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

Ann. I. H. Poincaré – AN 21 (2004) 807–826 www.elsevier.com/locate/anihpc 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-Gonzáleza, M.A. Rodríguez-Bellidob, M.A. Rojas-Medarc 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 withfree 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: [email protected] (F. Guillén-González), [email protected] (M.A. Rodríguez-Bellido), m[email protected]p.br (M.A. Rojas-Medar). 0294-1449/$ – see front matter 2004 Elsevier SAS. All rights reserved. doi:10.1016/j.anihpc.2003.11.002 808 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 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 hydrostaticNavier–Stokesproblem(also called Primitive Equations)is a modelthat appears in Geophysical fluid dynamics, in order to describe the general circulation in the Ocean or the Atmosphere [14]. This model has been extensivelystudied from a mathematicalpoint of view by severalauthors [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 physicalscales (because the domainof 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 to L2(∂Ω) (we recall that a weak solution has regularity H1(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. F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 809 – 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 hydrostaticStokesProblem,weakeningthe hypothesison the time derivativefor the windstress 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 Ω⊆R3given by Ω=(x,z)∈R3;x=(x, y) ∈S, −h(x)<z<0,(1) where Sis an open bounded domain of R2and h:S→R+is the depth function (which regularity assumptions will be precised later on). The boundary ∂Ω can be written as ∂Ω =Γs∪Γb∪Γlwhere: Γs=(x,0)|x∈S, Γb=x,−h(x)|x∈S, 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. h⩾hmin >0inS. Concretely, a vertical section of the domain can be viewed in Fig. 1. We start from the hydrostatic Stokes problem:      −νu−ν3∂2 zzu+∇p=fin Ω, ∇·u=0inS, ν3∂zu=Υon Γs, u=0on Γb∪Γl, (2) where u(x)=0 −h(x)u(x,z)dz. The unknowns are u:Ω→R2the horizontal components of the velocity, and a potential p:S→Rrepresenting the surface pressure stress (and the centripetal forces, see [14]). The data for (2)are the external forces, f:Ω→R2, the wind tension stress on the surface, Υ:Γs→R2, and the (eddy) horizontal and vertical viscosities, ν>0andν3>0, respectively. ,∇and ∇· denote the two-dimensional operators: ∂2 xx +∂2 yy,(∂x,∂ y)tand the horizontal divergence operator, respectively. We define the following dual problem:      −νΦ −ν3∂2 zzΦ+∇π=gin Ω, ∇·Φ=−ϕin S, ν3∂zΦ=0on Γs, Φ=0on Γb∪Γl, (3) where (Φ, π) are the unknowns and (g,ϕ)the data. 810 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 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=v∈L2(Ω)2;∇·v=0inS,v·n∂S =0, V=VH1=v∈H1(Ω)2;∇·v=0inS,v|Γb∪Γl=0, where n∂S denotes the outward normal vector to the boundary of Scontained in the plane z=0, V=ϕ∈C∞ b,l(Ω)2;∇·ϕ=0inS, and C∞ b,l(Ω) =ϕ∈C∞(Ω)2;supp(ϕ) is a compact set ⊆Ω\(Γb∪Γl). We denote H1 b,l(Ω) the space of functions of H1(Ω) vanishing on Γb∪Γl(i.e. H1 b,l(Ω) =C∞ b,l(Ω)H1). Respect to spaces for the pressure, let us introduce: L2 0(S) =q∈L2(S);S qdx=0, H=q∈H1(S);S qdx=0=H1(S) ∩L2 0(S). In the following, by Cwe will denote different positive constants. Now, we present the main result of this subsection: Theorem 2.1 (Strong solution of (3)). Suppose S⊆R2with ∂S ∈C3and h∈C3(S) with h⩾hmin >0in S, and the correspondingdomain Ω(definedin (1)).Ifg∈L2(Ω)2and ϕ∈H, then there exists a unique(strong)solution of (3) with Φ∈H2(Ω)2∩H1 b,l(Ω)2,π∈H1(S). Moreover, Φ2 H2(Ω) +π2 H1(S) ⩽Cg2 L2(Ω) +ϕ2 H1(S).(4) F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 811 First of all, we will obtain a weak solution of (3)using a mixed formulation for the problem. We introduce the notation: X=H1 b,l(Ω)2,M=L2 0(S), a(u,v)=ν Ω ∇u:∇vdΩ +ν3 Ω ∂zu·∂zvdΩ, ∀u,v∈X, b(u,p)=−S p∇·udx,∀u∈X, ∀p∈M, L,v= Ω g·vdΩ, ∀v∈X, R,q=S ϕqdx,∀q∈M. It is easy to verify that a(·,·)is a bilinear, symmetric, continuous and elliptic form on X×X,b(·,·)is a bilinear continuous form on X×M,Lis a linear continuous form on Xand Ris a linear continuous form on M. Then, we consider the (abstract) mixed problem: Find (Φ, π) ∈X×Msuch that: a(Φ,v)+b(v,π)=L,v∀v∈X, b(Φ,q) =R,q∀q∈M.(5) Proposition 2.2 (Existence and uniqueness of solution for (5) [7]). Suppose that: •a(·,·)is a bilinear continuous form on X×X, satisfying the V-elliptic condition i.e. there exists a0>0such that: a(v,v)⩾a0v2 X,∀v∈V, •b(·,·)is a bilinear continuous form on X×Msatisfying the inf-sup condition, i.e. there exists β0>0such that: inf p∈M\{0}sup v∈X\{0} b(v,p) vXpM ⩾β0. Then, for each pair (L, R) ∈X×M(the dual space of X×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),Φ∈H1 b,l(Ω)2and π∈L2 0(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 Ω⊂R3be a Lipschitz-continuous domain. The three-dimensional divergence operator, ∇3·:W⊥→L2 0(Ω) is an isomorphism, where W={v∈H1 0(Ω)3;∇ 3·v=0},W⊥is the orthogonal space of Wrespect to the H1 0(Ω)-norm, and L2 0(Ω) ={g∈L2(Ω), ΩgdΩ=0}. 812 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 Lemma 2.4. Assume Ω(defined in (1)) Lipschitz-continuous and h⩾hmin >0in S. The following inf-sup condition holds: sup v∈H1 0(Ω)2\{0}Sp∇·vdx vH1 0(Ω) ⩾Chmin pL2(S),∀p∈L2 0(S). Proof. Basically, we follow here an argument introduced in [5]. Let p∈L2 0(S). Easily, we can deduce 1 h(x)p∈ L2 0(Ω) (using that h⩾hmin ⩾0) and ∂zp=0. Indeed, we have: 1 h(x)pL2(Ω) ⩽1 √h(x)pL2(S) ⩽1 √hmin pL2(S). Then, applying Lemma 2.3, there exists a function U=(u,u 3)∈W⊥⊂H1 0(Ω)3such that ∇3·U=1 h(x)pand (in particular) ∇3UL2(Ω) ⩽C∇3·UL2(Ω) ⩽C1 h(x)pL2(Ω) ⩽C1 √hmin pL2(S).(6) Rewriting the bilinear form b(·,·), one has that for all v∈H1 0(Ω)2 b(v,p)=−S p∇·vdx=−S p∇ ·vdx=− Ω p∇·vdΩ (as ∂zp=0) =− Ω p∇3·(v,v 3)dΩ where v3is any function belonging to H1 0(Ω). Then, in particular b(u,p) =− Ω p∇3·(u,u 3)dΩ= Ω 1 h(x)p2dΩ =S p2dx=p2 L2(S).