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Generalized Differential Transform Method for
Fractional Differential Equations
By
Fahmi S. Barakat
Supervised by
Prof. Dr. Ayman H. Sakka
A thesis submitted in partial fulfillment of the requirements for the degree
of Master of Science in Mathematical Sciences
September/2018
زةــغب ةــالميــــــة اإلســـــــــامعـالج
حث العلمي والدراسات العلياـبال عمادة
العـلـــــــــــــــــــــومة ــــــــــــــــــــليـك
العلـــــــوم الـريـاضـيــةر ـــــــماجستي
The Islamic University of Gaza
Deanship of Research and Graduate studies
Faculty of Science
Master of Mathematical Science
i
Generalized Differential Transform Method for
Fractional Differential Equations
للمعادالت التفاضلية الكسريةطريقة التحويل التفاضلي المعممة
Declaration
I understand the nature of plagiarism, and I am aware of the University’s policy on
this.
The work provided in this thesis, unless otherwise referenced, is the researcher's own
work, and has not been submitted by others elsewhere for any other degree or
qualification.
Student's name: Fahmi S. Barakat
Signature:
Date:
Abstract
The Fractional Calculus is a useful mathematical tool for applied sciences. Nev-
ertheless, fractional calculus is somehow hard to tackle. Fractional calculus is a
generalization of classical calculus in which the order of the derivative can be any
complex number. In this thesis, we talked about the fractional calculus and we
review some examples and some basic properties and theorems for fractional calcu-
lus. Moreover we will study the generalized Taylor’s formula and we will talk about
the generalized differential transform method extensively because it is the most
important in this thesis. Then we explain some examples on fractional integro-
differential equations and solve it by generalized differential transform method and
sole the fractional first Painleve equation by the same method.
iii
Acknowledgements
Firstly, I thank Allah so much for this success and for finishing my studies.
I would like to thank my father, mother, brothers and my sisters who encouraged
me to complete my studies and their continued support for my Masters thesis.
Also, I would like to express my sincere gratitude to my advisor Prof. Ayman H.
Sakka for the continuous support of my Master’s study and related research, for
his patience, motivation, and immense knowledge. His guidance helped me in all
the time of research and writing of this thesis. I could not have imagined having a
better advisor and mentor for my Master’s study.
Finally, I extend my thanks to all my colleagues and friends at the university
for their endless love and support in good and bad circumstances.
iv
Contents
Abstract iii
Acknowledgements iv
Contents v
List of Figures vii
List of Figures vii
Introduction 1
1 Preliminaries 4
2 Fractional Integrals and Derivatives 8
2.1 Riemann-Liouville Fractional Integral . . . . . . . . . . . . . . . . . 8
2.2 Fractional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Examples of Fractional Integrals . . . . . . . . . . . . . . . . . . . . 10
2.4 Dirichlet’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5 Relation Between Fractional Integral and Fractional Derivatives . . 18
2.6 Leibniz’s Formula for Fractional Integrals and derivatives . . . . . . 22
v
3 Generalized Differential Transform Method 26
3.1 Generalized Taylor’s Formula . . . . . . . . . . . . . . . . . . . . . 26
3.2 Generalized Differential Transform Method . . . . . . . . . . . . . . 34
4 Applications of GDT Method 47
4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4.2 Fractional First Painleve equation . . . . . . . . . . . . . . . . . . 67
5 Conclusion 74
Bibliography 75
Bibliography 76
vi
List of Figures
4.1 y(t) for β=0 and α = 0.25 : (-)µ=1.75, (..)µ=1.5, (- -)µ=1.25 . . . 59
4.2 y(t) for µ=2 and α = 0.25 : (- -)β=0.75, (..)β=0.5, (-)β=0.25 . . . 61
4.3 y(t) for µ=2 and β2 = 0 : (-)β1=0.5, (..)β1=1, (- -)β1=1.5 . . . . . 64
4.4 y(t) for β=0.5 and α = 0.5 : (-)µ=1, (..)µ=1.5, (- -)µ=2 . . . . . . 66
vii
Introduction
Fractional Calculus started when LHopital, one of the founders of calculus, wrote
to Leibnitz about the meaning ofdnxdxn when n = 1
2. Leibnitz replied in 1695 saying
that ( it could be a paradox from which useful consequences would be drawn one
day).(Hans Joachim Haubold, 2017) This was the first question that open the door
to a new branch of mathematics namely the branch of fractional calculus. Then
the question was improved to be: Can n be any number (Miller and Ross, 1988).
The name fractional calculus does not mean the calculus of fraction nor a fraction
of any calculus. The fractional calculus is a branch of mathematics the studies the
integrals and derivatives of arbitrary order (Podlubny, 1998).
In 1812 P. S. Laplace used integrals to define a fractional derivative. In 1819
the derivative of arbitrary order has been appeared in a text by S. F. Lacriox. The
Lacroix’s method was weak because we can not use it for a derivative of arbitrary
order. After 3 years, Fourier mentioned derivative of arbitrary order where he
defined fractional operations in terms of his integral representation of a function
f(x). The first application of fractional operations was by Niel’s Henrik Able in
1823. Able applied the fractional calculus in the solution of an integral equation
that arise in the formulation of the tautochrone problem (Miller and Ross, 1988).
The goal of this thesis is to study fractional calculus and the generalized differ-
ential transform method (GDTM) to solve some integro-differential equation and
1
the fractional first Painleve equation.
The organization of this thesis is as follows: In Chapter one, we will give the
basic definitions and equations related to our study. In Chapter two, we will study
the fractional calculus and its popular definitions, theorems, also we will investigate
some examples on fractional calculus, then we will study the Dirichlet’s formula
and Leibniz’s formula for fractional integrals. In Chapter three, we will study the
generalized Taylor’s formula and the GDTM and its basic properties. In Chapter
four, we will use the GDTM to study some integro-differential examples, then we
will use the GDTM to study the fractional first Painleve equation.
2
Chapter 1
Preliminaries
Chapter 1
Preliminaries
Definition 1. (Bell, 2004) The gamma function Γ(t) is a meromorphic function of
t defined by
Γ(t) =
∫ ∞0
e−xxt−1dx, t > 0 (1.1)
If m is a non-negative integer then Γ(m + 1) = m!. Moreover Γ(t + 1) = tΓ(t)
for t > 0.
Definition 2. (Beals and Wong, 2010) The incomplete gamma function γ(α, x) is
defined by
γ(α, x) =
∫ x
0
e−ττα−1dτ. α > 0. (1.2)
and the function γ?(α, x) is defined as(Miller and Ross, 1988)
γ∗(α, x) =1
Γ(α)xαγ(α, x)
=1
Γ(α)xα
∫ x
0
e−ττα−1dτ, α > 0 (1.3)
Using the gamma function we can write the binomial coefficients as (Miller and
4
Ross, 1988)
(−zk
)=
Γ(1− z)
k!Γ(1− z − k)= (−1)k
Γ(z + k)
k!Γ(z)= (−1)k
(z + k − 1
k
). (1.4)
where z ∈ R/N and k is non-negative integer.
The Psi function is the derivative of logarithmic of gamma function (Andrews
et al., 1999)
ψ(t) =d
dt(ln(Γ(t)) =
ddt
(Γ(t))
Γ(t). (1.5)
The other name of the psi function is the digamma function. In particular we have
(Miller and Ross, 1988)
ψ(1) = −γ, (1.6)
ψ
(1
2
)= −γ − ln(2), (1.7)
where γ = 0.57721566 · · · is the Euler’s constant. The psi function satisfies the
recurrence relation (Miller and Ross, 1988)
ψ(t+ 1) = ψ(t) +1
t. (1.8)
Definition 3. (Bell, 2004) The beta function β(x, y) is defined as
β(x, y) =
∫ 1
0
tx−1(1− t)y−1 dt, x > 0, y > 0 (1.9)
The beta function satisfies β(x, y) = β(y, x), and it can be written in term of
5
the Gamma function as (Gorenflo and Mainardi, 1999)
β(x, y) =Γ(x)Γ(y)
Γ(x+ y). (1.10)
The incomplete beta function βτ (x, y) has the integral representation (Miller and
Ross, 1988)
βτ (x, y) =
∫ τ
0
tx−1(1− t)y−1 dt, x > 0, y > 0 (1.11)
If a(x) and b(x) are both differentiable, then(Abramowitz and Stegun, 1964)
d
dx
(∫ b(x)
a(x)
f(x, t)dt
)= f(x, b(x))
d
dxb(x)− f(x, a(x))
d
dxa(x) +
∫ b(x)
a(x)
∂
∂xf(x, t)dt
(1.12)
which called the Leibniz integral rule.
6
Chapter 2
Fractional Integrals andDerivatives
Chapter 2
Fractional Integrals and
Derivatives
2.1 Riemann-Liouville Fractional Integral
There are many definitions of fractional integral but the most popular definition is
the Riemann-Liouville fractional integral. The Riemann-Liouville fractional integral
is defined as follows:
Definition 4. (Miller and Ross, 1988) Let f be a piecewise continuous function on
the interval J ′ =(0,∞) and integrable on any finite subinterval of J=[0,∞). Let α
be a complex number with Re(α) > 0. Then for x > 0 and a ≥ 0 we define
Iαa f(x) =1
Γ(α)
∫ x
a
(x− t)α−1f(t)dt (2.1)
to be the Riemann-Liouville fractional integral of order α.
8
Definition 5. (Erturk and Momani, 2010) The space Cα is the space of all functions
f(x) which can be written of the form
f(x) = xλη(x)
or
f(x) = xλ(lnx)η(x),
where λ> -1 and η(x) is continuous on J=[0,∞). For m ∈ N, f ∈ Cmα if f (m) ∈ Cα.
2.2 Fractional Derivatives
Let D = ddx
be the differential operator. If n is a positive integer then Dnf(x)
means the nth order derivative of the function f(x).
Definition 6. (Kimeu, 2009) For any number ν ∈ R with ν>0, the fractional
derivative is defined as
Dνaf(x) = Dn[Iαa f(x)], (2.2)
where n = dνe is the least integer greater or equal to α and α = n− ν.
Definition 7. (Hans Joachim Haubold, 2017) The Riemann-Liouville fractional
derivative of order α is denoted and defined by
Dαa f(x) = (Dm Im−αf)(x)
=
1
Γ(m−α)
(ddx
)m ∫ xa
(x− t)m−α−1f(t)dt, m− 1 < α < m,(ddx
)mf(x), α = m.
Another popular definition of fractional calculus is the Caputo derivative.
9
Definition 8. (Odibat and Momani, 2008) For a function f(x), the Caputo frac-
tional derivative of order α>0 with a≥0, is defined by
Dαaf(x) = (Im−αa f (m))(x) =
1
Γ (m− α)
x∫a
(x− t)m−α−1f (m)(t)dt,
(2.3)
for m− 1<α ≤m, m ∈ N, x ≥ a, f(x) ∈ Cm−1.
2.3 Examples of Fractional Integrals
In this section we will consider some examples.
Example 2.1. First let us begin with the constant function f(x) = k. We get, by
Definition 4,
Iα0 f(x) =1
Γ(α)
∫ x
0
(x− t)α−1 k dt
=k
Γ(α)
∫ x
0
(x− t)α−1dt
=k
Γ(α)
−(x− t)α
α
∣∣∣∣x0
=k
αΓ(α)xα.
