New Application of Differential Transformation Method for Improved Boussinesq Equation

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Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014 doi: 10.14355/jmmf.2014.0301.02

New Application of Differential Transformation Method for Improved Boussinesq Equation S. Mahmoudvand1, B. Soltanalizadeh2 Department of Electrical Engineering, Mehrban@Branch,Islamic Azad University, Sarab, Iran Department of Mathematics, University of Houston, 4800 Calhoun Rd,Houston, TX 77204, USA

1 2

babak.soltanalizadeh@gmail.com

1

Abstract In this paper, the Differential Transformation Method (DTM) has been applied to find the exact solution of the improved Boussinesq equation by means of the known forms of the series solutions. In addition, a numerical test is presented to demonstrate the effectiveness and efficiency of the proposed method. The results prove that the DTM is one of the powerful techniques for linear and nonlinear equations.

Keywords Improved Boussinesq Equation; Method; Spectral Methods

Differential

Transformation

Introduction In the present work, we are dealing with the numerical approximation of the following second-order problem that, one of the important partial differential equations that commonly arise in problems of mathematical physics: utt − u xx − uu xx − ( u x ) − u xxtt = 0, 2

0 < x < L, 0 < t ≤ T ,

(1)

Our approach consists of reducing the problem to a set of algebraic equations by expanding the approximate solution as a arbitrary function with unknown coefficients.

The linear case of Boussinesq equation was discussed

by Whitham (Whitham 1974) and Zwillinger (Zwillinger 1997). Moreover the nonlinear Boussinesq equation has been discussed in (Zwillinger 1997; Calogero1982)and the modified Boussinesq equation has been derived in (Zwillinger 1997; Clarkson 1986). In (Bratsos 2007), a predictor corrector (P-C) scheme is applied successfully to a nonlinear Boussinesq equation. Authors of (Wang 2008) discussed the improved and generalized Boussinesq equation and some other schemes are presented in (Yildirim 2011; Sariaydin 2010).

The concept of the DTM was first proposed by Zhou (Zhou 1986), who solved linear and nonlinear problems 12

in electrical circuit problems. Chen and Ho (Chen 1999) developed this method for partial differential equations and Ayaz (Ayaz 2004) applied it to the system of differential equations. During recent years this method has been used to solve various types of equations. For example, this method has been used for differential algebraic equations (Cole 1951), partial differential equations (Chen 1999; Jang2001; Kangalgil 2009) fractional differential equations (Arikoglu 2007; Tari 2009) and Difference equations (Arikoglu 2006).

In (Soltanalizadeh 2011a; Soltanalizadeh 2011b; Soltanalizadeh 2012; Soltanalizadeh 2011c), this method has been utilized for Telegraph, KuramotoSivashinsky, RLW and Kawahara equations. S. Shahmorad et al. developed DTM to fractional-order integro-differential equations with nonlocal boundary conditions (Nazari 2010) and class of two-dimensional Volterra integral equations (Tari 2009). Abazari et. al. applied this method for Burgers (Abazari 2010) equation. Similar problems can be found in (Soltanalizadeh 2011; Ghehsareh 2011).

The Definitions and Operations of DT The One-dimensional Differential Transform The basic definitions and operations of onedimensional DT are introduced in (Zhou 1986; Chen 1999; Ayaz 2004) as follows: Definition 2.1 If u (t ) is analytic in the time domain T then d k u (t ) (2) = φ (t , k ), ∀t ∈ T . dt k for t = ti , φ (t , k ) = φ (ti , k ), where k belongs to the

non-negative integer, denoted as the K Therefore, Eq. ((2)) can be rewritten as d k u (t ) U i (k ) = φ (ti , k ) = [ ]t = t , ∀t ∈ T . i dt k

domain.

(3)

Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014

where U i (k ) is called the spectrum of u (t ) at t = ti in the K domain.

(t − ti ) k U (k ). k! k =0 ∞

(4)

Eq. ((4)) is known as the inverse transformation of U(k).

