# Stability Analysis for Travelling Wave Solutions of the Olver and Fifth-Order KdV Equations

Authors: A. R. Seadawy; W. Amer; A. Sayed
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

The Olver equation is governing a unidirectional model for describing long and small amplitude waves in shallow water waves. The solitary wave solutions of the Olver and fifth-order KdV equations can be obtained by using extended tanh and sech-tanh methods. The present results are describing the generation and evolution of such waves, their interactions, and their stability. Moreover, the methods can be applied to a wide class of nonlinear evolution equations. All solutions are exact and stable and have applications in physics.


---

## Body

## 1. Introduction

The research on the Korteweg-de Vries (KdV) equations attracted the interest of many scientists. The KdV equations describe nonlinear dispersive long waves; many other partial differential equations have been derived to model wave phenomena in diverse nonlinear systems. The KdV equation plays an important role in describing motions of long waves in shallow water under gravity, one-dimensional nonlinear lattice [1, 2], fluid mechanics [3, 4], quantum mechanics, plasma physics, nonlinear optics, and other areas. The KdV equation is a well-known model for the description of nonlinear long internal waves in a fluid stratified by both density and current. The steady-state version of this equation was produced by Long [5], while Benney [6] gave the integral expressions for calculation of the coefficients of the KdV equation for waves in a fluid with arbitrary stratification in the density and current.There are many classical methods proposed to solve the KdV equations, including direct integration, Lyapunov approach, Hirota’s dependent variable transformation, the inverse scattering transform, and the Bäcklund transformation [7–9]. A direct algebraic approach has also been developed by Parkes and Duffy [10] in which the solutions to the particular equation are represented by an automated tanh-function method [10]. Recently, Wazwaz considered the abundant solitons solutions, compactons and solitary patterns solutions, some new solitons, and periodic solutions of the fifth-order KdV equation [11, 12]. The adiabatic parameter dynamics of 1-soliton solution of the generalized fifth-order nonlinear KdV equation is obtained by virtue of the soliton perturbation theory [13, 14]. The authors present a Mathematica package that deals with complicated algebraic system and outputs directly the required solutions for particular nonlinear equations [15–21].Exact solutions to nonlinear evolution equations (NEEs) play an important role in nonlinear physical science, since the characteristics of these solutions may well simulate real-life physical phenomena [22–24]. The wave phenomena can be observed in fluid dynamics, plasma physics, elastic media, and so forth. The main task of this work is to show that our proposed methods, improved tanh and sech-tanh methods, are very efficient in solving the Olver equation and the fifth-order KdV equation by using extended tanh method and extended sech-tanh method [25–28].This paper is organized as follows. An introduction in is presented in Section1. In Section 2, an analysis of the extended tanh method and extended sech-tanh method is formulated. In Section 3, the travelling wave solutions of the Olver and the fifth-order KdV equations are obtained. Finally, the paper ends with a conclusion in Section 4.


## 2. An Analysis of the Methods

### 2.1. The sech-tanh Method

We suppose thatu
(
x
,
t
)
=
u
(
ξ
), where ξ
=
x
-
k
t, and u
(
ξ
) has the following formal travelling wave solution:
(1)
u
(
ξ
)
=
∑
i
=
1
n
‍
sec
h
i
-
1
ξ
(
A
i
sech
ξ
+
B
i
tanh
ξ
)
,
where A
0
,
A
1
,
…
,
A
n and B
1
,
…
,
B
n are constants to be determined.Step 1.
Equating the highest-order nonlinear term and the highest-order linear partial derivative in the ordinary differential equations yields the value ofn.Step 2.
By setting the coefficients ofsec
h
j
tanh
i for i
=
0,1 and j
=
1,2
,
… to zero, we have the following set of over determined equations in the unknowns A
0
,
A
i
,
B
i, and μ and k for i
=
1,2
,
…
,
n.Step 3.
By using Mathematica and Wu’s elimination methods, the algebraic equations in Step2 can be solved.


### 2.2. The Extendedtanh Method

Thetanh method developed and introduced an independent variable:
(2)
Y
=
tanh
(
μ
ξ
)
,
ξ
=
x
-
k
t
,
that is introduced and leads to the change of the following derivatives:
(3)
d
d
ξ
=
μ
(
1
-
Y
2
)
d
d
Y
,
d
2
d
ξ
2
=
-
2
μ
2
Y
(
1
-
Y
2
)
(
d
d
Y
)
2
+
μ
2
(
1
-
Y
2
)
2
d
2
d
Y
2
.The extendedtanh method admits the use of the finite expansion:
(4)
u
(
μ
ξ
)
=
S
(
Y
)
=
∑
i
=
0
m
‍
a
i
Y
i
+
∑
i
=
1
m
‍
b
i
Y
-
i
,
where m is a positive integer, in most cases, that will be determined. Expansion equation (4) reduces to the standard tanh method for b
i
,
1
⩽
i
⩽
m. The parameter m is usually obtained by balancing the linear terms of highest-order in the resulting equation with the highest-order nonlinear terms. Substituting of (4) into the ODE results in an algebraic system of equations in powers of Y that will lead to the determination of the parameters a
i
,
(
i
=
0
,
…
,
m
)
,
μ and c.Stability of Solution. Hamiltonian system for the momentum is given by
(5)
ν
=
1
2
∫
-
∞
∞
‍
∫
-
∞
∞
‍
u
2
(
x
,
y
)
d
x
d
y
,
at
t
=
0
.The sufficient condition for discussing the stability of solution is∂
ν
/
∂
k
>
0, where k is coefficient of time.


