Selçuk J. Appl. Math. Selçuk Journal of Vol. 10. No. 1. pp. 121-133, 2009 Applied Mathematics
An Application of Exp—Function Method to the Generalized Burger’s— Huxley Equation
M.M. Alipour2, D.D. Ganji1, A. G. Davodi3
1Department of Mechanical and Civil Engineering, Babol University of Technology,
Babol, Iran
e-mail: ddg_ davoo d@yaho o.com
2Department of Mechanical Engineering, K.N Toosi University of Technology, Tehran,
Iran
e-mail: m m alip our@ yaho o.com
3Department of Civil Engineering, Shahrood University of Technology, Shahrood, Iran
e-mail: a.g.davo di@ yahoo.com
Received: March 16, 2009
Abstract. In this work, we implement a relatively new analytical technique, the Exp—Function method, for solving nonlinear equations and absolutely special form of Generalized Burger’s—Huxley equation which may contain high nonlinear terms. This method can be used as an alternative to obtain analytical and approximate solutions of different types of fractional differential equations which applied in engineering mathematics. For more illustration of the efficiency and reliability of Exp method some numerical examples are presented. It is predicted that Exp—Function method can be found widely applicable in engineering. Key words: Exp—Function method; Generalized Burger’s—Huxley equation; Nonlinear Partial Differential equations.
2000 Mathematical Subject Classification: 35D10, 35D99, 35J70. 1. Introduction
Many problems in natural and engineering sciences are modeled by partial differ-ential equations. Nonlinear phenomena play important roles in applied mathe-matics, physics and also in engineering problems in which each parameter varies depending on different factors. Solving nonlinear equations may guide authors to know the described process deeply and sometimes leads them to know some facts which are not simply understood through common observations. More-over, obtaining exact solutions for these problems is a great purpose which has
been quite untouched. However, in recent years, numerical analysis [1] has con-siderably been developed to be used for nonlinear partial equations such as Gen-eralized Burger’s—Huxley equation that shows a prototype model for describing the interaction between reaction mechanisms, convection effects and diffusion transports [2]. The solitary waves and the existed property of a finite number of quantities (with physical interpretation) are conserved by the solutions. The Generalized Burger’s—Huxley equation is in the form of:
(1) +
− = (1 − )(− )
∀ 0 ≤ ≤ 1 ≥ 0 ( 0) = ()
When = 0 = 1 Eq. (1) is reduced to the Huxley equation which describes nerve pulse propagation in nerve fibers and wall motion in liquid crystals [3] and the subscripts t and x denote differentiation with respect to time and space, respectively.
In addition, in recent years, scientists have presented some new methods for solving nonlinear partial differential equations; for instance, Bäcklund transfor-mation method [4], Lie group method [5], Adomian’s decomposition method [6], inverse scattering method [7], Hirota’s bilinear method [8], homotopy analysis method [9] and He’s Homotopy perturbation method[10-11], He’s Variational iteration method[12-13] and Exp-Function method [14-15].
In this letter, we purpose to present implementation of Exp—Function method to generalized nonlinear Generalized Burger’s—Huxley equation. Having the available exact solution of special form of corresponding equations [16] would provide us to benefit an admissible comparison of the results which supports the applicability, accuracy and efficiency of the proposed methods.
2. Basic Idea of Exp-Function Method We first consider nonlinear equation form:
(2) ( ) = 0
Introduction a complete variation defines as:
(3) = ( + ) = ()
and therefore, the Eq(1) is the construction of ODE of form:
(4) ( 0 02002200 200 ) = 0 and then solution of () is form:
(5) () = P =− exp() P =− exp() =exp() + + −exp(−) exp() + + −exp(−)
where and are positive integers which are unknown to be further de-termined, and are unknown constants.
