 Applied Mathematics, 2011, 2, 258-263 doi:10.4236/am.2011.22030 Published Online February 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM The First Integral Method for Solving Maccari’s System Davood Rostamy1,2, Fatemeh Zabihi1, Kobra Karimi1, Siamak Khalehoghli2 1Department of Mat hem at ic s, Imam Khomeini International University, Qazvin, Iran 2Department of Mat hem at ic s, Islamic Azad University, South Tehran Branch, Tehran, Iran E-mail: rostamyd@yahoo.com Received November 13, 2010; revised December 31, 2010; accepted January 4, 2011 Abstract In this paper, we investigate the first integral method for solving the solutions of Maccari’s system. This idea can obtain some exact solutions of this system based on the theory of Commutative algebra. Keywords: First Integral Method, Maccari’s System, Exact Solution 1. Introduction The first integral method was first proposed for solving Burger-KdV equation  which is based on the ring theory of commutative algebra. This method was further developed by the same author in [2-10] and some other mathematicians [2,11,12,13]. The present paper investi-gates for the first time the applicability and effectiveness of the first integral method on the Maccari’s. We consid-er Maccari’s system: 2=0,=0.txxty xiQ QQRRR Q (1) For the first time, Maccari derived this system from the Kadomtsev-Petviash vili equ ation by using asymptoti- cally exact reduction me t hod based Fo uri er expansi o n an d spatiotemporal rescaling . Maccari’s system is a kind of nonlinear evolution equations that are often presented to describe the motion of the isolated waves, localized in a small part of space, in many fields such as hydrody-namic, plasma physics, nonlinear optic, etc. Zhang used Exp-function method for seeking exact solutions of Maccari’s system . The remaining structure of this article is organized as follows: Section 2 is a brief introduction to the first in- tegral method. In Section 3, implementing the first inte- gral method, some new exact solutions for Maccari’s sy- stem are reported. This describes ability and reliability of the method. A conclusion and future directions for re-search are all summarized in the last section. 2. The First Integral Method Consider a general nonlinear PDE in the form ,,,,,,, ,,,,=0.txyxxttyyxtxyytxxxPuuu uuuuuuuu (2) Using the wave variable =2xykt carries in-to the following ODE: ,, ,,=0,QUU UU (3) where prime denotes the derivative with respect to the same variable . Next, we introduce new independent variables =xu, =yu which change to a system of ODEs ==,.xyyfxy (4) According to the qualitative theory of differential equ-ations , if one can find two first integrals to system (4) under the same conditions, then analytic solutions to (4) can be solved directly. However, in general, it is difficult to realize this even for a single first integral, because for a given plane autonomous system, there is no general theo- ry telling us how to find it’s first integrals in a systematic way. A key idea of our approach here to find first inte- gral is to utilize the division theorem. For convenience, first let us recall the division theorem for two variables in the complex domain  . Theorem 2.1. Division theorem (see ) Suppose that ,Pxy and ,Qxy are polynomials of two va- riables x and y in ,xy and ,Pxy is irreducible in ,xy. If ,Qxy vanishes at all zero points of ,Pxy, then there exists a polynomial ,Gxy in ,xy such that ,= ,,.QxyPxyGxy 3. Exact Solutions for Maccari’ s System In order to seek exact solutions of system (1), we sup- D. ROSTAMY ET AL. Copyright © 2011 SciRes. AM 259pose ,,=,, exp,Qxytuxytikxytl (5) where ,k and  are constants to be determined later, l is an arbitrary constant. Substituting Equation (5) into system (1) and yields 222=0,=0,txxxty xiukuuku uRRR u (6) Using the transformation  =, =, =2,uuRRxykt (7) where  is a constant, system (6) become the follow-ing 22=0,2=0,ukuuRkR u   (8) where prime denotes the differential with respect to . Integrating the second segment of Equation (8) with re-spect to  and taking the integration constant as zero yields yields 21=.2Ruk (9) Substituting Equation (9) into the first segment of (8) yields 231=0.2uku uk   (10) Next, we introduce new independent variables =xu, =yu which change Equation (10) to a system of ODEs 23=1=.2xyykx xk (11) Now, we are applying the Division Theorem to seek the first integral to (9). Suppose that=xx and =yy are the nontrivial solutions to (9), and  =0,= ,miiipxya xy is an irreducible polynomial in ,xy such that  =0,= =0,miiipxya xy  (12) where  =0,1, ,iaxi m are polynomials of x and all relatively prime in ,xy, 0max. Equation (10) is also called the first integral to (9). We start our study by assuming =1m in (12). Note that dpd is a polynomial in x and y, and  ,=0px y implies 11=0dpd. By the