## 2.1. The sech-tanh Method

We suppose thatu
(
x
,
t
)
=
u
(
ξ
), where ξ
=
x
-
k
t, and u
(
ξ
) has the following formal travelling wave solution:
(1)
u
(
ξ
)
=
∑
i
=
1
n
‍
sec
h
i
-
1
ξ
(
A
i
sech
ξ
+
B
i
tanh
ξ
)
,
where A
0
,
A
1
,
…
,
A
n and B
1
,
…
,
B
n are constants to be determined.Step 1.
Equating the highest-order nonlinear term and the highest-order linear partial derivative in the ordinary differential equations yields the value ofn.Step 2.
By setting the coefficients ofsec
h
j
tanh
i for i
=
0,1 and j
=
1,2
,
… to zero, we have the following set of over determined equations in the unknowns A
0
,
A
i
,
B
i, and μ and k for i
=
1,2
,
…
,
n.Step 3.
By using Mathematica and Wu’s elimination methods, the algebraic equations in Step2 can be solved.


## 2.2. The Extendedtanh Method

Thetanh method developed and introduced an independent variable:
(2)
Y
=
tanh
(
μ
ξ
)
,
ξ
=
x
-
k
t
,
that is introduced and leads to the change of the following derivatives:
(3)
d
d
ξ
=
μ
(
1
-
Y
2
)
d
d
Y
,
d
2
d
ξ
2
=
-
2
μ
2
Y
(
1
-
Y
2
)
(
d
d
Y
)
2
+
μ
2
(
1
-
Y
2
)
2
d
2
d
Y
2
.The extendedtanh method admits the use of the finite expansion:
(4)
u
(
μ
ξ
)
=
S
(
Y
)
=
∑
i
=
0
m
‍
a
i
Y
i
+
∑
i
=
1
m
‍
b
i
Y
-
i
,
where m is a positive integer, in most cases, that will be determined. Expansion equation (4) reduces to the standard tanh method for b
i
,
1
⩽
i
⩽
m. The parameter m is usually obtained by balancing the linear terms of highest-order in the resulting equation with the highest-order nonlinear terms. Substituting of (4) into the ODE results in an algebraic system of equations in powers of Y that will lead to the determination of the parameters a
i
,
(
i
=
0
,
…
,
m
)
,
μ and c.Stability of Solution. Hamiltonian system for the momentum is given by
(5)
ν
=
1
2
∫
-
∞
∞
‍
∫
-
∞
∞
‍
u
2
(
x
,
y
)
d
x
d
y
,
at
t
=
0
.The sufficient condition for discussing the stability of solution is∂
ν
/
∂
k
>
0, where k is coefficient of time.


## 3. Application of the Methods

### 3.1. The Olver Equation

In this section, we will employ the proposed methods to solve Olver equation [23]:
(6)
v
t
+
(
1
-
q
6
2
4
q
2
)
v
x
+
q
1
v
x
x
x
x
x
+
q
2
v
2
v
x
+
q
3
v
v
x
x
x
+
q
4
v
x
v
x
x
+
(
q
5
-
q
3
q
6
2
q
2
)
v
x
x
x
=
0
,
where the coefficients q
i
,
(
i
=
1
,
…
,
6
) are real constants depending on surface tension. These coefficients are
(7)
q
1
=
(
19
360
-
τ
12
-
τ
2
8
)
ζ
2
,
q
2
=
-
3
8
χ
2
,
q
3
=
(
5
12
-
τ
4
)
χ
ζ
,
q
4
=
(
23
24
+
5
τ
8
)
χ
ζ
,
q
5
=
(
1
6
-
τ
2
)
ζ
,
q
6
=
3
2
χ
.
Here, τ represents a dimensionless surface tension coefficient, χ is the ratio of wave amplitude to undisturbed fluid depth, and ζ is the square of the ratio of fluid depth to wave length.