3. Application of Exp—Function Method to Generalized Burger’s— Huxley Equation
We begin with the transformation ( ) = ( ) or ( ) = 1( ), and
have: (6) = 1 1 −1 = 1 1 −1 = 1 [ 1 −1+ ( 1 − 1) 1 −22 ] (1 − )(− ) = 1(1 − )( − )
Substituting these results into Eq. (1), we have:
(7) 1 1 −1[+ − − ( 1 − 1) 2 −1] = 1 (1 − )( − )
Eq. (7) is simplified as follows:
(8) + − − (
1 − 1)
2
−1= (1 − )( − )
Introducing a complex variation defined as Eq.(3), Eq. (8) becomes an ordi-nary differential equation which reforms to:
(9) 0+ 02(00+ (1
− 1)
02−1) − (1 − )( − ) = 0
In order to determine values of and , we balance the linear term of the highest order 00with the highest order nonlinear term 3 in Eq. (9), we have:
(10) 00=1exp [( + 3) ] + 2exp [4] +
(11) 3= 3exp [(3 + ) ] + 4exp [4] +
where are determined coefficients only for simplicity. Balancing highest order
of Exp—function in Eq. (10) and (11), we have:
(12) + 3 = 3 +
which leads to the result:
(13) =
Similarly to determine values of and , we balance the linear term of lowest order in Eq. (9) (14) 00= + 1exp [− ( + 3) ] + 2exp [−4] and (15) 3= + 3exp [− (3 + ) ] + 4exp[−4]
where are determined coefficients only for simplicity. Balancing lowest order
of Exp—Function into Eqs. (14) and (15), we have:
(16) − ( + 3) = − (3 + )
This leads to the result:
(17) =
Case 1: = = 1, = = 1
According to this case, Eq. (5) reduces to:
(18) () = 1exp() + 0+ −1exp(−) exp() + 0+ −1exp(−)
Substituting Eq. (18) into Eq. (9), and by the help of MAPLE, we have:
(19) 1 {4 4+ 33+ 22+ 1+ 0+ −1− +−2−2+ −3−3+ −4−4)} = 0
and are coefficients of exp(). All mentioned relations and equations which
are unknown in this case are added to this paper in appendix.
Equating the coefficients of exp() must be zero, and therefore we have:
(20) 1= 0 2= 0 3= 0 4= 0 0= 0 −1= 0 −2= 0 −3= 0 −4= 0
Solving the system Eq. (20), simultaneously yields the following sets of nontriv-ial solutions: (21) 0= 0 −1= 0 −1= −1 0= 0 1= 1 1= = −[ − p 2+ 4(1 + )] 4(1 + ) = 2[ − 2(1 + )+ (1 + − ) + p2+ 4(1 + )] = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + ) (22) 0= 0+ −1 0 −1= 0 −1= −1 0= 0 1= 1 1= 1 = −[ − p 2+ 4(1 + )] 2(1 + ) = 2[ − 2(1 + )+ ( + − 1) +p2+ 4(1 + )] = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + ) (23) 0= 0 −1= 0 −1= − 0(0− 0) 2 0= 0 1= 1 1= = −[ − p 2+ 4(1 + )] 2(1 + ) = 2[ − 2(1 + )+ (1 + − ) + p2+ 4(1 + )] = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + ) (24) 0= 0 −1= 0 −1= 0 0= 0 1= 1 1= 1 = −[ − p 2+ 4(1 + )] 2(1 + ) = 2[ − 2(1 + )+ ( + − 1) +p2+ 4(1 + )] = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + )
(25) 0= 0 −1 = 0 −1= −1 0= 0 1= 1 1= 1 = −[ − p 2+ 4(1 + )] 4(1 + ) = 2[ − 2(1 + )+ ( + − 1) +p2+ 4(1 + )] = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + ) (26) 0= 0 −1= 0(0−0) −1= 0(0−0) 0= 0 1= 1 1= 0 = −[ + p 2+ 4(1 + )] 2(1 + ) = 2[ − 2(1 + )+ ( + − 1) −p2+ 4(1 + )] = − (1 + )− ( + − 1)( +p2+ 4(1 + )) 2(1 + ) (27) 0= 0 −1 = −1 −1= −1 0= 0 1= 1 1= 0 = [ − p 2+ 4(1 + )] 4(1 + ) = 2[ 2(1 + )− (1 + − ) − p2+ 4(1 + )] = = (1 + )+ (1 + − )( −p2+ 4(1 + )) 2(1 + ) (28) 0= 0 −1 = −1 −1= −1 0= 0 1= 1 1= 0 = [ + p 2+ 4(1 + )] 4(1 + ) = 2[ − 2(1 + )+ (1 + − ) − p2+ 4(1 + )] = − (1 + )− (1 + − )( +p2+ 4(1 + )) 2(1 + )
Substituting these results into Eq. (18), and = ( + )we obtain the follow-ing generalized solitonary solutions of Eq. (8) solution
(29) 1( ) = exp[( + )] exp[( + )] + −1exp[−( + )] = −[ − p 2+ 4(1 + )] 4(1 + ) = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + )
where, −1 is arbitrary constant parameter. Substituting −1 = 1 into Eq.(29), we have the following closed form of Eq. (8) and Eq. ( 1)
