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Jan 2014. Vol. 4, No. 7 ISSN2305-8269

International Journal of Engineering and Applied Science © 2012 - 2014 EAAS & ARF. All rights reserved

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18

EXACT TRAVELING WAVE SOLUTIONS FOR NONLINEAR

FRACTIONAL PARTIAL DIFFERENTIAL EQUATIONS USING THE

IMPROVED (G’/G) – EXPANSION METHOD

Elsayed M. E. Zayed1, Yasser. A. Amer and Reham M. A. Shohib

Mathematics Department, Faculty of Sciences, Zagazig University, Zagazig, Egypt

1Email of corresponding author: [email protected]

ABSTRACT

In this article, we apply the improved (G’/G)-expansion method to construct the exact solutions of the nonlinear

fractional partial differential equations (PDEs) in the sense of the modified Riemann-Liouville derivative. Based on

a nonlinear fractional complex transformation, certain fractional partial differential equations can be turned into

other nonlinear ordinary differential equations (ODEs) of integer orders. For illustrating the validity of this method,

we apply it to four nonlinear fractional PDEs equations namely, the space-time fractional Potential Kadomtsev-

Petviashvili (PKP) equation, the space-time fractional Symmetric Regularized Long wave (SRLW) equation, the

space-time fractional Sharma- Tasso-Olver (STO) equation and the space-time fractional Kolmogorov-Petrovskii-

Piskunov (KPP) equation. This method can be applied to many other nonlinear fractional PDEs in mathematical

physics.

Keywords: Nonlinear space-time fractional PDES, Improved ( / )G G - expansion method, Nonlinear

fractional complex transformation, Exact solutions; Modified Riemann- Liouville derivative.

INTRODCUTION

In recent years, nonlinear fractional differential

equations have been attracted great interest. It is

caused by both the development of the theory of

fractional calculus itself and by the applications of

such constructions in various sciences such as

physics, engineering and biology [11,15,20,

22,25,27,28]. For better understanding the

mechanisms of the complicated nonlinear physical

phenomena as well as further applying them in

practical life, the solutions of these equation

[2,16,19,26,30,31] are much involved. In the past,

many analytical and numerical method have been

proposed to obtain the solutions of nonlinear

fractional differential equations, such as the finite

difference method [17], the finite element method

[7], the differential transform method [3,21], the

Adomian decomposition method [4,5,12,23], the

variational iteration method [13,24,29], the homotopy

perturbation method [ 8], the improved ( / )G G -

expansion method [6,9,32], the fractional sub-

equation method [1,10,33,34] and so on.

In this article, we will apply the improved ( / )G G -

expansion method [26-28] for solving the nonlinear

fractional PDEs in the sense of the modified

Riemann-Liouville derivative obtained in [14,18].

The modified Riemann-Liouville derivative of order

is defined by the following expression:

0

( )( )

1( ) [ ( ) (0)] , 0 1,

(1 )( )

( ) , 1, 1

t

t

nn

dt f f d

dtD f t

f t n n n

(1)

We list some important properties for the modified

Riemann-Liouville derivative as follows:

mailto:[email protected]

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



19

(1 )

, 0(1 )

r r

t

rD t t r

r

(2)

( ) ( ) ( ) ( ) ( ) ( )t t t

D f t g t f t D g t g t D f t

(3)

( ( )) ( ( )) ( )

( ( ))[ ( )]

t g t

g

D f g t f g t D g t

D f g t g t

(4)

The rest of this article is organized as follows: In

Sec. 2, the improved ( / )G G -expansion method for

solving nonlinear fractional PDEs is given. In Sec.

3, we apply this method to establish the exact

solutions for the space-time nonlinear fractional PKP

equation, the space-time nonlinear fractional SRLW

equation, the space-time nonlinear fractional STO

equation and the space-time nonlinear fractional

KPP equation. In Sec. 4, some conclusions and

discussions are obtained.

DESCRIPTION PF THE IMPROVED

(G’/G)-EXPANSION METHOD FOR

SOLVING PDES

Suppose that we have the following nonlinear

fractional partial differential equation:

( , , ,...) 0, 0 1,t xF u D u D u (5)

where ,t xD u D u are the modified Riemann-

Liouville derivatives, and F is a polynomial in u(x,

t) and its partial fractional derivatives, in which the

highest order derivatives and the nonlinear terms are

involved. In the following, we give the main steps of

this method :

Step1. Using the nonlinear fractional complex

transformation [33,34]

0

( , ) ( ),

(1 ) (1 )

u x t U

kx ct

(6)

where k, c, 0 are constants with , 0k c , to reduce

Eq. (5) to the following ordinary differential equation

(ODE) with integer order:

( , ', '',...) 0,P U U U (7)

where, P is a polynomial in ( )U and its total

derivatives, while the dashes denote the derivatives

with respect to .

