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Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 125139 17 pageshttpdxdoiorg1011552013125139
Research ArticleTwo-Level Iteration Penalty Methods for the Navier-StokesEquations with Friction Boundary Conditions
Yuan Li and Rong An
College of Mathematics and Information Science Wenzhou University Wenzhou 325035 China
Correspondence should be addressed to Rong An anrong702gmailcom
Received 2 April 2013 Revised 12 June 2013 Accepted 19 June 2013
Academic Editor Stanislaw Migorski
Copyright copy 2013 Y Li and R An This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper presents two-level iteration penalty finite element methods to approximate the solution of the Navier-Stokes equationswith friction boundary conditionsThe basic idea is to solve the Navier-Stokes type variational inequality problem on a coarsemeshwith mesh size 119867 in combining with solving a Stokes Oseen or linearized Navier-Stokes type variational inequality problem forStokes Oseen or Newton iteration on a fine mesh with mesh size ℎ The error estimate obtained in this paper shows that if119867 ℎand 120576 can be chosen appropriately then these two-level iteration penalty methods are of the same convergence orders as the usualone-level iteration penalty method
1 Introduction
In this paper we consider a two-level iteration penaltymethod for the incompressible flows which are governed bythe incompressible Navier-Stokes equations
minus120583Δu + (u sdot nabla)u minus nabla119901 = f in Ω
div u = 0 in Ω
(1)
where Ω is a bounded domain in R2 assumed to have aLipschitz continuous boundary 120597Ω 120583 gt 0 represents theviscous coefficient u = (119906
1(119909) 1199062(119909)) denotes the velocity
vector 119901 = 119901(119909) the pressure and f = (1198911(119909) 1198912(119909)) the
prescribed body force vector The solenoidal conditiondiv u = 0means that the flows are incompressible
Instead of the classical whole homogeneous boundaryconditions here we consider the following slip boundaryconditions with friction type
u = 0 on Γ
u119899= 0 minus120590
120591(u) isin 119892120597 1003816100381610038161003816u120591
1003816100381610038161003816 on 119878
(2)
where Γ cap 119878 = 0 Γ cup 119878 = 120597Ω and 119892 is a scalar functionu119899= u sdot n and u
120591= u minus u
119899n are the normal and tangential
components of the velocity where n stands for the unit vector
of the external normal to 119878 120590120591(u) = 120590 minus 120590
119899n independent
of 119901 is the tangential components of the stress vector 120590which is defined by 120590
119894= 120590119894(u 119901) = (120583119890
119894119895(u) minus 119901120575
119894119895)119899119895with
119890119894119895(u) = (120597119906
119894120597119909119895) + (120597119906
119895120597119909119894) 119894 119895 = 1 2 The set 120597120595(a)
denotes a subdifferential of the function 120595 at a isin 1198712(119878)2
whose definition will be given in the next sectionThis type of boundary condition is firstly introduced
by Fujita [1] where some problems in hydrodynamics arestudied Some theoretical problems are also studied by manyscholars such as Fujita in [2ndash4] Y Li and K Li [5 6] andSaito and Fujita [7 8] and references cited in their work
The development of appropriate mixed finite elementapproximations is a key component in the search for efficienttechniques for solving the problem (1) quickly and efficientlyRoughly speaking there exist two main difficulties One isthe nonlinear term (u sdot nabla)u which can be processed by thelinearization method such as the Newton iteration methodStokes iteration method Oseen iteration method [9] or thetwo-level methods [10ndash17] The other is that the velocity andthe pressure are coupled by the solenoidal condition Thepopular technique to overcome the second difficulty is torelax the solenoidal condition in an appropriate method andto result in a pseudocompressible system such as the penaltymethod and the artificial compressiblemethod [18] Recentlyusing the Taylor-Hood element the authors [19] study the
2 Abstract and Applied Analysis
penalty finite elementmethod for the problem (1)-(2) Denote(uℎ120576 119901ℎ
120576) as the penalty finite element approximation solution
to (u 119901) isin (1198673(Ω)2 1198672(Ω)) The error estimate derived in
[19] is10038171003817100381710038171003817u minus uℎ120576
100381710038171003817100381710038171+10038171003817100381710038171003817119901 minus 119901ℎ
120576
10038171003817100381710038171003817le 119888 (120576 + ℎ
54) (3)
where 120576 gt 0 is the penalty parameter However thecondition number of the numerical discretization for thepenalty methods is 119874(120576minus1ℎminus2) which will result in an ill-conditioned problem when mesh size ℎ rarr 0 In order toavoid the choice of the small parameter 120576 Dai et al [20]have studied the iteration penalty finite element method andderive
10038171003817100381710038171003817u minus uℎ119896120576
100381710038171003817100381710038171+10038171003817100381710038171003817119901 minus 119901ℎ119896
120576
10038171003817100381710038171003817le 119888 (120576
119896+1+ ℎ54) (4)
where 119896 isin N+ is the iteration step numberIn this paper we combine the iteration penalty method
with the two-level method to approximate the solution ofthe problem (1)-(2) The iterative penalty method was firstintroduced by Cheng and Shaikh [21] for the Stokes equationsand further used to solve the pure Neumann problem [22]This iteration penalty method can be considered as the timediscretization of the artificial compressible method [23] Thetwo-level iteration penalty methods studied in this papercan be described as follows The first step and the secondstep are required to solve a small Navier-Stokes equationson the coarse mesh in terms of the iteration penalty method[20 21]The third step is required to solve a large linearizationproblem on the fine mesh in terms of the Stokes iterationOseen iteration or Newtonian iteration respectively Weprove that these two-level iteration penalty finite elementsolutions (u
120576ℎ 119901120576ℎ) are of the following error estimate
1003817100381710038171003817u minus u120576ℎ
10038171003817100381710038171+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817
le
119888 (ℎ54
+ 11986794
+ 12057611986754
+ 120576119896+1
)
(Stokes and Oseen iteration) 119888 (ℎ54
+ 11986752
+ 12057611986754
+ 120576119896+2
)
(Newtonian iteration)
(5)
Finally we propose an improved correction iteration schemefor (119906
120576ℎ 119901120576ℎ) in terms of the Newton iteration method We
prove that the correction finite element solutions (119906⋆120576ℎ 119901⋆
120576ℎ)
are of the following error estimates1003817100381710038171003817u minus u⋆
120576ℎ
10038171003817100381710038171+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le
119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(Stokes and Oseen iteration) 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(Newtonian iteration) (6)
Throughout this paper we will use 119888 to denote a positiveconstant whose valuemay change from place to place but thatremains independent of ℎ119867 and 120576 and that may depend on120583 Ω and the norms of u 119901 f and 119892
2 Preliminary
First we give the definition of the subdifferential property Let120595 be a given function possessing the properties of convexityand weak semicontinuity from below We say that the set120597120595(a) is a subdifferential of the function 120595 at a isin 119871
2(119878)2 if
and only if
120597120595 (a) = b isin 1198712(119878)2 120595 (h) minus 120595 (a)
ge b sdot (h minus a) forallh isin 1198712(119878)2 (7)
In what follows we employ the standard notation119867119897(Ω)(or 119867119897(Ω)2) and || sdot ||
119897 119897 ge 0 for the Sobolev spaces of all
functions having square integrable derivatives up to order 119897in Ω and the standard Sobolev norm When 119897 = 0 we write1198712(Ω) (or 1198712(Ω)2) and || sdot || instead of1198670(Ω) (or1198670(Ω)2) and
|| sdot ||0 respectively
For the mathematical setting we introduce the followingspaces
119881 = u isin 1198671(Ω)2 u|Γ = 0 u sdot n|119878 = 0
1198810= 1198671
0(Ω)2
119881120590= u isin 119881 div u = 0
119872 = 1198712
0(Ω) = 119902 isin 119871
2(Ω) int
Ω
119902119889119909 = 0
(8)
The space 119881 is equipped with the norm
v119881 = (intΩ
|nablav|2119889119909)12
(9)
It is well known that ||v||119881
is equivalent to ||v||1due to
Poincarersquos inequality Introduce two bilinear forms
119886 (u v) = 120583intΩ
nablau sdot nablav119889119909 forallu v isin 119881
119889 (V 119902) = intΩ
119902 div v119889119909 forallv isin 119881 119902 isin 119872
(10)
and a trilinear form
119887 (u vw) = intΩ
(u sdot nabla) v sdot w119889119909 minus 1
2intΩ
div uv sdot w119889119909
=1
2intΩ
(u sdot nabla) v sdot w119889119909 minus 1
2intΩ
(u sdot nabla)w sdot v119889119909(11)
It is easy to verify that this trilinear form satisfies the followingimportant properties [12 23]
119887 (u vw) = minus119887 (uw v) (12)
119887 (u vw) le 119873u119881v119881w119881 (13)
119887 (u vw) le 119873
2u12u12
119881
times (v119881w12w12119881
+ w119881v12v12119881)
(14)
Abstract and Applied Analysis 3
for all u vw isin 119881 and
|119887 (u vw)| + |119887 (v uw)| + |119887 (w u v)| le 119873u119881v2 w (15)
for all u isin 119881 v isin 1198672(Ω)2 and w isin 119871
2(Ω)2 where 119873 gt 0
depends only onΩGiven f isin 119871
2(Ω)2 and 119892 isin 119871
2(119878) with 119892 gt 0 on 119878
under the above notation the variational formulation of theproblem (1)-(2) reads as follows find (u 119901) isin (119881119872) suchthat for all (v 119902) isin (119881119872)
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591)
minus 119889 (v minus u 119901) ge (f v minus u)
119889 (u 119902) = 0
(16)
where 119895(120578) = int119878119892|120578|119889119904 for all 120578 isin 1198712(119878)2 Saito in [8] showed
that there exists some positive 120573 gt 0 such that
12057310038171003817100381710038171199021003817100381710038171003817 le sup
Visin119881
119889 (v 119902)v119881
(17)
then the variational inequality (16) is equivalent to thefollowing find u isin 119881
120590such that for all v isin 119881
120590
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591) ge (f v minus u)
(18)
The existence and uniqueness theoremof the solutionu to theproblem (18) has been shown in [19] Here we only recall it
Theorem 1 If the following uniqueness condition holds
41205811119873(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
1205832lt 1 (19)
then there exists a unique solution u isin 119881120590to the variational
inequality problem (18) such that
u119881 le21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt
120583
2119873 (20)
where 1205811gt 0 satisfies
1003816100381610038161003816(f v) minus 119895 (v120591)1003816100381610038161003816 le 1205811 (f +
100381710038171003817100381711989210038171003817100381710038171198712(119878)
) v119881 forallv isin 119881120590 (21)
3 Iteration Penalty FiniteElement Approximation
Suppose that Ω is a convex and polygon domain LetTℎbe a
family of quasi-uniform triangular partition ofΩ The corre-sponding ordered triangles are denoted by119870
1 1198702 119870
119899 Let
ℎ119894= diam(119870
119894) 119894 = 1 119899 and ℎ = maxℎ
1 ℎ2 ℎ
119899 For
every 119870 isin Tℎ let 119875119903(119870) denote the space of the polynomials
on 119870 of degree at most 119903 For simplicity we consider theconforming finite element spaces 119881
ℎand119872
ℎdefined by
119881ℎ= 119881 cap119882
ℎwith 119882
ℎ= vℎisin 119862(Ω)
2
vℎ
1003816100381610038161003816119870
isin [1198752(119870)]2
forall119870 isin Tℎ
119872ℎ=119902ℎisin119862 (Ω) 119902
ℎ
1003816100381610038161003816119870isin1198751(119870) forall119870isinT
ℎ intΩ
119902ℎ119889119909=0
(22)
Denote 1198810ℎ
= 1198810cap 119882ℎ It is well known that 119881
0ℎand 119872
ℎ
satisfy the Babuska-Brezzi condition [24 25]
1205811003817100381710038171003817119902ℎ
1003817100381710038171003817 le supwℎisin1198810ℎ
119889 (wℎ 119902ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
(23)
where 120581 gt 0 is a constant independent of ℎ Denote119877ℎand119876
ℎ
as the1198712 orthogonal projections onto119881ℎand119872
ℎ respectively
which satisfy1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817 + ℎ1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817119881le 119888ℎ119894v119894
forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(24)
1003817100381710038171003817119902 minus 119876ℎ1199021003817100381710038171003817 le 119888ℎ
11989510038171003817100381710038171199021003817100381710038171003817119895 forall119902 isin 119867
2(Ω) cap119872 119895 = 1 2 (25)
It follows from the trace inequality ||v||1198712(119878)
le 119888||v||12||v||12119881
[26] that1003817100381710038171003817v minus 119877ℎv
10038171003817100381710038171198712(119878)
le 1198881003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817
121003817100381710038171003817v minus 119877ℎv1003817100381710038171003817
12
119881
le 119888ℎ119894minus12
v119894 forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(26)
Let 120576 gt 0 be some small parameterThe one-level iterationpenalty finite element method for the problem (16) has beenstudied in [20] which can be described as follows
Step 1 Find (u0120576ℎ 1199010
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u0120576ℎ vℎminus u0120576ℎ) + 119887 (u0
120576ℎ u0120576ℎ vℎminus u0120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u0120576ℎ120591) minus 119889 (v
ℎminus u0120576ℎ 1199010
120576ℎ) ge (f v
ℎminus u0120576ℎ)
119889 (u0120576ℎ 119902ℎ) + 120576 (119901
0
120576ℎ 119902ℎ) = 0
(27)
Step 2 For 119896 = 1 2 find (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) such that
for all (vℎ 119902ℎ) isin (119881
ℎ119872ℎ)
119886 (u119896120576ℎ vℎminus u119896120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ vℎminus u119896120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u119896120576ℎ120591) minus 119889 (v
ℎminus u119896120576ℎ 119901119896
120576ℎ) ge (f v
ℎminus u119896120576ℎ)
119889 (u119896120576ℎ 119902ℎ) + 120576 (119901
119896
120576ℎ 119902ℎ) = 120576 (119901
119896minus1
120576ℎ 119902ℎ)
(28)
First we give the a priori estimate of the solution (u119896120576ℎ 119901119896
120576ℎ)
to the problem (28)
4 Abstract and Applied Analysis
Theorem 2 Suppose that (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) is the solution
to the problem (28) then it satisfies
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(29)
Proof Setting vℎ= 0 119902
ℎ= 1199010
120576ℎin (27) using (12) and Youngrsquos
inequality it yields that
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le (f u0120576ℎ) minus 119895 (u0
120576ℎ120591)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817119881
le120583
2
10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(30)
Then we have
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 2120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(31)
For 119896 = 1 2 setting vℎ= 0 119902
ℎ= 119901119896
120576ℎin (28) it yields that
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le (f u119896120576ℎ) minus 119895 (u119896
120576ℎ120591) + 120576 (119901
119896minus1
120576ℎ 119901119896
120576ℎ)
le120583
2
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+120576
2
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
(32)
Thus we obtain
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 12057610038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
le sdot sdot sdot le1198961205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 120576100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(33)
The next theorem gives the error estimate between thesolutions (u 119901) and (u119896
120576ℎ 119901119896
120576ℎ) to the problems (16) and (28)
respectively The proof can be found in [20]
Theorem 3 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
54+ 120576119896+1
) (34)
Next we will show the error estimate ||u minus u119896120576ℎ|| for
the penalty finite element approximation (28) This 1198712 erroranalysis is based on the regularity assumption that thefollowing linearized problem (35) is (1198672(Ω)2 1198671(Ω)) regular
Given z isin 1198712(Ω)2 find (w 120587) isin (119881119872) such that for all
(v 119902) isin (119881119872)
119886 (w v) + 119887 (u119896120576ℎ vw) + 119887 (v uw) minus 119889 (v 120587) = (z v)
119889 (w 119902) = 0(35)
According to (12) and (20) it is easy to verify that there existsa unique solution (w 120587) to the problem (35)The assumptionthat (35) is (1198672(Ω)2 1198671(Ω)) regular means that (w 120587) alsobelongs to (119867
2(Ω)2 1198671(Ω)) and the following inequality
holds
w2 + 1205871 le 119888 119911 (36)
Let 119868ℎbe the 1198712 orthogonal projections onto 119881
0ℎand satisfy
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881 le 119888ℎw2 (37)
Theorem 4 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
94+ 120576ℎ54
+ 120576119896+1
) (38)
Proof Setting z = uminusu119896120576ℎand v = uminusu119896
120576ℎin the first equation
of (35) we get
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w u minus u119896120576ℎ) + 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908)
+ 119887 (u minus u119896120576ℎ uw) minus 119889 (u minus u119896
120576ℎ 120587)
(39)
Taking v = uplusmn 119868ℎw 119902 = 119876
ℎ120587 in (16) and v
ℎ= u119896120576ℎplusmn 119868ℎw 119902ℎ=
119876ℎ120587 in (28) respectively we obtain
119886 (u 119868ℎw) + 119887 (u u 119868
ℎw) minus 119889 (119868
ℎw 119901) = (f 119868
ℎw)
119889 (u 119876ℎ120587) = 0
119886 (u119896120576ℎ 119868ℎw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119889 (119868
ℎw 119901119896120576ℎ) = (f 119868
ℎw)
119889 (u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896
120576ℎ 119876ℎ120587) = 120576 (119901
119896minus1
120576ℎ 119876ℎ120587)
(40)
Subtracting them we get
119886 (u minus u119896120576ℎ 119868ℎw) + 119887 (u u 119868
ℎw) minus 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw)
minus 119889 (119868ℎw 119901 minus 119901119896
120576ℎ) = 0
119889 (u minus u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587) = 0
(41)
Abstract and Applied Analysis 5
Substituting the previous equation into (39) it yields that
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w minus 119868ℎw u minus u119896
120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198691
+ 119887 (u119896120576ℎ u minus u119896
120576ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119887 (u minus u119896120576ℎ uw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119889 (119868ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198693
+ 120576 (119901119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198694
(42)
Using (34) (36) and (37) 1198691is estimated by
1198691= 119886 (w minus 119868
ℎw u minus u119896
120576ℎ)
le 12058310038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(43)
Similarly using (25) (34) (36) and (37) 1198693is estimated by
1198693= 119889 (119868
ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
le1003817100381710038171003817119868ℎw minus w1003817100381710038171003817119881
10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817120587 minus 119876ℎ1205871003817100381710038171003817
le 119888ℎ (ℎ54
+ 120576119896+1
) (w2 + 1205871)
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(44)
We rewrite 1198692as
1198692= 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎminus u 119868ℎw) minus 119887 (u minus u119896
120576ℎ u 119868ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw minus 119868
ℎw) + 119887 (u minus u119896
120576ℎ uw minus 119868
ℎw) (45)
Then from (13) (20) (29) (34) (36) and (37) it holds that
1198692le 119873(
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817119881+ u119881)
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119877ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(46)
Finally we estimate 1198694by
1198694= 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
= 120576 (119901119896minus1
120576ℎminus 119901119876
ℎ120587) + 120576 (119901 minus 119901
119896
120576ℎ 119876ℎ120587)
le 120576 (10038171003817100381710038171003817119901119896minus1
120576ℎminus 119901
10038171003817100381710038171003817+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817)1003817100381710038171003817119876ℎ120587
1003817100381710038171003817
le 119888120576 (ℎ54
+ 120576119896) 1205871
le 119888120576 (ℎ54
+ 120576119896)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(47)
Combining these estimates with (42) we conclude that (38)holds
4 Two-Level Iteration Penalty Methods
In this section based on the iteration penalty methoddescribed in the previous section the two-level iterationpenalty finite element methods for (16) are proposed in termsof the Stokes iteration Oseen iteration or Newtonian itera-tion From now on119867 and ℎ with ℎ lt 119867 are two real positiveparameters The coarse mesh triangulation T
119867is made as
in Section 3 And a fine mesh triangulation Tℎis generated
by a mesh refinement process to T119867 The conforming finite
element space pairs (119881ℎ119872ℎ) and (119881
119867119872119867) sub (119881
ℎ119872ℎ)
corresponding to the triangulationsTℎandT
119867 respectively
are constructed as in Section 3With the preavious notationswe propose the following two-level iteration finite elementmethods
41 Two-Level Stokes Iteration Penalty Method In Steps 1and 2 we solve (27) and (28) on the coarse mesh as in thefollwing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) such that for all
(v119867 119902119867) isin (119881
119867119872119867)
119886 (u0120576119867 v119867minus u0120576119867) + 119887 (u0
120576119867 u0120576119867 v119867minus u0120576119867) + 119895 (v
119867120591)
minus 119895 (u0120576119867120591
) minus 119889 (v119867minus u0120576119867 1199010
120576119867) ge (f v
119867minus u0120576119867)
119889 (u0120576119867 119902119867) + 120576 (119901
0
120576119867 119902119867) = 0
(48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) such that
for all (v119867 119902119867) isin (119881
119867119872119867)
119886 (u119896120576119867 v119867minus u119896120576119867) + 119887 (u119896
120576119867 u119896120576119867 v119867minus u119896120576119867) + 119895 (v
119867120591)
minus 119895 (u119896120576119867120591
) minus 119889 (v119867minus u119896120576119867 119901119896
120576119867) ge (f v
119867minus u119896120576119867)
119889 (u119896120576119867 119902119867) + 120576 (119901
119896
120576119867 119902119867) = 120576 (119901
119896minus1
120576119867 119902ℎ)
(49)
In Step 3 we solve a Stokes-type variational inequalityproblem on the fine mesh in terms of the Stokes iteration asin the following
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
penalty finite elementmethod for the problem (1)-(2) Denote(uℎ120576 119901ℎ
120576) as the penalty finite element approximation solution
to (u 119901) isin (1198673(Ω)2 1198672(Ω)) The error estimate derived in
[19] is10038171003817100381710038171003817u minus uℎ120576
100381710038171003817100381710038171+10038171003817100381710038171003817119901 minus 119901ℎ
120576
10038171003817100381710038171003817le 119888 (120576 + ℎ
54) (3)
where 120576 gt 0 is the penalty parameter However thecondition number of the numerical discretization for thepenalty methods is 119874(120576minus1ℎminus2) which will result in an ill-conditioned problem when mesh size ℎ rarr 0 In order toavoid the choice of the small parameter 120576 Dai et al [20]have studied the iteration penalty finite element method andderive
10038171003817100381710038171003817u minus uℎ119896120576
100381710038171003817100381710038171+10038171003817100381710038171003817119901 minus 119901ℎ119896
120576
10038171003817100381710038171003817le 119888 (120576
119896+1+ ℎ54) (4)
where 119896 isin N+ is the iteration step numberIn this paper we combine the iteration penalty method
with the two-level method to approximate the solution ofthe problem (1)-(2) The iterative penalty method was firstintroduced by Cheng and Shaikh [21] for the Stokes equationsand further used to solve the pure Neumann problem [22]This iteration penalty method can be considered as the timediscretization of the artificial compressible method [23] Thetwo-level iteration penalty methods studied in this papercan be described as follows The first step and the secondstep are required to solve a small Navier-Stokes equationson the coarse mesh in terms of the iteration penalty method[20 21]The third step is required to solve a large linearizationproblem on the fine mesh in terms of the Stokes iterationOseen iteration or Newtonian iteration respectively Weprove that these two-level iteration penalty finite elementsolutions (u
120576ℎ 119901120576ℎ) are of the following error estimate
1003817100381710038171003817u minus u120576ℎ
10038171003817100381710038171+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817
le
119888 (ℎ54
+ 11986794
+ 12057611986754
+ 120576119896+1
)
(Stokes and Oseen iteration) 119888 (ℎ54
+ 11986752
+ 12057611986754
+ 120576119896+2
)
(Newtonian iteration)
(5)
Finally we propose an improved correction iteration schemefor (119906
120576ℎ 119901120576ℎ) in terms of the Newton iteration method We
prove that the correction finite element solutions (119906⋆120576ℎ 119901⋆
120576ℎ)
are of the following error estimates1003817100381710038171003817u minus u⋆
120576ℎ
10038171003817100381710038171+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le
119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(Stokes and Oseen iteration) 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(Newtonian iteration) (6)
Throughout this paper we will use 119888 to denote a positiveconstant whose valuemay change from place to place but thatremains independent of ℎ119867 and 120576 and that may depend on120583 Ω and the norms of u 119901 f and 119892
2 Preliminary
First we give the definition of the subdifferential property Let120595 be a given function possessing the properties of convexityand weak semicontinuity from below We say that the set120597120595(a) is a subdifferential of the function 120595 at a isin 119871
2(119878)2 if
and only if
120597120595 (a) = b isin 1198712(119878)2 120595 (h) minus 120595 (a)
ge b sdot (h minus a) forallh isin 1198712(119878)2 (7)
In what follows we employ the standard notation119867119897(Ω)(or 119867119897(Ω)2) and || sdot ||
119897 119897 ge 0 for the Sobolev spaces of all
functions having square integrable derivatives up to order 119897in Ω and the standard Sobolev norm When 119897 = 0 we write1198712(Ω) (or 1198712(Ω)2) and || sdot || instead of1198670(Ω) (or1198670(Ω)2) and
|| sdot ||0 respectively
For the mathematical setting we introduce the followingspaces
119881 = u isin 1198671(Ω)2 u|Γ = 0 u sdot n|119878 = 0
1198810= 1198671
0(Ω)2
119881120590= u isin 119881 div u = 0
119872 = 1198712
0(Ω) = 119902 isin 119871
2(Ω) int
Ω
119902119889119909 = 0
(8)
The space 119881 is equipped with the norm
v119881 = (intΩ
|nablav|2119889119909)12
(9)
It is well known that ||v||119881
is equivalent to ||v||1due to
Poincarersquos inequality Introduce two bilinear forms
119886 (u v) = 120583intΩ
nablau sdot nablav119889119909 forallu v isin 119881
119889 (V 119902) = intΩ
119902 div v119889119909 forallv isin 119881 119902 isin 119872
(10)
and a trilinear form
119887 (u vw) = intΩ
(u sdot nabla) v sdot w119889119909 minus 1
2intΩ
div uv sdot w119889119909
=1
2intΩ
(u sdot nabla) v sdot w119889119909 minus 1
2intΩ
(u sdot nabla)w sdot v119889119909(11)
It is easy to verify that this trilinear form satisfies the followingimportant properties [12 23]
119887 (u vw) = minus119887 (uw v) (12)
119887 (u vw) le 119873u119881v119881w119881 (13)
119887 (u vw) le 119873
2u12u12
119881
times (v119881w12w12119881
+ w119881v12v12119881)
(14)
Abstract and Applied Analysis 3
for all u vw isin 119881 and
|119887 (u vw)| + |119887 (v uw)| + |119887 (w u v)| le 119873u119881v2 w (15)
for all u isin 119881 v isin 1198672(Ω)2 and w isin 119871
2(Ω)2 where 119873 gt 0
depends only onΩGiven f isin 119871
2(Ω)2 and 119892 isin 119871
2(119878) with 119892 gt 0 on 119878
under the above notation the variational formulation of theproblem (1)-(2) reads as follows find (u 119901) isin (119881119872) suchthat for all (v 119902) isin (119881119872)
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591)
minus 119889 (v minus u 119901) ge (f v minus u)
119889 (u 119902) = 0
(16)
where 119895(120578) = int119878119892|120578|119889119904 for all 120578 isin 1198712(119878)2 Saito in [8] showed
that there exists some positive 120573 gt 0 such that
12057310038171003817100381710038171199021003817100381710038171003817 le sup
Visin119881
119889 (v 119902)v119881
(17)
then the variational inequality (16) is equivalent to thefollowing find u isin 119881
120590such that for all v isin 119881
120590
