ISSN 1749-3889 (print), 1749-3897 (online) International Journal of Nonlinear Science Vol.7(2009) No.2,pp.174-181
The Application of He’s Exp-function Method for MKdV and Burgers’ Equations with Variable Coefficients K. R. Raslan ∗ Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr-City, Cairo, Egypt. (Received 15 April 2008, accepted 19 June 2008)
Abstract: In this paper, the Exp-function method with the aid of symbolic computational system is used to obtain generalized solitary solutions and periodic solutions of the MKdV equation, Burgers’ equation in one and two dimensional with variable coefficients. It is shown that the Exp-function method, with the help of symbolic computation, provides a powerful mathematical tool for solving other nonlinear evolution equations arising in mathematical physics. Keywords: Exp- function method; MKdV; Burgers’ equation with variable coefficients.
1
Introduction
Explicit solutions to the mathematical modeling of physical problems are fundamental importance. There are many methods in literature to solve the non-linear equations. Among them are Backlund transformation [1,5], Darboux transformation [2], Inverse scattering method [3], Hirota’s bilinear method [4], the tanh method [6], the sine-cosine method [7,10], the homogeneous balance method [8,9], variational iteration method [11-16], and the Riccati expansion method with constant coefficients [17]. The third order KdV equation has been the focus of considerable recent studies by [19-20]. Gardner [20] developed a variational and its Hamiltonian formulation to handle this problem and also Gardner et. Al. [21] introduced various methods for explicit solutions for the KdV equation. Soliman [22] used the collocation solution of the KdV equation using septic splines as element shape function. Ma [23] applied a special solution of square HopfCole type to an ordinary equation to obtain the bounded traveling wave solution to the two-dimensional Korteweg-de Vries Burgers equation. Very recently, He and Wu [24] proposed a straightforward and concise method, called Exp-function method, to obtain generalized solitary solutions and periodic solutions, applications of the method can be found in [25-27] for solving nonlinear evolution equations arising in mathematical physics. The solution procedure of this method, with the aid of Mathematica , is of utter simplicity and this method can easily extended to other kinds of nonlinear evolution equations. The aim of this paper is used Exp-function method to find the new exact solutions of the modified Korteweg-de Vries (MKdV) equation, the Burgers’ equation in one and two dimensional with variable coefficients which are believed to be useful in the numerical studies.
2
The Exp-function method
Consider the nonlinear PDE with variable coefficients as: F (u, ut , α(t) ux , β(t) uxx , γ(t) uxxx , ...) = 0, where u(x, t)is the solution of the Eq. (1). Using the transformations Z u(x, t) = f (η), η = k x + ω(t) dt, ∗
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(1)
(2)
K. R. Raslan: The Application of He’s Exp-function Method for · · ·
175
where k is constant and ω(t)is an integrable function of t. Based on then 2 ∂ d ∂ d ∂2 2 d (.) = ω(t) (.) , (.) = k (.) , (.) = k (.) , ... ∂t dη ∂x dη ∂ x2 d η2
(3)
We use Eq. (3) to change the PDE (1) to ODE with variable coefficients: G(f (η), ω(t) fη (η), k α(t) fη (η), k 2 β(t) fηη (η), k 3 γ(t) fηηη (η), ...) = 0,
(4)
next, a solution of Eq.(4) can be written in the form d P
f (η) =
n=−c q P n=−p
an exp(n η) ,
(5)
bn exp(n η)
where c, d, pand qare positive integers which are unknown to be further determined, an and bn are unknown constants. The balance of the linear term of highest order in Eq. (4) with the highest order nonlinear term then we can find the start of the summation c, p also the end d, q. Then using Eq. (5) in Eq. (4), and equating the coefficients of exp(n η) to be zero, a system of nonlinear algebraic equations is obtained. Solved it simultaneously, therefore, we obtain the solution of Eq.(4) and then the solution of Eq. (1) is obtained directly. The above analysis yields the following theorem. Theorem 1 The Exp-function solution of the nonlinear partial differential equation with variable coefficients Eq. (1) can be determined by Eq. (5) after determining all constant coefficients.
3
Applications
In this section, we discuss the problems which involving the nonlinear PDEs with variable coefficients by using the Exp-function method described in section 2.
