Derivation of a quasi-stationary coupled Darcy-Reynolds equation for incompressible viscous fluid flow through a thin porous medium with a fissure

Derivation of a quasi-stationary coupled Darcy-Reynolds equation for incompressible viscous fluid flow through a thin porous medium with a fissure Mar´ıa ANGUIANO Departamento de An´alisis Matem´atico Universidad de Sevilla, P. O. Box 1160, 41080-Sevilla (Spain) [email protected] Abstract We consider a non-stationary Stokes system in a thin porous medium of thickness εwhich is perforated by periodically distributed solid cylinders of size ε, and containing a fissure of width ηε. Passing to the limit when εgoes to zero, we find a critical size ηε≈ε2 3in which the flow is described by a 2D quasi-stationary Darcy law coupled with a 1D quasi-stationary Reynolds problem. AMS classification numbers: 75A05, 76A20, 76M50, 35B27. Keywords: Stokes equation; Darcy’s law; Reynolds equation; thin porous medium; fissure. 1 1 Introduction The aim of this work is to prove the convergence of the homogenization process for the non-stationary Stokes system in a thin porous medium Dεηεof thickness εwhich is perforated by periodically distributed solid cylinders of size εand contains a fissure {0≤x2≤ηε}of width ηε. We consider the fluid flow through a periodic distribution of vertical cylinders and a fissure. The periodic distribution of vertical cylinders and the fissure are confined between two parallel plates (see Figure 1). A representative elementary volume for the thin porous medium is a cube of lateral length εand vertical lentgth ε. The cube is repeated periodically in the space between the plates. Each cube can be divided into fluid part and a solid part, where the solid part has the shape of a vertical cylinder of height ε. " " " ⌘" x3 x2 x1 Figure 1: View of the domain Dεηε The question of a medium containing a fissure with properties different from those of the rest of the material has been the subject of many studies previously, see Ciarlet et al [1], Panasenko [2] and Chapter 13 of Sanchez-Palencia [3] among others. A similar problem of the one considered in this paper with a fixed height domain, but for the Laplace’s equation, was studied in Bourgeat and Tapiero [4]. The peculiar behavior observed for the Laplace’s equation when ηε≈ε2 3has motivated the analogous study for the Stokes system in Bourgeat et al [5] (see [6] for the Navier-Stokes system and [7] for a non-stationary Stokes system). In Anguiano [8], we consider a non-stationary Stokes system in a thin porous medium of thickness εwhich is perforated by periodically distributed solid cylinders of size aε. We apply an adaptation of the unfolding method in order to obtain rigorously quasi-stationary Darcy’s laws. The behavior observed when aε≈εhas motivated the fact of considering a thin porous medium containing a fissure. In this sense, our aim in the present paper is to extend the study of Bourgeat et al [5] to the case of a non-stationary Stokes system in a domain of small height ε, perforated by periodically distributed solid cylinders of size ε, containing a fissure of width ηε, which makes necessary to rescale in the height variable in order to work with a domain of height one. We find the same critical size as in Bourgeat et al [5], what means that the evolutive model and the thin thickness of the domain do not modify the critical size. However, the thin thickness of the domain leads us to use techniques of reduction of the dimension together with homogenization in order to obtain more simplified effective models than those obtained in Bourgeat et al [5]. More precisely, we obtain the following results corresponding to three characteristic situations depending on the parameter ηεwith respect to ε: •If ηεε2 3the fissure is not giving any contribution. In this case, in order to find the limit, we 2 use the results developed in Anguiano [8] and we obtain a 2D quasi-stationary Darcy’s law. •If ηεε2 3the fissure is dominant. We introduce a rescaling of the