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´

´

THEORIE

DES EQUATIONS

´

D’EVOLUTION

Jean-Yves CHEMIN

Laboratoire J.-L. Lions

Universit´e Paris 6, Case 187

75 232 Paris Cedex 05, France

adresse ´electronique: chemin@ann.jussieu.fr

2

Chapter 1

A first approach of evolution

equations

The purpose of this course is to provide some basic techniques in order to study evolution

partial differential equations. In such equations, one variable (namely the time variable) plays

a special role. Let us first present three examples of linear partial differential equations which

we shall meet later on in the course :

• the heat equation which is a model for so called parabolic equations

∂t u − ∆u = 0

• the wave wave equation, which is a model for so called parabolic equations

∂t2 u − ∆u = 0

• Schr¨odinger equation which is a model for so called dispersive equation

∂t u + i∆u = 0.

To each of this equation, will correspond a non linear model in the following.

The heat equation will appear in the following system which is a model for the description

of the evolution of an incompressible viscous fluid. A fluid is described by a time dependant

vector fields v which is supposed to describe the speed of a pointwise particle located in x at

time t. The system is the following

∂t v + v · ∇v − ν∆v = −∇p

div v = 0

v|t=0 = v0

Here, ν denotes a postive real number which represents the viscosity of the fluid. This system

is known as the incompressible Navier-Stokes system.

3

For the wave equation, the following system is related to gas dynamics. The unknown is

the couple (ρ, v) which satisfies

∂t ρ + v · ∇ρ + ρ div v = 0

1

∂t v + v · ∇v + ∇p = 0

ρ

with p = Aργ . Here, ρ is a scalar function with values in R+

? and represents the density of

the particules of the gas at time t in the point x and v a time dependant vector field which

describes 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

2

c =

γ−1

def

∂p

∂ρ

1

2

1

(2γA) 2 γ−1

=

ρ 2 .

γ−1

The first equation becomes

∂t c + v · ∇c +

γ−1

c div v = 0.

2

About the second one, let us observe that

1

γ−1

c∇c = ∇p.

2

ρ

The Euler system related to gas dynamics becomes

∂t c + v · ∇c + γ − 1 c div v = 0

2

γ−1

∂t v + v · ∇v +

c ∇c = 0.

2

(1.1)

Let us assume that the solution is ”small”, i.e. is a perturbation of magnitude ε of a stationnary flat state v = 0 and c = c, by an easy computation of the coefficients of the powers of ε,

we infer

∂t c + γ − 1 c div v = 0

2

(1.2)

γ−1

∂t v +

c ∇c = 0.

2

An obvious computation ensures that

∂t2 c

−

γ−1

2

2

c2 ∆c = 0.

This equation is called ”acoustic waves equation”.

Then, we shall study non linear Schr¨odinger equations of the type

i

∂t u + ∆u = ±|u|p−1 u

2

for a real number greater or equal to 1.

4

1.1

A review on ordinary differential equation

Before starting the study of evolution partial differential equation, let us have a look on basic

properties of ordinary differential equations.

1.1.1

The linear case

Let E be a Banach space, I an open interval of R and A a map from I to L(E), the set of

continuous linear maps from E into E. We want to solve the equation

(

def du

u˙ =

= A(t)u(t)

(ODE)

dt

u(0) = u0 .

The proof of the existence and uniqueness of solutions of this equation is very simple. Let λ

be a positive real number, let us introduce the space Eλ definied by

Z t

o

n

def

Eλ = u ∈ C(I, E) / kukλ = sup ku(t)k exp −λ

kA(t0 )kL(E) dt0 < ∞ .

t∈I

0

The solution of (ODE) are the same as the solutions of

Z t

def

A(t0 )u(t0 )dt0 .

Lu = u0 with Lu(t) = u(t) −

0

We have

Z

t

k(Lu − u)(t)k ≤

kA(t0 )kL(E) ku(t0 )kdt0 .

0

Thus we deduce that

Z t

k(Lu − u)(t)k exp −λ

kA(t0 )kL(E) dt0

0

t

Z

≤

0

Z t

Z

00

00

0

exp −λ

kA(t )kL(E) dt kA(t )kL(E) exp −λ

t0

t0

kA(t00 )kL(E) dt00 ku(t0 )kdt0 .

0

By definition of k · kλ , we infer that

Z t

1

k(Lu − u)(t)k exp −λ

kA(t0 )kL(E) dt0 ≤ kukλ

λ

0

and thus that kLu − ukλ ≤ λ−1 kukλ . This implies that, for λ greater than 1, L is invertible

in L(E). Then the proof of the existence and uniqueness of solutions is achieved.

1.1.2

The case of non linear equations with almost lipschitz vector field

We still work in a Banach space E and an interval I of R. Let F be a function of I × E into E.

In the whole of this section, µ will denote a functionfrom R+ into itsself, vanishing at 0,

positive outside 0, continuous and non decreasing.

Definition 1.1.1 Let (X, d) and (Y, δ) be two metric spaces. We denote by Cµ (X, Y ) the

set of the bounded functions from X into Y such that a constant C exists such that, for

any (x, y) ∈ X 2 , we have

δ(u(x), u(y)) ≤ Cµ(d(x, y)).

5

Remark If (Y, δ) is a Banach space (which we denote (E, k·k) in this case), the space Cµ (X, E)

is a Banach space equipped with the norm

kukµ = kukL∞ +

sup

(x,y)∈X×X, x6=y

ku(x) − u(y)k

·

µ(d(x, y))

The following theorem provides hypotheses for which there is existence and uniqueness of

integral curve for an ordinary differential equation.

Theorem 1.1.1 Let E be a Banach space, Ω an open subset of E , I an open interval of R

and (t0 , x0 ) an element of I × Ω. Let us consider a function F of L1loc (I; Cµ (Ω; E)). Let us

assume in addition that

Z 1

dr

= +∞.

(1.3)

0 µ(r)

Then an interval J exists such that t0 ∈ J ⊂ I and such that the equation

Z

t

(ODE) x(t) = x0 +

F (t0 , x(t0 ))dt0

t0

has a unique continuous solution defined on the interval J.

Remark If µ(r) = r, this theorem is nothing more than the familiar Cauchy-Lipschitz

theorem. But let us point out that other functions satisfy the hypotheses of the theorem; for

instance the function defined by µ(r) = −r log r for r ≤ e−1 and µ(r) = e−1 if not.

In order to prove, let us begin by the proof of uniqueness of trajectories. Let x1 (t) and x2 (t)

two solutions of (ODE) defined on a neighbourhoood Je of t0 with this the same initial data x0 .

Let us denote

def

ρ(t) = ||x1 (t) − x2 (t)||.

As F belongs to L1loc (I; Cµ (Ω, E)), we have

Z

t

0 ≤ ρ(t) ≤

γ(t0 )µ(ρ(t0 ))dt0

with γ ∈ L1loc (I)

and

γ ≥ 0.

(1.4)

t0

In the case when µ(r) = r, Gronwall lemma implies that ρ ≡ 0. Let us recall a version ogf

Gronwalll lemma which will be useful in the following.

Lemma 1.1.1 Let f and g be two C 0 (resp. C 1 ) non negative functions on [t0 , T ]. Let A be

a continuous functions on [t0 , T ]. Suppose that, for t in [t0 , T ],

1d 2

g (t) ≤ A(t) g 2 (t) + f (t)g(t).

2 dt

Then for any time t in [t0 , T ], we have

Z

t

g(t) ≤ g(t0 ) exp

0

0

Z

t

A(t ) dt +

t0

t0

6

Z t

A(t00 ) dt00 dt0 .

f (t ) exp

0

t0

(1.5)

Let us define

Z t

def

A(t0 ) dt0

gA (t) = g(t) exp −

Z t

def

A(t0 ) dt0 .

and fA (t) = f (t) exp −

t0

t0

Obviously, we have

1d 2

g ≤ fA gA so that for any positive ε,

2 dt A

1

d 2

gA

fA ≤ fA .

(gA + ε2 ) 2 ≤

2 + ε2 ) 12

dt

(gA

By integration, we get

2

(gA

(t)

2

1

2

2

(gA

(0)

+ε ) ≤

2

1

2

Z

+ε ) +

t

fA (t0 ) dt0 .

0

Having ε tend to 0 gives the result.

The key lemma for the proof of Theorem 1.1.1 is the following.

Lemma 1.1.2 Let ρ be a measurable non negative function, γ a non negative function locally

integrable and µ a continuous non decreasing function. Let us assume that, for a non negative

real number a, the function ρ satisfies

Z t

ρ(t) ≤ a +

γ(t0 )µ(ρ(t0 ))dt0 .

(1.6)

t0

If a is positive, then we have

Z

t

−M(ρ(t)) + M(a) ≤

0

γ(t )dt

0

Z

with M(x) =

x

t0

1

dr

·

µ(r)

(1.7)

If a = 0 and if µ satisfies (1.3), then the function ρ is identically 0.

In order to prove this lemma, let us assume that a is positive and let us define

Z t

def

Ra (t) = a +

γ(t0 )µ(ρ(t0 ))dt0 .

t0

The function Ra is a continuous non decreasing function. Thus, we have,

R˙ a (t) = γ(t)µ(ρ(t)).

As the function µ is non decreasing, we have

R˙ a (t) ≤ γ(t)µ(Ra (t)).

(1.8)

The function Ra is positive. As the function M is C 1 on ]0, ∞[, Inequality (1.8) implies that

−

d

R˙ a (t)

M(Ra (t)) =

≤ γ(t).

dt

µ(Ra (t))

By inegration, we get (1.7) using that the function −M is increasing and that ρ ≤ Ra .

Let us assume now that a = 0 and let us proceed by contraposition. Let us assume

that the function ρ is not identically 0 near t0 . As the function µ is non decreasing, we can

7

substitute supt0 ∈[t0 ,t] ρ(t0 ) to the function ρ (we continue to use the same notation ρ). A real

number t1 > t0 exists, such that ρ(t1 ) > 0. As the function ρ satisfies (1.6) for a = 0,it

satisfies also this inequality for any positive a0 . It comes from (1.7) that

Z t1

0

0

∀a > 0 , M(a ) ≤

γ(t0 )dt0 + M(ρ(t1 )).

t0

This implies that the integral

1

Z

0

dr

µ(r)

is convergent; the proof of the lemma is done.

Thanks to Inequality (1.4), the uniqueness of integral curve issued from a point is an

obvious consequence of Lemma 1.1.2.

Let us prove the existence. In order to do so, let us consider the classsical Picard scheme

Z t

xk+1 (t) = x0 +

F (t0 , xk (t0 ))dt0 .

t0

Let us skip the proof of the fact that, for a small enough interval J, the sequence (xk )k∈N is

bounded in L∞ (J). Let us prove that this sequence is a Cauchy one in the space of continuous

functions from J to E. In order to do so, let us define

def

ρk+1,n (t) = ||xk+1+n (t) − xk+1 (t)||.

It turns out that

Z

t

0 ≤ ρk+1,n (t) ≤

γ(t0 )µ(ρk,n (t0 ))dt0

t0

def

Let us define ρk (t) = supn ||xk+1+n (t) − xk+1 (t)||. As the function µ is non decreasing,

we have

Z t

0 ≤ ρk+1 (t) ≤

γ(t0 )µ(ρk (t0 ))dt0 .

t0

Thanks to Fatou lemma, we get, using that the function µ is non decreasing,

Z t

def

ρe(t) = lim sup ρk (t) ≤

γ(t0 )µ(e

ρ(t0 ))dt0 .

k→+∞

t0

Applying again Lemma 1.1.2, we find that ρe(t) is identically 0 near t0 ; this concludes the

proof of Theorem 1.1.1.

Let us point out that the concepts of iterative scheme and of Cauchy sequence plays a key

role.

1.1.3

Blow up criteria

The existence and uniqueness theorem for ordinary differential equations is a local theorem.

Let us investigate what can be necessary conditions for a blow up phenomena.

Proposition 1.1.1 Let F be a function of R ×E in E satisfying the hypothesis of Theorem 1.1.1 in any point x0 of E. Let us assume in addition that a locally bounded function M

from R+ into R+ and a locally integrable function β from R+ into R+ such that

kF (t, u)k ≤ β(t)M (kuk).

8

then, if the maximal interval of definition is ]T? , T ? [, then, if T ? is finite,

lim sup ku(t)k = ∞.

t→T ?

Let us first prove that, if we consider a time T > T0 such that ku(t)k is bounded on the

interval [T0 , 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 [T0 , T [, the hypothesis on F that, for any t of the

interval [T0 , T [, we have

kF (t, u(t))k ≤ Cβ(t).

The function β being integrable on the interval [T0 , T ], we have deduce que, for any ε strictement positive, it exists a positive real number η such that, pour tout t and t0 such that T −t < η

and T − t0 < η,

ku(t) − u(t0 )k < ε.

The space E being complete, an element u? of E exists such that

lim u(t) = u? .

t→T ?

Applying Theorem 1.1.1, we construct solution of (ODE) on some [T+? , T1 ] and the continuous

function defined by induction on the interval [T0 , T1 ] is a solution of the equation (ODE) on

the interval [T0 , T1 ].

Corollary 1.1.1 Under the hypothesis of Proposition 1.1.1, if we have in addition that

kF (t, u)k ≤ M kuk2 ,

then, if the interval ]T? , T ? [ is the maximal interval of definition of u and if T ? is finite, then

Z

T?

kx(t)kdt = ∞.

t0

The solution satisfies, for any t ≥ t0

Z

t

kx(t)k ≤ kx(t0 )k + M

kx(t0 )k2 dt0 .

(1.9)

t0

Gronwall’s Lemma implies that

Z t

kx(t)k ≤ kx0 k exp M

kx(t0 )kdt0 .

0

A more precise way of proving this result is the following.

def

Let T = sup{t ∈ [t0 , T ? [ / kx(t)k ≤ 2kx(t0 )k}. For any t ∈ [t0 , T ? [, we have, using (1.9),

kx(t)k ≤ kx(t0 )k + 4M (t − t0 )kx(t0 )k2 .