3.1
Application to the MKdV equation with variable coefficients
For illustration, we consider the MKdV equation with variable coefficients in the form : ut + α(t) u2 ux + β(t) uxxx = 0,
(6)
Using Eq. (2) and Eq. (3) then Eq.(6) becomes an ordinary differential equations, which reads ω(t)
df (η) df (η) d3 f (η) + k α(t) f 2 (η) + k 3 β(t) = 0. dη dη dη 3
(7)
According to the Exp-function method [24-27], we assume that the solution of Eq.(7) can be re-written in an alternative form as follows: f (η) =
ac exp(c η) + ... + a−d exp(−d η) , bp exp(p η) + ... + b−q exp(−q η)
(8)
In order to determine values of c and p, we balance the linear term of the highest order in Eq. (7) with the highest order nonlinear term. By simple calculation, we have f 000 (η) = f 2 (η) f 0 (η) =
c1 exp((c + 7 p)η) + .... , c2 exp(8 pη) + ....
c3 exp((3 c + 5 p)η) + .... c3 exp((3 c)η) + .... = , c4 exp(8 pη) + .... c4 exp(3 pη) + ....
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(9) (10)
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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 174-181
where ci are determined coefficients only simplicity. Balancing highest order of Exp-function in Eqs. (9) and (10), we have c + 7p = 3c + 5p, (11) which leads to the result p = c.
(12)
Similarly to determine values dand q, we balance the linear term of lowest order in Eq. (7) with the highest order nonlinear term, we have f 000 (η) =
... + d1 exp(−(7 q + d)η) , ... + d2 exp(−8qη)
and f 2 (η) f 0 (η) =
(13)
... + d3 exp(−(5 q + 3 d)η) , ... + d4 exp(−8qη)
(14)
where di are determined coefficients only for simplicity. Balancing highest order of Exp-function in Eqs. (13) and (14), we have −(7q + d) = −(5q + 3d), (15) This leads to q = d.
(16)
We can freely choose the values of cand d, but we will illustrate the final solution does not strangely depended upon the choice of values of cand d. For simplicity, we set p = c = 1andd = q = 1, the trial function, Eq. (5) becomes a1 exp(η) + a0 + a−1 exp(−η) f (η) = , (17) b1 exp(η) + b0 + b−1 exp(−η) Substituting Eq. (17) into Eq. (7), equating to zero the coefficients of all powers of exp(n η)to be zero yields a set of algebraic equations for a1 , a0 , a−1 , b0 , b−1 , b1 , k, α(t), β(t)and ω(t) as follows: −(kα(t) a2−1 + (k 3 β(t) + ω(t)) b2−1 ) (−a0 b−1 + a−1 b0 ) = 0,
(18a)
−2(k α(t) a2−1 (−a1 b−1 + a0 b0 ) − b2−1 ((4 k 3 β(t) + ω(t))a1 b−1 + (−2k 3 β(t) + ω(t))a0 b0 )+ kα(t)a3−1 b1 + a−1 b−1 (−kα(t)a20 + (−2k 3 β(t) + ω(t))b20 + (4k 3 β(t) + ω(t))b−1 b1 )) = 0,
(18b)
kα(t)a30 b−1 − kαa−1 a20 b0 + a0 (6kαa−1 a1 b−1 − 5kαa2−1 b1 + b−1 ((k 3 β(t) + ω(t))b20 +(−23 k 3 β(t) + ω(t))b−1 b1 )) − b0 (kα(t)a2−1 a1 − 5(k 3 β(t) + ω(t))a1 b2−1 3
