4_xxx_ibis.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 2, 2011, 129-141

ISSN 1307-5543 – www.ejpam.com

Applications of Fractional Differential Transform Method to

Fractional Differential-Algebraic Equations

Birol İbi̧s1,∗, Mustafa Bayram2, A. Göksel Ağargün3

1 Department of Main Sciences, Turkish Air Force Academy, Istanbul,Turkey
2 Mathematical Engineering, Faculty of Chemical and Metallurgical Engineering, Yıldız Technical

University, Istanbul,Turkey
3 Department of Mathematics, Faculty of Art and Sciences, Yıldız Technical University, Istanbul,Turkey

Abstract. In this paper, we implement fractional differential transform method (FDTM), which is a

semi analytical numerical technique, to fractional differential-algebraic equations (FDAEs). The frac-

tional derivatives are described in the Caputo sense. The method provides the solution in the form of

a rapidly convergent series. The method is illustrated by four examples of FDAEs and solutions are ob-

tained. Comparisons are made between fractional differential transform method (FDTM), Homotopy

Analysis Method (HAM) and the exact solutions. The results reveal that the proposed method is very

effective and simple.

2000 Mathematics Subject Classifications: 4A08,34K28,34B05,34B15,65L10,74S30

Key Words and Phrases: Fractional Differential Transform Method (FDTM), Fractional Differential-

Algebraic Equations (FDAEs), Caputo , Homotopy Analysis Method (HAM)

1. Introduction

Fractional differential equations (FDEs) have been succesfully modelled for many physical

and engineering phenomena such as seismic analysis, rheology, fluid flow, viscous damping,

viscoelastic materials and polymer physics [9, 2, 31, 16, 32, 3]. Most nonlinear FDEs don’t

have exact analytic solutions, therefore approximation and numerical techniques must be

used. Some of the recent analytic methods for solving nonlinear problems include the Ado-

mian decomposition method (ADM) [12, 11, 30, 38, 35, 14], variational iteration method

(VIM) [18, 19, 42, 36, 33], homotopy analysis method (HAM) [34, 13, 40, 37, 27] and frac-

tional method [17]. Among these solution techniques, the VIM and the ADM are the most

clear methods of solution of FDEs for providing instant and visible symbolic terms of ana-

lytic solutions, as well as numerical approximate solutions to nonlinear differential equations

without linearization or discretization.

∗Corresponding author.

Email address: bibis�hho.edu.tr (B. İbi̧s)

http://www.ejpam.com 129 c© 2011 EJPAM All rights reserved.



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 130

Many physical problems are governed by a system of differential-algebraic equations

(DAEs), and the solution of these equations has been a subject of many investigators in recent

years. Although many exact solutions for linear DAEs has been found, in general, there exists

no method that yields an exact solution for nonlinear DAEs. Numerical approaches for ap-

proximating solutions of DAEs have been presented [39, 29, 28, 6, 5, 4, 8, 24, 25, 7, 10, 15].

Recently, many important mathematical models can be expressed in terms of differential-

algebraic equations of fractional order. Homotopy analysis method was first introduced by

Liao [34], who employed the basic ideas of the homotopy in topology to propose a general

analytic method for nonlinear problems. Zurigat, Momani and Alawneh [26] applied this

method for fractional differential-algebraic equations (FDAEs).

The differential transform method (DTM) was first applied in the engineering domain in

[20]. The DTM is numerical method based on the Taylor series expansion which constructs

an analytical solution in the form of a polynomial. The traditional high order Taylor series

method requires symbolic computation. However, the DTM obtains a polynomial series so-

lution by means of an iterative procedure. Arikoglu and Ozkol implement a new analytical

technique for the field of fractional calculus, for solving fractional type differential equations

that will be named as Fractional Differential Transform Method (FDTM)[1]. In this paper,

fractional differential transform method (FDTM) is applied to solve fractional differential-

algebraic equations (FDAEs) of form

D
αi
∗ x i(t) = f (t, x1, x2, . . . , xn, x ′1, x ′2, . . . , x ′n), i = 1,2,3, . . . , n− 1, t ≥ 0, 0< αi ≤ 1 (1)

g(t, x1, x2, . . . , xn) = 0 (2)

subject to the initial conditions

x i(0) = ai , i = 1,2, . . . , n (3)

2. Basic Definitions

There are several definitions of a fractional derivative of order α > 0 [17, 22].e.g.Riemann-

Liouville, Grunwald-Letnikow, Caputo and Generalized Functions Approach. The most com-

monly used definitions are the Riemann-Liouville and Caputo. We give some basic definitions

and properties of the fractional calculus theory which are used further in this paper.

