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ISSN 0976-5727 (Print) ISSN 2319-8133 (Online) Abbr:J.Comp.&Math.Sci. 2014, Vol.5(3): Pg.296-304
Application of exp (−Φ(ξ )) -Expansion Method to Find Exact Solutions of Korteweg-de Vries and Boussinesq Equations Md. Abdus Salam1, M. M. Ali2 and Md. Abdul Aziz3 1,2
Department of Mathematics, Mawlana Bhashani Science and Technology University, Tangail, BANGLADESH. 3 Department of Electronics and Telecommunication Engineering, Prime University, Mirpur-1, BANGLADESH. (Received on: June 3, 2014) ABSTRACT In this work, exp (−Φ(ξ )) -expansion method is used to find generalized traveling wave solutions of nonlinear Korteweg-de Vries and Boussinesq equations. The validity of this method is checked by finding the solutions of these equations. Using mathematical software Maple-13 as a helping tool, some exact travelling wave solutions and their respective three-dimensional plots are drawn. Keywords: exp (−Φ(ξ )) -expansion method, traveling wave solutions, Korteweg-de Vries and Boussinesq equations.
1. INTRODUCTION Many problems of solid state physics, fluid mechanics, plasma physics, population dynamics, chemical kinetics, nonlinear optics etc. are frequently described by the nonlinear evolution equations. The basic strategy one may adopt to predict, control and quantify the underlying features of a system under investigation is to model the system in terms of some mathematical equations, which are generally nonlinear,
and then find exact analytic solutions of such model equations using a suitable method. In the last few decades, considerable efforts have been made to obtain exact analytical solutions of such nonlinear equations and a number of powerful and efficient methods have been developed for obtaining explicit traveling wave solutions. Some of these approaches are the tanhmethod1, (G ′ / G ) -expansion method2, the
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exp-function method3-4, modified simple equation method5-6, the homogeneous balance method7, generalized mapping method8, and so on. Recently, a new powerful technique called exp(−Φ(ξ )) -expansion method has been developed for a reliable treatment of nonlinear wave equations. This method is straightforward, concise and capable of producing new applications. The organization of the paper is as follows: In Section 2, a brief account of the exp(−Φ(ξ )) -expansion method is given. In Section 3, we will apply these methods to find exact traveling wave solutions of the equations. In Section 4, we sketch some graphs for various cases. Finally concluding remarks are presented in Section 5. 2. DESCRIPTION OF THE GENERALIZED EXP ( −Φ(ξ )) EXPANSION METHOD Suppose that a nonlinear equation, say in one or two independent variables x and t , is given by
