Commun Nonlinear Sci Numer Simulat 14 (2009) 2848–2852

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Commun Nonlinear Sci Numer Simulat

journal homepage: www.elsevier .com/locate /cnsns

Short communication

A simple analytical solution for the steady flow of a third gradefluid in a porous half space

Faiz Ahmad *

Centre for Advanced Mathematics and Physics, National University of Science and Technology, EME Campus, Peshawar Road, Rawalpindi, Pakistan

a r t i c l e i n f o a b s t r a c t

Article history:Received 24 October 2007Received in revised form 23 September2008Accepted 24 September 2008Available online 17 October 2008

PACS:47.50 Cd47.54 Bd47.56+r

Keywords:Third grade fluidPorous half spaceAnalytical solutionAsymptotic solution

1007-5704/$ - see front matter � 2008 Elsevier B.Vdoi:10.1016/j.cnsns.2008.09.029

* Tel.: +92 51 9278050; fax: +92 51 5473989.E-mail address: faizmath@hotmail.com

We solve the governing equations for the flow of a third grade fluid in a porous half space.We find a simple expression which describes the solution accurately over the wholedomain ½0;1Þ. The rate of exponential decay of the flow is independent of the parameterswhich characterize the nonlinear part of a third grade fluid. Coefficients in a series expan-sion depend only on a single material constant.

� 2008 Elsevier B.V. All rights reserved.

1. Introduction

The flow of non-Newtonian fluids has several technical applications, especially in the paper and textile industries. Manynonlinear problems involving Newtonian as well as non-Newtonian fluids have recently been solved by using the homotopyanalysis method (HAM), see for example [1–7] and references therein. In this note we shall focus on Hayat et al. [1], whohave discussed the flow of a third grade fluid in a porous half space. Also of interest are recent papers of Hayat et al. [8]and Tan et al. [9]. For unidirectional flow, they have generalized the relation� �

ðrpÞx ¼ �luk

1þ a1

l@

@tu; ð1Þ

for a second grade fluid to the following modified Darcy’s Law for a third grade fluid

ðrpÞx ¼ �uk

luþ a1@u@tþ 2b3

@u@y

� �2

u

" #: ð2Þ

In the above equations u;l and p, respectively denote the fluid velocity, dynamic viscosity and the pressure, a1; b3 are mate-rial constants and k and u, respectively represent the permeability and porosity of the porous half space which occupies theregion y > 0. Defining non dimensional fluid velocity f and the coordinate z

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F. Ahmad / Commun Nonlinear Sci Numer Simulat 14 (2009) 2848–2852 2849

z ¼ V0

my; f ðzÞ ¼ u

V0;

where V0 ¼ uð0Þ and m ¼ l=q represents the kinematic viscosity, the boundary value problem modelling the steady stateflow of a third grade fluid in a porous half space becomes

d2f

dz2 þ b1dfdz

� �2 d2f

dz2 � b2fdfdz

� �2

� cf ¼ 0; ð3Þ

f ð0Þ ¼ 1; f ðzÞ ! 0 as z!1: ð4Þ

Parameters appearing in (3) are defined as follows:

b1 ¼6b3V4

0

lm2 ;

b2 ¼2b3uV2

0

kl;

c ¼ um2

kV20

:

Note that the parameters are not independent, since

b2 ¼b1c3: ð5Þ

The HAM solution of the above problem found in [1] is of the form

XM

m¼0

fmðzÞ ¼X2Mþ1

n¼1

e�nzX2mþ1�2n

m¼n�1

akm;nzk: ð6Þ

Since akm;n itself, is a series containing 2m� kþ 1 terms, the above solution, if written explicitly, will cover a large number of

pages even for M ¼ 17, the level of approximation used by Hayat et al. [1]. Since they did not take into account relation (5)among the parameters, most of the graphical results presented by them are redundant.

