1. Introduction
Due to diversity of fluids in nature, a lot of models have been
proposed to describe their flow behavior in different circumstances.
A wide range of commonly encountered fluids includes Newtonian
fluids. But a large number of fluids appearing in industry differ
greatly from Newtonian fluids in their rheology. Newtonian fluids
are recognized by the linear relationship between stress and the
rate of strain. In many existing fluids with complex molecular
structure, the relation between stress and strain is found to be
non-linear. Therefore, the Newtonian fluids model can not be used to
predict, analyze and stimulate the behavior of many viscoelastic
fluids. Hence, in practical situations and applications in industry,
it is necessary to study of the flow behavior of non-Newtonian
fluids.
The inadequacy of the classical Navier-Stokes theory to describes
the behavior of the rheologically complex fluids such as polymer
solutions, heavy oils, blood and many emulsions, has led to the
development of models of non-Newtonian fluids. In particular, many
pastes, slurries, synovial, polymer solutions and suspensions
exhibit shear thinning behavior. The non-newtonian fluids are widely
used in chemical engineering, food industry, biological analysis,
petroleum industry, and many other fields. The academic workers and
engineers are very much interested in the geometry of flows of such
types of fluids. As compared to Newtonian fluids, the analysis of
the behavior of the motion of such fluids is much more complicated
and not easy to handle because of non-linear relationship between
stress and rate of strain.
In recent years many non-Newtonian models have been proposed. Among
these models, model of “fluids of differential type,” [1] have
received considerable attention. “Fluid of third grade” is a
subclass of of fluids of differential type, which has been studied
successfully in various types of flow situations [2, 3, 4] and is known
to capture the non-Newtonian affects such as shear thinning, shear
thickening as well as normal stress.
Most of the nonlinear differential equations do not have analytical
solution. However researchers used many numerical methods, but these
methods require much time and more efficient computing devices. Semi
analytical methods are more suitable than numerical methods to solve
nonlinear non-homogenous partial differential equations. The most
powerful tool for the calculation of analytical solutions of the
linear or nonlinear partial differential equation is Adomian’s
decomposition method (ADM), a method introduced by Adomian [5, 6].
The ADM provides analytical solution in the form of an infinite
convergent power series in which each term can be easily determined.
The ADM has been successfully applied to solve nonlinear
differential equations in studying interesting problems arising in
applied sciences and engineering [7, 8]. In 2009 Siddiqui, et al.
[8] studied parallel plate flow of a third grade fluid by means of
ADM and compare the results with num
erical schemes. Recently,
Siddiquie et al. [9] compare the ADM and HPM in solving the problem
of squeezing flow between two circular plates. Their comparison
shows that the ADM requires more computational efforts than the HPM,
but it yields more accurate results than the HPM.
In the past few years, magnetohydrodynamics (MHD) has gained
considerable importance because of its diverse applications in
physics and engineering. In astrophysical and geophysical
applications it is useful to study the stellar and solar structures
solar storms and flares, radio propagation through the ionosphere
etc. In engineering its applications are in MHD generators, MHD
pumps and MHD bearings. The use of a MHD fluid as lubricant is of
interest in industrial applications, because it prevents the
unexpected variation of lubricant viscosity with temperature under
certain extreme operating conditions.
Historically Rossow [10] initiated the boundary layer flow on a semi
infinite flat plate. Since then large amount of literature is
developed on this subject. The recent attempts in this directions
are made by Hayat et al. [11, 12, 13] , Khan et al. [14]. S. Islam et al.
[15], obtained series solutions for MHD flow between infinite
parallel plates.
The main aim of this work is to study the steady MHD flow of
incompressible third grade non-Newtonian fluid between two parallel
plates separated by a finite gap in the presence of a
transversal magnetic field and to study the steady flow of
incompressible third grade non-Newtonian fluid between two coaxial
cylinders of infinite length [16], using the ADM. Siddiqui,
et al. [8] solved the three fundamental problems of plane Couette
flow, fully developed plane Poiseuille flow and plane
Couette-Poiseuille flow by using the ADM. We extend the work of
Siddiqui [8] to the case of a MHD fluid.
We apply the ADM to solve the problems modeling the
MHD flow of a third grade fluid between two parallel plats separated
by a finite gab in the presence of a transversal magnetic field.
Specially, we study the three fundamental problems namely, plane
Couette flow (flow due to motion of either of the plates), Poiseulle
flow (flow due to application of external pressure gradient, while
both plates are stationary), and Generalized-Couette flow (flow due
to motion of one plate as well as applied external pressure
gradient). Also, the solution of ordinary third grade fluid are
recovered [8] as limiting cases of our obtained solutions. At the
end of this chapter, the influences of the parameters and other
material constants on the velocity field are analyzed graphically.
2.Mathematical formulation of the problem
The basic equations which govern the incompressible unidirectional
magnetohydrodynamic flow are:
\begin{equation}\label{1}
\nabla\cdot \textbf{V}=0,
\end{equation}
(1)
\begin{equation}\label{2}
\rho\bigg[\frac{\partial\textbf{V}}{\partial
t}+(\textbf{V}\cdot\nabla)\textbf{V}\bigg]=\rho{\textbf{f}}+\nabla\cdot
\textbf{T}+\textbf{J}\times \textbf{B},
\end{equation}
(2)
where, \(\textbf{J}\) is electric current density and \(\textbf{B}\) is
the total magnetic field, \(\textbf{B}=\textbf{B}_{0}+\textbf{b},\textbf{B}_{0}\) represents the imposed magnetic field and
\(\textbf{b}\) denotes the induced magnetic field. In the absence of
displacement currents, the Ohm’s Law and Maxwell’s equation [
4] are
$$\textbf{J}=\sigma[\textbf{E}+\textbf{V}\times \textbf{B}],$$
$$div\textbf{B}=0,\ \nabla\times \textbf{B}=\mu_{m}\textbf{J},\
curl\textbf{E}=-\frac{\partial \textbf{B}}{\partial t},$$
in which
\(\sigma\) is the electrical conductivity, \(\mu_{m}\) is the magnetic
permeability. The following assumptions are made in order to lead
our discussion,
- The density \(\rho,\) magnetic permeability
\(\mu_{m}\) and electrical conductivity \(\sigma\) are assumed to
be constant throughout the flow field region.
