Asymptotic behavior of a viscous fluid with slip boundary conditions on a slightly rough wall

Asymptotic behavior of a viscous fluid with slip boundary conditions on a slightly rough wall March 23, 2009 J. CASADO-D´ IAZ†M. LUNA-LAYNEZ†F.J. SU´ AREZ-GRAU† †Dpto. de Ecuaciones Diferenciales y An´alisis Num´erico, Facultad de Matem´aticas, C. Tarfia s/n 41012 Sevilla, Spain e-mail: jcasado[email protected], mlla[email protected], [email protected] Abstract For an oscillating boundary of period and amplitude ε, it is known that the asymptotic behavior when εtends to zero of a three-dimensional viscous fluid satisfying slip boundary conditions is the same as if we assume no-slip (adherence) boundary conditions. In the present paper we consider the case where the period is still εbut the amplitude is δεwith δε/ε converging to zero. We show that if δε/ε3 2tends to infinity, the equivalence between the slip and no-slip conditions still holds. If the limit of δε/ε3 2belongs to (0,+∞) (critical size) then we still have the slip boundary conditions in the limit but with a bigger friction coefficient. In the case where δε/ε3 2tends to zero the boundary behaves as a plane boundary. Besides the limit equation, we also obtain an approximation (corrector result) of the pressure and the velocity in the strong topology of L2and H1respectively. 1 Introduction For a viscous fluid in a three-dimensional domain such that its boundary is covered by microscopic periodic asperities, it has been proved that to impose slip or no-slip conditions on the boundary is asymptotically equivalent. From a physical point of view, this justifies that noslip (adherence) conditions on the boundary are usually imposed for viscous fluids. The above assertion was proved in [12] for a boundary described by the equation x3=−εΨ(x1 ε,x2 ε)∀(x1, x2)∈ω, (1.1) with ε > 0 devoted to converge to zero, ωa bounded open set of R2and Ψ a smooth periodic function such that Span({∇Ψ(y0) : y0∈R2}) = R2.(1.2) These results have been generalized in [6] to a non-periodic boundary x3= Φε(x1, x2)∀(x1, x2)∈ω, (1.3) assuming that Φεconverges ∗-weakly to zero in W1,∞(ω) and it is such that the support of the Young’s measure associated to ∇Φεcontains two non-linear independent vectors. 1 In the present paper, we are interested in the case of a weak rugosity described by Γε=nx= (x1, x2, x3)∈ω×R:x3=−δεΨ(x1 ε,x2 ε)o,(1.4) where δε>0 satisfies lim ε→0 δε ε= 0, and where Ψ is periodic and smooth. Remark that the assumptions imposed in [6] are not satisfied in this case. Indeed, if Φε(x1, x2) = −δεΨ(x1 ε,x2 ε), then ∇Φε(x1, x2) = −δε ε∇Ψ(x1 ε,x2 ε) converges strongly to zero in L∞(ω)2. Taking Ωε=nx= (x1, x2, x3)∈ω×R:−δεΨ³x1 ε,x2 ε´< x3<1o,Ω = ω×(0,1),(1.5) we will show that the following result proved in [12] for δε=εstill holds if δε/ε3 2tends to infinity Theorem 1.1 If uεis a sequence in H1(Ωε)3which satisfies the slip condition uεν= 0 on Γε (νdenotes the unit outward normal to Ωεon Γε) and it is such that kuεkH1(Ωε)3is bounded, then the weak limit u= (u1, u2, u3)of uεin H1(Ω)3(which exists at least for a subsequence) satisfies u3(x1, x2,0) = 0, 2 X i=1 ui(x1, x2,0)∂iΨ(y1, y2) = 0 a.e. (x1, x2, y1, y2)∈ω×R2. Applying this result to a viscous fluid in Ωεand assuming (1.2), we deduce that the slip and no-slip conditions on Γεare asymptotically equivalent in this case. However, we prove that this result does not hold if the limit λof δε/ε3 2belongs to (0,+∞). Indeed, if uεis a solution of the Navier-Stokes system, satisfying Navier’s law on Γε                −µ∆uε+∇pε+ (uε·∇)uε=fin Ωε div uε= 0 in Ωε uεν= 0 on Γε,∂Tuε ∂ν +γTuε= 0 on Γε uε= 0 on ∂Ωε\Γε (1.6) where Tis the orthogonal projection on the tangent space to Γε,µ > 0, γ≥0, and fbelongs to L6 5(ω×(−1,1))3, then the weak limits u= (u1, u2, u3) of uεin H1(Ω)3and pof pεin L2(Ω) (which exist at least for a subsequence) are also a solution of the corresponding Navier-Stokes system in Ω satisfying Navier’s law on {x3= 0} u3= 0 on {x3= 0},−∂(u1, u2) ∂x3 +γ(u1, u2) + λ2R(u1, u2) = 0 on {x3= 0},(1.7) where Ris a nonnegative symmetric squared matrix of order 2 which does not depend on λ. In this case we do not have the adherence condition in the limit but the rugosity is large enough to enlarge the friction coefficient in the limit from γI to γI +λ2R. If (1.2) is satisfied then Ris positive, and then, taking λconverging to infinity in (1.7) we recuperate the adherence condition u= 0 on {x3= 0}. When δε/ε3 2converges to zero we prove that the rugosity is so small that it has not any effect in the limit problem. In this case, the boundary condition on {x3= 0}satisfied in the limit problem is just u3= 0 on {x3= 0},−∂(u1, u2) ∂x3 +γ(u1, u2) = 0 on {x3= 0}. 