• Aucun résultat trouvé

Navier-Stokes Equations With Nonhomogeneous Boundary Conditions in a Convex Bi-Dimensional Domain

N/A
N/A
Protected

Academic year: 2022

Partager "Navier-Stokes Equations With Nonhomogeneous Boundary Conditions in a Convex Bi-Dimensional Domain"

Copied!
29
0
0

Texte intégral

(1)

Navier–Stokes equations with nonhomogeneous boundary conditions in a convex bi-dimensional domain

Vincent Girinon

Institut de Mathématiques, UMR CNRS 5219, Université Paul Sabatier, 31062 Toulouse Cedex 9, France Received 7 July 2008; accepted 22 December 2008

Available online 21 June 2009

Abstract

We prove the existence of a weak solution to Navier–Stokes equations describing the isentropic flow of a gas in a convex and bounded region,ΩR2, with nonhomogeneous Dirichlet boundary conditions on∂Ω. These results are also extended to flow domain surrounding an obstacle.

©2009 Elsevier Masson SAS. All rights reserved.

MSC:35Q30; 35A05

Keywords:Compressible Navier–Stokes equations; Inflow–outflow boundary conditions; Global existence of weak solutions

1. Introduction

1.1. Mathematical model

In the case of the isentropic flow of a gas in an open setΩR2, Navier–Stokes equations can be written in the form:

tρ+div(ρu)=0, (1.1)

t(ρu)+div(ρu⊗u)μu+μ)∇divu+ ∇p=ρf, (1.2)

where the densityρ=ρ(t,x), the velocityu=u(t,x)=(u1(t,x), u2(t,x))and the pressurep=p(t,x)are functions of the timet(0, T )and the spatial coordinatexΩ(T ∈R+). Moreover, the pressure only depends on the density according to the relation:

p=γ (a >0), (1.3)

where the adiabatic constantγsatisfies the condition:γ >1.

The viscosity coefficientsμandλare such thatμ >0 andλ+μ0.

Finally,fL×(0, T ))corresponds to external force density acting on the fluid.

E-mail address:girinon.vincent@neuf.fr.

0294-1449/$ – see front matter ©2009 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2008.12.007

(2)

The initial conditions for the density and the velocity are ρ(0,x)=ρ0(x) inΩ,

u(0,x)=u0(x) inΩ, (1.4)

with

ρ0Lγ(Ω), ρ00 a.e. onΩ, u0L2(Ω),

ρ0u0L2(Ω). (1.5)

The system is completed by the following boundary conditions

⎧⎨

ρ(t,x)=ρ(t,x) on

t(0,T )

{t} ×Γet, u(t,x)=a(t,x) on(0, T )×∂Ω,

(1.6) where

Γet:=

x∂Ω: a(t,x)·n(x) <0

, (1.7)

n(x)is the unit outward normal vector atx∂Ω, with aC1 [0, T] ×R2

and ρL

t(0,T )

{t} ×Γet

. (1.8)

1.2. Context

The first general proof giving the existence of a solution(ρ,u)to the system (1.1), (1.2), (1.4), (1.6) was obtained by P.-L. Lions [6,7] (in three-dimensional space) for homogeneous boundary conditions:a(t,x)=0on(0, T )×∂Ω (and soΓet= ∅). This approach was improved by E. Feireisl (see [2–4]) who released the constraint concerning the adiabatic exponentγin Lions’ work.

The case of nonhomogeneous boundary conditions was studied by S. Novo [8] in the particular case of an open set ΩR3defined byΩ=B(x0, R0)\S, whereB(x0, R0)is the open ball of centerx0and radiusR0>0, and where S is a compact set included inB(x0,R20)satisfying the cone property. Novo considers the case of constant boundary conditions on∂B(x0, R0):aR3\ {0},ρ>0 (Γet is then independent of time and corresponds to a half-sphere included in∂B(x0, R0)).

1.3. Our purpose

The goal of this article is to prove the existence of a solution(ρ,u)to the system (1.1), (1.2), (1.4), (1.6) under the following hypotheses:

(H1) Description ofΩ.

The problem is studied in an open convex and bounded setΩ, included inR2, of type Ω=g1 ]−∞,1[

,

whereg:R2Ris convex andC1onR2. We also suppose that0Ω.

(H2) Description of the incoming border area.

Γet is independent oft and is the intersection of∂Ω with a cone of vertex0. In the sequel, we simply writeΓe

rather thanΓet.

Notations.More precisely,Γeis defined by

Γe=∂ΩCθ12, (1.9)

whereθ1,θ2are real numbers such that 0θ1< θ2<2π, and where Cθ12:= rcos(θ ), rsin(θ )

, (r, θ )R+× ]θ1, θ2[

. (1.10)

(3)

Fig. 1.

