World Applied Sciences Journal 25 (3): 500-523, 2013
ISSN 1818-4952
© IDOSI Publications, 2013
DOI: 10.5829/idosi.wasj.2013.25.03.1193
Some New Exact Traveling Wave Solutions to
the (2+1)-Dimensional Breaking Soliton Equation
1
1
Md. Nur Alam, 2M. Ali Akbar and 1Kamruzzaman Khan
Department of Mathematics, Pabna University of Science and Technology, Bangladesh
2
Department of Applied Mathematics, University of Rajshahi, Bangladesh
Submitted: Jul 26, 2013;
Accepted: Sep 1, 2013;
Published: Oct 3, 2013
Abstract: In this article, the new generalized (G'/G)-expansion method is used to find the exact traveling wave
solutions of the (2+1)-dimensional breaking soliton equations. Abundant traveling wave solutions with
arbitrary parameters are successfully obtained by this method which are expressed in terms of hyperbolic,
trigonometric and rational functions. This shows that the new generalized (G'/G)-expansion method is a
powerful tool for constructing exact solution of nonlinear partial differential equations in physics, mathematics
and other applications. The method appears to be easier and faster by means of computer algebra.
Mathematics Subject Classification: 35C07 35C08 35P99
Keywords: The new generalized (G'/G)-expansion method
The (2+1)-dimensional breaking soliton
equations Nonlinear evolution equation Traveling wave solutions
INTRODUCTION
Recently, Naher and Abdullah [28] established a
highly effective extension of the (G'/G)-expansion
method, called the new generalized (G'/G)-expansion
method to obtain exact traveling wave solutions of
NLEEs. The objective of this article is to search for new
study relating to the new generalized (G'/G)-expansion
method for solving the (2+1)-dimensional breaking soliton
equations to demonstrate the correctness and
truthfulness of the method.
The rest of the article is organized as follows:
In Section 2, the description of the new generalized
(G'/G) expansion method is given. In Section 3, we will
apply this method to obtain the traveling wave
solution of the (2+1)-dimensional breaking soliton
equations. In Sections 4, we will give some conclusions.
In the recent years, the exact solutions of nonlinear
partial differential equations (NLPDEs) have been
investigated by many authors who are interested in
nonlinear phenomena which exist in all fields including
either the scientific works or engineering fields, such as
fluid mechanics,
chemical
physics, chemical
kinematics, plasma physics, elastic media, optical
fibers, solid state physics, biology, atmospheric and
oceanic phenomena and so on. The research of
traveling
wave solutions of some nonlinear
evolution equations derived from such fields played
an important role in the analysis of some phenomena.
To obtain traveling wave solutions, many effective
methods have been presented in the literature,
such as the (G'/G, 1/G)-expansion method [1, 2],
the (G'/G)-expansion
method [3-11], the novel
(G'/G)-expansion method [12], the modified simple
equation method [13, 14], the improved (G'/G)-expansion
method [15], the homogeneous balance method [16, 17],
the inverse scattering transform [18], the Exp-function
method [19-21], the Cole-Hopf transformation [22], the
Adomian decomposition method [23], the ansatz method
[24, 25], the tanh-function method [26], the tanh-coth
method [27] and so on.
Corresponding Author:
Description of the New Generalized (G'/G)-Expansion
Method: Let us consider a general nonlinear PDE in the
form
P (u , ut , u x , utt , ut x , u xx ,) = 0
(1)
where u = u(x, t) is an unknown function, P is a
polynomial in u(x, t) and its derivatives in which highest
order derivatives and nonlinear terms are involved and the
subscripts stand for the partial derivatives.
Md. Nur Alam, Department of Mathematics, Pabna University of Science and Technology, Bangladesh
500
World Appl. Sci. J., 25 (3): 500-523, 2013
Step 1: We combine the real variables x and t by a
complex variable ,
where G = G( ) satisfies the following auxiliary nonlinear
ordinary differential equation:
u ( x, t )= u ( ),
AGG′′ − BGG′ − E G 2 − C (G′)2 =
0
= x ± V t,
(2)
where the prime stands for derivative with respect to ; A,
B, C and E are real parameters.
where V is the speed of the traveling wave. The traveling
wave transformation (2) converts Eq. (1) into an ordinary
differential equation (ODE) for u = u( ):
Q(u, u′, u′′, u′′′,) = 0,
Step 4: To determine the positive integer N, taking the
homogeneous balance between the highest order
nonlinear terms and the derivatives of the highest order
appearing in Eq. (3).
(3)
where Q is a polynomial of u and it derivatives and the
superscripts indicate the ordinary derivatives with respect
to .
Step 5: Substitute Eq. (4) and Eq. (6) including Eq. (5)
into Eq. (3) with the value of N obtained in Step 4,
we obtain polynomials in (d + H)N (N = 0,1,2,...)
and (d + H) N (N = 0,1,2,...). Then, we collect each
coefficient of the resulted polynomials to zero
yields a set of algebraic equations for ai (i = 0,1,2,...,N)
and bi (i = 1,2,...,N), d and V.
Step 2: According to possibility, Eq. (3) can be integrated
term by term one or more times, yields constant(s) of
integration. The integral constant may be zero for
simplicity.
Step 3: Suppose the traveling wave solution of Eq. (3) can
be expressed as follows:
u( =
)
N
∑
ai (d + H )i +
Step 6: Suppose that the value of the constants
ai (i = 0,1,2,...,N), bi (i = 1,2,...,N), d and V can be found
by solving the algebraic equations obtained in
Step 5. Since the general solution of Eq. (6) is well
known to us, inserting the values of ai (i =
0,1,2,...,N), bi (i = 1,2,...,N), d and V into Eq. (4), we
obtain more general type and new exact traveling
wave solutions of the nonlinear partial differential
equation (1).
Using the general solution of Eq. (6), we have the
following solutions of Eq. (5):
N
∑ bi (d + H )−i ,
(4)
=i 0=i 1
where either aN or bN may be zero, but both aN and bN
could be zero at a time, (i = 0,1,2,...,N) and bi (i = 1,2,...,N)
and d are arbitrary constants to be determined later and
H( ) is given by
H( ) = (G'/G)
Family 1: When B
(5)
0,
= A – C and
 Ω
C1 sinh 
Ω
 G′  B
 2A
+
)  =
H (=

 Ω
2
G 2
C1 cosh 
 2A
Family 2: When B
0,
= B2 + 4E(A – C) > 0,


 + C2 cosh 




 + C2 sinh 


= A – C and

−C1 sin 
−Ω
 G′  B

+
)  =
H (=


2
G 2
C1 cos 

(6)



