# Comment on “New Exact Solutions to the KdV-Burgers-Kuramoto Equation with the Exp-Function Method”

Authors: Hong-Zhun Liu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We point out in this paper that the claims made by Kim et al. in the commented paper are incorrect and no new exact solution was obtained.


---

## Body

## 1. Introduction

In [1], Kim and Chun had investigated exact solutions to the following KdV-Burgers-Kuramoto equation:
(1)
u
t
+
ν
u
u
x
+
μ
u
u
x
x
x
+
α
u
x
x
+
γ
u
x
x
x
x
=
0
.
They took traveling wave transformation u
=
u
(
η
) with η
=
k
x
+
ω
t into account and transformed (1) into an ordinary differential equation:
(2)
ω
u
′
+
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
=
0
.
After implementing the Exp-function method [2] based on the truncated Painlevé, they had constructed the following four new generalized solitary wave solutions to (1).Case 1.
Consider the following:(3)
u
(
x
,
t
)
=
15
k
76
γ
ν
c
3
Y
6
+
c
2
Y
5
+
c
1
Y
4
+
c
0
Y
3
+
c
-
1
Y
2
+
c
-
2
Y
+
c
-
3
(
b
0
+
b
1
Y
)
6
,
where
(4)
c
-
3
=
(
α
2
-
16
μ
γ
)
b
1
6
,
c
-
2
=
(
6
α
2
-
96
μ
γ
)
b
1
5
b
0
,
c
-
1
=
(
15
α
2
-
240
μ
γ
)
b
1
4
b
0
2
,
c
0
=
(
20
α
2
-
320
μ
γ
)
b
1
3
b
0
3
,
c
1
=
(
15
α
2
-
240
μ
γ
)
b
1
2
b
0
4
,
c
2
=
(
6
α
2
-
96
μ
γ
)
b
1
b
0
5
,
c
3
=
(
α
2
-
16
μ
γ
)
b
0
6
,
Y
≜
exp
⁡
(
η
), as in the following, is an introduced variable, and k, ω, b
0, and b
1 are arbitrary constants.Other Three Cases. Consider
(5)
u
(
x
,
t
)
=
K
·
c
3
Y
6
+
c
2
Y
5
+
c
1
Y
4
+
c
0
Y
3
+
c
-
1
Y
2
+
c
-
2
Y
+
c
-
3
(
b
0
+
b
-
1
Y
)
6
,
where
(6)
c
-
3
=
(
α
2
-
16
μ
γ
)
b
0
6
,
c
-
2
=
(
6
α
2
-
96
μ
γ
)
b
0
5
b
-
1
,
c
-
1
=
(
15
α
2
-
240
μ
γ
)
b
0
4
b
-
1
2
,
c
0
=
(
20
α
2
-
320
μ
γ
)
b
0
3
b
-
1
3
,
c
1
=
(
15
α
2
-
240
μ
γ
)
b
0
2
b
-
1
4
,
c
2
=
(
6
α
2
-
96
μ
γ
)
b
0
b
-
1
5
,
c
3
=
(
α
2
-
16
μ
γ
)
b
-
1
6
,
and b
0, b
-
1 are arbitrary constants.Case 2.
K
=
-
15
k
/
76
γ
ν, k is an arbitrary constant, and ω subjects to
(7)
ω
=
k
2
(
15
α
2
-
76
α
k
γ
-
76
γ
2
k
2
-
316
μ
γ
)
76
γCase 3.
K
=
-
15
k
/
76
γ
ν, and k and ω subject to
(8)
k
=
±
-
μ
γ
γ
,
ω
=
μ
(
240
μ
γ
±
76
α
-
μ
γ
-
15
α
2
)
76
γ
2
.Case 4.
K
=
-
15
k
/
2888
γ
ν, and k and ω subject to
(9)
k
=
-
19
α
γ
±
437
α
2
γ
2
-
1216
γ
3
μ
152
γ
2
,
ω
=
20848
μ
α
2
γ
+
39936
μ
2
γ
2
+
707
α
4
-
184832
γ
3
-
(
1168
γ
α
μ
-
41
α
3
)
437
α
2
-
1216
μ
γ
184832
γ
3
.
The authors claimed that the above four solutions could not be directly constructed form the Exp-function method.They also claimed that they had obtained a new solitary wave solution in the case wherep
=
q
=
2 and d
=
q
=
2, which was given by
(10)
u
(
x
,
t
)
=
a
2
exp
⁡
⁡
(
2
η
)
+
a
2
b
1
exp
⁡
⁡
(
η
)
+
a
2
b
0
+
a
2
b
-
1
exp
⁡
⁡
(
-
η
)
exp
⁡
⁡
(
2
η
)
+
b
1
exp
⁡
⁡
(
η
)
+
b
0
+
b
-
1
exp
⁡
⁡
(
-
η
)
.


