# Assessment of the Exact Solutions of the Space and Time Fractional Benjamin-Bona-Mahony Equation via theG

Authors: Olusola Kolebaje; Oyebola Popoola
Journal: ISRN Mathematical Physics
 (2014)

---

## Abstract

Exact travelling wave solutions to the space and time fractional Benjamin-Bona-Mahony (BBM) equation defined in the sense of Jumarie’s modified Riemann-Liouville derivative via the(
G
′
/
G
) expansion and the modified simple equation methods are presented in this paper. A fractional complex transformation was applied to turn the fractional BBM equation into an equivalent integer order ordinary differential equation. New complex type travelling wave solutions to the space and time fractional BBM equation were obtained with Liu’s theorem. The modified simple equation method is not effective for constructing solutions to the fractional BBM equation.


---

## Body

## 1. Introduction

Nonlinear partial differential equations arise in a large number of physics, mathematics, and engineering problems. In the Soliton theory, the study of exact solutions to these nonlinear equations plays a very germane role, as they provide much information about the physical models they describe. In recent times, it has been found that many physical, chemical, and biological processes are governed by nonlinear partial differential equations of noninteger or fractional order [1–4].Various powerful methods have been employed to construct exact travelling wave solutions to nonlinear partial differential equations. These methods include the inverse scattering transform [5], the Backlund transform [6, 7], the Darboux transform [8], the Hirota bilinear method [9], the tanh-function method [10, 11], the sine-cosine method [12], the exp-function method [13], the generalized Riccati equation [14], the homogenous balance method [15], the first integral method [16, 17], the (
G
′
/
G
) expansion method [18, 19], and the modified simple equation method [20–22].In this paper, we apply the(
G
′
/
G
) expansion method and the modified simple equation method to construct travelling wave solutions to the space and time fractional Benjamin-Bona-Mahony (BBM) equation in the sense of Jumarie’s modified Riemann-Liouville derivative via a fractional complex transformation and we further get new complex type solutions to the equation by applying Liu’s theorem [23]. The Benjamin-Bona-Mahony equation is of the following form:
(1)
∂
γ
u
(
x
,
t
)
∂
t
+
∂
σ
u
(
x
,
t
)
∂
x
+
u
(
x
,
t
)
∂
σ
u
(
x
,
t
)
∂
x
-
∂
2
σ
∂
γ
u
(
x
,
t
)
∂
x
2
∂
t
=
0
.
This equation was introduced in [24] as an improvement of the Korteweg-de Vries equation (KdV equation) for modelling long waves of small amplitude in 1 + 1 dimensions. It is used in modelling surface waves of long wavelength in liquids, acoustic gravity waves in compressible fluids, and acoustic waves in anharmonic crystals.Jumarie’s modified Riemann-Liouville derivative of orderσ with respect to x is defined as [25] (2)
D
x
σ
f
(
x
)
=
{
1
Γ
(
1
-
σ
)
∫
0
x
(
x
-
ξ
)
-
σ
-
1
[
f
(
ξ
)
-
f
(
0
)
]
d
ξ
,
σ
<
0
,
1
Γ
(
1
-
σ
)
d
d
x
∫
0
x
(
x
-
ξ
)
-
σ
[
f
(
ξ
)
-
f
(
0
)
]
d
ξ
,
0
<
σ
<
1
,
[
f
(
σ
-
n
)
(
x
)
]
(
n
)
,
n
≤
σ
<
n
+
1
,
n
≥
1
.Some useful properties of the modified Riemann-Liouville derivative are listed below [25, 26]:
(3)
D
x
σ
x
k
=
Γ
(
1
+
k
)
Γ
(
1
+
k
-
σ
)
x
k
-
σ
,
D
x
σ
(
f
(
x
)
g
(
x
)
)
=
g
(
x
)
D
x
σ
f
(
x
)
+
f
(
x
)
D
x
σ
g
(
x
)
.


## 2. Description of the(
G
′
/
G
) Expansion Method

Here, we provide a brief explanation of the(
G
′
/
G
) expansion method for finding travelling wave solutions of nonlinear fractional partial differential equations. Suppose that the nonlinear evolution equation is in the following form:
(4)
℘
(
u
,
D
t
γ
u
,
D
x
σ
u
,
D
t
2
γ
u
,
D
x
2
σ
u
,
D
x
σ
D
t
γ
u
,
…
)
=
0
,
hhhhhhhhhhhhhhhhhhhhhhhh
0
<
σ
,
γ
≤
1
,
where ℘ is a polynomial of u
(
x
,
t
) and its derivatives (integer and fractional) with respect to x and t. σ and γ are parameters that describe the order of the space and time derivatives, respectively.Theorem 1.
Fractional Complex Transformation. To transform (4) into a nonlinear ordinary differential equation (ODE) of integer order by applying a fractional complex transformation proposed by Li and He [27],
(5)
u
(
x
,
t
)
=
U
(
ξ
)
,
ξ
=
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
,
where c is an arbitrary constant and (4) reduces to a nonlinear integer order ODE of the following form:
(6)
P
(
u
,
u
′
,
u
′′
,
u
′′′
,
…
)
=
0
.
The method assumes that the solution to (6) can be expressed as a polynomial in (
G
′
/
G
):
(7)
u
(
ξ
)
=
∑
i
=
0
m
α
i
[
G
′
(
ξ
)
G
(
ξ
)
]
i
,
α
m
≠
0
,
where α
0
,
α
1
,
…
,
α
m are constants to be determined and G
=
G
(
ξ
) is a solution of the linear ordinary differential equation of the following form:
(8)
G
′′
(
ξ
)
+
λ
G
′
(
ξ
)
+
μ
G
(
ξ
)
=
0
,
where λ and μ are arbitrary constants. The general solution of (8) gives [28] the following: (9)
G
′
(
ξ
)
G
(
ξ
)
=
{
λ
2
-
4
μ
2
[
C
1
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
C
1
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
]
-
λ
2
,
λ
2
-
4
μ
>
0
Hyperbolic
,
4
μ
-
λ
2
2
[
-
C
1
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
C
1
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
]
-
λ
2
,
λ
2
-
4
μ
<
0
Trigonometric
,
C
2
C
1
+
C
2
ξ
-
λ
2
,
λ
2
-
4
μ
=
0
Rational
.Equation (6) is integrated as long as all the terms contain derivatives, where integration constants are considered to be zero. To determine  m, one considers the homogenous balance between the highest order derivative and the highest order nonlinear term(s).
Substitute (7) with the determined value of m into (6), and collect all terms with the same order of (
G
′
/
G
) together. If the coefficients of (
G
′
/
G
)
i vanish separately, one has a set of algebraic equations in α
0
,
α
1
,
…
,
α
m, c, λ and μ that is solved with the aid of Mathematica.
Finally, substitutingα
0
,
α
1
,
…
,
α
m, c and the general solution to (8) into (7) yields the travelling wave solution of (4).Theorem 2 (Liu’s theorem [23]).
If a nonlinear evolution equation has a kink-type solution in the form of(10)
u
(
ξ
)
=
P
k
(
tanh
⁡
[
ψ
ξ
]
)
,
where P
k is a polynomial of degree  k, then it has a certain kink-bell-type solution in the following form:
(11)
u
(
ξ
)
=
P
k
(
tanh
⁡
[
2
ψ
ξ
]
±
i

