New Formal Solutions of Davey Stewartson Equation via Combined tanh Function Method with Symmetry Method

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1 Commun. Theor. Phys. Beijing China pp c International Academic Publishers Vol. 7 No. April New Formal Solutions of Davey Stewartson Equation via Combined tanh Function Method with Symmetry Method ZHI Hong-Yan and ZHANG Hong-Qing Department of Applied Mathematics Dalian University of Technology Dalian 60 China Received May 9 006; Revised September Abstract We combine the tanh function method with the symmetry group method to construct new type of solutions of Davey Stewartson equation and implemente it in a computer algebraic system. As a result some new types of solutions are obtained. This method is also applied to other differential equations if the nonlinear evolution equations admit nontrivial one-parameter group of transformation. PACS numbers: 0.30.Jr Key words: tanh function method symmetry group method Davey Stewartson equation exact solution Introduction With the development of solitary theory many effective methods were developed to seek the exact solutions of differential equations such as the inverse scattering method Bäcklund and Darboux transformation method Hirota bilinear method 5 the symmetry group method 6 9 the homogeneous balance method 0 the tanh function method and all kinds of auxiliary equation methods the Riccati method 3 6 projective Riccati equation method 78 the sine cosine function method 9 the Jacobi elliptic function method 0 Function rational expansion method and equation rational expasion method etc. Recently the tanh function method and auxiliary equation methods were attracting more and more researchers owing to the symbolic computation. And the symmetry method for the differential equations originally developed by Sophus Lie in the latter half of the nineteenth century was highly algorithmic and hence amendable to symbolic. With the symmetry group method we can obtain analytical results for partial differential equations such as the construction of similarity solutions or nonclassical solutions finding conservation laws equivalence mappings and linearization. 67 In fact the admitted symmetry group can transform the non-invariant solutions to the new formal ones. But it is impossible to construct the non-invariant solutions with the method. Recently Dong gave some new solutions of the -dimensional Burgers Korteweg-de-Vries equation by using the classical Lie method of infinitesimals and the direct method. 3 In this paper we use the direct ansätze method to construct the non-invariant solutions. Under the symmetry group of transformation we may get the new formal solutions. That is to say we give a method to construct new solutions of the nonlinear evolution equations which combines the two kinds of methods the tanh function method and symmetry group method to find new solutions of differential equations. In Sec. we introduce the combined method in details. And we give an example to illustrate the method in Sec. 3. We conclude the paper in the last section. Introduction of the New Method Consider a system of N nonlinear evolution equations N with n independent variables x = x x... x n and m dependent variables u = u u... u m given by A µ x u u u... r u = 0 µ =... N j u denotes the coordinates with components j u/ x i x i... x ij = u ii...i j i j =... n for j =... k corresponding to all j-th-order partial derivatives of u with respect to x. Our new method includes the following steps. Step Find all the infinitesimal generators of Lie group of transformation with the Infinitesimal Criterion. 