# A Viscous Fluid Flow through a Thin Channel with Mixed Rigid-Elastic Boundary: Variational and Asymptotic Analysis

Authors: R. Fares; G. P. Panasenko; R. Stavre
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the nonsteady Stokes flow in a thin tube structure composed by two thin rectangles with lateral elastic boundaries which are connected by a domain with rigid boundaries. After a variational approach of the problem which gives us existence, uniqueness, regularity results, and somea priori estimates, we construct an asymptotic solution. The existence of a junction region between the two rectangles imposes to consider, as part of the asymptotic solution, some boundary layer correctors that correspond to this region. We present and solve the problems for all the terms of the asymptotic expansion. For two different cases, we describe the order of steps of the algorithm of solving the problem and we construct the main term of the asymptotic expansion. By means of the a priori estimates, we justify our asymptotic construction, by obtaining a small error between the exact and the asymptotic solutions.


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## Body

## 1. Introduction

A few years ago we began to publish several papers dealing with an asymptotic analysis for an interaction problem between a viscous fluid and an elastic structure. Problems involving the interaction between a fluid and a deformable structure have been studied extensively in the last years due to their applications in many areas such as: engineering, biomechanics, biology, and hydroelasticity.In order to model the blood flow through a vessel we considered in [1] a non steady viscous flow in a thin rectangle with elastic walls, when at the ends of the flow domain periodicity conditions are imposed. The asymptotic approach was then extended for the nonperiodic case in [2], when the inflow and outflow velocities are given functions. Generalizations of the previous cited papers were obtained in [3–5]. In the first two papers we extended the results of [1] to a-three dimensional case, by considering a periodic, axisymmetric flow in a thin cylinder with elastic lateral boundary, while in [4] we studied two different cases: when the inertial term of the equation for the wall displacement is much smaller than the stress term and when these two terms are of the same order. Moreover, [5] deals with the asymptotic analysis for a fluid with variable viscosity.In all these papers, the fluid flow was described by the Stokes equations and the behavior of the elastic structure was simulated either by the Sophie Germain’s or by the Koiter’s equation. The fluid-structure interaction was mathematically expressed by the equality of the fluid velocity at the boundary and the time derivative of the wall displacement. We constructed an asymptotic solution and we proved that it represents a good approximation for the exact solution, by obtaining a small error between them.In this paper we consider the nonsteady Stokes flow in a thin tube structure. In two-dimensional case, a tube structure is some connected union of thin rectangles, having a dimension much smaller than the other one. The main difference between this paper and our previous works [2, 5] is that now the boundary of the flow domain contains elastic parts and rigid parts, as well.The flow domain consists of two thin rectangles with elastic boundary. The junction between the rectangles is realized by means of rigid boundaries. The interaction between the viscous fluid and the elastic boundaries produces normal displacements. The elastic boundaries behavior is described by the Sophie Germain’s equation.This domain models a vessel structure where a stent was placed. In fact, for treating arterial stenoses or occlusions, percutaneous angioplasty is indicated. It uses small inflatable balloons, single-use, that help dilate the artery at the site of narrowing. During angioplasty, a small wire mesh tube called a stent may be permanently placed in the newly opened artery or vein to help it remain open.There are two types of stents: bare stents (wire mesh) and covered stents (also commonly called stent grafts). The first are in use as part of the expansion of retrecissements vessels, the second impermeable and can be used to prevent or treat hemorrhage. Stents are used for arteries in the heart, the kidney, the arm, or the leg. They are also used for the aorta in the abdomen or chest (see Figures1 and 2).Figure 1
Bifurcation stenting.Figure 2
Stenting.We assimilate then the part of the vessel where the stent is placed to a domain with rigid boundaries.We suppose that the viscous fluid has a variable viscosity depending on a longitudinal variable for each rectangle. This situation models a blood flow in a vessel structure where the viscosity depends on the concentration of some substances diluted in blood or some blood cells. Indeed, the asymptotic analysis of the convection-diffusion equation set in such domains [6, 7] shows that in the case of the Neumann (impermeability) condition at the lateral boundary and small Reynolds numbers, the concentration is asymptotically close to the one-dimensional description, that is, the convection-diffusion equation set on the graph. The solution of the problem on the graph is the leading term of the asymptotic expansion, and it evidently depends on the longitudinal variable. On the other hand, the viscosity often depends on the concentration of the diluted substances or distributed cells, and so, it depends on the longitudinal variable. Of course, the fluid motion equation is coupled with the diffusion-convection equation in this case. However, if the velocity is small (in our case, it is of order ɛ2), then neglecting the convection, in comparison with the diffusion term or iterating with respect to the small term,we get the steady state diffusion equation; in absence of the source term in the right-hand side, it has a piecewise-linear asymptotic solution on the graph for the concentration. So, in this simplified situation, the diffusion equation can be solved before the fluid motion equation, and we obtain for the flow, the Stokes or Navier-Stokes equation with a variable viscosity depending (via concentration) on the longitudinal variable.There are, of course, many other practical problems involving fluids with variable viscosity. For example, the presence of bacteria in suspension (see [8]) may change locally the viscosity.The outline of the paper is as follows. In Section2 we give a description of the flow domain and of the coupled system which models our problem. The next section presents the variational formulation of the problem. The literature contains an important number of papers dealing with a variational approach of fluid-structure interaction problems. For instance, results concerning the existence of weak or strong solutions when the fluid domain is either fixed or depending on time can be found in [9–12]. We establish in this section results such as: existence, uniqueness, regularity and a priori estimates. In Section 4 we construct the asymptotic solution. As in our previous papers [1–5], the problem depends on two small parameters. The first small parameter, ɛ, is defined as the ratio of the dimensions of the thin rectangles; the second one, δ, corresponds to the softness of the wall. For various values of the small parameters ɛ and δ, an asymptotic expansion of the solution is constructed; the parameter δ is taken of the form δ=ɛγ, with γ∈ℕ, γ≥3. The asymptotic expansion is different for the cases: γ>3 and γ=3. The asymptotic solution contains three types of terms: the regular part, defined as in [1], the boundary layer correctors corresponding to the boundary conditions and the boundary layer correctors which realize the junction between the motion in the two rectangles. The first two types of correctors have already been introduced in [1, 2]. The third type is characteristic for structures with junction regions. The asymptotic solution is more complicated in this case since it contains also some truncation functions introduced in order to restrict the influence of the boundary layer correctors to the regions to which they correspond. We present and solve the problems for all the components of the asymptotic solution. For the two cases γ>3 and γ=3 the order of solving the problems is presented and the leading term of the asymptotic expansion is described. In the last section, we establish the error between the exact solution and the asymptotic one, by means of the a priori estimates obtained in Section 3. The small error between the two solutions justifies our asymptotic expansion.


## 2. The Physical Problem

We consider an incompressible, viscous fluid, with variable viscosity, occupying a thin domain,ℬɛ. The flow domain is a thin tube structure, composed by two thin rectangles with lateral elastic boundaries, connected by a region with rigid boundaries. We introduce the first small parameter of our problem, ɛ, ɛ=1/q, q∈ℕ in connection with the ratio of the two dimensions of the rectangles, as below. The thin rectangles are given by:(2.1)Dɛ1={(x1,x2)∈ℝ2:0<x1<1,-ɛ<x2<ɛ},Dɛ2={(x1,x2)∈ℝ2:-ɛ<x1<ɛ,0<x2<1},
and the junction region is(2.2)Dɛr={(x1,x2)∈ℝ2:-ɛ<x1<2ɛ,-ɛ<x2<2ɛ}∖({(x1,x2)∈ℝ2:-ɛ<x1<0,-ɛ<x2<0,x12+x22≥ɛ2}∪{(x1,x2)∈ℝ2:ɛ<x1<2ɛ,ɛ<x2<2ɛ,(x1-2ɛ)2+(x2-2ɛ)2≤ɛ2}).
The flow domain ℬɛ⊂ℝ2 is given by (Dɛ1∩{x1≥2ɛ})∪(Dɛ2∩{x2≥2ɛ})∪Dɛr, as shown in Figure 3.Figure 3
The domainℬɛ.The main difference between this paper and our previous works [2, 5] is that now the boundary of the flow domain ℬɛ contains elastic parts and rigid parts as well. Let A+εB+ε⌢ and A-εB-ε⌢ be the rigid parts of the boundary of ℬɛ defined as follows:(2.3)A+εB+ε⌢={(x1,x2)∈ℝ2:ɛ<x1<2ɛ,ɛ<x2<2ɛ,(x1-2ɛ)2+(x2-2ɛ)2=ɛ2},A-εB-ε⌢={(x1,x2)∈ℝ2:x1=-ɛ,0≤x2<2ɛ}∪{(x1,x2)∈ℝ2:0≤x1<2ɛ,x2=-ɛ}∪{(x1,x2)∈ℝ2:-ɛ<x1<0,-ɛ<x2<0,x12+x22=ɛ2};
the elastic parts of ∂ℬɛ are given by:(2.4)Γ±ɛ={(±ɛ,x2):2ɛ<x2<1}∪{(x1,±ɛ):2ɛ<x1<1}.
Let F1ɛ={(1,x2):-ɛ<x2<ɛ} and F2ɛ={(x1,1):-ɛ<x1<ɛ} be the inflow and outflow parts of the boundary of ℬɛ. Denoting F±ɛ=Γ±ɛ∪A±εB±ε⌢, we can write ∂ℬɛ=F+ɛ∪F-ɛ∪F1ɛ∪F2ɛ.We study the nonsteady, slow flow of the viscous fluid in the domainℬɛ previously described, when the fluid interacts with the elastic boundaries, Γ±ɛ. The interaction between the fluid and the elastic boundaries produces the normal displacements d±=d±(x1,x2,t). We neglect the tangential displacements and we consider that the elastic boundaries are clamped. We study the problem for t∈(0,T), with T an arbitrary positive constant independent on ɛ and we assume that the membranes are not very elastic so that the displacement of the boundaries is small enough. Consequently, at each time t, we approximate the position of the elastic membranes by its initial position and, hence, the fluid flow equations are considered in the initial configuration. We suppose that the displacements d±:F±ɛ↦ℝ have the following form:(2.5)d±(x1,x2,t)={d±(x1,t)onD¯ɛ1∩Γ±ɛ,d±(x2,t)onD¯ɛ2∩Γ±ɛ,0onF±ɛ∖Γ±ɛ.The problem described above, with nonhomogeneous boundary conditions for the velocity, is modeled by the following coupled system:(2.6)ρf∂u∂t-2divx(ν(x)Dxu)+∇xp=finℬɛ×(0,T),divxu=0inℬɛ×(0,T),ρh∂2d±∂t2+h3E12∂4d±∂xi4+μ∂5d±∂xi4∂t=g±±p|D¯ɛi∩Γ±ɛon(D¯ɛi∩Γ±ɛ)×(0,T),i=1,2,d+=0onA+εB+ε⌢×(0,T),d-=0onA-εB-ε⌢×(0,T),u=ψɛon(∂ℬɛ∖(Γ+ɛ∪Γ-ɛ))×(0,T),d±(1,±ɛ,t)=∂d±∂x1(1,±ɛ,t)=0in(0,T),d±(2ɛ,±ɛ,t)=∂d±∂x1(2ɛ,±ɛ,t)=0in(0,T),d±(±ɛ,1,t)=∂d±∂x2(±ɛ,1,t)=0in(0,T),d±(±ɛ,2ɛ,t)=∂d±∂x2(±ɛ,2ɛ,t)=0in(0,T),u·n=±∂d±∂t,u·τ=0onΓ±ɛ×(0,T),u(x,0)=0inℬɛ,d±(x,0)=∂d±∂t(x,0)=0onΓ±ɛ,
with n the outer unit normal on the boundary of ℬɛ and τ the unit tangent vector to ∂ℬɛ. The given data contained by the previous system are: some material constants and some given functions. ρf,ρ,μ,E are positive given constants in connection with the properties of materials representing the density of the fluid, the density of the elastic walls, a viscosity coefficient, and the two-dimensional Young’s modulus (The two dimensional Young’s modulus E is defined as E(3)/(1-ν̂2) where E(3) is the common three dimensional Young modulus and ν̂ is the Poisson ratio.), respectively, and the positive constant h stands for the thickness of the elastic walls.The given functions are:ν, the variable viscosity of the fluid, which satisfies ν∈C1(ℬɛ),ν(x)≥α>0 for all x∈ℬɛ, f, the exterior force applied to the fluid, g±, the exterior forces applied on the elastic walls, with g±:F±ɛ↦ℝ,(2.7)g±(x1,x2,t)={g±(x1,t)onD¯ɛ1∩Γ±ɛ,g±(x2,t)onD¯ɛ2∩Γ±ɛ,0onF±ɛ∖Γ±ɛ
and a small inflow-outflow velocity defined by means of the function ψɛ defined as follows:(2.8)ψɛ(x1,x2,t)=ɛ2ψ(ξ1,ξ2,t),∀x∈ℬɛ,
with (ξ1,ξ2)=(x1/ɛ,x2/ɛ). The function ψ is the trace of a function denoted also by ψ with the following properties:(2.9)ψ(ξ1,ξ2,t)=ψ(ξ2,t)e1,∀(ξ1,ξ2)∈(2,∞)×(-1,1),ψ(ξ1,ξ2,t)=ψ(ξ1,t)e2,∀(ξ1,ξ2)∈(-1,1)×(2,∞),divξψ=0,inDr,ψ=0onF+∪F-,ψ(ξ1,ξ2,0)=0,∫-11ψ(ξ2,t)dξ2+∫-11ψ(ξ1,t)dξ1=0,
where Dr=(1/ɛ)Dɛr,F±=F±ɛ. The unknowns of the system (2.6) are the velocity of the fluid, u, the pressure of the fluid, p, and the normal displacements of the elastic walls, d±.The fluid flow is described by the nonsteady Stokes equations. For the normal displacements we consider the Sophie-Germain’s equation. A “viscous” type term,μ(∂5d±i/∂xi4∂t), is added to the usual forth-order equation for the normal displacements. It corresponds to the viscoelastic behavior of the wall (the so called Kelvin-Voigt model). Usually, the Young’s modulus, E, has a value of 104–106 Pa. On the other hand, we assume that the characteristic longitudinal space scale for vessels is of order of cm and the time scale is of order of seconds. Let us use the S.I. system of units. This leads us to the necessity of scaling of every derivative is xi by the factor 102; that is, the fourth derivative will contain the additional factor 108. If h is of order 10-3 m or 10-2 m, then the coefficient ρh can be taken in the further analysis as a value of order of 1. The coefficient h3E/12 in (2.6)3 will be replaced (after scaling in xi) by a great coefficient δ-1 with the value δ of order of 10-7 to 10-4. If the ratio of thickness and the length of the vessel ɛ are of order 10-2, then δ is of order from ɛ2 to ɛ4. We assume that the “viscous” term is much smaller than the term with the coefficient δ-1 and hence the new coefficient denoted also by μ, obtained after scaling in xi, is O(1). More details concerning (2.6) can be found, for instance in [1].Due to the properties of the functionψɛ, the compatibility condition for the coupled system which describes the physical problem is(2.10)0=∫∂ℬɛu(x,t)·nds=ddt(∫2ɛ1(d+(x1,t)-d-(x1,t))dx1+∫2ɛ1(d+(x2,t)-d-(x2,t))dx2).
Using next the initial condition for the displacements, condition for the above coupled system becomes(2.11)∫2ɛ1(d+(x1,t)-d-(x1,t))dx1+∫2ɛ1(d+(x2,t)-d-(x2,t))dx2=0.
This condition states that the global area of the flow domain is preserved.


