perturbation solutions for asymmetric laminar flow in porous channel with expanding and contracting...

Appl. Math. Mech. -Engl. Ed., 35(2), 203–220 (2014)DOI 10.1007/s10483-014-1784-8c©Shanghai University and Springer-Verlag

Berlin Heidelberg 2014

Applied Mathematicsand Mechanics(English Edition)

Perturbation solutions for asymmetric laminar flow in porous channelwith expanding and contracting walls∗

Yan ZHANG (� �)1, Ping LIN (� �)1,2, Xin-hui SI (��)1

(1. Department of Mathematics and Mechanics, University of Science and

Technology Beijing, Beijing 100083, P. R. China;

2. Department of Mathematics, University of Dundee, Dundee DD1 4HN, U. K.)

Abstract The cases of large Reynolds number and small expansion ratio for the asym-metric laminar flow through a two-dimensional porous expanding channel are considered.The Navier-Stokes equations are reduced to a nonlinear fourth-order ordinary differentialequation by introducing a time and space similar transformation. A singular perturbationmethod is used for the large suction Reynolds case to obtain an asymptotic solution bymatching outer and inner solutions. For the case of small expansion ratios, we are ableto obtain asymptotic solutions by double parameter expansion in either a small Reynoldsnumber or a small asymmetric parameter. The asymptotic solutions indicate that theReynolds number and expansion ratio play an important role in the flow behavior. Nu-merical methods are also designed to confirm the correctness of the present asymptoticsolutions.

Key words singular perturbation method, regular perturbation method, porousexpanding channel, expansion ratio

Chinese Library Classification O175.8, O357.32010 Mathematics Subject Classification 76M45, 76D10

1 Introduction

The expanding and contracting walls have received considerable attentions in recent yearsdue to their wide applications, for example, in physiological pumps, peristaltic motion, problemsinvolving collapsible tubes, and lymphatics[1]. In general, there is the same permeability inbiological tissues. However, in biological organism, the difference of ischemia may lead todifferent permeability. Goto and Uchida[2] may be the first to examine the incompressiblelaminar flow in a semi-infinite porous pipe whose radius varies with time. The time dependentmovement 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 Dauenhauerand Majdalani[8] regarded the Reynolds number as a small parameter to obtain the perturbationsolutions for different Reynolds numbers and also discussed the same model numerically by using

∗ Received Dec. 17, 2012 / Revised Apr. 24, 2013Project supported by the Beijing Higher Education Young Elite Teacher Project (No.YETP0387),the Fundamental Research Funds for the Central Universities (Nos. FRF-TP-12-108A and FRF-BR-13-023), and the National Natural Science Foundation of China (Nos. 51174028 and 11302024)Corresponding author Xin-hui SI, E-mail: sixinhui [email protected]

204 Yan ZHANG, Ping LIN, and Xin-hui SI

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]. Theyinvestigated the flow through a porous channel with different permeability at both stationarywalls 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 paralleland uniformly porous walls of different permeability by using the finite difference based on thenumerical algorithm to solve the coupled pair of equations. Stephen[11] also examined multiplesolutions for the flow of Newtonian fluid in a parallel-walled channel with stationary porouswalls. Si et al.[12–13] proposed the models of the asymmetric laminar flow for the Newtonianand micropolar fluid flow in a porous expanding channel and obtained the series solutions bythe 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 asymmetricflow of fluid in a two-dimensional porous channel with different permeability velocities −v0and −v1 at the walls. We shall find that the expansion ratio has significant influence on thesolutions. A technique of matching the outer and inner expansions is adopted to obtain a higherorder asymptotic solution for the suction large Reynolds number case. We shall also considerthe case of small expansion ratio and try to construct the asymptotic solutions either in thecase of small Reynolds number or in the case of small asymmetric parameter. All asymmetricsolutions 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 thelength of the channel. Both walls have different permeability and expand or contract uniformlyat a time-dependent rate a(t). Let x and y be chosen as the co-ordinates and assume u and vto be the velocity components in the x- and y-directions, respectively.