## 2. Comment and Analysis

In this section, we will analyze the claims by Kim and Chun in [1].


### 2.1. Comment 1

It is not difficult to rewrite (3) and (5) in the following form:
(11)
u
(
x
,
t
)
=
A
6
Y
6
+
A
5
Y
5
+
A
4
Y
4
+
A
3
Y
3
+
A
2
Y
2
+
A
1
Y
+
A
0
B
6
Y
6
+
B
5
Y
5
+
B
4
Y
4
+
B
3
Y
3
+
B
2
Y
2
+
B
1
Y
+
B
0
,
where A
i and B
i are certain constants.Therefore, we have(12)
u
(
x
,
t
)
=
(
A
6
Y
6
+
A
5
Y
5
+
A
4
Y
4
+
A
3
Y
3
+
A
2
Y
2
+
A
1
Y
+
A
0
)
×
(
B
6
Y
6
+
B
5
Y
5
+
B
4
Y
4
+
B
3
Y
3
+
B
2
Y
2
+
B
1
Y
+
B
0
)
-
1
=
(
A
6
exp
⁡
(
6
η
)
+
A
5
exp
⁡
(
5
η
)
+
A
4
exp
⁡
(
4
η
)
+
A
3
exp
⁡
(
3
η
)
+
A
2
exp
⁡
(
2
η
)
+
A
1
exp
⁡
(
η
)
+
A
0
)
×
(
B
6
exp
⁡
(
6
η
)
+
B
5
exp
⁡
(
5
η
)
+
B
4
exp
⁡
(
4
η
)
vv
+
B
3
exp
⁡
(
3
η
)
+
B
2
exp
⁡
(
2
η
)
+
B
1
exp
⁡
(
η
)
+
B
0
)
-
1
=
(
A
6
exp
⁡
(
5
η
)
+
A
5
exp
⁡
(
4
η
)
+
A
4
exp
⁡
(
3
η
)
+
A
3
exp
⁡
(
2
η
)
+
A
2
exp
⁡
(
η
)
+
A
1
+
A
0
exp
⁡
(
-
η
)
)
×
(
B
6
exp
⁡
(
5
η
)
+
B
5
exp
⁡
(
4
η
)
+
B
4
exp
⁡
(
3
η
)
vv
+
B
3
exp
⁡
(
2
η
)
+
B
2
exp
⁡
(
η
)
+
B
1
+
B
0
exp
⁡
(
-
η
)
)
-
1
=
(
A
6
exp
⁡
(
4
η
)
+
A
5
exp
⁡
(
3
η
)
+
A
4
exp
⁡
(
2
η
)
+
A
3
exp
⁡
(
η
)
+
A
2
+
A
1
exp
⁡
(
-
η
)
+
A
0
exp
⁡
(
-
2
η
)
)
×
(
B
6
exp
⁡
(
4
η
)
+
B
5
exp
⁡
(
3
η
)
+
B
4
exp
⁡
(
2
η
)
vv
+
B
3
exp
⁡
(
η
)
+
B
2
+
B
1
exp
⁡
(
-
η
)
+
B
0
exp
⁡
(
-
2
η
)
)
-
1
=
(
A
6
exp
⁡
(
3
η
)
+
A
5
exp
⁡
(
2
η
)
+
A
4
exp
⁡
(
η
)
+
A
3
+
A
2
exp
⁡
(
-
η
)
+
A
1
exp
⁡
(
-
2
η
)
+
A
0
exp
⁡
(
-
3
η
)
)
×
(
B
6
exp
⁡
(
3
η
)
+
B
5
exp
⁡
(
2
η
)
+
B
4
exp
⁡
(
η
)
+
B
3
vv
+
B
2
exp
⁡
⁡
(
-
η
)
+
B
1
exp
⁡
⁡
(
-
2
η
)
+
B
0
exp
⁡
⁡
(
-
3
η
)
)
-
1
.
According to the Exp-function method [2], we can reobtain the solution (11) by assuming that the solution of (2) can be expressed in the form
(13)
u
(
η
)
=
M
-
d
exp
⁡
(
-
d
η
)
+
⋯
+
M
c
exp
⁡
(
c
η
)
N
-
q
exp
⁡
(
-
q
η
)
+
⋯
+
N
p
exp
⁡
(
p
η
)
,
where (i) p
=
c
=
5 and d
=
q
=
1, (ii) p
=
c
=
4 and d
=
q
=
2, (iii) p
=
c
=
d
=
q
=
3, respectively. It is worth to mention that the fact that the three cases of (i), (ii), and (iii) are equivalent has been emphasized in [3–5]. Thus the claim in the commented paper that “new generalized solitary wave solutions are constructed for the KdV-Burgers-Kuramoto equation, which cannot be directly constructed from the Exp-function method” is not true.