P(u, ut , u x , utt , u xt , u xx ,.........................) = 0, (2.1) where u = u ( x, t ) is an unknown function, P is a polynomial in u = u ( x, t ) and its various partial derivatives, in which the highest order derivatives and nonlinear terms are involved. The main steps of the exp(−Φ(ξ )) -expansion method are given below: Step-1:
The
traveling
wave
reducing Eq.(2.1) to an ODE for u = u (ξ ) in the form P(u, ωu ′, u ′, ω 2 u ′′, ωu ′′, u ′′,.........................) = 0 (2.2)
Step-2: We consider a solution of Eq.(2.2) of the form m
u (ξ ) = ∑ ai (exp(−Φ(ξ ))) i ,
where α i are constants, the positive integer m can be determined by considering the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in Eq. (2.2), and Φ = Φ(ξ ) satisfies the equation:
Φ′(ξ ) = exp( − Φ (ξ )) + µ exp( Φ(ξ )) + λ (2.4) Eq. (2.4) gives the following solutions: When λ 2 − 4 µ > 0 , µ ≠ 0 2 − λ 2 − 4 µ tanh λ − 4 µ (ξ + E ) − λ 2 Φ (ξ ) = ln 2µ
2 − λ2 − 4 µ coth λ − 4 µ (ξ + E ) − λ 2 Φ (ξ ) = ln 2µ
variable
u(x, t) = u(ξ ), where ξ = x + ω t , permits us
(2.3)
i =0
When λ 2 − 4 µ < 0 , µ ≠ 0
Md. Abdus Salam, et al., J. Comp. & Math. Sci. Vol.5 (3), 296-304 (2014) 2 4 µ − λ2 tan 4 µ − λ (ξ + E ) − λ 2 Φ (ξ ) = ln 2µ 2 4 µ − λ2 cot 4 µ − λ (ξ + E ) − λ 2 Φ (ξ ) = ln 2µ
298
we will illustrate the validity of the proposed method by applying it to solve several nonlinear evolution equations. 3. APPLICATION OF THE EXP ( −Φ(ξ )) -EXPANSION METHOD 3.1 Korteweg-de Vries equation: We know the Korteweg-de Vries equation is
ut + u xxx − 6uu x = 0
(3.1)
2
When λ − 4 µ > 0 , µ = 0 , λ ≠ 0 λ . Φ (ξ ) = − ln exp (λ (ξ + E ) ) − 1
When λ 2 − 4 µ = 0 , µ ≠ 0 , λ ≠ 0 2[ λ (ξ + E ) + 2 ] . Φ (ξ ) = ln − λ2 (ξ + E )
When λ 2 − 4 µ = 0 , µ = 0 , λ = 0 Φ (ξ ) = ln( ξ + E ) . Step-3: We substitute Eq. (2.3) into Eq.(2.2) and use Eq.(2.4) and then we account the function exp( − Φ (ξ )) . As a result of this substitution, we get a polynomial of exp( − Φ (ξ )) . We equate all the coefficients of same power of exp( − Φ (ξ )) to zero. This procedure yields a system of algebraic equations whichever can be solved to find
am , am −1 ,..........a0 .
Suppose the traveling-wave transformation is given by
u( x, t ) = u(ξ ) , ξ = x + ω t
(3.2)
where ω is a constant that to be determined later. Using Eq.(3.2), Eq.(3.1) is converted into the following ODE:
ω u ′ + u ′′′ − 6uu ′ = 0
(3.3)
Integrating Eq.(3.3), we get
ω u + u ′′ − 3u 2 + C = 0,
(3.4)
where C is an integrating constant. We suppose that Eq.(3.4) has the formal solution m
u (ξ ) = ∑ ai (exp( −Φ(ξ )))i
(3.5)
i =0
Step-4: Solving the algebraic equations system in Step-3, and by using the solutions of Eq. (2.4), we can construct the travelling wave solutions of the nonlinear evolution equation (2.1). In the subsequent sections,