Thus we infer

h

n

∀t ∈ t0 , min T, t0 +

oh

1

, kx(t)k ≤ 2kx0 k.

4M kx(t0 k)

9

Thanks to Proposition 1.1.1, we have

T ? − t0 ≥

c

·

kx0 k

Applying again this result at time t ∈ [t0 , T ? [, we find that

∀t ∈ [t0 , T ? [ ,

kx(t)k ≥

T?

c

·

−t

Exercice 1.1.1 Let F a function defined on R ×E such that

sup kF (t, x)k +

x∈E

sup

(x,y)∈E 2

0<kx−yk≤e−1

kF (t, x) − F (t, y)k

≤ β(t) with β ∈ L1loc (R).

−kx − yk log kx − yk

1) Prove that it existsd a map ψ from R ×E into E such that

Z t

ψ(t, x) = x +

F (t, ψ(t, x))dt

0

2) Prove that, for any t, ψ(t, ·) defines a homeomorphism of E such that

|x − y| ≤ e− exp

1.1.4

Rt

0

β(s)ds

⇒ |ψ(t, x) − ψ(t, y)|≤ |x − y|exp −

Rt

0

β(s)ds − exp −

e

Rt

0

β(s)ds

.

A compactness theorem : Peano’s theorem

The theorem is the following.

Theorem 1.1.2 (Peano) Let I be an open interval of R. Let us consider a function f

from I × Rd into Rd such that

• For any compact K of Rd , the function t 7→ kf (t)kL∞ (K) is locally integrable,

• For any t of I, the function x 7→ f (t, x) is continuous on Rd .

Then, for any point (t0 , x0 ) of I × Rd , an open interval J ⊂ I containing t0 and a continuous

function x on J exists such that

Z t

(ODE)

x(t) = x0 +

f (t0 , x(t0 ))dt0 .

t0

The structure of the proof is at least as intesting as the result itself. This proof will be

a model for the proof of existence of weak solutions for ther incompressible Navier-Stokes

equation we shall study in Chater 4.

There are three steps in the proof

• we regularize the function f and we apply Cauchy-Lipschitz’s Theorem to the sequence

of regularized functions; Proposition 1.1.1 ensures that the solutions of the regularized

problem have a commun interval of definition,

• then, we prove that the sequence of those solutions of the regularized problem are

relatively compact in the space C(J, Rd ),

• as a conclusion, we pass to the limit.

10

Let us procede to a classical regularization; let χ a non negative function of D(B(0, 1))

def

the integral of which is 1. Let us define χn (x) = nd χ(nx) and fn (t) = χn ? f (t). We have

kfn (t)kL∞ (K) ≤ kf (t)kL∞ (K+B(0,n−1 )) .

Morevoer, we have

k∂j fn (t)kL∞ (K) ≤ C(n + 1)kf (t)kL∞ (K+B(0,n−1 )) .

We can apply Cauchy-Lipschitz’s Theorem of to the function fn . Let Jn the maximal interval

of definition of xn . Let J an interval ouvert such that

Z

kf (t)kL∞ (B(x0 ,2) dt ≤ 1.

J

n

o

def

Let us define tn = sup t ∈ [t0 , ∞[∩J ∩ Jn / ∀t0 ≤ t , x(t0 ) ∈ B(x0 , 1) . For any t ≤ tn , we

have

Z

kfn (t)kL∞ (B(x0 ,1)) dt

kxn (t) − x0 k ≤

ZJ

≤

kf (t)kL∞ (B(x0 ,2)) dt

J

≤ 1.

Thus tn ≥ sup J ∩ Jn . working in the same way for the times less to t0 , we find, using

Proposition 1.1.1 that, for any n, J ⊂ Jn . This concludes the first part of the proof.

We have

d´ef

∀t ∈ J , X(t) = {xn (t), n ∈ N} ⊂ B(x0 , 1).

As we work on a finite dimensionnal space, X(t) is relatively compact. Moreover, we have

Z t0

0

kxn (t) − xn (t )k ≤

kfn (t00 )kL∞ (B(x0 ,1)) dt00

t

Z

≤

t0

t

kf (t00 )kL∞ (B(x0 ,2)) dt00 .

Thus, for any positive , it exists a positive real number α such that

∀(t, t0 ) ∈ J 2 , |t − t0 | < α =⇒ kxn (t) − xn (t0 )k < .

In other words, the family (xn )n∈N is equicontinuous on J. Ascoli’s Theorem ensures that

the set of functions xn is relatively compact in C(J; Rd ). Thus we can extract a subsequence

which converge uniformely on J to a function x of C(J; Rd ). Let omit 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)kL∞ (B(x0 ,1)) + kf (t, xn (t)) − f (t, x(t))k.

Thus for any t of J, we have

lim fn (t, xn (t)) = f (t, x(t)).

n→∞

Moreover, kfn (t, xn (t)k ≤ kf (t)kL∞ (B(x0 ,2)) . Lebesgue’s Theorem ensures that, for any t, we

have

Z t

Z t

0

0

0

lim

fn (t , xn (t ))dt =

f (t0 , x(t0 ))dt0 .

n→∞ t

0

t0

The theorem is proved.

11

Remarks

• All the theorems and all the proofs of this chapter must be known.

• To know more about ordinary differential equations and their historical aspect of Osgood’s theory, see the book by T.M. Fleet, Differential analysis, Cambridge University

Press, 1980.

• To know more about non lipschiztian vector fields satisfying Osgood condition, see the

book by J.-Y. Chemin, Fluides parfaits incompressibles, Ast´erisque, 230, 1995 or its

english version Incompressible perfect fluids, Oxford University Press, 1998.

12

Chapter 2

Sobolev spaces

Introduction

In this course, we shall restrict ourselves to Sobolev spaces modeled on L2 . These spaces

definitely play a crucial role in the study of partial differential equations, linear or not. The

key tool will be the Fourier transform.

2.1

Definition of Sobolev spaces on Rd

Definition 2.1.1 Let s be a real number, a tempered distribution u belongs to the Sobolev

space of index s, denoted H s (Rd ), or simply H s if no confusion is possible, if and only if

u

b ∈ L2loc (Rd ) and u

b(ξ) ∈ L2 (Rd ; (1 + |ξ|2 )s dξ).

and we note

kuk2H s

Z

(1 + |ξ|2 )s |b

u(ξ)|2 dξ.

=

Rd

Proposition 2.1.1 For any s real number, the space H s , equipped with the norm k · kH s , is

a Hilbert space.

The fact that the norm k · kH s comes from the scalar product

Z

def

(u|v)H s =

(1 + |ξ|2 )s u

b(ξ)b

v (ξ)dξ

Rd

is obvious. Let us prove that this space is complete. Let (un )n∈N a Cauchy sequence of

H s . By definition of the norm, the sequence (b

un )n∈N is a Cauchy sequence of the space

L2 (Rd ; (1 + |ξ|2 )s dξ). Thus, a function u

e exists in the space L2 (Rd ; (1 + |ξ|2 )s dξ) such that

lim kb

un − u

ekL2 (Rd ;(1+|ξ|2 )s dξ) = 0.

n→∞

(2.1)

In particular, the sequence(b

un )n∈N tends to u

e in the space S 0 of tempered distribtions. Let u =

F −1 u

e. As the Fourier transform is an isomorphism of S 0 , the sequence (un )n∈N tends to u in

the space S 0 , and also in H s thanks to (2.1).

Shortly said, this is nothing more than observing that the Fourier transform is an isometric

isomorphism from H s onto L2 (Rd ; (1 + |ξ|2 )s dξ).

13

Proposition 2.1.2 Let s be a non negative integer, the space H s (Rd ) is the space of functions u of L2 all the derivatives of which of order less or equal to m are distributions which

belongs to L2 . Moreover, the space H m equipped with the norm

def X

e

k∂ α uk2L2

kuk2H m =

|α|≤m

is a Hilbert space and this norm is equivalent to the norme k · kH s .

The fact that

e

kuk2H m = e(u|u)H m

Z

d´ef X

e

m

with (u|v)H =

|α|≤m R

∂ α u(x)∂ α v(x)dx.

d

ensures that the norm k · kH m comes from a scalar product. Moreover, a constant C exists

such that

X

X

(2.2)

|ξ|2|α| .

|ξ|2|α| ≤ (1 + |ξ|2 )s ≤ C 1 +

∀ξ ∈ Rd , C −1 1 +

0<|α|≤m

0<|α|≤m

As the Fourier transform is, up to a constant, an isometric isomorphism from L2 onto L2 , we

have

∂ α u ∈ L2 ⇐⇒ ξ α u

b ∈ L2 .

Thus, we have deduce that

u ∈ H m ⇐⇒ ∀α / |α| ≤ m , ∂ α u ∈ L2 .

Inequality (2.2) ensures the equivalence of the two norms using again the fact that the Fourier

transform is a isometric isomorphism up to a constant. The proposition is proved.

Exercice 2.1.1 Prove that the space S is continuously included in the space H s for any

real s.

d

Exercice 2.1.2 Prove that the mass of Dirac δ0 belongs to the space H − 2 −ε for any positive

d

real number ε. Prove that δ0 does not belong to the space H − 2 .

Exercice 2.1.3 Prove that, for any distribution to support compact u, it exists a real number s such that u belongs to the Sobolev space H s .

Exercice 2.1.4 Prove that the constant 1 does not belong to H s for any real number s.

Proposition 2.1.3 Let s a real number of the interval ]0, 1[. Prove that the space H s is the

space des functions u of L2 such that

Z

|u(x + y) − u(x)|2

dxdy.

|y|d+2s

Rd × Rd

Moreover, a constant C exists such that, for any function u of H s , we have

Z

|u(x + y) − u(x)|2

−1

2

2

C kukH s ≤ kukL2 +

dxdy ≤ Ckuk2H s .

|y|d+2s

Rd × Rd

14

Thanks to Fourier-Plancherel identity,we can write that

Z

Rd

|u(x + y) − u(x)|2

dx =

|y|d+2s

It turns out that

Z

Rd × Rd

Z

Rd

|u(x + y) − u(x)|2

dxdy

|y|d+2s

F (ξ)

|ei(y|ξ) − 1|2

|b

u(ξ)|2 dξ < ∞.

|y|d+2s

Z

=

F (ξ)|b

u(ξ)|2 dξ

with

Rd

def

Z

=

Rd

|ei(y|ξ) − 1|2 dy

.

|y|2s

|y|d

By an obvious change of variable, we see that the function F is radial and homogeneous of

degree 2s. Thus

Z iy1

|e − 1|2 dy

2s

F (ξ) = |ξ|

·

|y|2s |y|d

Let us prove now an interpolation inequality which will be very useful.

Proposition 2.1.4 If s = θs1 + (1 − θ)s2 with θ ∈ [0, 1], then, we have

kukH s ≤ kukθH s1 kuk1−θ

H s2 .

The proof consits in applying H¨older inequality with the measure |b

u(ξ)|2 dξ and the two

2

θs

2

(1−θ)s

1

2

functions (1 + |ξ| )

and (1 + |ξ| )

.

Theorem 2.1.1 Let s a real quelconque;

• the space D(Rd ) is dense in H s (Rd ),

• the multiplication by a function of S est a continuous function of H s into lui-mˆeme.

In order to prove the first point of this theorem, let us consider a distribution u of H s such

that, for any test function ϕ, we have (ϕ|u)H s = 0. This means that, for any test function ϕ,

we have

Z

ϕ(ξ)(1

b

+ |ξ|2 )s u

b(ξ)dξ = 0.

Rd

As S is continuously included in H s , as D is dense in S, and as the Fourier transform an

isomorphism of S, we have, for any function f of S,

Z

f (ξ)(1 + |ξ|2 )s u

b(ξ)dξ = 0.

Rd

As S is dense in L2 , this implies that (1 + |ξ|2 )s u

b(ξ) = 0, thus u

b = 0 and thus u = 0.

Let us prove now the second second point of the theorem. This proof is presented here

just for culture. We know that

ϕu

c = (2π)−d ϕ

b?u

b.

The point is to estimate the L2 norm of the function defined by

Z

2 2s

|ϕ(ξ

b − η)| × |b

u(η)|dη.

U (ξ) = (1 + |ξ |)

Rd

15

Let us define I1 (ξ) = {η / 2|ξ − η| ≤ |η|} and I2 (ξ) = {η / 2|ξ − η| ≥ |η|}. It is clear that we

have

U (ξ) = U1 (ξ) + U2 (ξ)

Z

2 2s

Uj (ξ) = (1 + |ξ| )

Ij (ξ)

with

|ϕ(ξ

b − η)| × |b

u(η)|dη.

Let us first observe that, if η ∈ I1 (ξ) , then

1

3

|η| ≤ |ξ| ≤ |η|.

2

2

We deduce that, for any real number s, a constant C exists such that, for any couple (ξ, η)

such that η belongs to I1 (ξ),

(1 + |ξ|2 )s ≤ C(1 + |η|2 )s .

Thus it turns out that

Z

U1 (ξ) ≤ C

s

Rd

u(η)|dη.

|ϕ(ξ

b − η)|(1 + |η|2 ) 2 |b

As ϕ

b belongs to S, it also belongs to L1 and we have

kU1 kL2 ≤ Ckϕk

b L1 kukH s .

But if η belongs to I2 (ξ), we have

1

3

5

|ξ − η| ≤ |ξ| ≤ |ξ − η| and |η| ≤ |ξ − η|.

2

2

2

Then we deduce that

2

|s|

2

Z

|s|

s

U2 (ξ) ≤ C(1 + |ξ| )

|ϕ(ξ

b − η)|(1 + |η|2 ) 2 (1 + |η|2 ) 2 |b

u(η)|dη

Rd

Z

s

≤ C

|ϕ(ξ

b − η)|(1 + |ξ − η|2 )|s| (1 + |η|2 ) 2 |b

u(η)|dη.

Rd

We know that ϕ

b belongs to S. A constant C exists such that

|ϕ(ζ)|

b

≤ C(1 + |ζ|2 )−

d+1

−|s|

2

.