Ω 

2 A 
Ω
2A
= B2 + 4E(A – C) < 0,


 + C2 cos 




−Ω
 + C2 sin 
2 A 

−Ω
2A
(7)




−Ω

2 A 
−Ω
2A
501
(8)
World Appl. Sci. J., 25 (3): 500-523, 2013
Family 3: When B
0,
= A – C and
= B2 + 4E(A – C) = 0,
C2
 G′  B
+
)  =
H (=

C1 + C2
G 2
Family 4: When B = 0,
 G′ 
( ) =
H=

G
∆
= A – C and
∆
) + C2 cosh(
A
∆
) + C2 sinh(
C1 cosh(
A
−∆
= A – C and
E > 0,
=
C1 sinh(
Family 5: When B = 0,
 G′ 
( ) =
H=

G
(9)
∆
)
A
∆
)
A
=
−∆
) + C2 cos(
A
−∆
) + C2 sin(
C1 cos(
A
−C1 sin(
(10)
E < 0,
−∆
)
A
−∆
)
A
(11)
Application of the Method: In this section, we will bring to bear the new generalized (G'/G)-expansion method to look
for the exact solutions and then the solitary wave solutions to the (2+1)-dimensional breaking soliton equations. Let us
consider the (2+1)-dimensional breaking soliton equations
ut + u xxy + 4 uvx + 4 u x=
v 0,
u=
y v x.
(12)
where is a constant.
Using the traveling wave variable u( ) = u(x, y, t), = x + y – Vt carries the (2+1)-dimensional breaking soliton equations
(12) into an ODE
′v 0,
−Vu ′ + u ′′′ + 4 uv′ + 4 u=
=
u ′ v′
(13)
Integrating the second equation in the system and neglecting constants of integration, we find
u=v
Substituting (14) into the first equation of the system and integrating, we find
(14)
−Vu + 4 u 2 + u′′ =
0
(15)
Taking the homogeneous balance between u2 and u'' in Eq. (15), we obtain N = 2. Therefore, the solution of Eq. (15)
is of the form:
u ( ) =a0 + a1(d + H ) + a2 (d + H )2 + b1( d + H ) − 1 + b2 (d + H ) −2 ,
(16)
where a0,a1,a2,b1,b2 and d are constants to be determined.
Substituting Eq. (16) along with Eqs. (5) and (6) into Eq. (15), the left-hand side is converted into polynomials
in (N+H)N (N = 0,1,2,..) and (d + H) N (N = 1,2,...). We collect each coefficient of these resulted polynomials to zero,
yields a set of simultaneous algebraic equations (for simplicity which are not presented here) for a0, a1, a2,b1,b2 d, K
and V. Solving these algebraic equations with the help of symbolic computation software Maple, we obtain the
following:
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World Appl. Sci. J., 25 (3): 500-523, 2013
Case 1:
2
1
B
−
a0 =
( B 2 + 4 E ), a1 = 0, a2 = − 3
,
, d= −
2
2
4A
2 A2
b1 = 0,
3m1
4
−
− 2 (4 E + B 2 )
,V =
b2 =
2 2
32 A
A
where
= A – C, m1 = (B4 + 8EB2
Case=
2: a0
3
4 A2
(17)
+ 16E2
2
), A, B, C and E are free parameters.
2
B
( B 2 + 4 E ), a1 = 0, a2 = − 3
,
, d= −
2
2 A2
b1 = 0,
3m1
4
−
,V =
(4 E + B 2 )
b2 =
32 A2 2
A2
where
= A – C, m1 = (B4 + 8EB2
3
Case 3: a0 =
−
2 A2
(− E
(18)
+ 16E2
2 2 ), a
+ dB + d=
1
d d=
, b1 0,=
b2 0,
V
=
=
2
A
3
2 A2
2
), A, B, C and E are free parameters.
2
(2 d 2 + B ), a2 = − 3
,
2 A2
(19)
(4 E + B2 )
where
= A – C, A, B, C and E are free parameters.
Case 4:
2
1
3
2 2 ), a
−
a0 =
( B 2 − 2 E + 6dB + 6d=
(2 d 2 + B ), a2 = − 3
,
1
2
4 A2
2 A2
2A
− 2 (4 E + B 2 )
d=
d , b1 =
0, b2 =
0, V =
A
where
(20)
= A – C, A, B, C and E are free parameters.
m2
B
Case 5: a0 =
, a1 =
0, a2 =
0, d =
, b1 =
−
0.
8 A2
2
3m2 2
−
− 2 (4 E + B 2 )
,V =
b2 =
2 2
32 A
A
where
(21)
= A – C, m2 = (B2 + 4E ), A, B, C and E are free parameters.
3m2
B
Case 6: a0 =
, a1 =
0, a2 =
0, d =
, b1 =
0,
−
8 A2
2
3m2 2
−
,V=
(4 E + B 2 )
b2 =
32 A2 2
A2
where
(22)
= A – C, m2 = (B2 + 4E ), A, B, C and E are free parameters.
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World Appl. Sci. J., 25 (3): 500-523, 2013
Case 7:
a0 =
m3
2 A2
where
, a1 = 0, a2 = 0, d = d , b1 =
m3 =−( − E
+ Bd
3m4
2 A2
, b2 =
+ d 2 2 ), m4 = (2d 3 2 − 2 Ed
3m5
2 A2
,V=
1
A2
( A2 − 4E − B 2 )
− EB + B 2 d + 3Bd 2 ), m5 =
−( d 4 2 − 2Ed 2 + 2 Bd 3 + B 2 d 2 − 2 BdE + E 2 ),
(23)
=A–
C A, B, C and E are free parameters.
Case 8:
3m7
3m8
m6
, a1 0, a2 =
0, d =
, b2
, V − 2 (4 E + B 2 )
a0 ==
d , b1 ===
2
2
2
4A
2A
2A
A
where
−( B 2 − 2 E + 6 Bd
m6 =
3 2 − 2 Ed
+ 6d 2 2 ), m
=
7 (2d
(24)
− EB + B 2 d + 3Bd 2 ), m8 =
−( d 4 2 − 2 Ed 2 + 2 Bd 3 + B 2 d 2 − 2 BdE + E 2 ),
= A – C, A, B, C and E are free parameters.
For case 1, substituting Eq. (17) into Eq. (16), together with Eq. (7) and simplifying, we obtain the following solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
u11 ( ) =−
{2( B 2 + 4 E ) + 3Ω coth 2 (
Ω
3m
Ω
)}.
) + 1 tanh 2 (
2A
2A
Ω
{2( B 2 + 4 E ) + 3Ω coth 2 (
Ω
3m
Ω
)}.
) + 1 tanh 2 (
2A
2A
Ω
{2( B 2 + 4 E ) + 3Ω tanh 2 (
Ω
3m
Ω
)}.
) + 1 coth 2 (
2A
2A
Ω
{2( B 2 + 4 E ) + 3Ω tanh 2 (
Ω
3m
Ω
)}.
) + 1 coth 2 (
2A
2A
Ω
1
8A
2
and
v11 ( ) =−
u12 ( ) =−
1
8A
2
1
8A
2
and
v12 ( ) =−
where
C2
1
8A
2
= x − {−
4
A2
(4 E + B 2 )}t.
Substituting Eq. (17) into Eq. (16), along with Eq. (8) and simplifying, we obtain following solutions (if C1 = 0 but
0; C2 = 0 but C1 0) respectively:
u13 ( ) =−
1
8A
2
{2( B 2 + 4 E ) − 3Ω cot 2 (
−Ω
3m
−Ω
)}.
) − 1 tan 2 (
2A
2A
Ω
{2( B 2 + 4 E ) − 3Ω cot 2 (
−Ω
3m
−Ω
)}
) − 1 tan 2 (
2A
2A
Ω
and
v13 ( ) =−
1
8A
2
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World Appl. Sci. J., 25 (3): 500-523, 2013
u14 ( ) =−
1
8A
2
{2( B 2 + 4 E ) − 3Ω cot 2 (
−Ω
3m
−Ω
)}.
) − 1 tan 2 (
2A
2A
Ω
{2( B 2 + 4 E ) − 3Ω cot 2 (
−Ω
3m
−Ω
) − 1 tan 2 (
)}
2A
Ω
2A
and
v14 ( ) =−
1
8A
2
Substituting Eq. (17) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
1
− 2 {( B 2 + 4 E ) + 6
u15 ( ) =
4A
2
(
3m
C2
C2
) 2 + 12 (
) −2 }.
C1 + C2
C1 + C2
8
2
(
3m
C2
C2
) 2 + 12 (
) −2 }.
C1 + C2
C1 + C2
8
and
1
− 2 {( B 2 + 4 E ) + 6
v15 ( ) =
4A
Substituting Eq. (17) into Eq. (16), along with Eq. (10) and simplifying, yields following exact traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
1
− 2 {( B 2 + 4 E ) + 6
u16 ( ) =
4A
2  −B

