The first integral method and traveling wave solutions to Davey Stewartson equation

Transcription

1 18 Nonlinear Analysis: Modelling Control 01 Vol. 17 No The first integral method traveling wave solutions to Davey Stewartson equation Hossein Jafari a1 Atefe Sooraki a Yahya Talebi a Anjan Biswas b a Department of Mathematics Computer Science University of Mazaran P.O. Box Babolsar Iran jafari@umz.ac.ir; atefe_sooraki@umz.ac.ir; talebi@umz.ac.ir b Department of Mathematical Sciences Delaware State University Dover DE USA biswas.anjan@gmail.com Received: 4 October 011 / Revised: 17 February 01 / Published online: 1 May 01 Abstract. In this paper the first integral method will be applied to integrate the Davey Stewartson s equation. Using this method a few exact solutions will be obtained using ideas from the theory of commutative algebra. Finally soliton solution will also be obtained using the traveling wave hypothesis. Keywords: first integral method Davey Stewartson equations exact solutions. 1 Introduction The first integral method was first proposed for solving Burger-KdV equation [1 which is based on the ring theory of commutative algebra. This method was further developed by the same author [ 3 some other mathematicians [ 4 5. The present paper investigates for the first time the applicability effectiveness of the first integral method on the Davey Stewartson equations. We consider the Davey Stewartson (DS equations [6 7: iq t + 1 σ( q xx + σ q yy + q q φ x q = 0 φ xx σ φ yy ( q x = 0. (1 The case σ = 1 is called the DS-I equation while σ = i is the DS-II equation. The parameter characterizes the focusing or defocusing case. The Davey Stewartson I II are two well-known examples of integrable equations in two space dimensions which arise 1 Corresponding author. c Vilnius University 01

2 Davey Stewartson equation 183 as higher dimensional generalizations of the nonlinear Schrodinger (NLS equation [8. They appear in many applications for example in the description of gravity-capillarity surface wave packets in the limit of the shallow water. Davey Stewartson first derived their model in the context of water waves from purely physical considerations. In the context q(x y t is the amplitude of a surface wave packet while φ(x y reperesents the velocity potential of the mean flow interacting with the surface wave [8. The remaining portion of this article is organized as follows: Section is a brief introduction to the first integral method. In Section 3 the first integral method will be implemented some new exact solutions for Davey Stewartson equation will be reported. Additionally in this section the traveling wave solution will also be obtained to retrieve soliton solution. A conclusion future directions for research will be summarized in the last section. The first integral method Consider a general nonlinear PDE in the form P (u u t u x u y u xx u tt u yy u xt u xy u yt u xxx... = 0. Initially we consider case σ = 1 (DS-I. So using the wave variable η = x αy + αt leads into the following ordinary differential equation (ODE: Q ( U U U U... = 0 ( where prime denotes the derivative with respect to the same variable η. For case σ = i (DS-II we use the wave variable η = x + αy αt into Eq. (. Next we introduce new independent variables x = u y = u η which change to a dynamical system of the following type: x = y y = f(x y. (3 According to the qualitative theory of differential equations [1 9 if one can find two first integrals to Eqs. (3 under the same conditions then analytic solutions to Eqs. (3 can be solved directly. However in general it is difficult to realize this even for a single first integral because for a given autonomous system in two spatial dimensions there does not exist any general theory that allows us to extract its first integrals in a systematic way. A key idea of our approach here to find first integral is to utilize the Division theorem. For convenience first let us recall the Division theorem for two variables in the complex domain C [10. Division theorem. (See [11. Suppose P (x y Q(x y are polynomials of two variables x y in C[x y P (x y is irreducible in C[x y. If Q(x y vanishes at all zero points of P (x y then there exists a polynomial G(x y in C[x y such that Q(x y = P (x yg(x y. Nonlinear Anal. Model. Control 01 Vol. 17 No

