california state university, northridge existence and
TRANSCRIPT
CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
EXISTENCE AND UNIQUENESS RESULTS FOR EULER AND
NAVIER-STOKES EQUATIONS
A thesis submitted in partial fulfillment of the requirementsfor the degree of Master of Science
in Mathematics
by
Nikolay Jacob Olshansky
December 2012
The thesis of Nikolay Jacob Olshansky is approved:
Dr. Jacek Polewczak Date
Dr. Vladislav Panferov Date
Dr. Stephen Breen, Chair Date
California State University, Northridge
ii
Dedication
To my wife Lyudmila.
iii
Acknowledgements
Special thanks to Dr. Stephen Breen. I would also like to thank Dr. Jacek Polewczak andDr. Vladislav Panferov for being on my committee and taking the time to help improve thiswork.
I would like to thank all the instructors I have had the pleasure of taking a course with atCSUN.
iv
Table of Contents
Signature page ii
Dedication iii
Acknowledgements iv
Abstract vi
1 An Introduction to Vortex Dynamics for Incompressible Fluid Flows. 11.1 The Euler and the Navier-Stokes Equations. . . . . . . . . . . . . . . . . . 11.2 Symmetry Groups for the Euler and the Navier-Stokes Equations. . . . . . 21.3 Particle Trajectories. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 The Vorticity, a Deformation Matrix, and Some Elementary Exact Solutions. 71.5 Some Remarkable Properties of the Vorticity in Ideal Fluid Flows. . . . . . 171.6 Some Conserved Quantities in Ideal and Viscous Fluid Flows. . . . . . . . 211.7 Leray’s Formulation of Incompressible Flows. . . . . . . . . . . . . . . . . 28
2 Technical Background. 332.1 The Gronwall’s Inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 Elementary Properties of the Fourier Transform. . . . . . . . . . . . . . . . 352.3 Elementary Properties of the Poisson Equation. . . . . . . . . . . . . . . . 452.4 Calculus Inequalities in the Sobolev Spaces. . . . . . . . . . . . . . . . . . 542.5 Properties of the Mollifiers. . . . . . . . . . . . . . . . . . . . . . . . . . . 722.6 The Hodge’s Decomposition of Vector Fields and Properties of the Leray’s
Projection Operator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3 Energy Methods for the Euler and the Navier-Stokes Equations. 913.1 Energy Methods: Elementary Concepts. . . . . . . . . . . . . . . . . . . . 91
3.1.1 A Basic Energy Estimate. . . . . . . . . . . . . . . . . . . . . . . 913.1.2 Approximation of Inviscid Flows by High Reynold’s Number Vis-
cous Flows. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.2 Local-in-Time Existence of Solutions by Means of Energy Methods. . . . . 98
3.2.1 Global Existence of Solution to a Regularization of the Euler andthe Navier-Stokes Equations. . . . . . . . . . . . . . . . . . . . . . 98
3.2.2 Local-in-Time Existence of Solutions to the Euler and the Navier-Stokes Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
v
ABSTRACT
EXISTENCE AND UNIQUENESS RESULTS FOR EULER AND NAVIER-STOKES
EQUATIONS
By
Nikolay Jacob Olshansky
Master of Science in Mathematics
We prove existence and uniqueness of a smooth solution to each of the initial value prob-lems for the Euler and the Navier-Stokes equations in 3-D case on some time interval[0, T ] applying classical energy methods. We use an approximation scheme with mollifi-cation and various a priori estimates on Sobolev norms of velocity v. At first, we justifyglobal-in-time existence and uniqueness of solutions to regularizations of the Euler and theNavier-Stokes equations. Then we show that there exists a time T such that a unique solu-tion exists on the time interval [0, T ], and subsequence of regularized solutions convergesto this solution. In our work, we follow the book ”Vorticity and incompressible flow” byA. J. Majda and A. L. Bertozzi. Numerous details are explained and several justificationsare added.
We consider some properties of the Euler and the Navier-Stokes equations and a reformu-lation of these equations due to Leray, which excludes the pressure term fom the equa-tions. Also, we prove some properties of the Fourier transform, the Poisson equation,Sobolev spaces, mollifiers, Leray’s projection operator into divergence-free vector space,and Hodge’s decomposition of vector fields, all of which we use in this work.
vi
Chapter 1
An Introduction to Vortex Dynamics for Incompressible Fluid Flows.
Studying vorticity, the curl of the velocity field, is an important aspect of the mathematicaltheory of incompressible flows. In this chapter we introduce the Euler and the Navier-Stokes equations for incompressible flows, inviscid and viscous respectively, present ele-mentary properties of these equations, and consider some elementary examples. At the endof the chapter, we derive a reformulation of the equations due to Leray, which excludes thepressure term from the equations. Our exposition of this material follows chapter 1 of [6].
1.1 The Euler and the Navier-Stokes Equations.
Incompressible flows of homogeneous fluids in RN , N = 2, 3, are determined by the fluidvelocity field v(x, t) = (v1(x, t), v2(x, t), . . . , vN(x, t))t onRN×[0,∞). The mathematicalmodel of such flows, that we study, constitutes the initial value problem including theNavier-Stokes equation (1.1), the incompressibility condition (1.2), and the initial velocityfield (1.3),
Dv
Dt= −∇p+ν∆v, (1.1)
div v = 0, (x, t) ∈ RN × [0,∞), (1.2)
v(x, 0) = v0, x ∈ RN , (1.3)
where scalar p(x, t) is the pressure, D/Dt is the convective derivative,
D
Dt=
∂
∂t+
N∑j=1
∂
∂xj
dxjdt
=∂
∂t+
N∑j=1
vj∂
∂xj, (1.4)
the gradient operator∇ is
∇ =
(∂
∂x1
,∂
∂x2
, . . . ,∂
∂xN
)t, (1.5)
the Laplace operator ∆ is
∆ = ∇2 =N∑j=1
∂2
∂x2j
, (1.6)
and the divergence of a velocity field divv is
divv =N∑j=1
∂vj
∂xj. (1.7)
The constant kinematic viscosity ν ≥ 0 and ν is inversely proportional to Reynolds numberRe. For ν = 0 (inviscid flow) the equation (1.1) is called the Euler equation. The Euler
1
equation is used in many applications to approximate viscous solution for high Reynoldsnumbers (very small ν).
The Navier-Stokes equation (1.1) follows from the law of conservation of momentum [7].
Note that the equations (1.1), (1.2) contain the time derivative of velocity v only. Laterin this chapter we show that the pressure p could be excluded being determined by thevelocity v(x.t).
1.2 Symmetry Groups for the Euler and the Navier-Stokes Equations.
Next proposition reviews some elementary symmetry groups for solutions to the Euler andthe Navier-Stokes equations.
Proposition 1.1 (Symmetry groups for the Euler and the Navier-Stokes equations.). Letv, p be a solution to the Euler or the Navier-Stokes equations. Then the following transfor-mations also yield solutions:
(i) Galilean invariance: For any constant velocity vector c ∈ RN ,
vc(x, t) = v(x− ct, t) + c, pc(x, t) = p(x− ct, t) (1.8)
is also a solution pair.
(ii) Rotation symmetry: for any rotation matrix Q (Qt = Q−1),
vQ(x, t) = Qtv(Qx, t), pQ(x, t) = p(Qx, t) (1.9)
is also a solution pair.
(iii) Scale invariance:for any λ, τ ∈ R,
vλ,τ (x, t) = λτv(xλ, tτ
), pλ,τ (x, t) = λ2
τ2p(xλ, tτ
), (1.10)
is a solution pair to the Euler equation, and for any τ ∈ R+,
vτ (x, t) = τ−1/2v(
xτ1/2
, tτ
), pτ (x, t) = τ−1p
(x
τ1/2, tτ
)(1.11)
is also a solution pair to the Navier-Stokes equation.
Proof. Let v(x, t) and p(x, t) be a solution pair to the Euler (ν = 0) or to the Navier-Stokes(ν > 0) equations.
(i) Show that vc = v(x− ct, t) + c and pc = p(x− ct, t) are solutions of these equationsas well ∀c ∈ RN . For the i-th component of v(x, t) and ∇xp(x, t) by the equation(1.1), we have
∂vi
∂t+
N∑j=1
vj∂vi
∂xj= − ∂p
∂xj+ ν
N∑j=1
∂2vi
∂x2j
(∗)
2
Substitute vi and p by vic and pc respectively. Since
∂vic∂t
=N∑j=1
∂vi
∂xj(− cj) +
∂vi
∂t,
∂kvic∂xkj
=∂kvi
∂xkj, k = 1, 2,
∂pc∂xi
=∂p
∂xi,
then we get
∂vi
∂t+
N∑j=1
∂vi
∂xj(− cj) +
N∑j=1
(vj + cj)∂vi
∂xj= − ∂p
∂xi+ ν
N∑j=1
∂2vi
∂x2j
.
After cancelation in the left hand side, the last equation coincides with the identity(∗). So, vc(x, t) and pc(x, t) satisfy the Navier-Stokes equation (1.1).
Also, since div vc =∑N
i=1∂vic∂xi
=∑N
i=1∂vi
∂xi= div v = 0, then vc satisfies the
incompressibility condition (1.2) too.
(ii) Show that vQ(x, t) = Q−1v(Qx, t), pQ(x, t) = p(Qx, t) is also a solution pair tothe Euler and the Navier=Stokes equations.
Because Q is a rotation matrix, Q−1 = Qt = Q and Q2 = I , the identity matrix.
Substitute vQ and pQ in the equations (1.1) and (1.2) instead of v and p respectively.We have
Q−1∂v
∂t+
N∑j=1
Q−1vjQ−1 ∂v
∂xjQ = −∇xpQ+ νQ−1∆vQ2
and Q−1 div vQ = 0 or div v = 0.Multiplying the first of these equations by Q, we get
∂v
∂t+
N∑j=1
vj ∂v∂xj
= (−∇xp+ ν∆v) I.
So, vQ and pQ is a solution pair to the Euler and Navier-Stokes equations.
(iii) Now substitute into the Euler equation and into the equation (1.2), vλ,τ = λτv(xλ, tτ
)and pλ,τ = λ2
τ2p(xλ, tτ
). We have
λ
τ 2
∂v
∂t+
N∑j=1
λ
τvjλ
τ
∂v
∂xj
1
λ= − λ2
τ 2∇xp
1
λand
λ
τdiv v
1
λ= 0.
Multiplying the first of these equations by τ2
λand the second one by τ , we get the
Euler equation and the incompressibility condition correspondingly which are iden-tities for the solution pair v and p. So, vλ,τ and pλ,τ is a solution pair to the Eulerequation.
3
Finally, substitute into the equations (1.1) and (1.2) vτ = 1√τv(
x√τ, tτ
)and pτ =
1τp(
x√τ, tτ
). Then we have
1√τ
∂v
∂t
1
τ+
N∑j=1
1√τvj
1√τ
∂v
∂xj
1√τ
= − 1
τ∇xp
1√τ
+ ν
(1√τ
)3
∆v
and
1√τ
div v1√τ
= 0.
Multiplying the first of these equations by τ 3/2 and the second one by τ , we get theNavier-Stokes equation and the incompressibility condition correspondingly whichare again identities for the solution pair v and p. Thus, vτ and pτ is a solution pair tothe Navier-Stokes equation.
Notice that inviscid flows require a two-parameter symmetry group for scaling transforma-tions while flows with viscosity need only a one parameter group for that.
1.3 Particle Trajectories.
Definition 1.1. Define the particle-trajectory mappingX( · , t) : α ∈ RN 7→ X(α, t) ∈ RNwhere α = (α1, α2, . . . , αN)t is the initial point of fluid particle location at time t = 0and X(α, t) = (X1, X2, . . . , XN)t is the location at time t. The parameter α is calledthe Lagrangian particle marker. For given fluid velocity v(x, t), the particle-trajectorymapping is determined by the following initial value problem:
dX(α, t)
dt= v (X(α, t), t) , X(α, 0) = α (1.12)
Note that in general the ODE (1.12) is nonlinear.
Definition 1.2. Consider an initial domain Ω ⊂ RN in fluid at time t = 0. In time t, fluidparticles move from the domain Ω to X(Ω, t) = X(α, t) : α ∈ Ω. Define the Jacobianof this transformation by
J(α, t) = det (∇αX(α, t)) , (1.13)
where∇α =[∂∂α1
, . . . , ∂∂αN
]Next we consider a relation between rate of change of J and fluid velocity v. To do this,we need the following lemma about derivative of determinant.
4
Lemma 1.1. Let A(t) = [aij(t)] with i, j = 1, 2, . . . , N be a matrix of order N and itselements aij ∈ C1 ((t1, t2)). Then ∀t ∈ (t1, t2), d
d tdet(A) =
∑Ni,j=1
d aijd t
Aij where Aij arecofactors of elements aij .
Proof. Consider the set of columns cj = [aij]Ni=1, j = 1, 2, . . . , N . ThenA = [c1 c2 . . . cN ].
Hence
d
d tdet (A(t)) = lim
h→0
1h[det (A(t+ h))− det (A(t))] =
limh→0
1hdet ([c1(t+ h) c2(t+ h) . . . cN(t+ h)])− det ([c1(t) c2(t) . . . cN(t)]) =
limh→0
1hdet([c1(t+ h) c2(t+ h) . . . cN(t+ h)])− det([c1(t) c2(t+ h) . . . cN(t+ h)])
+ det ([c1(t) c2(t+ h) c3(t+ h) . . . cN(t+ h)])−det ([c1(t) c2(t) c3(t+ h) . . . cN(t+ h)]) +
det ([c1(t) c2(t) c3(t+ h) . . . cN(t+ h)])− . . .− det ([c1(t) c2(t) . . . cN−1(t) cN(t+ h)]) +
det ([c1(t) c2(t) . . . cN−1(t) cN(t+ h)])− det ([c1(t) c2(t) . . . cN−1(t) cN(t)]) =
limh→0
det([
c1(t+h)−c1(t)h
c2(t+ h) . . . cN(t+ h)])
+
limh→0
det([c1(t) c2(t+h)−c2(t)
hc3(t+ h) . . . cN(t+ h)
])+ . . .
+ limh→0
det([c1(t) c2(t) . . . cN−1(t) cN (t+h)−cN (t)
h
])=
det([
d c1d t
c2 . . . cN])
+ det([c1
d c2d t
c3 . . . cN])
+ · · ·+ det([c1 c2 . . . cN−1
d cNd t
]),
using the multilinearity of the determinant.
Decompose each determinant in this sum by column consisting of derivatives. Then we get
d
d tdet [A(t)] =
N∑i=1
d
d t[ai1(t)] Ai1(t) +
N∑i=1
d
d t[ai2(t)] Ai2(t) + . . .
+N∑i=1
d
d t[aiN(t)] AiN(t) =
N∑i,j=1
d
d t[aij(t)] Aij(t).
Proposition 1.2 is an application of this lemma.
Proposition 1.2. Let X(·, t) be a particle-trajectory mapping of a smooth velocity fieldv ∈ RN . Then
∂J
∂ t= (divx v)|(X(α,t),t) J(α, t). (1.14)
5
Proof. Consider
∇αX(α, t) = [∇αX1(α, t) ∇αX2(α, t) . . . ∇αXN(α, t)] =
∂X1
∂α1
∂X1
∂α2. . . ∂X1
∂αN
∂X2
∂α1
∂X2
∂α2. . . ∂X2
∂αN
. . . . . . . . . . . .∂XN∂α1
∂XN∂α2
. . . ∂XN∂αN
.
Let Aij be cofactors of the elements ∂Xi∂αj
. Then decomposing det (∇αX(α, t)) by the k-thcolumn (1 ≤ k ≤ N) and using well-known properties of determinants, we have
N∑j=1
∂Xk
∂αjAij = δik det (∇αX(α, t)) = δik J,
where Kronecker’s symbol δik =
1, if k = i
0, if k 6= i.
Now by the definition of J (see (1.13)) and the Lemma 1.1, we obtain
∂J
∂t=∂
∂tdet
[∂Xi(α, t)
∂αj
]=
N∑i,j=1
∂
∂t
(∂Xi(α, t)
∂αj
)Aij =
N∑i,j=1
∂
∂αj
(∂Xi(α, t)
∂t
)Aij =
N∑i,j=1
∂vi (X(α, t))
∂αjAij =
N∑i,j=1
Aij
N∑k=1
viXk∂Xk
∂αj=
N∑i,k=1
viXk
N∑j=1
AijXk
∂αj=
N∑i,k=1
viXkδikJ =
(N∑i=1
viXi
)J = divX [v (X(α, t), t)] J.
As an immediate consequence of this proposition, we get that the incompressibility con-dition (1.2) implies J(α, t) is constant in time. Moreover, since at the time t = 0,X(α, 0) = α, then J(α, 0) = det(∇αα) = det(I) = 1, so J(α, t) = J(α, 0) = 1,∀t > 0.
In its turn, the fact that the Jacobian of the transformation is 1 implies that the flow preservesthe volume of a domain.
Definition 1.3. A flow X(·, t) is incompressible if for all subdomains Ω with smooth boun-daries and any t > 0, the flow is volume preserving:
volX(Ω, t) = vol Ω.
Next, a calculus formula determines the rate of change of any smooth function in domainX(Ω, t) moving with the fluid.
Proposition 1.3 (The Transport Formula). Let Ω ⊂ RN be an open, bounded domain with
6
a smooth boundary, and let X be a given particle-trajectory mapping of a smooth velocityfield v. Then for any smooth function f(x, t),
d
d t
∫X(Ω,t)
f dx =
∫X(Ω,t)
[ft + divx(fv)] dx. (1.15)
Proof. Use the change of variables, at first returning from X(Ω, t) to the initial domain Ωand then moving back to X(Ω, t).∫
X(Ω,t)
f(x, t) dx =
∫Ω
f (X(α, t), t) J(α, t) dα.
Then applying Proposition 1.2, we get
d
d t
[∫X(Ω,t)
f(x, t) dx
]=
∫Ω
[(∂f
∂ t+
N∑j=1
∂f
∂Xj
∂Xj
∂ t
)J(α, t)+f(X(α, t), t)
∂J
∂ t
]dx =
∫Ω
[(∂f
∂ t+
N∑j=1
∂f
∂Xj
vj
)J + f
∂J
∂ t
]dx =
∫Ω
(∂f
∂ t+∇Xf · v + f divxv
)Jdx =∫
Ω
[∂f
∂ t+ divx(fv)
]Jdx =
∫X(Ω,t)
[∂f
∂ t+ divx(fv)
]dx.
Note that for f ≡ 1, the Transport formula gives
d
d tvolX(Ω, t) =
∫X(Ω,t)
divx f dx,
which explains the equivalence of volume preservation and the incompressibility conditiondivx v = 0.
We summarize the results in the following proposition.
Proposition 1.4. For the smooth flow the following three conditions are equivalent:
(i) a flow is incompressible, i.e., ∀Ω ⊂ RN , t ≥ 0, volX(Ω, t) = vol Ω,
(ii) div v = 0,
(iii) J(α, t) = 1,∀t ≥ 0.
1.4 The Vorticity, a Deformation Matrix, and Some Elementary Exact Solutions.
In this section we consider some models of typical local behavior of incompressible flow.
7
Definition 1.4. Consider 3×3 matrix∇v = [vxj ]. Then∇v has the decomposition
∇v = D + Ω, (1.16)
where D is a symmetric matrix, and Ω is an antisymmetric matrix, given respectively by
D = 12
[∇v + (∇v)t
], (1.17)
Ω = 12
[∇v − (∇v)t
]. (1.18)
D is called the deformation or rate-of-strain matrix and Ω is called the rotation matrix.
Note that div v =∑3
i=1 vixi
= trD = tr(∇v). So, for the incompressible flow, tr D = 0.
Next, recall the definition of the vorticity.
Definition 1.5. Let v be a smooth vector field. The vorticity or rotor of the vector field v is
ω = curl v ≡ ∇× v =(∂v3
∂x2− ∂v2
∂x3, ∂v1
∂x3− ∂v3
∂x1, ∂v2
∂x1− ∂v1
∂x2
). (1.19)
Remark. The rotation matrix Ω and the vorticity ω of the vector field v satisfy the followingrelation:
Ωh = 12ω × h, ∀h ∈ R3. (1.20)
Proof. Let ω = (ω1, ω2, ω3)t be defined by the equation (1.19). Then Ω= 12[∇v−(∇v)t]
= 12
0
(∂v1
∂x2− ∂v2
∂x1
) (∂v1
∂x3− ∂v3
∂x1
)(∂v2
∂x1− ∂v1
∂x2
)0
(∂v2
∂x3− ∂v3
∂x2
)(∂v3
∂x1− ∂v1
∂x3
) (∂v3
∂x2− ∂v2
∂x3
)0
= 12
0 −ω3 ω2
ω3 0 −ω1
−ω2 ω1 0
.
Let h = (h1, h2, h3)t ∈ RN . Consider Ωh = 12
(−ω3h2 + ω2h3)( ω3h1 − ω1h3)(−ω2h1 + ω1h2)
= 12ω × h.
The following lemma provides a simple local description for an incompressible fluid flow.
Lemma 1.2. To linear order in (|x−x0|), every smooth incompressible velocity field v(x, t)is the unique sum of three terms:
v(x, t0) = v(x0, t0) + 12ω × (x− x0) +D(x− x0), (1.21)
where D is the symmetric deformation matrix with trD = 0 and ω is the vorticity.
8
Proof. Let x0, x ∈ RN and t0 ≥ 0. Using a Taylor series expansion for a smooth velocityfield v(x, t) at a fixed point (x0, t0), we get
v(x, t0) = v(x0, t0) +∇x [v(x0, t0)] (x− x0) +O(|x− x0|2
).
Then applying the definition (1.16) and the remark (1.20), we have
v(x, t0) = v(x0, t0) + Ω(x− x0) +D(x− x0) +O(|x− x0|2
)=
v(x0, t0) + 12ω × (x− x0) +D(x− x0) +O
(|x− x0|2
).
So, the equation (1.21) holds to linear order in (|x− x0|).
The next corollary gives a natural physical interpretation to each of three terms in theequation (1.21).
Corollary 1.1. To linear order in (|x− x0|), every smooth incompressible velocity fieldv(x, t) is the sum of infinitesimal translation, rotation, and deformation velocities.
Proof. Consider the initial value problem (1.12) for a particle-trajectory mapping corres-ponding to a fluid velocity field v(x, t).
1) Retaining the term v(x0, t0) only in the equation (1.21), we have
dX
d t= v(x0, t0), X|t=t0 = α.
Integrating from t0 to t > t0, we get
X(α, t) = α + v(x0, t0)(t− t0).
This equation describes an infinitesimal translation.
2) Now retain the second term 12ω × (x− x0) only in the equation (1.21). We obtain
dX
d t=
1
2ω × (X −X0), X|t=t0 = α.
Decompose vorticity ω = (ω1, ω2, ω3)t = (ω1, 0, 0)t + (0, ω2, 0)t + (0, 0, ω3)t
and consider a solution of the ODE above as the superposition of solutions of thisODE with ω replaced by each of its components separately. So for the third compo-nent of the ω, we have
dX
d t=
1
2
00ω3
×X1 −X1
0
X2 −X20
X3 −X30
=ω3
2
−(X2 −X20 )
(X1 −X10 )
0
.9
Consider the ODEs for each row of the matrices.dX1
d t= −ω3
2(X2 −X2
0 )dX2
d t= ω3
2(X1 −X1
0 ) X i|t=t0 = αi, i = 1, 2, 3.dX3
d t= 0,
Differentiating the first of these equations and applying the second one to the result,we get
d2(X1 −X10 )
d t2= −(ω3)2
4(X1 −X1
0 ).
Solve this equation. Then substituting the solution we get to the first equation fromthe system above, we obtain
X1(t)−X10 = C1 cos
(ω3t2
)+ C2 sin
(ω3t2
), X1|t=t0 = α1,
X2(t)−X20 = C1 sin
(ω3t2
)− C2 cos
(ω3t2
), X2|t=t0 = α2,
X3(t) = α3,
where C1, C2 ∈ R.
Introduce the following two-row matrices:
X ′ =
[X1
X2
], X ′0 =
[X1
0
X20
], α′ =
[α1
α2
], C ′ =
[C1
C2
],
and the rotation in the plane X1OX2 matrix Q(φ) =
[cosφ − sinφsinφ cosφ
], where φ is
an angle of rotation about axis parallel to the X3 axis.
Then we have:
X ′(t)−X ′0 = Q(ω3t2
)C ′, α′ −X ′0 = Q(ω3t02
)C ′.
Exclude C ′ and take in count that Q−1 = Qt and that
Q(ω3t2
)Q(ω3t02
) =
cos(ω3t2
)− sin
(ω3t2
)sin(ω3t2
)cos(ω3t2
) cos(ω3t0
2
)sin(ω3t0
2
)− sin
(ω3t0
2
)cos(ω3t0
2
) =
cos(ω3(t−t0)
2
)− sin
(ω3(t−t0)
2
)sin(ω3(t−t0)
2
)cos(ω3(t−t0)
2
) = Q(ω3(t−t0)
2
).
Hence C ′ = Qt(ω3t0
2
)(α′ −X ′0) and X ′(t) −X ′0 = Q
(ω3t2
)Qt(ω3t0
2
)(α′ −X ′0)
or X ′(t)−X ′0 = Q(ω3(t−t0)
2
)(α′ −X ′0).
Note that det(Q) = 1 and ‖X ′(t) − X ′0‖ = ‖α′ − X ′0‖, ∀t > 0. Therefore these
10
trajectories are circles centered atX0 on the planeX3 = α3 parallel to the coordinateplane X1OX2. So, replacing ω by its third component, we get a rotation in thedirection of (0, 0, ω3)t with angular velocity 1
2|ω3|.
Similarly, for the first and second components of ω, we get rotations in the directionsof (ω1, 0, 0)t and (0, ω2, 0)t with angular velocities 1
2|ω1| and 1
2|ω2| respectively.
Superposing these solutions for each components of ω, we conclude that the secondterm 1
2ω×(X−X0) in the equation (1.21) is an infinitesimal rotation in the direction
of ω with angular velocity 12|ω|.
3) Finally, retain the term D(X − X0) only in the particle-trajectory equation (1.21).Since D is a symmetric matrix, then there exists a rotation matrix Q such thatQtDQ = diag(γ1, γ2, γ3). Then tr(D) =tr(QtDQ) = γ1 + γ2 + γ3 = 0 becausetraces are invariant under similarity transformations. So, γ3 = −(γ1 + γ2). By therotation symmetry (the Proposition 1.1(ii)), we can replace v = D(X−X0) by vQ =QtDQ(X −X0) = diag(γ1, γ2, −(γ1 + γ2))(X −X0).
Thus, we have the following initial value problem:
dX
d t= diag(γ1, γ2, −(γ1 + γ2))(X −X0), X|t=t0 = α.
After integration, we get
X(t)−X0 = diag(eγ1t, eγ2t, e−(γ1+γ2)t
)C, where C ∈ R3.
Using the initial value X|t=t0 = α and the fact that diag−1(a1, a2, a3) =
diag(a−11 , a−1
2 , a−13 ) with a1, a2, a3 ∈ R and a2
1 + a22 + a2
3 6= 0, we have
C = diag(e−γ1t0 , e−γ2t0 , e(γ1+γ2)t0
)(α−X0).
Therefore
X(t) = X0 + diag(eγ1(t−t0), eγ2(t−t0), e−(γ1+γ2)(t−t0)
)(α−X0).
This solution describes a deformation of the fluid. For example, let X0 = 0 andγ1, γ2 < 0. Then the fluid particles create a jet along X3 axis (see Ex. 1.1).
In general, the third term D(X −X0) in the equation (1.21) represents an infinitesi-mal deformation velocity in the direction (X −X0).
So, we conclude that up to linear order in (|X −X0|), incompressible velocity fieldis the sum of infinitesimal translation, rotation, and deformation velocities.
In the next proposition, we consider a large class of exact solutions to the Euler and theNavier-Stokes equations. To prove this proposition, we will need the following lemma.
11
Lemma 1.3. Let A be a real, symmetric, 3×3 matrix with trA = 0, and b, c ∈ R3. Then
−A(b× c) = b× (Ac) + (Ab)× c. (1.22)
Proof. Here we provide a straightforward computational proof.
Let A =
a11 a12 a13
a12 a22 a23
a13 a23 a33
with a11 + a22 + a33 = 0, b =
b1
b2
b3
, and c =
c1
c2
c3
.
1) Compute L.H.S. of the equation (1.22).
b× c = (b2c3 − b3c2, b3c1 − b1c3, b1c2 − b2c1)t,
−A(b× c) = −
[a11(b2c3 − b3c2) + a12(b3c1 − b1c3) + a13(b1c2 − b2c1)][a12(b2c3 − b3c2) + a22(b3c1 − b1c3) + a23(b1c2 − b2c1)][a13(b2c3 − b3c2) + a23(b3c1 − b1c3) + a33(b1c2 − b2c1)]
=
[b1(a12c3 − a13c2) + b2(a13c1 − a11c3) + b3(a11c2 − a12c1)][b1(a22c3 − a23c2) + b2(a23c1 − a12c3) + b3(a12c2 − a22c1)][b1(a23c3 − a33c2) + b2(a33c1 − a13c3) + b3(a13c2 − a23c1)]
.2) Now compute the R.H.S. of the equation (1.22).
b× (Ac) =
[b2(a13c1 + a23c2 + a33c3)− b3(a12c1 + a22c2 + a23c3)][b3(a11c1 + a12c2 + a13c3)− b1(a13c1 + a23c2 + a33c3)][b1(a12c1 + a22c2 + a23c3)− b2(a11c1 + a12c2 + a13c3)]
(Ab)× c =
[(a12b1 + a22b2 + a23b3)c3 − (a13b1 + a23b2 + a33b3)c2][(a13b1 + a23b2 + a33b3)c1 − (a11b1 + a12b2 + a13b3)c3][(a11b1 + a12b2 + a13b3)c2 − (a12b1 + a22b2 + a23b3)c1]
=
[b1(a12c3 − a13c2) + b2(a22c3 − a23c2) + b3(a23c3 − a33c2)][b1(a13c1 − a11c3) + b2(a23c1 − a12c3) + b3(a33c1 − a13c3)][b1(a11c2 − a12c1) + b2(a12c2 − a22c1) + b3(a13c2 − a23c1)]
Hence the sum b× (Ac) + (Ab)× c =[b1(a12c3 − a13c2) + b2(a13c1 + a33c3 + a22c3) + b3(−a12c1 − a22c2 − a33c2)]
[b1(−a23c2 − a33c3 − a11c3) + b2(a23c1 − a12c3) + b3(a11c1 + a12c2 + a33c1)][b1(a22c2 + a23c3 + a11c2) + b2(−a11c1 − a13c3 − a22c1) + b3(a13c2 − a23c1)]
Finally, using that tr(A) = 0, we get b× (Ac) + (Ab)× c =[b1(a12c3 − a13c2) + b2(a13c1 − a11c3) + b3(a11c2 − a12c1)]
[b1(a22c3 − a23c2) + b2(a23c1 − a12c3) + b3(a12c2 − a22c1)][b1(a23c3 − a33c2) + b2(a33c1 − a13c3) + b3(a13c2 − a23c1)]
= −A(b× c).
12
Consider now the following important proposition.
Proposition 1.5. Let D(t) be a real, symmetric, 3× 3 matrix with trD(t) = 0. Determinethe vorticity ω(t) from the ODE in R3,
dω
d t= D(t)ω, ω|t=0 = ω0 ∈ R3, (1.23)
and the antisymmetric matrix Ω by means of the formula Ωh = 12ω × h, h ∈ R3. Then
v(x, t) = 12ω(t)× h+D(t)x, (1.24a)
p(x, t) = −12
[Dt(t) +D2(t) + Ω2(t)
]x · x (1.24b)
are exact solutions to the three-dimensional Euler and the Navier-Stokes equations.
Proof.
1) Consider componentwise the Navier-Stokes equation (1.1).
vit +3∑j=1
vjvixj = −pxi + ν∆xvi, i = 1, 2, 3.
Differentiating this equation by xk, k = 1, 2, 3, we have
(vixk)t +3∑j=1
vj(vixk)xj +3∑j=1
vjxkvixj = −pxixk + ν∆x(v
ixk
)
orDDt
(vixk) +3∑j=1
vjxkvixj
= −pxixk + ν∆x(vixk
).
Introduce the notation V ≡ ∆xv = [vixk ] and P ≡ [pxixk ], Hessian matrix of thepressure p.
Then from the last equation we get the following matrix equation for V.
DV
D t+ V 2 = −P + ν∆V. (1.25)
2) Represent V again as a sum of its symmetric partD and antisymmetric part Ω. HenceV 2 = (D + Ω)2 = (D2 + Ω2) + (DΩ + ΩD), where the first term is symmetric andthe second term is antisymmetric.
Note that P is a symmetric matrix.
So, the symmetric part of the equation (1.25) is
DDD t
+D2 + Ω2 = −P + ν∆D, (1.26)
13
and the antisymmetric part of the equation (1.25) is
DΩ
D t+DΩ + ΩD = ν∆Ω. (1.27)
To get an equation in terms of vorticity, use that Ωh = 12ω × h, ∀h ∈ R3.
Multiplying the matrix equation (1.27) by h from the right, we have
D(Ωh)
D t+D(Ωh) + Ω(Dh) = ν∆(Ωh)
orD
Dt(ω × h) +D(ω × h) + ω × (Dh) = ν∆(ω × h).
Applying the lemma 1.3, we get
D(ω × h) = −ω × (Dh)− (Dω)× h.
Then we obtainDω
Dt× h− (Dω)× h = ν(∆ω × h)
or (Dω
Dt−Dω − ν∆ω
)× h = 0, ∀h ∈ R3,
that givesDω
Dt= Dω + ν∆ω. (1.28)
The equations (1.26) and (1.28) have fundamental meaning in the incompressibleflow theory.
