i

SOLVING FRACTIONAL DIFFERENTIAL EQUATIONS BY USING

CONFORMABLE FRACTIONAL DERIVATIVES DEFINITION

By

Shadi Ahmad Al-Tarawneh

Supervisor

Dr. Khaled Jaber

This Thesis was Submitted in Partial Fulfillment of the Requirements for

the Master’s Degree of Science in Mathematics

Faculty of Graduate Studies

Zarqa University

May, 2016

ii

COMMITTEE DECISION

This Thesis/Dissertation (Solving Fractional Differential Equations by Using Conformable

Fractional Derivatives Definition) was Successfully Defended and Approved on

………………………..

Examination Committee

Signature

Dr. Khaled Jaber (Supervisor)

Assoc. Prof. of Mathematics ------------------------------

------------------------------

Dr. (Member)

------------------------------

Dr. (Member)

------------------------------

Dr. (Member)

------------------------------

iii

ACKNOWLEDGEMENT

In the name of Allah, the most Gracious, most Merciful.

First and foremost, I thank ALLAH for bestowing me with health, patience, and

knowledge to complete this thesis and without ALLAH’s grace, we couldn't have done it.

So to ALLAH returns all the praise and gratitude.

I would like to express my gratitude to Dr. Khaled Jaber, the supervisor of my thesis,

who was a generous and instructor. I was blessed to be supervised by him. Thanks go to

him for his guidance, suggestions and invaluable encouragement throughout the

development of this research.

Also, I should thank with great respect and honor all my professors, doctors and

instructors to be taught by them.

My great gratitude is due to my parents, beloved brothers, sisters and all friends for

their encouragement, support, prayers and being always there for me.

Last, but not least, I would like to thank my friend Omar Al Nasaan and my beloved

wife Ghosoun Al Hindi for their help, support, effort and encouragement was in the end

what made this thesis possible.

iv

Table of Contents

COMMITTEE DECISION ............................................................................................................... ii

ACKNOWLEDGEMENT ................................................................................................................ iii

Table of Contents ................................................................................................................................ iv

List of Symbols .................................................................................................................................. vi

List of Abbreviations ....................................................................................................................... vii

List of Figures and Tables .............................................................................................................. viii

ABSTRACT ........................................................................................................................................ ix

INTRODUCTION .............................................................................................................................. 1

Chapter one: Basic Concepts and Preliminaries ................................................................................. 3

1.1 History of Fractional Calculus ............................................................................................... 3

1.2 Some Special Functions .......................................................................................................... 3

1.2.1. Gamma Function ........................................................................................................... 4

1.2.2. The Beta Function ......................................................................................................... 8

1.2.3 Mittag-Leffler Function .............................................................................................. 10

1.3 The Popular Definitions of Fractional Derivatives/Integrals in Fractional Calculus .............. 12

1.3.1. Riemann-Liouville (RL) ................................................................................................ 13

1.3.2. M.Caputo (1967) .......................................................................................................... 13

1.3.3. Oldham and Spainer (1974) ....................................................................................... 13

1.3.4. Kolwanker and Gangel (1994) .................................................................................. 13

1.3.5. Conformable Fractional Derivative (2014) ............................................................. 13

1.4 Riemann-Liouville (R-L) Fractional Derivative and Integration ....................................... 14

1.4.1. Riemann-Liouville Fractional Integration .............................................................. 14

1.4.2. Riemann-Liouville Fractional Derivative ............................................................... 19

1.5 Caputo Fractional Operator .................................................................................................. 25

1.6 Comparison between Riemann-Liouville and Caputo Fractional Derivative Operators . 36

1.7 Ordinary Differential Equations ........................................................................................... 39

1.7.1. Bernoulli Differential Equation ................................................................................ 39

1.7.2 Second-Order Linear Differential Equations .......................................................... 39

Chapter Two: Conformable Fractional Definition ............................................................................ 41

2.1 Conformable Fractional Derivative ..................................................................................... 41

2.2. Conformable Fractional Integrals ....................................................................................... 52

2.3 Applications ............................................................................................................................ 54

v

2.4. Abel’s Formula and Wronskain for Conformable Fractional Differential Equation ...... 55

2.4.1. The Wronskain ............................................................................................................. 56

2.4.2. Abel’s Formula ............................................................................................................ 57

Chapter 3: Exact Solution of Riccati Fractional Differential Equation ............................................. 59

3.1 Fractional Riccati Differential Equation (FRDE) ............................................................... 59

3.2 Applications: .......................................................................................................................... 67

Future Work ...................................................................................................................................... 70

Conclusions ....................................................................................................................................... 71

REFERENCES.................................................................................................................................. 72

Abstract in Arabic ............................................................................................................................. 76

vi

List of Symbols

Symbol Denoted

ℕ The set of Natural Numbers

ℝ The set of Real Numbers

𝛾(𝑠, 𝑥) The Lower Incomplete Gamma Function

𝜓(𝑥) The Digamma Function

𝐵𝑥(𝑎, 𝑏) The Incomplete Beta Function

𝐸𝛼,𝛽(𝑧) The Two-Parameters Mittage-Leffler Function

𝐸𝛼,𝛽(𝑘)(𝑥) The k-th Derivative of Mittage-Leffler Function

𝐷𝑐−𝑝𝑓(𝑥) The Riemann-Liouville Fractional Integral

𝐷𝑐𝑝𝑓(𝑥) The Riemann-Liouville Fractional Derivative

𝐷𝑐 𝑎−𝑝𝑓(𝑥) The Caputo Fractional Derivative

𝑇𝛼𝑓(𝑥) The Conformable Fractional Derivative

𝐽𝛼𝑓(𝑥) The Conformable Fractional Integral

Γ(𝑥) Gamma Function

vii

List of Abbreviations

Abbreviation Denoted

R-L Riemann-Liouville

FDEs Fractional Differential Equations

FRDE Fractional Riccati Differential Equation

viii

List of Figures and Tables

Figure/Table Page

Figure 1 : Graph of Gamma Function Γ(𝑥) 5

Table 1: Comparison between Riemann-Liouville and Caputo

38

ix

SOLVING FRACTIONAL DIFFERENTIAL EQUATIONS BY USING

CONFORMABLE FRACTIONAL DERIVATIVES DEFINITION

By

Shadi Ahmad Al-Tarawneh

Supervisor

Dr. Khaled Jaber

ABSTRACT

Ordinary and partial fractional differential equations are very important in many fields

like Fluid Mechanics, Biology, Physics, Optics, Electrochemistry of Corrosion,

Engineering, Viscoelasticity, Electrical Networks and Control Theory of Dynamic Systems.

The fractional Ricatti equation is studied by many researchers by using different

numerical methods. Our interest in solving fractional differential equations began when

Prof. Khalil, et al., presented a new simpler and more efficient definition of fractional

derivative. The new definition reflects a nature extension of normal derivative which is

called “conformable fractional derivative”.

In this thesis, we found an exact solution to the fractional Ricatti differential equation,

and we introduced some theorems which lead us to find a second solution when we have a

given particular solution.

1

INTRODUCTION

The sense of differentiation operator 𝐷 = 𝑑 𝑑𝑥⁄ is known to all who went through

ordinary calculus. And for proper function 𝑓, the 𝑛’th derivative of 𝑓, namely 𝐷𝑛𝑓(𝑥) =

𝑑𝑛𝑓(𝑥)𝑑𝑥𝑛⁄ is well defined where 𝑛 is positive integer.

The beginning of derivative theory of non-integers order dates back to leibniz’s note in

his letter to L’Hopital, dated 30 September 1695 [4, 5]. He questioned that what would it

mean if the derivative of one half is discussed [4, 5, 10]. Ever after the fractional calculus

has got the interest, such as Euler, Laplace, Fourier, Abel, Liouville, Rieman, and Lauraut.

Since three centuries, fractional calculus became the traditional calculus but not very

common amongst science and engineering community. This field of applied mathematics

translates the reality of nature better! Therefore, to make this field ready as prevalent

subject to science and engineering community, add another dimension to understand or

describe basic nature in accessible way. Possibly factional calculus is what nature

comprehend and to talk with nature in this language is more effective [4].Fractional

calculus was a theoretical since till some economies and engineering applications involve

fractional differential equations [4].

Most fractional differential equations (FDEs) don’t have exact solution, so

approximate and numerical techniques [6, 24, 25] must be used. Various numerical and

approximate methods to solve the FDEs have been discussed as variational iteration

method [9], homotopy perturbation method [24], Adomain’s decomposition method [32],

2

homotopy analysis method [31], collocation method [12, 13, 28] and finite difference

method [26, 27, 29].

Riccati differential equation refers to the Italian Nobleman Count Jacopo Francesco

Riccati (1676-1754). The fractional Riccati equation is studied by many researchers using

different numerical methods [15, 18, 20].

Recently, Khalil, et al. [14] introduced a new definition of fractional calculus which is

simpler and more efficient. The new definition reflects a nature extension of normal

derivative which is called “conformable fractional derivative”.

The objective of the present thesis is to use conformable fractional derivative to solve

fractional differential equation, specifically, fractional Riccate differential equation.

The thesis is organized as follows, chapter one contains seven sections, and each

handles a preliminary concept of some important special functions and some basic

information about linear differential equation. Also this chapter gives the two familiar

operators of fractional calculus which are: Rieman-Liouville (R-L) and Caputo operators

and study several important rules, as well as, the differences between these operators.

