The solutions for the flow of micropolar fluid through an expanding or contracting channel with porous walls

Abstract

The unsteady, two-dimensional laminar flow of an incompressible micropolar fluid in a channel with expanding or contracting porous walls is investigated. The governing equations are transformed into a coupled nonlinear two-points boundary value problem by a suitable similarity transformation. Unlike the classic Berman problem (Berman in J. Appl. Phys. 24:1232-1235, 1953), three new solutions (totally six solutions) and no-solution interval, which is one of important characteristics for the laminar flow through porous pipe with stationary wall (Terrill and Thomas in Appl. Sci. Res. 21:37-67, 1969), are found numerically for the first time. The multiplicity of the solutions is strictly dependent on the expansion ratio. Furthermore, the asymptotic solutions are constructed by the Lighthill method, which eliminates the singularity of the similarity solution, for large injection and by the matching theorem for the suction Reynolds number, respectively. The analytical solutions also are compared with the numerical ones and the results agree well.

1 Introduction

The studies of laminar flow in an expanding or contracting channel or pipe with permeable wall have received considerable attentions due to its application in biophysical flows in recent years. The pioneering work may have been done by Uchida and Aoki [3] to simulate the successive peristaltic motion of the artery, who examined the viscous flow inside an impermeable tube with a contracting cross section. Later, Goto and Uchida [4] analyzed the incompressible laminar flow in a semi-infinite porous pipe whose radius varied with time. In addition, further experiments carried out by Wilens [5], Evans et al. [6], and Michel [7] proved the endothelial walls to be permeable with ultra-microscopic pores, through which the mass transfer takes place between blood, air, and tissue. It has also been pointed out that the deposition of cholesterol and the dilated damaged and inflamed arterial walls may cause an increase in its permeability. Recently, Majdalani et al. [8–10] investigated the flow of the fluid through an expanding channel with porous walls by numerical or asymptotical methods and analyzed the influence of the permeability and the expansion ratio on the velocity and pressure distribution.

In general, blood flow is regarded as a non-Newtonian fluid through a pipe or channel. The rheological properties of the blood are mainly due to the suspension of red blood cells, white cells, and platelets in the blood plasma. These properties strongly affect the dynamics of blood flow and render blood a non-Newtonian fluid [11]. The theory of micopolar fluids proposed by Eringen [12, 13] is one of the best theories of fluids to describe the deformation of such materials. Physically these fluids may represent the fluids consisting of rigid randomly oriented particles suspended in a viscous medium undergoing both translational and rotational motion. Furthermore, experimental studies have revealed that the blood’s viscosity decreases with the increase in shear rate, and that blood has a small yield stress. Airman et al. [14, 15] made excellent reviews about the applications of micropolar fluids. Pazanin [16] presented the result as regards an asymptotic approximation of the micropolar fluid flow through a thin or long straight pipe with variable cross section and explicitly addressed the effects of the microstructure on the flow. Ziabakhsh and Domairry [17] and Joneidi et al. [18] discussed the micropolar fluid in a porous channel by a homotopy analysis method (HAM). Ramachandran et al. [19] studied the heat transfer of a micropolar fluid past a curved surface with suction and injection using Van Dyke’s singular perturbation technique. Si et al. [20] investigated the mass and heat transfer of the micropolar fluid through an expanding channel with porous walls by HAM and analyzed the effects of the heat dissipation on the velocity and temperature distribution.

Apart from the above work, solution multiplicity is a classic problem for the equations describing the flow in channels or pipes with porous walls. A similar problem has been studied by Berman [1] and Terrill [21], who assumed that the wall is stationary. The multiplicity of the solutions has been addressed by Terrill [21], Terrill and Thomas [2], Robinson [22], MacGillivray and Lu [23], Lu et al. [24], and Zaturska et al. [25]. More recently, a theoretical treatment was done by Xu et al. [26], who revisited the Newtonian fluid in the channel with porous orthogonally moving walls by homotopy analysis method (HAM), and the results have shown that there exist dual or triple solutions corresponding to the cases reported by Zaturska et al. [25]. Almost at the same time, Si et al. [27, 28] also obtained the dual solutions of the flow in the expanding pipe and channel with porous walls, where they consider the small expansion ratio and large suction Reynolds number by singular perturbation methods. In addition, some other similar numerical and experimental studies also have been reported on multiple solutions. For example, Luo and Pedley [29] investigated two-dimensional limitation flow in a collapsible channel and multiple steady solutions were found for a range of physical parameters. Siviglia and Toffolon [30] discussed one-dimensional flow through a collapsible tube and found that geometrical alterations or variations of the mechanical properties of the tube wall affected the occurrence of multiple flow states. Lanzerstorfer and Kuhlmann [31] studied numerically the global stability of multiple solutions for the flow in a sudden expansion plane. Putkaradze and Vorobieff [32] made some experiments to show the multiple solutions of the flow in an expanding channel.

Motivated by the above-mentioned studies, the aim of this paper is to investigate the flow of an incompressible micropolar fluid in a channel with deforming porous walls. The paper is organized as follows. In Section 2 we introduce the laminar flow equations and the similarity transformation which reduce the governing equations to a system of high order nonlinear ordinary differential ones. In Section 3 we discuss the characteristics of different types of solution obtained by Bvp4c, which is a collocation method equivalent to the fourth order monoimplicit-Runge-Kutta method. In Section 4 we construct the asymptotic solutions with large injection or suction velocity, which are compared with the numerical ones obtained in Section 3. In Section 5 the conclusion is drawn.

2 Governing equations

We consider the unsteady, incompressible, laminar flow of a micropolar fluid in a semi-infinite channel with porous walls. The distance $2a(t)$ between the walls is much smaller than the width and length of the channel. One end of the channel is closed by a complicated solid membrane. Both walls have equal permeability $v_{w}$ and expand or contract uniformly in the transverse direction at a time-dependent rate $\dot{a}(t)$. As shown in Figure 1, a coordinate system is chosen with the origin at the center of the channel. Take x and y to be coordinate axes parallel and perpendicular to the channel walls, respectively.

Figure 1

The model of the porous channel with expanding or contracting walls.