Definition 1. A real function f (x), x > 0, is said to be in the space Cµ, µ ∈ R if there exists

a real number p > µ such that f (x) = x p f1(x), where f1(x) ∈ C[0,∞). Clearly Cµ < Cβ if

β < µ.

Definition 2. A function f (x), x > 0, is said to be in the space Cm
µ , m ∈ N ∪ {0} if f (m) ∈ Cµ.

Definition 3. The Riemann-Liouville fractional integral operator of order a α≥ 0 of a function,

f ∈ Cµ,µ ≥ −1 is defined as [31]

Jα f (x) =
1

Γ(α)

∫ x

0

(x − t)α−1 f (t)d t, α > 0, x > 0 (4)

J0 f (x) = f (x) (5)



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 131

Properties of the operator Jα can be found in [41, 23, 21], we mention only the following:

For f ∈ Cµ, µ ≥ −1, α,β ≥ 0 and γ > −1

JαJβ f (x) = Jα+β f (x) (6)

JαJβ f (x) = Jβ Jα f (x) (7)

Jαxγ =
Γ(γ+ 1)

Γ(α+ γ+ 1)
xα+γ (8)

The Riemann-Liouville derivative has certain disadvantages when trying to model real-world

phenomena using fractional differential equations. Therefore, we will introduce a modified

fractional differential operator proposed by Caputo’s work on the theory of viscoelasticity

[22].

Definition 4. The fractional derivative of f (x) in the Caputo sense is defined as

Dα∗ f (x) = J m−αDm f (x) =
1

Γ(m−α)

∫ x

0

(x − t)m−α−1 f (m)(t)d t, (9)

for m− 1< α≤ m, m ∈ N, x > 0, f ∈ Cm
−1.

Also, we need here two of its basic properties.

Lemma 1. If m− 1< α ≤ m, m ∈ N and f ∈ Cm
µ , m ≥ −1, then

Dα∗ J
α f (x) = f (x) (10)

JαDα∗ f (x) = f (x)−
m−1
∑

k=0

f (k)(0+)
x k

k!
, x > 0 (11)

3. Fractional Differential Transform Method (FDTM)

In this section, we introduce the fractional differential transform method used in this

paper to obtain approximate analytical solutions for FDAEs in Eq.(1). This method has been

developed in [1] as follows:

The fractional differentiation in Riemann-Liouville sense is defined by

Dα f (x) =
1

Γ(m−α)
Dm

�∫ x

0

(x − t)m−α−1 f (t)d t

�

(12)

for m− 1≤ α < m, m ∈ N , x > 0. Let us expand the analytical and continuous function f (x)

in terms of a fractional power series as follows:

f (x) =

∞
∑

k=0

F(k)x
k

β (13)



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 132

where β is the order of fraction and F(k) is the fractional differential transform of f (x) .

In order to avoid fractional initial and boundary conditions, we define the fractional deriva-

tive in the Caputo sense. The relation between the Riemann-Liouville operator and Caputo

operator is given by

Dα∗ f (x) = Dα

 

f (x)−
m−1
∑

k=0

f (k)(0+)
x k

k!

!

(14)

Setting f (x) = f (x)−
∑m−1

k=0 f (k)(0+) xk

k!
in Eq.(12)and using Eq.(14), we obtain fractional

derivative in the Caputo sense as follows:

Dα∗ f (x) =
1

Γ(m−α)
Dm





∫ x

0

(x − t)m−α−1

 

f (t)−
m−1
∑

k=0

f (k)(0+)
tk

k!

