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 Case 1, 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 Case 2, 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).
Case 3 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.