Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.2. pp. 41-53 , 2010 Applied Mathematics
Exact Travelling Solutions for the Generalized (1+1) Dimensional and the Generalized (2+1) Dimensional Ito Equations by0 -Expansion Method
M. Abdollahzadeh1, D. D. Ganji1∗, A. G. Davodi12, M. Barzegar13
1Babol University of Technology, Department of Mechanical Engineering, Babol, Iran
e-mail: ∗ddg_davood@yahoo.com
2Shahrood University of Technology, Department of Civil Engineering, Shahrood, Iran 3Iran’s University of Science and Technology, Department of Mechanical Engineering,
Tehran, Iran
Received Date: August 23, 2009 Accepted Date: October 19, 2010
Abstract. In this paper, we establish exact solutions for some nonlinear par-tial differenpar-tial integral equations (PDIE). The ³0´-expansion method was used to construct travelling wave solutions of the generalized (1+1) dimensional and the generalized (2+1) dimensional Ito equations. In this method we take the advantage of general solutions of second order linear ordinary differential equation (LODE) to solve effectively many nonlinear evolution equations. The ³
0 ´
-expansion method presents a wider applicability for handling nonlinear wave equations.
Key words: The 0-expansion method; Traveling wave solutions; the gener-alized (1+1) dimensional and the genergener-alized (2+1) dimensional Ito equations. 2000 Mathematics Subject Classification. 35G20.
1. Introduction
The investigation of travelling wave solutions for nonlinear partial differential equations plays an important role in the study of nonlinear physical phenomena. Nonlinear wave phenomena appear in various scientific and engineering fields, such as: solid state physics [1], fluid mechanics [2], chemical kinetics [3], plasma physics, population models [4], nonlinear optics .Analytical exact solutions to nonlinear partial differential equation play an important role in nonlinear sci-ence, especially they may provide us much physical information and more insight into the physical aspects of the problem and may lead to further applications.
In recent years, a variety of powerful methods, such as inverse scattering method [5, 6], bilinear transformation [7], Bäcklund and Darboux transformation [7-11], tanh-sech method [12-14], extended tanh method [15], Exp-function method[16-19], the sine-cosine method [20-22], the Jacobi elliptic function method [23-25], F-expansion method [26,27], Li group analysis[28], He’s Variational iteration method[29], He’s homotopy perturbation method[30-32] , homogeneous balance method [33,34] were used for obtaining explicit travelling and solitary wave so-lutions of nonlinear evoso-lutions equations .
The pioneer work Wang et al. [35] introduced the 0-expansion method for a reliable treatment of the nonlinear wave equations. The useful 0-expansion method is widely used by many authors in references [36-40].
In this paper, we will apply the 0-expansion method to obtain the exact travelling wave solution of the generalized (1+1) dimensional and generalized (2+1) dimensional Ito equations
(1.1) + + 3 (2+ ) + 3 Z 0 −∞ = 0 (1.2) + + 3 (2+ ) + 3 Z 0 −∞ + + = 0
which was recently derived by Wazwaz, by generalizing the bilinear forms of the standard KdV equation. Wazwaz obtained Multiple-soliton solutions of these equations using tanh method and Hirota bilinear method [41].
The rest of the paper is organized as follows. In Section 2, we present a method-ology of the generalized³0
´
-expansion method. In Section 3 and 4, we apply our method to the generalized (1+1) dimensional and the generalized (2+1) dimensional Ito equations. In Section 5, some conclusions are given.
2- The³0 ´
-Expansion Method
Wang has summarized the main steps for using ³0´-expansion method, as following:
1-We introduce the wave variables = − into the PDE, we get
(2.1) ( ) = 0
where ( ) is travelling wave solution. This enables us to use the following changes:
(2.2) ( ) = () (2.3) = − 2 2 = 2 2 2 = 2 2 = 2 2
And so on for the other derivates. Using (2.3) and (2.2), the nonlinear PDE (2.1) changes to a nonlinear ODE:
(2.4) Ψ¡ −0 0 200 00 −00 000 ¢= 0
If all terms of the resulting ODE contain derivatives in , then by integrating this equation, by considering the constant of integration to be zero, we obtain a simplified ODE.
2-Suppose that the solution of ODE (2.2) can be expressed by a polynomial in ³ 0 ´ as follows: (2.5) () = X =0 µ 0 ¶ = µ 0 ¶ + · · ·
Where = () satisfies the second order LODE in the form:
(2.6) 00+ 0+ = 0
where 0 = () , 00 =
2()
2 . ,and are constants to be deter-mined later, 6= 0, the unwritten part in (2.5) is also a polynomial
³ 0
´
, but the degree of which is generally equal to or less than − 1, the positive integer can be determined by considering the homogeneous balance the highest order derivatives and highest order nonlinear terms appearing in ODE (2.4).