+a−1 ((k β(t) +
ω(t))b20
(18c)
3
+ 6(−3k β(t) + ω(t))b−1 b1 ) = 0,
4(a1 b−1 − a−1 b1 )(kα(t)a20 + kα(t)a−1 a1 + k 3 β(t)b20
(18d)
+ω(t)b20 − 8k 3 β(t) b−1 b1 + ω(t)b−1 b1 = 0, kα(t)a20 a1 b0 − kα(t)a30 b1 + a0 (5kα(t)a21 b−1 − 6kα(t)a−1 a1 b1 + b1 (−(k 3 β(t) +ω(t))b20 + (23k 3 β(t) − ω(t))b−1 b1 )) + b0 (a1 ((k 3 β(t) + ω(t))b20 3
+6(−3k β(t) + ω(t))b−1 b1 ) +
a−1 (kα(t)a21
3
− 5(k β(t) +
ω(t))b21 ))
(18e)
= 0,
−2(−kα(t)a31 b−1 + kα(t)a21 (−a0 b0 + a−1 b1 ) + b21 ((−2k 3 β(t) + ω(t))a0 b0 +(4k 3 β(t) + ω(t))a−1 b1 ) − a1 b1 (−kα(t)a20 + (−2k 3 β(t) + ω(t))b20
(18f)
3
+(4k β(t) + ω(t))b−1 b1 )) = 0, (a1 b0 − a0 b1 )(kα(t)a21 + (k 3 β(t) + ω(t))b21 = 0,
(18g)
Solving the system of algebraic equations with the aid of Mathematica, we obtain k(α(t)a21 +k2 β(t)b21 ) b20 (2α(t)a21 +3k2 β(t)b21 ) ω(t) = − , a = , −1 2 2 b 8 α(t) a b 1 1
1
a0 =
α(t)a21 b0 +3k2 β(t)b0 b21 , α(t) a1 b1
b−1 =
b20 (2α(t)a21 +3k2 β(t)b21 ) , 8 α(t) a1 b21
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K. R. Raslan: The Application of He’s Exp-function Method for · · ·
177
q q 2 3 3 i 3 α(t) a1 i 23 α(t) a1 −a1 b0 ω(t) = p , a−1 = 0, a0 = , b−1 = 0, k = − p , b1 3 β(t)b31 β(t) b1 q q 2 i 23 α(t)3 a31 i α(t) a 1 a1 b0 3 p p ω(t) = − , b−1 = 0, k = , , a−1 = 0, a0 = b1 3 β(t)b31 β(t)b1
(19b)
(19c)
Inserting Eq. (19a-c) and Eq. (17) admits the solution of the MKdV equation in the forms: 2 2 2 R kx+ −
a1 e 2 u(x, t) =
k(α(t)a1 +k β(t)b1 ) dt b2 1
R
2 2 k(α(t)a2 1 +k β(t)b1 ) dt − b2 1
b0 (2α(t)a21 +3k2 β(t)b21 ) −kx− e 8α(t)a1 b21 2 R k(α(t)a1 +k2 β(t)b21 ) k x+ − dt b2 1 b1 e
+
β(t)b1
a1 e
−
b1 e i
+ b0 R
3
√2
!
α(t)a1 β(t) b1
√3
√2 √3
i
x+
! α(t) a1
β(t) b1
a1 e b1 e
i
2 2 k(α(t)a2 1 +k β(t)b1 ) dt − b2 1
√2 √3
x−
R
! α(t) a1
β(t) b1
x−
β(t)b1
√2
R
i
3
3 √
−
α(t)3 a3 1 dt β(t)b3 1
√2
3 3 i 3 α(t) a1 √ dt 3 β(t) b3 1
R
i
,
b20 (2α(t)a21 +3k2 β(t)b21 ) −k x− e 8 α(t) a1 b21 ! √2 √ R i 23 α(t)3 a31 i 3 α(t)a1 √ √ − x+ 3 dt
u(x, t) =
u(x, t) =
α(t)a21 b0 +3k2 β(t)b0 b21 + α(t)a1 b1
+
√2 3
3 √
,
(20b)
,
(20c)
+ b0
+
α(t)3 a3 1 dt β(t) b3 1
a1 b0 b1
(20a)
a1 b0 b1
+ b0
These are solutions for the MKdV equation with variable coefficients in the form (6). Also, we can find another new solutions if we take different cases of the parameters p = candd = q. For a special case we 1 take the values as, a1 = b1 = b0 = k = 1, α(t) = β(t) = 1+t 2 , then the exact solution (20) becomes in the form −1 −1 4 + ex−2 tan t + 85 e−x+2 tan t u(x, t) = (21) −1 −1 , 1 + ex−2 tan t + 58 e−x+2 tan t which plotted in the following Fig. 1 in three different cases.