Thus it tuns out that

Z

U (ξ) ≤ C

(1 + |ξ − η|2 )−

d+1

2

s

(1 + |η|2 ) 2 |b

u(η)|dη.

Rd

Thus kU2 kL2 ≤ CkukH s ; this concludes the proof of the theorem.

Exercice 2.1.5 Let FL1 = {u ∈ S 0 / u

b ∈ L1 }. Prove that, for any non negative real

number s, the product is a bilinear continuous map from FL1 ∩ H s × FL1 ∩ H s into FL1 ∩ H s .

What happens when s is greater than d/2?

16

Exercice 2.1.6 Let s a real number greater than 1/2. Prove that the map γ defined by

D(Rd ) −→ D(Rd−1 )

γ

ϕ

7−→ γ(ϕ) : (x2 , · · · , xd ) 7→ ϕ(0, x2 , · · · , xd )

1

can be extended in a continuous onto map from H s (Rd ) onto H s− 2 (Rd−1 ).

Hint : Write

Z

ϕ(ξ

b 1 , ξ2 , · · · , ξd )dξ1 .

FRd−1 ϕ(0, ξ2 , · · · , ξd ) = (2π)−1

R

and for the fact that the map in onto, observe that, if

−(n−1)

u = (2π)

Cs F

−1

!

1

(1 + |ξ 0 |2 )s− 2

0

vb(ξ ) ,

(1 + |ξ|2 )s

then u ∈ H s and γ(u) = v.

Let us prove a theorem which describes the dual of the space H s .

Theorem 2.1.2 The bilinear form B defined by

S ×S → C

Z

B

u(x)ϕ(x)dx

(u, ϕ) 7→

Rd

can b extended as a bilinear form continuous from H −s × H s to C. Moreover, the map δB

defined by

−s

H

−→ (H s )0

δB

u 7−→ δB (u) : (ϕ) 7→ B(u, ϕ)

is a linear and isometric isomorphism (up to a constant), which means that the bilinear form B

identifies the space H −s to the dual space of H s .

The important point of the proof of this theorem is inverse Fourier formula which ensures

that, for any couple (u, ϕ) of functions of S, we have

Z

B(u, ϕ) =

u(x)ϕ(x)dx

d

ZR

=

u(x)F(F −1 ϕ)(x)dx

Rd

Z

=

u

b(ξ)(F −1 ϕ)(ξ)dξ

Rd

Z

−d

= (2π)

u

b(ξ)ϕ(−ξ)dξ.

b

(2.3)

Rd

s

Multiplying and dividing by (1 + |ξ|2 ) 2 , we immediately get thanks to Cauchy-Schwarz inequality,

|B(u, ϕ)| ≤ (2π)−d kukH s kϕkH −s .

Thus the first point of the theorem. The fact that the map δB is one to one comes from the

fact that if, for any function ϕ ∈ S, we have B(u, ϕ) = 0, then u is 0. We have to prove

that the map is onto. In fact, we shall prove that δB is one to one and onto. For any real

17

number σ, the Fourier transform is an isometric (up to a constant) isomorphism from H σ

ˇ defined by

onto L2 (Rd , (1 + |ξ|2 )σ dξ). Let us now consider the bilinear form B

L2 (Rd , (1 + |ξ|2 )−s dξ) × L2 (Rd , (1 + |ξ|2 )s dξ) −→ C

Z

ˇ

B

−d

f (ξ)φ(−ξ)dξ.

(φ, f )

7−→ (2π)

Rd

If we prove that

δB = t FδBˇ F,

(2.4)

then Theorem 2.1.2 will be proved. Indeed, as F is an isomorphism from H s onto L2 (Rd , (1 +

|ξ|2 )s dξ), the map t F is an isomorphism from (L2 (Rd , (1 + |ξ|2 )s dξ))0 onto (H s )0 . We know

that δBˇ is an isomorphism from the space (L2 (Rd , (1 + |ξ|2 )s dξ))0 onto the space L2 (Rd , (1 +

|ξ|2 )−s dξ).

In order to prove Formula (2.4), let us write that

ht FδBˇ Fu, ϕi = hδBˇ Fu, Fϕi

= δBˇ (Fu, Fϕ).

Thanks to Identity (2.3), we have ht FδBˇ Fu, ϕi = hδB (u), ϕi. Thus the theorem is proved.

2.2

Sobolev embeddings

The purpose of this section is the study of embedding properties of Sobolev spaces H s (Rd )

into Lp spaces . Let us prove the following theorem.

Theorem 2.2.1 If s is greater than d/2, then the space H s is continuously included in the

space of continuous functions which tend to 0 at infinity. If s is a positive real number less

2d

than d/2, then the space H s is continuously included in L d−2s and we have

kf k

Lp

def

≤ Ckf kH˙ s

with kf kH˙ s =

Z

2s

R

d

|ξ| |fb(ξ)|2 dξ

1

2

.

The first point of this theorem is very easy to prove. Let us use the fact that

kukL∞ ≤ (2π)−d kb

ukL1

(2.5)

Indeed, if s is greater than d/2, we have,

|b

u(ξ)| ≤ (1 + |ξ|2 )−s/2 (1 + |ξ|2 )s/2 |b

u(ξ)|.

The fact that s is greater than d/2 implies that the function

ξ 7→ (1 + |ξ|2 )−s/2

belongs to L2 . Thus, we have

Z

kb

ukL1 ≤

2 −s

(1 + |ξ| )

The first point of the theorem is proved.

18

1

2

dξ

kukH s .

(2.6)

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 us

denote by vλ the function vλ (x) = v(λx). We have

kvλ kLp = λ

− dp

kvkLp

and also

Z

|ξ|2s |vbλ (ξ)|2 dξ = λ−2d

Z

|ξ|2s |b

v (λ−1 ξ)|2 dξ

= λ−d+2s kvk2H˙ s ,

with

def

kvk2H˙ s =

Z

|ξ|2s |b

v (ξ)|2 dξ.

Rd

The two quantities k · kLp and k · kH˙ s 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 kf kH˙ s = 1. On utilise then the fact that for any p de the interval ]1, +∞[, we have, for

any function measurable f ,

Z ∞

p

kf kLp = p

λp−1 m(|f | > λ)dλ.

0

Let us decompose f in a low and in a high frequencies by writing

f = f1,A + f2,A

with f1,A = F −1 (1B(0,A) fb)

and

f2,A = F −1 (1B c (0,A) fb).

(2.7)

As the support of the Fourier transform of f1,A is compact, the function f1,A is bounded and

more precisely,

kf1,A kL∞

≤ (2π)−d kfd

1,A kL1

Z

≤ (2π)−d

|ξ|−s |ξ|s |fb(ξ)|dξ

B(0,A)

!1

2

Z

≤ (2π)−d

|ξ|−2s dξ

B(0,A)

C

≤

d

(d − 2s)

1

2

A 2 −s .

(2.8)

The triangle inequality implies that, for any positive real number A,

(|f | > λ) ⊂ (2|f1,A | > λ) ∪ (2|f2,A | > λ).

Using Inequality (2.8), we have

1

def

A = Aλ =

λ(d − 2s) 2

4C

Thus we deduce that

kf kpLp

Z

=p

!p

d

λ

=⇒ m |f1,A | >

2

∞

λp−1 m(2|f2,Aλ | > λ)dλ.

0

19

= 0.

it is well known (this is Bienaim´e-Tchebychev inequality) that

Z

λ

=

m |f2,Aλ | >

dx

2

(|f2,Aλ |> λ2 )

Z

4|f2,Aλ (x)|2

dx

≤

λ2

(|f2,A |> λ )

2

λ

kf2,Aλ k2L2

≤ 4

·

λ2

For such a choice of A, we have

∞

Z

kf kpLp

λp−3 kf2,Aλ k2L2 dλ.

≤ 4p

0

(2.9)

As the Fourier transform is (up to a constant) an isometric isomorphism of L2 , we have

Z

2

−d

kf2,Aλ kL2 = (2π)

|fb(ξ)|2 dξ.

(|ξ|≥Aλ )

Thanks to Inequality (2.9), we get

Z

kf kpLp ≤ 4p(2π)−d

R+ × R

d

λp−3 1{(λ,ξ) /

b 2

|ξ|≥Aλ } (λ, ξ)|f (ξ)| dξdλ.

By definition of Aλ , we have

4C

d´ef

|ξ| ≥ Aλ ⇐⇒ λ ≤ Cξ =

d

(d − 2s)

1

2

|ξ| p .

Fubini’s theorem implies that

kf kpLp

−d

≤ 4p(2π)

λ

R

≤ 4

As 2s =

Cξ

Z

Z

d

p(2π)d

p−2

p−3

dλ |fb(ξ)|2 dξ

0

!p−2 Z

4C

(d − 2s)

|ξ|

1

2

R

d(p−2)

p

d

|fb(ξ)|2 dξ.

d(p − 2)

, the theorem is proved.

p

def

Corollary 2.2.1 Let p ∈]2, ∞[, and s > sp = d

1

kukLp ≤ Ckuk1−θ

kukθH˙ s

L2

2

−

1

. We have

p

with θ =

sp

·

s

The proof of this corollary is an application of the above theorem together with the fact

that

kukH˙ θs1 +(1−θ)s2 ≤ kukθH˙ s kuk1−θ

.

H˙ s2

1

20

2.3

Homogeneous Sobolev spaces

Definition 2.3.1 Let s be a real number, the homogeneous Sobolev space H˙ s is the space of

tempered distribtions such that u

b belongs to L1loc and satisfies

Z

2 def

|ξ|2s |b

u(ξ)|2 dξ < ∞.

kukH˙ s =

Rd

These spaces (or at least theirs norms) naturally appeared in the proof of Theorem 2.2.1.

The k · kH˙ s norm has the following scaling property

d

kf (λ·)kH˙ s = λ− 2 +s kf kH˙ s .

These spaces are different from the inhomogeneous H s spaces. Let us notice that if s is

positive, then H s is included in H˙ s but that if s is negative, then H˙ s is included in H s . The

inhomongeneous spaces is a decreasing family of spaces (with respect to the index s). The

homogeneous ones are not comparable together.

We shall only consider theses homogeneous spaces in the case when s is less than the half

dimension.

Proposition 2.3.1 If s < d/2, then the space H˙ s is a Banach space.

Let (un )n∈N a Cauchy sequence of H˙ s . The sequence (b

un )n∈N is a Cauchy one in the Banach

d

2

2s

space L (R \{0}; |ξ| dξ). Let f be its limit. It is clear that f belongs to L1loc (Rd \{0}).

Moreover,

Z

Z

1 Z

1

2

2

2s

2

|f (ξ)|dξ ≤

|ξ| f (ξ)| dξ

|ξ|−2s dξ

<∞

Rd

B(0,1)

B(0,1)

d´ef

because s is less than the half-dimension. Thus fb belongs to S 0 and to L1loc . Thus u = F −1 f

is well defined, belongs to H˙ s , and is the limit of the sequence (un )n∈N in the sense of the

norm H˙ s .

Exercice 2.3.1 1) Prove that the space

def

B = {u ∈ S 0 (Rd ) , u

b ∈ L1 (B(0, 1); dξ) ∩ L2 (Rd ; |ξ|2s dξ)}

def

equipped with the norme N (u) = kb

ukL1 (B(0,1)) + kukH˙ s is a Banach space.

2) Let s ≥ d/2. Give an example of a sequence (fn )n∈N of B, bounded in H˙ s (Rd ), such

that

lim N (fn ) = +∞.

n→∞

3) Then deduce that

(H˙ s , k

· kH˙ s ) is not a Banach space.

Exercice 2.3.2 Prove that, if k ∈ N, then we have

n

X

H˙ −k (Rd ) = u ∈ S 0 (Rd ) , u =

∂ α fα

o

with fα ∈ L2 .

|α|=k

Prove that a constant C exists such that

X

1

X

2

−1

2

α

C kukH˙ −k ≤ inf

kfα kL2 / u =

∂ fα ≤ CkukH˙ −k .

|α|=k

|α|=k

21

2.4

The spaces H01 (Ω) and H −1 (Ω)

Definition 2.4.1 Let Ω a domain of Rd , the space H01 (Ω) is defined as the closure of D(Ω)

in the sense of the norm H 1 (Rd ) .

The space H −1 (Ω) is the set of distributions u on Ω such that

def

kukH −1 (Ω) =

sup

|hu, f i| < ∞.

f ∈D(Ω)

kf kH 1 (Ω) ≤1

0

Proposition 2.4.1 The space H01 (Ω) is a Hilbert space equipped with the norm

kuk2L2 (Ω) + k∇uk2L2 (Ω)

1

2

.

The proof is an easy exercice left to the reader. The space H −1 (Ω) can be indentified to the

dual space of H01 (Ω) thanks to the following theorem.

Theorem 2.4.1 The bilinear map defined by

−1

H (Ω) × D(Ω) −→ C

B

(u, ϕ)

7−→ hu, ϕi

can be extended to a bilinear continuous map from H −1 (Ω) × H01 (Ω) into C, still denoted

by B. Moreover, the map δB defined by

(

H −1 (Ω) −→ (H01 (Ω))0

δB

def

u

7−→ δB (u)(ϕ) = B(u, ϕ)

is a linear isometric isomorphism between the space H −1 (Ω) and the dual space of H01 (Ω).

The fact that the bilinear map B can be extended because B is uniformely continuous. Let `

a linear form continuous on H01 (Ω). Its restriction on D(Ω) is a distribution u on Ω such that

∀ϕ ∈ D(Ω) , hu, ϕi = h`, ϕi.

By definition of the norm on (H01 (Ω))0 , the theorem is proved.

Theorem 2.4.2 (Poincar´

e Inequality) Let Ω be bounded open subset of Rd . A constant C exists such that

1

2

d

X

1

2

∀ϕ ∈ H0 (Ω) , kϕkL2 ≤ C

k∂j ϕkL2

.

j=1

Let R a positive real number such that Ω is included in ] − R, R[× Rd−1 .Then, for any test

function ϕ, we have

Z x1

∂ϕ

ϕ(x1 , · · · , xd ) =

(y1 , x2 , · · · , xd )dy1 .