#### 3.1.1. Using asech-tanh Method

Equation (6) was equivalently
(8)
(
1
-
k
-
q
6
2
4
q
2
)
v
′
+
q
1
v
′′′′
′
+
q
2
v
2
v
′
+
q
3
v
v
′′′
+
q
4
v
′
v
′′
+
(
q
5
-
q
3
q
6
2
q
2
)
v
′′′
=
0
,
obtained upon using the wave variable ξ
=
x
-
k
t. Balancing v
′′′′
′ with v
′
v
′′ in (8) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(9)
v
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (9) into (8) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1,2
,
3,4
,
5,6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.By solving the set of result equations by using Mathematica, we obtain the following solutions.Case I. Consider
(10)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
(
2
q
3
+
q
4
∓
E
)
q
2
,
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
×
(
-
19200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
2
q
3
q
4
+
q
4
(
q
4
±
E
)
)
+
q
3
2
q
4
(
2
q
3
+
q
4
∓
E
)
(
4
(
q
3
+
q
4
)
-
q
6
)
×
(
4
(
q
3
+
q
4
)
+
q
6
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
2
q
3
q
4
+
q
4
2
)
q
5
q
6
∓
q
4
q
5
q
6
E
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
20
q
2
+
96
q
3
2
+
176
q
3
q
4
+
5
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
-
4
q
1
q
2
(
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
20
q
2
2
q
5
2
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
+
q
3
(
48
q
3
3
+
256
q
3
2
q
4
+
q
3
(
288
q
4
2
∓
80
q
4
E
-
5
q
6
2
)
+
10
q
4
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
)
.
In this case, the generalized soliton solution can be written as
(11)
v
1
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
+
(
3
(
2
q
3
+
q
4
∓
E
)
q
2
)
sec
h
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure1(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 1(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (11) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (11) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(12)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
+
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(13)
v
2
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
2
q
3
3
)
+
(
3
(
2
q
3
+
q
4
+
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(14)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
-
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
,
k
=
1
(
8
q
2
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
-
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
-
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
-
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(15)
v
3
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
-
(
E
+
q
4
)
q
6
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
+
(
3
(
2
q
3
+
q
4
-
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


#### 3.1.2. Using the Extendedtanh Method

We have(16)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (16) into (8) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(17)
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
20
q
2
+
8
μ
4
(
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
)
)
.
In this case, the generalized soliton solution can be written as
(18)
u
1
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
tanh
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure2(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 2(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (18) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (18) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(19)
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
307200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
20
q
2
+
128
μ
4
(
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
256
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
32
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.
In this case, the generalized soliton solution can be written as
(20)
u
2
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
×
(
tanh
2
[
x
-
k
t
]
+
coth
⁡
2
[
x
-
k
t
]
)
.Case III. Consider
(21)
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
(
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
20
q
2
+
8
μ
4
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.In this case, the generalized soliton solution can be written as(22)
u
3
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
(
4
μ
2
(
-
20
q
1
q
2
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
coth
⁡
2
[
x
-
k
t
]
.


### 3.2. Solving the Fifth-Order Korteweg-de Vries Equation

In this section we will employ the proposed methods to solve the fifth-order Korteweg-de Vries equation:(23)
η
t
+
6
η
η
x
+
η
3
x
+
α
c
1
η
2
η
x
+
α
c
2
η
x
η
x
x
+
α
c
3
η
η
3
x
+
α
c
4
η
5
x
=
0
,
α
≪
1
.
Or equivalently
(24)
-
k
η
′
+
6
η
η
′
+
η
′′′
+
α
c
1
η
2
η
′
+
α
c
2
η
′
η
′′
+
α
c
3
η
η
′′′
+
α
c
4
η
′′′′
′
=
0
,
is obtained upon using the wave variable ξ
=
x
-
k
t, when the higher-order coefficients are given by
(25)
(
c
1
,
c
2
,
c
3
,
c
4
)
=
(
1
,
1
12
,
1
3
,
1
480
)
.


#### 3.2.1. Using asech-tanh Method

Balancingη
′′′′
′ with η
η
′′′ in (24) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(26)
η
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (27) into (24) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1
,
2
,
3
,
4
,
5
,
6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.Solve the set of result equations by using Mathematica; we obtain the following solutions.Case I. Consider
(27)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
c
2
+
6
c
3
∓
3
g
c
1
,
A
0
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
,
k
=
-
1
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
20
c
1
3
(
c
4
+
240
α
2
c
4
3
)
+
c
2
c
3
2
(
-
3
+
2
α
(
c
2
+
c
3
)
)
×
(
3
+
2
α
(
c
2
+
c
3
)
)
(
-
c
2
-
2
c
3
±
g
)
-
c
1
2
(
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
c
2
2
(
1
+
400
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
60
(
-
15
+
8
α
2
c
3
2
)
c
4
2
)
+
c
2
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
)
+
2
c
1
c
3
(
40
α
2
c
2
3
c
4
+
6
c
3
(
c
3
+
(
-
15
+
4
α
2
c
3
2
)
c
4
)
+
c
2
2
(
3
+
8
α
2
c
4
(
18
c
3
∓
5
g
)
)
+
c
2
(
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
3
(
-
60
c
4
±
g
)
+
2
c
3
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(28)
η
1
(
x
,
t
)
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
+
3
c
2
+
6
c
3
∓
3
g
c
1
sec
h
2
[
x
-
k
t
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure3(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (28) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (28) is plotted: stability contour of solitary waves.
(a)
(b)Figure3(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(29)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
+
g
)
2
c
1
,
B
2
=
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
-
6
(
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(30)
η
2
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
+
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(31)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
-
g
)
2
c
1
,
B
2
=
±
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
×
(
c
2
+
2
c
3
-
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
-
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
-
6
(
-
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
-
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(32)
η
3
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
-
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
±
3
2
c
1
.
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


