Asymptotic behavior of a non-Newtonian flow in a thin domain with Navier law on a rough boundary

Asymptotic behavior of a non-Newtonian flow in a thin domain with Navier law on a rough boundary F. J. Su´arez-Grau∗ Dpto. de Ecuaciones Diferenciales y An´alisis Num´erico Universidad de Sevilla 41012-Sevilla (Spain) Abstract We consider a non-Newtonian flow in a thin domain of thickness ε. The flow is described by the 3D incompressible Navier-Stokes (Stokes) system with a nonlinear viscosity, being a power of the shear rate (power law) of flow index p. The bottom of the domain is irregular by the present of slight roughness of amplitude εδand period εβ, satisfying the relation 1 < β < δ. Assuming pure slip or partial slip with a friction coefficient ε−γ, with γ > 0, on the rough boundary, we consider the limit when domain thickness tends to zero and we obtain different models depending on the magnitude δwith respect to 2p−1 pβ−p−1 p, and the magnitude γwith respect to 1 p−1. Keywords: non-Newtonian flow; thin fluid films; rough boundary; Navier condition; adherence condition; asymptotic behavior 2010 MSC: 76A05, 76A20, 35B27, 35Q30 1. Introduction Roughness of the solid surface as well as the rheological properties of the fluid have influence in the fluid-solid interface condition. For that reason, the choice of boundary conditions of fluid flow is a relevant problem to determine such influence. It is commonly accepted that viscous fluids adhere to rough5 ∗Corresponding author Email address: [email protected] (F. J. Su´arez-Grau ) Preprint submitted to Nonlinear Analysis: Theory, Methods & ApplicationsJanuary 18, 2015 surfaces, and so the no-slip condition at the rough surfaces of a domain is widely used. This does not seem always valid for the non-Newtonian fluids, indeed non- Newtonian fluids melt and solutions slip against the surface. This phenomenon has been related in many mechanical papers concerning non-Newtonian fluids (see [24], [34]).10 In this sense, it has been suggested that, in many cases, the velocity field of the fluid uεin a domain Ωεobeys a Navier condition at the rough surface Γε⊂∂Ωε:S ντ=−λ[uε]τon Γε, uεν= 0,on Γε, where Sis the deviatoric viscous stress tensor, νdenotes the outside unitary normal vector to Ωεon Γε, the subscript τdenotes the orthogonal projection on the tangent space of Γε, and λ > 0 is the friction coefficient. Notice that depending on the value of λ, we shall consider either pure slip, partial slip or no-slip at the boundary. This means that the friction coefficient15 λshall be either zero, positive or +∞. Recently, based on the so called rugosity effect, some results mathematically justify that viscous fluids adhere completely to the boundary of an impermeable domain. More precisely, these results accounts asymptotically for the transformation of pure slip boundary conditions on a rough surface in no-slip boundary20 conditions, as the amplitude of the roughness vanishes, provided that the energy of the solutions is uniformly bounded and there is enough roughness of the oscillating boundaries. We refer to the pioneering paper [15] for the case of periodic and very smooth boundaries, and to [6], [10], [11], [12] (see also [18], [21], [25], [31]) for having a quite complete understanding for arbitrary boundaries.25 In this paper, assuming Navier condition on a slightly rough surface of a thin domain, our goal is to identify, depending on the magnitudes of the amplitude of the roughness and the friction coefficient, the type of boundary condition in which the starting Navier condition is transformed in the limit.30 Next, let us set the position of the problem. We consider a smooth bounded 2 open set ω⊂R2and a function Ψ in W2,∞ loc (R2), periodic of period Z0= (−1/2,1/2)2, satisfying Span({∇Ψ(z0) : z0∈R2}) = R2,(1) which always holds except in the case where Ψ is constant in one direction (i.e. not included roughness of riblets-type: Ψ = Ψ(z1) or Ψ = Ψ(z2)).35 Then, we consider a spatial domain Ωεdetermined through Ωε=(x0, x3)∈R2×R:x0∈ω, −εδΨx0 εβ< x3< ε,(2) where εis a positive parameter devoted to tends to zero representing the characteristic thickness of the domain, and εδand εβrepresent the amplitud and the period, respectively, of the roughness satisfying 1 < β < δ < +∞, i.e. relation lim ε→0εδ−β= 0,lim ε→0εβ−1= 0.(3) By Γε⊂∂Ωεwe denote the rough boundary of Ωε, that is40 Γε=(x0, x3)∈R2×R:x0∈ω, x3=−εδΨx0 εβ.