NOTES DU COURS
”INTRODUCTION AUX EQUATIONS
AUX DERIVEES PARTIELLES
D’EVOLUTION”
Jean-Yves CHEMINLaboratoire J.-L. Lions
Universite Paris 6, Case 18775 232 Paris Cedex 05, France
adresse electronique: [email protected]
December 3, 2016
2
Contents
1 Ordinary Di↵erential Equations and transport equations 71.1 Linear ordinary di↵erential equations . . . . . . . . . . . . . . . . . . . . . . . . 81.2 Around Cauchy-Lipschitz theorem . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Non linear ordinary di↵erential equations solved by a compactness method . . 121.4 Transport equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 The heat flow and the Stokes flow in a bounded domain 232.1 The Dirichlet problem in a bounded domain . . . . . . . . . . . . . . . . . . . . 232.2 The stationnary Stokes’s problem . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 The Stokes’s flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3 Leray’s Theorem on Navier-Stokes equations in a bounded domain 353.1 The concept of turbulent solution . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 The proof of Leray’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.3 Stability of Leray solutions in dimension two . . . . . . . . . . . . . . . . . . . 43
4 Stability of Navier-Stokes equations in dimension 3 474.1 A su�cient condition of 3D stability . . . . . . . . . . . . . . . . . . . . . . . . 474.2 Existence of stable solutions in a bounded domain . . . . . . . . . . . . . . . . 50
4.2.1 Intermediate spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.2.2 The wellposedness result in V12� . . . . . . . . . . . . . . . . . . . . . . . 52
4.2.3 Some remarks about stable solutions . . . . . . . . . . . . . . . . . . . . 55
5 An example of a dispersive equation: the Schrodinger equation 575.1 The solution of some classical linear evolution PDE in Rd . . . . . . . . . . . . 575.2 The complex interpolation method in Lp space . . . . . . . . . . . . . . . . . . 615.3 The duality method and the TT ? argument . . . . . . . . . . . . . . . . . . . . 625.4 An example of application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.5 Refined convolution inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
6 Linear symmetric systems 696.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696.2 The wellposedness of linear symmetric systems . . . . . . . . . . . . . . . . . . 716.3 Finite propagation speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
A Sobolev spaces 81A.1 Definition of Sobolev spaces on Rd . . . . . . . . . . . . . . . . . . . . . . . . . 81A.2 Sobolev embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
3
A.3 Homogeneous Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88A.4 The spaces H1
0
(⌦) and H�1(⌦) . . . . . . . . . . . . . . . . . . . . . . . . . . . 89A.5 The Lindelhof principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4
Introduction
This text consists in notes of a series of lectures given in the second year of the Master”Mathematiques et applications fondamentales” of the University Pierre et Marie Curie. Thepurpose of this series of lectures is to expose some basic results and methods in the fieldevolution partial di↵erential equations. By basic, we mean that we stay far away from optimalor refined results for the the type of problems we look at.
We investigate four di↵erent class of problems:
• the first one is the linear transport equations associated with a non smooth vector fieldwhich is divergence free.
• the second is the problem is local and global existence of solutions of the incompressibleNavier-Stokes equations in a bounded domain
• the third one is the linear Schrodinger equation in the whole space. Thi is the opprtunityto put in light dispersive phenomena and to apply it to a simple exemple of semi linearSchrodinger equation.
• the fourth one concern linear and quasilinear symmetric systems of order 1. A basicexample of such a system is the system related to barotropic gases.
The structure of these notes is the following.
In the first chapter, we recal the basic theorem on ordinary di↵erential equations (linearand non linear Cauchy-Lipchitz theorem, Peano’s theorem). We explain the link betweenordinary di↵erential equations and transport equations in the smooth case. Then westudy transport equations associated with non smooth divergence free vector fields.
In the second chapter, we investigate the heat equation and the stationnary Stoke equa-tion. The concept of solution of Stokes problem is important. Then we study the heatequation and the evolution Stokes equation.
The third chapter is devoted to the proof of the global existence of weak solution for theincompressible Navier-Stokes equation in a bounded domain in dimension 2 or 3. Weprove the uniqueness of such solutions in dimension 2.
The fourth chapter studies the problem of uniqueness and more generally the problemof stability for the three dimensionnal incompressible Navier-Stokes system. We provealso the local existence for small data and the global existence for small data, the initialdata being in this case more regular than just of finite energy.
The fifth chapter is devoted to the study of the Schrodinger equation in the wholespace. We compute the fundamental solution and prove the Stricharz estimate whichis a consequence of some dispersive e↵ect. This is the opportunity to study complexinterpolation theory and also refined convolution inequality.
5
The sixth chapter is devoted to the study of linear symmetric system with boundedsmooth variable coe�cients.
The seventh chapter is an introduction to the Littlewood-Paley theory which is animportant tool of non linear partial di↵erential equation.
The aim of the eighth chapter is the proof of local wellposedness of quasilinear of quasi-linear systems
6
Chapter 1
Ordinary Di↵erential Equations andtransport equations
Introduction
The purpose of this chapter is twofold: first we want to recall very basic theorem, namelylinear and non linear Cauchy-Lipschitz theorem) and Peano’s theorem. Secondly, we presentthe basis of the Di-Perna-Lions theory about the resolution of transport equation associatedwith non smooth divergence free vector fields. The chapter is organized as follows.
In the first section, we prove the linear Cauchy-Lipschitz theorem using the method ofconjugation by an exponantial term involving a large parameter. This method provides avery short proof of global existence for a linear ordinary di↵erential equation.
The second section is devoted to the proof of the non linear Cauchy-Lipschitz theorem.Again, we use a conjugation with respect to an exponantial term involving a large parameterand the Lipschitz constant of the function defining the ordinary di↵erential equation. Themain point is that this method gives immediatly the existence on a time interval the lengthof which is estimated from below in a way which depends only on the supremum norm of thevector field which defines the ordinary di↵erential equation.
In the third section, we prove Peano’s theorem which claims the existence (and not theuniqueness which is false in general) for ordinary di↵erential equations associated with acontinuous vector field. The method consists in the regularization of the vector field. ThenAscoli’s theorem provides relative compactness of the family of solutions of the approximatedproblems and thus a solution.
In the forth section, we establish the basis of Di-Perna-Lions theory which claims thatfor divergence free vector fileds for which the first order derivatives belong to some Lp spacefor finite p, the associated transport equation as a unique solution for intial data in Lp0 .The existence part relies on the same method used to prove Peano’s theorem. The proof ofthe uniqueness follows a duality method which relies on the fact that the transport problem(which the same transport equation) satisfies the maximum principle.
As a conclusion of this introduction, let us point out that this chapter contains somefundamental techniques for studying evolution partial di↵erential equations:
• the conjugation by an exponantial term involving a large parameter;
• fixed point theorem which provides existecne and uniqueness together with an approxi-mation scheme;
7
• relative compactness of solutions of a family of approximated problems; the relativecompactness coming from some time regularity;
• duality method which provides uniqueness for linear problem without any idea of con-traction.
These methods will be used all along these notes.
1.1 Linear ordinary di↵erential equations
Theorem 1.1.1 Let E be a Banach space, I an open interval of R and A a continuous mapfrom I to L(E), the set of continuous linear maps from E into E. Let t
0
be in I, a unique C1
function x from I to E exists such that
(ODE)
8
>
<
>
:
du
dt= A(t)u(t)
u(0) = u0
.
Proof. Uniqueness is obvious because as it is a linear equation, it is enough to prove that if xis a solution of (ODE) which initial data 0, then it is identically zero. The set of t in I suchthat x(t) = 0 is a closed subset of I. It is open because if t
1
is sucht at x(t1
) = 0, then wehave by integration
supt2[t1�↵,t1+↵1]
kx(t)kE ↵ supt2[t1�↵,t1+↵1]
kA(t)kL(E)
supt2[t1�↵,t1+↵1]
kx(t)kE
Taking ↵ small enough implies that x is identically 0 near t1
. The proof of the existence ofsolutions of this equation is very simple. Let � be a positive real number, let us introduce thespace E� definied by
E� =n
x 2 C(I, E) / kxk� def= sup
t2Ikx(t)k exp
⇣
��Z t
0
A(t0)dt0⌘
< 1o
with A(t0) def= kA(t0)kL(E)
. It is an exercise left to the reader to prove that E� is a Banachspace. The solution of (ODE) are the same as the solutions of
(Id�PA)u = u0
with PAu(t)def=
Z t
0
A(t0)u(t0)dt0.
Let us prove that
kPAkL(E�) 1
�(1.1)
which will imply the theorem because the constant function with value x0
obviously belongsto E
⇤
for any positive real number �. By definition of A we have
kPAx(t)k exp⇣
��Z t
0
A(t0)dt0⌘
Z t
0
exp⇣
��Z t
t0A(t00)dt00
⌘
A(t0) exp⇣
��Z t0
0
A(t00)dt00⌘
kx(t0)kdt0.
8
By definition of k · k�, we infer that
kPAx(t)k exp⇣
��Z t
0
A(t0)dt0⌘
kxk�Z t
0
exp⇣
��Z t
t0A(t00)dt00
⌘
A(t0)dt0.
An obvious computation of integral implies that, for any t in I,
kPAx(t)k exp⇣
��Z t
0
A(t0)dt0⌘
1
�kxk�
which is exactly (1.1) ; the theorem is proved. 2
1.2 Around Cauchy-Lipschitz theorem
Let us state and prove the classical theorem about existence and uniqueness of solution of anordinary di↵erential equation associated with a Lipschitz function. This uses Picard’s fixedpoint theorem.
Theorem 1.2.1 (de Cauchy-Lipschitz) Let E be a Banach space and f a continuous func-tion on an open subset U of R⇥E with value in E. Let us consider a point (t
0
, x0
) of U suchthat an open interval J
0
containing t0
and a ball B0
centered in x0
and of radius R0
suchthat J
0
⇥B0
is included in U and such that the two functions
M0
(t)def= sup
x2B0
kf(t, x)kE and K0
(t)def= sup
(x1,x2)2B20
kf(t, x1
)� f(t, x2
)kEkx
1
� x2
kEare integrable on J
0
. Then, for any subinterval J of J0
containing t0
which satisfiesZ
JM
0
(t)dt R0
, (1.2)
a unique continuous function x on J with values in B0
exists such that, for any t in J
x(t) = x0
+ F (x)(t) with F (x)(t)def=
Z t
t0
f(t0, x(t0))dt0.
Proof. Let us denote by X the set of continuous functions on J with values in B0
. Let usconsider x in X. By definition of M
0
and using Condition (1.2), we get, for any t in J
�
�
�
Z t
t0
f(t0, x(t0))dt0�
�
�
E
�
�
�
Z t
t0
kf(t0, x(t0))kEdt0�
�
�
Z
JM
0
(t)dt R0
.
Thus the function F maps X into itself. Then for any positive �, let us consider the metricspace X equipped with the distance
d�(x1, x2)def= sup
t2J
✓
exp���
�
�
Z t
t0
K0
(t0)dt0�
�
�
◆
kx1
(t)� x2
(t)kE .
Let us estimate d�(F (x1
), F (x2
)). In order to do it, let us write that, for t � t0
,
kF (x1
)(t)� F (x2
)(t)kE Z t
t0
kf(t0, x1
(t0))� f(t0, x2
(t0))kE dt0
Z t
t0
K0
(t0)kx1
(t0)� x2
(t0)kE dt0.
9
From this we deduce that, by definition of d�,
K�(t)def= exp
⇣
��Z t
t0
K0
(t0)dt0⌘
kF (x1
)(t)� F (x2
)(t)kE
Z t
t0
K0
(t0) exp⇣
��Z t
t0K
0
(t00)dt00⌘
exp⇣
��Z t0
t0
K0
(t00)dt00⌘
kx1
(t0)� x2
(t0)kE dt0
d�(x1, x2)
Z t
t0
K0
(t0) exp⇣
��Z t
t0K
0
(t00)dt00⌘
dt0
1
�d�(x1, x2).
This proves that the function F is Lipschitz with constant 1/� on X. Thus it has a uniquefixed point in X and the theorem is proved. 2
Remark Let us point out that the concepts of iterative scheme and of Cauchy sequence playsa key role. Moreover, let us notice Condition (1.2) is the only restriction on the interval J .
The existence and uniqueness theorem for ordinary di↵erential equations is a local theorem.Let us first define the concept of maximal solution and then investigate what can be necessaryconditions for a blow up phenomena. For the sake of simplicity, we are going to define it inthe case when the function F is globally defined on the Banach space R⇥E. Because of theuniqueness part of Theorem 1.2.1, if x
1
and x2
are to solution of (ODE) define respectivelyon two interval J
1
and J2
for which t0
is an interior point, then they coincides on J1
\ J2
.Then it is an exercice of calculus to check that, if we are a collection (x�)�2⇤ are solution
of (ODE) defined on J�, then the function the function x on Jdef=[
�2⇤J� by
x|J� = x�
is a solution of (ODE) on J .
Definition 1.2.1 The solution x defined above the called the maximal solution of (ODE).
Proposition 1.2.1 Let F be a function of R⇥E in E satisfying the hypothesis of Theo-rem 1.2.1 in any point x
0
of E. Let us assume in addition that a locally bounded function Mfrom R+ into R+ and a locally integrable function � from R+ into R+ such that
kF (t, u)k �(t)M(kuk).
then, if the maximal interval of definition is ]T?, T ?[, then, if T ? is finite,
lim supt!T ?
ku(t)k = 1,
the same being true for T?.
Proof. Let us first prove that, if we consider a time T > T0
such that ku(t)k is bounded onthe interval [T
0
, T [, then we can define the solution on a larger interval [T0
, T1
] with T1
> T .As the function u is bounded on the interval [T
0
, T [, the hypothesis on F that, for any t ofthe interval [T
0
, T [, we havekF (t, u(t))k C�(t).
10
The function � being integrable on the interval [T0
, T ], we have deduce que, for any " stricte-ment positive, it exists a positive real number ⌘ such that, pour tout t and t0 such that T�t < ⌘and T � t0 < ⌘,
ku(t)� u(t0)k < ".
The space E being complete, an element u? of E exists such that
limt!T ?
u(t) = u?.
Applying Theorem 1.2.1, we construct solution of (ODE) on some [T ?+
, T1
] and the continuousfunction defined by induction on the interval [T
0
, T1
] is a solution of the equation (ODE) onthe interval [T
0
, T1
]. 2
Corollary 1.2.1 Under the hypothesis of Proposition 1.2.1, if we have in addition that
kF (t, u)k Mkuk2,
then, if the interval ]T?, T ?[ is the maximal interval of definition of u and if T ? is finite, then
Z T ?
t0
kx(t)kdt = 1.
Proof. The solution satisfies, for any t � t0
,
kx(t)k kx(t0
)k+M
Z t
t0
kx(t0)k2dt0. (1.3)
Gronwall’s Lemma implies that
kx(t)k kx0
k exp⇣
M
Z t
0
kx(t0)kdt0⌘
.
A more precise way of proving this result is the following. Let us define
Tdef= sup
�
t 2 [t0
, T ?[ / kx(t)k 2kx(t0
)k .
For any t in [t0
, T ?[, we have, using (1.3),
kx(t)k kx(t0
)k+ 4M(t� t0
)kx(t0
)k2.
Thus we infer
8t 2h
t0
,minn
T, t0
+1
4Mkx(t0
k)oh
, kx(t)k 2kx0
k.
Thanks to Proposition 1.2.1, we have
T ? � t0
� c
kx0
k ·
Applying again this result at time t 2 [t0
, T ?[, we find that
8t 2 [t0
, T ?[ , kx(t)k � c
T ? � t·
This proves the theorem. 2
11
1.3 Non linear ordinary di↵erential equations solved by a com-pactness method
In this section, we are going to work only in Rd. This point is important because we shall usethat bounded closed subsets are compact. The main point is the proof of Peano’s theorem.
Theorem 1.3.1 (Peano) Let I be an open interval of R. Let us consider a continuousfunction f from I⇥Rd into Rd. Then, for any point (t
0
, x0
) of I⇥Rd, an open interval J ⇢ Icontaining t
0
and a continuous function x on J exists such that
(ODE) x(t) = x0
+
Z t
t0
f(t0, x(t0)) dt0.
Proof. The structure of the proof is at least as intesting as the result itself. This proof willbe a model for the proof of existence of weak solutions for the incompressible Navier-Stokesequation we shall study in Chapter 3.
There are three steps in the proof:
• we regularize the function f and we apply Cauchy-Lipschitz’s Theorem to the sequenceof regularized functions; have a commun interval of definition,
• then, we prove that the sequence of those solutions of the regularized problem arerelatively compact in the space C(J,Rd),
• as a conclusion, we pass to the limit.
Let us proceed to a classical regularization; let � a non negative function of D(B(0, 1))
the integral of which is 1. Let us define �n(x)def= nd�(nx) and fn(t) = �n ? f(t). We have
kfn(t)kL1(K)
kf(t)kL1(K+B(0,n�1
))
.
Thus two positive real numbers ↵ and R independant of n, such that
8n 2 N ,
Z
J↵
supx2B(x0,R+1)
kf(t, x)kRd dt R. (1.4)
Thus, using Theorem 1.2.1 and the remark page 10, for all integer n, a unique solution xn of
(ODEn) xn(t) = x0
+
Z t
0
fn(t0, xn(t0)) dt0.
exists on J↵.
Now let us prove that the set Xdef= {xn, n 2 N} is relatively compact in C(J↵;Rd). First
of all, we have
8t 2 J , X(t)def= {xn(t), n 2 N} ⇢ B(x
0
, R).
As we work on a finite dimensionnal space, X(t) is relatively compact. Moreover, we have
kxn(t)� xn(t0)k
�
�
�
Z t0
tkfn(t00)kL1
(B(x0,R))
dt00�
�
�
�
�
�
Z t0
tkf(t00)kL1
(B(x0,R+1))
dt00�
�
�
.
12
Thus, for any positive ✏, it exists a positive real number ↵ such that
8(t, t0) 2 J2 , |t� t0| < ↵ =) kxn(t)� xn(t0)k < ✏.
As the function kf(t, ·)kL1(B(x0,R+1)
is integrable on J↵, the family (xn)n2N is equicontinuous
on J↵. Ascoli’s Theorem ensures that the set X is relatively compact in C(J↵;Rd). Thus wecan extract a subsequence which converge uniformely on J↵ to a function x of C(J ;Rd). Letomit to note the extraction in the following.
Now let us pass to the limit. For any t of J↵, we have
kfn(t, xn(t)� f(t, x(t))k kfn(t)� f(t)kL1(B(x0,R))
+ kf(t, xn(t))� f(t, x(t))k.
Thus for any t of J , we have
limn!1 fn(t, xn(t)) = f(t, x(t)).
Moreover, kfn(t, xn(t))k kf(t)kL1(B(x0,R+1))
. Lebesgue’s Theorem ensures that, for any t,we have
limn!1
Z t
t0
fn(t0, xn(t0))dt0 =
Z t
t0
f(t0, x(t0))dt0.
The theorem is proved. 2
1.4 Transport equations
Let us consider a regular time dependant vector field, say continuous in time and continuouslydi↵erentiable with respect to the variable x. Let us consider the flow of v, namely the map uniquely defined by
(EDO)
⇢
@t (t, x) = v(t, (t, x)) (0, x) = x.
For any time t, the map x 7! (t, x) is a C1 di↵eomorphism. Let us describe the link betweenthe (ODE) and the transport equation (T ) given by
(T )
⇢
@tf + v(t, x) ·rf(t, x) = 0f(0, x) = f
0
(x)
Indeed, the chain rule implies that
@t�
f(t, (t, x)�
= (@tf)(t, (t, x)) +dX
j=1
(@jf)(t, (t, x))@t j(t, x)
=�
@tf + v ·rf�
(t, (t, x)).
This implies that f is solution of (T ) if and only if
f(t, (t, x)) = f0
(x) which writes f(t, x) = f0
( �1(t, x)).
In the case when the vector field v is rough, the correspondance between flow and solutions ofthe transport equation is less simple. It turns out that in the case of incompressible flows, thetransport equation (T ) is easier to solve than the associated ordinary di↵erential equation.
13
About incompressible flow, let us remark that it is a classical exercise of calculus to provethat
@t det(D )(t, x) = (div v)(t, (t, x)) det(D )(t, x).
This implies that the incompressibily of the flow is equivalent to the fact that v is divergencefree. In this case of incompressibility, the equation (T ) requires only low regularity to be wellposed (i.e. to have a unique solution).
Let us define an appropriate space for the vector field v.
Definition 1.4.1 Let p be a real number in [1,1]. We define W 1,p(Rd) are the space offunctions a in Lp(Rd) such that the derivatives of a belong to Lp(Rd).
Let us also define the concept of weak continuity for functins f which are bounded in timewith value in Lp(Rd) which means that the function
t 7�! kf(t, ·)kLp(Rd
)
is bounded.
Definition 1.4.2 We denote by Cw,b([0, T ];Lq(Rq)) the set of functions which are boundedwith value in Lp(Rd) such that for any function ' in D(R1+d), the function
8
<
:
[0, T ] �! Rf 7�!
Z
Rdf(t, x)'(t, x)dx
is continuous.
The purpose of this section is the proof of the following theorem.
Theorem 1.4.1 Let q be in ]1,1]. Let us consider a time dependant vector divergencefree vector field v the components of which belong to L1([0, T ];W 1,q0(Rd)). Then for anyfunction f
0
in Lq(Rd), a unique solution f of (T ) exists in Cw,b([0, T ];Lq(Rq)) which meansthat the function f(t, ·) is well defined as a function of Lq(Rd), that kf(t, ·)kLq
(Rd)
is a bounded
function and that, for any ' in D(R1+d), the function
t 7�!Z
Rdf(t, x)'(t, x)dx
is continuous and we have
(Tw)
Z
Rdf(t, x)'(t, x)dx =
Z
Rdf0
(x)'(0, x)dx+
Z t
0
Z
Rdf(t0, x)
�
@t'+ v ·r'�(t0, x)dt0dx.
The proof of this theorem is can be splitted into two steps: the first one is an existencetheorem the proof of which has some analogy with the proof of Peano’s theorem and useAscoli’s theorem. The second step is the uniqueness theorem proved using a duality method.Let us state the existence theorem.
Theorem 1.4.2 Let q be in ]1,1]. Let us consider a time dependant vector divergence freevector field v the components of which belong to L1([0, T ];Lq0(Rd)). Then for any function f
0
in Lq(Rd), a solution f of (T ) exists in Cw([0, T ];Lq(Rq)).
14
Proof. As in the proof of Peano’s theorem, we proceed by regularization and compactness.Let us consider a sequence (vn)n2N a smooth vector fields in R⇥Rd such that
limn!1 kv � vnkL1
([0,T ];Lq(Rd
))
= 0 (1.5)
and a sequence (f0,n)n2N a smooth compactly supported function which tends to f
0
in Lq(Rd).Then let us denote by fn the solution of (T ) associated with the vector field vn and the initialdata f
0,n. Obviously, we have, for any ' in D(R1+d)Z
Rdfn(t, x)'(t, x)dx =
Z
Rdf0,n(x)'(0, x)dx+ In(')(t) with
In(')(t) def=
Z t
0
Z
Rdfn(t
0, x)�
@t'+ vn ·r'�(t0, x)dt0dx.(1.6)
The goal is to pass to the limit in the above expression. Let us first observe that
limn!1
Z
Rdf0,n(x)'(0, x)dx =
Z
Rdf0
(x)'(0, x)dx. (1.7)
As the vector field v is divergence free, we have
8t 2 [0, T ] , kfn(t, ·)kLq(Rd
)
= kf0,nkLq
(Rd)
.