(7) Therefore, using (6) and (7) |b(u,p)| ∇uL2(Ω)pL2(S) =pL2(S) ∇uL2(Ω) ⩾Chmin ∇3UL2(Ω) ∇uL2(Ω) ⩾Chmin, hence sup v∈H1 0(Ω)2\{0} |b(v,p)| vH1 0(Ω) ⩾Chmin pL2 0(S),∀p∈L2 0(S). 2 Applying Proposition 2.2 in our context, we deduce the following result: Theorem 2.5 (Weak solution of (3)). Suppose ΩLipschitz-continuousand h⩾hmin >0in S.Ifg∈H−1 b,l (Ω)2and ϕ∈L2 0(S), then there exists a unique weak solution of (3),(Φ, π) ∈H1 b,l(Ω)2×L2 0(S). Moreover, ΦH1(Ω) +πL2(S) ⩽CgH1 b,l (Ω)+ϕL2 0(S).(8) Remark 2.1. Taking into account that the space L2(S)/Ris isomorphic to L2 0(S)(see [7] for instance), we can replace in (8)the norm ϕL2 0(S)by ϕL2(S)/R. F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 813 In order to prove Theorem 2.1, we will use two auxiliary regularity results; a result for elliptic problems in Ω (see [16] and references therein cited) and Cattabriga’s result for the Stokes problem in S(see [15] for instance), that we recall here: Proposition 2.6 (Regularity for a elliptic problem in Ω, with mixed Neumann–Dirichlet boundary conditions). Assume h∈C3(S) with h⩾hmin >0in Sand ∂S ∈C3, and the corresponding domain Ω.Letube the unique solution of the problem:    −νw−ν3∂2 zzw=din Ω, ν3∂zw=Υon Γs, w=Ψl(respectively Ψb)on Γl(respectively Γb). Suppose d∈L2(Ω)2.IfΥ∈Hs−3/2 0(Γs)2,Ψl∈Hs−1/2 0(Γl)2,Ψb∈Hs−1/2 0(Γb)2with 3/2⩽s<2, then, w∈Hs(Ω)2. Moreover, if Υ∈H1/2+ε 0(Γs)2,Ψl∈H3/2+ε 0(Γl)2and Ψb∈H3/2+ε 0(Γb)2,forsomeε:0<ε<1/2, then w∈H2(Ω)2. Proposition 2.7 (Regularity for the Stokes problem in S). Let S⊆R2be an open set with ∂S ∈C3.Let(u,p) be the solution for the Dirichlet–Stokes problem in S: −v+∇π=ain S, ∇·v=bin S, v=con ∂S. If a∈Lα(S)2,b∈Ws−1,α(S) and c∈Ws−1 α,α(∂S)2with 1<α<+∞ and 1⩽s⩽2such that s−1 αis not an integer, and the compatibility condition Sbdx=∂S c·n∂S ds holds, then: v∈Ws,α(S)2and π∈Ws−1,α(S). This result was proved by Cattabriga [4], for any s⩾1 integer, and can be generalized to s∈Rusing 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).Sinceπ∈L2 0(S), in particular ∇π∈H−1(S). Therefore,we might consider the auxiliary function v∈H1 0(S)2 as the (unique) weak solution of the elliptic problem in S: νv=∇πin S, v=0on ∂S.(9) Now, we look at the elliptic problem verified by w=Φ−v(without pressure and free divergence restriction):      −νw−ν3∂2 zzw=gin Ω, ν3∂zw=0on Γs, w=0on Γl, w=−von Γb. (10) Here, we have made the identification v(x)=v(x,−h(x)) in order to consider vas a function defined in Γb. Due to the regularity of the data (g∈L2(Ω)2,v∈H1 0(Γb)2), we can apply Proposition 2.6 (for s=3/2) and deduce that w∈H3/2(Ω)2. Then, using Lemma B.1 (see Appendix B), w∈H3/2(S)2.Asvis independent of z, v=hv.Then −∇ ·(h v)=−∇·v=∇·w−∇·Φ=∇·w+ϕ∈H1/2(S). Now, as ∇·v=1 h∇·(h v)−∇h·v h, 814 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 then ∇·v∈H1/2(S) (using that h⩾hmin >0inS). In particular, ∇·v∈H1/2−ε(S) for all