3) Assume that both D and ω do not depend on the spatial variable x and that
v(x, t) = 12ω(t)× x+D(t)x.
Then ∇xω, ∆xω = 0, DωDt
= dωdt
+ v · ∇xω = dωdt
.
So, the equation for vorticity (1.28) reduces to the ODE dωdt
= Dω.
4) It remains to find the pressure p by solving the equation (1.26), the symmetric partof the Navier-Stokes equation.
Since under our assumption ∆xD(t) = 0, and Ω does not depend on x becauseΩh = 1
2ω(t)h, then the equation (1.26) gives −P (t) = dD
dt+D2 + Ω2.
Note that the right-hand side of this equation is a known symmetric matrix.
Because second order derivatives of p by spatial variable, elements of the Hessianmatrix P , do not depend on x, it is reasonable to represent p(x, t) as a quadraticfunction of x.
14
Let p(x, t) = ax21 + bx2
2 + cx23 + 2dx1x2 + 2ex1x3 + 2fx2x3, where a, b, c, d, e, f
are some functions of t.
Then a = 12px1x1 , b = 1
2px2x2 , c = 1
2px3x3 , d = 1
2px1x2 , e = 1
2px1x3 , and f = 1
2px2x3 .
So p(x, t) = 12
(px1x1x21 + px1x2x1x2 + px1x3x1x3 + px2x1x1x2 + px2x2x
22 +
px2x3x2x3 + px3x1x1x3 + px3x2x2x3 + px3x3x23) =
12
(px1x1x1 + px1x2x2 + px1x3x3)(px2x1x1 + px2x2x2 + px2x3x3)(px3x1x1 + px3x2x2 + px3x3x3)
·x1
x2
x3
= 12P x · x.
Thus, v(x, t) and p(x, t) determined by the equations (1.24), satisfy the Navier-Stokes equation by the construction above. Also, under our assumptions (v(x, t)is linear in x and ω = ω(t)) ∆xv, ∇xω = 0, that gives an independence on viscos-ity. So solutions to the equations (1.24) v(x, t) and p(x, t) satisfy the Euler equationas well.
At the end of this section, we consider some simple examples of the exact solutions to theequations (1.24) to illustrate the interaction between a rotation and a deformation.
Note that Corollary 1.1 is applicable up to the linear order in x while the pressure p in theequations (1.24) has a quadratic behavior. So, solutions to the equations (1.24) have directphysical meaning described in Corollary 1.1 locally in space and time only.
Example 1.1 (Jet Flows).In the equations (1.23), (1.24), take D = diag(−γ1, −γ2, γ1 + γ2), γi > 0, i = 1, 2 andω0 = 0. Then ω = diag[e−γ1t, e−γ2t, e(γ1+γ2)t]C, C ∈ R3, ω|t=0 = IC = 0. So. C = 0
and ω = 0, i.e., this flow is irrotational. Hence for the particle-trajectory mapping we havethe following initial value problem:
dX
dt= v(x, t) = DX, X|t=0 = α, α ∈ R3.
After integration, we get X(t) = diag[e−γ1t, e−γ2t, e(γ1+γ2)t]C1, C1 ∈ R3.
So, X(0) = IC1 = α and X(t) = diag[e−γ1t, e−γ2t, e(γ1+γ2)t]α.
Note that X21 (t) + X2
2 (t) = e−2(γ1+γ2)t(α21 + α2
2), i.e., the distance of a given fluid particleto the X3 axis decreases exponentially in time. So, this axisymmetric flow without swirlforms two jets along positive and negative directions of the X3 axis.
Example 1.2 (Strain Flows).Take again ω0 = 0 and D = diag(−γ, γ, 0), γ > 0. Then again ω = 0, i.e., flow isirrotational, and X(t) = diag(e−γt, eγt, 1)α. Hence X1(t)X2(t) = α1α2 and X3(t) = α3.So, this flow is independent on X3, and the fluid particle trajectoties are hyperbolas in theplane X3 = α3. It forms a strain flow.
15
Example 1.3 (A Vortex).Take now ω0 = (0, 0, k)t, k > 0 and D = 0 in the equations (1.23) and (1.24). Thenω(t) = ω0 and v(X, t) = 1
2ω0 × X = 1
2(−kX2, kX1, 0)t. We have already considered
these flows in details in the corollary 1.1(2). We got[X1(t)X2(t)
]= Q(
1
2kt)
[α1
α2
], X3(t) = α3,
where Q is 2× 2 rotation matrix, i. e., Q(12kt) =
[cos(1
2kt) sin(1
2kt)
sin(12kt) cos(1
2kt)
].
So, this flow forms a two-dimensional vortex in the plane X3 = α3.
Example 1.4 (A Rotating Jet).Here we will consider the superposition of a jet and a vortex taking ω0 = (0, 0, k)t, D =diag(−γ1, −γ2, γ1 + γ2), γ1, γ2, k > 0 in the equations (1.23) and(1.24).
Then ω(t) = (0, 0, ke(γ1+γ2)t)t. So, the rotation is around X3 axis, and the vorticityincreases exponentially in time.
From the equation (1.24a), we get the velocityv(x, t) = (−γ1X1 − k
2e(γ1+γ2)tX2, −γ2X2 + k
2e(γ1+γ2)tX1, (γ1 + γ2)X3)t.
Then coordinates of the particle-trajectories satisfy the following initial value problem:dX1
d t= −γ1X1 − k
2e(γ1+γ2)tX2,
dX2
d t= −γ2X2 + k
2e(γ1+γ2)tX1,
dX3
d t= (γ1 + γ2)X3,
Xi|t=0 = αi, i = 1, 2, 3.
So, X3(t) = α3e(γ1+γ2)t as for the jet flows in the example 1.1.
Now multiply the first from the equations above by X1, the second one by X2, and add theresults. We have
1
2
d
dt
(X2
1 +X22
)= −
(γ1X
21 + γ2X
22
).
Hence
−4 max(γ1, γ2)dt ≤ d(X21 +X2
2 )
X21 +X2
2
≤ −4 min(γ1, γ2)dt.
After integration we get
C1e−4 max(γ1, γ2)t ≤ X2
1 +X22 ≤ C2e
−4 min(γ1, γ2)t with C1, C2 > 0.
Hence for t = 0 we have C1 ≤ α21 + α2
2 ≤ C2.
To sharpen our estimate, choose C1 = α21 + α2
2 = C2. So we obtain
(α21 + α2
2) e−4 max(γ1, γ2)t ≤ X21 +X2
2 ≤ (α21 + α2
2) e−4 min(γ1, γ2)t
16
that shows for a given fluid particle the same exponential decreasing in time of minimaland maximal distances to the X3 axis as those for the jet without rotation in the example1.1.
Thus, the flow forms a jet in the negative and positive directions of the X3 axis, and theparticle trajectories spiral around the X3 axis with increasing angular velocity. This flow isa rotating jet, an axisymmetric flow with swirl.
1.5 Some Remarkable Properties of the Vorticity in Ideal Fluid Flows.
In this section we review some important properties of vorticity in inviscid flows (ν = 0).
In the case ν = 0, the vorticity equation (1.28) reduces to
Dω
Dt= Dω, (1.29)
where DDt
is a convective derivative, and D = ∇v − Ω.
ThenDω = ∇v ω,
since Ωω = 12ω × ω = 0.
One of the properties we will consider, the Vorticity-Transport Formula, is based on thefollowing lemma.
Lemma 1.4. Let v(x, t) be any smooth velocity field (not necessary divergence free) withassociated particle-trajectory mapping X(α, t) satifying
dX
d t= v(X(α, t), t), X(α, 0) = α. (1.30)
Let h(x, t) be a smooth vector field. Then h satisfies the initial value problem
Dh
Dt= (∇v)h, h(α, 0) = h0(α) (1.31)
if and only ifh (X(α, t), t) = ∇αX(α, t)h0(α). (1.32)
Proof. Take ∇α of both sides of the equation (1.30). We have
d
dt∇αX(α, t) = ∇x v (X(α, t), t)∇αX(α, t), ∇αX(α, 0) = ∇αα = I.
Recall that both ∇x v and ∇αX are 3 × 3 matrices. Now on this equation and the initial
17
value multiply from the right by h0(α) ∈ R3.
d
dt[∇αX(α, t)h0(α)]=∇xv(X(α, t), t)[∇αX(α, t)h0(α)] , ∇αX(α, 0)h0(α) = h0(α).
This ODE and initial condition are the same as in the initial value problem (1.31). So, bythe uniqueness of solutions to initial value problems, we conclude that
h (X(α, t), t) = ∇αX(α, t)h0(α).
The following proposition is an immediate application of the last lemma with h = ω.
Proposition 1.6 (Vorticity-Transport Formula). Let X(α, t) be the smooth particle trajec-tories corresponding to a divergence-free velocity field v(x, t). Then the solution ω to theinviscid vorticity equation (1.29), Dω
Dt= ∇v ω, is
ω (X(α, t), t) = ∇αX(α, t)ω0(α), where ω0(α) = ω|t=0 . (1.33)
Consider some interpretation of the formula (1.33). ∇αX(α, t) is 3 × 3 real matrix withdet (∇αX(α, t)) = 1 since the fluid is incompressible. Then the product of three complexeigenvalues is −1. If one of the eigenvalues is −1, then two others are λ and λ−1 with|λ| ≥ 1. In the case |λ| > 1, the formula (1.33) shows that the vorticity increases whenω0 aligns roughly with the complex eigenvector associated with λ and decreases whendirections of ω0 and the eigenvector associated with λ−1 are close.
In particular, the formula (1.33) leads to the following significant result for ideal fluid flowsin two dimensions.
Corollary 1.2. Let X(α, t) be the smooth particle trajectories corresponding to a diver-gence-free velocity field. Then the vorticity ω(x, t) satisfies
ω (X(α, t), t) = ω0(α), α ∈ R2, (1.34)
and the vorticity ω0(α) is conserved along particle trajectories for two-dimensional invis-cid fluid flows.
Proof. Let ∀t > 0, X3(α, t) = α3 =const, and X1α3
= X2α3
= 0. Also, for rotation in theplane parallel to X1oX2 the vorticity ω0(α) = [0, 0, k(α)]t, where k(α) is a real-valuedfunction.
Then by the formula (1.33), we have
ω (X(α, t), t) =
X1α1
X1α2
0X2α1
X2α2
00 0 1
00
k(α)
=
00
k(α)
= ω0(α), ∀t ≥ 0.
18
Another corollary of the Vorticity-Transport Formula shows one remarkable property ofvortex lines in inviscid flows.
At first, give a definition of vortex lines in fluid flows.
Definition 1.6. We say that a smooth curve C = y(s) ∈ RN : 0 < s < 1 is a vortex lineat fixed time t if it is tangent to the vorticity ω at each of its points, i. e., provided that
d y(s)
d s= λ(s)ω (y(s), t) for some λ(s) 6= 0. (1.35)
Proposition 1.7. In inviscid, smooth fluid flows, vortex lines move with the fluid.
Proof. Let a smooth curve C = y(s) ∈ RN : 0 < s < 1 be an initial vortex line.
Then dy(s)ds
= λ(s)ω0 (y(s)), where ω0(α) = ω|t=0 , α = y(s).
C evolves with fluid to the curve C(t) = X (y(s), t) ∈ RN : 0 < s < 1.
We show that C(t) is also a vortex line. By the definition 1.6, we have
dX (y(s), t)
ds= ∇αX (y(s), t)
dy(s)
ds= ∇αX (y(s), t)λ(s)ω0 (y(s))
The proposition 1.6 gives that∇αX (y(s), t)ω0 (y(s)) = ω (X (y(s), t) , t).
So dX(y(s),t)ds
= λ(s)ω (X (y(s), t) , t), i.e., C(t) is tangent to vortex ω at each of its points.
We conclude our review of some important properties of the vorticity in ideal fluid flowsdiscussing changing in time of their flux and velocity circulation.
Let S be a bounded, open, smooth surface with smooth, oriented boundary C on it, andS and C evolve with the fluid to S(t) = X(α, t) : α ∈ S and C(t) = X(α, t) :α ∈ C respectively, where X(α, t) are the smooth particle trajectories corresponding to adivergence-free velocity field v(x, t).
At first, we will get an analog of the transport formula (1.15) for circulation of velocity.
Proposition 1.8. Let C be a smooth, oriented, closed curve and let X(α, t) be the smoothparticle trajectories corresponding to a divergence-free velocity field v(x, t). Then
d
dt
∮C(t)
v · dl =
∮C(t)
Dv
Dt· dl (1.36)
19
Proof. Change variables α → X(α, t). Then C → C(t) and an infinitesimal element ofC, dα = (α+ dα)− α→ dl = X(α+ dα, t)−X(α, t). Using a Taylor series expansionto linear order of dα for smooth particle trajectories X(α + dα, t) at the point (α, t), weget that the infinitesimal element of the curve C(t), dl = ∇αX(α, t)dα.
So,d
dt
∮C(t)
v (X(α, t)t) · dl =d
dt
∮C
v(α, t) · ∇αX(α, t)dα =
∮C
[d
dtv(α, t) · ∇αX(α, t) + v(α, t) · ∇α
dX(α, t)
dt
]dα =∮
C
d
dtv(α, t) · ∇αX(α, t)dα +
∮C
v(α, t) · ∇αv(α, t)dα.
The second integral∮Cv(α, t) · ∇αv(α, t)dα =
∮Cv(α, t)dv(α, t) =
∮C
12dv2(α, t) =
12
∮Cd (|v(α, t)|2) = 0 as an integral of a perfect differential along a closed contour.
Hence reversing the change of variables in the first integral, we obtain
d
dt
∮C(t)
v · dl =
∮C(t)
d
dtv (X(α, t)t) · dl =
∮C(t)
(∂v
∂t+ v∇v
)· dl =
∮C(t)
Dv
Dt· dl.
As an immediate application of the formula (1.36), we get the Kelvin’s Conservation Lawof Circulation.
Proposition 1.9 (Kelvin’s Conservation of Circulation). For a smooth solution v to theEuler equation, the circulation ΓC(t) around a curve C(t) moving with the fluid,
ΓC(t) =
∮C(t)
v · dl, (1.37)
is constant in time.
Proof. Differentiate ΓC(t) by t. The formula (1.36) and the Euler equation DvDt
= −∇p give
d
dtΓC(t) =
∮C(t)
Dv
Dt· dl = −
∮C(t)
∇p · dl.
The contour C(t) moves with the fluid along particle trajectories X(α, t). Because flowis inviscid, there are no tangent forces acting on the fluid. Then ∇p · dl = 0, and soddt
ΓC(t) = 0. Thus, ΓC(t) =const. in time.
20
The conservation Law of Vorticity Flux is a consequence of this result.
Corollary 1.3 (Helmholtz’s Conservation of Vorticity Flux). For a smooth solution v to theEuler equation, the vorticity flux FS(t) through a surface S(t) moving with the fluid,
FS(t) =
∫S(t)
(ω · n) dS, (1.38)
where n is unit vector normal to S(t), is constant in time.
Proof. Let C(t) be a smooth boundary of the surface S(t). Apply Stoke’s formula to theprevious result
0 =d
dtΓC(t) =
d
dt
∮C(t)
v · dl =d
dt
∫S(t)
(curlv · n) dS =d
dt
∫S(t)
(ω · n) dS =d
dtFS(t).
So, FS(t) =const. in time.
Notice that if area of S(t) decreases, then by the Conservation Law of Vorticity Flux, ω|S(t)
must roughly increse as we have seen in the Example 1.4 for rotating jets.
Remark. In general, ΓC(t) is not conserved in viscous flows with ν > 0.
Proof. Applying the Navier-Stokes equation instead of the Euler equation, we have
d
dtΓC(t) =
∮C(t)
Dv
Dt· dl = −
∮C(t)
∇p · dl + ν
∮C(t)
∆v · dl.
The first integral is 0 because it is the integral of a conservative vector field arround a closedcontour.
Also note that curlω = ∇×ω = ∇× (∇× v) = (∇· v)∇− (∇·∇)v = (divv)∇−∆v =−∆v since divv = 0.
So, ddt
ΓC(t) = −ν∮C(t)
curlω · dl 6= 0 in general.
1.6 Some Conserved Quantities in Ideal and Viscous Fluid Flows.
We start this section with an important lemma.
Lemma 1.5. Let w be a smooth, divergence-free vector field in RN and let q be a smoothscalar such that
|w(x)| |q(x)| = o[(|x|1−N)
]as |x| → ∞. (1.39)
21
Then w and ∇q are orthogonal: ∫RNw · ∇q dx = 0. (1.40)
Proof. The proof is based on the Green’s formula:∫U
∇φ · ∇q dx = −∫U
q∆φ dx+
∫∂U
∂φ
∂nq dS,
where φ, q ∈ C1(U), U ⊂ RN , n is an outward unit vector, normal to ∂U , the boundaryof U .
Let U = x ∈ RN : |x| ≤ R for some R > 0.
Let ∇φ = w. Then ∆φ = ∇2φ = ∇· (∇φ) = ∇·w =divw = 0 and ∂φ∂n
= ∇φ ·n = w ·n.
So,∫|x|≤R w · ∇q dx =
∫|x|=R w · qn dS.
Estimate the flux on the right-hand side of the last equation. Let SR be the surface area ofthe sphere of radius R; SR = CNR
N−1 with CN , a coefficient depending on dimension N .∣∣∣∣∫|x|=R
w · qn dS∣∣∣∣ ≤ max|x|=R |w · qn|SR ≤ max|x|=R |w(x)| |q(x)|CNRN−1 =
o(R1−N)CNRN−1 → 0 if R→∞.
Thus,∫RN w · ∇q dx = 0, i.e., w and ∇q are orthogonal.
Recall that the Navier-Stokes and the Euler equations are obtained from the conservationlaw of momentum, and that the incompressibility condition div v = 0 is gotten from theconservation law of mass.
The next proposition lists some other quantities for solution of the Euler equation that areconserved globally in time.
Proposition 1.10. Let v (and ω = curl v) and p be a smooth solution to the Euler equationDvDt
= −∆p and div v = 0 in R3, vanishing sufficiently rapidly as |x| → ∞. Then thefollowing quantities are conserved for all time:
(i) the total flux V3 of velocity,
V3 =
∫R3
v dx (1.41)
(ii) the total flux Ω3 of vorticity,
Ω3 =
∫R3
ω dx (1.42)
22
(iii) the kinetic energy E3,
E3 = 12
∫R3
|v|2 dx (1.43)
(iv) the helicity H3,
H3 =
∫R3
v · ω dx (1.44)
(v) the fluid impulse I3,
I3 = 12
∫R3
x× ω dx (1.45)
Proof.
(i) Integrate over R3 the Euler equation ∂v∂t
+ (∇v)v = −∇p. We have componentwise:
d
dt
∫R3
vi(x, t) dx = −∫R3
∇vi(x, t) · v(x, t) dx−∫R3
pxi dx, i = 1, 2, 3.
Since v vanishes at infinity sufficiently rapidly such that |∇vi(x, t)| |v(x, t)| =o(|x|1−N
)as |x| → ∞, then by the Lemma 1.5, the first integral at the right-hade
side of the equation above is zero. Applying to the next integral in this equation theGauss-Green theorem, we get:∫
R3
pxi dx = limR→∞
∫|x|=R
pni dS,
where ni = xiR
is the ith component of the outward normal unit vector of the sphere|x| = R.
Since p vanishes sufficiently rapidly as |x| → ∞, then for large R, p must be aconstant
So, ddtV3 = limR→∞
pR
∫|x|=R xi dS = 0.
Hence V3 is constant in time.
(ii) Now integrate over R3 the vorticity equation.
dω
dt+∇ω v = ∇v ω.
We have componentwise:
d
dt
∫R3
ωi dx =
∫R3
ω · ∇vi dx−∫R3
v · ∇ωi dx, i = 1, 2, 3.
Apply the Green’s formula for both integrals at the right-hand side of this equation.
23
We get
d
dt
∫R3
ωi dx = −∫R3
vi(∇ · ω) dx+ limR→∞
∫|x|=R
(ω · n)vi dS +∫R3
ωi(∇ · v) dx− limR→∞
∫|x|=R
(v · n)ωi dS.
Because ω = ∇ × v, ∇ · ω = divω = ∇ · (∇ × v) = 0. ∇ · v = div v = 0 bythe incompressibility. Finally, take into account that v vanishes sufficiently rapidlyas |x| → ∞.
So, ddt
∫R3 ω
i dx = 0, i = 1, 2, 3, i.e., Ω3 is conserved for all time.
(iii) Apply the transport formula (1.15),
d
dt
∫X(Ω,t)
f dx =
∫X(Ω,t)
[ft + div(fv)]dx,
for Ω = R3 and f = 12v · v = 1
2|v|2. Then we have
d
dt
1
2
∫R3
|v|2 dx =
∫R3
[vt · v +∇
(1
2v · v
)· v +
(1
2v · v
)div v
]dx.
Now div v = 0 by the incompressibility. ∇(12v · v) = (v · ∇)v. Hence
d
dt
1
2
∫R3
|v|2 dx =
∫R3
[vt + (v · ∇)v] · v dx.
Using the Euler equation vt + (v · ∇)v = −∇p, we get
d
dt
1
2
∫R3
|v|2 dx = −∫R3
∇p · v dx.
Finally, since v and p vanish sufficiently rapidly as |x| → ∞, then by the Lemma1.5, the integral at the right-hand side of the last equation is zero.
So, ddtE3 = 0, and the kinetic energy E3 is constant in time.
(iv) Multiply the Euler equation by ω.
vt · ω + (v · ∇)v · ω = −∇p · ω.
Differentiating the products (v · ω)v and p ω, we have
div ((v · ω)v) = ∇(v ·ω) · v+ (v ·ω)div v = (v · ∇)v ·ω+ (v · ∇)ω · v+ (v ·ω)div v
and div(p ω) = ∇p · ω + p divω. Recall that divv = 0 by the incompressibility.
24
Hence we get
vt · ω + div ((v · ω)v)− (v · ∇)ω · v = −div(p ω) + p divω. (∗)
Now multiply the vorticity equation DωDt
= (ω · ∇)v by v to find
ωt · v + (v · ∇)ω · v = (ω · ∇)v · v = 12ω · ∇
(v2).
Using that div(v2ω) = ∇ (v2) · ω + v2divω, we obtain
ωt · v + (v · ∇)ω · v =1
2div(v2ω)− 1
2v2divω. (∗∗)
However, divω = 0 since divω = ∇ · ω = ∇ · (∇× v) = 0.
Adding the equations (∗) and (∗∗) together, we have
(v · ω)t + div((v · ω)v) =1
2div(v2ω)− div(p ω)
or
(v · ω)t + div[(v · ω)v +
(p− v2
2
)ω
]= 0.
Integrate this equation over R3.
d
dt
∫R3
v · ω dx+
∫R3
div[(v · ω)v +
(p− v2
2ω
)]= 0.
Apply the Gauss-Green’s formula to the second integral of this equation,
d
dtH3 + lim
R→∞
∫|x|=R
[(v · ω)v +
(p− v2
2
)ω
]· n dS = 0.
By the assumption, v and p vanish sufficiently rapidly when |x| → ∞, then ω =curl v must vanish sufficiently rapidly at infinity as well. Then the last limit is zero.So, d
dtH3 = 0, i.e., the helicity H3 is conserved for all time.
(v) To prove the conservation of the fluid impulse I3, apply the transport formula (1.15)to Ω = R3 and f = x× ω.
d
dt
∫R3
x× ω dx =
∫R3
[∂
∂t(x× ω) + v · ∇(x× ω) + (x× ω)div v
]dx,
and apply div v = 0. Then using a convective derivative, we have
d
dtI3 =
∫R3
D(x× ω)
Dtdx =
∫R3
(Dx
Dt× ω + x× Dω
Dt
)dx,
25
and
Dx
Dt=∂x
∂t+ v · ∇x = 0 + (∇x)v = I v = v.
By the vorticity equation, DωDt
= (ω · ∇)v, it follows that
d
dtI3 =
∫R3
[v × ω + x× (ω · ∇)v]dx,
and
v × ω + x× (ω · ∇)v =3∑j=1
v × (ωj ej) +3∑j=1
x× (ωj vxj) =
3∑j=1
ωj(v × ej + x× vxj) = ω · (v ×∇x+ x×∇v) = ω · ∇(x× v).
So,dI3
dt=
∫R3
ω · ∇(x× v)dx =
∫R3
3∑j=1
ωj(x× v)xj dx.
Adding∫R3
∑3j=1 ω
jxj
(x× v)dx =∫R3 divω(x× v)dx = 0 (since divω = 0) to the
right-hand side of the last equation, we get
dI3
dt=
∫R3
3∑j=1
[ωj(x× v)xj + ωjxj(x× v)]dx =
∫R3
3∑j=1
[ωj(x× v)]xjdx =3∑i=1
3∑j=1
∫R3
[ωj(x× v)i]xjdx ei.
Now apply the Gauss-Green’s formula. We obtain
dI3
dt=
3∑i,j=1
limR→∞
(∫|x|=R
ωj(x× v)inj dS
)ei.
Exploiting again the fact that both v and ω vanish sufficiently rapidly as |x| → ∞,we finally have dI3
dt= 0, i.e., the fluid impulse I3 is conserved for all time.
Notice that for 2D flows, the quantities V2, Ω2, E2, and I2 are also conserved for all time,while the helicityH2 ≡ 0 because v = (v1, v2, 0)t, ω = (0, 0, v2
x1−v1
x2)t, and so v ·ω ≡ 0.
26
At the end of this section, we will consider the effect that viscosity has on the conservationof energy. It is important for the study of the properties of general solution to the Navier-Stokes equation in chapter 3.
Proposition 1.11. Let v be a smooth solution to the Navier-Stokes equation, vanishingsufficiently rapidly as |x| → ∞. Then the kinetic energy E(t) satisfies the ODE
d
dtE(t) = −ν
∫RN|∇v|2 dx. (1.46)
Proof. This proof repeats the steps of the conservation energy proof for inviscid fluid (Prop.1.10(iii)) with using the Navier-Stokes equation instead of the Euler equation. So, applyingthe transport formula (1.15) to Ω = RN and f = 1
2v · v, we get
d
dt
∫RN
1
2|v|2dx =
∫RN
Dv
Dt· v dx,
or using the Navier-Stokes equation DvDt
== ∇p+ ν∆v, we have
d
dt
∫RN
1
2|v|2dx = −
∫RNv · ∇p dx+ ν
∫RNv ·∆v dx.
Again, the first integral at the right-hand side of this equation is 0 by the Lemma 1.5 sincev and p vanish sufficiently rapidly as |x| → ∞.
For the second integral at the right-hand side of this equation, we apply the Green’s formula∫U
φ ·∆ψ dx = −∫U
∇φ · ∇ψ dx+
∫∂U
∂ψ
∂n· φ dS,
where n is outward normal unit vector of ∂U , to φ = ψ = v.
Then we obtain
d
dtE(t) = ν
(−∫RN∇v · ∇v dx+ lim
R→∞
∫|x|=R
v(n · ∇v)dS
).
This limit is zero since v vanishes at infinity sufficiently rapidly. Hence we get the follow-ing rate of change of kinetic energy:
d
dtE(t) = −ν
∫RN|∇v|2 dx. (1.46)
Thus, for ν > 0, the kinetic energy is dissipated in viscous flows.
27
1.7 Leray’s Formulation of Incompressible Flows.
The goal of this section is to exclude the pressure p from the Navier-Stokes equation by useof the incompressibility condition divv = 0. The fact that the Navier-Stokes equation doesnot contain the time derivative of p makes this attempt possible.
The following proposition shows the resulting equation with v as the only unknown.
Proposition 1.12 (Leray’s Formulation of the Navier-Stokes Equation). Solving the Na-vier-Stokes equations
Dv
Dt= −∇p+ ν∆v (1.47)
anddiv v = 0 (1.48)
with smooth initial velocity v0, divv0 = 0, is equivalent to solving the evolution equation:
Dv
Dt= CN
∫RN
x− y|x− y|N
tr (∇v(y, t))2 dy + ν∆v, v|t=0 = v0. (1.49)
The pressure p(x, t) can be recovered from the velocity v(x, t) by the solution of the Poissonequation
−∆p = tr(∇v)2 =N∑
i,j=1
vixivjxj. (1.50)
Proof.
1) Exclude the pressure p from the equations (1.47), (1.48).
When we proved Proposition 1.5, we differentiated componentwise the Navier-Sto-kes equation by xj , j = 1, 2, . . . , N and got the matrix equation (1.25)
DV
Dt+ V 2 = −P + ν∆V,
where V ≡ ∇v = [vixj ] with trV = div v = 0 and P = [pxixj ] is the Hessian matrixof pressure.
Now take trace of this matrix equation.
tr(DV
Dt
)= tr
(∂V
∂t
)+ tr(v · ∇V ) =
∂
∂t(trV ) + v · ∇(trV ) =
D
Dt(trV ) =
D
Dt(div v),
tr (P ) = ∆p and tr(∆V ) = ∆(trV ) = ∆(div v).
So, we haveD
Dt(div v) + trV 2 = −∆p+ ν∆(div v). (1.51)
28
Alternatively, we could just take a divergence of the Navier-Stokes equation. Thenwe get
∇ ·[∂v
∂t+ (v · ∇)v
]= −∇ · ∇p+∇ · ν∆v;
∂
∂t(∇ · v) +∇(v · ∇)v + (v · ∇)(∇ · v) = −∆p+ ν∆(∇ · v);
∂
∂t(div v) +
N∑i=1
∂
∂xi
(N∑j=1
vj∂
∂xj
)vi + (v · ∇)div v = −∆p+ ν∆(div v);
D
Dt(div v) +
N∑j=1
N∑i=1
vjxivixj
= −∆p+ ν∆(div v).
Since V 2 = V · V = [vjxi ] · [vixk
] =[∑N
i=1 vjxivixk
], then trV 2 =
∑Nj=1
∑Ni=1 v
jxivixj
(j = k).
So, we get equation (1.51).
For incompressible flows divv = 0. Thus, the pressure p could be determined fromvelocity v(x, t) as a solution of the following Poisson equation:
−∆p = tr(∇v)2 =N∑
i,j=1
vixjvjxi. (1.50)
The right-hand side of this equation is smooth inRN and vanishes sufficiently rapidlyas |x| → ∞. Using elementary properties of the Poisson equation (see Theorem 2.3),we have
p(x, t) = Φ(x) ∗ tr (∇v(x, t))2 , (1.52)
where the fundamental solution of the Laplace’s equation in RN \ 0 is
Φ(x) =
−1
2πlog |x|, if N = 2,
Γ(N/2)
2πN/2(N−2)1
|x|N−2 , if N ≥ 3.(1.53)
Now to compute∇p, first see that
∇ (|x|) = ∇
(N∑j=1
x2j
)1/2
=N∑i=1
1
2|x|
(N∑j=1
2xjδij
)ei =
N∑i=1
xi|x|ei =
x
|x|,
where Kronecker symbol δij =
0, if i 6= j
1, if i = jand ei is the vector whose i-th com-
ponent is 1 and the other components are 0.
Then for N = 2, ∇Φ(x) = − 12π
1|x|∇ (|x|) = − x
2π|x|2 ,
29
and for N > 2, ∇Φ(x) = −Γ(N/2)
2πN/2|x|1−N∇ (|x|) = −Γ(N/2)
2πN/2x|x|N ,
i. e., ∇Φ(x) = −CN x|x|N with CN =
1
2π, if N = 2,
Γ(N/2)
2πN/2, if N > 2.
Hence∇p(x, t)=∇∫RNΦ(x− y)tr (∇v(y, t))2 dy=
∫RN∇Φ(x− y)tr (∇v(y, t))2 dy.
Finally,
−∇p(x, t) = CN
∫RN
x− y|x− y|N
tr (∇v(y, t))2 dy. (1.54)
Substitute∇p from this equation into the Navier-Stokes equation.
Dv
Dt= CN
∫RN
x− y|x− y|N
tr (∇v(y, t))2 dy + ν∆v. (1.55)
This closed evolution equation is of unknown v only.
2) Prove that the solution v to this equation is automatically divergence free, so incom-pressibility equation (1.48) can be dropped.
Actually, the evolution equation (1.55) is the Navier-Stokes equation when p(x, t)satisfies the equation (1.50). Applying the divergence operator to this (or the Navier-Stokes) equation, we got the equation (1.51) as we saw above. Using the equation(1.50), we reduce the equation (1.51) to the following one.
D
Dt(divv) = ν∆(divv). (1.56)
From the initial condition (1.3), to supply the incompressibility stipulation (1.2), wehave:
divv|t=0 = divv0 = 0. (1.57)
Then smooth solution divv must be bounded on a short time interval.
Denote divv = u. Now we will apply the energy method to prove the uniqueness ofsolution of the initial value problem
DuDt
= ν∆u, (x, t) ∈ RN × [0,∞),
u| = 0, x ∈ RN(1.58)
in the class of bounded functions.
Let u1 and u2 are two solutions of this IVP.
Then ∂(u1−u2)∂t
= −(v · ∇)(u1 − u2) + ν∆(u1 − u2).
Multiply this equation by (u1 − u2).
(u1 − u2) ∂∂t
(u1 − u2) = −(u1 − u2)(v · ∇)(u1 − u2) + (u1 − u2)ν∆(u1 − u2)
or12∂∂t
(u1 − u2)2 = −12(v · ∇)(u1 − u2)2 + ν(u1 − u2)∆(u1 − u2).
30
Integrate the last equation over RN .
12ddt
∫RN
(u1 − u2)2dx = −12
∫RN
(v · ∇)(u1 − u2)2dx+ ν
∫RN
(u1 − u2)∆(u1 − u2)dx.