This implies in particular that the sequence (fn)n2N is bounded in Lq([0, T ]⇥Rd). As q is di↵er-ent from 1, we can assume, up to an extraction we omit to note, that the sequence (fn)n2N con-vergence in the weak? sense to a function f in Lq([0, T ]⇥Rq). Moreover, as (fn)n2N is boundedin L1([0, T ];Lq(Rd)) this implies that the function f also belongs to L1([0, T ];Lq(Rd)). Letus prove that such a function f is a solution.
The first step is the proof of the following proposition.
Proposition 1.4.1 The family (In('))n2N is relatively compact in C([0, T ];R).Proof. Holder inequalities imply that
�
�
�
Z t
0
Z
Rdfn(t
0, x)@t'(t0, x)dt0dx�
�
�
kfnkL1([0,T ];Lq
(Rd))
k@t'kL1([0,T ];Lq0
(Rd))
and
�
�
�
Z t
0
Z
Rdfn(t
0, x)�
vn ·r'�(t0, x)dt0dx�
�
�
kfnkL1([0,T ];Lq
(Rd))
kvnkL1([0,T ];Lq0
(Rd))
⇥ kr'kL1([0,T ]⇥Rd
)
.
Thus (In('))n2N is a bounded family of L1([0, T ]). Now let us prove it is uniformely equicon-tinuous. Again Holder inequalities implies that
�
�
�
Z t2
t1
Z
Rdfn(t
0, x)@t'(t0, x)dt0dx�
�
�
kfnkL1([0,T ];Lq
(Rd))
k@t'kL1([0,T ];Lq0
(Rd))
|t1
� t2
| and�
�
�
Z t
0
Z
Rdfn(t
0, x)�
vn ·r'�(t0, x)dt0dx�
�
�
kfnkL1([0,T ];Lq
(Rd))
kvnkL1([t1,t2];Lq0
(Rd))
⇥ kr'kL1([0,T ]⇥Rd
)
.
As the sequence (vn)n2N tends to v in L1([0, T ];Lq(Rd)), then for any positive ", a positivereal number ↵" exists such that
|t1
� t2
| < ↵" =) 8n 2 N , kvnkL1([t1,t2];Lq0
(Rd))
< ".
Using Ascoli’s theorem, this proves the proposition. 2
15
Continuation of the proof of Theorem 1.4.2 . From Ascoli’s theorem, we deduce that, up to anextraction we omit to note, the sequence (In('))n2N converges to some continuous functionwe denote by I('). This implies that the sequence of functions
⇣
Z
Rdfn(t, x)'(t, x)dx
⌘
n2N
converge uniformely on [0, T ] to L(')(t) which satisfies, thanks to (1.6) and
L(')(t) =
Z
Rdf0
(x)'(0, x)dx+ I(')(t). (1.8)
Because on the weak convergence of (fn)n2N to f , this implies that, for any function ↵in D(]0, T [, we have
Z T
0
I(')(t)↵(t)dt = limn!1
Z T
0
⇣
Z
Rdfn(t, x)'(t, x)dx
⌘
↵(t)dt
= limn!1
Z
[0,T ]⇥Rdfn(t, x)'(t, x)↵(t)dtdx
=
Z
[0,T ]⇥Rdf(t, x)'(t, x)↵(t)dtdx
=
Z
[0,T ]
⇣
Z
Rdf(t, x)'(t, x)dx
⌘
↵(t)dt.
In order to compute I(') it is enough to worry about pointwise convergence. This impliesthat f belongs to Cw,b([0, T ];Lq(Rd) and that
limn!1
Z
Rdfn(t, x)'(t, x)dx =
Z
Rdf(t, x)'(t, x)dx uniformely on [0, T ]. (1.9)
Thanks to (1.8), this becomesZ
Rdf(t, x)'(t, x)dx =
Z
Rdf0
(x)'(0, x)dx+ I(')(t). (1.10)
Now let us compute I('). In order to get to result, it is enough now to think interm topointwise convergence. Holder estimates imply that, for any t in [0, T ],
�
�
�
Z t
0
Z
Rdfn(t
0, x)�
(vn � v) ·r'�(t0, x)dt0dx�
�
�
kfnkL1([0,T ];Lq
(Rd))
kvn � vkL1([0,T ];Lq0
(Rd))
⇥ kr'kL1([0,T ]⇥Rd
)
.
Thus
limn!1 sup
t2[0,T ]
�
�
�
In(')(t)�Z t
0
Z
Rdfn(t
0, x)�
@t'+ v ·r'�(t0, x)dt0dx�
�
�
= 0.
As the sequence (fn)n2N tends weakly ? to f as a sequence of functions of Lq([0, T ]⇥Rd), thesequence (In('))n2N tends to
Z t
0
f(t0, x)�
@t'+ v ·r'�(t0, x)dt0dx.
This proves the theorem. 2
16
Then we are going to prove a uniqueness theorem with an additionnal hypohesis of regu-larity about the vector field v.
Theorem 1.4.3 Let q be in ]1,1] and v a time depending divergence free vector field thecomponent of which belongs to L1([0, T ];W 1,q0(Rd)). Let is f is a solution of (Tw) with initialdata 0 i.e. for all ' in D(R1+d),
(Tw)
Z
Rdf(t, x)'(t, x)dx =
Z t
0
Z
Rdf(t0, x)
�
@t'+ v ·r'�(t0, x)dt0dx,
then f = 0.
Proof. The strategy is the following: we regularize f in space by convolution by a smoothcompactly supported function, we establish the equation (Tw,") satisfied by the family ofapproximated solutions and then in this equation, we use as a test function '� which is thesolution of
(T �)
⇢
@t'� + v�(t, x) ·r'�(t, x) = 0'�(t
0
, x) = �(x)
where � is a given smooth compactly supported function on Rd and v� a suitable family ofapproximation of v. Let us define
a"(x)def=
1
"d⇢⇣ ·"
⌘
? a.
where ⇢ is a smooth compactly supported function on Rd, even, non negative and with in-tegral 1. For any compactly supported function ' on D(R1+d), we get, using '" as a testfunction in (Tw) that
(Tw)
Z
Rdf(t, x)'"(t, x)dx =
Z t
0
Z
Rdf(t0, x)
�
@t'" + v ·r'"
�
(t0, x)dt0dx,
As the function ⇢ is even we haveZ
Rdf(t, x)'"(t, x)dx =
Z
Rdf"(t, x)'(t, x)dx and
Z t
0
Z
Rdf(t0, x)@t'"(t
0, x)dt0dx =
Z t
0
Z
Rdf"(t
0, x)@t'(t0, x)dt0dx.(1.11)
By definition of the convolution and because the function ⇢ is even, we have
E def=
Z t
0
Z
Rdf(t0, x)
�
v ·r'"
�
(t0, x)dt0dx�Z t
0
Z
Rdf"(t
0, x)�
v ·r'�(t0, x)dt0dx
=1
"d
dX
j=1
Z t
0
Z
Rdf(t0, x)⇢
⇣x� y
"
⌘
�
vj(t0, x)@j'(t0, y)dt0dx� vj(t0, y)@j'(t0, y)�
dt0dxdy .
Using that the vector field v is divergence free, an integration by parts gives
E =
Z t
0
Z
RdC"(v, f)(t0, y)'(t0, y)dt0dy with
C"(v, f)(t, y)def=
1
"d+1
dX
j=1
Z
Rd(@j⇢)
⇣x� y
"
⌘
�
vj(t0, x)� vj(t0, y)�
f(t0, x)dx.(1.12)
17
Now (Tw) writes
Z
Rdf"(t, x)'(t, x)dx =
Z t
0
Z
Rdf"(t
0, x)�
@t'+ v ·r'�(t0, x)dt0dx
+
Z t
0
Z
RdC"(v, f)(t0, x)'(t0, x)dx.
(1.13)
The main step of the proof of the following proposition
Proposition 1.4.2 For any positive real number ", we have
kC"(v, f)kL1([0,T ]⇥Rd
)
��| · |⇢��L1
(Rd)
kfkL1([0,T ];Lq
(Rd))
krvkL1([0,T ];Lq0
(Rd))
. (1.14)
Moreover, for any function f in the space L1([0, T ];Lq(Rd)) and any divergence free vectorfield v in L1([0, T ];Lq0(Rd)), we have
lim"!0
kC"(v, f)kL1([0,T ]⇥Rd
)
= 0.
Proof. The proof of the first inequality relies on the following lemma, which can be understoodas a Lp version of finite increments inequality.
Lemma 1.4.1 For any p in [1,1[, then for any a in W 1,p the map
⇢
Rd �! Lp(Rd)z 7�! a(·+ z)
is lipschitzian.
Proof. By density, we can assume that a is a smooth compactly supported function. Then wecan write
a(x+ z0)� a(x+ z) =dX
j=1
(z0j � zj)
Z
1
0
@ja�
x+ z + s(z0 � z)�
ds.
Taking the Lp norm in x in the above identity gives
ka(·+ z0)� a(·+ z)kLp dX
j=1
|z0j � zj | k@jakLp .
This proves the lemma. 2
Continuation of the proof of Proposition 1.4.2 Changing variable x = y + "z gives
C"(v, f)(t, y) =1
"
dX
j=1
Z
Rd@j⇢(z)(v
j(t0, y + "z)� vj(t0, y)�
f(t0, y + "z)dz. (1.15)
Holder inequality implies that
kC"(v, f)(t, )kL1(Rq
)
1
"
dX
j=1
Z
Rd|@j⇢(z)| kvj(t0, ·+"z)�vj(t0, ·)kLq0
(Rd)
kf(t0, ·+"z)kL1(Lq
(Rq))
dz.
18
Lemma 1.4.1 implies that
kC"(v, f)(t, )kL1(Rq
)
dX
j=1
Z
Rd|zj | |@j⇢(z)| krvj(t0, ·)kLq0
(Rd)
kf(t0, ·)kLq(Rq
)
dz.
This gives (1.14). Now let us prove the second part of the proposition. Because of (1.14),it is enough to prove it for smooth divergence free vector field. Writing a Taylor formula atorder 2 in (1.15) gives
C"(v, f) =dX
j=1
dX
k=1
C",1j,k (v, f) + C",2
j,k (f) with
C",1j,k (v, f)(t, y)
def= @kv
j(t, y)
Z
Rdzk@j⇢(z)f(t
0, y + "z)dz and
C",2j,k (v, f)(t, y)
def= "
X
`=1
Z
Rdzkz`@j⇢(z)
Z
1
0
(1� t)(@k@`vj)(t, y + �"z)f(t0, y + "z)dz.
Let us observe that we have
kC",2j,k (v, f)kL1
([0,T ]⇥Rd)
"
Z T
0
Z
Rd
Z
1
0
kf(t, ·+ "z)kLq(Rd
)
|z|2|⇢(z)|⇥ kr2vj(t, ·+ s+ "z)kLq0
(Rd)
dsdzdt
Cv,⇢"kfkL1([0,T ];Lq0
(Rd))
. (1.16)
We know that, for almost every t in [0, T ], we have
lim"!0
�
�
�
�
Z
Rdzk@j⇢(z)f(t, ·+ "z)dz �
⇣
Z
Rdzk@j⇢(z)dz
⌘
f(t, ·)�
�
�
�
Lq(Rd
)
= 0.
Because ⇢ is of integral 1, we haveZ
Rdzk@j⇢(z)dz = 0 if j 6= k and
Z
Rdzj@j⇢(z)dz = 1.
We get, for almost every t,
lim"!0
�
�
�
�
dX
j=1
dX
k=1
C",1j,k (f)(t, ·)� div v(t, ·)f(t, ·)
�
�
�
�
L1(Rd
)
= 0
As, for any positive ",
kC",1j,k (v, f)(t, ·)kL1
(Rd)
k@kvj(t, ·)kLq0(Rd
)
kf(t, ·)kLq(Rd
)
,
Lebesgue theorem and the fact that that v is divergence free together with (1.16) implies theconclusion of Proposition 1.4.2. 2
Continuation of the proof of Theorem 1.4.2 Passing to the limit in (Tw) is not su�cient.Indeed, we have, using Proposition 1.4.2 and (1.11)
Z
Rdf(t, x)'(t, x)dx =
Z t
0
Z
Rdf(t0, x)
�
@t'+ v ·r'�(t0, x)dt0dx (1.17)
19
for any function ' in D(R1+d). But we cannot solve
@t'+ v ·r' = 0
with smooth solution '. Thus we need to regularize the vector field v. Let us consider thefamily '� given by (T �) in page 17. Applying (Tw) with '� gives thanks to (1.11) and (1.13),for some t
0
in [0, T ] and for any positive �,
Z
Rdf(t
0
, x)�"(x)dx =
Z t0
0
Z
Rdf(t0, x)
�
@t'�" + v ·r'�
"
�
(t0, x)dt0dx
=
Z t0
0
Z
Rdf(t0, x)
�
(v � v�) ·r'�"
�
(t0, x)dt0dx
+
Z t0
0
Z
Rdf(t0, x)
�
@t'�" + v� ·r'�
"
�
(t0, x)dt0dx
=
Z t0
0
Z
Rdf"(t
0, x)�
(v � v�) ·r'��
(t0, x)dt0dx
+
Z t
0
Z
RdC"(v, f)(t0, y)'�(t0, y)dt0dy.
Now for fixed postive ", let us take the limit when � tends to 0. By integration by parts, weget because v and v� are divergence free,
Z t0
0
Z
Rdf"(t
0, x)�
(v � v�) ·r'��
(t0, x)dt0dx = �Z t0
0
Z
Rd'�(t0, x)
�
(v � v�) ·rf"�
(t0, x)dt0dx.
Maximum principle for transport equation ensures that
8� > 0 , k'�kL1([0,T ]⇥Rd
)
k�kL1(Rd
)
. (1.18)
Thus, using the fact that the vector field v (and thus all the vector fields v�) is divergencefree, we get
�
�
�
Z t0
0
Z
Rdf"(t
0, x)�
(v � v�) ·r'��
(t0, x)dt0dx�
�
�
k�kL1(Rd
)
kv � v�kL1([0,T ];Lq0
(Rd))
krf"kL1([0,T ];Lq
(Rd))
.
By definition of the regularization process of f , we have
krf"kL1([0,T ];Lq
(Rd))
C"�1kfkL1([0,T ];Lq
(Rd))
.
This implies that, for any positive ",
lim�!0
supt2[0,T ]
�
�
�
Z t
0
Z
Rdf"(t
0, x)�
(v � v�) ·r'��
(t0, x)dt0dx�
�
�
= 0. (1.19)
Let (�n)n2N be a sequence of positive real numbers which tends to 0. Thanks to (1.18), we canassume (up to an extraction we omit to note) that the sequence ('�n)n2N converges weakly-?to some function ' in L1([0, T ]⇥ Rd). Thus we get, for any t in [0, T ], and any fixed ",
limn!1
Z t
0
Z
RdC"(v, f)(t0, y)'�n(t0, y)dt0dy =
Z t
0
Z
RdC"(v, f)(t0, y)'(t0, y)dt0dy.
20
Together with (1.19), this gives
Z
Rdf(t
0
, x)�"(x)dx =
Z t
0
Z
RdC"(v, f)(t0, y)'(t0, y)dt0dy.
Proposition 1.4.2 implies that
lim"!0
Z
Rdf(t
0
, x)�"(x)dx =
Z
Rdf(t
0
, x)�(x)dx = 0.
Theorem 1.4.3 is proved. 2
21
22
Chapter 2
The heat flow and the Stokes flowin a bounded domain
Introduction
In this chapter, we first describe the structure of the Laplace operator in a bounded domainwith Dirichlet boundary condition. Then we solve the heat flow with Dirichlet boundarycondition.
In the second section, the investigate the equivalent of the Dirichlet problem for divergencefree vector field still in a bounded domain: the stationnary Stokes problem. The main pointis that the divergence free constrain gives rice a new concept of solution.
In the third section, we solve the Stokes flow which the equivalent of the heat flow in thecontext of divergence free vector field. Again the concept of solutions will be di↵erent of theone of the heat flow; it takes into account the divergence free condition.
2.1 The Dirichlet problem in a bounded domain
We are going to present the solution of classical linear partial di↵erential equations in abounded domain with Dirichlet boudary condition. In order to do so, let us recall the basictheorem about the Laplacian on a bounded domain with Dirichlet boundary condition.
Theorem 2.1.1 For any f in H�1(⌦), a unique solution of the equation ��u = f existsin H1
0
(⌦) in the following sense:
8v 2 H1
0
(⌦) , (rv|r')L2(⌦)
= hf, viH�1(⌦)⇥H1
0 (⌦)
. (2.1)
It also exists a non decreasing sequence (�k)k2N of positive real numbers which tends to infinityand a hilbertian basis of L2(⌦) denoted by (ek)k2N such that
��ek = �kek.
Moreover, the sequence (��1
k ek)k2N is an orthonormal basis of H1
0
(⌦). Finally, if f belongsto H�1(⌦), then
kfk2H�1(⌦)
=X
k
��2
k (hf, eki)2.
Remark Thus, the space H�1(⌦) is a Hilbert space and (�kek)k2N is a hilbertian basisof H�1(⌦).
23
Proof of Theorem 2.1.1. The first assertion comes simply from the fact that the dual spaceof H1
0
(⌦) is described in two di↵erent ways. As H1
0
(⌦) is a Hilbert space, a unique u existsin H1
0
(⌦) such that8v 2 H1
0
(⌦) , (u|v)H10= hf, vi
As we have (u|v)H10= h��u, viH1⇥H1
0, we get that ��u = f in H�1.
Now let us observe that as the space L2 is continuously included in H�1(⌦), we can definean operator B as follows:
B
⇢
L2 �! H1
0
(⌦) ⇢ L2(⌦)f 7�! u
such that u is the solution in H1
0
(⌦) of ��u = f . The operator B is of course continuousfrom L2(⌦) into H1
0
(⌦). Thanks to Rellich’s theorem (see A.4.3 page 91 for the statement andthe proof), the operator B is compact from L2(⌦) into L2(⌦). Let us prove it is self adjointand positive. Let us write that for any couple of functions (f, g) in L2(⌦), we have
(Bf |g)L2 = hg,BfiH�1⇥H10.
By definition of B, we have for any g in L2(⌦), g = �Bg. Thus we infer that
(Bf |g)L2 = h�Bg,BfiH�1⇥H10.
As for any u and v in H1
0
(⌦), we have
h�u, viH�1⇥H10= (u|v)H1
0,
we deduce that(Bf |g)L2 = (Bf |Bg)H1
0.
If f = g, this gives in particular (Bf |f)L2 = kBfk2H1
0. Thus the operator B is compact,
selfadjoint and positive. The spectral theorem about compact selfadjoint operator appliedto B implies the existence of a non increasing sequence (µk)k2N of positive real numberswhich tends to 0 and a hilbertian basis of L2(⌦) denoted (ek)k2N such that, for any k, thefunction ek belongs to L2(⌦) and such that Bek = µkek. This implies that ��ek = µ�1
k ek.We have,
kfkH�1(⌦)
= sup(ck)2Bf
hf,X
k
��1
k ckeki
where Bf denotes the set of sequences having only a finite number of non zero terms and of `2
norm less or equal to to 1. Thus
kfkH�1(⌦)
= sup(ck)2Bf
X
k
��1
k hf, ekick = k(��1
k hf, eki)k2Nk`2(N).
Theorem 2.1.1 is proved. 2
Let us first study the case of the heat equation in a bounded domain with Dirichletboundary condition (which means that the solution ”vanishes on the boundary”). As we arefar away from smooth function, we need a definiton of ”weak solution”.
Definition 2.1.1 Let ⌦ be a bounded domain of Rd. Let us consider a function u continuouson R with value in H�1(⌦) which moreover belongs to L1
loc
(R+;H1
0
(⌦)), a distribution u0
in H�1(⌦) and a function f which belongs to L1
loc
(R+;H�1(⌦)).
24
We say that u is the solution of the heat equation with initial data u0
and external force fwith Dirichlet boundary condition if, for any function � continuously di↵erentiable with valuein H1
0
(⌦),
hu(t),�(t)iH�1⇥H10= hu
0
,�(0)iH�1⇥H10+
Z t
0
h��(t0) + @t�(t0), u(t0)iH�1⇥H1
0dt0
+
Z t
0
hf(t0),�(t0)iH�1⇥H10dt0.
The symbolic writing of this equation is
(HE)
8
<
:
@tu��u = fu|t=0
= u0
u|@⌦ = 0
This type of solution, if it exists, is unique because of the following proposition.
Proposition 2.1.1 Let us consider a solution u of the heat equation with initial data 0 andexternal force 0 with Dirichlet boundary condtion. Then u is identically 0.
Proof. This hypothesis means exacly that, for any function � continuously di↵erentiable withvalue in H1
0
(⌦), we have
hu(t),�(t)iH�1⇥H10=
Z t
0
h��(t0) + @t�(t0), u(t0)iH�1⇥H1
0dt0.
Let us consider the hilbertian basis (ek)k2N given by Theorem 2.1.1. Let us apply the aboveidentity with a the test function �(t) = ek. This gives
hu(t), ekiH�1⇥H10=
Z t
0
h�ek, u(t0)iH�1⇥H1
0dt0.
As �ek = ��2kek, we gives
hu(t), ekiH�1⇥H10= ��2k
Z t
0
hek, u(t0)iH�1⇥H10dt0.
As for almost every t, u(t) belongs to H1
0
(⌦), we get
hek, u(t)iH�1⇥H10= hu(t), ekiH�1⇥H1
0= (u(t)|ek)L2 .
This implies that hu(t), eki = 0 for any t and any k. The proposition is proved. 2
In order to prove the existence, we need some additionnal regularity properties on theinitial data u
0
and the external force f . We have the following theorem.
Theorem 2.1.2 Let u0
be in L2(⌦) and f in L2
loc
(R+;H�1(⌦)). A (unique) solution exists(in the sense of Definition 2.1.1) which belongs to the space
C(R+;L2(⌦)) \ L2
loc
(R+;H1
0
(⌦))
and which satisfies
1
2ku(t)k2L2 +
Z t
0
kru(t0)k2L2dt0 =1
2ku
0
k2L2 +
Z t
0
hf(t0), u(t0)idt0.
25
Proof. Because a uniqueness, we can prove that the solution exists on [0, T ] for any positive T .By time cuto↵, we are reduced to the case when f belongs to L2(R+;H�1(⌦)).
The method used is a regularization method. Let us denote by Ek the orthogonal pro-jection on Vk the space generated by the first k + 1 eigenvectors e`. Let us first prove thefollowing regularization lemma.
Lemma 2.1.1 For any force f in L2(R+;H�1(⌦)), a sequence (fk)k2N of functions of thespace L2(R+;H1
0
(⌦)) exists which converges to f in L2(R+;H�1(⌦)) and morevoer satisfies
8k 2 N , fk 2 C(R1;Vk).
Proof. Thanks to Theorem 2.2.2 and to the Lebesgue Theorem,
limk!1
Ek f = f in L2(R+;H�1(⌦)).
