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Applicable Analysis
An International Journal
Volume 99, 2020 - Issue 12
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Articles

Asymptotic analysis of the nonsteady micropolar fluid flow through a curved pipe

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Pages 2045-2092 | Received 11 Aug 2018, Accepted 20 Nov 2018, Published online: 12 Dec 2018

ABSTRACT

We consider the nonsteady flow of a micropolar fluid in a thin (or long) curved pipe via rigorous asymptotic analysis. Germano's reference system is employed to describe the pipe's geometry. After writing the governing equations in curvilinear coordinates, we construct the asymptotic expansion up to a second order. Obtained in the explicit form, the asymptotic approximation clearly demonstrates the effects of pipe's distortion, micropolarity and the time derivative. A detailed study of the boundary layers in space is provided as well as the construction of the divergence correction. Finally, a rigorous justification of the proposed effective model is given by proving the error estimates.

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1. Introduction

The model of micropolar fluids was first introduced in 1960s by Eringen in his well-known paper [Citation1], providing a generalization of the classical Navier–Stokes model. The main advantage of the micropolar fluid model is due to the fact that it takes into account the microstructure of the fluid particles and effects such as rotation and shrinking. Consequently, numerous non-Newtonian fluids such as liquid crystals, animal blood, muddy fluids and even water in small scales can be described in a more precise and realistic manner. In order to describe the rotation of the particles, a new vector field is introduced – the angular velocity field of rotation (microrotation), and accordingly a new equation coming from the conservation of the angular momentum. In this way, we obtain a coupled system of partial differential equations with four new viscosity coefficients. The model of micropolar fluids has been a subject of research in both the mathematical and engineering community: a comprehensive survey of the underlying mathematical theory can be found in the monograph by Lukaszezwicz [Citation2], whereas recent results concerning engineering applications in biomedicine and blood flow modelling can be found in [Citation3–8].

In last two decades, one can also find a number of papers providing a rigorous derivation of new asymptotic models for steady-state flows in thin domains. The model describing the flow of Newtonian fluid in a three-dimensional curved pipe has been formally derived and justified via error estimate by Marušić–Paloka [Citation9]. The asymptotic behaviour of a steady flow in a system of thin pipes has been considered by the same author in [Citation10]. The steady flow of a micropolar fluid has been investigated by Dupuy, Panasenko and Stavre [Citation11,Citation12] in two-dimensional domains such as a periodically constricted tubes and curvilinear channels. A three-dimensional setting including an undeformed pipe and a curved pipe has been addressed by Pažanin [Citation13,Citation14], while multiple pipe system has been treated by Beneš and Pažanin [Citation15].

Most recently, new asymptotic models for the nonsteady flows have been proposed and mathematically justified in case of classical, Newtonian fluid by Panasenko and Pileckas [Citation16–18] in an infinite cylinder and a system of thin pipes. The asymptotic behaviour of nonsteady Newtonian fluid flow in a curved three-dimensional thin pipe with moving walls has been considered by Castineira et al. [Citation19,Citation20]). A higher-order model describing the nonsteady flow of a micropolar fluid through a straight pipe has been rigorously derived in [Citation21,Citation22] by the authors of this paper. In view of that and inspired by the applications, our aim here is to investigate a nonsteady micropolar flow through a pipe with an arbitrary (generic) central curve.

The paper is organized as follows. In Section 2, we formally describe the geometry of the curved pipe using Germano's frame of reference (introduced in [Citation23,Citation24]) and consider the equations describing the nonsteady micropolar fluid flow. In Section 3, we introduce the curvilinear coordinates and rewrite the governing problem in an undeformed pipe. In Section 4, using two-scale expansion technique, we compute the asymptotic approximation up to a second-order in order to capture the effects of the pipe's geometry, micropolarity as well as the time derivative. It should be emphasized that the asymptotic solution is provided in the form of the explicit formulae, which is particularly important with regards to numerical simulations. We also conduct a detailed boundary layer analysis in the vicinity of pipe's ends and construct the divergence correction to improve the order of accuracy of our model. In Section 5, we prove the error estimates in suitable norms evaluating the difference between the exact solution of the problem (which cannot be found due to the complexity of the flow domain and the equations) and our second-order asymptotic approximation. By doing that, we justify the usage of the proposed effective model and that represents our main contribution. We strongly believe that the presented result can be useful in the practice, especially one related to blood flow modelling. Finally, in the appendix, we provide complete expressions computed for the higher-order regular part correctors as well as for the exponentially decreasing functions appearing in the boundary layers problems.

2. The setting of the problem

2.1. Description of the pipe's geometry

In the following, we define our thin (or long) curved pipe Ωεα with a smooth central curve γ and a circular cross-section. We suppose that the curve γ is a generic curve in R3 parameterized by its arc length x1[0,l]. We denote by πC3([0,l];R3) its natural parametrization such that π(x1) 0, for every x1[0,l]. At each point, π(x1) of the curve γ we define its curvature (flexion) as κ(x1)=|π(x1)| and introduce Frenet's basis by t=π,n=1κt,b=t×n, where t is the tangent, n the normal and b the binormal. The torsion of γ is denoted by τ(x1)=|b(x1)|. It is known that the Frenet's basis (t,n,b) satisfies the system (1) t=κn,n=κt+τb,b=τn.(1) Let 0<ε1 be a small parameter and B=B(0,1)R2 a unit circle. We define the undeformed thin pipe as Ωε={x=(x1,x2,x3)R3:x10,l, x=(x2,x3)εB}=0,l×Bε. Next, we introduce the mapping Φεα:ΩεR3 by (2) Φεα(x)=π(x1)+x2nα(x1)+x3bα(x1),(2) where (3) nα(x1)=cosα(x1)n(x1)+sinα(x1)b(x1),bα(x1)=sinα(x1)n(x1)+cosα(x1)b(x1),(3) and (4) α(x1)=x0x1τ(ξ)dξ+α0,(4) with x0 and α0 being arbitrary constants. The local injectivity of Φεα can be easily established assuming ε is sufficiently small (see [Citation14, Section 3.1] for details).

Figure 1. Germano's frame of reference.

Figure 1. Germano's frame of reference.

We can now define the curved pipe with the central curve γ and circular cross-section Bε by putting Ωεα=Φεα(Ωε). We also denote by Σεi=Φεα({i}×Bε), i=0,l the ends of the pipe and by Γεα=Φεα(0,l×Bε) its lateral boundary.

2.2. The flow equations

The equations describing the nonsteady flow of a micropolar fluid in a thin curved pipe Ωεα read: (5) uεtμΔuε+pε=arotwε+F,divuε=0,inΩεα,wεtδΔwεβdivwε+2awε=arotuε+G.(5) Here uε is the velocity field, pε is the pressure and wε is the microrotation field. The external sources of linear and angular momentum are given by the functions F(t,x~)=f(t,x1) and G(t,x~)=g(t,x1), where x~=Φεα(x). We employ the notation μ=ν+νr, a=2νr, δ=ca+cd and β=c0+cdca, where ν is the Newtonian viscosity, νr is the microrotation viscosity, while c0,ca and cd are the coefficients of angular viscosities.

We prescribe the following boundary and initial condition: (6) uε=0 on Γεα,uε=ε2hiε on Σεi,i=0,l,wε=0 on Ωεα,uε(0,x~)=0,wε(0,x~)=0,(6) where we suppose that the prescribed velocities h0ε,hlε are given in the form (7) h0ε(t,x~)=h0t(t,xε)t(0)+h0n(t,xε)n(0)+h0b(t,xε)b(0),(7) where x~=Φ(0,x), and (8) hlε(t,x~)=hlt(t,xε)t(l)+hln(t,xε)n(l)+hlb(t,xε)b(l),(8) where x~=Φ(l,x). To assure the well-posedness of the governing problem, the following compatibility condition needs to be fulfilled: (9) F(t)=Bεhit(t,xε)=ε2Bhit(t,y)=ε2F0(t),i=0,l.(9) Since the initial velocity equals to zero, there also needs to hold (10) F(0)=0.(10) In the following, we suppose that h0ε,hlε,F and G vanish in the neighbourhood of t=0, i.e. (11) h0ε(t,),hlε(t,),F(t,),G(t,),t[0,T],T>0.(11) Let jh0εtjL2(0,T;H1/2(Σε0)) and jhlεtjL2(0,T;H1/2(Σεl)) for j=0,1 and let hεL2(0,T;W1,2(Ωεα)) with hεtL2(0,T;W1,2(Ωεα)) be the solution of the following problem (see Lemma 5.2): divhε=0 in Ωεα,hε=0 on Γεα,hε=ε2hiε on Σεi, i=0,l. A weak solution of our problem (Equation5)–(Equation11) is a pair (uε,wε)=(vε+hε,wε) such that divvε=0,vε(0,x~)=0,wε(0,x~)=0,vε,wεL2(0,T;W1,2(Ωεα))L(0,T;L2(Ωεα)),vt,wtL2(0,T;L2(Ωεα)), and (vε,wε) satisfy the integral identities (12) Ωεαvεtϕ+μΩεαvεϕ=aΩεαrotwεϕ+ΩεαFϕΩεαhεtϕΩεαhεϕ,ϕW1,2(Ωεα),divϕ=0,Ωεαwεtψ+δΩεαwεψ+βΩεαdivwεdivψ+2aΩεαwεψ=aΩεαrot(vε+hε)ψ+ΩεαGψ,ψW1,2(Ωεα).(12) Let F, GL2(0,T;L2(Ωεα)). There then exists a unique weak solution (vε,wε) satisfying (12) for a.e. t[0,T]. The proof follows the same arguments as the proof of Theorem 2.1.1 from [Citation2, Chapter III, Section 2] which can be straighforwardly adapted to our setting.

3. Rewriting the problem

In this section, we will provide the geometric tools necessary to write the problem (Equation5)–(Equation11) on the reference domain Ωε, allowing us to perform the asymptotic analysis.

3.1. Covariant and contravariant basis

Let us consider the following mapping: Φ~εα:[0,T]×Ωε[0,T]×Ωεα,Φ~εα(t,x)=(t~,x~)=(t~,Φεα(x)), We introduce the covariant basis as the gradient of the mapping Φεα consisting of vectors ai(x)=Φεα(x)xi,i=1,2,3. Using (Equation2)–(Equation3) and Frenet's relations (Equation1), one can easily obtain a1=(1κ(eαx))t(α+τ)(eαx)n+(α+τ)(eαx)b,a2=cosαn+sinαb,a3=sinαn+cosαb, where (13) eα=(cosα,sinα),eα=(sinα,cosα).(13) From (Equation4), it directly follows that α=τ, so we have a1=(1κ(eαx))t. The contravariant basis is the dual basis to the covariant basis defined by the relation aiaj=δij,i,j=1,2,3. It can be easily verified that a1=11κ(eαx)t,a2=cosαn+sinαb,a3=sinαn+cosαb. More details can be found in [Citation14, Section 3.1].

3.2. Micropolar equations in the reference domain

Let us denote the scalar field Pε=pεΦ~εα, where Φ~εα:[0,T]×Ωε[0,T]×Ωεα,pε:[0,T]×ΩεαR and Pε:[0,T]×ΩεR.

We also introduce the vector fields Uε=uεΦ~εα=Vε1a1+Vε2a2+Vε3a3,Wε=wεΦ~εα=Wε1a1+Wε2a2+Wε3a3, where uε,wε:[0,T]×ΩεαR3 and Uε,Wε:[0,T]×ΩεR3. By Vεi=Uεai,Wεi=Wεai, we denote the corresponding covariant components.

Using the expressions for the differential operators in curvilinear coordinates derived in [Citation14, Section 3.3], we can write the equations (Equation5) on the reference domain Ωε.