- The electrical conductivity \(\sigma\) of the
fluid considers being finite.
- The total magnetic field B is
perpendicular to the velocity field V and the induced
magnetic field b is negligible compared with the applied magnetic
field \(B_{0}\) so that the magnetic Reynolds number is small
[6].
- We assume a situation where no energy is
added or extracted from the fluid by the electric
field, which implies that there is no electric
field present in the fluid flow region.
Under these assumptions, the magnetohydrodynamic force involved in
(2) takes the following form
\begin{equation}
\textbf{J}\times \textbf{B}=\sigma[\textbf{B}_{0}(\textbf{V}\cdot
\textbf{B}_{0})-\textbf{V}(\textbf{B}_{0}\cdot\textbf{B}_{0})]=-\sigma
\textbf{B}_{0}^{2}\textbf{V}.
\end{equation}
(3)
For incompressible third grade fluid, the constitutive equation for
extra stress tensor is given in [
17].
For the problem under consideration we assume a velocity field for
one dimensional flow and stress tensor of the form
\begin{equation}\label{4}
\textbf{V}=(u(y), 0, 0),\ \ \ \textbf{S}=S(y),
\end{equation}
(4)
By using Eq. (4), the continuity equation (1) is identically
satisfied and the equation of motion (2), in the absence of
gravitational effect becomes
\begin{equation}
-\frac{\partial p}{\partial x}+\mu\frac{\partial^{2}u}{\partial
y^{2}}+6(\beta_{2}+\beta_{3})\bigg(\frac{\partial u}{\partial
y}\bigg)^{2}\frac{\partial^{2} u}{\partial y^{2}}-\sigma
B_{0}^{2}u=0,
\end{equation}
(5)
\begin{equation}
-\frac{\partial p}{\partial y}+\frac{\partial}{\partial
y}\bigg\{(2\alpha_{1}+\alpha_{2})\bigg(\frac{\partial u}{\partial
y}\bigg)^{2}\bigg\}=0.
\end{equation}
(6)
\begin{equation}
\frac{\partial p}{\partial z}=0.
\end{equation}
(7)
Introducing the generalized pressure \(\widehat{p}\)
\begin{equation}
\widehat{p}=-p(x,y)+(2\alpha_{1}+\alpha_{2})\bigg(\frac{\partial
u}{\partial y}\bigg)^{2}
\end{equation}
(8)
and substituting \(\widehat{p}\) in Eq. (6), we find that
\begin{equation}\label{9}
\frac{d\widehat{p}}{dy}=0,
\end{equation}
(9)
showing that \(\widehat{p}=\widehat{p}(x)\). Consequently, Eq.(5)
reduces to the single equation
\begin{equation}\label{10}
-\frac{d\widehat{p}}{dx}+\mu\frac{d^{2}
u}{dy^{2}}+6\beta\bigg(\frac{d u}{dy}\bigg)^{2}\frac{d^{2}
u}{dy^{2}}-\sigma B_{0}^{2}u=0,
\end{equation}
(10)
where for simplicity we have introduced
\(\beta=\beta_{2}+\beta_{3}\). Eq. (10) is a second-order
nonlinear ordinary differential equation. This equations governs the
unidirectional flow of a non-Newtonian third grade fluid between two
infinite parallel plates.
3. Solution of the problem using Adomian Decomposition Method
In this section, we use the ADM to solve Eq. (10) with boundary
conditions corresponding to different problems of Couette flow.
3.1 Plane Couette Flow
We consider the steady laminar flow of an incompressible third grade
fluid between two infinite parallel plate at \(y=0,\) and \(y=h\). The
plates are separated by distance \(h\). The upper plate moves parallel
to itself with uniform velocity \(U\), while the lower one is at rest.
The plates are non conducting and a transversal magmatic field is
applied in the vertical upward direction. Let x-axis be taken along
the direction of flow and \(y\) in the direction normal to the flow.
The pressure \(p\) is constant and fluid properties vary along y-axis
only. Thus, governing equation (10) for such a flow, in the
absence of pressure gradient, reduces to
\begin{equation}\label{11}
\frac{d^{2} u}{d y^{2}}+\frac{6\beta}{\mu}\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}-m^{2}u=0,
\end{equation}
(11)
where, \(m^{2}={\sigma B_{0}^{2}}/{\mu}\), subject to the boundary
conditions
\begin{equation*}
u(y)=0\ \ \text{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{12}
u(y)=U\ \ \textrm{at}\ \ y=h.
\end{equation}
(12)
Introducing the following non-dimensional parameters
\begin{equation}\label{13}
u^{*}=\frac{u}{U},\ \ y^{*}=\frac{y}{h},\ \ \beta^{*}=\frac{\beta
}{\mu h^{2}/U^{2}},\ \ {m^{*2}}=\frac{\sigma
B_{0}^{2}}{{\mu}/{h^{2}}}.
\end{equation}
(13)
Dropping the \(‘*’\) the boundary value problem (1) becomes
\begin{equation} \frac{d^{2} u}{d y^{2}}+6\beta\bigg(\frac{d
u}{d y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}-m^{2}u=0,
\end{equation}
(14)
with the boundary conditions
\begin{equation*}
u(y)=0 \ \ \textrm{at} \ \ y=0,
\end{equation*}
\begin{equation}
u(y)=1 \ \ \textrm{at} \ \ y=1.