where ai are constants, and Φ = Φ(ξ ) satisfies the equation: Φ ′(ξ ) = exp( − Φ (ξ )) + µ exp( Φ (ξ )) + λ
(3.6)
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Balancing the order of u ′′ and u 2 in Eq.(3.4), we have m = 2 . So Eq.(3.5) can be rewritten as u (ξ ) = a2 (exp( −Φ(ξ ))) 2 + a1 exp( −Φ(ξ )) + a0 , a2 ≠ 0
(3.7) to be
where a2 , a1 , a0 are constants determined later. Substituting Eq.(3.7) into Eq.(3.4) and collecting all the terms with the same power of exp( − Φ (ξ )) together, equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows:
e −4 Φ (ξ ) : 6a2 − 3a22 = 0 e −3 Φ ( ξ ) : e −2Φ (ξ ) :
− 6a1a2 + 10 a2 λ + 2a1 = 0
− 6a0 a2 − 3a12 + 3a1λ + 4a2 λ2 − 40a2 µ + ωa2 + 8a2 µ = 0
u3,1 (ξ ) =
1 2 λ + 8µ + ω − 6
u3,2 (ξ ) =
1 2 λ + 8µ + ω − 6
(
(
)
)
e − Φ (ξ ) :
ωa1 − 6a0 a1 + 6a2 µλ + a1λ2 + 2a1µ = 0
e0 : ω a0 + a1µλ + 2a2 µ 2 − 3a02 + C = 0 Solving the algebraic equations above, yields: 1 6 1 C = 12 a0 =
8 µ + λ2 + ω , a = 2λ , a = 2, 1 2 λ 4 − ω 2 + 16 µ 2 − 8 λ 2 µ
(3.8)
Substituting Eq.(3.8) into Eq.(3.7), we get u(ξ ) =
1 8µ + λ2 + ω + 2λ exp(−Φ(ξ )) + 2 exp(−2Φ(ξ )) , 6
(
)
where ξ = x + ω t
(3.9)
When λ 2 − 4 µ > 0 , µ ≠ 0
4µλ λ2 − 4µ λ2 − 4µ tanh (ξ + E) + λ 2
4µλ λ2 − 4µ λ − 4µ coth (ξ + E) + λ 2
+
8µ 2 2
+
λ2 − 4µ 2 (ξ + E) + λ λ − 4µ tanh 2 2 8µ
+
8µ 2
+
4µ − λ2 2 (ξ + E) − λ 4µ − λ tan 2 2 8µ
2
2
λ2 − 4µ 2 (ξ + E) + λ λ − 4µ coth 2
When λ 2 − 4 µ < 0 , µ ≠ 0 u3,3 (ξ ) =
1 2 λ + 8µ + ω + 6
u3,4 (ξ ) =
1 2 λ + 8µ + ω + 6
(
(
)
)
4µλ 4µ − λ2 4µ − λ2 tan (ξ + E) − λ 2
4µλ 4µ − λ2 4µ − λ cot (ξ + E) − λ 2 2
When λ2 − 4 µ > 0, µ = 0, λ ≠ 0
2
4µ − λ2 2 (ξ + E ) − λ 4µ − λ cot 2
2
Md. Abdus Salam, et al., J. Comp. & Math. Sci. Vol.5 (3), 296-304 (2014)
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1 2 2λ2eλ (ξ + E ) u3,5 (ξ ) = λ + 8µ + ω + 2 6 eλ (ξ + E ) − 1 When λ2 − 4 µ = 0, µ ≠ 0, λ ≠ 0
(
)
(
)
1 2 λ3 (ξ + E )(λξ + λE + 4) λ + 8µ + ω − 6 2(λξ + λE + 2) 2 When λ 2 − 4 µ = 0 , µ = 0 , λ = 0 1 2λ 2 u3,7 (ξ ) = (λ2 + 8µ + ω ) + + 6 ξ + E (ξ + E )2 u3,6 (ξ ) =
(
)
3.2 SOLUTION OF BOUSSINESQ EQUATION The Boussinesq equation is
u tt − u xx − u xxxx − 3( u 2 ) xx = 0
(3.10)
The transformation Eq.(3.2) reduces the above equation to an ODE
(ω 2 − 1) u ′′ − u ′′′′ − 3 ( u 2 )′′ = 0
(3.11)
Balancing the order of u ′′′′ and (u 2 )′′ in Eq.(3.11), we have m = 2 , hence we substitute Eq.(3.7) into eq.(3.11) and collect all the terms with the same power of exp( − Φ (ξ )) together, equate each coefficient to zero, yields a set of simultaneous algebraic equations as follows:
e −6 Φ (ξ ) : − 120 a 2 − 60 a 22 = 0 e −5Φ (ξ ) : − 336a2λ − 108a22λ − 24a1 − 72a1a2 = 0 e − 4 Φ ( ξ ) : − 96 a 22 µ − 48 a 22 λ 2 − 240 a 2 µ − 6 a1 − 18 a12 − 126 a1a 2 λ + 6ω 2 a 2 − 36 a 0 a 2 − 330 a 2 λ2 − 60 a1λ = 0