∂y

1

−R

22

Cauchy-Schwarz Inequality implies that

2

∂ϕ

dy1 .

(y

,

x

,

·

·

·

,

x

)

|ϕ(x1 , · · · , xd )| ≤ 2R

1

2

d

−R ∂y1

x1

Z

2

By integration in x1 , we get

Z

Z

2

|ϕ(x1 , · · · , xd )| dx1 ≤ 2R

2

∂ϕ

dy1 .

(y

,

x

,

·

·

·

,

x

)

1

2

d

Ω×]−R,R[ ∂y1

Ω

Then, integrating with respect to the other d − 1 variables, we find

2

Z

Z

∂ϕ

2

|ϕ(x1 , · · · , xd )| dx ≤ 2R

∂y1 (y1 , x2 , · · · , xd ) dy1 dx2 · · · dxd

Ω×]−R,R[

Ω

≤ 4R2

d

X

k∂j ϕk2L2 .

j=1

As D(Ω) is dense in H01 (Ω), the theorem is proved. It obviously implies the following corollary.

Corollary 2.4.1 The space H01 (Ω) equipped with the norm

u 7−→

X

d

k∂j uk2L2

1

2

def

= k∇ukL2

j=1

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 compactness

theorem.

Theorem 2.4.3 Let K be a compact of Rd and (s, s0 ) a couple of real numbers such that s0 <

s the space of distributions of H s (Rd ) the support of which is included

s. Let us denote by HK

s in H s0 is compact.

in K. The embedding of HK

K

Before proving this theorem, let us give some immediate corollaries.

Theorem 2.4.4 Let Ω be a bounded open subset of Rd with d ≥ 2. If p is a real number less

than

def 2d ,

pc =

d−2

1

p

then the embedding of H0 (Ω) in L (Ω) is compact.

Let us prove Theorem 2.4.3. Without any loss on generalitry, we can assume that the

compact K is included in the interior of the cube [0, 2π]d . Now let us introduce the linear

map defined by

∞ (Td )

DK −→ CX

P

u(x − k)

u 7−→

k∈2π Zd

where DK denotes the set of smooth compactly supported functions the support of which

s in H s (Td )

is included in K. This map can be extended to a linear continuous map of HK

defined by

def X

H s (Td ) = {u ∈ D(Td ) / kuk2H s (Td ) =

(1 + |n|2 )s |b

u(n)|2 < +∞}.

n∈Z

23

This comes from the fact that a constant C exists such that

∀ϕ ∈ DK , kP ϕkH s (Td ) ≤ CkϕkH s .

(2.10)

In order to prove this, let us consider a function χ of D]0, 2π[d ) with value 1 near K. Then,

let us write

Z

ϕ(n)

b

=

ϕ(x)e−inx dx

Z

=

ϕ(x)χ(x)e−inx dx

Z

−inx

= (2π)−d ϕ(ξ)F(χe

b

)(ξ)dξ

Z

= (2π)−d ϕ(ξ)b

b χ(n − ξ)dξ.

As χ is a smooth compactly supported function, for any integer N , a constant CN exists such

that

Z

|ϕ(ξ)|

b

|ϕ(n)|

b

≤ CN

dξ.

−N

Rd (1 + |n − ξ|)

The result is an obvious consequence of the following two lemmas.

Lemma 2.4.1 For any (a, b) ∈ Rd , for any s ∈ R, we have

s

(1 + |a + b|2 ) 2 ≤ 2

|s|

2

(1 + |a|2 )

|s|

2

s

(1 + |b|2 ) 2 .

Lemma 2.4.2 Let (Xj,µj ) two measured spaces and k a function measurable from X1 × X2

into R such that

Z

Z

n

o

d´ef

M = max sup

|k(x1 , x2 )|dµ1 (x1 ), sup

|k(x1 , x2 )|dµ2 (x2 ) < ∞.

x2 ∈X2

x1 ∈X1

X1

X2

Then the map defined by

Z

(Kf )(x2 ) =

k(x1 , x2 )f (x1 )dµ1 (x1 ),

X1

maps Lp (X1 , dµ1 ) into Lp (X2 , dµ2 ) for any p ∈ [1, ∞]. More precisely, we have

kKf kLp (X2 ,dµ2 ) ≤ M kf kLp (X1 ,dµ1 ) .

Proof of Lemma 2.4.1 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 case

when s is negative, we have

s

s

s

s

(1 + |b|2 )− 2 ≤ 2− 2 (1 + |a + b|2 )− 2 (1 + |a|2 )− 2 .

Thus the result is proved.

24

0

Proof of the Lemma 2.4.2 Let g be an element of norm 1 of Lp (X2 , dµ2 ). We have

Z

Z

|k(x1 , x2 )| |f (x1 )| |g(x2 )|dµ1 (x1 )dµ2 (x2 ).

|Kf (x2 )| |g(x2 )|dµ2 (x2 ) ≤

X1 ×X2

X2

H¨older Inequality for the measure |k(x1 , x2 )|dµ1 (x1 )dµ2 (x2 ) implies that

Z

1

Z

p

p

|Kf (x2 )| |g(x2 )|dµ2 (x2 ) ≤

|f (x1 )| |k(x1 , x2 )|dµ1 (x1 )dµ2 (x2 )

X1 ×X2

X2

10

p

.

|g(x2 )| |k(x1 , x2 )|dµ1 (x1 )dµ2 (x2 )

Z

p0

×

X1 ×X2

Then Fubini’s Theorem ensures the result.

0

Now let us prove that the embedding of H s (Td ) in H s (Td ) is compact. In order to do so,

let us observe that, if we define

X

def

i(n|x)

iN (ϕ) = (2π)−d

ϕ(n)e

b

,

n≤|N |

where N denotes any integer, then we have

X

0

2

|ϕ − iN (ϕ)|2H s0 (Td ) =

(1 + |n|2 )s |ϕ(n)|

b

n>|N |

0

= (1 + |N |2 )s −s

X

2

(1 + |n|2 )s |ϕ(n)|

b

n>|N |

s0 −s

= (1 + |N |2 )

|ϕ|2H s0 (Td ) .

0

Thus, the embedding i of H s (Td ) in H s (Td ) is compact as a limit of finite rank operators.

Let us consider the map defined by

s0 d

0

H (T ) −→ H s (Rd )

Mχ

u

7−→ χu

Using the Fourier transform of u, we get

Z

F(χu)(ξ) =

e−i(x|ξ) χ(x)u(x)dx

R

Z

X

u

b(n)

e−i(x|ξ−n) χ(x)dx

d

=

Rd

n∈Zd

=

X

u

b(n)b

χ(n − ξ).

d

n∈Z

The proof of the continuity of Mχ is strictly analogous to the proof of the continuity of P ,.

s in H s0 is equal to P ◦ i ◦ M ; the theorem is

Moreover, it is clear that the embedding of HK

χ

K

proved.

Remarks

The proofs of this chapter must be known except the proof of the second point of Theorem 2.1.1.

To know more about Sobolev spaces, the reader can conslut the classical book

R. A. Adams, Sobolev spaces, Pure and Applied Mathematics, Vol. 65, Academic Press,

1975.

25

26

Chapter 3

Extrema problem and the least

action principle

3.1

The problem of Dirichlet vu comme a problem d’extremum

In this section and also in the following one, Ω denotes a bounded domain of Rd . Let f be an

element of H −1 (Ω), let us consider the functionnal F defined par

( 1

H0 (Ω) −→ R

F

1

u

7−→

k∇uk2L2 − hf, ui.

2

Dirichlet Theorem is the following:

Theorem 3.1.1 The functionnal F has a unique minimum which is the unique solution

in H01 (Ω) of −∆u = f in the distribution sense in Ω.

In order to prove this theorem, let us observe that the functionnal F bounded from below

because

1

F (u) ≥

k∇uk2L2 − k∇ukL2 kf kH −1 (Ω)

2

2 1

1

≥

k∇ukL2 − kf kH −1 (Ω) − kf k2H −1 (Ω) .

(3.1)

2

2

The functionnal F has a lower bound m. Let us consider a minimizing sequence (un )n∈N i.e.

a sequence (un )n∈N such that lim F (un ) = m. Using lnequality (3.1), we have

n→∞

k∇un kL2 ≤ 2F (un ) + kf kH −1 (Ω)

1

2

+ kf kH −1 (Ω) .

The sequence (un )n∈N is a bounded sequence of the space H01 (Ω) which is complete. Thus it

exists a function u in H01 (Ω) and a subsequence of (un )n∈N (which we still denote by (un )n∈N )

such that (un )n∈N tends weakly to u. Moreover, we know that the sequence (k∇un kL2 )n∈N

converges to m + hf, ui. Thanks to the properties of the weak limit we have

lim k∇un kL2 ≥ k∇ukL2 .

n→∞

Let us assume that ∇ukL2 < lim k∇un kL2 . Then, we have F (u) < m which is in contradicn→∞

tion with the fact that m is the infimum of F . Thus

lim k∇un kL2 = k∇ukL2

n→∞

27

and then the lower bound is a minimum and the sequence (un )n∈N converges strongly to u

in H01 (Ω)).

Now let us prove that u is a solution of Laplace Equation. The functionnal F is the sum

of the quadratic functionnal (the norm to the square) and of of a linear functionnal (both

continuous). We have, for any function h of H01 (Ω),

F (u + h) = F (u) + 2(∇u|∇h)L2 − hf, hi + k∇hk2L2 .

(3.2)

If u is a minumum, then the differential vanishes at u and thus u is a solution of Laplace

Equation. Moreover, Relation (3.2) implies that the minimum is unique and it is characterised

by the fact that, for any h in H01 (Ω), we have (∇u|∇h)L2 − hf, hi = 0. Thus the theorem is

proved.

Exercice 3.1.1 Let Ω a bounded domain of Rd and f a distribution of H −1 (Ω). Prove that

a vector field v exists in L2 (Ω) such that div v = f .

Let us prove now a result about the spectral strucutre of the Laplacian in a bounded

domain.

Theorem 3.1.2 It exists a non decreasing sequence (λk )k∈N of positive real numbers which

thends to infinity and a hilbertian basis of L2 (Ω) denoted by (ek )k∈N such that the se1

quence (λ−1

k ek )k∈N is an orthonormal basis of H0 (Ω) such that

−∆ek = λk ek .

Moreover, if f belongs to H −1 (Ω), then

kf k2H −1 (Ω) =

X

2

λ−2

k (hf, ek i) .

k

Remark Thus, the space H −1 (Ω) is a Hilbert space and (λk ek )k∈N is a hilbertian basis

of H −1 (Ω).

Proof of Theorem 3.1.2

As the space L2 is continuously included in H −1 (Ω), we can define an operator B as

follows:

2

L −→ H01 (Ω) ⊂ L2 (Ω)

B

f 7−→ u

such that u

from L2 (Ω)

into L2 (Ω).

functions of

is the solution in H01 (Ω) of −∆u = f . The operator B is of course continuous

into H01 (Ω). Thanks to Theorem 2.4.3, the operator B is compact from L2 (Ω)

Moreover, the operator B is selfadjoint and positive, i.e. that, for any couple of

L2 (Ω) (f, g), we have

(Bf |g)L2 = (f |Bg)L2

and (Bf |f )L2 > 0

if f 6= 0.

By definition of B, it exists a couple of functions in H01 (Ω) (u, v) such that we have,

(Bf |g)L2 = −(Bf |∆Bg)L2 = (∇Bf |∇Bg)L2 .

Thus the operator B is compact, selfadjoint and positive. The spectral theorem applied to B

implies the existence of a non increasing sequence (µk )k∈N of positive real numbers which tends

28

to 0 and a hilbertian basis of L2 (Ω) denoted (ek )k∈N such that , for any k, the function ek

belongs to L2 (Ω) and such that Bek = µk ek . This implies that −∆ek = µ−1

k ek . We have,

kf kH −1 (Ω) = sup hf,

(ck )∈Bf

X

λ−1

k ck ek i

k

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

X

−1

kf kH −1 (Ω) = sup

λ−1

k hf, ek ick = k(λk hf, ek i)k∈N k`2 (N) .

(ck )∈Bf

k

Theorem 3.1.2 is proved.

3.2

The problem of Stokes

This problem is analogous to the Dirichlet problem, but we work on the set of divergence free

vector field. Nevertheless, the fact that we impose a constrain (even a linear one) of the space

on which we search the minimum will introduce an important change. The Laplace eqaution

will become the Stokes equation. Let us first define of the space we are going to work with.

Definition 3.2.1 Let us denote by Vσ (Ω) the set of divergence free vector fields whose componants are in H01 (Ω) and by H(Ω) the closure in (L2 (Ω))d de Vσ (Ω).

Let us state the analogous of Dirichlet theorem in this framework. As in the preceeding

section, let us consider a vector feld f whose componants are in H −1 (Ω); then we define the

functionnal F

(

Vσ (Ω) −→ R

F

1

k∇uk2L2 − hf, ui.

u

7−→

2

Theorem 3.2.1 Let f ∈ V 0 (Ω). It exists a unique minimum of the functionnal F which is

also the unique solution of following equation

−∆u − f ∈ (Vσ (Ω))◦

which means that, for any vector field v of Vσ (Ω), we have

−h∆u, vi = hf, vi.

(3.3)

The existence and the uniqueness of a minimum u for the functionnal F can be proved following

exactly the same lines as in the case of Dirichlet problem. The fact that the differential of F

vanishes at point u implies the relation (3.3).

Remarks

• The fact that a vector field g of H −1 (Ω) belongs the polar set (in the sense of the duality)

of H01 (Ω) implies in particular that, for any function ϕ of D(Ω), we have

hg i , −∂j ϕi + hg j , ∂i ϕi = 0

which implies that ∂j g i − ∂i g j = 0, i.e. the curl of g is identically 0.

29

• Very simple domains exist sucht that it exists a vector field of H −1 (Ω) which are of

divergence and of curl identically 0 and which are not gradients of functions.

def

Let us consider the domain of the plan Ω = {x ∈ R2 / 0 < R1 < |x| < R2 } and the vector

field f defined by (−∂2 log |x|, ∂1 log |x|). We have the following lemma.