The equation (Equation5)1 expressing the balance of momentum in the reference domain Ωε takes the following form: Vε1t(1+κ(eαx)+)t+Vε2ta2+Vε3ta3μ(ΔVε1+κcosα(Vε1x2Vε2x1)+κsinα(Vε3x1Vε1x3))t+μ(x1(Vε2κcosαVε3κsinακ(Vε2cosα+Vε3sinα))+κ2Vε1κ(eαx)(ΔVε1κcosαVε1x2))tμ(κ2(eαx)sinαVε1x3+κ2(eαx)2ΔVε1)tμ(ΔVε2κcosαVε2x2+κsinαVε2x3)a2μ(2κ(eαx)(κcosαVε2x2+κsinαVε2x3)+(2κVε1x1+κVε1κ2Vε2)cosα+κτVε1sinα)a2μ(κ2cosα(eαx)Vε2x2κ2sinα(eαx)Vε2x3)a2μ(ΔVε3+κsinαVε3x3κcosαVε3x2)a3μ(2κ(eαx)(κsinαVε3x3κcosαVε3x2)+κτVε1cosα(2κVε1x1+κVε1κ2Vε2)sinα)a3 (14) μ(κ2cosα(eαx)Vε3x2κ2sinα(eαx)Vε3x3)a3+Pεx1(1+κ(eαx)+)t+Pεx2a2+Pεx3a3=a(Wε3x2Wε2x3)t+a(1+κ(eαx)+)(Wε1x3Wε3x1)a2+a(1+κ(eαx)+)(Wε2x1Wε1x2)a3+f1t+(f2cosα+f3sinα)a2+(f3cosαf2sinα)a3.(14) The equation (Equation5)2 expressing the balance of mass can be written in the reference domain Ωε as: (15) [11κ(eαx)]2(Vε1x1κ(Vε2cosαVε3sinα)+(κ(eαx)+κτ(eαx))Vε1+κ(eαx)(κ(eαx)+κτ(eαx))Vε1+κ2(eαx)cosαVε2κ2(eαx)sinαVε3)+Vε2x2+Vε3x3=Vε1x1κ(Vε2cosαVε3sinα)+(κ(eαx)+κτ(eαx))Vε1+κ(eαx)(κ(eαx)+κτ(eαx))Vε1+κ2(eαx)cosαVε2κ2(eαx)sinαVε3+Vε2x2+Vε3x3+2κ(eαx)(Vε1x1κ(Vε2cosαVε3sinα)+(κ(eαx)+κτ(eαx))Vε1+κ2(eαx)cosαVε2κ2(eαx)sinαVε3)+3κ2(eαx)2(Vε1x1κ(Vε2cosαVε3sinα))+4κ3(eαx)3Vε1x1+O(ε3)=0.(15) Finally, the equation (Equation5)3 describing the balance of angular momentum in the reference domain Ωε reads: Wε1t(1+κ(eαx)+)t+Wε2ta2+Wε3ta3δ(ΔWε1+κcosα(Wε1x2Wε2x1)+κsinα(Wε3x1Wε1x3))t+δ(x1(Wε2κcosαWε3κsinακ(Wε2cosαWε3sinα))+κ2Wε1κ(eαx)(ΔWε1κcosαWε1x2))tδ(κ2(eαx)sinαWε1x3+κ2(eαx)2ΔWε1)tδ(ΔWε2κcosαWε2x2+κsinαWε2x3)a2δ(2κ(eαx)(κcosαWε2x2+κsinαWε2x3)+(2κWε1x1+κWε1κ2Wε2)cosα+κτWε1sinα))a2δ(κ2cosα(eαx)Wε2x2κ2sinα(eαx)Wε2x3)a2 δ(ΔWε3+κsinαWε3x3κcosαWε3x2)a3δ(2κ(eαx)(κsinαWε3x3κcosαWε3x2)+κτWε1cosα(2κWε1x1+κWε1κ2Wε2)sinα)a3δ(κ2cosα(eαx)Wε3x2κ2sinα(eαx)Wε3x3)a3β((1+κ(eαx))(2Wε1x12(κcosα)Wε2)tβ(1+κ(eαx))(κcosαWε2x1+(κsinα)Wε3+κsinαWε3x1+2Wε2x1x2+2Wε3x1x3)tβ(2Wε1x1x2κ(Wε2x2cosαWε3x2sinα)+(κWε1+κ2cosαWε2κ2sinαWε3)cosα+2κcosαWε1x1)a2β(2κ2(Wε2cosαWε3sinα)cosα+κτWε1sinα+(κ(eαx)+κτ(eαx))Wε1x2+κ2(eαx)Wε2x2cosα)a2β(κ2(eαx)Wε3x2sinα+2Wε2x22+2Wε3x2x3+2κ(eαx)(2Wε1x1x2κ(Wε2x2cosαWε3x2sinα)))a2 (16) β(2Wε1x1x3κ(Wε2x3cosαWε3x3sinα)+κτWε1cosα(κWε1+κ2cosαWε2κ2sinαWε3)sinα)a3β(2κ(Wε1x1κ(Wε2cosαWε3sinα))sinα+κ(eαx)Wε1x3+κτ(eαx)Wε1x3+κ2(eαx)cosαWε2x3)a3β(κ2(eαx)Wε3x3sinα+2Wε2x2x3+2Wε3x32+2κ(eαx)(2Wε1x1x3κ(Wε2x3cosαWε3x3sinα))]a3+2a(1+κ(eαx)+)Wε1t+2aWε2a2+2aWε3a3=a(Vε3x2Vε2x3)t+a((1+κ(eαx)+)(Vε1x3Vε3x1))a2+a((1+κ(eαx)+)(Vε2x1Vε1x2))a3+g1t+(g2cosα+g3sinα)a2+(g3cosαg2sinα)a3.(16) Note that f=f1t+f2n+f3b,g=g1t+g2n+g3b.

4. Asymptotic expansion

4.1. Regular part

In this section, we construct the regular part of the asymptotic approximation of the solution (uε,wε,pε). In view of that, let us introduce Uε=uεΦ~εα=Vε1a1+Vε2a2+Vε3a3,Wε=wεΦ~εα=Wε1a1+Wε2a2+Wε3a3,Pε=pεΦ~εα and formally expand (for i=1,2,3) (17) Vεi(t,x)=ε2V0i(t,x1,x2ε,x3ε)+ε3V1i(t,x1,x2ε,x3ε)+ε4V2i(t,x1,x2ε,x3ε)+Wεi(t,x)=ε2W0i(t,x1,x2ε,x3ε)+ε3W1i(t,x1,x2ε,x3ε)+ε4W2i(t,x1,x2ε,x3ε)+Pε(t,x)=P0(t,x1)+εP1(t,x1,x2ε,x3ε)+ε2P2(t,x1,x2ε,x3ε)+ε3P3(t,x1,x2ε,x3ε)+(17)

4.1.1. Zero-order approximation

We first compute the zero-order approximation for the velocity V0=(V01,V02,V03) and the microrotation W0=(W01,W02,W03). Plugging the expansions (Equation17) into the equations (Equation14)–(Equation15) and collecting the ε0 terms, we obtain the following: μ(ΔyV01t+ΔyV02a2+ΔyV03a3)+P0x1t+P1y2a2+P1y3a3=f1t+(cosαf2+sinαf3)a2+(sinαf2+cosαf3)a3. where y=xε. In this way, we obtain the problem for the velocity and pressure zero-order approximation posed in Ω=0,l×B: (18) μ(ΔyV01,ΔyV02,ΔyV03)+(P0x1,P1y2,P1y3)=(f1,f2cosα+f3sinα,f3cosαf2sinα) in Ω,divyV0=0 in Ω,V0=0 on Γ=0,l×B.(18) The system (Equation18) can be solved by taking (19) V01(t,x1,y)=14μ(1|y|2)(f1(t,x1)P0(x1,t)x1),V02=V03=0,P1(t,x1,y)=f2(t,x1)(eα(x1)y)+f3(t,x1)(eα(x1)y),(19) where eα(x1) and eα(x1) are given by (Equation13).

To determine the zero-order pressure approximation, we collect the ε2 terms in the divergence equation to obtain: (20) V01x1+divyV1=0in Ω.(20) Integrating (Equation20) over B and using the divergence theorem yields x1BV01=0. The boundary conditions give BV01(t,0,y)=BV01(t,l,y)=F0(t), (justified in Section 4.2) implying (21) BV01=F0(t).(21) From (Equation19) and (Equation21), we now get π8μ(f1(t,x1)P0(t,x1)x1)=F0(t) to obtain (22) P0(t,x1)=8μπF0(t)x1+0x1f1(t,ξ)dξ+p0(t),(22) where p0(t) is an arbitrary function of time. It now follows from (Equation19)1 and (Equation22) that (23) V01(t,y)=2π(1|y|2)F0(t).(23) We now turn our attention to computing the microrotation zero-order approximation W0=(W01,W02,W03). Plugging the expansions (Equation17) into (Equation16), we arrive at the following problem: δΔyW0βy(divyW0)=(g1,g2cosα+g3sinα,g3cosαg2sinα) in Ω,W0=0 on Γ leading to (24) W01(t,x1,y)=14δ(1|y|2)g1(t,x1),W02(t,x1,y)=12(2δ+β)(1|y|2)(g2(t,x1)cosα(x1)+g3(t,x1)sinα(x1)),W03(t,x1,y)=12(2δ+β)(1|y|2)(g3(t,x1)cosα(x1)g2(t,x1)sinα(x1)).(24) As expected, the zero-order velocity and microrotation approximation given by (Equation23) and (Equation24) do not feel the effects of curvature, torsion, micropolarity or the time derivative so we need to compute the first- and second-order correctors to capture those effects.

4.1.2. First-order corrector

Again, substituting (Equation17) into the Equation (Equation14) and collecting the terms of order ε, we get the following equation for the first component of the velocity first-order corrector V11: (25) μ(ΔyV11+κcosαV01y2κsinαV01y3+κ(eαy)ΔyV01)+κ(eαy)P0x1+P1x1=a(W03y2W02y3)inΩ.(25) Substituting the expressions for V01,P0,P1,W02 and W03 derived in the previous section and given by (Equation19), (Equation22), (Equation23) and (Equation24) into (Equation25), we obtain the following problem for V11: (26) μΔyV11=y2H1(t,x1)+y3H2(t,x2) in Ω,V11=0 on Γ,(26) where the functions H1(t,x1) and H2(t,x1) are provided in the appendix (see (EquationA1)).

It can be easily shown that the solution of problem (Equation26) is given by (27) V11(t,x1,y)=18μ(1|y|2)(y2H1(t,x1)+y3H2(t,x1)).(27) In a similar manner, we obtain the equations for the two other velocity components V1=(V12,V13): (28) μ(ΔyV12κcosαV02y2+κsinαV02y3,ΔyV13+κsinαV03y3κcosαV03y2)+(P2y2,P2y3)=a(W01y3,W01y2) in Ω,V01x1κV02+V12y2+V13y3=0 in Ω,V1=0 on Γ.(28) Applying the expressions for V01,V02,V03 and W01 given by (Equation19), (Equation23) and (Equation24) into the equation (28), we arrive at: (29) μΔyV1+yP2=a2δg1(y3,y2) in Ω,divyV1=0 in Ω,V1=0 on Γ.(29) The solution of the problem (Equation29) reads: (30) V12(t,x1,y)=a16μδg1(t,x1)(1|y|2)y3,V13(t,x1,y)=a16μδg1(t,x1)(1|y|2)y2,P2(t,x1)=x1p1(t)+12x12(p2(t)p1(t)l),(30) where p1(t) and p2(t) will be determined in Section 4.2.1 from the compatibility conditions related to the second-order boundary layer correctors for the velocity and pressure.

We will now determine the expressions for the microrotation first-order correctors. Plugging (Equation17) into the Equation (Equation16) and collecting the terms of order ε, we obtain: (31) δ(ΔyW11+κcosαW01y2κsinW01y3+κ(eαy)ΔyW01)β(2W02x1y2+2W03x1y3)=a(V03y2V02y3).(31) Employing the expressions for V02,V03,W01,W02 and W03 given by (Equation19) and (Equation24) in the Equation (Equation31), we get the following problem for W11: (32) δΔyW11=y2H3(t,x1)+y3H4(t,x1) in Ω,W11=0 on Γ,(32) where H3(t,x1) and H4(t,x1) are given in the appendix (see (EquationA2)).

It follows (33) W11(t,x1,y)=18μ(1|y|2)(y2H3(t,x1)+y3H4(t,x1)).(33) Substituting the expansions (Equation17) into the Equation (Equation16), we deduce the equations for (W12,W13): (34) δ(ΔyW12κcosαW02y2+κsinαW02y3)β(2W01x1y2κcosαW02y2+κsinαW03y2+2W12y22+2W13y2y3)=aV01y3,δ(ΔyW13+κsinαW03y3κcosαW03y2)β(2W01x1y3κcosαW02y3+κsinαW03y3+2W13y32+2W12y2y3)=aV01y2.(34) In view of the expressions for V01,W01,W02 and W03 given by (Equation19), (Equation23) and (Equation24), we obtain the following problem for (W12,W13): (35) δΔyW12β(2W12y22+2W13y2y3)=y2H5+y6H6 in Ω,δΔyW13β(W13y32+W12y2y3)=y2H7+y3H8 in Ω,W12=W13=0 on Γ,(35) where H5(t,x1),,H8(t,x1) can be found in the appendix (see (EquationA3)).

The solution of the problem (Equation35) can be written as: (36) W12(t,x1,y)=(1|y|2)(y2H9(t,x1)+y3H10(t,x1)),W13(t,x1,y)=(1|y|2)(y2H11(t,x1)+y3H12(t,x1)),(36) where H9(t,x1),,H12(t,x1) are given in the appendix (see (EquationA4)).

Looking at the obtained expressions for the velocity and microrotation first-order correctors (see (Equation27), (Equation30), (Equation33), (Equation36) with (EquationA1)–(EquationA4)), we can clearly see the effects of the curvature, torsion and micropolarity, as in the stationary case (see [Citation14, Section 4.2]). In order to capture the effects of the time derivative as well, we need to continue the computation and construct the second-order correctors.

4.1.3. Second-order corrector

Plugging the expansions (Equation17) into the Equation (Equation14)–(Equation15) and collecting the terms of order ε2, we get the equation for the first component of the velocity second-order corrector V21: (37) V01tμ(2V01x12+ΔyV21+κcosαV11y2κsinαV11y3κ2V01+κ(eαy)(ΔyV11κcosαV01y2+κsinαV01y3))μκ2(eαy)2ΔyV01+κ2(eαy)2P0x1+κ(eαy)P1x1+P2x1=a(W13y2W12y3).(37) Inserting the expressions for V01,V11,P0,P1,P2,W12 and W13 given by (Equation19) (Equation22), (Equation23), (Equation27), (Equation30) and (Equation36) into (Equation37) yields (38) ΔyV21=A1y22+A2y2y3+A3y32+A4 in Ω,V21=0 on Γ,(38) where the functions A1(t,x1),,A4(t,x1) are provided in the appendix (see (EquationA5)).

