International Journal of Analysis and Applications

Volume 18, Number 5 (2020), 738-747

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-18-2020-738

NEW APPROACH OF MHD BOUNDARY LAYER FLOW TOWARDS A POROUS

STRETCHING SHEET VIA SYMMETRY ANALYSIS AND THE GENERALIZED

EXP-FUNCTION METHOD

A.A. GABER1,2∗, M.H. SHEHATA1

1Department of Mathematics, College of Science, and Human Studies at Hotat Sudair, Majmaah

University, Saudi Arabia

2Department of Mathematics, Faculty of Education, Ain Shams University, Roxy, Hiliopolis, Cairo, Egypt

∗Corresponding author: aagaber6@gmail.com, a.gaber@mu.edu.sa

Abstract. Due to importance of the slip effect on modeling the boundary layer flows, symmetries and

exact solution investigations have been introduced in this paper for studying the effect of a slip boundary

layer on the stretching sheet through a porous medium. The exact solution of the investigating model

is obtained in term of exponential via the generalized Exp-Function method. This solution satisfies the

boundary conditions. Finally, the effect of parameters on the velocity field is studied.

1. Introduction

Symmetry group analysis based on the transformation groups, now known as Lie groups, is the most

important solution method for the nonlinear problems in the literature. This approach is used to analysis

the symmetries of the differential equations. Then, the corresponding symmetry groups can be used to

simplify the analysis of the problems governing by the differential equations in the engineering science,

mathematical physics, and mechanics. Lie groups characterize the symmetry of the differential equations

and may be a point, a contact, and a potential or a nonlocal symmetry. It has also been verified that these

kinds of groups can be represented by their infinitesimals that contain dependent variables, independent

Received March 8th, 2020; accepted May 12th, 2020; published June 24th, 2020.

1991 Mathematics Subject Classification. 35Q35, 34B15.

Key words and phrases. boundary layer; exact solutions; generalized exp-function method.

©2020 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

738

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-738


Int. J. Anal. Appl. 18 (5) (2020) 739

variables and the derivatives of dependent variables as arguments. In the last century, the application of the

Lie groups has been developed by a number of mathematicians. Ovsiannikov [1], Olver [2], Ibragimov [3],

and Bluman and Kumei [4] are some of the mathematicians who have huge number of studies in that field

[5-9].

The boundary layer [10-13] equations are especially interesting from a physical point of view because they

have the capacity to admit a large number of invariant solutions i.e. basically closed-form solutions. In the

present context, invariant solutions are meant to be a reduction to a simpler equation such as an ordinary

differential equation (ODE). Prandtl’s boundary layer equations admit more and different symmetry groups.

Symmetry groups or simply symmetries are invariant transformations which do not alter the structural form

of the equation under investigation (Bluman and Kumei [1]).

This work is organized as follows. The problem is formulated in Section 2 and in Section 3 we calculate

the symmetries of the thermal boundary layer equations. All invariant solutions of the thermal boundary

layer equations in Section 4. Finally, we show the effect of parameters on the velocity field.

2. Formulation of the problem

We consider the steady state 2D magnetohydrodynamic (MHN) boundary layer, incompressible and vis-

cous flow on stretching sheet through a porous medium, where M is the magnetic parameter, kp is the

permeability parameter and fw is the mass transfer parameter, which is positive for suction and negative for

injection.

∂u

∂x
+
∂υ

∂y
= 0,

u
∂u

∂x
+ υ

∂u

∂y
= v

∂2u

∂y2
−

v

k0
u−

α0B
2
0

ρ
u. (1)

In (1) u and υ are the components of velocity respectively in the x and y directions, k0 is the permeability

of the porous medium, B0 is magnetic field of uniform strength and σ0 is electrical conductivity, v =
µ
ρ

is the

kinematic viscosity, µ is the coefficient of fluid viscosity and ρ is the fluid density. By using the boundary

layer approximations and neglecting viscous dissipation.