sech

[
2
ψ
ξ
]
)
,
where i is the imaginary number unit.


## 3. Description of the Modified Simple Equation Method

Here, we highlight the basic ideas of the modified simple equation method. Suppose that the fractional nonlinear evolution equation is in the following form:(12)
℘
(
u
,
D
t
γ
u
,
D
x
σ
u
,
D
t
2
γ
u
,
D
x
2
σ
u
,
D
x
σ
D
t
γ
u
,
…
)
=
0
,
where ℘ is a polynomial of u and its fractional partial derivatives with respect to x and t. Suppose a fractional complex transformation (see (5)) which allows us to reduce (12) to an ordinary differential equation (ODE) of the following form:
(13)
P
(
u
,
u
′
,
u
′′
,
…
)
=
0
.
We suppose that the solution to (13) can be expressed as a polynomial of (
ϕ
′
/
ϕ
) in the following form:
(14)
u
(
ξ
)
=
∑
i
=
0
N
A
i
[
ϕ
′
(
ξ
)
ϕ
(
ξ
)
]
i
,
where N is a positive integer obtained by balancing the highest order derivatives and the highest order nonlinear terms in (13). A
i
(
i
=
0,1
,
2
,
…
,
N
) are arbitrary constants to be determined such that A
N
≠
0 and ϕ
(
ξ
) is an unknown function to be determined and is not a solution of any predefined differential equation.We substitute (14) and its derivatives into (13) taking into account the function  ϕ
(
ξ
). As a result of this substitution we obtain a polynomial of  ϕ
-
j
(
ξ
) with the derivatives of ϕ
(
ξ
). Equating all the coefficients of ϕ
-
j
(
ξ
) to zero yields a system of equations which can be solved to obtain A
i and ϕ
(
ξ
). Finally, substituting the values of A
i, ϕ
(
ξ
) and its derivative ϕ
′
(
ξ
) into (14) gives the exact solution of (12).


## 4. Application

### 4.1. Solutions for Space and Time Fractional BBM Equation via(
G
′
/
G
) Expansion Method

In this section, we apply the(
G
′
/
G
) expansion method to construct travelling wave solution of the space and time fractional Benjamin-Bona-Mahony equation. Consider the space and time fractional BBM equation (1) which can be written in subscript notation as
(15)
D
t
γ
u
+
D
x
σ
u
x
+
u
D
x
σ
u
x
-
D
x
2
σ
D
t
γ
u
=
0
.
We make the transformation u
(
x
,
t
)
=
u
(
ξ
), ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). Equation (15) becomes
(16)
(
1
-
c
)
u
′
+
u
u
′
+
c
u
′′′
=
0
.
Integrating (16) once with respect to ξ and setting the integration constant to zero yields
(17)
(
1
-
c
)
u
+
u
2
2
+
c
u
′′
=
0
.
To get m, we consider the homogenous balance between the highest order derivative u
′
′ and the highest order nonlinear term u
2:
(18)
2
m
=
m
+
2
⟹
m
=
2
.
Then (7) becomes
(19)
u
(
ξ
)
=
α
2
(
G
′
G
)
2
+
α
1
(
G
′
G
)
+
α
0
,
where
α
2
≠
0
,
where α
0, α
1, and α
2 are constants to be determined later. From (19),
(20)
u
′
(
ξ
)
=
-
2
α
2
(
G
′
G
)
3
+
(
-
2
α
2
λ
-
α
1
)
(
G
′
G
)
2
+
(
-
2
α
2
μ
-
α
1
λ
)
(
G
′
G
)
-
α
1
μ
,
u
′′
(
ξ
)
=
6
α
2
(
G
′
G
)
4
+
(
10
α
2
λ
+
2
α
1
)
(
G
′
G
)
3
+
(
8
α
2
μ
+
4
α
2
λ
2
+
3
α
1
λ
)
(
G
′
G
)
2
+
(
6
α
2
λ
μ
+
2
α
1
μ
+
α
1
λ
2
)
(
G
′
G
)
+
2
α
2
μ
2
+
α
1
λ
μ
.
Substituting (19) and its derivatives into (17) and collecting all terms with the same power of (
G
′
/
G
) together yields a simultaneous set of nonlinear algebraic equations as follows:
(21)
(
G
′
G
)
0
:
2
(
1
-
c
)
α
0
+
α
0
2
+
4
α
2
c
μ
2
+
2
α
1
c
λ
μ
=
0
,
(
G
′
G
)
1
:
(
1
-
c
)
α
1
+
α
0
α
1
+
6
α
2
c
λ
μ
+
2
α
1
c
μ
+
α
1
c
λ
2
=
0
,
(
G
′
G
)
2
:
2
(
1
-
c
)
α
2
+
2
α
0
α
2
+
α
1
2
+
16
α
2
c
μ
+
6
α
1
c
λ
+
8
α
2
c
λ
2
=
0
,
(
G
′
G
)
3
:
α
1
α
2
+
2
α
1
c
+
10
α
2
c
λ
=
0
,
(
G
′
G
)
4
:
α
2
2
+
12
α
2
c
=
0
.
Solving this algebraic system of equations with the aid of Mathematica yields the following solution:
(22)
α
2
=
-
12
c
,
α
1
=
-
12
c
λ
,
α
0
=
-
1
+
c
-
c
λ
2
-
8
c
μ
.
Substituting the solution to the algebraic equation and the general solution to (8) into (19), we obtain three types of travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation.Case 1.
Whenλ
2
-
4
μ
>
0, we obtain the following hyperbolic function solutions: (23)
u
(
ξ
)
=
-
12
c
(
λ
2
-
4
μ
2
[
C
1
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
C
1
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
]
-
λ
2
)
2
-
12
c
λ
(
λ
2
-
4
μ
2
[
C
1
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
C
1
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
]
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
whereξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
)
. If we set C
1
=
0 and C
2
≠
0 in (23), we obtain
(24)
u
1
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
coth
⁡
2
⁡
[
λ
2
-
4
μ
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
If we set C
1
≠
0 and C
2
=
0 in (23), we obtain
(25)
u
2
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
tanh
2
⁡
[
λ
2
-
4
μ
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
Applying Theorem 1 to (24) yields two further solutions:
(26)
u
3
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
+
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
-
2
,
u
4
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
-
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
-
2
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
).
Applying Theorem1 to (25) yields two further solutions:
(27)
u
5
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
+
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
2
,
u
6
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
-
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
2
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
).Case 2.
Whenλ
2
-
4
μ
<
0, we obtain the following trigonometric function solutions: (28)
u
(
ξ
)
=
-
12
c
(
4
μ
-
λ
2
2
[
-
C
1
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
C
1
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
]
-
λ
2
)
2
-
12
c
λ
(
4
μ
-
λ
2
2
[
-
C
1
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
C
1
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
]
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
,where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). If we set C
1
=
0 and C
2
≠
0 in (28), we obtain
(29)
u
7
(
ξ
)
=
-
1
+
c
-
2
c
(
4
μ
-
λ
2
)
-
3
c
(
4
μ
-
λ
2
)
×
cot
2
⁡
[
4
μ
-
λ
2
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
If we set C
1
≠
0 and C
2
=
0 in (28), we obtain
(30)
u
8
(
ξ
)
=
-
1
+
c
-
2
c
(
4
μ
-
λ
2
)
-
3
c
(
4
μ
-
λ
2
)
×
tan
2
⁡
[
4
μ
-
λ
2
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.Case 3.
Whenλ
2
-
4
μ
=
0, we obtain a rational function solution:
(31)
u
(
ξ
)
=
-
12
c
(
C
2
C
1
+
C
2
ξ
-
λ
2
)
2
-
12
c
λ
(
C
2
C
1
+
C
2
ξ
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). If we set C
1
=
0 and C
2
≠
0 in (31), we obtain
(32)
u
9
(
ξ
)
=
-
1
+
c
-
12
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
.
If we set C
1
≠
0 and C
2
=
0 in (31), we obtain
(33)
u
10
(
ξ
)
=
-
1
+
c
.


