Chapter 7

Comparison of FDTM andMFDTM for Solving Nonlinear

Fractional Klein-Gordon Equation

In this chapter, Two-dimensional fractional differential transform method and mod-

ified fractional differential transform method applied successfully for finding the

approximate analytical solution of the nonlinear fractional Klein-Gordon equation.

The plotted graph illustrates the behavior of the solution for different values of frac-

tional order α. Three test examples are given to demonstrate the ability of both the

methods for solving nonlinear fractional Klein-Gordon equation.

7.1 Introduction

In recent past, the glorious developments have been envisaged in the field of frac-

tional calculus and fractional differential equations. Differential equations involving

fractional order derivatives are used to model a variety of systems, of which the im-

portant applications lie in field of viscoelasticity, electrode-electrolyte polarization,

heat conduction, electromagnetic waves, diffusion equations and so on (Hilfer, 2000;

Caputo, 1967). Due to its tremendous scope and applications in several disciplines, a

considerable attention has been given to exact and numerical solutions of fractional

differential equations. A great deal of researches has shown the advantageous use

of the fractional calculus in the modeling and control of many dynamical systems

(Podlubny, 1999; Laroche and Knittel, 2005; Calderon et al., 2006; Sabatier et al.,

2006; Vinagre et al., 2007; Monje et al., 2008). Other than modeling aspects of these

differential equations, the solution techniques and their reliability are rather more

important aspects. It is also equally important to handle critical points which cause

98

sudden divergence, convergence and bifurcation of the solutions of the model. In

order to achieve the goal of highly accurate and reliable solutions, several methods

have been proposed to solve the fractional order differential equations. Some of

the recent analytic/numerical methods are Adomian decomposition method (Ray

and Bera, 2005; Momani and Odibat, 2007; Odibat and Momani, 2008), finite dif-

ference method (Meerschaert and Tadjeran, 2006), fractional differential transform

method (Arikoglu and Ozkol, 2007; Erturk and Momani, 2008), variational iteration

method (Mustafa Inc, 2008; Das, 2009), operational matrix method (Saadatmandi

and Dehghan, 2010), homotopy analysis method (Pandey et al., 2011), generalized

differential transform method (Odibat et al., 2008; Liu and Hou, 2011), finite ele-

ment method (Jiang and Ma, 2011) and the references therein.

In this chapter,we consider the fractional Klein-Gordon equation

∂αu(x, t)

∂tα=∂2u(x, t)

∂x2+ au(x, t) + bu2(x, t) + cu3(x, t) (7.1.1)

with initial condition

u(x, 0) = f(x), x ∈ R (7.1.2)

where a, b and c are real constants. Recently,Golmankhaneh et al. (2011) pre-

sented the application of homotopy perturbation method for solving fractional Klein-

Gordon equation. It is well known that linear and nonlinear Klein-Gordon equa-

tions model many problems in classical and quantum mechanics, solitons and con-

densed matter physics. For instance, nonlinear sine Klein-Gordon equation models

a Josephson junction, the motion of rigid pendula attached to a stretched wire, and

dislocations in crystals (Barone et al., 1971; El-Sayed, 2003; Wazwaz, 2005; Odibat

and Momani, 2007; Yusufoglu, 2008).

99

7.2 Description of the method

In this section, we present the comparison of fractional differential transform method

(FDTM) and modified fractional differential transform method (MFDTM) for the

nonlinear fractional Klein-Gordon equation (7.1.1)-(7.1.2).

7.2.1 FDTM

First, we consider the two-dimensional fractional differential transform method for

the equation (7.1.1) with respect to (x, t) according to Theorem 1.5.1 is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(k, h+ 1) = (k + 1)(k + 2)Uα,1(k + 2, h)

+ aUα,1(k, h) + bk∑

r=0

h∑s=0

Uα,1(r, h− s)Uα,1(k − r, s)

+ c

k∑r=0

k−r∑q=0

h∑s=0

h−s∑p=0

Uα,1(r, h− s− p)Uα,1(q, s)Uα,1(k − r − q, p) (7.2.1)

where a, b and c are unknown constants.

