Basis of Mathematical Methods in Fluid
Mechanics
Jean-Pierre Puel 1
November 30., 2011
1J.-P. Puel: (jppuel@bcamath.org) Ikerbasque and BCAM, Bizkaia Technology Park,
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) withT >0 is the time variable.
The so-called “state variables” or quantities which determine the flow are
• The fluid velocityu.
• The pressurep.
• The density of the fluidρ.
• Sometimes other quantitites like the temperatureθ, ...
Eulerian description.
u(x, t) is the velocity of the particle of fluid which is at positionxat time t. There-fore, ift′ 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 velocityv(x, t) in order to avoid confusions. If we call (x(t))t≥t0
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
dt(t) =u(x(t), t), x(t0) =x0.
Letϕbe a function of x andt. The particle derivative of ϕis
d
dtϕ(x0, t0) = d
dt(ϕ(x(t), t))/t=t0
= ∂ϕ
∂t(x0, t0) +
N
X
i=1
ui(x0, t0) ∂ϕ ∂xi
(x0, t0)
=∂ϕ ∂t +
N
X
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 the velocity 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 quantityϕ, which is a function of x and tisconserved if
∀V(t), d dt
Z
V(t)ϕ(x, t)dx= 0.
Therefore for everyV(t) andδt(small) we have
Z
V(t+δt)
ϕ(y, t+δt)dy=
Z
V(t)
V(t)
V(t+δt)
u x
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 inδt. We have
y(x, δt) =x+δt.u(x, t) +l.o.t.
so that
dyi=dxi+δt N
X
j=1 ∂ui
∂xj
(x, t)dxj+l.o.t.
Therefore
+δt
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
Another way to obtain this equation is the following. LetS(t) be the surface limiting the volumeV(t) and letνbe the outward pointing unit normal vector to this surface. We first have to show that
d dt
Z
V(t)
ϕ(x, t)dx=
Z
V(t) ∂ϕ
∂t(x, t)dx+
Z
S(t)
ϕ(x, t)u(x, t)ν(x)dS.
Then applying Stokes-Ostrogradskii formula we have
Z
S(t)
ϕ(x, t)u(x, t)ν(x)dS=
Z
V(t)
div (u(x, t).ϕ(x, t))dx,
which shows that
d dt
Z
V(t)ϕ(x, t)dx=
Z
V(t)
∂ϕ
∂t(x, t) + div (u(x, t).ϕ(x, t))
dx.
1.2.1 Conservation of mass
Here we take
ϕ(x, t) =ρ(x, t)
whereρ is the density of the fluid. We obtain the equation of conservation of mass which 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 remains constant during the flow. Therefore we here take
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 satisfiesρ0(x) =ρ0 = Cst (independent ofx), 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 each element stays constant during the flow (dtdρ= 0).
1.2.3 Stress tensor. Conservation of momentum
n
dS
F1
F2
The force exerted byF1 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 a scalar quantity at each point. This quantity is thehydrostatic pressure.
For a moving fluid, it appears tangential stresses : friction between the fluid layers gliding along each other, due to theviscosityof 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 byF1 onF2 is given by
σ.n= ( N
X
j=1
σijnj)i=1,···,N = (σijnj)i=1,···,N
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′ ,
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 that usually, a fluid at rest is compressed).
(σij′ ) is independent of translations and local rotations, and therefore independent ofu itself and ofωij = 12(∂x∂uij −∂u∂xji) (ω= curlu is the vorticity of the fluid). Therefore,σ′
ij only depends on the symmetric part of the tensor of velocity gradients
e= (eij) where
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
whereη is the shear viscosity andζ is the volumic velocity. For an incompressible fluid, we have divu= 0 so thatell= 0 and then
σij′ = 2ηeij.
Conservation of momentum.
LetV(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 ofV(t) exerted on the surface S(t), the fundamental law of dynamics takes the (vectorial) form
d
Here f is the volumic density of external forces per unit of mass. For example f can represent
• 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). • ....
From the equation expressing conservation of mass we have ∂ρ On the other hand, from Stokes-Ostrogradskii formula we have
Z
Therefore we obtain for every volume V(t)
Z
This gives us the equation for conservation of momentum
ρ∂ui
which can also be written as
1.3
Basic equations : Navier-Stokes, Euler, Stokes,....
If we neglect the spatial variations of viscosities ∂x∂η
j and
∂ζ
∂xj we obtain : •For a Newtonian viscous compressible fluid,
∂ρ
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 equa-tions
•For a perfect (inviscid) incompressible fluid we obtain the Euler equations
divu= 0,
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
wherep0 is the hydrostatic pressure (in absence of flow). We obtain
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
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 =u2.
Equilibrium between the stresses in each of the fluids and the stresses localized on the interface (nis the normal vector andτ are tangent vectors)
(σ1.n)τ = (σ2.n)τ (equality of tangential stresses),
(σ1.n)n−(σ2.n)n=γ(1 R +
1 R′),
whereγ 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 for the energy and the coupling will appear throughρ or σ in the fluid equation.
