Selçuk J. Appl. Math. Selçuk Journal of Vol. 9. No.1. pp. 3 - 10 , 2008 Applied Mathematics
Exact Traveling Wave Solutions of the (2+1) Dimensional Boussinesq Equation
Mohammed Khalfallah
Mathematics Department, Faculty of Science, South Valley University, Qena, Egypt; e-mail:m m _ kalf@ yaho o.com
Received : April 10, 2007
Summary. The repeated homogeneous balance method is used to construct exact traveling wave solutions of the (2+1) dimensional Boussinesq equation, in which the homogeneous balance method is applied to solve the Riccati equation and the reduced nonlinear ordinary differential equation, respectively. Many new exact traveling wave solutions are successfully obtained. This method is straightforward and concise, and it can be also applied to other nonlinear evo-lution equations.
Key words:2000 Mathematical Subject Classification 15A24, 15A69 1. Introduction
The investigation of the exact traveling wave solutions of nonlinear evolutions equations plays an important role in the study of nonlinear physical phenomena. For example, the wave phenomena observed in fluid dynamics, elastic media, optical fibers, etc.
In recent years Wong et al. presented a useful homogeneous balance (HB) method [1-3] for finding exact solutions of a given nonlinear partial differential equations. Fan [4] used HB method to search for Backlund transformation and similarity reduction of nonlinear partial differential equations. Also, he showed that there is a close connection among the HB method, Wiess, Tabor, Carnevale(WTC)method and Clarkson, Kruskal(CK)method.
In this paper, we use the HB method to solve the Riccati equation 0 = 2+ and the reduced nonlinear ordinary differential equation for the (2+1) dimen-sional Boussinesq equation, respectively. It makes the HB method use more extensively.
For the (2+1) dimensional Boussinesq equation [5] (1) − − − (2)− = 0
let us consider the traveling wave solutions
(2) ( ) = () = + + + where and are constants.
Substituting (2)into (1), then (1) is reduced to the following nonlinear ordinary differential equation
(3) (2− 2− 2)00− 22¡02+ 00¢− 40000= 0
The aim of this paper is to present other new traveling wave solutions of Eq.(1). Then we introduce the following auxiliary ordinary equation (Riccati equation)
(4) 02+
where are constants. The homogeneous balance method admits the use of the finite expansion in the form
(5) =
X
=0
where are constants to be determined later and satisfy the Eq.(4). It is
easy to show that = 2 if balancing 00 with 0. Therefore use the ansatz
(6) = 0+ 1 + 22
Substituting Eq.(4),and (6) into Eq.(3), and equating the coefficients of like powers of ( = 0 1 2 3 4 5) to zero yields the system of algebraic equations to 0 1 2 and −16324− 22122− 2222− 42022+ 2222= 0 −162214− 212− 4012− 122122+ 212= 0 −1362224− 8122− 822− 16022− 122222 +822= 0 −40314− 2212− 42012− 36122+ 2212= 0 −240324− 62122− 6222− 122022− 32222 +62 22= 0 (7) −24414− 242122= 0
for which, with the aid of "Mathematica", we get the following solution
(8) 0=
1 22(
2
− 2− 84) 1= 0 2= −622
(I)Let = Σ=0tanh Balancing 0with 2leads to
(9) = 0+ 1tanh
Substituting Eq.(9) into Eq.(4),we obtain the following solution of Eq.(4)
(10) = tanh = −1
tanh = −1
From Eq.(6), (8) and (10), we get the following traveling wave solutions of (2+1) dimensional Boussinesq equation (1)
(11) ( ) = 1 22(
2
− 2− 84) − 62tanh2( + + + ) Similarly, let = Σ
=0coth, then we obtain the following traveling wave
solutions of (2+1) dimensional Boussinesq equation (1) (12) ( ) = 1
22( 2
− 2− 84) − 62coth2( + + + ) (II) From [11], when = 1 the Riccati equation (4) has the following solutions
(13) = ⎧ ⎨ ⎩ −√− tanh(√−) 0 −1 = 0 √ tan(√) 0
From (6), (8) and (13), we have the following traveling wave solutions of (2+1) dimensional Boussinesq equation (1).
