IJST (2012) A4: 431-452
Iranian Journal of Science & Technology
http://www.shirazu.ac.ir/en
Solutions of the perturbed Klein-Gordon equations
A. Biswas1*, G. Ebadi2, M. Fessak1, A. G. Johnpillai3, S. Johnson1, 4,
E. V. Krishnan5 and A. Yildirim6
1
Department of Mathematical Sciences, Delaware State University, Dover, DE 19901-2277, USA
2
Department of Mathematical Sciences, University of Tabriz Tabriz, 51666-14766, Iran
3
Department of Mathematics, Eastern University, Sri lanka
4
Lake Forest High School, 5407 Killens Pond Road, Felton, DE-19943, USA
5
Department of Mathematics and Statistics, Sultan Qaboos University, P.O. Box 36, Al Khod 123, Muscat, Oman
6
4146 Sk. No: 16 Zeytinalani Mah. Urla, Izmir, Turkey
E-mail: [email protected]
Abstract
This paper studies the perturbed Klein-Gordon equation by the aid of several methods of integrability. There are
six forms of nonlinearity that are considered in this paper. The parameter domains are thus identified.
Keywords: The perturbed Klein-Gordon equation; integrability; nonlinearity
1. Introduction
The nonlinear Klein-Gordon equation (KGE)
appears in Theoretical Physics in the context of
relativistic quantum mechanics. There have been
several studies conducted with this equation by
Physicists and Applied Mathematicians across the
globe [1-10]. One of the most important tasks is to
carry out the integration of the perturbed KGE. This
paper will focus on obtaining the solution of the
the nonlinear function will now be related to the
nonlinear function of the sine-Gordon equation
(SGE).This will allow us to justify the study of
perturbation terms that arise in the theory of long
Josephson junctions that are modeled by the SGE.
2.1. Six Forms of Nonlinearity
perturbed KGE by the aid of G /G method, expfunction method and, finally, the traveling wave
solution will be obtained. There are six types of
nonlinearities of the KGE that will be considered in
this paper.
There are six types of nonlinearity to be
considered. They are labeled as Forms I through VI.
For each of these forms, the connection to SGE will
be illustrated in the next six subsections. In each of
the following forms of nonlinearity, a , b and c
are real valued constants. The exponent n is a
positive integer.
2. Mathematical Analysis
2.1.1. Form-I
The dimensionless form of the nonlinear KGE that
studied in this paper is given by [4]
In this case, the nonlinear function
given by
'
qtt  k 2 qxx  F (q) = 0 .
(1)
Here, in (1), the dependent variable q represents
the wave profile, while x and t are the
independent variables that represent the spatial and
temporal variables respectively. Also, k is a real
valued constant.The function F ( q ) represents the
nonlinear function. The mathematical analysis of
*Corresponding author
Received: 18 February 2012 / Accepted: 16 May 2012
F (q) = aq  bq 2
F (q ) is
(2)
Equation (2) can be approximated by
qtt  k 2 qxx  a sin q  b1 cos q  b2 cos 2q = 0 (3)
for small values of q , where b1 and b2 must be
chosen such that there is no constant term and the
quadratic term has coefficient b . For better
approximations, higher terms such as sin 2q must
be added.
IJST (2012) A4: 1-12
432
2.1.4. Form-IV
2.1.2. Form-II
Here,
In this case,
F (q) = aq  bq 3
(4)
and KGE with this form is sometimes known as the
 equation . This form can be approximated by
4
qtt  k 2 qxx  a1 sin q  a2 sin 2q = 0
(5)
for small values of q , where a1 and a2 must be
F (q) = aq  bq n  cq 2 n 1
(9)
In this case, if n is odd, then this can be
approximated by
qtt  k 2 qxx  a1 sin q
 a2 sin 2q    an sin nq = 0
(10)
q , where a1 through an must
be chosen to exactly satisfy the coefficients of q ,
q n , and q 2 n 1 . For better approximations, higher
terms must be added. If n is even, then this can be
chosen such that the linear coefficient is a and the
cubic coefficient is b . For better approximations,
higher terms such as sin 3q must be added.
for small values of
2.1.3. Form-III
approximated by
Here,
qtt  k 2 qxx  a1 sin q  a2 sin 2q
   an sin nq  b1 cos q  b2 cos 2q  
F (q) = aq  bq n
(6)
This encompasses the previous two forms in a
generalized format. The special case is when n = 1
and n = 2 collapses to the previous two forms. If
n is odd, then this can be approximated by
qtt  k 2 qxx  a1 sin q  a2 sin 2q
(7)
  n  1 
   a n 1 sin  
 q  = 0
2



2
for small values of q , where a1 through a( n 1)/2
must be chosen such that only the linear term and
n  
bn cos    1 q  = 0
1
2  
2
(11)
q , where a1 through an must
be chosen to exactly satisfy the coefficients of q
2 n 1
and q
, and b1 through bn/2 1 must be chosen
for small values of
n
such that there are no even terms until the q term
and it has coefficient b . For better
approximations, higher terms must be added.
q n survive and have coefficients a and b
2.1.5. Form-V
respectively. For better approximations, higher
terms must be added. If n is even, then this can be
approximated by
For this form,
qtt  k 2 qxx  a1 sin q  a2 sin 2q  
This is similar to the other forms, and can be
approximated by a series of sine and cosine terms.
n  
 a n sin    1 q   b1 cos q 
1
2  
2
n  
b2 cos 2q    bn cos    1 q  = 0 (8)
1
2  
2
for small values of q , where a1 through an/2 1
must be chosen such that only the linear term
survives with coefficient a , and b1 through bn/2 1
F (q) = aq  bq1 n  cq n 1.
(12)
2.1.6. Form-VI
This is the logarithmic form of nonlinearity. In
this case [6],
F (q ) = aq  bq ln q
(13)
Define
f ( x) = x log x 2
(14)
must be chosen such that there are no even terms
until the q term and it has coefficient b . For
better approximations, higher terms must be added.
n
It is clear that f is an odd function, thus the
Fourier series collapses to just the sine terms
433
IJST (2012) A4: 1-12

