Int. J. Anal. Appl. (2023), 21:91 Transmission Problem Between Two Herschel-Bulkley Fluids in a Three Dimensional Thin Layer Salim Saf1,∗, Farid Messelmi2 1Laboratory of Pure and Applied Mathematics, Amar Telidji University of Laghouat, Laghouat 03000, Algeria 2Department of Mathematics and LDMM Laboratory Ziane Achour University, Djelfa 17000, Algeria ∗Corresponding author: s.saf@lagh-univ.dz Abstract. The paper is devoted to the study of steady-state transmission problem between two Herschel-Bulkley fluids in a three dimensional thin layer. 1. Introduction The rigid viscoplastic and incompressible fluid of Herschel-Bulkley has been studied and used by many mathematicians, physicists and engineers, to model the flow of metals, plastic solids and a variety of polymers. Due to the existence of the yield limit, the model can capture phenomena connected with the development of discontinuous stresses. A particularity of Herschel-Bulkley fluid lies in the presence of rigid zones located in the interior of the flow and as yield limit increases, the rigid zones become larger and may completely block the flow, this phenomenon is known as the blockage property. The literature concerning this topic is extensive; see e.g. [4, 11, 12, 14, 15]. The purpose of this paper is to study the asymptotic behavior of the steady flow of Herschel-Bulkley fluid in a three-dimensional thin layer. The paper is organized as follows. In section 2 we present the mechanical problem of the steady flow of Herschel-Bulkley fluid in a three-dimensional thin layer. We introduce some notations and preliminaries. Moreover, we define some function spaces and we recall the variational formulation. In Section 3, we are interested in the asymptotic behavior, to this aim we prove some convergence Received: Jun. 22, 2023. 2020 Mathematics Subject Classification. 76A05, 49J40, 76B15. Key words and phrases. Herschel-Bulkley fluid; transmission; asymptotic behaviour; thin layer. https://doi.org/10.28924/2291-8639-21-2023-91 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-91 2 Int. J. Anal. Appl. (2023), 21:91 results concerning the velocity and pressure when the thickness tends to zero. Besides, the uniqueness of a limit solution has been also established. 2. Problem statement Denoting by ω the fixed region in plan x = (x1,x2) ∈R2. Introducing the function h : ω → R such that 0 < h0 ≤ h(x,y) ≤ h1 for all (x,y) ∈R2, where h0 and h1 are constants. Considering the following domains Ω1 = { (x,y,z) ∈R3/ (x,y) ∈ ω and 0 < z < h(x,y) } , Ωε1 = { (x1,x2,x3) ∈R3/ (x1,x2) ∈ ω and 0 < x3 < εh(x1,x2) } , Ω2 = { (x,y,z) ∈R3/ (x,y) ∈ ω and −h(x,y) < z < 0 } Ωε2 = { (x1,x2; x3) ∈R3/ (x1,x2) ∈ ω and −εh(x1,x2) < x3 < 0 } , where ε > 0. Remark that if (x1,x2; x3) ∈ Ωεi then (x,y,z) = (x1,x2, x3 ε ) ∈ Ωi. This permits us to define, for every function ϕεi : Ω ε i →R, the function ϕ̂ ε i : Ωi →R given by ϕ̂εi (x,y,z) = ϕ ε i (x1,x2,x3), i = 1, 2. Let 1 < p ≤ 2, p′ the conjugate p, ( 1 p + 1 p′ = 1 ) and fi = (fi1, fi2, fi3) ∈ Lp ′ i (Ωi ) 3 a given functions. We define the functions f εi ∈ L p′(Ωεi ) 3 such that f̂ ε i = fi, i = 1, 2. We consider a mathematical problem modeling the steady flow of a rigid viscoplastic and incompressible Herschel-Bulkley fluid. We suppose