2 Besides obtaining the limit problem of (1.6) we get a corrector (i.e. a strong approximation) of uεand pεin H1(Ωε)3and L2(Ωε) respectively. The homogenization of problem (1.6) has also been studied in [9] for a very general choice of Ωε, in particular it is not imposed a periodic structure for ∂Ωε. These results of [9] can be applied for Ωεgiven by (1.8), giving the existence of a family of vector spaces {V(x1, x2)}(x1,x2)∈ω⊂ {x3= 0}, a finite positive Borel measure τin {x3= 0}, absolutely continuous respect to the capacity, and a symmetric positively Borel function Ain {x3= 0}valuated in the symmetric nonnegatives matrices, such that, up to a subsequence, uεconverges weakly in H1(Ω)3to a solution uof the variational problem                    u∈H1(Ω)3,div u= 0 in Ω, u(x1, x2,0) ∈V(x1, x2) in ω, Z{x3=0}|u|2dτ < +∞ µZΩ Du :Dv dx +ZΩ [(u·∇)u]v dx +γZ{x3=0} uv dx +Z{x3=0} Auv dτ =ZΩ fv dx, ∀v∈H1(Ω)3,div v= 0 in Ω, v(x1, x2,0) ∈V(x1, x2) in ω, Z{x3=0}|v|2dτ < +∞. The results of the present paper give the value of the spaces V(x1, x2), the measure τand the matrix Aaccording to the value of λ= limε→0δε/ε3 2: •If λ= +∞, then V(x1, x2) = Span({∇Ψ(y0) : y0∈R2})⊥×{0},A≡0. •If λ∈(0,+∞), then V(x1, x2) = R2×{0},τis the usual surface measure corresponding to the plane boundary {x3= 0}and A(x1, x2,0)ξ=λ2R(x1, x2,0)ξfor ξ∈R2×{0}. •If λ= 0, then V(x1, x2) = R2×{0},A≡0. So, our work provides in particular an example where the measure τand the matrix A, whose existence is proved in [9], are different of zero. Another example giving Aand τdifferent of zero has been obtained in [7] and [8], where analogously to (1.8), Ωεis defined by Ωε=nx= (x1, x2, x3)∈ω×R:−εΨ³x1 ε´< x3<1o,(1.8) with Ψ nonnegative, periodic of period 1, and not constant. In this case V(x1, x2) = {0}×R× {0},τis the usual surface measure corresponding to {x3= 0}and A(x1, x2,0)ξ=γµZ1 0p1 + |Ψ0(y)|2dy −1¶ξ, for ξ∈ {0}×R×{0}. We remark that this example has a different nature of the one given in the present paper. Indeed, in [7] and [8] the terms Aand µwhich appear in the limit problem of (1.6) are due to the fact that the surface element measure p1 + |Ψ0(x1 ε)|2dx1dx2, corresponding to the oscillating boundary in Ωε, does not converge to the surface element dx1dx2on {x3= 0}but to R1 0p1 + |Ψ0(y)|2dy dx1dx2. However, in the case where Γεis described by (1.4), the surface element q1 + δ2 ε ε2|Ψ0(x1 ε)|2dx1dx2converges to dx1dx2. The proof of the results corresponding to the present paper is based on an original adaptation of the unfolding method ([5], [10], [14]) which is very related to the two-scale convergence method ([1], [17], [18]). The unfolding method is a very efficient tool to study periodic homogenization problems where the size of the periodic cell tends to zero. The idea is to introduce suitable changes of variables which transform every periodic cell into a simpler reference set by using a supplementary variable (microscopic variable). Although our main interest in the present paper is to study the asymptotic behavior of (1.6), with Ωεdescribed by (1.8) and δε/ε converging to zero, in the last section of the paper we complete the work showing that if the limit of δε/ε is strictly positive (and possibly +∞), 3 then Theorem 1.1 still holds. The proof of this result is also obtained by using the unfolding method. The results of the present paper can be extended for dimension 2 using essentially the same proof. In fact, the results obtained for the Stokes problem hold in arbitrary dimension (taking the second members converging in the convenient spaces). It is important to mention that the exponent 3/2 which appears in the critical case does not depend on the dimension. As we have mentioned above, the present paper is devoted to study the asymptotic behavior of a viscous fluid near a periodic oscillating boundary on which we consider slip conditions. A related problem was considered in [2] assuming that the solution is periodic with the same period of the boundary. We also refer to [3] for the case of Fourier’s conditions. In both works [2] and [3] the boundary is supposed to be described by (1.1). 