Proposition 3.2.1 The vector field f is of divergence free 0, but it is not the gradient of a

function.

The fact that f is of divergence free is obvious. The fact that its curl is 0 follows from the

fact that that the function x 7→ log |x| is harmonic on Ω. Let us assume that f is a gradient of

some distribution −p. As f is smooth, p is also smooth. Let us consider the flow of f = −∇p.

By definition of f , its trajectories are periodic. Let us consider a trajectory γ from of a point

of Ω such that f 6= 0 (here all points are like this). We have

dγ

d

(p ◦ γ)(t) =

∇p(γ(t)) 2 = −|∇p(γ(t))|2 ≤ 0.

dt

dt

L

The fact that the derivative en t = 0 is n´egative is contradictoire with la p´eriodicit´e of the

trajectoire γ. La proposition 3.2.1 is provede.

As shown by the following proposition, belonging to the polar space (in the sens of the

duality H −1 , H01 (Ω)) of (Vσ (Ω))◦ is stronger than being curl free. Let us admit the following

proposition .

Proposition 3.2.2 Let f in V 0 (Ω). If f belongs `a (Vσ (Ω))◦ i.e. if

∀v ∈ Vσ (Ω) ,

d

X

hf j , v j i = 0,

j=1

then it exists p in D0 (Ω) such that f = −∇p. If the boundary of Ω is a C 1 , hypersurface,

then p ∈ L2 (Ω).

Asinto the cas of Dirichlet problem, nous allons appliquer un result of theory spectrale on

les operators audoadjoints compacts pour obtenir the following theorem.

Theorem 3.2.2 Il exists a non decreasing sequence (λk )k∈N of reals strictement positives tendant to l’infini and a hilbertian basis of H(Ω) denoted (ek )k∈N such that the sequence (λ−1

k ek )k∈N

soit une base hilbertienne of Vσ (Ω) and such that

−∆ek − λ2k ek ∈ (Vσ (Ω))◦ .

Moreover, if f ∈ V 0 (Ω), alors

kf k2Vσ0 (Ω) =

X

2

λ−2

k (hf, ek i)

n

X

lim f −

hf, ek iek

and

n→∞

k∈N

k=0

(Vσ (Ω))0

= 0.

The proof is very close to the proof of Theorem 3.1.2. As the space H(Ω) is continuously

included in H −1 (Ω), we can define the operator B

H(Ω) −→ Vσ (Ω) ⊂ H(Ω)

B

f

7−→ u

such that u is the solution in H01 (Ω) of −∆u − f ∈ (Vσ (Ω))◦ . The following of the proof is

strictement analogous to the one of Dirichlet problem.

The orthogonal projection of L2 (Ω) on H(Ω), denoted by P, is the Leray projection.

30

3.3

How to modelize fluids using the least action principle

In this section, we shall always assume that the fluid extends to the whole space Rd , which

means that we neglect boundary effects. We want to describe the evolution of a perfect fluid

between two times t0 and t1 . A particule of fluid located at point x to l’time t0 sera located

at point ψ1 (x) at the time t1 . The possible incompressibility of the fluid will be described by

the fact that the map ψ1 , assumed to be a diffeomorphism, will preserved the measure, i.e.

its jacobian is of determinant 1.

Let us precise the functionnal spaces we are going to work with. In all this section, we

consider a diffeomorphism ψ1 which preserves the volume if the fluid is assumed incompressible.

Definition 3.3.1 Let us denote by L the space of C 1 functions from [t0 , t1 ] × Rd in Rd such

that ψ(t0 ) = Id and ψ(t1 ) = ψ1 , such that, each time t, the function ψ(t) is a diffeomorphism

of Rd and then such that ∂t ψ(t) is continuous from [t0 , t1 ] into L2 .

Let us denote by L0 the space des functions of L such that, for any time t, the diffeomorphism ψ(t) preserves the measure.

A possible evolution of a compressible fluid between the situation at time t0 and the one

at time t1 by the diffeomorphism ψ1 , is modelized by a function ψ of the space L. A possible

evolution of an incompressible fluid between the situation at time t0 and the situation at

time t1 described by the diffeomorphism ψ1 , is modelized by a function ψ of the space L0 .

This is the point of view is called the lagragian one.

Let us defined a functionnal the extremal points of which will decribes the evolutions of

the fluid.

Definition 3.3.2 Let F a C ∞ function and ρ0 a C ∞ positive function; Let us define the

action by the map A defined from L into R

A(ψ)

def

A1 (ψ) + A2 (ψ) with

Z t1 Z

def 1

A1 (ψ) =

|∂t ψ(t, x)|2 dxdt and

2 t0 Rd

Z Z

1 t1

def

A2 (ψ) = =

F ((Jψ(t, x))−1 ρ0 (x))Jψ(t, x)dxdt.

2 t0 Rd

=

def

where Jψ denotes the jacobian determinant of ψ, i.e. Jψ = det Dψ.

Remark The term A1 modelize cinetic energy of the system, the term A2 of internal energy.

Proposition 3.3.1 The functionnals A1 and A2 are differentiables of L in R and we have

Z t1 Z

DA1 (ψ)h =

ρ0 (x)∂t ψ(t, x) · ∂t h(t, x)dxdt and

t0

t1

Z

DA2 (ψ)h

=

Rd

G((Jψ(t, x))−1 ρ0 (x))(div τ )(t, ψ(t, x))Jψ(t, x)dtdx with

t0

G(y)

div h(t, x)

def

def

=

F (y) − yF 0 (y) and τ (t, x) = h(t, ψ −1 (t, x)) and

=

d

X

∂ j

h (t, x) ;

∂xj

j=1

31

la differential being defined sur

− def

→

L = {h ∈ C ∞ ([0, 1] × Rd ; Rd ) / ∀x ∈ Rd , h(0, x) = h(1, x) = 0}.

As the functionnal A1 is quadratic, the computation of DA1 is trivial. The computation

of DA2 comes from the chain rule. The proof of the following lemma is left to the reader as

an exercise.

Lemma 3.3.1 We have the following formula

DJ(ψ) · h =

d

X

det Dx ψ 1 , · · · , Dx ψ j−1 , Dx hj , Dx ψ j+1 · · · , Dx ψ d .

j=1

def

If we assume that τ (t, x) = h(t, ψ −1 (t, x)), then we have

(DJ(ψ) · h)(t, x) = (div τ )(t, ψ(t, x))J(ψ)(t, x).

The chain rule implies that

D(Jψ)−1 h(t, x) = −(Jψ(t, x))−1 (div τ )(t, ψ(t, x)) ;

the proposition is proved.

Now we can define perfect compressible fluids .

Definition 3.3.3 A perfect compressible fluid is a a fluid whose evolution follows extremals

of the functional A, i.e. following an element ψ of L such that

→

−

∀h ∈ L , DA(ψ)h = 0.

3.4

The eulerian point of view in the compressible case

The above definition seems rather implicit. The purpose of this section is to from a lagrangian

description , i.e. a description which is based on the evolution of each pointwise particule x

to a eulerian description, i.e. to a description of the fluid based on the knowledge of the speed

of the fluid fluid in whole of its points.

Mathematically, the link between these two points of view is the theory of ordinary differential equations. Indeed, let us consider an element ψ of L. Then the associated vector field

is defined by

v(t, x) = ∂t ψ(t, ψ −1 (t, x)).

(3.4)

→

−

It is clear that v belongs to L and that ψ is solution of

∂t ψ(t, x) = v(t, ψ(t, x))

(3.5)

ψ(0, x) = x.

→

−

Conversely, if v belongs to L , Cauchy-Lipschitz Theorem allows to define a flow ψ belonging

to L by the above system (3.5).

Now let us state the theorem which justifies this approach and derives the Euler system

of an incompressible fluid.

32

Theorem 3.4.1 Let ψ be an extremal of the functionnal A, i.e. an element of L such that

→

−

∀h ∈ L , DA(ψ)h = 0.

Let us consider time dependant vector field vdefined by the above relation (3.4); let us define

def

ρ(t, x) = ρ0 (ψ −1 (t, x))Jψ(t, ψ −1 (t, x))−1 ,

then the couple (ρ, v) satisfies the following system, called compressible Euler system

(Ecomp )

∂t ρ + v · ∇ρ + ρ div v = 0

ρ(∂t v + v · ∇v) + ∇p = 0

def

with v · ∇a =

d

X

vj

j=1

∂a

∂xj

and p = G(ρ).

The proof is very simple. First, the equation on ρ, which comes form the mass conservation,

is nothing more than the translation in terms of partial differential equations of the fact that,

by definition of ρ, we have

ρ(t, ψ(t, x)) = ρ0 (x)(Jψ(t, x))−1 .

By time derivation of this formula, it comes frome the chain rule and from Lemma 3.3.1

(∂t ρ + v · ∇ρ)(t, ψ(t, x)) = −ρ0 (x)(Jψ(t, x))−2 DJ(ψ)(t, ψ(t, x)) · ∂t ψ(t, x)

= −ρ0 (x)(Jψ(t, x))−2 DJ(ψ) · v(t, ψ(t, x))

= −ρ0 (x)(Jψ(t, x))−1 (div v)(t, ψ(t, x))

= −(ρ div v)(t, ψ(t, x)).

For the second equation, we use the hypothesis DA(ψ)h = 0. Using an integration by parts

and the definition of v and ρ, we find that

Z t1 Z

DA1 (ψ)h = −

ρ0 (x)∂t2 ψ(t, x)h(t, x)dxdt

t0

t1

Z

Rd

Z

= −

t0

Z t1

Rd

ρ0 (x)∂t v(t, ψ(t, x))h(t, x)dxdt

Z

= −

Rd

t0

ρ0 (x)(∂t v + v · ∇v)(t, ψ(t, x))τ (t, ψ(t, x))dxdt

Performing the change of variable y = ψ(t, x), we find, by definition of ρ,

Z t1 Z

DA1 (ψ)h =

ρ(x)(∂t v + v · ∇v)(t, x)τ (t, x)dxdt.

t0

Performing the change of variable y = ψ(t, x) into the formula of DA2 , it comes

Z t1 Z

DA2 (ψ)h = −

∇p(t, x)τ (t, x)dtdx.

t0

Rd

Applying these two formulas (3.6) and (3.7), we find that

Z t1 Z

ρ(x)(∂t v + v · ∇v)(t, x) + ∇p(t, x) τ (t, x)dtdx

DA(ψ)h = −

t0

(3.6)

Rd

Rd

Thus the theorem is proved.

33

(3.7)

3.5

The incompressible case

As above, the idea is to define a perfect incompressible fluid as an incompressible fluid whose

evolution follows extremals of the functionnal action A1 restricted to the space L0 . In order

to define the notion of infinitesimal variation on the space L0 which is included in an affine

space. Following the classical definition of the tangent space of a submanifold of Rd , we give

the following definition.

Definition 3.5.1 An infinitesimal variation at a point ψ of the space L0 is the derivative

at 0 of any function Θ continuously differentiable from [0, 1] into L0 such that Θ(0) = ψ.

Because of some regularity problem, it is not very easy to describe exactly the set of

infinitesimal variations. Moreover, the intuition of finite dimensionnal spaces must be used

with a lot of care always because of regularity problems. The following proposition will be

enough for our purpose here.

→

−

Proposition 3.5.1 Let us denote by L 0 the set of vector fields τ whose coefficients are

continuously differentiables on [t0 , t1 ] × Rd and such that

τ (t0 ) = τ (t1 ) = 0

and ∀ t ∈ [t0 , t1 ] , div τ (t) = 0.

→

−

Let θ be an infinitesimal variation at a point ψ of L0 ; it exists a vector field τ of the space L 0

such that θ(t, x) = τ (t, ψ(t, x)).

Conversely, let α a smooth compactly supported function on ]t0 , t1 [ and τ a divergence

free vector field whose componants belong to the space S ; if θ(t, x) = α(t)τ (ψ(t, x)) then θ

is an infinitesimal variation at point ψ.

Let θ an infinitesimal variation at point ψ. By definition, it exists a function Θ continuously

differentiable of [0, 1] in L0 such that

∂s Θ(s, t, x)|s=0 = θ(t, x)

et

Θ(0, t, x) = ψ(t, x).

(3.8)

As for any s and any t, Θ(s, t) is a diffeomorphism, we can define a vector field τe(s, t, x) by

τe(s, t, x) = ∂s Θ(s, t, Θ−1 (s, t, x)).

This means that

∂s Θ(s, t, x) = τe(s, t, Θ(s, t, x))

Thanks to Lemma 3.3.1, it is enough to apply the chain rule, which gives

∂s det Dx Θ(s, t, x) = div τe(s, t, Θ(s, t, x)) × det(Dx θ(s, x)).

As det Dx Θ(s, t, x) = 1, we have

∀(s, t) ∈ [0, 1] × [t0 , t1 ] , div τe(s, t) = 0.

As ∂s θ(0, t, x) = τe(0, t, ψ(t, x)), we have the first point of the proposition defining

def

τ (t, x) = τe(0, t, x).

The second point is very easy. It is enough to solve the following differential equation

∂s Θ(s, t, x) = α(t)τ (t, Θ(s, t, x)),

Θ(0, t, x) = ψ(t, x).

Now let us define mathematically what a perfect incompressible fluid is.

34

(3.9)

Definition 3.5.2 A fluid is perfect and incompressible if its evolution between the time t0

and situation ψ1 at time t1 if its follows an element ψ of L0 such that, for any infinitesimal

variation θ at point ψ, we have DA1 (ψ) · θ = 0.

The above definition can be formulated as the fact perfect incompressible fluids follows

extremales of the action functionnal action, which is defined on the space of curves of measure

preserving diffeomorphisms.

The above description of the evolution of an incompressible fluid by a curve on the spaceof

measure preserving diffeomorphisms; it is the lagrangien point of view. Let us take now the

Eulerian point of view.