At any pointx of∂Ω, letn(x)be the unit outward normal vector toΩ andτ (x)be the unit tangent vector (image ofn(x)under rotation of angle+π2). Letx1andx2be the points of∂Ω belonging to the boundary ofCθ12. These points are of type:

x1= |x1|cos(θ1),|x1|sin(θ1)

and x2= |x2|cos(θ2),|x2|sin(θ2)

. (1.11)

Then, forx∈ {x1,x2}andt∈ [0, T], the vectora(t,x)is colinear toτ (x). In addition, we suppose that:

(H3) Sufficient condition for no reflux.

(See Fig. 1.) For allt∈ [0, T],

a(t,x1)= −a(t,x1)τ (x1) and a(t,x2)= +a(t,x2)τ (x2).

Remark 1.1.The boundary condition on density is adataof our problem: the flow inΩ depends on this data but the opposite is false. From this point of view, we can understand assumption(H3)as a way to express the “independence”

ofΓefrom the flow inΩ(intuitively,(H3)forbids fluid’s particles which leaveΩto follow its boundary untilΓeand next to go back inΩ). In our work, hypothesis(H3)appears in the proof of Lemma 2.5. This lemma gives the essential argument which enables us to construct an interesting extension of the initial density in order to recoverρ=ρon (0, T )×Γe(see Lemma 2.7 and Section 4.1).

In accordance with the terminology of [9, p. 413], we calla bounded energy renormalized weak solutionof system (1.1), (1.2), (1.4), (1.6), any pair(ρ,u)such that:

(i) Regularity and initial conditions.

⎧⎪

⎪⎪

⎪⎪

⎪⎩

ρL 0, T;Lγ(Ω)

and ρ0 a.e. onΩ×(0, T ), uL2 0, T;W1,2(Ω)

withu(t,x)=a(t,x)on∂Ω×(0, T ), ρC [0, T];Lγw(Ω)

and ρuC [0, T];L

γ+1

w (Ω) with

ρ(0)=ρ0, (ρu)(0)=ρ0u0.

(1.12)

(ii) Equations’ interpretation. The continuity equation (1.1) and the momentum equation (1.2) hold in D×(0, T )).

(iii) Boundary conditions for density. For any functionηD(R2×(0, T ))such thatη=0 onQs, where Qs:=

t(0,T )

x∂Ω: a(t,x)·n(x) >0

× {t}, (1.13)

we have

(4)

T 0

Ω

(ρ∂tη+ρu· ∇η) dxdt= T 0

Γe

ρa·nη dSdt. (1.14)

Moreover, for any functionbC(R+)C1(R+)such that b(s)csλ0 ifs∈ ]0,1],

b(s)csλ1 ifs∈ [1,+∞[, (1.15)

with

c >0, λ0∈ ]−1,+∞[, λ1

−∞ 2 −1

, (1.16)

we have T 0

Ω

b(ρ)∂tη+b(ρ)u· ∇η

T

0

Ω

ρb(ρ)b(ρ)

div= T 0

Γe

b(ρ)a·nη. (1.17)

(iv) Energy inequality. For almost everyt(0, T ),

E(t )+μ t 0

Ω

(uu)2

dxds++μ) t 0

Ω

div(u−u)2

dxds

E0+ t 0

Ω

ρf·(uu) dxdst 0

Ω

ρ∂tu·(uu) dxdsμ t 0

Ω

u: ∇(uu) dxds

+μ) t 0

Ω

divudiv(u−u) dxdst 0

Ω

(ρu)· (uu)· ∇ u

dxds

a t 0

Ω

ργdivudxdst 0

Γe

a

γ−1ργ u·ndSds, (1.18)

with

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

E(t ):=1 2

Ω

ρ(t )(uu)(t )2dx+

Ω

a

γ−1ρ(t )γdx, E0:=1

2

Ω

ρ0 u0u(0)2

dx+

Ω

a

γ−1ρ0γdx,

whereuis a vector field defined on[0, T] ×R2that is equal toaon[0, T] ×∂Ω (it will be introduced in Section 2.3).

We are ready to state the main result of this paper:

Theorem 1.1.For initial conditionsρ0andu0which fulfil(1.5), for boundary conditionsaandρwhich fulfil(1.8) and under hypotheses(H1),(H2),(H3), there exists a bounded energy renormalized weak solution to the problem (1.1),(1.2),(1.4),(1.6).