2
+
∆
2
−2
coth(
∆ 
3m  − B
+
)  + 12 
A
8 2

coth(
∆ 
∆
∆ 
3m  − B
+
)  + 12 
coth(
)  }.
A
A
8 2


∆
coth(
∆ 
)  }.
A

and
1
− 2 {( B 2 + 4 E ) + 6
v16 ( ) =
4A
2  −B
1
− 2 {( B 2 + 4 E ) + 6
u17 ( ) =
4A
2  −B

2

2
+
+
∆
2
−2
2
−2
2
−2
∆
∆ 
∆
∆ 
3m  − B
+
tanh(
)  + 12 
tanh(
)  }.
2
A
A
 8 

∆
∆ 
∆
∆ 
3m  − B
+
tanh(
)  + 12 
tanh(
)  }.
A
A
 8 2

and
1
− 2 {( B 2 + 4 E ) + 6
v17 ( ) =
4A
2  −B

2
+
Substituting Eq. (17) into Eq. (16), along with Eq. (11) and simplifying, we obtain the following solutions (if C1 = 0
but C2 0; C2 = 0 but C1 0) respectively:
1
− 2 {( B 2 + 4 E ) + 6
u18 ( ) =
4A
2  −B
 2

+
−∆
cot(
−∆ 
) 
A

2
−2
+
−∆
−∆ 
3m1  − B
+
cot(
)  }.
2  2
A
8 

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World Appl. Sci. J., 25 (3): 500-523, 2013
and
1
v18 ( ) =
− 2 {( B 2 + 4 E ) + 6
4A
2  −B
+
 2

−∆
2
cot(
−∆ 
)

A

tan(
−∆ 
)

A

tan(
−∆ 
) 
A

−2
−∆
−∆ 
3m1  − B
+
cot(
)  }.
2  2
A
8 

+
1
u19 ( ) =
− 2 {( B 2 + 4 E ) + 6
4A
2  −B

2
−
−∆
2
−2
+
−∆
−∆ 
3m1  − B
−
tan(
)  }.
2  2
A
8 

and
1
− 2 {( B 2 + 4 E ) + 6
v19 ( ) =
4A
2  −B

2
−
−∆
2
−2
−∆
−∆ 
3m  − B
+ 12 
−
tan(
)  }.
2
A
8 

Again for case 2, substituting Eq. (18) into Eq. (16), together with Eq. (7) and simplifying, our obtained exact
solutions becomes (if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
=
u21 ( )
3
8A
2
{2( B 2 + 4 E ) − Ω coth 2 (
Ω
Ω
m
)}.
) − 1 tanh 2 (
2A
2A
Ω
{2( B 2 + 4 E ) − Ω coth 2 (
Ω
Ω
m
)}.
) − 1 tanh 2 (
2A
2A
Ω
{2( B 2 + 4 E ) − Ω tanh 2 (
Ω
Ω
m
)}.
) − 1 coth 2 (
2A
2A
Ω
{2( B 2 + 4 E ) − Ω tanh 2 (
Ω
Ω
m
)}.
) − 1 coth 2 (
2A
2A
Ω
and
=
v21 ( )
=
u 22 ( )
3
8A
2
3
8 A2
and
=
v22 ( )
where
C2
3
8 A2
4
=
(4 E + B 2 )}t.
x −{
A2
Substituting Eq. (18) into Eq. (16), along with Eq. (8) and simplifying, we obtain following solutions (if C1 = 0 but
0; C2 = 0 but C1 0) respectively:
=
u 23 ( )
3
8 A2
{2( B 2 + 4 E ) + Ω cot 2 (
−Ω
−Ω
m
)}.
) + 1 tan 2 (
2A
2A
Ω
and
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World Appl. Sci. J., 25 (3): 500-523, 2013
=
v23 ( )
3
8A
2
3
=
u 24 ( )
8A
2
{2( B 2 + 4 E ) + Ω cot 2 (
−Ω
−Ω
m
)}.
) + 1 tan 2 (
2A
2A
Ω
{2( B 2 + 4 E ) + Ω cot 2 (
−Ω
−Ω
m
)}.
) + 1 tan 2 (
2A
2A
Ω
{2( B 2 + 4 E ) + Ω cot 2 (
−Ω
−Ω
m
)}.
) + 1 tan 2 (
2A
2A
Ω
and
=
v24 ( )
3
8A
2
Substituting Eq. (18) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
3
)
u 25 ( =
4A
2
{( B 2 + 4 E ) − 2
2
(
C2
m
C2
) 2 − 12 (
) −2 }.
C1 + C2
C1 + C2
8
{( B 2 + 4 E ) − 2
2
(
C2
m
C2
) 2 − 12 (
) −2 }.
C1 + C2
C1 + C2
8
and
3
)
v25 ( =
2
4A
Substituting Eq. (18) into Eq. (16), along with Eq. (10) and simplifying, yields following exact traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
3
{( B 2 + 4 E ) − 2
2
4A
u 26 ( =
)
2  −B

