Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 784134, 5 pages http://dx.doi.org/10.1155/2013/784134 Research Article Symmetry Reduced and New Exact Nontraveling Wave Solutions of (2+1)-Dimensional Potential Boiti-Leon-Manna-Pempinelli Equation Chen Han-Lin and Xian Da-Quan School of Sciences, Southwest University of Science and Technology, Mianyang, Sichuan 621010, China Correspondence should be addressed to Xian Da-Quan; daquanxian@163.com Received 21 September 2012; Accepted 8 December 2012 Academic Editor: Zhengde Dai Copyright © 2013 C. Han-Lin and X. Da-Quan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. With the aid of Maple symbolic computation and Lie group method, (2+1)-dimensional PBLMP equation is reduced to some (1+1)dimensional PDE with constant coefficients. Using the homoclinic test technique and auxiliary equation methods, we obtain new exact nontraveling solution with arbitrary functions for the PBLMP equation. 1. Introduction 2. Symmetry of (1) In this paper, we will consider the potential Boiti-LeonManna-Pempinelli (PBLMP) equation This section is devoted to Lie point group symmetries of (1). Let π’π¦π‘ + π’π₯π₯π₯π¦ − 3π’π₯π₯ π’π¦ − 3π’π₯ π’π₯π¦ = 0, π = π (π₯, π¦, π‘, π’, π’π₯ , π’π¦ , π’π‘ ) (1) where π’ : π π₯ × π π¦ × π π‘+ → π . By some transformations, the PBLMP equation (1) can be equivalent to the asymmetric Nizhnik-Novikov-Veselov (ANNV) system. In fact, the ANNV equation can be obtained from the inner parameterdependent symmetry constraint of the KP equation [1] and may be considered as a model for an incompressible fluid [2]. The PainleveΜ analysis, Lax pair, and some exact solutions have been studied for the PBLMP equation [3]. Tang and Lou obtained the bilinear form of (1) and variable separation solutions including two arbitrary functions by the multilinear variable separation approach [4, 5]. In this paper, by means of Maple symbolic computation, we will use the Lie group method [6, 7], homoclinic test technique [8, 9] and so forth to reduce and solve the PBLMP equation. First, we will derive symmetry of (1). Then we use the symmetry to reduce (1) to some (1 + 1)-dimensional PDE with constant coefficients. Finally, solving the reduced PDE by Homoclinic test technique and auxiliary equation methods [10, 11] implies abundant exact nontraveling wave periodic solutions for the PBLMP equation. (2) be the symmetry of (1). Based on Lie group theory [6], π satisfies the following symmetry equation: ππ¦π‘ + ππ¦π₯π₯π₯ − 3π’π₯π₯ ππ¦ − 3π’π¦ ππ₯π₯ − 3π’π₯ ππ₯π¦ − 3π’π₯π¦ ππ₯ = 0. (3) To get some symmetries of (1), we take the function π in the form π = π (π₯, π¦, π‘) π’π₯ + π (π₯, π¦, π‘) π’π¦ + π (π₯, π¦, π‘) π’π‘ +π (π₯, π¦, π‘) π’ + π (π₯, π¦, π‘) , (4) where π, π, π, π, π are functions of π₯, π¦, π‘ to be determined, and π’(π₯, π¦, π‘) satisfies (1). Substituting (4) and (1) into (3), one can get 1 π = π1 π₯ + π (π‘) , 3 π = π1 π‘ + π2 , 1 π = π1 , 3 π = π (π¦) , 1 π = πσΈ (π‘) π₯ + π (π‘) , 3 (5) 2 Abstract and Applied Analysis where π1 , π2 are arbitrary constants. π(π‘), π(π‘) are arbitrary functions of t. π(π¦) is a arbitrary function of y. Substituting (5) into (4), we obtain the symmetries of (1) as follows: Substituting (15) into (1) yields a reduced PDE of (1) with constant coefficients: 1 1 π = ( π1 π₯ + π (π‘)) π’π₯ + π (π¦) π’π¦ + (π1 π‘ + π2 ) π’π‘ + π1 π’ 3 3 π€ππππ − 6π€ππ π€π + π€ππ‘ = 0. 