3.1. Unsteady Flow of Third-Grade Fluid over a Flat Rigid Plate with Porous Medium
By following the methodology of References. [20, 21], the equation of motion for the unsteady flow of third-grade fluid over the rigid plate with porous medium is
(1)
ρ
∂
U
∂
t
=
μ
∂
2
U
∂
y
2
+
α
1
∂
3
U
∂
y
2
∂
t
+
6
β
3
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
ϕ
κ
[
μ
+
α
1
∂
∂
t
+
2
β
3
(
∂
U
∂
y
)
2
]
U
.
Here
U
is the velocity component,
ρ
is the density,
μ
the coefficient of viscosity,
α
1
and
β
3
are the material constants (for details on these material constants and the conditions that are satisfied by these constants, the reader is referred to [8]),
ϕ
the porosity and
κ
the permeability of the porous medium.
In order to solve (1) mentioned above, boundary conditions are specified as follows:
(2)
U
(
0
,
t
)
=
U
0
V
(
t
)
,
t
>
0
,
(3)
U
(
∞
,
t
)
=
0
,
t
>
0
,
(4)
U
(
y
,
0
)
=
0
,
y
>
0
,
where
U
0
is the reference velocity. The first boundary condition (2) is the no-slip condition and the second boundary condition (3) says that the main stream velocity is zero. This is not a restrictive assumption since we can always measure velocity relative to the main stream. The initial condition (4) indicates that the fluid is initially at rest.
On introducing the nondimensional quantities
(5)
U
-
=
U
U
0
,
y
-
=
U
0
y
ν
,
t
-
=
U
0
2
t
ν
,
α
-
=
α
1
U
0
2
ρ
ν
2
,
β
-
=
2
β
3
U
0
4
ρ
ν
3
,
1
K
-
=
ϕ
ν
2
κ
U
0
2
.
Equation (1) and the corresponding initial and the boundary conditions take the form
(6)
∂
U
∂
t
=
∂
2
U
∂
y
2
+
α
∂
3
U
∂
y
2
∂
t
+
3
β
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
1
K
[
U
+
α
∂
U
∂
t
+
β
U
(
∂
U
∂
y
)
2
]
,
(7)
U
(
0
,
t
)
=
V
(
t
)
,
t
>
0
,
(8)
U
(
y
,
t
)
⟶
0
as
y
⟶
∞
,
t
>
0
,
(9)
U
(
y
,
0
)
=
0
,
y
>
0
.
For simplicity, we ignore the bars of the nondimensional quantities. We rewrite (6) as
(10)
∂
U
∂
t
=
μ
*
∂
2
U
∂
y
2
+
α
*
∂
3
U
∂
y
2
∂
t
+
3
γ
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
γ
*
U
(
∂
U
∂
y
)
2
-
1
K
*
U
,
where
(11)
μ
*
=
1
(
1
+
α
/
K
)
,
α
*
=
α
(
1
+
α
/
K
)
,
γ
=
β
(
1
+
α
/
K
)
,
γ
*
=
β
/
K
(
1
+
α
/
K
)
,
1
K
*
=
1
/
K
(
1
+
α
/
K
)
.
We know that from the principal of Lie group theory that if a differential equation is explicitly independent of any dependent or independent variable, then this particular differential equation remains invariant under the translation symmetry corresponding to that particular variable. We notice that (10) is independent of
t
, so it is invariant under the Lie point symmetry generator
X
=
∂
/
∂
t
. So, by the theory of invariants, the solution of (10) corresponding to operator
X
is obtained by
(12)
d
t
1
=
d
y
0
=
d
U
0
,
implyingthe steady-state solution given by
(13)
U
=
F
(
η
)
,
where
η
=
y
.
Inserting (13) into (10) yields
(14)
μ
*
∂
2
F
∂
η
2
+
3
γ
(
∂
F
∂
η
)
2
∂
2
F
∂
η
2
-
γ
*
F
(
∂
F
∂
η
)
2
-
1
K
*
F
=
0
,
and the transformed boundary conditions are given by
(15)
F
(
0
)
=
l
1
,
where
l
1
is a constant
F
(
η
)
⟶
0
,
as
y
⟶
∞
.
We have now transformed the governing nonlinear PDE (10) into a nonlinear ODE (14) as well as the boundary conditions (7)–(9) to the boundary conditions (15).
In order to solve (14) for
F
(
η
)
, we assume a solution of the form
(16)
F
(
η
)
=
A
exp
(
B
η
)
,
where
A
and
B
are the constants to be determined. Substituting (16) into (14), we obtain
(17)
(
μ
*
B
2
-
1
K
*
)
+
e
2
B
η
(
3
γ
A
2
B
4
-
γ
*
A
2
B
2
)
=
0
.
Separating (17) in the powers of
e
0
and
e
2
B
η
, we deduce
(18)
e
0
:
μ
*
B
2
-
1
K
*
=
0
,
(19)
e
2
B
η
:
3
γ
A
2
B
4
-
γ
*
A
2
B
2
=
0
.