## 3. Variational Formulation: Existence, Uniqueness, Regularity, andA Priori Estimates

In order to obtain the above properties for the solution of the physical problem, we introduce the variational framework.To simplify the computations, we consider first (2.6) with homogeneous boundary conditions on F1ɛ and F2ɛ, that is, the problem for ψɛ=0. Then, the same properties for the solution of (2.6) follow with the usual technique for nonhomogeneous problems.Taking into account the conditions for the velocity and for the displacements and the condition (2.11) we introduce the following spaces:(3.1)Vɛ={v∈(H1(ℬɛ))2:divv=0inℬɛ,v=0on∂ℬɛ∖(Γ+ɛ∪Γ-ɛ),v·τ=0onΓ±ɛ},Wɛ={∫2ɛ1(β+,β-)∈H1(F+ɛ)×H1(F-ɛ):(β+,β-)∈H02(Γ+ɛ)×H02(Γ-ɛ),(β+,β-)=(0,0)onA+εB+ε⌢×A-εB-ε⌢,∫2ɛ1(β+1(x1)-β-1(x1))dx1+∫2ɛ1(β+2(x2)-β-2(x2))dx2=0}.
Choosing for the data the regularity f∈L2(0,T;(L2(ℬɛ))2) and g=(g+,g-)∈L2(0,T;F+ɛ)×L2(0,T;F-ɛ) we consider the following variational problem:(3.2)Find(u,d)∈L2(0,T;Vɛ)×H1(0,T;Wɛ),with(u̇,d̈)∈L2(0,T;(Vɛ)′)×L2(0,T;(Wɛ)′)whichsatisfiesa.e.in(0,T):ρfddt∫ℬɛu·φdx+2∫ℬɛνDu:Dφdx+∑i=12{ρhddt∫2ɛ1∂d∂t·βdxi+h3E12∫2ɛ1∂2d∂xi2·∂2β∂xi2dxi+μ∫2ɛ1∂3d∂xi2∂t·∂2β∂xi2dxi}=∫ℬɛf·φdx+∑i=12∫2ɛ1g·βdxi∀φ∈Vɛ,∀β∈Wɛ,withφ·n=±β±onΓ±ɛ,u·n=±∂d±∂tonΓ±ɛ,u(0)=0,d(0)=ḋ(0)=0.
Here and below d=(d+,d-) and β=(β+,β-).For the nonhomogeneous boundary conditions we still obtain the variational formulation (3.2) with u replaced by u-ψɛ and f replaced by f-ρf(∂ψɛ/∂t)+2div(νDψɛ).Theorem 3.1.
The variational problem (3.2) has the unique solution (u,d) with (u̇,d̈)∈L2(0,T;(L2(ℬɛ))2)×L2(0,T;L2(F+ɛ)×L2(F-ɛ)).Proof.
Let us start with the proof of the uniqueness for the solution of (3.2). Consider (u1,d1) and (u2,d2) two solutions of (3.2) and define (u,d)=(u1-u2,d1-d2). Subtracting the two relations (3.2)1 and taking as test function (φ,β)=(u,d) we get:
(3.3)ρf2ddt∫ℬɛu2dx+2∫ℬɛνDu:Dudx+∑i=12(ρh2ddt∫2ɛ1(∂d∂t)2dxi+h3E24ddt∫2ɛ1(∂2d∂xi2)2dxi+μ∫2ɛ1(∂3d∂xi2∂t)2dxi)=0.
Integrating from 0 to t this equality and taking into account the initial conditions, we obtain: u=0 a.e. in (0,T) and d=0 a.e. in (0,T). Hence the problem (3.2) has a unique solution.
For proving the existence and the regularity of the functionsu and d, we will use the Galerkin’s method.
We begin with the construction of a basis for the spaceWɛ. Let {ζj}j∈ℕ be a basis of H02(2ɛ,1) chosen by considering the eigenfunctions of the following problem:
(3.4)ζj(iv)=αjζjin(2ɛ,1),ζj(2ɛ)=ζj(1)=0,ζj′(2ɛ)=ζj′(1)=0,
where ζj(iv) is the fourth derivative of ζj and αj>0, for all j∈ℕ. We define {βj}j∈ℕ as follows: βj=(β+j,β-j) where
(3.5)β±j(x1,x2)={ζj(x1)for(x1,x2)∈(2ɛ,1)×{±ɛ},0for(x1,x2)∈A±εB±ε⌢,ζj(x2)for(x1,x2)∈{±ɛ}×(2ɛ,1).
It is easy to check that {βj}j∈ℕ is a basis for Wɛ. We choose the functions of the basis {ζj}j∈ℕ such that
(3.6)∫ℬɛβi·βkdx=δik.
As a consequence of the previous relation we also get:
(3.7)∫ℬɛβi′′·βk′′=αkδik.
We consider now {ψi}i∈ℕ a basis of the space V0ɛ={u∈(H01(ℬɛ))2:divu=0inℬɛ}, constructed with the eigenfunctions of the following Stokes problem:
(3.8)-2div(νDψi)+∇qi=λiψiinℬɛ,divψi=0inℬɛ,ψi=0on∂ℬɛ,
with λi>0 for all i∈ℕ. The functions ψi,i∈ℕ are uniquely determined from the condition
(3.9)∫ℬɛψj·ψkdx=δjk,
which implies
(3.10)2∫ℬɛνDψi:Dψkdx=λkδjk.
Next, for any βj we consider the following problem:
(3.11)Find(φj,pj)∈Vɛ×L2(ℬɛ)suchthat-2div(νD(φj))+∇pj=0in(H-1(ℬɛ))2,φj·n=±β±jonΓ±ɛ.
Following the classical results of [13] for nonhomogeneous Stokes problems, we obtain a unique φj and a function pj unique up to an additive constant. Moreover, for any j,k∈ℕ(3.12)∫ℬɛνDφj:Dψkdx=0.
By means of the functions {βj}j∈ℕ,{ψi}i∈ℕ,{φj}j∈ℕ, we are now in a position to define, for each n,m∈ℕ, an approximate solution (unm,dn) of (3.2) as follows:
(3.13)unm(x,t)=∑i=1mai(t)ψi(x)+∑j=1nḃj(t)φj(x),dn(x,t)=∑j=1nbj(t)βj(x),
with ai,bj:[0,T]↦ℝ,i=1,…,m,j=1,…,n scalar unknown functions. These functions are determined below from the condition that (unm,dn) is the solution for the problem:
(3.14)ρf∫ℬɛ∂unm∂t·ψidx+2∫ℬɛνDunm:Dψidx=∫ℬɛf·ψidx,fori∈{1,…,m},ρf∫ℬɛ∂unm∂t·φjdx+2∫ℬɛνDunm:Dφjdx+∑l=12(ρh∫2ɛ1∂2dn∂t2·βjdxl+h3E12∫2ɛ1∂2dn∂xl2·∂2βj∂xl2dxl+μ∫2ɛ1∂3dn∂xl2∂t·∂2βj∂xl2dxl)=∫ℬɛf·φjdx+∑l=12∫2ɛ1g·βjdxl,forj∈{1,…,n},unm·n=±∂dn±∂tonΓ±ɛ,unm(0)=0,dn(0)=ḋn(0)=0.
We introduce the notations:
(3.15)pik=∫ℬɛφk·ψidx,qkj=∫ℬɛφk·φjdx,rik=2∫ℬɛνDφi:Dφkdx.
Taking into account the previous notations and the relations (3.6), (3.7), (3.9), (3.10), (3.12) we get from (3.14) the following m+n linear differential system for the unknown functions ai,bj,i=1,…,m,j=1,…,n:
(3.16)ρfȧi(t)+λiai(t)+∑k=1npikb̈k(t)=∫ℬɛf·ψidx,ρf∑k=1mpkjȧk(t)+ρf∑k=1mqkjb̈k(t)+∑k=1nrkjḃk(t)+ρhb̈j(t)+h3E12αjbj(t)+μαjḃj(t)=∫ℬɛf·φjdx+∑l=12∫2ɛ1g·βjdxl,ai(0)=bj(0)=ḃj(0)=0,i=1,…,m,j=1,…,n.
The previous system uniquely determines the unknown functions ai,bj,i=1,…,m,j=1,…,n. For more details see [1, Section  3].
In the sequel we obtain somea priori estimates which give the regularity of the solution for (3.2). Computing ∑i=1mai(t)·(3.14)1+∑j=1nḃj(t)·(3.14)2 and using (3.13) we get:(3.17)ρf2ddt∫ℬɛ(unm)2dx+∫ℬɛν(Dunm)2+∑i=12(ρh2ddt∫2ɛ1(∂dn∂t)2dxi+h3E24ddt∫2ɛ1(∂2dn∂xi2)2dxi+μ∫2ɛ1(∂3dn∂xi2∂t)2dxi)=∫ℬɛf·unmdx+∑i=12∫2ɛ1g·∂dn∂tdxi.
Integrating from 0 to t, using the property of ν and the initial conditions we obtain, as in [1], the first estimates:
(3.18)∥umn∥L∞(0,T;(L2(ℬɛ))2)≤C(f,g),∥Dumn∥L2(0,T;(L2(ℬɛ))4)≤C(f,g),∥∂d±n∂t∥L∞(0,T;L2(Γ±ɛ))≤C(f,g),∥∂2d±n∂s2∥L∞(0,T;L2(Γ±ɛ))≤C(f,g),∥∂3d±n∂s2∂t∥L2(Γ±ɛ×(0,T))≤C(f,g),
with C(f,g)=C(T,μ,ρf,ρ,E,h)(∥f∥L2(0,T;(L2(ℬɛ))2)+∥g∥L2((0,T);L2(F+ɛ×F-ɛ))) and s is the variable of the parametrisation for Γ+ɛ or on Γ-ɛ.
The second estimates are obtained computing∑i=1mȧi(t)·(3.14)1+∑j=1nb̈j(t)·(3.14)2:(3.19)∥∂umn∂t∥L2(0,T;(L2(ℬɛ))2)≤(h3E12)1/2C(f,g),∥Dumn∥L∞((0,T);(L2(ℬɛ))4)≤(h3E12)1/2C(f,g),∥∂2d±n∂t2∥L2(Γ±ɛ×(0,T))≤(h3E12)1/2C(f,g),∥∂3d±n∂s2∂t∥L∞(0,T;L2(Γ±ɛ))≤(h3E12)1/2C(f,g).
From (3.18)1,2 and (3.19)1,2 we get the boundedness of {unm}m,n∈ℕ in L∞(0,T;Vɛ)∩H1(0,T;(L2(ℬɛ))2). Consequently, we can pass to the limit in (3.14) which yields the existence result of the theorem. To achieve the proof, we notice that the regularity stated in (3.2) follows from the estimates (3.18) and (3.19).As a consequence of Theorem3.1 we introduce the pressure which appears in (2.6)1,3.Corollary 3.2.
There exists a unique functionp∈L2(0,T;H1(ℬɛ)) such that (u,p,d) satisfies (2.6)1,3 a.e. in ℬɛ×(0,T) and on (D¯ɛi∩Γ±ɛ)×(0,T), respectively.To obtain this result we follow the ideas of [1, Section  3].The last result of this section presents the estimates for(u,p,d), the unique solution of (2.6).Corollary 3.3.
Let(u,p,d) be the solution of the problem (2.6) corresponding to the data f,g, with f∈L2(0,T;(L2(ℬɛ))2) and g∈L2(0,T;F+ɛ)×L2(0,T;F-ɛ). Then the following estimates hold:
(3.20)∥u∥L∞(0,T;(L2(ℬε))2)≤C(f,g),∥Du∥L2(0,T;(L2(ℬε))4)≤C(f,g),∥∂u∂t∥L2(0,T;(L2(ℬε))2)≤(h3E12)1/2C(f,g),∥∂d±∂t∥L∞(0,T;L2(Γ±ε))≤C(f,g),∥∂2d±∂s2∥L∞(0,T;L2(Γ±ε))≤C(f,g),∥∂2d±∂t2∥L2(Γ±ε×(0,T))≤(h3E12)1/2C(f,g),∥∂3d±∂s2∂t∥L∞(0,T;L2(Γ±ε))≤(h3E12)1/2C(f,g),∥∇p∥L2(0,T;(L2(ℬε))2)≤(h3E12)1/2C(f,g).Remark 3.4.
If we consider the nonhomogeneous problem (which corresponds toψɛ) we obtain for the corresponding solution the same estimates (3.20) with a different f, but which depends in the same way on ɛ.


## 4. Asymptotic Analysis

In order to approximate the solution of (2.6), we consider more regular data than in Section 3. We suppose that:(4.1)ψɛ∈(𝒞∞(ℬɛ×(0,T)))2,f∈(𝒞∞(ℬɛ×(0,T)))2,g±∈𝒞∞(F±ɛ×(0,T)),ν∈𝒞∞(ℬɛ),∃t*<Tsuchthatf(x,t)=g±(x,t)=ψɛ(x,t)=0∀(x,t)∈ℬɛ×(0,t*),
and f, g± and ν are chosen as follows:(4.2)f(x1,x2,t)={f(x1,t)e1in(ℬɛ∩{x1>13})×(0,T),f(x2,t)e2in(ℬɛ∩{x2>13})×(0,T),0in(ℬɛ∩{x1≤13,x2≤13})×(0,T),g±(x1,x2,t)={g±(x1,t)on(F±ɛ∩{x1>13})×(0,T),g±(x2,t)on(F±ɛ∩{x2>13})×(0,T),0on(F±ɛ∩{x1≤13,x2≤13})×(0,T),ν(x1,x2)={ν(x1)inℬɛ∩{13<x1<23},ν(x2)inℬɛ∩{13<x2<23},ν0inℬɛ∩({x1≤13,x2≤13}∪{x1≥23}∪{x2≥23}),
with ν0 constant.


### 4.1. Construction of the Asymptotic Solution

In the sequel we introduce the second small parameterδ=12/h3E and we take δ=ɛγ, with γ∈ℕ, γ≥3. The asymptotic solution approximating the periodic flow in an infinite rectangle (see [1, Section  5]) is modified by using two types of correctors: the first type corresponds to the boundary conditions on x1=1 and x2=1 and the second one represents the boundary layer functions in a neighborhood of (x1,x2)=0. So, the asymptotic solution contains the regular part of the solution (which has two terms, due to the form of the flow domain), two boundary layer functions corresponding to x1=1 and to x2=1, respectively, and the correctors in (x1,x2)=0. Since the terms of the asymptotic solution should vanish in different parts of the flow domain, we multiply them with the truncation functions defined as follows: χ, η:ℝ↦ℝ, χ,η∈𝒞∞(ℝ),(4.3)χ(τ)={0,|τ|<2,1,|τ|>3,η(τ)={0,|τ|<34,1,|τ|>78.We construct the asymptotic solution as below:(4.4)ûa(k)(x1,x2,t)=u1(k)(x1,x2ɛ,t)χ(x1ɛ)+u2(k)(x1ɛ,x2,t)χ(x2ɛ)+ubl(k)1(x1-1ɛ,x2ɛ,t)η(x1)+ubl(k)2(x1ɛ,x2-1ɛ,t)η(x2)+ubl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2),p̂a(k)(x1,x2,t)=p1(k)(x1,x2ɛ,t)χ(x1ɛ)+p2(k)(x1ɛ,x2,t)χ(x2ɛ)+pbl(k)1(x1-1ɛ,x2ɛ,t)η(x1)+pbl(k)2(x1ɛ,x2-1ɛ,t)η(x2)+pbl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2),d̂±a(k)(x1,x2,t)=d±1(k)(x1,t)χ(x1ɛ)+d±2(k)(x2,t)χ(x2ɛ)+d±bl(k)1(x1-1ɛ,t)η(x1)+d±bl(k)2(x2-1ɛ,t)η(x2)+d±bl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2).Due to the definition of the truncation functions, we notice that inℬɛ∩{1/4≤xi≤3/4}, that is, in a neighborhood of the region of variable viscosity, the asymptotic solution reduces to its regular part, (ui(k),pi(k),d±i(k)), in a neighborhood of {xi=1} it reduces to (ui(k)+ubl(k)i,pi(k)+pbl(k)i,d±i(k)+d±bl(k)i), i=1,2, while in Dɛr the asymptotic solution is equal to the corrector in x=0,(ubl(k)0,pbl(k)0,d±bl(k)0). This means that the regular part of the asymptotic solution (which can be computed and has a very simple expression) represents an approximation for the exact solution in all the flow domain except some neighborhoods of x1=1, x2=1 and x=0.We give next the expressions of the three components of the asymptotic solution and we explain their role in this construction.


#### 4.1.1. The Regular Part of the Asymptotic Solution

For each rectangleDɛ1 and Dɛ2 we define the corresponding regular part of the asymptotic expansion as in [1]. The regular part corresponding to Dɛ1 has the expression:(4.5)u1(k)(x1,x2ɛ,t)=∑j=0kɛj+2u1,j1(x1,x2ɛ,t)e1+∑j=0kɛj+3u2,j1(x1,x2ɛ,t)e2,p1(k)(x1,x2ɛ,t)=∑j=0kɛj+1pj1(x1,x2ɛ,t)+∑j=0kɛjqj1(x1,t),d±1(k)(x1,t)=∑j=0kɛj+γd(±)j1(x1,t).
It represents the solution of the problem (2.6)1,2,3,11 set in the infinite rectangle in Ox1 direction, (-∞,∞)×(-ɛ,ɛ); the functions which appear in (4.5) can be computed explicitly and represent a good approximation of the exact solution of (2.6) in a neighborhood of the region with variable viscosity of Dɛ1 (as we will prove in the last section).In a similar way we introduce the regular part of the asymptotic solution corresponding toDɛ2.Hence, the sum of the first two terms of the asymptotic solution represents a good approximation of the exact solution of (2.6) in some neighborhood of the region of variable viscosity of the flow domain, but it is not close to it on the inflow/outflow boundaries and in Dɛr. Since the purpose of the asymptotic construction is to approximate the exact solution with a small error between the exact and asymptotic solution, we modify and complete the regular part by two types of boundary layer correctors.


#### 4.1.2. The Boundary Layer Correctors forx1=1 and x2=1

These boundary layer functions are introduced in order to repair the traces of the regular part of the asymptotic solution onx1=1 and x2=1. They are given by:(4.6)ubl(k)i(xɛ,t)=∑j=0kɛj+2uj(i)(xɛ,t),pbl(k)i(xɛ,t)=∑j=0kɛj+1pj(i)(xɛ,t),d±bl(k)i(xiɛ,t)=∑j=0kɛj+γdj(i)(xiɛ,t),i∈{1,2}.
The corrector with i=1 corresponds to the end x1=1 and that with i=2 corresponds to x2=1.From the definition of the truncation functions and of the asymptotic solution we notice that the influence of each corrector defined above is significant only near the corresponding end of the flow domain.


#### 4.1.3. The Boundary Layer Corrector inx=0

This corrector is necessary in order to realize the junction between the two parts of the asymptotic solution, corresponding to the two branches of the flow domain and to obtain the conditions on the rigid boundaries of the domain. The expressions of the correctors inx=0 are given by:(4.7)ubl(k)0(xɛ,t)=∑j=-1kɛj+2uj(0)(xɛ,t),pbl(k)0(xɛ,t)=∑j=-1kɛj+1pj(0)(xɛ,t),d±bl(k)0(xɛ,t)=∑j=0kɛj+γd(±)j(0)(xɛ,t).
From the definition of the truncation function η it follows that the corrector in x=0 appears in the expression of the asymptotic solution (4.4) only in ℬɛ∩{x1≤1/4,x2≤1/4}, which represents a neighborhood of Dɛr with constant viscosity and where the forces are equal to zero.Remark 4.1.
The boundary layer method is close to the well-known method of matching of asymptotic expansions (see [14]). However there is a difference between these two methods. In the boundary layer method the regular expansion is reexpanded with respect to the fast variable (as in the matching method), but then the new expansion is multiplied by a cutting function. This product being inserted into the equation gives a discrepancy in the right-hand side which is then compensated by an appropriate boundary layer corrector. So, we get finally a unique expression for the asymptotic approximation in each point of the domain. In the matching approach the reexpanded regular ansatz (outer expansion) coexists with the so-called inner expansion defined near the boundary, so that there are several overlapping expressions for an asymptotic approximation in different parts of the domain. We emphasize that this difference is not too important and corresponds more to the form of presentation of the result.


### 4.2. The Determination of the Asymptotic Solution

This subsection is devoted to the resolution of the problems satisfied by the three different components of the asymptotic solution presented in the previous subsection. Since some computations are different with respect to the values ofγ, we will analyse the problems and the order of solving them for γ>3 and for γ=3. Moreover, in each case we will specify the leading term of the asymptotic expansion.We begin the approach with the problems for the boundary layer correctors corresponding tox1=1 and x2=1 since the study of these problems is the same both for γ>3 and for γ=3. To fix the ideas, we obtain in the sequel the problems for the corrector corresponding to the end x1=1. As we noticed before, the term containing this corrector is not equal to zero only in a neighborhood of the boundary x1=1. So, the problems and the other relations corresponding to this corrector are obtained substituting the asymptotic solution in (2.6)1,2,3,6,7,11. We obtain two separate problems: one for the velocity-pressure correctors and the other one for the displacements correctors.Since in this neighborhood the viscosity is constant, the problem for(uj(1),pj(1)) has constant coefficients. Denoting by Π1 the semi-infinite rectangle (-∞,0)×(-1,1) and imposing for the velocity and for the pressure the condition of decay at -∞, we obtain for (uj(1),pj(1)) the problem:(4.8)-ν0Δξξuj(1)+∇ξpj(1)=-ρf∂uj-2(1)∂tinΠ1×(0,T),divξuj(1)=0inΠ1×(0,T),uj(1)(ξ1,±1,t)=∂d(±)j+2-γ(1)∂t(ξ1,t)e2in(-∞,0)×(0,T),uj(1)(0,ξ2,t)=ψ(ξ2,t)δj0e1-u1,j1(1,ξ2,t)e1-u2,j-11(1,ξ2,t)e2in(-1,1)×(0,T),uj(1)→0,pj(1)→0,uniformlywhenξ1→-∞.
The compatibility condition for (4.8) reads:(4.9)∫-11u1,j1(1,ξ2,t)dξ2=ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1+δj0∫-11ψ(ξ2,t)dξ2.
For any γ≥3 the right-hand side of (4.8)3 is known; so the boundary layer correctors for the velocity and for the pressure corresponding to x1=1 are uniquely determined from (4.8) (see [2, Section  4]). The condition (4.9) represents a relation for the regular part of the asymptotic solution.The boundary layer correctors for the displacements exponentially stabilizing to zero at-∞ are obtained as the unique solution of the following problems:(4.10)∂4d(±)j(1)∂ξ14=-ρh∂2d(±)j-4-γ(1)∂t2-μ∂5d(±)j-γ(1)∂ξ14∂t±pj-5(1)|ξ2=±1in(-∞,0)×(0,T),∂ad(±)j(1)∂ξ1a→0,uniformly,whenonlyξ1→-∞,a∈{0,1,2,3}.Since at the stepj the problem (4.10) gives both d(±)j(1) and d(±)j+1(1), introducing the asymptotic solution into (2.6)7 we obtain two boundary conditions for the regular part of the asymptotic solution for the displacements:(4.11)d(±)j1(1,t)=-d(±)j(1)(0,t),∂d(±)j1∂x1(1,t)=-∂d(±)j+1(1)∂ξ1(0,t).In a similar way, we obtain the boundary layer correctors corresponding to the endx2=1. The boundary layers for the velocity-pressure are defined on Π2×(0,T), with Π2=(-1,1)×(-∞,0), and the boundary layers for the displacements are defined also on (-∞,0)×(0,T).We study next the problems for the regular parts of the asymptotic solution. The results are obtained for the regular part corresponding toDɛ1; the regular part corresponding to Dɛ2 may be obtained from the previous with some obvious changes.Introducing (4.5) into (2.6)1,2,3,11 and collecting together the terms of the same order with respect to ɛ we are leaded to consider the following problem for (u1,j1,u2,j1,pj1,qj1,d(±)j1):(4.12)-ν(x1)∂2u1,j1∂ξ22+∂qj1∂x1=fδj0-ρf∂u1,j-21∂t+2∂∂x1(ν(x1)∂u1,j-21∂x1)+ν(x1)∂2u2,j-21∂ξ2∂x1-∂pj-11∂x1,∂pj1∂ξ2=-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22,∂u1,j1∂x1+∂u2,j1∂ξ2=0in(0,1)×(-1,1)×(0,T),∂4d(+)j1∂x14-qj1=g+δj0+pj-11|ξ2=1-ρh∂2d(+)j-γ1∂t2-μ∂5d(+)j-γ1∂x14∂tin(0,1)×{-1}×(0,T),∂4d(-)j1∂x14+qj1=g-δj0-pj-11|ξ2=-1-ρh∂2d(-)j-γ1∂t2-μ∂5d(-)j-γ1∂x14∂tin(0,1)×{1}×(0,T),uj1(x1,±1,t)=∂d(±)j-γ+31∂t(x1,t)e2in(0,1)×{±1}×(0,T).
The two cases, γ>3 and γ=3, appear because of the last relation of the previous system. We can see, indeed, that for γ>3 the unknown of this relation is uj1, while for γ=3, (4.12)6 contains two unknowns. From this point, the computations are different with respect to the values of γ.We introduce the functions:(4.13)N1(ξ2)=12(ξ22-1),N2(ξ2)=∫-1ξ2N1(τ)dτ,
with the properties: N1′′=1,N1(±1)=0 and N2(1)=-2/3.We also use the notations:(4.14)Uj-11(x1,ξ2,t)=ρf∂u1,j-21∂t-2∂∂x1(ν(x1)∂u1,j-21∂x1)-ν(x1)∂2u2,j-21∂ξ2∂x1+∂pj-11∂x1,𝒟(±)j-11(x1,t)=±pj-11|ξ2=1-ρh∂2d(±)j-γ1∂t2-μ∂5d(±)j-γ1∂x14∂t,D-1:F→∫-1ξ2F(x1,τ)dτ,D-2:F→∫-1ξ2∫-1θF(x1,τ)dτdθ-12(ξ2+1)∫-11∫-1θF(x1,τ)dτdθ,Ix1-k:F→∫0x1∫0ak-1⋯∫0a1F(s,ξ2,t)ds,Jx1:F→Ix1-4(F)+x13(2∫01Ix1-3(F)-∫01Ix1-2(F))-x2(3∫01Ix1-3(F)-∫01Ix1-2(F)).