fissure in order to work with a domain with size one, and then we prove that the limit of the velocity is a Dirac measure concentrated on the line {x2= 0} ∩ {x3= 0}representing the corresponding tangential line flow. Meanwhile in the porous medium the effective velocity is equal to zero. •If ηε≈ε2 3with ηε/ε2 3→λ, 0 < λ < +∞, it appears a coupling effect and the effective flow behaves as 2D quasi-stationary Darcy flow in the porous medium coupled with the tangential flow of the line {x2= 0}∩{x3= 0}. Compared to the first case ηεε2 3, the effective velocity has now an additional tangential component concentrated on {x2= 0}∩{x3= 0}. Moreover, the limit problem is now given by a new variational equation, in which appears the parameter λ, and consists of a 2D quasi-stationary Darcy law in the porous medium coupled with a 1D quasi-stationary Reynolds problem on the line {x2= 0}∩{x3= 0}. 2 The domain and some notations 2.1 The domain Let ω⊂R2be smooth bounded connected open set and Ω = ω×(0,1) ⊂R3. We define Ω+= Ω ∩ {x2>0},Ω−= Ω ∩ {x2<0},Σ=Ω∩ {x2= 0},Σ1= Σ ∩ {x3= 0}. For some η0>0 we define the domains D= Ω−∪(η0e2+ Ω+)∪(Σ ×[0, η0]e2), D0=D∩ {x3= 0}, with e2= (0,1,0). Let ε > 0 be a small parameter devoted to tend to zero and 0 < ηε< η0be a small parameter devoted to tend to zero with ε. A periodic porous medium is defined by a domain ωand an associated microstructure, or periodic cell Y0= [0,1]2, which is made of two complementary parts: the fluid part Y0 f, and the solid part Y0 s (Y0 fSY0 s=Y0and Y0 fTY0 s=∅). More precisely, we assume that Y0 sis a smooth and connected set strictly included in Y0. For k0= (k1, k2)∈Z2, each cell Y0 k0=k0+Y0is divided in a fluid part Y0 fk0and a solid part Y0 sk0. We define Y=Y0×(0,1) ⊂R3, and is divided in a fluid part Yfand a solid part Ys. We also denote Y− s=[ k0∈Z2 − Ysk0, Y + s=[ k0∈Z2 + Ysk0, all the solid parts in R2×(0,1), where Z2 −={k0∈Z2, k2<0}and Z2 +={k0∈Z2, k2>0}. It is obvious that Ef=(R2×(0,1)) \(Y− s∪Y+ s)∩Ω is the fluid part in Ω. Following [9], we make the following assumptions on Yf,Ef,Ysand Y∗ s=Y+ s∪Y− s: i) Yfis an open connected set of strictly positive measure, with a locally Lipschitz boundary. ii) Yshas strictly positive measure in Y. 3 iii) Efand the interior of Y∗ sare open sets with boundaries of class C0,1and are locally located on one side of their boundaries. Moreover Efis connected. We also define Y− s,ε =εY 0− s×(0,1), Y + s,εηε= (ηεe2+εY 0+ s)×(0,1),e Sεηε=∂(Y− s,ε ∪Y+ s,εηε). We denote by e Aεηε= (Y− s,ε ∪Y+ s,εηε)∩Dthe solid part of the domain D, e Dεηε=D\e Aεηεthe fluid part of the domain D(including the fissure), e Iηε= Σ ×(0, ηε)e2the fissure in D, e Ωεηε=e Dεηε\e Iηεthe fluid part of the porous medium in D. Let us define a domain with thickness ε, given by Ωε= Ω ∩ {0< x3< ε} ⊂ R3. We also define Ωε += Ω+∩ {0< x3< ε},Ωε −= Ω−∩ {0< x3< ε},Σε= Ωε∩ {x2= 0}, and Dε= Ωε −∪η0e2+ Ωε +∪(Σε×[0, η0]e2). The microscale of a porous medium is the small positive number ε. The domain ωis covered by a regular mesh of size ε: for k0= (k1, k2)∈Z2, each cell Y0 k0,ε =εk0+εY 0is divided in a fluid part Y0 fk0,ε and a solid part Y0 sk0,ε, i.e. is similar to the unit cell Y0rescaled to size ε. We define Yk0,ε =Y0 k0,ε ×(0,1) ⊂R3, which is also divided in a fluid part Yfk0,ε and a solid part Ysk0,ε. Now, we denote by Aεηε,Dεηε,Iηεand Ωεηεthe sets e Aεηε,e Dεηε,e Iηεand e Ωεηε, respectively, with thickness ε, i.e., Aεηε=e Aεηε∩ {0< x3< ε}- the solid part of the domain Dε, Dεηε=e Dεηε∩ {0< x3< ε}- the fluid part of the domain Dε(including the fissure), Iηε=e Iηε∩ {0< x3< ε}- the fissure in Dε, Ωεηε=e Ωεηε∩ {0< x3< ε}- the fluid part of the porous medium in Dε. Finally we define Ω+ εηε=Dεηε∩ {x2> ηε},Ω− εηε=Dεηε∩ {x2<0},Γηε=∂Σε×(0, ηε)e2, and D+=D∩ {x2>0}, D−= Ω−. 