A note on Abundant new exact solutions for the (3+1)-dimensional Jimbo-Miwa equation

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1 A note on Abundant new exact solutions for the (3+1)-dimensional Jimbo-Miwa equation Nikolay A. Kudryashov, Dmitry I. Sinelshchikov Deartment of Alied Mathematics, National Research Nuclear University MEPhI, 31 Kashirskoe Shosse, Moscow, Russian Federation Abstract We demonstrate that eight from twenty seven solutions obtained by Li and Dai [Abundant new exact solutions for the (3+1)-dimensional Jimbo- Miwa equation, J. Math. Anal. Al. 361 (010) ] are wrong and do not satisfy the equation. The other nineteen exact solutions are not new solutions and can be found from the well known solution. Li and Dai [1] looked for exact solutions of the (3 + 1) Jimbo Miwa equation u xxxy + 3 u x u xy + 3 u y u xx + u yt 3 u xz = 0. (1) Authors [1] used the traveling wave ansatz u(x, y, z, t) = U(), = k x + l y + m z + ω t in Eq.(1). After integrating with resect to they obtained the nonlinear ordinary differential equation in the form k 3 l U + 3 k l (U ) + ( l ω 3 k m)u = C. () Here C is an constant of integration. Authors [1] alied the generalized Riccati equation method to look for exact solution of Eq.(). However we know very well the general solution of Eq.(). Let us demonstrate this fact. Denoting U = V () in Eq.() we have the following equation k 3 l V + 3 k l V + ( l ω 3 k m)v = C (3) Multilying Eq. (3) on V and integrating the equation with resect to we obtain the equation k 3 l (V ) + k l V 3 + l ω 3 k m V = C V + C 1, (4) where C 1 is an arbitrary constant. Using transformation ( V = k + l ω 3 k m ) 1 k 3 l from Eq. (4) we obtain (5) ( ) = 4 3 g g 3, (6) 1

2 where g = ( l ω 3 k m) + 1 k l C 1 k 6 l, g 3 = ( l ω 3 k m)3 108 k 4 l C k l C( l ω 3 k m) 0 k 9 l 3. We can see that the general solution of Eq.(6) and consequently Eq.() is exressed via the Weierstrass ellitic function []. As for the rational, eriodic and solitary wave solutions of () they can be found from Eq. (6) or Eq. (4) [3 8]. In the case of C 1 = C = 0 solution of Eq. (4) takes the form { V () = l ω 3 k m }) (1 k + tan 1 l ω 3 k m l k 3 ( + C ), (8) l where C in an arbitrary constant. Integrating solution (8) once with resect to, we obtain the solution of Eq. () { } lω 3 k m 1 l ω 3 k m U() = C 3 k k 3 tan l k 3 ( + C ). (9) l Here C 3 in an constant of integration. If C 1 = C = 0 and ω = 3 k m l we have rational solution of Eq. (4) (7) k V () = ( + C ), (10) where C in an arbitrary constant. Integrating (10) once with resect to, we obtain the rational solution of Eq. () U() = C 3 + k. (11) + C Here C 3 in an integration constant. The aer [1] contains the imortant misrint: the value of ω resented by Li and Dai in formula (.8) is wrong. The right value of ω is the following ω = 4 k3 l q r + 3 k m k 3 l. (1) l We checked all solutions by Li and Dai [1] with value of ω (1) and obtained that exressions u 7, u 10, u 11, u 1, u 17 u 19, u, u 3 and u 5 with lower sign do not satisfy Eq. (). Solutions u 1 u 4, u 6, u 8, u 9, u 13 u 16, u 0, u 1, u 4 u 6 and u 5 with uer sign are the same and can be obtained from formula (9). Substituting (1) in solution (9) we have U(z) = C 3 k { 1 } 4 q r tan 4 q r ( + C ). (13)

3 Taking into account formula at 4 q r < 0 from Eq. (13) we have U(z) = C 3 + k 4 q r tanh i tan iα = tanh α (14) { 1 } 4 q r( + C ). (15) We can see that (15) coincides with u 1 at C 3 = a 0 + k, C = 0. Using the identity tanh( + iπ/) = coth (16) in the case of C 3 = a 0 + k, C = iπ from (15) we obtain solution u by Li and Dai. Taking into account the identity tanh ± isech = tanh( ± iπ 4 ) (17) one can see that u 3 is equal to (15) at C 3 = a 0 + k, C = ± iπ. In the case of C 3 = a 0 + k, C = iπ and using formula coth + csch = tanh( + iπ ) (18) we can see that (15) is equal to u 4 with uer sign. Using formula coth csch = tanh (19) and suose that C 3 = a 0 + k, C = 0 in (15) we have solution u 4 with lower sign. Let us show that solution u 8 is articular case of (15). Denoting C 3 = a 0 + k, C = iπ + φ, φ = φ, φ = arccosh 4 q r from (15) we 4 q r 4 q r 3