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591) ge (f v minus u)
(18)
The existence and uniqueness theoremof the solutionu to theproblem (18) has been shown in [19] Here we only recall it
Theorem 1 If the following uniqueness condition holds
41205811119873(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
1205832lt 1 (19)
then there exists a unique solution u isin 119881120590to the variational
inequality problem (18) such that
u119881 le21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt
120583
2119873 (20)
where 1205811gt 0 satisfies
1003816100381610038161003816(f v) minus 119895 (v120591)1003816100381610038161003816 le 1205811 (f +
100381710038171003817100381711989210038171003817100381710038171198712(119878)
) v119881 forallv isin 119881120590 (21)
3 Iteration Penalty FiniteElement Approximation
Suppose that Ω is a convex and polygon domain LetTℎbe a
family of quasi-uniform triangular partition ofΩ The corre-sponding ordered triangles are denoted by119870
1 1198702 119870
119899 Let
ℎ119894= diam(119870
119894) 119894 = 1 119899 and ℎ = maxℎ
1 ℎ2 ℎ
119899 For
every 119870 isin Tℎ let 119875119903(119870) denote the space of the polynomials
on 119870 of degree at most 119903 For simplicity we consider theconforming finite element spaces 119881
ℎand119872
ℎdefined by
119881ℎ= 119881 cap119882
ℎwith 119882
ℎ= vℎisin 119862(Ω)
2
vℎ
1003816100381610038161003816119870
isin [1198752(119870)]2
forall119870 isin Tℎ
119872ℎ=119902ℎisin119862 (Ω) 119902
ℎ
1003816100381610038161003816119870isin1198751(119870) forall119870isinT
ℎ intΩ
119902ℎ119889119909=0
(22)
Denote 1198810ℎ
= 1198810cap 119882ℎ It is well known that 119881
0ℎand 119872
ℎ
satisfy the Babuska-Brezzi condition [24 25]
1205811003817100381710038171003817119902ℎ
1003817100381710038171003817 le supwℎisin1198810ℎ
119889 (wℎ 119902ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
(23)
where 120581 gt 0 is a constant independent of ℎ Denote119877ℎand119876
ℎ
as the1198712 orthogonal projections onto119881ℎand119872
ℎ respectively
which satisfy1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817 + ℎ1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817119881le 119888ℎ119894v119894
forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(24)
1003817100381710038171003817119902 minus 119876ℎ1199021003817100381710038171003817 le 119888ℎ
11989510038171003817100381710038171199021003817100381710038171003817119895 forall119902 isin 119867
2(Ω) cap119872 119895 = 1 2 (25)
It follows from the trace inequality ||v||1198712(119878)
le 119888||v||12||v||12119881
[26] that1003817100381710038171003817v minus 119877ℎv
10038171003817100381710038171198712(119878)
le 1198881003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817
121003817100381710038171003817v minus 119877ℎv1003817100381710038171003817
12
119881
le 119888ℎ119894minus12
v119894 forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(26)
Let 120576 gt 0 be some small parameterThe one-level iterationpenalty finite element method for the problem (16) has beenstudied in [20] which can be described as follows
Step 1 Find (u0120576ℎ 1199010
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u0120576ℎ vℎminus u0120576ℎ) + 119887 (u0
120576ℎ u0120576ℎ vℎminus u0120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u0120576ℎ120591) minus 119889 (v
ℎminus u0120576ℎ 1199010
120576ℎ) ge (f v
ℎminus u0120576ℎ)
119889 (u0120576ℎ 119902ℎ) + 120576 (119901
0
120576ℎ 119902ℎ) = 0
(27)
Step 2 For 119896 = 1 2 find (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) such that
for all (vℎ 119902ℎ) isin (119881
ℎ119872ℎ)
119886 (u119896120576ℎ vℎminus u119896120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ vℎminus u119896120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u119896120576ℎ120591) minus 119889 (v
ℎminus u119896120576ℎ 119901119896
120576ℎ) ge (f v
ℎminus u119896120576ℎ)
119889 (u119896120576ℎ 119902ℎ) + 120576 (119901
119896
120576ℎ 119902ℎ) = 120576 (119901
119896minus1
120576ℎ 119902ℎ)
(28)
First we give the a priori estimate of the solution (u119896120576ℎ 119901119896
120576ℎ)
to the problem (28)
4 Abstract and Applied Analysis
Theorem 2 Suppose that (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) is the solution
to the problem (28) then it satisfies
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(29)
Proof Setting vℎ= 0 119902
ℎ= 1199010
120576ℎin (27) using (12) and Youngrsquos
inequality it yields that
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le (f u0120576ℎ) minus 119895 (u0
120576ℎ120591)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817119881
le120583
2
10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(30)
Then we have
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 2120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(31)
For 119896 = 1 2 setting vℎ= 0 119902
ℎ= 119901119896
120576ℎin (28) it yields that
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le (f u119896120576ℎ) minus 119895 (u119896
120576ℎ120591) + 120576 (119901
119896minus1
120576ℎ 119901119896
120576ℎ)
le120583
2
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+120576
2
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
(32)
Thus we obtain
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 12057610038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
le sdot sdot sdot le1198961205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 120576100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(33)
The next theorem gives the error estimate between thesolutions (u 119901) and (u119896
120576ℎ 119901119896
120576ℎ) to the problems (16) and (28)
respectively The proof can be found in [20]
Theorem 3 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
54+ 120576119896+1
) (34)
Next we will show the error estimate ||u minus u119896120576ℎ|| for
the penalty finite element approximation (28) This 1198712 erroranalysis is based on the regularity assumption that thefollowing linearized problem (35) is (1198672(Ω)2 1198671(Ω)) regular
Given z isin 1198712(Ω)2 find (w 120587) isin (119881119872) such that for all
(v 119902) isin (119881119872)
119886 (w v) + 119887 (u119896120576ℎ vw) + 119887 (v uw) minus 119889 (v 120587) = (z v)
119889 (w 119902) = 0(35)
According to (12) and (20) it is easy to verify that there existsa unique solution (w 120587) to the problem (35)The assumptionthat (35) is (1198672(Ω)2 1198671(Ω)) regular means that (w 120587) alsobelongs to (119867
2(Ω)2 1198671(Ω)) and the following inequality
holds
w2 + 1205871 le 119888 119911 (36)
Let 119868ℎbe the 1198712 orthogonal projections onto 119881
0ℎand satisfy
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881 le 119888ℎw2 (37)
Theorem 4 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
94+ 120576ℎ54
+ 120576119896+1
) (38)
Proof Setting z = uminusu119896120576ℎand v = uminusu119896
120576ℎin the first equation
of (35) we get
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w u minus u119896120576ℎ) + 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908)
+ 119887 (u minus u119896120576ℎ uw) minus 119889 (u minus u119896
120576ℎ 120587)
(39)
Taking v = uplusmn 119868ℎw 119902 = 119876
ℎ120587 in (16) and v
ℎ= u119896120576ℎplusmn 119868ℎw 119902ℎ=
119876ℎ120587 in (28) respectively we obtain
119886 (u 119868ℎw) + 119887 (u u 119868
ℎw) minus 119889 (119868
ℎw 119901) = (f 119868
ℎw)
119889 (u 119876ℎ120587) = 0
119886 (u119896120576ℎ 119868ℎw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119889 (119868
ℎw 119901119896120576ℎ) = (f 119868
ℎw)
119889 (u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896
120576ℎ 119876ℎ120587) = 120576 (119901
119896minus1
120576ℎ 119876ℎ120587)
(40)
Subtracting them we get
119886 (u minus u119896120576ℎ 119868ℎw) + 119887 (u u 119868
ℎw) minus 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw)
minus 119889 (119868ℎw 119901 minus 119901119896
120576ℎ) = 0
119889 (u minus u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587) = 0
(41)
Abstract and Applied Analysis 5
Substituting the previous equation into (39) it yields that
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w minus 119868ℎw u minus u119896
120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198691
+ 119887 (u119896120576ℎ u minus u119896
120576ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119887 (u minus u119896120576ℎ uw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119889 (119868ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198693
+ 120576 (119901119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198694
(42)
Using (34) (36) and (37) 1198691is estimated by
1198691= 119886 (w minus 119868
ℎw u minus u119896
120576ℎ)
le 12058310038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(43)
Similarly using (25) (34) (36) and (37) 1198693is estimated by
1198693= 119889 (119868
ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
le1003817100381710038171003817119868ℎw minus w1003817100381710038171003817119881
10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817120587 minus 119876ℎ1205871003817100381710038171003817
le 119888ℎ (ℎ54
+ 120576119896+1
) (w2 + 1205871)
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(44)
We rewrite 1198692as
1198692= 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎminus u 119868ℎw) minus 119887 (u minus u119896
120576ℎ u 119868ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw minus 119868
ℎw) + 119887 (u minus u119896
120576ℎ uw minus 119868
ℎw) (45)
Then from (13) (20) (29) (34) (36) and (37) it holds that
1198692le 119873(
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817119881+ u119881)
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119877ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(46)
Finally we estimate 1198694by
1198694= 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
= 120576 (119901119896minus1
120576ℎminus 119901119876
ℎ120587) + 120576 (119901 minus 119901
119896
120576ℎ 119876ℎ120587)
le 120576 (10038171003817100381710038171003817119901119896minus1
120576ℎminus 119901
10038171003817100381710038171003817+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817)1003817100381710038171003817119876ℎ120587
1003817100381710038171003817
le 119888120576 (ℎ54
+ 120576119896) 1205871
le 119888120576 (ℎ54
+ 120576119896)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(47)
Combining these estimates with (42) we conclude that (38)holds
4 Two-Level Iteration Penalty Methods
In this section based on the iteration penalty methoddescribed in the previous section the two-level iterationpenalty finite element methods for (16) are proposed in termsof the Stokes iteration Oseen iteration or Newtonian itera-tion From now on119867 and ℎ with ℎ lt 119867 are two real positiveparameters The coarse mesh triangulation T
119867is made as
in Section 3 And a fine mesh triangulation Tℎis generated
by a mesh refinement process to T119867 The conforming finite
element space pairs (119881ℎ119872ℎ) and (119881
119867119872119867) sub (119881
ℎ119872ℎ)
corresponding to the triangulationsTℎandT
119867 respectively
are constructed as in Section 3With the preavious notationswe propose the following two-level iteration finite elementmethods
41 Two-Level Stokes Iteration Penalty Method In Steps 1and 2 we solve (27) and (28) on the coarse mesh as in thefollwing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) such that for all
(v119867 119902119867) isin (119881
119867119872119867)
119886 (u0120576119867 v119867minus u0120576119867) + 119887 (u0
120576119867 u0120576119867 v119867minus u0120576119867) + 119895 (v
119867120591)
minus 119895 (u0120576119867120591
) minus 119889 (v119867minus u0120576119867 1199010
120576119867) ge (f v
119867minus u0120576119867)
119889 (u0120576119867 119902119867) + 120576 (119901
0
120576119867 119902119867) = 0
(48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) such that
for all (v119867 119902119867) isin (119881
119867119872119867)
119886 (u119896120576119867 v119867minus u119896120576119867) + 119887 (u119896
120576119867 u119896120576119867 v119867minus u119896120576119867) + 119895 (v
119867120591)
minus 119895 (u119896120576119867120591
) minus 119889 (v119867minus u119896120576119867 119901119896
120576119867) ge (f v
119867minus u119896120576119867)
119889 (u119896120576119867 119902119867) + 120576 (119901
119896
120576119867 119902119867) = 120576 (119901
119896minus1
120576119867 119902ℎ)
(49)
In Step 3 we solve a Stokes-type variational inequalityproblem on the fine mesh in terms of the Stokes iteration asin the following
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
for all u vw isin 119881 and
|119887 (u vw)| + |119887 (v uw)| + |119887 (w u v)| le 119873u119881v2 w (15)
for all u isin 119881 v isin 1198672(Ω)2 and w isin 119871
2(Ω)2 where 119873 gt 0
depends only onΩGiven f isin 119871
2(Ω)2 and 119892 isin 119871
2(119878) with 119892 gt 0 on 119878
under the above notation the variational formulation of theproblem (1)-(2) reads as follows find (u 119901) isin (119881119872) suchthat for all (v 119902) isin (119881119872)
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591)
minus 119889 (v minus u 119901) ge (f v minus u)
119889 (u 119902) = 0
(16)
where 119895(120578) = int119878119892|120578|119889119904 for all 120578 isin 1198712(119878)2 Saito in [8] showed
that there exists some positive 120573 gt 0 such that
12057310038171003817100381710038171199021003817100381710038171003817 le sup
Visin119881
119889 (v 119902)v119881
(17)
then the variational inequality (16) is equivalent to thefollowing find u isin 119881
120590such that for all v isin 119881
120590
119886 (u v minus u) + 119887 (u u v minus u) + 119895 (v120591) minus 119895 (u
120591) ge (f v minus u)
(18)
The existence and uniqueness theoremof the solutionu to theproblem (18) has been shown in [19] Here we only recall it
Theorem 1 If the following uniqueness condition holds
41205811119873(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
1205832lt 1 (19)
then there exists a unique solution u isin 119881120590to the variational
inequality problem (18) such that
u119881 le21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt
120583
2119873 (20)
where 1205811gt 0 satisfies
1003816100381610038161003816(f v) minus 119895 (v120591)1003816100381610038161003816 le 1205811 (f +
100381710038171003817100381711989210038171003817100381710038171198712(119878)
) v119881 forallv isin 119881120590 (21)
3 Iteration Penalty FiniteElement Approximation
Suppose that Ω is a convex and polygon domain LetTℎbe a