#### 3.2.2. Using the Extendedtanh Method

We have(33)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (33) into (23) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(34)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.
In this case, the generalized soliton solution can be written as
(35)
η
1
(
x
,
t
)
=
(
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
tanh
2
[
μ
(
x
-
k
t
)
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure4(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (35) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (35) is plotted: stability contour of solitary waves.
(a)
(b)Figure4(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(36)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
32
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
384
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(37)
η
2
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
×
(
tanh
2
[
μ
(
x
-
k
t
)
]
+
coth
⁡
2
[
μ
(
x
-
k
t
)
]
)
.Figure5(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (37) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (37) is plotted: stability contour of solitary waves.
(a)
(b)Figure5(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case III. Consider
(38)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(39)
η
3
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
coth
⁡
2
[
μ
(
x
-
k
t
)
]
.Figure6(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
]. Figure 6(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (39) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (39) is plotted: stability contour of solitary waves.
(a)
(b)


## 3.1. The Olver Equation

In this section, we will employ the proposed methods to solve Olver equation [23]:
(6)
v
t
+
(
1
-
q
6
2
4
q
2
)
v
x
+
q
1
v
x
x
x
x
x
+
q
2
v
2
v
x
+
q
3
v
v
x
x
x
+
q
4
v
x
v
x
x
+
(
q
5
-
q
3
q
6
2
q
2
)
v
x
x
x
=
0
,
where the coefficients q
i
,
(
i
=
1
,
…
,
6
) are real constants depending on surface tension. These coefficients are
(7)
q
1
=
(
19
360
-
τ
12
-
τ
2
8
)
ζ
2
,
q
2
=
-
3
8
χ
2
,
q
3
=
(
5
12
-
τ
4
)
χ
ζ
,
q
4
=
(
23
24
+
5
τ
8
)
χ
ζ
,
q
5
=
(
1
6
-
τ
2
)
ζ
,
q
6
=
3
2
χ
.
Here, τ represents a dimensionless surface tension coefficient, χ is the ratio of wave amplitude to undisturbed fluid depth, and ζ is the square of the ratio of fluid depth to wave length.


### 3.1.1. Using asech-tanh Method

Equation (6) was equivalently
(8)
(
1
-
k
-
q
6
2
4
q
2
)
v
′
+
q
1
v
′′′′
′
+
q
2
v
2
v
′
+
q
3
v
v
′′′
+
q
4
v
′
v
′′
+
(
q
5
-
q
3
q
6
2
q
2
)
v
′′′
=
0
,
obtained upon using the wave variable ξ
=
x
-
k
t. Balancing v
′′′′
′ with v
′
v
′′ in (8) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(9)
v
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (9) into (8) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1,2
,
3,4
,
5,6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.By solving the set of result equations by using Mathematica, we obtain the following solutions.Case I. Consider
(10)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
(
2
q
3
+
q
4
∓
E
)
q
2
,
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
×
(
-
19200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
2
q
3
q
4
+
q
4
(
q
4
±
E
)
)
+
q
3
2
q
4
(
2
q
3
+
q
4
∓
E
)
(
4
(
q
3
+
q
4
)
-
q
6
)
×
(
4
(
q
3
+
q
4
)
+
q
6
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
2
q
3
q
4
+
q
4
2
)
q
5
q
6
∓
q
4
q
5
q
6
E
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
20
q
2
+
96
q
3
2
+
176
q
3
q
4
+
5
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
-
4
q
1
q
2
(
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
20
q
2
2
q
5
2
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
+
q
3
(
48
q
3
3
+
256
q
3
2
q
4
+
q
3
(
288
q
4
2
∓
80
q
4
E
-
5
q
6
2
)
+
10
q
4
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
)
.
In this case, the generalized soliton solution can be written as
(11)
v
1
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
+
(
3
(
2
q
3
+
q
4
∓
E
)
q
2
)
sec
h
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure1(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 1(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (11) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (11) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(12)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
+
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(13)
v
2
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
2
q
3
3
)
+
(
3
(
2
q
3
+
q
4
+
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(14)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
-
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
,
k
=
1
(
8
q
2
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
-
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
-
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
-
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(15)
v
3
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
-
(
E
+
q
4
)
q
6
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
+
(
3
(
2
q
3
+
q
4
-
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


### 3.1.2. Using the Extendedtanh Method

We have(16)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (16) into (8) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(17)
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
20
q
2
+
8
μ
4
(
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
)
)
.
In this case, the generalized soliton solution can be written as
(18)
u
1
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
tanh
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure2(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 2(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (18) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (18) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(19)
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
307200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
20
q
2
+
128
μ
4
(
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
256
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
32
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.
In this case, the generalized soliton solution can be written as
(20)
u
2
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
×
(
tanh
2
[
x
-
k
t
]
+
coth
⁡
2
[
x
-
k
t
]
)
.Case III. Consider
(21)
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
(
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
20
q
2
+
8
μ
4
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.In this case, the generalized soliton solution can be written as(22)
u
3
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
(
4
μ
2
(
-
20
q
1
q
2
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
coth
⁡
2
[
x
-
k
t
]
.