### 4.2. Solutions for Space and Time Fractional BBM Equation via MSE Method

Now, we apply the modified simple equation method to construct travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation. The procedure for transforming the nonlinear fractional PDE into an ODE in the MSE method is the same as that of the(
G
′
/
G
) expansion method. Hence, from (17), we have
(34)
(
1
-
c
)
u
+
u
2
2
+
c
u
′′
=
0
.
We get N
=
2 by considering the homogenous balance between the highest order derivative and the highest order nonlinear term. Then (14) becomes
(35)
u
(
ξ
)
=
A
2
(
ϕ
′
(
ξ
)
ϕ
(
ξ
)
)
2
+
A
1
(
ϕ
′
(
ξ
)
ϕ
(
ξ
)
)
+
A
0
,
where A
0, A
1, and A
2 are constants to be determined such that A
2
≠
0 and ϕ
(
ξ
) is an unidentified function to be determined. It is easy to find that
(36)
u
′
=
A
1
(
ϕ
′′
ϕ
)
-
A
1
(
ϕ
′
ϕ
)
2
+
2
A
2
(
ϕ
′
ϕ
′′
ϕ
2
)
-
2
A
2
(
ϕ
′
ϕ
)
3
,
u
′′
=
A
1
(
ϕ
′′′
ϕ
)
-
3
A
1
(
ϕ
′
ϕ
′′
ϕ
2
)
+
2
A
1
(
ϕ
′
ϕ
)
3
+
2
A
2
(
ϕ
′
ϕ
′′′
ϕ
2
)
+
2
A
2
(
ϕ
′′
ϕ
)
2
-
10
A
2
(
ϕ
′
2
ϕ
′′
ϕ
3
)
+
6
A
2
(
ϕ
′
ϕ
)
4
,
u
2
=
A
0
2
+
2
A
0
A
1
(
ϕ
′
ϕ
)
+
(
2
A
0
A
2
+
A
1
2
)
(
ϕ
′
ϕ
)
2
+
2
A
1
A
2
(
ϕ
′
ϕ
)
3
+
A
2
2
(
ϕ
′
ϕ
)
4
.
Substituting these values into (17) and equating the coefficients of ϕ
0, ϕ
-
1, ϕ
-
2, ϕ
-
3, and ϕ
-
4, respectively, to zero, we obtain
(37)
2
(
1
-
c
)
A
0
+
A
0
2
=
0
,
(38)
2
(
1
-
c
)
A
1
ϕ
′
+
2
A
0
A
1
ϕ
′
+
2
c
A
1
ϕ
′′′
=
0
,
(39)
2
(
1
-
c
)
A
2
ϕ
′
2
+
2
A
0
A
2
ϕ
′
2
+
A
1
2
ϕ
′
2
-
6
c
A
1
ϕ
′
ϕ
′′
+
4
c
A
2
ϕ
′
ϕ
′′′
+
4
c
A
2
ϕ
′′
2
=
0
,
(40)
2
A
1
A
2
ϕ
′
3
+
4
c
A
1
ϕ
′
3
-
20
c
A
2
ϕ
′
2
ϕ
′′
=
0
,
(41)
A
2
2
ϕ
′
4
+
12
c
A
2
ϕ
′
4
=
0
.
Solving (37) and (41), respectively, yields
(42)
A
0
=
0,2
(
c
-
1
)
,
A
2
=
-
12
c
h
h
h
2
(
where
c
≠
0
)
since
A
2
≠
0
.
From (38), (39), and (40), we have
(43)
A
1
=
12
c
ϕ
′′
ϕ
′
,
or
A
1
=
0
,
ϕ
′′
=
0
,
ϕ
′′′
ϕ
′
=
c
-
1
-
A
0
c
.Case 1.
WhenA
1
=
12
c
ϕ
′
′
/
ϕ
′, we get A
1
=
±
12
c
(
c
-
1
). This yields an absurd solution and hence this case is discarded.Case 2.
WhenA
1
=
0 and ϕ
′
′
=
0, we have
(44)
ϕ
′
(
ξ
)
=
α
1
,
where α
1 is a constant of integration. Integrating (44) with respect to ξ, we obtain
(45)
ϕ
(
ξ
)
=
α
1
ξ
+
α
2
,
where α
2 is a constant of integration. Substituting the value of ϕ
′ and ϕ from (44) and (45), respectively, into (35) yields
(46)
u
(
ξ
)
=
-
12
c
(
α
1
α
1
ξ
+
α
2
)
2
+
A
0
,
when A
0
=
0, we obtain the exact solution of (15) as
(47)
u
(
x
,
t
)
=
-
12
c
[
α
1
α
1
ξ
+
α
2
]
2
,
when A
0
=
2
(
c
-
1
), we obtain the exact solution of (15) as
(48)
u
(
x
,
t
)
=
-
12
c
[
α
1
α
1
ξ
+
α
2
]
2
+
2
(
c
-
1
)
.
Since α
1 and α
2 are arbitrary constants, therefore, if we set α
1
=
1 and α
2
=
0, we have the solutions (47) and (48) as
(49)
u
11
=
-
12
c
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
,
u
12
=
-
12
c
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
+
2
(
c
-
1
)
.
Bulent and Erdal [29] used direct algebraic method to get new complex travelling wave solutions to the BBM equation (15) by employing a transformation given by u
(
x
,
t
)
=
u
(
ξ
), ξ
=
i
k
(
x
-
c
t
). The results reported by [29] are
(50)
u
1
,
1
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
-
α
tanh
⁡
(
i
k
-
α
(
x
-
c
t
)
)
]
2
,
α
<
0
,
(51)
u
1
,
2
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
-
α
coth
⁡
⁡
(
i
k
-
α
(
x
-
c
t
)
)
]
2
,
α
<
0
,
(52)
u
1
,
3
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
α
tan
⁡
(
i
k
α
(
x
-
c
t
)
)
]
2
,
α
>
0
,
(53)
u
1
,
4
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
α
cot
⁡
(
i
k
α
(
x
-
c
t
)
)
]
2
,
α
>
0
,
(54)
u
1
,
5
=
-
1
i
k
(
x
-
c
t
)
,
α
=
0
.
A generalized (
G
′
/
G
) expansion method which uses the Klein-Gordon equation as the auxiliary equation was applied by Yanhong and Baodan [30] to a form of BBM equation given by
(55)
u
t
+
u
x
+
u
u
x
-
k
u
x
x
t
=
0
.
The travelling wave solutions to the BBM equation obtained by [30] using a transformation given by u
(
x
,
t
)
=
u
(
ξ
), ξ
=
x
-
c
t are
(56)
u
2
,
1
=
-
12
k
c
ξ
2
+
c
-
1
,
(57)
u
2
,
2
=
48
k
c
B
[
e
ξ
+
B
e
-
ξ
]
2
-
4
k
c
+
c
-
1
,
(58)
u
2
,
3
=
-
12
k
c
cosech
2
⁡
ξ
-
4
k
c
+
c
-
1
,
(59)
u
2
,
4
=
12
k
c
sech
2
⁡
ξ
-
4
k
c
+
c
-
1
,
(60)
u
2
,
5
=
-
12
k
c
tanh
2
⁡
ξ
-
12
k
c
coth
⁡
2
⁡
ξ
+
8
k
c
+
c
-
1
,
(61)
u
2
,
6
=
-
12
k
c
(
1
+
B
2
)
[
B
sin
⁡
ξ
+
cos
⁡
⁡
ξ
]
2
+
4
k
c
+
c
-
1
,
(62)
u
2
,
7
=
-
12
k
c
m
2
·
s
n
2
(
ξ
)
-
12
k
c
·
n
s
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
(63)
u
2
,
8
=
12
k
c
m
2
·
c
n
2
(
ξ
)
-
12
k
c
·
d
c
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
(64)
u
2
,
9
=
12
k
c
m
2
·
d
n
2
(
ξ
)
-
12
k
c
·
c
d
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
where s
n
(
ξ
), c
n
(
ξ
), d
n
(
