Exact Jacobian Elliptic Function Solutions to sine-gordon Equation

Transcription

1 Commun. Theor. Phys. (Beijing, China 44 (005 pp c International Academic Publishers Vol. 44, No., July 5, 005 Exact Jacobian Elliptic Function Solutions to sine-gordon Equation FU Zun-Tao,,, YAO Zhen-Hua, LIU Shi-Kuo, and LIU Shi-Da, School of Physics & Laboratory for Severe Storm and Flood Disaster, Peking University, Beijing 0087, China State Key Laboratory for Turbulence and Complex Systems, Peking University, Beijing 0087, China (Received November 30, 004 Abstract In this paper, four transformations are introduced to solve single sine-gordon equation by using the knowledge of elliptic equation and Jacobian elliptic functions. It is shown that different transformations are required in order to obtain more kinds of solutions to the single sine-gordon equation. PACS numbers: Ge Key words: Jacobian elliptic function, periodic wave solution, transformation, sine-gordon equation Introduction The sine-gordon-type equations, including single sine- Gordon (SSG for short equation u xt = sin u, ( double sine-gordon (DSG for short equation u xt = sin u + β sin(u, ( and triple sine-gordon (TSG for short equation u xt = sin u + β sin(u + γ sin(3u (3 are widely applied in physics and engineering. For example, DSG equation is a frequent object of study in numerous physical applications, such as Josephson arrays, ferromagnetic materials, charge density waves, smectic liquid crystal dynamics. [ 5] Actually, SSG equation and DSG equation also arise in nonlinear optics 3 He spin waves and other fields. In a resonant fivefold degenerate medium, the propagation and creation of ultra-short optical pulses, the SSG and DSG models are usually used. However, in some cases, one has to consider other sine-gordon equations. For instance, TSG equation is used to describe the propagation of strictly resonant sharp line optical pulses. [6] Due to the wide applications of sine-gordon type equations, many solutions to them, such as tan coth ξ, tan tanh ξ, tan sech ξ, tan sn ξ and so on, have been obtained in different functional forms by different methods. [7 ] Due to the special forms of the sine- Gordon-type equations, it is very difficult to solve them directly, so it needs some transformations. In this paper, based on the introduced transformations, we will show systematical results about solutions for SSG equation ( by using the knowledge of elliptic equation and Jacobian elliptic functions. [3 9] The First Kind of Transformation and Solutions to SSG Equation In order to solve the sine-gordon-type equations, certain transformations must be introduced. For example, the transformation u = tan v or v = tan u, (4 has been introduced in Refs. [7] and [9] to solve DSG equation. When the transformation (4 is considered, the SSG equation ( can be rewritten as ( + v v tx vv t v x v v 3 = 0. (5 Equation (5 can be solved in the frame v = v(ξ, ξ = k(x ct, (6 where k and c are wave number and wave speed, respectively. Substituting Eq. (6 into Eq. (5, we have ( + v d v ( dv dξ v + v + v 3 = 0, dξ = k c. (7 And we suppose that equation (7 has the following solution: j=n v = v( b j y j, b n 0, y(ξ, (8 j=0 where y satisfies elliptic equation, [3 0] with y = dy/dξ, y = a 0 + a y + a 4 y 4, a 4 0 (9 y = a y + a 4 y 3. (0 The project supported by National Natural Science Foundation of China under Grant No Correspondence author, fuzt@pku.edu.cn Correspondence address

