Some New Exact Traveling Wave Solutions for the
Generalized Benney-Luke (GBL) Equation
with Any Order
Jun-Jie Wang1,2, Lian-Tang Wang2, Kuan-De Yang1 1
Mathematics Department, Simao Teachers’ College, Pu’er, China 2
Mathematics Department, Northwest University, Xi’an, China Email: [email protected]
Received December 5,2011; revised February 22, 2012; accepted February 29, 2012
ABSTRACT
In the paper, a auxiliary equation expansion method and its algorithm is proposed by studying a second order nonlinear ordinary differential equation with a four-degree term. The method is applied to the generalized Benney-Luke (GBL) equation with any order. As a result, some new exact traveling wave solutions are obtained which singular solutions, triangular periodic wave solutions and jacobian elliptic function solutions. This algorithm can also be applied to other nonlinear wave equations in mathematical physics.
Keywords: Generalized Benney-Luke (GBL) Equation; Jacobian Elliptic Function; Singular Solution
1. Introduction
Nonlinear phenomena that appear in many areas of sci- entific fields such as solid state physics, plasma physics, fluid dynamics, mathematical biology and chemical ki- netics can be modeled by partial differential equation. A broad class of analytical solutions methods and numeri- cal solutions methods were used in handle these prob- lems. The investigation of exact traveling wave solution to nonlinear equations plays an important role in the study of nonlinear physical phenomena. Various methods for seeking traveling wave solutions to nonlinear partial differential equations are proposed such as inverse scat-tering transform method [1], BÄacklund and Darboux transform [2-6], Hirota method [7], Lie group method [8,9] and so on.
In the paper, we shall consider the following general- ized Benney-Luke (GBL) equation [10]:
2 1
( ) 2
p
tt tt t x
p
xx yy x xt y yt
a b p
0 (1.1)
The paper is organized as follows. In Section 2, we present the auxiliary equation method and its algorithm. In Section 3, we present some exact traveling wave solu- tions of system (1.1). Finally some conclusion are given.
2. The Auxiliary Equation Algorithm
We outline our auxiliary equation algorithm:
Step I. For a given nonlinear wave equation with one physical field u x y t
, ,
; in three variables x y t, , ,
, ,t x, y, tx, ty, xy, xxx,
0F u u u u u u u u (2.1) We seek its special solution, traveling wave solution, in the form of
, ,
,u x y t u x y ct (2.2)
where is constant. Substitution (2.2) into (2.1) gives rise to a nonlinear ordinary differential equation
c
, , , ,
0G u u u u (2.3)
Step II. To seek the traveling wave solution of (2.3), we assume that (2.3) has the solution in the form of
0
n i i i
u a
(2.4)with the new variable
satisfying:
20 2 4
h h h
42
(2.5)
j
h are constants and is the integer to be determined
later.
n
Step III. Determined the parameter . Substituing Equa- tion (2.4) along with Equation (2.5) into Equation (2.3) and balancing the highest derivative term with the non- linear terms in the Equation (2.3), we then obtain the va- lue of .
n
n
Step IV. Determine the parameters
, ;i j 0,1, 2, , , 0, 2, 4 ,
Substitution Equation (2.4) along with Equation (2.5) into Equation (2.3) and setting the coefficients of all
powers to zero, we will
obtain a system of nonlinear algebraic equations (NAEs) with respect to the parameters
0,1, 2, , 0,1i j
i n j
, ;i j 0,1, 2, , , 0, 2, 4 ,
c a h i n j
By solving the NAEs if available, we can determine those parameters explicitly.
