Solving the Generalized Kaup Kupershmidt Equation

Transcription

1 Adv Studies Theor Phys, Vol 6, 2012, no 18, Solving the Generalized Kaup Kupershmidt Equation Alvaro H Salas Department of Mathematics Universidad de Caldas, Manizales, Colombia Universidad Nacional de Colombia, Manizales FIZMAKO Research Group asalash2002@yahoocom Abstract In this paper we apply the Cole-Hopf transformation to find soliton solutions and the simplified Hirota s method to find one and two soliton solutions for the generalized Kaup Kupershmidt equation Mathematics Subject Classification: 35C05 Keywords: Kaup Kupershmidt equation, KK equation, Cole-Hopf transformation, nonlinear pde, fifth order dv, nonlinear evolution equation, simplified Hirota s method, exp function method 1 Introduction Hirota s method has been one of the most successful direct techniques for constructing exact solutions of various nonlinear PDEs from soliton theory In this paper we mae use of a simplified version of this method [2] without ever using the bilinear forms Furthermore, the simplified method can easily be implemented in any symbolic manipulation pacage This papaer is organized as follows : We first find a soliton solution for the general fkdv by using a Cole-Hopf transformation [1][3] In the next section we apply the simplified Hirota s method [2] to find multisoliton solutions for the generalized Kaup Kupershmidt equation 2 Soliton solutions to the general fkdv by the Cole-Hopf transformation The general fifth order KdV equation reads 21 u t + ωu xxxxx + αuu xxx + βu x u xx + γu 2 u x =0,

2 880 A H Salas where α, β, γ and ω are arbitrary real parameters To obtain a soliton solution, we apply the generalized Cole-Hopf transformation 22 u = A 2 ln fx, t+b, x x where A and B are some constants and fx, t = 1 + expθ, θ = x ct + δ Upon substitution of 22 into 21, we obtain the equation Aω 7 + ABα 5 + AB 2 γ 3 Ac 2 e θ + 57Aω 7 + A 2 α 7 + A 2 β 7 9ABα 5 +2A 2 Bγ 5 +3AB 2 γ 3 3Ac 2 e 2θ + 302Aω 7 11A 2 α 7 5A 2 β 7 + A 3 γ 7 10ABα 5 +2A 2 Bγ 5 +2AB 2 γ 3 2Ac 2 e 3θ + 302Aω 7 +11A 2 α 7 +5A 2 β 7 A 3 γ 7 +10ABα 5 2A 2 Bγ 5 2AB 2 γ 3 +2Ac 2 e 4θ + 57Aω 7 A 2 α 7 A 2 β 7 +9ABα 5 2A 2 Bγ 5 3AB 2 γ 3 +3Ac 2 e 5θ + Aω 7 ABα 5 AB 2 γ 3 + Ac 2 e 6θ =0 Equating the coefficients of e θ, e 2θ,, e 6θ to zero, we obtain an algebraic system Solving this system yields : First solution : A = 6α +3β 3 2α + β 2 40ωγ, B = γ 2α 2 β + 2α + β 2 40ωγ, 12ω 5 γ+ β 2α β + 2α + β 2 40ωγ c = 23 ux, t = 2 R 2α β x + 5 R 2α ββ +12ωγt + δ 5, x + 5 R 2α ββ +12ωγt + δ +1 where R = 2α + β 2 40ωγ Second solution : A = 3 2α + β + 2α + β 2 40γω 2α 2 + β + 2α + β 2 40γω, B =, γ

3 24 ux, t = Solving the generalized Kaup Kupershmidt equation 881 β 5 2α + β + 2α + β 2 40γω 12γω c = 2 R +2α + β where R = 2α + β 2 40ωγ x 5 βr +2α + β 12ωγt + δ 5, x 5 βr +2α + β 12ωγt + δ +1 3 The generalized Kaup Kupershmidt equation This equation reads 31 u t +10abuu xxx +25abu x u xx + bu 2 u x +20a 2 bu xxxxx =0, where a 0 and b 0 The well-nown Kaup Kupershmidt equation [5][2] KK equation 32 u t +10uu xxx +25u x u xx +20u 2 u x + u xxxxx =0, is obtained from 31 for the values a =1/20 and b =20 From Eqs 23 and 24 we obtain the following solutions of 31 : u 1 x, t = a2 x 5 4 a2 b 5 t + δ +1 u 2 x, t =20a x 220a 2 b 5 t + δ+1 Solution 33 has the form 22 and it corresponds to values A =30a, B = 5a 2 /2 and c = 5 4 a2 b 5 From 33 and 34 we obtain periodic solutions if we change by 1 and δ by 1δ These solutions are : 35 u 3 x, t = 5 2 a2 1 6 cos x 5 4 a2 b 5 t + δ +1 u 4 x, t = 20a cos x 220a 2 b 5 t + δ+1 31 Simplified Hirota s method We will apply this method to find multisolitons solutions for Eq 31 Using the Cole-Hopf transformation 37 u =30a 2 ln f, x x we get a fourth degree equation in f = fx, t and its derivatives, f 4 20a 2 bf xxxxxxx + f xxt + f 3 50a 2 bf xxx f xxxx 120a 2 bf xx f xxxxx 140a 2 bf x f xxxxxx 2f x f xt f t f xx +

