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Hindawi Publishing CorporationInternational Journal of Differential EquationsVolume 2013 Article ID 532987 13 pageshttpdxdoiorg1011552013532987
Research ArticleAnalysis of Mixed Elliptic and Parabolic BoundaryLayers with Corners
Gung-Min Gie1 Chang-Yeol Jung2 and Roger Temam3
1 Department of Mathematics University of California Riverside 900 University Avenue Riverside CA 92521 USA2Ulsan National Institute of Science and Technology San 194 Banyeon-ri Eonyang-eup Ulju Gun Ulsan 689-798 Republic of Korea3The Institute for Scientific Computing and Applied Mathematics Indiana University 831 East Third StreetBloomington IN 47405 USA
Correspondence should be addressed to Gung-Min Gie ggiemathucredu
Received 27 January 2013 Accepted 1 April 2013
Academic Editor Norio Yoshida
Copyright copy 2013 Gung-Min Gie et al 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
We study the asymptotic behavior at small diffusivity of the solutions 119906120576 to a convection-diffusion equation in a rectangular domainΩ The diffusive equation is supplemented with a Dirichlet boundary condition which is smooth along the edges and continuousat the corners To resolve the discrepancy on 120597Ω between 119906
120576 and the corresponding limit solution 1199060 we propose asymptoticexpansions of 119906120576 at any arbitrary but fixed order In order to manage some singular effects near the four corners ofΩ the so-calledelliptic and ordinary corner correctors are added in the asymptotic expansions as well as the parabolic and classical boundary layerfunctionsThen performing the energy estimates on the difference of 119906120576 and the proposed expansions the validity of our asymptoticexpansions is established in suitable Sobolev spaces
1 Introduction
We consider a singularly perturbed convection-diffusionequation in a rectangular domain Ω = (0 1) times (0 1)
119871120576119906120576
= minus120576Δ119906120576
minus 120597119909119906120576
= 119891 in Ω
119906120576
= 119892 on 120597Ω
(1)
Here 120576 is a small but strictly positive diffusivity parameterand 119891 = 119891(119909 119910) is a given smooth data with 119863
120572
1198911198712(Ω)
le
120581120572 independent of 120576 for some 120572rsquos as needed in the analysis
below The function 119892 = 119892(119909 119910) is assumed to be continuouson 120597Ω and smooth on each edge of 120597Ω Namely defining therestriction of 119892 to the edges of 120597Ω as follows
1198921003816100381610038161003816119909=0
= 1198921(119910) 119892
1003816100381610038161003816119909=1
= 1198922(119910)
1198921003816100381610038161003816119910=0
= 1198923(119909) 119892
1003816100381610038161003816119910=1
= 1198924(119909)
(2)
we assume that
1198921(0) = 119892
3(0) 119892
1(1) = 119892
4(0)
1198922(0) = 119892
3(1) 119892
2(1) = 119892
4(1)
119892119894is smooth for 1 le 119894 le 4
(3)
If these compatibility conditions were not satisfied someadditional considerations would be necessary which we donot address here
In what follows we study the asymptotic behavior of thesolutions to (1) at small diffusivity 120576
In a very nice related earlier work [1] the asymptoticbehavior of the solutions of a convection-diffusion equationsimilar to (1) was discussed More precisely in a rectangulardomain Ω = (0 119897
0) times (0 119897
1) the authors considered
minus1205761015840
Δ1199061205761015840
minus 1198871205971199091199061205761015840
+ 1198881199061205761015840
=119891 in Ω
1199061205761015840
= 119892 on 120597Ω
(4)
2 International Journal of Differential Equations
where 119887 gt 0 119888 ge 0 are constants and 119891 is a given smooth
function in ΩThe function 119892 satisfying the analogue versionof (3) in Ω is assumed to be continuous and piecewisesmooth on 120597Ω By constructing in [1] asymptotic expansionsof119906120576
1015840
with respect to a small diffusivity 1205761015840 the boundary layers
of (4) were analyzed in a systematic way The validity of theirasymptotic expansions was established using the maximumprinciple
Using a simple change of variables which maps Ω toΩ and setting 119906
1205761015840
= 119906120576
119890120582119909 with a suitable 120582 one can
transform (4) to (1) where the two diffusivity parameters120576 and 120576
1015840 respectively in (1) and (4) are compatible thatis 1205761205761015840 is of order one Hence via this transformation ouranalysis of (1) in this paper is applicable to (4) as well Ourmotivations for conducting the present study appear below Inparticular we significantly simplify the proofs of [1] andmakethe study suitable for more general equations or systemsin particular those which do not satisfy the maximumprinciple
In the boundary layer analysis associated with theconvection-diffusion equation (1) (or (4)) one of the mostimportant and difficult points is to resolve any possiblesingularity near the four corners of the rectangular domainΩTowards this end we simplify and improve the method usedin [1] More precisely concerning the asymptotic expansionof 119906120576 we introduce the so-called elliptic boundary layer
corrector appearing in Section 32 near the inflow boundaryat 119909 = 1 only while in [1] it was used near both theinflow and outflow boundaries at 119909 = 1 and 119909 = 0This simplification in the construction of the asymptoticexpansions is mainly based on an observation that the cornersingularities at (0 0) and (0 1) (where the flows go out)can be handled by the ordinary corner corrector appearingin Section 34 (which is much easier to analyze than theelliptic corrector) Using the energy estimates rather thanthe maximum principle as in [1] we prove the validityof our proposed asymptotic expansions in suitable Sobolevspaces Here we make use of the Hardy inequality (seeeg [2]) in the estimates As we said above our energyestimation approach can be easily extended to some higherorder equations or systems which do not admit a maximumprinciple
This paper is organized as follows In Section 2 weintroduce a formal expansion of119906120576 as the sumof the outer andinner expansionsThe outer expansion (outside of the bound-ary layers) which is easy to obtain appears in Section 2Thenin Section 3 by performing the boundary layer analysis in asystematic way we construct the inner expansion (inside theboundary layers) In Section 4 we state and prove the mainconvergence result Theorem 9 concerning the differencebetween 119906
120576 and the proposed asymptotic expansion at a givenorder In addition in the appendix we prove Lemma 1 thatcontains some delicate pointwise estimates on the parabolicboundary layer correctors introduced in Section 31 Thestudy of the elliptic corrector and its approximation alsoappears in the appendix
Throughout this paper we use the notation conventionsbelow
Notation 1 120581 = 120581(119891 119892 Ω 119899) is a constant depending onthe data but independent of 120576 here 119899 ge 0 is the fixed (butarbitrary) order of the asymptotic expansion as it appears in(6)
Notation 2 An est is a function (or a constant) whose normin all Sobolev spaces119867119904 (and thus spaces119862119904) is exponentiallysmall with for each 119904 a bound of the form 119888
1119890minus1198882120576120574
1198881 1198882 120574 gt
0 119888119894 120574 depending possibly on 119904
Notation 3 For a fixed 120576 gt 0 we define the energy norm of1198671
(Ω)
sdot120576= radic120576 sdot
1198671(Ω)
+ sdot1198712(Ω)
(5)
2 Asymptotic Expansions
To study the asymptotic behavior of 119906120576 solution of (1) we
propose an asymptotic expansion 119906120576 of the following type
119906120576
cong
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) (6)
Here at each order of 120576119895 119895 ge 0 119906119895 corresponds to the outerexpansion (outside of the boundary layer) of 119906120576 To balancethe discrepancy on the boundary 120597Ω of 119906120576 and of the 119906119895 119895 ge 0we introduce the correctors Θ
119895 119895 ge 0 which will contributemainly inside of the boundary layerΘ119895 will be itself the sumof several boundary layer functions as we will see below Theasymptotic expansion (6) is made at order 119899 ge 0 As we willsee below the expansion itself depends on 119899 but 119899 which isset at the beginning of the study can be chosen arbitrary large
To determine the asymptotic expansion (6) we start withthe outer expansion for 119906
120576 119906120576
cong sum119899
119895=0120576119895
119906119895 Inserting this
expansion into (1) we formally find the equations for the 119906119895
minus1205971199091199060
= 119891
minus120597119909119906119895
= Δ119906119895minus1
1 le 119895 le 119899
(7)
We supplement these equations with the following inflowboundary conditions (see (1)ndash(3))
1199060
= 1198922(119910) at 119909 = 1
119906119895
= 0 at 119909 = 1 1 le 119895 le 119899
(8)
While (7) are ldquonaturalrdquo the choice of the boundary conditionsin (8) is not obvious it will be eventually justified bythe convergence theorem below Then by integrating theequations in 119909 we find the smooth outer solutions 119906
119895 in thefollowing form
1199060
= int
1
119909
119891 (1199091 119910) 119889119909
1+ 1198922(119910) (9)
119906119895
= int
1
119909
Δ119906119895minus1
(1199091 119910) 119889119909
1 1 le 119895 le 119899 (10)
International Journal of Differential Equations 3
Under this construction of the outer expansion we noticethat119906120576 cong sum
119899
119895=0120576119895
119906119895 satisfies the boundary condition (1)
2along
the edge 119909 = 1 only and not on the three other sides of120597Ω 119910 = 0 119910 = 1 and 119909 = 0 (in general) Hence weexpect boundary layers to occur near those edges and the so-called correctors (corresponding to the Θ
119895) will be necessaryto account for these discrepancies
In Section 3 performing the boundary layer analysis wedefine for each 0 le 119895 le 119899 the corrector Θ
119895 appearing in(11) belowThen in Section 4 using (6) (9) (10) and (11) weprove our main result which is stated as Theorem 9
3 Construction of the CorrectorsBoundary Layers Analysis
Wewant the boundary value of 119906120576 tomatch that of its approx-imation In general the boundary values of the diffusivesolution 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 match only at
119909 = 1 but not on the other sides of 120597Ω 119910 = 0 119910 = 1 and119909 = 0 Hence we expect that some boundary layers will occurnear those three edges To resolve this inconsistency we willdefine a number of boundary layer functions on the sides andcorners of 120597Ω following an analysis reminiscent of the theoryof the Prandtl boundary layer in [3ndash6]
For the moment comparing 119906120576 with a finite sum
sum119899
119895=0120576119895
119906119895 we infer from the definition of the 119906
119895 in (7)ndash(10)that the values of these functions coincide on the segment119909 = 1 In general they will not coincide anywhere else on120597Ω First to correct the inconsistency on the sides 119910 = 0 1we will define the so-called parabolic correctors 120593
119895
= 120593119895
119897+
120593119895
119906 Then to correct the inconsistency at 119909 = 0 we will
introduce the (classical) corrector 120579119895 However this will not
be enough because additional inconsistencies appear at thecorners ofΩ and as we shall see below it will be necessary tointroduce additional correctors 120585
119895 120577119895 and 120578119895 to handle these
inconsistenciesIn summary in the subsections below we will define the
Θ119895 in the form
Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
0 le 119895 le 119899 (11)
where the role and location of each corrector are explainedrespectively in Box 1 and Figure 1 The stretched variables ofthe corrector functions which will be introduced and usedin the following subsections are indicated in the followingequation
120593119895
119897= 120593119895
119897(119909
119910
radic120576
) 120593119895
119906= 120593119895
119906(119909
(1 minus 119910)
radic120576
)
120585119895
119897= 120585119895
119897(
(1 minus 119909)
120576
119910
120576
) 120585119895
119906= 120585119895
119906(
(1 minus 119909)
120576
(1 minus 119910)
120576
)
120579119895
= 120579119895
(
119909
120576
119910)
120577119895
119897= 120577119895
119897(
119909
120576
119910
radic120576
) 120577119895
119906= 120577119895
119906(
119909
120576
(1 minus 119910)
120576
)
120578119895
119897= 120578119895
119897(
119909
120576
119910) 120578119895
119906= 120578119895
119906(
119909
120576
119910)
(12)
31 Parabolic Boundary Layers (PBL) At the bottom (or top)boundary that is at 119910 = 0 (or 119910 = 1) to balance thediscrepancy of 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 (see
the 119906119895 as defined in (9) and (10)) we construct below the
parabolic boundary layer correctors 120593119895 0 le 119895 le 119899We formally insert the asymptotic expansion 119906
120576
cong
sum119899
119895=0120576119895
120593119895 into (1) Using the arguments similar to those of
the Prandtl theory we see that the thickness of the boundarylayer near 119910 = 0 or 119910 = 1 should be of order 120576
12 Then bycollecting the terms at the same order of 120576119895 0 le 119895 le 119899 we findthe equations of the parabolic boundary layer correctors
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω 0 le 119895 le 119899 (13)
where we set 120593minus1 = 0Along the boundaries 119910 = 0 119910 = 1 and 119909 = 1 comparing
(1)2and the boundary values of the functions in (9) and (10)
we find the boundary conditions of 120593119895 0 le 119895 le 119899 which read
120593119895
= 0 at 119909 = 1 0 le 119895 le 119899
1205930
= 1198923(119909) minus 119906
0
(119909 0) at 119910 = 0
120593119895
= minus119906119895
(119909 0) at 119910 = 0 1 le 119895 le 119899
1205930
= 1198924(119909) minus 119906
0
(119909 1) at 119910 = 1
120593119895
= minus119906119895
(119909 1) at 119910 = 1 1 le 119895 le 119899
(14)
We will show below how to actually construct 120593119895 For themoment comparing the functions 119906
120576 and sum119899
119895=0120576119895
(119906119895
minus 120593119895
)