Theorem 3.5.1 Let ψ an evolution of a pecfect imcompressible fluid and v the divergence

free vector field associated

with ψ by (3.4). It exists then a distribution temp´er´ee p such that,

P

if l’on pose v · ∇ = di=1 v i ∂i , we have

∂t v + v · ∇v = −∇p.

Let us suppose that ψ is an evolution of a perfect incompressible fluid. Definition 3.5.1 and

Proposition 3.5.1 imply that, for any α ∈ D(]t0 , t1 [) and any divergence free vector field, we

have τ ∈ C ∞ ([t0 , t1 ]; S(Rd ; Rd )),

Z t1 Z

∂t ψ(t, x)∂t (α(t)τ (t, ψ(t, x)))dtdx = 0.

Rd

t0

Form (3.5), it comes

Z

t1

t0

Z

Rd

v(t, ψ(t, x))∂t (α(t)τ (t, ψ(t, x)))dtdx = 0.

An integration by parts with respect the time variable ensures that

Z t1 Z

∂t (v(t, ψ(t, x)))α(t)τ (t, ψ(t, x))dtdx = 0.

t0

Rd

As ∂t (v(t, ψ(t, x)) = (∂t v + v · ∇v)(t, ψ(t, x)), and as ψ(t) is a measure preserving diffeomorphism, we have, for any α ∈ D(]t0 , t1 [) and any divergence free vector field τ ,

Z t1 Z

(∂t v + v · ∇v)(t, x)α(t)τ (t, x)dtdx = 0.

(3.10)

t0

Rd

As ∂t v + v · ∇v is a continuous function of the time, we have, for any divergence free vector

field τ and for any time t,

Z

(∂t v + v · ∇v)(t, x)τ (t, x)dx = 0.

(3.11)

Rd

The first part of the theorem will be a consequence of the following proposition, which is

classical in the theory ofdistributions.

Proposition 3.5.2 Let w a vector field the coefficients of which are tempered distribtions.

The existence of a tempered distribution p such that w = ∇p is equivalent to the fact that

the curl of w is 0, i.e. ∂j wi = ∂i wj .

35

The proof of this proposition is a exercice of theory of distributions. Let us consider f0 a

smooth compactly supported function ofone real variable of integral 1. Let us assume that

the dimension d is equal to 1. Then let us define, for φ ∈ D(R),

Z x

Z ∞

0

0

φ(y )dy f0 (y) dy.

φ(y) −

Φ(φ)(x) =

−∞

−∞

It is clear that Φ(∂x φ) = φ and that Φ(φ) is a smooth compactly supported function. Morever,

the map Φ defined above is a linear continuous map from S(R) into itsself. Indeed, if x ≤ −A,

with A strictement positif, then

Z x

Z ∞

Z x

0

0

|f0 (y)|dy.

φ(y )dy

|φ(y)|dy +

|Φ(φ)(x)| ≤

−∞

−∞

−∞

If −x is strictement positif, we have, for any integer N greater that 2,

Z

|Φ(φ)(x)| ≤

+ |y|)N +1 |φ(y)|

supy∈R (1 + |y|)N +1 |f0 (y)|

+ kφkL1

dy 0 .

(1 + |y 0 |)N +1

(1 + |y 0 |)N +1

∞ sup

y∈R (1

−x

Let us define NN (φ) = kφkL1 + supy∈Rd |(1 + |y|)N +1 φ(y)|. We have

|x|N |Φ(φ)(x)| ≤ CNN (φ).

As

Z

∞

Z

∞

φ(y) −

0

φ(y )dy

−∞

0

f0 (y) dy = 0,

−∞

the same thing holds for positive x. The fact that

Z

∂x Φ(φ) = φ − f0

∞

φ(y 0 )dy 0

−∞

concludes the proof of the continuity of Φ on S(R).

Let w a tempered distribution on R. Let us define p by

< p, φ >= − < w, Φ(φ) > .

As Φ is a linear continuous map from S(R) into itsself, p is a tempered distribution. Moreover,

< ∂x p, φ >= − < p, ∂x φ >=< w, Φ(∂x φ) >=< w, φ > .

Thus the result holds in dimension d = 1.

Let us do an induction by assuming that the result holds in dimension d ≤ k − 1. Let w a

vector field the coefficients of which are tempered distribtions on Rk and satisfies the hypothesis on the proposition.Thanks to the induction hypothesis, a tenpered distribution π on Rk−1

exists such that

< ∂i π, θ >=< wi , f0 ⊗ θ > .

Then let us define the tempered distribution on Rk by

Z

1

∞

< p, φ >= − < w , Φk (φ) > + < π,

φ(y1 )dy1 >,

−∞

36

with Φk (φ)(x) = Φ(φ(·, x2 , . . . , xk ))(x1 ). Now it is enough to check that ∂i p = wi . First, let

us observe that

Z ∞

< ∂1 p, φ >=< w1 , Φk (∂1 φ) > − < π,

∂1 φ(y1 )dy1 >=< w1 , φ > .

−∞

If i ≥ 2, we have

< ∂i p, φ > = + < w1 , Φk (∂i φ) > − < π,

Z

∞

∂i φ(y1 )dy1 >

−∞

Z ∞

= − < ∂i w1 , Φk (φ) > + < ∂i π,

φ(y1 )dy1 >

Z ∞

= − < ∂1 wi , Φk (φ) > + < wi , f0 ⊗

φ(y1 )dy1 >

−∞

Z ∞

i

i

φ(y1 )dy1 >

= < w , ∂1 Φk (φ) > + < w , f0 ⊗

−∞

−∞

i

= < w ,φ >

Thus the proposition.

We can easily deduce the two following corollaries :

Corollary 3.5.1 Let w a vector fields the coefficients of which are tempered distribtions such

that, for any divergence free vector field u of S(Rd ; Rd ), we have

< w, u >=

X

< wi , ui >= 0.

i

A tempered distribution p exists such that w = ∇p.

Corollary 3.5.2 Let w a divergence free vector field on R2 whose coefficients are tempered

distributions. It exists a tempered distribution f such that

def

w = ∇⊥ f = (−∂2 f, ∂1 f ).

In order to prove the first corollary, let us consider a function φ of the space S(Rd ). In the

case when i and j are two distinct positive integers less or equal to d, let us consider the vector

field u whose componant of index j is ∂i φ, whose componant of index i is −∂j φ the others

being identically 0. It is clear that u is a divergence free vector field. Thus by hypothesis

< w, u >=< wj , ∂i φ > − < wi , ∂j φ >=< ∂j wi − ∂i wj , φ >= 0.

It turns out that ∂j wi − ∂i wj = 0 and Corollary 3.5.1 is proved.

In order to prove the second corollary, let us consider the divergence free vector field

w

e = (−w2 , w1 ). it is clear that we have ∂1 w

e2 − ∂2 w

e1 = div w = 0. It exists donc a tempered

distribution f such that w

e = (∂1 f, ∂2 f ). Thus le corollary 3.5.2.

37

Let us go back to the proof of Theorem 3.5.1. It is clear from Corollary 3.5.1 and from

(3.10) that, if ψ is a evolution of a perfect incompressible fluid, then the vector field associated

with ψ(t) throug (3.4) satisfies

∂t v + v · ∇v = −∇p.

(3.12)

Conversely, if v is a vector field satisfying (3.12). Let us consider the flow ψ of v defined

by (3.5) et θ an infinitesimal variation in ψ. Then, we have

t1

Z

t0

Z

Z

Rd

t1

∂t ψ(t, x)∂t θ(t, x)dtdx = −

t0

Z

Rd

∂t2 ψ(t, x)θ(t, x)dtdx.

Proposition 3.5.1 ensures the existence of a divergence free vector field τ such that θ =

τ (t, ψ(t, x)). It turns out that

Z

t1

t0

Z

Rd

Z

t1

∂t ψ(t, x)∂t θ(t, x)dtdx =

Z

(∇p)(ψ(t, x))τ (t, ψ(t, x))dtdx.

t0

Rd

For any time t, the diffeomorphism ψ(t) preserves the measure; this concludes the proof the

theorem.

Now let us give a weak formulation to Equation (3.12).This formulation is equivalent

to the one of Relation (3.12) in the case when the vector field solution is smooth enough.

Nevertheless, it can be important to have a weak formulation. If v is a divergence free vector

field continuously differentiable, we have v · ∇a = div(av) for any function a continuously

differentiable. This gives

∂t v + v · ∇v = ∂t v + div v ⊗ v,

where div v ⊗ v denotes the vector field whose coordiante of order j is

d

X

∂j (v i v j ). Thus we

j=1

get the following formulation of incompressible Euler equations.

∂t v + div v ⊗ v = −∇p

div v = 0

(E)

v|t=0 = v0 .

Remarks

• The proofs of sections 3.1 and 3.2 must be known. The sections 3.3, 3.4 and 3.5 are

important in the point of view of modelization. It is not necessary to know the proofs

of these sections.

38

Chapter 4

Leray’s Theorem on Navier-Stokes

equations

In this chapitre, we shall prove the existence of global solutions for the incompressible NavierStokes system in a bounded domain with Dirichlet boundary conditions which means

∂t v + v · ∇v − ν∆v = −∇p

div v = 0

v|t=0 = v0

v|∂Ω = 0.

Before studying this problem, we study a simplier one, called ”time dependent Stokes problem”.

4.1

The time dependent Stokes problem

Given a positive viscosity ν, the evolution Stokes

∂t u − ν∆u

div u

(ESν )

u|∂Ω

u|t=0

problem reads as follows:

=

=

=

=

f − ∇p

0

0

u0 ∈ H.

Let us define what a solution of this problem is.

Definition 4.1.1 Let u0 be in H and f in L2loc (R+ ; V 0 ). We shall say that u is a solution

of (ESν ) with initial data u0 and external force f if and only if u belongs to the space

+

2

+

C(R+ ; Vσ0 ) ∩ L∞

loc (R ; H) ∩ Lloc (R ; Vσ )

and satisfies, for any Ψ in C 1 (R+ ; Vσ ),

Z

hu(t), Ψ(t)i +

ν∇u : ∇Ψ − u · ∂t Ψ (t0 , x) dxdt0

[0,t]×Ω

Z

u0 (x) · Ψ(0, x) dx +

=

Ω

The following theorem holds.

39

Z

0

t

hf (t0 ), Ψ(t0 )i dt0 .

Theorem 4.1.1 The problem (ESν ) has a unique solution in the sense of the above definition.

Moreover this solution belongs to C(R+ ; H) and satisfies

Z t

Z t

1

1

k∇u(t0 )k2L2 dt0 = ku0 k2L2 +

hf (t0 ), u(t0 )idt0 .

ku(t)k2L2 + ν

2

2

0

0

Proof of Theorem 4.1.1. In order to prove uniqueness, let us consider some function u

in C(R+ ; Vσ0 ) ∩ L2loc (R+ ; Vσ ) such that, for all Ψ in C 1 (R+ ; Vσ ),

Z tZ

hu(t), Ψ(t)i +

ν∇u : ∇Ψ − u · ∂t Ψ (t0 , x) dxdt0 = 0.

0

Ω

This is valid in particular for the time independent function Ψ(t) ≡ ek where the family vector

fields (ek )k∈N is given by Theorem 3.2.2. This gives

Z tZ

hu(t), ek i = −ν

∇u(t0 , x) : ∇ek (x)dxdt0

0

Ω

Z t

= ν

hu(t0 ), ∆ Pk Ψi.

0

Thanks to the spectral Theorem 3.2.2 together with the fact that, for almost every t0 , u(t)

belongs to Vσ , we have

Z

− ∇u(t0 , x) : ∇ek (x)dx = h∆ek , u(t0 )i = λ2k hek , u(t0 )i.

Ω

This gives

t

Z

hu(t), ek i =

λ2k hek , u(t0 )idt0 .

0

The fact that hu(0), ek i = 0 implies that, for any k, hu(t), ek i = (u|ek )H = 0. Thus u ≡ 0.

In order to prove existence, let us consider a sequence (fk )k∈N associated with f by

Lemma 4.3.1 page 43 and then the approximated problem

∂t uk − ν Pk ∆uk = fk

def X

(ESν,k )

with Pk f =

hf, ej iej .

(4.1)

uk|t=0 = Pk u0

j≤k

Again thanks to Theorem 3.2.2 page 30, it is a linear ordinary differential equation on Hk

which has a global solution uk which is C 1 (R+ ; Hk ). By an energy estimate in (ESν,k ) we get

that

1d

kuk (t)k2L2 + νk∇uk (t)k2L2 = hfk (t), uk (t)i.

2 dt

A time integration gives

Z t

Z t

1

1

2

0 2

0

2

kuk (t)kL2 + ν

k∇uk (t )kL2 dt = k Pk u(0)kL2 +

hfk (t0 ), uk (t0 )idt0 .

(4.2)

2

2

0

0

In order to pass to the limit, we write an energy estimate for uk − uk+` , which gives

Z t

def 1

2

δk,` (t) =

k(uk − uk+` )(t)kL2 + ν

k∇(uk − uk+` )(t0 )k2L2 dt0

2

0

Z t

1

2

k(Pk − Pk+` )u(0)kL2 +

h(fk − fk+` )(t0 ), uk (t0 )idt0

=

2

0

Z t

1

C

2

≤

k(Pk − Pk+` )u(0)kL2 +

k(fk − fk+` )(t0 )k2Vσ0 dt0

2

ν 0

Z

Z

ν t

ν t

0 2

0

+

k∇(uk − uk+` )(t )kL2 dt +

k(uk − uk+` )(t0 )k2L2 dt0 .

2 0

2 0

40

Using Poincar´e’s inequality, this implies that

δk,` (t) ≤

k(Pk − Pk+` )u(0)k2L2

Z

C

+

ν

0

t

k(fk − fk+` )(t0 )k2Vσ0 dt0 .