If U(k) is defined as d k q (t )u (t ) (5) U (k ) = M (k )[ ]t = t , k = 0,1, 2,.... i dt k then the function u (t ) can be described as 1 ∞ (t − ti ) k U (k ) . u (t ) = ∑ q (t ) k =0 k! M (k )

(6)

where M (k ) ≠ 0 , q (t ) ≠ 0. M (k ) is called the weighting factor and q (t ) is regarded as a kernel corresponding to u (t ) . If M (k ) = 1 and q (t ) = 1 then Eqs. ((4)) and ((6)) are equivalent. In this paper, the transformation with M (k ) = 1/ k! and q (t ) = 1 is applied. Thus from Eq. ((6)), we have U (k ) =

1 d k u (t ) [ ]t = t , k = 0,1, 2,.... i k! dt k

1 ∂k +h [ k h w( x, t )]x = x , t = t , 0 0 k!h! ∂x ∂t

W (h, k ) =

(7)

formed function, which is also called the T-function.

Let w( x, y ) be the original function while the uppercase W (k , h) stands for the transformed function.

Then the differential inverse transform of W (k , h) is defined as: ∞

∞

w( x, t ) = ∑ ∑ W (k , h)( x − x0 ) k (t − t0 ) h .

Using Eq. ((23)) in Eq. ((9)), we have ∞ ∞ 1 ∂k +h [ k h w( x, t )]x = x , t = t x k t h w( x, t ) = ∑ ∑ 0 0 k =0 h =0 k!h! ∂x ∂t ∞

∞

(12)

= ∑ ∑ W ( k , h) x k t h . k =0 h =0

Afterward, from the above definitions and Eq. ((11)) and ((12)), we can obtain some of the fundamental mathematical operations performed by twodimensional differential transform in Table 1 and 2 which table 1 is the original functions and table 2 is the transformed functionof table 1.

TABLE 1

Original function w( x, t ) = u ( x, t ) ± v( x, t ) w( x, t ) = cu ( x, t )

obtained by finite-term Taylor series plus a remainder, as

w( x, t ) =

∂ u ( x, t ) ∂x

∂r + s u ( x, t ) ∂x r ∂t s w( x, t ) = u ( x, t )v( x, t )

(8)

w( x, t ) =

n

= ∑ (t − t0 ) k U (k ) + Rn +1 (t ). k =0

In order to speed up the convergence rate and improve the accuracy of calculation, the entire domain of t needs to be split into sub-domains.

w( x, t ) =

∂u ( x, t ) ∂v( x, t ) ∂x ∂x

TABLE 2

W ( k , h) = U ( k , h) ± V ( k , h)

The Two-dimensional Differential Transform

W (k , h) = cU (k , h)

W (k , h) = (k + 1)U (k + 1, h)

Consider a function of two variables w( x, t ) , and suppose that it can be represented as a product of two single-variable functions, i.e., w( x, t ) = f ( x) g (t ) . Based on the properties of one-dimensional differential transform, function w( x, t ) can be represented as ∞ ∞

(9)

w( x, t ) = ∑ ∑ W (i, j ) xi t j . i =0 j =0

where W (i, j ) is called the spectrum of w( x, y ) . Now the basic definitions and operations dimensional DT are introduced as follows.

(11)

k =0 h =0

Using the differential transform, a differential equation in the domain of interest can be transformed to an algebraic equation in the K domain and u (t ) can be

1 n (t − ti ) k U (k ) + Rn +1 (t ) u (t ) = ∑ q (t ) k =0 k! M (k )

(10)

where the spectrum function W (k , h) is the trans-

Definition 2.2 If ut is analytic, then it can be shown as

u (t ) = ∑

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of

two-

Definition 2.3 If w( x, t ) is analytic and continuously differentiable with respect to time t in the domain of interest, then

W ( k , h) =

(k + r )!(h + s )!

U (k + r , h + s) k!h!

