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Hydrostatic Stokes equations with non-smooth data for mixed boundary conditions

Équations de Stokes hydrostatiques avec conditions aux limites mixtes à données peu régulières

F. Guillén-González

a

, M.A. Rodríguez-Bellido

b

, M.A. Rojas-Medar

c

aDpto. de Ecuaciones Diferenciales y Análisis Numérico, Universidad de Sevilla, Aptdo. 1160, 41080 Sevilla, Spain bDpto. de Matemática Aplicada I, E. T. S. de Arquitectura, Universidad de Sevilla, Avda. Reina Mercedes, 2, 41012 Sevilla, Spain

cDpto. de Matemática Aplicada, IMECC-UNICAMP, C. P. 6065, 13083-970, Campinas-SP, Brazil Received 30 June 2003; received in revised form 10 October 2003; accepted 21 November 2003

Available online 8 June 2004

Abstract

The main subject of this work is to study the concept of very weak solution for the hydrostatic Stokes system with mixed boundary conditions (non-smooth Neumann conditions on the rigid surface and homogeneous Dirichlet conditions elsewhere on the boundary). In the Stokes framework, this concept has been studied by Conca [Rev. Mat. Apl. 10 (1989)] imposing non-smooth Dirichlet boundary conditions.

In this paper, we introduce the dual problem that turns out to be a hydrostatic Stokes system with non-free divergence condition. First, we obtain strong regularity for this dual problem (which can be viewed as a generalisation of the regularity results for the hydrostatic Stokes system with free divergence condition obtained by Ziane [Appl. Anal. 58 (1995)]). Afterwards, we prove existence and uniqueness of very weak solution for the (primal) problem.

As a consequence of this result, the existence of strong solution for the non-stationary hydrostatic Navier-Stokes equations is proved, weakening the hypothesis over the time derivative of the wind stress tensor imposed by Guillén-González, Masmoudi and Rodríguez-Bellido [Differential Integral Equations 50 (2001)].

2004 Elsevier SAS. All rights reserved.

Résumé

Le but principal de ce travail est d’étudier le concept de solution très faible pour le système de Stokes hydrostatique avec conditions aux limites mixtes (condition de Neumann non régulière sur la surface rigide et condition de Dirichlet homogène dans le reste). Dans le cas du problème de Stokes, ce sujet a été étudié par Conca [Rev. Mat. Apl. 10 (1989)] en imposant une condition aux limites de Dirichlet non homogène et peu régulière.

E-mail addresses: guillen@us.es (F. Guillén-González), angeles@us.es (M.A. Rodríguez-Bellido), marko@ime.unicamp.br (M.A. Rojas-Medar).

0294-1449/$ – see front matter 2004 Elsevier SAS. All rights reserved.

doi:10.1016/j.anihpc.2003.11.002

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Dans ce papier, on introduit le problème dual qui est aussi un système de Stokes hydrostatique mais avec une condition de divergence non nulle. D’abord, on obtient la régularité forte pour le problème dual (ce résultat peut être consideré comme une généralisation des résultats de régularité pour le système de Stokes hydrostatique avec la condition de divergence nulle, obtenus par Ziane [Appl. Anal. 58 (1995)]). On montre ensuite l’existence et unicité de solution très faible pour le problème primal.

Comme conséquence de ce résultat, on montre l’existence de solution forte pour le problème de Navier-Stokes hydrostatique non-stationnaire, avec une hypothèse sur la dérivée par rapport au temps du tenseur du vent plus faible que celle qui était imposée par Guillén-González, Masmoudi et Rodríguez-Bellido [Differential Integral Equations 50 (2001)].