\end{equation}
(15)
In the operator form, Eq. (4) becomes
\begin{equation}
Lu=-{6\beta}\bigg(\frac{d u}{d y}\bigg)^{2}\frac{d^{2} u}{d
y^{2}}+m^{2}u,
\end{equation}
(16)
where \(L={d^{2}}/{d y^{2}}\) and inverse operator is given by
\(\tilde{L}^{-1}=\int\int(\cdot) dydy\). Applying \(\tilde{L}^{-1}\)
on both sides of Eq. (6), we have
\begin{equation}
\tilde{L}^{-1}Lu=-\tilde{L}^{-1}\bigg[{6\beta}\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}\bigg]+\tilde{L}^{-1}[m^{2}u],
\end{equation}
(17)
\begin{equation}
u(y)=Ay+B-{6\beta}\tilde{L}^{-1}\bigg[\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}\bigg]+\tilde{L}^{-1}[m^{2}u],
\end{equation}
(18)
where \(A\) and \(B\) are constants of integration.
To solve Eq. (8) by the ADM, we let
\begin{equation}
u=\sum_{n=0}^{\infty}u_{n}(y),
\end{equation}
(19)
\begin{equation}
Nu=\sum_{n=0}^{\infty}A_{n},
\end{equation}
(20)
where
\begin{equation}
Nu=\bigg(\frac{d u}{d y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}
\end{equation}
(21)
In view of Eqs. (9) and (10), Eq. (8) becomes
\begin{equation}
\sum_{n=0}^{\infty}u_{n}(y)=Ay+B- 6\beta
\tilde{L}^{-1}\sum_{n=0}^{\infty}A_{n}+m^{2}\tilde{L}^{-1}
\sum_{n=0}^{\infty}u_{n}(y).
\end{equation}
(22)
We identify the zeroth component as
\begin{equation}
u_{0}(y)=Ay+B,
\end{equation}
(23)
and the remaining components as the recurrence relation
\begin{equation}
u_{n+1}(y)=-6\beta
\tilde{L}^{-1}[A_{n}]+m^{2}\tilde{L}^{-1}[u_{n}(y)],\ \ n\geq0,
\end{equation}
(24)
where \(A_{n}\)’s are the Adomian polynomials that represent the non
linear term in (11).
The first few Adomian polynomials as follows:
\begin{equation}
A_{0}=\bigg(\frac{d u_{0}}{d y}\bigg)^{2}\frac{d^{2} u_{0}}{d y^{2}},
\end{equation}
\begin{equation}
A_{1}=2\frac{d u_{0}}{d y}\frac{d u_{1}}{d y}\frac{d^{2} u_{0}}{d
y^{2}}+\bigg(\frac{d u_{0}}{d y}\bigg)^{2}\frac{d^{2} u_{1}}{d
y^{2}},
\end{equation}
\begin{equation}\label{25}
A_{2}=2\frac{d u_{0}}{d y}\frac{d^{2} u_{0}}{d y^{2}}\frac{d
u_{2}}{d y}+\bigg(\frac{d u_{1}}{d y}\bigg)^{2}\frac{d^{2} u_{0}}{d
y^{2}}+2\frac{d u_{0}}{d y}\frac{d u_{1}}{d y}\frac{d^{2} u_{1}}{d
y^{2}}+\bigg(\frac{d u_{0}}{d y}\bigg)^{2}\frac{d^{2} u_{2}}{d
y^{2}}
\end{equation}
(25)
$$\vdots$$
Using expression (9) in (5), we have the following boundary
conditions
\begin{equation}
u_{0}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation}
\begin{equation}
u_{0}(y)=1\ \ \textrm{at}\ \ y=1,
\end{equation}
(26)
and
\begin{equation}
u_{n}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation}
\begin{equation}\label{27}
u_{n}(y)=0\ \ \textrm{at}\ \ y=1,\ \ n\geq 1.
\end{equation}
(27)
From Eq. (23) and boundary conditions (26), we obtain the
zeroth-order solution as
\begin{equation}\label{28}
u_{0}(y)=y,
\end{equation}
(28)
From (24), (25) and (27), we have the first-order
problem as
\begin{equation}
u_{1}(y)=-6\beta \tilde{L}^{-1}[A_{0}]+m^{2}\tilde{L}^{-1}[u_{0}],
\end{equation}
(29)
subject to the boundary conditions
\begin{equation*}
u_{1}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{30}
u_{1}(y)=0\ \ \textrm{at}\ \ y=1.
\end{equation}
(30)
Solving (29) and (30), the first-order solution is given by
\begin{equation}\label{31}
u_{1}(y)=\frac{m^{2}}{6}[y^{3}-y].
\end{equation}
(31)
In view of (24), (25) and (27), the second-order
problem is
\begin{equation}
u_{2}(y)=-6\beta \tilde{L}^{-1}[A_{1}]+m^{2}\tilde{L}^{-1}[u_{1}],
\end{equation}
(32)
subject to boundary conditions
\begin{equation*}
u_{2}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{33}
u_{2}(y)=0\ \ \textrm{at}\ \ y=1.
\end{equation}
(33)
The corresponding solution is
\begin{equation}\label{34}
u_{2}(y)={-\beta}m^{2}[y^{3}-y]+\frac{m^{4}}{6}\bigg[\frac{y^{5}}{20}-
\frac{y^{3}}{6}+\frac{7y}{60}\bigg].