family of quasi-uniform triangular partition ofΩ The corre-sponding ordered triangles are denoted by119870
1 1198702 119870
119899 Let
ℎ119894= diam(119870
119894) 119894 = 1 119899 and ℎ = maxℎ
1 ℎ2 ℎ
119899 For
every 119870 isin Tℎ let 119875119903(119870) denote the space of the polynomials
on 119870 of degree at most 119903 For simplicity we consider theconforming finite element spaces 119881
ℎand119872
ℎdefined by
119881ℎ= 119881 cap119882
ℎwith 119882
ℎ= vℎisin 119862(Ω)
2
vℎ
1003816100381610038161003816119870
isin [1198752(119870)]2
forall119870 isin Tℎ
119872ℎ=119902ℎisin119862 (Ω) 119902
ℎ
1003816100381610038161003816119870isin1198751(119870) forall119870isinT
ℎ intΩ
119902ℎ119889119909=0
(22)
Denote 1198810ℎ
= 1198810cap 119882ℎ It is well known that 119881
0ℎand 119872
ℎ
satisfy the Babuska-Brezzi condition [24 25]
1205811003817100381710038171003817119902ℎ
1003817100381710038171003817 le supwℎisin1198810ℎ
119889 (wℎ 119902ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
(23)
where 120581 gt 0 is a constant independent of ℎ Denote119877ℎand119876
ℎ
as the1198712 orthogonal projections onto119881ℎand119872
ℎ respectively
which satisfy1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817 + ℎ1003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817119881le 119888ℎ119894v119894
forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(24)
1003817100381710038171003817119902 minus 119876ℎ1199021003817100381710038171003817 le 119888ℎ
11989510038171003817100381710038171199021003817100381710038171003817119895 forall119902 isin 119867
2(Ω) cap119872 119895 = 1 2 (25)
It follows from the trace inequality ||v||1198712(119878)
le 119888||v||12||v||12119881
[26] that1003817100381710038171003817v minus 119877ℎv
10038171003817100381710038171198712(119878)
le 1198881003817100381710038171003817v minus 119877ℎv
1003817100381710038171003817
121003817100381710038171003817v minus 119877ℎv1003817100381710038171003817
12
119881
le 119888ℎ119894minus12
v119894 forallv isin 1198673(Ω)2 cap 119881 119894 = 1 2 3
(26)
Let 120576 gt 0 be some small parameterThe one-level iterationpenalty finite element method for the problem (16) has beenstudied in [20] which can be described as follows
Step 1 Find (u0120576ℎ 1199010
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u0120576ℎ vℎminus u0120576ℎ) + 119887 (u0
120576ℎ u0120576ℎ vℎminus u0120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u0120576ℎ120591) minus 119889 (v
ℎminus u0120576ℎ 1199010
120576ℎ) ge (f v
ℎminus u0120576ℎ)
119889 (u0120576ℎ 119902ℎ) + 120576 (119901
0
120576ℎ 119902ℎ) = 0
(27)
Step 2 For 119896 = 1 2 find (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) such that
for all (vℎ 119902ℎ) isin (119881
ℎ119872ℎ)
119886 (u119896120576ℎ vℎminus u119896120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ vℎminus u119896120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u119896120576ℎ120591) minus 119889 (v
ℎminus u119896120576ℎ 119901119896
120576ℎ) ge (f v
ℎminus u119896120576ℎ)
119889 (u119896120576ℎ 119902ℎ) + 120576 (119901
119896
120576ℎ 119902ℎ) = 120576 (119901
119896minus1
120576ℎ 119902ℎ)
(28)
First we give the a priori estimate of the solution (u119896120576ℎ 119901119896
120576ℎ)
to the problem (28)
4 Abstract and Applied Analysis
Theorem 2 Suppose that (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) is the solution
to the problem (28) then it satisfies
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(29)
Proof Setting vℎ= 0 119902
ℎ= 1199010
120576ℎin (27) using (12) and Youngrsquos
inequality it yields that
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le (f u0120576ℎ) minus 119895 (u0
120576ℎ120591)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817119881
le120583
2
10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(30)
Then we have
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 2120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(31)
For 119896 = 1 2 setting vℎ= 0 119902
ℎ= 119901119896
120576ℎin (28) it yields that
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le (f u119896120576ℎ) minus 119895 (u119896
120576ℎ120591) + 120576 (119901
119896minus1
120576ℎ 119901119896
120576ℎ)
le120583
2
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+120576
2
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
(32)
Thus we obtain
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 12057610038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
le sdot sdot sdot le1198961205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 120576100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(33)
The next theorem gives the error estimate between thesolutions (u 119901) and (u119896
120576ℎ 119901119896
120576ℎ) to the problems (16) and (28)
respectively The proof can be found in [20]
Theorem 3 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
54+ 120576119896+1
) (34)
Next we will show the error estimate ||u minus u119896120576ℎ|| for
the penalty finite element approximation (28) This 1198712 erroranalysis is based on the regularity assumption that thefollowing linearized problem (35) is (1198672(Ω)2 1198671(Ω)) regular
Given z isin 1198712(Ω)2 find (w 120587) isin (119881119872) such that for all
(v 119902) isin (119881119872)
119886 (w v) + 119887 (u119896120576ℎ vw) + 119887 (v uw) minus 119889 (v 120587) = (z v)
119889 (w 119902) = 0(35)
According to (12) and (20) it is easy to verify that there existsa unique solution (w 120587) to the problem (35)The assumptionthat (35) is (1198672(Ω)2 1198671(Ω)) regular means that (w 120587) alsobelongs to (119867
2(Ω)2 1198671(Ω)) and the following inequality
holds
w2 + 1205871 le 119888 119911 (36)
Let 119868ℎbe the 1198712 orthogonal projections onto 119881
0ℎand satisfy
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881 le 119888ℎw2 (37)
Theorem 4 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
94+ 120576ℎ54
+ 120576119896+1
) (38)
Proof Setting z = uminusu119896120576ℎand v = uminusu119896
120576ℎin the first equation
of (35) we get
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w u minus u119896120576ℎ) + 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908)
+ 119887 (u minus u119896120576ℎ uw) minus 119889 (u minus u119896
120576ℎ 120587)
(39)
Taking v = uplusmn 119868ℎw 119902 = 119876
ℎ120587 in (16) and v
ℎ= u119896120576ℎplusmn 119868ℎw 119902ℎ=
119876ℎ120587 in (28) respectively we obtain
119886 (u 119868ℎw) + 119887 (u u 119868
ℎw) minus 119889 (119868
ℎw 119901) = (f 119868
ℎw)
119889 (u 119876ℎ120587) = 0
119886 (u119896120576ℎ 119868ℎw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119889 (119868
ℎw 119901119896120576ℎ) = (f 119868
ℎw)
119889 (u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896
120576ℎ 119876ℎ120587) = 120576 (119901
119896minus1
120576ℎ 119876ℎ120587)
(40)
Subtracting them we get
119886 (u minus u119896120576ℎ 119868ℎw) + 119887 (u u 119868
ℎw) minus 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw)
minus 119889 (119868ℎw 119901 minus 119901119896
120576ℎ) = 0
119889 (u minus u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587) = 0
(41)
Abstract and Applied Analysis 5
Substituting the previous equation into (39) it yields that
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w minus 119868ℎw u minus u119896
120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198691
+ 119887 (u119896120576ℎ u minus u119896
120576ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119887 (u minus u119896120576ℎ uw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119889 (119868ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198693
+ 120576 (119901119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198694
(42)
Using (34) (36) and (37) 1198691is estimated by
1198691= 119886 (w minus 119868
ℎw u minus u119896
120576ℎ)
le 12058310038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(43)
Similarly using (25) (34) (36) and (37) 1198693is estimated by
1198693= 119889 (119868
ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
le1003817100381710038171003817119868ℎw minus w1003817100381710038171003817119881
10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817120587 minus 119876ℎ1205871003817100381710038171003817
le 119888ℎ (ℎ54
+ 120576119896+1
) (w2 + 1205871)
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(44)
We rewrite 1198692as
1198692= 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎminus u 119868ℎw) minus 119887 (u minus u119896
120576ℎ u 119868ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw minus 119868
ℎw) + 119887 (u minus u119896
120576ℎ uw minus 119868
ℎw) (45)
Then from (13) (20) (29) (34) (36) and (37) it holds that
1198692le 119873(
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817119881+ u119881)
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119877ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(46)
Finally we estimate 1198694by
1198694= 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
= 120576 (119901119896minus1
120576ℎminus 119901119876
ℎ120587) + 120576 (119901 minus 119901
119896
120576ℎ 119876ℎ120587)
le 120576 (10038171003817100381710038171003817119901119896minus1
120576ℎminus 119901
10038171003817100381710038171003817+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817)1003817100381710038171003817119876ℎ120587
1003817100381710038171003817
le 119888120576 (ℎ54
+ 120576119896) 1205871
le 119888120576 (ℎ54
+ 120576119896)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(47)
Combining these estimates with (42) we conclude that (38)holds
4 Two-Level Iteration Penalty Methods
In this section based on the iteration penalty methoddescribed in the previous section the two-level iterationpenalty finite element methods for (16) are proposed in termsof the Stokes iteration Oseen iteration or Newtonian itera-tion From now on119867 and ℎ with ℎ lt 119867 are two real positiveparameters The coarse mesh triangulation T
119867is made as
in Section 3 And a fine mesh triangulation Tℎis generated
by a mesh refinement process to T119867 The conforming finite
element space pairs (119881ℎ119872ℎ) and (119881
119867119872119867) sub (119881
ℎ119872ℎ)
corresponding to the triangulationsTℎandT
119867 respectively
are constructed as in Section 3With the preavious notationswe propose the following two-level iteration finite elementmethods
41 Two-Level Stokes Iteration Penalty Method In Steps 1and 2 we solve (27) and (28) on the coarse mesh as in thefollwing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) such that for all
(v119867 119902119867) isin (119881
119867119872119867)
119886 (u0120576119867 v119867minus u0120576119867) + 119887 (u0
120576119867 u0120576119867 v119867minus u0120576119867) + 119895 (v
119867120591)
minus 119895 (u0120576119867120591
) minus 119889 (v119867minus u0120576119867 1199010
120576119867) ge (f v
119867minus u0120576119867)
119889 (u0120576119867 119902119867) + 120576 (119901
0
120576119867 119902119867) = 0
(48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) such that
for all (v119867 119902119867) isin (119881
119867119872119867)
119886 (u119896120576119867 v119867minus u119896120576119867) + 119887 (u119896
120576119867 u119896120576119867 v119867minus u119896120576119867) + 119895 (v
119867120591)
minus 119895 (u119896120576119867120591
) minus 119889 (v119867minus u119896120576119867 119901119896
120576119867) ge (f v
119867minus u119896120576119867)
119889 (u119896120576119867 119902119867) + 120576 (119901
119896
120576119867 119902119867) = 120576 (119901
119896minus1
120576119867 119902ℎ)
(49)
In Step 3 we solve a Stokes-type variational inequalityproblem on the fine mesh in terms of the Stokes iteration asin the following
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
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4 Abstract and Applied Analysis
Theorem 2 Suppose that (u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) is the solution
to the problem (28) then it satisfies
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(29)
Proof Setting vℎ= 0 119902
ℎ= 1199010
120576ℎin (27) using (12) and Youngrsquos
inequality it yields that
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le (f u0120576ℎ) minus 119895 (u0
120576ℎ120591)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817119881
le120583
2
10038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(30)
Then we have
12058310038171003817100381710038171003817u0120576ℎ
10038171003817100381710038171003817
2
119881+ 2120576
100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(31)
For 119896 = 1 2 setting vℎ= 0 119902
ℎ= 119901119896
120576ℎin (28) it yields that
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le (f u119896120576ℎ) minus 119895 (u119896
120576ℎ120591) + 120576 (119901
119896minus1
120576ℎ 119901119896
120576ℎ)
le120583
2
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+120576
2
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
(32)
Thus we obtain
12058310038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901119896
120576ℎ
10038171003817100381710038171003817
2
le1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 12057610038171003817100381710038171003817119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
le sdot sdot sdot le1198961205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+ 120576100381710038171003817100381710038171199010
120576ℎ
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(33)
The next theorem gives the error estimate between thesolutions (u 119901) and (u119896
120576ℎ 119901119896
120576ℎ) to the problems (16) and (28)
respectively The proof can be found in [20]
Theorem 3 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
54+ 120576119896+1
) (34)
Next we will show the error estimate ||u minus u119896120576ℎ|| for
the penalty finite element approximation (28) This 1198712 erroranalysis is based on the regularity assumption that thefollowing linearized problem (35) is (1198672(Ω)2 1198671(Ω)) regular
Given z isin 1198712(Ω)2 find (w 120587) isin (119881119872) such that for all
(v 119902) isin (119881119872)
119886 (w v) + 119887 (u119896120576ℎ vw) + 119887 (v uw) minus 119889 (v 120587) = (z v)
119889 (w 119902) = 0(35)
According to (12) and (20) it is easy to verify that there existsa unique solution (w 120587) to the problem (35)The assumptionthat (35) is (1198672(Ω)2 1198671(Ω)) regular means that (w 120587) alsobelongs to (119867
2(Ω)2 1198671(Ω)) and the following inequality
holds
w2 + 1205871 le 119888 119911 (36)
