# Homotopy Analysis Solution for Magnetohydrodynamic Squeezing Flow in Porous Medium

Authors: Inayat Ullah; M. T. Rahim; Hamid Khan; Mubashir Qayyum
Journal: Advances in Mathematical Physics
 (2016)

---

## Abstract

The aim of the present work is to analyze the magnetohydrodynamic (MHD) squeezing flow through porous medium using homotopy analysis method (HAM). Fourth-order boundary value problem is modeled through stream functionψ
(
r
,
z
) and transformation ψ
(
r
,
z
)
=
r
2
f
(
z
). Absolute residuals are used to check the efficiency and consistency of HAM. Other analytical techniques are compared with the present work. It is shown that results of good agreement can be obtained by choosing a suitable value of convergence control parameter h in the valid region R
h. The influence of different parameters on the flow is argued theoretically as well as graphically.


---

## Body

## 1. Introduction

The squeezing movement normal to two plates is observed in many hydromechanical devices such as motors, engines, and hydraulic lifters, where compression/injection processes using pistons and clutches are found. Due to the utility of these devices, significant research effort is being made for their improvement. Other industrial applications include polymer processing, while medical applications include the modeling of synthetics transportation inside living bodies. As such, the study of squeezing effect, in addition to other properties such as magnetohydrodynamics (MHD) and porosity, has become one of the most active topics in fluid mechanics. The study of porosity effects after introduction of the modified Darcy Law [1] specifically contributed to oil and gas production industry, detection of ground water pollution due to leakage of chemicals from tanks and oil pipelines, ground water hydrology, and recovery of crude oil from pores of reservoir rocks [2–5]. These contributions and others [6, 7] are generally found in reservoir, chemical, civil, environmental, agricultural, and biomedical engineering.These studies are typically modeled using small parameters in nonlinear differential equations which are expressed as series expansions. The exact solution using perturbation methods is therefore not always possible and this poses a considerable challenge to researchers. A recently developed analytic method known as homotopy analysis method (HAM) by Liao in 1992, however, has given promising results as it does not require modeling of the small parameter [8]. In fact, in quite contrast, HAM provides a way to accelerate the series solution conversion in the form of auxiliary parameter. The equations are reduced by HAM to a set of linear ordinary differential equations based on the homotopy of topology. These sets of equations can then be computed by mathematical software like Mathematica, Maple, MATLAB, Octave, SageMath, or Maxima. Applications are found in various problems of science and engineering such as expression of skin friction coefficient and reduced Nusselt and Sherwood numbers [9–12]. Analytical tools like homotopy perturbation method (HPM), δ-expansion method, and Adomian decomposition method (ADM) are special cases of HAM [13].Mabood and Khan [14, 15] successfully applied homotopy analysis method for the study of heat transfer on MHD stagnation point flow in porous medium and boundary layer flow and heat transfer over a permeable flat plate in a Darcian porous medium. Analytic solutions for unsteady two-dimensional and axisymmetric flows were presented by Rashidi et al. [16]. The study of heat and mass transfer in the context of squeezing flow was performed by Mustafa et al. [17]. Kirubhashankar and Ganesh [18] analyzed electrically conducting MHD viscous flows. A two-dimensional MHD problem was approximated using homotopy perturbation method (HPM) by Siddiqui et al. [19, 20]. Unsteady flow of viscous fluid between porous plates is studied by Ganesh and Krishnambal [21] and Mohamed Ismail et al. [22]. Shevanian et al. [23, 24] successfully used HAM to study singular linear vibrational BVPs and MHD squeezing flow between two parallel disks. The same authors [25, 26] used predictor homotopy analysis method (PHAM) to investigate nonlinear reactive transport model and nanoboundary layer flows with nonlinear Navier boundary condition.This work is an effort to investigate MHD squeezing flow of Newtonian fluid between two parallel plates passing through porous medium by homotopy analysis method. Using similarity transforms, the governing partial differential equations are converted to equivalent nonlinear ordinary differential equation and then solved using the mentioned scheme. Velocity profile of fluid is argued by varying various parameters involved.