(30) 1( ) = exp[( + )] exp[( + )] + exp[−( + )] = 2[1 + tanh(( + )] 1( ) = [ 2+ 2tanh(( + )] 1 = −[ − p 2+ 4(1 + )] 4(1 + ) = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + ) 2( ) = exp[( + )] + 0 exp[( + )] + 0+−10 + −1exp[−( + )] (31) 2( ) = ( exp[( + )] + 0 exp[( + )] + 0+−10 + −1exp[−( + )] )1 = −[ − p 2+ 4(1 + )] 2(1 + ) = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + ) (32) 3( ) = exp[( + )] + 0 exp[( + )] + 0−0(0−2 0)exp[−( + )] 3( ) = ( exp[( + )] + 0 exp[( + )] + 0−0(0−2 0)exp[−( + )] )1 = −[ − p 2+ 4(1 + )] 2(1 + ) = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + ) (33) 4( ) = exp[( + )] exp[( + )] + 0] 4( ) = ( exp[( + )] exp[( + )] + 0] )1 = −[ − p 2+ 4(1 + )] 2(1 + ) = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + )
(34) 5( ) = exp[( + )] exp[( + )] + −1exp[−( + )] 5( ) = ( exp[( + )] exp[( + )] + −1exp[−( + )]) 1 = −[ − p 2+ 4(1 + )] 4(1 + ) = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + )
where, −1 is arbitrary constant parameter. Substituting −1 = 1 into Eq.(34), we have the following closed form of Eq. ( 8) and Eq. ( 1)
(35) 5( ) = exp[( + )] exp[( + )] + exp[−( + )] = 1 2+ 1 2tanh(( + )) 5( ) = [ 1 2+ 1 2tanh(( + )] 1 = −[ − p 2+ 4(1 + )] 4(1 + ) = − (1 + )− ( + − 1)( −p2+ 4(1 + )) 2(1 + ) (36) 6( ) = 0+ 0(0− 0) exp[( + )] exp[( + )] + 0+ 0(0− 0) exp[−( + )] 6( ) = ( 0+ 0(0− 0) exp[( + )] exp[( + )] + 0+ 0(0− 0) exp[−( + )] )1 = −[ + p 2+ 4(1 + )] 2(1 + ) = − (1 + )− ( + − 1)( +p2+ 4(1 + )) 2(1 + ) (37) 7( ) = 0+ −1exp[−( + )] exp[( + )] + −1exp[−( + )] 7( ) = ( 0+ −1exp[−( + )] exp[( + )] + −1exp[−( + )]) 1 =[ − p 2+ 4(1 + )] 4(1 + ) = (1 + )+ (1 + − )( −p2+ 4(1 + )) 2(1 + )
where, −1 0are arbitrary constant parameters. Substituting −1= 1 0= 0
(38) 7( ) = exp[−( + )] exp[( + )] + exp[−( + )] = 2[1 + tanh(−( + )] 7( ) = [ 2+ 2tanh(−( + )] 1 =[ − p 2+ 4(1 + )] 4(1 + ) = (1 + )+ (1 + − )( −p2+ 4(1 + )) 2(1 + ) (39) 8( ) = 0+ −1exp[−( + )] exp[( + )] + −1exp[−( + )] 8( ) = ( 0+ −1exp[−( + )] exp[( + )] + −1exp[−( + )]) 1 =[ + p 2+ 4(1 + )] 4(1 + ) = − (1 + )− (1 + − )( +p2+ 4(1 + )) 2(1 + )
where, −1 0are arbitrary constant parameters. Substituting −1= 1 0= 0
into Eq.(39), we have the following closed form of Eq. ( 8) and Eq. (1)
(40) 8( ) = exp[−( + )] exp[( + )] + exp[−( + )] = 2[1 + tanh(−( + )] 8( ) = [ 2+ 2tanh(−( + )] 1 =[ + p 2+ 4(1 + )] 4(1 + ) = − (1 + )− (1 + − )( +p2+ 4(1 + )) 2(1 + ) Case2: = = 2, = = 2
As mentioned above the values of and can be freely chosen, we set = = 2 and = = 2, then the trial function, Eq. (4) becomes:
(41) () = 2exp(2) + 1exp() + 0+ −1exp(−) + −2exp(−2) 2exp(2) + 1exp() + 0+ −1exp(−) + −2exp(−2)
There are some free parameters in Eq. (41), we set 2= 1, 1= 0 −1= 0 and
(42) () = 2exp(2) + 0+ −2exp(−2) exp(2) + 0+ −1exp(−) + −2exp(−2)
By the same manipulation as illustrated above, we obtain:
(43) 0= 0 −2= 0 −2= −2 0= 0 2= 1 2= −1= 0 = −[ − p 2+ 4(1 + )] 8(1 + ) = [ − 2(1 + )+ (1 + − ) + p2+ 4(1 + )] = − 2(1 + )− (1 + − )( −p2+ 4(1 + )) 4(1 + )
where, −2is arbitrary constant parameter. Substituting Eq. (43) into (42) yields the following solution:
(44) () = exp(2) exp(2) + −2exp(−2) = −[ − p 2+ 4(1 + )] 8(1 + ) = − 2(1 + ) − (1 + − )( −p2+ 4(1 + )) 4(1 + ) Or (45) ( ) = exp[2( + )] exp[2( + )] + −2exp[−2( + )] = −[ − p 2+ 4(1 + )] 8(1 + ) = − 2(1 + ) − (1 + − )( −p2+ 4(1 + )) 4(1 + )
Substituting −2 = 1 into Eq.(45), we have the following closed form of Eq. ( 8) and Eq. (1) (46) ( ) = exp[( + )] exp[( + )] + exp[−( + )] = 2[1 + tanh(( + )] ( ) = [ 2+ 2tanh(( + )] 1 = −[ − p 2+ 4(1 + )] 4(1 + ) = − (1 + )− (1 + − )( −p2+ 4(1 + )) 2(1 + )
4. Conclusion
In this Letter, the Exp—Function method was used for finding solutions of Gen-eralized Burger’s—Huxley equation which may contain high nonlinear terms. It can be concluded that the Exp—Function method is very powerful and efficient technique in finding exact solutions for wide class of problems.
