The Stokes equations with Fourier boundary conditions on a wall with asperities

THE STOKES EQUATIONS WITH FOURIER BOUNDARY CONDITIONS ON A WALL WITH ASPERITIES Youcef AMIRAT∗ , Blanca CLIMENT† , Enrique FERN ´ ANDEZ-CARA†and Jacques SIMON∗ Abstract We study the effect of the rugosity of a wall on the solution of the Stokes system complemented with Fourier boundary conditions. We consider the case of small periodic asperities of size ε. We prove that the velocity field, pressure and drag respectively converge to the velocity field, pressure and drag of a homogenized Stokes problem, where a different friction coefficient appears. This shows that, contrarily to the case of Dirichlet boundary conditions, rugosity is dominant here. ∗Laboratoire de Math´ematiques Appliqu´ees, Universit´e Blaise Pascal (Clermont-Ferrand 2), 63177 Aubi`ere Cedex, France, E-mails: [email protected], [email protected] bpclermont.fr. †Dpto. de Ecuaciones Diferenciales y An´alisis Num´erico, Universidad de Sevilla, Aptdo. 1160, 41080 Sevilla, Spain, E-mails: [email protected], car[email protected]. 0 1 Introduction Let us consider a fluid in a domain Oε, limited at the bottom by a plane wall P and at the top by a wall Rε. We assume that Pmoves at a constant velocity while Rεis at rest. The latter is assumed to consist of a smooth wall Rcovered with periodically distributed asperities of small size ε. We are then concerned with the asymptotic behaviour, as ε→0, of the velocity and the pressure in the fluid. The case in which the fluid adheres to the walls has been considered in [?], [?] and [?]. In [?], [?], the wall Ris a plate. Using boundary layer correctors, it is proved that, outside a neighbourhood of the rugose zone, the flow behaves asymptotically as a Couette flow, up to an exponentially small error. An accurate approximation of the drag is given which shows that there is no palpable drag reduction. These results have been then extended to the case of a flow governed by Navier-Stokes equations, see [?]. Let us also mention a recent paper by W. J¨ager and A. Mikeli´c [?] on the laminar viscous channel flow, with the lateral surface of the channel containing surface irregularities. The fluid satisfies a no-slip boundary condition on the rugose surface and it is supposed that a uniform pressure gradient is maintened in the longitudinal direction in the channel. So the limit flow is a Hagen-Poiseuille flow. Using the correponding boundary layers, the authors derive a wall law which gives an approximation of the tangential drag force at order O(ε3/2). As appears from the quoted works, if the fluid is assumed to adhere to the walls, that is to say, if Dirichlet conditions are imposed, then the effect of rugosity is negligible. More precisely, the drag, the velocity field, and the pressure related to Rεconverge to those related to Ras ε→0. In this paper, we