Apply an integration by parts to the first integral on the right-hand side of this equa-tion,
− 12
∫RN
(v · ∇)(u1 − u2)2dx = −12
N∑j=1
∫RNvj ∂
∂xj(u1 − u2)2dx =
12
N∑j=1
∫RN
(u1 − u2)2vjxjdx−12
N∑j=1
limR→∞
∫|x|=R
(u1 − u2)2vjnj0 dS =
12
∫RN
(u1 − u2)2divv dx.
For the second integral, apply the Green’s formula
ν
∫RN
(u1 − u2) ·∆(u1 − u2) dx = −ν∫RN∇(u1 − u2) · ∇(u1 − u2) dx
+ limR→∞
∫|x|=R
(u1 − u2) · ∇(u1 − u2)n0 dS = −ν∫RN|∇(u1 − u2)|2 dx.
Here n0 is the outward normal unit vector, and both limits are 0 since v (and sou =divv) vanishes sufficiently rapidly at infinity.
Then we have:
12ddt
∫RN
(u1 − u2)2dx = 12
∫RN
divv (u1 − u2)2dx− ν∫RN|∇(u1 − u2)|2dx
≤ 12
∫RN
divv (u1 − u2)2dx.
Now we use the boundness of divv. Hence there exists a constant C > 0 such that
12ddt
∫RN
(u1 − u2)2dx ≤ C
∫RN
(u1 − u2)2dx.
After that we will use Gronwall’s lemma (see section 2.1) for φ(t) = 2C,η(t) =
∫RN (u1 − u2)2dx, and ψ(t) ≡ 0.
So, η′(t) ≤ φ(t)η(t) + ψ(t) on [0, T ] for some T > 0 implies that ∀t ∈ [0, T ],
η(t) ≤ e∫ T0 φ(s)ds
[η(0) +
∫ T0ψ(s)ds
].
31
Therefore we get:∫RN
(u1 − u2)2dx ≤ e2CT
[∫RN
(u1 − u2)2dx
]∣∣∣∣t=0
.
Since u1|t=0 = u2|t=0, then u1 ≡ u2, ∀t ∈ [0, T ], i.e., the solution u of the IVP(1.58) is unique. So, the trivial solution u ≡ 0 is the only solution of this problem inclass of bounded functions.
Thus, divv ≡ 0 for v satisfying the evolution equation (1.55) with the initial condition
v|t=0 = v0, divv0 = 0. (1.59)
We have proved that the IVP with the Navier-Stokes equation (1.2)-(1.3) is equivalentto the IVP (1.55), (1.59).
By the Lemma 1.5, a divergence-free vector field v and the gradient of scalar p are ortho-gonal in L2(RN) if both v and p vanish sufficiently rapidly as |x| → ∞,i.e.,
(v,∇p)0 =
∫RNv · ∇p dx = 0.
Denote the projection operator on the space of divergence-free vector fields by P .
So
Pv = w, divw = 0;
Pv = v if and only if divv = 0,
P (∇p) = 0.
A general concept of decomposing a vector field into the divergence free part and thegradient part, orthogonal to it, is very convenient for studying incompressible flows andwill be considered in detail in the next chapter in the Hodge’s Decomposition section.
From this point of view, the equation (1.55) is a projection of the Navier-Stokes equationon the space of divergence-free vector fields. Using the projection operator P , the equation(1.55) can be written as following:
vt = P (−v · ∇v) + ν∆v. (1.60)
This Leray’s formulation of the Navier-Stokes equation and the energy method will be usedin Chapter 3.
32
Chapter 2
Technical Background.
In this chapter, we review some properties of the Fourier transform, the Poisson equa-tion, Sobolev spaces, mollifiers, and Leray’s projection operator. Also, we will prove hereHodge’s decomposition of vector fields and Gronwall’s inequality we have used in the pre-vious chapter.
2.1 The Gronwall’s Inequality.
Gronwall’s inequality is useful to derive various estimates in energy methods. This inequal-ity often is used in differential or integral forms.
Lemma 2.1 (Gronwall’s inequality in differential form.). If:
1) η(t) ≥ 0 and is differentiable on [0, T ],
2) ϕ(t), ψ(t) ≥ 0 and are integrable on [0, T ],
3) η′(t) ≤ ϕ(t)η(t) + ψ(t),
then ∀t ∈ [0, T ],
η(t) ≤ e∫ t0 ϕ(s) ds
[η(0) +
∫ t
0
ψ(s) ds
]. (2.1)
Proof. Perform a substitution η(t) = µ(t)e∫ t0 ϕ(s) ds. Note that η(0) = µ(0).
Then from the inequality 3), we have
µ′(t)e∫ t0 ϕ(s) ds + µ(t)e
∫ t0 ϕ(s) dsϕ(t) ≤ ϕ(t)µ(t)e
∫ t0 ϕ(s) ds + ψ(t)
orµ′(t) ≤ ψ(t)e−
∫ t0 ϕ(s) ds ≤ ψ(t) since e−
∫ t0 ϕ(s) ds ≤ 1.
Integrating the last inequality on [0, t], 0 ≤ t ≤ T , we get∫ t
0
µ′(s) ds ≤∫ t
0
ψ(s) ds.
So, µ(t) ≤∫ t
0
ψ(s) ds+ µ(0).
Multiplying this inequality by e∫ t0 ϕ(s) ds, we obtain (2.1).
η(t) ≤ e∫ t0 ϕ(s) ds
[η(0) +
∫ t
0
ψ(s) ds
].
33
Lemma 2.2 (Gronwall’s inequality in integral form.). If:
1) c(t) ≥ 0 and c(t) ∈ C1([0, T ]),
2) u(t), q(t) ∈ C([0, T ]),
3) q(t) ≤ c(t) +∫ t
0u(s)q(s) ds,
then
q(t) ≤ c(t) +
∫ t
0
c(s)u(s)e∫ ts u(τ) dτds (2.2)
or equivalently,
q(t) ≤ c(0)e∫ t0 u(s) ds +
∫ t
0
c′(s)e∫ ts u(τ) dτds. (2.2a)
Proof.
1) Define v(s) := e−∫ s0 u(τ) dτ
∫ s0u(τ)q(τ) dτ . Note that v ∈ C1([0, T ]) and v(0) = 0.
Then v′(s) = u(s)e−∫ s0 u(τ) dτ
[q(s)−
∫ s0u(τ)q(τ) dτ
]. Since by the hypothesis 3,)
q(t)−∫ t
0u(s)q(s) ds ≤ c(t), then v′(s) ≤ u(s)c(s)e−
∫ s0 u(τ) dτ .
Integrating this inequality on [0, t], t ∈ [0, T ] and taking in count that v(0) = 0, we
have v(t) ≤∫ t
0u(s)c(s)e−
∫ s0 u(τ) dτ ds.
2) By the definition of v,∫ t
0u(s)q(s) ds = v(t)e
∫ t0 u(s) ds. Applying the estimate for v
obtained above, we get∫ t
0
u(s)q(s) ds ≤ e∫ t0 u(τ) dτ
∫ t
0
u(s)c(s)e−∫ s0 u(τ) dτds =
∫ t
0
u(s)c(s)e∫ ts u(τ) dτds.
Use this bound along with the hypothesis 3)
q(t) ≤ c(t) +
∫ t
0
u(s)c(s)e∫ ts u(τ) dτ ds. (2.2)
Integrating by parts at the right-hand side of this inequality gives us the followingresult,
q(t) ≤ c(t) +
∫ t
0
c(s) dds
(−e
∫ ts u(τ dτ)
)ds
= c(t)− c(s)e∫ ts u(τ) dτ
∣∣∣s=ts=0
+
∫ t
0
c′(s)e∫ ts u(τ) dτ ds
= c(t)− c(t) + c(0)e∫ t0 u(τ) dτ +
∫ t
0
c′(s)e∫ ts u(τ) dτ ds. (2.2a)
34
2.2 Elementary Properties of the Fourier Transform.
In this section, we review some basic properties of the Fourier transform.
Definition 2.1. The Fourier transform f for a function f ∈ L1(RN) is
f(ξ) =
∫RNe−2πix·ξf(x) dx, ξ ∈ RN . (2.3)
The inverse transform for f is
f(ξ) =
∫RNe2πix·ξf(x) dx, ξ ∈ RN . (2.4)
Note that f(x) = f(−x). Also, |e±2πix·ξ| = 1, and |∫RN e
±2πix·ξf(x) dx|≤∫RN |e
±2πix·ξ| |f(x)| dx = ‖f‖L1 <∞ since f ∈ L1(RN).
So, f , f ∈ L∞(RN), and ‖f‖L∞ , ‖f‖L∞ ≤ ‖f‖L1 .
The Fourier transform is very convenient to apply to derivatives and convolutions. The lastone is described by the following lemma.
Lemma 2.3. If f, g ∈ L1(RN), then the Fourier transform of their convolution is theproduct of their Fourier transforms:
f ∗ g = f g. (2.5)
Proof. We will use Fubini’s theorem to prove (2.5).
f ∗ g(ξ) =
∫RNe−2πix·ξ
[∫RNf(x− y)g(y) dy
]dx
=
∫∫RN×RN
e−2πi(x−y)·ξe−2πiy·ξf(x− y)g(y) dy dx.
Make substitution z(x) = x− y, so dz=dx. Then
f ∗ g(ξ) =
∫∫RN×RN
e−2πiz·ξe−2πiy·ξf(z)g(y) dydz
=
∫RNe−2πiz·ξf(z) dz
∫RNe−2πiy·ξg(y) dy = f(ξ)g(ξ).
Next three lemmas will help us to prove some properties of the Fourier transform.
35
Lemma 2.4. Let f, g ∈ L1(RN). Then∫RNf(x)g(x) dx =
∫RNf(y)g(y) dy. (2.6)
Proof. First, show that both integrals in the equation (2.6) are finite.∣∣∣∣∫RNf(x)g(x) dx
∣∣∣∣ ≤ supx∈RN
|g(x)|∫RN|f(x)| dx = ‖g‖L∞ ‖f‖L1 ≤ ‖g‖L1 ‖f‖L1 <∞.
Similarly∣∣∣∫RN f(y)g(y) dy
∣∣∣ ≤ ‖f‖L1 ‖g‖L1 <∞. Now use Fubini’s theorem.∫RNf(x)g(x) dx =
∫RNf(x)
[∫RNe−2πiy·xg(y) dy
]dx =
∫∫RN×RN
e−2πiy·xf(x)g(y) dydx∫RNg(y)
[∫RNe−2πix·yf(x) dx
]dy =
∫RNg(y)f(y) dy.
Similarly, it could be shown that∫RNf(x)g(x) dx =
∫RNf(y)g(y) dy. (2.7)
Here we consider one example of directly finding the Fourier transform. We will use thisexample later in proving Plancherel’s theorem.
Example 2.1. Let fε := e−ε|x|2, ε > 0. Then fε(ξ) =
(πε
)N/2e−(π|ξ|)2/ε.
Proof. First, show that fε ∈ L1(RN). Using the Fubini’s theorem, we have∫RN|fε(x)| dx =
∫RNe−ε|x|
2
dx =
∫RNe−ε
∑Nn=1 x
2n dx1dx2 . . . dxn =
N∏n=1
∫ ∞−∞
e−εx2n dxn =
N∏n=1
1√ε
∫ ∞−∞
e−(√ε xn)2 d(
√ε xn) =
N∏n=1
√πε
= (πε)N/2 <∞.
36
Then the Fourier transform
fε(ξ) =
∫RNe−2πix·ξ e−ε|x|
2
dx =
∫RNe−
∑Nn=1(εx2n+2
√εxnπiξn/
√ε) dx1dx2 . . . dxn =∫
RNe−∑Nn=1
[(√εxn+πiξn/
√ε)
2−(πiξn)2/ε
]dx1dx2 . . . dxn =
e−(π|ξ|)2/εN∏n=1
∫ ∞−∞
e−(√εxn+πiξn/
√ε)2 dxn =
e−(π|ξ|)2/εN∏n=1
1√ε
∫ ∞−∞
e−(√εxn+πiξn/
√ε)2 d
(√εxn + πiξn√
ε
)=
e−(π|ξ|)2/εN∏n=1
√πε
=(πε
)N/2e(π|ξ|)2/ε,
where we have omitted the details of the contour shift arising from the substitution xn →(√εxn + πi√
εξn
).
Note that applying the Lemma 2.3 to fε(x) and to some function g ∈ L1(RN), we get∫RNg(x)e−ε|x|
2
dx =(πε
)N/2 ∫RNg(y)e−(π|y|)2/ε dy. (2.8)
Lemma 2.5. Let f ∈ Lp(RN), 1 ≤ p < ∞. Then limh→0 ‖fh − f‖Lp = 0, wherefh(x) = f(x+ h).
Proof.
1) Since C∞0 (RN) is dense in Lp(RN), then for each p ∈ [1,∞), ∀ε > 0 and ∀f ∈Lp(RN), ∃g ∈ C∞0 such that ‖f − g‖Lp < ε
3.
2) g(x) is uniformly continuous because it is continuous on its compact support.So gh → g uniformly as h→ 0 where gh(x) : = g(x+ h).
Let g(x) has support in some compact V ⊂ RN . For sufficiently small |h| > 0,gh has support in the compact V . Then ‖gh − g‖Lp(RN ) = ‖gh − g‖Lp(V ) ≤supx∈V |gh(x)− g(x)| [vol(V )]1/p → 0 as h → 0 by the uniform convergence of gon V .
So, ∀ε > 0, ∃r > 0 s.t. ∀h ∈ Br(0), ‖gh − g‖Lp < ε3.
3) Note that by the substitution y = x+ h, we have
‖fh − gh‖Lp(RN ) = ‖f − g‖Lp(RN ) <ε3, ∀ε > 0.
37
4) Thus ∀ε > 0 and ∀f ∈ Lp(RN), ∃r > 0 and ∃g ∈ C∞0 (RN) s.t. ∀h ∈ Br(0),‖fh−f‖Lp = ‖fh−gh+gh−g+g−f‖Lp ≤ ‖fh−gh‖Lp +‖gh−g‖Lp +‖f −g‖Lp< ε
3+ ε
3+ ε
3= ε, i.e., limh→0 ‖fh − f‖Lp = 0.
Lemma 2.6. Let f ∈ L1(RN)∩L2(RN). Define g(x) : = f(x)∗f(−x) where f is complexconjugate of f . Then g ∈ L1(RN) ∩ C(RN).
Proof. Note that since f ∈ L1(RN), then f(−x), f(x), and so f(−x) ∈ L1(RN).Moreover, their L1 norms in RN are the same:
‖f(x)‖L1 = ‖f(−x)‖L1 = ‖f(x)‖L1 = ‖f(−x)‖L1 .
1) Show that g ∈ L1(RN). Use again Fubini’s theorem.∫RN|g(x)| dx ≤
∫RN
(∫RN|f(x− y)|
∣∣f(−y)∣∣ dy) dx
=
∫∫RN×RN
|f(−y)| |f(x− y)| dy dx =
∫RN|f(−y)|
(∫RN|f(x− y)| dx
)dy.
Use substitution z = −y and w(x) = x− y = x+ z.∫RN|g(x)| dx ≤ (−1)N
∫ ∞−∞
. . .
∫ ∞−∞|f(z)|
(∫RN|f(w)| dw
)(−1)N dz
=
∫RN|f(z)| dz ‖f‖L1 = (‖f‖L1)2 .
So, g ∈ L1(RN) and ‖g‖L1 ≤ (‖f‖L1)2.
2) Show that g ∈ C(RN).
Remind that by the lemma 2.5 for p = 1, limh→0 ‖fh − f‖L1 = 0, i.e., ∀ε > 0,∃r1 > 0 s.t. ∀h ∈ Br1(0), ‖fh − f‖L1 < ε
2‖f‖L∞.
Similarly limh→0
∥∥f [−(x+ h)]− f(−x)∥∥L1 = 0, i.e., ∀ε > 0, ∃r2 > 0 s.t.
∀h ∈ Br2(0),∥∥f [−(x+ h)]− f(−x)
∥∥L1 <
ε2‖f‖L∞
.
38
Now ∀x ∈ RN and ∀ε > 0, ∃r = minr1, r2 > 0 s.t. ∀h ∈ Br(0),
|g(x+ h)− g(x)| =∣∣f(x+ h) ∗ f(−x− h)− f(x) ∗ f(−x)
∣∣ ≤∣∣f(x+ h) ∗ f(−x− h)− f(x) ∗ f(−x− h)∣∣
+∣∣f(x) ∗ f(−x− h)− f(x) ∗ f(−x)
∣∣ =∣∣[f(x+ h)− f(x)] ∗ f(−x− h)∣∣+∣∣f(x) ∗
[f(−x− h)− f(−x)
]∣∣ =∣∣∣∣∫RN
[f(x+ h− y)− f(x− y)] f(−y − h) dy
∣∣∣∣+
∣∣∣∣∫RNf(x− y)
[f(−y − h)− f(−y)
]dy
∣∣∣∣≤
supy∈RN
∣∣f(−y − h)∣∣ ∫
RN|f(z + h)− f(z)| dz
+ supz∈RN
|f(z)|∫RN
∣∣f(−y − h)− f(−y)∣∣ dy
where z(y) = x− y.
So, |g(x+ h)− g(x)| ≤ ‖f‖L∞‖fh − f‖L1 + ‖f‖L∞‖f(−y − h)− f(−y)‖L1 <
‖f‖L∞(
ε2‖f‖L∞
+ ε2‖f‖L∞
)= ε, ∀ε > 0, ∀x ∈ RN .
Thus, g ∈ C(RN).
An important property of the Fourier transform is that it keeps the L2 norm unchanged asthe following theorem shows.
Theorem 2.1 (Plancherel’s Theorem). Let f ∈ L1(RN) ∩ L2(RN). Then f , f ∈ L2(RN)and
‖u‖L2 = ‖u‖L2 = ‖u‖L2 . (2.9)
Proof.
1) Define again g(x) := f(x) ∗ ϕ(x) with ϕ(x) := f(−x).
By the Lemma 2.6, g ∈ L1(RN) with L1 norm bounded by ‖f‖2L1 . The Lemma 2.3
gives that g(ξ) = f(ξ)ϕ(ξ).
Show that f(ξ) and ϕ(ξ) are complex conjugate.
ϕ(ξ) =
∫RNe−2πix·ξf(−x) dx.
Perform the substitution y = −x. So, the components yn = −xn, 1 ≤ n ≤ N and
39
dx = dx1dx2 . . . dxn = (−1)Ndy1dy2 . . . dyn = (−1)Ndy. Then
ϕ(ξ) = (−1)N∫ ∞−∞
∫ ∞−∞
. . .
∫ ∞−∞
e2πiy·ξf(y)(−1)Ndy =∫ ∞−∞
∫ ∞−∞
. . .
∫ ∞−∞
e−2πiy·ξf(y)dy =
∫RNe−2πiy·ξf(y) dy =
¯f(ξ).
Hence g(ξ) = f(ξ)¯f(ξ) = |f(ξ)|2, i.e., g(ξ) is real-valued non-negative function.
2) Show that g ∈ L1(RN) and then f ∈ L2(RN) since
‖g‖L1 =
∫RN|g(ξ)| dξ =
∫RN|f(ξ)|2 dξ = ‖f‖2
L2 .
We will use the function fε(x) = e−ε|x|2 , ε > 0 from example 2.1.∫
RN|g(ξ)| dξ =
∫RNg(ξ) dξ =
∫RNg(ξ) lim
ε→0+e−ε|ξ|
2
dξ since limε→0+
e−ε|ξ|2
= 1.
Note that fε(ξ) is increasing as ε→ 0+.Apply the Monotone Convergence theorem and then equation (2.8).∫
RN|g(ξ)| dξ = lim
ε→0+
∫RNg(ξ)e−ε|ξ|
2
dξ = limε→0+
(πε)N/2
∫RNg(x)e−π
2|x|2/ε dx.
Perform substitution y = πx√ε. So, xj =
√επyj , j = 1, 2, . . . , N , and
dx = dx1dx2 . . . dxN = εN/2
πNdy.
Then ∫RN|g(ξ)| dξ = lim
ε→0+
(πε
)N/2 εN/2
πN
∫RNg(√
εyπ
)e−|y|
2
dy.
Since e−|y|2 ≤ 1, and so g(√
εyπ
)e−|y|
2 ≤ g(√
εyπ
)∈ L1(RN), then we can apply
now the Dominated Convergence theorem and the Fubini’s theorem after that.∫RN|g(ξ)| dξ = π−N/2
∫RN
limε→0+
g(√
εyπ
)e−|y|
2
dy =
π−N/2g(0)
∫RNe−
∑Nj=1 y
2j dy = π−N/2g(0)
N∏j=1
∫ ∞−∞
e−y2j dyj =
π−N/2g(0)πN/2 = g(0).
Thus, g ∈ L1(RN), f ∈ L2(RN), and ‖g‖L1 = ‖f‖2L2 = g(0).
Following the same arguments, we can also show that f ∈ L2(RN).
40
3) The relation (2.9) follows from the last result.
‖f‖2L2 = g(0) = (f ∗ ϕ)(0) =
∫RNf(0− x)ϕ(x)dx
=
∫RNf(−x)f(−x)dx =
∫RN|f(−x)|2 dx.
After substitution y = −x, we have
‖f‖2L2 =
∫RN|f(y)|2 dy = ‖f‖2
L2 .
Then‖f‖L2 = ‖f‖L2 in RN . (2.9a)
By the same way, we can get
‖f‖L2 = ‖f‖L2 in RN . (2.9b)
Based on Plancherel’s theorem, we can define the Fourier transform in L2(RN). SinceL1(RN)∩L2(RN) is dense in L2(RN), given f ∈ L2(RN), there exists a sequence fk ⊂L1(RN)∩L2(RN) such that ‖fk − f‖L2 → 0 as k →∞. Then fk is a Cauchy sequencein L2(RN). Hence, the sequence fk ⊂ L1(RN) ∩ L2(RN) is a Cauchy sequence inL2(RN) as well, because by Plancherel’s theorem, ‖fk1 − fk2‖L2 = ‖ fk1 − fk2‖L2 =
‖fk1 − fk2‖L2 , k1, k2 ∈ N. However, L2(RN) is complete, so, ∃f ∈ L2(RN) such thatf := limk→∞ fk. Also, ‖f‖L2 = limk→∞ ‖fk‖L2 = limk→∞ ‖fk‖L2 = ‖f‖L2 .
In the same way, we can define the inverse Fourier transform in L2(RN) as f(ξ) :=limk→∞ fk(ξ).
Lemma 2.4 could be extended to L2(RN).
Lemma 2.7. Let f, g ∈ L2(RN). Then∫RNf(x)g(x) dx =
∫RNf(y)g(y) dy. (2.7)
Proof. Recall again that L1(RN) ∩ L2(RN) is dense in L2(RN). So, there exist sequencesfk, gk ∈ L1(RN) ∩ L2(RN) such that fk → f and gk → g as k → ∞, i.e., ∀ε >0, ∃k0(ε) ∈ N s.t. ∀k ≥ k0 ‖fk − f‖L2 , ‖gk − g‖L2 < ε.Note that by the Plancherel’s theorem, ‖fk − f‖L2 = ‖fk − f‖L2 and ‖gk − g‖L2 =‖gk − g‖L2 .
41
1) Show that ‖fk‖L2 is finite.
‖fk‖L2 = ‖fk−f+f‖L2 ≤ ‖fk−f‖L2+‖f‖L2 < ε+‖f‖L2 , ∀k > k0(ε), ε > 0.
2) Show that for the L2 inner product, (fk, gk)→ (f , g) as k →∞.Consider∣∣∣∣∫
RNfkgk dx−
∫RNf g dx
∣∣∣∣ =∣∣∣∣∫RNfkgk dx−
∫RNfkg dx+
∫RNfkg dx−
∫RNf g dx
∣∣∣∣≤∣∣∣∣∫RNfk(gk − g) dx
∣∣∣∣+
∣∣∣∣∫RN
(fk − f
)g dx
∣∣∣∣ .Applying the Cauchy-Schwartz inequality, we get∣∣(fk, gk)− (f , g)
∣∣ ≤ ‖fk‖L2‖gk − g‖L2 + ‖fk − f‖L2‖g‖L2 <(‖fk‖L2 + ‖g‖L2
)ε < (ε+ ‖f‖L2 + ‖g‖L2) ε, ∀k ≥ k0, ∀ε > 0.
So, (fk, gk)→ (f , g) as k →∞.
Similarly, it could be shown that (fk, gk)→ (f, g) as k →∞.
3) Using the Lemma 2.4 for fk, gk ∈ L1(RN) ∩ L2(RN), we obtain the equation (2.7)for f, g ∈ L2(RN).∫
RNf(x)g(x) dx = lim
k→∞
∫RNfk(x)gk(x) dx =
limk→∞
∫RNfk(x)gk(x) dx =
∫RNf(x)g(x) dx.
We conclude our review of the Fourier transform by the following theorem.
Theorem 2.2 (Properties of the Fourier transform). Let f, g,∈ L2(RN). Then:
(i) ∫RNf(x)g(x) dx =
∫RNf(ξ)¯g(ξ) dξ, (2.10)
(ii) ∀ multiindices α such that Dαf ∈ L2(RN),
Dαf = (2πiξ)αf(ξ), (2.11)
42
(iii) if f, g ∈ L1(RN) ∩ L2(RN), then
f ∗ g = f g (2.5)
and
(f ∗ g) = f g (2.12)
(iv)
(f ) = f (2.13)
and
(f ) = f. (2.14)
So, the Fourier transform is a bijection L2(RN)→ L2(RN).
Proof.
(i) Apply the Plancherel’s theorem for (f + αg), α ∈ C, in RN .
‖f + αg‖2L2 = ‖f + αg‖2
L2 = ‖f + αg‖2L2 .
Then ∫RN
(f + αg)(f + αg) dx =
∫RN
(f + αg)(f + αg) dξ
or∫RN
(|f |2 + |α|2|g|2 + αgf + αf g
)dx =
∫RN
(|f |2 + |α|2|g|2 + αg¯f + αf
¯g) dξ,
‖f‖L2 + |α|2‖g‖L2 +
∫RN
(αgf + αf g) dx = ‖f‖L2 + |α|2‖g‖L2 +
∫RN
(αg¯f + αf
¯g) dξ.
Hence ∫RN
(αgf + αf g) dx =
∫RN
(αg¯f + αf
¯g) dξ.
For α = 1, we have:∫RN
(gf + fg) dx =
∫RN
(g¯f +f
¯g) dξ.
For α = i, we have: i∫RN
(gf − fg) dx = i∫RN
(g¯f −f ¯
g) dξ.
43
Subtracting the last equation from the previous one, we obtain
2∫RNf g dx = 2
∫RNf
¯g dξ. (2.10)
(ii) At first, we will prove the property (ii) for C∞0 (RN) by the mathematical induction.
Let f ∈ C∞0 (RN), and R > 0 is large enough such that spt(f) ⊂ B(0, R). Use thedefinition of the Fourier transform and then integrate by parts, to find the transformof ∂f
∂xj, 1 ≤ j ≤ N .
∂f
∂xj(ξ) =
∫RNe−2πix·ξ ∂f(x)
∂xjdx =
∫B(0,R)
e−2πix·ξ ∂f(x)
∂xjdx =∫
∂B(0,R)
e−2πix·ξf(x)νj(x) dS(x)−∫B(0,R)
∂
∂xj
(e−2πix·ξ) f(x) dx =
0− (−2πiξj)
∫B(0,R)
e−2πix·ξf(x) dx = 2πiξj
∫RNe−2πix·ξf(x) dx = 2πiξj f(ξ).
Here νj is the j-th component of the unit normal outward vector of ∂B(0, R).
For another step of the mathematical induction proof, assume that the property (ii)holds for Dβf , that is, for all multiindices β such that |β| = |α| − 1, i.e., Dβf(ξ) =(2πiξ)β f(ξ).
It follows that Dαf(x) =(Dβf
)xj
(x) for some 1 ≤ j ≤ N . Then
Dαf(ξ) = (2πiξj)Dβf(ξ) = (2πiξj)(2πiξ)β f(ξ) = (2πiξ)αf(ξ).
So, the property (ii) has been proved on C∞0 (RN).Now we prove that this property extends to any function f ∈ L2(RN) such thatDαf ∈ L2(RN) for multiindex α.C∞0 (RN) is dense in L2(RN), i.e., ∀f ∈ L2(RN), ∃ fk ⊂ C∞0 (RN) such thatfk → f as k →∞.fk converges uniformly on compact subsets of RN .So, Dαfk → Dαf as k →∞.Then using the definition of the Fourier transform in L2(RN) and the property (ii) inC∞0 (RN), we get
Dαf(ξ) = limk→∞
Dαfk(ξ) = limk→∞
(2πiξ)αfk(ξ) = (2πiξ)α limk→∞
fk(ξ) = (2πiξ)αf(ξ).
(2.11)
(iii) By the Lemma 2.3, the equation (2.5) holds in L1(RN) ∩ L2(RN) that is dense inL2(RN). Applying the density argument by the same way, as we did above proving
44
(ii), we get the equation (2.5) in L2(RN).Similarly, we can prove that
(f ∗ g) = f g. (2.12)
(iv) Show that ∀g ∈ L2(RN), ¯g(ξ) = g(ξ).
¯g(ξ) =
∫RNe−2πix·ξ g(x) dx =
∫RNe−2πix·ξ ¯g(x) dx =
∫RNe2πix·ξg(x) dx = g(ξ).
Next, using the Lemma 2.7, the last relation, and the property (i), we have∫RN
(f ) g dx =
∫RNf g dx =
∫RNf
¯g dx =
∫RNf ¯g dx =
∫RNfg dx.
Hence∫RN
((f )− f
)g dx = 0, ∀g ∈ L2(RN).
Therefore (f ) = f. (2.13)
The same proof works to show that
(f ) = f. (2.14)
Thus, the Fourier transform is a bijection L2(RN)→ L2(RN).
The Fourier transform is a powerful tool in solution of PDE. In particular, it could beapplied to solve the Poisson equation.
2.3 Elementary Properties of the Poisson Equation.
In this section, we review solution of the Poisson equation in RN using the Fourier trans-form.
Lemma 2.8. Let f ∈ C20(RN). Then for N ≥ 3, (Φ ∗ f)(x) is the solution of the Poisson
equation −∆u = f where the fundamental solution of Laplace’s equation is
Φ(x) =Γ(N/2)
2πN/2(N − 2)
1
|x|N−2, x ∈ RN \ 0. (2.15)
Proof.
1) Find the Fourier transform of ∆u using the property (2.11) of the Fourier transform,
∆u(ξ) = (N∑j=1
D2ju) (ξ) =
N∑j=1
(2πiξj)2u(ξ) = −4π2|ξ|2u(ξ).
45
Then applying the Fourier transform to both sides of the Poisson equation, we have
4π2|ξ|2u(ξ) = f(ξ) or u(ξ) = 14π2|ξ|2 f(ξ).
Take the inverse Fourier transform from both sides of the last equation and apply theproperties of the inverse Fourier transfom (2.13) and (2.12).We obtain
u(x) = 14π2 ( 1
|ξ|2 ) (x) ∗ f(x) = Φ(x) ∗ f(x),
where the fundamental solution of Laplace’s equation is
Φ(x) = 14π2 ( 1
|ξ|2 ) (x).
Note that 1|ξ|2 is not integrable for N = 2 at ξ = 0. So, we consider here the case
N > 2 only.
2)
(a) To find this inverse Fourier transform, we will use identity:
1
|ξ|2=
∫ ∞0
e−t|ξ|2
dt.
Then(1
|ξ|2
)(x) =
∫ ∞0
(∫RNe2πix·ξe−t|ξ|
2
dξ
)dt =∫ ∞
0
(∫RNe2πi
∑Nj=1 xjξje−t
∑Nj=1 ξ
2j dξ1dξ2 . . . dξN
)dt =∫ ∞
0
(N∏j=1
∫ ∞−∞
e2πixjξj−tξ2j dξj
)dt =
∫ ∞0
N∏j=1
Ij(t) dt
where Ij(t) =
∫ ∞−∞
e−(√tξj)
2+2πixjξj dξj.
(b)
Ij(t) =
∫ ∞−∞
e−(√tξj)
2+2πixj√
t(√tξj)+
π2x2jt−π2x2jt dξj =
e−π2x2jt
∫ ∞−∞
e−(√
tξj−πixj√t
)2
dξj.
46
Make the substitution z(ξj) =√tξj − πixj√
t. Then we have
Ij(t) = 1√te−
π2x2jt
∫Λj
e−z2
dz where Λj := z ∈ C|Imz = −πxj√t.
Without lost of generality, assume that xj ≤ 0.
Now define the counterclockwise oriented contour λ = ∪4k=1λk with
λ1 = (x, y)| −R ≤ x ≤ R, y = 0, λ2 = (x, y)|x = R, 0 ≤ y ≤ −πxj√t,
λ3 = (−x, y)| −R ≤ x ≤ R, y = −πxj√t,
λ4 = (x,−πxj√t− y)| x = −R, 0 ≤ y ≤ −πxj√
t for some R > 0.
e−z2 is analytic on C. So, by the Cauchy integral theorem,
∮λe−z
2dz = 0.
Then4∑
k=1
∫λk
e−z2
dz = 0, and∫λ1
e−z2
dz =
−∫λ2
e−z2
dz −∫λ3
e−z2
dz −∫λ4
e−z2
dz.
Now take the limit when R→∞.
limR→∞
∫λ1
e−z2
dz =
∫ ∞−∞
e−x2
dx =√π,
limR→∞
−∫λ3
e−z2
dz = limR→∞
∫−λ3
e−z2
dz =
∫Λj
e−z2
dz.