#### 4.2.1. The Order of Solving the Problems forγ>3

The regular part corresponding toDɛ1 is computed by integrating (4.12), as stated below.Proposition 4.2.
The unknownsuj1,qj1,pj1,d(±)j1 are determined from (4.12), up to nine functions of t.Proof.
Integrating twice (4.12)1 from −1 to ξ2 and using the boundary conditions (4.12)6 we get:
(4.15)u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),
which contains as unknowns u1,j1 and qj1. The other functions contained by this relation are either known from previous computations or equal to zero.
We integrate next the incompressibility condition (4.12)3 with respect to ξ2 with the boundary condition (4.12)6 for ξ2=-1 and we obtain(4.16)u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)).
The previous two relations give the components of the velocity uj1 with respect to qj1.
The pressure approximations are determined from (4.12)2, supposing that the integration functions, depending on x1, t are equal to zero, since we consider that any function depending only on x1, t could be contained in qj+11. 											(4.17)pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
Taking ξ2=1 in (4.16) and using the boundary condition (4.12)6 for ξ2=1 we obtain the following second order differential equation for the function qj1:
(4.18)∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))=32∂∂t(d(+)j-γ+31-d(-)j-γ+31)+32∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2.
Integrating (4.18) from x1 to 1, we express ∂qj1/∂x1 by means of (∂qj1/∂x1)(1,t), which represents the only unknown of this expression. This function of t is obtained as follows: we take x1=1 in (4.15) and we introduce the result into (4.9). Hence, we determined the expression of ∂qj1/∂x1 in (0,1)×(0,T), which is:
(4.19)1ν(x1)(∂qj1∂x1-fδj0)=-32δj0∫-11ψ(ξ2,t)dξ2-32ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1-32∂∂t∫x11(d(+)j-γ+31-d(-)j-γ+31)(s,t)ds+32∫-11D-2(1ν(x1)Uj-11)dξ2.
Introducing (4.19) into (4.15) and (4.18) into (4.16) we determine uj1 in (0,1)×(-1,1)×(0,T).
We integrate nextν(x1) · (4.19) from 0 to x1 and we get qj1 determined up to the function of t, qj1(0,t):
(4.20)qj1(x1,t)=δj0∫0x1f(s,t)ds-32∫0x1ν(θ)∫θ1∂∂t(d(+)j-γ+31-d(-)j-γ+31)(s,t)dsdθ-32(ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1+δj0∫-11ψ(ξ2,t)dξ2)(∫0x1ν(s)ds)+32∫0x1∫-11D-2(Uj-11(s,ξ2,t))dξ2ds+qj1(0,t).
The functions d(±)j1 satisfy the fourth-order differential equations:
(4.21)∂4d(±)j1∂x14=±qj1+g±δj0+𝒟(±)j-11,
with qj1 given by (4.20). Writting qj1 as Qj1(x1,t)+qj1(0,t) and integrating four times (4.21) with respect to x1 we obtain the following expressions for d(±)j1:
(4.22)d(+)j1(x1,t)=d(+)j1(0,t)(1-3x12+2x13)+∂d(+)j1∂x1(0,t)(x1-2x12+x13)+d(+)j1(1,t)(3x12-2x13)+∂d(+)j1∂x1(1,t)(-x12+x13)+qj1(0,t)(x1224-x1312+x1424)+Jx1(Qj1)+Jx1(g+)δj0+Jx1(𝒟(+)j-11),d(-)j1(x1,t)=d(-)j1(0,t)(1-3x12+2x13)+∂d(-)j1∂x1(0,t)(x1-2x12+x13)+d(-)j1(1,t)(3x12-2x13)+∂d(-)j1∂x1(1,t)(-x12+x13)-qj1(0,t)(x1224-x1312+x1424)-Jx1(Qj1)+Jx1(g-)δj0+Jx1(𝒟(-)j-11).
Hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined up to the functions qj1(0,t), d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t), which achieves the proof.In a similar way we express the regular part of the asymptotic solution corresponding toDɛ2 depending on 9 unknown functions of t.We continue the construction with the problems for the corrector inx=0. As we previously said, the term of the asymptotic solution containing this corrector is not equal to zero only in a neighborhood of Dɛr. In this neighborhood, the expression of the asymptotic solution reduces to (u1(k)(x1,ξ2,t)χ(ξ1)+u2(k)(ξ1,x2,t)χ(ξ2)+ubl(k)0(ξ,t),p1(k)(x1,ξ2,t)χ(ξ1)+p2(k)(ξ1,x2,t)χ(ξ2)+pbl(k)0(ξ,t),d±1(k)(x1,t)χ(ξ1)+d±2(k)(x2,t)χ(ξ2)+d±bl(k)0(ξ,t)). For obtaining the problems satisfied by the correctors in x=0, we introduce the previous expression of the asymptotic solution in (2.6)1,2,3,4,5,6,11, with ν=ν0 and f=g=0 in (2.6)1,3; for derivating the terms which contain the two types of variables xi and ξi we proceed as follows: we replace ∂/∂xi by (1/ɛ)(∂/∂ξi), we replace xi by ɛξi and we expand the functions as a Taylor expansion with respect to xi=ɛξi. We introduce the notations ℬ=Dr∪{(ξ1,ξ2):ξ1≥2,ξ2∈(-1,1)}∪{(ξ1,ξ2):ξ1∈(-1,1),ξ2≥2},A±B±⌢=(1/ɛ)A±εB±ε⌢ and we obtain for (uj(0),pj(0)) the following nondivergence free problem:(4.23)-ν0Δξξuj(0)+∇ξpj(0)=-ρf∂uj-2(0)∂t+Fj+(1-χ(ξ1))ξ1′qj+11(0,t)e1+(1-χ(ξ2))ξ2′qj+12(0,t)e2inℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0uniformlywhenξ1→∞,uj(0)→0uniformlywhenξ2→∞,
for j≥0 and(4.24)-ν0Δξξu-1(0)+∇ξp-1(0)=(1-χ(ξ1))ξ1′q01(0,t)e1+(1-χ(ξ2))ξ2′q02(0,t)e2inℬ,divu-1(0)=0inℬ,u-1(0)=0onA+B+⌢∪A-B-⌢,u-1(0)(ξ1,±1,t)=0forξ1≥2,u-1(0)(±1,ξ2,t)=0forξ2≥2,u-1(0)→0uniformlywhenξ1→∞,u-1(0)→0uniformlywhenξ2→∞.
Unlike the problem (4.8) which give the correctors for x1=1, the problems (4.23), (4.24) have unknown right hand sides. The functions Fj are known, but qj+11(0,t) and qj+12(0,t) are elements from the next approximation, so they are unknown. The function θ̃j is given by:(4.25)θ̃j=-divξ(χ(ξ1)∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2)+χ(ξ2)∑l=0jξ2ll!(∂lu1,j-1-l2∂x2l(ξ1,0,t)e1+∂lu2,j-l2∂x2l(ξ1,0,t)e2)),
and it has to satisfy a compatibility condition.The problem (4.24) has the unique solution (with the pressure unique up to an additive function of t)(4.26)u-1(0)=0,p-1(0)={(1-χ(ξ1))q01(0,t)forξ1≥2,q01(0,t)=q02(0,t)inDr,(1-χ(ξ2))q02(0,t)forξ2≥2.
We obtain next the problems for the correctors in x=0 corresponding to the displacements. We first notice that(4.27)d(±)j(0)=0onA±B±⌢.
Let us denote by d̂(±)j-1(0) the unique solution of(4.28)∂4d̂(±)j-1(0)∂ξ14=ρh(1-χ(ξ1))+∑l=0jξ1ll!∂l+2d(±)j-γ-4-l1∂t2∂x1l(0,t)+∑l=2j∂4ξ14((1-χ(ξ1))ξ1ll!)∂ld(±)j-l1∂x1l(0,t)+μ∑l=0j∂4ξ14((1-χ(ξ1))ξ1ll!)∂l+1d(±)j-γ-l1∂t∂x1l(0,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lpj-5-l1∂x1l(0,±1,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lqj-4-l1∂x1l(0,t),∂ad̂(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3},
and by d̃(±)j-1(0) the unique solution of(4.29)∂4d̃(±)j-1(0)∂ξ14=∂4d̂(±)j-1(0)∂ξ14-ρh∂2d(±)j-γ-4(0)∂t2-μ∂5d(±)j-γ(0)∂ξ14∂t+pj-5/ξ2=±1(0),∂ad̃(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3}.
Introducing the asymptotic solution into (2.6)3 and identifying the coefficients of the power of ɛ we obtain:(4.30)d(±)j(0)(ξ1,±1,t)=d̃(±)j-1(0)(ξ1,±1,t)-(1-χ(ξ1))×(d̃(±)j-1(0)(2,±1,t)-(2-ξ1)∂d̃(±)j-1(0)∂ξ1(2,±1,t))ξ1>2,(4.31)d(±)j1(0,t)=2∂d̃(±)j-1(0)∂ξ1(2,±1,t)-d̃(±)j-1(0)(2,±1,t),∂d(±)j1∂x1(0,t)=-∂d̃(±)j(0)∂ξ1(2,±1,t).
The relation (4.30) gives the corrector in x=0 for the displacements corresponding to one branch of the flow domain.We notice that the right-hand side of (4.31) is known, since the problems (4.28) and (4.29) can be solved also if we replace j-1 by j.We are now in a position to return to the regular part of the asymptotic solution. For each regular part of the asymptotic solution, we need 9 relations, in order to determine the 9 unknown functions of integration. For the regular part corresponding toDɛ1, (4.11) and (4.31) give 8 functions: d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t). It remains undetermined qj1(0,t). We show next that at the jth step this function is already determined at the previous step. For this purpose, we return to the problem (4.23). As usual, we should construct boundary layer functions stabilizing to zero at infinity. As we imposed this condition for the velocity in (4.23)6,7, it is known that the pressure stabilizes to some functions of t. At the jth step, we determine qj+11(0,t) and qj+12(0,t) from the condition of the exponential decay of pj(0) at infinity. To this end, we consider instead of (4.23) and (4.24), the following problem with known right-hand side for j≥-1:(4.32)-ν0Δξξuj(0)+∇ξp¯j(0)=-ρf∂uj-2(0)∂t+Fjinℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0,p¯j(0)→αj1(t)uniformlywhenξ1→∞,uj(0)→0,p¯j(0)→αj2(t)uniformlywhenξ2→∞
with αj1(t),αj2(t) unknown. Supposing for the time being that the function θ̃j satisfies the compatibility condition, it follows that (4.32) has a unique solution (the uniqueness for the pressure being understood up to an additive function of t). This means that the function αj(t), with αj(t)=αj1(t)-αj2(t) is fixed. We define for j≥-1(4.33)pj(0)=p¯j(0)+χ(ξ1)qj+11(0,t)+χ(ξ2)qj+12(0,t).
Standard computations show that (uj(0),p¯j(0)) is solution for (4.32) iff (uj(0),pj(0)) satisfies (4.23) or (4.24) for j=-1 with pj(0)→αji(t)+qj+1i(0,t) uniformly when ξi→∞,i=1,2. To obtain the desired behavior for pj(0) when ξi→∞,i=1,2, it suffices to take αji(t)=-qj+1i(0,t).A first equation for the unknownsqj+11(0,t) and qj+12(0,t) is(4.34)qj+11(0,t)-qj+12(0,t)=αj(t),
with α-1(t)=0. The other equation is obtained introducing the asymptotic solution into the compatibility condition (2.11):(4.35)∫01(d(+)j+11(x1,t)-d(-)j+11(x1,t))dx1+∫01(d(+)j+12(x2,t)-d(-)j+12(x2,t))dx2=-∑l=0j(∂l(d(+)j-l1-d(-)j-l1)∂x1l(0,t)+∂l(d(+)j-l2-d(-)j-l2)∂x2l(0,t))∫03τll!(χ(τ)-1)dτ-∫-∞0(d(+)j(1)(ξ1,1,t)-d(+)j(1)(ξ1,-1,t))dξ1-∫-∞0(d(+)j(2)(1,ξ2,t)-d(+)j(2)(-1,ξ2,t))dξ2-∫2∞(d(+)j(0)(ξ1,1,t)-d(+)j(0)(ξ1,-1,t))dξ1-∫2∞(d(+)j(0)(1,ξ2,t)-d(+)j(0)(-1,ξ2,t))dξ2.
The right-hand side of (4.35) is known from the jth approximation. Replacing in the left hand side of (4.35) d(±)j+11 given by (4.22) (for j replaced by j+1) and d(±)j+12 given by a similar relation, we are leaded to the second equation for the unknowns qj+11(0,t) and qj+12(0,t):(4.36)qj+11(0,t)+qj+12(0,t)=βj(t),
with βj a known function for j≥-1, with(4.37)β-1(t)=-6!(∫01Ix1-1(f)dx1+∫01Ix2-1(f)dx2)+3×6!2(∫-11ψ(ξ2,t)dξ2∫01Ix1-1(ν)dx1+∫-11ψ(ξ1,t)dξ1∫01Ix2-1(ν)dx2)-6!2(∫01Ix1-4(g+-g-)dx1+∫01Ix2-4(g+-g-)dx2)+6!4(∫01Ix1-3(g+-g-)dx1+∫01Ix2-3(g+-g-)dx2)-6!4!(∫01Ix1-2(g+-g-)dx1+∫01Ix2-2(g+-g-)dx2).
From (4.34) and (4.36) we determine qj+11(0,t) and qj+12(0,t), which means that the right-hand side of (4.23) is now known, that allows to solve (4.23), if we show that the problem (4.23) is well posed, that is, the compatibility condition is fulfilled. Moreover, this justifies the previous assertion that qji(0,t),i=1,2 are computed at the (j-1)th step. We can say that the entire jth approximation was obtained for γ>3 if we prove.Proposition 4.3.
For anyj∈ℕ, the function θ̃j satisfies the compatibility condition:
(4.38)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫∂ℬuj(0)·nds.Proof.
From (4.23) and (4.35) it follows:
(4.39)∫∂ℬuj(0)·nds=-ddt(∫-∞0(d(+)j+2-γ(1)(ξ1,t)-d(-)j+2-γ(1)(ξ1,t))dξ1-∫-∞0(d(+)j+2-γ(2)(ξ2,t)-d(-)j+2-γ(2)(ξ2,t))dξ2)-ddt∑l=0j+2-γ(∂l(d(+)j+2-γ-l1-d(-)j+2-γ-l1)∂x1l(0,t)+∂l(d(+)j+2-γ-l2-d(-)j+2-γ-l2)∂x2l(0,t))∫03sll!(χ(s)-1)ds-ddt(∫01(d(+)j+3-γ1(x1,t)-d(-)j+3-γ1(x1,t))dx1+∫01(d(+)j+3-γ2(x2,t)-d(-)j+3-γ2(x2,t))dx2).
We express next the three terms of the right-hand side of the previous equality with respect to the components of the velocity corresponding to the regular parts of the asymptotic solution.
From (4.9), the corresponding relation for x2=1 and (2.9)6 the first term of the right-hand side is equal to(4.40)T1=-∫-11u1,j1(1,ξ2,t)dξ2-∫-11u2,j2(ξ1,1,t)dξ1.
The third term is expressed by means of (4.12)6 and its analogous for uj2 and it is equal to
(4.41)T3=-∫01(u2,j1(x1,1,t)-u2,j1(x1,-1,t))dx1-∫01(u1,j2(1,x2,t)-u1,j2(-1,x2,t))dx2.
Computing now T1+T3 and using (4.12)3 and the similar one for uj2 we get
(4.42)T1+T3=-∫-11u1,j1(0,ξ2,t)dξ2-∫-11u2,j2(ξ1,0,t)dξ1.
The second term is given by
(4.43)T2=∑l=0j(∂l(u2,j-1-l1(0,1,t)-u2,j-1-l1(0,-1,t))∂x1l+∂l(u1,j-1-l2(1,0,t)-u1,j-1-l2(-1,0,t))∂x2l)∫03sll!(χ(s)-1)ds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ2,0,t)∂x2le2))dξ+∫-11u1,j1(0,ξ2,t)dξ2+∫-11u2,j2(ξ1,0,t)dξ2.
Finally, computing T1+T2+T3, we obtain the following expression of the right-hand side of (4.38):
(4.44)∫∂ℬuj(0)·nds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ1,0,t)∂x2le2))dξ.
We compute next the left-hand side of (4.38). Since χ(ξ1)=χ(ξ2)=0 on ℬ∩{ξ1<2,ξ2<2}, we can write:
(4.45)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ+∫(-1,1)×(2,3)θ̃j(ξ1,ξ2,t)dξ.
We express the first term of the right-hand side of the previous relation taking into account that for (ξ1,ξ2)∈(2,3)×(-1,1), we have χ(ξ2)=0:
(4.46)∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ=∫(0,3)×(-1,1)divξ((1-χ(ξ1))∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))dξ.
For obtaining the above equality we also used the property:
(4.47)divξ(∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))=0.
A similar result is obtained for the second term of the right-hand side and, comparing the expression of ∫ℬθ̃j(ξ1,ξ2,t)dξ with the right-hand side of (4.44), the proof is complete.The last result of this subsection gives the leading term of the regular part of the asymptotic solution corresponding toDɛ1.Proposition 4.4.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.48)u1,01(x1,ξ2,t)=-32(∫-11ψ(ξ2,t)dξ2)N1(ξ2),u2,01(x1,ξ2,t)=0,p01(x1,ξ2,t)=0,q01(x1,t)=∫0x1f(s,t)ds-32(∫-11ψ(ξ2,t)dξ2)∫0x1ν(s)ds+12β-1(t),
with β-1(t) defined by (4.37) and
(4.49)∂4d(±)01∂x14=g±±q01,d(±)01(0,t)=d(±)01(1,t)=∂d(±)01∂x1(0,t)=∂d(±)01∂x1(1,t)=0.