ε>0. Therefore, if we consider the Stokes problem in Sthat satisfies (v,π): −νv+∇π=0in S, ∇·v∈H1/2−ε(S) in S, v=0on ∂S,(11) from Proposition 2.7 (for s=1/2−εand α=2) one has v∈H3/2−ε 0(S)2and π∈H1/2−ε(S),forallε>0. Returning to system (10)(and (11)), using again Proposition 2.6fors=2−ε(and Proposition 2.7), one has w∈H2−ε(Ω)2(and v∈H2−ε 0(S)2and π∈H1−ε(S)). Notice that, following with this “bootstrap” argument, it is not possible to obtain w∈H2(Ω)2by means of Proposition 2.6, since v/∈H3/2+ε 0(Ω)2because in general ∂v ∂n=0on∂S. Therefore,we changethe argument.In orderto increase the regularity ofthe pressure to π∈H1(S),weintegrate (3)1in the z-variable, arriving at: −νΦ+h(x)∇π=Gin S, ∇·Φ=−ϕin S, Φ=0on ∂S, where G=g+ν3∂zΦ(x,0)−ν3∂zΦx,−h(x)+(∇Φ)x,−h(x)∇h(x). The last term of Gis the vector whose ith componentis the scalar product (∇Φi)|Γb·∇h, coming from the equality Φi=Φi−(∇Φi)|Γb·∇h(i=1,2)(here, Φ|Γb=0is used). We will see that G∈L2(S)2. Since w∈H2−ε(Ω)2, in particular, ∂zΦ=∂zw∈H1−ε(Ω)2.(12) From g∈L2(Ω)2and (12),wehavethatg+ν3∂zΦ(x,0)−ν3∂zΦ(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 equality Φ|Γb= Φ(x,−h(x)) =0, we obtain: (∇Φi)|Γb=(∂zΦi)|Γb∇h(x). Then, (∇Φi)|Γb·∇h=(∂zΦi)|Γb∇h(x)2. Therefore, it suffices to analyze the regularity of |∇h(x)|2(∂zΦ)|Γb. Since h∈H2(S), in particular |∇h|2∈Lp(S) for all p>1. From (12),(∂zΦ)|Γb∈H1/2−ε(Γb)2→ L4/(1+2ε)(Γb)2. In particular, (∂zΦ)|Γb∈Lq(S)2for a certain q>2. Then |∇h(x)|2(∂zΦ)|Γb∈L2(S)2, hence we can conclude that G∈L2(S)2. Notice that, using the equality h∇π=∇(h π) −π∇h, the previous problem can be rewritten as the Stokes problem in S: −νΦ+∇(hπ) =G+π∇hin S, ∇·Φ=−ϕin S, Φ=0on ∂S. Since G+π∇h∈L2(S)2(recall that π∈H1/2−ε(S))andϕ∈H1(S) such that Sϕdx=0, by Proposition 2.7, Φ∈H2(S)2and hπ ∈H1(S). In particular, π∈H1(S) (using once again that h⩾hmin >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 unknown Φ. Then, using the new regularity π∈H1(S) and applying Proposition 2.6, one has Φ∈H2(Ω)2. Finally, inequality (4)can be deduced by construction, thanks to the continuous dependence of the auxiliary problems (9),(10)and (11).2 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 815 2.3. Very weak regularity for the primal problem Suppose the following regularity hypothesis for the data: f∈H2(Ω)2∩H1 b,l(Ω)2,Υ∈H−3/2(Γs). (H) Denote by ·,·Sthe duality between H1(S)and H1(S),by·,·Ωthe duality between (H 2(Ω) ∩H1 b,l(Ω)) and H2(Ω) ∩H1 b,l(Ω) and by ·,·Γsthe duality between H−3/2(Γs)and H3/2 0(Γs). Definition 2.8. Apair(u,p) is said a very weak solution of (2)if u∈L2(Ω)2,p∈H1(S)/Rand satisfies: Ωu·(−νΦ −ν3∂2 zzΦ+∇π)dΩ −p,∇·ΦS=f,ΦΩ+Υ,ΦΓs ∀Φ∈H2(Ω)2∩H1 b,l(Ω)2with ∂zΦ|Γs=0,∀π∈H1(S). (13) Remark 2.2. Notice that p∈H1(S)/Rmeans that p∈H1(S)is defined up to an additive constant. From Φ∈H2(Ω)2∩H1 b,l(Ω)2one has ∇·Φ∈H1(S) and S∇·Φ=0 (therefore, p,∇·ΦS=p+c,∇·ΦS for all c∈R). Remark 2.3. It is easy to see that if the data (f,Υ) are regular and (u,p) ∈H1 b,l(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). Let l:L2(Ω)2×H→Rdefined by: l(g,ϕ)=f,ΦΩ+Υ,ΦΓs, 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 that lis a linear and continuous operator from L2(Ω)2×Hinto R. Indeed, from (4) one has l(L2(Ω)×H)⩽Cf(H2(Ω)∩H1 b,l(Ω))+ΥH−3/2(Γs). 