### 2.2. Comment 2

In this section, we show that the solutions in the mentioned four cases are incorrect. Here, we should point out that it is difficult for us to solve original algebra system appearing in [1] and therefore we verify the four cases in an ad hoc way.


#### 2.2.1.  Solution Analysis

At the beginning, substituting (4) into (3), we find
(14)
u
(
x
,
t
)
=
15
k
(
α
2
-
16
μ
γ
)
76
γ
ν
·
(
b
0
Y
+
b
1
)
6
(
b
0
+
b
1
Y
)
6
,
and substituting (6) into (5), we find
(15)
u
(
x
,
t
)
=
K
(
α
2
-
16
μ
γ
)
·
(
b
0
+
b
-
1
Y
)
6
(
b
0
Y
+
b
-
1
)
6
.So we assume that the solution of (2) can be expressed in the form
(16)
u
(
η
)
=
L
·
(
k
0
+
k
1
Y
)
6
(
k
1
+
k
0
Y
)
6
,
where L, k
0, and k
1 are constants to be determined.We emphasize that (16) needs only three undetermined parameters. Unlike (5), there are ten undetermined parameters. Hence assumption (16) can reduce the computation burden.Substituting (16) into (2) and setting the coefficients of all powers of Y
i to zero yield a system of algebraic equations for L, k
0, k
1, k, and ω. Solving these algebraic equations, we can determine certain solutions to (2). Obviously, these solutions cover (14) and (15). However, it is to our surprise that with the aid of Maple we determine none of nontrivial solutions after solving the above system equations. Hence further verification should be made.