3 184 H. Jafari et al. 3 Exact solutions of Davey Stewartson equation In order to seek exact solutions of Eqs. (1 we first consider case σ = 1. So Eqs. (1 reduces to iq t + 1 (q xx + q yy + q q φ x q = 0 φ xx φ yy ( q x = 0. (4 Now in order to seek exact solutions of Eqs. (4 we assume q(x y t = u(x y t exp [ i(αx + βy + kt + l (5 where α β k are constants to be determined later l is an arbitrary constant. We assume β = 1. Substitute Eq. (5 into Eqs. (4 to yield i(u t + αu x + u y + 1 (u xx + u yy 1 ( α + k + 1 u + u 3 φ x u = 0 φ xx φ yy ( u x = 0. (6 Using the transformation u = u(η φ = φ(η η = x αy + αt where α is a constant Eqs. (6 further reduces to α ( u 1 ( α + k + 1 u + u 3 φ u = 0 φ 4α φ ( u (7 = 0 where prime denotes the differential with respect to η. Integrating the second part of Eq. (7 with respect to η taking the integration constant as zero yields φ = Substituting Eq. (8 into the first part of (7 yields α ( u 1 ( α + k + 1 u Application of Division theorem 1 4α u. (8 ( 1 4α u 3 = 0. (9 In order to apply the Division theorem we introduce new independent variables x = u y = u η which change Eq. (9 to the dynamical system given by x = y y = α + k α x + 1 4α x3. (10

4 Davey Stewartson equation 185 Now we are going to apply the Division theorem to seek the first integral to (10. Suppose that x = x(η y = y(η are the nontrivial solutions to (10 P (x y = m a i (xy i is an irreducible polynomial in C[x y such that P ( x(η y(η = m ( a i x(η y(η i = 0 (11 where a i (x (i = m are polynomials in x all relatively prime in C[x y a m (x 0. Equation (11 is also called the first integral to (10. We start our study by assuming m = 1 in (11. Note that dp dη is a polynomial in x y P [x(η y(η = 0 implies dp dη = 0. By the Division theorem there exists a polynomial H(x y = h(x + g(xy in C[x y such [ dp P dη = x x η + P y y η (1 = 1 a i(xy i ia i (xy i 1 [ α + k α x + 1 4α x3 = ( h(x + g(xy ( 1 a i (xy i (1 where prime denotes differentiating with respect to the variable x. On equating the coefficients of y i (i = 1 0 on both sides of (1 we have a 1(x = g(xa 1 (x (13 a 0(x = h(xa 1 (x + g(xa 0 (x (14 [ α + k + 1 a 1 (x 1 + 4α x + 1 4α x3 = h(xa 0 (x. (15 Since a 1 (x is a polynomial in x from (13 we conclude that a 1 (x is a constant g(x = 0. For simplicity we take a 1 (x = 1 balancing the degrees of h(x a 0 (x we conclude that deg h(x = 1. Now suppose that h(x = Ax+B then from (14 we find a 0 (x = 1 Ax + Bx + D where D is an arbitrary integration constant. Substituting a 0 (x a 1 (x h(x in (15 setting all the coefficients of powers x to be zero we obtain a system of nonlinear algebraic equations by solving it we obtain A = 1 4α B = 0 D = α + k α 1 4α (16 Nonlinear Anal. Model. Control 01 Vol. 17 No

5 186 H. Jafari et al. A = 1 4α B = 0 D = + k α α 1 + 4α. (17 Using (16 (17 in (11 we obtain y + 1 4α x + α + k α 1 + 4α = 0 y 1 4α x α + k α 1 + 4α = 0 respectively where 0. Combining this equations with (10 where we obtain the exact solutions of Eq. (10 as follows: u 1 (η = A 1 tan ( B 1 η A 1 c 1 where A 1 = u (η = A 1 tan ( B 1 η A 1 c 1 (α + k + 1(1 4α (1 + 4α B 1 = 1 (α + k α where 0 c 1 is an arbitrary constant. Therefore the exact solutions to (10 can be written as u 1 (x y t = A 1 tan [ B 1 (x αy + αt A 1 c 1 u (x y t = A 1 tan [ B 1 (x αy + αt A 1 c 1 where 0. Then exact solutions for Eqs. (4 are q 1 = e i(αx+y+kt+l A 1 tan [ B 1 (x αy + αt A 1 c 1 φ 1 = 3(1 4α A3/ 1 tan 3[ B 1 (x αy + αt A 1 c 1 q = e i(αx+y+kt+l A 1 tan [ B 1 (x αy + αt A 1 c 1 φ 1 = 3(1 4α A3/ 1 tan 3[ B 1 (x αy + αt A 1 c 1 where 0.

