Pramana – J. Phys. (2018) 91:9 © Indian Academy of Sciences

https://doi.org/10.1007/s12043-018-1583-4

Exact solutions to (2+1)-dimensional Chaffee–Infante equation

YUANYUAN MAO

Department of Mathematics, Northeast Petroleum University, Daqing 163318, China

E-mail: [email protected]

MS received 28 November 2017; revised 4 January 2018; accepted 8 January 2018

Abstract. In this paper, the canonical-like transformation method and trial equation method are applied to(2+1)-

dimensional Chaffee–Infante equation, and some exact solutions are obtained. In particular, a new solution in terms

of elliptic functions is given.

Keywords. Chaffee–Infante equation; travelling wave solutions; canonical-like transformation method; trial

equation method.

PACS Nos 05.45.Yv; 02.30.Jr; 03.65.Vf

1. Introduction

The construction of exact solutions to nonlinear

differential equations is an important and difcult task.

It also plays a signicant role in mathematics and

physics. Various powerful methods such as inverse

scattering method 1, direct method 2, symmetri-

cal method 3, the complete discrimination system for

polynomial method 4–14 and so on have been pro-

posed for obtaining approximate and exact solutions for

various nonlinear equations.

In this paper, we consider (2+1)-dimensional

Chaffee–Infante equation 15

uxt+ (?uxx+au3?au)x+?uyy=0,(1)

where aand?are arbitrary constants. Assume that a

substance spreads in a region with the concentration of

the diffusion as u(x,y,z).LetD(x,y,z,t)be the diffu-

sion coefcient, then according to the law of diffusion,

we have

d m=?D ? u

?n d sdt,(2)

where mrepresents the amount of diffusion material and

D >0.Accordingtoeq.(2) and the law of conservation

of mass yields

m =

t2

t1

D ? u

?n d s

d t

=

u(x,y,z,t2)?u(x,y,z,t1)dxdydz,(3)

that is,

?

? x

D ? u

?x

+ ?

? x

D ? u

?y

+ ?

? z

D ? u

?x

= ? u

?t . (4)

Assume that D=1 and the inuential factor isf(u)=

u3? u.Weget

? u

?t = ?2u

? x2+ ?2u

? y2+ ?2u

? z2+?(u?u3),(5)

where parameter ?adjusts the relative balance of the

diffusion term and the nonlinear term. So the (1+1)-

dimensional Chaffee–Infante equation is

ut? uxx=?(u?u3),(6)

and the derivation process of (2+1)-dimensional

Chaffee–Infante equation is similar to it.

The (2+1)-dimensional Chaffee–Infante equation is

a well-known reaction diffusion equation. It describes

the physical process of mass transport and particle dif-

fusion, and has been widely used in environmental

science, uid dynamics, high-energy physics, electronic

science, and so on. Therefore, building exact solutions

to this equation has great scientic signicance and

a broad application background (see 16 and refer-

ences therein). In 15, the authors use exp-function

method to obtain solutions to eq. ( 1), and by this method,

we cannot get solutions of other types such as ellip-

tic function solutions. It is also difcult to get elliptic

function solutions to eq. ( 1) by some other methods,

but canonical-like transformation method can be used

9 Page 2 of 4Pramana – J. Phys. (2018) 91:9

to get elliptic function solutions. The canonical-like

transformation method was proposed by Liu 17to

obtain exact travelling wave solutions to some diffusion-

reaction equations such as Fisher equation, Burgers–

KdV equation and Newell–Whitehead–Kwahara

equation and so on. In this paper, we use Liu’s canonical-

like transformation method to nd solutions to eq. (1)

and get a new solution in terms of elliptic functions. In

addition, we use trial equation method 18–23toget

exact solutions of other forms. Further studies for exact

solutions and integrability of differential equations in

mathematical physics can be found in many papers (see,

for example, 24–30).