#### 4.2.2. The Solution of the Problems forγ=3

Unlike in the previous case, forγ=3 we cannot solve the problems corresponding to Dɛ1 and to Dɛ2 separately. We begin with the problems corresponding to the functions with the index 1. The relations (4.15)–(4.17) are still true. Hence, pj1 is also given by (4.17). The difference with respect to the previous case is that now we cannot express the other unknowns only in function of qj1, since in the right-hand side of (4.16) d(-)j1 is also unknown.We introduce the notations:(4.50)dj1(x1,t)=d(+)j1(x1,t)+d(-)j1(x1,t),𝒟j-11=𝒟(+)j-11+𝒟(-)j-11,
with 𝒟(±)j-11 given by (4.14)2. The function dj1 is obtained as the unique solution of a fourth-order differential problem:(4.51)∂4dj1∂x14=(g++g-)δj0+𝒟j-11,dj1(1,t)=-(d(+)j(1)(0,t)+d(-)j(1)(0,t)),∂dj1∂x1(1,t)=-(∂d(+)j+1(1)∂ξ1(0,t)+∂d(-)j+1(1)∂ξ1(0,t)),dj1(0,t)=2(∂d̃(+)j-1(0)∂ξ1(2,1,t)+∂d̃(-)j-1(0)∂ξ1(2,-1,t))-(d̃(+)j-1(0)(2,1,t)+d̃(-)j-1(0)(2,-1,t)),∂dj1∂x1(0,t)=-(∂d̃(+)j(0)∂ξ1(2,1,t)+∂d̃(-)j(0)∂ξ1(2,-1,t)),
the boundary conditions being obtained by means of (4.11) and (4.31). This means that d(-)j1 can also be expressed via d(+)j1. We show that all the unknowns of the regular part of the asymptotic solution can be expressed in function of d(+)j1; hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined if we obtain and solve a problem for d(+)j1. As we cannot obtain all the necessary conditions for d(+)j1, we have to consider the problem for the couple of unknowns, d(+)j1, d(+)j2.Theorem 4.5.
The approximations of the displacement,d(+)j1, d(+)j2, are obtained as the unique solution of the following system of two sixth-order parabolic equations:
(4.52)∂d(+)j1∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)j1∂x15(x1,t))=Gj1(x1,t)-∂H1∂x1(x1,t)δj0,∂d(+)j2∂t(x2,t)-13∂∂x2(1ν(x2)∂5d(+)j2∂x25(x2,t))=Gj2(x2,t)-∂H2∂x2(x2,t)δj0,d(+)j1(1,t)=-d(+)j(1)(0,t),∂d(+)j1∂x1(1,t)=-∂d(+)j+1(1)∂ξ1(0,t),d(+)j1(0,t)=2∂d̃(+)j-1(0)∂ξ1(2,1,t)-d̃(+)j(0)(2,1,t),∂d(+)j1∂x1(0,t)=-∂d̃(+)j(0)∂ξ1(2,1,t),d(+)j2(1,t)=-d(+)j(2)(0,t),∂d(+)j2∂x2(1,t)=-∂d(+)j+1(2)∂ξ2(0,t),d(+)j2(0,t)=2∂d̃(+)j-1(0)∂ξ2(1,2,t)-d̃(+)j(0)(1,2,t),∂d(+)j2∂x2(0,t)=-∂d̃(+)j(0)∂ξ2(1,2,t),∂5d(+)j1∂x15(1,t)=Lj1(t),∂5d(+)j2∂x25(1,t)=Lj2(t),∂4d(+)j1∂x14(0,t)-∂4d(+)j2∂x24(0,t)=Mj(t),∂5d(+)j1∂x15(0,t)+∂5d(+)j2∂x25(0,t)=Nj(t),d(+)j1(x1,0)=d(+)j2(x1,0)=0,
where Hi, Gji, Lji(t), Mj(t), and Nj(t) are known functions, given below.Proof.
The first relation between the unknownsd(+)j1 and qj1 is given by (4.12)4, that is,
(4.53)∂4d(+)j1∂x14-qj1=g+δj0+𝒟(+)j-11.
The second one is obtained from (4.16) for γ=3 and ξ2=1, (4.12)6 for ξ2=1 and (4.50)1:
(4.54)2∂d(+)j1∂t=∂dj1∂t+23∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))-∫-11D-2(∂∂x1(1ν(x1)Uj-11)dξ2.
Eliminating qj1 from (4.53) and (4.54) we get (4.52)1 with
(4.55)H1(x1,t)=13ν(x1)(∂∂x1g++f),Gj1(x1,t)=12(∂dj1∂t-∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2)-13∂∂x1(1ν(x1)∂∂x1𝒟(+)j-11).
As we can see,H1 depends only on the data and Gj1 depends on some functions determined at the previous approximations. Similar expressions are obtained for H2 and Gj2.
Relations (4.52)3,4 are in fact (4.11) and (4.31), respectively, and the same remark holds for (4.52)5,6. We obtain next (4.52)7 as follows: we derivate (4.53) with respect to x1 and we take x1=1; we also take x1=1 in (4.15) and we eliminate (∂qj1/∂x1)(1,t) from these two relations, by means of (4.9). The right-hand side of (4.52)7 is(4.56)Lj1(t)=(∂∂x1g+(1,t)+f(1,t)-32ν0∫-11ψdξ2)δj0+32∫-11D-2(Uj-11)(1,ξ2,t)dξ2+∂∂x1𝒟(+)j-11-3ν02ddt∫∞0(d(+)j-1(1)-d(-)j-1(1))(ξ1,t)dξ1
and a similar expression we obtain for Lj2(t). Relation (4.52)9 is a consequence of (4.34) for j+1 replaced by j, of (4.53) for x1=0 and of the corresponding relation for the index 2. The right hand side of (4.52)9 is
(4.57)Mj1(0,t)=αj-1(t)+𝒟(+)j-11(0,t)-𝒟(+)j-12(0,t).
Finally, (4.52)10 is a consequence of the compatibility condition for (4.32), obtained as follows: (4.25) can be written as
(4.58)θ̃j=θ̂j-1-divξ(χ(ξ1)u1,j1(ξ1,0,t)e1+χ(ξ2)u2,j2(0,ξ2,t)e2),
where
(4.59)θ̂j-1=-divξ(χ(ξ1)∑l=1jξ1l-1(l-1)!(ξ1l∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂l-1u2,j-l1∂x1l-1(0,ξ2,t)e2)+χ(ξ2)∑l=1jξ2l-1(l-1)!(∂lu1,j-l2∂x2l(ξ1,0,t)e1+ξ2l∂lu2,j-l2∂x2l(ξ1,0,t)e2))
from (4.32) we get
(4.60)γj(t):=∂qj1∂x1(0,t)+∂qj2∂x2(0,t)=-3ν02∫ℬθ̂j-1+3ν02∫2∞(d(+)j-1(0)(ξ1,1,t)-d(-)j-1(0)(ξ1,-1,t))dξ1+3ν02∫2∞(d(+)j-1(0)(1,ξ2,t)-d(-)j-1(0)(-1,ξ2,t))dξ2+32∫-11D-2(Uj-11)(0,ξ2,t)dξ2+32∫-11D-2(Uj-12)(ξ1,0,t)dξ1.
Next, we derivate (4.53), we take x1=0 and we add with the same computation for the index  2. Replacing the previous relation, we obtain (4.52)10, with(4.61)Nj(t)=γj(t)+∂∂x1(𝒟(+)j-1(0,t)+𝒟(-)j-1(0,t))
which completes the proof.The other components of the regular part of the asymptotic solution corresponding toDɛ1 have the following expressions:(4.62)d(-)j1(x1,t)=dj1(x1,t)-d(+)j1(x1,t),qj1(x1,t)=∂4d(+)j1∂x14(x1,t)-g+(x1,t)δj0+𝒟(+)j-11(x1,t),u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)),pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
As in the previous subsection, we discuss the expression of the leading term of the asymptotic solution.Proposition 4.6.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.63)u1,01(x1,ξ2,t)=1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t))N1(ξ2),u2,01(x1,ξ2,t)=-∂∂x1(1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t)))N2(ξ2),q01(x1,t)=∂4d(+)01∂x14(x1,t)-g+(x1,t),p01(x1,ξ2,t)=0,d(-)01(x1,t)=d01(x1,t)-d(+)01(x1,t),
with d01,d(+)01 satisfying the problems presented in the proof, (4.64) and (4.65).Proof.
The relations (4.63)1,2,3,4 are consequences of (4.15)–(4.17) and of (4.53) for j=0. It remains to show that d01,d(+)01,d(+)02 can be determined as unique solutions of some problems.
Forj=0 the problem for d01 becomes:(4.64)∂4d01∂x14(x1,t)=g+(x,t)+g-(x1,t),d01(1,t)=∂d01∂x1(1,t)=0,d01(0,t)=∂d01∂x1(0,t)=0.
Finally, the problem for d(+)01,d(+)02 is the following:
(4.65)∂d(+)01∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)01∂x15(x1,t))=12∂d01∂t-13∂∂x1(1ν(x1)(∂∂x1g++f)),∂d(+)02∂t(x2,t)-13∂∂t(1ν(x2)∂5d(+)02∂x25(x2,t))=12∂d02∂x2-13∂∂x2(1ν(x2)(∂∂x2g++f)),d(+)01(1,t)=∂d(+)01∂x1(1,t)=d(+)01(0,t)=∂d(+)01∂x1(0,t)=0,d(+)02(1,t)=∂d(+)02∂x2(1,t)=d(+)02(0,t)=∂d(+)02∂x2(0,t)=0,∂5d(+)01∂x15(1,t)=-3ν02∫-11ψ(ξ2,t)dξ2+f(1,t)+∂∂x1g+(1,t),∂5d(+)02∂x25(1,t)=-3ν02∫-11ψ(ξ1,t)dξ1+f(1,t)+∂∂x2g+(1,t),∂4d(+)01∂x14(0,t)-∂4d(+)02∂x24(0,t)=0,∂5d(+)01∂x15(0,t)+∂5d(+)02∂x25(0,t)=0,d(+)01(x1,0)=d(+)02(x2,0)=0.At the end of this subsection we notice that the system (4.52) can be solved also for j replaced by j+1 which gives, by means of (4.53), qj+11(0,t), qj+12(0,t). Hence, the system (4.23) can be solved, and all the jth approximation is determined for γ=3.


#### 4.2.3. Conclusions

In the previous two subsections we determined thejth approximation of the asymptotic solution for two different cases: γ>3 and γ=3. For γ>3 we can separate the problems for the two branches of the elastic tube and we obtain the regular part of the asymptotic solution (for D1ɛ) as follows:(i)
the macroscopic variableqj1 is the unique solution of the problem:
(4.66)∂∂x1(1ν(x1)(∂qj1∂x1))=ajq(x1,t),∂qj1∂x1(1,t)=bjq(t),qj1(0,t)=cjq(t),
with ajq,bjq,cjq known functions;(ii)
the macroscopic variablesd(±)j1 are the unique solutions of the fourth-order differential problems:
(4.67)∂4d(+)j1∂x14=ajd(x1,t),d(+)j1(0,t),d(+)j1(1,t),∂d(+)j1∂x1(0,t),∂d(+)j1∂x1(1,t)given,
and a similar one for d(-)j1;(iii)
u1,j1,u2,j1,pj1 given by (4.15)–(4.17). The main term of the regular part of the asymptotic solution is presented in (4.48)-(4.49). Unlike in the case γ>3, for γ=3 the problems must be solved for the two branches together, since they cannot be uncoupled;(iv)
we first introduce the auxiliary functionsdj1,dj2, with dj1 the unique solution of (4.51); then the regular part of the asymptotic solution is obtained as follows;(v)
the macroscopic variablesd(+)j1,d(+)j2 are obtained as the unique solution of the sixth-order parabolic system (4.52);(vi)
the macroscopic variablesqj1,qj2 are given by (4.53) and the similar equation for the index 2;(vii)
u1,j1,u2,j1,pj1 given also by (4.15)–(4.17), but we obtain different expressions.The main term of the regular part of the asymptotic solution is given by (4.63) and (4.65).


## 4.1. Construction of the Asymptotic Solution

In the sequel we introduce the second small parameterδ=12/h3E and we take δ=ɛγ, with γ∈ℕ, γ≥3. The asymptotic solution approximating the periodic flow in an infinite rectangle (see [1, Section  5]) is modified by using two types of correctors: the first type corresponds to the boundary conditions on x1=1 and x2=1 and the second one represents the boundary layer functions in a neighborhood of (x1,x2)=0. So, the asymptotic solution contains the regular part of the solution (which has two terms, due to the form of the flow domain), two boundary layer functions corresponding to x1=1 and to x2=1, respectively, and the correctors in (x1,x2)=0. Since the terms of the asymptotic solution should vanish in different parts of the flow domain, we multiply them with the truncation functions defined as follows: χ, η:ℝ↦ℝ, χ,η∈𝒞∞(ℝ),(4.3)χ(τ)={0,|τ|<2,1,|τ|>3,η(τ)={0,|τ|<34,1,|τ|>78.We construct the asymptotic solution as below:(4.4)ûa(k)(x1,x2,t)=u1(k)(x1,x2ɛ,t)χ(x1ɛ)+u2(k)(x1ɛ,x2,t)χ(x2ɛ)+ubl(k)1(x1-1ɛ,x2ɛ,t)η(x1)+ubl(k)2(x1ɛ,x2-1ɛ,t)η(x2)+ubl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2),p̂a(k)(x1,x2,t)=p1(k)(x1,x2ɛ,t)χ(x1ɛ)+p2(k)(x1ɛ,x2,t)χ(x2ɛ)+pbl(k)1(x1-1ɛ,x2ɛ,t)η(x1)+pbl(k)2(x1ɛ,x2-1ɛ,t)η(x2)+pbl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2),d̂±a(k)(x1,x2,t)=d±1(k)(x1,t)χ(x1ɛ)+d±2(k)(x2,t)χ(x2ɛ)+d±bl(k)1(x1-1ɛ,t)η(x1)+d±bl(k)2(x2-1ɛ,t)η(x2)+d±bl(k)0(x1ɛ,x2ɛ,t)η(1-x1)η(1-x2).Due to the definition of the truncation functions, we notice that inℬɛ∩{1/4≤xi≤3/4}, that is, in a neighborhood of the region of variable viscosity, the asymptotic solution reduces to its regular part, (ui(k),pi(k),d±i(k)), in a neighborhood of {xi=1} it reduces to (ui(k)+ubl(k)i,pi(k)+pbl(k)i,d±i(k)+d±bl(k)i), i=1,2, while in Dɛr the asymptotic solution is equal to the corrector in x=0,(ubl(k)0,pbl(k)0,d±bl(k)0). This means that the regular part of the asymptotic solution (which can be computed and has a very simple expression) represents an approximation for the exact solution in all the flow domain except some neighborhoods of x1=1, x2=1 and x=0.We give next the expressions of the three components of the asymptotic solution and we explain their role in this construction.


### 4.1.1. The Regular Part of the Asymptotic Solution

For each rectangleDɛ1 and Dɛ2 we define the corresponding regular part of the asymptotic expansion as in [1]. The regular part corresponding to Dɛ1 has the expression:(4.5)u1(k)(x1,x2ɛ,t)=∑j=0kɛj+2u1,j1(x1,x2ɛ,t)e1+∑j=0kɛj+3u2,j1(x1,x2ɛ,t)e2,p1(k)(x1,x2ɛ,t)=∑j=0kɛj+1pj1(x1,x2ɛ,t)+∑j=0kɛjqj1(x1,t),d±1(k)(x1,t)=∑j=0kɛj+γd(±)j1(x1,t).
It represents the solution of the problem (2.6)1,2,3,11 set in the infinite rectangle in Ox1 direction, (-∞,∞)×(-ɛ,ɛ); the functions which appear in (4.5) can be computed explicitly and represent a good approximation of the exact solution of (2.6) in a neighborhood of the region with variable viscosity of Dɛ1 (as we will prove in the last section).In a similar way we introduce the regular part of the asymptotic solution corresponding toDɛ2.Hence, the sum of the first two terms of the asymptotic solution represents a good approximation of the exact solution of (2.6) in some neighborhood of the region of variable viscosity of the flow domain, but it is not close to it on the inflow/outflow boundaries and in Dɛr. Since the purpose of the asymptotic construction is to approximate the exact solution with a small error between the exact and asymptotic solution, we modify and complete the regular part by two types of boundary layer correctors.


### 4.1.2. The Boundary Layer Correctors forx1=1 and x2=1

These boundary layer functions are introduced in order to repair the traces of the regular part of the asymptotic solution onx1=1 and x2=1. They are given by:(4.6)ubl(k)i(xɛ,t)=∑j=0kɛj+2uj(i)(xɛ,t),pbl(k)i(xɛ,t)=∑j=0kɛj+1pj(i)(xɛ,t),d±bl(k)i(xiɛ,t)=∑j=0kɛj+γdj(i)(xiɛ,t),i∈{1,2}.
The corrector with i=1 corresponds to the end x1=1 and that with i=2 corresponds to x2=1.From the definition of the truncation functions and of the asymptotic solution we notice that the influence of each corrector defined above is significant only near the corresponding end of the flow domain.


### 4.1.3. The Boundary Layer Corrector inx=0

This corrector is necessary in order to realize the junction between the two parts of the asymptotic solution, corresponding to the two branches of the flow domain and to obtain the conditions on the rigid boundaries of the domain. The expressions of the correctors inx=0 are given by:(4.7)ubl(k)0(xɛ,t)=∑j=-1kɛj+2uj(0)(xɛ,t),pbl(k)0(xɛ,t)=∑j=-1kɛj+1pj(0)(xɛ,t),d±bl(k)0(xɛ,t)=∑j=0kɛj+γd(±)j(0)(xɛ,t).
From the definition of the truncation function η it follows that the corrector in x=0 appears in the expression of the asymptotic solution (4.4) only in ℬɛ∩{x1≤1/4,x2≤1/4}, which represents a neighborhood of Dɛr with constant viscosity and where the forces are equal to zero.Remark 4.1.
The boundary layer method is close to the well-known method of matching of asymptotic expansions (see [14]). However there is a difference between these two methods. In the boundary layer method the regular expansion is reexpanded with respect to the fast variable (as in the matching method), but then the new expansion is multiplied by a cutting function. This product being inserted into the equation gives a discrepancy in the right-hand side which is then compensated by an appropriate boundary layer corrector. So, we get finally a unique expression for the asymptotic approximation in each point of the domain. In the matching approach the reexpanded regular ansatz (outer expansion) coexists with the so-called inner expansion defined near the boundary, so that there are several overlapping expressions for an asymptotic approximation in different parts of the domain. We emphasize that this difference is not too important and corresponds more to the form of presentation of the result.


## 4.1.1. The Regular Part of the Asymptotic Solution

For each rectangleDɛ1 and Dɛ2 we define the corresponding regular part of the asymptotic expansion as in [1]. The regular part corresponding to Dɛ1 has the expression:(4.5)u1(k)(x1,x2ɛ,t)=∑j=0kɛj+2u1,j1(x1,x2ɛ,t)e1+∑j=0kɛj+3u2,j1(x1,x2ɛ,t)e2,p1(k)(x1,x2ɛ,t)=∑j=0kɛj+1pj1(x1,x2ɛ,t)+∑j=0kɛjqj1(x1,t),d±1(k)(x1,t)=∑j=0kɛj+γd(±)j1(x1,t).
It represents the solution of the problem (2.6)1,2,3,11 set in the infinite rectangle in Ox1 direction, (-∞,∞)×(-ɛ,ɛ); the functions which appear in (4.5) can be computed explicitly and represent a good approximation of the exact solution of (2.6) in a neighborhood of the region with variable viscosity of Dɛ1 (as we will prove in the last section).In a similar way we introduce the regular part of the asymptotic solution corresponding toDɛ2.Hence, the sum of the first two terms of the asymptotic solution represents a good approximation of the exact solution of (2.6) in some neighborhood of the region of variable viscosity of the flow domain, but it is not close to it on the inflow/outflow boundaries and in Dɛr. Since the purpose of the asymptotic construction is to approximate the exact solution with a small error between the exact and asymptotic solution, we modify and complete the regular part by two types of boundary layer correctors.


## 4.1.2. The Boundary Layer Correctors forx1=1 and x2=1

These boundary layer functions are introduced in order to repair the traces of the regular part of the asymptotic solution onx1=1 and x2=1. They are given by:(4.6)ubl(k)i(xɛ,t)=∑j=0kɛj+2uj(i)(xɛ,t),pbl(k)i(xɛ,t)=∑j=0kɛj+1pj(i)(xɛ,t),d±bl(k)i(xiɛ,t)=∑j=0kɛj+γdj(i)(xiɛ,t),i∈{1,2}.
The corrector with i=1 corresponds to the end x1=1 and that with i=2 corresponds to x2=1.From the definition of the truncation functions and of the asymptotic solution we notice that the influence of each corrector defined above is significant only near the corresponding end of the flow domain.


## 4.1.3. The Boundary Layer Corrector inx=0

This corrector is necessary in order to realize the junction between the two parts of the asymptotic solution, corresponding to the two branches of the flow domain and to obtain the conditions on the rigid boundaries of the domain. The expressions of the correctors inx=0 are given by:(4.7)ubl(k)0(xɛ,t)=∑j=-1kɛj+2uj(0)(xɛ,t),pbl(k)0(xɛ,t)=∑j=-1kɛj+1pj(0)(xɛ,t),d±bl(k)0(xɛ,t)=∑j=0kɛj+γd(±)j(0)(xɛ,t).
From the definition of the truncation function η it follows that the corrector in x=0 appears in the expression of the asymptotic solution (4.4) only in ℬɛ∩{x1≤1/4,x2≤1/4}, which represents a neighborhood of Dɛr with constant viscosity and where the forces are equal to zero.Remark 4.1.
The boundary layer method is close to the well-known method of matching of asymptotic expansions (see [14]). However there is a difference between these two methods. In the boundary layer method the regular expansion is reexpanded with respect to the fast variable (as in the matching method), but then the new expansion is multiplied by a cutting function. This product being inserted into the equation gives a discrepancy in the right-hand side which is then compensated by an appropriate boundary layer corrector. So, we get finally a unique expression for the asymptotic approximation in each point of the domain. In the matching approach the reexpanded regular ansatz (outer expansion) coexists with the so-called inner expansion defined near the boundary, so that there are several overlapping expressions for an asymptotic approximation in different parts of the domain. We emphasize that this difference is not too important and corresponds more to the form of presentation of the result.


## 4.2. The Determination of the Asymptotic Solution

This subsection is devoted to the resolution of the problems satisfied by the three different components of the asymptotic solution presented in the previous subsection. Since some computations are different with respect to the values ofγ, we will analyse the problems and the order of solving them for γ>3 and for γ=3. Moreover, in each case we will specify the leading term of the asymptotic expansion.We begin the approach with the problems for the boundary layer correctors corresponding tox1=1 and x2=1 since the study of these problems is the same both for γ>3 and for γ=3. To fix the ideas, we obtain in the sequel the problems for the corrector corresponding to the end x1=1. As we noticed before, the term containing this corrector is not equal to zero only in a neighborhood of the boundary x1=1. So, the problems and the other relations corresponding to this corrector are obtained substituting the asymptotic solution in (2.6)1,2,3,6,7,11. We obtain two separate problems: one for the velocity-pressure correctors and the other one for the displacements correctors.Since in this neighborhood the viscosity is constant, the problem for(uj(1),pj(1)) has constant coefficients. Denoting by Π1 the semi-infinite rectangle (-∞,0)×(-1,1) and imposing for the velocity and for the pressure the condition of decay at -∞, we obtain for (uj(1),pj(1)) the problem:(4.8)-ν0Δξξuj(1)+∇ξpj(1)=-ρf∂uj-2(1)∂tinΠ1×(0,T),divξuj(1)=0inΠ1×(0,T),uj(1)(ξ1,±1,t)=∂d(±)j+2-γ(1)∂t(ξ1,t)e2in(-∞,0)×(0,T),uj(1)(0,ξ2,t)=ψ(ξ2,t)δj0e1-u1,j1(1,ξ2,t)e1-u2,j-11(1,ξ2,t)e2in(-1,1)×(0,T),uj(1)→0,pj(1)→0,uniformlywhenξ1→-∞.
The compatibility condition for (4.8) reads:(4.9)∫-11u1,j1(1,ξ2,t)dξ2=ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1+δj0∫-11ψ(ξ2,t)dξ2.
For any γ≥3 the right-hand side of (4.8)3 is known; so the boundary layer correctors for the velocity and for the pressure corresponding to x1=1 are uniquely determined from (4.8) (see [2, Section  4]). The condition (4.9) represents a relation for the regular part of the asymptotic solution.The boundary layer correctors for the displacements exponentially stabilizing to zero at-∞ are obtained as the unique solution of the following problems:(4.10)∂4d(±)j(1)∂ξ14=-ρh∂2d(±)j-4-γ(1)∂t2-μ∂5d(±)j-γ(1)∂ξ14∂t±pj-5(1)|ξ2=±1in(-∞,0)×(0,T),∂ad(±)j(1)∂ξ1a→0,uniformly,whenonlyξ1→-∞,a∈{0,1,2,3}.Since at the stepj the problem (4.10) gives both d(±)j(1) and d(±)j+1(1), introducing the asymptotic solution into (2.6)7 we obtain two boundary conditions for the regular part of the asymptotic solution for the displacements:(4.11)d(±)j1(1,t)=-d(±)j(1)(0,t),∂d(±)j1∂x1(1,t)=-∂d(±)j+1(1)∂ξ1(0,t).In a similar way, we obtain the boundary layer correctors corresponding to the endx2=1. The boundary layers for the velocity-pressure are defined on Π2×(0,T), with Π2=(-1,1)×(-∞,0), and the boundary layers for the displacements are defined also on (-∞,0)×(0,T).We study next the problems for the regular parts of the asymptotic solution. The results are obtained for the regular part corresponding toDɛ1; the regular part corresponding to Dɛ2 may be obtained from the previous with some obvious changes.Introducing (4.5) into (2.6)1,2,3,11 and collecting together the terms of the same order with respect to ɛ we are leaded to consider the following problem for (u1,j1,u2,j1,pj1,qj1,d(±)j1):(4.12)-ν(x1)∂2u1,j1∂ξ22+∂qj1∂x1=fδj0-ρf∂u1,j-21∂t+2∂∂x1(ν(x1)∂u1,j-21∂x1)+ν(x1)∂2u2,j-21∂ξ2∂x1-∂pj-11∂x1,∂pj1∂ξ2=-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22,∂u1,j1∂x1+∂u2,j1∂ξ2=0in(0,1)×(-1,1)×(0,T),∂4d(+)j1∂x14-qj1=g+δj0+pj-11|ξ2=1-ρh∂2d(+)j-γ1∂t2-μ∂5d(+)j-γ1∂x14∂tin(0,1)×{-1}×(0,T),∂4d(-)j1∂x14+qj1=g-δj0-pj-11|ξ2=-1-ρh∂2d(-)j-γ1∂t2-μ∂5d(-)j-γ1∂x14∂tin(0,1)×{1}×(0,T),uj1(x1,±1,t)=∂d(±)j-γ+31∂t(x1,t)e2in(0,1)×{±1}×(0,T).
The two cases, γ>3 and γ=3, appear because of the last relation of the previous system. We can see, indeed, that for γ>3 the unknown of this relation is uj1, while for γ=3, (4.12)6 contains two unknowns. From this point, the computations are different with respect to the values of γ.We introduce the functions:(4.13)N1(ξ2)=12(ξ22-1),N2(ξ2)=∫-1ξ2N1(τ)dτ,
with the properties: N1′′=1,N1(±1)=0 and N2(1)=-2/3.We also use the notations:(4.14)Uj-11(x1,ξ2,t)=ρf∂u1,j-21∂t-2∂∂x1(ν(x1)∂u1,j-21∂x1)-ν(x1)∂2u2,j-21∂ξ2∂x1+∂pj-11∂x1,𝒟(±)j-11(x1,t)=±pj-11|ξ2=1-ρh∂2d(±)j-γ1∂t2-μ∂5d(±)j-γ1∂x14∂t,D-1:F→∫-1ξ2F(x1,τ)dτ,D-2:F→∫-1ξ2∫-1θF(x1,τ)dτdθ-12(ξ2+1)∫-11∫-1θF(x1,τ)dτdθ,Ix1-k:F→∫0x1∫0ak-1⋯∫0a1F(s,ξ2,t)ds,Jx1:F→Ix1-4(F)+x13(2∫01Ix1-3(F)-∫01Ix1-2(F))-x2(3∫01Ix1-3(F)-∫01Ix1-2(F)).


