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EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 2, 2011, 174-185

ISSN 1307-5543 – www.ejpam.com

On the Solution of Fractional Order Nonlinear Boundary Value

Problems By Using Differential Transformation Method

Che Haziqah Che Hussin1 Adem Kılıçman2,∗

1 Department of Mathematics, Faculty of Science, Universiti Putra Malaysia, 43400 UPM, Serdang,

Selangor, Malaysia
2 Department of Mathematics and Institute for Mathematical Research, Faculty of Science, Universiti

Putra Malaysia, 43400 UPM, Serdang, Selangor, Malaysia

Abstract. In this research, we study about fractional order for nonlinear of fifth-order boundary value

problems and produce a theorem for higher order of fractional of nth-order boundary value problems.

The aim of this study was to evaluate and validate the theorem and provide several numerical exam-

ples to test the performance of our theorem. We also make comparison between exact solutions and

differential transformation method(DTM) by calculating the error between them. It is shown that DTM

has very small error and suitable in several numerical solutions since it is effective and provide high

accuracy.

2000 Mathematics Subject Classifications: 35C10,74S30, 65L10, 34B05, 34B15

Key Words and Phrases: Differential transformation method, Taylor series, fractional order of nonlin-

ear boundary value problems

1. Introduction

Recently, there has been an increasing interest in differential transformation method to

solve ordinary differential equations, partial differential equations as well as the integral

equations. For details, see [1, 3, 4, 6] and [7]. It is also known that the DTM concept

was introduced by Zhou in [14] in order to solve the related problems in electrical circuit

analysis for linear and nonlinear initial value problems. Since then DTM was applied to sev-

eral different problems in linear and nonlinear boundary value problems and also seems that

the method is easy to perform to solve problem numerically, for example, see [9, 8, 11] and

[12].

The DTM is approximation to exact solutions which are differentiable and it has very high

accuracy with minor error. This method is different with the traditional high order Taylor

∗Corresponding author.

Email addresses: akili
man�putra.upm.edu.my (A. Kılıçman), haziqah�math.upm.edu.my (C. Hussin)

http://www.ejpam.com 174 c© 2011 EJPAM All rights reserved.



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 175

series since high order of the Taylor series needs a long time in computation and it requires

the computation of the necessary derivatives [13]. The DTM can be applied in the high order

differential equations and it is an alternative way to get Taylor series solution for the given

differential equations. This method finally gives series solution but truncated series solution

in practice. In addition, the series of the method always coincides with the Taylor expan-

sion of true solution because it has very small error. Ayaz [2], was studied in application

of two-dimensional DTM in partial differential equations and Borhanifar and Abazari [5] was

also studied for two-dimensional and three-dimensional DTM in partial differential equations.

Recently, the authors studied the higher-order boundary value problems for higher-order

nonlinear differential equation and further made a comparison among differential transfor-

mation method and Adomian decomposition method, and exact solutions, see [10].

The basic definitions and operations of differential transformation method is discussed

in Section 2. The proposed theorems and methodology will be described in Section 3. In

Section 4, we provide several numerical examples to prove that the DTM has high accuracy.In

addition, result is displayed in Section 5 and finally the conclusion had made in Section 6.

We introduced new theorem and proved the theorem. This theorem is about fractional order

of nonlinear function. By using this theorem, we can solve the higher order of fractional

order for nonlinear function easily, more efficient and the result is more accurate because we

generate general form of high fractional order for nth order boundary value problems.

2. The Differential Transformation Method (DTM)

It is necessary here to clarify exactly what is meant by the differential transform of the

function y(x) for the kth derivative. It is defined like the following [9]:

Y (k) =
1

k!

�

dk y(x)

d x k

�

x=x0

(1)

where y(x) is the original function and Y (k) is the transformed function. The inverse differ-

ential transform of Y (k) is defined as

y(x) =

∞
∑

k=0





�

x − x0

�k

k!



Y (k) (2)

Substitute (1) into (2), we will get

y(x) =

∞
∑

k=0

�

x − x0

�k 1

k!

�

dk y(x)

d x k

�

x=x0

(3)

which is the Taylor’s series for y(x) at x = x0.