assume that, on the walls, the fluid satisfies conditions of Fourier kind. In particular, on the rugose wall we impose σ·n+ku = 0 on Rε,(1) where σis the usual stress tensor and kis a friction coefficient. We prove that these conditions, contrarily to Dirichlet (no-slip) conditions, bring a leading part to rugosity : as ε→0, the drag experienced by the rugose wall Rεconverges to the drag associated with the smooth wall Rprovided with an homogenized friction coefficient Kwhich is not constant and depends on the profile of asperities. We calculate the limit flow and we give estimates of the deviations of the drag, velocity field and pressure, in terms of the size εof the asperities. In the particular case of a plate, the limit drag is larger than the drag of the smooth wall, see Remark 2.1. In the case of Laplace or Poisson equation with Fourier or Neumann boundary conditions, similar results to those of the present paper have been obtained by O.A. Oleinik, A.S. Shamaev and G.A. Yosifian [?] and by G.A. Chechkin, A. Friedman and A.L. Piatnitski [?]. See also E. Sanchez-Palencia [?]. 1 In fact, it is not completely realistic to assume that a fluid satisfies (??), unless the wall Rεhas some kind of porosity, see R.L. Panton [?, p. 149–152]. It would be more appropriate, instead, to impose the slip conditions u·n= 0,(σ·n)tang +ku = 0 on Rε,(2) where (σ·n)tang denotes the tangential component of σ·n. However, (??) may be used for large kas a formal approximation of the Dirichlet condition u= 0. It would be very interesting to extend the present results to the case in which conditions like (??) are considered, but there are some technical difficulties to do this. Our results may be considered as a first step in this direction. Let us mention here that slip boundary conditions have been considered by G. Allaire [?], and D. Cioranescu, P. Donato and H. I. Ene [?] for homogenization of Stokes or Navier-Stokes equations in domains containing periodically distributed obstacles. For conditions of Fourier kind, we refer to C. Conca [?], [?]. 2 The main result The smooth wall Ris assumed to be the graph of a function ron IR2, with ris Lipschitz-continuous, positive and (l1, l2)-periodic (3) (the latter means that ris periodic with respect to xiwith period lifor i= 1 or 2). For each ε > 0, the rugose wall Rεis assumed to be the graph of the function rε defined on IR2as follows : for any x′= (x1, x2), rε(x′) = r(x′)(1 + εη(x′,x′ ε)).(4) Here, η=η(x′, y′) is a function on IR2×IR2satisfying ηis Lipschitz-continuous and l1, l2-periodic with respect to x′and y′. In order to ensure that rεsatisfies (??), we assume that εkηkL∞(S2)≤1 2,1 εis an integer number. (5) The fluid occupies the unbounded domain Oε={x∈IR3:x′∈IR2,0< x3< rε(x′)}. Setting S= (0, l1)×(0, l2), Oεcan be viewed as generated by periodic translations of the bounded domain Ωε={x∈IR3:x′∈S, 0< x3< rε(x′)}. 