From (19), we find
(20)
B
=
±
γ
*
3
γ
.
We choose
(21)
B
=
-
γ
*
3
γ
,
so that our solution would satisfy the second boundary condition at infinity. Using the value of
B
in (18), we obtain
(22)
γ
*
3
γ
μ
*
-
1
K
*
=
0
.
Thus, the solution for
F
(
η
)
can be written as
(23)
U
=
F
(
η
)
=
l
1
exp
(
-
γ
*
3
γ
η
)
,
provided that condition (22) holds, where
A
=
l
1
by using the first boundary condition.
3.2. Unsteady Flow of Third-Grade Fluid over a Flat Porous Plate with Porous Medium
By employing the same geometry as we have explained in Section 2, we extend the previous problem by incorporating the effects of plate suction or injection. We provide the closed-form solution of the problem by reducing the governing nonlinear PDE into an ODE with the aid of Lie groups.
The present analysis deals with a porous plate with suction or injection and thus the velocity field is given by
(24)
V
=
[
U
(
y
,
t
)
,
-
W
0
,
0
]
,
where
W
0
>
0
denotes the suction velocity and
W
0
<
0
indicates blowing velocity. One can see that the incompressibility constraint is identically satisfied; that is,
(25)
div
V
=
0
.
So, the unsteady flow through a porous medium and over a porous plate in the absence of the modified pressure gradient takes the form
(26)
ρ
[
∂
U
∂
t
-
W
0
∂
U
∂
y
]
=
μ
∂
2
U
∂
y
2
+
α
1
∂
3
U
∂
y
2
∂
t
-
α
1
W
0
∂
3
U
∂
y
3
+
6
β
3
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
ϕ
κ
×
[
μ
+
α
1
∂
∂
t
-
α
1
W
0
∂
∂
y
+
2
β
3
(
∂
U
∂
y
)
2
]
U
.
The boundary conditions remain the same as given in (2)–(4). Defining the nondimensional parameters as
(27)
U
-
=
U
U
0
,
y
-
=
U
0
y
ν
,
t
-
=
U
0
2
t
ν
,
α
-
=
α
1
U
0
2
ρ
ν
2
,
β
-
=
2
β
3
U
0
4
ρ
ν
3
,
1
K
-
=
ϕ
ν
2
κ
U
0
2
,
W
-
=
W
0
U
0
.
Equation (28) becomes
(28)
[
∂
U
∂
t
-
W
∂
U
∂
y
]
=
∂
2
U
∂
y
2
+
α
∂
3
U
∂
y
2
∂
t
-
α
W
∂
3
U
∂
y
3
+
3
β
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
1
K
×
[
U
+
α
∂
U
∂
t
-
α
W
∂
U
∂
y
+
β
U
(
∂
U
∂
y
)
2
]
,
with the boundary conditions taking the form as given in (7)–(9). We rewrite (28) as
(29)
∂
U
∂
t
=
μ
*
∂
2
U
∂
y
2
+
α
*
∂
3
U
∂
y
2
∂
t
-
α
*
W
∂
3
U
∂
y
3
+
3
γ
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
γ
*
U
(
∂
U
∂
y
)
2
+
W
∂
U
∂
y
-
1
K
*
U
,
where
μ
*
,
α
*
,
γ
*
,
γ
, and
1
/
K
*
are defined in (11). Now, we have to solve (29) subject to the boundary conditions (7)–(9).
As it can be seen, (29) is invariant under the time-translation symmetry generator
X
=
∂
/
∂
t
. The invariant solution corresponding to
∂
/
∂
t
is the steady-state solution given by
(30)
U
=
G
(
η
)
,
where
η
=
y
.
Using (30) into (29) yields
(31)
μ
*
d
2
G
d
η
2
-
α
*
W
d
3
G
d
η
3
+
3
γ
(
d
G
d
η
)
2
d
2
G
d
η
2
-
γ
*
G
(
d
G
d
η
)
2
+
W
d
G
d
η
-
1
K
*
G
=
0
,
with the transformed boundary conditions
(32)
G
(
0
)
=
l
2
,
where
l
2
is a constant,
G
(
η
)
⟶
0
,
d
G
d
η
⟶
0
as
y
⟶
∞
.
Following the same methodology as adopted in solving the previous problem, (31) admits an exact solution of the form (which we require to be zero at infinity due to the condition
G
(
∞
)
=
0
)
(33)
U
=
G
(
η
)
=
l
2
exp
(
-
γ
*
3
γ
η
)
,
provided that
(34)
γ
*
3
γ
μ
*
+
α
W
γ
*
3
γ
γ
*
3
γ
-
γ
*
3
γ
W
-
1
K
*
=
0
.
Note that, if we set
W
=
0
, we can recover the condition given in (22).