(4) From the other side it is well-known that for non-Newtonian flow through a thin domain the non linear Poiseuille law is used. For that reason, in this study we deal with the case where the viscosity is not constant. Thus, we consider that the viscosity satisfies the non linear power law, which is widely used for melted polymers, oil, mud,... Denoting the shear rate by D[uε] = 1 2(Duε+Dtuε), the viscosity as a function of the shear rate is given by ηp(D[uε]) = µ|D[uε]|p−2, p > 1, where the two material parameters µ > 0 and 1 <p<+∞are called the consistency and the flow index, respectively. For simplicity we suppose µ= 1. Recall that p= 2 yields the Newtonian fluid. For p < 2 the fluid is pseudoplastic (shear thinning), which is characteristic of high polymers, polymer solutions, and many suspension, whereas for p > 2 the fluid is dilatant (shear45 thickening), whose behavior is reported for certain slurries, like mud, clay, or 3 cement, and implies an increased resistance to flow with intensified shearing. Therefore, assuming the fluid incompressible and in a stationary state, and depending on the value of p, the velocity uεand the pressure πεsatisfy:50 - For 9/5≤p < +∞, the non-Newtonian Navier-Stokes system −div (ηp(D[uε])D[uε]) + (uε∇)uε+∇πε=fεin Ωε, div uε= 0 in Ωε. (5) - For 1 <p<9/5, due to the known technical difficulties with inertial term for p < 9/5, the non-Newtonian Stokes system −div (ηp(D[uε])D[uε]) + ∇πε=fεin Ωε, div uε= 0 in Ωε, (6) where the right-hand side fεis of the form fε(x) = e fx0,x3 ε,a.e. x∈Ωε,(7) with e fassumed in Lp0(ω×(−1,1))3.55 In order to study the roughness effects on the flow, we assume Navier condition on Γε, hηp(D[uε])D[uε]νiτ=−λ ε−γuετon Γε, uεν= 0 on Γε,(8) where λ ε−γis the friction coefficient expressed in terms of the magnitude of the domain. Here λ=O(1), and to cover interesting cases γ∈ {−∞} ∪ [0,+∞);60 namely γ=−∞ yields pure slip, and γ∈[0,+∞) yields partial slip. Notice that when γ > 1 p−1the friction coefficient is so big that we must consider no-slip condition instead of Navier one (see (96) for how to reach that conclusion). For that reason, from now on we assume γ∈ {−∞} ∪ [0,1 p−1]. 65 4 For simplicity, we assume no-slip condition on the rest of the boundary, uε= 0 on ∂Ωε\Γε.(9) The goal of this work is to study the asymptotic behavior of the solutions of the above problems and to know in which way the irregular boundary affects the flow. To do this, it is required to study the asymptotic study of the problem by using monotonicity arguments together with an appropriate combination of two changes of variables: the usual rescaling to transform the thin domain into one with fixed height providing the macroscopic behavior of the fluid, and the unfolding method (see e.g. [2], [13], [16], [20], [26]) to capture the microscopic behavior of the fluid near the rough boundary. Then, denoting by eu0and eπ the limit velocity and pressure, we recover the simplified non-Newtonian Stokes system in Ω = ω×(0,1): −∂y3|∂y3eu0|p−2∂y3eu0= 2p 2e f0− ∇y0eπin Ω, The roughness effects apppear into the boundary condition on Γ = ω× {0}; we show that the boundary condition on Γ satisfied by eu0includes the roughness effects, which depend on the values related with the amplitude of the roughness and the friction coefficient of the Navier condition, i.e. δand γrespectively:70 - Concerning δ∈(β, +∞); there is one critical regime, namely δ=e βp, and therefore three different regimes, namely δ∈(β, e βp), δ=e βp, and δ > e βp. The parameter e βpis defined by e βp:= 2p−1 pβ−p−1 p,(10) which satisfies e βp∈(β, +∞), ∀p > 1. - Concerning γ∈ {−∞} ∪ [0,1 p−1]; three regimes are addressed, namely the75 critical regime γ=1 p−1, and thus γ∈[0,1 p−1) and γ=−∞. Below every cases are discussed: (i) Case δ∈(β, e βp). We show that eu0satisfies no-slip: eu0= 0 on Γ, 5 for every γ∈ {−∞} ∪ [0,1 p−1]. This means that no matter if we start with pure or partial slip, actually the roughness is so strong that the limit velocity of the fluid vanishes at the bottom.80 (ii) Case δ=e βp. This depends on γ: - If γ=1 p−1, then eu0satisfies partial slip: −|∂y3eu0(y0,0)|p−2∂y3eu0(y0,0) = −λ2p 2eu0(y0,0)−2p 2R(eu0(y0,0)) on Γ, which means that, starting from partial slip with friction coefficient of magnitude of order ε−1 p−1, we also obtain partial slip in the limit, but with a higher friction coefficient coming from the effects of the roughness (see Theorem 3.2 for more details of R).85 - If γ∈[0,1 p−1), then eu0satisfies partial slip: −|∂y3eu0(y0,0)|p−2∂y3eu0(y0,0) = −2p 2R(eu0(y0,0)) on Γ, which means that, starting from partial slip with a friction coefficient of magnitude of order smaller than ε−1 p−1, we obtain partial slip with only the friction coefficient coming from the