2
+
∆
2
−2
2
−2
2
−2
2
−2
coth(
∆ 
∆
∆ 
m  −B
+
)  − 12 
coth(
)  }.
2
A
A
 8 

coth(
∆ 
∆
∆ 
m  −B
+
)  − 12 
coth(
)  }.
A
A
 8 2

tanh(
∆ 
∆
∆ 
m  −B
+
)  − 12 
tanh(
)  }.
A
A
 8 2

tanh(
∆ 
∆
∆ 
m  −B
+
)  − 12 
tanh(
)  }.
A
A
8 2


and
3
v26 ( =
)
4A
{( B 2 + 4 E ) − 2
2
3
u27 ( =
)
4A
{( B 2 + 4 E ) − 2
2
2  −B

2
2  −B

2
+
+
∆
∆
and
3
)
v27 ( =
4A
{( B 2 + 4 E ) − 2
2
2  −B

2
+
∆
Substituting Eq. (18) into Eq. (16), along with Eq. (11) and simplifying, our obtained exact solutions become (if C1
= 0 but C2 0; C2 = 0 but C1 0) respectively:
3
)
u28 ( =
4A
{( B 2 + 4 E ) − 2
2
2  −B
 2

+
−∆
cot(
−∆ 
) 
A

2
−2
−
−∆
−∆ 
m1  − B
+
cot(
)  }.
2  2
A
8 

507
World Appl. Sci. J., 25 (3): 500-523, 2013
and
3
v28 ( =
)
4A
{( B 2 + 4 E ) − 2
2
2  −B
 2

+
−∆
2
cot(
−∆ 
)

A

tan(
−∆ 
)

A

tan(
−∆ 
)

A

−2
m1  − B
−∆
−∆ 
+
cot(
)  }.
2  2
A
8 

−
3
u29 ( =
)
4A
{( B 2 + 4 E ) − 2
2
2  −B

2
−
−∆
2
−2
−
m1  − B
−∆
−∆ 
−
tan(
)  }.
2  2
A
8 

and
v29 ( =
)
3
4A
{( B 2 + 4 E ) − 2
2
2  −B

2
−
−∆
2
−2
m  −B
−∆
−∆ 
− 12 
−
tan(
)  }.
A
8 2

Again for case 3, substituting Eq. (19) into Eq. (16) along with Eq. (7) and simplifying, the traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively become:
=
u31 ( )
Ω 
1 
).
3(4 E + B 2 ) − 3Ω coth 2 (
2
2 A 
8 A 
and
v31 ( )
=
Ω 
1 
).
3(4 E + B 2 ) − 3Ω coth 2 (
2
2 A 
8 A 
u32 ( )
=
Ω 
1 
).
3(4 E + B 2 ) − 3Ω tanh 2 (
2
2 A 
8 A 
and
v32 ( )
=
where
Ω 
1 
) .
3(4 E + B 2 ) − 3Ω tanh 2 (
2
2 A 
8 A 


=
(4 E + B 2 )  t .
x−
 A2

Substituting Eq. (19) into Eq. (16), along with Eq. (8) and simplifying, yields exact solutions (if C1 = 0 but C2
C2 = 0 but C1 0) respectively:
508
0;
World Appl. Sci. J., 25 (3): 500-523, 2013
u33 ( )
=
−Ω 
1 
) .
3(4 E + B 2 ) − 3Ω cot 2 (
2
2A
8 A 

and
v33 ( )
=
−Ω 
1 
) .
3(4 E + B 2 ) − 3Ω cot 2 (
2
2A
8 A 

u34 ( )
=
−Ω 
1 
) .
3(4 E + B 2 ) + 3Ω tan 2 (
2
2A
8 A 

and
v34 ( )
=
−Ω 
1 
) .
3(4 E + B 2 ) + 3Ω tan 2 (
2
2A
8 A 

Substituting Eq. (19) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
=
u35 ( )
1 
2
3(4 E + B ) − 12
8 A2 
2
(

C2
)2  ,
C1 + C2

2
(

C2
)2  ,
C1 + C2

and
=
v35 ( )
1 
2
3(4 E + B ) − 12
8 A2 
Substituting Eq. (19) into Eq. (16), together with Eq. (10) and simplifying, yields following traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
=
u36 ( )

3 
∆
∆  
)  .
E + ∆  B coth(
) − ∆ coth 2 (
2
A
A
2 A 

 
and
=
v36 ( )

3 
∆
∆  
E + ∆  B coth(
) − ∆ coth 2 (
)  
2
A
A
2 A 

 
=
u37 ( )

3 
∆
∆  
E + ∆  B tanh(
) − ∆ tanh 2 (
)  
2
A
A
2 A 

 
and
=
v37 ( )
C2

3 
∆
∆  
)  .
E + ∆  B tanh(
) − ∆ tanh 2 (
2
A
A
2 A 

 
Substituting Eq. (19) into Eq. (16), along with Eq. (11) and simplifying, our exact solutions become (if C1 = 0 but
0; C2 = 0 but C1 0) respectively:
509
World Appl. Sci. J., 25 (3): 500-523, 2013
=
u38 ( )

3 
−∆
−∆ 
)  .
E + ∆  iB cot(
) − ∆ cot 2 (
2
A
A
2 A 


and
=
v38 ( )

3 
−∆
−∆ 
2
+
∆
−
∆
) .
E
iB
cot(
)
cot
(



A
A
2 A2 


=
u39 ( )

3 
−∆
−∆  
)   ,
E − ∆  iB tan(
) + ∆ tan 2 (
2
A
A
2 A 

 
and
=
v39 ( )