2 Notation The elements x∈R3will be decomposed as x= (x0, x3) with x0∈R2,x3∈R. By Y0, we denote the unitary cube of R2,Y0= (−1 2,1 2)2, and by b Qthe set b Q=Y0×(0,+∞). For every M > 0 we write b QM=Y0×(0, M). We use the index ]to mean periodicity with respect Y0, for example L2 ](Y0) denotes the space of functions u∈L2 loc(R2) which are Y0-periodic, while L2 ](b Q) denotes the space of functions bu∈L2 loc(R2×(0,+∞)) such that Zb Q|bu|2dy < +∞,bu(y0+k0, y3) = bu(y),∀k0∈Z2,a.e. y∈R2×(0,+∞). For a bounded measurable set Θ ⊂RN, we denote by L2 0(Θ) the space of functions of L2(Θ) with zero mean value in Θ. We denote by εand δεtwo positive parameters devoted to tend to zero such that lim ε→0δε= 0,lim ε→0 δε ε= 0. For a fixed bounded Lipschitz open set ω⊂R2and a function Ψ ∈W2,∞ ](Y0), Ψ ≥0 in Y0, we denote Ω = ω×(0,1),Ωε=½x∈R3:x0∈ω, −δεΨ(x0 ε)< x3<1¾ Γ = ω×{0},Γε=½x∈R3:x0∈ω, x3=−δεΨ(x0 ε)¾. We denote by νthe outside unitary orthogonal vector to Ωεon ∂Ωεand by Tthe orthogonal projection from R3to {ν}⊥(or equivalently, the tangent projection on Γ). Our aim in the present paper is to study the asymptotic behavior of a viscous fluid in Ωε, which satisfies a slip boundary condition on Γε. For this purpose we will use an adaptation of the method introduced in [5] to the study of periodic homogenization problems. It is now known as the unfolding method. We refer to [10], [13], [14], [17] for different applications of this method and its relation with the two-scale convergence method of G. Neguetseng and G. Allaire ([1], [18]). In order to apply the unfolding method, we will need the following notation. For k0∈Z2, we denote Ck0 ε=εk0+εY 0,Ωk0 ε= Ωε∩(Ck0 ε×(−∞,1)). We define κ:R2→Z2by κ(x0) = k0⇔x0∈Ck0 1. 4 Remark that κis well defined up to a set of zero measure in R2(the set ∪k0∈Z2∂Ck0 1). Moreover, for every ε > 0, we have κ(x0 ε) = k0⇔x0∈Ck0 ε. For a.e. x0∈R2we define Cε(x0) = Ck0 εsuch that x0∈Ck0 ε. For every ρ > 0, we take ωρ={x∈ω: dist(x, ∂ω)> ρ} and Iρ,ε =©k0∈Z2:Ck0 ε∩ωρ6=∅ª. We define by Vthe space of functions bv:R2×(0,+∞)→Rsuch that bv∈H1 ](b QM),∀M > 0,∇bv∈L2 ](b Q)3. It is a Hilbert space endowed with the norm k·kVdefined by kbvk2 V=kbvk2 L2(Y0×{0})+k∇bvk2 L2(b Q)3. We denote by Oεa generic real sequence which tends to zero with εand can change from line to line. We denote by Ca generic positive constant which can change from line to line. 3 Main Results In this section we describe the asymptotic behavior of a viscous fluid in the geometry Ωεdescribed in Section 2 and satisfying slip conditions on Γε. The proof of the corresponding results will be given in the next section. Our first result is referred to the Stokes system. Namely, let us consider a sequence (uε, pε)∈ H1(Ωε)3∩L2(Ωε), which satisfies          −∆uε+∇pε=fε−div Gεin Ωε div uε= 0 in Ωε uεν= 0 on Γε,∂Tuε ∂ν −TGεν−Tgε= 0 on Γε, (3.1) The second members fε∈L6 5(Ωε)3,Gε∈L2(Ωε)3×3,gε∈L2(Γε)3are assumed to satisfy kfεkL6 5(Ωε)3+kGεkL2(Ωε)3×3+kgεkL2(Γε)3≤C, ∀ε > 0,(3.2) and lim ε→0Z{x3<sε}³|fε|6 5+|Gε|2´dx = 0,∀s > 0.(3.3) Remark 3.1 Condition (3.2) implies that, up to a subsequence, there exist f∈L6 5(Ω)3,G∈ L2(Ω)3×3,g∈L2(Γ)3such that fε* f in L6 5(Ω)3, Gε* G in L2(Ω)3×3, gε¡x0,−δεΨ(x0 ε)¢* g in L2(Γ)3.(3.4) Observe that we are not imposing any boundary conditions on ∂Ωε\Γεin (3.1). We are interested in the behavior of (uε, pε) near Γεand this does not depend on these boundary conditions. However these conditions are necessary to obtain an existence and uniqueness result of solution of problem (3.1). In this way, we have 5 Theorem 3.2 We consider fε∈L6 5(Ωε)3,Gε∈L2(Ωε)3×3and gε∈L2(Γε)3which satisfy (3.2). Then, adding one of the following conditions i) uε= 0 on ∂Ωε\Γε.(3.5) ii)          ωis a rectangle, uε,∂uε ∂ν −pεν−Gενare periodic of period ωwith respect to x0, uε= 0 on {1}×ω, (3.6) problem (3.1) has a unique solution (uε, pε)in H1(Ωε)3×L2 0(Ωε). Moreover, there exists C > 0, which does not depend on ε, such that kuεkH1(Ωε)3+kpεkL2(Ωε)≤C. (3.7) Instead of supposing some boundary conditions on ∂Ωε\Γε, Theorem 3.4 below describes the asymptotic behavior of a solution (uε, pε) of (3.1) such that (3.7) holds. Indeed, next Proposition asserts that it is enough to assume the existence of uε∈H1(Ωε)3, with kuεkH1(Ωε)3bounded such that              ZΩε (Duε−Gε) : Dv dx =ZΩε fεv dx, ∀v∈H1 0(Ωε)3,with div v= 0 in Ωε div uε= 0 in Ωε uεν= 0 on Γε,∂Tuε ∂ν −TGεν−Tgε= 0 on Γε. Proposition 3.3 We consider fε∈L6 5(Ωε)3,Gε∈L2(Ωε)3×3and gε∈L2(Γε)3which satisfy (3.2) and a sequence uε∈H1(Ωε)3such that kuεkH1(Ωε)3is bounded and ZΩε (Duε−Gε) : Dv dx =ZΩε fεv dx, ∀v∈H1 0(Ωε)3,with div v= 0 in Ωε. Then, there exists a unique pε∈L2 0(Ωε), such that uεsatisfies −∆uε+∇pε=fε−div Gεin Ωε. Moreover kpεkL2 0(Ωε)is bounded. Our main result referred to the asymptotic behavior of the solution of (3.1) is given by the following Theorem. Theorem 3.4 We consider fε∈L6 5(Ωε)3,Gε∈L2(Ωε)3×3and gε∈L2(Γε)3satisfying (3.2), (3.3) and such that there exist f∈L6 5(Ω)3,G∈L2(Ω)3×3,g∈L2(Γ)3, which satisfy (3.4). We also assume (uε, pε)is a solution of (3.1) such that (3.7) holds. Then, there exists (u, p)∈ H1(Ω)3×L2 0(Ω) such that, up to a subsequence, uε* u in H1(Ω)3, pε* p in L2(Ω).(3.8) The pair (u, p)satisfies the Stokes system (−∆u+∇p=f−div G in Ω div u= 0 in Ω.