3. Exact Traveling Wave Solutions for the
Generalized Benney-Luke (GBL)
Equation with Any Order
By considering that the traveling wave solutions of (1.1) propagate in the direction of the positive x-axis, i.e.
x y t, ,
u x ct y, Ref. [10] gave the following trav-
eling equation:
2 2
1
1 2
2 2
2
0
xx xxxx yy xxyy
p p p
x xx x y yy y xy
c u a bc u u a bc u
c p u u pu u u u u
(3.1)
In this paper, we will study the traveling wave solu- tions of (1.1) with the form:
, ,
,u x y t x y ct
x y ct
where c is the wave speed. Substituting it into (1.1) and noting (3.1), we have
2 2
2 2 2 4
1 1
1 2 2 1
0 1
p p
c
a bc a bc a
c p
p
(3.2)
Integrating (3.2) once with respect to and taking
the integral constant as zero, we obtain
2 2 2 2 2 4
1 1
1 2
2 1
0 1
p p
c a bc a bc a
c p
p
(3.3)
Letting
d
,
d d
(3.4)and writing
2 2
2 2 2 4
1
2 2 2
1
, 2
1
1 2
p
c A
a bc a bc a
c B
p a bc a bc a
then from (3.4), we have that
2
1 2
d
2 d
p
A p B
(3.5)
Let
p
We have
2 2
1 1
2
p
A p B
p p
3
0
(3.6)
We expand the solution of Equation (3.6) in form of Equation (2.4). Substitution Equation (2.4) along with Equation (2.5) and balancing the highest order derivative term.
2
with nonlinear term 3 in Equation (3.6) gives
n . Hence, we have
20 1 2
a a a
(3.7)
wherea a a0, 1, 2are constant to be determined later;
satifiyed Equation (2.5).
We can get the following equation by Step IV
2 3 4 5 6
0 1 2 3 4 5 6 0
f f f f f f f (3.8)
If
2 2
1
pm n
p l
, we can get the following solution:
3 4
2 3
2 3 2 2 2
h p p
a
p B p
1 0 a
3
0 3
3 2 2
2 2 1
A p p
a
p B p p
3
2 3
2 1
Ap h
p p
6 2
0 2
3 4
16 1
p A h
h p p
3
3
1 3
4 2
3
3 2 2
2 2 1
2 3 2 2
2
p
A p p
H
p B p p
h p p
p B p
(3.9)
4
where
6 2 3
2 2
2 3
3 4
4 4
2 1 16 1
p A Ap
p p
h p p
h
Th 3.1. Suppose that
6 2 3
4 4
2 3
3 4
,
2 1 16 1
p A Ap
h h
p p
h p p
If , that Equation (1.1) has a kink profile solu-
tion
4 0 h
3
1 3
1 3
4 2
4 0
3
3 2 2
2 2 1
2 3 2 2
tanh d
2
p
A p p
p B p p
h p p
h
p B P
Th 3.2. 1) If
2
6 2 3
4 4
2 2 3 2
3 4
1 ,
2 1 16 1
k h h
p A Ap
k p p k
h p p
that Equation (1.1) has two Jacobian elliptic function solutions
3
2 3
1 3
4 2 4
0
3 2
3 2 2
2 2 1
2 3 2 2
d
2
p
A p p
p B p p
h p p h
sn
p B P k
3
3 3
1
4
3 2 0
4 3
4 0 2
3 2 2
2 2 1
2 3 2 2
d
2
p
A p p
p B p p
h cn
h p p k
p B P h
dn k
2) If
2 6 2
4
2 2
3 4
2 3
4 2 3
1 ,
16 1
2
2 1
k h
p A
k
h p p
k h Ap
k
p p
that Equation (1.1) has a Jacobian elliptic function solution
3
4 3
1 3
4 4
0
3 2
3 2 2
2 2 1
2 3 2 2
d
2
p
A p p
p B p p
h p p h
cn
p B P k
