NEW EXACT SOLUTIONS OF SOME NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS VIA THE IMPROVED EXP-FUNCTION METHOD

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1 NEW EXACT SOLUTIONS OF SOME NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS VIA THE IMPROVED EXP-FUNCTION METHOD M. F. El-Sabbagh, R. Zait and R. M. Abdelazeem Mathematics Department, Faculty of Science, MiniaUniversity, Egypt. Corresponding ABSTRACT In this paper, we establish new exact solutions of some nonlinear partial differential equations (PDEs) of interest such as the Kaup Kupershmidt, the generalized shallow water, the Boussinesq equations via the improved Exp function method. Also the method is used to construct periodic and solitary wave solutions for the considered equations as well. Keywords: Nonlinear PDEs, Exact solutions, The improved Exp function method.. INTRODUCTION The nonlinear evolution equations (NLEEs) are widely used as models to describe complex physical phenomena in various field of science, particularly in fluid mechanics, solid state physics, plasma waves and chemical physics. Nonlinear equations covers also the following subjects: surface wave in compressible fluid, hydro magnetic waves in cold plasma, acoustic waves in un harmonic crystal, ect.. The wide applicability of these equations is the main reason for they have attracted so much attention from mathematicians in the last decades. The investigation of the exact solutions of non linear partial differential equations (PDEs) plays an important role in the study of non-linear physical phenomena. When we want to understand the physical mechanism of phenomena in nature, described by non linear PDEs, exact solutions have to be explored. The study of nonlinear PDEs becomes one of the most important topics in mathematical physics. Recently there are many new methods to obtain exact solutions of nonlinear PDEs such as sine-cosine function method [-5], tanh function method [6-8], ( G/ G )-expansion method [9-3 ], extended Jacobi elliptic function method [4, 5]. He and Wu [6], proposed a straightforward and concise method, called Exp-function method [7-24], to obtain generalized solitary wave solutions of nonlinear PDEs. Ali [25] improved this method and obtained new exp-function solutions and periodic solutions as well. In this paper, we use the improved Exp-function method [25, 26 ] to search for new solitary wave solutions, compact like solutions and periodic solutions of some nonlinear PDEs, such as the Kaup Kupershmidt equation [27-35], the generalized shallow water equation [36, 37], and the Boussinesq equation [38-42]. 2. THE IMPROVED EXP-FUNCTION METHOD: We present the improved Exp-function method [25] in the following steps: - Consider the following nonlinear PDE with two independent variables x, t and dependant variable u: N(u, u t, u x, u xx, u xt, u tt, ) = 0 (), where N is in general a polynomial function of its argument and the subscripts denote the partial derivatives. 2- We seek a traveling wave solution of Eq. () in the form u x, t = u ξ, ξ = kx + ωt 2, where k and ω are constants to be determined 3- Using the transformation (2), Eq. () can be reduced to an ordinary differential equation (ODE): G u, u, u, = 0 3, where G is a polynomial of u and its derivatives. 4- Through this method, we express the solution of the nonlinear PDE () in the form: 32

2 u η = m j =0 n i=0 A j exp (jξ) B i exp (iξ) 4, where m and n are positive integers that could be freely chosen. 5- To determine u x, t explicitly, one may apply the following producer: (i) Substitute Eq. (4) into Eq. (3), then the lift hand side of Eq. (3) is converted into a polynomial in exp (ξ). Setting all coefficients of exp (ξ) to zero yield a system of algebraic equations for A 0, A, A 2,. A m, B 0, B, B 2,. B n, k and ω. (ii) Solve these algebraic equations to obtain A 0, A, A 2,, A m, B 0, B, B 2,, B n, k and ω. 3. APPLICATIONS In order to illustrate the effectiveness of the above method, examples of mathematical and physical interests are chosen as follows: 3.The Kaup Kupershmidt equation [27-35] The Kaup Kupershmidt equation is the nonlinear fifth-order partial differential equation; It is the first equation in a hierarchy of integrable equations with Lax operator u. It has properties similar (but not identical) to x + u x x those of the better known KdV hierarchy in which the Lax operator has order two. In the present paper we introduce new exact solutions of the Kaup Kupershmidt equation via the improved Expfunction method as follows: Consider the Kaup Kupershmidt equation is given as: u t = u xxxxx 20uu xxx 50u x u xx + 80u 2 u x (5) Using the transformation: u = u η, η = x ct 6, where c is a constant to be determined later. Substituting Eq. (6) into Eq. (5) we get cu u + 20uu + 50u u 80u 2 u = 0 7, where primes denote derivatives with respect to η. Now we study the following cases: Case : m = 2, n = 2: According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written as: u x, t = A 0 + A exp (η) + A 2 exp (2η) B 0 + B exp (η) + B 2 exp (2η) (8) In case B 2 0, Eq. (8) can be simplified as: u x, t = A 0 + A exp (η) + A 2 exp (2η) B 0 + B exp (η) + exp (2η) (9) Substituting Eq. (9) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of exp (η)yields a set of algebraic equations for A 0, A, A 2, B 0, B and c. Solving these system of algebraic equations, with the aid of Maple, we obtain families of solutions as the following: 33

