Exact Solutions for Three Fractional PartialDifferential Equations by the (G’/G) Method

Nina Shang, Bin Zheng∗

Abstract—In this paper, based on certain variable transfor-mation, we apply the known (G’/G) method to seek exactsolutions for three fractional partial differential equations:the space fractional (2+1)-dimensional breaking soliton equa-tions, the space-time fractional Fokas equation, and the space-time fractional Kaup-Kupershmidt equation. The fractionalderivative is defined in the sense of modified Riemann-liouvillederivative. With the aid of mathematical software Maple, anumber of exact solutions including hyperbolic function solu-tions, trigonometric function solutions, and rational functionsolutions for them are obtained.

Index Terms—(G’/G) method, fractional partial differentialequation, exact solution, variable transformation.

I. INTRODUCTION

In the literature, research on the theory of differentialequations, integral equations and matrix equations includevarious aspects, such as the existence and uniqueness ofsolutions [1,2], seeking for exact solutions [3,4], numericalmethod [5-7]. Among these investigations, research on thetheory and applications of fractional differential and integralequations has been the focus of many studies due to theirfrequent appearance in various applications in physics, biol-ogy, engineering, signal processing, systems identification,control theory, finance and fractional dynamics, and hasattracted much attention of more and more scholars. Forexample, Bouhassoun [8] extended the telescoping decompo-sition method to derive approximate analytical solutions offractional differential equations. Bijura [9] investigated thesolution of a singularly perturbed nonlinear system fractionalintegral equations. Blackledge [10] investigated the applica-tion of a certain fractional Diffusion equation, and applied itfor predicting market behavior.

Among the investigations for fractional differential equa-tions, research for seeking exact solutions and approximatesolutions of fractional differential equations is a hot topic.Many powerful and efficient methods have been proposedso far (for example, see [11-24]) including the fractionalvariational iteration method, the Adomian’s decompositionmethod, the homotopy perturbation method, the fractionalsub-equation method, the finite difference method, the finiteelement method and so on. Using these methods, solutionswith various forms for some given fractional differentialequations have been established.

In [25], we extended the known (G’/G) method [26-28] tosolve fractional partial differential equations, and obtainedsome exact solutions for the space-time fractional general-ized Hirota-Satsuma coupled KdV equations and the time-fractional fifth-order Sawada-Kotera equation successfully.

Manuscript received March 16, 2013; revised 15 May, 2013.N. Shang and B. Zheng are with the School of Science, Shandong

University Of Technology, Zibo, Shandong, 255049 China ∗e-mail: zheng-bin2601@126.com

In this paper, we will furthermore test the validity of the(G’/G) method by applying it to solve other fractional partialdifferential equations. The fractional derivative is definedin the sense of modified Riemann-liouville derivative. Welist the definition and some important properties for themodified Riemann-Liouville derivative of order α as follows[22-24,29]:

Dαt f(t) =

1Γ(1− α)

ddt

∫ t

0(t− ξ)−α(f(ξ)− f(0))dξ,

0 < α < 1,(f (n)(t))(α−n), n ≤ α < n + 1, n ≥ 1.

Dαt tr =

Γ(1 + r)Γ(1 + r − α)

tr−α, (1)

Dαt (f(t)g(t)) = g(t)Dα

t f(t) + f(t)Dαt g(t), (2)

Dαt f [g(t)] = f ′g[g(t)]Dα

t g(t) = Dαg f [g(t)](g′(t))α. (3)

The rest of this paper is organized as follows. In Section2, we give the description of the (G’/G) method for solvingfractional partial differential equations. Then in Section 3we apply this method to seek exact solutions for the spacefractional (2+1)-dimensional breaking soliton equations, thespace-time fractional Fokas equation, and the space-timefractional Kaup-Kupershmidt equation. Some conclusionsare presented at the end of the paper.

II. DESCRIPTION OF THE (G’/G) METHOD FOR SOLVINGFRACTIONAL PARTIAL DIFFERENTIAL EQUATIONS

In this section we give the description of the (G′G ) method

for solving fractional partial differential equations.Suppose that a fractional partial differential equation, say

in the independent variables t, x1, x2, ..., xn, is given by

P (u1, ...uk, Dαt u1, ..., D

αt uk, Dβ

x1u1, ...,

Dβx1

uk, ..., Dγxn

u1, ..., Dγxn

uk, ...) = 0, (4)

where ui = ui(t, x1, x2, ..., xn), i = 1, ..., k are unknownfunctions, P is a polynomial in ui and their various partialderivatives including fractional derivatives.