roughness. - If γ= +∞, then, starting from pure slip, eu0satisfies partial slip with the same friction coefficient as the previous case.90 (iii) Case δ > e βp. This depends on γ: - If γ=1 p−1, then eu0satisfies partial slip: −|∂y3eu0(y0,0)|p−2∂y3eu0(y0,0) = −λ2p 2eu0(y0,0) on Γ, which means that, starting from partial slip with friction coefficient of magnitude of order ε−1 p−1, we obtain partial slip with only λas the friction coefficient. Here the roughness is so slight that no roughness effects appear in the limit.95 6 - If γ∈ {−∞} ∪ [0,1 p−1), then eu0satisfies pure slip: −|∂y3eu0(y0,0)|p−2∂y3eu0(y0,0) = 0 on Γ, which means that, starting from pure or partial slip with a friction coefficient of magnitude of order smaller than ε−1 p−1, we obtain pure slip. To conclude with this discussion, notice that the usual case δ=β(amplitude and period of the same order) is not included in this work, because we have considered 1 < β < δ which is necessary to apply the adaptation of the unfolding100 method. However, as for δ∈(β, e βp) the roughness effects are so strong that we get no-slip in the limit, then for δ≤βthe no-slip condition in the limit must be verified. In this sense, due to the wide variety of choices of parameters δand βconcerning the roughness, and γrelative to the friction coefficient, this study may be instrumental for understanding the effects of the rough boundary and105 fluid microstructure on the lubrication process. Moreover, by means of the limit equation, as usual in the asymptotic study of fluids in thin domains, we obtain the corresponding non-linear Reynolds type equation assuming the pure slip, partial slip or no-slip boundary conditions ob-110 tained in each case. Related to this, for the rigorous mathematical justification of the linear Reynolds equation for a flow between two plain surfaces, we refer to [3] and [32], whereas for the justification of the non-linear Reynolds type equation we give the references [9] and [29]. Remark that the last reference has been later extended in [7] and [8], assuming stick-slip conditions given by Tresca115 law on the boundary. Finally, comment that approximate the behavior of flows in domains containing singular boundaries is critical for the description of a number of phenomena related to the behavior of flows over rough surfaces. In this sense, it worth to re-120 fer the following papers concerning fluid flows and domains containing corners, [4], [22] and [30]. 7 2. Notation The elements x∈R3will be decomposed as x= (x0, x3) with x0∈R2, x3∈R.125 By Z0, we denote the unitary cube of R2,Z0= (−1 2,1 2)2, and by b Qthe set b Q=Z0×(0,+∞). For every M > 0 we write b QM=Z0×(0, M). We assume p, with p > 1, and p0conjugate exponents, i.e. 1/p + 1/p0= 1. We use the index # to mean periodicity with respect Z0, for example Lp #(Z0), q > 1 denotes the space of functions u∈Lp loc(R2) which are Z0-periodic, while Lp #(b Q) denotes the space of functions bu∈Lp loc(R2×(0,+∞)) such that Zb Q |bu|pdz < +∞,bu(z0+k0, z3) = bu(z),∀k0∈Z2,a.e. z∈R2×(0,+∞). For a bounded measurable set Θ ⊂RN, we denote by Lp 0(Θ) the space of functions of Lp(Θ) with null integral. We denote by ε,εβand εδthree positive parameters which tend to zero and satisfy 1 < β < δ < +∞, i.e. lim ε→0εδ−β= 0,lim ε→0εβ−1= 0. Then, for a function Ψ ∈W2,∞ #(Z0), Ψ ≥0 in Z0, we define the open set Λε⊂R3 130 by Λε=x∈R3:−εδΨx0 εβ< x3< ε,(11) and for a Lipschitz bounded connected open set ω⊂R2, we take Ωε= Λε∩(ω×R),(12) Ω− ε= Ωε∩(ω×(−∞,0)),Ω+ ε= Ωε∩(ω×(0,+∞)),(13) Γε=x∈R3:x0∈ω, x3=−εδΨx0 εβ,(14) 135 e Ωε=y∈R3:y0∈ω, −εδ−1Ψy0 εβ< y3<1,(15) 8 e Γε=y∈R3:y0∈ω, y3=−εδ−1Ψy0 εβ,(16) Ω = ω×(0,1),Γ = ω× {0}.(17) We denote by νthe outside unitary normal vector to Ωεon ∂Ωε. The orthogonal projection on the tangent space of ∂Ωεwill be denoted by T, i.e. Tξ =ξ−(ξν)ν, ∀ξ∈R3,a.e. on ∂Ωε. For k0∈Z2and ρ > 0, we denote Ck0 ρ=ρk0+ρZ0, Qk0 ρ= Λε∩(Ck0 ρ×R). We define κ:R2→Z2by κ(x0) = k0⇔x0∈Ck0 1. Remark that κis well defined up to a set of zero measure in R2(the set ∪k0∈Z2∂Ck0 1). Moreover, for every ρ > 0, we have κx0 ρ=k0⇔x0∈Ck0 ρ. For a.e. x0∈R2we define Cεβ(x0) as the square Ck0 εβsuch that x0belongs to Ck0 εβ.140 We denote by Vthe space of functions bv:R2×(0,+∞)→Rsuch that bv∈W1,p #(b QM),∀M > 0,∇bv∈Lp #(b Q)3. It is a Banach space endowed with the norm k·kVdefined by kbvkp V=kbvkp Lp(Z0×{0})+k∇bvkp Lp( 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. 