f ( x) = bn sin(nx)
(15)
n =1
where the coefficients are defined by
bn =
1

f ( x) sin(nx) dx
 

(16)
Solving these coefficients leads to the sine
expansion of x log x

x log x 2 = 4[
n =1

2
(1)
n
n 1
log( )
(17)
0
The perturbed KGE with general form of
nonlinearity is given by Equation (19). For
traveling wave solution the hypothesis
sin t
dt
t
(20)
is substituted into the (19) where v represents the
velocity of the wave and g ( x, t ) represents the
wave profile. Thus equation (19) reduces to
v
where Si is the sine integral defined by
Si ( x) = 
3.1. Description of the method
q( x, t ) = g ( x  vt )
1
Si (n )]sin(nx)
 n2
x
the solution of the perturbed KGE. The saerach will
be for soliton solutions only, in this section. After a
general description of the method, the six forms of
nonlinearity will be considered in separate
subsections.
2
 k 2   v   v 2  g ''  F ( g )   g
   v    g '   vg '''  g '''' = 0
(18)
where
Here,
(21)
g ' = dg/ds , g '' = d 2 g/ds 2 and so on.
2.2. Perturbation Terms
s = x  vt
The perturbed KGE that will be studied in this
paper is given by [4]
In order to solve (21) by the traveling wave
hypothesis, it is necessary to set
qtt  k 2 qxx  F (q) =  q   qt   qx
 qxt   qtt   qxxt   qxxxx
 =  =  = = 0
(19)
These perturbation terms typically arise in the
study of long Josephson junction in the context of
sine-Gordon equation (SGE). Since SGE can be
approximated by KGE as seen in the previous subsection, an exact solution of the perturbed KGE will
make sense in this context [9].
For the perturbation terms,  represents losses
across the junction,  accounts for dissipative
losses in Josephson junction theory due to tunneling
of normal electrons across the dielectric barrier, 
is generated by a small inhomogeneous part of the
local inductance,  represents diffusion and  is
the capacity inhomogeneity. Finally,  accounts
for losses due to a current along the barrier [9].
In this paper, therefore the perturbed KGE, given
by (19) for the six forms of nonlinearity as defined
in (2)-(13), will be integrated in the subsequent
three sections.
3. Traveling Wave Solutions
In this section, the traveling wave solutions will be
obtained. This is the most fundamental approach to
(22)
(23)
Thus equation of study given by (19) reduces to
qtt  k 2 qxx  F (q) =  q   qxt   qtt
(24)
and hence (21) reduces to
v
2
 k 2   v   v 2  g ''  F ( g )   g = 0. (25)
'
Multiplying both sides by g and choosing the
integration constant to be zero, since the search is
for soliton solution, gives
v
2
 k 2   v   v 2  g ' 
2
2g ' F ( g )ds   g 2 = 0.
(26)
Separating variables and integrating leads to
x  vt
v2  k 2   v  v2
dg
=
 g  2 g F ( g )ds 



2
'
1
2
.
(27)
IJST (2012) A4: 1-12
434
This expression will be evaluated for the six
different functions F that will be studied in the
following subsections.
which is from the nonlinearity that is given by (4).
Thus, in this case, using (27) we obtain
x  vt
In this case, the perturbed KGE that will be
studied is given by
v2  k 2   v  v2

1
bg 2 
1 
=
tanh  1 

2(  a) 
 a

qtt  k 2 qxx  aq  bq 2 =  q   qxt   qtt
which leads to the soliton solution
3.1.1. Form-I
(28)
(36)
for the nonlinearity that is given by (2). This is the
quadratic form of nonlinearity and hence (28) is the
g ( x  vt ) = q ( x, t ) = Asech[ B ( x  vt )], (37)
perturbed  equation. In this case (27) reduces to
where the amplitude of the soliton is given by
4
x  vt
v  k   v  v
2
2
2
= 3
dg
g 3(  a)  2bg
(29)
1
2
x  vt
B=
(30)
g ( x  vt ) = q ( x, t ) = Asech 2[ B ( x  vt )], (31)
where the amplitude A and the inverse width
of the soliton are given by
3(a   )
2b
B
(32)
1
 a
2
2
2 v  k   v  v2
(39)
is the width of the soliton. Hence the constraint (34)
is still valid and it is additionally necessary that
(40)
must also hold.
3.1.3. Form-III
In this case, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq n =  q   qxt   qtt . (41)
Here equation (27) yields
and
B=
(38)
b(  a) < 0
which leads to the soliton solution
A=
2a  
b
and
which gives, upon integration
v2  k 2   v  v2

2
2bg 
1 
=
tanh  1 

3(  a) 
 a

A=
1
 a
,
2
2
2 v  k   v  v2
x  vt
(33)
respectively. The width of the soliton forces the
constraint
(  a )  v 2  k 2   v   v 2  > 0
(34)
3.1.2. Form-II
After simplification, this implies that the 1-soliton
solution to (41) is given by
2
In this case, the perturbed KGE that will be
studied is given by
qtt  k qxx  aq  bq =  q   qxt   qtt
2
2

(n  1)   a
v2  k 2   v  v2


2bg n 1
1 
(42)
tanh  1 
.
(n  1)(  a) 

=
3
(35)
q ( x, t ) = g ( x  vt ) = Asech n 1[ B( x  vt )], (43)
where the amplitude of the soliton is given by
1
 (n  1)(a   )  n 1
A=

2b

(44)
435
IJST (2012) A4: 1-12
3.1.5. Form-V
and the width B is
B=
For this form, the perturbed KGE that will be
studied is given by
(n  1)
 a
2
2
2
v  k   v  v2
(45)
which again requires the same constraint in (34).
qtt  k 2 qxx  aq  bq1 n  cq n 1
=  q   qxt   qtt
Here, equation (27) reduces to give
3.1.4. Form-IV
In this case, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq n  cq 2 n 1 =  q   qxt   qtt . (46)
x  vt
v2  k 2   v  v2
A = 2i g n 
x  vt
v  k 2   v  v2
 g n 1 
1
=
ln 
,
(n  1)   a  G 
2
 gn 
(47)
(48)
2b
c
2 1 
g n 1 
g 2n2
(n  1)(  a )
n(  a)
which finally leads to
A
1  C cosh[ B( x  vt )]
1
n 1
, (49)
C=g
C=
2J
1  1  4 JK
1
n 1
 a
v  k   v  v2
2
(57)
2
2(n  1)(  a )
b
(50)
(51)
2J
1  1  4 JK
(58)
(59)
1  1  4 JK
1  1  4 JK
(60)
J=
2b
(2  n)(  a )
(61)
K=
2c
(2  n)(  a)
(62)
and  is the elliptic integral of the third kind that
is defined as
 (n;  | k ) = 
sin 
1  nt 
2
dt
(1  t 2 )(1  k 2t 2 )
.
(52)
3.1.6. Form-VI
For this form, the perturbed KGE that will be
studied is given by
(53)
2
4b
4c
= 1.

2
2
(n  1) (  a ) n(  a )
n
2
0
with the constraints (34) and
b(  a) < 0

n
D=
where
B = (n  1)

2 Jg n
B = n 1  Jg  Kg 
1  1  4 JK
2b
g n 1  2
(n  1)(  a)
 (n  1)(  a ) 
A = 

b


A
 isinh 1C D , (56)
B
2J
1  1  4 JK
n
where
q( x, t ) = g ( x  vt ) =
=
where
In this case, equation (27) gives
G=
(55)
(54)
qtt  k 2 qxx  aq  bq ln q =  q   qxt   qtt . (63)
Here, from (27) we get
IJST (2012) A4: 1-12
436
1
x  vt
2 
b
2
=    a    b ln g  (64)
2
2
2
b 
2
v  k   v  v

Upon simplifying (64), the Gaussons for the
perturbed KGE are given by
q( x, t ) = g ( x  vt ) = Ae  B
2 ( x  vt )2
,
(65)
v 2 B 2Q'' ,) = 0
(71)
and restricts the general solutions on travelling
wave solutions, where U = U ( ) , and prime
denotes the derivative with respect to
  a
A = exp 
 b
1
 
2
(66)
b
1
2
2
2 v  k   v  v2
(67)
(68)
to hold in order for the Gaussons to exist.
4.
G' /G Method
'
In this section, the G /G method will be adopted
to carry out the integration of the perturbed KGE.
This study will be divided into the following six
subsections based on the types of nonlinearity that
are being studied.
4.1. Description of the method
'
of the G /G -expansion method for solving certain
nonlinear partial differential equations (PDEs).
Suppose that we have a nonlinear PDE for u ( x, t ) ,
in the form of
(69)
where P is a polynomial in its arguments, which
includes nonlinear terms and the highest order
derivatives. The wave transformation
q( x, t ) = Q( ),  = B( x  vt ),
(70)
reduces Eq. (69) to the ordinary differential
equation (ODE)
P (Q, BQ' , vBQ' , B 2Q'' , vB 2Q'' ,
(72)
are constants to be
d 2G ( )
dG ( )