3) If
6 2
2 4 2
3 4
3
2 4 3
1 ,
16 1
2
2 1
p A
k h
h p p
Ap
k h
p p
that Equation (1.1) has a Jacobian elliptic function solu- tion
3
5 3
1 3
4 4
0
3 2
3 2 2
2 2 1
2 3 2 2
d
2
p
A p p
p B p p
h p p h
dn
p B P k
4) If
6 2 3
2 2
4 4
2 3
3 4
, 1
2 1 16 1
p A Ap
k h k h
p p
h p p
that Equation (1.1) has two Jacobian elliptic function solutions
3
6 3
1 3
4 3
4 0
3 2 2
2 2 1
2 3 2 2 1
d
2
p
A p p
p B p p
h p p
p B P sn h
3
7 3
1 3
4 0
4 3
4 0
3 2 2
2 2 1
2 3 2 2
d
2
p
A p p
p B p p
dn h
h p p
p B P cn h
5) If
2 2
6 2 3
4 4
2 2 3
3 4
2 1 ,
1 2 1 1
16 1
k h
k h
p A Ap
k p p k
h p p
2
that Equation (1.1) has a Jacobian elliptic function solu- tion
3
8 3
1
3 4
3
4
0 2
3 2 2
2 2 1
2 3 2 2 1
d
2
1
p
A p p
p B p p
h p p
p B P h
cn k
6) If
2
6 2 3
4 4
2 2 3 2
3 4
2 ,
1 2 1 1
16 1
k h h
p A Ap
k p p k
h p p
that Equation (1.1) has a Jacobian elliptic function so - tion
lu
3
9 3
1
3 4
3
4
d
2
p B P h
0 2
3 2 2
2 2 1
2 3 2 2 1
1
p
A p p
p B p p
h p p
dn k
7) If
6 2
2 4 2
3 4
3
2 4 3
1 , 16 1
2 2 1
p A
k h
h p p
Ap
k h
p p
that Equation (1.1) has a Jacobian elliptic function solu- tion
3
10 3
1 3
4 0
4 3
4 0
(3 2 2 ) 2( 2) ( 1 )
2 3 2 2
d
2
p
A p p
p B p p
cn h
h p p
p B P sn h
8) If
2
6 2 3
4 4
2 2 3 2
3 4
2 ,
1 2 1 1
16 1
k h h
p A Ap
k p p k
h p p
that Equation (1.1) has a Jacobian elliptic function solu- tion
3
11 3
1
4
3 2 0
4
0 2
3 2 2
2 2 1
2 3 2 2 1
1
p
A p p
p B p p
h sn
h p p k
k
3
4
d
2
p B P h
cn
9) If
6 2
4
2 2 2
3 4
2 3
4
3 2 2
, 1 16 1
2 1
2 1 1
h p A
k k
h p p
k h
Ap
p p k k
that Equation (1.1) has a Jacobian elliptic function solution
3 3
1 4
3 2 2 0
4
0 2 2
3 2 2
2 2 1
2 3 2 2 ( 1)
1
p
A p p
p B p p
h sn
h p p k k
dn k k
12
3
4
d
2
p B P h
10) If
6 2
2 2
4 2
3 4
3
2 4 3
1 , 16 1
2 1 2 1
p A
k k h
h p p
Ap
k h
p p
that Equation (1.1) has a Jacobian elliptic function solution
3
13 3
1 3
4 0
4 3
4 0
3 2
2 2 1
2 3 2 2
d
2
p
A p
p B p p
dn h
h p p
p B P sn h
2p
11) If
3
2 4,
3
2 1 2 2
42 1
16 1
h k h
p p
h p p
6 2 3
4
p A Ap
that Equation (1.1) has four Jacobian elliptic function solutions
3 3
4
14 3 3
1
4 0
4 0 4 0
3 2 2 2 3 2 2
2
2 2 1
4 1
A
d
4 4
p
p p h p p
p B P
p B p p
cn h
sn h sn h
3
15 3
4 4 0
3
4 0
3 2 2
2 2 1
d
2 1 4
A p p
p B p p
p B P cn h
1 3
2 3 2 2 4
p
h p p sn h
3 3
4
16 3 3
1
4 0 4 0
3 2 2 2 3 2 2
2
2 2 1
4 4
p d
A p p h p
p B P
p B p p
ksn h idn h
p
3 3
4
17 3 3
1
4 0
2
4 0
3 2 2 2 3 2 2
2
2 2 1
4
d
4 1
p
A p p h p
p B P
p B p p
dn h
kcn h i m
p12) If
2
6 2 3
4 4
2 3 2
3 4
4 1 ,
2 1 2 1
16 1
k h
p A Ap
