Mathematical methods and fluid mechanics1

Basis of Mathematical Methods in Fluid
Mechanics
Jean-Pierre Puel

1

November 30., 2011

1

J.-P. Puel: (jppuel@bcamath.org) Ikerbasque and BCAM, Bizkaia Technology Park,
48160, Derio, Spain and Laboratoire de Mathématiques de Versailles, Université de Versailles
Saint-Quentin

These notes correspond to a course taught in BCAM in november 2011 with
some complements. Most of them also follow a DEA course taught in University of
Versailles St Quentin some time before. They are very classical basis for equations
of fluid mechanics (for viscous fluids) and for the mathematical analysis of these
equations and they intend to be essentially self contained. The first chapter can be
found in the introductions of many books related to fluid mechanics, for example
[1] and the next chapters follow closely notes written by L. Tartar in [3] which is

difficult to find. But of course, most of the results and the proofs can be found
(sometimes presented differently) in classical books where Navier-Stokes equations
are studied like [4] or [5] and the references therein.

1

Chapter 1

Equations of viscous fluid flows
1.1

Introduction

The physical domain is determined by
• An open subset Ω of IR2 or IR3 (say in general IRN ).
• x ∈ Ω is the sapce variable.
• t ∈ (0, T ) with T > 0 is the time variable.
The so-called “state variables” or quantities which determine the flow are
• The fluid velocity u.
• The pressure p.

• The density of the fluid ρ.
• Sometimes other quantitites like the temperature θ, .....
Eulerian description.
u(x, t) is the velocity of the particle of fluid which is at position x at time t. Therefore, if t′ 6= t, u(x, t′ ) is the velocity of a different particle. The observer is located
at position x and looks at particles passing at this position. Here we will focus
essentially on this Eulerian description.
Lagrangian description.
Let us call here the velocity v(x, t) in order to avoid confusions. If we call (x(t))t≥t0
the trajectory of a particle emanating from point x0 at time t0 (in fact this trajectory
is an unknown of the problem). Then v(x0 , t) (t ≥ t0 ) is the velocity, at time t of
the particle which was at position x0 at time t0 and therefore, the velocity at time
2

t of the particle which is at position x(t). The observer is here transpoted by the
flow and follows the particle.
Particle derivative.
Let (t, x(t)) the trajectory of a fluid particle starting from (t0 , x0 ). We have with
the Eulerian velocity,
dx
(t) = u(x(t), t),

dt
x(t0 ) = x0 .
Let ϕ be a function of x and t. The particle derivative of ϕ is
d
d
ϕ(x0 , t0 ) = (ϕ(x(t), t))/t=t0
dt
dt
N
X
∂ϕ
∂ϕ
=
ui (x0 , t0 )
(x0 , t0 ) +
(x0 , t0 )
∂t
∂x
i
i=1

1.2

N
X

=

 ∂ϕ

+

=

 ∂ϕ

+ u.∇ϕ

∂t
∂t

ui

i=1

∂ϕ 
∂xi /(x0 ,t0 )


/(x0 ,t0 )

.

Conservation laws

Let V (t) be a volume of fluid which, when t varies, contains the same particles
(which have been moving). That is to say V (t) is a volume transported by the
velocity field u(x, t) or for t′ ≥ t
V (t′ ) = {x(t′ ),

dx
(s) = u(x(s), s), t ≤ s ≤ t′ ,
ds

x(t) ∈ V (t)}.

We say that a quantity ϕ, which is a function of x and t is conserved if
∀V (t),

d
dt

Z

ϕ(x, t)dx = 0.
V (t)

Therefore for every V (t) and δt (small) we have
Z

ϕ(y, t + δt)dy =
V (t+δt)

3

Z

ϕ(x, t)dx.
V (t)

y
V(t+δt)

u

x
V(t)

We can write
V (t + δt) = {y(δt),

dy
(s) = u(y(s), s),
dt

, y(0) = x,

x ∈ V (t)}.

Let us expand every quantity at the first order in δt. We have
y(x, δt) = x + δt.u(x, t) + l.o.t.
so that
dyi = dxi + δt

N
X
∂ui

j=1

∂xj

(x, t)dxj + l.o.t.

Therefore
dy = dy1 ∧ dy2 ∧ · · · ∧ dyN = dx1 ∧ dx2 ∧ · · · ∧ dxN
4

N
X
∂u1

+δt

j=1

∂xj





dxj ∧ dx2 ∧ · · · ∧ dxN + · · ·

+δt dx1 ∧ dx2 ∧ · · · ∧

N
X
∂uN 

j=1

∂xj

+ l.o.t.

so that
dy = (1 + δt(div u) + l.o.t.)dx.
Now we have
ϕ(y(x, δt), t + δt) = ϕ(x + δtu(x, t) + l.o.t., t + δt)

= ϕ(x, t) + δt

N
X
∂ϕ
∂ϕ
uj (x, t)
(x, t) + δt
(x, t) + l.o.t.
∂t
∂xj
j=1

Therefore
Z

ϕ(y, t + δt)dy
V (t+δt)

Z

∂ϕ
(x, t) + u(x, t).∇ϕ(x, t) + o(δt))(1 + δtdiv u(x, t) + o(δt))dx
∂t
V (t)
Z
Z

 ∂ϕ
(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)div u(x, t) dx + o(δt).
ϕ(x, t)dx + δt
=
V (t) ∂t
V (t)
=



ϕ(x, t) + δt(

As we have

Z

ϕ(y, t + δt)dy =
V (t+δt)

Z

ϕ(x, t)dx,
V (t)

we obtain
∀V (t),

Z

∀δt > 0,

V (t)

 ∂ϕ

∂t



(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)div u(x, t) dx = o(δt),

and therefore
∀V (t),

Z

V (t)

 ∂ϕ

∂t



(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)div u(x, t) dx = 0,

that is to say the equation
∂ϕ
(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)div u(x, t) = 0
∂t
or

∂ϕ
(x, t) + div (u(x, t).ϕ(x, t)) = 0.
∂t
5

Another way to obtain this equation is the following. Let S(t) be the surface limiting
the volume V (t) and let ν be the outward pointing unit normal vector to this surface.
We first have to show that
d
dt

Z

ϕ(x, t)dx =
V (t)

Z

V (t)

∂ϕ
(x, t)dx +
∂t

Z

ϕ(x, t)u(x, t)ν(x)dS.
S(t)

Then applying Stokes-Ostrogradskii formula we have
Z

ϕ(x, t)u(x, t)ν(x)dS =
S(t)

Z

div (u(x, t).ϕ(x, t))dx,
V (t)

which shows that
d
dt

1.2.1

Z

ϕ(x, t)dx =
V (t)

Z

V (t)

 ∂ϕ

∂t



(x, t) + div (u(x, t).ϕ(x, t)) dx.

Conservation of mass

Here we take
ϕ(x, t) = ρ(x, t)
where ρ is the density of the fluid. We obtain the equation of conservation of mass
which can take different forms.
(1.2.1)

∂ρ
(x, t) + div (u(x, t)ρ(x, t)) = 0 in Ω × (0, T ),
∂t

or
(1.2.2)

∂ρ
(x, t) + u(x, t).∇ρ(x, t) + ρ(x, t)div u(x, t) = 0 in Ω × (0, T ),
∂t
or even using the particle derivative

(1.2.3)

1.2.2

d
ρ(x, t) + ρ(x, t)div u(x, t) = 0 in Ω × (0, T ).
dt

Conservation of volume. Incompressibility

A fluid is incompressible if the volume occupied by a group of fluid particles remains
constant during the flow. Therefore we here take
ϕ(x, t) = 1
6

which gives the incompressibility condition
(1.2.4)

div u(x, t) = 0 in Ω × (0, T ).

In that case, the conservation of mass becomes
(1.2.5)
(1.2.6)

∂ρ
(x, t) + u(x, t).∇ρ(x, t) = 0 in Ω × (0, T ),
∂t
ρ(x, 0) = ρ0 (x) in Ω.

When the inital density satisfies ρ0 (x) = ρ0 = Cst (independent of x), this implies
ρ(x, t) = ρ0 .
Also if ρ(x, t) = ρ0 equation for conservation of mass implies that div u(x, t) = 0.
In fact we see that
div u = 0 ⇒

d
ρ = 0 ⇒ ρ(x(t), t) = Cst
dt

and

d
ρ = 0 ⇒ div u = 0.
dt
Therefore the fluid is incompressible (div u = 0) if and only if the density of each
d
ρ = 0).
element stays constant during the flow ( dt
ρ(x(t), t) = Cst ⇒

1.2.3

Stress tensor. Conservation of momentum

Let dS be a surface element in the fluid which separates the fluid in two parts F 1
and F 2 and let n be the unit normal vector to dS pointing towards the exterior of
F 2.