(3.9) Moreover, denoting (this limit exists at least for a subsequence) λ= lim ε→0 δε ε3 2∈[0,+∞],(3.10) it also satisfies the following boundary condition on Γ 6 i) If λ= 0, then (u3= 0 on Γ −∂3u0+ (Ge3)0−g0= 0 on Γ.(3.11) ii) If λ∈(0,+∞), then defining (b φi,bqi)∈V3×L2 ](b Q),i= 1,2as a solution of                −∆b φi+∇bqi= 0 in R2×R+ div b φi= 0 in R2×R+ b φi 3=∂iΨon R2×{0} −∂3(b φi)0= 0 on R2×{0}, (3.12) and R∈R2×2by Ri,j =Zb Q Db φi:Db φjdy, ∀i, j ∈ {1,2}(3.13) we have (u3= 0 on Γ −∂3u0+λ2Ru0+ (Ge3)0−g0= 0 on Γ.(3.14) iii) If λ= +∞, then defining W=Span({(∇Ψ(y0),0) : y0∈Y0}∪{e3}),(3.15) and Qthe orthogonal projection from R3to W⊥, we have (u∈W⊥on Γ −∂3Qu +QGe3−Qg = 0 on Γ.(3.16) The matrix Rwhich appears in problem (3.14) is defined throughout a solution (b φi,bqi) of (3.12). The following Proposition ensure the existence and uniqueness of solution of this problem. It also gives some smoothness properties for (b φi,bqi). In particular it shows that Db φiis uniquely defined and then that the matrix Ris well defined. Proposition 3.5 There exists a unique solution (b φi,bqi)of problem (3.14) in ¡(V/R)2×V¢× L2 ](b Q)¢. Moreover, for every r∈[2,+∞), one has kDb φikLr(b Q)3×3+kbqikLr(b Q)<+∞.(3.17) Remark 3.6 From Lemma 4 in [4] (see also [16] for related results) we can easily show that every solution (b φi,bqi),i= 1,2, of (3.12) is in C∞ ](ˆ Q)3×C∞ ](ˆ Q)and there exists a unique solution in ¡(V/R)2×V¢×L2 ](b Q)¢. Moreover, for every α∈NNand every z > 0there exist two positive constants Cz,α,τ(the last one does not depend on αor z) such that |Dαb φi|(y) + |Dαbqi|(y)≤Cz,αe−τy3,∀y∈R2×(z, +∞). Remark 3.7 For λ= 0, the boundary conditions on Γ, (3.11), in the limit problem of (3.1) are the same we would find if Γεdoes not vary with ε, i.e. in this case the rugosity of Γεis very slight and the solution (uε, pε)of (3.1) behaves as if Γεcoincides with the plane boundary Γ. For 0<λ<+∞(critical size), the boundary condition satisfied by the limit uof uεon the tangent space to Γcontains the new term λ2Ru. The effect of the rugosity of the wall Γεis not worthless in this case. Finally, for λ= +∞the rugosity of Γεis so strong that the limit uof uε does not only satisfies the condition u3= 0 on Γ, but it is also such that its tangent velocity on 7 Γ,u0, is orthogonal to the vectors ∇Ψ(y0), for every y0∈Y0. In particular, if the linear space spanned by {∇Ψ(y0) : y0∈Y0}has dimension 2 (this holds if and only if Ψis not constant in any straight line of R2, see [12]), we get that usatisfies the adherence condition u= 0 on Γ, i.e. although we have imposed a slip condition on Γε, the rugosity forces uto satisfy a no-slip (adherence) condition on Γ. This result extends to the case where lim ε→0 δε ε= 0,lim ε→0 δε ε3 2 = +∞, the results obtained in [12] for δε=ε(see also [6] for the nonperiodic case). The limit equation (3.14) corresponding to the critical size λ∈(0,+∞)can be considered as the general one. In fact, if λis tending to zero or +∞in (3.14) we get (3.11) and (3.16) respectively. Remark 3.8 If in Theorem 3.4, we also assume that one of the conditions (3.5) or (3.6) holds, then, assuming that there exists the limit λgiven by (3.10), we deduce that (3.8) holds without extracting any subsequence. Moreover, if (3.5) is satisfied, then u= 0 on ∂Ω\Γ, while if (3.6) is satisfied, then u,∂u ∂ν −pν −Gν are periodic with respect to x0and u= 0 on {1}×ω. Theorem 3.4 gives an approximation of (uε, pε) in the weak topology of H1(Ω)3×L2(Ω). Assuming that the second members fε,Gεand gεof (3.1) satisfy |fε|6 5is equiintegrable in Ω,(3.18) Gεconverges strongly to Gin L2(Ω)3×3,(3.19) let us now obtain an asymptotic expansion of (DuεχΩε, pεχΩε) which converges in the strong topology of R2(corrector result). Theorem 3.9 We consider fε∈L6 5(Ωε)3,Gε∈L2(Ωε)3×3and gε∈L2(Γε)3which satisfy (3.2),(3.3) and (3.18) and are such that there exist f∈L6 5(Ω)3,G∈L2(Ω)3×3and g∈L2(Γ)3 which satisfy (3.4) and (3.19). Let (uε, pε)∈H1(Ωε)3×L2(Ωε)be a solution of (3.1) which satisfies (3.7) and it is such that there exists (u, p)∈H1(Ω)3×L2(Ω) which satisfies (3.8). We also assume that there exists the limit λgiven by (3.10). Then, we have i) lim ε→0ÃZΩε\Ω|uε|2dx +ZΩ|uε−u|2dx!= 0.(3.20) ii) If λ= 0 or +∞, then, for every ϕ∈C1 c(ω×(−1,1)) we have lim ε→0ÃZΩε\Ω|Duε|2ϕ dx +ZΩ|D(uε−u)|2ϕ dx!= 0,(3.21) lim ε→0ÃZΩε\Ω|pε|2ϕ dx +ZΩ|pε−p|2ϕ dx!= 0.(3.22) iii) If λ∈(0,+∞), then, taking (b φi,bqi),i= 1,2as a solution of (3.12) and defining bu: ω×(R2×(0,+∞)) →R3and bp:ω×(R2×(0,+∞)) →Rby bu(x0, y) = −λu1(x0,0)b φ1(y)−λu2(x0,0)b φ2(y),(3.23) bp(x0, y) = −λu1(x0,0)bq1(y)−λu2(x0,0)bq2(y),(3.24) 8 for a.e. (x0, y)∈ω×(R2×(0,+∞)), we have for every ϕ∈C1 c(ω×(−1,1)) lim ε→0ÃZΩε\Ω|Duε|2ϕ dx +ZΩ¯¯¯¯Duε−Du −1 √εDybu(x0,x ε)¯¯¯¯ 2 ϕ dx!