Then, let us consider a sequence (✏n)n2N of positive real numbers which tends to 0 and afunction � smooth and compactly supported from R into R with integral 1. As
limn!1
1
"n�⇣ ·"n
⌘
? (1R+fk) = 1R+fk in L2(R+;Vk),
the lemma is proved. 2
Continuation of the proof of Theorem 2.1.2 Let us observe that the spaces Vk are stable underthe action of �. Thus the heat equation with initial data EN u
0
can be written
d
dtUN = �UN + fN (2.2)
which is a linear ordinary di↵erential equation on Vk the solution of which is given by
UN (t) =NX
k=0
✓
e��2kt(u
0
|ek)L2 +
Z t
0
e��2k(t�t0)hfN (t0), ekidt0
◆
ek. (2.3)
From (2.2), we infer that
1
2
d
dtkUN (t)k2L2 = (�UN |UN (t))L2 + (fN (t)|UN (t))
= �(rUN |rUN (t))L2 + hfN (t), UN (t)i.A time integration gives
1
2kUN (t)k2L2 +
Z t
0
krUN (t0)k2L2dt0 =1
2kEN u(0)k2L2 +
Z t
0
hfN (t0), UN (t0)i dt0. (2.4)
Now let us prove that (UN )N2N is a Cauchy sequence in L1(R+;L2(⌦))\L2(R+;H1
0
(⌦)). Bylinearity, we have
d
dt(UN+P � UN ) = �(UN+P � UN ) = fN+P � fN .
Using again the above energy method, we infer that
1
2kUN+P (t)� UN (t)k2L2 +
Z t
0
krUN+P �rUN (t0)k2L2dt0
=1
2kEN+P u
0
� EN u(0)k2L2 +
Z t
0
hfN+P � fN (t0), UN+P (t0)� UN (t0)i dt0.
26
Using that
hg, vi kgkH�1(⌦)
kvkH10 (⌦)
and that ab 1
2a2 +
1
2b2,
we infer that
kUN+P (t)� UN (t)k2L2 +
Z t
0
krUN+P �rUN (t0)k2L2dt0
kEN+P u0
� EN u(0)k2L2 +
Z t
0
kfN+P (t0)� fN (t0)k2H�1
(⌦)
dt0.
As (EN u0
)N2N is a Cauchy sequence in L2(⌦) and (fN )N2N is a Cauchy sequence in thespace L2(R+;H�1(⌦)), the theorem is proved. 2
One of the important point to notice is that the solution u is in some sense explicite,because given by the following formula which comes immediatly from (2.3) once noticed thatthe sequence (UN )N2N is a Cauchy one.
U(t) =X
k2N
✓
e��2kt(u
0
|ek)L2 +
Z t
0
e��2k(t�t0)hfN (t0), ekidt0
◆
ek. (2.5)
2.2 The stationnary Stokes’s problem
This problem is analogous to the Dirichlet problem, but we work on the set of divergence freevector field. The Laplace equation will become the Stokes equation. Let us first define of thespace we are going to work with.
Definition 2.2.1 Let us denote by V(⌦) (resp. V�(⌦)) the set of vector fields (resp. of diver-gence free vector fields) the componants of which are in H1
0
(⌦). Theses spaces are equippedwith the norm
kuk2V def=
dX
j,k=1
k@jukk2L2 .
Let us denote by H(⌦) the closure of V�(⌦) in the space (L2(⌦))d. Let us denote by V 0(⌦)the space (H�1(⌦))d. For f in V 0 and v in V, we note
hf, viV 0⇥V =dX
j=1
hf j , vjiH�1⇥H10.
We denote by (V�(⌦))� the polar set of V�(⌦), i.e. the set of elements of V 0(⌦) such that,for any v in V�(⌦), we have hf, viV 0⇥V = 0. When no confusion is possible, we omitted tomention the domain ⌦.
When no confusion is possible, we drop ⌦ in the notations.
Remark The space H is stricly included in the space of divergence free vector fields thecomponents fo which are in L2. In fact, when the boundary is regular, it is possible to provethat, if v belongs to the space of vector fields the components of which are in L2(⌦), thenthe value of v · n where n denotes a (local) unit normal vector of the boundary of ⌦ (denotedby @⌦) at the boundary @⌦ makes sense because of the divergence free condition. After asuitable localization, it can be recuded to the following exercise.
27
Exercice 2.2.1 Let us consider L2
�(R3) the set of vector fields of R3 the components of whichare in L2(R3). Let us prove that, for any x
3
and y3
in R, we have
�
�v3(·, x3
)� v3(·, y3
)�
�
H�1(R2
)
C|x3
� y3
| 12 kvkL2(R3
)
.
Let us state the analogous of Dirichlet theorem in this framework. We have the followingtheorem, which is the analogue of the classical Dirichlet theorem.
Theorem 2.2.1 Let f be in V 0. It exists a unique solution u in the space V� of the followingequation
��u� f 2 V��
which means that, for any vector field v of V�, we have
�h�u, viV 0⇥V = hf, viV 0⇥V . (2.6)
This also means that ��u = f in V 0�.
Proof. Let us consider f|V�. It is of course a continuous linear form on V�. Riesz theorem
applied to the Hilbert space (V�, k · kV) implies that a unique vector u exists in V� such that
8h 2 V� , (u|h)V� = hf, hiV 0⇥V .
In fact the scalar product on V� is simply the scalar product on V restricted to V�. Thus,we get
8h 2 V� , (u|h)V� = �h�u, hiV 0⇥V = �h�h, uiV 0⇥V . (2.7)
This gives the theorem. 2
The existence and the uniqueness of a minimum u for the functionnal F can be proved fol-lowing exactly the same lines as in the case of Dirichlet problem. The fact that the di↵erentialof F vanishes at point u implies the relation (2.6).
Remarks
• The fact that a vector field g ofH�1(⌦) belongs the polar set (in the sense of the duality)of V�(⌦) implies in particular that, for any function ' of D(⌦), we have
hgi,�@j'iH�1⇥H10+ hgj , @i'iH�1⇥H1
0= 0
which implies that @jgi � @igj = 0, i.e. the curl of g is identically 0.
• Very simple domains exist such that it exists a smooth vector field which are of diver-gence and of curl identically 0 and which is not the gradient of a function on ⌦.
Let us consider the domain of the plan ⌦def= {x 2 R2 / 0 < R
1
< |x| < R2
} and the vectorfield f defined by (�@
2
log |x|, @1
log |x|). We have the following lemma.
Proposition 2.2.1 The vector field f is curl free, but it is not the gradient of a distribution.
28
Proof. The fact that f is of divergence free is obvious. The fact that its curl is 0 follows from thefact that that the function x 7! log |x| is harmonic on ⌦. Let us assume that f is a gradient of somedistribution �p. As f is a smooth function, so is p is also smooth. Let us consider the flow of f = �rp.By definition of f , its trajectories are periodic. Let us consider a trajectory � from of a point of ⌦such that f 6= 0 (here all points are like this). Thanks to the chain rule, we have
d
dt(p � �)(t) = hDp(�(t)),
d�
dti.
By definition of the gradient, we get
hDp(�(t)),d�
dti =
⇣
rp(�(t))|d�dt
⌘
By definition of �, we infer thatd
dt(p � �)(t) = �|rp(�(t)|2.
The fact that the derivative is negative is in contradiction with the periodicity of the trajectory �.
Proposition 2.2.1 is proved. 2
As shown by the following proposition, belonging to V�� is stronger than being curl free.
Let us admit the following proposition.
Proposition 2.2.2 If f belongs to V��, then it exists p in L2
loc
(⌦) such that f = �rp. If theboundary of ⌦ is a C1 hypersurface, then p belongs to L2(⌦).
Remarks Let us note that the above theorem implies that the vector field f defined inProposition 2.2.1 does not belongs to V�
� in spite of the fact that it is curl free. The twonotions coincide in so called ”simply connected” domain which means essentially that thereno hole in them. Often a generic element of V�
� is denoted by rp.
As in the case of Dirichlet problem, the spectral structure of selfadjoint compact operatorswill give the following result.
Theorem 2.2.2 A non decreasing sequence (�k)k2N of positive numbers which tends to in-finity and an orthonormal basis (ek)k2N of H exist such that the sequence (��1
k ek)k2N is anorthonormal basis of V� and such that
��ek � �2kek 2 V�� which means ��ek = �2kek in V 0
�.
Moreover, if f is in V 0, then
kfk2V 0�=X
k2N��2
k (hf, eki)2 and limn!1
�
�
�
f �nX
k=0
hf, ekiek�
�
�
V 0�
= 0.
Proof. It is very close to the proof of analogous theorem for the Laplacian. As the space H iscontinuously included in V 0, we can define the operator B
B
⇢ H �! V� ⇢ Hf 7�! u
such that u is the solution in V� of ��u = f in V 0�.
The operator B is of course continuous from H into V�. Thanks to Rellich’s theorem, theoperator B is compact from H into H. Let us prove it is self adjoint and positive. Let uswrite that for any couple of functions (f, g) in H, we have
(Bf |g)L2 = hg,BfiV 0⇥V .
29
By definition of B, we have for any g in H, g = �Bg in V 0�. Thus we infer that
(Bf |g)H = hg,BfiV 0�⇥V� = h�Bg,BfiV 0
�⇥V� .
As for any u and v in V�, we have
h�u, viV 0�⇥V� = (u|v)V� ,
we deduce that
(Bf |g)H = (Bf |Bg)V� .
Thus the operator B is compact, selfadjoint and positive. The spectral theorem applied to Bimplies the existence of a non increasing sequence (µk)k2N of positive real numbers whichtends to 0 and a Hilbertian basis of H denoted (ek)k2N such that, for any k, the function ekbelongs to H and such that Bek = µkek. This implies that ��ek = µ�1
k ek in V 0�. We have,
kfkV 0�= sup
(ck)2B0
⌦
f,X
k
��1
k ckek↵
V 0⇥V
where B0
denotes the set of sequences having only a finite number of non zero terms and of `2
norm less or equal to to 1. Thus
kfkV 0�= sup
(ck)2B0
X
k
��1
k hf, ekick = k(��1
k hf, eki)k2Nk`2(N).
The theorem is proved. 2
Let us define some orthogonal projection which will play a key role in that follows.
Definition 2.2.2 Let us denote by Hk the k+1 dimensionnal vector space generated by thefamily (ej)
0jk. For f in V 0, let us state
Pk fdef=
X
jk
hf, ejiV 0⇥V ej and (2.8)
P fdef=
X
j2Nhf, ejiV 0⇥V ej . (2.9)
The map P is called the Leray projection on divergence free vector field.
Let us notice that from Theorem 2.2.2, we have, for all non negative integer k,
8f 2 V 0 kPk fkV 0�
kP fkV 0� kfkV 0 , (2.10)
8f 2 (L2)d kPk fkH kP fkH kfk(L2
)
d and (2.11)
8f 2 V kPk fkV� kP fkV� kfkV . (2.12)
The operator Pk (resp. P) is the orthogonal projection of (L2)d on Hk (resp. H) and also theorthogonal projection of V on Hk (resp. V�).
30
2.3 The Stokes’s flow
Given a positive viscosity ⌫, the evolution Stokes problem reads as follows:
(ES⌫)
8
>
>
<
>
>
:
@tu� ⌫�u = f �rpdiv u = 0u|@⌦ = 0u|t=0
= u0
2 H.
Let us define what a solution of this problem is.
Definition 2.3.1 Let u0
be in H and f in L2
loc(R+;V 0). We shall say that u is a solution
of (ES⌫) with initial data u0
and external force f if and only if u belongs to the space
C(R+;V 0�) \ L1
loc(R+;H) \ L2
loc(R+;V�)
and satisfies, for any in C1(R+;V�),
hu(t), (t)iV 0�⇥V� = (u
0
| (0))H +
Z t
0
⌦
⌫� (t0) + @t (t0), u(t0)
↵
V 0�⇥V�
dt0
+
Z t
0
hf(t0), (t0)iV 0�⇥V� dt
0.
The following theorem holds.
Theorem 2.3.1 The problem (ES⌫) has a unique solution in the sense of the above definition.Moreover this solution belongs to C(R+;H) and satisfies the following energy equality:
1
2ku(t)k2H + ⌫
Z t
0
ku(t0)k2V�dt0 =
1
2ku
0
k2H +
Z t
0
hf(t0), u(t0)iV 0�⇥V� dt
0.
Proof. In order to prove uniqueness, let us consider some function u in C(R+;V 0�)\L2
loc(R+;V�)
such that, for all in C1(R+;V�),
hu(t), (t)iV 0�⇥V� =
Z t
0
⌦
⌫� (t0) + @t (t0), u(t0)
↵
V 0�⇥V�
dt0.
This is valid in particular for the time independent function (t) ⌘ ek where the family vectorfields (ek)k2N is given by Theorem 2.2.2. This gives
hu(t), eki = ⌫
Z t
0
h�ek, u(t0)idt0
Thanks to the spectral Theorem 2.2.2, we get
hu(t), eki = �⌫�2kZ t
0
hek, u(t0)idt0.
This implies that, for any k, hu(t), eki = 0. Thus u ⌘ 0. 2
In order to prove existence, we use the following approximation lemma, the proof of whichis exactly analogous to the proof of Lemma 2.1.1, and is thus omitted.
31
Lemma 2.3.1 For any force f in L2
loc(R+;V 0), a sequence (fk)k2N exists in C1(R+;V�) such
that for any integer k and for any positive t, the vector field fk(t) belongs to Hk, and
limk!1
kfk � fkL2([0,T ];V 0
�)= 0.
Let us consider a sequence (fk)k2N associated with f given by the above lemma and thenlet us consider approximated problem
(ES⌫,k)
⇢
@tuk � ⌫ Pk�uk = fkuk |t=0
= Pk u0.(2.13)
Again thanks to Theorem 2.2.2, it is a linear ordinary di↵erential equation on Hk which hasa global solution uk which is C1(R+;Hk). By an energy estimate in (ES⌫,k) we get that
1
2
d
dtkuk(t)k2H + ⌫kuk(t)k2V�
= hfk(t), uk(t)iV 0�⇥V� .
A time integration gives
1
2kuk(t)k2H + ⌫
Z t
0
kuk(t0)k2V�dt0 =
1
2kPk u(0)k2H +
Z t
0
hfk(t0), uk(t0)iV 0�⇥V� dt
0. (2.14)
In order to pass to the limit, we write an energy estimate for uk � uk+`, which gives
�k,`(t)def=
1
2k(uk � uk+`)(t)k2H + ⌫
Z t
0
kuk(t0)� uk+`)(t0)k2V�
dt0
=1
2k(Pk �Pk+`)u(0)k2H +
Z t
0
h(fk � fk+`)(t0), uk(t0)iV 0⇥V dt0
1
2k(Pk �Pk+`)u(0)k2H +
1
2⌫
Z t
0
kfk(t0)� fk+`(t0)k2V 0
�dt0
+⌫
2
Z t
0
k(uk � uk+`)(t0)k2V�
dt0.
This implies that
�k,`(t) k(Pk �Pk+`)u(0)k2L2 +1
2⌫
Z t
0
k(fk � fk+`)(t0)k2V 0
�dt0.
This implies that the sequence (uk)k2N is a Cauchy sequence in C(R+;H)\L2
loc(R+;V�). Let
us denote by u the limit and prove that u is a solution in the sense of Definition 2.3.1. As uk isa C1 solution of the ordinary di↵erential equation (ES⌫,k), we have, for any in C1(R+;V�),
d
dthuk(t), (t)iV 0
�⇥V� = ⌫h�uk(t), (t)iV 0�⇥V� + hfk(t), (t)iV 0
�⇥V� + huk(t), @t (t)iV 0�⇥V� .
Using Relation (2.7), we get by time integration,
huk(t), (t)iV 0�⇥V� = (Pk u0| (0))H +
Z t
0
⌦
⌫� (t0) + @t (t0), uk(t0)
↵
V 0�⇥V�
dt0
+
Z t
0
hfk(t0), (t0)iV 0�⇥V� dt
0.
Passing to the limit in the above equality gives the existence part of the theorem. Let usprove the energy equality. As (uk)k2N is a Cauchy sequence in L1([0, T ];V�)\L2([0, T ];V�),we get the energy equality passing in the limite into (2.14). 2
32
Remark. The solution is given by the explicit formula
u(t) =X
j2NUj(t)ej with
Uj(t)def= e�⌫µ2
j t(u0
|ej)L2 +
Z t
0
e�⌫µ2j (t�t0)hf(t0), eji dt0. (2.15)
The formula shows that the solution to (ES⌫) depends only on the initial data (of course)and of f|V�
.
33
34
Chapter 3
Leray’s Theorem on Navier-Stokesequations in a bounded domain
In this chapter, we shall prove the existence of global solutions for the incompressible Navier-Stokes system in a bounded domain with Dirichlet boundary conditions which can be formallywritten
(NS)
8
>
>
<
>
>
:
@tv + v ·rv � ⌫�v = �rpdiv v = 0v|t=0
= v0
v|@⌦ = 0.
In all this chapter, ⌦ will denoted a bounded domain of Rd with d in {2, 3}. Let us point outthat no hypothesis is made on regularity of the boundary of ⌦. We want to solve the abovesystem for initial data of finite kinetic energy.
The first section consists in defining the concept of ”turbulent” solution, which is a nonlinear generalization of Definition 2.1.1 page 24, called also Leray solution.
The purpose of the the second section is the proof of the existence of a solution in thespirit of the proof of Peano’s Theorem 1.3.1 page 12. Here we project the equation (NS) onthe space Hk defined in Definition 2.2.2 page 30 This approximated problem is an ordinarydi↵erential equation in the finite dimensionnal space Hk. Then we pass to the limit using thesmoothing e↵ect of the Stokes operator and the time regularity given by the equation (in factwe use Ascoli’s theorem).
In the third section, we prove that the Leray solution is unique and (even stable) in thecase when the space dimension is 2. The main point is that Sobolev embeddings related to theenergy spaces are better in dimension 2. The problem of uniqueness in the three dimensionnalcase is the purpose of the next chapter.
3.1 The concept of turbulent solution
Let us state now the weak formulation of the incompressible Navier–Stokes system (NS⌫).
Definition 3.1.1 We shall say that u is a weak solution of the Navier–Stokes equationson [0, T [⇥⌦ (T can be 1) with Dirichlet boundary condition, with an initial data u
0
in Hand an external force f in L2
loc([0, T [R+;V 0) if and only if u belongs to the space
C([0, T [;V 0�) \ L1
loc([0, T [;H) \ L2
loc([0, T [;V�)
35
and for any function in C1([0, T [;R+;V�), the vector field u satisfies the following condition:
hu(t), (t)iV 0�⇥V� = hu
0
, (0)iV 0�⇥V� +
Z t
0
⌦
⌫� (t0) + @t (t0), u(t0)
↵
V 0�⇥V�
dt0
+
Z t
0
�
u(t0)⌦ u(t0)|r (t0))L2dt0 +Z t
0
hf(t0), (t0)iV 0�⇥V� dt
0 with (u⌦ u)`,k = u`uk
for any t in [0, T [.
Let us point out that this definition make sense because, using Gagliardo–Nirenberg’s inequal-ity stated in Corollary A.2.1 page 88 we get
8u 2 V , kukL4 krukd4L2kuk1�
d4
L2 .
Thus the energy space is included in the space L8dloc
(R+;L4) and as d is less than 4, productof coordinates of u belongs to L1
loc
(R+;L2) The non linear term
�Z t
0
�
u(t0)⌦ u(t0)|r (t0)�L2dt
0
is well defined for u in the energy space E def= L1
loc
(R+;H) \ L2
loc
(R+;V�).
Let us remark that the above relation means that the equality in (NS⌫) must be understoodas an equality in the sense of V 0
�. Now let us state the Leray theorem.
Theorem 3.1.1 Let ⌦ be a domain of Rd and u0
a vector field in H. Then, there existsa global (i.e. defined on [0,1[) weak solution u to (NS⌫) in the sense of Definition 3.1.1.Moreover, this solution satisfies the energy inequality for all t � 0,
1
2ku(t)k2H + ⌫
Z t
0
ku(t0)k2V�dt0 1
2ku
0
k2H +
Z t
0
hf(t0), u(t0)iV 0�⇥V�dt
0. (3.1)
It is convenient to state the following definition.
Definition 3.1.2 A solution of (NS⌫) in the sense of the above Definition 3.1.1 which more-over satisfies the energy inequality (3.1) is called a Leray solution (or a turbulent solutionof (NS⌫).
Let us remark that the energy inequality implies a control on the energy.
Proposition 3.1.1 Any Leray solution u of (NS⌫) satisfies
ku(t)k2H + ⌫
Z t
0
ku(t0)k2V�dt0 ku
0
k2H +1
⌫
Z t
0
kf(t0)k2V 0�dt0.
Proof. By definition of the norm k · kV 0�, we have, as almost every positive t0, u(t0) belongs
to V�, we
hf(t0), u(t)0iV 0�⇥V� kf(t0)kV 0
�ku(t0)kV� . (3.2)
Inequality (3.1) becomes
ku(t)k2H + 2⌫
Z t
0
ku(t0)k2V�dt0 ku
0
k2H + 2
Z t
0
kf(t0)kV 0�ku(t0, ·)kV�dt
0.
Thus, we get, using the fact that 2ab a2 + b2,
ku(t)k2H + ⌫
Z t
0
ku(t0)k2V�dt0 ku
0
k2H +1
⌫
Z t
0
kf(t0)k2V 0�dt0.
Thus the proposition is proved.
36
3.2 The proof of Leray’s theorem
The structure of the proof is the following:
• first a family of approximated solutions (uk)k2N is built in spaces with finite frequenciesby using simple ordinary di↵erential equations results in L2–type spaces.
• Next, thank energy estimates, we prove that, for any positive T , this family is weaklycompact in L2([0, T ];V�) and strongly compact in L1([0, T ];V 0
�)
• Finally the conclusion is obtained by passing to the limit in the weak formulation, takingespecially care of the nonlinear terms.
The study of the non linearity is based on the following simple (but key) lemma.
Lemma 3.2.1 Let us define the bilinear map
Q
⇢ V ⇥ V �! V 0
(u, v) 7�! � div(u⌦ v).(3.3)
For any u and v in V, the following estimates hold. For d in {2, 3}, a constant C exists suchthat, for any ' in V,
hQ(u, v),'i Ckrukd4L2krvk
d4L2kuk1�
d4
L2 kvk1�d4
L2 kr'kL2 .
Moreover for any u in V� and any v in V, hQ(u, v), vi = 0.
Proof. The first two inequalities follow directly from Gagliardo–Nirenberg’s inequality statedin Corollary A.2.1 page 88, once noticed that
hQ(u, v),'i ku⌦ vkL2kr'kL2
kukL4kvkL4kr'kL2 .
In order to prove the third assertion, let us assume that u and v are two vector fields thecomponents of which belong to D(⌦). Then we deduce from integrations by parts that
hQ(u, v), vi = �Z
⌦
(div(u⌦ v) · v)(x) dx
= �dX
`,m=1
Z
⌦
@m(um(x)v`(x))v`(x) dx
=dX
`,m=1
Z
⌦
um(x)v`(x)@mv`(x) dx
= �Z
⌦
|v(x)|2 div u(x) dx� hQ(u, v), vi.
Thus, we have
hQ(u, v), vi = �1
2
Z
⌦
|v(x)|2 div u(x) dx.
The two expressions are continuous on V and by definition, D is dense in V. Thus the aboveformula is true for any (u, v) in V ⇥ V, which completes the proof. 2
37
Now let us proceed to the proof of Leray’s theorem step by step.
Construction of approximated solutions
We use the projections Pk introduced in Definition 2.2.2. Let us introduce the function
Fk
⇢ Hk �! Hk
v 7�! Pk(div(v ⌦ v)).
The properties of Fk will be the consequence of the following lemma which will be very useful.