4 obtain U() = a 0 + k + k 4 q r 4 q r coth φ = a q r cosh 4 q r φ + sinh 4 q r φ +k = sinh 4 q r φ ( ) 4qr cosh 4 q r φ cosh φ + sinh 4 q r φ sinh φ = a 0 + k = sinh 4 q r φ 4qr cosh 4 q r = a 0 + k = a 0 sinh 4 q r φ 4kqr cosh 4 q r 4 q r sinh 4 q r cosh 4 q r (0) In the same way one can show that u 9 is equal to (15). Using the same the double angle formulas for the hyerbolic functions we can see that u 1 is equal to u 9 and consequently to Eq. (15). By means of formulae tanh + coth = tanh( iπ ) (1) and in the case of C 3 = a 0 + k, C = 0 from (15) we have solution u 5 with uer sign. In the case of C 3 = a 0 + k, C = i π from (13) we obtain solution u 13 from [1]. Denoting C 3 = a 0 + k, C = π in (13) we have solution u 14 from [1]. Using formula tan ± sec = tan( ± π 4 ) () we obtain that u 15 is equal to (13) at C 3 = a 0 + k, C = ± π. Let us show that solution u 0 is articular case of Eq. (13). Denoting C 3 = a0 + k, C = π + φ, φ = φ, φ = arccos 4qr 4 q r 4qr from Eq. (13) 4

5 we obtain U() = a 0 + k + k 4 q r 4 q r cot + φ = a q r 4 q r 4 q r cos + φ + sin + φ +k = 4 q r sin + φ ( ) 4 q r 4 q r 4qr cos + φ cos φ + sin + φ sin φ = a 0 + k = 4 q r sin + φ 4 q r 4qr cos + = a 0 + k = a q r sin + φ 4 q r 4kqr cos 4 q r 4 q r 4 q r sin + cos (3) In the same way one can obtain that solution u 1 is articular case of Eq. (13). Using the double angle formulas for trigonometric functions one can see that u 4 is equal to u 1 and consequently to Eq. (15). By means of following identities k d u 5 = a 0 + d + cosh( ) sinh( ) = a d 0 + k d + e = ( ) { } 1 e +φ φ = a 0 + k + 1 = a 1 + e +φ 0 + k + k tanh (4) φ = ln d we have that u 5 is equal to Eq. (15) at r = 0, C 3 = a0 + k, C = φ. Using identity ( ) (cosh + sinh ) + φ d + cosh + sinh = tanh + 1, φ = ln 1 d (5) we obtain that u 6 is equal to Eq. (15) at r = 0, C 3 = a0 + k, C = φ. One can see that u 7 coincide with Eq. (10) if we take C 3 = a 0 and C = c1 q. So Li and Dai [1] found twenty seven solutions of Eq. (). However as we have seen above there is the general solution of Eq. (). We can also see that Li and Dai in aer [1] made the second common error from the list of errors given by Kudryashov in [9]: Some authors do not use the known general solutions of ordinary differential equations. 5

6 References [1] Zitian Li, Zhenge Dai, Abundant new exact solutions for the (3+1)- dimensional Jimbo-Miwa equation, J. Math. Anal. Al. 361 (010) [] E. T. Whittaker, G. N. Watson A Course of Modern Analysis (4th edition), Cambridge, Cambridge University Press: 008 [3] N.A. Kudryashov Exact soliton solutions of the generalized evolution equation of wave dynamics, Journal of Alied Mathematics and Mechanics 1988; 5 (3): [4] Kudryashov N.A. Exact solutions of generalized Kuramoto-Sivashinsky equation. Phys Lett A 1990; 147: [5] Kudryashov N.A., Loguinova N.B. Be careful with the Ex-function method. Commun Nonlinear Sci Numer Simul 009; 14: [6] Kudryashov N.A. On new travelling wave solutions of the KdV and the KdV-Burgers equations. Commun Nonlinear Sci Numer Simul 009; 14: [7] Parkes, E.J., Duffy, B.R. Travelling solitary wave solutions to a comound KdV-Burgers equation. Phys Lett A 1997; 9: 17-0 [8] E.J. Parkes A note on solitary-wave solutions to comound KdV-Burgers equations. Phys Lett A 003; 317: 44-8 [9] Kudryashov N.A. Seven common errors in finding exact solutions of nonlinear differential equations. Commun Nonlinear Sci Numer Simul 009; 14: