the numerical solutions of differential transform method and the laplace transform method for a...

Nonlinear Analysis: Hybrid Systems 4 (2010) 425–431

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Nonlinear Analysis: Hybrid Systems

journal homepage: www.elsevier.com/locate/nahs

The numerical solutions of differential transform method and theLaplace transform method for a system of differential equationsMontri Thongmoon a,∗, Sasitorn Pusjuso ba Department of Mathematics, Faculty of Science, Mahasarakham University, Mahasarakham, 44150, Thailandb Department of Mathematics, Statistics and Computer, Faculty of Science, Ubonrajathanee University, Ubonrajathanee, 34190, Thailand

a r t i c l e i n f o

Article history:Received 29 September 2009Accepted 7 October 2009

Keywords:Differential transform methodLaplace transform methodSystem of differential equations

a b s t r a c t

The differential transform method is one of the approximate methods which can be eas-ily applied to many linear and nonlinear problems and is capable of reducing the size ofcomputational work. Exact solutions can also be achieved by the known forms of the seriessolutions. In this paper, we present the definition and operation of the one-dimensionaldifferential transform and investigate the particular exact solutions of system of ordinarydifferential equations that usually arise in mathematical biology by a one-dimensional dif-ferential transformmethod. The numerical results of the presentmethod are presented andcompared with the exact solutions that are calculated by the Laplace transform method.

Published by Elsevier Ltd

1. Introduction

The differential transformation method is a numerical method based on a Taylor expansion. This method constructs ananalytical solution in the form of a polynomial. The concept of differential transformmethodwas first proposed and appliedto solve linear and nonlinear initial value problems in electric circuit analysis by [1]. Chen and Liu have applied this methodto solve two-boundary-value problems [2]. Jang, Chen and Liu apply the two-dimensional differential transform methodto solve partial differential equations [3]. Yu and Chen apply the differential transformation method to the optimization ofthe rectangular fins with variable thermal parameters [4,5]. Unlike the traditional high order Taylor series method whichrequires a lot of symbolic computations, the differential transform method is an iterative procedure for obtaining Taylorseries solutions. This method will not consume too much computer time when applying to nonlinear or parameter varyingsystems. This method gives an analytical solution in the form of a polynomial. But, it is different from Taylor series methodthat requires computation of the high order derivatives. The differential transform method is an iterative procedure that isdescribed by the transformed equations of original functions for solution of differential equations.In this paper, three systems of the ordinary differential equation problem are considered by a differential transformation

technique, a closed form series solution or an approximate solution can be obtained and the numerical solutions arecompared with the exact solutions that are calculated from the Laplace transform method.

2. Basic definitions

As in Refs. [6,7,2,3,8], the basic definition of the differential transformation are introduced as follows:

∗ Corresponding author. Tel.: +66 043754244; fax: +66 043754244.E-mail addresses:[email protected] (M. Thongmoon), [email protected] (S. Pusjuso).

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426 M. Thongmoon, S. Pusjuso / Nonlinear Analysis: Hybrid Systems 4 (2010) 425–431

Definition 1. The one-dimensional differential transform of function c(x) is defined as follows:

C(k) =1k!

[∂k

∂xkc(x)

]x=x0

(1)

where c(x) is analytic and differentiated continuouslywith respect to x in the domain of interest and C(k) is the transformedfunction, which is called the T -function in brief.

Definition 2. The differential inverse transform of C(k) is defined as follows:

c(x) =∞∑k=0

C(k)(x− x0)k. (2)

When (x0 = 0) then (1) and (2) can be written as:

Definition 3. The one-dimensional differential transform of function c(x) is defined as follows:

C(k) =1k!

[∂k

∂xkc(x)

]x=0

(3)

where c(x) is the original function and C(k) is the transformed function, which is called the T -function.

Definition 4. The differential inverse transform of C(k) is defined as follows:

c(x) =∞∑k=0

C(k)xk. (4)

Substituting (3) into (4) we have

c(x) =∞∑k=0

1k!

[∂k

∂xkc(x)

]x=0xk. (5)

In real applications, the function c(x) by a finite series of (4) can be written as

c(x) =n∑k=0

C(k)xk (6)

and (4) implies that

c(x) =∞∑

k=n+1

C(k)xk

is neglected as it is small. Usually, the values of n are decided by a convergency of the series coefficients.From the definition of (3) and (4), it is readily proved that the transformed functions complywith the basic mathematical

operations [6,7,9,2,3,8]:

Theorem 5. If c(t) = u(t)± v(t), then

C(k) = U(k)± V (k). (7)

Theorem 6. If c(t) = αu(t), then

C(k) = αU(k) (8)

where α is a constant.