  G ( ) = 0.
2
d
d
where

and

(73)
are arbitrary constants. Using the
general solutions of Eq.
(73) , we have


 2  4
 2  4 
 2
 )  c2 sinh(
)  
 c1 cosh(
2
2
   4 
  ,  2  4  > 0,

2

 2  4
 2  4  2
 )  c2 cosh(
) 

 c1 sinh(
2
2




2
2


4
4







 )  c2 sinh(
)  
 c cosh(
G'  4    2  1
2
2
  ,  2  4  < 0,
=
2
G 

4   2
4   2  2
sinh(
)
cosh(
)

c
c




1
2

2
2



c1
 2

 ,   4 = 0

c1  c2 2




where c1 and c2 are arbitrary constants.
The objective of this section is to outline the use
P(q, qx , qt , qxx qxt , qtt ,) = 0
l
determined later and G ( ) satisfies a second order
linear ordinary differential equation (LODE):
which forces the constraint condition given by
b  v2  k 2   v  v2  > 0
G' /G as follows:
 G' 
Q( ) = a   , am  0
l
l =0
G
m
where al , l = 0,1, , m
and
B=
We
assume that the solution of Eq. (71) can be
expressed by a polynomial in
where the amplitude A and width B of the
Gausson are given by
.
To determine
four steps:
q explicitly, we take the following
Step 1. Determine the integer m by substituting
Eq. (72) along with Eq. (73) into Eq. (71) , and
balance the highest order nonlinear term(s) and the
highest order partial derivative.
Step 2. Substitute Eq. (72) with the value of m
determined in Step 1, along with Eq.
(73) into Eq.
(71) and collect all terms with the same order of
G' /G together, the left-hand side of Eq. (71) is
'
converted into a polynomial in G /G . Then set
each coefficient of this polynomial to zero to derive
a set of algebraic equations for v , B ,  ,  and
al for l = 0,1, , m.
437
IJST (2012) A4: 1-12
Step 3. Solve the system of algebraic equations
obtained in Step 2, for v , B ,  ,  and al , for
l = 0,1, , m by using Maple.
Step 4. Use the results obtained in the above
mentioned steps to derive a series of fundamental
solutions Q ( ) of Eq. (71) depending on G' /G ,
since the solutions of Eq. (73) are well known to
us, then we can obtain the exact solutions of Eq.
(69) .
6 B 2 y
b
a2 =
 2 (a   )   2 2 z
=
4B2 z
In this section, we will demonstrate the G /G expansion method on several forms of the perturbed
KGE.
4.2.1. Form-I
z =  2 k 2     2   2 , y = k 2  v 2  v  v 2.
When  =   4  =
qtt  k 2 qxx  aq  bq 2 =  q   qt   qx
 qxt   qtt
(74)
2
> 0 we obtain
6 B (  v   )
,
5b
(75)
and
(83)
15B 2  2   2 (5(  a )  6 B (  v   ))
,
10  2b
f ( ) =
''
(82)
6 B 2 y
d2 =
,
b
(70) is
(v  k   v   v ) B Q  (a   )Q
bQ 2  (  v   ) BQ' = 0,
2
B2 z
where
d0 =
2
 2 (a   )
q( x, t ) = Q( ) = d 0  d1 f ( )  d 2 f ( ) 2 , (81)
d1 =
In this case, the perturbed KGE that will be
studied is given by
2
(80)
hyperbolic function solutions:
'
which by the wave transformation
converted to
.
where
2
4.2. Applications to Perturbed KGE
(79)
 (c1 sinh( )  c2 cosh( ))
,
(c1 cosh( )  c2 sinh( ))
(84)
(85)
 = (  1)( 2   2 v 2 )   2v   ,
where k , a , b ,  ,  ,  ,  and  are
constants. According to step 1, we get
m  2 = 2m, hence m = 2 .
We then assume that Eq. (21) has the following
formal solutions:
1  2 (a   )

and  = x  vt , v = .
2

2
B z
2
 (a   )
2
< 0 we obtain
When  =   4  =
B2 y
 G 
 G 
Q = a2    a1    a0 ,
G
G
trigonometric function solutions:
2
(76)
where a0 , a1 , and a2 are constants which are
unknown to be determined by solving the set of
algebraic equations obtained by step 2 by use of
Maple, we get the following results:
a0 =
1
(3 2 B 2 z   2 (a   ))
2
2b
6
a1 =
B(5 By   v   )
5b
(77)
=
q( x, t ) = Q( ) = d 0  d1 f ( )  d 2 f ( ) 2 , (86)
where
f ( ) =
When
 (c1 sin( )  c2 cos( ))
.
(c1 cos( )  c2 sin( ))
 =  2  4 =
 2 (a   )
B2 y
= 0,
(87)
the
rational function solutions obtained as:
(78)
q( x, t ) = Q ( ) = d 0  d1 f ( )  d 2 f ( ) 2 ,
where
(88)
IJST (2012) A4: 1-12
438
c2
f ( ) =
.
c1  c2
(89)

in (86) and (88) have the same
and d 0 , d1 , d 2 ,
values as Eq. (82), (83), (84) and
 , c1 , c2 , B
are
arbitrary constants.
4.2.2. Form-II
In this case, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq 3 =  q   qt   qx
 qxt   qtt
(90)
where by the wave transformation
converted to
(70) is
yB 2Q''  (a   )Q  bQ 3  (  v   ) BQ' = 0. (91)
''
Balancing Q
and Q
3
gives m  2 = 3m,
hence m = 1 . Then assuming
Q = a1
G
 a0
G
(92)
2by (3By  v   )
where a0 = 
6by
 2 (a   )
2 zB
2
,  = x

t.

4.2.3. Form-III
In this case, the perturbed KGE to be studied is
given by
qtt  k 2 qxx  aq  bq n =  q   qt   qx
 qxt   qtt
(96)
and, by the wave transformation (94) is reduced
to
yB 2Q''  ( a   )Q  bQ n  (  v   ) BQ' = 0, (97)
(93)
where y = v  k   v   v .
2
(94)
Here, y = v  k  v  v
(95)

z =  2  k 2  2     2.
2
2
and
Therefore,
q( x, t ) = Q( ) = 

=
2
n
B 2 2 z  2 2 (a   )
.
4 zB 2
2
and
2
Balancing Q and Q yields m  2 = nm and
2byB
a1 = 
b
=
c1 sinh( )  c2 cosh( )

,

c1 cosh( )  c2 sinh( )


2  2 (a   ) zB 2 > 0
2


4

=

,


c1 sin( )  c2 cos( )
,
f ( ) = 
c1 cos( )  c2 sin( )