The solution of (Equation38) is given in the form V21(t,x1,y)=(|y|21)[B1y22+B2y2y3+B3y32+B4], where B1(t,x1)=7A1A396, B2(t,x1)=112A2, B3(t,x1)=7A3A196, B4(t,x1)=14A4+132(A1+A3). It is important to notice that the effects of the time derivative appear in the second-order corrector as the derivative of the flux function F0(t) (see (EquationA5)).

In a similar way, we now obtain the equations for the second and third component of the velocity second-order corrector V2=(V22,V23): (39) μ(ΔyV22κcosαV12y2+κsinαV12y3+(2κV01x1+κV01)cosα+κτV01sinα)+P3y2=a(W11y3+κ(eαy)W01y3W03x1) in Ω,μ(ΔyV23+κsinαV13y3κcosαV13y2+κτV01cosα(2κV01x1+κV01)sinα)+P3y3=a(W02x1W11y2κ(eαy)W01y2) in Ω,V22y2+V23y3=V11x1+κ(V12cosαV13sinα)(κ(eαy)+κτ(eαy))V010 in Ω,V22=V23=0 on Γ.(39) The system of equations (Equation39) admits a unique solution since B(V11x1+κ(V12cosαV13sinα)(κ(eαy)+κτ(eαy))V01)dB=0 (see [Citation25]). The explicit solution is constructed below.

Inserting the expressions for V01,W01,W02,W03,V11V12,V13,W11 given by (Equation23), (Equation24), (Equation27), (Equation30) and (Equation33) into (Equation39), we get: (40) μΔyV22+P3y2=A5y22+A6y2y3+A7y32+A8,μΔyV23+P3y3=A9y22+A10y2y3+A11y32+A12,V22y2+V23y3=A13y23+A14y33+A13y2y32+A14y22y3A13y2A14y3,V22=V23=0 on Γ,(40) where A5(t,x1),,A14(t,x1) are given in the appendix (see (EquationA6)).

We seek the solution of (Equation40) in the following form: (41) V22(t,x1,y)=(1|y|2)(B5y22+B6y2y3+B7y23+B8),V23(t,x1,y)=(1|y|2)(B9y22+B10y2y3+B11y23+B12),P3(t,x1,y)=M1y23+M2y33+M3y2y32+M4y22y3+M5y2+M6y3.(41) Plugging (Equation41) into the system (Equation40), we obtain a system of equations relating B5(t,x1),,B12(t,x1) and M1(t,x1),,M6(t,x1) with A5(t,x1),,A14(t,x1). Solving this system, we deduce B5(t,x1),,B12(t,x1) and M1(t,x1),,M6(t,x1) (see (EquationA7)–(EquationA8) in the appendix).

In this way, we have obtained the explicit expressions for the velocity second-order corrector (V22,V23) and pressure third-order corrector P3 and we now compute the microrotation second-order corrector in the following.

Substituting (Equation17) into (Equation16) and collecting the ε2 terms, we obtain the following problem for the first component of the microrotation second-order corrector W21: (42) W01tδ(2W01x12+ΔyW21+κcosα(W11y2W02x1)+κsinα(W03x1W11y3))δx1(W02κcosαW03κsinακ(W02cosαW03sinα))+δκ2W01δκ(eαy)(ΔyW11κcosαW01y2+κsinαW01y3)δκ2(eαy)2ΔyW01β(2W01x12(κcosα)W02κcosαW02x1+(κsinα)W03+κsinαW03x1+2W12x1y2+2W13x1y3)β(κ(eαy)2W02x1y2+κ(eαy)W03x1y3)+2aW01=a(V13y2V12y3) in Ω,W21=0 on Γ.(42) Inserting the expressions for W01,W02,W03,V12,V13,W11,W12,W13 given by (Equation24), (Equation30), (Equation33) and (Equation36) into (Equation42) yields (43) ΔyW21=C1y22+C2y2y3+C3y32+C4 in Ω,W21=0 on Γ,(43) where the functions C1(t,x1),,C4(t,x1) are given in the appendix (see (EquationA9)).

We find the solution of the problem (Equation43) in the form (44) W21(t,x1,y)=(|y|21)(D1y22+D2y2y3+D3y32+D4),(44) where (45) D1=7C1C396,D2=112C2,D3=7C3C196,D4=14C4+132(C1+C3).(45)

Plugging expansions (Equation17) into equation (Equation16) and collecting the ε2 terms, we finally obtain the problem for the second and third component of the microrotation second-order corector (W22,W23) as: W02tδ(2W02x12+ΔyW22κcosαW12y2+κsinαW12y3+2κ(eαy)(κcosαW02y2+κsinαW02y3))δ((2κW01x1+κW01κ2W02)cosα+κτW01sinα+κ2cosα(eαy)W02y2κ2sinα(eαy)W02y3)β(2W11x1y2κ(W12y2cosαW13y2sinα)+(κW01+κ2cosαW02κ2sinαW03)cosα)2βκ(W01x1κ(W02cosαW03sinα))cosαβ(κτW01sinα+κ(eαy)W01y2+κτ(eαy)W01y2)βκ2(eαy)(W02y2cosαW03y3sinα)β(2W22y22+2W23y2y3)2βκ(eαy)(2W01x1y2κ(W02y2cosαW03y2sinα))+2aW02=a(V11y3+κ(eαy)V01y3V03x1) in Ω,W03tδ(2W03x12+ΔyW23+κsinαW13y3κcosαW13y2+2κ(eαy)(κsinαW03y3κcosαW03y2)) (46) δ(κτW01cosα(2κW01x1+κW01κ2W02)sinα+κ2cosα(eαy)W03y2κ2sinα(eαy)W03y3)β(2W11x1y3κ(W12y3cosαW13y3sinα)+κτW01cosα(κW01+κ2cosαW02κ2sinαW03))2βκ(W01x1κ(W02cosαW03sinα))sinαβ(κ(eαy)W01y3+κτ(eαy)W01y3)βκ2(eαy)(cosαW02y3sinαW03y3)β(2W23y32+2W22y2y3)2βκ(eαy)(2W01x1y3κ(W02y3cosαW03y3sinα))+2aW03=a(V02x1V11y2κ(eαy)V01y2) in Ω,W22=W23=0 on Γ.(46) Plugging the expressions for V01,V02,V03,V11,W01,W02,W03,W11,W12,W13 given by (Equation19), (Equation23), (Equation24), (Equation27), (Equation33) and (Equation36) into (Equation46) yields: (47) δ(2W22y22+W22y32)β(2W22y22+W23y2y3)=C5y22+C6y2y3+C7y32+C8 in Ω,δ(2W23y22+2W23y32)β(2W22y2y3+2W23y32)=C9y22+C10y2y3+C11y32+C12 in Ω,W22=W23=0 on Γ.(47) The functions C5(t,x1),,C12(t,x1) are given in the appendix (see (EquationA10)–(EquationA11)).

The solution of the problem (Equation47) takes the following form (48) W22(t,x1,y)=(|y|21)(D5y22+D6y2y3+D7y32+D8),W23(t,x1,y)=(|y|21)(D9y22+D10y2y3+D11y32+D12),(48) where D5(t,x1),,D12(t,x1) can be found in the appendix (see (EquationA12)).

To conclude this subsection, we notice that the expressions for (W21,W22,W23) given by (Equation44)–(Equation45) and (48) contain the effects of the time derivative of the external force function g.

4.2. Boundary layers

The regular part of our expansion (49) Uε,reg[2](t,x)=ε2V0(t,xε)+ε3V1(t,x1,xε)+ε4V2(t,x1,xε),Vj=Vj1a1+Vj2a2+Vj3a3,j=0,1,2,Wε,reg[2](t,x)=ε2W0(t,x1xε)+ε3W1(t,x1,xε)+ε4W2(t,x1,xε),Wj=Wj1a1+Wj2a2+Wj3a3,j=0,1,2,Pε,reg[3](t,x)=P0(t,x1)+εP1(t,x1,xε)+ε2P2(t,x1,xε)+ε3P3(t,x1,xε)(49) has been computed to satisfy the governing equations and the lateral boundary conditions, while the boundary conditions at the ends of the pipe have been neglected in the process. To fix that, we introduce the boundary layer correctors at x1=0: (50) Uε,bl,0[2](t,x)=ε2V0(t,x1ε,xε)+ε3V1(t,x1ε,xε)+ε4V2(t,x1ε,xε),Vj=Vj1a1+Vj2a2+Vj3a3,j=0,1,2,Wε,bl,0[2](t,x)=ε2W0(t,x1ε,xε)+ε3W1(t,x1ε,xε)+ε4W2(t,x1ε,xε),Wj=Wj1a1+Wj2a2+Wj3a3,j=0,1,2,Pε,bl,0[2](t,x)=εP0(t,x1ε,xε)+ε2P1(t,x1ε,xε)+ε3P2(t,x1ε,xε),(50) and the boundary layer correctors on the opposite side x1=l: (51) Uε,bl,l[2](t,x)=ε2Y0(t,x1lε,xε)+ε3Y1(t,x1lε,xε)+ε4Y2(t,x1lε,xε),Yj=Yja1+Yja2+Yja3,j=0,1,2,Wε,bl,l[2](t,x)=ε2Z0(t,x1lε,xε)+ε3Z1(t,x1lε,xε)+ε4Z2(t,x1lε,xε),Zj=Zj1a1+Zj2a2+Zj3a3,j=0,1,2,Pε,bl,l[2](t,x)=εQ0(t,x1lε,xε)+ε2Q1(t,x1lε,xε)+ε3Q2(t,x1lε,xε).(51) The boundary layer corrector at x1=0 given by (Equation50) are defined on the semi-infinite cylinder G0=0,×B, whereas the boundary layer correctors at x1=l given by (Equation51) are defined on Gl=,0×B.

It should be emphasized that, due to (Equation11), the homogeneous initial conditions (Equation6)3 are automatically satisfied and for this reason the boundary layer in time does not appear.

4.2.1. Boundary layer for the velocity and pressure

Zero-order approximation: Substituting the boundary layer correctors (Equation50) into the Equations (Equation14)–(Equation15) and collecting the ε0 terms, we obtain the problem for the velocity and pressure zero-order boundary layer approximation at x1=0: (52) μΔV0+P0=0in G0,divV0=0 in G0,V0=0 on ω=0,×B,V0(t,0,)=(h0tV01(t,0,),h0ncosα(0)+h0bsinα(0),h0bcosα(0)h0nsinα(0)).(52) System (Equation52) admits a unique (up to an additive constant in the pressure) solution V0(t,)W1,2(G0),P0(t,)Lloc2(G0) since the necessary compatibility conditions holds B(h0tV01(t,0,))=F0(t)F0(t)=0. It is also well known that these functions in some sense exponentially decay to zero as y1 (see e.g. [Citation25]).

The problem for the velocity and pressure zero-order boundary layer approximations at x1=l given by (Equation51) is analogous as for (V0,P0) : (53) μΔY0+Q0=0in Gl,divY0=0in Gl,Y0=0 on σ=,0×B,Y0(t,0,)=(hltV01(t,l,),hlncosα(l)+hlbsinα(l),hlbcosα(l)hlnsinα(l)).(53) The well-posedness and the exponential decay of the solution (Y0,Q0) follow analogously as for (V0,P0).

First-order corrector: The first-order boundary layer correctors at x1=0 for the velocity and pressure (V1,P1) are given as the solution of the following problem: (54) μΔV1+P1=Ξ(t,0,V0,W0,P0) in G0,divV1=κ(0)cosα(0)V02κ(0)sinα(0)V03(2κ(0)cosα(0)y2V01y12κ(0)sinα(0)y3V01y1), V1=0 on ω=0,×B,V1(t,0,)=(V11(t,0,),V12(t,0,),V13(t,0,))(κ(0)(eα(0)y)V01(t,0,),0,0)(κ(0)(eα(0)y)V01(t,0,),0,0),(54) where Ξ(t,x1,J,K,j) is an exponentially decreasing function provided in the appendix (see (EquationA13)).

To ensure the well-posedness of the problem (Equation54), we need to verify the compatibility condition (55) G0divV1=BV11(t,0,).(55) Since div(y2V0)=V02,div(y3V0)=V03, we obtain G0V02=By2V01(t,0,)=By2(V01h0t)=By2h0t,G0V03=By3V01(t,0,)=By3(V01h0t)=By3h0t yielding G0divV1=κ(0)cosα(0)G0V02κ(0)sinα(0)G0V032κ(0)cosα(0)G0y2V01y1+2κ(0)sinα(0)G0y3V01y1=κ(0)cosα(0)By2h0t+κ(0)sinα(0)By3h0t2κ(0)cosα(0)G0y2V01y1+2κ(0)sinα(0)G0y3V01y1. On the other hand, we have BV11(t,0,)=BV11(t,0,)κ(0)cosα(0)By2V01(t,0,)+κ(0)sinα(0)By3V01(t,0,)κ(0)cosα(0)By2(h0tV01(t,0,))+κ(0)sinα(0)By3(h0tV01(t,0,))=κ(0)cosα(0)By2h0t+κ(0)sinα(0)By3h0t, and, in order to satisfy the compatibility condition (Equation55), we need to verify the following: (56) 2κ(0)cosα(0)G0y2V01y1+2κ(0)sinα(0)G0y3V01y1=2κ(0)cosα(0)By2h0t2κ(0)sinα(0)By3h0t.(56) Using the fact that V01 exponentially decays as y1, we have: G0y2V01y1=By2lima0aV01y1=By2lima(V01(t,a,)V0(t,0,))=By2(h0tV01(t,0,))=By2h0t and G0y3V01y1=By3h0t, thus verifying the compatibility condition (Equation56). Consequently, the well-posedness of the problem (Equation54) is established. Notice that (V1,P1) exponentially decay to zero as y1 since Ξ is an exponentially decreasing function in y1 (see [Citation25]).