As a corolllary, we get that Fk is locally Lipschitz in Hk and that Fk satisfies
kFk(v)kHk. �
d4k kvk2Hk
. (3.4)
Thanks to Theorem 2.2.2 and to the above lemma, we can solve the following ordinarydi↵erential equation
(NS⌫,k)
( dukdt
(t) = ⌫ Pk�uk(t) + Fk(uk(t)) + fk(t)
uk(0) = Pk u0
where (fk)k2N a the sequence of approximation of f given by Lemma 2.3.1. Theorem 2.2.2implies that Pk� is a linear map from Hk into itself. Thus the continuity properties of Qand Pk allow to apply the Cauchy–Lipschitz theorem. This gives the existence of Tk in ]0,+1]and a unique maximal solution uk of (NS⌫,k) in C1([0, Tk[;Hk). To prove that Tk = +1, letus use the energy estimate
1
2
d
dtkuk(t)k2L2 = �⌫kruk(t)k2L2 + (Fk(uk)|uk)L2 + (fk(t
0)|uk(t))L2 .
Because of Lemma 3.2.1, we get
1
2
d
dtkuk(t)k2H = �⌫kruk(t)k2L2 + (fk(t
0)|uk(t))L2
= �⌫kuk(t)k2V�+ hfk(t), uk(t)iV 0⇥V .
By integration in time, we infer that
1
2kuk(t)k2H + ⌫
Z t
0
kuk(t0)k2V�dt0 =
1
2kuk(0)k2H +
Z t
0
hfk(t0), uk(t0)iV 0⇥Vdt0. (3.5)
Using (3.2) and the (well known) fact that 2ab a2 + b2, we get
kuk(t)k2H + ⌫
Z t
0
kuk(t0)k2V�dt0 kuk(0)k2H +
1
⌫
Z t
0
kfk(t0)k2V 0�dt0 . (3.6)
Using Corollary 1.2.1 page 11, we get that, for all k, Tk is infinite and the above estimate isvalid for any time t.
Now let us write the ordinary di↵erential equation on Hk in terms of Definition 3.1.1. Forany function in C1(R+;V�), the function
t 7�! (uk(t)| (t))L2 = huk(t), (t)iV 0�⇥V�
is a C1 function and we have
d
dthuk(t), (t)iV 0
�⇥V� =Dduk
dt, (t)
E
V 0�⇥V�
+D
uk(t),d
dt
E
V 0�⇥V�
38
Thanks to (NS⌫,k), we get
d
dthuk(t), (t)iV 0
�⇥V� =⌦
⌫�uk(t) + Fk(u(t)) + fk(t), (t)↵
V 0�⇥V�
+D
uk(t),d
dt
E
V 0�⇥V�
=⌦
⌫� (t) + @t (t), uk(t)↵
V 0�⇥V�
+�
uk(t0)⌦ uk(t
0)|rPk (t0)�
L2 + hfk(t0), (t0)iV 0⇥V .
By time integration, this gives,
huk(t), (t)iV 0�⇥V� = (Pk u0| (0))H +
Z t
0
⌦
⌫� (t0) + @t (t0), uk(t0)
↵
V 0�⇥V�
dt0
= �Z t
0
�
uk(t0)⌦ uk(t
0)|rPk (t0)�
L2dt0 +Z t
0
hfk(t0), (t0)iV 0⇥V dt0 .(3.7)
The problem consists now to pass to the limit in the above formula.
The relative compactness of the family (uk)k2NThis is described by the following proposition.
Proposition 3.2.1 It exists an extraction � and a function u belonging to the space
C(R+;V 0�) \ L1
loc
(R+;H) \ L2
loc
(R+;V�)
such that for any positive real number T ,
8 2 L2([0, T ];V�) , limk!1
Z T
0
(u�(k)(t)| (t))V�dt =
Z T
0
(u(t)| (t))V�dt ,
limk!1
ku�(k) � ukL1([0,T ];V 0
�)= 0 and lim
k!1ku�(k) � ukL2
([0,T ];L4)
= 0 .
Proof. The key point is the proof of the following lemma.
Lemma 3.2.2 For any positive time T , the set U(T )def=�
uk |[0,T ]
, k 2 N
is relativelycompact in the set of continuous fonction on [0, T ] with value in V 0
�.
Proof. We proof that for any positive ", the set U(T ) can be covered by a finite number of ballof radius ". In order to do it, we use Theorem 2.2.1 which implies that for any couple (k
0
, k) ofpositive integers such that k is greater than or equal to k
0
, for any non negative real number t,we get
kuk(t)� Pk0 uk(t)k2V 0�=
X
j�k0+1
��2jhuk(t), eji2.
Using that the sequence (�j)j is non decreasing, we get, by Theorem 2.2.2,
kuk(t)� Pk0 uk(t)k2V 0�
��2
k0+1
X
j
�2j huk(t), eji)2
��2
k0+1
kukk2L1([0,T ];H)
.
As limj!1
�j = +1 and as (uk)k2N is a bounded sequence of L1([0, T ];H), we get that
8" , 9k0
/ 8k , kuk � Pk0 ukkL1([0,T ];V 0
�)<"
2· (3.8)
39
Now, let us prove that the sequence (Pk0 uk)k2N is relativement compact in C([0, T ];Hk0).As all the norm are equivalent on the finite dimensionnal vector space Hk0 , let us con-sider Hk0equipped with the norm V 0
�. It turns out that
�
�
�
Pk0
dukdt
(t)�
�
�
V 0�
kuk(t)kV� + kuk(t)k2�d2
H kuk(t)kd2V�.
Using the energy estimate (3.6), we infer that⇣
Pk0
dukdt
⌘
k2Nis a bounded sequence of the
space L4d ([0, T ];V 0
�) which means that
8k 2 N ,�
�
�
Pk0
dukdt
�
�
�
L4d([0,T ];V 0
�)
T 1� 2d ku
0
kH + Cku0
k2H.
Thus, by integration and Holder estimate, we get, for any (t, t0) in [0, T ]2,
kPk0 uk(t)� Pk0 uk(t)kV 0� |t� t0|1� 4
d�
T 1� 2d ku
0
kH + Cku0
k2H�
. (3.9)
Moreover, for any t in [0, T ], the set {Pk0 uk(t), k 2 N} in bounded in the finite dimensionnalspace Hk0 . Thus it is relatively compact in C([0, T ];Hk0). Together with (3.9), this allowsto apply Ascoli’s compactness theorem. Thus, in particular, the set {Pk0 uk, k 2 N} canbe recovered by a finite number of balls of radius "/2 in the norm L1([0, T ];V 0
�). Togetherwith (3.8), this proves the lemma. 2
Continuation of the proof of Proposition 3.2.1 Now let is observe that as (uk)k2N is boundedin the Hilbert space L2([0, T ];V�), it is weakly convergent in this Hilbert space to some v(up to an omitted extraction). Let us prove that u = v on ]0, T [⇥⌦. Let us consider anyfonction ↵ in D(]0, T [) and any non negative integer k. The weak convergence in L2([0, T ;V�)implies that
limn!1
Z T
0
↵(t)�2khun(t), ekidt =Z T
0
↵(t)�2khv(t), ekidt.
The fact that (un)n2N converges to u in C([0, T ];V 0�) implies that
limn!1
Z T
0
↵(t)��1
k hun(t), ekidt =Z T
0
↵(t)��1
k hu(t), ekidt.
This implies that, for any fonction ↵ in D(]0, T [ and any non negative integer k,
Z T
0
↵(t)hu(t), ekidt =Z T
0
↵(t)hv(t), ekidt
which gives that, for any t, hu(t), eki = hv(t), eki and thus u ⌘ v.
Now let use a diagonal process. Let us consider an increasing sequence (Tp)p2N of positivereals numbers tending to infinity. The above two compactness result imply the existence ofa sequence of extraction (�p)p2N, of a sequence (eup)p2N of functions such that eup belongsto C([0, Tp];V 0
�), and a sequence (evp)p2N such that ev belongs to L2([0, Tp];V�) which satisfies,for all p,
limk!1
ku�1�···��p(k) � eupkL1([0,Tp];V 0
�)= 0 and
8 2 L2([0, Tp];V�) , limk!1
Z Tp
0
(u�1�···��p(k)(t)| (t))V�dt =
Z Tp
0
(evp(t)| (t))V�dt
40
Moreover, as for any t, the sequence (uk(t))k2N is bounded in H, so, for any t in Tp, eup(t)belongs to H. Moreover, because of the uniqueness of the limit we have for any p0 less than p,
eup0 = eup|[0,Tp]and evp0 = evp|[0,Tp]
.
Thus we can define u and v on R+ by
u|[0,Tp]= eup and v|[0, Tp] = evp.
As both convergence imply convergence in the sense of distributions we get that u = v. Nowlet us define �(k) = �
1
� · · · � �k(k). It is obvious that, for any p,
limk!1
ku�(k) � ukL1([0,Tp];V 0
�)= 0 and
8 2 L2([0, Tp];V�) , limk!1
Z Tp
0
�
u�(k)(t)| (t)�
V�dt =
Z Tp
0
(u(t)| (t))V�dt.
Let us prove the last convergence of the proposition. As d 4, using Gagiardo-Nirenberginequality (see Corollary A.2.1 page 88) we get
ku�(k)(t)� u(t)k2L4 Cku�(k)(t)� u(t)k2�d2
L2 kru�(k)(t)�ru(t)kd2L2 (3.10)
Moreover, using Cauchy Schwarz inequality, we can write that
kvk2H =1X
j=0
��1
j hv, eji�jhv, eji
✓ 1X
j=0
��2
j hv, eji2◆
12✓ 1X
j=0
�2j hv, eji2◆
12
= kvkV 0�kvkV� .
Plugging this in (3.10) gives
ku�(k)(t)� u(t)k2L4 Cku�(k)(t)� u(t)k1�d4
V 0�
kru�(k)(t)�ru(t)k1+d4
L2
By time integration this gives
ku�(k) � uk2L2([0,T ];L4
)
Cku�(k) � uk1�d4
L1([0,T ];V 0
�)kru�(k) �ruk1+
d4
L1+ d4([0,T ];L2
)
.
As d is less than 4, Holder estimate implies that
ku�(k) � uk2L2([0,T ];L4
)
CT12� d
8 ku�(k) � uk1�d4
L1([0,T ];V 0
�)kru�(k) �ruk1+
d4
L2([0,T ];L2
)
.
As (uk)k2N is bounded in L2([0, T ];V�), We infer using Proposition 3.2.1 that
limk!1
ku�(k) � ukL2([0,T ];L4
)
= 0.
The proposition is proved. 2
41
Conclusion of the proof the proof Theorem 3.1.1 The problem is to pass to the limit inRelation (3.7). We skip the notation of the extraction ' in that follows.
First of all, because of Proposition 3.2.1 and Relation (2.7), we have
limk!1
huk(t), (t)iV 0�⇥V� = hu(t), (t)iV 0
�⇥V� and (3.11)
limk!1
Z t
0
h@t (t0) +� (t0), uk(t0)iV 0�⇥V�dt
0 =Z t
0
h@t (t0) +� (t0), u(t0)iV 0�⇥V�dt
0. (3.12)
Now the main problem consists in passing to the limit in the non linear term
NLk(t)def=
Z t
0
�
uk(t0)⌦ uk(t
0)|rPk (t0)�
L2dt0.
This implies that
limk!1
⇣
NLk(t)�Z t
0
�
u(t0)⌦ u(t0)|rPk (t0)�
L2dt0⌘
= 0. (3.13)
Let us observe that, for any d 4, u ⌦ u belongs to L1
loc
(R+;L2). Now let us observe that,for have function a in L2 and any function b in H1
0
, we haveZ
⌦
a(x)@jb(x)dx = �h@ja, biH�1⇥H10.
Thus, for almost every t0, we have�
u(t0)⌦ u(t0)|rPk (t0)�
L2 = hdiv(u(t0)⌦ u(t0),Pk (t0)iV 0⇥V .
For any t0, we have
limk!1
kPk (t0)� (t0)kV� = 0 and
�
�hdiv(u(t0)⌦ u(t0),Pk (t0)iV 0⇥V
�
� ku(t0)k2L4k (t0)kV .
Lebesgue theorem implies that
limk!1
Z t
0
�
u(t0)⌦ u(t0)|rPk (t0)�
L2dt0 =
Z t
0
�
u(t0)⌦ u(t0)|r (t0)�L2dt
0.
Then using (3.11)–(3.13) and observing that Pk u0 tends to u0
in H, we infer that
hu(t), (t)i = (u0
)| (0))H �Z t
0
h@t (t0) +� (t0), u(t0)iV 0�⇥V�dt
0
= �Z t
0
�
u(t0)⌦ u(t0)|r (t0)�L2dt
0 +Z t
0
hf(t0), (t0)iV 0�⇥V� dt
0(3.14)
which claims that u is a (weak) solution of the incompressible Navier-Stokes system.
It remains to prove the energy inequality (3.1). Let us start from Relation (3.5) which canbe written
1
2kuk(t)k2L2 + ⌫
Z t
0
kruk(t0)k2L2dt0 =
1
2kuk(0)k2L2 +
Z t
0
hfk(t0), uk(t0)idt0. (3.15)
Let us notice that�
�hfk(t)� f(t), uk(t)iV 0�⇥V�
�
� kfk(t)� f(t)kV 0�kuk(t)kV� .
42
As the sequence (uk)k2N is bounded in L2([0, T ];V�) and as the sequence (fk)k2N tends to fin L2
loc
(R+;V 0�), we infer that
limk!1
⇣
Z t
0
hfk(t0), uk(t0)iV 0�⇥V�dt
0 �Z t
0
hf(t0), uk(t0)iV 0�⇥V�dt
0⌘
= 0.
As V� is dense in V 0� and as Pk u0 tends to u
0
in H, we infer that
limk!1
⇣1
2kuk(0)k2L2+
Z t
0
hfk(t0), uk(t0)iV 0�⇥V�dt
0⌘
=1
2ku(0)k2L2 +
Z t
0
hf(t0), u(t0)iV 0�⇥V�dt
0.(3.16)
Proposition 3.2.1 implies in particular that for any time t � 0 and any v in V�,
limk!1
(uk(t)|v)H = limk!1
huk(t), v)iV 0�⇥V� = hu(t), v)iV 0
�⇥V� = (u(t)|v)H.
As V� is dense in H, we get that for any t � 0, the sequence (uk(t))k2N converges weaklytowards u(t) in the Hilbert space H. Hence
ku(t)k2H lim infk!1
kuk(t)k2H for all t � 0.
On the other hand, (uk)k2N converges weakly to u in L2
loc([0, T ];V), so that for all non nega-tive t, we have
Z t
0
kru(t0)k2L2 dt0 lim infk!1
Z t
0
kruk(t0)k2L2 dt0.
The use of (3.15) and of (3.16) gives energy inequality (3.1) and Leray theorem is proved.
3.3 Stability of Leray solutions in dimension two
In a two dimensional domain, the Leray weak solutions are unique and even stable. Moreprecisely, we have the following theorem.
Theorem 3.3.1 For any data u0
in H and f in L2
loc(R+;V 0), the Leray weak solution is
unique. Moreover, it belongs to C(R+;H) and satisfies, for any (s, t) such that 0 s t,
1
2ku(t)k2H + ⌫
Z t
sku(t0)k2V�
dt0 =1
2ku(s)k2H +
Z t
shf(t0), u(t0)iV 0
�⇥V� dt0. (3.17)
Furthermore, the Leray solutions are stable in the following sense. Let u (resp. v) be theLeray solution associated with u
0
(resp. v0
) in H and f (resp. g) in the space L2
loc(R+;V 0)
then,
k(u� v)(t)k2H + ⌫
Z t
0
ku(t0)� v(t0)k2V�dt0
✓
ku0
� v0
k2H +1
⌫
Z t
0
k(f � g)(t0)k2V 0�dt0◆
exp⇣CE2(t)
⌫4
⌘
with
E(t)def= min
⇢
ku0
k2H +1
⌫
Z t
0
kf(t0)k2V 0�dt0 , kv
0
k2H +1
⌫
Z t
0
kg(t0)k2V 0�dt0�
.
43
Proof. The main point is that because of the Gagliardo-Nirenberg inequality in dimension 2,the non linear term Q(u, u) can be thought as an good external force for the linear Stokesproblem. Indeed, as u belongs to L1
loc(R+;H)\L2
loc(R+;V�), thanks to Lemma 3.2.1 page 37,
the non linear term Q(u, u) belongs to L2
loc(R+;V 0). Thus u is the solution of (ES⌫) with
initial data u0
and external force f + Q(u, u). Theorem 2.3.1 immediately implies that ubelongs to C(R+;H) and satisfies, for any (s, t) such that 0 s t,
1
2ku(t)k2H + ⌫
Z t
sku(t0)k2V�
dt0 =1
2ku(s)k2H
+
Z t
shf(t0), u(t0)iV 0
�⇥V� dt0 +Z t
shQ(u(t0), u(t0)), u(t0)iV 0
�⇥V� dt0.
Using Lemma 3.2.1, we get the energy equality (3.17).
To prove the stability, let us observe that, by di↵erence wdef= u�v is the solution of (ES⌫)
with data u0
� v0
and external force f � g +Q(u, u)�Q(v, v), Theorem 2.3.1 implies that
kw(t)k2H + 2⌫
Z t
0
kw(t0)k2V�dt0 = kw(0)k2H
+ 2
Z t
0
h(f � g)(t0), w(t0)iV 0�⇥V� dt
0 + 2
Z t
0
h(Q(u, u)�Q(v, v))(t0), w(t0)iV 0�⇥V� dt
0.
The non linear term is estimated thanks to the following lemma.
Lemma 3.3.1 In two dimensional domains, if a and b belong to V�, we have
|hQ(a, a)�Q(b, b), a� bi| Ckr(a� b)k32L2ka� bk
12L2krak
12L2kak
12L2 .
Proof. Lemma 3.2.1 implies that the quantity hQ(a, b), ai is well defined. AshQ(b, b� a), b� ai = 0,
we have
hQ(a, a)�Q(b, b), a� bi = hQ(a, a)�Q(b, a), a� bi= hQ(a� b, a), a� bi. (3.18)
Using again Lemma 3.2.1, we get the result. 2
Continuation of the proof of Theorem 3.3.1. Using that 2ab a2 + b2, we get
kw(t)k2H +3
2⌫
Z t
0
kw(t0)k2V�dt0 kw(0)k2L2 +
2
⌫
Z t
0
k(f � g)(t0)k2V 0�dt0
+ C
Z t
0
kw(t0)k32V�kw(t0)k
12Hku(t0)k
12V�ku(t0)k
12H dt0.
Using (with ✓ = 1/4) the convexity inequality
ab ✓a1✓ + (1� ✓)b1�
1✓ (3.19)
we infer that
kw(t)k2H + ⌫
Z t
0
kw(t0)k2V�dt0 kw(0)k2L2 +
2
⌫
Z t
0
k(f � g)(t0)k2V 0�dt0
+C
⌫3
Z t
0
kw(t0)k2H�ku(t0)k2V�
ku(t0)k2H�
dt0.
44
Gronwall’s lemma implies that
kw(t)k2H + ⌫
Z t
0
kw(t0)k2V�dt0
⇣
kw(0)k2H +2
⌫
Z t
0
k(f � g)(t0)k2V 0�dt0⌘
⇥ exp
✓
C
⌫3sup
t02[0,t]ku(t0)k2H
Z t
0
ku(t0)k2V�dt0◆
.
The energy estimate tells us that
supt02[0,t]
ku(t0)k2L2
Z t
0
ku(t0)k2V�dt0 1
⌫
✓
ku0
k2H +2
⌫
Z t
0
kf(t0)k2V 0�dt0◆
2
.
As u and v play the same role, the theorem is proved. 2
Remarks
• This chapter must be known.
• If you want to know more about the basis of the subject, you can read the seminal paperof J. Leray ”Essai on the mouvement of a liquide visqueux emplissant the space, ActaMatematica, 63, 1933, pages 193–248.
• To have a more recent review of results on incompressible Navier-Stokes, we can see thebooks of P. Constantin and C. Foias Navier-Stokes equations, Chicago University Press,1988 and of P.-G. Lemarie-Rieusset, Recent developments in the Navier-Stokes problem.Chapman & Hall/CRC, Research Notes in Mathematics, 431, 2002.
• If you are interested in developments related to geophysical fluids, you can see the bookof J.-Y. Chemin, B. Desjardins, I. Gallagher and E. Grenier, Mathematical Geophysics;an introduction to rotating fluids and Navier-Stokes equations, Oxford Lecture series inMathematics and its maps, 32, Oxford University Press, 2006.
45
46
Chapter 4
Stability of Navier-Stokes equationsin dimension 3
Introduction
This chapter investigates the stability of the turbulent solutions constructed in the previouschapter in the case when ⌦ is a bounded domain of R3. This question remains open ingeneral ans is one of the most challeging question in the field of non linear paertial di↵erentialequations.
In the first section, we establish a regularity criteria for a turbulent solution to be stableamong all turbulent solution.
The purose of the second section is the proof of the existence of a stable solution in the casewhen the initial data is more regular (roughly speaking belongs to the Sobolev space H
12 ).
We obtain local in time exsitence of such a solution which becomes global when the intialdata is small enough in a suitable sense.
4.1 A su�cient condition of 3D stability
The stability result is the following.
Theorem 4.1.1 Let u be a Leray solution associated with initial velocity u0
inH and externalforce f in L2([0, T ];V 0). We assume that u belongs to the space L4([0, T ];V�) for somepositive T . Then u is unique, belongs to C([0, T ];H) and satisfies, for any (s, t) such that 0 s t T ,
1
2ku(t)k2H + ⌫
Z t
sku(t0)k2V�
dt0 =1
2ku(s)k2H +
Z t
shf(t0), u(t0)iV 0
�⇥V� dt0. (4.1)
Moreover, let v be any Leray solution associated with v0
in H and g in L2
loc([0, T ];V 0). Then,for all t in [0, T ],
k(u� v)(t)k2H + ⌫
Z t
0
ku� v(t0)k2V�dt0
✓
ku0
� v0
k2H +2
⌫
Z t
0
k(f � g)(t0)k2V 0�dt0◆
exp⇣C
⌫3
Z t
0
ku(t0)k4V�dt0⌘
.
47
Proof. Thanks to Lemma 3.2.1, the fact that u belongs to L4([0, T ];V�) implies that
kQ(u, u)kL2([0,T ];V 0
)
Ckuk12L1
([0,T ];L2)
kuk32
L3([0,T ];H1
0 )
CT18 kuk
12L1
([0,T ];L2)
kuk32
L4([0,T ];H1
0 ). (4.2)
Hence the non linear term Q(u, u) belongs to L2([0, T ];V 0). Thus, exactly as in the twodimensional case, u is the solution of (ES⌫) with initial data u
0
and external force f+Q(u, u).Theorem 2.3.1 immediately implies that u belongs to C([0, T ];H) and satisfies, for any (s, t)such that 0 s t,
1
2ku(t)k2H + ⌫
Z t
sku(t0)k2V�
dt0 =1
2ku(s)k2H
+
Z t
shf(t0), u(t0)iV 0⇥V dt0 +
Z t
shQ(u(t0), u(t0)), u(t0)iV 0
�⇥V� dt0.
Using Lemma 3.2.1, we get the energy equality (4.1).
As u and v are two Leray solutions, we can write that
�⌫(t)def= k(u� v)(t)k2H + 2⌫
Z t
0
ku(t0)� v(t0)k2V�dt0
= ku(t)k2H + 2⌫
Z t
0
ku(t0)k2V�dt0 + kv(t)k2H + 2⌫
Z t
0
kv(t0)k2V�dt0 � 2P (t) with
P (t)def= (v(t)|u(t))H + 2⌫
Z t
0
(v(t0)|u(t0))V� dt0
This can be written
�⌫(t) ku0
k2H + kv0
k2H � 2P (t) + 2
Z t
0
hg(t0), v(t0)i dt0 + 2
Z t
0
hf(t0), u(t0)i dt0. (4.3)
The idea is to use the solution u as a test function in Definition 3.1.1 page 35 in order toestimate �2P (t). The following lemma makes it possible.