2  2 (a   )
 2  4 =
< 0,

zB 2

 c1
2  2 (a   )
,  2  4 =
= 0,

zB 2
 c1  c2
2byB
f ( )
b
''
2
2
n 1
hence m =
. We use Q = W
, so we have
n 1
(n  1) 2 bW 4  (a   )W 2 
2 y (n  1) B 2W ''W 2 y (n  3) B 2 W '  = 0. (98)
2
According to step 1, we get
2m  2 = 4m,
hence m = 1 .We therefore assume that
2by (v   )
6by
 G 
W = a1    a0 ,
G
where a0 = 
a1 = 
2by (3By  v   )
6by
b( a   )
b
(99)
(100)
(101)
439
v=
IJST (2012) A4: 1-12
   2  4k 2  4k 2
,  = 0,
2(  1)
Q = [
 = 0.
(102)
In this case, the perturbed KGE that will be
studied is given by
2by (3By  v   )
6by
qtt  k 2 qxx  aq  bq 2 =  q   qt   qx
 qxt   qtt
(104)
2
b( a   )  
 n 1

  f ( ) ] and

b
 2

 c1 sinh( )  c2 cosh( )
2
 c cosh( )  c sinh( ) ,   4  > 0,
2
 1


 c sin( )  c2 cos( )
,  2  4  < 0,
f ( ) =  1
cos(

)
sin(

)
c

c
2
 1


c1

,
 2  4 = 0,

c1  c2
with
=
 2  4
2
,
The exp-function method is based on the
assumption that traveling wave solutions of Eq.
(21) can be expressed in the following form:
d
a e 
n
m=  p
 k 2   v   v 2  B 2Q''  (a   )Q
bQ 2  (  v   ) BQ' = 0,
(105)
where k , a , b ,  ,  ,  and  are arbitrary
constants. Using the ansatz (101), for the linear
2
term of highest order Q by simple calculation, we
have
c1e( c 3 p )  
c2 e 4 p 
(106)
c3e(2 c  2 p )  
,
c4 e 4 p  
(107)
where ci are determined coefficients only for
simplicity. Balancing highest order of exp-function
2
''
in Q and Q , we have
2c  2 p = c  3 p,
5.1. Description of the method
n
b e
2
Q2 =
This section will integrate the perturbed KGE by
the exponential function technique that is
abbreviated as exp-function approach. The details
are in the following subsection below.
,
v
and
5. Exp-Function Method
Q( ) =
which by the wave transformation, (102) is
converted to
Q'' =
   2  4k 2  4k 2
 = x
t.
2(  1)
n=c
q
5.1.1. Form-I
(103)
which leads to the result
p = c.
(109)
Similarly, to determine values of d and q , we
balance the linear term of lowest order in Eq. (105)
m
m
Q'' =
d1e  (3q  d )  
d 2 e 4 q  
where c , d , p , and q are positive integers
which are unknown to be further determined, an
and bm are unknown constants. To determine the
values of c and p , we balance the linear term of
highest order in Eq. (21) with the highest order
nonlinear term. Similarly, to determine the values
of d and q , we balance the linear term of lowest
order in Eq. (21) with the lowest order nonlinear
term.
(108)
and
Q2 =
  d3e  (2 d  2 q )
,
  d 4 e 4 q
(110)
(111)
where di are determined coefficients only for
simplicity. Balancing lowest order of exp-function
in Eqs. (110) and (111), we have
(3q  d ) = (2d  2q ),
(112)
IJST (2012) A4: 1-12
440
which leads to the result
d = q.
(113)
Since the final solution does not strongly depend
upon the choice of values of c and d , for
simplicity, we set p = c = 1 and d = q = 1 , the
trial function, Eq. (101) becomes
Q( ) =
a1e   a0  a1e
.
b1e   b0  b1e
where in the second, third and fourth set of
solutions h is the root of
(1   ) Z 2   Z  k 2 = 0
(120)
b0 , b1 are free parameters with b  0 and
   h  0 conditions.
and
Set 3.
(114)
a1 = a0 = b1 = 0, a1 =
Substituting Eq. (114) into Eq. (105), and
equating to zero the coefficients of all powers of
e n yields a set of algebraic equations for a1 , a0 ,
a1 , b1 , b0 , v , B and b1 . Solving the system of
algebraic equations with the aid of Maple, we
obtain the following several sets of solutions.
b1 (a   )
, v = h, (121)
b
so that
q( x, t ) = Q( ) =
b1 (a   )e 
,
b(b1e  b0 )
(122)
where
b0 , b1 are free parameters and b  0 .
Set 4.
a1 = a1 = b1 = 0, v = h,
a0 =
b0 (a   )
.
b
Set 1.
3b (a   )
a1 = 0, a1 = 0, a0 = 0
,
b
b2

b1 = 0 , v = , B = h ,

4b1
(115)
So
so
q ( x, t ) = Q( ) =
3b0 (a   )
,
1 b02 e 

b(b1e  b0 
)
4 b1

t

(116)
(124)
b0 , b1 are free parameters and b  0 .
5.1.2. Form-II
In this case, the perturbed KGE that will be
studied is given by
   k 2  2   2  Z 2  a   = 0. (117)
qtt  k 2 qxx  aq  bq 3 =  q   qt   qx
 qxt   qtt   qxxt   qxxxx
(125)
 , b1  0 ,  = x 
root of
2
where
b0 (a   )
,
b(b1e  b0 )
and h are the
where b ,
 
q( x, t ) = Q( ) =
(123)
(70) is
and
b0 , b1 are free parameters, Here and the
following cases a , b , k ,  ,  ,  ,  are free
which by the wave transformation
converted to
parameters.
yB 2Q''  (a   )Q  bQ 3  (  v   ) BQ'
v B 3Q'''   B 4Q'''' = 0,
(126)
a1 = a1 = b1 = 0, v = h,
b (a   )
a 
a0 = 0
,B =
,
   h
b
Set 2.
(118)
y = v 2  k 2   v  v 2
so
q( x, t ) = Q( ) =
b0 (a   )
,
b(b1e  b0 )
where
(119)
and k , a , b ,  ,  ,  ,  ,  and  are all
arbitrary constants. Using the ansatz (101) and
441
IJST (2012) A4: 1-12
balancing highest order of exp-function in
Q 3 and
''
2
p = c.
(127)
Similarly, balancing the linear term of lowest
3
Q'' leads to the result which leads
a1e  a0  a1e
.
b1e  b0  b1e
(129)
Substituting Eq. (129) into Eq. (126), and
equating to zero the coefficients of all powers of
en yields a set of algebraic equations for a1 , a0 ,
a1 , b1 , b0 , and b1 . Solving the system of
algebraic equations with the aid of Maple, we
obtain the following several sets of solutions.
Set 1.
a1 = a1 , b0 = b0 , v =
h
.
B
q( x, t ) = Q( ) =
b

a0  b1e 
a 
a1 = a1 = b0 = 0, b1 =
q( x, t ) = Q( ) =
where b ,
a1e
,
b0
where
2
(138)
(139)
  2     2  Z 2  a   = 0,
(140)
and
B=
a1 , b0 are free parameters.
a0
. (141)
ba02


b1e 
e
8b1 (a   )
a1 = a1 = b1 =  = c = 0, v = h,
b
a0
a 
(142)
3(a   )
,
2(  h   )
(143)
b0 = 
2
ba02
,  = 0,  = 0
8b1 ( a   )
of
Set 4.
(  1) Z  ( B   B   ) Z  a
 B 2 k 2    B  B 4 = 0, (132)
2
(137)
a0 , b1 are free parameters and h is the root
(131)
b0  0 and h is a root of
.