## 2. Mathematical Model

In two dimensions, if cylindrical coordinates of the velocityu of moving plates are [
u
r
,
0
,
u
z
], ∇ is defined to be (
∂
/
∂
r
,
1
/
r
∂
/
∂
θ
,
∂
/
∂
z
), ω
=
∇
×
u, Ω
(
r
,
z
)
=
∂
u
z
/
∂
r
-
∂
u
r
/
∂
z, v
=
u
r
2
+
u
z
2, and ρ, p, and κ denote the density, pressure, and permeability, respectively, then, from momentum equation of steady squeezing flow in porous medium with MHD effect, r, θ, and z components are as follows.r component is as follows:(1)
∂
∂
r
ρ
v
+
p
-
ρ
Ω
u
z
=
-
μ
∂
Ω
∂
z
+
A
u
r
.θ component is as follows:(2)
1
r
∂
p
∂
θ
=
0
⟹
∂
p
∂
θ
=
0
⟹
p
=
f
r
,
z
.z component is as follows:(3)
∂
∂
r
ρ
v
+
p
+
ρ
Ω
u
r
=
μ
r
∂
∂
r
r
Ω
-
μ
κ
u
z
.Here, A
=
μ
/
κ
+
σ
B
0
2 with B
0 as imposed magnetic field. Introducing stream function [27] ψ
(
r
,
z
) defined by(4)
u
r
r
,
z
=
1
r
∂
ψ
∂
z
,
u
z
r
,
z
=
-
1
r
∂
ψ
∂
r
,it can be easily proved that the continuity equation ∇
·
u
=
0 is satisfied. With the help of (4) and eliminating ∂
/
∂
r
(
ρ
v
+
p
), (1) and (3) reduce to single PDE as follows:(5)
ρ
∂
ψ
∂
r
∂
∂
z
η
2
ψ
r
2
-
∂
ψ
∂
z
∂
∂
r
η
2
ψ
r
2
=
-
μ
r
η
4
ψ
+
μ
κ
η
2
ψ
+
η
B
0
2
r
∂
2
ψ
∂
z
2
.Here, η
=
∂
2
/
∂
r
2
-
1
/
r
∂
/
∂
r
+
∂
2
/
∂
z
2. If the moving plates are separated by distance 2
d, then(6)
u
r
=
0
,
u
z
=
-
v
,
a
t
z
=
d
,
u
z
=
0
,
∂
u
r
∂
z
=
0
,
a
t
z
=
0
.Using transformation ψ
(
r
,
z
)
=
r
2
f
(
z
), (5) reduces to single ODE as follows:(7)
d
4
d
z
4
f
z
+
2
ρ
μ
f
z
d
3
d
z
3
f
z
-
1
κ
d
2
d
z
2
f
z
-
σ
B
0
2
μ
d
2
d
z
2
f
z
=
0
.The boundary conditions in (6) are transformed to(8)
d
2
d
z
2
f
0
=
0
,
f
0
=
0
,
d
d
z
f
d
=
0
,
f
d
=
-
u
2
.Using nondimensional parameters 2
/
u
f
∗
=
f, d
z
∗
=
z, u
/
μ
R
m
p
=
ρ
d, m
h
=
σ
d
B
0
/
μ, and m
p
=
h
/
κ and omitting ∗ sign, (7) and (8) become(9)
d
4
d
z
4
f
z
+
R
m
p
f
z
d
3
d
z
3
f
z
-
m
p
d
2
d
z
2
f
z
-
m
h
d
2
d
z
2
f
z
=
0
,
(10)
d
2
d
z
2
f
0
=
0
,
f
0
=
0
,
d
d
z
f
1
=
0
,
f
1
=
1
,where R
m
p is Reynold and m
h, m
p are Hartmann numbers.


## 3. Application of HAM to Squeezing Flow

HAM logically contains some analytic techniques such as Adomian’s decomposition method, Lyapunov’s artificial small parameter method, andδ-expansion method. Thus, this technique can be regarded as a unified or generalized theory of these analytical techniques. Unlike other analytic techniques, the homotopy analysis method provides a simple way to control and adjust the convergence region and rate of solution series of nonlinear problems. Thus, this method is valid for nonlinear problems with strong nonlinearity. Homotopy analysis method provides great freedom to use base functions to express solutions of a nonlinear problem so that one can approximate a nonlinear problem more efficiently by means of better base functions [13].In the present section, this technique is applied on (9) using boundary conditions (10). For solution expression polynomial base function {
z
2
n
+
1
∣
n
=
0,1
,
2,3
,
…
} is used to determine f
(
z
) as follows:(11)
f
z
=
∑
m
=
0
∞
b
n
z
2
n
+
1
,where b
n are constants. By rule of solution expression and according to conditions in (10), the initial guess of the problem is(12)
f
0
z
=
3
z
-
z
3
2
.The auxiliary linear operator is chosen as(13)
L
f
z
=
d
4
d
z
4
f
zwith the property(14)
L
C
1
z
3
6
+
C
2
z
2
2
+
C
3
z
+
C
4
=
0
.Here, C
i
(
i
=
1,2
,
3,4
) are integration constants whose obtained values are(15)
C
2
=
C
4
=
0
,
C
1
=
-
3
,
C
3
=
1.5
.By rule of solution expression in (2.51) and by (2.39) in [13], the auxiliary function H
(
z
) is chosen to be H
(
z
)
=
1.