W (k , h) = ∑ r =0∑ s =0U (r , h − s )V (k − r , s ) k

h

W (h, k ) = ∑ r =0∑ s =0(r + 1)(k − r + 1) × U (r + 1, h − s )V (k − r + 1, s ) k

h

Application of the DTM In this section, the DTM is applied to solve the improved Boussinesq equation.

Remark 3.1 The symbol ⊗ is used to denote the differential transform version of multiplication.

Consider equation (1) utt − u xx − uu xx − ( u x ) − u xxtt = φ ( x, t ) , 2

(12)

13

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Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014

with the initial conditions u ( x, 0) = f ( x),

h

(13)

s =0 r =0

U (k − r + 1, s ) − φ (k .h)},

and

∂u ( x, 0) = g ( x), ∂t and boundary conditions u (0, t ) = p (t ),

(14) (15)

k = 0,1, 2,..., N , h = 0,1, 2,..., N .

Example 3.1. Consider the the equations (12) - (16) with φ ( x, t ) = −2exp(2 x + 2t ) − exp( x + t ),

and

∂u (0, t ) = q (t ).

(16) ∂x Let U (k , h) be the differential transform of u ( x, t ) .

Applying Table 1 and Definition 2.3 when x0 = t0 = 0 , we get the differential transform version of Eq. ((13)) as follows: (h + 1)(h + 2)U (k , h + 2)

−(k + 1)(k + 2)U (k + 2, h) ∂u ∂u ∂u |x = k , t = h − |x = k , t = h ⊗ |x = k , t = h −u ⊗ ∂x ∂x ∂x (k + 2)!(h + 2)!

U (k + 2, h + 2) = φ (k , h). − k!h! then we have (h + 1)(h + 2)U (k , h + 2) − (k + 1)(k + 2)U (k + 2, h) h

k

− ∑ ∑ (r + 1)(k − r + 1)U (r + 1, h − s )U (k − r + 1, s )

g ( x) = exp ( x),

p (t ) = exp (t ),

q (t ) = exp (t ).

− U (1, 0)U (1, 0) − φ (0, 0)) =

U (3, 2) =

1 ((1)(2)U (1, 2) − (2)(3)U (3, 0) 3!2! k =1

r =0 k =1

(17)

r =0

U (2 − r , s ) − φ (1, 0)) = −

−

U (4, 2) = (18)

and

U (0, h) ,

and

U (0, h) ,

k

− ∑ ∑ (r + 1)(r + 2)U (r + 2, h − s )U (k − r , s ) s =0 r =0

14

∑ (r + 1)(r + 2)U (r + 2, 0)U (2 − r , 0) k =2

− ∑ (r + 1)(3 − r )U (r + 1, 0) r =0

(20) (21)

for

1 , 2!4! and for h = 1 and k = 0,1, 2,..., N , we get 1 ((2)(3)U (0,3) − (1)(2)U (2,1) U (2,3) = 2!3!

U (3 − r , s ) − φ (2, 0)) = −

h =1

− ∑ (1)(2)U (2,1 − s )U (0, s ) s =0 h =1

− ∑ U (1,1 − s )U (1, s ) − φ (0,1)) = s =0

U (3,3) =

1 , 2!3!

1 ((2)(3)U (1,3) − (2)(3)U (3,1) 3!3!

h =1 k =1

− ∑ ∑ (r + 1)(r + 2)U (r + 2,1 − s )U (1 − r , s )

k!h! {(h + 1)(h + 2)U (k , h + 2) (k + 2)!(h + 2)! h

2! ((1)(2)U (2, 2) − (3)(4)U (4, 0) 4!2! k =2

(19)

h = 2,3,...N ,can be obtained from Eqs. ((19)) - ((22)).

Using equation Eq. ((18)), we find the remainder values of U as follows: U (k + 2, h + 2) =

−(k + 1)(k + 2)U (k + 2, h)

1 , 2!3!

r =0

Therefore, the value of U (k , 0) , and U (k ,1) , for k = 0,1, 2,...N ,

(24)

− ∑ (r + 1)(2 − r )U (r + 1, 0)

s =0 r =0

(k + 2)!(h + 2)!