a1 = b1 = b0 = k = 1, 1 α(t) = β(t) = 1+t 2
a1 = b1 = b0 = k = 1, α(t) = β(t) = sin(t)
a1 = b1 = b0 = k = 1, α(t) = β(t) = tanh(t)
Fig. 1: Solutions of the MKdV equations with variable coefficients in different form
3.2
Application to the Burgers’ equation with variable coefficients
Consider the nonlinear equation of Burgers’ equation with variable coefficients in the form ut + α(t) u ux − β(t) uxx = 0, IJNS homepage:http://www.nonlinearscience.org.uk/
(22)
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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 174-181
Using Eq. (2) and Eq. (3) then Eq.(22) becomes an ordinary differential equations, which reads ω(t)
df (η) df (η) d2 f (η) + k α(t) f (η) − k 2 β(t) = 0, dη dη dη 2
(23)
In view of the Exp-function method [24-27], the highest linear term is now given by f 00 (η) = and f (η) f 0 (η) =
c1 exp((c + 3 p)η) + .... , c2 exp(4 pη) + ....
(24)
c3 exp((2 c + 2 p)η) + .... c3 exp((2 c)η) + .... = , c4 exp(4 pη) + .... c4 exp(2 pη) + ....
(25)
Balancing highest order of Exp-function in Eq. (24) and (25), we get p = c.
(26)
By a similar derivation as illustrated above, we have q = d.
(27)
We can freely choose the values of c and d, but we will illustrate the final solution does not strangely depended upon the choice of values of c and d. For simplicity, we set p = c = 1andd = q = 1, the trial function becomes a1 exp(η) + a0 + a−1 exp(−η) f (η) = , (28) b1 exp(η) + b0 + b−1 exp(−η) Equating to zero the coefficients of all powers of exp(n η) to be zero yields a set of algebraic equations for a1 , a0 , a−1 , b0 , b−1 , b1 , k, α(t), β(t) and ω(t) solving this system, we obtain √ α(t)a1 b0 +kβ(t)b0 b1 +kβ(t)b1 b20 −4b−1 b1 k(α(t)a1 +kβ(t)b1 ) b−1 (α(t)a1 +2kβ(t)b1 ) (29) ω(t) = − , a−1 = , a0 = b1 α(t)b1 α(t)b1 Then the solution of the Burgers equation can be written in the form √ R k(α(t)a1 +kβ(t)b1 ) α(t)a1 b0 +kβ(t)b0 b1 +kβ(t)b1 b20 −4b−1 b1 k x+ − dt b 1 a e + + α(t)b1 1 R k(α(t)a1 +kβ(t)b1 ) dt b−1 (α(t)a1 +2kβ(t)b1 ) −k x− − b 1 e α(t)b1 µ u(x, t) = , R k(α(t)a1 +kβ(t)b1 ) R k(α(t)a1 +kβ(t)b1 ) ¶ k x+ − dt −k x− − dt b b 1 1 b1 e + b0 + b−1 e For a special case we take the values as a1 = b1 = b−1 = k = 1, b0 = 2, α(t) = the exact solution (30) becomes in the form
1 , and β(t) 1+t2
=
(30)
1 ,then 1+t2
−1
u(x, t) =
ex + 3 e2 tan t , −1 ex + e2 tan t
(31)
which plotted in the following Fig. 2 in three different cases
3.3
Application to the Burgers’ equation in two dimensional with variable coefficients.