Also,
∣∣∣∣∣∫λ2,4
e−z2
dz
∣∣∣∣∣ =
∣∣∣∣∣∫ −πxj√
t
0
e−(±R+iy)2 dy
∣∣∣∣∣ ≤∫ −πxj√t
0
∣∣∣e−R2
e∓2Riy ey2∣∣∣ dy = e−R
2
∫ −πxj√t
0
ey2
dy → 0 as R→∞.
Therefore∫
Λje−z
2dz =
√π and Ij(t) =
√πte−
π2x2jt .
(c) Thus,
(1
|ξ|2
)(x) =
∫ ∞0
N∏j=1
√πte−
π2x2jt dt = πN/2
∫ ∞0
1tN/2
e−∑Nj=1
π2x2jt dt
= πN/2∫ ∞
0
1tN/2
e−π2|x|2t dt.
47
Note that applying L’Hopital rule [[(N + 1)/2]] times, we have
limt→0+
e−π2|x|2t
t= lim(
1t
)→∞
(1t)N/2
eπ2|x|2
(1t
) = 0 for N ∈ N.
However, when t→∞, e−π2|x|2t → 1. So, our integrand is integrable at
infinity only if N/2 > 1, i.e., for N ≥ 3.
Make the change of variable s = π2|x|2t
.(1
|ξ|2
)(x) = πN/2
∫ 0
∞
sN/2
πN |x|Ne−s(−π
2|x|2
s2
)ds =
π2−N/2
|x|N−2
∫ ∞0
s(N/2)−2e−s ds.
So, Φ(x) = 14π2
(1
|ξ|2
)(x) =
Γ(N2− 1)
4πN/2|x|N−2
where the gamma function Γ(z) =
∫ ∞0
tz−1e−t dt, Re z > 0.
Since Γ(z + 1) = zΓ(z), then Γ(N2− 1) = Γ(N/2)
(N/2)− 1. Then
Φ(x) =Γ(N/2)
2πN/2(N − 2)
1
|x|N−2, x ∈ RN \ 0. (2.15)
Next, we consider the case N = 2 separately.
Lemma 2.9. Let f ∈ C20(R2). The (Φ ∗ f)(x) is the solution of the Poisson equation
−∆u = f where the fundamental solution of the Laplace’s equation
Φ(x) = − 12π
log |x|, x ∈ R2 \ 0 (2.16)
Proof.
1) Assume that for N = 2, like in the case N ≥ 3, Φ(x) is a radial function satisfyingthe Laplace’s equation, i.e.,
Φ(x) = v(r) where r = |x| and ∆Φ(x) = 0.
Then
Φxj = v′(r)rxj = v′(r)xjr, j = 1, 2, r > 0.
48
Hence
Φxjxj = v′′(r)r2xj
+ v′(r)rxjxj = v′′(r)(xjr
)2+ v′(r)
r−xjxjr
r2.
Therefore
∆Φ = Φx1x1 + Φx2x2 = v′′(r)x21+x22r2
+ v′(r)(
2r− x21+x22
r3
)= v′′(r) + 1
rv′(r) = 0.
For w = v′(r), we get a separable ODE of the first order
w′ + 1rw = 0.
After separation variables and integration, we have
log |w| = − log r + log |C1|, C1 ∈ R \ 0, or v′ = C1
r.
After a second integration, we obtain
v(r) = C1 log r + C2, C2 ∈ R.
So, choosing C1 = − 12π
and C2 = 0, we get
Φ(x) = − 12π
log |x|, x ∈ R2 \ 0. (2.16)
Φ(x) from (2.16) is defined as the fundamental solution of the Laplace’s equation inR2 \ 0.
2) Show that
u(x) = (Φ ∗ f)(x) =
∫R2
Φ(x− y)f(y) dy ∈ C2(R2).
(a) Note that Φ has a singularity when x = y. To avoid this problem, use the factthat a convolution is commutative. Indeed, changing the variable of integrationby z(y) = x− y, we get
u(x) =
∫R2
Φ(z)f(x− z) dz = (f ∗ Φ)(x).
Take some h ∈ R, e1 = (1, 0), e2 = (0, 1), j = 1, 2 and consider the ratio
u(x+hej)−u(x)
h=
∫R2
Φ(z)f(x+hej−z)−f(x−z)
hdz.
49
Recall that f ∈ C20(R2). Then by Taylor’s theorem, we have
f(x+ hej − z) = f(x− z) + ∂f(x−z)∂xj
h+∂2f(x+cej−z)∂x2j
h2 with c between 0 and h.
So,
f(x+hej−z)−f(x−z)h
→ ∂f(x−z)∂xj
as h→ 0.
(b) Show that pointwise convergence above is uniform convergence.Take h = 1
nand define
gn(x) :=f(x+
ejn− z)− f(x− z)
1/n, M(x) :=
∣∣∣gn(x)− ∂f(x−z)∂xj
∣∣∣ .By the previous result, gn → ∂f(x−z)
∂xjas n→∞.
M(x) is continuous in R2 since f ∈ C2(R2). Take R > 0 so big that
B(0, R) ⊃ spt(gn), spt(∂f(x−z)∂xj
). Then ∀n ∈ N, ∃xn ∈ B(0, R)-compact,
such that M(xn) =∣∣∣gn(xn)− ∂f(xn−z)
∂xj
∣∣∣ = supx∈B(0,R) M(x).
Since gn(x)→ ∂f(x−z)∂xj
pointwise as n→∞, then M(x)→ 0 as n→∞.
We get that supx∈B(0,R)
∣∣∣gn(x)− ∂f(x−z)∂xj
∣∣∣→ 0 as n→∞.
So, gn(x)→ ∂f(x−z)∂xj
uniformly as n→∞.
Therefore, f(x+hej−z)−f(x−z)h
→ ∂f(x−z)∂xj
.
(c) Now, employing the uniform convergence of the integrand, we obtain
∂u(x)∂xj
= limh→0
u(x+hej)−u(x)
h= lim
h→∞
∫R2
Φ(z)f(x+hej−z)−f(x−z)
hdz
=
∫R2
Φ(z) ∂f(x−z)∂xj
dz =(∂f∂xj
)(x).
Repeating the same argument, we can get
∂2u(x)∂xj∂xk
=
∫R2
Φ(z)∂2f(x−z)∂xj∂xk
dz =(∂2f(x)∂xj∂xk
∗ Φ)
(x), j, k = 1, 2.
So, u(x) = (Φ ∗ f)(x) ∈ C2(R2).
3) It remains to show that u = Φ ∗ f satisfies the Poisson equation −∆u = f in R2.According to the previous part of our proof,
∆u(x) = (Φ ∗∆f)(x) =
∫R2
Φ(z)∆xf(x− z) dz.
50
Choose some ε > 0. Note that the integrand is C∞ function in R2 \B(0, ε).Represent
∆u(x) = Iε + Jε,
where Iε :=
∫B(0,ε)
Φ(z)∆xf(x− z) dz and Jε :=
∫R2\B(0,ε)
Φ(z)∆xf(x− z) dz
and we will take the limit as ε→ 0+ afterwards.
(a) Show that Iε → 0 as ε→ 0+.Estimate Iε.
|Iε| ≤ supx∈R2
|∆f(x)|∫B(0,ε)
|Φ(z)| dz ≤ C
∫B(0,ε)
|Φ(z)| dz, C ∈ R,
since f ∈ C20(R2), and ∆f is continuous on compact set and so is bounded.
Switch to polar coordinates because Φ is a radial function.
|Iε| ≤ C
∫B(0,ε)
12π| log r|r drdθ = C
2π
∫ 2π
0
dθ
∫ ε
0
| log r|r dr = C
∫ ε
0
| log r|r dr.
Assume that 0 < ε < 1. So, | log r| = − log r. Integrate by parts taking intoaccount that the integrand is singular at the origin.
|Iε| ≤ −C limδ→0+
(r2
2log r
∣∣∣εδ− 1
2
∫ ε
δ
r dr
)=
C2
[−ε2 log ε+ ε2
2+ lim
δ→0+
(δ2 log δ − δ2
2
)]≤ C
2ε2(− log ε) = C
2ε2| log ε|.
Here, to find the limit, we applied L’Hopital’s rule.
limδ→0+
δ2 log δ = limδ→0+
log δδ−2 =
(−∞∞
)= lim
δ→0+
δ−1
−2δ−3 = −12
limδ→0+
δ2 = 0.
So, |Iε| → 0 as ε→ 0+.
(b) Show that Jε → −f(x) as ε→ 0+.
i. Since ∆xf(x− z) = ∆zf(x− z)(−1)2, then
Jε =∫R2\B(0,ε)
Φ(z)∆zf(x− z) dz.
Let again R > ε > 0 be so big that B(0, R) ⊃ spt(f).
So, Jε =∫B(0,R)\B(0,ε)
Φ(z)∆zf(x− z) dz.
51
Now integrate by parts. We have
Jε = limR→∞
∫B(0,R)\B(0,ε)
Φ(z)∇z[∇zf(x− z)] dz =
limR→∞
[−∫B(0,R)\B(0,ε)
∇Φ(z)∇zf(x− z) dz +∫∂B(0,ε)
Φ(z)∂f(x−z)∂ν
dS(z) +
∫∂B(0,R)
Φ(z)∂f(x−z)∂ν
dS(z)
].
Note that ν is an exterior normal unit vector for the annulus B(0, R)\B(0, ε), i.e., ν is the outward unit vector normal to the outer circle∂B(0, R), and it is the inward unit vector normal to the inner circle∂B(0, ε).
If R→∞, then the last integral is 0 since f vanishes at infinity.Hence
Jε = Kε + Lε
where Kε := −∫B(0,R)\B(0,ε)
∇Φ(z)∇zf(x− z) dz
and Lε :=
∫∂B(0,ε)
Φ(z)∂f(x−z)∂ν
dS(z).
ii. Show that Lε → 0 as ε→ 0+.At first, estimate
∣∣∣∂f(x−z)∂ν
∣∣∣ on ∂B(0, ε) using the Cauchy-Schwartz inequa-lity.∣∣∣∂f(x−z)
∂ν
∣∣∣ = |ν · ∇f(x− z)| ≤ |ν| |∇f(x− z)|
≤ 1 sup |∇f(x− z)| ≤ ‖∇f‖L∞(R2) <∞ since f ∈ C20(R2).
Then |Lε| ≤ ‖∇f‖L∞∫∂B(0,ε)
|Φ(z)| dS(z).
Since |Φ(z)| = 12π| log ε| =const on ∂B(0, ε) as a radial function, then
|Lε| ≤ ‖∇f‖L∞ 12π| log ε|2πε because
∫∂B(0,ε)
dS(z) = 2πε, the length of
the circumference ∂B(0, ε).Applying L’Hopital’s rule, we have
limε→0+
ε| log ε| = − limε→0+
log ε1/ε
=(∞∞
)= − lim
ε→0+
ε−1
−ε−2 = limε→0+
ε = 0.
So, |Lε| → 0 as ε→ 0+.iii. Finally, show that Kε → −f(x) as ε → 0+. Perform integration by parts
52
for Kε and use again the fact that f vanishes at infinity.
Kε = −∫R2\B(0,ε)
∇Φ(z) · ∇zf(x− z) dz =∫R2\B(0,ε)
∆Φ(z)f(x− z) dz −∫∂B(0,ε)
∂Φ(z)∂ν
f(x− z) dS(z).
Since Φ(z) is a solution of the Laplace’s equation inR2\0, then ∆Φ(z)≡0 in R2 \ 0, and so
Kε = −∫∂B(0,ε)
∂Φ(z)∂ν
f(x− z) dS(z).
Now calculate ∂Φ(z)∂ν
on the circumference ∂B(0, ε).
∂Φ(z)∂ν
= ν · ∇Φ(z).
∂Φ(z)∂zj
= − 12π
∂
∂zjlog |z| = − 1
2π1|z|
∂
∂zj
[(z2
1 + z22)1/2
]=
− 12π|z|
12(z2
1 + z22)−1/2 2zj = − zj
2π|z|2 , j = 1, 2.
Then ∇Φ(z) = − 12π|z|2 z.
Since the vector z is an outward pointing vector normal to the circum-ference ∂B(0, ε), then ν = − z
|z| is an inward pointing vector normal to the∂B(0, ε).
So, ∂Φ(z)∂ν
= (− z|z|)(−
z2π|z|2 ) = |z|2
2π|z|3 = 12π|z| = 1
2πεon B(0, ε).
Therefore Kε = − 12πε
∫∂B(0,ε)
f(x− z) dS(z).
The integrand f is continuous on ∂B(0, ε). Apply the Mean Value theoremto get
Kε = − 1(2πε) f(x− z∗)(2πε) = f(x− z∗),
where z∗ is some point on the ∂B(0, ε).
Note that a choice of the constant for Φ(x) in the equation (2.16) is ex-plained by the necessity of cancellation by 2π to obtain −f(x) as ε→ 0+.
Exploiting the continuity of f once more, we attain that Kε → −f(x) asε→ 0+ because z∗ → 0 as ε→ 0+.
Thus, ∆u(x) = −f(x).
We summarize both lemmas in the following theorem.
53
Theorem 2.3. The fundamental solution of the Laplace’s equation in R2 \ 0 is
Φ(x) =
−1
2πlog |x|, if N = 2,
Γ(N/2)
2πN/2(N−2)1
|x|N−2 , if N ≥ 3.(2.15-2.16)
If f ∈ C20(RN), then there exists a solution u ∈ C2(RN) to the Poisson equation−∆u = f
in RN given by the convolution u(x) = (Φ ∗ f)(x).
In the next section, we introduce Sobolev spaces we will need in the chapter 3.
2.4 Calculus Inequalities in the Sobolev Spaces.
Here we review some useful properties of Sobolev spaces.
Definition 2.2. Let p ∈ [1,∞) and k ∈ Z+ ∪ 0.
The Sobolev space W k,p(RN) := f ∈ LP (RN) |Dαf ∈ Lp(RN) for |α| ≤ k
where Dα is the weak derivative. Dα = ∂α11 ∂α2
2 . . . ∂αNN , D0f = f, |α| =∑N
j=1 αj .
The norm ‖f‖Wk,p(RN ) :=
∑0≤|α|≤k
‖Dαf‖pLp(RN )
1/p
.
In particular, for the case p = 2, we use notation Hk(RN) := W k,2(RN), and so the norm
‖f‖Hk(RN ) :=
∑0≤|α|≤k
‖Dαf‖2L2(RN )
1/2
. (2.17)
We will use the following notation for the derivatives of k-th order:
Dkf = Dαf | |α| = k; ‖Dkf‖Lp(RN ) =∑|α|=k
‖Dαf‖Lp(RN ). (2.18)
Together with the norm defined by the equation (2.17), we will use else two other normsdefined by
‖f‖Hk(RN ) =
[∫RN|f(ξ)|2(1 + |ξ|2)k dξ
]1/2
(2.19)
where f(ξ) denotes the Fourier transform of f(x), and
‖f‖Hk(RN ) =∑
0≤|α|≤k
‖Dαf‖L2(RN ). (2.20)
The Sobolev spaceHk generalizes to the case k = s ∈ R with norm defined by the equation
54
(2.19).
The following lemma justifies an equivalence of all these norms in Hk(RN).
Lemma 2.10. The norms defined by the equations (2.17), (2.19), and (2.20) are equivalentin Hk(RN), k ∈ Z+ ∪ 0.
Proof.
1) Prove the equivalency in Hk(RN) of the norms defined by the equations (2.17) and(2.19).
(a) Starting with the norm defined by (2.17) and applying Plancherel’s theorem andthe property of the Fourier transform (2.11), we have
‖f‖2Hk(RN ) =
∑0≤|α|≤k
‖Dαf‖2L2(RN ) =
∑0≤|α|≤k
‖Dαf‖2L2(RN ) =
∑0≤|α|≤k
‖N∏j=1
(2πiξj)αj f‖2
L2(RN ) =∑
0≤|α|≤k
∫RN|2πi|2|α||ξα|2|f(ξ)|2 dξ =
∫RN
( ∑0≤|α|≤k
(4π2)|α||ξα|2)|f(ξ)|2 dξ where ξα = ξα1
1 ξα22 . . . ξαNN =
N∏j=1
ξαjj .
Hence we have the following estimate:∫RN
( ∑0≤|α|≤k
|ξα|2)|f(ξ)|2 dξ ≤ ‖f‖2
Hk(RN ) ≤
(4π2)k∫RN
( ∑0≤|α|≤k
|ξα|2)|f(ξ)|2 dξ
Comparing the bounds above with the Hk-norm in (2.19), we can see that it issufficient to show that
∑0≤|α|≤k |ξα|2 and (1 + |ξ|2)k are comparable.
(b) Show that ∑0≤|α|≤k
|ξα|2 =k∑
m=0
(|ξ|2)m
.
55
Indeed,
∑0≤|α|≤k
|ξα|2 = 1 +N∑j=1
ξ2j +
∑|α|=2
|ξα|2 + · · ·+∑|α|=k
|ξα|2.
where∑|α|=2
|ξα|2 =N∑j=1
(ξ2j )
2 + 2∑l 6=j
(ξlξj)2 =
(N∑j=1
ξ2j
)2
=(|ξ|2)2.
Here products ξlξj and ξjξl correspond to derivatives Dlj and Djl respectively.Assume that ∑
|α|=k−1
|ξα|2 = (|ξ|2)k−1
Then ∑|α|=k
|ξα|2 =N∑j=1
ξ2j
∑|α|=k−1
|ξα|2 = |ξ|2(|ξ|2)k−1 = (|ξ|2)k.
So, by mathematical induction,∑0≤|α|≤k
|ξα|2 =k∑
m=0
(|ξ|2)m
.
(c) Now consider
(1 + |ξ|2)k =k∑
m=0
(k
m
)(|ξ|2)m
The coefficients(k
m
)≥ 1. Then
∑0≤|α|≤k
|ξα|2 =k∑
m=0
(|ξ|2)m ≤ (1 + |ξ|2)k.
On the other hand,
(1 + |ξ|2)k ≤(
k
[[k/2]]
) ∑0≤|α|≤k
|ξα|2,
where(
k[[k/2]]
)is the biggest of binomial coefficients
(km
), 0 ≤ m ≤ k.
Thus, the Hk-norm defined by the equations (2.17) and (2.19) are equivalent.
2) Show that the equations (2.17) and (2.20) define equivalent norms in Hk(RN).
56
(a) Enumerate and denote all ‖Dαf‖L2(RN ), 0 ≤ |α| ≤ k as am, 0 ≤ m ≤M
with M =k∑l=0
N l = Nk+1−1N−1
. Then ‖Dαf‖L2(RN ) =M∑m=0
am,
and
∑0≤|α|≤k
‖Dαf‖2L2(RN )
1/2
=
(M∑m=0
a2m
)1/2
where am ≥ 0, m ∈ [0,M ].
It follows that (M∑m=0
a2m
)1/2
≤M∑m=0
am
since
M∑m=0
a2m ≤
(M∑m=0
am
)2
=M∑m=0
a2m + 2
∑m6=l
amal.
(b) On the other hand, since 2amal ≤ a2m + a2
l ,(M∑m=0
am
)2
≤M∑m=0
a2m +
∑m6=l
(a+m2 + a2l ) =
M∑m=0
a2m +M
M∑m=0
a2m =
(M + 1)M∑m=0
a2m.
So,M∑m=0
am ≤√M + 1
(M∑m=0
a2m
)1/2
.
Therefore, the two Hk-norms defined by the equations (2.17) and (2.20) are equiva-lent.
The next lemma asserts that an L2 bound on a higher derivative implies a pointwise boundon lower derivative.
Lemma 2.11 (Sobolev Inequality). If s > N/2, k ∈ Z+ ∪ 0, then Hs+k ⊂ Ck(RN),and there exists C > 0 such that
|v|Ck(RN ) ≤ C‖v‖Hs+k(RN ), ∀v ∈ Hs+k(RN). (2.21)
57
Proof.
1) Using the property (2.13) of the Fourier transform and the definition (2.4) of theinverse Fourier transform, we have, ∀x ∈ RN , ∀|α| ≤ s+ k,
|Dαv(x)| = |(Dαv) (x)| =∣∣∣∣∫RNe2πix·ξ(Dαv)(ξ) dξ
∣∣∣∣ ≤∫RN|e2πix·ξ| |(Dαv)(ξ)| dξ =
∫RN|(Dαv)(ξ)| dξ = ‖(Dαv)‖L1(RN ).
So, supx∈RN
|Dαv(x)| ≤ ‖(Dαv)‖L1(RN ).
2) In proving the previous lemma about an equivalence of the norms in Sobolev space,we got that there exists C1 > 0 such that
∑0≤|α|≤k
|ξα| ≤ C1
∑0≤|α|≤k
|ξα|21/2
, and∑
0≤|α|≤k
|ξα|2 ≤ (1 + |ξ|2)k
so,∑
0≤|α|≤k
|ξα| ≤ C1(1 + |ξ|2)k/2.
Then applying the property of the Fourier transform (2.11), the last inequality, andthen the Cauchy-Schwartz inequality after that, we have
∑0≤|α|≤k
‖Dαv‖L1(RN ) =∑
0≤|α|≤k
∫RN|2πi||α| |ξα| |v(ξ)| dξ =
∫RN
∑0≤|α|≤k
(2π)|α| |ξα|
|v(ξ)| dξ ≤ (2π)k∫RN
∑0≤|α|≤k
|ξα|
|v(ξ)| dξ ≤
C1(2π)k∫RN
(1 + |ξ|2)k/2 |v(ξ)| dξ = C1(2π)k∫RN
(1 + |ξ|2)(s+k)/2
(1 + |ξ|2)s/2|v(ξ)| dξ ≤
C1(2π)k∥∥(1 + |ξ|2)(s+k)/2|v(ξ)|
∥∥L2(RN )
∥∥(1 + |ξ|2)−s/2∥∥L2(RN )
=
C1(2π)k[∫
RN(1 + |ξ|2)s+k|v(ξ)|2 dξ
]1/2 [∫RN
(1 + |ξ|2)−s dξ
]1/2
.
So, using the norm in Sobolev space defined by the equation (2.19), we get
∑0≤|α|≤k
‖Dαv‖L1(RN ) ≤ C1(2π)k ‖v‖Hs+k(RN )
[∫RN
(1 + |ξ|2)−s dξ
]1/2
.
58
3) Need to show that∫RN (1 + |ξ|2)−s dξ <∞.
Turn to spherical coordinates in RN .∫RN
(1 + |ξ|2)−s dξ = ωN
∫ ∞0
(1 + r2)−srN−1 dr
where ωN is the surface area of a unit ball in RN .
Denote C2 =
∫ 1
0
(1 + r2)−srN−1 dr <∞.
Then∫RN
(1 + |ξ|2)−s dξ = ωN(C2 +
∫ ∞1
(1 + r2)−srN−1 dr).
Now using that r2 ≥ 1 in the last integral, we have∫RN
(1 + |ξ|2)−s dξ ≤ ωN(C2 +
∫ ∞1
2−sr−2s+N−1 dr).
Since by the hypothesis s > N2
, then −2s < −N , and −2s+N − 1 < −1.Hence the integral at the right-hand side of the last inequality is convergent and isequal to (2s−N)−1.
So,∫RN
(1 + |ξ|2)−s dξ ≤ C3 <∞ where C3 = ωN(C2 + 2−s
2s−N ).
4) We obtained∑0≤|α|≤k
‖Dαv‖L1(RN ) ≤ C ‖v‖Hs+k(RN ) where C = C1(2π)kC3.
Then ∑0≤|α|≤k
supx∈RN
|Dαv(x)| ≤ C ‖v‖Hs+k(RN ).
Hence
sup0≤|α|≤k
supx∈RN
|Dαv(x)| = |v|Ck(RN ) ≤ C ‖v‖Hs+k(RN ). (2.21)
Before we consider important calculus inequalities in Sobolev spaces, we will prove sev-eral useful propositions known as Gagliardo-Nirenberg estimates, following Chapter 13,Section 3 of [10].
59
Up to the end of this section, we will assume that all functional spaces are considered overRN without mentioning that explicitly.
Proposition 2.1. For real k ≥ 1, 1 ≤ p ≤ k, 1 ≤ j ≤ N , we have
‖Dju‖2L2k/p ≤ C ‖u‖Lq2 ‖D2
jju‖Lq1 ∀u ∈ C∞0 , (2.22)
hence ∀u ∈ Lq2 ∩Hq1 , where q1 = 2kp+1
, q2 = 2kp−1
.
Proof.
1) Let v ∈ C∞0 , real q ≥ 2.Show that v|v|q−2 ∈ C1
0 .
(a) For v 6= 0, Dj(|v|) = Dj(√v2) = 2v
2|v|Djv = sign(v)Djv.
Then Dj(v|v|q−2) = Djv |v|q−2 + (q − 2)v|v|q−3 v|v|Djv =
|v|q−2Djv + (q − 2)|v|q−4v2Djv = (q − 1)|v|q−2Djv for v 6= 0.
(b) Let v(x0) = 0 at some x0 ∈ RN . Show that Dj(v|v|q−2)|x=x0= 0.
limh→0
(v|v|q−2)(x0 + ejh)− (v|v|q−2)(x0)
h
v(x0 + ejh)
v(x0 + ejh)=
limh→0
(v|v|q−2)(x0 + ejh)
v(x0 + ejh)
v(x0 + ejh)
h=
limh→0|v(x0 + ejh)|q−2 lim
h→0
v(x0 + ejh)− v(x0)
h= |v(x0)|q−2Djv(x0) = 0.
So, ∀v ∈ C∞0 , (v|v|q−2) ∈ C10 .
2) Let v = Dju. Consider
Dj(u v|v|q−2) = Dju v|v|q−2 + u (q − 1)|v|q−2Djv = |v|q + u(q − 1)|v|q−2Djv.
Hence |v|q = Dj(u v|v|q−2)− u(q − 1)|v|q−2Djv
or |Dju|q = Dj(uDju|Dju|q−2)− u(q − 1)|Dju|q−2D2jju.
3) Integrating the last equality, we have
‖Dju‖qLq =
∫RN|Dju|q dx =∫
RNDj(uDju|Dju|q−2) dx− (q − 1)
∫RNu|Dju|q−2D2
jju dx.
60
The first integral∫RNDj(uDju|Dju|q−2) dx =
∫RN−1
(uDju|Dju|q−2)∣∣∞−∞ dN−1x = 0,
since u vanishes at infinity.Then applying the Generalized Holder’s inequality for 3 factors in the integrand ofthe second integral, we have
‖Dju‖qLq ≤ |q − 1|∣∣∣∣∫RNu|Dju|q−2D2
jju dx
∣∣∣∣ ≤|q − 1|
∫RN
∣∣u|Dju|q−2D2jju∣∣ dx ≤ |q − 1|‖u‖Lq2 ‖D2
jju‖Lq1∥∥|Dju|q−2
∥∥Lq/(q−2)
where 1q1
+ 1q2
+ q−2q
= 1 or 1q1
+ 1q2
= 2q
= pk
or q = 2kp.
Note that∥∥|Dju|q−2∥∥Lq/(q−2) =
(∫RN|Dju|
(q−2) qq−2 dx
)1q
(q−2)
= ‖Dju‖q−2Lq .
So, for q = 2kp
, we obtained
‖Dju‖qLq ≤ |q − 1| ‖u‖Lq2 ‖D2jju‖Lq1 ‖Dju‖q−2
Lq .
Hence,
‖Dju‖2Lq ≤ C ‖u‖Lq2 ‖D2
jju‖Lq1 with C = 2kp− 1. (2.22)
The next proposition is an application of the estimate (2.22).
Proposition 2.2. Let l,m, n ∈ Z+ and k, p ∈ R. If n ≤ p ≤ k+ 1−m and l ≥ n, then forsufficiently small ε > 0
‖Dlu‖L2k/p ≤ C ε ‖Dl−nu‖L2k/(p−n) + C(ε) ‖Dl+mu‖L2k/(p+m) . (2.23)
Proof. Let k, p ∈ R, k ≥ 1, 1 ≤ p ≤ k.
1) SubstituteDl−1u instead of u in the inequality (2.22) and take the square root of bothsides of the inequality
‖DjDl−1u‖L2k/p ≤
√C ‖Dl−1u‖1/2
L2k/(p−1) ‖D2jjD
l−1u‖1/2
L2k(p+1) .
61
Summing on j and recalling (2.18), we have
N∑j=1
∑|α|=l−1
‖DjDαu‖L2k/p ≤
√C
N∑j=1
∑|α|=l−1
‖Dαu‖L2k/(p−1)
1/2 ∑|α|=l−1
‖D2jjD
αu‖L2k/(p+1)
1/2
.
Note that the left-hand side of the last inequality is∑|β|=l
‖Dβu‖L2k/p = ‖Dlu‖L2k/p .
Denoting
aj =
∑|α|=l−1
‖Dαu‖L2k/(p−1)
1/2
, bj =
∑|α|=l−1
‖D2jjD
αu‖L2k/(p+1)
1/2
and applying the Cauchy-Schwartz inequality to the right-hand side of the last in-equality, we get
N∑j=1
ajbj ≤
(N∑j=1
a2j
)1/2( N∑j=1
b2j
)1/2
.
Then we obtain
‖Dlu‖L2k/p ≤√C
N ∑|α|=l−1
‖Dαu‖L2k/(p−1)
1/2 N∑j=1
∑|α|=l−1
‖D2jjD
αu‖L2k/(p+1)
1/2
.
Rewrite the last factor on the right-hand side by adding the norms of all mixed deriva-tives of the second order of Dαu, |α| = l − 1, to obtain
‖Dlu‖L2k/p ≤√C√N
∑|α|=l−1
‖Dαu‖L2k/(p−1)
1/2 ∑|α|=l+1
‖Dαu‖L2k/(p+1)
1/2
or
‖Dlu‖L2k/p ≤√CN ‖Dl−1u‖1/2
L2k/(p−1) ‖Dl+1u‖1/2
L2k/(p+1) .
62
After squaring, we have
‖Dlu‖2L2k/p ≤ CN ‖Dl−1u‖L2k/(p−1) ‖Dl+1u‖L2k/(p+1) . (2.24)
Return to the previous inequality and apply the Cauchy inequality with ε:
ab ≤ εa2 + b2
4ε, a, b ∈ R, ε > 0.
We get
‖Dlu‖L2k/p ≤ C1ε ‖Dl−1u‖L2k/(p−1) + C(ε) ‖Dl+1u‖L2k/(p+1) , ∀ε > 0 (2.25)
where C1 =√CN and C(ε) = C1
4ε.
2) Take l ≥ 2 and 2 ≤ p ≤ k. Using in the inequality (2.25), (p− 1) and (l− 1) insteadof p and l correspondingly, we get
‖Dl−1u‖L2k/(p−1) ≤ C1ε1 ‖Dl−2u‖L2k/(p−2) + C(ε1) ‖Dlu‖L2k/p , ε1 > 0. (2.26)
Now replace ‖Dl−1u‖L2k/(p−1) in the inequality (2.25) by the right-hand side of theinequality (2.26).
‖Dlu‖L2k/p ≤ C21εε1‖Dl−2u‖L2k/(p−2) + C1εC(ε1)‖Dlu‖L2k/p + C(ε)‖Dl+1u‖L2k/(p+1)
or
[1− C1εC(ε1)] ‖Dlu‖L2k/p ≤ C21εε1 ‖Dl−2u‖L2k/(p−2) + C(ε) ‖Dl+1u‖L2k/(p+1) .
Fix ε1 and pick ε so small that C1εC(ε1) ≤ 12.
Then
‖Dlu‖L2k/p ≤ C2ε ‖Dl−2u‖L2k/(p−2) + C1(ε) ‖Dl+1u‖L2k/(p+1) (2.27)
where C2 = 2C21ε1 and C1(ε) = 2C(ε).
3) Repeat this procedure. Take l ≥ 3, 3 ≤ p ≤ k and use (p− 1) and (l− 1) instead ofp and l correspondingly in the inequality (2.27).
‖Dl−1u‖L2k/(p−1) ≤ C2ε2 ‖Dl−3u‖L2k/(p−3) + C1(ε2) ‖Dlu‖L2k/p , ε2 > 0.
Replace ‖Dl−1u‖L2k/(p−1) in the inequality (2.25) by the right-hand side of the lastinequality to obtain
‖Dlu‖L2k/p ≤ C1C2εε2‖Dl−3‖L2k/(p−3) + C1εC1(ε2)‖Dlu‖L2k/p + C(ε)‖Dl+1u‖L2k/(p+1) .
63
Fix ε2 and pick ε so small that C1εC1(ε2) ≤ 12. Then
‖Dlu‖L2k/p ≤ C3ε ‖Dl−3u‖L2k/(p−3) + C2(ε) ‖Dl+1u‖L2k/(p+1)
where C3 = 2C1C2ε2 and C2(ε) = 2C1(ε).
4) Assume that l ≥ n and n ≤ p ≤ k.Continue to repeat the procedure described above. After repeating (n− 1) times, wewill have
‖Dlu‖L2k/p ≤ Cε ‖Dl−nu‖L2k/(p−n) + C(ε) ‖Dl+1u‖L2k/(p+1) , ∀ε > 0. (2.28)
5) Now we will work on the second term on the right-hand side of the inequality (2.28).Let l ≥ n, n ≤ p ≤ k− 1. Apply the inequality (2.25) using now (p+ 1) and (l+ 1)instead of p and l correspondingly. We get
‖Dl+1u‖L2k/(p+1) ≤ C1ε′ ‖Dlu‖L2k/p + C(ε′) ‖Dl+2‖L2k/(p+2) , ∀ε′ > 0 (∗)
Substituting the right-hand side of this inequality into the inequality (2.28) for‖Dl+1u‖L2k/(p+1) , we have
‖Dlu‖L2k/p ≤ Cε‖Dl−nu‖L2k/(p−n) + C(ε)C1ε′‖Dlu‖L2k/p + C(ε)C(ε′)‖Dl+2‖L2k/(p+2)
.