!

d t



 (15)

Since the initial conditions are implemented for the integer order derivatives, the transforma-

tion of the initial conditions are defined as follows:

F(k) =

(

1

(k/β)!

h

dk/β f (x)

d xk/β

i

x=0
for k = 0,1,2, . . . , (αβ − 1) , k/β ∈ N+

0 , k/β /∈ N+
(16)

where, α is the order of fractional differential equation considered. The following theorems

that can be deduced from Eqs.(12) and (13) are given below,for proofs and details see [1].

Theorem 1. if f (x) = g(x)± h(x), then F(k) = G(k)±H(k)

Theorem 2. if f (x) = g(x)h(x), then F(k) =
∑k

l=0 G(l)H(k− l)

Theorem 3. if f (x) = g1(x)g2(x) · · · gn−1(x)gn(x), then

F(k) =

k
∑

kn−1=0

kn−1
∑

kn−2=0

· · ·
k3
∑

k2=0

k2
∑

k1=0

G1(k1)G2(k2− k1) · · ·Gn−1(kn−1 − kn−2)Gn(k− kn−1) (17)

Theorem 4. if f (x) = x p, then F(k) = δ(k− βp)

where, δ(k) =

¨

1 if k = 0

0 if k 6= 0
(18)

Theorem 5. if f (x) = Dαg(x), then F(k) =
Γ(α+1+k/β)

Γ(1+k/β)
G(k+αβ)

Theorem 6. if f (x) = Dα1 g1(x)D
α2 g2(x) · · ·D

αn−1 gn−1(x)D
αn gn(x), then

F(k) =

k
∑

kn−1=0

kn−1
∑

kn−2=0

· · ·
k3
∑

k2=0

k2
∑

k1=0

Γ(α1 + 1+ k1/β)

Γ(1+ k1/β)

Γ(α2 + 1+ (k2− k1)/β)

Γ(1+ (k2− k1)/β)
· · ·

×
Γ(αn+ 1+ (k− kn−1/β)

Γ(1+ (k− kn−1/β)
G1(k1 +α1β)G2(k2− k1 +α2β) · · ·Gn(k− kn−1 +αnβ)

where βαi ∈ Z+ for i = 1,2, . . . , n.



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 133

4. Numerical Examples

In order to demonstrate the effectiveness of the fractional differential transform method,

we consider the following FDAEs. All the results are calculated by using the symbolic calculus

software Maple.

Example 1. We consider the following fractional differential-algebraic equations.

Dα∗ x(t)− t y ′(t) + x(t)− (1+ t)y(t) = 0, 0< α≤ 1 (19)

y(t)− sin(t) = 0 (20)

with initial conditions as

x(0) = 1, y(0) = 0 (21)

For the special case when α = 1 the exact solution is x(t) = e−t + tsin(t), y(t) = sin(t).

Eqs.(19)-(20) are transformed by using Theorems 1, 2, 4, 5 and Eq.(13) as follows:

X (k+αβ) =
Γ(1+ k/β)

Γ(α+ 1+ k/β)





k
∑

l=0

�

Γ(2+ l/β)

Γ(1+ l/β)
Y (l + β) + Y (l)

�

δ(k− l − β)− X (k)+ Y (k)





(22)

Y (k) =

∞
∑

i=0

(−1)i

(2i+ 1)!
δ(k− β(2i+ 1)) (23)

where β is the unknown value of the fractions. Initial conditions in Eq.(21) are transformed

by using Eq.(16) as follows:

X (0) = 1, Y (0) = 0, X (k) = Y (k) = 0 for k = 1,2, . . . ,αβ − 1 (24)

From Eqs.(22)-(24), X (k) and Y (k) are obtained for different values of α and using the inverse

transformation in Eq.(13), x(t) and y(t) are evaluated.Numerical results with comparison to

Ref. [26] is given in Table 1.

Example 2. Consider the following fractional differential-algebraic equations.