3 
−∆
−∆  
)  .
E − ∆  iB tan(
) + ∆ tan 2 (
2
A
A
2 A 

 
Again for case 4, substituting Eq. (20) into Eq. (16) along with Eq. (7) and simplifying, the traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively become:
u41 ( )
=
Ω 
1 
) .
(4 E + B 2 ) − 3Ω coth 2 (
2
2
A 
8 A 
and
v41 ( )
=
Ω 
1 
) .
(4 E + B 2 ) − 3Ω coth 2 (
2
2
A 
8 A 
u4 2 ( )
=
Ω 
1 
).
(4 E + B 2 ) − 3Ω tanh 2 (
2
2 A 
8 A 
and
v42 ( )
=
where
Ω 
1 
) .
(4 E + B 2 ) − 3Ω tanh 2 (
2
2 A 
8 A 


(4 E + B 2 )  t .
x−
=
 A2

Substituting Eq. (20) into Eq. (16), along with Eq. (8) and simplifying, yields exact solutions (if C1 = 0 but C2
= 0 but C1 0) respectively:
u 43 ( )
=
−Ω 
1 
).
(4 E + B 2 ) − 3Ω cot 2 (
2
2A
8 A 

and
510
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World Appl. Sci. J., 25 (3): 500-523, 2013
v43 ( )
=
−Ω 
1 
).
(4 E + B 2 ) − 3Ω cot 2 (
2
2A
8 A 

u 44 ( )
=
−Ω 
1 
).
(4 E + B 2 ) + 3Ω tan 2 (
2
2A
8 A 

and
v44 ( )
=
−Ω 
1 
).
(4 E + B 2 ) + 3Ω tan 2 (
2
2A
8 A 

Substituting Eq. (20) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
u=
45 ( )
1 
2
(4 E + B ) − 12
8 A2 
2
(

C2
)2  ,
C1 + C2

2
(

C2
)2  ,
C1 + C2

and
v=
45 ( )
1 
2
(4 E + B ) − 12
8 A2 
Substituting Eq. (20) into Eq. (16), together with Eq. (10) and simplifying, yields following traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
u46 =
( )

1 
∆
∆  
)   .
(− B 2 + 2 E ) + 6 ∆  B coth(
) − ∆ coth 2 (
2
A
A
4 A 

 
and
v46 =
( )

1 
∆
∆  
(− B 2 + 2E ) + 6 ∆  B coth(
) − ∆ coth 2 (
)  
2
A
A
4 A 

 
u47 =
( )

1 
∆
∆  
)   .
(− B 2 + 2 E ) + 6 ∆  B tanh(
) − ∆ tanh 2 (
2
A
A
4 A 

 
and
v47 =
( )

1 
∆
∆  
(− B 2 + 2E ) + 6 ∆  B tanh(
) − ∆ tanh 2 (
)  
2
A
A
4 A 

 
Substituting Eq. (20) into Eq. (16), along with Eq. (11) and simplifying, our exact solutions become (if C1 = 0 but C2
0; C2 = 0 but C1 0) respectively:
u48 =
( )

1 
−∆
−∆  
)  .
(− B 2 + 2E ) + 6 ∆  iB cot(
) − ∆ cot 2 (
2
A
A
4 A 

 
and
511
World Appl. Sci. J., 25 (3): 500-523, 2013
v48 =
( )

1 
−∆
−∆  
(− B 2 + 2E ) + 6 ∆  iB cot(
) − ∆ cot 2 (
)  
2
A
A
4 A 

 
u49 =
( )

1 
−∆
−∆  
)   ,
(− B 2 + 2 E ) − 6 ∆  iB tan(
) + ∆ tan 2 (
2
A
A
2 A 

 
and
v49 (=
)

1 
−∆
−∆  
2
2
−
+
−
∆
+
∆
)  .
B
E
iB
(
2
)
6
tan(
)
tan
(



A
A
2 A2 

 
Again for case 5, substituting Eq. (21) into Eq. (16) along with Eq. (7) and simplifying, the traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively become:
u=
51 ( )
Ω 
m2  3m2
) .
tanh 2 (
1 −
2 A 
Ω
8 A2 
and
v=
51 ( )
Ω 
m2  3m2
) .
1−
tanh 2 (
2
2 A 
Ω
8 A 
u=
52 ( )
Ω 
m2  3m2
).
1−
coth 2 (
2
2
Ω
A 
8 A 
and
v=
52 ( )
Ω 
m2  3m2
).
1−
coth 2 (
2
2 A 
Ω
8 A 
where

(4 E
= x − −
 A2

+ B 2 )  t.

Substituting Eq. (21) into Eq. (16), along with Eq. (8) and simplifying, yields exact solutions (if C1 = 0 but C2
= 0 but C1 = 0) respectively:
u=
53 ( )
−Ω 
m2  3m2
) .
1+
tan 2 (
2
Ω
2A
8 A 

and
v=
53 ( )
−Ω 
m2  3m2
).
1+
tan 2 (
2
2
Ω
A
8 A 

u=
54 ( )
−Ω 
m2  3m2
).
1+
cot 2 (
2
2A
Ω
8 A 

512
0; C2
World Appl. Sci. J., 25 (3): 500-523, 2013
and
v=
54 ( )
−Ω 
m2  3m2
) .
1+
cot 2 (
2
2
Ω
A
8 A 

Substituting Eq. (21) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
u=
55 ( )
m2  3m2
C2

) −2  ,
1−
(
2
2 C +C
8 A  4

1
2
and
v=
55 ( )
m2  3m2
C2

) −2  ,
1−
(
2
2 C +C
8 A  4

1
2
Substituting Eq. (21) into Eq. (16), together with Eq. (10) and simplifying, yields following traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 = 0) respectively:
−2 

∆
∆  
m2  3m2  B
+
u56 ( =
)
coth(
)   .
−
1 −
A
8 A2  4 2  2
 

and
−2 

∆
∆  
m2  3m2  B
+
v56 ( =
)
coth(
)   .
−
1 −
A
8 A2  4 2  2
 

u57 ( =
)
−2 

∆
∆  
m2  3m2  B
−
−
+
1
tanh(
)

  .

A
8 A2  4 2  2
 

and
v57 ( =
)
−2 

∆
∆  
m2  3m2  B
−
−
+
1
tanh(
)

 .