The governing equations of the flow are expressed as follows [18, 20]:

$$\begin{aligned}& \frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}=0, \end{aligned}$$

(1)

$$\begin{aligned}& \frac{\partial{u}}{\partial{t}}+u\frac{\partial{u}}{\partial{x}}+v\frac {\partial{u}}{\partial{y}}=- \frac{1}{\rho}\frac{\partial{p}}{\partial {x}}+\biggl(\nu+\frac{\kappa}{\rho}\biggr) \biggl( \frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\biggr)+\frac{\kappa}{\rho}\frac{\partial N}{\partial y}, \end{aligned}$$

(2)

$$\begin{aligned}& \frac{\partial{v}}{\partial{t}}+u\frac{\partial{v}}{\partial{x}}+v\frac {\partial{v}}{\partial{y}}=- \frac{1}{\rho}\frac{\partial{p}}{\partial {y}}+\biggl(\nu+\frac{\kappa}{\rho}\biggr) \biggl( \frac{\partial^{2}v}{\partial x^{2}}+\frac{\partial^{2}v}{\partial y^{2}}\biggr)-\frac{\kappa}{\rho}\frac{\partial N}{\partial x}, \end{aligned}$$

(3)

$$\begin{aligned}& \rho j\biggl(\frac{\partial N}{\partial t}+u\frac{\partial N}{\partial x}+v \frac{\partial N}{\partial y}\biggr)=-\kappa\biggl(2N+\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x} \biggr)+\gamma\biggl(\frac{\partial^{2}N}{\partial y^{2}}+\frac{\partial^{2}N}{\partial x^{2}}\biggr), \end{aligned}$$

(4)

where u, v is the velocity components in the x and y directions and N is the microrotation, respectively. ρ, p, μ, κ, j, and γ are the density, pressure, dynamic viscosity, vortex viscosity coefficient, micro-inertia, and the spin gradient viscosity coefficient, respectively. Here γ is assumed to be [33]

$$ \gamma=\biggl(\mu+\frac{\kappa}{2}\biggr)j. $$

(5)

The corresponding boundary conditions are [8, 20]

$$\begin{aligned}& u(y)=0,\qquad v(y)=v_{w}=A\dot{a},\qquad N(y)=0,\quad y=a(t), \end{aligned}$$

(6)

$$\begin{aligned}& \frac{\partial u}{\partial y}=0, \qquad v(y)=0,\qquad N(y)=0,\quad y=0, \end{aligned}$$

(7)

where A is the injection coefficient, which is a measure of the wall permeability.

Motivated by the definition of the stream function [8, 20], in this paper we define

$$ u=-\nu a^{-2}xF_{\eta}(\eta,t),\qquad v=\nu a^{-1}F(\eta,t),\qquad N=\nu a^{-3}xG(\eta,t),\quad \eta=y/a(t). $$

(8)

Substituting equation (8) into equations (1), (2), (3), (4), one obtains the following equations:

$$\begin{aligned}& (1+K)F_{\eta\eta\eta\eta}-KG_{\eta\eta}+\alpha(\eta F_{\eta\eta\eta }+3F_{\eta\eta})+(F_{\eta} F_{\eta\eta}-FF_{\eta\eta\eta})-a^{2} \nu ^{-1}F_{\eta\eta t}=0, \end{aligned}$$

(9)

$$\begin{aligned}& \biggl(1+\frac{K}{2}\biggr)G_{\eta\eta}+\alpha(\eta G_{\eta}+3G)+\mathit{Re}(F_{\eta }G-FG_{\eta})+N_{1}(F_{\eta\eta}-2G)-a^{2} \nu^{-1}G_{t}=0, \end{aligned}$$

(10)

where $\alpha=\frac{a\dot{a}}{\upsilon}$ is the expansion ratio, and $K=\frac{\kappa}{\mu}$ and $N_{1}=\frac{\kappa a^{2}}{j\mu}$ are micropolar parameters. In accordance with the physical meaning, the positive α represents the expanding walls.

The boundary conditions become

$$\begin{aligned}& F(0)=0,\qquad F''(0)=0,\qquad G(0)=0, \end{aligned}$$

(11)

$$\begin{aligned}& F(1)=\mathit{Re},\qquad F'(1)=0,\qquad G(1)=0, \end{aligned}$$

(12)

where $\mathit{Re}=\frac{a v_{w}}{\upsilon}$ is the permeability Reynolds number.

A similar solution with respect to both space and time can be developed following the transformation described by Uchida and Aoki [4], Dauenhauer and Majdalani [10], and Si et al. [20], respectively. This can be accomplished by considering the case: α is a constant, and $F_{\eta\eta t}=0$ and $G_{t}=0$. Similar to [10], the normal pressure gradient and axial pressure gradient can be obtained by substituting equation (8) into equations (2) and (3), which are listed as follows:

$$ \triangle{p}_{n}=\rho\nu^{2}a^{-2} \int^{\eta}_{0}\bigl( \alpha\bigl(F+\eta F'\bigr)-FF'+(1+K)F''-KG \bigr)\, d\eta, $$

(13)

and

$$ \triangle{p}_{x}=-\rho\nu^{2} a^{-4} \frac{x^{2}}{2} \bigl(2\alpha F'+\alpha\eta F''+F^{\prime 2}-FF''+(1+K)F'''-K F'\bigr). $$

(14)

Equations (9), (10), (11), (12) can be normalized by the following transformations:

$$ v^{*}=\frac{v}{\dot{a}},\qquad u^{*}=\frac{u}{\dot{a}},\qquad x^{*}= \frac{x}{a},\qquad f=\frac {F}{\mathit{Re}},\qquad g=\frac{G}{\mathit{Re}}, $$

(15)

and so

$$\begin{aligned}& (1+K)f^{(4)}-Kg''+\alpha\bigl( \eta f'''+3f'' \bigr)+\mathit{Re}\bigl(f' f''-ff''' \bigr)=0, \end{aligned}$$

(16)

$$\begin{aligned}& \biggl(1+\frac{K}{2}\biggr)g''+ \alpha\bigl(\eta g'+3g\bigr)+\mathit{Re}\bigl(f'g-fg' \bigr)+N_{1}\bigl(f''-2g\bigr)=0. \end{aligned}$$

(17)

The boundary conditions become

$$\begin{aligned}& f(0)=0,\qquad f''(0)=0,\qquad g(0)=0, \end{aligned}$$

(18)

$$\begin{aligned}& f(1)=1,\qquad f'(1)=0,\qquad g(1)=0. \end{aligned}$$

(19)

In physical meaning, this model is based on a decelerating wall dilation rate that follows a plausible assumption [3, 10]:

$$ \frac{a\dot{a}}{\nu}=\frac{a_{0}\dot{a}_{0}}{\nu}=\alpha, $$

(20)

where $a_{0}$ and $\dot{a}_{0}$ are the initial height and expansion rate of the channel. Then the expressions for the height and velocity at time t of the wall can be obtained,

$$ \frac{a(t)}{a_{0}}=\bigl(1+2\nu\alpha t a_{0}^{-2} \bigr)^{1/2},\qquad \frac{\dot {a}(t)}{\dot{a}_{0}}=\bigl(1+2\nu\alpha t a_{0}^{-2}\bigr)^{-1/2}. $$

(21)

3 Numerical results and discussions

As mentioned before, multiple solutions associated with fluid mechanics have been attracted many researchers’ attention. Recently, Yao [34] claimed that the Reynolds number is insufficient to determine a flow field uniquely for a given geometry. The non-uniqueness of fluid flow means that the different frequencies of flows can exist on each stable bifurcation branch. Here the multiple solutions are obtained numerically by Bvp4c [35] and we set the maximum residual error as 10⁻⁴ during the computation.

