Similarity Reductions of (2+1)-Dimensional Multi-component Broer Kaup System

Commun. Theor. Phys. Beijing China 50 008 pp. 803 808 c Chinese Physical Society Vol. 50 No. 4 October 15 008 Similarity Reductions of +1-Dimensional Multi-component Broer Kaup System DONG Zhong-Zhou 1 CHEN Yong 1 and WANG Ling 3 1 Shanghai Key Laboratory of Trustworthy Computing East China Normal University Shanghai 0006 China Key Laboratory of Mathematics Mechanization Chinese Academy of Sciences Beijing 100080 China 3 Department of Mathematics and Computer Science Chizhou University Chizhou 47000 Anhui Province China Received November 6 007 Abstract Painlevé property of the +1-dimensional multi-component Broer Kaup BK system is considered by using the standard Weiss Kruskal approaches. Applying the Clarkson and Kruskal CK direct method to the +1- dimensional multi-component BK system some types of similarity reductions are obtained. By solving the reductions one can get the solutions of the +1-dimensional multi-component BK system. PACS numbers: 0.30.Jr 03.40.Kf 04.0.Jb Key words: multi-component Broer Kaup system similarity reduction direct method 1 Introduction Nonlinear partial differential equations widely describe complex phenomena in various fields of sciences such as physics chemistry biology etc. Then solving nonlinear problems gets important and significant in nonlinear science. To find some similarity solutions of a nonlinear physics problem the Lie group approach is one of the most important methods. One may use the classical Lie group approach [1 and the nonclassical Lie group approach. [ However both the classical Lie group approach and the nonclassical Lie group approach are quite complicated in practical calculations. The direct method which was presented first by Clarkson and Kruskal [3 has been widely used to find the similarity solutions for many important mathematical and physical models. On the one hand the direct method can simplify the calculation. On the other hand one believes that the results obtained by the CK direct method contain those obtained by the classical Lie group approach and the results of the nonclassical Lie approach include those of the direct method. [4 It is known that possessing Painlevé property is also one of the most important integrable properties which means that the solutions of the model are single-valued about an arbitrary singularity manifold and various other integrable properties are linked with the Painlevé property. [5 7 There are some kinds of approaches to study the Painlevé property of a nonlinear evolution equation such as the ARS algorithm [5 the standard Weiss Kruskal approaches [6 and the Conte s invariant method. [7 In Ref. [8 Lou and Hu used the Darboux-transformation-related symmetry constraints of the Kadomtsev Petviashvili KP equation to get some integrable 1+1- dimensional and +1-dimensional multi-component BK system. In this paper first using the standard Weiss Kruskal approaches we try to investigate the Painlevé property of the +1-dimensional