1 Introduction

The expanding and contracting walls have received considerable attentions in recent years due to their wide applications,for example,in physiological pumps,peristaltic motion,problems involving collapsible tubes,and lymphatics^[1]. In general,there is the same permeability in biological tissues. However,in biological organism,the difference of ischemia may lead to different permeability. Goto and Uchida^[2] may be the first to examine the incompressible laminar flow in a semi-infinite porous pipe whose radius varies with time. The time dependent movement in a long porous expanding channel was obtained by Ma et al.^{[3, 4]} and Barron et al.^[5] experimentally. Subsequently,Majdalani et al.^{[6, 7]},Majdalani and Zhou^[7],and Dauenhauer andMajdalani^[8] regarded the Reynolds number as a small parameter to obtain the perturbation solutions for different Reynolds numbers and also discussed the same model numerically by using the shooting method coupled with the Runge-Kutta integration scheme.

Above works considered that there is the same permeability on the wall for the channel or the pipe. The asymmetric case may be tracked back to Terrill and Shrestha’s work^[9]. They investigated the flow through a porous channel with different permeability at both stationary walls and obtained the asymptotic solution for large Reynolds numbers. Muhammad et al.^[10] investigated the steady,laminar,and incompressible flow of a micropolar fluid through parallel and uniformly porous walls of different permeability by using the finite difference based on the numerical algorithm to solve the coupled pair of equations. Stephen^[11] also examined multiple solutions for the flow of Newtonian fluid in a parallel-walled channel with stationary porous walls. Si et al.^{[12, 13]} proposed the models of the asymmetric laminar flow for the Newtonian and micropolar fluid flow in a porous expanding channel and obtained the series solutions by the homotopy analysis method for some values of Reynolds numbers and expansion ratios, respectively. However,they did not consider the case of larger Reynolds numbers.

Motivated by the above mentioned work,in this paper,we shall consider the asymmetric flow of fluid in a two-dimensional porous channel with different permeability velocities −v₀ and −v₁ at the walls. We shall find that the expansion ratio has significant influence on the solutions. A technique of matching the outer and inner expansions is adopted to obtain a higher order asymptotic solution for the suction large Reynolds number case. We shall also consider the case of small expansion ratio and try to construct the asymptotic solutions either in the case of small Reynolds number or in the case of small asymmetric parameter. All asymmetric solutions will be verified by numerical solutions.

2 Formulation of problem

Consider the asymmetric laminar flow through a two-dimensional porous channel with ex- panding and contracting walls. The distance between the porous walls is much smaller than the length of the channel. Both walls have different permeability and expand or contract uniformly at a time-dependent rate ˙a(t). Let x and y be chosen as the co-ordinates and assume u and v to be the velocity components in the x- and y-directions,respectively.

The Navier-Stokes equations with the corresponding boundary conditions describing the flow through the expanding porous channel are

Introduce the stream function

with η = y/a .

Substituting (6) into (1)-(3) yields the following differential equation:

where α is the wall expansion ratio defined by The negative of α corresponds to the contraction of the two walls. <>The boundary conditions (4)-(5) then become Assume |v₁| > |v₀|. Let where R is the permeation Reynolds number,R is positive for injection and negative for suction, and α₂ is an asymmetric parameter.

A similar solution with respect to both space and time was developed by Uchida and Aoki^[14] by assuming that α is constant and f = f(η).

Under these assumptions,(7) and (9) become

An integration of (11) produces

where K is an integration constant. 3 Solution for large suction Reynolds number

For the large suction Reynolds number,(13) may be written as

where

Since later we shall look for a linear outer solution (as shown in (17)),the second- and third-order derivatives of f(η) in the region away from the boundary layer will be almost zero. That is why we can write the right-hand side of (14) as β² + 2εαβ. 3.1 Outer solution

In order to obtain the outer solution of (14) with the boundary conditions (12),we neglect the conditions f′(−1) = 0,f′(1) = 0 when there are two boundary layers at each wall. Since the large suction leads to the boundary layer at each wall,the outer solution would be of the form f(η) ~ η in physical meaning. Then,we suggest that the solution of (14) satisfying the conditions

is where 3.2 Inner solution

Since there is a boundary layer at each wall and there are two inner regions,two inner solutions will have to be obtained,respectively.

(i) Solution near the wall η = −1

Introduce an appropriate stretching variable

and a solution of the following form is sought:

Substituting (20) into (14) yields

From (19),the boundary conditions in (12) at the wall η = −1 become Equating coefficients of "i (i = 0,1,2) produce

The solutions of (23)-(25) subjected to (22) are

The second-order inner solution near η = −1 is

(ii) Solution near the wall η = 1

The stretching variable is chosen to be

and the form of the solution is assumed to be

Substituting (31) into (14) produces

The boundary conditions at the wall η = 1 become The process is similar to the case above. We can get The second-order inner solution near η = 1 is 3.3 Match of outer and inner solutions

The constants β_n(ε) and γ_n(ε) will be determined by matching the outer and inner solutions in a region where both f^o(η) and fⁱ(η) are valid.