Fix ε′ and pick ε so small that C(ε)C1ε′ ≤ 1
2. Then
‖Dlu‖L2k/p ≤ C ′ε ‖Dl−n‖L2k/(p−n) + C ′(ε) ‖Dl+2u‖L2k/(p+2) (∗∗)
where C ′ = 2C and C ′(ε) = 2C(ε)C1ε′.
6) Take l ≥ n, n ≤ p ≤ k − 2 and use now (p + 2) and (l + 2) instead of p and lcorrespondingly in the inequality (2.25).
‖Dl+2u‖L2k/(p+2) ≤ Cε′′ ‖Dl+1u‖L2k/(p+1) + C(ε′′) ‖Dl+3u‖L2k/(p+3) , ∀ε′′ > 0.
In the last inequality replace ‖Dl+1u‖L2k/(p+1) by the right-hand side of the inequality(∗).
‖Dl+2u‖L2k/(p+2) ≤CC1ε
′ε′′‖Dlu‖L2k/p + Cε′′C(ε′)‖Dl+2u‖L2k/(p+2) + C(ε′′)‖Dl+3u‖L2k/(p+3) .
Fix ε′ and pick ε′′ so small that Cε′′C(ε′) ≤ 12. Then
‖Dl+2u‖L2k/(p+2) ≤ C ′′ε′′ ‖Dlu‖L2k/p + C ′′(ε′′) ‖Dl+3u‖L2k/(p+3)
64
where C ′′ = 2CC1ε′ and C ′′(ε′′) = 2C(ε′′).
Use this inequality to replace ‖Dl+2u‖L2k/(p+2) in the inequality (∗∗),
‖Dlu‖L2k/p ≤ C ′ε ‖Dl−nu‖L2k/(p−n) +
C ′(ε)C ′′ε′′ ‖Dlu‖L2k/p + C ′(ε)C ′′(ε′′) ‖Dl+3u‖L2k/(p+3) .
After that, fix ε′′ and pick ε so small that C ′(ε)C ′′ε′′ ≤ 12. Then
‖Dlu‖L2k/p ≤ C ′′′ε ‖Dl−nu‖L2k/(p−n) + C ′′′(ε) ‖Dl+3‖L2k/(p+3) .
where C ′′′ = 4C and C ′′′(ε) = 8C(ε)C1ε′C(ε′′).
7) Assume n ≤ p ≤ k− (m−1), l ≥ n. Repeat (m−1) times the procedure describedin part 6). For sufficiently small ε, we will obtain the following inequality:
‖Dlu‖L2k/p ≤ Cε ‖Dl−nu‖L2k/(p−n) + C(ε) ‖Dl+mu‖L2k/(p+m) . (2.23)
We underline a special case of Proposition 2.2 in the following corollary.
Corollary 2.1. If positive integers are satisfying l ≤ p ≤ k − 1, then
‖Dlu‖L2k/p ≤ Cε ‖u‖L2k/(p−l) + C(ε) ‖Dl+k−pu‖L2 (2.29)
In particular, if p = l < k, then
‖Dlu‖L2k/p ≤ Cε‖u‖L∞ + C(ε) ‖Dku‖L2 , (2.30)
for all u ∈ C∞0 .
Proof. If l = n, l ≤ p ≤ k − (m − 1), then the inequality (2.23) gives for sufficientlysmall ε > 0
‖Dlu‖L2k/p ≤ Cε ‖u‖L2k/(p−l) + C(ε) ‖Dl+mu‖L2k/(p+m) (2.31)
Hence for p = k −m or m = k − p, we obtain
‖Dlu‖L2k/p ≤ Cε ‖u‖L2k/(p−l) + C(ε) ‖Dl+k−pu‖L2 . (2.29)
The inequality (2.30) follows for p = l.
65
Next, we need two propositions to estimate ‖Dlu‖L2k/p by a product that is often moreconvenient than a sum like in the inequality (2.29).
Proposition 2.3. Let l,m, µ be nonnegative integers satisfying l ≤ max(µ,m), and p1, q,r ∈ [1,∞]. Suppose the estimate
‖Dlu‖Lq ≤ C1 ‖Dµu‖Lr + C2 ‖Dmu‖Lp1 (2.32)
is valid ∀u ∈ C∞0 . Then
‖Dlu‖Lq ≤ (C1 + C2) ‖Dµu‖β/(α+β)Lr ‖Dmu‖α/(α+β)
Lp1 (2.33)
with α = Nq− N
r+ µ− l, β = −N
q+ N
p1−m+ l. (2.34)
provided α, β 6= 0. In this case, α and β have the same sign.
Proof. Use the following notations:
Q = ‖Dlu‖Lq , R = ‖Dµu‖Lr , P = ‖Dmu‖Lp1 .
So, the inequality (2.32) is
Q ≤ C1R + C2P. (2.32a)
Replace u(x) in the inequality (2.32) by u(sx) for some s > 0.
Q = ‖Dlu‖Lq =∑|γ|=l
‖Dγu‖Lq =∑|γ|=l
(∫RN|Dγu(sx)|q dx
)1/q
.
Change to the variable t = sx. So dtj = sdxj and dt = SNdx.
Dγu(sx) = Dγu(t)(dtdx
)γ= Dγu(t)s|γ| = Dγu(t)sl.
Then
Q =∑|γ|=l
(∫RN|Dγu(t)|qslq−N dt
)1/q
= sl−(N/q)∑|γ|=l
(∫RN|Dγu(t)|q dt
)1/q
=
sl−(N/q)∑|γ|=l
‖Dγu‖Lq = sl−(N/q)‖Dlu‖Lq = sl−(N/q)Q.
Similarly, we get R = sµ−(N/r)R and P = sm−(N/p1)P .
66
So, the inequality (2.32a) gives
sl−(N/q)Q ≤ C1sµ−(N/r)R + C2s
m−(N/p1)P, ∀s > 0
or
Q ≤ C1sµ−(N/r)−l+(N/q)R + C2s
m−(N/p1)−l+(N/q)P = C1sαR + C2S
−βP.
If α > 0 and β < 0, then for s → 0+, the last inequality implies Q = 0. If α < 0 andβ > 0, then when s → ∞, we get Q = 0 as well. However, Q 6= 0, ∀u ∈ C∞0 . So, α andβ must have the same sign.Now choose s such that sαR = s−βP , i.e., sα+β = P
Ror s =
(PR
)1/(α+β).
Then s−βP = sαR =(PR
)α/(α+β)R = Pα/(α+β)R1−[α/(α+β)] = Pα/(α+β)Rβ/(α+β).
So, Q ≤ (C1 + C2)Pα/(α+β)Rβ/(α+β).
Thus, we got the estimate (2.33).
Proposition 2.4. If l, p, and k are positive integers satisfying l ≤ p ≤ k − 1, then
‖Dlu‖L2k/p ≤ C ‖u‖(k−p)/(k+l−p)L2k/(p−l) ‖Dk+l−pu‖l/(k+l−p)
L2 . (2.35)
In particular, if p = l < k, then
‖Dlu‖L2k/l ≤ C ‖u‖1−(p/k)L∞ ‖Dku‖l/kL2 . (2.36)
Proof. Apply the Proposition 2.3 to the estimate (2.29). Comparing that estimate with thehypothesis (2.32), we take
µ = 0, m = k + l − p, q = 2kp, r = 2k
p−l , p1 = 2.
Then the relations (2.34) give
α = Np2k− N(p−l)
2k+ 0− l = l(N
2k− 1) = l(N−2k)
2k
and β = −Np2k
+ N2− k − l + p+ l = −Np+Nk+(p−k)2k
2k= (N−2k)(k−p)
2k.
Hence α + β = N−2k2k
(l + k − p), βα+β
= k−pl+k−p , and α
α+β= l
l+k−p .
So, the estimate (2.33) implies
‖Dlu‖L2k/p ≤ C ‖u‖(k−p)/(k+l−p)L2k/(p−l) ‖Dk+l−pu‖l/(k+l−p)
L2 . (2.35)
For p = l, this inequality modifies to (2.36).
To prove the next proposition, we need to consider the following small lemma.
67
Lemma 2.12. If k, l,m ∈ Z+ and k = l +m, then ∀a, b ≥ 0, ∃C ∈ (0, 1) such that
am/k bl/k ≤ C(a+ b).
Proof. Let 0 ≤ m ≤ k and l = k−m. Keeping only one term in the binomial formula, wehave
(a+ b)k ≥(k
m
)am bl.
Raising both sides of this inequality to the power 1/k, we get
a+ b ≥(k
m
)1/k
am/k bl/k.
So,
am/k bl/k ≤ C(a+ b) with C =
(k
m
)−1/k
< 1.
Proposition 2.5. If |β|+ |γ| = k, then
‖(Dβf)(Dγg)‖L2 ≤ C ‖f‖L∞ ‖Dkg‖L2 +C ‖Dkf‖L2 ‖g‖L∞ , ∀f, g ∈ C0∩Hk. (2.37)
Proof. Let |β| = l, |γ| = m, and so l +m = k. Then l2k
+ m2k
= 12, and using the Holder’s
inequality, we have
‖(Dβf)(Dγg)‖L2 ≤ ‖Dβf‖L2k/l ‖Dγg‖L2k/m .
Apply the estimate (2.36) for each factor of the last inequality to find
‖(Dβf)(Dγg)‖L2 ≤(C1 ‖f‖1−(l/k)
L∞ ‖Dkf‖l/kL2
) (C2 ‖g‖1−(m/k)
L∞ ‖Dkg‖m/kL2
)=
C1C2
(‖f‖L∞ ‖Dkg‖L2
)m/k (‖Dkf‖L2 ‖g‖L∞)l/k
.
Now use Lemma 2.12. We obtain
‖(Dβf)(Dγg)‖L2 ≤ C ‖f‖L∞ ‖Dkg‖L2 + C ‖Dkf‖L2 ‖g‖L∞ (2.37)
with C = C1C2C3 where C1 and C2 are the constants from the estimate (2.36), and C3 isthe constant from the Lemma 2.12.
Beginning from this point and thereafter, we will use an abbreviated notation ‖u‖k instead
68
of ‖u‖Hk(RN ). So, ‖u‖0 means ‖u‖L2(RN ).
The following theorem is an application of the Gagliardo-Nirenberg estimates proved inthe Propositions 2.1-2.5.
Theorem 2.4 (Calculus Inequalities in the Sobolev Space).
(i) For all m ∈ Z+ ∪ 0, there exists C > 0 such that, for all u, v ∈ L∞(RN) ∩Hm(RN),
‖uv‖m ≤ C(‖u‖L∞ ‖v‖m + ‖u‖m ‖v‖L∞), (2.38)∑0≤|α|≤m
‖Dα(uv)− uDαv‖0 ≤ C(‖∇u‖L∞ ‖v‖m−1 + ‖u‖m ‖v‖L∞) (2.39)
(ii) For all s > N2
, Hs(RN) is a Banach algebra. That is, there exists C > 0 such that,for all u, v ∈ Hs(RN),
‖uv‖s ≤ C ‖u‖sv‖s. (2.40)
Proof.
(i)
(a) Let α, β, γ be multi-indices, and |β|+ |γ| = |α| = m.By definition
‖Dm(uv)‖0 :=∑|α|=m
‖Dα(uv)‖0.
Use the Leibnitz’s formula in RN , so that
‖Dm(uv)‖0 =∑|α|=m
∥∥∥ ∑β+γ=α
(α
β
)(Dβu)(Dγv)
∥∥∥0
with(α
β
)=
N∏j=1
(αjβj
).
Apply the triangle-inequality and the estimate (2.37) after that to obtain
‖Dm(uv)‖0 ≤∑|α|=m
∑β+γ=α
(α
β
)‖(Dβu)(Dγv)‖0 ≤
∑|α|=m
∑|β|≤|α|
(α
β
)Cβ
(‖u‖L∞ ‖D|α|v‖0 + ‖D|α|u‖0 ‖v‖L∞
)=
Cm
(‖u‖L∞ ‖Dmv‖0 + ‖Dmu‖0 ‖v‖L∞
)with Cm =
∑|α|=m
∑|β|≤|α|
(α
β
)Cβ.
69
(b) Similarly, we can get the following estimate for ‖Dm−1(uv)‖0.
‖Dm−1(uv)‖0 ≤ Cm−1
(‖u‖L∞ ‖Dm−1v‖0 + ‖Dm−1u‖0 ‖v‖L∞
).
Continuing in descending order of m, we finally have
‖D(uv)‖0 ≤ C1
(‖u‖L∞ ‖Dv‖0
).
Also,
‖uv‖0 ≤ ‖u‖L∞ ‖v‖0 ≤ ‖u‖L∞ ‖v‖0 + ‖u‖0 ‖v‖L∞ .
Adding all these inequalities, we obtain
‖uv‖m :=∑|α|≤m
‖Dα(uv)‖0 ≤ C(‖u‖L∞ ‖v‖m + ‖u‖m ‖v‖L∞
). (2.38)
(c) Prove now the inequality (2.39).Use again the Leibnitz’s formula in RN .
Dα(uv) =∑
β+γ=α(β≥0)
(α
β
)(Dβu)(Dγv) = u(Dαv) +
∑β+γ=α(β>0)
(α
β
)(Dβu)(Dγv).
Let ej be a vector with 1 as its j-th component and with 0s as all other compo-nents, and jβ be the number of the first nonzero component of multi-index β,i.e.,
jβ = infi∈I
i where I = i ∈ Z+ | i ≤ N and βi > 0.
Denote β′ = β − ejβ and Djβu = ujβ .Then
Dα(uv)− u(Dαv) =∑
β+γ=α(β>0)
(α
β
)(Dβ−ejβ (Djβu)
)(Dγv) =
∑|β′|+|γ|=|α|−1
(α
β′ + ejβ
)(Dβ′ujβ) (Dγv)
Hence
∥∥Dα(uv)− u(Dαv)∥∥
0≤
∑|β′|+|γ|=|α|−1
(α
β′ + ejβ
)∥∥(Dβ′ujβ) (Dγv)∥∥
0.
70
Apply now the estimate (2.37).
‖Dα(uv)− u(Dαv)‖0 ≤∑|β|≤|α|
C
(α
β
)(‖ujβ‖L∞ ‖D|α−1|v‖0 + ‖D|α−1|ujβ‖0 ‖v‖L∞
).
Since
‖ujβ‖L∞ ≤ max1≤j≤N
‖uj‖L∞ = ‖∇u‖L∞
and
‖D|α−1|ujβ‖0 ≤ ‖D|α−1|∇u‖0 = ‖D|α|u‖0,
then
‖Dα(uv)− u(Dαv)‖0 ≤ C1
(‖∇u‖L∞ ‖D|α−1|v‖0 + ‖D|α|u‖0 ‖v‖L∞
)where C1 =
∑|β|≤|α|C
(αβ
).
Now, taking a sum by all α with 0 ≤ |α| ≤ m, we have∑0≤|α|≤m
‖Dα(uv)− u(Dαv)‖0 ≤ C1
(‖∇u‖L∞ ‖v‖m−1 + ‖u‖m ‖v‖L∞
).
(2.39)
(ii) The calculus inequality (2.40) is an implication of the inequality (2.38).Let s ∈ Z+, s > N/2. By the Sobolev inequality (2.21) for k = 0, there exists suchC > 0 that
|v|C = supx∈RN
|v(x)| ≤ C ‖v‖s
Then there exists C1 > 0 and C2 > 0 such that
‖u‖L∞ ≤ C1‖u‖s and ‖v‖L∞ ≤ C2‖v‖s.
Applying these last inequalities to enforce the calculus inequality (2.38), we obtain
‖uv‖s ≤ C(C1 ‖u‖s ‖v‖s + C2 ‖u‖s ‖v‖s
)= C3 ‖u‖s ‖v‖s (2.40)
where C3 = C (C1 + C2).
71
These calculus inequalities conclude our review of properties of Sobolev space.
2.5 Properties of the Mollifiers.
Energy methods we will apply in the chapter 3, involves regularization of equations. As aregularizing tool, we will use a mollifier.
Definition 2.3. Given any radial function
ρ(|x|) ∈ C∞0 (RN), ρ ≥ 0,
∫RNρ dx = 1. (2.41)
The mollification Jεv of functions v ∈ Lp(RN), 1 ≤ p ≤ ∞ is defined by
(Jεv) (x) = ε−N∫RNρ(x−y
ε)v(y) dy, ε > 0. (2.42)
Note that since ρ(x−yε
) ∈ C∞0 (RN), then by the Leibnitz’s rule
Dα(Jεv)(x) = ε−N∫RNDαρ(x−y
ε)v(y) dy
for all multi-indices α, i.e., (Jεv)(x) ∈ C∞0 (RN).
Theorem 2.5 (Properties of Mollifiers).
(i) For all v ∈ C0(RN), Jεv → v uniformly on any compact set Ω ⊂ RN and
‖Jεv‖L∞ ≤ ‖v‖L∞ . (2.43)
(ii) Mollifiers commute with weak derivatives,
DαJεv = JεDαv, ∀|α| ≤ m, v ∈ Hm(RN). (2.44)
(iii) For all u ∈ Lp(RN), v ∈ Lq(RN), 1p
+ 1q
= 1,∫RN
(Jεu)v dx =
∫RNu(Jεv) dx, i.e., (Jεu, v)L2 = (u, Jεv)L2 . (2.45)
(iv) For all v ∈ Hs(RN), Jεv converges to v in Hs(RN), and the rate of convergence inHs−1-norm is linear in ε.
limε→0+
‖Jεv − v‖s = 0, (2.46)
‖Jεv − v‖s−1 ≤ Cε ‖v‖s. (2.47)
72
(v) For all v ∈ Hm(RN), k ∈ Z+ ∪ 0, and ε > 0,
‖Jεv‖m+k ≤Cmkεk‖v‖m, (2.48)
‖JεDkv‖L∞ ≤Ck
ε(N/2)+k‖v‖0. (2.49)
Proof.
(i) (a) First, prove estimate (2.43).
|Jεv| ≤ ε−N∫RNρ(x−y
ε)|v(y)| dy ≤ ‖v‖L∞ ε−N
∫RNρ(x−y
ε) dy.
Let u(y) = x−yε
. Then du(y) = (−1)N
εNdy.
Hence
ε−N∫RNρ(x−y
ε) dy = ε−N
∫ −∞∞
∫ −∞∞
. . .
∫ −∞∞
ρ(u) εN(−1)N
du =
∫RNρ(u) du = 1.
So, |Jεv| ≤ ‖v‖L∞ .
Then
‖Jεv‖L∞ ≤ ‖v‖L∞ . (2.43)
(b) Show that Jεv → v uniformly on any compact set Ω ⊂ RN .Since ρ(|x|) ∈ C∞0 (RN), then ∃ε > 0 such that ρ(x
ε) = 0 for |x| > ε.
Given compact Ω ⊂ RN . Define Ωε := x ∈ RN : dist(x,Ω) ≤ ε. Then Ωε
is compact as well. Hence v is uniformly continuous on Ωε:
∀ε′ > 0, ∃δ > 0 such that ∀x, y ∈ Ωε, |x− y| ≤ δ ⇒ |v(x)− v(y)| < ε′.
Now for x ∈ Ω and for ε > δ, we have
|(Jεv)(x)− v(x)| =∣∣∣ε−N ∫
RNρ(x−y
ε)v(y) dy − v(x) · 1
∣∣∣ =
∣∣∣ε−N ∫RNρ(x−y
ε)v(y) dy − v(x)ε−N
∫RNρ(x−y
εdy)∣∣∣ =
ε−N∣∣∣∫RNρ(x−y
ε)[v(y)− v(x)] dy
∣∣∣ ≤ ε−N∫RNρ(x−y
ε)|v(y)− v(x)| dy
< ε−Nε′∫RNρ(x−y
ε) dy = ε′ since ε−N
∫RNρ(x−y
ε) dy = 1 by the part (i(a)).
73
So, ∀ε′ > 0, ∃ε > 0 such that ∀x ∈ Ω, |(Jεv)(x) − v(x)| < ε′, i.e., Jεv → vuniformly on Ω.
(ii) Apply the Leibnitz rule. Then use the fact that
Dαxρ(x−y
ε) = (−1)|α|Dα
y ρ(x−yε
).
After that, integrate by parts |α| times taking in count that both ρ and Dαv,0 ≤ |α| ≤ m vanish in infinity.
Dα(Jεv)(x) = Dαx ε−N∫RNρ(x−y
ε)v(y) dy = ε−N
∫RNDαx [ρ(x−y
ε)]v(y) dy =
(−1)|α|ε−n∫RNDαy [ρ(x−y
ε)]v(y) dy = (−1)|α| ε−N(−1)|α|
∫RNρ(x−y
ε)Dα
y v(y) dy
= (JεDαv) (x).
We got the property (2.44).
(iii) Let u ∈ Lp(RN), v ∈ Lq(RN), and 1p
+ 1q
= 1.
By the Holder’s inequality, ‖uv‖L1 ≤ ‖u‖Lp ‖v‖Lq . Then (uv) ∈ L1(RN).
So, (ρuv) ∈ L1(RN) since ρ ∈ C∞0 (RN).Changing an order of integration in integral over RN × RN , we have∫
RN(Jεu)(x)v(x) dx =
∫RNεN[∫
RNρ(x−y
ε)u(y) dy
]v(x) dx =∫
RN×RNε−Nρ(x−y
ε)u(y)v(x) dxdy =
∫RNu(y)
[ε−N
∫RNρ(x−y
ε)v(x) dx
]dy
=
∫RNu(y) (Jεv)(y) dy.
We obtained the property (2.45).
(iv) (a) Let v ∈ Cs0(RN) and let compact Ω ⊃ ∪|α|≤s sptDαv.
By the property (i), (Jεv − v)→ 0 uniformly on compact Ω as ε→ 0+.By the property (ii), DαJεv = JεD
αv. So, applying the property (i), we getthat (Jε(D
αv)−Dαv)→ 0 uniformly on Ω as ε→ 0+, ∀α satisfying |α| ≤ s.Apply now the triangle inequality , the commutation property of mollifiers withderivatives (ii), and the estimate (2.43) to the difference [Dα(Jεv)−Dαv].
|Dα(Jεv)−Dαv| ≤ |Jε(Dαv)|+ |Dαv| ≤ 2‖Dαv‖L∞ .
Define g(x) := 2‖Dαv‖L∞ . g(x) is integrable on RN since∫RNg(x) dx =
∫Ω
2 ‖Dαv‖L∞ dx = 2 ‖Dαv‖L∞Vol(Ω) <∞.
74
So, we can apply the Lebesgue Dominated theorem. Then we get
∀|α| ≤ s, ∃ limε→0‖Jε(Dαv)−Dαv‖0 = 0
or after summation by α, we obtain
limε→0‖Jεv − v‖Hs = 0. (2.46)
(b) Since spaces of continuous functions are dense in L2, then the equation (2.46)holds ∀Dαv ∈ L2(RN), ∀|α| < s or ∀v ∈ Hs(RN).
(c) For the proof of (2.44), recall that the radial function ρ(|x|) ∈ C∞0 (RN), ; ρ ≥ 0,and
∫RN ρ dx = 1.
Since ρ(|x|) = ρ
((∑Nj=1 x
2j
)1/2)
, then ρ is an even function of xj, ∀1 ≤ j ≤
N .
(1) The Fourier thansform ρ(εξ) =∫RN e
−2πix·εξρ(|x|) dx.Hence ρ(0) =
∫RN ρ(|x|) dx = 1.
Next, show that∇ρ(0) = 0.The j-th component(∇ρ(εξ)
)j
=∂
∂ξjρ(εξ) =∫
RN
∂
∂ξje−2πiε
∑Nk=1 xkξkρ(|x|) dx = (−2πiε)
∫RNe−2πix·ξεxjρ(|x|) dx.
Then (∇ρ(0)
)j
= (−2πiε)
∫RN−1
(∫ ∞−∞
xjρ(|x|) dxj)∏k 6=j
dxk = 0,
since xjρ(|x|) is an odd function of xj that makes the internal integral overR equal to 0.Using the Taylor extension of the first order at ξ = 0, we get
ρ(εξ) = ρ(0) +∇ρ(0) εξ +O((εξ)2
)= 1 +O
(|εξ|2
)or ∣∣ρ(εξ)− 1
∣∣ ≤ Cε2|ξ|2 for some constant C.
75
Estimate the following ratio:∣∣∣∣∣[ρ(εξ)− 1
]21 + |ξ|2
∣∣∣∣∣ ≤∣∣ρ(εξ)− 1
∣∣2|ξ|2
≤ C2ε4|ξ|4
|ξ|2= C2ε4|ξ|2. (∗)
Also ∣∣ρ(εξ)∣∣ ≤ ∫
RN
∣∣e−2πiεξ·x∣∣ ρ(|x|) dx =
∫RNρ(|x|) dx = 1.
Then
|ρ(εξ)− 1| ≤∣∣ρ(εξ)
∣∣+ 1 ≤ 2.
Hence ∣∣∣∣∣[ρ(εξ)− 1
]21 + |ξ|2
∣∣∣∣∣ ≤∣∣ρ(εξ)− 1
∣∣2|ξ|2
≤ 4
|ξ|2. (∗∗)
For |ξ| < 1ε, by the estimate (∗), we have∣∣∣∣∣
[ρ(εξ)− 1
]21 + |ξ|2
∣∣∣∣∣ ≤ C2ε4 1ε2
= C2ε2.
For |ξ| ≥ 1ε, the estimate (∗∗) gives∣∣∣∣∣
[ρ(εξ)− 1
]21 + |ξ|2
∣∣∣∣∣ ≤ 4ε2.
So, ∀ξ ∈ RN , we obtain∣∣∣∣∣[ρ(εξ)− 1
]21 + |ξ|2
∣∣∣∣∣ ≤ C1ε2 with C1 = maxC2, 4. (∗∗∗)
(2) Note that mollification of v(x) is a convolution,
Jεv(x) = ε−N∫RNρ(x−y
ε)v(y) dy = ρε(x) ∗ v(x), where ρε(x) = ε−Nρ(x
ε).
Consider the Fourier transform
ρε(ξ) =
∫RNe−2πiξ·xε−Nρ(x
ε) dx.
76
Perform the substitution of x = εy. Then
dx =N∏k=1
dxk =N∏k=1
d(εyk) = εNN∏k=1
dyk = εN dy.
Hence
ρε(ξ) =
∫RNe−2πi(εξ)·yρ(y) dy = ρ(εξ).
Now, applying the Fourier transform property (2.5) about convolutions, weobtain
Jεv(ξ) = ρε(ξ) v(ξ) = ρ(εξ) v(ξ).
(3) Finally, using the definition of the Sobolev norm (2.19), the last equality,and the bound (∗∗∗), we get the estimate (2.47).
‖Jεv − v‖2s−1 =
∫RN
(1 + |ξ|2)s−1∣∣∣ (Jεv − v)(ξ)
∣∣∣2 dξ=
∫RN
(1 + |ξ|2)s−1∣∣∣Jεv(ξ)− v(ξ)
∣∣∣2 dξ=
∫RN
(1 + |ξ|2)s−1∣∣∣ρ(εξ)v(ξ)− v(ξ)
∣∣∣2 dξ=
∫RN
(1 + |ξ|2)s−1∣∣∣ρ(εξ)− 1
∣∣∣2 ∣∣∣v(ξ)∣∣∣2 dξ
=
∫RN
∣∣∣ρ(εξ)− 1∣∣∣2
1 + |ξ|2(1 + |ξ|2)s
∣∣∣v(ξ)∣∣∣2 dξ
≤ C1ε2
∫RN
(1 + |ξ|2)s∣∣∣v(ξ)
∣∣∣2 dξ = C1ε2‖v‖2
s.
(v) (a) Prove the important estimate (2.48).First, note that since ρ(x) ∈ C∞0 (RN), then ρβ(x) := Dβρ(x) ∈ C∞0 (RN).So,∫RN |ρβ(x)| dx = Cβ <∞.
Let m, k ∈ Z+ ∪ 0, ε > 0, |α| ≤ m, and |β| ≤ k. Compute
DβDαJεv(x) = DβDαε−N∫RNρ(x−y
ε)v(y)dy = ε−NDα
∫RNDβxρ(x−y
ε)v(y)dy.
Let u = x−yε
. Then
Dβxρ(u) = Dβ
uρ(u)(Dxu)|β| = ρβ(u)ε−|β|.
77
So,
DβDαJεv(x) = ε−|β|−NDα
∫RNρβ(x−y
ε)v(y)dy = ε−|β|−N
∫RNDαxρβ(x−y
ε)v(y)dy.
Since Dαx ρβ(x−y
ε) = (−1)|α|Dy ρβ(x−y
ε), then
DβDαJεv(x) = (−1)|α| ε−|β|−N∫RNDαy ρβ(x−y
ε) v(y) dy.
Now integrate by parts |α| times. We get
DβDαJεv(x) = ε−|β|−N∫RNρβ(x−y
ε)Dα
y v(y) dy.
Hence
|DβDαJεv(x)|2 = ε−2|β|−2N
∣∣∣∣∫RNρβ(x−y
ε)Dα
y v(y) dy
∣∣∣∣2≤ ε−2|β|−2N
[∫RN|ρβ(x−y
ε)| |Dα
y v(y)| dy]2
= ε−2|β|−2N
[∫RN
√|ρβ(x−y
ε)|√|ρβ(x−y
ε)|∣∣Dα
y v(y)∣∣ dy]2
.
Applying the Schwartz inequality, we have
∣∣DβDαJεv(x)∣∣2 ≤
ε−2|β|[ε−N
∫RN|ρβ(x−y
ε)| dy
] [ε−N
∫RN|ρ(x−y
ε)||Dα
y v(y)|2 dy]
≤ Cβ ε−2|β|−N
∫RN|ρβ(x−y
ε)||Dα
y v(y)|2 dy
since ε−N∫RN|ρβ(x−y
ε)| dy =
∫RN|ρβ(u)| du = Cβ after substitution u = x−y
ε.
Estimate now the L2-norm of DβDαJεv by using the Fubini theorem.
‖DβDαJεv‖20 ≤
Cβε2|β|
∫RNε−N
(∫RN|ρβ(x−y
ε)| |Dα
y v(y)|2 dy)dx =
Cβε2|β|
∫RN|Dα
y v(y)|2(∫
RNε−N |ρβ(x−y
ε)| dx
)dy ≤ C2
β
ε2|β|‖Dαv(y)‖2
0.
78
So,
‖Jεv‖m+k =
(∑|β|≤k|α|≤m
‖DβDαJεv‖20
)1/2
≤
(∑|β|≤k|α|≤m
C2β
ε2|β|‖Dαv‖2
0
)1/2
≤
(∑|α|≤m
C2kα
ε2k‖Dαv‖2
0
)1/2
≤
(C2mk
ε2k
∑|α|≤m
‖Dαv‖20
)1/2
= Cmkεk‖v‖m,
where C2mk = max|α|≤mC
2kα, C2
kα = max|β|≤k C2β .
Also, we used here that ε−2|β| ≤ ε−2k since 0 < ε < 1.
(b) It remains to prove the estimate (2.49).Let ρk(x) = Dkρ(x). Since mollifiers commute with derivatives by the property(2.44), then we have
|JεDkv(x)| = |DkJεv(x)| = ε−N∣∣∣∫RNDkxρ(x−y
ε) v(y) dy
∣∣∣.Because
Dkx ρ(x−y
ε) = Dk
u ρ(u) ε−k = ε−kρk(x−yε
) with u = x−yε,
we get using the Schwartz inequality,
|JεDkv(x)| = ε−N−k∣∣∣∫RNρk(
x−yε
)v(y) dy∣∣∣
≤ ε−(N/2)−kε−N/2[∫
RN|ρk(x−yε )|2 dy
]1/2 [∫RN|v(y)|2 dy
]1/2
≤ Ckε(N/2)+k
‖v‖0,
where C2k = ε−N
∫RN|ρk(x−yε )|2 dy =
∫RN|ρk(u)|2 du since |ρk(u)|2 ∈ C0(RN).
So,
‖JεDkv(x)‖L∞ ≤Ck
ε(N/2)+k‖v‖0. (2.49)
79
2.6 The Hodge’s Decomposition of Vector Fields and Properties of the Leray’sProjection Operator.
At first, prove Hodge’s decomposition in RN .
Proposition 2.6 (Hodge’s Decomposition inRN ). Every vector field v∈L2(RN)∩C∞(RN)has a unique decomposition:
v = w +∇q, divw = 0, (2.50)
where w is a divergence free vector field and q is a scalar in RN with the followingproperties:
(i) w,∇q ∈ L2(RN) ∩ C∞(RN),
(ii) w ⊥ ∇q in L2, i.e., (w,∇q)L2 = 0,
(iii) for any multi-index β of the derivative Dβ, |β| ≥ 0,
‖Dβv‖20 = ‖Dβw‖2
0 + ‖∇Dβq‖20. (2.51)
Proof.(i) (a) Show that w, ∇q ∈ C∞(RN).
From the equations (2.50), we get
div v = div(w +∇q) = divw + div(∇q) = 0 + ∆q = ∆q.
So,
∆q = div v (2.52)
Thus, q ∈ C∞(RN) as a solution of the Poisson equation (2.52). Then w ∈C∞(RN) as well since w = v − ∇q. This part also proves the existence ofq and w such that (2.50) holds.
(b) Prove that w,∇q ∈ L2(RN).Assume v ∈ C∞(RN) with bounded spt v ⊂ x ∈ RN | |x| < R for someR > 0.As a solution of the Poisson equation (2.52),
q(x) = −Φ(x) ∗ div v(x)
where Φ(x) is the fundamental solution of the Laplace’s equation given by(2.15) and (2.16).