Thus if f(x) = k, then
Iα0 f(x) =k xα
Γ(α + 1).
For Riemann-Liouville fractional derivative, by Definition 7, consider m = 1 we
10
have
Dα0 f(x) =
1
Γ(1− α)D
∫ x
0
(x− t)−α (k)dt
=k
Γ(1− α)D
[−(x− t)−α+1
(1− α)
]x0
=k
Γ(1− α)D
[(x)1−α
(1− α)
]
Thus if f(x) = k, then
Dα0 f(x) =
k x−α
Γ(1− α). (2.4)
If we use the Caputo fractional derivative for f(x) = k, by Definition 8 we will find
Dα0 f(x) = 0
If we take α = 12
and k = 1, then
Iα0 (1) =x
12
Γ(112)
=2x
12
√π.
Dα0 (1) =
x−12
Γ(12)
=x−12
√π.
Example 2.2. Consider f(x) = xµ with µ>− 1. Then, by Definition 4, we have
Iα0 f(x) =1
Γ(α)
∫ x
0
(x− t)α−1 tµdt.
11
Let t = xτ . Then dt = xdτ , and we get
Iα0 xµ =
1
Γ(α)
∫ 1
0
(x− xτ)α−1 (xτ)µ x dτ.
Take x as a common factor we have
Iα0 xµ =
xµ+α
Γ(α)
∫ 1
0
(1− τ)α−1 (τ)µ dτ. α > 0, µ+ 1 > 0
By Equation (1.9), we obtain
Iα0 xµ =
xµ+α
Γ(α)β(µ+ 1, α).
By (1.10) we conclude that
Iα0 xµ =
Γ( µ+ 1) xµ+α
Γ( µ+ α + 1).
For Riemann-Liouville fractional derivative, by Definition 7, consider m = 1 we
have
Dα0 f(x) =
1
Γ(1− α)D
∫ x
0
(x− t)−α (t)µdt
12
Let t = xτ . Then dt = xdτ , and we get
Dα0 f(x) =
1
Γ(1− α)D
∫ 1
0
x−α (1− τ)−αxµτµx dτ
=D(xµ−α+1)
Γ(1− α)
∫ 1
0
(1− τ)−ατµdτ
=(µ− α + 1)(xµ−α)
Γ(1− α)β(1− α, µ+ 1)
=(µ− α + 1)(xµ−α)
Γ(1− α)
Γ(1− α)Γ(µ+ 1)
Γ(2 + µ− α)
Thus if f(x) = xµ, then
Dα0 f(x) =
Γ(µ+ 1)xµ−α
Γ(1 + µ− α). (2.5)
If we use the Caputo fractional derivative for f(x) = xµ, by Definition 8, we have
Dα0 f(x) =
1
Γ(1− α)
∫ x
0
(x− t)−α µ (t)µ−1dt
Let t = xτ . Then dt = xdτ , and we get
Dα0 f(x) =
µ
Γ(1− α)
∫ 1
0
x−α (1− τ)−αxµ−1τµ−1x dτ
=µ xµ−α
Γ(1− α)
∫ 1
0
(1− τ)−ατµ−1dτ
=µ xµ−α
Γ(1− α)β(1− α, µ)
=µ xµ−α
Γ(1− α)
Γ(1− α)Γ(µ)
Γ(1 + µ− α)
Thus if f(x) = xµ, then
Dα0 f(x) =
Γ(µ+ 1)xµ−α
Γ(1 + µ− α). (2.6)
Which is the same rustle as Riemann-Liouville fractional derivative.
13
Example 2.3. Suppose f(x) = eax, where a is a constant. Then, by Definition 4,
we obtain
Iα0 f(x) =1
Γ(α)
∫ x
0
(x− t)α−1 eatdt.
Using τ = x− t, we have dτ = −dt, and hence
Iα0 eax =
1
Γ(α)
∫ 0
x
−(τ)α−1 ea(x−τ)dτ,
=eax
Γ(α)
∫ x
0
(τ)α−1 e−aτdτ.
Using the incomplete gamma function (1.3) we obtained
Iα0 eax = eaxxαγ∗(α, ax).
If we use the Caputo fractional derivative for f(x) = eat, by Definition 8, we have
Dα0 f(x) =
1
Γ(1− α)
∫ x
0
(x− t)−α a (e)atdt
Let τ = x− t. Then dt = −dτ , we get
Dα0 f(x) =
aeax
Γ(1− α)
∫ 0
x
−τ−α (e)−aτdτ
=aeax
Γ(1− α)
∫ x
0
τ−α (e)−aτdτ
Using the incomplete gamma function (1.2) we obtained
Dα0 (e)at =
aeax
Γ(1− α)γ(1− α, ax). (2.7)
14
Example 2.4. Consider f(x) = (x− a)λ, where 0 < x < a. Then
Iαa0f(x) =1
Γ(α)
∫ x
a
(x− t)α−1 (t− a)λ dt.
Using the transformation t = a+ (x− a)τ , then dt = (x− a)dτ , we obtain
Iαa (x− ax)λ =1
Γ(α)
∫ 1
0
(x− a)α−1 (1− τ)α−1 (x− a)λ τλ (x− a)dτ
=(x− a)α+λ
Γ(α)
∫ 1
0
(1− τ)α−1τλdτ,
The integral can be written in terms of the beta function, so we have
Iαa (x− a)λ =(x− a)α+λ
Γ(α)β(α, λ+ 1)
=(x− a)α+λ
Γ(α)
Γ(α)Γ(λ+ 1)
Γ(α + λ+ 1)
Thus
Iαa (x− a)λ =Γ(λ+ 1)
Γ(α + λ+ 1)(x− a)α+λ (2.8)
If we use the Caputo fractional derivative for f(x) = (x− a)λ, by Definition 8, we
have
Dαa f(x) =
1
Γ(1− α)
∫ x
a
(x− t)−α λ (t− a)λ−1dt.
Let t = a+ (x− a)τ , then dt = (x− a)dτ , we get
Dαa (x− a)λ =
λ
Γ(1− α)
∫ 1
0
(x− a)−α (1− τ)−α (x− a)λ−1 τλ−1 (x− a)dτ
=λ(x− a)λ−α
Γ(1− α)
∫ 1
0
(1− τ)−α τλ−1dτ.
15
The integral can be written in terms of the beta function, so we have
Dαa (x− a)λ =
λ(x− a)λ−α
Γ(1− α)β(1− α, λ)
=λ(x− a)λ−α
Γ(1− α)
Γ(1− α) Γ(λ)
Γ(1− α + λ)
Thus
Dαa (x− a)λ =
λ Γ(λ)
Γ(1− α + λ)(x− a)λ−α (2.9)
2.4 Dirichlet’s Formula
If L(x, y) is jointly continuous on [a, b]× [a, b], we know from the elementary theory
of functions that
∫ b
a
∫ x
a
L(x, y)dydx =
∫ b
a
∫ b
y
L(x, y)dxdy.
Definition 9. (Miller and Ross, 1988) Let F be jointly continuous function on the
Euclidean plane, and let λ, µ, and α be positive numbers. Then the Dirichlet’s
formula reads
∫ t
a
(t−x)µ−1dx
∫ x
a
(y−a)λ−1(x−y)α−1)F (x, y)dy =
∫ t
a
(y−a)λ−1dy
∫ t
y
(t−x)µ−1(x−y)α−1F (x, y)dx.
We will use The Dirichlet’s formula to proof the following theorem.
Theorem 2.4.1. (Kimeu, 2009) Let f be a continuous function on J = [0,∞), and
let µ and α be positive numbers. Then for all x
Iα0 [Iµ0 f(x)] = Iα+µ0 f(x) = Iµ0 [Iα0 f(x)].
16
Proof. Firstly, by Definition 4,
Iα0 [Iµ0 f(x)] = Iα0
(1
Γ(µ)
∫ t
0
(t− y)µ−1f(y)dy
)=
1
Γ(α)
∫ x
0
(x− t)α−1
(1
Γ(µ)
∫ t
0
(t− y)µ−1f(y)dy
)dt
=1
Γ(α)Γ(µ)
∫ x
0
(x− t)α−1
∫ t
0
(t− y)µ−1f(y)dydt.
By Dirichlet’s formula, we get
Iα0 [Iµf(x)] =1
Γ(α)Γ(µ)
∫ x
0
f(y)dy
∫ x
y
(x− t)α−1(t− y)µ−1dt.
Using t = y + (x− y)τ , we have
Iα0 [Iµ0 f(x)] =1
Γ(α)Γ(µ)
∫ x
0
f(y)dy
∫ 1
0
(x− y)α−1(1− τ)α−1(x− y)µ−1τµ−1(x− y)dτ
=1
Γ(α)Γ(µ)
∫ x
0
(x− y)α+µ−1f(y)dy
∫ 1
0
(1− τ)α−1τµ−1dτ.
Using Equation (1.9), we have
Iα0 [Iµ0 f(x)] =β(α, µ)
Γ(α)Γ(µ)
∫ x
0
(x− y)α+µ−1f(y)dy.
By Equation (1.10), we obtained
Iα0 [Iµ0 f(x)] =1
Γ(α + µ)
∫ x
0
(x− y)α+µ−1f(y)dy.
From the Definition 4, we conclude that
Iα0 [Iµ0 f(x)] = Iα+µ0 f(x).
17
By the same way can show that
Iµ0 [Iα0 f(x)] = Iα+µ0 f(x).
2.5 Relation Between Fractional Integral and Frac-
tional Derivatives
Theorem 2.5.1. (Miller and Ross, 1988) Let f be a continuous function on J =
[0,∞), α>0, and Df of class C. Then
(a) Iα+10 [Df(x)] = Iα0 f(x)− f(0)
Γ(α+1)xα,
(b) D[Iα0 f(x)] = Iα−10 f(x) = Iα0 [Df(x)] + f(0)
Γ(α)xα−1.
Proof. From Definition 4, part (a) can be proved as follows :
Iα+10 [Df(x)] =
1
Γ(α + 1)
∫ x
0
(x− t)αDf(t)dt.
Integration by parts yields
Iα+10 [Df(x)] =
1
Γ(α + 1)
[(x− t)α f(t)
∣∣∣∣x0
+
∫ x
0
α(x− t)α−1f(t)dt
]=
1
Γ(α + 1)[(0− xαf(0)) + αΓ(α)Iα0 [f(x)]]
=αΓ(α)
Γ(α + 1)Iα0 [f(x)]− f(0)
Γ(α + 1)xα.
18
It follows that
Iα+10 [Df(x)] = Iα0 [f(x)]− f(0)
Γ(α + 1)xα.
From Definition 4, part (b) can be proved as follows:
Iα0 [f(x)] =1
Γ(α)
∫ x
0
(x− t)αf(t)dt.
Using t = x− τλ, where λ = 1α
, we get
Iα0 [f(x)] =1
Γ(α)
∫ 0
xα(τλ)α−1f(x− τλ) (−λτλ−1)dτ,
=λ
Γ(α)
∫ xα
0
τ 1−λ τλ−1f(x− τλ)dτ,
=1
αΓ(α)
∫ xα
0
f(x− τλ)dτ,
=1
Γ(α + 1)
∫ xα
0
f(x− τλ)dτ.