2004 Elsevier SAS. All rights reserved.

MSC: 35A05; 35Q35; 35B40; 35B65; 76D03

Keywords: Hydrostatic Stokes equations; Mixed boundary conditions; Non-smooth boundary data; Transposition method

Mots-clés : Équations de Stokes hydrostatiques ; Conditions aux limites mixtes ; Données aux limites non régulières ; Méthode de transposition

1. Introduction

The hydrostatic Navier–Stokes problem (also called Primitive Equations) is a model that appears in Geophysical fluid dynamics, in order to describe the general circulation in the Ocean or the Atmosphere [14]. This model has been extensively studied from a mathematical point of view by several authors [12,13,3,2,10,8,9], who in particular have established existence and uniqueness results for the stationary and non-stationary problems.

Let us recall that the Primitive Equations are variants of the Navier–Stokes system, where some simplifications have been made based on the analysis of physical scales (because the domain of study have a depth scale negligible in comparison to horizontal scales). Concretely, rigid-lid hypothesis and hydrostatic pressure are imposed [10].

These simplifications reduce the dimension of the system from a numerical point of view. However, it does not make any easier its mathematical analysis. For instance, this system is no longer parabolic for the vertical velocity, which depends upon derivatives for the horizontal velocity, loosing basically an order of regularity.

As far as we know, all the results concerning strong solutions for the Primitive Equations are based on Ziane’s results for the hydrostatic Stokes problem [16].

In [9], Guillén-González, Masmoudi and Rodríguez-Bellido used Ziane’s results in order to obtain existence of strong solutions (global in time for small data or local in time for any data) for the non-stationary hydrostatic Navier–Stokes problem. This fact forced to impose some regularity hypothesis on the data (more precisely, on the time derivative for the Neumann boundary condition) that we consider as not optimal.

In [6], Conca defines the very weak solution concept for the Stokes problem, analysing what kind of regularity can be obtained for a Stokes system when Dirichlet boundary data only belong toL2(∂Ω)(we recall that a weak solution has regularityH1(Ω), therefore one has to impose Dirichlet data in H1/2(∂Ω)). See also [1] for the non-hilbertian case.

In the present paper, we study a very weak solution concept for the hydrostatic Stokes problem, with mixed Dirichlet–Neumann boundary conditions. Moreover, we apply this research as an auxiliary problem in order to obtain strong regularity for the non-stationary case.

By sections, the main contributions of this paper are the following:

– In Section 2 we set up the formulation of the (stationary) hydrostatic Stokes problem (2), and we define the dual problem associated (3). Using a mixed formulation, we obtain, first, a weak solution for the dual problem (where the “divergence condition" does not vanish). Afterwards, we obtain strong solution for the dual problem. Finally, we define the very weak solution concept for the hydrostatic Stokes problem, by means of a transposition argument (using the strong regularity for the dual problem) and we prove the existence (and uniqueness) of very weak solutions.

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– In Section 3 we give a differential interpretation of the very weak solution and a meaning for the boundary conditions; Dirichlet on the bottom and Neumann on the surface.

– In Section 4 we apply the results obtained in Section 2 in order to obtain strong solutions for the non-stationary hydrostatic Stokes Problem, weakening the hypothesis on the time derivative for the wind stress tensor imposed in [9] (here, a very weak solution will be used to lift the time derivative of the Neumann data). Finally, the application to the nonlinear non-stationary problem (or non-stationary Primitive Equations) is sketched, rewriting the arguments used in [9].

2. Existence of very weak solutions 2.1. Formulation of the problem

We consider an open, bounded and Lipschitz-continuous domain⊆R3given by =

(x, z)∈R3;x=(x, y)S,h(x) < z <0

, (1)

whereS is an open bounded domain ofR2andh:S→R+is the depth function (which regularity assumptions will be precised later on). The boundary∂Ω can be written as∂Ω=ΓsΓbΓlwhere:

Γs=

(x,0)|xS , Γb=

x,h(x)

|xS , and

Γl=

(x, z)/x∂S,h(x) < z <0 .

In the main results of this work, we are going to impose the hypothesis of minimal depth strictly positive, i.e.

hhmin>0 inS.

Concretely, a vertical section of the domain can be viewed in Fig. 1.