2 4 FU Zun-Tao, YAO Zhen-Hua, LIU Shi-Kuo, and LIU Shi-Da Vol. 44 There n in Eq. (8 can be determined by the partial balance between the highest order derivative terms and the highest degree nonlinear term in Eq. (7. Here we know that the degree of v is and from Eqs. (9 and (0, one has O(v = O(y n = n, ( O(y = O(y 4 = 4, O(y = O(y 3 = 3, ( and actually one can have So one has O(y (l = l +. (3 O(v = n, O(v = n +, O(v = n +, O(v (l = n + l. (4 For SSG equation (, we have n =, so the ansatz solution of Eq. (7 can be rewritten as v = b 0 + b y, b 0. (5 Substituting Eq. (5 into Eq. (7 results in an algebraic equation for y, which can be used to determine expansion coefficients in Eq. (5 and some constraints can also be obtained. Here we have b 0 = 0, b = a + a 0 = a 4 a, (6 from which the constraints can be determined as a + a 4 > 0, > 0, a > 4a 0 a 4. (7 a 0 a Considering the constraints (7, the solutions to the elliptic equation (9 can be used to derive the final results, here eleven cases can be obtained. Case If a 0 = 0, a =, a 4 =, csch ξ, b 0 = 0, b = ±, k, (8 Eq. ( is u = tan csch ξ. (9 Case If a 0 =, a = m, a 4 = m, where 0 m is called modulus of Jacobian elliptic functions, [0 3] m cn ξ, b 0 = 0, b = ±, k, 0 < m <, (0 where k is an arbitrary constant, and cn ξ is Jacobian elliptic cosine function. [0 3] So the solution to SSG Eq. ( is m u = tan cn ξ. ( Case 3 If a 0 = m, a = m, a 4 =, nc ξ cn ξ, b 0 = 0, b = ± m, k, 0 < m <, ( u 3 = tan nc ξ. (3 m Case 4 If a 0 =, a = m, a 4 = (m m, sd ξ sn ξ, b 0 = 0, b = ±m k, 0 < m <, (4 where k is an arbitrary constant, and sn ξ is Jacobian elliptic sine function and is Jacobian elliptic function of the third kind. [0 3] So the solution to SSG Eq. ( is u 4 = tan m sd ξ. (5 Case 5 If a 0 =, a = m, a 4 =, cs ξ cn ξ sn ξ, b 0 = 0, b = ±, m k, 0 < m <, (6 u 5 = tan cs ξ. (7 Case 6 If a 0 =, a = m, a 4 =, cs ξ cn ξ sn ξ, b 0 = 0, b = ±, m k, 0 < m <, (8 u 6 = tan cs ξ. (9 Case 7 If a 0 = m (m, a = m, a 4 =, ds ξ sn ξ, b 0 = 0, b = ± m, k, 0 < m, (30 u 7 = tan ( ± m ds ξ. (3 Actually, when m, u 7 recovers u.

3 No. Exact Jacobian Elliptic Function Solutions to sine-gordon Equation 5 Case 8 If a 0 =, a = m, a 4 =, sc ξ sn ξ cn ξ, b 0 = 0, b = ±, m k, 0 < m <, (3 ( u 8 = tan ± sc ξ. (33 Case 9 If a 0 =, a = m, a 4 =, sc ξ sn ξ cn ξ, b 0 = 0, b = ±, m k, 0 < m <, (34 u 9 = tan sc ξ. (35 Case 0 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, cn ξ ± sn ξ, b 0 = 0, b = ± (, mk, 0 < m <, (36 u 0 = tan cn ξ ( ± sn ξ. (37 Case If a 0 = ( /4, a = ( + m /, a 4 = ( /4, cn ξ ± sn ξ, b 0 = 0, b = ± ( + m, mk, 0 < m <, (38 u = tan ( + m cn ξ ± sn ξ. (39 Remark The solutions u 0 and u in terms of rational functions have not been