The Exp—Function method has many merits and more advantages than exact solutions. Calculations in the Exp—Function method are simple and straight-forward. The reliability of the method and reduction in the size of computa-tional domain gives this method a wider applicability. The results show that Exp—Function method is a powerful mathematical tool for solving systems of nonlinear partial differential equations having wide applications in engineering. References
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Appendix:
Case 1, Generalized Burger’s—Huxley equation: = (+ 0+ −1−)4 4= −231+ 2 21− 231+ 241= 0 3= 2210− 2310+ 22120− 210+ 120− 32210 −322 10+ 2201+ 42310− 01− 2310− 201 +3 10= 0 2= 622120− 3 2 2 1−1+ 4231−1− 42−11− 322100 +22 100+ 221−1+ 4212−1+ 2120+ 231−1 −2−11− 2120− 20− 221−1+ 22100 +220− 3212−1+ 42100− 231−1− 322100 −22 0+ 2221−1+ 22−11+ 21220− 32120− 32120 −23 1−122120= 0 1= −322010− 322010− 6210−1− 62101 +122210−1− 32210−1− 230− 42−10+ 42130+ 5210−1 −200+ 21−10− 610−1+ 4220−1+ 1200− 3−10 −2−10+ 2120−1+ 621−10− 6210−1+ 2010− 30 −2−110− 42210−1+ 421−10+ 2200− 21200+ 21−10 +32 10−1+ 221200+ 22210−1− 230+ 22−10+ 22200 +421−10− 3221−10− 32210−1+ 42−110− 3221−10= 0 0= 22120−1− 32120−1− 422−1− 3 2 2 1−1−1 −6210−10+ 2212−1− 3220−1+ 821−1−1+ 22−1+ 622 12−1− 42120−1+ 240− 3212−1+ 122120−1− 22100−1 +222 0−1− 3221−1−1− 22−1+ 42−100+ 42100−1 −23 00− 321−12 + 22020− 2 3 00+ 2212−1+ 2220−1 −32120−1+ 421−1−1− 3220−1+ 4220−1− 6210−10 −4−100− 402−1− 22−100+ 22120−1+ 4100−1 −42212−1− 420−1+ 4120−1+ 421−1−1= 0
−1= 122102−1+ 4230−1− 621−1−10− 212−10− 230−1 +42 1−1−10+ 62−10−1− 02−10+ 30−1+ 61−10−1 −42 12−10− 422−10+ 42−10−1− 2−10−1− 621−10−1 −2 −1200+ 220−10+ 21−1−10− 32−10+ 22−10 −12−10+ 3102−1− −1200+ 20−10− 502−1 −322 0−10+ 421−1−10− 3212−10− 320−12 + 22−1200 +2212−10− 621−10−1− 3202−1+ 22−12 0− 230−1 −3212−10+ 2220−10+ 42−10−1− 3220−10= 0 −2= 4213−1+ 62202−1+ 22−100−1− 2−12 −1+ 202−1 −22 02−1− 2 3 −1+ 220−1−1− 23−1− 4212−1−1− 2−120 +422−1−1+ 212−1−1− 2002−1+ 212−1−1− 32∗ 02−10 +42−100−1+ 22−120− 3202−1−1− 322−11−1+ 2202−1 +2212−1−1+ 2 ∗ 22−1−1− 3202−10− 3220−1−1 −322 −11−1− 2−13 − 22−120= 0 −3= −2 02−1−1− 322−10−1− 23−10− 2−10−1 +02−1−1− 23−10+ 202−1−1+ 2−10−1− 3−10 +222 −10−1+ 4203−1+ 22−10−1− 322−10−1= 0 −4 = −23−1−1− 23−1−1+ 22−12−1+ 24−1= 0