Step 1. Execute certain variable transformation

ui(t, x1, x2, ..., xn) = Ui(ξ), ξ = ξ(t, x1, x2, ..., xn), (5)

such that Eq. (4) can be turned into the following ordinarydifferential equation of integer order with respect to thevariable ξ:

P (U1, ..., Uk, U ′1, ..., U

′k, U ′′

1 , ..., U ′′k , ...) = 0. (6)

Step 2. Suppose that the solution of (6) can be expressedby a polynomial in (G′

G ) as follows:

Uj(ξ) =mj∑

i=0

aj,i(G′

G)i, j = 1, 2, ..., k, (7)

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where G = G(ξ) satisfies the second order ODE in the form

G′′ + λG′ + µG = 0, (8)

and λ, µ, aj,i, i = 0, 1, ..., mj , j = 1, 2, ..., k are constantsto be determined later, aj,m 6= 0. The positive integer mj

can be determined by considering the homogeneous balancebetween the highest order derivatives and nonlinear termsappearing in (6).

By the generalized solutions of Eq. (8) we have

(G′

G) =

−λ2 +

√λ2 − 4µ

2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

),

λ2 − 4µ > 0,

−λ2 +

√4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

),

λ2 − 4µ < 0,

−λ2 + C2

C1 + C2ξ, λ2 − 4µ = 0,

(9)where C1, C2 are arbitrary constants.

Step 3. Substituting (7) into (6) and using (8),collecting all terms with the same order of (G′

G )together, the left-hand side of (6) is converted intoanother polynomial in (G′

G ). Equating each coefficient ofthis polynomial to zero, yields a set of algebraic equationsfor λ, µ, aj,i, i = 0, 1, ..., mj , j = 1, 2, ..., k.

Step 4. Solving the equations system in Step 3, and using(9), we can construct a variety of exact solutions for Eq. (4).

III. APPLICATION OF THE PROPOSED METHOD TO SPACEFRACTIONAL (2+1)-DIMENSIONAL BREAKING SOLITON

EQUATIONS

We consider the space fractional (2+1)-dimensional break-ing soliton equations

ut + a∂2α+βu∂x2αyβ + 4au∂αv

∂xα + 4a∂αu∂xα v = 0,

∂βu∂yβ = ∂αv

∂xα ,, (10)

where 0 < α, β ≤ 1. Eqs. (10) are variation of the following(2+1)-dimensional breaking soliton equations equations [30-33] {

ut + auxxy + 4auvx + 4auxv = 0,uy = vx.

(11)

For Eqs. (11), some periodic wave solutions, non-travelingwave solutions, and Jacobi elliptic function solutions werefound in [20-23]. But we notice no research has been paidfor Eqs. (10) so far. In the following, we will apply thedescribed method in Section 2 to Eqs. (10).

To begin with, we suppose u(x, y, t) = U(ξ), v(x, y, t) =V (ξ), where ξ = ct + k1

Γ(1 + α)xα + k2Γ(1 + β)yβ + ξ0,

k1, k2, c, ξ0 are all constants with k1, k2, c 6= 0.Then by use of (1) and the first equality in Eq. (3), we

obtain Dαx u = Dα

x U(ξ) = U ′(ξ)Dαx ξ = k1U

′(ξ), Dβy u =

Dβy U(ξ) = U ′(ξ)Dβ

y ξ = k2U′(ξ), ut = cU ′(ξ), and then

Eqs. (10) can be turned into{

cU ′ + ak21k2U

′′′ + 4ak1UV ′ + 4ak1V U ′ = 0,k2U

′ = k1V′. (12)

Suppose that the solution of Eqs. (12) can be expressedby

U(ξ) =m1∑i=0

ai(G′G )i,

V (ξ) =m2∑i=0

bi(G′G )i.

(13)

Balancing the order of U ′′′ and UV ′, U ′ and V ′ in (12) wehave m1 = m2 = 2. So

U(ξ) = a0 + a1(G′G )1 + a2(G′

G )2,

V (ξ) = b0 + b1(G′G )1 + b2(G′

G )2.(14)

Substituting (14) into (12), using Eq. (8) and collecting allthe terms with the same power of (G′

G ) together, equatingeach coefficient to zero, yields a set of algebraic equations.Solving these equations yields:

a0 = a0, a1 = −32k21λ, a2 = −3

2k21,

b0 = −8ak21k2µ + ak2

1k2λ2 + c + 4aa0k2

4ak1,

b1 = −32k1k2λ, b2 = −3

2k1k2,

where λ, µ, a0 are arbitrary constants.Substituting the result above into Eqs. (14), and combining

with (9) we can obtain the following exact solutions for Eqs.(10).

