pplied Mathematical Sciences, Vol. 6, 202, no. 27, 325-334 Exact Solutions for a BBM(m,n) Equation with Generalized Evolution Wei Li Yun-Mei Zhao Department of Mathematics, Honghe University Mengzi, Yunnan, 6600, China wellars@63.com bstract In this paper, employed exp-function method F-expansion method, we study the BBM(m,n) equation with generalized evolution, four families of exact solutions of exp-function type are obtained. Under different parametric conditions, every family of solution can be reduced to some solitary wave solutions periodic wave solutions. The results presented in this paper improve the previous results. Mathematics Subject Classification: 35Q5; 35Q58; Keywords: Exp-function method; F-expansion method; BBM(m,n) equation; Riccati equation; Exact solution Introduction The research work of nonlinear evolution equations in pplied Mathematics Theoretical Physics has been going on for the past forty years. By dint of some new methods, many new results have been published in this area for a long time. Here, it is worth to mention that the two methods, the Expfunction method [,2] F-expansion method [3] can be combined to form one method [4-6]. In this paper, by using Exp-function method combined with F-expansion method, we will study the BBM(m,n) equation with generalized evolution [7] (u l ) t a(u l ) x k(u m ) x b(u n ) xxt =0, () where a, b, k are arbitrary constants l, m, n are integers. In [7], using a solitary wave ansatz in the form of sech p tanh p functions, respectively, Triki Ismail obtained exact bright dark soliton solutions for Eq. (). In this work, we will explore more types of exact solutions of Eq.().
326 Wei Li Yun-Mei Zhao 2 Description of the method In this section, we review the combining the Exp-function method with F-expansion method [5,6] at first. Given a nonlinear partial differential equation, for instance, in two variables, as follows: P (u, u x,u t,u xx,u xt, )=0, (2) where P is in general a nonlinear function of its variables. We firstly use the Exp-function method to obtain new exact solutions of the following Riccati equation φ (ξ) = d dξ φ(ξ) = φ2 (ξ), (3) where are arbitrary constants, then using the Riccati equation (3) as auxiliary equation its exact solutions, we obtain exact solutions of the nonlinear partial differential equation(2). Seeking for the exact solutions of Eq. (3), we introducing a complex variable η, defined by η = μξ ξ 0, (4) where μ is a constant to be determined later, ξ 0 is an arbitrary constant, Riccati equation (3) converts to μφ φ 2 =0, (5) where prime denotes the derivative with respect to η. ccording to the Exp-function method, we assume that the solution of Eq. (5) can be expressed in the following form φ(η) = a e exp(eη) a d exp( dη) b g exp(gη) b f exp( fη), (6) where e, d, g f are positive integers which are given by the homogeneous balance principle, a e,..., a d, b g,..., b f are unknown constants to be determined. To determine the values of e g, we usually balance the linear term of the highest order in Eq. (5) with the highest order nonlinear term. Similarly, we can determine d f by balancing the linear term of the lowest order in Eq. (5) with the lowest order nonlinear term, we obtain e = g, d = f. For simplicity, we set e = g = d = f =, then Eq. (6) becomes φ(η) = a exp(η)a 0 a exp( η) b exp(η)b 0 b exp( η). (7) Substituting Eq. (7) into Eq. (5), equating to zero the coefficients of all powers of exp(nη) (n = 2,, 0,, 2) yields a set of algebraic equations for a, a 0,
