Basis of Mathematical Methods in FluidMechanics

Jean-Pierre Puel 1

November 30., 2011

1J.-P. Puel: (jppuel@bcamath.org) Ikerbasque and BCAM, Bizkaia Technology Park,48160, Derio, Spain and Laboratoire de Mathematiques de Versailles, Universite de VersaillesSaint-Quentin

These notes correspond to a course taught in BCAM in november 2011 withsome complements. Most of them also follow a DEA course taught in University ofVersailles St Quentin some time before. They are very classical basis for equationsof fluid mechanics (for viscous fluids) and for the mathematical analysis of theseequations and they intend to be essentially self contained. The first chapter can befound 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 isdifficult to find. But of course, most of the results and the proofs can be found(sometimes presented differently) in classical books where Navier-Stokes equationsare 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. There-fore, if t = t, u(x, t) is the velocity of a different particle. The observer is locatedat position x and looks at particles passing at this position. Here we will focusessentially 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 trajectoryis an unknown of the problem). Then v(x0, t) (t ≥ t0) is the velocity, at time t ofthe 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 theflow and follows the particle.Particle derivative.Let (t, x(t)) the trajectory of a fluid particle starting from (t0, x0). We have withthe Eulerian velocity,

dx

dt(t) = u(x(t), t),

x(t0) = x0.

Let ϕ be a function of x and t. The particle derivative of ϕ is

d

dtϕ(x0, t0) =

d

dt(ϕ(x(t), t))/t=t0

=∂ϕ

∂t(x0, t0) +

N

i=1

ui(x0, t0)∂ϕ

∂xi(x0, t0)

=∂ϕ

∂t+

N

i=1

ui∂ϕ

∂xi

/(x0,t0)

=∂ϕ

∂t+ u.∇ϕ

/(x0,t0).

1.2 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 thevelocity field u(x, t) or for t ≥ t

V (t) = x(t), dx

ds(s) = u(x(s), s), t ≤ s ≤ t

, 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

V (t)ϕ(x, t)dx = 0.

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

V (t+δt)ϕ(y, t+ δt)dy =

V (t)ϕ(x, t)dx.

3

V(t)

V(t+δt)

ux

y

We can write

V (t+ δt) = y(δt), dy

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

j=1

∂ui

∂xj(x, t)dxj + l.o.t.

Therefore

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

4

+δt

N

j=1

∂u1

∂xjdxj ∧ dx2 ∧ · · · ∧ dxN

+ · · ·

+δt

dx1 ∧ dx2 ∧ · · · ∧

N

j=1

∂uN

∂xj

+ l.o.t.

so thatdy = (1 + δt(divu) + 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∂ϕ

∂t(x, t) + δt

N

j=1

uj(x, t)∂ϕ

∂xj(x, t) + l.o.t.

Therefore

V (t+δt)ϕ(y, t+ δt)dy

=

V (t)

ϕ(x, t) + δt(

∂ϕ

∂t(x, t) + u(x, t).∇ϕ(x, t) + o(δt))(1 + δtdivu(x, t) + o(δt))dx

=

V (t)ϕ(x, t)dx+ δt

V (t)

∂ϕ

∂t(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)divu(x, t)

dx+ o(δt).

As we have

V (t+δt)ϕ(y, t+ δt)dy =

V (t)ϕ(x, t)dx,

we obtain

∀V (t), ∀δt > 0,

V (t)

∂ϕ

∂t(x, t)+u(x, t).∇ϕ(x, t)+ϕ(x, t)divu(x, t)

dx = o(δt),

and therefore

∀V (t),

V (t)

∂ϕ

∂t(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)divu(x, t)

dx = 0,

that is to say the equation

∂ϕ

∂t(x, t) + u(x, t).∇ϕ(x, t) + ϕ(x, t)divu(x, t) = 0

or∂ϕ

∂t(x, t) + div (u(x, t).ϕ(x, t)) = 0.

5

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

d

dt

V (t)ϕ(x, t)dx =

V (t)

∂ϕ

∂t(x, t)dx+

S(t)ϕ(x, t)u(x, t)ν(x)dS.

Then applying Stokes-Ostrogradskii formula we have

S(t)ϕ(x, t)u(x, t)ν(x)dS =

V (t)div (u(x, t).ϕ(x, t))dx,

which shows that

d

dt

V (t)ϕ(x, t)dx =

V (t)

∂ϕ

∂t(x, t) + div (u(x, t).ϕ(x, t))

dx.

1.2.1 Conservation of mass

Here we takeϕ(x, t) = ρ(x, t)

where ρ is the density of the fluid. We obtain the equation of conservation of masswhich can take different forms.

∂ρ

∂t(x, t) + div (u(x, t)ρ(x, t)) = 0 in Ω× (0, T ),(1.2.1)

or∂ρ

∂t(x, t) + u(x, t).∇ρ(x, t) + ρ(x, t)divu(x, t) = 0 in Ω× (0, T ),(1.2.2)

or even using the particle derivative

d

dtρ(x, t) + ρ(x, t)divu(x, t) = 0 in Ω× (0, T ).(1.2.3)

1.2.2 Conservation of volume. Incompressibility

A fluid is incompressible if the volume occupied by a group of fluid particles remainsconstant during the flow. Therefore we here take

ϕ(x, t) = 1

6

which gives the incompressibility condition

divu(x, t) = 0 in Ω× (0, T ).(1.2.4)

In that case, the conservation of mass becomes

∂ρ

∂t(x, t) + u(x, t).∇ρ(x, t) = 0 in Ω× (0, T ),(1.2.5)

ρ(x, 0) = ρ0(x) in Ω.(1.2.6)

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 divu(x, t) = 0.In fact we see that

divu = 0 ⇒ d

dtρ = 0 ⇒ ρ(x(t), t) = Cst

and

ρ(x(t), t) = Cst ⇒ d

dtρ = 0 ⇒ divu = 0.

Therefore the fluid is incompressible (divu = 0) if and only if the density of eachelement stays constant during the flow ( d

dtρ = 0).

1.2.3 Stress tensor. Conservation of momentum

Let dS be a surface element in the fluid which separates the fluid in two parts F1and F2 and let n be the unit normal vector to dS pointing towards the exterior ofF2.

7

n

dS

F1

F2

The force exerted by F1 on F2 per surface unit is called stress.For a fluid at rest (u = 0), this force is normal to dS, so it is characterized by ascalar quantity at each point. This quantity is the hydrostatic pressure.For a moving fluid, it appears tangential stresses : friction between the fluid layersgliding 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 suchthat

• σ is symmetric : σij = σji. This comes from an equilibrium equation.

• The force exerted by F1 on F2 is given by

σ.n = (N

j=1

σijnj)i=1,···,N = (σijnj)i=1,···,N

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 thefluid deformation, that is to say which exist when the fluid is at rest, and thosewhich are due to the fluid deformation. We set

σij = −pδij + σij ,

where p is the pressure, δij is the Kronecker symbol and (σij) is the tensor of

viscosity stresses. (Sign − in front of the pressure is just a choice indicating thatusually, a fluid at rest is compressed).(σ

ij) is independent of translations and local rotations, and therefore independent

of u itself and of ωij =12(

∂ui

∂xj− ∂uj

∂xi) (ω = curlu is the vorticity of the fluid).

Therefore, σijonly depends on the symmetric part of the tensor of velocity gradients

e = (eij) where

eij =1

2(∂ui

∂xj+

∂uj

∂xi).

Newtonian fluids.For Newtonian fluids (only case which will be considered here), the relation between(σ

ij) and (eij) is linear and (after some remarks on isotropy etc) the relation can be

written

σij = η. (2eij −

2

3δijell)

deformation without volume changes

+ζ. (δijell) isotropic dilation

,

where η is the shear viscosity and ζ is the volumic velocity.For an incompressible fluid, we have divu = 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 andlimited by a surface S(t). In the presence of external forces represented by f andthe action of the exterior of V (t) exerted on the surface S(t), the fundamental lawof dynamics takes the (vectorial) form

d

dt

V (t)ρudx =

V (t)ρfdx+

S(t)σ.ndS.

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

V (t)ρuidx =

V (t)

∂

∂t(ρui) + div (ρuiu)

dx

=

V (t)ui

∂ρ

∂t+ div (ρu)

dx+

V (t)ρ

∂ui

∂t+ u.∇ui

dx.

From the equation expressing conservation of mass we have

∂ρ

∂t+ div (ρu) = 0,

so thatd

dt

V (t)ρuidx =

V (t)ρ

∂ui

∂t+ u.∇ui

dx =

V (t)ρdui

dtdx.

On the other hand, from Stokes-Ostrogradskii formula we have

S(t)(σ.n)idS =

V (t)

N

j=1

σijnjdS =

V (t)

N

j=1

∂σij

∂xjdx.

Therefore we obtain for every volume V (t)

V (t)ρ

∂ui

∂t+ u.∇ui

dx =

V (t)

N

j=1

∂σij

∂xjdx+

V (t)ρfidx.