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:  Ω u·gdΩ +˜p,ϕH,H=l(g,ϕ), ∀g∈L2(Ω)2,∀ϕ∈H.(14) Moreover, one has: uL2(Ω) +˜pH⩽Cf(H2∩H1 b,l )+ΥH−3/2(Γs).(15) In order to rewrite the linear form of (14) in terms of (Φ, π), one needs the following result Proposition 2.10. The space H1(S)/Ris isomorphic to H. Proof. In Appendix A. 2 822 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 separate the regularity in xand z, considering the anisotropic spaces and estimates that were introduced in [9] and that we overview here: Definition 4.5. Given p,q ∈[1,+∞], we say that a function ubelongs to Lq zLp x(Ω) if: u(·,z)∈Lq(Sz)and u(·,z) Lq(Sz)∈Lp(−hmax,0), where Sz={x∈S/(x,z)∈Ω}for a fixed z∈(−hmax,0), and its norm is given by the expression: u(·,z) Lq(Sz)Lp(−hmax,0). Lemma 4.6 (Interpolation inequalities [9]). (a) Let v∈L2(Ω) be a function such that ∂zv∈L2(Ω) and (vnz)|Γb=0. Then, v∈L∞ zL2 x(Ω) and satisfy the estimate: v2 L∞ zL2 x⩽2vL2(Ω)∂zvL2(Ω).(34) More generally, if v∈H1(Ω) then v∈L∞ zL2 x(Ω), and there exists a constant C=C(Ω) > 0such that: v2 L∞ zL2 x⩽C(Ω)vL2(Ω)vH1(Ω),∀v∈H1(Ω). (35) (b) Let v∈L2(Ω) be a function such that ∇v∈L2(Ω)2and (vnxi)|Γb∪Γl=0(i=1,2). Then, v∈L2 zL4 x(Ω) and verifies the estimate: v2 L2 zL4 x⩽4vL2(Ω)∇vL2(Ω).(36) More generally, if v∈H1(Ω) then v∈L2 zL4 x, and there exists a constant C=C(Ω) > 0such that: v2 L2 zL4 x⩽C(Ω)vL2(Ω)vH1(Ω).(37) (c) Let v∈L2(Ω)2be a function such that ∇·v∈H1(Ω). Then, if we consider v3defined as in (31), we have that v3∈L∞ zL4 x(Ω) and satisfies the following estimate: v3L∞ zL4 x⩽C(Ω)∇ ·v1/2 L2(Ω)∇ ·v1/2 H1(Ω).(38) By using the above inequalities, we can bound the terms I1and I2as follows: I2⩽C(w3)m+(e3)m2 L∞ zL4 x∂zwm+∂zem2 L2 zL4 x ⩽C ν3/2Awm2 L2(Ω)wmV+C ν3wm2 Vem2 H1(Ω)em2 H2(Ω) +Cem2 H1(Ω)em2 H2(Ω) for a constant C=C(Ω) > 0. For the I1-term, we bound in the usual (isotropic) form: I1⩽Cwm+em2 L4(Ω)∇wm+∇em2 L4(Ω) ⩽C∇wm3/2 L2(Ω)wm1/2 L2(Ω) +em3/2 H1(Ω)em1/2 L2(Ω) ×∇wm3/2 H1(Ω)∇wm1/2 L2(Ω) +em3/2 H2(Ω)em1/2 H1(Ω) ⩽C1 ν11 wm8 V+1 ν7em8 H1(Ω) +1 νem1/2 H1(Ω)em3/2 H2(Ω)wm2 V +1 2Awm2 L2(Ω) +Cem5/2 H1(Ω)em3/2 H2(Ω). F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 823 Plugging the above estimates in (33),wehave: d dtwm2 V+Awm2 L2(Ω) ⩽C ν3/2Awm2 L2(Ω)wmV+C ν11 wm10 V+a(t)wm2 V+b(t), (39) where a(t) and b(t) are bounded functions in L1(0,T), depending on νand the data. From here, we can argue in two ways: supposing small data (to obtain global in time strong regularity) or for any data (to obtain local in time strong regularity). In order to obtain global in time strong solution for small data, we reason as follows: First, supposing the estimate: wm(t)V<γν 