The following theorems are easy to prove and considered the fundamental operations of

differential transforms Method (DTM).



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 176

Theorem 1. If t(x) = r(x)± p(x) then T (k) = R(k)± P(k).

Theorem 2. If t(x) = αr(x) then, T (k) = αR(k).

Theorem 3. If t(x) =
dr(x)

d x
then, T (k) = (k+ 1)R (k+ 1).

Theorem 4. If t(x) =
d2r(x)

d x2 then, T (k) = (k+ 1) (k+ 2)R (k+ 2).

Theorem 5. If t(x) =
d b r(x)

d x b then, T (k) = (k+ 1) (k+ 2) . . . (k+ b)R (k+ b).

Theorem 6. If t(x) = r(x)p(x) then T (k) =
∑k

l=0 P (l)R (k− l).

Theorem 7. If t(x) = x b then T (k) = δ (k− b) where, δ (k− b) =

¨

1 if k = b

0 if k 6= b

Theorem 8. If t(x) = ex p (λx) then, T (k) = λ
k

k!

Theorem 9. If t(x) = (1+ x)b then, T (k) =
b(b−1)...(b−k+1)

k!
.

Theorem 10. If t(x) = sin
�

j x +α
�

then, T (k) =
jk

k!
sin
�

πk

2
+α
�

.

Theorem 11. If t(x) = cos
�

j x +α
�

then, T (k) =
jk

k!
cos
�

πk

2
+α
�

.

2.1. Two-dimensional DTM

We note that the differential transform methods can easily be extended to the multiple

dimensional cases, For example if we take a function with two variables, for instance y(x , t)

having a transform Y (k, j) then two-dimensional differential transformation method can be

applied several partial differential equations. Thus the two dimensional form of the differen-

tial transform methods is defined as the following:

Y (k, j) =
1

k! j!

�

∂ k+ j

∂ x k∂ y j
y(x , t)

�

x=0
y=0

(4)

Differential equation in form of y(x , t) is like the following:

y(x , t) =

∞
∑

k=0

∞
∑

j=0

Y (k, j)x k y j . (5)

From Eq. (4) and (5) we can demonstrate as follows:

Y (k, j) =

∞
∑

k=0

∞
∑

j=0

1

k! j!

�

∂ k+ j

∂ x k∂ y j
y(x , t)

�

x=0
y=0

. (6)

It is clear that Eq (6) implies the two dimensional of Taylor series expansion. One easily

deduce several similar results as in the Section 1.



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 177

3. General Solution for nth-order Boundary Value Problems for mth-order

Nonlinear Functions

Consider the following problem. If y(x) is transformable then on using the equation (1)

we consider the solution to the high order differential equation y(n)(x) = e−x y
1

m (x). Now we

can consider several cases as follows:

If n= 5 and p = 1

m
= 1

2
then on using the equation (1) one can easily prove that,

Y (k+ 5) =
k!

(k+ 5)!













1
�

−1

2

�5







�

(−1)k

k!

�

k!Y (k)







k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!













1
�

1

2

�5







�

(−1)k

k!

�

k!Y (k)







k=1,3,5,...

.

Now, if p = 1

3
then,

Y (k+ 5) =
k!

(k+ 5)!













1
�

−2

3

�5







�

(−1)k

k!

�

k!Y (k)







k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!













1
�

2

3

�5







�

(−1)k

k!

�

k!Y (k)







k=1,3,5,...

Similarly, if p = 1

m
then,

Y (k+ 5) =
k!

(k+ 5)!





 

1

( 1

m
− 1)5

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!





 

1

(1− 1

m
)5

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

If p = 1

m+1
then,

Y (k+ 5) =
k!

(k+ 5)!





 

1

( 1

m+1
− 1)5

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!





 

1

(1− 1

m+1
)5

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 178

Thus, if n= 1 and p = 1

m
then,

Y (k+ 1) =
k!

(k+ 1)!





 

1

( 1

m
− 1)

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ 1) =
k!

(k+ 1)!





 

1

(1− 1

m
)

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

If n= 2 then,

Y (k+ 2) =
k!

(k+ 2)!





 

1

( 1

m
− 1)2

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ 2) =
k!