3 Family : B 0 =, A = 5, A 0 = 2, c =, A 2 = 2, B = 2. u x, t = 2 5exp (x + t) + exp (2x + 22t) 2 + 2exp (x + t) + exp (2x + 22t) (0) Figure. Traveling wave solution of Eq. 5 for solution 0. Family 2: B 0 =, A = 5 8, A 0 =, c = 6 6, A 2 = 6, B = 2. u 2 x, t = 6 5 t exp (x ) + 6 exp (2x + t 8 ) + 2exp (x + t 6 ) + exp (2x + t () 8 ) Figure 2. Traveling wave solution of Eq. 5 for solution. Case 2: m = 3, n = 3: According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written as: 34

4 u x, t = A 0 + A exp (η) + A 2 exp (2η) + A 3 exp (3η) B 0 + B exp (η) + B 2 exp (2η) + B 3 exp (3η) (2) In case B 3 0, Eq. (2) can be simplified as: u x, t = A 0 + A exp (η) + A 2 exp (2η) + A 3 exp (3η) B 0 + B exp (η) + B 2 exp (2η) + exp (3η) 3 Substituting Eq. (3) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of ex p η yields a set of algebraic equations for A 0, A, A 2, A 3, B 0, B, B 2 and c. Solving the system of algebraic equations given above, with the aid of Maple, we obtain: Family : A 0 = 0, A = B 8 2 2, A 2 = 5 B 2 2, A 3 =, B 2 0 = 0, B = B 4 2 2, B 2 = B 2, c =. u 5 x, t = 4 B 2 2 exp (x + t) 5B 2 exp (2x + 22t) + exp (3x + 33t) 2 B 2 2 exp (x + t) + 2B 2 exp (2x + 22t) + 2exp (3x + 33t) (4) If B 2 =, then exp (x + t) 5exp (2x + 22t) + exp (3x + 33t) u 5 x, t = 4 (5) 2 exp (x + t) + 2exp (2x + 22t) + 2exp (3x + 33t) Figure 3. Traveling wave solution of Eq. 5 for solution 5, B 2 =. Family 2: A 0 = 0, A = B , A 2 = 5 B 6 2, A 3 =, B 6 0 = 0, B = B 4 2 2, B 2 = B 2, c = u 6 x, t = 4 B 2 2 exp (x + t 6 ) 5B 2exp (2x + t 3t ) + exp (3x ) 4(B 2 2 exp (x + t 6 ) + 4B 2exp (2x + t (6) 3t ) + 4exp (3x )) 6. 35

5 If B 2 =, then u 6 x, t = t exp (x ) 5exp (2x + t 3t ) + exp (3x ) 4(ex p x + t 6 + 4ex p 2x + t 8 + 4ex p 3x + 3t 6 ) (7) Figure 4. Traveling wave solution of Eq. 5 for solution 7, B 2 =. 3.2 The generalized shallow water equation: [36, 37] The shallow water wave equations describe the evolution of incompressible flow, neglecting density change along the depth. The shallow water wave equations are applicable to cases where the horizontal scale of the flow is much bigger than the depth of fluid. The shallow water equations have been extensively used for a wide variety of coastal phenomena, such as tide-currents, pollutant- dispersion storm-surges, tsunami-wave propagation, etc.. In the present paper we introduce new exact solutions of the generalized shallow water equation via the improved Exp-function Method as follows: Consider the generalized shallow water equation: u xxxt + αu x u xt + βu t u xx u xt u xx = 0 8, where α and β are arbitrary nonzero constants. Using the transformation: u = u η, η = x ct 9, where c is a constant to be determined later. Substituting Eq. (9) into Eq. (8) we get cu cαu u cβu u cu u = 0 (20), where the prime denotes the differential with respect to η. Now we study the following cases: Case : m = 2, n = 2: According to the improved Exp-function method, the travelling wave solution of Eq. (8) in this case can be written as: u x, t = A 0 + A exp (η) + A 2 exp (2η) B 0 + B exp (η) + B 2 exp (2η) (2) 36