3/2for γsmall enough, (40) we can control globally in time the terms Awm2 L2(Ω)wmVand C ν11 wm10 Vappearing on the right hand side of (39), and finish the proof in a standard way. Secondly, we have to prove the estimate (40), which is obtained using the homogeneous initial data and adequate smallness hypothesis on the data (see [9] for more details). In order to obtain local in time strong solution for any data,usingthatwm(0)=0and wm:[0,T]→H1(Ω)2 is a continuous function, we can chose a time Tmsuch that: wm(t)V⩽ν3/2 2C,∀t∈[0,T m].(41) Then, we can prove that Tmis bounded from below by a time T∗⩾0 independent on m. Indeed, integrating (39) between 0 and t,t∈[0,T m], and using (41)we obtain ([9]): wm(t)2 V+ t 0Awm(s)2 L2(Ω) ds ⩽K(ν)t +Cν2 t 0em(s)2 H2(Ω) ds. Therefore, choosing now Tsuch that: K(ν)T+Cν2em2 L2(0,T;H2(Ω)) <ν3 4C2, we verify that Tmcan be chosen equal to Tfor each m. Now, the proof of the existence of strong solution in (0,T ) follows in a standard manner. 2 Acknowledgements The first author has been partially supported by the project BFM2000-1317,and the third author by the projects CNPq-Brasil 3013541-03-0and Fapesp-Brasil 01/07557-3. Appendix A. The isomorphism Proposition A.1 (before cited as Proposition 2.10). The space H1(S)/Ris isomorphic to H. Proof. Recall that H={ϕ/ϕ∈H1(S), Sϕdx=0}. (a) H1(S) =H⊕H⊥(identifying H⊥with the space spanned by the constant functions over S). In fact, any constant function ω=αin S(α∈R), verifies: (ω, ϕ)1=S ∇ω·∇ϕdx+S ωϕdx=αS ϕdx=0∀ϕ∈H, 824 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 where ((·,·))1denotes the inner product in H1(S). Therefore, ω=α∈H⊥. On the other hand, every function v∈H1(S) can be written in a unique manner as: v=ϕ+α, with α=1 |S|Svdx∈Rand ϕ=v−α∈H. (b) The isomorphism: We define the operator T:H1(S)/R→H, q→Tq as Tq,ϕH,H=q,ϕH1(S),H1(S),∀ϕ∈H. This operator is well-defined because if αis a constant, T(q+α) =Tq. Indeed, T(q+α),ϕ=q+α, ϕ= q,ϕ, because of Sϕdxfor all ϕ∈H. Therefore, it suffices to prove that Tis a continuous bijection. Tis one-to-one: Suppose that Tq =0, i.e. Tq,ϕ=0forallϕ∈H.Letv∈H1(S),thenv=ϕ+αwith α=1 |S|Svdxand ϕ∈H. Then, q,vH1(S),H1(S) =q,ϕ +αH1(S),H1(S) =Tq,ϕH,H+1 |S|q,1H1(S),H1(S)) S vdx =1 |S|q,1H1(S),H1(S) S vdx, therefore, defining β=1 |S|q,1H1(S),H1(S) ∈R, one has q−β,vH1(S),H 1(S) =0forallv∈H1(S), hence q−β belongs to the zero equivalent class in H1(S)/R, and therefore q=0inH1(S)/R. Tis onto:Foranyl∈H,wehavetoprovethatthereexistsanelementq∈H1(S)/Rsuch that Tq=l, i.e., q,ϕH1(S),H1(S) =l,ϕH,H∀ϕ∈H. Indeed, it suffices to define q,vH1(S),H1(S) =l,ϕH,Hif v=ϕ+α. Tis continuous: Using the standard norm definitions, qH1(S)/R=inf c∈Rq+cH1(S)=inf c∈Rsup v∈H1(S) q+c,vH1(S),H 1(S) vH1(S) ⩾inf c∈Rsup ϕ∈H q+c,ϕH1(S),H1(S) ϕH1(S) =sup ϕ∈H q,ϕH1(S),H1(S) ϕH1(S) =sup ϕ∈H Tq,ϕH,H ϕH1(S) =TqH.2 Appendix B. Regularity in S Lemma B.1. Suppose h∈H2(S).Ifw∈Hs(Ω) for any s:1⩽s⩽2,thenw∈Hs(S). Proof. If w∈L2(Ω), then integrating in (−h(x), 0)we obtain: w2= 0  −h(x) w(x,z)dz 2 ⩽0  −h(x)w(x,z) 2dzh(x). Integrating now in S, we obtain: F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 825 Sw2dx⩽S h(x)0  −h(x)w(x,z) 2dsdx⩽hL∞(S)  Ωw(x,z) 2dΩ =hL∞(S)w2 L2(Ω), that implies: wL2(S) ⩽h1/2 L∞(S) wL2(Ω).