The transformed version of (7.1.2) is

Uα,1(k, 0) =1

k!

dkf(x)

dxk

∣∣∣x=0

, k = 0, 1, 2, · · · (7.2.2)

From Eq. (7.2.1), we get the following recurrence equation

Uα,1(k, h+ 1) =Γ(αh+ 1)

Γ(α(h+ 1) + 1)

[(k + 1)(k + 2)Uα,1(k + 2, h)

+ aUα,1(k, h) + bk∑

r=0

h∑s=0

Uα,1(r, h− s)Uα,1(k − r, s)

+ c

k∑r=0

k−r∑q=0

h∑s=0

h−s∑p=0

Uα,1(r, h− s− p)Uα,1(q, s)Uα,1(k − r − q, p)],

k, h = 0, 1, · · · (7.2.3)

100

We can obtain all values of Uα,1(k, h) by using Eq. (7.2.2) in Eq. (7.2.3). By

substituting the quantities Uα,1(k, h) in (1.5.4), We get the following series solution

u(x, t) =∞∑k=0

∞∑h=0

Uα,1(k, h)(x− x0)k(t− t0)

hα. (7.2.4)

7.2.2 MFDTM

The modified fractional differential transform for the Eq. (7.1.1) with respect to the

variable t according to Theorem 1.5.2 is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(x, h+ 1) =

∂2Uα,1(x, h)

∂x2

+ aUα,1(x, h) + b∑m=0

hUα,1(x,m)Uα,1(x, h−m)

+ ch∑

m=0

m∑l=0

Uα,1(x, h−m)Uα,1(x, l)Uα,1(x,m− l) (7.2.5)

where a, b and c are unknown constants. The modified transformed version of (7.1.2)

with respect to the variable t is

Uα,1(x, 0) = f(x) (7.2.6)

We get the following recurrence equation from Eq. (7.2.5),

Uα,1(x, h+ 1) =Γ(αh+ 1)

Γ(α(h+ 1) + 1)

[∂2Uα,1(x, h)

∂x2+ aUα,1(x, h)

+ b∑m=0

hUα,1(x,m)Uα,1(x, h−m)

+ ch∑

m=0

m∑l=0

Uα,1(x, h−m)Uα,1(x, l)Uα,1(x,m− l)],

h = 0, 1, · · · . (7.2.7)

101

We can obtain all values of Uα,1(x, h) by substituting the Eq. (7.2.6) in Eq. (7.2.7).

Therefore, by substituting the quantities Uα,1(x, h) in (1.5.9), we get the following

series solution

u(x, t) =∞∑h=0

Uα,1(x, h)(t− t0)αh. (7.2.8)

7.3 Numerical Results

In this section, three numerical examples are presented to authenticate the fractional

differential transform method and modified fractional differential transform method.

Example 7.3.1. First, we consider the linear fractional Klein-Gordon equation

∂αu(x, t)

∂tα− ∂2u(x, t)

∂x2− u(x, t) = 0 (7.3.1)

subject to the initial conditions

u(x, 0) = 1 + sin(x) (7.3.2)

Case I: FDTM

The transformed version of (7.3.1) is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(k, h+ 1)− (k + 1)(k + 2)Uα,1(k + 2, h)− Uα,1(k, h) = 0

(7.3.3)

The transformed version of (7.3.2) is

102

Uα,1(k, 0) =

1, k = 0

1k!, k = 1, 5, 9 · · ·

− 1k!, k = 3, 7, 11 · · ·

(7.3.4)

From Eq. (7.3.3), we get the following recurrence equation as

Uα,1(k, h+ 1) =Γ(αh+ 1)