BBM(m,n) equation with generalized evolution 327 a, b, b 0, b μ. Solving the system of algebraic equations by using Maple, we obtain the new exact solution of Eq. (3), which read φ = b exp( ξ ξ 0)a exp( ξ ξ 0) b exp( ξ ξ 0) a exp( ξ ξ, (8) 0) where a b are free parameters ; φ 2 = (a 2 0 b2 0 ) 4 b exp(2 ξ ξ 0)a 0 b exp( 2 ξ ξ 0) (a 2 0 b2 0 ) 4b exp(2 ξ ξ 0)b 0 b exp( 2 ξ ξ 0) (9) where a 0, b 0 b are free parameters. By choosing properly values of a 0, a, b 0, b, we find many kinds of hyperbolic function solutions triangular periodic solutions of Eq. (3), which are listed as follows: (i) When ξ 0 =0,b =,a = ±, tanh( < 0, the solution (8) becomes ξ), (0), coth( ξ). () (ii) When ξ 0 =0,b = i, a =, φ = > 0, the solution (8) becomes tan( ξ), (2) cot( ξ). (3) (iii) When ξ 0 =0,b 0 =0,b =,a 0 = ±2, < 0, the solution (9) becomes [coth(2 ξ) ± csch(2 ξ)]. (4) (iv) When ξ 0 =0,b 0 =0,b = i, a 0 = ±2, < 0, the solution (9) becomes [tanh(2 ξ) ± isech(2 ξ)]. (5)
328 Wei Li Yun-Mei Zhao (v) When ξ 0 =0,b 0 =0,b =,a 0 = ±2, > 0, the solution (9) becomes φ = [tan(2 ξ) ± sec(2 ξ)]. (6) (vi) When ξ 0 =0,b 0 =0,b = i, a 0 = ±2, > 0, the solution (9) becomes [cot(2 ξ) csc(2 ξ)]. (7) For simplicity, in the rest of the paper, we consider ξ 0 =0. 3 Solutions for the BBM(m,n) equation In order to obtain new exact travelling wave solutions for Eq. (), we use u(x, t) =u(ξ), ξ = B(x ωt), (8) where B ω are constants, substituting the (8) into Eq. (), integrating two sides of the equation with respect to ξ setting the constants of integration equal to zero, we obtain (a ω)u l ku m bωb 2 (u n ) =0. (9) 3.. Case I: l = n, m n In this case, balancing the order of the nonlinear term u m with the term (u n ) in (9), we obtain mp = np 2, (20) so that P = 2 m n. (2) To get a closed form solution, it is natural to use the transformation Eq. (9) becomes u = v, (22) (a ω)(m n) 2 v 2 k(m n) 2 v 3 ωbb [ 2 n( m)(v ) 2 n(m n)vv ] =0. (23) Now, we assume that the solution of Eq. (23) can be expressed in the following form N v = v(ξ) = α j φ j (ξ), (24) j=0
BBM(m,n) equation with generalized evolution 329 where N is positive integers which are given by the homogeneous balance principle, φ(ξ) is a solution of Eq. (3). Balancing vv term with v 3 term in (23) gives N = 2. Therefore, we obtain v = α 0 α φ(ξ)α 2 φ 2 (ξ). (25) Substituting Eq. (25) into (23) using the Riccati equation (3), collecting the coefficients of φ(ξ), we have [ C0 C φ(ξ)c 2 φ 2 (ξ) C 5 φ 5 (ξ)c 6 φ 6 (ξ) ] =0. (26) D Because the expresses to these coefficients D, C 0 =0, C =0, C 2 =0, C 3 = 0, C 4 =0,C 5 =0,C 6 =0ofφ(ξ) in Eq. (26) are too lengthiness, so we omit them. But we can directly use the comm solve in mathematical software Maple to solve the following set of algebraic equations C 0 =0, C =0, C 2 =0, C 3 =0, C 4 =0, C 5 =0, C 6 =0. (27) Solved the above algebraic equations, we obtain α 0 = form: (m n)(ω a), α =0, α 2 = (m n)(ω a), B = (m n) a ω ωb (28) Thus from Eqs. (25) (28) we obtain the solution of Eq. (23) in the v = (m n)(ω a) (m n)(ω a) φ 2 (ξ), (29) where φ(ξ) is a solution of Eq. (3). Substituting new solutions (8) (9) of Riccati equation into solution (29), using the transformation (22), we have the following two families of solutions to Eq. (). Family. u (x, t) = { (m n)(ω a) (m n)(ω a) 2 b exp( ξ)a exp( ξ b exp( ξ) a exp( ξ), (30) where ξ = () a ω ωb (x ωt), l = n.