We start from the hydrostatic Stokes problem:





νuν3zz2u+ ∇p=f inΩ,

∇ · u =0 inS, ν3zu=Υ onΓs,

u=0 onΓbΓl,

(2)

whereu(x)= 0h(x)u(x, z) dz. The unknowns are u : →R2 the horizontal components of the velocity, and a potentialp:S→Rrepresenting the surface pressure stress (and the centripetal forces, see [14]). The data for (2)are the external forces, f : →R2 , the wind tension stress on the surface,Υ:Γs →R2, and the (eddy) horizontal and vertical viscosities, ν >0 and ν3>0, respectively. ,∇ and ∇· denote the two-dimensional operators:xx2 +yy2 ,(∂x, ∂y)t and the horizontal divergence operator, respectively.

We define the following dual problem:





νΦν3zz2Φ+ ∇π=g in,

∇ · Φ = −ϕ inS, ν3zΦ=0 onΓs,

Φ=0 onΓbΓl,

(3)

where(Φ, π )are the unknowns and(g, ϕ)the data.

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Fig. 1. A vertical section of the domain.

2.2. Weak and strong regularity for the dual problem

The following functional spaces will be used for the velocities:

H=VL2=

vL2(Ω)2; ∇ · v =0 inS,v ·n∂S=0 , V =VH1=

vH1(Ω)2; ∇ · v =0 inS, v|ΓbΓl =0 ,

where n∂S denotes the outward normal vector to the boundary ofScontained in the planez=0, V=

ϕCb,l(Ω)2; ∇ · ϕ =0 inS , and

Cb,l(Ω)=

ϕC(Ω)2;supp(ϕ)is a compact set⊆\bΓl) .

We denoteHb,l1 (Ω)the space of functions ofH1(Ω)vanishing onΓbΓl(i.e.Hb,l1 (Ω)=Cb,l(Ω)H1).

Respect to spaces for the pressure, let us introduce:

L20(S)=

qL2(S);

S

q dx=0

,

H=

qH1(S);

S

q dx=0

=H1(S)L20(S).

In the following, byCwe will denote different positive constants.

Now, we present the main result of this subsection:

Theorem 2.1 (Strong solution of(3)). SupposeS⊆R2with∂SC3andhC3(S)withhhmin>0 inS, and the corresponding domainΩ(defined in(1)). If gL2(Ω)2andϕH, then there exists a unique (strong) solution of (3) withΦH2(Ω)2Hb,l1 (Ω)2H1(S). Moreover,

Φ2H2(Ω)+ π2H1(S)C

g2L2(Ω)+ ϕ2H1(S)

. (4)

(5)

First of all, we will obtain a weak solution of(3)using a mixed formulation for the problem. We introduce the notation:

X=Hb,l1 (Ω)2, M=L20(S), a(u,v)=ν

u :vdΩ+ν3

zu·zvdΩ,u,vX,

b(u, p)= −

S

p

∇ · u

dx,uX,pM,

L,v =

g·vdΩ,vX,

R, q =

S

ϕ q dx,qM.

It is easy to verify thata(·,·)is a bilinear, symmetric, continuous and elliptic form onX×X,b(·,·)is a bilinear continuous form onX×M,Lis a linear continuous form onXandRis a linear continuous form onM.

Then, we consider the (abstract) mixed problem: Find(Φ, π )X×Msuch that:

a(Φ,v)+b(v, π )= L,vvX,

b(Φ, q)= R, qqM. (5)

Proposition 2.2 (Existence and uniqueness of solution for (5) [7]). Suppose that:

a(·,·)is a bilinear continuous form onX×X, satisfying theV-elliptic condition i.e. there existsa0>0 such that:

a(v,v)a0v2X,vV ,

b(·,·)is a bilinear continuous form onX×Msatisfying the inf-sup condition, i.e. there existsβ0>0 such that:

pinfM\{0} sup

vX\{0}

b(v, p)

vXpM β0.