Then
∇q(x) = −∫|y|≤R
∇xΦ(x− y) div v(y) dy
80
Now
∇Φ(x) =
− 1
2π
x
|x|2, if N = 2
−Γ(N/2)
2πN/2x
|x|N, if N ≥ 3
= − 1
ωN
x
|x|N.
where ωN is the surface area of a unit sphere in RN .
So,
∇q(x) =1
ωN
∫|y|≤R
x− y|x− y|N
divv(y) dy.
Using the Taylor expansion, we have for large |x|
|x− y|−N =
(|x|∣∣∣∣ x|x| − y
|x|
∣∣∣∣)−N = |x|−N(x
|x|− y
|x|
)2(−N/2)
=
|x|−N(x2
|x|2− 2
x · y|x|2
+y2
|x|2
)−N/2= |x|−N
[1 +
(−2
x · y|x|2
+y2
|x|2
)]−N/2=
|x|−N[
1− N
2
(−2
x · y|x|2
+y2
|x|2
)+
N2
(N2
+ 1)
2
(−2
x · y|x|2
+y2
|x|2
)2
+ . . .
].
This binomial series converges if∣∣∣−2 x·y
|x|2 + y2
|x|2
∣∣∣ < 1.
Assume that ∣∣∣∣−2x · y|x|2
+y2
|x|2
∣∣∣∣ ≤ 2|x| |y||x|2
+|y|2
|x|2< 1
Then |y||x| <√
2− 1 or |x| > (√
2 + 1)|y|.
So, for |y| ≤ R and |x| > (√
2 + 1)R, we have∣∣∣∣−2x · y|x|2
+y2
|x|2
∣∣∣∣ ≤ 2|y||x|
+|y|2
|x|2≤ 2R
|x|+R2
|x|2≤ C
|x|= O
(1
|x|
),
81
and
|x− y|−N = |x|−N[1 +O
(−2
x · y|x|2
+y2
|x|2
)]= |x|−N
[1 +O
(1
|x|
)].
Then
x− y|x− y|N
=x− y|x|N
[1 +O
(1
|x|
)]=x− y|x|N
+x− y|x|N
O(
1
|x|
)=
x
|x|N− y
|x|N+ (x− y)O
(1
|x|N+1
).
Also, since v ∈ C∞(RN), then ∃C0 > 0 such that |divv(y)| ≤ C0 on |y| ≤ R.
Then
∇q(x) =1
ωN
[x
|x|N
∫|y|≤R
divv(y) dy − 1
|x|N
∫|y|≤R
y divv(y) dy+
O(
1
|x|N+1
)∫|y|≤R
(x− y)divv(y) dy
].
Applying the Gauss-Green theorem, we get∫|y|≤R
divv(y) dy =
∫|y|=R
v(y)ν(y) dS = 0 since spt v ⊂ x ∈ RN | |x| < R.
Estimate the remaining two integrals.∣∣∣∫|y|≤R
ydivv(y) dy∣∣∣ ≤ ∫
|y|≤R|y| |divv(y)| dy ≤ RC0Vol (B(0, R)) ,
∣∣∣∫|y|≤R
(x− y)divv(y) dy∣∣∣ ≤ ∫
|y|≤R|x− y| |divv(y)| dy ≤
C0
∫|y|≤R
(|x|+ |y|) dy ≤ 2C0|x|Vol (B(0, R)) .
So,
∇q(x) =1
ωN
[0 +O(|x|−N) + |x|O(|x|−N−1)
]= O(|x|−N),
i.e., |∇q(x)| ≤ C1|x|−N for some C1 > 0 and |x| > (√
2 + 1)R.
82
Now it is convenient to use polar coordinates to check if∇q ∈ L2(RN).∫|x|≥3R
|∇q(x)|2 dx ≤ C21
∫|x|≥3R
|x|−2N dx = C21ωN
∫ ∞3R
r−2NrN−1 dr =
C21ωN lim
a→∞
(− 1
NrN
)∣∣∣∣a3R
=C2
1ωNN
lima→∞
(− 1
aN+
1
(3R)N
)=
C21ωN
N3NRN<∞.
Thus, ∇q ∈ L2(RN). Hence w = v −∇q ∈ L2(RN) as well.
(ii) Show that w ⊥ ∇q in L2(RN).
(a) Recall that v vanishes sufficiently rapidly at infinity. Let v = O(|x|−N) as|x| → ∞. From the part(i), we have that ∇q(x) = O(|x|−N) as |x| → ∞.Then w = v −∇q = O(|x|−N).
(b) Estimate q(x) for large |x|.
q(x) = −∫|y|≤R
Φ(x− y) divv(y) dy since spt v ⊂ B(0, R),
and Φ(x) is given by the equations (2.15) for N > 2 and (2.16) for N=2.
In case N > 2,
q(x) =1
ωN(2−N)
∫|y|≤R
|x− y|−N+2 divv(y) dy.
Similarly as we did in the part (i), we get for large |x|
|x− y|−N+2 = |x|−N+2
[1 +O
(−2
x · y|x|2
+y2
|x|2
)].
Then
|q(x)| ≤ |x|−N+2
ωN |2−N |
[∫|y|≤R
divv(y) dy+
C2
∫|y|≤R
∣∣∣−2x · y|x|2
+y2
|x|2∣∣∣ ∣∣∣divv(y)
∣∣∣ dy] ≤|x|−N+2
ωN(N − 2)
[0 + 2C2
|x||x|2
∫|y|≤R|y| |divv(y)| dy +
C2
|x|2
∫|y|≤R|y|2|divv(y)| dy
]
≤ |x|−N+2
ωM(N − 2)
(2C2
|x|RC0 +
C2
|x|2R2C0
)VolB(0, R)
≤ C3|x|−N+2
|x|= O(|x|−N+1).
83
In case N = 2,
q(x) =1
2π
∫|y|≤R
ln |x− y| divv(y) dy.
|x− y| ≤ |x|+ |y| ≤ 2|x| implies ln |x− y| ≤ ln |x|+ ln 2 ≤ C3 ln |x|
for large |x|. Then
|q(x)| ≤ 1
2π
∫|y|≤R| ln |x− y|| |divv(y)| dy ≤
C3
2πC0Vol
(B(0, R)
)ln |x| = O(ln |x|).
Thus,
q(x) =
O(ln |x|), if N = 2,
O(|x|−N+1), if N > 2.
(c) So, as |x| → ∞,
|w(x)| |q(x)| =
o(|x|−1), if N = 2,
O(|x|−2N+1), if N > 2.
Here for N = 2, we used L’Hopital’s rule.
lim|x|→∞
|w(x)| |q(x)||x|−1
≤ lim|x|→∞
C|x|−2 ln |x||x|−1
= C lim|x|→∞
ln |x||x|
= C lim|x|→∞
1
|x|= 0.
So, |w(x)| |q(x)| = o(|x|−N+1) as |x| → ∞. Then the conditions of Lemma1.5 are satisfied, and applying this lemma, we get that w and∇q are orthogonal:∫
RNw · ∇q dx = 0.
Hence
‖v‖20 =
∫RNv2 dx =
∫RN
(w +∇q)2 dx =∫RNw2 dx+ 2
∫RNw · ∇q dx+
∫RN
(∇q)2 dx = ‖w‖20 + ‖∇q‖2
0.
Note that the orthogonality of the decomposition (2.50) implies its uniqueness.In detail, if v = w +∇q = w′ +∇q′, then the previous estimates apply to w′
and ∇q′ so that (w′,∇q′)L2 = (w′,∇q)L2 = (w,∇q′)L2 = 0, and this willimply that ‖w − w′‖L2 = ‖∇q −∇q′‖L2 = 0.
84
(iii) Differentiating the equations (2.50) and (2.52) for any multi-index β, we have
Dβv = Dβw +Dβ(∇q), divDβw = 0,
∆Dβq = div (Dβv).
Applying for Dβv parts (i) and (ii) of the current proposition, we get that Dβw andDβ(∇q) are orthogonal, and consequently, we have
‖Dβv‖20 = ‖Dβw‖2
0 + ‖∇Dβq‖20 (2.51)
For now, we assumed in our proof that v ∈ C∞0 (RN).
(iv) Finally, show that all statements hold for v ∈ L2(RN) ∩ C∞(RN).
(a) For this purpose, consider a function ρ(x) ∈ C∞0 (RN) such that ρ(x) ≡ 1 for|x| ≤ 1 and ρ(x) ≡ 0 for |x| ≥ 2.
Define vn(x) := v(x)ρ( |x|n
), n ∈ N.
vn(x) ⊂ C∞0 (RN), ∀v ∈ C∞(RN) that implies this proposition holds forvn(x) ∀n.
Also, vn(x) ⊂ L2(RN).
Clearly, vn → v as n→∞.
(b) Since vn ∈ C∞0 (RN), then by the previous proof, ∃ wn,∇qn ∈ L2(RN) ∩C∞(RN) such that Proposition 2.6 holds for vn = wn +∇qn.Need to show that ∃w, q such that w,∇q ∈ L2(RN), wn → w, and
∇qn → ∇q in L2(RN).Let n,m ∈ Z+. vn = wn +∇qn and vm = wm +∇qm.Subtracting these equations, we get the following decomposition:
vn − vm = (wn − wm) + (∇qn −∇qm) where div (wn − wm) = 0.
By Proposition 2.6, (wn − wm) ⊥ (∇qn −∇qm) and
‖vn − vm‖20 = ‖wn − wm‖2
0 + ‖∇qn −∇qm‖20.
The last equation leads to the following two inequalities:
‖vn − vm‖0 ≥ ‖wn − wm‖0 and ‖vn − vm‖0 ≥ ‖∇qn −∇qm‖0.
Therefore, vn is Cauchy in L2(RN) implies that both wn and ∇qn areCauchy in L2(RN).So, ∃w, α ∈ L2(RN) such that wn → w and ∇qn → α as n→∞.
(c) Need to check that w is divergence-free, i.e., divw = 0 in sense of weak deriva-
85
tives.From Proposition 2.6, we have divwn = 0, ∀n ∈ Z+ in sense of weak deriva-tives.Let β ∈ C∞0 (RN). Consider
|(wn − w,∇β)L2(RN )| ≤ ‖wn − w‖0 ‖∇β‖0.
As n→∞, ‖wn − w‖0 → 0 implies (wn − w, ∇β)L2(RN ) → 0.
Integrating by parts, we get
(wn − w, ∇β)L2(RN ) = −(∇wn −∇w, β)L2(RN ) = (∇w, β)L2(RN ) = 0,
∀β ∈ C∞0 (RN).
This shows that divw = 0 in L2(RN).(d) Show that w ⊥ α in L2((RN)). Let n,m ∈ Z+. Applying integration by parts
again, we have
(wn,∇qm)L2 = −(∇wn, qm)L2 = 0 since divwn = 0.
Then (wn,∇qm)L2 → (w, α)L2 = 0 as n,m→∞.(e) Need to point out that
(w, α)L2 = 0, ∀w such that divw = 0,
is a sufficient condition for there to exist q ∈ L2 such that ∇q = α [9, Lemma2.2.1].
Thus, this proposition has proved for every vector field v ∈ L2(RN) ∩ C∞(RN).
Next theorem generalizes Proposition 2.6 and surveys some useful properties of the Leray’sprojection operator defined in Chapter 1.
Theorem 2.6 (The Hodge Decomposition in Hm). Every vector field v ∈ Hm(RN),m ∈ Z+ ∪ 0, has the unique orthogonal decomposition
v = w +∇ϕ
wherew is divergence-free vector field inRN , and ϕ is a scalar inRN , such that the Leray’sprojection operator P v = w satisfies
(i) P v, ∇ϕ ∈ Hm,∫RN P v · ∇ϕdx = 0, divP v = 0, and
‖P v‖2m + ‖∇ϕ‖2
m = ‖v‖2m, (2.53)
86
(ii) P commutes with weak derivatives,
P Dαv = DαP v, ∀v ∈ Hm, |α| ≤ m, (2.54)
(iii) P commutes with mollifiers Jε,
P (Jεv) = Jε(P v), ∀v ∈ Hm, ε > 0, (2.55)
(iv) P is symmetric,(P u, v)m = (u,P v)m. (2.56)
Proof.(i) (a) By the part (iii) of the Proposition 2.6, Dβw and Dβ(∇ϕ) are orthogonal in
L2(RN) ∩ C∞(RN) for any multi-index β, |β| ≤ m.
(w,∇ϕ)Hm(RN ) :=
∫RN
∑0≤|β|≤m
Dβw ·Dβ(∇ϕ) dx =
∑0≤|β|≤m
(Dβw,Dβ(∇ϕ)
)L2RN = 0.
So,w and∇ϕ are orthogonal inHm(RN)∩C∞(RN). Hence this decompositionof v is unique.Also, by the part (i) of the Proposition 2.6, P v, ∇ϕ ∈ Hm(RN) ∩ C∞(RN)since v ∈ Hm(RN) ∩ C∞(RN).To get the equation (2.53), we apply equation (2.51) from Proposition 2.6.
‖v‖2m :=
∑0≤|β|≤m
‖Dβv‖20 =
∑0≤|β|≤m
(‖Dβw‖2
0 + ‖Dβ(∇ϕ)‖20
)=
∑0≤|β|≤m
‖Dβw‖20 +
∑0≤|β|≤m
‖Dβ(∇ϕ)‖20 := ‖w‖2
m + ‖∇ϕ‖2m.
So,‖v‖2
m = ‖P v‖2m + ‖∇ϕ‖2
m (2.53)
(b) (1) Extend the previous to all v ∈ Hm(RN) using the argument thatHm(RN) ∩ C∞(RN) is dense in Hm(RN) [3, p.251].
So, for given v ∈ Hm(RN), ∃ vn ⊂ Hm(RN) ∩ C∞(RN) such that
‖vn − v‖m → 0 as n→∞.By Proposition 2.6, ∀n, ∃ wn, qn such that wn,∇qn ∈ L2(RN) ∩C∞(RN), wn ⊥ ∇qn in L2(RN),
vn = wn +∇qn ‖vn‖2m = ‖wn‖2
m + ‖∇qn‖2m.
87
Since ‖vn‖m → ‖v‖m as n→∞,we have
limn→∞
(‖wn‖2
m + ‖∇qn‖2m
)= ‖v‖2
m.
(2) Similarly as we did in Proposition 2.6, we can show that ∃w, α∈Hm(RN),wn → w, and ∇qn → α in Hm(RN); w ⊥ α in L2(RN); and that w isdivergence-free, i.e., divw = 0 in sense of weak derivatives.Also, since α ∈ Hm(RN) ⇒ α ∈ L2(RN), it follows from previous argu-ment in Proposition 2.6 that ∃ ϕ ∈ L2(RN) such that∇ϕ = α.
(3) Need to clarify that P v = w.∀n ∈ Z+, P vn = wn, and wn → w in Hm(RN) as n→∞.Consider
‖wn − P v‖m = ‖P vn − P v‖m = ‖P (v − vn)‖m ≤ 1 · ‖vn − v‖m → 0
as n→∞.
So,w = lim
n→∞wn = P v.
(4) Thus,v = P v +∇ϕ where P v,∇ϕ ∈ Hm(RN),
divP v = 0, and (P v, ∇ϕ)L2(RN ) =
∫RNP v · ∇ϕdx = 0.
The orthogonality of P v and∇ϕ implies uniqueness of the decompositionand that
‖v‖2m = ‖P v +∇ϕ‖2
m = ‖P v‖2m + ‖∇ϕ‖2
m. (2.53)
(ii) Note that div(Dαw) = Dα(divw) = 0, |α| ≤ m, since divw = 0.Then P (Dαw) = Dαw.By Proposition 2.6, part (iii),
Dαv = Dαw +Dα∇ϕ and Dαw ⊥ Dα(∇ϕ) in L2(RN).
HenceP(Dα∇ϕ) = 0 and
P(Dαv) = P(Dαw) + P(Dα∇ϕ) = Dαw + 0 = Dα(P w). (2.54)
(iii) w ∈ Hm(RN) and divw = 0. Since by the Proposition 2.6, part (ii), mollifierscommute with weak derivatives, then
div(Jεw) = Jε(divw) = 0.
88
So, P(Jεw) = Jεw.Show that P(Jε∇ϕ) = 0.Let u ∈ Hm(RN) and divu = 0 that implies div(Jεu) = 0.Applying the symmetry property of mollifiers (Th.2.5(iii)) and then integrating byparts, we have
(u, Jε∇ϕ)L2(RN ) = (Jεu,∇ϕ)L2(RN ) = −(∇(Jεu), ϕ
)L2(RN )
= 0.
So, Jε(∇ϕ) ⊥ u, ∀u from the divergence-free space in Hm(RN).Hence P(Jε∇ϕ) = 0.Now we get
P(Jεv) = P [Jε(w +∇ϕ)] = P(Jεw + Jε∇ϕ) =
P(Jεw) + P(Jε∇ϕ) = Jεw + 0 = Jε(P v). (2.55)
Thus, the Leray’s projection operator commutes with mollifiers.
(iv) Let u, v ∈ Hm(RN), u = Pu+∇q, v = Pv +∇ϕ, ∇Pu = 0, ∇Pv = 0.Using results from the Proposition 2.6,part (iii), we have for any multi-index β,|β| ≤ m,
Dβu = Dβ(Pu) +Dβ(∇q), Dβv = Dβ(Pv) +Dβ(∇ϕ),
∇Dβ(Pu) = 0, ∇Dβ(Pv) = 0.
Applying an integration by parts, we get(Dβ(Pu), Dβ(∇ϕ)
)L2(RN )
=(Dβ(Pu),∇(Dβϕ)
)L2(RN )
=
−(∇Dβ(Pu), Dβϕ
)L2(RN )
= 0.
Hence
(Pu,∇ϕ)m :=∑
0≤|β|≤m
(Dβ(Pu), Dβ(∇ϕ))L2(RN ) = 0.
Similarly, we can obtain
(Pv,∇q)m = (∇q,Pv)m = 0.
Now we have
(Pu, v)m = (Pu,Pv +∇ϕ)m = (Pu,Pv)m + (Pu,∇ϕ)m = (Pu,Pv)m + 0 =
(Pu,Pv)m + (∇q,Pv)m = (Pu+∇q,Pv)m = (u,Pv)m.
89
So, we get(Pu, v)m = (u,Pv)m. (2.56)
This theorem concludes our review of the technical background necessary to work on theinitial value problems for the Euler and the Navier-Stokes equations.
90
Chapter 3
Energy Methods for the Euler and the Navier-Stokes Equations.
In this chapter, we prove existence and uniqueness of a smooth solution of the initial valueproblems for the Euler and the Navier-Stokes equations on some time interval [0, T ) apply-ing classical energy methods. These methods use an approximation scheme and various apriori estimates on Sobolev norms of velocity v allowing to replace the unbounded opera-tors∇ and ∆ by some operators bounded in a Banach space.
3.1 Energy Methods: Elementary Concepts.
The kinetic energy of the fluid, E = 12
∫RN |v|
2 dx, is a half of the square of L2 norm of thevelocity.In Section 1.7, we got that
dE(t)
dt= −ν
∫RN|∇v|2 dx < 0. (3.1)
So, the kinetic energy dissipates for viscous flows (case ν > 0). In the case ν = 0, forinviscid flows, kinetic energy is conserved.Thus, a priori, we have boundness of the velocity in L2 norm.
3.1.1 A Basic Energy Estimate.
Here we will derive a basic energy estimate for a smooth solution of the Euler or the Navier-Stokes equations.
Proposition 3.1 (Basic Energy Estimate). Let v1 and v2 be two smooth solutions to theNavier-Stokes equations with external forces F1 and F2 correspondingly and the same vis-cosity ν ≥ 0. Suppose that these solutions exist on a common time interval [0, T ] and, forfixed time, decay fast enough at infinity to belong to L2(RN). Then
sup0≤t≤T
‖v1 − v2‖0 ≤[‖ (v1 − v2)|t=0 ‖0 +
∫RN‖F1 − F2‖0 dt
]exp(∫
RN‖∇v2‖L∞ dt
). (3.7)
Proof. Since v1 and v2 are solutions to the Navier- Stokes equations with respective exter-nal forces F1 and F2, then we have
DviDt
= −∇pi + ν∆vi + Fi,
div vi = 0,
vi|t=0 = v0i(i = 1, 2).
(3.2)
We assume that these solutions exist on a common time interval [0, T ] and that they vanish
91
sufficiently rapidly as |x| → ∞, so that vi ∈ L2(RN).Take a difference of corresponding equations (3.2) and add and subtract v1 · ∇v2 to andfrom respectively the left-hand sides of the first equations.
v1t + v1 · ∇v1 = −∇p1 + ν∆v1 + F1,−v2t + v2 · ∇v2 = −∇p2 + ν∆v2 + F2,
(v1 − v2)t + v1 · ∇v1 − v1 · ∇v2 + v1 · ∇v2 − v2 · ∇v2 = −∇(p1 − p2) + ν∆(v1 − v2)
+ (F1 − F2).
So, we get
vt + v1 · ∇v + v · ∇v2 = −∇p+ ν ∆v + F , div v = 0, v|t=0 = v0 (3.2a)
where v = v1 − v2, p = p1 − p2, F = F1 − F2, and v0 = v01 − v02 .
Take the L2 inner product ( , ) of both sides of the first equation (3.2a) with v.
(vt, v) + (v1 · ∇v, v) + (v · ∇v2, v) = −(∇p, v) + ν(∆v, v) + (F , v). (3.3)
Next, integrate by parts and use the divergence theorem and the fact that vi, i = 1, 2, andso v, are divergence free and vanish sufficiently rapidly at infinity.Note that a multiplication by the unknown quantity (or a function of it) and an integrationby parts are common for all energy methods.Show that (v1 · ∇v, v) = −(∇p, v) = 0.
1) (v1 · ∇v, v) =
∫RN
(v1 · ∇v) · v dx =
∫RN
N∑i=1
( N∑j=1
v1j∂
∂xjvi
)vi dx
=
∫RN
N∑j=1
v1j
N∑i=1
∂vi∂xjvi dx =
∫RN
N∑j=1
v1j
N∑i=1
12
∂
∂xj(v2i )dx =
∫RN
N∑j=1
v1j∂
∂xj(1
2v2)dx
=
∫RNv1 · ∇(1
2v2) dx =
∫RN
∇ ·[(1
2v2)v1
]− 1
2v2(∇ · v1)
dx
=
∫RN
div[(1
2v2)v1
]− (1
2v2)divv1
dx =
∫RN
div[(1
2v2)v1
]− 0dx
= limR→∞
∫|x|=R
(12v2)v1 · n dS = 0
where n is the outward normal unit vector of the sphere |x| = R.
92
2) − (∇p, v) = −∫RN∇p · v dx = −
N∑i=1
∫RNpxi vi dx
= −N∑i=1
(−∫RNp(vi)xi dx+ lim
R→∞
∫|x|=R
p vi ni dS)
(after integration by parts)
=
∫RNp
N∑i=1
(vi)xi dx+ 0 =
∫RNp∇ · v dx =
∫RNp div v dx = 0.
Also
− (∆v, v) = −N∑i=1
∫RN
∆vi vi dx = −N∑i=1
∫RN
(∇ · ∇vi)vi dx
=N∑i=1
∫RN
(∇vi · ∇vi) dx−N∑i=1
limR→∞
∫|x|=R
(∇vivi) · ni dS (after integration by parts)
=N∑i=1
(∇vi,∇vi)−0 = (∇v,∇v) =
∫RN∇v·∇v dx =
∫RN|∇v|2 dx = ‖∇v(, t)‖2
0 ≥ 0.
Then from equation (3.3), we get the following basic energy identity:
(vt, v) + ν‖∇v‖20 = −(v · v2, v) + (F , v). (3.4)
Note that
(vt, v) =
∫RNvt · v dx =
∫RN
N∑i=1
(vi)tvi dx =N∑i=1
∫RN
(12v2i )t dx =
12ddt
∫RN
N∑i=1
v2i dx = 1
2ddt
∫RN|v|2 dx = 1
2ddt‖v(, t)‖2
0 = 12 2 ‖v(, t)‖0
ddt
(‖v(, t)‖0
).
Using the Cauchy-Schwartz and Schwartz inequalities, find an upper estimate for the right-hand side of the equation (3.4).
−(v · ∇v2, v) = −∫RN
(v · ∇v2) · v dx ≤∣∣∣∫RN
(v · ∇v2) · v dx∣∣∣ ≤∫
RN|(v · ∇v2) · v| dx ≤∫
RN|v · ∇v2| |v| dx ≤
∫RN|v| |∇v2| |v| dx ≤ ‖∇v2‖L∞
∫RN|v|2 dx = ‖∇v2‖L∞ ‖v(, t)‖2
0;
(F , v) =
∫RNF · v dx ≤
∣∣∣∫RNF · v dx
∣∣∣ ≤ ‖F (, t)‖0 ‖v(, t)‖0.
93
Then from equation (3.4), we get the following inequality:
‖v(, t)‖0ddt
(‖v(, t)‖0
)+ ν‖∇v(, t)‖2
0 ≤‖∇v2‖L∞ ‖v(, t)‖2
0 + ‖F (, t)‖0 ‖v(, t)‖0. (3.5)
Drop ν‖∇v(, t)‖20 in the left-hand side of the last inequality and divide this inequality by
‖v(, t)‖0. Then we have
ddt
(‖v(, t)‖0
)≤ ‖∇v2‖L∞ ‖v(, t)‖0 + ‖F (, t)‖0. (3.6)
Recall Gronwall’s lemma:If η(t), ϕ(t), ψ(t) ≥ 0, η(t) is absolutely continuous and ϕ(t), ψ(t) are integrable on [0, T ]for some T > 0, and η′(t) ≤ ϕ(t)η(t) + ψ(t), then
∀t ∈ [0, T ], η(t) ≤ e∫ T0 ϕ(s) ds
[η(0) +
∫ T
0
ψ(s) ds].
Suppose that
η(t) = ‖v(, t)‖0, ϕ(t) = ‖∇v2‖L∞ , and ψ(t) = ‖F (, t)‖0.
Hence from the inequality (3.6), applying the Gronwall’s lemma, we get
‖v(, t)‖0 ≤ e∫ T0 ‖∇v2‖L∞ dt
(‖v(, t)‖0 +
∫ T
0
‖F (, t)‖0 dt).
Thus, we derived the Basic Energy estimate.
sup0≤t≤T
‖v1 − v2‖0 ≤[‖ (v1 − v2)|t=0 ‖0 +
∫RN‖F1 − F2‖0 dt
]exp(∫ T
0
‖∇v2‖L∞ dt). (3.7)
Note that the Basic Energy estimate (3.7) does not depend explicitly on the viscosity. So,the same estimate is applicable for solutions to both The Euler and the Navier-Stokes equa-tions.
The Basic Energy estimate (3.7) immediately implies uniqueness of smooth solutions tothe Euler and to the Navier-Stokes equations.
Corollary 3.1 (Uniqueness of Solutions). Let v1 and v2 be two smooth L2 solutions to theNavier-Stokes equations on [0, T ] with the same initial data and forcing F . Then v1 = v2.
Proof. Indeed, with v1|t=0 = v2|t=0 and F1 = F2, the Basic Energy estimate (3.7) gives
94
that v1 = v2, ∀t ∈ [0, T ].
Another consequence of the Basic Energy estimate is the following useful estimate on thegradient of v = v1 − v2 that we will apply in the next subsection.
Corollary 3.2. Under the assumptions of Proposition 3.1,
ν
∫ T
0
‖∇v(, t)‖20 dt ≤ C(v2, T )
[‖v(, 0)‖2
0 +
(∫ T
0
‖F (, t)‖0 dt
)2](3.9)
where the constant C(v2, T ) depends on∫ T
0‖∇v2‖L∞ dt.
Proof. Integrate the inequality (3.5) in time on [0, T ] taking into account that∫ T
0
‖v(, t)‖0ddt
(‖v(, t)‖0
)dt = 1
2‖‖2
0
∣∣T0.
We get
12‖v(, T )‖2
0 − 12‖v(, 0)‖2
0 + ν
∫ T
0
‖∇v(, t)‖20 dt ≤
sup0≤t≤T
‖v(, t)‖20
∫ T
0
‖∇v2‖L∞ dt+ sup0≤t≤T
‖v(, t)‖0
∫ T
0
‖F (, t)‖0 dt. (3.8)
Drop 12‖v(, T )‖2
0 ≥ 0 at the left-hand side of the last inequality and use the estimate (3.7)for sup0≤t≤T ‖v(, t)‖0 at the right-hand side of this inequality.Denote a =
∫ T0‖∇v2‖L∞ dt. Then we have
ν
∫ T
0
‖∇v(, t)‖20 dt ≤ 1
2‖v(, 0)‖2
0+
ae2a[‖v(, 0)‖2
0 + 2‖v(, 0)‖0
∫ T
0
‖F (, t)‖0 dt+(∫ T
0
‖F (, t)‖0 dt)2]
+∫ T
0
‖F (, t)‖0 dt ea[‖v(, 0)‖0 +
∫ T
0
‖F (, t)‖0 dt]
or
ν
∫ T
0
‖∇v(, t)‖20 dt ≤ ‖v(, 0)‖2
0(12
+ ae2a)+
‖v(, 0)‖0
∫ T
0
‖F (, t)‖0 dt ea(2aea + 1) +
(∫ T
0
‖F (, t)‖0 dt)2
ea(aea + 1).
Note that a =∫ T
0‖∇v2‖L∞ dt ≥ 0. So, ea ≥ 1 > 1
2.
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Also, by the elementary inequality
‖v(, 0)‖0
∫ T
0
‖F (, t)‖0 dt ≤ 12
[‖v(, 0)‖2
0 +(∫ T
0
‖F (, t)‖0 dt)2]
.
Applying that to the last estimate, we finally get
ν
∫ T
0
‖∇v(, t)‖20 dt ≤ C(v2, T )
[‖v(, 0)‖2
0 +
(∫ T
0
‖F (, t)‖0 dt
)2](3.9)
where C(v2, T ) = ea(2aea + 32) with a =
∫ T0‖∇v2‖L∞ dt.
3.1.2 Approximation of Inviscid Flows by High Reynold’s Number Viscous Flows.
We continue to assume here that both the smooth inviscid solution v0 to the Euler equationand the smooth viscous solution vν to the Navier-Stokes equations exist on a commontime interval. The Basic Energy estimate allows to show that for the same initial data, thesolution vν converges to the solution v0 as the viscosity ν → 0.
Proposition 3.2 (Estimate of Closeness of Smooth Solutions to the Euler and to the Navier-Stokes Equations). Given fixed initial data, let vν , ν ≥ 0, be a smooth solution to the Euler(for ν = 0) or the Navier-Stokes (for ν > 0) equations. Suppose that for 0 ≤ ν ≤ ν0 thesesolutions exist on a common time interval [0, T ] and vanish sufficiently rapidly as |x| → ∞.Then
sup0≤t≤T
‖vν − v0‖0 ≤ C1(v0, T ) νT (3.10)
and ∫ T
0
‖∇(vν − v0)‖0 dt ≤ C2(v0, T ) ν1/2T 3/2 (3.11)
where C1(v0, T ) = sup0≤t≤T
‖∆v0(, t)‖0 e∫ T0 ‖∇v
0‖L∞ dt,
C2(v0, T ) = C1/2(v0, T ) sup0≤t≤T
‖∆v0(, t)‖0 with C(v0, T ) the same as in (3.9).
Proof. Suppose that in the Navier-Stokes equation
Dvν
Dt= −∇p+ ν∆vν + F,
the external force F = −ν∆vν . Then we get
Dvν
Dt= −∇p,
96
the Euler equation. So, vν = v0 in this case.Consider now two smooth solutions to the Navier-Stokes equation: v1 = vν for externalforce F1 = 0 and v2 = v0 for F2 = −ν∆v0. Note that (vν − v0)|t=0 = 0.Applying to these solutions the Basic Energy estimate (3.7),we get
sup0≤t≤T
‖vν − v0‖0 ≤∫ T
0
‖ν∆v0‖0 dt e∫ T0 ‖∇v
0‖L∞ dt ≤ ν sup0≤t≤T
‖∆v0‖0 Te∫ T0 ‖∇v
0‖L∞ dt
or
sup0≤t≤T
‖vν − v0‖0 ≤ C1(v0, T ) νT (3.10)
where C1(v0, T ) = sup0≤t≤T
‖∆v0(, t)‖0 e∫ T0 ‖∇v
0‖L∞ dt.
From the estimate (3.9), we obtain
ν
∫ T
0
‖∇(vν − v0)‖20 dt ≤ C(v0, T )
(∫ T
0
‖ν∆v0(, t)‖0 dt
)2
.
To derive the second estimate of this proposition, apply the Schwartz inequality to theleft-hand side of the previous inequality.∫ T
0
‖∇(vν − v0)‖0 dt =
∫ T
0
1 ‖∇(vν − v0)‖0 dt ≤
(∫ T
0
12 dt
)1/2(∫ T
0
‖∇(vν − v0)‖20 dt
)1/2
=
(Tν
)1/2(ν
∫ T
0
‖∇(vν − v0)‖20 dt
)1/2
.
Now upper bound the last term using the previous estimate. We get
∫ T
0
∇(vν − v0)‖0 dt ≤(TνC(v0, T )
)1/2[(∫ T
0
‖ν∆v0(, t)‖0 dt
)2]1/2
≤
(TνC(v0, T )
)1/2
ν sup0≤t≤T
‖∆v0(, t)‖0 T
or ∫ T
0
‖∇(vν − v0)‖0 dt ≤ C2(v0, T ) ν1/2T 3/2 (3.11)
where C2(v0, T ) =√C(v0, T ) sup
0≤t≤T‖∆v0(, t)‖0.