### 3.1. Zeroth-Order Deformation Equation

Using the homotopy introduced in [13], zeroth-order deformation equation is given by(16)
1
-
q
L
ϕ
z
;
q
-
f
0
z
=
q
h
H
z
N
ϕ
z
;
q
,where N is a nonlinear operator defined by(17)
N
ϕ
z
;
q
=
d
4
ϕ
z
;
q
d
z
4
+
R
m
p
ϕ
z
;
q
d
3
ϕ
z
;
q
d
z
3
-
m
p
d
2
ϕ
z
;
q
d
z
2
-
m
h
d
2
ϕ
z
;
q
d
z
2
.From here, the zeroth-order problem obtained is(18)
1
-
q
L
ϕ
z
;
q
-
f
0
z
=
q
h
H
z
N
ϕ
z
;
q
,
(19)
d
2
ϕ
0
;
q
d
z
2
=
0
,
ϕ
0
;
q
=
0
,
d
ϕ
1
;
q
d
z
=
0
,
ϕ
1
;
q
=
1
.Here, q is an embedding parameter and h is a nonzero auxiliary parameter. It is observed that,atq
=
0,(20)
ϕ
0
;
q
=
f
0
z
;
atq
=
1,(21)
ϕ
1
;
q
=
f
z
.Hence, when q varies from 0 to 1, then ϕ
(
z
;
q
) varies from f
0
(
z
) to f
(
z
). By Maclaurin’s expansion, ϕ
(
z
;
q
) can be expressed as(22)
ϕ
z
;
q
=
f
0
z
+
∑
n
=
1
∞
f
n
z
q
n
.The value of the auxiliary parameter h is chosen in such a way that the series in (18) converges at q
=
1; that is,(23)
f
z
=
f
0
z
+
∑
n
=
1
∞
f
n
z
,where f
n
(
z
)
=
1
/
n
!
∂
n
f
(
z
;
q
)
/
∂
z
n
q
=
0.


### 3.2.nth-Order Deformation Equation

Differentiate (18) and (19) n-times with respect to q and put q
=
0 to get nth-order deformation equation as follows:(24)
L
f
n
z
-
χ
n
f
n
-
1
z
=
h
H
z
R
n
f
n
-
1
,
(25)
d
2
f
n
0
d
z
2
=
0
,
f
n
0
=
0
,
d
f
n
1
d
z
=
0
,
f
n
1
=
0
.Equation (24) reduces to(26)
f
n
z
=
χ
n
f
n
-
1
z
+
h
L
-
1
H
z
R
n
f
n
-
1
,where(27)
L
-
1
=
C
1
x
3
6
+
C
2
x
2
2
+
C
3
x
+
C
4
+
⨌
⁡
·
d
z
d
z
d
z
d
z
,
(28)
R
n
f
n
-
1
=
f
n
-
1
i
v
+
R
m
p
∑
k
=
0
n
-
1
f
k
f
n
-
1
-
k
′
′
′
-
m
h
+
m
p
f
n
-
1
′
′
,
(29)
χ
n
=
0
,
n
≤
0
1
,
n
>
0
.