3- By substituting Eq. (2.5) into Eq. (2.4) and using second order LODE (2.6), collecting all terms with the same order of³0´ together, the left-hand side of Eq. (2.4) is converted into another polynomial in³0´. Equating each coefficient of this polynomial to zero, yield a set of algebraic equations involving and .
4-Assuming that the constants and can be obtained by solving the algebraic equations in Step 3, since the general solutions of the second order LODE (2.6) have been well known for us, then substituting and general solutions of Eq. (2.6) into Eq.(2.5) we have more travelling wave solutions of the nonlinear evolution equation (1.1) and (1.2).
3. The (1+1)-Dimensional Ito Equation
In this section, we will demonstrate the proposed method on (1+1) dimensional Ito equation [41]: (3.1) + + 3 (2+ ) + 3 Z 0 −∞ = 0
By using the potential = , Eq. (3.1) can be reduced to:
(3.2) + + 6+ 3+ 3= 0
As described in Section 2 we perform a traveling wave reduction ( ) = () with the argument = − , generating the reduced nonlinear ODE:
(3.3) 2000− (5)− 6 (00)2− 60000= 0 Or equivalently:
(3.4) 2000(5)− 3¡(02¢¨= 0 Integrating Eq. (3.4) twice, we will obtain:
(3.5) 20− 000− 3(02= 0
Setting = 0 as new depend variable, a second order nonlinear ODE results:
(3.6) 2 − 002= 0
Suppose that the solution of ODE (3.6) can be expressed by
(3.7) () = µ 0 ¶ +
(3.8) 00+ 0+ = 0 By using Eq. (3.7) and Eq. (3.8) it is easily derived that
2() = 2 µ 0 ¶2 + · · · (3.9) 0() = − µ 0 ¶+1 + · · · 00() = ( + 1) µ0 ¶+2 + · · ·
Considering the homogeneous balance between 2 and 00 in Eq. (3.6), based on Eq. (3.9), we required that 2 = + 2 ⇒ = 2 so we can write Eq. (3.7) as: (3.10) () = 0+ 1 µ 0 ¶ + 2 µ 0 ¶2 26= 0 By using Eq. (3.8) and Eq. (3.10), it is derived that
(3.11) 2() = 2 0+ 201 µ 0 ¶ + (202+ 21) µ 0 ¶2 + 212 µ 0 ¶3 +22 µ 0 ¶4 and (3.12) 00() = 222+ 1 + (21 + 12+ 62) µ 0 ¶ +(422+ 31 + 82) µ 0 ¶2 + (21+ 102) µ 0 ¶3 +62 µ 0 ¶4
By substituting Eqs. (3.10)— (3.12) into Eq. (3.6) and collecting all terms with the same power of³0´ together, the left-hand side of Eq. (3.6) is converted into another polynomial in³0´. Equating each coefficient of this polynomial
to zero, yields a set of simultaneous algebraic equations for 0 1 2 and As follows: (3.13a) µ 0 ¶0 : 20− 222− 320− 1 = 0 (3.13b) µ 0 ¶1 : 21− 21 − 12− 62 − 610= 0 (3.13c) µ 0 ¶2 : −82 − 321− 31 − 620+ 22− 422= 0 (3.13d) µ 0 ¶3 : −21− 102 − 612= 0 (3.13e) µ 0 ¶4 : −322− 62= 0 Solving the algebraic equations above, yields:
(3.14) 0= −2 1= −2 2= −2 = 2− 4 or (3.15) 0= − 2 3 − 1 3 2 1= −2 2= −2 = 4 − 2 where and are arbitrary constants.
By using Eqs. (3.14) and (3.15), expression (3.7) can be written as
(3.16) 1() = −2 − 2 µ 0 ¶ − 2 µ 0 ¶2 where = − , or (3.17) 2() = − 2 3 − 1 3 2 − 2 µ 0 ¶ − 2 µ 0 ¶2 where = − (4 − 2).