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 the flow does not vary with time. Therefore it corresponds to cancel all terms containing
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
−ν∆u=−∇p+f in Ω, (2.1.1)
divu= 0 in Ω, (2.1.2)
u= 0 on Γ, (2.1.3)
whereν >0 is the given viscosity andf is the given external force andu andp are the velocity and the pressure.
For each component ui,i= 1,· · ·, N we have
−ν∆ui=−
∂p ∂xi
+fi.
Letw= (w1,· · ·, wN) be a vector function which is “regular” and which is zero on the boundary, with divw = 0. If we multiply the equation for ui by wi, integrate on Ω and sum up fori= 1 to N we obtain
−ν
N
X
i=1
Z
Ω∆uiwidx=− N
X
i=1
Z
Ω ∂p ∂xi
widx+ N
X
i=1
Z
Ωfiwidx.
Asw/Γ= 0, by integration by parts we have
−
N
X
i=1
Z
Ω ∂p ∂xi
widx=
Z
Ω
and
Thereforeu must satisfy for allw regular, vanishing on Γ such that divw= 0,
ν
This suggests a variational formulation of Stokes system.
Precise mathematical formulation.
Let us define the space
V ={w∈H01(Ω)N, divw= 0} we look foru such that
a(u, w) = also solution to the minimization problem
J(u) = min w∈V J(w), (2.1.8)
where
First of all, as Ω is bounded, from Poincar´e inequality, we can equipH1
0(Ω) with the norm v→(RΩ|∇v|2dx)12 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)
Now the mapping w→ divw is linear continuous from H1
0(Ω)N toL2(Ω), and the spaceV is the kernel of this mapping. Therefore,V is a closed subspace ofH01(Ω)N and therefore, it is a Hilbert space for the norm (and the scalar product) induced by the one inH1
0(Ω)N.
It is now immediate to see that
(v, w)∈V ×V →a(v, w)
is a continuous bilinear form onV ×V which is obviously coercive when ν >0. On the other hand, whenf ∈H−1(Ω)N, the mappingw→PN
i=1 < fi, wi >is linear continuous fromV to IR.
We can then apply Lax-Milgram Theorem which shows the first part of Theorem 2.1.1.
Thatuis the solution to the minimization problem for functionalJ comes from the fact that the bilinear forma(., .) is symmetric.
Interpretation. Relation with Stokes problem.
What is the relation with Stokes problem and in particular what about the pressure pwhich seems to have disappeared?
Let us writeV the space
V ={ϕ∈ D(Ω)N, divϕ= 0},
whereD(Ω) =C0∞(Ω). Of courseV ⊂V and it can be shown thatV is dense inV. Using distribution theory we have forϕ∈ V
Therefore we obtain N
X
i=1
<−ν∆ui−fi, ϕi>D′
,D= 0 ∀ϕ∈ V.
The distribution T = (T1,· · ·, TN)∈ D′(Ω)N withTi=−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 which vanish onV are of the form T =−∇p.
So here we see that there existsp∈ D′(Ω) such that for i= 1,· · ·, N −ν∆ui=fi−
∂p ∂xi
in D′(Ω), divu= 0, in Ω,
u= 0 on Γ.
In fact we can see that for everyi= 1,· · ·, N
∂p ∂xi
=fi+ν∆ui ∈H−1(Ω).
Thereforep∈ D′(Ω) and ∇p∈H−1(Ω) and this implies that p∈L2(Ω). Of course pis 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 uniqueu∈V and a unique
p∈L2(Ω)/IR (p is unique up to the addition of a constant) such that
−ν∆ui=fi−
∂p ∂xi
in H−1(Ω)N, (2.1.10)
divu= 0, in Ω, (2.1.11)
u= 0 on Γ. (2.1.12)
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∈H1(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 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.
LetF,G,H be three Hilbert spaces and letA be a continuous linear operator from H toF. 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}. ThenN is also a closed linear subspace ofH.
Ifh ∈H, we can consider h1 = ProjKerAh 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 everyh∈H we have the decomposition
h=h1+h2, h1 ∈KerA, h2 ∈N.
Moreover, ifh∈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
which says that the projections are continuous. (It is clear here in the case of Hilbert spaces that the constantsC1 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 |A˜hn|F ≤ n1 (therefore|A˜hn|F →0 in F).
As|h˜n|H = 1 we can extract a subsequence, still denoted (˜hn) such that
˜
hn⇀ h0 in H weakly.
AsN is a closed subspace (then closed for the weak topology) and ˜hn∈N,h0 ∈N. AsAis linear continuous fromH toF, it is continuous for the weak topologies and therefore,
A˜hn⇀ Ah0 in F weakly.
But we already know thatA˜hn→0 in F strongly. Therefore Ah0 = 0 and h0 ∈N. This says thath0∈KerA∩N so that h0 = 0.