(5)

1.4. Sketch of the proof

For proving Theorem 1.1, we follow a method similar to the one used by S. Novo in [8]. The strategy consists in extending the domain of study (we work on an open setD which containsΩ) and, consequently, in extending the data (initial conditions, densityf) in order to come back to a problem with Dirichlet homogeneous condition for the velocity (on the boundary ofD). Then the problem is solved by the method of Lions and Feireisl. It consists in working with a “perturbed version” of Eqs. (1.1) and (1.2) on which Novo added a penalization’s term. More precisely, the system is:

tρ+div(ρu)=ερ,

t(ρu)+div(ρu⊗u)μu+μ)∇divu+ ∇ γ+δρβ

+ερ· ∇u+E(uu)=ρf.

(1.19) Here,ε >0,δ >0,m >0 andβ >max{6, γ}.χE is the characteristic function of the setE:=D\Ω. The equations are studied onD×(0, T )and we consider the following boundary conditions:

u(t,x)=0 and nρ(t,x)=0 on∂D×(0, T ). (1.20)

The system is completed by suitable initial conditions:

ρ(0,x)=ρ0,ε,δ(x) and u(0,x)=u0,ε,δ(x) inD, (1.21)

whereρ0andu0 are extensions ofρ0 andu0(u0,ε,δ,ρ0,ε,δ denote some “regularized versions” ofu0 andρ0). We extend simplyu0byu(0)outsideΩ. The extension ofρ0is more complex and depends onρin order to take the boundary conditions on density (1.6) into account (see Section 4.1). Let us present the main points of the proof of Theorem 1.1.

Step A: Construction of approximate solutions.

By a Faedo–Galerkin method, we prove that, for anynN, there exists an approximate solutionn,un)to the problem (1.19)–(1.21). The integernis relative to the dimension of the vector space which belongs the vector field un(t,·)fort∈ [0, T]. By adding the diffusive termερ to the continuity equation, we obtain a parabolic equation which “regularizes” the density. Among other things, this regularization allows calculations which provide an energy estimate for the approximate solutions. Indeed, fort∈ [0, T], we have

E[ρn,un](t )+ε t 0

D

δβρnβ2|∇ρn|2+μ t 0

D

|∇un|2++μ) t 0

D

|divun|2

E[ρn,un](0)+ t

0

D

ρnf·um t 0

E

(unu)·un, (1.22)

whereE[ρn,un](t ):=12

Dρnu2n(t )+

D(γa1ρnγ(t )+βδ1ρnβ(t )).

Step B: Passage to the limitn→ +∞.

Choosing a suitable subsequence (still indexed byn), we prove that{n,un)}converges to a pairm,um)which is an “exact solution” of (1.19)–(1.21). Actually, the “perturbed continuity equation” is satisfied in the strong sense onD×(0, T ), whereas the momentum equation holds inD(D×(0, T )). At last, replacingρnbyρm,unbyumand E[ρn,un](0)by 12

Dρ0,ε,δu0,ε,δ2 +

D(γa1ρ0,ε,δγ +βδ1ρ0,ε,δβ ), the energy inequality (1.22) holds a.e. on(0, T ).

Step C: Passage to the limitm→ +∞.

This step consists in getting rid of the penalization termE(uu)in order to “force” the velocity to take boundary conditions into account (recall thatucoincides withaon∂Ω). More precisely, we want to prove that an extracted sequence of{m,um)}converges to a pairε,uε)such thatuε=uonE×(0, T ). The first difficulty comes from the energy inequality (1.22) satisfied byρmandum: it does not permit to obtain estimates independent ofmbecause of the term−mt

0

E(umu)·um. It is solved by working with a new estimate deduced from (1.22) and from two new integral relations established by using some suitable test functions (depending onu) in Eqs. (1.19).

(6)

Step D: Passage to the limitε→0.

Here, we prove that{ε,uε)}converges to a couple of functionsδ,uδ). In this step, the difficulty comes from the lack of suitable estimates on the sequence {ρε}: whereas the strong convergence of the density was deduced from results of parabolic regularity in steps B and C, it comes here from properties of the effective viscous flux εγ+δρεβ+μ)divuεdiscovered by Lions. Remark that the passage to the limit in the energy inequality is also a problem in view of the lack of information about{ρε}. In our case, this matter will be solved thanks to the specific construction of the vector fieldu(see Section 3.5).

At the end of step C, we haveuε=uonE×(0, T ), so we can deduce thatuδ=uonE×(0, T ). This last result is used in order to “recover” the boundary condition for the density.

Step E: Passage to the limitδ→0.

The purpose of this final step is to show that the sequence{δ,uδ)}converges to a couple(ρ,u)which is a bounded energy renormalized weak solution of system (1.1), (1.2), (1.4), (1.6). This part uses ideas developed by E. Feireisl, mainly the essential result about the oscillations amplitude of the density, that is

sup

k>0

lim sup

δ0

Tkδ)Tk(ρ)

Lγ+1×(0,T ))

C, (1.23)

whereC >0 is a constant (independent ofδandk) andTk (k∈N) a “cut-off” function defined by Tk(s):=kT

s k

, where

TC(R), T is increasing and concave onR,

T (s)=s fors1, T (s)=2 fors3. (1.24)

This result enables us to end the proof of Theorem 1.1.