Substituting general solution of Eq. (2.6) into Eqs. (3.16) and (3.17) we have the solutions of Eq. (3.6) as follows:
When 2− 4 0, (3.18) 1() = −2 + 2 2 − 1 2( 2 − 4) µ 1sinh12 √ 2−4+ 2cosh12 √ 2−4 1cosh12 √ 2 −4+2sinh12 √ 2 −4 ¶2 where = − , or (3.19) 2() = − 2 3+ 2 6 − 1 2( 2 −4) µ 1sinh12 √ 2−4+ 2cosh12 √ 2−4 1cosh12√2−4+2sinh12√2−4 ¶2 where = − (4 − 2). When 2− 4 0, (3.20) 3() = −2 + 2 2 − 1 2(4 − 2 ) µ −1sin12 √ 4−2+ 2cos12 √ 4−2 1cos12 √ 4−2+ 2sin12 √ 4−2 ¶2 where = − , or (3.21) 4() = − 2 3 + 2 6 − 1 2(4 − 2 ) µ −1sin12 √ 4−2+ 2cos12 √ 4−2 1cos12 √ 4−2+ 2sin12 √ 4−2 ¶2 where = − (4 − 2).
Recalling that = 0 and using the potential transformation ( ) = ( ) and the travelling wave reduction ( ) = () and ( ) = (), we have: When 2− 4 0, (3.22) 1( ) = 2 − 2 Ã 1sinh( √ 2 ( − )) + 2cosh( √ 2 ( − )) 1cosh( √ 2 ( − )) + 2sinh( √ 2 ( − )) !2 where = 2− 4, or (3.23) 2( ) = 6 − 1 2 Ã 1sinh( √ 2 ( + )) + 2cosh( √ 2 ( + )) 1cosh( √ 2 ( + )) + 2sinh( √ 2 ( + )) !2
where = 2− 4. And when 2− 4 0,
(3.24) 3( ) = − 1 2 − 1 2 ³ −1sin(12 √ (+))+2cos(12 √ (+)) 1cos(12 √ (+))+2sin(12 √ (+)) ´2
where = 4 − 2, or (3.25) 4( ) = − 1 6 − 1 2 ³ −1sin(12 √ (−))+2cos(12 √ (−)) 1cos(12 √ (−))+2sin(12 √ (−)) ´2 where = 4 − 2.
1 2are arbitrary constants .If 1and 2are taken as special values, the various results can be rediscovered. For instance, if 1 6= 0 and 2 = 0, then 1( ) can be written as:
(3.26) 1( ) = 2 − 2 Ã 1sinh( √ 2 ( − )) 1cosh( √ 2 ( − )) !2 = 2sec 2( √ 2 ( − )) Which is equal to the exact solution of Eq (3.1) obtained by Wazwaz [41]. 4. The (2+1)-Dimensional Ito Equation
In this section we apply the³0´-expansion method to the (2+1)-dimensional Ito equation [41]: (4.1) + + 3(2+ ) + 3 Z −∞ 0+ + = 0
That can be reduced to:
(4.2) + + 6+ 3+ 3+ + = 0 Upon using the potential = , where and are arbitrary constants. We seek travelling wave solutions for Eq. (4.3) in the form:
(4.3) ( ) = () = + − Using Eq. (4.3), Eq. (4.2) is carried to a nonlinear ODE:
(4.4) 2000(5)− 6(002− 60000− 000− 000= 0 or equivalently
(4.5) ( − ( + ))000(5)− 3¡(02¢¨= 0 Integrating Eq. (4.5) twice, we will obtain:
Introducing = 0 as a new depend variable, we obtain the reduced nonlinear ODE:
(4.7) ( − ( + )) − 002= 0
Considering the homogeneous balance between 2 and 00 in Eq. (4.7), we required that 2 = + 2 ⇒ = 2 so we suppose that the solution of Eq. (4.7) is of the form (4.8) () = 0+ 1 µ 0 ¶ + 2 µ 0 ¶2 26= 0
Where = () satisfies the second order LODE (2.6) and 0 1 2 and gare constants to be determined later.
By using Eq. (4.8) and Eq. (2.6), it is derived that
(4.9) 2() = 20+201 ³ 0 ´ +(202+21) ³ 0 ´2 +212 ³ 0 ´3 +22³0´ 4 and (4.10) 00() = 2 22+ 1 + (21 + 12+ 62) µ 0 ¶ +(422+ 31 + 82) µ 0 ¶2 + (21+ 102) µ 0 ¶3 +62 ³ 0 ´4
By substituting Eqs. (4.8)— (4.10) into Eq. (4.7) and collecting all terms with the same power of³0
´
together, the left-hand side of Eq. (4.7) is converted into another polynomial in³0´. Equating each coefficient of this polynomial to zero, yields a set of simultaneous algebraic equations for 0 1 2 and As follows: (4.11a) µ0 ¶0 : 20− 320− 0− 222− 1 − 0= 0 (4.11b) µ 0 ¶1 : −62 − 610− 21 − 12+ 21− 1− 1= 0 (4.11c) µ 0 ¶2 : 22−82−620−2−422−31−2−321= 0
(4.11d) µ 0 ¶3 : −612− 21− 102 = 0 (4.11e) µ 0 ¶4 : −322− 62= 0 Solving the algebraic equations above, yields:
(4.12) 0= −2 1= −2 2= −2 = 2− 4 + + or (4.13) 0= − 2 3 − 1 3 2 1= −2 2= −2 = 4 − 2+ + where and are arbitrary constants.