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

Fig. 1 Model for porous expanding channel with different permeability

Perturbation solutions for asymmetric laminar flow in porous channel 205

∂u

∂x+∂v

∂y= 0, (1)

∂u

∂t+ u

∂u

∂x+ v

∂u

∂y= −1

ρ

∂p

∂x+ υ

(∂2u

∂x2+∂2u

∂y2

), (2)

∂v

∂t+ u

∂v

∂x+ v

∂v

∂y= −1

ρ

∂p

∂y+ υ

(∂2v

∂x2+∂2v

∂y2

), (3)

u = 0, v = −v0 = −A0a, y = −a(t), (4)

u = 0, v = −v1 = −A1a, y = a(t), (5)

where A0 = v0a and A1 = v1

a represent the measures of the lower and upper wall permeability,respectively.

Introduce the stream function

ψ =υx

aF (η, t) (6)

with η = ya .

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

Fηηηη + α(ηFηηη + 3Fηη) + FFηηη − FηFηη − a2υ−1Fηηt = 0, (7)

where α is the wall expansion ratio defined by

α =aa

υ. (8)

The negative of α corresponds to the contraction of the two walls.The boundary conditions (4)–(5) then become

F (−1) =av0υ, Fη(−1) = 0, F (1) =

av1υ, Fη(1) = 0. (9)

Assume |v1| � |v0|. Let

R =av1υ, f =

F

R, α2 = 1 − v0

v1, l =

x

a, (10)

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

f IV + α(ηf ′′′ + 3f ′′) +R(ff ′′′ − f ′f ′′) = 0, (11)

f(−1) = 1 − α2, f ′(−1) = 0, f(1) = 1, f ′(1) = 0. (12)

An integration of (11) produces

f ′′′ + α(ηf ′′ + 2f ′) +R(ff ′′ − f ′2) = K, (13)

where K is an integration constant.


3 Solution for large suction Reynolds number

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

εf ′′′ + αε(ηf ′′ + 2f ′) + f ′2 − ff ′′ = β2 + 2εαβ, (14)

where

ε = − 1R, −K

R= β2 + 2εαβ. (15)

Since later we shall look for a linear outer solution (as shown in (17)), the second- andthird-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 neglectthe conditions f ′(−1) = 0, f ′(1) = 0 when there are two boundary layers at each wall. Sincethe large suction leads to the boundary layer at each wall, the outer solution would be of theform f(η) ∼ η in physical meaning. Then, we suggest that the solution of (14) satisfying theconditions

f(−1) = 1 − α2, f(1) = 1 (16)

is

fo(η) = β(ε)η + γ(ε), (17)

where

β(ε) =∞∑

n=0

εnβn, γ(ε) =∞∑

n=0

εnγn. (18)

3.2 Inner solutionSince 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 = −1Introduce an appropriate stretching variable

ξ = ε−1(η + 1), (19)

and a solution of the following form is sought:

f i(η) = a+ εω(ξ, ε), (20)

where a = 1 − α2, ω =∞∑

n=0εnωn(ξ).

Substituting (20) into (14) yields

∞∑n=0

εnω′′′n (ξ) + αε

((ξε− 1)

∞∑n=0

εnω′′n(ξ) + 2ε

∞∑n=0

εnω′n(ξ)

)

+ ε( ∞∑

n=0

εnω′n(ξ)

)2

−(a+ ε

∞∑n=0

εnωn(ξ))( ∞∑

n=0

εnω′′n(ξ)

)

= ε( ∞∑

n=0

εnβn

)2

+ 2ε2α∞∑

n=0

εnβn. (21)


From (19), the boundary conditions in (12) at the wall η = −1 become

ωn(0) = ω′n(0) = 0, ω′

n(∞) = βn. (22)

Equating coefficients of εi (i = 0, 1, 2) produce

ω′′′0 − aω′′

0 = 0, (23)

ω′′′1 − aω′′

1 = αω′′0 − ω′2

0 + ω0ω′′0 + β2

0 , (24)

ω′′′2 − aω′′

2 = ω0ω′′1 + ω1ω

′′0 − 2ω′

0ω′1 − αξω′′

0 + αω′′1 − 2αω′

0 + 2β0β1 + 2αβ0. (25)

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

ω0 =β0

a(1 + aξ − eaξ), (26)