U (k + 2, h + 2) = φ (k , h). k!h!

By the initial conditions we have f k (0) , k = 0,1, 2,..., N , U (k , 0) = k! g k (0) , k = 0,1, 2,..., N .

U (k ,1) = k!

Similarly, from the boundary conditions we have p h (0) , h = 2,3,..., N , U (0, h) = h! q h (0) U (1, h) = , h = 2,3,..., N . h!

1 , 2!2!

− ∑ (r + 1)(r + 2)U (r + 2, 0)U (1 − r , 0)

− ∑ ∑ (r + 1)(r + 2)U (r + 2, h − s )U (k − r , s ) h

f ( x) = exp ( x),

From equations (20)-(23) we have 1 U (k , 0) = U (k ,1) = , k = 0,1, 2,..., N , k!

1 U (0, h) = U (1, h) = , h = 2,3,..., N . h!

Using (24) for h = 0 and k = 0,1, 2,...N , we have 1 U (2, 2) = ((1)(2)U (0, 2) − (1)(2)U (2, 0) 2!2! − (1)(2)U (2, 0)U (0, 0)

k

s =0 r =0

k

− ∑ ∑ (r + 1)(k − r + 1)U (r + 1, h − s )

s =0 r =0 h =1 k =1

(22)

− ∑ ∑ (r + 1)(2 − r )U (r + 1,1 − s ) s =0 r =0

U (2 − r , s ) − φ (1,1)) =

1 , 3!31

(25)

Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014

U (4,3) =

1 ((2)(3)U (2,3) − (3)(4)U (4,1) 3!3!

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U (4, 2) = 0. and for h = 1 we have 1 U (2,3) = ((2)(3)U (0,3) − (1)(2)U (2,1) 2!3!

h =1k =2

− ∑ ∑ (r + 1)(r + 2)U (r + 2,1 − s )U (2 − r , s ) s =0 r =0 h =1k =2

h =1

− ∑ ∑ (r + 1)(3 − r )U (r + 1,1 − s )U (3 − r , s )

− ∑ (1)(2)U (2,1 − s )U (0, s )

s =0 r =0

s =0

1 −φ (2,1)) = 4!31 .

By continuing this process, we obtain 1 1 1 u ( x, t );(1 + t + t 2 + t 3 ... + t n + ...)

2!

3! n!

1 2 1 3 1 n +( x + xt + xt + xt ... + xt + ...)

2!

3! n!

1 2 1 2 1 2 2 1 23 x t + x t +( x + x t + 2!

2!

2!2!

2!3!

1 2 n x t + ...) + ..., +... + 2!n! which is the Taylor expansion of the

h =1

− ∑ U (1,1 − s )U (1, s ) − φ (0,1)) = 1, s =0

U (3,3) = h =1 k =1

− ∑ ∑ (r + 1)(r + 2)U (r + 2,1 − s )U (1 − r , s ) s =0 r =0 h =1 k =1

− ∑ ∑ (r + 1)(2 − r )U (r + 1,1 − s ) s =0 r =0

U (2 − r , s ) − φ (1,1)) = 0, U (4,3) = 1 Hence we obtain u ( x, t );(−2 x − 4 xt − 2 xt 3 ) + ( x 2 + 2 x 2 t + x 2 t 3 )

+( x 4 + 2 x 4 t + x 4 t 3 ) = (1 + 2t + t 3 )(−2 x)

u ( x, t ) = exp ( x + t ) 0 ≤ x ≤ L,

+(1 + 2t + t 3 ) x 2 + (1 + 2t + t 3 ) x 4

which is the exact solution of this example.

Example 3.2.Consider the equations (12) - (16) with

φ ( x, t ) = (−6 − 32t − 16t 2 − 10t 3 − 16t 4 − 4t 6 ) +(−18 − 114t − 24t 2 − 24t 3 − 24t 4 − 6t 6 ) x 2 +(40 + 160t + 160t + 80t + 160t + 40t ) x 3

4

6

3

+(−30 − 114t − 120t 2 − 60t 3 − 120t 4 − 30t 6 ) x 4 +(28 + 112t + 112t 2 + 56t 3 + 112t 4 + 28t 6 ) x 6

f ( x ) = x 4 + x 2 − 2 x, p (t ) = 0,

= (1 + 2t + t 3 )(−2 x + x 2 + x 4 ). which is the exact solution of this example.

Conclusions

+(12 + 36t + 48t 2 + 24t 3 + 48t 4 + 12t 6 ) 2

1 ((2)(3)U (1,3) − (2)(3)U (3,1) 3!3!

g ( x) = 2( x 4 + x 2 − 2 x),

q (t ) = −2(t 3 + 2t + 1).

Applying equations (20)-(23) in initial and boundary conditions of this problem, we have U (0, 0) = 0, U (1, 0) = −2, U (2, 0) = 1,

U (3, 0) = 0, U (4, 0) = 1, U (0,1) = 0, U (1,1) = −4, U (2,1) = 2, U (3,1) = 0, U (4,1) = 2, U (0, 2) = U (0,3) = U (0, 4) = 0, U (1, 2) = 0, U (1,3) = −2, U (1, 4) = 0.

From equation (25) for h = 0 we have 1 U (2, 2) = ((1)(2)U (0, 2) − (1)(2)U (2, 0) 2!2! −(1)(2)U (2, 0)U (0, 0) − U (1, 0)U (1, 0) − φ (0, 0)) = 0, 1 ((1)(2)U (1, 2) − (2)(3)U (3, 0) U (3, 2) = 3!2!

The basic goal of this paper is to employ DTM method to obtain the solution of the improved Boussinesq equation with some supplementally conditions. The analytical solutions can be obtained by applying this method with a simple iterative process. This method reduces the computational difficulties of the other methods and all the calculations can be made with simple iterative process. Therefore, this method can be applied to many complicated linear and nonlinear PDEs and systems of PDEs and does not require linearization, discretization or perturbation. It is observed that the method is an effective and reliable tool for the solution of such problems.

REFERENCES

Abazari, R., and A. Borhanifar, "Numerical study of the solution of the Burgers and coupled Burgers equations by a differential transformation method", Comput. Math.

Appl., 59 (2010) 2711–2722.

Arikoglu, A., and I. Ozkol, "Solution of difference equations

k =1

by using differential transformation method", Appl. Math.

r =0

Comput. 174 (2006) 1216–1228.

− ∑ (r + 1)(r + 2)U (r + 2, 0)U (1 − r , 0) k =1

− ∑ (r + 1)(2 − r )U (r + 1, 0)U (2 − r , s ) − φ (1, 0)) = 0, r =0

Arikoglu, A., and I. Ozkol, "Solution of fractional differential

15

www.sjmmf.org

Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014

equations by using differential transformation method",

dimensional Kadomtsev-Petviashvili equation", Z. fï‫½؟‬r

Chaos Solitons Fractals. 34 (2007) 1473–1481.

Naturforschung A, 65 (2010) 411–417

Ayaz, F., "Solutions of the systems of differential equations

Soltanalizadeh, B., "An Approximation Method Based on

by differential transform method", Appl. Math. Comput.,

Matrix Formulated Algorithm for the Numerical Study of

147 (2004) 547ï567-½‫؟‬.

a Biharmonic Equation", Australian Journal of Basic and

Bratsos, A. G., "A second order numerical scheme for the solution of the one-dimensional Boussinesq equation", Numer Algor., 46 (2007) 45-58.

Solitons: Tools to Solve and Investigate Nonlinear Evolution Equations." New York: North-Holland, 1982.

Chen, C. K., "Solving partial differantial equations by two dimensional differential transformation method", Appl.

Math. Copmut. 106 (1999) 171–179. equations by two dimensional differential transform method", Appl. Math. Comput., 106 (1999) 171–179.