for some 119899 ge 0 we see that the boundary values of thesefunctions coincide at 119909 = 1 and 119910 = 0 1 However thishas been achieved at the price of introducing some newinconsistencies Indeed we notice in (14) that for each 119895 ge 0the boundary conditions at119910 = 0 and119910 = 1 are not consistentwith that at 119909 = 1 That is in general for any 119894 ge 1
120597119894
1199091199060
(1 0) minus 120597119894
1199091198923(1) = 0 120597
119894
1199091199060
(1 1) minus 120597119894
1199091198924(1) = 0
120597119894
119909119906119895
(1 0) = 0 120597119894
119909119906119895
(1 1) = 0 119895 ge 1
(15)
Due to this inconsistency of the boundary data if we define120593119895 0 le 119895 le 119899 as a solution of (13) with (14) some derivatives
of 120593119895 0 le 119895 le 119899 get singular at the two corners (1 119896) 119896 = 0 1Therefore to define smooth (smoother) correctors120593119895 0 le 119895 le
119899 we will modify the boundary conditions (14) as introducedin (23)
4 International Journal of Differential Equations
120593119895
= 120593119895
119897+ 120593119895
119906is the parabolic boundary layer corrector near 119910 = 0 and 119910 = 1
120585119895
= 120585119895
119897+ 120585119895
119906is the elliptic corrector which resolves the compatibility issues in the construction of 120593119895 at the corners (1 0) and (1 1)
120579119895 is the ordinary boundary layer corrector near 119909 = 0
120577119895
= 120577119895
119897+ 120577119895
119906is the ordinary corner layer corrector which manages the effect of 120593119895 along 119909 = 0
120578119895
= 120578119895
119897+ 120578119895
119906is the supplementary corrector that manages the effects of 120579119895 and 120577
119895 along the edges 119910 = 0 and 119910 = 1
Box 1 The different correctors
119910
1
0 1 119909
120577119895
119906 120578119895
119906
120577119895
119897 120578119895
119897
120579119895
119906119895
120593119895
119906
120593119895
119897
120585119895
119906
120585119895
119897
Figure 1 Location of the outer solution 119906119895 and the boundary layer
correctors
311 On Modification of Boundary Conditions for SmoothCorrectors at 119910 = 0 1 Equation (13) with initial and bound-ary conditions (14) is a parabolic problem (heat equation)in which minus119909 is the (positive) time like variable so that theboundary conditions at 119909 = 1 are the analogue of theinitial conditions for an evolution problem and the possibleinconsistencies in (15) correspond to the (analogue of ) thecompatibility conditions between the initial and boundaryvalues for a parabolic equation see [7ndash9] and the referencesthereinDue to the lack of consistency at the corners (1 0) and(1 1) the correctors constructed as in (13) and (14) might getsingular derivatives at those corners We will overcome thisdifficulty by considering some corner functions at 119909 = 1 and119910 = 0 1 similar to what is done in [10 11] or [12] for thecompatibility issue in parabolic problems see [13] as well
Defining 120575(119909) as a smooth cut-off function independentof 120576 such that
120575 (119909) =
0 for 0 le 119909 le
1
2
1 for 119909 ge
3
4
(16)
we set
120574119895
119896(119909) = 120574
119895
119896(119909) 120575 (119909) 119896 = 0 1 (17)
where
1205740
119896(119909) =
2119899+1
sum
119894=1
(119909 minus 1)119894
119894
[120597119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896)]
120574119895
119896(119909) = minus
2119899+1minus2119895
sum
119894=1
(119909 minus 1)119894
119894
120597119894
119909119906119895
(1 119896) 1 le 119895 le 119899
(18)
Note that the 1205740
119896 119896 = 0 1 depend on 119899 but this dependency
is not made explicit to make the notations (slightly) simplerNote also (comparing to (15)) that at 119909 = 1 for each 119896 = 0 1
120574119895
119896(1) = 0 0 le 119895 le 119899
120597119894
1199091205740
119896(1) = 120597
119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896) 1 le 119894 le 2119899 + 1
120597119894
119909120574119895
119896(1) = minus120597
119894
119909119906119895
(1 119896) 1 le 119895 le 119899 1 le 119894 le 2119899 + 1 minus 119895
(19)
We introduce the stretched variables
119910 =
119910
radic120576
119910 =
(1 minus 119910)
radic120576
(20)
Then using (13) (14) (17) and (20) we define the paraboliccorrector 120593119895 0 le 119895 le 119899 as
120593119895
= 120593119895
119897+ 120593119895
119906= 120593119895
119897(119909 119910) + 120593
119895
119906(119909 119910) (21)
where 120593119895
119897and 120593
119895
119906(lower and upper parts of 120593
119895 resp) aredefined as the solutions of the 1D heat equations below
For 0 le 119895 le 119899
minus1205972
119910120593119895
119897minus 120597119909120593119895
119897= 1205972
119909120593119895minus1
119897for (119910 119909) isin R
+
times (0 1)
120593119895
119897= ℎ119895
0(119909) at 119910 = 0
120593119895
119897997888rarr 0 as 119910 997888rarr infin
120593119895
119897= 0 at 119909 = 1
minus1205972
119910120593119895
119906minus 120597119909120593119895
119906= 1205972
119909120593119895minus1
119906for (119910 119909) isin R
+
times (0 1)
120593119895
119906= ℎ119895
1(119909) at 119910 = 0
120593119895
119906997888rarr 0 as 119910 997888rarr infin
120593119895
119906= 0 at 119909 = 1
(22)
Here the ℎ119895
119896(119909) 0 le 119895 le 119899 119896 = 0 1 are defined by
ℎ0
0(119909) = 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909)
ℎ119895
0(119909) = minus119906
119895
(119909 0) minus 120574119895
0(119909) 1 le 119895 le 119899
ℎ0
1(119909) = 119892
4(119909) minus 119906
0
(119909 1) minus 1205740
1(119909)
ℎ119895
1(119909) = minus119906
119895
(119909 1) minus 120574119895
1(119909) 1 le 119895 le 119899
(23)
International Journal of Differential Equations 5
Note that the smooth boundary conditions ℎ119895
119896(119909) 0 le 119895 le 119899
119896 = 0 1 along 119910 or 119910 = 0 are compatible with the other 0boundary (initial) conditions at 119909 = 1 in the sense that foreach 0 le 119895 le 119899
120597119894
119909ℎ119895
119896(1) = 0 119896 = 0 1 0 le 119894 le 2119899 + 1 minus 2119895 (24)
From [14ndash16] or [1] we recall the explicit expressions of120593119895
119897= 120593119895
119897(119909 119910) 0 le 119895 le 119899
1205930
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ0
0(119909 +
1199102
21199102
1
) 1198891199101 (25)
120593119895
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ119895
0(119909 +
1199102
21199102
1
)1198891199101
+
1
2radic120587
int
1minus119909
0
int
infin
0
1
radic1199091
exp(minus
(119910 minus 1199101)2
41199091
)
minus exp(minus
(119910 + 1199101)2
41199091
)
times 1205972
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
(26)
The expressions of 120593119895
119906= 120593119895
119906(119909 119910) 0 le 119895 le 119899 are identical
to (25) and (26) with ℎ0
0 ℎ1198950 and 119910 respectively replaced by
ℎ0
1 ℎ119895
1and 119910
312 Estimates on the Parabolic Boundary Layers We nowstate some pointwise estimates for the 120593
119895
119897and 120593
119895
119906 0 le 119895 le 119899
which are proved in the appendix
Lemma 1 For each 0 le 119895 le 119899 and 119904 ge 0 one has the followingpointwise estimates
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
yradic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(27)
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119906
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
1 minus 119910
radic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(28)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
From Lemma 1 it is easy to deduce the following 119871119901
estimates
Lemma 2 For each 0 le 119895 le 119899 and 119904 ge 0 one has for 0 le
119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119906
10038171003817100381710038171003817119871119901(Ω)
120581120576(119904minus119898)2+12119901
(29)
Using (21) and (22) we find the equations for120593119895 = 120593119895
119897+120593119895
119906
0 le 119895 le 119899
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω
120593119895
= ℎ119895
0(119909) + 120593
119895
119906
10038161003816100381610038161003816119910=0
at 119910 = 0
120593119895
= ℎ119895
1(119909) + 120593
119895
119897
10038161003816100381610038161003816119910=1
at 119910 = 1
120593119895
= 0 at 119909 = 1
(30)
Thanks to Lemma 1 we notice that
120593119895
119906
10038161003816100381610038161003816119910=0
= est 120593119895
119897
10038161003816100381610038161003816119910=1
= est (31)
where est denotes an exponentially small term with respectto the (small) parameter 120576 as defined in Notation 2
32 Elliptic Boundary Layers (EBL) In Section 31 to con-struct the consistent parabolic boundary layer correctors 120593
1198950 le 119895 le 119899 we considered the modified boundary conditions(23) including 120574
119895
119896 instead of the natural boundary conditions
(14) Here on the two sides 119910 = 0 and 119910 = 1 ofΩ to cancel 120574119895119896
which is exactly the difference of (14) and (23) we introduceelliptic boundary layer correctors 120585119895 0 le 119895 le 119899 which like 120574
119895
119896
or 120574119895
119896 depend on the order 119899 of the asymptotic expansion in
(6)Using (17) we define
120585119895
= 120585119895
119897+ 120585119895
119906 0 le 119895 le 119899 (32)
where 120585119895
119897and 120585119895
119906are the solutions of the elliptic systems below
For 0 le 119895 le 119899
minus120576 1205972
119909120585119895
119897minus 1205761205972
119910120585119895
119897minus 120597119909120585119895
119897= 0 in Ω
120585119895
119897= 120574119895
0(119909) at 119910 = 0
120585119895
119897= 0 at 119909 = 0 1 or 119910 = 1
minus120576 1205972
119909120585119895
119906minus 120576 1205972
119910120585119895
119906minus 120597119909120585119895
119906= 0 in Ω
120585119895
119906= 120574119895
1(119909) at 119910 = 1
120585119895
119906= 0 at 119909 = 0 1 or 119910 = 0
(33)
The systems (33) being elliptic equations their well-posedness are easy to verify and hence we omit the proofhere
The equations satisfied by the elliptic correctors 120585119895 = 120585119895
119897+
120585119895
119906read as followsFor 0 le 119895 le 119899
minus1205761205972
119909120585119895
minus 1205761205972
119910120585119895
minus 120597119909120585119895
= 0 in Ω
120585119895
= 120574119895
119896(119909) at 119910 = 119896 119896 = 0 1
120585119895
= 0 at 119909 = 0 1
(34)
6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 International Journal of Differential Equations
where 119887 gt 0 119888 ge 0 are constants and 119891 is a given smooth
function in ΩThe function 119892 satisfying the analogue versionof (3) in Ω is assumed to be continuous and piecewisesmooth on 120597Ω By constructing in [1] asymptotic expansionsof119906120576
1015840
with respect to a small diffusivity 1205761015840 the boundary layers
of (4) were analyzed in a systematic way The validity of theirasymptotic expansions was established using the maximumprinciple
Using a simple change of variables which maps Ω toΩ and setting 119906
1205761015840
= 119906120576
119890120582119909 with a suitable 120582 one can
transform (4) to (1) where the two diffusivity parameters120576 and 120576
1015840 respectively in (1) and (4) are compatible thatis 1205761205761015840 is of order one Hence via this transformation ouranalysis of (1) in this paper is applicable to (4) as well Ourmotivations for conducting the present study appear below Inparticular we significantly simplify the proofs of [1] andmakethe study suitable for more general equations or systemsin particular those which do not satisfy the maximumprinciple
In the boundary layer analysis associated with theconvection-diffusion equation (1) (or (4)) one of the mostimportant and difficult points is to resolve any possiblesingularity near the four corners of the rectangular domainΩTowards this end we simplify and improve the method usedin [1] More precisely concerning the asymptotic expansionof 119906120576 we introduce the so-called elliptic boundary layer
corrector appearing in Section 32 near the inflow boundaryat 119909 = 1 only while in [1] it was used near both theinflow and outflow boundaries at 119909 = 1 and 119909 = 0This simplification in the construction of the asymptoticexpansions is mainly based on an observation that the cornersingularities at (0 0) and (0 1) (where the flows go out)can be handled by the ordinary corner corrector appearingin Section 34 (which is much easier to analyze than theelliptic corrector) Using the energy estimates rather thanthe maximum principle as in [1] we prove the validityof our proposed asymptotic expansions in suitable Sobolevspaces Here we make use of the Hardy inequality (seeeg [2]) in the estimates As we said above our energyestimation approach can be easily extended to some higherorder equations or systems which do not admit a maximumprinciple
This paper is organized as follows In Section 2 weintroduce a formal expansion of119906120576 as the sumof the outer andinner expansionsThe outer expansion (outside of the bound-ary layers) which is easy to obtain appears in Section 2Thenin Section 3 by performing the boundary layer analysis in asystematic way we construct the inner expansion (inside theboundary layers) In Section 4 we state and prove the mainconvergence result Theorem 9 concerning the differencebetween 119906
120576 and the proposed asymptotic expansion at a givenorder In addition in the appendix we prove Lemma 1 thatcontains some delicate pointwise estimates on the parabolicboundary layer correctors introduced in Section 31 Thestudy of the elliptic corrector and its approximation alsoappears in the appendix
Throughout this paper we use the notation conventionsbelow
Notation 1 120581 = 120581(119891 119892 Ω 119899) is a constant depending onthe data but independent of 120576 here 119899 ge 0 is the fixed (butarbitrary) order of the asymptotic expansion as it appears in(6)
Notation 2 An est is a function (or a constant) whose normin all Sobolev spaces119867119904 (and thus spaces119862119904) is exponentiallysmall with for each 119904 a bound of the form 119888
1119890minus1198882120576120574
1198881 1198882 120574 gt
0 119888119894 120574 depending possibly on 119904
Notation 3 For a fixed 120576 gt 0 we define the energy norm of1198671
(Ω)
sdot120576= radic120576 sdot
1198671(Ω)
+ sdot1198712(Ω)
(5)
2 Asymptotic Expansions
To study the asymptotic behavior of 119906120576 solution of (1) we
propose an asymptotic expansion 119906120576 of the following type