Compared with the classic works obtained by Robinson [22], MacGillivray and Lu [23], Zaturska et al. [25], which are related with the Newtonian flow with stationary porous walls, the numerical solutions obtained by us lead to the new conclusions:

• (i) As is shown in Figures 2, 3, 4, the solutions of types I, II, III exist for all the values of the expansion ratio. In addition, the interval of existence for the solutions types II, III varies with the expansion ratio.

  Figure 2

  

  The effects of the negative expansion ratio α on the solutions as $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$ .

  Figure 3

  

  The effects of the positive expansion ratio α on the solutions as $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$ .

  Figure 4

  

  Axial velocity for type IV as $\pmb{K=0.1}$ and $\pmb{N_{1}=0.1}$ .

• (ii) The positive expansion ratio leads to the appearance of new types of solutions, which are illustrated in Figures 5-7. However, these solutions do not exist for every value of the expansion ratio.

  Figure 5

  

  Axial velocity for type IV as $\pmb{K=0.1}$ , $\pmb{\alpha=5}$ , and $\pmb{N_{1}=0.1}$ .

  Figure 6

  

  Axial velocity for type V as $\pmb{K=0.1}$ , $\pmb{\alpha=5}$ , and $\pmb{N_{1}=0.1}$ .

  Figure 7

  

  Axial velocity for type VI as $\pmb{K=0.1}$ , $\pmb{\alpha=5}$ , and $\pmb{N_{1}=0.1}$ .

• (iii) For some values of the positive expansion ratio (i.e. $\alpha=5$), the interval $(2.72,9.77)$ of no solution appears. This phenomenon is similar to the case of the Newtonian flow through a stationary pipe with porous wall.

• (iv) The micropolar parameter has influence on the solutions, but it does not lead to the appearance of the new types of solutions, which is shown in Figures 8, 9.

  Figure 8

  

  The effects of K on solutions as $\pmb{\alpha=0.5}$ , $\pmb{N_{1}=0.1}$ .

  Figure 9

  

  The effects of $\pmb{N_{1}}$ on solutions as $\pmb{\alpha=0.5}$ , $\pmb{K=0.1}$ .

4 Perturbation analysis for this problem

4.1 Solution for the large injection Reynolds number

The singularity is due to large blowing at each wall which pushes the shear layer to the core region [36]. In order to eliminate the singular behavior, we employ the Lighthill method to obtain the asymptotic solution which is valid to arbitrary order as regards the derivative of the axial velocity.

One treats $\varepsilon=1/\mathit{Re}$ as the perturbation parameter, then equations (16), (17) become

$$\begin{aligned}& \varepsilon(1+K)f'''- \varepsilon Kg'+\varepsilon\alpha\bigl(\eta f''+2f' \bigr)+\bigl(f^{\prime (2)}-ff''\bigr)= \lambda, \end{aligned}$$

(22)

$$\begin{aligned}& \varepsilon(1+K/2)g''+\varepsilon\alpha \bigl(\eta g'+3g\bigr)-\varepsilon N_{1} \bigl(2g-f''\bigr)+f'g-fg'=0, \end{aligned}$$

(23)

where λ is an integer constant. First of all, we introduce a variable transformation of η,

$$ \eta=\xi+\varepsilon X_{1}(\xi)+\varepsilon^{2} X_{2}(\xi)+O\bigl(\varepsilon^{3}\bigr), $$

(24)

where the functions $X_{1}$, $X_{2}$ will be determined next.

Assume the function f, g, and the constant λ can be expanded in the following forms:

$$ f(\xi)=\sum_{i=0}^{\infty}\varepsilon^{i}f_{i}(\xi),\qquad g(\xi)=\sum _{i=0}^{\infty}\varepsilon^{i}g_{i}( \xi),\qquad \lambda=\sum_{i=0}^{\infty}\varepsilon^{i}\lambda_{i}. $$

(25)

Substituting (25) into (22), (23), and collecting the same powers of ε, one can obtain the leading order of the solution

$$ \dot{f}_{0}^{2}-f_{0} \ddot{f}_{0}=\lambda_{0} ,\qquad f_{0} \dot{g}_{0}-\dot{f}_{0}g_{0}=0, $$

(26)

and the first order of the solution

$$\begin{aligned}& (1+K)\dddot{f}_{0}-K\dot{g}_{0}+\alpha\xi \ddot{f}_{0}+2\alpha\dot {f}_{0}-f_{0} \ddot{f}_{1}-\ddot{f}_{0}f_{1} +2 \dot{f}_{0} \dot{f}_{1}=\lambda_{1}+2 \lambda_{0}\dot{X}_{1}, \end{aligned}$$

(27)

$$\begin{aligned}& (1+K/2)\ddot{g}_{0}+\alpha(\xi\dot{g}_{0}+3g_{0})+( \dot{f}_{1}g_{0}-f_{1}\dot{g}_{0}) +N_{1}(\ddot{f}_{0}-2g_{0})+\dot{f}_{0}g_{1}-f_{0} \dot{g}_{1}=0. \end{aligned}$$

(28)

Here the overdot $\dot{\hphantom{a}}$ denotes the derivation with respect to ξ.

4.1.1 A. The transformed boundary condition at the wall

We assume ξ̃ is the root of (24) at $\eta=1$, then

$$\begin{aligned} \tilde{\xi} =& 1-\varepsilon X_{1}(\tilde{\xi})- \varepsilon^{2}X_{2}(\tilde{\xi}) +O\bigl( \varepsilon^{3}\bigr) \\ =& 1-\varepsilon\bigl\{ X_{1}(1)+\dot{X}_{1}(1)\bigl[- \varepsilon X_{1}(\tilde{\xi })-\varepsilon^{2} X_{2}(\tilde{\xi})\bigr]+\cdots\bigr\} \\ &{}-\varepsilon^{2} \bigl\{ X_{2}(1)+\dot{X}_{2}(1) \bigl[-\varepsilon X_{1}(\tilde{\xi })-\varepsilon^{2} X_{2}(\tilde{\xi})\bigr]+\cdots\bigr\} +\cdots \\ =& 1-\varepsilon X_{1}(1)-\varepsilon^{2} \bigl[X_{2}(1)-\dot{X}_{1}(1)X_{1}(1)\bigr]+ O\bigl( \varepsilon^{3}\bigr), \end{aligned}$$