1.5. Outline of this paper

In Section 2, we specify the “extended domain”D, the extensions of the data as well as the vector fieldu. We do not detail the steps A, B and C because they can be found in [8] or in [9]. Sections 3 and 4 are respectively dedicated to steps D and E. Finally, Section 5 deals with other particular configurations as the presence of an internal obstacle or the case of a rectangular domain. In these cases, we indicate the modifications in the proof to obtain the existence of a weak solution.

2. Departure model

2.1. Extension of domain

According to hypothesis (H1), there existsR0>0 such that

ΩB(0, R0), (2.1)

whereB(0, R0)denotes the open disk of center0and radiusR0. We define

D:=B(0, R0+3). (2.2)

2.2. Extension of data

We extendfby0outsideΩ (and we still notefthis extension) so that:fL(D×(0, T )).

The regularity of the initial conditions plays an important role in the proof. For clarity, we will specify in the beginning of each step with what type of initial conditions we work.

– So, in the statement of Theorem 3.1,u0andρ0denote the initial conditions for the “extended model” (they are defined onD).

– In Section 3.3, where we start the description of the step D, initial conditions are still defined onD.

– Finally, in Section 4,u0andρ0will designate the initial conditions defined by (1.5). Then we specify the suitable extensions onDthat permit to prove Theorem 1.1.

(7)

2.3. Vector fieldu

The vector field u, which appears in the energy inequality (1.18), plays an essential role in our proof: it is constructed in order to coincide withaon(0, T )×∂Ω and to obtain Lemmas 2.2 and 2.5. Indeed:

– Lemma 2.2 will be used in order to permit the passage to the limit in energy inequality at steps D and E (see Proposition 3.1),

– Lemma 2.5 conducts to Lemma 2.6 which enables to define an interesting extension of the initial density ρ0. Thanks to this extension, we will be able to take the integral formulas (1.14) and (1.17) into account (see Sec- tion 4.1).

2.3.1. An auxiliary function

Lemma 2.1.Under hypothesis(H1):

(i) SetΩc=R2\Ω. Then, for everyxΩc,g(x)·x1−g(0) >0.

(ii) There exists a mappingk:R2\ {0} →R+of classC1onR2\ {0}such that, for anyxR2\ {0},g(k(x)x)=1.

(iii) The mappingkis(−1)-homogeneous.

Proof. (i) LetxΩcandϕx:R+Rbe the function defined byϕx(t ):=g(tx). The functionϕxisC1and convex onR+, thus:ϕx(1)ϕx(1)ϕx(0), that is,g(x)·xg(x)g(0). SincexΩc,0Ω, we haveg(x)1 and g(0) <1. Therefore∇g(x)·x1−g(0) >0 and the proof of (i) is complete.

(ii) LetxR2\ {0}and let us prove that the equation(Ex): ϕx(t )=1 admits a unique solution inR+.

Existence. On the one hand, sinceΩ is bounded, there existst0>0 such thatt0x, henceϕx(t0)1. On the other hand, due to(H1), we haveϕx(0) <1. Thus, by continuity ofϕx, there existstx>0 such thatϕx(tx)=1.

Uniqueness. Assume the existence of two solutions of(Ex), tx andsx such that 0< sx< tx. Thus, there exists λ(0,1)such thatsx=λtx. But, due to the convexity ofϕx,

ϕx(sx)λϕx(tx)+(1λ)ϕx(0)=λ+(1λ)g(0) <1, which gives a contradiction.

Conclusion. For allxR2\ {0}, there exists a unique realk(x) >0 such thatg(k(x)x)=1.

Regularity of mappingk. Consider theC1-functionG:R+×R2Rdefined byG(t,x)=g(tx)−1. Let(t0,x0)R+×(R2\ {0})such thatG(t0,x0)=0 (thust0=k(x0)). We can write∂tG(t0,x0)=t10g(t0x0)·t0x0 and, since t0x0Ωc, due to (i), we havetG(t0,x0) >0. According to the Implicit Mapping Theorem,kisC1on a neighborhood ofx0.

(iii) LetxR2\ {0}andλ >0. Sinceϕλx(t )=ϕx(λt )for alltR+, we deduce thatλk(λx)is a solution of(Ex).

Due to the uniqueness of the solution, we haveλk(λx)=k(x). Consequently,kis(−1)-homogeneous. 2 Remark 2.1.