Step2. We assume that Eq. (7) has the formal

solution :

( )

( ) ,( )

im

i

i m

GU a

G

(8)

where ( ,..., )ia i m m are constants to be

determined later , and ( )G satisfies the following

linear ODE:

( ) ( ) ( ) 0,G G G (9)

where and are constants.

Step3. The positive integer m in Eq. (8) can be

determined by balancing the highest-order derivatives

with the nonlinear terms appearing in Eq. (7).

Step4. We substitute (8) along with Eq. (9) into Eq.

(7) to obtain polynomials in , ( 0, 1, 2, ...)

i

Gi

G

.

Equating all the coefficients of these polynomials to

zero, yields a set of algebraic equations.

Step5. We solve the algebraic equations of step 4,

using the Maple or Mathematica to find the values of

, , , ,ia k c . Substituting these values into (8) and

using the ratios:

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



20

2

1 2

1 2

2

1 2

1 2

22

1 2

1 1

1 1

2 2

2 2

cosh sinh4,

2 2 sinh cosh

sin cos( ) 4,

( ) 2 2 cos sin

4 0,2

c c

c c

c cG

G c c

cif

c c

(10)

where 2 2

14 , 4 0 and

2 2

2 4 4, 0. We obtain the exact

solutions of Eq. (5),where c1 and c2 are arbitrary

constants.

APPLICATIONS

In this section we construct the exact solutions of the

following four nonlinear fractional PDEs using the

proposed method of Sec.2 as follows:

Example1. The space-time nonlinear fractional

PKP equation

This equation is well-known [1] and has the form:

4 2 21 3 3

( ) 0,4 2 4

x x x y t xD u D uD u D u D D u

(11)

where 0 1 . Eq. (11) has been investigated in

[1] using the fractional sub-equation method. Let us

now solve Eq. (11) using the proposed method of

Sec. 2. To this end, we use the nonlinear fractional

complex transformation

1 2

0

( , , ) ( ),

(1 ) (1 ) (1 ),

u x y t U

k x k y ct

(12)

where 1 2 0, , ,k k c are constants, to reduce Eq. (11)

to the following ODE with integer order:

4 3 2 2

1 1 2 13 (3 4 ) 0,k U k U k ck U (13)

By balancing 2U with U , we have m=1.

Consequently, Eq. (13) has the formal solutions:

1

1 0 1( ) ,G G

U a a aG G

(14)

where 1 0 1, ,a a a

are constants to be determined

later, such that 1 0a or

1 0a . Substituting

(14) along with Eq. (9) into Eq. (13), collecting all

the terms of the same orders , 0, 1, 2, ...

i

Gi

G

and setting each coefficient to zero, we have the

following set of algebraic equations:

4

4 2 3

1 1 1 1

3

4 2 3

1 1 1 1

2

4 2 3 2 2 2

1 1 1 1 1 1 1 1

2

1 2 1

4 3 3 2

1 1 1 1 1 1 1

2

1 2 1

: 2 0,

: 2 0,

: (8 7 ) 3 ( 2 2 )

(3 4 ) 0,

: (8 ) 3 (2 4 )

(3 4 ) 0,

Ga k a k

G

Ga k a k

G

Gk a a k a a a a

G

a k k c

Gk a a k a a a

G

a k k c

0

4 2 2 2

1 1 1 1 1

3 2 2 2 2 2

1 1 1 1 1 1 1 1 1 2 1

1

4 3 3 2

1 1 1 1 1 1 1

2

1 2 1

: (2 2 )

3 ( 4 2 ) ( )(3 4 ) 0

: (8 ) 3 (2 4 )

(3 4 ) 0,

Gk a a a a

G

k a a a a a a a a k k c

Gk a a k a a a

G

a k k c

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



21

2

4 2 2 3 2 2 2 2

1 1 1 1 1 1 1 1

2

1 2 1

3

2 4 2 3

1 1 1 1

4

3 4 2 2 3

1 1 1 1

: (7 8 ) 3 ( 2 2 )

(3 4 ) 0,

: 2 0,

: 2 0.