This gives us the equation for conservation of momentum

ρ

∂ui

∂t+ u.∇ui

=

N

j=1

∂σij

∂xj+ ρfi,(1.2.7)

which can also be written as

ρ

∂ui

∂t+ u.∇ui

= − ∂p

∂xi+

N

j=1

∂σij

∂xj+ ρfi,(1.2.8)

or in vectorial form

ρ

∂u

∂t+ (u.∇)u

= −∇p+ divσ + ρf.(1.2.9)

10

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

If we neglect the spatial variations of viscosities ∂η

∂xjand ∂ζ

∂xjwe obtain :

• For a Newtonian viscous compressible fluid,

∂ρ

∂t+ div (ρu) = 0,(1.3.10)

ρ

∂u

∂t+ (u.∇)u

= −∇p+ η∆u+ (ζ +

η

3)∇(divu) + ρf.(1.3.11)

In order to complete this system we need to give the pressure law, or the energyequation.• For a Newtonian viscous incompressible fluid we obtain the Navier-Stokes equa-tions

divu = 0,(1.3.12)

ρ

∂u

∂t+ (u.∇)u

= −∇p+ η∆u+ ρf.(1.3.13)

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

divu = 0,(1.3.14)

ρ

∂u

∂t+ (u.∇)u

= −∇p+ ρf.(1.3.15)

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

ρf = ∇p,(1.3.16)

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 =

x

L, u

=u

U, t

=t

L/U, p

=p− p012ρU

2,

where p0 is the hydrostatic pressure (in absence of flow). We obtain

∂u

∂t+ (u.∇)u = −∇

p +

η

ρUL∆

u + f

.(1.3.17)

Setting

ν =η

ρ

1

Re=

ν

UL,

11

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

divu = 0,(1.3.18)∂u

∂t= −∇p+ ν∆u+ f.(1.3.19)

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 = u

2.

Equilibrium between the stresses in each of the fluids and the stresses localized onthe interface (n is the normal vector and τ are tangent vectors)

(σ1.n)τ = (σ2

.n)τ (equality of tangential stresses),

(σ1.n)n− (σ2

.n)n = γ(1

R+

1

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

u/t=0 = u0, ρ/t=0 = ρ0, · · ·

Remark 1.3.1 • If we want to consider an interaction between a fluid and a struc-ture, the coupling must occur in the boundary conditions.• If we want to consider the thermal effects, we must have to add an equation forthe 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 thefluid 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 pre-vious equations. This does not say that the fluid is at rest, but says only that theflow does not vary with time. Therefore it corresponds to cancel all terms containing∂

∂t.

13

Chapter 2

Stokes equations. Mathematicalformulation

2.1 Stationnary Stokes equations

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

−ν∆u = −∇p+ f in Ω,(2.1.1)

divu = 0 in Ω,(2.1.2)

u = 0 on Γ,(2.1.3)

where ν > 0 is the given viscosity and f is the given external force and u and p arethe velocity and the pressure.For each component ui, i = 1, · · · , N we have

−ν∆ui = − ∂p

∂xi+ fi.

Let w = (w1, · · · , wN ) be a vector function which is “regular” and which is zero onthe boundary, with divw = 0. If we multiply the equation for ui by wi, integrateon Ω and sum up for i = 1 to N we obtain

−ν

N

i=1

Ω∆uiwidx = −

N

i=1

Ω

∂p

∂xiwidx+

N

i=1

Ωfiwidx.

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

−N

i=1

Ω

∂p

∂xiwidx =

Ωp(divw)dx = 0,

14

and

−ν

N

i=1

Ω∆uiwidx = ν

N

i=1

Ω∇ui∇widx.

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

ν

N

i=1

Ω∇ui∇widx =

N

i=1

Ωfiwidx.

This suggests a variational formulation of Stokes system.Precise mathematical formulation.Let us define the space

V = w ∈ H10 (Ω)

N, divw = 0(2.1.4)

and the bilinear form defined for all v ∈ V and w ∈ V by

a(v, w) = ν

N

i=1

Ω∇vi∇widx.(2.1.5)

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

a(u,w) =N

i=1

Ωfiwidx (or

N

i=1

< fi, wi > ), ∀w ∈ V,(2.1.6)

u ∈ V.

Here < ., . > denotes the duality pairing between H−1(Ω) and H

10 (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

a(u,w) =N

i=1

< fi, wi >, ∀w ∈ V,(2.1.7)

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

J(u) = minw∈V

J(w),(2.1.8)

u ∈ V,

15

where

J(w) =1

2a(w,w)−

N

i=1

< fi, wi > .(2.1.9)

Proof.First of all, as Ω is bounded, from Poincare inequality, we can equip H

10 (Ω) with the

norm v → (Ω |∇v|2dx)

12 and the associated scalar product and it is then a Hilbert

space. Then H10 (Ω)

N is also a Hilbert space for the norm (and the correspondingscalar product)

w = (w1, · · · , wN ) → (N

i=1

Ω|∇wi|2dx)

12 .

Now the mapping w → divw is linear continuous from H10 (Ω)

N to L2(Ω), and the

space V is the kernel of this mapping. Therefore, V is a closed subspace of H10 (Ω)

N

and therefore, it is a Hilbert space for the norm (and the scalar product) inducedby the one in H

10 (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.On the other hand, when f ∈ H

−1(Ω)N , the mapping w →

N

i=1 < fi, wi > is linearcontinuous from V to IR.We can then apply Lax-Milgram Theorem which shows the first part of Theorem2.1.1.That u is the solution to the minimization problem for functional J comes from thefact that the bilinear form a(., .) is symmetric.Interpretation. Relation with Stokes problem.

What is the relation with Stokes problem and in particular what about thepressure p which seems to have disappeared?Let us write V the space

V = ϕ ∈ D(Ω)N , divϕ = 0,

where D(Ω) = C∞0 (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

i=1

Ω∇ui∇ϕidx = ν

N

i=1

< ∇ui,∇ϕi >D,D= −ν

N

i=1

< ∆ui,ϕi >D,D .

16

Therefore we obtain

N

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, divϕ >D,D= 0.

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

−ν∆ui = fi −∂p

∂xiin D(Ω),

divu = 0, in Ω,

u = 0 on Γ.

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

∂xi= fi + ν∆ui ∈ H

−1(Ω).

Therefore p ∈ D(Ω) and ∇p ∈ H−1(Ω) and this implies that p ∈ L

2(Ω). Of coursep 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

−ν∆ui = fi −∂p

∂xiin H

−1(Ω)N ,(2.1.10)

divu = 0, in Ω,(2.1.11)

u = 0 on Γ.(2.1.12)

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

17

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 ∈ H2(Ω)N ∩ V, p ∈ H

1(Ω)/IR

and we have||u||H2(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 whichenable to prove most of the properties which have been accepted so far withoutproof. This section follows very closely lecture notes given by L. Tartar on thesubject, 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 fromH to F . Then

KerA = h ∈ H, Ah = 0is 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 = ProjKerAh and h2 = h− h1. From the definition ofprojection we have

∀l ∈ KerA, (h− h1, l)H = 0

which impliesh2 = 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 abovedecomposition is then unique and moreover there exist two constants C1 > 0 andc2 > 0 such that

∀h ∈ H, |h1|H ≤ C1|h|H , |h2|H ≤ C2|h|H

18

which says that the projections are continuous. (It is clear here in the case of Hilbertspaces 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∃C > 0, ∀h ∈ H, |h|H ≤ C(|Ah|F + |Bh|G).(2.2.13)

Lemma 2.2.1 Under the above hypotheses

∃CN , ∀h ∈ N, |h|H ≤ CN |Ah|F .(2.2.14)

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, |hn|H ≥ n|Ahn|F .

Let us define

hn =hn

|hn|H∈ N.

Then |hn|H = 1 and |Ahn|F ≤ 1n(therefore |Ahn|F → 0 in F ).

As |hn|H = 1 we can extract a subsequence, still denoted (hn) such that

hn h0 in H weakly.

As N is a closed subspace (then closed for the weak topology) and hn ∈ N , h0 ∈ N .As A is linear continuous from H to F , it is continuous for the weak topologies andtherefore,

Ahn Ah0 in F weakly.

But we already know that Ahn → 0 in F strongly. Therefore Ah0 = 0 and h0 ∈ N .This says that h0 ∈ KerA ∩N so that h0 = 0.Now we have

hn 0 in H weakly and Ahn → 0 in F strongly.

As B is compact we have

Bhn → B0 = 0 in G strongly

so that |Bhn|G → 0.But from (2.2.13) we have

|hn|H ≤ C(|Ahn|F + |Bhn|G) → 0.

19

This gives a contradiction with the fact that |hn|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 hn ∈ H such that Ahn = yn. We have the decompsition

hn = hn + hn, hn ∈ N, hn ∈ 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 thatthere 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 suchthat 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)

andA = ∇ (Gradient operator)

in order to show that ∇ has a closed image in H−1(Ω)N .

Let us define the space

X(Ω) = g ∈ H−1(Ω), ∇g ∈ H

−1(ΩN.(2.2.15)

Lemma 2.2.2X(IRN ) = L

2(IRN ).(2.2.16)

20

Proof.It is clear that L

2(IRN ) ⊂ X(IRN ). Let us show that if g ∈ X(IRN ), then g ∈L2(IRN ) by using the Fourier transform.

We know that

g ∈ L2(IRN ) ⇐⇒ g ∈ L

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

12 g ∈ L

2(IRN ).