Lemma 4.1.1 Let v be a weak solution of (NS) in the sense of Definition 3.1.1 page 35.Then for any function such that
@t 2 L2
loc(R+;V 0�) , 2 C(R+;H) and 2 L4
loc(R+;V�) (4.4)
we have
hv(t), (t)iV 0�⇥V� = (v
0
| (0))H +
Z t
0
h@t (t0) +� (t0), v(t0)iV 0�⇥V�dt
0
�Z t
0
hdiv(v(t0)⌦ v(t0)), (t0)iV 0�⇥V� dt
0 +Z t
0
hg(t0), (t0)iV 0�⇥V� dt
0 .
Proof. It relies on density argument. If a function satisfies (4.4), then, by Lebesgueconvergence theorem, we get, for any time T ,
limk!1
k@t Pk � @t kL2([0,T ];V 0
�)= lim
k!1kPk � kL1
([0,T ];H)
= limk!1
kPk � kL4([0,T ];V�)
= 0.
48
Then the following time regularization
",kdef=
1
"�⇣ ·"
⌘
? 1[0,T ]
Pk
approximate by C1 functions with value in V�. Then, we have to check that we can passto the limit in all terms of the definition of weak solution. All the terms are obvious exceptthe term
Z t
0
hdiv(v(t0)⌦ v(t0)), (t0))V 0�⇥V� dt
0
From Gagliardo–Nirenberg’s inequality stated in Corollary A.2.1 page 88, we get that v be-
longs L83loc(R
+, L4) and thus v ⌦ v belongs to L43loc(R
+;L2). As we have convergence of thefamily ",k in L4
loc
(R+;V�), we can pass to the limit. 2
Continuation of the proof of Theorem 4.1.1 The idea is to use u as a test function for the(weak) solution v. This gives
(v(t)|u(t))H = (v0
|u0
)H +
Z t
0
⇣
�⌫(v(t0)|u(t0))V� + h@tu(t0), v(t0)iV 0⇥V⌘
dt0
�Z t
0
hdiv v ⌦ v, u(t0)idt0 +Z t
0
hg(t0), u(t0)i dt0 .
The fact that u is a solution of Navier-Stokes with initial data u0
and external force f meansexactly that, for any t0,
@tu(t0) = ⌫�u(t0)� div u⌦ u(t0) + f(t0) in the space L2
loc
(R+;V 0�).
As for almost every t0, v(t0) belongs to V�, we infer that
Z t
0
h@tu(t0), v(t0)iV 0�⇥V�dt
0 =Z t
0
⌦
⌫�u(t0)� div(u(t0)⌦ u(t0)) + f(t0), v(t0)↵
V 0�⇥V�
dt0.
Moreover, for almost every t0, we have (v(t0)|u(t0))V� = �h�u(t0), v(t0)i. This gives
(v(t)|u(t))L2 = (v0
|u0
)L2 � 2
Z t
0
(v(t0)|u(t0))V�dt0
�Z t
0
�
⌫hdiv v ⌦ v, ui+ hdiv u⌦ u, vi�dt0 +Z t
0
�hf(t0), v(t0)i+ hg(t0), u(t0)i�dt0 .
Using (4.3), this gives
�⌫(t) ku0
� v0
k2L2�Z t
0
�hdiv v ⌦ v, ui+ hdiv u⌦ u, vi�dt0
+ 2
Z t
0
h(f � g)(t0), (u� v)(t0)i dt0 .(4.5)
As hdiv v ⌦ v, vi = hdiv u⌦ u, ui = 0, we have
hdiv v ⌦ v, ui+ hdiv u⌦ u, vi = hdiv v ⌦ v � u⌦ u, u� vi= hdiv v ⌦ (v + u� v)� u⌦ u, u� vi= hdiv(v � u⌦ u, u� vi.
49
Using the divergence free condition, we
�
�hdiv v ⌦ v, ui+ hdiv u⌦ u, vi�� ku� vk2L4krukL2 .
Using Gagliardo–Nirenberg’s inequality stated in Corollary A.2.1 page 88 and the convexityinequality (3.19) page 44, we infer
hdiv v ⌦ v, ui+ hdiv u⌦ u, vi ⌫34 kr(u� v)k
32L2C⌫
� 34 krukL2ku� vk
14L2
1
2⌫kruk2L2 +
C
⌫3kru(t)k4L2ku� vk2L2 .
Moreover, we have
h(f � g)(t0), (u� v)(t0)i 1
2⌫kr(u� v)(t0)k2L2 +
1
2⌫kf(t0)� g(t0)k2V 0 .
Plugging the above two inequalities in (4.5) gives
k(u� v)(t)k2L2 + ⌫
Z t
0
kr(u� v)(t0)k2L2 dt0 ku0
� v0
k2L2
+C
⌫3
Z t
0
kru(t0)k4L2k(u� v)(t0)k2L2 +1
2⌫
Z t
0
kf(t0)� g(t0)k2V 0 dt0 .
Gronwall lemma gives the result. 2
4.2 Existence of stable solutions in a bounded domain
The purpose of this section is the proof of the existence of solutions of the system (NS⌫)which are L4 in time with values in V�. In order to state (and prove) a sharp theorem, weshall introduce intermediate spaces between the spaces V 0
� and V�. Then, we shall prove aglobal existence theorem for small data and then a local in time theorem for large data.
4.2.1 Intermediate spaces
We shall define a family of intermediate spaces between the spaces V 0� and V�. This can be
done by abstract interpolation theory but we prefer to do it here in an explicit way.
Definition 4.2.1 Let s be in [�1, 1]. We shall denote by Vs� the space of vector fields u in V 0
such that
kuk2Vs�
def=X
j2Nµ2sj hu, eji2 < +1.
Theorem 2.2.2 implies that V0
� = H and V1
� = V�. Moreover, it is obvious that, when s is nonnegative, Vs
� endowed with the norm k · kVs�is a Hilbert space.
The following proposition will be important in the following two paragraphs.
Proposition 4.2.1 The space V12� is embedded in L3 and the space PL
32 is embedded in V� 1
2� .
Moreover precilsey, we have
kakL3 . kakV
12�
and kP akV� 1
2�
. kakL
32.
50
Proof. This proposition can be proved using abstract interpolation theory. We prefer topresent here a self contained proof in the spirit of the proof of Theorem A.2.1. Let us consider a
in V12� . Without any loss of generality, we can assume that kak
V12�
1. Let us define, for a
positive real number ⇤,
a⇤
def=
X
j / µj<⇤
ha, ejiej and b⇤
def= a� a
⇤
.
Using the fact that�
x 2 ⌦/ |a(x)| > ⇤ ⇢ �x 2 ⌦/ |a⇤
(x)| > ⇤/2 [�x 2 ⌦/ |b⇤
(x)| > ⇤/2 ,we can write
kak3L3 3
Z
+1
0
⇤2meas��
x 2 ⌦ / |a⇤
(x)| > ⇤/2 � d⇤
+ 3
Z
+1
0
⇤2meas��
x 2 ⌦ / |b⇤
(x)| > ⇤/2 � d⇤
3⇥ 26Z
+1
0
⇤�4ka⇤
k6L6 d⇤+ 3⇥ 22Z
+1
0
kb⇤
k2L2 d⇤. (4.1)
Thanks to Theorem A.2.1, we have, by definition of the k · kVs�norm,
ka⇤
k2L6 Cka⇤
k2V�
CX
j / µj<⇤
µ2
j ha, eji2
C⇤X
j / µj<⇤
µjha, eji2 C⇤.
Plugging this estimate into (4.1) gives
kak3L3 C
Z
+1
0
⇤�2ka⇤
k2V�d⇤+ C
Z
+1
0
kb⇤
k2L2 d⇤.
Using Theorem 2.2.2 and the definition of a⇤
gives
kak3L3 X
j2N
Z
+1
µj
⇤�2µ2
j (a|ej)2H d⇤+ CX
j2N
Z µj
0
(a|ej)2 d⇤
CX
j2Nµjha, eji2
C.
This proves the first part of the proposition.
The second part is obtained by a duality argument. By definition, we have, for any ain V 0,
kP akV� 1
2�
= k(µ� 12
j ha, eji)j2Nk`2
= sup↵2B
X
j2N↵jµ
� 12
j ha, eji (4.2)
51
where B denotes the sequence which are 0 outside a finite set and of `2(N) norm of size lessthan 1. Let us write that, for any ↵ in B, for N large enough, we have
NX
j=0
µ� 1
2j ↵jha, eji = ha,
NX
j=0
↵jµ� 1
2j eji (4.3)
As a is in L32 (⌦) which is included in V 0 , we have, for any � in V� because ' is in L3,
ha,'i =Z
⌦
a(x) · '(x) dx.
Applying Holder inequality in (4.3) gives
�
�
�
�
NX
j=0
µ� 1
2j ↵jha, eji
�
�
�
�
kakL
32
�
�
�
�
NX
j=0
µ� 1
2j ↵jej
�
�
�
�
L3
.
The embedding of V12� into L3 and the definition of the V� norm implies
�
�
�
�
NX
j=0
µ� 1
2j ↵jha, eji
�
�
�
�
CkakL
32
�
�
�
�
NX
j=0
µ� 1
2j ↵jej
�
�
�
�
V12�
CkakL
32k↵k`2
CkakL
32.
This completes the proof of Proposition 4.2.1. 2
4.2.2 The wellposedness result in V12�
The aim of this paragraph is the proof of the following existence theorem with data in V12� .
Theorem 4.2.1 If the initial data u0
belongs to V12� and the external force f belongs to the
space L2
loc(R+
;V� 12
� ), then a positive time T exists such that a solution u of (NS⌫) exists
in L4([0, T ];V�). This solution is unique and belongs to C([0, T ];V12� ).
Moreover, a constant c exists (which can be chosen independent of the domain ⌦) suchthat, if
ku0
kV
12�
+1
⌫kfk
L2(R+
;V� 12
� )
c⌫,
then the above solution is global.
Proof. For the sake of simplicity, we shall ignore the external force in the proof. Let us observethat the map
Q(
V� ⇥ V� �! V� 12
�
(u, v) 7�! P div(u⌦ v)
is a bilinear continuous map. Indeed, as u is divergence free, we can write
div(u⌦ v)` =3
X
k=1
uk@kv`.
52
Then Holder inequality gives
k div(u⌦ v)`kL
32. kukL6krvkL2 .
Sobolev embedding V� ,! L6 and dual Sobolev embedding of Proposition 4.2.1 implies that
kP div(u⌦ v)kV� 1
2�
. kukV�kvkV� . (4.4)
Now let us define the followong bilienar operator
B
⇢
L4([0, T ];V�)⇥ L4([0, T ];V�) �! L4([0, T ];V�)(u, v) 7�! B(u, v)
where B(u, v) is the solution of the linear Stokes problem with initial data 0 and external
force1
2
�Q(u, v) +Q(v, u)�
. Using (4.4), it turns out that
�
�P�Q(u, v) +Q(v, u)
�
�
�
L2([0,T ];V� 1
2� )
. kukL4([0,T ];V�)
kvkL4([0,T ];V�)
.
As V� 12
� is by construction a subspace of V 0�, the term Q(u, v) +Q(v, u) can be considered as
an external force for the Stokes problem of evolution. Thus the bilinear operator B is welldefined. The fact that it maps continuously L4([0, T ];V�) ⇥ L4([0, T ];V�) into L4([0, T ];V�)will be a consequence of the following lemma.
Lemma 4.2.1 Let us consider f in L2([0, T ];V� 12
� ). A constant C exists, such that, for any pin [4,1], the solution Lf of the Stokes problem with external force f and initial data 0satisfies
X
j
µ1+
4p
j k��hLf(t), eki�
�
2
Lp([0,T ])
. kfk2L2
([0,T ];V� 12
� )
.
Proof. Using Formula (2.15) page 33, we get
hLf(t), eji =Z t
0
e�⌫µ2j (t�t0)hf(t0), eji dt0.
Using the Young’s inequality, we get
�
�hLf(t), eki�
�
Lp([0,T ]
1
⌫12+
1p
µ� 1
2� 2p
j
�
�hf(t), eji�
�
L2([0,T ])
.
By definition of the V� 12
� norm, we get the result. 2
Continuation of the proof of Theorem 4.2.1 Let us denote by S(t)u0
the solution of the linearStokes problem with initial data u
0
and external force 0. Using again Formula (2.15) page 33,we get
hS(t)u0
, eji = hu0
, ejie�⌫µ2j t.
Thus, for any p in [4,1], we get
X
j
µ1+
4p
j khS(t)u0
(t), ejik2Lp([0,T ])
. 1
⌫2p
X
j
µj(hu0, eji2 (4.5)
53
The idea of the Kato theory is the following: u is a solution of incompressible Navier-Stokesequation in the space L4([0, T ];V�) if and only if u satisfies
u = S(t)u0
+B(u, u).
In other terms, u is a fixed point of the map
u 7�! S(t)u0
+B(u, u).
Now let us observe that, thanks to the Cauchy–Schwarz inequality, for any a in `2(L4[0, T ]),
Z T
0
kaj(t)k4`2(N)dt =
Z T
0
⇣
X
j2Na2j (t)
⌘
2
dt
=X
j2N,k2N
Z T
0
a2j (t)a2
k(t)dt
X
j2N,k2Nkajk2L4
([0,T ])
kakk2L4([0,T ])
�
�
�
(kajkL4([0,T ])
)j2N�
�
�
4
`2
Let us notice that this is a particular case of the Minkowski inequality. Then we deduce fromLemma 4.2.1 that
kB(u, u)kL4([0,T ];V�)
C
⌫34
kuk2L4([0,T ];V�)
.
Using (4.5) with p = 4 and Minkowski inequality gives
kS(t)u0
kL4([0,T ];V�)
1
⌫14
ku0
kV
12�
. (4.6)
Using Picard fixed point theorem, if we prove that limT!0
kuLkL4([0,T ];V�)
= 0 we conclude the
proof of the theorem up to the continuity of u. Let us observe that, for any positive ", aninterger j" exsits such that
k(Id�Pj")u0kV� ⌫14 "
2with Pk a
def=X
k0k
ha, ek0iek0 .
Then Inequality (4.6) implies that
kS(t)Pj" u0kL4([0,T ];V�)
"
2· (4.7)
Then we have
kS(t)Pj")u0kL4([0,T ];V�)
T14 kS(t)Pj")u0kL1
([0,T ];V�)
T14µ
12j"ku
0
kV
12�
.
Together with (4.7), this implies that
limT!0
kuLkL4([0,T ];V�)
= 0.
54
In order to prove the continuity in time with value in V12� of u, let us observe that, as
@tu = �u+ P div(u⌦ u)
we have that @tu belongs to
L4([0, T ];V 0�) + L2([0, T ];V� 1
2� ) ,! L2([0, T ];V 0).
Thus u is continuous with value in V 0�. Using Lemma 4.2.1 and Inequality (4.2.1) we infer
thatX
j
µjkhu(t), ejik2L1([0,T ])
< 1
Thus for any positive real number ", an integer j" exists such that
X
j>j"
µjkhu(t), ejik2L1([0,T ])
<"2
4·
Now, it turns out that for all (t1
, t2
) in [0, T ]2, one has
ku(t1
)�u(t2
)kV
12�
✓
X
j>j"
µjkhu(t), ejik2L1([0,T ])
◆
12
+
✓
X
jj"
µjhu(t1)� u(t2
), eji2◆
12
"
2+ µj"ku(t1)� u(t
2
)kV 0�.
As u is continuous in time with value in V 0�, the whole Theorem 4.2.1 is proved. 2
4.2.3 Some remarks about stable solutions
In this paragraph, we shall assume that the external force f is identically 0. We shall establishsome results about the maximal existence time of the solution constructed in the precedingparagraph.
Proposition 4.2.2 Let us assume that the initial data u0
belongs to V�. Then the maximal
time of existence T ? of the solution u in the space C([0, T ?[;V12� ) \ L4
loc([0, T?[;V�) satisfies
T ? � c⌫3
kru0
k4L2
·
Proof. Let us observe that , if u0
belongs to V�, we have
kS(t)u0
kL4([0,T ];V�)
T14 ku
0
kV�
This ensures the proposition. 2
From this proposition, we infer the following corollary.
Corollary 4.2.1 Let T ? be the maximal time of existence for a solution u of the system (NS⌫)
in the space C([0, T ?);V12� ) \ L4
loc([0, T?);V�). If T ? is finite, then
Z T ?
0
kru(t)k4L2dt = +1 and T ? c
⌫5ku
0
k4L2 .
55
Proof. For almost every t, u(t) belongs to V�. Then, thanks to the above proposition, themaximal time of existence of the solution starting at time t, which is of course T ?� t, satisfies
T ? � t � c⌫3
kru(t)k4L2
·
This can be written as
kru(t)k4L2 � c⌫3
T ? � t·
This gives the first part of the corollary. Taking the square root of the above inequality gives,thanks to the energy estimate,
c⌫52
Z T ?
0
dt
(T ? � t)12
1
2ku
0
k2L2 .
The corollary is proved.
Remarks
• Sections 3.3 and 4.2 must be known.
• Again books of P. Constantin et C. Foias Navier-Stokes equations, Chigago Univer-sity Press, 1988, de P.-G. Lemarie-Rieusset, Recent developments in the Navier-Stokesproblem. Chapman & Hall/CRC, Research Notes in Mathematics, 431, 2002 and ofJ.-Y. Chemin, B. Desjardins, I. Gallagher and E. Grenier, Mathematical Geophysics;an introduction to rotating fluids and Navier-Stokes equations, Oxford Lecture series inMathematics and its maps, 32, Oxford University Press, 2006 give more details.
56
Chapter 5
An example of a dispersiveequation: the Schrodinger equation
Introduction
The chapter is devoted to study of the Schrodinger equation in the whole space Rd. Thisequation writes
(LS) i@tu��u = 0 and u|t=0
= u0
.
and represents the free evolution of a system of quantum particules in the whole space Rd.Here,the preserved L2 norm does not represent a kinetic energy like in the preceeding chapters butthe total mass. Indeed if the function u is a solution of (LS), the quantity |u(t, x)|2 representsthe density of particules at time t and point x.
The purpose of the first section is to define the concept of weak solution of (LS) possibilywith an external force f . Then we compute explicitely the solution of (LS) in term of Fouriertransform and also in term in convolution. In other term, we compute the fundamental solu-tion. From this formula, we deduce a so called dispersive estimates which is a mathematicaltranslation of the fact that free particules spread in the whole space.
The purpose of the second section is to explain the procedure of complex interpolationwhich allow to extend the two inequalities about an operator between Lp spaces in a familyof inequalities.
In the third section, we expose a general method called TT ? method which allows totranslate a dispersive estimate in space time estimates known as Strichartz estimates. Basedon duality arguments, this proof requires also refined Young inequalities.
In the last section, we use Strichartz estimates to prove the existence ans the uniquenessof solution to a semi-linear Schrodinger equation.
5.1 The solution of some classical linear evolution PDE in Rd
The case of Schrodinger equation follows the same lines.
Definition 5.1.1 Let u be a continuous function from R with value in S 0(Rd), which meansexactly that for any � in S(Rd), that maps defined by
t 7�! hu(t),�i
57
is continuous. Let f be a locally integrale map from R into S 0(Rd), which means exactly thatfor any � in S(Rd), that maps defined by
t 7�! hf(t),�i
is locally integrable on R. Let u0
be in S 0(Rd). We say that u is a solution of (LS) if andonly if, for any function � of S(R⇥Rd), we have, for any t in R,
hu(T ),�(T )i � hu0
,�(0)i =Z T
0
⌦
u(t), i��(t) + @t�(t)↵
dt� i
Z T
0
hf(t),�(t)idt.
Proposition 5.1.1 If u0
belongs to S 0(Rd) and f is locally integrale form R into S 0(Rd),there is a unique solution of (LS) with a given by
u(t) = F�1
✓
e�it|⇠|2bu0
(⇠)� i
Z t
0
e�i(t�t0)|⇠|2bf(t0, ⇠)dt0
◆
. (5.1)
Proof. Let us first prove the result using a duality method. As the equation is linear, unique-ness consists only in proving that if, for all � of S(R⇥Rd), if
hu(T ),�(T )i =Z T
0
⌦
u(t), i��(t) + @t�(t)↵
dt (5.2)
then u(T ) = 0 in S 0(Rd) for all time T . Let us consider any function �T in S 0(Rd). Let usobserve that
�(t)def= F�1
�
e�i(T�t)|⇠|2b�T�
satisfies i��(t) + @t�(t) = 0. Then, Identity (5.2) implies that hu(T ),�(T )i = 0 and theuniqueness is proved.
Let us proved that the function
u(t)def= F�1
�
e�it|·|2bu0
�
.
is a solution of (LS) with initial data u0
and external force 0. In order to do it, let us compute,for any � in C1(R;S(Rd)),
U(T )def= �i
Z T
0
⌦
u(t), i�b�(t) + @tb�(t)↵
dt.
We have, because of the properties of the Fourier transform,
U(T ) =
Z T
0
⌦
u(t),�iF�| · |2�(t)�↵ dt+Z t
0
hu(t), @tb�(t)i dt
=
Z T
0
⌦
bu(t),�i| · |2�(t)↵ dt+Z T
0
hbu(t), @t�(t)i dt.
By definition of u, we get
U(t) =
Z T
0
⌦
e�it|·|2bu0
,�i| · |2�(t)↵ dt+Z T
0
he�it|·|2bu0
, @t�(t)i dt
= �D
bu0
,
Z T
0
e�it|·|2��i| · |2�(t) + @t�(t)�
E
dt.
58
As we have
e�it|⇠|2��i|⇠|2�(t, ⇠) + @t�(t, ⇠)�
= @t�
e�it|⇠|2�(t, ⇠)�
, (5.3)
we infer that
U(t) =D
bu0
,
Z T
0
@t�
e�it|·|2�(t)�
E
dt
= hbu0
, e�iT |·|2�(t)i � hbu0
,�(0)i.By definition of u, this means exactly that
U(T ) = hbu(T ),�(t)i � hbu0
,�(0)i= hu(T ), b�(t)i � hbu
0
, b�(0)i.In the case when the external force f is identically 0, Formula (5.1) is established. In the casewhen u
0
= 0, let us compute
V (T )def=
Z T
0
⌦
F (t), i��(t) + @t�(t)↵
dt with F (t)def=
Z t
0
S(t� t0)f(t0)dt0,
where S(t)fdef= F�1
�
e�it|·|2f�
. By definiton of S, we get
V (T ) =
Z T
0
⇣
Z t
0
⌦
f(t0)dt0, S(t� t0)�
i��(t) + @t�(t)↵
dt0⌘
dt.
Again by definition of S, we have, using (5.3),⇣
FS(t� t0)�
i��(t) + @t�(t)�
⌘
(⇠) = e�i(t�t0)|⇠|2��i|⇠|2b�(t, ⇠) + @tb�(t, ⇠)�
= @t�
e�i(t�t0)|⇠|2b�(t, ⇠)
�
= @tFS(t� t0)�(t).
By definition of V , we infer that
V (T ) = �i
Z T
0
Z t
0
⇣
@t⌦
f(t0), S(t� t0)�(t)↵
dt0⌘
dt.