Theorem 7. If c(t) = ∂∂t u(t), then

C(k) = (k+ 1)U(k+ 1). (9)

Theorem 8. If c(t) = ∂r

∂tr u(t), then

C(k) = (k+ 1)(k+ 2) · · · (k+ r)U(k+ r). (10)

M. Thongmoon, S. Pusjuso / Nonlinear Analysis: Hybrid Systems 4 (2010) 425–431 427

Table 1Differential transformation values of Example 1, for k = 0, 1, 2, 3, . . . .

k k+ 1 X(k+ 1) Y (k+ 1)

0 1 −1 11 2 1 −1/22 3 −5/6 1/23 4 5/8 −7/244 5 −7/120 23/1205 6 −19/144 1/80...

.

.

....

.

.

.

Theorem 9. If c(t) = u(t)v(t), then

C(k) =k∑r=0

U(r)V (k− r). (11)

Theorem 10. If c(t) = eλt , then C(k) = λk

k! .

Theorem 11. If c(t) = (1+ t)m, then C(k) = m(m−1)···(m−k+1)k! .

Theorem 12. If c(t) = sin(ωt + α), then C(k) = ωk

k! sin(πk2! + α).

Theorem 13. If c(t) = cos(ωt + α), then C(k) = ωk

k! cos(πk2! + α).

3. Numerical experiments

3.1. Example 1

Consider the following non-linear differential system:

dx(t)dt+dydt+ x(t)+ y(t) = 1

dy(t)dt= 2x(t)+ y(t)

(12)

with the initial conditions

x(0) = 0; y(0) = 1. (13)

Taking the differential transform method to Eqs. (12) and (13), we obtain

(k+ 1)X(k+ 1)+ (k+ 1)Y (k+ 1)+ X(k)+ Y (k) = 1(k+ 1)Y (k+ 1) = 2X(k)+ Y (k) (14)

rearranging Eq. (14) can be written as:

X(k+ 1) =1k+ 1

[1− (k+ 1)Y (k+ 1)− X(k)− Y (k)]

Y (k+ 1) =1k+ 1

[2X(k)+ Y (k)](15)

with initial conditions

X(0) = 0; Y (0) = 1. (16)

The numerical results for the differential transformation method are presented in Table 1.From (6), the approximate solution when n = 6 is

x(t) =6∑k=0

X(k)tk (17)


1 2 3 4 5 6 7 8 9 10 11-1

-0.5

0

0.5

1

1.5

2Numerical solutions

x values from DTM

y values from DTM

x values from Exact solution

y values from Exact solution

valu

e o

f x

and

y

value of t

Fig. 1. Numerical solutions of differential transform method and exact solutions when 0 ≤ t ≤ 1 and1t = 0.1.

and

y(t) =6∑k=0

Y (k)tk. (18)

Substituting all values of X(k) and Y (k) into Eqs. (17) and (18), we obtain

x(t) = −t + t2 − 5/6t3 + 5/8t4 − 7/120t5 − 19/144t6 (19)

y(t) = 1+ t − 1/2t2 + 1/2t3 − 7/24t4 + 23/120t5 + 1/80t6. (20)

Using the Laplace transform method, the exact solution of this example is the form

x(t) = e−t − 1y(t) = 2− e−t .

The numerical solutions are presented in Fig. 1:

3.2. Example 2

Consider the following non-homogenous differential system:

dx(t)dt= z(t)− cos t

dy(t)dt= z(t)− et

dz(t)dt= x(t)− y(t)

(21)


x(0) = 1; y(0) = 0; z(0) = 2. (22)


(k+ 1)X(k+ 1) = Z(k)−1k!cos

(πk2

)(k+ 1)Y (k+ 1) = Z(k)−

1k!

(k+ 1)Z(k+ 1) = X(k)− Y (k)

(23)


Table 2Differential transformation values of Example 2, for k = 0, 1, 2, . . . .

k k+ 1 X(k+ 1) Y (k+ 1) Z(k+1)

0 1 1 1 11 2 1/2 0 02 3 1/6 −1/6 1/63 4 1/24 0 1/124 5 1/120 1/120 1/1205 6 1/720 0 0...

.

.

....

.

.

....

rearranging Eq. (23) can be written as:

X(k+ 1) =1k+ 1

[Z(k)−

1k!cos

(πk2

)]Y (k+ 1) =

1k+ 1

[Z(k)−

1k!

]Z(k+ 1) =

1k+ 1

[X(k)− Y (k)]

(24)

with initial conditions

X(0) = 1; Y (0) = 0; Z(0) = 2. (25)

Then the numerical solutions for X(k+ 1), Y (k+ 1), Z(k+ 1) for k = 0, 1, 2, . . . are presented as in Table 2:From (6), the approximate solution when n = 6 is

x(t) =6∑k=0

X(k)tk; (26)

y(t) =6∑k=0

Y (k)tk; (27)

and

z(t) =6∑k=0

Z(k)tk. (28)

Substituting all values of X(k), Y (k) and Z(k) for all k = 0, 1, 2, . . . , 6 into Eqs. (26)–(28), we have the numerical solution:

x(t) = 1+ t +12t2 +

16t3 +

124t4 +

1120t5 +

1720t6

y(t) = t −16t3 +

1120t5

z(t) = 2+ t +16t3 +

112t4 +

1120t5.