9 Let us introduce some notation which will be useful in the following. Associated to the change of variables (19), we introduce operators: Dy0, divy0, and245 Dε,y,Dε,y, divε,y by (Dε,yev)i,j = (Dy0ev)i,j =∂yjevifor i= 1,2,3, j = 1,2, (Dε,yev)i,3=1 ε∂y3evifor i= 1,2,3. Dε,y[ev] = 1 2(Dε,yev+Dt ε,yev) divε,yev= divy0ev0+1 ε∂y3ev3= 2 X i=1 ∂yievi+1 ε∂y3ev3. Taking into account the definition of rescaled functions euε,eπε, it is easily observed that the Navier-Stokes problem (5) and the Stokes problem (6) are equivalent, after rescaling, to −divε|Dε,y[euε]|p−2Dε,y[euε]+ (euε∇ε,y)euε+∇ε,y epε=e fin e Ωε, divε,y euε= 0 in e Ωε, (38) 250 −divε|Dε,y[euε]|p−2Dε,y[euε]+∇ε,y epε=e fin e Ωε, divε,y euε= 0 in e Ωε. (39) Proposition 4.1. There exists C > 0, such that for every wε∈W1,p(Ωε), 1<p<+∞, with wε= 0 on ω× {ε}, we have the following: (i) (Poincare’s inequality) kwεkLp(Ωε)≤Cεk∂x3wεkLp(Ωε).(40) (ii) (Korn’s inequality)255 kDwεkLp(Ωε)3×3≤CkD[wε]kLp(Ωε)3×3.(41) Proof. Statement (i) follows using that wε(x) = −Zε x3 ∂x3wε(x0, t)dt, a.e. x∈Ωε. 16 In order to prove (41), we extend wε(x) by zero for x3> ε. Then wεbelongs to W1,p(Ω∗ ε), with Ω∗ ε=x∈R3:x0∈ω, −εδΨx0 εβ< x3<1. and thus the result follows from Sec. 5.2 in[11].  Proposition 4.2 (Estimates for velocity). Let uεbe a solution of (5) (9/5≤ p < +∞) or the unique solution of (6) (1<p<9/5), and euεgiven by (20) the corresponding rescaled solutions of (38) or (39). Then we have260 kuεkLp(Ωε)3≤Cε 2p−1 p(p−1) +1,kDuεkLp(Ωε)3×3≤Cε 2p−1 p(p−1) ,kD[uε]kLp(Ωε)3×3≤Cε 2p−1 p(p−1) , (42) keuεkLp( e Ωε)3≤Cε p p−1,kDε,yuεkLp( e Ωε)3×3≤Cε 1 p−1,kDε,y[uε]kLp( e Ωε)3×3≤Cε 1 p−1, kDy0euεkLp( e Ωε)3×2≤Cε 1 p−1,k∂y3euεkLp( e Ωε)3≤Cε p p−1 (43) Proof. Using energy equality corresponding to the momentum equation (5) and (6), hypothesis (7), Korn inequality (41), and Poincar´e inequality (40) give (42). Finally, estimates (42) combined to the change of variables (19) easily imply265 that euεsatisfies (43).  Proposition 4.3 (Estimates for pressure). Let πεbe a solution of (5) (9/5≤ p < +∞) or the unique solution of (6) (1<p<9/5), and eπεgiven by (20) the corresponding rescaled solutions of (38) or (39). Then we have270 kπεkLp0(Ωε)≤Cε 1 p0,keπεkLp0( e Ωε)≤C, (44) k∇ε,yeπεkW−1,p0 0( e Ωε)3≤C, k∇y0eπεkW−1,p0 0( e Ωε)3≤C, k∂y3eπεkW−1,p0 0( e Ωε)3≤Cε. (45) 17 Proof. Rescaled momentum equation (38) gives h∇εeπε,eϕiW−1,p0( e Ωε),W 1,p 0( e Ωε)=h∇y0eπε,eϕ0i+1 εh∂y3eπε,eϕ3i =−Ze Ωε |Dεeuε|p−2Dεeuε:Dεeϕ dy +Ze Ωεe feϕ dy −Ze Ωε (euε∇ε)euεeϕ dy, (46) for every eϕ∈W1,p 0(e Ωε)3. By (43) we have that for every p > 1, ZΩε |Dε[uε]|p−2Dε[uε] : Dεϕ dy≤ kDε[uε]kp−1 Lp0(Ωε)3×3kDεeϕkLp( e Ωε) ≤1 εkDε[uε]kp−1 Lp0(Ωε)3×3keϕkW1,p 0( e Ωε) ≤CkeϕkW1,p 0( e Ωε), Ze Ωεe feϕ dy≤CkϕkW1,p 0. (47) Hence, to derive estimates for pressure eπεfrom (46), we just need to consider the initial terms. It can be written275 Ze Ωε (euε∇ε,y)euεeϕ dy =−Ze Ωεeuεe ⊗euε∇y0eϕ dy +1 εZe Ωε ∂y3euε,3euεϕ dy +Ze Ωεeuε,3∂y3euεϕ dx, (48) where (ue ⊗v)i,j =uivj, i = 1,2; j= 1,2,3. We consider separately the two terms in the right-hand side of (48). (i) First term of (48): Applying Holder’s inequality, we have Ze Ωεeuεe ⊗euε∇y0eϕ dy≤Ckuεk2 Lq0( e Ωε)3kϕkW1,p 0( e Ωε)3, with 2/q0+ 1/p = 1, i.e. q0= 2p/(p−1). The Sobolev embedding theorem and (43) imply that keuεkLp∗( e Ωε)≤Cε 1 p−1,(49) 18 where p∗= 3p/(3 −p) if 9/5≤p < 3, p∗∈[p, +∞) if p= 3 and p∗= [p, +∞] if280 p > 3. In order to interpolate between Lpand W1,p we introduce interpolation parameter θsuch that 1 q0=θ p+1−θ p∗ in order to have the interpolation (use (43) and (49)) keuεkLq0( e Ωε)3≤ keuεkθ Lp( e Ωε)3keuεk1−θ Lp∗( e Ωε)3≤Cεθp p−1+(1−θ)1 p−1. We choose θfrom inequality θp p−1+ (1 −θ)1 p−1≥0, that is 1≥θ≥θ0= max 0,−1 p−1. Then, θ0= 0, and we have Ze Ωεeuεe ⊗euε∇y0eϕ dy≤Ckuεk2 Lq0( e Ωε)3kϕkW1,p 0( e Ωε)3≤Cε2(θ+1 p−1)kϕkW1,p 0( e Ωε)3, which implies Ze Ωεeuεe ⊗euε∇y0eϕ dy≤CεαkϕkW1,p 0( e Ωε)3,(50) wich α > 0. (ii) Estimate of the second part of the right-hand side of (48) has the form C εk∂y3euε,3kLp( e Ωε)keuεkLq0( e Ωε)3kϕkLq0( e Ωε)3. Working as in item (i), and taking into account last estimate in (43), we get 285 1 εZe Ωε ∂y3euε,3euεϕ dy +Ze Ωεeuε,3∂y3euεϕ dy≤CεαkϕkW1,p 0( e Ωε)3,(51) with α=θ+ 2/(p−1) >0. Putting together (50) and (51) we obtain the estimate Ze Ωε (euε∇ε,y)euεeϕ dy ≤CεαkϕkW1,p 0( e Ωε)3, 19 with α > 0. Now, previous estimate and (47) give estimate (45), i.e. h∇ε,yeπε,eϕiW−1,p0( e Ωε),W 1,p 0( e Ωε)≤CkϕkW1,p 0( e Ωε)3, which, using the Bogovskii operator (see [5]), gives the second estimate in (44). Applying the rescaling x3=εy3, this estimate implies the first one. 290 For the Stokes case, proceeding similarly by using the rescaled momentum equation (39), the result follows from (46) and (47).  Proof of Theorem 3.1. The classical theory (see e.g.