– ForxR2\ {0},k(x)is the ratio (strictly positive) of the homothety of center0that transformsxinto some point of∂Ωand k(x)1 is the gauge of the convexΩ atx.

– Settingk(0)= +∞, we haveΩ= {x∈R2: k(x) >1}and∂Ω= {x∈R2: k(x)=1}.

2.3.2. Definition ofu

Letkb∈ ]1,32[and chooser0>0 such that B(0, r0)Ω\Ωb whereΩb:=

xR2: 1< k(x) < kb

. (2.3)

For(t,x)∈ [0, T] ×R2, we set u(t,x)=

a t, k(x)x +C

1 k(x)−1

x

ψ(x), (2.4)

(8)

whereCis a strictly positive real constant andψis aC-function onR2such that 0ψ(x)1 and ψ(x)=

1 ifr0|x|R0+2,

0 if|x|r20 or|x|R0+3. (2.5)

Due to the regularity ofa(see (1.8)) andk, we can claim thatuisC1on[0, T] ×R2and has a compact support inD. We complete these properties.

Lemma 2.2.We can chooseCsuch that:

(i) For(t,x)∈ [0, T] ×Ωcsatisfying|x|R0+2,divu(t,x)0.

(ii) For(t,x)∈ [0, T] ×Ωb,divu(t,x)0.

(iii) For(t,x)∈ [0, T] ×Ωcsuch thatR0+1|x|R0+2,u(t,x)·x0.

Proof. According to Lemma 2.1(iii), we know that∇k(x)·x= −k(x). Consequently, for(t,x)∈ [0, T] ×R2satis- fyingr0|x|R0+2, we have

divu(t,x)=divA(t,x)+C

−∇k(x)·x k(x)2 +

2 k(x)−2

=divA(t,x)+C 3

k(x)−2

, whereA(t,x):=a(t, k(x)x).

IfxΩcΩb, thenk(x)∈ ]0, kb[. Thus we have divu(t,x)divA(t,x)+C(k3

b −2). SetC0:= {xR2: r0|x|R0+2}. ChoosingCsuch that

C kb

3−2kbdivAL((0,T )×C0), (2.6)

then we get (i) and (ii).

On the other hand, for(t,x)∈ [0, T] ×R2such thatr0|x|R0+2, u(t,x)·x=A(t,x)·x+C

1 k(x)−1

|x|2C 1

k(x)−1

|x|2A(t,x)|x|. Butk(x)xΩB(0, R0)and, consequently,k(x)1 |Rx0|. Hence

u(t,x)·x|x|

C

|x|

R0 −1

|x| −A(t,x) .

Moreover, if we haveR0+1|x|R0+2, then:u(t,x)·x|x|(C− |A(t,x)|).

Therefore, if in addition to condition (2.6), we chooseCso that C|A|

L((0,T )×C0), (2.7)

we obtain (iii). 2 2.4. Invariance

SinceuisC1with compact support in[0, T]×R2, it is globally Lipschitz on[0, T]×R2. Thus, for any(t0,x0)∈ [0, T] ×R2, the Cauchy problem:

dy

ds(s)=u s,y(s)

and y(t0)=x0,

admits a unique solutionY(·;t0,x0)on[0, T]. Furthermore, the mapping

Y:[0, T] × [0, T] ×R2R2, (s, t,x)Y(s;t,x), (2.8) isC1on[0, T] × [0, T] ×R2.

(9)

2.4.1. Generalities

The results of this section are inspired from [1, pp. 211–220] but we repeat them under a form adapted to our purpose.

Definition 2.1.A subsetMofR2is callednegatively invariant with respect toYif, and only if, for everyxM, for everyt∈ [0, T],Y(s;t,x)Mas soon ass∈ [0, t].

Here is a sufficient condition so that an open region is negatively invariant.

Lemma 2.3. Let mN and let f1, . . . , fm:R2R be C1-functions on R2. Consider the open set O = j=m

j=1 fj1(]−∞,0[)and suppose that, for everyx∂O, u(t,x)=0 for allt∈ [0, T], or

there existsj∈ {1, . . . , m}such thatxfj1({0})andfj(x)·u(t,x) >0for allt∈ [0, T]. ThenOis negatively invariant.

The following result will permit us to prove that the mappingGeintroduced in Lemma 2.7 is a diffeomorphism.

Lemma 2.4. Let mN and let f1, . . . , fm:R2R be C1-functions on R2. Assume the closed set F = m

j=1fj1(]−∞,0])is negatively invariant. If we set Γ :=

x∂F: ∃j ∈ {1, . . . , m}fj(x)=0andt∈ [0, T],fj(x)·u(t,x) >0 , then, for allxΓ and fors, t∈ [0, T]such thats > t,Y(s;t,x) /F.