67 This can be realized in Maple with liesymm package. We assume that the infinitesimal generator is V = ξ i x u η µ x u i =... n µ =... m. x i u µ Step Calculate the corresponding one-parameter group of nontrivial transformations by solving the ordinary differential equationsodes d x i dɛ = dũ j ξi x ũ i =... n dɛ = ηj x ũ j =... m with x = x ũ = u at ɛ = 0. We assume that the results are x i = X i x u ɛ ũ j = U j x u ɛ i =... n; j =... m. The project partially supported by the State Key Basic Research Program of China under Grant No. 00CB zhidlut@yahoo.com.cn

2 588 ZHI Hong-Yan and ZHANG Hong-Qing Vol. 7 Step 3 Seek the simple solutions of the given nonlinear evolution equations u i = u i x i =... m with the tanh function method the auxiliary equation methods etc. Step i If n i= ξ i x u u j η j x u 0 j =... m x i uj=u jx under the obtained Lie group of transformations we obtain the new type of solutions ii If ũ j = U j x u ɛ = U j x ux ɛ = Ũj x ũ ɛ j =... m. n i= ξ i x u u j η j x u = 0 j =... m x i uj=u jx that is to say u j = u j x is invariant under the transformation. Then we cannot obtain new solutions from the known one. In the following we choose the coupled -dimensional nonlinear system of Davey Stewartson equations to illustrate the key step of the combined method. 3 Exact Solutions of the -Dimensional Davey Stewartson Equation Now we consider the coupled -dimensional nonlinear system of Davey Stewartson equation 5 iu t pu xx u yy r u u uv = 0 v xx pv yy r u xx = 0 a b the ux y t and vx y t are complex valued functions and p = ± r is a constant. Equation with p = or p = are called the DS I and DS II equations respectively. We apply the standard infinitesimal procedure to find the symmetry group of transformations. We find it more convenient to rewrite the corresponding system in a real form by separating u = a bi v = e f i into real and imaginary parts. Then we obtain a system of four real partial differential equations: a t pb xx b yy ra b b 3 af be = 0 a b t pa xx a yy ra 3 ab bf ae = 0 b e xx pe yy raa xx bb xx a x b x = 0 f xx pf yy = 0. c d We write Eqs. as a system i x y t a b e f = 0 i = 3. The point symmetry x = x ɛξx y t a b e f Oɛ ỹ = y ɛηx y t a b e f Oɛ t = t ɛτx y t a b e f Oɛ ã = a ɛax y t a b e f Oɛ b = b ɛbx y t a b e f Oɛ ẽ = e ɛex y t a b e f Oɛ f = f ɛf x y t a b e f Oɛ 3 is admitted by Eqs. if and only if it satisfies the determining equations pr V i x y t a b e f = 0 i=0 i = 3 for any a b e f solving Eqs. V = ξ x η y τ t A a B b E e F f is the infinitesimal generator of the point symmetry 3 and pr X is the second prolongation of the infinitesimal generator. This condition provides us with quite a complicated system of determining equations a system of linear partial differential equations