(29)

thus the condition at the wall can be obtained

$$\begin{aligned}& f|_{\eta=1}=1\quad \Longrightarrow\quad 1 =f|_{\xi=\tilde{\xi}} = f|_{\xi =1}+\dot{f}|_{\xi=1}\bigl\{ -\varepsilon X_{1}(1)- \varepsilon^{2}\bigl[X_{2}(1)-\dot {X}_{1}(1)X_{1}(1) \bigr]+ \cdots\bigr\} \\ & \hphantom{f|_{\eta=1}=1\quad \Longrightarrow\quad 1}=f_{0}|_{\xi=1}+\varepsilon[f_{1}-X_{1} \dot{f}_{0}]|_{\xi=1}+O\bigl(\varepsilon^{2} \bigr), \end{aligned}$$

(30)

$$\begin{aligned}& f'|_{\eta=1}=0 \quad \Longrightarrow\quad 0 = \dot{f}|_{\xi=\tilde{\xi}} = \dot {f}|_{\xi=1}+\ddot{f}|_{\xi=1}\bigl\{ -\varepsilon X_{1}(1)-\varepsilon ^{2}\bigl[X_{2}(1)- \dot{X}_{1}(1)X_{1}(1)\bigr]+ \cdots\bigr\} \\& \hphantom{f'|_{\eta=1}=0 \quad \Longrightarrow\quad 0}=\dot{f}_{0}|_{\xi=1}+\varepsilon[ \dot{f}_{1}-X_{1} \ddot{f}_{0}]|_{\xi=1} +O \bigl(\varepsilon^{2}\bigr), \end{aligned}$$

(31)

$$\begin{aligned}& g|_{\eta=1}=0\quad \Longrightarrow\quad 0=g|_{\xi=\tilde{\xi}} = g|_{\xi =1}+\dot{g}|_{\xi=1}\bigl\{ -\varepsilon X_{1}(1)- \varepsilon^{2}\bigl[X_{2}(1)-\dot {X}_{1}(1)X_{1}(1) \bigr]+ \cdots\bigr\} \\& \hphantom{g|_{\eta=1}=0\quad \Longrightarrow\quad 0}=g_{0}|_{\xi=1}+\varepsilon[g_{1}-X_{1} \dot{g}_{0}]|_{\xi=1}+ O\bigl(\varepsilon^{2} \bigr). \end{aligned}$$

(32)

Hence, the boundary conditions of $f_{i}$ and $g_{i}$ at $\eta=1$ are

$$\begin{aligned}& f_{0}|_{\xi=1}=1 ,\qquad (f_{1}-X_{1} \dot{f}_{0})|_{\xi=1}=0,\qquad \ldots, \end{aligned}$$

(33)

$$\begin{aligned}& \dot{f}_{0}|_{\xi=1}=0 ,\qquad ( \dot{f}_{1}-X_{1} \ddot{f}_{0})|_{\xi=1}=0, \qquad \ldots, \end{aligned}$$

(34)

$$\begin{aligned}& g_{0}|_{\xi=1}=0,\qquad (g_{1}-X_{1} \dot{g}_{0})|_{\xi=1}=0,\qquad \ldots. \end{aligned}$$

(35)

4.1.2 B. The transformed boundary condition at the core

One supposes that ξ̂ is the root of (24) at $\eta=0$, then

$$ \hat{\xi}=-\varepsilon X_{1}(\hat{\xi})- \varepsilon^{2} X_{2}(\hat{\xi})+ O\bigl( \varepsilon^{3}\bigr)=-\varepsilon X_{1}(0)- \varepsilon^{2} \bigl[X_{2}(0)-\dot{X}_{1}(0) X_{1}(0)\bigr]+ O\bigl(\varepsilon^{3}\bigr), $$

(36)

thus we can induce

$$\begin{aligned}& f|_{\eta=0}=0\quad \Longrightarrow\quad 0 = f|_{\xi=\hat{\xi}} =f|_{\xi=0}+\dot{f}|_{\xi=0}\bigl\{ -\varepsilon X_{1}(0)-\varepsilon ^{2}\bigl[X_{2}(0)- \dot{X}_{1}(0)X_{1}(0)\bigr]+ \cdots\bigr\} \\& \hphantom{f|_{\eta=0}=0\quad \Longrightarrow\quad 0}=f_{0}|_{\xi=0}+\varepsilon[f_{1}-X_{1} \dot{f}_{0}]|_{\xi=0}+ O\bigl(\varepsilon ^{2}\bigr), \end{aligned}$$

(37)

$$\begin{aligned}& f''|_{\eta=0}=0\quad \Longrightarrow\quad 0 = \ddot{f}+\dot{f}\bigl[\bigl(-\varepsilon\ddot{X}_{1}- \varepsilon^{2}\ddot {X}_{2}\bigr)/\bigl(1+\varepsilon \dot{X}_{1}+\varepsilon^{2}\dot{X}_{2}\bigr) \bigr]|_{\xi=\hat {\xi}} \\& \hphantom{f''|_{\eta=0}=0\quad \Longrightarrow\quad 0} =\bigl\{ \ddot{f}|_{\xi=0} +\dddot{f}|_{\xi=0} \bigl[-\varepsilon X_{1}(0)-\varepsilon^{2} \bigl(X_{2}(0)-\dot{X}_{1}(0)X_{1}(0)\bigr)+ \cdots \bigr]\bigr\} \\& \hphantom{f''|_{\eta=0}=0\quad \Longrightarrow\quad 0 ={}}{}+\bigl\{ \dot{f}|_{\xi=0}+\ddot{f}|_{\xi=0}\bigl[- \varepsilon X_{1}(0) -\varepsilon^{2}\bigl(X_{2}(0)- \dot{X}_{1}(0)X_{1}(0)\bigr)+ \cdots\bigr]\bigr\} \\& \hphantom{f''|_{\eta=0}=0\quad \Longrightarrow\quad 0 ={}}{}\cdot \bigl[\bigl(-\varepsilon\ddot{X}_{1}- \varepsilon^{2}\ddot{X}_{2}\bigr)/\bigl(1+\varepsilon\dot {X}_{1}+\varepsilon^{2}\dot{X}_{2}\bigr) \bigr]|_{\xi=\hat{\xi}} \\& \hphantom{f''|_{\eta=0}=0\quad \Longrightarrow\quad 0} =\ddot{f}_{0}|_{\xi=0}+\varepsilon[ \ddot{f}_{1}-X_{1} \dddot{f}_{0}-\ddot {X}_{1}\dot{f}_{0}]|_{\xi=0}+ O\bigl( \varepsilon^{2}\bigr), \end{aligned}$$

(38)

$$\begin{aligned}& g|_{\eta=0}=0\quad \Longrightarrow\quad 0 = g|_{\xi=\hat{\xi}} =g|_{\xi=0}+\dot{g}|_{\xi=0}\bigl\{ -\varepsilon X_{1}(0)- \varepsilon ^{2}\bigl[X_{2}(0)-\dot{X}_{1}(0)X_{1}(0) \bigr]+ \cdots\bigr\} \\& \hphantom{g|_{\eta=0}=0\quad \Longrightarrow\quad 0} =g_{0}|_{\xi=0}+\varepsilon[g_{1}-X_{1} \dot{g}_{0}]|_{\xi=0}+ O\bigl(\varepsilon^{2} \bigr). \end{aligned}$$