The appropriate boundary conditions for the problem are given by

u = Bx, υ = υw at y = 0,

u −→ 0 at y = ∞. (2)

where B is the stretching rate, υw is the wall velocity and the velocity components along x, y coordinates,

respectively, are

u =
∂ψ

∂y
and υ =

−∂ψ
∂x

(3)



Int. J. Anal. Appl. 18 (5) (2020) 740

where ψ is the stream function.

Using the relations (3) in the boundary layer (2) and in the energy (1) we get the following equations

∂ψ

∂y

∂2ψ

∂y∂x
−
∂ψ

∂x

∂2ψ

∂y2
= v

∂3ψ

∂y3
−
v

k

∂ψ

∂y
−
α0B

2
0

ρ

∂ψ

∂y
, (4)

The boundary conditions (2) then become

∂ψ

∂y
= Bx,

∂ψ

∂x
= υw at y = 0,

∂ψ

∂y
−→ 0 at y = ∞. (5)

3. Symmetry analysis for the boundary layer equations

Firstly, we shall derive the similarity solutions using the Lie-group method [11] under which (1) is invariant.

Consider the one-parameter (ε) Lie group of infinitesimal transformations in (x, y, ψ) given by Lie point

symmetries

x∗ = x∗(x,t,ψ; ε), y∗ = y∗(x,y,ψ; ε)

ψ∗ = ψ∗(x,y,ψ; ε). (6)

With associated infinitesimal form

x∗ = x + εη(x,y,ψ; ε) + o(ε2),

y∗ = y + εζ(x,y,ψ; ε) + o(ε2),

ψ∗ = ψ + εψ(x,y,ψ; ε) + o(ε2), (7)

where “ε” is a small parameter.

If we set:

∆1 =
∂ψ

∂y

∂2ψ

∂y∂x
−
∂ψ

∂x

∂2ψ

∂y2
−v

∂3ψ

∂y3
+ (

v

k
+
α0B

2
0

ρ
)
∂ψ

∂y
, (8)

The invariance conditions[1-4]

Γ(3)(∆α) = 0 whenever ∆α = 0, α = 1, 2, (9)

where Γ(3) is given by

Γ(3) = χ + gx
∂

∂ψx
+ gxx

∂

∂ψxx
+ gxt

∂

∂ψxt
+ gxxx

∂

∂ψxxx
. (10)

where

χ = ζ
∂

∂x
+ τ

∂

∂y
+ g

∂

∂ψ
(11)



Int. J. Anal. Appl. 18 (5) (2020) 741

The components ζx, ζy, τx, τy, gx, gxx, gxxy....can be determined from the following expressions:

gs = Dsg −ψtDsζ −ψxDsτ, g
sj = Djg −ψtsDjζ −ψxsDjτ (12)

Equation (9) gives the following system of linear partial differential equations:

ζy = 0, ζψ = 0,

gy = 0, gx = 0,

τy = 0, τψ = 0,

gψψ = 0, ζx −gψ = 0, (13)

Solving the system (13), after substitution from (12) into(13), and using the invariance of the boundary

conditions (6), yields

ζ = λ1x + λ2

τ = λ3(x)

g = λ1ψ + λ4 (14)

In order to study the group theoretic structure, the vector field operator V is written as

V = V1(λ1) + V2(λ2) + V3(λ3) + V4(λ4), (15)

where

V1 = x
∂

∂x
+ ψ

∂

∂ψ
, V2 =

∂

∂x
,

V3 = λ3(x)
∂

∂y
, V4 =

∂

∂ψ
. (16)

It is easy to verify, that the vector fields are closed under the Lie bracket as follows

[V1,V1] = [V2,V2] = [V3,V3] = [V4,V4] = [V1,V3] = 0

[V2,V3] = [V2,V4] = 0, [V3,V1] = [V3,V2] = [V3,V4] = 0

[V4,V2] = [V4,V3] = 0, [V1,V2] = −[V2,V1] = −V2

[V1,V4] = −[V4,V1] = −V4

Further, from the symmetries given in (16) the following possibilities exist for the solution of (9).