#### 2.2.2. Solution Check

In what follows, after careful numerical inspection, we show that the four cases are incorrect.Firstly, we check the solution in Case1, namely, (3) and (4) (or (14)) with arbitrary constants b
0, b
1, k, and ω. We observe that the nontrivial solution (3) with (4) is independent of ω, which is impossible. Indeed, given ϕ
(
η
) is a nontrivial solution of Case 1, we rewrite (2) as
(17)
-
ω
u
′
=
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
.
Substituting ϕ
(
η
) into (17), we have
(18)
-
ω
ϕ
′
(
η
)
=
k
ν
ϕ
(
η
)
ϕ
′
(
η
)
+
k
2
α
ϕ
′′
(
η
)
+
k
3
μ
ϕ
′′′
(
η
)
11
+
k
4
γ
ϕ
′′′′
(
η
)
.
Fixing b
0, b
1, and k and leaving ω free, we can find that the right-hand side of (18) is determined, while the left-hand side is not. This is a contradiction.Secondly, we take Case2, namely, (5) with (6) and (7), into account. Setting b
0
=
k
=
1, b
-
1
=
2, γ
=
μ
=
1
/
4, α
=
1
/
2, and ν
=
45
/
19 for simplicity, we have
(19)
ω
=
-
121
76
,
u
=
1
4
(
1
+
2
Y
2
+
Y
)
6
=
1
4
(
1
+
2
exp
⁡
(
η
)
2
+
exp
⁡
(
η
)
)
6
.
Substituting above values into the left-hand side of (2), we obtain
(20)
ω
u
′
+
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
=
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
,
where
(21)
F
(
η
)
=
-
945
+
23246
exp
⁡
(
η
)
+
233632
exp
⁡
(
2
η
)
+
562784
exp
⁡
(
3
η
)
+
290396
exp
⁡
(
4
η
)
-
108644
exp
⁡
(
5
η
)
-
77299
exp
⁡
(
6
η
)
+
34514
exp
⁡
(
7
η
)
+
32332
exp
⁡
(
8
η
)
+
7256
exp
⁡
(
9
η
)
.
And by taking η
=
0, we have
(22)
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
|
η
=
0
=
1
.
Since the right-hand side of (20) is not zero for all value of η, we conclude that the solution in Case 2 is not admitted by the original ordinary differential equation (2) and KdV-Burgers-Kuramoto equation (1).Case3 and Case 4 can be checked in a way similar to Case 2; here we omit the details.At the end of this section, we should point out that (10) can be exactly simplified to the constant a
2 as follows:(23)
u
(
x
,
t
)
=
a
2
exp
⁡
(
2
η
)
+
a
2
b
1
exp
⁡
(
η
)
+
a
2
b
0
+
a
2
b
-
1
exp
⁡
(
-
η
)
exp
⁡
(
2
η
)
+
b
1
exp
⁡
(
η
)
+
b
0
+
b
-
1
exp
⁡
(
-
η
)
=
a
2
,
which is trivial.So, we conclude that not any new exact solution was obtained.