multi-component BK system. Second applying the CK direct method we construct similarity solutions of the +1-dimensional multicomponent BK system: G jy = G jxx G j H j x 1 H jy =H jxx H j H jx z 1 G ixx j = 1...N. Differentiating Eq. with respect to z one can get H jyz = H jxxz H jx H jz H j H jxz G ixx. 3 Now we consider Eqs. 1 and 3. Painlevé Property of +1-Dimensional Multi-component Broer Kaup System Using the standard Weiss Kruskal approaches we give the form of the expansion about the singular manifold H j = φ α j h jk φ k G j = φ g jk φ k 4 k=0 k=0 where h jk = h jk x y z g jk = g jk x y z and φ = φx y z are analytic functions of x y z. Substituting Eq. 4 into Eqs. 1 and 3 leads to α j = 1 = h j0 = φ x g j0 = φ x φ z j=1 and the recursion relations have the forms i 3φ x g j0 h ji + ii 3φ xg ji = f 1 h jl g jl 5 i i 3 φ xφ z h ji + i i 3φ x g ji = f h jl g jl 6 j=1 The project supported by National Natural Science Foundation of China under Grant No. 10735030 Shanghai Leading Academic Discipline Project under Grant No. B41 Natural Science Foundation of Zhejiang Province of China under Grant No. Y604056 and the Doctoral Foundation of Ningbo City under Grant No. 005A61030 E-mail address: ychen@sei.ecnu.edu.cn

804 DONG Zhong-Zhou CHEN Yong and WANG Ling Vol. 50 where f 1 and f are complicated functions of h jl g jl l i 1 and the derivatives of the singularity φ. From Eqs. 5 and 6 the resonance values of i are given by a 1 0 0 b 0 0 0 a 0 0 b 0 0 0 a N 0 0 b c 0 0 d d d 0 c 0 d d d 0 0 c d d d = i + 1i N 1 i N i 3 3N 1 i 4φ 4N x φ N z = 0 where a k = i 3φ x g k0 b = ii 3φ x c = i i 3 φ xφ z d = i i 3φ x i.e. i = 1 0...0... 3...3 4...4. Resonances are those values of i at which it is possible to introduce arbitrary functions into the expansions 4. For each nontrivial resonance there occurs a compatibility condition that must be satisfied if the solution has a single-valued expansion. After detailed calculation we find the compatibility conditions at i = 0...0... 3...3 4...4 are satisfied identically for Eqs.5 and 6. The resonance at i = 1 corresponds to the arbitrary function φ defining the singularity manifold for the +1-dimensional multi-component BK system. According to the Weiss Kruskal approach equations 1 and 3 possess Painlevé properties. It is said that the +1-dimensional multi-component BK system is integrable under the meaning that it possesses Painlevé property. 3 Similarity Reduction and Solutions of +1-Dimensional Multi-component Broer Kaup System To Eqs. 1 and 3 suppose that the solution has the following form H j = α j x y z + x y zp j ξ η 7 G j = a j x y z + b j x y zq j ξ η 8 where α j x y z x y z ξ = ξx y z η = ηx y z a j x y z and b j x y z are functions of x y z to be determined. Substituting Eqs. 7 and 8 into Eqs. 1 and 3 yields δ 0 + δ 1 P j + δ Q j + δ 3 P jξ + δ 4 P jη + δ 5 Q jξ + δ 6 Q jη + δ 7 Q jξξ + δ 8 Q jξη + δ 9 Q jηη + δ 10 P j Q j + δ 11 P jξ Q j + δ 1 P jη Q j + δ 13 P j Q jξ + δ 14 P j Q jη = 0 9 γ 0i + γ 1i P j + γ i P jξ + γ 3i P jη + γ 4i P jξξ i=0 + γ 5i P jξη + γ 6i P