2. Elliptic functions solutions by canonical-like

transformation method

According to ref. 17, the canonical-like transforma-

tion method is as follows. We consider the following

ordinary differential equation:

u

(? )?Au(? )=Bu(? )+Du(? ).(7)

If we take?=u

,eq.(7) becomes

d?

du=Bu

+Du+A?

?.(8)

In order to solve the rst-order nonlinear ODE (8), we

re-parametriseuand?by a new parameters,thatis,we

take the canonical-like transformations

u=a

11(s)v(s)+a12(s)v(s),(9)

?=a

21(s)v(s)+a22(s)v(s),(10)

wherea

ij(s)(i,j=1,2)andv(s)are functions to be

determined. In order to obtain the parameters, we set

a

12=0. After a lot of calculations, we get

a

11=

+3(?1)A

1/2

(s?s0)2/(?1),(11)

a

22=

(?1)A+3

1/2

(s?s0)2/(?1),(12)

a

21=A?h2

+3(?1)A

1/2

(s?s0)2/(?1).(13)

Then eq. (7) becomes

v

(s)=Mv(s),(14)

andsis determined by

d?

ds=1?(s),(15)

whereM=Bh

?(+3)/2,h=±?4D+A2and?(s)=

h(s?s

0).The general solution of eq. (14)isgivenby

±(s?s

0)=

dv 2M+1v+1+c,(16)

wherecis an integral constant. Ifc=0, we have

v(s)=

±1?

2

2M

+1(s?s0)

2/(1?)

.(17)

And in the special case ofc=1and=3, we get

v(s)=

2

M

1/4asn(?(s?s0),m)+bcn(?(s?s0),m)

csn(?(s?s0),m)+dcn(?(s?s0),m,

(18)

where

a=3?2?

2,b=2?2?1,

c=2?

2?4,d=?2,

m=16+12?

2

17+12?2,?=

19+12?2

52?32?2.

So, we can get solutions of eq. (7) from the formulas

(17)and(18)

u(? )=6

B

(?1)A+3

(+2)/2e2A+3(???0)

f

e(?1)A+3(???0)?s0

2?1

(19)

and

u(? )=

3

A

1/2

(eh(???0)?s0)

2M

1/4

×asn(?(e

h(???0)?s0),m)+bcn(?(eh(???0)?s0),m)

csn(?(eh(???0)?s0),m)+dcn(?(eh(???0)?s0),m).

(20)

Now, we use the canonical-like transformation

method to obtain the solutions of eq. (1). Letu=u(? )

and?=kx+ly+wt, then eq. (1) becomes

(kw+?l

2)u?k3u+3aku2u?aku=0,(21)

and integrating it yields

(kw+?l

2)u?k3u+aku3?aku=c0.(22)

Ifc

0=0, by (19)and(20), the solutions of eq. (1)

can be written as

u(x,y,t)=

6k2

a

fkw+?l2

3k3

5/2ekw+?l23k3(kx+ly+?t??0)

ekw+?l23k3(kx+ly+?t??0)?s0

(23)

Pramana – J. Phys. (2018) 91:9 Page 3 of 4 9

and

u(x,y,t)=

3

A

1/2

(eh(kx+ly+?t??0)?s0)

2M

1/4

×a1sn(?(eh(kx+ly+?t??0)?s0),m)+b1cn(?(eh(kx+ly+?t??0)?s0),m)

c1sn(?(eh(kx+ly+?t??0)?s0),m)+d1cn(?(eh(kx+ly+?t??0)?s0),m),(24)

where

A=kw+?l

2

k3,h=±(kw+?l2)2+4ak4

k3,

M=a

k2h?3,a1=3?2?2,b1=2?2?1,

c

1=2?2?4,d1=?2,

m=16+12?

2

17+12?2,?=

19+12?2

26?16?2.

Furthermore, according to ref. 17, we have

2(kw+?

2)2=?9ak4,(25)

from which we have

h=kw+?l

2

3k3.

Formula (24) is a new solution represented by elliptic

functions with double period.

3. Other solutions by trial equation method

Ifc

0=0, by taking

u=a

0+a1?+b1

?(26)

and

?=?