ω1 =β1

a(1 + aξ) +

3β20

a3− αβ0

a2+

(β20

a3

(− 1

2a2ξ2 + 3aξ − 3

)− β1

a+αβ0

a2(1 − aξ)

)eaξ, (27)

ω2 =β2

a(1 + aξ) +

(89β30

4a5+

6β0β1

a3

)− 9αβ2

0

a4+α2β0

a3− αβ1

a2

+(β3

0

a5

(− 1

8a4ξ4 +

32a3ξ3 − 8a2ξ2 + 22aξ − 45

2

)+

2β0β1

a3

(− a2ξ2

2+ 3aξ − 3

)

− β2

a+αβ2

0

a4

(− 1

2a3ξ3 +

72a2ξ2 − 9aξ + 9

)+α2β0

a3

(− 1

2a2ξ2 + aξ − 1

)

+αβ1

a2(1 − aξ) +

αβ0a2ξ2

2a3+

)eaξ +

β30e2aξ

4a5. (28)

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

f i(η) = a+ ε(ω0 + εω1 + ε2ω2). (29)

(ii) Solution near the wall η = 1The stretching variable is chosen to be

λ = ε−1(η − 1), (30)

and the form of the solution is assumed to be

f(η) = 1 + εϕ(λ, ε), (31)

where ϕ =∞∑

n=0εnϕn(λ).

Substituting (31) into (14) produces

ϕ′′′(λ) + αε((λε+ 1)ϕ′′ + 2εϕ′) + εϕ′2 − (1 + εϕ)ϕ′′ = εβ2 + 2ε2αβ. (32)

The boundary conditions at the wall η = 1 become

ϕ(0) = 0, ϕ′(0) = 0, ϕ′(∞) = β(ε). (33)

The process is similar to the case above. We can get

ϕ0 = β0(1 + λ− eλ), (34)

ϕ1 = β1(1 + λ) + 3β20 + αβ0 +

(β2

0

(− 1

2λ2 + 3λ− 3

)− β1 + αβ0(λ− 1)

)eλ, (35)


ϕ2 =β2(1 + λ) +894β3

0 + 6β0β1 + 3αβ20 + αβ1 + α2β0

+(β3

0

(− 1

8λ4 +

32λ3 − 8λ2 + 22λ− 45

2

)+ 2β0β1

(− λ2

2+ 3λ− 3

)− β2

+ αβ20

(16λ3 − 3

2λ2 + 3λ− 3

)+ α2β0

(− 1

2λ2 + λ− 1

)+ αβ1(λ− 1)

+αβ0λ

2

2

)eλ +

β30e2λ

4. (36)

The second-order inner solution near η = 1 is

f i(η) = 1 + ε(ϕ0 + εϕ1 + ε2ϕ2). (37)

3.3 Match of outer and inner solutionsThe constants βn(ε) and γn(ε) will be determined by matching the outer and inner solutions

in a region where both fo(η) and f i(η) are valid.The outer solution expressed in terms of the inner variable ξ is

fo(η) =β(ε)η + γ(ε)

= (β0 + εβ1 + ε2β2 + ε3β3 + · · · )(εξ − 1)

+ γ0 + εγ1 + ε2γ2 + ε3γ3 + · · · . (38)

The inner solution given by (29) is

f i(η) = a+ ε(ω0 + εω1 + ε2ω2)

= a+ ε(β0

a(1 + aξ − eaξ) + ε

β1

a(1 + aξ) + ε

3β20

a3− ε

αβ0

a2

+ ε(β2

0

a3

(− 1

2a2ξ2 + 3aξ − 3

)− β1

a+αβ0

a2(1 − aξ)

)eaξ

+ ε2β2

a(1 + aξ) + ε2

(89β30

4a5+

6β0β1

a3

)− ε2

9αβ20

a4+ ε2

α2β0

a3− ε2

αβ1

a2

+ ε2(β3

0

a5

(− 1

8a4ξ4 +

32a3ξ3 − 8a2ξ2 + 22aξ − 45

2

)+

2β0β1

a3

(− a2ξ2

2+ 3aξ − 3

)