that the consistency and yield limit of the fluid are respectively µiε p, giε where µi, gi > 0, i = 1, 2 and p represents the power-law index. The first fluid occupies a bounded domain Ωε1 ⊂ R 3 with the boundary ∂Ωε1 of class C 1. The second one occupies a bounded domain Ωε2 ⊂ R 3 with the boundary ∂Ωε2 of class C 1. We denote by Ωε the domain Ωε1 ∪ Ω ε 2 and we suppose that ∂Ω ε 1 = ω∪ Γ ε 1 and ∂Ωε2 = ω ∪ Γ ε 2 the velocity is known and equal to zero, where ω, Γ ε 1, Γ ε 2 are measurable domains and meas(Γε1), meas(Γ ε 2) > 0. The fluids are acted upon by given volume forces of densities f ε 1 , f ε 2 respectively. We denote by S3 the space of symmetric tensors on R3. We define the inner product and the Euclidean norm on R3 and S3, respectively, by u.υ = ulυl ∀u,υ ∈R3 and σ.τ = σlmτ lm ∀σ,τ ∈ S3. |u| = (u.u) 1 2 ∀u ∈R3 and |σ| = (σ.σ) 1 2 ∀σ ∈ S3. Here and below, the indices l and m run from 1 to 3 and the summation convention over repeated indices is used. We denote by σ̃ε i the deviator of σεi given by σεi = −p ε i I3 + σ̃ ε i , where pεi , i = 1, 2 represents the hydrostatic pressure and I3 denotes the identity matrix of size 2. We consider the rate of deformation operator defined for every vεi ∈ W 1, pi (Ωεi ) 3 by D(vεi ) = (Dlm(v ε i )), Dlm(v ε i ) = 1 2 ((υεi )l, m + (υ ε i )m, l), i = 1, 2. Int. J. Anal. Appl. (2023), 21:91 3 We denote by n the unit outward normal vector on the boundary ω oriented to the exterior of Ωε1 and to the interior of Ωε2, see the figure below. For every vector field v ε i ∈ W 1, pi (Ωεi ) 3 we also write vεi for its trace on ∂Ωεi , i = 1, 2. The steady-state transmission problem for the Herschel-Bulkley fluids in thin layer is given by the following mechanical problem. Problem Pε. Find the velocity field uεi = (u ε i1,u ε i2,u ε i3) : Ω ε i → R 3, the stress field σεi = (σεi1,σ ε i2,σ ε i3) : Ω ε i → S3 and the pressure p ε i : Ω ε i →R 2, i = 1, 2 such that div σε1 + f ε 1 = 0 in Ω ε 1. (2.1) div σε2 + f ε 2 = 0 in Ω ε 2. (2.2) σ̃ε1 = µ1ε p ∣∣D(uε1)∣∣p−2 D(uε1) + g1ε D(uε1 )|D(uε1 )| if ∣∣D(uε1)∣∣ 6= 0∣∣∣σ̃ε1∣∣∣ ≤ g1ε if ∣∣D(uε1)∣∣ = 0   in Ωε1, (2.3) σ̃ε2 = µ2ε p ∣∣D(uε2)∣∣p−2 D(uε2) + g2ε D(uε2 )|D(uε2 )| if ∣∣D(uε2)∣∣ 6= 0∣∣∣σ̃ε2∣∣∣ ≤ g2ε if ∣∣D(uε2)∣∣ = 0   in Ωε2, (2.4) div uε1 = 0 in Ω ε 1, (2.5) div uε2 = 0 in Ω ε 2 , (2.6) uε1 = 0 on Γ ε 1 , (2.7) uε2 = 0 on Γ ε 2, (2.8) uε1 −u ε 2 = 0 on ω, (2.9) σε1.n−σ ε 2.n = 0 on ω. (2.10) Here, the flow is given by the equations (2.1) and (2.2). Equations (2.3) and (2.4) represents, respectively, the constitutive laws of Herschel-Bulkley fluids where and are the consistencies and yield limits of the two fluids, respectively, 1 < p ≤ 2 are the power law index of the two fluids, respectively. Equations (2.5) and (2.6) represents the incompressibility condition. (2.7), (2.8) give the velocities on the boundaries Γε1 and Γ ε 2, respectively. Finally, on the boundary part ω, (2.9) and (2.10) represents the transmission condition for liquid-liquid interface. Let us define now the following Banach spaces W 1, p Γi (Ωεi ) = { vi ∈ W 1, p(Ωεi ) : vi = 0 on Γ ε i } , (2.11) W p, ε div (Ωεi ) = { vi ∈ W 1, p(Ωεi ) 3 : div(vi ) = 0 in Ω ε i } , (2.12) W p div (Ωi ) = { vi ∈ W 1, p(Ωi )3 : div(vi ) = 0 in Ωi } , (2.13) L p 0 (Ω ε i ) = { ϕεi ∈ L p(Ωεi ) : ∫ Ωε i ϕεi (x1,x2,x3)dx1dx2dx3 = 0 } , (2.14) 4 Int. J. Anal. Appl. (2023), 21:91 L p 0 (Ωi ) = { ϕi ∈ L p(Ωi ) : ∫ Ωi ϕi (x,y,z)dxdydz = 0 } , (2.15) Wz (Ωi ) = { ϕi ∈ L p(Ωi ) : ∂ϕi ∂z ∈ Lp(Ωi ) } , i = 1, 2. (2.16) Wz = Wz (Ω1) 2 ×Wz (Ω2)2 , (2.17) Wεdiv = { (v1,v2) ∈ W ,p, ε div (Ωε1) ×W p, ε div (Ωε2) : v1 = v2 on ω, v1 = 0 on Γε1 , v2 = 0 on Γ ε 2 } , (2.18) Wε = { (v1,v2) ∈ W 1, p Γε1 (Ωε1) 3 ×W 1, p Γε2 (Ωε2) 3 : v1 = v2 on ω } . (2.19) For the rest of this article, we will denote by c possibly different positive constants depending only on the data of the problem. The use of Green’s formula permits us to derive the following variational formulation of the me- chanical problem (Pε), see [4,13,15]. Problem PVε . For prescribed data (f ε1 , f ε 2 ) ∈ L p′(Ωε1) 3 × Lp ′ (Ωε2) 3. Find (uε1,u ε 2) ∈ W ε div and (pε1,p ε 2) ∈ L p′ 0 (Ω ε 1) ×L p′ 0 (Ω ε 2) satisfying the variational inequality µ1ε p ∫ Ωε1 |D(uε1)| p−2 D(uε1).D(v1 −u ε 1)dx1dx2dx3 + g1ε ∫ Ωε1 |D(v1)|dx1dx2dx3 −g1ε ∫ Ωε1 |D(uε1)|dx1dx2dx3 + µ2ε p ∫ Ωε2 |D(uε2)| p−2 D(uε2).D(v2 −u ε 2)dx1dx2dx3 +g2ε ∫ Ωε2 |D(v2)|dx1dx2dx3 −g2ε ∫ Ωε2 |D(uε2)|dx1dx2dx3 ≥ ∫ Ωε1 f ε1 .(v1 −u ε 1)dx1dx2dx3 + ∫ Ωε1 pε1 div(v1 −u ε 1)dx1dx2dx3 + ∫ Ωε2 f ε2 .(v2 −u ε 2)dx1dx2dx3 + ∫ Ωε2 pε2 div(v2 −u ε 2)dx1dx2dx3, ∀(v1,v2) ∈ W ε. (2.20) It is known that this variational problem has a unique solution (uε1,u ε 2) ∈ W ε div and (p ε 1,p ε 2) ∈ L p′ 0 (Ω ε 1)× L p′ 0 (Ω ε 2), see for more details [11,13,15]. 3. Asymptotic behavior In this section, we establish some results concerning the asymptotic behavior of the solution when ε tends to zero. We begin by recalling the following lemmas see [1,3,4,8,16]. Lemma 3.1. (1) Poincaré’s inequality. For every vi ∈ W 1,p Γi (Ωεi ) 3 we have ‖vεi ‖Lp(Ωε i )3 ≤ ε ∥∥∥∥∂vεi∂x2 ∥∥∥∥ Lp(Ωε i )3 , i = 1, 2. (3.1) (2) Korn’s inequality. For every vi ∈ W 1,p Γi (Ωεi ) 3 there exists a positive constant C0 independent on ε, such that ‖∇vεi ‖Lp(Ωε i )9 ≤ C0 ‖D(v ε i ) ‖Lp(Ωε i )9 , i = 1, 2. (3.2) Int. J. Anal. Appl. (2023), 21:91 5 Lemma 3.2 (Minty). Let E be a banach spaces, A : E → E′ a monotone and hemi-continuous operator, J : E → ]−∞, +∞] a proper and convex functional. Let u ∈ E and f ∈ E′ . the following assertions are equivalent: 1.〈Au; v −u〉E′×E + J(v) −J(u) ≥ 〈f ; v −u〉E′×E, ∀v ∈ E 2.〈Av; v −u〉E′×E + J(v) −J(u) ≥ 〈f ; v −u〉E′×E. ∀v ∈ E The main results of this section are stated by the following proposition. Proposition 3.1. Let (uε1,u ε 2) ∈ W ε div and (p ε 1,p ε 2) ∈ L p′ 0 (Ω ε 1) ×L p′ 0 (Ω ε 2) be the solution of variational problem (PVε). Then, there exists (û1 , û2) ∈Wz (Ω1)3×Wz (Ω2)3 and (p̂1 , p̂2) ∈ L p′ 0 (Ω1)×L p′ 0 (Ω2) such that (ûε1, û ε 2) → (û1 , û2) in Wz (Ω1) 3 ×Wz (Ω2)3 weakly, (3.3)( ∂ûε13 ∂z , ∂ûε23 ∂z ) → (0, 0) in Lp(Ω1) ×Lp(Ω2) weakly, (3.4) (p̂ε1, p̂ ε 2) → (p̂1 , p̂2) in L p′ 0 (Ω1) ×L p′ 0 (Ω2) weakly. (3.5) Proof. choosing (v1,v2) = (0, 0) as test function in inequality (2.20), we deduce that µ1ε p ‖D(uε1)‖ p Lp(Ωε1) 9 + µ2ε p ‖D(uε2)‖ p Lp(Ωε2) 9 ≤ ∫ Ωε1 f ε1 .u ε 1dx1dx2dx3 + ∫ Ωε2 f ε2 .u ε 2dx1dx2dx3, this permits us to obtain, making use of Poincaré’s and Korn’s inequalities and by passage to variables x , y and z ∥∥∥ûε1∥∥∥ Lp(Ω1)3 + ∥∥∥ûε2∥∥∥ Lp(Ω2)3 ≤ c, (3.6)∥∥∥∥∥∂û ε 1 ∂z ∥∥∥∥∥ Lp(Ω1)3 + ∥∥∥∥∥∂û ε 2 ∂z ∥∥∥∥∥ Lp(Ω2)3 ≤ c, (3.7) ∥∥∥∥∥∂û ε 1 ∂x ∥∥∥∥∥ Lp(Ω1)3 + ∥∥∥∥∥∂û ε 2 ∂x ∥∥∥∥∥ Lp(Ω2)3 ≤ c ε , (3.8) ∥∥∥∥∥∂û ε 1 ∂y ∥∥∥∥∥ Lp(Ω1)3 + ∥∥∥∥∥∂û ε 2 ∂y ∥∥∥∥∥ Lp(Ω2)3 ≤ c ε . (3.9) Moreover, we get using the incompressibility condition (2.5), (2.6) and Green’s formula, for any function (ϕε1,ϕ ε 2) ∈ W 1,p1 Γ1 (Ωε1) ×W 1,p2 Γ2 (Ωε2)∫ Ω1 ∂ûε13 ∂z ϕ̂ε1dxdydz + ∫ Ω2 ∂ûε23 ∂z ϕ̂ε2dxdydz = ε ∫ Ω1 ( ûε11 ∂ϕ̂ε1 ∂x + ûε12 ∂ϕ̂ε1 ∂y ) dxdydz + ε ∫ Ω2 ( ûε21 ∂ϕ̂ε2 ∂x + ûε22 ∂ϕ̂ε2 ∂y ) dxdydz. 6 Int. J. Anal. Appl. (2023), 21:91 Which gives, making use (2.16)∥∥∥∥∥∂û ε 13 ∂z ∥∥∥∥∥ W−1,p ′ (Ω1) + ∥∥∥∥∥∂û ε 23 ∂z ∥∥∥∥∥ W−1,p ′ (Ω2) ≤ cε. (3.10) We can then extract a subsequences still denoted by (ûε1, û ε 2) such that (ûε1, û ε 2) → (û1, û2) in L p(Ω1) 3 ×Lp(Ω2)3 weakly, (3.11)( ∂ûε1 ∂z , ∂ûε2 ∂z ) → ( ∂û1 ∂z , ∂û1 ∂z ) in Lp(Ω1) 3 ×Lp(Ω2)3 weakly, (3.12)( ∂ûε13 ∂z , ∂ûε23 ∂z ) → (0, 0) in Lp(Ω1) ×Lp(Ω2) weakly. (3.13) Let now (vε1,v ε 2 ) ∈ W 1,p Γ1 (Ωε1) 3 ×W 1,p Γ2 (Ωε2) 3 , we obtain by setting (uε1 −v ε 1,u ε 2 −v ε 2 ) as test function in inequality (2.20), using the incompressibility conditions (2.5) and (2.6) as well as the Green formula and Holder’s inequality ∫ Ωε1 ∇pε1.v ε 1dx1dx2dx3 + ∫ Ωε2 ∇pε2.v ε 2dx1dx2dx3 ≤ µ1ε p (∫ Ωε1 |D(uε1)| p dx1dx2dx3 ) 1 p′ (∫ Ωε1 |D(vε1 )| p dx1dx2dx3 )1 p +g1ε 1 p′ +1Meas(Ωε1) 1 p′ (∫ Ωε1 |D(vε1 )| p dx1dx2dx3 )1 p +ε ∥∥∥f̂ ε1 ∥∥∥ Lp ′ (Ωε1) 3 ∥∥∥v̂ε1∥∥∥ W 1, p Γ1 (Ω1)3 + ε ∥∥∥f̂ ε2 ∥∥∥ Lp ′ (Ωε2) 3 ∥∥∥v̂ε2∥∥∥ W 1,p Γ2 (Ω2)3 +µ2ε p (∫ Ωε2 |D(uε2)| p dx1dx2dx3 ) 1 p′ (∫ Ωε2 |D(vε2 )| p dx1dx2dx3 )1 p +g2ε 1 p′ +1Meas(Ωε2) 1 p′ (∫ Ωε2 |D(vε2 )| p dx1dx2dx3 )1 p . (3.14) On the other hand, it is easy to check that, after some algebraic manipulations, we find (∫ Ωε i |D(vεi )| p dx1dx2dx3 )1 p ≤ ε 1 p −1 ∥∥∥v̂εi ∥∥∥ W 1, p Γi (Ωi ) 3 , i = 1, 2. (3.15) Hence, from (3.7), (3.8), (3.9), (3.14) and (3.15) if follows that∫ Ωε1 ∇pε1.v ε 1dx1dx2dx3 + ∫ Ωε2 ∇pε2.v ε 2dx1dx2dx3 ≤ cε (∥∥∥v̂ε1∥∥∥ W 1, p Γ1 (Ω1)3 + ∥∥∥v̂ε2∥∥∥ W 1, p Γ2 (Ω2)3 ) . (3.16) Int. J. Anal. Appl. (2023), 21:91 7 Passing to the variables x, y and z in the left hand side of (3.16 ) we find the following estimates∥∥∥p̂ε1∥∥∥ L p′ 0 (Ω1) + ∥∥∥p̂ε2∥∥∥ L p′ 0 (Ω2) ≤ c, (3.17)∥∥∥∥∥∂p̂ ε 1 ∂x ∥∥∥∥∥ W−1, p ′ (Ω1) + ∥∥∥∥∥∂p̂ ε 2 ∂x ∥∥∥∥∥ W−1, p ′ (Ω2) ≤ c, (3.18) ∥∥∥∥∥∂p̂ ε 1 ∂y ∥∥∥∥∥ W−1, p ′ (Ω1) + ∥∥∥∥∥∂p̂ ε 2 ∂y ∥∥∥∥∥ W−1, p ′ (Ω2) ≤ c, (3.19) ∥∥∥∥∥∂p̂ ε 1 ∂z ∥∥∥∥∥ W−1, p ′ (Ω1) + ∥∥∥∥∥∂p̂ ε 2 ∂z ∥∥∥∥∥ W−1, p ′ (Ω2) ≤ εc. (3.20) Consequently, we can extract a subsequence still denoted by ( p̂ε1, p̂ ε 2 ) such that( p̂ε1, p̂ ε 2 ) → (p̂1, p̂2) in L p′ 0 (Ω1) ×L p′ 0 (Ω2) weakly, (3.21) which achieves the proof. This proof permits also to deduce that limit pressure verify( p̂ε1 (x,y,z) , p̂ ε 2(x,y,z) ) = (p̂1(x,y), p̂2(x,y)). � Proposition 3.2. The velocity limit given by (3.3) verifies∫ h(x,y) 0 ( ∂û11(x,y,z) ∂x + ∂û12(x,y,z) ∂y ) dz + ∫ 0 −h(x,y) ( ∂û21(x,y,z) ∂x + ∂û22(x,y,z) ∂y ) dz = 0 ∀(x,y) ∈ ω (3.22) Proof. We know from incompressibility conditions (2.5) and (2.6) that∫ Ωε1 div(uε1(x1,x2,x3))ϕ1(x1,x2)dx1dx2dx3 + ∫ Ωε2 div(uε2(x1,x2,x3))ϕ2(x1,x2)dx1dx2dx3 = 0 for all (ϕ1,ϕ2) ∈ D(ω) 2. This implies, using Green’s formula∫ Ωε1 uε11 dϕ1 dx1 (x1,x2)dx1dx2dx3 + ∫ Ωε1 uε12 dϕ1 dx2 (x1,x2)dx1dx2dx3 + ∫ Ωε2 uε21 dϕ2 dx1 (x1,x2)dx1dx2dx3 + ∫ Ωε2 uε22 dϕ2 dx2 (x1,x2)dx1dx2dx3 = ∫ Ωε1 ∂uε13 ∂x3 ϕ1(x1,x2)dx1dx2dx3 + ∫ Ωε2 ∂uε33 ∂x3 ϕ2(x1,x2)dx1dx2dx3. Hence, by passage to the variables x, y and z using Geen’s formula, we can infer∫ ω ϕ(x,y) (∫ h(x,y) 0 ( ∂ûε11 ∂x + ∂ûε12 ∂y ) dz + ∫ 0 −h(x,y) ( ∂ûε21 ∂x + ∂ûε22 ∂y ) dz ) dxdy = 0 ϕ ∈ D(ω). 