= 0,(3.25) lim ε→0ÃZΩε\Ω|pε|2ϕ dx +ZΩ¯¯¯¯pε−p−1 √εbp(x0,x ε)¯¯¯¯ 2 ϕ dx!= 0.(3.26) Remark 3.10 If in Theorem 3.9, we also assume that (uε, pε)satisfies one of the assumptions (3.5) or (3.6), then in (3.21), (3.22), (3.25) and (3.26), we can take ϕ= 1. As a consequence of the results given above for the Stokes system, we can now describe the asymptotic behavior of the Navier-Stokes system posed in Ωε. To simplify the exposition we assume that the second member are fixed functions. The case of varying second members fε, Gεand gεis analogous. Theorem 3.11 For µ > 0,γ≥0and f∈L6 5(ω×(−1,1))3, we consider a solution (uε, pε)∈ H1(Ωε)3×L2(Ωε)of the Navier-Stokes system                −µ∆uε+∇pε+ (uε·∇)uε=fin Ωε div uε= 0 in Ωε uεν= 0 on Γε ∂Tuε ∂ν +γTuε= 0 on Γε, (3.27) which satisfies (3.7). Then, up to a subsequence, we have (3.8) with (u, p)solution of the Navier-Stokes system (−µ∆u+∇p+ (u·∇)u=fin Ω div u= 0 in Ω.(3.28) Defining λby (3.10) (this limit exists up to a subsequence), we also have i) If λ= 0, then (u3= 0 on Γ −∂3u0+γu0= 0 on Γ.(3.29) Moreover (3.21), (3.22) hold for every ϕ∈C1 c(ω×(−1,1)). ii) If λ∈(0,+∞), then (u3= 0 on Γ −∂3u0+γu0+λ2Ru0= 0 on Γ,(3.30) with Rdefined by (3.13). Moreover, defining bu,bpby (3.23), (3.24) we have (3.25), (3.26) for every ϕ∈C1 c(ω×(−1,1)). iii) If λ= +∞, then defining Wby (3.15) and Qas the ortogonal projection from R3onto W⊥we have (u∈W⊥on Γ −∂3Qu +γQu = 0 on Γ.(3.31) Moreover, (3.21), (3.22) hold for every ϕ∈C1 c(ω×(−1,1)). Remark 3.12 Assuming also that (uε, pε)satisfies one of the boundary conditions given by (3.5) or (3.6) there exists at least a solution of the Navier-Stokes system (3.27). Moreover, this solution satisfies (3.7) and thus, Theorem 3.11 can be applied. In this case, we can take ϕ= 1 in (3.21), (3.22), (3.25), (3.26). 9 Step 1. We start with the most difficult case λ∈(0,+∞) (critical size), which we will carry out more in detail. We consider v∈C1 c(ω×(−1,1))3,bv∈C1 c(ω;C1 ](b Q)3), with Dybv(x0, y) = 0 a.e. in {y3> M}, for some M > 0, such that (v(x0, x3) = v(x0,0) if x3≤0 v3(x0,0) = 0,(bv(x0, y0, y3) = bv(x0, y0,0) if y3≤0 bv3(x0, y0,0) = −λ∇Ψ(y0)v0(x0,0).(4.28) Besides, we take ζ∈C∞(R) such that ζ(x3) = 1 if x3<1 3, ζ(x3) = 0 if x3>2 3,(4.29) and Rε>0 such that Rε→ ∞, Rε"µδε ε3 2−λ¶2 +ε#→0.(4.30) Then, we define vε∈H1(Ωε)3by                      v0 ε(x) = v0(x) + √εbv0(x0,x ε)ζ(x3) vε,3(x) = v3(x) + √ε"bv3(x0,x ε)ζ(x3)+ +ζ(x3 εRε )∇Ψ(x0 ε)µµλ−δε ε3 2¶v0(x0,0) −δε εbv0(x0,x0 ε,0)¶#. Since v∈C1 c(ω×(−1,1))3,bv∈C1 c(ω;C1 ](b Q)3) and ζ(x3) = 0 if x3>2/3, the sequence vε satisfies vε= 0 on ∂Ωε\Γε.(4.31) Moreover, taking into account the properties (4.28) of vand bv, it is not difficult to check that vεν= 0 on Γε.(4.32) Properties (4.31) and (4.32) of vεallow us to take it as test function in (4.27). To simplify the calculus we will first estimate the derivative of vε. Taking into account that Dybv= 0 a.e. in {y3> M}and that ζ= 1 a.e. on {x3<1/3}, we have Dvε(x) = Dv(x) + 1 √εDybv(x0,x ε) + hε(x),(4.33) where, using that v,bvand ζare bounded and have bounded derivatives, the function hε∈ C0(¯ Ωε)3×3satisfies |hε| ≤ C√ε+C·µ 1 √εRε +1 √嶵¯¯¯¯ δε ε3 2−λ¯¯¯¯+δε ε¶+√ε¯¯¯¯ δε ε3 2−λ¯¯¯¯+δε ε3 2¸χ{x3<εRε}, a.e. in Ωε. Using that Rεtends to infinity and that δε/ε3 2is bounded, we get |hε| ≤ C√ε+C·1 √ε¯¯¯¯ δε ε3 2−λ¯¯¯¯+ 1¸χ{x3<2 3εRε},a.e. in Ωε. Therefore, by (4.30), we have ZΩε|hε|2dx ≤Oε+CRε"µδε ε3 2−λ¶2 +ε#=Oε.(4.34) 16 Taking vεas test function in (4.27) and using that kuεkH1(Ωε)3,kpεkL2(Ωε)are bounded, kvε−vkC0(¯ Ωε)3tends to zero, (4.33) and (4.34), we get ZΩε (Duε:Dv −pεdiv v)dx +1 √εZΩε³Duε:Dybv(x0,x ε)−pεdivybv(x0,x ε)´dx =ZΩε fεv dx +ZΩε Gε:µDv +1 √εDybv(x0,x ε)¶dx +ZΓε gεv dσ +Oε. (4.35) In this equality we use that ZΩε\Ω¯¯¯¯Dv +1 √εDybv(x0,x ε)¯¯¯¯ 2 dx ≤C ε|Ωε\Ω| ≤ Cδε ε≤C√ε, and that (3.3) and Dybv= 0 a.e. in {y3> M}imply 1 √εZΩε|GεDybv(x0,x ε)|dx ≤C √εÃZ{x3<Mε}|Gε|2dx!1 2 |{x3< Mε}|1 2=Oε. Therefore, (4.35) can be written as ZΩ (Duε:Dv −pεdiv v)dx +1 √εZΩ³Duε:Dybv(x0,x ε)−pεdivybv(x0,x ε)´dx =ZΩ fεv dx +ZΩ Gε:Dv dx +ZΓε gεv dσ +Oε, which taking into account (3.4), (3.8) and (4.28) proves ZΩ (Du :Dv −pdiv v)dx +1 √εZΩ³Duε:Dybv(x0,x ε)−pεdivybv(x0,x ε)´dx =ZΩ fv dx +ZΩ G:Dv dx +ZΓ gv dσ +Oε. (4.36) In order to estimate the second term in (4.36), we introduce the sequences buε,bpεrespectively defined by (4.5) and (4.6). By (3.7) and Lemmas 4.4 and 4.5, we can assume that there exist bp∈L2(ω×b Q) and bu∈L2(ω;V3) which satisfy (4.8), (4.11), (4.12) and (4.13), 1 √εZΩ³Duε:Dybv(x0,x ε)−pεdivybv(x0,x ε)´dx =ZωZb QMµ1 √εDybuε:Dybv−√εbpεdivybv¶dy dx0 =Zω×b Q¡Dybu:Dybv−bpdivybv¢dx0dy +Oε. Substituting in (4.36) we set ZΩ (Du :Dv −pdiv v)dx +Zω×b Q¡Dybu:Dybv−bpdivybv¢dx0dy =ZΩ fv dx +ZΩ G:Dv