Let 119868ℎbe the 1198712 orthogonal projections onto 119881
0ℎand satisfy
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881 le 119888ℎw2 (37)
Theorem 4 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u119896120576ℎ 119901119896
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16) and
(28) respectively then they satisfy
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817le 119888 (ℎ
94+ 120576ℎ54
+ 120576119896+1
) (38)
Proof Setting z = uminusu119896120576ℎand v = uminusu119896
120576ℎin the first equation
of (35) we get
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w u minus u119896120576ℎ) + 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908)
+ 119887 (u minus u119896120576ℎ uw) minus 119889 (u minus u119896
120576ℎ 120587)
(39)
Taking v = uplusmn 119868ℎw 119902 = 119876
ℎ120587 in (16) and v
ℎ= u119896120576ℎplusmn 119868ℎw 119902ℎ=
119876ℎ120587 in (28) respectively we obtain
119886 (u 119868ℎw) + 119887 (u u 119868
ℎw) minus 119889 (119868
ℎw 119901) = (f 119868
ℎw)
119889 (u 119876ℎ120587) = 0
119886 (u119896120576ℎ 119868ℎw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119889 (119868
ℎw 119901119896120576ℎ) = (f 119868
ℎw)
119889 (u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896
120576ℎ 119876ℎ120587) = 120576 (119901
119896minus1
120576ℎ 119876ℎ120587)
(40)
Subtracting them we get
119886 (u minus u119896120576ℎ 119868ℎw) + 119887 (u u 119868
ℎw) minus 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw)
minus 119889 (119868ℎw 119901 minus 119901119896
120576ℎ) = 0
119889 (u minus u119896120576ℎ 119876ℎ120587) + 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587) = 0
(41)
Abstract and Applied Analysis 5
Substituting the previous equation into (39) it yields that
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w minus 119868ℎw u minus u119896
120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198691
+ 119887 (u119896120576ℎ u minus u119896
120576ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119887 (u minus u119896120576ℎ uw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119889 (119868ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198693
+ 120576 (119901119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198694
(42)
Using (34) (36) and (37) 1198691is estimated by
1198691= 119886 (w minus 119868
ℎw u minus u119896
120576ℎ)
le 12058310038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(43)
Similarly using (25) (34) (36) and (37) 1198693is estimated by
1198693= 119889 (119868
ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
le1003817100381710038171003817119868ℎw minus w1003817100381710038171003817119881
10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817120587 minus 119876ℎ1205871003817100381710038171003817
le 119888ℎ (ℎ54
+ 120576119896+1
) (w2 + 1205871)
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(44)
We rewrite 1198692as
1198692= 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎminus u 119868ℎw) minus 119887 (u minus u119896
120576ℎ u 119868ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw minus 119868
ℎw) + 119887 (u minus u119896
120576ℎ uw minus 119868
ℎw) (45)
Then from (13) (20) (29) (34) (36) and (37) it holds that
1198692le 119873(
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817119881+ u119881)
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119877ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(46)
Finally we estimate 1198694by
1198694= 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
= 120576 (119901119896minus1
120576ℎminus 119901119876
ℎ120587) + 120576 (119901 minus 119901
119896
120576ℎ 119876ℎ120587)
le 120576 (10038171003817100381710038171003817119901119896minus1
120576ℎminus 119901
10038171003817100381710038171003817+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817)1003817100381710038171003817119876ℎ120587
1003817100381710038171003817
le 119888120576 (ℎ54
+ 120576119896) 1205871
le 119888120576 (ℎ54
+ 120576119896)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(47)
Combining these estimates with (42) we conclude that (38)holds
4 Two-Level Iteration Penalty Methods
In this section based on the iteration penalty methoddescribed in the previous section the two-level iterationpenalty finite element methods for (16) are proposed in termsof the Stokes iteration Oseen iteration or Newtonian itera-tion From now on119867 and ℎ with ℎ lt 119867 are two real positiveparameters The coarse mesh triangulation T
119867is made as
in Section 3 And a fine mesh triangulation Tℎis generated
by a mesh refinement process to T119867 The conforming finite
element space pairs (119881ℎ119872ℎ) and (119881
119867119872119867) sub (119881
ℎ119872ℎ)
corresponding to the triangulationsTℎandT
119867 respectively
are constructed as in Section 3With the preavious notationswe propose the following two-level iteration finite elementmethods
41 Two-Level Stokes Iteration Penalty Method In Steps 1and 2 we solve (27) and (28) on the coarse mesh as in thefollwing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) such that for all
(v119867 119902119867) isin (119881
119867119872119867)
119886 (u0120576119867 v119867minus u0120576119867) + 119887 (u0
120576119867 u0120576119867 v119867minus u0120576119867) + 119895 (v
119867120591)
minus 119895 (u0120576119867120591
) minus 119889 (v119867minus u0120576119867 1199010
120576119867) ge (f v
119867minus u0120576119867)
119889 (u0120576119867 119902119867) + 120576 (119901
0
120576119867 119902119867) = 0
(48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) such that
for all (v119867 119902119867) isin (119881
119867119872119867)
119886 (u119896120576119867 v119867minus u119896120576119867) + 119887 (u119896
120576119867 u119896120576119867 v119867minus u119896120576119867) + 119895 (v
119867120591)
minus 119895 (u119896120576119867120591
) minus 119889 (v119867minus u119896120576119867 119901119896
120576119867) ge (f v
119867minus u119896120576119867)
119889 (u119896120576119867 119902119867) + 120576 (119901
119896
120576119867 119902119867) = 120576 (119901
119896minus1
120576119867 119902ℎ)
(49)
In Step 3 we solve a Stokes-type variational inequalityproblem on the fine mesh in terms of the Stokes iteration asin the following
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Substituting the previous equation into (39) it yields that
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
2
= 119886 (w minus 119868ℎw u minus u119896
120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198691
+ 119887 (u119896120576ℎ u minus u119896
120576ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119887 (u minus u119896120576ℎ uw) + 119887 (u119896
120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198692
+ 119889 (119868ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198693
+ 120576 (119901119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198694
(42)
Using (34) (36) and (37) 1198691is estimated by
1198691= 119886 (w minus 119868
ℎw u minus u119896
120576ℎ)
le 12058310038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119868ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(43)
Similarly using (25) (34) (36) and (37) 1198693is estimated by
1198693= 119889 (119868
ℎw minus w 119901 minus 119901119896
120576ℎ) minus 119889 (u minus u119896
120576ℎ 120587 minus 119876
ℎ120587)
le1003817100381710038171003817119868ℎw minus w1003817100381710038171003817119881
10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817120587 minus 119876ℎ1205871003817100381710038171003817
le 119888ℎ (ℎ54
+ 120576119896+1
) (w2 + 1205871)
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(44)
We rewrite 1198692as
1198692= 119887 (u119896
120576ℎ u minus u119896
120576ℎ 119908) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎ 119868ℎw) minus 119887 (u u 119868
ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw) + 119887 (u minus u119896
120576ℎ uw)
+ 119887 (u119896120576ℎ u119896120576ℎminus u 119868ℎw) minus 119887 (u minus u119896
120576ℎ u 119868ℎw)
= 119887 (u119896120576ℎ u minus u119896
120576ℎw minus 119868
ℎw) + 119887 (u minus u119896
120576ℎ uw minus 119868
ℎw) (45)
Then from (13) (20) (29) (34) (36) and (37) it holds that
1198692le 119873(
10038171003817100381710038171003817u119896120576ℎ
10038171003817100381710038171003817119881+ u119881)
10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817119881
1003817100381710038171003817w minus 119877ℎw1003817100381710038171003817119881
le 119888ℎ (ℎ54
+ 120576119896+1
) w2
le 119888ℎ (ℎ54
+ 120576119896+1
)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(46)
Finally we estimate 1198694by
1198694= 120576 (119901
119896minus1
120576ℎ 119876ℎ120587) minus 120576 (119901
119896
120576ℎ 119876ℎ120587)
= 120576 (119901119896minus1
120576ℎminus 119901119876
ℎ120587) + 120576 (119901 minus 119901
119896
120576ℎ 119876ℎ120587)
le 120576 (10038171003817100381710038171003817119901119896minus1
120576ℎminus 119901
10038171003817100381710038171003817+10038171003817100381710038171003817119901 minus 119901119896
120576ℎ
10038171003817100381710038171003817)1003817100381710038171003817119876ℎ120587
1003817100381710038171003817
le 119888120576 (ℎ54
+ 120576119896) 1205871
le 119888120576 (ℎ54
+ 120576119896)10038171003817100381710038171003817u minus u119896120576ℎ
10038171003817100381710038171003817
(47)
Combining these estimates with (42) we conclude that (38)holds
4 Two-Level Iteration Penalty Methods
In this section based on the iteration penalty methoddescribed in the previous section the two-level iterationpenalty finite element methods for (16) are proposed in termsof the Stokes iteration Oseen iteration or Newtonian itera-tion From now on119867 and ℎ with ℎ lt 119867 are two real positiveparameters The coarse mesh triangulation T
119867is made as
in Section 3 And a fine mesh triangulation Tℎis generated
by a mesh refinement process to T119867 The conforming finite
element space pairs (119881ℎ119872ℎ) and (119881
119867119872119867) sub (119881
ℎ119872ℎ)
corresponding to the triangulationsTℎandT
119867 respectively
are constructed as in Section 3With the preavious notationswe propose the following two-level iteration finite elementmethods
41 Two-Level Stokes Iteration Penalty Method In Steps 1and 2 we solve (27) and (28) on the coarse mesh as in thefollwing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) such that for all
(v119867 119902119867) isin (119881
119867119872119867)
119886 (u0120576119867 v119867minus u0120576119867) + 119887 (u0
120576119867 u0120576119867 v119867minus u0120576119867) + 119895 (v
119867120591)
minus 119895 (u0120576119867120591
) minus 119889 (v119867minus u0120576119867 1199010
120576119867) ge (f v
119867minus u0120576119867)
119889 (u0120576119867 119902119867) + 120576 (119901
0
120576119867 119902119867) = 0
(48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) such that
for all (v119867 119902119867) isin (119881
119867119872119867)
119886 (u119896120576119867 v119867minus u119896120576119867) + 119887 (u119896
120576119867 u119896120576119867 v119867minus u119896120576119867) + 119895 (v
119867120591)
minus 119895 (u119896120576119867120591
) minus 119889 (v119867minus u119896120576119867 119901119896
120576119867) ge (f v
119867minus u119896120576119867)
119889 (u119896120576119867 119902119867) + 120576 (119901
119896
120576119867 119902119867) = 120576 (119901
119896minus1
120576119867 119902ℎ)
(49)
In Step 3 we solve a Stokes-type variational inequalityproblem on the fine mesh in terms of the Stokes iteration asin the following
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(50)
As a direct consequence of Theorem 2 the solution(u119896120576119867 119901119896
120576119867) to the problem (49) satisfies
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881le radic
2119896 + 1
2
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) (51)
12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
le(2119896 + 1) 120581
2
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(52)
Next we estimateu120576ℎ Taking v
ℎ= 0 119902ℎ= 119901120576ℎin (50) it yields
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(53)
That is
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881leradic21205811
120583(f+1003817100381710038171003817119892
10038171003817100381710038171198712(119878))+
radic2119873
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881+radic120576
radic120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
(54)
Suppose that the initial data satisfies
71198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (55)
then using (51)-(52) we can estimate 119906120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
leradic21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
radic2
7
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
7119873
le (radic2 +radic2119896 + 1
7)1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) +
120583
12119873
le (radic2 +radic2119896 + 1
7)
radic2120583
7radic2119896 + 1119873
+120583
12119873
lt (2
7+2
49+1
7)120583
119873=23120583
49119873lt
120583
2119873
(56)
By the classical existence theorem for the variational inequal-ity problem of the second kind in the finite dimension [27]we have the following
Theorem 5 Under the uniqueness condition (55) there existsa unique solution (u
120576ℎ 119901120576ℎ) to the problem (50) Moreover u
120576ℎ
satisfy (56)
It follows fromTheorems 3 and 4 that (u119896120576119867 119901119896
120576119867) is of the
following error estimates10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817le 119888 (119867
54+ 120576119896+1
) (57)
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817le 119888 (119867
94+ 12057611986754
+ 120576119896+1
) (58)
Next we begin to prove the following error estimate for thesolution (u
120576ℎ 119901120576ℎ) to the problem (50)
Theorem 6 Suppose that the uniqueness condition (55) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (50)
respectively then they satisfy1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(59)
Proof Define a generalized bilinear form on (119881ℎ119872ℎ) times
(119881ℎ119872ℎ) by
B120576ℎ(uℎ 119901ℎ vℎ 119902ℎ) = 119886 (u
ℎ vℎ) minus 119889 (v
ℎ 119901ℎ)
+ 119889 (uℎ 119902ℎ) + 120576 (119901
ℎ 119902ℎ)
(60)
Taking vℎ= 119877ℎu 119902ℎ= 119901120576ℎminus 119876ℎ119901 in (50) we have
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= B120576ℎ(u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
= 119886 (u120576ℎ u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901120576ℎ)