In this note we use an alternative approach to find an analytical solution of the problem. We determine the asymptoticform of the solution and utilize this information to develop a series solution. We find that an approximate expression con-taining only two terms describes the solution accurately over the whole domain ½0;1Þ. A comparison with the numericalsolution, for some typical values of the parameters, shows that the two solutions match with relative error less than 0.5%.More significantly, the approximate solution clearly demonstrates how various physical parameters play their part in deter-mining properties of the flow.

2. Approximate solution

Write Eq. (3) in the form

d2f

dz2 ¼b2

dfdz

� �2þ c

� �f

1þ b1dfdz

� �2 : ð7Þ

It is clear that since b2 < b1c, f 00ðzÞ < cf ðzÞ. On multiplying with f 0ðzÞ, which is less than zero on ½0;1Þ, we get f 0f 00 > cff 0. Anintegration from 0 to z gives

½f 0ðzÞ�2 > ½f 0ð0Þ�2 � c þ c½f ðzÞ�2: ð8Þ

If f 0ð0Þ < �ffiffifficp

then ½f 0ð0Þ�2 � c > 0 and the inequality (8) gives f 0ðzÞ < �ffiffifficp

f ðzÞ. Finally an integration from 0 to z givesf ðzÞ < expð�

ffiffifficp

zÞ on ½0;1Þ.For large z; f ðzÞ is small and changes slowly, otherwise it should change its sign. Therefore, for large z; j f 0ðz j� 1 and Eq.

(7) reduces to

d2f

dz2 � cf ¼ 0 ð9Þ

This indicates that asymptotically

f ðzÞ � ae�ffifficp

z ð10Þ

Define a new independent variable w ¼ e�ffifficp

z. Then

dfdz¼ �

ffiffifficp

wdfdw

;d2f

dw2 ¼ c w2 d2f

dw2 þwdfdw

!;

2850 F. Ahmad / Commun Nonlinear Sci Numer Simulat 14 (2009) 2848–2852

and the boundary value problem 3,4 transforms to

�f þwdfdwþw2 d2f

dw2 � b2fdfdw

� �2( )

þ b1cw3 dfdw

� �3

þ b1cw4 d2f

dw2

dfdw

� �2

¼ 0; f ð0Þ ¼ 0; f ð1Þ ¼ 1: ð11Þ

The original problem had an infinite domain while the transformed problem is defined on the interval [0,1]. This makes theproblem easily tractable. We assume a series solution

f ðwÞ ¼X1n¼0

anwn: ð12Þ

In view of the condition f ð0Þ ¼ 0, a0 ¼ 0. Substitute (12) in (11) and equate to zero coefficients of w;w2; . . . We obtain

3a2 ¼ 0;

8a3 � a31b2 þ a3

1b1c ¼ 0;

15a4 � 5a21a2b2 þ 8a2

1a2b1c ¼ 0;

24a5 � 7a21a3b2 þ 15a2

1a3b1c ¼ 0;

ð13Þ

35a6 � 4a32b2 � 22a1a2a3b2 � 9a2

1a4b2 þ 16a32b1c þ 72a1a2a3b1c þ 24a2

1a4b1c ¼ 0, and so on. Solving the above equations suc-cessively, we find

a2 ¼ 0; a3 ¼b2 � b1c

8a3

1; a4 ¼ 0; a5 ¼ðb2 � b1cÞð7b2 � 15b1cÞ

192a5

1; a6 ¼ 0; ð14Þ

where a1 is, as yet, arbitrary and is fixed by the second boundary condition f ð1Þ ¼ 1, which requires

a1 þb2 � b1c

8

� �a3

1 þðb2 � b1cÞð7b2 � 15b1cÞ

192a5

1 � 1 ¼ 0 ð15Þ

Once a1 has been found, as a root of Eq. (15), an approximate analytical solution of the problem, to the fifth order in w, whichbecomes, in the original coordinates

f ðzÞ ¼ a1e�ffifficp

z þ a3e�3ffifficp

z þ a5e�5ffifficp

z ð16Þ

It is clear that the above procedure can be carried to any desired order. However, due to the presence of exponentially decay-ing terms in (16) any additional term is not likely to significantly alter the accuracy of the approximate solution (16) withonly three terms.