jηη + γ 7i P jξξξ + γ 8i P jξξη + γ 9i P jξηη + γ 10i P jηηη + γ 11i P j + γ 1i P j P jξ + γ 13i P j P jη + γ 14i P jξ + γ 15i P jξ P jη + γ 16i P jη + γ 17i P j P jξξ + γ 18i P j P jξη + γ 19i P j P jηη + γ 0i Q i = 0 10 where δ i i = 0 1...14 and γ ji j = 0 1...0; i = 0 1...N are given by δ 0 = a jy + a jxx + α j a jx + α jx a j δ 1 = a jx + a j x δ = b jy + b jxx + α j b jx + α jx a j δ 3 = a j ξ x δ 4 = a j η x δ 5 = b j ξ y + b jx ξ x + b j ξ xx + α j b j ξ x δ 6 = b j η y + b jx η x + b j η xx + α j b j η x δ 7 = b j ξ x δ 8 = b j ξ x η x δ 9 = b j η x δ 10 = b jx + x b j δ 11 = b j ξ x δ 1 = b j η x δ 13 = b j ξ x δ 14 = b j η x. 11 For γ ji if j is fixed we can take γ j = γ j0 γ j1...γ jn. So γ 0 = γ 00 γ 01... γ 0N = α jyz α jxxz + α jx α jz + α j α jxz a 1xx a xx...a Nxx γ 1 = γ 10 γ 11... γ 1N = yz xxz + α jx z + α jz x + α j xz + α jxz 1 0 0...0 γ = γ 0 γ 1... γ N = [y ξ z + z ξ y + ξ yz xx ξ z xz ξ x x ξ xz z ξ xx ξ xxz + α jx ξ z + α jz ξ x + α j x ξ z + z ξ x + ξ xz 1 0 0...0 γ 3 = γ 30 γ 31... γ 3N = [y η z + z η y + η yz xx η z xz η x x η xz z η xx η xxz + α jx η z + α jz η x + α j x η z + z η x + η xz 1 0 0...0 γ 4 = γ 40 γ 41... γ 4N = [ ξ y x ξ x ξ xx ξ z z ξ x + ξ xz α j ξ z ξ x 1 0 0...0 γ 5 = γ 50 γ 51... γ 5N = [ ξ y η z + ξ z η y ξ xx η z + ξ z η xx + ξ xz η x + ξ x η xz x ξ x η z + ξ z η x z ξ x η x + α j ξ x η z + ξ z η x 1 0 0...0 γ 6 = γ 60 γ 61... γ 6N = [ η y η z η xx η z η x η xz x η x η z z η x + α j η x η z 1 0 0...0 γ 7 = γ 70 γ 71... γ 7N = ξ xξ z 1 0 0...0 γ 8 = γ 80 γ 81... γ 8N = ξ x ξ z η x + ξ x η z 1 0 0...0 γ 9 = γ 90 γ 91... γ 9N = η x ξ x η z + ξ z η x 1 0 0...0 γ 10 = γ 100 γ 101...γ 10N = η xη z 1 0 0...0 γ 11 = γ 110 γ 111...γ 11N = x z + xz 1 0 0...0

No. 4 Similarity Reductions of +1-Dimensional Multi-component Broer Kaup System 805 γ 1 = γ 10 γ 11...γ 1N = x ξ z + z ξ x + z ξ x + ξ xz 1 0 0...0 γ 13 = γ 130 γ 131...γ 13N = x η z + z η x + z µ x + η xz 1 0 0...0 γ 14 = γ 140 γ 141...γ 14N = βj ξ x ξ z 1 0 0...0 γ 15 = γ 150 γ 151...γ 15N = βj ξ z η x + ξ x η z 1 0 0...0 γ 16 = γ 160 γ 161...γ 16N = βj η x η z 1 0 0...0 γ 17 = γ 170 γ 171...γ 17N = βj ξ x ξ z 1 0 0...0 γ 18 = γ 180 γ 181...γ 18N = βj ξ z η x + ξ x η z 1 0 0...0 γ 19 = γ 190 γ 191...γ 19N = βj η xη z 1 0 0...0 γ 0 = γ 00 γ 01...γ 0N = 0 b 1xx b xx...b Nxx γ 1 = γ 10 γ 11...γ 1N = 0 b 1x ξ x + b 1 ξ xx b x ξ x + b ξ xx...b Nxξ x + b N ξ xx γ = γ 0 γ 1...γ N = 0 b 1x η x + b 1 η xx b x η x + b η xx...b Nxη x + b N η xx γ 3 = γ 30 γ 31...γ 3N = 0 b 1 ξx b ξx...b N ξx γ 4 = γ 40 γ 41...γ 4N = 0 4b 1 ξ x η x 4b ξ x η x...4b N ξ x η x γ 5 = γ 50 γ 51...