− β2

a+αβ2

0

a4

(− 1

2a3ξ3 +

72a2ξ2 − 9aξ + 9

)+α2β0

a3

(− 1

2a2ξ2 + aξ − 1

)

+αβ1

a2(1 − aξ) +

αβ0a2ξ2

2a3

)eaξ + ε2

β30e2aξ

4a5

). (39)

Keeping ε fixed as ξ → ∞ and matching the inner solution (39) with the outer solution (38)give

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

− β0 + γ0 = a,

− β1 + γ1 =β0

a,

− β2 + γ2 =β1

a+

3β20

a3− αβ0

a2,

− β3 + γ3 =β2

a+

89β30

4a5+

6β0β1

a3− 9αβ2

0

a4+α2β0

a3− αβ1

a2.

(40)

Similarly, using the inner solution (37) valid near η = 1 and the outer solution given by(17), the functions fo(η) is obtained by substituting η = λε+ 1 into (17). Matching the outer


and inner solutions, we can get⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

β0 + γ0 = 1,β1 + γ1 = β0,

β2 + γ2 = β1 + 3β20 + αβ0,

β3 + γ3 = β2 +89β3

0

4+ 6β0β1 + 3αβ2

0 + αβ1 + α2β0.

(41)

Using equations (40) and (41), β0, β1, β2, β3, γ0, γ1, γ2, and γ3 are given by

β0 =1 − a

2,

β1 = − (a− 1)2

4a,

β2 =(a− 1)(−4a3 + 2a2 + 3a4 − 2αa3 − 4a+ 3 − 2αa)

8a3,

β3 = − 164a5

((a− 1)(214a− 165a2 + 48a3 − 48a4 + 16αa3 + 72αa+ 165a5 + 40αa5

− 16αa4 + 16α2a5 − 16α2a2 − 24αa6 − 88αa2 − 89 − 214a6 + 89a7)),

γ0 =1 + a

2,

γ1 =1 − a2

4a,

γ2 =(a− 1)(−4a3 + 3a4 − 2αa3 + 4a− 3 + 2αa)

8a3,

γ3 = − 164a5

((a− 1)(−214a+ 141a2 − 16a3 − 16a4 − 72αa+ 141a5 + 40αa5

+ 16α2a5 + 16α2a2 − 24αa6 + 88αa2 + 89 − 214a6 + 89a7)). (42)

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

f(η) = f0 + αf1(η) +O(α2), (43)

where fi(η) is independent of the parameter α. Substituting (43) into (11), the equation andrelevant conditions for f0 are

f IV0 +Rf0f

′′′0 −Rf ′

0f′′0 = 0, (44)

f0(−1) = 1 − α2, f ′0(−1) = 0, f0(1) = 1, f ′

0(1) = 0. (45)

Assume R is also small, it can be used as a secondary perturbation parameter. Then, f0(η) canbe expanded in the form

f0 = f00 +Rf01 +O(R2). (46)


Substituting (46) into (44) yields two sets of equations emerge at leading order in R

f IV00 = 0, (47)

f00(−1) = 1 − α2, f ′00(−1) = 0, f00(1) = 1, f ′

00(1) = 0. (48)

The solution of (47) subjected to (48) is

f00 = −14α2η

3 + 1 +34α2η − 1

2α2. (49)

The first-order in R is

f IV01 = f ′

00f′′00 − f00f

′′′00 =

34α2

2η3 +

32α2 − 3

4α2

2, (50)

f01(−1) = 0, f ′01(−1) = 0, f01(1) = 0, f ′

01(1) = 0. (51)

The solution of (50) satisfying (51) is

f01 =1

1 120α2

2η7 +

116η4α2 − 1

32α2

2η4 − 3

1 120α2

2η3 +

12

(18α2

2 −14α2

)η2

+1

560α2

2η −132α2

2 +116α2. (52)

Combining f00 and f01, the leading order solution becomes

f0 = f00 +Rf01

= − 14α2η

3 + 1 +34α2η − 1

2α2

+R( 1

1 120α2

2η7 +

116η4α2 − 1

32α2

2η4 − 3

1 120α2

2η3

+12

(18α2

2 −14α2

)η2 +

1560

α22η −

132α2

2 +116α2

). (53)