ξ
), n
s
(
ξ
), d
c
(
ξ
), and c
d
(
ξ
) are Jacobi elliptic functions and 0
<
m
<
1 gives the modulus of the Jacobi elliptic functions. These functions generate hyperbolic functions when m
→
1 and trigonometric functions when m
→
0 as follows:(65)
m
⟶
1
s
n
(
ξ
)
⟶
tanh
⁡
(
ξ
)
c
n
(
ξ
)
⟶
sech
⁡
(
ξ
)
d
n
(
ξ
)
⟶
sech
⁡
(
ξ
)
m
⟶
0
s
n
(
ξ
)
⟶
sin
⁡
(
ξ
)
c
n
(
ξ
)
⟶
cos
⁡
⁡
(
ξ
)
d
n
(
ξ
)
⟶
1
n
s
(
ξ
)
=
1
s
n
(
ξ
)
d
c
(
ξ
)
=
d
n
(
ξ
)
c
n
(
ξ
)
c
d
(
ξ
)
=
c
n
(
ξ
)
d
n
(
ξ
)
.Remark 3.
When the arbitrary constantsC
1 and C
2 are taken to be zero separately in (23), (28), and (31), we get (24), (25), (29), (30), and (32) which are the travelling wave solutions to the space and time fractional BBM equation obtained via the (
G
′
/
G
) expansion method. Equations (49) are the travelling wave solution to the space and time fractional BBM equation obtained through the modified simple equation method. We see that the modified simple equation is not very effective for constructing travelling wave solutions to the space and time fractional BBM equation because the integer N obtained from considering the homogenous balance between the highest nonlinear terms and the highest derivative is greater than 1.Remark 4.
Applying Liu’s theorem to (24) and (25), we obtain new complex travelling wave solutions in the form of (26) and (27).Remark 5.
All the travelling wave solutions to the space and time fractional Benjamin-Bona-Mahony equation obtained were checked by putting them back into (1) with the aid of Mathematica.Remark 6.
Forσ
=
γ
=
1, k
=
-
i, and α
=
(
4
μ
-
λ
2
)
/
4, the complex travelling wave solutions u
1
,
1, u
1
,
2, u
1
,
3, and u
1
,
4 in (50)–(54) by [29] reduce to some of the solutions we obtained via the (
G
′
/
G
) expansion method; that is, u
1
,
1 (see (50)), u
1
,
2 (see (51)), u
1
,
3 (see (52)), and u
1
,
4 (see (53)) are the same as u
2 (see (25)), u
1 (see (24)), u
8 (see (30)), and u
7 (see (29)), respectively. Hence, the complex direct algebra method does not produce new complex solutions as suggested in [29]. Solution u
1
,
5 (see (54)) reported by [29] does not satisfy the space and time fractional BBM equation for σ
=
γ
=
1.Remark 7.
We also examine the solutions to the BBM equation from [30] for k
=
1. For σ
=
γ
=
1 and c
=
1, u
2
,
1 (see (56)) reduces to u
9 (see (32)). For λ
2
-
4
μ
=
4, u
2
,
3 (see (58)) and u
2
,
4 (see (59)) reduce to u
1 (see (24)) and u
2 (see (25)), respectively. When the modulus of the Jacobi elliptic functions is m
=
1, then both u
2
,
8 (see (63)) and u
2
,
9 (see (64)) reduce to u
2 (see (25)) for λ
2
-
4
μ
=
4. We observe that u
7 (see (29)) and u
8 (see (30)) are, respectively, the same as u
2
,
7 (see (62)) and u
2
,
8 (see (63)) when λ
2
-
4
μ
=
-
4 and m
=
0. It should be noted that u
2
,
2 (see (57)), u
2
,
5 (see (60)), and u
2
,
6 (see (61)) are not valid solutions to the space and time fractional BBM equation for σ
=
γ
=
1. Also u
2
,
7 (see (62)) and u
2
,
9 (see (64)) for m
=
1 and m
=
0, respectively, do not satisfy the space and time fractional BBM equation for σ
=
γ
=
1.The shapes of (24), (25), (29), and (30) for selected values ofc, λ, and μ at different σ and γ values are presented in Figures 1–4. The solution u
1 (see (24)) is a singular kink shaped travelling wave solution of (15). Figure 1 shows the shape of (24) with c
=
1, λ
=
3, and μ
=
2 for the interval 0
≤
x, t
≤
10 at different σ and γ values. From Figure 1 and u
(
x
,
t
)
=
u
(
x
-
c
t
), we observe that, for c
>
0, the propagation of the wave will be in the positive x-direction. If we take c
<
0, then the propagation of the wave will be in the negative x-direction. The solution u
2 (see (25)) is a bell-shaped Soliton solution of (15). Figure 2 shows the shape of (25) with c
=
1, λ
=
3, and μ
=
2 for the interval 0
≤
x
,
t
≤
10 at different σ and γ values. From Figure 2 and u
(
x
,
t
)
=
u
(
x
-
c
t
), we also observe that, for c
>
0, the propagation of the wave will be in the positive x-direction. If we take c
<
0, then the propagation of the wave will be in the negative x-direction. Figures 3 and 4 show, respectively, the shape of u
3 (see (29)) and u
4 (see (30)) with c
=
1 and λ
=
μ
=
2 for the interval 0
≤
x, t
≤
10 at different σ and γ values. u
3 (see (29)) and u
4 (see (30)) are exact periodic travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation.Shape of (24) with c
=
1, λ
=
3,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (25) with c
=
1, λ
=
3, μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (29) with c
=
1, λ
=
2,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (30) with c
=
1, λ
=
2,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)