When 0 we have (14) ( ) = 1 22( 2 − 2− 84) + 62 tanh2(√−( + + + )) When = 0 we have (15) ( ) = 1 22( 2 − 2− 84) − 6 2 ( + + + )2 When 0 we have (16) ( ) = 1 22( 2 − 2− 84) − 62 tan2(√−( + + + )) Solutions (11), (12), (14), and (16) obtained by this method is presented in [12]. (III) We suppose that the Riccati equation (4) have the following solutions of the form (17) = 0+ X =1 (+ −1)
with = 1 cosh + = sinh cosh + which satisfy 0() = −()() 02() − () 2() = 1 − 2() + (2− 1)2() Balancing 0 with 2 leads to
(18) = 0+ 1 + 1
Substituting Eq.(18) into (4), collecting the coefficient of the same power of
( = 0 1 2; = 0 1) and setting each of the obtained coefficients to zero yield the following set of algebraic equations
21+ (2− 1)12+ (2− 1)1= 0 211+ 1= 0 201− 212− 1= 0 201= 0 (19) 2 0+ 12+ = 0
which have solutions
(20) 0= 0 1= ± r (2− 1) 42 1= − 1 2 where 4 = −1 From Eqs.(17-20), we have
(21) =−1
2(
sinh ∓p(2− 1)
cosh + ) From Eqs.(6), (8) and (21), we obtain
(22) ( ) = 1 2( 1 2( 2 − 2− 84) − 32(± √ 2− 1 − sinh() + cosh() ) 2) where = + + +
(IV) We take in the Riccati equation (4) being of the form
(23) = 1() +
4()
where
(24) = 2+
where 1 2and 3 are constants to be determined.
Substituting (23), (24) into (4), we have
(25) 2(1+2)0212+ (1− 24)1 + 04− 24− = 0
Setting 1+ 2= 21 we get 1= 2 if we assume that 4= 21 and = −
2 1
4
then Eq.(25) becomes
(26) 202= 0
By solving Eq.(26), we have
(27) = −1 = − 1 1+ 3 Substituting (27) and 4= 21 into (23), we have
(28) = − 1 1 (1+ 3) +1 2 If 3= 1 in (28), we get (29) = −1 2tanh( 1 21) If 3= −1 in (28), we get (30) = −1 2coth( 1 21)
From (6), (8) and (28), we obtain the following traveling wave solutions of (2+1) dimensional Boussinesq equation (1)
(31) ( ) = 1 2( 1 2( 2 − 2− 84) − 3212( 21(+++)− 1 1(+++)+ 3 )2) When 3= 1 we have from (29)
(32) ( ) = 1 2( 1 2( 2 − 2− 84) − 3212tanh 2 (1 2( + + + ))) Clearly, (11) is the special case of (32) with 1 = 2 When 3 = −1 we have
from (30) (33) ( ) = 1 2( 1 2( 2 − 2− 84) − 3212coth 2 (1 2( + + + ))) Clearly, (12) is the special case of (33) with 1= 2
(V) We suppose that the Riccati equation (4) have the following solutions of the form
= 0+
X
=1
where = sinh or = cosh It is easy to find that = 1 by balancing 0 and 2. So we choose
(34) = 0+ 1sinh + 1cosh
when = sinh We substitute (34) and = sinh into (4) and set the coefficient of sinh cosh ( = 0 1 2; = 0 1) to zero. A set of algebraic equations is obtained as follows
20+ 12+ = 0 201= 0
21+ 12= 1
201= 0
(35) 211= 1
for which, we have the following solutions
(36) 0= 0 1= 0 1= 1 where =−1 and (37) 0= 0 1= ± 1 2 1= 1 2 where = −41
When = sinh we have
(38) sinh = − cosh = − coth From (35)-(38), we obtain (39) = − where = −1 and (40) =coth ± 2 where = −1 4.
Clearly, (39) is the special case of (31)with 1= 2.
From (6), (8), (39) and (40), we get the exact traveling wave solutions of (2+1) dimensional Boussinesq equation (1) in the following form
(41) ( ) = 1 22(
2
which is identical with (12). (42) ( ) = 1 2( 1 2( 2 − 2− 84) −3 2 2 (() ± ()) 2)
Where = + + + These solutions of (1) are solitary wave solutions. They are linear combinations of kink solitary and bell solitary wave solutions. Similarly, when = cosh we obtain the following exact traveling wave solutions of (2+1) dimensional Boussinesq equation equation (1) in the following form (43) ( ) = 1 22( 2 − 2− 84) − 62cot2() (44) ( ) = 1 2( 1 2( 2 − 2− 84) −3 2 2 (cot() ± csc()) 2) where = + + +
In this paper new solitons and periodic solutions were formally derived. These solutions may be helpful to describe waves features in plasma physics. More-over, the obtained results in this work clearly demonstrate the reliability of the repeated homogeneous balance method.
In summary we have used the HB balance method to obtain many traveling wave solutions of (2+1) dimensional Boussinesq equation.
We now summarize the key steps as follows Step1: For a given nonlinear evolution equation (45) ( ) = 0
we consider its traveling wave solutions ( ) = () = + + + then Eq.(44) is reduced to a nonlinear ordinary differential equation
(46) ( 0 00 000 ) = 0 where a prime denotes
Step2: For a given ansatz equation (for example, the ansatz equation is 02+ in this paper), the form of is decided and the HB method is used on Eq.(46) to find the coefficients of
Step3: The HB method is used to solve the ansatz equation.
Step4: Finally, the traveling wave solutions of Eq.(45) are obtained by combining step2 and 3.
From the above procedure, it is easy to find that the HB method is more effective and simple than other methods and a lot of solutions can also be applied to other nonlinear evolution equations.
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