3. Comparison with numerical solution

We check the accuracy of Eq. (16) by comparing it with a numerical solution for some typical values of parameters. Wetake b1 ¼ 0:6; b2 ¼ 0:1; c ¼ 0:5. Substituting these values in (15) and solving for a1 we find a1 ¼ 1:02229 and Eq. (16)becomes

f ðzÞ ¼ 1:02229e�0:70711z � 0:0267093e�2:12132z þ 0:0044196e�3:53553z: ð17Þ

For the numerical solution, we need to determine f 0ð0Þ. We accomplish this by the shooting method and find, correct to sixdecimal positions, f 0ð0Þ ¼ �0:678301. In comparison, the slope, at the origin, of the approximate solution (17) is �0.681835.

3.1. Pade approximation

Sometimes a Pade approximation is helpful in improving the accuracy of an approximate solution available in the form ofa polynomial. Also it may be judiciously employed to enlarge the interval of convergence of such a solution [10]. A [3,2] Padeapproximant for the solution (17) can be written as

1:02229wþ 0:14245w3

1þ 0:165471w2 ; ð18Þ

where w ¼ e�0:70711z. In Table 1, we compare the analytical solution (17) with the numerical solution as well as the valuesgiven by the Pade approximant (18). Entries in the lower half of the second column have been left blank because they merelyreplicate the values in the first column. If the results were to be rounded off to three positions after the decimal, then theapproximate solution is in complete agreement with the exact. The maximum error over the interval [0,5] is less thanone part in 900.

The solution is presented graphically in Fig. 1. Difference between the approximate solution and the exact (i.e. numerical)is so small as to be invisible on this scale.

Fig. 1. Variation of the fluid speed as a function of the dimensionless coordinate z. Approximate and exact solutions are indistinguishable on this scale.

Table 1Comparison of analytical and numerical results.

z Analytical, Eq. (17) Pade approximation (18) Numerical

0.0 1.0 0.9994 1.00.2 0.8722 0.8719 0.87260.4 0.7601 0.7600 0.76060.6 0.6619 0.6618 0.66240.8 0.5760 0.5760 0.57651.0 0.5010 0.5010 0.50141.2 0.4356 0.4356 0.43591.6 0.3289 0.3289 0.32922.0 0.2482 0.2482 0.24842.5 0.1744 0.17452.7 0.1514 0.15163.0 0.1225 0.12263.4 0.09234 0.092423.6 0.08016 0.080244.0 0.06042 0.060474.2 0.05245 0.052504.4 0.04553 0.045584.6 0.03953 0.039574.8 0.03432 0.034355.0 0.02979 0.02982

F. Ahmad / Commun Nonlinear Sci Numer Simulat 14 (2009) 2848–2852 2851

4. Conclusions

If we make use of the relation between the parameters then the coefficients in Eq. (11) will depend only on the singleparameter b2. This indicates that the solution depends on the kinematic viscosity only through the exponential term. Therate of decay is inversely proportional to V0 and directly proportional to the kinematic viscosity. The first term in the solutioncorresponds to the Newtonian flow and the second term includes the nonlinear effects inherent in the flow of a third gradefluid. If b2 is small, the second term represents a small correction to the Newtonian flow, otherwise we have to take intoaccount more terms of the series (12).

In contrast to perturbation series, accuracy of solution (16) increases as z becomes larger. It seems that a transformationof a boundary value problem, such as the one applied in this paper, can prove advantageous even if the problem is eventuallytackled by HAM or some other method.

Acknowledgment

The author is grateful to an anonymous referee for helpful suggestions.

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