Then for t>0, and by Leibniz integral rule(1.12) we have
D[Iα0 f(x)] =1
Γ(α + 1)
[f(0)(αxα−1 +
∫ xα
0
∂f(x− τλ)∂x
dτ
].
Now reversing the transformation t = x− τλ, we have
D[Iαf(x)] =αxα−1
Γ(α + 1)f(0) +
1
Γ(α + 1)
∫ 0
x
D[f(t)](−α(x− t)α−1)dt,
=xα−1
Γ(α)f(0) +
1
Γ(α)
∫ x
0
(x− t)α−1D[f(t)]dt.
19
Therefore, by Definition 4, we get
D[Iα0 f(x)] = Iα0 [Df(x)] +xα−1
Γ(α)f(0).
and by applying part (a), we get
D[Iα0 f(x)] = Iα−10 f(x)
Theorem 2.5.2. (Miller and Ross, 1988) Let p be a positive integer , let Dp−1f be
continuous on J = [0,∞) and let α>0. Then
(a) Iα0 f(x) = Iα+p0 [Dpf(x)] +Qp(x, α),
(b) Dp[Iα0 f(x)] = Iα0 [Dpf(x)] +Qp(x, α− p),
where
Q√ =
p−1∑k=0
xα+k
Γ(α + k + 1)Dkf(0).
Proof. To prove part (a), we put p = 1. As a result we obtain
Iα0 f(x) = Iα+10 [Df(x)] +Q1(x, α),
= Iα+10 [Df(x)] +
xα
Γ(α + 1)f(0). (2.10)
Replacing α by α + 1 and f by Df in Equation (2.10), we get
Iα+10 Df(x) = Iα+2
0 [D2f(x)] +xα+1
Γ(α + 2)Df(0).
20
Also from Equation (2.10), we get
Iα0 f(x)− xα
Γ(α + 1)f(0) = Iα+2
0 [D2f(x)] +xα+1
Γ(α + 2)Df(0)
Iα0 f(x) = Iα+20 [D2f(x)] +
xα+1
Γ(α + 2)Df(0) +
xα
Γ(α + 1)f(0)
= Iα+20 [D2f(x)] +
1∑k=0
xα+k
Γ(α + k + 1)Dkf(0)
= Iα+20 [D2f(x)] +Q2(x, α).
Repeated iteration establishes part(a).
Now for part (b), if we differentiate part (b) of Theorem 2.5.1, we get
D2[Iα0 f(x)] = D{Iα0 [Df(x)]}+f(0)
Γ(α− 1)xα−2. (2.11)
Replacing f by Df in part(b) of Theorem 2.5.1, we have
D[Iα0 (Df(x))] = Iα0 [D2f(x)] +Df(0)
Γ(α)xα−1. (2.12)
Now substitute Equation (2.12) into Equation (2.11) to get
D2[Iα0 f(x)] = Iα0 [D2f(x)] +Df(0)
Γ(α)xα−1 +
f(0)
Γ(α− 1)xα−2
= Iα0 [D2f(x)] +1∑
k=0
xα+k−2
Γ(α + k − 1)Dkf(0)
= Iα0 [D2f(x)] +Q2(x, α− 2).
Repeated iteration establishes part (b).
21
2.6 Leibniz’s Formula for Fractional Integrals and
derivatives
In this section we will talk about fractional integral of the product of two functions.
In the elementary calculus the derivative of two functions can be written as a sum
of products of derivative as
Dn[f(x)g(x)] =n∑k=0
(n
k
)[Dkg(x)][Dn−kf(x)],
where f and g are assumed to have derivatives of order n on some interval. Now
we want to produce a similarly formula as above for fractional integral.
Theorem 2.6.1. Suppose that f and g are continuous on [0,∞). Then
Iα0 [f(x)g(x)] =∞∑k=0
(−αk
)Dkg(x) Iα+k
0 f(x). (2.13)
Proof.
Iα0 [f(x)g(x)] =1
Γ(α)
∫ x
0
(x− t)α−1[f(t)g(t)]dt. (2.14)
Using Taylor series g(t) can be written as
g(t) =∞∑k=0
Dkg(x)
k!(t− x)k
=∞∑k=0
(−1)kDkg(x)
k!(x− t)k
= g(x) +∞∑k=1
(−1)kDkg(x)
k!(x− t)k. (2.15)
22
Now substituting g from Equation (2.15) into Equation (2.14) gives
Iα0 [f(x)g(x)] =1
Γ(α)
∫ x
0
(x− t)α−1f(t)
[(g(x) +
∞∑k=1
(−1)kDkg(x)
k!(x− t)k)
]dt
=1
Γ(α)
∫ x
0
(x− t)α−1f(t)g(x)dt
+1
Γ(α)
∫ x
0
(x− t)α−1f(t)×∞∑k=1
(−1)kDkg(x)
k!(x− t)kdt
=g(x)
Γ(α)
∫ x
0
(x− t)α−1f(t)dt
+1
Γ(α)
∫ x
0
(x− t)α−1f(t)×∞∑k=1
(−1)kDkg(x)
k!(x− t)kdt.
By definition of fractional integral we get
Iα0 [f(x)g(x)] = g(x)Iα0 [f(x)] +1
Γ(α)
∫ x
0
(x− t)α−1f(t)×∞∑k=1
(−1)kDkg(x)
k!(x− t)kdt.
Now interchange the order of integration and summation, we have
Iα0 [f(x)g(x)] = g(x) Iα0 [f(x)] +∞∑k=1
(−1)kDkg(x)
k!Γ(α)
∫ x
0
(x− t)α+k−1f(t)dt.
Also by Definition 4, the integral∫ x
0(x− t)α+k−1 f(t) dt = Γ(α + k)Iα+k
0 [f(x)]. As
a result we get
Iα0 [f(x)g(x)] = g(x) Iα0 [f(x)] +∞∑k=1
(−1)kDkg(x)
k!Γ(α)Γ(α + k) Iα+k
0 f(x)
= g(x) Iα0 [f(x)] +∞∑k=1
(−1)kΓ(α + k)
k!Γ(α)Dkg(x) Iα+k
0 [f(x)].
23
By Equation (1.4), we get
Iα0 [f(x)g(x)] = g(x) Iα0 [f(x)] +∞∑k=1
(−αk
)Dkg(x) Iα+k
0 f(x)
=∞∑k=0
(−αk
)Dkg(x) Iα+kf(x).
Similarly, we can be proved that
Theorem 2.6.2. Let f and g be continuous on [0,∞), then
Dα0 [f(x)g(x)] =
∞∑k=0
(α
k
)Dkg(x) Dα−k
0 f(x). (2.16)
for α > 0
24
Chapter 3
Generalized DifferentialTransform Method
Chapter 3
Generalized Differential
Transform Method
3.1 Generalized Taylor’s Formula
Theorem 3.1.1. (Erturk et al., 2008) Suppose that f(x) ∈ C[a,b] and Dαa f(x) ∈
C(a,b], where Dαa is the Caputo fractional derivative of order α, for 0 < α ≤ 1.
Then we have
f(x) = f(a) +1
Γ(α)(Dα
a f)(t)(x− a)α , (3.1)
where a≤ t≤x, ∀ x ∈ (a,b].
Proof. From the definition of Caputo fractional derivative of f(x) (8), if we put m
= 1, we get
(Dαa f)(x) = I1−α
a f (1)(x).
26
Applying Riemann-Liouville integral of order α to both sides we have
Iαa (Dαa f)(x) = Iαa (I1−α
a f (1))(x)
= I1af
(1)(x)
= f(x)− f(a). (3.2)
The left hand side can be written as
Iαa (Dαa f)(x) =
1
Γ(α)
∫ x
a
(x− ξ)α−1Dαa f(ξ)dξ.
By the mean value theorem for integrals we get
Iαa (Dαa f)(x) =
Dαa f(t)
Γ(α)
∫ x
a
(x− ξ)α−1dξ , a ≤ t ≤ x
=Dαa f(t)
Γ(α)
(−(x− ξ)α
α
∣∣∣∣xa
)=
1
Γ(α + 1)Dαa f(t) (x− a)α. (3.3)
So that, from Equation (3.2) and Equation (3.3), we have
f(x)− f(a) =1
Γ(α)(Dα
a f)(t)(x− a)α
f(x) = f(a) +1
Γ(α)(Dα
a f)(t)(x− a)α.
For the special case of α = 1, we have the classical mean value theorem as
f ′(c) =f(x)− f(a)
x− a.
27
Theorem 3.1.2. (Erturk et al., 2008) Suppose that (Dα)nf(x), (Dα)n+1f(x)∈ C(a,b]
for 0<α≤1. Then we have
(Inαa (Dαa )nf)(x)− (I(n+1)α
a (Dαa )(n+1)f)(x) =
(x− a)nα
Γ(nα + 1)((Dα
a )nf)(a), (3.4)
where (Dαa )n =Dα
a .Dαa · · ·Dα
a f(x)︸ ︷︷ ︸n-times
.
Proof. We will start with the second term in the left side which can be written as
(I(n+1)αa (Dα
a )n+1f)(x) = (Inαa Iαa (Dαa )nDα
a f)(x)
= (Inαa (Dαa )nIαaD
αa f)(x).
From Equation (3.2), we get
(I(n+1)αa (Dα
a )n+1f)(x) = (Inαa (Dαa )n)(f(x)− f(a))
= (Inαa (Dαa )nf)(x)− (Inαa (Dα
a )nf)(a).
Rearranging the last equation we have
(Inαa (Dαa )nf)(x)− (I(n+1)α
a (Dαa )n+1f)(x) = (Inαa (Dα
a )nf)(a)
=1
Γ(nα)
∫ x
a
(x− t)nα−1((Dα)nf)(a)dt
=((Dα)nf)(a)
Γ(nα + 1)(x− t)nα.
Theorem 3.1.3. (Bansal and Jain, 2016) (Generalized Taylor’s Formula)
Suppose that (Dαa )kf(x) ∈ C(a, b] for k=0,1,2,· · · ,n+1, where 0<α≤1. Then we
28
have
f(x) =n∑i=0
(x− a)iα
Γ(iα + 1)((Dα
a )if)(a) +(Dα
a )n+1f(t)
Γ((n+ 1)α + 1)(x− a)(n+1)α, (3.5)
where a<t≤ x and x ∈ (a,b].
Proof. We will prove the theorem by induction. From Equation (3.4), if we take
n = 0, we have
f(x) = f(a) + (IαaDαa f)(x). (3.6)
For n = 1, Equation (3.4) become
(Iαa (Dαa )f)(x)− (I2α
a (Dαa )2f)(x) =
(x− a)α
Γ(α + 1)(Dα
a f)(a).
Then by Equation (3.6), we have
f(x)− f(a) =(x− a)α
Γ(α + 1)(Dα
a f)(a) + (I2αa (Dα
a )2f)(x).
