Invariant Sets and Exact Solutions to Higher-Dimensional Wave Equations

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1 Commun. Theor. Phys. Beijing, China) ) pp c Chinese Physical Society Vol. 49, No. 5, May 5, 2008 Invariant Sets and Exact Solutions to Higher-Dimensional Wave Equations QU Gai-Zhu, ZHANG Shun-Li,,2, and ZHU Chun-Rong,3 Center for Nonlinear Studies, Department of Mathematics, Northwest University, Xi an 70069, China 2 Center of Nonlinear Science, Ningbo University, Ningbo 352, China 3 Department of Mathematics, Anhui Normal University, Wuhu 24000, China Received June 8, 2007) Abstract The invariant sets and exact solutions of the 2)-dimensional wave equations are discussed. It is shown that there exist a class of solutions to the equations which belong to the invariant set E 0 = {u : u x = v x u), u y = v y u)}. This approach is also developed to solve N)-dimensional wave equations. PACS numbers: Jr, Sv Key words: wave equation, invariant set, exact solution Introduction It has been the highlights to seek exact solutions of nonlinear partial differential equations PDEs) in mathematical physics. Constructing exact solutions to nonlinear PDEs has theoretical significance as well as practical value in mechanics, physics and engineering, etc. Since 970s, various ways to solve nonlinear systems have been developed, which include the inverse scattering transformation method, ] the Darboux transformation method, 2] the Bäcklund transformation method, 3] the bilinear method and multi-linear method, 4] the classical and non-classical Lie group approaches, 5] the CK s direct method, 6] the deformation mapping method, 7] the geometrical method, 8] the truncated Painlevé expansion method, 9] the function expansion method, 0] the homogeneous balance method, ] the variable separation approaches, 2] and so on. In Ref. 3], Galaktionov introduced an extension to the scaling group for the equation, u t = Kx, u x, u xx,...,u k) ), which is governed by the invariance of the set S 0 = {u : u x = x u)}, where uk) denotes the k-th-order derivative of u with respect to x, is a smooth function to be determined. It is also valid to solve the nonlinear evolution equations of the KdV-type. 4] Qu and Estevez 5] extended the scaling group to a more general form, which is governed by the invariant set S = { u : u x = x expn ) u) ɛu) ]} z) dz. This approach has been used successfully to construct solutions to a number of evolution equations. 5,6] In Refs. 7] and 8], we further extended the Galaktionov s approach to discuss the 2)-dimensional reactiondiffusion equations and the generalized thin film equation with respect to the invariant set E 0 = {u : u x = v x u), u y = v y u)}. In this paper, we develop the approach 3,4] to study the higher-dimensional wave equations and their solutions in terms of the invariant set E 0. In Sec. 2, by way of the invariant set E 0, we construct some interesting solutions of the wave equations. In Sec. 3, we discuss the invariant set and solutions of N)-dimensional wave equations. Section 4 is the concluding remarks. 2 2)-Dimensional Wave Equations Consider the 2)-dimensional wave equation, u tt = Au)u xx Bu)u yy Cu)u 2 xdu)u 2 y Qu), ) where Au), Bu), Cu), Du), and Qu) are smooth functions of u. Equation ) has a wide range of physical applications in fluid dynamics, plasma, and elastic media, etc. Here we introduce the invariant set E 0 = {u : u x = v x u), u y = v y u)}, where v is some smooth function of x and y, is a smooth function of u to be determined from the invariant condition ux, y, 0) E 0 = ux, y, t) E 0, for t 0, ]. or u E 0, we obtain solutions of Eq. ) given by dz = vx, y) ht). z) In the set E 0, we have the following expressions u t = h, u tt = h h 2, u x = v x, u xx = v xx v 2 x, u y = v y, u yy = v yy v 2 y. Substituting them into Eq. ), we obtain The project supported by National Natural Science oundation of China under Grant Nos and and the Natural Science oundation of Shaanxi Province of China under Grant No. 2005A3 zhangshunli@nwu.edu.cn