## 3.1.1. Using asech-tanh Method

Equation (6) was equivalently
(8)
(
1
-
k
-
q
6
2
4
q
2
)
v
′
+
q
1
v
′′′′
′
+
q
2
v
2
v
′
+
q
3
v
v
′′′
+
q
4
v
′
v
′′
+
(
q
5
-
q
3
q
6
2
q
2
)
v
′′′
=
0
,
obtained upon using the wave variable ξ
=
x
-
k
t. Balancing v
′′′′
′ with v
′
v
′′ in (8) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(9)
v
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (9) into (8) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1,2
,
3,4
,
5,6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.By solving the set of result equations by using Mathematica, we obtain the following solutions.Case I. Consider
(10)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
(
2
q
3
+
q
4
∓
E
)
q
2
,
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
×
(
-
19200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
2
q
3
q
4
+
q
4
(
q
4
±
E
)
)
+
q
3
2
q
4
(
2
q
3
+
q
4
∓
E
)
(
4
(
q
3
+
q
4
)
-
q
6
)
×
(
4
(
q
3
+
q
4
)
+
q
6
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
2
q
3
q
4
+
q
4
2
)
q
5
q
6
∓
q
4
q
5
q
6
E
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
20
q
2
+
96
q
3
2
+
176
q
3
q
4
+
5
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
-
4
q
1
q
2
(
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
20
q
2
2
q
5
2
+
20
q
2
q
3
(
2
(
q
3
+
q
4
)
-
q
5
q
6
)
)
+
q
3
(
48
q
3
3
+
256
q
3
2
q
4
+
q
3
(
288
q
4
2
∓
80
q
4
E
-
5
q
6
2
)
+
10
q
4
(
8
q
4
(
q
4
∓
E
)
-
q
6
2
)
)
)
.
In this case, the generalized soliton solution can be written as
(11)
v
1
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
×
(
8
q
3
3
+
4
q
3
2
(
3
q
4
∓
E
)
+
40
q
1
q
2
(
-
2
q
3
-
q
4
±
E
)
+
2
q
2
q
5
(
q
4
±
E
)
+
q
3
(
4
q
4
2
∓
E
q
6
-
q
4
(
q
6
±
4
E
)
)
)
+
(
3
(
2
q
3
+
q
4
∓
E
)
q
2
)
sec
h
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure1(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 1(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (11) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (11) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(12)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
+
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
q
4
(
E
-
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(13)
v
2
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
2
q
3
3
-
10
q
1
q
2
(
E
+
2
q
3
+
q
4
)
+
q
3
2
(
E
+
3
q
4
)
+
2
q
2
(
-
E
+
q
4
)
q
5
+
q
3
(
q
4
(
E
+
q
4
)
+
(
E
-
q
4
)
q
6
)
(
2
q
3
3
)
+
(
3
(
2
q
3
+
q
4
+
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
-
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(14)
A
0
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
(
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
)
,
A
1
=
B
1
=
0
,
A
2
=
3
(
2
q
3
+
q
4
-
E
)
2
q
2
,
B
2
=
∓
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
,
k
=
1
(
8
q
2
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
-
1200
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
+
q
4
(
E
+
2
q
3
)
+
q
4
2
)
q
5
q
6
)
+
20
q
1
2
q
2
2
(
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
5
q
4
(
-
E
+
q
4
)
+
40
q
2
+
12
q
3
2
+
22
q
3
q
4
-
10
q
6
2
)
+
q
3
2
q
4
(
-
E
+
2
q
3
+
q
4
)
(
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
20
q
2
2
q
5
2
+
20
q
2
q
3
(
(
2
q
3
+
q
4
)
-
q
5
q
6
)
+
q
3
(
-
5
q
6
2
(
q
3
+
2
q
4
)
)
(
q
3
+
q
4
)
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
-
E
+
q
4
)
)
-
5
q
6
2
(
q
3
+
2
q
4
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(15)
v
3
(
x
,
t
)
=
1
4
q
2
(
10
q
1
q
2
-
q
3
(
q
3
+
q
4
)
)
·
(
-
(
E
+
q
4
)
q
6
)
-
(
E
+
q
4
)
q
6
)
)
2
q
3
3
-
10
q
1
q
2
(
-
E
+
2
q
3
+
q
4
)
+
q
3
2
(
-
E
+
3
q
4
)
+
2
q
2
(
E
+
q
4
)
q
5
+
q
3
(
q
4
(
-
E
+
q
4
)
-
(
E
+
q
4
)
q
6
)
)
+
(
3
(
2
q
3
+
q
4
-
E
)
2
q
2
)
sec
h
2
[
x
-
k
t
]
∓
(
3
2
q
2
20
q
1
q
2
-
(
2
q
3
+
q
4
)
2
+
(
2
q
3
+
q
4
)
E
)