Then, for each pair(L, R)X×M (the dual space ofX×M) the mixed problem(5)has a unique solution (Φ, π )X×M. Moreover, the following mapping is an isomorphism:

(L, R)X×M(Φ, π )X×M.

Therefore, in order to prove existence and uniqueness of weak solution of(3),ΦHb,l1 (Ω)2andπL20(S), we only have to prove that the inf-sup condition holds. For this purpose, we use the following result:

Lemma 2.3 [7]. Let ⊂R3 be a Lipschitz-continuous domain. The three-dimensional divergence operator,

3·:WL20(Ω)is an isomorphism, whereW = {vH01(Ω)3; ∇3·v=0},W is the orthogonal space of Wrespect to theH01(Ω)-norm, andL20(Ω)= {gL2(Ω), g dΩ=0}.

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Lemma 2.4. Assume (defined in (1)) Lipschitz-continuous and hhmin>0 in S. The following inf-sup condition holds:

sup

vH01(Ω)2\{0}

Sp∇ · vdx vH1

0(Ω)

C

hminpL2(S),pL20(S).

Proof. Basically, we follow here an argument introduced in [5]. LetpL20(S). Easily, we can deduce h(x)1 pL20(Ω)(using thathhmin0) andzp=0. Indeed, we have:

1 h(x)p

L2(Ω)

1

h(x)p L2(S)

1

hminpL2(S).

Then, applying Lemma 2.3, there exists a function U=(u, u3)WH01(Ω)3such that∇3·U=h(x)1 pand (in particular)

3UL2(Ω)C3·UL2(Ω)C 1

h(x)p L2(Ω)

C 1

hminpL2(S). (6) Rewriting the bilinear formb(·,·), one has that for all vH01(Ω)2

b(v, p)= −

S

p

∇ · v dx= −

S

p∇ ·vdx= −

p∇ ·vdΩ

(aszp=0)= −

p3·(v, v3) dΩ

wherev3is any function belonging toH01(Ω). Then, in particular b(u, p)=

p3·(u, u3) dΩ =

1

h(x)p2dΩ=

S

p2dx= p2L2(S). (7) Therefore, using (6) and (7)

|b(u, p)|

uL2(Ω)pL2(S)

= pL2(S)

uL2(Ω)

C

hmin

3UL2(Ω)

uL2(Ω)

C

hmin,

hence sup

vH01(Ω)2\{0}

|b(v, p)| vH1

0(Ω)

C

hminpL2

0(S),pL20(S). 2

Applying Proposition 2.2 in our context, we deduce the following result:

Theorem 2.5 (Weak solution of (3)). Suppose Lipschitz-continuous andhhmin>0 inS. If gHb,l1(Ω)2and ϕL20(S), then there exists a unique weak solution of(3),(Φ, π )Hb,l1 (Ω)2×L20(S). Moreover,

ΦH1(Ω)+ πL2(S)C gH1

b,l(Ω)+ ϕL2

0(S)

. (8)

Remark 2.1. Taking into account that the spaceL2(S)/Ris isomorphic toL20(S)(see [7] for instance), we can replace in(8)the normϕL2

0(S)byϕL2(S)/R.

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In order to prove Theorem 2.1, we will use two auxiliary regularity results; a result for elliptic problems in (see [16] and references therein cited) and Cattabriga’s result for the Stokes problem inS(see [15] for instance), that we recall here:

Proposition 2.6 (Regularity for a elliptic problem in, with mixed Neumann–Dirichlet boundary conditions).

AssumehC3(S)withhhmin>0 inS and∂SC3, and the corresponding domainΩ. Let u be the unique solution of the problem:



νwν3zz2w=d inΩ, ν3zw=Υ onΓs,

w=Ψl (respectivelyΨb) onΓl(respectivelyΓb).

Suppose dL2(Ω)2. If ΥH0s3/2s)2, ΨlH0s1/2l)2, ΨbH0s1/2b)2 with 3/2 s <2, then, wHs(Ω)2. Moreover, ifΥH01/2+εs)2lH03/2+εl)2andΨbH03/2+εb)2, for someε: 0< ε <1/2, then wH2(Ω)2.