97
The estimates (3.10) and (3.11) give the following corollary.
Corollary 3.3. For the assumptions of Proposition 3.2, vν → v0 as ν → 0 in theL∞
([0, T ]
);L2(RN) norm with rate first order of ν, and the gradients∇vν → ∇v0 with
rate square root of ν.
Thus, for small viscosity (high Reynold’s number), solutions to the Euler equation v0 andto the Navier-Stokes equation vν are close, and the viscous solution vν can be a goodapproximation for the inviscid solution v0.
Recall that we assumed that both vν and v0 are smooth solutions on a common time interval.It is not necessarily true for solutions that are not smooth. It is worth mention, that forsmall viscosity, the time interval of existence of solution to the Navier-Stokes equation vν
includes the time interval of existence of solution to the Euler equation[2].
3.2 Local-in-Time Existence of Solutions by Means of Energy Methods.
Here we will prove the local-in-time existence of a solution to the Navier-Stokes equationin 3-D case without forcing. Note that the results of this section can be extended to the caseof sufficiently smooth L2 forcing.
We will use independent of viscosity ν estimates. So, the proof holds for both the Navier-Stokes and Euler equations.
At first, we will find an approximate equation whose solution satisfies an energy estimateand exists for all time. Then we obtain the solution to the Navier-Stokes equation as a limitin the approximation scheme.
3.2.1 Global Existence of Solution to a Regularization of the Euler and the Navier-Stokes Equations.
In this subsection, we will use the Picard theorem for ODEs on a Banach space.
Theorem 3.1 (Picard Theorem on a Banach Space). Let O ⊆ B be an open subset of aBanach space B and let F : O → B be a locally Lipschitz continuous mapping, i.e., forany X ∈ O there exists L > 0 and an open neighborhood UX ⊂ O of X such that
‖F (X)− F (X)‖B ≤ L‖X − X‖B for all X, X ∈ UX .
Then for any X0 ∈ O, there exists a time T such that the ODE
dX
dt= F (X), X|t=0 = X0
has a unique (local) solution X ∈ C1[(−T, T );O].
The Navier-Stokes equations contain operators ∇ and ∆ unbounded in Banach space, thatdoes not permit to apply directly the existence theory for ODEs. Therefore we will regu-
98
larize the Navier-Stokes equations in such way that regularized equations will satisfy theconditions of the Picard theorem.
To construct the approximate equation, we will use a mollification that allows us to convertunbounded differential operators into bounded operators.
We regularize the Navier-Stokes equations with initial data
vt + v · ∇v = −∇p+ ν∆v,
div v = 0,
v|t=0 = v0,
(3.12)
replacing them with the following approximate equations with unknowns vε and pε forsome parameter ε > 0:
vεt + Jε[(Jεv
ε) · ∇(Jεvε)]
= −∇pε + νJε(Jε∆vε),
div vε = 0,
vε|t=0 = v0.
(3.13)
where Jε is defined in the equations (2.41)-(2.42).
To exclude the pressure pε and the incompressibility condition div vε = 0, apply to theequations (3.13) Leray’s operator P projecting onto the space of divergence-free functions:
V s = v ∈ Hs(RN) : div v = 0. (3.14)
If vε ∈ V s, then P vε = vε and P(vεt) =(Pvε
)t
= vεt . Also, P(∇pε) = 0.Since P commutes with derivatives and mollifiers, then
P[νJε(Jε∆v
ε)]
= νJ2ε
[∆(P vε)
]= νJ2
ε (∆vε).
Hence we have
vεt + PJε[(Jεv
ε) · ∇(Jεvε)]
= νJ2ε (∆vε) (3.15)
ordvε
dt= Fε(v
ε),
vε|t=0 = v0(3.16)
where Fε(vε) = F (1)
ε (vε)− F (2)ε (vε), F (1)
ε (vε) = νJ2ε (∆vε),
F (2)(vε) = PJε[(Jεv
ε) · ∇(Jεvε)].
(3.17)
We reduced the regularized Navier-Stokes and Euler equations (3.13) to the ODE (3.17) inthe Banach space V s. Here vε ∈ C1
([0, Tε);V
s), i.e., vε(t) is in the space of C1 functions
on the interval [0, Tε) for some Tε > 0 with values in V s-Banach space equiped with theSobolev norm ‖ ‖s.
99
At first, prove that C1([0, Tε);V
s)
is a Banach space.
Lemma 3.1. Let B be a Banach space and C1([0, T );B
):= f, df
dt: [0, T ) → B | f ∈
C1([0, T )
) for some T > 0. Then C1
([0, T );B
)is a Banach space with the norm
‖f‖C1 := sup0≤t<T
‖f(t)‖B + sup0≤t<T
‖df(t)dt‖B.
Proof.
1) Show that ‖ ‖C1 is a norm.
(a) ∀t ∈ [0, T ), ‖f(t)‖B ≥ 0 and ‖df(t)dt‖B ≥ 0 as ‖‖B is a norm. Then ‖f‖C1 ≥ 0.
(b) If f(t) ≡ 0 on [0, T ), then df(t)dt≡ 0 on [0, T ). Hence ‖f(t)‖B = ‖df(t)
dt‖B = 0
∀t ∈ [0, T ) since ‖ ‖B is a norm. So, ‖f‖C1 = 0.
Let ‖f‖C1 = 0. Then ‖f‖C1 ≥ ‖f(t)‖B ≥ 0 and ‖f‖C1 ≥ ‖df(t)dt‖B ≥ 0,∀t ∈
[0, T ) imply ‖f(t)‖B = ‖df(t)dt‖B ≡ 0, ∀t ∈ [0, T ).
So, f(t) ≡ 0 ∀t ∈ [0, T ) since ‖ ‖B is a norm.Thus, ‖f‖C1 = 0 iff f(t) ≡ 0 on [0, T ).
(c) ∀α ∈ R, ‖αf‖C1 = sup0≤t<T
‖αf(t)‖B + sup0≤t<T
‖αdf(t)dt‖B =
sup0≤t<T
(|α| ‖f(t)‖B
)+ sup
0≤t<T
(|α| ‖df(t)
dt‖B)
=
|α|(
sup0≤t<T
‖f(t‖B + sup0≤t<T
‖df(t)dt‖B))
= |α| ‖f‖C1 .
(d) By the Triangle inequality for norm ‖ ‖B, we have
∀f(t), g(t) ∈ B, ‖f(t) + g(t)‖B ≤ ‖f(t)‖B + ‖g(t)‖B, ∀t ∈ [0, T ).
Then
sup0≤t<T
‖f(t) + g(t)‖B ≤ sup0≤t<T
‖f(t)‖B + sup0≤t<T
‖g(t)‖B.
Similarly, we can get the following inequality for derivatives:
sup0≤t<T
‖ddt
[f(t) + g(t)
]‖B ≤ sup
0≤t<T‖df(t)dt‖B + sup
0≤t<T‖dg(t)dt‖B.
Adding the last two inequalities, we obtain the Triangle inequality for ‖ ‖C1:
‖f + g‖C1 ≤ ‖f‖C1 + ‖g‖C1 .
Thus, ‖ ‖C1 is a norm on C1([0, T ); B
).
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2) Prove that C1([0, T ); B
)is complete in ‖ ‖C1 norm.
(a) Assume un∞n=1 ⊂ C1([0, T ); B
)is a Cauchy sequence. So, given ε > 0,
∃N ∈ N such that ∀m > n > N, ‖un − um‖C1 < ε, that implies
‖un(t)− um(t)‖B ≤ ‖un − um‖C1 < ε, ∀t ∈ [0, T ).
Then un(t)∞n=1 is a Cauchy in B, ∀t ∈ [0, T ).Since B is complete, then ∃ u(t) ∈ B, such that ∀t ∈ [0, T ),‖un(t)− u(t)‖B → 0 as n→∞.
(b) Show that un(t) converges to u(t) uniformly on [0, T ).Since un(t) is a Cauchy sequence in ‖ ‖C1 norm, so given ε > 0,
‖un − um‖C1 < ε3∀m > n > N .
Now consider
‖um(t)−un(t)‖B ≤ sup0≤t<T
‖um(t)−un(t)‖B ≤ ‖um−un‖C1 < ε3, ∀m,n > N,
so un(t) is Cauchy in ‖ ‖B norm for each t ∈ [0, T ). Since B is complete,∃ u(t) ∈ B such that ‖un(t)− u(t)|B → 0, as n→∞, t ∈ [0, T ).Now for any t0 ∈ [0, T ),
‖um(t0)− un(t0)‖B ≤ sup0≤t<T
‖um(t)− un(t)‖B < ε3.
Let n→∞, so‖um(t0)− u(t0)‖B ≤ ε
3.
Since this is true for arbitrary t0 ∈ [0, T ) for the some fixed ε, so
sup0≤t<T
‖um(t)− u(t)‖B ≤ ε3. (∗)
Therefore, un(t)→ u(t) uniformly on [0, T ).Similarly, it could be shown that Cauchy sequence dun(t)
dt∞n=1 converges uni-
formly on [0, T ).Hence u(t) is differentiable and limn→∞
dun(t)dt
= du(t)dt
on [0, T ).[8]
(c) Show that u(t) ∈ C1([0, T ); B
).
Given any s, t ∈ [0, T ),
‖u(t)− u(s)‖B ≤ ‖u(t)− un(t)‖B + ‖un(t)− un(s)‖B + ‖un(s)− u(s)‖B.
Applying the inequality (∗) to the first and to the third terms at the right-handside of the last inequality, we have
‖u(t)− u(s)‖B ≤ 2ε3
+ ‖un(t)− un(s)‖B.
un(t) is continuous on [0, T ). So, for given ε > 0, ∃ δ > 0, such that∀s, t ∈ [0, T ) with |s− t| < δ, we have ‖un(t)− un(s)‖B < ε
3.
101
Combine this inequality with the previous one to get
∀ε > 0, ‖u(t)− u(s)‖B < ε ∀s, t ∈ [0, T ) such that |s− t| < δ.
So, u : [0, T )→ B is continuous.Similarly, we can show that du(t)
dtis continuous on [0, T ).
Then u ∈ C1([0, T ); B
), and so C1
([0, T ); B
)is complete.
Thus, C1([0, T ); B
)is a Banach space with norm ‖ ‖C1 .
Next, we prove the following proposition.
Proposition 3.3. Consider an initial condition v0 ∈ V m, m ∈ Z+ ∪ 0. Then
i) for any ε > 0 there exists the unique solution vε ∈ C1([0, Tε); V
m)
to the ODE(3.16), where Tε = T (‖v0‖m, ε);
ii) on any time interval [0, T ] on which this solution belongs to C1([0, T ];V 0
),
sup0≤t≤T
‖vε‖0 ≤ ‖v0‖0. (3.18)
Proof.
i) Our goal is to show that the right-hand side of the ODE (3.16) Fε(vε) = F(1)ε (vε) −
F(2)ε (vε) satisfies the conditions of the Picard theorem (Theorem 3.1).
(a) At first, clarify that Fε maps V m to V m.Let vε ∈ V m, so div vε = 0. Since the mollifier Jε commutes with derivatives,then
divF (1)ε (vε) = div(νJ2
ε ∆vε) = νJ2ε ∆(div vε) = 0.
Also, P is a projection into divergence-free vector fields, so
F (2)ε (vε) = PJε
[(Jεv
ε) · (∇Jεvε)]
impliesdivF (2)
ε (vε) = 0.
ThendivFε(vε) = divF (1)
ε (vε)− divF (2)ε (vε) = 0.
Thus, Fε : V m → V m.
102
(b) Prove that F (1)ε (vε) is Lipschitz continuous.
Let v1, v2 ∈ C1([0, Tε]; V
m). Consider
‖F (1)ε (v1)− F (1)
ε (v2)‖m = ν ‖J2ε ∆(v1 − v2)‖m = ν ‖∆J2
ε (v1 − v2)‖m
since the mollifier Jε commutes with derivatives.Then by the definition of Sobolev norm, we have
‖F (1)ε (v1)− F (1)
ε (v2)‖m = ν[∫RN
(1 + |ξ|2)m∣∣[∆J2
ε (v1 − v2)]∣∣2 dξ]1/2
=
ν[∫RN
(1 + |ξ|2)m(2π)4|ξ|4∣∣[J2
ε (v1 − v2)]∣∣2 dξ]1/2
≤
16π4ν[∫RN
(1 + |ξ|2)m+2∣∣[J2
ε (v1 − v2)]∣∣2 dξ]1/2
=
16π4ν‖J2ε (v1 − v2)‖m+2 ≤ 16π4ν Cm2
ε2‖J2
ε (v1 − v2)‖m ≤
16π4ν Cm2
ε2Cm0
ε0‖v1 − v2‖m = C3ν
ε2‖v1 − v2‖m with C3 = 16π4νCm2Cm0.
Here we consequently used the property of the Fourier transform
∂βf(ξ) = (2πiξ)β f(ξ) with multi-index β, (2.11)
the elementary fact that |ξ|4 ≤ (1 + |ξ|2)2, and twice the estimate for mollifiers
‖Jεv‖m+k ≤ Cmkεk‖v‖m ∀v ∈ Hm, ∀ε > 0, ∀k ∈ Z+ ∪ 0,
for k = 2 and k = 0.So, F (1)
ε (vε) is Lipschitz continuous.
(c) Now show that F (2)ε (vε) is Lipschitz continuous too.
Let again v1, v2 ∈ C1([0, Tε); V
m). Consider
‖F (2)ε (v1)− F (2)
ε (v2)‖m =
‖F (2)ε (v1)− PJε
[(Jεv1) · ∇(Jεv2)
]+ PJε
[(Jεv1) · ∇(Jεv2)
]− F (2)
ε (v2)‖m=∥∥PJε[(Jεv1) · ∇(Jεv1)
]− PJε
[(Jεv1) · ∇(Jεv2)
]+
PJε[(Jεv1) · ∇(Jεv2)
]− PJε
[(Jεv2) · ∇(Jεv2)
]∥∥m≤
‖PJε[(Jεv1) · ∇
(Jε(v1 − v2)
)]‖m + ‖PJε
[(Jε(v1 − v2)
)· ∇(Jεv2)
]‖m.
By the Hodge decomposition in Hm, we have
‖P v‖2m + ‖∇ϕ‖2
m = ‖v‖2m. (2.53)
So, omitting the Leray’s projection operator P , we will use that ‖P v‖m ≤‖v‖m in the next inequality.
103
We will also continue using the estimates for mollifiers (2.48) and (2.49), theCalculus inequality in the Sobolev space (2.38), and the property of mollifiers(2.44). Then we get
‖F (2)ε (v1)− F (2)
ε (v2)‖m ≤∥∥Jε[(Jεv1) · ∇(Jε(v1 − v2)
)]∥∥m
+∥∥Jε[(Jε(v1 − v2)
)· ∇(Jεv2)
]∥∥m
(2.48)≤
Cm0
∥∥(Jεv1) · ∇(Jε(v1 − v2)
)∥∥m
+∥∥Jε(v1 − v2) · ∇(Jεv2)
∥∥m
(2.38),(2.44)≤
CCm0 [ ‖Jεv1‖L∞ ‖DmJε∇(v1 − v2)‖0 + ‖DmJεv1‖0 ‖Jε∇(v1 − v2)‖L∞ +
‖Jε(v1 − v2)‖L∞‖DmJε∇v2‖0 + ‖DmJε(v1 − v2)‖0 ‖Jε∇v2‖L∞ ](2.48),(2.49)≤
CCm0
[C0
εN/2‖v1‖0 ‖
=Dm+1
Dm∇ Jε(v1 − v2)‖0 + ‖Jεv1‖m C1
ε(N/2)+1 ‖v1 − v2‖0 +
C0
εN/2‖v1 − v2‖0 ‖
=Dm+1
Dm∇ Jεv2‖0 + ‖Jε(v1 − v2)‖m C1
ε(N/2)+1 ‖v2‖0
]≤
CCm0
[C0
εN/2‖v1‖0 ‖Jε(v1 − v2)‖m+1 + ‖Jεv1‖m C1
ε(N/2)+1 ‖v1 − v2‖0 +
C0
εN/2‖v1 − v2‖0 ‖Jεv2‖m+1 + ‖Jε(v1 − v2)‖m C1
ε(N/2)+1 ‖v2‖0
] (2.48)≤
CCm0
[C0
εN/2‖v1‖0
Cm1
ε‖v1 − v2‖m + C0m
εm‖v1‖0
C1
ε(N/2)+1 ‖v1 − v2‖0 +
C0
εN/2‖v1 − v2‖0
C0(m+1)
εm+1 ‖v2‖0 + Cm0
ε0‖v1 − v2‖m C1
ε(N/2)+1 ‖v2‖0
]≤
CCm0‖v1 − v2‖mε(N/2)+1+m
(C0Cm1‖v1‖0ε
m + C1Cm0‖v1‖0 + C0C(m+1)0‖v2‖0 +
C1Cm0εm‖v2‖0) ≤ C2‖v1 − v2‖m
ε(N/2)+1+m
(‖v1‖0 + ‖v2‖0
)with C2 = CCm0 max(C0Cm1 + C1Cm0), (C0C(m+1)0 + C1Cm0) andε ∈ (0, 1).Here we used the facts that
‖Dm∇f‖0 ≤ ‖f‖m+1, ‖Dmf‖0 ≤ ‖f‖m, ‖v1 − v2‖0 ≤ ‖v1 − v2‖m, εm < 1.
Also, we twice applied the estimate (2.48) not for k = 0 and k = 1 as earlier,but for k = m and k = m+ 1, i.e.
‖Jεv1‖0+m ≤ Cm0
εm‖v1‖0
and
‖Jε(v1 − v2)‖m+1 ≤C(m+1)0
εm+1 ‖v1 − v2‖0.
So, both F (1)ε (vε) and F (2)
ε (vε) are Lipschitz continuous.
104
(d) Then for Fε(vε) = F(1)ε (vε)− F (2)
ε (vε), we get
‖Fε(v1)− Fε(V2)‖m ≤ ‖F (1)ε (v1)− F (1)
ε (v2)‖m + ‖F (2)ε (v1)− F (2)
ε (v2)‖m
≤ νC3
ε2‖v1 − v2‖m +
C2‖v1‖0 + ‖v2‖0
ε(N/2)+m+1‖v1 − v2‖m = C‖v1 − v2‖m (3.19)
where C = C(‖v1‖0, ‖v2‖0, ε, N,m).Let OM = v ∈ V m | ‖v‖m < M.So, Fε is locally Lipschitz continuous on any such open set OM .Hence the Picard theorem (Theorem 3.1) gives us that for any given initial con-dition v0 ∈ Hm, there exists the unique solution to the IVP (3.16)vε ∈ C1
([0, Tε); OM
), m ∈ Z+ ∪ 0, for some Tε > 0 (locally in time).
ii) Next, prove the energy bound (3.18) for any vε ∈ C1([0, Tε); OM
).
Take the L2 inner product of the equation (3.16) with vε.(vε, dv
ε
dt
)L2(RN )
=(vε, Fε(v
ε))L2(RN )
,
i.e., ∫RNvε · dvε
dtdx =
∫RNvε · Fε(vε) dx
or
12ddt
∫RN
(vε)2 dx = ν
∫RNvε · J2
ε (∆vε) dx−∫RNvε · PJε
[(Jεv
ε) · ∇(Jεvε)]dx. (3.20)
Now use the properties of mollifiers and the properties of the projection operator Pfrom Theorem 2.5(iii) and Theorem 2.6(iii) respectively and after that, integrate byparts both of the integrals on the right-hand side of the last equation. Then we have
ν
∫RNvε · J2
ε (∆vε) dxTh.2.5(iii)= ν
∫RN
(Jεvε) · Jε(∇2vε) dx
int.byparts=
− ν∫RN∇(Jεv
ε) · ∇(Jεvε) dx = −ν
∫RN
(∇Jεvε)2 dx = −ν∫RN
[Jε(∇vε)
]2dx
and
−∫RNvε · PJε
[(Jεv
ε) · ∇(Jεvε)]dx
Th.2.6(iii)= −
∫RN
=vε
Pvε · Jε[(Jεv
ε) · ∇(Jεvε)]dx
Th.2.5(iii)= −
∫RN
(Jεvε) ·[(Jεv
ε) · ∇(Jεvε)]dx = −
∫RN
(Jεvε) · 1
2∇(Jεv
ε)2 dxint.byparts=
12
∫RN∇ · (Jεvε) (Jεv
ε)2 dx = 12
∫RNJε(=divvε=0
∇ · vε)(Jεv
ε)2 dx = 0.
105
Combining these results with the equation (3.20), we obtain
12ddt
∫RN
(vε)2 dx = −ν∫RN
(∇Jεvε)2 dx
orddt‖vε‖2
0 = −2ν‖∇Jεvε‖20 ≤ 0 for ν ≥ 0.
Thus, ‖vε‖0 is not increasing in time, i.e.,
‖vε‖0 ≤ ‖v0‖0, ∀t ≥ 0
orsup
0≤t≤T‖vε‖0 ≤ ‖v0‖0. (3.18)
Note that the energy estimate (3.18) does not contain explicitly ε and ν. So, it could beapplied for any positive ε to regularize both the Navier-Stokes or the Euler equations.
We emphasize here the importance of the special choice of the regularization in the equation(3.17) that makes possible to obtain this valuable estimate.
We have proved the existence and the uniqueness of solution vε to the regularized Navier-Stokes and Euler equations locally in time. Now we want to show that this equation hasglobal-in-time solutions. For this purpose, using the fact that the functional F in the re-gularized equation (3.16) is time independent, we will apply the following continuationproperty of ODEs on a Banach space.
Theorem 3.2 (Continuation of an Autonomous ODE on a Banach Space). Let O ⊂ Bbe an open set of Banach space B, and let F : O → B be a locally Lipschitz continuousoperator. Then the unique solution X ∈ C1
([0, T ); O
)to the autonomous ODE,
dX
dt= F (X), X|t=0 = X0 ∈ O,
either exists globally in time, or T <∞ and X(t) leaves the open space O as t→ T−.[4]
Now we ready to get the main result of this subsection.
Theorem 3.3 (Global Existence of Regularized Solutions). Given an initial condition v0 ∈V m, m ∈ Z+ ∩ 0, for any ε > 0, there exists for all time a unique solution vε ∈C1([0,∞); V m
)to the regularized equation (3.16).
Proof. By the Proposition 3.6, there exists the unique solution vε ∈ C1([0, Tε); V
m)
to theregularized equation (3.16). Now show that there exists an a priori bound on ‖vε(, t)‖m.
106
Take the weak derivative Dα with multi-index α in the spatial variables of equation (3.16).We have
Dα(dvε
dt
)= Dα
(Fε(v
ε)).
Then take the L2 inner product of the last equation with Dα(vε). We obtain∫RNDα(vε) · d
dt
(Dα(vε)
)dx =
∫RNDα(vε) ·Dα
(Fε(v
ε))dx.
For the left-hand side of this equality, we have∫RNDα(vε) · d
dt
(Dα(vε)
)dx = 1
2
∫RN
ddt
(Dα(vε)
)2dx = 1
2ddt
∫RN
∣∣Dα(vε)∣∣2 dx =
12ddt‖Dα(vε)‖2
0 = ‖Dα(vε)‖0ddt
(‖Dα(vε)‖0
).
Bound the right-hand side of that equality using the Schwartz inequality,∫RNDα(vε) ·Dα
(Fε(v
ε))dx ≤
∣∣∣∫RNDα(vε) ·Dα
(Fε(v
ε))dx∣∣∣ ≤ ‖Dα(vε)‖0 ‖Dα
(Fε(v
ε))‖0.
Then dividing both sides by ‖Dα(vε)‖0, we get
ddt‖Dα(vε)‖0 ≤ ‖Dα
(Fε(v
ε))‖0.
Take a sum by 0 ≤ |α| ≤ m and use the Sobolev norm ‖vε‖m =∑
0≤|α|≤m ‖Dα(vε)‖0.We have
ddt‖vε(, t)‖m ≤ ‖Fε
(vε(, t)
)‖m (∗)
Recall now the relation (3.19).
‖Fε(v1)− Fε(v2)‖m ≤ C(‖v1‖0, ‖v2‖0, ε, N,m) ‖v1 − v2‖m (3.19)
where C =ν2C4
ε2+C2(‖v1‖0 + ‖v2‖0)
ε(N/2)+m+1and C2, C4 are positive constants.
Choose v1 = vε and v2 = 0. We get
‖Fε(vε)‖m ≤ C(‖vε‖0, ε, N,m) ‖vε‖m.
So, applying this inequality to the inequality (∗), we obtain
ddt‖vε(, t)‖m ≤ C(‖vε‖0, ε, N,m) ‖vε‖m. (∗∗)
107
By the energy estimate (3.18), we have
‖vε(, t)‖0 ≤ ‖v0‖0 on [0, T ].
Applying that to the inequality (∗∗) and taking into account that C is monotone increasingin ‖vε‖0, we get the following bound:
ddt‖vε(, t)‖m ≤ C(‖v0‖0, ε, N,m) ‖vε‖m.
Now use Gronwall’s inequality in differential form:if η(t) ≥ 0 and absolutely continuous on [0, T ], ϕ(t), ψ(t) ≥ 0 and summable on [0, T ],and almost everywhere η′(t) ≤ ϕ(t)η(t) + ψ(t), then
η(t) ≤ e∫ T0 ϕ(s) ds
[η(0) +
∫ T
0
ψ(s) ds], t ∈ [0, T ].
Take η(t) = ‖vε(, t)‖m, ϕ(t) ≡ C, and ψ(t) ≡ 0. We get the following bound:
‖vε(, t)‖m ≤ ‖v0‖m eCt ≤ ‖v0‖m eCT .
So, ∀T ≥ 0, ‖vε‖m is bounded.Hence vε never leaves set O where it is bounded, and Fε is Lipschitz continuous.Therefore by the Theorem 3.2, there exists a unique solution vε to the regularized equation(3.16) for all time t ∈ [0,∞).
Thus, the solution vε exists and is unique globally-in-time.
3.2.2 Local-in-Time Existence of Solutions to the Euler and the Navier-Stokes Equa-tions.
The goal of this subsection is to show that for any viscosity 0 ≤ ν <∞, there exists a timeinterval [0, T ] of existence of the unique solution vν∈C
([0, T ];C2(R3)
)∩C1
([0, T ];C(R3)
)to the Euler and the Navier-Stokes equations, and that a subsequence of regularized solu-tions (vε) converges to vν .
At first, we need to prove the following a priori higher-order energy estimate that is inde-pendent of ε.
Proposition 3.4 (The Hm Energy Estimate). Let v0 ∈ V m. Then the unique regularizedsolution vε ∈ C1
([0,∞); V m
)to the equation (3.16) satisfies
12ddt‖vε‖2
m + ν‖Jε∇vε‖2m ≤ Cm ‖∇Jεvε‖L∞ ‖vε‖2
m. (3.21)
Proof. Let vε ∈ C1([0,∞); V m
)be a smooth solution to the equation (3.16), i.e.,
vεt = νJ2ε ∆vε − PJε
[(Jεv
ε) · ∇(Jεvε)].
108
Take the derivative Dα, |α| ≤ m of this equation.
Dαvεt = νDαJ2ε ∆vε −DαPJε
[(Jεv
ε) · ∇(Jεvε)].
Consider the L2 inner product of both sides of this equation with Dαvε.(Dαvεt , D
αvε)L2 = ν
(DαJ2
ε ∆vε, Dαvε)L2 −
(DαPJε
[(Jεv
ε) · ∇(Jεvε)], Dαvε
)L2 . (∗)
Note that the left-hand side of this equation can be written in the following form:
(Dαvεt , D
αvε)L2 = 1
2
∫RN
ddt
(DαV ε)2 dx = 12ddt‖Dαvε‖2
0.
To the first term at the right-hand side of the equation (∗), apply the properties of themollifiers (ii) and (iii) from Theorem 2.5 and an integration by parts with the divergencetheorem. Then we get
(DαJ2ε ∆vε, Dαvε)L2
Th.2.5(ii)= (J2
εDα∆vε, Dαvε)L2 = (J2
ε ∆Dαvε, Dαvε)L2
Th.2.5(ii)=
(∇ · ∇J2εD
αvε, Dαvε)L2
Int.byparts= −(∇J2
εDαvε, ∇Dαvε)L2
Th.2.5(ii)= −(J2
ε∇Dαvε, ∇Dαvε)L2
Th.2.5(iii)= −(Jε∇Dαvε, Jε∇Dαvε)L2 = −‖Jε∇Dαvε‖2
0
Th.2.5(ii)= −‖DαJε∇vε‖2
0.
Consider now the L2 inner product(PJε
[(Jεv
ε) ·∇(DαJεvε)], Dαvε
)L2 and apply the pro-
perties of Leray’s projection operator (ii) and (iv) from Theorem 2.6 and then all operationswe just have mentioned above.
(PJε
[(Jεv
ε) · ∇(DαJεvε)], Dαvε
)L2
Th.2.6(iv)=
(Jε[(Jεv
ε) · ∇(DαJεvε)], PDαvε
)L2
Th.2.6(ii)=
(Jε[(Jεv
ε) · ∇(DαJεvε)], Dα(Pvε)
)L2
Th.2.5(iii)(Pvε=vε)
=([
(Jεvε) · ∇(DαJεv
ε)], JεD
αvε)L2 =∫
RN(Jεv
ε) · ∇(DαJεvε) · (JεDαvε) dx
Th.2.5(ii)=
(Jεv
ε,[∇(JεD
αvε) · (JεDαvε)])L2 =
(Jεv
ε, 12∇(JεD
αvε)2)L2
Int.byparts= −1
2
(∇ · (Jεvε), (JεD
αvε)2)L2
Th.2.5(ii)=
− 12
(Jε(∇ · vε), (JεD
αvε)2)L2 = 0 since ∇ · vε = div vε = 0.
Add this inner product (zero value) to the right-hand side of the equation (∗) and then take
109
a sum of both sides of the resulting equation by α, 0 ≤ |α| ≤ m. We have
12ddt‖vε‖2
m = −ν ‖Jε∇vε‖2m −
∑0≤|α|≤m
(DαPJε[(Jεvε) · ∇(Jεv
ε)], Dαvε)L2
+
∑0≤|α|≤m
(PJε[(Jεvε) · ∇(DαJεv
ε)], Dαvε)L2
or12ddt‖vε‖2
m + ν ‖Jε∇vε‖2m =
−∑
0≤|α|≤m
(DαPJε[(Jεvε) · ∇(Jεv
ε)]− PJε[(JεV ε) · ∇(DαJεvε)], Dαvε
)L2. (∗∗)
Work now to find a bound for the right-hand side of this equation.
−∑
0≤|α|≤m
(DαPJε[(Jεvε) · ∇(Jεv
ε)]− PJε[(JεV ε) · ∇(DαJεvε)], Dαvε
)L2
Th.2.5(ii)Th.2.6(iv)
≤
∣∣∣ ∑0≤|α|≤m
PJεDα[(Jεv
ε) · ∇(Jεvε)]− [(Jεv
ε) · ∇(DαJεvε)], Dαvε
∣∣∣ Th.2.6(iv)Th.2.5(iii)
≤
∑0≤|α|≤m
∣∣∣(Dα[(Jεvε) · ∇(Jεv
ε)]− [(Jεvε) · ∇(DαJεv
ε)], DαJε(Pvε)
)L2
∣∣∣ Pvε=vε,Schwartz ineq.≤
Th.2.5(ii,iii)
∑0≤|α|≤m
‖Dα[(Jεvε) · ∇(Jεv
ε)]− [(Jεvε) · ∇(DαJεv
ε)]‖0 ‖DαJεvε‖0 ≤(
for 0 ≤ |α| ≤ m, ‖DαJεvε‖0 ≤ ‖Jεvε‖m
Th.2.5(v)for k=0≤ Cm0 ‖vε‖m
)Cm0 ‖vε‖0
∑0≤|α|≤m
‖Dα[(Jεvε) · ∇(Jεv
ε)]− [(Jεvε) ·Dα(∇Jεvε)]‖0
(2.39)
≤(Applying the Calculus inequality (2.39), we choose u = Jεv
ε and v = ∇(JεVε).)
Cm0C ‖vε‖m(‖∇Jεvε‖L∞ ‖Dm−1∇Jεvε‖0 + ‖DmJεv
ε‖0 ‖∇Jεvε‖L∞)≤(
Use that ‖Dm−1∇Jεvε‖0 ≤ ‖DmJεvε‖0
)2Cm0C ‖vε‖m ‖∇Jεvε‖L∞ ‖DmJεv
ε‖0 ≤ 2Cm0C ‖vε‖m ‖∇Jεvε‖L∞ ‖Jεvε‖mTh.2.5(v)
for k=0≤
Th.2.5(ii)
Cm ‖vε‖2m ‖Jε∇vε‖L∞ with Cm = 2C2
m0C.
Utilizing this bound, from the equation (∗∗), we obtain the following estimate:
12ddt‖vε‖2
m + ν ‖Jε∇vε‖2m ≤ Cm ‖Jε∇vε‖L∞ ‖vε‖2
m. (3.21)
110
Now show that the solutions to the regularized equation (3.16) form a contraction in thelow norm C
([0, T ]; L2(R3)
).
Lemma 3.2. The bounded family vε forms a Cauchy sequence in C([0, T ]; L2(R3)
). In
particular, there exists a constant C, such that, for all ε and ε′,
sup0<t<T
‖vε − vε′‖0 ≤ C maxε, ε′. (3.22)
Proof. Let vε and vε′ be two solutions to the equation (3.16) with parameters ε and ε′
respectively, i.e.,
dvε
dt= νJ2
ε ∆vε − PJε[(Jεvε) · ∇(Jεvε)], vε|t=0 = v0
anddvε′
dt= νJ2
ε′∆vε′ − PJε′ [(Jε′vε
′) · ∇(Jε′v
ε′)], vε′∣∣∣t=0
= v0.