4 " " ⌘" x3 x2 ⌦+ "⌘" ⌦ "⌘" I⌘" " " x2=0 x2=⌘" x1 x2 Figure 2: View of the domain Dεηεfrom above (left) and lateral (right) 2.2 Some notations Let us introduce some notations which will be useful in the following. For a vectorial function v= (v1, v2, v3) and a scalar function w, we introduce the operators: Dε,∇εand divεby (Dεv)i,j =∂xjvifor i= 1,2,3, j = 1,2, (Dεv)i,3=1 ε∂y3vifor i= 1,2,3, ∇εw= (∇x0w, 1 ε∂y3w)t, divεv= divx0v0+1 ε∂y3v3, and moreover the operators Dηε,∇ηεand divηεby (Dηεv)i,1=∂x1vifor i= 1,2,3, (Dηεv)i,2=1 ηε ∂y2vifor i= 1,2,3, (Dηεv)i,3=1 ε∂y3vifor i= 1,2,3, ∇ηεw= (∂x1w, 1 ηε ∂y2w, 1 ε∂y3w)t, divηεv=∂x1v1+1 ηε ∂y2v2+1 ε∂y3v3. 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. 5 3 Setting and main results Hereinafter, the points x∈R3will be decomposed as x= (x0, x3) with x0∈R2,x3∈R. We also use the notation x0to denote a generic vector of R2. In this section, we describe the asymptotic behavior of an incompressible viscous fluid in a thin porous medium with a fissure. The proof of the corresponding results will be given in the next sections. Our results are referred to the non-stationary Stokes system. Namely, for f∈C([0, T]×D)3let us consider a sequence (uε, pε)∈L2(0, T;H1 0(Dεηε))3×L2(0, T;L2(Dεηε)), which satisfies      ∂uε ∂t −µ∆uε+∇pε=fin (0, T)×Dεηε, div uε= 0 in (0, T)×Dεηε, uε(0, x)=0, x ∈Dεηε, (3.1) where T > 0, µ > 0 is the viscosity and Dεηεis defined in Section 2. The right-hand side fis of the form f(t, x) = (f0(t, x0),0),a.e. x∈D, (3.2) where f0∈C([0, T]×D)2.(3.3) This choice of fis usual when we deal with thin domains. Since the thickness of the domain εis small then the vertical component of the force can be neglected and, moreover the force can be considered independent of the vertical variable. Finally, we may consider Dirichlet boundary conditions without altering the generality of the problem under consideration, uε= 0 on (0, T)×∂Dεηε.(3.4) For any fixed ε, under the assumptions of fand u0 ε, a classical result (see Temam [10]) shows that (3.1)-(3.4) has at least one weak solution (uε, pε)∈L2(0, T;H1 0(Dεηε))3×L2(0, T;L2(Dεηε)), where pεis uniquely defined up to an additive constant, that is, it is uniquely defined if we consider the corresponding equivalence class: pε∈L2(0, T;L2(Dεηε)/R). Our aim is to study the asymptotic behavior of uεand pεwhen εtends to zero. For this purpose, we use the dilatation in the variable x3 y3=x3 ε,(3.5) in order to have the functions defined in an open set with fixed height e Dεηεgiven in Section 2. Namely, we define ˜uε∈L2(0, T;H1 0(e Dεηε))3, ˜pε∈L2(0, T;L2(e Dεηε)/R) by ˜uε(t, x0, y3) = uε(t, x0, εy3),˜pε(t, x0, y3) = pε(t, x0, εy3), a.e. (t, x0, y3)∈(0, T )×e Dεηε. Using the transformation (3.5), the system (3.1) can be rewritten as        ∂˜uε ∂t −µ∆ε˜uε+∇ε˜pε=fin (0, T)×e Dεηε, divε˜uε= 0 in (0, T)×e Dεηε, ˜uε(0, x0, y3)=0,(x0, y3)∈e Dεηε, (3.6) 6 with Dirichlet boundary conditions ˜uε= 0 on (0, T)×∂e Dεηε,(3.7) where we set ∆εw= ∆x0w+ε−2∂2 y3wand e Dεηεis defined in Section 2. Our goal then is to describe the asymptotic behavior of this new sequence (˜uε, ˜pε). Moreover, in order to study the behavior of ˜uε, ˜pεin the fissure we rewrite our equations in the unit cylinder e I1= Σ ×(0,1)e2by introducing the change of variable y2=x2 ηε ,(3.8) which transform e Iηεin a fixed domain e I1. We define the new functions ˜ Uε(t, x1, y2, y3) = ˜uε(t, x1, ηεy2, y3),˜ Pε(t, x1, y2, y3) = ˜pε(t, x1, ηεy2, y3)−cεηε,(3.9) with cεηε=1 |e Iηε|Ze Iηε ˜pε(t, x0, y3)dx0dy3.(3.10) Using the transformation (3.8), the system (3.6) can be rewritten as        ∂˜ Uε ∂t −µ∆ηε˜ Uε+∇ηε˜ Pε=f(t, x1, ηεy2) in (0, T)×e I1, divηε˜ Uε= 0 in (0, T)×e I1, ˜ Uε(0, x1, ηεy2, y3)=0,(x1, ηεy2, y3)∈e I1, (3.11) with Dirichlet boundary conditions ˜ Uε= 0 on (0, T)×∂e I1,(3.12) where we set ∆ηεw=∂2 x1w+η−2 ε∂2 y2w+ε−2∂2 y3w. Our main result referred to the asymptotic behavior of the solution of (3.6) is given by the following theorem. Theorem 3.1. We distingue three cases depending on the relation between the parameter ηεwith respect to ε: i) if ηεε2 3, then there exists (˜v, ˜p)∈L2((0, T)×D)3×L2(0, T ;L2(D)/R), with ˜v3= 0 and ˜p independent of y3, such that the solution (ε−2˜uε,˜pε)of problem (3.6)-(3.7) satisfies ε−2˜uε*˜vin L2((0, T )×D)3,˜pε→˜pin L2(0, T;L2(D)/R).