\end{equation}
(34)
Similarly, third-order problem is
\begin{equation}\label{35}
u_{3}(y)=-6\beta \tilde{L}^{-1}[A_{2}]+m^{2}\tilde{L}^{-1}[u_{2}],
\end{equation}
(35)
with boundary conditions
\begin{equation*}
u_{3}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{36}
u_{3}(y)=0\ \ \textrm{at}\ \ y=1.
\end{equation}
(36)
The corresponding third-order solution is given by
\begin{equation}
u_{3}(y)=6m^{2}\beta^{2}[y^{3}-y]
{-\beta}m^{4}\bigg[\frac{2y^{5}}{5}-
\frac{2y^{3}}{3}+\frac{4y}{15}\bigg]+\frac{m^{6}}{6}\bigg[\frac{y^{7}}{840}
-\frac{y^{5}}{120} + \frac{7y^{3}}{360}-\frac{31y}{2520}\bigg].
\end{equation}
(37)
Inserting (18), (21), (24) and (27) in (9), the
solution of differential equation (1) takes the form
\begin{equation*}
u(y)=\sum_{n=0}^{\infty}u_{n}(y)=u_{0}(y)+u_{1}(y)+u_{2}(y)+u_{3}(y)+\cdots
\end{equation*}
or, equivalently
\begin{equation*}
u(y)=y+\frac{m^{2}}{6}[y^{3}-y] {-\beta}m^{2}[y^{3}-y]
+6{\beta}^{2}m^{2}[y^{3}-y]
\end{equation*}
\begin{equation}\label{38}
{-\beta}m^{4}\bigg[\frac{2y^{5}}{5}-
\frac{2y^{3}}{3}+\frac{4y}{15}\bigg]+\frac{m^{4}}{6}\bigg[\frac{y^{5}}{20}-
\frac{y^{3}}{6}+\frac{7y}{60}\bigg]
+\frac{m^{6}}{6}\bigg[\frac{y^{7}}{840}-
\frac{y^{5}}{120}+\frac{7y^{3}}{360}-\frac{31y}{2520}\bigg].
\end{equation}
(38)
It represents the velocity field for the MHD flow of a non-Newtonian
third grade fluid between two parallel plates. By taking $m=0$ in
Eq. (38) we recover the same solution as obtained in [
8] .
3.2. Fully developed plane Poiseuille Flow
For fully developed Poiseuille Flow, we consider the steady laminar
flow of third grade fluid between two stationary infinite parallel
plats under constant pressure gradient. Let the separation between
two plates is $2h$ and the plates are at $y=h$ and $y=-h.$ Thus,
the governing equation (10) in the presence of the constant
pressure gradient and transversal magnetic field takes the form
\begin{equation}\label{39}
\frac{d^{2} u}{d y^{2}}+\frac{6\beta}{\mu}\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d
y^{2}}^{2}-m^{2}u=\frac{1}{\mu}\frac{d\widehat{p}}{dx},
\end{equation}
(39)
with the boundary conditions
\begin{equation*}
u(y)=0\ \ \textrm{at}\ \ y=h,
\end{equation*}
\begin{equation}\label{40}
\ u(y)=0\ \ \textrm{at}\ \ y=-h,
\end{equation}
(40)
Introducing the non-dimensional parameters
\begin{equation}\label{41}
u^{*}=\frac{u}{U},\ \ y^{*}=\frac{y}{h},\ \ \beta^{*}=\frac{\beta
U^{2}}{\mu h^{2}},\ \ x^{*}=\frac{x}{h},\ \
p^{*}=\frac{\widehat{p}}{\mu U/h},\ \ {m^{*2}}=\frac{\sigma
B_{0}^{2}}{{\mu}/{h^{2}}},
\end{equation}
(41)
Eqs. (39) and (40) after dropping \(‘\ast’\) take the
following form:
\begin{equation}\label{42}
\frac{d^{2} u}{d y^{2}}+6\beta\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}-m^{2}u=\frac{dp}{dx},
\end{equation}
(42)
\begin{equation*}
u(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{43}
u(y)=0 \ \ \textrm{at} \ \ y=-1.
\end{equation}
(43)
Let us apply the ADM to solve the above boundary value problem.
Accordingly, in operator form Eq. (42) becomes
\begin{equation}
Lu=\frac{dp}{dx}-{6\beta}\bigg[\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}\bigg]+[m^{2}u],
\end{equation}
(44)
where \(L={d^{2}}/{dy^{2}}\), and \(Nu=({d u}/{d y})^{2}{d^{2}
u}/{d y^{2}}\) are linear and nonlinear terms respectively.
Applying the inverse operator \(\tilde{L}^{-1}=\int\int(\cdot) dydy\)
on both sides to equation (44), we get
\begin{equation}
\tilde{L}^{-1}Lu=\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg)-
{6\beta}\tilde{L}^{-1}\bigg[\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}\bigg]+\tilde{L}^{-1}[m^{2}u].
\end{equation}
(45)
In view of Eqs. (9) and (10), Eq. (45) becomes
\begin{equation}
\sum_{n=0}^{\infty}u_{n}(y)=Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg)-
{6\beta}\tilde{L}^{-1}\sum_{n=0}^{\infty}A_{n}+m^{2}\tilde{L}^{-1}
\sum_{n=0}^{\infty}u_{n}(y).
\end{equation}
(46)
From expression (46), we have the following recurrence relations
for zeroth and higher order solutions respectively
\begin{equation}\label{47}
u_{0}(y)=Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg),
\end{equation}
(46)
\begin{equation}\label{48}
u_{n+1}(y)=-{6\beta}\tilde{L}^{-1}[A_{n}]+m^{2}\tilde{L}^{-1}[u_{n}(y)],\
n\geq0,
\end{equation}
(46)
Using expression (9), the boundary conditions (43) becomes
\begin{equation*}
u_{n}(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{49}
u_{n}(y)=0\ \ \textrm{at}\ \ y=-1.