Chapter two focuses on a new definition of “conformable fractional derivatives” and

studies the rules of differentiation and integration.

In chapter three we found an exact solution of fractional Riccati differential equation

and introduced some theorems which lead us to find a second solution when we have a

given particular solution.

3

Chapter one: Basic Concepts and Preliminaries

This chapter shows popular fractional derivatives presented by Riemann-liouville

(R-L) and Caputo’s fractional differential operators and their properties. At first, it is

needed to introduce some special functions like Gamma function, Beta function and

Mittage-Leffler and their properties, then we will introduce some basic differential

equations of first order.

1.1 History of Fractional Calculus

The history of “fractional derivative” started in 1695 by L’Hopetal, when he

questioned Leibniz what would it mean 𝐷𝑛𝑥

𝐷𝑥𝑛 if 𝑛 =

1

2 in his letter, Leibniz answered that

would be a paradox. This was the beginning of “fractional derivative” and influence on

this new concept to a number of mathematicians like Laplace, Euler, Fourier, Lacroix,

Riemann, Abel and Liouville. Lacroix was the first mathematician who released a

paper mentioning fractional derivatives in it. He began with the polynomial 𝑓(𝑥) =

𝑥𝑚, where m is a positive integer, and differentiated it n times where 𝑚 ≥ 𝑛 to get

𝐷𝑓

𝐷𝑥=

𝑚!

(𝑚−𝑛)! 𝑥𝑚−𝑛 ,then he used Legendres symbol Γ to have

𝐷𝑛𝑓

𝐷𝑥𝑛=

Γ(𝑚+1)

Γ(𝑚−𝑛+1)𝑥𝑚−𝑛 .

Using this formula when 𝑚 = 1 𝑎𝑛𝑑 𝑛 =1

2 he obtained 𝐷

1

2𝑓 =2√𝑥

√𝜋.

1.2 Some Special Functions

In this section we are going to introduce the basic definitions and properties of

the upcoming special functions: Gamma, Beta and Mittag-Leffler which are the corner

stone in fractional calculus.

4

1.2.1. Gamma Function

The Gamma function is considered as an extension to the factorial function to

real and complex numbers not only integers. It plays an important role in many fields

of applied science. It has many equivalent definitions, from those, one can prove that

the Gamma function is defined for all real numbers except at 𝑥 = 0,−1 ,−2 ,…, also

Γ(𝑥) has an integral representation for complex number 𝑍, where the real part of the

complex number Z is positive[17], and it can be presented in many formulas as we

will discuss below.

Definition 1.2.1. [23] (Euler, 1730) Let 𝑥 > 0 The Gamma function is defined by

Γ(𝑥) = ∫(− log(𝑡))𝑥−1𝑑𝑡

1

0

, (1.1)

by elementary changes of variables these historical definitions take the more usual

forms:

Theorem 1.2.1. [17, 23] For 𝑥 > 0,

Γ(𝑥) = ∫ 𝑡𝑥−1𝑒−𝑡𝑑𝑡 ,

∞

0

(1.2)

or sometimes

Γ(𝑥) = 2∫ 𝑡2𝑥−1𝑒−𝑡2𝑑𝑡

∞

0

. (1.3)

Proof: Use respectively the changes of variable 𝑢 = −log (𝑡) and 𝑢2 = − log(𝑡) in

(1.1)

5

Figure 1 : Graph of Gamma Function Γ(𝑥)

Another definition of the Gamma function was written in a letter from Euler to

his friend Gold bach in October 13, 1729 is shown below.

Definition 1.2.2. [23] (Euler, 1729 and Gauss, 1811) Let 𝑥 > 0 , 𝑝 ∈ 𝑁, define:

Γ𝑝(𝑥) =

𝑝!. 𝑝𝑥

𝑥(𝑥 + 1)… (𝑥 + 𝑝)

= 𝑝𝑥

𝑥(1 + 𝑥 1⁄ )…(1 + 𝑥 𝑝⁄ )

(1.4)

Theorem 1.2.2. [23] (Weierstrass) For any real number, except the non-positive

integers {0,-1 …} we have the infinite product

6

1

Γ(𝑥)= 𝑥𝑒𝛾𝑥∏ (1 +

𝑥

𝑝) 𝑒−𝑥 𝑝⁄∞

𝑝=1 . (1.5)

where γ is the Euler’s constant γ =0.5772156649015328606065120900824024310421...

which is defined by: 𝛾 = 𝑙𝑖𝑚𝑝→∞ (1 +1

2+⋯+

1

𝑝− 𝑙𝑜𝑔 (𝑝)) .

Below are two important properties of Gamma function.

Theorem 1.2.3. [16, 17, 23] let 𝑥 ≠ 0, 𝑛 ∈ ℕ, then:

1. Γ(n + 1) = n! (1.6)

2. Γ(𝑥) = Γ(𝑥+1)

𝑥, for negative value of x . (1.7)

3. Γ(𝑥)Γ(1 − 𝑥) = 𝜋

𝑠𝑖𝑛(𝜋𝑥) . (1.8)

4. 𝑑𝑛

𝜕𝑥𝑛Γ(𝑥) = ∫ 𝑡𝑥−1𝑒−𝑡(𝑙𝑛 𝑡)𝑛𝑑𝑡

∞

0

, 𝑥 > 0 . (1.9)

5. Γ(𝑥) = 𝑥−1∏ (1 +1

𝑛)𝑥

(1 +𝑥

𝑛)−1

∞𝑛=1 . (1.10)

6. Γ (1

2+ 𝑧)Γ (

1

2− 𝑧) = 𝜋 sec 𝜋𝑧. (1.11)

7. 1

Γ(𝑧)= 𝑧 lim

𝑛→∞{𝑛−𝑧∏ (1 +

𝑧

𝑘)

𝑛

𝑘=1

} (1.12)

From the above we can get:

(a) Γ (1

2) = √𝜋

(b) Γ (5

2) =

3

2Γ (3

2) =

3

2.1

2Γ (1

2) =

3

4√𝜋

7

(c) Γ (−3

2) =

Γ(−3 2⁄ + 1)

−32⁄

=Γ(−12 )

−32⁄=

Γ(12)

−32 ∙

−12

=4

3√𝜋

Definition 1.2.3. The lower incomplete Gamma function is defined by [17, 19]:

𝛾(𝑠, 𝑥) = ∫𝑡𝑠−1𝑒−𝑡𝑥

0

. 𝑑𝑡 (1.13)

and the upper incomplete Gamma function

𝛤(𝑠, 𝑥) = ∫ 𝑡𝑠−1𝑒−𝑡. 𝑑𝑡

∞

𝑥

(1.14)

The Relation between Gamma function and incomplete Gamma function is given by

[17].

(a) 𝛾(𝑠, 𝑥) =∑

𝑥𝑠𝑒−𝑥𝑥𝑘

𝑠(𝑠 + 1)…(𝑠 + 𝑘)

∞

𝑘=0

= 𝑥𝑠𝛤(𝑠)𝑒−𝑥∑𝑥𝑘

𝛤(𝑠 + 𝑘 + 1)

∞

𝑘=0

(1.15)

(b) 𝑙𝑖𝑚𝑥→∞

𝛾(𝑠, 𝑥) = Γ(𝑠) (1.16)

(c) 𝛾(𝑠, 𝑥) + Γ(𝑠, 𝑥) = Γ(𝑠) (1.17)

Definition 1.2.4. The Digamma function ψ(x) is defined by [17]

𝜓(𝑥) =

𝑑

𝑑𝑥𝑙𝑛 Γ(𝑥) =

𝛤′(𝑥)

𝛤(𝑥) (1.18)

Here are some properties of Digamma functions:

8

1) 𝜓(𝑧 + 𝑛) = 𝜓(𝑧) +1

𝑧+

1

𝑧 + 1+⋯+

1

𝑧 + 𝑛 − 1 (1.19)

2) 𝜓(𝑧) − 𝜓(1 − 𝑧) =−𝜋

tan(𝜋𝑧) (1.20)

1.2.2. The Beta Function

The Beta function is useful function related to the Gamma functions. It is defined

for 𝑥 > 0 and 𝑦 > 0 by the two equivalent identities:

Definition 1.2.5. [23] The Beta function (or Eulerian integral of the first kind) is given

by

𝐵(𝑥, 𝑦) = ∫ 𝑡𝑥−1(1 − 𝑡)𝑦−1𝑑𝑡1

0; 0 ≤ 𝑡 ≤ 1 (1.21)

= 2 ∫ 𝑠𝑖𝑛(𝑡)2𝑥−1 𝑐𝑜𝑠(𝑡)2𝑦−1 𝑑𝑡

𝜋 2⁄

0

; 0 ≤ 𝑡 ≤𝜋

2

This definition is also applicable for complex numbers 𝑥 and 𝑦 such as 𝑅𝑒(𝑥) > 0

and 𝑅𝑒(𝑦) > 0, and Euler gave (1.22) in 1730. The name of Beta function was

introduced for the first time by Jacques Binet (1786-1856) in (1839) [23] and he

provided many achievements on the subject.