2 20 QU Gai-Zhu, ZHANG Shun-Li, and ZHU Chun-Rong Vol. 49 h h 2 = Av xx Bv yy A C)v 2 x B D)v 2 y Q. 2) Differentiating Eq. 2) with respect to x and y respectively yields Q h 2 v x = Av xxx Bv yyx A v xx B v yy v x A C) vx 3 2A C)v x v xx B D) v x vy 2 2B D)v y v yx, 3) Q h 2 v y = Av xxy Bv yyy A v xx B v yy v y A C) v 2 xv y 2A C)v x v xy B D) v 3 y 2B D)v y v yy. 4) It is very hard to determine the coefficient functions in Eq. ) by Eqs. 3) and 4). To illustrate our approach, we just consider several special cases for different choices of vx, y) hereafter Case v xx = v yy = 0. rom v xx = v yy = 0, we simply take v = xy. In this case, E 0 becomes the set E = {u : u x = yu), u y = xu)}. If 0, substituting v = xy into Eqs. 3) and 4) respectively, we obtain B D) x 2 A C) y 2 2B D) x Q y h 2 = A C) y 2 B D) x 2 2A C) y Q x Differentiating Eq. 5) with respect to x and Eq. 6) with respect to y respectively yields B D x { 2B 2 D) ] 2B D) } y x A C y 3 2B D) Q/) ] y y = 0, A C xy { 2A 2 C) ] 2A C) } y B D x 3 2A C) Q/) ] x x = 0. h 2 =, 5). 6) rom them, we get the coefficient functions in Eq. ) satisfying A C = 0, B Q D = 0, = c, and ht) = c t c 2, where and hereafter c i denote arbitrary constants and c 0. In this case, solutions of Eq. ) are given by z) dz = xy c t c 2. If = 0, i.e. = u, we also have the above formulas. Letting A = B = u m, = u k, we have C = D = ku m, Q = c ku 2k. Thus we have shown that the equation u tt = u m u xx u yy ) ku m u 2 x u 2 y) c ku 2k { x c0 exp 2 y 2 ] c t, k =, u = 2 k)xy c t c 3 )] / k), k. 7) Case 2 v xy = 0. rom v xy = 0, we can see that vx, y) = fx) gy), and E 0 turns into E = {u : u x = f x)u), u y = g y)u)}. Now we distinguish three subcases: Subcase 2. fx) = ln x, gy) = ln y. When fx) = ln x, gy) = ln y, vx, y) = ln x ln y, v x = /x, v y = /y, and E = {u : u x = x u), u y = } y u). If 0, equation 3) becomes h 2 = x 2 2A A C) A A C y 2 B B D) Q and equation 4) becomes ), 8) h 2 = y 2 2B B D) B B D x 2 A A C) Q ). 9) Differentiating Eq. 8) with respect to x and Eq. 9) with respect to y respectively yields ] B B G 2G x 3 D) Q/) ] x = 0, ] xy 2

3 No. 5 Invariant Sets and Exact Solutions to Higher-Dimensional Wave Equations 2 where ] A A H 2H y 3 C) Q/) ] y = 0, ] yx 2 G = 2A A C) A A C, H = 2B B D) B B D. These give the constraints for the coefficient functions in Eq. ) A A C = 0, B B Q D = 0, = c, and ht) = c tc 2. Equation ) z) dz = ln xy c t c 2. If = 0, i.e. = u, then the above expressions also hold. Letting A = B = u m, = u k, then C = D = u m k ku m, Q = c ku 2k, and we can see that the equation u tt = u m u xx u yy )u m k ku m )u 2 xu 2 y)c ku 2k { c0 xy e c t, k =, u = k)ln xy c t c 3 )] / k), k. 0) Subcase 2.2 fx) = ln cos x, gy) = ln cos y. Taking fx) = ln cos x, gy) = ln cosy, then v x = tanx, v y = tany, and E = {u : u x = tanx)u), u y = tany)u)}. If 0, equation 3) becomes while equation 4) becomes h 2 = { A A C) 2A A C)]tanx) 2 B B D) tany) 2 h 2 = { B B D) 2B B D)]tany) 2 A A C) tanx) 2 A B Q ) } 2A A C), ) A B Q ) } 2B B D). 2) Differentiating Eq. ) with respect to x and Eq. 2) with respect to y yields B B G 2G]tanx) 3 D tanxtany) 2 2G M ) = 0, A A H 2H]tany) 3 C tan ytanx) 2 2H N ) = 0, where G = A A C) 2A A C)], H = B B D) 2B B D)], M = A B Q ) 2A A C)], N = A B Q ) 2B B D)]. Solving them, we obtain A A C = 0, B B D = 0, A B Q = c, and ht) = c t c 2. But if = 0, i.e. = u, we still have the above formulas. Letting A = B = u m, = u k, then C = D = u m k ku m ), Q = c ku 2k 2u mk. Thus the equation u tt = u m u xx u yy ) u m k ku m )u 2 x u 2 y) c ku 2k 2u mk c 0 e c t u = cos x cosy, k =, k) ln cos x cos y c t c 3 )] / k), k. Subcase 2.3 fx) = x 2 /2, gy) = y 2 /2. If we take fx) = x 2 /2, gy) = y 2 /2, then v x = x, v y = y, E = {u : u x = xu), u y = yu)}. If 0, we can verify that equations 3) and 4) respectively turn to be h 2 = A C) x 2 B D) y 2 2A C) 3) A B Q, 4)