dx +ZΓ gv dσ, (4.37) for every v∈C1 c(ω×(−1,1))3,bv∈C1 c(ω;C1 ](b Q)3), with Dybv(x0, y) = 0 a.e. in {y3> M}, for some M > 0, and such that (4.28) is satisfied. By density, this equality holds true for every v∈H1(Ω)3, and every bv∈L2(ω;V3) such that v= 0 on ∂Ω\Γ, 17 v3(x0,0) = 0,bv3(x0, y0,0) = −λ∇Ψ(y0)v0(x0,0),a.e. (x0, y0)∈ω×Y0. Let us now obtain an equation for ueliminating buand bpin (4.37). For this purpose, we take v= 0 in (4.37). This proves that (bu, bp) (extended by periodicity to ω×R2×(0,+∞)) is a solution of                    −∆ybu+∇ybp= 0 in R2×R+ divybu= 0 in R2×(0,+∞) (bu, bp)∈V3×L2 ](b Q) bu3(x0, y0,0) = −λ∇Ψ(y0)u0(x0,0) on R2×{0} −∂y3bu0= 0 on R2×{0}, (4.38) a.e. in ω. Defining (b φi,bqi), i= 1,2, by (3.12), we deduce by linearity and uniqueness Dybu(x0, y) = −λ(u1(x0,0)Dyb φ1(y) + u2(x0,0)Dyb φ2(y)) a.e. in R2×(0,+∞),(4.39) bp(x0, y) = λ(u1(x0,0)bq1(y) + u2(x0,0)bq2(y)) a.e. in R2×(0,+∞).(4.40) Now, for v∈H1(Ω)3, with v= 0 on ∂Ω\Γ, v3= 0 on Γ, we take vand ˆv(x0, y) = −λ¡v1(x0,0)b φ1(y) + v2(x0,0)b φ2(y)¢, as test functions in (4.37). Taking into account (4.39) we get ZΩ Du :Dv dx −ZΩ pdiv v dx +λ2ZΓ Ru0v0dx0 =ZΩ fv dx +ZΩ G:Dv dx +ZΓ gv dx0. (4.41) By the arbitrariness of v, this proves that (u, p) is a solution of (3.14). Step 2. The case λ= 0. As in Step 1, we consider v∈C1 c(ω×(−1,1))3, with v(x0, x3) = v(x0,0) if x3≤0, v3= 0 on Γ. Then, for ζ∈C∞(R) which satisfies (4.29), we define vε∈H1(ω×(−1,1))3by    v0 ε(x) = v0(x) vε,3(x) = v3(x)−δε εζ(x3 ε)∇Ψ(x0 ε)v0(x). The sequence vεsatisfies vεν= 0 on Γεand vε= 0 on ∂Ωε\Γε. We take vεas test function in (4.27). Using (3.3), (3.4), (3.8) and that λ= 0 implies ZΩε|D(vε−v)|2dx →0,kvε−vkL∞(Ωε)3→0,(4.42) we can pass to the limit in (4.27) to get     ZΩ Du :Dv dx −ZΩ pdiv v dx =ZΩ fv dx +ZΩ G:Dv dx +ZΓ gv dx0 ∀v∈H1(Ω)3, v3= 0 on Γ, v = 0 on ∂Ω\Γ. (4.43) This is equivalent to (3.11). Step 3. The case λ= +∞. We consider v∈C1 c(ω×(−1,1))3, with v(x0, x3) = v(x0,0) if x3≤0, v3= 0 on Γ and such that v0(x0,0)∇Ψ(y0) = 0 a.e. (x0, y0)∈ω×Y0. 18 Observe that the properties of vimply that vν = 0 on Γε,v= 0 on ∂Ωε\Γε. Taking vε=vin (4.27), passing to the limit in εand reasoning by density we get     ZΩ Du :Dv dx −ZΩ pdiv v dx =ZΩ fv dx +ZΩ G:Dv dx +ZΓ gv dx0 ∀v∈H1(Ω)3, v3= 0 on Γ, v0(x0,0)∇Ψ(y0) = 0 a.e. (x0, y0)∈ω×Y0, v = 0 on ∂Ω\Γ. This is equivalent to (3.16). ¤ Proof of Proposition 3.5. To simplify the notation we just prove the result for the pair (b φ1,bq1) which will be just denoted by (b φ, bq) (the case i= 2 is completely analogous). Step 1. Existence of b φ. The function b φcan be defined equivalently as the solution of the variational problem            b φ∈(V/R)2×V,b φ3=∂1Ψ on Y0×{0},div b φ= 0 in b Q Zb Q Db φ:Dbv dy = 0 ∀bv∈(V/R)2×V,bv3= 0 on Y0×{0},div bv= 0 in b Q, (4.44) and then its existence and uniqueness follows from the Lax-Milgram theorem. Step 2. Extension of b φto R3and existence of the pressure bq. For ζ∈C∞((−∞,0]) with ζ(0) = 1, dζ dt (0) = d2ζ dt2(0) = 0 and ζ(t) = 0 if t < −1, we extend b φto Y0×Rby            b φ1(y) = b φ1(y0,−y3)−2dζ dy3 (y3)Ψ(y0) b φ2(y) = b φ2(y0,−y3) b φ3(y) = −b φ3(y0,−y3) + 2ζ(y3)∂1Ψ(y0), a.e. in Y0×(−∞,0). Then, denoting by Wthe space of functions bw:R3→Rsuch that bw∈H1 ](Y0×(−M, M)),∀M > 0,∇bw∈L2 ](b Y0×R)3, we have that b φsatisfies            b φ∈W3,div b φ= 0 in Y0×R Zb Q (Db φ+H) : Dbw dy = 0 ∀bw∈W3,div bw= 0 in Y0×R, (4.45) where His a matrix function defined by zero in Y0×((−∞,−1) ∪(0,+∞)) and by H(y) = 2        dζ dy3 (y3)∇Ψ(y0)d2ζ dy2 3 (y3)Ψ(y0) 0 0 −ζ(y3)∇∂1Ψ(y0)−dζ dt (y3)∂1Ψ(y0)        ,∀y∈Y0×(−1,0). This implies (reasoning similarly to Lemma A.1 in [11], see also [15]) that for every bϕ∈H1(R3)3 with compact support and such that div bϕ= 0 in R3, one has ZR3 (Db φ+H) : Dbϕ dy = 0, 19 and so, there exists bq∗∈L2 loc(R3) such that ZR3bq∗div bϕ=ZR3 (Db φ+H) : Dbϕ dy, (4.46) for every bϕ∈H1(R3)3with compact support. Step 3. Let us prove that bq∗is periodic with respect to y0with period Y0. Since Db φ+His periodic with respect to y0with period Y0, equation (4.46) implies that for every bϕ∈H1(R3)3with compact support we have ZR3 (bq∗(y1+ 1, y2, y3)−bq∗(y))div bϕ(y)dy = 0. Using that for every bh∈L2(R3) with compact support and mean value zero there exists bϕ∈ H1(R3)3with compact support and div bϕ=bhin R3, we deduce that bq∗satisfies ZR3 (bq∗(y1+ 1, y2, y3)−bq∗(y))bh(y)dy = 0, for every bh∈L2(R3) with compact support and mean value zero. This implies that there exists c∈Rsatisfying bq∗(y1+ 1, y2, y3) = bq∗(y) + ca.e. y∈R3.