Exp-Function Method and Fractional Complex Transform for Space-Time Fractional KP-BBM Equation

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1 Commun. Theor. Phys. 68 ( Vol. 68, No. 2, August 1, 2017 Ex-Function Method and Fractional Comlex Transform for Sace-Time Fractional KP-BBM Equation Ozkan uner Cankiri Karatekin University, Faculty of Economics and Administrative Sciences, Deartment of International Trade, Cankiri, Turkey (Received March 16, 2017; revised manuscrit received Aril 12, 2017 Abstract In the resent article, He s fractional derivative, the ansatz method, the ( /-exansion method, and the ex-function method are used to construct the exact solutions of nonlinear sace-time fractional Kadomtsev Petviashvili Benjamin Bona Mahony (KP-BBM. As a result, different tyes of exact solutions are obtained. Also we have examined the relation between the solutions obtained from the different methods. These methods are an efficient mathematical tool for solving fractional differential equations (FDEs and it can be alied to other nonlinear FDEs. PACS numbers: Jr, Wz, Yv DOI: / /68/2/149 Key words: ansatz method, ex-function method, He s fractional derivative, ( /-exansion method, sacetime fractional KP-BBM equation 1 Introduction It is worth nothing that the standard mathematical models of integer-order derivatives, including nonlinear models, do not work adequately in many cases. For this reason, many henomena in alied mathematics and other science can be described successfully by using the theory of derivatives of fractional order. The analysis and alications of fractional differential equations are object of an increasing interest in many different fields such as control theory, mechanics, engineering, hysics, electricity, chemistry, biology, economics, and signal image rocessing. [13] A variety of owerful methods have been used to solve FDEs, such as Adomian s decomosition method (ADM, the ( /-exansion method, the first integral method, the kudryashov method, the ex-function method, the rojective Riccati equation method, the fractional subequation method, the functional variable method, the modified trial equation method, the simlest equation method and so on. [423] In this aer, our aim is to aly the ex-function method, the ansatz method and the ( / exansion method for obtaining solution of the KP-BBM equation for the first time. These methods are an effective and owerful mathematical tool in finding exact solutions of nonlinear sace time fractional KP-BBM equation, which has very imortant alication areas in mathematical hysics. For examle, in the case of an offshore structure, the fluid flow is considered, relatively, unbounded. So, it will exert ressure forces around this structure. For this reason researchers are interested to one of the bidirectional traveling waves models, such as KP-BBM equation. [24] Corresonding author, ozkanguner@karatekin.edu.tr c 2017 Chinese Physical Society and IOP Publishing Ltd The He s fractional derivative of order α is given as [25] Dwu α 1 d n = Γ(n α dw n w t 0 (ξ w nα1 (u 0 (ξ u(ξdξ, (1 u 0 (ξ is a known function. The He s fractional derivative rovides the below roerties, which we will use in Sec. 3. D α wc = 0, (2 D α w{αf(w + βg(w} = αd α wf(w + βd α wg(w, (3 α, β, and c are constants. 