(29)

Using the Laplace transform method, the exact solution of this example is the form

x(t) = et

y(t) = sin tz(t) = et + cos t.

(30)

From Eqs. (29) and (30), we have the numerical solutions in Fig. 2: We see that the numerical solution in Eq. (29) is closedto the exact solution in Eq. (30).

3.3. Example 3

Consider the differential equation:

d2x(t)dt2

+ y(t) = 1

d2y(t)dt2

+ x(t) = 0(31)


1 2 3 4 5 6 7 8 9 10 110

0.5

1

1.5

2

2.5

3

3.5Numerical solutions

x values from DTMy values from DTMz values from DTMx values from Exact solutiony values from Exact solutionz values from Exact solution

val

ue

of

x, y

an

d z

value of t


Table 3Differential transformation values of Example 3, for k = 0, 1, 2, . . . .

k k+ 2 X(k+ 2) Y (k+ 2)

0 2 1/2 01 3 1/6 02 4 1/12 −1/243 5 1/20 −1/1204 6 5/144 −1/3605 7 121/5040 −1/840...

.

.

....

.

.

.


x(0) = y(0) = x′(0) = y′(0) = 0. (32)


(k+ 1)(k+ 2)X(k+ 2)+ Y (k) = 1(k+ 1)(k+ 2)Y (k+ 2)+ X(k) = 0 (33)


X(0) = Y (0) = X(1) = Y (1) = 0. (34)

Then the numerical solutions for X(k+2), Y (k+2) for k = 0, 1, 2, . . . are presented as in Table 3: From (6), the approximatesolution when n = 7 is

x(t) =7∑k=0

X(k)tk (35)

and

y(t) =7∑k=0

Y (k)tk. (36)

Substituting all values of X(k) and Y (k) into Eqs. (35) and (36), we have the numerical solution:

x(t) =t2

2−t3

6+t4

12+t5

20+5t6

144+121t7

5040

y(t) = −t4

24−t5

120−t6

360−t7

840.

(37)


1 2 3 4 5 6 7 8 9 10 11-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6 Numerical solutions

x values from DTM

y values from DTM

x values from Exact solution

y values from Exact solution

value of t

valu

e o

f x

and

y


Using the Laplace transform method, the exact solution of this example is in the form

x(t) =et

4+e−t

4−cos t2

y(t) = 1−et

4−e−t

4−cos t2.

(38)

From Eqs. (37) and (38), we have the numerical solutions in Fig. 3: We see that the numerical solution in Eq. (37) is closedthe exact solution in Eq. (38).

4. Conclusion and discussion

One-dimensional differential transforms have been applied to linear and non-linear systems of ordinary differentialequations. Numerical examples have been presented to show that the approach is promising and the research is worthcontinuing in this direction. Using the differential transform method, the solution of the system of ordinary differentialequations can be obtained in Taylor’s series form. All the calculations in the method are very easy. The calculated results arequite reliable. Therefore, this method can be applied to many complicated linear and non-linear ODEs.

Acknowledgement

The authors would like to thank the referee for his valuable suggestions that improved the presentation of the paper.

References

[1] X. Zhou, Differential Transformation and its Applications for Electrical Circuits, Huazhong University Press, Wuhan, China, 1986 (in Chinese).[2] C.L. Chen, Y.C. Liu, Differential transformation technique for steady nonlinear heat conduction problems, Appl. Math. Comput. 95 (1998) 155164.[3] M.J. Jang, C.L. Chen, Analysis of the response of a strongly nonlinear damped system using a differential transformation technique, Appl. Math. Comput.88 (1997) 137–151.

[4] L.T. Yu, C.K. Chen, The solution of the Blasius equation by the differential transformation method, Math. Comput. Modelling 28 (1) (1998) 101–111.[5] L.T. Yu, C.K. Chen, Application of taylor transformation to optimize rectangular fins with variable thermal parameters, Appl. Math. Model. 22 (1998)11–21.

[6] F. Ayas, On the two-dimensional differential transform method, Appl. Math. Comput. 143 (2003) 361–374.[7] F. Ayas, Solutions of the system of differential equations by differential transform method, Appl. Math. Comput. 147 (2004) 547–567.[8] I.H. Abdel-Halim Hassan, Differential transformation technique for solving higher-order initial value problems, Appl. Math. Comput. 154 (2004)299–311.

[9] Y. Duan, R. Liu, Y. Jiang, Lattice Boltzmann model for the modified Burger’s equation, Appl. Math. Comput. 202 (2008) 489–497.

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