6 In case B 2 0, Eq.(2) can be simplified as: u x, t = A 0 + A exp (η) + A 2 exp (2η) B 0 + B exp (η) + exp (2η) 22 Substituting Eq. (22) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of exp (η)yields a set of algebraic equations for A 0, A, A 2, B 0, B and c. Solving the system of algebraic equation given above, with the aid of Maple, we obtain: A 0 = B 0( 24+A 2 β +A 2 α), A α+β = 0, A 2 = A 2, B 0 = B 0, B = 0, c =. 3 u x, t = B 0 ( 24 + A 2 β + A 2 α) + A α + β 2 exp (2x + 2t 3 ) B 0 + exp (2x + 2t (23) 3 ) If A 2 = B 0 =, α = β = 2, then u x, t = 5 + exp (2x + 2t 3 ) + exp (2x + 2t 3 ) (24) Figure 5. Traveling wave solution of Eq. 8 for solution Eq. 24, A 2 = B 0 =, α = β = 2. Case 2: m = 3, n = 3: According to the improved Exp-function method, the travelling wave solution of Eq. (8) in this case can be written as: u x, t = A 0 + A exp (η) + A 2 exp (2η) + A 3 exp (3η) B 0 + B exp (η) + B 2 exp (2η) + B 3 exp (3η) (25) In case B 3 0, Eq.(25) can be simplified as: u x, t = A 0 + A exp (η) + A 2 exp (2η) + A 3 exp (3η) B 0 + B exp (η) + B 2 exp (2η) + exp (3η) 26 Substituting Eq. (26) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of exp (η) yields a set of algebraic equations fora 0, A, A 2, A 3, B 0, B, B 2 and c. Solving the system of algebraic equation given above, with the aid of Maple, we obtain: 37

7 Family : A 0 = 0, A = B ( 24 + A 3 β + A 3 α), A α + β 2 = 0, A 3 = A 3, B 0 = 0, B = B, B 2 = B 2, c = 3 u 2 x, t = B ( 24 + A 3 β + A 3 α) exp (x + t α + β 3 ) + A 3exp (3x + t) B exp (x + t (27) 3 ) + exp (3x + t) If B = A 3 = 2, α = β =, then u 2 x, t = 20exp (x + t ) + 2exp (3x + t) 3 2exp (x + t (28) 3 ) + exp (3x + t) Figure 6. Traveling wave solution of Eq. 8 for solution Eq. 28, A 3 = B = 2, α = β =. Family 2: A 0 = B 0(A 3 β + A 3 α 36), A α + β = 0, A 2 = 0, A 3 = A 3, B 0 = B 0, B = 0, B 2 = 0, c = 8 u 3 x, t = B 0 (A 3 β + A 3 α 36) + A α + β 3 exp (3x + 3t 8 ) B 0 + exp (3x + 3t (29) 8 ) If B 0 = A 3 = 2, α = β =, then u 3 x, t = exp (3x + 3t 8 ) 2 + exp (3x + 3t 8 ) (30) 38

8 Figure 7. Traveling wave solution of Eq. 8 for solution Eq. 30, A 3 = B 0 = 2, α = β =. 3.3 The Boussinesq equation: [38-42] The Boussinesq-type equations, which include the lowest-order effects of nonlinearity and frequency dispersion as additions to the simplest non-dispersive linear long wave theory, provide a sound and increasingly well-tested basis for the simulation of wave propagation in coastal regions. The standard Boussinesq equations for variable water depth were first derived by Peregrine (967), who used depth-averaged velocity as a dependent variable. In the present paper we introduce new exact solutions of the Boussinesq equation via the improved Exp-function method as follows: Consider the Boussinesq equation: u tt u xx u xxxx 6(u x ) 2 6uu xx = 0 (3) Using the transformation: u = u η, ξ = kx + ωt 32, where k, ω are constants to be determined later. Substituting Eq. (32) into Eq. (3) we get ω 2 u k 2 u k 4 u 6k 2 (u ) 2 6k 2 uu = 0 33, where the prime denotes the differential with respect to ξ. Now we study the following cases: Case : m = 2, n = 3: According to the improved Exp-function method, the travelling wave solution of Eq. (3) in this case can be written as: u x, t = A 0 + A exp (ξ) + A 2 exp (2ξ) B 0 + B exp (ξ) + B 2 exp (2ξ) + B 3 exp (3ξ) (34) In case B 3 0, Eq.(34) can be simplified as: u x, t = A 0 + A exp (ξ) + A 2 exp (2ξ) B 0 + B exp (ξ) + B 2 exp (2ξ) + exp (3ξ) 35 Substituting Eq. (35) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of ex p ξ yields a set of algebraic equations for A 0, A, A 2, B 0, B, B 2, k and ω. Solving the system of algebraic equation given above, with the aid of Maple, we obtain: 39