2 Observe that ∂Ωεconsists of the following parts of the walls Rεand P Rε={x∈IR3:x′∈S, x3=rε(x′)}, P ={x∈IR3:x′∈S, x3= 0} and the lateral immaterial boundary L={x∈IR3:x′∈∂S, 0≤x3≤rε(x′)}. For each m≥0, let us introduce the space Hm per(Ωε) = {v∈Hm loc(Oε) : v∈Hm(Ωε), v(x+ (l1,0,0)) = v(x+ (0, l2,0)) = v(x) for a.e. x∈ Oε}. In other words, Hm per(Ωε) is formed by all (l1, l2)-periodic functions on Oεwhich are Hmin any bounded subset (and not only in compact subsets, as Hm loc means). As usual, for m= 0, we write L2 per(Ωε). The velocity field and the pressure are assumed to satisfy                uε∈(H1 per(Ωε))3, pε∈L2 per(Ωε), −ν∆uε+∇pε= 0,∇·uε= 0 in Oε, σε·n+ku = 0 on Rε, σε·n+k(u−g) = 0 on P, (6) where ν > 0 is the viscosity, σε=σ(uε, pε) = −pεId + ν(∇uε+t∇uε),(7) nis the outwards unit normal vector field, k > 0 is a friction coefficient and g= (g′,0) is the velocity of the wall P. The existence and uniqueness of a solution is proved in proposition 3.1. Remark that the boundary conditions on Rεand Pare meaningful, since (??) gives σε∈ (L2 per(Ωε))3×3and ∇·σε=−∇pε+ν∆uε+t∇(∇·uε) = 0.(8) This allows to define the normal trace σε·nin (H−1/2 loc (∂Oε))3. The hydrodynamical drag Tεassociated with the bounded part Rεof the wall Rε(the drag of the whole wall is infinite) is by definition the projection of the force exerted on Rεby the fluid, that is Tε=−g·ZRε σε·n ds. (9) 3 Thanks to the boundary condition on Rε, it reads as well Tε=g·ZRε uεds. Let us now define the homogenized friction coefficient K=K(x′, r(x′)). First, we introduce the following function m∈L∞(IR2×IR2) : for any (x′, y′)∈IR2×IR2, m(x′, y′) = 1 + |∇r(x′) + r(x′)∇y′η(x′, y′)|2 1 + |∇r(x′)|21 2.(10) Then we put, for all x′∈IR2, hmi(x′) = 1 |S|ZSm(x′, y′)dy′. Obviously, we have hmi ∈ L∞(IR2). The homogenized friction coefficient is given R as follows : for all x′∈IR2, K(x′, r(x′)) = khmi(x′).(11) It is therefore a (l1, l2)-periodic function which belongs to L∞(R). We will prove that (uε, pε) converges in an appropriate sense to the unique solution of the system                u0∈(H1 per(Ω))3, p0∈L2 per(Ω), −ν∆u0+∇p0= 0,∇·u0= 0 in O, σ0·n+Ku0= 0 on R, σ0·n+k(u0−g) = 0 on P, (12) where σ0=−p0Id + ν(∇u0+t∇u0), Ris the graph of r,Ois the domain bounded by Rand Pand Ω = {x∈IR3:x′∈S, 0< x3< r(x′)}. Furthermore, we will prove that the limit drag is T0=−g·ZRσ0·n ds =g·ZRKu0ds. Notice that, in general, u0is not the velocity field related to the smooth wall R because K6=k, and T0is not the drag experienced by R. Since the domain Ωεvaries with ε, the convergence of uεand pεcannot hold in the whole domain Ω. We will obtain convergence outside a neighbourhood of Rof arbitrary small size δ > 0, that is, in all subdomains of the form ωδ={x∈IR3:x′∈S, 0< x3< r(x′)−δ}. Our main result is the following. 