(I)V1

(II)V2 + V3

(III)V2 + V3 + V4



Int. J. Anal. Appl. 18 (5) (2020) 742

Having determined the infinitessimals, the symmetry variables are found by solving the auxiliary equation

dx

ζ
=
dy

τ
=
dψ

g
. (17)

4. Reductions and exact solutions

Now we look the similarity solutions with respect to the generators V1

η∗ = y, ψ = xF(η∗), (18)

The reduced system of ODEs is

F 82 −FF ′′ −vF ′′′ + (
v

k
+
α0B

2
0

ρ
)F ′ = 0, (19)

The boundary condition take the following forms

F ′ = B, F = υw at η
∗ = 0

F ′ = 0 at η∗ →∞. (20)

We look for a similarity solution of (19) ,and boundary condition (20) as the following form:

F =
√
Bv f(η) and η =

√
B

v
η∗ (21)

Using (21) we obtain the following self-similar equations

f′′′ −f82 + ff′′ − (kp + M)f′ = 0, (22)

subject to the boundary conditions

f(0) = fw, f
′(0) = 1 f′(∞) = 0 (23)

where M =
α0B

2
0

ρB
is the magnetic field, kp =

v
k0B

is the permeability of the porous medium and fw =
υw√
Bv

where fw > 0 corresponds to suction and fw < 0 for injection. Equation (22) is nonlinear differential equation

which can be solved by the generalized He’s Exp-Function method. In view of the generalized Exp-Function

method [14-16], we assume that the solution of (22) can be expressed in the form

f(η) =
a−c[φ(η)

−c] + ... + ap[φ(η)
p]

r−d[φ(η)−d] + ... + rq[φ(τ)q]
, (24)

where c, d, p and q are positive integers which are unknown to be further determined, an and rm are unknown

constants. In addition, φ(η) satisfies Riccati equation,

φ′(η) = A + Bφ(η) + Cφ(η)2. (25)



Int. J. Anal. Appl. 18 (5) (2020) 743

In order to determine values of c and p, we balance the linear term of the highest order in Eq. (24) with

the highest order nonlinear term f′′′ and f82, we have

f′′′(η) =
a1φ
−c−8d−3 + ... + a2φ

p+8q+3

r1φ
−9d + ... + r2φ

9q
, (26)

f82(η) =
a3φ
−2c−6d−2 + ... + a4φ

2p+6q+2

r3φ
−9d + ... + r4φ

9q
, (27)

where ai and ri are determined coefficients only for simplicity. From balancing the lowest order and highest

order of φ (26) and (27), we obtain −7d − c − 3 = −6d − 2c − 2, which leads to the limit c = d + 1,and

7q + p + 3 = 6q + 2p + 2,which leads to the limit p = q + 1, for simplicity d = q = 0, the function in Eq. (24),

becomes

f(η) = γ−1φ
−1 + γ0 + γ1φ (28)

Substituting (28) into (22), equating to zero the coefficients of all powers of φ(η) yields a set of algebraic

equations for γ0, γ1 and γ−1, we obtain the following system

γ21BC + 12γ1BC
2 + 2γ0γ1C

2 = 0,

− 6γ−1A
3 + γ2−1A

2, γ21C
2 + 6γ1C

3 = 0,

− 12γ−1A
2B + γ2−1AB + 2γ0γ−1A

2 = 0,

− 6γ−1A
3 + γ2−1A

2 = 0,

γ21BC + 12γ1BC
2 + 2γ0γ1C

2 = 0,

− 12γ−1A
2B + γ2−1AB + 2γ0γ−1A

2 −Mγ1B + 5γ−1γ1BC − (2(−γ−1C + γ1A))γ1B

+ γ1(8ABC + B
3) + γ1(γ−1BC + γ1AB) + γ0γ1(2AC + B

2) = 0,

− (2(−γ−1C + γ1A))γ1C −γ
2
1B

2 + 2γ−1γ1C
2 + 3γ0γ1BC + γ

2
1(2AC + B2) + γ1(8AC

2 + 7B2C)