## 3.1. Zeroth-Order Deformation Equation

Using the homotopy introduced in [13], zeroth-order deformation equation is given by(16)
1
-
q
L
ϕ
z
;
q
-
f
0
z
=
q
h
H
z
N
ϕ
z
;
q
,where N is a nonlinear operator defined by(17)
N
ϕ
z
;
q
=
d
4
ϕ
z
;
q
d
z
4
+
R
m
p
ϕ
z
;
q
d
3
ϕ
z
;
q
d
z
3
-
m
p
d
2
ϕ
z
;
q
d
z
2
-
m
h
d
2
ϕ
z
;
q
d
z
2
.From here, the zeroth-order problem obtained is(18)
1
-
q
L
ϕ
z
;
q
-
f
0
z
=
q
h
H
z
N
ϕ
z
;
q
,
(19)
d
2
ϕ
0
;
q
d
z
2
=
0
,
ϕ
0
;
q
=
0
,
d
ϕ
1
;
q
d
z
=
0
,
ϕ
1
;
q
=
1
.Here, q is an embedding parameter and h is a nonzero auxiliary parameter. It is observed that,atq
=
0,(20)
ϕ
0
;
q
=
f
0
z
;
atq
=
1,(21)
ϕ
1
;
q
=
f
z
.Hence, when q varies from 0 to 1, then ϕ
(
z
;
q
) varies from f
0
(
z
) to f
(
z
). By Maclaurin’s expansion, ϕ
(
z
;
q
) can be expressed as(22)
ϕ
z
;
q
=
f
0
z
+
∑
n
=
1
∞
f
n
z
q
n
.The value of the auxiliary parameter h is chosen in such a way that the series in (18) converges at q
=
1; that is,(23)
f
z
=
f
0
z
+
∑
n
=
1
∞
f
n
z
,where f
n
(
z
)
=
1
/
n
!
∂
n
f
(
z
;
q
)
/
∂
z
n
q
=
0.


## 3.2.nth-Order Deformation Equation

Differentiate (18) and (19) n-times with respect to q and put q
=
0 to get nth-order deformation equation as follows:(24)
L
f
n
z
-
χ
n
f
n
-
1
z
=
h
H
z
R
n
f
n
-
1
,
(25)
d
2
f
n
0
d
z
2
=
0
,
f
n
0
=
0
,
d
f
n
1
d
z
=
0
,
f
n
1
=
0
.Equation (24) reduces to(26)
f
n
z
=
χ
n
f
n
-
1
z
+
h
L
-
1
H
z
R
n
f
n
-
1
,where(27)
L
-
1
=
C
1
x
3
6
+
C
2
x
2
2
+
C
3
x
+
C
4
+
⨌
⁡
·
d
z
d
z
d
z
d
z
,
(28)
R
n
f
n
-
1
=
f
n
-
1
i
v
+
R
m
p
∑
k
=
0
n
-
1
f
k
f
n
-
1
-
k
′
′
′
-
m
h
+
m
p
f
n
-
1
′
′
,
(29)
χ
n
=
0
,
n
≤
0
1
,
n
>
0
.


## 4. Exact Solution in Case of Zero Reynold Number

In this section, a special case is studied when the Reynold number is zero and hence (9) becomes a linear differential equation. The exact solution obtained using boundary conditions in (10) is given by(30)
f
z
=
-
e
-
M
z
M
z
-
e
M
z
-
M
z
e
M
z
+
2
M
+
e
2
M
z
+
M
-
e
M
M
e
2
M
+
M
-
e
2
M
+
1
.Here, M
=
m
h
+
m
p. Homotopy analysis solution is also derived in this case. The operator N in (17) becomes(31)
N
ϕ
z
;
q
=
d
4
ϕ
z
;
q
d
z
4
-
m
p
d
2
ϕ
z
;
q
d
z
2
-
m
h
d
2
ϕ
z
;
q
d
z
2and for nth-order deformation equation R
n in (28) becomes(32)
R
n
f
n
-
1
=
f
n
-
1
i
v
-
m
h
+
m
p
f
n
-
1
′
′
.In Table 4, comparison of exact solution with fifth- and tenth-order HAM solutions is made with the help of absolute error.


## 5. Convergence of HAM Solution

Solution obtained by homotopy analysis method in (23) contains auxiliary parameter h which adjusts and controls the convergence. There is great freedom to choose the auxiliary parameter. For influence of h on the solution, the convergence of f
(
n
)
(
0
), where n is odd, is considered. The valid region R
h of h for which f
′
(
0
) converges is shown for different order solutions in Figure 1. The curve f
′
(
0
) versus h is said to be h-curve. From Figure 1, it is observed that R
h increases with the increase of approximation order. For fifth-order solution, the valid region for h is -
1.6
≤
h
≤
-
0.3. It is obvious from Figure 2 that when m
h, m
p increase further, the valid region moves towards the right. Figures 3 and 4 are constructed to examine R
h for increasing m
h, m
p, and R
m
p.Figure 1
h-curve of different order solutions for m
h
=
0.3, m
p
=
0.7, and R
m
p
=
1.Figure 2
h-curve of fifth-order solution for various m
h and m
p and R
m
p
=
1.Figure 3
h-curve of fifth-order solution for various R
m
p and m
h
=
0.6 and m
p
=
0.4.Figure 4
h-curve of tenth-order solution for various m
h, m
p, and R
m
p.