h
p p k
h p p
that Equation (1.1) has a Jacobian elliptic function solution
3
18 3
1 4
3
0 2 4
3
4
0 2
3 2 2
2 2 1
4
2 3 2 2 1
d
2 4
1 1
p
A p p
p B p p
h cn
h p p k
p B P h
sn k
13) If
6 2
4 2
3 ,
p A
h
4
2 3
4 2 3
16 1
2 2
2 1
h p p
k h
Ap
k
p p
that Equation (1.1) has three Jacobian elliptic function solutions
3 3
4
19 3 3
1
4 4
0 0
2 2
3 2 2 2 3 2 2
2
2 2 1
4 4
d
p
A p p h p
p B P
p B p p
h h
sn icn
k k
3 3
4
20 3 3
1 4
0 2
d 1
k
i 2 4 4
0 0
2 2
3 2 2 2 3 2 2
2
2 2 1
4
4 4
p
A p p h p p
p B P
p B p p
h dn
h h
k sn cn
k k
3
21 3
1 4
3
0 2 4
3
4 0 2
3 2 2
2 2 1
4
2 3 2 2
d
2 4
1
p
A p p
p B p p
h ksn
h p p k
p B P h
dn k
14) If
2
6 2 3
4 4
2 3 2
3 4
2 1
,
1
2 1
16 1
k h
p A Ap
h
k
p p
h p p
that Equation (1.1) has a Jacobian elliptic function solu- tion
3
22 3
1 4
3
0 2
4 3
4
0 2
3 2 2
2 2 1
4
2 3 2 2 1
d
2 4
1
1
p
A p p
p B p p
h dn
h p p k
p B P h
ksn k
15) If
6 2
2 2
4 2
3 4
16h p 1 p
3
2 4 3
1 ,
2 1
2 1
p A
k h
Ap
k h
p p
that Equation (1.1) has a Jacobian elliptic function solu- tion
p
3 3
4
23 3 3
1
4 0 4 0
3 2 2 2 3 2 2
2
2 2 1
4 4
p
A p p h p p
p B P
p B p p
kcn h dn h d
16) If
6 2
4
2 2
3 2
4
2 3
4 2
3 2
,
16 1 1
2 1
2 1 1
h p A
h p p k
k h
Ap
p p k
that Equation (1.1) has a Jacobian elliptic function solu- tion
3 3
4
24 3 3
1 4
0 2 2
4 4
0
2 2
2 2
3 2 2 2 3 2 2
2
2 2 1
4
1
d
4 4
1 1
p
A p p h p p
p B P
P B p p
h sn
k
h h
dn cn
k k
0
17) If
2
6 2 3
4 4
2 4 3 4
2 2
, 16
k h
h
p A Ap
h p
3
4 1 p k 2 p 1 p k
that Equation (1.1) has a Jacobian elliptic function solu- tion
3 3
4
25 3 3
1
4 0 4
2 4
0 4
3 2 2 2 3 2 2 2
2 2 1
4
d
4 1
p
A p p h p p
p B P
p B p p
h cn
k h k dn
k
4. Conclusion
In the paper, we apply the auxiliary equation method t study Equation (1.1), and get some new conclusions. So
dic wave solutions and Jacobian elliptic function
solutions. These solutions may be useful for describing certain nonlinear physical phenomena of Equation (1.1). The method which we have propose in this paper is standard, direct and computerized method ,which allow us to do complicated and tedious algebraic calculation.It is shown that the algorithm can be also applied to other nonlinear wave equations such as generalized BBM equa- tion,
2
0
p p
t x xxt
u au u bu u
and generalized Klein-Gordon equation:
2 41 2 3
tt xx yy 0
p p
b u b u u b u u
u k u u
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new exact traveling wave solution of Equation (1.1) are obtain which include new singular solution, triangular perio