Exact Solutions For Fractional Partial Differential Equations By A New Generalized Fractional Sub-equation Method

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1 Exact Solutions For Fractional Partial Differential Equations y A New eneralized Fractional Sub-equation Method QINHUA FEN Shandong University of Technology School of Science Zhangzhou Road 12, Zibo, CHINA fqhua@sina.com Abstract: In this paper, combining with a fractional complex transformation, we propose a new generalized fractional sub-equation method named fractional ( D ) method to seek exact solutions for fractional partial differential equations. This method is the fractional version of the improved ( /) method. ased on this method, some new exact solutions for the space-time fractional (2+1)-dimensional breaking soliton equations and the space-time fractional Fokas equation are successfully found. Under some special cases, we get solitary wave solutions for them. Key Words: Fractional sub-equation method; Fractional partial differential equations; Exact solutions; Fractional complex transformation; Fractional breaking soliton equations; Fractional Fokas equation. 1 Introduction Fractional differential equations are generalizations of classical differential equations of integer order. The main advantage of fractional differential equations in comparison with classical differential equations of integer order mainly lies in that fractional derivative is more useful in describing the memory and hereditary properties of materials and processes. In recent decades, fractional differential equations have gained much attention as they are widely used to describe various complex phenomena in many fields such as the fluid flow, signal processing, control theory, systems identification, biology and other areas. The mathematical modeling and simulation of systems and processes leads to the research of the theory for fractional differential equations. Many authors have investigated some aspects of fractional differential equations so far. Among these investigations for fractional differential equations, research for seeking numerical solutions and exact solutions of fractional differential e- quations has been a hot topic, which can also provide valuable reference for other related research. Many powerful and efficient methods have been proposed to obtain numerical solutions and exact solutions of fractional differential equations so far. For example, these methods include the ( /) method [1-3 the Homo- Separation of Variables method [4 the variational iterative method [5-7 the Adomian decomposition method [8,9 the fractional sub-equation method [10-12 the homotopy perturbation method [13-15] and so on. ased on these methods, a variety of fractional differential equations have been investigated. In this paper, using a new fractional sub-equation, we propose a new generalized fractional sub-equation method named fractional ( D ) method to seek exact solutions for fractional partial differential equations in the sense of modified Riemann-Liouville derivative. This method can be seen as the fractional version of the improved ( /) method [16]. The modified Riemann-Liouville fractional derivative, defined by Jumarie in [17-20 has many excellent characters in handling with many fractional calculus problems. Many authors have investigated various applications of the modified Riemann-Liouville fractional derivative (for example, see [21-23]). We now list the definition for it as follows. Definition 1 The modified Riemann-Liouville derivative of order is defined by the following expression: d t (t ) dt 0 Γ(1 ) (f() f(0))d, 0 < < 1, Dt f(t) = (f (n) (t)) ( n), n < n+1, n 1. Definition 2 The Riemann-Liouville fractional integral of order on the interval [0, t] is defined by I f(t) = t 0 (t s) 1 f(s)ds. Γ() E-ISSN: Volume 15, 2016