Now∇g ∈ H

−1(IRN )N ⇐⇒ (1 + |ξ|2)|12 |ξ|g| ∈ L

2(IRN )

so that

g ∈ X(IRN ) ⇐⇒ (1 + |ξ|2)(1 + |ξ|2)|g|2 ∈ L1(IRN ) ⇐⇒ g ∈ L

2(IRN ).

Lemma 2.2.3X(IRN

+ ) = L2(IRN

+ ).(2.2.17)

Proof.It is clear that L

2(IRN

+ ) ⊂ X(IRN

+ ). Let us show that X(IRN

+ ) ⊂ L2(IRN

+ ). To thisend, we are going to exhibit a continuous extension from X(IRN

+ ) to X(IRN ).Let us first define a restriction operator from H

1(IRN ) to H10 (IR

N

+ ). For u ∈D(IRN ) = C

∞0 (IRN ) we set

Qu(x1, · · · , xN ) =0, if xN < 0,u(x1, · · · , xN ) +

2j=1 aju(x1, · · · , xN−1,−jxN ) if xN > 0.

We impose that

1 +2

j=1

aj = 0

which impliesQu(x1, · · · , xN−1, 0) = 0.

It is clear that for i = 1, · · · , N − 1 we have

Q(∂u

∂xi) =

∂Qu

∂xi.

For i = N we have

Q(∂u

∂xN) =

∂

∂xNRu

where

Ru(x1, · · · , xN ) =0, if xN < 0,u(x1, · · · , xN ) +

2j=1

aj

−ju(x1, · · · , xN−1,−jxN ) if xN > 0.

21

We also want R to be continuous from H1(IRN ) to H

10 (IR

N

+ ) which imposes

1 +2

j=1

aj

−j= 0.

Thus we choose the coefficients aj such that

1 +2

j=1

aj = 0, 1 +2

j=1

aj

−j= 0

and they are well defined by these relations. Now we extend Q and R by continuityto H

1(IRN ) and we defineP =t

Q.

As Q is linear continuous from H1(IRN ) to H

10 (IR

N

+ ), P is linear continuous fromH

−1(IRN

+ ) to H−1(IRN ). Moreover we have for i = 1, · · · , N − 1,

∀ϕ ∈ H1(IRN ), <

∂

∂xiPu,ϕ >= − < Pu,

∂ϕ

∂xi>= − < u,Q(

∂ϕ

∂xi) >

= − < u,∂

∂xiQϕ >=<

∂u

∂xi, Qϕ >=< P (

∂u

∂xi),ϕ >,

so that∂

∂xiPu = P (

∂u

∂xi).

So, for i = 1, · · · , N − 1, as ∂u

∂xi∈ H

−1(IRN

+ ) we have P ( ∂u

∂xi) = ∂

∂xiPu ∈ H

−1(IRN ).Now for i = N we have in the same way

∂

∂xNPu =t

R∂u

∂xN

and tR is linear continuous from H

−1(IRN

+ ) to H−1(IRN ).

Therefore, if u ∈ X(IRN

+ ), then Pu ∈ X(IRN ) and P is a continuous linear operator.It remains to prove that P is an extension.For u ∈ H

10 (IR

N

+ ) let us define by Eu the extension by 0 outside IRN

+ . It is wellknown that u → Eu is linear continuous from H

10 (IR

N

+ ) to H1(IRN ). We denote by

Π the transposition of this operator E which is the restriction to IRN

+ and which isa linear continuous operator from H

−1(IRN ) to H−1(IRN

+ ). We have to show thatΠ.P = Id.But Π.P =t

E.tQ =t (Q.E) and it is clear that Q.E = Id so that Π.P = Id.

If u ∈ X(IRN

+ ), Pu ∈ X(IRN ) = L2(IRN ) from Lemma 2.2.2. Then Π.Pu = 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,

I

i=1

θi = 1.

We write

u =I

i=1

θiu.

In the case θi ∈ C∞0 (Ω), θiu can be extended by zero to have θiu ∈ X(IRN ).

When θi ∈ C∞(Ω), we can find a C

2 diffeomorphism η such that θiuη−1 ∈ X(IRN ).Then θiu η−1 ∈ L

2(IRN ) which implies θiu ∈ 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|2

H−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

∀g ∈ L2(Ω), |g|2

L2(Ω) ≤ C|g|2X(Ω).(2.2.19)

Proof.When g ∈ L

2(Ω) we have|g|2

H−1(Ω) ≤ C|g|2L2(Ω)

and|∇g|2

H−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 twoHilbert 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|2

L2(Ω) ≤ C|g|2X(Ω).

23

This gives Lemma 2.2.5.Now, as Ω is bounded, H1

0 (Ω) is compactly embedded in L2(Ω) and then L

2(Ω) isalso 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 defineKer∇ = g ∈ L

2(Ω), ∇g = 0

as Ω is a connected set, we see that the elements of Ker∇ are constants in Ω.Therefore we have

Ker∇⊥ = g ∈ L2(Ω),

Ωgdx = 0.

Lemma 2.2.7 If (pn) is a sequence in L2(Ω) such that

Ω pndx = 0 and |∇pn|H−1(Ω)N

is bounded, then |pn|L2(Ω) is bounded.

Proof.We have pn ∈ Ker∇⊥. Then from Lemma 2.2.1, there exists a constant C > 0independent 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 ∈ H10 (Ω)

N, < y, w >= 0, ∀y ∈ Y = V.

24

Actually, if w ∈ H10 (Ω)

N and < ∇p, w >= 0 ∀p ∈ L2(Ω) we have < w,∇ϕ >= 0

∀ϕ ∈ C∞0 (Ω) so that < divw,ϕ >= 0 ∀ϕ ∈ C

∞0 (Ω) and therefore divw = 0 so that

w ∈ V .On the other hand we have

∀p ∈ L2(Ω), ∀w ∈ V, < ∇p, w >= 0.

Therefore V = Y⊥ 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 thisproves Lemma 2.2.8.Lemma 2.2.8 enables us to give a completely correct interpretation of Stokes problemin Theorem 2.1.3. Indeed, if gi = −ν∆ui − fi and g = (g1, · · · , gN ) we have fromthe 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 inter-

pretation of Stokes problem.This result can also be proved by another method.For > 0 let us consider the problem

a(u, w) +1

Ωdivudivwdx =

N

i=1

< fi, wi >, ∀w ∈ H10 (Ω)

N,

u ∈ H10 (Ω)

N.

This problem has a unique solution u which satisfies

−ν∆u −∇(1

divu) = f

u ∈ H10 (Ω)

N.

If we write

p = −1

divu

then

p ∈ L2(Ω),

Ωpdx = 0.

On the other hand we have

a(u, u) +1

Ω|divu|2dx =

N

i=1

< fi, u,i >

25

so that there exists a constant M independent of such that

||u||H10 (Ω)N ≤ M, | 1√

divu|L2(Ω) ≤ M.

Then, after extraction of a subsequence, we have

u u in H10 (Ω)

N weakly,

divu → 0 in L2(Ω) strongly.

Then divu = 0 so that u ∈ V and for every w ∈ V (divw = 0) we have

a(u, w) =N

i=1

< fi, wi >

so that

a(u,w) =N

i=1

< fi, wi >

and u is solution of the variational form of Stokes problem.Now p satisfies

∇p = f + ν∆u,

Ωpdx = 0.

Then p is bounded in H−1(Ω)N and satisfies

Ω pdx = 0 so that p is bounded in

L2(Ω). After extraction of a subsequence,

p p in L2(Ω) weakly

and we have

∀w ∈ H10 (Ω)

N, a(u,w) =

N

i=1

< fi, wi > +

Ωpdivwdx

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 haveF = ImA⊕ L.

Here, F = H−1(Ω)N = W

where W = H10 (Ω)

N so that F = W .

26

Lemma 2.2.9W = (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, < u− v,Ag >= 0

which says that u− v ∈ (ImA)⊥.We then have

u = v + w, v ∈ L⊥, w ∈ ImA

⊥.

Lemma 2.2.10 If tA : W → H is defined by

∀u ∈ W, ∀g ∈ H, (tAu, g)H =< u,Ag >

then we haveKertA = ImA

⊥, and Imt

A = KerA⊥.

27

Proof.If u ∈ KertA, we have ∀g ∈ H, < u,Ag >= 0 and then u ∈ ImA

⊥.If u ∈ ImA

⊥, we have ∀g ∈ H, < u,Ag >= 0 and therefore (tAu, g)H = 0 so thattAu = 0 and u ∈ KertA.Now if g ∈ Imt

A, there exists u ∈ W such that g =tAu. We have for f ∈ KerA,

(g, f)H = (tAu, f)H =< u,Af >= 0, so that g ∈ KerA⊥. This shows that ImtA ⊂

KerA⊥ and as KerA⊥ is closed, we have

ImtA ⊂ KerA⊥

.

Let us suppose that ImtA = KerA⊥. Then there exists g ∈ H

= H and there existsf0 ∈ KerA⊥ such that

∀f ∈ ImtA, (g, f)H = 0, and (g, f0)H = 1.

Then for every u ∈ W , (g,tAu)H = 0 so that < u,Ag >= 0 and therefore Ag = 0in W

. Therefore, g ∈ KerA which implies (g, f0)H = 0. This gives a contradiction.Therefore we have

ImtA = KerA⊥

.