Gk a a k a a a a

G

a k k c

Ga k a k

G

Ga k a k

G

On solving these algebraic equations with the aid of

Maple or Mathematica we have the following cases:

Case 1.

1 1 2 2

4 2

1 2 1 1 1 1

1

0, , , ,

1(16 2 ), 2 , 2

4

k k k k

c k k a k a kk

.

Case 2.

1 1 2 2

4 2

1 2 1 1 1

1

0, , , ,

1(4 2 ), 2 , 0

4

k k k k

c k k a k ak

.

Case 3.

1 1 2 2

4 2 2

1 2 1 1 1

1

0, , , ,

1( ( 4 ) 2 ), 2 , 0.

4

k k k k

c k k a k ak

Let us now write down the following exact solutions

of the space-time fractional PKP equation (13) for

case 1 (Similarly for cases 2,3 which are omitted here

for simplicity):

(i) If 0 (Hyperbolic function solutions)

In this case, we have the exact solution:

1

1 2

1

1 2

1 2

0 1

1 2

sinh( ) cosh( )( , , ) 2

cosh( ) sinh( )

sinh cosh2 ,

cosh sinh

c cu x y t k

c c

c ca k

c c

(15)

If we set c1=0 and2 0c in (15) we have the

solitary solution:

1 1

0 1

( , , ) 2 coth( )

2 tanh( ),

u x y t k

a k

(16)

while if we set 2

0c and1

0c , in (15) we have the

solitary wave solution:

2 1( , , ) ( , , ),u x y t u x y t (17)

(ii)If 0 (Trigonometric function solutions)

In this case we have the exact solution:

1

1 2

1

1 2

1 2

0 1

1 2

sin( ) cos( )( , , ) 2

cos( ) sin( )

sin( ) cos( )2 ,

cos( ) sin( )

c cu x y t k

c c

c ca k

c c

(18)

If we set c1=0 and 2 0c in (18) we have the

periodic solution:

3 1 0 1( , , ) 2 tan( ) 2 cot( )u x y t k a k (19)

while if we set c2=0 and 1 0c , in (18) we have the

periodic solution:

4 3( , , ) ( , , )u x y t u x y t (20)

where

4 21 2

1 2 0

1

1(16 2 )

(1 ) (1 ) 4 (1 )

k x k y tk k

k

(iii) If 0 (Rational function solutions)

1

2 2

1 0 1

1 2 1 2

( , , ) 2 2c c

u x y t k a kc c c c

(21)

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



22

where

2

1 2 20

1(1 ) (1 ) 2 (1 )

k x k y k t

k


SRLW equation


2 2 2 2

( ) ( ) 0,t x t x t x t x

D u D u uD D u D u D u D D u

(22)



solve Eq. (22) using the proposed method of Sec. 2.

To this end , we use the nonlinear fractional complex

transformation

0

( , ) ( ), ,(1 ) (1 )

kx ctu x t U

(23)

where 0, ,k c are constants, to reduce Eq. (22) to

the following ODE with integer order:

2 2 2 2 2( ) 0,

2

kck c U k c U U (24)



2 1 2

2 1 0 1 2( ) ,

G G G GU a a a a a

G G G G

(25)

where 2 1 0 1 2, , , ,a a a a a are constants to be

determined later, such that 2 0a or 2 0a .

Substituting Eq.(25) along with Eq. (9) into Eq. (24),

collecting all the terms of the same orders

, 0, 1, 2,...

iG

iG

and setting each

coefficient to zero, we have the following set of

algebraic equations:

4

2 2 2

2 2

3

2 2

1 2 2 1

2

2 2 2 2 2 2

2 1 0 2 2 1 2

2 2 2 2 2

1 1 2 0 1 1 2 1

: 6 0,2

: (10 2 ) 0,

: ( ) ( 2 ) (8 3 4 ) 0,2

: ( ) ( ) (2 6 ) 0,

G kca a k c

G

Ga a kc k c a a

G

G kca k c a a a k c a a a

G

Ga k c kc a a a a k c a a a

G

0

2 2 2

0 0 2 2 1 1

2 2 2

2 1 1 2

: ( ) ( 2 2 )2

(2 2 ) 0,

G kca k c a a a a a

G

k c a a a a

1

2 2

1 1 2 0 1

2 2 2

1 2 1

: ( ) ( )

(2 6 ) 0,

Ga k c kc a a a a

G

k c a a a

2

2 2 2

2 1 0 2

2 2 2

2 1 2

3

2 2 2

1 2 2 1

4

2 2 2 2

2 2

: ( ) ( 2 )2

(8 3 4 ) 0,

: (10 2 ) 0,

: 6 0,2

G kca k c a a a

G

k c a a a

Ga a kc k c a a

G

G kca a k c

G

On solving the above set of algebraic equations with

the aid of Maple or Mathematica we have the

following cases:

Case 1.