4 Theorem 2.1 Assume r∈W3,∞(IR2). There exists C > 0such that, for any ε > 0 satisfying (??), we have : |Tε−T0| ≤ C√ε. (13) Moreover, for any δ > 0there exists Cδ>0such that, for any εsatisfying (??), we have : kuε−u0kH1(ωδ)≤Cδ√ε, (14) kpε−p0kL2(ωδ)≤Cδ√ε. (15) ⊔⊓ Remark 2.1 The drag Tεof a rugose plate is strictly greater than the drag Tof the corresponding (homogenized) smooth plate. Indeed, assume that r(x′)≡l3(a positive real number) and ηdepends only on y′. Then hmiis independent of x′and is greater than 1 unless ηis a constant. Accordingly, we have lim ε→0Tε=T0=νl1l2khmi|g|2 ν(1 + hmi) + l3khmi> T =νl1l2k|g|2 2ν+l3k. ⊔⊓ Remark 2.2 Notice that 1 1 + |∇r(x′)|2≤ hmi(x′)≤1 + (r(x′))2 1 + |∇r(x′)|2 1 |S|ZS|∇y′η(x′, y′)|dy′1/2 . The last quantity is bounded from above by the local asymptotic ratio of the Rε-area and the R-area. In other words, for any x′∈S, we have hm′i(x′)≤lima→0limε→0|RεSB(x′;a)| lima→0|RSB(x′;a)| where B(x′;a) is the ball centered at (x′, r(x′)) of radius a.⊔⊓ Remark 2.3 The drag Tεcan also be written in the form Tε=g·ZPσε·n ds =−kg ·ZP(uε−g)ds. (16) Indeed, the following equalities hold : −g·ZRε σε·n ds =−g·ZRε∪Pσε·n ds +g·ZPσε·n ds =−g·ZΩε∇·σεds +g·ZPσε·n ds. A similar equality holds for T0.⊔⊓ 5 3 Existence, uniqueness and estimates We will prove that there exists exactly one solution (uε, pε) to (??) and exactly one solution (u0, p0) to (??). Notice that (??) is similar to (??) with a varying friction coefficient, since both walls Rεand Rare the graphs of periodic Lipschitz functions. Therefore, in order to put these two problems in the same framework, we will assume in this Section that k∈L∞ per(Ωε), k ≥κ > 0,(17) where κis a real number. We will use the following variational formulation :            uε∈(H1 per(Ωε))3, pε∈L2 per(Ωε), 2νZΩε e(uε) : e(ϕ)−ZΩε pε∇·ϕ+ZRε∪Pkuε·ϕ=ZPkg ·ϕ∀ϕ∈(H1 per(Ωε))3, ∇·uε= 0, (18) where e(ϕ) = 1 2(∇ϕ+t∇ϕ) and e(u) : e(ϕ) = Pi,j eij(u)eij(ϕ). Proposition 3.1 Problem (??)is equivalent to (??)and possesses exactly one solution. Furthermore, one has kuεkH1(Ωε)+kpεkL2(Ωε)≤C, (19) where Cis independent of ε.⊔⊓ Obviously, this result also provides the existence and uniqueness of a solution (u0, p0) to (??). In order to prove this proposition, we need some previous results. In particular, we need a Korn inequality for a special class of star-shaped domains. By definition, Dis star-shaped, with respect to a ball Bif the segment connecting any two points x∈Band y∈Dlies in D. Lemma 3.1 There exists C > 0such that, for any bounded domain D⊂IR3of diameter Rwhich is star-shaped with respect to a ball Bof radius ρand for any v∈(H1(D))3, the following inequality holds k∇vk2 L2(D)≤CR ρ3ke(v)k2 L2(D)+k∇vk2 L2(B).(20) ⊔⊓ 6 For the proof see O.A. Oleinik, A.S. Shamaev and G.A. Yosifian, [?, Theorem 2.10, p. 23]. Let us put D⋆={x∈IR3:x′∈S, 0< x3<l3 2}. Lemma 3.2 There exists C > 0, only depending on Sand l3, such that, for all v∈(H1(D⋆))3, k∇vk2 L2(D⋆)≤Cke(v)k2 L2(D⋆)+ZP|v|2ds.