2 Descrition of the Methods for FDEs Firstly we consider the following (2+1-dimensional nonlinear sace-time FDE of the tye H(u, D α t u, D α x u, D α y u, D 2α t u, D α t D α x u, D α t D α y u, D 2α x u,... = 0, 0 < α 1, (4 u is an unknown function, and H is a olynomial of u and its artial fractional derivatives. The fractional comlex transform was roosed by He and Li to convert a fractal sace (time to its continous artner. The hysical basis was illustrated by his and other colleageus recent review articles. [2633] The fractional comlex transform [34] u(x, y, t = U(ξ, (5 Γ(1 + α + byα Γ(1 + α ctα Γ(1 + α, (6 a, b, and c are nonzero constants reduce (4 to an integer ordered ODE. htt:// htt://ct.it.ac.cn

2 150 Communications in Theoretical Physics Vol. 68 as One should note that the chain rule can be calculated D α t u = σ t du dξ Dα t ξ, D α x u = σ x du dξ Dα x ξ, D α y u = σ y du dξ Dα y ξ, (7 σ t, σ x and σ y are called the sigma indexes see Refs. [35 36], and we can take σ t = σ x = σ y = L, L is a constant. Substituting Eq. (5 with Eqs. (1 and (7 into Eq. (4, we can rewrite Eq. (4 in the following nonlinear ODE; P (U, U, U, U,... = 0, (8 the rime denotes the derivation with resect to ξ. Now we consider three different methods. 2.1 Ansatz Method The ansatz method is given as follows: u(x, y, t = Asech ξ (9 is the form the solution in looked for and Γ(1 + α + byα Γ(1 + α ctα Γ(1 + α (10 is the fractional comlex transform ; A, a, b are free arameters of the soliton and c is the velocity of the soliton. The exonent is unknown at this oint and will be determined later. From the ansatz given above with two equalities, it is ossible to obtain necessary derivatives. Then, the obtained derivatives are substituted in the equation and we collect all terms with the same order of necessary terms. Then by equating each coefficient of the resulting olynomial to zero, we obtain a set of algebraic equations for; A, a, b and c. Finally solving the system of equations we can get exact solution of Eq. (4. Likewise, is obtained balancing between the exonents of the sech and tanh functions or multilications of these functions. [3742] 2.2 ( /-Exansion Method We assume that the solution of Eq. (8 can be exressed in the finite series exansion n ( i U( a i, an 0, (11 i=0 a i (i = 0, 1, 2,..., n and n are constants, (ξ is the general solution of the the following second order LODE (ξ + λ (ξ + µ( 0, (12 in which λ and µ are constants to be secified later. The ositive integer n can be determined easily by considering the homogeneous balance between the highest order derivative term and the nonlinear term aearing in Eq. (8. By substituting Eq. (11 into Eq. (8 and using Eq. (12 then setting the coefficients of ( / to zero, we obtain a set of equations system for a i (i = 0, 1, 2,..., n, λ, µ, a, b, and c. Next ste by solving the equations system and substituting a i (i = 0, 1, 2,..., n, λ, µ, a, b, c and the general solutions of Eq. (12 into Eq. (11, we can find traveling wave solutions of Eq. (4. [4344] 2.3 Ex-Function Method By the ex-function method, [45] we assume that the wave solution of Eq. (8 can be exressed as U( a c ex[cξ] + + a d ex[dξ] b ex[ξ] + + b q ex[qξ], (13, q, c, and d are ositive integers, which will be further determined, a n and b m are unknown constants. [4647] The ex-function method was orginally roosed to solve PDEs. When this method was successfully extended to fractional calculus [48] by He in 2013, it became an effective tool for fractional differential equations, see Refs. [49 52]. 3 Exact Solutions of Sace-Time Fractional KP-BBM Equation In this section, we aly the above methods to seek exact solutions of the following sace-time fractional KP- BBM equation: [53] Dx α [Dt α u + Dx α u Dx α u 2 qdt α (Dx 2α u] + rdy 2α u = 0, 0 < α 1, (14 α is a arameter describing the order of the fractional sace-time derivative and, q, r are constants. This roblem is within context of PDEs in multile sace dimension. Feng and Meng used the fractional Jacobi ellitic equation-based sub-equation method to obtain a variety of exact solutions in the forms of the Jacobi ellitic functions. When α = 1 Eq. (14 becomes the known KP- BBM equation of integer order. For our urose, we use the transformations Eqs. (5 and (6 and by twice integrating and setting the constants of integration to zero, Eq. (14 reduced into following ODE (a 2 ac + rb 2 U a 2 U 2 + qa 3 cl 2 U = 0, (15 U = du/dξ. 3.1 Exact Solutions by Ansatz Method To obtain the bright soliton solution of Eq. (15, the solitary wave ansatz admits the use of the assumtion, U( Asech s ξ, (16 Γ(1 + α + byα Γ(1 + α ctα Γ(1 + α, (17 which a, b and A are constant coefficients and c is the velocity of the soliton. From the ansatz (16 (17, we obtain d 2 U(ξ dξ 2 = As 2 sech s ξ As(s + 1sech s+2 ξ, (18

3 No. 2 Communications in Theoretical Physics 151 U 2 ( A 2 sech 2s ξ. (19 Thus, substituting the ansatz (16 (19 into Eq. (15, yields to (a 2 ac + rb 2 Asech s ξ a 2 A 2 sech 2s ξ + qa 3 cl 2 As 2 sech s ξ qa 3 cl 2 As(s + 1sech s+2 0. (20 Now, from Eq. (20, equating the exonents s + 2 and 2s leads to s + 2 = 2s, s = 2. (21 From Eq. (20, setting the coefficients of sech s+2 ξ and sech 2s ξ terms to zero, we obtain a 2 A 2 qa 3 cl 2 As(s + 1 = 0, (22 by using Eq. (21 and after some calculations, we have A = 6qacL2. (23 We find, from setting the coefficients of sech ξ terms in Eq. (20 to zero (a 2 ac + rb 2 A + qa 3 cl 2 As 2 = 0, (24 also we get c = a2 + rb 2 a 4qa 3 L 2. (25 Thus finally, the 1-soliton solution of Eq. (14 is given by: u(x, y, t = 6qacL2 sech 2( ax α Γ(1 + α + byα Γ(1 + α (a 2 + rb 2 t α (a 4qa 3 L 2 Γ(1 + α. ( Exact Solutions by ( /-Exansion Method Balancing the linear term of highest order U with the highest order nonlinear term U 2 we can obtain n + 2 = 2n, n = 2. (27 So solutions of Eq. (15 can be exressed by a olynomial in ( / as follows: ( ( 2 U( a 0 + a 1 + a 2, a2 0. (28 By using Eqs. (12 