《Another New ( G0 )-expansion Method and its Applications.doc》由会员分享，可在线阅读，更多相关《Another New ( G0 )-expansion Method and its Applications.doc（21页珍藏版）》请在三一文库上搜索。
1、精品论文GAnother New ( G0 )-expansion Method and its ApplicationsHouxia Shi, Rongfen Shi, Chao LiSchool of Mathematics and Statistics, Lan-zhou University, Lan-zhou 730000, China0GAbstract: In this paper, the ( G )-expansion method is improved not only by making extension to the expression of the soluti
2、ons of the NLEEs, but also by substituting the linear subsidiary equation in30the G -expansion method for a nonlinear one in form of F 0 2 = Pnai F i . Making use of the extendedG i=0method, we study the compound KdV-Burgers equation, the Variant Boussinesq equations and the Hirota-Satsuma coupled K
3、dV equations and obtain rich new families of exact solutions, including the solution-like solutions, period-form solutions.PACS : 02.30.Jr, 04.20.JbGKeywords : extended ( G0 )-expansion method, homogeneous balance principle, compound KdV-Burgers equation, Variant Boussinesq equations, Hirota-Satsuma
4、 coupled KdV equations.1 IntroductionAs mathematical models of the complex physics phenomena, the investigation of explicit solutions of the nonlinear evolution equations (NLEEs) will help one to understand these phenomena better. e-mail: Over last years, much work has been done by mathematicians an
5、d physicists to find special solutions of NLEEs, such as the inverse scattering transform 1, the Backlund/Darboux transform 2-4, the Hirotas bilinear operators 5, the truncated Painleve expansion 6, the tanh-function expansion and its various extension 7-9, the Jacobi elliptic function expansion 10-
6、11, the F-expansion 12-15, the sub-ODE method 16-19, the homogeneous balance method 20-22, the exp-function expansion method and so on.GRecently, M.L. Wang, X.Zh Li and J.L. Zhang 24 present a ( G0 )-expansion method to seek0G0Gmore travelling wave solutions of NLEEs. In this paper, we develop the (
7、 G )-expansion method by changing the linear auxiliary equation of the ( G )-expansion method into a nonlinear one in form ofF =0 2 Pni=0ai F i , and we also make an extension to the expression of the solution. Then, we apply0the extended method to the compound KdV-Burgers equation, the Variant Bous
8、sinesq equations and the Hirota-Satsuma coupled KdV equations, and derive rich new families of exact solutions, including the solution-like solutions, period-form solutions.GThe rest of the paper is organized as follows. In Section 2, we describe the extended ( G )-expansionmethod, and give the main
9、 steps of the method. In the subsequent sections, from Section 3 to Section5, we illustrate the method in detail with the compound KdV-Burgers equation, Variant Boussinesq equations and Hirota-Satsuma coupled KdV equations. In Section 6, we briefly make a summary to the extended method.G2 Summary of
10、 the improved ( G0 )-expansion methodGIn this section, we describe the general theory of the improved ( G0 )-expansion method. For a givenNLEE, say, in two variablesP (u, ut , ux, utt , uxx, uxt , ) = 0, (2.1) where u = u(x, t) is an unknown function, P is a polynomial in u = u(x, t) and its various
11、 partial derivatives, in which the highest order derivatives and nonlinear terms are involved. In the followingGwe give the main steps of the modified ( G0 )-expansion method.step 1. We suppose thatu(x, t) = u(), = x V t, (2.2)the travelling wave variable (2.2) permits us reducing Eq.(2.1) to an ODE
12、 for u = u()GP (u, V u0, u0, V 2u00 , V u00 , u00 , ) = 0. (2.3) Step 2. We assume that the solution of Eq.(2.3) can be expressed by a polynomial in ( G0 ) asfollows:u() = 0 +NXi=NG0i ( G)i , N = 0, (2.4)where 0, i , (i = N, , N ) are constants to be determined later, and G = G() satisfies a nonline
13、arODE in the formG0 2 + qG2 = p, p = 0, (2.5)精品论文http:/where q and p are constants, and N in Eq.(2.4) is a positive integer that can be determined by considering the homogeneous balance principle in Eq.(2.3).Step 3. Substituting Eq.(2.4) into Eq.(2.3) and using Eq.(2.5), the left-hand side of Eq.(2.