The equation of f1 and the corresponding boundary conditions are

f IV1 + ηf ′′′

0 + 3f ′′0 +Rf0f

′′′1 +Rf1f

′′′0 −Rf ′

0f′′1 −Rf ′

1f′′0 = 0, (54)

f1(−1) = 0, f ′1(−1) = 0, f1(1) = 0, f ′

1(1) = 0. (55)

To obtain f1, the above process is repeated. Assume R is small, f1 can be written as

f1 = f10 +Rf11 +O(R2). (56)

We can get

f10 = − 110 080

Rα22η

9 +1

192Rα2

2η6 − 1

96Rα2η

6 +120α2η

5 +3

5 600Rα2

2η5

− 164Rα2

2η4 +

132Rα2η

4 − 1725 200

Rα22η

3 − 110α2η

3 +164Rα2

2η2 − 1

32Rα2η

2

+1

4 200Rα2

2η +120α2η − 1

192Rα2

2 +196Rα2. (57)


Substituting (53) and (57) into (43), f becomes

f(η) = − 14α2η

3 + 1 +34α2η − 1

2α2 +R

( 11 120

α22η

7 +116η4α2 − 1

32α2

2η4 − 3

1 120α2

2η3

+12

(18α2

2 −14α2

)η2 +

1560

α22η −

132α2

2 +116α2

)+ α

(− 1

10 080Rα2

2η9 +

1192

Rα22η

6

− 196Rα2η

6 +120α2η

5 +3

5 600Rα2

2η5 − 1

64Rα2

2η4 +

132Rα2η

4 − 1725 200

Rα22η

3

− 110α2η

3 +164Rα2

2η2 − 1

32Rα2η

2 +1

4 200Rα2

2η +120α2η − 1

192Rα2

2 +196Rα2

). (58)

4.2 Perturbation solution for small α2

f(η) can be expanded in the form

f(η) = 1 + α2f1(η) +O(α22), (59)

where fi(η) is independent of the parameter α2. Substituting (59) into (11) yields the equationfor f1

f IV1 + αηf ′′′

1 + 3αf ′′1 +Rf ′′′

1 = 0, (60)

f1(−1) = −1, f ′1(−1) = 0, f1(1) = 0, f ′

1(1) = 0. (61)

Assume α is also small, it can be used as a secondary perturbation parameter. Then, f1(η) canbe written as

f1 = f10 + αf11 +O(α2). (62)

Substituting (62) into (61) and equating equal powers of α, we can get

f10 =−ReR − 3Re−R − 4e−R + 2RηeR + 2Re−R +Rη2e−R −Rη2eR + 4e−Rη

4D1, (63)

where

D1 =ReR +Re−R + e−R − eR, (64)

f11 = − 14R2D2

1

(6R3 + 10ηR2 − 6η2R3 + η2R2e2R + 2e−Rηη2R2D1 + e−Rη3R2D1

− eRη3R2D1 − 12e−RηD1 + ηR3e2R − 5ηR2e−2R − ηR3e−2R − 5ηR2e2R

− η2R2e−2R + η2R3e2R + η2R3e−2R + 6e−Rη−RR2 − 2e−Rη+RR3

+ 12e−Rη−RR− 2e−Rη−RR3 − 12e−Rη+R + 12e−Rη−R − 6e−Rη+RR2

+ 12e−Rη+RR−R3e2R + 3R2e2R −R3e−2R − 3R2e−2R). (65)


Combining f10 and f11, the leading order solution becomes

f1 =−ReR − 3Re−R − 4e−R + 2RηeR + 2Rηe−R +Rη2e−R −Rη2eR + 4e−Rη

4D1

+ α(− 1

4R2D21

(6R3 + 10ηR2 − 6η2R3 + η2R2e2R + 2e−Rηη2R2D1 + e−Rη3R2D1

− eRη3R2D1 − 12e−RηD1 + ηR3e2R − 5ηR2e−2R − ηR3e−2R − 5ηR2e2R − η2R2e−2R

+ η2R3e2R + η2R3e−2R + 6e−Rη−RR2 − 2e−Rη+RR3 + 12e−Rη−RR− 2e−Rη−RR3

− 12e−Rη+R + 12e−Rη−R − 6e−Rη+RR2 + 12e−Rη+RR−R3e2R

+ 3R2e2R −R3e−2R − 3R2e−2R)). (66)