Lemma 2.2.11 ImtA 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

Aun with un ∈ W . Moreover we know from the previous lemma that gn ∈ KerA⊥.From Lemma 2.2.9, we can decompose un as un = un + vn with un ∈ L

⊥ andvn ∈ ImA

⊥ = KertA. Then tAun =t

Aun so that gn =tAun with un ∈ L

⊥.If h ∈ W

we have h = Ag + l with g ∈ KerA⊥ = N and l ∈ L. (the fact that wecan take g ∈ KerA⊥ is immediate since H = KerA⊕KerA⊥.) Then

< um − up, h >=< um − up, Ag >=< um − up, l >=< um − up, Ag >

= (tA(um − up), g)H = (gm − gp, g)H .

Then for every h ∈ W

| < um − up, h > | ≤ |gm − gp|H |g|H≤ |gm − gp|HCN ||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 . NowtAun = gn →t

Au in H and therefore g =tAu so that g ∈ Imt

A and ImtA is closed

in H.

28

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

tAu = f and ||u||W ≤ C|f |H .

Proof.We have W = ImA

⊥⊕L⊥ = KertA⊕L

⊥. Now it is immediate to see that tA : L⊥ →

H is injective. But as ImtA = KerA⊥ we see that the mapping t

A : L⊥ → KerA⊥

is one-to-one (bijective). On the other hand we have |tAu|H ≤ C||u||W which saysthat t

A is continuous.Now l

⊥ and KerA⊥ are Banach spaces as they are closed subspaces of Banachspaces. Then from Banach Theorem, t

A : L⊥ → KerA⊥ has a continuous inversewhich 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) =

g ∈ L2(Ω),

Ω gdx = 0, there exists u ∈ H

10 (Ω)

N such that

divu = g

and||u||

H10 (Ω)N ≤ C|g|L2(Ω).

Proof.Take

A = ∇ : L2(Ω) = H → H−1(Ω)N = W

.

Then KerA = g ∈ L2(Ω), g = Cst and KerA⊥ = L

20(Ω).

If g ∈ L20(Ω), there exists u ∈ L

⊥ ⊂ W such that tAu = −g and ||u||W ≤ C|g|H .

Then u ∈ H10 (Ω)

N and for every h ∈ H we have

(−g, h)H = (tAu, h)H =< u,Ah >=< u,∇h > .

Taking h ∈ C∞0 (Ω) we see that

(tAu, h)H =< u,∇h >D,D= − < divu, h >D,D=< −g, h >D,D .

Then we have

divu = g, u ∈ H10 (Ω)

N, and ||u||

H10 (Ω)N ≤ C|g|L2(Ω).

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 ∈ C

∞0 (Ω)N = D(Ω)N , divϕ = 0.

Let us consider an increasing sequence of open sets (Ωn) such that Ωn ⊂ Ω and∪nΩn = Ω. If u ∈ H

10 (Ωn)N and divu = 0, we can extend u by 0 outside Ωn. If (ρ)

is a regularizing sequence (mollifiers), for small enough we have ρ ∗ u ∈ C∞0 (Ω)N

and div (ρ ∗ u) = 0. Then < f, u >= lim→0 < f, ρ ∗ u >= 0. Therefore

∀u ∈ H10 (Ωn)

N, divu = 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. Thenf = ∇p where p ∈ L

2loc(Ω).

Now if Ω is starshaped (for example with respect to 0), for 0 < θ < 1 we haveθΩ ⊂ Ω. If u ∈ H

10 (Ω)

N, divu = 0, then uθ defined by uθ(x) = u(x

θ) has a compact

support in Ω and divuθ = 0. Then we can approximate uθ by ρ ∗ uθ ∈ C∞0 (Ω)N

with div (ρ ∗ uθ) = 0 so that

< f, u >= limθ→1

lim→0

< f, ρ ∗ uθ >= 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 regularand strictly starshaped. Then there exists q ∈ L

2(ω) such that f/ω = ∇q.But p/ω − q is constant on ω. Therefore p/ω ∈ L

2(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 thereexists p ∈ L

2(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 ∈ L

2(Ω).

Lemma 2.2.17 The space C∞(Ω)N is dense in X and the mapping ϕ → ϕ.n de-

fined on C∞(Ω)N can be extended to a linear continuous map from X to H

− 12 (Γ).

30

Proof.Let v ∈ X. For every ψ ∈ H

12 (Γ) and ϕ ∈ H

1(Ω) such that ϕ/Γ = ψ (we take acontinuous extension), we define

L(ψ) =

Ωdiv vϕdx+

Ωv.∇ϕdx.

If ϕ ∈ H10 (Ω), then the right hand side is 0 so that L depends only on ψ. Moreover

we have|L(ψ)| ≤ ||v||X ||ϕ||H1(Ω) ≤ C||v||X ||ψ||

H12 (Γ)

.

Therefore, there exists an element of H12 (Γ) which corresponds to v.n for regular

functions v such that

∀ψ ∈ H12 (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 iscontained in H. Now if f ∈ L

2(Ω)N and < f, v >= 0 for every v ∈ V . Then weknow that there exists p ∈ L

2(Ω) such that f = ∇p. Then p ∈ H1(Ω). Therefore,

for every v ∈ H, we have

< f, v >=< ∇p, v >=

Ωpdiv vdx+ < v.n, p >Γ= 0.

Then < f, v >= 0 for every v ∈ H and this shows that the closure of V contains Hand 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 thevelocity and p the pressure) such that

∂u

∂t− ν∆u = f −∇p in Ω× (0, T ),(2.3.20)

divu = 0 in Ω× (0, T ),(2.3.21)

u = 0 on Σ = Γ× (0, T ),(2.3.22)

u(0) = u0 in Ω.(2.3.23)

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

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

beH = w ∈ L

2(Ω)N , divw = 0, w.n = 0(2.3.24)

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, forf ∈ H (we denote by (., .) the scalar product in H which is the L

2 scalar product)

a(u,w) = (f, w) ∀w ∈ V,

u ∈ V.

As the injection V ⊂ H is compact continuous it is immediate to see that themapping T : f ∈ H → u ∈ V → u ∈ H is linear continuous and compact. Moreoverif we write Tf = u and Tg = v, we have as a(., .) is symmetric

(Tf, g) = (u, g) = (g, u) = a(v, u) = a(u, v) = (f, v) = (f, Tg),

so that T is selfadjoint.Therefore, (see for example [2]), there exists a decreasing sequence (µn)n of strictlypositive real numbers with µn → 0 when n → +∞ and a sequence (wn)n of elementsof V which form an orthonormal basis in H such that

Twn = µnwn.

Setting now λn = 1µn

we have

0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · · , λn → +∞

and

a(wn, w) = λn(wn, w) ∀w ∈ V,

wn ∈ V,

(wm, wp) = δmp.

If we equip V with the (equivalent) scalar product a(v, w), we can see that ( wn√λn

)nform an orthonormal basis in V .Let v ∈ H. Then we have

v =+∞

n=1

(v, wn)wn with+∞

n=1

|(v, wn)|2 = |v|2H < +∞.

32

If v ∈ V we have

v =+∞

n=1

(v, wn)wn with+∞

n=1

λn|(v, wn)|2 = ||v||2V < +∞.

If v ∈ V we have

v =+∞

n=1

< v,wn > wn with+∞

n=1

| < v,wn > |2

λn

= ||v||2V < +∞.

2.3.2 Existence and uniqueness result

First of all, at least formally, if we multiply the Stokes evolution system by a functionw ∈ V , as for the stationnary case, the pressure p disappear and we can write theproblem as

d

dt(u(t), w) + a(u(t), w) =< f(t), w > on (0, T ), ∀w ∈ V,(2.3.25)

u(0) = u0.(2.3.26)

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

there exists a unique solution u ∈ C([0, T ];H) ∩ L2(0, T ;V ) with du

dt∈ L

2(0, T ;V )of problem (2.3.25),(2.3.26). Moreover, there exists p ∈ D(]0, T [;L2(Ω)) (in factp ∈ H

−1(0, T ;L2(Ω))) such that (u, p) satisfies the Stokes evolution equation

∂u

∂t− ν∆u = f −∇p in Ω× (0, T ),(2.3.27)

divu = 0 in Ω× (0, T ),(2.3.28)

u = 0 on Σ = Γ× (0, T ),(2.3.29)

u(0) = u0 in Ω.(2.3.30)

Proof.we have

u0 =+∞

n=1

(u0, wn)wn with+∞

n=1

|(u0, wn)|2 = |u0|2H < +∞

and

f(t) =+∞

n=1

< f(t), wn > wn with

T

0

+∞

n=1

| < f(t), wn > |2

λn

dt = ||f ||2

L2(0,T ;V ) < +∞.

33

Let us write

uM

0 =M

n=1

(u0, wn)wn

and

fM (t) =

M

n=1

< f(t), wn > wn.

We know that uM0 converges to u0 in H and fM converges to f in L

2(0, T ;V ) whenM → +∞. First of all, calling V

M = spanw1, · · · , wM, we look for

uM (t) =

M

n=1

un(t)wn

such that

d

dt(uM (t), w) + a(uM (t), w) =< f

M (t), w > on (0, T ), ∀w ∈ VM,

uM (0) = u

M

0 .