(3.13) Moreover, ˜p∈L2(0, T;H1(D)/R)and (˜ V , ˜p)is the unique solution of the 2D quasi-stationary Darcy law (where tis only a parameter)        ˜ V0(t, x0) = 1 µKf0(t, x0)− ∇x0˜p(t, x0)in (0, T )×D0, divx0˜ V(t, x0)=0in (0, T)×D0, ˜ V(t, x0)·n= 0 in (0, T)×∂D0, (3.14) 7 where ˜ V(t, x0) = R1 0˜v(t, x0, y3)dy3and K∈R2×2is a symmetric, positive, tensor defined by its entries Kij =ZYf Dywi(y) : Dywj(y)dy, i, j = 1,2,(3.15) where wi(y),i= 1,2, with RYfwi 3dy = 0, denotes the unique solution in H1 #(Yf)3of the local stationary Stokes problems in 3D        −∆ywi+∇yqi=eiin Yf, divywi= 0 in Yf, wi= 0 in ∂(Y\Yf), wi, qiY0−periodic. (3.16) ii) if ηεε2 3and let (˜ Uε,˜ Pε)be a solution of (3.11)-(3.12). Then there exist ˜ U ∈ L2((0, T)×e I1)3, independent of y3, with ˜ U2=˜ U3= 0, and ˜ P∈L2(0, T;L2(e I1)/R)only depending on tand x1, such that for a subsequence, ηε−2˜ Uε*˜ Uin L2((0, T)×e I1)3,˜ Pε*˜ Pin L2(0, T;L2(e I1)/R), where ˜ U1(t, x1, y2) = y2(1 −y2) 2f1(t, x1,0) −∂x1˜ P(t, x1).(3.17) Moreover, it holds that ηε−3˜uε? *˜ VδΣ1in L2(0, T;M(D))3,(3.18) where ˜ V ∈ L2((0, T)×Σ1)3, with ˜ V2=˜ V3= 0, such that ˜ V1(t, x1) = Z1 0 ˜ U1(t, x1, y2)dy2=1 12 f1(t, x1,0) −∂x1˜ P(t, x1),(3.19) and, in fact ˜ P∈L2(0, T;H1(Σ1)/R)is the unique solution of the 1D quasi-stationary Reynolds problem on Σ1(where tis only a parameter)    ∂x1f1(t, x1,0) −∂x1˜ P(t, x1))= 0 in (0, T)×Σ1, f1(t, x1,0) −∂x1˜ P(t, x1)·n= 0 on (0, T)×∂Σ1.(3.20) iii) if ηε≈ε2 3, with ηε/ε2 3→λ,0<λ<+∞, then there exist a Darcy velocity ˜v, a Reynolds velocity ˜ Vand a pressure field ˜psuch that ε−2˜uε? *˜v+λ3˜ VδΣ1in L2(0, T;M(D))3, ˜pε→˜pin L2(0, T ;L2(D)/R),(3.21) where δΣ1is the Dirac measure concentrated on Σ1, and M(D)3is the space of Radon meaures on D. The velocities ˜vand ˜ Vare linked with the pressure ˜pthrough the 2D Darcy law (3.14) in (0, T)×D0and the 1D Reynolds problem (3.20) on (0, T)×Σ1. The pressure field ˜p∈ L2(0, T;H1(D0)/R)with ˜p(·,0) ∈L2(0, T;H1(Σ1)/R), is the unique solution of the variational problem ZT 0ZD0 1 µKf0(t, x0)− ∇x0˜p(t, x0)· ∇x0ϕ(t, x0)dx0dt +λ3 12 ZT 0ZΣ1 (f1(t, x1,0) −∂x1˜p(t, x1)) ∂x1ϕ(t, x1,0) dx1dt = 0, (3.22) for every ϕ∈L2(0, T;H1(D0)) with ϕ(·,0) ∈L2(0, T;H1(Σ1)). 8 Remark 3.2. The coupled problem (3.22) corresponding to the critical case ηε≈ε2 3, with ηε/ε2 3→λ, 0< λ < +∞, can be considered as the general one. In fact, if λtends to infinity in (3.22) we recover the 1D quasi-stationary Reynolds problem (3.20), meanwhile if λtends to zero we recover the 2D quasi-stationary Darcy law (3.14). 4 A Priori Estimates Let us begin with a lemma on Poincar´e inequality in the porous medium e Ωεηε, which will be very useful (see for example Lemma 4.1 in [8]). Lemma 4.1. There exists a constant Cindependent of ε, such that, for any function v∈H1(e Dεηε)3 and v= 0 on e Sεηε, one has kvkL2(e Ωεηε)3≤Cε kDεvkL2(e Ωεηε)3×3.(4.23) Next, we give an useful estimate in the fissure e Iηε. Lemma 4.2. There exists a constant Cindependent of ε, such that, for any function v∈H1(e Dεηε)3 and v= 0 on e Sεηε, one has kvkL2(e Iηε)3≤Cηε 1 2(ηε+ε)1 2kDεvkL2(e Dεηε)3×3.(4.24) Proof. For any function w(y)∈H1(e I1)3with w= 0 in ∂e I1, the Poincar´e inequality in e I1states that Ze I1 |w|2dz ≤CZe I1 |∂z2w|2dz, (4.25) where the constant Cdepends only on e I1. For every k0∈Z2, by the change of variable z1=x1, z2=x2 ηε , z3=x3 ε, dz =dx εηε , ∂z2=ηε∂x2,(4.26) we rescale (4.25) from e I1to Iηε. This yields that, for any function w(x)∈H1(Iηε)3with w= 0 in ∂Iηε, one has ZIηε |w|2dx ≤Cη2 εZIηε |∂x2w|2dx ≤Cη2 εZIηε |Dxw|2dx, (4.27) with the same constant Cas in (4.25). Finally, applying the dilatation (3.5) in (4.27), we obtain Z˜ Iηε |w|2dx0dy3≤Cη2 εZ˜ Iηε |Dεw|2dx0dy3, which gives kvkL2(e Iηε)3≤CηεkDεvkL2(e Iηε)3×3.(4.28) Next, if we choose a point y∈Aεηε, which is close to the point x∈Iηε, then we have v(x)−v(y) = Dv(ξ)(x−y)≤(ε+ηε)|Dv|. 