We get

f(η) = 1 + α2f1(η)

= 1 + α2

(−ReR − 3Re−R − 4e−R + 2RηeR + 2Rηe−R +Rη2e−R −Rη2eR + 4e−Rη

4D1

+ α(− 1

4R2D21

(6R3 + 10ηR2 − 6η2R3 + η2R2e2R + 2e−Rηη2R2D1

+ e−Rη3R2D1 − eRη3R2D1 − 12e−RηD1 + ηR3e2R − 5ηR2e−2R

− ηR3e−2R − 5ηR2e2R − η2R2e−2R + η2R3e2R + η2R3e−2R

+ 6e−Rη−RR2 − 2e−Rη+RR3 + 12e−Rη−RR− 2e−Rη−RR3 − 12e−Rη+R

+ 12e−Rη−R − 6e−Rη+RR2 + 12e−Rη+RR−R3e2R + 3R2e2R −R3e−2R

− 3R2e−2R))). (67)

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

R1 =av0υ, f =

F

R1, α1 =

v1v0

− 1, l =x

a. (68)

Equation (7) and the corresponding condition (9) become

f IV + α(ηf ′′′ + 3f ′′) +R1(f ′′′ − f ′f ′′) = 0, (69)

f(−1) = 1, f ′(−1) = 0, f(1) = 1 + α1, f ′(1) = 0. (70)


In a very similar process to the above small α2 case, we can obtain

f(η) =1 + α1f1(η)

=1 + α1

·(3R1eR1 +R1e−R1 − 4eR1 + 2R1ηeR1 + 2R1ηe−R1 −R1η

2eR1 +R1η2e−R1 + 4e−R1η

4E1

+ α(− 1

4E21

(6R1 − 2η3 + 10η − 6R1η2 − 2e−R1η−R1R1 − 2e−R1η+R1R1

+ 2e−R1η−R1η2 − 2e−R1η+R1η2 − e2R1η3R1 − R1e2R1 −R1e−2R1 + 3e2R1

− 3e−2R1 +R1η2e2R1 + η2R1e−2R1 +R1ηe2R1 + e−2R1η3R1 + 6e−R1η−R1

− 6e−R1η+R1 − 5ηe2R1 + e2R1η3 + η2e2R1 + e−2R1η3 − η2e−2R1 − 5ηe−2R1

− ηR1e−2R1 + 2e−R1η+R1R1η2 + 2e−R1η−R1R1η

2))), (71)

where E1 = R1eR1 +R1e−R1 + e−R1 − eR1 .

5 Comparison of numerical and analytical solutions

The accuracy of the analytical solutions for the suction case will be determined by comparingwith 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

f ′′(−1) = β0aR+2β2

0

a− β1a− αβ0

−(13β3

0

2a5+

4β0β1

a− β2a− 2αβ2

0

a2+αβ0

a− αβ1

) 1R

(72)

and

f ′′(1) =β0R + 2β20 − β1 + αβ0

−(13β3

0

2+ 4β0β1 − β2 + αβ1 + αβ0

) 1R. (73)

5.2 Case of small expansion ratio α and small RFor this case, f ′(−1) and f ′(1) can be, respectively, given by

f ′′(−1) =32α2 − 19

70Rα2

2 +12Rα2 +

12 100

Rαα22 −

25αα2 (74)

and

f ′′(1) = −32α2 − 8

35Rα2

2 +12Rα2 − 1

2 100Rαα2

2 +25αα2. (75)

5.3 Case of small expansion ratio α and small α2

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

f ′′(−1) =α2

2D21

(RD1e−R −ReRD1 + 2eRR2D1 − αeRR2D1 − 4αReRD1

+ αeRD1 + 3αe−RD1 + αe−2R + 5αe2R − αRe−2R − 7αRe2R

− 6α− 3αR2 + αR3e2R + αR3 + 3αR2e2R) (76)


and

f ′′(1) = − α2

2D21

(−RD1e−R +ReRD1 − 2e−RR2D1 + αe−RR2D1 − 4αRD1e−R

− αe−RD1 − 3αeRD1 + 5αe−2R + αe2R + 7αRe−2R + αRe2R − 6α

+ 3αR2e−2R − αR3 − αR3e−2R − 3αR2). (77)