This gives us for every n = 1, · · · ,M

d

dtun(t) + λnun(t) =< f(t), wn >,

un(0) = (u0, wn),

and we even have an explicit formula for un

un(t) = (u0, wn)e−λnt +

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

n=P+1

un(t)wn,

|(uM − uP )(t)|2H =

M

n=P+1

|un(t)|2,

||uM − uP ||2

L2(0,T ;V ) =

T

0

M

n=P+1

λn|un(t)|2dt.

34

From the equation giving un we obtain

1

2

d

dt|un(t)|2 + λn|un(t)|2 =< f(t), wn > un(t) ≤

| < f(t), wn > |√λn

√λn|un(t)|.

For t ∈ [0, T ] we now integrate on (0, t) to get

|un(t)|2 +

t

0λn|un(s)|2ds ≤ |(u0, wn)|2 +

t

0

| < f(s), wn > |2

λn

ds.

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

||uM − uP ||2

C([0,T ;H) + ||uM − uP ||2

L2(0,T ;V ) ≤ 2|uM0 − u

P

0 |2H + ||fM − fP ||2

L2(0,T ;V )

.

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) =+∞

n=1

un(t)wn.

The regularity of du

dtcomes immediately from the equation for un for example.

Taking test functions w in ∪MVM first then in its closure which is V it is now easy

to see that u satisfies

d

dt(u(t), w) + a(u(t), w) =< f(t), w > on (0, T ), ∀w ∈ V,

u(0) = u0.

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

U(t) =

t

0u(s)ds and F (t) =

t

0f(s)ds,

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),

divU(t) = 0.

35

As F (·)− u(·) + u0 is continuous in time, we clearly have P ∈ C([0, T ];L2(Ω)/IR).

Now taking the time derivative we obtain with p = d

dtP

∂u

∂t− ν∆u = f −∇p in Ω× (0, T ),

divu = 0 in Ω× (0, T ),

u = 0 on Σ = Γ× (0, T ),

u(0) = u0 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 corre-

sponding Stokes equation satisfies

u ∈ C([0, T ], V ) ∩ L2(0, T ;H2(Ω)N ),

∂u

∂t∈ L

2(0, T ;H),(2.3.31)

p ∈ L2(0, T ;H1(Ω)/IR).(2.3.32)

Proof.Multiplying the equation by ∂u

∂twe have

Ω|∂u∂t

(t)|2dx+ν

2

d

dt

Ω|∇u(t)|2dx =

Ωf(t)

∂u

∂t(t)dx ≤ |f(t)|H |∂u

∂t(t)|H .

This gives an estimate on ∂u

∂tin L

2(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 Stokesequation with right hand side in L

2(0, T ;H), which gives (from the regularity resultfor the stationnary problem) v ∈ L

2(0, T,H2(Ω)N ) and p ∈ L2(0, T ;H1(Ω)/IR).

36

Chapter 3

Navier-Stokes equations. Theevolution case.

3.1 Notations and preliminaries

All along this chapter we will assume that Ω is bounded and regular, except if it isspecially mentionned. We will also take the density ρ = 1 in order to simplify thepresentation. The Navier-Stokes equations then take the form

∂u

∂t+ (u.∇)u− ν∆u = f −∇p, in Ω× (0, T ),(3.1.1)

divu = 0, in Ω× (0, T ),(3.1.2)

u = 0, on Γ× (0, T ),(3.1.3)

u(0) = u0 in Ω.(3.1.4)

We introduce the notations

a(v, w) =N

i,j=1

Ω

∂vi

∂xj.∂wi

∂xjdx =

Ω∇v.∇wdx, ∀v, w ∈ V,(3.1.5)

b(u, v, w) =N

i,j=1

Ωuj

∂vi

∂xjwidx, ∀u, v, w ∈ V.(3.1.6)

Lemma 3.1.1 For N ≤ 4, b(., ., .) is a trilinear continuous form on H1(Ω)N ×

H1(Ω)N ×H

1(Ω)N . Moreover we have

∀u ∈ V, ∀v, w ∈ H1(Ω)N , b(u, v, w) + b(u,w, v) = 0.(3.1.7)

37

In particular we have

∀u ∈ V, ∀v ∈ H1(Ω)N , b(u, v, v) = 0.(3.1.8)

Proof.For N ≤ 4 we have H1(Ω) ⊂ L

4(Ω) with continuous injection. Therefore, if u, v, w ∈V we have

uj ∈ L4(Ω,

∂vi

∂xj∈ L

2(Ω), wi ∈ L4(Ω),

and the term

Ωuj

∂vi

∂xjwidx

makes perfect sense with

|

Ωuj

∂vi

∂xjwidx| ≤ C|uj |L4(Ω)||vi||H1(Ω)|wi|L4(Ω) ≤ C||uj ||H1(Ω)||vi||H1(Ω)||wi||H1(Ω).

Therefore, b(., ., .) is well defined, is trilinear and

|b(u, v, w)| ≤ C||u||H1(Ω)N ||v||H1(Ω)N ||w||H1(Ω)N ,

which says that it is continuous.Now we have

b(u, v, w) + b(u,w, v) =N

i,j=1

Ωuj(

∂vi

∂xjwi +

∂wi

∂xjvi)dx

=N

i,j=1

Ωuj

∂

∂xj(viwi)dx = −

N

i,j=1

Ω(∂uj

∂xj)viwidx = 0

= −N

i=1

Ω(divu)viwidx = 0.

We now give a key lemma in dimension N = 2 which is no longer valid in higherdimension and which makes the crucial difference between the study of Navier-Stokesequations 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

∀ϕ ∈ H1(IR2), |ϕ|

L4(IR2) ≤ C||ϕ||12

H1(IR2)|ϕ|

12

L2(IR2).(3.1.9)

In particular we have

∀ϕ ∈ H10 (Ω), |ϕ|L4(Ω) ≤ C||ϕ||

12

H10 (Ω)

|ϕ|12L2(Ω).(3.1.10)

38

Proof.Of course it is sufficient to prove the above inequalities for ϕ ∈ C

∞0 (IR2). We can

write

ϕ2(x1, x2) =

x2

−∞

∂

∂x2ϕ2(x1, y)dy = 2

x2

−∞ϕ(x1, y)

∂ϕ

∂x2(x1, y)dy

and then

|ϕ(x1, x2)|2 ≤ 2

x2

−∞ϕ(x1, y)

2dy

12

x2

−∞(∂ϕ

∂x2(x1, y))

2dy

12

≤ 2 +∞

−∞ϕ(x1, y)

2dy

12 +∞

−∞(∂ϕ

∂x2(x1, y))

2dy

12.

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

|ϕ(x1, x2)|2 ≤ 2 +∞

−∞ϕ(y, x2)

2dy

12 +∞

−∞(∂ϕ

∂x1(y, x2))

2dy

12.

Multiplying these two inequalities we obtain

|ϕ(x1, x2)|4 ≤ 4 +∞

−∞ϕ(x1, y)

2dy

12 +∞

−∞(∂ϕ

∂x2(x1, y))

2dy

12

.

+∞

−∞ϕ(y, x2)

2dy

12 +∞

−∞(∂ϕ

∂x1(y, x2))

2dy

12

≤ λ(x1).µ(x2).

Therefore

IR2|ϕ(x1, x2)|4dx1dx2 ≤ 4

+∞

−∞λ(x1)dx1

+∞

−∞µ(x2)dx2

.

But we have +∞

−∞λ(x1)dx1 ≤

IR2|ϕ(x1, y)|2dx1dy

12

IR2| ∂ϕ∂x2

(x1, y)|2dx1dy 1

2

≤ |ϕ|L2(IR2)|

∂ϕ

∂x2|L2(IR2).

In the same way we have +∞

−∞µ(x2)dx2 ≤ |ϕ|

L2(IR2)|∂ϕ

∂x1|L2(IR2).

Therefore

|ϕ|4L4(IR2) ≤ 4|ϕ|

L2(IR2)|∂ϕ

∂x2|L2(IR2)|ϕ|L2(IR2)|

∂ϕ

∂x1|L2(IR2) ≤ 2|ϕ|2

L2(IR2)||ϕ||2H1(IR2).

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

10 (Ω)N is linear continuous on

V and therefore defines (in a non unique way) an element of V that we will stilldenote by f . We will also denote the duality between H

−1(Ω)N and H10 (Ω)

N by< ·, · >.We will write A the operator defined by

∀u, v ∈ V, a(u, v) =< Au, v > .

Then it is immediate to see that

A ∈ L(V, V ).

Now for N ≤ 4 we can write

∀u, v, w ∈ V, b(u, v, w) =< B(u, v), w >

where(u, v) → B(u, v)

is bilinear continuous from V × V in V and we have

||B(u, v)||V ≤ C||u||V ||v||V .

When u ∈ L2(0, T ;V ) and v ∈ L

2(0, T ;V ) it is clear that B(u, v) ∈ L1(0, T ;V ).

Lemma 3.1.3 In dimension N = 2, when u, v ∈ L2(0, T ;V ) ∩ L

∞(0, T ;H), then

B(u, v) ∈ L2(0, T ;H−1(Ω)N ).