Subtract these equations and take L2 inner product from their difference with (vε − vε′).Take into account that(
ddt
(vε − vε′), vε − vε′)L2 = 1
2
∫R3
ddt
(vε − vε′)2 dx = 12ddt‖vε − vε′‖2
0.
We will have12ddt‖vε − vε′‖2
0 = T1 + T2
where T1 = ν(J2ε ∆vε − J2
ε′∆vε′ , vε − vε′)L2 and
T2 =(PJε′ [(Jε′vε
′) · ∇(Jε′v
ε′)]− PJε[(Jεvε) · ∇(Jεvε)], vε − vε′
)L2 .
At first, find a bound for T1.After adding and subtracting J2
ε′∆vε and grouping, we get
T1 = ν(J2ε ∆vε − J2
ε′∆vε + J2
ε′∆vε − J2
ε′∆vε′ , vε − vε′)L2 = T11 + T12
with T11 = ν((J2ε − J2
ε′)∆vε, vε − vε′
)L2 and T12 = ν
(J2ε′∆(vε − vε′), vε − vε′
)L2 .
For T11, consequently apply the Schwartz inequality, add and subtract Jε∆vε, Jε′∆vε, and∆vε, apply the Triangle inequality, and the properties of mollifiers (iv,v) from Theorem 2.5.
111
Then we have
T11
Schwartzineq.
≤ ‖(J2ε − J2
ε′)∆vε‖0 ‖vε − vε
′‖0 =
‖(J2ε ∆vε−Jε∆vε)+(Jε∆v
ε−∆vε)− (J2ε′∆v
ε−Jε′∆vε)− (Jε′∆vε−∆vε)‖0 ‖vε−vε
′‖0
≤ [‖Jε(Jε∆vε)− (Jε∆vε)‖0 + ‖Jε∆vε −∆vε‖0 + ‖Jε′(Jε′∆vε)− (Jε′∆v
ε)‖0 +
‖Jε′∆vε −∆vε‖0] ‖vε − vε′‖0
Th.2.5(iv)
≤
(C1ε‖Jε∆vε‖1 + C2ε‖∆vε‖1 + C3ε
′‖Jε′∆vε‖1 + C4ε′‖∆vε‖1
)‖vε − vε′‖0
Th.2.5(v)
≤(C1C10ε‖∆vε‖1 + C2ε‖∆vε‖1 + C3C10ε
′‖∆vε‖1 + C4ε′‖∆vε‖1
)‖vε − vε′‖0 ≤
C maxε, ε′ ‖∆vε‖1 ‖vε − vε′‖0 ≤ C maxε, ε′ ‖vε‖3 ‖vε − vε
′‖0.
Here C = maxC1C10, C2, C3C10, C4, and we used that ‖∆vε‖1 ≤ ‖vε‖3.
For T12, apply the properties of mollifiers (ii,iii) from the Theorem 2.5 and then integrateby parts.
T12 = ν(∆J2
ε′(vε − vε′), vε − vε′
)L2
Th.(ii,iii)= ν
(∇2Jε′(v
ε − vε′), Jε′(vε − vε′))L2
Int.byparts=
−ν(∇Jε′(vε − vε
′), ∇Jε′(vε − vε
′))L2 = −ν‖∇Jε′(vε − vε
′)‖2
0 ≤ 0.
So,
T1 = T11 + T12 ≤ T11 ≤ C maxε, ε′ ‖vε‖3 ‖vε − vε′‖0.
Now estimate the second term, T2. Begin from applying the symmetry property of theprojection operator P (Th.2.6(iv)).
T2 =(PJε′[(Jε′v
ε′) · ∇(Jε′vε′)]− Jε
[(Jεv
ε) · ∇(Jεvε)], vε − vε′
)L2
=(Jε′[(Jε′v
ε′) · ∇(Jε′vε′)]− Jε
[(Jεv
ε) · ∇(Jεvε)], Pvε − Pvε′
)L2.
Recall that Pvε = vε and Pvε′ = vε′ since vε and vε′ are divergence-free.
In the following grouping, subtract and add Jε′[(Jε′v
ε′) · ∇(Jε′vε)], Jε′
[(Jε′v
ε′) · ∇(Jεvε)],
112
Jε′[(Jε′v
ε) · ∇(Jεvε)], and Jε′
[(Jεv
ε) · ∇(Jεvε)]. Then we have
T2 =(Jε′[(Jε′v
ε′) · ∇(Jε′vε′)]− Jε′
[(Jε′v
ε′) · ∇(Jε′vε)]
+ Jε′[(Jε′v
ε′) · ∇(Jε′vε)]−
Jε′[(Jε′v
ε′) · ∇(Jεvε)]
+ Jε′[(Jε′v
ε′) · ∇(Jεvε)]− Jε′
[(Jε′v
ε) · ∇(Jεvε)]
+
Jε′[(Jε′v
ε) · ∇(Jεvε)]− Jε′
[(Jεv
ε) · ∇(Jεvε)]
+ Jε′[(Jεv
ε) · ∇(Jεvε)]−
Jε[(Jεv
ε) · ∇(Jεvε)], vε − vε′
)L2
=
R1 +R2 +R3 +R4 +R5
where
R1 =((Jε′ − Jε)[(Jεvε) · ∇(Jεv
ε)], vε − vε′)L2 ,
R2 =(Jε′ [(Jε′ − Jε)vε · ∇(Jεv
ε)], vε − vε′)L2 ,
R3 =(Jε′ [Jε′(v
ε′ − vε) · ∇(Jεvε)], vε − vε′
)L2 ,
R4 =(Jε′ [(Jε′v
ε′) · ∇(Jε′ − Jε)vε], vε − vε′)L2 ,
R5 =(Jε′ [(Jε′v
ε′) · ∇Jε′(vε′ − vε)], vε − vε′
)L2 .
Estimate separately each part of T2. To do that, we will use the Schwartz and triangle in-equalities for L2 norm, properties of mollifiers (Th.2.5), the Calculus inequality in Sobolevspace (Th.2.4(i)), and the Sobolev inequality. Also we will prove a small auxiliary claimthat we will apply to R3, and integrate by parts evaluating R5.
|R1|SchwartzIneq.
≤ ‖(Jε′ − Jε)[(Jεvε) · ∇(Jεvε)]‖0 ‖vε − vε
′‖0 =
‖Jε′ [(Jεvε) · ∇(Jεvε)]− [(Jεv
ε) · ∇(Jεvε)] + [(Jεv
ε) · ∇(Jεvε)] −
Jε[(Jεvε) · ∇(Jεv
ε)]‖0 ‖vε − vε′‖0
TriangleIneq.
≤(‖Jε′ [(Jεvε) · ∇(Jεv
ε)]− [(Jεvε) · ∇(Jεv
ε)]‖0 +
‖Jε[(Jεvε) · ∇(Jεvε)]− [(Jεv
ε) · ∇(Jεvε)]‖0) ‖vε − vε′‖0
Th.2.5(iv)(2.47)≤(
C1ε′‖(Jεvε) · ∇(Jεv
ε)‖1 + C2ε‖(Jεvε) · ∇(Jεvε)‖1
)‖vε − vε′‖0 ≤
(C1 + C2) maxε, ε′‖(Jεvε) · ∇(Jεvε)‖1 ‖vε − vε
′‖0
Th.2.4(i)(2.38)≤
C3(C1 + C2) maxε, ε′(‖Jεvε‖L∞‖∇2(Jεv
ε)‖0 + ‖∇(Jεvε)‖0‖∇(Jεv
ε)‖L∞)
‖vε − vε′‖0
Th.2.5(i,ii)
≤
113
C3(C1 + C2) maxε, ε′(C4‖vε‖L∞‖Jεvε‖2 + ‖Jεvε‖1‖Jε(∇vε)‖L∞
)‖vε − vε′‖0
Th.2.5(i,ii,v)Sobolev In. for k=0
≤
C3(C1 +C2) maxε, ε′(C4C5‖vε‖sC20‖vε‖2 +C10‖vε‖1 ‖∇vε‖L∞
)‖vε − vε′‖0
Sobolev In.for k=1≤
C3(C1 + C2) maxε, ε′(C4C5C20‖vε‖s ‖vε‖2 + C10‖vε‖1C6‖vε‖s+1
)‖vε − vε′‖0 ≤
C1 maxε, ε′‖vε‖2m ‖vε − vε
′‖0
with C1 = (C1 + C2)C3 maxC4C5C20, C10C6, and we choose s = m− 1 so that
m > N2
+ 1.
Here we used that ‖vε‖1 ≤ ‖vε‖2 ≤ ‖vε‖s+1 where s > N2
.
Now estimate R2.
∣∣R2∣∣ =
∣∣(Jε′ [(Jε′ − Jε)vε · ∇(Jεvε)], vε − vε′
)L2
∣∣ Th.2.5(iii)=
∣∣([(Jε′−Jε)vε ·∇(Jεvε)], Jε′(v
ε−vε′))L2
∣∣SchwartzIneq.≤∥∥[(Jε′−Jε)vε ·∇(Jεv
ε)]∥∥
0
∥∥Jε′(vε−vε′)∥∥0
Th.2.5(v)(2.48)≤
Th.2.4(i)(2.38)
C1
(‖(Jε′ − Jε)vε‖L∞ ‖∇(Jεv
ε)‖0 + ‖(Jε′ − Jε)vε‖0 ‖∇Jεvε)‖L∞)C00‖vε − vε
′‖0
Th.2.5(i,v)
≤Sobolev In.for k=0,1
C1C00
(C2‖(Jε′ − Jε)vε‖s ‖Jε(∇vε)‖0 + ‖Jε′vε − Jεvε − vε + vε‖0C3‖Jεvε‖s+1
)‖vε − vε′‖0
Th.2.5(v)Triangle In.≤
C1C00 [C2‖Jε′vε − vε − Jεvε + vε‖sC00‖∇vε‖0 +(‖Jε′vε − vε‖0 + ‖Jεvε − vε‖0
)C3C(s+1)0‖vε‖s+1
]‖vε − vε′‖0
Triangle In.≤
Th.2.5(iv)
C1C00
[C2C00
(‖Jε′vε − vε‖s + ‖Jεvε − vε‖s
)‖vε‖1 +(
C4ε′‖vε‖1 + C5ε‖vε‖1
)C3C(s+1)0‖vε‖s+1
]‖vε − vε′‖0
Th.2.5(iv)
≤
C1C00 [C2C00(C6ε′ + C7ε)‖vε‖s+1 ‖vε‖1 +
(C4ε′ + C5ε)C3C(s+1)0‖vε‖1 ‖vε‖s+1
]‖vε − vε′‖0 ≤
114
C1C00
[C2C00(C6 + C7) + C3C(s+1)0(C4 + C5)
]maxε, ε′‖vε‖s+1‖vε‖1‖vε − vε
′‖0 ≤C2 maxε, ε′‖vε‖2
m ‖vε − vε′‖0
where C2 = C1C00 maxC2C00(C6 + C7), C3C(s+1)0(C4 + C5), and we choose s =
m− 1 so that m > N2
+ 1.
Also, we used here that ‖vε‖1 ≤ ‖vε‖s+1 with s > N2.
To estimate R3, we need the following claim.
Claim. Let a, b : RN → RN and ai, bi ∈ Hm(RN), i = 1, 2, . . . , N, m ∈ Z+ ∪ 0. Then
|(b · ∇a, b)L2| ≤ N3‖b‖20 ‖∇a‖L∞ .
Proof.∣∣∣(b · ∇a, b)L2
∣∣∣ =∣∣∣∫RN
(b · ∇a) · b dx∣∣∣ ≤ ∫
RN
∣∣∣[(b · ∇)a] · b∣∣∣ dx
=
∫RN
∣∣∣ N∑i=1
(N∑j=1
bj∂ai∂xj
)bi
∣∣∣dx ≤∫RN
N∑i,j=1
∣∣bi∣∣ ∣∣bj∣∣ ∣∣ ∂ai∂xj
∣∣dx ≤ N∑i,j=1
supx∈RN
∣∣ ∂ai∂xj
∣∣ ∫RN
∣∣bibj∣∣dx=
N∑i,j=1
∥∥ ∂ai∂xj
∥∥L∞
∫RN
∣∣bibj∣∣ dx ≤ N2 max1≤i,j≤N
∥∥ ∂ai∂xj
∥∥L∞
∫RN
N∑i,j=1
∣∣bibj∣∣ dx= N2‖∇a‖L∞
∫RN
(N∑i=1
b2i + 2
∑i<j
∣∣bibj∣∣+ (N − 1)N∑i=1
b2i − (N − 1)
N∑i=1
b2i
)dx
= N2‖∇a‖L∞∫RN
(N
N∑i=1
b2i −
∑i<j
(|bi| − |bj|)2
)dx ≤ N3‖∇a‖L∞
∫RN
N∑i=1
b2i dx
= N3‖∇a‖L∞‖b‖20.
We used the following definition of L∞-norm:
If a = (a1, a2, . . . , aN), ai ∈ C1(X) for some open set X,
then ‖∇a‖L∞(X) := max1≤i,j≤N
‖ ∂ai∂xj‖L∞(X).
Alternatively, we could use an equivalent norm
‖∇a‖L∞ :=N∑
i,j=1
‖ ∂ai∂xj‖L∞ .
115
They are equivalent since
max1≤i,j≤N
‖ ∂ai∂xj‖L∞ ≤
N∑i,j=1
‖ ∂ai∂xj‖L∞ ≤ N2 max
1≤i,j≤N‖ ∂ai∂xj‖L∞ .
Return to our estimate of R3. Choose a = Jεvε and b = Jε′(v
ε′ − vε).
∣∣R3∣∣ =
∣∣(Jε′ [Jε′(vε′ − vε) · ∇(Jεvε)], vε − vε′
)L2
∣∣ Th.2.5(iii)=∣∣([Jε′(vε′ − vε) · ∇(Jεv
ε)], −Jε′(vε′ − vε)
)L2
∣∣ =∣∣−(b · ∇a, b)
L2
∣∣ Claim≤N3 ‖b‖2
0 ‖∇a‖L∞ = N3 ‖Jε′(vε′ − vε)‖2
0 ‖∇(Jεvε)‖L∞
Th.2.5(ii,v)
≤
N3C200‖vε
′ − vε‖20 ‖Jε(∇vε)‖L∞
Th.2.5(i)
≤ N3C200‖vε − vε
′‖20 ‖∇vε‖L∞
Sobolev Ineq.k=1, s>N/2
≤
N3C200 ‖vε − vε
′‖20C1 ‖V ε‖s+1 = C3 ‖vε‖m ‖vε − vε
′‖20
with C3 = N3C200C1, and we choose s = m− 1 so that m > N
2+ 1.
∣∣R4∣∣ =
∣∣(Jε′ [(Jε′vε′) · ∇(Jε′ − Jε)vε], vε − vε′)L2
∣∣ Th.2.5(iii)=
∣∣([(Jε′vε′) · ∇(Jε′ − Jε)vε], Jε′(vε − vε′))L2
∣∣ SchwartzIneq.
≤
‖(Jε′vε′) · ∇(Jε′ − Jε)vε‖0, ‖Jε′(vε − vε
′)‖0
Calculus In.Th.2.4(i)
≤Th.2.5
(v)
C1
(‖Jε′vε
′‖L∞ ‖∇(Jε′−Jε)vε‖0 +‖Jε′vε′‖0 ‖∇(Jε′−Jε)vε‖L∞
)C00 ‖vε−vε
′‖0
Th.2.5(i,v)
≤Sobolev Ineq.k=1, s>N/2
C1C00
(‖vε′‖L∞ ‖(Jε′ − Jε)vε‖1 + C00 ‖vε
′‖0 ‖(Jε′ − Jε)vε‖s+1
)‖vε − vε′‖0
Sobolev Ineq.k=0, s>N/2
≤
C1C00
[C2‖vε
′‖s ‖(Jε′vε − vε)− (Jεvε − vε)‖1 +
C00 ‖vε′‖0 ‖(Jε′vε − vε)− (Jεv
ε − vε)‖s+1
]‖vε − vε′‖0
TriangleIneq.≤
C1C00
[C2 ‖vε
′‖s(‖Jε′vε − vε‖1 + ‖Jεvε − vε‖1
)+
C00 ‖vε′‖0
(‖Jε′vε − vε‖s+1 + ‖Jεvε − vε‖s+1
)]‖vε − vε′‖0
Th.2.5(iv)
≤
116
C1C00
[C2 ‖vε
′‖s(C3ε′ + C4ε)‖vε‖2 + C00 ‖vε
′‖0(C5ε′ + C6ε)‖vε‖s+2
]‖vε − vε′‖0 ≤
C1C00 maxε, ε′[C2(C3 + C4)‖vε′‖s‖vε‖2 + C00(C5 + C6)‖vε′‖0‖vε‖s+2
]‖vε − vε′‖0
≤ C4 maxε, ε′ ‖vε′‖s ‖vε‖s+2 ‖vε − vε′‖0 ≤ C4 maxε, ε′ ‖vε′‖m ‖vε‖m ‖vε − vε
′‖0
where C4 = C1C00 maxC2(C3+C4), C00(C5+C6), and we choose s = m−2 so that
m > N2
+ 2.
R5 =(Jε′ [(Jε′v
ε′)·∇Jε′(vε′−vε)], vε−vε′
)L2
Th.2.5(iii)=([(Jε′v
ε′)·∇Jε′(vε′−vε)], Jε′(vε−vε
′))L2
=
∫RN
[(Jε′vε′) · ∇Jε′(vε
′ − vε)] · [−Jε′(vε′ − vε)] dx.
Denote Jε′vε′= a and Jε′(vε
′ − vε) = b where a, b : RN → RN .Then
R5 = −∫RN
(a · ∇b) · b dx = −∫RN
N∑j=1
( N∑i=1
ai∂bj∂xi
)bj dx = −
∫RN
N∑i=1
ai( N∑j=1
∂bj∂xibj)dx =
− 12
∫RN
N∑i=1
ai
N∑j=1
∂(b2j )
∂xidx = −1
2
∫RN
N∑i=1
ai∂∂xi
( N∑j=1
b2j
)dx =
− 12
∫RN
N∑i=1
ai∂∂xi
(|b|2)dx = −1
2
∫RNa · ∇
(|b|2)dx
Int. byparts= 1
2
∫RN
(∇ · a
)|b|2 dx = 0
since vε, vε′ and so a, b vanish at infinity, and
∇ · a = div a = div(Jε′vε′)
Th.2.5(ii)= Jε′(div vε
′) = 0
because vε′ ∈ V m(RN).
Putting T1 and T2 together, we get for m > N2
+ 2.
12ddt‖vε − vε′‖2
0 = T1 + T2 = T1 +5∑i=1
Ri ≤ T1 +5∑i=1
|Ri| ≤
C maxε, ε′‖vε‖3 ‖vε − vε′‖0 + C1 maxε, ε′‖vε‖2
m ‖vε − vε′‖0+
C2 maxε, ε′‖vε‖2m ‖vε − vε
′‖0 + C3‖vε‖m ‖vε − vε′‖2
0+
C4 maxε, ε′‖vε′‖m ‖vε‖m ‖vε − vε′‖0 + 0 ≤
C5‖vε‖m[maxε, ε′
(1 + 2‖vε‖m + ‖vε′‖m
)+ ‖vε − vε′‖0
]‖vε − vε′‖0 ≤
C5‖vε‖m(1 + 2‖vε‖m + ‖vε′‖m
)(maxε, ε′+ ‖vε − vε′‖0
)‖vε − vε′‖0
where C5 = maxC, C1 + C2, C3, C4 and ‖vε‖3 ≤ ‖vε‖m for m > N2
+ 2.
117
Now use that the family (vε) is bounded. This will be proved as part 1) of the proof ofTheorem 3.4.Let vε, vε′ ∈ OM = v ∈ V m(RN) | ‖v‖m < M.Then 1 + 2‖vε‖m + ‖vε′‖m < 1 + 3M . Denote C(M) = C5M(1 + 3M) > 0.We obtain the following IVP for linear OD inequality:
12 2 ‖vε − vε′‖0 d
dt‖vε − vε′‖0 ≤ C(M)[maxε, ε′+ ‖vε − vε′‖0] ‖vε − vε′‖0 ,
‖vε − vε′‖0
∣∣t=0
= ‖vε0 − vε′
0 ‖0 = ‖v0 − v0‖0 = 0.
Take‖vε − vε′‖0 = ϕ(t)eC(M)t with ϕ(0) = 0.
Then we have
dϕdteC(M)t ≤ C(M) maxε, ε′.
Integrating in t, we get ∫ t
0
dϕ ≤ C(M) maxε, ε′∫ t
0
e−C(M)t dt
orϕ(t) ≤ maxε, ε′
(1− e−C(M)t
).
So,
‖vε − vε′‖0 ≤ maxε, ε′(eC(M)t − 1
)≤ maxε, ε′eC(M)t.
Thereforesup
0≤t≤T‖vε − vε′‖0 ≤ C maxε, ε′ (3.22)
where C = eC(M)T depends on some selected M > 0 and T .
Thus, (vε) is a Cauchy sequence in C ([0, T ];L2(R3)).
We now prove the main theorem of this section (and entire chapter).
Theorem 3.4 (Local-in-Time Existence of Solutions to the Euler and the Navier-StokesEquations). Given an initial condition v0 ∈ V m, m ≥ [[N
2]] + 2, then the following results
are true.
i) There exists a time T with the rough upper bound
T <1
C ′m ‖v0‖m(3.23)
such that for any viscosity 0 ≤ ν < ∞ there exists the unique solution vν ∈C ([0, T ];C2(R3)) ∩ C1 ([0, T ];C(R3)) to the Euler or the Navier-Stokes equations.
118
The solution vν is the limit of subsequence of approximate solutions vε to the equa-tions (3.16) and (3.17).
ii) The approximate solution vε and the limit vν satisfy the higher-order energy estimates
sup0≤t≤T
‖vε‖m ≤‖v0‖m
1− C ′mT‖v0‖m, (3.24)
sup0≤t≤T
‖vν‖m ≤‖v0‖m
1− C ′mT‖v0‖m.
Proof.
1) Show that (vε), family of solutions of the regularized equation, is uniformly boundedin Hm, ∀m > N
2+ 1.
Apply the energy estimate (3.21) and the Sobolev inequality (2.21) to bound the timederivative of ‖vε‖m.
ddt‖vε‖m
(3.21)≤ Cm‖Jε∇vε‖L∞ ‖vε‖m − ν‖Jε∇vε‖2
m ≤ Cm‖Jε∇vε‖L∞ ‖vε‖mTh.2.5
(i)
≤
Cm‖∇vε‖L∞ ‖vε‖m(2.21)≤ C ′m ‖vε‖2
m.
Thus,ddt‖vε‖m ≤ C ′m ‖vε‖2
m. (3.25)
Here we applied the Sobolev inequality to ‖∇vε‖L∞ (k = 1). So, it must be thatm > N
2+ 1.
Integrate the inequality (3.25) in time.∫ ‖vε‖m‖v0‖m
d‖vε‖m‖vε‖2
m
≤∫ t
0
C ′m dt,
− 1
‖vε‖m+
1
‖v0‖m≤ C ′mt or ‖vε‖m ≤
1
‖v0‖−1m − C ′mt
, ∀t ∈ [0, T ].
The last upper bound makes sense when ‖v0‖−1m − C ′mt > 0 only. Therefore
T < 1C′m‖v0‖m
.The bound 1
‖v0‖−1m −C′mt
is monotonically increasing on [0, T ]. Then, for all ε,
sup0≤t≤T
‖vε‖m ≤‖v0‖m
1− C ′mT‖v0‖m(3.24)
Thus, for T < 1C′m‖v0‖m
, ∀m > N2
+ 1, the family (vε) is uniformly bounded inC([0, T ];Hm(RN)
).
2) Show that the family of time derivatives(dvε
dt
)is uniformly bounded in Hm−2(RN).
119
From the regularized equation (3.16), we have∥∥dvεdt
∥∥m−2≤∥∥F (1)
ε (vε)∥∥m−2
+∥∥F (2)
ε (vε)∥∥m−2
.
(a) ‖F (1)ε (vε)‖m−2 = ν‖J2
ε ∆vε‖m−2
Th.2.5(ii)= ν‖∆J2
ε vε‖m−2 ≤ ν‖J2
ε vε‖m
Th.2.5(v)
≤
νCm0‖Jεvε‖mTh.2.5
(v)
≤ νC2m0‖vε‖m = νC(1)(m)‖vε‖m with C(1)(m) = C2
m0.
(b) ‖F (2)ε (vε)‖m−2 = ‖PJε[(Jεvε) · ∇(Jεv
ε)]‖m−2
Th.2.6(i)
≤ ‖Jε[(Jεvε) · ∇(Jεvε)]‖m−2
Th.2.5(v)
≤ Cm0‖(Jεvε) · ∇(Jεvε)‖m−2
(2.38)≤
Cm0C[‖Jεvε‖L∞ ‖Dm−2∇(Jεv
ε)‖0 + ‖Dm−2(Jεvε)‖0 ‖∇(Jεv
ε)‖L∞] Th.2.5(v)
≤
Cm0C[C0
εN/2‖vε‖0 ‖Dm−1(Jεv
ε)‖0 + ‖Jεvε‖m−2C1
ε(N/2)+1‖vε‖0
] Th.2.5(v)
≤
Cm0C[C0
εN/2‖vε‖0
C(m−1)0
1‖vε‖m−1 +
C(m−2)0
1‖vε‖m−2
C1
ε(N/2)+1‖vε‖0
]=
Cm0CεN/2‖vε‖0
(C0C(m−1)0‖vε‖m−1 +
C(m−2)0C1
ε‖vε‖m−2
)≤
Cm0Cε(N/2)+1
(C0C(m−1)0ε+ C(m−2)0C1
)‖vε‖2
m = C(2)(ε,N,m) ‖vε‖2m
with C(2)(ε,N,m) = Cm0Cε(N/2)+1
(C0C(m−1)0ε+ C(m−2)0C1
).
Here we used that ‖vε‖0 ≤ ‖vε‖m and ‖vε‖m−2 ≤ ‖vε‖m−1 ≤ ‖vε‖m.
Also, we assumed here that m− 2 > N2
. So, m > N2
+ 2.
(c) Putting together the estimates in Hm−2 of F (1)ε and F (2)
ε , we obtain∥∥dvεdt
∥∥m−2≤ νC(1)(m) ‖vε‖m + C(2)(ε,N,m) ‖vε‖2
m
Since the family (vε) is uniformly bounded in Hm(RN) for m > N2
+ 1, then
the family(dvε
dt
)is uniformly bounded in Hm−2(RN) for m > N
2+ 2.
3) Let vε, vε′ ∈ OM = v ∈ V m | ‖v‖m < M and M = ‖v0‖m1−C′mT ‖v0‖m
.
By the Lemma 3.2, sup0≤t≤T ‖vε − vε′‖0 ≤ C
(‖v0‖m, T
)maxε, ε′, i.e., the fa-
mily (vε) ⊂ OM forms a Cauchy sequence in C ([0, T ];L2(R3)), that is a Ba-nach space (Lemma 3.1). Then as ε→ 0+, (vε) strongly converges to somev ∈ C ([0, T ];L2(R3)), i.e., ∃ v ∈ C ([0, T ];L2(R3)) such that
sup0≤t≤T
‖vε − v‖0 ≤ Cε, ∀ε > 0. (3.26)
120
4) To this point, we proved that (vε) is uniformly bounded in a high norm, and it stronglyconverges in L2-norm (in H0).Next, show that (vε) has strong convergence in all intermediate norms. For thispurpose, we will use the following interpolation lemma for Sobolev spaces.
Lemma 3.3 (Interpolation in Sobolev Spaces). [1] Given ε > 0, there exists a con-stant Cs so that for all v ∈ Hs(RN) and 0 < s′ < s,
‖v‖s′ ≤ Cs ‖v‖1−(s′/s)0 ‖v‖s′/ss . (3.27)
Let s = m and 0 < m′ < m. Apply the Interpolation lemma to the difference vε − vand take the supremum by t ∈ [0, T ]. We have
sup0≤t≤T
‖vε − v‖m′ ≤ Cm sup0≤t≤T
‖vε − v‖1−(m′/m)0 sup
0≤t≤T‖vε − v‖m′/mm .
Since (vε) is uniformly bounded inHm(RN) by (3.24), we know ∃v ∈ Hm(RN) anda subsequence (vεj) such that vεj → v weakly in Hm(RN) [5, Theorem 3.7].It follows that [5, Proposition 3.6]
‖v‖m ≤ limj→∞‖vεj‖m ≤
‖v0‖m1− C ′mT‖v0‖m
.
Hence using the triangle inequality, we obtain
sup0≤t≤T
‖vε − v‖m ≤ sup0≤t≤T
‖vε‖m + sup0≤t≤T
‖v‖m ≤2‖v0‖m
1− C ′mT‖v0‖m.
Putting this estimate together with the bound (3.26), we have
sup0≤t≤T
‖vε − v‖m′ ≤ Cm[C(‖v0‖m, T )ε
]1−(m′/m)( 2‖v0‖m
1− C ′mT‖v0‖m
)m′/m=
C1(‖v0‖m, T )ε1−(m′/m).
Therefore (vε) strongly converges to v as ε→ 0 in C([0, T ];Hm′(R3)
),
∀ 0 ≤ m′ < m.
5) Employing the Sobolev inequality for (vε − v) with m′ = s+ k, k = 2, and s > N2
,we get
‖vε − v‖C2(RN ) ≤ C ‖vε − v‖m′ with some constant C.
So,sup
0≤t≤T‖vε − v‖C2(RN ) ≤ C2ε
1−(m′/m) with C2 = CC1.
121
Thus, (vε) strongly converges to v in C([0, T ];C2(RN)
)with N
2+ 2 < m′ < m or
dN2e+ 2 ≤ m′ < m where dN
2e is a ceiling function of N
2, the least integer greater or
equal to N2
.
6) Note that strong convergence in C([0, T ];C2(RN)
)implies an uniform convergence
for the sequence (vε) and for sequences of spatial derivatives of vε up to the secondorder since from the last inequality, we have
sup0≤t≤T
∑|α|≤2
supx∈RN
|Dαx (vε − v)| < C2ε
1−(m′/m) with dN2e+ 2 ≤ m′ < m.
Alternatively, we can apply the following corollary of the Arzela-Ascoli theorem:
If a sequence of differentiable functions and sequence of their derivatives are uni-formly bounded, then there exists a subsequence that converges uniformly.
Hence subsequences of (∇vε) and (∆vε) converge uniformly to ∇v and ∆v respec-tively as ε → 0+. The projection operator P is uniformly continuous. Also, usingthe properties of mollifiers (i,iv) from the Theorem 2.5, we get
limε→0+
Jεvε = lim
ε→0+vε = v (strong and uniform convergence).
Then Fε(vε) = νJ2ε ∆vε − PJε
[(Jεv
ε) · ∇(Jεvε)]
converges uniformly inC([0, T ];C(R3)
)to w = [ν∆v − P(v · ∇v)] ∈ C
([0, T ];C(R3)
)as ε→ 0+.
Note that vε → v in C([0, T ];C2(R3)
)implies that this is true in C
([0, T ];C(R3)
)since C2(R3) ⊆ C(R3).By the equation (3.16), vεt = Fε(v
ε).Hence there exists subsequence (v
εjt ) (induced by uniformly convergent subsequence
(vεj)) that uniformly converges to w in C([0, T ];C(RN)
).
Justify that w = vt in the Banach space C([0, T ];C(R3)
).
Using Fundamental Theorem of Calculus, we can write
vε(t, x) =
∫ t
0
vεs(s, x) ds
with vεs(s, x) continuous in both variables.Now, as ε→ 0+, vεs(s, x)→ w(s, x) uniformly in s by previous, and so∫ t
0
vεs(s, x) ds→∫ t
0
w(s, x)ds as ε→ 0+.
Also, we know that as ε→ 0+, vε(t, x)→ v(t, x) uniformly in t.
122
It follows that
v(t, x) =
∫ t
0
w(s, x) ds
Consider
v(t+h,x)−v(t,x)h
= 1h
[∫ t+h
0
w(s, x) ds−∫ t
0
w(s, x) ds
]= 1
h
∫ t+h
t
w(s, x) ds.
Then in C(R3) norm we have∥∥∥∥v(t+h,x)−v(t,x)h
− w(t, x)
∥∥∥∥C(R3)
=
∥∥∥∥ 1h
∫ t+h
t
w(s, x) ds− 1h
∫ t+h
t
w(t, x) ds
∥∥∥∥C(R3)
= 1h
∥∥∥∥∫ t+h
t
[w(s, x)− w(t, x)] ds
∥∥∥∥C(R3)
≤ 1h
∫ t+h
t
‖w(s, x)− w(t, x)‖C(R3) ds
≤ 1h h sup
t≤s≤t+h‖w(s, x)− w(t, x)‖C(R3) .
For the last term, w(s, ) continuous on [0, T ] implies w(s, ) uniformly continuouson [0, T ], so given ε > 0, ∃ h = h(ε) > 0 such that ∀ s, t ∈ [0, h],
‖w(s, x)− w(t, x)‖C(R3) < ε.
Therefore,
supt≤s≤t+h
‖w(s, x)− w(t, x)‖C(R3) ≤ ε.
So, vt = w = ν∆v − P(v · ∇v).Thus, v = vν , a classical solution to the Euler or the Navier-Stokes equations.Note that then vν ∈ C
([0, T ];C2(RN)
)∩ C1
([0, T ];C(RN)
).
This concludes chapter 3 and the present paper.
123
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