Γ(α(h+ 1) + 1)

[(k + 1)(k + 2)Uα,1(k + 2, h) + Uα,1(k, h)

],

k, h = 0, 1, · · · . (7.3.5)

substituting Eq. (7.3.4) in (7.3.5), yields the Uα,1(k, h) values and some values of

Uα,1(k, h) are listed in Table 7.1. Using Uα,1(k, h) values in (7.2.4), we obtained the

series solution as

u(x, t) =∞∑k=0

∞∑h=0

Uα,1(k, h)xktαh

= 1 +[x− x3

3!+x5

5!− x7

7!+ · · ·

]+[ tα

Γ(α+ 1)

+t2α

Γ(2α+ 1)+

t3α

Γ(3α + 1)+ · · ·

]+ · · · (7.3.6)

Case II: MFDTM

The modified transformed version of (7.3.1) with respect to the variable t is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(x, h+ 1)− ∂2Uα,1(x, h)

∂x2− Uα,1(x, h) = 0 (7.3.7)

The modified transformed version of (7.3.2) with respect to the variable t is

Uα,1(x, 0) = 1 + sin(x) (7.3.8)

103

Tab. 7.1: Some values of Uα,1(k, h) for Example 7.3.1

Uα,1(k, h)

k h = 0 h = 1 h = 2 h = 3 h = 4 h = 5

0 1 1Γ(α+1)

1Γ(2α+1)

1Γ(3α+1)

1Γ(4α+1)

1Γ(5α+1)

1 1 0 0 0 0 0

2 0 0 0 0 0 0

3 − 13!

0 0 0 0 0

4 0 0 0 0 0 0

5 15!

0 0 0 0 0

From Eq. (7.3.7), we get the following recurrence equation as

Uα,1(x, h+ 1) =Γ(αh+ 1)

Γ(α(h+ 1) + 1)

[∂2Uα,1(x, h)

∂x2+ Uα,1(x, h)

],

h = 0, 1, · · · (7.3.9)

substituting (7.3.8) into (7.3.9) gives Uα,1(k, h) values

Uα,1(x, 1) =1

Γ(α+ 1), Uα,1(x, 2) =

1

Γ(2α+ 1), Uα,1(x, 3) =

1

Γ(3α+ 1), · · ·

(7.3.10)

Using Uα,1(x, h) values in (7.2.8), we obtained the series solution as

u(x, t) =∞∑h=0

Uα,1(x, h)tαh

=1 + sin x+tα

Γ(α + 1)+

t2α

Γ(2α + 1)+

t3α

Γ(3α + 1)+ · · · (7.3.11)

In the limit of infinitely many terms, equations (7.3.6) and (7.3.11) yields the

104

(a) FDTM

(b) MFDTM

(c) Exact

Fig. 7.1: u(x, t) obtained by using (a) FDTM, (b) MFDTM and (c) Exact solutionwhen α = 1.5.

105

(a) FDTM

(b) MFDTM

(c) Exact

Fig. 7.2: u(x, t) obtained by using (a) FDTM, (b) MFDTM and (c) Exact solutionwhen α = 2.5.

106

(a) FDTM

(b) MFDTM

(c) Exact

Fig. 7.3: u(x, t) obtained by using (a) FDTM, (b) MFDTM and (c) Exact solutionwhen α = 3.15.

107

solution,

u(x, t) = sin x+ Eα (tα) (7.3.12)

where Eα (tα) is the Mittag -Leffler function defined as Mittag-Leffler (1904)

u(x, t) = 1 + sin x+∞∑h=1

(tα)h

Γ(hα+ 1)(7.3.13)

When α→ 2, the approximate solution (7.3.6) and (7.3.11) takes the following form

uapp(x, t) = 1 + sin x+t2

Γ(3)+

t4

Γ(5)+

t6

Γ(7)+ · · ·

= sin x+ 1 +t2

2!+t4

4!+t6

6!+ · · · (7.3.14)

Using Taylor series expansion near t = 0, we get

uex(x, t) = sinx+ cosh t (7.3.15)

which is exactly same as the solution obtained in Ravi Kanth and Aruna (2009).