Proof.We have < B(u, v), w >= − < B(u,w), v > so that

| < B(u, v), w > | ≤ C|u|L4(Ω)|v|L4(Ω)||w||H10 (Ω)N .

Then

||B(u, v)||H−1(Ω)N ≤ C|u|L4(Ω)|v|L4(Ω) ≤ C|u|12L2(Ω)||u||

12V|v|

12L2(Ω)||v||

12V.

This implies

T

0||B(u, v)||2

H−1(Ω)Ndt ≤ C||u||L∞(0,T ;H)||v||L∞(0,T ;H)||u||L2(0,T ;V )||v||L2(0,T ;V ),

which proves the Lemma 3.1.3.

40

Lemma 3.1.4 In dimension N = 3, when u, v ∈ L2(0, T ;V )∩L

∞(0, T ;H) we have

B(u, v) ∈ L43 (0, T ;H−1(Ω)N ).

Proof.We still have

||B(u, v)||H−1(Ω)N ≤ C|u|L4(Ω)|v|L4(Ω).

Then T

0||B(u, v)||

43H−1(Ω)Ndt ≤ C

T

0|u|

43L4(Ω)|v|

43L4(Ω)dt.

But

|u|4L4(Ω) =

Ω|u|4dx =

Ω|u||u|3dx ≤ (

Ω|u|2dx)

12 (

Ω|u|6dx)

12

≤ |u|L2(Ω)|u|3L6(Ω) ≤ |u|L2(Ω)||u||3V .

Therefore

T

0||B(u, v)||

43H−1(Ω)Ndt ≤

T

0|u|

13L2(Ω)|v|

13L2(Ω)||u||V ||v||V dt

≤ ||u||13L∞(0,T ;H)||v||

13L∞(0,T ;H)||u||L2(0,T ;V )||v||L2(0,T ;V )

and

||B(u, v)||L

43 (0,T ;H−1(Ω)N )

≤ C||u||14L∞(0,T ;H)||v||

14L∞(0,T ;H)||u||

34L2(0,T ;V )||v||

34L2(0,T ;V ).

It is now natural to formulate Navier-Stokes equations in the following way. Welook for u ∈ L

2(0, T ;V ) ∩ L∞(0, T ;H) such that (with f ∈ L

2(0, T ;V ))

d

dt(u(t), w) + a(u(t), w) + b(u(t), u(t), w) =< f(t), w >, ∀w ∈ V.(3.1.11)

In order to give a sense to the initial condition we will need u to be continuous intime with values in some suitable space in order to impose

u(0) = u0 ∈ H.(3.1.12)

We have for the two dimensional case

Proposition 3.1.5 In dimension N = 2, if u ∈ L2(0, T ;V )∩L

∞(0, T ;H) and if usatisfies (3.1.11) with f ∈ L

2(0, T ;V ), then

u ∈ C([0, T ], H),du

dt∈ L

2(0, T ;V )

41

and we can impose initial condition (3.1.12) with u0 ∈ H. These solutions will becalled weak solutions of Navier-Stokes equations.We then have

du

dt+Au+B(u, u) = f in L

2(0, T ;V ),(3.1.13)

u(0) = u0,(3.1.14)

and, using the results for Stokes equation, when f ∈ L2(0, T ;H−1(Ω)N ), there exists

a pressure p ∈ H−1(0, T ;L2(Ω)) such that

∂u

∂t+ (u.∇)u− ν∆u = f −∇p, in Ω× (0, T ),(3.1.15)

divu = 0, in Ω× (0, T ),(3.1.16)

u = 0, on Γ× (0, T ),(3.1.17)

u(0) = u0.(3.1.18)

Proof.If u ∈ L

2(0, T ;V )∩L∞(0, T ;H) we have, from Lemma 3.1.3, B(u, u) ∈ L2(0, T ;H−1(Ω)N ).

We are then lead to the interpretation of Stokes problem with right hand sidef −B(u, u) ∈ L

2(0, T ;H−1(Ω)N ) and the result follows immediately.

3.2 Existence and uniqueness in dimension N = 2

Uniqueness of weak solutions for Navier-Stokes equations in dimension N = 3 is oneof the major open problems nowadays in mathematics. Of course we will not discussthis question here but we will give the uniqueness result in dimension N = 2 below.

Theorem 3.2.1 If f ∈ L2(0, T ;V ) and u0 ∈ H, there exists at most one solution

u to the Navier-Stokes equations satisfying u ∈ L2(0, T ;V ) ∩ L

∞(0, T ;H).

Proof.Let us suppose that we have two solutions u

1 and u2 in the admissible class. We

then have

d

dt(u1 − u

2) +A(u1 − u2) +B(u1, u1)−B(u2, u2) = 0,

(u1 − u2)(0) = 0.

Taking the scalar product with (u1 − u2) we obtain

1

2

d

dt|u1 − u

2|2H + ν||u1 − u2||2V + (B(u1, u1)−B(u2, u2), u1 − u

2) = 0 a.e. in t.

42

But we have

(B(u1, u1)−B(u2, u2), u1−u2) = (B(u1−u

2, u

1), u1−u2)+(B(u2, u1−u

2), u1−u2)

and(B(u2, u1 − u

2), u1 − u2) = b(u2, u1 − u

2, u

1 − u2) = 0.

Then

|(B(u1, u1)−B(u2, u2), u1 − u2)| ≤ C|u1 − u

2|L4(Ω)||u1||V |u1 − u2|L4(Ω)

≤ C||u1||V |u1 − u2|H ||u1 − u

2||V .

We then obtain

1

2

d

dt|u1 − u

2|2H + ν||u1 − u2||2V ≤ C||u1||V |u1 − u

2|H ||u1 − u2||V

≤ ν||u1 − u2||2V +

C2

4ν||u1||2V |u1 − u

2|H ,

|(u1 − u2)(0)|H = 0.

Then

1

2

d

dt|u1 − u

2|2H ≤ C2

4ν||u1||2V |u1 − u

2|H ,

|(u1 − u2)(0)|H = 0,

and as u1 ∈ L2(0, T ;V ), ||u1||2

V∈ L

1(0, T ).

Lemma 3.2.2 (Gronwall inequality.) If ϕ ≥ 0 satisfies

dϕ

dt(t) ≤ θ(t)ϕ(t) on (0, T )

with θ ∈ L1(0, T ), then we have

∀t ∈ [0, T ], ϕ(t) ≤ ϕ(0)e(

t

0θ(s)ds)

.

Proof.we have

d

dt

ϕ(t)e−(

t

0θ(s)ds)

≤ 0.

We can now use Gronwall inequality in our context and as |(u1 − u2)(0)|H = 0 we

obtain∀t ∈ (0, T ), |(u1 − u

2)(t)|H = 0.

This finishes the proof of the uniqueness result in dimension N = 2. We can nowgive the existence result.

43

Theorem 3.2.3 If f ∈ L2(0, T, V ) and u0 ∈ H, there exists a (weak) solution

u ∈ L2(0, T ;V ) ∩ L

∞(0, T ;H) of the Navier-Stokes equations.

Proof.We will give the proof in the case of dimension N = 2 but the existence theoremfor weak solutions is still true in dimension N = 3 and the proof requires the use ofsome interpolation spaces.First step. The proof uses the Galerkin method with a special basis. Let usconsider the special basis (wn) of eigenfunctions of operator A as described in theprevious chapter. Then (wn) is orthonormal in H and orthogonal in V . Let usfirst consider the following “approximate problem” . First of all we define the finitedimensional space

Vm = spanw1, · · · , wm.

Then we look for a function um defined on (0, T ) with values in V

m such that almosteverywhere in t and for every j = 1, · · · ,m

d

dt(um(t), wj) + a(um(t), wj) + b(um(t), um(t), wj) =< f(t), wj >,(3.2.19)

um(0) = u

m

0 ,(3.2.20)

where

um

0 =m

j=1

(u0, wj)wj ∈ Vm

andum

0 → u0 in H when m → +∞.

Then

um(t) =

m

i=1

gim(t)wi,

and, as the (wi) are orthonormal in H and orhtogonal in V , the system can bewritten as

dgjm

dt+ λjgjm +Gj(t, g1m, · · · , gmm) = 0, ∀j = 1, · · · ,m,(3.2.21)

gjm(0) = (u0, wj),(3.2.22)

where

Gj(t, g1m, · · · , gmm) = b(m

i=1

gim(t)wi,

m

k=1

gkm(t)wk, wj)− < f(t), wj > .

44

Using Cauchy-Lipschitz theorem, it is easy to show that this nonlinear system ofordinary differential equations has a unique solution locally in time and thereforeon an interval of time [0, Tm] ⊂ [0, T ].Second step. A priori estimates (I).We will argue indifferently on the (gjm) or on u

m as it gives the same thing. Mul-tiplying equation for um by gjm and summing up in j we get

1

2

d

dt|um|2H + a(um(t), um(t)) + b(um(t), um(t), um(t)) =< f(t), um(t) > .

As b(um(t), um(t), um(t)) = 0, this gives

1

2

d

dt|um|2H + ν||um(t)||2V ≤ ||f(t)||V ||um(t)||V ≤ ν

2||um(t)||2V +

1

2ν||f(t)||2V ,

and thend

dt|um|2H + ν||um(t)||2V ≤ 1

ν||f(t)||2V .

