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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



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 =



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



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[

≤ 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 )

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


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 +

=


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



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

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.


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

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

0 Ω
Z
Z t
=
u0 (x) · Ψ(0, x) dx +
hf (t0 ), Ψ(t0 )i dt0 with


∇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


0





(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



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


= −

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.



43

Thus, we have
1
hQ(u, v), vi = −
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 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.


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



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





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



0



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


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