119906120576
cong
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) (6)
Here at each order of 120576119895 119895 ge 0 119906119895 corresponds to the outerexpansion (outside of the boundary layer) of 119906120576 To balancethe discrepancy on the boundary 120597Ω of 119906120576 and of the 119906119895 119895 ge 0we introduce the correctors Θ
119895 119895 ge 0 which will contributemainly inside of the boundary layerΘ119895 will be itself the sumof several boundary layer functions as we will see below Theasymptotic expansion (6) is made at order 119899 ge 0 As we willsee below the expansion itself depends on 119899 but 119899 which isset at the beginning of the study can be chosen arbitrary large
To determine the asymptotic expansion (6) we start withthe outer expansion for 119906
120576 119906120576
cong sum119899
119895=0120576119895
119906119895 Inserting this
expansion into (1) we formally find the equations for the 119906119895
minus1205971199091199060
= 119891
minus120597119909119906119895
= Δ119906119895minus1
1 le 119895 le 119899
(7)
We supplement these equations with the following inflowboundary conditions (see (1)ndash(3))
1199060
= 1198922(119910) at 119909 = 1
119906119895
= 0 at 119909 = 1 1 le 119895 le 119899
(8)
While (7) are ldquonaturalrdquo the choice of the boundary conditionsin (8) is not obvious it will be eventually justified bythe convergence theorem below Then by integrating theequations in 119909 we find the smooth outer solutions 119906
119895 in thefollowing form
1199060
= int
1
119909
119891 (1199091 119910) 119889119909
1+ 1198922(119910) (9)
119906119895
= int
1
119909
Δ119906119895minus1
(1199091 119910) 119889119909
1 1 le 119895 le 119899 (10)
International Journal of Differential Equations 3
Under this construction of the outer expansion we noticethat119906120576 cong sum
119899
119895=0120576119895
119906119895 satisfies the boundary condition (1)
2along
the edge 119909 = 1 only and not on the three other sides of120597Ω 119910 = 0 119910 = 1 and 119909 = 0 (in general) Hence weexpect boundary layers to occur near those edges and the so-called correctors (corresponding to the Θ
119895) will be necessaryto account for these discrepancies
In Section 3 performing the boundary layer analysis wedefine for each 0 le 119895 le 119899 the corrector Θ
119895 appearing in(11) belowThen in Section 4 using (6) (9) (10) and (11) weprove our main result which is stated as Theorem 9
3 Construction of the CorrectorsBoundary Layers Analysis
Wewant the boundary value of 119906120576 tomatch that of its approx-imation In general the boundary values of the diffusivesolution 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 match only at
119909 = 1 but not on the other sides of 120597Ω 119910 = 0 119910 = 1 and119909 = 0 Hence we expect that some boundary layers will occurnear those three edges To resolve this inconsistency we willdefine a number of boundary layer functions on the sides andcorners of 120597Ω following an analysis reminiscent of the theoryof the Prandtl boundary layer in [3ndash6]
For the moment comparing 119906120576 with a finite sum
sum119899
119895=0120576119895
119906119895 we infer from the definition of the 119906
119895 in (7)ndash(10)that the values of these functions coincide on the segment119909 = 1 In general they will not coincide anywhere else on120597Ω First to correct the inconsistency on the sides 119910 = 0 1we will define the so-called parabolic correctors 120593
119895
= 120593119895
119897+
120593119895
119906 Then to correct the inconsistency at 119909 = 0 we will
introduce the (classical) corrector 120579119895 However this will not
be enough because additional inconsistencies appear at thecorners ofΩ and as we shall see below it will be necessary tointroduce additional correctors 120585
119895 120577119895 and 120578119895 to handle these
inconsistenciesIn summary in the subsections below we will define the
Θ119895 in the form
Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
0 le 119895 le 119899 (11)
where the role and location of each corrector are explainedrespectively in Box 1 and Figure 1 The stretched variables ofthe corrector functions which will be introduced and usedin the following subsections are indicated in the followingequation
120593119895
119897= 120593119895
119897(119909
119910
radic120576
) 120593119895
119906= 120593119895
119906(119909
(1 minus 119910)
radic120576
)
120585119895
119897= 120585119895
119897(
(1 minus 119909)
120576
119910
120576
) 120585119895
119906= 120585119895
119906(
(1 minus 119909)
120576
(1 minus 119910)
120576
)
120579119895
= 120579119895
(
119909
120576
119910)
120577119895
119897= 120577119895
119897(
119909
120576
119910
radic120576
) 120577119895
119906= 120577119895
119906(
119909
120576
(1 minus 119910)
120576
)
120578119895
119897= 120578119895
119897(
119909
120576
119910) 120578119895
119906= 120578119895
119906(
119909
120576
119910)
(12)
31 Parabolic Boundary Layers (PBL) At the bottom (or top)boundary that is at 119910 = 0 (or 119910 = 1) to balance thediscrepancy of 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 (see
the 119906119895 as defined in (9) and (10)) we construct below the
parabolic boundary layer correctors 120593119895 0 le 119895 le 119899We formally insert the asymptotic expansion 119906
120576
cong
sum119899
119895=0120576119895
120593119895 into (1) Using the arguments similar to those of
the Prandtl theory we see that the thickness of the boundarylayer near 119910 = 0 or 119910 = 1 should be of order 120576
12 Then bycollecting the terms at the same order of 120576119895 0 le 119895 le 119899 we findthe equations of the parabolic boundary layer correctors
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω 0 le 119895 le 119899 (13)
where we set 120593minus1 = 0Along the boundaries 119910 = 0 119910 = 1 and 119909 = 1 comparing
(1)2and the boundary values of the functions in (9) and (10)
we find the boundary conditions of 120593119895 0 le 119895 le 119899 which read
120593119895
= 0 at 119909 = 1 0 le 119895 le 119899
1205930
= 1198923(119909) minus 119906
0
(119909 0) at 119910 = 0
120593119895
= minus119906119895
(119909 0) at 119910 = 0 1 le 119895 le 119899
1205930
= 1198924(119909) minus 119906
0
(119909 1) at 119910 = 1
120593119895
= minus119906119895
(119909 1) at 119910 = 1 1 le 119895 le 119899
(14)
We will show below how to actually construct 120593119895 For themoment comparing the functions 119906
120576 and sum119899
119895=0120576119895
(119906119895
minus 120593119895
)
for some 119899 ge 0 we see that the boundary values of thesefunctions coincide at 119909 = 1 and 119910 = 0 1 However thishas been achieved at the price of introducing some newinconsistencies Indeed we notice in (14) that for each 119895 ge 0the boundary conditions at119910 = 0 and119910 = 1 are not consistentwith that at 119909 = 1 That is in general for any 119894 ge 1
120597119894
1199091199060
(1 0) minus 120597119894
1199091198923(1) = 0 120597
119894
1199091199060
(1 1) minus 120597119894
1199091198924(1) = 0
120597119894
119909119906119895
(1 0) = 0 120597119894
119909119906119895
(1 1) = 0 119895 ge 1
(15)
Due to this inconsistency of the boundary data if we define120593119895 0 le 119895 le 119899 as a solution of (13) with (14) some derivatives
of 120593119895 0 le 119895 le 119899 get singular at the two corners (1 119896) 119896 = 0 1Therefore to define smooth (smoother) correctors120593119895 0 le 119895 le
119899 we will modify the boundary conditions (14) as introducedin (23)
4 International Journal of Differential Equations
120593119895
= 120593119895
119897+ 120593119895
119906is the parabolic boundary layer corrector near 119910 = 0 and 119910 = 1
120585119895
= 120585119895
119897+ 120585119895
119906is the elliptic corrector which resolves the compatibility issues in the construction of 120593119895 at the corners (1 0) and (1 1)
120579119895 is the ordinary boundary layer corrector near 119909 = 0
120577119895
= 120577119895
119897+ 120577119895
119906is the ordinary corner layer corrector which manages the effect of 120593119895 along 119909 = 0
120578119895
= 120578119895
119897+ 120578119895
119906is the supplementary corrector that manages the effects of 120579119895 and 120577
119895 along the edges 119910 = 0 and 119910 = 1
Box 1 The different correctors
119910
1
0 1 119909
120577119895
119906 120578119895
119906
120577119895
119897 120578119895
119897
120579119895
119906119895
120593119895
119906
120593119895
119897
120585119895
119906
120585119895
119897
Figure 1 Location of the outer solution 119906119895 and the boundary layer
correctors
311 On Modification of Boundary Conditions for SmoothCorrectors at 119910 = 0 1 Equation (13) with initial and bound-ary conditions (14) is a parabolic problem (heat equation)in which minus119909 is the (positive) time like variable so that theboundary conditions at 119909 = 1 are the analogue of theinitial conditions for an evolution problem and the possibleinconsistencies in (15) correspond to the (analogue of ) thecompatibility conditions between the initial and boundaryvalues for a parabolic equation see [7ndash9] and the referencesthereinDue to the lack of consistency at the corners (1 0) and(1 1) the correctors constructed as in (13) and (14) might getsingular derivatives at those corners We will overcome thisdifficulty by considering some corner functions at 119909 = 1 and119910 = 0 1 similar to what is done in [10 11] or [12] for thecompatibility issue in parabolic problems see [13] as well
Defining 120575(119909) as a smooth cut-off function independentof 120576 such that
120575 (119909) =
0 for 0 le 119909 le
1
2
1 for 119909 ge
3
4
(16)
we set
120574119895
119896(119909) = 120574
119895
119896(119909) 120575 (119909) 119896 = 0 1 (17)
where
1205740
119896(119909) =
2119899+1
sum
119894=1
(119909 minus 1)119894
119894
[120597119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896)]
120574119895
119896(119909) = minus
2119899+1minus2119895
sum
119894=1
(119909 minus 1)119894
119894
120597119894
119909119906119895
(1 119896) 1 le 119895 le 119899
(18)
Note that the 1205740
119896 119896 = 0 1 depend on 119899 but this dependency
is not made explicit to make the notations (slightly) simplerNote also (comparing to (15)) that at 119909 = 1 for each 119896 = 0 1
120574119895
119896(1) = 0 0 le 119895 le 119899
120597119894
1199091205740
119896(1) = 120597
119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896) 1 le 119894 le 2119899 + 1
120597119894
119909120574119895
119896(1) = minus120597
119894
119909119906119895
(1 119896) 1 le 119895 le 119899 1 le 119894 le 2119899 + 1 minus 119895
(19)
We introduce the stretched variables
119910 =
119910
radic120576
119910 =
(1 minus 119910)
radic120576
(20)
Then using (13) (14) (17) and (20) we define the paraboliccorrector 120593119895 0 le 119895 le 119899 as
120593119895
= 120593119895
119897+ 120593119895
119906= 120593119895
119897(119909 119910) + 120593
119895
119906(119909 119910) (21)
where 120593119895
119897and 120593
119895
119906(lower and upper parts of 120593
119895 resp) aredefined as the solutions of the 1D heat equations below
For 0 le 119895 le 119899
minus1205972
119910120593119895
119897minus 120597119909120593119895
119897= 1205972
119909120593119895minus1
119897for (119910 119909) isin R
+
times (0 1)
120593119895
119897= ℎ119895
0(119909) at 119910 = 0
120593119895
119897997888rarr 0 as 119910 997888rarr infin
120593119895
119897= 0 at 119909 = 1
minus1205972
119910120593119895
119906minus 120597119909120593119895
119906= 1205972
119909120593119895minus1
119906for (119910 119909) isin R
+
times (0 1)
120593119895
119906= ℎ119895
1(119909) at 119910 = 0
120593119895
119906997888rarr 0 as 119910 997888rarr infin
120593119895
119906= 0 at 119909 = 1
(22)
Here the ℎ119895
119896(119909) 0 le 119895 le 119899 119896 = 0 1 are defined by
ℎ0
0(119909) = 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909)
ℎ119895
0(119909) = minus119906
119895
(119909 0) minus 120574119895
0(119909) 1 le 119895 le 119899
ℎ0
1(119909) = 119892
4(119909) minus 119906
0
(119909 1) minus 1205740
1(119909)
ℎ119895
1(119909) = minus119906
119895
(119909 1) minus 120574119895
1(119909) 1 le 119895 le 119899
(23)
International Journal of Differential Equations 5
Note that the smooth boundary conditions ℎ119895
119896(119909) 0 le 119895 le 119899
119896 = 0 1 along 119910 or 119910 = 0 are compatible with the other 0boundary (initial) conditions at 119909 = 1 in the sense that foreach 0 le 119895 le 119899
120597119894
119909ℎ119895
119896(1) = 0 119896 = 0 1 0 le 119894 le 2119899 + 1 minus 2119895 (24)
From [14ndash16] or [1] we recall the explicit expressions of120593119895
119897= 120593119895
119897(119909 119910) 0 le 119895 le 119899
1205930
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ0
0(119909 +
1199102
21199102
1
) 1198891199101 (25)
120593119895
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ119895
0(119909 +
1199102
21199102
1
)1198891199101
+
1
2radic120587
int
1minus119909
0
int
infin
0
1
radic1199091
exp(minus
(119910 minus 1199101)2
41199091
)
minus exp(minus
(119910 + 1199101)2
41199091
)
times 1205972
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
(26)
The expressions of 120593119895
119906= 120593119895
119906(119909 119910) 0 le 119895 le 119899 are identical
to (25) and (26) with ℎ0
0 ℎ1198950 and 119910 respectively replaced by
ℎ0
1 ℎ119895
1and 119910
312 Estimates on the Parabolic Boundary Layers We nowstate some pointwise estimates for the 120593
119895
119897and 120593
119895
119906 0 le 119895 le 119899
which are proved in the appendix
Lemma 1 For each 0 le 119895 le 119899 and 119904 ge 0 one has the followingpointwise estimates
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
yradic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(27)
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119906
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
1 minus 119910
radic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(28)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
From Lemma 1 it is easy to deduce the following 119871119901
estimates
Lemma 2 For each 0 le 119895 le 119899 and 119904 ge 0 one has for 0 le
119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119906