9 On the other hand, we have ZT 0 |<∇ε˜pε(t), ϕ(t)(wε−w)>D|dt =ZT 0<∇x0˜pε(t), ϕ(t)Rε(w0 ε−w0)>e Dεηεdt =ZT 0hµ∆x0˜v0 ε(t), ϕ(t)Rε(w0 ε−w0)ie Dεηε+hf0(t), ϕ(t)Rε(w0 ε−w0)ie Dεηε−h∂˜v0 ε(t) ∂t , ϕ(t)Rε(w0 ε−w0)ie Dεηεdt, and using Cauchy-Schwarz’s inequality, estimate (4.32), the first estimate in (4.34), the estimates of the restricted operator Rεapplied to Dx0instead of Dε, and taking into account that ηεε2 3and ε1, we get ZT 0 |<∇ε˜pε(t), ϕ(t)(wε−w)>D|dt ≤C ZT 0 ϕ(t)2kw0 ε−w0k2 L2(D)2dt1/2 +εZT 0 ϕ(t)2kDx0w0 ε−Dx0w0k2 L2(D)2×2dt1/2!→0 as ε→0, by virtue of (5.45) and the Rellich Theorem. This implies that ∇ε˜pε→ ∇x0˜pstrongly in L2(0, T;H−1(D))3, which implies the strong convergence of the pressure given in (5.43). Lemma 5.2. Let ηεε2 3and let (˜vε,˜pε)be the extended solution of (3.6)-(3.7). Let (˜v, ˜p)∈L2((0, T )× D)3×L2(0, T;L2(D)/R)be given by Lemma 5.1. Then, ˜p∈L2(0, T ;H1(D)/R)and (˜v, ˜p)is the unique solution of Darcy’s law (3.14). Proof. We apply Theorem 3.1-(i) in [8], because in the present paper aε≈εin the porous part, in order to obtain that (˜v, ˜p) is the unique solution of Darcy’s law (3.14). Finally, the classical theory of the elliptic equation implies existence of the unique solution ˜pbelongs to L2(0, T;H1(D)/R). Proof of Theorem 3.1-i).It remains to prove convergence (3.13) of the whole velocity ˜uε, i.e. to prove ε−2k˜uεkL2((0,T )×e Iηε)3→0.(5.46) For this, it is sufficient to prove that ε−2k˜uεkL2((0,T )×e Iηε)3→0 for ηεε, (5.47) and ε−2k˜uεkLq((0,T )×e Iηε)3→0 for εηεε1 α,1< α < 3 2,(5.48) for a qwhich will be defined below. Using (4.31) and using ηεε, we have ε−2k˜uεkL2((0,T )×e Iηε)3≤C ηε 5 2 ε2+ηε ε+ηε ε1 2!, 16 so that (5.47) easily holds. Using H¨older’s inequality with the conjugate exponents 2 qand 2 2−qwe obtain ε−2k˜uεkLq((0,T )×e Iηε)3≤C ηε 1 q+2 ε2+ηε 1 q+1 2 ε+ηε 1 q ε1 2!. Now we take ηε=ε1 α. Then we find that ε−2k˜uεkLq((0,T )×e Iηε)3≤Cε1 α1 q+2−2+ε1 α1 q+1 2−1+ε1 qα −1 2.(5.49) We seek an optimal qsuch that the right hand side in (5.49) tends to zero. It is easy to prove that we have a convergence to zero for any q∈1,2 2(α−1)+1. Therefore, (5.48) holds and so we have (5.46). 5.2 Problem in the fissure part ηεε 2 3 The proof of Theorem 3.1-ii) will be developed in different lemmas. Lemma 5.3. Let ηεε2 3and let (˜ Uε,˜ Pε)be the solution of (3.11)-(3.12). Then there exist subsequences of ˜ Uεand ˜ Pεstill denoted by the same, and functions ˜ U ∈ L2((0, T)×e I1)3, independent of y3, with ˜ U2=˜ U3= 0,˜ P∈L2(0, T;L2(e I1)/R)such that ηε−2˜ Uε*˜ Uin L2((0, T)×e I1)3,˜ Pε*˜ Pin L2(0, T;L2(e I1)/R).(5.50) Moreover, ˜ P=˜ P(x1)and ˜ U1is given by expression (3.17). Proof. Taking into account ηεε2 3and estimates (4.31), (4.32), (4.33), (4.41) with the change of variable (3.8), we have k˜ UεkL2((0,T)×e I1)3≤Cηε2,(5.51) k∂x1˜ UεkL2((0,T)×e I1)3≤Cηε,k∂y2˜ UεkL2((0,T)×e I1)3≤Cηε2,(5.52) k∂y3˜ UεkL2((0,T)×e I1)3≤Cε ηε,(5.53) k˜ UεkL∞(0,T;L2(e I1))3≤Cηε,(5.54) k˜ PεkL2(0,T;L2(e I1)/R)≤C. (5.55) From the estimates (5.51) and (5.55), there exist ˜ U ∈ L2((0, T)×e I1)3,˜ P∈L2(0, T;L2(e I1)/R) such that convergence (5.50) holds. Moreover ηε−2∂y2˜ Uε* ∂y2˜ Uin L2((0, T)×e I1)3,(5.56) and from (5.54), there exists ˜ W ∈ L∞(0, T;L2(e I1))3such that ηε−1˜ Uε∗ *˜ Win L∞(0, T;L2(e I1))3.(5.57) The estimate (5.53) implies that ε−1η−1 ε∂y3˜ Uεis bounded in L2((0, T)×e I1)3. This together with ηεε2 3implies that η−2 ε∂y3˜ Uεtends to ∂y3˜ U= 0. This implies that ˜ Udoes not depend on y3. 17 As ˜ Udoes not depend on y3, let ϕ∈C∞ 0((0, T)×e I1)3independent of y3. Taking into account that divηε˜ Uε= 0 in (0, T )×e I1, we have ηε−1ZT 0Ze I1∂x1˜ Uε 1+ηε−1∂y2˜ Uε 2+ε−1∂y3˜ Uε 3ϕ dx1dy2dy3dt =−ηε−1ZT 0Ze I1 ˜ Uε 1∂x1ϕ dx1dy2dy3dt −ηε−2ZT 0Ze I1 ˜ Uε 2·∂y2ϕ dx1dy2dy3dt = 0. Taking the limit ε→0 we obtain ZT 0Ze I1 ˜ U2∂y2ϕ dx1dy2dy3dt = 0, so that ˜ U2=˜ U2(t, x1). Since ˜ U,∂y2˜ U ∈ L2((0, T)×e I1)3the traces ˜ U(t, x1,0), ˜ U(t, x1,1) are well defined in L2((0, T)×Σ)3. Analogously to the proof of Lemma 4.2 we choose a point β(x1,y3)∈e Aεηε, which is close to the point α(x1,y3)∈Σ, then we have ZT 0ZΣ |˜ Uε(t, x0,0, y3)|2dx1dy3dt =ZT 0ZΣ |˜uε(t, x1,0, y3)|2dx1dy3dt ≤CZT 0ZΣ Z(β(x1,y3),α(x1,y3)) Dε˜uε·(α(x1,y3)−β(x1,y3))d`!2 