Let us observe that
d
dthF (t),�(t)i =
d
dt
Z t
0
⌦
f(t0), S(t� t0)�(t)↵
dt0
= hf(t),�(t)i+Z t
0
@t⌦
f(t0), S(t� t0)�(t)↵
dt0.
Thus, we infer that
V (T ) = �i
Z T
0
d
dthF (t),�(t)i dt0 + i
Z T
0
hf(t),�(t)i dt
= �ihF (T ),�(T )i+ i
Z T
0
hf(t),�(t)i dt.
The result is proved. 2
59
Let us now give a formula for the solution of the Schrodinger equation.
Proposition 5.1.2 Let u0
a function in L1(Rd). The solution of the linear equation Schro-dinger
i@tu��u = 0 and u|t=0
= u0
.
is given by
u(t, x) =1
(4⇡it)d2
Z
Rdei
|x�y|24t u
0
(y)dy.
with, for z�d2def= |z|� d
2 e�i d2 ✓ if z = |z|ei✓ with ✓ in [�⇡/2,⇡/2].Proof. Proposition 5.1.1 claims that
u(t) = F�1(e�it|·|2bu0
).
Thus the point is to compute the Fourier transform of the purely imaginary gaussian. In orderto do it, let us remark that, for any ⇠ in Rd, the two functions
z 7�!Z
Rde�ih⇠,xie�z|x|2 dx and z 7�!
⇣⇡
z
⌘
d2e�
|⇠|24z (5.4)
are holomorphic on the domain D of complex numbers with positive real part. As we havefor any positive real number a, we have
Z
Rde�ih⇠,xie�a|x|2 dx =
✓
⇡
a
◆
d2
e�|⇠|24a .
This claims that the two functions of (5.4) coincide on the intersection of the real line with D.Thus they also coincide on the whole domain D. Now let (zn)n2N be a sequence of elementsof D which converges to it for t 6= 0. For any function � in S, we have by virtue of Lebesgue’sdominated convergence theorem,
limn!1
Z
Rde�zn|x|2�(x) dx =
Z
Rde�it|x|2�(x) dx and
limn!1
Z
Rde�
|⇠|24zn �(⇠) d⇠ =
Z
Rde�
|⇠|24it �(⇠) d⇠.
As we have
F⇣
e�zn|·|2⌘
=⇣ ⇡
zn
⌘
d2e�
|⇠|24zn ,
passing to the limit in S 0(Rd) when n tends to 0 gives
F�e�it|·|2�(⇠) =⇣⇡
z
⌘
d2e�
|⇠|24it . (5.5)
This gives the result. 2
Corollary 5.1.1 If the denote by eit�u0
the solution of (LS) with f ⌘ 0, we have
keit�u0
kL1 ⇣ 1
4⇡|t|⌘
d2 ku
0
kL1
60
5.2 The complex interpolation method in Lp space
We present here the theory of complex interpolation in the particular case of Lp space. Itallows to extend inequalities between Lp space The basic theorem is the following.
Theorem 5.2.1 Let us consider (Xk, µk)1k2
two measured spaces and (pj , qj)j2{0,1} twoelements of [1,1]2. Let us consider an operator A which maps continuously the space Lpj (X
1
)into the space Lqj (X
2
) for j in {0, 1}. For any ✓ in [0, 1], if
✓
1
p✓, 1
q✓
◆
def= (1� ✓)
✓
1
p0
, 1
q0
◆
+ ✓
✓
1
p1
, 1
q1
◆
,
then A maps continuously Lp✓(X1
) into Lq✓(X2
) and
kAkL(Lp✓(X1);L
q✓(X2))
A✓ with A✓def= kAk1�✓
L(Lp0(X1);Lq0
(X2))kAk✓L(Lp1
(X1);Lq1(X2))
.
Proof. Let us first point out that there is nothing to prove if ✓ is equal to 0 or 1. Fromnow on, we assume that ✓ is in ]0, 1[. This implies that (p✓, q✓) belongs to ]1,1[2. Thus weconsider only (f,') in L1 \ L1(X
1
)⇥ L1 \ L1(X2
) such that kfkLp✓(X1)
k = k'kLq0✓(X2)
= 1.
It is enough to prove thatZ
X2
(Af)(x2
)'(x2
)dµ2
(x2
) A✓. (5.6)
Let us consider a complex number z in the strip S of complex numbers the real part of whichis between 0 and 1. Let us define
fz(x1)def=
f(x1
)
|f(x1
)| |f(x1)|p✓
⇣1�zp0
+
zp1
⌘
and
'z(x2)def=
'(x2
)
|'(x2
)| |'(x2)|q0✓
✓1�zq00
+
zq01
◆ (5.7)
Obviously, we have f✓ = f and '✓ = '. Moreover, for any t in R, we have, for j in {0, 1},
|fj+it(x1)| = |f(x1
)|p✓pj and |'j+it(x2)| = |'(x
2
)|q0✓q0j . (5.8)
It can be checked that the function defined by
F (z)def=
Z
X2
(Afz)(x2)'z(x2)dµ2
(x2
)
is holomorphic and bounded on S and continuous on the closure of S. Using Lindelof’sprinciple (see Lemme A.5.1 page 92), we infer that
F (✓) M1�✓0
M ✓1
with Mjdef= sup
t2R|F (j + it)|. (5.9)
Using (5.8), we observe that that fj+it belongs to Lpj and kfj+itkLpj(Lqj
)
= 1. We infer that
Mj kAkL(Lpj(X1);L
qj(X2))
.
and the lemma is proved. 2
61
Let us give some immediate applications of this result.
Corollary 5.2.1 For any p in [1, 2], the Fourier transform maps Lp(Rd) into Lp0(Rd) with
norm less or equal to (2⇡)dp0 .
Corollary 5.2.2 In Rd, we have, for any p in [1, 2],
keit�kL(Lp;Lp0
)
⇣ 1
4⇡|t|⌘d
⇣1p� 1
2
⌘
.
5.3 The duality method and the TT ? argument
This section describes the so-called TT ? argument which is the standard method for convertingthe dispersive estimates (presented in the previous section) into inequalities involving suitablespace-time Lebesgue norms of the solution.
In all this section, we denote by k · kLp(Lr
)
the norm in Lp(R;Lr(Rd)). Let us now statethe “abstract” Strichartz estimates.
Theorem 5.3.1 Let (U(t))t2R be a bounded family of continuous operators on L2(Rd) suchthat, for some positive real numbers � and C
0
, we have
kU(t)U?(t0)fkL1 C0
|t� t0|� kfkL1 . (5.10)
Then, for any (p, q) in [2,1]2 such that
2
p+
2�
r= � and (p, r,�) 6= (2,1, 1) , (5.11)
we have for some positive constant C
kU(t)u0
kLp(Lr
)
Cku0
kL2 .
Proof. It is based on a duality argument together with the Hardy-Littlewood-Sobolev inequal-ity stated in Theorem 5.5.2. Let us first notice that
kU(t)u0
kLp(Lq
)
= sup'2Bp,r
�
�
�
�
Z
R⇥RdU(t)u
0
(x)'(t, x) dt dx
�
�
�
�
= sup'2Bp,r
�
�
�
�
Z
R(U(t)u
0
|'(t))L2 dt
�
�
�
�
whereBp,r
def=�
� 2 D(R1+d;C) / k�kLp0(Lr0
)
1
.
By definition of the adjoint operator, we have
kU(t)u0
kLp(Lq
)
= sup'2Bp,r
�
�
�
�
⇣
u0
�
�
�
Z
RU?(t)'(t) dt
⌘
L2
�
�
�
�
.
By virtue of the Cauchy-Schwarz inequality, we deduce that
kU(t)u0
kLp(Lq
)
ku0
kL2 sup'2Bp,r
�
�
�
Z
RU?(t)'(t) dt
�
�
�
L2. (5.12)
62
Let us write that�
�
�
Z
RU?(t)'(t) dt
�
�
�
2
L2=
Z
R2
�
U?(t0)'(t0)�
�U?(t)'(t)�
L2 dt0 dt
=
Z
R2
�
U(t)U?(t0)'(t0)�
�'(t)�
L2 dt0 dt
=
Z
R2
⌦
U(t)U?(t0)'(t0),'(t)↵
dt0 dt. (5.13)
Now, let us observe that, using Theorem 5.2.1, we infer that
8r 2 [2,1] , kU(t)U?(t0)fkLr C
|t� t0|�(1� 2r )kfkLr0 . (5.14)
Thanks to (5.11) we infer that�
�
�
Z
RU?(t)'(t) dt
�
�
�
2
L2 C
Z
R2
1
|t� t0| 2pk'(t0)kLr0k'(t)kLr0 dt0 dt.
Because p > 2, the Hardy-Littlewood-Sobolev inequality page 65 gives�
�
�
Z
RU?(t)'(t) dt
�
�
�
2
L2 Ck'k2
Lp0(Lq0
)
.
Thanks to (5.12), the theorem is proved. 2
Form this theorem, we can deduce Strichatz estimates for the linear Schrodinger equation.
Theorem 5.3.2 Let us consider (p, q)in [2,1] such that
2
p+
d
r=
d
2with (p, r) 6= (2,1). (5.15)
A constant C exists such that if u be a solution of (LS) (in the sense of Definition 5.1.1)with u
0
in L2(Rd) and f in L1(R;L2(Rd)) given by Proposition 5.1.1 page 58, then
kukLp(R;Lq
(Rd))
C�ku
0
kL2 + kfkL1(R;L2
(Rd)
�
.
Proof. It is a clear consequence of Theorem 5.3.1 that
keit�u0
kLp(Lr
)
Cku0
kL2 . (5.16)
Now, we can to prove the theorem in the case when u0
= 0. In this case, let us write thesolution as
u(t) =
Z t
0
ei(t�t0)�f(t0)dt0.
We have, for any t,
ku(t)kLr(Rd
)
C
Z t
0
kei(t�t0)�f(t0)kLr(Rd
)
dt0
C
Z
Rkei(t�t0)�f
+
(t0)kLr(Rd
)
dt0
with f+
(t)def= 1R+(t)f(t). Taking the Lp norm in time in the above inequality, using (5.16)
and the translation invariance of the Lebesgue measure on R gives
kukLp(R;Lr
(Rd))
C
Z
Rkf
+
(t0)kL2(Rd
)
dt0
CkfkL1(R;L2
(Rd))
.
This concluded the proof of Theorem 5.3.2. 2
63
5.4 An example of application
As an application of the results of the previous section, we here solve the initial boundaryvalue problem for the cubic semilinear Schrodinger equation in R2:
(NLS3
)
(
i@tu� 1
2�u = P
3
(u, u)
u|t=0
= u0
where P3
is some given homogeneous polynomial of degree 3.
Theorem 5.4.1 There exists a constant c such that, for any initial data u0
in L2(R2) satis-fying ku
0
kL2 c, Equation (NLS3
) has a unique solution u in the space L3(R;L6(R2)) whichin addition belongs to L1(R;L2(R2)).
Remark Let us first have a look on the scaling properties of Equation (NLS3
). If u is a
solution of (NLS3
), then u�(t, x)def= �u(�2t,�x) is also a solution of the same equation. In
the scale of Sobolev spaces, L2(R2) is the only invariant space.
Proof of Theorem 5.4.1. Let Q be the nonlinear functional defined by
(
i@tQ(u)� 1
2�Q(u) = P
3
(u, u)
Q(u)|t=0
= 0.
The functional Q maps continuously the space L3(R;L6(R2)) into the space L1(R;L2(R2))\L3(R;L6(R2)). Indeed Theorem 5.3.2 leads to
kQ(u)kL3(R;L6
(R2))
CkP3
(u, u)kL1(R;L2
(R2))
Ckuk3L3
(R;L6(R2
))
.
As Q(u)�Q(v) satisfies
⇣
i@t +1
2�⌘
(Q(u)�Q(v)) = P3
(u, u)� P3
(v, v),
we get, using Theorem 5.3.2 again,
kQ(u)�Q(v)kL1(R;L2
(R2))\L3
(R;L6(R2
))
Cku� vkL3(R;L6
(R2))
⇥⇣
kuk2L3
(R;L6(R2
))
+ kvk2L3
(R;L6(R2
))
⌘
.(5.17)
Now, it is obvious that u is a solution of (NLS3
) if and only if u is a fixed point of the map
F (u)def= U(t)u
0
+Q(u).
Applying Theorem 5.3.2 and Estimate (5.17) with v = 0, we get that
kF (u)kL3(R;L6
)
. ku0
kL2 + kuk3L3(R;L6
)
.
Thus, if 8C2ku0
k2L2 1, then the ball B(0, 2Cku0
kL2) of center 0 and radius 2Cku0
kL2 of theBanach space L3(R;L6(R2)) is invariant by the map F . Using again Inequality (5.17), we get,for any u and v in B(0, 2Cku
0
kL2),
kF (u)� F (v)kL3(R;L6
)
8C3ku0
k2L2ku� vkL3(R;L6
)
.
64
Thus, if in addition
8C3ku0
k2L2 1
2,
then Picard’s fixed point theorem implies that a unique solution u exists in some neighborhoodof 0 in L3(R;L6). Clearly, Inequality (5.17) implies that uniqueness holds true in L3(R;L6)without any smallness condition.
Finally, the energy estimate entails that this solution belongs to L1(R;L2). Indeed,multiplying Equation (NLS
3
) by u, integrating over R2 then taking the real part, we discoverthat
1
2
d
dtkuk2L2 = =m
Z
uP3
(u, u) dx,
whence, for all t in R,
ku(t)kL2 ku0
kL2 + C
�
�
�
�
Z t
0
ku(t0)k3L6 dt0�
�
�
�
.
This completes the proof of the theorem. 2
5.5 Refined convolution inequalities
The purpose of this section is to improve the classical Young inequalities which clains that, if
1
p+
1
q= 1 +
1
r(5.18)
then kf ? gkLr kfkLpkgkLq . In order to state these refined inequalities, let us introduce theso caled weak Lp spaces.
Definition 5.5.1 For p in [1,1[, we denote by Lpw(X,µ)) space the set of measure functions
such thatkfkp
Lpw(X,µ)
def= sup
�>0
�pµ(|f | > �) < 1,
Let us notice that we have
µ(|f | > �) Z
(|fk�)
⇣ |f(x)|�
⌘pdµ(x) 1
�pkfkpLp .
Thus Lp is a subset of Lpw. Typical examples of weak Lp functions. The function | · |�↵ belongs
for Ld↵w (Rd, dx).
Theorem 5.5.1 Let (p, q, r) be in ]1,1[3 and satisfy (5.18). A constant C exists such that
kf ? gkLr CkfkLpkgkLqw
Let us notice that the above theorem implies the well-known Hardy-Littlewood-Sobolevinequalities on Rd .
Theorem 5.5.2 (Hardy-Littlewood-Sobolev inequality) Let ↵ be in ]0, d[ and (p, r)in ]1,1[2 satisfy
1
p+↵
d= 1 +
1
r· (5.19)
Then a constant C exists such that
k | · |�↵ ? fkLr(Rd
)
CkfkLp(Rd
)
.
65
The proof of Theorem 5.5.1 relies on the atomic decomposition which is decribe in thefollowing proposition.
Proposition 5.5.1 Let (X,µ) be a measure space and p be in [1,1[. Let f be a nonnegativefunction in Lp. Then a sequence of positive real numbers (ck)k2Z and a sequence of nonnegativefunctions (fk)k2Z (the atoms) exist such that
f =X
k2Zckfk
where the support of the functions fk are pairwise disjoint and
µ(Supp fk) 2k+1, (5.20)
kfkkL1 2�kp , (5.21)
1
2kfkpLp
X
k2Zcpk 2kfkpLp . (5.22)
Remarks
• Let us notice that, because of Inequalities (5.20) and (5.21), the functions fk belongsto La or any a in [1,1] and satisfies
kfkkLa 21a 2
k⇣
1a� 1
p
⌘
. (5.23)
• The fact that f belongs to Lp is given by the behavior of the sequence (ck)k2Z.
• As inferred by the definition given below, the sequence (ckfk)k2Z is independent of pand depends only on f .
Proof of Proposition 5.5.1. If µ(f > 0) is not finite, let us define, for any k in Z,
�kdef= inf
n
� /µ(f > �) < 2ko
.
If µ(f > 0) is finite, let us define by k0
the smallest integer such that
µ(f > 0) 2k0
Then, for k k0
, let us define
�kdef= inf
n
� /µ(f > �) < 2ko
.
In all that follows in this proof, we shall implicitely consider that, if µ(f > 0) is finite, then,all the sequence define are 0 for k � k
0
. Let us notice that (�k)k2Z is a decreasing sequence.By definition of the sequence (�k)k2Z, we get
8� < �k , µ(f > �) � 2k. (5.24)
The monotonic convergence theorem and the definition of (�k)k2Z, we have that
µ(f > �k) 2k (5.25)
66
Then, let us define
ckdef= 2
kp�k and fk
def= c�1
k 1(�k+1<f�k)
f.
It is obvious that kfkkL1 2�kp . The point is now the proof of (5.22). As the support of the
functions (fk)k2Z are pairwise disjoint, one may write
kfkpLp =X
k2ZcpkkfkkpLp .
Taking advantage of Inequalities (5.20) and (5.21), we claim that
kfkkpLp 2 for all k 2 Z .
which claims exactly thatX
k2Zcpk 2kfkpLp .
Moreoverwhich, owing to the fact that f is a nonnegative function of Lp, converges to 0when k tends to +1.
Thanks to (5.25), we have µ(Supp fk) 2k+1. This givesX
k2Zcpk =
X
k2Z2k�pk
= pX
k2Z
Z 1
0
2k1]0,�k[
(�)�p�1 d�.
Using Fubini’s theorem, we get
X
k2Zcpk = p
Z 1
0
�p�1
✓
X
k /�k>�
2k◆
d�.
By definition of sequence (�k)k2Z, having � < �k implies that µ(f > �) � 2k. Thus we inferthat
X
k2Zcpk p
Z 1
0
�p�1
✓
X
k / 2kµ(f>�)
2k◆
d�
2p
Z 1
0
�p�1µ(f > �) d�.
Now, the right inequality in (5.22) follows from the fact that, owing to Fubini’s theorem, wehave
kfkpLp = p
Z 1
0
�p�1µ(|f | > �) d� (5.26)
which implies the desired inequality. 2
Proof of Theorem 5.5.1 Let f and g be two nonnegative measurable functions on Rd. Let usconsider a nonnegative function h in Lr0 and let us define
I(f, g, h)def=
Z
R2df(y)g(x� y)h(x) dµ(x) dµ(y).
Arguing by homogeneity, one can assume that kfkLp = kgkLqw= khkLr0 = 1. Defining
Cjdef=�
y 2 Rd , 2�j+1q < g(y) 2�
jq
and gjdef= 1Cjg,
67
we can write
I(f, g, h) X
j2ZI(f, gj , h).
Because kgkLqw= 1, we have kgjkLs 2
1s 2
�j⇣
1q� 1
s
⌘
for all s in [1,1]. Thus if we directly apply
Young’s inequality with p, q and r, we find that I(f, gj , h) 21q so that series
�
(I(f, gj , h)�
j2Zhas no reason to converge. In order to bypass this di�culty, one may introduce the atomicdecomposition for f and h given by Proposition 5.5.1 which leads to
I(f, g, h) =X
j,k,`
ckd`I(fk, gj , h`).
Using Young’s inequalities, we get for any (a, b) in [1,1]2 such that b a0 and for any ( ef,eh)in La ⇥ Lb,
I(fk, gj , h`) kfkkLakgjkLbkh`kLc0 with1
a+
1
b= 1 +
1
c·
Using Proposition 5.5.1, this gives
I(fk, gj , h`) 4⇥ 2j⇣
1b� 1
q
⌘
2k⇣
1a� 1
p
⌘
2`(1c0� 1
r0 ) = 4⇥ 2j⇣
1q� 1
b
⌘
2k⇣
1a� 1
p
⌘
2`(1r� 1
c ).
Using the condition (5.18) on (p, q, r) and (a, b, c) implies
I(fk, gj , h`) 4⇥ 2j⇣
1q� 1
b
⌘
2k⇣
1a� 1
p
⌘
2`⇣
1p+
1q� 1
a� 1
b
⌘
4⇥ 2(j�`)
⇣1b� 1
q
⌘
2(k�`)
⇣1a� 1
p
⌘
. (5.27)
As (p, q, r) is in ]1,1[2, a positive real number " exists, if
1
adef=
1
p� " sg(k � `) and
1
bdef=
1
q� " sg(j � `)
then (a, b) is in [1,1]2 and1
a+
1
b� 1. With this choice of a and b, (5.27) becomes
I(fk, gj , h`) 4⇥ 2�"|j�`|�"|k�`|.
Now, using Young’s inequality for Z equipped with the counting measure, we deduce that
I(f, g, h) 4X
k,`
ckd`2�"|k�`|X
j
2�"|j�`|
C
"
X
k,`
ckd`2�"|k�`|
C
"2k(ck)k`pkk(d`)k`p0 .
Condition (5.18) implies that r0 p0 and thus
I(f, g, h) C
"2k(ck)k`pk(d`)k`r0 .
The theorem is proved. 2
68
Chapter 6
Linear symmetric systems
6.1 Definition and examples
Let us first define the concept in the framework of linear system with variable coe�cients.Let I be a closed interval of R which has 0 as an interior point. Let us consider a family Aof smooth bounded functions (Ak)0kd from I ⇥ Rd into the space of N ⇥N matrices withreal coe�cients. Let us assume that all their derivatives in the x variable are bounded. Weconsider the system
(LS)
8
>
<
>
:
@tU +dX
k=1
Ak@kU +A0
U = F
U|t=0
= U0
.
Let us introduce the following notations. If U in D0(Rd) and ' in D(Rd),
hU,'i =NX
i=1
hU i,'ii
and if U and V belongs to (L2(Rd))N ,
(U |V )L2 =NX
i=1
Z
RdU i(x)V i(x)dx.
Definition 6.1.1 A function U in (C(I;L2))N is a solution of (LS) if and only if for any 'in D(R⇥Rd), for any t in I,
hU(t),'(t)i =Z t
0
hU(t0), @t'(t0)idt0 + hU0
,'(0)i
+
Z t
0
X
k=1
h(AkU)(t0), @k'(t0)idt0 +Z t
0
h(divAU)(t0),'(t0)idt0 +Z t
0
hF (t0),'(t0)idt0.
Let us define the concept of symmetric system.
Definition 6.1.2 The above system (LS) is symmetric if and only if for any k 2 {1, . . . , d}and any (t, x) in I ⇥Rd the matrices Ak(t, x) are symmetric, which means that for any k, wehave Ak,i,j(t, x) = Ak,j,i(t, x).
69
Let us of a look to an example coming from the gas dynamics. The unknown is thecouple (⇢, v) which satisfies
8
<
:
@t⇢+ v ·r⇢+ ⇢ div v = 0
@tv + v ·rv +1
⇢rp = 0
with p = A⇢� . Here, ⇢ is a scalar function with values in R+
? and represents the density ofthe particules of the gas at time t in the point x and v a time dependant vector field whichdescribes the speed of a particule located in x at time t.
It will be clear later on that we have to change the unknowns defining
cdef=
2
� � 1
✓
@p
@⇢
◆
12
=(4�A)
12
� � 1⇢
��12 .
The first equation becomes
@tc+ v ·rc+� � 1
2c div v = 0.
About the second one, let us observe that
� � 1
2crc =
1
⇢rp.
The Euler system related to gas dynamics becomes8
>
<
>
:
@tc+ v ·rc+� � 1
2c div v = 0
@tv + v ·rv +� � 1
2crc = 0.