\end{equation}
(49)
From equation (47) and (49), we have the Zeroth-order
problem as
\begin{equation}\label{50}
u_{0}(y)=Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg),
\end{equation}
(50)
with the boundary conditions
\begin{equation*}
u_{0}(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{51}
u_{0}(y)=0\ \ \textrm{at}\ \ y=-1.
\end{equation}
(51)
The corresponding zeroth-order solution is given by
\begin{equation}\label{52}
u_{0}(y)=\frac{dp}{dx}\bigg[\frac{y^{2}}{2}-\frac{1}{2}\bigg].
\end{equation}
(51)
In view of Eqs. (25), (48) and (49), we get following
problems of different orders with corresponding boundary conditions
as: The first-order problem is
\begin{equation}\label{53}
u_{1}(y)=-6\beta
\tilde{L}^{-1}[A_{0}]+m^{2}\tilde{L}^{-1}[u_{0}(y)],
\end{equation}
(53)
subject to boundary conditions
\begin{equation*}
u_{1}(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{54}
u_{1}(y)=0\ \ \textrm{at}\ \ y=-1.
\end{equation}
(54)
The corresponding solution is
\begin{equation}\label{55}
u_{1}(y)={-2\beta}\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[\frac{y^{4}}{4}-
\frac{1}{4}\bigg]+
\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{2}}{2}\bigg(\frac{y^{4}}{12}-
\frac{y^{2}}{2}+\frac{5}{12}\bigg)\bigg].
\end{equation}
(55)
The second-order problem is
\begin{equation}\label{56}
u_{2}(y)=-{6\beta}\tilde{L}^{-1}[A_{1}]+m^{2}\tilde{L}^{-1}[u_{1}(y)],
\end{equation}
(56)
with the boundary conditions
\begin{equation*}
u_{2}(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{57}
u_{2}(y)=0\ \ \textrm{at}\ \ y=-1.
\end{equation}
(56)
The corresponding second-order solution is
\begin{equation*}
u_{2}(y)=3({-2\beta})^{2}\bigg(\frac{dp}{dx}\bigg)^{5}\bigg[\frac{y^{6}}{6}-\frac{1}{6}\bigg]+
({-2\beta})\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[m^{2}\bigg(\frac{11y^{6}}{120}
-\frac{3y^{4}}{8}-\frac{y^{2}}{8}+\frac{49}{120}\bigg)\bigg]
\end{equation*}
\begin{equation}\label{58}
+\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{4}}{2}\bigg(\frac{y^{6}}{360}-
\frac{y^{4}}{24}+\frac{5y^{2}}{24}-\frac{61}{360}\bigg)\bigg].
\end{equation}
(56)
The third-order problem is
\begin{equation}\label{59}
u_{3}(y)=-{6\beta}\tilde{L}^{-1}[A_{2}]+m^{2}\tilde{L}^{-1}[u_{2}(y)],
\end{equation}
(59)
with the boundary conditions
\begin{equation*}
u_{3}(y)=0\ \ \textrm{at}\ \ y=1,
\end{equation*}
\begin{equation}\label{60}
u_{3}(y)=0\ \ \textrm{at}\ \ y=-1.
\end{equation}
(59)
The corresponding third-order solution is given by
\begin{equation*}
u_{3}(y)=12({-2\beta})^{3}\bigg(\frac{dp}{dx}\bigg)^{7}
\bigg[\frac{y^{8}}{8}-\frac{1}{8}\bigg]+
3({-2\beta})^{2}\bigg(\frac{dp}{dx}\bigg)^{5}
\bigg[m^{2}\bigg(\frac{127y^{8}}{1120}-
\frac{25y^{6}}{60}-\frac{y^{4}}{16}
\end{equation*}
\begin{equation*}
-\frac{y^{2}}{12}+\frac{503}{1120}\bigg)\bigg]+({-2\beta})
\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[m^{4}\bigg
(\frac{17y^{8}}{1120}-\frac{11y^{6}}{80}+\frac{y^{4}}{3}+
\frac{49y^{2}}{240}-\frac{93}{224}\bigg)\bigg]
\end{equation*}
\begin{equation}
+\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{6}}{2}\bigg(\frac{y^{8}}{20160}
-\frac{y^{6}}{720}+\frac{5y^{4}}{288}-
\frac{61y^{2}}{720}+\frac{277}{4032}\bigg)\bigg].
\end{equation}
(61)
Inserting (52), (55), (58) and (61) in (9) we
obtain the fourth order approximate solution for a fully developed
plan poiseuille MHD flow of a non-Newtonian third grade fluid.