D
α1
∗ x(t)− x(t)− z(t)x(t) = 1 (25)

D
α2
∗ z(t)− y(t) + x2(t) + z(t) = 0, 0< α1,α2 ≤ 1 (26)

y(t)− x2(t) = 0 (27)

with initial conditions as

x(0) = y(0) = z(0) = 1 (28)

For α1 = α2 = 1 the exact solution is x(t) = et , y(t) = e2t , z(t) = e−t . By using Theorems

1, 2, 4, 5 and Eq.(13), Eqs.(25)-(27) are transformed to,

X (k+α1β1) =
Γ(1+ k/β1)

Γ(α1 + 1+ k/β1)



X (k)−
k
∑

l=0

Z(l)X (k− l) + δ(k)



 (29)



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 134Table 1: Numeri
al results with 
omparison to Ref. [26℄ in Example 1
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 0.7642925 0.7642925 0.8492995 0.8492996 0.9148208 0.9148208 0.9148208

0.2 0.7545097 0.7545096 0.8016696 0.8016697 0.8584646 0.8584646 0.8584646

0.3 0.7903162 0.7903162 0.7979000 0.7978999 0.8294743 0.8294743 0.8294743

0.4 0.8524950 0.8524950 0.8250873 0.8250871 0.8260874 0.8260874 0.8260874

0.5 0.9323247 0.9323247 0.8760146 0.8760144 0.8462434 0.8462434 0.8462434

0.6 1.0242052 1.0242052 0.9454582 0.9454582 0.8875971 0.8875971 0.8875971

0.7 1.1237906 1.1237906 1.0290755 1.0290757 0.9475377 0.9475377 0.9475377

0.8 1.2273291 1.2273291 1.1229592 1.1229595 1.0232138 1.0232138 1.0232138

0.9 1.3313916 1.3313915 1.2234363 1.2234368 1.1115639 1.1115639 1.1115639

1.0 1.4327552 1.4327552 1.3269757 1.3269767 1.2093505 1.2093504 1.2093504

Z(k+α2β2) =
Γ(1+ k/β2)

Γ(α2 + 1+ k/β2)



Y (k)−
k
∑

l=0

X (l)X (k− l)− Z(k)



 (30)

Y (k) =

k
∑

l=0

X (l)X (k− l) (31)

where β1 and β2 are the unknown values of the fractions and β = LC M(β1,β2). From

Eq.(16), initial conditions in Eq.(28) can be transformed as follows:

X (0) = Z(0) = 1, X (k) = 0, k = 1,2, . . . ,α1β1 − 1, Z(k) = 0, k = 1,2, . . . ,α2β2− 1 (32)

From Eqs.(29)-(32), X (k), Y (k) and Z(k) are calculated and using the inverse transformation

rule in Eq.(13), x(t),y(t) and z(t) are calculated for different values of α1 and α2. Numerical

comparisons are given in Table 2-3-4.Table 2: Numeri
al results of x(t) with 
omparison to HAM in Example 2
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 1.4678849 1.4678849 1.2187069 1.2187069 1.1051709 1.1051709 1.1051709

0.2 1.7411322 1.7411322 1.4000280 1.4000280 1.2214028 1.2214028 1.2214028

0.3 1.9927891 1.9927891 1.5841270 1.5841270 1.3498588 1.3498588 1.3498588

0.4 2.2392557 2.2392557 1.7769089 1.7769089 1.4918247 1.4918247 1.4918247

0.5 2.4871415 2.4871415 1.9813870 1.9813870 1.6487213 1.6487213 1.6487213

0.6 2.7401183 2.7401183 2.1997453 2.1997453 1.8221188 1.8221188 1.8221188

0.7 3.0006469 3.0006469 2.4338838 2.4338838 2.0137527 2.0137527 2.0137527

0.8 3.2706054 3.2706054 2.6856249 2.6856249 2.2255409 2.2255409 2.2255409

0.9 3.5515666 3.5515666 2.9568131 2.9568125 2.4596031 2.4596031 2.4596031

1.0 3.8450351 3.8450346 3.2493750 3.2493684 2.7182818 2.7182818 2.7182818



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 135Table 3: Numeri
al results of y(t) with 
omparison to HAM in Example 2
α = 0.5 α = 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 2.1546862 2.1546862 1.4852465 1.4852465 1.2214028 1.2214028 1.2214028