8 Int. J. Anal. Appl. (2023), 21:91 Then, ∫ h(x,y) 0 ( ∂ûε11 ∂x + ∂ûε12 ∂y ) dz + ∫ 0 −h(x,y) ( ∂ûε21 ∂x + ∂ûε22 ∂y ) dz = 0. Moreover, the fact that û1 = (û ε 11, û ε 12) ∈ L p (Ω1) 2 , û2 = (û ε 21, û ε 22) ∈ L p (Ω2) 2 and∫h(x,y) 0 û1(x,y,z)dz + ∫ 0 −h(x,y) û2(x,y,z)dz is continuous and linear, it is weakly continuous. Thus, by passage to the limit when ε tends to zero, taking into account the boundaries conditions (2.7) ,(2.8) and (2.9) it follows that∫ h(x,y) 0 ( ∂û11 ∂x + ∂û12 ∂y ) dz + ∫ 0 −h(x,y) ( ∂û21 ∂x + ∂û22 ∂y ) dz = 0, ∀(x,y) ∈ ω. � We derive in the proposition below the strong equation verified by the limit solution (û1, û2) = ((û11, û12), (û21, û22)) ∈ Wz, and (p̂1, p̂2) ∈ L p′ 0 (Ω1) ×L p′ 0 (Ω2). Proposition 3.3. if ( ∂û1 ∂z , ∂û2 ∂z ) 6= (0, 0) then the limit point (û1, û2) and (p̂1, p̂2) given by (3.3) and (3.5) verify the limit problem − ∂ ∂z  µ1 2 p 2 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z + √ 2 2 g1 ∂û1 ∂z∣∣∣∂û1∂z ∣∣∣ + µ2 2 p 2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z + √ 2 2 g2 ∂û2 ∂z∣∣∣∂û2∂z ∣∣∣   = f̂ 1 −∇p1(x,y) + f̂ 2 −∇p2(x,y) in W −1, p′(Ω)2. (3.23) Proof. Introducing the operator Φ defined as follows Φ : W ε → W ε′, 〈Φ(uε1,u ε 2), (v ε 1,v ε 2 )〉W ε′×W ε = µ1ε p ∫ Ωε1 |D(uε1)| p−2 D(uε1).D(v ε 1 )dx1dx2dx3 +µ2ε p ∫ Ωε2 |D(uε2)| p−2 D(uε2).D(v ε 2 )dx1dx2dx3. It is easy to verify that Φ is monotone and hemi-continuous (see for more details the reference [2,5,15] ). Moreover, we know that the functional (vε1,v ε 2 ) ∈ W ε → g1ε ∫ Ωε1 |D(vε1 )|dx1dx2dx3 + g2ε ∫ Ωε2 |D(vε2 )|dx1dx2dx3, is proper and convex. Then, the use of Minty’s lemma permits us to affirm that (2.20) is equivalent to the following inequality Int. J. Anal. Appl. (2023), 21:91 9 µ1ε p ∫ Ωε1 |D(vε1 )| p−2 D(vε1 ).D(v ε 1 −u ε 1)dx1dx2dx3 + g1ε ∫ Ωε1 |D(vε1 )|dx1dx2dx3 −g1ε ∫ Ωε1 |D(uε1)|dx1dx2dx3 + µ2ε p ∫ Ωε2 |D(vε2 )| p−2 D(vε2 ).D(v ε 2 −u ε 2)dx1dx2dx3 +g2ε ∫ Ωε2 |D(vε2 )|dx1dx2dx3 −g2ε ∫ Ωε1 |D(uε2)|dx1dx2dx3 ≥ ∫ Ωε1 f ε1 .(v ε 1 −u ε 1)dx1dx2dx3 + ∫ Ωε1 pε1 div(v1 −u ε 1)dx1dx2dx3 + ∫ Ωε2 f ε2 .(v ε 2 −u ε 2)dx1dx2dx3 + ∫ Ωε2 pε2 div(v2 −u ε 2)dx1dx2dx3, ∀(v ε 1,v ε 2 ) ∈ W ε. Our goal now is to pass to the limit when ε tends to zero.To this aim, we use Proposition (3.3) and the weak lower semi-continuity of the convex and continuous functional (vε1,v ε 2 ) ∈ W ε → g1ε ∫ Ωε1 |D(vε1 )|dx1dx2dx3 + g2ε ∫ Ωε2 |D(vε2 )|dx1dx2dx3, We find the following limit inequality µ1 ∫ Ω1   12 ∣∣∣∂v̂11∂z ∣∣∣2 + 12 ∣∣∣∂v̂12∂z ∣∣∣2 + ∣∣∣∂v̂13∂z ∣∣∣2   p−2 2 × [ 1 2 ∂v̂11 ∂z ∂(v̂11−û11) ∂z + 1 2 ∂v̂12 ∂z ∂(v̂12−û12) ∂z +∂v̂13 ∂z ∂(v̂13−û13) ∂z ] dxdydz + + g1 ∫ Ω1   12 ∣∣∣∂v̂11∂z ∣∣∣2 + 12 ∣∣∣∂v̂12∂z ∣∣∣2 + ∣∣∣∂v̂13∂z ∣∣∣2   1 2 dxdydz −g1 ∫ Ω1   12 ∣∣∣∂û11∂z ∣∣∣2 + 12 ∣∣∣∂û12∂z ∣∣∣2 + ∣∣∣∂û13∂z ∣∣∣2   1 2 dxdydz +µ2 ∫ Ω2   12 ∣∣∣∂v̂21∂z ∣∣∣2 + 12 ∣∣∣∂v̂22∂z ∣∣∣2 + ∣∣∣∂v̂23∂z ∣∣∣2   p−2 2 × [ 1 2 ∂v̂21 ∂z ∂(v̂21−û21) ∂z + 1 2 ∂v̂22 ∂z ∂(v̂22−û22) ∂z +∂v̂23 ∂z ∂(v̂23−û23) ∂z ] dxdydz +g2 ∫ Ω2   12 ∣∣∣∂v̂21∂z ∣∣∣2 + 12 ∣∣∣∂v̂22∂z ∣∣∣2 + ∣∣∣∂v̂23∂z ∣∣∣2   1 2 dxdydz −g2 ∫ Ω2   12 ∣∣∣∂û21∂z ∣∣∣2 + 12 ∣∣∣∂û22∂z ∣∣∣2 + ∣∣∣∂û23∂z ∣∣∣2   1 2 dxdydz ≥ ∫ Ω1 f̂1.