The existence of the unique solution exponentially decaying to zero corresponding to the problem on the opposite side x1=l for the first-order boundary layer correctors (Y1,Q1) follows in a similar manner.

Second-order corrector: The second-order boundary layer correctors at x1=0 for the velocity and pressure (V2,P2) are given by: (57) μΔV2+P2=Π(t,0,V0,V1,W0,W1,P0,P1) in G0,divV2=κ(0)cosα(0)V12κ(0)sinα(0)V13(κ(0)(eα(0)y)+κ(0)τ(0)(eα(0)y))V01κ2(0)(eα(0)y)cosα(0)V02+κ2(0)(eα(0)y)sinα(0)V03+2κ(0)(eα(0)y)(κ(0)cosα(0)V02κ(0)sinα(0)V03)2κ(0)(eα(0)y)V11y13κ2(0)(eα(0)y)2V01y1 in G0,V2=0 in ω=0,×B,V2(t,0,)=V2(t,0,)(κ(eαy)V11(t,0,),0,0)(κ2(eαy)2V01(t,0,),0,0)(κ(eαy)V11(t,0,),0,0)(κ2(eαy)2V01(t,0,),0,0),(57) where Π(t,x1,J0,J1,K0,K1,j0,j1) is an exponentially decreasing function given in the appendix (see (EquationA14)).

To ensure the well-posedness of (Equation57), we check the compatibility condition (58) G0divV2=BV21(t,0,).(58) Using the relations V12=div(y2V1)κ(0)cosα(0)y2V02+κ(0)sinα(0)y2V03+(2κ(0)cosα(0)y22V01y12κ(0)sinα(0)y2y3V01y1),V13=div(y3V1)κ(0)cosα(0)y3V02+κ(0)sinα(0)y3V03+(2κ(0)cosα(0)y2y3V01y12κ(0)sinα(0)y32V01y1), and integrating (Equation57)2 over G0, the first two terms can be written as: κ(0)cosα(0)G0V12κ(0)sinα(0)G0V13=B(κ(0)cosα(0)y2κ(0)sinα(0)y3)V11(t,0,)+2κ2(0)cos2α(0)G0y22V01y14κ(0)sinαcosα(0)G0y2y3V01y1+2κ2(0)sin2(0)y32V01y1=B(κ(0)cosα(0)y2κ(0)sinα(0)y3)(V11(t,0,)+κ(0)(eα(0)y)h0t)+B(2κ2(0)cos2α(0)y224κ(0)sinα(0)cosα(0)y2y3+2κ2(0)sin2(0)y32)(V01(t,0,)h0t). while other terms in the divergence equations can be written in the following way: G0((κ(0)cosα(0)+κ(0)τ(0)sinα(0))y2+(κ(0)sinα(0)+κ(0)τ(0)cosα(0))y3)V01=B(((κ(0)cosα(0)+κ(0)τ(0)sinα(0))y2+(κ(0)sinα(0)+κ(0)τ(0)cosα(0))y3)0V01),G0(κ2(0)cos2α(0)y2V02κ2(0)sinαcosα(0)(y3V02+y2V03)+κ2(0)sin2α(0)y3V03)=B(κ2(0)cos2α(0)2y22κ2(0)sinα(0)cosα(0)y2y3+κ2(0)sin2α(0)2y32)(V01(t,0,)h0t). Using the fact that V01,V11 exponentially decay to zero as y1, we get: G02κ(0)(cosα(0)y2sinα(0)y3)V11y1G03κ2(0)(cos2α(0)y222sinα(0)cosα(0)y2y3+sin2(0)y32)V01y1=B(2κ(0)cosα(0)y2+2κ(0)sinα(0)y3)(V11(t,0,)+κ(0)(eα(0)y)h0t)+B(3κ2(0)cos2α(0)y2+6κ2sinα(0)cosα(0)y2y33κ2(0)sin2(0)y32)(V01(t,0,)h0t). On the other hand, we have BV21(t,0,)=BV21(t,0,)=B(|y|21)(B1(t,0)y22+B2(t,0)y2y3+B3(t,0)y32+B~4(t,0)+14μp1(t)). To ensure that the compatibility condition (Equation58), we choose p1(t)=8μπ(G0divV2+B(|y|21)(B1(t,0)y22+B2(t,0)y2y3+B3(t,0)y32+B~4(t,0))), where B~4(t,x1)=14(2μπddtF0κcosα8μH1+κsinα8μH2+2κ2πF0aμ(H11H10))+132(A1+A3).

The existence of a unique solution exponentially decaying to zero corresponding to the problem on the opposite side for the second-order boundary layer correctors (Y2,Q2) follows analogously, choosing the corresponding p2(t) such that the compatibility condition holds.

Remark 4.1

Simplified model

Let us consider a simplified model by assuming the following: κ,α are constants, f,g vanish in the neighbourhood of x1=0 and x1=l, 0lg=0 and there hold the following relations: By22h0t=By22hlt, By2y3h0t=By2y3hlt, By32h0t=By32hlt. We also take P2(t,x1)=x1pbl(t)+pdiv(t,x1), where pbl(t) will be obtained from the compatibility conditions for the second-order boundary layer correctors and pdiv will be chosen such that the ε4 term in the divergence equation vanishes, thus improving our final estimate (see Section 4.3 and Section 5).

The following compatibility conditions needs to be fulfilled: (59) G0divV2=BV21(t,0,),GldivY2=BY21(t,0,).(59) One can easily verify that the Equations (Equation59)1 and (Equation59)2 are equivalent due to the assumptions of the simplified model, so choosing pbl(t)=8μπ(G0divV2+B(|y|21)(B1(t,0)y22+B2(t,0)y2y3+B3(t,0)y32+B~4(t,0))), where B~4(t,x1)=14(2μπddtF0κcosα8μH1+κsinα8μH2+2κ2πF0aμ(H11H10))+132(A1+A3), both the compatibility conditions (Equation59) are satisfied.

We will address these considerations in Section 4.3 as they are essential in order to improve the divergence estimate.

4.2.2. Boundary layer for the microrotation

As for the velocity and pressure, we need to correct the microrotation approximation to satisfy the conditions on the end of the pipe Σεi, i=0,l.

Plugging the boundary layer correctors given by (Equation50) into the angular momentum equation (Equation16) and collecting the ε0 terms, we obtain the problem for the microrotation zero-order boundary layer approximation at x1=0: (60) δΔW0βdivW0=0 in G0,W0=0 on ω=0,×B,W0(t,0,)=W0(t,0,).(60) The existence of the solution W0(t,)W1,2(G0) exponentially decaying to zero as y1 can be easily proven (see [Citation25]).

The first-order boundary layer microrotation corrector W1 is the solution of the following system: (61) δW1βdivW1=Θ(t,0,V0,W0) in G0,W1=0 on ω=0,×B,W1(t,0,)=W1(t,0,)(κ(eαy)W01(t,0,),0,0)(κ(eαy)W01(t,0,),0,0),(61) where Θ(t,x1,J,K) is an exponentially decreasing function given in the appendix (see (EquationA15)). The existence of the unique exponentially decreasing solution W1(t,)W1,2(G0) is established in the same manner as for W0.

Finally, the system for the second-order boundary layer microrotation corrector W2 is given with (62) δW2βdivW2=Λ(t,0,V0,V1,W0,W1) in G0,W2=0 on ω=0,×B,W2(t,0,)=W2(t,0,)(κ(eαy)W11(t,0,),0,0)(κ2(eαy)2W01(t,0,),0,0)(κ(eαy)W11(t,0,),0,0)(κ2(eαy)2W01(t,0,),0,0),(62) where Θ(t,x1,J0,J1,K0,K1) is an exponentially decreasing function given in the appendix (see (EquationA16)). The existence of a unique exponentially decreasing solution W2(t,)W1,2(G0) is established as for W0 and W1.

The construction of the boundary layer correctors Zj,j=0,1,2 on the opposite side x1=l can be done in the same way and the exponential decay easily follows.

4.3. Divergence correction

Collecting the regular part of the approximation given by (Equation49) and the boundary layer correctors at x1=0 and x1=l given by (Equation50)–(Equation51), we define our approximation as: (63) Uε[2](t,x)=Uε,reg[2](t,x)+Uε,bl,0[2](t,x)+Uε,bl,l[2](t,x)=ε2(V0(t,xε)+V0(t,x1ε,xε)+Y0(t,x1lε,xε))+ε3(V1(t,x1,xε)+V1(t,x1ε,xε)+Y1(t,x1lε,xε))+ε4(V2(t,x1,xε)+V2(t,x1ε,xε)+Y2(t,x1lε,xε)),(63) (64) Wε[2](t,x)=Wε,reg[2](t,x)+Wε,bl,0[2](t,x)+Wε,bl,l[2](t,x)=ε2(W0(t,x1xε)+W0(t,x1ε,xε)+Z0(t,x1lε,xε))+ε3(W1(t,x1,xε)+W1(t,x1ε,xε)+Z1(t,x1lε,xε))+ε4(W2(t,x1,xε)+W2(t,x1ε,xε)+Z2(t,x1lε,xε)),(64) (65) Pε[2](t,x)=Pε,reg[3](t,x)+Pε,bl,0[2](t,x)+Pε,bl,l[2](t,x)=P0(t,x1)+ε(P1(t,x1,xε)+P0(t,x1ε,xε)+Q0(t,x1lε,xε))+ε2(P2(t,x1,xε)+P1(t,x1ε,xε)+Q1(t,x1lε,xε)),+ε3(P3(t,x1,xε)+P2(t,x1ε,xε)+Q2(t,x1lε,xε)),(65) (66) uε[2](t,x~)=Uε[2](t,x),wε[2](t,x~)=Wε[2](t,x),pε[2](t,x~)=Pε[2](t,x), x~=Φεα(x).(66) where (67) Vj=Vj1a1+Vj2a2+Vj3a3,Vj=Vj1a1+Vj2a2+Vj3a3,Yj=Yj1a1+Yj2a2+Yj3a3,j=0,1,2,Wj=Wj1a1+Wj2a2+Wj3a3,Wj=Wj1a1+Wj2a2+Wj3a3,Zj=Zj1a1+Zj2a2+Zj3a3,j=0,1,2.(67)

It is well-known that the incompresibility equation need one more corrector than the momentum and angular momentum equation (see e.g. [Citation9]). Therefore, we need to correct the residual in the divergence equation. Plugging the expansion (Equation63) into the divergence Equation (Equation15), we obtain: divUε[2]=ε(V02y2+V03y3)+ε(V01y1+V02y2+V03y3)+ε(Y01y1+Y02y2+Y03y3)+ε2(V01x1+V12y2+V13y3)+ε2(V11y1+V12y2+V13y3κ(V02cosαV03sinα)+2κ(y2cosαV01y1y3sinαV01y1))+ε2(Y11y1+Y12y2+Y13y3κ(Y02cosαY03sinα)+2κ(y2cosαY01y1y3sinαY01y1))+ε3(V11x1κ(V12cosαV13sinα)+(κ(eαy)+κτ(eαy))V01+V22y2+V23y3)+ε3(V21y1+V22y2+V23y3κ(V12cosαV13sinα)+(κ(eαy)+κτ(eαy))V01+κ2(eαy)cosαV02κ2(eαy)sinαV03+2κ(eαy)(V11y1κ(V02cosαV03sinα))+3κ2(eαy)2V01y1) +ε3(Y21y1+Y22y2+Y23y3κ(Y12cosαY13sinα)+(κ(eαy)+κτ(eαy))Y01+κ2(eαy)cosαY02κ2(eαy)sinαY03+2κ(eαy)(Y11y1κ(Y02cosαY03sinα))+3κ2(eαy)2Y01y1)+ε4(V21x1κ(V22cosαV23sinα)+(κ(eαy)+κτ(eαy))V11+κ(eαy)(κ(eαy)+κτ(eαy))V01+κ2(eαy)(cosαV12sinαV13)+2κ(eαy)(V11x1κ(V12cosαV13sinα))+2κ(eαy)(κ(eαy)+κτ(eαy))V01)+ε4(κ(V22cosαV23sinα)+(κ(eαy)+κτ(eαy))V11+κ(eαy)(κ(eαy)+κτ(eαy))V01+κ2(eαy)cosαV12κ2(eαy)sinαV13+2κ(eαy)(V21y1κ(V12cosαV13sinα)) (68) +2κ(eαy)((κ(eαy)+κτ(eαy))V01+κ2(eαy)cosαV02κ2(eαy)sinαV03)+3κ2(eαy)2(V11y1κ(V02cosαV03sinα))+4κ3(eαy)3V01y1)+ε4(κ(Y22cosαY23sinα)+(κ(eαy)+κτ(eαy))Y11+κ(eαy)(κ(eαy)+κτ(eαy))Y01+κ2(eαy)cosαY12κ2(eαy)sinαY13+2κ(eαy)(Y21y1κ(Y12cosαY13sinα))+2κ(eαy)((κ(eαy)+κτ(eαy))Y01+κ2(eαy)cosαY02κ2(eαy)sinαY03)+3κ2(eαy)2(Y11y1κ(Y02cosαY03sinα))+4κ3(eαy)3Y01y1)+O(ε5)(68) Taking into account (Equation18)2, (Equation29)2, (Equation39)3, (Equation52)2, (Equation53)2, (Equation54)2, (Equation57)2, we get from (Equation68) the following equation: divUε[2]=ε2(κ(0)(V02cosα(0)V03sinα(0))κ(x1)(V02cosα(x1)V03sinα(x1)))+ε2(2κ(0)(y2cosα(0)y3sinα(0))V01y1+2κ(x1)(y2cosα(x1)y3sinα(x1))V01y1)+ε2(κ(l)(Y02cosα(l)Y03sinα(l))κ(x1)(Y02cosα(x1)Y03sinα(x1)))+ε2(2κ(l)(y2cosα(l)y3sinα(l))Y01y1+2κ(x1)(y2cosα(x1)y3sinα(x1))Y01y1)+ε3(κ(0)(V12cosα(0)V13sinα(0))κ(x1)(V12cosα(x1)V13sinα(x1)))+ε3((κ(0)(y2cosα(0)y3sinα(0))κ(0)τ(0)(y2sinα(0)+y3cosα(0)))V01)+ε3((κ(x1)(y2cosα(x1)y3sinα(x1))+κ(x1)τ(x1)(y2sinα(x1)+y3cosα(x1)))V01)+ε3(κ2(0)(y2cosα(0)y3sinα(0))(cosα(0)V02sinα(0)V03)) (69) +ε3(κ2(x1)(y2cosα(x1)y3sinα(x1))(cosα(x1)V02sinα(x1)V03))+ε3(2κ(0)(y2cosα(0)y3sinα(0))V11y1+2κ(x1)(y2cosα(x1)y3sinα(x1))V11y1)+ε3(3κ2(0)(cosα(0)sinα(0))2V01y1+3κ2(x1)(cosα(x1)sinα(x1))2V01y1)+ε4(V21x1κ(V22cosαV23sinα)+(κ(eαy)+κτ(eαy))V11+κ(eαy)(κ(eαy)+κτ(eαy))V01+κ2(eαy)(cosαV12sinαV13)+2κ(eαy)(V11x1κ(V12cosαV13sinα))+2κ(eαy)(κ(eαy)+κτ(eαy))V01)+πε,res,(69) where πε,resL2(Ωεα)=O(ε11/2). We write the above equation as (70) πε=πε,1+πε,2+πε,3+πε,res(70) where πε,1 is ε2 term (first six lines), πε,2 the ε3 term (seventh to fourteenth line) and πε,3 the ε4 term (last four lines) in equation (Equation69).