+ 119889 (u120576ℎ 119901120576ℎminus 119876ℎ119901) + 120576 (119901
120576ℎ 119901120576ℎminus 119876ℎ119901)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
le (f u120576ℎminus 119877ℎu) + 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
+ 120576 (119901119896
120576119867 119901120576ℎminus 119876ℎ119901) + 119895 (119877
ℎu120591) minus 119895 (u
120576ℎ120591)
minusB120576ℎ(119877ℎu 119876ℎ119901 u120576ℎminus 119877ℎu 119901120576ℎminus 119876ℎ119901)
(61)
Let v = u120576ℎand v = 2u minus 119877
ℎu in the first inequality of (16)
then
119886 (u u120576ℎminus u) + 119887 (u u u
120576ℎminus 119906) + 119895 (u
120576ℎ120591)
minus 119895 (u120591) minus 119889 (u
120576ℎminus u 119901) ge (f u
120576ℎminus u)
119886 (u u minus 119877ℎu) + 119887 (u u u minus 119877
ℎu) + 119895 (2u
120591minus 119877ℎu120591)
minus 119895 (u120591) minus 119889 (u minus 119877
ℎu 119901) ge (f u minus 119877
ℎu)
(62)
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
Adding the above two inequalities gives
(f u120576ℎminus 119877ℎu) le 119886 (u u
120576ℎminus 119877ℎu)
+ 119887 (u u u120576ℎminus 119877ℎu) minus 119889 (u
120576ℎminus 119877ℎu 119901)
+ 119895 (2u120591minus 119877ℎu120591) + 119895 (u
120576ℎ120591) minus 2119895 (u
120591)
(63)
Substituting the above inequality into (61) it yields that
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198682
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(64)
It follows fromHolderrsquos inequality andYoungrsquos inequality that
1198681= 119886 (u minus 119877
ℎu u120576ℎminus 119877ℎu) le 1205831003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 2120583
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
1198683= 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901) minus 119889 (u
120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901)
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 +
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817119881
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
le120583
8
1003817100381710038171003817u120576ℎ minus 119877ℎu1003817100381710038171003817
2
119881+ 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1
120583
1003817100381710038171003817119901 minus 119876ℎ1199011003817100381710038171003817
2
+1
41205782
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817
2
119881
(65)
where 120578 gt 0 is some small constant determined later Werewrite 119868
2as
1198682= 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ) + 119887 (u u119896
120576119867minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u119896
120576119867minus 119906 119877
ℎu minus u120576ℎ)
(66)
Then using (13) (15) and Youngrsquos inequality we can estimate1198682by
1198682le 119873u2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
+ 11987310038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038174
119881)
(67)
It is easy to show that
1198684= 120576 (119901
119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
le 12057610038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
41205782
10038171003817100381710038171003817119901119896
120576119867minus 119876ℎ11990110038171003817100381710038171003817
2
le 12057821003817100381710038171003817119901120576ℎ minus 119876ℎ119901
1003817100381710038171003817
2
+1205762
21205782(10038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817
2
+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
2
)
(68)
Finally from triangle inequality 1198685is estimated by
1198685= 119895 (2u
120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)
le 2int119878
1198921003816100381610038161003816u120591 minus 119877ℎu120591
1003816100381610038161003816 119889119904
le 2100381710038171003817100381711989210038171003817100381710038171198712(119878)
1003817100381710038171003817u120591 minus 119877ℎu12059110038171003817100381710038171198712(119878)
(69)
Substituting (65)ndash(69) into (64) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881
le1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817119881
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +1003817100381710038171003817u120591 minus 119877ℎu120591
1003817100381710038171003817
12
1198712(119878)
+ 12057610038171003817100381710038171003817119901 minus 119901119896
120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
le2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817 + 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(70)
where we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all w
ℎisin 1198810ℎ let v = u plusmn w
ℎin (16) and
vℎ= u120576ℎplusmn wℎin (50) respectively Then we get
119886 (uwℎ) + 119887 (u uw
ℎ) minus 119889 (w
ℎ 119901) = (f w
ℎ)
119886 (u120576ℎwℎ) + 119887 (u119896
120576119867 u119896120576119867wℎ) minus 119889 (w
ℎ 119901120576ℎ) = (f w
ℎ)
(71)
Subtracting them and using (13) (15) we obtain
119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ) minus 119887 (u119896
120576119867 u119896120576119867wℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u u minus u119896
120576119867wℎ)
+ 119887 (u minus u119896120576119867 uwℎ) minus 119887 (u minus u119896
120576119867 u minus u119896
120576119867wℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(72)
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
Therefore it follows from (23) that ||119876ℎ119901 minus 119901
120576ℎ|| can be
estimated by
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 + 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119873u210038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881
le 1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(73)
If we choose 120578 = 1205814radic120583 such that (2120578radic120583)sdot(120583120581) = 12 thensubstituting (73) into (70) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (74)
From (73) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (75)
Thus we complete the proof of (59)
42 Two-Level Oseen Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve an Oseen type variational inequalityproblem on the fine mesh in terms of the Oseen iteration asin the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ) + 119895 (v
ℎ120591)
minus 119895 (u120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ) ge (f v
ℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(76)
From (12) it is easy to show that the solution (u120576ℎ 119901120576ℎ) to
the problem (76) satisfies
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881+ 120576
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
2120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
(77)
Suppose that the initial data satisfies
2radic4119896 + 21198731205811
1205832(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (78)
then using (52) we can estimate u120576ℎby
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le radic120576
120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le (radic2119896 + 1
2+ 1)
1205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))
le
radic4119896 + 21205811
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) le
120583
2119873
(79)
For two-level Oseen iteration penalty method the solu-tion (u
120576ℎ 119901120576ℎ) is of the following error estimate
Theorem 7 Suppose that the uniqueness condition (78) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (76)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(80)
Proof Proceeding as in the proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198686
minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
(81)
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198686 Using
(12) (13) (15) and Youngrsquos inequality we have
1198686= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 u120576ℎminus 119877ℎu 119877ℎu minus u120576ℎ)
+ 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
= 119887 (u119896120576119867 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (u119896120576119867minus u u 119877
ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
+ 11987311990621003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
10038171003817100381710038171003817u119896120576119867minus u10038171003817100381710038171003817
le120583
8
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
10038171003817100381710038171003817u119896120576119867minus u100381710038171003817100381710038172
+1003817100381710038171003817119877ℎu minus u1003817100381710038171003817
2
119881)
(82)
Then substituting (65) (68) (69) and (82) into (81) it yieldsthat
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
2120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(83)
In (83) we use (24)ndash(26) and (57)-(58) Next we estimate||119901120576ℎminus 119876ℎ119901|| For all 119908
ℎisin 1198810ℎ proceeding as in the proof
of (72) from (51) and (78) we can show119889 (wℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u uw
ℎ)
minus 119887 (u119896120576119867 u120576ℎwℎ)
= 119886 (u minus u120576ℎwℎ) + 119887 (u119896
120576119867 u minus u
120576ℎwℎ)
+ 119887 (u minus u119896120576119867 uwℎ)
le (1205831003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
+11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817wℎ
1003817100381710038171003817119881
le (5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817)1003817100381710038171003817wℎ
1003817100381710038171003817119881
(84)
It follows from (23) and (84) that1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119889 (wℎ 119876ℎ119901 minus 119901) + 119889 (w
ℎ 119901 minus 119901
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817 +5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119873u2
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
le5120583
4
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+ 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
)
(85)
If we choose 120578 = 1205815radic120583 such that (2120578radic120583) sdot (51205834120581) = 12then substituting (85) into (83) we show
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (86)
From (85) again we obtain
1003817100381710038171003817119901 minus 119901120576ℎ1003817100381710038171003817 le 119888 (ℎ
54+ 12057611986754
+ 11986794
+ 120576119896+1
) (87)
Thus we complete the proof of (80)
43 Two-Level Newton Iteration Penalty Method In Steps 1and 2 we solve (48) and (49) on the coarse mesh as in thefollowing
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
In Step 3 we solve a linearized Navier-Stokes typevariational inequality problem on the fine mesh in terms ofthe Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u120576ℎ vℎminus u120576ℎ) + 119887 (u119896
120576119867 u120576ℎ vℎminus u120576ℎ)
+ 119887 (u120576ℎ u119896120576119867 vℎminus u120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u
120576ℎ120591) minus 119889 (v
ℎminus u120576ℎ 119901120576ℎ)
ge (f vℎminus u120576ℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎminus u120576ℎ)
119889 (u120576ℎ 119902ℎ) + 120576 (119901
120576ℎ 119902ℎ) = 120576 (119901
119896
120576119867 119902ℎ)
(88)
In this section we will suppose that the initial datasatisfies
81198731205811
1205832radic2119896 + 1
2(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878)) lt 1 (89)
Then from (51) u119896120576119867
satisfies ||u119896120576119867||119881
le 1205838119873 Let vℎ=
0 119902ℎ= 119901120576ℎin (88) Using (12) we obtain
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ) + 120576
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
le (f u120576ℎ) minus 119895 (u
120576ℎ120591) minus 119887 (u119896
120576119867 u119896120576119867 u120576ℎ) + 120576 (119901
119896
120576119867 119901120576ℎ)
le 1205811(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))1003817100381710038171003817u120576ℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
+ 12057610038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
le120583
2
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881+120576
2
1003817100381710038171003817119901120576ℎ1003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817
2
+1205812
1
120583(f + 1003817100381710038171003817119892
10038171003817100381710038171198712(119878))2
+1198732
120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
4
119881
(90)
Since
1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119887 (u119896
120576119867 u120576ℎ u120576ℎ)
ge 1205831003817100381710038171003817u120576ℎ
1003817100381710038171003817
2
119881minus 119873
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881ge7120583
8
1003817100381710038171003817u120576ℎ1003817100381710038171003817
2
119881
(91)
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
then
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
le21205811
radic3120583
(f + 100381710038171003817100381711989210038171003817100381710038171198712(119878)
) +2radic120576
radic3120583
10038171003817100381710038171003817119901119896
120576119867
10038171003817100381710038171003817+2119873
radic3120583
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817
2
119881
leradic2
4radic3
1
radic2119896 + 1
120583
119873+
1
4radic3
10038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881+
120583
4radic3119873
lt (1
4+1
55+1
6)120583
119873=
574120583
1320119873lt
120583
2119873
(92)
Theorem 8 Suppose that the uniqueness condition (89) holdsLet (u 119901) isin 119867
3(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and (u
120576ℎ 119901120576ℎ) isin
(119881ℎ119872ℎ) be the solutions to the problems (16) and (88)
respectively then they satisfy
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(93)Proof Proceeding as the in proof of (64) we can get
1205831003817100381710038171003817u120576ℎ minus 119877ℎu
1003817100381710038171003817
2
119881+ 120576
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
2
= 119886 (u minus 119877ℎu u120576ℎminus 119877ℎu)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198681
+ minus119889 (u120576ℎminus 119877ℎu 119901 minus 119876
ℎ119901) + 119889 (u minus 119877
ℎu 119901120576ℎminus 119876ℎ119901)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198683
+ 120576 (119901119896
120576119867minus 119876ℎ119901 119901120576ℎminus 119876ℎ119901)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198684
+ 119895 (2u120591minus 119877ℎu120591) minus 2119895 (u
120591) + 119895 (119877
ℎu120591)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198685
+ 119887 (u119896120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ) minus 119887 (u u 119877
ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198687
(94)
In the above equation 1198681 1198683 1198684 and 119868
5have been estimated