(21) Proof: Suppose the assertion in this lemma is false. Then, for each m≥1, there exists wm∈(H1(D⋆))3such that k∇wmk2 L2(D⋆)> mke(wm)k2 L2(D⋆)+ZP|wm|2ds. Let us put vm=wm/k∇wmkL2(D⋆). Then k∇vmkL2(D⋆)= 1 and ke(vm)k2 L2(D⋆)+ZP|vm|2<1 m, whence we obtain the following as m→ ∞ : e(vm)→0 in (L2(D⋆))3×3,ZP|vm|2ds →0.(22) On the other hand, the following estimate holds for all v∈(H1(D⋆))3: ZD⋆|v|2ds ≤l3ZP|v|2+l3 2ZD⋆|∇v|2.(23) Indeed, for any regular vand any x= (x′, x3) in D⋆, one has v(x) = v(x′,0) + Zx3 0∂x3v(x′, y3)dy3. Consequently, |v(x)|2≤2|v(x′,0)|2+ 2Zl3/2 0|∂x3v(x′, y3)|dy32, whence Zl3/2 0|v(x′, x3)|2dx3≤l3|v(x′,0)|2+Zl3/2 0|∂x3v(x′, y3)|dy32 7 and ZSZl3/2 0|v(x′, x3)|2dx3dx′ ≤l3ZP|v(x′,0)|2dx′+1 2ZSZl3/2 0|∂x3v(x′, y3)|2l3dy3dx′. This proves (??), at least when vis regular enough. By density, (??) holds for all v in (H1(D⋆))3. It follows from (??) that vmis uniformly bounded in (H1(D⋆))3. From the compactness of the embedding H1(D⋆)֒→L2(D⋆), there exists a subsequence, still denoted vm, that converges strongly in (L2(D⋆))3×3to some v∈(H1(D⋆))3. In view of Korn inequality in Lipschitz domains, one has kvm′−vmk2 H1(D⋆)≤C(ke(vm′)−e(vm)k2 L2(D⋆)+kvm′−vmk2 L2(D⋆)), where the constant Cdepends only on Sand l3, see [?], [?]. These inequalities and (??) show that vmconverges strongly in H1(D⋆) to vand, also, that k∇vkL2(D⋆)= 1,ke(v)kL2(D⋆)= 0,ZP|v|2ds = 0.(24) But the equality e(v) = 0 implies that vis a rigid displacement, i.e. v=Ax +b where Ais a skew-symmetric constant matrix and bis a constant vector. This fact, together with the third equality in (??), implies v= 0. This leads to a contradiction and proves the lemma. ⊔⊓ Lemma 3.3 Let εsatisfy (??). Then, for any v∈(H1(Ωε))3, kvk2 H1(Ωε)≤Cke(v)k2 L2(Ωε)+ZP|v|2,(25) where Cdepends only of l1,l2,rand η. Proof: The function rεdefined by (??) is Lipschitz-continous, with a Lipschitz constant independent of ε. It is also bounded from below by a positive number independent of ε. Therefore, if l1and l2are small enough, there exists a ball B independent of εsuch that Ωεis star-shaped with respect to B. Moreover, Bcan be chosen in D⋆. For arbitrarily given l1and l2, by dividing Sin sufficiently small squares, it follows that Ωεis the union of mdomains Ωi εwhich are respectively starshaped with respect to the balls Bi, with mand Biindependent of ε. Lemma 3.1 yields the following for each iand for all v∈(H1(Ωε))3: k∇vk2 L2(Ωi ε)≤Cke(v)k2 L2(Ωi ε)+k∇vk2 L2(Bi). Adding these inequalities for i= 1, ..., m, we find that k∇vk2 L2(Ωε)≤Cke(v)k2 L2(Ωε)+k∇vk2 L2(D⋆) 8 5 Some technical lemmas Lemma 5.1 Let γ=γ(x′, y′)be a Lipschitz function on S2, periodic with respect to y′and satisfying ZSγ(x′, y′)dy′= 0 for any x′∈S. There exists a real number Csuch that : i)For all functions ϕand ψin H1(Ω) and 0< ε ≤1, we have ZΩγ(x′,x′ ε)ψ(x)ϕ(x)dx≤CεkψkH1(Ω)kϕkH1(Ω).