γ 5N = 0 b 1 ηx b ηx...b N ηx. 1 Equations 9 and 10 are PDEs of P and Q with respect to ξ and η only for the coefficients of different derivatives and with the powers of P and Q being functions of ξ and η. That is to say the constrained conditions δ i = δ j i ξ η i = 0 1...14 13 γ ji = γ j i Γ jiξ η; i = 1...N; j = 1...5.14 For some fixed non-zero δ j and γ j i must be satisfied where i and Γ ji are some functions of ξ and η to be determined later. To determine the non-fixed functions {α i β i P i a i b i Q i ξ η i Γ ji } we can use some remarks to simplify the calculations Remark 1 If α i x y z or a i x y z has the form α i = β i x y zξ i ξ η+α 0i x y z or a i = b i x y zξ i ξ η+ a 0i x y z then we can take Ξ i 0. Remark If β i x y z or b i x y z has the form β i = β 0i x y zξ i ξ η or b i = b 0i x y z Ξ i ξ η then we can take Ξ i C i = constant. Remark 3 If ξ = ξξ 0 x y z η or η = ηξ η 0 x y z then we can take ξ ξ 0 or η η 0. Remark 4 If ξx y z or ηx y z has the form Ξξ = ξ 0 x y z or Hη = η 0 x y z where Ξξ or Hη is any invertible function then we can take Ξξ = ξ or Hη = η. To discuss further three cases should be considered: i ξ x 0 ii ξ x = η x = 0 ξ y 0 and iii ξ x = η x = 0 ξ y = η y = 0. Only the first two cases will be discussed in detail because the last one is quite trivial. i ξ x 0. In this case we can choose δ j = δ 11 = b j ξ x as common factors. As far as the choice of γ j i we will consider the following three cases. Using Remarks 1 4 to fix the freedoms in the determination of {α i β i P i a i b i Q i ξ η i Γ ji } and analyzing Eqs. 13 and 14 with Eqs. 11 and 1 carefully one can obtain three possible solutions of Eqs. 1 and 3. a η z 0. From Substituting Eqs. 11 into Eq. 9 we obtain we can have 0 = 1 = = 4 = 5 = 0 6 = 11 = 13 = 1 15 3 = a j b j 9 = η x ξ x 7 = ξ x 8 = η x 10 = b jx + x b j b j ξ x 1 = 14 = η x ξ x 16 α j = ξ xx ξ y ξ x η y = ξ x 17 b jy + b jxx + α j b jx + α jx b j = 0. 18 3 = a j b j a j = 3 b j 7 = ξ x 19 = ξ x 7. 0 Using Remarks 1 and with Eq. 0 yields So one can arrive at a j = 0 = ξ x. 1 8 = η x. From Eq. we can have ξ x η x = 8 η = 1 ξ 8 ξ 1 dξ 1 + C 1 y z. 3 Using Remark 3 with Eq. 3 yields η = C 1 y z. 4 In the meantime we also get 9 = 1 = 14 = 0.

806 DONG Zhong-Zhou CHEN Yong and WANG Ling Vol. 50 From we can have b j = C jy z 10 = b jx + x b j b j ξ x 5 Using Remark with Eq. 6 yields ξ exp 10 ξ 1 dξ 1. 6 b j = C jy z = C jy z ξ x. 7 Here we choose γ j i = γ 80 = ξ xη z. Substituting Eqs. 1 17 1 4 and 7 into Eq. 10 we arrive at Γ 0 = Γ 1 = Γ = Γ 3 = Γ 4 = Γ 5 = Γ 9 = Γ 10 = Γ 13 = Γ 16 = Γ 19 = Γ = Γ 4 = Γ 5 = 0 0...0 8 Γ 6 = Γ 15 = Γ 18 = 1 0 0...0 9 ξz Γ 7 = 0 0...0 Γ 8 = 1 0 0...0 η z Γ 11 = ξ xxξ xz + ξ x ξ xxz ξxc 3 0 0...0 30 Γ 1 = ξ xxξ z + 3ξ x ξ xz ξxη 0 0...0 z Γ 14 = Γ 17 = ξ z 0 0...0 31 η z Γ 15 = Γ 18 = 1 0 0...0 Γ 0 = 0 4b 1xx ξxc 3 4b xx ξxc 3... 4b Nxx ξxc 3 3 Γ 1 = 0 4b 1xξ x + b 1 ξ xx ξxc 3 4b xξ x + b ξ xx ξxc 3... 4b Nxξ x + b N ξ xx ξxc 3 33 Γ 3 = 0 1 1...1 34 ξ xx = 0 C = 4C iy z ξ x 35 y + α jx = 0 36 ξ y + α j ξ x = 0. 37 From Eq. 30 we can have ξ z = Γ 70 η z ξ = η Γ 70 η 1 dη 1 + C 3 x y. 38 Using Remark 3 with Eq. 38 yields So we can obtain ξ = C 4 yx + C 5 y = C 4y P j = a 3 + a a a 6 tan [ a a 6 a 1 + a ξ + a 3 η/a ξ = C 3 x y. 39 α j = C 4y C 4 y x C 5y C 4 y b j = 1 C C 4 y 40 C 1y = C 4y C = 4C iy z C 4 y 41 Γ 11 = Γ 1 = Γ 14 = Γ 17 = Γ 0 = Γ 1 = 0 0 0...0.4 Solving Eqs. 18 36 and 37 with Eqs. 40 41 and 4 one can arrive at α j = C 4y C 4 y x C 5y C 4 y = C 4y a j = 0 b j = C 4 yc 6 z 43 ξ = C 4 yx + C 5 y η = C4ydy + 4 C 6 zdz + c 0 44 where c 0 is an arbitrary constant C 4 y C 5 y and C 6 z are arbitrary functions of the corresponding variables. From the above analysis the first type of similarity reduction of Eqs. 1 and 3 is simplified to Q jη + Q jξξ + P jξ Q j + P j Q jξ = 0 45 P jηη + P jξξη P jξ P jη P j P jξη Q iξξ = 0 46 with ξ and η satisfying Eq. 44. By solving Eqs. 45 and 46 one can obtain a Q j = c j a 4 { tan [ a a 6 a 1 + a ξ + a 3 η/a / a a 6 a 1 /a } + a5 1 + tan [ a a 6 a 1 + a ξ + a 3 η/a where c j j = 1...N a i i = 1...6 are arbitrary constants and N j=1 c j = 0. So the corresponding solution of Eqs. 1 and 3 reads H j = C 4 y C 4 y x C 5 y C 4 y a 3C 4 y + a a a 6 C 4 y tan [ a a 6 a 1 + a ξ + a 3 η/a a a 47 G j = c j C 4 yc 6 z a { 4 tan[ a a 6 a 1 + a ξ + a 3 η/a / } a a 6 a 1 /a + a5 1 + tan [ a a 6 a 1 + a ξ + a 3 η/a 48

No. 4 Similarity Reductions of +1-Dimensional Multi-component Broer Kaup System 807 with ξ and η satisfying Eq. 44. If η z = 0 and ξ z 0 we can consider the following case. b η z = 0 ξ z 0. In this case the corresponding solution of Eqs. 1 and 3 has the form H j = C 1 y 4C 1 y x + 1 C1 y P jξ η 49 G j = C z C1 y Q jξ η 50 where C 1 y and C z are arbitrary functions of the corresponding variables. The reduction equations then read Q jη + Q jξξ + P jξ Q j + P j Q jξ = 0 51 P jξη + P jξξξ P jξ P j P jξξ with ξ = η = C1 y x + Q iξξ = 0 5 C zdz + c 0 4 C 1 y dy + c 1 53 where c i i = 0 1 are arbitrary constants. Integrating Eq. 49 with respect to ξ we can obtain P jη + P jξξ P j P jξ Q iξ = 0. 54 By taking P j = H j and Q j = G j one can arrive at the same equation as Eqs. 5 and 6 in Ref. [8 as follows: G jη + G jξξ + H jξ G j + H j G jξ = 0 55 H jη + H jξξ H j H jξ G iξ = 0. 56 It is easy to see that equations 55 and 56 have the following solution H j = a 1 4 1 + 1 + exp ξ a 1 η a 57 G j = 4c j exp ξ a 1 η a k[1 + exp ξ a 1 η a 58 where a 1 a and c j j = 1...N are arbitrary constants and k = N c i. So equations 1 and 3 have the solution H j = C 1 y 4C 1 y x + C1 y G j = [ a 1 4 1 + 1 + exp ξ a 1 η a 59 8c j C z exp ξ a 1 η a k C 1 y[1 + exp ξ a 1 η a 60 with ξ and η satisfying Eq. 53. c ξ z = η z = 0. In this simple case the corresponding solution of Eqs. 1 and 3 can be written as H j = 1 4 x C 1y exp 1 c 0 y + c 0 1 exp y P j ξ η 61 1 G j = c 0 exp y C j zq j ξ η 6 where C 1 y and C j z are arbitrary functions of the corresponding variables and c 0 is an arbitrary constant. The reduction equations read with Q jη + Q jξξ + P jξ Q j + P j Q jξ = 0 63 Q iξξ = 0 64 ξ = c 0 exp 1 yx + C 1y η = c 0 expy. 65 By solving Eqs. 63 and 64 one can arrive at P j = F 1j η/ξ F j ηξ + F 3jη F 1j ηξ + F j η 66 Q j = F 1j ηξ + F j η 67 where F ij η i = 1 3 are arbitrary functions. So the solution of Eqs. 1 and 3 can be written as H j = 4x C 1y exp 1 c 0 y + c 0 exp y [ F 1j η/ξ F j ηξ + F 3jη 68 [F 1j ηξ + F j η 1 G j = c 0 exp y C j z[f 1j ηξ + F j η 69 with ξ and η satisfying Eq. 65. ii ξ x = η x = 0 ξ y 0. In this case we can choose δ j = δ 5 = b j ξ y and γ j i = γ 50 = ξ y η z + ξ z η y. Using Remarks 1 4 we obtain two possible independent solutions of Eqs. 1 and 3. a x 0. The reduction equations in this case have the forms with P j + Q jξ + P j Q j = 0 70 P jξη + P j P jη = 0 71 ξ = 4ay + b η = C z 7 while a and b are arbitrary constants. The corresponding special solution of Eqs. 1 and 3 reads H j = C 3 y C [ 4y x + C 3 y +a [ G j = C 1 z x x C 3 ydy P j ξ η 73 C 3 ydy [1 + Q j ξ η 74

808 DONG Zhong-Zhou CHEN Yong and WANG Ling Vol. 50 where C 1 z and C i y i = 3 4 are arbitrary functions of the corresponding variables. P j ξ η and Q j ξ η satisfy Eqs. 70 and 71. It is easy to know that equations 70 and 71 have the solution P j = ξ + F 1j η Q j = ξ 4F 1j ηξ + F j η [ξ + F 1j η where F ij η i = 1 are arbitrary functions. So equations 73 and 74 can be rewritten as H j = C 3 y C 4y x + C 3 y [ + a x C 3 ydy ξ + F 1j η [ G j = C 1 z x C 3 ydy [ 1 + ξ 4F 1j ηξ + F j η [ξ + F 1j η with ξ and η satisfying Eq. 7. b x = 0 The reduction equations in this case can be written as P j + Q jξ = 0 75 P jξη = 0 76 with ξ = c 0 c 1 y + c η = C 5 z. The reduction Eqs. 75 and 76 can be easily solved. The results read P j = F 1jξ + F j η Q j = F 1j ξ F j ηξ + F 3j η where F 1j ξ F j η and F 3j η are arbitrary functions of the corresponding variables. The corresponding special solution of Eqs. 1 and 3 is as follows: H j = C 1jy + C j z + c 0 [F 1jc 0 c 1 y + c + F j C 5 z G j = c 1 C 4j zx c 1 C 4j zc 1j y c 1 C j zc 4j zy + C 3j z + C 4j z[ F 1j c 0 c 1 y + c F j C 5 z c 0 c 1 y + c + F 3j C 5 z where c i i = 0 1 are arbitrary constants C 1j y C j z C 3j z C 4j z and C 5 z are arbitrary functions of the corresponding variables. 4 Conclusions It is proved that the +1-dimensional multicomponent BK system possesses Painlevé properties. The CK direct method is a useful tool of solving partial differential equations. But the application to the multicomponent equations is seldom reported. In this paper we apply the CK direct method to the +1-dimensional multi-component BK system and obtain some explicit solutions. All the solutions obtained in this paper have not been reported in other journals. Acknowledgments We would like to thank Prof. Lou Sen-Yue for his helpful discussions. References [1 P.J. Olver Application of Lie Group to Differential Equations Springer Berlin 1986. [ G.W. Bluman and J.D. Cole J. Math. Mech. 18 1969 105. [3 P.A. Clarkson and M.D. Kruskal J. Math. Phys. 30 1989 01; S.Y. Lou Phys. Lett. A 151 1990 133; S.Y. Lou X.Y. Tang and J. Lin J. Math. Phys. 41 000 886. [4 S.Y. Lou J. Phys. A 3 1990 L649; Sci. China Ser. A: Math. Phys. Astron. Technol. Sci. 34 1991 1098; J. Math. Phys. 33 199 4300; Phys. Lett. A 176 1993 96. [5 M.J. Ablowitz A. Ramani H. Segur Lett. Nuovo Cimento 3 1978 333. [6 J. Weiss M. Tabor J. Carnevale J. Math. Phys. 4 1983 55; S.Y. Lou and Q.X. Wu Phys. Lett. A 6 1999 344. [7 R. Conte Phys. Lett. A 140 1989 383; S.Y. Lou Phys. Rev. Lett. 80 1998 507. [8 S.Y. Lou and X.B. Hu Commun. Theor. Phys. Beijing China 9 1998 145.