2 Some important properties for the modified Riemann-Liouville derivative and fractional integral are listed as follows (see [17, Eqs. (3.10)-(3.13) and see also [10-12,21-23]) (the interval concerned below is always defined by [0,t]): Dt t r = Γ(1+r) Γ(1+r ) tr ; (1) Dt (f(t)g(t))=g(t)d t f(t)+f(t)d t g(t);(2) Dt f[g(t)] = f g [g(t)]d t g(t) = Dg f[g(t)](g (t)) ; (3) I (Dt f(t)) = f(t) f(0) (4) I (g(t)dt f(t)) = f(t)g(t) f(0)g(0) I (f(t)dt g(t)). (5) The rest of this paper is organized as follows. In Section 2, we give the description of the fractional ( D ) method for solving fractional partial differential equations. Then in Section 3 we apply this method to establish exact solutions for the space-time fractional (2+1)-dimensional breaking soliton equations and the space-time fractional Fokas equation. Some conclusions are presented at the end of the paper. 2 Description of the fractional ( D ) method In this section we give the description of the ( D ) method for solving fractional partial differential equations. Suppose that a fractional partial differential equation, say in the independent variables t,x 1,x 2,...,x n, is given by P(u 1,..u k,dt u 1,..,Dt u k,dx 1 u 1,..,Dx 1 u k,... Dx n u 1,..,Dx n u k,dt 2 u 1,...,Dt 2 u k,dx 2 1 u 1,...) = 0, (6) where u i = u i (t,x 1,x 2,...,x n ), i = 1,...,k are unknown functions, P is a polynomial in u i and their various partial derivatives including fractional derivatives. Step 1. Suppose that u i (t,x 1,x 2,...,x n ) = U i (), xi = ct+k 1 x 1 +k 2 x k n x n + 0. (7) Then by the second equality in Eq. (3), Eq. (7) can be turned into the following fractional ordinary differential equation with respect to the variable : P(U 1,...,U k,c D U 1,...,c D U k,k 1 D U 1,..., k 1 D U k,...,k n D U 1,...,k n D U k,c 2 D 2 U 1,...,c 2 D 2 U k,k 2 1 D 2 U 1,...) = 0. (8) Step 2. Suppose that the solution of (8) can be expressed by a polynomial in ( D ) as follows: m j U j () = a j,i ( D )i, j = 1,2,...,k, (9) i=0 where a j,i, i = 0,1,...,m, j = 1,2,...,k are constants to be determined later with a j,m 0, the positive integer m can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (8), = () satisfies the following fractional ordinary differential equation: AD 2 () D () C(D ())2 E 2 () = 0, (10) where D () denotes the modified Riemann- Liouville derivative of order for () with respect to, and A,, C, E are real parameters. In order to obtain the general solutions for Eq. (10), we suppose () = H(η), and a nonlinear fractional complex transformation η = Γ(1+). Then by Eq. (1) and the first equality in Eq. (3), Eq. (10) can be turned into the following second ordinary differential equation: AHH (η) HH (η) C(H (η)) 2 EH 2 (η) = 0. (11) Denote 1 = 2 +4E(A C), 2 = E(A C). y the general solutions of Eq. (11) [16] we have the following expressions for H (η) H(η) : When 0, 1 > 0: H (η) H(η) = [ C 1 η +C 1 η 1 η h +C 1 η h When 0, 1 < 0: H (η) H(η) = + 1 [ C 1 η +C 1 η 1 η +C 1 η When 0, 1 = 0: (12) (13) H (η) H(η) = + C 2 C 1 +C 2 η, (14) E-ISSN: Volume 15, 2016

3 When = 0, 2 > 0: H η (η) H(η) = (A C) [C (A C) +C η (A C) η h (A C) +C η h (A C) When = 0, 2 < 0: H η (η) H(η) = (A C) [ C (A C) +C η (A C) η (A C) +C η (A C) (15) (16) Since D () = D H(η) = H (η)d η = H (η), we obtain the following expressions for D () () : When 0, 1 > 0: D () () = Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) When 0, 1 < 0: D () () = + 1 Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 Γ(1+) When 0, 1 = 0: (17) (18) D () () = + C 2Γ(1+) C 1 Γ(1+)+C 2, (19) When = 0, 2 > 0: D () () = (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) When = 0, 2 < 0: D () () = (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) (A C)Γ(1+) +C (A C)Γ(1+) (20) (21) Step 