Thus
f(x) =(x− a)α
Γ(α + 1)(Dα
a f)(a) + f(a) + (I2αa (Dα
a )2f)(x). (3.7)
For n = 2, Equation (3.4) become
(I2αa (Dα
a )2f)(x)− (I3αa (Dα
a )3f)(x) =(x− a)2α
Γ(2α + 1)((Dα
a )2f)(a),
and hence
(I2αa (Dα
a )2f)(x) =(x− a)2α
Γ(2α + 1)((Dα
a )2f)(a) + (I3αa (Dα
a )3f)(x).
29
Thus by Equation (3.7), we get
f(x) =(x− a)2α
Γ(2α + 1)((Dα
a )2f)(a)+(x− a)α
Γ(α + 1)(Dα
a f)(a)+f(a)+(I3αa (Dα
a )3f)(x). (3.8)
If we repeat the previous iteration for n, we conclude
f(x) =(x− a)nα
Γ(nα + 1)((Dα
a )nf)(a)+· · ·+ (x− a)α
Γ(α + 1)(Dα
a f)(a)+f(a)+(I(n+1)αa (Dα
a )(n+1)f)(x).
So
f(x) =n∑i=0
(x− a)iα
Γ(iα + 1)((Dα
a )if)(a)) + (I(n+1)αa (Dα
a )(n+1)f)(x). (3.9)
Now from Equation (3.3) and replacing α by (n+ 1)α, we get
I(n+1)αa (D(n+1)α
a f)(x) =1
Γ((n+ 1)α + 1)D(n+1)αa f(t) (x− a)(n+1)α. (3.10)
Substituted Equation (3.9) into Equation (3.8), we have
f(x) =n∑i=0
(x− a)iα
Γ(iα + 1)((Dα
a )if)(a)) +D
(n+1)αa f(t)
Γ((n+ 1)α + 1)(x− a)(n+1)α.
In the special case α = 1, Equation (3.5) reduces to
f(x) =n∑i=0
(x− a)i
i!f (i)(a) +
f (n+1)(t)
(n+ 1)!(x− a)(n+1),
which is the classical Taylor’s formula, and the second term in the right hand side is
the remainder term known as the Lagrange remainder. The radius of convergence
30
R, for the generalized Taylor’s formula
n∑i=0
(x− b)iα
Γ(iα + 1)((Dα
b )if)(b),
depends on f(x) and b, and is given by
R = |x− b|α limn→∞
∣∣∣∣∣ Γ(nα + 1)
Γ((n+ 1)α + 1)
(D(n+1)αb f)(b)
(Dnαb f)(b)
∣∣∣∣∣ .Theorem 3.1.4. (Erturk et al., 2008) Suppose that (Dα
a )kf(x) ∈ C(a,b] for k=0,1,2,· · · ,n+1,
where 0<α≤1. If x ∈ [a,b], then
f(x) ∼= PαN(x) =
N∑i=0
(x− a)iα
Γ(iα + 1)((Dα
a )if)(a). (3.11)
Furthermore, the error term, denoted by RαN , is given by
RαN =
(Dαa )N+1f(t)
Γ((N + 1)α + 1)(x− a)(N+1)α,
where a<t≤x.
To increase the accuracy of the approximation f(x) ∼= PαN(x) we must choose
large N so that the error does not exceed a specified bound.
Theorem 3.1.5. (Garg and Manohar, 2015) Suppose that f(x) = (x − a)λg(x),
where a, λ > and g(x) has the generalized power series expansion
g(x) =∞∑n=0
an(x− a)nα,
31
with a radius of converge R>0, 0 < α ≤1. Then
DγaD
βaf(x) = Dγ+β
a f(x) (3.12)
∀ (x-a) ∈ (0,R), the coefficients an = 0 for n given by nα+ λ− β = 0 and either :
(a) λ > µ, µ = max(β + [γ], [β + γ]) or
(b) λ ≤ µ, ak = 0, for k = 0, 1, ·,[µ−λα
]here [x] denotes the greatest integer less than or equal to x.
Proof. For the first case of λ > µ, µ = max(β+[γ], [β+γ]), from Caputo fractional
differential definition with m = [β] + 1, we have
Dβaf(x) = I [β]+1−β
a D[β]+1a
∞∑n=0
an (x− a)nα+λ.
Performing term by term differentiation which is justified since the series involving
derivatives up to the order [β] + 1 of the term (x− a)nα+λ are uniformly convergent
for (x− a) ∈ (0, R), we obtain
Dβaf(x) = I [β]+1−β
a
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− [β])(x− a)nα+λ−[β]−1.
Let we change the order of integration and summation, then using Definition 4 and
equation (2.8)
Dβaf(x) =
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− [β])I [β]+1−βa (x− a)nα+λ−[β]−1.
=∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β + 1)(x− a)nα+λ−β. (3.13)
32
Applying the same argument as above, we now have
DγaD
βaf(x) = Dγ
a
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β + 1)(x− a)nα+γ−β
= I [γ]+1−γa D[γ]+1
a
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β + 1)(x− a)nα+γ−β
= I [γ]+1−γa
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β − [γ])(x− a)nα+λ−β−[γ]−1
=∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β − γ + 1)(x− a)nα+λ−β−γ (3.14)
forλ− β − [γ] > 0
Next
Dγ+βa f(x) = Dγ+β
a
∞∑n=0
an (x− a)nα+γ
= I [γ+β]+1−γ−βa D[γ+β]+1
a
∞∑n=0
an (x− a)nα+γ
= I [γ+β]+1−γ−βa
∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− [γ + β])(x− a)nα+λ−[γ+β]−1
=∞∑n=0
anΓ(nα + λ+ 1)
Γ(nα + λ− β − γ + 1)(x− a)nα+λ−β−γ (3.15)
forλ− [β + γ] > 0
Which is precisely DγaD
βaf(x) as in equation (3.14).
The conditions mentioned with (3.14) and (3.15) can be combined and written
as condition given in part (a).
for part (b), λ ≤ µ, we take ak = 0 for k = 0, 1, · · · , l − 1, where l − 1 =[µ−λα
]
33
we have due to the uniform convergence of derived series up to the order [β] + 1,
Dβaf(x) = Dβ
a
∞∑n=0
an (x− a)nα+λ
=∞∑n=l
anΓ(nα + λ+ 1)
Γ(nα + λ− β + 1)(x− a)nα+λ−β
=∞∑r=0
ar+lΓ((r + l)α + λ+ 1)
Γ((r + l)α + λ− β + 1)(x− a)(r+l)α+λ−β (3.16)
If we let λ′ = lα + λ, then (3.16) becomes as
Dβaf(x) =
∞∑r=0
ar+lΓ(rα + λ′ + 1)
Γ(rα + λ′ − β + 1)(x− a)rα+λ′−β (3.17)
Which is the same as equation (3.13) and the proof proceeds as in part (a).
3.2 Generalized Differential Transform Method
The Generalized Differential Transform Method (GDTM) is an application of the
Generalized Taylor’s formula and it is used to obtain numerical solutions of integro-
differential equations of fractional order.
Definition 10. (Erturk and Momani, 2010) The generalized differential transform
of a function f(x) is defined as
Fα(k) =1
Γ(αk + 1)[(Dα
x0)kf(x)]x=x0 , k = 0, 1, · · · , (3.18)
where (Dαx0
)k=Dαx0
.Dαx0
.Dαx0· · · Dα
x0, k-times, is the sequential derivative of order
αk.
34
The inverse transform of Fα(k) is defined as
f(x) =∞∑k=0
Fα(k)(x− x0)αk. (3.19)
We will use Equation (3.19) to approximate a function f(x) by a finite sum. The
special case when α = 1, the generalized differential transform reduces to the clas-
sical differential transform. Next we study some basic properties of the generalized
differential transform.
Theorem 3.2.1. (Erturk and Momani, 2010)
(a) If f(x) = g(x) ± h(x), then Fα(k) = Gα(k)±Hα(k).
(b) If f(x) = ag(x), then Fα(k) = aGα(k), where a is a constant.
Proof. We can prove Theorem (3.2.1) by linearity of fractional derivative.
Theorem 3.2.2. (Erturk and Momani, 2010) If f(x) = g(x)h(x), then
Fα(k) =k∑l=0
Gα(l)Hα(k − l).
35
Proof. From Equation (3.19), we get
f(x) = g(x)h(x)
=
(∞∑k=0
Gα(k)(x− x0)αk
)(∞∑k=0
Hα(k)(x− x0)αk
)= [Gα(0) +Gα(1)(x− x0)α +Gα(2)(x− x0)2α + · · · ]×
[Hα(0) +Hα(1)(x− x0)α +Hα(2)(x− x0)2α + · · · ]
= Gα(0)Hα(0) + (x− x0)α(Gα(0)Hα(1) +Gα(1)Hα(0)) +
(x− x0)2α(Gα(0)Hα(2) +Gα(1)Hα(1) +Gα(2)Hα(0)) + · · ·
=∞∑k=0
[k∑l=0
Gα(l)Hα(k − l)
](x− x0)αk.
Then by Equation (3.19) we have
Fα(k) =k∑l=0
Gα(l)Hα(k − l).
Theorem 3.2.3. (Erturk and Momani, 2010) If f(x)=g1(x)g2(x) · · · gn−1(x)gn(x),
then
Fα(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).
(3.20)
Proof. We will prove this theorem by induction. First it true for n=2, by Theorem
(3.2.2). Assume is true for n. Now for n+ 1, let
f(x) = g1(x)g2(x) · · · gn(x)gn+1(x),
36
and
h(x) = g1(x)g2(x) · · · gn−1(x)gn(x).
Then
f(x) = h(x)gn+1(x).
By Theorem (3.2.2), we get
Fα(x) =l∑
k=0
H(k)Gn+1(l − k).
Now by induction hypothesis, we have
Hα(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).
As result, we get
Fα(k) =l∑
k=0
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)Gn+1(l−k).
Theorem 3.2.4. (Erturk and Momani, 2010) If f(x) = Dαx0g(x), then
Fα(k) =Γ(α(k + 1) + 1)
Γ(αk + 1)Gα(k + 1).
Proof. From the definition of generalized differential transform method (3.18), we
37
get
Fα(k) =1
Γ(αk + 1)
[(Dαx0
)kf(x)
]x=x0
=1
Γ(αk + 1)
[(Dαx0
)kDαx0g(x)
]x=x0
=1
Γ(αk + 1)
[(Dαx0
)k+1g(x)
]x=x0
=Γ(α(k + 1) + 1)
Γ(αk + 1)Γ(α(k + 1) + 1)
[(Dαx0
)k+1g(x)
]x=x0
=Γ(α(k + 1) + 1)
Γ(αk + 1)Gα(k + 1).
If a function f(x) satisfies the conditions of Theorem 3.1.5, then the generalized
differential transform becomes
Fα(k) =1
Γ(αk + 1)
[Dαkx0f(x)
]x=x0
. (3.21)
Theorem 3.2.5. (Erturk and Momani, 2010) If f(x) = (x− x0)γ, then
Fα(k) = δ(k − γ
α), (3.22)
where δ(k) =
1, k = 0,
0, k 6= 0.