When λ2 − 4µ > 0, we obtain the hyperbolic functionsolutions:

u1(x, y, t) = a0 − 32k2

1λ[−λ2 +

√λ2 − 4µ

2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]

−32k2

1[−λ2 +

√λ2 − 4µ

2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]2,

v1(x, y, t) = −8ak21k2µ + ak2

1k2λ2 + c + 4aa0k2

4ak1

−32k1k2λ[−λ

2 +√

λ2 − 4µ2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]

−32k1k2[−λ

2 +√

λ2 − 4µ2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]2,

(15)

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where ξ = ct + k1Γ(1 + α)xα + k2

Γ(1 + β)yβ + ξ0.

In particular, if we take C2 = 0, then we obtain thefollowing solitary wave solutions, which are shown in Figs.1-2.

u2(x, y, t) = a0 + 38k2

1λ2

−38k2

1(λ2 − 4µ)[tanh

√λ2 − 4µ

2 ξ]2,

v2(x, y, t) = −8ak21k2µ + ak2

1k2λ2 + c + 4aa0k2

4ak1

+38k1k2λ

2 − 38k1k2(λ2 − 4µ)[tanh

√λ2 − 4µ

2 ξ]2,

When λ2 − 4µ < 0, we obtain the periodic functionsolutions:

u3(x, y, t) = a0 − 32k2

1λ[−λ2 +

√4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]

−32k2

1[−λ2 +

√4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]2,

v3(x, y, t) = −8ak21k2µ + ak2

1k2λ2 + c + 4aa0k2

4ak1

−32k1k2λ[−λ

2 +√

4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]

−32k1k2[−λ

2 +√

4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]2,

(16)where ξ = ct + k1

Γ(1 + α)xα + k2Γ(1 + β)yβ + ξ0.

When λ2 − 4µ = 0, we obtain the rational functionsolutions:

u4(x, y, t) = a0 − 32k2

1λ[−λ2 + C2

C1 + C2ξ]

−32k2

1[−λ2 + C2

C1 + C2ξ]2,

v4(x, y, t) = −8ak21k2µ + ak2

1k2λ2 + c + 4aa0k2

4ak1

−32k1k2λ[−λ

2 + C2C1 + C2ξ

]

−32k1k2[−λ

2 + C2C1 + C2ξ

]2,(17)

where ξ = ct + k1Γ(1 + α)xα + k2

Γ(1 + β)yβ + ξ0.

Remark 1. The established solutions above for the spacefractional (2+1)-dimensional breaking soliton equations arenew exact solutions so far in the literature.

IV. APPLICATION OF THE PROPOSED METHOD TOSPACE-TIME FRACTIONAL FOKAS EQUATION

We consider the space-time fractional Fokas equation

4∂2αq

∂tα∂xα1

− ∂4αq

∂x3α1 ∂xα

2

+∂4αq

∂x3α2 ∂xα

1

+ 12∂αq

∂xα1

∂αq

∂xα2

+12q∂2αq

∂xα1 ∂xα

2

− 6∂2αq

∂yα1 ∂yα

2

= 0, 0 < α ≤ 1. (18)

In [24], the authors solved Eq. (18) by a fractional Riccatisub-equation method, and obtained some exact solutions forit. Now we will apply the described method in Section 3 toEq. (18).

Suppose q(x, y, t) = U(ξ), where ξ = cΓ(1 + α) tα +

k1Γ(1 + α)xα

1 + k2Γ(1 + α)xα

2 + l1Γ(1 + α)yα

1 + l2Γ(1 + α)yα

2 +ξ0, k1, k2, l1, l2, c, ξ0 are all constants withk1, k2, l1, l2, c 6= 0. Then by use of (1) and the firstequality in Eq. (3), Eq. (18) can be turned into

4ck1U′′ − k3

1k2U(4) + k3

2k1U(4) + 12k1k2(U ′)2

+12k1k2UU ′′ − 6l1l2U′′ = 0. (19)

Suppose that the solution of Eq. (19) can be expressed by

U(ξ) =m∑

i=0

ai(G′

G)i, (20)

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where G = G(ξ) satisfies Eq. (8). By Balancing the orderbetween the highest order derivative term and nonlinear termin Eq. (19) we can obtain m = 2. So we have

U(ξ) = a0 + a1(G′

G) + a2(

G′

G)2. (21)

Substituting (21) into (19) and collecting all the terms withthe same power of (G′

G ) together, equating each coefficientto zero, yields a set of algebraic equations. Solving theseequations, yields:

a0 =k31k2λ

2 − k1k32λ

2 + 8k31k2µ− 8k1k

32µ− 4ck1 + 6l1l2

12k1k2,

a1 = λ(k21 − k2

2), a2 = k21 − k2

2.