of the coefficients. This step is entirely algorithmic and is implemented on the computer algebra package Maple. The final step of integrating the determining equations is less algorithmic. Solving these huge numbers of determining equations we find ξ = c xt c x c 6t c 7 η = c yt c y c t c 5 τ = c t c t c 3 E = c t c e F = c t c f A = c at c a c 8 by x p c by c 6 p bx bc 8 B = c bt c b c 8 ay x p c ay c 6 p ax ac 8. So we can obtain eight functional independent infinitesimal generators of Eqs. as follows: V = x V = y V 3 = t V = b a a b V 5 = t x bx p a ax p b V 6 = t y by a ay b V 7 = x x y y t t a a b b e e f f

3 No. New Formal Solutions of Davey Stewartson Equation via Combined tanh Function Method 589 V 8 = xt x yt y t t at by bx p a ay ax p bt b et e ft f. From u = a bi v = e f i we can get ũ = a bi ɛa Bi Oɛ ṽ = e f i ɛe F i Oɛ. Then we can obtain the infinitesimal generator admitted by Eq. Y = x Y = y Y 3 = t Y = iu u Y 6 = t y iyu u Y 5 = t x p ixu u Y 7 = x x y y t t u u v v Y 8 = xt x yt y t t t i tv v. y x p u u Now we find the one-parameter group of transformation admitted by Eq.. From Y 8 we have d x dɛ = x t dỹ dɛ = ỹ t ỹ x p d t dɛ = t dũ t dɛ = ũ i dṽ dɛ = ṽ t 5 with x = x t = t ỹ = y ũ = u ṽ = v at ɛ = 0. Solving the ordinary differential equations 5 we have the group of transformation: x = x ɛt ỹ = y ɛt t = t ɛt ɛiy x /p ũ = u ɛt exp ɛt ṽ = v ɛt. 6 Obviously the above transformation is nontrivial. Similarly we can get the transformations: x = x ɛ ỹ = y t = t ũ = u ṽ = v ; 7 x = x ỹ = y ɛ t = t ũ = u ṽ = v ; 8 x = x ỹ = y t = t ɛ ũ = u ṽ = v ; 9 x = x ỹ = y t = t ũ = u e iɛ ṽ = v ; 0 x = x ɛt ỹ = y t = t iɛx ɛt ũ = u exp ṽ = v ; p x = x ỹ = y ɛt t = t iɛy ɛt ũ = u exp ṽ = v ; x = x e ɛ ỹ = y e ɛ t = t e ɛ ũ = u e ɛ ṽ = v e ɛ ; 3 by solving the ODEs corresponding to Y Y 7 respectively. In the following we seek the simple solutions of Eq.. The transformation ax y t = az bx y t = bz ex y t = ez fx y t = fz z = kx ly λt transforms Eqs. into λa pk l b ra b b 3 af be = 0 5a λb pk l a ra 3 ab bf ae = 0 5b k pl e rk a b = 0 5c k pl f = 0. 5d From Eqs. 5c and 5d we may assume e = rk a b k pl α = Ma b α f = α. 6 Throughout this paper we write rk /k pl = M. Substituting Eq. 6 into Eqs. 5a and 5b we obtain λa pk l b ra b b 3 aα bma b α = 0 λb pk l a bα rab a 3 ama b α = 0. 7a 7b Then we assume that equations 7 have the following formal solutions: a = β 0 β tanhz β sechz b = δ 0 δ tanhz δ sechz. 8 With the aid of Maple substituting Eq. 8 into Eqs. 7 yields two polynomial equations in sechξ and tanhξ. And the degree of tanhξ is 0 or. Setting the coefficients of tanh i ξsech j ξ i = 0 ; j = 0... to be zero yields a set of over-determined algebraic equations with respect to β 0 β β δ 0 δ δ k l λ α α. Solving the over-determined algebraic equations we get the following results There are other results we only list some of them for simplicity. Case δ = β 0 = δ = α = β = 0 pk β = l pl k r 3 k pl α = rδ 0 pl 3 k pl k pk l and λ k l δ 0 are arbitrary constants. Case δ = α = δ 0 = β = 0 pk β 0 = l k pl r 3 k pl pk λ = l k pl r 3 k pl rδ pl 3 k k pl 3pk l δ