4 882 A H Salas f 2 200a 2 bf x f 2 xxx 150a2 bf 2 xx f xxx+450a 2 bf x f xx f xxxx +540a 2 bf 2 x f xxxxx+2f t f 2 x + f 1 150a 2 b3f x f 3 xx 8f 3 xf xxxx + 300a 2 bf 3 x4f x f xxx 3f 2 xx =0 Obviously, this last equation can be written as 38 f 4 Lf+f 3 N 1 f,f+f 2 N 2 f,f,f+fn 3 f,f,f,f+n 4 f,f,f,f,f =0, where the linear operator L and the nonlinear operators N 1, N 2, N 3, N 4 are defined as 39 Lf =20a 2 bf xxxxxxx + f xxt, N 1 f,g =50a 2 bf xxx g xxxx 120a 2 bf xx g xxxxx 140a 2 bf x g xxxxxx 310 2f x g xt f t g xx, N 2 f,g,h = 200a 2 bf x g xxx h xxx 150a 2 bf xx g xx h xxx + 450a 2 bf x g xx h xxxx + 540a 2 bf x g x h xxxxx +2f t g x h x, 311 N 3 f, g, h, φ = 150a 2 b3f x g xx h xx φ xx 8f x g x h x φ xxxx, 312 N 4 f, g, h, φ, ϕ = 300a 2 b4f x g x h x φ x ϕ xxx 3f x g x h x φ xx ϕ xx, for auxiliary functions form f, g, h, φ and ϕ We see a solution of 38 in the 313 fx, t =1+ ɛ n f n x, t n=1 We substitute 313 into 38 and equate the coefficients of different powers of ɛ to zero The following perturbation scheme follows: vspace05cm

5 Solving the generalized Kaup Kupershmidt equation Oɛ 1 :Lf 1 =0 315 Oɛ 2 :Lf 2 = N 1 f 1,f 1 Oɛ 3 :Lf 3 =f 1 N 1 f 1,f 1 N 1 f 1,f 2 N 1 f 2,f N 2 f 1,f 1,f 1 Oɛ 4 :Lf 4 = N 1 f 1,f 1 f N 1 f 1,f 2 f 1 + N 1 f 2,f 1 f 1 + 2N 2 f 1,f 1,f 1 f 1 + f 2 N 1 f 1,f 1 N 1 f 1,f 3 N 1 f 2,f 2 N 1 f 3,f 1 N 2 f 1,f 1,f 2 N 2 f 1,f 2,f 1 N 2 f 2,f 1,f N 3 f 1,f 1,f 1,f 1 Oɛ 5 :Lf 5 = N 1 f 2,f 1 f 1 2 3N 2 f 1,f 1,f 1 f N 1 f 1,f 3 f 1 + N 1 f 2,f 2 f 1 + N 1 f 3,f 1 f 1 + 2N 2 f 1,f 1,f 2 f 1 +2N 2 f 1,f 2,f 1 f 1 +2N 2 f 2,f 1,f 1 f 1 + 3N 3 f 1,f 1,f 1,f 1 f 1 +f 1 3 2f 2 f 1 + f 3 N 1 f 1,f 1 + f 2 f 1 2 N 1 f 1,f 2 N 1 f 1,f 4 +f 2 N 1 f 2,f 1 N 1 f 2,f 3 N 1 f 3,f 2 N 1 f 4,f 1 +2f 2 N 2 f 1,f 1,f 1 N 2 f 1,f 1,f 3 N 2 f 1,f 2,f 2 N 2 f 1,f 3,f 1 N 2 f 2,f 1,f 2 N 2 f 2,f 2,f 1 N 2 f 3,f 1,f 1 N 3 f 1,f 1,f 1,f 2 N 3 f 1,f 1,f 2,f 1 N 3 f 1,f 2,f 1,f N 3 f 2,f 1,f 1,f 1 N 4 f 1,f 1,f 1,f 1,f 1 Noticeably, the number of terms in the right hand side RHS of the equations grows rapidly as the order in ɛ increases 311 The One-Soliton solution To find the one-soliton solution, tae 319 f 1 = expθ, with θ = x ct + δ Equation 314 gives the dispersion law c =20a 2 b 5 To solve 315, we first compute its RHS, 320 N 1 f 1,f 1 = 150a 2 7 exp2θ