in the proof of Theorem 6 Here we only estimate 1198687 We
rewrite 1198687as
1198687= 119887 (u119896
120576119867 u120576ℎ 119877ℎu minus u120576ℎ) + 119887 (u
120576ℎ u119896120576119867 119877ℎu minus u120576ℎ)
minus 119887 (u u 119877ℎu minus u120576ℎ) minus 119887 (u119896
120576119867 u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus u u 119877
ℎu minus u120576ℎ) + 119887 (u
120576ℎ u120576ℎminus u 119877
ℎu minus u120576ℎ)
minus 119887 (u120576ℎminus u119896120576119867 u120576ℎminus u119896120576119867 119877ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198688
+ 119887 (u120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
1198689
+ 119887 (119877ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986810
+ 119887 (u119896120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
11986811
(95)
From (13) (20) and Youngrsquos inequality 1198688is estimated by
1198688= 119887 (u
120576ℎminus 119877ℎu u 119877
ℎu minus u120576ℎ) + 119887 (119877
ℎu minus u u 119877
ℎu minus u120576ℎ)
le 119873u1198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+ 119873u119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
le120583
2
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 4120583
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(96)
Using (13) and (24) we estimate 1198689by
1198689= 119887 (u
120576ℎ 119877ℎu minus u 119877
ℎu minus u120576ℎ)
= 119887 (u120576ℎminus 119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
+ 119887 (119877ℎu 119877ℎu minus u 119877
ℎu minus u120576ℎ)
le 1198731003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
+ 1198731003817100381710038171003817119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u10038171003817100381710038171198811003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817119881
le 1198621ℎ21003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881+120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888
1003817100381710038171003817119877ℎu minus u10038171003817100381710038172
119881
(97)
where 1198621gt 0 is independent of ℎ 119867 and 120576 Similarly it
follows from (13) (24) and (57) that
11986810= 119887 (119877
ℎu minus u120576ℎ 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
le 119873(1003817100381710038171003817119877ℎu minus u1003817100381710038171003817119881 +
10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
le 1198622(ℎ2+ 11986754
+ 120576119896+1
)1003817100381710038171003817119877ℎu minus u
120576ℎ
1003817100381710038171003817
2
119881
(98)
where 1198622gt 0 is independent of ℎ 119867 and 120576 Finally we can
estimate 11986811by
11986811= 119887 (u119896
120576119867minus 119877ℎu 119877ℎu minus u119896120576119867 119877ℎu minus u120576ℎ)
le 11987310038171003817100381710038171003817119877ℎu minus u119896120576119867
10038171003817100381710038171003817
2
119881
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817119881
le120583
32
1003817100381710038171003817119877ℎu minus u120576ℎ
1003817100381710038171003817
2
119881+ 119888 (
1003817100381710038171003817119877ℎu minus u10038171003817100381710038174
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
4
119881)
(99)
For sufficiently small ℎ 119867 and 120576 such that 1198621ℎ2+ 1198622(ℎ2+
11986754
+ 120576119896+1
) = 132 substituting (65) (68) (69) and (95)ndash(99) into (94) it yields that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881le
4120578
radic120583
1003817100381710038171003817119901120576ℎ minus 119876ℎ1199011003817100381710038171003817
+ 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(100)
For all 119908ℎisin 1198810ℎ proceeding as in the proof of (72) we can
show119889 (vℎ 119901 minus 119901
120576ℎ)
= 119886 (u minus u120576ℎ vℎ) + 119887 (u u v
ℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ)
minus 119887 (u120576ℎ u119896120576119867 vℎ) + 119887 (u119896
120576119867 u119896120576119867 vℎ)
(101)
Since
119887 (u u vℎ) minus 119887 (u119896
120576119867 u120576ℎ vℎ) minus 119887 (119906
120576ℎ u119896120576119867 vℎ)
+ 119887 (u119896120576119867 u119896120576119867 vℎ)
= 119887 (u minus u120576ℎ u vℎ) + 119887 (u u minus 119877
ℎu Vh)
minus 119887 (u minus u120576ℎ u minus 119877
ℎu vℎ)
+ 119887 (u120576ℎminus u119896120576119867 119877ℎu minus u119896120576119867 vℎ)
minus 119887 (u119896120576119867 u120576ℎminus 119877ℎu vℎ)
le (119873u1198811003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+ 119873u119881
1003817100381710038171003817u minus 119877ℎu1003817100381710038171003817119881
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ (1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)
times (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576119867
10038171003817100381710038171003817119881(1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
le 1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
1003817100381710038171003817vℎ1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)1003817100381710038171003817vℎ
1003817100381710038171003817119881
(102)
where 1198623gt 0 is independent of ℎ 119867 and 120576 then from (23)
we have
1205811003817100381710038171003817119876ℎ119901 minus 119901120576ℎ
1003817100381710038171003817 le 1198623 (1 + ℎ2+ 1198672+ 120576119896+1
)1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881
+ 119888 (1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119876ℎ119901
1003817100381710038171003817
+1003817100381710038171003817u minus 119877ℎu
1003817100381710038171003817
2
119881+10038171003817100381710038171003817u minus u119896120576119867
10038171003817100381710038171003817
2
119881)
(103)
Thus for sufficiently small ℎ 119867 120576 and 120578 such that
1198623(1 + ℎ
2+ 1198672+ 120576119896+1
)
120581
4120578
radic120583lt1
2 (104)
substituting (103) into (100) we obtain1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881le 119888 (ℎ
54+ 12057611986754
+ 11986752
+ 120576119896+2
) (105)
From (103) again we obtain1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
) (106)
Thus we complete the proof of (93)
Remark 9 In terms of Theorems 6 7 and 8 if we choose120576 = 119874(119867) 119867 = 119874(ℎ
59) for the two-level Stokes or Oseen
iteration penalty methods and 120576 = 119874(11986754) 119867 = 119874(ℎ12) for
the two-level Newton iteration penalty method then1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (107)
44 An Improved Scheme In this section we will propose ascheme to improve the error estimates derived in Theorems6ndash8 which is described as follows
In Steps 1 and 2 we solve (48) and (49) on the coarsemesh as in the following
Step 1 Find (u0120576119867 1199010
120576119867) isin (119881
119867119872119867) by (48)
Step 2 For 119896 = 1 2 find (u119896120576119867 119901119896
120576119867) isin (119881
119867119872119867) by (49)
At Step 3 we solve a linearized problem (50) or (76) or(88) on the fine mesh in terms of Stokes iteration or Oseeniteration or Newton iteration as in the following
Step 3 Find (u120576ℎ 119901120576ℎ) isin (119881
ℎ119872ℎ) by (50) or (76) or (88)
At Step 4 we solve a Newton correction of (u120576ℎ 119901120576ℎ)
on the fine mesh in terms of Newton iteration as in thefollowing
Step 4 Find (u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) such that for all (v
ℎ 119902ℎ) isin
(119881ℎ119872ℎ)
119886 (u⋆120576ℎ vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u⋆120576ℎ vℎminus u⋆120576ℎ)
+ 119887 (u⋆120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
+ 119895 (vℎ120591) minus 119895 (u⋆
120576ℎ120591) minus 119889 (v
ℎminus u⋆120576ℎ 119901⋆
120576ℎ)
ge (f vℎminus u⋆120576ℎ) + 119887 (u
120576ℎ u120576ℎ vℎminus u⋆120576ℎ)
119889 (u⋆120576ℎ 119902ℎ) + 120576 (119901
⋆
120576ℎ 119902ℎ) = 120576 (119901
120576ℎ 119902ℎ)
(108)
First we show the following theorem
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Abstract and Applied Analysis
Theorem 10 Let (u119896120576ℎ 119901119896
120576ℎ) and (u⋆
120576ℎ 119901⋆
120576ℎ) be the solutions of
(28) and (108) respectively Then there holds that10038171003817100381710038171003817u119896120576ℎminus u⋆120576ℎ
10038171003817100381710038171003817119881+10038171003817100381710038171003817119901119896
120576ℎminus 119901⋆
120576ℎ
10038171003817100381710038171003817
le 119888 (10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881+ 12057612 10038171003817100381710038171003817
119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817)
(109)
where (u120576ℎ 119901120576ℎ) is the solution to the problem (50) or (76) or
(88)
Proof Under the uniqueness condition (89) the solution u120576ℎ
satisfies ||u120576ℎ||119881le 1205832119873 Taking v
ℎ= u⋆120576ℎ 119902ℎ= 119901⋆
120576ℎminus 119901119896
120576ℎin
(28) and vℎ= u119896120576ℎ 119902ℎ= 119901119896
120576ℎminus 119901⋆
120576ℎin (108) and adding them
yield
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎminus 119901119896
120576ℎ) + 119887 (u119896
120576ℎ u119896120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u120576ℎ u⋆120576ℎ u⋆120576ℎminus u119896120576ℎ)
minus 119887 (u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ) + 119887 (u
120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
= 120576 (119901120576ℎminus 119901119896minus1
120576ℎ 119901⋆
120576ℎ) + 119887 (u
120576ℎminusu119896120576ℎ u120576ℎminusu119896120576ℎ u⋆120576ℎminusu119896120576ℎ)
+ 119887 (u119896120576ℎminus u⋆120576ℎ u120576ℎ u⋆120576ℎminus u119896120576ℎ)
(110)
Using (13) Holderrsquos inequality and Youngrsquos inequality weobtain
12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+ 120576
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
le 12057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817+ 119873
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
times10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
le120576
2
10038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
2
+120576
2
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817
2
+120583
2
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881
+120583
4
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881+1198732
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
4
119881
(111)
That is
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881le radic
2120576
120583
10038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+2119873
120583
10038171003817100381710038171003817u120576ℎminus u119896120576ℎ
10038171003817100381710038171003817
2
119881(112)
For all wℎisin 1198810ℎ taking v
ℎ= u119896120576ℎplusmn wℎin the first inequality
of (28) and vℎ= u⋆120576ℎplusmn wℎin the first inequality of (108) it
yields that
119886 (u119896120576ℎwℎ) + 119887 (u119896
120576ℎ u119896120576ℎwℎ) minus 119889 (w
ℎ 119901119896
120576ℎ) = (f w
ℎ)
119886 (u⋆120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ) + 119887 (u⋆
120576ℎ u120576ℎwℎ)
minus 119889 (wℎ 119901⋆
120576ℎ) = (f w
ℎ) + 119887 (u
120576ℎ u120576ℎwℎ)
(113)
1
10
Γ
Γ
x
y
S2
S1
Figure 1 Domain Ω
Subtracting them and using (23) (112) we obtain
12058110038171003817100381710038171003817119901⋆
120576ℎminus 119901119896
120576ℎ
10038171003817100381710038171003817
le supwℎisin1198810ℎ
119889 (wℎ 119901⋆
120576ℎminus 119901119896
120576ℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u
120576ℎ u⋆120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u⋆120576ℎ u120576ℎwℎ)minus119887 (u119896
120576ℎ u119896120576ℎwℎ)minus119887 (u
120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
= supwℎisin1198810ℎ
119886 (u⋆120576ℎminus u119896120576ℎwℎ) + 119887 (u⋆
120576ℎminus u119896120576ℎ u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
+ supwℎisin1198810ℎ
119887 (u120576ℎ u⋆120576ℎminus u119896120576ℎwℎ)minus119887 (u119896
120576ℎminus u120576ℎ u119896120576ℎminus u120576ℎwℎ)
1003817100381710038171003817wℎ1003817100381710038171003817119881
le 12058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 2119873
1003817100381710038171003817u120576ℎ1003817100381710038171003817119881
10038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881
+ 11987310038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 212058310038171003817100381710038171003817u⋆120576ℎminus u119896120576ℎ
10038171003817100381710038171003817119881+ 119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
le 2radic212058312057610038171003817100381710038171003817119901120576ℎminus 119901119896minus1
120576ℎ
10038171003817100381710038171003817+ 5119873
10038171003817100381710038171003817u119896120576ℎminus u120576ℎ
10038171003817100381710038171003817
2
119881
(114)
From (34) andTheorems 6ndash10 we get the following errorestimates
Theorem 11 Let (u 119901) isin 1198673(Ω)2cap 119881 times 119867
2(Ω) cap 119872 and
(u⋆120576ℎ 119901⋆
120576ℎ) isin (119881
ℎ119872ℎ) be the solutions to the problems (16)
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 13
IsoValue IsoValue IsoValueIsoValue000054865600016460500027434500038408500049382400060356400071330400082304400093278300104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
005015025035045055065075085095
minus085minus095
minus075minus065minus055minus045minus035minus025minus015minus005
Figure 2 Streamline of flow and pressure contour for exact solution
and (108) respectively Then for the two-level Stokes or Oseeniteration penalty methods they satisfy
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 11986792
+ 1205763211986754
+ 1205761211986794
+ 120576119896+12
)
(115)
And for the two-level Newton iteration penalty method theysatisfy1003817100381710038171003817u minus u⋆
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817
le 119888 (ℎ54
+ 1198675+ 1205763211986754
+ 1205761211986752
+ 120576119896+12
)
(116)
Remark 12 If we choose 119867 = 119874(ℎ518
) 120576 = 119874(ℎ54) in (115)
for two-level Stokes or Oseen iteration penalty methods and119867 = 119874(ℎ
14) 120576 = 119874(ℎ
54) in (116) for the two-level Newton
iteration penalty method then we obtain
1003817100381710038171003817u minus u⋆120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901
⋆
120576ℎ
1003817100381710038171003817 le 119888ℎ54 (117)
5 Numerical Results
In this section we will give numerical results to confirm theerror analysis obtained in Section 4 Since these two-levelStokesOseenNewton iteration penalty methods are given in
the form of the variational inequality problems which arenot directly solved the appropriate iteration algorithm mustbe constructed Here we use the Uzawa iteration algorithmintroduced in [28]