[23], [27], [36]) gives295 the existence of at least one weak solution (uε, πε)∈W1,p(Ωε)3×Lp0 0(Ωε), p0=p/(p−1) for (5), under assumption 9/5≤p < +∞, and the existence of a unique one for (6), under assumption 1 < p < 9/5. The estimates for velocity and pressure given in (18) follow from Proposition 4.2 and 4.3. 300 5. Some compactness results In this section we obtain some compactness results about the behavior of a sequence (uε, πε) satisfying a priori estimates (18) together with boundary conditions uε= 0 on ω× {ε},uεν= 0 on ∂Ωε\(ω× {ε}), but where (uε, πε) is not necessarily the solution of any PDE.305 Lemma 5.1. Let uεbe in W1,p(Ωε)3,p > 1, with uε= 0 on ω× {ε},uεν= 0 on ∂Ωε\(ω× {ε}),div uε= 0 in Ωε, and such that there exists a constant C independent of εsatisfying ZΩε |Duε|pdx ≤Cε2p−1 p−1.(52) Let us define euε∈W1,p(e Ωε)3by (20). Then, for a subsequence of εstill denoted by ε, there exist eu0∈W1,p(0,1; Lp(ω))2and ew∈W2,p(0,1; W−1,p(ω)) such that310 20 eu0(1) = 0 in Lp(ω),ew(0) = ew(1) = 0 in W−1,p(ω),(53) divy0eu0+∂y3ew= 0 in W1,p(0,1; W−1,p(ω)),(54) divy0Z1 0eu0(y0, t)dt = 0 in Lp(ω),(55) Z1 0eu0(y0, t)dt ν = 0 in W−1 p,p(∂ω),(56) 315 ε−1 p−1euε*0in W1,p(Ω)3,(57) ε−p p−1˜uε*(eu0,0) in W1,p(0,1; Lp(ω))3,(58) ε−2p−1 p−1˜uε,3*˜win W2,p(0,1; W−1,p(ω)),(59) ε−p p−1divy0eu0 ε+ε−2p−1 p−1∂y3euε,3*0in W1,p(0,1; W−1,p(ω)).(60) Proof. Since uεvanishes on ω× {ε}, estimates (40) and (52) imply that uε also satisfies ZΩε |uε|pdx ≤Cε2p−1 p−1+p. This inequality combined to the change of variables (19) and inequality (52) imply that euεsatisfies320 Ze Ωε |euε|pdy ≤Cε p2 p−1,Ze Ωε|∇y0euε|p+1 εp|∂y3euε|pdy ≤Cε p p−1,(61) Therefore, up to a subsequence, there exist eu∈W1,p(0,1; Lp(ω))3, with eu(1) = 0, such that ε−p p−1euε*euin W1,p(0,1; Lp(ω))3,(62) 21 such that (58) and (60) hold. By (62), we also have that ε−p p−1divy0eu0 ε*divy0eu0,in W1,p(0,1; W−1,p(ω)),(63) and together with div uε= 0 implies that ∂y3euε,3/ε2p−1 p−1is bounded in W1,p(0,1; W−1,p(ω)). Using then that euε,3= 0 on ω× {1}, we deduce that euε,3/ε2p−1 p−1is325 bounded in W2,p(0,1; W−1,p(ω)) and therefore, up to a subsequence, there exists ew∈W2,p(0,1; W−1,p(ω)), with ˜w(1) = 0 in W−1,p(ω), such that (59) holds. By (62), we get that eu3= 0 which finishes the proof of (58). From (59), (60) and (63), we also deduce (54).330 Now, we consider η∈C∞(ω). Integrating by parts in e Ωεand taking into account that uεν= 0 on ∂Ωε, we get 0 = Ze Ωεε−p p−1divy0eu0 ε+ε−2p−1 p−1∂y3euε,3η(y0)dy =−Ze Ωε ε−p p−1eu0 ε(y)∇y0η(y0)dy. Since (61) implies Ze Ωε\Ωε−p p−1eu0 εdy →0, we can write the previous equality as ZΩ ε−p p−1eu0 ε(y)∇y0η(y0)dy +Oε= 0. Passing to the limit in this equality by means of (58), we get ZωZ1 0eu0(y0, y3)dy3∇y0η(y0)dy0= 0, which implies (55) and (56). Integrating (54) with respect to y3in (0,1), we now deduce that ew(0) = 0, which concludes the proof of (53).  The change of variables (19) does not provide the information we need about335 the behavior of uεin the part of Ωεclose to Γε. To solve this difficulty, we introduce an adaptation of the unfolding method (see e.g. [2], [13], [16], [20], [26]), which is strongly related to the two-scale convergence method ([1], [33]). 22 For this purpose, given uε∈W1,p(Ωε)3,uε= 0 on ∂Ωε\Γε, and assuming uε extended by zero to the set Λεgiven by (11), we define buεby340 buε(x0, z) = uεεβκx0 εβ+εβz0, εβz3,a.e. (x0, z0)∈R2×b Zε,(64) with b Zε=z∈Z0×R:−εδ−βΨ(z0)< z3< ε1−β. Remark 5.2. For k0∈Z2the restriction of buεto Ck0 εβ×b Zεdoes not depend on x0, while as function of zit is obtained from uεby using the change of variables z0=x0−εβk0 εβ, z3=x3 εβ,(65) which transforms Qk0 εβinto b Zε. Therefore, the idea in the definition of the func-345 tion buεis to apply a dilatation in order to study the behavior of uεat a very small distance of Γε. In addition, we observe that the change of variables (65), with x0fixed, transforms Γεinto the surface {z3=−εδ−βΨ(z0)}which, thanks to the assumption εδ−βconverging to zero, almost agrees with the flat boundary {z3= 0}.350 We will use the following lemma, whose proof is elementary and thus omitted. Lemma 5.3. Let vε∈Lp(R2),p > 1, be a sequence which converges weakly to a function vin Lp(R2). We define ¯vε∈Lp(R2)by ¯vε(x0) = − ZCεβ(x0) vε(η0)d η0,a.e. x0∈R2. Then we have: (i) The sequence ¯vεconverges weakly to vin Lp(R2). Moreover, if the convergence of vεis strong in Lp(R2), then the convergence of ¯vεis also strong355 in Lp(R2). 