Proposition 2.7 (Regularity for the Stokes problem inS). LetS⊆R2be an open set with∂SC3. Let(u, p)be the solution for the Dirichlet–Stokes problem inS:

v+ ∇π=a inS,

∇ ·v=b inS,

v=c on∂S.

If aLα(S)2,bWs1,α(S)and cWsα1(∂S)2with 1< α <+∞and 1s2 such thatsα1 is not an integer, and the compatibility condition Sb dx= ∂Sc·n∂Sdsholds, then:

vWs,α(S)2 and πWs1,α(S).

This result was proved by Cattabriga [4], for any s 1 integer, and can be generalized to s ∈R using interpolation techniques, as in Lions and Magenes [11].

Proof of Theorem 2.1. Let us consider(Φ, π )the weak solution of the dual problem (3) (obtained in Theorem 2.5). SinceπL20(S), in particularπH1(S). Therefore, we might consider the auxiliary function vH01(S)2 as the (unique) weak solution of the elliptic problem inS:

νv= ∇π inS,

v=0 on∂S. (9)

Now, we look at the elliptic problem verified by w=Φv (without pressure and free divergence restriction):





νwν3zz2w=g in, ν3zw=0 onΓs,

w=0 onΓl,

w= −v onΓb.

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Here, we have made the identification v(x)=v(x,h(x))in order to consider v as a function defined inΓb. Due to the regularity of the data (gL2(Ω)2,vH01b)2), we can apply Proposition 2.6 (fors=3/2) and deduce that wH3/2(Ω)2. Then, using Lemma B.1 (see Appendix B),wH3/2(S)2. As v is independent ofz, v =hv. Then

−∇ ·(hv)= −∇ · v = ∇ · w − ∇ · Φ = ∇ · w +ϕH1/2(S).

Now, as

∇ ·v=1

h∇ ·(hv)−∇h·v h ,

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then∇ ·vH1/2(S)(using thathhmin>0 inS). In particular,∇ ·vH1/2ε(S)for allε >0. Therefore, if we consider the Stokes problem inSthat satisfies(v, π ):

νv+ ∇π=0 inS,

∇ ·vH1/2ε(S) inS,

v=0 on∂S,

(11) from Proposition 2.7 (fors=1/2−εand α=2) one has vH03/2ε(S)2 andπH1/2ε(S), for all ε >0.

Returning to system(10)(and(11)), using again Proposition 2.6 fors=2−ε (and Proposition 2.7), one has wH2ε(Ω)2(and vH02ε(S)2andπH1ε(S)).

Notice that, following with this “bootstrap” argument, it is not possible to obtain wH2(Ω)2by means of Proposition 2.6, since v∈/H03/2+ε(Ω)2because in general∂n∂v=0 on∂S.

Therefore, we change the argument. In order to increase the regularity of the pressure toπH1(S), we integrate (3)1in thez-variable, arriving at:

νΦ +h(x)π=G inS,

∇ · Φ = −ϕ inS, Φ =0 on∂S, where

G= g +ν3zΦ(x,0)−ν3zΦ

x,h(x)

+(Φ)

x,h(x)

h(x).

The last term of G is the vector whoseith component is the scalar product(Φi)|Γb· ∇h, coming from the equality Φi =Φi(Φi)|Γb· ∇h (i=1,2)(here,Φ|Γb =0 is used). We will see that GL2(S)2.

Since wH2ε(Ω)2, in particular,

zΦ=zwH1ε(Ω)2. (12)

From gL2(Ω)2 and(12), we have thatg +ν3zΦ(x,0)−ν3zΦ(x,h(x))L2(S)2. Therefore, we focus our attention on the term(Φi)(x,h(x))· ∇h(x). Deriving with respect to the x-variables the equalityΦ|Γb= Φ(x,h(x))=0, we obtain:

(Φi)|Γb=(∂zΦi)|Γbh(x).