10038171003817100381710038171003817119871119901(Ω)
120581120576(119904minus119898)2+12119901
(29)
Using (21) and (22) we find the equations for120593119895 = 120593119895
119897+120593119895
119906
0 le 119895 le 119899
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω
120593119895
= ℎ119895
0(119909) + 120593
119895
119906
10038161003816100381610038161003816119910=0
at 119910 = 0
120593119895
= ℎ119895
1(119909) + 120593
119895
119897
10038161003816100381610038161003816119910=1
at 119910 = 1
120593119895
= 0 at 119909 = 1
(30)
Thanks to Lemma 1 we notice that
120593119895
119906
10038161003816100381610038161003816119910=0
= est 120593119895
119897
10038161003816100381610038161003816119910=1
= est (31)
where est denotes an exponentially small term with respectto the (small) parameter 120576 as defined in Notation 2
32 Elliptic Boundary Layers (EBL) In Section 31 to con-struct the consistent parabolic boundary layer correctors 120593
1198950 le 119895 le 119899 we considered the modified boundary conditions(23) including 120574
119895
119896 instead of the natural boundary conditions
(14) Here on the two sides 119910 = 0 and 119910 = 1 ofΩ to cancel 120574119895119896
which is exactly the difference of (14) and (23) we introduceelliptic boundary layer correctors 120585119895 0 le 119895 le 119899 which like 120574
119895
119896
or 120574119895
119896 depend on the order 119899 of the asymptotic expansion in
(6)Using (17) we define
120585119895
= 120585119895
119897+ 120585119895
119906 0 le 119895 le 119899 (32)
where 120585119895
119897and 120585119895
119906are the solutions of the elliptic systems below
For 0 le 119895 le 119899
minus120576 1205972
119909120585119895
119897minus 1205761205972
119910120585119895
119897minus 120597119909120585119895
119897= 0 in Ω
120585119895
119897= 120574119895
0(119909) at 119910 = 0
120585119895
119897= 0 at 119909 = 0 1 or 119910 = 1
minus120576 1205972
119909120585119895
119906minus 120576 1205972
119910120585119895
119906minus 120597119909120585119895
119906= 0 in Ω
120585119895
119906= 120574119895
1(119909) at 119910 = 1
120585119895
119906= 0 at 119909 = 0 1 or 119910 = 0
(33)
The systems (33) being elliptic equations their well-posedness are easy to verify and hence we omit the proofhere
The equations satisfied by the elliptic correctors 120585119895 = 120585119895
119897+
120585119895
119906read as followsFor 0 le 119895 le 119899
minus1205761205972
119909120585119895
minus 1205761205972
119910120585119895
minus 120597119909120585119895
= 0 in Ω
120585119895
= 120574119895
119896(119909) at 119910 = 119896 119896 = 0 1
120585119895
= 0 at 119909 = 0 1
(34)
6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 3
Under this construction of the outer expansion we noticethat119906120576 cong sum
119899
119895=0120576119895
119906119895 satisfies the boundary condition (1)
2along
the edge 119909 = 1 only and not on the three other sides of120597Ω 119910 = 0 119910 = 1 and 119909 = 0 (in general) Hence weexpect boundary layers to occur near those edges and the so-called correctors (corresponding to the Θ
119895) will be necessaryto account for these discrepancies
In Section 3 performing the boundary layer analysis wedefine for each 0 le 119895 le 119899 the corrector Θ
119895 appearing in(11) belowThen in Section 4 using (6) (9) (10) and (11) weprove our main result which is stated as Theorem 9
3 Construction of the CorrectorsBoundary Layers Analysis
Wewant the boundary value of 119906120576 tomatch that of its approx-imation In general the boundary values of the diffusivesolution 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 match only at
119909 = 1 but not on the other sides of 120597Ω 119910 = 0 119910 = 1 and119909 = 0 Hence we expect that some boundary layers will occurnear those three edges To resolve this inconsistency we willdefine a number of boundary layer functions on the sides andcorners of 120597Ω following an analysis reminiscent of the theoryof the Prandtl boundary layer in [3ndash6]
For the moment comparing 119906120576 with a finite sum
sum119899
119895=0120576119895
119906119895 we infer from the definition of the 119906
119895 in (7)ndash(10)that the values of these functions coincide on the segment119909 = 1 In general they will not coincide anywhere else on120597Ω First to correct the inconsistency on the sides 119910 = 0 1we will define the so-called parabolic correctors 120593
119895
= 120593119895
119897+
120593119895
119906 Then to correct the inconsistency at 119909 = 0 we will
introduce the (classical) corrector 120579119895 However this will not
be enough because additional inconsistencies appear at thecorners ofΩ and as we shall see below it will be necessary tointroduce additional correctors 120585
119895 120577119895 and 120578119895 to handle these
inconsistenciesIn summary in the subsections below we will define the
Θ119895 in the form
Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
0 le 119895 le 119899 (11)
where the role and location of each corrector are explainedrespectively in Box 1 and Figure 1 The stretched variables ofthe corrector functions which will be introduced and usedin the following subsections are indicated in the followingequation
120593119895
119897= 120593119895
119897(119909
119910
radic120576
) 120593119895
119906= 120593119895
119906(119909
(1 minus 119910)
radic120576
)
120585119895
119897= 120585119895
119897(
(1 minus 119909)
120576
119910
120576
) 120585119895
119906= 120585119895
119906(
(1 minus 119909)
120576
(1 minus 119910)
120576
)
120579119895
= 120579119895
(
119909
120576
119910)
120577119895
119897= 120577119895
119897(
119909
120576
119910
radic120576
) 120577119895
119906= 120577119895
119906(
119909
120576
(1 minus 119910)
120576
)
120578119895
119897= 120578119895
119897(
119909
120576
119910) 120578119895
119906= 120578119895
119906(
119909
120576
119910)
(12)
31 Parabolic Boundary Layers (PBL) At the bottom (or top)boundary that is at 119910 = 0 (or 119910 = 1) to balance thediscrepancy of 119906
120576 and the outer expansion sum119899
119895=0120576119895
119906119895 (see
the 119906119895 as defined in (9) and (10)) we construct below the
parabolic boundary layer correctors 120593119895 0 le 119895 le 119899We formally insert the asymptotic expansion 119906
120576
cong
sum119899
119895=0120576119895
120593119895 into (1) Using the arguments similar to those of
the Prandtl theory we see that the thickness of the boundarylayer near 119910 = 0 or 119910 = 1 should be of order 120576
12 Then bycollecting the terms at the same order of 120576119895 0 le 119895 le 119899 we findthe equations of the parabolic boundary layer correctors
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω 0 le 119895 le 119899 (13)
where we set 120593minus1 = 0Along the boundaries 119910 = 0 119910 = 1 and 119909 = 1 comparing
(1)2and the boundary values of the functions in (9) and (10)
we find the boundary conditions of 120593119895 0 le 119895 le 119899 which read
120593119895
= 0 at 119909 = 1 0 le 119895 le 119899
1205930
= 1198923(119909) minus 119906
0
(119909 0) at 119910 = 0
120593119895
= minus119906119895
(119909 0) at 119910 = 0 1 le 119895 le 119899
1205930
= 1198924(119909) minus 119906
0
(119909 1) at 119910 = 1
120593119895
= minus119906119895
(119909 1) at 119910 = 1 1 le 119895 le 119899
(14)
We will show below how to actually construct 120593119895 For themoment comparing the functions 119906
120576 and sum119899
119895=0120576119895
(119906119895
minus 120593119895
)
for some 119899 ge 0 we see that the boundary values of thesefunctions coincide at 119909 = 1 and 119910 = 0 1 However thishas been achieved at the price of introducing some newinconsistencies Indeed we notice in (14) that for each 119895 ge 0the boundary conditions at119910 = 0 and119910 = 1 are not consistentwith that at 119909 = 1 That is in general for any 119894 ge 1
120597119894
1199091199060
(1 0) minus 120597119894
1199091198923(1) = 0 120597
119894
1199091199060
(1 1) minus 120597119894
1199091198924(1) = 0
120597119894
119909119906119895
(1 0) = 0 120597119894
119909119906119895
(1 1) = 0 119895 ge 1
(15)
Due to this inconsistency of the boundary data if we define120593119895 0 le 119895 le 119899 as a solution of (13) with (14) some derivatives
of 120593119895 0 le 119895 le 119899 get singular at the two corners (1 119896) 119896 = 0 1Therefore to define smooth (smoother) correctors120593119895 0 le 119895 le
119899 we will modify the boundary conditions (14) as introducedin (23)
4 International Journal of Differential Equations
120593119895
= 120593119895
119897+ 120593119895
119906is the parabolic boundary layer corrector near 119910 = 0 and 119910 = 1
120585119895
= 120585119895
119897+ 120585119895
119906is the elliptic corrector which resolves the compatibility issues in the construction of 120593119895 at the corners (1 0) and (1 1)
120579119895 is the ordinary boundary layer corrector near 119909 = 0
120577119895
= 120577119895
119897+ 120577119895
119906is the ordinary corner layer corrector which manages the effect of 120593119895 along 119909 = 0
120578119895
= 120578119895
119897+ 120578119895
119906is the supplementary corrector that manages the effects of 120579119895 and 120577
119895 along the edges 119910 = 0 and 119910 = 1
Box 1 The different correctors
119910
1
0 1 119909
120577119895
119906 120578119895
119906
120577119895
119897 120578119895
119897
120579119895
119906119895
120593119895
119906
120593119895
119897
120585119895
119906
120585119895
119897
Figure 1 Location of the outer solution 119906119895 and the boundary layer
correctors
311 On Modification of Boundary Conditions for SmoothCorrectors at 119910 = 0 1 Equation (13) with initial and bound-ary conditions (14) is a parabolic problem (heat equation)in which minus119909 is the (positive) time like variable so that theboundary conditions at 119909 = 1 are the analogue of theinitial conditions for an evolution problem and the possibleinconsistencies in (15) correspond to the (analogue of ) thecompatibility conditions between the initial and boundaryvalues for a parabolic equation see [7ndash9] and the referencesthereinDue to the lack of consistency at the corners (1 0) and(1 1) the correctors constructed as in (13) and (14) might getsingular derivatives at those corners We will overcome thisdifficulty by considering some corner functions at 119909 = 1 and119910 = 0 1 similar to what is done in [10 11] or [12] for thecompatibility issue in parabolic problems see [13] as well
Defining 120575(119909) as a smooth cut-off function independentof 120576 such that
120575 (119909) =
0 for 0 le 119909 le
1
2
1 for 119909 ge
3
4
(16)
we set
120574119895
119896(119909) = 120574
119895
119896(119909) 120575 (119909) 119896 = 0 1 (17)
where
1205740
119896(119909) =
2119899+1
sum
119894=1
(119909 minus 1)119894
119894
[120597119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896)]
120574119895
119896(119909) = minus
2119899+1minus2119895
sum
119894=1
(119909 minus 1)119894
119894
120597119894
119909119906119895
(1 119896) 1 le 119895 le 119899
(18)
Note that the 1205740
119896 119896 = 0 1 depend on 119899 but this dependency
is not made explicit to make the notations (slightly) simplerNote also (comparing to (15)) that at 119909 = 1 for each 119896 = 0 1
120574119895
119896(1) = 0 0 le 119895 le 119899
120597119894
1199091205740
119896(1) = 120597
119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896) 1 le 119894 le 2119899 + 1
120597119894
119909120574119895
119896(1) = minus120597
119894
119909119906119895
(1 119896) 1 le 119895 le 119899 1 le 119894 le 2119899 + 1 minus 119895
(19)
We introduce the stretched variables
119910 =
119910
radic120576
119910 =
(1 minus 119910)
radic120576
(20)
Then using (13) (14) (17) and (20) we define the paraboliccorrector 120593119895 0 le 119895 le 119899 as
120593119895
= 120593119895
119897+ 120593119895
119906= 120593119895
119897(119909 119910) + 120593
119895
119906(119909 119910) (21)
where 120593119895
119897and 120593
119895
119906(lower and upper parts of 120593
119895 resp) aredefined as the solutions of the 1D heat equations below
For 0 le 119895 le 119899
minus1205972
119910120593119895
119897minus 120597119909120593119895
119897= 1205972
119909120593119895minus1
119897for (119910 119909) isin R
+
times (0 1)
120593119895
119897= ℎ119895
0(119909) at 119910 = 0
120593119895
119897997888rarr 0 as 119910 997888rarr infin
120593119895
119897= 0 at 119909 = 1
minus1205972
119910120593119895
119906minus 120597119909120593119895
119906= 1205972
119909120593119895minus1
119906for (119910 119909) isin R
+
times (0 1)
120593119895
119906= ℎ119895
1(119909) at 119910 = 0
120593119895
119906997888rarr 0 as 119910 997888rarr infin
120593119895
119906= 0 at 119909 = 1
(22)
Here the ℎ119895
119896(119909) 0 le 119895 le 119899 119896 = 0 1 are defined by
ℎ0
0(119909) = 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909)
ℎ119895
0(119909) = minus119906
119895
(119909 0) minus 120574119895
0(119909) 1 le 119895 le 119899
ℎ0
1(119909) = 119892
4(119909) minus 119906
0
(119909 1) minus 1205740
1(119909)
ℎ119895
1(119909) = minus119906
119895
(119909 1) minus 120574119895
1(119909) 1 le 119895 le 119899
(23)
International Journal of Differential Equations 5
Note that the smooth boundary conditions ℎ119895
119896(119909) 0 le 119895 le 119899
119896 = 0 1 along 119910 or 119910 = 0 are compatible with the other 0boundary (initial) conditions at 119909 = 1 in the sense that foreach 0 le 119895 le 119899
120597119894
119909ℎ119895
119896(1) = 0 119896 = 0 1 0 le 119894 le 2119899 + 1 minus 2119895 (24)
From [14ndash16] or [1] we recall the explicit expressions of120593119895
119897= 120593119895
119897(119909 119910) 0 le 119895 le 119899
1205930
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ0
0(119909 +
1199102
21199102
1
) 1198891199101 (25)
120593119895
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ119895
0(119909 +
1199102
21199102
1
)1198891199101
+
1
2radic120587
int
1minus119909
0
int
infin
0
1
radic1199091
exp(minus
(119910 minus 1199101)2
41199091
)
minus exp(minus
(119910 + 1199101)2
41199091
)
times 1205972
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
(26)
The expressions of 120593119895
119906= 120593119895
119906(119909 119910) 0 le 119895 le 119899 are identical