, B = h ,

v=
q ( x, t ) = Q( ) =
So
where
a0
Set 3.
2
(130)
(136)
So,
k 
a1 = a0 = 0, b1 = b1 = 0,

b
 0,  h    0.
a 
(128)
Since the final solution does not strongly depend
upon the choice of values of c and d , for
simplicity, we set p = c = 1 , d = q = 1 , the trial
function, Eq. (101) becomes
2
(4  9 (  a)) Z  2 2
9k 2 (a   ) = 0 (135)
and
d = q.
Q( ) =
b1 , a0  0 and h is the root of
9(  1)(a   )  2  Z
Q leads to the result
order in Q and
to the result
where
where
Set 2.
a1 = a1 = b1 =  = c = 0, v = h,
b0 = 
b
3(a   )
a0 , B =
,
a 
2(  h   )
(133)
(134)
h = 9(  1)(a   )  2 2  Z 2 
(4  9 (  a )) Z  2 2  9k 2 (a   )
and
a0 , b0 , b1 are free parameters. So
(144)
IJST (2012) A4: 1-12
442
q ( x, t ) = Q( ) =
a0
b
a0  b1e 

a 
. (145)
W 4 leads to the result
a0b
b
(a   )
a 
,
(146)
(147)
(1   ) Z  ( B   ) Z
 B  k 2 B 2  2  2a = 0
2
(148)
and a0 , a1 are free parameters and
a0  a1e 
ba0
order in WW  and W
leads to the result
4
leads to the result which
(156)
Since the final solution does not strongly depend
upon the choice of values of c and d , for
simplicity we set p = c = 1 , d = q = 1 , the trial
function, Eq. (101) becomes
Q( ) =
q( x, t ) = Q( )
b
a1e  
a 
(155)
d = q.
Here h is the root of

p = c.
Similarly, balancing the linear term of lowest
b
h
b1 = 
a1 , v = .
a 
B
=
(154)
and k , a, b,  ,  ,  ,  and  are arbitrary
constants. Using the ansatz (101) and balancing
highest order of exp-function in WW  and
Set 5. a1 = b1 =  = c = 0,
b0 =
y = k 2   v  v 2  v 2 .
.
(149)
a1e   a0  a1e
.
b1e   b0  b1e
(157)
Substituting Eq. (157) into Eq. (152), and
equating to zero the coefficients of all powers of
en yields a set of algebraic equations for a1 , a0 ,
a1 , b1 , b0 , and b1. Solving the system of
b
(a   )
a 
algebraic equations with the aid of Maple, we
obtain the following several sets of solutions.
5.1.3. Form-III
In this case, the perturbed KGE that will be
studied is given by
Set 1. a1 = a1 = b0 = 0,
qtt  k 2 qxx  aq  bq n =  q   qt   qx
 qxt   qtt
(150)
ba02
b1 =
2(n  1)(a   )b1
(158)
and by q  Q ( ),  = B ( x  vt )
v=

, B = h(n  1)  .

(159)
(151)
is reduced to
v
2
 k   v  v  B Q  aQ  bQ
2
2
2
''
= Q  v  BQ   BQ .
'
By q = W
2
n 1
'
Here h is root of
n
(152)
, this is converted to
2 y (n  1) B 2W W  2 y (3  n) B 2W 2
2(   v)(n  1) BWW   bW 4 (n  1) 2
W 2 (n  1) 2 (a   ) = 0, (153)
here
 4
2
 4  4 2  4k 2  2  Z 2
  a = 0.
(160)
2
So, q ( x, t ) = W n 1
443
IJST (2012) A4: 1-12




a0


=

2
ba0
 
 b e 
e
 1

2(n  1)(a   )b1
2
n 1
and a0 and a1 are free parameters.
(161)
In this case, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq n  cq 2 n 1 =  q 
 qt   qx   qxt   qtt
where a0 and b1 are arbitrary constants.
Set 2. a1 = b1 = 0, v =
a 

, a0 = 
b0 ,
b

 a 
(a   )b 1  
b0 
e 

b
a




b


b
=


b0  b 1e








b1 = 
h
, b0 =
B
1
2
n 1
(n  1) 2 (a   )W 2  bW 3  cW 4 
 B 2W 2 (n  2)  BW W ' (  v )
(n  1)  y (n  1)B 2W W '' = 0,
''
4
in WW and W leads to the result
p = c.
(169)
Similarly, balancing the linear term of lowest
''
order in WW and
leads to the result
ba0
b
(a   )
a 
,
(164)









(170)
Since the final solution does not strongly depend
upon the choice of values of c and d , so for
simplicity, we set p = c = 1 , d = q = 1 , the trial
function, Eq. (101) becomes
Q( ) =
2
n 1
W 4 leads to the result which
d = q.
and



a0  a1e 

=
ba0
b

a1e 


a 
b
 (a   )
a 

(168)
here y = v 2  k 2   v   v 2 and k , a, b,  ,  ,  , c, 
and  are arbitrary constants. Using the ansatz
(101) and balancing highest order of exp-function
(163)
b
a1 ,
a 
q ( x, t ) = W
(167)
which, by q = W n 1 is reduced to
2
 (a   )b1
n 1
a1 =
, (162) and q ( x, t ) = W
a 
b
b
Set 3. a1 = b1 = 0, v =
5.1.4. Form-IV
a1e   a0  a1e
b1e   b0  b1e
(171)
Substituting Eq. (171) into Eq. (168), and
equating to zero the coefficients of all powers of
exp(n ) yields a set of algebraic equations for
2
n 1
a1 , a0 , a1 , b1 , b0 , and b1. Solving the system of
(165)
algebraic equations with the aid of Maple, we
obtain the following several sets of solutions.
Set 1.
where h is the root of
a1 = a1 = 0, v =
(  1) Z 2  ( B   ) Z  (  a )(n  1)
 k 2 B 2   B = 0,
(166)
a0 =

,

(a   )(n  1)b0
,
b
(172)
IJST (2012) A4: 1-12
b1 = 
444
 W '   4n3 (  v   ) BW 3W '
(c(n  1) 2 (a   )  b 2 n)b02
,
4(nb 2b1 )
So q ( x, t ) = W
2
=

(a   )(n  1)b0


b

2



(
c
(
n
1)
(
a

)
b 2 n)b02 

e  b0  b1e
2

4(nb b1 )
12 n 2 (n  4) B 4 W ''  W 2 = 0.
2
a 
.
   2    k 2  2
B = (n  1)  
1
n 1
2






1
n 1
Here k , a , b ,  ,  ,  , c ,  ,  and 
are arbitrary constants. Using the ansatz (101) and
balancing highest order of exp-function in WW 
and
(173)
W 4 leads to the result
p = c.
(180)
Similarly, balancing the linear term of lowest
''
order in WW and W
leads to the result
where b0 and b1 are free parameters.
4
leads to the result, which
d = q.
5.1.5. Form-V
For this form, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq1 n  c1q n 1 =  q   qt (174)
 qx   qxt   qtt   qxxt   qxxxx
(175)
q ( x, t ) = Q( ),  = B( x  vt )
(176)
by
Since the final solution does not strongly depend
upon the choice of values of c and d , for
simplicity, we set p = c = 1 , d = q = 1, the trial
function, Eq. (101) becomes
Q( ) =
a1e  a0  a1e
b1e  b0  b1e
B 2  v 2  k 2   v   v 2  Q''  aQ
bQ1 n  cQ n 1   Q  B(  v   )Q'
v B 3Q'''   B 4Q'''' = 0.
en yields a set of algebraic equations for a1 , a0 ,
a1 , b1 , b0 and b1 . Solving the system of
(177)
By
algebraic equations with the aid of Maple, we
obtain the following set of solutions.
a1 = a1 = b1 = b1 = 0, b0 = ha0 ,
4
n
is reduced to 12n
(178)
16 (n  4)n 2 B 4W 'W '''W 2  an 4W 4
 n 4W 4  bn 4  cn 4W 8
3
5.1.6. Form-VI
For this form, the perturbed KGE that will to be
studied is given by
4 vn3 B3W 3W ''  4 n3 B 4W ''''W 3
48 n(n  2)(n  4) B W
WW
' 2
(184)
where a0 is an arbitrary constant.
4n3  v 2  k 2   v  v 2  B 2W 3W ''
4
(183)
where h is root of
bZ 8  c 2  (  a ) Z 4 = 0,
 v(n  4) B 3W 2W W 
2
8 vn(n  2)(n  4) B 3W W ' 
(182)
Substituting Eq. (182) into Eq. (177), and
equating to zero the coefficients of all powers of
is converted to
Q =W
(181)
''
8 (n  2)(3n  4)(n  4) B 4 W ' 
(179)
4
4n 2  v 2  k 2   v   v 2  (n  4) B 2W 2
qtt  k 2 qxx  aq  bq ln q =  q   qt
 qx   qxt   qtt
(185)
with Q( x, t ) = Q( ),  = B( x  vt ) is converted to
B 2  v 2  k 2   v   v 2  Q''  (a   )Q
445
IJST (2012) A4: 1-12
bQ ln Q  (  v   ) BQ' = 0
and by
(186)
Set 2.
ln q = u it is reduced to