\begin{eqnarray*}
u(y)&=&\frac{dp}{dx}\bigg[\frac{y^{2}}{2}-\frac{1}{2}\bigg]
{-2\beta}\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[\frac{y^{4}}{4}-\frac{1}{4}\bigg]
+3({-2\beta})^{2}\bigg(\frac{dp}{dx}
\bigg)^{5}\bigg[\frac{y^{6}}{6}-\frac{1}{6}\bigg]\\
&&+12({-2\beta})^{3}
\bigg(\frac{dp}{dx}\bigg)^{7}
\end{eqnarray*}
\begin{equation*}
\bigg[\frac{y^{8}}{8}-\frac{1}{8}\bigg]+\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{2}}{2}\bigg(\frac{y^{4}}{12}-
\frac{y^{2}}{2}+\frac{5}{12}\bigg)\bigg]+
({-2\beta})\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[m^{2}\bigg(\frac{11y^{6}}{120}-\frac{3y^{4}}{8}
\end{equation*}
\begin{equation*}
-\frac{y^{2}}{8}+\frac{49}{120}\bigg)\bigg]+3({-2\beta})^{2}\bigg(\frac{dp}{dx}\bigg)^{5}\bigg[m^{2}\bigg(\frac{127y^{8}}{1120}-
\frac{25y^{6}}{60}
-\frac{y^{4}}{16}-\frac{y^{2}}{12}+\frac{503}{1120}\bigg)\bigg]
\end{equation*}
\begin{equation*}
+\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{4}}{2}\bigg(\frac{y^{6}}{360}-
\frac{y^{4}}{24}+\frac{5y^{2}}{24}-\frac{61}{360}\bigg)\bigg]+({-2\beta})\bigg(\frac{dp}{dx}\bigg)^{3}\bigg[m^{4}\bigg
(\frac{17y^{8}}{1120}-\frac{11y^{6}}{80}
\end{equation*}
\begin{equation}\label{62}
+\frac{y^{4}}{3}+
\frac{49y^{2}}{240}-\frac{93}{224}\bigg)\bigg]+\bigg(\frac{dp}{dx}\bigg)\bigg[\frac{m^{6}}{2}\bigg(\frac{y^{8}}{20160}-
\frac{y^{6}}{720}+\frac{5y^{4}}{288}-
\frac{61y^{2}}{720}+\frac{277}{4032}\bigg)\bigg].
\end{equation}
(61)
By taking \(m=0\) in the above expression we recover the same solution
as in [Eq. 2.2.52, 8]. When we take \(\beta=0\) and \(m=0\) we have the
exact solution for the viscous Newtonian fluid.
3.3. MHD Generalized-Couette Flow
We now again consider the steady laminar fully developed flow of a
third grade fluid between two infinite horizontal parallel plates at
a distance $h$ apart. The plates are non conducting and a
transversal magmatic field is applied in the vertical upward
direction. The motion of the fluid is due to the motion of upper
plate of uniform velocity $U$ and constant pressure gradient along
$x$ direction, while the lower plate is at rest. The resulting
differential equation (39) and the corresponding boundary
conditions for this flow are given as respectively
\begin{equation}\label{63}
\frac{d^{2} u}{d y^{2}}+\frac{6\beta}{\mu}\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d
y^{2}}^{2}-m^{2}u=\frac{1}{\mu}\frac{d\widehat{p}}{dx},
\end{equation}
(63)
\begin{equation*}
u(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{64}
u(y)=U\ \ \textrm{at}\ \ y=h.
\end{equation}
(64)
Introducing the non-dimensional parameters (41), Eqs. (63)
and (54) becomes
\begin{equation}\label{65}
\frac{d^{2} u}{d y^{2}}+{6\beta}\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}-m^{2}u=\frac{dp}{dx},
\end{equation}
(65)
\begin{equation*}
u(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{66}
u(y)=1\ \ \textrm{at}\ \ y=1.
\end{equation}
(66)
Follow the same procedure as in previous section, we define linear
operator \(L={d^{2}}/{dy^{2}}\) for Eq. (65) and apply the inverse
operator \(\tilde{L}^{-1}=\int\int(\cdot) dydy\) on both sides to
equation (55) we get
\begin{equation}\label{67}
u(y)=Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg)-
{6\beta}\tilde{L}^{-1}\bigg[\bigg(\frac{d u}{d
y}\bigg)^{2}\frac{d^{2} u}{d y^{2}}\bigg]+\tilde{L}^{-1}[m^{2}u],
\end{equation}
(67)
where A and B are constants of integration.
In view of decomposition series (9) and (10), Eq. (67)
gives
\begin{equation}
\sum_{n=0}^{\infty}u_{n}(y)=
Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg)-
{6\beta}\tilde{L}^{-1}\sum_{n=0}^{\infty}A_{n}
+m^{2}\tilde{L}^{-1}\sum_{n=0}^{\infty}u_{n}(y),
\end{equation}
(68)
From (68) one obtains the recurrence relations
\begin{equation}\label{69}
u_{0}(y)=Ay+B+\tilde{L}^{-1}\bigg(\frac{dp}{dx}\bigg),
\end{equation}
(69)
\begin{equation}
u_{n+1}(y)=-{6\beta}\tilde{L}^{-1}[A_{n}]+m^{2}\tilde{L}^{-1}[u_{n}(y)],\
n\geq0.
\end{equation}
(70)
Using decomposition series (9) into (66), we have the
following boundary conditions
\begin{equation*}
u_{0}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{71}
u_{0}(y)=1\ \ \textrm{at}\ \ y=1.
\end{equation}
(71)
and
\begin{equation*}
u_{n}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}
u_{n}(y)=0\ \ \textrm{at}\ \ y=1,\ \ n\geq1.
\end{equation}
(72)
From Eq. (69) and boundary conditions (71), we obtain the
zeroth-order solution as
\begin{equation}
u_{0}(y)=y+\alpha[y^{2}-y].
\end{equation}
(73)
In view of recurrence relations (70) along with the Adomian
polynomials (25), we get the following problems of different
orders with corresponding boundary conditions (72) as: The first-order problem is
\begin{equation}\label{74}
u_{1}(y)=-6\beta \tilde{L}^{-1}[A_{0}]+m^{2}\tilde{L}^{-1}[u_{0}],
\end{equation}
(74)
subject to the boundary conditions
\begin{equation*}
u_{1}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}
u_{1}(y)=-6\beta \tilde{L}^{-1}[A_{0}]+m^{2}\tilde{L}^{-1}[u_{0}],
\end{equation}
(75)
\begin{equation}\label{75}
u_{1}(y)=0\ \ \textrm{at}\ \ y=1.