to (25) and (26) with ℎ0
0 ℎ1198950 and 119910 respectively replaced by
ℎ0
1 ℎ119895
1and 119910
312 Estimates on the Parabolic Boundary Layers We nowstate some pointwise estimates for the 120593
119895
119897and 120593
119895
119906 0 le 119895 le 119899
which are proved in the appendix
Lemma 1 For each 0 le 119895 le 119899 and 119904 ge 0 one has the followingpointwise estimates
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
yradic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(27)
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119906
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
1 minus 119910
radic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(28)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
From Lemma 1 it is easy to deduce the following 119871119901
estimates
Lemma 2 For each 0 le 119895 le 119899 and 119904 ge 0 one has for 0 le
119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119906
10038171003817100381710038171003817119871119901(Ω)
120581120576(119904minus119898)2+12119901
(29)
Using (21) and (22) we find the equations for120593119895 = 120593119895
119897+120593119895
119906
0 le 119895 le 119899
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω
120593119895
= ℎ119895
0(119909) + 120593
119895
119906
10038161003816100381610038161003816119910=0
at 119910 = 0
120593119895
= ℎ119895
1(119909) + 120593
119895
119897
10038161003816100381610038161003816119910=1
at 119910 = 1
120593119895
= 0 at 119909 = 1
(30)
Thanks to Lemma 1 we notice that
120593119895
119906
10038161003816100381610038161003816119910=0
= est 120593119895
119897
10038161003816100381610038161003816119910=1
= est (31)
where est denotes an exponentially small term with respectto the (small) parameter 120576 as defined in Notation 2
32 Elliptic Boundary Layers (EBL) In Section 31 to con-struct the consistent parabolic boundary layer correctors 120593
1198950 le 119895 le 119899 we considered the modified boundary conditions(23) including 120574
119895
119896 instead of the natural boundary conditions
(14) Here on the two sides 119910 = 0 and 119910 = 1 ofΩ to cancel 120574119895119896
which is exactly the difference of (14) and (23) we introduceelliptic boundary layer correctors 120585119895 0 le 119895 le 119899 which like 120574
119895
119896
or 120574119895
119896 depend on the order 119899 of the asymptotic expansion in
(6)Using (17) we define
120585119895
= 120585119895
119897+ 120585119895
119906 0 le 119895 le 119899 (32)
where 120585119895
119897and 120585119895
119906are the solutions of the elliptic systems below
For 0 le 119895 le 119899
minus120576 1205972
119909120585119895
119897minus 1205761205972
119910120585119895
119897minus 120597119909120585119895
119897= 0 in Ω
120585119895
119897= 120574119895
0(119909) at 119910 = 0
120585119895
119897= 0 at 119909 = 0 1 or 119910 = 1
minus120576 1205972
119909120585119895
119906minus 120576 1205972
119910120585119895
119906minus 120597119909120585119895
119906= 0 in Ω
120585119895
119906= 120574119895
1(119909) at 119910 = 1
120585119895
119906= 0 at 119909 = 0 1 or 119910 = 0
(33)
The systems (33) being elliptic equations their well-posedness are easy to verify and hence we omit the proofhere
The equations satisfied by the elliptic correctors 120585119895 = 120585119895
119897+
120585119895
119906read as followsFor 0 le 119895 le 119899
minus1205761205972
119909120585119895
minus 1205761205972
119910120585119895
minus 120597119909120585119895
= 0 in Ω
120585119895
= 120574119895
119896(119909) at 119910 = 119896 119896 = 0 1
120585119895
= 0 at 119909 = 0 1
(34)
6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Differential Equations
120593119895
= 120593119895
119897+ 120593119895
119906is the parabolic boundary layer corrector near 119910 = 0 and 119910 = 1
120585119895
= 120585119895
119897+ 120585119895
119906is the elliptic corrector which resolves the compatibility issues in the construction of 120593119895 at the corners (1 0) and (1 1)
120579119895 is the ordinary boundary layer corrector near 119909 = 0
120577119895
= 120577119895
119897+ 120577119895
119906is the ordinary corner layer corrector which manages the effect of 120593119895 along 119909 = 0
120578119895
= 120578119895
119897+ 120578119895
119906is the supplementary corrector that manages the effects of 120579119895 and 120577
119895 along the edges 119910 = 0 and 119910 = 1
Box 1 The different correctors
119910
1
0 1 119909
120577119895
119906 120578119895
119906
120577119895
119897 120578119895
119897
120579119895
119906119895
120593119895
119906
120593119895
119897
120585119895
119906
120585119895
119897
Figure 1 Location of the outer solution 119906119895 and the boundary layer
correctors
311 On Modification of Boundary Conditions for SmoothCorrectors at 119910 = 0 1 Equation (13) with initial and bound-ary conditions (14) is a parabolic problem (heat equation)in which minus119909 is the (positive) time like variable so that theboundary conditions at 119909 = 1 are the analogue of theinitial conditions for an evolution problem and the possibleinconsistencies in (15) correspond to the (analogue of ) thecompatibility conditions between the initial and boundaryvalues for a parabolic equation see [7ndash9] and the referencesthereinDue to the lack of consistency at the corners (1 0) and(1 1) the correctors constructed as in (13) and (14) might getsingular derivatives at those corners We will overcome thisdifficulty by considering some corner functions at 119909 = 1 and119910 = 0 1 similar to what is done in [10 11] or [12] for thecompatibility issue in parabolic problems see [13] as well
Defining 120575(119909) as a smooth cut-off function independentof 120576 such that
120575 (119909) =
0 for 0 le 119909 le
1
2
1 for 119909 ge
3
4
(16)
we set
120574119895
119896(119909) = 120574
119895
119896(119909) 120575 (119909) 119896 = 0 1 (17)
where
1205740
119896(119909) =
2119899+1
sum
119894=1
(119909 minus 1)119894
119894
[120597119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896)]
120574119895
119896(119909) = minus
2119899+1minus2119895
sum
119894=1
(119909 minus 1)119894
119894
120597119894
119909119906119895
(1 119896) 1 le 119895 le 119899
(18)
Note that the 1205740
119896 119896 = 0 1 depend on 119899 but this dependency
is not made explicit to make the notations (slightly) simplerNote also (comparing to (15)) that at 119909 = 1 for each 119896 = 0 1
120574119895
119896(1) = 0 0 le 119895 le 119899
120597119894
1199091205740
119896(1) = 120597
119894
1199091198923+119896
(1) minus 120597119894
1199091199060
(1 119896) 1 le 119894 le 2119899 + 1
120597119894
119909120574119895
119896(1) = minus120597
119894
119909119906119895
(1 119896) 1 le 119895 le 119899 1 le 119894 le 2119899 + 1 minus 119895
(19)
We introduce the stretched variables
119910 =
119910
radic120576
119910 =
(1 minus 119910)
radic120576
(20)
Then using (13) (14) (17) and (20) we define the paraboliccorrector 120593119895 0 le 119895 le 119899 as
120593119895
= 120593119895
119897+ 120593119895
119906= 120593119895
119897(119909 119910) + 120593
119895
119906(119909 119910) (21)
where 120593119895
119897and 120593
119895
119906(lower and upper parts of 120593
119895 resp) aredefined as the solutions of the 1D heat equations below
For 0 le 119895 le 119899
minus1205972
119910120593119895
119897minus 120597119909120593119895
119897= 1205972
119909120593119895minus1
119897for (119910 119909) isin R
+
times (0 1)
120593119895
119897= ℎ119895
0(119909) at 119910 = 0
120593119895
119897997888rarr 0 as 119910 997888rarr infin
120593119895
119897= 0 at 119909 = 1
minus1205972
119910120593119895
119906minus 120597119909120593119895
119906= 1205972
119909120593119895minus1
119906for (119910 119909) isin R
+
times (0 1)
120593119895
119906= ℎ119895
1(119909) at 119910 = 0
120593119895
119906997888rarr 0 as 119910 997888rarr infin
120593119895
119906= 0 at 119909 = 1
(22)
Here the ℎ119895
119896(119909) 0 le 119895 le 119899 119896 = 0 1 are defined by
ℎ0
0(119909) = 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909)
ℎ119895
0(119909) = minus119906
119895
(119909 0) minus 120574119895
0(119909) 1 le 119895 le 119899
ℎ0
1(119909) = 119892
4(119909) minus 119906
0
(119909 1) minus 1205740
1(119909)
ℎ119895
1(119909) = minus119906
119895
(119909 1) minus 120574119895
1(119909) 1 le 119895 le 119899
(23)
International Journal of Differential Equations 5
Note that the smooth boundary conditions ℎ119895
119896(119909) 0 le 119895 le 119899
119896 = 0 1 along 119910 or 119910 = 0 are compatible with the other 0boundary (initial) conditions at 119909 = 1 in the sense that foreach 0 le 119895 le 119899
120597119894
119909ℎ119895
119896(1) = 0 119896 = 0 1 0 le 119894 le 2119899 + 1 minus 2119895 (24)
From [14ndash16] or [1] we recall the explicit expressions of120593119895
119897= 120593119895
119897(119909 119910) 0 le 119895 le 119899
1205930
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ0
0(119909 +
1199102
21199102
1
) 1198891199101 (25)
120593119895
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ119895
0(119909 +
1199102
21199102
1
)1198891199101
+
1
2radic120587
int
1minus119909
0
int
infin
0
1
radic1199091
exp(minus
(119910 minus 1199101)2
41199091
)
minus exp(minus
(119910 + 1199101)2
41199091
)
times 1205972
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
(26)
The expressions of 120593119895
119906= 120593119895
119906(119909 119910) 0 le 119895 le 119899 are identical
to (25) and (26) with ℎ0
0 ℎ1198950 and 119910 respectively replaced by
ℎ0
1 ℎ119895
1and 119910
312 Estimates on the Parabolic Boundary Layers We nowstate some pointwise estimates for the 120593
119895
119897and 120593
119895
119906 0 le 119895 le 119899
which are proved in the appendix
Lemma 1 For each 0 le 119895 le 119899 and 119904 ge 0 one has the followingpointwise estimates
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
yradic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(27)
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119906
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
1 minus 119910
radic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(28)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
From Lemma 1 it is easy to deduce the following 119871119901
estimates
Lemma 2 For each 0 le 119895 le 119899 and 119904 ge 0 one has for 0 le
119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119906
10038171003817100381710038171003817119871119901(Ω)
120581120576(119904minus119898)2+12119901
(29)
Using (21) and (22) we find the equations for120593119895 = 120593119895
119897+120593119895
119906
0 le 119895 le 119899
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω
120593119895
= ℎ119895
0(119909) + 120593
119895
119906
10038161003816100381610038161003816119910=0
at 119910 = 0
120593119895
= ℎ119895
1(119909) + 120593
119895
119897
10038161003816100381610038161003816119910=1
at 119910 = 1
120593119895
= 0 at 119909 = 1
(30)
Thanks to Lemma 1 we notice that
120593119895
119906
10038161003816100381610038161003816119910=0
= est 120593119895
119897
10038161003816100381610038161003816119910=1
= est (31)
where est denotes an exponentially small term with respectto the (small) parameter 120576 as defined in Notation 2
32 Elliptic Boundary Layers (EBL) In Section 31 to con-struct the consistent parabolic boundary layer correctors 120593
1198950 le 119895 le 119899 we considered the modified boundary conditions(23) including 120574
119895
119896 instead of the natural boundary conditions
(14) Here on the two sides 119910 = 0 and 119910 = 1 ofΩ to cancel 120574119895119896
which is exactly the difference of (14) and (23) we introduceelliptic boundary layer correctors 120585119895 0 le 119895 le 119899 which like 120574
119895
119896
or 120574119895
119896 depend on the order 119899 of the asymptotic expansion in
(6)Using (17) we define
120585119895
= 120585119895
119897+ 120585119895
119906 0 le 119895 le 119899 (32)
where 120585119895
119897and 120585119895
119906are the solutions of the elliptic systems below
For 0 le 119895 le 119899
minus120576 1205972
119909120585119895
119897minus 1205761205972
119910120585119895
119897minus 120597119909120585119895
119897= 0 in Ω
120585119895
119897= 120574119895
0(119909) at 119910 = 0
120585119895
119897= 0 at 119909 = 0 1 or 119910 = 1
minus120576 1205972
119909120585119895
119906minus 120576 1205972
119910120585119895
119906minus 120597119909120585119895
119906= 0 in Ω
120585119895
119906= 120574119895
1(119909) at 119910 = 1
120585119895
119906= 0 at 119909 = 0 1 or 119910 = 0
(33)
The systems (33) being elliptic equations their well-posedness are easy to verify and hence we omit the proofhere
The equations satisfied by the elliptic correctors 120585119895 = 120585119895
119897+
120585119895
119906read as followsFor 0 le 119895 le 119899
minus1205761205972
119909120585119895
minus 1205761205972
119910120585119895
minus 120597119909120585119895
= 0 in Ω
120585119895
= 120574119895
119896(119909) at 119910 = 119896 119896 = 0 1
120585119895
= 0 at 119909 = 0 1
(34)
6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
Submit your manuscripts athttpwwwhindawicom
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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
International Journal of Differential Equations 5
Note that the smooth boundary conditions ℎ119895
119896(119909) 0 le 119895 le 119899
119896 = 0 1 along 119910 or 119910 = 0 are compatible with the other 0boundary (initial) conditions at 119909 = 1 in the sense that foreach 0 le 119895 le 119899
120597119894
119909ℎ119895
119896(1) = 0 119896 = 0 1 0 le 119894 le 2119899 + 1 minus 2119895 (24)
From [14ndash16] or [1] we recall the explicit expressions of120593119895
119897= 120593119895
119897(119909 119910) 0 le 119895 le 119899
1205930
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ0
0(119909 +
1199102
21199102
1
) 1198891199101 (25)
120593119895
119897= radic
2
120587
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) ℎ119895
0(119909 +
1199102
21199102
1
)1198891199101
+
1
2radic120587
int
1minus119909
0
int
infin
0
1
radic1199091
exp(minus
(119910 minus 1199101)2
41199091
)
minus exp(minus
(119910 + 1199101)2
41199091
)
times 1205972
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
(26)
The expressions of 120593119895