×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


## 3.1.2. Using the Extendedtanh Method

We have(16)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (16) into (8) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(17)
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
20
q
2
+
8
μ
4
(
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
2
μ
4
)
)
.
In this case, the generalized soliton solution can be written as
(18)
u
1
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
tanh
2
[
x
-
k
t
]
,
where E
=
-
40
q
1
q
2
+
(
2
q
3
+
q
4
)
2, and (
2
q
3
+
q
4
)
2
>
40
q
1
q
2.Figure2(a) shows the stability dark solitary wave solutions with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
]. Figure 2(b) shows the stability contour of solitary wave solution with (χ
=
0.25, ζ
=
0.025, and τ
=
0.5) in the interval [
-
10,10
] and time in the interval [
0,5
].(a) Travelling waves solutions of (18) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (18) is plotted: stability contour of solitary waves.
(a)
(b)Case II. Consider
(19)
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
-
307200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
)
+
40
q
1
2
q
2
2
(
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
20
q
2
+
128
μ
4
(
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
12
q
3
2
+
22
q
3
q
4
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
256
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
5
q
2
+
32
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.
In this case, the generalized soliton solution can be written as
(20)
u
2
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
×
(
tanh
2
[
x
-
k
t
]
+
coth
⁡
2
[
x
-
k
t
]
)
.Case III. Consider
(21)
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
,
a
0
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
)
,
k
=
1
8
q
2
(
-
10
q
1
q
2
+
q
3
(
q
3
+
q
4
)
)
2
·
(
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
-
19200
μ
4
q
1
3
q
2
3
+
4
q
2
2
q
5
2
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
+
4
q
2
q
3
(
2
q
3
(
q
3
+
q
4
)
2
-
(
2
q
3
2
-
(
±
E
-
2
q
3
)
q
4
+
q
4
2
)
q
5
q
6
(
2
q
3
(
q
3
+
q
4
)
2
)
+
40
q
1
2
q
2
2
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
-
5
q
6
2
)
20
q
2
+
8
μ
4
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
(
+
22
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
12
q
3
2
)
-
5
q
6
2
)
+
q
3
2
q
4
(
±
E
+
2
q
3
+
q
4
)
(
16
μ
4
(
q
3
+
q
4
)
2
-
q
6
2
)
-
4
q
1
q
2
(
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
8
q
3
(
q
3
+
q
4
)
×
(
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
5
q
2
+
2
μ
4
×
(
3
q
3
2
+
13
q
3
q
4
+
5
q
4
(
±
E
+
q
4
)
)
)
+
20
q
2
2
q
5
2
-
20
q
2
q
3
q
5
q
6
-
5
q
3
(
q
3
+
2
q
4
)
q
6
2
)
)
.In this case, the generalized soliton solution can be written as(22)
u
3
(
x
,
t
)
=
1
q
2
(
q
4
±
E
)
·
(
4
μ
2
(
-
20
q
1
q
2
+
(
q
3
+
q
4
)
(
2
q
3
+
q
4
)
)
±
4
μ
2
E
(
q
3
+
q
4
)
+
2
q
2
q
5
-
q
3
q
6
(
4
μ
2
(
-
20
q
1
q
2
)
-
3
μ
2
(
2
q
3
+
q
4
±
E
)
q
2
coth
⁡
2
[
x
-
k
t
]
.


## 3.2. Solving the Fifth-Order Korteweg-de Vries Equation

In this section we will employ the proposed methods to solve the fifth-order Korteweg-de Vries equation:(23)
η
t
+
6
η
η
x
+
η
3
x
+
α
c
1
η
2
η
x
+
α
c
2
η
x
η
x
x
+
α
c
3
η
η
3
x
+
α
c
4
η
5
x
=
0
,
α
≪
1
.
Or equivalently
(24)
-
k
η
′
+
6
η
η
′
+
η
′′′
+
α
c
1
η
2
η
′
+
α
c
2
η
′
η
′′
+
α
c
3
η
η
′′′
+
α
c
4
η
′′′′
′
=
0
,
is obtained upon using the wave variable ξ
=
x
-
k
t, when the higher-order coefficients are given by
(25)
(
c
1
,
c
2
,
c
3
,
c
4
)
=
(
1
,
1
12
,
1
3
,
1
480
)
.


### 3.2.1. Using asech-tanh Method

Balancingη
′′′′
′ with η
η
′′′ in (24) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(26)
η
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (27) into (24) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1
,
2
,
3
,
4
,
5
,
6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.Solve the set of result equations by using Mathematica; we obtain the following solutions.Case I. Consider
(27)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
c
2
+
6
c
3
∓
3
g
c
1
,
A
0
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
,
k
=
-
1
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
20
c
1
3
(
c
4
+
240
α
2
c
4
3
)
+
c
2
c
3
2
(
-
3
+
2
α
(
c
2
+
c
3
)
)
×
(
3
+
2
α
(
c
2
+
c
3
)
)
(
-
c
2
-
2
c
3
±
g
)
-
c
1
2
(
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
c
2
2
(
1
+
400
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
60
(
-
15
+
8
α
2
c
3
2
)
c
4
2
)
+
c
2
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
)
+
2
c
1
c
3
(
40
α
2
c
2
3
c
4
+
6
c
3
(
c
3
+
(
-
15
+
4
α
2
c
3
2
)
c
4
)
+
c
2
2
(
3
+
8
α
2
c
4
(
18
c
3
∓
5
g
)
)
+
c
2
(
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
3
(
-
60
c
4
±
g
)
+
2
c
3
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(28)
η
1
(
x
,
t
)
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
+
3
c
2
+
6
c
3
∓
3
g
c
1
sec
h
2
[
x
-
k
t
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure3(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (28) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (28) is plotted: stability contour of solitary waves.
(a)
(b)Figure3(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(29)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
+
g
)
2
c
1
,
B
2
=
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
-
6
(
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(30)
η
2
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
+
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(31)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
-
g
)
2
c
1
,
B
2
=
±
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
×
(
c
2
+
2
c
3
-
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
-
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
-
6
(
-
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
-
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(32)
η
3
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
-
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
±
3
2
c
1
.
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