The Beta function is symmetric as will be shown in the next theorem:

Theorem 1.2.5. let 𝑅𝑒(𝑥) > 0 and 𝑅𝑒(𝑦) > 0 , Then

𝐵(𝑥, 𝑦) =

Γ(𝑥)Γ(𝑦)

Γ(𝑥 + 𝑦)= 𝐵(𝑦, 𝑥) (1.22)

Proof: by using the definite integral (1.3)

9

Γ(𝑥)Γ(𝑦) = 4∫ 𝑢2𝑥−1𝑒−𝑢2𝑑𝑢∫ 𝑣2𝑦−1𝑒−𝑣

2𝑑𝑣

∞

0

∞

0

= 4∫ ∫ 𝑒−(𝑢2+𝑣2) 𝑢2𝑥−1𝑣2𝑦−1

∞

0

𝑑𝑢𝑑𝑣

∞

0

Now by using the polar variables 𝑢 = 𝑟 cos𝜃 and 𝑣 = 𝑟 sin 𝜃 so that,

Γ(𝑥)Γ(𝑦) = 4∫ ∫ 𝑒−𝑟2

𝜋 2⁄

0

𝑟2(𝑥+𝑦)−1 cos2x−1 𝜃 sin2y−1 𝜃 𝑑𝑟𝑑𝜃

∞

0

= 2∫ 𝑒−𝑟2

∞

0

𝑟2(𝑥+𝑦)−1𝑑𝑟. 2 ∫ cos2x−1 𝜃 sin2y−1 𝜃𝑑𝜃

𝜋 2⁄

0

= Γ(x + y)B(x, y) ∎

From relation (1.23) follows

𝐵(𝑥 + 1, 𝑦) =Γ(𝑥 + 1)Γ(𝑦)

Γ(𝑥 + 𝑦 + 1)=

xΓ(𝑥)Γ(𝑦)

(x + y)Γ(𝑥 + 𝑦)=

𝑥

𝑥 + 𝑦𝐵(𝑥, 𝑦)

This is the beta function functional equation

𝐵(𝑥 + 1, 𝑦) =𝑥

𝑥 + 𝑦𝐵(𝑥, 𝑦) (1.23)

Definition 1.2.6. The incomplete Beta function 𝐵𝜏(𝑥, 𝑦)is defined by:

𝐵𝜏(𝑥, 𝑦) = ∫ 𝑡𝑥−1(1 − 𝑡)𝑦−1. 𝑑𝑡 ,

𝜏

0

0 < 𝜏 < 1 (1.24)

Note that from the above:

𝐵 (1

2,1

2) = 𝜋

10

𝐵 (1

3,2

3) =

2 √3

3𝜋

𝐵 (1

4,3

4) = 𝜋 √2

𝐵(𝑥, 1 − 𝑥) = 𝜋

sin (𝜋𝑥)

𝐵(𝑥, 1) = 1

𝑥

𝐵(𝑥, 𝑛) = (𝑛 − 1)!

𝑥. (𝑥 + 1)… (𝑥 + 𝑛 − 1) 𝑛 ≥ 1

𝐵(𝑚, 𝑛) = (𝑚 − 1)! (𝑛 − 1)!

(𝑚 + 𝑛 − 1)! 𝑚 ≥ 1 , 𝑛 ≥ 1

1.2.3 Mittag-Leffler Function

The Mittag-Leffler function is a generalization of the exponential function and it is

one of the most important functions that are related to fractional differential equations.

Definition 1.2.7. [3, 5, 17] The one and two-parameter Mittag-Leffler functions are

defined, respectively, by:

𝐸𝑎(𝑥) = ∑

𝑥𝑛

Γ(𝑎𝑛 + 1) , 𝑎 > 0

∞

𝑛=0

(1.25)

𝐸𝑎,𝑏(𝑥) = ∑

𝑥𝑛

Γ(𝑎𝑛 + 𝑏) , 𝑎 > 0, 𝑏 > 0

∞

𝑛=0

(1.26)

If 𝑎 = 1 and 𝑏 ∈ ℕ

11

𝐸1,1(𝑥) = ∑

𝑥𝑛

Γ(𝑛 + 1)=

∞

𝑛=0

∑𝑥𝑛

n!

∞

𝑛=0

= 𝑒𝑥 (1.27)

𝐸1,2(𝑥) = ∑

𝑥𝑛

Γ(𝑛 + 2)

∞

𝑛=0

=∑𝑥𝑛

(𝑛 + 1)!

∞

𝑛=0

=1

𝑥∑

𝑥𝑛+1

(n + 1)!

∞

𝑛=0

=𝑒𝑥 − 1

𝑥

(1.28)

𝐸1,3(𝑥) = ∑

𝑥𝑛

Γ(𝑛 + 3)

∞

𝑛=0

=∑𝑥𝑛

(𝑛 + 2)!

∞

𝑛=0

=1

𝑥2∑

𝑥𝑛+2

(n + 2)!

∞

𝑛=0

=𝑒𝑥 − 1− 𝑥

𝑥2

(1.29)

In general,

𝐸1,𝑚 =1

𝑥𝑚−1{𝑒𝑥 − ∑

𝑥𝑛

n!

𝑚−2

𝑛=0

} (1.30)

Easily, we can obtain the following:

(a) 𝐸2,1(𝑥2) = ∑

𝑥2𝑛

𝛤(2𝑛 + 1)

∞

𝑛=0

=∑𝑥2𝑛

(2𝑛)!

∞

𝑛=0

= 𝑐𝑜𝑠ℎ(𝑥) (1.31)

(b) 𝐸2,2(𝑥2) = ∑

𝑥2𝑛

𝛤(2𝑛 + 2)

∞

𝑛=0

=∑𝑥2𝑛+1

𝑥(2𝑛 + 1)!

∞

𝑛=0

=𝑠𝑖𝑛ℎ(𝑥)

𝑥 (1.32)

(c) 𝐸2,1(−𝑥2) = ∑

(−𝑥2)𝑛

Γ(2𝑛 + 1)

∞

𝑛=0

=∑(−1)𝑛𝑥2𝑛

(2𝑛)!

∞

𝑛=0

= 𝑐𝑜𝑠(𝑥) (1.33)

(d) 𝐸2,2(−𝑥2) = ∑

(−𝑥2)𝑛

Γ(2𝑛 + 2)

∞

𝑛=0

=∑(−1)𝑛𝑥2𝑛+1

𝑥(2𝑛 + 1)!

∞

𝑛=0

=𝑠𝑖𝑛(𝑥)

𝑥 (1.34)

The Mittage-Leffler function has the following relations :

12

𝐸𝑎,𝑏(𝑥) = 𝑥 𝐸𝑎,𝑎+𝑏(𝑥) +

1

𝛤(𝑏) (1.35)

𝐸𝑎,𝑏(𝑥) = 𝑏𝐸𝑎,𝑏+1(𝑥) + 𝑎𝑥

𝑑

𝑑𝑥 𝐸𝑎,𝑏+1(𝑥) (1.36)

Obviously, from (1.36) we have

𝑑

𝑑𝑥 𝐸𝑎,𝑏(𝑥) =

1

𝑎𝑥 [𝐸𝑎,𝑏−1(𝑥) − (𝑏 − 1) 𝐸𝑎,𝑏(𝑥) ] (1.37)

The 𝑚-th derivative of Mittage-Leffler function is given as follows:

𝑑𝑚

𝑑𝑥𝑚[𝑥𝑏−1 𝐸𝑎,𝑏(𝑥

𝑎)] = 𝑥𝑏−𝑚−1𝐸𝑎,𝑏−𝑚(𝑥𝑎) , 𝑏 − 𝑚 > 0 , 𝑚 = 0, 1 ,⋯ (1.38)

The integration of the Mittage-Leffler function is given as follows:

∫ 𝐸𝑎,𝑏(𝜆 𝑡

𝑎)𝑡𝑏−1𝑥

0

𝑑𝑡 = 𝑥𝑏𝐸𝑎,𝑏+1(𝜆 𝑥𝑎) (1.39)

The relation (1.39) is a special case and the following relation is more general:

1

Γ(𝑣)∫ (𝑥 − 𝑡)𝑣−1𝑥

0

𝐸𝑎,𝑏(𝜆 𝑡𝑎)𝑡𝑏−1𝑑𝑡 = 𝑥𝑏+𝑣−1 𝐸𝑎,𝑏+𝑣(𝜆 𝑥

𝑎) , 𝑣 > 0 (1.40)

From (1.40) we obtain the following important formulas:

1

Γ(𝑝)∫ (𝑥 − 𝑡)𝑝−1𝑒𝑎𝑡 𝑑𝑡 = 𝑥𝑝 𝐸1,𝑝+1(𝑎𝑥) , 𝑝 > 0 𝑥

0

(1.41)

1

Γ(𝑝)∫ (𝑥 − 𝑡)𝑝−1 cosh(𝑎𝑡) 𝑑𝑡 = 𝑥𝑝 𝐸2,𝑝+1((𝑎𝑥)

2) , 𝑝 > 0 𝑥

0

(1.42)

1

𝛤(𝑝)∫ (𝑥 − 𝑡)𝑝−1 𝑠𝑖𝑛ℎ(𝑎𝑡) 𝑑𝑡 = 𝑎 𝑥𝑝+1 𝐸2,𝑝+2((𝑎𝑥)

2) , 𝑝 > 0 𝑥

0

(1.43)

1.3 The Popular Definitions of Fractional Derivatives/Integrals in

Fractional Calculus

In this section we listed the popular definition of fractional calculus [3]:

13

1.3.1. Riemann-Liouville (RL) [3, 4, 5]:

𝐷𝑡𝛼

𝑎 𝑓(𝑡) =1

𝛤(𝑛 − 𝛼)(𝑑

𝑑𝑡)𝑛

∫𝑓(𝑥)

(𝑡 − 𝑥)𝛼−𝑛+1𝑑𝑥

𝑡

𝑎

(1.44)

(𝑛 − 1) ≤ 𝛼 < 𝑛 ,where 𝛼 is a real number, 𝑛 is integer.