14、3) isconverted into another polynomial in ( G0 ), collecting all terms with the same order of ( G0 )i together.G GGEquating each coefficient of the ( G0 )i to zero, yields a set of algebraic equations for 0 , i , (i =N, , N ) and V .Step 4. Assume that the constants 0, i , (i = N, , N ) and V can be
15、 obtained by solving the algebraic equations in Step 3. Since the solutions of the Eq.(2.5) can be obtained, we substitute 0, i , (i = N, , N ), V and the general solutions of Eq.(2.5) into (2.4), then we have travelling wave solutions of Eq.(2.1).We find that Eq.(2.5) admits many fundamental soluti
16、ons which are listed as follows. Case 1. If q 0, p 0, say, the Eq.(2.5) is equivalent toG0 2 + qG2 = p, q 0, p 0, (2.6)then the Eq.(2.6) possesses three triangular type solutions18G11 () = r psin(qr p q( c), G12 () = G13 () = 1cos(qr psin(2qq( c),q( c) + 2r pcos(2q q( c),(2.7)where c is an arbitrary
17、 constant, and 1 = 1, 2 = 1.Case 2. If q 0, the Eq.(2.5) is equivalent toG0 2 qG2 = p, q 0, p 0(2.8)精品论文then the Eq.(2.8) possesses a hyperbolic type solutionG20 () = r psinh(q q( c), (2.9)where c is an arbitrary constant.Case 3. If q 0, p 0, p 0, (2.10)then the Eq.(2.10) possesses a hyperbolic type
18、 solutionG30 () = r pcosh(q q( c), (2.11)where c is an arbitrary constant.Case 4. If q = 0, p = 0, the Eq.(2.5) possesses a polynomial type solutionG40 () = p + c, (2.12)where c is an arbitrary constant.Remark : When p = 0, q = 0, the Eq(2.5) possesses the exponential type solution, however, theNLEE
19、s only has trivial solutions, so we omit this case.3 compound KdV-Burgers equationWe start with the well-known compound KdV-Burgers equation 26ut + Auux + Bu2ux + uxx suxxx = 0, (3.1)where A, B, , s are constants, which is a composition of the KdV, mKdV and Burgers equations, involving nonlinear, di
20、spersion and dissipation effect.The travelling wave variable belowu(x, t) = u(), = x V t, (3.2)permits us converting Eq.(3.1) into an ODE0V u+ Auu0+ Bu2 u0+ u00 su000= 0, (3.3)Considering the homogeneous balance principle, we suppose the solution of Eq.(3.3) in the form:G G0u() = 1( G0 ) + 0 + 1( G
21、), 1 = 0, (3.4)where G = G() satisfies the Eq.(2.5).0GUsing Eq.(2.5), substituting (3.4) into Eq.(3.3) and eliminating ( G )i , yields a set of simultaneous algebraic equations for 1, 0 , 1 and V . Solving the algebraic equations and yields(1)1 = 0, 0 = 3As 6B6Bs , 1 = 6sB, V1 =B2B2 24qs2B 3sA212Bs
22、,(3.5)B(2)1 = q6sB, 0 = 3As 6B6Bs , 1 = 6sB, V2 =B2B2 96qs2B 3sA212Bs .By using (3.5), expression (3.4) can be written as:6sB G03As 6Bu1 () = B ( G )6Bs ,(3.6)q6sB G3As 6B6sB G0u2 () = ( )B G06Bs ( ), B GSubstituting the general solutions of Eq.(2.5) into (3.6), we have exact solutions of Eq.(3.1).C
23、ase 1. When q 0, p 0,6sBq3As 6Bu111 () = u121 () = cot B6sBqtan B6Bs ,3As 6B6Bs ,(3.7)6sBq1 cos 2 sin 3As 6Bu131 () = (B 1sin + 2)cos 6Bs ,where = q( c), = x V1t, 1 = 1, 2 = 1, c is an arbitrary constant.u112 () = u122 () = 6sBqtan B6sBqtan B6sBqcot B6sBqcot Bq3As 6B6Bs ,3As 6B6Bs ,(3.8)6sBq1 sin +
24、2 cos 6sBq1 cos 2 sin 3As 6Bu132 () = (B 1cos 2)sin (B 1sin + 2)cos 6Bs ,where = q( c), = x V2t, 1 = 1, 2 = 1, c is an arbitrary constant.Case 2. When q 0,6sBq3As 6Bu201 () = coth B6Bs , V = V1 ,(3.9)u202 () = 6sBqtanh B3As 6B6Bs 6sBqcoth , V = V2, Bwhere = q( c), = x V t, c is an arbitrary constant
25、.Case 3. When q 0, p 0, p 0,H111 () = 2q cot2 2q, u111 () = 2q cot + 0 ,H121 () = 2q tan2 2q, u121 () = 2q tan + 0 , 1 cos 2 sin 2(4.9)H131 () = 2q(1sin + 2)cos 2q,131 q( sin + cos u () =2 1 cos 2 sin 1 2) + 0 ,where 0 and c are arbitrary constants, = q( c), = x 0 t.H112 () = H122 () = 2q(cot2 + tan
26、2 ), u112 () = u122 () = 2q(cot + tan ) + 0,(4.10) 1 cos 2 sin 2 1 sin + 2 cos 2H132 () = 2q(1sin + 2)cos 2q(1cos 2) ,sin 1 cos 2 sin 1 sin + 2 cos u132 () = 2q(1 sin + 2 cos ) 2q(1 cos 2 sin ) + 0 ,where 0 and c are arbitrary constants, = q( c), = x 0 t.H113 () = 2q(cot2 + tan2 ) 4q, u113 () = 2q(c
27、ot tan ) + 0, H123 () = 2q(tan2 + cot2 ) 4q, u123 () = 2q(cot tan ) + 0,(4.11) 1 cos 2 sin 2 1 sin + 2 cos 2H133 () = 2q(1sin + 2)cos 2q(1cos 2)sin 4q, 1 cos 2 sin 1 sin + 2 cos u133 () = 2q(1 sin + 2 cos ) 2q(1 cos 2 sin ) + 0 ,where 0 and c are arbitrary constants, = q( c), = x 0 t.Case 2. When q
28、0,H201 () = 2q coth2 2q, u201 () = 2q coth + 0,H202 () = 2q coth2 2q tanh2 ,u202 () = 2q coth 2q tanh + 0, H203 () = 2q coth2 2q tanh2 4q, u203 () = 2q coth 2q tanh + 0,where 0 and c are arbitrary constants, = q( c), = x 0 t.Case 3. Whenq 0, p 0,H301 () = 2q tanh2 2q, u301 () = 2q tanh + 0,H302 () = 2q tanh2 2q coth2 ,u302 () = 2q tanh 2q coth + 0, H303 () = 2q tanh2 2q coth2 4q, u303 () = 2q tanh 2q coth + 0,where 0 and c are arbitrary constants, = q