A
8 A2  4 2  2
 

Substituting Eq. (21) into Eq. (16), along with Eq. (11) and simplifying, our exact solutions become (if C1 = 0 but C2
0; C2 = 0 but C1 0) respectively:
−2 

−∆
−∆  
m2  3m2  B
+
)
cot(
)  .
u58 ( =
−
1 −
A
8 A2  4 2  2
 

and
)
v58 ( =
−2 

−∆
−∆  
m2  3m2  B
−
−
+
1
cot(
)



 .
A
8 A2  4 2  2
 

513
World Appl. Sci. J., 25 (3): 500-523, 2013
u59 ( =
)
−2 

−∆
−∆  
m2  3m2  B
−
−
−
1
tan(
)



 ,
A
8 A2  4 2  2
 

and
v59 ( =
)
−2 

−∆
−∆  
m2  3m2  B
−
−
−
1
tan(
)



 .
A
8 A2  4 2  2
 

Again for case 6, substituting Eq. (22) into Eq. (16) along with Eq. (7) and simplifying, the traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively become:
u=
61 ( )
Ω 
3m2  m2
) .
1−
tanh 2 (
2
2 A 
Ω
8 A 
and
v=
61 ( )
Ω 
3m2  m2
) .
1−
tanh 2 (
2
2 A 
Ω
8 A 
u=
62 ( )
Ω 
3m2  m2
) .
1−
coth 2 (
2
2 A 
Ω
8 A 
and
v=
62 ( )
where
Ω 
3m2  m2
) .
1−
coth 2 (
2
Ω
2 A 
8 A 


(4 E + B 2 )  t .
x−
=
2
 A

Substituting Eq. (22) into Eq. (16), along with Eq. (8) and simplifying, yields exact solutions (if C1 = 0 but C2
= 0 but C1 0) respectively:
u=
63 ( )
−Ω 
3m2  m2
).
1+
tan 2 (
2
Ω
2A
8 A 

and
v=
63 ( )
−Ω 
3m2  m2
).
1+
tan 2 (
2
Ω
2A
8 A 

u=
64 ( )
−Ω 
3m2  m2
) .
1+
cot 2 (
2
2
Ω
A
8 A 

and
v=
64 ( )
−Ω 
3m2  m2
) .
1+
cot 2 (
2
Ω
2
A
8 A 

514
0; C2
World Appl. Sci. J., 25 (3): 500-523, 2013
Substituting Eq. (22) into Eq. (16), along with Eq. (9) and simplifying, our obtained solution becomes:
u=
65 ( )

3m2 
m
C2
) −2  ,
1 − 22 (
2
+
C
C
8 A  4

1
2
and
v=
65 ( )

3m2 
m
C2
) −2  ,
1 − 22 (
2
C1 + C2
8 A  4

Substituting Eq. (22) into Eq. (16), together with Eq. (10) and simplifying, yields following traveling wave solutions
(if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
−2 

∆
∆  
m2  B
3m2 
+
u66 ( =
)
coth(
)   .
−
1 −
A
8 A2  4 2  2
 

and
v66 ( =
)
−2 

m2  B
∆
∆  
3m2 
−
−
+
1
coth(
)



 
A
8 A2  4 2  2
 

−2 

∆
∆  
m2  B
3m2 
+
u67 ( =
)
tanh(
)   .
−
1 −
A
8 A2  4 2  2
 

and
v67 ( =
)
−2 

∆
∆  
m2  B
3m2 
−
−
+
1
tanh(
)



 .
A
8 A2  4 2  2
 

Substituting Eq. (22) into Eq. (16), along with Eq. (11) and simplifying, our exact solutions become (if C1 = 0 but C2
0; C2 = 0 but C1 0) respectively:
)
u68 ( =
−2 

−∆
−∆  
3m2 
m2  B
−
−
+
1
cot(
)



 .
A
8 A2  4 2  2
 

and
v68 ( =
)
−2 

−∆
−∆  
m2  B
3m2 
−
−
+
1
cot(
)



 .
A
8 A2  4 2  2
 

u69 ( =
)
−2 

−∆
−∆  
m2  B
3m2 
−
−
−
1
tan(
)



 ,
A
8 A2  4 2  2
 

515
World Appl. Sci. J., 25 (3): 500-523, 2013
and
)
v69 ( =
−2 

−∆
−∆  
3m2 
m2  B
−
−
−
1
tan(
)



 
A
8 A2  4 2  2
 

Again for case 7, substituting Eq. (23) into Eq. (16), along with Eq. (7) and simplifying yields following traveling wave
solutions (if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
−1
−2 



Ω
Ω 
Ω
Ω  
B
B
3 
+
+
u71=
( )
coth(
)  + m5  d +
coth(
)  .
 m3 + m4  d +

2
2
2 A 
2
2
2 A  
2 A2 




and
( )
v71=
−1
−2 



Ω
Ω 
Ω
Ω  
3 
B
B
+
+
+
+
+
+
coth(
)
coth(
)
m
m
d
m
d





.
3
4
5
2
2
2 A 
2
2
2 A  
2 A2 




( )
u7 2 =
−1
−2 



Ω
Ω 
Ω
Ω  
3 
B
B
+
+
+
+
+
+
tanh(
)
tanh(
)
m
m
d
m
d

 .
 3
4 
5 
2
2
2 A 
2
2
2 A  
2 A2 




and
−1
−2 



Ω
Ω 
Ω
Ω  
B
B
3 
+
+
v72=
( )
tanh(
)  + m5  d +
tanh(
)  .
m3 + m4  d +

2
2
2 A 
2
2
2 A  
2 A2 




where


=
(4 E + B 2 )  t .
x−
 A2

Substituting Eq. (23) into Eq. (16), along with Eq. (8) and simplifying, our exact solutions become (if C1 = 0 but C2
0; C2 = 0 but C1 0) respectively:
u73=
( )
and
v73=
( )
−1


B
−Ω
−Ω 
3 
m
m
d
+
+
+
cot(
)



3
4


2
2
2A
2 A2 



−2 

B
−Ω
−Ω  
+ m5  d +
+
cot(
) 


2
2
2A

 
−1


B
−Ω
−Ω 
3 
m
m
d
+
+
+
cot(
)

 3
4 
2
2
2A
2 A2 



−2 

B
−Ω
−Ω  
+ m5  d +
+
cot(
) 


2
2
2A

 
u7 4 =
( )
−1


B
−Ω
−Ω 
3 
m
m
d
+
+
+
tan(
)



3
4


2
2
2A
2 A2 



−2 

B
−Ω
−Ω  
+ m5  d +
+
tan(
) 


2
2
2A

 
516
World Appl. Sci. J., 25 (3): 500-523, 2013
and
v74=
( )
−1


B
−Ω
−Ω 
3 
m
m
d
+
+
+
tan(
)


 3
4

2
2
2A
2 A2 



−2 

B
−Ω
−Ω  
tan(
) 
+ m5  d +
+


2
2
2A

 
Substituting Eq. (23) into Eq. (16) together with Eq. (9) and simplifying, we obtain
u75=
( )
−1
−2