23 (ii) For every τ0∈R2, we have ¯vε(x0+εβτ0)−¯vε(x0) εβ*∇v τ0in W−1,p(R2). Lemma 5.4. We consider a sequence uε∈W1,p(Ωε)3,p > 1, satisfying (52), uε= 0 on ω×{ε},uεν= 0 on ∂Ωε\(ω×{ε}). We define euε∈W1,p(e Ωε)3by (20) and suppose there exists eu0∈W1,p(0,1; Lp(ω))2such that (58) holds. Taking into account the definition of e βpgiven in (10), then we have the following:360 (i) If δ∈(β, e βp), then eu0(x0,0)∇Ψ(z0)=0 a.e. (x0, z0)∈ω×Z0.(66) (ii) If δ=e βp, then there exists bu∈Lp(ω;V3)with bu3(x0, z0,0) = −∇Ψ(z0)eu0(x0,0),a.e. (x0, z0)∈ω×Z0,(67) such that for every M > 0, the sequence buεdefined by (64) satisfies ε−βp−1 p−2p−1 p(p−1) Dzbuε* Dzbuin Lp(ω×b QM)3×3.(68) Besides, if div uε= 0 in Ωε, then divzbu= 0 in ω×b Q. (69) Proof. We divide the proof in four steps.365 Step 1. Let us obtain some estimates for the sequence buεdefined by (64). For M > 0, definition of (64) of buεand (52) prove that for every ε > 0 small enough (depending on M), we have ZR2× b QM |Dzbuε(x0, z)|pdx0dz ≤εβ(p+2) X k0∈Z2Zb QM |Duε(εβ(k0+z0), εβz3)|pdz0 ≤X k0∈Z2 εβ(p−1) ZQk0 εβ |Duε|pdx ≤εβ(p−1) ZΩε |Duε|pdx ≤Cεβ(p−1)+ 2p−1 p−1. (70) On the other hand, defining ¯uε(x0) = − ZCεβ(x0) uε(τ0,0) dτ =− ZCεβ(x0)euε(τ0,0) dτ =ZZ0buε(x0, z0,0) dz0,(71) 24 a.e. x0∈R2, using the inequality370 ZωZb QM |buε(x0, z)−¯uε(x0)|pdzdx0≤CMZωZb QM |Dzbuε|pdzdx0,(72) where CMdoes not depend on ε, and taking into account (70), we deduce that b Uε=buε(x0, y)−¯uε εβp−1 p+2p−1 p(p−1) is bounded in Lp(R2;W1,p(b QM))3∀M > 0.(73) Thus, there exists bu∈Lp(ω;W1,p(b QM))3for every M > 0, such that, up to a subsequence, b Uε*buin Lp(ω;W1,p(b QM))3∀M > 0,(74) and then ε−βp−1 p−2p−1 p(p−1) Dzbuε* Dzbuin Lp(ω×QM)3×3,∀M > 0.(75) By semicontinuity, inequality (70) proves Zω× b QM |Dzbu|pdx0dz ≤C∀M > 0. Once we prove the Z0-periodicity of buin z0(Step 2), the arbitrariness of Mwill375 then imply that bubelongs to Lp(ω;V3). Moreover, if we also assume that div uε= 0 in Ωε, then by definition (64) of buε, we have divzbuε= 0 in R2×b QM, which combined with (75) proves (69). Step 2. Let us prove that buis Z0-periodic in the variable z0.380 We observe that by definition (64) of buε, for every M > 0, we have buεx1+εβ, x2,−1 2, z2, z3=buεx1, x2,1 2, z2, z3, a.e. (x0, z2, z3)∈R2×(−1 2,1 2)×(0, M), which implies b Uεx1+εβ, x2,−1 2, z2, z3−b Uεx0,1 2, z2, z3=−¯uε(x1+εβ, x2) + ¯uε(x0) εβp−1 p+2p−1 p(p−1) (76) Since ε−p p−1euε(x0,0) is bounded in Lp(R2)3, we can apply Lemma 5.3-(ii) to deduce that the right-hand side of this equality tends to zero in W−1,p(R2). 25 However, integrating in the x3variable, we have455 Zωuεx0,−δεΨx0 rε−uε(x0,0) p dx0≤Cεδ(p−1) ZΩε |Duε|pdx =Oε,(95) and so, using that uε(x0,0) = euε(x0,0) in ω, we get λε−γZΓε uεϕεdσ=λε−γ−1Zωeu0 ε(y0,0)eϕ0(y0,0)dy0+Oε. By convergence (58), in order to pass to the limit in this term it must be verified the following relation −γ−1 = −p p−1, which is true for γ=1 p−1. Let us discuss the pass of the limit of this term depending on the parameter γ: - If γ=−∞, then pure slip is considered, and so this term is zero. - If γ∈[0,1 p−1), then λε−γZΓε uεϕεdσ =λε−γ−1Zωeu0 ε(y0,0)eϕ0(y0,0)dy0+Oε =λε 1 p−1−γZω ε−p p−1eu0 ε(y0,0)eϕ0(y0,0)dy0+Oε =Oε, because 1 p−1−γ > 0. - If γ=1 p−1, then460 λε−1 p−1ZΓε uεϕεdσ =λZω ε−p p−1eu0 ε(y0,0)eϕ0(y0,0)dy0+Oε =λZωeu0(y0,0)eϕ0(y0,0)dy0. (96) - If γ∈(1 p−1,+∞), then λε−γZΓε uεϕεdσ =λε−γ−1Zωeu0 ε(y0,0)eϕ0(y0,0)dy0+Oε =λε 1 p−1−γZω ε−p p−1eu0(y0,0)eϕ0(y0,0)dy0+Oε, 32 and this goes to infinity, because 1 p−1−γ < 0. For that reason, we drop this case; it must be required no-slip condition on Γε, instead of Navier condition. •Fifth term in (81). Thanks to (84), (86) and the change of variables (19) we get465 ZΩε fε(x)ϕε(x)dx =ε−1ZΩ+ εe f0x0,x3 εeϕ0x0,x3 εdx +Oε =ZΩe f0(y)eϕ0(y)dy +Oε. (97) From (92)-(97), we then deduce that the limit system reads as ZΩe ξ: 2 X i=1 ∂y3eϕi(y)(ei⊗e3) !dy+ZΩ ∇y0eπ(y0)eϕ0(y)dy + ZωZb Qb ξ:Dzbϕ(x0, z)dzdx0+λZΓeu0eϕ0dσ =ZΩe f0(y)eϕ0(y)dy, (98) for every eϕ0∈C1 c(ω×(−1,1))2,bϕ∈C1 c(ω;C1 ](b Q))3such that (82) is satisfied, with λ= 0 in the case γ∈ {−∞} ∪ [0,1 p−1) and λ6= 0 in the case γ=1 p−1. By density, this holds true for every eϕ0∈W1,p(0,1; L2(ω))2, and every bϕ∈ Lp(ω;V)3such that eϕ0(x0,1) = 0 a.e. x0∈ω, ∇Ψ(z0)eϕ0(x0,0) + bϕ3(x0, z0,0) = 0 a.e. (x0, z0)∈ω×Z0. Step 2. Previous property. Let us prove lim ε→0ε−2p−1 p−1ZΩε S(D[uε]) : Duεdx +λε−2p p−1ZΓε |uε|2dσ =ZΩe ξ: 2 X i=1 ∂y3euiei⊗e3 !dy +λ ZΓ |eu0|2dσ +ZωZb Qb ξ:Dzbu(x0, z)dzdx0, (99) with λ= 0 in the case γ∈ {−∞} ∪ [0,1 p−1) and λ6= 0 in the case γ=1 p−1. For this purpose we take ε−2p−1 p−1uεas test function in (5). Using div uε= 0 in Ωε, this gives470 ε−2p−1 p−1ZΩε S(D[uε]) : Duεdx +λε−2p p−1ZΓε |uε|2dσ =ε−2p−1 p−1ZΩε fεuεdx. (100) 33 In order to pass to the limit for the last term we use the change of variables (19) and (58), which easily gives ε−2p−1 p−1ZΩε fεuεdx =ε−p p−1Ze Ωεe feuεdy =ZΩe f0eu0dy +Oε. So, (100) can be written as ε−2p−1 p−1ZΩε S(D[uε]) : Duεdx +λε−2p p−1ZΓε |uε|2dσ =ZΩe f0eu0dy +Oε(101) Taking now eu0and buas test functions in (98), and taking into account that divy0eu0= 0 in Ω and divzbu= 0 in ω×b Q, we then deduce (99) from (101). Step 3. Corrector result for D[uε].In order to prove (31), we define b ϑε(x0, z) = − ZCεβ(x0) Dz[bu(s0, z)]ds0,a.e. (x0, z)∈ω×b Q. It is well known that it holds475 b ϑε→Dz[bu] in Lp(ω×b Q)3×3,(102) and that using the properties of the p-Laplace operator Swe get S(b ϑε)→S(Dz[bu]) in Lp0(ω×b Q)3×3.