4 590 ZHI Hong-Yan and ZHANG Hong-Qing Vol. 7 pk β = l pl k r pl 3 k δ α = rδ pl 3 k pl k pk l and l k δ are arbitrary constants. Case 3 δ = α = β 0 = δ = β = 0 k β = pl pk l 3 rk rpl α = rδ 0 pl 3 k pl k pk l and l k λ δ 0 are arbitrary constants. Case β = δ = β = 0 δ 0 = α k pl rβ 0 pl 3 k δ = λ k pl rβ 0 pl 3 k α Ω = rβ 0 pl 3 k pl k Ω = 9 r β 0 λ α k λ pl 6 r β 0 pl α pl k α p l r β 0 p l λ p l and l k β 0 α λ are arbitrary constants. Case 5 and l k β λ are arbitrary constants. Case 6 α = β 0 = δ = β = 0 δ 0 = λ k pl rβ pl 3 k δ N = 3 rk rpl α = λ k λ k pl λ p l β r p l β r pl k 36 β r k 8β r pl 3 k k pl N = β rpl 3 β rk pk l k l p p k l δ = β = 0 β 0 = β α = k p l α β 3 rk rpl λ = 6 β pk l 3 rk rpl δ 0 = 3 rk rpl δ = 3 rk rpl and l k β and α are arbitrary constants. According to Cases 6 Eq. 8 and u = a bi v = e f i = Ma b α α i we can obtain the simple solutions of the original Eqs. as follows: pk u = l pl k r 3 k pl sech kx ly λt δ 0 i v = M pk l pl k r 3 k pl sech kx ly λt Mδ0 rδ 0 pl 3 k pl k λ k l and δ 0 are arbitrary constants; pk l u = pk l k pl pk r 3 k pl l pl k r pl 3 k δ sech z iδ tanh z v = rδ pl 3 k pl k pk l pk l k pl M r 3 k pl pk l pl k r pl 3 k sech z pk l k pl pk l pl k r 3 k pl r pl 3 k δ sech z δ z = kx ly t pk l k pl r 3 k pl rδ pl 3 k k pl 3pk l δ

5 No. New Formal Solutions of Davey Stewartson Equation via Combined tanh Function Method 59 l k δ are arbitrary constants; k u 3 = pl pk l 3 rk rpl tanh kx ly λt δ 0 i k pl pk l v 3 = M 3 rk rpl tanh kx ly λt δ0 rδ 0 pl 3 k pl k l k λ δ 0 are arbitrary constants; u = β 0 i k pl rβ 0 pl 3 k α λ tanh kx ly λt k v = M β0 pl r β0 pl 3 k α λ tanh kx ly λt pk l Ω rβ 0 pl 3 k pl k α i Ω = 9 r β 0 λ α k λ pl 6 r β 0 pl α pl k α p l r β 0 p l λ p l and l k β 0 α λ are arbitrary constants; λ k pl u 5 = β tanh kx ly λt i rβ pl 3 k N 3 rk sech kx ly λt rpl λ k v 5 = α M β tanh pl kx ly λt rβ pl 3 k N 3 rk sech kx ly λt rpl α = λ k λ k pl λ p l β r p l β r pl k 36 β r k 8β r pl 3 k k pl N = β rpl 3 β rk pk l k l p p k l and l k β and λ are arbitrary constants. u 6 = tanh z β i r pl 3 k v 6 = M β N r pl 3 k tanh z k p l α β 3 rk rpl α i z = kx ly 6 β t pk l 3 rk rpl and l k β and α are arbitrary constants. Remark i We can also assume that Eqs. 7 have the other formal solutions like β β sechz a = β 0 µ µ tanhz µ 3 sechz b = δ δ δ sechz 0 µ µ tanhz µ 3 sechz etc. ii In Refs. and 5 the authors introduced u = uξ e iη v = vξ ξ = kx py λt η = αx βy γt to seek the explicit solutions of Eqs. uξ and vξ were real functions. So they could not obtain the solutions like u v u 6 v 6. If u = ux y t v = vx y t is a solution of Eqs. then under the transformation 6 ũ = x u tɛ ỹ tɛ t ɛipỹ x exp tɛ p ṽ = v x/ tɛ ỹ/ tɛ t/ tɛ tɛ is also a solution of the system. Then from u u 6 v v 6 we can obtain the new type of solutions of the original Eqs. ũ = pk l pl k k x lỹ λ t ɛipỹ x r 3 k pl sech δ 0 i exp p pk ṽ = tɛ l pl k k x lỹ λ t M r 3 k pl sech δ0 rδ 0 pl 3 k pl k pk l