B
C2 
B
C2  
3 
+
+
+
+
+
+
m
m
d
m
d
 3

 .
4
5
2
2
C1 + C2 
C1 + C2  
2 A2 




and
v75=
( )
−1
−2


B
C2 
B
C2  
3 
+
+
+
+
+
+
m
m
d
m
d
 3

 .
4
5
2
2
C1 + C2 
C1 + C2  
2 A2 




Substituting Eq. (23) into Eq. (16), along with Eq. (10) and simplifying, we obtain following traveling wave but
solutions (if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
)
u7 6 ( =
and
−1
−2 



∆
∆ 
∆
∆  
3 
+
+
+
+
coth(
)
coth(
m
m
d
m
d
)




 3
4
5

 .
A
A
2 A2 






)
v76 ( =
−1
−2 



∆
∆ 
∆
∆  
3 
+
+
+
+
coth(
)
coth(
m
m
d
m
d
)





3
4
5



 .
A
A
2 A2 



 

)
u7 7 ( =
−1
−2 



∆
∆ 
∆
∆  
3 
+
+
+
+
tanh(
)
tanh(
m
m
d
m
d
)

  .
 3
4 
5 
A
A
2 A2 



 

and
)
v77 ( =
−1
−2 



∆
∆ 
∆
∆  
3 
+
+
+
+
tanh(
)
tanh(
m
m
d
m
d
)




 3
4
5

 .
A
A
2 A2 



 

Substituting Eq. (23) into Eq. (16), together with Eq. (11) and simplifying, our obtained exact solutions become (if
C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
)
u78 ( =
−1
−2



 
−∆
−∆ 
−∆
−∆
3 
+
+
+
+
cot(
)
(
cot(
))
m
m
d
m
d





3
4
5

 .
A
A
2 A2 






and
517
World Appl. Sci. J., 25 (3): 500-523, 2013
)
v78 ( =
−1
−2



 
−∆
−∆ 
−∆
−∆
3 
+
+
+
+
cot(
)
(
cot(
))
m
m
d
m
d





3
4
5

 .
A
A
2 A2 






−1
−2 



−∆
−∆ 
−∆
−∆  
3 
)
tan(
)  + m5  d −
tan(
u7 9 ( =
)  .
 m3 + m4  d −


A
A
2 A2 



 

and
v79 ( =
)
−1
−2 



−∆
−∆ 
−∆
−∆  
3 
+
−
+
−
m
m
d
tan(
)
m
d
tan(
)
 3
4 
5 

 
A
A
2 A2 



 

Finally for case 8, substituting Eq. (24) into Eq. (16), along with Eq. (7) and simplifying yields following traveling
wave solutions (if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
( )
u81=
−1
−2 



Ω
Ω 
Ω
Ω  
1 
B
B
+
+
+
+
+
+
6
coth(
)
6
coth(
)
m
m
d
m
d

 .
 6
7 
8 
2
2
2 A 
2
2
2 A  
4 A2 




and
( )
v81=
−1
−2 



Ω
Ω 
Ω
Ω  
1 
B
B
+
+
+
+
+
+
6
coth(
)
6
coth(
)
m
m
d
m
d




 6
.
7
8
2
2
2 A 
2
2
2 A  
4 A2 




−1
−2 



Ω
Ω 
Ω
Ω  
1 
B
B
+
+
( )
tanh(
)  + 6 m8  d +
tanh(
)  .
u82=
m6 + 6 m7  d +
2
2
2 A 
2
2
2 A  
4 A2 




and
v82=
( )
where
−1
−2 



Ω
Ω 
Ω
Ω  
B
B
1 
+
+
+
+
+
+
m
m
d
m
d
6
tanh(
)
6
tanh(
)

 .
 6
7 
8 
2
2
2 A 
2
2
2 A  
4 A2 





= x − −
(4 E
 A2

+ B 2 )  t.

Substituting Eq. (24) into Eq. (16), along with Eq. (8) and simplifying, our exact solutions become (if C1 = 0 but C2
0; C2 = 0 but C1 0) respectively:
u83=
( )
−1


B
−Ω
−Ω 
1 
m
m
d
+
+
+
6
cot(
)

 6
7 
2
2
2A
4 A2 



−2 

B
−Ω
−Ω  
+ 6m8  d +
+
cot(
) 


2
2
2A

 
518
World Appl. Sci. J., 25 (3): 500-523, 2013
and
−1


B
−Ω
−Ω 
1 
v83=
+
( )
cot(
)
 m6 + 6m7  d +

2
2
2A
4 A2 



−2 

B
−Ω
−Ω  
+ 6m8  d +
+
cot(
) 


2
2
2A

 
u84=
( )
−1


B
−Ω
−Ω 
1 
m
m
d
+
+
+
6
tan(
)

 6
7 
2
2
2A
4 A2 



−2 

B
−Ω
−Ω  
+ 6m8  d +
+
tan(
) 


2
2
2A

 
and
−1


B
−Ω
−Ω 
1 
v84=
+
( )
tan(
)
 m6 + 6m7  d +

2
2
2A
4 A2 



−2 

B
−Ω
−Ω  
+ 6m8  d +
+
tan(
) 


2
2
2A

 
Substituting Eq. (24) into Eq. (16) together with Eq. (9) and simplifying, we obtain
( )
u85=
−1
−2


 
1 
B
C2 
B
C2
+
+
+
+
+
+
6
6
m
m
d
m
d

.




6
7
8
2
2
C1 + C2 
C1 + C2  
4 A2 




and
v85=
( )
−1
−2


B
C2 
B
C2  
1 
+
+
+
+
+
+
m
m
d
m
d
6
6
 6

 .
7
8
2
2
C1 + C2 
C1 + C2  
4 A2 




Substituting Eq. (24) into Eq. (16), along with Eq. (10) and simplifying, we obtain following traveling wave but
solutions (if C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
)
u86 ( =
−1
−2 



∆
∆ 
∆
∆  
1 
+
+
+
+
coth(
)
coth(
m
m
d
m
d
)





6
7
8

 .
A
A
4 A2 



 

and
−1
−2 



∆
∆ 
∆
∆  
1 
)
coth(
)  + m8  d +
coth(
v86 ( =
)   .
m6 + m7  d +
A
A
4 A2 



 

519
World Appl. Sci. J., 25 (3): 500-523, 2013
)
u87 ( =
−1
−2 



∆
∆ 
∆
∆  
1 
+
+
+
+
tanh(
)
tanh(
m
m
d
m
d
)





6
7
8

 .
A
A
4 A2 



 

and
−1
−2 



∆
∆ 
∆
∆  
1 
)
tanh(
)  + m8  d +
tanh(
v87 ( =
)  .
m6 + m7  d +
A
A
4 A2 



 