### 4.2.1. The Order of Solving the Problems forγ>3

The regular part corresponding toDɛ1 is computed by integrating (4.12), as stated below.Proposition 4.2.
The unknownsuj1,qj1,pj1,d(±)j1 are determined from (4.12), up to nine functions of t.Proof.
Integrating twice (4.12)1 from −1 to ξ2 and using the boundary conditions (4.12)6 we get:
(4.15)u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),
which contains as unknowns u1,j1 and qj1. The other functions contained by this relation are either known from previous computations or equal to zero.
We integrate next the incompressibility condition (4.12)3 with respect to ξ2 with the boundary condition (4.12)6 for ξ2=-1 and we obtain(4.16)u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)).
The previous two relations give the components of the velocity uj1 with respect to qj1.
The pressure approximations are determined from (4.12)2, supposing that the integration functions, depending on x1, t are equal to zero, since we consider that any function depending only on x1, t could be contained in qj+11. 											(4.17)pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
Taking ξ2=1 in (4.16) and using the boundary condition (4.12)6 for ξ2=1 we obtain the following second order differential equation for the function qj1:
(4.18)∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))=32∂∂t(d(+)j-γ+31-d(-)j-γ+31)+32∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2.
Integrating (4.18) from x1 to 1, we express ∂qj1/∂x1 by means of (∂qj1/∂x1)(1,t), which represents the only unknown of this expression. This function of t is obtained as follows: we take x1=1 in (4.15) and we introduce the result into (4.9). Hence, we determined the expression of ∂qj1/∂x1 in (0,1)×(0,T), which is:
(4.19)1ν(x1)(∂qj1∂x1-fδj0)=-32δj0∫-11ψ(ξ2,t)dξ2-32ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1-32∂∂t∫x11(d(+)j-γ+31-d(-)j-γ+31)(s,t)ds+32∫-11D-2(1ν(x1)Uj-11)dξ2.
Introducing (4.19) into (4.15) and (4.18) into (4.16) we determine uj1 in (0,1)×(-1,1)×(0,T).
We integrate nextν(x1) · (4.19) from 0 to x1 and we get qj1 determined up to the function of t, qj1(0,t):
(4.20)qj1(x1,t)=δj0∫0x1f(s,t)ds-32∫0x1ν(θ)∫θ1∂∂t(d(+)j-γ+31-d(-)j-γ+31)(s,t)dsdθ-32(ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1+δj0∫-11ψ(ξ2,t)dξ2)(∫0x1ν(s)ds)+32∫0x1∫-11D-2(Uj-11(s,ξ2,t))dξ2ds+qj1(0,t).
The functions d(±)j1 satisfy the fourth-order differential equations:
(4.21)∂4d(±)j1∂x14=±qj1+g±δj0+𝒟(±)j-11,
with qj1 given by (4.20). Writting qj1 as Qj1(x1,t)+qj1(0,t) and integrating four times (4.21) with respect to x1 we obtain the following expressions for d(±)j1:
(4.22)d(+)j1(x1,t)=d(+)j1(0,t)(1-3x12+2x13)+∂d(+)j1∂x1(0,t)(x1-2x12+x13)+d(+)j1(1,t)(3x12-2x13)+∂d(+)j1∂x1(1,t)(-x12+x13)+qj1(0,t)(x1224-x1312+x1424)+Jx1(Qj1)+Jx1(g+)δj0+Jx1(𝒟(+)j-11),d(-)j1(x1,t)=d(-)j1(0,t)(1-3x12+2x13)+∂d(-)j1∂x1(0,t)(x1-2x12+x13)+d(-)j1(1,t)(3x12-2x13)+∂d(-)j1∂x1(1,t)(-x12+x13)-qj1(0,t)(x1224-x1312+x1424)-Jx1(Qj1)+Jx1(g-)δj0+Jx1(𝒟(-)j-11).
Hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined up to the functions qj1(0,t), d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t), which achieves the proof.In a similar way we express the regular part of the asymptotic solution corresponding toDɛ2 depending on 9 unknown functions of t.We continue the construction with the problems for the corrector inx=0. As we previously said, the term of the asymptotic solution containing this corrector is not equal to zero only in a neighborhood of Dɛr. In this neighborhood, the expression of the asymptotic solution reduces to (u1(k)(x1,ξ2,t)χ(ξ1)+u2(k)(ξ1,x2,t)χ(ξ2)+ubl(k)0(ξ,t),p1(k)(x1,ξ2,t)χ(ξ1)+p2(k)(ξ1,x2,t)χ(ξ2)+pbl(k)0(ξ,t),d±1(k)(x1,t)χ(ξ1)+d±2(k)(x2,t)χ(ξ2)+d±bl(k)0(ξ,t)). For obtaining the problems satisfied by the correctors in x=0, we introduce the previous expression of the asymptotic solution in (2.6)1,2,3,4,5,6,11, with ν=ν0 and f=g=0 in (2.6)1,3; for derivating the terms which contain the two types of variables xi and ξi we proceed as follows: we replace ∂/∂xi by (1/ɛ)(∂/∂ξi), we replace xi by ɛξi and we expand the functions as a Taylor expansion with respect to xi=ɛξi. We introduce the notations ℬ=Dr∪{(ξ1,ξ2):ξ1≥2,ξ2∈(-1,1)}∪{(ξ1,ξ2):ξ1∈(-1,1),ξ2≥2},A±B±⌢=(1/ɛ)A±εB±ε⌢ and we obtain for (uj(0),pj(0)) the following nondivergence free problem:(4.23)-ν0Δξξuj(0)+∇ξpj(0)=-ρf∂uj-2(0)∂t+Fj+(1-χ(ξ1))ξ1′qj+11(0,t)e1+(1-χ(ξ2))ξ2′qj+12(0,t)e2inℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0uniformlywhenξ1→∞,uj(0)→0uniformlywhenξ2→∞,
for j≥0 and(4.24)-ν0Δξξu-1(0)+∇ξp-1(0)=(1-χ(ξ1))ξ1′q01(0,t)e1+(1-χ(ξ2))ξ2′q02(0,t)e2inℬ,divu-1(0)=0inℬ,u-1(0)=0onA+B+⌢∪A-B-⌢,u-1(0)(ξ1,±1,t)=0forξ1≥2,u-1(0)(±1,ξ2,t)=0forξ2≥2,u-1(0)→0uniformlywhenξ1→∞,u-1(0)→0uniformlywhenξ2→∞.
Unlike the problem (4.8) which give the correctors for x1=1, the problems (4.23), (4.24) have unknown right hand sides. The functions Fj are known, but qj+11(0,t) and qj+12(0,t) are elements from the next approximation, so they are unknown. The function θ̃j is given by:(4.25)θ̃j=-divξ(χ(ξ1)∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2)+χ(ξ2)∑l=0jξ2ll!(∂lu1,j-1-l2∂x2l(ξ1,0,t)e1+∂lu2,j-l2∂x2l(ξ1,0,t)e2)),
and it has to satisfy a compatibility condition.The problem (4.24) has the unique solution (with the pressure unique up to an additive function of t)(4.26)u-1(0)=0,p-1(0)={(1-χ(ξ1))q01(0,t)forξ1≥2,q01(0,t)=q02(0,t)inDr,(1-χ(ξ2))q02(0,t)forξ2≥2.
We obtain next the problems for the correctors in x=0 corresponding to the displacements. We first notice that(4.27)d(±)j(0)=0onA±B±⌢.
Let us denote by d̂(±)j-1(0) the unique solution of(4.28)∂4d̂(±)j-1(0)∂ξ14=ρh(1-χ(ξ1))+∑l=0jξ1ll!∂l+2d(±)j-γ-4-l1∂t2∂x1l(0,t)+∑l=2j∂4ξ14((1-χ(ξ1))ξ1ll!)∂ld(±)j-l1∂x1l(0,t)+μ∑l=0j∂4ξ14((1-χ(ξ1))ξ1ll!)∂l+1d(±)j-γ-l1∂t∂x1l(0,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lpj-5-l1∂x1l(0,±1,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lqj-4-l1∂x1l(0,t),∂ad̂(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3},
and by d̃(±)j-1(0) the unique solution of(4.29)∂4d̃(±)j-1(0)∂ξ14=∂4d̂(±)j-1(0)∂ξ14-ρh∂2d(±)j-γ-4(0)∂t2-μ∂5d(±)j-γ(0)∂ξ14∂t+pj-5/ξ2=±1(0),∂ad̃(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3}.
Introducing the asymptotic solution into (2.6)3 and identifying the coefficients of the power of ɛ we obtain:(4.30)d(±)j(0)(ξ1,±1,t)=d̃(±)j-1(0)(ξ1,±1,t)-(1-χ(ξ1))×(d̃(±)j-1(0)(2,±1,t)-(2-ξ1)∂d̃(±)j-1(0)∂ξ1(2,±1,t))ξ1>2,(4.31)d(±)j1(0,t)=2∂d̃(±)j-1(0)∂ξ1(2,±1,t)-d̃(±)j-1(0)(2,±1,t),∂d(±)j1∂x1(0,t)=-∂d̃(±)j(0)∂ξ1(2,±1,t).
The relation (4.30) gives the corrector in x=0 for the displacements corresponding to one branch of the flow domain.We notice that the right-hand side of (4.31) is known, since the problems (4.28) and (4.29) can be solved also if we replace j-1 by j.We are now in a position to return to the regular part of the asymptotic solution. For each regular part of the asymptotic solution, we need 9 relations, in order to determine the 9 unknown functions of integration. For the regular part corresponding toDɛ1, (4.11) and (4.31) give 8 functions: d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t). It remains undetermined qj1(0,t). We show next that at the jth step this function is already determined at the previous step. For this purpose, we return to the problem (4.23). As usual, we should construct boundary layer functions stabilizing to zero at infinity. As we imposed this condition for the velocity in (4.23)6,7, it is known that the pressure stabilizes to some functions of t. At the jth step, we determine qj+11(0,t) and qj+12(0,t) from the condition of the exponential decay of pj(0) at infinity. To this end, we consider instead of (4.23) and (4.24), the following problem with known right-hand side for j≥-1:(4.32)-ν0Δξξuj(0)+∇ξp¯j(0)=-ρf∂uj-2(0)∂t+Fjinℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0,p¯j(0)→αj1(t)uniformlywhenξ1→∞,uj(0)→0,p¯j(0)→αj2(t)uniformlywhenξ2→∞
with αj1(t),αj2(t) unknown. Supposing for the time being that the function θ̃j satisfies the compatibility condition, it follows that (4.32) has a unique solution (the uniqueness for the pressure being understood up to an additive function of t). This means that the function αj(t), with αj(t)=αj1(t)-αj2(t) is fixed. We define for j≥-1(4.33)pj(0)=p¯j(0)+χ(ξ1)qj+11(0,t)+χ(ξ2)qj+12(0,t).
Standard computations show that (uj(0),p¯j(0)) is solution for (4.32) iff (uj(0),pj(0)) satisfies (4.23) or (4.24) for j=-1 with pj(0)→αji(t)+qj+1i(0,t) uniformly when ξi→∞,i=1,2. To obtain the desired behavior for pj(0) when ξi→∞,i=1,2, it suffices to take αji(t)=-qj+1i(0,t).A first equation for the unknownsqj+11(0,t) and qj+12(0,t) is(4.34)qj+11(0,t)-qj+12(0,t)=αj(t),
with α-1(t)=0. The other equation is obtained introducing the asymptotic solution into the compatibility condition (2.11):(4.35)∫01(d(+)j+11(x1,t)-d(-)j+11(x1,t))dx1+∫01(d(+)j+12(x2,t)-d(-)j+12(x2,t))dx2=-∑l=0j(∂l(d(+)j-l1-d(-)j-l1)∂x1l(0,t)+∂l(d(+)j-l2-d(-)j-l2)∂x2l(0,t))∫03τll!(χ(τ)-1)dτ-∫-∞0(d(+)j(1)(ξ1,1,t)-d(+)j(1)(ξ1,-1,t))dξ1-∫-∞0(d(+)j(2)(1,ξ2,t)-d(+)j(2)(-1,ξ2,t))dξ2-∫2∞(d(+)j(0)(ξ1,1,t)-d(+)j(0)(ξ1,-1,t))dξ1-∫2∞(d(+)j(0)(1,ξ2,t)-d(+)j(0)(-1,ξ2,t))dξ2.
The right-hand side of (4.35) is known from the jth approximation. Replacing in the left hand side of (4.35) d(±)j+11 given by (4.22) (for j replaced by j+1) and d(±)j+12 given by a similar relation, we are leaded to the second equation for the unknowns qj+11(0,t) and qj+12(0,t):(4.36)qj+11(0,t)+qj+12(0,t)=βj(t),
with βj a known function for j≥-1, with(4.37)β-1(t)=-6!(∫01Ix1-1(f)dx1+∫01Ix2-1(f)dx2)+3×6!2(∫-11ψ(ξ2,t)dξ2∫01Ix1-1(ν)dx1+∫-11ψ(ξ1,t)dξ1∫01Ix2-1(ν)dx2)-6!2(∫01Ix1-4(g+-g-)dx1+∫01Ix2-4(g+-g-)dx2)+6!4(∫01Ix1-3(g+-g-)dx1+∫01Ix2-3(g+-g-)dx2)-6!4!(∫01Ix1-2(g+-g-)dx1+∫01Ix2-2(g+-g-)dx2).
From (4.34) and (4.36) we determine qj+11(0,t) and qj+12(0,t), which means that the right-hand side of (4.23) is now known, that allows to solve (4.23), if we show that the problem (4.23) is well posed, that is, the compatibility condition is fulfilled. Moreover, this justifies the previous assertion that qji(0,t),i=1,2 are computed at the (j-1)th step. We can say that the entire jth approximation was obtained for γ>3 if we prove.Proposition 4.3.
For anyj∈ℕ, the function θ̃j satisfies the compatibility condition:
(4.38)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫∂ℬuj(0)·nds.Proof.
From (4.23) and (4.35) it follows:
(4.39)∫∂ℬuj(0)·nds=-ddt(∫-∞0(d(+)j+2-γ(1)(ξ1,t)-d(-)j+2-γ(1)(ξ1,t))dξ1-∫-∞0(d(+)j+2-γ(2)(ξ2,t)-d(-)j+2-γ(2)(ξ2,t))dξ2)-ddt∑l=0j+2-γ(∂l(d(+)j+2-γ-l1-d(-)j+2-γ-l1)∂x1l(0,t)+∂l(d(+)j+2-γ-l2-d(-)j+2-γ-l2)∂x2l(0,t))∫03sll!(χ(s)-1)ds-ddt(∫01(d(+)j+3-γ1(x1,t)-d(-)j+3-γ1(x1,t))dx1+∫01(d(+)j+3-γ2(x2,t)-d(-)j+3-γ2(x2,t))dx2).
We express next the three terms of the right-hand side of the previous equality with respect to the components of the velocity corresponding to the regular parts of the asymptotic solution.
From (4.9), the corresponding relation for x2=1 and (2.9)6 the first term of the right-hand side is equal to(4.40)T1=-∫-11u1,j1(1,ξ2,t)dξ2-∫-11u2,j2(ξ1,1,t)dξ1.
The third term is expressed by means of (4.12)6 and its analogous for uj2 and it is equal to
(4.41)T3=-∫01(u2,j1(x1,1,t)-u2,j1(x1,-1,t))dx1-∫01(u1,j2(1,x2,t)-u1,j2(-1,x2,t))dx2.
Computing now T1+T3 and using (4.12)3 and the similar one for uj2 we get
(4.42)T1+T3=-∫-11u1,j1(0,ξ2,t)dξ2-∫-11u2,j2(ξ1,0,t)dξ1.
The second term is given by
(4.43)T2=∑l=0j(∂l(u2,j-1-l1(0,1,t)-u2,j-1-l1(0,-1,t))∂x1l+∂l(u1,j-1-l2(1,0,t)-u1,j-1-l2(-1,0,t))∂x2l)∫03sll!(χ(s)-1)ds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ2,0,t)∂x2le2))dξ+∫-11u1,j1(0,ξ2,t)dξ2+∫-11u2,j2(ξ1,0,t)dξ2.
Finally, computing T1+T2+T3, we obtain the following expression of the right-hand side of (4.38):
(4.44)∫∂ℬuj(0)·nds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ1,0,t)∂x2le2))dξ.
We compute next the left-hand side of (4.38). Since χ(ξ1)=χ(ξ2)=0 on ℬ∩{ξ1<2,ξ2<2}, we can write:
(4.45)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ+∫(-1,1)×(2,3)θ̃j(ξ1,ξ2,t)dξ.
We express the first term of the right-hand side of the previous relation taking into account that for (ξ1,ξ2)∈(2,3)×(-1,1), we have χ(ξ2)=0:
(4.46)∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ=∫(0,3)×(-1,1)divξ((1-χ(ξ1))∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))dξ.
For obtaining the above equality we also used the property:
(4.47)divξ(∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))=0.
A similar result is obtained for the second term of the right-hand side and, comparing the expression of ∫ℬθ̃j(ξ1,ξ2,t)dξ with the right-hand side of (4.44), the proof is complete.The last result of this subsection gives the leading term of the regular part of the asymptotic solution corresponding toDɛ1.Proposition 4.4.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.48)u1,01(x1,ξ2,t)=-32(∫-11ψ(ξ2,t)dξ2)N1(ξ2),u2,01(x1,ξ2,t)=0,p01(x1,ξ2,t)=0,q01(x1,t)=∫0x1f(s,t)ds-32(∫-11ψ(ξ2,t)dξ2)∫0x1ν(s)ds+12β-1(t),
with β-1(t) defined by (4.37) and
(4.49)∂4d(±)01∂x14=g±±q01,d(±)01(0,t)=d(±)01(1,t)=∂d(±)01∂x1(0,t)=∂d(±)01∂x1(1,t)=0.