7

F1
n

dS

F2

The force exerted by F 1 on F 2 per surface unit is called stress.
For a fluid at rest (u = 0), this force is normal to dS, so it is characterized by a
scalar quantity at each point. This quantity is the hydrostatic pressure.
For a moving fluid, it appears tangential stresses : friction between the fluid layers
gliding along each other, due to the viscosity of the fluid.
One can show (or it is commonly admitted) that there exists a tensor σ = (σij )
(represented by a (2, 2) or (3, 3) or (N, N ) matrix called the stress tensor such
that
• σ is symmetric : σij = σji . This comes from an equilibrium equation.
• The force exerted by F 1 on F 2 is given by
σ.n = (

N
X

σij nj )i=1,···,N = (σij nj )i=1,···,N

j=1

with the convention of summation for repeated indices.
8

Tensor of viscosity stresses.
Among the stresses, it is convenient to separate those which do not depend of the
fluid deformation, that is to say which exist when the fluid is at rest, and those
which are due to the fluid deformation. We set
′
σij = −pδij + σij
,
′ ) is the tensor of
where p is the pressure, δij is the Kronecker symbol and (σij
viscosity stresses. (Sign − in front of the pressure is just a choice indicating that
usually, a fluid at rest is compressed).
′ ) is independent of translations and local rotations, and therefore independent
(σij
∂u

∂ui
− ∂xji ) (ω = curlu is the vorticity of the fluid).
of u itself and of ωij = 21 ( ∂x
j
′ only depends on the symmetric part of the tensor of velocity gradients
Therefore, σij
e = (eij ) where
∂uj
1 ∂ui
+
).
eij = (
2 ∂xj
∂xi

Newtonian fluids.
For Newtonian fluids (only case which will be considered here), the relation between
′ ) and (e ) is linear and (after some remarks on isotropy etc) the relation can be
(σij
ij
written
2
′
σij
= η.
(2eij − δij ell )
+ζ.
(δij ell ) ,
| {z }
|
{z3
}
isotropic dilation

deformation without volume changes

where η is the shear viscosity and ζ is the volumic velocity.
For an incompressible fluid, we have div u = 0 so that ell = 0 and then
′
σij
= 2ηeij .

Conservation of momentum.
Let V (t) be any volume transported by the flow asociated with the velocity u and
limited by a surface S(t). In the presence of external forces represented by f and
the action of the exterior of V (t) exerted on the surface S(t), the fundamental law
of dynamics takes the (vectorial) form
d
dt

Z

ρudx =
V (t)

Z

ρf dx +
V (t)

Z

σ.ndS.
S(t)

Here f is the volumic density of external forces per unit of mass. For example f
can represent
• Gravity forces.
9

• An electrostatic force for a fluid with electrical charges.
• A Coriolis force for a fluid in a rotating reference frame.
• A magnetic force for a fluid containing magnetic particles (ferrofluid).
• ....
For each component i = 1, · · · , N we have
d
dt

Z

V (t)

Z

ρui dx =

∂

∂t

V (t)

=

Z

V (t)



(ρui ) + div (ρui u) dx

ui

 ∂ρ

∂t



+ div (ρu) dx +

Z

 ∂u

ρ
V (t)

i

∂t

From the equation expressing conservation of mass we have
∂ρ
+ div (ρu) = 0,
∂t
so that

Z

Z

Z

(σ.n)i dS =

Z

 ∂u

d
dui
i
ρ
ρ
ρui dx =
+ u.∇ui dx =
dx.
dt V (t)
∂t
dt
V (t)
V (t)
On the other hand, from Stokes-Ostrogradskii formula we have

S(t)

Z

N
X

V (t) j=1

σij nj dS =

Z

N
X
∂σij

V (t) j=1

∂xj

dx.

Therefore we obtain for every volume V (t)
Z

ρ
V (t)

 ∂u

i

∂t



+ u.∇ui dx =

Z

N
X
∂σij

V (t) j=1

∂xj

dx +

Z

V (t)

This gives us the equation for conservation of momentum
ρ

(1.2.7)

 ∂u

i

∂t



+ u.∇ui =

N
X
∂σij

j=1

∂xj

+ ρfi ,

which can also be written as
(1.2.8)

ρ

 ∂u

i

∂t

+ u.∇ui



N
′
X
∂σij
∂p
+
+ ρfi ,
=−
∂xi j=1 ∂xj

or in vectorial form
(1.2.9)

 ∂u

ρ

∂t



+ (u.∇)u = −∇p + div σ ′ + ρf.

10



+ u.∇ui dx.

ρfi dx.

1.3

Basic equations : Navier-Stokes, Euler, Stokes,....

If we neglect the spatial variations of viscosities
• For a Newtonian viscous compressible fluid,
(1.3.10)
(1.3.11)

∂η
∂xj

and

∂ζ
∂xj

we obtain :

∂ρ
+ div (ρu) = 0,
∂t
 ∂u

η
ρ
+ (u.∇)u = −∇p + η∆u + (ζ + )∇(div u) + ρf.
∂t
3

In order to complete this system we need to give the pressure law, or the energy
equation.
• For a Newtonian viscous incompressible fluid we obtain the Navier-Stokes equations
(1.3.12)
(1.3.13)

div u = 0,
 ∂u

ρ
+ (u.∇)u = −∇p + η∆u + ρf.
∂t

• For a perfect (inviscid) incompressible fluid we obtain the Euler equations
(1.3.14)
(1.3.15)

div u = 0,

 ∂u
+ (u.∇)u = −∇p + ρf.
ρ
∂t

• For a steady fluid (u = 0) we obtain
(1.3.16)

ρf = ∇p,

which is the fundamental principle of hydrostatics.
A-dimensional form of Navier-Stokes equations.
Let L and U be the respective reference scales of length and velocity of the flow.
We write
x
u
t
p − p0
x′ = , u′ = , t′ =
, p′ = 1 2 ,
L
U
L/U
2 ρU
where p0 is the hydrostatic pressure (in absence of flow). We obtain
(1.3.17)

∂u′
η
+ (u′ .∇′ )u′ = −∇′ p′ +
∆′ u ′ + f ′ .
′
∂t
ρU L

Setting
ν=

η
ρ

1
ν
=
,
Re
UL
11

this defines the Reynolds number Re.
Stokes equations.
For small Reynolds numbers, or large viscosity, for laminar flows, we can neglect
the convective terms (u.∇)u to obtain the Stokes equations
(1.3.18)
(1.3.19)

div u = 0,
∂u
= −∇p + ν∆u + f.
∂t

Boundary conditions.
• Boundary conditions on the surface of a solid body.
No penetration of the fluid
usol .n = ufluid .n.
For a viscous fluid : no-slip boundary condition
ufluid = usol .
• Boundary condition at the (fixed) interface of two fluids.
Continuity of velocities
u1 = u2 .
Equilibrium between the stresses in each of the fluids and the stresses localized on
the interface (n is the normal vector and τ are tangent vectors)
(σ 1 .n)τ = (σ 2 .n)τ

(equality of tangential stresses),

1
1
+ ′ ),
R R
where γ is the surface tension coefficient between fluid 1 and fluid 2 and R and R′
are the principal curvature radii of the interface.
Initial conditions.
They must describe the flow at initial time (t = 0) by the datas
(σ 1 .n)n − (σ 2 .n)n = γ(

u/t=0 = u0 ,

ρ/t=0 = ρ0 , · · ·

Remark 1.3.1 • If we want to consider an interaction between a fluid and a structure, the coupling must occur in the boundary conditions.
• If we want to consider the thermal effects, we must have to add an equation for
the energy and the coupling will appear through ρ or σ in the fluid equation.
• We can also consider coupling with other phenomena like transport of a specy
(salinity in an ocean, physico-chemical elements of a fluid, ...). The coupling in the
fluid equation will then have to be defined in a consistent way.
12

Remark 1.3.2 We can consider the stationnary problems corresponding to the previous equations. This does not say that the fluid is at rest, but says only that the
flow does not vary with time. Therefore it corresponds to cancel all terms containing
∂
∂t .