(42) Deriving w, we get that ∇w=∇w+w|Γb∇h.TakingtheL2(S)-norm and using (42), we obtain: ∇wL2(S) ⩽∇wL2(S) +w|Γb∇hL2(S) ⩽h1/2 L∞(S)∇wL2(Ω) +w|Γb1/2 L4(S)∇h1/2 L4(S). Let us focus our attention on the terms of L4(S)-type: w|ΓbL4(S) =Swx,−h(x)4dx1/4 = Γbwx,−h(x)41+∇h(x)2−1/2dσ1/4 ⩽wL4(Γb)⩽CwH1(Ω), where we have used the continuity of the trace function from H1(Ω) in L4(∂Ω). On the other hand, using the Sobolev embedding H2(S) →W1,4(S), one has: ∇hL4(S) ⩽ChH2(S). Therefore, ∇wL2(S) ⩽h1/2 L∞(S) ∇wL2(Ω) +Cw1/2 H1(Ω)h1/2 H2(Ω).(43) Estimates (42)and (43)let us deduce that if w∈H1(Ω) and h∈H2(S),thenw∈H1(S). Now, we study the case where w∈H2(Ω) and h∈H2(S). From the previous estimates, we only need to estimate second order derivatives for w. Without loss of generality, we will only reason for ∂2 xxw.Wehave: ∂2 xxw=∂2 xxw+2(∂xw)|Γb∂xh(x)−(∂zw)|Γb∂xh(x)2+w|Γb∂2 xxh(x). Therefore, using the same arguments as before: ∂2 xxwL2(S) ⩽h1/2 L∞(S) ∂2 xxwL2(Ω) +C∂xw1/2 H1(Ω)h1/2 H2(S) +∂zw1/2 H1(Ω)hH2(S) +w|Γb1/2 L∞(S)h1/2 H2(S) ⩽h1/2 L∞(S) ∂2 xxwL2(Ω) +Cw1/2 H2(Ω)h1/2 H2(S)1+h1/2 H2(S),(44) where we have used in the last term that if w∈H2(Ω),thenw|∂Ω ∈H3/2(∂Ω) →L∞(∂Ω). Expression (44) together with (42)and (43), let us deduce that if w∈H2(Ω) and h∈H2(S),thenw∈H2(S). We have just proved that, if w∈H1(Ω) then w∈H1(S),andifw∈H2(Ω) then w∈H2(S).In the case Hs(Ω) for s∈(1,2), interpolation results [11] let us see Hs(Ω) =[H1(Ω), H 2(Ω)]θand Hs(S) = [H1(S), H 2(S)]θwith θ=s−1, hence we can deduce that if w∈Hs(Ω),thenw∈Hs(S).2 Appendix C. Interpolation results Lemma C.1. Let S⊆R2be a bounded open set with ∂S ∈C∞.IfΥ∈L2(0,T;H1/2+ε 0(Γs)) for some ε>0and ∂tΥ∈L2(0,T;H−3/2(Γs)),thenΥ∈C0([0,T];H−1/2(Γs)). 826 F. Guillén-González et al. / Ann. I. H. Poincaré – AN 21 (2004) 807–826 In order to prove this lemma, we will use some interpolation results appearing in Lions and Magenes [11]: Theorem C.2 [11, p. 79]. Suppose that Γs∈C∞.Lets1,s2⩾0such that si=λ+1/2(λinteger, i=1,2).Let θ∈[0,1]such that: (1−θ)s1−θs2=µ+1/2and =−µ−1/2(µ integer⩾0). (45) Then, Hs1 0(Γs), H −s2(Γs)θ=H(1−θ)s1−θs2 0(Γs)if (1−θ)s1−θs2⩾0, H(1−θ)s1−θs2(Γs)if (1−θ)s1−θs2⩽0.(46) Proposition C.3 [11, p. 53]. Let Xand Ytwo separable Hilbert spaces such that X⊂Y,Xdense in Ywith continuous embedding. Then: Ht1(Ω;X),Ht2(Ω;Y) θ=H(1−θ)t1+θt2Ω;[X, Y ]θ. Proof of Lemma C.1. In our case, taking in Theorem C.2, µ=0, s1=1/2+ε,s2=3/2+εwith ε<εone has H1/2+ε 0(Γs), H −3/2−ε(Γs)θ=H−1/2+δ(Γs), if one imposes that (1−θ)s1−θs2=−1/2+δ<0 with δis small enough, then: θ=1+ε−δ 2+ε+ε>1 2(taking ε>ε +2δ). From hypothesis Υ∈H1(0,T;H−3/2(Γs)) ∩H0(0,T;H1/2+ε 0(Γs)). 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