### 3.2.2. Using the Extendedtanh Method

We have(33)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (33) into (23) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(34)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.
In this case, the generalized soliton solution can be written as
(35)
η
1
(
x
,
t
)
=
(
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
tanh
2
[
μ
(
x
-
k
t
)
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure4(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (35) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (35) is plotted: stability contour of solitary waves.
(a)
(b)Figure4(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(36)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
32
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
384
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(37)
η
2
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
×
(
tanh
2
[
μ
(
x
-
k
t
)
]
+
coth
⁡
2
[
μ
(
x
-
k
t
)
]
)
.Figure5(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (37) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (37) is plotted: stability contour of solitary waves.
(a)
(b)Figure5(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case III. Consider
(38)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(39)
η
3
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
coth
⁡
2
[
μ
(
x
-
k
t
)
]
.Figure6(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
]. Figure 6(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (39) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (39) is plotted: stability contour of solitary waves.
(a)
(b)


## 3.2.1. Using asech-tanh Method

Balancingη
′′′′
′ with η
η
′′′ in (24) gives m
=
2. sech-tanh method equation (1) admits the use of the finite expansion:
(26)
η
(
ξ
)
=
A
0
+
A
1
sech
ξ
+
B
1
tanh
ξ
+
A
2
sec
h
2
ξ
+
B
2
sech
ξ
tanh
ξ
,
and by substituting from (27) into (24) and setting the coefficients of sec
h
j
tanh
i for i
=
0,1 and j
=
1
,
2
,
3
,
4
,
5
,
6
,
7 to zero, we have the following set of overdetermined equations in the unknowns A
0, A
1, A
2, B
1, B
2, and k.Solve the set of result equations by using Mathematica; we obtain the following solutions.Case I. Consider
(27)
A
1
=
B
1
=
B
2
=
0
,
A
2
=
3
c
2
+
6
c
3
∓
3
g
c
1
,
A
0
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
,
k
=
-
1
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
20
c
1
3
(
c
4
+
240
α
2
c
4
3
)
+
c
2
c
3
2
(
-
3
+
2
α
(
c
2
+
c
3
)
)
×
(
3
+
2
α
(
c
2
+
c
3
)
)
(
-
c
2
-
2
c
3
±
g
)
-
c
1
2
(
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
c
2
2
(
1
+
400
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
60
(
-
15
+
8
α
2
c
3
2
)
c
4
2
)
+
c
2
(
±
g
(
1
-
400
α
2
c
4
2
)
+
2
c
3
(
1
+
880
α
2
c
4
2
)
)
)
+
2
c
1
c
3
(
40
α
2
c
2
3
c
4
+
6
c
3
(
c
3
+
(
-
15
+
4
α
2
c
3
2
)
c
4
)
+
c
2
2
(
3
+
8
α
2
c
4
(
18
c
3
∓
5
g
)
)
+
c
2
(
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
3
(
-
60
c
4
±
g
)
+
2
c
3
×
(
3
+
4
α
2
c
4
(
16
c
3
∓
5
g
)
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(28)
η
1
(
x
,
t
)
=
1
2
α
c
1
(
-
c
3
(
c
2
+
c
3
)
+
10
c
1
c
4
)
·
(
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
c
3
(
3
+
2
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
∓
g
)
+
c
1
(
c
2
(
1
-
20
α
c
4
)
±
g
+
20
c
4
(
-
3
-
2
α
c
3
±
α
g
)
)
)
+
3
c
2
+
6
c
3
∓
3
g
c
1
sec
h
2
[
x
-
k
t
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure3(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (28) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (28) is plotted: stability contour of solitary waves.
(a)
(b)Figure3(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(29)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
+
g
)
2
c
1
,
B
2
=
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
-
6
(
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(30)
η
2
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
+
g
)
-
2
c
1
(
g
+
5
(
12
+
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
+
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
∓
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
-
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.Case III. Consider
(31)
A
1
=
B
1
=
0
,
A
2
=
3
(
c
2
+
2
c
3
-
g
)
2
c
1
,
B
2
=
±
3
2
c
1
·
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
,
A
0
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
,
k
=
1
8
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
·
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
-
80
c
1
3
c
4
(
1
+
15
α
2
c
4
2
)
+
c
2
c
3
2
(
-
6
+
α
(
c
2
+
c
3
)
)
(
6
+
α
(
c
2
+
c
3
)
)
×
(
c
2
+
2
c
3
-
g
)
-
4
c
1
c
3
(
5
α
2
c
2
3
c
4
+
c
2
2
(
6
+
α
2
(
-
5
g
+
18
c
3
)
c
4
)
+
3
c
3
(
4
c
3
+
(
-
60
+
α
2
c
3
2
)
c
4
)
+
c
2
(
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
-
6
(
-
g
+
60
c
4
)
+
c
3
(
12
+
α
2
(
-
5
g
+
16
c
3
)
c
4
)
)
)
+
4
c
1
2
(
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
c
2
2
(
1
+
25
α
2
c
4
2
)
+
2
(
c
3
2
+
60
c
3
c
4
+
30
(
-
30
+
α
2
c
3
2
)
c
4
2
)
+
c
2
(
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
-
g
(
-
1
+
25
α
2
c
4
2
)
+
2
c
3
(
1
+
55
α
2
c
4
2
)
)
)
)
.
In this case, the generalized soliton solution can be written as
(32)
η
3
(
x
,
t
)
=
-
1
4
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
·
(
+
c
2
(
-
1
+
5
α
c
4
)
)
c
3
(
6
+
α
(
c
2
+
c
3
)
)
(
c
2
+
2
c
3
-
g
)
-
2
c
1
(
-
g
+
5
(
12
-
g
α
+
2
α
c
3
)
c
4
+
c
2
(
-
1
+
5
α
c
4
)
)
)
+
3
(
c
2
+
2
c
3
-
g
)
2
c
1
sec
h
2
[
x
-
k
t
]
±
3
2
c
1
.
-
(
c
2
+
2
c
3
)
2
+
20
c
1
c
4
+
(
c
2
+
c
3
)
g
×
sech
[
x
-
k
t
]
tanh
[
x
-
k
t
]
.