By using Eqs. (4.12) and (4.13), expression (4.8) can be written as
(4.14) () = −2 − 2 µ 0 ¶ − 2 µ 0 ¶2 where = + − (2− 4 + + ), or (4.15) () = −2 3 − 1 3 2 − 2 µ 0 ¶ − 2 µ 0 ¶2 where = + − (4 − 2+ + ).
Substituting general solution of Eq. (2.6) into Eqs. (4.14) and (4.15), we have solutions of Eq (4.7) as follows: (4.16) 1() = −2 + 2 2 − 1 2( 2 − 4) µ 1sinh12 √ 2 −4+2cosh12 √ 2 −4 1cosh12 √ 2−4+ 2sinh12 √ 2−4 ¶2 Where = + − (2− 4 + + ) (4.17) 2() = − 2 3+ 2 6 − 1 2( 2 −4) µ 1sinh12√2−4+2cosh12√2−4 1cosh12 √ 2−4+ 2sinh12 √ 2−4 ¶2 where = + − (4 − 2+ + ). And when 2− 4 0,
(4.18) 3() = −2 + 2 2 − 1 2(4 − 2 ) µ −1sin12 √ 4−2+ 2cos12 √ 4−2 1cos12√4−2+2sin12√4−2 ¶2 where = + − (2− 4 + + ), or (4.19) 4() = − 2 3 + 2 6 − 1 2(4 − 2 ) µ −1sin12 √ 4−2+ 2cos12 √ 4−2 1cos12 √ 4−2+ 2sin12 √ 4−2 ¶2 where = + − (4 − 2+ + ).
Recalling that = 0and using the potential transformation ( ) = ( ) and the travelling wave reduction ( ) = ()and( ) = () , we have: When 2− 4 0, (4.20) 1( ) = 2 −2 µ 1sinh( √ 2 (+−(++)))+2cosh( √ 2 (+−(++))) 1cosh( √ 2 (+−(++)))+2sinh( √ 2 (+−(++))) ¶2 where = 2− 4 + + , or (4.21) 2( ) = 6 −2 µ 1sinh( √ 2 (+−(+−)))+2cosh( √ 2 (+−(+−))) 1cosh( √ 2 (+−(+−)))+2sinh( √ 2 (+−(+−))) ¶2 where = 2− 4. When 2− 4 0, (4.22) 3( ) = − 1 2 −2 ³−1sin(12 √ (+−(+−)))+2cos(12 √ (+−(+−))) 1cos(12 √ (+−(+−)))+2sin(12 √ (+−(+−))) ´2 where = 4 − 2,or (4.23) 4( ) = − 1 6 −12³−1sin(12 √ (+−(++)))+2cos(12 √ (+−(++))) 1cos(12√(+−(++)))+2sin(12√(+−(++))) ´2 where = 4 − 2.
1 2are arbitrary constants .If 1and 2are taken as special values, the various results can be rediscovered. For instance, if 16= 0, 2= 0 and we take + + = then 1( ) can be written as:
(4.24) 1( ) = − ( + ) 2 − − ( + ) 2 µ 1sinh( √ −(+) 2 (+−)) 1cosh( √ −(+) 2 (+−)) ¶2 = − ( + ) 2 sec 2( p − ( + ) 2 ( + − ))
Which is equal to the exact solution of Eq. (4.1) obtained by Wazwaz [41]. 5- Conclusions
The³0´-expansion method was successfully used to establish travelling wave solutions of the generalized (1+1) dimensional and the generalized (2+1) dimen-sional Ito equations. Many well known nonlinear wave equations were handled by this method. The performance of this method is reliable and effective and gives more solutions. This method has more advantages .it is direct and concise .It is elementary that the general solutions of the second order LODE have been well known for the researchers and effective that it can be used in many other nonlinear evolution equations. The availability of computer systems like Math-ematica or Maple facilitates the tedious algebraic calculations. The method which we have proposed in this letter is also a standard, direct and comput-erizable method, which allows us to solve complicated and tedious algebraic calculation.
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