38
Proof. First f(x) satisfies the condition in Theorem 3.1.5. Thus
Fα(k) =1
Γ(αk + 1)
[Dαkf(x)
]x=x0
=1
Γ(αk + 1)
1
Γ(m− αk)
∫ x
x=x0
(x− t)m−αk−1 f (m)(t) dt
=1
Γ(αk + 1)Γ(m− αk)
∫ x
x=x0
(x− t)m−αk−1 Γ(γ + 1)
Γ(γ −m+ 1)(t− x0)γ−m dt
=Γ(γ + 1)
Γ(αk + 1)Γ(m− αk)Γ(γ −m+ 1)
∫ x
x=x0
(x− t)m−αk−1 (t− x0)γ−m dt.
Using t = x0 + (x− x0)τ , we get
Fα(k) =Γ(γ + 1)
Γ(αk + 1)Γ(m− αk)Γ(γ −m+ 1)∫ 1
0
(x− x0)m−αk−1(1− τ)m−αk−1 (x− x0)γ−mτ γ−m (x− x0)dτ
=Γ(γ + 1) (x− x0)γ−αk
Γ(αk + 1)Γ(m− αk)Γ(γ −m+ 1)
∫ 1
0
(1− τ)m−αk−1 τ γ−m dτ.
By the definition of Beta function Equation (1.9), we have
Fα(k) =Γ(γ + 1) (x− x0)γ−αk
Γ(αk + 1)Γ(m− αk)Γ(γ −m+ 1)β(m− αk, γ −m+ 1)
=Γ(γ + 1) (x− x0)γ−αk
Γ(αk + 1)Γ(m− αk)Γ(γ −m+ 1)
Γ(m− αk)Γ(γ −m+ 1)
Γ(γ − αk + 1)
∣∣∣∣x=x0
=Γ(γ + 1) (x− x0)γ−αk
Γ(αk + 1)Γ(γ − αk + 1)
∣∣∣∣x=x0
=
1, k = γα,
0, k 6= γα.
= δ(k − γ
α).
39
Theorem 3.2.6. (a) If f(x) = e(x−x0), then
Fα(k) = E1(k) =
1
(αk)!, αk ∈ Z+,
0, αk /∈ Z+.(3.23)
(b) If f(x) = e−(x−x0), then
Fα(k) = E2(k) =
(−1)αk
(αk)!, αk ∈ Z+,
0, αk /∈ Z+.(3.24)
Proof. We can prove Proposition 3.2.6 by the Taylor series expansion for ex.
Theorem 3.2.7. (Erturk and Momani, 2010) If f(x) = Dβx0g(x), m − 1 < β ≤ m
and the function g(x) satisfies the conditions in Theorem (3.1.5), then
Fα(k) =Γ(αk + β + 1)
Γ(αk + 1)Gα
(k +
β
α
). (3.25)
Proof. From the definition of generalized differential transform (3.18), we have
Fα(k) =1
Γ(αk + 1)
[Dαkx0f(x)
]x=x0
=1
Γ(αk + 1)
[Dαkx0Dβx0g(x)
]x=x0
.
40
Since g(x) satisfies the conditions of Theorem 3.1.5, we have
Fα(k) =1
Γ(αk + 1)
[Dαk+βx0
g(x)]x=x0
=1
Γ(αk + 1)
Γ(αk + β + 1)
Γ(αk + β + 1)
[Dαk+βx0
g(x)]x=x0
=Γ(αk + β + 1)
Γ(αk + 1)
1
Γ(α(k + βα
) + 1)
[Dα(k+ β
α)
x0 g(x)
]x=x0
=Γ(αk + β + 1)
Γ(αk + 1)Gα(k +
β
α).
Theorem 3.2.8. (Erturk and Momani, 2010) If f(x)=∫ xx0g(t) dt, then
Fα(k) =Gα(k − 1
α)
αk, (3.26)
where k ≥ 1α
.
Proof. From Equation (3.19) we can write g(x) as
g(x) =∞∑k=0
Gα(k)(x− x0)αk.
Thus
f(x) =
∫ x
x0
∞∑k=0
Gα(k)(t− x0)αkdt.
41
Interchanging the order of summation and integration, we have
f(x) =∞∑k=0
∫ x
x0
Gα(k)(t− x0)αkdt
=∞∑k=0
Gα(k)
∫ x
x0
(t− x0)αkdt
=∞∑k=0
Gα(k)
[(t− x0)αk+1
αk + 1
]xx0
=∞∑k=0
Gα(k)(x− x0)αk+1
αk + 1.
Now let k = m− 1α
. Then
f(x) =∞∑
m= 1α
Gα(m− 1
α)(x− x0)αm
αm.
There for, by Equation (3.19), we have
Fα(k) =Gα(k − 1
α)
αk,
where k ≥ 1α
.
Theorem 3.2.9. (Erturk and Momani, 2010) If f(x) = g(x)∫ xx0h(t)dt, then
Fα(k) =k∑
k1= 1α
Hα(k1 − 1α
)
αk1
Gα(k − k1), (3.27)
where k1 ≥ 1α
.
Proof. First let c(x) =∫ xx0h(t)dt. Then from Theorem 3.2.8, we have Cα(k) =
42
Hα(k− 1α
)
αk, where k1 ≥ 1
α. Using Theorem 3.2.1, we get
Fα(k) =k∑
k1=0
Cα(k1)Gα(k − k1)
=k∑
k1= 1α
Hα(k − 1α
)
αkGα(k − k1).
Theorem 3.2.10. (Erturk and Momani, 2010) If f(x)=∫ xx0h1(t)h2(t) · · ·hn−1(t)hn(t)dt
then
Fα(k) =1
αk
k− 1α∑
kn−1=0
kn−1∑kn−2=0
· · ·k3∑k2=0
k2∑k1=0
H1(k1)H2(k2−k1) · · ·Hn−1(kn−1−kn−2)Hn(k−kn−1−1
α),
(3.28)
where k≥ 1α
.
Proof. First let
c(t) = h1(t)h2(t) · · ·hn−1(t)hn(t).
Then f(x)=∫ xx0c(t)dt, by Theorem (3.2.8), we obtain
Fα(k) =Cα(k − 1
α)
αk, k ≥ 1
α,
where Cα(k) is the transform of c(t). Thus, by Theorem 3.2.3, we have
Cα(k) =k∑
kn−1=0
kn−1∑kn−2=0
· · ·k3∑k2=0
k2∑k1=0
H1(k1)H2(k2−k1) · · ·Hn−1(kn−1−kn−2)Hn(k−kn−1).
43
Then we get
Fα(k) =1
αk
k− 1α∑
kn−1=0
kn−1∑kn−2=0
· · ·k3∑k2=0
k2∑k1=0
H1(k1)H2(k2−k1) · · ·Hn−1(kn−1−kn−2)Hn(k−kn−1−1
α),
where k≥ 1α
.
Theorem 3.2.11. (Erturk and Momani, 2010) If
f(x) = [g1(x)g2(x) · · · gm−1(x)gm(x)]
∫ x
x0
h1(t)h2(t) · · ·hn−1(t)hn(t)dt,
then
Fα(k) =k∑
k1= 1α
1
αk
k1− 1α∑
jn−1=0
jn−1∑jn−2=0
· · ·j3∑j2=0
j2∑j1=0
k−k1∑im−1=0
im−1∑im−2=0
G1(i1)G2(i2 − i1) · · ·Gm−1(im−1 − im−2)
Gm(k − im−1 − k1)×H1(j1)H2(j2 − j1) · · ·Hn−1(jn−1 − jn−2)Hn(k − jn−1 −1
α), (3.29)
where k≥ 1α
.
Proof. First let c(x) = g1(x)g2(x) · · · gm−1(x)gm(x) and z(t) = h1(t)h2(t) · · ·hn−1(t)hn(t).
Then f(x) = c(x)∫ xx0z(t)dt and by Theorem 3.2.9, we have
Fα(k) =
k1= 1α∑
k
Zα(k1 − 1α
)
αk1
Cα(k − k1), k ≥ 1
α.
Now by Theorem (3.2.3), we have
Cα(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),
44
and
Zα(k) =k∑
jn−1=0
jn−1∑jn−2=0
· · ·j3∑j2=0
j2∑j1=0
H1(j1)H2(j2−j1) · · ·Hn−1(jn−1−jn−2)Hn(k−jn−1).
Therefore we obtain
Fα(k) =k∑
k1= 1α
1
αk
k1− 1α∑
jn−1=0
jn−1∑jn−2=0
· · ·j3∑j2=0
j2∑j1=0
k−k1∑im−1=0
im−1∑im−2=0
G1(i1)G2(i2 − i1) · · ·Gm−1(im−1 − im−2)
Gm(k − im−1 − k1)×H1(j1)H2(j2 − j1) · · ·Hn−1(jn−1 − jn−2)Hn(k − jn−1 −1
α),
where k≥ 1α
.
45
Chapter 4
Applications of GDTM
Chapter 4
Applications of GDT Method
In this chapter we will study the application of GDT to some fractional integro-
differential equations. Moreover we will apply the GDT to fractional first Painleve
equation. I used the Mathematica program to solve the examples in this chapter.
4.1 Examples
Example 4.1. (Erturk and Momani, 2010) Consider the linear fractional integro-
differential equation
(D0.75y)(t) =−t2et
5y(t) +
6t2.25
Γ(3.25)+ et
∫ t
0
sy(s)ds (4.1)
subject to the initial condition y(0) = 0. If we set α = 14, then, by Theorem 3.2.7,
we get the GDT of (D0.75y) as
GDT (D0.75y) =Γ(k
4+ 7
4)
Γ(k4
+ 1)Y 1
4(k + 3).
47
By Theorem 3.2.3 and Proposition 3.2.5, we have the GDT of −t2et
5y(t) as
GDT
(−t2et
5y(t)
)=−1
5
k∑k2=0
k2∑k1
δ(k1 − 8)E1(k2 − k1)Y 14(k − k2),
where the E1(k) is the GDT of et. Using Theorem 3.2.5, we can find the GDT of
6t2.25
Γ(3.25)as
GDT
(6t2.25
Γ(3.25)
)=
6
Γ(3.25)δ(k − 9).
Finally from Theorem 3.2.11, we find the GDT of et∫ t
0sy(s)ds to be
GDT
(et∫ t
0
sy(s)ds
)=
k∑k1=4
4
k1
k1−4∑j1=0
δ(j1 − 4)Y 14(k1 − j1 − 4)E1(k − k1).
Thus the GDT of Equation (4.1), becomes
Y 14(k + 3) =
Γ(k4
+ 1)
Γ(k4
+ 74)
[−1
5
k∑k2=0
k2∑k1
δ(k1 − 8)E1(k2 − k1)Y 14(k − k2) +
6
Γ(3.25)δ(k − 9)
+k∑
k1=4
4
k1
k1−4∑j1=0
δ(j1 − 4)Y 14(k1 − j1 − 4)E1(k − k1)
]. (4.2)
Therefore. we have Y 14(k) = 0 for k=0,1,2,· · · ,11, and Y 1
4(12) = 1, and for k ≥ 13
Y 14(k) = 0. Now from the inverse transformation equation (3.19), we obtain
y(t) = Y 14(12)t
124
= t3.
48
Let as show that y(t) is the exact solution. The left hand side in Equation (4.1), by
the definition of Riemann-Liouville derivative (7), become
(D0.75y)(t) =1
Γ(0.25)
∫ t
0
3(t− s)−0.75s2ds.