## 4.1. Solutions for Space and Time Fractional BBM Equation via(
G
′
/
G
) Expansion Method

In this section, we apply the(
G
′
/
G
) expansion method to construct travelling wave solution of the space and time fractional Benjamin-Bona-Mahony equation. Consider the space and time fractional BBM equation (1) which can be written in subscript notation as
(15)
D
t
γ
u
+
D
x
σ
u
x
+
u
D
x
σ
u
x
-
D
x
2
σ
D
t
γ
u
=
0
.
We make the transformation u
(
x
,
t
)
=
u
(
ξ
), ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). Equation (15) becomes
(16)
(
1
-
c
)
u
′
+
u
u
′
+
c
u
′′′
=
0
.
Integrating (16) once with respect to ξ and setting the integration constant to zero yields
(17)
(
1
-
c
)
u
+
u
2
2
+
c
u
′′
=
0
.
To get m, we consider the homogenous balance between the highest order derivative u
′
′ and the highest order nonlinear term u
2:
(18)
2
m
=
m
+
2
⟹
m
=
2
.
Then (7) becomes
(19)
u
(
ξ
)
=
α
2
(
G
′
G
)
2
+
α
1
(
G
′
G
)
+
α
0
,
where
α
2
≠
0
,
where α
0, α
1, and α
2 are constants to be determined later. From (19),
(20)
u
′
(
ξ
)
=
-
2
α
2
(
G
′
G
)
3
+
(
-
2
α
2
λ
-
α
1
)
(
G
′
G
)
2
+
(
-
2
α
2
μ
-
α
1
λ
)
(
G
′
G
)
-
α
1
μ
,
u
′′
(
ξ
)
=
6
α
2
(
G
′
G
)
4
+
(
10
α
2
λ
+
2
α
1
)
(
G
′
G
)
3
+
(
8
α
2
μ
+
4
α
2
λ
2
+
3
α
1
λ
)
(
G
′
G
)
2
+
(
6
α
2
λ
μ
+
2
α
1
μ
+
α
1
λ
2
)
(
G
′
G
)
+
2
α
2
μ
2
+
α
1
λ
μ
.
Substituting (19) and its derivatives into (17) and collecting all terms with the same power of (
G
′
/
G
) together yields a simultaneous set of nonlinear algebraic equations as follows:
(21)
(
G
′
G
)
0
:
2
(
1
-
c
)
α
0
+
α
0
2
+
4
α
2
c
μ
2
+
2
α
1
c
λ
μ
=
0
,
(
G
′
G
)
1
:
(
1
-
c
)
α
1
+
α
0
α
1
+
6
α
2
c
λ
μ
+
2
α
1
c
μ
+
α
1
c
λ
2
=
0
,
(
G
′
G
)
2
:
2
(
1
-
c
)
α
2
+
2
α
0
α
2
+
α
1
2
+
16
α
2
c
μ
+
6
α
1
c
λ
+
8
α
2
c
λ
2
=
0
,
(
G
′
G
)
3
:
α
1
α
2
+
2
α
1
c
+
10
α
2
c
λ
=
0
,
(
G
′
G
)
4
:
α
2
2
+
12
α
2
c
=
0
.
Solving this algebraic system of equations with the aid of Mathematica yields the following solution:
(22)
α
2
=
-
12
c
,
α
1
=
-
12
c
λ
,
α
0
=
-
1
+
c
-
c
λ
2
-
8
c
μ
.
Substituting the solution to the algebraic equation and the general solution to (8) into (19), we obtain three types of travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation.Case 1.
Whenλ
2
-
4
μ
>
0, we obtain the following hyperbolic function solutions: (23)
u
(
ξ
)
=
-
12
c
(
λ
2
-
4
μ
2
[
C
1
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
C
1
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
]
-
λ
2
)
2
-
12
c
λ
(
λ
2
-
4
μ
2
[
C
1
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
C
1
cosh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
+
C
2
sinh
⁡
⁡
(
(
λ
2
-
4
μ
/
2
)
ξ
)
]
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
whereξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
)
. If we set C
1
=
0 and C
2
≠
0 in (23), we obtain
(24)
u
1
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
coth
⁡
2
⁡
[
λ
2
-
4
μ
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
If we set C
1
≠
0 and C
2
=
0 in (23), we obtain
(25)
u
2
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
tanh
2
⁡
[
λ
2
-
4
μ
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
Applying Theorem 1 to (24) yields two further solutions:
(26)
u
3
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
+
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
-
2
,
u
4
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
-
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
-
2
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
).
Applying Theorem1 to (25) yields two further solutions:
(27)
u
5
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
+
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
2
,
u
6
(
ξ
)
=
-
1
+
c
+
2
c
(
λ
2
-
4
μ
)
-
3
c
(
λ
2
-
4
μ
)
×
[
tanh
⁡
(
λ
2
-
4
μ
ξ
)
-
i
sech
⁡
(
λ
2
-
4
μ
ξ
)
]
2
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
).Case 2.
Whenλ
2
-
4
μ
<
0, we obtain the following trigonometric function solutions: (28)
u
(
ξ
)
=
-
12
c
(
4
μ
-
λ
2
2
[
-
C
1
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
C
1
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
]
-
λ
2
)
2
-
12
c
λ
(
4
μ
-
λ
2
2
[
-
C
1
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
C
1
cos
⁡
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
+
C
2
sin
⁡
(
(
4
μ
-
λ
2
/
2
)
ξ
)
]
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
,where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). If we set C
1
=
0 and C
2
≠
0 in (28), we obtain
(29)
u
7
(
ξ
)
=
-
1
+
c
-
2
c
(
4
μ
-
λ
2
)
-
3
c
(
4
μ
-
λ
2
)
×
cot
2
⁡
[
4
μ
-
λ
2
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.
If we set C
1
≠
0 and C
2
=
0 in (28), we obtain
(30)
u
8
(
ξ
)
=
-
1
+
c
-
2
c
(
4
μ
-
λ
2
)
-
3
c
(
4
μ
-
λ
2
)
×
tan
2
⁡
[
4
μ
-
λ
2
2
(
x
σ
Γ
(
1
+
σ
)
-
c
t
γ
Γ
(
1
+
γ
)
)
]
.Case 3.
Whenλ
2
-
4
μ
=
0, we obtain a rational function solution:
(31)
u
(
ξ
)
=
-
12
c
(
C
2
C
1
+
C
2
ξ
-
λ
2
)
2
-
12
c
λ
(
C
2
C
1
+
C
2
ξ
-
λ
2
)
-
1
+
c
-
c
λ
2
-
8
c
μ
,
where ξ
=
x
σ
/
Γ
(
1
+
σ
)
-
c
t
γ
/
Γ
(
1
+
γ
). If we set C
1
=
0 and C
2
≠
0 in (31), we obtain
(32)
u
9
(
ξ
)
=
-
1
+
c
-
12
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
.
If we set C
1
≠
0 and C
2
=
0 in (31), we obtain
(33)
u
10
(
ξ
)
=
-
1
+
c
.