B 2  v 2  k 2   v   v 2  u''   u' 
2
a1 = b1 = b0 = 0, v = h, B =

(a   )  bu  B(  v   )u = 0,
'
(187)
here k , a , b ,  ,  ,  ,  and  are arbitrary
constants. Using the ansatz (101) and balancing
highest order of exp-function in WW
leads to the result
''
and W
p = c.
(a   )b1
.
b
Here a0 , b1 are arbitrary constants and h is the
roots of
(1   ) Z 2  Z   k 2 = 0. (195)
(188)
''
order in u and
u 
' 2
leads to the result, which
leads to the result
d = q.
(189)
Since the final solution does not strongly depend
upon the choice of values of c and d , for
simplicity, we set p = c = 1 , d = q = 1 , the trial
function, Eq. (101) becomes
a1e  a0  a1e
.
b1e  b0  b1e
(190)
Substituting Eq. (190) into Eq. (186), and
equating to zero the coefficients of all powers of
en yields a set of algebraic equations for a1 , a0 ,
a1 , b1 , b0 and b1 . Solving the system of
algebraic equations with the aid of Maple, we
obtain the following set of solutions.
Set 1.
(a   )b0
,
a1 = b1 = b1 = 0, v = h, a0 =
b
b
B=
.
(191)
h 
Here h is the root of
(192)
and a1 and b0 are arbitrary constants. So
 ( a  ) b0
b
q ( x, t ) = Q( ) = e
a0 
So q ( x, t ) = Q ( ) = eu ( ) = e
where b  0 and
 ( a  ) b1 
e
b


b1e
, (196)
h  .
6. Mapping Methods
Now, we solve eq. (19) by a mapping method and a
modified mapping method which generates a
variety of periodic wave solutions (PWSs) in terms
of squared Jacobi elliptic functions (JEFs) and we
subsequently derive their infinite period
counterparts in terms of hyperbolic functions which
are solitary wave solutions (SWSs), shock wave
solutions or singular solutions. For solving by these
methods, we set in eq. (21)  = = = =0 . Thus
eq. (21) reduces to
 v k
2
2
 v  v 2  g ''  F ( g )  g =0.
(197)
6.1. Form-I
In this case, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq 2 =  q   qxt   qtt . (198)
Using the travelling wave hypotheses (20) and
(22), eq. (198) reduces to
Ag ''  Bg C g 2 =0,
(199)
where
(1   ) Z 2  Z   k 2 = 0
u ( )
(194)
4
Similarly, balancing the linear term of lowest
Q( ) =
a1 =
b
,
h 
=e
(200)
6.1.1. Mapping Method
 a1e
b0
A=v 2  k 2  v  v 2 , B = a  , C = b.
. (193)
Here, we assume that eq. (199) has a solution in
the form
IJST (2012) A4: 1-12
446
g = A0  A1 f ,
(201)

where
' 2
2
2
 R f 3.
3
(202)
f ( s )=cn 2 ( s ) . Thus we obtain the PWS of eq.
(199) as
Eq. (201) is the mapping relation between the
solution to eq. (199) and that of eq. (202). We
substitute eq. (201) into eq. (199), use eq. (202) and
equate the coefficients of like powers of f to zero
so that we arrive at the set of equations
AA1 R C A12 =0,
(203)
AA1 Q  B A1  2C A0 A1 =0,
(204)
AA1 P  B A0  2C A02 =0.
(205)
From eqs. (203) and (204), we obtain A1 = 
AR
C
AQ  B
. Eq. (205) gives rise to the
and A0 = 
2C
a    4(2m 2  1)(v 2  k 2   v   v 2 )
g (s) =
2b
2
2
2
2
6m (v  k   v   v ) 2
cn ( s ).

(210)
b
When
(209).
A2 (Q 2  4 PR)= B 2 .
P =  2(1 m 2 ), Q =4(2m 2 ), R = 6.
2
Here, eq. (202) has the solution f ( s )=dn ( s ) .
So, we obtain the PWS of eq. (199) as
2
2
2
2
g ( s) = a    4(2  m )(v  k   v   v )
2b
6(v  k   v   v ) 2
dn ( s ).
b
2
2
(206)
When
P =2, Q =  4(1 m ), R =6m . Now eq.
2
(202) has two solutions f ( s )=sn ( s ) and
f ( s )=cd 2 ( s ) . So, we obtain the PWSs of eq.
Case 1.
2
(211)
m1 , the SWS (209) is retained.
2
(199) as
a    4(1  m 2 )(v 2  k 2   v   v 2 )
g ( s )=
2b
2
2
2
6m (v  k   v   v 2 ) 2

(207)
sn ( s),
b
and
a    4(1  m 2 )(v 2  k 2   v   v 2 )
2b
2
2
2
6m (v  k   v   v 2 ) 2
cd ( s ).

(208)
b
g ( s )=
When
m1 , the eq. (210) gives rise to the SWS
Case3.

constraint relation
2
(209)
Case 2. P =2(1 m 2 ), Q =  4(2m 2 1), R = 6m 2 .
In this case, eq. (202) has the solution
f '' = P Q f  R f 2 ,
 f  =2 P f Q f
6(v 2  k 2  v  v 2 )
tanh 2 (s ).
b
m1 , the eq. (207) gives rise to the SWS
a    8(v 2  k 2   v   v 2 )
g ( s )=
2b
P =2m 2 , Q =  4(1 m 2 ), R =6. Now, eq.
2
(202) has two solutions f ( s )=ns ( s ) and
f ( s )=dc 2 ( s ) .In this case, we obtain the two
Case 4.
PWSs of eq. (199) as
g ( s) =

a    4(1  m 2 )(v 2  k 2   v   v 2 )
2b
6(v 2  k 2   v   v 2 ) 2
ns ( s), (212) and
b
2
2
2
2
g ( s) = a    4(1  m )(v  k   v   v )
2b

6(v  k   v   v ) 2
dc ( s ).
b
2
2
2
(213)
When m1 , eq. (212) leads us to the singular
solution
g ( s )=
a    8(v 2  k 2   v   v 2 )
2b
447

IJST (2012) A4: 1-12
6(v 2  k 2   v   v 2 )
coth 2 ( s ).
b
(214)
P =  2m 2 , Q =4(2m 2 1), R =6(1 m 2 ).
2
So, eq. (202) has the solution f ( s )=nc ( s ) .
Case5.
Thus we obtain the PWS of eq. (199) as
a    4(2m  1)(v  k   v   v )
2b
2
2
2
6(1  m )(v  k   v   v 2 ) 2
nc ( s ). (215)

b
2
2
2
2
g (s) =
P =  2, Q =4(2 m 2 ), R = 6(1m 2 ).
2
Here, eq. (202) has the solution f ( s )=nd ( s ) .
Case 6.
So, we obtain the PWS of eq. (199) as
a    4(2  m )(v  k   v   v )
2b
2
2
2
6(1  m )(v  k   v   v 2 ) 2
nd ( s ). (216)

b
2
2
2
2
g ( s )=
P =2, Q =4(2 m 2 ), R =6(1 m 2 ). Here,
2
eq. (202) has the solution f ( s )=sc ( s ) . Thus the
Case 7.