This implies that for every t ∈ (0, Tm)

|um(t)|2H+ν

t

0||um(s)||2V ds ≤ |um0 |2H+

1

ν

t

0||f(s)||2V ds ≤ |um0 |2H+

1

ν

T

0||f(s)||2V ds.

This gives an a priori bound on the norm |um(t)|2Hand therefore we can take Tm = T

for every m and we obtain

∀t ∈ (0, T ), |um(t)|2H ≤ |um0 |2H +1

ν

T

0||f(s)||2V ds ≤ M

and

ν

T

0||um(s)||2V ds ≤ |um0 |2H +

1

ν

T

0||f(s)||2V ds ≤ M.

This shows that (um) stays bounded in L∞(0, T ;H) and in L

2(0, T, V ). After extrac-tion of a subsequence still denoted by (um), there exists u ∈ L

∞(0, T ;H)∩L2(0, T ;V )such that when m → +∞

um

u in L2(0, T ;V ) weakly,

um

u in L∞(0, T,H) weak ∗ .

As we work in dimension N = 2, from Lemma 3.1.3 we see that B(um, um) stays

bounded in L2(0, T, V ) and we can suppose that

B(um, um) g in L

2(0, T, V ) weakly.

45

The question is now : how to prove that g = B(u, u)?Third step. A priori estimates (II).Let us show that du

m

dtstays bounded in L

2(0, T, V ). We have

<du

m

dt(t), wj > + < Au

m(t), wj > + < B(um(t), um(t)), wj >=< f(t), wj > .

Let us write hm = f − B(um, um) − Au

m. We know from the previous estimatesthat hm stays bounded in L

2(0, T, V ) and we have

<du

m

dt(t), wj >=< hm(t), wj > .

Therefore

||dum

dt||2L2(0,T,V ) =

T

0(m

j=1

| < dum

dt(t), wj > |2

λj

)dt ≤ ||hm||2L2(0,T ;V ),

and dum

dtstays bounded in L

2(0, T, V ).We now use the following compactness lemma.

Lemma 3.2.4 Let V be compactly embedded in H and let (um) be a sequence ofL2(0, T ;V ) such that

• (um) is bounded in L2(0, T ;V ).

• (dum

dt) is bounded in L

2(0, T ;V ).Then the sequence (um) is relatively compact in L

2(0, T ;H) and therefore, thereexists a subsequence which converges (strongly) in L

2(0, T ;H).

Applying this lemma in our context, we see that (denoting again the subsequenceby (um))

um → u in L

2(0, T ;H).

Let us then show that B(um, um) converges to B(u, u) in L

2(0, T ;V ) weakly. It issufficient to show that g = B(u, u). This amounts to show that

∀w ∈ L2(0, T ;V ),

T

0< B(um, u

m), w > dt →

T

0< B(u, u), w > dt.

We have

T

0< B(um, u

m), w > dt =

T

0b(um, u

m, w)dt = −

T

0b(um, w, u

m)dt

= −N

i,j=1

T

0

Ωum

j

∂wi

∂xjum

i dxdt.

46

As ∂wi

∂xjis a fixed function in L

2(0, T ;L2(Ω)) we only have to show that for every

i, j = 1, · · · , N ,um

j um

i → ujui in L2(0, T ;L2(Ω)) weakly.

From Lemma 3.1.2, we know that (because we work in dimensionN = 2) L2(0, T ;V )∩L∞(0, T ;H) ⊂ L

4(0, T ;L4(Ω)) with continuous injection. As (um) is bounded inL2(0, T ;V )∩L

∞(0, T ;H) we see that (um) is bounded in L4(0, T ;L4(Ω)) and there-

foreum → u in L

4(0, T ;L4(Ω)) weakly.

Then um

jum

iis bounded in L

2(0, T ;L2(Ω)) and this implies that it converges weaklyin this space. But we know that

um → u in L

2(0, T ;H) strongly

so thatum

j um

i → ujui in L1(0, T ;L1(Ω)) strongly.

Thereforeum

j um

i ujui in L2(0, T ;L2(Ω)) weakly,

andB(um, u

m) B(u, u) in L2(0, T ;V ) weakly.

Fourth step. Passage to the limit.Let ϕ ∈ C

∞([0, T ]) such that ϕ(T ) = 0. After integration by parts we have

−

T

0(um(t), wj)ϕ

(t)dt− (um0 , wj)ϕ(0) +

T

0a(um(t), wj)ϕ(t)dt

+

T

0< B(um(t), um(t)), wj > ϕ(t)dt =

T

0< f(t), wj > ϕ(t)dt.

For fixed j let m tend to +∞. All terms have a limit and we obtain for every j

−

T

0(u(t), wj)ϕ

(t)dt− (u0, wj)ϕ(0) +

T

0a(u(t), wj)ϕ(t)dt

+

T

0< B(u(t), u(t)), wj > ϕ(t)dt =

T

0< f(t), wj > ϕ(t)dt.

As the (wj) form a basis in V we then have for every w ∈ V and every ϕ ∈ C∞([0, T ])

with ϕ(T ) = 0

−

T

0(u(t), w)ϕ(t)dt− (u0, w)ϕ(0) +

T

0a(u(t), w)ϕ(t)dt(3.2.23)

+

T

0< B(u(t), u(t)), w > ϕ(t)dt =

T

0< f(t), w > ϕ(t)dt.

47

Now if we choose first ϕ ∈ C∞0 (]0, T [) we obtain

d

dt(u(t), w) + a(u(t), w)+ < B(u(t), u(t), w >=< f(t), w > in D(]0, T [),

with f ∈ L2(0, T ;V ) and u ∈ L

2(0, T ;V )∩L∞(0, T ;H). Therefore du

dt∈ L

2(0, T ;V )and u ∈ C([0, T ];H) and the above equation holds in L

2(0, T ).Now we take a general ϕ ∈ C

∞([0, T ]) with ϕ(T ) = 0 and ϕ(0) = 0. We multiplythe above equation by ϕ, integrate on (0, T ) and integrate by parts and we obtainby comparison with the relation (3.2.23) that

∀w ∈ V, (u(0), w)ϕ(0) = (u0, w)ϕ(0)

so thatu(0) = u0.

Therefore u is a weak solution to the Navier-Stokes equations and Theorem 3.2.3 isproved.

3.3 Complements in dimension N = 2

First of all we will give a result of dependence with respect to the datas.

Theorem 3.3.1 In dimension N = 2, let u1 be the weak solution to Navier-Stokesequations corresponding to the datas u

10 and f

1 and u2 be the weak solution corre-

sponding to the datas u20 and f

2. Then there exists a constant C = C(u2) > 0 suchthat

||u1 − u2||L2(0,T ;V ) + |u1 − u

2|L∞(0,T ;H) ≤ C(u2)(|u10 − u20|H + ||f1 − f

2||L2(0,T ;V )).

Proof.We have by difference

d

dt(u1 − u

2) +A(u1 − u2) +B(u1, u1)−B(u2, u2) = f

1 − f2,

(u1 − u2)(0) = u

10 − u

20.

Multiplying by u1 − u

2 we get

1

2

d

dt|u1(t)− u

2(t)|2H + ν||u1(t)− u2(t)||2V ≤ ||f1(t)− f

2(t)||V ||u1(t)− u2(t)||V

+| < B(u1, u1)(t)−B(u2, u2)(t), u1(t)− u2(t) > |

48

As we have already seen,

| < B(u1, u1)(t)−B(u2, u2)(t), u1(t)−u2(t) > | ≤ C|u1(t)−u

2(t)|H ||u1(t)−u2(t)||V ||u2(t)||V .

From this we obtain

d

dt|u1(t)− u

2(t)|2H + ν||u1(t)− u2(t)||2V ≤ 2

ν||f1(t)− f

2(t)||2V

+2C

ν||u2(t)||2V |u1(t)− u

2(t)|2H .

Now, as ||u2||2V∈ L

1(0, T ), using Gronwall inequality we get

maxt∈[0,T ]

|u1(t)− u2(t)|2H ≤ C(u2)(|u10 − u

20|2H + ||f1 − f

2||2L2(0,T ;V )).

Then we also have

ν

T

0||u1(t)− u

2(t)||2V dt ≤ C(u2)(|u10 − u20|2H + ||f1 − f

2||2L2(0,T ;V )).

This finishes the proof of Theorem 3.3.1.Next we give a regularity result again in dimension N = 2.

Theorem 3.3.2 In dimension N = 2, if f ∈ L2(0, T ;H) and u0 ∈ V , then the

weak solution (u, p) of Navier-Stokes equations satisfies

u ∈ L2(0, T ;H2(Ω)N ) ∩ L

∞(0, T ;V ),∂u

∂t∈ L

2(0, T ;H), p ∈ L2(0, T,H1(Ω)/IR).

Proof.We use again the special basis (wn) and we go back to the approximate problem setin the finite dimensional space V

m where we look for um such that

(du

m

dt+Au

m +B(um, um)− f, wi) = 0, i = 1, · · · ,m.

As Awi = λiwi, we see that Aum(t) ∈ Vm. So we have

(du

m

dt+Au

m +B(um, um)− f,Au

m) = 0.