### 4.2.2. The Solution of the Problems forγ=3

Unlike in the previous case, forγ=3 we cannot solve the problems corresponding to Dɛ1 and to Dɛ2 separately. We begin with the problems corresponding to the functions with the index 1. The relations (4.15)–(4.17) are still true. Hence, pj1 is also given by (4.17). The difference with respect to the previous case is that now we cannot express the other unknowns only in function of qj1, since in the right-hand side of (4.16) d(-)j1 is also unknown.We introduce the notations:(4.50)dj1(x1,t)=d(+)j1(x1,t)+d(-)j1(x1,t),𝒟j-11=𝒟(+)j-11+𝒟(-)j-11,
with 𝒟(±)j-11 given by (4.14)2. The function dj1 is obtained as the unique solution of a fourth-order differential problem:(4.51)∂4dj1∂x14=(g++g-)δj0+𝒟j-11,dj1(1,t)=-(d(+)j(1)(0,t)+d(-)j(1)(0,t)),∂dj1∂x1(1,t)=-(∂d(+)j+1(1)∂ξ1(0,t)+∂d(-)j+1(1)∂ξ1(0,t)),dj1(0,t)=2(∂d̃(+)j-1(0)∂ξ1(2,1,t)+∂d̃(-)j-1(0)∂ξ1(2,-1,t))-(d̃(+)j-1(0)(2,1,t)+d̃(-)j-1(0)(2,-1,t)),∂dj1∂x1(0,t)=-(∂d̃(+)j(0)∂ξ1(2,1,t)+∂d̃(-)j(0)∂ξ1(2,-1,t)),
the boundary conditions being obtained by means of (4.11) and (4.31). This means that d(-)j1 can also be expressed via d(+)j1. We show that all the unknowns of the regular part of the asymptotic solution can be expressed in function of d(+)j1; hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined if we obtain and solve a problem for d(+)j1. As we cannot obtain all the necessary conditions for d(+)j1, we have to consider the problem for the couple of unknowns, d(+)j1, d(+)j2.Theorem 4.5.
The approximations of the displacement,d(+)j1, d(+)j2, are obtained as the unique solution of the following system of two sixth-order parabolic equations:
(4.52)∂d(+)j1∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)j1∂x15(x1,t))=Gj1(x1,t)-∂H1∂x1(x1,t)δj0,∂d(+)j2∂t(x2,t)-13∂∂x2(1ν(x2)∂5d(+)j2∂x25(x2,t))=Gj2(x2,t)-∂H2∂x2(x2,t)δj0,d(+)j1(1,t)=-d(+)j(1)(0,t),∂d(+)j1∂x1(1,t)=-∂d(+)j+1(1)∂ξ1(0,t),d(+)j1(0,t)=2∂d̃(+)j-1(0)∂ξ1(2,1,t)-d̃(+)j(0)(2,1,t),∂d(+)j1∂x1(0,t)=-∂d̃(+)j(0)∂ξ1(2,1,t),d(+)j2(1,t)=-d(+)j(2)(0,t),∂d(+)j2∂x2(1,t)=-∂d(+)j+1(2)∂ξ2(0,t),d(+)j2(0,t)=2∂d̃(+)j-1(0)∂ξ2(1,2,t)-d̃(+)j(0)(1,2,t),∂d(+)j2∂x2(0,t)=-∂d̃(+)j(0)∂ξ2(1,2,t),∂5d(+)j1∂x15(1,t)=Lj1(t),∂5d(+)j2∂x25(1,t)=Lj2(t),∂4d(+)j1∂x14(0,t)-∂4d(+)j2∂x24(0,t)=Mj(t),∂5d(+)j1∂x15(0,t)+∂5d(+)j2∂x25(0,t)=Nj(t),d(+)j1(x1,0)=d(+)j2(x1,0)=0,
where Hi, Gji, Lji(t), Mj(t), and Nj(t) are known functions, given below.Proof.
The first relation between the unknownsd(+)j1 and qj1 is given by (4.12)4, that is,
(4.53)∂4d(+)j1∂x14-qj1=g+δj0+𝒟(+)j-11.
The second one is obtained from (4.16) for γ=3 and ξ2=1, (4.12)6 for ξ2=1 and (4.50)1:
(4.54)2∂d(+)j1∂t=∂dj1∂t+23∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))-∫-11D-2(∂∂x1(1ν(x1)Uj-11)dξ2.
Eliminating qj1 from (4.53) and (4.54) we get (4.52)1 with
(4.55)H1(x1,t)=13ν(x1)(∂∂x1g++f),Gj1(x1,t)=12(∂dj1∂t-∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2)-13∂∂x1(1ν(x1)∂∂x1𝒟(+)j-11).
As we can see,H1 depends only on the data and Gj1 depends on some functions determined at the previous approximations. Similar expressions are obtained for H2 and Gj2.
Relations (4.52)3,4 are in fact (4.11) and (4.31), respectively, and the same remark holds for (4.52)5,6. We obtain next (4.52)7 as follows: we derivate (4.53) with respect to x1 and we take x1=1; we also take x1=1 in (4.15) and we eliminate (∂qj1/∂x1)(1,t) from these two relations, by means of (4.9). The right-hand side of (4.52)7 is(4.56)Lj1(t)=(∂∂x1g+(1,t)+f(1,t)-32ν0∫-11ψdξ2)δj0+32∫-11D-2(Uj-11)(1,ξ2,t)dξ2+∂∂x1𝒟(+)j-11-3ν02ddt∫∞0(d(+)j-1(1)-d(-)j-1(1))(ξ1,t)dξ1
and a similar expression we obtain for Lj2(t). Relation (4.52)9 is a consequence of (4.34) for j+1 replaced by j, of (4.53) for x1=0 and of the corresponding relation for the index 2. The right hand side of (4.52)9 is
(4.57)Mj1(0,t)=αj-1(t)+𝒟(+)j-11(0,t)-𝒟(+)j-12(0,t).
Finally, (4.52)10 is a consequence of the compatibility condition for (4.32), obtained as follows: (4.25) can be written as
(4.58)θ̃j=θ̂j-1-divξ(χ(ξ1)u1,j1(ξ1,0,t)e1+χ(ξ2)u2,j2(0,ξ2,t)e2),
where
(4.59)θ̂j-1=-divξ(χ(ξ1)∑l=1jξ1l-1(l-1)!(ξ1l∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂l-1u2,j-l1∂x1l-1(0,ξ2,t)e2)+χ(ξ2)∑l=1jξ2l-1(l-1)!(∂lu1,j-l2∂x2l(ξ1,0,t)e1+ξ2l∂lu2,j-l2∂x2l(ξ1,0,t)e2))
from (4.32) we get
(4.60)γj(t):=∂qj1∂x1(0,t)+∂qj2∂x2(0,t)=-3ν02∫ℬθ̂j-1+3ν02∫2∞(d(+)j-1(0)(ξ1,1,t)-d(-)j-1(0)(ξ1,-1,t))dξ1+3ν02∫2∞(d(+)j-1(0)(1,ξ2,t)-d(-)j-1(0)(-1,ξ2,t))dξ2+32∫-11D-2(Uj-11)(0,ξ2,t)dξ2+32∫-11D-2(Uj-12)(ξ1,0,t)dξ1.
Next, we derivate (4.53), we take x1=0 and we add with the same computation for the index  2. Replacing the previous relation, we obtain (4.52)10, with(4.61)Nj(t)=γj(t)+∂∂x1(𝒟(+)j-1(0,t)+𝒟(-)j-1(0,t))
which completes the proof.The other components of the regular part of the asymptotic solution corresponding toDɛ1 have the following expressions:(4.62)d(-)j1(x1,t)=dj1(x1,t)-d(+)j1(x1,t),qj1(x1,t)=∂4d(+)j1∂x14(x1,t)-g+(x1,t)δj0+𝒟(+)j-11(x1,t),u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)),pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
As in the previous subsection, we discuss the expression of the leading term of the asymptotic solution.Proposition 4.6.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.63)u1,01(x1,ξ2,t)=1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t))N1(ξ2),u2,01(x1,ξ2,t)=-∂∂x1(1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t)))N2(ξ2),q01(x1,t)=∂4d(+)01∂x14(x1,t)-g+(x1,t),p01(x1,ξ2,t)=0,d(-)01(x1,t)=d01(x1,t)-d(+)01(x1,t),
with d01,d(+)01 satisfying the problems presented in the proof, (4.64) and (4.65).Proof.
The relations (4.63)1,2,3,4 are consequences of (4.15)–(4.17) and of (4.53) for j=0. It remains to show that d01,d(+)01,d(+)02 can be determined as unique solutions of some problems.
Forj=0 the problem for d01 becomes:(4.64)∂4d01∂x14(x1,t)=g+(x,t)+g-(x1,t),d01(1,t)=∂d01∂x1(1,t)=0,d01(0,t)=∂d01∂x1(0,t)=0.
Finally, the problem for d(+)01,d(+)02 is the following:
(4.65)∂d(+)01∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)01∂x15(x1,t))=12∂d01∂t-13∂∂x1(1ν(x1)(∂∂x1g++f)),∂d(+)02∂t(x2,t)-13∂∂t(1ν(x2)∂5d(+)02∂x25(x2,t))=12∂d02∂x2-13∂∂x2(1ν(x2)(∂∂x2g++f)),d(+)01(1,t)=∂d(+)01∂x1(1,t)=d(+)01(0,t)=∂d(+)01∂x1(0,t)=0,d(+)02(1,t)=∂d(+)02∂x2(1,t)=d(+)02(0,t)=∂d(+)02∂x2(0,t)=0,∂5d(+)01∂x15(1,t)=-3ν02∫-11ψ(ξ2,t)dξ2+f(1,t)+∂∂x1g+(1,t),∂5d(+)02∂x25(1,t)=-3ν02∫-11ψ(ξ1,t)dξ1+f(1,t)+∂∂x2g+(1,t),∂4d(+)01∂x14(0,t)-∂4d(+)02∂x24(0,t)=0,∂5d(+)01∂x15(0,t)+∂5d(+)02∂x25(0,t)=0,d(+)01(x1,0)=d(+)02(x2,0)=0.At the end of this subsection we notice that the system (4.52) can be solved also for j replaced by j+1 which gives, by means of (4.53), qj+11(0,t), qj+12(0,t). Hence, the system (4.23) can be solved, and all the jth approximation is determined for γ=3.


### 4.2.3. Conclusions

In the previous two subsections we determined thejth approximation of the asymptotic solution for two different cases: γ>3 and γ=3. For γ>3 we can separate the problems for the two branches of the elastic tube and we obtain the regular part of the asymptotic solution (for D1ɛ) as follows:(i)
the macroscopic variableqj1 is the unique solution of the problem:
(4.66)∂∂x1(1ν(x1)(∂qj1∂x1))=ajq(x1,t),∂qj1∂x1(1,t)=bjq(t),qj1(0,t)=cjq(t),
with ajq,bjq,cjq known functions;(ii)
the macroscopic variablesd(±)j1 are the unique solutions of the fourth-order differential problems:
(4.67)∂4d(+)j1∂x14=ajd(x1,t),d(+)j1(0,t),d(+)j1(1,t),∂d(+)j1∂x1(0,t),∂d(+)j1∂x1(1,t)given,
and a similar one for d(-)j1;(iii)
u1,j1,u2,j1,pj1 given by (4.15)–(4.17). The main term of the regular part of the asymptotic solution is presented in (4.48)-(4.49). Unlike in the case γ>3, for γ=3 the problems must be solved for the two branches together, since they cannot be uncoupled;(iv)
we first introduce the auxiliary functionsdj1,dj2, with dj1 the unique solution of (4.51); then the regular part of the asymptotic solution is obtained as follows;(v)
the macroscopic variablesd(+)j1,d(+)j2 are obtained as the unique solution of the sixth-order parabolic system (4.52);(vi)
the macroscopic variablesqj1,qj2 are given by (4.53) and the similar equation for the index 2;(vii)
u1,j1,u2,j1,pj1 given also by (4.15)–(4.17), but we obtain different expressions.The main term of the regular part of the asymptotic solution is given by (4.63) and (4.65).


## 4.2.1. The Order of Solving the Problems forγ>3

The regular part corresponding toDɛ1 is computed by integrating (4.12), as stated below.Proposition 4.2.
The unknownsuj1,qj1,pj1,d(±)j1 are determined from (4.12), up to nine functions of t.Proof.
Integrating twice (4.12)1 from −1 to ξ2 and using the boundary conditions (4.12)6 we get:
(4.15)u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),
which contains as unknowns u1,j1 and qj1. The other functions contained by this relation are either known from previous computations or equal to zero.
We integrate next the incompressibility condition (4.12)3 with respect to ξ2 with the boundary condition (4.12)6 for ξ2=-1 and we obtain(4.16)u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)).
The previous two relations give the components of the velocity uj1 with respect to qj1.
The pressure approximations are determined from (4.12)2, supposing that the integration functions, depending on x1, t are equal to zero, since we consider that any function depending only on x1, t could be contained in qj+11. 											(4.17)pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
Taking ξ2=1 in (4.16) and using the boundary condition (4.12)6 for ξ2=1 we obtain the following second order differential equation for the function qj1:
(4.18)∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))=32∂∂t(d(+)j-γ+31-d(-)j-γ+31)+32∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2.
Integrating (4.18) from x1 to 1, we express ∂qj1/∂x1 by means of (∂qj1/∂x1)(1,t), which represents the only unknown of this expression. This function of t is obtained as follows: we take x1=1 in (4.15) and we introduce the result into (4.9). Hence, we determined the expression of ∂qj1/∂x1 in (0,1)×(0,T), which is:
(4.19)1ν(x1)(∂qj1∂x1-fδj0)=-32δj0∫-11ψ(ξ2,t)dξ2-32ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1-32∂∂t∫x11(d(+)j-γ+31-d(-)j-γ+31)(s,t)ds+32∫-11D-2(1ν(x1)Uj-11)dξ2.
Introducing (4.19) into (4.15) and (4.18) into (4.16) we determine uj1 in (0,1)×(-1,1)×(0,T).
We integrate nextν(x1) · (4.19) from 0 to x1 and we get qj1 determined up to the function of t, qj1(0,t):
(4.20)qj1(x1,t)=δj0∫0x1f(s,t)ds-32∫0x1ν(θ)∫θ1∂∂t(d(+)j-γ+31-d(-)j-γ+31)(s,t)dsdθ-32(ddt∫-∞0(d(+)j+2-γ(1)-d(-)j+2-γ(1))(ξ1,t)dξ1+δj0∫-11ψ(ξ2,t)dξ2)(∫0x1ν(s)ds)+32∫0x1∫-11D-2(Uj-11(s,ξ2,t))dξ2ds+qj1(0,t).
The functions d(±)j1 satisfy the fourth-order differential equations:
(4.21)∂4d(±)j1∂x14=±qj1+g±δj0+𝒟(±)j-11,
with qj1 given by (4.20). Writting qj1 as Qj1(x1,t)+qj1(0,t) and integrating four times (4.21) with respect to x1 we obtain the following expressions for d(±)j1:
(4.22)d(+)j1(x1,t)=d(+)j1(0,t)(1-3x12+2x13)+∂d(+)j1∂x1(0,t)(x1-2x12+x13)+d(+)j1(1,t)(3x12-2x13)+∂d(+)j1∂x1(1,t)(-x12+x13)+qj1(0,t)(x1224-x1312+x1424)+Jx1(Qj1)+Jx1(g+)δj0+Jx1(𝒟(+)j-11),d(-)j1(x1,t)=d(-)j1(0,t)(1-3x12+2x13)+∂d(-)j1∂x1(0,t)(x1-2x12+x13)+d(-)j1(1,t)(3x12-2x13)+∂d(-)j1∂x1(1,t)(-x12+x13)-qj1(0,t)(x1224-x1312+x1424)-Jx1(Qj1)+Jx1(g-)δj0+Jx1(𝒟(-)j-11).
Hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined up to the functions qj1(0,t), d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t), which achieves the proof.In a similar way we express the regular part of the asymptotic solution corresponding toDɛ2 depending on 9 unknown functions of t.We continue the construction with the problems for the corrector inx=0. As we previously said, the term of the asymptotic solution containing this corrector is not equal to zero only in a neighborhood of Dɛr. In this neighborhood, the expression of the asymptotic solution reduces to (u1(k)(x1,ξ2,t)χ(ξ1)+u2(k)(ξ1,x2,t)χ(ξ2)+ubl(k)0(ξ,t),p1(k)(x1,ξ2,t)χ(ξ1)+p2(k)(ξ1,x2,t)χ(ξ2)+pbl(k)0(ξ,t),d±1(k)(x1,t)χ(ξ1)+d±2(k)(x2,t)χ(ξ2)+d±bl(k)0(ξ,t)). For obtaining the problems satisfied by the correctors in x=0, we introduce the previous expression of the asymptotic solution in (2.6)1,2,3,4,5,6,11, with ν=ν0 and f=g=0 in (2.6)1,3; for derivating the terms which contain the two types of variables xi and ξi we proceed as follows: we replace ∂/∂xi by (1/ɛ)(∂/∂ξi), we replace xi by ɛξi and we expand the functions as a Taylor expansion with respect to xi=ɛξi. We introduce the notations ℬ=Dr∪{(ξ1,ξ2):ξ1≥2,ξ2∈(-1,1)}∪{(ξ1,ξ2):ξ1∈(-1,1),ξ2≥2},A±B±⌢=(1/ɛ)A±εB±ε⌢ and we obtain for (uj(0),pj(0)) the following nondivergence free problem:(4.23)-ν0Δξξuj(0)+∇ξpj(0)=-ρf∂uj-2(0)∂t+Fj+(1-χ(ξ1))ξ1′qj+11(0,t)e1+(1-χ(ξ2))ξ2′qj+12(0,t)e2inℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0uniformlywhenξ1→∞,uj(0)→0uniformlywhenξ2→∞,
for j≥0 and(4.24)-ν0Δξξu-1(0)+∇ξp-1(0)=(1-χ(ξ1))ξ1′q01(0,t)e1+(1-χ(ξ2))ξ2′q02(0,t)e2inℬ,divu-1(0)=0inℬ,u-1(0)=0onA+B+⌢∪A-B-⌢,u-1(0)(ξ1,±1,t)=0forξ1≥2,u-1(0)(±1,ξ2,t)=0forξ2≥2,u-1(0)→0uniformlywhenξ1→∞,u-1(0)→0uniformlywhenξ2→∞.
Unlike the problem (4.8) which give the correctors for x1=1, the problems (4.23), (4.24) have unknown right hand sides. The functions Fj are known, but qj+11(0,t) and qj+12(0,t) are elements from the next approximation, so they are unknown. The function θ̃j is given by:(4.25)θ̃j=-divξ(χ(ξ1)∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2)+χ(ξ2)∑l=0jξ2ll!(∂lu1,j-1-l2∂x2l(ξ1,0,t)e1+∂lu2,j-l2∂x2l(ξ1,0,t)e2)),
and it has to satisfy a compatibility condition.The problem (4.24) has the unique solution (with the pressure unique up to an additive function of t)(4.26)u-1(0)=0,p-1(0)={(1-χ(ξ1))q01(0,t)forξ1≥2,q01(0,t)=q02(0,t)inDr,(1-χ(ξ2))q02(0,t)forξ2≥2.
We obtain next the problems for the correctors in x=0 corresponding to the displacements. We first notice that(4.27)d(±)j(0)=0onA±B±⌢.
Let us denote by d̂(±)j-1(0) the unique solution of(4.28)∂4d̂(±)j-1(0)∂ξ14=ρh(1-χ(ξ1))+∑l=0jξ1ll!∂l+2d(±)j-γ-4-l1∂t2∂x1l(0,t)+∑l=2j∂4ξ14((1-χ(ξ1))ξ1ll!)∂ld(±)j-l1∂x1l(0,t)+μ∑l=0j∂4ξ14((1-χ(ξ1))ξ1ll!)∂l+1d(±)j-γ-l1∂t∂x1l(0,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lpj-5-l1∂x1l(0,±1,t)-(1-χ(ξ1))∑l=0jξ1ll!∂lqj-4-l1∂x1l(0,t),∂ad̂(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3},
and by d̃(±)j-1(0) the unique solution of(4.29)∂4d̃(±)j-1(0)∂ξ14=∂4d̂(±)j-1(0)∂ξ14-ρh∂2d(±)j-γ-4(0)∂t2-μ∂5d(±)j-γ(0)∂ξ14∂t+pj-5/ξ2=±1(0),∂ad̃(±)j-1(0)∂ξ1a→0,uniformly,whenξ1→-∞,a∈{0,1,2,3}.
Introducing the asymptotic solution into (2.6)3 and identifying the coefficients of the power of ɛ we obtain:(4.30)d(±)j(0)(ξ1,±1,t)=d̃(±)j-1(0)(ξ1,±1,t)-(1-χ(ξ1))×(d̃(±)j-1(0)(2,±1,t)-(2-ξ1)∂d̃(±)j-1(0)∂ξ1(2,±1,t))ξ1>2,(4.31)d(±)j1(0,t)=2∂d̃(±)j-1(0)∂ξ1(2,±1,t)-d̃(±)j-1(0)(2,±1,t),∂d(±)j1∂x1(0,t)=-∂d̃(±)j(0)∂ξ1(2,±1,t).
The relation (4.30) gives the corrector in x=0 for the displacements corresponding to one branch of the flow domain.We notice that the right-hand side of (4.31) is known, since the problems (4.28) and (4.29) can be solved also if we replace j-1 by j.We are now in a position to return to the regular part of the asymptotic solution. For each regular part of the asymptotic solution, we need 9 relations, in order to determine the 9 unknown functions of integration. For the regular part corresponding toDɛ1, (4.11) and (4.31) give 8 functions: d(±)j1(0,t), (∂d(±)j1/∂x1)(0,t), d(±)j1(1,t), (∂d(±)j1/∂x1)(1,t). It remains undetermined qj1(0,t). We show next that at the jth step this function is already determined at the previous step. For this purpose, we return to the problem (4.23). As usual, we should construct boundary layer functions stabilizing to zero at infinity. As we imposed this condition for the velocity in (4.23)6,7, it is known that the pressure stabilizes to some functions of t. At the jth step, we determine qj+11(0,t) and qj+12(0,t) from the condition of the exponential decay of pj(0) at infinity. To this end, we consider instead of (4.23) and (4.24), the following problem with known right-hand side for j≥-1:(4.32)-ν0Δξξuj(0)+∇ξp¯j(0)=-ρf∂uj-2(0)∂t+Fjinℬ,divuj(0)=θ̃jinℬ,uj(0)=0onA+B+⌢∪A-B-⌢,uj(0)(ξ1,±1,t)=∂d(±)j+2-γ(0)∂t(ξ1,±1,t)e2forξ1≥2,uj(0)(±1,ξ2,t)=∂d(±)j+2-γ(0)∂t(±1,ξ2,t)e1forξ2≥2,uj(0)→0,p¯j(0)→αj1(t)uniformlywhenξ1→∞,uj(0)→0,p¯j(0)→αj2(t)uniformlywhenξ2→∞
with αj1(t),αj2(t) unknown. Supposing for the time being that the function θ̃j satisfies the compatibility condition, it follows that (4.32) has a unique solution (the uniqueness for the pressure being understood up to an additive function of t). This means that the function αj(t), with αj(t)=αj1(t)-αj2(t) is fixed. We define for j≥-1(4.33)pj(0)=p¯j(0)+χ(ξ1)qj+11(0,t)+χ(ξ2)qj+12(0,t).
Standard computations show that (uj(0),p¯j(0)) is solution for (4.32) iff (uj(0),pj(0)) satisfies (4.23) or (4.24) for j=-1 with pj(0)→αji(t)+qj+1i(0,t) uniformly when ξi→∞,i=1,2. To obtain the desired behavior for pj(0) when ξi→∞,i=1,2, it suffices to take αji(t)=-qj+1i(0,t).A first equation for the unknownsqj+11(0,t) and qj+12(0,t) is(4.34)qj+11(0,t)-qj+12(0,t)=αj(t),
with α-1(t)=0. The other equation is obtained introducing the asymptotic solution into the compatibility condition (2.11):(4.35)∫01(d(+)j+11(x1,t)-d(-)j+11(x1,t))dx1+∫01(d(+)j+12(x2,t)-d(-)j+12(x2,t))dx2=-∑l=0j(∂l(d(+)j-l1-d(-)j-l1)∂x1l(0,t)+∂l(d(+)j-l2-d(-)j-l2)∂x2l(0,t))∫03τll!(χ(τ)-1)dτ-∫-∞0(d(+)j(1)(ξ1,1,t)-d(+)j(1)(ξ1,-1,t))dξ1-∫-∞0(d(+)j(2)(1,ξ2,t)-d(+)j(2)(-1,ξ2,t))dξ2-∫2∞(d(+)j(0)(ξ1,1,t)-d(+)j(0)(ξ1,-1,t))dξ1-∫2∞(d(+)j(0)(1,ξ2,t)-d(+)j(0)(-1,ξ2,t))dξ2.
The right-hand side of (4.35) is known from the jth approximation. Replacing in the left hand side of (4.35) d(±)j+11 given by (4.22) (for j replaced by j+1) and d(±)j+12 given by a similar relation, we are leaded to the second equation for the unknowns qj+11(0,t) and qj+12(0,t):(4.36)qj+11(0,t)+qj+12(0,t)=βj(t),
with βj a known function for j≥-1, with(4.37)β-1(t)=-6!(∫01Ix1-1(f)dx1+∫01Ix2-1(f)dx2)+3×6!2(∫-11ψ(ξ2,t)dξ2∫01Ix1-1(ν)dx1+∫-11ψ(ξ1,t)dξ1∫01Ix2-1(ν)dx2)-6!2(∫01Ix1-4(g+-g-)dx1+∫01Ix2-4(g+-g-)dx2)+6!4(∫01Ix1-3(g+-g-)dx1+∫01Ix2-3(g+-g-)dx2)-6!4!(∫01Ix1-2(g+-g-)dx1+∫01Ix2-2(g+-g-)dx2).
From (4.34) and (4.36) we determine qj+11(0,t) and qj+12(0,t), which means that the right-hand side of (4.23) is now known, that allows to solve (4.23), if we show that the problem (4.23) is well posed, that is, the compatibility condition is fulfilled. Moreover, this justifies the previous assertion that qji(0,t),i=1,2 are computed at the (j-1)th step. We can say that the entire jth approximation was obtained for γ>3 if we prove.Proposition 4.3.
For anyj∈ℕ, the function θ̃j satisfies the compatibility condition:
(4.38)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫∂ℬuj(0)·nds.Proof.
From (4.23) and (4.35) it follows:
(4.39)∫∂ℬuj(0)·nds=-ddt(∫-∞0(d(+)j+2-γ(1)(ξ1,t)-d(-)j+2-γ(1)(ξ1,t))dξ1-∫-∞0(d(+)j+2-γ(2)(ξ2,t)-d(-)j+2-γ(2)(ξ2,t))dξ2)-ddt∑l=0j+2-γ(∂l(d(+)j+2-γ-l1-d(-)j+2-γ-l1)∂x1l(0,t)+∂l(d(+)j+2-γ-l2-d(-)j+2-γ-l2)∂x2l(0,t))∫03sll!(χ(s)-1)ds-ddt(∫01(d(+)j+3-γ1(x1,t)-d(-)j+3-γ1(x1,t))dx1+∫01(d(+)j+3-γ2(x2,t)-d(-)j+3-γ2(x2,t))dx2).
We express next the three terms of the right-hand side of the previous equality with respect to the components of the velocity corresponding to the regular parts of the asymptotic solution.
From (4.9), the corresponding relation for x2=1 and (2.9)6 the first term of the right-hand side is equal to(4.40)T1=-∫-11u1,j1(1,ξ2,t)dξ2-∫-11u2,j2(ξ1,1,t)dξ1.
The third term is expressed by means of (4.12)6 and its analogous for uj2 and it is equal to
(4.41)T3=-∫01(u2,j1(x1,1,t)-u2,j1(x1,-1,t))dx1-∫01(u1,j2(1,x2,t)-u1,j2(-1,x2,t))dx2.
Computing now T1+T3 and using (4.12)3 and the similar one for uj2 we get
(4.42)T1+T3=-∫-11u1,j1(0,ξ2,t)dξ2-∫-11u2,j2(ξ1,0,t)dξ1.
The second term is given by
(4.43)T2=∑l=0j(∂l(u2,j-1-l1(0,1,t)-u2,j-1-l1(0,-1,t))∂x1l+∂l(u1,j-1-l2(1,0,t)-u1,j-1-l2(-1,0,t))∂x2l)∫03sll!(χ(s)-1)ds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ2,0,t)∂x2le2))dξ+∫-11u1,j1(0,ξ2,t)dξ2+∫-11u2,j2(ξ1,0,t)dξ2.
Finally, computing T1+T2+T3, we obtain the following expression of the right-hand side of (4.38):
(4.44)∫∂ℬuj(0)·nds=∑l=0j∫(0,3)×(-1,1)divξ((1-χ(ξ1))ξ1ll!(∂lu1,j-l1(0,ξ2,t)∂x1le1-∂lu2,j-1-l1(0,ξ2,t)∂x1le2))dξ+∑l=0j∫(-1,1)×(0,3)divξ((1-χ(ξ2))ξ2ll!(∂lu1,j-1-l2(ξ1,0,t)∂x2le1-∂lu2,j-l2(ξ1,0,t)∂x2le2))dξ.
We compute next the left-hand side of (4.38). Since χ(ξ1)=χ(ξ2)=0 on ℬ∩{ξ1<2,ξ2<2}, we can write:
(4.45)∫ℬθ̃j(ξ1,ξ2,t)dξ=∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ+∫(-1,1)×(2,3)θ̃j(ξ1,ξ2,t)dξ.
We express the first term of the right-hand side of the previous relation taking into account that for (ξ1,ξ2)∈(2,3)×(-1,1), we have χ(ξ2)=0:
(4.46)∫(2,3)×(-1,1)θ̃j(ξ1,ξ2,t)dξ=∫(0,3)×(-1,1)divξ((1-χ(ξ1))∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))dξ.
For obtaining the above equality we also used the property:
(4.47)divξ(∑l=0jξ1ll!(∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂lu2,j-1-l1∂x1l(0,ξ2,t)e2))=0.
A similar result is obtained for the second term of the right-hand side and, comparing the expression of ∫ℬθ̃j(ξ1,ξ2,t)dξ with the right-hand side of (4.44), the proof is complete.The last result of this subsection gives the leading term of the regular part of the asymptotic solution corresponding toDɛ1.Proposition 4.4.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.48)u1,01(x1,ξ2,t)=-32(∫-11ψ(ξ2,t)dξ2)N1(ξ2),u2,01(x1,ξ2,t)=0,p01(x1,ξ2,t)=0,q01(x1,t)=∫0x1f(s,t)ds-32(∫-11ψ(ξ2,t)dξ2)∫0x1ν(s)ds+12β-1(t),
with β-1(t) defined by (4.37) and
(4.49)∂4d(±)01∂x14=g±±q01,d(±)01(0,t)=d(±)01(1,t)=∂d(±)01∂x1(0,t)=∂d(±)01∂x1(1,t)=0.