Division Theorem, there exists a polynomial ,=Hxyhxgxy in ,xy such that    11 11111123=0 =01=0=1=2=,iiiiiiiiidpp xp ydxyaxyiaxyk xxkhxgxya xy   (13) where prime denotes differentiating with respect to the variable x. On equating the coefficients of =2,1,0iyi on both sides of (13), we have 11=,ax gxax (14) 010=,axhxax gxax (15)   23101=.2axk xxhxaxk (16) Since, 1ax is a polynomial of x, from (12) we con-clude that 1ax is a constant and =0.gx For sim-plicity, we take 1=1ax , and balancing the degrees of hx and 0ax we conclude that deg =1hx , only. Now suppose that =hxAxB, then From (13), we find 201=,2axAx BxD where D is an arbitrary integration constant. Substi-tuting 0ax, 1ax and hx in (14) and setting all the coefficients of powers x to be zero, we obtain a system of nonlinear algebraic equations and by solving it, we obtain the following solutions: 222=,=0,= 2,22ABDk kk (17) and 222=,=0,=2.22ABDk kk (18) Using (15) and (16) in (10), we obtain 22222=0,222kkyxk and 22222=0,222kkyxk respectively. Combining this equations with (9), we ob-tain the exact solutions of Equation (10) as follows: D. ROSTAMY ET AL. Copyright © 2011 SciRes. AM 260 2221 12=2tanh 2,2ukk ikkkc   22 22 12=2tanh 2,2ukk ikkkc   where 1c is an arbitrary constant. Therefore, the exact solutions to (10) can be written as  22 21 12,,=2tanh 22,2uxytkkik xyktkkc   22 22 12,,=2tanh 22.2uxytkkikxyktkkc   Then exact solutions for system (1) are 22 221 122 21 12= 22,tanh 22=2tanh 22.2ikxyt lRk ikxyktkkcQe kk ikxyktkkc   (19) and 22 222 122 22 12= 22,tanh 22=2tanh (2)2.2ikxyt lRk ikxyktkkcQekkikx yktkkc   (20) Now we assume that m = 2 in (10). By the Division Theor em, there exists a polynomial  ,=Hxyhx gxy in ,xy such that    11 11221123=0 =02=0=1=2=.iiiiiiiiidpp xp ydxyaxyiaxyk xxkhxgxyaxy   (21) On equating the coefficients of =3,2,1,0iyi on both sides of (21), we have 22=,ax gxax (22)  121=,axhxax gxax (23)   2302101=2 2,ax axkxxkhxaxgxax (24)   23101=.2axk xxhxaxk (25) Since, 2ax is a polynomial of x, from (20) we conclude that 2ax is a constant and =0gx . For simplicity, we take 2=1ax , and balancing the de-grees of ,hx0ax and 1ax we conclude that deg =1hx or 0, therefore we have two cases: Case1: Suppose that deg =1hx and =hxAxB, then from (21) we f ind 211=,2a xAxBxD where D is an arbitrary integration constant. From (22) we find  24302221=82 2212,2AABaxx xkADBkx BDxE  where E is an arbitrary integration constant. Substituting 0ax, 1ax and hx in (25) and setting all the co- efficients of powers x to be zero, we obtain a system of nonlinear algebraic equations and by solving it, we obtain 2222=,=0,2=22,22=,2kkEBDkkAk (26) D. ROSTAMY ET AL. Copyright © 2011 SciRes. AM 261and 2222=,2=0,=22,22=.2kkEBDk kAk  (27) Using (26) and (25) in (10), we obtain 2221=0,222kkyxk and 2221=0,222kkyxk respectively. Combining this Equations with (11), we obtain two exact solutions to Equation (10) which was obtained in case m = 1, i.e. 2221 12=2tanh 2,2ukkik kkc   22 22 12=2tanh 2.2ukkik kkc   where 1c is an arbitrary constant. Therefore, the exact solutions to (10) can be written as  22 21 12,,=2tanh 22,2uxytkkikxyktkkc   22 22 12,,=2tanh 22.2uxytkk ikxyktkkc   Then the exact solutions for system (1) are: 22 221 122 21 12= 22,tanh 22=2tanh22 .2ikxyt lRk ikxyktkkcQe kk ikxyktkkc   (28) and 22 222 122 22 12=22,tanh22=2tanh 22.2ikxytlRk ikxyktkkcQekkikx yktkkc    (29) Case 2: In this case suppose that deg =0hx and =hxA, then from (21) we find 1=,ax AxB where B is an arbitrar y integration constant. From (22) we find  242201=,22 2AaxxkxABxDk  where D is an arbitrary integration constant. Substi-tuting 0ax, 1ax and hx in (23) and setting all the coefficients of powers x to be zero, we obtain a sys-tem of nonlinear algebraic equations and by solving it, we obtain =0, =0.AB (30) Using (28) in (10), we obtain 22241=0.22ykx xk  Combining this equations with (9), we obtain the exact solutions to Equation (10) as follows: 2212212223222242 2=,4ckk kckk kkkeuee   221221222422 2242 2=,14ckk kckk kkkeue  where 1c is an arbitrary constant. Then the exact solu-tions to (10) can be written as: D. ROSTAMY ET AL. Copyright © 2011 SciRes. AM 262 22122122( 2)232222 242 2,, =,4ckkxyktkxykt kckkkkeuxyt ee    22122122( 2)2422 22242 2,, =,14ckkxyktkckkxyktkkkeuxyte     Then solutions of system (1) are 22122122122122 2(2)23222222222232222 232=,442 2=.4ckkxyktkxykt kckkckkxyktkikxytlxykt kckkkeReekkeQee       (31) and 22122122122122 2224222 2222222422 22232=,1442 2=.14ckkxyktkckkxyktkckkxyktkikxytlckkxyktkkeRekkeQe         (32) 4. Conclusions We described this method for finding some new exact solutions for the Maccari’s system. We have obtained four exact solutions to the Maccari’s system. The solu-tions obtained are expressed by the hyperbolic and ex-ponential functions. 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