e −3Φ (ξ ) : 2ω 2 a1 − 84a22 µλ − 180a1a2 µ − 440a2 µλ − 130a2λ3 − 30a1λ2 − 12a0 a1 − 10a2λ − 54a1a2λ2 − 2a1 − 60a0 a2λ + 10ω 2 a2λ − 50a1λ2 − 40a1µ = 0 e −2 Φ (ξ ) : − 8a2 µ − 4a2 λ2 − 602a1µλ − 36a22 µ − 136a2 µ 2 − 90a1a2 µλ − 12a12λ2 + 8ω 2 a2 µ + 4ω 2 a2λ2 + 3ω 2 a1λ − 48a0 a2 µ − 16a2λ4 − 24a0 a2λ2 − 24a12 µ − 3a1λ − 232a2 µλ2 − 15a1λ3 − 18a0 a1λ = 0
e − Φ (ξ ) : − 120a2 µ 2λ + 6ω 2 a2 µλ − a1λ2 − 6a2 µλ − 36a0 a2 µλ − 6a0 a1λ2 − 12a0 a1µ + ω 2 a1λ2 − 30a2 µλ3 − a1λ4 − 22a1λ2 µ − 18a1µλ2 − 16a1µ 2 − 36a1a2 µ 2 − 2a1µ + 2ω 2 a1µ = 0
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e0 :
Md. Abdus Salam, et al., J. Comp. & Math. Sci. Vol.5 (3), 296-304 (2014)
− 8a1µ 2λ − 12a0 a2 µ 2 − 14a2 µ 2λ2 − a1λµ − 2a2 µ 2 − 6a12 µ 2 + 2ω 2 a2 µ 2 + ω 2 a1λµ − 16a2 µ 3 − a1λ3 µ − 6a0 a1λµ = 0
Solving the algebraic equations above, yields:
a0 = −
1 8µ + λ2 + 1 − ω 2 , a1 = −2λ , a2 = −2 6
(
)
(3.12)
Substituting Eq.(3.12) into Eq.(3.7), we get
1 u(ξ ) = − 8µ + λ2 + 1 − ω 2 − 2λ exp(−Φ(ξ )) − 2 exp(−2Φ(ξ )), where ξ = x + ω t 6
(
)
(3.13)
When λ 2 − 4 µ > 0 , µ ≠ 0 u3,8 (ξ ) = −
1 8µ + λ2 + 1 − ω 2 + 6
(
)
1 u3,9 (ξ ) = − 8µ + λ2 +1− ω2 + 6
(
)
4µλ λ2 − 4µ λ2 − 4µ tanh (ξ + E) + λ 2
−
4µλ λ2 − 4µ λ − 4µ coth (ξ + E) + λ 2
−
2
8µ 2 2
λ2 − 4µ 2 (ξ + E) + λ λ − 4µ tanh 2
8µ2 2
λ2 − 4µ 2 (ξ + E) + λ λ − 4µ coth 2
When λ 2 − 4 µ < 0 , µ ≠ 0 1 u3,10(ξ ) = − 8µ + λ2 +1−ω2 − 6
(
)
1 u3,11(ξ ) = − 8µ + λ2 +1−ω2 − 6
(
)
4µλ 4µ − λ2 4µ − λ tan (ξ + E) − λ 2
−
2
4µλ 4µ −λ2 4µ −λ cot (ξ + E) −λ 2 2
−
8µ2 2
4µ − λ2 2 4 µ − λ tan (ξ + E) − λ 2
8µ2
When λ 2 − 4 µ > 0 , µ = 0 , λ ≠ 0 u 3,12 (ξ ) = −
1 2 λ2 e λ (ξ + E ) 8 µ + λ2 + 1 − ω 2 − 2 6 e λ (ξ + E ) − 1
(
)
(
)
2
When λ − 4 µ = 0 , µ ≠ 0 , λ ≠ 0
u 3,13 (ξ ) = −
1 λ3 (ξ + E )(λξ + λ E + 4 ) 8 µ + λ2 + 1 − ω 2 + 6 2 (λξ + λ E + 2 ) 2
(
)
2
4µ −λ2 2 4 µ − λ cot (ξ + E) −λ 2
Md. Abdus Salam, et al., J. Comp. & Math. Sci. Vol.5 (3), 296-304 (2014)
When λ 2 − 4 µ = 0 , µ = 0 , λ = 0
u 3,14 (ξ ) = −
1 2λ 2 8µ + λ 2 + 1 − ω 2 − − 6 ξ + E (ξ + E )2
(
)
4. GRAPHICAL REPRESENTATION
Fig. 1: Profile for
u 3,1 when λ = 3, µ = 1, ω = 1, E = 1.
Fig. 2: Profile for
u 3,3 when λ = 1, µ = 1, ω = 1, E = 1
Fig. 3: Profile for
u 3,5 when λ = 2, µ = 0, ω = 1, E = 5.
Fig. 4: Profile for
u3,8 when λ = 3, µ = 1, ω = 1, E = 1.
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equations. Three-dimensional plots of some of the investigated solutions visualize the underlying dynamics that make the method more reliable and effective. The method will be applied in further works to establish more entirely new solutions for other kinds of nonlinear equations. REFERENCES
Fig. 5: Profile for
u 3,10 when λ = 1, µ = 1, ω = 1, E = 2.
Fig. 6: Profile for
u 3,13 when λ = 2, µ = 2, ω = 1, E = 2. 5. CONCLUSION These solutions may be important for explaining some practical physical phenomena which are modeled by these
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