(4.47) We define bb(y) = (∂2b φ1(y) + H1,2(y), ∂3b φ1(y) + H1,3(y)),¯ b(y2, y3) = Z1 0bb(y)dy1 and, for n∈N,bρn:R→Rby bρn(s) =              sif 0 ≤s≤1 1 if 1 ≤s≤n n+ 1 −sif n≤s≤n+ 1 0 if s≤0 or s≥n+ 1. Then, for bη∈C∞ c(R2), we take bϕ(y)=(bρn(y1)bη(y2, y3),0,0) as test function in (4.46). Using the periodicity of Db φ+Hand (4.47), we get −nc ZR2bη dy2dy3=ZR2bη·Z1 0 (bq∗(y)−bq∗(y1+n, y2, y3)) dy1¸dy2dy3=ZR3bq∗div bϕ dy =Zn+1 0bρn(y1)ZR2 ˆ b∇bη dy2dy3dy1=nZR2 ¯ b∇bη dy2dy3, for every bη∈C∞ c(R2). This implies c= div ¯ bin R2. Integrating this equality in (−1 2,1 2)×(t,t+1), with t > 0 and taking into account that ¯ b= (¯ b2,¯ b3) is periodic of period 1 with respect to y2 we deduce c=Z1 2 −1 2 (¯ b3(y2, t + 1) −¯ b3(y2, t))dy2, Integrating again for t∈(s, s + 1) with s > 0, this implies |c| ≤ Z(−1 2,1 2)×(s,s+2) |¯ b3(y2, y3)|dy2dy3≤√2ÃZ(−1 2,1 2)×(s,s+2) |¯ b3(y2, y3)|2dy2dy3!1 2 , 20 for every s > 0. Since k¯ b3(y2, y3)kL2((−1/2,1/2)×R)<+∞,we conclude that c= 0 and then by (4.47) it follows bq∗(y1+1, y2, y3) = bq∗(y), for a.e. y∈R3. Analogously, bq∗(y1, y2+1, y3) = bq∗(y), for a.e. y∈R3. Thus bqis periodic with respect to y0with period Y0. Step 4. To finish the proof of Proposition 3.5, let us prove that we can choose a representative bqof bq∗(remark that bq∗is defined up to a constant) such that for every r≥2 kDb φkLr(Y0×R)3×3+k∇bqkLr(Y0×R)<+∞.(4.48) This will imply that the pair (b φ, bq) satisfies the thesis of Proposition 3.5 in R2×(0,+∞) (the uniqueness of (b φ, bq) in ¡(V/R)2×V¢×L2 ](b Q) is straightforward). First we prove that for every r≥2 there exists Cr>0 such that for any n∈N,      kDb φkLr(nY 0×(−n,n))3×3+kbq∗−¯qnkLr(nY 0×(−n,n)) ≤Cr n3(r−2) 2rkDb φkL2(2nY 0×(−2n,2n))3×3+CrkHkLr(2nY 0×(−2n,2n))3×3,(4.49) where we have denoted ¯qn=1 2n3ZnY 0×(−n,n)bq∗dy. For n= 1, the proof of (4.49) follows from Theorem 4.4, chapter 4, in [16], the general case follows using a dilatation which transforms Y0×(−1,1) in nY 0×(−n, n). Using the periodicity with respect to y0of b φ,bq∗and Hwe can write (4.49) as      kDb φkLr(Y0×(−n,n))3×3+kbq∗−¯qnkLr(Y0×(−n,n)) ≤2Cr n(r−2) 2rkDb φkL2(Y0×(−2n,2n))3×3+ 2CrkHkLr(Y0×(−2n,2n))3×3.(4.50) This inequality implies that, up to a subsequence, there exists the limit bqof bq∗−¯qn. Passing to the limit in nin (4.50) we get (4.48). ¤ Proof of Theorem 3.9. Step 1. Let us first prove (3.20). The Rellich-Kondrachov theorem and (3.8) give that uεconverges to ustrongly in L2(Ω)3. On the other hand, from (4.3) and H¨older’s inequality we have lim ε→0ZΩε\Ω|uε|2dx ≤lim sup ε→0³kuεk2 L6(Ωε)3|Ωε\Ω|2 3´≤lim sup ε→0 Cδ 2 3 εkuεk2 H1(Ωε)3= 0.(4.51) This proves that (3.20)) holds (for any value of λ∈[0,+∞]). In Steps 2, 3 and 4, let us prove the corrector result for Duεand pε. Step 2. We consider vε∈H1(Ωε)3such that vεν= 0 on Γε,kvεkH1(Ωε)3≤C, (4.52) and such that there exists v∈H1(Ω)3with vεconverging weakly to vin H1(Ω)3. By Lemma 4.5, the third component v3of vvanishes on Γ, and if λ= +∞it also holds v0(x0,0)∇Ψ(y0) = 0 a.e. (x0, y0)∈ω×Y0. Let us prove that for any ϕ∈C1 c(ω×(−1,1)), we have          lim ε→0µZΩε Duε:Dvεϕ dx −ZΩε pεdiv vεϕ dx¶ =ZΩ Du :Dv ϕ dx −ZΩ pdiv v ϕ dx, if λ= 0,+∞,(4.53) 21          lim ε→0µZΩε Duε:Dvεϕ dx −ZΩε pεdiv vεϕ dx¶ =ZΩ Du :Dv ϕ dx −ZΩ pdiv v ϕ dx +λ2ZΓ Ru0v0ϕ dx0, if λ∈(0,+∞).(4.54) where Ris defined by (3.13). For this purpose, given ϕ∈C1 c(ω×(−1,1)), we take vεϕas test function in (3.1). This gives ZΩε Duε:D(vεϕ)dx −ZΩε pεdiv (vεϕ)dx =ZΩε fεvεϕdx +ZΩε Gε:D(vεϕ)dx +ZΓε gεvεϕdσ. (4.55) Let us pass to the limit in each term of this equality. Using that fεconverges weakly to fin L6 5(Ω)3,|fε|6is equiintegrable, vεconverges to vin measure in Ω and vεconverges weakly to vin L6(Ω)3, we have that fεvεconverges strongly to fv in L1(Ω). Therefore ZΩ fεvεϕ dx =ZΩ fvϕ dx +Oε.(4.56) On the other hand, by (3.3), (4.3), we get ZΩε\Ω fεvεϕ dx =Oε.(4.57) Reasoning analogously with the second term of (4.55), thanks to the strong convergence of Gεin L2(Ω)3×3, we have ZΩε Gε:D(vεϕ)dx =ZΩ G:D(vϕ)dx +Oε.(4.58) For the las term in (4.55), we use ZΓε gεvεϕ dσ =Zω (gεvεϕ)¡x0,−δεΨ(x0 ε)¢s1 + µδε ε¶2¯¯¯¯∇Ψ(x0 ε)¯¯¯¯ 2 dx0. The inequality            Zω¯¯¯¯vε¡x0,−δεΨ(x0 ε)¢−vε(x0,0)¯¯¯¯ 2 dx0=Zω¯¯¯¯¯Z0 −δεΨ(x0 ε) ∂3vε(x0, s)ds¯¯¯¯¯ 2 dx0 ≤CδεZΩε|∂3vε|2dx =Oε, (4.59) and the compact imbedding of H1(Ω) into L2(Γ) give that vε(x0,−δεΨ(x0 ε)) converges strongly to v(x0,0) in L2(ω)3. Thus, the weak convergence of gε(x0,−δεΨ(x0 ε)) to g(x0) in L2(ω)3, Ψ in W2,∞ ](Y0) and δε/ε tends to zero, imply ZΓε gεvεϕ dσ =ZΓ gvϕ dσ +Oε.(4.60) As vεconverges to vstrongly in L2(Ω)3and pεconverges to pweakly in L2(Ω), we have ZΩ pεvε∇ϕ dx =ZΩ pv∇ϕ dx +Oε.(4.61) On the other hand, thanks to (4.3) and (3.7) ¯¯¯¯¯ZΩε\Ω pεvε∇ϕ dx¯¯¯¯¯≤ kpεkL2(Ωε)kvεkL6(Ωε)3k∇ϕkL3(Ωε\Ω)3=Oε, 22 which, together with (4.61), gives ZΩε pεdiv (vεϕ)dx =ZΩε pεdiv vεϕ dx +ZΩ p v∇ϕ dx +Oε.