Then,

(Φi)|Γb· ∇h=(∂zΦi)|Γbh(x)2.

Therefore, it suffices to analyze the regularity of|∇h(x)|2(∂zΦ)|Γb.

Since hH2(S), in particular |∇h|2Lp(S) for all p >1. From (12), (∂zΦ)|ΓbH1/2εb)2L4/(1+2ε)b)2. In particular,(∂zΦ)|ΓbLq(S)2 for a certainq >2. Then |∇h(x)|2(∂zΦ)|ΓbL2(S)2, hence we can conclude that GL2(S)2.

Notice that, using the equalityhπ= ∇(h π )πh, the previous problem can be rewritten as the Stokes problem inS:

νΦ + ∇(hπ )=G+πh inS,

∇ · Φ = −ϕ inS,

Φ =0 on∂S.

Since G+πhL2(S)2(recall thatπH1/2ε(S)) andϕH1(S)such that Sϕ dx=0, by Proposition 2.7, ΦH2(S)2andH1(S). In particular,πH1(S)(using once again thathhmin>0).

Now, we go back to the dual problem(3). Moving the pressure term to the right hand side, we consider the corresponding elliptic problem with the unknownΦ. Then, using the new regularityπH1(S) and applying Proposition 2.6, one has ΦH2(Ω)2. Finally, inequality (4) can be deduced by construction, thanks to the continuous dependence of the auxiliary problems(9),(10)and(11). 2

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2.3. Very weak regularity for the primal problem

Suppose the following regularity hypothesis for the data:

f

H2(Ω)2Hb,l1 (Ω)2

, ΥH3/2s). (H)

Denote by·,·S the duality betweenH1(S)andH1(S), by·,· the duality between(H2(Ω)Hb,l1 (Ω)) andH2(Ω)Hb,l1 (Ω)and by·,·Γs the duality betweenH3/2s)andH03/2s).

Definition 2.8. A pair(u, p)is said a very weak solution of (2)if uL2(Ω)2,pH1(S)/Rand satisfies:

u·(νΦν3zz2Φ+ ∇π ) dΩp,∇ · ΦS= f, Φ+ Υ, ΦΓs

ΦH2(Ω)2Hb,l1 (Ω)2 with zΦ|Γs =0, ∀πH1(S). (13) Remark 2.2. Notice that pH1(S)/R means that pH1(S) is defined up to an additive constant. From ΦH2(Ω)2Hb,l1 (Ω)2one has∇ · ΦH1(S)and S∇ · Φ =0 (therefore,p,∇ · ΦS= p+c,∇ · ΦS

for allc∈R).

Remark 2.3. It is easy to see that if the data (f, Υ ) are regular and (u, p)Hb,l1 (Ω)2×L2(Ω)/Ris a weak solution of(2), then(u, p)is also a very weak solution of(2)(i.e., the previous definition is a generalization of the variational formulation).

Letl:L2(Ω)2×H→Rdefined by:

l(g, ϕ)= f, Φ+ Υ, ΦΓs,

with(Φ, π )the strong solution for the dual problem(3)with data(g, ϕ)(given in Theorem 2.1). It is easy to prove thatlis a linear and continuous operator fromL2(Ω)2×HintoR. Indeed, from (4) one has

l(L2(Ω)×H)C

f(H2(Ω)H1

b,l(Ω))+ ΥH3/2s)

.

Applying the classical Riesz’ identification, one can easily prove the following:

Lemma 2.9. Assuming (H) and the hypothesis of Theorem 2.1, there exists a unique pair(u,p)˜ ∈L2(Ω)2×H verifying:

u·gdΩ+ ˜p, ϕH,H=l(g, ϕ),gL2(Ω)2,ϕH. (14) Moreover, one has:

uL2(Ω)+ ˜pHC

f(H2Hb,l1 )+ ΥH3/2s)

. (15)

In order to rewrite the linear form of (14) in terms of(Φ, π ), one needs the following result Proposition 2.10. The spaceH1(S)/Ris isomorphic toH.