6 Davey Stewartson equation 187 Notice that integrating of Eq. (8 with respect to η taking the integration constant as zero yields φ = 3(1 4α u3. Now we assume that m = in (11. By the Division theorem there exists a polynomial H(x y = h(x + g(xy in C[x y such that [ dp P dη = x x η + P y y η = a i(xy i+1 + (1 [ α ia i (xy i 1 + k α x + 1 4α x3 = ( h(x + g(xy ( a i (xy i (18 On equating the coefficients of y i (i = from both sides of (18 we have a (x = g(xa (x (19 a 1(x = h(xa (x + g(xa 1 (x (0 ( α a + k + 1 0(x = a (x 1 + 4α x + 1 4α x3 + h(xa 1 (x + g(xa 0 (x (1 [ α + k + 1 a 1 (x 1 + 4α x + 1 4α x3 = h(xa 0 (x. ( Since a (x is a polynomial of x from (19 we conclude that a (x is a constant g(x = 0. For simplicity we take a (x = 1 balancing the degrees of h(x a 0 (x a 1 (x we conclude that deg h(x = 1 or 0 therefore we have two cases: Case 1. Suppose that deg h(x = 1 h(x = Ax + B then from (0 we find a 1 (x = 1 Ax + Bx + D where D is an arbitrary integration constant. From (1 we find [ A a 0 (x = 8 1 4α x 4 + AB [ 1 x3 + ( B +AD α + k α x +BDx+E where E is an arbitrary integration constant. Substituting a 0 (x a 1 (x h(x in ( setting all the coefficients of powers x to be zero we obtain a system of nonlinear algebraic equations by solving it we obtain: A = α + k α ( α (3 + k α B = 0 D = 1 4α E = 1 4α α Nonlinear Anal. Model. Control 01 Vol. 17 No

7 188 H. Jafari et al. A = 4 1 4α B = 0 D = 1 4α E = 1 ( 4α α + k α. Using (3 (4 in (11 we obtain y + 1 4α x + α + k α 1 + 4α = 0 α + k α y 1 4α x α + k α 1 + 4α = 0 respectively where 0. Combining this equations with (10 we obtain two exact solutions to Eq. (10 which was obtained in case m = 1. Case. In this case suppose that deg h(x = 0 h(x = A then from (0 we find a 1 (x = Ax + B where B is an arbitrary integration constant. From (1 we find a 0 (x = [ A 1 4α x4 + α + k α x + ABx + D where D is an arbitrary integration constant. Substituting a 0 (x a 1 (x h(x in ( setting all the coefficients of powers x to be zero we obtain a system of nonlinear algebraic equations by solving it we obtain: Using (5 in (11 we obtain (4 A = 0 B = 0 D = 0. (5 y α + k α x 1 4α x4 = 0. Combining this equations with (10 we obtain the exact solutions to Eq. (10 as follows: where u 3 (η = u 4 (η = A e B(η+A/4c1 e B + 4(1 + 4α e A/ A e B(η+A/4c (1 + 4α e B(η+A/c1 A = 4 (4α 1 ( α + k + 1 B = (4α 1(α + k + 1 (4α + 1(16α 1