(103) We also define e ϑ(y) = 1 2 2 X i=1 ∂y3eui(y) (ei⊗e3+e3⊗ei),a.e. y∈e Ωε.(104) We consider a new parameter εs>0, with 1 < s < β, which implies lim ε→0εs−1= 0,lim ε→0εs−β= +∞.(105) Using the H¨older inequality and applying the change of variables y=x/ε, we prove480 Z{x3>εs} |ε−β pb ϑε(x0,x εβ)|pdx =ε2βX k0∈Z2ZZ0Zε1−β εs−β− ZCk0 εβ Dz[bu](s0, z)ds0 p dz3dz0 ≤ZωZZ0Z∞ εs−β |Dz[bu](s0, z)|2dz3dz0ds0=Oε. (106) 34 and on the contrary, we get ε−1Z{x3<sε} |e ϑ(x0,x3 ε)|pdx =Z{0<y3<sε ε} |e ϑ(y)|pdy =Oε.(107) Since properties of the p-Laplace operator S(i.e. p-coercivity, (p−1)-growth and monotonicity) , proving the strong convergence (31) is equivalent to prove the following statement Eε:= ε−2p−1 p−1ZΩ− ε S(D[uε]) : D[uε]dx +ε−2p−1 p−1ZΩ+ εS(D[uε]) −Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) :D[uε]−ε1 p−1e ϑ(x0,x3 ε)−ε−β p+2p−1 p(p−1) b ϑε(x0,x rε )dx +ε−2p p−1λZΓε |uε−εp p−1(˜u0(x0,0),0)|2dσ→0, with λ= 0 in the case γ∈ {−∞} ∪ [0,1 p−1) and λ6= 0 in the case γ=1 p−1. Developing the expression of Eε, we have Eε=ε−2p−1 p−1ZΩε S(D[uε]) : D[uε]dx +ε−2p p−1λZΓε |uε|2dσ −ε−2ZΩ+ ε S(D[uε]) : e ϑ(x0,x3 ε)dx −ε−β p−2p−1 pZΩ S(D[uε]) : b ϑε(x0,x εβ)dx −ε−2p−1 p−1ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx +ε−2ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):e ϑ(x0,x3 ε)dx +ε−β p−2p−1 pZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):b ϑε(x0,x εβ)dx −2ε−p p−1λZΓε u0 εeu0(x0,0)dσ +λZΓε |u(x0,0)|2dσ. (108) Let us pass to the limit in the different terms of this expression. •First and second terms. They can be estimated using the previous property 35 (99) given in Step 2, which gives ε−2p−1 p−1ZΩε S(D[uε]) : D[uε]dx +ε−2p p−1ZΓε |uε|2dσ =ZΩe ξ:e ϑ(y)dy +λZΓ |eu0|2dσ +Zω× b Qb ξ:Dz[bu]dx0dz +Oε. •Third term. We use the change of variables (19) together with (90) prove −ε−2ZΩ+ ε S(D[uε]) : e ϑ(x0,x3 ε)dx =−Ze Ω+ ε S(ε−1 p−1Dε,y[euε]) : e ϑ(y)dy =−ZΩe ξ:e ϑ(y)dy +Oε. •Fourth term. Using the sequence sεgiven in (105), we get −ε−β p−2p−1 pZΩ+ ε S(D[uε]) : b ϑε(x0,x εβ)dx =−ε−β p−2p−1 pZ{x3>εs} S(D[uε]) : b ϑε(x0,x εβ)dx −ε−β p−2p−1 pZ{0<x3<εs} S(D[uε]) : b ϑε(x0,x εβ)dx Taking into account property (106), the first integral converges to zero, and by485 applying the change of variables (65) in the second one together with the weak convergence of S(ε−βp−1 p−2p−1 p(p−1) Dz[buε]) to b ξin Lp0(ω×b Q)3×3and (102), we have −ε−β p−2p−1 pZΩ+ ε S(D[uε]) : b ϑε(x0,x rε )dx =−Zω× b Qε1−β S(ε−βp−1 p−2p−1 p(p−1) Dz[buε]) : b ϑεdx0dz =−Zω× b Qb ξ:Dz[bu]dx0dz +Oε. •Fifth term. We use the sequence εsdefined in (105) to split this term as follows −ε−2p−1 p−1ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx =−ε−2p−1 p−1Z{x3>εs} Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx −ε−2p−1 p−1Z{0<x3<εs} Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx 36 By using property (106) and (18), we prove for p > 2: ε−2p−1 p−1Z{x3>εs}Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ)−Sε1 p−1e ϑ(x0,x3 ε):D[uε]dx ≤ε−2p−1 p−1Z{x3>εs}ε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) p−2+ε1 p−1e ϑ(x0,x3 ε) p−2· ·ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ)|D[uε]|dx ≤Cε−2p−1 p−1  Z{x3>εs}ε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) pdx!p−2 p + Z{x3>εs}ε1 p−1e ϑ(x0,x3 ε) pdx!p−2 p · · Z{x3>εs}ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) pdx!1 p Z{x3>εs} |D[uε]|pdx!1 p =Oε, and for 9/5≤p < 2 (Navier-Stokes case), and for 1 <p<9/5 (Stokes case): ε−2p−1 p−1Z{x3>εs}Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ)−Sε1 p−1e ϑ(x0,x3 ε):D[uε]dx ≤Cε−2p−1 p−1Z{x3>εs}ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) p−1|D[uε]|dx ≤Cε−2p−1 p−1 Z{x3>εs}ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) pdx!1 p Z{x3>εs} |D[uε]|pdx!1 p =Oε. Analogously, by using (107), we can prove ε−2p−1 p−1Z{0<x3<εs}Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ) −Sε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx =Oε. 