(53) ii)For all functions ψ∈H2(Ω) and ϕ∈H1(Ω) and 0< ε ≤1, we have ZSγ(x′,x′ ε)ϕ(x′, r(x′))ψ(x′, r(x′)) dx′≤C√εkψkH2(Ω)kϕkH1(Ω).(54) ⊔⊓ For the proof, see O.A. Oleinik, A.S. Shamaev and G.A. Yosifian [?, Lemma 1.6, p. 8]. Lemma 5.2 Assume r∈W3,∞(S). There exists a positive number C, independent of u0and ε, with the following properties : i)For 0< ε ≤1, we have k∇·(t Nεu)kL2(Ω) ≤Cεku0kH2(Ω).(55) ii)For any function ϕ∈(H1(Ω))3and 0< ε ≤1, we have ZΩ(b eε(u) : b eε(ϕ))jε−ZΩe(u0) : e(ϕ)≤C√εku0kH3(Ω)kϕkH1(Ω).(56) Proof: Notice that u0∈(H3(Ω))3, since r∈W3,∞(S). Let us first prove (??). Since by definition Nε= (1 + εηε)Mε, we see that ∇·(t Nεw) = ∂x1((1 + εηε)w1) + ∂x2((1 + εηε)w2) +∂x3(−x3ε(∂x1ηε)w1−x3ε(∂x2ηε)w2+w3) =∇·w+εηε(∂x1w1+∂x2w2)−x3(ε(∂x1ηε)∂x3w1+ε(∂x2ηε)∂x3w2), 15 for any function w. With the particular choice w=u=u0+x3εηε∂x3u0, since ∇·u0= 0, we obtain ∇·u=x3(ε(∂x1ηε)∂x3u01 +ε(∂x2ηε)∂x3u02) + εηε∂x3u03 and εηε∂x1u1−x3ε(∂x1ηε)∂x3u1=εηε∂x1u01 −x3ε(∂x1ηε)∂x3u01 +x3(εηε)2∂2 x1x3u01 −x2 3ε2ηε(∂x1ηε)∂2 x3x3u01. Using the corresponding similar equality for u2, we deduce that ∇·(t Nεu) = x3(εηε)2(∂2 x1x3u01 +∂2 x2x3u02) −x2 3εηε(ε(∂x1ηε)∂2 x3x3u01 +ε(∂x2ηε)∂2 x3x3u02). The inequality (??) follows, since |ηε(x′)| ≤ kηkL∞(S2)and |ε∂xiηε(x′)|=ε∂xiη(x′,x′ ε) + ∂yiη(x′,x′ ε)≤2kηkW1,∞(S2). Let us now prove (??). The definition (??) of b eεyields the following for all u and ϕ: (b eε(u) : b eε(ϕ)) = 1 2Mε∇u:Mε∇ϕ+1 2Mε∇u:t (Mε∇ϕ). Moreover, the definition of Mεleads to the identities, for 1 ≤k≤3, (Mε∇ϕ)ik =            ∂xiϕk−1 jε x3ε(∂xiηε)∂x3ϕki= 1,2, ∂x3ϕk−1 jε εηε∂x3ϕki= 3. For i= 3, we have used jε= 1 + εηεand thus 1/jε= 1 −εηε/jε. In particular, we obtain (Mε∇u)ik =            ∂xiu0k+x3εηε∂2 xix3u0k−1 jε ε2ηε∂xiηε∂2 x3x3u0ki= 1,2, ∂x3u0k+1 jε x3εηε∂2 x3x3u0ki= 3. Therefore, b eε(u)jε:b eε(ϕ)−e(u0) : e(ϕ) = 2 X i=1 3 X k=1 aik +X α εηεbα, where aik =−1 2x3ε(∂xiηε)∂x3ϕk(∂xiu0k+∂xku0i) 16 and each bαis the product of a first or second derivative of u0, a derivative of ϕand some of the following functions : x3,jε, (jε)−1,εηεand ε∂xiηε. Since |εηε| ≤ Cε, we see that ZΩ|εηεbα| ≤ Cεku0kH2(Ω)kϕkH1(Ω). Since ε∂xiηε(x′) = ε∂xiη(x′,x′ ε) + ∂yiη(x′,x′ ε), we can also put aik =−1 2x3∂yiηε∂x3ϕk(∂xiu0k+∂xku0i) + ε∂xiηεb′ ik,(57) where the b′ ik have the same structure as bα. Hence, ZΩ|ε∂xiηεb′ ik| ≤ Cεku0kH2(Ω)kϕkH1(Ω). The integrals of the other terms in the right-hand side of (??) are as follows : ZΩ−x3∂yjηε(∂x3ϕk)(∂xlu0m) =ZΩ∂yjηε∂x3(x3∂xlu0m)ϕk−Z∂Ωx3∂yjηεϕk(∂xlu0m)n3ds. Since 1/ε is an integer and ηis periodic with respect to all its variables, we have ZS∂yjηε=ZS∂yiη(x′,x′ ε)dx′= 0.(58) Therefore, thanks to lemma 5.1, part i), ZΩ∂yjηε∂x3(x3∂xlu0m)ϕkdx≤Cεku0mkH2(Ω)kϕkkH1(Ω). On the other hand, from lemma 5.1, part ii), we have Z∂Ω∂yjηεx3(∂xlu0m)ϕkn3ds=ZS∂yjη(x′,x′ ε)(x3(∂xlu0m)ϕk)(x′, r(x′)) dx′ ≤√εku0mkH3(Ωε)kϕkkH1(Ωε). Here, we have used that n3ds =dx′. This proves that ZΩai,k≤Cεku0kH3(Ω)kϕkkH1(Ω). This completes the proof of (??). 