Now we have
˜
hn⇀0 in Hweakly and A˜hn→0 in F strongly.
AsB is compact we have
B˜hn→B0 = 0 in Gstrongly
so that|B˜hn|G →0. But from (2.2.13) we have
This gives a contradiction with the fact that |˜hn|H = 1. Therefore (2.2.14) is true. Let us now show that ImAis closed in F.
Let us take a sequence (yn) such that
yn∈ImA, yn→y in F.
Asyn∈ImA, there exists ¯hn∈H such that A¯hn=yn. We have the decompsition
¯
hn=hn+ ˆhn, hn∈N, hˆn∈KerA.
ThenAhn=yn and hn∈N.
Asyn→y inF, (yn) is a Cauchy sequence inF. 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 inH which is complete. This shows that there existsh∈H such thathn→h inH 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 thaty=Ahand y∈ImAso that ImAis 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 inH−1(Ω)N. Let us define the space
X(Ω) ={g∈H−1(Ω), ∇g∈H−1(ΩN}. (2.2.15)
Lemma 2.2.2
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
g∈L2(IRN) ⇐⇒ ˆg∈L2(IRN) and thatg∈H−1(IRN) ⇐⇒ (1+|ξ|2)−12ˆg∈L2(IRN).
Now
∇g∈H−1(IRN)N ⇐⇒ (1 +|ξ|2)|12|ξ|gˆ| ∈L2(IRN)
so that
g∈X(IRN) ⇐⇒ (1 +|ξ|2)(1 +|ξ|2)|ˆg|2∈L1(IRN) ⇐⇒ gˆ∈L2(IRN).
Lemma 2.2.3
X(IRN+) =L2(IRN+). (2.2.17)
Proof.
It is clear that L2(IRN+) ⊂ X(IRN+). Let us show that X(IRN+) ⊂L2(IRN+). To this end, we are going to exhibit a continuous extension fromX(IRN+) to X(IRN).
Let us first define a restriction operator from H1(IRN) to H01(IRN+). For u ∈ D(IRN) =C0∞(IRN) we set
Qu(x1,· · ·, xN) =
0, if x
N <0,
u(x1,· · ·, xN) +P2j=1aju(x1,· · ·, xN−1,−jxN) if xN >0. We impose that
1 + 2
X
j=1 aj = 0
which implies
Qu(x1,· · ·, xN−1,0) = 0. It is clear that for i= 1,· · ·, N −1 we have
Q(∂u ∂xi
) = ∂Qu ∂xi
.
Fori=N we have
Q( ∂u ∂xN
) = ∂ ∂xN
Ru
where
Ru(x1,· · ·, xN) =
0, if x
N <0,
u(x1,· · ·, xN) +P2j=1 aj
We also wantR to be continuous from H1(IRN) to H01(IRN+) which imposes
Thus we choose the coefficientsaj such that
1 +
and they are well defined by these relations. Now we extendQandR by continuity toH1(IRN) and we define It remains to prove thatP is an extension.
For u ∈ H1
0(IRN+) let us define by Eu the extension by 0 outside IRN+. It is well known thatu→Eu is linear continuous from H01(IRN+) to H1(IRN). 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
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
X
i=1
θi= 1.
We write
u= I
X
i=1 θiu.
In the caseθi∈C0∞(Ω), θiu can be extended by zero to have θiu∈X(IRN).
Whenθi ∈C∞(Ω), we can find aC2 diffeomorphismηsuch thatθiu◦η−1 ∈X(IRN). Thenθiu◦η−1∈L2(IRN) which impliesθiu∈L2(Ω).
Now we have, algebraically, X(Ω) =L2(Ω). Let us take onX(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 thatX(Ω) is a Hilbert space.
Lemma 2.2.5 There exists a constant C >0 such that
∀g∈L2(Ω), |g|L22(Ω) ≤C|g|2X(Ω).
(2.2.19)
Proof.
Wheng∈L2(Ω) we have
|g|2H−1(Ω) ≤C|g|2L2(Ω)
and
|∇g|2H−1(Ω)N ≤C|g|2L2(Ω).
Let us consider the identity map fromL2(Ω) toX(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 constantC >0 such that
This gives Lemma 2.2.5. Now, as Ω is bounded,H1
0(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
Ker∇⊥={g∈L2(Ω),
Z
Ωgdx= 0}.
Lemma 2.2.7 If(pn)is a sequence inL2(Ω)such thatRΩpndx= 0and|∇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 > 0 independent ofnsuch 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 thatf =−∇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)}.
ThenY is a closed subspace ofH−1(ΩN. Let us show that
Actually, if w ∈ H01(Ω)N and < ∇p, w >= 0 ∀p ∈ L2(Ω) we have < w,∇ϕ >= 0 ∀ϕ∈C∞
0 (Ω) so that <divw, ϕ >= 0∀ϕ∈C0∞(Ω) and therefore divw= 0 so that w∈V.