## 2.1. Comment 1

It is not difficult to rewrite (3) and (5) in the following form:
(11)
u
(
x
,
t
)
=
A
6
Y
6
+
A
5
Y
5
+
A
4
Y
4
+
A
3
Y
3
+
A
2
Y
2
+
A
1
Y
+
A
0
B
6
Y
6
+
B
5
Y
5
+
B
4
Y
4
+
B
3
Y
3
+
B
2
Y
2
+
B
1
Y
+
B
0
,
where A
i and B
i are certain constants.Therefore, we have(12)
u
(
x
,
t
)
=
(
A
6
Y
6
+
A
5
Y
5
+
A
4
Y
4
+
A
3
Y
3
+
A
2
Y
2
+
A
1
Y
+
A
0
)
×
(
B
6
Y
6
+
B
5
Y
5
+
B
4
Y
4
+
B
3
Y
3
+
B
2
Y
2
+
B
1
Y
+
B
0
)
-
1
=
(
A
6
exp
⁡
(
6
η
)
+
A
5
exp
⁡
(
5
η
)
+
A
4
exp
⁡
(
4
η
)
+
A
3
exp
⁡
(
3
η
)
+
A
2
exp
⁡
(
2
η
)
+
A
1
exp
⁡
(
η
)
+
A
0
)
×
(
B
6
exp
⁡
(
6
η
)
+
B
5
exp
⁡
(
5
η
)
+
B
4
exp
⁡
(
4
η
)
vv
+
B
3
exp
⁡
(
3
η
)
+
B
2
exp
⁡
(
2
η
)
+
B
1
exp
⁡
(
η
)
+
B
0
)
-
1
=
(
A
6
exp
⁡
(
5
η
)
+
A
5
exp
⁡
(
4
η
)
+
A
4
exp
⁡
(
3
η
)
+
A
3
exp
⁡
(
2
η
)
+
A
2
exp
⁡
(
η
)
+
A
1
+
A
0
exp
⁡
(
-
η
)
)
×
(
B
6
exp
⁡
(
5
η
)
+
B
5
exp
⁡
(
4
η
)
+
B
4
exp
⁡
(
3
η
)
vv
+
B
3
exp
⁡
(
2
η
)
+
B
2
exp
⁡
(
η
)
+
B
1
+
B
0
exp
⁡
(
-
η
)
)
-
1
=
(
A
6
exp
⁡
(
4
η
)
+
A
5
exp
⁡
(
3
η
)
+
A
4
exp
⁡
(
2
η
)
+
A
3
exp
⁡
(
η
)
+
A
2
+
A
1
exp
⁡
(
-
η
)
+
A
0
exp
⁡
(
-
2
η
)
)
×
(
B
6
exp
⁡
(
4
η
)
+
B
5
exp
⁡
(
3
η
)
+
B
4
exp
⁡
(
2
η
)
vv
+
B
3
exp
⁡
(
η
)
+
B
2
+
B
1
exp
⁡
(
-
η
)
+
B
0
exp
⁡
(
-
2
η
)
)
-
1
=
(
A
6
exp
⁡
(
3
η
)
+
A
5
exp
⁡
(
2
η
)
+
A
4
exp
⁡
(
η
)
+
A
3
+
A
2
exp
⁡
(
-
η
)
+
A
1
exp
⁡
(
-
2
η
)
+
A
0
exp
⁡
(
-
3
η
)
)
×
(
B
6
exp
⁡
(
3
η
)
+
B
5
exp
⁡
(
2
η
)
+
B
4
exp
⁡
(
η
)
+
B
3
vv
+
B
2
exp
⁡
⁡
(
-
η
)
+
B
1
exp
⁡
⁡
(
-
2
η
)
+
B
0
exp
⁡
⁡
(
-
3
η
)
)
-
1
.
According to the Exp-function method [2], we can reobtain the solution (11) by assuming that the solution of (2) can be expressed in the form
(13)
u
(
η
)
=
M
-
d
exp
⁡
(
-
d
η
)
+
⋯
+
M
c
exp
⁡
(
c
η
)
N
-
q
exp
⁡
(
-
q
η
)
+
⋯
+
N
p
exp
⁡
(
p
η
)
,
where (i) p
=
c
=
5 and d
=
q
=
1, (ii) p
=
c
=
4 and d
=
q
=
2, (iii) p
=
c
=
d
=
q
=
3, respectively. It is worth to mention that the fact that the three cases of (i), (ii), and (iii) are equivalent has been emphasized in [3–5]. Thus the claim in the commented paper that “new generalized solitary wave solutions are constructed for the KdV-Burgers-Kuramoto equation, which cannot be directly constructed from the Exp-function method” is not true.


## 2.2. Comment 2

In this section, we show that the solutions in the mentioned four cases are incorrect. Here, we should point out that it is difficult for us to solve original algebra system appearing in [1] and therefore we verify the four cases in an ad hoc way.


### 2.2.1.  Solution Analysis

At the beginning, substituting (4) into (3), we find
(14)
u
(
x
,
t
)
=
15
k
(
α
2
-
16
μ
γ
)
76
γ
ν
·
(
b
0
Y
+
b
1
)
6
(
b
0
+
b
1
Y
)
6
,
and substituting (6) into (5), we find
(15)
u
(
x
,
t
)
=
K
(
α
2
-
16
μ
γ
)
·
(
b
0
+
b
-
1
Y
)
6
(
b
0
Y
+
b
-
1
)
6
.So we assume that the solution of (2) can be expressed in the form
(16)
u
(
η
)
=
L
·
(
k
0
+
k
1
Y
)
6
(
k
1
+
k
0
Y
)
6
,
where L, k
0, and k
1 are constants to be determined.We emphasize that (16) needs only three undetermined parameters. Unlike (5), there are ten undetermined parameters. Hence assumption (16) can reduce the computation burden.Substituting (16) into (2) and setting the coefficients of all powers of Y
i to zero yield a system of algebraic equations for L, k
0, k
1, k, and ω. Solving these algebraic equations, we can determine certain solutions to (2). Obviously, these solutions cover (14) and (15). However, it is to our surprise that with the aid of Maple we determine none of nontrivial solutions after solving the above system equations. Hence further verification should be made.