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



23

2 2 2 2 2

2 2

2 2 2 2 2

1 1 2

2 2 2 2 2 2 2 2 2 2 2

2 03 3

1, , , [ ( )],

4

3[ ( )], 0,

3 1[ ( )] , [3 ( )]

4

c c k k k c c kk c

a k c c k a akc

a k c c k a k c c kk c kc

Case 2.

2 2 2 1

12 2

2 2 2

0 1 1 2 1 1 2

1, , [144( ) ], ,

576 12

1[48( ) ], 0, , 12

48

ac c k k c k a

k c kc

a c k a a a a a a kckc


of the space-time fractional PKP equation (22) for

case 1 (Similarly for case 2 which is omitted here for

simplicity):

(i) If 2 4 0 (Hyperbolic function solutions)


2 2 2 2 2

2 2 2

2 2 2 2

1 22 2

2 2

2 2 2 2

1 2

2 2 2 2 2 2

3

1

3 ( ) 3( , ) [

cosh sinh

2 2( )]

2 2

sinh cosh

2 2

3[ ( )]

4

k c k cu x t k c

kc kc

k c k cc c

kc kck cc k

kc k c k cc c

kc kc

k c c k

k

2 2 2 2

1 22 2

32 2 2 2

1 2

2

cosh sinh

2 2

2 2

sinh cosh

2 2

k c k cc c

kc kck c

c kc k c k cc c

kc kc

(26)


solitary solution:

2 2 2 2 2

1

2 2 2 2

2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

3 3

1

2

3 ( )( , )

3[ ( )] tanh

2 2 2

3[ ( )]tanh ,

4 2 2 2

k c k cu x t

kc

k c k ck c c k

kc kc kc

k c c k k c k c

k c kc kc

(27)

while if we set c2=0 and1 0c , in (26) we have the

solitary wave solution: 2 2 2 2 2

2

2 2 2 2

2 2 2 2 2

2 2 2 2 2 2 2 2 2 2

3 3

1

2

3 ( )( , )

3[ ( )] coth

2 2 2

3[ ( )]coth ,

4 2 2 2

k c k cu x t

kc

k c k ck c c k

kc kc kc

k c c k k c k c

k c kc kc

. (28)

If 2 0c and

2 2

1 2c c , then we have the solitary

wave solution: 2 2 2 2 2

3

1

2 2 2 2

2 2 2 2 2

1

2

2 2 2 2 2 2 2 2 2 2

13 3

3 ( )( , )

3[ ( )] coth

2 2 2

3[ ( )]coth ,

4 2 2 2

k c k cu x t

kc

k c k ck c c k

kc kc kc

k c c k k c k c

k c kc kc

(29)

where1 1

1

2

tanhc

c

, while if 1 0c and

2 2

2 1 ,c c then we have the solitary wave solution:

2 2 2 2 2

4

2 2 2 2

2 2 2 2 2

1

2 2 2 2 2 2 2 2 2 2

13 3

2

1

3 ( )( , )

3[ ( )] tanh

2 2 2

3[ ( )]tanh ,

4 2 2 2

k c k cu x t

kc

k c k ck c c k

kc kc kc

k c c k k c k c

k c kc kc

(30)

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



24

where 1 2

1

1

cothc

c

and

0.(1 ) (1 )

kx ct

Exampe3. The space-time nonlinear fractional

STO equation


2 2 2 2

3 ( ) 3 3 0,t x x x x

D u D u u D u uD u D u

(31)



now solve Eq. (31) using the proposed method of

Sec. 2. To this end, we use the transformation (6) to

reduce Eq. (31) to the following ODE with integer

order:

2 3 33 0,cU k UU kU k U (32)

By balancing3U with U , we have m=1.