1.3.2. M.Caputo (1967) [3,4]:

𝐷𝑡𝛼

𝑎𝑐 𝑓(𝑡) =

1

𝛤(𝑛 − 𝛼)∫

𝑓(𝑛)(𝑥)

(𝑡 − 𝑥)𝛼+1−𝑛𝑑𝑥

𝑡

𝑎

(1.45)

(𝑛 − 1) ≤ 𝛼 < 𝑛 , where 𝛼 is a real number and 𝑛 is integer

1.3.3. Oldham and Spainer (1974) [4]:

The scaling property for fractional derivatives

𝑑𝑞𝑓(𝛽𝑥)

𝑑𝑥𝑞= 𝛽𝑞

𝑑𝑞𝑓(𝛽𝑥)

𝑑(𝛽𝑥)𝑞 (1.46)

1.3.4. Kolwanker and Gangel (1994) [4]:

Kolwanker and Gangel (KG) defined a local fractional derivative to explain the

behavior of “continuous but nowhere differentiable” function for 0 < 𝑝 < 1 , the local

fractional derivative at point 𝑥 = 𝑦 , for 𝑓: [0,1] → ℝ is:

𝐷𝑝𝑓(𝑦) =

𝑑𝑝(𝑓(𝑥) − 𝑓(𝑦))

𝑑(𝑥 − 𝑦)𝑝 (1.47)

1.3.5. Conformable Fractional Derivative (2014) [4]:

let 𝑓: [0,∞) → 𝑅 , 𝑡 > 0 , then the Conformable fractional derivative of 𝑓 of order 𝛼 is

defined by

14

𝑇𝛼(𝑓)(𝑡) = 𝑙𝑖𝑚

𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀 (1.48)

for all 𝑡 > 0 , 𝛼 ∈ (0,1)

1.4 Riemann-Liouville (R-L) Fractional Derivative and Integration

In this section, we listed some presentations, rules and properties of Riemann-

Liouville integration and differentiation and their proofs.

1.4.1. Riemann-Liouville Fractional Integration

We need to use the following fact to define the fractional integration of Riemann-

Liouville:

If 𝑓 is an integrable function on[𝑎, 𝑏], then for 𝑛 ∈ ℕ and for 𝑥 ∈ [𝑎, 𝑏], we have

𝐷𝛼−𝑛𝑓(𝑥) =

1

(𝑛 − 1)!∫(𝑥 − 𝑡)𝑛−1𝑓(𝑡)

𝑥

𝑎

𝑑𝑡. (1.49)

By using (𝑛 − 1)! = Γ(𝑛) and if we replace the order (𝑛) by the order(𝑝),

where 𝑝 ∈ 𝑅, then we get the following definition:-

Definition 1.4.1. [3, 4, 19, 22] let 𝑓(𝑥) be a piecewise continuous on 𝜇 = (0,∞) and

intergrable on any finite subinterval of 𝜇′ = [0,∞) and for 𝑝 > 0 , 𝑥 > 0 we call

𝐷−𝑝𝑓(𝑥) =1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑓(𝑡) 𝑑𝑡

𝑥

0

(1.50)

The Riemann-Liouville fractional integral of order 𝑝 of 𝑓

Remark 1.4.1. [19, 22] The Riemann’s definition is given by:

15

𝐷𝑐−𝑝𝑓(𝑥) =

1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑓(𝑡) 𝑑𝑡.

𝑥

𝑐

(1.51)

where 𝑝 > 0 , 𝑥 > 𝑐

The Liouville’s definition is given by:

𝐷−∞−𝑝 𝑓(𝑥) =

1

Γ(𝑝) ∫(𝑥 − 𝑡)𝑝−1𝑓(𝑡)𝑑𝑡.

𝑥

−∞

where p > 0

Note that we use the symbol 𝐷−𝑝𝑓(𝑥) instead of 𝐷0−𝑝𝑓(𝑥) when the lower limit of

the integral equals zero.

Properties 1.4.1. [19,22] If 𝑓(𝑥) and ℎ(𝑥) are continuous functions a, 𝑏 ∈ 𝑅 , and

𝑛,𝑚 > 0 , then:

𝐷𝑎−𝑛(𝐷𝑎

−𝑚𝑓(𝑥)) = 𝐷𝑎−𝑚(𝐷𝑎

−𝑛𝑓(𝑥)) = 𝐷𝑎−(𝑛+𝑚)𝑓(𝑥) (1.52)

𝐷𝛼−𝑛(𝑎𝑓(𝑥) + 𝑏ℎ(𝑥)) = 𝑎𝐷𝛼

−𝑛𝑓(𝑥) + 𝑏𝐷𝛼−𝑛ℎ(𝑥) (1.53)

Theorem 1.4.1. [19,22] (Basic rules of Riemann-Liouville fractional integral)

Let 𝑝 > 0 , 𝑥 > 0, then

1. 𝐷−𝑝𝑥𝜇 =Γ(𝜇 + 1)

Γ(𝑝 + 𝜇 + 1)𝑥𝑝+𝜇 , 𝜇 > −1 (1.54)

2. 𝐷−𝑝𝑐 =𝑐

Γ(𝑝 + 1)𝑥𝑝 , 𝑐 is a constant (1.55)

16

3. 𝐷−𝑝𝑒𝑐𝑥 =

𝑒𝑐𝑥

𝑐𝑝𝛤(𝑝)𝛾(𝑝, 𝑐𝑥) , 𝑎 > 0

where γ(p,cx) is the lower incomplete Gamma functions

(1.56)

4. 𝐷−𝑝(sin 𝑐𝑥) = 𝑐𝑥𝑝+1𝐸2,𝑝+2 (−(𝑐𝑥 )2) (1.57)

5. 𝐷−𝑝(cos 𝑐𝑥) = 𝑥𝑝+1𝐸2,𝑝+1 (−(𝑐𝑥 )2) (1.58)

6. 𝐷−𝑝(cosh 𝑐𝑥) = 𝑥𝑝𝐸2,𝑝+1 ((𝑐𝑥 )2) (1.59)

7. 𝐷−𝑝(sinh 𝑐𝑥) = 𝑐𝑥𝑝+1𝐸2,𝑝+2 ((𝑐𝑥 )2) (1.60)

8. 𝐷−𝑝 𝑙𝑛 𝑥 =𝑥𝑝

Γ(𝑝 + 1)[𝑙𝑛 𝑥 − 𝛾 − 𝜓(𝑝 + 1)] (1.61)

where 𝜓 is the digamma function and 𝛾 = −𝜓(1) = −Γ′(1) ≈

0.5772157 is Euler constant.

Proof:

(1) 𝐷−𝑝𝑥𝜇 = 1

Γ(𝑝)∫ (𝑥 − 𝑡)𝑝−1𝑡𝜇𝑥

0𝑑𝑡

=1

Γ(𝑝)∫(1 −

𝑡

𝑥)𝑝−1

𝑥𝑝−1𝑡𝜇𝑥

0

𝑑𝑡

By substituting 𝑢 = 𝑡 𝑥⁄

= 1

Γ(𝑝)∫(1 − 𝑢)𝑝−1𝑥𝑝−1(𝑢𝑥)𝜇1

0

𝑥𝑑𝑢

=1

Γ(𝑝)∫(1 − 𝑢)𝑝−1𝑢𝜇𝑥𝜇+𝑝1

0

𝑑𝑢

17

= 1

Γ(𝑝)𝑥𝜇+𝑝𝛣(𝜇 + 1, 𝑝)

=Γ(𝜇 + 1)

Γ(𝑝 + 𝜇 + 1)𝑥𝑝+𝜇 ∎

(2) If we set 𝜇 = 0 in (1.54), then the proof is complete

(3)

𝐷−𝑝𝑒𝑐𝑥 =1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑥

0

𝑒𝑐𝑡𝑑𝑡

=1

Γ(𝑝)∫(

𝑐(𝑥 − 𝑡)

𝑐)

𝑝−1𝑥

0

𝑒𝑐𝑡𝑑𝑡

=1

Γ(𝑝)∫𝑢𝑝−1

𝑐𝑝−1

𝑐𝑥

0

𝑒𝑐𝑥−𝑢𝑑𝑢

𝑐=

𝑒𝑐𝑥

𝑐𝑝𝛤(𝑝)∫ 𝑢𝑝−1𝑐𝑥

0

𝑒−𝑢𝑑𝑢 ,

By substituting 𝑢 = 𝑐(𝑥 − 𝑡)

=𝑒𝑐𝑥

𝑐𝑝Γ(𝑝)𝛾(𝑝, 𝑐𝑥) ∎

(4) 𝐷−𝑝(𝑠𝑖𝑛 𝑐𝑥) =1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑥

0

𝑠𝑖𝑛(𝑐𝑡) 𝑑𝑡

=1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑥

0

𝑠𝑖𝑛(𝑐𝑡)

𝑐𝑡𝑐𝑡 𝑑𝑡

Simply by using (1.34) and (1.40),

𝐷−𝑝(𝑠𝑖𝑛 𝑐𝑥) = 𝑐𝑥𝑝+1𝐸2,𝑝+2(−(𝑐𝑥)2) ∎

18

(5) Follows by using (1.33) and (1.40) and the same as (4).