6 884 A H Salas Thus, f 2 will be of the form f 2 = λ exp2θ Calculating the left hand side LHS of 315, 321 Lf 2 = 2400a 2 b 7 e 40a2 bt 5 +2x+2δ = 2400λa 2 b 7 exp2θ and equating it with 320 we get λ =1/16 so 322 f 2 = 1 16 exp2θ It is straightforward to chec that f n = 0 for n 3 Therefore, using 37 and 313 with ɛ = 1, the one-soliton solution of 31 generated by f = 1 + expθ exp2θ = 1 + expx 20a2 b 5 t + δ exp2x 20a2 b 5 t + δ is 480a 4+e 2 x 20a2 b 5t+δ +16e x 20a2 b 5 t+δ 323 ux, t = 16 + e x 20a2 b 5 t+δ +16e x 20a2 b 5 t+δ The Two-Soliton Solution For the two-soliton solution, we start with 324 f 1 = expθ 1 + expθ 2, where θ i = i x c i t + δ i, i =1, 2 From 314 we get c i =20a 2 bi 5, i =1, 2 To find f 2, the RHS of 315 has to be calculated We obtain N 1 f 1,f 1 = 150a 2 b1 7 e2θ a 2 b2 7 e2θ a 2 be θ 1+θ e θ 1 +θ 2 Obviously, f 2 must be of the form 326 f 2 = p exp2θ 1 +q exp2θ 2 +r expθ 1 + θ 2 In contrast to what happened for the KdV, the Lax, and SK equations, the terms in exp2θ i no longer drop out We now substitute 326 into 315 Computation of the LHS yields Lf 2 = 2400a 2 b1 7 pe2θ a 2 b2 7 qe2θ a 2 b 1 2 r e θ 1 +θ 2 Equating 325 and 327 gives p = q = 1 16 and 328 Therefore, 329 r = f 2 = 1 16 expθ expθ expθ 1 + θ 2

7 Solving the generalized Kaup Kupershmidt equation 885 Proceeding in a similar way with 316 we find 330 f 3 = s exp2θ 1 + θ 2 + expθ 1 +2θ 2, s = To find f 4, equation 317 has to be solved, which leads to 332 f 4 = s 2 exp2θ 1 +2θ 2 = exp2θ 1 +2θ 2 After verifying that all f n will be zero, for n 5, the form of f for ɛ =1 will be f = expθ 1 + expθ exp2θ exp2θ 2+rexpθ 1 + θ s exp2θ 1 + θ 2 + expθ 1 +2θ 2 + s 2 exp2θ 1 +2θ 2, where r and s are given by 328 and 331, respectively The two-soliton solution of Eq 31 could be obtained by substituting 333 into 37 In a similar fashion, we may find three and four soliton solutions For additional details, see [2] 4 Conclusions We have obtained many solutions of the generalized KK equation by using three distinct methods The Exp-function method is a promising method because it can establish a variety of solutions of distinct physical structures This method allows us to obtain many new solutions for evolution equations References [1] Cole J D, A quasi-linear parabolic equation occuring in aerodynamics,quart Appl Math, 1951, 9, pages [2] Hereman W & Nuseir A, Symbolic methods to find exact solutions of nonlinear partial differential equations, 14th IMACS World Congress Atlanta, July, 1994 [3] Hopf E, The partial differential equation u t + uu x = u xx, Comm Pure Appl Math, 3, 1950, [4] He JH & Wu XH, Exp-function method for nonlinear wave equations, Chaos, Solitons and Fractals, Volume 30, Issue 3, Nov 2006, Pages: [5] Satsuma J & Kaup DJ, J Phys Soc Japan 43, Received: March, 2012