(39)

Hence, the boundary conditions of $f_{i}$ and $g_{i}$ at $\eta=0$ are

$$\begin{aligned}& f_{0}|_{\xi=0}=0, \qquad (f_{1}-X_{1} \dot{f}_{0})|_{\xi=0}=0,\qquad \ldots, \end{aligned}$$

(40)

$$\begin{aligned}& \ddot{f}_{0}|_{\xi=0}=0 ,\qquad ( \ddot{f}_{1}-X_{1} \dddot{f}_{0} - \ddot{X}_{1}\dot{f}_{0})|_{\xi=0}=0,\qquad \ldots, \end{aligned}$$

(41)

$$\begin{aligned}& g_{0}|_{\xi=0}=0 ,\qquad (g_{1}-X_{1} \dot{g}_{0})|_{\xi=0}=0,\qquad \ldots. \end{aligned}$$

(42)

Using (40), (41), and (42), the solution for (26) can be obtained,

$$ f_{0}(\xi)=\sin\biggl(\frac{\pi}{2} \xi\biggr), \qquad g_{0}(\xi)=0, $$

(43)

and then $\lambda_{0}=\frac{\pi^{2}}{4}$.

Substituting the above results into (27), (28) yields the following equations for $f_{1}$, $g_{1}$:

$$\begin{aligned}& \sin\biggl(\frac{\pi}{2} \xi\biggr)\ddot{f}_{1}-\pi \cos\biggl(\frac{\pi}{2} \xi\biggr)\dot{f}_{1}-\frac{\pi^{2}}{4} \sin\biggl(\frac{\pi}{2} \xi\biggr)f_{1} \\& \quad =\lambda_{1}+\frac{\pi^{2}}{2}\dot{X}_{1}+\biggl( \frac{(1+K)\pi^{3}}{8}-\alpha\pi \biggr)\cos\biggl(\frac{\pi}{2}\xi\biggr)+ \frac{\alpha\pi^{2}}{4}\xi\sin\biggl(\frac{\pi}{2}\xi\biggr), \end{aligned}$$

(44)

and

$$ \sin\biggl(\frac{\pi}{2} \xi\biggr)\dot{g}_{1}- \frac{\pi}{2}\cos\biggl(\frac{\pi}{2} \xi\biggr)g_{1}+N_{1} \frac{\pi^{2}}{4}\sin\biggl(\frac{\pi}{2} \xi\biggr)=0. $$

(45)

It is noted that using the method of variation of parameters to solve equation (44) will cause a singularity term in the solution [36]. In order to eliminate the singularity and to obtain $f_{1}$, we can set

$$ \lambda_{1}+\frac{\pi^{2}}{2}\dot{X}_{1}+ \biggl(\frac{(1+K)\pi^{3}}{8}-\alpha\pi\biggr)\cos \biggl(\frac{\pi}{2}\xi \biggr)+\frac{\alpha\pi^{2}}{4}\xi\sin\biggl(\frac{\pi}{2}\xi\biggr)=0, $$

(46)

then we have

$$ X_{1}(\xi)=\biggl(\frac{2\alpha}{\pi^{2}}-\frac{(1+K)}{2} \biggr)\sin\biggl(\frac{\pi}{2}\xi \biggr)+\frac{\alpha}{\pi}\xi\cos\biggl( \frac{\pi}{2}\xi\biggr)-\frac{2\lambda_{1}}{\pi ^{2}}\xi+C_{1}. $$

(47)

Thus, the equation for $f_{1}$ becomes

$$ \sin\biggl(\frac{\pi}{2} \xi\biggr)\ddot{f}_{1}-\pi \cos\biggl(\frac{\pi}{2} \xi\biggr)\dot{f}_{1}-\frac{\pi^{2}}{4} \sin\biggl(\frac{\pi}{2}\xi\biggr)f_{1}=0, $$

(48)

whose solution is

$$ f_{1}(\xi)=K_{1}\cos\biggl(\frac{\pi}{2} \xi\biggr)+K_{2}\biggl(\frac{2}{\pi}\sin\biggl(\frac{\pi}{2} \xi\biggr)-\xi\cos\biggl(\frac{\pi}{2} \xi\biggr)\biggr), $$

(49)

where $K_{1}$, $K_{2}$ can be determined by boundary conditions, and the results are $K_{1}=\frac{\pi}{2}C_{1}$, $K_{2}=0$, $\lambda_{1}=\alpha-\frac{(1+K)\pi^{2}}{4}$. For obtaining the nontrivial solution of $f_{1}$, we set $C_{1} = 1$. Thus, one can obtain $f_{1}$, $g_{1}$:

$$ f_{1}=\frac{\pi}{2}\cos\biggl(\frac{\pi}{2} \xi\biggr), $$

(50)

and

$$ g_{1}=\frac{\pi}{2}N_{1}\sin \biggl( \frac{\pi}{2}\xi\biggr)\log\biggl[\cot\biggl(\frac{\pi}{4}\xi\biggr) \biggr]. $$

(51)

Finally, we obtain the asymptotic solution of f, g in terms of ξ as follows:

$$\begin{aligned}& f(\eta) =\sin\biggl(\frac{\pi}{2}\xi\biggr) +\varepsilon \frac{\pi}{2}\cos\biggl(\frac{\pi }{2}\xi\biggr), \end{aligned}$$

(52)

$$\begin{aligned}& g(\eta) =\varepsilon \frac{\pi}{2}N_{1}\sin\biggl( \frac{\pi}{2}\xi\biggr)\log\biggl[\cot\biggl(\frac{\pi}{4}\xi\biggr) \biggr], \end{aligned}$$

(53)

where

$$ \eta=\xi +\varepsilon\biggl[\biggl(\frac{2\alpha}{\pi^{2}}- \frac{1+K}{2}\biggr)\sin\biggl(\frac{\pi}{2}\xi \biggr)+ \frac{\alpha}{\pi}\xi\cos\biggl(\frac{\pi}{2}\xi\biggr)-\biggl( \frac{2\alpha}{\pi ^{2}}-\frac{1+K}{2}\biggr)\xi+1\biggr]. $$

(54)

Tables 1 and 2 give the comparison between the asymptotic results of $-f''(1)$, $g'(1)$ with the numerical ones and the results agree well. The axial velocity and microrotation velocity profiles are showed in Figures 10, 11.

Figure 10

The effects of the expansion ratio α on the axial velocity as $\pmb{\mathit{Re}=-100}$ , $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$ .

Figure 11

The effects of the expansion ratio α on microrotation as $\pmb{\mathit{Re}=-100}$ , $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$ .