6(v 2  k 2   v   v 2 ) 2
cs ( s ).
b
(219)
When m1, the PWS (219) will give rise to
the singular solution (214).
P =  2m 2 (1 m 2 ), Q =4(2m 2 1), R =6.
2
Thus eq. (202) has the solution f ( s )=ds ( s ) . So,
Case10.
the PWS of eq. (199) is
a    4(2m 2  1)(v 2  k 2   v   v 2 )
2b
2
2
2
6(v  k   v   v ) 2

(220)
ds ( s).
b
g (s) =
When m1, the PWS (220) leads to the same
singular solution (214).
It is evident from the constraint relation (206) that
Q 2  4 PR should always be positive with our
choices of P, Q and R for real solutions to exist.
In all the cases considered, we can see that
Q 2  4 PR is equal to 16m 4 16m 2 16 which is
always positive for 0 m1. Thus all our solutions
are valid with the constraint relation (206).
PWS of eq. (199) is
2
2
2
2
g ( s) = a    4(2  m )(v  k   v   v )
2b

6(1  m )(v  k   v   v ) 2
sc ( s). (217)
b
2
2
2
2
Case8. P =2, Q =4(2m 1), R = 6m
In this case, eq. (202) has the
2
2
(1 m 2 ).
solution
2
f ( s)=sd ( s ) . Thus the PWS of eq. (199) is
a    4(2m 2  1)(v 2  k 2   v   v 2 )
2b
2
2
2
2
6m (1  m )(v  k   v   v 2 ) 2
g (s) =

b
sd ( s ) (218)
P =2(1m 2 ), Q =4(2 m 2 ), R =6. Here,
2
eq. (202) has the solution f ( s )=cs ( s ) . Thus the
Case 9.
PWS of eq. (199) is
g ( s)=
a    4(2  m 2 )(v 2  k 2   v   v 2 )
2b
6.1.2. Modified Mapping Method
In this case, we assume that eq. (199) has a
solution in the form
g = A0  A1 f  B1 f 1 ,
(221)
where f satisfies eq. (202). Eq. (221) is the
mapping relation between the solution to eq. (202)
and that of eq. (199).
We substitute eq. (221) into eq. (199), use eq.
(202) and equate the coefficients of like powers of
f to zero so that we will obtain a set of equations
giving rise to the solutions
A0 = 
AQ  B
AR
3P A
, A1 = 
, B1 = 
, (222)
2C
C
C
and the constraint relation
A2 (Q 2 16 PR)= B 2 .
(223)
Case 1. P =2, Q =  4(1 m ), R =6m
Here, eq. (202) has two solutions
2
2
.
IJST (2012) A4: 1-12
448
In this case, the perturbed KGE that will be
studied is given by
f ( s)=sn 2 ( s) and f ( s )=cd 2 ( s ) .
The PWSs of eq. (199) are
qtt  k 2 qxx  aq  bq3 =  q   qxt   qtt  qxxxx .
a   (v  k   v   v )

2b
2b
2
2
2
2
2
[4(1  m )  12m sn ( s ) 12m ns ( s )], (224)
2
2
g (s) =
and
Using the travelling wave hypotheses (20) and
(22), eq. (230) reduces to
Ag ''  Bg C g 3 =0,
(231)
Where,
a   (v  k   v   v )

(225)
2b
2b
  4(1  m 2 )  12m 2 cd 2 ( s )  12m 2 dc2 ( s )  .
2
2
2
g (s) =
When m1 , eq. (224) degenerates to the
solution
a   (v 2  k 2   v   v 2 )
g (s) =

2b
2b
2
 8  12 tanh (s )  coth 2 (s ) .
2
(226)
a   (v 2  k 2   v   v 2 )
g (s) =

2b
2b
[4(2  m )  12dn ( s ) 12(1  m )nd ( s )]. (227)
2
(232)
6.2.1. Mapping Method
Here, we assume that eq. (231) has a solution in
the form
g = A0  A1 f ,
(233)
1
f '' = P f Q f 3 , f ' 2 = P f 2  Q f 4  r.
2
(234)
2
f ( s)=dn 2 ( s) . So, the PWS of eq. (199) is
2
A= v 2  k 2  v  v 2 , B = a  , C = b.
where
Case 2. P =  2(1 m ), Q =4(2 m ), R = 6.
In this case, eq. (202) has the solution
When
(230)
2
2
2
m1 , eq. (227) leads us to the SWS
a   2(v 2  k 2   v   v 2 )
g (s) =

2b
b
2
 1  3sech ( s)  .
Eq. (233) is the mapping relation between the
solution to eq. (231) and that of eq. (234).We
substitute eq. (233) into eq. (231), use eq. (234) and
equate the coefficients of like powers of f to zero
so that we arrive at the set of equations
Q AA1 C A13 =0,
(235)
3C A0 A12 =0,
(236)
( P A B) A1 3C A02 A1 =0,
(237)
B A0 C A03 =0,
(238)
(228)
P =2, Q =4(2 m 2 ), R =6(1 m 2 ).
2
Thus eq. (202) has the solution f ( s )=sc ( s ) . So,
Case 3.
from which we obtain
A0 =0, A1 = 
QB
, P A B =0.
PC
(239)
the PWS of eq. (199) is
  4(2  m 2 )  12(1  m 2 )sc 2 ( s )  12cs 2 ( s )  . (229)
Case 1. P =  2, Q =2, R =1.
In this case, the solution of eq. (234) is
f ( s)=tanh( s ) . So, we have the shock wave
solution of eq. (231) as
When m1 , eq. (229) degenerates to the
singular solution (214).
g ( s )= 
g (s) =
a   (v 2  k 2   v   v 2 )

2b
2b
6.2. Form-II
Case 2.
a 
tanh( s ).
b
P =1, Q =  2, R =0.
(240)
449
IJST (2012) A4: 1-12
So, the solution of eq. (234) is
Thus the SWS of eq. (231) is
g ( s )= 
f ( s)=sech( s ) .
Now, we use the modified mapping method in
which we assume a solution of eq. (231) in the form
2(a   )
sech( s ).
b
(241)
P = (1 m 2 ), Q =2m 2 , R =1.
Here, the solutions of eq. (234) are f ( s )=sn( s )
and f ( s )=cd( s ) . So, the PWSs of eq. (231) are
Case 3.
g ( s )= 
6.2.2. Modified Mapping Method
2(a   )
msn( s ),
b(1 m 2 )
(242)
g = A0  A1 f  B1 f 1
(248)
where f satisfies eq. (234).
We substitute eq. (248) into eq. (231), use eq. (234)
and equate the coefficients of like powers of f to
zero to arrive at a set of equations fro which it can
be found that
A0 =0, A1 =  
QA
C
(249)
and
2(a   )
g ( s )= 
mcd( s ).
b(1 m 2 )
(243)
B1 =  
2R A
, P A B 3C A1 B1 =0.
C
(250)
Thus for real solutions of eq. (231) to exist, when
When m1 , eq. (242) reduces to the shock
wave solution (240).
Case 4. P =2 m , Q =  2, R = m 1.
In this case, the solution of eq. (234) is
f ( s)=dn( s ) . Here, the PWS of eq. (231) is
2
g ( s )= 
2
2(a   )
dn( s ).
b(2 m 2 )
(244)
When m1 , the SWS (241) is recovered from
eq. (244).
P = (1 m ), Q =2, R = m .
The solutions of eq. (234) are f ( s )=ns( s ) and
f ( s )=dc( s ) . So, the PWSs of eq. (231) are
Case 5.
g ( s )= 
2
2(a   )
ns( s ),
b(1 m 2 )
2
(245)
and
g ( s )= 
Case 1. P =  2, Q =2, R =1.
In this case, the solution of eq. (234) is
f ( s)=tanh( s ) . So, we have the solution of eq.
2(v 2  k 2  v  v 2 )
b
 tanh( s )  coth( s ) .
(231) as g ( s )=
(251)
P = (1 m 2 ), Q =2m 2 , R =1.
Here, the solutions of eq. (234) are f ( s )=sn( s )
and f ( s )=cd( s ) . Thus the PWSs of eq. (231) are
Case 2.
2(v 2  k 2  v  v 2 )
g ( s )=
b
 msn( s )  ns( s ) ,
(252)
and
2(a   )
dc( s ).
b(1 m 2 )
(246)
When m1 , eq. (245) degenerates to the
singular solution
g ( s )= 
Q and R are both positive, A and C should be
of opposite signs and when Q and R are both
negative, A and C should be of the same signs.
a 
coth( s ).
b
(247)
2(v 2 k 2  v  v 2 )
b
 mcd( s )  dc( s ) .
g (s) =
(253)
When m1 , eq. (252) gives the same solution
(251).
IJST (2012) A4: 1-12
450
P =2m 2 , Q =2, R =1m 2 .
So, the solution of eq. (234) is f ( s )=cs( s ) . Thus
Case 3.
 i , i i
are
12
determined
prolongation formulae
we have the PWS of eq. (231) as
g (s) =