13

Chapter 2

Stokes equations. Mathematical
formulation
2.1

Stationnary Stokes equations

Let Ω be a connected open subset of IRN that we will suppose to be bounded and
regular and let Γ be its boundary. The Stokes equation with no-slip boundary
condition can be written in the following (vectorial) form
(2.1.1)

−ν∆u = −∇p + f

(2.1.2)

div u = 0

(2.1.3)

u=0

in

on

in

Ω,

Ω,

Γ,

where ν > 0 is the given viscosity and f is the given external force and u and p are
the velocity and the pressure.
For each component ui , i = 1, · · · , N we have
∂p
−ν∆ui = −
+ fi .
∂xi
Let w = (w1 , · · · , wN ) be a vector function which is “regular” and which is zero on
the boundary, with div w = 0. If we multiply the equation for ui by wi , integrate
on Ω and sum up for i = 1 to N we obtain
−ν

N Z
X

i=1 Ω

∆ui wi dx = −

N Z
X

i=1 Ω

N
X
∂p
wi dx +
∂xi
i=1

Z

Ω

As w/Γ = 0, by integration by parts we have
−

N Z
X

i=1 Ω

∂p
wi dx =
∂xi

Z

14

p(div w)dx = 0,
Ω

fi wi dx.

and
−ν

N Z
X

i=1 Ω

∆ui wi dx = ν

N Z
X

i=1 Ω

∇ui ∇wi dx.

Therefore u must satisfy for all w regular, vanishing on Γ such that div w = 0,
ν

N Z
X

i=1 Ω

∇ui ∇wi dx =

N Z
X

i=1 Ω

fi wi dx.

This suggests a variational formulation of Stokes system.
Precise mathematical formulation.
Let us define the space
V = {w ∈ H01 (Ω)N ,

(2.1.4)

div w = 0}

and the bilinear form defined for all v ∈ V and w ∈ V by
a(v, w) = ν

(2.1.5)

N Z
X

i=1 Ω

∇vi ∇wi dx.

If f = (f1 , · · · , fN ) ∈ L2 (Ω)N (for example but we could also take f ∈ H −1 (Ω)N ),
we look for u such that
(2.1.6)

a(u, w) =

N Z
X

i=1 Ω

fi wi dx

(or

N
X

< fi , wi > ),

i=1

∀w ∈ V,

u ∈ V.
Here < ., . > denotes the duality pairing between H −1 (Ω) and H01 (Ω). We then
have
Theorem 2.1.1 For every f ∈ H −1 (Ω)N , there exists a unique solution u ∈ V to
the problem
(2.1.7)

a(u, w) =

N
X

< fi , wi >,

i=1

∀w ∈ V,

u ∈ V.
The mapping f → u is linear continuous from H −1 (Ω)N to V and moreover u is
also solution to the minimization problem
(2.1.8)

J(u) = min J(w),
w∈V

u ∈ V,
15

where

N
X
1
< fi , w i > .
J(w) = a(w, w) −
2
i=1

(2.1.9)

Proof.
First of all, as Ω is bounded, from Poincaré inequality, we can equip H01 (Ω) with the
R
1
norm v → ( Ω |∇v|2 dx) 2 and the associated scalar product and it is then a Hilbert
space. Then H01 (Ω)N is also a Hilbert space for the norm (and the corresponding
scalar product)
w = (w1 , · · · , wN ) → (

N Z
X

i=1 Ω

1

|∇wi |2 dx) 2 .

Now the mapping w → div w is linear continuous from H01 (Ω)N to L2 (Ω), and the
space V is the kernel of this mapping. Therefore, V is a closed subspace of H01 (Ω)N
and therefore, it is a Hilbert space for the norm (and the scalar product) induced
by the one in H01 (Ω)N .
It is now immediate to see that
(v, w) ∈ V × V → a(v, w)
is a continuous bilinear form on V × V which is obviously coercive when ν > 0.
P
On the other hand, when f ∈ H −1 (Ω)N , the mapping w → N
i=1 < fi , wi > is linear
continuous from V to IR.
We can then apply Lax-Milgram Theorem which shows the first part of Theorem
2.1.1.
That u is the solution to the minimization problem for functional J comes from the
fact that the bilinear form a(., .) is symmetric.
Interpretation. Relation with Stokes problem.
What is the relation with Stokes problem and in particular what about the
pressure p which seems to have disappeared?
Let us write V the space
V = {ϕ ∈ D(Ω)N ,

div ϕ = 0},

where D(Ω) = C0∞ (Ω). Of course V ⊂ V and it can be shown that V is dense in V .
Using distribution theory we have for ϕ ∈ V
a(u, ϕ) = ν

N Z
X

i=1 Ω

∇ui ∇ϕi dx = ν

N
X
i=1

< ∇ui , ∇ϕi >D′ ,D = −ν

16

N
X
i=1

< ∆ui , ϕi >D′ ,D .

Therefore we obtain
N
X
i=1

< −ν∆ui − fi , ϕi >D′ ,D = 0 ∀ϕ ∈ V.

The distribution T = (T1 , · · · , TN ) ∈ D′ (Ω)N with Ti = −ν∆ui − fi is such that
∀ϕ ∈ V,

< T, ϕ >D′ ,D = 0.

Lemma 2.1.2 (de Rham’s Lemma) Let T ∈ D′ (Ω)N such that
∀ϕ ∈ V,

< T, ϕ >D′ ,D = 0.

Then there exists a distribution p ∈ D′ (Ω) such that
T = −∇p.
In fact it is easy to see that
∀p ∈ D′ (Ω),

∀ϕ ∈ V,

< −∇p, ϕ >D′ ,D =< p, div ϕ >D′ ,D = 0.

de Rham’s Lemma, which is very difficult, shows that the only distributions which
vanish on V are of the form T = −∇p.
So here we see that there exists p ∈ D′ (Ω) such that for i = 1, · · · , N

div u = 0,

∂p
in
∂xi
in Ω,

u=0

Γ.

−ν∆ui = fi −
on

D′ (Ω),

In fact we can see that for every i = 1, · · · , N

∂p
= fi + ν∆ui ∈ H −1 (Ω).
∂xi

Therefore p ∈ D′ (Ω) and ∇p ∈ H −1 (Ω) and this implies that p ∈ L2 (Ω). Of course
p is defined up to the addition of a constant. We then obtain
Theorem 2.1.3 Let f ∈ H −1 (Ω)N . Then there exists a unique u ∈ V and a unique
p ∈ L2 (Ω)/IR (p is unique up to the addition of a constant) such that
(2.1.10)

−ν∆ui = fi −

(2.1.11)

div u = 0,

∂p
in
∂xi
in Ω,

(2.1.12)

u=0

Γ.

on

((u, p) is solution to Stokes problem).
17

H −1 (Ω)N ,

We can also obtain a regularity result.
Theorem 2.1.4 Let f ∈ L2 (Ω)N . The solution (u, p) of Stokes problem satisfies
u ∈ H 2 (Ω)N ∩ V,

p ∈ H 1 (Ω)/IR

and we have
||u||H 2 (Ω)N + |p|L2 (Ω)/IR ≤ C|f |L2 (Ω)N .

2.2

Complements of functional analysis and applications

This section will provide some complementary results of functional analysis which
enable to prove most of the properties which have been accepted so far without
proof. This section follows very closely lecture notes given by L. Tartar on the
subject, see [3].
First of all we start with some abstract results.
Let F , G, H be three Hilbert spaces and let A be a continuous linear operator from
H to F . Then
KerA = {h ∈ H, Ah = 0}

is a closed linear subspace of H. Let us define
N = KerA⊥ = {h ∈ H,

∀l ∈ KerA,

(h, l)H = 0}.

Then N is also a closed linear subspace of H.
If h ∈ H, we can consider h1 = ProjKerA h and h2 = h − h1 . From the definition of
projection we have
∀l ∈ KerA, (h − h1 , l)H = 0
which implies

h2 = h − h1 ∈ N.

Therefore for every h ∈ H we have the decomposition
h = h1 + h2 ,

h1 ∈ KerA,

h2 ∈ N.