In this example, we solve the Burgers’ equation in two dimensional with variable coefficients which can be written as ut + α(t) u(ux + uy ) − β(t) (uxx + uyy ) = 0, (32) where u(x, y, t)is the solution of the equation (32). We use the transformations Z u(x, y, t) = f (η), η = kx + m y + ω(t) dt,
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K. R. Raslan: The Application of He’s Exp-function Method for · · ·
a1 = b1 = b−1 = k = 1, b0 = 2 1 α(t) = β(t) = 1+t 2
a1 = b1 = b−1 = k = 1, b0 = 2 α(t) = β(t) = sin(t)
179
a1 = b1 = b−1 = k = 1, b0 = 2 α(t) = β(t) = tanh(t)
Fig. 2: Solutions of the Burgers’ equations with variable coefficients in different form where k , m are constants and ω(t)is an integrable function of t. Then the equation (32) becomes ω(t)
df (η) df (η) d2 f (η) + (k + m) α(t) f (η) − (k 2 + m2 ) β(t) = 0, dη dη dη 2
(34)
In view of the Exp-function method [24-27], we assume that the solution of equation (34) can be expressed in the form ac exp(c η) + ... + a−d exp(−d η) , (35) f (η) = bp exp(p η) + ... + b−q exp(−q η) By the same in the last example we find that p = candd = q. For simplicity, we set p = c = 1andd = q = 1, the trial function, Eq. (35) becomes f (η) =
a1 exp(η) + a0 + a−1 exp(−η) , b1 exp(η) + b0 + b−1 exp(−η)
(36)
Substituting Eq. (36) into Eq. (34), equating to zero the coefficients of all powers of exp(n η)to be zero yields a set of algebraic equations for a1 , a0 , a−1 , b0 , b−1 , b1 , k, m, α(t), β(t) and ω(t) solving this system, we obtain 2 2 −(k+m)α(t)a1 −(k2 +m2 )β(t)b1 1 +2(k +m )β(t)b1 , a−1 = b−1 (k+m)α(t)a , b1 (k+m)α(t)b 1 √ (k+m)α(t)a1 b0 +(k2 +m2 )β(t)b1 (b0 + b20 −4b−1 b1 ) (k+m)α(t)b1
ω(t) = a0 =
(37)
then the solution of the Burgers equation can be written in the form R −(k+m)α(t)a1 −(k2 +m2 )β(t)b1 k x+my+ dt b 1 + a1 e √ (k+m)α(t)a1 b0 +(k2 +m2 )β(t)b1 (b0 + b20 −4b−1 b1 ) + (k+m)α(t)b1 R k(α(t)a1 +kβ(t)b1 ) 2 2 dt b−1 (k+m)α(t)a1 +2(k +m )β(t)b1 −k x−my− − b 1 e (k+m)α(t)b1 u(x, y, t) = µ R k(α(t)a1 +kβ(t)b1 ) R k(α(t)a1 +kβ(t)b1 ) ¶ , k x+my+ − dt −k x−my− − dt b1 b1 b1 e + b0 + b−1 e For a special case we take the values as a1 = b1 = b−1 = k = m = 1, b0 = 2, 1 , then the exact solution (38) becomes in the form 1+t2
α(t) =
(38)
1 andβ(t) 1+t2
=
−1
ex+y + 3 e4 tan t u(x, y, t) = x+y , (39) −1 e + e4 tan t which plotted in the following Fig. 3 in three different cases at time t = 1. From the last two examples then we can solved Burgers’ equation in three dimensional with variable coefficients in the form ut + α(t) u(ux + uy + uz ) − β(t) (uxx + uyy + uzz ) = 0,
(40)
where u(x, y, z, t)is the solution of the Eq.(40) using the same method Exp-function method and done the changes required. IJNS homepage:http://www.nonlinearscience.org.uk/
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International Journal of Nonlinear Science,Vol.7(2009),No.2,pp. 174-181
a1 = b1 = b−1 = k = m = 1, 1 b0 = 2, α(t) = β(t) = 1+t 2
a1 = b1 = b−1 = k = m = 1, b0 = 2, α(t) = sin(t) β(t) = sin(t)
a1 = b1 = b−1 = k = m = 1, b0 = 2, α(t) = tanh(t) β(t) = tanh(t)
Fig. 3: Solutions of the Burgers’ equations with variable coefficients in different form.
4
Conclusion
In this paper, the Exp-function method was applied successfully for finding nonlinear partial differential equations with variable coefficients in one, two and three dimensional. Nonlinear partial differential equations with variable coefficients which are the MKdV equation, the Burgers’ equation in one and two dimensional have been solved. It can be concluded that the Exp-function method is very powerful and efficient technique in finding exact solutions for wide classes of problems. The Exp-function method has got many merits and much more advantages than the exact solutions. Calculations in Exp-function method are simple and straightforward. The reliability of the method and the reduction in the size of computational domain give this method a wider applicability. The results show that the Exp-function method is a powerful mathematical tool for solving systems of nonlinear partial differential equations have wide applications in physics and engineering.
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