(k+ 2)!





 

1

(1− 1

m
)2

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

If n= q then,

Y (k+ q) =
k!

(k+ q)!





 

1

( 1

m
− 1)q

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ q) =
k!

(k+ q)!





 

1

(1− 1

m
)q

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

If n= q+ 1 then,

Y (k+ q+ 1) =
k!

(k+ q+ 1)!





 

1

( 1

m
− 1)q+1

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

and

Y (k+ q+ 1) =
k!

(k+ q+ 1)!





 

1

(1− 1

m
)q+1

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

The above calculations suggest that the general case by using the induction method can be

proved and is given as the following theorem.

Theorem 12. Let y(x) is transformable then solution to the high order differential equation

y(n)(x) = e−x y
1

m (x) is given by

Y (k+ n) =
k!

(k+ n)!





 

1

( 1

m
− 1)n

!
�

(−1)k

k!

�

k!Y (k)





k=0,2,4,...

Y (k+ n) =
k!

(k+ n)!





 

1

(1− 1

m
)n

!
�

(−1)k

k!

�

k!Y (k)





k=1,3,5,...

.

In this research, we study nth-order boundary value problems having fractional-order

with nonlinear functions. We also compute error between exact solution and differential

transformation method (DTM).



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 179

4. Numerical Examples

In this section, we provide three examples to make better understanding to the theorem.

Example 1

First, we take the case of fractional in form 1

2
order for fifth order BVP.

y(5) = e−x
p

y (x) 0< x < 1 (7)

subject to the boundary conditions

y(0) = 1, y ′(0) = −2, y ′′(0) = 4, y(1) = e−2, y ′(1) = −2e−2. (8)

Applying Eq. (8) to Eq. (1) at x = 0, the following transformed boundary conditions can be

obtained

Y (0) = 1, Y (1) = −2, Y (2) = 2. (9)

By using differential transform properties in theorem (13) to Eq. (7) then the transformed

equation is given by

Y (k+ 5) =
k!

(k+ 5)!













1
�

−1

2

�5







�

(−1)k

k!

�

k!Y (k)







k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!













1
�

1

2

�5







�

(−1)k

k!

�

k!Y (k)







k=1,3,5,...

(10)

which is based on Eq. (1), t =
y′′′(0)

3!
= Y (3) and w =

y(4)(0)

4!
= Y (4). Using the transformed

boundary conditions in Eq. (9) and transformed equation in Eq. (10), we can get the solution

for Y (k), k ≥ 5 easily. The values of t and w can be evaluated by using boundary conditions

in Eq. (8) at x = 1 for N = 20 by solving two equations such as:

21
∑

k=0

Y (k) = e−2 and

21
∑

k=0

kY (k) = −2e−2.

These two equations give t = −1.333333333 and w = 0.6666666668.

As a result the following series solution can be formed by applying the inverse transfor-

mation equation in Eq. (2) up to N = 20.

y(x) =1.0− 2.0x + 2.0x2− 1.333333333x3+ 0.6666666668x4− 0.2666666667x5

+ 0.08888888889x6− 0.02539682540x7+ 0.006349206348x8

− 0.001410934745x9+ 0.0002821869489x10− 0.00005130671797x11



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 180

+ 0.000008551119662x12− 0.000001315556871x13+ 0.0000001879366960x14

− 0.00000002505822612x15+ 0.000000003132278265x16

− 0.0000000003685033252x17+ 4.094481391× 10−11 x18

− 4.309980415× 10−12 x19 + 4.309980412× 10−13 x20.

Example 2

Then, for Example 2 we consider fractional in form 1

3
order for fifth order boundary value

problems. For instance:

y(5)(x) = ex 3
p

(y(x)) 0< x < 1 (11)

subject to the boundary conditions

y(0) = 1, y ′(0) = −
3

2
, y ′(0) =

9

4
, y(1) = e−

3

2 , y ′(1) = (−3/2)e−
3

2 . (12)

Applying Eq. (12) to Eq. (1) at x = 0, the following transformed boundary conditions can be

found

Y (0) = 1, Y (1) =
−3

2
, Y (2) =

9

8
. (13)

Next, by using differential transform properties in theorem (13) to Eq. (11) then the trans-

formed equation is given by

Y (k+ 5) =
k!