On the other hand we have
∀p∈L2(Ω), ∀w∈V, <∇p, w >= 0. ThereforeV =Y⊥ which implies
V⊥= (Y⊥)⊥=Y =Y.
Then iff ∈H−1(Ω)N is such that ∀w∈V < f, w >= 0 thenf ∈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 inter-pretation of Stokes problem.
This result can also be proved by another method. Forǫ >0 let us consider the problem
a(uǫ, w) + 1 ǫ
Z
Ωdivuǫdivwdx= N
X
i=1
< fi, wi>, ∀w∈H01(Ω)N,
uǫ∈H01(Ω)N.
This problem has a unique solution uǫ which satisfies
−ν∆uǫ− ∇( 1
ǫdivuǫ) =f uǫ∈H01(Ω)N.
If we write
pǫ=− 1 ǫdivuǫ then
pǫ∈L2(Ω),
Z
Ω
pǫdx= 0.
On the other hand we have
a(uǫ, uǫ) + 1 ǫ
Z
Ω|
divuǫ|2dx= N
X
i=1
so that there exists a constantM independent ofǫsuch that
||uǫ||H1
0(Ω)N ≤M, |
1 √
ǫdivuǫ|L2(Ω)≤M.
Then, after extraction of a subsequence, we have
uǫ ⇀ u in H01(Ω)N weakly,
divuǫ→0 in L2(Ω) strongly.
Then divu= 0 so that u∈V and for everyw∈V (divw= 0) we have
a(uǫ, w) = N
X
i=1
< fi, wi >
so that
a(u, w) = N
X
i=1
< fi, wi >
and uis solution of the variational form of Stokes problem. Nowpǫ satisfies
∇pǫ =f+ν∆uǫ,
Z
Ω
pǫdx= 0.
Then pǫ is bounded in H−1(Ω)N and satisfiesRΩpǫdx= 0 so that pǫ is bounded in
L2(Ω). After extraction of a subsequence,
pǫ ⇀ p in L2(Ω) weakly
and we have
∀w∈H01(Ω)N, a(u, w) = N
X
i=1
< fi, wi >+
Z
Ω
pdivwdx
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 inF. 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 =H1
Lemma 2.2.9
W = (ImA)⊥⊕L⊥. Proof.
Ifu∈(ImA)⊥∩L⊥, then
∀g∈H, < u, Ag >= 0, ∀l∈L, < u, l >= 0.
As∀h∈W′ we have the decomposition h=Ag+lwithg∈H and l∈L, we have ∀h∈W′, < u, h >= 0.
Let nowu∈W. Let us show that there existsv∈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 everyl∈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 thatu−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 have
Proof.
Ifu∈KertA, we have ∀g∈H, < u, Ag >= 0 and thenu∈ImA⊥.
If u∈ ImA⊥, we have ∀g ∈H, < u, Ag >= 0 and therefore (tAu, g)
H = 0 so that tAu= 0 and u∈KertA.
Now if g∈ImtA, there exists u∈W such that g=t Au. We have for f ∈ KerA, (g, f)H = (tAu, f)H =< u, Af >= 0, so thatg ∈KerA⊥. This shows that ImtA⊂ KerA⊥ and as KerA⊥ is closed, we have
ImtA⊂KerA⊥.
Let us suppose that ImtA6= KerA⊥. Then there existsg∈H′=H and there exists f0∈KerA⊥ such that
∀f ∈ImtA, (g, f)H = 0, and (g, f0)H = 1.
Then for everyu ∈W, (g,tAu)H = 0 so that < u, Ag >= 0 and therefore Ag = 0 inW′. 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 ImtAsuch that gn→g inH. We have gn=t
Au˜n 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⊥ and
vn∈ImA⊥= KertA. Then tAu˜n=tAun so that gn=tAun withun∈L⊥.
If h ∈W′ we have h =Ag+l with g∈ KerA⊥ =N and l ∈L. (the fact that we can takeg∈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 everyh∈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. Now tAu
Lemma 2.2.12 For every f ∈KerA⊥, there existsu∈L⊥ such that
tAu=f and
||u||W ≤C|f|H.
Proof.
We haveW = ImA⊥⊕L⊥ = KertA⊕L⊥. Now it is immediate to see thattA:L⊥→ H is injective. But as ImtA = KerA⊥ we see that the mapping tA :L⊥ →KerA⊥ is one-to-one (bijective). On the other hand we have|tAu|
H ≤C||u||W which says thattA is continuous.
Now l⊥ and KerA⊥ are Banach spaces as they are closed subspaces of Banach spaces. Then from Banach Theorem, tA : 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) = {g∈L2(Ω), RΩgdx= 0}, there exists u∈H01(Ω)N such that
divu=g
and
||u||H1
0(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⊥=L2
0(Ω).