reported in the literature, they are new solutions to SSG Eq. (. Remark The solutions from u to u in terms of Jacobian elliptic functions have not been given in Ref. []. Remark 3 form of In Ref. [0], Peng solved DSG equation in u xt = sin u + λ sin(u, (40 and obtained some solutions in terms of Jacobian elliptic functions. He pointed out that he can obtain solutions to SSG Eq. ( with = when λ = 0. However, his conclusion is wrong, for the coefficients of solutions he obtained in terms of sn (Eq. (40 in Ref. [0], dn (Eq. (46 in Ref. [0], ns (Eq. (5 in Ref. [0] and dc (Eq. (5 in Ref. [0] are imaginary, but in fact they should be real. For example, from the constraint (4 in Ref. [0] ( k ω 4λ( + m kω + 4λ = 0, (4 if λ = 0, we have kω = ±. (4 Substituting Eq. (4 into solution (40 in Ref. [0], ( ( + m kω + λ + u = arctan ± kω sn (kx ωt, (43 we can derive the coefficient [ ( + m kω + ]/kω to be i m or i, with i. So solutions given by Peng in terms of sn (Eq. (40 in Ref. [0], dn (Eq. (46 in Ref. [0], ns (Eq. (5 in Ref. [0] and dc (Eq. (5 in Ref. [0] are not real solutions, which is contrary to the origin of SSG Eq. (. Remark 4 Based on the above results, we can see that when the auxiliary equation, such as elliptic equation (9, is applied to solve nonlinear evolution equations, the constraints must be involved, otherwise, the obtained solutions may be trivial. 3 The Second Kind of Transformation and Solutions to SSG Equation The second transformation under consideration is u = tan ( or v v = tanu, (44 which has been introduced in Ref. [7] to solve DSG equation. When the transformation (44 is considered, the SSG Eq. ( can be rewritten as ( + v v tx vv t v x + v + v 3 = 0. (45 We can see that the difference between Eq. (5 and Eq. (45 is that the in Eq. (5 is replaced by in Eq. (45, so the solutions to Eq. ( under the transformation (44 can be easily obtained by replacing by and v by /v in solutions from u to u. Case If a 0 = 0, a =, a 4 =, the solution to SSG Eq. ( is u = tan sinh ξ. (46

4 6 FU Zun-Tao, YAO Zhen-Hua, LIU Shi-Kuo, and LIU Shi-Da Vol. 44 Case If a 0 =, a = m, a 4 = m, the solution to SSG Eq. ( is u 3. Case 3 If a 0 = m, a = m, a 4 =, the solution to SSG Eq. ( is u. Case 4 If a 0 =, a = m, a 4 = (m m, the solution to SSG Eq. ( is u 7. Case 5 If a 0 =, a = m, a 4 =, the solution to SSG Eq. ( is u 8. Case 6 If a 0 = m (m, a = m, a 4 =, the solution to SSG Eq. ( is u 4. Case 7 If a 0 =, a = m, a 4 =, the solution to SSG Eq. ( is u 9. Case 8 If a 0 =, a = m, a 4 =, the solution to SSG Eq. ( is u 5. Case 9 If a 0 =, a = m, a 4 =, the solution to SSG Eq. ( is u 6. Case 0 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, the solution to SSG Eq. ( is u 3 = tan ( ± sn ξ cn ξ. (47 Case If a 0 = ( /4, a = ( + m /, a 4 = ( /4, the solution to SSG Eq. ( is ( + m u 4 = tan ± sn ξ. (48 cn ξ Remark Most of the solutions from u to u 4 in terms of Jacobian elliptic functions have not been given in the literature. 