By integration by parts, we have
(D0.75y)(t) =3
Γ(0.25)
[− 4(t− s)0.25s2 − 32
5(t− s)1.25s− 128
45(t− s)2.25
]t0
,
=3
Γ(0.25)× 128
45t2.25,
=128
15 Γ(0.25)t2.25.
Now for the right hand side, we get
−t2et
5y(t) +
6t2.25
Γ(3.25)+ et
∫ t
0
sy(s)ds =−t5et
5+
6t2.25
Γ(3.25)+ et
∫ t
0
s4ds,
=6t2.25
Γ(3.25).
Since Γ(3.25) = 4564
Γ(0.25), the right hand side becomes 12815 Γ(0.25)
t2.25 which is the left
hand side. So that the function y(t) = t3 is the exact solution of Equation (4.1).
As another case, if we set α = 34, then applying the GDT to both sides of Equation
(4.1), we find
Y 34(k + 1) =
Γ(3k4
+ 1)
Γ(3k4
+ 74)
[−1
5
k∑k2=0
k2∑k1
δ(k1 −8
3)E1(k2 − k1)Y 3
4(k − k2) +
6
Γ(3.25)δ(k − 3)
+k∑
k1= 43
4
3k1
k1− 43∑
j1=0
δ(j1 −4
3)Y 3
4(k1 − j1 −
4
3)E1(k − k1)
]. (4.3)
49
Then from the initial condition we have Y 34(k) = 0, for k = 0,1,2,3, Y 3
4(4) = 1, and
Y 34(k) = 0, for k ≥ 5. By the inverse transformation equation (3.19), we have
y(t) = Y 34(4)t3
= t3.
Thus we get the same result as the case α = 14.
Example 4.2. (Yıldırım et al., 2010) Consider the following nonlinear fractional
integro-differential equation
(Dµy)(t) = 1 +
∫ t
0
e−xy2(x)dx, 0 < t ≤ 1, 3 < µ ≤ 4, (4.4)
with the boundary conditions
y(0) = 1, y(1) = e, y′′(0) = 1, y′′(1) = e.
Using GDT, Theorem 3.2.1, Theorem 3.2.3, Theorem 3.2.7, Theorem 3.2.10, Propo-
sitions 3.2.5 and Propositions 3.2.6, the GDT of Equation (4.4) leads
Yα(k+µ
α) =
Γ(αk + 1)
Γ(αk + µ+ 1)
[δ(k) +
k− 1α∑
k2=0
1
αk
k2∑k1=0
E2(k1)Yα(K2− k1)Yα(k− k2−1
α)
],
(4.5)
where E2 is the transform of e−t. The boundary conditions are transformed as
Yα(0) = 1, Yα(1
α) = A, Yα(
2
α) =
1
2, Yα(
3
α) =
B
6, (4.6)
50
where A = y′(0) and B = y(3)(0). To obtain the other boundary conditions we use
(3.19). As a result we have
e =n∑k=0
Yα(k), e =n∑k=0
αk(αk − 1)Yα(k). (4.7)
We use the recurrence relation (4.5) and the transform of the boundary condition
(4.6) to calculate Yα(k), 1 ≤ k ≤ 41. After that we solve Equation (4.7). We get A
and B for different value of µ and α in Table 1.
Table 1: The value of A and B for Example 4.2
µ = 3.25, α = 0.25 µ = 3.5, α = 0.5 µ = 3.75, α = 0.25 µ = 4, α = 1
A 0.92464246386389 0.94771552784705 0.95897518300426 0.96252966411221
B 1.09091770795158 1.14134184158935 1.21523661695289 1.29760132651330
By the inverse transformation equation (3.19), we find the values of y(t) ≈41∑k=0
Yα(k)(x−
x0)αk as given in Table 2.
51
Table 2: Numerical results for Example 4.2
µ = 3.25, α = 0.25 µ = 3.5, α = 0.5 µ = 3.75, α = 0.25 µ = 4, α = 1
t y(t)
0 1 1 1 1
0.1 1.09771390815997 1.09998895160648 1.10111077507635 1.10147339851847
0.2 1.20702751427101 1.21137193386889 1.21355937178814 1.21430263800801
0.3 1.32970257155423 1.33571680913262 1.33881851248205 1.33993445700234
0.4 1.46758385027066 1.47471349572387 1.47847983973151 1.47991287402151
0.5 1.62255658631268 1.63013829846679 1.63423500963543 1.63587548073304
0.6 1.79651375935473 1.80382615087404 1.80785869357487 1.80954744965809
0.7 1.99132268690975 1.99764282022245 2.00118998644618 2.00273258925801
0.8 2.20878663175633 2.21345355858212 2.21611023030709 2.21730060428921
0.9 2.45059825481466 2.45308558108126 2.45451549407104 2.45516950309659
1 2.71828182845905 2.71828182845905 2.71828182845905 2.71828182845905
Example 4.3. (Yıldırım et al., 2010) Consider the linear fractional integro-differential
equation
Dµy(x) = x(1 + ex) + 3ex + y(x)−∫ x
0
y(t)dt, 0 < t ≤ 1, 3 < µ ≤ 4, (4.8)
where 0 < x < 1, 3 < µ ≤ 4, with the following boundary conditions:
y(0) = 1, y(1) = 1 + e, y′′(0) = 2, y′′(1) = 3e.
Using GDT Theorem 3.2.1, Theorem 3.2.2, Theorem 3.2.7, Theorem 3.2.8, Propo-
sitions 3.2.5 and Propositions 3.2.6, GDT of Equation (4.8) is the following recur-
52
rence relation
Yα(k+µ
α) =
Γ(αk + 1)
Γ(αk + µ+ 1)
[ k∑k1=0
E1(k−k1)δ(k1−1
α)+3E1(k)+δ(k− 1
α)+Yα(k)−
Yα(k − 1α
)
αk
].
(4.9)
From the boundary conditions we find
Yα(0) = 1, Yα(1
α) = A, Yα(
2
α) = 1 and Yα(
3
α) =
1
6B, (4.10)
where A = y′(0) and B = y(3)(0). For the other boundary conditions from inverse
Equation (3.19) we have
1 + e =N∑k=0
Yα(k), 3e =N∑k=0
αk(αk − 1)Yα(k). (4.11)
We use the recurrence relation (4.9) and the transformed boundary conditions (4.10)
to calculate Yα(k), 1 ≤ k ≤ 41, we take N = 41 because the values of Yα(k) is so
small when k > 41. After that we solve Equation (4.11). We get A and B for
different value of µ and α in Table 3.
Table 3: The value of A and B for Example 4.3
µ = 3.25, α = 0.25 µ = 3.5, α = 0.5 µ = 3.75, α = 0.25 µ = 4, α = 1
A 0.98974472947971 0.94439050993174 0.89781890023719 0.85462784878015
B 2.7287260222733 3.52144157408021 4.16577542401644 4.67695672374404
By the inverse transformation equation (3.19), we find the value of y(t) ≈41∑k=0
Yα(k)(x−
x0)αk and the values of y(t) at t = 0.1j, j = 0, 1, 2, · · · , 10 are given in Table 4.
53
Table 4: Numerical results for Example 4.3
µ = 3.25, α = 0.25 µ = 3.5, α = 0.5 µ = 3.75, α = 0.25 µ = 4, α = 1
t y(t)
0 1 1 1 1
0.1 1.10950369262754 1.10505559538800 1.10950369262754 1.09624677475153
0.2 1.24236144190876 1.23393798983689 1.24236144190876 1.21723902720088
0.3 1.40235570035610 1.39079904097706 1.40235570035610 1.36785682218008
0.4 1.59374044947830 1.58016553421027 1.59374044947830 1.55317211770043
0.5 1.82130448531571 1.80700048872745 1.82130448531571 1.77850569745885
0.6 2.09045323896625 2.07678069159722 2.09045323896625 2.04949307216779
0.7 2.40730385514182 2.39558688058562 2.40730385514182 2.37216041364104
0.8 2.77879325812683 2.77020652425362 2.77879325812683 2.75301167323103
0.9 3.21280008133921 3.20825005479628 3.21280008133921 3.19912814126617
1 3.71828182845905 3.71828182845905 3.71828182845905 3.71828182845905
Example 4.4. (Odibat and Momani, 2008) Consider the homogeneous linear frac-
tional differential equation
Dαy(t) = −y(t), (4.12)
where y(0)=1, t>0 and 0< α≤1. By GDT we have
Yα(k + 1) = − Γ(αk + 1)
Γ(α(k + 1) + 1)Yα(k). (4.13)
54
The initial condition is transformed as Yα(0) = 1. Now for α=0.25, we can find
Y0.25(k) =(−1)k
Γ(0.25k + 1). Thus, by inverse transform equation (3.19), we have
y(t) =∞∑k=0
Y0.25(k)t0.25k
=∞∑k=0
(−1)k
Γ(0.25k + 1)t0.25k.
By the same way for α = 0.5, we find Y0.5(k) = (−1)k
Γ(0.5k+1), and hence
y(t) =∞∑k=0
Y0.5(k)t0.5k
=∞∑k=0
(−1)k
Γ(0.5k + 1)t0.5k.
For α = 0.75, we find Y0.75(k) = (−1)k
Γ(0.75k+1), and as a result we get
y(t) =∞∑k=0
Y0.75(k)t0.75k
=∞∑k=0
(−1)k
Γ(0.75k + 1)t0.75k.
Also for α = 1, we find Y1(k) = (−1)k
Γ(k+1), and
y(t) =∞∑k=0
Y1(k)tk
=∞∑k=0
(−1)k
Γ(k + 1)tk
=∞∑k=0
(−1)k
k!tk
= e−t.
55
The last case agree with the classical derivative case.
Example 4.5. (Odibat and Momani, 2008) Consider the inhomogeneous linear
fractional differential equation
Dαy(t) =2
Γ(3− α)t2−α − 1
Γ(2− α)t1−α − y(t) + t2 − t, (4.14)
where y(0)=0, t>0, 0< α ≤1. By the GDTM we have
Yα(k+1) =Γ(αk + 1)
Γ(α(k + 1) + 1)
[2δ(k − 2−α
α)
Γ(3− α)−δ(k − 1−α
α)
Γ(2− α)− Yα(k) + δ(k − 2
α)− δ(k − 1
α)
].
(4.15)
The initial condition is transformed as Yα(0) = 0. First if we take α=0.25, we get
Y0.25(4) = −1, Y0.25(8) = 1, Y0.25(k) = 0 ∀k ∈ N-{4,8}. From the inverse transform
equation (3.19), we have
y(t) = t2 − t.
For α=0.5, also we get Y0.5(2) = −1, Y0.5(4) = 1, Y0.5(k) = 0 ∀k ∈ N-{2,4}, from
the inverse transform equation (3.19), we have
y(t) = t2 − t.
Finally if α=0.75, we get Y0.75(k) = 0 ∀k ∈ N. Using the inverse transform equation
(3.19), we have the zero solution. If α=1, also we get Y1(1) = −1, Y1(2) = 1,
Y1(k) = 0 ∀k ∈ N-{1,2}, and from the inverse transform equation (3.19), we have
y(t) = t2 − t.