Ifg∈L20(Ω), there existsu∈L⊥⊂W such that tAu=−g and ||u||W ≤C|g|H. Thenu∈H01(Ω)N and for everyh∈H we have
(−g, h)H = (tAu, h)H =< u, Ah >=< u,∇h > .
Takingh∈C0∞(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∈H01(Ω)N, and ||u||H1
0(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
Proof.
We recall thatV ={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. Ifu∈H01(Ωn)N and divu= 0, we can extenduby 0 outside Ωn. If (ρǫ) is a regularizing sequence (mollifiers), forǫsmall 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, divu= 0, < f/Ωn, u >= 0.
Then there existspn∈L2(Ωn) such thatf/Ωn =∇pn.
But pn+1−pn is constant on Ωn and we can assume that this constant is 0. Then
f =∇p wherep∈L2loc(Ω).
Now if Ω is starshaped (for example with respect to 0), for 0 < θ < 1 we have θΩ⊂Ω. If u∈H1
0(Ω)N, divu= 0, thenuθ defined by uθ(x) =u(xθ) has a compact support in Ω and divuθ = 0. Then we can approximate uθ by ρǫ∗uθ ∈ C0∞(Ω)N with div (ρǫ∗uθ) = 0 so that
< f, u >= lim
θ→1limǫ→0< f, ρǫ∗uθ>= 0. Then, in this case, there existsp∈L2(Ω) such thatf =∇p.
In the general case, every point of Γ has a connected neighborhoodωwhich is regular and strictly starshaped. Then there existsq ∈L2(ω) such that f
/ω =∇q. Butp/ω−q is constant onω. Thereforep/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 inV.
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. ThereforeV ⊃V, but it is clear thatV ⊂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, divv = 0, v.n= 0 onΓ}.
Proof. Let us define
X={v∈L2(Ω)N, divv∈L2(Ω)}.
Proof.
Let v ∈ X. For every ψ ∈ H12(Γ) and ϕ ∈ H1(Ω) such that ϕ/Γ = ψ (we take a
continuous extension), we define
L(ψ) =
Z
Ωdivvϕdx+
Z
Ωv.∇ϕdx.
Ifϕ∈H01(Ω), then the right hand side is 0 so thatL depends only onψ. 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
functionsv 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 ofV 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∈ H1(Ω). Therefore, for everyv∈H, we have
< f, v >=<∇p, v >=
Z
Ω
pdivvdx+< v.n, p >Γ= 0.
Then< f, v >= 0 for everyv∈H and this shows that the closure of V containsH 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 andp 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)
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
H ={w∈L2(Ω)N, divw= 0, w.n= 0} (2.3.24)
wherenis 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 inH which is theL2 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 the mappingT :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 thatT is selfadjoint.
Therefore, (see for example [2]), there exists a decreasing sequence (µn)n of strictly positive real numbers withµn→0 whenn→+∞and a sequence (wn)nof elements ofV which form an orthonormal basis inH such that
T wn=µnwn.
Setting nowλn= µ1n 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 producta(v, w), we can see that (√wn
λn)n
form an orthonormal basis inV. Letv∈H. Then we have
v= +X∞
n=1
(v, wn)wn with +X∞
n=1
Ifv∈V we have
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 pressurep disappear and we can write the problem as
We obtain the following result
Let us write
uM0 = M
X
n=1
(u0, wn)wn
and
fM(t) = M
X
n=1
< f(t), wn> wn.
We know thatuM
0 converges tou0 inH andfM converges tof inL2(0, T;V′) when M →+∞. First of all, callingVM =span{w1,· · ·, wM}, we look for
uM(t) = M
X
n=1
un(t)wn
such that
d dt(u
M(t), w) +a(uM(t), w) =< fM(t), w > on (0, T), ∀w∈VM,
uM(0) =uM0 .
This gives us for everyn= 1,· · ·, M
d
dtun(t) +λnun(t) =< f(t), wn>, un(0) = (u0, wn),
and we even have an explicit formula forun
un(t) = (u0, wn)e−λnt+
Z t
0 < fn(s), wn> e
−λn(t−s)ds.
Let us show that (uM) is a Cauchy sequence inC([0, T;H) and inL2(0, T;V). For M ≥P + 1 we have
(uM −uP)(t) = M
X
n=P+1
un(t)wn,
|(uM −uP)(t)|2H = M
X
n=P+1
|un(t)|2,
||uM −uP||2L2(0,T;V) = Z T
0 M
X
n=P+1
From the equation givingun we obtain converges to a functionu in these spaces which are complete. We can also write
u(t) = +X∞
n=1
un(t)wn.
The regularity of dudt comes immediately from the equation forun for example. Taking test functionswin∪MVM first then in its closure which isV it is now easy to see that usatisfies
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 equationw=wn and seeing that (u(t), wn) satisfies the same equation asun(t).