6 59 ZHI Hong-Yan and ZHANG Hong-Qing Vol. 7 λ k l δ 0 are arbitrary constants; ũ = tɛ pk l k pl pk r 3 k pl iδ tanh z l pl k r pl 3 k δ sech z ɛipỹ x exp p rδ ṽ = tɛ pl 3 k pk l k pl pl k M r 3 k pl pk l pl k r pl 3 k sech z pk l pk l k pl pk l pl k r 3 k pl r pl 3 k δ sech z δ z = k x lỹ t pk l k pl rδ pl 3 k r 3 k pl k pl 3pk l δ and l k δ are arbitrary constants; ũ 3 = tɛ k pl pk l k x lỹ λ t 3 rk rpl tanh k ṽ 3 = tɛ pl pk l k x lỹ λ t M 3 rk rpl tanh l k λ δ 0 are arbitrary constants; ũ = β 0 i k pl tɛ rβ 0 pl 3 k k ṽ = tɛ M β0 pl r β0 pl 3 k Ω rβ 0 pl 3 k pl k α i α λ tanh δ 0 i exp δ0 ɛipỹ x p rδ 0 pl 3 k pl k pk l k x lỹ λ t ɛipỹ x exp p k x lỹ λ t α λ tanh Ω = 9 r β 0 λ α k λ pl 6 r β 0 pl α pl k α p l r β 0 p l λ p l and l k β 0 α and λ are arbitrary constants; ũ 5 = k x lỹ λ t β tanh iλ k pl tɛ rβ pl 3 k N k x lỹ λ t ɛipỹ i 3 rk rpl sech x exp p k x lỹ λ t λ k ṽ 5 = tɛ α M β tanh pl rβ pl 3 k N k x lỹ λ t 3 rk rpl sech α = λ k λ k pl λ p l β r p l β r pl k 36 β r k 8β r pl 3 k k pl N = β rpl 3 β rk pk l k l p p k l and l k β λ are arbitrary constants; ũ 6 = ɛipỹ x tanh z β i tɛ r pl 3 k exp p

7 No. New Formal Solutions of Davey Stewartson Equation via Combined tanh Function Method 593 ṽ 6 = tɛ Mβ N r pl 3 k tanh z k p l α β k x lỹ z = 6 β t pk l 3 rk rpl 3 rk rpl α i and l k β and α are arbitrary constants. Similarly under the transformations and we can also obtain other new solutions. It is easy to see that ũ ũ 6 ṽ ṽ 6 are different from u u 6 v v 6 when ɛ 0 respectively. It is impossible to obtain them by the tanh function method directly. As far as we know u v u 6 v 6 ũ ṽ ũ 6 ṽ 6 have not been presented by others. Conclusion In this paper we combine the direct ansätze method with the symmetry group method to obtain new type of solutions for the coupled two-dimensional Davey Stewartson equation. When one system has a nontrivial group of transformation we can use the combined method to obtain new formal solutions. Of course we can use the auxiliary equation methods to construct simple solutions. In fact we can reduce the original system by the symmetry group but it is troublesome to do so and the reduced system is always complicated and difficult to be solved. So our new combined method seems to be more useful to seek the exact solutions of the nonlinear evolution equations. References M.J. Ablowitz and P.A. Clarkson Solitons Nonlinear Evolution Equations and Inverse Scattering Cambridge University Press Cambridge 99. M. Wadati H. Sanuki and K. Konno Prog. Theor. Phys K. Konno and M. Wadati Prog. Theor. Phys C.H. Gu et al. Soliton Theory and Its Application Springer Berlin R. Hirota Phys. Rev. Lett P.J. Olver Applications of Lie Groups to Differential Equations Springer Berlin G. Bluman and S. Kumei Symmetries and Differential Equations Springer New York P.A. Clarkson and M.D. Kruskal J. Math. Phys S.Y. Lou X.Y. Tang and J. Lin J. Math. Phys M.L. Wang Phys. Lett. A W. Hereman Comput. Phys. Commun E.J. Parkes and B.R. Duffy Comput. Phys. Commun E.G. Fan Phys. Lett. A Z.Y. Yan Phys. Lett. A B. Li Y. Chen and H.Q. Zhang Z. Naturforsch. A H.Y. Zhi et al. Commun. Theor. Phys. Beijing China R. Conte and M. Musette J. Phys. A T.C. Bountis V. Papageorgiou and P. Winternitz J. Math. Phys C. Yan Phys. Lett. A S.K. Liu Z.T. Fu S.D. Liu and Q. Zhao Phys. Lett. A Y. Chen Q. Wang and B. Li Z. Naturforsch. A Y. Chen Q. Wang and B. Li Chaos Soliton and Fractals Z.Z. Dong X.Q. Liu and C.L. Bai Commun. Theor. Phys. Beijing China E.G. Fan J. Phys. A: Math. Gen Z.Y. Yan Commun. Theor. Phys. Beijing China