and (28 we have ( U 4 3 ( 6a 2 + (2a1 + 10a 2 λ( + (8a2 µ + 3a 1 λ + 4a 2 λ 2 ( ( + (6a 2 λµ + 2a 1 µ + a 1 λ 2 ( U 2 ( a 2 4 ( 3 ( 2 ( 2 + 2a1 a 2 + 2a0 a 2 + a 2 2 ( 1 + 2a0 a 1 + 2a 2 µ 2 + a 1 λµ, ( a 2 0. (30 Substituting Eqs. (28 (30 into Eq. (15, collecting the coefficients of ( / i (i = 0,..., 4 and setting it to zero we obtain the algebraic system and solving this system by Male yields Case 1 a 0 = a 2 = 6(a 2 + rb 2 ql 2 µ [1 + a 2 ql 2 (4µ λ 2 ], a 6(a 2 + rb 2 ql 2 λ 1 = [1 + a 2 ql 2 (4µ λ 2 ], 6(a 2 + rb 2 ql 2 [1 + a 2 ql 2 (4µ λ 2 ], a = a, b = b, c = a 2 + rb 2 a[1 + a 2 ql 2 (4µ λ 2 ], (31 λ and µ are arbitrary constants. By using Eq. (31, exression (28 can be written as 6(a 2 + rb 2 ql 2 µ U( [1 + a 2 ql 2 (4µ λ 2 ] + 6(a 2 + rb 2 ql 2 λ ( 6(a 2 + rb 2 ql 2 ( 2 [1 + a 2 ql 2 (4µ λ 2 +, (32 ] [1 + a 2 ql 2 (4µ λ 2 ] Case 2 λ and µ are arbitrary constants and λ 2 4µ 0. a 0 = cl(λ2 + 2µ q 4µ λ 2, a 1 = 6λcL q 4µ λ 2, a 2 = 6cL q 4µ λ 2, a = 1 q ql 4µ λ 2, b = 1 q ql 4µ λ 2, c = c, (33

4 152 Communications in Theoretical Physics Vol. 68 By using Eq. (33, we get U( cl(λ2 + 2µ q 4µ λ 2 6λcL q ( 4µ λ 2 6cL q ( 2. (34 4µ λ 2 Substituting general solutions of Eq. (12 into Eq. (32 we get solutions of this equation as follows: When λ 2 4µ > 0, U 1 ( 3qL2 (a 2 + rb 2 (4µ λ 2 { ( C1 sinh 1 [1 + a 2 ql 2 (4µ λ λ2 4µξ + C 2 cosh 1 2 λ2 4µξ 2 }, (35 ] C 1 cosh 2 1 λ2 4µξ + C 2 sinh 2 1 λ2 4µξ When λ 2 4µ < 0, U 2 ( 3qL2 (a 2 + rb 2 (4µ λ 2 [1 + a 2 ql 2 (4µ λ 2 ] Γ(1 + α + byα Γ(1 + α (a 2 + rb 2 t α a[1 + a 2 ql 2 (4µ λ 2 ]Γ(1 + α. { ( C1 sin µ λ2 ξ + C 2 cos 1 2 4µ λ2 ξ C 1 cos 2 1 4µ λ2 ξ + C 2 sin 2 1 4µ λ2 ξ Γ(1 + α + byα Γ(1 + α (a 2 + rb 2 t α a[1 + a 2 ql 2 (4µ λ 2 ]Γ(1 + α. 2 }, (36 In articular, if C 1 0, C 2 = 0, λ > 0, µ = 0, then U 1 and U 2 become u 1 (x, y, t = 3qL2 λ 2 (a 2 + rb 2 (a 2 qλ 2 L 2 1 sech2( λax α 2Γ(1 + α + λbxα 2Γ(1 + α λ(a 2 + rb 2 t α 2a[1 a 2 qλ 2 L 2, (37 ]Γ(1 + α Substituting general solutions of Eq. (12 into Eq. (34 also we get solutions When λ 2 4µ > 0, U 3 ( ± cl q(4µ λ2 3cL ( C1 sinh 1 q(4µ 2 λ2 4µξ + C 2 cosh 1 2 λ2 4µξ 2 λ2, (38 C 1 cosh 2 1 λ2 4µξ + C 2 sinh 2 1 λ2 4µξ 1 q x α ql 4µ λ 2 Γ(1 + α 1 q y α ql 4µ λ 2 Γ(1 + α ctα Γ(1 + α. When λ 2 4µ < 0, U 4 ( ± cl q(4µ λ2 3cL ( C1 sin 1 q(4µ 2 4µ λ2 ξ + C 2 cos 1 2 4µ λ2 ξ 2 λ2, (39 C 1 cos 2 1 4µ λ2 ξ + C 2 sin 2 1 4µ λ2 ξ 1 q x α ql 4µ λ 2 Γ(1 + α 1 q y α ql 4µ λ 2 Γ(1 + α ctα Γ(1 + α. In articular, if C 1 0, C 2 = 0, λ > 0, µ = 0, then U 3 becomes { ( 1 3tanh 2 and U 4 becomes u 2 (x, y, t = ± clλ q u 3 (x, y, t = ± clλ q {1 + 3tanh 2 ( x α 2 qlγ(1 + α y α 2 qlγ(1 + α λctα 2Γ(1 + α }, (40 x α 2 qlγ(1 + α y α } 2 qlγ(1 + α λctα. (41 2Γ(1 + α 3.3 Exact Solutions by Ex-Function Method Balancing the order of U and U 2 in Eq. (15, we obtain U = c 1 ex[(c + 3ξ] +, (42 c 2 ex[4ξ] + U 2 = c 3 ex[2cξ] + c 4 ex[ξ] +, (43 c i are determined coefficients only for simlicity. Balancing highest order of Ex-function in Eqs. (42 and (43 we get (c + 3 = (2c +, (44 = c. (45 Similar rocess we balance the linear term of lowest order in Eq. (15, we obtain q = d. (46