This gives

ν

2

d

dt||um||2V + |Aum|2

L2(Ω)N ≤ |B(um, um)− f |L2(Ω)N |Aum|L2(Ω)N .

49

But if v ∈ H2(Ω)N ∩H

10 (Ω)

N we have

|vj∂vi

∂xj|L2(Ω) ≤ |vj |L4(Ω)|

∂vi

∂xj|L4(Ω).

As vj ∈ H10 (Ω) we have seen that

|vj |L4(Ω) ≤ C|vj |12L2(Ω)||vj ||

12H1(Ω).

It can also be shown (again in dimension N = 2) that when z ∈ H1(Ω)

|z|L4(Ω) ≤ C(Ω)|z|12L2(Ω)||z||

12H1(Ω).

In order to see that we have to use an extension of z to H1(IRN ) which is continuous

and also continuous from L2(Ω) to L

2(IRN ). We can then use this inequality forz = ∂vi

∂xj. We then have

|vj∂vi

∂xj|L2(Ω) ≤ |vj |L4(Ω)|

∂vi

∂xj|L4(Ω)

≤ C|vj |12L2(Ω)||vj ||

12H1(Ω)|

∂vi

∂xj|12L2(Ω)||

∂vi

∂xj||

12H1(Ω)

≤ |vj |12L2(Ω)||vj ||

12H1(Ω)||vi||

12H1(Ω)||vi||

12H2(Ω).

Then

|B(v, v)|L2(Ω)N ≤ C|w|12H||w||V |Aw|

12L2(Ω)N .

Using this inequality we have

ν

2

d

dt||um||2V + |Aum|2

L2(Ω)N ≤ C|Aum|32L2(Ω)N |u

m|12H||um||V + |f |L2(Ω)N |Aum|L2(Ω)N

≤ 1

2|Aum|2

L2(Ω)N + C|f |2L2(Ω)N + C|um|2H ||um||4V .

This implies

d

dt||um||2V + |Aum|2

L2(Ω)N ≤ C|f |2L2(Ω)N + C|um|2H ||um||2V ||um||2V

||um(0)||V ≤ C.

We already know that |um|2H||um||2

Vis bounded in L

1(0, T ). Then, from Gronwallinequality, we can say that um stays bounded in L

∞(0, T ;V ) and Aum stays bounded

in L2(0, T, L2(Ω)N ). As we already know that um converges to u, this implies

u ∈ L∞(0, T ;V ) and Au ∈ L

2(0, T ;L2(Ω)N ).

50

From this we have B(u, u) ∈ L4(0, T ;L2(Ω)N ). Using now the regularity result for

Stokes equation with right hand side f −B(u, u) we obtain

∂u

∂t∈ L

2(0, T ;H) and p ∈ L2(0, T ;H1(Ω)/IR).

This gives Theorem 3.3.2.

51

Chapter 4

Stationnary Navier-Stokesequations

4.1 Existence results and some cases of uniqueness

Stationnary Navier-Stokes equations can be written as follows.

−ν∆ui +N

j=1

uj∂ui

∂xj= fi −

∂p

∂xiin Ω, i = 1, · · · , N,(4.1.1)

divu = 0 in Ω,(4.1.2)

u = 0 on Γ.(4.1.3)

It is natural to set the problem in the following variational form. Let f ∈ H−1(Ω)N .

We look for u ∈ V such that

a(u,w) + b(u, u, w) =< f,w > ∀w ∈ V,(4.1.4)

u ∈ V.(4.1.5)

We will always here suppose that N ≤ 4.

Remark 4.1.1 As N ≤ 4, for every v ∈ V we have B(v, v) ∈ V (and B(v, v) ∈

H−1(Ω)N ) and

||B(v, v)||V ≤ C0||v||2Vor

||B(v, v)||H−1(Ω)N ≤ C0||v||2V .

Let us start with the case of small datas f (or of large viscosity ν).

52

Theorem 4.1.2 If f ∈ H−1(Ω)N and if ||f ||H−1(Ω)N <

ν2

C0, then there exists a

unique solution u of Navier-Stokes equations with

||u||V ≤ 1

ν||f ||H−1(Ω)N .

Proof.Let v ∈ V . Let us set

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

Then a(., .) is a bilionear continuous form on V × V and

a(w,w) = a(w,w) + b(v, w,w) ≥ ν||w||2V .

From Lax-Milgram theorem, for any F ∈ V (we take < F,w >=< f,w > ∀w ∈ V )

there exists a unique u such that

a(u,w) =< F,w > ∀w ∈ V,

u ∈ V.

Moreover we have

||u||V ≤ 1

ν||F ||V ≤ 1

ν||f ||H−1(Ω)N .

Let us setu = T (v, f).

The problem is then to show that T (., f) has a fixed point. We notice that for fixedf , v → T (v, f) maps the ball of V or radius 1

ν||f ||H−1(Ω)N into itself. Let us assume

that ||f ||H−1(Ω)N <ν2

C0and let us show then that T (., f) is a strict contraction. Let

us setu = T (v, f) u = T (v, f).

We then haveA(u− u) +B(v, u)−B(v, u) = 0

so that

ν||u− u||2V =< B(v, u)−B(v, u), u− u >

=< B(v − v, u), u− u >≤ C0||v − v||V ||u||V ||u− u||V .

Therefore

||u− u||V ≤ C0

ν||u||V ||v − v||V .

53

So T (., f) will be a strict contraction if C0||u||Vν

< 1 but we have

C0||u||Vν

≤C0||f ||H−1(Ω)N

ν2< 1

and therefore T (., f) has a unique fixed point in the ball of radius 1ν||f ||H−1(Ω)N

which is a solution to the Navier-Stokes equations.Let us now consider the general case N ≤ 4.

Theorem 4.1.3 For every f ∈ H−1(Ω)N , there exists u ∈ V solution of the Navier-

Stokes equations.In the general case there is no uniqueness result.

Let us consider the special basis (wn) of H and the finite dimensional space Vm

already defined. We first look for um ∈ Vm such that

a(um, wj) + b(um, um, wj) =< f,wj >, j = 1, · · · ,m.

In order to solve this problem we define Tm(zm) = vm where

a(vm, wj) + b(zm, vm, wj) =< f,wj >, j = 1, · · · ,m,

vm ∈ Vm.

This last problem has a unique solution and we have (as above)

||vm||V ≤||f ||H−1(Ω)N

ν.

So Tm maps this ball of V of radius||f ||

H−1(Ω)N

νinto itself and Tm is Lipschitz (see

above) so it is continuous. From Brouwer fixed point theorem (we are here workingin finite dimension), Tm has (at least) a fixed point which we call um which satisfies

a(um, wj) + b(um, um, wj) =< f,wj >, j = 1, · · · ,m.

As b(um, um, u

m) = 0 we see that

||um||V ≤||f ||H−1(Ω)N

ν.

Therefore, we can extract a subsequence, still denoted (um) such that

um

u in V weakly.

54

From the compactness of the embedding V ⊂ H we have

um → u in H strongly.

Now for every v ∈ V we have

b(um, um, v) = −b(um, v, u

m) = −N

i,j=1

Ωum

j

∂vi

∂xjum

i dx.

We know that umjum

i→ ujui in L

1(Ω) strongly and that umjum

iis bounded in L

2(Ω).Therefore we have

um

j um

i ujui in L2(Ω) weakly

and therefore, for every v ∈ V we have

b(um, um, v) = −b(um, v, u

m) → −b(u, v, u) = b(u, u, v).

Now, for fixed j we can pass to the limit in m → +nfty and obtain

a(u,wj) + b(u, u, wj) =< f,wj > ∀j,u ∈ V,

and u is a solution to Navier-Stokes equations.

4.2 Some particular flows

4.2.1 Couette flows

We consider a flow between two parallel planes z = 0 and z = a with velocity0 on the lower plane z = 0 and velocity vz = 0 and vx = v0 on the plane z = aand under the action of gravity.We have the solution given by

vz = 0, vx = vx(z)

with

ν∂2vx

∂z2= 0, −ρg =

∂p

∂z

which gives

vx = v0z

a, vz = 0, ∇xp = 0.

55

4.2.2 Poiseuille flows

In dimension N = 2 we have a flow between two parallel planes y = −a andy = a with no-slip boundary condition under the effect of a constant pressuregradient in the variable x, ∂p

∂x= −K.

We have the solutionvx = vx(y), vy = 0,

and

ν∂2vx

∂y2+K = 0, vx(±a) = 0

which gives

vx =K

ν(−y

2

2+

a2

2).

56

Bibliography

[1] G.K. Batchelor An Introduction to Fluid Dynamics, 2nd paperback ed. Cam-bridge Univ. Press, Cambridge, U.K., 1999.

[2] H. Brezis Analyse Fonctionnelle, Masson, Paris, 1983.

[3] L. Tartar Nonlinear partial differential equations using compactness method,M.R.C. Report, University of Wisconsin, 1977.

[4] R. Temam Navier-Stokes Equations, 2nd ed., North-Holland, Amsterdam, 1986.

[5] J.-L.Lions Quelques Methodes de Resolutions de Problemes aux Limites nonLineaires, Dunod, Paris, 1969.

57