6 Results and discussion

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

Table 1 Comparisons of f ′′(−1) and f ′′(1) for large suction Reynolds number

α2 R α

f ′′(−1) f ′′(1)

Numerical Perturbation Numerical Perturbation

solutions solutions solutions solutions

1.943 –72.826 1 64.519 795 37 64.531 759 85 –68.758 114 73 –68.800 819 81

1.916 –77.476 1 65.777 176 12 65.782 375 32 –72.312 855 07 –72.349 752 86

1.931 –77.387 1 67.364 938 39 67.372 578 36 –72.766 741 75 –72.804 879 45

1.905 –79.968 1 66.719 636 90 66.721 393 52 –74.294 088 66 –74.328 947 65

1.900 –72.000 –1 61.327 753 98 61.316 694 32 –68.507 555 29 –68.489 126 37

1.930 –77.300 –1 69.200 307 23 69.196 009 56 –74.656 558 62 –74.638 757 67

1.916 –75.300 –1 65.878 878 46 65.871 175 66 –72.221 577 53 –72.202 985 38

1.400 –120.000 –3 33.556 284 57 32.104 172 00 –86.255 752 12 –86.402 217 71

1.600 –130.000 –4 64.125 319 06 63.920 546 00 –107.038 295 7 –106.998 263 3

1.500 –110.000 4 35.916 001 74 35.520 596 59 –79.594 935 77 –79.528 764 20

In order to analysis the effect of α2 on the velocity fields, Figs. 2 and 3 are illustrated forsome fixed values of negative R and α. The maximum values of velocity increase with the

Fig. 2 Variation of f ′(η) for different α2 asR = −75.3 and α = −1

Fig. 3 Variation of f ′(η) for different α2 asR = −75.3 and α = 4


increase of α2. Furthermore, the profile becomes symmetric as α2 increases. Figure 3 showsthat 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 anincreasing 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 isexpanding. The reversal flow occurs as α = −10 and R = −75.3, which shows that theexpansion ratio has important influence on the flow velocity.

Fig. 4 Variation of f ′(η) for different α as R = −75.3 and α2 = 1.2

Figures 5 and 6 illustrate the effects of expansion ratio α and larger Reynolds number R onvelocity 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 decreaseof the magnitude of R, the reversal flow disappears gradually and the velocity becomes moresymmetric. Figure 6 shows the influence of R on f ′(η) as α = 10, the phenomenon of flowreversal is more obvious as the magnitude of R increases. It shows that the larger suctionReynolds numbers have important influence on the flow behavior.

Fig. 5 Variation of f ′(η) for different R asα2 = 1.2 and α = −10

Fig. 6 Variation of f ′(η) for different R as α2 =1.2 and α = 10

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


Table 2 Comparisons of f ′′(−1) and f ′′(1) for small expansion ratio and small R

α R α2

f ′′(−1) f ′′(1)



0.01 0.02 0.4 0.601 540 237 0.601 531 444 –0.595 146 100 –0.595 131 4440.02 0.03 1.6 2.390 393 011 2.390 355 017 –2.380 846 124 –2.380 755 0170.06 0.04 0.9 1.337 707 703 1.337 606 640 –1.317 990 644 –1.317 806 640

–0.03 –0.06 0.8 1.196 093 806 1.196 023 406 –1.225 051 289 –1.224 823 4060.03 0.06 0.7 1.054 696 811 1.054 620 420 –1.027 528 914 –1.027 320 4200.03 –0.05 0.6 0.882 713 982 0.882 685 457 –0.903 786 828 –0.903 685 4570.07 –0.09 1.1 1.599 023 315 1.599 254 941 –1.644 088 629 –1.643 804 941

–0.11 0.12 1.3 2.029 438 920 2.030 143 663 –1.975 786 227 –1.975 543 6630.12 –0.14 1.5 2.157 170 248 2.158 482 000 –2.210 994 608 –2.210 982 000