The outer solution expressed in terms of the inner variable ~ is

The inner solution given by (29) is Keeping ε fixed as ξ and matching the inner solution (39) with the outer solution (38) give

Similarly,using the inner solution (37) valid near η = 1 and the outer solution given by (17),the functions f^o(η) is obtained by substituting η = λε + 1 into (17). Matching the outer and inner solutions,we can get

Using equations (40) and (41),β₀,β₁,β₂,β₃,γ₀, γ₁, γ₂,and γ₃ are given by 4 Solution for small expansion ratio

In the following section,we shall consider two cases of small expanding ratio for the equation (11) with the conditions (12). 4.1 Perturbation solution for small R

f(η) can be expanded in the form

where f_i(η) is independent of the parameter α. Substituting (43) into (11),the equation and relevant conditions for f₀ are Assume R is also small,it can be used as a secondary perturbation parameter. Then,f₀(η) can be expanded in the form Substituting (46) into (44) yields two sets of equations emerge at leading order in R The solution of (47) subjected to (48) is The first-order in R is The solution of (50) satisfying (51) is Combining f₀₀ and f₀₁,the leading order solution becomes The equation of f₁ and the corresponding boundary conditions are To obtain f₁,the above process is repeated. Assume R is small,f1 can be written as We can get Substituting (53) and (57) into (43),f becomes 4.2 Perturbation solution for small α₂

f(η) can be expanded in the form

where f_i(η) is independent of the parameter α₂. Substituting (59) into (11) yields the equation for f₁ Assume α is also small,it can be used as a secondary perturbation parameter. Then,f₁(η) can be written as Substituting (62) into (61) and equating equal powers of α,we can get where Combining f₁₀ and f₁₁,the leading order solution becomes We get

Remark 1 Similarly,we can consider a mixed injection case with small α and α₁. As- suming |v₀| > |v₁|,let

Equation (7) and the corresponding condition (9) become In a very similar process to the above small α₂ case,we can obtain 5 Comparison of numerical and analytical solutions

The accuracy of the analytical solutions for the suction case will be determined by comparing with the corresponding numerical solutions. 5.1 Case of large suction Reynolds number

For this case,f′′(−1) and f′′(1) can be,respectively,given by

5.2 Case of small expansion ratio α and small R

For this case,f′(−1) and f′(1) can be,respectively,given by

5.3 Case of small expansion ratio α and small α₂

For this case,f′′(−1) and f′′(1) can be,respectively,given by

and 6 Results and discussion

In Table 1,the results are for suction at both walls (R < 0,and 1 < α₂ 6 2),and the values of f′′(−1) and f′′(1) given by (72) and (73) are tabulated. A comparison between the numerical solutions and asymptotic solutions shows that decreasing values of α₂ diminish the accuracy of the solution,but altering the values of R have small effects on the accuracy of the solution. It can be seen that the analytical solution agrees reasonably well with the numerical one for large R.

In order to analysis the effect of α₂ on the velocity fields,Figs. 2 and 3 are illustrated for some fixed values of negative R and α. The maximum values of velocity increase with the increase of α₂. Furthermore,the profile becomes symmetric as α₂ increases. Figure 3 shows that the similar trend of the velocity is observed in comparison with Fig. 2.

In Fig. 4,the results for different α are shown as α₂ and R are fixed. The velocity is an increasing function of α,and the profile becomes more symmetric when the wall contracts. It indicates that the velocity is much greater sensitive to wall regression when the wall is expanding. The reversal flow occurs as α = −10 and R = −75.3,which shows that the expansion ratio has important influence on the flow velocity.

Figures 5 and 6 illustrate the effects of expansion ratio α and larger Reynolds number R on velocity fields. For fixed α₂ and α,when the suction velocity is imposed at the upper wall (i.e., R < 0) and the wall is contracting,the profile becomes more asymmetric. With the decrease of the magnitude of R,the reversal flow disappears gradually and the velocity becomes more symmetric. Figure 6 shows the influence of R on f′(η) as α = 10,the phenomenon of flow reversal is more obvious as the magnitude of R increases. It shows that the larger suction Reynolds numbers have important influence on the flow behavior.

In Table 2,the results are for the case of small α and R (with 0 < α₂ 6 2). The values of f′′(−1) and f′′(1) are given by (74) and (75),respectively. A comparison between the numerical solutions and asymptotic solutions is shown. They agree well.

The influence of the expansion ratio α on the velocity is illustrated for fixed values of R and α₂ in Figs. 7 and 8. The profile is symmetric and the maximum value of the velocity increases with the increase of α. Figures 9 and 10 represent the influence of suction Reynolds number R on lower or upper wall on the velocity fields,respectively. We can find that the profile becomes more asymmetric when the absolute value of R increases.

The effect of α₂ on f′(η) is given in Figs. 11-14 for fixed α and R. In Fig. 11,the maximum value of the suction velocity increases by increasing of α₂ when the wall is expanding and the figure is symmetric. Figures 12-14 show the similar trend in comparison with Fig. 11. It shows that α₂ has significant influence on the velocity.

In Table 3,the results are for the case of small α and α₂. The values of f′′(−1) and f′′(1) are given by (76) and (77). A comparison between the numerical solutions and asymptotic solutions is shown. It can be seen that the asymptotic solution agrees reasonably well with the numerical one for small α and α₂.

In Figs. 15 and 16,the influence of R on the velocity is illustrated for fixed values of positive α₂ and α. The maximum values of the velocity shift towards to the upper wall with the increase of R. Figure 16 shows the influence of R on the velocity for fixed values of α₂ and negative α. It shows the similar trend of the velocity in comparison with Fig. 16.

In Figs. 17 and 18,the influence of different α₂ with fixed α and R is shown. It shows that α₂ has significant influence on the velocity with the increase of α₂. Furthermore,there occurs a suction boundary layer on the upper wall.

The effect of α on the velocity is given in Fig. 19 for fixed α₂ and R. Because of the influence of large injection velocity,the velocity almost has no changes near the wall. In another word, the small expansion ratio almost has no influence on the velocity.

7 Conclusions

In this paper,the asymptotic solutions of asymmetric laminar flow through a two-dimensional porous channel are obtained and compared with the numerical ones. The Berman problem for the symmetric case is a special case corresponding to parameter α₂ = 2. Then,this work can be seen as an extension of the previous works. The results show that the Reynolds numbers, expansion ratio,and α₂ have much important influence on the velocity.