0.2 3.0315412 3.0315412 1.9600784 1.9600784 1.4918247 1.4918247 1.4918247

0.3 3.9712084 3.9712084 2.5094586 2.5094586 1.8221188 1.8221188 1.8221188

0.4 5.0142660 5.0142660 3.1574052 3.1574052 2.2255409 2.2255409 2.2255409

0.5 6.1858731 6.1858732 3.9258942 3.9258942 2.7182815 2.7182818 2.7182815

0.6 7.5082482 7.5082482 4.8388794 4.8388794 3.3201150 3.3201169 3.3201169

0.7 9.0038821 9.0038821 5.9237902 5.9237903 4.0551908 4.0552000 4.0552000

0.8 10.6968505 10.696866 7.2125789 7.2125809 4.9529970 4.9530324 4.9530324

0.9 12.6037326 12.603625 8.7427133 8.7427398 6.0495302 6.0496475 6.0496475

1.0 14.7830711 14.784077 10.558121 10.558395 7.3887125 7.3890561 7.3890561Table 4: Numeri
al results of z(t) with 
omparison to HAM in Example 2
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 0.7235784 0.7235784 0.8282505 0.8282505 0.9048374 0.9048374 0.9048374

0.2 0.6437883 0.6437883 0.7325847 0.7325847 0.8187308 0.8187308 0.8187308

0.3 0.5920184 0.5920184 0.6603375 0.6603375 0.7408182 0.7408182 0.7408182

0.4 0.5536063 0.5536063 0.6021211 0.6021211 0.6703201 0.6703201 0.6703201

0.5 0.5231566 0.5231566 0.5536026 0.5536026 0.6065307 0.6065307 0.6065307

0.6 0.4980246 0.4980246 0.5122851 0.5122851 0.5488116 0.5488116 0.5488116

0.7 0.4767027 0.4767027 0.4765549 0.4765549 0.4965853 0.4965853 0.4965853

0.8 0.4582460 0.4582460 0.4452924 0.4452924 0.4493290 0.4493290 0.4493290

0.9 0.4420214 0.4420214 0.4176820 0.4176821 0.4065697 0.4065697 0.4065697

1.0 0.4275836 0.4275836 0.3931083 0.3931083 0.3678795 0.3678794 0.3678794

Example 3. Consider the following fractional differential- algebraic equations.

x(t) + y(t) = e−t + sin(t) (33)

Dα∗ x(t) + x(t)− y(t) + sin(t) = 0, 0< α ≤ 1 (34)

with initial conditions as

x(0) = 1, y(0) = 0 (35)

For the special case when α = 1 the exact solution is x(t) = e−t , y(t) = sin(t). Eqs.(33)-

(34) are transformed by using Theorems 1, 4 and 5 as follows:

Y (k) = −X (k) + E(k) + S(k) (36)

X (k+αβ) =
Γ(1+ k/β)

Γ(α+ 1+ k/β)
[−X (k)+ Y (k)− S(k)] (37)



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 136

where β is the unknown value of the fractions, E(k) and S(k) are the fractional differential

transform of x(t) = e−t and y(t) = sin(t) that can be evaluated using Eq.(13) as

E(k) =

(

(−1)k/β

(k/β)!
if k/β ∈ Z+

0 if k/β /∈ Z+
(38)

S(k) =

∞
∑

i=0

(−1)i

(2i+ 1)!
δ(k− β(2i+ 1)) (39)

From Eq.(16), initial conditions in Eq.(35) can be transformed as follows:

X (0) = 1, Y (0) = 0, X (k) = Y (k) = 0, for k = 1,2, . . . ,αβ − 1 (40)

From Eqs.(38)-(40), X (k) and Y (k) are calculated and using the inverse transformation rule

in Eq.(13), x(t) and y(t) are calculated for different values of. Numerical comparisons are

given in Table 5-6.Table 5: Numeri
al results of x(t) with 
omparison to HAM in Example 3
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 0.7608910 0.7608910 0.8373931 0.8373931 0.9048374 0.9048374 0.9048374