4 22 QU Gai-Zhu, ZHANG Shun-Li, and ZHU Chun-Rong Vol. 49 h 2 = B D) y 2 A C) x 2 2B D) A B Q. 5) Likewise, differentiating Eq. 4) with respect to x and Eq. 5) with respect to y, we have A C x B 3 D xy { 2A 2 C) ] 2A C) A B Q ) ] } x = 0, B D y A 3 C x { 2B 2 D) ] 2B D) A B Q y ) ] } y = 0. rom them, we can get the constraints for the coefficient functions in Eq. ) A C = 0, B D = 0, A B Q = c, and ht) = c t c 2. But if = 0, i.e. = u, we can also obtain the above expressions. Letting A = B = u m, = u k, then we have C = D = ku m, Q = c ku 2k 2u mk. Then the equation u tt = u m u xx u yy ) ku m u 2 x u 2 y) c ku 2k 2u mk u = x 2 y 2 c 0 exp 2 ] c t, k =, x 2 y 2 k) )] 6) / k) c t c 3, k. 2 Case 3 v xx v yy = 0 and v xx 0, v yy 0. A solution of v xx v yy = 0 is v = arctanx/y). In this case, E 0 becomes the set E = {u : u x = y/x 2 y 2 )]u), u y = x/x 2 y 2 )]u)}. If 0, then equations 3) and 4) respectively become h 2 = x 3 2B D)] yx 2 y 2 ) 2 6A B) B D x 2 y 2 ) 2 2A B) 4A C) 2B xy D)] x 2 y 2 ) 2 y 2 2A B) A C x 2 y 2 ) 2 Q/), 7) y 3 h 2 = 2A C)] xx 2 y 2 ) 2 6A B) A C x 2 y 2 ) 2 2A B) 4B D) 2A xy C)] x 2 y 2 ) 2 x 2 2A B) B D x 2 y 2 ) 2 Q/). 8) Differentiating Eq. 7) with respect to x and Eq. 8) with respect to y, we have 2B x 4 B D)] y 2 x 2 y 2 ) 2 D 2x 3 G yx 2 y 2 ) 2 G 6B D) ] where x 2 x 2 y 2 ) 2 G M 2xy ) x 2 y 2 ) 2 y 2 Q/) ] I 4I 2M) x 2 y 2 ) 2, 2A y 4 A C)] x 2 x 2 y 2 ) 2 C 2y 3 H xx 2 y 2 ) 2 H N 2y 2 ) x 2 y 2 ) 2 H 6A C) ] xy x 2 y 2 ) 2 J x 2 Q/) ] 4J 2N) x 2 y 2 ) 2, G = 6A B) B D, H = 6A B) A C, I = 2A B) A C, J = 2A B) B D, M = A B) 2A C) B D)], N = A B) 2B D) A C)]. y 2 x 2

5 No. 5 Invariant Sets and Exact Solutions to Higher-Dimensional Wave Equations 23 They give the constraints for the coefficient functions in Eq. ) A = B, A C = 0, B Q D = 0, = c, and ht) = c t c 2. If = 0, i.e. = u, the above formulas are still gained. It is easy to show that the equation u tt = u m u xx u yy ) ku m u 2 x u 2 y) c ku 2k u = c 0 exp arctan x y ] c t, k =, k) arctan x y )] / k) 9) c t c 3, k. 3 N)-Dimensional Wave Equations More generally, by using the same method, we now discuss the N)-dimensional wave equation u tt = A i u)u xi x i B i u)u 2 x i Qu). 20) i= i= Now we introduce the extended-invariant set Ẽ 0 = {u : u xi = v xi u), i =,...,N}, where v is some smooth function of x i i =,...,N), u) is a smooth function of u to be determined by the invariant condition ux, x 2,..., x N, 0) Ẽ0 = ux, x 2,...,x N, t) Ẽ0, for t 0, ]. It follows that the corresponding solution of Eq. 20) is given by z) dz = vx, x 2,...,x N ) ht), where ht) satisfies h h 2 = A i v xi x i A i B i )vx 2 i Q. i= i= Similarly, we distinguish two cases as follows: Case v xi x i = 0 i =,...,N). N Ẽ = {u : u xi = x j u), i =,..., N}, j=,i j and the coefficient functions in Eq. 20) satisfy A i Q B i = 0 i =,...,N), = c. The corresponding solution of Eq. 20) is given by z) dz = x x 2 x N c t c 2. Case 2 v xi x j = 0 i j, i, j =,...N). In this case, vx, x 2,...,x N ) = f x )f 2 x 2 ) f N x N ). In the same way as in Sec. 2, we have the following results. Subcase 2. v = ln x ln x 2... ln x N. Ẽ = {u : u xi = /x i )u), i =,...,N}, and the coefficient functions in Eq. 20) satisfy A i A i Q B i = 0 i =,...,N), = c. The corresponding solution of Eq. 20) is given by z) dz = ln x x 2... x N c t c 2. Subcase 2.2. v = ln cos x ln cosx 2 ln cosx N ). Ẽ = {u : u xi = tanx i u), i =,...,N}, and the coefficient functions in Eq. 20) satisfy A i A i B i = 0 i =,...,N), A i Q = c. i= The corresponding solution of Eq. 20) is given by z) dz = ln cosx cos x 2 cos x N ) c t c 2. Subcase 2.3 v = x 2 x 2 2 x 2 N)/2. Ẽ = {u : u xi = x i u), i =,...,N}, and the coefficient functions in Eq. 20) satisfy A i B i = 0 i =, 2,...,N), A i Q = c. i= The corresponding solution of Eq. 20) is given by z) dz = x 2 i c t c 2. 2 i= 4 Concluding Remarks In this paper, we have extended the scaling and rotation group to construct some interesting solutions of 2)-dimensional wave equations. We have also applied this approach to solve N)-dimensional wave equations. We shall pay much attention to studying some soliton-type equations via the approach in future.

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