5 No. 2 Communications in Theoretical Physics 153 For simlicity, we set = c = 1 and q = d = 1, so Eq. (13 reduces to U( a 1 ex(ξ + a 0 + a 1 ex(ξ b 1 ex(ξ + b 0 + b 1 ex(ξ. (47 Substituting Eq. (47 into Eq. (15, and solving this system of algebraic equations by using Male, we get the following results Case 1 a 1 = 0, a 0 = 3b 0qL 2 (a 2 + rb 2 (qa 2 L 2 1, a 1 = 0, b 1 = b2 0 4b 1, b 0 = b 0, b 1 = b 1, a = a, b = b, c = a2 + rb 2 a(qa 2 L 2 1, (48 b 0 and b 1 are free arameters. Substituting these results into Eq. (47, we get the following exact solution Case 2 u 1 (x, y, t = ( 3qL 2 b 0 (a 2 +rb 2 /(qa 2 L 2 1 ( ( (b 2 0 /4b1 ex ax α Γ(1+α + byα Γ(1+α (a 2 +rb 2 t α a(qa 2 L 2 ax +b 1Γ(1+α 0+b 1 ex α Γ(1+α + byα Γ(1+α (a 2 +rb 2 t α a(qa 2 L 2 1Γ(1+α a 1 = 0, a 0 = ± 3Lcb 0 q, a 1 = 0, b 1 = b2 0 4b 1, b 0 = b 0, b 1 = b 1, b 0 and b 1 are free arameters. Similar rocess we get u 2,3 (x, y, t =. (49 a = 1 ql, b = 1 rql, c = c, (50 ±(3Lcb ( 0 q/ (b 2 0 /4b1 ex x α y qlγ(1+α α ctα rqlγ(1+α Γ(1+α +b 0+b 1 ex ( x ( α y qlγ(1+α α ctα rqlγ(1+α Γ(1+α. (51 If we take b 0 = 2, b 1 = 1, a 2 + rb = qa 2 L 2, and = 3qL 2 Eq. (49 becomes u 1,2 (x, y, t = 1 2 sech2( ax α 2Γ(1 + α + by α 2Γ(1 + α t α. (52 2aΓ(1 + α Similarly; if we take b 0 = 2, b 1 = 1, and = 3cL q Eq. (51 becomes u 3,4 (x, y, t = 1 2 sech2( 3cx α Γ(1 + α 3cy α rγ(1 + α ct α. (53 2Γ(1 + α When the established solutions are comared with each other, it can be seen that: Remark 1 If we take a = 3c/, b = a/ r, and ac = 1 in Eq. (52, this solution gives solution (53. Remark 2 If we take λ = 1, a 2 + rb 2 1 = qa 2 L 2, and = 3qL 2 in Eq. (37, we get solution (52. Remark 3 Also, our solutions (26, (40, and (41 are different and never been obtained. Remark 4 Moreover, we note that the hiyerbolic function solutions are new exact solutions that have not been reorted by other authors for the sace-time fractional KP-BBM equation. [53] 4 Conclusion In this aer, exact solutions of sace time fractional KP-BBM equation is obtained by using ansatz, ( /- exansion and ex-function methods. The obtained solutions are different from the Feng and Meng solutions. [53] The result (37 is in agreement with Eqs. (52 and (53, which are obtained by ex-function method. The solutions obtained in subsection (3.1, (3.2, and (3.3 have not been reorted in the literature before. The obtained solutions namely single soliton solution would be imortant significance for the exlanation of some ractical hysical henomena. According to the results of subsection (3.1, (3.2, (3.3, we conclude that the ( /-exansion method gives more solution than the other methods. These methods have a owerful tool rather than the other methods due to the fact that the methods are useful in FDEs of giving single soliton solutions. By choosing roer values, we have examined the relation between the solutions obtained by different methods. These methods can be alied to other FDEs. The comutations associated in this work were erformed with the aid of the software Male.

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