1 + πσΈ (π‘) π₯ + π (π‘) . 3 (6) (16) Integrating (16) once with respect to π and taking integration constant to zero yield π€πππ − 3π€π2 + π€π‘ = 0. (17) 3. Symmetry Reduction of (1) Based on the integrability of reduced equation of the symmetry (6), we consider the following three cases. Case 1. Let π1 = π2 = 0, π(π‘) = π, π(π‘) = 1, π(π¦) = −1/π(π¦) in (6), then −1 π = π(π¦) (ππ (π¦) π’π₯ − π’π¦ + π (π¦)) , (7) where π is an arbitrary nonzero constant, π(π¦) =ΜΈ 0. Solving the differential equation for π = 0, one gets Combining the above results, we obtain some reduced equations of (1) expressed by (10), (13), and (17), respectively. Meanwhile many new explicit solutions of (1) from these reduced Equations. can be achieved. We omit other cases based on symmetries (6) here. 4. Solve Reduced PDE and Get Exact Nontraveling Wave Solutions of (1) (8) In this section, we seek exact nontraveling wave solutions of (1) by using some appropriate methods to solve reduced equations (10), (13), and (17). Substituting (8) into (1), we get the following (1 + 1)-dimensional nonlinear PDE with constant coefficients: 4.1. Solve Reduced PDE (10). Now, we seek solutions of (10) by auxiliary equation method. Make transformation as follows: π’ = ∫ π (π¦) ππ¦ + π€ (π, π‘) , π = π₯ + ∫ ππ (π¦) ππ¦. ππ€ππππ − 6ππ€π π€ππ + ππ€ππ‘ − 3π€ππ = 0. (9) Integrating (9) once with respect to π and taking integration constant to zero yield ππ€πππ − 3ππ€π2 + ππ€π‘ − 3π€π = 0. π = π(π¦) (π’π¦ + ππ (π¦) π’π‘ ) . π = π‘ − ∫ π (π¦) ππ¦. (12) (13) Case 3. Let π1 = π2 = 0, π(π‘) = 1, π(π‘) = 0, π(π¦) = −1/π(π¦) in (6), then −1 π = π(π¦) (π (π¦) π’π₯ − π’π¦ ) . (14) Solving the equation for π = 0, we obtain π’ = π€ (π, π‘) , π = π₯ + ∫ π (π¦) ππ¦. where π, π are nonzero constants. Substituting (18) into (10) obtains an ordinary differential equation for π(π) as follows: π3 ππσΈ σΈ σΈ − 3ππ2 πσΈ 2 + (ππ − 3π) πσΈ = 0, (19) where πσΈ = ππ/ππ. Let πσΈ = π, then (19) can be written as π3 ππσΈ σΈ − 3ππ2 π2 + (ππ − 3π) π = 0. Substituting (12) into (1), we have the function π€(π₯, π) which must satisfy the following PDE: π€π₯π₯π₯π − 3π€π₯π₯ π€π − 3π€π₯ π€π₯π + π€ππ = 0. (18) (11) Solving the differential equation for π = 0, one gets π’ = π€ (π₯, π) , π = ππ + ππ‘, (10) Case 2. Taking π1 = 0, π2 = 1, π(π‘) = 0, π(π‘) = 0, π(π¦) = 1/π(π¦) in (6) yields −1 π€ (π, π‘) = π (π) , (15) (20) This is the fourth type of ellipse equation (12), its solutions are as follows: 3π − ππ 3π − ππ { { sech2 [√ (π − π0 )] , − { 2 { 4π3 π { { 2π π { { { ππ (3π − ππ) > 0, { { { { { { { { { 3π − ππ 3π − ππ { { csch2 [√ (π − π0 )] , 2 π = { 2π π 4π3 π { { { ππ (3π − ππ) > 0, { { { { { { { { { 3π − ππ 3π − ππ 2 { { − sec [√− (π − π0 )] , { 2π { { 2π 4π3 π { { ππ (3π − ππ) < 0, { (21) Abstract and Applied Analysis 3 where π0 is the integration constant. From the result of (21), some new exact solutions π’1 through π’3 of (1) can be obtained: 1.5 u 3π − ππ π’1 = ∫ π (π¦) ππ¦ − √ ππ 0.5 −0.5 −1.5 −15 −10 3π − ππ × tanh [√ (π (π₯ + π ∫ π (π¦) ππ¦) + ππ‘ − π0 )] , 4π3 π −5 t 0 5 10 × coth [√ 3π − ππ ππ 3π − ππ (π (π₯ + π ∫ π (π¦) ππ¦) + ππ‘ − π0 )] , 4π3 π ππ (3π − ππ) > 0, π’3 = ∫ π (π¦) ππ¦ − √ ππ − 3π ππ ππ (ππ − 3π) < 0. (22) Particularly, we assume π = π = 1, π = 2, π(π¦) = sin(π¦), π0 = 0, π₯ = sech(π‘), then the solution π’1 can be depicted by Figure 1(a). If π = −1, π = 1, π = −1, π(π¦) = β cos(π¦), π0 = 0, π₯ = sin(t), then π’3 can be depicted by Figures 1(b) and 2(a). 4.2. Solve Reduced PDE (13). Make transformation to (13) as follows: π€ (π₯, π) = π (π) , π = ππ₯ + ππ, −1.5 −1 −0.5 0 0.5 1 1.5 −15 −10 −5 0 5 10 15 y (b) Figure 1: (a) The figure of π’1 as π = 1, π = 1, π = 2, π(π¦) = sin(π¦), π0 = 0, π₯ = sech(π‘). (b) The figure of π’3 as π = −1, π = 1, π = −1, π(π¦) = − cos(π¦), π0 = 0, π₯ = sin(π‘). 0.2 0.1 0 u −0.1 −0.2 −0.1 −0.05 t 0 0.05 0.1 (23) where π, π are non-zero constants. Substituting (23) into (13) then we have ππσΈ + π3 πσΈ σΈ σΈ − 3πΎ2 πσΈ 2 = 0. y 6 4 2 u 0 −2 −4 −6 t ππ − 3π × tan [√ (π (π₯ + π ∫ π (π¦) ππ¦) + ππ‘ − π0 )] , 4π3 π −5 (a) ππ (3π − ππ) > 0, π’2 = ∫ π (π¦) ππ¦ − √ −10 15 10 5 0 −6 −4 −2 0 2 4 y 6 8 10 (a) (24) It is equivalent to (19). Based on the above accordant idea, we can get −0.16 −0.12 π π’4 = √− 2π π ×tanh [√− 3 (ππ₯+π (π‘−∫π (π¦) ππ¦)− π0)] , 2π π’5 = √− u −0.08 −0.04 ππ < 0, π 2π π ×coth [√− 3 (ππ₯+π(π‘−∫π (π¦) ππ¦)−π0)] , 2π 0 −10 −5 0 y 5 −1 −2 −3 t −4 10 −5 (b) ππ < 0, Figure 2: (a) The figure of π’3 as π = −1, π = 1, π = −1, π(π¦) = cos(π¦), π0 = 0, π₯ = sin(π‘). (b) The figure of π’9 as π1 = 1, π1 = 1, π2 = 1, π(π¦) = sin(π¦), π₯ = sin(π‘). 4 Abstract and Applied Analysis π’6 = √ π 2π ×tan[√ where π2 = −1. Substituting (30)–(33) into (28) yields the solutions π’7 through π’11 of (1) as follows: π (ππ₯+π (π‘−∫ π (π¦) ππ¦)−π0 )] , 2π3 π’7 = −2π1 tanh (π1 (π₯ + ∫ π (π¦) ππ¦) − 4π12 π‘ + ππ > 0. (25) 4.3. Solve Reduced PDE (17). In this section, we use homoclinic test technique [8, 9] to (17) and transform the unknown function as follows: π€ (π, π‘) = −2(ln π (π, π‘))π . (π·π π·π‘ + π·π4 ) (π when π2 > 0 in (30); π’8 = −2π1 coth (π1 (π₯ + ∫ π (π¦) ππ¦) − 4π12 π‘ + (26) Substituting (26) into (17) and using the bilinear form, we can get ⋅ π) = 0, π’9 = −2π1 π2 (27) π=π π1 (π−π1 π‘) + π1 cos (π2 (π + π2 π‘)) + π2 π , × (coth (π1 (π₯ + ∫ π (π¦) ππ¦) − (π12 − 3π22 ) π‘ + ln (28) π1 π2 ) π1 + sin (π2 (π₯ + ∫ π (π¦) ππ¦) − (3π12 − π22 ) π‘)) where π2 , π1 , π2 , π1 , and π2 are real constants to be determined. Substituting (28) into (27) yields a set of algebraic equations as follows: × (π2 sinh (π1 (π₯ + ∫ π (π¦) ππ¦) π1 π1 π2 (4 (π12 − π22 ) + π2 − π1 ) = 0, − (π12 − 3π22 ) π‘ + ln π1 ((π14 + π44 − 6π12 π22 ) − π12 π1 − π22 π2 ) = 0, π1 π2 π1 π2 (4 (π12 − π22 ) + π2 − π1 ) = 0, 1 ln (−π2 )) , 2 (35) when π2 < 0 in (30); where the Hirota operator π· is defined in [12]. In this case we choose extended homoclinic test function −π1 (π−π1 π‘) 1 ln π2 ) , 2 (34) π1 π2 ) π1 −1 (29) + π1 cos (π2 (π₯ + ∫ π (π¦) ππ¦) − (3π12 − π22 ) π‘)) , π1 π2 ((π14 + π44 − 6π12 π22 ) − π12 π1 − π22 π2 ) = 0, (36) when π1 π1 π2 > 0 in (31) (see Figure 2(b)); 4 (4π14 π2 + π12 π24 ) − 4π12 π1 π2 − π12 π22 π2 = 0. π’10 (π₯, π¦, π‘) = π2 tan (π2 (π₯ + ∫ π (π¦) ππ¦) + 4π22 π‘) , Solving the above equations (29) yields (37) π = π1 , (1) { 1 π1 = 4π12 , π2 = π2 , π2 = π2 , π1 = 0, when π2 = 1 in (32); π2 = π2 , π’11 (π₯, π¦, π‘) (30) { {π = π1 , π2 = π2 , π1 = π1 , π2 = (2){ 1 { 2 2 π2 = −3π12 + π22 , {π1 = −3π2 + π1 , π = π2 π, π2 = π2 , π1 = π1 , (3) { 1 π1 = −4π22 , π2 = 4π22 , π2 π2 − 1 22 , π1 (31) π2 = π2 , × sin (π2 (π₯+∫ π (π¦) ππ¦)+(8π22 −π2 ) π‘)+sin (π2 (π₯+∫ π (π¦) ππ¦)+π2 π‘) , cos (π2 (π₯+∫ π (π¦) ππ¦)+(8π22 −π2 ) π‘)+cos (π2 (π₯+∫ π (π¦) ππ¦)+π2 π‘) (38) when π1 = 2 in (33). Remark 1. If one lets π€π = π£ in (16), then (16) can be written as π£π‘ − 6π£π£π + π£πππ = 0. (32) {π = π2 π, π2 = π2 , π1 = π1 , (4) { 1 2 π2 = π2 , {π1 = π2 − 8π2 , = −2π2 1 π2 = π12 , 4 (39) This is the famous KdV equation. 5. Conclusions (33) In this paper, a combination of Lie group method and homoclinic test technique and so forth is applied and thus the Abstract and Applied Analysis symmetries (6) are obtained. The (2+1)-dimensional potential Boiti-Leon-Manna-Pempinelli equation (1) is reduced to (1 + 1)-dimensional nonlinear PDE of constant coefficients (10), (13), and (17). Further auxiliary equation method and homoclinic test technique are used and some new exact nontraveling wave solutions are obtained. And they include some special and strange structures to be further studied and other relevant solutions about symmetry (6) will be discussed later in another paper. Our results show that combining the Lie group method with homoclinic test technique and so forth is effective in finding nontraveling wave exact solutions of nonlinear evolution equations. Acknowledgments The authors would like to thank Professor Dai Zhengde for his helpful discussions. This work was supported by the Chinese Natural Science Foundation Grant no. 10971169 and the key research projects of Sichuan Provincial Educational Administration, no. 10ZA021. References [1] S.-Y. Lou and X.-B. Hu, “Infinitely many Lax pairs and symmetry constraints of the KP equation,” Journal of Mathematical Physics, vol. 38, no. 12, pp. 6401–6427, 1997. [2] M. J. Ablowitz and P. A. Clarkson, Solitons, Nonlinear Evolution Equations and Inverse Scattering, vol. 149, Cambridge University Press, Cambridge, Mass, USA, 1991. [3] P. G. EsteΜvez and S. Leble, “A wave equation in 2 + 1 PainleveΜ analysis and solutions,” Inverse Problems, vol. 11, no. 4, pp. 925– 937, 1995. [4] X. Y. 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