## 3.2.2. Using the Extendedtanh Method

We have(33)
u
(
ξ
)
=
a
0
+
a
1
Y
+
a
2
Y
2
+
b
1
Y
+
b
2
Y
2
.
By substituting (33) into (23) and collecting the coefficient of Y, we obtain a system of algebraic equations for a
0, a
1, a
2, b
1, b
2, and k. Solving this system gives the following solution.Case I. Consider
(34)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
b
2
=
0
,
a
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.
In this case, the generalized soliton solution can be written as
(35)
η
1
(
x
,
t
)
=
(
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
tanh
2
[
μ
(
x
-
k
t
)
]
,
where g
=
(
c
2
+
2
c
3
)
2
-
40
c
1
c
4 and (
c
2
+
2
c
3
)
2
>
40
c
1
c
4.Figure4(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (35) is plotted: stability bright solitary waves. (b) Travelling waves solutions of (35) is plotted: stability contour of solitary waves.
(a)
(b)Figure4(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case II. Consider
(36)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
,
a
1
=
b
1
=
0
,
a
2
=
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
32
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
384
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(37)
η
2
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
×
(
tanh
2
[
μ
(
x
-
k
t
)
]
+
coth
⁡
2
[
μ
(
x
-
k
t
)
]
)
.Figure5(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (37) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (37) is plotted: stability contour of solitary waves.
(a)
(b)Figure5(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].Case III. Consider
(38)
a
0
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
a
1
=
a
2
=
b
1
=
0
,
b
2
=
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
,
k
=
-
9
+
2
μ
4
α
2
c
2
(
±
g
+
c
2
+
2
c
3
)
c
1
α
-
24
μ
4
α
c
4
-
(
±
g
-
c
2
)
c
2
(
c
1
-
3
c
3
)
2
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
2
+
(
c
1
-
3
c
3
)
2
α
c
1
(
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
.In this case, the generalized soliton solution can be written as(39)
η
3
(
x
,
t
)
=
(
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
-
6
±
4
g
μ
2
α
+
4
μ
2
α
(
c
2
+
2
c
3
)
+
(
±
g
-
c
2
)
(
c
1
-
3
c
3
)
c
3
(
c
2
+
c
3
)
-
10
c
1
c
4
)
×
(
2
α
c
1
)
-
1
-
3
μ
2
(
±
g
+
c
2
+
2
c
3
)
c
1
coth
⁡
2
[
μ
(
x
-
k
t
)
]
.Figure6(a) shows the stability bright solitary wave solutions with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
]. Figure 6(b) shows the stability contour of solitary wave solution with (
α
=
-
1
) in the interval [
-
5,5
] and time in the interval [
0,0.5
].(a) Travelling waves solutions of (39) is plotted: stability dark solitary waves. (b) Travelling waves solutions of (39) is plotted: stability contour of solitary waves.
(a)
(b)


## 4. Conclusion

Contour plots produced by Mathematica are drawn shaded, in such a way that regions with higher values ofu
i
(
x
,
t
), v
i
(
x
,
t
), and η
i
(
x
,
t
), i
=
1
,
2
,
3, are drawn lighter. As with all Mathematica graphics commands, options allow you to control the appearance of the graph. Contours plot allows you to determine the number of contours to be drawn. The default is ten equally spaced curves. An analytic study was conducted on the Olver and fifth-order KdV equations. We formally derived travelling wave solutions for the Olver and fifth-order KdV equations. However, by using another distinct approach, we derived one traveling wave solutions for each Olver and fifth-order KdV equations. The structures of the obtained solutions are distinct and stable.



---

*Source: 839485-2014-03-24.xml*