2.4.2. A negatively invariant region Lemma 2.5.We define the open setDeby

De:=

xCθ12D: g(x) >1

. (2.9)

ThenDeis negatively invariant with respect toY.

Proof. Step 1. We prove thatM=Cθ12Ωcis negatively invariant with respect toY.

First of all, we define an open setOby O=f11 ]−∞,0[

f21 ]−∞,0[

f31 ]−∞,0[

, (2.10)

where the functionsf1,f2andf3are defined onR2by

f1(x)= −x·x1, f2(x)=x·x2, f3(x)=1−g(x). (2.11) (vis the image ofvunder the rotation of angle+π2 andx1,x2are still given by (1.11).)

Forε >0, letuεbe the vector field defined on[0, T] ×R2by uε(t,x)=u(t,x)εx+ε2

x2

|x2|− x1

|x1|

. (2.12)

This vector field isC1on[0, T] ×R2and globally Lipschitz on[0, T] ×R2. Thus, the associated flowYε:[0, T] × [0, T] ×R2R2,(s, t,x)Yε(s;t,x), is alsoC1.

Forx∂Of11({0})andt∈ [0, T], we have

f1(x)·uε(t,x)= −x1 ·a(t,x1(x)ε2

|x2|x1·x2+ε2|x1|,

= −x1 ·a(t,x1(x)+ε2|x1| 1−cos(θ2θ1)

. (2.13)

(10)

But, in view of(H3),a(t,x1)= −|a(t,x1)|τ (x1)= −|a(t,x1)|n(x1), thus

−x1 ·a(t,x1)=a(t,x1)x1 ·n(x1)=a(t,x1)x1·n(x1)=|a(t,x1)|

|∇g(x1)| x1· ∇g(x1).

According to Lemma 2.1(i),x1· ∇g(x1) >0, therefore−x1 ·a(t,x1)0. Furthermore, sinceψ0 andθ2θ1∈ ]0,2π[, we deduce from (2.13):

f1(x)·uε(t,x) >0. (2.14)

An analogous calculation shows that, forx∂Of21({0})andt∈ [0, T],

f2(x)·uε(t,x) >0. (2.15)

Now, for everyx∂Of31({0})andt∈ [0, T],

f3(x)·uε(t,x)= −∇g(x)·a(t,x)+εg(x)·xε2g(x)· x2

|x2|− x1

|x1|

−∇g(x)·a(t,x)+εg(x)·x−2ε2g(x).

But ∇g(x) · a(t,x) 0 (indeed xΓe) and ∇g(x)· x 1 −g(0) (according to Lemma 2.1(i)), thus:

f3(x)·uε(t,x)ε(1g(0))−2ε2|∇g|L(∂Ω). Consequently, forε >0 small enough,

f3(x)·uε(t,x) >0. (2.16)

Due to (2.14)–(2.16) and Lemma 2.3, O is negatively invariant with respect to Yε (for ε >0 small enough).

Thus, for all xO and for every s, t ∈ [0, T] such that st, Yε(s;t,x)O. Since uε uniformly converges touon[0, T] ×R2, according to theorems dealing with differential equations with parameters, we can claim that Yε(s;t,x)−−−→ε0 Y(s;t,x). We deduce that, for allxO,

Y(s;t,x)O (0stT ). (2.17)

SinceM=Oand due to the continuity ofY, the above result is still valid forxM. This provesMis negatively invariant with respect toY.

Step 2. We show thatDeis negatively invariant with respect toY.

Sinceuis equal to zero on∂D× [0, T], it is obvious thatDis negatively invariant with respect toY. Conse- quently,De=MDis negatively invariant (as intersection of two negatively invariant sets). 2

2.4.3. A diffeomorphism

Let us start with some classical properties of the flowYassociated withu. Lemma 2.6.

(i) The mappingY:[0, T]×[0, T]×R2R2,(s, t,x)Y(s;t,x)isC1and the following partial derivatives exist and are continuous on[0, T] × [0, T] ×R2:

2Y

∂s∂t = 2Y

∂t ∂s and, fori=1,2, 2Y

∂s∂xi = 2Y

∂xi∂s.

(ii) The mappingX:[0, T] ×R2R2,(t,x)X(t,x):=Y(t;0,x)isC1on[0, T] ×R2and, for allt ∈ [0, T], X(t,·)is aC1-diffeomorphism onR2onto itself whose inverse mapping isX(t,·)1:R2R2,xY(0;t,x).

Moreover, for allxR2,

det ∇xX(t,x)

=exp t

0

divu s,X(s,x) ds

.

(11)

Now, we are able to prove the following result:

Lemma 2.7.The mappingGe:(0, T )×ΓeR2,(t,x)Y(0;t,x)is aC1-diffeomorphism from(0, T )×Γeonto an open setGeincluded inDe.