119906= 120593119895
119906(119909 119910) 0 le 119895 le 119899 are identical
to (25) and (26) with ℎ0
0 ℎ1198950 and 119910 respectively replaced by
ℎ0
1 ℎ119895
1and 119910
312 Estimates on the Parabolic Boundary Layers We nowstate some pointwise estimates for the 120593
119895
119897and 120593
119895
119906 0 le 119895 le 119899
which are proved in the appendix
Lemma 1 For each 0 le 119895 le 119899 and 119904 ge 0 one has the followingpointwise estimates
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
yradic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(27)
10038161003816100381610038161003816119910119904
120597119894
119909120597119898
119910120593119895
119906
10038161003816100381610038161003816le 120581120576(119904minus119898)2 exp(minus119888
1 minus 119910
radic120576
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(28)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
From Lemma 1 it is easy to deduce the following 119871119901
estimates
Lemma 2 For each 0 le 119895 le 119899 and 119904 ge 0 one has for 0 le
119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119910119904
120597119894
119909120597119898
119910120593119895
119906
10038171003817100381710038171003817119871119901(Ω)
120581120576(119904minus119898)2+12119901
(29)
Using (21) and (22) we find the equations for120593119895 = 120593119895
119897+120593119895
119906
0 le 119895 le 119899
minus1205761205972
119910120593119895
minus 120597119909120593119895
= 1205972
119909120593119895minus1 in Ω
120593119895
= ℎ119895
0(119909) + 120593
119895
119906
10038161003816100381610038161003816119910=0
at 119910 = 0
120593119895
= ℎ119895
1(119909) + 120593
119895
119897
10038161003816100381610038161003816119910=1
at 119910 = 1
120593119895
= 0 at 119909 = 1
(30)
Thanks to Lemma 1 we notice that
120593119895
119906
10038161003816100381610038161003816119910=0
= est 120593119895
119897
10038161003816100381610038161003816119910=1
= est (31)
where est denotes an exponentially small term with respectto the (small) parameter 120576 as defined in Notation 2
32 Elliptic Boundary Layers (EBL) In Section 31 to con-struct the consistent parabolic boundary layer correctors 120593
1198950 le 119895 le 119899 we considered the modified boundary conditions(23) including 120574
119895
119896 instead of the natural boundary conditions
(14) Here on the two sides 119910 = 0 and 119910 = 1 ofΩ to cancel 120574119895119896
which is exactly the difference of (14) and (23) we introduceelliptic boundary layer correctors 120585119895 0 le 119895 le 119899 which like 120574
119895
119896
or 120574119895
119896 depend on the order 119899 of the asymptotic expansion in
(6)Using (17) we define
120585119895
= 120585119895
119897+ 120585119895
119906 0 le 119895 le 119899 (32)
where 120585119895
119897and 120585119895
119906are the solutions of the elliptic systems below
For 0 le 119895 le 119899
minus120576 1205972
119909120585119895
119897minus 1205761205972
119910120585119895
119897minus 120597119909120585119895
119897= 0 in Ω
120585119895
119897= 120574119895
0(119909) at 119910 = 0
120585119895
119897= 0 at 119909 = 0 1 or 119910 = 1
minus120576 1205972
119909120585119895
119906minus 120576 1205972
119910120585119895
119906minus 120597119909120585119895
119906= 0 in Ω
120585119895
119906= 120574119895
1(119909) at 119910 = 1
120585119895
119906= 0 at 119909 = 0 1 or 119910 = 0
(33)
The systems (33) being elliptic equations their well-posedness are easy to verify and hence we omit the proofhere
The equations satisfied by the elliptic correctors 120585119895 = 120585119895
119897+
120585119895
119906read as followsFor 0 le 119895 le 119899
minus1205761205972
119909120585119895
minus 1205761205972
119910120585119895
minus 120597119909120585119895
= 0 in Ω
120585119895
= 120574119895
119896(119909) at 119910 = 119896 119896 = 0 1
120585119895
= 0 at 119909 = 0 1
(34)
6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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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
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6 International Journal of Differential Equations
In the error analysis below in Section 4 we will see thatno extra terms related to the 120585
119895 appear see (65) below Thisis because the 120585
119895 as appearing in (34) satisfies the sameequation as for 119906
120576 supplemented with the (exact) boundaryconditions that we need From this observation as it willbe justified in Section 4 we notice that our main result inTheorem 9 does not require any estimate on the 120585
119895As an extra informationwhichmight be useful elsewhere
but not in this paper by performing the energy estimate on(34) we find that
1003817100381710038171003817100381712058511989510038171003817100381710038171003817120576
le 12058112057614
0 le 119895 le 119899 (35)
In the appendix we introduce an explicit approximate 120585 of120585 Then thanks to the estimates on 120585 one can obtain somepointwise estimates on 120585 as well see Appendix B
At this stage the functions 119906120576 and sum
119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
)
match along the sides 119909 = 1 and 119910 = 0 1 (with singularderivatives of high orders for the 120593
119895) More precisely from(1) (8) (30) (31) and (34) we see that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
) =
1198921+
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
) at 119909 = 0
0 at 119909 = 1
119899
sum
119895=0
120576119895
120593119895
119906= est at 119910 = 0
119899
sum
119895=0
120576119895
120593119895
119897= est at 119910 = 1
(36)
We will now deal with the side 119909 = 0 and construct theordinary boundary layer 120579
119895 and the ordinary corner layerfunctions (correctors) 120577119895
33 Ordinary Boundary Layers (OBL) To handle the dis-crepancy of 119906
120576 and sum119899
119895=0120576119895
119906119895 at 119909 = 0 which is a part of
119906120576
minus sum119899
119895=0120576119895
(119906119895
+ 120593119895
+ 120585119895
) appearing in (36) the so-calledordinary boundary layer corrector 120579
119895 is introduced in thissection We insert a formal expansion 119906
120576
cong suminfin
119895=0120576119895
120579119895 into
the diffusive equation (1) Then using the stretched variable119909 = 119909120576 we find the equations of 120579119895
minus1205761205972
119909120579119895
minus 120597119909120579119895
= 120576minus1
1205972
119910120579119895minus2 in Ω 0 le 119895 le 119899 (37)
where we set 120579minus1 = 120579minus2
= 0Themain role of the 120579
119895 is at each order of 120576119895 to cancel theerror 119906120576 minussum
119899
119895=0120576119895
119906119895 at 119909 = 0 Hence we supplement (37) with
the boundary conditions
1205790
= 1198921(119910) minus 119906
0
(0 119910) at 119909 = 0
120579119895
= minus119906119895
(0 119910) at 119909 = 0 1 le 119895 le 119899
120579119895
997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899
(38)
The explicit expressions of 120579119895 0 le 119895 le 119899 are inductively
shown to be of the form
120579119895
= 119875119896
(
119909
120576
119910) exp(
minus119909
120576
) 119895 = 2119896 2119896 + 1 (39)
where 119875119896
(119909120576 119910) is a polynomial in 119909120576 of degree 119896 whosecoefficients independent of 120576 are linear combinations of the120597119904
119906119895minus119904
(0 119910)120597119910119904 0 le 119904 le 119895 and 119875
119896
(0 119910) = minus119906119895
(0 119910)According to (39) we deduce the following lemmas
Lemma3 For each 0 le 119895 le 119899 one has the pointwise estimates10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057911989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 119904 119894 119898 ge 0 (40)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 4 For each 0 le 119895 le 119899 one has10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057911989510038171003817100381710038171003817119871119901(Ω)
le 120581120576119904minus119894+(1119901)
119904 119894 119898 ge 0 (41)
Thanks to Lemma 3 we notice that the effect of 120579119895 nearthe boundary 119909 = 1 is exponentially small
120579119895
(1 119910) = est 0 le 119895 le 119899 (42)
From (36) (38) and (42) we infer that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
)
=
119899
sum
119895=0
120576119895
120593119895 at 119909 = 0
119899
sum
119895=0
120576119895
120579119895
= est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
) =
119899
sum
119895=0
120576119895
120579119895
+ est at 119910 = 1
(43)
34 Ordinary Corner Layers (OCL) To account for the valuesum119899
119895=0120576119895
120593119895 at 119909 = 0 which is exactly the difference of 119906120576 and
sum119899
119895=0120576119895
(119906119895
+120593119895
+ 120585119895
+ 120579119895
) at 119909 = 0 (see (43)) we introduce theordinary corner layer correctors 120577119895 0 le 119895 le 119899 in the form
120577119895
= 120577119895
119897+ 120577119895
119906 (44)
where the 120577119895
119897(or 120577119895
119906) are the correctors near the corner (0 0)
(or (0 1)) as constructed belowTo define 120577
119895
119897(or 120577119895
119906) we insert a formal expansion 119906
120576
cong
suminfin
119895=0120576119895
120577119895 into (1) Then using the stretched variables 119909 = 119909120576
and 119910 = 119910radic120576 near (0 0) and using 119909 and 119910 = (1 minus 119910)radic120576
near (1 1) we collect the terms of order 120576119895 0 le 119895 le 119899 As a
result we find the equations for 120577119895
119897(or 120577119895
119906)
minus1205761205972
119909120577119895
119896minus 120597119909120577119895
119896= 1205972
119910120577119895minus1
119896in Ω 0 le 119895 le 119899 119896 = 119906 or 119897
(45)
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 7
Here we set 120577minus1119896
= 0 for 119896 = 119897 119906At each order of 120576
119895 0 le 119895 le 119899 to cancel the error 120593119895
near the corner (0 0) or (0 1) we supplement (45) with theboundary conditions
120577119895
119896= minus120593119895
119896(0 119910) at 119909 = 0 0 le 119895 le 119899 119896 = 119897 119906
120577119895
119896997888rarr 0 as 119909 997888rarr infin 0 le 119895 le 119899 119896 = 119897 119906
(46)
The explicit solutions 120577119895
119897of (45) and (46) 0 le 119895 le 119899 are
inductively found to be of the following form
120577119895
119897= 119875119895
(
119909
120576
119910
radic120576
) exp(
minus119909
120576
) (47)
where119875119895
(119909120576 119910radic120576) is a polynomial in 119909120576 of degree 119895whosecoefficients independent of 120576 are linear combinations of thequantities 120576
119904
1205972119904
120593119895minus119904
119897(0 119910radic120576)120597119910
2119904 0 le 119904 le 119895 In the samemanner the explicit forms of the 120577
119895
119906 0 le 119895 le 119899 can be found
as wellUsing Lemma 1 and (47) we obtain the lemmas below
Lemma 5 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has thepointwise estimates
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038161003816100381610038161003816119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038161003816100381610038161003816le 120581120576119896minus119894+((119904minus119898)2) exp(minus119888(
119909
120576
+
1 minus 119910
radic120576
))
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(48)
for a generic constant 119888 gt 0 independent of 119909 119910 and 120576
Lemma 6 For each 0 le 119895 le 119899 and 119896 119904 ge 0 one has for0 le 119894 + 119898 le 2119899 + 2 minus 2119895
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119897
10038171003817100381710038171003817119871119901(Ω)
+
10038171003817100381710038171003817119909119896
119910119904
120597119894
119909120597119898
119910120577119895
119906
10038171003817100381710038171003817119871119901(Ω)
le 120581120576119896minus119894+((119904minus119898)2)+(32119901)
(49)
Thanks to (46) and Lemma 5 we estimate the values of120577119895 0 le 119895 le 119899 along the sides 119909 = 1 119910 = 0 and 119910 = 1
120577119895
=
est at 119909 = 1
120577119895
119897(119909 0) + est at 119910 = 0
120577119895
119906(119909 0) + est at 119910 = 1
(50)
Now from (43) (46) and (50) we observe that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897) + est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
) =
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906) + est at 119910 = 1
(51)
35 Supplementary Correctors Now our task is to handle ateach order 0 le 119895 le 119899 the effects of 120579119895 and 120577
119895 along the topand bottom boundaries 119910 = 0 1 To this end using (51) weconstruct the supplementary correctors 120578119895 in the form
120578119895
= 120578119895
119897+ 120578119895
119906 0 le 119895 le 119899 (52)
where
120578119895
119897= minus (1 minus 119910) (120579
119895
+ 120577119895
119897) (119909 119910 = 0)
120578119895
119906= minus119910 (120579
119895
+ 120577119895
119906) (119909 119910 = 1)
(53)
From Lemmas 3 and 5 it is easy to verify the lemmasbelow
Lemma 7 For each 0 le 119895 le 119899 and 119904 ge 0 one has the pointwiseestimates
10038161003816100381610038161003816119909119904
120597119894
119909120597119898
11991012057811989510038161003816100381610038161003816le 120581120576119904minus119894 exp (minus119888
119909
120576
) 0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(54)
for a constant 119888 independent of 119909 119910 and 120576
Lemma 8 For each 0 le 119895 le 119899 and 119904 ge 0 one has
10038171003817100381710038171003817119909119904
120597119894
119909120597119898
11991012057811989510038171003817100381710038171003817119871119901(Ω)
le 120576119904minus119894+(1119901)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895
(55)
Using (3) (14) (17) (22) (23) (38) and (46) we find thatnear the corner (0 0)
(1205790
+ 1205770
119897) (0 0) = 119892
1(0) minus 119906
0
(0 0) minus 1205930
119897(0 0)
= 1198921(0) minus 119906
0
(0 0) minus (1198923(0) minus 119906
0
(0 0)) = 0
(56)
More generally one can verify that
(120579119895
+ 120577119895
119897) (0 0) = 0 (120579
119895
+ 120577119895
119906) (0 1) = 0 0 le 119895 le 119899
(57)
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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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