17 Lemma 5.3 Let us assume that r∈W3,∞(S). Then, for any εsatisfying (??)and any ϕ∈(H1 per(Ω))3, we have ZΩp0∇·(t Nεϕ−ϕ)≤C√εkp0kH2(Ω)kϕkH1(Ω),(59) where Cis independent of ε,p0and ϕ. Proof: Let us first notice that p0∈H2(Ω), since r∈W3,∞(S). We have ∇·(t Nεϕ−ϕ) = εηε(∂x1ϕ1+∂x2ϕ2)−x3 2 X j=1 (ε∂xjηε+∂yjηε)∂x3ϕj. Then, multiplying by p0and integrating by parts the last term, we see that ZΩp0∇·(t Nεϕ−ϕ) = ZΩεp0 2 X j=1 (ηε∂xjϕj−x3∂xjηε∂x3ϕj) +ZΩ∂x3(x3p0) 2 X j=1 ∂yjηεϕj−ZSr(x′)p0(x′, r(x′)) 2 X j=1 ∂yjηε(x′)ϕj(x′, r(x′)) dx′. In the right-hand side, the first integral is bounded by ZΩεp0 2 X j=1 (ηε∂jϕj−εx3∂xjηε∂x3ϕj)≤Cεkp0kL2kϕjkH1(Ω). Using lemma 5.1, part i) and (??), the second integral is bounded by ZΩ∂x3(x3p0) 2 X j=1 ∂yjηεϕj≤Cεkp0kH2(Ω)kϕjkH1(Ω). On the other hand, from lemma 5.1, part ii), the boundary integral satisfies ZSr(x′)p0(x′, r(x′)) 2 X j=1 ∂yjηε(x′)ϕj(x′, r(x′)) dx′≤C√εkp0kH2(Ω)kϕkH1(Ω). Hence, lemma 5.3 is proved. ⊔⊓ Lemma 5.4 Assume that ψ∈L2 per(Ω) is such that ZΩψ= 0. Then there exists ϕ∈(H1 0(Ω))3such that ∇· t(Nεϕ) = ψin Ω,kϕkH1(Ω) ≤CkψkL2(Ω),(60) where Cis independent of ε,ψand ϕ. 18 Proof: Let us put ψε=1 1 + εηε(ψ◦L−1 ε). Clearly, ψε∈L2 per(Ωε). From lemma 3.4 with w0= 0, there exists wε∈(H1 0(Ωε))3 such that ∇·wε=ψεin Ωεand kwεkH1(Ωε)≤CkψεkL2(Ωε), where Cdoes not depend of ε,ψεand wε. In accordance with (??), the function ϕ=b wεsatisfies ∇·(t Nεϕ) = ψin Ω. Moreover, ZΩε|ψε|2=ZΩ 1 1 + εηε|ψ|2≤ZΩ|ψ|2. ⊔⊓ 6 The proofs of proposition 4.1 and theorem 2.1 Proof of proposition 4.1: We will first proof the estimates (??). Let us write the weak formulation of the problem satisfied by (u0, p0) : (u0, p0)∈(H1 per(Ω))3×L2 per(Ω),(61) 2νZΩe(u0) : e(ϕ)−ZΩp0∇·ϕ+ZRKuε·ϕ+kZPuε·ϕ=kZPg·ϕ ∀ϕ∈(H1 per(Ω))3, (62) ∇·u0= 0 in Ω.(63) Substracting (??) from (??), we obtain, for all ϕ∈(H1 per(Ω))3, 2νZΩb eε(b uε) : b eε(ϕ)jε−e(u0) : e(ϕ)−ZΩb pε∇·(t Nεϕ)−p0∇·ϕ +ZR(kµεb uε−Ku0)·ϕ ds +kZP(b uε−u0)·ϕ ds = 0. (64) Let us set b uε=u+zε, where u=u0+x3εηε∂x3u0. Then (??) reads kzεkH1(Ω) ≤C√ε. (65) Since u=u0on P, (??) gives : 2νZΩ(b eε(zε) : b eε(ϕ))jε+kZRzε·ϕµεds +kZPzε·ϕ ds =−2νZΩb eε(u) : b eε(ϕ)jε−e(u0) : e(ϕ)+ZΩ(b pε−p0)∇·(t Nεϕ) +ZΩp0∇·(t Nεϕ−ϕ)−ZR(kµεu−Ku0)·ϕ ds. (66) 19 Let us choose ϕ=zε. Then the left-hand side is larger than Ckzεk2 (H1(Ω))3thanks to Korn inequality (??). Using (??) to estimate the first integral in the right-hand side of (??), we see that kzεk2 H1(Ω) ≤C√εku0kH3(Ω)kzεkH1(Ω) +A+B+ZR(kµεu−Ku0)·zεds,(67) where Cis independent of εand Aand Bare respectively given by A=ZΩ(b pε−p0)∇·(t Nεzε), B =ZΩp0∇·(t Nεzε−zε). Thanks to (??), ∇·(t Nεzε) = ∇·(t Nεb uε)−∇·(t Nεu) = −∇·(t Nεu). Thus, (??) and (??) give A≤Cε(kb pεkL2(Ω) +kp0kL2(Ω))ku0kH2(Ω) ≤Cε. (68) On the other hand, (??) implies B≤C√εkp0kH2(Ω)kzεkH1(Ω) ≤C√εkzεkH1(Ω).