## 6. Results and Discussion

Analytic solution, using homotopy analysis method, of magnetohydrodynamics squeezing flow through porous medium is studied. Four figures (Figures1–4) are constructed, for various values of Reynold and Hartmann numbers, to examine the valid region R
h which has a vital role in convergence of analytic solution. Table 1 shows different order absolute residuals for HAM solutions and it is clear to see that as the order of approximation increases further, the solution converges to exact solution. Fifth-order absolute residuals for various values of R
m
p and for fixed values of m
h, m
p are displayed in Table 2. Table 3 is constructed to display tenth-order absolute residuals for different values of m
h, m
p, and h while keeping Reynold number fixed. Exact solution is obtained in case of zero Reynold number when the differential equation in (9) becomes linear. Fifth- and tenth-order HAM solutions are compared with this exact solution in Table 4 using the concept of absolute error. Keeping h and Reynold and Hartmann numbers fixed, Table 5 displays different order HAM solutions. Table 6 shows important information about the consistency and efficiency of HAM by means of average absolute residuals for fifth-order approximation. Convergence of the present technique is given in Table 7 for different order approximations by means of f
′
(
0
), |
f
′
′
′
(
0
)
|, f
v
(
0
), and |
f
v
i
i
(
0
)
|. As f
n
(
0
) becomes zero for even n, odd order derivatives are taken to study the convergence of the technique used. Comparison of different analytical techniques and one numerical Mathematica command NDSolve with the present work is displayed in Table 8 which shows that results obtained by HAM are in high agreement. One can refine these results by selecting suitable h in the valid region R
h. The rapid convergence of HAM can also be seen in Figure 5 which shows the residuals of various analytical schemes.Table 1
Absolute residuals keepingm
h
=
m
p
=
0.5 and R
m
p
=
1.
Absolute residuals for different order HAM solutions
z
5th order
10th order
15th order
20th order
25th order
0.0
0
0
0
0
0
0.1
9.5367
×
1
0
-
9
7.7605
×
1
0
-
14
0
5.5511
×
1
0
-
17
0
0.2
3.2067
×
1
0
-
8
3.0953
×
1
0
-
13
2.2210
×
1
0
-
16
2.2210
×
1
0
-
16
2.2210
×
1
0
-
16
0.3
1.6171
×
1
0
-
7
9.2082
×
1
0
-
13
4.4410
×
1
0
-
16
2.2210
×
1
0
-
16
2.2210
×
1
0
-
16
0.4
5.3675
×
1
0
-
7
2.1336
×
1
0
-
12
0
0
0
0.5
1.3172
×
1
0
-
6
3.5714
×
1
0
-
12
0
4.4410
×
1
0
-
16
0
0.6
1.8889
×
1
0
-
6
2.8302
×
1
0
-
12
4.4410
×
1
0
-
16
8.8820
×
1
0
-
16
4.4410
×
1
0
-
16
0.7
2.0711
×
1
0
-
7
5.1026
×
1
0
-
12
4.4410
×
1
0
-
16
4.4410
×
1
0
-
16
4.4410
×
1
0
-
16
0.8
5.7686
×
1
0
-
6
2.1931
×
1
0
-
11
0
4.4410
×
1
0
-
16
1.3323
×
1
0
-
15
0.9
1.3705
×
1
0
-
5
3.2338
×
1
0
-
11
4.4410
×
1
0
-
16
8.8820
×
1
0
-
16
0
1.0
1.2902
×
1
0
-
5
2.8715
×
1
0
-
12
2.2205
×
1
0
-
15
4.4410
×
1
0
-
16
4.4410
×
1
0
-
16Table 2
Absolute residuals for differentR
m
p keeping m
h and  m
p fixed.
Fifth-order absolute residuals
z
R
m
p
=
0.1
R
m
p
=
0.3
R
m
p
=
0.5
R
m
p