4 The Third Kind of Transformation and Solutions to SSG Equation The third transformation is introduced in the form u = sin v or v = sin u, (49 and the SSG equation ( can be rewritten as ( v v tx + vv t v x v( v = 0. (50 In the travelling wave frame (6, the formal solution of Eq. (50 by the elliptic equation expansion method (8 can be written as Eq. (5. Substituting Eq. (5 into Eq. (50, we have a4 k b 0 = 0, b = ± c, k a ± a 4a 0 a 4 (5 with constraints a 4 k c > 0, a 4a 0 a 4 0. (5 Similarly, there are some cases to consider. For example, Case If a 0 =, a = ( + m, a 4 = m, sn ξ, b 0 = 0, b = ±m, k, 0 < m, (53 Case u 5 = sin m sn ξ. (54 If a 0 =, a = ( + m, a 4 = m, sn ξ, b 0 = 0, b = ±, k m, 0 < m, (55 u 6 = sin sn ξ. (56 Actually, when m, u 5 and u 6 all recover, u 7 = sin tanh ξ, k. (57 Case 3 If a 0 =, a = m, a 4 = m, cn ξ, b 0 = 0, b = ±, k m, 0 < m, (58 u 8 = sin cn ξ. (59 Actually, when m, u 8 recovers Case 4 u 9 = sin sech ξ, k. (60 If a 0 = m, a = m, a 4 =,, b 0 = 0, b = ±, k, 0 < m, (6 u 0 = sin. (6 Actually, when m, u 0 recovers u 9. Case 5 If a 0 = m, a = m, a 4 =,, b 0 = 0, b = ±, (m k, 0 < m <, (63 u = sin. (64

5 No. Exact Jacobian Elliptic Function Solutions to sine-gordon Equation 7 Case 6 If a 0 = m, a = ( + m, a 4 =, ns ξ, b 0 = 0, b = ± m, k m, 0 < m, (65 u = sin ( ± m ns ξ. (66 Case 7 If a 0 = m, a = ( + m, a 4 =, ns ξ, b 0 = 0, b = ±, k, 0 < m, (67 u 3 = sin ns ξ. (68 Actually, when m, u and u 3 all recover, u 4 = sin coth ξ, k. (69 Case 8 If a 0 = m, a = m, a 4 =, nc ξ, b 0 = 0, b = ±, ( k, 0 < m <, (70 Case 9 u 5 = sin nc ξ. (7 If a 0 =, a = m, a 4 = m, nd ξ, b 0 = 0, b = ±, (m k, 0 < m <, (7 Case 0 u 6 = sin nd ξ. (73 If a 0 =, a = m, a 4 = m, nd ξ, b 0 = 0, b = ±, k, 0 < m <, (74 u 7 = sin nd ξ. (75 Case If a 0 =, a = m, a 4 = (m m, sd ξ, b 0 = 0, b = ±, m k, 0 < m <, (76 Case u 8 = sin sd ξ. (77 If a 0 =, a = ( + m, a 4 = m, cd ξ, b 0 = 0, b = ±m, k, 0 < m, (78 Case 3 u 9 = sin m cd ξ. (79 If a 0 =, a = ( + m, a 4 = m, cd ξ, b 0 = 0, b = ±, m k, 0 < m, (80 u 30 = sin cd ξ. (8 Case 4 If a 0 = m (m, a = m, a 4 =, ds ξ, b 0 = 0, b = ±, ( k, 0 < m <, (8 u 3 = sin ds ξ. (83 Case 5 If a 0 = m, a = ( + m, a 4 =, dc ξ, b 0 = 0, b = ± m, m k, 0 < m, (84 u 3 = sin ( ± m dc ξ. (85 Case 6 If a 0 = m, a = ( + m, a 4 =, dc ξ, b 0 = 0, b = ±, k, 0 < m, (86 u 33 = sin dc ξ. (87 Apart from these Jacobian elliptic solutions, y also has some rational solutions in terms of Jacobian elliptic functions. For example,

6 8 FU Zun-Tao, YAO Zhen-Hua, LIU Shi-Kuo, and LIU Shi-Da Vol. 44 Case 7 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, ± m sn ξ, b 0 = 0, b = ± + m, 4 ( + m k, 0 < m <, (88 u 34 = sin. (89 + m ± m sn ξ Case 8 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, ± m sn ξ, b 0 = 0, b = ±, 4 ( k, 0 < m <, (90 u 35 = sin. (9 ± m sn ξ Case 9 If a 0 = m /4, a = ( m /, a 4 = m /4, m sn ξ ±, b m 0 = 0, b = ± m +, 4 ( m + k, 0 < m, (9 u 36 = sin m m + m sn ξ. (93 ± Case 0 If a 0 = m /4, a = ( m /, a 4 = m /4, m sn ξ ±, b m 0 = 0, b = ± m, 4 ( m k, 0 < m, (94 u 37 = sin m m m sn ξ. (95 ± When m, u 36 and u 37 all recover ( u 38 = sin ± tanh ξ, ± sech ξ 4 k. (96 Case If a 0 = /4, a = ( m /, a 4 = m 4 /4, sn ξ ±, b 0 = 0, b = ± m +, m 4 ( m + k, 0 < m, (97 u 39 = sin m m + sn ξ. (98 ± Case If a 0 = /4, a = ( m /, a 4 = (m 4 /4, sn ξ ±, b 0 = 0, b = ± m, m 4 ( m k, 0 < m, (99 u 40 = sin m m sn ξ.