Modified extended Tanh method

1 July 2002Physics Letters A 299 (2002) 179–188 www.elsevier.com/locate/pla Modified extended tanh-function method for solving nonlinear partial differential equations S.A. Elwakil a , S.K. El-labany b , M.A. Zahran a,∗ , R. Sabry b a Theoretical Physics Group, Physics Department, Faculty of Science, Mansoura University, Mansoura, Egypt b Theoretical Physics Group, Physics Department, Faculty of Science, Mansoura University, New Damietta 34517, Damietta, Egypt Received 20 February 2002; received in revised form 2 May 2002; accepted 7 May 2002 Communicated by A.R. Bishop Abstract Based on an extended tanh-function method, a general method is suggested to obtain multiple travelling wave solutions for nonlinear partial differential equations (PDEs). The validity and reliability of the method is tested by its application to some nonlinear PDEs. The obtained results are compared with that of an extended tanh-function method and hyperbolic-function method. New exact solutions are found.  2002 Elsevier Science B.V. All rights reserved. Keywords: Nonlinear evolution equation; Traveling wave solution; Symbolic computation 1. Introduction Nonlinear PDEs are widely used to describe complex phenomena in various fields of sciences, especially in physics. Various methods for obtaining explicit travelling solitary wave solutions to nonlinear evolution equations have been proposed. Among these are Hirota’s bilinear methods [1], the inverse scattering transform [1], Painlevé expansions [2], the real exponential method [3], homogeneous balance method [4,5], rank analytical method [6] and several “ansatz” methods (see, for example, Ref. [7]). Recently, tanh-function method [8–13] has been proposed to construct exact solutions to PDEs. Later, Fan [14,15] has proposed an extended tanh-function method and obtained new travelling wave solutions that cannot be obtained by tanh-function method [8–13]. Now, let us simply describe the modified extended tanh-function. Consider a given PDE, say in two variables (1) H(u, u x , u t , u xx , . . .) =0. * Corresponding author. E-mail addresses: [email protected] (M.A. Zahran), [email protected] (R. Sabry). 0375-9601/02/$ – see front matter  2002 Elsevier Science B.V. All rights reserved. PII: S0375- 9601( 02) 00669- 2 180 S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 We first consider its travelling solutions u(x, t ) =u(ζ ), ζ =x +λt , then Eq. (1) becomes an ordinary differential equation. The next crucial step is that the solution we are looking for is expressed in the form (2) u(ζ ) = m i=0 a i ω i + m i=0 b i ω −i and (3) ω =b +ω 2 , where b is a parameter to be determined, ω = ω(ζ ), ω = dω/dζ . The parameter m can be found by balancing the highest-order linear term with the nonlinear terms [9,14]. Substituting (2) and (3) into the relevant ordinary differential equation will yield a system of algebraic equations with respect to a i , b i , b, λ (where i = 0, . . . , m) because all the coefficients of ω i have to vanish. With the aid of Mathematica, one can determine a i , b i , b and λ. The Riccati equation (3) has the general solutions (4) ω = _ − √ −b tanh √ −bζ, with b < 0, − √ −b coth √ −bζ, with b < 0, (5) ω =− 1 ζ , with b =0, and (6) ω = _ √ b tan √ bζ, with b > 0, − √ b cot √ bζ, with b > 0. Since coth- and cot-type solutions appear in pairs with tanh- and tan-type solutions, respectively, they are omitted in this Letter. In the next section, we study some nonlinear equations to illustrate this method. 2. Examples Example 1. Consider the nonlinear evolution equation (7) u t t +αu xx +βu +γ u 3 =0, where α, β, and γ are constants. Eq. (7) contains some particular important equations such as Duffing, Klein– Gordon, Landau–Ginsburg–Higgs, and φ 4 equation [16]. To look for the travelling wave solution of Eq. (7), we use the transformation u(x, t ) =u(ζ ), ζ =x +λt . Then Eq. (7) is reduced to the following ordinary differential equation: (8) _ λ 2 +α _ u +βu +γ u 3 =0. Balancing u with u 3 yields m=1. Therefore, we have (9) u =a 0 +a 1 ω +b 0 +b 1 ω −1 . Substituting Eq. (9) into Eq. (8) and making use of Eq. (3), with the help of Mathematica we get a system of algebraic equations for a 0 , a 1 , b 0 , b 1 , b, and λ: (a 0 +b 0 ) _ β +γ _ (a 0 +b 0 ) 2 +6a 1 b 1 __ =0, 2bαa 1 +βa 1 +2bλ 2 a 1 +3γ a 2 0 a 1 +6γ a 0 a 1 b 0 +3γ a 1 b 2 0 +3γ a 2 1 b 1 =0, S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 181 γ a 0 a 2 1 +γ a 2 1 b 0 =0, 2αa 1 +2λ 2 a 1 +γ a 3 1 =0, 2bαb 1 +βb 1 +2bλ 2 b 1 +3γ a 2 0 b 1 +6γ a 0 b 0 b 1 +3γ b 2 0 b 1 +3γ a 1 b 2 1 =0, γ a 0 b 2 1 +γ b 0 b 2 1 =0, 2b 2 αb 1 +2b 2 λ 2 b 1 +γ b 3 1 =0, from which we find (10) b =− β 2(α +λ 2 ) , a 0 =−b 0 , a 1 =0, b 1 =±b _ −2(α +λ 2 ) γ , (11) b =− β 2(α +λ 2 ) , a 0 =−b 0 , a 1 =± _ −2(α +λ 2 ) γ , b 1 =0, (12) b =− β 8(α +λ 2 ) , a 0 =−b 0 , a 1 =± _ −2(α +λ 2 ) γ , b 1 =−ba, (13) b = β 4(α +λ 2 ) , a 0 =−b 0 , a 1 =± _ −2(α +λ 2 ) γ , b 1 =ba 1 . According to Eq. (10), it is clear that (α +λ 2 )/γ < 0 for b 1 to be real. Then, for b > 0 the solution to Eq. (7) reads (14) u(x, t ) =± _ β γ cot __ −β 2(α +λ 2 ) (x +λt ) _ , where (α +λ 2 )/γ < 0 . While for b < 0 the solution to Eq. (7) is (15) u(x, t ) =± _ −β γ coth __ β 