=
0.7
R
m
p
=
0.9
0.1
1.2301
×
1
0
-
10
1.3902
×
1
0
-
9
2.2210
×
1
0
-
9
9.1597
×
1
0
-
11
8.9372
×
1
0
-
9
0.2
9.2250
×
1
0
-
10
2.2645
×
1
0
-
9
2.2210
×
1
0
-
9
5.9589
×
1
0
-
10
6.5040
×
1
0
-
9
0.3
2.8005
×
1
0
-
9
2.1661
×
1
0
-
9
4.4410
×
1
0
-
10
4.7810
×
1
0
-
10
3.1478
×
1
0
-
8
0.4
5.5909
×
1
0
-
9
6.2185
×
1
0
-
10
2.2210
×
1
0
-
9
1.2061
×
1
0
-
9
9.3540
×
1
0
-
8
0.5
8.2828
×
1
0
-
9
2.9842
×
1
0
-
9
2.2210
×
1
0
-
8
2.0999
×
1
0
-
9
3.0340
×
1
0
-
7
0.6
8.8277
×
1
0
-
9
8.9656
×
1
0
-
9
4.4410
×
1
0
-
8
3.2913
×
1
0
-
9
6.0391
×
1
0
-
7
0.7
3.9394
×
1
0
-
9
1.6855
×
1
0
-
8
4.4410
×
1
0
-
8
1.4133
×
1
0
-
8
4.0297
×
1
0
-
7
0.8
1.1760
×
1
0
-
8
2.4190
×
1
0
-
8
2.2210
×
1
0
-
10
3.2937
×
1
0
-
8
1.1930
×
1
0
-
7
0.9
4.9318
×
1
0
-
8
2.7364
×
1
0
-
8
4.4410
×
1
0
-
9
1.0928
×
1
0
-
8
4.0659
×
1
0
-
7
1.0
1.3750
×
1
0
-
7
2.5064
×
1
0
-
8
2.2205
×
1
0
-
8
1.3853
×
1
0
-
7
5.5660
×
1
0
-
6Table 3
Absolute residuals for different values ofm
h, m
p and  h keeping R
m
p
=
1.
Tenth-order absolute residuals
m
h
=
0.5
m
h
=
2.5
m
h
=
3.5
m
h
=
4.5
m
h
=
5.5
z
m
p
=
2.5
m
p
=
3.5
m
p
=
5.5
m
p
=
7.5
m
p
=
9.5
h
=
-
0.905
h
=
-
0.81
h
=
-
0.76
h
=
-
0.702
h
=
-
0.652
0.1
1.7770
×
1
0
-
12
5.2593
×
1
0
-
11
2.0127
×
1
0
-
8
3.2010
×
1
0
-
7
1.7200
×
1
0
-
6
0.2
2.2357
×
1
0
-
12
2.7173
×
1
0
-
10
3.7920
×
1
0
-
8
6.4820
×
1
0
-
7
3.9832
×
1
0
-
6
0.3
6.6970
×
1
0
-
12
6.7927
×
1
0
-
10
5.0219
×
1
0
-
8
8.4460
×
1
0
-
7
5.8732
×
1
0
-
6
0.4
1.2603
×
1
0
-
11
1.0666
×
1
0
-
9
5.3527
×
1
0
-
8
5.7542
×
1
0
-
7
4.2189
×
1
0
-
6
0.5
2.1863
×
1
0
-
11
1.0404
×
1
0
-
9
4.8000
×
1
0
-
8
4.9288
×
1
0
-
7
5.2428
×
1
0
-
6
0.6
3.3222
×
1
0
-
11
2.0734
×
1
0
-
10
4.8612
×
1
0
-
8
2.0817 ×1
0
-
6
2.2440
×
1
0
-
5
0.7
3.7364
×
1
0
-
11
1.7351
×
1
0
-
9
1.1400
×
1
0
-
7
2.0580
×
1
0
-
6
2.9649
×
1
0
-
5
0.8
2.6140
×
1
0
-
11
5.6690
×
1
0
-
9
4.1453
×
1
0
-
7
6.0343
×
1
0
-
6
3.7432
×
1
0
-
5
0.9
3.5145
×
1
0
-
12
1.6370
×
1
0
-
8
1.4760
×
1
0
-
6
3.9871
×
1
0
-
5
3.6410
×
1
0
-
4
1.0
2.8584
×
1
0
-
11
5.7388
×
1
0
-
8
5.2801
×
1
0
-
6
1.5630
×
1
0
-
4
1.5182
×
1
0
-
3Table 4
Absolute errors of fifth- and tenth-order HAM solutions whenR
m
p
=
0.
Absolute errors whenR
m
p
=
0
z
Exact solution
Fifth-order AE
Tenth-order AE
0.0
0
0
0
0.1
0.147171
7.58785506649317
×
1
0
-
10
0
0.2
0.291618
1.39478356642186
×
1
0
-
9
3.33066907387546
×
1
0
-
16
0.3
0.430586
1.80948145356296
×
1
0
-
9
8.32667268468867
×
1
0
-
16
0.4
0.561271
1.94815263920844
×
1
0
-
9
7.77156117237609
×
1
0
-
16
0.5
0.680780
1.81075676675135
×
1
0
-
9
4.44089209850062
×
1
0
-
16
0.6
0.786113
1.45204859247627
×
1
0
-
9
2.22044604925031
×
1
0
-
16
0.7
0.874125
9.70818092582703
×
1
0
-
10
5.55111512312578
×
1
0
-
16
0.8
0.941500
4.90335771985428
×
1
0
-
10
7.77156117237609
×
1
0
-
16
0.9
0.984713
1.33847599670389
×
1
0
-
10
1.11022302462515
×
1
0
-
16
1.0
1.000000
2.22044604925031
×
1
0
-
16
2.22044604925030
×
1
0
-
16
A
E
=
|
e
x
a
c
t
s
o
l
u
t
i
o
n
-
H
A
M
s
o
l
u
t
i
o
n
|.Table 5
Different order solutions keepingm
h
=
m
p
=
0.5, R
m
p
=
1, and h
=
-
0.97.