Remark 2.2.Geis the set of all the initial positions of fluid’s particles which, under the action of the flow generated byu, crossΓeat one time.

Proof of Lemma 2.7. Remark thatΓeadmits aC1parametrization:

Φe:]θ1, θ2[ →R2, θΦe(θ )=k m(θ )

m(θ ), (2.18)

wherem(θ ):=(cos(θ ),sin(θ )).

Φeis injective and regular on]θ1, θ2[(for allθ∈ ]θ1, θ2[,Φe(θ )=0). Indeed, a quick calculation yields for every θ∈ ]θ1, θ2[,Φe(θ )= −(k(m(θ ))). Now, consider the mapping

Ge:(0, T )×1, θ2)R2, (t, θ )Ge t, Φe(θ ) .

Step 1. We prove thatGeis aC1local diffeomorphism from(0, T )×1, θ2)onto an open setGeDe.

TheC1regularity ofGeresults immediately from Lemma 2.6(i) and from the regularity ofΦe. Furthermore, for all(t, θ )(0, T )×1, θ2),

JacGe(t, θ )=det ∂Y

∂t 0;t, Φe(θ )

,xY 0;t, Φe(θ ) Φe(θ )

.

But, observing that for every(t, ξ )∈ [0, T] ×R2, we haveY(0;t,X(t, ξ ))=ξ and differentiating this relation with respect tot, one obtains

∂Y

∂t 0;t,X(t, ξ )

+ ∇xY 0;t,X(t, ξ )

tX(t, ξ )=0.

Since tX(t, ξ )= u(t,X(t, ξ )), we have ∂Y∂t(0;t,X(t, ξ )) = −∇xY(0;t,X(t, ξ ))u(t,X(t, ξ )). When ξ de- scribesR2,x:=X(t, ξ )describes alsoR2. Thus, for all(t,x)∈ [0, T] ×R2,

∂Y

∂t (0;t,x)= −∇xY(0;t,x)u(t,x).

Consequently, thanks to Lemma 2.6(ii), for(t, θ )(0, T )×1, θ2), we get JacGe(t, θ )=det

xY 0;t, Φe(θ )

×det

Φe(θ ),u t, Φe(θ )

= det[Φe(θ ),u(t, Φe(θ ))] exp(t

0divu(s,X(s, Φe(θ ))) ds).

But det[Φe(θ ),u(t, Φe(θ ))] =Φe(θ )·u(t, Φe(θ ))so, for all(t, θ )(0, T )×1, θ2), det

Φe(θ ),u t, Φe(θ )

= −Φe(θ )n Φe(θ )

·u t, Φe(θ )

>0, and therefore JacGe(t, θ )=0.

Due to the Inverse Mapping Theorem, Ge is a C1 local diffeomorphism from (0, T )×1, θ2) onto Ge :=

Ge((0, T )×1, θ2)). SinceΓe is included inDe which is negatively invariant relatively toY, we can claim than Geis an open set included inDe, henceGeDe.

Step 2. We prove thatGeis a(global)diffeomorphism from(0, T )×ΓeontoGe. At this point, it is sufficient to show thatGeis injective.

Let(t,x)and(t,x)(0, T )×Γe such that Ge(t,x)=Ge(t,x)=y0. Therefore, the functionsY(·;t,x)and Y(·;t,x)coincide with the solutionyof the following Cauchy problem:

dy

ds(s)=u s,y(s)

on[0, T], y(0)=y0.

(12)

Suppose thatt=t and, for example, let us deal with the caset < t. Sincey(t )=xΓe, according to Lemma 2.4 applied toDe, for alls∈ ]t, T],y(s) /De. Thus, in particular,y(t) /Γe, that isxewhich is absurd. Consequently, we havet=t, and thusx=y(t )=y(t)=x. 2

3. Step D: Passage to the limit onε

3.1. Conclusion of step C

The details about steps A, B and C, can be found in [8] or [9, pp. 415–418]. The following theorem sums up the results of step C.

Theorem 3.1.For anyε >0,δ >0and for every couple(ρ0,u0)such that u0L2(D),

ρ0W1,(D)W2,2(D) with infDρ0>0 and nρ0(x)=0 on∂D, (3.1) there exists a pair of functions(ρ,u)such that

⎧⎪

⎪⎪

⎪⎪

⎪⎩

ρL 0, T;Lβ(D)

and ρ0 a.e. onD×(0, T ), uL2 0, T;W1,20 (D)

withu=uonE×(0, T ), ρC [0, T];Lβ(D)

and ρuC [0, T];L

β+1

w (Ω) with

ρ(0)=ρ0, (ρu)(0)=ρ0u0. These functions satisfy in the strong sense

tρ+div(ρu)=ερ inD×(0, T ),

nρ(t,x)=0 on∂D×(0, T ), ρ(0,x)=ρ0(x) inD,

and in the sense of distributions inΩ×(0, T ):

t(ρu)+div(ρu⊗u)μu+μ)∇divu+ ∇ γ+δρβ

+ε(ρ· ∇)u=ρf.