Substituting the result above into Eq. (21), and combiningwith (9) we can obtain the following exact solutions to Eq.(18).

When λ2 − 4µ > 0,

q1(t, x1, x2, y1, y2) =

k31k2λ

2 − k1k32λ

2 + 8k31k2µ− 8k1k

32µ− 4ck1 + 6l1l2

12k1k2

+λ(k21 − k2

2)[−λ

2+

√λ2 − 4µ

2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]

+(k21 − k2

2)[−λ

2+

√λ2 − 4µ

2

(C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

)]2, (22)

where ξ = cΓ(1 + α) tα + k1

Γ(1 + α)xα1 + k2

Γ(1 + α)xα2 +

l1Γ(1 + α)yα

1 + l2Γ(1 + α)yα

2 + ξ0.

When λ2 − 4µ < 0,

q2(t, x1, x2, y1, y2) =

k31k2λ

2 − k1k32λ

2 + 8k31k2µ− 8k1k

32µ− 4ck1 + 6l1l2

12k1k2

+λ(k21 − k2

2)[−λ

2+

√4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]

+(k21 − k2

2)[−λ

2+

√4µ− λ2

2

(−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

)]2, (23)

where ξ = cΓ(1 + α) tα + k1

Γ(1 + α)xα1 + k2

Γ(1 + α)xα2 +

l1Γ(1 + α)yα

1 + l2Γ(1 + α)yα

2 + ξ0.In Fig. 3, the periodic function solution (23) is demon-

strated with some certain parameters.When λ2 − 4µ = 0,

q3(t, x1, x2, y1, y2) =

k31k2λ

2 − k1k32λ

2 + 8k31k2µ− 8k1k

32µ− 4ck1 + 6l1l2

12k1k2+

λ(k21−k2

2)[−λ

2+

C2

C1 + C2ξ]+(k2

1−k22)[−

λ

2+

C2

C1 + C2ξ]2,

(24)where ξ = c

Γ(1 + α) tα + k1Γ(1 + α)xα

1 + k2Γ(1 + α)xα

2 +l1

Γ(1 + α)yα1 + l2

Γ(1 + α)yα2 + ξ0.

Remark 2. As one can see, the established solutionsfor the space-time fractional Fokas equation above aredifferent from the results in [24], and are new exactsolutions so far to our best knowledge.

V. APPLICATION OF THE PROPOSED METHOD TOSPACE-TIME FRACTIONAL KAUP-KUPERSHMIDT

EQUATION

We consider the following space-time fractional Kaup-Kupershmidt equation

Dαt u + D5β

x u + 45u2Dβxu− 75

2Dβ

xuD2βx u

−15uD3βx u = 0, 0 < α, β ≤ 1, (25)

which is a variation of the following Kaup-Kupershmidtequation [34-36]:

ut + uxxxxx + 45uxu2 − 752

uxuxx − 15uuxxx = 0, (26)

To begin with, we suppose u(x, t) = U(ξ), where ξ =c

Γ(1 + α) tα + kΓ(1 + β)xβ + ξ0, k, c, ξ0 are all constants

with k, c 6= 0. Then by use of (1) and the first equality inEq. (3), Eq. (25) can be turned into

cU ′ + k5U (5) + 45kU2U ′ − 752

k3U ′U ′′ − 15k3UU ′′′ = 0.

(27)

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Suppose that the solution of Eq. (27) can be expressed by

U(ξ) =m∑

i=0

ai(G′

G)i. (28)

Balancing the order of U (5) and UU ′′′ in Eq. (27) we havem = 2. So

U(ξ) = a0 + a1(G′

G) + a2(

G′

G)2. (29)

Substituting (29) into (27), using Eq. (8) and collecting allthe terms with the same power of (G′

G ) together, equatingeach coefficient to zero, yields a set of algebraic equations.Solving these equations, yields:

Case 1:

a2 = 8k2, a1 = 8k2λ, a0 =23k2(λ2 + 8µ), k = k,

c = −11k5(−8λ2µ + 16µ2 + λ4),

g =1664

9k7µ3 +

1043

k7λ4µ− 4163

k7λ2µ2. (30)

Substituting the result above into Eq. (29) and combiningwith (9) we can obtain the following exact solutions to Eq.(25).