We now obtain for πε,1: (71) πε,1=ε2(κ(0)(cosα(0)cosα(x1))+(κ(0)κ(x1))cosα(x1))V02ε2(κ(0)(sinα(0)sinα(x1))+(κ(0)κ(x1))sinα(x1))V03ε2(2κ(0)(cosα(0)cosα(x1))+(2κ(0)2κ(x1))cosα(x1))y2V01y1+ε2(2κ(0)(sinα(0)sinα(x1))+2(κ(0)κ(x1))sinα(x1))y3V01y1+ε2(κ(l)(cosα(l)cosα(x1))+(κ(l)κ(x1))cosα(x1))Y02ε2(κ(l)(sinα(l)sinα(x1))+(κ(l)κ(x1))sinα(x1))Y03ε2(2κ(l)(cosα(l)cosα(x1))+(2κ(l)2κ(x1))cosα(x1))y2Y01y1+ε2(2κ(l)(sinα(l)sinα(x1))+2(κ(l)κ(x1))sinα(x1))y3Y01y1(71) Using a simple change of variables and the fact that V0 is concentrated near y1=0 and Y0 near x1=l, we deduce the estimate (72) πε,1L2(Ωεα)=O(ε9/2),(72) and it easily follows from (Equation69)–(Equation70) that πεL2(Ωεα)=O(ε9/2). It is important to observe that, since πε,i,i=1,2,3 do not vanish when integrated over the cross-section B, we cannot define the divergence corrector in the general case. For that reason, we look at the simplified model (see Remark 4.1) and complete the construction of the divergence correction in the following.

If we assume that κ and α are constants, it easily follows from (Equation69) that πε,1, πε,2=0, so we have (73) πε=πε,3+πε,res.(73) As in the general case, Bπε,3 0, so we obtain πεL2(Ωεα)=O(ε5).

If we additionally assume that f,g vanish in the neighbourhood of x1=0 and x1=l and 0lg=0, the following relations hold (74) By22h0t=By22hlt, By2y3h0t=By2y3hlt, By32h0t=By32hlt(74) and take P2=x1pbl(t)+pdiv(t,x1), we can choose pdiv such that the Bπε,3=0 (note that we have obtained pbl from the compatibility conditions for the second-order velocity and pressure boundary layer correctors).

The ε4 divergence term that needs to be corrected reads: (75) πε,3=V21x1κ(V22cosαV23sinα)+κ2(eαy)(cosαV12sinαV13)+2κ(eαy)(V11x1κ(V12cosαV13sinα)),(75) while (76) pdiv(x1,t)=μBdB0x1(0ζB(|y|21)(B1x1y22+B2x1y2y3+B3x1y32+14A~4x1+132(A1x1+A3x1))dξ)dζ+μB0x1(0ζκ(cosαBV22sinαBV23)dξ)dζμB0x1(0ζBκ2(eαy)(cosαV12sinαV13)dξ)dζμB0x1(0ζB2κ(eαy)(18μ(1|y|2)(y2H1x1+y3H2x1)κ(V12cosαV13sinα))dξ)dζ,(76) with A4=(2/μπ)(d/dt)F0(κcosα/8μ)H1+(κsinα/8μ)H2(a/μ)(H11H10). Observe that pdiv(x1,t)/x1 vanishes at x1=0 and x1=l due to our assumptions and does not appear in the compatibility conditions for the velocity and pressure second-order boundary layer correctors at x1=0 and x1=l.

Finally, we can now define our divergence corrector as (77) Ψε(t,x)=ε5i=23Ψi(x1,xε)ai,(77) where Ψε is the solution of the problem (78) divyΨε=πε,3,Ψε=0 on 0,l×B.(78) In this case, we define u~ε[2](t,x~)=U~ε[2](t,x), x~=Φεα(x), where U~ε[2]=Uε[2]Ψε.

In this way, the divergence estimate is improved, namely: (79) divu~ε[2]=π~ε,π~εL2(Ωεα)=O(ε11/2).(79)

5. Error analysis

5.1. A priori estimates

To derive the a priori estimates, we first provide two technical results, namely the Poincaré's inequality and the estimates for the divergence equation.

The following result is a direct consequence of [Citation9, Lemma 7].

Lemma 5.1

Poincaré's inequality

Let T0, and t[0,T]. There then exists a constant C>0, independent of ε, such that (80) ϕε(t,)L2(Ωεα)Cεϕε(t,)L2(Ωεα),(80) for any ϕε(t,)W1,2(Ωεα) such that ϕε=0 on Γεα.

We will also need to consider the following result on the divergence equation, which can be straightforwardly deduced from [Citation9, Lemma 9].

Lemma 5.2

Divergence equation

Let T0, and t[0,T]. Futhermore, let jKtjL2(0,T;L2(Ωεα)),jg0tjL2(0,T;H1/2(Σε0)),jgltjL2(0,T;H1/2(Σεl)) for j=0,1,2. There then exists ϕεL2(0,T;W1,2(Ωεα)) such that jϕεtjL2(0,T;W1,2(Ωεα)),j=1,2 satisfying div ϕε=K in Ωεα,ϕε=0 on Γεα,ϕ=g0 on Σε0,ϕ=gl on Σεl, where ΩεαK=Σε0g0t(0)+Σεlglεt(l). Moreover, if git=(git)t,ginb=(Itt)gi,i=0,l, then ϕε satisfies the following estimate (81) ϕεL2(0,T;L2(Ωεα))C(1εKL2(0,T;L2(Ωεα))+1ε(g0tL2(0,T;H1/2(Σε0))+gltL2(0,T;H1/2(Σεl))))+C(g0nbL2(0,T;H1/2(Σε0))+glnbL2(0,T;H1/2(Σεl))),||ϕεt||L2(0,T;L2(Ωεα))C(1ε||Kt||L2(0,T;L2(Ωεα))+1ε(||g0tt||L2(0,T;H1/2(Σε0))+||gltt||L2(0,T;H1/2(Σεl)))+C(||g0nbt||L2(0,T;H1/2(Σε0))+||glnbt||L2(0,T;H1/2(Σεl))),||2ϕεt2||L2(0,T;L2(Ωεα))C(1ε||2Kt2||L2(0,T;L2(Ωεα))+1ε(||2h0tt2||L2(0,T;H1/2(Σε0))+||2hltt2||L2(0,T;H1/2(Σεl)))+C(||2h0nbt2||L2(0,T;H1/2(Σε0))+||2hlnbt2||L2(0,T;H1/2(Σεl))),(81) where the constant C>0 is independent of ε.

Remark 5.3

It can be easily shown that we need higher time regularity conditions for the given data to ensure that jKtjL2(0,T;L2(Ωεα)),jg0tjL2(0,T;H1/2(Σε0)),jgltjL2(0,T;H1/2(Σεl)),j=0,1,2, namely: jFtj,jGtj,h0εtj,hlεtjL2(0,T;L2(Ωεα)), j=0,1,2,3.

The next step is to derive the a priori estimates for the velocity, pressure and microrotation.

Proposition 5.4

Apriori estimates

Let T0, and let (uε,pε,wε)=(vε+hε,pε,wε) be the solution of the problem (Equation5)–(Equation11). Then there exists a constant C>0, independent of ε, such that (82) supt[0,T]vεL2(Ωεα)2+0TvεL2(Ωεα)2Cε4,supt[0,T]||vεt||L2(Ωεα)2+0T||vεt||L2(Ωεα)2Cε4,supt[0,T]wεL2(Ωεα)2+0TwεL2(Ωεα)2Cε4,supt[0,T]||wεt||L2(Ωεα)2+0T||wεt||L2(Ωεα)2Cε4,supt[0,T]pεL2(0,T;L2(Ωεα))2Cε2.(82)

Proof.

Using wε as a test function in the integral identity (12)2 yields (83) 12ddtΩεα|wε|2+αΩεα|wε|2+βΩεα(divwε)2+2aΩεα|wε|2=aΩεαrotvεwε+aΩεαrothεwε+ΩεαGwε,(83) Using inequality (Equation80), we estimate the terms on the right–hand side of (Equation83) to obtain: (84) |Ωεαrotvεε|rotvεL2(Ωεα)wεL2(Ωεα)CεvεL2(Ωεα)wεL2(Ωεα),|Ωεαrothεwε|rothεL2(Ωεα)wεL2(Ωεα)CεhεL2(Ωεα)wεL2(Ωεα),|ΩεαGwε|GL2(Ωεα)wεL2(Ωεα)Cε2wεL2(Ωεα).(84) Using estimates (Equation84), Young's inequality, integrating over t in equation (Equation83) and applying estimate (Equation81)1 we have: (85) supt[0,T]wεL2(Ωεα)2+0TwεL2(Ωεα)2Cε20TvεL2(Ωεα)2+Cε4.(85) Differentiating the integral identity (12)2 and taking wεt as a test function, we obtain: (86) 12ddtΩεα|wεt|2+αΩεα|wεt|2+βΩεα(divwt)2+2aΩεα|wεt|2=aΩεαrotvεtwεt+aΩεαrothεtwεt+ΩεαGtwεt.(86) Again, estimating the terms on the right–hand side as above, using Young's inequality, integrating over t in (Equation86) and using (Equation81)2, we obtain : (87) supt[0,T]||wεt||L2(Ωεα)2+0T||wεt||L2(Ωεα)2Cε20T||vεt||L2(Ωεα)2+Cε4.(87)

We now multiply the integral identity (12)1 with vε and integrate over Ωεα to obtain (88) 12ddtΩεα|vε|2+μΩεα|vε|2=aΩεαrotwεvε+ΩεαFvεΩεhεtvεΩεαhεvε.(88) We carefully estimate the terms on the right–hand side of (Equation88) as follows: (89) |Ωεαrotwεvε|CεvεL2(Ωεα)wεL2(Ωεα),|ΩεαFvε|FL2(Ωεα)vεL2(Ωεα)Cε2vεL2(Ωεα),|Ωεαhεtvε|||hεt||L2(Ωεα)vεL2(Ωεα)Cε2||hεt||L2(Ωεα)vεL2(Ωεα),|Ωεαhεvε|hεL2(Ωεα)vεL2(Ωεα).(89) Applying (Equation89) and Young's inequality on the right–hand side of (Equation88), integrating over t and using estimates (Equation81)1–(Equation81)2, we have (90) supt[0,T]vεL2(Ωεα)2+0TvεL2(Ωεα)2Cε20TwεL2(Ωεα)2+Cε4.(90) Finally, combining the estimates (Equation85) and (Equation90), for sufficiently small ε, we obtain: (91) supt[0,T]vεL2(Ωεα)2+0TvεL2(Ωεα)2Cε4.(91) and (92) supt[0,T]wεL2(Ωεα)2+0TwεL2(Ωεα)2Cε4(92) proving (Equation82)1 and (Equation82)3.