330 Wei Li Yun-Mei Zhao If we set b =,a = ±, < 0 in Eq. (30), we obtain { (m n)(ω a) u () (x, t) = sech 2 ( } ξ), (3) { (m n)(ω a) u (2) (x, t) = csch 2 ( } ξ). (32) Setting b = i, a =, > 0 in Eq. (30), we get { } (m n)(ω a) u (3) (x, t) = sec 2 ( ξ), (33) { } (m n)(ω a) u (4) (x, t) = csc 2 ( ξ). (34) Family 2 { (m n)(ω a) (m n)(ω a) u 2 (x, t) = where ξ = () 4 a ω ωb (a 2 0 b2 0 ) b (a 2 0 b2 0 ) 4b exp(2 ξ)a 0 (x ωt), l = n. 2 b exp( 2 ξ) exp(2 ξ)b 0b exp( 2 ξ), (35) If we set b 0 =0,b =,a 0 = ±2, < 0 in Eq. (35), we obtain u 2() (x, t) = { (m n)(ω a) (m n)(ω a) [ coth(2 ξ) ± csch(2 ξ)] 2 } (36) (m n)(ω a) = kn [ cosh(2 ξ)]. (37) Setting b 0 =0,b = i, a 0 = ±2, < 0 in Eq. (35), we get u 2(2) (x, t) = { (m n)(ω a) (m n)(ω a) [ tanh(2 ξ) ± isech(2 ξ)] 2 }. (38)
BBM(m,n) equation with generalized evolution 33 Setting b 0 =0,b =,a 0 = ±2, > 0 in Eq. (35), we have u 2(3) (x, t) = { (m n)(ω a) (m n)(ω a) [ tan(2 ξ) ± sec(2 ξ)] 2 }. (39) Setting b 0 =0,b = i, a 0 = ±2, > 0 in Eq. (35), we have u 2(4) (x, t) = { (m n)(ω a) (m n)(ω a) [ cot(2 ξ) csc(2 ξ)] 2 } (40) (m n)(ω a) = kn [ ± cos(2 ξ)]. (4) 3.2. Case II: m = n, l n In this case, balancing the order of the nonlinear term u l with the term (u n ) in (9), we obtain lp = np 2, (42) so that P = 2 l n. (43) To get a closed form solution, it is natural to use the transformation Eq. (9) becomes u = v, (44) (a ω)() 2 v 3 k() 2 v 2 ωbb 2 [ n( l)(v ) 2 n(l n)vv ] =0. (45) By the same manipulation as is illustrated in Case I, we obtain the solution of Eq. (45) in the form: v = k(l n) (ω a) k(l n) (n l) (ω a) φ2 (ξ), ξ = k ωb (x ωt), (46) where φ(ξ) is a solution of Eq. (3). Substituting new solutions (8) (9) of Riccati equation into solution (46), using the transformation (44), we have the following two families of solutions
332 Wei Li Yun-Mei Zhao to Eq. (). Family 3 u 3 (x, t) = { k(l n) (ω a) k(l n) (ω a) b exp( b exp( ξ) a 2 ξ)a exp( ξ exp( ξ), (47) where ξ = (n l) k ωb (x ωt), m = n. If we set b =,a = ±, < 0 in Eq. (47), we obtain u 3() (x, t) = { k(l n) (ω a) sech2 ( } ξ), (48) { k(l n) u 3(2) (x, t) = (ω a) csch2 ( } ξ). (49) Setting b = i, a =, u 3(3) (x, t) = > 0 in Eq. (47), we get { } k(l n) (ω a) sec2 ( ξ), (50) Family 4 u 3(4) (x, t) = { } k(l n) (ω a) csc2 ( ξ). (5) u 4 (x, t) = where ξ = (n l) { k(l n) (ω a) k(l n) (ω a) (a 2 0 b2 0 ) exp(2 4 b ξ)a 0 b exp( 2 (a 2 0 b2 0 ) exp(2 ξ)b 0b exp( 2 ξ) k ωb 4b (x ωt), m = n. 2 ξ), (52)
BBM(m,n) equation with generalized evolution 333 If we set b 0 =0,b =,a 0 = ±2, < 0 in Eq. (52), we obtain { k(l n) k(l n) u 4() (x, t) = (ω a) (ω a) [ coth(2 ξ) ± csch(2 ξ)] 2 } (53) k(l n) = n(ω a) [ cosh(2 ξ)]. (54) Setting b 0 =0,b = i, a 0 = ±2, < 0 in Eq. (52), we get u 4(2) (x, t) = { k(l n) k(l n) (ω a) (ω a) [ tanh(2 ξ) ± isech(2 ξ)] 2 }. Setting b 0 =0,b =,a 0 = ±2, > 0 in Eq. (52), we have { k(l n) k(l n) u 4(3) (x, t) = (ω a) (ω a) [ tan(2 ξ) ± sec(2 ξ)] 2 }. (55) (56) Setting b 0 =0,b = i, a 0 = ±2, > 0 in Eq. (52), we have { k(l n) k(l n) u 4(4) (x, t) = (ω a) (ω a) [ cot(2 ξ) csc(2 ξ)] 2 } 4 Conclusions k(l n) = n(ω a) [ ± cos(2 ξ)] (57). (58) In this paper, by using Exp-function method combined with F-expansion method, we studied the BBM(m,n) equation with generalized evolution, four families of exact solutions of exp-function type are obtained. Under different parametric conditions, every family of solution can be reduced to some solitary wave solutions periodic wave solutions, the majority of these results are very different to those in Ref. [7]. This shows that some solutions with different wave forms can be expressed by the same one solution of exp-function type. From these abundant results, it is easy to know that the Exp-function method combined F-expansion method is useful to many nonlinear partial equations.
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