,(27)

where?is a function to be determined, eq. (22) becomes

a

1?b1

?2

??

?

Aa1?Ab1

?2

?+2b1

?3?2

=B

a0+a1?+b1

?

3

?

a0+a1?+b1

?

+c

0,(28)

where

A=k?+?l

2

k3,B=ak2,

and the prime indicates differential with respect to?,

e.g.,?

=d?/d?. Without loss of generality, we take

the solution of eq. (28) as follows:

?=?

2+b0,(29)whereb

0is a constant to be determined. Substituting

(29)into(28) yields a polynomial of?. Setting all coef-

cients of this polynomial to zero, we get a system of

algebraic equations

2b

1b2

0?Bb3

1=0,(30)

Ab

1b0?Ba0b2

1=0,(31)

4b

1b0?2b1b0?Ba1b2

1?Ba2

0b1+Bb1=0,(32)

?Aa

1b0+Ab1?Ba3

0?Ba0a1b1+Ba0?c=0,

(33)

2a

1b0?2b1+2b1?Ba2

0a1?Ba2

1b1+Ba1=0,

(34)

?Aa

1?Ba0a2

1=0,(35)

2a

1?Ba3

1=0.(36)

Solving this algebraic equation system, we obtain a

family of values of parameters

a

0=?A?2B,a1=

2

B,b0=?2B?A

2

8,

b

1=?A

2?2B

4?2B.(37)

So we have

?=?

2B?A2

8tanh?

?

2B?A2

8???0

?

?

(38)

and

?=?

2B?A2

8coth?

?

2B?A2

8???0

?

?

,(39)

where?

0is an arbitrary constant. Therefore, the solu-

tions of eq. (1)aregivenby

u(x,y,t)=?kw+?l

2

k2?2a

?m

2k2?atanh

m2?2k3(kx+ly+wt)??0

+m

2k2?acoth

m2?2k3(kx+ly+wt)??0

(40)

9 Page 4 of 4Pramana – J. Phys. (2018) 91:9

and

u(x,y,t)

=?kw+?l

2

k2?2a

?m

2k2?acoth

m2?2k3(kx+ly+wt)??0

+m

2k2?atanh

m2?2k3(kx+ly+wt)??0

,

(41)

wherem=?

2ak4?k2?2?2k??l2??2l4. Expres-

sions (40)and(41) are solutions represented by hyper-

bolic tangent function and hyperbolic cotangent

function.

4. Conclusion

In this paper, the canonical-like transformation method

and trial equation method are applied to obtain exact

solutions to(2+1)-dimensional Chaffee–Infante equa-

tion. Among those, some new solutions are given. The

results show that canonical-like transformation method

and trial equation method are powerful for solving non-

linear problems arising in mathematical physics.

References

1 M J Ablowitz and P A Clarkson,Solitons, nonlinear

evolutions and inverse scattering(Cambridge Univer-

sity Press, Cambridge, 1991)2 R Hirota,Direct methods in soliton theory

(Springer-Verlag, Berlin, 1980)

3 M Wadati,Stud. Appl. Math.59(2), 153 (1978)

4 C Liu,Commun. Theor. Phys.44(5), 799 (2005)

5 C Liu,Commun. Theor. Phys.54(6), 991 (2006)

6 C Liu,Commun. Theor. Phys.43(4), 787 (2005)

7 C Liu,Commun. Theor. Phys.48(4), 601 (2007)

8 C Liu,Chin. Phys.14(9), 1710 (2005)

9 C Liu,Chin. Phys.16(7), 1832 (2007)

10 C Liu,Chin. Phys. Lett.21(12), 2369 (2004 )

11 C Liu,Commun. Theor. Phys.49(2), 291 (2008)

12 C Liu,Commun. Theor. Phys.49(1), 153 (2008)

13 C Liu,Comput. Phys. Commun.181(2), 317 (2010)

14 Y Kai,Pramana – J. Phys.87(4), 59 (2016)

15 R Sakthivel and C Chun,Z. Naturforsch. A65(3), 197

(2010)

