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Authors requiring further information regarding Elsevier’s archiving and manuscript policies are encouraged to visit: http://www.elsevier.com/copyright Available online at www.sciencedirect.com Chaos, Solitons and Fractals 41 (2009) 587–593 www.elsevier.com/locate/chaos A new integrable equation with no smooth solitons Zhijun Qiao *, Liping Liu Department of Mathematics, The University of Texas Pan-American, 1201 West University Drive, Edinburg, TX 78541, USA Accepted 20 November 2007 Communicated by Prof. M. Wadati Abstract In this paper, we propose a new completely integrable equation: 1 1 1 1 m ; t ¼ 2 À 2 2 m xxx 2 m x which has no smooth solitons. This equation is shown to have bi-Hamiltonian structure and Lax pair, which imply inte- grability of the equation. Studying this new equation, we develop two new kinds of soliton solutions under the inho- mogeneous boundary condition limjxj!1m ¼ B where B is nonzero constant. One is continuous and piecewise smooth ‘‘W/M”-shape-peaks solitary solution and the other one-single-peak soliton. The two new kinds of peaked solitons can not be written as the regular type peakon: cejxÀctj, where c is a constant. We will provide graphs to show those new kinds of peaked solitons. Ó 2008 Elsevier Ltd. All rights reserved. 1. Introduction Recently, the study of peaked and cusped soliton equations has arisen lot of attractive attention. The typical repre- sentative of such equations is the well-known Harry–Dym (HD) equation [8] 1 ut ¼ pffiffiffi : u xxx Wadati et al. [18] generalized the HD equation to an integrable hierarchy. In their paper [19,20], Wadati et al. first time proposed the cusp soliton, which is a kind of peaked soliton whose left and right derivatives equal infinities, for the HD equation. Later, there are several authors studying the cusp and peaked soliton solutions for the integrable equations [2–5,9,11,13,15–17]. In this paper, we propose a new peaked soliton equation: 1 1 1 1 m ; t ¼ 2 À 2 ð1Þ 2 m xxx 2 m x * Corresponding author. E-mail address: [email protected] (Z. Qiao). 0960-0779/$ - see front matter Ó 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2007.11.034 588 Z. Qiao, L. Liu / Chaos, Solitons and Fractals 41 (2009) 587–593 where m is a scalar function and subscripts denote the partial derivatives. This equation is shown to have bi-Hamilto- nian structure, and Lax pair that implies its integrability. Through studying equation (1), we develop two new kinds of soliton solutions under the inhomogeneous boundary condition limjxj!1m ¼ B, where B is nonzero constant. One is continuous and piecewise smooth ‘‘W/M”-shape-peaks solitary solution and the other one-single-peak soliton. The two new kinds of peaked solitons cannot be equivalent to the regular peakon: cejxÀctj, where c is a constant. There is no smooth soliton found for the new Eq. (1). We will take some graphs to show how these three peaks soltions and one-single-peak solitons look like. 2. Hamiltonian structure and integrability Eq. (1) can be cast in the following Hamiltonian structure: 1 1 1 1 dH þ dH þ J 1 K 0 mt ¼ 2 À 2 ¼ ¼ ; ð2Þ 2 m xxx 2 m x dm dm where J ¼omoÀ1mo; ð3Þ o K ¼ o3 À o; o ¼ ; ð4Þ Z ox þ 1 1 H 0 ¼ dx; 2 ZX m 1 1 4 4 H þ m2 x; 1 ¼ 3 þ 5 þ 7 x Þd 2 X 4m 5m 7m X ¼ðx0; x0 þ T Þ or X ¼ ð1; þ1Þ is the domain of m that needs to be periodic with T or to approach the same con- þ þ stant as x goes to 1, and H 0 , H 1 are two Hamiltonian functions. Both operator K and operator J are Hamiltonian, and furthermore our Eq. (1) is bi-Hamiltonian (see Remark 1). Remark 1. Apparently, the operator K ¼ o3 À o is Hamiltonian (see [10], chapter 7) because of constant coefficients and skew-symmetric property. From Ref. [10], we also know that the operator J is Hamiltonian if and only if PrV JhðAJ Þ¼0, where Z 1 AJ ¼ ðh ^ JhÞdx 2 oÀ1 is the associated bi-vectorR of J, and h is a basic uni-vector corresponding to m. Let P ¼ mhx, then P x ¼ mhx, 1 Jh ¼ðmPÞx, A ¼2 h ^ðmPÞx dx, and Z Z 1 ÀÁ1 ÀÁ PrV JhðAJ Þ¼ h ^ Jh ^ðmPÞx À hx ^ Jh ^ P dx ¼ h ^ðmPÞx ^ðmPÞx þ hx ^ðmxP þ mP xÞ^P dx Z2 2 1 ÀÁ ¼ h ^ðm P þ m2h Þ^P dx ¼ 0: 2 x x x So, J is Hamiltonian. In a similar way, we can prove that K þ J is also Hamiltonian. Therefore, K and J form a Ham- iltonian pair. In order to show the integrability of this equation, let us consider the following spectral problem ! 