"Application

of

Differential

Kawahara Equation", Australian Journal of Basic and Applied Sciences, 5(12) (2011) 490–495 Soltanalizadeh, B., "Differential Transformation method for solving

one-space-dimensional

Telegraph

equation",

Comp. Appl. Math., 30(3) (2011) 639–653.

Modified

equation with an integral condition", J. Appl. Math.

Informatics, In Press.

Soltanalizadeh,

Clarkson, P. A., "The Painlevï Property, a Modified and a

B.,

Soltanalizadeh, B., "Numerical study of the Telegraph

Chen, C.K., and S.H. Ho, "Solving partial differential

Equation

Soltanalizadeh,

Transformation Method for Numerical Analysis of

Calogero, F., and A. "Degasperis, Spectral Transform and

Boussinesq

Applied Sciences, 5(12) (2011) 501–506.

Kadomtsev-

Petviashvili Equation", Physica D 19 (1986) 447-450.

Differential

B.,

and

A.

Transformation

Yildirim, method

"Application for

of

numerical

computation of Regularized Long Wave equation", Zeitschrift fuer Naturforschung A., 67a (2012) 160–166.

Ghehsareh, H.R., B. Soltanalizadeh, S. Abbasbandy, "A

Soltanalizadeh, B., and M. Zarebnia, "Numerical analysis of

matrix formulation to the wave equation with nonlocal

the linear and nonlinear Kuramoto-Sivashinsky equation

boundary condition", Inter. J. Comput. Math., 88 (2011)

by using Differential Transformation method", Inter. J.

1681–1696.

Appl. Math. Mechanics, 7(12) (2011) 63-ï72½‫؟‬.

J. D. Cole, "On a quasilinear parabolic equation occurring in aerodynamics", Quart. Appl. Math. 9 (1951) 225–236.

Jang, M. J., and C. K. Chen, "Two-dimensional differential transformation method for partial differantial equations",

two-dimensional linear and nonlinear Volterra integral equations by the differential transform method", J. Comput. Appl. Math. 228 (2009) 70–76 Wang, S., and H. Xue, "Global solution for a generalized

Appl. Math. Copmut. 121 (2001) 261–270.

Kangalgil, F., and F. Ayaz, "Solitary wave solutions for the KDV and mKDV equations by differential transformation

Boussinesq equation", Appl. Math. Comput. 204 (2008) 130-136.

Whitham, G. B., "Linear and Nonlinear Waves." New York:

method", Choas Solitons Fractals. 41 (2009) 464–472.

Momani, S., Z. Odibat, I. Hashim, "Algorithms for nonlinear

Wiley, PP. 9, 1974 .

fractional partial differantial equations: A selection of

Yildirim, A., S. Sefa Anil, K. Yasemin, "Analytical approach

numerical methods", Topol. Method Nonlinear Anal. 31

to Boussinesq equation with space- and time-fractional

(2008) 211-226.

derivatives" Inter. j. numer. meth. fluids, 66 (2011)

Nazari, D., and S. Shahmorad, "Application of the fractional

1315–1324.

fractional-order

Zhou, J.K., "Differential Transformation and Its Applications

integro-differential equations with nonlocal boundary

for Electrical Circuits", Huazhong University Press,

conditions", J. Comput. Appl. Math. 234 (2010) 883–891.

Wuhan, China, 1986.

differential

transform

method

to

Sariaydin, S., and A. Yildirim, "Analytical approach to (2+1)-dimensional

16

Tari, A., M.Y. Rahimib, S. Shahmorad, "Solving a class of

Boussinesq

equation

and

(3+1)-

Zwillinger, D., "Handbook of Differential Equations, 3rd ed.

Boston, MA: Academic Press", pp. 129-130, 1997.

Journal of Modern Mathematics Frontier Volume 3 Issue 1, March 2014

Samad Mahmoudvand finished his Master degree in the applied mathematics in the University of Tabriz.

www.sjmmf.org

Babak Soltanalizadeh is a PhD student in department of Mathematics, university of Houston.

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