(k+ 5)!













1
�

−2

3

�5







�

(−1)k

k!

�

k!Y (k)







k=0,2,4,...

and

Y (k+ 5) =
k!

(k+ 5)!













1
�

2

3

�5







�

(−1)k

k!

�

k!Y (k)







k=1,3,5,...

(14)

which is based on Eq. (1), t =
y′′′(0)

3!
= Y (3) and w =

y(4)(0)

4!
= Y (4). On using the transformed

boundary conditions in Eq. (13) and transformed equation in Eq. (14), we can get the solution

for Y (k), k ≥ 5 easily. The values of t and w can be evaluated by using boundary conditions

in Eq. (12) at x = 1 for N = 20 by solving two equations such as:

20
∑

k=0

Y (k) = e−
3

2 and

20
∑

k=0

kY (k) = (−3/2)e−
3

2 .

These two equations give values for t = −0.5625000003 and w = 0.2109375003.

As a result the following series solution can be formed by applying the inverse transfor-

mation equation in Eq. (2) up to N = 20.

y(x) =1.0− 1.500000000x+ 1.125000000x2− 0.5625000003x3



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 181

+ 0.2109375003x4− 0.06328125000x5+ 0.01582031250x6

− 0.003390066964x7+ 0.0006356375561x8− 0.0001059395928x9

+ 0.00001589093890x10− 0.000002166946213x11

+ 0.0000002708682766x12− 0.00000003125403193x13

+ 0.000000003348646282x14− 0.0000000003348646277x15

+ 3.139355885× 10−11 x16 − 2.770019898× 10−12 x17

+ 2.308349916× 10−13 x18 − 1.822381515× 10−14 x19

+ 1.366786134× 10−15 x20.

Example 3

Finally we perform fractional-order in form 1

4
for sixth-order boundary value problems

such as:

y(6)(x) = ex 4
p

(y(x)) 0< x < 1 (15)

subject to the boundary conditions

y(0) = 1, y ′(0) =
−4

3
, y ′′(0) =

16

9
, y(1) = e−

4

3 , y ′(1) =

�

4

3

�

e−
4

3

y ′′(1) =

�

16

9

�

e−
4

3 .

(16)

Applying Eq. (16) to Eq. (1) at x = 0, the following transformed boundary conditions can be

found

Y (0) = 1, Y (1) =
−4

3
, Y (2) =

8

9
. (17)

Then, by using differential transform properties in theorem (13) to Eq. (15) then the trans-

formed equation is given by

Y (k+ 6) =
k!

(k+ 6)!

















1
�

−3

4

�6









�

(−1)k

k!

�

k!Y (k)









k=0,2,4,...

and (18)

Y (k+ 6) =
k!

(k+ 6)!

















1
�

3

4

�6









�

(−1)k

k!

�

k!Y (k)









k=1,3,5,...

(19)

which is based on Eq.(1), t =
y′′′(0)

3!
= Y (3), w =

y(4)(0)

4!
= Y (4) and z =

y(5)(0)

5!
= Y (5). Using

the transformed boundary conditions in Eq. (17) and transformed equation in Eq. (18), we



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 182

can get the solution for Y (k), k ≥ 6 easily. The values of t,w and z can be evaluated by using

boundary conditions in Eq. (16) at x = 1 for N = 20 by solving three equations such as:

20
∑

k=0

Y (k) = e−
4

3 ,

20
∑

k=0

kY (k) =

�

4

3

�

e−
4

3 and

20
∑

k=0

k(k− 1)Y (k) =

�

16

9

�

e−
4

3 . (20)

These three equations give values for t = −0.4047692491, 0.1573589866 and

z = −0.5413042140e− 1. Consequently the following series solution can be formed by ap-

plying the inverse transformation equation in Eq. (2) up to N = 20.

y(x) =1.0− 1.333333333 x+ 0.8888888889 x2− 0.4047692491 x3+

0.1573589866 x4− 0.05413042140 x5+ 0.007803688462 x6+

0.001486416850 x7+ 0.0002477361417 x8+ 0.00003760348951 x9+

0.000005847526229 x10+ 0.0000009143223916 x11+

0.00000006590644330 x12− 0.000000006759635211 x13+

0.0000000006437747820 x14− 5.863055289× 10−11 x15+

5.698335785× 10−12 x16 − 5.765261068× 10−13 x17+

2.770487778× 10−14 x18 + 1.944201950× 10−15 x19+

1.296134633× 10−16 x20.