\end{equation}
(75)
Solving (74) and (75), the first-order solution is given by
\begin{equation*}
u_{1}(y)=\frac{-\beta}{4\alpha}\bigg[1+\alpha(2y-1)\bigg]^{4}+{y}
\bigg[\frac{\beta}{4\alpha}(\bar{A}-\bar{B})\bigg]+
\bigg(\frac{\beta}{4\alpha}\bar{B}\bigg)
\end{equation*}
\begin{equation}\label{79}
+m^{2}\bigg[\alpha\bigg(\frac{y^{4}}{12}-\frac{y^{3}}{6}+\frac{y}{12}\bigg)
+\frac{y^{3}}{3}-\frac{y}{3}\bigg],
\end{equation}
(75)
The second-order problem is
\begin{equation}\label{80}
u_{2}(y)=-{6\beta}\tilde{L}^{-1}[A_{1}]+m^{2}\tilde{L}^{-1}[u_{1}(y)],
\end{equation}
(75)
subject to
\begin{equation*}
u_{2}(y)=0\ \ \textrm{at}\ \ y=0,
\end{equation*}
\begin{equation}\label{81}
u_{2}(y)=0\ \ \textrm{at}\ \ y=1.
\end{equation}
(75)
The second-order solution is
\begin{equation*}
u_{2}(y)=\frac{\beta^{2}}{\alpha}[1+\alpha(2y-1)]^{6}
-\bigg(\frac{\beta}{2\alpha}\bigg)^{2}(\bar{A}-
\bar{B})[1+\alpha(2y-1)]^{3}-
\bigg(\frac{{m^{2}\beta}}{480\alpha^{3}}\bigg)
\end{equation*}
\begin{equation*}
[1+\alpha(2y-1)]^{6}+
\bigg(\frac{m^{2}\beta}{24\alpha}\bigg)[(\bar{A}-\bar{B})y^{3}]+
\bigg(\frac{m^{2}\beta}{8\alpha}\bigg)[\bar{B}y^{2}]+
6m^{2}({-\beta})\bigg[\frac{y^{3}}{6}
\end{equation*}
\begin{equation*}
+\frac{5\alpha{y^{4}}}{12}-\frac{\alpha
{y^{3}}}{2}+\frac{2{\alpha^{2}y^{5}}}{5}+
\frac{2{\alpha^{3}y^{6}}}{15}-\frac{2{\alpha^{3}y^{5}}}{5}+\frac{\alpha^{2}
y^{3}}{2}+\frac{5{\alpha^{3}y^{4}}}{12}-\frac{\alpha^{3}
y^{3}}{6}-\frac{5{\alpha^{2}y^{4}}}{6}\bigg]
\end{equation*}
\begin{equation*}
+24m^{2}({-\alpha\beta})\bigg[\frac{y^{4}}{24}+\frac{\alpha
y^{5}}{15}-\frac{\alpha y^{4}}{12}-\frac{y^{2}}{12}+\frac{\alpha
y^{2}}{8}+\frac{\alpha^{2}y^{6}}{45}-\frac{\alpha^{2}y^{5}}{15}-\frac{\alpha
y^{3}}{18}+\frac{\alpha^{2}y^{3}}{36}
\end{equation*}
\begin{equation*}
+\frac{\alpha^{2}y^{4}}{24}
-\frac{\alpha^{2}y^{2}}{24}\bigg]+m^{4}\bigg[\frac{y^{5}}{120}+\frac{\alpha
y^{6}}{360}-\frac{\alpha y^{5}}{120}-\frac{y^{3}}{36}+\frac{\alpha
y^{3}}{72}\bigg]+\frac{\beta^{2}}{\alpha}{y}[(1-\alpha )^{6}-(1
\end{equation*}
\begin{equation*}
+\alpha)^{6}]+\bigg(\frac{\beta}{2\alpha}\bigg)^{2}{y}
(\bar{A}-\bar{B})[(1+\alpha)^{3}-(1-\alpha)^{3}]
+\bigg(\frac{m^{2}\beta}{480\alpha^{3}}\bigg){y}[(1+\alpha)^{6}
\end{equation*}
\begin{equation*}
-(1-\alpha)^{6}]-6m^{2}({-\beta}){y}\bigg[\frac{1}{6}-\frac{\alpha}
{12}+\frac{9{\alpha^{2}}}{10}-\frac{5{\alpha^{2}}}{6}-\frac{\alpha^{3}}{60}\bigg]
-24m^{2}({-\alpha\beta})
\end{equation*}
\begin{equation*}
{y}\bigg[\frac{-1}{24}+\frac{19\alpha
}{360}-\frac{\alpha^{2}}{60}\bigg]-m^{4}{y}\bigg[\frac{-7}{360}+\frac{\alpha}{120}\bigg]
-\bigg(\frac{m^{2}\beta}{8\alpha}\bigg)[\bar{B}y]
-\bigg(\frac{m^{2}\beta}{24\alpha}\bigg)y(\bar{A}-\bar{B})
\end{equation*}
\begin{equation}
-\bigg(\frac{\beta^{2}}{\alpha}\bigg)
[(1-\alpha)^{6}]+\bigg(\frac{\beta}{2\alpha}\bigg)^{2}(\bar{A}-\bar{B})
(1-\alpha)^{3}
+\bigg(\frac{{m^{2}}\beta}{480\alpha^{3}}\bigg)(1-\alpha )^{6}.