3.3. Unsteady Magnetohydrodynamic (MHD) Flow of Third-Grade Fluid over a Flat Porous Plate with Porous Medium
In this problem, we extend the previous one by considering the fluid to be electrically conducting under the influence of a uniform magnetic field applied transversely to the flow.
The unsteady MHD flow of a third-grade fluid in a porous half space with plate suction/injection is governed by
(35)
ρ
[
∂
U
∂
t
-
W
0
∂
U
∂
y
]
=
μ
∂
2
U
∂
y
2
+
α
1
∂
3
U
∂
y
2
∂
t
-
α
1
W
0
∂
3
U
∂
y
3
+
6
β
3
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
ϕ
κ
×
[
μ
+
α
1
∂
∂
t
-
α
1
W
0
∂
∂
y
+
2
β
3
(
∂
U
∂
y
)
2
]
×
U
-
σ
B
0
2
U
,
where
σ
is the electrical conductivity and
B
0
the uniform applied magnetic field. Again the boundary conditions remain the same as given in (2)–(4). We define the dimensionless parameters as
(36)
U
-
=
U
U
0
,
y
-
=
U
0
y
ν
,
t
-
=
U
0
2
t
ν
,
α
-
=
α
1
U
0
2
ρ
ν
2
,
β
-
=
2
β
3
U
0
4
ρ
ν
3
,
1
K
-
=
ϕ
ν
2
κ
U
0
2
,
M
-
2
=
σ
B
0
2
ν
ρ
U
0
2
,
W
-
=
W
0
U
0
.
Equation (26) takes the form
(37)
[
∂
U
∂
t
-
W
∂
U
∂
y
]
=
∂
2
U
∂
y
2
+
α
∂
3
U
∂
y
2
∂
t
-
α
W
∂
3
U
∂
y
3
+
3
β
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
1
K
[
U
+
α
∂
U
∂
t
-
α
W
∂
U
∂
y
+
β
U
(
∂
U
∂
y
)
2
]
-
M
2
U
.
We rewrite (37) as
(38)
∂
U
∂
t
=
μ
*
∂
2
U
∂
y
2
+
α
*
∂
3
U
∂
y
2
∂
t
-
α
*
W
∂
3
U
∂
y
3
+
3
γ
(
∂
U
∂
y
)
2
∂
2
U
∂
y
2
-
γ
*
U
(
∂
U
∂
y
)
2
+
W
∂
U
∂
y
-
(
1
K
*
+
M
*
2
)
U
,
where
μ
*
,
α
*
,
γ
*
,
γ
, and
1
/
K
*
are defined in (11) and
(39)
M
*
2
=
M
2
(
1
+
α
/
K
)
.
Now, we need to solve (38) subject to the boundary conditions (7)–(9). Since (38) is invariant under the time-translation symmetry generator
X
=
∂
/
∂
t
, the invariant solution corresponding to
∂
/
∂
t
is the steady-state solution. Consider
(40)
U
=
H
(
η
)
,
where
η
=
y
.
Invoking (40) in (38) yields
(41)
μ
*
d
2
H
d
η
2
-
α
*
W
d
3
H
d
η
3
+
3
γ
(
d
H
d
η
)
2
d
2
H
d
η
2
-
γ
*
H
(
d
H
d
η
)
2
+
W
d
H
d
η
-
(
1
K
*
+
M
*
2
)
H
=
0
,
with the transformed boundary condition
(42)
H
(
0
)
=
l
3
,
where
l
3
is a constant
,
H
(
η
)
⟶
0
,
d
H
d
η
⟶
0
as
y
⟶
∞
.
Utilizing the same method adopted to solve the first problem, (41) admits an exact solution of the form (which we require to be zero at infinity due to the second boundary condition)
(43)
U
=
H
(
η
)
=
l
3
exp
(
-
γ
*
3
γ
η
)
,
provided that
(44)
γ
*
3
γ
μ
*
+
α
W
γ
*
3
γ
γ
*
3
γ
-
γ
*
3
γ
W
-
1
K
*
-
M
*
2
=
0
.
Note that, if we set
W
=
M
*
=
0
, we can recover the previous two conditions given in (22) and (34).
Remark 1.
We note that the similarity solutions (23), (33), and (43) are the same, but the imposing conditions (22), (34), and (44) under which these solutions are valid are different. These solutions do show the effects of the third-grade fluid parameters
γ
and
γ
*
on the flow. However, they do not directly contain the term which is responsible for showing the effects of suction/blowing and magnetic field on the flow. The imposing conditions do contain the magnetic field and suction/blowing parameters. Thus, these closed-form solutions are valid for the particular values of the suction/blowing and the magnetic field parameters. To show the effects of these interesting phenomena, we have solved the reduced (14), (31), and (41) with the boundary conditions (15), (32), and (42) numerically using the powerful numerical solver NDSolve in Mathematica. These numerical solutions are plotted in Figures 2–4 against the emerging parameters of the flow.