Proof. In Appendix A. 2

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Now, we are able to show the result about existence (and uniqueness) of very weak solutions.

Theorem 2.11. Under conditions of Lemma 2.9, there exists a unique very weak solution (u, p) of (2) in L2(Ω)2×(H1(S))/R. Moreover, one has:

uL2(Ω)+ pH1(S)/RC

f(H2(Ω)Hb,l1 (Ω))+ ΥH−3/2s)

. (16)

Proof. From Lemma 2.9, there exists a unique pair(u,p )˜ ∈L2(Ω)2×Hverifying(14).

Using the isomorphism between(H1(S))/RandH, we can identifyp˜ with a distributionp in(H1(S))/R such that ˜p, ϕH,H= p, ϕH1(S),H1(S),∀ϕH. Therefore, we conclude that(u, p)is a solution of(13), i.e. a very weak solution of(2). The uniqueness is deduced from the linearity of the problem. Finally,(16)is deduced from(15)and the isomorphism betweenH1(S)/RandH. 2

3. Interpretation for the differential problem 3.1. Differential equation

From now on, we fix fL2(Ω)2instead of f(H2(Ω)Hb,l1 )(in order to have a space of distributions). Let us consider(u, p)the very weak solution of(2). Takingπ=0 andΦD(Ω)in(13), one has:

u·(νΦν3zz2Φ) dΩ

p,∇ · Φ

S=

f·Φ dΩ,

hence we can deduce the system(2)1in the distributional sense. TakingΦ=0 andπD(S)in(13),

u· ∇π dΩ =0

hence we can deduce∇ · u =0 inD(S). Therefore, we have:

Proposition 3.1. Assume fL2(Ω)2. Let(u, p)L2(Ω)2×(H1(S))/Rbe the unique very weak solution of(2).

Then u andpverify Eqs.(2)1−2in the distributional sense inΩandS, respectively.

3.2. Sense for the boundary conditions

From uL2(Ω)2, we have thatuL2(S)2. On the other hand, since∇ ·u =0, in particular∇ ·uL2(S).

Then we can conclude thatu ·n∂SH1/2(S). Moreover, takingΦ=0 andπH1(S)in(13), we get:

u· ∇π dΩ=0, hence

0=

S

u · ∇π dS=

u ·n∂S, π

∂S,πH1(S)

where·,·∂S denotes the dualityH1/2(∂S), H1/2(∂S). Therefore,

u ·n∂S=0 inH1/2(∂S). (17)

(11)

3.2.1. Dirichlet boundary conditions

In order to give a sense at the Dirichlet boundary conditions(2)4onΓbΓl, we define the operator D(u,p):H1/2bΓl)2→R,

D(u,p)(Ψ )=

u·(νΦν3zz2Φ) dΩ

p,∇ · Φ

Sf, Φ,ΨH1/2bΓl)2, whereΦ=Φ(Ψ )is the unique weak solution,ΦH2(Ω)2H01(Ω)2, of the problem:



2Φ=0 in, Φ=0 on∂Ω,

∂Φ

∂n =Ψ onΓbΓl, zΦ=0 onΓs.

It is easy to show that D(u,p)is a linear continuous operator, i.e.:

D(u,p)(H1/2bΓl)2).

Notice that if(u, p)is a weak solution of(2), then:

D(u,p)(Ψ )=

ΓbΓl

u·Ψ dσ,Ψ,

therefore we can denote this operator as the generalized trace overΓbΓl. Replacingπ=0 andΦ=Φ(Ψ )in(13), we obtain that:

D(u,p)(Ψ )=0, ∀ΨH1/2bΓl)2. Thus, it follows that D(u,p)=0 in(H1/2bΓl)2). 3.2.2. Neumann boundary condition

In this case, in order to give a meaning to the Neumann boundary conditions(2)3onΓs, we define the operator N(u,p):H03/2(S)2→R,

N(u,p)(Ψ )=

u·(νΦν3zz2Φ) dΩ

p,∇ · Φ

Sf, Φ, whereΦ=Φ(Ψ )is the unique weak solution,ΦH2(Ω)2, of the problem:



2Φ=0 in,

Φ=Ψ onΓs, Φ=0 onΓbΓl,

∂Φ

∂n =0 on∂Ω.