16 P Constantin,Integral manifolds and inertial manifolds

for disspative partial equation(Springer-Verlag, New

York, 1989)

17 C Liu,Chaos Solitons Fractals42, 441 (2009)

18 C Liu,Far East J. Appl. Math.40(1

), 49 (2010)

19 C Liu,Acta Phys. Sinica54(6), 2505 (2005)

20 C Liu,Acta Phys. Sinica54(10), 4506 (2005)

21 C Liu,Commun. Theor. Phys.45(2), 219 (2006)

22 C Liu,Commun. Theor. Phys.54, 3395 (2006)

23 C Liu,Found. Phys.41(5), 793 (2011)

24 C Liu,Chaos Solitons Fractals40, 708 (2009)

25 H L Fan,Appl. Math. Comput.219, 748 (2012)

26 H L Fan and X Li,Pramana – J. Phys.81(6), 925 (2013)

27 C S Liu,Rep. Math. Phys.67(1), 109 (2011)

28 D Y Dai and Y P Yuan,Appl. Math. Comput.242, 729

(2014)

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(2016)

30 A Saha, B Talukdar, U Das and S Chaterjee,Pramana –

J. Phys.88, 28 (2017)

https://doi.org/10.1007/s12043-018-1583-4

Exact solutions to (2+1)-dimensional Chaffee–Infante equation

YUANYUAN MAO

Department of Mathematics, Northeast Petroleum University, Daqing 163318, China

E-mail: [email protected]

MS received 28 November 2017; revised 4 January 2018; accepted 8 January 2018

Abstract. In this paper, the canonical-like transformation method and trial equation method are applied to(2+1)-

dimensional Chaffee–Infante equation, and some exact solutions are obtained. In particular, a new solution in terms

of elliptic functions is given.

Keywords. Chaffee–Infante equation; travelling wave solutions; canonical-like transformation method; trial

equation method.

PACS Nos 05.45.Yv; 02.30.Jr; 03.65.Vf

1. Introduction

The construction of exact solutions to nonlinear

differential equations is an important and difcult task.

It also plays a signicant role in mathematics and

physics. Various powerful methods such as inverse

scattering method 1, direct method 2, symmetri-

cal method 3, the complete discrimination system for

polynomial method 4–14 and so on have been pro-

posed for obtaining approximate and exact solutions for

various nonlinear equations.

In this paper, we consider (2+1)-dimensional

Chaffee–Infante equation 15

uxt+ (?uxx+au3?au)x+?uyy=0,(1)

where aand?are arbitrary constants. Assume that a

substance spreads in a region with the concentration of

the diffusion as u(x,y,z).LetD(x,y,z,t)be the diffu-

sion coefcient, then according to the law of diffusion,

we have

d m=?D ? u

?n d sdt,(2)

where mrepresents the amount of diffusion material and

D >0.Accordingtoeq.(2) and the law of conservation

of mass yields

m =

t2

t1

D ? u

?n d s

d t

=

u(x,y,z,t2)?u(x,y,z,t1)dxdydz,(3)

that is,

?

? x

D ? u

?x

+ ?

? x

D ? u

?y

+ ?

? z

D ? u

?x

= ? u

?t . (4)

Assume that D=1 and the inuential factor isf(u)=

u3? u.Weget

? u

?t = ?2u

? x2+ ?2u

? y2+ ?2u

? z2+?(u?u3),(5)

where parameter ?adjusts the relative balance of the

diffusion term and the nonlinear term. So the (1+1)-

dimensional Chaffee–Infante equation is

ut? uxx=?(u?u3),(6)

and the derivation process of (2+1)-dimensional

Chaffee–Infante equation is similar to it.