37 Then, taking into account the above, (58), (68), and (103), we get −ε−2p−1 p−1ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx =−ε−2p−1 p−1Z{x3>εs} Sε1 p−1e ϑ(x0,x3 ε):D[uε]dx −ε−2p−1 p−1Z{0<x3<εs} Sε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):D[uε]dx +Oε =−Z{y3>εs−1} Se ϑ(y): ε−p p−1 2 2 X i=1 ∂y3euε,i(y)(ei⊗e3+e3⊗ei)!dy −Zω× b Qεs−β Sb ϑε(x0, z):ε−βp−1 p−2p−1 p(p−1) Dz[buε]dx0dz +Oε =−ZΩ Se ϑ(y):e ϑ(y)dy −Zω× b Q S(Dz[bu]) : (Dz[bu]) dx0dz +Oε. (109) •Sixth and seventh terms. Reasoning as in the fifth term, we have490 ε−2ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):e ϑ(x0,x3 ε)dx =ZΩ Se ϑ(y):e ϑ(y)dy +Oε, (110) ε−β p−2p−1 pZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑε(x0,x εβ):b ϑε(x0,x εβ)dx =Zω× b Q S(Dz[bu]) : Dz[bu]dx0dz +Oε. (111) •Eighth term. Using ZΓε |uε−uε(x0,0)|pdx =Oε, and (21), we have −2λε−p p−1ZΓε u0 ε˜u0(x0,0)dσ =−2λZΓ ε−p p−1˜u0 ε˜u0dσ+Oε=−2λZΓ |˜u0|2dσ+Oε. •Ninth term. We have λZΓε |u(x0,0)|2dσ =λZΓ |u(x0,0)|2dσ +Oε. 38 The estimates obtained for the different terms on the right-hand side of (108) prove that Eεtends to zero and then (31). Step 4. Identification of e ξand b ξ.By using the corrector result of D[uε],495 obtained in Step 3, we will prove that e ξ= 2−p 2|∂y3eu0|p−2 2 X i=1 ∂y3eui(ei⊗e3+e3⊗ei) in Ω,(112) b ξ=S(Dz[bu]) in ω×b Q. (113) On the one hand, using a sequence εssatisfying (105), the weak convergences of S(ε−1 p−1Dε,y[euε]) to ˜ ξin Lp0(Ω)3×3and S(ε−βp−1 p−2p−1 p(p−1) Dz[buε]) to b ξ in Lp0(ω×b QM)3×3, for some M > 0, then for every e φ∈C1 c(ω×(−1,1))3and500 b φ∈C1 c(ω;C1 #(b Q)3), with Dzφ2(x0, y) = 0 a.e. in {z3> M}, for some M > 0, we prove that ε−2p−1 p−1ZΩ+ ε S(D[uε]) : ε1 p−1 2 X i=1 ∂y3e φi(x0,x3 ε)ei⊗e3!+ε−β p+2p−1 p(p−1) Dzb φ(x0,x εβ)!dx =ε−2Z{x3>εs} S(D[uε]) : 2 X i=1 ∂y3e φi(x0,x3 ε)ei⊗e3!dx +Z{0<x3<εs} ε−β p−2p−1 pS(D[uε]) : Dzb φ(x0,x εβ)dx =Z{y3>εs−1} S(ε−1 p−1Dy,ε[euε]) : 2 X i=1 ∂y3e φi(y)ei⊗e3!dy +Zω× b Qεs−r S(ε−βp−1 p−2p−1 p(p−1) Dz[buε]) : Dzb φ(x0, z)dx0dz =ZΩe ξ: 2 X i=1 ∂y3e φi(y)ei⊗e3!dy +Zω× b Qb ξ:Dzb φ(x0, z)dx0dz +Oε (114) On the other hand, taking into account the sequence εssatisfying (105), using the corrector result (31) for D[uε], properties (106), (107), and proceeding as in 39 (109), we prove505 ε−2p−1 p−1ZΩ+ ε S(D[uε]) : ε1 p−1 2 X i=1 ∂y3e φi(x0,x3 ε)ei⊗e3!+ε−β p+2p−1 p(p−1) Dzb φ(x0,x εβ)!dx =ε−2p−1 p−1ZΩ+ ε Sε1 p−1e ϑ(x0,x3 ε) + ε−β p+2p−1 p(p−1) b ϑ(x0,x εβ) : ε1 p−1 2 X i=1 ∂y3e φi(x0,x3 ε)ei⊗e3!+ε−β p+2p−1 p(p−1) Dzb φ(x0,x εβ)!dx +Oε =ZΩ Se ϑ(y): 2 X i=1 ∂y3e φi(x0,x3 ε)ei⊗e3!dy +Zω× b Q S(Dz[bu]) : Dzb φ(x0, z)dx0dz +Oε. (115) Thus, (114) and (115) implies ZΩ ˜ ξ: 2 X i=1 ∂y3e φi(y)ei⊗e3!dy =ZΩ Se ϑ(y): 2 X i=1 ∂y3e φi(y)ei⊗e3!dy, Zω× b Qb ξ:Dzb φ(x0, z)dx0dz =Zω× b Q S(Dz[bu]) : Dzb φ(x0, z)dx0dz for every e φand b φas above. By density this implies (113). To get (112), it remains to prove S(e ϑ(y)) = 2−p 2|∂y3eu0|p−2 2 X i=1 ∂y3eui(y)(ei⊗e3+e3⊗ei) in Ω. This follows just taking into account that S(e ϑ) can be expressed in term of the second invariant of the strain tensor DII (e ϑ) = D(e ϑ)D(e ϑ)tby |D(e ϑ)|p−2=|DII (e ϑ)|p/2−1. Step 5. Boundary layer system satisfied by bu.From the previous steps, the limit problem (98) reads as ZΩ 2−p 2|∂y3eu0|p−2 2 X i=1 ∂y3eui(y)(ei⊗e3+e3⊗ei) : 2 X i=1 ∂y3eϕi(y)ei⊗e3!dy +ZΩ ∇y0eπeϕ0dy +Zω× b Q S(Dz[bu]) : Dzbϕ dzdx0+λZΓ ˜u0eϕ0dσ =ZΩe f0eϕ0dy, (116) 40 for every ϕ∈W1,p(Ω)3, and every bϕ∈Lp(ω;V3) such that eϕ= 0 on ∂Ω\Γ, eϕ3(x0,0) = 0,bϕ3(x0, z0,0) = −λ∇Ψ(z0)eϕ0(x0,0),a.e. (x0, y0)∈ω×Z0, with λ= 0 in the case γ∈ {−∞} ∪ [0,1 p−1) and λ6= 0 in the case γ=1 p−1. Now, we will obtain an equation for (eu0,eπ) eliminating buin (116). For this purpose, we take eϕ0= 0 in (116) we prove that bu∈ V3satisfies510                        Zω× b Q S(Dz[bu]) : Dzbϕ dx0dz = 0 divzbu= 0, bu3(x0, z0,0) = −∇Ψ(z0)eu0(x0,0) on R2× {0}, S(Dzbu)i,3= 0, i = 1,2,on R2× {0}, (117) a.e. in ω. Defining (b φξ0,bqξ0), for every ξ0∈R2, by (24), we deduce that bu(x0, z) = b φ eu0(x0,0)(z),a.e. (x0, z)∈ω×b Q. (118) Now, for ϕ∈W1,p(Ω)3, with ϕ= 0 on ∂Ω\Γ, ϕ3= 0 on Γ, we take eϕ0and bϕ(x0, z) = eϕ1(x0,0)b φe1(z) + eϕ2(x0,0)b φe2(z), as test functions in (116). Taking into account (118) and definition (25) we get ZΩ 2−p 2|∂y3eu0|p−2 2 X i=1 ∂y3eui(y)(ei⊗e3+e3⊗ei) : 2 X i=1 ∂y3eϕi(y)ei⊗e3!dy − ZΩeπdivy0˜ϕ0dy +λZΓeu0˜ϕ0dσ +Zω R(˜u0(y0,0)) ˜ϕ0(y0,0)dy0=ZΩ ˜ f0˜ϕ0dy. (119) By the arbitrariness of ϕ, this proves that (eu0,eπ) is a solution of (23)-515 (26) with λ6= 0 (case γ=1 p−1), or a solution of (23)-(27) with λ= 0 (case γ={−∞} ∪ [0,1 p−1)). 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