(v̂1 − û1)dxdydz + ∫ Ω1 p̂1 div(v̂1 − û1)dxdydz + ∫ Ω2 f̂2.(v̂2 − û2)dxdydz + + ∫ Ω2 p̂2 div(v̂1 − û1)dxdydz. (3.24) Furthermore, from (3.3) and (3.4) we find( ∂û13 ∂z , ∂û23 ∂z ) = (0, 0) in Ω1 × Ω2. It follows, keeping in mind (3.22), that û1(x,y,z) = (û11(x,y,z), û12(x,y,z), 0) , and û2(x,y,z) = (û21(x,y,z), û22(x,y,z), 0) . This permits also to choose (v̂13, v̂23) = (0, 0) in ( 3.24). 10 Int. J. Anal. Appl. (2023), 21:91 Considering now the operator Φ such that Φ : Wz → W ′z , 〈Φ (û1, û2) , (v̂1, v̂2)〉W ′z ×Wz = µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂v̂1∂z dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂v̂2∂z dxdydz. It is clear that the operator Φ is monotone and hemi-continuous and the functional (v̂1, v̂2) ∈ Wz → √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂v̂1∂z ∣∣∣∣dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂v̂2∂z ∣∣∣∣dxdydz, is proper and convex. Hence, we deduce using again Minty’s lemma µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂(v̂1 − û1)∂z dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂v̂1∂z ∣∣∣∣dxdydz − √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂(v̂2 − û2)∂z dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂v̂2∂z ∣∣∣∣dxdydz − √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣dxdydz ≥ ∫ Ω1 ( f̂ 1 −∇p̂1 ) .(v̂1 − û1)dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .(v̂2 − û2)dxdydz ∀(v̂1, v̂2) ∈ Wz . (3.25) Setting (v̂1, v̂2) = (2û1, 2û2) and (v̂1, v̂2) = (0, 0) in (3.25), to obtain the following inequalities µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣dxdydz ≥ ∫ Ω1 ( f̂ 1 −∇p̂1 ) .û1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .û2dxdydz, and − µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p dxdydz − √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣dxdydz − µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p dxdydz − √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣dxdydz ≥ − ∫ Ω1 ( f̂ 1 −∇p̂1 ) .û1dxdydz − ∫ Ω2 ( f̂ 2 −∇p̂2 ) .û2dxdydz. Int. J. Anal. Appl. (2023), 21:91 11 Consequently, we get by combining these two inequalities µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣dxdydz = ∫ Ω1 ( f̂ 1 −∇p̂1 ) .û1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .û2dxdydz. (3.26) Due to the fact that W 1, p Γi (Ωi ) is dense in Wz (Ωi ) , see [1, 6], we can take w1 = v̂1 − û1 and w2 = v̂2 − û2 in (3.25) we obtain µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂w1∂z dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂w1∂z ∣∣∣∣dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂w2∂z dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂w2∂z ∣∣∣∣dxdydz ≥ ∫ Ω1 ( f̂ 1 −∇p̂1 ) .w1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz ∀(w1,w2) ∈ W 1, p Γ1 (Ω1) 2 ×W 1, p Γ2 (Ω2) 2 . (3.27) Changing (w1,w2) to (−w1,−w2) in (3.27), we obtain |F (w1,w2)| ≤ √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂w1∂z ∣∣∣∣dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂w2∂z ∣∣∣∣dxdydz , where F (w1,w2) = µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂w1∂z dxdydz + µ22 