(6.1)
Let us assume that the solution is ”small”, i.e. is a perturbation of magnitude " of a station-nary flat state v = 0 and c = c, by an easy computation of the coe�cients of the powers of ",we infer
8
>
<
>
:
@tc+� � 1
2c div v = 0
@tv +� � 1
2crc = 0.
(6.2)
An obvious computation ensures that
@2t c�✓
� � 1
2
◆
2
c2�c = 0.
This equation is called ”acoustic waves equation”.
The reason why this definition is fundamental is the following. Let us consider a solutionof (LS) and let us look to the evolution of its energy. This question leads to the followingformal computation which will be made rigourous in the following section.
1
2
d
dtkU(t)k2L2 = �
dX
k=1
⇣
Ak@kU�
�U⌘
L2� (A
0
U |U)L2 + (F |U)L2
70
By integration by part, we get that
�⇣
Ak@kU�
�U⌘
L2= �
X
i,j
Z
RdAk,i,j@kU
iU `dx
=X
i,j
Z
RdAk,i,jU
i@kU`dx+
X
i,j
Z
Rd@kAk,i,jU
iU `dx.
If the system (LS) is symmetric, then we have
�dX
k=1
⇣
Ak@kU�
�U⌘
L2=
1
2((divA)U |U)L2 with (divA)i,j
def=
dX
k=1
@kAk,i,j
This implies that
�
�
�
�
dX
k=1
⇣
Ak(t)@kU(t)�
�U(t)⌘
L2
�
�
�
�
1
2k divA(t)kL1kU(t)k2L2 .
Thus we get that
d
dtkU(t)k2L2 a
0
(t)kUk2L2 +(F |U)L2 with a0
(t)def= k divA(t, ·)kL1 +2kA
0
(t, ·)kL1 . (6.3)
The purpose of this section is to study linear symmetric systems. First, we want to solvethem and then to study basic properties of their solutions. In this section, for s 2 N we shallstate
|U(t)|2s def=
X
1jN1|↵|d
k@↵xU `(t)k2L2 .
6.2 The wellposedness of linear symmetric systems
The goal of this paragraph is the proof of the following wellposedness theorem.
Theorem 6.2.1 Let (LS) be a linear symmetric system. Then, if U0
belongs to Hs andif F is a continuous function with value in Hs, then a unique solution of (S) exists in thespace C0(I,Hs) \ C1(I;Hs�1).
Proof. It requires four steps:
• We first prove a-priori estimates for smooth enough solutions of the system (S).
• Then we apply Friedrichs method.
• Then we pass to the limit in the case of smooth enough initial data and we get existencein any case by smoothing of the initial data.
• Finally, we get uniqueness using existence for the adjoint system.
A priori estimates use in a crucial way the symmetry hypothesis and are true only forsmooth enough solutions.
71
Lemma 6.2.1 For any non negative integer s, a locally bounded function as exists such thatfor any function U in C0(I,Hs+1) \ C1(I,Hs), we have for any t in I,
|U(t)|s |U(0)|s expZ t
0
as(t0)dt0 +
Z t
0
|F (t0)|s exp✓
Z t
t0as(t
00)dt00◆
dt0,
with
F = @tU +dX
k=1
Ak@kU +A0
u .
Proof. To start with, let us prove this lemma for s = 0. Let us consider a function U in thespace C0(I;H1) \ C1(I;L2). By definition of F , we have
1
2
d
dt|U(t)|2
0
= (@tU |U)L2
= (F |U)L2 � (A0
U |U)L2 �dX
k=1
(Ak@kU |U)L2 .
As the system (LS) is symmetric and U belongs to C0(I;H1)\C1(I;L2), computations donepage 71 which lead to (6.3) are rigourous. Thus we have
d
dt|U(t)|2
0
a0
(t)|U(t)|20
+ 2|F (t)|0
|U(t)|0
(6.4)
with a0
(t)def= k divA(t, ·)kL1 + 2kA
0
(t, ·)kL1 . By Gronwall lemma, we get
|U(t)|0
|U(0)|0
exp
Z t
0
a0
(t0)dt0 +Z t
0
|F (t0)|0
exp
✓
Z t
t0a0
(t00)dt00◆
dt0 (6.5)
and the lemma is proved is the case when s = 0.
Remark Let us point out that the above computations are also valid when the matrices (Ak)have C1 coe�cients and A
0
has C0 coe�cients.
Let us study the case when s is any non negative integer. To do so, we shall proceed byinduction on the integer s. Let us assume that Lemma 6.2.1 is proved for some s. Let Ube a function in C0(I,Hs+2) \ C1(I;Hs+1). Let us introduce the function (with N(d + 1)components) eU defined by
eU = (U, @1
U, · · · , @dU) .
As, for any j in {1, . . . , d},
F = @tU +dX
k=1
Ak@kU +A0
U,
we obtain by di↵erentation of the equation,
@t (@jU) = �dX
k=1
Ak@k@jU �dX
k=1
(@jAk) · @kU � @j(A0
U)� @jF.
72
We may write
@t eU = �dX
k=1
Bk @k eU � B0
eU + eF with (6.6)
eF = (F, @1
F, · · · , @dF ) and (6.7)
Bk =
0
@
Ak 0Ak
0 Ak
1
A . (6.8)
The coe�cients of B0
are computed from those of Ak (k = 0, . . . , d) and their first orderderivatives. The induction hypothesis allows to conclude the proof of Lemma 6.2.1. 2
Remark Let us point out that the proof of the inequalities of Lemma 6.2.1 done abovedemands exactly one more derivative than in the statement of Theorem 6.2.1.
Second step of the proof of Theorem 6.2.1 . This leads us to use a smoothing method, theFriedrich method. It consists in smoothing both the initial data and the system itself. Moreprecisely, let us consider the system (LSn) defined by
(LSn)
8
>
<
>
:
@tUn +dX
k=1
En (Ak@kUn) + En(A0
Un) = En F
En U|t=0
= En U0
where En is the cuto↵ operator defined on L2 by
En udef= F�1(1B(0,n)bu). (6.9)
This is nothing else that the orthogonal projection of L2 on the closed space L2
n of the L2
functions the Fourier transform of which are supported in the ball of center 0 and radius n.Fourier Plancherel tells us in particular that the operator @k is a continuous on L2
n. As thefunctions Ak are bounded it turns out that the linear operator
V 7�!dX
k=1
En (Ak@kV ) + En(A0
V )
is continuous on L2
n. Thus the system (LSn) is a linear sytstem of ordinary di↵erentialequations on L2
n. This implies the existence of a unique function Un continuous on I withvalue in L2
n which is a solution of (LSn). Moreover as the functions Ak are smooth functionsin (t, x), using the equation (LSn), we get that Un is a smooth function on I with value in Hs
for any integer s.
Let us prove that the functions Un satisfy the energy estimates of Lemma 6.2.1. Moreprecisely, we have the following lemma.
Lemma 6.2.2 For any non negative integer s, a locally bounded function as exists such thatfor any n 2 N and any t in I we have,
|Un(t)|s |En U(0)|s expZ t
0
as(t0)dt0 +
Z t
0
|En F (t0)|s exp✓
Z t
t0as(t
00)dt00◆
dt0,
73
Proof. Taking the scalar product of (LSn) with Un in L2, we get using the fact that theoperator En is selfadjoint on L2 and that En Un = Un,
d
dt|Un(t)|2
0
= �2dX
k=1
(Ak@kUn|Un)L2 � 2(A0
Un)|Un)L2 � 2(En F |Un)L2 .
We proceed exactly as in the proof of Lemma 6.2.1. As the system (LS) is symmetric and Un
belongs to C0(I;H1)\C1(I;L2), computations done page 71 which lead to (6.3) are rigourous.Thus we have
d
dt|Un(t)|2
0
a0
(t)|Un(t)|20
+ 2|En F (t)|0
|Un(t)|0 (6.10)
with a0
(t)def= k divA(t, ·)kL1 + 2kA
0
(t, ·)kL1 . Gronwall Lemma implies that
|Un(t)|0 |En U0
|0
exp
Z t
0
a0
(t0)dt0 +Z t
0
|En F (t0)|0
exp
✓
Z t
t0a0
(t00)dt00◆
dt0.
The proof of the lemma for any integer s works exactly as the one of Lemma 6.2.1 and isomitted. 2
Third step of the proof of Theorem 6.2.1 . The third step consists in the proof of the followingwellposedness result.
Proposition 6.2.1 Let s � 3. We consider the linear symmetric system
(LS)
8
>
<
>
:
@tU +dX
k=1
Ak @kU +A0
U = F
U(0) = U0
with F in C(I;Hs) and U0
in Hs. A unique solution U exists in C(I,Hs�2) \ C1(I;Hs�3)which moreover satisfies the energy estimate
8� s , 8 t 2 I , |U(t)|� |U0
|� exp
Z t
0
as(t0)dt0 +
Z t
0
|F (t0)|� exp✓
Z t
t0as(t
00)dt00◆
dt0.
Proof. Let us consider the sequence (Un)n2N of solution of (LSn). We shall prove that (Un)n2Nis a Cauchy sequence in L1(I;Hs�2). In order to do so, let us state Vn,p
def= Un+p � Un. We
have8
>
<
>
:
@tVn,p +dX
k=1
En+p (Ak @kVn,p) + En+p(A0
Vn,p) = Fn,p
Vn,p(0) = (En+p�En)U0
(6.11)
with
Fn,pdef=
dX
k=1
(En+p�En) (Ak @kUn) � (En+p�En)(A0
Un) + (En+p�En)F.
Lemma 6.2.2 tells us that the sequence (Un)n2N is bounded in L1(I;Hs). Moreover we have
|(En+p�En)a|��1
C
n|a|�.
74
Thus we have
|(En+p�En) (Ak @kUn(t))|s�2
C
nsupk
|(En+p�En) (Ak @kUn(t))|s�1
C
n|Un(t)|s.
The sames arguments give�
�
�
(En+p�En)(A0
Un(t)) + (En+p�En)F (t)�
�
�
s�2
C
n2
(|Un(t)|s + |F (t)|s). (6.12)
Energy estimate implies that
|Vn,p(t)|s�2
C
nexp⇣
t
Z t
0
as(t0)dt0
⌘
.
Thus the sequence (Un)n2N is a Cauchy one in L1(I,Hs�2). Moreover, using (6.11) and (6.12),we infer that (@tUn)n2N is a Cauchy sequence in L1(I;Hs�3). Let us denote by U thelimit of (Un)n2N. Of course, U belongs to C(I;Hs�2) \ C1(I;Hs�3). Let us check that thisfunction U is solution of (LS). As F belongs to C(I;Hs), we have that
limn!1En F = F in L1(I;Hs). (6.13)
As the sequence (Un)n2N is bounded in L1(I;Hs), we have that
k(En� Id)Ak(U)@kUnkL1(I;Hs�2
)
C
n·
Thus U is a solution of (LS). To conclude, let us point out that the sequence (Un)n2Nis bounded in L1(I;Hs). Using interpolation inequality, we get that for any s0 < s, thesequence (Un)n2N is a Cauchy one in C(I,Hs0). Thus U belongs to C(I,Hs0). Using thefact that U is a solution of (LS), we get that U belongs to C(I,Hs0) \ C1(I;Hs0�1). Butas (Un)n2N is bounded in L1(I;Hs), it weakly converges to U in L1(I;Hs). Using (6.13),the fact that (En U0
)n2N converges to U0
in Hs and that
kUkL1([0,t];Hs
)
lim supn!1
kUnkL1([0,t];Hs
)
we get, passing to the limit in Lemma 6.2.2 that
|Un(t)|s |En U(0)|s expZ t
0
as(t0)dt0 +
Z t
0
|En F (t0)|s exp✓
Z t
t0as(t
00)dt00◆
dt0.
Proposition 6.2.1 is proved. 2
Last step of the proof of Theorem 6.2.1 . Let us consider the sequence (eUn)n2N of solutions of8
>
<
>
:
@ eUn
@t+
dX
k=1
Ak@k eUn +A0
eUn = En F
eUn|t=0
= En U0
.
Thanks to Proposition 6.2.1 this solution does exist in C1(I,Hs) for any positive real number s.
Let us state Vn,pdef= Un+p � Un. It satisfies
8
>
<
>
:
@t eVn,p +dX
k=1
Ak@k eVn,p +A0
eVn,p = (En+p�En)F
eVn,p|t=0
= (En+p�En)U0
.
75
Lemma 6.2.1 implies that
|eVn,p(t)|s |(En+p�En)U(0)|s expZ t
0
as(t0)dt0
+
Z t
0
|(En+p�En)F (t0)|s exp✓
Z t
t0as(t
0)dt0◆
dt.
As the function F is continuous from I into Hs, the sequence (En F )n2N converges to F inthe space L1([0, T ];Hs). As U
0
belongs to Hs, the sequence (En U0
)n2N converges to U0
in Hs. Thus the sequence (eUn)n2N is a Cauchy sequence in L1(I;Hs). It converges to afunction U of C(I;Hs) which is of course solution of the system (LS). The fact that @tUbelongs to C(I;Hs�1) comes immediately from the fact that U is solution of (S).
The existence part of Theorem 6.2.1 and also the uniqueness when s � 1 is now proved.The following proposition will conclude the proof of Theorem 6.2.1.
Proposition 6.2.2 Let U be a solution C0(I;L2) of the symmetric system (LS).
(LS)
8
>
<
>
:
@tU +dX
k=1
Ak @kU +A0
U = 0
U|t=0
= 0.
Then U ⌘ 0.
Proof. In order to prove this proposition, we shall use a duality method. Let be a functionof D(]0, T [⇥Rd). Let us consider the solution of
(tLS)
8
>
<
>
:
�@t'�dX
k=1
@k(Ak') +tA
0
' =
'|t=T = 0.
The system (tLS) can be understood as the adjoint system of the system (LS). As we have
@k(Ak') = Ak@k'+ @kAk',
the system (tLS) becomes
(tS)
8
>
<
>
:
�@t'�dX
k=1
Ak@k'+ eA0
' =
'|t=T = 0
with eA0
def= tA
0
�dX
k=1
@kAk.
This is obviously a linear symmetric system. The existence part of Theorem 6.2.1 tells usthat a solution ' of (tLS) exists in C1(I,Hs) for any s in N. Thus we have
hU, i =D
U,�@t'�dX
k=1
Ak@k'+ eA0
'E
= �Z
IhU(t, ·), @t'(t, ·idt�
dX
k=1
Z
IhU(t), @k(Ak')(t)idt
+
Z
I⇥RdU(t, x)tA
0
'(t, x)dtdx.
76
Considering the weak regularity of U , each integration by part must be justified. UsingTheorem 6.3.2 page 79 below (the proof of which is totally independant of Proposition 6.2.2),we have that for any t in I, the function '(t, ·) belongs to D(Rd). By definition of the derivativeof distributions, we have
hU(t), @k(Ak')(t)i =X
i,j
hU i(t), @k(Ak,i,j'j)(t)i
=X
i,j
h@kU i(t),Ak,i,j'j(t)i.
Because the matrices Ak are symmetric, we have for any t in I,
hU(t), @k(Ak')(t)i = hAk@U(t)
@xk,'(t)i.
It turns ot that
hU, i = �Z
IhU(t, ·), @t'(t, ·idt�
D
dX
k=1
Ak@kU � eA0
U,'E
.
In order to justify the time integration by part, let us observe that U belongs to C1(I,H�1).Indeed, as for smooth function we have
hAk@kV,'i = �hV, @ktAk'i � hV, tAk@k'i (kAk(t, ·)kL1 + a
0
(t))kV kL2k'kH1 .
This implies that @tU belongs to C0(I;H�1). Now, let us use the smoothing operator En
defined by (6.9). The function En U belongs to C1(I;Hs) for any s. Using this with s greaterthan d/2 implies that for any x, the function
t 7! En U(t, x)
exists and is a C1 function on I. This impies that
�Z
IEn U(t, x)
@'
@t(t, x)dt = �En U(T, x)'(T, x) + En U(0, x)'(0, x)
+
Z
I
@ En U
@t(t, x)'(t, x)dt.
Using the fact that U0
= 0 and that '(T, ·) = 0, we get that
�Z
IEn U(t, x)@t'(t, x)dt =
Z
I@t(En U)(t, x)'(t, x)dt.
By integration in the variable x and interchanging time and space integration, we get that
�Z
IhEn U(t, ·), @t'(t, ·)idt =
Z
Ih@t(En U)(t, ·),'(t, ·)idt.
As U is a function of C(I;L2) \ C1(I;H�1), we have
limn!1En U = U in L1(I, L2) and lim
n!1En @tU = @tU in L1(I,H�1).
77
Passing to the limit in the above equality gives
�hU, @t'i =Z
Ih@tU(t, ·),'(t, ·)idt
and thus
hU, i =Z
I
D
@tU(t, ·) +dX
k=1
Ak@kU(t, ·) +A0
U(t, ·),'(t, ·)E
.
Ass U is solution of (LS) with F = 0, then U ⌘ 0 which ends the proof of the proposition 2
The whole Theorem 6.2.1 is nowproved. 2
6.3 Finite propagation speed
The phenomena of finite propagation speed is describe by the following theorem.
Theorem 6.3.1 Let (LS) be a symmetric system. A constant C0
exists such that, for anypositive real number R and any data F 2 C0(I, L2) and U
0
2 L2 such that
F (t, x) ⌘ 0 when |x| < R� C0
t and U0
(x) ⌘ 0 when |x| < R. (6.14)
then the unique solution U of the system (LS) in C0(I, L2) with data F and U0
satisfies
U(t, x) ⌘ 0 when |x| < R� C0
t.
An other form of this statement is given by the following corollary.
Corollary 6.3.1 If the data F and U0
satisfy
F (t, x) ⌘ 0 for |x| > R+ C0
t and U0
(x) ⌘ 0 for |x| > R,
then the solution U satisfies
U(t, x) ⌘ 0 when |x| > R+ C0
t.
Proof of Theorem 6.3.1. To start with, let us regularize the data U0
and F perturbing theirsupport as less as possible. Let � be a function of D(B(0, 1)) the integral of which is 1. Forany positive ✏, we state
�"(x)def=
1
"d�⇣x
"
⌘
·Now let us consider the data
U0,"
def= �" ? U0
and F"(t, ·) def= �" ? F (t, ·).
Of course, we have
Supp U0," ⇢ Supp U
0
+B(0, ") and F"(t, ·) ⇢ Supp F (t, ·) +B(0, ").
The support hypothesis are satisfies for U0," and F" with R+" instead of R and the associated
solution U" is C1(I;Hs) for any s 2 N. Thus it is enough to prove Theorem 6.3.1 with thoseregular solutions, namely the following statement.
78
Theorem 6.3.2 Let (LS) be a symmetric system. A constant C0
exists such that, for anypositive real number R and any data F 2 C0(I,H1) \ C1(I, L2) and U
0
2 H1 such that
F (t, x) ⌘ 0 when |x| < R� C0
t and U0
(x) ⌘ 0 when |x| < R. (6.15)
then the unique solution U of the system (LS) in C0(I,H1) \ C1(I, L2) with data F and U0
satisfiesU(t, x) ⌘ 0 when |x| < R� C
0
t.
Proof. The method used is weighted energy estimates. More precisely, for ⌧ greater than 1,let us introduce
U⌧ (t, x)def= e⌧�(t,x)U(t, x).
with �(t, x) = �t + (x). The function is a smooth real valued function on Rd which willbe choosen later on.
@tU⌧ +dX
k=1
Ak @kU⌧ + B⌧U⌧ = F⌧
with
F⌧ (t, x)def= e⌧�(t,x)F (t, x) and
B⌧def= A
0
+ ⌧
@t�Id +dX
k=1
@k�Ak
!
Considering the form of the function �, we have
B⌧ = A0
� ⌧
Id�dX
k=1
@k Ak
!
Thus a constant K > 0 exists such that for any (t, x) 2 I ⇥ Rd, any vector W 2 RN and anypositive real number ⌧ , we have
kr kL1 K ) (B⌧ (t, x)W |W ) (A0
(t, x)W |W ).
Then let us write the energy estimate and use the above inequality and relation (6.3); we get
d
dt|U⌧ (t)|2
0
= �2dX
k=1
(Ak@kU |U⌧ )L2 � 2(B⌧U⌧ |U⌧ )L2 + 2(F⌧ |U⌧ )L2
a0
(t)|U⌧ |20
+ (F⌧ |U⌧ )L2
Using Gronwall Lemma, we get
|U⌧ (t)|0 |U⌧ (0)|0 expZ t
0
a0
(t0)dt0 +Z t
0
|F⌧ (t0)|
0
exp⇣
Z t
t0a0
(t00)dt00⌘
dt0. (6.16)
Let us point out that the above inequality is independant of ⌧ . Now let us state C0
= 1/Kand let us pick up a smooth function = (|x|) such that
�2"+K(R� |x|) (x) �"+K(R� |x|) and kr kL1 K. (6.17)
79
Then we have8(t, x) 2 I ⇥ Rd , |x| � R� C
0
t =) �t+ (x) �".When ⌧ tends to +1 in the inequality (6.16), we get that
8t 2 I , lim⌧!1
Z
Rde2⌧�(t,x)|u(t, x)|2dx = 0.
Thus U(t, x) ⌘ 0 on the open set t < (x). But, if (t0
, x0
) satisfies |x0
| < R � C0
t0
, it ispossible to pick up a function satisfying (6.17) and such that t
0
< (x0
). This proves thetheorem. 2
80
Appendix A
Sobolev spaces
Introduction
In this course, we shall restrict ourselves to Sobolev spaces modeled on L2. These spacesdefinitely play a crucial role in the study of partial di↵erential equations, linear or not. Thekey tool will be the Fourier transform. This chapter is not treated during the lectures. It ishere as a convenient reference for the members of the audience of the lectures.
A.1 Definition of Sobolev spaces on Rd
Definition A.1.1 Let s be a real number, a tempered distribution u belongs to the Sobolevspace of index s, denoted Hs(Rd), or simply Hs if no confusion is possible, if and only if
bu 2 L2
loc(Rd) and bu(⇠) 2 L2(Rd; (1 + |⇠|2)sd⇠).
and we note
kuk2Hs =
Z
Rd(1 + |⇠|2)s|bu(⇠)|2d⇠.
Proposition A.1.1 For any s real number, the space Hs, equipped with the norm k · kHs ,is a Hilbert space.
Proof. The fact that the norm k · kHs comes from the scalar product
(u|v)Hsdef=
Z
Rd(1 + |⇠|2)sbu(⇠)bv(⇠)d⇠
is obvious. Let us prove that this space is complete. Let (un)n2N a Cauchy sequence of Hs.By definition of the norm, the sequence (bun)n2N is a Cauchy sequence of L2(Rd; (1+ |⇠|2)sd⇠).Thus, a function eu exists in the space L2(Rd; (1 + |⇠|2)sd⇠) such that
limn!1 kbun � eukL2
(Rd;(1+|⇠|2)sd⇠) = 0. (A.1)
In particular, the sequence(bun)n2N tends to eu in the space S 0 of tempered distribtions. Letus define u = F�1
eu. As the Fourier transform is an isomorphism of S 0, the sequence (un)n2Ntends to u in the space S 0, and also in Hs thanks to (A.1).
Shortly said, this is nothing more than observing that the Fourier transform is an isometricisomorphism from Hs onto L2(Rd; (1 + |⇠|2)sd⇠). 2
81
Proposition A.1.2 Let s be a non negative integer, the space Hs(Rd) is the space of func-tions u of L2 all the derivatives of which of order less or equal to m are distributions whichbelongs to L2. Moreover, the space Hm equipped with the norm
ekuk2Hmdef=
X
|↵|m
k@↵uk2L2
is a Hilbert space and this norm is equivalent to the norme k · kHs .