Moreover, if h ∈ KerA ∩ N , then it is clear that h = 0. This says that
H = KerA ⊕ N
and KerA and N are closed, so that they are topological supplements. The above
decomposition is then unique and moreover there exist two constants C1 > 0 and
c2 > 0 such that
∀h ∈ H,

|h1 |H ≤ C1 |h|H ,
18

|h2 |H ≤ C2 |h|H

which says that the projections are continuous. (It is clear here in the case of Hilbert
spaces that the constants C1 and C2 are less or equal to 1.)
Now we will assume that there exists a compact linear operator B from H to G
such that
(2.2.13)
∃C > 0, ∀h ∈ H, |h|H ≤ C(|Ah|F + |Bh|G ).
Lemma 2.2.1 Under the above hypotheses
(2.2.14)

∃CN ,

∀h ∈ N,

|h|H ≤ CN |Ah|F .

Moreover, the image of A denoted ImA is closed in F .
Proof.
Let us suppose that (2.2.14) is not true. Then
∀n ∈ IN,

∃hn ∈ N,

Let us define
h̃n =

|hn |H ≥ n|Ahn |F .

hn
∈ N.
|hn |H

Then |h̃n |H = 1 and |Ah̃n |F ≤ n1 (therefore |Ah̃n |F → 0 in F ).
As |h̃n |H = 1 we can extract a subsequence, still denoted (h̃n ) such that
h̃n ⇀ h0

in

H weakly.

As N is a closed subspace (then closed for the weak topology) and h̃n ∈ N , h0 ∈ N .
As A is linear continuous from H to F , it is continuous for the weak topologies and
therefore,
Ah̃n ⇀ Ah0 in F weakly.
But we already know that Ah̃n → 0 in F strongly. Therefore Ah0 = 0 and h0 ∈ N .
This says that h0 ∈ KerA ∩ N so that h0 = 0.
Now we have
h̃n ⇀ 0

in

H weakly

and

Ah̃n → 0

in

As B is compact we have
B h̃n → B0 = 0

in

G strongly

so that |B h̃n |G → 0.
But from (2.2.13) we have
|h̃n |H ≤ C(|Ah̃n |F + |B h̃n |G ) → 0.
19

F strongly.

This gives a contradiction with the fact that |h̃n |H = 1. Therefore (2.2.14) is true.
Let us now show that ImA is closed in F .
Let us take a sequence (yn ) such that
yn ∈ ImA,

yn → y

in

F.

As yn ∈ ImA, there exists h̄n ∈ H such that Ah̄n = yn . We have the decompsition
h̄n = hn + ĥn ,

hn ∈ N,

ĥn ∈ KerA.

Then Ahn = yn and hn ∈ N .
As yn → y in F , (yn ) is a Cauchy sequence in F . and
A(hm − hp ) = ym − yp ,

(hm − hp ) ∈ N.

From (2.2.14) we see that
|hm − hp |H ≤ CN |ym − yp |F
and therefore, (hn ) is a Cauchy sequence in H which is complete. This shows that
there exists h ∈ H such that hn → h in H and as Ais continuous,
Ahn → Ah

in

F.

But we know that Ahn = yn converges to y in F . Then there exists h ∈ H such
that y = Ah and y ∈ ImA so that ImA is closed.
This finishes the proof of Lemma 2.2.1.
We are going to show that hese abstract results can be applied to the case
H = L2 (Ω),

F = H −1 (ΩN ,

G = H −1 (Ω)

and
A=∇

(Gradient operator)

in order to show that ∇ has a closed image in H −1 (Ω)N .
Let us define the space
(2.2.15)
Lemma 2.2.2
(2.2.16)

X(Ω) = {g ∈ H −1 (Ω),

∇g ∈ H −1 (ΩN }.

X(IRN ) = L2 (IRN ).

20

Proof.
It is clear that L2 (IRN ) ⊂ X(IRN ). Let us show that if g ∈ X(IRN ), then g ∈
L2 (IRN ) by using the Fourier transform.
We know that
1

g ∈ L2 (IRN ) ⇐⇒ ĝ ∈ L2 (IRN ) and that g ∈ H −1 (IRN ) ⇐⇒ (1+|ξ|2 )− 2 ĝ ∈ L2 (IRN ).
Now

1

∇g ∈ H −1 (IRN )N ⇐⇒ (1 + |ξ|2 )| 2 |ξ|ĝ| ∈ L2 (IRN )
so that
g ∈ X(IRN ) ⇐⇒ (1 + |ξ|2 )(1 + |ξ|2 )|ĝ|2 ∈ L1 (IRN ) ⇐⇒ ĝ ∈ L2 (IRN ).
Lemma 2.2.3
(2.2.17)

2
N
X(IRN
+ ) = L (IR+ ).

Proof.
N
N
N
2
It is clear that L2 (IRN
+ ) ⊂ X(IR+ ). Let us show that X(IR+ ) ⊂ L (IR+ ). To this
N
end, we are going to exhibit a continuous extension from X(IR+ ) to X(IRN ).
Let us first define a restriction operator from H 1 (IRN ) to H01 (IRN
+ ). For u ∈
D(IRN ) = C0∞ (IRN ) we set
Qu(x1 , · · · , xN ) =



We impose that

0, if xN < 0,
P
u(x1 , · · · , xN ) + 2j=1 aj u(x1 , · · · , xN −1 , −jxN )
1+

2
X

if

xN > 0.

if

xN > 0.

aj = 0

j=1

which implies
Qu(x1 , · · · , xN −1 , 0) = 0.
It is clear that for i = 1, · · · , N − 1 we have
Q(
For i = N we have
Q(

∂Qu
∂u
)=
.
∂xi
∂xi

∂u
∂
)=
Ru
∂xN
∂xN

where
Ru(x1 , · · · , xN ) =



0, if xN < 0,
P
u(x1 , · · · , xN ) + 2j=1
21

aj
−j u(x1 , · · · , xN −1 , −jxN )

We also want R to be continuous from H 1 (IRN ) to H01 (IRN
+ ) which imposes
1+

2
X
aj

j=1

−j

= 0.

Thus we choose the coefficients aj such that
1+

2
X

aj = 0,

1+

2
X
aj

j=1

j=1

−j

=0

and they are well defined by these relations. Now we extend Q and R by continuity
to H 1 (IRN ) and we define
P =t Q.
As Q is linear continuous from H 1 (IRN ) to H01 (IRN
+ ), P is linear continuous from
−1 (IRN ). Moreover we have for i = 1, · · · , N − 1,
H −1 (IRN
)
to
H
+
∂ϕ
∂ϕ
∂
P u, ϕ >= − < P u,
>= − < u, Q(
)>
∂xi
∂xi
∂xi
∂u
∂u
∂
Qϕ >=<
, Qϕ >=< P (
), ϕ >,
= − < u,
∂xi
∂xi
∂xi
∀ϕ ∈ H 1 (IRN ),

so that

<

∂u
∂
Pu = P(
).
∂xi
∂xi

∂u
∂u
∈ H −1 (IRN
So, for i = 1, · · · , N − 1, as ∂x
+ ) we have P ( ∂xi ) =
i
Now for i = N we have in the same way

∂
∂xi P u

∈ H −1 (IRN ).

∂
∂u
P u =t R
∂xN
∂xN
−1 (IRN ).
and t R is linear continuous from H −1 (IRN
+ ) to H
N
N
Therefore, if u ∈ X(IR+ ), then P u ∈ X(IR ) and P is a continuous linear operator.
It remains to prove that P is an extension.
N
For u ∈ H01 (IRN
+ ) let us define by Eu the extension by 0 outside IR+ . It is well
N
N
1
1
known that u → Eu is linear continuous from H0 (IR+ ) to H (IR ). We denote by
Π the transposition of this operator E which is the restriction to IRN
+ and which is
a linear continuous operator from H −1 (IRN ) to H −1 (IRN
).
We
have
to show that
+
Π.P = Id.
But Π.P =t E.t Q =t (Q.E) and it is clear that Q.E = Id so that Π.P = Id.
N
N
2
If u ∈ X(IRN
+ ), P u ∈ X(IR ) = L (IR ) from Lemma 2.2.2. Then Π.P u = u ∈
L2 (IRN
+ ) and this finishes the proof of Lemma 2.2.3.

22

Lemma 2.2.4 If Ω is a bounded regular open set in IRN , then
X(Ω) = L2 (Ω).