(6) By using (1.42), we get:

𝐷−𝑝(𝑐𝑜𝑠ℎ(𝑐𝑥)) =1

𝛤(𝑝)∫(𝑥 − 𝑡)𝑝−1 𝑐𝑜𝑠ℎ(𝑐𝑡) . 𝑑𝑡

𝑥

0

= 𝑥𝑝𝐸2,𝑝+1((𝑐𝑥)2) ∎

(7) Follows by using (1.43) and the same as (6).

(8) The proof can be found in [19].

Remark 1.4.2. [19,22] The fractional integral of 𝑐𝑜𝑠(𝑐𝑥) , 𝑠𝑖𝑛(𝑐𝑥) can be expressed in

generalized 𝑠𝑖𝑛 and 𝑐𝑜𝑠 functions as:

𝐷−𝑝 cos(𝑐𝑥) =1

Γ(𝑝)∫(𝑥 − 𝑡)𝑝−1𝑥

0

cos(𝑐𝑡) 𝑑𝑡 = 𝐶𝑥(𝑝, 𝑎) (1.62)

𝐷−𝑝 sin(𝑐𝑥) =1

Γ(𝑝)∫(𝑥 − 𝑡)

𝑥

0

sin(𝑐𝑡) 𝑑𝑡 = 𝑆𝑥(𝑝, 𝑎) (1.63)

Remark 1.4.3 [19, 22] We can express the fractional integral function 𝑒𝑐𝑥 by using

Mittage-Leffler function as

𝐷−𝑝𝑒𝑐𝑥 = 𝑥𝑝𝐸1,𝑝+1(𝑐𝑥) (1.61)

Proof:

By using (1.15) and (1.27), then

D−p𝑒𝑐𝑥 =𝑒𝑐𝑥

𝑐𝑝Γ(𝑝)𝛾(𝑝, 𝑐𝑥) =

𝑒𝑐𝑥

𝑐𝑝Γ(𝑝)(𝑐𝑥)𝑝Γ(𝑝)𝑒−𝑐𝑥𝐸1,𝑝+1(𝑐𝑥)

19

= 𝑥𝑝𝐸1,𝑝+1(𝑐𝑥)

1.4.2. Riemann-Liouville Fractional Derivative

The most important approaches to define the fractional derivative is using the

integration of fractional order in the same as the following fact:

𝐷𝛼𝑝𝑓 = 𝐷𝛼

𝑞(𝐷𝑝−𝑞𝑓) , 𝑝, 𝑞 ∈ ℕ, 𝑞 > 𝑝

Riemann-Liouville use the later fact to introduce the following definition:

Definition 1.4.2. [3, 5, 19,22] (Riemann-Liouville fractional derivative)

The Riemann-Liouville fractional derivative of 𝑓(𝑥) of order 𝛼, 𝑛 − 1 < 𝛼 < 𝑛,

𝑛 ∈ ℕ is defined by:

𝐷𝑎𝛼𝑓(𝑥) = 𝐷𝑛 (𝐷𝑎

−(𝑛−𝛼)𝑓(𝑥))

=𝑑𝑛

𝑑𝑥𝑛1

Γ(𝑛 − 𝛼)∫(𝑥 − 𝑡)𝑛−𝛼+1𝑥

𝑎

𝑓(𝑡). 𝑑𝑡

(1.62)

Definition 1.4.3. [3, 5, 19, 22] Let 𝑓(𝑥) be a function defined on the closed interval

[𝑎, 𝑏] and let 𝛼 ∈ [0,1), then the left Riemann-Liouville 𝛼 derivative of 𝑓(𝑥) is:

𝐷𝑎𝛼𝑓(𝑥) =

1

Γ(1 − 𝛼)

𝑑

𝑑𝑥∫

𝑓(𝑡)

(𝑥 − 𝑡)𝛼

𝑥

𝑎

. 𝑑𝑡 (1.63)

The right Riemann-Liouville 𝛼 derivative of 𝑓(𝑥) is

𝐷𝑏𝛼𝑓(𝑥) =

−1

Γ(1 − 𝛼)

𝑑

𝑑𝑥∫

𝑓(𝑡)

(𝑡 − 𝑥)𝛼

𝑏

𝑥

. 𝑑𝑡 (1.64)

20

But when 𝛼 is any number greater than 1. Then the definition will be as the

following

Definition 1.4.4. [3, 5, 19, 22]

Let 𝑓(𝑥) be a function defined on the closed interval [𝑎, 𝑏] and let 𝛼 ∈

[𝑛 − 1, 𝑛), 𝑛 ∈ ℕ. Then the left Riemann-Liouville 𝛼 derivative of 𝑓(𝑥) is:

𝐷𝑎𝛼𝑓(𝑥) =

1

Γ(𝑛 − 𝛼)

𝑑𝑛

𝑑𝑥𝑛∫

𝑓(𝑡)

(𝑥 − 𝑡)𝛼−𝑛+1

𝑥

𝑎

. 𝑑𝑡 (1.65)

and the right Riemann-Liouville 𝛼 derivative of 𝑓(𝑥) is

𝐷𝑏𝛼𝑓(𝑥) =

(−1)𝑛

Γ(𝑛 − 𝛼)

𝑑𝑛

𝑑𝑥𝑛∫

𝑓(𝑡)

(𝑡 − 𝑥)𝛼−𝑛+1

𝑏

𝑥

. 𝑑𝑡 (1.66)

Note that the required condition required in the definitions is to be 𝑛-times

continuously differentiable.

The relationship between integration and differentiation of Riemann-Liouville

operators for the arbitrary order 𝑝 are shown as follows:

The Derivative of fractional integral could be shown as:

𝐷𝛼𝑝 (𝐷𝛼

−𝑞𝑓(𝑥)) = 𝐷𝑝−𝑞𝑓(𝑥), (1.67)

where 𝑓(𝑥) is continuous also 𝑝 ≥ 𝑞 ≥ 0

precisely, when 𝑞 ≥ 0 then 𝐷𝛼𝑞 (𝐷𝛼

−𝑞𝑓(𝑥)) = 𝑓(𝑥) (1.68)

21

Preposition 1.4.2. [19,22] Let 𝑓1(𝑥), 𝑓2(𝑥) be two functions defined on[𝑎, 𝑏], and let

𝛼 ∈ [𝑛 − 1, 𝑛), 𝑛 ∈ ℕ, 𝜆, 𝛽 ∈ ℂ and 𝐷𝑎𝛼𝑓1(𝑥),𝐷𝑎

𝛼𝑓2(𝑥) exist, then

𝐷𝑎𝛼[𝜆𝑓1(𝑥) + 𝛽𝑓2(𝑥)] = 𝜆𝐷𝑎

𝛼𝑓1(𝑥) + 𝛽𝐷𝑎𝛼𝑓2(𝑥) (1.69)

Proof:

𝐷𝑎𝛼[𝜆𝑓1(𝑥) + 𝛽𝑓2(𝑥)] =

1

Γ(𝑛 − 𝛼)

𝑑𝑛

𝑑𝑥𝑛∫[𝜆𝑓1(𝑥) + 𝛽𝑓2(𝑥)]

(𝑥 − 𝑡)𝛼−𝑛+1

𝑥

𝑎

𝑑𝑡

=𝜆

Γ(𝑛 − 𝛼)

𝑑𝑛

𝑑𝑥𝑛∫

𝑓1(𝑥)

(𝑥 − 𝑡)𝛼−𝑛+1

𝑥

𝑎

𝑑𝑡 +𝛽

Γ(𝑛 − 𝛼)

𝑑𝑛

𝑑𝑥𝑛∫

𝑓2(𝑥)

(𝑥 − 𝑡)𝛼−𝑛+1

𝑥

𝑎

𝑑𝑡

= 𝜆𝐷𝑎𝛼𝑓1(𝑥) + 𝛽𝐷𝑎

𝛼𝑓2(𝑥) ∎

Preposition 1.4.3. (Interpolation Property)

Let 𝑔(𝑥) be a function defined on [𝑎, 𝑏] and let 𝛼 ∈ [0,1). Let 𝑔(𝑥) have a

continuous derivative of sufficient order and 𝐷𝑎𝛼𝑔(𝑥) exists, then

lim𝛼→1

𝐷𝑎𝛼𝑔(𝑥) = 𝑔′(𝑥) (1.70)

and lim𝛼→0

𝐷𝑎𝛼𝑔 (𝑥) = 𝑔(𝑥) (1.71)

Proof: see [22]

We can generalize the above equalities in preposition 1.4.3 for any positive

number 𝛼 to be

lim𝛼→𝑛

𝐷𝑎𝛼𝑔(𝑥) = 𝑔(𝑛)(𝑥) (1.72)

22

and lim𝛼→𝑛−1

𝐷𝑎𝛼𝑔 (𝑥) = 𝑔(𝑛−1)(𝑥) (1.73)

where 𝛼 ∈ [𝑛 − 1, 𝑛) , 𝑛 ∈ ℕ and with the same condition of the preposition (1.4.3).