3. Substituting (9) into (8) and using (10), collecting all terms with the same order of ( D ) together, the left-hand side of (8) is converted into another polynomial in( D ). Equating each coefficient of this polynomial to zero, yields a set of algebraic equations for a j,i, i = 0,1,...,m, j = 1,2,...,k. Step 4. Solving the equations system in Step 3, and using (17)-(21), we can construct a variety of exact solutions for Eq. (6). Remark 3 If we set = 1 in Eq. (10), then it becomes A () () C( ()) 2 E 2 () = 0, which is the foundation of the improved ( /) method [16] for solving partial differential e- quations (PDEs). So the fractional ( D ) method is the extension of the improved ( /) method to fractional case. 3 Applications of the fractional ( D ) method to some fractional partial differential equations 3.1 Space-time fractional (2+1)-dimensional breaking soliton equations Consider the space-time fractional (2+1)-dimensional breaking soliton equations [24] u t +a 3 u +4au v x 2 y x +4a u x v = 0, u y = v x, (22) where 0 < 1. In [24 the authors obtained some new exact solutions for Eqs. (22) by use of a fractional sub-equation method, which is based on the following fractional sub-equation: D 2 ()+λd ()+µ() = 0, (23) In the following, we will apply the fractional ( D ) method described in Section 3 to solve E- qs. (22). To begin with, we suppose u(x,y,t) = U(), v(x,y,t) = V(), where = k 1 x+k 2 y+ct+ 0, k 1, k 2, c, 0 are all constants with k 1, k 2, c 0. Then by use of the second equality in Eq. (4), Eqs. (22) can be turned into c D U +ak2 1 k 2 D3 U +4ak 1 UD V +4ak 1 VD U = 0, k 2 D U = k 1 D V. (24) Suppose that the solution of Eqs. (24) can be expressed by U() = m 1 i=0 V() = m 2 i=0 a i ( D )i, b i ( D )i., (25) E-ISSN: Volume 15, 2016

4 alancing the order of D 3 U and UD V, D U and D V in (24) we have m 1 = m 2 = 2. So U() = a 0 +a 1 ( D )1 +a 2 ( D )2, V() = b 0 +b 1 ( D )1 +b 2 ( D )2. (26) Substituting (26) into (24), using the properties (1)-(3) and Eq. (10), collecting all the terms with the same power of ( D ) together, equating each coefficient to zero, yields a set of algebraic equations. Solving these equations yields: a 0 = a 0, a 1 = 3(A C)k2 1, a 2 = 2, b 0 = ak2 1 k ak1 2k 2 ae(c A)+c A 2 +4aa 0 A 2 k2, b 1 = 3(A C)k 1 k 2, b 2 = k 2 2A, 2 where a 0 is an arbitrary constant. Substituting the result above into Eqs. (26), and combining with (17)-(21) we can obtain the following exact solutions to Eqs. (22). Family 1: when 0, 1 > 0: u 1 (x,y,t) = a 0 + 3(A C)k1 2 Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) 2 Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) v 1 (x,y,t) = 2, ak1 2k ak1 2k 2 ae(c A)+c A 2 +4aa 0 A 2 k2 + 3(A C)k 1 k 2 Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) k 2 Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) 2, (27) (28) wherec 1, C 2 are arbitrary constants, = k 1 x+k 2 y+ ct + 0. In particular, if we set C 2 = 0 in (27)-(28), then we obtain the following solitary solutions: u 2 (x,y,t) = a 0 + 3(A C)k tanh tanh v 2 (x,y,t) = 1 Γ(1+) } 1 Γ(1+) }2, ak2 1 k ak 2 1 k 2 ae(c A)+c A 2 +4aa 0 A 2 k 2 4aA 2 k 1 + 3(A C)k 1 k tanh 3(C2 +A 2 2CA)k 1 k tanh 1 Γ(1+) } Family 2: when 0, 1 < 0: 1 Γ(1+) }2. (29) (30) u 3 (x,y,t) = a 0 + 3(A C)k Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 Γ(1+) Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 2, Γ(1+) v 3 (x,y,t) = ak2 1 k ak1 2k 2 ae(c A)+c A 2 +4aa 0 A 2 k2 + 3(A C)k 1 k Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 Γ(1+) k 2 2A Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 2, Γ(1+) (31) (32) where C 1, C 2 are arbitrary constants, = k 1 x+k 2 y +ct+ 0. Family 3: when 0, 1 = 0: u 4 (x,y,t) = a 0 + 3(A C)k2 1 + C 2Γ(1+) C 1 Γ(1+)+C 2 } 2 + C 2Γ(1+) C 1 Γ(1+)+C 2 } 2, (33) E-ISSN: Volume 15, 2016