0.2 0.6909262 0.6909262 0.7494391 0.7494391 0.8187308 0.8187308 0.8187308

0.3 0.6396502 0.6396502 0.6816129 0.6816129 0.7408182 0.7408182 0.7408182

0.4 0.5970878 0.5970877 0.6250322 0.6250322 0.6703201 0.6703201 0.6703201

0.5 0.5599926 0.5599926 0.5760122 0.5760122 0.6065307 0.6065307 0.6065307

0.6 0.5268894 0.5268894 0.5326238 0.5326238 0.5488116 0.5488116 0.5488116

0.7 0.4969640 0.4969640 0.4937128 0.4937128 0.4965853 0.4965853 0.4965853

0.8 0.4697022 0.4697024 0.4585197 0.4585198 0.4493290 0.4493290 0.4493290

0.9 0.4447448 0.4447444 0.4265076 0.4265076 0.4065697 0.4065697 0.4065697

1.0 0.4218207 0.4218206 0.3972736 0.3972738 0.3678794 0.3678795 0.3678794

Example 4. We consider the following fractional differential-algebraic equations.

D
α1
∗ x(t)− t2 x(t) + y(t)− 2t = 0 (41)

D
α2
∗ y(t)− 2z(t) + 2(t + 1) = 0, 0< α1,α2 ≤ 1 (42)

z(t)− y(t)− 2t x(t) + t4 − t − 1= 0 (43)

with initial conditions as

x(0) = 0, y(0) = 0, z(0) = 1 (44)

For α1 = α2 = 1 the exact solution is x(t) = t2, y(t) = t4, z(t) = 2t3 + t + 1. By using

Theorems 1, 2, 4, 5 and Eq.(13), Eqs.(41)-(43) are transformed to,

X (k+α1β1) =
Γ(1+ k/β1)

Γ(α1 + 1+ k/β1)





k
∑

l=0

δ(l − 2β1)X (k− l)− Y (k) + 2δ(k− β1)



 (45)



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 137Table 6: Numeri
al results of y(t) with 
omparison to HAM in Example 3
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000

0.1 0.2437798 0.2437798 0.1672777 0.1672777 0.0998334 0.0998334 0.0998334

0.2 0.3264739 0.3264739 0.2679610 0.2679610 0.1986693 0.1986693 0.1986693

0.3 0.3966883 0.3966883 0.3547256 0.3547256 0.2955202 0.2955202 0.2955202

0.4 0.4626507 0.4626507 0.4347062 0.4347062 0.3894183 0.3894183 0.3894183

0.5 0.5259636 0.5259636 0.5099441 0.5099441 0.4794255 0.4794255 0.4794255

0.6 0.5865647 0.5865647 0.5808303 0.5808303 0.5646425 0.5646425 0.5646425

0.7 0.6438391 0.6438391 0.6470902 0.6470902 0.6442177 0.6442177 0.6442177

0.8 0.6969830 0.6969827 0.7081653 0.7081653 0.7173561 0.7173561 0.7173561

0.9 0.7451519 0.7451522 0.7633890 0.7633890 0.7833269 0.7833269 0.7833269

1.0 0.7875300 0.7875299 0.8120769 0.8120766 0.8414710 0.8414710 0.8414710

Y (k+α2β2) =
Γ(1+ k/β2)

Γ(α2 + 1+ k/β2)

�

2Z(k)− δ(k− β2)− 2δ(k)
�

(46)

Z(k) = Y (k) + 2

k
∑

l=0

δ(l − β)X (k− l)− 4δ(k− 4β) + δ(k− β)+ δ(k) (47)

where β1 and β2 are the unknown value of the fractions and β = LC M(β1,β2).From Eq.(16),

initial conditions in Eq.(44) can be transformed as follows:

X (0) = 0, k = 0,1,2, . . . ,α1β1 − 1, Y (0) = 0, k = 0,1,2, . . . ,α2β2 − 1, Z(0) = 1 (48)

From Eqs.(45)-(48), X (k), Y (k) and Z(k) are obtained up to and using the inverse transfor-

mation rule in Eq.(13), x(t) ,y(t) and z(t) are calculated for different values of α1 and α2.