Preposition1.4.4. (Some properties of Riemann-Liouville fractional derivative)

1) The integral of (Riemann-Liouville) derivative is given by

𝐷𝑎−𝑝(𝐷𝑎

𝑞𝑓(𝑥)) = 𝐷𝑎

𝑞−𝑝𝑓(𝑥) −∑[𝐷𝛼

𝑞−𝑘−1𝑓(𝑥)]

𝑥=𝑎

(𝑥 − 𝑎)𝑝−𝑘−1

Γ(𝑝 − 𝑘)

𝑛−1

𝑘=0

(1.74)

where 𝑛 − 1 < 𝑞 < 𝑛 , 𝑛 ∈ ℕ

2) 𝐷𝑎−𝛼(𝐷𝑎

𝛼𝑓(𝑥)) = 𝑓(𝑥) −∑[𝐷𝛼𝛼−𝑘−1𝑓(𝑥)]𝑥=𝑎

(𝑥 − 𝑎)𝛼−𝑘−1

Γ(𝛼 − 𝑘)

𝑛−1

𝑘=0

(1.75)

3) The fractional derivative of fractional derivative is shown as:-

𝐷𝑎𝑝(𝐷𝑎

𝛼𝑓(𝑥)) = 𝐷𝑝+𝛼𝑓(𝑥) − ∑[𝐷𝛼𝛼−𝑘−1𝑓(𝑥)]𝑥=𝑎

(𝑥 − 𝑎)−𝑝−𝑘−1

Γ(−p − k)

𝑚−1

𝑘=0

where 𝑛 − 1 < 𝑝 < 𝑛 , 𝑚 − 1 < 𝛼 < 𝑚, 𝑛,𝑚 ∈ ℕ

(1.76)

Remark 1.4.4.

𝐷𝑝𝐷𝑞𝑓(𝑥) = 𝐷𝑝+𝑞𝑓(𝑥) = 𝐷𝑞𝐷𝑝𝑓(𝑥) (1.77)

if and only if

𝑓(𝑘)(0) = 0 , 𝑘 = 0,1,… , 𝑟 where 𝑟 = max(𝑛,𝑚),

where 𝑚− 1 ≤ 𝑝 < 𝑚 and 𝑛 − 1 ≤ 𝑞 < 𝑛

23

Theorem 1.4.2. [19,22] The Riemann-Liouville 𝑝 derivative does not satisfy the

following

1) 𝐷𝛼𝑝(𝑓ℎ) = 𝑓𝐷𝑎

𝑝(ℎ) + ℎ𝐷𝑎𝑝(𝑓)

2) 𝐷𝑎𝑝(𝑓 ∘ ℎ) = 𝑓(𝑝)(ℎ(𝑥))ℎ(𝛼)(𝑥)

3) 𝐷𝛼𝑝 (𝑓ℎ⁄ ) =

ℎ𝐷𝑎𝑝(𝑓) − 𝑓𝐷𝑎

𝑝(ℎ)

ℎ2

Theorem 1.4.2. [19,22] The Riemann-Liouville 𝑝 derivative of known functions:

Let 𝑝 > 0 , 𝑥 > 0 , 𝑘, 𝑐 ∈ ℝ , then

1) 𝐷𝑝𝑥𝜇 =Γ(𝜇 + 1)

Γ(𝜇 − 𝑝 + 1)𝑥𝜇−𝑝 , 𝜇 > −1 (1.78)

2) 𝐷𝑝𝑐 =𝑐

Γ(1 − 𝑝)𝑥−𝑝 (1.79)

3) 𝐷𝑝𝑒𝑐𝑥 = 𝑥−𝑝𝐸1,1−𝑝(𝑐𝑥) (1.80)

4) 𝐷𝑝 cos(𝑐𝑥) = 𝑥−𝑝𝐸2,1−𝑝(−(𝑐𝑥)2) (1.81)

5) 𝐷𝑝 sin(𝑐𝑥) = 𝑐𝑥1−𝑝𝐸2,2−𝑝((𝑐𝑥)2) (1.82)

6) 𝐷𝑝 cosh(𝑐𝑥) = 𝑥−𝑝𝐸2,1−𝑝((𝑐𝑥)2) (1.83)

7) 𝐷𝑝 sinh(𝑐𝑥) = 𝑐𝑥1−𝑝𝐸2,2−𝑝((𝑐𝑥)2) (1.84)

8) 𝐷𝑝 ln(𝑥) =𝑥−𝑝

Γ(1 − 𝑝)[ln(𝑥) − 𝛾 − 𝜓(1 − 𝑝)] (1.85)

Proof:

(1) Let 𝑛 − 1 < 𝑝 < 𝑛 , 𝑛 ∈ ℕ, then

𝐷𝑝𝑥𝜇 = 𝐷𝑛[𝐷−(𝑛−𝑝)𝑥𝜇]

24

= 𝐷𝑛 [Γ(𝜇 + 1)

Γ(𝜇 + 𝑛 − 𝑝 + 1)𝑥𝜇+𝑛−𝑝]

=Γ(𝜇 + 1)

Γ(𝜇 + 𝑛 − 𝑝 + 1).Γ(𝜇 + 𝑛 − 𝑝 + 1)

Γ(𝜇 + 𝑛 − 𝑝 − 𝑛 + 1)𝑥𝜇+𝑛−𝑝−𝑛

=Γ(μ + 1)

Γ(μ − p + 1)𝑥𝜇−𝑝 ∎

(2) It follows by substituting 𝜇 = 0 in (1.78)

(3) Using 𝐷−𝑝𝑒𝑐𝑥 = 𝑥𝑝𝐸1,𝑝+1(𝑐𝑥)

= 𝑥𝑝∑(𝑐𝑥)𝑘

Γ(𝑘 + 𝑝 + 1)

∞

𝑘=0

=∑𝑐𝑘𝑥𝑘+𝑝

Γ(𝑘 + 𝑝 + 1)

∞

𝑘=0

=∑𝑐𝑘

Γ(𝑘 + 1)

∞

𝑘=0

.Γ(𝑘 + 1)

Γ(𝑘 + 𝑝 + 1)𝑥𝑘+𝑝

=∑𝑐𝑘

Γ(𝑘 + 1)

∞

𝑘=0

𝐷−𝑝𝑥𝑘

(1.86)

Now, by using (1.86) we have

𝐷𝑝𝑒𝑐𝑥 = 𝐷𝑛[𝐷−(𝑛−𝑝)𝑒𝑐𝑥] = 𝐷𝑛 [∑𝑐𝑘

Γ(k + 1)

∞

𝑘=0

𝐷−(𝑛−𝑝)𝑥𝑘]

=∑𝑐𝑘

Γ(k + 1)

∞

𝑘=0

𝐷𝑝𝑥𝑘 =∑𝑐𝑘

Γ(k + 1).Γ(𝑘 + 1)𝑥𝑘−𝑝

Γ(𝑘 − 𝑝 + 1)

∞

𝑘=0

= 𝑥−𝑝∑(𝑐𝑥)𝑘

Γ(𝑘 − 𝑝 + 1)

∞

𝑘=0

= 𝑥−𝑝𝐸1,1−𝑝(𝑐𝑥) ∎

(4) 𝐷𝑝 cos(𝑐𝑥) = 𝐷𝑛[𝐷−(𝑛−𝑝) cos(𝑐𝑥)]

25

= 𝐷𝑛[𝑥𝑛−𝑝𝐸2,𝑛−𝑝+1(−(𝑐𝑥)2)]

= 𝑥𝑛−𝑝+1−𝑛+1𝐸2,𝑛−𝑝+1−𝑛(−(𝑐𝑥)2)

= 𝑥−𝑝𝐸2,1−𝑝(−(𝑐𝑥)2) ∎

(5) Similarly of (4).

(6) 𝐷𝑝 cosh(𝑐𝑥) = 𝐷𝑛[𝐷−(𝑛−𝑝) cosh(𝑐𝑥)]

= 𝐷𝑛[𝑥𝑛−𝑝𝐸2,𝑛−𝑝+1((𝑐𝑥)2)]

= 𝑥𝑛−𝑝+1−𝑛−1𝐸2,𝑛−𝑝+1−𝑛(−(𝑐𝑥)2)

= 𝑥−𝑝𝐸2,1−𝑝((𝑐𝑥)2) ∎

(7) Similarly of (6)

(8) To proof see [19]

1.5 Caputo Fractional Operator

In 1967 M.Caputo published a paper[11]. He put a new definition of fractional

derivative. In this section we introduced Caputo fractional derivative and some

properties of this definition.

Definition 1.5.1. [3, 4, 5, 11, 19, 22] let 𝑓 be 𝑛 −times differentiable function,

𝑥, 𝑎 ∈ ℝ , 𝑥 > 𝑎 and 𝛼 ∈ [0,1). Then the Caputo fractional differential operator of

order 𝛼 of 𝑓 is defined by:

𝐷𝑎𝛼𝑐 𝑓(𝑥) =

1

Γ(1 − 𝛼)∫

𝑓′(𝑡)

(𝑥 − 𝑡)𝛼

𝑥

𝑎

𝑑𝑡

26

Definition 1.5.2. [3, 4, 5, 11, 19,22] let 𝑓 be 𝑛-times differentiable function, 𝑥, 𝑎 ∈

ℝ , 𝑥 > 𝑎 and 𝛼 ∈ (𝑛, 𝑛 − 1). Then the caputo fractional differential operator of 𝛼 is

defined as:

𝐷𝑎𝛼𝑓(𝑥)𝑐 =

1

Γ(𝑛 − 𝛼)∫

𝑓(𝑛)(𝑡)

(𝑥 − 𝑡)𝛼−𝑛+1

𝑥

𝑎

𝑑𝑡

Remark 1.5.1.

Because of similarity between (R-L) and Caputo fractional integration, the symbol

𝐷𝑎−𝛼𝑓(𝑥) will be indicated to (R-L) and Caputo fractional integral.