Different order HAM solutions
z
1st order
3rd order
5th order
15th order
25th order
0.0
0
0
0
0
0
0.1
0.15034713893
0.15029407026
0.15029422756
0.15029422727
0.15029422694
0.2
0.29758357211
0.29748028081
0.29748058054
0.29748058002
0.29748057936
0.3
0.43861285278
0.43846608333
0.43846649504
0.43846649440
0.43846649341
0.4
0.57036617920
0.57018808886
0.57018856454
0.57018856394
0.57018856267
0.5
0.68981403459
0.68962312433
0.68962359743
0.68962359707
0.68962359568
0.6
0.79397520822
0.79379541675
0.79379580965
0.79379580971
0.79379580853
0.7
0.87992232433
0.87977913949
0.87977938815
0.87977938864
0.87977938832
0.8
0.94478300617
0.94469588175
0.94469597496
0.94469597563
0.94469597737
0.9
0.98573580197
0.98570687775
0.98570688141
0.98570688178
0.98570688792
1.0
0.99999999999
1.00000000000
1.00000000000
1.00000000000
1.00000001530Table 6
Average absolute residuals for various values ofm
h, m
p, R
m
p, and h.
Fifth-order average absolute residuals
⁢m
h
⁢m
p
R
m
p
⁢h
Solution
Residual
0.3
0.7
1.0
-
0.972
0.6896235970708491
3.385181024
×
10
-
12
1.2
0.8
0.7
-
0.685
0.5614010061254293
5.195453594
×
1
0
-
7
3.0
2.0
0.5
-
0.773
0.8625404743084730
1.845019426
×
1
0
-
8
3.5
4.5
0.9
-
0.851
0.6537531470383126
9.278417177
×
1
0
-
8
6.0
4.0
1.0
-
0.597
0.7562438682896610
5.552400535
×
1
0
-
5
10.0
5.0
1.5
-
0.656
0.7456786847598219
1.257868035
×
1
0
-
5
12.0
8.0
2.0
-
0.590
0.6266880364547981
2.172601809
×
1
0
-
5
15.0
10.0
2.5
-
0.520
0.7309895729856828
4.155602273
×
1
0
-
4Table 7
Convergence of HAM for different order approximations.
Approximation order
f
′
(
0
)
|
f
′
′
′
(
0
)
|
f
v
(
0
)
|
f
(
v
i
i
)
(
0
)
|
1
0.00866
3.11432
1.45499
8.729999
3
0.00812
3.11177
1.58119
15.22983
5
0.00812
3.11178
1.58118
15.25585
7
0.00812
3.11178
1.58118
15.25593
10
0.00812
3.11178
1.58118
15.25593
15
0.00812
3.11178
1.58118
15.25593
20
0.00812
3.11178
1.58118
15.25593
25
0.00812
3.11178
1.58118
15.25595Table 8
Comparison of HAM with numerical and other analytical techniques.
Absolute residuals for numerical and different analytical schemes
z
NDSolve
DTM
DJM
ADM
OHAM
HAM
0
3.47
×
1
0
-
3
0
0
0
0
0
0.1
3.20
×
1
0
-
4
2.36
×
1
0
-
13
2.86
×
1
0
-
12
3.20
×
1
0
-
5
1.51
×
1
0
-
7
0
0.2
2.02
×
1
0
-
4
1.21
×
1
0
-
10
3.56
×
1
0
-
10
2.51
×
1
0
-
4
2.72
×
1
0
-
7
2.22
×
1
0
-
16
0.3
4.57
×
1
0
-
5
4.62
×
1
0
-
9
5.81
×
1
0
-
9
8.30
×
1
0
-
4
3.31
×
1
0
-
7
4.44
×
1
0
-
16
0.4
5.67
×
1
0
-
5
6.11
×
1
0
-
8
4.09
×
1
0
-
8
2.00
×
1
0
-
3
3.05
×
1
0
-
7
0
0.5
1.51
×
1
0
-
5
4.51
×
1
0
-
7
1.79
×
1
0
-
7
3.51
×
1
0
-
3
1.75
×
1
0
-
7
0
0.6
3.45
×
1
0
-
5
2.31
×
1
0
-
6
5.78
×
1
0
-
7
5.70
×
1
0
-
3
6.62
×
1
0
-
8
4.44
×
1
0
-
16
0.7
5.07
×
1
0
-
7
9.12
×
1
0
-
6
1.50
×
1
0
-
6
8.33
×
1
0
-
3
4.05
×
1
0
-
7
4.44
×
1
0
-
16
0.8
4.56
×
1
0
-
5
3.00
×
1
0
-
5
3.25
×
1
0
-
6
1.12
×
1
0
-
2
8.08
×
1
0
-
7
4.44
×
1
0
-
16
0.9
1.20
×
1
0
-
4
8.50
×
1
0
-
5
6.17
×
1
0
-
6
1.41
×
1
0
-
2
1.23
×
1
0
-
6
0
1.0
6.99
×
1
0
-
4
2.20
×
1
0
-
4
1.03
×
1
0
-
5
1.64
×
1
0
-
2
1.61
×
1
0
-
6
2.22
×
1
0
-
15Figure 5
Comparison of HAM with other analytical schemes through residual.