Differentiating the integral identity (12)1 and using vεt as a test function, we get (93) 12ddtΩεα|vεt|2+μ0t|vεt|2=aΩεαrotwεtvεt+ΩεαFtvεtΩεα2hεt2vεtΩεαhεtvεt.(93) In much the same way as for (Equation88), estimating the terms on right–hand side using Young's inequality in equation (Equation93), integrating over t and using (Equation81)2–(Equation81)3, we arrive at: (94) ||vεt||L2(Ωεα)2+0t||vεt||L2(Ωεα)2Cε20t||wεt||L2(Ωεα)2+Cε4.(94) Combining the estimates (Equation87) and (Equation94), for sufficiently small ε we have: (95) supt[0,T]||vεt||L2(Ωεα)2+0T||vεt||L2(Ωεα)2Cε4,(95) and (96) supt[0,T]||wεt||L2(Ωεα)2+0T||wεt||L2(Ωεα)2Cε4(96) leading to (Equation82)2 and (Equation82)4.

We now need to estimate the pressure. Let dε be the solution of the problem: (97) divdε=pε,dε=0 on Ωεα.(97) Supposing Ωεαpε=0, for all t, it follows from Lemma 5.2 that (Equation97) has at least one solution satisfying (98) dεL2(0,T;L2(Ωεα))CεpεL2(0,T;L2(Ωεα)).(98) Multiplying the momentum equation (Equation5)1 with dε and integrating over Ωεα yields (99) pεL2(Ωεα)2Ωεαvεtdε+Ωεαhεtdε+μΩεαvεdε+μΩεαhεdεaΩεαrotwεdεΩεαfεdε.(99) We estimate the terms on the right–hand side of (Equation99) using Poincaré's inequality (Equation80) to obtain (100) |Ωεαvεtdε|||vεt||L2(Ωεα)dεL2(Ωεα)Cε2||vεt||L2(Ωεα)dεL2(Ωεα),|Ωεαhεtdε|||hεt||L2(Ωεα)dεL2(Ωεα)Cε2||hεt||L2(Ωεα)dεL2(Ωεα),|Ωεαvεdε|vεL2(Ωεα)dεL2(Ωεα),|Ωεαhεdε|hεL2(Ωεα)dεL2(Ωεα),|Ωεαrotwεdε|CεwεL2(Ωεα)dεL2(Ωεα),|Ωεαfεdε|CfεL2(Ωεα)dεL2(Ωεα)Cε2dεL2(Ωεα).(100)

Using the estimates (Equation100) and Young's inequality on the right–hand side of equation (Equation99), integrating over t and using (Equation81)1–(Equation81)2, (Equation91)–(Equation92), (Equation95) and (Equation98), we deduce (101) pεL2(0,T;L2(Ωεα))Cε(101) completing the proof.

5.2. Error estimate

In the following, we provide the main result of this section.

Theorem 5.5

Error estimate

Let T0, and let (uε[2],wε[2],pε[2]) be the asymptotic solution defined by (Equation63)–(Equation67) and (uε,wε,pε) the solution of the governing problem (Equation5)–(Equation11). Then the following estimates hold: (102) supt[0,T]uε(t,)uε[2](t,)L2(Ωεα)+(uεuε[2])||L2(0,T;L2(Ωεα))Cε7/2,supt[0,T]||uε(t,)tuε[2](t,)t||L2(Ωεα)+||(uεtuε[2]t)||L2(0,T;L2(Ωεα))Cε7/2,supt[0,T]wε(t,)wε[2](t,)L2(Ωεα)+(wεwε[2])||L2(0,T;L2(Ωεα))Cε9/2,supt[0,T]||wε(t,)twε[2](t,)t||L2(Ωεα)+||(wεtwε[2]t)||L2(0,T;L2(Ωεα))Cε9/2,pεpε[2]L2(0,T;L2(Ωεα))Cε5/2,(102) where the constant C>0 is independent of ε.

Proof.

The asymptotic approximation wε[2] for the microrotation satisfies the following equation: (103) wε[2]tδΔwε[2]βdivwε[2]+2awε[2]=arotuε[1]+G+ξε in Ωεα,wε[2]=0 on Γεα,wε[2]=η0ε on Σε0,wε[2]=ηlε on Σεl,wε[2](0,)=0.(103) where ξεL2(0,T;L2(Ωεα))=O(ε4) and ηiεL2(0,T;H1/2(Σεi)),=O(ε11/2), i=0,l.

Denoting sε=wεwε[2] and subtracting (5)3,(6)2,(6)3,2 and (Equation103), we obtain (104) sεtαΔsεβdivsε+2asε=arot(uεuε[1])ξε,sε=0 on Γεα,sε=η0 on Σε0,sε=ηl on Σεl.(104) It follows from Lemma 5.2 that we can construct a function Dε with the same trace as sε on Ωεα such that (105) DεL2(0,T;L2(Ωεα))+||Dεt||L2(0,T;L2(Ωεα))+||2Dεt2||L2(0,T;L2(Ωεα))Cε5.(105) Multiplying the equation (Equation104) by s=sεDε and integrating over Ωεα yield (106) 12ddtΩεα|sε|2+αΩεα|sε|2+βΩεα(divsε)2+2aΩεα|sε|2=aΩεαrot(uεuε[1])sεΩεαDεtsεδΩεαsεDεβΩεαdivDεdivsε2aΩεαsεDεΩεαξεsε(106) We now estimate the terms on the right–hand side of (Equation106) using Poincare's inequality (Equation80) in the following way: (107) |Ωεαrot(uεuε)sε|C(uεuε[1])L2(Ωεα)sεL2(Ωεα)C((uεuε[2])L2(Ωεα)+(uε[2]uε[1])L2(Ωεα))sεL2(Ωεα)Cε(uεuε[2])L2(Ωεα)sεL2(Ωε)+Cε5sεL2(Ωεα),|ΩεαDεtsε|||Dεt||L2(Ωεα)sεL2(Ωεα)ε2||Dεt||L2(Ωεα)sεL2(Ωεα),|ΩεαsεDε|sεL2(Ωεα)DεL2(Ωεα),|ΩεαdivDεdivsε|DεL2(Ωεα)sεL2(Ωεα),|ΩεαsεDε|sεL2(Ωεα)DεL2(Ωεα)Cε2sεL2(Ωεα)DεL2(Ωεα),|Ωεαξεsε|ξεL2(Ωεα)sεL2(Ωεα)CεξεL2(Ωεα)sεL2(Ωε).(107) Applying the estimates (Equation107) and Young's inequality on the right hand–side of equation (Equation106), integrating over t and using (Equation105), we get (108) supt[0,T]sεL2(Ωεα)2+0TsεL2(Ωεα)2Cε20t(uεuε[2])L2(Ωεα)2+Cε10.(108) Differentiating identity (Equation104)1 with respect to t and using sεt=sεtDεt as a test function leads to: (109) 12ddtΩεα|sεt|2+δΩεα|sεt|2+βΩεα(divsεt)2+2aΩεα|sεt|2=aΩεαrot(uεtuε[1]t)sεtΩεα2Dεt2sεtαΩεαsεtDεtβΩεαdivDεtdivsεt2aΩεαsεtDεtΩεαξεtsεt.(109) Similarly as above, estimating the terms on the right–hand side of (Equation109), applying Young's inequality and integrating over t we obtain: (110) supt[0,T]||sεt||L2(Ωεα)2+0T||sεt||L2(Ωεα)2Cε20T||(uεtuε[2]t)||L2(Ωεα)2+Cε10.(110) The problem satisfied by the asymptotic approximation uε[2] reads: (111) uε[2]tμΔuε[2]+pε[2]=arotwε[1]+F+Eε in Ωεα,divuε[2]=πε in Ωεα,uε[2]=0 on Γεα,uε[2]=ε2h0+r0 on Σε0, uε[2]=ε2hl+rl on Σεl,uε[2](0,)=0,(111) where EεL2(0,T;L2(Ωεα))=O(ε4),πεL2(0,T;L2(Ωεα))=O(ε9/2),riεL2(0,T;H1/2(Σεi))=O(ε11/2),i=0,l.

Denoting Rε=uεuε[2], rε=pεpε[2] and subtracting (5)1,(5)2,(6)1,(6)3,1 and (Equation111), we obtain: (112) RεtμΔRε+rε=arot(wεwε[1])Eε in Ωεα,divRε=πε in Ωεα,Rε=0 on Γεα,Rε=r0 on Σε0Rε=rl on Σεl,Rε(0,)=0.(112) It is important to emphasize that the divergence estimate πεL2(0,T;L2(Ωεα))=O(ε9/2) cannot be improved in the general case, due to its structure (see (Equation68)–(Equation72)). However, under certain assumptions on κ,α and h0ε,hlε,F and G, this estimate can, in fact, be improved providing better overall estimates. We refer the reader to (Equation73)–(Equation79) in Section 4.3 and Remark 5.6 at the end of this Section.

Let us define ζε as the solution of the problem divζε=πε in Ωεα,ζε=0 on Γεα,ζε=r0 on Σε0Rε=rl on Σεl satisfying the following estimates (see Lemma 5.2): (113) ζεL2(0,T;L2(Ωεα))+||ζεt||L2(0,T;L2(Ωεα))+||2ζεt2||L2(0,T;L2(Ωεα))Cε7/2.(113) Multiplying the equation (Equation112)1 with function Rε=Rεζε and integrating over Ωεα, we obtain: (114) 12ddtRεL2(Ωεα)2+μRεL2(Ωεα2ΩεαζεtRε+μΩεαζεRε+aΩεαrot(wεwε[1])RεΩεαEεRε.(114) Estimating the terms on the right–hand side of Equation (Equation114) using Poincaré's inequality (Equation80) yields (115) |ΩεαζεtRε|||ζεt||L2(Ωεα)RεL2(Ωεα)Cε2||ζεt||L2(Ωεα)RεL2(Ωεα),|ΩεαζεRε|ζεL2(Ωεα)RεL2(Ωεα),|Ωεα)rot(wεwε[1])Rε|C(wεwε[1])L2(Ωεα)RεL2(Ωεα)C((wεwε[2])L2(Ωεα)+(wε[2]wε[1])L2(Ωεα))RεL2(Ωεα)CεsεL2(Ωεα)RεL2(Ωεα)+Cε5RεL2(Ωεα),|ΩεαEεRε|EεL2(Ωεα)RεL2(Ωεα)CεEεL2(Ωεα)RεL2(Ωεα).(115) Applying the estimates (Equation115) and Young's inequality on the right–hand side of (Equation114), integrating over t and using (Equation108), (Equation113), for sufficiently small ε we get: (116) supt[0,T]RεL2(Ωεα)2+0TRεL2(Ωεα)2Cε7,(116) and from (Equation108) we now easily get (117) supt[0,T]sεL2(Ωεα)2+0TsεL2(Ωεα)2Cε9,(117) implying (Equation102)1 and (Equation102)3.

Differentiating equation (Equation112)1 with respect to t, multiplying by Rεt=Rεtζεt and integrating over Ωεα, we get: (118) 12ddtΩεα|Rεt|L2(Ωεα)2+μΩεα|Rεt|L2(Ωεα)2=aΩεαrot(wεtwε[1]t)RεtΩεαEεtRεtΩεα2ζεt2RεtαΩεαζεtRεt.(118) In much the same way, using Poincaré's and Young's inequality to estimate the terms on the right–hand side of (Equation118), integrating with respect to t and using (Equation110), (Equation113), for sufficiently small ε we obtain: (119) supt[0,T]||Rεt||L2(Ωεα)2+0T||Rεt||L2(Ωεα)2Cε7.(119) From (Equation110), it follows (120) supt[0,T]||sεt||L2(Ωεα)2+0T||sεt||L2(Ωεα)2Cε9,(120) thus obtaining the estimates (Equation102)2 and (Equation102)4.

Let now dε be the solution of the problem (121) divdε=rε in Ωεα,dε=0 on Ωεα.(121) Supposing that Ωεαrε=0, for all t, it follows from Lemma (5.2) that the problem (Equation121) has at least one solution satisfying the estimate (122) dεL2(0,T;L2(Ωεα))CεrεL2(0,T;L2(Ωεα)).(122) Multiplying the equation (Equation112)1 by dε and integrating over Ωεα yields (123) rεL2(Ωεα)2=ΩεαRεtdε+μΩεαRεdεaΩεαrot(wεwε[1])dε+ΩεαEεdε+Ωεαζεtdε+μΩεαζεdε.(123) Estimating the terms on the right–hand of (Equation123) and using Poincaré's inequality yields: (124) |ΩεαRεtdε|||Rεt||L2(Ωεα)dεL2(Ωεα)Cε2||Rεt||L2(Ωε)dεL2(Ωεα),|ΩεαRεdε|RεL2(Ωεα)dεL2(Ωεα),|Ωεαrot(wεwε[1])dε|ε(wεwε[1])L2(Ωεα)dεL2(Ωεα)Cε((wεwε[2])L2(Ωεα)+(wε[2]wε[1])L2(Ωεα))dεL2(Ωεα)CεsεL2(Ωεα)dεL2(Ωεα)+Cε5dεL2(Ωεα),|ΩεαEεdε|EεL2(Ωεα)dεL2(Ωεα)CεEεL2(Ωεα)dεL2(Ωεα),|Ωεαζεtdε|||ζεt||L2(Ωεα)dεL2(Ωεα)Cε2||∂∇ζεt||L2(Ωεα)dεL2(Ωεα),|Ωεαζεdε|ζεL2(Ωεα)dL2(Ωεα).(124) Applying (Equation124) and Young's inequality on the right–hand side of (Equation123), integrating over t and using (Equation113), (Equation116), (Equation117), (Equation119) and (Equation122), we finally obtain: (125) rεL2(0,T;L2(Ωεα))Cε5/2,(125) completing the proof.