For simplicity we only give the Uzawa iterationmethod for solving the variational inequality problem(16) Similar schemes can be used to solve the two-level StokesOseenNewton iteration penalty schemes inSection 4 First there exists a multiplier 120582 isin Λ such thatthe variational inequality problem (16) is equivalent to thefollowing variational identity problem
119886 (u v) + 119887 (u u v) minus 119889 (v 119901) + int119878
120582119892v120591119889119904 = (f v)
forallv isin 119881
119889 (u 119902) = 0 forall119902 isin 119872
120582119906120591=1003816100381610038161003816119906120591
1003816100381610038161003816 ae on 119878
(118)
where 120582 isin Λ = 120574 isin 1198712(119878) |120574(119909)| le 1 ae on 119878 In this
case we can solve the problem (16) by the following Uzawaiteration scheme
1205820isin Λ is given (119)
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Abstract and Applied Analysis
000054869600016460900027434800038408700049382700060356600071330500082304400093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499787014997802499780349977044997705499760649975074997508499740949974
IsoValue IsoValue IsoValueIsoValueminus0950016minus0850017minus0750017minus0650018minus0550018minus0450019minus0350019minus025002minus015002minus00500208
Figure 3 Streamline of flow and pressure contour by Stokes method
then 120582119899 is known we compute (u119899 119901119899) and 120582119899+1 by
119886 (u119899 v) + 119887 (u119899 u119899 v) minus 119889 (v 119901119899)
= (f v) minus int119878
120582119899119892v120591119889119904 forallv isin 119881
119889 (u119899 119902) = 0 forall119902 isin 119872
120582119899+1
= 119875Λ(120582119899+ 120588119892u119899
120591) 120588 gt 0
(120)
where
119875Λ(120574) = sup (minus1 inf (1 120574)) forall120574 isin 119871
2(119878) (121)
Consider the problems (1)-(2) in the fixed square domain(0 1) times (0 1) (see Figure 1) Let 120583 = 01 The external force 119891is chosen such that the exact solution (u 119901) is
u (119909 119910) = (1199061(119909 119910) 119906
2(119909 119910))
119901 (119909 119910) = (2119909 minus 1) (2119910 minus 1)
1199061(119909 119910) = minus119909
2119910 (119909 minus 1) (3119910 minus 2)
1199062(119909 119910) = 119909119910
2(119910 minus 1) (3119909 minus 2)
(122)
It is easy to verify that the exact solution u satisfies u = 0
on Γ u sdotn = 1199061= 0 119906
2= 0 on 119878
1and 1199061= 0 u sdotn = 119906
2= 0 on
1198782 Moreover the tangential vector 120591 on 119878
1and 119878
2are (0 1)
and (minus1 0) Thus we have
120590120591= 4120583119910
2(119910 minus 1) on 119878
1
120590120591= 4120583119909
2(119909 minus 1) on 119878
2
(123)
On the other hand from the nonlinear slip boundary condi-tions (2) there holds that
10038161003816100381610038161205901205911003816100381610038161003816 le 119892 (124)
then the function 119892 can be chosen as 119892 = minus120590120591ge 0 on 119878
1and
1198782In all experiments we choose 120583 = 01 iteration initial
value 1205820 = 1 and 120588 = 1205832 In terms of Theorems 6 and 7 forthe two-level StokesOseen penalty iteration methods thereholds that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986794
+ 120576119896+1
)
(125)
Then we choose 120576 = 119874(119867) 119867 = 119874(ℎ59) 119896 = 2 such that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (126)
We pick eight coarse mesh size values that is 119867 =
14 16 18 118 In Table 1 the scaling between 1119867
and 1ℎ = (1119867)95 is given
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 15
000054869600016460900027434800038408800049382700060356600071330500082304500093278400104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
00499791014997902499780349977044997705499760649976074997508499750949974
IsoValue IsoValue IsoValue IsoValueminus0950015minus0850016minus0750017minus0650017minus0550018minus0450018minus0350019minus0250019minus015002minus00500204
Figure 4 Streamline of flow and pressure contour by Oseen method
000054869700016460900027434900038408800049382800060356700071330700082304600093278500104252
00115226001262001371740014814800159122001700960018107001920440020301800213992
0049993601499930249992034999104499905499890649988074998708499860949986
IsoValue IsoValue IsoValue IsoValueminus0949997minus0849998minus0749999minus065minus0550001minus0450002minus0350003minus0250004minus0150005minus00500055
Figure 5 Streamline of flow and pressure contour by Newton method
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Abstract and Applied Analysis
Table 1 Comparison of the scaling between 1119867 and 1ℎ
1119867 4 6 8 10 12 14 16 181ℎ 12125 25157 42224 63095 87604 115619 147033 181756
Table 2 Numerical relative error for velocity with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 104374119890 minus 03 184283119890 minus 04 152977119890 minus 04 152631119890 minus 04
Oseen 104345119890 minus 03 178281119890 minus 04 145638119890 minus 04 145275119890 minus 04
Table 3 Numerical relative error for pressure with 119867 = 114 andℎ = 1115
1205760
001 0001 00001 000001Stokes 185222119890 minus 04 620827119890 minus 05 596578119890 minus 05 596412119890 minus 05
Oseen 185112119890 minus 04 613519119890 minus 05 588604119890 minus 05 588398119890 minus 05
Table 4 Numerical relative error for Stokes method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 151517119890 minus 02 541171119890 minus 03 2 02816 25 330825119890 minus 03 125062119890 minus 03 2 08898 42 115862119890 minus 03 444992119890 minus 04 2 230210 63 525908119890 minus 04 199251119890 minus 04 2 503812 87 291391119890 minus 04 105862119890 minus 04 2 980614 115 184283119890 minus 04 620827119890 minus 05 2 1762116 147 130602119890 minus 04 396161119890 minus 05 2 4054918 181 101944119890 minus 04 277123119890 minus 05 2 53445Order 1727 1913
Setting 120576 = 1205760119867 the comparison of relative error
||u minus u120576ℎ||119881||u||119881
and ||119901 minus 119901120576ℎ||||119901|| for different 120576
0gt
0 is shown in Tables 2 and 3 when we use the two-levelStokesOseen penalty iteration methods with 1119867 = 14 and1ℎ = 115 We can see that for our present testing case itsuffices to set 120576 = 0001119867 if it is hoped to be as large aspossible
Thus set 120576 = 0001119867 and 1ℎ asymp (1119867)95 Tables
4 and 5 display the relative 1198671 errors of the velocity andthe relative 119871
2 errors of the pressure and their averageconvergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseeniteration penalty method respectively Based on Tables 4 and5 the two-level StokesOseen iteration penalty methods canreach the convergence orders of 119874(ℎ54) for both velocityand pressure in1198671- and 1198712-norms respectively as shown in(126)
Next we give the numerical results by using the two-levelNewton iteration penalty method In terms of Theorem 8there holds that
1003817100381710038171003817u minus u120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888 (ℎ54
+ 12057611986754
+ 11986752
+ 120576119896+2
)
(127)
Table 5 Numerical relative error for Oseen method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 12 149411119890 minus 02 538449119890 minus 03 2 03166 25 323052119890 minus 03 124245119890 minus 03 2 10068 42 112567119890 minus 03 441332119890 minus 04 2 261310 63 508693119890 minus 04 193710119890 minus 04 2 572812 87 281455119890 minus 04 104715119890 minus 04 2 1121314 115 178281119890 minus 04 613519119890 minus 05 2 2004516 147 126872119890 minus 04 391207119890 minus 05 2 4224218 181 995916119890 minus 05 273676119890 minus 05 2 59391Order 1728 1915
Table 6 Numerical relative error for Newton method
1119867 1ℎ u minus u120576ℎ119881u119881
119901 minus 119901120576ℎ119901 Iteration CPU(s)
4 16 810332119890 minus 03 302474119890 minus 03 2 05356 36 151165119890 minus 03 598186119890 minus 04 2 22568 64 476536119890 minus 04 189983119890 minus 04 2 699110 100 208297119890 minus 04 802333119890 minus 05 2 1697712 144 110901119890 minus 04 389471119890 minus 05 2 3780414 196 720875119890 minus 05 221184119890 minus 05 2 78811Order 1811 1947
Then we choose 120576 = 00111986754 and 1ℎ = (1119867)2 such that1003817100381710038171003817u minus u
120576ℎ
1003817100381710038171003817119881+1003817100381710038171003817119901 minus 119901120576ℎ
1003817100381710038171003817 le 119888ℎ54 (128)
Because when 119867 = 116 and ℎ = 1256 this method doesnot work and the computer displays ldquoout of memoryrdquo Thusin this experiment we pick six coarse mesh size values thatis 119867 = 14 16 114 Table 6 displays the relative 1198671errors of the velocity and the relative 1198712 errors of the pressureand their average convergence orders and CPU time whenwe use the two-level Newton iteration penalty method Basedon Tables 4 and 5 we can see that the two-level Newtoniteration penalty method also reaches the convergence ordersof 119874(ℎ54) for both velocity and pressure in 119867
1- and 1198712-
norms respectively as shown in (128)Figures 2 3 4 and 5 show the streamline of flow and
the pressure contour of the numerical solution by the two-level StokesOseenNewton iteration penalty methods andthe exact solution respectively
Acknowledgments
This work is supported by the National Natural ScienceFoundation ofChina underGrant nos 10901122 11001205 andby Zhejiang Provincial Natural Science Foundation of Chinaunder Grant no LY12A01015
References
[1] H Fujita ldquoFlow Problems with Unilateral Boundary condi-tionsrdquo Leccons Colleege de France 1993
[2] H Fujita ldquoA mathematical analysis of motions of viscousincompressible fluid under leak or slip boundary conditionsrdquoRIMS Kokyuroku no 888 pp 199ndash216 1994
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 17
[3] H Fujita ldquoNon-stationary Stokes flows under leak boundaryconditions of friction typerdquo Journal of Computational Mathe-matics vol 19 no 1 pp 1ndash8 2001
[4] H Fujita ldquoA coherent analysis of Stokes flows under boundaryconditions of friction typerdquo Journal of Computational andApplied Mathematics vol 149 no 1 pp 57ndash69 2002
[5] Y Li and K Li ldquoExistence of the solution to stationary Navier-Stokes equations with nonlinear slip boundary conditionsrdquoJournal of Mathematical Analysis and Applications vol 381 no1 pp 1ndash9 2011
[6] Y Li and K Li ldquoGlobal strong solutions of two-dimensionalNavier-Stokes equations with nonlinear slip boundary condi-tionsrdquo Journal of Mathematical Analysis and Applications vol393 no 1 pp 1ndash13 2012
[7] N Saito and H Fujita ldquoRegularity of solutions to the Stokesequation under a certain nonlinear boundary condition TheNavier-Stokes Equationsrdquo in Proceedings of the 2nd Interna-tional Conference on the Navier Stokes-Equations Theory andNumericalMethods vol 223 ofDekker Lecture Notes in Pure andApplied Mathematics pp 73ndash86 2001
[8] N Saito ldquoOn the Stokes equation with the leak and slipboundary conditions of friction type regularity of solutionsrdquoPublications of the Research Institute for Mathematical SciencesKyoto University vol 40 no 2 pp 345ndash383 2004
[9] Y He and J Li ldquoConvergence of three iterative methods basedon the finite element discretization for the stationary Navier-Stokes equationsrdquo Computer Methods in Applied Mechanics andEngineering vol 198 no 15-16 pp 1351ndash1359 2009
[10] R An and H L Qiu ldquoTwo-Level Newton iteration methodsfor Navier-Stokes type variational inequality problemrdquo TheAdvances in Applied Mathematics and Mechanics vol 5 no 1pp 36ndash54 2013
[11] V Girault and J-L Lions ldquoTwo-grid finite-element schemesfor the steady Navier-Stokes problem in polyhedrardquo PortugaliaeMathematica vol 58 no 1 pp 25ndash57 2001
[12] Y He and K Li ldquoTwo-level stabilized finite element methodsfor the steady Navier-Stokes problemrdquo Computing vol 74 no4 pp 337ndash351 2005
[13] W Layton and H W J Lenferink ldquoA multilevel mesh indepen-dence principle for the Navier-Stokes equationsrdquo SIAM Journalon Numerical Analysis vol 33 no 1 pp 17ndash30 1996
[14] Y Li and R An ldquoTwo-level pressure projection finite ele-ment methods for Navier-Stokes equations with nonlinear slipboundary conditionsrdquo Applied Numerical Mathematics vol 61no 3 pp 285ndash297 2011
[15] K Li and Y Hou ldquoAn AIM and one-step Newton methodfor the Navier-Stokes equationsrdquo Computer Methods in AppliedMechanics and Engineering vol 190 no 46-47 pp 6141ndash61552001
[16] M Marion and J Xu ldquoError estimates on a new nonlinearGalerkin method based on two-grid finite elementsrdquo SIAMJournal onNumerical Analysis vol 32 no 4 pp 1170ndash1184 1995
[17] J Xu ldquoTwo-grid discretization techniques for linear and non-linear PDEsrdquo SIAM Journal on Numerical Analysis vol 33 no5 pp 1759ndash1777 1996
[18] R Temam Navier-Stokes Equations Theory and Numericalanalysis AMS Chelsea Providence RI USA 2001
[19] Y Li and R An ldquoPenalty finite element method for Navier-Stokes equations with nonlinear slip boundary conditionsrdquoInternational Journal for Numerical Methods in Fluids vol 69no 3 pp 550ndash566 2012
[20] X X Dai P P Tang and M H Wu ldquoAnalysis of an iterativepenalty method for Navier-Stokes equations with nonlinearslip boundary conditionsrdquo International Journal for NumericalMethods in Fluids vol 72 no 4 pp 403ndash413 2013
[21] X L Cheng and A W Shaikh ldquoAnalysis of the iterative penaltymethod for the Stokes equationsrdquo Applied Mathematics Lettersvol 19 no 10 pp 1024ndash1028 2006
[22] X Dai ldquoFinite element approximation of the pure Neumannproblem using the iterative penalty methodrdquo Applied Mathe-matics and Computation vol 186 no 2 pp 1367ndash1373 2007
[23] J Shen ldquoOn error estimates of the penalty method for unsteadyNavier-Stokes equationsrdquo SIAM Journal on Numerical Analysisvol 32 no 2 pp 386ndash403 1995
[24] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquo Numerische Mathematik vol 20 pp 179ndash192 197273
[25] F Brezzi ldquoOn the existence uniqueness and approximationof saddle-point problems arising from Lagrangian multipli-ersrdquo Revue Francaise drsquoAutomatique Informatique RechercheOperationnelle vol 8 no 2 pp 129ndash151 1974
[26] S C Brenner and L R Scott The Mathematical Theory ofFinite Element Methods vol 15 of Texts in Applied MathematicsSpringer New York NY USA 2nd edition 2002
[27] R Glowinski Numerical Methods for Nonlinear VariationalProblems Scientific Computation Springer Berlin Germany2008
[28] Y Li and K Li ldquoUzawa iteration method for Stokes typevariational inequality of the second kindrdquo Acta MathematicaeApplicatae Sinica vol 27 no 2 pp 303ndash315 2011
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of