The five term fractional differential transform method, modified fractional differen-

tial transform method and exact solution for the three different values fractional

order α are shown in Fig. 7.1- 7.3. It can be seen that the modified fractional dif-

ferential transform method results are far better approximations than the fractional

differential transform method results.

Example 7.3.2. Next, we consider the nonlinear fractional Klein-Gordon equation

∂αu(x, t)

∂tα− ∂2u(x, t)

∂x2+ u2(x, t) = 0 (7.3.16)

subject to the initial conditions

108

u(x, 0) = 1 + sin x (7.3.17)

Case I: FDTM

The transformed version of (7.3.16) is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(k, h+ 1)− (k + 1)(k + 2)Uα,1(k + 2, h)

+k∑

r=0

h∑s=0

Uα,1(r, h− s)Uα,1(k − r, s) = 0,

k, h = 0, 1, · · · . (7.3.18)

The transformed version of (7.3.17) is

Uα,1(k, 0) =

1, k = 0

1k!, k = 1, 5, 9 · · ·

− 1k!, k = 3, 7, 11 · · ·

(7.3.19)

We obtain the Uα,1(k, h) values by substituting (7.3.19) in (7.3.18)

Uα,1(0, 1) =1

Γ(α+ 1), Uα,1(0, 2) = 0,

Uα,1(1, 1) =3

Γ(α+ 1), Uα,1(1, 2) =

11

Γ(2α + 1),

Uα,1(2, 1) =1

Γ(α+ 1), Uα,1(2, 2) =

12

Γ(2α + 1),

Uα,1(3, 1) =1

2Γ(α + 1), Uα,1(3, 2) =

1

6Γ(2α + 1)

...... (7.3.20)

Using Uα,1(k, h) values in (7.2.4), we obtained the series solution as

109

u(x, t) =∞∑k=0

∞∑h=0

Uα,1(k, h)xktαh

= 1 + x− x3

3!+x5

5!− x7

7!+ · · ·

+

(1 + 3x+ x2 − x3

2− x4

3+ · · ·

)tα

Γ(α + 1)

+

(11x+ 12x2 +

x3

6− 4x4 + · · ·

)t2α

Γ(2α + 1)+ · · · (7.3.21)

Case II: MFDTM

The modified transformed version of (7.3.16) with respect to the variable t is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(x, h+ 1)− ∂2Uα,1(x, h)

∂x2+

h∑m=0

Uα,1(x,m)Uα,1(x, h−m) = 0,

h = 0, 1, · · · (7.3.22)

The modified transformed version of (7.3.17) with respect to the variable t is

Uα,1(x, 0) = 1 + sin x (7.3.23)

substituting (7.3.23) into (7.3.22) yields the values of Uα,1(x, h) as

Uα,1(x, 1) =1

2Γ(α + 1)(−3 + cos 2x− 6 sinx) ,

Uα,1(x, 2) =1

Γ(2α + 1)(12− 12 cos 2x+ 25 sin x− sin 3x) , · · ·

Using Uα,1 values in (7.2.8), we obtained the series solution

u(x, t) =∞∑h=0

Uα,1(x, h)tαh = 1 + sin x+ (−3 + cos 2x− 6 sin x)

tα

2Γ(α + a)+ · · ·

(7.3.24)

110

It is interesting to note that, the solution of (7.3.21) and (7.3.24) obtained above

are same as that of solution obtained by Golmankhaneh et al. (2011) using the

homotopy perturbation method.