dx1dy3dt, so that, by Cauchy-Schwarz’s inequality, k˜ Uε(t, x1,0, y3)k2 L2((0,T)×Σ)3≤CεkDε˜uεk2 L2((0,T)×e Dεηε)3×3. Taking into account estimate (4.32) and ηεε2 3, we have ηε−2k˜ Uε(t, x1,0, y3)k2 L2((0,T)×Σ)3≤Cεηε→0 as ε→0, which implies that ˜ U(t, x1,0) = 0 , and analogously ˜ U(t, x1,1) = 0 . Consequently ˜ U2= 0 . It remains to prove that ˜ U3= 0. In order to do that, as ˜ Udoes not depend on y3, we take a test function v= (0,0, v3(x1, y2)) in (3.11), d dt Ze I1 ˜ Uε 3(t)v3dx1dy2dy3+Ze I1 ∂2 x1˜ Uε 3(t)v3dx1dy2dy3+1 η2 εZe I1 ∂2 y2˜ Uε 3(t)v3dx1dy2dy3= 0, in D0(0, T). We consider ϕ∈C1 c([0, T]) such that ϕ(T) = 0 and ϕ(0) 6= 0. Multiplying by ϕand integrating between 0 and T, we have −ZT 0 d dtϕ(t)Ze I1 ˜ Uε 3(t)v3dx1dy2dy3dt +ZT 0 ϕ(t)Ze I1 ∂2 x1˜ Uε 3(t)v3dx1dy2dy3dt +1 η2 εZT 0 ϕ(t)Ze I1 ∂2 y2˜ Uε 3(t)v3dx1dy2dy3dt = 0. 18 We pass to the limit when εtends to zero, and using the convergences (5.56) and (5.57) with v3ϕ(t)∈L2((0, T)×e I1), v3 d dtϕ(t)∈L1(0, T;L2(e I1)), we can deduce that ˜ U3= 0. Finally, we compute the expression of ˜ Ugiven in (3.17). First, we take a test function v= (0,0, εv3) in (3.11), and we obtain εd dt Ze I1 ˜ Uε 3(t)v3dx1dy2dy3+εZe I1 ∂2 x1˜ Uε 3(t)v3dx1dy2dy3+ε η2 εZe I1 ∂2 y2˜ Uε 3(t)v3dx1dy2dy3 +1 εZe I1 ∂2 y3˜ Uε 3(t)v3dx1dy2dy3−Ze I1 ˜ Pε∂y3v3dx1dy2dy3= 0, in D0(0, T). Multiplying by ϕand integrating between 0 and T, we have −εZT 0 d dtϕ(t)Ze I1 ˜ Uε 3(t)v3dx1dy2dy3dt +εZT 0 ϕ(t)Ze I1 ∂2 x1˜ Uε 3(t)v3dx1dy2dy3dt +ε η2 εZT 0 ϕ(t)Ze I1 ∂2 y2˜ Uε 3(t)v3dx1dy2dy3dt +1 εZT 0 ϕ(t)Ze I1 ∂2 y3˜ Uε 3(t)v3dx1dy2dy3dt −ZT 0 ϕ(t)Ze I1 ˜ Pε∂y3v3dx1dy2dy3dt = 0. We pass to the limit when εtends to zero, and using the estimate (5.53), the convergences (5.50) and (5.57) with v3ϕ(t)∈L2((0, T)×e I1), v3 d dtϕ(t)∈L1(0, T;L2(e I1)), we can deduce that ˜ Pdoes not depend on y3. We take a test function v= (0, ηεv2,0), independent of y3, in (3.11), and we obtain ηε d dt Ze I1 ˜ Uε 2(t)v2dx1dy2dy3+ηεZe I1 ∂2 x1˜ Uε 2(t)v2dx1dy2dy3+1 ηεZe I1 ∂2 y2˜ Uε 2(t)v2dx1dy2dy3 −Ze I1 ˜ Pε∂y2v2dx1dy2dy3=ηεZe I1 f2v2dx1dy2dy3, in D0(0, T). Multiplying by ϕand integrating between 0 and T, we have −ηεZT 0 d dtϕ(t)Ze I1 ˜ Uε 2(t)v2dx1dy2dy3dt +ηεZT 0 ϕ(t)Ze I1 ∂2 x1˜ Uε 2(t)v2dx1dy2dy3dt +1 ηεZT 0 ϕ(t)Ze I1 ∂2 y2˜ Uε 2(t)v2dx1dy2dy3dt −ZT 0 ϕ(t)Ze I1 ˜ Pε∂y2v2dx1dy2dy3dt =ηεZT 0 ϕ(t)Ze I1 f2v2dx1dy2dy3dt. We pass to the limit when εtends to zero, and using the convergences (5.50) and (5.57) with v2ϕ(t)∈L2((0, T)×e I1), v2 d dtϕ(t)∈L1(0, T;L2(e I1)), 19 we can deduce that ˜ P=˜ P(t, x1). Now, taking into account that ˜ Udoes not depend on y3and ˜ U2=˜ U3= 0, we take a test function v= (v1(x1, y2),0,0) in (3.11), d dt Ze I1 ˜ Uε 1(t)v1dx1dy2dy3+Ze I1 ∂2 x1˜ Uε 1(t)v1dx1dy2dy3+1 η2 εZe I1 ∂2 y2˜ Uε 1(t)v1dx1dy2dy3 −Ze I1 ˜ Pε∂x1v1dx1dy2dy3=Ze I1 f1(t, x1, ηεy2)v1dx1dy2dy3, in D0(0, T). Multiplying by ϕand integrating between 0 and T, we have −ZT 0 d dtϕ(t)Ze I1 ˜ Uε 1(t)v1dx1dy2dy3dt +ZT 0 ϕ(t)Ze I1 ∂2 x1˜ Uε 1(t)v1dx1dy2dy3dt +1 η2 εZT 0 ϕ(t)Ze I1 ∂2 y2˜ Uε 1(t)v1dx1dy2dy3dt −ZT 0 ϕ(t)Ze I1 ˜ Pε∂x1v1dx1dy2dy3dt =ZT 0 ϕ(t)Ze I1 f1(t, x1, ηεy2)v1dx1dy2dy3dt. We pass to the limit when εtends to zero, and using the convergences (5.50) and (5.57) with v1ϕ(t)∈L2((0, T)×e I1), v1 d dtϕ(t)∈L1(0, T;L2(e I1)), we obtain the ODE    −∂2 y2˜ U1(t, x1, y2) = f1(t, x1,0) −∂x1˜ P(t, x1), ˜ U1(t, x1,0) = ˜ U1(t, x1,1) = 0, which gives the expression (3.17) for ˜ U1. Proof of Theorem 3.1-ii).It remains to prove the convergence (3.18) of the whole velocity to the function Vgiven by (3.19), and also prove that ˜ P∈L2(0, T;H1(Σ)/R) is the unique solution of the Reynolds problem (3.20). Taking as test function ϕ∈C∞((0, T)×D), independent of y3, in the equation divε˜uε= 0 in (0, T)×D, we obtain ZT 0ZD divε˜uεϕ dx0dy3dt =−ZT 0ZD ˜v0 ε·∇x0ϕ dx0dy3dt−ηεZT 0Ze I1 (˜ Uε)0·∇x0ϕ(t, x1, ηεy2)dx1dy2dy3dt = 0, so that multiplying by ηε−3, ZT 0Ze I1 ηε−2˜ Uε 1∂x1ϕ(t, x1, ηεy2)dx1dy2dy3dt (5.58) =−ZT 0ZD ηε−3˜vε· ∇x0ϕ dx0dy3dt −ZT 0Ze I1 ηε−2˜ Uε 2∂x2ϕ(t, x1, ηεy2)dx1dy2dy3dt. 20 Using (4.30) and taking into account ηεε2 3, we obtain ηε−3k˜vεkL2((0,T )×D)3≤C ε ηε 3 2 +ε2 ηε3!