Remark 1.5.2

The symbol 𝐷𝑐 𝑎𝛼𝑓(𝑥)is used to denote Caputo fractional derivative of order 𝛼

with lower limit 𝑎 and the symbol 𝐷𝛼𝑓(𝑥)𝑐 is used to denote caputo fractional

derivative of order 𝛼 with lower limit 0.

Preposition 1.5.1. [11, 19,22] let 𝑓(𝑥), 𝑔(𝑥) be two functions such that both

𝐷𝑐 𝑎𝛼𝑓(𝑥), 𝐷𝑎

𝛼𝑐 𝑔(𝑥) exist for 𝛼 ∈ [0,1) and let 𝑎, 𝑏 ∈ ℂ.

Then

𝐷𝑎𝛼(𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)) = 𝑎 𝐷𝑎

𝛼𝑐 𝑓(𝑥) +𝑐 𝑏 𝐷𝑎𝛼𝑐 𝑔(𝑥) (1.87)

Proof: using the definition of Caputo fractional 𝛼 derivative

𝐷𝑎𝛼𝑐 (𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)) =

1

Γ(1 − 𝛼)∫(𝑎𝑓(𝑡) + 𝑏𝑔(𝑡))′

(𝑥 − 𝑡)𝛼

𝑥

𝑎

𝑑𝑡

27

=1

Γ(1 − 𝛼)[𝑎∫

𝑓′(𝑥)

(𝑥 − 𝑡)𝛼

𝑥

𝑎

𝑑𝑡 + 𝑏∫𝑔′(𝑥)

(𝑥 − 𝑡)𝛼𝑑𝑡

𝑥

𝑎

]

=1

Γ(1−𝛼)𝑎 ∫

𝑓′(𝑥)

(𝑥−𝑡)𝛼

𝑥

𝑎𝑑𝑡 +

1

Γ(1−𝛼)𝑏 ∫

𝑔′(𝑥)

(𝑥−𝑡)𝛼𝑑𝑡

𝑥

𝑎

= 𝑎 𝐷𝑎𝛼𝑐 𝑓(𝑥) + 𝑏 𝐷𝑎

𝛼𝑐 𝑔(𝑥) ∎

We can generalize the previous result for any 𝛼 ∈ [𝑛 − 1, 𝑛)

The Relation between integration and differentiation of Caputo operator of order

𝛼 are given as shown:

The Caputo derivative of fractional integral is

𝐷𝑎𝛼𝑐 (𝐷𝑎

−𝛼𝑓(𝑥)) = 𝑓(𝑥) (1.88)

The fractional integral of Caputo derivative is

𝐷𝑎−𝛼( 𝐷𝑐 𝑎

𝛼𝑓(𝑥)) = 𝑓(𝑥) − ∑(𝑥 − 𝑎)𝑚

𝑚!

𝑛−1

𝑚=0

𝑓(𝑚)(𝑎) (1.89)

From (1.88) and (1.89) we have

𝐷𝑎𝛼𝑐 (𝐷𝑎

−𝛼𝑓(𝑥)) ≠ 𝐷𝑎−𝛼( 𝐷𝑐 𝑎

𝛼𝑓(𝑥)) (1.90)

Generally, we can conclude:

𝐷𝑛[𝐷−(𝑛−𝛼)𝑓(𝑥)] ≠ 𝐷−(𝑛−𝛼)[𝐷𝑛𝑓(𝑥)]

28

Thus

𝐷𝑎𝛼𝑓(𝑥) ≠ 𝐷𝑎

𝛼𝑐 𝑓(𝑥) (1.91)

which implies that the Caputo derivative is not equivalent with (Riemann-

Liouville) derivative.

Preposition 1.5.2. [11, 19,22] let 𝑛 ∈ ℕ , 𝛼 ∈ [𝑛 − 1, 𝑛). Let the function 𝑓(𝑥) be an n-

times differentiable function. Then the representation of the Caputo 𝛼 derivative:

𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝐷𝑎

−(𝑛−𝛼) 𝐷𝑎𝑛𝑐 𝑓(𝑥) (1.92)

where 𝐷𝑎−𝛼𝑓(𝑥) =

1

Γ(𝛼)∫

𝑓(𝑡)

(𝑥−𝑡)1−𝛼

𝛼

𝑎𝑑𝑡

is the Riemann-Liouville 𝛼 integral

Theorem1.5.1. [11, 19,22] (Relation between Caputo 𝛼 derivative and Riemann-

Liouville 𝛼 derivative).

Let 𝑛 ∈ ℕ,𝛼 ∈ [𝑛 − 1, 𝑛). And let 𝑓(𝑥) be a function such that 𝐷𝑎𝛼𝑐 𝑓(𝑥) and

𝐷𝑎𝛼𝑓(𝑥) exist. Then the relation between the (R-L) and the Caputo derivatives is given

by:

𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝐷𝑎

𝛼𝑓(𝑥) −∑(𝑥 − 𝑎)𝑘−𝛼

Γ(𝑘 + 1 − 𝛼)

𝑛−1

𝑘=0

𝑓(𝑘)(𝑎) (1.93)

Proof: The well-known Taylor Series expansion of 𝑓 about 𝑥 = 0 is

29

𝑓(𝑥) = 𝑓(0) + 𝑥𝑓′(0) +𝑥2

𝑥!𝑓′′(0) +⋯+

𝑥𝑛−1

(𝑛 − 1)!𝑓(𝑛−1)(0) + 𝑅𝑛−1

=∑𝑥𝑘

Γ(𝑘 + 1)

𝑛−1

𝑘=0

𝑓(𝑘)(0) + 𝑅𝑛−1

(1.94)

where, considering the following

𝐷−𝑛𝑓(𝑡) = ∫∫ …

𝑡1

𝑎

𝑡

𝑎

∫ 𝑓(𝜆)

𝑡𝑛−1

𝑎

𝑑𝜆…𝑑𝜆2 𝑑𝜆1

=1

(𝑛 − 1)!∫𝑓(𝜆)

𝑡

𝑎

(𝑡 − 𝜆)(𝑛−1)𝑑𝜆

(1.95)

The previous formula is called cauchy’s formula for repeated integration.

𝑅𝑛−1 = ∫𝑓(𝑛)(𝑡)(𝑥 − 𝑡)𝑛−1

(𝑛 − 1)!

𝑥

0

𝑑𝑡 =1

Γ(𝑛)∫𝑓(𝑛)(𝑡)

𝑥

0

(𝑥 − 1)𝑛−1𝑑𝑡 (1.96)

Now, by using linearity of Riemann-Liouville, the (Riemann-Liouville) derivative

of power function, the properties of Riemann-Liouville integrals and the representation

formula.

𝐷𝑎𝛼𝑓(𝑡) = 𝐷𝑎

𝛼 [∑𝑥𝑘

Γ(𝑘 + 1)𝑓(𝑘)(0) + 𝑅𝑛−1

𝑛−1

𝑘=0

] = ∑𝐷𝑎𝛼

𝑛−1

𝑘=0

𝑥𝑘

Γ(𝑘 + 1)𝑓(𝑘)(0) + 𝐷𝑎

𝛼𝑅𝑛−1

=∑Γ(𝑘 + 1)

Γ(𝑘 − 𝛼 + 1)

𝑛−1

𝑘=0

𝑥𝑘−𝛼

Γ(𝑘 + 1)𝑓(𝑘)(0) + 𝐷𝑎

𝛼𝐷−𝑛𝑓(𝑛)(𝑥)

30

=∑𝑥𝑘−𝛼

Γ(𝑘 − 𝛼 + 1)

𝑛−1

𝑘=0

𝑓(𝑘)(0) + 𝐷−(𝑛−𝛼)𝑓(𝑛)(𝑥)

= ∑𝑥𝑘−𝛼

Γ(𝑘 − 𝛼 + 1)

𝑛−1

𝑘=0

𝑓(𝑘)(0) + 𝐷𝑎𝛼𝑐 𝑓(𝑥)

∴ 𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝐷𝑎

𝛼𝑓(𝑡) −∑𝑥𝑘−𝛼

Γ(𝑘 − 𝛼 + 1)

𝑛−1

𝑘=0

𝑓(𝑘)(0)

Preposition 1.5.3. [11, 19, 22] Let 𝛼 ∈ [0,1], let 𝑓(𝑥) be a function with second

continuous bounded derivative in [𝑎, 𝑇] for every 𝑇 > 𝑎 and 𝐷𝑎𝛼𝑐 𝑓(𝑥) exist, then:

1) lim𝑎→1

𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝑓′(𝑥) (1.97)

2) lim𝑎→0

𝐷𝑐 𝑎𝛼𝑓(𝑥) = 𝑓(𝑥) − 𝑓(𝑎) (1.98)

To proof see [11].

We can generalize the above equations in preposition 1.5.3 for any positive 𝛼 to be:

lim𝑎→𝑛

𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝑓(𝑛)(𝑥) (1.99)

and lim𝑎→𝑛−1

𝐷𝑐 𝑎𝛼𝑓(𝑥) = 𝑓(𝑛−1)(𝑥) − 𝑓(𝑛−1)(0) (1.100)

where 𝛼 ∈ [𝑛 − 1, 𝑛), 𝑛 ∈ ℕ and with the same condition of the preposition.