(2.2.18)

Proof.
Let (θi ) be a partition of unity so that
θi ∈ C ∞ (Ω),

0 ≤ θi ≤ 1,

We write
u=

I
X

I
X

θi = 1.

i=1

θi u.

i=1

In the case θi ∈ C0∞ (Ω), θi u can be extended by zero to have θi u ∈ X(IRN ).
When θi ∈ C ∞ (Ω), we can find a C 2 diffeomorphism η such that θi u◦η −1 ∈ X(IRN ).
Then θi u ◦ η −1 ∈ L2 (IRN ) which implies θi u ∈ L2 (Ω).
Now we have, algebraically, X(Ω) = L2 (Ω).
Let us take on X(Ω) the norm defined by
∀g ∈ X(Ω),

|g|2X(Ω) = |∇g|2H −1 (Ω)N + |g|2H −1 (Ω) .

Equipped with this norm it is clear that X(Ω) is a Hilbert space.
Lemma 2.2.5 There exists a constant C > 0 such that
(2.2.19)

∀g ∈ L2 (Ω),

Proof.
When g ∈ L2 (Ω) we have

|g|2L2 (Ω) ≤ C|g|2X(Ω) .

|g|2H −1 (Ω) ≤ C|g|2L2 (Ω)

and
|∇g|2H −1 (Ω)N ≤ C|g|2L2 (Ω) .
Let us consider the identity map from L2 (Ω) to X(Ω), i.e.








Id : L2 (Ω), | · |L2 (Ω) → X(Ω), | · |X(Ω) .
From Lemma 2.2.4 this is a one-to-one mapping which is continuous between two
Hilbert spaces. From Banach Theorem, it is bi-continuous (the inverse is continuous)
which says that there exists a constant C > 0 such that
∀g ∈ L2 (Ω),

|g|2L2 (Ω) ≤ C|g|2X(Ω) .
23

This gives Lemma 2.2.5.
Now, as Ω is bounded, H01 (Ω) is compactly embedded in L2 (Ω) and then L2 (Ω) is
also compactly embedded in H −1 (Ω).
Let us set
H = L2 (Ω),

F = H −1 (Ω)N ,

G = H −1 (Ω),

A = ∇,

B = Id.

All hypotheses of Lemma 2.2.1 are fulfilled. We then have the following results.
Lemma 2.2.6 The operator ∇ : L2 (Ω) → H −1 (Ω)N has a closed image.
If we define
Ker∇ = {g ∈ L2 (Ω),

∇g = 0}

as Ω is a connected set, we see that the elements of Ker∇ are constants in Ω.
Therefore we have
Z
⊥
2
gdx = 0}.
Ker∇ = {g ∈ L (Ω),
Ω

Lemma 2.2.7 If (pn ) is a sequence in L2 (Ω) such that
is bounded, then |pn |L2 (Ω) is bounded.

R

Ω pn dx

= 0 and |∇pn |H −1 (Ω)N

Proof.
We have pn ∈ Ker∇⊥ . Then from Lemma 2.2.1, there exists a constant C > 0
independent of n such that
|pn |L2 (Ω) ≤ C|∇pn |H −1 (Ω)N
and this implies the lemma.
Lemma 2.2.8 Let f ∈ H −1 (Ω)N such that
∀w ∈ V,

< f, w >= 0.

Then there exists p ∈ L2 (Ω)/IR such that f = −∇p.
Moreover, there exists a constant C > 0 such that
|p|L2 (Ω)/IR ≤ C|f |H −1 (Ω)N .
Proof.
Let us write
Y = {∇p,

p ∈ L2 (Ω)}.

Then Y is a closed subspace of H −1 (ΩN . Let us show that
{w ∈ H01 (Ω)N ,

< y, w >= 0, ∀y ∈ Y } = V.
24

Actually, if w ∈ H01 (Ω)N and < ∇p, w >= 0 ∀p ∈ L2 (Ω) we have < w, ∇ϕ >= 0
∀ϕ ∈ C0∞ (Ω) so that < div w, ϕ >= 0 ∀ϕ ∈ C0∞ (Ω) and therefore div w = 0 so that
w ∈V.
On the other hand we have

Therefore V = Y

⊥

∀p ∈ L2 (Ω),

∀w ∈ V,

< ∇p, w >= 0.

which implies
V ⊥ = (Y ⊥ )⊥ = Y = Y.

Then if f ∈ H −1 (Ω)N is such that ∀w ∈ V < f, w >= 0 then f ∈ V ⊥ = Y and this
proves Lemma 2.2.8.
Lemma 2.2.8 enables us to give a completely correct interpretation of Stokes problem
in Theorem 2.1.3. Indeed, if gi = −ν∆ui − fi and g = (g1 , · · · , gN ) we have from
the variational formulation
g ∈ H −1 (Ω)N

and

∀w ∈ V,

< g, w >= 0.

Then there exists p ∈ L2 (Ω)/IR such that g = −∇p, which gives the correct interpretation of Stokes problem.
This result can also be proved by another method.
For ǫ > 0 let us consider the problem
1
a(uǫ , w) +
ǫ

Z

Ω

div uǫ div wdx =

N
X
i=1

< fi , wi >, ∀w ∈ H01 (Ω)N ,

uǫ ∈ H01 (Ω)N .
This problem has a unique solution uǫ which satisfies
1
−ν∆uǫ − ∇( div uǫ ) = f
ǫ
N
1
uǫ ∈ H0 (Ω) .
If we write

1
pǫ = − div uǫ
ǫ

then
pǫ ∈ L2 (Ω),

Z

Ω

pǫ dx = 0.

On the other hand we have
1
a(uǫ , uǫ ) +
ǫ

Z

Ω

|div uǫ |2 dx =
25

N
X
i=1

< fi , uǫ,i >

so that there exists a constant M independent of ǫ such that
1
| √ div uǫ |L2 (Ω) ≤ M.
ǫ

||uǫ ||H 1 (Ω)N ≤ M,
0

Then, after extraction of a subsequence, we have
uǫ ⇀ u

H01 (Ω)N weakly,

in

div uǫ → 0

L2 (Ω) strongly.

in

Then div u = 0 so that u ∈ V and for every w ∈ V (div w = 0) we have
a(uǫ , w) =

N
X

< fi , w i >

N
X

< fi , w i >

i=1

so that
a(u, w) =

i=1

and u is solution of the variational form of Stokes problem.
Now pǫ satisfies
Z
∇pǫ = f + ν∆uǫ ,

Ω

Then pǫ is bounded in H −1 (Ω)N and satisfies
L2 (Ω). After extraction of a subsequence,

R

pǫ dx = 0.

Ω pǫ dx

= 0 so that pǫ is bounded in

L2 (Ω) weakly

pǫ ⇀ p in
and we have
∀w ∈

H01 (Ω)N ,

a(u, w) =

N
X

< fi , w i > +

i=1

Z

pdiv wdx
Ω

so that
−ν∆u = f − ∇p

in

H −1 (Ω)N .

Let us now go back to the abstract formulation. We know that ImA is closed in F .
Then writing
L = {l ∈ F, ∀u ∈ H, (Au, l)F = 0}
we have
F = ImA ⊕ L.

Here, F = H −1 (Ω)N = W ′ where W = H01 (Ω)N so that F ′ = W .
26

Lemma 2.2.9
W = (ImA)⊥ ⊕ L⊥ .
Proof.
If u ∈ (ImA)⊥ ∩ L⊥ , then
∀g ∈ H, < u, Ag >= 0,

∀l ∈ L, < u, l >= 0.

As ∀h ∈ W ′ we have the decomposition h = Ag + l with g ∈ H and l ∈ L, we have
∀h ∈ W ′ ,

< u, h >= 0.

Let now u ∈ W . Let us show that there exists v ∈ W such that
∀h ∈ W ′ ,

< v, h >=< u, Ag > .

The mapping h ∈ W ′ →< u, Ag > is linear. Moreover
| < u, Ag > | ≤ ||u||W ||Ag||W ′ ≤ C||u||W ||h||W ′
and so the mapping is continuous. Therefore, there exists v ∈ W such that
∀h ∈ W ′ ,

< v, h >=< u, Ag > .

Now for every l ∈ L we can write l = A0 + l and then
∀l ∈ L,

< v, l >=< u, A0 >= 0

and v ∈ L⊥ .
This implies
∀g ∈ H,

which says that u − v ∈ (ImA)⊥ .
We then have
u = v + w,

< u − v, Ag >= 0
v ∈ L⊥ ,

w ∈ ImA⊥ .

Lemma 2.2.10 If t A : W → H is defined by
∀u ∈ W,

∀g ∈ H,

(t Au, g)H =< u, Ag >

then we have
Kert A = ImA⊥ ,

and

27

Imt A = KerA⊥ .