## 4.2. Solutions for Space and Time Fractional BBM Equation via MSE Method

Now, we apply the modified simple equation method to construct travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation. The procedure for transforming the nonlinear fractional PDE into an ODE in the MSE method is the same as that of the(
G
′
/
G
) expansion method. Hence, from (17), we have
(34)
(
1
-
c
)
u
+
u
2
2
+
c
u
′′
=
0
.
We get N
=
2 by considering the homogenous balance between the highest order derivative and the highest order nonlinear term. Then (14) becomes
(35)
u
(
ξ
)
=
A
2
(
ϕ
′
(
ξ
)
ϕ
(
ξ
)
)
2
+
A
1
(
ϕ
′
(
ξ
)
ϕ
(
ξ
)
)
+
A
0
,
where A
0, A
1, and A
2 are constants to be determined such that A
2
≠
0 and ϕ
(
ξ
) is an unidentified function to be determined. It is easy to find that
(36)
u
′
=
A
1
(
ϕ
′′
ϕ
)
-
A
1
(
ϕ
′
ϕ
)
2
+
2
A
2
(
ϕ
′
ϕ
′′
ϕ
2
)
-
2
A
2
(
ϕ
′
ϕ
)
3
,
u
′′
=
A
1
(
ϕ
′′′
ϕ
)
-
3
A
1
(
ϕ
′
ϕ
′′
ϕ
2
)
+
2
A
1
(
ϕ
′
ϕ
)
3
+
2
A
2
(
ϕ
′
ϕ
′′′
ϕ
2
)
+
2
A
2
(
ϕ
′′
ϕ
)
2
-
10
A
2
(
ϕ
′
2
ϕ
′′
ϕ
3
)
+
6
A
2
(
ϕ
′
ϕ
)
4
,
u
2
=
A
0
2
+
2
A
0
A
1
(
ϕ
′
ϕ
)
+
(
2
A
0
A
2
+
A
1
2
)
(
ϕ
′
ϕ
)
2
+
2
A
1
A
2
(
ϕ
′
ϕ
)
3
+
A
2
2
(
ϕ
′
ϕ
)
4
.
Substituting these values into (17) and equating the coefficients of ϕ
0, ϕ
-
1, ϕ
-
2, ϕ
-
3, and ϕ
-
4, respectively, to zero, we obtain
(37)
2
(
1
-
c
)
A
0
+
A
0
2
=
0
,
(38)
2
(
1
-
c
)
A
1
ϕ
′
+
2
A
0
A
1
ϕ
′
+
2
c
A
1
ϕ
′′′
=
0
,
(39)
2
(
1
-
c
)
A
2
ϕ
′
2
+
2
A
0
A
2
ϕ
′
2
+
A
1
2
ϕ
′
2
-
6
c
A
1
ϕ
′
ϕ
′′
+
4
c
A
2
ϕ
′
ϕ
′′′
+
4
c
A
2
ϕ
′′
2
=
0
,
(40)
2
A
1
A
2
ϕ
′
3
+
4
c
A
1
ϕ
′
3
-
20
c
A
2
ϕ
′
2
ϕ
′′
=
0
,
(41)
A
2
2
ϕ
′
4
+
12
c
A
2
ϕ
′
4
=
0
.
Solving (37) and (41), respectively, yields
(42)
A
0
=
0,2
(
c
-
1
)
,
A
2
=
-
12
c
h
h
h
2
(
where
c
≠
0
)
since
A
2
≠
0
.
From (38), (39), and (40), we have
(43)
A
1
=
12
c
ϕ
′′
ϕ
′
,
or
A
1
=
0
,
ϕ
′′
=
0
,
ϕ
′′′
ϕ
′
=
c
-
1
-
A
0
c
.Case 1.
WhenA
1
=
12
c
ϕ
′
′
/
ϕ
′, we get A
1
=
±
12
c
(
c
-
1
). This yields an absurd solution and hence this case is discarded.Case 2.
WhenA
1
=
0 and ϕ
′
′
=
0, we have
(44)
ϕ
′
(
ξ
)
=
α
1
,
where α
1 is a constant of integration. Integrating (44) with respect to ξ, we obtain
(45)
ϕ
(
ξ
)
=
α
1
ξ
+
α
2
,
where α
2 is a constant of integration. Substituting the value of ϕ
′ and ϕ from (44) and (45), respectively, into (35) yields
(46)
u
(
ξ
)
=
-
12
c
(
α
1
α
1
ξ
+
α
2
)
2
+
A
0
,
when A
0
=
0, we obtain the exact solution of (15) as
(47)
u
(
x
,
t
)
=
-
12
c
[
α
1
α
1
ξ
+
α
2
]
2
,
when A
0
=
2
(
c
-
1
), we obtain the exact solution of (15) as
(48)
u
(
x
,
t
)
=
-
12
c
[
α
1
α
1
ξ
+
α
2
]
2
+
2
(
c
-
1
)
.
Since α
1 and α
2 are arbitrary constants, therefore, if we set α
1
=
1 and α
2
=
0, we have the solutions (47) and (48) as
(49)
u
11
=
-
12
c
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
,
u
12
=
-
12
c
(
x
σ
/
Γ
(
1
+
σ
)
-
c
(
t
γ
/
Γ
(
1
+
γ
)
)
)
2
+
2
(
c
-
1
)
.
Bulent and Erdal [29] used direct algebraic method to get new complex travelling wave solutions to the BBM equation (15) by employing a transformation given by u
(
x
,
t
)
=
u
(
ξ
), ξ
=
i
k
(
x
-
c
t
). The results reported by [29] are
(50)
u
1
,
1
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
-
α
tanh
⁡
(
i
k
-
α
(
x
-
c
t
)
)
]
2
,
α
<
0
,
(51)
u
1
,
2
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
-
α
coth
⁡
⁡
(
i
k
-
α
(
x
-
c
t
)
)
]
2
,
α
<
0
,
(52)
u
1
,
3
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
α
tan
⁡
(
i
k
α
(
x
-
c
t
)
)
]
2
,
α
>
0
,
(53)
u
1
,
4
=
-
1
+
c
+
8
c
k
2
α
+
12
c
k
2
×
[
-
α
cot
⁡
(
i
k
α
(
x
-
c
t
)
)
]
2
,
α
>
0
,
(54)
u
1
,
5
=
-
1
i
k
(
x
-
c
t
)
,
α
=
0
.
A generalized (
G
′
/
G