\end{equation}
(74)
Summarizing these solutions, we get final form of solution as
\begin{equation*}
u(y)=y+\alpha[y^{2}-y]+
\frac{-\beta}{4\alpha}[1+\alpha(2y-1)]^{4}+{y}
\bigg[\frac{\beta}{4\alpha}(\bar{A}-\bar{B})\bigg]+
\bigg(\frac{\beta}{4\alpha}\bar{B}\bigg)
\end{equation*}
\begin{equation*}
+m^{2}\bigg[\alpha\bigg(\frac{y^{4}}{12}-\frac{y^{3}}{6}+\frac{y}{12}\bigg)+\frac{y^{3}}{3}-\frac{y}{3}\bigg]
+\frac{{\beta}^{2}}{\alpha}[1+\alpha(2y-1)]^{6}
\end{equation*}
\begin{equation*}
-\bigg(\frac{\beta}{2\alpha}\bigg)^{2}(\bar{A}-\bar{B})[1+\alpha(2y-1)]^{3}
– \bigg(\frac{m^{2}\beta}{480\alpha^{3}}\bigg)[1+\alpha(2y-1)]^{6}
+\bigg(\frac{m^{2}\beta}{24\alpha}\bigg)[(\bar{A}-\bar{B})y^{3}]
\end{equation*}
\begin{equation*}
+\bigg(\frac{m^{2}\beta}{8\alpha}\bigg)[\bar{B}y^{2}]+
6m^{2}({-\beta})
\end{equation*}
\begin{equation*}
\bigg[\frac{y^{3}}{6}+\frac{5\alpha{y^{4}}}{12}-\frac{\alpha
y^{3}}{2}+\frac{2{\alpha^{2}y^{5}}}{5}+\frac{2{\alpha^{3}y^{6}}}{15}-\frac{2{\alpha^{3}y^{5}}}{5}+\frac{\alpha^{2}
y^{3}}{2}+\frac{5{\alpha^{3}y^{4}}}{12}-\frac{\alpha^{3}
y^{3}}{6}-\frac{5{\alpha^{2}y^{4}}}{6}\bigg]
\end{equation*}
\begin{equation*}
+24m^{2}({-\alpha\beta})
\end{equation*}
\begin{equation*}
\bigg[\frac{y^{4}}{24}+\frac{\alpha y^{5}}{15}-\frac{\alpha
y^{4}}{12}-\frac{y^{2}}{12}+\frac{\alpha
y^{2}}{8}+\frac{\alpha^{2}y^{6}}{45}-\frac{\alpha^{2}y^{5}}{15}-\frac{\alpha
y^{3}}{18}+\frac{\alpha^{2}y^{3}}{36}+\frac{\alpha^{2}y^{4}}{24}
-\frac{\alpha^{2}y^{2}}{24}\bigg]
\end{equation*}
\begin{equation*}
+m^{4}\bigg[\frac{y^{5}}{120}+\frac{\alpha
y^{6}}{360}-\frac{\alpha y^{5}}{120}-\frac{y^{3}}{36}+\frac{\alpha
y^{3}}{72}\bigg]
\end{equation*}
\begin{equation*}
+ \frac{\beta^{2}}{\alpha}{y}[(1-\alpha
)^{6}-(1+\alpha)^{6}]+\bigg(\frac{\beta}{2\alpha}\bigg)^{2}{y}
(\bar{A}-\bar{B})[(1+\alpha)^{3}-(1-\alpha )^{3}]
\end{equation*}
\begin{equation*}
+\bigg(\frac{{m^{2}\beta}}{480\alpha^{3}}\bigg)
{y}[(1+\alpha)^{6}-(1-\alpha)^{6}]-6m^{2}({-\beta}){y}\bigg[\frac{1}{6}-\frac{\alpha
}{12}+\frac{9{\alpha^{2}}}{10}-\frac{5{\alpha^{2}}}{6}-\frac{\alpha^{3}}{60}\bigg]-24m^{2}({-\alpha\beta})
\end{equation*}
\begin{equation*}
{y}\bigg[\frac{-1}{24}+\frac{19\alpha
}{360}-\frac{\alpha^{2}}{60}\bigg]-m^{4}{y}\bigg[\frac{-7}{360}+\frac{\alpha
}{120}\bigg]-\bigg(\frac{m^{2}\beta}{8\alpha}\bigg)[\bar{B}y]-
\bigg(\frac{m^{2}\beta}{24\alpha}\bigg)y(\bar{A}-\bar{B})
\end{equation*}
\begin{equation}
-\frac{{\beta}^{2}}{\alpha}(1-\alpha)^{6}+\bigg(\frac{\beta}{2\alpha}\bigg)^{2}(\bar{A}-\bar{B})
(1-\alpha)^{3}
+\bigg(\frac{m^{2}\beta}{480\alpha^{3}}\bigg)(1-\alpha )^{6}.
\end{equation}
(74)
where,
$\alpha=\bigg(\frac{1}{2}\frac{dp}{dx}\bigg)$, and
$\bar{A}=\bigg(1+\frac{1}{2}\frac{dp}{dx}\bigg)^{4},$
$\bar{B}=\bigg(1-\frac{1}{2}\frac{dp}{dx}\bigg)^{4}.$
By taking $m=0$ in the above expression we recover the same solution
as obtained in [Eq. 2.2.52, 8]. When we take $\beta=0$ and $m=0$ we
have the exact solution for the viscous Newtonian fluid.
4. Conclusion
This work is mainly concerned with the flows of third grade fluid in Cartesian and Cylindrical frame. We have discussed the velocity field corresponding to the steady, incompressible unidirectional flow of the third grade fluid between to infinite parallel plates under the influence of MHD force term have been determined by mean of Adomain decomposition method.
After lengthy computations the solution obtained in series form satisfy all the imposed initial and boundary conditions. Graphically, results shown that by increasing the strength of magnitude field. Fluid velocity gradually decreases and by varying values of parameters velocity also varies. By taking $m=0$, we get the similar results as obtained in [8].