→0 as ε→0.(5.59) Taking the limit in (5.58) as ε→0, using convergence (5.50), ˜ U2= 0 and ˜ U1independent of y3, we have ZT 0ZΣ ˜ U1∂x1ϕ(x1,0) dx1dy2dt = 0, and by definition (3.19), we get ZT 0ZΣ1f1(t, x1,0) −∂x1˜ P(t, x1)∂x1ϕ(t, x1,0) dx1dt = 0. Consequently, ˜ P∈L2(0, T;H1(Σ1)/R) and is the unique solution of (3.20). Finally, we consider ϕ∈C0((0, T)×D)3, independent of y3, and so we have ZT 0ZD ηε−3˜uε·ϕ dx0dy3dt =ZT 0ZD ηε−3˜vε·ϕ dx0dy3dt +ZT 0Ze I1 ηε−2˜ Uε·ϕ(t, x1, ηεy2)dx1dy2dy3dt. Using (5.59), convergence (5.50) and ˜ U2=˜ U3= 0, we obtain ZT 0ZD ηε−3˜uε·ϕ dx0dy3dt →ZT 0ZΣ ˜ U1(t, x1, y2)ϕ1(t, x1,0) dx1dy2dt =ZT 0ZΣ1 ˜ V1(t, x1)ϕ1(t, x1,0) dx1=ZT 0 h˜ V1(t, x1)δΣ1, ϕiM(D)3,C0(D)3dt, which implies (3.18). 5.3 Effects of coupling ηε≈ε 2 3 The conclusion of the previous two subsections is that for any sequence of solutions (˜vε,˜pε) with ηεε2 3and ( ˜ Uε,˜ Pε) with ηεε2 3, and letting ε→0, we can extract subsequences still denoted by ˜vε,˜pε,˜ Uε,˜ Pεand find functions ˜v∈L2(0, T ;H1(0,1; L2(ω)3)) with ˜v3= 0, ˜p∈L2(0, T;H1(D)/R), ˜ U ∈ L2((0, T)×e I1)3, independent of y3, with ˜ U2=˜ U3= 0, ˜ P∈L2(0, T;H1(Σ)/R) such that ε−2˜vε*(˜v0,0) in L2(0, T;H1(0,1; L2(ω)3)),˜pε→˜pin L2(0, T;L2(D)/R), ηε−2˜ Uε*(˜ U1,0,0) in L2((0, T)×e I1)3,˜ Pε*˜ Pin L2(0, T;L2(e I1)/R). (5.60) Moreover such limit functions ˜v, ˜p, ˜ U,˜ Pnecessarily satisfy the equations ˜ V0(t, x0) = 1 µKf0(t, x0)− ∇x0˜p(t, x0)in (0, T )×D0, ˜ U1(t, x1, y2) = y2(1 −y2) 2f1(t, x1,0) −∂x1˜ P(t, x1)in (0, T)×e I1, (5.61) where ˜ V0(t, x0) = R1 0˜v0(t, x0, y3)dy3. We are going to find the connection between the functions ˜pand ˜ P, i.e. to find the coupling effects between the solution in the porous part and in the fissure. 21 Lemma 5.4. Let ηε≈ε2 3, with ηε/ε2 3→λ,0< λ < +∞, and let ˜pε∈L2(0, T;L2(D)/R),˜p∈ L2(0, T;H1(D)/R),˜ P∈L2(0, T;H1(Σ)/R)be such that (5.60) and (5.61) hold. Then, ZT 0ZD0 1 µKf0(t, x0)− ∇x0˜p(t, x0)· ∇x0ϕ(t, x0)dx0dt +λ3 12 ZT 0ZΣ1f1(t, x1,0) −∂x1˜ P(t, x1)∂x1ϕ(t, x1,0) dx1dt = 0, (5.62) for every ϕ∈L2(0, T;H1(D0)) with ϕ(t, ·,0) ∈L2(0, T;H1(Σ1)). Proof. Let ϕε(t, x0, y3) = ϕ(t, x0, εy3)∈L2(0, T;H1(D)) with ϕ∈L2(0, T;H1(D)) and ϕ(t, ·,0) ∈ L2(0, T;H1(Σ)). Taking into account the definitions (5.42) of ˜vεand (3.9) of ˜ Uε, and from divε˜uε= 0 in (0, T)×Dwe have ZT 0ZD ε−2˜uε· ∇εϕεdx0dy3dt =ZT 0ZD ε−2˜vε· ∇εϕεdx0dy3dt +ηε ε2 33ZT 0Ze I1 ηε−2˜ Uε· ∇εϕε(t, x1, ηεy2, y3)dx1dy2dy3dt = 0, and by the definition of ϕε, we can deduce ZT 0ZD ε−2˜vε· ∇ϕ(t, x0, εy3)dx0dy3dt +ηε ε2 33ZT 0Ze I1 ηε−2˜ Uε· ∇ϕ(t, x1, ηεy2, εy3)dx1dy2dy3dt = 0. Taking the limit as ε→0, using (5.60), ˜v3=˜ U2=˜ U3= 0, ηε/ε2 3→λ, and taking into account that ˜ U1does not depend on y3, we obtain ZT 0ZD ˜v0(t, x0, y3)· ∇x0ϕ(t, x0,0) dx0dy3dt +λ3ZT 0ZΣ ˜ U1(t, x1, y2)∂x1ϕ(t, x1,0,0) dx1dy2dt = 0, and taking into account expressions (5.61) and (3.19), we get (5.62). We are going to prove the relation ˜p(t, x1,0) = ˜ P(t, x1) + C, with C∈R. Then (3.22) follows from (5.62). Lemma 5.5. Let ηε≈ε2 3,ηε/ε2 3→λ,0< λ < +∞, and let ˜p,˜ Pbe the limit pressures from (5.60). Then, there exists C∈Rsuch that ˜p(t, x1,0) = ˜ P(t, x1) + C, (5.63) and ˜p∈L2(0, T ;H1(D0)/R)with ˜p(t, ·,0) ∈L2(0, T;H1(Σ1)/R)is the unique solution of the variational problem (3.22). Proof. We need to extend the test functions considered in the proof of Lemma 5.2 to the fissure e Iηε. To do this, we define I0 ηε=e Iηε∩ {x3= 0},Bηε=D0 −∪Σ1∪I0 ηεand Y1=Yf∩ {x2= 0}, and we consider φ(y0)∈C∞ #(Bηε)3be such that φ(y0) = 0 in Y0\Y0 f. We define φε(x0) =      φx0 εin D0 −, K2e2in I0 ηε,where K2=ZY1 φ2(y1,0)dy1. 22 Let ϕ∈C∞ 0(B1), with B1=D−∪Σ∪e I1be such that ZΣ ϕ(x1,0, y3)dx1dy3= 0.(5.64) Taking in (3.6) as test function wε(x0, y3) =    ϕ(x0, y3)φx0 εin D−, ϕx1,x2 ηε, y3K2e2in e Iηε, we obtain d dt ZBηε ˜uε(t)·wεdx0dy3!