–0.17 –0.13 1.6 2.497 117 203 2.495 158 370 –2.539 668 524 –2.536 758 3700.24 –0.18 1.9 2.669 172 864 2.672 900 023 –2.687 006 724 –2.690 000 0230.28 0.23 2.0 2.766 550 030 2.756 408 381 –2.766 550 029 –2.756 408 381

The influence of the expansion ratio α on the velocity is illustrated for fixed values of R andα2 in Figs. 7 and 8. The profile is symmetric and the maximum value of the velocity increaseswith the increase of α. Figures 9 and 10 represent the influence of suction Reynolds number R

Fig. 7 Variation of f ′(η) for different α asR = 0.03 and α2 = 0.8

Fig. 8 Variation of f ′(η) for different α as R =−0.03 and α2 = 0.8

Fig. 9 Variation of f ′(η) for different R asα = 0.06 and α2 = 0.6

Fig. 10 Variation of f ′(η) for different R as α =−0.08 and α2 = 1.1


on lower or upper wall on the velocity fields, respectively. We can find that the profile becomesmore 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 maximumvalue of the suction velocity increases by increasing of α2 when the wall is expanding and thefigure is symmetric. Figures 12–14 show the similar trend in comparison with Fig. 11. It showsthat α2 has significant influence on the velocity.

Fig. 11 Variation of f ′(η) for different α2 asα = −0.03 and R = −0.08

Fig. 12 Variation of f ′(η) for different α2 asα = 0.03 and R = 0.08

Fig. 13 Variation of f ′(η) for different α2 asα = −0.03 and R = 0.08

Fig. 14 Variation of f ′(η) for different α2 asα = 0.03 and R = −0.08

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 asymptoticsolutions is shown. It can be seen that the asymptotic solution agrees reasonably well with thenumerical one for small α and α2.

In Figs. 15 and 16, the influence of R on the velocity is illustrated for fixed values of positiveα2 and α. The maximum values of the velocity shift towards to the upper wall with the increaseof 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.


Table 3 Comparisons of f ′′(−1) and f ′′(1) for small expansion ratio and small α2

α2 R α

f ′′(−1) f ′′(1)



0.040 0 –13.88 0.010 0.021 367 5 0.021 524 2 –0.577 880 –0.576 724

0.007 8 –43.76 0.004 0.003 985 0 0.003 990 5 –0.345 660 –0.345 319

0.173 0 –64.78 0.035 0.084 750 0 0.087 762 8 –11.617 301 –11.294 703

0.193 2 –32.46 0.120 0.094 867 8 0.098 957 1 –6.546 507 –6.370 229

0.213 2 –57.83 0.150 0.103 065 5 0.107 922 9 –12.891 087 –12.437 279

0.031 2 –63.74 –0.080 0.015 802 1 0.015 887 8 –2.013 418 –2.004 576

0.092 3 –13.76 –0.094 0.049 577 2 0.050 394 8 –1.326 489 –1.320 443

0.126 4 –67.87 –0.110 0.062 839 0 0.064 349 9 –8.814 086 –8.643 118

0.183 7 –79.34 –0.124 0.089 973 8 0.093 309 5 –15.126 285 –14.668 068

0.212 7 –87.39 –0.140 0.103 349 2 0.107 921 7 –19.401 544 –18.695 774

Fig. 15 Variation of f ′(η) for different R asα2 = 0.06 and α = 0.04

Fig. 16 Variation of f ′(η) for different R asα2 = 0.06 and α = −0.04

In Figs. 17 and 18, the influence of different α2 with fixed α and R is shown. It shows thatα2 has significant influence on the velocity with the increase of α2. Furthermore, there occurs

Fig. 17 Variation of f ′(η) for different α2 asR = −50 and α = −0.03

Fig. 18 Variation of f ′(η) for different α2 asR = −50 and α = 0.03


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.

Fig. 19 Variation of f ′(η) for different α as R = −50 and α2 = 0.06

7 Conclusions

In this paper, the asymptotic solutions of asymmetric laminar flow through a two-dimensionalporous channel are obtained and compared with the numerical ones. The Berman problem forthe symmetric case is a special case corresponding to parameter α2 = 2. Then, this work canbe 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.

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