−Mγ1C, 5γ1γ−1AB + 2γ−1B(−γ−1C + γ1A) + Mγ−1B −γ−1(8ABC + B
3) + γ0γ−1(2AC + B

2)

+ γ−1(γ−1BC + γ1AB) + 24γ−1ABC + 6γ−1B(2AC + B
2) − 6γ−1(4ABC + B(2AC + B

2)) = 0 (29)

Solving the system of algebraic equations with the aid of Maple, we obtain the following results:

γ−1 =
((kp + M) −B2 + γ0B)

C
, γ1 = 0 at A = 0. (30)

γ−1 = −6C, γ1 = 6A at B = 0. (31)



Int. J. Anal. Appl. 18 (5) (2020) 744

Substituting (30) into (28), the solutions of (1) can be written as:

f(η) = γ0 −
((kp + M) −B2 + γ0B)

C

C exp(Bη) − 1
B exp(Bη)

, (32)

where γ =
a−1
r−1

. Now we have to apply the boundary conditions to the solution (19), noting that the third

one is already satisfied. On using the first two boundary conditions we then need to solve the system:

γ0 −
((kp + M) −B2 + γ0B)

C

C − 1
B

= fw,

−Bγ + B2 −M +
((kp + M) −B2 + γ0B)(C − 1)

C
= 1. (33)

By solving Eq. (33) then substituting in Eq. (32), we obtain the closed form solution

f(η) = γ0 −
((kp + M) −B2 + γ0B)

B
+

((kp + M) −B2 + γ0B)
CB exp(Bη)

, (34)

where C 6= 0, γ0 =
−C+1+fw( 12 fw+

1
2

√
f2w+4+4(kp+M))

1
2
fw+

1
2

√
f2w+4+4(kp+M)

and B = 1
2
fw +

1
2

√
f2w + 4 + 4(kp + M).

5. Results and discussion

Figs. 1–3 have been made in order to see the effects of the permeability of the porous medium kp,

suction/injection parameter fw and the MHD parameter M on the velocity field.

Fig. (1) From this figure, rise in M indicates the raise of magnetic field which acts like a resistive force

and consequently fluid flow slowdowns relatively and hence boundary layer thickness increases.

Fig. (2) The effect of the influence of the porous medium on horizontal velocity. It is found that the

horizontal velocity decreases with the increase of k i.e. Increased permeability parameter (kp) caues an

increase in resistance to fluid along the surface, and this leads to increase the thickness of the boundary

layer.

Fig. (3) show the effects of suction (fw > 0) and injection (fw < 0) on the horizontal velocity f
/(η) the

effect of suction is to decrease the horizontal velocity whereas the effect of injection is to increase this.

6. Conclusion

In this paper, the couple system of MHD boundary layer flow towards a porous stretching sheet have been

reduced by symmetry method to ordinary differential equations. the exact solutions of ordinary differential

equations is obtained by the generalized Exp-Function method. Finally, some plots have been given for study

the effects of various parameters on velocity of fluid .

7. Acknowledgements

The authors would like to thank the deanship of scientific research of Majmaah niversity for the financial

grant received for conducting this research.



Int. J. Anal. Appl. 18 (5) (2020) 745

Figure 1. Velocity profile for different values of magnatic field M

Figure 2. Velocity profile for different values of porous medium Kp

Figure 3. Velocity profile for several values of fw



Int. J. Anal. Appl. 18 (5) (2020) 746

APPENDIX A: The exact solutions for Riccati Equation.

cases A B C solutions of Ricati equation

1 6=0 6= 0 6= 0 −B+
√
4AC−B2 tan( 1

2
(
√
4AC−B2(ζ+d0)))

2C

2 free 6= 0 0 −A
B

+ exp(Bζ)

3 0 6= 0 6= 0 −B exp(Bζ+Bd0)
C exp(Bζ+Bd0)−1

4 0 -1 -1 −d0
exp(ζ)+d0

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.

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	1. Introduction
	2. Formulation of the problem
	3. Symmetry analysis for the boundary layer equations
	4. Reductions and exact solutions
	5. Results and discussion
	6. Conclusion
	7. Acknowledgements
	References