5 v 4 (x,y,t) = ak2 1 k ak 2 1 k 2 ae(c A)+c A 2 +4aa 0 A 2 k 2 4aA 2 k 1 + 3(A C)k 1 k 2 3(C2 +A 2 2CA)k 1 k C 2Γ(1+) C 1 Γ(1+)+C 2 } 2, C 2Γ(1+) C 1 Γ(1+)+C 2 } (34) where C 1, C 2 are arbitrary constants, and = k 1 x+k 2 y +ct+ 0. Family 4: when = 0, 2 > 0: u 5 (x,y,t) = a 0 + 3(A C)k2 1 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) v 5 (x,y,t) = 2, ak2 1 k ak1 2k 2 ae(c A)+c A 2 +4aa 0 A 2 k2 + 3(A C)k 1 k 2 2A 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) k 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) 2, (35) (36) where C 1, C 2 are arbitrary constants, = k 1 x+k 2 y +ct+ 0. Family 5: when = 0, 2 < 0: u 6 (x,y,t) = a 0 + 3(A C)k2 1 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) (A C)Γ(1+) +C (A C)Γ(1+) 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) (A C)Γ(1+) +C (A C)Γ(1+) 2, (37) v 6 (x,y,t) = ak2 1 k ak1 2k 2 ae(c A)+c A 2 +4aa 0 A 2 k2 + 3(A C)k 1 k 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) (A C)Γ(1+) +C (A C)Γ(1+) k 2 2A 2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) 2, (A C)Γ(1+) +C (A C)Γ(1+) (38) wherec 1, C 2 are arbitrary constants, = k 1 x+k 2 y+ ct Space-time fractional Fokas equation Consider the space-time fractional Fokas equation 4 2 q t x 1 +12q 2 q x 1 x 2 4 q x 3 1 x q y 1 y q q x 1 x 2 = 0, 0 < q x 3 2 x 1 (39) In [25 the authors solved Eq. (37) by a fractional Riccati sub-equation method. ased on the this method, some exact solutions for it were obtained. Now we will apply the fractional ( D ) method described in Section 3 to solve Eq. (37). Suppose q(t,x 1,x 2,y 1,y 2 ) = U(), where = ct+k 1 x 1 +k 2 x 2 +l 1 y 1 +l 2 y 2 + 0,k 1, k 2, l 1, l 2, c, 0 are all constants with k 1, k 2, l 1, l 2, c 0. Then by use of the second equality in Eq. (3), Eq. (37) can be turned into 4c k 1 D2 U k3 1 k 2 D4 U +k3 2 k 1 D4 U+ 12k 1 k 2 (D U)2 +12k 1 k 2 UD2 U 6l 1 l 2 D2 U = 0. (40) Suppose that the solution of Eq. (38) can be expressed by U() = m i=0 a i ( D )i, (41) where = () satisfies Eq. (10). y alancing the order between the highest order derivative term and nonlinear term in Eq. (38), we can obtain m = 2. So we have U() = a 0 +a 1 ( D )+a 2( D )2. (42) Substituting (40) into (38), using the properties (1)-(3) and Eq. (10), collecting all the terms with the E-ISSN: Volume 15, 2016

6 same power of ( D ) together, equating each coefficient to zero, yields a set of algebraic equations. Solving these equations, yields: a 0 = 4A2 c k 1 6A2 l 1 l 2 +8AEk 2 k3 12k 1 k 2 A2 1 8AEk3 2 k 1 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k, 1 k 2 A2 a 1 = (Ak3 a 2 = A2 k 3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C), k1 k 2 A2 1 k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3k. k1 k 2 A2 Substituting the result above into Eq. (40), and combining with (17)-(21) we can obtain the following exact solutions to Eq. (37). Family 1: when 0, 1 > 0: q 1 (t,x 1,x 2,y 1,y 2 ) = 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k1 k 2 A2 Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3k k 1 k 2 A2 } Γ(1+) +C 1 Γ(1+) 1 h Γ(1+) +C 1 h Γ(1+) 2, (43) wherec 1, C 2 are arbitrary constants, = ct+k 1 x 1 + k 2 x 2 +l 1 y 1 +l 2 y In particular, if we setc 2 = 0 in (41), then we obtain the following solitary solutions: q 2 (t,x 1,x 2,y 1,y 2 ) = 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k1 k 2 A2 tanh 1 Γ(1+) } + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3k k 1 k 2 A2 } + 1 tanh Family 2: when 0, 1 < 0: q 3 (t,x 1,x 2,y 1,y 2 ) = 1 Γ(1+) }2. 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k1 k 2 A2 + 1 Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 Γ(1+) + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3k k 1 k 2 A2 } + 1 Γ(1+) +C 1 Γ(1+) 1 Γ(1+) +C 1 Γ(1+) 2, (44) where C 1, C 2 are arbitrary constants, = ct+k 1 x 1 +k 2 x 2 +l 1 y 1 +l 2 y Family 3: when 0, 1 = 0: q 4 (t,x 1,x 2,y 1,y 2 ) = 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k1 k 2 A2 + C 2Γ(1+) C 1 Γ(1+)+C 2 } + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3k k 1 k 2 A2 } + C 2Γ(1+) C 1 Γ(1+)+C 2 } 2, (45) where C 1,C 2 are arbitrary constants, and = ct+k 1 x 1 +k 2 x 2 +l 1 y 1 +l 2 y E-ISSN: Volume 15, 2016