Proof.
If u ∈ Kert A, we have ∀g ∈ H, < u, Ag >= 0 and then u ∈ ImA⊥ .
If u ∈ ImA⊥ , we have ∀g ∈ H, < u, Ag >= 0 and therefore (t Au, g)H = 0 so that
t Au = 0 and u ∈ Kert A.
Now if g ∈ Imt A, there exists u ∈ W such that g =t Au. We have for f ∈ KerA,
(g, f )H = (t Au, f )H =< u, Af >= 0, so that g ∈ KerA⊥ . This shows that Imt A ⊂
KerA⊥ and as KerA⊥ is closed, we have
Imt A ⊂ KerA⊥ .

Let us suppose that Imt A 6= KerA⊥ . Then there exists g ∈ H ′ = H and there exists
f0 ∈ KerA⊥ such that
∀f ∈ Imt A, (g, f )H = 0,

and

(g, f0 )H = 1.

Then for every u ∈ W , (g,t Au)H = 0 so that < u, Ag >= 0 and therefore Ag = 0
in W ′ . Therefore, g ∈ KerA which implies (g, f0 )H = 0. This gives a contradiction.
Therefore we have
Imt A = KerA⊥ .
Lemma 2.2.11 Imt A is closed in H.
Proof.
Let (gn ) be a sequence of elements of Imt A such that gn → g in H. We have gn =t
Aũn with ũn ∈ W . Moreover we know from the previous lemma that gn ∈ KerA⊥ .
From Lemma 2.2.9, we can decompose ũn as ũn = un + vn with un ∈ L⊥ and
vn ∈ ImA⊥ = Kert A. Then t Aũn =t Aun so that gn =t Aun with un ∈ L⊥ .
If h ∈ W ′ we have h = Ag + l with g ∈ KerA⊥ = N and l ∈ L. (the fact that we
can take g ∈ KerA⊥ is immediate since H = KerA ⊕ KerA⊥ .) Then
< um − up , h >=< um − up , Ag >=< um − up , l >=< um − up , Ag >
= (t A(um − up ), g)H = (gm − gp , g)H .

Then for every h ∈ W ′
| < um − up , h > | ≤ |gm − gp |H |g|H
≤ |gm − gp |H CN ||Ag||W ′

≤ C.CN |gm − gp |H ||h||W ′ .

This implies
||um − up ||W ≤ C.CN |gm − gp |H .

Therefore, (un is a Cauchy sequence in W which converges to some u ∈ W . Now
t
t
t
t
n = gn → Au in H and therefore g = Au so that g ∈ Im A and Im A is closed
in H.

t Au

28

Lemma 2.2.12 For every f ∈ KerA⊥ , there exists u ∈ L⊥ such that
t

Au = f

and

||u||W ≤ C|f |H .

Proof.
We have W = ImA⊥ ⊕L⊥ = Kert A⊕L⊥ . Now it is immediate to see that t A : L⊥ →
H is injective. But as Imt A = KerA⊥ we see that the mapping t A : L⊥ → KerA⊥
is one-to-one (bijective). On the other hand we have |t Au|H ≤ C||u||W which says
that t A is continuous.
Now l⊥ and KerA⊥ are Banach spaces as they are closed subspaces of Banach
spaces. Then from Banach Theorem, t A : L⊥ → KerA⊥ has a continuous inverse
which shows the lemma.
As an application we have the following result concenring operator div .
Theorem 2.2.13
There exists a constant C > 0 such that for every g ∈ L20 (Ω) =
R
2
{g ∈ L (Ω), Ω gdx = 0}, there exists u ∈ H01 (Ω)N such that
div u = g
and
||u||H 1 (Ω)N ≤ C|g|L2 (Ω) .
0

Proof.
Take
A = ∇ : L2 (Ω) = H → H −1 (Ω)N = W ′ .

Then KerA = {g ∈ L2 (Ω), g = Cst} and KerA⊥ = L20 (Ω).
If g ∈ L20 (Ω), there exists u ∈ L⊥ ⊂ W such that t Au = −g and ||u||W ≤ C|g|H .
Then u ∈ H01 (Ω)N and for every h ∈ H we have
(−g, h)H = (t Au, h)H =< u, Ah >=< u, ∇h > .
Taking h ∈ C0∞ (Ω) we see that
(t Au, h)H =< u, ∇h >D′ ,D = − < div u, h >D′ ,D =< −g, h >D′ ,D .
Then we have
div u = g,

u ∈ H01 (Ω)N ,

and

||u||H 1 (Ω)N ≤ C|g|L2 (Ω) .
0

Lemma 2.2.14 If f ∈ H −1 (Ω)N and < f, ϕ >= 0 ∀ϕ ∈ V, then there exists
p ∈ L2 (Ω) such that f = ∇p.
29

Proof.
We recall that V = {ϕ ∈ C0∞ (Ω)N = D(Ω)N , div ϕ = 0}.
Let us consider an increasing sequence of open sets (Ωn ) such that Ωn ⊂ Ω and
∪n Ωn = Ω. If u ∈ H01 (Ωn )N and div u = 0, we can extend u by 0 outside Ωn . If (ρǫ )
is a regularizing sequence (mollifiers), for ǫ small enough we have ρǫ ∗ u ∈ C0∞ (Ω)N
and div (ρǫ ∗ u) = 0. Then < f, u >= limǫ→0 < f, ρǫ ∗ u >= 0. Therefore
∀u ∈ H01 (Ωn )N ,

div u = 0,

< f/Ωn , u >= 0.

Then there exists pn ∈ L2 (Ωn ) such that f/Ωn = ∇pn .
But pn+1 − pn is constant on Ωn and we can assume that this constant is 0. Then
f = ∇p where p ∈ L2loc (Ω).
Now if Ω is starshaped (for example with respect to 0), for 0 < θ < 1 we have
θΩ ⊂ Ω. If u ∈ H01 (Ω)N , div u = 0, then uθ defined by uθ (x) = u( xθ ) has a compact
support in Ω and div uθ = 0. Then we can approximate uθ by ρǫ ∗ uθ ∈ C0∞ (Ω)N
with div (ρǫ ∗ uθ ) = 0 so that
< f, u >= lim lim < f, ρǫ ∗ uθ >= 0.
θ→1 ǫ→0

Then, in this case, there exists p ∈ L2 (Ω) such that f = ∇p.
In the general case, every point of Γ has a connected neighborhood ω which is regular
and strictly starshaped. Then there exists q ∈ L2 (ω) such that f/ω = ∇q.
But p/ω − q is constant on ω. Therefore p/ω ∈ L2 (ω) and therefore, p ∈ L2 (Ω).
Theorem 2.2.15 The space of regular functions V (which has been recalled above)
is dense in V .
Proof.
Let f ∈ H −1 (Ω)N such that for every ϕ ∈ V we have < f, ϕ >= 0. Then there
exists p ∈ L2 (Ω) such that f = ∇p. This implies that for every w ∈ V we have
< f, w >= 0. Therefore V ⊃ V , but it is clear that V ⊂ V and the theorem follows.
Theorem 2.2.16 The closure of V in L2 (Ω)N is the space
H = {v ∈ L2 (Ω)N , div v = 0, v.n = 0 onΓ}.
Proof.
Let us define
X = {v ∈ L2 (Ω)N , div v ∈ L2 (Ω)}.
Lemma 2.2.17 The space C ∞ (Ω)N is dense in X and the mapping ϕ → ϕ.n de1
fined on C ∞ (Ω)N can be extended to a linear continuous map from X to H − 2 (Γ).
30

Proof.
1
Let v ∈ X. For every ψ ∈ H 2 (Γ) and ϕ ∈ H 1 (Ω) such that ϕ/Γ = ψ (we take a
continuous extension), we define
L(ψ) =

Z

div vϕdx +
Ω

Z

v.∇ϕdx.
Ω

If ϕ ∈ H01 (Ω), then the right hand side is 0 so that L depends only on ψ. Moreover
we have
|L(ψ)| ≤ ||v||X ||ϕ||H 1 (Ω) ≤ C||v||X ||ψ|| 12 .
H (Γ)

1
2

Therefore, there exists an element of H (Γ) which corresponds to v.n for regular
functions v such that
1

∀ψ ∈ H 2 (Γ),

L(ψ) =< v.n, ψ >Γ .