Interpretaton of this equation and the relation with Stokes problem are more diffi-cult. We need to consider
U(t) =
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 timet. Therefore, there existsP(t)∈L2(Ω) such that
AsF(·)−u(·) +u0 is continuous in time, we clearly haveP ∈C([0, T];L2(Ω)/IR). Now taking the time derivative we obtain withp= dtdP
∂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∂t we have
Z
Ω| ∂u ∂t(t)|
2dx+ν 2
d dt
Z
Ω|∇
u(t)|2dx=
Z
Ω
f(t)∂u
∂t(t)dx≤ |f(t)|H| ∂u ∂t(t)|H.
This gives an estimate on ∂u∂t in L2(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 inL2(0, T;H), which gives (from the regularity result for the stationnary problem)v∈L2(0, T, H2(Ω)N) and p∈L2(0, T;H1(Ω)
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
∂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
X
i,j=1
Z
Ω ∂vi
∂xj
.∂wi ∂xj
dx=
Z
Ω∇
v.∇wdx, ∀v, w ∈V, (3.1.5)
b(u, v, w) = N
X
i,j=1
Z
Ω uj
∂vi
∂xj
widx, ∀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 ×H1(Ω)N. Moreover we have
In particular we have
which says that it is continuous. Now we have
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 dimensionN = 2 and in dimensionN = 3.
Lemma 3.1.2 In dimension N = 2, there exists a constant C >0 such that
Proof.
Of course it is sufficient to prove the above inequalities for ϕ∈ C∞
0 (IR2). We can
In the same way we can write (exchanging the roles of x1 and x2)
|ϕ(x1, x2)|2 ≤2
Multiplying these two inequalities we obtain
|ϕ(x1, x2)|4 ≤4
In the same way we have
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,H1
0(Ω)N is linear continuous on
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 Athe operator defined by
∀u, v∈V, a(u, v) =< Au, v > .
Then it is immediate to see that
A∈ L(V, V′). Now forN ≤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 fromV ×V inV′ and we have
||B(u, v)||V′ ≤C||u||V||v||V.
Whenu∈L2(0, T;V) and v∈L2(0, T;V) it is clear thatB(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||H1 0(Ω)N.
Then
||B(u, v)||H−1(Ω)N ≤C|u|L4(Ω)|v|L4(Ω) ≤C|u| 1 2
L2(Ω)||u|| 1 2
V|v|
1 2
L2(Ω)||v|| 1 2
V.
This implies
Z T
0 ||
B(u, v)||2H−1
(Ω)Ndt≤C||u||L∞(0,T;H)||v||L∞(0,T;H)||u||L2(0,T;V)||v||L2(0,T;V),
Lemma 3.1.4 In dimension N = 3, whenu, v∈L2(0, T;V)∩L∞(0, T;H)we have
It is now natural to formulate Navier-Stokes equations in the following way. We look foru∈L2(0, T;V)∩L∞(0, T;H) such that (with f ∈L2(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 in time 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, ifu∈L2(0, T;V)∩L∞(0, T;H) and ifu
satisfies (3.1.11) withf ∈L2(0, T;V′), then
u∈C([0, T], H), du dt ∈L
and we can impose initial condition (3.1.12) with u0 ∈H. These solutions will be
called 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, whenf ∈L2(0, T;H−1(Ω)N), there exists a pressurep∈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.
Ifu∈L2(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 side f−B(u, u)∈L2(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 dimensionN = 3 is one of the major open problems nowadays in mathematics. Of course we will not discuss this question here but we will give the uniqueness result in dimensionN = 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 satisfyingu∈L2(0, T;V)∩L∞(0, T;H).
Proof.
Let us suppose that we have two solutions u1 and u2 in the admissible class. We then have
d dt(u
1
−u2) +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|u
1−u2|2
But we have
(B(u1, u1)−B(u2, u2), u1−u2) = (B(u1−u2, u1), u1−u2) + (B(u2, u1−u2), u1−u2)
and
(B(u2, u1−u2), u1−u2) =b(u2, u1−u2, u1−u2) = 0.
Then
|(B(u1, u1)−B(u2, u2), u1−u2)| ≤C|u1−u2|L4(Ω)||u1||V|u1−u2|L4(Ω)
≤C||u1||V|u1−u2|H||u1−u2||V.
We then obtain 1 2
d dt|u
1−u2|2
H+ν||u1−u2||2V ≤C||u1||V|u1−u2|H||u1−u2||V
≤ν||u1−u2||2V +C 2
4ν||u 1||2
V|u1−u2|H,
|(u1−u2)(0)|H = 0.
Then
1 2
d dt|u
1−u2|2 H ≤
C2 4ν||u
1||2
V|u1−u2|H,
|(u1−u2)(0)|H = 0,
and as u1 ∈L2(0, T;V), ||u1||2V ∈L1(0, T).