1 1 w1 À 2 2 km w1 w1 ¼ 1 1 Uðm; kÞ ; ð5Þ w2 x À 2 km 2 w2 w2 where k is a spectral parameter, m is a scalar potential function periodic or approaching the same constant at both infin- T ities, and w ¼ðw1; w2Þ is the spectral function corresponding to the spectral parameter k. Then, we have Krk ¼ k2Jrk; ð6Þ k 2 2 where rk ¼ 2 ðw1 þ w2Þ. Z. Qiao, L. Liu / Chaos, Solitons and Fractals 41 (2009) 587–593 589 Remark 2. Eq. (6) plays a very important role in the discussions of the periodic solutions of the new wave equation (1), which we will deal with in a subsequent paper [16]. Actually, on the basis of those two operators, following our earlier method [12,14] we are able to generate a new integrable hierarchy. A direct calculation leads to the following statement. Eq. (1) has the following Lax pair: w w 1 ¼ Uðm; kÞ 1 ; ð7Þ w w 2 x 2 w w 1 ¼ V ðm; kÞ 1 ; ð8Þ w2 t w2 where ! À 1 1 km Uðm; kÞ¼ 2 2 ; À 1 km 1 0 2 2 1 2 k 2 mðmxÀmxxÞþ3mx k @ À m k þ m4 A V ðm; kÞ¼ 2 : 2 k2 mðmxþmxxÞ3mx k À þ m4 m w w In fact, one can use mathematical software Maple to check that the compatibility condition 1 ¼ 1 , namely w2 xt w2 tx U t À V x þ½U; V ¼0 generates equation (1). So the wave equation (1) is accordingly completely integrable by the Inverse Scattering Transformation [1]. 3. W/M-shape-peaks solitons and new one-single-peak solitons 3.1. Traveling wave setting Let mðx; tÞ¼pffiffiffiffiffiffiffi1 , then Eq. (1) becomes vðx;tÞ o vðx; tÞ o3 o À ot ¼ vðx; tÞ vðx; tÞ: ð9Þ vðx; tÞð3=2Þ ox3 ox Let us consider the traveling wave solutions of the Eq. (9) through a generic setting vðx; tÞ¼UðnÞ, where n ¼ x À ct, and c is the wave speed. Substituting it into Eq. (9) yields the following ODE: À3=2 U nnn À U n ¼ cU U n: ð10Þ Apparently U ¼ constant is a solution, which is not interesting for us. Let us find non-trivial solutions. Taking indef- inite integral twice on both sides of the ODE (10), we obtain 2c pffiffiffiffi þ U nn À U þ C1 ¼ 0; ð11Þ U pffiffiffiffi U 2 1 4c U þ C U À þ U 2 þ C ¼ 0; ð12Þ 1 2 2 n 2 where C1 and C2 are two constants to be determined. To have solitary traveling wave solutions, we set U ¼ V 2 and impose the boundary condition lim V ¼ A; A > 0; ð13Þ n!1 1 which implies m ! A as x approaches 1 (see paper [15,17] for more details). Substituting the boundary condition (13) into the ODEs (11) and (12) generates the following two constants 2c C ¼ A2 À ; ð14Þ 1 A 1 C ¼ A4 À 2cA: ð15Þ 2 2 590 Z. Qiao, L. Liu / Chaos, Solitons and Fractals 41 (2009) 587–593 So the ODE (12) becomes 2ð2c À A3Þ pffiffiffiffi U 02 ¼ U 2 þ U À 8c U þ A4 þ 4cA: ð16Þ A 3.2. W/M-shape-peaks solitons Setting U ¼ V 2 and taking integral on both sides of the ODE (16), we arrive at 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 3 B A3 þ 2c þ A2V þ ðA3 þ cÞðAV 2 þ 2A2V þ A3 þ 4cÞ C 2 4c A B C 2lnðA þ V þ ðA þ V Þ þ Þpffiffiffiffiffiffiffiffiffiffiffiffiffi @2ln2þ ln A A A3 þ c AðV À AÞ ¼jnj: pffiffiffiffi ffiffiffiffiffiffiffiffiA3 3 In general, we can not get an explicit form of V. But, if p ¼ 2, namely, c ¼ 3, then we have A3þc pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4A pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi AðA þ 2V þ V 2 þ 2AV À 2A2Þ 2lnðA þ V þ V 2 þ 2AV À 2A2Þ2ln2À 2ln ¼jnj; V À A which implies qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 þ 2X þ 3X 2 À 3ð3 þ 2X þ 3X 2ÞðX À 1Þ2 V ¼ A ; 4X À1jnjþln 2 X ¼ e 2 ; 3 n ¼ x þ A3t: 4 Since m ¼ 1, we denote B ¼ 1 –0, then m ! B as n !1, therefore we obtain the following explicit solution of Eq.