The continuity follows immediately.
Back to the proof of Theorem 2.2.16.
Of course, for elements of V this normal trace is 0 and therefore the closure of V is
contained in H. Now if f ∈ L2 (Ω)N and < f, v >= 0 for every v ∈ V. Then we
know that there exists p ∈ L2 (Ω) such that f = ∇p. Then p ∈ H 1 (Ω). Therefore,
for every v ∈ H, we have
< f, v >=< ∇p, v >=

Z

Ω

pdiv vdx+ < v.n, p >Γ = 0.

Then < f, v >= 0 for every v ∈ H and this shows that the closure of V contains H
and the proof of Theorem 2.2.16 is complete.

2.3

Evolution Stokes equations

We consider the following evolution problem : we look for (u, p) (u will be the
velocity and p the pressure) such that
(2.3.20)
(2.3.21)

∂u
− ν∆u = f − ∇p in Ω × (0, T ),
∂t
div u = 0 in Ω × (0, T ),

(2.3.22)

u=0

on

(2.3.23)

u(0) = u0

Σ = Γ × (0, T ),

in

Ω.

When Ω is a bounded open set we can use a Fourier method which will be presented
below.
31

2.3.1

Special basis

First of all we define the space H as the closure in L2 (Ω)N of V which is shown to
be
(2.3.24)
H = {w ∈ L2 (Ω)N , div w = 0, w.n = 0}
where n is the unit outward pointing normal vector on the boundary Γ.
We go back to the stationnary Stokes problem which can be written in the form, for
f ∈ H (we denote by (., .) the scalar product in H which is the L2 scalar product)
a(u, w) = (f, w)
u ∈ V.

∀w ∈ V,

As the injection V ⊂ H is compact continuous it is immediate to see that the
mapping T : f ∈ H → u ∈ V → u ∈ H is linear continuous and compact. Moreover
if we write T f = u and T g = v, we have as a(., .) is symmetric
(T f, g) = (u, g) = (g, u) = a(v, u) = a(u, v) = (f, v) = (f, T g),
so that T is selfadjoint.
Therefore, (see for example [2]), there exists a decreasing sequence (µn )n of strictly
positive real numbers with µn → 0 when n → +∞ and a sequence (wn )n of elements
of V which form an orthonormal basis in H such that
T w n = µn w n .
Setting now λn =

1
µn

we have

0 < λ1 ≤ λ 2 ≤ · · · ≤ λ n ≤ · · ·

,

λn → +∞

and
a(wn , w) = λn (wn , w)
wn ∈ V,

∀w ∈ V,

(wm , wp ) = δmp .
If we equip V with the (equivalent) scalar product a(v, w), we can see that ( √wλn )n
n
form an orthonormal basis in V .
Let v ∈ H. Then we have
v=

+∞
X

n=1

(v, wn )wn

with

+∞
X

n=1

32

|(v, wn )|2 = |v|2H < +∞.

If v ∈ V we have
v=

+∞
X

(v, wn )wn

+∞
X

with

n=1

n=1

λn |(v, wn )|2 = ||v||2V < +∞.

′

If v ∈ V we have
v=

+∞
X

< v, wn > wn

+∞
X

| < v, wn > |2
= ||v||2V ′ < +∞.
λ
n
n=1

with

n=1

2.3.2

Existence and uniqueness result

First of all, at least formally, if we multiply the Stokes evolution system by a function
w ∈ V , as for the stationnary case, the pressure p disappear and we can write the
problem as
(2.3.25)
(2.3.26)

d
(u(t), w) + a(u(t), w) =< f (t), w >
dt
u(0) = u0 .

on

(0, T ),

∀w ∈ V,

We obtain the following result
Theorem 2.3.1 Let us assume that u0 ∈ H and f ∈ L2 (0, T ; H −1 (Ω)N ). Then
2
′
there exists a unique solution u ∈ C([0, T ]; H) ∩ L2 (0, T ; V ) with du
dt ∈ L (0, T ; V )
′
2
of problem (2.3.25),(2.3.26). Moreover, there exists p ∈ D (]0, T [; L (Ω)) (in fact
p ∈ H −1 (0, T ; L2 (Ω))) such that (u, p) satisfies the Stokes evolution equation
∂u
− ν∆u = f − ∇p in Ω × (0, T ),
∂t
div u = 0 in Ω × (0, T ),

(2.3.27)
(2.3.28)
(2.3.29)

u=0

on

(2.3.30)

u(0) = u0

Σ = Γ × (0, T ),

in

Ω.

Proof.
we have
u0 =

+∞
X

(u0 , wn )wn

with

+∞
X

n=1

n=1

|(u0 , wn )|2 = |u0 |2H < +∞

and
f (t) =

+∞
X

n=1

< f (t), wn > wn

with

Z T  +∞
X | < f (t), wn > |2 
0

n=1

33

λn

dt = ||f ||2L2 (0,T ;V ′ ) < +∞.

Let us write
uM
0 =

M
X

(u0 , wn )wn

n=1

and
f M (t) =

M
X

< f (t), wn > wn .

n=1
M converges to f in L2 (0, T ; V ′ ) when
We know that uM
0 converges to u0 in H and f
M → +∞. First of all, calling V M = span{w1 , · · · , wM }, we look for

uM (t) =

M
X

un (t)wn

n=1

such that
d M
(u (t), w) + a(uM (t), w) =< f M (t), w >
dt
uM (0) = uM
0 .

on

(0, T ),

∀w ∈ V M ,

This gives us for every n = 1, · · · , M
d
un (t) + λn un (t) =< f (t), wn >,
dt
un (0) = (u0 , wn ),
and we even have an explicit formula for un
un (t) = (u0 , wn )e−λn t +

Z t
0

< fn (s), wn > e−λn (t−s) ds.

Let us show that (uM ) is a Cauchy sequence in C([0, T ; H) and in L2 (0, T ; V ). For
M ≥ P + 1 we have
(uM − uP )(t) =

M
X

n=P +1

|(uM − uP )(t)|2H =
||uM − uP ||2L2 (0,T ;V ) =

un (t)wn ,

n=P +1

Z T
0

34

M
X

M
X

|un (t)|2 ,

n=P +1

λn |un (t)|2 dt.

From the equation giving un we obtain
| < f (t), wn > | √
1d
√
|un (t)|2 + λn |un (t)|2 =< f (t), wn > un (t) ≤
λn |un (t)|.
2 dt
λn
For t ∈ [0, T ] we now integrate on (0, t) to get
|un (t)|2 +

Z t
0

λn |un (s)|2 ds ≤ |(u0 , wn )|2 +

Z t
| < f (s), wn > |2

λn

0

ds.

Summing up from n = P + 1 to M we immediately obtain




P 2
M
||uM − uP ||2C([0,T ;H) + ||uM − uP ||2L2 (0,T ;V ) ≤ 2 |uM
− f P ||2L2 (0,T ;V ′ ) .
0 − u0 |H + ||f

Therefore, (uM ) is a Cauchy sequence in C([0, T ]; H) ∩ L2 (0, T ; V ) and consequently
converges to a function u in these spaces which are complete. We can also write
u(t) =

+∞
X

un (t)wn .

n=1

The regularity of du
dt comes immediately from the equation for un for example.
Taking test functions w in ∪M V M first then in its closure which is V it is now easy
to see that u satisfies
d
(u(t), w) + a(u(t), w) =< f (t), w >
dt
u(0) = u0 .

on

(0, T ),

∀w ∈ V,

Uniqueness comes immediately by taking in the above equation w = wn and seeing
that (u(t), wn ) satisfies the same equation as un (t).
Interpretaton of this equation and the relation with Stokes problem are more difficult. We need to consider
U (t) =

Z t

u(s)ds

and

F (t) =

Z t

f (s)ds,

0

0

and to integrate our equation in time, which gives
(u(t) − u0 , w) + a(U (t), w) =< F (t), w >,

∀w ∈ V.

This is a stationnary Stokes problem at time t. Therefore, there exists P (t) ∈ L2 (Ω)
such that
−ν∆U (t) + u(t) − u0 = F (t) − ∇P (t),

div U (t) = 0.