## 4.2.2. The Solution of the Problems forγ=3

Unlike in the previous case, forγ=3 we cannot solve the problems corresponding to Dɛ1 and to Dɛ2 separately. We begin with the problems corresponding to the functions with the index 1. The relations (4.15)–(4.17) are still true. Hence, pj1 is also given by (4.17). The difference with respect to the previous case is that now we cannot express the other unknowns only in function of qj1, since in the right-hand side of (4.16) d(-)j1 is also unknown.We introduce the notations:(4.50)dj1(x1,t)=d(+)j1(x1,t)+d(-)j1(x1,t),𝒟j-11=𝒟(+)j-11+𝒟(-)j-11,
with 𝒟(±)j-11 given by (4.14)2. The function dj1 is obtained as the unique solution of a fourth-order differential problem:(4.51)∂4dj1∂x14=(g++g-)δj0+𝒟j-11,dj1(1,t)=-(d(+)j(1)(0,t)+d(-)j(1)(0,t)),∂dj1∂x1(1,t)=-(∂d(+)j+1(1)∂ξ1(0,t)+∂d(-)j+1(1)∂ξ1(0,t)),dj1(0,t)=2(∂d̃(+)j-1(0)∂ξ1(2,1,t)+∂d̃(-)j-1(0)∂ξ1(2,-1,t))-(d̃(+)j-1(0)(2,1,t)+d̃(-)j-1(0)(2,-1,t)),∂dj1∂x1(0,t)=-(∂d̃(+)j(0)∂ξ1(2,1,t)+∂d̃(-)j(0)∂ξ1(2,-1,t)),
the boundary conditions being obtained by means of (4.11) and (4.31). This means that d(-)j1 can also be expressed via d(+)j1. We show that all the unknowns of the regular part of the asymptotic solution can be expressed in function of d(+)j1; hence, the regular part of the asymptotic solution corresponding to Dɛ1 is determined if we obtain and solve a problem for d(+)j1. As we cannot obtain all the necessary conditions for d(+)j1, we have to consider the problem for the couple of unknowns, d(+)j1, d(+)j2.Theorem 4.5.
The approximations of the displacement,d(+)j1, d(+)j2, are obtained as the unique solution of the following system of two sixth-order parabolic equations:
(4.52)∂d(+)j1∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)j1∂x15(x1,t))=Gj1(x1,t)-∂H1∂x1(x1,t)δj0,∂d(+)j2∂t(x2,t)-13∂∂x2(1ν(x2)∂5d(+)j2∂x25(x2,t))=Gj2(x2,t)-∂H2∂x2(x2,t)δj0,d(+)j1(1,t)=-d(+)j(1)(0,t),∂d(+)j1∂x1(1,t)=-∂d(+)j+1(1)∂ξ1(0,t),d(+)j1(0,t)=2∂d̃(+)j-1(0)∂ξ1(2,1,t)-d̃(+)j(0)(2,1,t),∂d(+)j1∂x1(0,t)=-∂d̃(+)j(0)∂ξ1(2,1,t),d(+)j2(1,t)=-d(+)j(2)(0,t),∂d(+)j2∂x2(1,t)=-∂d(+)j+1(2)∂ξ2(0,t),d(+)j2(0,t)=2∂d̃(+)j-1(0)∂ξ2(1,2,t)-d̃(+)j(0)(1,2,t),∂d(+)j2∂x2(0,t)=-∂d̃(+)j(0)∂ξ2(1,2,t),∂5d(+)j1∂x15(1,t)=Lj1(t),∂5d(+)j2∂x25(1,t)=Lj2(t),∂4d(+)j1∂x14(0,t)-∂4d(+)j2∂x24(0,t)=Mj(t),∂5d(+)j1∂x15(0,t)+∂5d(+)j2∂x25(0,t)=Nj(t),d(+)j1(x1,0)=d(+)j2(x1,0)=0,
where Hi, Gji, Lji(t), Mj(t), and Nj(t) are known functions, given below.Proof.
The first relation between the unknownsd(+)j1 and qj1 is given by (4.12)4, that is,
(4.53)∂4d(+)j1∂x14-qj1=g+δj0+𝒟(+)j-11.
The second one is obtained from (4.16) for γ=3 and ξ2=1, (4.12)6 for ξ2=1 and (4.50)1:
(4.54)2∂d(+)j1∂t=∂dj1∂t+23∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))-∫-11D-2(∂∂x1(1ν(x1)Uj-11)dξ2.
Eliminating qj1 from (4.53) and (4.54) we get (4.52)1 with
(4.55)H1(x1,t)=13ν(x1)(∂∂x1g++f),Gj1(x1,t)=12(∂dj1∂t-∫-11D-2(∂∂x1(1ν(x1)Uj-11))dξ2)-13∂∂x1(1ν(x1)∂∂x1𝒟(+)j-11).
As we can see,H1 depends only on the data and Gj1 depends on some functions determined at the previous approximations. Similar expressions are obtained for H2 and Gj2.
Relations (4.52)3,4 are in fact (4.11) and (4.31), respectively, and the same remark holds for (4.52)5,6. We obtain next (4.52)7 as follows: we derivate (4.53) with respect to x1 and we take x1=1; we also take x1=1 in (4.15) and we eliminate (∂qj1/∂x1)(1,t) from these two relations, by means of (4.9). The right-hand side of (4.52)7 is(4.56)Lj1(t)=(∂∂x1g+(1,t)+f(1,t)-32ν0∫-11ψdξ2)δj0+32∫-11D-2(Uj-11)(1,ξ2,t)dξ2+∂∂x1𝒟(+)j-11-3ν02ddt∫∞0(d(+)j-1(1)-d(-)j-1(1))(ξ1,t)dξ1
and a similar expression we obtain for Lj2(t). Relation (4.52)9 is a consequence of (4.34) for j+1 replaced by j, of (4.53) for x1=0 and of the corresponding relation for the index 2. The right hand side of (4.52)9 is
(4.57)Mj1(0,t)=αj-1(t)+𝒟(+)j-11(0,t)-𝒟(+)j-12(0,t).
Finally, (4.52)10 is a consequence of the compatibility condition for (4.32), obtained as follows: (4.25) can be written as
(4.58)θ̃j=θ̂j-1-divξ(χ(ξ1)u1,j1(ξ1,0,t)e1+χ(ξ2)u2,j2(0,ξ2,t)e2),
where
(4.59)θ̂j-1=-divξ(χ(ξ1)∑l=1jξ1l-1(l-1)!(ξ1l∂lu1,j-l1∂x1l(0,ξ2,t)e1+∂l-1u2,j-l1∂x1l-1(0,ξ2,t)e2)+χ(ξ2)∑l=1jξ2l-1(l-1)!(∂lu1,j-l2∂x2l(ξ1,0,t)e1+ξ2l∂lu2,j-l2∂x2l(ξ1,0,t)e2))
from (4.32) we get
(4.60)γj(t):=∂qj1∂x1(0,t)+∂qj2∂x2(0,t)=-3ν02∫ℬθ̂j-1+3ν02∫2∞(d(+)j-1(0)(ξ1,1,t)-d(-)j-1(0)(ξ1,-1,t))dξ1+3ν02∫2∞(d(+)j-1(0)(1,ξ2,t)-d(-)j-1(0)(-1,ξ2,t))dξ2+32∫-11D-2(Uj-11)(0,ξ2,t)dξ2+32∫-11D-2(Uj-12)(ξ1,0,t)dξ1.
Next, we derivate (4.53), we take x1=0 and we add with the same computation for the index  2. Replacing the previous relation, we obtain (4.52)10, with(4.61)Nj(t)=γj(t)+∂∂x1(𝒟(+)j-1(0,t)+𝒟(-)j-1(0,t))
which completes the proof.The other components of the regular part of the asymptotic solution corresponding toDɛ1 have the following expressions:(4.62)d(-)j1(x1,t)=dj1(x1,t)-d(+)j1(x1,t),qj1(x1,t)=∂4d(+)j1∂x14(x1,t)-g+(x1,t)δj0+𝒟(+)j-11(x1,t),u1,j1(x1,ξ2,t)=1ν(x1)D-2(Uj-11)+1ν(x1)(∂qj1∂x1-fδj0)N1(ξ2),u2,j1(x1,ξ2,t)=∂d(-)j-γ+31∂t-∂∂x1(1ν(x1)(∂qj1∂x1-fδj0))N2(ξ2)-D-1D-2(∂∂x1(1ν(x1)Uj-11)),pj1=D-1(-ρf∂u2,j-31∂t+∂∂x1(ν(x1)∂u2,j-31∂x1)+∂∂x1(ν(x1)∂u1,j-11∂ξ2)+2ν(x1)∂2u2,j-11∂ξ22).
As in the previous subsection, we discuss the expression of the leading term of the asymptotic solution.Proposition 4.6.
Forj=0 the regular part of the asymptotic solution corresponding to Dɛ1 is given by
(4.63)u1,01(x1,ξ2,t)=1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t))N1(ξ2),u2,01(x1,ξ2,t)=-∂∂x1(1ν(x1)(∂5d(+)01∂x15(x1,t)-∂∂x1g+(x1,t)-f(x1,t)))N2(ξ2),q01(x1,t)=∂4d(+)01∂x14(x1,t)-g+(x1,t),p01(x1,ξ2,t)=0,d(-)01(x1,t)=d01(x1,t)-d(+)01(x1,t),
with d01,d(+)01 satisfying the problems presented in the proof, (4.64) and (4.65).Proof.
The relations (4.63)1,2,3,4 are consequences of (4.15)–(4.17) and of (4.53) for j=0. It remains to show that d01,d(+)01,d(+)02 can be determined as unique solutions of some problems.
Forj=0 the problem for d01 becomes:(4.64)∂4d01∂x14(x1,t)=g+(x,t)+g-(x1,t),d01(1,t)=∂d01∂x1(1,t)=0,d01(0,t)=∂d01∂x1(0,t)=0.
Finally, the problem for d(+)01,d(+)02 is the following:
(4.65)∂d(+)01∂t(x1,t)-13∂∂x1(1ν(x1)∂5d(+)01∂x15(x1,t))=12∂d01∂t-13∂∂x1(1ν(x1)(∂∂x1g++f)),∂d(+)02∂t(x2,t)-13∂∂t(1ν(x2)∂5d(+)02∂x25(x2,t))=12∂d02∂x2-13∂∂x2(1ν(x2)(∂∂x2g++f)),d(+)01(1,t)=∂d(+)01∂x1(1,t)=d(+)01(0,t)=∂d(+)01∂x1(0,t)=0,d(+)02(1,t)=∂d(+)02∂x2(1,t)=d(+)02(0,t)=∂d(+)02∂x2(0,t)=0,∂5d(+)01∂x15(1,t)=-3ν02∫-11ψ(ξ2,t)dξ2+f(1,t)+∂∂x1g+(1,t),∂5d(+)02∂x25(1,t)=-3ν02∫-11ψ(ξ1,t)dξ1+f(1,t)+∂∂x2g+(1,t),∂4d(+)01∂x14(0,t)-∂4d(+)02∂x24(0,t)=0,∂5d(+)01∂x15(0,t)+∂5d(+)02∂x25(0,t)=0,d(+)01(x1,0)=d(+)02(x2,0)=0.At the end of this subsection we notice that the system (4.52) can be solved also for j replaced by j+1 which gives, by means of (4.53), qj+11(0,t), qj+12(0,t). Hence, the system (4.23) can be solved, and all the jth approximation is determined for γ=3.


## 4.2.3. Conclusions

In the previous two subsections we determined thejth approximation of the asymptotic solution for two different cases: γ>3 and γ=3. For γ>3 we can separate the problems for the two branches of the elastic tube and we obtain the regular part of the asymptotic solution (for D1ɛ) as follows:(i)
the macroscopic variableqj1 is the unique solution of the problem:
(4.66)∂∂x1(1ν(x1)(∂qj1∂x1))=ajq(x1,t),∂qj1∂x1(1,t)=bjq(t),qj1(0,t)=cjq(t),
with ajq,bjq,cjq known functions;(ii)
the macroscopic variablesd(±)j1 are the unique solutions of the fourth-order differential problems:
(4.67)∂4d(+)j1∂x14=ajd(x1,t),d(+)j1(0,t),d(+)j1(1,t),∂d(+)j1∂x1(0,t),∂d(+)j1∂x1(1,t)given,
and a similar one for d(-)j1;(iii)
u1,j1,u2,j1,pj1 given by (4.15)–(4.17). The main term of the regular part of the asymptotic solution is presented in (4.48)-(4.49). Unlike in the case γ>3, for γ=3 the problems must be solved for the two branches together, since they cannot be uncoupled;(iv)
we first introduce the auxiliary functionsdj1,dj2, with dj1 the unique solution of (4.51); then the regular part of the asymptotic solution is obtained as follows;(v)
the macroscopic variablesd(+)j1,d(+)j2 are obtained as the unique solution of the sixth-order parabolic system (4.52);(vi)
the macroscopic variablesqj1,qj2 are given by (4.53) and the similar equation for the index 2;(vii)
u1,j1,u2,j1,pj1 given also by (4.15)–(4.17), but we obtain different expressions.The main term of the regular part of the asymptotic solution is given by (4.63) and (4.65).