56
Example 4.6. (Erturk et al., 2008) Consider the following linear fractional differ-
ential equation
dµy
dtµ+ ωµ−β
dβy
dtβ= 0, 1 < µ ≤ 2, 0 ≤ β < 1. (4.16)
with initial conditions y(0) = 0, y′(0) = 1. The initial conditions is transformed as
Yα(0) = 0, Yα( 1α
) = 1. First we will fix β = 0, and we will consider the cases µ=2,
1.75, 1.5 and 1.25. Equation (4.16) becomes
dµy
dtµ+ ωµ y = 0. (4.17)
Now for µ=2, we have the simple harmonic oscillator which is
d2y
dt2+ ω2 y = 0.
For α=1, we have by GDT
Y1(k + 2) = −Γ(k + 1)
Γ(k + 3)
[ω2 Y1(k)
].
Then we calculated Y1(k) for k = 2, 3, · · · and we find that Yα(k) can be written as
Y1(k) =
0, if k is even
(−1)k−12 ωk−1
k!, if k is odd.
57
By the inverse transform equation (3.19), we get
y(t) =∞∑k=0
Y1(k)tk
= t− ω2
3!t3 +
ω4
5!t5 − · · ·
=1
ω
[ωt− ω3
3!t3 +
ω5
5!t5 − · · ·
]=
1
ωsin(ωt).
This is the exact solution of simple harmonic oscillator.
Now for α=0.25, by GDT Equation (4.17) becomes as
Y 14(k + 4µ) = −
Γ(14k + 1)
Γ(14k + µ+ 1)
[ωµY 1
4(k)]. (4.18)
We use the recurrence relation (4.18) and the transform of boundary conditions to
calculate Y 14(k), 0 < k ≤ 296, we take N = 296 because the values of Yα(k) is so
small when k > 296. Then inverse transform equation (3.19), we find the numerical
result of y(t) ≈296∑k=0
Y 14(k)(x− x0)
14k and the values of y(t) at t = 0, 1, 2, · · · , 10 are
given in Table 5 and Figure 4.1.
58
Table 5: Numerical results for Example 4.6 when α=0.25 and β=0
µ = 1.25 µ = 1.5 µ = 1.75
t y(t)
0 0 0 0
1 0.68228514566433 0.73748224790189 0.7921721503389
2 0.83851676635498 0.82993969202460 0.85140682890250
3 0.77629730669340 0.57534600869266 0.35415666341335
4 0.67609971811335 0.31295156839240 -0.15087740469876
5 0.59839122154365 0.18202084109387 -0.29846033557419
6 0.54879160372066 0.16239554720194 -0.11598355408457
7 0.51799360603922 0.18496380339671 0.13626358099416
8 0.49720508207815 0.20177973265536 0.24113963468050
9 0.48129512610695 0.19998315963671 0.16987130045728
10 0.46791064769444 0.18672750847812 0.03546892784447
Figure 4.1: y(t) for β=0 and α = 0.25 : (-)µ=1.75, (..)µ=1.5, (- -)µ=1.25
59
As another case, fix µ=2 and change β between 0.25, 0.5 and 0.75. Then let α=0.25.
By GDT, we get
Y 14(k + 8) = −
Γ(14k + 1)
Γ(14k + 3)
[ω2−βΓ(1
4k + β + 1)
Γ(14k + 1)
Y 14(k + 4β)
]= −
Γ(14k + β + 1)
Γ(14k + 3)
[ω2−βY 1
4(k + 4β)
]. (4.19)
We use the recurrence relation (4.19) and the transform of boundary conditions to
calculate Y 14(k), 0 < k ≤ 200, we take N = 200 because the values of Yα(k) is so
small when k > 200. Then inverse transform equation (3.19), we find the numerical
result of y(t) ≈200∑k=0
Y 14(k)(x− x0)
14k and the values of y(t) at t = 0, 1, 2, · · · , 10 are
given in Table 6 and Figure 4.2.
Table 6: Numerical results for Example 4.6 when α=0.25 and µ=2
β = 0.25 β = 0.5 β = 0.75
t y(t)
0 0 0 0
1 0.79217215033897 0.73748224790189 0.68228514566433
2 0.85140682890250 0.82993969202460 0.83851676635498
3 0.35415666341335 0.57534600869266 0.77629730669340
4 -0.15087740469876 0.31295156839240 0.67609971811335
5 -0.2984603355742 0.18202084109387 0.59839122154365
6 -0.11598355408457 0.16239554720194 0.54879160372066
7 0.13626358099416 0.18496380339671 0.51799360603922
8 0.24113963468050 0.20177973265536 0.49720508207815
9 0.16987130045728 0.19998315963671 0.48129512610695
10 0.03546892784447 0.18672750847812 0.46791064769444
60
Figure 4.2: y(t) for µ=2 and α = 0.25 : (- -)β=0.75, (..)β=0.5, (-)β=0.25
Example 4.7. (Bansal and Jain, 2016) Consider the Bagley-Torvik equation
D2y(t) + AD32y(t) +By(t) = g(t). (4.20)
We will take two special cases. The first case is
D2y(t) +D32y(t) + y(t) = t2 + 4
√t
π+ 2, (4.21)
withe initial conditions y(0) = 0, y′(0) = 0. The second case is
D2y(t) +D32y(t) + y(t) = t+ 1, (4.22)
61
withe initial conditions y(0) = 1, y′(0) = 1. Solving Equation (4.21) by GDT and
taking α=0.5, we have
Y 12(k+4) =
Γ(12k + 1)
Γ(12k + 3)
[2δ(k) +
4√πδ(k − 1) + δ(k − 4)−
Γ(12k + 21
2)
Γ(12k + 1)
Y 12(k + 3)− Y 1
2(k)
].
(4.23)
The initial conditions are transformed as Y 12(0) = 0 and Y 1
2(2) = 0. For other
values of Y 12(k), we get Y 1
2(4) = 1 and Y 1
2(k) = 0 ∀k 6= 4. By the inverse transform
equation (3.19) we get
y(t) = t2.
Now for the Equation (4.22) by GDTM and take α=0.5 we have
Y 12(k + 4) =
Γ(12k + 1)
Γ(12k + 3)
[δ(k) + δ(k − 2)−
Γ(12k + 21
2)
Γ(12k + 1)
Y 12(k + 3)− Y 1
2(k)
]. (4.24)
The initial conditions are transformed as Y 12(0) = 1 and Y 1
2(2) = 1. By calculating
the Y 12(k), ∀k 6= 0, 2 we find Y 1
2(k) = 0 . By the inverse transform equation (3.19)
we get
y(t) = t+ 1.
Example 4.8. (Erturk et al., 2008) Consider the following initial value problem
Dµy(t)− ADβ1y(t)−BDβ2y(t) = g(t), (4.25)
with initial conditions
y(i)(0) = ci i = 0, 1, 2, · · ·
where 0< β2 < β1 < µ, m − 1 < µ ≤ m, b − 1 < β1 ≤ b, s − 1 < β2 ≤ s and m,
b and s ∈ N. If we put µ = 2, β1 = 32
and β2 = 0,then we get the Bagley-Torvik
62
equation. We will fix µ = 2, β2 = 0, A = B = −1 and g(t) = 0, we have
D2y(t) +Dβ1y(t) + y(t) = 0, (4.26)
with initial condition y(0) = 1 and y′(0) = 1. If we take α = 0.5, then by GDT, we
get
Y 12(k + 4) = −
Γ(12k + 1)
Γ(12k + 3)
[Γ(1
2k + β1 + 1)
Γ(12k + 1)
Y 12(k + 2β1) + Y 1
2(k)
]. (4.27)
And the initial conditions transform as
Y 12(0) = 1, Y 1
2(1) = 0, Y 1
2(2) = 1, and Y 1
2(3) = 0. (4.28)
We will change β1 between 0.5, 1 and 1.5. Using Equations (4.27) and (4.28) to
calculate Y 12(k) for k = 4, 5, 6, · · · , 250, we take N = 250 because the values of
Yα(k) is so small when k > 250. Then by inverse transform equation (3.19) we can
find the numerical result for y(t) ≈250∑k=0
Y 12(k)(x − x0)
12k and the values of y(t) at
t = 0, 1, 2, · · · , 10 are given in Table 7 and Figure 4.3.
63
Table 7: Numerical results for Example 4.8 when α=0.5, µ=2 and β2 = 0
β1 = 0.5 β1 = 1 β1 = 1.5
t y(t)
0 1 1 1
1 1.22409741741251 1.19320734850639 1.60547009109590
2 0.39160568690478 0.56985399481222 1.49225256629752
3 -0.0833154908998 0.00888787660963 0.91246117978678
4 0.13309480777701 -0.20265264815597 0.23223689734488
5 0.39685609216470 -0.16253298732755 -0.24722231144687
6 0.33807337130083 -0.05318181220415 -0.39471576502379
7 0.16572806220600 0.01799732467266 -0.25325246315076
8 0.12603210067992 0.03370846884274 0.02881428761821
9 0.18620710897531 0.01987040772525 0.28977622906387
10 0.21184752310356 0.00321536387689 0.42494519415940
Figure 4.3: y(t) for µ=2 and β2 = 0 : (-)β1=0.5, (..)β1=1, (- -)β1=1.5
64
Example 4.9. (Erturk et al., 2008) Consider the following non-linear initial value
problem
Dµy(t)− y(t)Dβy(t)− 1 = 0, m− 1 < µ ≤ m, 0 ≤ β < µ. (4.29)
Subject to the initial condition y(i)(0) = 0, i = 0, 1, 2, · · · ,m− 1. Selecting β = 0.5,
α = 0.5, then Equation (4.29) become as
Dµy(t) = y(t)D12y(t) + 1, 1 < µ ≤ 2.
Then change µ between 1, 1.5 and 2. By GDTM we have
Y 12(k + 2µ) =
Γ(12k + 1)
Γ(12k + µ+ 1)
[k∑l=0
Γ(12l + 3
2)
Γ(12l + 1)
Y 12(l + 1)Y 1
2(k − l) + δ(k)
].(4.30)
The initial conditions transform as
Y 12(k) = 0, k = 0, 1, · · · , 2µ− 1. (4.31)
We will change µ between 1, 1.5 and 2. Using Equations (4.30) and (4.31) to
calculated Y 12(k) for k = 0, 1, 2, · · · , 200, we take N = 200 because the values of
Yα(k) is so small when k > 200. Then by inverse transform equation (3.19) we can
find the numerical result for y(t) ≈200∑k=0
Y 12(k)(x − x0)
12k and the values of y(t) at
t = 0, 0.1, 0.2, · · · , 1 are given in Table 8 and Figure 4.4.