35

As F (·) − u(·) + u0 is continuous in time, we clearly have P ∈ C([0, T ]; L2 (Ω)/IR ).
d
P
Now taking the time derivative we obtain with p = dt
∂u
− ν∆u = f − ∇p in Ω × (0, T ),
∂t
div u = 0 in Ω × (0, T ),

u=0

on

u(0) = u0

Σ = Γ × (0, T ),

in

Ω,

and p ∈ H −1 (0, T ; L2 (Ω)/IR ).
Regularity result.
We also have the regularity result.
Theorem 2.3.2 If u0 ∈ V and f ∈ L2 (0, T ; H), then the solution u of the corresponding Stokes equation satisfies
(2.3.31)

u ∈ C([0, T ], V ) ∩ L2 (0, T ; H 2 (Ω)N ),

(2.3.32)

p ∈ L2 (0, T ; H 1 (Ω)/IR ).

Proof.
Multiplying the equation by
Z

ν d
∂u
| (t)|2 dx +
2 dt
Ω ∂t

Z

∂u
∂t

we have
2

Ω

∂u
∈ L2 (0, T ; H),
∂t

|∇u(t)| dx =

Z

f (t)
Ω

∂u
∂u
(t)dx ≤ |f (t)|H | (t)|H .
∂t
∂t

2
∞
This gives an estimate on ∂u
∂t in L (0, T ; H) and of u in L (0, T, V ). Then from
the equation, for almost every fixed t we can consider it as a stationnary Stokes
equation with right hand side in L2 (0, T ; H), which gives (from the regularity result
for the stationnary problem) v ∈ L2 (0, T, H 2 (Ω)N ) and p ∈ L2 (0, T ; H 1 (Ω)/IR ).

36

Chapter 3

Navier-Stokes equations. The
evolution case.
3.1

Notations and preliminaries

All along this chapter we will assume that Ω is bounded and regular, except if it is
specially mentionned. We will also take the density ρ = 1 in order to simplify the
presentation. The Navier-Stokes equations then take the form
(3.1.1)
(3.1.2)
(3.1.3)
(3.1.4)

∂u
+ (u.∇)u − ν∆u = f − ∇p,
∂t
div u = 0, in Ω × (0, T ),
u = 0,

on

u(0) = u0

in

Ω × (0, T ),

Γ × (0, T ),

in

Ω.

We introduce the notations
(3.1.5)

a(v, w) =

N Z
X

i,j=1 Ω

(3.1.6)

b(u, v, w) =

∂vi ∂wi
.
dx =
∂xj ∂xj
N Z
X

i,j=1 Ω

uj

Z

Ω

∇v.∇wdx,

∂vi
wi dx,
∂xj

∀v, w ∈ V,

∀u, v, w ∈ V.

Lemma 3.1.1 For N ≤ 4, b(., ., .) is a trilinear continuous form on H 1 (Ω)N ×
H 1 (Ω)N × H 1 (Ω)N . Moreover we have
(3.1.7)

∀u ∈ V, ∀v, w ∈ H 1 (Ω)N ,

b(u, v, w) + b(u, w, v) = 0.

37

In particular we have
∀u ∈ V, ∀v ∈ H 1 (Ω)N ,

(3.1.8)

b(u, v, v) = 0.

Proof.
For N ≤ 4 we have H 1 (Ω) ⊂ L4 (Ω) with continuous injection. Therefore, if u, v, w ∈
V we have
∂vi
∈ L2 (Ω), wi ∈ L4 (Ω),
uj ∈ L4 (Ω,
∂xj
and the term

Z

Ω

uj

∂vi
wi dx
∂xj

makes perfect sense with
|

Z

Ω

uj

∂vi
wi dx| ≤ C|uj |L4 (Ω) ||vi ||H 1 (Ω) |wi |L4 (Ω) ≤ C||uj ||H 1 (Ω) ||vi ||H 1 (Ω) ||wi ||H 1 (Ω) .
∂xj

Therefore, b(., ., .) is well defined, is trilinear and
|b(u, v, w)| ≤ C||u||H 1 (Ω)N ||v||H 1 (Ω)N ||w||H 1 (Ω)N ,
which says that it is continuous.
Now we have
b(u, v, w) + b(u, w, v) =

N Z
X

i,j=1 Ω

=

N Z
X

i,j=1 Ω

=−

N Z
X

uj

i=1 Ω

uj (

N
X
∂
(vi wi )dx = −
∂xj
i,j=1

∂vi
∂wi
wi +
vi )dx
∂xj
∂xj

Z

(
Ω

∂uj
)vi wi dx = 0
∂xj

(div u)vi wi dx = 0.

We now give a key lemma in dimension N = 2 which is no longer valid in higher
dimension and which makes the crucial difference between the study of Navier-Stokes
equations in dimension N = 2 and in dimension N = 3.
Lemma 3.1.2 In dimension N = 2, there exists a constant C > 0 such that
(3.1.9)

∀ϕ ∈ H 1 (IR2 ),

1

1

2
2
|ϕ|L4 (IR2 ) ≤ C||ϕ||H
1 (IR2 ) |ϕ|L2 (IR2 ) .

In particular we have
(3.1.10)

∀ϕ ∈ H01 (Ω),

1

1

2
2
|ϕ|L4 (Ω) ≤ C||ϕ||H
1 (Ω) |ϕ|L2 (Ω) .
0

38

Proof.
Of course it is sufficient to prove the above inequalities for ϕ ∈ C0∞ (IR2 ). We can
write
Z x2
Z x2
∂ϕ
∂ 2
ϕ(x1 , y)
ϕ (x1 , y)dy = 2
(x1 , y)dy
ϕ2 (x1 , x2 ) =
∂x
∂x
−∞
−∞
2
2
and then

 Z x2

2

|ϕ(x1 , x2 )| ≤ 2
≤2

 Z +∞
−∞

−∞

2

ϕ(x1 , y) dy

ϕ(x1 , y)2 dy

 1  Z x2
2

−∞

 1  Z +∞ ∂ϕ
2

(

∂x2

−∞

(

1
∂ϕ
(x1 , y))2 dy 2
∂x2

(x1 , y))2 dy

1

2

.

In the same way we can write (exchanging the roles of x1 and x2 )
 Z +∞

|ϕ(x1 , x2 )|2 ≤ 2

−∞

ϕ(y, x2 )2 dy

 1  Z +∞ ∂ϕ
2

(y, x2 ))2 dy

 1  Z +∞ ∂ϕ
2

(x1 , y))2 dy

(

∂x1

−∞

1
2

.

Multiplying these two inequalities we obtain
|ϕ(x1 , x2 )|4 ≤ 4

 Z +∞

ϕ(x1 , y)2 dy

(

1
2

−∞
−∞ ∂x2
 Z +∞
 1  Z +∞ ∂ϕ
1
(
.
ϕ(y, x2 )2 dy 2
(y, x2 ))2 dy 2
−∞ ∂x1
−∞
≤ λ(x1 ).µ(x2 ).

Therefore
Z

IR2

 Z +∞

|ϕ(x1 , x2 )|4 dx1 dx2 ≤ 4

−∞

λ(x1 )dx1

 Z +∞
−∞



µ(x2 )dx2 .

But we have
Z +∞
−∞

λ(x1 )dx1 ≤

Z

2

IR2

|ϕ(x1 , y)| dx1 dy

≤ |ϕ|L2 (IR2 ) |

1  Z
2

IR2

|

1
∂ϕ
(x1 , y)|2 dx1 dy 2
∂x2

∂ϕ
| 2 2.
∂x2 L (IR )

In the same way we have
Z +∞
−∞

µ(x2 )dx2 ≤ |ϕ|L2 (IR2 ) |

∂ϕ
| 2 2.
∂x1 L (IR )

Therefore
|ϕ|4L4 (IR2 ) ≤ 4|ϕ|L2 (IR2 ) |

∂ϕ
∂ϕ
|L2 (IR2 ) |ϕ|L2 (IR2 ) |
| 2 2 ≤ 2|ϕ|2L2 (IR2 ) ||ϕ||2H 1 (IR2 ) .
∂x2
∂x1 L (IR )
39

This finishes the proof of Lemma 3.1.2.
Notations.
We will denote by < ., . > the duality between V ′ and V . Notice that when f ∈
H −1 (Ω)N , then the mapping w →< f, w >H −1 (Ω)N ,H 1 (Ω)N is linear continuous on
0
V and therefore defines (in a non unique way) an element of V ′ that we will still
denote by f . We will also denote the duality between H −1 (Ω)N and H01 (Ω)N by
< ·, · >.
We will write A the operator defined by
∀u, v ∈ V,

a(u, v)