5. Result

There are numerical results for differential transformation method and comparison to

exact solution of fifth-order BVPs for fractional order 1

2
and 1

3
. We also provided sixth-order

boundary value problems for fractional-order of degree 1

4
. They are in Table 1, Table 2 and

Table 3 respectively. Table 1: Comparison numeri
al result for Example 1.
x Exact solution DTM (N=21) Error

0.0 1 1 0

0.1 0.8187307532 0.8187307532 0

0.2 0.6703200461 0.6703200461 0

0.3 0.5488116361 0.5488116361 0

0.4 0.4493289640 0.4493289640 0

0.5 0.3678794413 0.3678794413 0

0.6 0.3011942120 0.3011942120 0

0.7 0.2465969642 0.2465969642 0

0.8 0.2018965183 0.2018965183 0

0.9 0.1652988883 0.1652988884 0.1× 10−9

1.0 0.1353352833 0.1353352834 0.1× 10−9



C. Hussin, A. Kılıçman / Eur. J. Pure Appl. Math, 4 (2011), 174-185 183Table 2: Comparison numeri
al result for Example 2.
x Exact solution DTM (N=21) Error

0.0 1 1 0

0.1 0.8607079765 0.8607079765 0

0.2 0.7408182206 0.7408182206 0

0.3 0.6376281517 0.6376281517 0

0.4 0.5488116361 0.5488116361 0

0.5 0.4723665528 0.4723665528 0

0.6 0.4065696598 0.4065696597 0.1× 10−9

0.7 0.3499377491 0.3499377490 0.1× 10−9

0.8 0.3011942119 0.3011942118 0.1× 10−9

0.9 0.2592402607 0.2592402606 0.1× 10−9

1.0 0.2231301601 0.2231301601 0Table 3: Comparison numeri
al result for Example 3.
x Exact solution DTM (N=21) Error

0.0 1 1 0

0.1 0.8751733191 0.8751659889 0.0000073302

0.2 0.7659283385 0.7658857067 0.426318× 10−4

0.3 0.6703200462 0.6702203323 0.997139× 10−4

0.4 0.5866462197 0.5864923238 0.0001538959

0.5 0.5134171191 0.5132373530 0.1797661× 10−3

0.6 0.4493289642 0.4491646634 0.0001643008

0.7 0.3932407212 0.3931270548 0.0001136664

0.8 0.3441537873 0.3441018869 0.519004× 10−4

0.9 0.3011942119 0.3011846781 0.0000095338

1.0 0.2635971382 0.2635971384 0.2× 10−9

From the tables, we can see that Differential Transformation Method has minor error. By

the way, for Table 1 it has no error from points 0.0 to 0.8. For points 0.9 and 1.0, the error

is 0.1e− 9. The maximum error for Example 1 is
∑1

i=0 x i = 2.0× 10−10. That means DTM

is very accurate with exact solution for fractional order 1

2
of fifth-order BVP. By comparison

to fractional order 1

3
, the maximum error is

∑1

i=0 x i = 4.0× 10−10. From point x = 0.0 to

x = 0.5 it has no error but at points x = 0.6 to 0.9, the error is 1× 10−10 and it declines to 0

for point 1.0 while for sixth-order BVPs of 1

4
fractional order function, the maximum error is

1
∑

i=0

x i = 0.9364059e− 3. From Table 3, we can see it has no error only for point x = 0.0.



REFERENCES 184

6. Conclusion

The results of this research support the idea that DTM has high degree of accuracy in

numerical solution. We introduced general form of nth-order BVPs for fractional order func-

tion. This research will serve as a base for future studies in nonlinear function especially in

fractional order function. Therefore, we proved the DTM method very successful and power-

ful in numerical solution for the bounded domains. The computations in the examples were

computed by using Maple 9.

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