The (2+1)-dimensional Chaffee–Infante equation is

a well-known reaction diffusion equation. It describes

the physical process of mass transport and particle dif-

fusion, and has been widely used in environmental

science, uid dynamics, high-energy physics, electronic

science, and so on. Therefore, building exact solutions

to this equation has great scientic signicance and

a broad application background (see 16 and refer-

ences therein). In 15, the authors use exp-function

method to obtain solutions to eq. ( 1), and by this method,

we cannot get solutions of other types such as ellip-

tic function solutions. It is also difcult to get elliptic

function solutions to eq. ( 1) by some other methods,

but canonical-like transformation method can be used

9 Page 2 of 4Pramana – J. Phys. (2018) 91:9

to get elliptic function solutions. The canonical-like

transformation method was proposed by Liu 17to

obtain exact travelling wave solutions to some diffusion-

reaction equations such as Fisher equation, Burgers–

KdV equation and Newell–Whitehead–Kwahara

equation and so on. In this paper, we use Liu’s canonical-

like transformation method to nd solutions to eq. (1)

and get a new solution in terms of elliptic functions. In

addition, we use trial equation method 18–23toget

exact solutions of other forms. Further studies for exact

solutions and integrability of differential equations in

mathematical physics can be found in many papers (see,

for example, 24–30).

2. Elliptic functions solutions by canonical-like

transformation method

According to ref. 17, the canonical-like transforma-

tion method is as follows. We consider the following

ordinary differential equation:

u

(? )?Au(? )=Bu(? )+Du(? ).(7)

If we take?=u

,eq.(7) becomes

d?

du=Bu

+Du+A?

?.(8)

In order to solve the rst-order nonlinear ODE (8), we

re-parametriseuand?by a new parameters,thatis,we

take the canonical-like transformations

u=a

11(s)v(s)+a12(s)v(s),(9)

?=a

21(s)v(s)+a22(s)v(s),(10)

wherea

ij(s)(i,j=1,2)andv(s)are functions to be

determined. In order to obtain the parameters, we set

a

12=0. After a lot of calculations, we get

a

11=

+3(?1)A

1/2

(s?s0)2/(?1),(11)

a

22=

(?1)A+3

1/2

(s?s0)2/(?1),(12)

a

21=A?h2

+3(?1)A

1/2

(s?s0)2/(?1).(13)

Then eq. (7) becomes

v

(s)=Mv(s),(14)

andsis determined by

d?

ds=1?(s),(15)

whereM=Bh

?(+3)/2,h=±?4D+A2and?(s)=

h(s?s

0).The general solution of eq. (14)isgivenby

±(s?s

0)=

dv 2M+1v+1+c,(16)

wherecis an integral constant. Ifc=0, we have

v(s)=

±1?

2

2M

+1(s?s0)

2/(1?)

.(17)

And in the special case ofc=1and=3, we get

v(s)=

2

M

1/4asn(?(s?s0),m)+bcn(?(s?s0),m)

csn(?(s?s0),m)+dcn(?(s?s0),m,

(18)

where

a=3?2?

2,b=2?2?1,

c=2?

2?4,d=?2,

m=16+12?

2

17+12?2,?=

19+12?2

52?32?2.

So, we can get solutions of eq. (7) from the formulas

(17)and(18)

u(? )=6

B

(?1)A+3

(+2)/2e2A+3(???0)

f

e(?1)A+3(???0)?s0

2?1

(19)

and

u(? )=

3

A

1/2

(eh(???0)?s0)

2M

1/4

×asn(?(e

h(???0)?s0),m)+bcn(?(eh(???0)?s0),m)

csn(?(eh(???0)?s0),m)+dcn(?(eh(???0)?s0),m).