Table 1 The comparison between asymptotic and numerical values of $\pmb{-f''(1)}$ for large injection Reynolds number as $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$

Table 2 The asymptotic and numerical values of $\pmb{g'(1)}$ for large injection Reynolds number at $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$

4.2 Solution for large suction Reynolds number

In order to capture an analytical solution satisfying all the boundary conditions and valid in the whole channel, the method of matched asymptotic expansion are used. Then equations (16), (17) can be written as

$$\begin{aligned}& \varepsilon(1+K)f'''- \varepsilon Kg'+\varepsilon\alpha\bigl(\eta f''+2f' \bigr)+\bigl(f^{\prime 2}-ff''\bigr)= \frac{k}{\mathit{Re}}, \end{aligned}$$

(55)

$$\begin{aligned}& \varepsilon\biggl(1+\frac{K}{2}\biggr)g''+ \varepsilon\alpha\bigl(\eta g'+3g\bigr)-\varepsilon N_{1} \bigl(2g-f''\bigr)+f'g-fg'=0, \end{aligned}$$

(56)

$$\begin{aligned}& \varepsilon=\frac{1}{\mathit{Re}},\quad \beta^{2}+\bigl((1+K) \sigma+2\alpha\beta-K\delta \bigr)\varepsilon=\frac{k}{\mathit{Re}}, \end{aligned}$$

(57)

where k is a constant of integration. Here

$$\begin{aligned}& \beta=f'(0)=\beta_{0}+\varepsilon \beta_{1}+\varepsilon^{2}\beta_{2}+\cdots ,\qquad \sigma=f'''(0)=\sigma_{0}+ \varepsilon\sigma_{1}+\varepsilon^{2}\sigma _{2}+ \cdots, \end{aligned}$$

(58)

$$\begin{aligned}& \delta=g'(0)=\delta_{0}+\varepsilon \delta_{1}+\varepsilon^{2}\delta_{2}+\cdots, \end{aligned}$$

(59)

where the coefficients $\beta_{i}$, $\sigma_{i}$, and $\delta_{i}$ are constants determined by matching with the inner solutions. We assume that the forms of the outer solution are

$$ f=f_{0}+\varepsilon f_{1}+ \varepsilon^{2}f_{2}+\varepsilon^{3}f_{3}+O \bigl(\varepsilon^{3}\bigr),\qquad g=g_{0}+\varepsilon g_{1}+\varepsilon^{2}g_{2}+\varepsilon^{3}g_{3}+O \bigl(\varepsilon^{3}\bigr). $$

(60)

Substituting (60) into (55), (56) and equating the same power of the coefficient ε, one obtains

$$\begin{aligned}& \varepsilon^{0}\mbox{: }\ f_{0}^{\prime 2}-f_{0}f_{0}''= \beta_{0}^{2}, \end{aligned}$$

(61)

$$\begin{aligned}& \hphantom{\varepsilon^{0}\mbox{: }}\ f_{0}'g_{0}-f_{0}g_{0}'=0, \end{aligned}$$

(62)

$$\begin{aligned}& \varepsilon^{1}\mbox{: } 2f_{0}'f_{1}'-f_{0}f_{1}''-f_{1}f_{0}''=Kg_{0}'-(1+K)f_{0}'''- \alpha\bigl(\eta f_{0}''+2f_{0}' \bigr) \\& \hphantom{\varepsilon^{1}\mbox{: } 2f_{0}'f_{1}'-f_{0}f_{1}''-f_{1}f_{0}''={}}{}+2\beta_{0}\beta_{1}+(1+K)\sigma_{0}+2\alpha \beta_{0}-K\delta_{0}, \end{aligned}$$

(63)

$$\begin{aligned}& \hphantom{\varepsilon^{1}\mbox{: }}f_{1}'g_{0}+f_{0}'g_{1}-f_{0}g_{1}'-f_{1}g_{0}'=N_{1} \bigl(2g_{0}-f_{0}''\bigr)-\biggl(1+ \frac {K}{2}\biggr)g_{0}''-\alpha\bigl( \eta g_{0}'+3g_{0}\bigr), \end{aligned}$$

(64)

$$\begin{aligned}& \varepsilon^{2}\mbox{: }2f_{0}'f_{2}'-f_{0}f_{2}''-f_{2}f_{0}''=Kg_{1}'- \alpha\eta f_{1}''-2\alpha f_{1}'-f_{1}^{\prime 2}+f_{1}f_{1}''-(1+K)f_{1}'''+2 \beta_{0}\beta_{2} \\& \hphantom{\varepsilon^{2}\mbox{: }2f_{0}'f_{2}'-f_{0}f_{2}''-f_{2}f_{0}''={}}{}+\beta_{1}^{2}+(1+K)\sigma_{1}+2 \alpha\beta_{1}-K\delta_{1}, \end{aligned}$$

(65)

$$\begin{aligned}& \hphantom{\varepsilon^{2}\mbox{: }}f_{0}'g_{2}+f_{2}'g_{0}-f_{0}g_{2}'-f_{2}g_{0}'=f_{1}g_{1}'-f_{1}'g_{1}-N_{1}f_{1}''+2N_{1}g_{1}-3 \alpha g_{1} \\& \hphantom{\varepsilon^{2}\mbox{: }f_{0}'g_{2}+f_{2}'g_{0}-f_{0}g_{2}'-f_{2}g_{0}'={}}{}-\alpha\eta g_{1}'-\biggl(1+ \frac{K}{2}\biggr)g_{1}'', \end{aligned}$$

(66)

$$\begin{aligned}& \varepsilon^{3}\mbox{: }2f_{0}'f_{3}'-f_{0}f_{3}''-f_{3}f_{0}''=Kg_{2}'- \alpha\eta f_{2}''-2\alpha f_{2}'+f_{1}f_{2}''-(1+K)f_{2}'''+f_{2}f_{1}''-2f_{1}'f_{2}' \\& \hphantom{\varepsilon^{3}\mbox{: }2f_{0}'f_{3}'-f_{0}f_{3}''-f_{3}f_{0}''={}}{}+2\beta_{0}\beta_{3} +(1+K) \sigma_{2}+2(\alpha+\beta_{1})\beta_{2}-K \delta_{2}, \end{aligned}$$

(67)

$$\begin{aligned}& \hphantom{\varepsilon^{3}\mbox{: }}f_{0}'g_{3}+f_{3}'g_{0}-f_{0}g_{3}'-f_{3}g_{0}'=f_{1}g_{2}'-f_{2}'g_{1}-N_{1}f_{2}''+2N_{1}g_{2}-3 \alpha g_{2}-\alpha\eta g_{2}' \\& \hphantom{\varepsilon^{3}\mbox{: }f_{0}'g_{3}+f_{3}'g_{0}-f_{0}g_{3}'-f_{3}g_{0}'={}}{}+f_{2}g_{1}'-g_{2}f_{1}'- \biggl(1+\frac{K}{2}\biggr)g_{2}'', \\& \ldots. \end{aligned}$$

(68)

The corresponding boundary conditions for the outer solutions are

$$ f_{i}(0)=0,\qquad f_{i}''(0)=0, \qquad g_{i}(0)=0,\quad i=0,1,2,\ldots. $$

(69)

According to the boundary conditions (69), it is not difficult to find the solutions of (61), (62),

$$ f_{0}=\beta_{0}\eta,\qquad g_{0}= \omega_{0}\eta, $$

(70)

then $\sigma_{0}=f_{0}'''(0)=0$, $\delta_{0}=g_{0}'(0)=\omega_{0}$.