1

1 0 1( ) ,G G

U a a aG G

(33)

where 1 0 1, ,a a a are constants to be determined

later, such that 1 0a or 1 0a . Substituting

(33) along with Eq. (9) into (32), collecting all the

terms of the same order , 0, 1, 2, 3

iG

iG

and setting each coefficient to zero, we have the

following set algebraic equations:

3

2 2 3 3

1 1 1

2

2 2 2 3

1 0 1 0 1 1

2 2 2 2

1 1 0 1 0 1 1 1

3 2

1 1

: 3 2 0,

: 3 ( ) 3 3 0,

: 3 ( ) 3 ( )

(2 ) 0,

Ga k a k a k

G

Gk a a a a a k a k

G

Gca k a a a k a a a a

G

k a a

0

2 3

0 0 1 1 0 0 1 1

3 2

1 1

: 3 ( ) ( 6 )

( ) 0,


G

k a a

1

2 2 2 2

1 1 0 1 0 1 1 1

3 2

1 1

2

2 2 2 3

1 0 1 0 1 1

3

2 2 3 2 3

1 1 1

: 3 ( ) 3 ( )

(2 ) 0

: 3 ( ) 3 3 0,

: 3 2 0,


G

k a a

Gk a a a a a k a k

G

Ga k a k a k

G

By solving these algebraic equations with the aid of

Maple or Mathematica we have the following cases:

Case 1.

3 2

1 0 1

, , , , ( 4 ),

2 , , 0

k k c k

a k a k a

Case 2.

3 2

1 1 0

, , , , ( 4 ),

0, 2 ,

k k c k

a a k a k

Case 3.

3 2

1 1 0

, , , , ( 4 ),

0, 2 ,

k k c k

a a k a k


of the space-time fractional STO equation (31) for

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



25

case 1 (Similarly for cases 2, 3 which are omitted

here for simplicity) :

(i) If 2 4 0 (Hyperbolic function solutions)


2 2

1 2

2

2 2

1 2

cosh 4 sinh 42 2

( , ) 4

sinh 4 cosh 42 2

,

c c

u x t k

c c

(34)


solitary solution:

2

2

1

4( , ) 4 tanh ,

2u x t k

(35)

while if we set c2=0 and1 0c , in (34) we have the


2

2

2

4( , ) 4 coth .

2u x t k

(36)

If 2 0c and2 2


wave solution:

2 2

3 1 2( , ) 4 coth 4 ,u x t k

(37)

where1 1

1

2

tanhc

c

, while if 1 0c and

2 2

2 1 ,c c then we have the solitary wave solution:

2 2

4 1 2( , ) 4 tanh 4 ,u x t k

(38)

where 1 2

1

1

cothc

c

.

(ii)If 2 4 0 (Trigonometric function

solutions)


2 2

1 2

2

2 2

1 2

sin 4 cos 42 2

( , ) 4 ,

cos 4 sin 42 2

c c

u x t k

c c

(39)


periodic solution:

2

2

1

4( , ) 4 cot ,

2u x t k

(40)


periodic solution:

2

2

2

4( , ) 4 tan ,

2u x t k

(41)

If 2 0c and

2 2

1 2c c , then we have the periodic

solution:

2 2

3 1 2( , ) 4 cot 4 ,u x t k

(42)

where1 1

1

2

tanc

c

, while if 1 0c and

2 2

2 1 ,c c then we have the periodic solution:

2 2

4 1 2( , ) 4 tan 4 ,u x t k

(43)

where 1 2

1

1

cotc

c

and

3 2

0( 4 ) .(1 ) (1 )

kx tk


KPP equation


2 2 3

1 0,t xD u D u u u u (44)

where 0 1 and , , are non zero

constants . This equation is important in the physical

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



26

fields, and it includes the Fisher equation. Huxlay

equation, Burgers equation, Chaffee- Infanfe equation

and Fitzhugh-Nagumo equation. When 1 Eq.

(44) has been discussed in [6] using the ( / )G G

expansion method. Let us solve Eq. (44) using the

proposed method of Sec. 2. To this end , we use the

nonlinear fractional complex transformation (6), to

reduce Eq. (44) to the following ODE with integer

order:

2 2 3

1 0,cU k U U U U (45)


Consequently, Eq. (45) has the same formal solution

(44). Substituting (44) along with Eq. (9) into Eq.