### 2.2.2. Solution Check

In what follows, after careful numerical inspection, we show that the four cases are incorrect.Firstly, we check the solution in Case1, namely, (3) and (4) (or (14)) with arbitrary constants b
0, b
1, k, and ω. We observe that the nontrivial solution (3) with (4) is independent of ω, which is impossible. Indeed, given ϕ
(
η
) is a nontrivial solution of Case 1, we rewrite (2) as
(17)
-
ω
u
′
=
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
.
Substituting ϕ
(
η
) into (17), we have
(18)
-
ω
ϕ
′
(
η
)
=
k
ν
ϕ
(
η
)
ϕ
′
(
η
)
+
k
2
α
ϕ
′′
(
η
)
+
k
3
μ
ϕ
′′′
(
η
)
11
+
k
4
γ
ϕ
′′′′
(
η
)
.
Fixing b
0, b
1, and k and leaving ω free, we can find that the right-hand side of (18) is determined, while the left-hand side is not. This is a contradiction.Secondly, we take Case2, namely, (5) with (6) and (7), into account. Setting b
0
=
k
=
1, b
-
1
=
2, γ
=
μ
=
1
/
4, α
=
1
/
2, and ν
=
45
/
19 for simplicity, we have
(19)
ω
=
-
121
76
,
u
=
1
4
(
1
+
2
Y
2
+
Y
)
6
=
1
4
(
1
+
2
exp
⁡
(
η
)
2
+
exp
⁡
(
η
)
)
6
.
Substituting above values into the left-hand side of (2), we obtain
(20)
ω
u
′
+
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
=
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
,
where
(21)
F
(
η
)
=
-
945
+
23246
exp
⁡
(
η
)
+
233632
exp
⁡
(
2
η
)
+
562784
exp
⁡
(
3
η
)
+
290396
exp
⁡
(
4
η
)
-
108644
exp
⁡
(
5
η
)
-
77299
exp
⁡
(
6
η
)
+
34514
exp
⁡
(
7
η
)
+
32332
exp
⁡
(
8
η
)
+
7256
exp
⁡
(
9
η
)
.
And by taking η
=
0, we have
(22)
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
|
η
=
0
=
1
.
Since the right-hand side of (20) is not zero for all value of η, we conclude that the solution in Case 2 is not admitted by the original ordinary differential equation (2) and KdV-Burgers-Kuramoto equation (1).Case3 and Case 4 can be checked in a way similar to Case 2; here we omit the details.At the end of this section, we should point out that (10) can be exactly simplified to the constant a
2 as follows:(23)
u
(
x
,
t
)
=
a
2
exp
⁡
(
2
η
)
+
a
2
b
1
exp
⁡
(
η
)
+
a
2
b
0
+
a
2
b
-
1
exp
⁡
(
-
η
)
exp
⁡
(
2
η
)
+
b
1
exp
⁡
(
η
)
+
b
0
+
b
-
1
exp
⁡
(
-
η
)
=
a
2
,
which is trivial.So, we conclude that not any new exact solution was obtained.


## 2.2.1.  Solution Analysis

At the beginning, substituting (4) into (3), we find
(14)
u
(
x
,
t
)
=
15
k
(
α
2
-
16
μ
γ
)
76
γ
ν
·
(
b
0
Y
+
b
1
)
6
(
b
0
+
b
1
Y
)
6
,
and substituting (6) into (5), we find
(15)
u
(
x
,
t
)
=
K
(
α
2
-
16
μ
γ
)
·
(
b
0
+
b
-
1
Y
)
6
(
b
0
Y
+
b
-
1
)
6
.So we assume that the solution of (2) can be expressed in the form
(16)
u
(
η
)
=
L
·
(
k
0
+
k
1
Y
)
6
(
k
1
+
k
0
Y
)
6
,
where L, k
0, and k
1 are constants to be determined.We emphasize that (16) needs only three undetermined parameters. Unlike (5), there are ten undetermined parameters. Hence assumption (16) can reduce the computation burden.Substituting (16) into (2) and setting the coefficients of all powers of Y
i to zero yield a system of algebraic equations for L, k
0, k
1, k, and ω. Solving these algebraic equations, we can determine certain solutions to (2). Obviously, these solutions cover (14) and (15). However, it is to our surprise that with the aid of Maple we determine none of nontrivial solutions after solving the above system equations. Hence further verification should be made.