8 International Journal of Differential Equations
Using (42) (50) and (57) we infer from (52) and (53) thatfor 0 le 119895 le 119899
120578119895
=
0 at 119909 = 0
est at 119909 = 1
minus (120579119895
+ 120577119895
119897) at 119910 = 0
minus (120579119895
+ 120577119895
119906) at 119910 = 1
(58)
Now we finally obtain from (51) and (58) that
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895
)
=
0 at 119909 = 0
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
+ 120578119895
) = est at 119909 = 1
119899
sum
119895=0
120576119895
(120593119895
119906+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119897+ 120578119895
) + est = est at 119910 = 0
119899
sum
119895=0
120576119895
(120593119895
119897+ 120579119895
+ 120577119895
+ 120578119895
)
=
119899
sum
119895=0
120576119895
(120579119895
+ 120577119895
119906+ 120578119895
) + est = est at 119910 = 1
(59)
The est on the right hand side of (59)234
are causedrespectively by (120579
119895
+ 120577119895
+ 120578119895
) at 119909 = 1 (120593119895119906+ 120577119895
119906) at 119910 = 0
and (120593119895
119897+ 120577119895
119897) at 119910 = 1
4 Asymptotic Error Analysis The Main Result
We define
120588 = (119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
))
10038161003816100381610038161003816100381610038161003816100381610038161003816120597Ω
= (the right-hand side of (59))
(60)
where Θ119895
= 120593119895
+ 120585119895
+ 120579119895
+ 120577119895
+ 120578119895 as defined in (11)
We construct an extension 120588 of 120588 in the form
120588 (119909 119910) = 119903 + (1 minus 119909) (120588 minus 119903)1003816100381610038161003816119909=0
+ 119909 (120588 minus 119903)1003816100381610038161003816119909=1
+(1 minus 119910) (120588 minus 119903)1003816100381610038161003816119910=0
+ 119910 (120588 minus 119903)1003816100381610038161003816119910=1
(61)
where
119903 (119909 119910) = (1 minus 119909) (1 minus 119910) 1205881003816100381610038161003816119909=119910=0
+ 119909 (1 minus 119910) 1205881003816100381610038161003816119909=1119910=0
+ (1 minus 119909) 1199101205881003816100381610038161003816119909=0119910=1
+ 1199091199101205881003816100381610038161003816119909=119910=1
(62)
Then one can verify that 120588|120597Ω
= 120588 and
120588 is an est in Ω (63)
We set
119908120576119899
= 119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
) minus 120588 (64)
Then using (1) (7) (11) (30) (34) (37) (45) (53) and (60)the equation of 119908
120576119899reads
119871120576119908120576119899
= 120576119899+1
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)
+ 120576119899
(1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) + est in Ω
119908120576119899
= 0 on 120597Ω
(65)
where
Ξ1= (1 minus 119910) 120597
2
119910120579119899
(119909 119910 = 0) + 119910 1205972
119910120579119899
(119909 119910 = 1)
Ξ2= (1 minus 119910) (120597
2
119910120579119899minus1
+ 1205761205972
119910120577119899
119897) (119909 119910 = 0)
+ 119910 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
119906) (119909 119910 = 1)
(66)
We multiply (65) by 119890119909
119908120576119899
and integrate over Ω Thenusing Lemmas 1 3 and 5 and the Hardy inequality (see eg[2]) we find that
1205761003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671 +
1 minus 120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712
le 120581120576119899+1
1003816100381610038161003816100381610038161003816
int
Ω
(Δ119906119899
+ 1205972
119909120593119899
+ 1205972
119910120579119899
+ Ξ119899
1)119908120576119899119889119909 119889119910
1003816100381610038161003816100381610038161003816
+ 120581120576119899
1003816100381610038161003816100381610038161003816
int
Ω
119909 (1205972
119910120579119899minus1
+ 1205761205972
119910120577119899
+ Ξ119899
2) (
119908120576119899
119909
)119889119909119889119910
1003816100381610038161003816100381610038161003816
le 1205811205762119899+2
+
1
4
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198712 +
120576
2
1003816100381610038161003816119908120576119899
1003816100381610038161003816
2
1198671
(67)
Thanks to the elliptic regularity theory and (67) we alsofind that
1003817100381710038171003817119908120576119899
10038171003817100381710038171198672(Ω)
le 1205811003817100381710038171003817Δ119908120576119899
10038171003817100381710038171198712(Ω)
le 120581120576minus11003817
100381710038171003817120597119909119908120576119899
+ lower order terms10038171003817100381710038171198712(Ω)
le 120581120576119899minus(12)
(68)
Finally from (63) (64) (67) and (68) we obtain theconvergence result on the difference of 119906120576 and the proposedasymptotic expansions that we summarize in Theorem 9
Theorem 9 For each fixed 119899 ge 0 as the diffusivity parameter120576 vanishes the difference between the solution 119906
120576 of (1) andits asymptotic expansion (6) converges to zero in the followingsense
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
10038171003817100381710038171003817100381710038171003817100381710038171003817120576
le 120581120576119899+1
(69)
10038171003817100381710038171003817100381710038171003817100381710038171003817
119906120576
minus
119899
sum
119895=0
120576119895
(119906119895
+ Θ119895
)
100381710038171003817100381710038171003817100381710038171003817100381710038171198672(Ω)
le 120581120576119899minus(12)
(70)
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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Stochastic AnalysisInternational Journal of
International Journal of Differential Equations 9
Remark 10 Due to the continuity of 119892 as appearing in (3) 1205850can be omitted in (11) for 119895 = 0 at order 1205760Then one can verifythat
10038171003817100381710038171003817119906120576
minus (1199060
+ 1205930
+ 1205790
+ 1205770
+ 1205780
)
10038171003817100381710038171003817120576
le 12058112057634
(71)
which gives a convergence estimate that a factor of 12057614 is
worse than what is stated in (69) for 119899 = 0
Appendix
A Proof of Lemma 1
To prove (27) for 119910 = 119910radic120576 since 119910119904 exp(minus119888119910) le
1205811205761199042 exp(minus119888
0119910) for all 119910 0 lt 119888
0lt 119888 it suffices to prove that
10038161003816100381610038161003816120597119894
119909120597119898
119910120593119895
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119898 le 2119899 + 2 minus 2119895 119894 119898 ge 0
(A1)
Note that (28) would follow similarly Since ℎ119895
(119909) = ℎ119895
0(119909) =
minus119906119895
(119909 119894) minus 120574119895
0(119909) for 119895 ge 1 and 119892
3(119909) minus 119906
0
(119909 0) minus 1205740
0(119909) for
119895 = 0 thanks to (3) we note that 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le
2119899 + 1 minus 2119895Now to prove (A1) we use an induction on 119895
(1) We begin with 119895 = 0
From (25) since 120597119896
119909ℎ0
(1) = 0 for 0 le 119896 le 2119899 + 1differentiating 120593
0
119897in 119909 we find that
10038161003816100381610038161003816120597119894
1199091205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894
119909ℎ0
(119909 +
1199102
21199102
1
)1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101 0 le 119894 le 2119899 + 2
(A2)
Using exp(minus1199102
12) le 120581 exp(minus119888119910
1) for all 119910
1ge 0 for some
119888 gt 0 (A1) for 119898 = 0 is proved Since minus1205930
119897119910 119910minus 1205930
119897119909= 0 for
119898 = 2(119896 + 1) 119896 ge 0 from (A2) we easily find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
1199101205930
119897
100381610038161003816100381610038161003816
=
10038161003816100381610038161003816120597119894+119896+1
1199091205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A3)
thus (A3) holds for 0 le 119894 + 119898 = 119894 + 2(119896 + 1) le 119896 + 1 + 2119899 + 2and this implies that (A1) holds for119898 = 2(119896+1) 119896 ge 0 119895 = 0
Since 120597119894+1
119909ℎ0
(1) = 0 0 le 119894+1 le 2119899+1 we now notice that
10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816= radic
2
120587
100381610038161003816100381610038161003816100381610038161003816
int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
) 120597119894+1
119909ℎ0
times (119909 +
1199102
21199102
1
)
119910
1199102
1
1198891199101
100381610038161003816100381610038161003816100381610038161003816
le 120581119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)119889 (minus119910minus1
1)
0 le 119894 + 1 le 2119899 + 2
(A4)
Integrating by parts in the last integral we deduce that10038161003816100381610038161003816120597119894
1199091205971199101205930
119897
10038161003816100381610038161003816
le 120581(radic1 minus 119909 exp(minus
1199102
4 (1 minus 119909)
)
+119910int
infin
119910radic2(1minus119909)
exp(minus
1199102
1
2
)1198891199101)
le 120581(exp(minus119888
119910
radic1 minus 119909
) + 119910 exp(minus119888
119910
radic1 minus 119909
))
le (since 119905 exp (minus119888119905) le 120581 exp (minus1198880119905)
forall119905 ge 0 for some 0 lt 1198880lt 119888)
le 120581 exp(minus1198880
119910
radic1 minus 119909
) 0 le 119894 + 1 le 2119899 + 2
(A5)
Hence for119898 = 2119896 + 1 119896 ge 0 using minus1205930
119897119910119910minus 1205930
119897119909= 0 again we
note from (A5) that10038161003816100381610038161003816120597119894
1199091205972119896+1
1199101205930
119897
10038161003816100381610038161003816=
10038161003816100381610038161003816120597119894+119896
1199091205971199101205930
119897
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
0 le 119894 + 119896 + 1 le 2119899 + 2
(A6)
thus (A6) holds for 0 le 119894 + 119898 = 119894 + 2119896 + 1 le 119896 + 2119899 + 2 andthis proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 = 0 Hence (A1) isproved for 119895 = 0
(2) We now assume that (A1) holds at any order less thanor equal to 119895 minus 1 and we want to prove it at the order119895
At order 119895 ge 1 we perform the analysis as followsWe first note that the homogeneous solution of 120593119895
119897 that is
the first integral of (26) for 120593119895119897 can be similarly estimatedWe
just replace ℎ0
0(119909) by ℎ
119895
0(119909) in the above analysis
Since 120597119896
119909ℎ119895
(1) = 0 for 0 le 119896 le 2119899+1minus2119895 the first integralis then estimated as in (A1) for 119895 = 0 We just replace theabove 2119899 + 2 by 2119899 + 2 minus 2119895 We now estimate the particularsolution of 120593119895
119897 that is the second integral of (26) denoted by
119868 For simplicity in the analysis below we write that
119869 = exp(minus
(119910 minus 1199101)2
41199091
) minus exp(minus
(119910 + 1199101)2
41199091
) (A7)
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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Differential EquationsInternational Journal of
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 International Journal of Differential Equations
Using the induction assumption at order 119895 minus 1 from (A1) for119895 minus 1 we note that for 0 le 2 + 119894 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 2 minus 2119895
1205972+119894
119909120593119895minus1
119897(1minus
1199101) = 0 (A8)
100381610038161003816100381610038161205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909 minus 1199091
)
le 120581 exp(minus119888
1199101
radic1 minus 119909
)
(A9)
Thanks to (A8) differentiating 119868 the second integral of (26)we find that
120597119894
119909119868 =
1
2radic120587
int
1minus119909
0
int
infin
0
119869
radic1199091
1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
0 le 119894 le 2119899 + 2 minus 2119895
(A10)
Since 0 le 1199091
le 1 minus 119909 le 1 |119869| le 2 exp(minus(119910 minus 1199101)2
41199091) le
2 exp(minus(119910minus1199101)2
4(1minus119909)) from (A9) we find that for 0 le 119894 le
2119899 + 2 minus 2119895
10038161003816100381610038161003816120597119894
119909119868
10038161003816100381610038161003816le 120581int
1minus119909
0
1
radic1199091
1198891199091
times int
infin
0
exp(minus
(119910 minus 1199101)2
4 (1 minus 119909)
) exp(minus119888
1199101
radic1 minus 119909
)1198891199101
le (setting 120573 = 2119888radic1 minus 119909)
le 120581int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
minus
120573
2 (1 minus 119909)
119910 +
1205732
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)int
infin
0
exp(minus
(1199101minus 119910 + 120573)
2
4 (1 minus 119909)
)1198891199101
le 120581 exp(minus119888
119910
radic1 minus 119909
)
(A11)
This proves (A1) for 119898 = 0 119895 ge 1For 119898 = 2(119896 + 1) 119896 ge 0 thanks to the induction
assumption on 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 +
2 + 2(119896 minus 119904) le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A12)
from (A11) we also note that
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 + 1 le 2119899 + 2 minus 2119895
(A13)
Since minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909 we thus find that
100381610038161003816100381610038161003816
120597119894
1199091205972(119896+1)
119910119868
100381610038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896+1
119909119868
10038161003816100381610038161003816+
119896
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 2 (119896 + 1) le 2119899 + 2 minus 2119895
(A14)
This proves (A1) for 119898 = 2(119896 + 1) 119896 ge 0 119895 ge 1Since 120597
119910exp(minus(119910minus119910
1)2
41199091) = minus120597
1199101
exp(minus(119910minus1199101)2
41199091)
differentiating (A10) in 119910 we observe that10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
120597119910
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
= 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
1205971199101
119869 sdot 1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
le (integrating by parts in 1199101)
le 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
exp(minus
1199102
41199091
)1205972+119894
119909120593119895minus1
119897(119909 + 119909
1 0) 119889119909
1
100381610038161003816100381610038161003816100381610038161003816
+ 120581
100381610038161003816100381610038161003816100381610038161003816
int
1minus119909
0
1
radic1199091
int
infin
0
119869 sdot 1205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101) 11988911991011198891199091
100381610038161003816100381610038161003816100381610038161003816
(A15)
where 119869 = exp(minus(119910 minus 1199101)2
41199091) + exp(minus(119910 + 119910