Substituting (70) into (63), one obtains

$$ 2f_{1}'-\eta f_{1}''=2 \beta_{1}, $$

(71)

then the solution of (71) satisfying the boundary conditions (69) is

$$ f_{1}=\beta_{1}\bigl(C_{1} \eta^{3}+\eta\bigr), $$

(72)

where $C_{1}$ is an integration constant, which will be determined next.

Substituting (70), (72) into (64), one obtains

$$ g_{1}-\eta g_{1}'= \frac{\omega_{0}}{\beta_{0}}(2N_{1}-4\alpha)\eta, $$

(73)

then the solution of (73) satisfying the boundary conditions (69) is

$$ g_{1}=\biggl(\omega_{1}-\frac{\omega_{0}}{\beta_{0}}(2N_{1}-4 \alpha)\ln{\eta}\biggr)\eta. $$

(74)

Because $g_{1}$ means the microrotation velocity of particles, $g'_{1}(0)\rightarrow\infty$ leads to $\omega_{0}=0$. Then

$$ g_{1}=\omega_{1}\eta. $$

(75)

Substituting (70), (72), (75) into (65), one obtains

$$\begin{aligned} f_{2} =&4C_{1}\alpha\eta^{3}\ln\eta+ \frac{3}{10}\beta_{0}C_{1}^{2} \eta^{5}+\biggl(C_{2}-\frac {4}{3}C_{1}\alpha \biggr)\eta^{3} \\ &{}+\biggl(\beta_{2}+\frac{\sigma_{1}+K\sigma_{1}+K\omega _{1}-K\delta_{1}}{2\beta_{0}}-3C_{1}(1+K) \biggr)\eta, \end{aligned}$$

(76)

where $C_{2}$ is an integration constant. Similarly, $f'''_{2}(0)\rightarrow\infty$ leads to $C_{1}=0$, $\sigma_{1}=f_{1}'''(0)=0$, thus

$$ f_{2}=C_{2}\eta^{3}+\biggl( \beta_{2}+\frac{K\omega_{1}-K\delta_{1}}{2\beta_{0}}\biggr)\eta, $$

(77)

and then $\delta_{1}=g_{1}'(0)=\omega_{1}$, which will be determined in the next section.

In order to obtain the inner solution in the viscous layer, we introduce the stretching transformation $\tau=(1-\eta)/\varepsilon$. Then (16) can be written as

$$\begin{aligned}& (1+K){\dddot{f}}-\varepsilon^{2}K \dot{g}-\alpha\bigl( \varepsilon \ddot{f}-\varepsilon^{2}\tau\ddot{f}-2\varepsilon^{2} \dot{f}\bigr)-\bigl(\dot{f}^{2}-f\ddot{f}\bigr) \\& \quad = - \varepsilon^{2}\beta^{2} -\bigl((1+K)\sigma+2\alpha\beta-K \delta\bigr)\varepsilon^{3}, \end{aligned}$$

(78)

$$\begin{aligned}& \biggl(1+\frac{K}{2}\biggr)\ddot{g}+\alpha\bigl(3 \varepsilon^{2}g-\varepsilon \dot{g}+\varepsilon^{2}\tau \dot{g}\bigr)+(f\dot{g}-\dot{f}g)- \varepsilon^{2} 2N_{1}g+N_{1} \ddot{f}=0. \end{aligned}$$

(79)

Here the overdot denotes the derivative with respect to τ. According to the boundary conditions (19), we assume the inner solutions near the wall to be

$$ f(\tau)=1+\sum_{i=1}^{\infty} \varepsilon^{i}\phi_{i}(\tau),\qquad g(\tau)=\sum _{i=1}^{\infty}\varepsilon^{i} \psi_{i}(\tau). $$

(80)

Substituting (80) into (78), (79) yields the following equations:

$$\begin{aligned}& \varepsilon^{1}\mbox{: }(1+K)\dddot{\phi}_{1}+ \ddot{\phi}_{1}=0, \end{aligned}$$

(81)

$$\begin{aligned}& \hphantom{\varepsilon^{1}\mbox{: }}\biggl(1+\frac{K}{2}\biggr)\ddot{ \psi}_{1}+\dot{\psi}_{1}+N_{1}\ddot{ \phi}_{1}=0, \end{aligned}$$

(82)

$$\begin{aligned}& \varepsilon^{2}\mbox{: }(1+K)\dddot{ \phi}_{2}+\ddot{\phi}_{2}=\alpha\ddot{\phi}_{1}+\dot {\phi}_{1}^{2}-\phi_{1}\ddot{\phi}_{1}- \beta_{0}^{2}, \end{aligned}$$

(83)

$$\begin{aligned}& \hphantom{\varepsilon^{2}\mbox{: }}\biggl(1+\frac{K}{2}\biggr)\ddot{ \psi}_{2}+\dot{\psi}_{2}=\alpha\dot{\psi}_{1}+\dot{ \phi} _{1}\psi_{1}-\phi_{1}\dot{ \psi}_{1}-N_{1}\ddot{\phi}_{2}, \\& \ldots. \end{aligned}$$

(84)

The boundary conditions corresponding to the inner solution are

$$ \phi_{i}(0)=0,\qquad \phi_{i}'(0)=0, \qquad \psi_{i}(0)=0,\quad i=1,2,3,\ldots. $$

(85)

The solution of (81) is

$$ \phi_{1}=D_{1}\bigl(e^{-A_{0}\tau}+A_{0} \tau-1\bigr), $$

(86)

where $A_{0}=\frac{1}{1+K}$. Then the first two inner solution of f can be expressed as

$$ f(\tau)=1+D_{1}\bigl(e^{-A_{0}\tau}+A_{0} \tau-1\bigr)\varepsilon. $$

(87)

The outer solution of f expressed in terms of the inner variable τ is

$$ f(\eta)=\beta_{0}+(\beta_{1}- \beta_{0} \tau)\varepsilon+\cdots. $$

(88)

As $\tau\rightarrow\infty$, matching the inner solution (87) with the outer solution (88) gives

$$ \beta_{0}=1,\qquad D_{1}=-\frac{1}{A_{0}}, \qquad \beta_{1}=\frac{1}{A_{0}}. $$