(00 ± When m, u 39 and u 40 all recover u 38. Remark Transformation (49 and the solutions from u 5 to u 40 in terms of Jacobian elliptic functions have not been given in the literature. 5 The Fourth Kind of Transformation and Solutions to SSG Equation The fourth transformation is introduced in the form u = cos v or v = cos u, (0 and the SSG equation ( can be rewritten as ( v v tx + vv t v x + v( v = 0. (0 We can see that the difference between Eq. (0 and Eq. (50 is that the in Eq. (50 is replaced by in Eq. (0, so the solutions to Eq. ( under the transformation (0 can be easily obtained by replacing by and sin by cos in solutions from u 5 to u 40. So we have b 0 = 0, b = ± a 4k c, k a ± with constraints a 4a 0 a 4, (03 a 4k c > 0, a 4a 0 a 4 0. (04 Similarly, there are some cases to be considered. For example,

7 No. Exact Jacobian Elliptic Function Solutions to sine-gordon Equation 9 Case If a 0 =, a = ( + m, a 4 = m, the solution to SSG u 4 = cos m sn ξ. (05 Case If a 0 =, a = ( + m, a 4 = m, the solution to SSG u 4 = cos sn ξ. (06 Actually, when m, u 4 and u 4 all recover u 43 = cos tanh ξ, k. (07 with Case 3 If a 0 =, a = m, a 4 = m, the solution to SSG u 44 = cos cn ξ. (08 Actually, when m, u 44 recovers u 45 = cos sech ξ, k. (09 Case 4 If a 0 = m, a = m, a 4 =, the solution to SSG u 46 = cos. (0 Actually, when m, u 46 recovers u 45. Case 5 If a 0 = m, a = m, a 4 =, the solution to SSG u 47 = cos ( ±. ( Case 6 If a 0 = m, a = ( + m, a 4 =, the solution to SSG u 48 = cos ( ± m ns ξ. ( Case 7 If a 0 = m, a = ( + m, a 4 =, the solution to SSG u 49 = cos ns ξ. (3 Actually, when m, u 48 and u 49 all recover u 50 = cos coth ξ, k. (4 Case 8 If a 0 = m, a = m, a 4 =, the solution to SSG u 5 = cos nc ξ. (5 Case 9 If a 0 =, a = m, a 4 = m, the solution to SSG u 5 = cos nd ξ. (6 Case 0 If a 0 =, a = m, a 4 = m, the solution to SSG ( u 53 = cos ± nd ξ. (7 Case If a 0 =, a = m, a 4 = (m m, the solution to SSG equation ( ( is u 54 = cos ± sd ξ. (8 Case If a 0 =, a = ( + m, a 4 = m, the solution to SSG u 55 = cos m cd ξ. (9 Case 3 If a 0 =, a = ( + m, a 4 = m, the solution to SSG u 56 = cos cd ξ. (0 Case 4 If a 0 = m (m, a = m, a 4 =, the solution to SSG u 57 = cos ds ξ. ( Case 5 If a 0 = m, a = ( + m, a 4 =, the solution to SSG u 58 = cos ( ± m dc ξ. ( Case 6 If a 0 = m, a = ( + m, a 4 =, the solution to SSG u 59 = cos dc ξ. (3 Apart from these Jacobian elliptic solutions, y also has some rational solutions in terms of Jacobian elliptic functions, for example, Case 7 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, the solution to SSG equation ( is u 60 = cos. (4 + m ± m sn ξ Case 8 If a 0 = ( /4, a = ( + m /, a 4 = ( /4, the solution to SSG equation ( is u 6 = cos. (5 ± m sn ξ Case 9 If a 0 = m /4, a = ( m /, a 4 = m /4, the solution to SSG u 6 = cos m m + m sn ξ. (6 ± Case 0 If a 0 = m /4, a = ( m, a 4 = m /4, the solution to SSG u 63 = cos m m m sn ξ. (7 ± When m, u 64 and u 65 all recover ( u 64 = cos ± tanh ξ, 4 ± sech ξ k. (8