When λ2 − 4µ > 0,

u1(x, t) = −2k2λ2 + 2k2(λ2 − 4µ)

C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

2

+23k2(λ2 + 8µ), (31)

where ξ = −11k5(−8λ2µ + 16µ2 + λ4)Γ(1 + α) tα+ k

Γ(1 + β)xβ +ξ0.

In particular, when λ > 0, µ = 0, C1 6= 0, C2 = 0, wecan deduce the soliton solutions of the Kaup-Kupershmidtequation as follows:

u2(x, t) = 2k2λ2sech2(λξ

2) +

23k2λ2. (32)

When λ2 − 4µ < 0,

u3(x, t) = −2k2λ2 + 2k2(4µ− λ2)−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

2

+23k2(λ2 + 8µ), (33)

where ξ = −11k5(−8λ2µ + 16µ2 + λ4)Γ(1 + α) tα+ k

Γ(1 + β)xβ +ξ0.

When λ2 − 4µ = 0,

u4(x, t) = −2k2λ2 +8k2C2

2

(C1 + C2ξ)2+

23k2(λ2 + 8µ), (34)

where ξ = −11k5(−8λ2µ + 16µ2 + λ4)Γ(1 + α) tα+ k

Γ(1 + β)xβ +ξ0.

Case 2:

a2 = k2, a1 = k2λ, a0 =112

k2(λ2 + 4µ), k = k,

c = − 116

k5(−8λ2µ + 16µ2 + λ4),

g = −29k7µ3 − 1

24k7λ4µ +

16k7λ2µ2 +

1288

k7λ6.

Substituting the result above into Eq. (29) and combiningwith (9) we can obtain the following exact solutions to Eq.(25).

When λ2 − 4µ > 0,

u5(x, t) = −14k2λ2 +

14k2(λ2 − 4µ)

C1 sinh

√λ2 − 4µ

2ξ + C2 cosh

√λ2 − 4µ

2ξ

C1 cosh

√λ2 − 4µ

2ξ + C2 sinh

√λ2 − 4µ

2ξ

2

+112

k2(λ2 + 4µ), (35)

where ξ =− 1

16k5(−8λ2µ + 16µ2 + λ4)

Γ(1 + α) tα+ kΓ(1 + β)xβ+

ξ0.

In particular, when λ > 0, µ = 0, C1 6= 0, C2 = 0, wecan deduce the soliton solutions of the Kaup-Kupershmidtequation as follows:

u6(x, t) =14k2λ2sech2(

λξ

2) +

112

k2λ2. (36)

When λ2 − 4µ < 0,

u7(x, t) = −14k2λ2 +

14k2(4µ− λ2)

−C1 sin

√4µ− λ2

2ξ + C2 cos

√4µ− λ2

2ξ

C1 cos

√4µ− λ2

2ξ + C2 sin

√4µ− λ2

2ξ

2

+112

k2(λ2 + 4µ), (37)

where ξ =− 1

16k5(−8λ2µ + 16µ2 + λ4)

Γ(1 + α) tα+ kΓ(1 + β)xβ+

ξ0.

When λ2 − 4µ = 0,

u8(x, t) = −14k2λ2 +

k2C22

(C1 + C2ξ)2+

112

k2(λ2 +4µ), (38)

IAENG International Journal of Applied Mathematics, 43:3, IJAM_43_3_04

(Advance online publication: 16 August 2013)

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where ξ =− 1

16k5(−8λ2µ + 16µ2 + λ4)

Γ(1 + α) tα+ kΓ(1 + β)xβ+

ξ0.

Remark 3. The established solutions in Eqs. (31)-(38) are new exact solutions for the space-time fractionalKaup-Kupershmidt equation.

VI. CONCLUSION

We have applied the known (G’/G) method to solve thespace fractional (2+1)-dimensional breaking soliton equa-tions, the space-time fractional Fokas equation, and thespace-time fractional Kaup-Kupershmidt equation. Based oncertain fractional transformation, such fractional partial dif-ferential equations can be turned into ordinary differentialequations of integer order, the solutions of which can beexpressed by a polynomial in (G′

G ), where G satisfies theODE G′′(ξ) + λG′(ξ) + µG(ξ) = 0. With the aid ofmathematical software, a variety of exact solutions for thesefractional partial differential equations are obtained. Beingconcise and powerful, we note that this approach can also beapplied to solve other fractional partial differential equations.

ACKNOWLEDGMENT

The author would like to thank the anonymous reviewersvery much for their valuable suggestions on improving thispaper.

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IAENG International Journal of Applied Mathematics, 43:3, IJAM_43_3_04

(Advance online publication: 16 August 2013)

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