Lemma 3.2.2 (Gronwall inequality.) If ϕ≥0 satisfies
dϕ
dt(t)≤θ(t)ϕ(t) on (0, T)
withθ∈L1(0, T), then we have
∀t∈[0, T], ϕ(t)≤ϕ(0)e(
Rt
0θ(s)ds).
Proof. we have
d dt
ϕ(t)e−(
Rt
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−u2)(t)|H = 0.
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 theorem for weak solutions is still true in dimensionN = 3 and the proof requires the use of some interpolation spaces.
First step. The proof uses the Galerkin method with a special basis. Let us consider the special basis (wn) of eigenfunctions of operator A as described in the previous chapter. Then (wn) is orthonormal in H and orthogonal in V. Let us first consider the following “approximate problem” . First of all we define the finite dimensional space
Vm = span{w1,· · ·, wm}.
Then we look for a functionum defined on (0, T) with values inVmsuch that almost everywhere intand for everyj = 1,· · ·, m
d dt(u
m(t), w
j) +a(um(t), wj) +b(um(t), um(t), wj) =< f(t), wj >, (3.2.19)
um(0) =um0 , (3.2.20)
where
um0 = m
X
j=1
(u0, wj)wj ∈Vm
and
um0 →u0 in H when m→+∞.
Then
um(t) = m
X
i=1
gim(t)wi,
and, as the (wi) are orthonormal in H and orhtogonal in V, the system can be written 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
X
i=1
gim(t)wi, m
X
k=1
Using Cauchy-Lipschitz theorem, it is easy to show that thisnonlinear system of ordinary differential equations has a unique solution locally in time and therefore on an interval of time [0, Tm]⊂[0, T].
Second step. A priori estimates (I).
We will argue indifferently on the (gjm) or onum as it gives the same thing. Mul-tiplying equation forum by g
jm and summing up inj we get for everym and we obtain
The question is now : how to prove thatg=B(u, u)?
Third step. A priori estimates (II).
Let us show that dudtm stays bounded inL2(0, T, V′). We have
We now use the following compactness lemma.
Lemma 3.2.4 Let V be compactly embedded in H and let (um) be a sequence of
As ∂wi in this space. But we know that
um→u in L2(0, T;H) strongly
Fourth step. Passage to the limit.
Now if we choose firstϕ∈C0∞(]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[),
withf ∈L2(0, T;V′) andu∈L2(0, T;V)∩L∞(0, T;H). Therefore du
dt ∈L2(0, T;V′) and u∈C([0, T];H) and the above equation holds in L2(0, T).
Now we take a general ϕ∈C∞([0, T]) with ϕ(T) = 0 and ϕ(0) 6= 0. We multiply the above equation byϕ, integrate on (0, T) and integrate by parts and we obtain by comparison with the relation (3.2.23) that
∀w∈V, (u(0), w)ϕ(0) = (u0, w)ϕ(0)
so that
u(0) =u0.
Thereforeuis a weak solution to the Navier-Stokes equations and Theorem 3.2.3 is proved.
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-Stokes equations corresponding to the datas u10 and f1 and u2 be the weak solution corre-sponding to the datas u2
0 and f2. Then there exists a constantC =C(u2)>0 such
that
||u1−u2||L2(0,T;V)+|u1−u2|L∞(0,T;H) ≤C(u2)(|u1
0−u20|H +||f1−f2||L2(0,T;V′)).
Proof.
We have by difference
d dt(u
1−u2) +A(u1−u2) +B(u1, u1)−B(u2, u2) =f1−f2,
(u1−u2)(0) =u10−u20.
Multiplying by u1−u2 we get
1 2
d dt|u
1(t)−u2(t)|2
H+ν||u1(t)−u2(t)||2V ≤ ||f1(t)−f2(t)||V′||u1(t)−u2(t)||V
As we have already seen,
|< B(u1, u1)(t)−B(u2, u2)(t), u1(t)−u2(t)>| ≤C|u1(t)−u2(t)|H||u1(t)−u2(t)||V||u2(t)||V.
From this we obtain
d dt|u
1(t)
−u2(t)|2H +ν||u1(t)−u2(t)||2V ≤ 2 ν||f
1(t)
−f2(t)||2V′
+2C ν ||u
2(t)||2
V|u1(t)−u2(t)|2H.
Now, as ||u2||2V ∈L1(0, T), using Gronwall inequality we get
max t∈[0,T]|u
1(t)
−u2(t)|2H ≤C(u2)(|u10−u20|2H +||f1−f2||2L2(0,T;V′
)).
Then we also have
ν
Z T
0 ||
u1(t)−u2(t)||2Vdt≤C(u2)(|u10−u20|2H +||f1−f2||2L2(0,T;V′)).
This finishes the proof of Theorem 3.3.1.
Next we give a regularity result again in dimensionN = 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 set in the finite dimensional space Vm where we look for um such that
(du m
dt +Au
m+B(um, um)−f, w
i) = 0, i= 1,· · ·, m.