This implies immediately that the sequence (uk )k∈N is a Cauchy one in the space C(R+ ; H) ∩

L2loc (R+ ; Vσ ). Let us denote by u the limit and prove that u is a solution in the sense of

Definition 4.1.1. As uk is a C 1 solution of the ordinary differential equation (ESν,k ), we have,

for a Ψ in C 1 (R+ ; Vσ ),

d

huk (t), Ψ(t)i = νh∆uk (t), Ψ(t)i + hfk (t), Ψ(t)i + huk (t), ∂t Ψ(t)i.

dt

By time integration, we get

Z tZ

∇uk (t0 , x) : ∇Ψ(t0 , x) dxdt0

huk (t), Ψ(t)i = −ν

0

Ω

Z t

Z t

+

hfk (t0 ), Ψ(t0 )idt0 + hPk u(0), Ψ(0)i +

huk (t0 ), ∂t Ψ(t0 )idt0 .

0

0

Passing to the limit in the above equality and in (4.2) gives the theorem.

Remark. The solution is given by the explicit formula

X

u(t) =

Uj (t)ej with

j∈N

Uj (t)

def

=

2

e−νµj t (u0 |ej )L2 +

t

Z

2

0

e−νµj (t−t ) hf (t0 ), ej i dt0 .

(4.3)

0

In the case of the whole space Rd , we have the following analogous formula

Z

−d

eix·ξ u

b(t, ξ)dξ with

u(t, x) = (2π)

d

R

Z t

2

0

def

−ν|ξ|2 t

u

b(t, ξ) = e

u

b0 (ξ) +

e−ν|ξ| (t−t ) fb(t0 , ξ) dt0 .

(4.4)

0

4.2

The concept of weak solution

Let us state now the weak formulation of the incompressible Navier–Stokes system (N Sν ).

Definition 4.2.1 Given a domain Ω in Rd , we shall say that u is a weak solution of the

Navier–Stokes equations on R+ ×Ω with an initial data u0 in H and an external force f

in L2loc (R+ ; V 0 ) if and only if u belongs to the space

+

2

+

C(R+ ; Vσ0 ) ∩ L∞

loc (R ; H) ∩ Lloc (R ; Vσ )

and for any function Ψ in C 1 (R+ ; Vσ ), the vector field u satisfies the following condition:

Z

Z tZ

(u · Ψ)(t, x) dx +

ν∇u : ∇Ψ − u ⊗ u : ∇Ψ − u · ∂t Ψ (t0 , x) dxdt0

Ω

0 Ω

Z

Z t

=

u0 (x) · Ψ(0, x) dx +

hf (t0 ), Ψ(t0 )i dt0 with

Ω

∇u : ∇Ψ =

0

d

X

∂j uk ∂j Ψk

and u ⊗ u : ∇Ψ =

j,k=1

d

X

j,k=1

41

uj uk ∂j Ψk .

Let us remark that the above relation means that the equality in (N Sν ) must be understood

as an equality in the sense of Vσ0 . Now let us state the Leray theorem.

Theorem 4.2.1 Let Ω be a domain of Rd and u0 a vector field in H. Then, there exists

a global weak solution u to (N Sν ) in the sense of Definition 4.2.1. Moreover, this solution

satisfies the energy inequality for all t ≥ 0,

1

2

Z

Z tZ

2

|∇u(t0 , x)|2 dxdt0

Z t

Z

1

hf (t0 , ·), u(t0 , ·)idt0 .

|u0 (x)|2 dx +

2 Ω

0

|u(t, x)| dx + ν

Ω

0

≤

Ω

(4.5)

It is convenient to state the following definition.

Definition 4.2.2 A solution of (N Sν ) in the sense of the above Definition 4.2.1 which moreover satisfies the energy inequality (4.5) is called a Leray solution of (N Sν ).

Let us remark that the energy inequality implies a control on the energy.

Proposition 4.2.1 Any Leray solution u of (N Sν ) satisfies

t

Z

ku(t)k2L2

k∇u(t

+ν

0

0

)k2L2 dt0

≤

C

+

ν

ku0 k2L2

t

Z

0

kf (t0 )k2Vσ0 dt0 .

Proof of Proposition 4.2.1 By definition of the norm k · kVσ0 , we have

hf (t, ·), u(t, ·)i ≤ kf (t, ·)kVσ0 ku(t, ·)kVσ .

Inequality (4.5) becomes

ku(t)k2L2 + 2ν

Z

t

0

k∇u(t0 )k2L2 dt0 ≤ ku0 k2L2 +

Z

0

t

kf (t0 )kVσ0 ku(t0 , ·)k2V dt0 .

As ku(t0 , ·)k2V = k∇u(t0 , ·)k2L2 , we get, using the fact that 2ab ≤ a2 + b2 ,

ku(t)k2L2

Z

t

k∇u(t

+ 2ν

0

0

)k2L2 dt0

≤

ku0 k2L2

C

+

ν

Z

0

t

kf (t0 )k2Vσ0 dt0 .

Thus the proposition is proved.

The outline of this section is now the following:

• first approximate solutions are built in spaces with finite frequencies by using simple

ordinary differential equations results in L2 –type spaces.

• Next, a compactness result is derived.

• Finally the conclusion is obtained by passing to the limit in the weak formulation, taking

especially care of the nonlinear terms.

42

4.3

Construction of approximate solutions

In this section, we intend to build approximate solutions of the Navier–Stokes equations. We

use the projections Pk defined in (4.1) and denote by Hk the space Pk H = Pk V 0 .

Lemma 4.3.1 For any bulk force f in L2loc (R+ ; V 0 ), a sequence (fk )k∈N exists in C 1 (R+ ; Vσ )

such that for any k ∈ N and for any t > 0, the vector field fk (t) belongs to Hk , and

lim kfk − f kL2 ([0,T ];Vσ0 ) = 0.

k→∞

Proof of Lemma 4.3.1: Thanks to Theorem 3.2.2 and to the Lebesgue Theorem, a sequence (fek )k∈N exists in L2loc (R+ ; Vσ ) such that for any positive integer k and for almost all

positive t, the vector field fek (t) belongs to Hk and

∀T > 0 , lim kfek − f kL2 ([0,T ];Vσ0 ) = 0.

k→∞

A standard (and omitted) time regularization procedure concludes the proof of the lemma.

In order to construct the approximate solution, let us study the non linear term.

Definition 4.3.1 Let us define the bilinear map

V × V → V0

Q

(u, v) 7→ − div(u ⊗ v).

Sobolev embeddings stated in Theorem 2.2.1 ensure that Q is continuous: in the sequel, the

following lemma will be useful.

Lemma 4.3.2 For any u and v in V, the following estimates hold. For d in {2, 3}, a constant C

exists such that, for any ϕ ∈ V,

d

d

1− d

1− d

hQ(u, v), ϕi ≤ Ck∇ukL4 2 k∇vkL4 2 kukL2 4 kvkL2 4 k∇ϕkL2 .

Moreover for any u in Vσ and any v in V, hQ(u, v), vi = 0.

Proof of Lemma 4.3.2. The first two inequalities follow directly from Gagliardo–Nirenberg’s

inequality stated in Corollary 2.2.1, once noticed that

hQ(u, v), ϕi ≤ ku ⊗ vkL2 k∇ϕkL2

≤ kukL4 kvkL4 k∇ϕkL2 .

In order to prove the third assertion, let us assume that u and v are two vector fields the

components of which belong to D(Ω). Then we deduce from integrations by parts that

Z

hQ(u, v), vi = − (div(u ⊗ v) · v)(x) dx

Ω

= −

d Z

X

∂m (um (x)v ` (x))v ` (x) dx

`,m=1 Ω

=

d Z

X

um (x)v ` (x)∂m v ` (x) dx

`,m=1 Ω

Z

= −

|v(x)|2 div u(x) dx − hQ(u, v), vi.

Ω

43

Thus, we have

1

hQ(u, v), vi = −

2

Z

|v(x)|2 div u(x) dx.

Ω

The two expressions are continuous on V and by definition, D is dense in V. Thus the formula

is true for any (u, v) ∈ Vσ × V, which completes the proof.

def

Thanks to Theorem 3.2.2 and to the above lemma, we can define Fk (u) = Pk Q(u, u).

Now let us introduce the following ordinary differential equation

u˙ k (t) = ν Pk ∆uk (t) + Fk (uk (t)) + fk (t)

(N Sν,k )

uk (0) = Pk u0 .

Theorem 3.2.2 implies that Pk ∆ is a linear map from Hk into itself. Thus the continuity

properties on Q and Pk allow to apply the Cauchy–Lipschitz theorem. This gives the existence

of Tk ∈]0, +∞] and a unique maximal solution uk of (N Sν,k ) in C ∞ ([0, Tk [; Hk ). In order to

prove that Tk = +∞, let us observe that, thanks to Lemma 4.3.2 and Theorem 3.2.2

d

ku˙ k (t)kL2 ≤ νλk kuk (t)kL2 + Cλk4 kuk (t)k2L2 + kfk (t)kL2 .

If kuk (t)kL2 remains bounded on some interval [0, T [, so does ku˙ k (t)kL2 . Thus, for any k, the

function uk satisfies the Cauchy criteria when t tends to T . Thus the solution can be extended

beyond T . It follows that a uniform bound on kuk (t)kL2 will imply that Tk = +∞.

4.4

A priori bounds

The purpose of this paragraph is the proof of the following proposition.

Proposition 4.4.1 The sequence (uk )k∈N in bounded in the space

8

+

+

2

+ 4

d

L∞

loc (R ; H) ∩ Lloc (R ; Vσ ) ∩ Lloc (R L (Ω)).

Moreover, the sequence (∆uk )k∈N is bounded in the space L2loc (R+ ; Vσ0 ).

Proof of Proposition 4.4.1 Let us now estimate the L2 norm of uk (t). Taking the L2

scalar product of equation (N Sν,k ) with uk (t), we get

1d

kuk (t)k2L2 = ν(∆uk (t)|uk (t))L2 + (Fk (uk (t))|uk (t))L2 + (fk (t)|uk (t))L2 .

2 dt

By definition of Fk , Lemma 4.3.2 implies that

(Fk (uk (t))|uk (t))L2 = hQ(uk (t), uk (t)), uk i = 0.

Thus we infer that

1d

kuk (t)k2L2 + ν(∇uk (t)|∇uk (t))L2 = (fk (t)|uk (t))L2 .

2 dt

By time integration, we get the fundamental

Stokes system: for all t ∈ [0, Tk )

Z t

1

2

kuk (t)kL2 + ν

k∇uk (t0 )k2L2 dt0 =

2

0

(4.6)

energy estimate for the approximate Navier–

1

kuk (0)k2L2 +

2

44

Z

0

t

(fk (t0 )|uk (t0 ))L2 dt0 .

(4.7)

Using the (well known) fact that 2ab ≤ a2 + b2 , we get

kuk (t)k2L2 + ν

Z

t

0

k∇uk (t0 )k2L2 dt0 ≤ kuk (0)k2L2 +

C

ν

Z

t

0

kfk (t0 )k2Vσ0 dt0

Gronwall’s lemma implies that (uk )k∈N remains uniformly bounded in H for all time, hence

+

that Tk = +∞. In addition, the sequence (uk )k∈N is bounded in the space L∞

loc (R ; H) ∩

L2loc (R+ ; Vσ ). Using Gagliardo-Nirenberg inequalities (see Corollary 2.2.1 page 20), we deduce

that the sequence (uk )k∈N is bounded in the space

8

+

2

+

4

d

L∞

loc (R ; H) ∩ Lloc (R ; Vσ ) ∩ Lloc (L (Ω)).

Moreover, we have, for any v ∈ Vσ ,

h−∆uk , vi = (∇uk |∇v)L2 ≤ kuk kH01 kvkV .

By definition of the norm k·kVσ0 , we infer that the sequence (∆uk )k∈N is bounded in L2loc (R+ ; Vσ0 ).

The whole proposition is proved.

4.5

Compactness properties

Let us now prove the following fundamental result.

+

+

2

Proposition 4.5.1 A vector field u exists in L∞

loc (R ; H) ∩ Lloc (R ; Vσ ) such that up to an

extraction (which we omit to note) we have for any positive real number T , for all vector

fields Ψ ∈ L2 ([0, T ]; V),

Z

lim

(∇uk (t, x) − ∇u(t, x)) : ∇Ψ(t, x) dtdx = 0.

(4.8)

k→∞ [0,T ]×Ω

Z

lim

k→∞ [0,T ]×Ω

|uk (t, x) − u(t, x)|2 dtdx = 0.

(4.9)

lim kuk − ukL∞ ([0,T ];Vσ0 ) = 0.

(4.10)

k→∞

Proof of Proposition 4.5.1. A standard diagonal process (omitted) with an increasing

sequence of positive real numbers Tn reduces the proof of (4.8)–(4.10) to the proof of the

same result for any time interval [0, T ]. The relative weak compactness of the sequence (uk )k∈N

in the Hilbert space L2 ([0, T ]; Vσ )is obvious. Thus (4.8) is true with u in L2 ([0, T ]; Vσ ).

In order to prove (4.9), let us establish th at

∀ε , ∃k0 / ∀k , kuk − Pk0 uk kL2 ([0,T ]×Ω) + kuk − Pk0 uk kL∞ ([0,T ];Vσ0 ) <

ε

·

2

The proof of the claim is based on Theorem 3.2.2. Using this result, we can write that

kuk −

Pk0 uk k2L2 ([0,T ];H)

Z

T

=

0

Z

X

huk (t), ej i)2 dt

j≥k0 +1

T

=

0

45

X

j≥k0 +1

2

2

λ−2

j λj huk (t), ej i) dt.

(4.11)

Using that the sequence (λj )j is increasing, we get, by Theorem 3.2.2,

kuk −

Pk0 uk k2L2 ([0,T ];H)

≤

≤

λ−2

k0 +1

T

Z

X

λ2j huk (t), ej i)2 dt

0

j

−2

2

λk0 +1 kuk kL2 ([0,T ];Vσ ) .

Following the same lines we get

2

kuk − Pk0 uk k2L2 ([0,T ];Vσ0 ) ≤ λ−2

k0 +1 kuk (t)kH .