Now, assuming that(u, p)is a weak solution of(2), one has:

N(u,p)(Ψ )=

Γs

zu·Ψ dσ,Ψ,

therefore we can denote this operator as generalized normal trace overΓs. Replacingπ=0 and this newΦ=Φ(Ψ )in(13), we obtain that:

N(u,p)(Ψ )= Υ, ΨΓs,ΨH03/2(S)2

Thus, it follows that N(u,p)=Υ in(H03/2(S)2)H3/2(S)2.

(12)

4. Application to non-stationary problems 4.1. The linear case

In this subsection, we are going to apply the result obtained for the stationary case in order to get strong solutions for the linear non-stationary Primitive Equations (also called non-stationary hydrostatic Stokes problem):









tuνuν3zz2u+ ∇p=F in(0, T )×Ω,

∇ · u =0 in(0, T )×S,

u|t=0=u0 in,

ν3zu=Υ on(0, T )×Γs, u=0 on(0, T )×bΓl).

(18)

The following result was given in [9]:

Theorem 4.1. SupposeS⊆R2with∂SC3andhC3(S) withhhmin>0 in S. If FL2((0, T )×Ω)2, u0V,ΥL2(0, T;H01/2+εs)2)for someε >0, with∂tΥL2(0, T;H1/2s)2), then there exists a unique strong solution u of problem(18)in(0, T ). Moreover,

u2L(V )+ u2L2(H2(Ω))+ tu2L2(H )

C

u02V + Υ (0)2H−1/2s)+ F2L2(L2(Ω))+ Υ2

L2(H01/2+εs))+ tΥ2L2(H−1/2s))

. (19)

The proof of this theorem is based on the following two results about the stationary problem(2); a weak regularity result given by Lions, Temam and Wang in [13], and a strong regularity result due to M. Ziane [16], respectively:

Lemma 4.2. Suppose⊆R3 be Lipschitz-continuous. If fHb,l−1(Ω)2andΥH1/2s)2, then the problem (2)has a unique weak solution uV. Moreover,

u2V C

Υ2H1/2s)+ f2H1 b,l(Ω)

. (20)

Lemma 4.3. Suppose S ⊆R2 with ∂SC3 and hC3(S) with h hmin>0 in S. If fL2(Ω)2 and ΥH01/2+εs)2, for someε >0, the unique solution u of the problem(2)belongs toH2(Ω)2V. Moreover,

u2H2(Ω)C Υ2

H01/2+εs)+ f2L2(Ω)

. (21)

In this section, using Theorem 2.11 instead of Lemma 4.2, we will obtain strong solutions of(19)imposing less regularity overtΥ (replacingH1/2s)byH3/2s)). More precisely, the new result is:

Theorem 4.4. Suppose S ⊆R2 with ∂SC3 domain and hC3(S) with h hmin >0 in S. If FL2((0, T )×Ω)2, u0V, ΥL2(0, T;H01/2+εs)2)L(0, T;H1/2s)2) for some ε >0 with tΥL2(0, T;H3/2s)2)andΥ (0)H1/2s)2, then there exists a unique strong solution u of the problem(18)in (0, T ). Moreover,

u2L(V )+ u2L2(H2(Ω))+ tu2L2(H )

C

u02V + Υ (0)2H1/2s)

+ F2L2(L2(Ω))+ Υ2

L2(H01/2+εs))L(H−1/2s))+ tΥ2L2(H−3/2s))

. (22)

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