Substituting Eq. (24) into Eq. (16), together with Eq. (11) and simplifying, our obtained exact solutions become (if
C1 = 0 but C2 0; C2 = 0 but C1 0) respectively:
−1
−2



 
−∆
−∆ 
−∆
−∆
1 
cot(
)  + 6m8  (d +
cot(
))  .
u88 ( ) =
 m6 + 6 m7  d +



A
A
4 A2 



 

and
v88 ( )=
−1
−2



 
−∆
−∆ 
−∆
−∆
1 
+
+
+
+
6
cot(
)
6
(
cot(
))
m
m
d
m
d

 .
 6
7 
8 
A
A
4 A2 



 

u89 ( )=
−1
−2 



−∆
−∆ 
−∆
−∆  
1 
+
−
+
−
6
tan(
)
6
tan(
m
m
d
m
d
)





6
7
8

 .
A
A
4 A2 



 

and
v89 ( )=
−1
−2 



−∆
−∆ 
−∆
−∆  
1 
+
−
+
−
6
tan(
)
6
tan(
m
m
d
m
d
)





6
7
8



 .
A
A
4 A2 



 

Bekir and Uygun [29] investigated solutions of the (2+1)-dimensional breaking soliton equations by the (G'/G)expansion method and obtained only twelve solutions (A. 1)-(A. 12) (see appendix). Moreover, in this article one hundred
forty four solutions of the (2+1)-dimensional breaking soliton equations are constructed by applying the new approach
of generalized (G'/G)-expansion method. But by means of the new approach of generalized (G'/G)-expansion method we
obtained solutions are different to Bekir and Uygun [29] solutions. Furthermore, we obtain solutions u1 ( ) − u1 ( ),
u2 ( ) − u2 ( ),
1
9
u3 ( ) − u3 ( ),
1
9
u4 ( ) − u4 ( ),
1
9
u5 ( ) − u5 ( ),
1
9
u6 ( ) − u6 ( ),
1
9
u7 ( ) − u7 ( ), ,
1
9
v2 ( ) − v2 ( ), v3 ( ) − v3 ( ), v4 ( ) − v4 ( ), v5 ( ) − v5 ( ), v6 ( ) − v6 ( ), v7 ( ) − v7 ( )
1
9
1
9
1
9
1
9
1
9
1
9
,
u8 ( ) − u8 ( )
1
9
,
1
9
v1 ( ) − v1 ( ),
1
9
v8 ( ) − v8 ( ). . These solutions are new
1
9
and were not obtained by Bekir and Uygun [29]. On the other hand, the auxiliary equation used in this paper is different,
so obtained solutions is also different.
CONCLUSION
trigonometric functions and the rational functions.
This study shows that the new generalized (G'/G)expansion method is quite efficient and practically well
suited to be used in finding exact solutions of NLEEs.
Also, we observe that the new generalized (G'/G)expansion method is straightforward and can be applied
to many other nonlinear evolution equations.
The new generalized (G'/G)-expansion method is
used to obtain the exact solutions of the (2+1)dimensional breaking soliton equations. The solution
method is very simple and effective. The solutions are
expressed in the form of the hyperbolic functions, the
520
World Appl. Sci. J., 25 (3): 500-523, 2013
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Appendix: Bekir and Uygun’s solution [29]
Bekir and Uygun [29] established exact solutions of the well-known (2+1)-dimensional breaking soliton equations by
using the (G'/G)-expansion method. They found the following solutions of the form,
When
2
– 4µ > 0

1
 C1 sinh 
3
2
2 
u1 = (4 − )
8
1

 C1 cosh  2



1
 C1 sinh 
3
2
v1 = (4 − 2 ) 
1
8

 C1 cosh  2


where = x + y – (
2

1
 C1 sinh 
3
2
v2= (4 − 2 ) 
8
1

 C1 cosh  2


When
2
2
−4
2
−4
2
−4
2
−4

1
 + C2 cosh 

2

1
 + C2 sinh 

2

1
 + C2 cosh 

2

1
 + C2 sinh 

2
2
2
−4
2
−4
2
−4
2
−4
2
−4
2
−4
2
−4
2
−4

2

 + 3 − 3 ,
8
2



(A.1)
2


2
 + 3 − 3 ,

8
2


(A.2)
– 4µ)t, or

1
 C1 sinh 
3
2
u2= (4 − 2 ) 
1
8

 C1 cosh  2


where = x + y + (
2
2
−4
2
−4
2
−4
2
−4

1
 + C2 cosh 

2

1
 + C2 sinh 

2

1
 + C2 cosh 

2

1
 + C2 sinh 

2
2


2
1
 +
−
,

8 2


(A.3)
2

2

1
 +
−
,
8
2



(A.4)
– 4µ)t
– 4µ < 0

1
4 −
 −C1 sin 
3 2
2
u3 = ( − 4 ) 
1
8

 C1 cos  2 4 −



1
4 −
 −C1 sin 
3 2
2
v3 = ( − 4 ) 
1
8

 C1 cos  2 4 −


2
2
2
2

1
4 −
 + C2 cos 

2

1
4 −
 + C2 sin 

2

1
4 −
 + C2 cos 

2

1
4 −
 + C2 sin 

2
2
2
2
2
2


2
 + 3 − 3 ,
 
8
2
 
 
2


2
 + 3 − 3 ,
 
8
2
 
 
522
(A.5)
(A.6)
World Appl. Sci. J., 25 (3): 500-523, 2013
where = x + y – (
2
– 4µ)t, or

1
4 −
 −C1 sin 
3 2
2
u4= ( − 4 ) 
8
1

 C1 cos  2 4 −



1
4 −
 −C1 sin 
3 2
2


v4= ( − 4 )
1
8

 C1 cos  2 4 −


where = x + y + (
When
=
u5
=
v5
2
2
2
2
2
2

1
4 −
 + C2 cos 

2

1
4 −
 + C2 sin 

2

1
4 −
 + C2 cos 

2

1
4 −
 + C2 sin 

2
2

2

1
 +
−
,
8 2
 
 
 
2
2
2
2
(A.7)
2


2
1
 +
−
,
 
8 2
 
 
(A.8)
– 4µ)t.
– 4µ = 0,
3C2 2
2[C1 + C2 ( x + y ))
3C2 2
2[C1 + C2 ( x + y ))
where = x + y – (
2
2
2
+
+
2
3
8
2
3
8
−
3
,
2
(A.9)
−
3
,
2
(A.10)
– 4µ)t, or
2
3C2 2
1
−
+
−
,
u6 =
2
8 2
2[C1 + C2 ( x + y ))
(A.11)
2
3C22
1
−
+
−
,
v6 =
2
8
2
2[C1 + C2 ( x + y ))
(A.12)
523