7 Family 4: when = 0, 2 > 0: q 5 (t,x 1,x 2,y 1,y 2 ) = 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k1 k 2 A2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3 k k } 1 k 2 A2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) h (A C)Γ(1+) +C h (A C)Γ(1+) 2, (46) where C 1, C 2 are arbitrary constants, and = ct+k 1 x 1 +k 2 x 2 +l 1 y 1 +l 2 y Family 5: when = 0, 2 < 0: q 6 (t,x 1,x 2,y 1,y 2 ) = 4A2 c k1 6A2 l1 l 2 +8AEk 2 k3 1 8AEk3 2 k 1 12k1 k 2 A2 k3 1 k 2 2 8k1 3k 2 CE+2 k2 3k 1 +8Ek3 2 k 1 C 12k1 k 2 A2 (Ak3 1 k 2 Ak3 2 k 1 k3 1 k 2 C+k3 2 k 1 C) k 1 k 2 A2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) (A C)Γ(1+) +C (A C)Γ(1+) + A2 k1 3k 2 A2 k2 3k 1 2Ak3 1 k 2 C k1 k 2 A2 + 2Ak3 2 k 1 C+k3 1 k 2 C2 k2 3 k } k1 k 2 A2 (A C) [ C (A C)Γ(1+) +C (A C)Γ(1+) 2, (A C)Γ(1+) +C (A C)Γ(1+) (47) wherec 1, C 2 are arbitrary constants, = ct+k 1 x 1 + k 2 x 2 +l 1 y 1 +l 2 y Conclusions y use of a new fractional sub-equation, we have proposed a new fractional ( D ) method to seek exact solutions for fractional partial differential equations. As for applications, we apply this method to solve the space-time fractional (2+1)-dimensional breaking soliton equations and the space-time fractional Fokas equation. Abundant new exact solutions including hyperbolic function solutions, trigonometric function solutions, and rational function solutions for the two e- quations have been successfully found. This method is based on the homogeneous balancing principle and the fractional complex transformation. eing concise and powerful, it can be applied to solve other fractional partial differential equations. Acknowledgements This work is partially supported by Natural Science Foundation of Shandong Province (China) (ZR2013AQ009). The author would like to thank the referees very much for their valuable suggestions on improving this paper. References: [1] N. Shang,. Zheng, Exact Solutions for Three Fractional Partial Differential Equations by the ( /) Method, IAEN International Journal of Applied Mathematics, 43(3), 2013, pp [2]. Zheng, Exact Solutions for Some Fractional Partial Differential Equations by the ( /) Method, Math. Pro. Engi., 2013, article ID: , 2013, pp [3]. Zheng, ( /)-Expansion Method for Solving Fractional Partial Differential Equations in the Theory of Mathematical Physics, Commun. Theor. Phys., 58, 2012, pp [4] A. Karbalaie and M. M. Montazeri, Application of Homo-separation of variables method on nonlinear system of PDEs, WSEAS Transactions on Mathematics, 13, 2014, pp [5] J.H. He, A new approach to nonlinear partial differential equations, Commun. Nonlinear Sci. Numer. Simul., , pp [6] E. M. E. Zayed and H. M. Abdel Rahman, On using the He s polynomials for solving the nonlinear coupled evolution equations in mathematical physics, WSEAS Transactions on Mathematics, 11(4), 2012, pp [7] S. uo and L. Mei, The fractional variational iteration method using He s polynomials, Phys. Lett. A, 375,2011, pp [8] A. M. A. El-Sayed and M. aber, The Adomian decomposition method for solving partial differential equations of fractal order in finite domains, Phys. Lett. A, 359, 2006, pp [9] A. M. A. El-Sayed, S. H. ehiry and W. E. Raslan, Adomian s decomposition method for solving an intermediate fractional advectiondispersion equation, Comput. Math. Appl. 59, 2010, pp E-ISSN: Volume 15, 2016

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