AsAwi=λiwi, we see thatAum(t)∈Vm. So we have
(du m
dt +Au
m+B(um, um)−f, Aum) = 0.
This gives
ν 2
d dt||u
m||2
But ifv∈H2(Ω)N ∩H01(Ω)N we have
Using this inequality we have
From this we have B(u, u) ∈L4(0, T;L2(Ω)N). Using now the regularity result for Stokes equation with right hand sidef −B(u, u) we obtain
∂u ∂t ∈L
2(0, T;H) and p
∈L2(0, T;H1(Ω)/IR).
Chapter 4
Stationnary Navier-Stokes
equations
4.1
Existence results and some cases of uniqueness
Stationnary Navier-Stokes equations can be written as follows.
−ν∆ui+ N
X
j=1 uj
∂ui
∂xj
=fi−
∂p ∂xi
in 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. Letf ∈H−1(Ω)N. We look foru∈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||2
V
or
||B(v, v)||H−1(Ω)N ≤C0||v||2V.
Theorem 4.1.2 If f ∈ H−1(Ω)N and if ||f||H−1(Ω)N < ν
2
C0, then there exists a
unique solutionu of Navier-Stokes equations with
||u||V ≤ 1
ν||f||H−1(Ω)N.
Proof.
Letv∈V. Let us set ˜
a(u, w) =a(u, w) +b(v, u, w).
Then ˜a(., .) is a bilionear continuous form onV ×V and
˜
a(w, w) =a(w, w) +b(v, w, w)≥ν||w||2V.
From Lax-Milgram theorem, for anyF ∈V′ (we take< F, w >=< f, w > ∀w∈V) there exists a unique usuch 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 set
u=T(v, f).
The problem is then to show thatT(., f) has a fixed point. We notice that for fixed f,v→T(v, f) maps the ball ofV or radius ν1||f||H−1
(Ω)N into itself. Let us assume
that||f||H−1
(Ω)N < ν
2
C0 and let us show then that T(., f) is a strict contraction. Let
us set
u=T(v, f) uˆ=T(ˆv, f).
We then have
A(u−uˆ) +B(v, u)−B(ˆv,uˆ) = 0
so that
ν||u−uˆ||2V =< B(ˆv,uˆ)−B(v, u), u−u >ˆ
=< B(ˆv−v, u), u−u >ˆ ≤C0||vˆ−v||V||u||V||u−uˆ||V.
Therefore
||u−uˆ||V ≤ C0
SoT(., f) will be a strict contraction if C0||u||V
ν <1 but we have
C0||u||V
ν ≤
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 caseN ≤4.
Theorem 4.1.3 For everyf ∈H−1(Ω)N, there existsu∈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 defineTm(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 ofV of radius
||f||H−1(Ω)N
ν into itself andTm is Lipschitz (see above) so it is continuous. From Brouwer fixed point theorem (we are here working in finite dimension),Tm has (at least) a fixed point which we callum which satisfies
a(um, wj) +b(um, um, wj) =< f, wj >, j= 1,· · ·, m.
Asb(um, um, um) = 0 we see that
||um||V ≤ ||
f||H−1(Ω)N
ν .
Therefore, we can extract a subsequence, still denoted (um) such that
From the compactness of the embeddingV ⊂H we have
um →u in H strongly.
Now for everyv∈V we have
b(um, um, v) =−b(um, v, um) =− N
X
i,j=1
Z
Ω umj ∂vi
∂xj
umi dx.
We know thatumj umi →ujuiinL1(Ω) strongly and thatumj umi is bounded inL2(Ω). Therefore we have
umj umi ⇀ ujui in L2(Ω) weakly
and therefore, for everyv∈V we have
b(um, um, v) =−b(um, v, um)→ −b(u, v, u) =b(u, u, v).
Now, for fixedj we can pass to the limit inm→+nf tyand obtain
a(u, wj) +b(u, u, wj) =< f, wj > ∀j,
u∈V,
and uis 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 velocity 0 on the lower plane{z= 0} and velocityvz = 0 and vx=v0 on the plane{z=a} and under the action of gravity.
We have the solution given by
vz = 0, vx=vx(z)
with
ν∂ 2v
x
∂z2 = 0, −ρg = ∂p ∂z which gives
vx=v0 z
4.2.2 Poiseuille flows
In dimension N = 2 we have a flow between two parallel planes {y = −a} and {y = a} with no-slip boundary condition under the effect of a constant pressure gradient in the variable x, ∂p∂x =−K.
We have the solution
vx =vx(y), vy = 0,
and
ν∂ 2v
x
∂y2 +K = 0, vx(±a) = 0 which gives
vx = K
ν (− y2
2 + a2
Bibliography
[1] G.K. Batchelor An Introduction to Fluid Dynamics, 2nd paperback ed. Cam-bridge Univ. Press, CamCam-bridge, 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. TemamNavier-Stokes Equations, 2nd ed., North-Holland, Amsterdam, 1986.