(69) In order to estimate the integral over Rin (??), let us put kµεu−Ku0= (kmε−K)u0+k(µε−mε)u0+kµε(u−u0),(70) where mε(x′) = m(x′, x′/ε). From the definition of K, we see that (kmε−K)(x′) = kγ(x′, x′/ε), where γ(x′, y′) = m(x′, y′)−hmi(x′). Thus, lemma 5.1, part ii), gives : ZR(kmε−K)u0·zεds≤C√εku0kH2(Ω)kzεkH1(Ω).(71) From (??), (??) and (??), we deduce that µ2 ε−(mε)2=|εr∇x′ηε+εηε∇r|2+ 2(εr∇x′ηε+εηε∇r)(r∇y′ηε+∇r) 1 + |∇r|2. This implies that |µε−mε| ≤ Cε and ZRk(µε−mε)u0·zεds≤Cεku0kH1(Ω)kzεkH1(Ω).(72) 20 Finally, |u−u0|=|x3ηε∂x3u0| ≤ C|∂x3u0|and ZRkµε(u−u0)·zεds≤Cεku0kH2(Ω)kzεkH1(Ω).(73) From (??), (??), (??) and (??), we see that ZR(kµεu−Ku0)·zεds≤C√εkzεkH1(Ω).(74) Using (??), (??) and (??) in (??), we obtain kzεk2 H1(Ω) ≤C√εkzεkH1(Ω) +Cε. This implies (??) and, therefore, (??) holds. Let us now prove (??). Choosing ϕ∈(H1 0(Ω))3in (??), we find 2νZΩ(b eε(zε) : b eε(ϕ))jε=ZΩ(b pε−p0)∇·(t Nεϕ) + ZΩp0∇·(t Nεϕ−ϕ) −2νZΩ(b eε(u) : b eε(ϕ))jε−e(u0) : e(ϕ). (75) Let ψbe a function in L2 per(Ω) and let ϕ∈(H1 0(Ω))3be the function furnished by lemma 5.4. Then (??) reads ZΩ(b pε−p0)ψ= 2νZΩ(b eε(zε) : b eε(ϕ))jε +2νZΩ(b eε(u) : b eε(ϕ))jε−e(u0) : e(ϕ)−ZΩp0∇·(t Nεϕ−ϕ). Using (??), (??), (??) and (??), it is not hard to see that ZΩ(b pε−p0)ψ≤C√εkψkL2. Since ψis arbitrary, this implies (??). This completes the proof. ⊔⊓ Proof of theorem 2.1: First, let us prove that (??) follows from (??). Let δ∗= 1/(2kηkL∞(S2)). Thanks to (??), we have |Tε−T0|=kg ·ZP(u0−uε)≤k|g|Cδ∗ku0−uεkH1(ωδ∗). Therefore, (??) implies (??). Let us now prove (??). It suffices to assume δ≤δ∗, since (??) for δ=δ∗implies (??) for any δ≥δ∗. Using (??) and putting b ωδ=L−1 ε(ωδ), we have kϕk2 H1(ωδ)=Zbωδ jε(|b ϕ|2+|Mε∇b ϕ|2)≤Ckb ϕkH1(bωδ). 21 Then, it suffices to prove that kb uε−b u0kH1(bωδ)≤C√ε (where Ccan depend of δ). From (??), we see that it is sufficient to prove that ku−b u0kH1(bωδ)≤C√ε. (76) Since u=u0=b u0on P, we only have to estimate the L2norms of the first derivatives ∂xj(u−b u0). In fact, we are going to prove that, for all x∈b ωδ, |∂xj(u−u0)(x)| ≤ C√ε. (77) Recall that u=u0+εx3ηε∂x3u0. Then, ∂xju=   ∂xju0+x3(∂yjηε+ε∂xjηε)∂x3u0+εx3ηε∂2 x3xju0j= 1,2, ∂x3u0+εηε∂x3u0+εx3ηε∂2 x3x3u0j= 3. Thanks to (??), ∂xjb u0=  d ∂xju0+x3(∂yjηε+ε∂xjηε)d ∂x3u0j= 1,2, (1 + εηε)d ∂x3u0j= 3. Since ηε,∂xjηεand ∂yjηεare bounded, it follows that |∂xj(u−b u0)| ≤ C 3 X k=1 |∂xku0−d ∂xku0|+Cε(|d ∂x3u0|+|∂x3u0|+|∂x3xju0|)). Since r∈W3,∞(S), we have u0∈(H3(Ω))3and thus Morrey-Sobolev theorem gives u0∈(C1,1/2(Ω))3, with a norm in this space bounded by Cku0kH3(Ω). Therefore, for all x∈b ωδ, |(∂xku0−d ∂xku0)(x)| ≤ Cδku0kH3(Ω)|x−Lε(x)|1/2≤Cδ√ε and (??) is established. Thanks to Poincar´e inequality, (??) holds too. This completes the proof of (??). Finally, let us prove (??). Arguing as before, it suffices to show that kb pε−b p0kL2(bωδ)≤C√ε. From (??), it suffices to prove that kp0−b p0kL2(bωδ)≤C√ε. 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