8 Davey Stewartson equation while c 1 is an arbitrary constant. Then the exact solutions to (10 can be written as: u 3 (x y t = u 4 (x y t = Then solutions of Eqs. (4 are q 3 = u 3 (x y te i(αx+y+kt+l A e B(x αy αt+a/4c1 e A/(x αy+αt + 4(1 + α e A/c1 A e B(x αy+αt+a/4c1 1 + e B(x αy+αt+a/4c1. A 3 φ 3 = e 3B(x αy+αt+3/4ac1 (1 4α e 6B(x αy+αt (1 + 4α 3 e 3/Ac1 q 4 = u 4 (x y te i(αx+y+kt+l A 3 φ 4 = e 3B(x αy+αt+3/4ac1 (1 4α 1 + A 3 e6b(x αy+αt+3/ac1 where 0. Now we consider the case when σ = i where Eqs. (1 transforms the following: iq t 1 (q xx q yy + q q φ x q = 0 φ xx + φ yy ( q x = 0. (6 We will solve equation (6 similarly as in the case where σ = 1 with the only difference being η = x + αy αt φ = 1+4α u x = y y = α k 1 1 4α x + 1 4α x3. By applying the Division theorem as in the case where σ = 1 by assuming m = 1 the exact solutions of Eqs. (10 as follows: u 1 (η = A 3 tan ( B 3 η A 3 c 1 u (η = A 3 tan ( B 3 η A 3 c 1 where A 3 = (α α 1( 1 4α (1 4α B 3 = 1 (α k α Nonlinear Anal. Model. Control 01 Vol. 17 No

9 190 H. Jafari et al. 0. Then the exact solutions of (6 are q 1 = e i(αx+y+kt+l A 1 tan [ B 1 (x αy + αt A 1 c 1 φ 1 = 3(1 4α A3/ 1 tan 3 [ B 1 (x αy + αt A 1 c 1 q 1 = e i(αx+y+kt+l A 3 tan [ B 3 (x αy + αt A 3 c 1 φ 1 = 3(1 4α A3/ 3 tan 3 [ B 3 (x αy + αt A 3 c 1 where 0 for case m = in Case 1 we obtain two exact solutions to Eqs. (10 which was obtained in case m = 1. As for case m = in Case we obtain exact solutions for Eqs. (10 as follows: where u 3 (η = u 4 (η = A 4 e B4(η+A4/4c1 e B4 + 4(1 + 4α e A4/ A 4 e B4(η+A4/4c (1 + 4α e B4(η+A4/c1 A 4 = 4 (1 4α ( α k 1 B 4 = (1 4α (α k 1 ( 4α 1(1 16α where 0 c 1 is an arbitrary constant. Hence the exact solutions to (6 are q 3 = A 4 e B(x αy αt+a4/4c1 e A4/(x αy+αt + 4(1 + α e A4/c1 e i(αx+y+kt+l A 3 φ 3 = 4e 3B4(x αy+αt+3/4a4c1 (1 4α e 6B(x αy+αt (1 + 4α 3 e 3/A4c1 q 4 = A 4e B4(x αy+αt+a4/4c1 e i(αx+y+kt+l 1 + e B4(x αy+αt+a4/4c1 A 3 φ 4 = 4e 3B4(x αy+αt+3/4a4c1 (1 4α 1 + A 3. 4 e6b4(x αy+αt+3/a4c1 3. Traveling wave solutions The traveling wave hypothesis will be used to obtain the 1-soliton solution to the Davey Stewartson equation (1. It needs to be noted that this equation was already studied by

10 Davey Stewartson equation 191 traveling wave hypothesis in 011 [1 where the power law nonlinearity was considered. Additionally the ansatz method was used to extract the exact 1-soliton solution to the Davey Stewartson equation in 011 that too was studied with power law nonlinearity [13. In this subsection the starting point is going to be Eq. (9 which can now be re-written as u = α + k + 1 4α u α u3. (7 Now multiplying both sides of (7 by u integrating while choosing the integration constant to be zero since the search is for soliton solutions yields ( u = bu au 4 (8 where a = 4α 1 b = α + k + 1 4α. + 1 Separating variables in (8 integrating gives x αy + αt = 1 a sech 1 u b b that yields the 1-soliton solution as u(x αy + αt = A sech [ B(x αy + αt where the amplitude A of the soliton is given by [ b (4α A = a = 1(α 1/ + k + 1 (4α B = α + k + 1 b = + 1 4α + 1 this leads to the constraints α + k + 1 > 0 ( 4α 1 ( α + k + 1 > 0. Hence by virtue of (5 q(x y t = A sech [ B(x αy + αt e i(αx+βy+kt+l. (9 Finally from (8 the topological 1-soliton solution is given by φ(x y t = B tanh [ B(x αy + αt. (30 Thus (9 (30 together constitute the 1-soliton solution of the Davey Stewartson equation given by (1. Nonlinear Anal. Model. Control 01 Vol. 17 No