9 Family : A 0 = 0, A = 0, A 2 = k 2 B 2, B 0 = 0, B = B 2 2 4, B 2 = B 2, ω = k + k 2 u x, t = k 2 B 2 exp (2kx + 2kt + k 2 ) B exp (kx + kt + k2 ) + B 2 exp (2kx + 2kt + k 2 ) + exp (3kx + 3kt + k 2 ) (36) If k =, B 2 = 2, then 2exp (2x + 2t 2) u x, t = exp (x + t 2) + 2exp (2x + 2t 2) + exp (3x + 3t 2) (37) Figure 8. Traveling wave solution of Eq. 3 for solution Eq. 37, B 2 = 2, k =. Family 2: A 0 = 0, A = 0, A 2 = k 2 B 2, B 0 = 0, B = B 2 2 4, B 2 = B 2, ω = k + k 2 u 2 x, t = k 2 B 2 exp (2kx 2kt + k 2 ) B exp (kx kt + k2 ) + B 2 exp (2kx 2kt + k 2 ) + exp (3kx 3kt + k 2 ) (38) If B 2 = 2, k =, then 2ex p 2x 2t 2 u 2 x, t = exp (x t 2) + 2exp (2x 2t 2) + exp (3x 3t 2) (39) 40

10 Figure 9. Traveling wave solution of Eq. 3 for solution Eq. 39, B 2 = 2, k =. Case 2: m = 2, n = 4: According to the improved Exp-function method, the travelling wave solution of Eq. (3) in this case can be written as: u x, t = A 0 + A ex p ξ + A 2 ex p 2ξ B 0 + B ex p ξ + B 2 ex p 2ξ + B 3 ex p 3ξ + B 4 ex p 4ξ (40) In case B 4 0, Eq.(40) can be simplified as: u x, t = A 0 + A ex p ξ + A 2 ex p 2ξ B 0 + B ex p ξ + B 2 ex p 2ξ + B 3 ex p 3ξ + ex p 4ξ 4 Substituting Eq. (4) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of ex p ξ yields a set of algebraic equations for A 0, A, A 2, B 0, B, B 2, B 3, k and ω. Solving the system of algebraic equation given above, with the aid of Maple, we obtain: Family : A 0 = 0, A = 0, A 2 = A 2, B 0 = A k 4, B = 0, B 2 = A 2 4k 2, B 3 = B 3, ω = k + 4k 2 u 3 x, t = A 2 exp (2kx + 2kt + 4k 2 ) A k 4 + A 2 4k 2 exp (2kx + 2kt + 4k2 ) + exp (4kx + 4kt + 4k 2 ) (42) If A 2 = 2, k =, then 2exp (2x + 2t 5) u 3 x, t = 6 + (43) 2 exp (2x + 2t 5) + exp (4x + 4t 5) 4

11 Figure 0. Traveling wave solution of Eq. 3 for solution Eq. 43, A 2 = 2, k =. Family 2: A 0 == 0, A = 0, A 2 = A 2, B 0 = A k 4, B = 0, B 2 = A 2 4k 2, B 3 = B 3, ω = k + 4k 2 u 4 x, t = A 2 exp (2kx 2kt + 4k 2 ) A k 4 + A 2 4k 2 exp (2kx 2kt + 4k2 ) + exp (4kx 4kt + 4k 2 ) (44) If A 2 = 2, k =, then 2exp (2x 2t 5) u 4 x, t = 6 + (45) 2 exp (2x 2t 5) + exp (4x 4t 5) Figure. Traveling wave solution of Eq. 33 for solution Eq. 45, A 2 = 2, k =. 42

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