Proof. The fact that
ekuk2Hm = e(u|u)Hm with e(u|v)Hmdef=
X
|↵|m
Z
Rd@↵u(x)@↵v(x)dx.
ensures that the norm k · kHm comes from a scalar product. Moreover, a constant C existssuch that
8⇠ 2 Rd , C�1
⇣
1 +X
0<|↵|m
|⇠|2|↵|⌘
(1 + |⇠|2)s C⇣
1 +X
0<|↵|m
|⇠|2|↵|⌘
. (A.2)
As the Fourier transform is, up to a constant, an isometric isomorphism from L2 onto L2, wehave
@↵u 2 L2 () ⇠↵bu 2 L2.
Thus, we have deduce that
u 2 Hm () 8↵ / |↵| m, @↵u 2 L2.
Inequality (A.2) ensures the equivalence of the two norms using again the fact that the Fouriertransform is a isometric isomorphism up to a constant. The proposition is proved. 2
Exercice A.1.1 Prove that the space S is continuously included in the space Hs for anyreal s.
Exercice A.1.2 Prove that the mass of Dirac �0
belongs to the space H� d2�" for any positive
real number ". Prove that �0
does not belong to the space H� d2 .
Exercice A.1.3 Prove that, for any distribution to support compact u, it exists a real num-ber s such that u belongs to the Sobolev space Hs.
Exercice A.1.4 Prove that the constant 1 does not belong to Hs for any real number s.
Proposition A.1.3 Let s a real number of the interval ]0, 1[. Prove that the space Hs is thespace des functions u of L2 such that
Z
Rd ⇥Rd
|u(x+ y)� u(x)|2|y|d+2s
dxdy.
Moreover, a constant C exists such that, for any function u of Hs, we have
C�1kuk2Hs kuk2L2 +
Z
Rd ⇥Rd
|u(x+ y)� u(x)|2|y|d+2s
dxdy Ckuk2Hs .
82
Proof. Thanks to Fourier-Plancherel identity,we can write that
Z
Rd
|u(x+ y)� u(x)|2|y|d+2s
dx =
Z
Rd
|ei(y|⇠) � 1|2|y|d+2s
|bu(⇠)|2d⇠ < 1.
It turns out thatZ
Rd ⇥Rd
|u(x+ y)� u(x)|2|y|d+2s
dxdy =
Z
RdF (⇠)|bu(⇠)|2d⇠ with
F (⇠)def=
Z
Rd
|ei(y|⇠) � 1|2|y|2s
dy
|y|d ·
By an obvious change of variable, we see that the function F is radial and homogeneous ofdegree 2s. Thus
F (⇠) = |⇠|2sZ |eiy1 � 1|2
|y|2sdy
|y|d ·
This concludes the proof of the proposition. 2
Let us prove now an interpolation inequality which will be very useful.
Proposition A.1.4 If s = ✓s1
+ (1� ✓)s2
with ✓ in [0, 1], then, we have
kukHs kuk✓Hs1kuk1�✓Hs2 .
The proof consits in applying Holder inequality with the measure |bu(⇠)|2d⇠ and the twofunctions (1 + |⇠|2)✓s1 and (1 + |⇠|2)(1�✓)s2 .
Theorem A.1.1 Let s a real quelconque;
• the space D(Rd) is dense in Hs(Rd),
• the multiplication by a function of S is a continuous function of Hs into itself.
Proof. In order to prove the first point of this theorem, let us consider a distribution u of Hs
such that, for any test function ', we have ('|u)Hs = 0. This means that, for any testfunction ', we have
Z
Rdb'(⇠)(1 + |⇠|2)sbu(⇠)d⇠ = 0.
which means that (1 + | · |2)sbu = 0 as a tempered distribution. As the multiplication by thefunction (1 + | · |2)�s is continuous in S 0, we have that bu = 0 as a tempered distribution.Thus u ⌘ 0.
Let us prove now the second second point of the theorem. This proof is presented herejust for culture. We know that
c'u = (2⇡)�db' ? bu.
The point is to estimate the L2 norm of the function defined by
U(⇠) = (1 + |⇠2|) s2
Z
Rd|b'(⇠ � ⌘)|⇥ |bu(⌘)|d⌘.
We shall use the following lemma.
83
Lemma A.1.1 For any (a, b) in Rd, for any s 2 R, we have
(1 + |a+ b|2) s2 2
|s|2 (1 + |a|2) |s|
2 (1 + |b|2) s2 .
Proof. Let us first observe that
1 + |a+ b|2 1 + 2(|a|2 + |b|2) 2(1 + |a|2)(1 + |b|2).Taking the power s/2 of this inequality, we find the result for non negative s � 0. In the casewhen s is negative, we have
(1 + |b|2)� s2 2�
s2 (1 + |a+ b|2)� s
2 (1 + |a|2) s2 .
Thus the result is proved. 2
Continuation of the proof of Theorem A.1.1 Lemma A.1.1 implies that
U(⇠) Z
Rd(1 + |⇠ � ⌘|2| |s|2 |b'(⇠ � ⌘)|⇥ (1 + |⌘|2| s2 |bu(⌘)|d⌘.
Young’s law implies that kU2
kL2 CkukHs ; this concludes the proof of the theorem. 2
Exercice A.1.5 Let FL1 = {u 2 S 0 / bu 2 L1}. Prove that, for any non negative realnumber s, the product is a bilinear continuous map from FL1\Hs⇥FL1\Hs into FL1\Hs.What happens when s is greater than d/2?
Exercice A.1.6 Let s a real number greater than 1/2. Prove that the map � defined by
�
⇢ D(Rd) �! D(Rd�1)' 7�! �(') : (x
2
, · · · , xd) 7! '(0, x2
, · · · , xd)
can be extended in a continuous onto map from Hs(Rd) onto Hs� 12 (Rd�1).
Hint : Write
FRd�1'(0, ⇠2
, · · · , ⇠d) = (2⇡)�1
Z
Rb'(⇠
1
, ⇠2
, · · · , ⇠d)d⇠1.and for the fact that the map in onto, observe that, if
u = (2⇡)�(n�1)CsF�1
(1 + |⇠0|2)s� 12
(1 + |⇠|2)s bv(⇠0)
!
,
then u 2 Hs and �(u) = v.
Let us prove a theorem which describes the dual of the space Hs.
Theorem A.1.2 The bilinear form B defined by
B
8
<
:
S ⇥ S ! C(u,') 7!
Z
Rdu(x)'(x)dx
can b extended as a bilinear form continuous from H�s ⇥ Hs to C. Moreover, the map �Bdefined by
�B
⇢
H�s �! (Hs)0
u 7�! �B(u) : (') 7! B(u,')
is a linear and isometric isomorphism (up to a constant), which means that the bilinear form Bidentifies the space H�s to the dual space of Hs.
84
Proof. The important point of the proof of this theorem is inverse Fourier formula whichensures that, for any couple (u,') of functions of S, we have
B(u,') =
Z
Rdu(x)'(x)dx
=
Z
Rdu(x)F(F�1')(x)dx
=
Z
Rdbu(⇠)(F�1')(⇠)d⇠
= (2⇡)�d
Z
Rdbu(⇠)b'(�⇠)d⇠. (A.3)
Multiplying and dividing by (1 + |⇠|2) s2 , we immediately get thanks to Cauchy-Schwarz in-
equality,|B(u,')| (2⇡)�dkukHsk'kH�s .
Thus the first point of the theorem. The fact that the map �B is one to one comes from thefact that if, for any function ' in S, we have B(u,') = 0, then u is 0. We shall prove that �Bis one to one and onto.
Let � and ' be in S0
. One can write that�
�
�
�
Z
Rd�(x)'(x) dx
�
�
�
�
=
�
�
�
�
Z
Rd(F�1�)(⇠)(F')(⇠) d⇠
�
�
�
�
= (2⇡)�d
�
�
�
�
Z
Rd|⇠|�s
b�(�⇠)|⇠|s b'(⇠) d⇠�
�
�
�
(2⇡)�dk�k˙H�sk'k
˙Hs .
As S0
is dense in H� when |�| < d/2, then one can extend B to H�s⇥ Hs. Of course, if (u,�)is in H�s ⇥ S then B(u,�) = hu,�i.
Let L be a linear functional on Hs. Consider the linear functional Ls defined by
Ls :
⇢
L2(Rd) �! Cf 7�! hL,F�1
�
(1 + | · |2)� s2 f�i.
It is obvious that
supkfkL2=1
|hLs, fi| = supkfkL2=1
|hL,F�1
�
(1 + | · |2)� s2 f�i|
= supk�kHs=1
|hL,�i|
= kLk(
˙Hs)
0 .
Riesz representation theorem implies that a function g exists in L2 such that
8h 2 L2 , hLs, hi =Z
Rdg(⇠)h(⇠) d⇠.
Stating �(⇠)def= F�1
�
(1 + | · |2)� s2h�
(�⇠), we infer that, for any � in Hs
hL,�i = hLs, hi=
Z
Rdg(⇠)(1 + |⇠|2) s
2 b�(�⇠)d⇠.
85
Then stating udef= F�1
�
(1 + | · |2) s2 g�
, we infer that
hL,�i =Z
Rdbu(⇠)b�(�⇠)d⇠
and the theorem is proved. 2
A.2 Sobolev embeddings
The purpose of this section is the study of embedding properties of Sobolev spaces Hs(Rd)into Lp spaces . Let us prove the following theorem.
Theorem A.2.1 If s is greater than d/2, then the space Hs is continuously included in thespace of continuous functions which tend to 0 at infinity. If s is a positive real number less
than d/2, then the space Hs is continuously included in L2d
d�2s and we have
kfkLp Ckfk˙Hs with kfk
˙Hs
def=⇣
Z
Rd|⇠|2s| bf(⇠)|2d⇠
⌘
12.
Proof. The first point of this theorem is very easy to prove. Let us use the fact that
kukL1 (2⇡)�dkbukL1 (A.4)
Indeed, if s is greater than d/2, we have,
|bu(⇠)| (1 + |⇠|2)�s/2(1 + |⇠|2)s/2|bu(⇠)|. (A.5)
The fact that s is greater than d/2 implies that the function
⇠ 7! (1 + |⇠|2)�s/2
belongs to L2. Thus, we have
kbukL1 ✓
Z
(1 + |⇠|2)�sd⇠
◆
12
kukHs .
The first point of the theorem is proved.
The proof of the second point is more delicate. A way to unterstand the index p =2d/(d � 2s) is the use of a scaling argument. Let us consider a function v on Rd and let usdenote by v� the function v�(x) = v(�x). We have
kv�kLp = ��dp kvkLp
and alsoZ
|⇠|2s| bv�(⇠)|2d⇠ = ��2d
Z
|⇠|2s|bv(��1⇠)|2d⇠= ��d+2skvk2
˙Hs ,
with
kvk2˙Hs
def=
Z
Rd|⇠|2s|bv(⇠)|2d⇠.
86
The two quantities k · kLp and k · k˙Hs have the same scaling, which means that they have the
same behaviour with respect to changes of unit. Thus, it make sense to compare them.Multiplying f by a positive real number, it is enough to prove the inequality in the case
when kfk˙Hs = 1. On utilise then the fact that for any p de the interval ]1,+1[, we have, for
any function measurable f ,
kfkpLp = p
Z 1
0
�p�1m(|f | > �)d�.
Let us decompose f in a low and in a high frequencies by writing
f = f1,A + f
2,A with f1,A = F�1(1B(0,A)
bf) and f2,A = F�1(1Bc
(0,A)
bf). (A.6)
As the support of the Fourier transform of f1,A is compact, the function f
1,A is bounded andmore precisely,
kf1,AkL1 (2⇡)�dkdf
1,AkL1
(2⇡)�d
Z
B(0,A)
|⇠|�s|⇠|s| bf(⇠)|d⇠
(2⇡)�d
Z
B(0,A)
|⇠|�2sd⇠
!
12
C
(d� 2s)12
Ad2�s. (A.7)
The triangle inequality implies that, for any positive real number A,
(|f | > �) ⇢ (2|f1,A| > �) [ (2|f
2,A| > �).
Using Inequality (A.7), we have
A = A�def=
�(d� 2s)12
4C
!
pd
=) m
✓
|f1,A| > �
2
◆
= 0.
Thus we deduce that
kfkpLp = p
Z 1
0
�p�1m(2|f2,A�
| > �)d�.
it is well known (this is Bienaime-Tchebychev inequality) that
m
✓
|f2,A�
| > �
2
◆
=
Z
(|f2,A�|>�
2 )dx
Z
(|f2,A�|>�
2 )
4|f2,A�
(x)|2�2
dx
4kf
2,A�k2L2
�2·
For such a choice of A, we have
kfkpLp 4p
Z 1
0
�p�3kf2,A�
k2L2d�. (A.8)
87
As the Fourier transform is (up to a constant) an isometric isomorphism of L2, we have
kf2,A�
k2L2 = (2⇡)�d
Z
(|⇠|�A�)
| bf(⇠)|2d⇠.
Thanks to Inequality (A.8), we get
kfkpLp 4p(2⇡)�d
Z
R+ ⇥Rd�p�31{(�,⇠) / |⇠|�A�}(�, ⇠)| bf(⇠)|2d⇠d�.
By definition of A�, we have
|⇠| � A� () � C⇠def=
4C
(d� 2s)12
|⇠| dp .
Fubini’s theorem implies that
kfkpLp 4p(2⇡)�d
Z
Rd
✓
Z C⇠
0
�p�3d�
◆
| bf(⇠)|2d⇠
4p(2⇡)d
p� 2
4C
(d� 2s)12
!p�2
Z
Rd|⇠|
d(p�2)p | bf(⇠)|2d⇠.
As 2s =d(p� 2)
p, the theorem is proved. 2
Corollary A.2.1 Let p be in ]2,1[, and s gretar than spdef= d
⇣1
2� 1
p
⌘
. We have
kukLp Ckuk1�✓L2 kuk✓
˙Hs with ✓ =sps·
Proof. It is an application of the above theorem together with the fact that
kuk˙H✓s1+(1�✓)s2 kuk✓
˙Hs1kuk1�✓
˙Hs2.
A.3 Homogeneous Sobolev spaces
Definition A.3.1 Let s be a real number, the homogeneous Sobolev space Hs is the spaceof tempered distribtions such that bu belongs to L1
loc and satisfies
kuk2˙Hs
def=
Z
Rd|⇠|2s|bu(⇠)|2d⇠ < 1.
These spaces (or at least theirs norms) naturally appeared in the proof of Theorem A.2.1.The k · k
˙Hs norm has the following scaling property
kf(�·)k˙Hs = ��
d2+skfk
˙Hs .
These spaces are di↵erent from the inhomogeneous Hs spaces. Let us notice that if s ispositive, then Hs is included in Hs but that if s is negative, then Hs is included in Hs. Theinhomongeneous spaces is a decreasing family of spaces (with respect to the index s). Thehomogeneous ones are not comparable together.
We shall only consider theses homogeneous spaces in the case when s is less than the halfdimension.
88
Proposition A.3.1 If s < d/2, then the space Hs is a Banach space.
Proof. Let (un)n2N a Cauchy sequence of Hs. The sequence (bun)n2N is a Cauchy one in the Ba-nach space L2(Rd \{0}; |⇠|2sd⇠). Let f be its limit. It is clear that f belongs to L1
loc(Rd \{0}).
Moreover,Z
B(0,1)|f(⇠)|d⇠
⇣
Z
Rd|⇠|2sf(⇠)|2d⇠
⌘
12⇣
Z
B(0,1)|⇠|�2sd⇠
⌘
12< 1
because s is less than the half-dimension. Thus bf belongs to S 0 and to L1
loc. Thus udef= F�1f
is well defined, belongs to Hs, and is the limit of the sequence (un)n2N in the sense of thenorm Hs. 2
Exercice A.3.1 1) Prove that the space
B def= {u 2 S 0(Rd) , bu 2 L1(B(0, 1); d⇠) \ L2(Rd; |⇠|2sd⇠)}
equipped with the norme N(u)def= kbukL1
(B(0,1)) + kuk˙Hs is a Banach space.
2) Let s � d/2. Give an example of a sequence (fn)n2N of B, bounded in Hs(Rd), suchthat
limn!1N(fn) = +1.
3) Then deduce that (Hs, k · k˙Hs) is not a Banach space.
Exercice A.3.2 Prove that, if k 2 N, then we have
H�k(Rd) =n
u 2 S 0(Rd) , u =X
|↵|=k
@↵f↵ with f↵ 2 L2
o
.
Prove that a constant C exists such that
C�1kuk˙H�k inf
⇢
⇣
X
|↵|=k
kf↵k2L2
⌘
12/ u =
X
|↵|=k
@↵f↵
�
Ckuk˙H�k .
A.4 The spaces H10(⌦) and H�1(⌦)
Definition A.4.1 Let ⌦ a domain of Rd, the space H1
0
(⌦) is defined as the closure of D(⌦)in the sense of the norm H1(Rd) .
The space H�1(⌦) is the set of distributions u on ⌦ such that
kukH�1(⌦)
def= sup
f2D(⌦)
kfkH10(⌦)1
|hu, fi| < 1.
Proposition A.4.1 The space H1
0
(⌦) is a Hilbert space equipped with the norm
⇣
kuk2L2(⌦)
+ kruk2L2(⌦)
⌘
12.
The proof is an easy exercice left to the reader. The space H�1(⌦) can be indentified to thedual space of H1
0
(⌦) thanks to the following theorem.
89
Theorem A.4.1 The bilinear map defined by
B
⇢
H�1(⌦)⇥D(⌦) �! C(u,') 7�! hu,'i
can be extended to a bilinear continuous map from H�1(⌦) ⇥ H1
0
(⌦) into C, still denotedby B. Moreover, the map �B defined by
�B
(
H�1(⌦) �! (H1
0
(⌦))0
u 7�! �B(u)(')def= B(u,')
is a linear isometric isomorphism between the space H�1(⌦) and the dual space of H1
0
(⌦).
Proof. The fact that the bilinear map B can be extended because B is uniformely continuous.Let ` a linear form continuous on H1
0
(⌦). Its restriction on D(⌦) is a distribution u on ⌦ suchthat
8' 2 D(⌦) , hu,'i = h`,'i.By definition of the norm on (H1
0
(⌦))0, the theorem is proved. 2
Remark The spacen H�1(⌦) is exactly the set of the resctrictions to ⌦ of the distributionswhich belongs to H�1(Rd). Indeed, let us observe that any linear form on H1
0
(⌦) can beextended to H1(Rd) simply by the extention which is 0 on the orthogonal of the closedsubspace H1
0
(⌦) is the space H1(Rd).
Theorem A.4.2 (Poincare Inequality) Let ⌦ be bounded open subset of Rd. A con-stant C exists such that
8' 2 H1
0
(⌦) , k'kL2 C
0
@
dX
j=1
k@j'k2L2
1
A
12
.
Proof. Let R a positive real number such that ⌦ is included in ]�R,R[⇥Rd�1.Then, for anytest function ', we have
'(x1
, · · · , xd) =Z x1
�R
@'
@y1
(y1
, x2
, · · · , xd)dy1.
Cauchy-Schwarz Inequality implies that
|'(x1
, · · · , xd)|2 2R
Z x1
�R
�
�
�
�
@'
@y1
(y1
, x2
, · · · , xd)�
�
�
�
2
dy1
.
By integration in x1
, we getZ
⌦
|'(x1
, · · · , xd)|2dx1 2R
Z
⌦⇥]�R,R[
�
�
�
�
@'
@y1
(y1
, x2
, · · · , xd)�
�
�
�
2
dy1
.
Then, integrating with respect to the other d� 1 variables, we findZ
⌦
|'(x1
, · · · , xd)|2dx 2R
Z
⌦⇥]�R,R[
�
�
�
�
@'
@y1
(y1
, x2
, · · · , xd)�
�
�
�
2
dy1
dx2
· · · dxd
4R2
dX
j=1
k@j'k2L2 .
As D(⌦) is dense in H1
0
(⌦), the theorem is proved. 2
90
It obviously implies the following corollary.
Corollary A.4.1 The space H1
0
(⌦) equipped with the norm
u 7�!✓ dX
j=1
k@juk2L2
◆
12 def= krukL2
is a Hilbert space and the this norm is equivalent to the previous one.
In order to conclude this chapter, let us prove the following very important compactnesstheorem.
Theorem A.4.3 For any positive real number s, M and R, the set
BsM,R
def=�
u 2 Hs(Rd) / kukHs M and u|B(0,R)
c = 0
is relativement compact in L2.
Proof. It is enough to prove that for any positive real number ", the set BsM,R can be covered
by a finite number of L2 balls of radius ". The first step consits in a cut o↵ in Fourierspace, the second step in a cuto↵ in physical space and then the theorem follows from Ascoli’scompacness theorem.
Let us consider a function � in S(Rd) such that the Fourier transform has value 1 onthe ball centered at 0 and of radius 1 and has values between 0 and 1. Let us consider thefamily (�↵)↵>0
defined by
�↵(x)def=
1
↵d�⇣x
↵
⌘
·Let us observe that, for any u in Hs the Hs norm of which is less than or equal to M satisfies
ku� �↵ ? uk2L2 = (2⇡)dZ
Rd
�
1� b�(↵⇠)�2(1 + |⇠|2)�2s(1 + |⇠|2)2s|bu(⇠)|2d⇠
⇣
1 +1
↵
⌘�2skuk2Hs
↵2sM2.
The we get that
↵ ⇣ "
3M
⌘
1s=) 8u 2 Bs
M,R , ku� �↵" ? ukL2 "
3· (A.9)
Let us notice that this part does not use the hypothesis of compact support.
For the cuto↵ in the physical space, let us observe that
|x| � 2R =) 8y 2 B(0, R) , |x� y| � 1
2|x|.
Thus we infer that, for any ↵ and any u in BsM,R,
|x| ���↵ ? u(x)�
� 2↵1�d
Z
B(0,R)
|x� y|↵
�⇣x� y
↵
⌘
u(y)dy.
We deduce from law of convolution that
8u 2 BsM,R ,
�
� (| · |�↵) ? u�
�
L2 C↵M
91
↵ "R
3CM=) 8u 2 Bs
M,R , ku� �↵" ? ukL2(B(0,R)
c)
"
3· (A.10)
Now let us notice that, for any positive ↵
k�↵ ? ukL1 C↵� d2M and k@j(�↵ ? u)kL1 C↵� d
2�1M.
Thus, because of Ascoli’s theorem, for any positve ↵ the set {(�↵ ? u)|B(0,2R)
, u 2 BsM,R} can
be recovered by a family number of balls of radius "/(3VolB(0, 2R)12 ) in L1(B(0, 2R). But
balls of radius "/(3VolB(0, 2R)12 ) are included in balls of radius "/3 in L2(B(0, 2R)). Then
choosing
↵"def= min
n⇣ "
3M
⌘
1s , "R
3CM
o
we get the results thanks to Assertions (A.9) and (A.10). 2
A.5 The Lindelhof principle
Lemma A.5.1
92
Liste des questions de cours pour l’examen
• Theoreme de Cauchy-Lipschiz (le theoreme 1.2.1 page 9 et sa demonstration)
• Theoreme de Peano (le theoreme 1.3.1 page 12 et sa demonstration)
• Theoreme d’unicite pour les equations de transport (le theoreme 1.4.3 page 17 et sademonstration)
• Le chapitre 3 dans son entier
• Le theoreme de Fujita-Kato (le theoreme 4.2.1 page 52 et sa demonstration)
• Le theoreme 5.3.1 page 62 et sa demonstration.
93