) expansion method which uses the Klein-Gordon equation as the auxiliary equation was applied by Yanhong and Baodan [30] to a form of BBM equation given by
(55)
u
t
+
u
x
+
u
u
x
-
k
u
x
x
t
=
0
.
The travelling wave solutions to the BBM equation obtained by [30] using a transformation given by u
(
x
,
t
)
=
u
(
ξ
), ξ
=
x
-
c
t are
(56)
u
2
,
1
=
-
12
k
c
ξ
2
+
c
-
1
,
(57)
u
2
,
2
=
48
k
c
B
[
e
ξ
+
B
e
-
ξ
]
2
-
4
k
c
+
c
-
1
,
(58)
u
2
,
3
=
-
12
k
c
cosech
2
⁡
ξ
-
4
k
c
+
c
-
1
,
(59)
u
2
,
4
=
12
k
c
sech
2
⁡
ξ
-
4
k
c
+
c
-
1
,
(60)
u
2
,
5
=
-
12
k
c
tanh
2
⁡
ξ
-
12
k
c
coth
⁡
2
⁡
ξ
+
8
k
c
+
c
-
1
,
(61)
u
2
,
6
=
-
12
k
c
(
1
+
B
2
)
[
B
sin
⁡
ξ
+
cos
⁡
⁡
ξ
]
2
+
4
k
c
+
c
-
1
,
(62)
u
2
,
7
=
-
12
k
c
m
2
·
s
n
2
(
ξ
)
-
12
k
c
·
n
s
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
(63)
u
2
,
8
=
12
k
c
m
2
·
c
n
2
(
ξ
)
-
12
k
c
·
d
c
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
(64)
u
2
,
9
=
12
k
c
m
2
·
d
n
2
(
ξ
)
-
12
k
c
·
c
d
2
(
ξ
)
+
4
k
c
m
2
+
4
k
c
+
c
-
1
,
where s
n
(
ξ
), c
n
(
ξ
), d
n
(
ξ
), n
s
(
ξ
), d
c
(
ξ
), and c
d
(
ξ
) are Jacobi elliptic functions and 0
<
m
<
1 gives the modulus of the Jacobi elliptic functions. These functions generate hyperbolic functions when m
→
1 and trigonometric functions when m
→
0 as follows:(65)
m
⟶
1
s
n
(
ξ
)
⟶
tanh
⁡
(
ξ
)
c
n
(
ξ
)
⟶
sech
⁡
(
ξ
)
d
n
(
ξ
)
⟶
sech
⁡
(
ξ
)
m
⟶
0
s
n
(
ξ
)
⟶
sin
⁡
(
ξ
)
c
n
(
ξ
)
⟶
cos
⁡
⁡
(
ξ
)
d
n
(
ξ
)
⟶
1
n
s
(
ξ
)
=
1
s
n
(
ξ
)
d
c
(
ξ
)
=
d
n
(
ξ
)
c
n
(
ξ
)
c
d
(
ξ
)
=
c
n
(
ξ
)
d
n
(
ξ
)
.Remark 3.
When the arbitrary constantsC
1 and C
2 are taken to be zero separately in (23), (28), and (31), we get (24), (25), (29), (30), and (32) which are the travelling wave solutions to the space and time fractional BBM equation obtained via the (
G
′
/
G
) expansion method. Equations (49) are the travelling wave solution to the space and time fractional BBM equation obtained through the modified simple equation method. We see that the modified simple equation is not very effective for constructing travelling wave solutions to the space and time fractional BBM equation because the integer N obtained from considering the homogenous balance between the highest nonlinear terms and the highest derivative is greater than 1.Remark 4.
Applying Liu’s theorem to (24) and (25), we obtain new complex travelling wave solutions in the form of (26) and (27).Remark 5.
All the travelling wave solutions to the space and time fractional Benjamin-Bona-Mahony equation obtained were checked by putting them back into (1) with the aid of Mathematica.Remark 6.
Forσ
=
γ
=
1, k
=
-
i, and α
=
(
4
μ
-
λ
2
)
/
4, the complex travelling wave solutions u
1
,
1, u
1
,
2, u
1
,
3, and u
1
,
4 in (50)–(54) by [29] reduce to some of the solutions we obtained via the (
G
′
/
G
) expansion method; that is, u
1
,
1 (see (50)), u
1
,
2 (see (51)), u
1
,
3 (see (52)), and u
1
,
4 (see (53)) are the same as u
2 (see (25)), u
1 (see (24)), u
8 (see (30)), and u
7 (see (29)), respectively. Hence, the complex direct algebra method does not produce new complex solutions as suggested in [29]. Solution u
1
,
5 (see (54)) reported by [29] does not satisfy the space and time fractional BBM equation for σ
=
γ
=
1.Remark 7.
We also examine the solutions to the BBM equation from [30] for k
=
1. For σ
=
γ
=
1 and c
=
1, u
2
,
1 (see (56)) reduces to u
9 (see (32)). For λ
2
-
4
μ
=
4, u
2
,
3 (see (58)) and u
2
,
4 (see (59)) reduce to u
1 (see (24)) and u
2 (see (25)), respectively. When the modulus of the Jacobi elliptic functions is m
=
1, then both u
2
,
8 (see (63)) and u
2
,
9 (see (64)) reduce to u