12

(254)
1
2(v 2  k 2   v   v 2 )
csch( s ). (255)
b
Case 4. P =2 m , Q =  2, R = (1 m ).

2
(260)
12
i
Dt
and
Dx
are
given
by
D1 = Dt =  t  qt  q  and D2 = Dx =  x  qx  q  .
7.1.1. Form-I
In this case, the perturbed KGE that will be
studied is given by
2
Here, the solution of eq. (234) is f ( s )=dn( s ) .
So, we have the PWS of eq. (231) is
g (s) = 
 i = Di (W )   j qij ,
defined by W =    qi . Here the differential
operators
When m1 , eq. (254) leads to the singular
solution
2
the
in which W is the Lie characteristic function
  cs(s )  1 m sc(s ) .
g ( s )= 
by
 i i = Di Di (W )   j q ji i , i, j = 1, 2,
2(v 2  k 2   v   v 2 )
b
2
uniquely
2(v 2  k 2   v   v 2 )
b

  dn(s )  1m 2 nd(s ) .
qtt  k 2 qxx  aq  bq 2   qt   qx 
 qxt = 0.
(261)
The equation (261) admits the following two Lie
point symmetries
(256)
When m1 , eq. (256) gives rise to the SWS
2(v 2  k 2   v   v 2 )  sech( s ). (257)
g ( s)=  
b
7. Lie Group Analysis
In this section, we study some forms of the
perturbed KGE by the aid of Lie group analysis.
Initially, the method will be described in a succinct
manner and will subsequently be applied to solve a
couple of forms of the perturbed KGE.
X1 = t , X 2 =  x .
(262)
The group-invariant solution corresponding to the
combination of symmetries X 1  cX 2 , where c is
a constant is given by
q = h(  ),
(263)
where  = x  ct . The group-invariant solution
(263) reduces the equation (261), choosing
 =  c , to the following nonlinear second-order
ordinary differential equation (ODE)
h  Ah  Bh 2 = 0,
(264)
where
A := a/(c 2  k 2   c), B := b/(c 2  k 2   c)
7.1. Description of the method
The
partial
differential
equation
admits the
E (t , x, q, qt , qx , qtt , qtx , qxx ) = 0
symmetry generator X in the form of prolongation
given by
X =  1 (t , x, q ) t   2 (t , x, q ) x   (t , x, q ) q
 i  q   i i  q ,
(258)
if XE |E =0 = 0,
(259)
i
12
i1i2
where the summation convention is used whenever
appropriate. In (258), the additional coefficients
and `prime' denotes differentiation with respect to
 . Integrating the equation (264) with respect to
 again and letting the constant of integration be
zero we obtain the following nonlinear first-order
ODE
h2 =  Ah 2 
2B 3
h.
3
(265)
The equation (265) has the following solution
after substituting the values of A and B
451
h(  ) =
IJST (2012) A4: 1-12


3a
a
  , (266)
sech 2  d 
2
2
2b
4(k  c   c) 

where d is a constant of integration. Hence we
obtain the following solitary wave exact groupinvariant solutions for the equation (261), in which
 =  c , given by
Further integration of the equation (253) and then
substituting the values of A , B and C we obtain
the following solution for the equation (272)
D
h(  ) =
[ E  cosh(e  F  )]
1
n 1
,
(273)
where e is a constant of integration,
3a
a
q( x, t ) = sech 2 (d 
2
2b
4(k  c 2   c)
 ( x  ct )).
(267)
7.1.2. Form-IV
1

 n 1
an(n  1)
 ,
D=
 n[nb 2  ad (n  1) 2 ] 


E=
For this form, the perturbed KGE that will be
studied is given by
qtt  k 2 qxx  aq  bq n  dq 2 n 1
 qxt   qtt = 0, n > 1.
The
(268)
(269)
X 1  cX 2 -group-invariant
solution
corresponding to the combination of symmetries
X 1 and X 2 , where c is a constant is given by
q = h(  ),
(270)
where  = x  ct .
The reduced nonlinear second-order ODE
resulting from substituting the group-invariant
solution (270) into the equation (268) is given by
h  Ah  Bh n  Ch 2 n 1 = 0,
(271)
where
A := a/( c 2  c 2  k 2   c),
B := b/( c 2  c 2  k 2   c),
2
2
and `prime' denotes differentiation with respect to
 . Integrating the equation (271) with respect to
 and letting the constant of integration be zero
yields the following nonlinear first-order ODE
h2 = Ah 2 
2 B n 1 C 2 n
h  h .
(n  1)
n
F = (n  1)
,
(275)
a
.
(c  c  k 2   c )
2
(272)
(276)
2
Thus we get the following solitary travelling
wave exact group-invariant solutions for the
equation (268) given by
D
q ( x, t ) =
[ E  cosh(e  F ( x  ct )]
1
n 1
,
(277)
where the constants D, E and F are given by the
equations (274), (275) and (276), respectively.
8. Conclusions
This paper studied the KGE with six forms of
nonlinearity including the log law nonlinearity. The
perturbation terms are taken from the theory of long
Josephson junctions, modeled by the sine-Gordon
equation and its type, and we were justified in
studying them in the context of KGE. Thus, the
perturbed KGE was integrated in the presence of
these strong perturbation terms by the aid of several
integration tools. In particular, the traveling wave
'
C := d/( c  c  k   c)
2
n[nb  ad (n  1) 2 ]
and
Again the equation (268) admits the following
two Lie point symmetries
X1 = t , X 2 =  x .
bn
2
(274)
solutions were obtained, G /G method approach
was used, exp-function method was carried out,
mapping method and its versions were also applied
and finally the Lie symmetry approach was also
utilized to extract several forms of solution to the
perturbed KGE.
In addition to soliton solutions, cnoidal waves,
snoidal waves, other PWS and rational solutions
were obtained. The limiting cases of these PWS are
also given. These solutions are going to be
IJST (2012) A4: 1-12
extremely useful in the context of relativistic
quantum mechanics where KGE arises.
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