(4.62) Finally, from the equality ZΩε Duε:D(vεϕ)dx =ZΩε Duε:Dvεϕ dx +ZΩε Duε:¡vε⊗∇ϕ¢dx, the strong convergence of vεto vin L2(Ω)3, the weak convergence of Duεto Du in L2(Ω)3×3 and that by (4.3) ¯¯¯¯¯ZΩε\Ω Duε:¡vε⊗∇ϕ¢dx¯¯¯¯¯≤ kDuεkL2(Ωε)3×3kvεkL6(Ωε)3k∇ϕkL3(Ωε\Ω)3=Oε, we derive ZΩε Duε:D(vεϕ)dx =ZΩε Duε:Dvεϕ dx −ZΩ Du : (v⊗∇ϕ)dx +Oε.(4.63) By (4.55), (4.56), (4.57), (4.58), (4.60), (4.62) and (4.63), we have then proved lim ε→0µZΩε Duε:Dvεϕ dx −ZΩε pεdiv (vε)ϕ dx¶=−ZΩ Du : (v⊗∇ϕ)dx +ZΩ p v∇ϕ dx +ZΩ fvϕ dx +ZΩ G:D(vϕ)dx +ZΓ gvϕ dx. But using vϕ as test function in the equation satisfied by the pair (u, p) (Theorem 3.4) we have that the second member of the above equality is equal to        ZΩ Du :Dv ϕ dx −ZΩ pdiv v ϕ dx if λ= 0,+∞ ZΩ Du :Dv ϕ dx −ZΩ pdiv v ϕ dx +λ2ZΓ Ru0v0ϕ dx0if λ∈(0,+∞). This proves (4.53) and (4.54). Step 3. Let us prove (3.21), (3.25). Using vε=uεin Step 2 and taking into account that div uε= 0 in Ωε, equalities (4.53) and (4.54) give for every ϕ∈C1 c(ω×(−1,1)), ϕ≥0 in ω×(−1,1), lim ε→0ZΩε|Duε|2ϕ dx =ZΩ|Du|2ϕ dx, if λ= 0,+∞,(4.64) lim ε→0ZΩε|Duε|2ϕ dx =ZΩ|Du|2ϕ dx +λ2ZΓ Ru0u0ϕ dx0,if λ∈(0,+∞).(4.65) Since uεconverges weakly to uin H1(Ω)3, equality (4.64) proves (3.21). In order to prove (3.25), we take rε>0 such that lim ε→0rε= 0,lim ε→0 rε ε= +∞,lim ε→0 rε εZ{x3<ε} |Du|2dx = 0.(4.66) Then we decompose ZΩε|Duε|2ϕ dx =ZΩε\Ω|Duε|2ϕ dx +Z{x3>rε} |Duε|2ϕ dx +Z{0<x3<rε} |Duε|2ϕ dx. (4.67) 23 Let us estimate each term in the right hand side of (4.67). Clearly lim inf ε→0ZΩε\Ω|Duε|2ϕ dx ≥0.(4.68) For the second term in the right side of (4.68), we use that for every ρ > 0 the weak convergence of Duεto Du in L2(ω×(ρ, 1))3×3gives lim inf ε→0Z{x3>rε} |Duε|2ϕ dx ≥lim inf ε→0Z{x3>ρ} |Duε|2ϕ dx ≥Z{x3>ρ} |Du|2ϕ dx. So we have lim inf ε→0Z{x3>rε} |Duε|2ϕ dx ≥sup ρ>0Z{x3>ρ} |Du|2ϕ dx =ZΩ|Du|2ϕ dx. (4.69) For the third term on the right hand side of (4.67), we take M > 0 and ε > 0 small enough such that M < rε/ε. Defining buεby (4.5) and using the change of variables (4.7) and the uniform continuity of ϕ, we get          ZΩ∩{x3<rε} |Duε|2ϕ dx =Zω×b Qrε ε¯¯Dy(buε √ε)¯¯2ϕ(x0,0) dx0dy +Oε ≥Zω×b QM ¯¯Dy(buε √ε)¯¯2ϕ(x0,0) dx0dy +Oε, (4.70) On the other hand, we saw in Step 1 in the proof of Theorem 3.4 that buε/√εconverges weakly to b u, defined by (3.23), in L2(ωρ;H1(b QM)3), for every ρ > 0. Therefore lim inf ε→0Z{0<x3<rε} |Duε|2ϕ dx ≥sup M>0Zω×b QM |Dybu|2ϕ(x0,0) dx0dy =Zω×b Q|Dybu|2ϕ(x0,0) dx0dy =λ2ZΓ Ru0u0ϕ dx0. (4.71) By (4.65), statements (4.67), (4.68), (4.69), (4.70) and (4.71) imply lim ε→0ZΩε\Ω|Duε|2ϕ dx = 0,(4.72) lim ε→0ZΩ∩{x3>rε} |Duε|2ϕ dx =ZΩ|Du|2ϕ dx (4.73) and lim ε→0ZΩ∩{x3<rε} |Duε|2ϕ dx = lim ε→0Zω×b Qrε ε ¯¯Dy(buε √ε)¯¯2ϕ(x0,0) dx0dy=Zω×b Q |Dybu|2ϕ(x0,0) dx0dy. (4.74) From (4.73), (4.74) and the weak convergence of uεto uin H1(Ω)3and of 1 √εDybuεto Dybu in L2(ω×b QM)3×3, for every M > 0, we obtain lim ε→0ZΩ∩{x3>rε}|D(uε−u)|2ϕ dx = 0,(4.75) lim ε→0Zω×b Qrε ε¯¯¯¯Dyµbuε √ε−bu¶¯¯¯¯ 2 ϕ(x0,0) dx0dy = 0.(4.76) 24 Therefore, taking ρ > 0 such that ϕ(x) = 0 if x06∈ ωρand using that buε(x0, y) does not depend on x0in Ck0 ε×Y, for every k0∈Iρ,ε, we get                            Z{0<x3<rε}¯¯¯¯¯Duε−1 ε2ZCε(x0) Dyµbu √ε¶(z0,x ε)d z0¯¯¯¯¯ 2 ϕdx =ε3X k0∈Iρ,ε Zb Qrε ε¯¯¯¯¯ 1 ε5 2ZCk0 ε Dyµbuε(z0, y) √ε−bu(z0, y)¶dz0¯¯¯¯¯ 2 ϕ(x0,0) dy +Oε ≤Zω×b Qrε ε¯¯¯¯ 1 √εDybuε−Dybu¯¯¯¯ 2 ϕ(x0,0) dz0dy +Oε=Oε. (4.77) This gives a corrector result for Duεin Ω ×{x3< rε}(this type of correctors is usual when we apply the unfolding method, see e.g. [10], [13], [14]). Let us improve (4.77) using the smoothness properties of bu. By (3.23), Holder’s inequality, (3.17) with r= 4, Lemma 4.6 below and (4.66) we have Z{x3<rε}¯¯¯¯¯Dyµbu √ε¶(x0,x ε)−1 ε2ZCε(x0) Dyµbu √ε¶(z0,x ε)d z0¯¯¯¯¯ 2 ϕ dx ≤C ε 2 X i=1 Z{x3<rε}¯¯¯¯¯ui(x0,0) −1 ε2ZCε(x0) ui(z0,0) dz0¯¯¯¯¯ 2¯¯¯Dyb φi(x ε)¯¯¯2ϕ dx ≤C ε 2 X i=1  Z{x3<rε}¯¯¯¯¯ui(x0,0) −1 ε2ZCε(x0) ui(z0,0) dz0¯¯¯¯¯ 4 ϕ dx  1 2ÃZ{x3<rε}¯¯¯Dyb φi(x ε)¯¯¯4ϕ dx!1 2 ≤Crrε ε 2 X i=1  Zωρ¯¯¯¯¯ui(x0,0) −1 ε2ZCε(x0) ui(z0,0) dz0¯¯¯¯¯ 4 dx0  1 2µZb Q¯¯¯Dyb φi¯¯¯4dy¶1 2 ≤Crrε ε X k0∈Iρ,ε ZCk0 ε¯¯¯¯¯u0(x0,0) −1 ε2ZCk0 ε u0(z0,0) dz0¯¯¯¯¯ 4 dx0  1 2 ≤Crrε ε X k0∈Iρ,ε ÃZCk0 ε×(0,ε)¯¯Du0¯¯2dx0!2  1 2 ≤Crrε εÃZ{x3<ε}¯¯Du0¯¯2dx0!2 =Oε. Thus, (4.77) implies ZΩ∩{x3<rε}¯¯¯¯Duε−1 √εDybu(x0,x ε)¯¯¯¯ 2 ϕdx = 0.(4.78) By (4.72), (4.75), (4.78) and lim ε→0ZΩ∩{x3<rε}|Du|2dx = lim ε→0 1 εZΩ∩{x3>rε}¯¯¯Dybu(x0,x ε)¯¯¯2dx = 0, (the second equality is immediate using the change of variables y=x/ε) we deduce (3.25). Step 4. Let us now prove that (3.22) and (3.26) hold. For every ε > 0, let Lε:L2 0(Ωε)−→ H1 0(Ωε)3be the linear continuous operator defined in Proposition 4.1-i), and let us denote vε=Lε(pε). Then div vε=pεin Ωε, and thanks to (3.7) and Proposition 4.1-i), we have kvεkH1 0(Ωε)3≤ kLεkkpεkL2(Ωε)≤C, ∀ε > 0.(4.79) 25