8 30 FU Zun-Tao, YAO Zhen-Hua, LIU Shi-Kuo, and LIU Shi-Da Vol. 44 Case If a 0 = /4, a = ( m /, a 4 = m 4 /4, the solution to SSG u 65 = cos m m + sn ξ.(9 ± Case If a 0 = /4, a = ( m /, a 4 = m 4 /4, the solution to SSG u 66 = cos ( m ± m sn ξ. (30 ± When m, u 65 and u 66 all recover u 64. Remark Transformation (0 and the solutions from u 4 to u 66 in terms of Jacobian elliptic functions have not been given in the literature. 6 Conclusion In this paper, four transformations are introduced to solve single sine-gordon equation by using the knowledge of elliptic equation and Jacobian elliptic functions. It is shown that different transformations are required in order to obtain more kinds of solutions to the single sine- Gordon equation. Here some new solutions have not been reported in the literature. It is shown that different transformations play different roles in obtaining exact solutions, some transformations may not work for a specific parameter of SSG equation (. Of course, there are still more efforts needed to explore what kinds of transformations are more suitable to solving sine-gordon equation. Because different transformations result in different partial balances for sine-gordon equation, which will lead to different expansion truncations in the elliptic equation expansion method. Finally, these will result in different solutions of the sine-gordon equation. References [] E. Goldobin, A. Sterck, and D. Koelle, Phys. Rev. E 63 ( [] K.M. Leung, D.L. Mills, P.S. Riseborough, and S.E. Trullinger, Phys. Rev. B 7 ( [3] M. Salerno and N.R. Quintero, Phys. Rev. E 65 ( [4] S.Y. Lou and G.J. Ni, Phys. Lett. A 40 ( [5] V.A. Gani and A.E. Kudryavtsev, Phys. Rev. E 60 ( [6] R.K. Bullough, P.J. Caudrey, and H.M. Gibbs, Solitons, eds. R.K. Bullough and P.J. Caudrey, Spinger, Berlin (980. [7] M. Panigrahi and P.C. Dash, Phys. Lett. A 3 ( [8] J.S. Yang and S.Y. Lou, Chin. Phys. Lett. ( [9] E.G. Fan and J. Zhang, Phys. Lett. A 305 ( [0] Y.Z. Peng, Phys. Lett. A 34 ( [] Sirendaoreji and J. Sun, Phys. Lett. A 98 ( [] K.W. Chow, Wave Motion 35 (00 7. [3] Z.T. Fu, S.D. Liu, and S.K. Liu, Commun. Theor. Phys. (Beijing, China 39 ( [4] Z.T. Fu, S.K. Liu, and S.D. Liu, Commun. Theor. Phys. (Beijing, China 40 ( [5] Z.T. Fu, Z. Chen, S.K. Liu, and S.D. Liu, Commun. Theor. Phys. (Beijing, China 4 ( [6] Z.T. Fu, Z. Chen, S.D. Liu, and S.K. Liu, Commun. Theor. Phys. (Beijing, China 4 ( [7] Z.T. Fu, S.K. Liu, and S.D. Liu, Commun. Theor. Phys. (Beijing, China 4 ( [8] Z.T. Fu, S.K. Liu, S.D. Liu, and Q. Zhao, Phys. Lett. A 90 (00 7. [9] S.K. Liu, Z.T. Fu, S.D. Liu, and Q. Zhao, Phys. Lett. A 89 ( [0] S.K. Liu and S.D. Liu, Nonlinear Equations in Physics, Peking University Press, Beijing (000. [] F. Bowman, Introduction to Elliptic Functions with Applications, London Universities, London (959. [] V. Prasolov and Y. Solovyev, Elliptic Functions and Elliptic Integrals, American Mathematical Society, Providence, R.I. (997. [3] Z.X. Wang and D.R. Guo, Special Functions, World Scietific, Singapore (989.