+µZBηε Dε˜uε(t) : Dεwεdx0dy3=ZBηε f0(t)·w0 εdx0dy3+ZBηε ˜pε(t) divεwεdx0dy3. We consider ψ∈C1 c([0, T]) such that ψ(T) = 0 and ψ(0) 6= 0. Multiplying by ψand integrating between 0 and T, we have −ZT 0 d dtψ(t)ZBηε ˜uε(t)·wεdx0dy3dt +µZT 0 ψ(t)ZBηε Dε˜uε(t) : Dεwεdx0dy3dt (5.65) =ZT 0 ψ(t)ZBηε f0(t)·w0 εdx0dy3dt +ZT 0 ψ(t)ZBηε ˜pε(t) divεwεdx0dy3dt. Using (5.51), we have K2ZT 0 d dtψ(t)Ze Iηε ˜ Uε 2(t)·ϕx1,x2 ηε , y3dx0dy3dt =K2ηεZT 0 d dtψ(t)Ze Iηε ˜ Uε 2(t)·ϕ(x1, y2, y3)dx1dy2dy3dt≤Cη3 ε→0 as ε→0. We observe that K2ZT 0 ψ(t)Ze Iηε f0(t)·ϕ0x1,x2 ηε , y3e2dx0dy3dt =ηεK2ZT 0 ψ(t)Ze I1 f0(t)·ϕ0(x1, y2, y3)e2dx1dy2dy3dt →0 as ε→0, and by the definition of wεin e Iηεand using estimates (5.52), (5.53), we deduce K2ZT 0 ψ(t)Ze Iηε Dε˜ Uε(t)∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt =K2ZT 0 ψ(t)Ze I1 Dηε˜ Uε(t)∂y2ϕ(x1, y2, y3)dx1dy2dy3dt≤Cηε→0 as ε→0, 23 Then, from (5.65), we can deduce that −ZT 0 d dtψ(t)ZD− ˜uε(t)·wεdx0dy3dt +ZT 0 ψ(t)ZD− Dε˜vε(t) : Dεwεdx0dy3dt (5.66) =ZT 0 ψ(t)ZD− f0(t)·w0 εdx0dy3dt +ZT 0 ψ(t)ZD− ˜pε(t)divεwεdx0dy3dt +K2ZT 0 ψ(t)Ze Iηε ˜pε(t)∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt +Oε. For the last term on the right hand side, we have K2ZT 0 ψ(t)Ze Iηε ˜pε(t)∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt =K2ZT 0 ψ(t)Ze Iηε cεηε(t)∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt +K2ZT 0 ψ(t)Ze Iηε (˜pε(t)−cεηε(t))∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt, where cεηεis defined in (3.10). Using (5.60), we obtain K2ZT 0 ψ(t)Ze Iηε (˜pε(t)−cεηε(t))∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt =K2ZT 0 ψ(t)Ze I1 ˜ Pε(t)∂y2ϕ(x1, y2, y3)dx1dy2dy3dt →K2ZT 0 ψ(t)Ze I1 ˜ P(t, x1)∂y2ϕ(x1, y2, y3)dx1dy2dy3dt =−K2ZT 0 ψ(t)ZΣ ˜ P(t, x1)ϕ(x1,0, y3)dx1dy3dt, (5.67) as ε→0, where ˜ Pεis given by (3.9), and using (5.64), we have K2ZT 0 ψ(t)cεηε(t)Ze Iηε ∂x2ϕ(x1,x2 ηε , y3)dx0dy3dt =K2ZT 0 ψ(t)cεηε(t)Ze I1 ∂y2ϕ(x1, y2, y3)dx1dy2dy3dt = 0. Passing to the limit in (5.66) similarly as in the proof of Theorem 6.1-(i) in [8] by using an adaptation of the unfolding method, and taking into account (5.67) and ZT 0 ψ(t)ZD0 −×Y ˜p(t, x0) divx0(ϕ(x0, y3)φ(y0)) dx0dydt =−ZT 0 ψ(t)ZD0 −×Y ∇x0˜p(t, x0)ϕ(x0, y3)φ(y0)dx0dydt +ZT 0 ψ(t)ZΣ×Y1 ˜p(t, x1,0)ϕ(x1,0, y3)φ2(y1,0) dx1dy1dy3dt =−ZT 0 ψ(t)ZD0 −×Y ∇x0˜p(t, x0)ϕ(x0, y3)φ(y0)dx0dydt +K2ZT 0 ψ(t)ZΣ ˜p(t, x1,0)ϕ(x1,0, y3)dx1dy3dt, then we have ZT 0 ψ(t)ZΣ˜p(t, x1,0) −˜ P(t, x1)ϕ(x1,0, y3)dx1dy3dt = 0, 24 so that Z(0,T)×Σ1˜p(t, x1,0) −˜ P(t, x1)ϑ(t, x1)dx1dt = 0, for every ϑ∈C∞ 0((0, T)×Σ1) such that RΣϑ dx1= 0, a.e. t∈(0, T). Finally we conclude that there exists a constant C∈Rsuch that (5.63) holds and ˜p(t, x1,0) ∈L2(0, T ;H1(Σ1)/R). Using (5.63) into (5.62), we obtain the variational formulation (3.22) for the limit pressure ˜pin the Banach space of functions v∈L2(0, T;H1(D0)) such that v(t, x1,0) ∈L2(0, T ;H1(Σ1)). Since K∈R2×2is a symmetric, positive, tensor given by (3.15), it can be proved that (3.22) has a unique solution in that Banach space with the norm |v|L2(0,T ;H1(D0)) +|v(x1,0)|L2(0,T;H1(Σ1)). Proof of Theorem 3.1-iii).It remains to prove the convergence (3.21) of the whole velocity. Let ϕ∈C0((0, T)×D)3. Then ZT 0ZD ε−2˜uε·ϕ dx0dy3dt =ZT 0ZD ε−2˜vε·ϕ dx0dy3dt +ηε ε2 33ZT 0Ze I1 ηε−2˜ Uε·ϕ(t, x1, ηεy2, y3)dx1dy2dy3dt = 0. Taking the limit as ε→0, using (5.60), ˜v3=˜ U2=˜ U3= 0 and ηε/ε2 3→λ, we obtain ZT 0ZD ε−2˜uε·ϕ dx0dy3dt →ZT 0ZD ˜v0·ϕ0dx0dy3dt +λ3ZT 0Ze I1 ˜ U1(t, x1, y2)ϕ(t, x1,0, y3)dx1dy2dy3dt. Taking into account that ZT 0Ze I1 ˜ U(t, x1, y2)ϕ(t, x1,0, y3)dx1dy2dy3dt =ZT 0ZΣ1 V(t, x1)Z1 0 ϕ(t, x1,0, y3)dy3dx1dt =ZT 0 hVδΣ1, ϕiM(D)3,C0(D)3dt, where V(t, x1) is given by (3.19), we get (3.21). Acknowledgments The author would like to thank the referees for the detailed remarks which allowed to improve this paper. The author has been supported by Junta de Andaluc´ıa (Spain), Proyecto de Excelencia P12FQM-2466, and in part by European Commission, Excellent Science-European Research Council (ERC) H2020-EU.1.1.-639227. References [1] Ciarlet P-G, Ledret H, Nzwenga R. Mod´elisation de la jonction entre un corps ´elastique tridimensionnel et une plaque. C. R. Acad. Sci., Paris, S´erie I. 1987; 305: 55-58. 25