(43), collecting all the terms of the same orders

, 0, 1, 2, 3

iG

iG

and setting each

coefficient to zero, we have the following set

algebraic equations:

3

2 3

1 1

2

2 2 2

1 1 1 0 1

: 2 0,

: 3 3 0,

Ga k a

G

Ga c a k a a a

G

2 2

1 1 1 1 1 1 0

2 2

0 1 1 1

0

2

1 1 1 1 1 0

2 3

0 1 1 0 0 1 1

: (2 ) 2

3 ( ) 0,

: ( ) ( )

( 2 ) ( 6 ) 0,

Ga c k a a a a a

G

a a a a

Gc a a k a a a

G

a a a a a a a

1

2 2

1 1 1 1 1 1 0

2 2

0 1 1 1

2

2 2 2

1 1 1 0 1

3

2 2 3

1 1

: (2 ) 2

3 ( ) 0,

: 3 3 0,

: 2 0,

Ga c k a a a a a

G

a a a a

Ga c a k a a a

G

Ga k a

G

By solving the above set of algebraic equations with

the aid of Maple or Mathematica we have the

following cases:

Case 1.

2

12

2

1

1 1 0

10, , , , ( 4 ),

32

( 4 ) 2 2, , , .

32 22

k kk

ka a k c a

k

Case 2.

2 2 2

1

1 0 1

1, , , , ( ( 4 ) ),

4

1 1 10, ( ), 2 ,

2 2 2 2

k k k

a a k a k c k


of the space-time fractional KPP equation (44) for

case 1 (Similarly for case 2 which is omitted here for

simplicity):

(i) If 0 (Hyperbolic function solutions)


1 2

1 2

1

2

1 21

1 2

cosh sinh2( , )

2sinh cosh

cosh sinh( 4 ) 1,

16 2 sinh cosh

c cu x t k

c c

c c

k c c

(46)


solitary solution:

1

2

2( , ) tanh

2

( 4 )coth ,

16 2

u x t k

k

(47)



Jan 2014. Vol. 4, No. 7 ISSN2305-8269



27

2

2

2( , ) coth

2

( 4 )tanh ,

16 2

u x t k

k

(48)

If 1 0c and

2 2


wave solution:

4 1

2

1

2( , ) tanh

2

( 4 )coth ,

16 2

u x t k

k

(49)

where 1 2

1

1

cothc

c

.

(ii) If 0 (Trigonometric function solutions)


1 2

1 2

1

2

1 2

1 2

sin cos2( , )

2cos sin

sin cos( 4 ),

16 2 cos sin

c cu x t k

c c

c c

k c c

(50)


periodic solution:

1

2

2( , ) cot

2

( 4 )tan ,

16 2

u x t k

k

(51)


periodic solution:

2

2

2( , ) tan

2

( 4 )cot ,

16 2

u x t k

k

(52)

If 2 0c and

2 2

1 2c c , then we have the periodic

solution:

3 1

2

1

2( , ) cot

2

( 4 )tan ,

16 2

u x t k

k

(53)

where1 1

1

2

tanc

c

. If 1 0c and

2 2

2 1 ,c c

then we have the periodic solution:

4 1

2

1

2( , ) tan

2

( 4 )cot ,

16 2

u x t k

k

(54)

where 1 2

1

1

cotc

c

and

0(1 ) (1 )2

kx k t

.

(iii) If 0 (Rational function solution)

2 2

1 2 1 2

1

2 2( , )

2 32

,c ck

u x tc c c ck

(55)

where 1 02(1 ) (1 )

kx tk

Figures of some exact solutions

In this section we give some figures to illustrate the

solutions of the equations (13) ,(24), (33) and (46) as

follows:

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



28

Fig. 1 Numerical simulation solution of the space-

time fractional PKP equation.


time fractional SRLW equation.

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



29

Fig.3 Numerical simulation solution of the space-

time fractional STO equation.

Jan 2014. Vol. 4, No. 7 ISSN2305-8269



30


time fractional KPP equation.

Some conclusions and discussions

In this article, we have extended successfully the

improved ( / )G G - expansion method to solve

four nonlinear fractional partial differential

equations. As applications, abundant new exact

solutions for the space-time nonlinear fractional

PKP equation, the space-time nonlinear fractional

SRLW equation, the space-time nonlinear fractional

STO equation and the space-time nonlinear

fractional KPP equation have been successfully

found. As one can see, the nonlinear fractional

complex transformation for is very important,

which ensures that certain nonlinear fractional PDEs

can be turned into other nonlinear ODEs of integer

orders, whose solutions can be expressed in the form

(8) where ( )G satisfies the linear ODE (9). Some

numerical examples with diagrams have been given

for fractional and non fractional orders to illustrate

our results. Besides, as this method is based on the

homogeneous balancing principle, so it can also be

applied to other nonlinear fractional partial

differential equations, where the homogeneous

balancing principle is satisfied.

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