## 5. Error Estimates

In the last section, we estimate the error between the exact solution and the asymptotic one, in order to justify the asymptotic expansion. Introducing the asymptotic solution of orderk, (ûa(k),p̂a(k),d̂±a(k)), into (2.6) we want to obtain for it a problem of the same type as (2.6).The next computations hold for any integerγ≥3. We begin with the continuity equation satisfied by the asymptotic velocity of order k. Standard computations give(5.1)divxûa(k)=dɛ(k)inℬɛ×(0,T),
with(5.2)dɛ(k)(x,t)={0inℬɛ∩{x1<18}∪{x2<18},-η′(1-xi)∑j=0kɛj+2u1,j(0)(xɛ,t)inℬɛ∩{18<xi<14},i=1,2,η′(x1)∑j=0kɛj+2u1,j(1)(x1-1ɛ,x2ɛ,t)inℬɛ∩{14<x1<1},η′(x2)∑j=0kɛj+2u2,j(2)(x1ɛ,x2-1ɛ,t)inℬɛ∩{14<x2<1}.
We notice that the asymptotic velocity of order k is not a divergence free function.We present next the equations for the asymptotic solution, which correspond to (2.6)1 and (2.6)3:(5.3)ρf∂ûa(k)∂t-2divx(ν(x)Dxûa(k))+∇xp̂a(k)=f+Fɛ(k)inℬɛ×(0,T),ρh∂2d̂±a(k)∂t2+1ɛγ∂4d̂±a(k)∂xi4+μ∂5d̂±a(k)∂xi4∂t=g±±p̂(k)a|D¯ɛ∩Γ±ɛ+G±ɛ(k)on(D¯ɛi∩Γ±ɛ)×(0,T),i=1,2,
with Fɛ(k),G±ɛ(k) residual known functions. Since the expressions of these residuals are rather complicated, but not too technical to obtain, let us give below only the estimates for their norms in the corresponding spaces: ∥Fɛ(k)∥L2(0,T;(L2(ℬɛ))2)=O(ɛk+3/2),
(5.4)∥G±ɛ(k)∥L2((D¯ɛi∩Γ±ɛ)×(0,T))={O(ɛ3/2)fork∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ})fork≥4.
These estimates will be used to derive finally the error estimate between the exact and the asymptotic solution. More details concerning these estimates can be found in [2, Theorem  5.1].We derive in the sequel the conditions on the elastic boundaries. Introducing the asymptotic solution of orderk into (2.6)11 we get(5.5)ûa(k)·n=±∂d̂±a(k)∂t±A±ɛk,ûa(k)·τ=0onΓ±ɛ×(0,T),
where the expression of A±ɛk on Γ±ɛ∩∂Dɛ1 is given by:(5.6)A±ɛk(x1,±1,t)={-(ɛk+4∂d±k+4-γ1∂t+⋯+ɛk+γ∂d±k1∂t)(x1,t)-η(x1)(ɛk+3∂d±k+3-γ(1)∂t+⋯+ɛk+γ∂d±k(1)∂t)(x1-1ɛ,±1,t)onΓ±ɛ∩{x1>14},-χ(x1ɛ)(ɛk+4∂d±k+4-γ1∂t+⋯+ɛk+γ∂d±k1∂t)(x1,t)-η(1-x1)(ɛk+3∂d±k+3-γ(0)∂t+⋯+ɛk+γ∂d±k(0)∂t)(x1ɛ,±1,t)onΓ±ɛ∩{2ɛ<x1<14},
and an analogous expression can be written on Γ±ɛ∩∂Dɛ2. We notice that for γ=3 the first terms in the above definition are zero.We complete the problem (5.1), (5.3), (5.5) with the following boundary conditions:(5.7)ûa(k)=0,d̂±a(k)=0on(A+εB+ε⌢∪A-εB-ε⌢)×(0,T),ûa(k)(1,x2,t)=ɛ2ψɛ(x2,t)e1+ɛk+3u2,k1(1,ξ2,t)e2onF1ɛ×(0,T),ûa(k)(x1,1,t)=ɛ2ψɛ(x1,t)e2+ɛk+3u1,k2(ξ1,1,t)e1onF2ɛ×(0,T),d̂±a(k)(1,±ɛ,t)=0,∂d̂±a(k)∂x1(1,±ɛ,t)=ɛk+γ∂d±k1∂x1(1,t),d̂±a(k)(2ɛ,±ɛ,t)=∂d̂±a(k)∂x1(2ɛ,±ɛ,t)=0,d̂±a(k)(±ɛ,1,t)=0,∂d̂±a(k)∂x1(±ɛ,1,t)=ɛk+γ∂d±k2∂x1(1,t),d̂±a(k)(±ɛ,2ɛ,t)=∂d̂±a(k)∂x1(±ɛ,2ɛ,t)=0t∈(0,T),
and initial conditions:(5.8)ûa(k)(x,0)=0inℬɛ,d̂±a(k)(x,0)=∂d̂±a(k)∂t(x,0)=0onΓ±ɛ.
The compatibility condition for the problem (5.1), (5.3) with the boundary and initial conditions (5.5), (5.7), (5.8) is given by:(5.9)∫ℬɛdɛ(k)dx=∫2ɛ1∂∂t(d̂+a(k)(x1,ɛ,t)-d̂-a(k)(x1,-ɛ,t))dx1+∫2ɛ1∂∂t(d̂+a(k)(ɛ,x2,t)-d̂-a(k)(-ɛ,x2,t))dx2+∫2ɛ1(A+ɛk(x1,ɛ,t)-A-ɛk(x1,-ɛ,t))dx1+∫2ɛ1(A-ɛk(ɛ,x2,t)-A+ɛk(-ɛ,x2,t))dx2.
We notice that the boundary conditions for ûa(k) and d̂±a(k) are of different type from those for u and d±, respectively; so we cannot apply directly the estimates (3.20) which hold for different right-hand sides in the equations but for the same boundary conditions. Moreover, the asymptotic velocity of order k is not a divergence free function. To overcome this difficulty, we define new functions Ua(k) and D±a(k) in order to obtain for (Ua(k),p̂±a(k),D±a(k)) a problem of the same type as (2.6).Construction ofD±a(k)
For this construction we use some ideas from [5]. We first define D±(k):F±ɛ×(0,T)→ℝ, satisfying:
(5.10)D±(k)(1,±ɛ,t)=0,∂D±(k)∂x1(1,±ɛ,t)=ɛk+γ∂d±k1∂x1(1,t),D±(k)(2ɛ,±ɛ,t)=∂D±(k)∂x1(2ɛ,±ɛ,t)=0,D±(k)(±ɛ,1,t)=0,∂D±(k)∂x1(±ɛ,1,t)=ɛk+γ∂d±k2∂x2(1,t),D±(k)(±ɛ,2ɛ,t)=∂D±(k)∂x2(±ɛ,2ɛ,t)=0,D±(k)(x,t)=0(x1,x2)∈A±εB±ε⌢,D±(k)(x,0)=∂D±(k)∂t(x,0)=0,(5.11)∫2ɛ1∂∂t(D+(k)(x1,ɛ,t)-D-(k)(x1,-ɛ,t))dx1+∫2ɛ1∂∂t(D+(k)(ɛ,x2,t)-D-(k)(-ɛ,x2,t))dx2=∫ℬɛdɛ(k)dx-∫2ɛ1(A+ɛk(x1,ɛ,t)-A-ɛk(x1,-ɛ,t))dx1-∫2ɛ1(A+ɛk(ɛ,x2,t)-A-ɛk(-ɛ,x2,t))dx2=:Wɛk(t).
Let us consider
(5.12)D±(k)(x,t)={(x1-1)(x1-2ɛ)2(a±(k)1(t)(x1-1)+1(1-2ɛ)2ɛk+γ∂d±k1∂x1(1,t))in(2ɛ,1)×{±ɛ},(x2-1)(x2-2ɛ)2(a±(k)2(t)(x2-1)+1(1-2ɛ)2ɛk+γ∂d±k2∂x2(1,t))in{±ɛ}×(2ɛ,1),0onA±εB±ε⌢,
with a±(k)i,i=1,2 four functions which are determined by imposing some conditions, as we show below. It is obvious that the functions D±(k) satisfy (5.10)1,2,3,4 without any conditions on a±(k)i.We introduce next the following notations:(5.13)Wɛk,1(t)=∫Dɛ1dɛ(k)dx-∫2ɛ1(A+ɛk(x1,ɛ,t)-A-ɛk(x1,-ɛ,t))dx1,Wɛk,2(t)=∫Dɛ2dɛ(k)dx-∫2ɛ1(A+ɛk(ɛ,x2,t)-A-ɛk(-ɛ,x2,t))dx2.
We notice that from (5.2) we obtain Wɛk(t)=Wɛk,1(t)+Wɛk,2(t). A relation between the functions a+(k)1 and a-(k)1 follows from (d/dt)∫2ɛ1(D+(k)(x1,ɛ,t)-D-(k)(x1,-ɛ,t))dx1=Wɛk,1(t), with the initial conditions (5.10)6. In a similar way we establish a relation between a+(k)2 and a-(k)2. So we proved the possibility of construction of functions D±(k) which satisfy (5.10) and (5.11).We define next(5.14)D̂±a(k)=d̂±a(k)-D±(k),
and we verify that it satisfies the same boundary conditions as d±.Construction ofÛa(k)
We look for a functionU(k):ℬɛ→ℝ2 satisfying the following problem, for t∈(0,T):
(5.15)U(k)(t)∈(H1(ℬɛ))2,divxU(k)(t)=dɛ(k)(t)inℬɛ,U(k)(t)=0inℬɛ∩{x1<2ɛ,x2<2ɛ},U(k)(1,x2,t)=ɛk+3u2,k1(1,ξ2,t)e2onF1ɛ,U(k)(x1,1,t)=ɛk+3u1,k2(ξ1,1,t)e1onF2ɛ,U(k)(x1,±ɛ,t)=(∂D±(k)∂t±A±ɛk)(x1,±ɛ,t)e2onΓ±ɛ∩∂Dɛ1,U(k)(±ɛ,x2,t)=(∂D±(k)∂t±A±ɛk)(±ɛ,x2,t)e1onΓ±ɛ∩∂Dɛ2.Remark 5.1.
SinceU(k) vanishes in ℬɛ∩{x1<2ɛ,x2<2ɛ}, (5.15) represents in fact two separate problems, one in Dɛ1 and the second one on Dɛ2. The compatibility condition for each problem is fulfilled due to the construction of D±(k).We can prove that.Proposition 5.2.
The problem (5.15) has at least one solution, with the property
(5.16)∥U(k)(t)∥(H1(ℬɛ))2=O(ɛk+3/2).Proof.
Due to (5.15)3, we can consider the problem (5.15) as two separate problems: one in Dɛ1 and the second in Dɛ2, each problem with its compatibility condition. So, we can obtain the estimate (5.16) only for Dɛ1. For any (x1,x2)∈Dɛ1∩{x1>2ɛ} we introduce the new variable (y1,y2)=((x1-2ɛ)/(1-2ɛ),x2/ɛ); obviously (y1,y2)∈(0,1)×(-1,1). We define a new function ηɛ1(k):(0,1)×(-1,1)×(0,T)→ℝ2, by ηɛ1(k)(y1,y2,t)=(1/(1-2ɛ))U1(k)(x1,x2,t)e1+(1/ɛ)U2(k)(x1,x2,t)e2. Obvious computations lead to the following problem for ηɛ1(k)(t):
(5.17)divyηɛ1(k)(t)=dɛ(k)((1-2ɛ)y1+2ɛ,ɛy2,t)in(0,1)×(-1,1),ηɛ1(k)(0,y2,t)=0in(-1,1),ηɛ1(k)(1,y2,t)=ɛk+2u2,k1(1,y2,t)e2in(-1,1),ηɛ1(k)(y1,±1,t)=1ɛ(∂D±(k)∂t±A±ɛk)((1-2ɛ)y1+2ɛ,±ɛ,t)e2in(0,1).
The right-hand sides of (5.17)1,4 contain either correctors (which are of order O(exp(-σ/ɛ)), with σ independent on ɛ), or terms of order O(ɛk+2). Applying a result of [15], Chapter III, page 127 we get:
(5.18)∥ηɛ1(k)(t)∥(H1((0,1)×(-1,1)))2=O(ɛk+2).
Expressing next the norm of U(k) with respect to the norm of ηɛ1(k) we obtain ∥U(k)(t)∥(H1(Dɛ1∩{x1>2ɛ}))2≤(1/ɛ1/2)∥ηɛ1(k)(t)∥(H1((0,1)×(-1,1)))2, that is, ∥U(k)(t)∥(H1(Dɛ1∩{x1>2ɛ}))2=O(ɛk+(3/2)). With similar computations we get the same estimate for ∥U(k)(t)∥(H1(Dɛ2∩{x2>2ɛ}))2, which completes the proof.We notice that from (5.16) and from the regularity of the right-hand sides of (5.17) it follows that ∥U(k)∥L2(0,T;(H1(ℬɛ))2)=O(ɛk+(3/2)).The function(5.19)Ûa(k)=ûa(k)-U(k)
satisfies the same type of boundary conditions as u on ∂ℬɛ. Moreover Ûa(k)(x,0)=0, due to (4.1)4. As a consequence of the previous constructions, we obtain for (Ûa(k),p̂a(k),D̂±a(k)) the following problem:(5.20)ρf∂Ûa(k)∂t-2divx(ν(x)DxÛa(k))+∇xp̂a(k)=f+ℱɛ(k)inℬɛ×(0,T),divxÛa(k)=0inℬɛ×(0,T),ρh∂2D̂±a(k)∂t2+1ɛγ∂4D̂±a(k)∂xi4+μ∂5D̂±a(k)∂xi4∂t=g±±p̂(k)a|D¯ɛi∩Γ±ɛ+𝒢±ɛ(k)onΓ±ɛ×(0,T),D̂±a(k)=0onA±εB±ε⌢×(0,T),Ûa(k)=ψɛon(∂ℬɛ∖(Γ+ɛ∪Γ-ɛ))×(0,T),D̂±a(k)(1,±ɛ,t)=∂D̂±a(k)∂x1(1,±ɛ,t)=0in(0,T),D̂±a(k)(2ɛ,±ɛ,t)=∂D̂±a(k)∂x1(2ɛ,±ɛ,t)=0in(0,T),D̂±a(k)(±ɛ,1,t)=∂D̂±a(k)∂x2(±ɛ,1,t)=0in(0,T),D̂±a(k)(±ɛ,2ɛ,t)=∂D̂±a(k)∂x2(±ɛ,2ɛ,t)=0in(0,T),Ûa(k)·n=±∂D̂±a(k)∂t,Ûa(k)·τ=0onΓ±ɛ×(0,T),Ûa(k)(x,0)=0inℬɛ,D̂±a(k)(x,0)=∂D̂±a(k)∂t(x,0)=0onΓ±ɛ,
with ℱɛ(k)=Fɛ(k)-ρf(∂U(k)/∂t)+2divx(ν(x)DxU(k)),𝒢±ɛ(k)=G±ɛ(k)-ρh(∂2D±(k)/∂t2)-(1/ɛγ)(∂4D±(k)/∂xi4)-μ(∂5D±(k)/∂xi4∂t).We are now in a position to apply (3.20) in order to establish the error between the exact solution of (2.6) and its approximation given by (4.4).Theorem 5.3.
Let(ûa(k),p̂a(k),d̂±a(k)) be the asymptotic solution of order k given by (4.4) and (u,p,d) the exact solution of the physical problem (2.6). Then the following estimates hold:
(5.21)∥u-ûa(k)∥L∞(0,T;(L2(ℬɛ))2)={O(ɛ3/2)forK∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ})forK≥4,∥D(u-ûa(k))∥L2(0,T;(L2(ℬɛ))4)={O(ɛ3/2)forK∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ})forK≥4,∥∂∂t(d±-d̂±a(K))∥L∞(0,T;L2(Γ±ɛ))={O(ɛ3/2)forK∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ})forK≥4,∥∂2∂s2(d±-d̂±a(K))∥L∞(0,T;L2(Γ±ɛ))={O(ɛ3/2+γ/2)forK∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ}+γ/2)forK≥4,∥∇(p-p̂a(k))∥L∞(0,T;(H-1(ℬɛ))2)={O(ɛ5/2-γ/2)forK∈{0,1,2,3},O(ɛmin{k-5/2,k+4-γ}-γ/2)forK≥4.Proof.
Let us obtain the estimate (5.21)2. From (5.16) and (5.19) it follows that
(5.22)∥D(u-ûa(k))∥L2(0,T;(L2(ℬɛ))4)≤∥D(u-Ûa(k))∥L2(0,T;(L2(ℬɛ))4)+O(ɛk+3/2).
It remains to estimate the first term of the right-hand side of the previous inequality. For this we first estimate min(∥ℱɛ(k)∥L2(0,T;(L2(ℬɛ))2),∥𝒢±ɛ(k)∥L2((D¯ɛi∩Γ±ɛ)×(0,T))) and then we apply (3.20)2. This estimate is given either by G±ɛ(k) or by (1/ɛγ)(∂4D±(k)/∂xi4), the other terms having greater order than these. Following the ideas of [2, Theorem  5.1] we obtain (5.21)2; the other estimates of (5.21) are proved with the same technique and the proof is achieved.In order to improve the estimates (5.21) we analyze the order of the leading term of the asymptotic solution. We prove that.Proposition 5.4.
For the leading term of the asymptotic solution one has the following estimates:(5.23)∥ûa(0)∥L∞(0,T;(L2(ℬɛ))2)=O(ɛ5/2),∥Dûa(0)∥L2(0,T;(L2(ℬɛ))4)=O(ɛ3/2),∥∂∂td̂±a(0)∥L∞(0,T;L2(Γ±ɛ))=O(ɛγ),∥∂2∂s2d̂±a(0)∥L∞(0,T;L2(Γ±ɛ))=O(ɛγ),∥∇p̂a(0)∥L∞((0,T);(H-1(ℬɛ))2)=O(ɛ1/2).Proof.
Let us establish (5.23)1. From (4.48)1,2 (for γ>3) and from (4.63)1,2 (for γ=3) we obtain for the regular velocity corresponding to Dɛ1(5.24)∥u1(0)χ∥L∞(0,T;(L2(ℬɛ))2)=O(ɛ5/2).
This means that the regular part of the asymptotic velocity in the norm of (5.24) is O(ɛ5/2). It remains to estimate in the same norm the boundary layer functions. For the corrector corresponding to x1=1 we have
(5.25)∥ubl(0)1η∥(L2(ℬɛ))2≤ɛ3(∫-1/4ɛ0∫-11(u0(1)(ξ1,ξ2,t))2dξ)1/2≤ɛ3(∫-∞0∫-11(u0(1)(ξ1,ξ2,t))2dξ)1/2=O(ɛ3),
which yields
(5.26)∥ubl(0)1η∥L∞(0,T;(L2(ℬɛ))2)=O(ɛ3).
Finally, for the corrector in x=0 we have
(5.27)∥ubl(0)0η·η∥(L2(ℬɛ))2≤ɛ3(∫ℬ∩{ξ1<1/ɛ4,ξ2<1/ɛ4}(u0(0)(ξ1,ξ2,t))2dξ)1/2=O(ɛ3),
that leads to an estimate of the same type as (5.26). Hence, we obtain (5.23)1 as a consequence of (5.24) and (5.26). More details concerning this type of computations can be found in [5, Theorem  5.1].The previous Proposition allows us to improve the estimates given by Theorem5.3 in the following sense.Theorem 5.5.
Let(ûa(j),p̂a(j),d̂±a(j)) be the asymptotic solution of order j and (u,p,d) the exact solution of the physical problem. Then the following estimates hold:
(5.28)∥u-ûa(j)∥L∞(0,T;(L2(ℬɛ))2)=O(ɛj+7/2),∥D(u-ûa(j))∥L2((0,T);(L2(ℬɛ))4)=O(ɛj+5/2),∥∂∂t(d±-d̂±a(j))∥L∞(0,T;L2(Γ±ɛ))=O(ɛj+γ+1),∥∂2∂s2(d±-d̂±a(j))∥L∞(0,T;L2(Γ±ɛ))=O(ɛj+γ+1),∥∇(p-p̂a(j))∥L∞((0,T);(H-1(ℬɛ))2)=O(ɛj+3/2).Proof.
Letj≥0 be a fixed integer and k≫j. Then from (5.21)1 and (5.23)1 we get
(5.29)∥u-ûa(j)∥L∞(0,T;(L2(ℬɛ))2)≤∥u-ûa(k)∥L∞(0,T;(L2(ℬɛ))2)+∥ûa(k)-ûa(j)∥L∞(0,T;(L2(ℬɛ))2)=O(ɛmin{k-5/2,k+4-γ})+O(ɛj+7/2)=O(ɛj+7/2).
The other estimates of (5.28) are obtained in a similar way, and the proof is achieved.



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