65
Table 8: Numerical results for Example 4.9 when α=0.5 and β=2
µ = 1 µ = 1.5 µ = 2
t y(t)
0 0 0 0
0.1 0.10145957446229 0.03164119704450 0.01000019223216
0.2 0.20861544471241 0.08973839419417 0.04000870097289
0.3 0.32517494106533 0.16582396255616 0.09008097180471
0.4 0.45572819130117 0.25780070611355 0.16039455041540
0.5 0.60707161536039 0.36551858650512 0.25134970565133
0.6 0.79052583933837 0.49015616590979 0.36369492270078
0.7 1.02745815705366 0.63419091111478 0.49868280927614
0.8 1.36574799258948 0.80167938199360 0.65826890432093
0.9 1.94870667696565 0.99887592260523 0.84537557600682
1 3.63876987167381 1.23538864948199 1.06425814323943
Figure 4.4: y(t) for β=0.5 and α = 0.5 : (-)µ=1, (..)µ=1.5, (- -)µ=2
66
4.2 Fractional First Painleve equation
In this section we will study the solutions of the fractional first Painleve equation
(FPI)
Dβy = 6y2 + t, (4.32)
where 1 < β ≤ 2, by the GDTM. Equation (4.32) with the initial conditions
y(0) = 0, y′(0) = 1 has been solved by iterative reproducing kernel algorithm(Arqub
and Maayah, 2018). We will apply the GDTM to equation (4.32) with the following
four cases of initial conditions:
1. y(0) = 0, y′(0) = 0,
2. y(0) = 0, y′(0) = 1,
3. y(0) = 1, y′(0) = 0,
4. y(0) = 1, y′(0) = 1.
Now if we choose β=1.25 and α = 14. Then by GDT we have
Y 14(k + 5) =
Γ(14k + 1)
Γ(14k + 9
4)
[6
k∑l=0
Y 14(l)Y 1
4(k − l) + δ(k − 4)
](4.33)
The four cases of the initial conditions are transformed as
1. Y 14(0) = 0, Y 1
4(1) = 0, Y 1
4(2) = 0, Y 1
4(3) = 0 and Y 1
4(4) = 0,
2. Y 14(0) = 0, Y 1
4(1) = 0, Y 1
4(2) = 0, Y 1
4(3) = 0 and Y 1
4(4) = 1,
3. Y 14(0) = 1, Y 1
4(1) = 0, Y 1
4(2) = 0, Y 1
4(3) = 0 and Y 1
4(4) = 0,
4. Y 14(0) = 1, Y 1
4(1) = 0, Y 1
4(2) = 0, Y 1
4(3) = 0 and Y 1
4(4) = 1.
67
Using Equations (4.33) and the initial conditions to calculating Y 14(k) for k =
0, 1, 2, · · · , 330, we take N = 330 because the values of Yα(k) is so small when
k > 330. Then by inverse transform equation (3.19) we can find the numerical
result for y(t) ≈330∑k=0
Y 14(k)(x− x0)
14k and the values of y(t) at t = 0, 0.1, 0.2, · · · , 1
are given in Table 9.
Table 9: Numerical results for FPI with β=1.25 and α=0.25
Case 1 Case 2 Case 3 Case 4
t y(t)
0 0. 0. 1. 1.
0.1 0.002206092587085 0.10305118820628 1.3884666377310564 1.5277913871352868
0.2 0.010503321665671 0.21913222374815 2.5384952410430417 3.1584036463264225
0.3 0.026233726176523 0.36213063231879 9.322170024226963 20.731784374578634
0.4 0.050470001377871 0.55731934189941 660873.0783039257 2.61749×108
0.5 0.084500445377701 0.86094088636104 3.48195×1013 1.85537×1016
0.6 0.130237568733330 1.43368994496577 1.07146×1020 6.20991×1022
0.7 0.190810870476772 2.97920758566714 3.61691×1025 2.20094×1028
0.8 0.271631543885182 15.9636859162489 2.31703×1030 1.46142×1033
0.9 0.382600295284582 7906.33255655693 4.10978×1034 2.66947×1037
1. 0.543345484268809 2.45492×107 2.63007×1038 1.75241×1041
For β=1.5 and α=0.5, then by GDT we have
Y 12(k + 3) =
Γ(12k + 1)
Γ(12k + 5
2)
[6
k∑l=0
Y 12(l)Y 1
2(k − l) + δ(k − 2)
](4.34)
The four cases of the initial conditions are transformed as
68
1. Y 12(0) = 0, Y 1
2(1) = 0, and Y 1
2(2) = 0,
2. Y 12(0) = 0, Y 1
2(1) = 0, and Y 1
2(2) = 1,
3. Y 12(0) = 1, Y 1
2(1) = 0, and Y 1
2(2) = 0,
4. Y 12(0) = 1, Y 1
2(1) = 0, and Y 1
2(2) = 1.
Using Equations (4.34) and the initial conditions to calculating Y 12(k) for k =
0, 1, 2, · · · , 330, we take N = 330 because the values of Yα(k) is so small when
k > 330. Then by inverse transform equation (3.19) we can find the numerical
result for y(t) ≈330∑k=0
Y 12(k)(x− x0)
12k and the values of y(t) at t = 0, 0.1, 0.2, · · · , 1
are given in Table 10.
Table 10: Numerical results for FPI with β=1.5 and α=0.5
Case 1 Case 2 Case 3 Case 4
t y(t)
0 0. 0. 1. 1.
0.1 0.000951543878665 0.10128219898238 1.15678282198616 1.270090377883161
0.2 0.00538367999743 0.20924750962310 1.53406639514928 1.834973311450455
0.3 0.01484685204048 0.33168992277226 2.31206231227754 3.057040660218881
0.4 0.03053955457452 0.48038377099985 4.21963979134045 6.677461905017484
0.5 0.05357987396060 0.67498205287182 11.9394687957457 37.1021437218602
0.6 0.08518494172773 0.95223365862394 1388.41752178538 8.1494×108
0.7 0.12688052598810 1.39086383434693 1.35851×1013 5.31358×1019
0.8 0.18079867958226 2.19328174624819 3.90656×1022 1.85056×1029
0.9 0.25014660700778 4.05272023869296 1.05342×1031 5.38707×1037
1 0.34001066304450 11.2058798974724 3.95159×1038 2.10762×1045
69
For β=1.7 and α=0.1, then by GDT we have
Y 110
(k + 17) =Γ( 1
10k + 1)
Γ( 110k + 2.7)
[6
k∑l=0
Y 110
(l)Y 110
(k − l) + δ(k − 10)
](4.35)
The four cases of the initial conditions are transformed as
1. Y 110
(k) = 0, for k = 0, 1, · · · , 16,
2. Y 110
(k) = 0, for k = 0, 1, · · · , 9, Y 14(10) = 1, and Y 1
10(k) = 0, fork =
11, 12, · · · , 16,
3. Y 110
(0) = 1, and Y 110
(k) = 0, for k = 1, 2, · · · , 16,
4. Y 110
(0) = 1, Y 110
(k) = 0, for k = 1, 2, · · · , 9, Y 110
(10) = 1, and Y 110
(k) = 0, for
k = 11, 12, · · · , 16.
Using Equations (4.35) and the initial conditions to calculating Y 110
(k) for k =
0, 1, 2, · · · , 330, we take N = 330 because the values of Yα(k) is so small when
k > 330. Then by inverse transform equation (3.19) we can find the numerical
result for y(t) ≈330∑k=0
Y 110
(k)(x−x0)110k and the values of y(t) at t = 0, 0.1, 0.2, · · · , 1
are given in Table 11.
70
Table 11: Numerical results for FPI with β=1.7 and α=0.1
Case 1 Case 2 Case 3 Case 4
t y(t)
0 0. 0. 1. 1.
0.1 0.00047840646478 0.10063447499985 1.10198982777833 1.20827246166575
0.2 0.00310883416491 0.20517143443723 1.3416572867382 1.59005381985771
0.3 0.00929277699505 0.31881865567949 1.76413671832011 2.25052151066583
0.4 0.02022045695566 0.44912578077147 2.52047335869762 3.50665910431151
0.5 0.03699741383226 0.60741129440363 4.03230797400287 6.40853856403214
0.6 0.06072765067536 0.81173972049523 7.81443312020797 16.3898199303591
0.7 0.09261021014764 1.09333037186162 23.0850227186997 100.422810484206
0.8 0.13407371253566 1.51152787347315 176.728040487152 2118.11180561518
0.9 0.18697546214563 2.19343148784927 3418.36834157114 65721.9010900269
1 0.25390758881515 3.45898299673972 80285.5823115865 1.79217×106
Comparing the results for case 2 with the results obtained in (Arqub and Maayah,
2018) we see that the results agree to four significant digits.
For β=2 and α=1, then by GDT we have
Y1(k + 2) =Γ(k + 1)
Γ(k + 3)
[6
k∑l=0
Y1(l)Y1(k − l) + δ(k − 1)
](4.36)
The four cases of the initial conditions are transformed as
1. Y1(0) = 0 and Y1(1) = 0,
2. Y1(0) = 0 and Y1(1) = 1,
3. Y1(0) = 1 and Y1(1) = 0,
71
4. Y1(0) = 1 and Y1(1) = 1.
Using Equations (4.36) and the initial conditions to calculating Y1(k) for k =
0, 1, 2, · · · , 330, we take N = 330 because the values of Yα(k) is so small when
k > 330. Then by inverse transform equation (3.19) we can find the numerical
result for y(t) ≈330∑k=0
Y1(k)(x − x0)k and the values of y(t) at t = 0, 0.1, 0.2, · · · , 1
are given in Table 12.
Table 12: Numerical results for FPI with β=2 and α=1
Case 1 Case 2 Case 3 Case 4
t y(t)
0 0 0 1 1
0.1 0.000166666696429 0.10021674767740 1.03047070993338 1.13255215511016
0.2 0.001333340952412 0.20213945271664 1.1263664313749 1.3442471565806
0.3 0.004500195273941 0.30863074916750 1.30145354657786 1.66856550740472
0.4 0.010668617398949 0.42398628948907 1.5830549490782 2.16745379377078
0.5 0.020844963736745 0.55434011899828 2.02276285430256 2.96255321026678
0.6 0.036050038449190 0.70846208804717 2.72124554649151 4.31635683790429
0.7 0.057338608507627 0.89924993800221 3.89089292018484 6.87983988404422
0.8 0.085834760549389 1.14653172641296 6.03835199271656 12.6511997392176
0.9 0.122790915116629 1.48252443049193 10.6226501190303 30.489014837526
1 0.169681440907945 1.96312822372003 23.3937131859602 152.049523303308
When β = 2 the fractional derivative reduces to the classical derivative and this
case was the subject of many studies. In particular our results in the case 2 are
in good agreement with the results in (Arqub and Maayah, 2018), also its in good
agreement with the results in (A. H. Sakka, pear)
72
Chapter 5
Conclusion
Chapter 5
Conclusion
In this thesis, we studied the fractional calculus and mentioned some examples
and some basic properties and theorems. Fractional calculus is more generalized
than classical calculus. In classical calculus the order of derivative is a positive
integer, but in fractional calculus the order of the derivative may be any real number
α > 0. Then we studied the generalized Taylor’s formula and generalized differential
transform method. In addition we applied the generalized differential transform
method to solve the fractional integro-differential equation and the fractional first
Painleve equation. As recommendations for future study, i propose the following:
1. To consider the initial value problems for fractional partial differential equa-
tions.
2. To extend the work to nonlinear partial differential equations of fractional
order and both initial and boundary value problems.
3. To apply the generalized differential transform method to get solutions for
other fractional order Painleve equations.
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4. To study and investigate if it is possible to modify GDT method to be applied
in case Riemann-Liouville fractional derivatives is used.
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