Numerical comparisons are given in Table 7-8-9.Table 7: Numeri
al results of x(t) with 
omparison to HAM in Example 4
α = 0.5 α = 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000

0.1 0.0468839 0.0468839 0.0220868 0.0220868 0.0100000 0.0100000 0.0100000

0.2 0.1256709 0.1256709 0.0738823 0.0738823 0.0400000 0.0400000 0.0400000

0.3 0.2069285 0.2069285 0.1483883 0.1483883 0.0900000 0.0900000 0.0900000

0.4 0.2635406 0.2635406 0.2403223 0.2403223 0.1600000 0.1600000 0.1600000

0.5 0.2692381 0.2692381 0.3435494 0.3435494 0.2500000 0.2500000 0.2500000

0.6 0.2064374 0.2064374 0.4503451 0.4503451 0.3600000 0.3600000 0.3600000

0.7 0.0771395 0.0771393 0.5510811 0.5510811 0.4900000 0.4900000 0.4900000

0.8 −0.0874721 −0.0874740 0.6342822 0.6342821 0.6400000 0.6400000 0.6400000

0.9 −0.2250269 −0.2249716 0.6872107 0.6872098 0.8100000 0.8100000 0.8100000

1.0 −0.2550524 −0.2536490 0.6972563 0.6972492 1.0000000 1.0000000 1.0000000



B. İbi̧s, M. Bayram and A. Ağargün / Eur. J. Pure Appl. Math, 4 (2011), 129-141 138Table 8: Numeri
al results of y(t) with 
omparison to HAM in Example 4
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000

0.1 0.0047962 0.0047962 0.0006640 0.0006640 0.0001000 0.0001000 0.0001000

0.2 0.0446486 0.0446486 0.0079990 0.0079990 0.0016000 0.0016000 0.0016000

0.3 0.1654479 0.1654479 0.0347313 0.0347313 0.0081000 0.0081000 0.0081000

0.4 0.4099255 0.4099255 0.0988115 0.0988115 0.0256000 0.0256000 0.0256000

0.5 0.7956628 0.7956628 0.2220253 0.2220253 0.0625000 0.0625000 0.0625000

0.6 1.2918230 1.2918231 0.4277910 0.4277910 0.1296000 0.1296000 0.1296000

0.7 1.8063925 1.8063925 0.7376692 0.7376692 0.2401000 0.2401000 0.2401000

0.8 2.1992173 2.1992062 1.1662343 1.1662344 0.4096000 0.4096000 0.4096000

0.9 2.3319582 2.3316970 1.7141421 1.7141424 0.6561000 0.6561000 0.6561000

1.0 2.1509580 2.1479845 2.3596212 2.3596233 1.0000000 1.0000000 1.0000000Table 9: Numeri
al results of z(t) with 
omparison to HAM in Example 4
α = 0.5 α= 0.75 α = 1

t xHAM xF DT M xHAM xF DT M xHAM xF DT M xExact

0.0 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.1 1.1140730 1.1140730 1.1049813 1.1049813 1.1020000 1.1020000 1.1020000

0.2 1.2933169 1.2933169 1.2359520 1.2359520 1.2160000 1.2160000 1.2160000

0.3 1.5815050 1.5815050 1.4156643 1.4156643 1.3540000 1.3540000 1.3540000

0.4 1.9951580 1.9951580 1.6654692 1.6654692 1.5280000 1.5280000 1.5280000

0.5 2.5024009 2.5024009 2.0030745 2.0030746 1.7500000 1.7500000 1.7500000

0.6 3.0099479 3.0099480 2.4386051 2.4386051 2.0320000 2.0320000 2.0320000

0.7 3.3742876 3.3742875 2.9690827 2.9690826 2.3860000 2.3860000 2.3860000

0.8 3.4496824 3.4496478 3.5714858 3.5714850 2.8240000 2.8240000 2.8240000

0.9 3.1713371 3.1706481 4.1950213 4.1950119 3.3580000 3.3580000 3.3580000

1.0 2.6477139 2.6406864 4.7541339 4.7540532 4.0000000 4.0000000 4.0000000

5. Conclusion

In this paper, fractional differential transform method (FDTM) is extended to solve frac-

tional differential-algebraic equations (FDAEs). The results of this method are in good agree-

ment with those obtained by using the homotopy analysis method (HAM). The study em-

phasized our belief that the method is a reliable technique to handle fractional differential-

algebraic equations and the FDTM offer significant advantages in terms of its straightforward

applicability, its computational effectiveness and its accuracy. In general, FDTM can be used

as a powerful solver for the solution of fractional differential-algebraic equations.



REFERENCES 139

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