(20)

Now, we use the canonical-like transformation

method to obtain the solutions of eq. (1). Letu=u(? )

and?=kx+ly+wt, then eq. (1) becomes

(kw+?l

2)u?k3u+3aku2u?aku=0,(21)

and integrating it yields

(kw+?l

2)u?k3u+aku3?aku=c0.(22)

Ifc

0=0, by (19)and(20), the solutions of eq. (1)

can be written as

u(x,y,t)=

6k2

a

fkw+?l2

3k3

5/2ekw+?l23k3(kx+ly+?t??0)

ekw+?l23k3(kx+ly+?t??0)?s0

(23)

Pramana – J. Phys. (2018) 91:9 Page 3 of 4 9

and

u(x,y,t)=

3

A

1/2

(eh(kx+ly+?t??0)?s0)

2M

1/4

×a1sn(?(eh(kx+ly+?t??0)?s0),m)+b1cn(?(eh(kx+ly+?t??0)?s0),m)

c1sn(?(eh(kx+ly+?t??0)?s0),m)+d1cn(?(eh(kx+ly+?t??0)?s0),m),(24)

where

A=kw+?l

2

k3,h=±(kw+?l2)2+4ak4

k3,

M=a

k2h?3,a1=3?2?2,b1=2?2?1,

c

1=2?2?4,d1=?2,

m=16+12?

2

17+12?2,?=

19+12?2

26?16?2.

Furthermore, according to ref. 17, we have

2(kw+?

2)2=?9ak4,(25)

from which we have

h=kw+?l

2

3k3.

Formula (24) is a new solution represented by elliptic

functions with double period.

3. Other solutions by trial equation method

Ifc

0=0, by taking

u=a

0+a1?+b1

?(26)

and

?=?

,(27)

where?is a function to be determined, eq. (22) becomes

a

1?b1

?2

??

?

Aa1?Ab1

?2

?+2b1

?3?2

=B

a0+a1?+b1

?

3

?

a0+a1?+b1

?

+c

0,(28)

where

A=k?+?l

2

k3,B=ak2,

and the prime indicates differential with respect to?,

e.g.,?

=d?/d?. Without loss of generality, we take

the solution of eq. (28) as follows:

?=?

2+b0,(29)whereb

0is a constant to be determined. Substituting

(29)into(28) yields a polynomial of?. Setting all coef-

cients of this polynomial to zero, we get a system of

algebraic equations

2b

1b2

0?Bb3

1=0,(30)

Ab

1b0?Ba0b2

1=0,(31)

4b

1b0?2b1b0?Ba1b2

1?Ba2

0b1+Bb1=0,(32)

?Aa

1b0+Ab1?Ba3

0?Ba0a1b1+Ba0?c=0,

(33)

2a

1b0?2b1+2b1?Ba2

0a1?Ba2

1b1+Ba1=0,

(34)

?Aa

1?Ba0a2

1=0,(35)

2a

1?Ba3

1=0.(36)

Solving this algebraic equation system, we obtain a

family of values of parameters

a

0=?A?2B,a1=

2

B,b0=?2B?A

2

8,

b

1=?A

2?2B

4?2B.(37)

So we have

?=?

2B?A2

8tanh?

?

2B?A2

8???0

?

?

(38)

and

?=?

2B?A2

8coth?

?

2B?A2

8???0

?

?

,(39)

where?

0is an arbitrary constant. Therefore, the solu-

tions of eq. (1)aregivenby

u(x,y,t)=?kw+?l

2

k2?2a

?m

2k2?atanh

m2?2k3(kx+ly+wt)??0

+m

2k2?acoth

m2?2k3(kx+ly+wt)??0

(40)

9 Page 4 of 4Pramana – J. Phys. (2018) 91:9

and

u(x,y,t)

=?kw+?l

2

k2?2a

?m

2k2?acoth

m2?2k3(kx+ly+wt)??0

+m

2k2?atanh

m2?2k3(kx+ly+wt)??0

,

(41)

wherem=?

2ak4?k2?2?2k??l2??2l4. Expres-

sions (40)and(41) are solutions represented by hyper-

bolic tangent function and hyperbolic cotangent

function.

4. Conclusion

In this paper, the canonical-like transformation method

and trial equation method are applied to obtain exact

solutions to(2+1)-dimensional Chaffee–Infante equa-

tion. Among those, some new solutions are given. The

results show that canonical-like transformation method

and trial equation method are powerful for solving non-

linear problems arising in mathematical physics.

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