p2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂w2∂z dxdydz − ∫ Ω1 ( f̂ 1 −∇p̂1 ) .w1dxdydz − ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz. (3.28) Now, utilising the Hahn-Banach theorem, ∃(m1,m2) ∈ L∞ (Ω1) 2 × L∞ (Ω2)2 , with ‖|m1|‖∞ ,‖|m2|‖∞ ≤ 1, such that F ((w1,w2)) = − √ 2 2 g1 ∫ Ω1 m1. ∂w1 ∂z dxdydz − √ 2 2 g2 ∫ Ω2 m2. ∂w2 ∂z dxdydz. (3.29) In particular, from (3.26) and (3.28), we get ∫ Ω1 m1. ∂u1 ∂z dxdydz + ∫ Ω2 m2. ∂u2 ∂z dxdydz = ∫ Ω1 ∣∣∣∣∂u1∂z ∣∣∣∣dxdydz + ∫ Ω2 ∣∣∣∣∂u2∂z ∣∣∣∣dxdydz. (3.30) 12 Int. J. Anal. Appl. (2023), 21:91 Rewriting (3.29) as µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂w1∂z dxdydz + µ22 p2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂w2∂z dxdydz + √ 2 2 g1 ∫ Ω1 m1. ∂w1 ∂z dxdydz + √ 2 2 g2 ∫ Ω2 m2. ∂w2 ∂z dxdydz − ∫ Ω1 ( f̂ 1 −∇p̂2 ) .w1dxdydz − ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz = 0. (3.31) Next using (3.30), we have∫ ∣∣∣∂û1 ∂z ∣∣∣6=0 (∣∣∣∣∂û1∂z ∣∣∣∣−m1.∂û1∂z ) dxdydz + ∫ ∣∣∣∂û2 ∂z ∣∣∣6=0 (∣∣∣∣∂û2∂z ∣∣∣∣−m2.∂û2∂z ) dxdydz = 0. As ‖|m1|‖∞ ,‖|m2|‖∞ ≤ 1, we deduce∣∣∣∣∂û1∂z ∣∣∣∣ = m1.∂û1∂z and ∣∣∣∣∂û2∂z ∣∣∣∣ = m2.∂û2∂z . Hence, if (∣∣∣∂û1∂z ∣∣∣ ,∣∣∣∂û2∂z ∣∣∣) 6= (0, 0) , we get ∫ Ω1  µ1 2 p 2 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z + √ 2 2 g1 ∂û1 ∂z∣∣∣∂û1∂z ∣∣∣   .∂w1 ∂z dxdydz ∫ Ω2  µ2 2 p 2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z + √ 2 2 g2 ∂û2 ∂z∣∣∣∂û2∂z ∣∣∣   .∂w2 ∂z dxdydz = ∫ Ω1 ( f̂ 1 −∇p̂1 ) .w1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz, ∀(w1,w2) ∈ W 1, p Γ1 (Ω1) 2 ×W 1, p Γ2 (Ω2) 2 . Consequenty, we get by using a simple integration by parts − ∫ Ω1 ∂ ∂z  µ1 2 p 2 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z dxdydz + √ 2 2 g1 ∂û1 ∂z∣∣∣∂û1∂z ∣∣∣   .w1dxdydz − ∫ Ω2 ∂ ∂z  µ2 2 p 2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z dxdydz + √ 2 2 g2 ∂û2 ∂z∣∣∣∂û2∂z ∣∣∣   .w2dxdydz = ∫ Ω1 ( f̂ 1 −∇p̂1 ) .w1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz, ∀(w1,w2) ∈ W 1, p Γ1 (Ω1) 2 ×W 1, p Γ2 (Ω2) 2 . Let us consider w ∈ W 1,p0 (Ω) 2 : w = { w1 in Ω1 w2 in Ω2 , Int. J. Anal. Appl. (2023), 21:91 13 and ã1 =   − ∂ ∂z ( µ1 2 p 2 ∣∣∣∂û1∂z ∣∣∣p−2 ∂û1∂z + √22 g1 ∂û1∂z∣∣∣∂û1 ∂z ∣∣∣ ) in Ω1 0 in Ω2 , ã2 =   0 in Ω1 − ∂ ∂z ( µ2 2 p 2 ∣∣∣∂û2∂z ∣∣∣p−2 ∂û2∂z + √22 g2 ∂û2∂z∣∣∣∂û2 ∂z ∣∣∣ ) in Ω2 , b̃1 = { f̂ 1 −∇p̂1 in Ω1 0 in Ω2 , b̃2 = { 0 in Ω1 f̂ 2 −∇p̂2 in Ω2 . Then, ∫ Ω (ã1 + ã2).wdxdydz = ∫ Ω1 (ã1 + ã2).w1dxdydz + ∫ Ω2 (ã1 + ã2).w2dxdydz, = ∫ Ω1 ã1.w1dxdydz + ∫ Ω2 ã2.w2dxdydz, = ∫ Ω1 − ∂ ∂z  µ1 2 p 2 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z + √ 2 2 g1 ∂û1 ∂z∣∣∣∂û1∂z ∣∣∣   .w1dxdydz + ∫ Ω2 − ∂ ∂z  µ2 2 p 2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z + √ 2 2 g2 ∂û2 ∂z∣∣∣∂û2∂z ∣∣∣   .w2dxdydz, = ∫ Ω1 ( f̂ 1 −∇p̂1 ) .w1dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .w2dxdydz, = ∫ Ω1 b̃1.w1dxdy dz + ∫ Ω2 b̃2.w2dxdydz, = ∫ Ω (b̃1 + b̃2).wdxdydz ∀w ∈ W 1,p 0 (Ω) 2. Which eventually gives (3.23). From now on we will denote by (û1, û2) ∈ Wz and (p̂1, p̂2) ∈ L p′ 0 (Ω1)×L p′ 0 (Ω2) the solution of the limit problem (3.23). � The following proposition shows the uniqueness of the limit solution (û1, p̂1) and (û2, p̂2). Proposition 3.4. The limit strong problem (3.23) has a unique, solution (û1, û2) ∈ Wz and (p̂1, p̂2) ∈ L p′ 0 (Ω1) ×L p′ 0 (Ω2) with the condition ( 3.22) . Proof. suppose that the limit problem (3.23) has at least two solutions (û1, û2) ∈ Wz, (p̂1, p̂2) ∈ L p′ 0 (Ω1) ×L p′ 0 (Ω2) and ( û1, û2 ) ∈ Wz, ( p̂1, p̂2 ) ∈ Lp ′ 0 (Ω1) ×L p′ 0 (Ω2). In particular, (û1, p̂1), (û2, p̂2) 14 Int. J. Anal. Appl. (2023), 21:91 and ( û1, p̂1 ) , ( û2, p̂2 ) are solutions of the weak formulation ( 3.25). Then µ1 2 p 2 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z .