## 7. Conclusion

All tables (Tables1–8) show the efficiency and consistency of the mentioned scheme. The valid region R
h can be more refined to choose such value of h for which the obtained solution converges rapidly which is the beauty of the present analytic technique. It is also observed that while increasing the value of Hartmann numbers (Figures 2 and 4), the valid region moves towards the right. Table 3 also tells the same story. The influences of different parameters on the velocity profile are displayed in Figures 6–9 and the following observations are made:(1)
Velocity of fluid in porous medium and the Reynold numberR
m
p are inversely proportional to each other; that is, increasing Reynold number (when density of fluid increases) results in the decrease in velocity of fluid while keeping Hartmann number fixed as shown in Figure 6.
(2)
Figure7 shows that while increasing m
h (i.e., increasing the imposed magnetic effect) and keeping m
p, R
m
p fixed, the velocity of fluid increases.
(3)
The same effect is studied in Figure8 by increasing m
p (i.e., when the permeability decreases) and keeping m
h and R
m
p fixed.
(4)
Increasing Hartmann and Reynold numbers together, it is observed from Figure9 that the velocity of fluid increases. It is concluded that Hartmann number is more influential as compared to Reynold number.Figure 6
Velocity profile for different values ofR
m
p with m
h
=
m
p
=
0.5.Figure 7
Velocity profile for different values ofm
h with m
p
=
R
m
p
=
1.Figure 8
Velocity profile for different values ofm
p with m
h
=
R
m
h
=
1.Figure 9
Velocity profile for equal values ofm
p, m
h, and R
m
p.



---

*Source: 3541512-2016-06-22.xml*