Moreover, setting

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

Pδ(ρ):= a

γ−1ργ + δ β−1ρβ, E(t ):=

D

1

2ρ(t ) u(t )u(t )2

+Pδ ρ(t ) dx,

E0:=

D

1

2ρ0 u0u(0)2

+Pδ0)

dx, we have for almost everyt(0, T ),

E(t )+ t 0

D

μ

(uu)2

++μ)

div(u−u)2 dxds

+ε t 0

D

δβρβ2|∇ρ|2dxds+ t 0

D

γ+δρβ

divudx

E0+ t 0

D

ρf·(uu) dxdst 0

D

ρ∂tu·(uu) dxds

(13)

μ t 0

D

u: ∇(uu) dxds+μ) t 0

D

divudiv(u−u) dxds

+ε t 0

D

(ρ· ∇)(uu)

·udxdst 0

D

ρu· (uu)· ∇ u

dxds. (3.2)

3.2. Aim of step D

The purpose of this section is to establish the following result:

Theorem 3.2.For anyδ >0and for every couple(ρ0,u0)such that

⎧⎨

u0L2(D) withu0=u(0)onE,

ρ0Lβ(D) withρ00onD and ρ0|DeC(De),

ρ0u0L2(D),

(3.3) there exists a pair(ρ,u)such that

⎧⎪

⎪⎪

⎪⎪

⎪⎩

ρL 0, T;Lβ(D)

and ρ0 a.e. onD×(0, T ), uL2 0, T;W1,20 (D)

and u=u a.e. onE×(0, T ), ρC [0, T];Lβw(D)

and ρuC [0, T];L

β+1

w (Ω) with

ρ(0)=ρ0, (ρu)(0)=ρ0u0.

(3.4)

Moreover, the pair(ρ,u)satisfies the continuity equation in the sense of distributions inD×(0, T )and the momentum equation is satisfied in the sense of distributions inΩ×(0, T ):

tρ+div(ρu)=0,

t(ρu)+div(ρu⊗u)μu+μ)∇divu+ ∇ γ+δρβ

=ρf. (3.5)

These functions also have the following properties ρC [0, T];Lp(D)

(1p < β), ρLβ+1 0, T;Lβloc+1(Ω) , ρuL2 0, T;Lm(D)

(1m < β), ρ|u|2L2 0, T;Lr(D) 1r <β+1

. (3.6)

Moreover, for any function ηD(R2×(0, T )) such that η|Qs =0 (where Qs is defined by (1.13)), and for any functionbC(R+)C1(R+)satisfying(1.15)with

c >0, λ0∈ ]−1,+∞[, λ1

−∞ 2 −1

, (3.7)

we have T 0

Ω

b(ρ)∂tη+b(ρ)u· ∇η dxdt

T 0

Ω

ρb(ρ)b(ρ)

divuη dxdt

= T 0

Γe

b

ρ0(Y(0;t,x)) J (t,x)

η(t,x)u(t,x)·n(x) dS(x) dt, (3.8) whereJ (t,x)is defined by

J (t,x)=exp t

0

divu s,Y(s;t,x) ds

. (3.9)

Références

Documents relatifs

However in all these approaches [12, 9, 29, 30], the pair (A, C) is completely detectable, and the feedback control laws determined in [6, 12], in the case of an internal control,

In section 3, we study the regularity of solutions to the Stokes equations by using the lifting method in the case of regular data, and by the transposition method in the case of

The proof is based on the weighted estimates of both pressure and kinetic energy for the approximate system which result in some higher integrability of the density, and the method

We also study the existence of weak solutions to the three-dimensional instationary Navier–Stokes equations for more regular data, but without any smallness assumption on the

L’accès aux archives de la revue « Annali della Scuola Normale Superiore di Pisa, Classe di Scienze » ( http://www.sns.it/it/edizioni/riviste/annaliscienze/ ) implique l’accord

— We proved in Section 2 of [7] a local existence and uniqueness theorem (Theorem 2.2) for solutions of the Navier- Stokes equations with initial data in a space 7-^1 ^2,&lt;5s ^

We present in this note the existence and uniqueness results for the Stokes and Navier-Stokes equations which model the laminar flow of an incompressible fluid inside a

[12] Marius Mitrea and Sylvie Monniaux, On the analyticity of the semigroup generated by the Stokes operator with Neumann-type boundary conditions on Lipschitz subdomains of