∂(v̂1 − û1)∂z dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂v̂1∂z ∣∣∣∣dxdydz − √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂û1∂z ∣∣∣∣dxdydz + µ2 2 p 2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣p−2 ∂û2∂z .∂(v̂2 − û2)∂z dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂v̂2∂z ∣∣∣∣dxdydz − √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂û2∂z ∣∣∣∣dxdydz ≥ ∫ Ω1 ( f̂ 1 −∇p̂1 ) .(v̂1 − û1)dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .(v̂2 − û2)dxdydz (v̂1, v̂2) ∈ Wz , (3.32) and µ1 2 p 2 ∫ Ω1 ∣∣∣∣∣∂û1∂z ∣∣∣∣∣ p−2 ∂û1 ∂z . ∂(v̂1 − û1) ∂z dxdydz + √ 2 2 g1 ∫ Ω1 ∣∣∣∣∂v̂1∂z ∣∣∣∣dxdydz − √ 2 2 g1 ∫ Ω1 ∣∣∣∣∣∂û1∂z ∣∣∣∣∣dxdydz + µ22 p2 ∫ Ω2 ∣∣∣∣∣∂û2∂z ∣∣∣∣∣ p−2 ∂û2 ∂z . ∂(v̂2 − û2) ∂z dxdydz + √ 2 2 g2 ∫ Ω2 ∣∣∣∣∂v̂2∂z ∣∣∣∣dxdydz − √ 2 2 g2 ∫ Ω2 ∣∣∣∣∣∂û2∂z ∣∣∣∣∣dxdydz ≥ ∫ Ω1 ( f̂ 1 −∇p̂1 ) .(v̂1 − û1)dxdydz + ∫ Ω2 ( f̂ 2 −∇p̂2 ) .(v̂2 − û2)dxdydz (v̂1, v̂2) ∈ Wz. (3.33) Setting (v̂1, v̂2) = ( û1, û2 ) and (v̂1, v̂2) = (û1, û2) as test functions in (3.32) and (3.33), respectively. Subtracting the tow obtained inequalities, we can infer µ 2 p 2 ∫ Ω1  ∣∣∣∣∣∂û1∂z ∣∣∣∣∣ p−2 ∂û1 ∂z − ∣∣∣∣∂û1∂z ∣∣∣∣p−2 ∂û1∂z   .∂(û1 − û1) ∂z dxdydz + µ2 2 p 2 ∫ Ω2  ∣∣∣∣∣∂û2∂y ∣∣∣∣∣ p−2 ∂û2 ∂y − ∣∣∣∣∂û2∂y ∣∣∣∣p−2 ∂û2∂y   .∂(û2 − û2) ∂y dxdydz ≤ ∫ Ω1 ∇ ( p̂1 − p̂1 ) . ( û1 − û1 ) dxdydz + ∫ Ω2 ∇ ( p̂2 − p̂2 ) . ( û2 − û2 ) dxdydz. (3.34) Observe that for every x,y ∈Rn, ( |x|p−2 x −|y|p−2 y ) .(x −y) ≥ (p− 1) |x −y|2 (|x| + |y|)2−p , 1 < p ≤ 2. Int. J. Anal. Appl. (2023), 21:91 15 This leads, making use (3.34), to µ1 (p− 1) 2 p 2 ∫ Ω1 (∣∣∣∣∣∂û1∂z ∣∣∣∣∣ + ∣∣∣∣∂û1∂z ∣∣∣∣ )p−2 ∣∣∣∣∣∂(û1 − û1)∂z ∣∣∣∣∣ 2 dxdydz + µ2 (p− 1) 2 p 2 ∫ Ω2 (∣∣∣∣∣∂û2∂y ∣∣∣∣∣ + ∣∣∣∣∂û2∂y ∣∣∣∣ )p−2 ∣∣∣∣∣∂(û2 − û2)∂z ∣∣∣∣∣ 2 dxdydz ≤ ∫ ω (( p̂1 − p̂1 )∫ h(x,y) 0 ( ∂(û11 − û11) ∂x + ∂(û12 − û12) ∂y ) dz ) dxdy + ∫ ω (( p̂2 − p̂2 )∫ 0 −h(x,y) ( ∂(û21 − û21) ∂x + ∂(û22 − û22) ∂y ) dz ) dxdy. This use of (3.22) gives µ1 (p− 1) 2 p 2 ∫ Ω1 (∣∣∣∣∣∂û1∂z ∣∣∣∣∣ + ∣∣∣∣∂û1∂z ∣∣∣∣ )p−2 ∣∣∣∣∣∂(û1 − û1)∂z ∣∣∣∣∣ 2 dxdydz + µ2 (p− 1) 2 p 2 ∫ Ω2 (∣∣∣∣∣∂û2∂y ∣∣∣∣∣ + ∣∣∣∣∂û2∂y ∣∣∣∣ )p−2 ∣∣∣∣∣∂(û2 − û2)∂z ∣∣∣∣∣ 2 dxdydz = 0. (3.35) On the other hand, the application of Hölder’s inequality leads to∫ Ω1 ∣∣∣∣∣∂(û1 − û1)∂z ∣∣∣∣∣ p dxdydz + ∫ Ω2 ∣∣∣∣∣∂(û2 − û2)∂z ∣∣∣∣∣ p dxdydz ≤ c  ∫ Ω1 ∣∣∣∂(û1−û1)∂z ∣∣∣2(∣∣∣∂û1∂z ∣∣∣ + ∣∣∣∂û1∂z ∣∣∣)2−pdxdydz   p 2 × (∫ Ω1 (∣∣∣∣∣∂û1∂z ∣∣∣∣∣ + ∣∣∣∣∣∂û1∂z ∣∣∣∣∣ )p dxdydz )2−p 2 +c  ∫ Ω2 ∣∣∣∂(û2−û2)∂z ∣∣∣2(∣∣∣∂û2∂z ∣∣∣ + ∣∣∣∂û2∂z ∣∣∣)2−pdxdydz   p 2 × (∫ Ω2 (∣∣∣∣∣∂û2∂z ∣∣∣∣∣ + ∣∣∣∣∂û2∂z ∣∣∣∣ )p dxdydz )2−p 2 . Which gives, keeping in mind (3.35)∥∥∥∥∥∂(û1 − û1)∂z ∥∥∥∥∥ Lp(Ω1) 2 = 0 and ∥∥∥∥∥∂(û2 − û2)∂z ∥∥∥∥∥ Lp(Ω2) 2 = 0, using Poincare’s inequality, we deduce∥∥∥û1 − û1∥∥∥ Lp(Ω1) 2 = 0 and ∥∥∥û2 − û2∥∥∥ Lp(Ω2) 2 = 0, we deduce that (û1, û2) = ( û1, û2 ) a.e. in Ω1 × Ω2.. 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