Remark 5.6

If we assume κ and α to be constant, due to consideration in Section 4.3 (see (Equation73)), following the proof of Theorem 5.5, we can improve the estimates (Equation102) to obtain: supt[0,T]uε(t,)uε[2](t,)L2(Ωεα)+(uεuε[2])||L2(0,T;L2(Ωεα))Cε4,supt[0,T]||uε(t,)tuε[2](t,)t||L2(Ωεα)+||(uεtuε[2]t)||L2(0,T;L2(Ωεα))Cε4,supt[0,T]wε(t,)wε[2](t,)L2(Ωεα)+(wεwε[2])||L2(0,T;L2(Ωεα))Cε5,supt[0,T]||wε(t,)twε[2](t,)t||L2(Ωεα)+||(wεtwε[2]t)||L2(0,T;L2(Ωεα))Cε5,pεpε[2]L2(0,T;L2(Ωεα))Cε3. If we, additionally, assume that f,g vanish in the neighbourhood of x1=0 and x1=l, 0lg=0 and take P2(x1,t)=x1pbl(t)+pdiv(x1,t), following the procedure given in Section 4.3 (see (Equation75)–(Equation79)), we can even further improve the error estimates. Namely, we get: supt[0,T]uε(t,)uε[2](t,)L2(Ωεα)+(uεuε[2])||L2(0,T;L2(Ωεα))Cε9/2,supt[0,T]||uε(t,)tuε[2](t,)t||L2(Ωεα)+||(uεtuε[2]t)||L2(0,T;L2(Ωεα))Cε9/2,supt[0,T]wε(t,)wε[2](t,)L2(Ωεα)+(wεwε[2])||L2(0,T;L2(Ωεα))Cε5,supt[0,T]||wε(t,)twε[2](t,)t||L2(Ωεα)+||(wεtwε[2]t)||L2(0,T;L2(Ωεα))Cε5,pεpε[2]L2(0,T;L2(Ωεα))Cε7/2.

Acknowledgments

The authors of this work have been supported by the Croatian Science Foundation (scientific project: Asymptotic analysis of the boundary value problems in continuum mechanics). The authors are grateful to the Referees for their valuable remarks and comments that helped to correct and improve the paper.

Disclosure statement

No potential conflict of interest was reported by the authors.

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Appendix

Regular part of the expansion

In the following, we write down the functions (depending on t and x1) appearing in the computations for the first- and second-order correctors (see (Equation26), (Equation32), (Equation35), (Equation36), (Equation38), (Equation40), (Equation41), (Equation43), (Equation47) and (Equation48)).

First-order corrector

The functions H1(t,x1) and H2(t,x1) in the problem (Equation26) for the first component of the velocity first-order corrector V11 are given by: (A1) H1(t,x1)=cosα(4μκπF0κf1f2x1+τf3a2δ+βg3)+sinα(τf2f3x1+a2δ+βg2),H2(t,x1)=sinα(4μκπF0+κf1+f2x1τf3a2δ+βg3)cosα(τf2+f3x1+a2δ+βg2).(A1) The functions H3(t,x1) and H4(t,x1) in the problem (Equation32) for the first component of the microrotation first-order corrector W11 are given by: (A2) H3(t,x1)=3κ2g1cosα+β2δ+β((τg3g2x1)cosα(τg2+g3x1)sinα),H4(t,x1)=3κ2g1sinαβ2δ+β((τg2+g3x1)cosα+(τg3g2x1)sinα).(A2) The functions H5(t,x1),,H8(t,x1) in the problem (Equation35) for the second and third component of the microrotation first-order corrector (W12,W13) are given by: (A3) H5(t,x1)=δκcosα2δ+β(g2(t,x1)cosα+g3(t,x1)sinα)+βκ2δ+βg2(t,x1)β2δg1(t,x1)x1,H6(t,x1)=δκsinα2δ+β(g2cosα+g3sinα)4aπF0(t),H7(t,x1)=κδcosα2δ+β(g3cosαg2sinα)+4aπF0(t),H8(t,x1)=δκsinα2δ+β(g2(t,x1)sinαg3(t,x1)cosαsinα)+βκ2δ+βg2(t,x1)β2δg1(t,x1)x1.(A3) The functions H9(t,x1),,H12(t,x1) in the expressions (Equation36) for the second and third component of the microrotation first-order corrector (W12,W13) are given by: (A4) H9(t,x1)=βH8(4δ+3β)H52[β2(4δ+3β)2],H10(t,x1)=βH7(4δ+β)H62[β2(4δ+β)2],H11(t,x1)=12βH64δ+βββH7(4δ+β)H62[β2(4δ+β)2],H12(t,x1)=12βH54δ+βββH8(4δ+β)H52[β2(4δ+β)2].(A4)

Second-order corrector

The functions A1(t,x1),,A4(t,x1) in the problem (Equation38) for the first component of the velocity second-order corrector V21 are given as follows: (A5) A1(t,x1)=2πμddtF0+3κcosα8μH1κsinα8μH22κ2πF0+κμcosαH1+4κ2cos2απF0κ2μcos2α(8μπF0f1)+κμ(cos2αf2x1sinαcosαf2+cosαsinαf3x1+cos2αf3)aμ(3H11+H10),A2(t,x1)=κcosα4μH2κsinα4μH1+κcosαμH2κsinαμH18κ2cosαsinαπF0+2κ2sinαcosαμ(8μπF0+f1)κμ(2sinαcosαf2x1(cos2αsin2α)f2+(cos2αsin2α)f3x12f3sinαcosα)+aμ(2H122H9),A3(t,x1)=2πμddtF0(t)+κcosα8μH13κsinα8μH22κ2πF0(t)κsinαμH2+4κ2sin2απF0(t)κ2sin2αμ(8πF0f1)+κμ(f2x1sin2αsinαcosαf2+f3x1sinαcosα+f3sin2α)aμ(H11+3H10),A4(t,x1)=2μπddtF0(t)κcosα8μH1+κsinα8μH2+2κ2πF0(t)+1μp1(t)+1μx1(p2(t)p1(t)l)aμ(H11H10).(A5) The functions A5(t,x1),,A14(t,x1) in the problem (Equation40) for the second and third component of the velocity second-order corrector and third-order pressure corrector (V22,V23,P3) are given as follows: A5(t,x1)=ag1μκsinα16μδ+2μ(κcosα+κτsinα)F0πaH4+a2(2δ+β)(g3x1cosα+g3τsinαg2x1sinα+τg2cosα),A6(t,x1)=ag1μκcosα8μδ2aH3aκg1cosα2δ,A7(t,x1)=3ag1μκsinα16μδ2μ(κcosα+κτsinα)F0π3aH4+aκsinαg12δ+a2(2δ+β)(g3x1cosα+g3τsinαg2x1sinα+τg2cosα),A8(t,x1)=ag1μκsinα16μδ+2μ(κcosα+κτsinα)F0π+aH4a2(2δ+β)(g3x1cosα+g3τsinαg2x1sinα+τg2cosα),A9(t,x1)=3aκcosαg116μδ2μ(κτcosακsinα)F0πa2(2δ+β)(g2x1cosα+g2τsinα+g3x1sinαg3τsinα)+3aH3+ag1κcosα2δ, (A6) A10(t,x1)=ag1μκsinα8μδ+2aH4ag1κsinα2δ,A11(t,x1)=aκcosαg116μδ2μ(κτcosακsinα)F0πa2(2δ+β)(g2x1cosα+g2τsinα+g3x1sinαg3τsinα)+aH3,A12(t,x1)=ag1μκcosα16μδ+2μ(κτcosακsinα)F0π+a2(2δ+β)(g2x1cosα+g2τsinα+g3x1sinαg3τsinα)aH3,A13(t,x1)=18μH1x1+aκg1sinα16μδ+2κcosα+2κτsinαπF0,A14(t,x1)=18μH2x1+aκg1cosα16μδ+2κsinα+2κτcosαπF0.(A6) The functions M1(t,x1),,M6(t,x1) and B5(t,x1),,B12(t,x1) in the expressions (Equation41) for the second and third component of the velocity second-order corrector and third-order pressure corrector (V22,V23,P3) are given by: (A7) M1(t,x1)=μA13+13A5112A7+124A10,M2(t,x1)=144147μA14+4147A68147A9+8147A11,M3(t,x1)=μA13+14A738A10,M4(t,x1)=96105μA14821A6+521A9+17105A11,M5(t,x1)=43μA13+16A7A8524A10,M6(t,x1)=998735μA14191882A6+79441A9+5491311640A11+A12,(A7) and (A8) B5(t,x1)=2196μM1+196μM3+796μA5196μA7,B6(t,x1)=16μM4+112μA6,B7(t,x1)=396μM1796μM3196μA5+796μA7,B8(t,x1)=996μM1396μM3+396μA5+396μA714μM5+14μA8,B9(t,x1)=396μM2796μM4+796μA9196μA11,B10(t,x1)=16μM3+112μA10,B11(t,x1)=2196μM2+196μM4+796μA11196μA9,B12(t,x1)=996μM2396μM4+396μA9+696μA1114μM6+14μA12.(A8) The functions C1(t,x1),,C4(t,x1) in the problem (Equation43) for the first component of the microrotation second-order corrector W21 are given as follows: C1(t,x1)=14δ2g1t14δ2g1x12+3κcosα8μH312δ(2δ+β)(g2x1cosα+τg2sinα+g3x1sinατg3cosα)+κsinα2(2δ+β)(g3x1cosα+τg3sinαg2x1sinα+τg2cosα)κsinα8μH4+12(2δ+β)x1((κκ)cosα(g2cosα+g3sinα)(κκ)sinα(g3cosαg2sinα))κ24δg1+κμcosαH3+κ22δcos2αg1+β4δ22g1x12+β2δ(2δ+β)((κcosα)(g2cosα+g3sinα)κcosα(g2cosα+g3sinα)+(κsinα)(g3cosαg2sinα)+κsinα(g3cosαg2sinα))+3βδH9x1+βδH12x1+βκδ(2δ+β)cosα(g2x1cosα+τg2sinα+g3x1sinατg3cosα)+4a216μδ2g1a2δ2g1,C2(t,x1)=κcosα4μH4κsinα4μH3κsinαμH3+κμcosαH4+κ2sinαcosαδg1+2βδH11x1βκsinαδ(2δ+β)(g2x1cosα+τg2sinα+g3x1sinατg3cosα)+βκcosαδ(2δ+β)(g3x1cosα+τg3sinαg2x1sinατg2cosα), C3(t,x1)=14δ2g1t14δ2g1x12+κcosα8μH3κcosα2(2δ+β)(g2x1cosα+τg2sinα+g3x1sinατg3sinα)+κsinα12(2δ+β)(g3x1cosα+τg3sinαg2x1sinα+τg2cosα)3κsinα8μH4+12δ(2δ+β)(x1((κκ)cosα(g2cosα+g3sinα)(κκ)sinα(g3cosαg2sinα)))κ214δg1κμsinαH4+κ2sin2α2δg1+β4δ22g1x12+β2δ(2δ+β)((κcosα)(g2cosα+g3sinα)κcosα(g2cosα+g3sinα)+(κsinα)(g3cosαg2sinα)+κsinα(g3cosαg2sinα))βδH9x1+3βδH12x1βκsinαδ(2δ+β)(g3x1cosα+τg3sinαg2x1sinα+τg2cosα)+a2g14μδ2ag12δ2, (A9) C4(t,x1)=14δ2g1t+14δ2g1x12κcosα8μH3+κcosα2δ(2δ+β)(g2x1cosα+τg2sinα+g3x1sinατg3sinα)κsinα2(2δ+β)(g3x1cosα+τg3sinαg2x1sinα+τg2cosα)+κsinα8μH412(2δ+β)x1((κκ)cosα(g2cosα+g3sinα)(κκ)sinα(g3cosαg2sinα))+κ24δg1β4δ22g1x12β2δ(2δ+β)((κcosα)(g2cosα+g3sinα)κcosα(g2cosα+g3sinα))+(κsinα)(g3cosαg2sinα)+κsinα(g3cosαg2sinα))+βδH9x1βδH12x1a28μδ2g1+a2δ2g1.(A9) The functions C5(t,x1),,C12(t,x1) in the problem (Equation47) for the second and third component of the microrotation second-order corrector (W22,W23) read: C5(t,x1)=12δ+β(g2tcosα+g3tsinα)δ2(2δ+β)(2g2x12cosα+g2x1τsinα+(τg2+τg2x1)sinατ2g2cosα+2g3x12sinαg3x1τcosα(τg3+τg3x1)cosατ2g3sinα)+3δκcosαH9δκsinαH10+2δκ22δ+βcos2α(g2cosα+g3sinα)14((2κg1x1+κg12δκ22δ+β(g2cosα+g3sinα))cosα+κτg1sinα)cos2αδκ22δ+β(g2cosα+g3sinα)3β8μH3x1+3βκcosαH93βκsinαH11βcosα(κ4δg1+κ2cosα2(2δ+β)(g2cosα+g3sinα)κ2sinα2(2δ+β)(g3sinαg2cosα)+2κ(14δg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)sinα12(2δ+β)(g3cosαg2sinα)))βκτ4δsinαg1βκ2δcosαg1βκτ2δsinαg1βκ2cos2α12δ+β(g2cosα+g3sinα)βκcosα(1δg1x1κ(12δ+βcosα(g2cosα+g3sinα)+12δ+βsinα(g3cosαg2sinα)))a8μH2+a2δ+β(g2cosα+g3sinα), C6(t,x1)=2δκcosαH102δκsinαH92δκ22δ+βsinαcosα(g2cosα+g3sinα)β4μH4x1+2βκcosαH102βκsinαH12+sinαβκg12δβκτ2δcosαg1+βκ2sinαcosα12δ+β(g2cosα+g3sinα)+βκ2sinαcosα(g3cosαg2sinα)βκcosα1δg1x1+2βκ2sinαcosα12δ+β(g2cosα+g3sinα)2βκ2sin2α12δ+β(g3cosαg2sinα)a4μH14aκπcosαF0, C7(t,x1)=12δ+β(g2tcosα+g3tsinα)δ2(2δ+β)(2g2x12cosα+g2x1τsinα+(τg2+τg2x1)sinατ2g2cosα+2g3x12sinαg3x1τcosα(τg3+τg3x1)cosατ2g3sinα)+δκcosαH93δκsinαH10+2δκ22δ+βsin2α(g2cosα+g3sinα)14((2κg1x1+κg12δκ22δ+β(g2cosα+g3sinα))cosα)+κτg1sinα)+κ2sin2αδ2δ+β(g2cosα+g3sinα)β8μH3x1+βκcosαH9βκsinαH11βcosα(κ4δg1+κ2cosα2(2δ+β)(g2cosα+g3sinα)κ2sinα2(2δ+β)(g2cosαg3sinα)+2κ(14δg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)sinα12(2δ+β)(g3cosαg2sinα))))sinαβκτ4δg1βκ2sin2α12δ+β(g3cosαg2sinα)2βκsin2α12δ+β(g3cosαg2sinα)3a8μH2+4aκπsinαF0+a2δ+β(g2cosα+g3sinα), (A10) C8(t,x1)=12δ+β(g2tcosα+g3tsinα)+δ2(2δ+β)(2g2x12cosα+g2x1τsinα+(τg2+τg2x1)sinατ2g2cosα+2g3x12sinαg3x1τcosα(τg3+τg3x1)cosατ2g3sinα)δκcosαH9+δκsinαH10+14((2κg1x1+κg12δκ22δ+β(g2cosα+g3sinα))cosα+κτg1sinα)+β8μH3x1βκcosαH9+βκsinαH11+βcosα(κ4δg1+κ2cosα12(2δ+β)(g2cosα+g3sinα)κ2sinα12(2δ+β)(g2cosα+g3sinα)+2κ(14αg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)sinα12(2δ+β)(g3cosαg2sinα))))+βκτsinαg14δ+a8μH2a2δ+β(g2cosα+g3sinα),(A10) and C9(t,x1)=12(2δ+β)(g3tcosαg2tsinα)δ2(2δ+β)(2g3x12cosα+τg3x1sinα+(τg3+τg3x1)sinατ2g3cosα2g2x12sinα+τg2x1cosα+(τg2+τg2x1)cosα+τ2g2sinα)δκsinαH12+3δκcosαH11+2δκ2cos2α2δ+β(g3cosαg2sinα)(κτcosα4g1(κ2g1x1+κ4g1κ22(2δ+β)(g2cosα+g3sinα))sinα)δκ2cos2α2δ+β(g3cosαg2sinα)β8μH4x1+βκcosαH10βκsinαH12βκτcosα14δg1+βκ(κ14δg1+κ2cosα12(2δ+β)(g2cosα+g3sinα)κ2sinα12(2δ+β)(g3cosαg2sinα))2βκ(14δg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)12(2δ+β)(g3cosαg2sinα)sinα))sinα+3a8μH1+4aκπF0cosα+a2δ+β(g3cosαg2sinα), C10(t,x1)=2δκsinαH11+2δκcosαH124δκ22δ+βsinαcosα(g3cosαg2sinα)+2δκ2cosαsinα12δ+β(g3cosαg2sinα)β4μH3x1+2βκcosαH9βκsinαH11βκcosα12δg1βκτsinα12δg1βκ2cos2α12δ+β(g2cosα+g3sinα)+βκ2cosαsinα12δ+β(g3cosαg2sinα)βκcosα1δg1x1+2βκ2cosα12δ+β(cosα(g2cosα+g3sinα)+sinα(g3cosαg2sinα))+a4μH24aκπsinαF0, C11(t,x1)=12(2δ+β)(g3tcosαg2tsinα)δ2(2δ+β)(2g3x12cosα+τg3x1sinα+(τg3+τg3x1)sinατ2g3cosα2g2x12sinα+τg2x1cosα+(τg2+τg2x1)cosα+τ2g2sinα)3δκsinαH12+δκcosαH11+2δκ2sin2α12δ+β(g3cosαg2sinα)(cosακτ4g1(κ2g1x1+κ4g1)sinα)+κ22(2δ+β)(g2cosα+g3sinα)sinαδκ2sin2α12δ+β(g3cosαg2sinα)3β8μH4x1+3βκcosαH103βκsinαH12βκτcosα14δg1+βκ(κ14δg1+κ2cosα12(2δ+β)(g2cosα+g3sinα)κ2sinα12(2δ+β)(g3cosαg2sinα))2βκ(14δg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)12(2δ+β)(g3cosαg2sinα)sinα))sinα+βκ2δsinαg1βκτcosα12δg1βκ2sinα(cosα12δ+β(g2cosα+g3sinα)+sinα12δ+β(g3cosαg2sinα))2βκsinα(12δg1x1+κ12δ+β(cosα(g2cosα+g3sinα)+sinα(g3cosαg2sinα)))+a8μH1+a2δ+β(g3cosαg2sinα), (A11) C12(t,x1)=22(2δ+β)(g3tcosαg2tsinα)+δ2(2δ+β)(2g3x12cosα+τg3x1sinα+(τg3+τg3x1)sinατ2g3cosα2g2x12sinα+τg2x1cosα+(τg2+τg2x1)cosα+τ2g2sinα)+δκsinαH12δκcosαH11+(cosακτ4g1(κ2g1x1+κ4g1κ22(2δ+β)(g2cosα+g3sinα))sinα)+β8μH4x1βκcosαH10+βκsinαH12+βκτcosα14δg1βκ(κ14δg1+κ2cosα12(2δ+β)(g2cosα+g3sinα)κ2sinα12(2δ+β)(g3cosαg2sinα))+2βκ(14δg1x1κ(cosα12(2δ+β)(g2cosα+g3sinα)12(2δ+β)sinα(g3cosαg2sinα)))sinαa8μH1a2δ+β(g3cosαg2sinα).(A11) The functions D5(t,x1),,D12(t,x1) in the expressions (48) for the second and third component of the microrotation second-order corrector (W22,W23) are given as follows: (A12) D5(t,x1)=(2δ12β212δ+6β)D7+C53βC1012δ+6β14δ12β+12β212δ+6β,D6(t,x1)=4β(D9+D11)C612δ+6β,D7(t,x1)=(C73βC1012δ+6β)(14δ12β+12β212δ+6β)(2δ2β+12β212δ+6β)(C53βC1012δ+6β)(2δ2β+12β212δ+6β)(2δ12β212δ+6β)+(14δ2β+12β212δ+6β)2,D8(t,x1)=2δ(D5+D7)+2βD5+βD10C84α+2β,D9(t,x1)=C93βC612δ+6β+(2δ+2β12β212δ+6β)D1114δ2β+12β212δ+6β,D10(t,x1)=4β(D5+D7)C1012δ+6β,D11(t,x1)=(C113βC612δ+6β)(14δ2β+12β212δ+6β)+(2δ12β212δ+6β)(C93βC612δ+6β)(2δ+12β212δ+6β)(2δ+2β12β212δ+6β)+(14δ12β+12β212δ+6β)(14δ2β+12β212δ+6β),D12(t,x1)=2δ(D9+D11)+2βD11+βD6C124δ+2β.(A12)