11 19 H. Jafari et al. 4 Conclusions We described the first integral method for finding some new exact solutions for the Davey Stewartson equation. We have obtained four exact solutions to the Davey Stewartson equation. The solutions obtained are expressed in terms of trigonometric exponential functions. In addition the traveling wave hypothesis is used to obtain the 1-soliton solution of the equation where a topological non-topological soliton pair is retrieved. These new solutions may be important for the explanation of some practical problems. One context where this equation is studied from a practical stpoint is in the study of water waves with finite depth which moves in one direction [14. However in a twodimensional scenario this equation models both short waves long waves. Many explicit solutions are given in this reference which are all applicable to the study of finite depth water waves. This paper also gives a substantial set of solution set that is also meaningful. Addditionally this equation can also be generalized to a fractional derivative case that has been recently touched (15. References 1. Z. Feng On explicit exact solutions to the compound Burgers-KdV equation Phys. Lett. A 93(1 pp X. Deng Traveling wave solutions for the generalized Burgers Huxley equation Appl. Math. Comput. 04( pp B. Lu H. Zhang F. Xie Traveling wave solutions of nonlinear partial equations by using the first integral method Appl. Math. Comput. 16(4 pp A. Maccari The Kadomtsev Petviashvili equation as a source of integrable model equations J. Math. Phys. 37 pp L.F. Tascan A. Bekir M. Koparan Traveling wave solutions of nonlinear evolution equations by using the first integral method Commun. Nonlinear Sci. Numer. Simul. 14(5 pp M. McConnell A.S. Fokas B. Pelloni Localised coherent solutions of the DSI DSII equations a numerical study Math. Comput. Simul. 64 pp H.A. Zedan S.Sh. Tantawy Solutions of Davey Stewartson equations by homotopy perturbation method Comput. Math. Phys. 49 pp A. Davey K. Stewartson On three-dimensional packets of surfaces waves Proc. R. Soc. Lond. Ser. A 338 pp T.R. Ding C.Z. Li Ordinary Differential Equations Peking Univ. Press Peking China Z. Feng Traveling wave behavior for a generalized fisher equation Chaos Solitons Fractals 38( pp T.S. Zhang Exp-function for solving Maccari s system Phys. Lett. A 371(1 pp

12 Davey Stewartson equation G. Ebadi E.V. Krishnan M. Labidi E. Zerrad A. Biswas Analytical numerical solutions for Davey Stewartson equation with power law nonlinearity Waves Rom Complex Media 1(4 pp G. Ebadi A. Biswas The G /G method 1-soliton solution of the Davey Stewartson equation Math. Comput. Modelling 53(5 6 pp R.-J. Wang Y.-C. Huang Exact solutions excitations for the Davey Stewartson equations with nonlinear gain terms Eur. Phys. J. D 57 pp H. Jafari A. Kadem D. Baleanu T. Yilmaz Solutions of the fractional Davey Stewartson equations with variational iteration method Rom. Rep. Phys. 64( 01 (in press. 16. D. Rostamy F. Zabihi K. Karimi S. Khalehoghli The first integral method for solving Maccari s system Appl. Math. ( pp H. Jafari M. Alipour Solutions of Davey Stewartson equation using homotopy analysis method Nonlinear Anal. Model. Control 15(4 pp Nonlinear Anal. Model. Control 01 Vol. 17 No