## 2.2.2. Solution Check

In what follows, after careful numerical inspection, we show that the four cases are incorrect.Firstly, we check the solution in Case1, namely, (3) and (4) (or (14)) with arbitrary constants b
0, b
1, k, and ω. We observe that the nontrivial solution (3) with (4) is independent of ω, which is impossible. Indeed, given ϕ
(
η
) is a nontrivial solution of Case 1, we rewrite (2) as
(17)
-
ω
u
′
=
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
.
Substituting ϕ
(
η
) into (17), we have
(18)
-
ω
ϕ
′
(
η
)
=
k
ν
ϕ
(
η
)
ϕ
′
(
η
)
+
k
2
α
ϕ
′′
(
η
)
+
k
3
μ
ϕ
′′′
(
η
)
11
+
k
4
γ
ϕ
′′′′
(
η
)
.
Fixing b
0, b
1, and k and leaving ω free, we can find that the right-hand side of (18) is determined, while the left-hand side is not. This is a contradiction.Secondly, we take Case2, namely, (5) with (6) and (7), into account. Setting b
0
=
k
=
1, b
-
1
=
2, γ
=
μ
=
1
/
4, α
=
1
/
2, and ν
=
45
/
19 for simplicity, we have
(19)
ω
=
-
121
76
,
u
=
1
4
(
1
+
2
Y
2
+
Y
)
6
=
1
4
(
1
+
2
exp
⁡
(
η
)
2
+
exp
⁡
(
η
)
)
6
.
Substituting above values into the left-hand side of (2), we obtain
(20)
ω
u
′
+
k
ν
u
u
′
+
k
2
α
u
′′
+
k
3
μ
u
′′′
+
k
4
γ
u
′′′′
=
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
,
where
(21)
F
(
η
)
=
-
945
+
23246
exp
⁡
(
η
)
+
233632
exp
⁡
(
2
η
)
+
562784
exp
⁡
(
3
η
)
+
290396
exp
⁡
(
4
η
)
-
108644
exp
⁡
(
5
η
)
-
77299
exp
⁡
(
6
η
)
+
34514
exp
⁡
(
7
η
)
+
32332
exp
⁡
(
8
η
)
+
7256
exp
⁡
(
9
η
)
.
And by taking η
=
0, we have
(22)
27
exp
⁡
(
η
)
(
1
+
2
exp
⁡
(
η
)
)
2
152
(
2
+
exp
⁡
(
η
)
)
13
·
F
(
η
)
|
η
=
0
=
1
.
Since the right-hand side of (20) is not zero for all value of η, we conclude that the solution in Case 2 is not admitted by the original ordinary differential equation (2) and KdV-Burgers-Kuramoto equation (1).Case3 and Case 4 can be checked in a way similar to Case 2; here we omit the details.At the end of this section, we should point out that (10) can be exactly simplified to the constant a
2 as follows:(23)
u
(
x
,
t
)
=
a
2
exp
⁡
(
2
η
)
+
a
2
b
1
exp
⁡
(
η
)
+
a
2
b
0
+
a
2
b
-
1
exp
⁡
(
-
η
)
exp
⁡
(
2
η
)
+
b
1
exp
⁡
(
η
)
+
b
0
+
b
-
1
exp
⁡
(
-
η
)
=
a
2
,
which is trivial.So, we conclude that not any new exact solution was obtained.


## 3. Conclusion

In this paper, we emphasize that the paper [1] contains some errors. We have to point out that similar mistakes had been analyzed in some published papers (see, e.g., [6, 7]). We hope that the results will help people have a good understanding of the work made by Kim et al.



---

*Source: 240784-2014-02-24.xml*