(89)

The solution of (82) is

$$ \psi_{1}=-\frac{2N_{1}}{KA_{0}}\bigl(e^{-A_{0}\tau}-1 \bigr)-\Omega_{1}\bigl(e^{-B_{0}\tau}-1\bigr), $$

(90)

where $B_{0}=\frac{2}{2+K}$ and $\Omega_{1}$ will be determined in the following matching process. Then the first two inner solution of g can be expressed as follows:

$$ g(\tau)=-\varepsilon\frac{2N_{1}}{KA_{0}}\bigl(e^{-A_{0}\tau}-1 \bigr)-\varepsilon \Omega_{1}\bigl(e^{-B_{0}\tau}-1\bigr). $$

(91)

The outer solution of g expressed in terms of the inner variable τ is

$$ g(\eta)=\omega_{1}-\tau\omega_{1}\varepsilon+ \cdots. $$

(92)

As $\tau\rightarrow\infty$, matching the inner solution (91) with the outer solution (92) gives

$$ \omega_{1}=0,\qquad \Omega_{1}=- \frac{2N_{1}}{KA_{0}},\qquad \delta_{1}=g'_{1}(0)=0. $$

(93)

According to (66), one can obtain

$$ g_{2}=\omega_{2}\eta. $$

(94)

Then we obtain $f_{3}$,

$$\begin{aligned} f_{3} =&4C_{2}\alpha\eta^{3}\ln\eta+ \biggl(C_{3}-\frac{4C_{2}\alpha}{3}\biggr)\eta^{3} \\ &{}+\biggl[ \frac {2\alpha-6C_{2}+1}{2A_{0}}+\biggl(\frac{A_{0}\alpha-A_{0}+1}{A_{0}}\biggr)\beta_{2}+ \frac {K\omega_{2}}{2}\biggr]\eta. \end{aligned}$$

(95)

Similarly, $f'''_{3}(0)\rightarrow\infty$ leads to $C_{2}=0$, $\sigma_{2}=f_{2}'''(0)=0$, thus

$$ f_{3}=C_{3}\eta^{3}+\biggl[ \frac{2\alpha+1}{2A_{0}}+\biggl(\frac{A_{0}\alpha-A_{0}+1}{A_{0}}\biggr)\beta _{2}+ \frac{K\omega_{2}}{2}\biggr]\eta, $$

(96)

then

$$ f_{2}=\beta_{2}\eta, $$

(97)

and from (83), one can obtain

$$\begin{aligned} \phi_{2} =&\frac{5}{A^{2}_{0}}+\frac{2\alpha}{A_{0}}- \frac{D_{2}}{A^{2}_{0}}+\biggl(\frac {D_{2}-2}{A_{0}}-\alpha\biggr)\tau \\ &{}+e^{-A_{0}\tau} \biggl(\frac{D_{2}-2\alpha-3\tau }{A_{0}}-\frac{\tau^{2}}{2}-\alpha \tau-\frac{5}{A^{2}_{0}} \biggr). \end{aligned}$$

(98)

As $\tau\rightarrow\infty$, matching the inner and outer solution, one obtains

$$ D_{2}=A_{0}\alpha+1, \qquad \beta_{2}=\frac{4}{A^{2}_{0}}+\frac{\alpha}{A_{0}}. $$

(99)

Then we obtain $g_{3}$,

$$ g_{3}=\bigl(\omega_{3}-\omega_{2}(2N_{1}-4 \alpha)\ln\eta\bigr)\eta, $$

(100)

thus $g'_{3}(0)\rightarrow\infty$ leads to $\omega_{2}=0$. From (84), we obtain

$$\begin{aligned} \psi_{2} =&\frac{N_{1}}{K^{2}(K+2)(3K+4)}\bigl(2e^{-B_{0}\tau}(K+1) \bigl(2K^{4}+(3\alpha -8)K^{3} \\ &{}+\bigl(3\tau^{2}+6( \alpha+1)\tau+10(\alpha-5)\bigr)K^{2}+4\bigl(\tau^{2}+2(\alpha+1) \tau+2(\alpha-9)\bigr)K-32\bigr) \\ &{}-e^{-A_{0}\tau }\bigl(8+10K+3K^{2}\bigr) \bigl(2(3\tau+\alpha-4)K^{2} \\ &{}+\bigl(\tau^{2}+2(\alpha+3)\tau+2(\alpha -8)\bigr)K-8\bigr)-4K^{2}(1+K)^{3}e^{-\frac{(3K+4)\tau}{(K+1)(K+2)}}\bigr), \end{aligned}$$

(101)

then $\delta_{2}=g'_{2}(0)=-\frac{2(K+1)(7K+10)N_{1}}{(K+2)(3K+4)}$.

Similarly, the comparison between the asymptotic and numerical solutions for different Reynolds numbers and expansion ratios is given in Tables 3 and 4. The axial velocity profiles and microrotation velocity profiles are presented in Figures 12, 13, and the results agree well.

Figure 12

The axial velocity as $\pmb{\mathit{Re}=100}$ , $\pmb{K=0.1}$ , $\pmb{\alpha=5}$ , $\pmb{N_{1}=0.1}$ .

Figure 13

The microrotation as $\pmb{\mathit{Re}=100}$ , $\pmb{K=0.1}$ , $\pmb{\alpha=5}$ , $\pmb{N_{1}=0.1}$ .

Table 3 The comparison between asymptotic and numerical solutions of $\pmb{-f''(1)}$ for large suction Reynolds number as $\pmb{K=0.1}$ , $\pmb{N_{1}=0.1}$

Table 4 The asymptotic and numerical values of $\pmb{g'(1)}$ for large suction Reynolds number at $\pmb{K=0.2}$ , $\pmb{N_{1}=0.1}$

5 Conclusions

The micropolar fluid through a deforming channel with porous walls is investigated numerically and analytically. In addition, the perturbation solutions are constructed for large injection and suction permeability Reynolds number. Some conclusions can be drawn:

• (i) The multiplicity of the solutions is influenced by the Reynolds number and the expansion ratio of the channel. New types of solutions are captured when the walls are expanding. Moreover, the non-existence of solutions for a specified range of Reynolds numbers with a particular positive expanding ratio is discovered.

• (ii) In order to eliminate the singularity of the solution for the case of large injection, the analytical solution is constructed by the Lighthill method.

• (iii) For the case of a large suction, the matched asymptotic expansion is used to obtain the analytical solution by the matching of the inner solution with the outer solution.

In this paper, we only construct two solutions for large injection or suction Reynolds number by singular perturbation methods. Some other multiple asymptotic solutions may be constructed to complement the above results in the future, which will help us to better understanding the stability of the micropolar flow through a deforming channel with porous walls.