2(α +λ 2 ) (x +λt ) _ , where (α +λ 2 )/γ < 0. Due to Eq. (11), it is clear that (α +λ 2 )/γ < 0 for a 1 to be real. Then, for b > 0 the solution to Eq. (7) reads (16) u(x, t ) =± _ β γ tan _ _ −β 2(α +λ 2 ) (x +λt ) _ , where (α +λ 2 )/γ < 0. While for b < 0 it is (17) u(x, t ) =∓ _ −β γ tanh _ _ β 2(α +λ 2 ) (x +λt ) _ , where (α +λ 2 )/γ < 0. From Eq. (12), it is clear that γ (α +λ 2 ) < 0 for a 1 and b 1 to be real. Then, for b > 0 we get (18) u(x, t ) =∓ _ β 4γ _ tan __ −β 8(α +λ 2 ) (x +λt ) _ −cot __ −β 8(α +λ 2 ) (x +λt ) __ , 182 S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 where γ (α +λ 2 ) < 0. Also, the solution of Eq. (7) due to Eq. (12) for the case b < 0 is (19) u(x, t ) =± _ −β 4γ _ tanh _ _ β 8(α +λ 2 ) (x +λt ) _ +coth _ _ β 8(α +λ 2 ) (x +λt ) __ , where γ (α +λ 2 ) < 0. Eq. (13) indicates that γ (α +λ 2 ) < 0 for a 1 and b 1 to be real. Then, for b < 0, (20) u(x, t ) =± _ β 2γ csch __ −β 4(α +λ 2 ) (x +λt ) _ sech __ −β 4(α +λ 2 ) (x +λt ) _ , while for b > 0, (21) u(x, t ) =± _ − 2β γ csc _ _ β (α +λ 2 ) (x +λt ) _ . We have recovered some new exact solutions that are not obtained by the hyperbolic-function method [16]. Example 2. Consider the two-dimensional KdV–Burgers equation (22) (u t +uu x +pu xxx −qu xx ) x +ru yy =0, where p, q are real constants and r =±1 [13,14]. On using u(x, t ) =u(ζ ), ζ =x +dy +ct , Eq. (22) reduces to (23) (cu +uu +pu −qu ) +rd 2 u =0. Balancing u with uu gives m=2, hence (24) u =a 0 +a 1 ω +a 2 ω 2 +b 0 +b 1 ω −1 +b 2 ω −2 . Substitute Eq. (24) into Eq. (23) and make use of Eq. (3). With the help of Mathematica, we get a system of algebraic equations, for a 0 , a 1 , a 2 , b 0 , b 1 , b 2 , b, c, and d: −2b 2 qa 1 +b 2 a 2 1 +2b 2 a 2 _ c +8bp +d 2 r +a 0 +b 0 _ +b 1 (2bq +b 1 ) +2b 2 _ c +8bp +d 2 r +a 0 +b 0 _ =0, bca 1 +8b 2 pa 1 +bd 2 ra 1 +ba 0 a 1 −8b 2 qa 2 +3b 2 a 1 a 2 +ba 1 b 0 +ba 2 b 1 =0, −8bqa 1 +4ba 2 1 +8bca 2 +136b 2 pa 2 +8bd 2 ra 2 +8ba 0 a 2 +6b 2 a 2 2 +8ba 2 b 0 =0, 2ca 1 +40bpa 1 +2d 2 ra 1 +2a 0 a 1 −40b 2 qa 2 +18ba 1 a 2 +2a 1 b 0 +2a 2 b 1 =0, −6qa 1 +3a 2 1 +6ca 2 +240bpa 2 +6d 2 ra 2 +6a 0 a 2 +16ba 2 2 +6a 2 b 0 =0, 2pa 1 −2qa 2 +a 1 a 2 =0, 12pa 2 +a 2 2 =0, bcb 1 +8b 2 pb 1 +bd 2 rb 1 +ba 0 b 1 +bb 0 b 1 +8bqb 2 +ba 1 b 2 +3b 1 b 2 =0, 8b 2 qb 1 +4bb 2 1 +8bcb 2 +136b 2 pb 2 +8bd 2 rb 2 +8ba 0 b 2 +8bb 0 b 2 +6b 2 2 =0, 2b 2 cb 1 +40b 3 pb 3 1 +2b 2 d 2 rb 1 +2b 2 a 0 b 1 +2b 2 b 0 b 1 +40b 2 qb 2 +2b 2 a 1 b 2 +18bb 1 b 2 =0, 6b 3 qb 1 +3b 2 b 2 1 +6b 2 cb 2 +240b 3 pb 2 +6b 2 d 2 rb 2 +6b 2 a 0 b 2 +6b 2 b 0 b 2 +16bb 2 2 =0, 2b 4 pb 1 +2b 3 qb 2 +b 2 b 1 b 2 =0, 12b 4 pb 2 +b 2 b 2 2 =0. S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 183 Solving the above system using Mathematica, we get a 0 =− _ c +8bp − q 2 25p +d 2 r +b 0 _ , a 1 = 12q 5 , a 2 =−12p, (25) b 1 =0, b 2 =0, b =− q 2 100p 2 , a 0 =− _ c +8bp − q 2 25p +d 2 r +b 0 _ , a 1 =0, a 2 =0, (26) b 1 =− 12bq 5 , b 2 =−12b 2 p, b =− q 2 100p 2 , where b =0, a 0 =− _ c +8bp − q 2 25p +d 2 r +b 0 _ , a 1 = 12q 5 , a 2 =−12p, (27) b 1 =− 12bq 5 , b 2 =−12b 2 p, b =− q 2 400p 2 , where b =0. Due to Eq. (25), it is clear that b 0 and we have (28) u(x, t ) =−c −d 2 r − 12p ζ 2 , b =0 and q =0, (29) u(x, t ) = 3q 2 25p − _ c +d 2 r _ − 6q 2 25p tanh _ qζ 10p _ − 3q 2 25p tanh 2 _ qζ 10p _ , ζ =x +dy +ct . In Eq. (26) b < 0, thus we get (30) u(x, t ) = 3q 2 25p − _ c +d 2 r _ − 6q 2 25p coth _ qζ 10p _ − 3q 2 25p coth 2 _ qζ 10p _ , ζ =x +dy +ct . Finally, as a result of Eq. (27), one gets (31) u(x, t ) =− _ c +d 2 r _ − 3q 2 25p csch 2 _ qζ 10p __ 1 +sinh _ qζ 5p __ , b < 0, ζ =x +dy +ct . In fact, the solution obtained by Fan [14] which is u 1 =−c + q 25p −d 2 r − 12q 5z − 12p z 2 , z =x +dy +ct , is not correct and holds only under the condition that b and q must be equal to zero. The correct solution is given above by Eq. (28). Solution in (28) is a rational-type solution. Solution (29) is identical to that obtained by Fan [14]. As shown in [14], since c and d are left arbitrary, we could make the transformation d →(5p/q)d, c +d 2 r →−c, then solution (29) is exactly the same with that in [17]. Solution (30) is guaranteed by replacing the tanh-type solution of the Riccati equation by coth-type in (29). Solution (31) is a new one. We have recovered all the solutions that obtained by an extended tanh-function method and new solutions are obtained. Example 3. Consider the system (32) u t +ϑ x +uu x +pu xxt =0, 184 S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 (33) ϑ t +(uϑ) x +qu xxx =0, which is the variant Boussinesq equation [14,18]. Using u(x, t ) = U(ζ ), ϑ(x, t ) = V (ζ ), ζ = x + ct , Eqs. (32) and (33) become (34) cU +V +UU +cpU =0, (35) cV +(UV ) +qU =0. Balancing U with UU , and U with (UV) leads to the following ansatz: (36) U = 2 i=0 a i ω i + 2 i=0 b i ω −i , V = 2 i=0 α i ω i + 2 i=0 β i ω −i , where ω satisfies Eq. (3). Substituting Eq. (36) into Eqs. (34) and (35), and making use of Eq. (3) gives ba 2 1 +2bca 2 +16b 2 cpa 2 +2ba 0 a 2 +2ba 2 b 0 +2bα 2 =0, ca 1 +8bcpa 1 +a 0 a 1 +3ba 1 a 2 +a 1 b 0 +a 2 b 1 +α 1 =0, a 2 1 +2ca 2 +40bcpa 2 +2a 0 a 2 +2ba 2 2 +2a 2 b 0 +2α 2 =0, 2cpa 1 +a 1 a 2 =0, 12cpa 2 +a 2 2 =0, b 2 1 +2cb 2 +16bcpb 2 +2a 0 b 2 +2b 0 b 2 +2β 2 =0, bcb 1 +8b 2 cpb 1 +ba 0 b 1 +bb 0 b 1 +ba 1 b 2 +3b 1 b 2 +bβ 1 =0, bb 2 1 +2bcb 2 +40b 2 cpb 2 +2ba 0 b 2 +2bb 0 b 2 +2b 2 2 +2bβ 2 =0, 2b 3 cpb 1 +bb 1 b 2 =0, 12b 3 cpb 2 +bb 2 2 =0, −(c +2bcp +a 0 −ba 2 +b 0 )b 1 +bα 1 −β 1 +a 1 _ b(c +2bcp +a 0 +b 0 ) −b 2 _ =0, 8b 2 qa 2 +ba 2 α 0 +ba 1 α 1 +bcα 2 +ba 0 α 2 +bb 0 α 2 +ba 2 β 0 =0, 8bqa 1 +a 1 α 0 +cα 1 +a 0 α 1 +3ba 2 α 1 +b 0 α 1 +3ba 1 α 2 +b 1 α 2 +a 1 β 0 +a 2 β 1 =0, 20bqa 2 +a 2 α 0 +a 1 α 1 +cα 2 +a 0 α 2 +2ba 2 α 2 +b 0 α 2 +a 2 β 0 =0, 2qa 1 +a 2 α 1 +a 1 α 2 =0, 6qa 2 +a 2 α 2 =0, 8bqb 2 +b 2 α 0 +b 2 β 0 +b 1 β 1 +cβ 2 +a 0 β 2 +b 0 β 2 =0, 8b 2 qb 1 +bb 1 α 0 +bb 2 α 1 +bb 1 β 0 +bcβ 1 +ba 0 β 1 +bb 0 β 1 +3b 2 β 1 +ba 1 β 2 +3b 1 β 2 =0, 20b 2 qb 2 +bb 2 α 0 +bb 2 β 0 +bb 1 β 1 +bcβ 2 +ba 0 β 2 +bb 0 β 2 +2b 2 β 2 =0, 2b 3 qb 1 +bb 2 β 1 +bb 1 β 2 =0, 6b 3 qb 2 +bb 2 β 2 =0, _ b(c +a 0 +b 0 ) −b 2 _ α 1 −(2bp +α 0 −bα 2 +β 0 )b 1 −(c +a 0 −ba 2 +b 0 )β 1 + _ b(2bp +α 0 +β 0 ) −β 2 _ a 1 =0, which with the help of Mathematica yields three different cases. S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 185 Case 1. a 0 =−c −8bcp − q 2cp −b 0 , a 1 =0, a 2 =−12cp, b 1 =0, b 2 =0, α 0 =−4bq + q 2 4c 2 p 2 −β 0 , α 1 =0, α 2 =−6q, (37) β 1 =0, β 2 =0, where c =0 and b are left arbitrary. Case 2. a 0 =−c −8bcp − q 2cp −b 0 , a 1 =0, a 2 =0, b 1 =0, b 2 =−12b 2 cp, α 0 =−4bq + q 2 4c 2 p 2 −β 0 , α 1 =0, α 2 =0, (38) β 1 =0, β 2 =−6b 2 q, where b =0 and c =0. Case 3. a 0 =−c −8bcp − q 2cp −b 0 , a 1 =0, a 2 =−12cp, b 1 =0, b 2 =−12b 2 cp, α 0 =−4bq + q 2 4c 2 p 2 −β 0 , α 1 =0, α 2 =−6q, (39) β 1 =0, β 2 =−6b 2 q, where b =0 and c =0. According to case 1, we have three different types of travelling wave solutions for u and ϑ: Type 1: for b =0, (40) u =−c − q 2cp − 12cp (x +ct ) 2 , ϑ = q 2 4c 2 p 2 − 6q (x +ct ) 2 . Type 2: for b < 0, u =−c −8bcp − q 2cp +12bcptanh 2 _√ −b(x +ct ) _ , (41) ϑ =−4bq + q 2 4c 2 p 2 +6bq tanh 2 _√ −b(x +ct ) _ . Type 3: for b > 0, u =−c −8bcp − q 2cp −12bcptan 2 _√ b(x +ct ) _ , (42) ϑ =−4bq + q 2 4c 2 p 2 −6bq tan 2 _√ b(x +ct ) _ . The solutions for u and ϑ in Eqs. (40)–(42) are identical to those obtained by Fan [14]. The solutions obtained from case 2 can be obtained directly from Eqs. (37) and (38) by replacing tanh- and tan-functions by coth- and cot-functions, respectively, so there is no need to list them here. Finally, case 3 admits the following two types: 186 S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 Type 1: for b < 0, u =−c −8bcp − q 2cp +12bcpcoth 2 _√ −b(x +ct ) _ +12bcptanh 2 _√ −b(x +ct ) _ , (43) ϑ =−4bq + q 2 4c 2 p 2 +6bq coth 2 _√ −b(x +ct ) _ +6bq tanh 2 _√ −b(x +ct ) _ . Type 2: for b > 0, u =−c −8bcp − q 2cp −12bcpcot 2 _√ b(x +ct ) _ −12bcptan 2 _√ −b(x +ct ) _ , (44) ϑ =−4bq + q 2 4c 2 p 2 −6bq cot 2 _√ b(x +ct ) _ −6bq tan 2 _√ b(x +ct ) _ . The solutions for u and ϑ in Eqs. (43) and (44) cannot be obtained by extended tanh-function method [14]. Example 4. Consider the nonlinear equation (45) u t − _ u 2 _ xx −pu +qu 2 =0, where p, q =0 are constants. Eq. (45) is known as the nonlinear reaction–diffusion equation [8,14]. Applying the transformation u(x, t ) =U(ζ ), ζ =x +ct to Eq. (45) will result in (46) cU − _ U 2 _ −pU +qU 2 =0. Balancing U with (U 2 ) yields m=−1, which is not a positive integer. Following the idea of [8,9,14], we can introduce the transformation U =V −1 . Thus we get (47) −pV 3 −6(V ) 2 +(q −cV )V 2 +2VV =0. Balancing V 2 V with V V yields the required balancing number, m=1.Therefore, V becomes (48) V =(a 0 +b 0 ) +a 1 ω +b 1 ω −1 . After substitution of Eqs. (48) and (3) in Eq. (47), we get 4ba 0 a 1 +2qa 0 a 1 −3pa 2 0 a 1 −2bca 0 a 2 1 +4ba 1 b 0 +2qa 1 b 0 −6pa 0 a 1 b 0 −2bca 2 1 b 0 −3pa 1 b 2 0 −3pa 2 1 b 1 =0, −ca 2 0 a 1 −8ba 2 1 +qa 2 1 −3pa 0 a 2 1 −bca 3 1 −2ca 0 a 1 b 1 −3pa 2 1 b 1 −ca 1 b 2 1 +16a 1 b 1 −ca 2 1 b 1 =0, 4a 0 a 1 −2ca 0 a 3 1 −pa 3 1 +4a 1 b 0 −2ca 2 1 b 0 =0, −2a 2 1 −ca 3 1 =0, 4ba 0 b 1 +2qa 0 b 1 −3pa 2 0 b 1 +4bb 0 b 1 +2qb 0 b 1 −6pa 0 b 0 b 1 −3pb 2 0 b 1 +2ca 0 b 2 1 −3pa 1 b 2 1 +2cb 0 b 2 1 =0, bca 2 0 b 1 +16b 2 a 1 b 1 +2bca 0 b 0 b 1 +bcb 2 0 b 1 −8bb 2 1 +qb 2 1 −3pa 0 b 2 1 +bca 1 b 2 1 −3pb 0 b 2 1 +cb 3 1 =0, 4b 2 a 0 b 1 +4b 2 b 0 b 1 +2bca 0 b 2 1 +2bcb 0 b 2 1 −pb 3 1 =0, −2b 2 b 2 1 +bcb 3 1 =0, −pa 3 0 +b 2 0 (q −cb 0 ) +cb 2 0 b 1 −6b 2 1 −ba 2 1 (6b +cb 1 ) +a 2 0 (q −bca 1 −3pb 0 +cb 1 ) +a 1 _ −bcb 2 0 +2b 1 (16b +q −3pb 0 ) +cb 2 1 _ +a 0 _ −3pb 2 0 −6pa 1 b 1 +2b 0 (q −bca 1 +cb 1 ) _ =0, which yields three different cases: S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 187 Case 1. (49) b =− p 2 16c 2 , c =∓ p √ q , a 0 = p 2c 2 −b 0 , a 1 =− 2 c , b 1 =0. Case 2. (50) b =− p 2 16c 2 , c =∓ p √ q , a 0 = p 2c 2 −b 0 , a 1 =0, b 1 =− p 2 8c 3 . Case 3. (51) b =− q 64 , c =∓ p √ q , a 0 = p 2c 2 −b 0 , a 1 =− 2 c , b 1 = b 2c . According to case 1, b < 0 and we have (52) u = 2p q _ 1 ∓tanh _√ q 4 _ x ∓ p √ q t __ _ −1 . In case 3, b < 0 and hence (53) u =− p q _ −1 +e ( √ q/2)(±x−(p/ √ q)t ) _ . The solutions in Eq. (52) are identical to those obtained in [14]. The solutions obtained from case 2 can be obtained directly from Eq. (52) by replacing tanh-function by coth-function, respectively. So, there is no need to list them. The solutions given by Eq. (53) appear