Boundary layers

In the following, we write down the exponentially decreasing functions Σ,Π,Θ and Λ appearing in the equations for the velocity, pressure and microrotation first- and second-order boundary layers correctors in Sections 4.2.1 and 4.2.2 (see (Equation54), (Equation57), (Equation61) and (Equation62)).

Boundary layers for the velocity

The function Ξ(t,x1,J,K,j) appearing in the equation (Equation54) is given as (A13) Ξ(t,x1,J,K,j)=μ(κcosα(J1y2J2y1)+κsinα(J3y1V1y3)κ(J2y1cosαJ3y1sinα)+κJ2y1cosακJ3y1sinα+κ(eαy)ΔJ1,κcosαJ2y2+κsinαJ2y3+2κcosαJ1y1,κsinαJ3y3κcosαJ3y22κsinαJ1y1)(κ(eαy)jy1,0,0)+a(K3y2K2y3,K1y3K3y1,K2y1K1y2),(A13) where J=(J1,J2,J3),K=(K1,K2,K3) are vector functions, j is a scalar function, x1[0,l] and t[0,T].

The function Π(t,x1,J0,J1,K0,K1,j0,j1) from the equation (Equation57) reads: (A14) Π(t,x1,J0,J1,K0,K1,j0,j1)=(J01t,J02t,J03t)+μ(κcosα(J11y2J12y1)+κsinα(J13y1J11y3)(κcosαJ12y1κsinαJ13y1)+κ(J12y1cosαJ13y1sinα)(J02(κcosα)J03(κsinα)J02(κcosα)+J03(κsinα))κ2J01+κ(eαy)(ΔJ11κcosαJ01y2+κsinαJ01y3)+κ2(eαy)2ΔJ01,κcosαJ12y2+κsinαJ12y3+2κ(eαy)(κcosαJ02y2+κsinαJ02y3)+(2κJ11y1+κJ01κ2J02)cosα+κτJ01sinα+κ2cosα(eαy)J02y2κ2sinα(eαy)J02y3,κsinαJ13y3κcosαJ13y2+2κ(eαy)κsinαJ03y32κ(eαy)κcosαJ03y2+κτJ01cosα(2κJ11y1+κJ01κ2J02)sinα+κ2cosα(eαy)J03y2κ2sinα(eαy)J03y3)(κ(eαy)j1y1+κ2(eαy)2j0y1,0,0)+a(K13y2K12y3,K11y3K13y1+κ(eαy)(K01y3K03y1),K12y1K11y1+κ(eαy)(K02y1K01y2),(A14) where J0=(J01,J02,J03),J1=(J12,J12,J13),K0=(K01,K02,K03),K1=(K11,K12,K13) are vector functions, j0,j1 are scalar functions, x1[0,l] and t[0,T].

Boundary layers for the microrotation

The function Θ(t,x1,J,K) in the equation (Equation61) takes the following form: (A15) Θ(t,x1,J,K)=δ(κcosα(K1y2K2y1)+κsinα(K3y1K1y3)κ(K2y1cosαK3y1sinα)+κ(K2y1cosαK3y1sinα)+κ(eαy)ΔK1,κcosαK2y2+κsinαK2y3+2κcosαK1y1,κsinαK3y3κcosαK3y22κsinαK1y1)+β(κcosαK2y1+κsinαK3y1,κ(K2y2cosαK3y2sinα)+2κcosαK1y1,κ(K2y3cosαK3y3sinα)2κsinαK1y1)+a(J3y2J2y3,J1y3J3y1,J2y1J1y2),(A15) where J=(J1,J2,J3), ,K=(K1,K2,K3) are vector functions, x1[0,l] and t[0,T].

The function Λ(t,x1,J0,J1,K0,K1) from the equation (Equation62) reads: Λ(t,x1,J0,J1,K0,K1)=(K01t,K02t,K03t)+δ(κcosα(K11y2K12y1)+κsinα(K13y1K11y3)(κcosαK12y1κsinαK13y1)+κ(J12y1cosαJ13y1sinα)(K02(κcosα)K03(κsinα)J02(κcosα)+J03(κsinα))κ2K02+κ(eαy)(ΔK11κcosαK01y2+κsinαK01y3)+κ2(eαy)2ΔK01,κcosαK12y2+κsinαK12y3+2κ(eαy)(κcosαK02y2+κsinαK02y3)+(2κK11y1+κK01κ2K02)cosα+κτK01sinα+κ2cosα(eαy)K02y2κ2sinα(eαy)K02y3,κsinαK13y3κcosαK13y2+2κ(eαy)κsinαK03y32κ(eαy)κcosαK03y2+κτK01cosα(2κK11y1+κK01κ2K02)sinα+κ2cosα(eαy)K03y2κ2sinα(eαy)K03y3) +β((κcosα)K02κcosαK12y1+(κsinα)K03+κsinαK13y1+κ(eαy)(κcosαK02y1+κsinαK03y1),κ(K12y2cosαK13y2sinα)+(κK01+κ2K02κ2sinαK03+2κ(K11y1κ(K02cosαK03sinα)))cosα+κτK01sinα+κ(eαy)K11y2+κτ(eαy)K11y2+κ2(eαy)K02y2cosακ2(eαy)K03y2sinα+2κ(eαy)(2K11y1y2κ(K02y2cosαK03y2sinα)),κ(K12y3cosαK13y3sinα)+κτK01cosα(κK01+κ2cosαK02κ2sinαK03+2κ(K11y1κ(K02cosαK03sinα)))sinα+κ(eαy)K01y3+κτ(eαy)K01y3 (A16) +κ2(eαy)(cosαK02y3sinαK02y3)+2κ(eαy)2K11y1y32κ2(eαy)(K02y3cosαK03y3sinα))+2a(K01,K02,K03)+a(J13y2J12y3,J11y3J13y1+κ(eαy)(J01y3J03y1),J12y1J11y2+κ(eαy)(J02y1J01y2)),(A16) where J0=(J01,J02,J03),J1=(J12,J12,J13),K0=(K01,K02,K03),K1=(K11,K12,K13) are vector functions, x1[0,l] and t[0,T].