The fact that lim λk = +∞ ensures (4.11).

k→∞

Now, let us prove that the sequence (Pk0 uk )k is relativement compact in Hk0 . Let us

notice that Hk0 = Pk0 H is a finite dimensionnal vector space. Using Theorem 3.2.2, it turns

out that

d

2− d

k Pk0 u˙ k (t)kL2 ≤ λk0 k Pk0 uk (t)kVσ + λk0 kuk (t)kL2 2 k∇uk (t)kL2 2 .

4

Using energy estimate (4.8), we infer that (Pk0 u˙ k )k is a bounded sequence of L d ([0, T ]; L2 )

which means that

∀k , k Pk0 u˙ k k 4

≤ Cu0 ,f,k0 .

2

L d ([0,T ];L )

Thus, by integration and H¨older estimate, we get, for any (t, t0 ) ∈ [0, T ]2 ,

4

k Pk0 uk (t) − Pk0 uk (t)kL2 ≤ |t − t0 |1 d Cu0 ,f,k0 .

(4.12)

Moreover, for any t in [0, T ], the set {Pk0 uk (t), k ∈ N} in bounded in the finite dimensionnal

space Hk0 . Thus it is relatively compact. Together with (4.12), this allows to apply Ascoli’s

compactness theorem. Thus, in particular, the set {Pk0 uk , , k ∈ N} can be recovered by a

finite number of balls of radius ε/2. Together with (4.11) , this proves (4.9) and (4.10).

4.6

End of the proof of the Leray Theorem

The local strong convergence of (uk )k∈N will be crucial to pass to the limit in (N Sν,k ) to

obtain solutions of (N Sν ).

According to the definition of a weak solution of (N Sν ), let us consider a test function Ψ

in C 1 (R+ ; Vσ ). Because uk is a solution of (N Sν,k ), we have

d

˙

huk (t), Ψ(t)i = hu˙ k (t), Ψ(t)i + huk (t), Ψ(t)i

dt

= νhPk ∆uk (t), Ψ(t)i + hPk Q(uk (t), uk (t)), Ψ(t)i

˙

+ hfk (t), Ψ(t)i + huk (t), Ψ(t)i.

We have after integration by parts

hPk ∆uk (t), Ψ(t)i = −ν(uk (t)| Pk Ψ(t))Vσ = −ν(uk (t)|Ψ(t))Vσ

Z

hPk Q(uk (t), uk (t)), Ψ(t)i =

uk (t, x) ⊗ uk (t, x) : ∇ Pk Ψ(t, x) dx and

ZΩ

˙

huk (t), Ψ(t)i

=

uk (t, x) · ∂t Ψ(t, x) dx.

Ω

46

By time integration between 0 and t, we infer that

Z t

huk (t), Ψ(t)i +

ν(∇uk (t0 )|∇Ψ(t0 ))Vσ − (uk (t0 )|∂t Ψ(t0 ))H dt0

0

Z tZ

Z t

−

uk ⊗ uk : ∇ Pk Ψ dxdt0 = huk (0), Ψ(0)i +

hfk (t0 ), Ψ(t0 )i dt0 .

0

0

Ω

We now have to pass to the limit. Using (4.10), we deduce that, for any t ∈ [0, T ],

lim huk (t), Ψ(t)i = hu(t), Ψ(t)i.

k→∞

(4.13)

Now, using (4.9) gives

Z

lim

k→∞ 0

t

(uk (t0 )|∂t Ψ(t0 ))H dt0 =

t

Z

(u(t0 )|∂t Ψ(t0 ))H dt0 .

(4.14)

0

Thanks to Theorem 3.2.2, we have

Z t

Z t

lim

hfk (t0 ), Ψ(t0 )i dt0 =

hf (t0 ), Ψ(t0 )i dt0 .

k→∞ 0

(4.15)

0

Now, we have to treat the non linear term. Let us start by proving the following preliminary lemma.

Lemma 4.6.1 Let H be a Hilbert space, and let (An )n∈N be a bounded sequence of linear

operators on H such that

∀h ∈ H, lim kAn h − hkH = 0.

n→∞

Then if ψ ∈ C([0, T ]; H) we have lim sup kAn ψ(t) − ψ(t)kH = 0.

n→∞ t∈[0,T ]

Proof of Lemma 4.6.1. The function ψ is continuous in time with values in H, so for all

positive ε, the compact ψ([0, T ]), can be covered by a finite number of balls of radius

ε

2(A + 1)

def

with A = sup kAn kL(H) .

n

and center (ψ(t` ))0≤`≤N . Then we have, for all t in [0, T ] and ` in {0, · · · , N },

kAn ψ(t) − ψ(t)kH ≤ kAn ψ(t) − An ψ(t` )kH + kAn ψ(t` ) − ψ(t` )kH + kψ(t` ) − ψ(t)kH .

The assumption on An implies that for any `, the sequence (An ψ(t` ))n∈N tends to ψ(t` ). Thus,

an integer nN exists such that, if n ≥ nN ,

ε

∀` ∈ {0, · · · N } , kAn ψ(t` ) − ψ(t` )kH < ·

2

We infer that, if n ≥ nN , for all t ∈ [0, T ] and all ` ∈ {0, · · · , N },

ε

kAn ψ(t) − ψ(t)kH ≤ kAn ψ(t) − An ψ(t` )kH + kψ(t` ) − ψ(t)kH + ·

2

For any t, let us choose ` such that

kψ(t) − ψ(t` )kH ≤

The lemma is proved.

47

ε

·

2(A + 1)

Now let us pass to the limit in the non linear term. As above the sequence uk is bounded

8

in L 3 ([0, T ]; L4 (Ω)), we have in fact

Z tZ

Z tZ

(uk ⊗ uk : ∇Ψ)(t0 , x) dxdt0 .

(uk ⊗ uk : ∇Pk Ψ)(t0 , x) dxdt0 = lim

lim

k→∞ 0

k→∞ 0

Ω

Ω

So it is enough to prove that

Z tZ

Z tZ

0

0

(uk ⊗ uk : ∇Ψ)(t , x) dxdt =

(u ⊗ u : ∇Ψ)(t0 , x) dxdt0 .

lim

k→∞ 0

Ω

0

Ω

It is enough to prove that lim kuk ⊗ uk − u ⊗ ukL1 ([0,T ];L2 (Ω)) = 0 which will be implied by

k→∞

lim kuk − ukL2 ([0,T ];L4 (Ω)) = 0.

(4.16)

k→∞

Using (2.2.1), we have

1− d

d

4

kuk − ukL2 ([0,T ];L4 (Ω)) ≤ Ckuk − ukL2 ([0,T

k∇(uk − u)kL4 2 ([0,T ]×Ω) .

]×Ω)

Proposition 4.5.1 allows to conclude the proof of the fact that u is a solution of (N Sν ) in the

sense of Definition 4.2.1.

It remains to prove the energy inequality (4.5). Assertion (4.10) of Proposition 4.5.1

implies in particular that for any time t ≥ 0 and any v ∈ Vσ ,

lim (uk (t)|v)H = lim huk (t), v)i = hu(t), v)i = (u(t)|v)H .

k→∞

k→∞

As Vσ is dense in H, we get that for any t ≥ 0, the sequence (uk (t))k∈N converges weakly

towards u(t) in the Hilbert space H. Hence

ku(t)k2L2 ≤ lim inf kuk (t)k2L2

for all

k→∞

t ≥ 0.

On the other hand, (uk )k∈N converges weakly to u in L2loc (R+ ; V), so that for all non negative t,

we have

Z

Z

t

0

k∇u(t0 )k2L2 dt0 ≤ lim inf

k→∞

t

0

k∇uk (t0 )k2L2 dt0 .

Taking the lim inf in the energy equality for approximate solutions (4.7) yields the energy

k→∞

inequality (4.5).

Remarks

• Ce chapitre is to savoir.

• Si vous ˆetes curieux, vous pouvez consulter l’article fondateur of J. Leray ”Essai on

the mouvement of a liquide visqueux emplissant the space, Acta Matematica, 63, 1933,

pages 193–248.

• Pour en savoir more on the equation of Navier-Stokes incompressible, vous pouvez consulter by exemple les livres of P. Constantin and C. Foias Navier-Stokes equations,

Chigago University Press, 1988 et of P.-G. Lemari´e-Rieusset, Recent developments in

the Navier-Stokes problem. Chapman & Hall/CRC, Research Notes in Mathematics,

431, 2002.

• Si vous ˆetes int´eress´es by des d´eveloppements li´es `a la g´eophysique, vous pouvez consulter the livre of J.-Y. Chemin, B. Desjardins, I. Gallagher and E. Grenier, Mathematical

Geophysics; an introduction to rotating fluids and Navier-Stokes equations, Oxford Lecture series in Mathematics and its maps, 32, Oxford University Press, 2006.

48

Chapter 5

Stability of Navier-Stokes equations

In this chapter we intend to investigate the stability of the Leray solutions constructed in the

previous chapter. It is useful to start by analyzing the linearised version of the Navier-Stokes

equations, so the first section of the chapter is devoted to the proof of the wellposedness

of the time dependent Stokes system. The study will be applied in Section 5.1 to the two

dimensional Navier-Stokes equations, and the more delicate case of three space dimensions

will be dealt with in Sections 5.2 to 5.3.

5.1

Stability in dimension two

In a two dimensional domain, the Leray weak solutions are unique and even stable. More

precisely, we have the following theorem.

Theorem 5.1.1 For any data u0 in H and f in L2loc (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,

Z t

Z t

1

1

0 2

0

2

2

k∇u(t )kL2 dt = ku(s)kL2 +

hf (t0 ), u(t0 )i dt0 .

(5.1)

ku(t)kL2 + ν

2

2

s

s

Furthermore, the Leray solutions are stable in the following sense. Let u (resp. v) be the

Leray solution associated with u0 (resp. v0 ) in H and f (resp. g) in the space L2loc (R+ ; V 0 )

then,

Z t

2

k(u − v)(t)kL2 + ν

k∇(u − v)(t0 )k2L2 dt0

0

Z

CE 2 (t)

1 t

≤ ku0 − v0 k2L2 +

k(f − g)(t0 )k2Vσ0 dt0 exp

with

ν 0

ν4

Z

Z

1 t

1 t

def

2

0 2

0

2

0 2

0

E(t) = min ku0 kL2 +

kf (t )kVσ0 dt , kv0 kL2 +

kg(t )kVσ0 dt .

ν 0

ν 0

+

+

2

Proof of Theorem 5.1.1. As u belongs to L∞

loc (R ; H)∩Lloc (R ; Vσ ), thanks to Lemma 4.3.2

+

2

0

page 43, the non linear term Q(u, u) belongs to Lloc (R ; V ). Thus u is the solution of (ESν )

with initial data u0 and external force f + Q(u, u). Theorem 4.1.1 immediately implies that u

belongs to C(R+ ; H) and satisfies, for any (s, t) such that 0 ≤ s ≤ t,

Z t

1

1

2

ku(t)kL2 + ν

k∇u(t0 )k2L2 dt0 = ku(s)k2L2

2

2

s

Z t

Z t

0

0

0

+

hf (t ), u(t )i dt +

hQ(u(t0 ), u(t0 )), u(t0 )i dt0 .

s

s

49

Using Lemma 4.3.2, we get the energy equality (5.1).

def

To prove the stability, let us observe that, by difference w = u − v is the solution of (ESν )

with data u0 − v0 and external force f − g + Q(u, u) − Q(v, v), Theorem 4.1.1 implies that

Z t

2

k∇w(t0 )k2L2 dt0 = kw(0)k2L2

kw(t)kL2 + 2ν

0

Z t

Z t

+ 2 h(f − g)(t0 ), w(t0 )i dt0 + 2 h(Q(u, u) − Q(v, v))(t0 ), w(t0 )i dt0 .

0

0

The non linear term is estimated thanks to the following lemma.

Lemma 5.1.1 In two dimensional domains, if a and b belong to Vσ , we have

3

1

1

1

|h(Q(a, a) − Q(b, b)), a − bi| ≤ Ck∇(a − b)kL2 2 ka − bkL2 2 k∇akL2 2 kakL2 2 .

Proof of Lemma 5.1.1. It is an exercise of algebra to deduce from Lemma 4.3.2 that

hQ(a, a) − Q(b, b), a − bi = hQ(a − b, a), a − bi.

(5.2)

Using again Lemma 4.3.2, we get the result.

Let us go back to the proof of Theorem 5.1.1. Using that 2ab ≤ a2 + b2 , we get

Z t

Z

3

2 t

2

0 2

0

2

kw(t)kL2 + ν

k∇w(t )kL2 dt ≤ kw(0)kL2 +

k(f − g)(t0 )k2Vσ0 dt0

2 0

ν 0

Z t

3

1

1

1

+C

k∇w(t0 )kL2 2 kw(t0 )kL2 2 k∇u(t0 )kL2 2 ku(t0 )kL2 2 dt0 .

0

Using (with θ = 1/4) the convexity inequality

1

1

ab ≤ θa θ + (1 − θ)b1− θ

(5.3)

we infer that

kw(t)k2L2

Z

t

k∇w(t

+ν

0

0

)k2L2

0

dt ≤

kw(0)k2L2

2

+

ν

Z t

C

+ 3

ν

0

Z

0

t

k(f − g)(t0 )k2Vσ0 dt0

kw(t0 )k2L2 k∇u(t0 )k2L2 ku(t0 )k2L2 dt0 .

Gronwall’s lemma implies that

Z t

Z

2 t

2

0 2

0

2

kw(t)kL2 + ν

k∇w(t )kL2 dt ≤

k(f − g)(t0 )k2Vσ0 dt0

kw(0)kL2 +

ν 0

0

Z t

C

k∇u(t0 )k2L2 dt0 .

× exp 3 sup ku(t0 )k2L2

ν t0 ∈[0,t]

0

The energy estimate tells us that

2

Z t

Z

1

2 t

0 2

0 2

0

2

0 2

0

sup ku(t )kL2

k∇u(t )kL2 dt ≤

ku0 kL2 +

kf (t )kVσ0 dt .

ν

ν 0

0

t0 ∈[0,t]

As u and v play the same role, the theorem is proved.

50

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