to be new. It has been shown that Eq. (45) have explicit (nonpropagating) similarity solutions that are explosively unstable, i.e., the instabilities tend to grow to infinite amplitude in a limited period of time. This typical behaviour is shown by solution (52) which is singular and blows up for x →−∞ [8]. There is much current interest in the formation of so-called “hot spots” or “blow-up” of solutions [14,19–21]. 3. Conclusion We have developed a modification of an extended tanh-function method, which is named modified extended tanh-function method. The validity and reliability of the method is tested by applying it to some nonlinear PDEs. We have successfully recovered the previously known solutions that had been found by an extended tanh-function method. We also found new exact solutions that are not obtained by both an extended tanh- and a hyperbolic- function methods. The travelling wave solutions derived in this Letter include soliton, periodical, rational and singular solutions. Soliton and periodical solutions are physically clear. The particle system describing the motion of pole of rational solutions of KdV and Boussinesq equation was analyzed in [14,18] which confirms the applicability of our obtained rational solutions. Acknowledgement We would like to express our sincere thanks to referees for helpful advice and suggestions. 188 S.A. Elwakil et al. / Physics Letters A 299 (2002) 179–188 References [1] P.G. Drazin, R.S. Johnson, Solitons: An Introduction, Cambridge University Press, Cambridge, 1989. [2] F. Cariello, M. Tabor, Physica D 39 (1989) 77. [3] W. Hereman, M. Takaoka, J. Phys. A 23 (1990) 4805. [4] M. Wang, Phys. Lett. A 213 (1996) 279. [5] E. Fan, Phys. Lett. A 265 (2000) 353. [6] X. Feng, Int. J. Theor. Phys. 39 (1) (2000) 207. [7] B.Q. Lu, B.Z. Xiu, Z.L. Pang, X.F. Jiang, Phys. Lett. A 175 (1993) 113. [8] W. Malfliet, Am. J. Phys. 60 (1992) 650. [9] E.J. Parkes, B.R. Duffy, Comput. Phys. Commun. 98 (1996) 288. [10] B.R. Duffy, E.J. Parkes, Phys. Lett. A 214 (1996) 271. [11] E.J. Parkes, B.R. Duffy, Phys. Lett. A 229 (1997) 217. [12] Y.T. Gao, B. Tian, Comput. Math. Appl. 33 (1997) 115. [13] V.V. Gudkov, J. Math. Phys. 38 (1997) 4794. [14] E. Fan, Phys. Lett. A 277 (2000) 212. [15] E. Fan, Z. Naturforsch. 56A (2001) 312. [16] C. Bai, Phys. Lett. A 288 (2001) 191. [17] E.J. Parkes, J. Phys. A: Math. Gen. 27 (1994) L497. [18] R.L. Sachs, Physica D 30 (1988) 1. [19] N.A. Kudryashov, D. Zargaryan, J. Phys. A: Math. Gen. 29 (1996) 8067. [20] N.F. Smyth, J. Austral. Math. Soc. Ser. B 33 (1992) 403. [21] P.A. Clarkson, E.L. Mansfield, Physica D 70 (1993) 250.