1)2
41199091) le
2 exp(minus(119910 minus 1199101)2
4(1 minus 119909)) Here we note that 120597119910119869 = minus120597
1199101
119869Thanks to the induction assumption on 119895 minus 1 from (A1) wethus note that for 0 le 2 + 119894 + 1 le 2119899 + 2 minus 2(119895 minus 1) that is0 le 119894 le 2119899 + 1 minus 2119895
100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 1199101)
10038161003816100381610038161003816le 120581 exp(minus119888
1199101
radic1 minus 119909
) (A16)
and for 0 le 2+ 119894 le 2119899+2minus2(119895minus1) that is 0 le 119894 le 2119899+2minus2119895100381610038161003816100381610038161205972+119894
1199091205971199101
120593119895minus1
119897(119909 + 119909
1 0)
10038161003816100381610038161003816le 120581 (A17)
As in (A11) we similarly find that for 0 le 119894 le 2119899 + 1 minus 2119895
10038161003816100381610038161003816120597119894
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
) (A18)
Hence for119898 = 2119896+1 119896 ge 0 using the induction assumptionat order 119895 minus 1 from (A1) we note that for 0 le 119894 + 119904 + 2 + 2(119896 minus
119904) minus 1 le 2119899 + 2 minus 2(119895 minus 1)
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) (A19)
from (A18) we also note that
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816le 120581 exp(minus119888
119910
radic1 minus 119909
)
for 0 le 119894 + 119896 le 2119899 + 1 minus 2119895
(A20)
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi 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
International Journal of Differential Equations 11
Using minus119868119910119910
minus 119868119909= 120593119895minus1
119897119909119909again we thus find that
10038161003816100381610038161003816120597119894
1199091205972119896+1
119910119868
10038161003816100381610038161003816
le
10038161003816100381610038161003816120597119894+119896
119909120597119910119868
10038161003816100381610038161003816+
119896minus1
sum
119904=0
100381610038161003816100381610038161003816
120597119894+119904+2
1199091205972(119896minus119904)minus1
119910120593119895minus1
119897
100381610038161003816100381610038161003816
le 120581 exp(minus119888
119910
radic1 minus 119909
) for 0 le 119894 + 2119896 + 1 le 2119899 + 2 minus 2119895
(A21)
This proves (A1) for 119898 = 2119896 + 1 119896 ge 0 119895 ge 1 Hence thelemma is proved
B Study of the Elliptic CornerCorrectors 120585
119895
= 120585119895
119897+ 120585119895
119906
For each 0 le 119895 le 119899 using the stretched variables
119883 =
1 minus 119909
2120576
119884 =
119910
2120576
(B1)
one can define an approximation 120585
119895
119897of 120585119895119897as a solution of the
system
minus1205972
119883120585
119895
119897minus 1205972
119884120585
119895
119897+ 2120597119883120585
119895
119897= 0
120585
119895
119897= 0 at 119883 = 0
120585
119895
119897= 120574119895
0(1 minus 2120576119883) at 119884 = 0
120585
119895
119897997888rarr 0 at 119883
2
+ 1198842
997888rarr infin 119884 gt 0
(B2)
The explicit expression of 120585119895
is given as (see [1])
120585
119895
119897=
119884
120587
int
infin
0
[
1198701(1199041)
1199041
minus
1198701(1199042)
1199042
]
times 120574119895
0(1 minus 2120576119904) exp (minus (119904 minus 119883)) 119889119904
(B3)
where 1198701is the modified Bessel function of the second kind
of the first order and
1199041= radic(119883 minus 119904)
2
+ 1198842 119904
2= radic(119883 + 119904)
2
+ 1198842 (B4)
Using an appropriate barrier function and the maximumprinciple (see [1]) one can verify that
1003816100381610038161003816100381610038161003816
120585
119895
119897(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581 exp (minus119888 (radic1198832+ 1198842minus 119883)) (B5)
for a constant 119888 independent of 120576 An approximation 120585
119895
119906of 120585119895119906
can be constructed in the same mannerUsing (33) and (B5) we notice that 120585119895 minus (120585
119895
119897+ 120585
119895
119906) = est
at 119910 = 0 1 We deduce from (17) (18) and (B2)2that 120574119895119896(1) =
0 and hence 120585119895
minus (120585
119895
119897+ 120585
119895
119906) = 0 at 119909 = 1 Moreover using
Lemma B1 below we see that 120585119895minus(120585
119895
119897+120585
119895
119906) = 0 at 119909 = 0Then
using these observations thanks to the maximum principlewe find that
1003816100381610038161003816100381610038161003816
120585119895
minus (120585
119895
119897+ 120585
119895
119906)
1003816100381610038161003816100381610038161003816
= est 0 le 119895 le 119899 (B6)
which implies that the elliptic corrector 120585119895 0 le 119895 le 119899
satisfies the pointwise estimates similar to (B5) as for itsapproximation 120585
119895
= 120585
119895
119897+ 120585
119895
119906
Lemma B1 There exists a positive constant 120581 independent of120576 such that
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le 120581min 119880 (119883) 119880 (119884) 0 le 119895 le 119899 (B7)
where
119880 (119885) =
1 if 0 le 119885 le 119886
1 minus 4(119885 minus 119886)2 if 119886 lt 119885 le 119887
0 if 119885 gt 119887
(B8)
and 119886 = 38120576 119887 = 119886 + (12)
To prove Lemma B1 we will recall below the weakmaximum principle (see eg [17])
We consider an operator 119871 of the form
119871119906 =
119898
sum
119894119895=1
119863119894(119886119894119895
119863119895119906) +
119898
sum
119894=1
(119863119894(119887119894
119906) + 119888119894
119863119894119906) + 119889119906 (B9)
where the coefficients 119886119894119895 119887119894 119888119894 and 119889 are assumed to be
measurable functions in a domainΩ119898
sub R119898 and119863119894denotes
120597120597119909119894 1 le 119894 le 119898
We assume that 119871 is strictly elliptic in Ω119898 that is there
exists 120582 gt 0 such that119898
sum
119894119895=1
119886119894119895
120585119894120585119895ge 120582
10038161003816100381610038161205851003816100381610038161003816
2
forall120585 isin R119898
(B10)
and that 119871 has bounded coefficients that is for someconstants Λ and ] ge 0 we have119898
sum
119894119895=1
1003816100381610038161003816100381611988611989411989510038161003816100381610038161003816
2
le Λ2
120582minus2
119898
sum
119894=1
(
1003816100381610038161003816100381611988711989410038161003816100381610038161003816
2
+
1003816100381610038161003816100381611988811989410038161003816100381610038161003816
2
) + 120582minus1
|119889| le ]2
(B11)
We also assume that
int
Ω119898
(119889V minus
119898
sum
119894=1
119887119894
119863119894V) le 0 forallV ge 0 V isin C
1
119888(Ω119898) (B12)
We say that a weakly differentiable function 119906 satisfies119871119906 = 0 (ge0 or le0) in the weak sense if
int
Ω119898
(
119898
sum
119894119895=1
119886119894119895
119863119895119906119863119894V +
119898
sum
119894=1
(119887119894
119906119863119894V minus 119888119894
119863119894119906V) minus 119889119906V) = 0
(le 0 or ge 0)
(B13)
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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
12 International Journal of Differential Equations
for all nonnegative functions V isin C1119888(Ω119898) provided that
119886119894119895
119863119895119906 + 119887119894
119906 and 119888119894
119863119894119906 + 119889119906 1 le 119894 le 119898 are locally integrable
Now we state the following maximum principle
Lemma B2 Under the conditions (B10) (B11) and (B12) if119906 isin 119867
1
(Ω119898) satisfies 119871119906 ge 0 or 119871119906 le 0 in the weak sense
then one has respectively
supΩ119898
119906 le sup120597Ω119898
119906+ or inf
Ω119898
119906 ge inf120597Ω119898
119906minus
(B14)
Thanks to Lemma B2 we can now prove Lemma B1
Proof of Lemma B1 Given 1205760
gt 0 thanks to (B2)4 we may
choose a regionΩ0= (119883 119884) isin [0infin)times[0infin) | 119883
2
+1198842
le
1198772
with 119877 sufficiently large so that |120585
119895
| lt 1205760on the circle
1198832
+ 1198842
= 1198772 119883119884 gt 0 We then define the barrier function
= (119883 119884) = 119862119880 (119883) + 1205760 (119883 119884) isin [0infin) times [0infin)
(B15)
where 119880(119883) is given in (B8) Here 119862 is a positive constantand it will be determined below Note that (119883 119884) isin 119867
1
(Ω0)
Going back to the elliptic problem (B2) and writing theelliptic operator 119871 as
119871120585
119895
= 120585
119895
119883119883+ 120585
119895
119884119884minus 2120585
119895
119883 (B16)
we find that the operator 119871 satisfies (B10) (B11) and (B12)We also find that for V isin C1
119888(Ω0) V ge 0
int
Ω0
(119883V119883
+ 119884V119884+ 2119883V)
= intint
119877
0
119862 (119880119883V119883
+ 2119880119883V) 119889119883119889119884 ge 0
(B17)
Indeed if 0 le 119883 le 119886 or119883 gt 119887 (119886 119887 as in (B8)) since119880119883
= 0we only have to consider the case 119886 lt 119877 le 119887 For this 119877 since119880 isin 119867
2
(Ω0) we note that int119877
0
(119880119883V119883+2119880119883V)119889119883 = int
119877
0
(minus119880119883119883
+
2119880119883)V119889119883 = int
119877
119886
(minus119880119883119883
+ 2119880119883)V119889119883 ge 0 for all non-negative
V isin C1119888(Ω0)
Since 119871(plusmn120585
119895
) = 0 we thus find that 119871( plusmn 120585
119895
) le 0 in theweak sense
Thanks to the cut-off function 120575(119909) choosing 119862 =
max119909isin[121]
|120574119895
(119909)| we find that
(119883 0) = 119862119880 (119883) + 1205760ge
10038161003816100381610038161003816120574119895
(119909)
10038161003816100381610038161003816120594[121]
(119909)
=
10038161003816100381610038161003816120574119895
(1 minus 2120576119883)
10038161003816100381610038161003816120594[014120576]
(119883) ge
10038161003816100381610038161003816119875119895
(119883)
10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816
120585
119895
(119883 0)
1003816100381610038161003816100381610038161003816
(0 119884) = 119862 + 1205760ge 0 =
1003816100381610038161003816100381610038161003816
120585
119895
(0 119884)
1003816100381610038161003816100381610038161003816
(119883 119884) ge 1205760gt
1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
on 120597Ω0with 119883
2
+ 1198842
= 1198772
119883 119884 gt 0
(B18)
Thus we note that plusmn 120585
119895
ge 0 on 120597Ω0 Using Lemma B2 we
conclude that for (119883 119884) isin Ω0
( plusmn 120585
119895
) (119883 119884) ge infΩ0
( plusmn 120585
119895
) ge inf120597Ω0
( plusmn 120585
119895
)
minus
= 0
(B19)
Hence we have1003816100381610038161003816100381610038161003816
120585
119895
(119883 119884)
1003816100381610038161003816100381610038161003816
le (119883 119884) = 119862119880 (119883) + 1205760 (B20)
Letting 119877 rarr infin we deduce that |120585119895
(119883 119884)| le 119862119880(119883) Sim-ilarly we can conclude that |120585
119895
(119883 119884)| le 119862119880(119884) This provesthe lemma
Acknowledgments
This work was supported in part by the Research Fund ofIndiana University by NSF Grants DMS 0906440 DMS1206438 and DMS 1212141 and by Basic Science ResearchProgram through theNational Research Foundation of Korea(NRF) funded by the Ministry of Education Science andTechnology (2012001167)
References
[1] S-D Shih and R B Kellogg ldquoAsymptotic analysis of a singularperturbation problemrdquo SIAM Journal onMathematical Analysisvol 18 no 5 pp 1467ndash1511 1987
[2] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press Cambridge UK 1988
[3] L Prandtl ldquoVerber Flussigkeiten bei sehr kleiner Reibungrdquo inProceedings of the Verk III Intem Math Kongr Heidelberg pp484ndash491 Teuber Leibzig Germany 1905
[4] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleiner Rei-bungrdquo in Verhandlung des 3 Internat Math Kongr Heidelbergpp 484ndash491 Teubner Leipzig Germany 1905
[5] L Prandtl Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik edited by W Tollmien HSchlichting and H Gortler Springer Berlin Germany 1961
[6] L Prandtl ldquoUber Flussigkeitsbewegung bei sehr kleinerReibungrdquo in Gesammelte Abhandlungen zur angewandtenMechanik Hydro- und Aerodynamik W Tollmien H Schlicht-ing and H Gortler Eds Springer Berlin Germany 1961
[7] J B Rauch and F J Massey III ldquoDifferentiability of solutionsto hyperbolic initial-boundary value problemsrdquo Transactions ofthe American Mathematical Society vol 189 pp 303ndash318 1974
[8] S Smale ldquoSmooth solutions of the heat and wave equationsrdquoCommentarii Mathematici Helvetici vol 55 no 1 pp 1ndash12 1980
[9] R Temam ldquoBehaviour at time 119905 = 0 of the solutions of semilin-ear evolution equationsrdquo Journal of Differential Equations vol43 no 1 pp 73ndash92 1982
[10] Q Chen Z Qin and R Temam ldquoNumerical resolution near119905 = 0 of nonlinear evolution equations in the presence ofcorner singularities in space dimension 1rdquo Communications inComputational Physics vol 9 no 3 pp 568ndash586 2011
[11] N Flyer and B Fornberg ldquoAccurate numerical resolution oftransients in initial-boundary value problems for the heatequationrdquo Journal of Computational Physics vol 184 no 2 pp526ndash539 2003
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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
International Journal of Differential Equations 13
[12] N Flyer and B Fornberg ldquoOn the nature of initial-boundaryvalue solutions for dispersive equationsrdquo SIAM Journal onApplied Mathematics vol 64 no 2 pp 546ndash564 2004
[13] G-M Gie ldquoAsymptotic expansion of the Stokes solutions atsmall viscosity the case of non-compatible initial datardquo Com-munications in Mathematical Sciences to appear
[14] J R Cannon The One-Dimensional Heat Equation vol 23of Encyclopedia of Mathematics and Its Applications Addison-Wesley Publishing Company Advanced Book Program Read-ing Mass USA 1984
[15] C-Y Jung and R Temam ldquoNumerical approximation oftwo-dimensional convection-diffusion equations with multipleboundary layersrdquo International Journal of Numerical Analysisand Modeling vol 2 no 4 pp 367ndash408 2005
[16] G-M Gie M Hamouda and R Temam ldquoBoundary layers insmooth curvilinear domains parabolic problemsrdquo Discrete andContinuous Dynamical Systems Series A vol 26 no 4 pp 1213ndash1240 2010
[17] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order vol 224 of Fundamental Principles ofMathematical Sciences Springer Berlin Germany 2nd edition1983
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