2 (see (25)) for λ
2
-
4
μ
=
4. We observe that u
7 (see (29)) and u
8 (see (30)) are, respectively, the same as u
2
,
7 (see (62)) and u
2
,
8 (see (63)) when λ
2
-
4
μ
=
-
4 and m
=
0. It should be noted that u
2
,
2 (see (57)), u
2
,
5 (see (60)), and u
2
,
6 (see (61)) are not valid solutions to the space and time fractional BBM equation for σ
=
γ
=
1. Also u
2
,
7 (see (62)) and u
2
,
9 (see (64)) for m
=
1 and m
=
0, respectively, do not satisfy the space and time fractional BBM equation for σ
=
γ
=
1.The shapes of (24), (25), (29), and (30) for selected values ofc, λ, and μ at different σ and γ values are presented in Figures 1–4. The solution u
1 (see (24)) is a singular kink shaped travelling wave solution of (15). Figure 1 shows the shape of (24) with c
=
1, λ
=
3, and μ
=
2 for the interval 0
≤
x, t
≤
10 at different σ and γ values. From Figure 1 and u
(
x
,
t
)
=
u
(
x
-
c
t
), we observe that, for c
>
0, the propagation of the wave will be in the positive x-direction. If we take c
<
0, then the propagation of the wave will be in the negative x-direction. The solution u
2 (see (25)) is a bell-shaped Soliton solution of (15). Figure 2 shows the shape of (25) with c
=
1, λ
=
3, and μ
=
2 for the interval 0
≤
x
,
t
≤
10 at different σ and γ values. From Figure 2 and u
(
x
,
t
)
=
u
(
x
-
c
t
), we also observe that, for c
>
0, the propagation of the wave will be in the positive x-direction. If we take c
<
0, then the propagation of the wave will be in the negative x-direction. Figures 3 and 4 show, respectively, the shape of u
3 (see (29)) and u
4 (see (30)) with c
=
1 and λ
=
μ
=
2 for the interval 0
≤
x, t
≤
10 at different σ and γ values. u
3 (see (29)) and u
4 (see (30)) are exact periodic travelling wave solutions of the space and time fractional Benjamin-Bona-Mahony equation.Shape of (24) with c
=
1, λ
=
3,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (25) with c
=
1, λ
=
3, μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (29) with c
=
1, λ
=
2,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)Shape of (30) with c
=
1, λ
=
2,μ
=
2, and 0
≤
x, t
≤
10 for (a) σ
=
γ
=
1, (b) σ
=
0.5 and γ
=
1, (c) σ
=
1 and γ
=
0.5, and (d) σ
=
γ
=
0.5.
(a)
(b)
(c)
(d)


## 5. Conclusion

Hyperbolic, trigonometric, and rational function travelling wave solutions to the space and time fractional Benjamin-Bona-Mahony equation defined in the sense of Jumarie’s modified Riemann-Liouville derivative have been obtained using the(
G
′
/
G
) expansion method and modified simple equation method. Liu’s theorem was also applied to obtain new complex travelling wave solutions to the space and time fractional BBM equation. The results were compared to those obtained by Bulent and Erdal [29] using the direct algebraic method for complex travelling wave solution and those by Yanhong and Baodan [30] using the generalized (
G
′
/
G
) expansion method with the Klein-Gordon equation as an auxiliary equation. The (
G
′
/
G
) expansion method is equivalent to the direct algebraic method and the generalized (
G
′
/
G
) expansion method with some of their reported solutions found not to satisfy the BBM equation. Also, the modified simple equation method is not effective for constructing travelling wave solutions to the space and time fractional BBM equation because the index N obtained from considering the homogenous balance between the highest nonlinear term and the highest derivative is greater than 1. The improvement of the modified simple equation scheme for evolution equations with N
>
1 is still an open problem.



---

*Source: 217184-2014-03-06.xml*