Example 7.3.3. Finally, we consider the nonlinear fractional Klein-Gordon equa-

tion

∂αu(x, t)

∂tα− ∂2u(x, t)

∂x2− u+ u3(x, t) = 0 (7.3.25)

subject to the initial condition

u(x, 0) = − sechx (7.3.26)

Case I: FDTM

The transformed version of (7.3.25) is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(k, h+ 1)− (k + 1)(k + 2)Uα,1(k + 2, h)− Uα,1(k, h)

+k∑

r=0

k−r∑q=0

h∑s=0

h−s∑p=0

Uα,1(r, h− s− p)Uα,1(q, s)Uα,1(k − r − q, p) = 0,

k, h = 0, 1, · · · . (7.3.27)

The transformed version of (7.3.26) is

Uα,1(k, 0) =dk sechx

dxk(7.3.28)

We obtain the Uα,1(k, h) values by substituting (7.3.28) in (7.3.27) are

Uα,1(0, 0) = −1, Uα,1(0, 1) =1

Γ(α + 1), Uα,1(0, 2) = − 1

9Γ(2α + 1),

Uα,1(1, 0) = 0, Uα,1(1, 1) = 0, Uα,1(1, 2) = 0,

Uα,1(2, 0) =1

2, Uα,1(2, 1) = − 7

2Γ(α + 1), Uα,1(2, 2) =

109

2Γ(2α + 1),

111

Uα,1(3, 0) = 0, Uα,1(3, 1) = 0, Uα,1(3, 2) = 0,

Uα,1(4, 0) = − 5

24, Uα,1(4, 1) =

89

24Γ(α + 1), Uα,1(4, 2) =

−225

24Γ(2α + 1),

......

...

Using Uα,1(k, h) values in (7.2.4), we obtained the series solution

u(x, t) =∞∑k=0

∞∑h=0

Uα,1(k, h)xktαh

=− 1 +x2

2− 5x4

24+

61x6

720− · · ·+

(1 +

x2

2+

89x4

24+ · · ·

)tα

Γ(α+ 1)

+

(−9 +

109x2

2− 225x4

24+ · · ·

)t2α

Γ(2α+ 1)+ · · · (7.3.29)

Case II: MFDTM

The modified transformed version of (7.3.25) with respect to the variable t is

Γ(α(h+ 1) + 1)

Γ(αh+ 1)Uα,1(x, h+ 1)− ∂2Uα,1(x, h)

∂x2− Uα,1(x, h)

+h∑

m=0

m∑l=0

Uα,1(x,m)Uα,1(x, h−m) = 0,

h = 0, 1, · · · . (7.3.30)

The modified transformed version of (7.3.26) with respect to the variable t is

Uα,1(x, 0) = − sechx (7.3.31)

substituting (7.3.31) into (7.3.30) gives Uα,1(x, h) values are

Uα,1(x, 1) =1

Γ(α + 1)

(2− cosh 2x(sechx)2

),

Uα,1(x, 2) = − 1

2Γ(2α + 1)(53− 36 cosh 2x+ cosh 4x) (sechx)5, · · · (7.3.32)

112

Using Uα,1 values in (7.2.8), we obtained the series solution

u(x, t) =∞∑h=0

Uα,1(x, h)tαh

= − sechx+(2− cosh 2x(sechx)2

) tα

Γ(α + 1)+ · · · (7.3.33)

Five term solution of u(x, t) using fractional differential transform method and

modified fractional differential transform method corresponding to the values of

α = 1.5, 2.5, 3.15 are plotted in Fig 7.4- 7.6 respectively.

113

(a) FDTM

(b) MFDTM

Fig. 7.4: u(x, t) obtained by using (a) FDTM and (b) MFDTM when α = 1.5.

114

(a) FDTM

(b) MFDTM

Fig. 7.5: u(x, t) obtained by using (a) FDTM and (b) MFDTM when α = 2.5.

115

(a) FDTM

(b) MFDTM

Fig. 7.6: u(x, t) obtained by using (a) FDTM and (b) MFDTM when α = 3.15.