Preposition 1.5.4. [11, 19, 22]

The Caputo differential operator does not satisfy the following:

31

1) 𝐷𝑎𝛼(𝑓ℎ)𝑐 = 𝑓 𝐷𝑎

𝛼𝑐 (ℎ) + ℎ 𝐷𝑎𝛼𝑐 (𝑓)

2) 𝐷𝑎𝛼𝑓

ℎ𝑐 =

ℎ 𝐷𝑎𝛼𝑐 (𝑓) + 𝑓 𝐷𝑎

𝛼𝑐 (ℎ)

ℎ2

3) 𝐷𝑎𝛼(𝑓 ∘ ℎ)𝑐 = 𝑓(𝛼)(ℎ(𝑥))ℎ(𝛼)(𝑥)

where 𝑓(𝛼)(𝑥), ℎ(𝛼)(𝑥) are the Caputo 𝛼 derivative.

Now, I will give counter example to show that the above rule does not satisfy for

Caputo Operator considers that:

𝐷𝛼𝑐 (𝑡) =1

Γ(2 − 𝛼)𝑡1−𝛼 ∴ 𝐷

13𝑐 (𝑡) = 1.1077𝑡2 3⁄

𝐷𝛼𝑐 (𝑡2) =2

Γ(3 − 𝛼)𝑡2−𝛼 ∴ 𝐷

13𝑐 (𝑡2) = 1.3293𝑡5 3⁄

𝐷𝛼𝑐 (𝑡3) =6

Γ(4 − 𝛼)𝑡3−𝛼 ∴ 𝐷

13𝑐 (𝑡3) = 1.4954𝑡8 3⁄

Let 𝑓(𝑥) = 𝑡 , 𝑔(𝑥) = 𝑡2 , ℎ(𝑡) = 𝑡3

𝐷13𝑐 (𝑓𝑔) = 𝐷

13𝑐 (𝑡3) = 1.4954𝑡8 3⁄

𝑓 𝐷13𝑐 (𝑔) + 𝑔 𝐷

13𝑐 (𝑓) = 𝑡(1.3293𝑡5 3⁄ ) + 𝑡2(1.1077𝑡2 3⁄ )

= 1.3293𝑡8 3⁄ + 1.1077𝑡8 3⁄ = 2.4370 𝑡8 3⁄

Obviously

𝐷13(𝑓𝑔)𝑐 ≠ 𝑓 𝐷

13𝑐 (𝑔) + 𝑔 𝐷

13𝑐 (𝑓)

Also

𝐷13𝑐 (ℎ(𝑡)

𝑓(𝑡)) = 𝐷

13𝑐 (𝑡3

𝑡) = 𝐷

13𝑐 (𝑡2) = 1.3293𝑡5 3⁄

But

32

𝑓(𝑡) 𝐷13𝑐 ℎ(𝑡) − ℎ(𝑡) 𝐷

13𝑐 𝑓(𝑡)

𝑓2(𝑡)=𝑡(1.4954𝑡5 3⁄ ) − 𝑡3(1.1077𝑡2 3⁄ )

𝑡2

= 1.4954𝑡2 3⁄ − 1.1077𝑡5 3⁄

Thus

𝐷13𝑐 (ℎ

𝑓) ≠

𝑓 𝐷13𝑐 (ℎ) − ℎ 𝐷

13𝑐 (𝑓)

ℎ2

It is easy to show that the composition Rule does not satisfy.

Preposition 1.5.5. [11, 19, 22] suppose that 𝛼 ∈ [𝑛 − 1, 𝑛),𝑚, 𝑛 ∈ ℝ, and 𝐷𝑎𝛼𝑓(𝑥)𝑐

exist. Then

𝐷𝑎𝛼𝑐 𝐷𝑚𝑐 𝑓(𝑥) = 𝐷𝛼+𝑚𝑐 𝑓(𝑥) ≠ 𝐷𝑚𝑐 𝐷𝑎

𝛼𝑐 𝑓(𝑥) (1.101)

Now we give counter example to show that the Caputo derivative is not commute

Example:

𝐷𝑎𝛼𝑐 𝑥𝑝 = {

Γ(𝑝 + 1)

Γ(𝑝 − 𝛼 + 1)𝑥𝑝−𝛼 𝑖𝑓 𝑛 − 1 ≤ 𝛼 < 𝑛 , 𝑝 > 𝑛 − 1, 𝑛 ∈ ℕ

0 𝑖𝑓 𝑛 − 1 ≤ 𝛼 < 𝑛 , 𝑝 ≤ 𝑛 − 1, 𝑝, 𝑛 ∈ ℕ

Now if we take 𝛼 =1

2 , 𝑚 = 3 , 𝑝 = 2. Then

𝐷1 2⁄ 𝐷3𝑐𝑐 [𝑥2] = 0

But 𝐷3𝑐 𝐷1 2⁄𝑐 [𝑥2] = 𝐷3𝑐 [Γ(3)

Γ(5

2)𝑥3

2]

= −38 Γ(3)Γ(52)𝑥−

32

33

Corollary 1.5.1. let 𝑛 ∈ ℕ,𝛼 ∈ [𝑛 − 1, 𝑛), 𝜇 = 𝛼 − (𝑛 − 1). Let 𝑓(𝑥) be a function

such that 𝐷𝑎𝛼𝑓(𝑥)𝑐 exist, then

𝐷𝑎𝛼𝑐 𝑓(𝑥) = 𝐷𝜇𝑐 𝐷𝑛−1𝑐 𝑓(𝑥) (1.102)

Theorem 1.5.2. [11, 19,22] (Some basic rules of Caputo fractional derivative):

Let 𝛼 ∈ [𝑛 − 1, 𝑛), 𝑛 ∈ ℕ,

1) 𝐷𝑎𝛼𝑐 𝑐 = 0 , 𝑐 is constant (1.103)

2) 𝐷𝑎𝛼𝑐 𝑥𝜇 = {

Γ(𝜇 + 1)

Γ(𝜇 − 𝛼 + 1)𝑥𝜇−𝛼 𝑖𝑓 𝜇 > 𝑛 − 1, 𝑥 > 0, 𝜇 ∈ ℝ

0 𝑖𝑓 𝜇 ≤ 𝑛 − 1, 𝑥 > 0, 𝜇 ∈ ℕ

(1.104)

3) 𝐷𝑎𝛼𝑐 𝑒𝑐𝑥 = 𝑐𝑛𝑥𝑛−𝛼𝐸1,𝑛−𝛼+1(𝑐𝑥) (1.105)

4) 𝐷𝑎𝛼𝑐 sin(𝑐𝑥) = −

1

2𝑖(𝑖𝑐)𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑖𝑐𝑥) − (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑖𝑐𝑥)] (1.106)

5) 𝐷𝑎𝛼𝑐 cos(𝑐𝑥) =

1

2(𝑖𝑐)𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑖𝑐𝑥) + (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑖𝑐𝑥)] (1.107)

6) 𝐷𝑎𝛼𝑐 cos(𝑐𝑥) =

1

2(𝑖𝑐)𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑖𝑐𝑥) + (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑖𝑐𝑥)] (1.108)

7) 𝐷𝑎𝛼𝑐 sinh(𝑐𝑥) = −

1

2𝑐𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑐𝑥) − (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑐𝑥)] (1.109)

8) 𝐷𝑎𝛼𝑐 cosh(𝑐𝑥) =

1

2𝑐𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑐𝑥) + (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑐𝑥)] (1.110)

Proof:

34

1) By applying the Caputo definition and because of the n’th derivative 𝑐(𝑛), (𝑛 ∈

ℕ , 𝑛 ≥ 1) of constant equals 0, then

𝐷𝑎𝛼𝑐 𝑐 =

1

Γ(𝑛 − 𝛼)∫

𝑐(𝑛)

(𝑥 − 𝑐)𝛼−𝑛+1

𝑥

𝑎

𝑑𝑡 = 0

2) The second case has an easy proof

( 𝐷𝑎𝛼𝑐 𝑡𝜇 = 0, 𝛼 ∈ (𝑛 − 1, 𝑛), 𝜇 ≤ 𝑛 − 1, 𝑛 ∈ ℕ)

It follows from the pattern of the proof of (1). But the first case is more

interesting. We can prove it by two ways. Directly by using Caputo definition.

Firstly, let α ∈ (𝑛 − 1, 𝑛), 𝜇 > 𝑛 − 1, 𝜇 ∈ ℝ

𝐷𝑎𝛼𝑥𝜇 =

1

Γ(𝑛 − 𝛼)𝑐 ∫

𝐷𝑛𝑡𝜇

(𝑥 − 𝑡)𝛼+1−𝑛

𝑥

0

𝑑𝑡

=1

Γ(𝑛 − 𝛼)∫

Γ(𝜇 + 1) 𝑡𝜇−𝑛

(𝑥 − 𝑡)𝛼+1−𝑛Γ(𝜇 − 𝑛 + 1)

𝑥

0

𝑑𝑡

Now by plugging t= 𝑥𝑢 ; 0 ≤ 𝑢 ≤ 1

=Γ(𝜇 + 1)

Γ(𝑛 − 𝛼)Γ(𝜇 − 𝑛 + 1)∫(𝑥𝑢)𝜇−𝑛((1 − 𝑢)𝑥)𝑛−𝛼−11

0

𝑥𝑑𝑢

=Γ(𝜇 + 1)

Γ(𝑛 − 𝛼)Γ(𝜇 − 𝑛 + 1)𝑥𝜇−𝑛𝛽(𝜇 − 𝑛 + 1, 𝑛 − 𝛼)

=Γ(𝜇 + 1)

Γ(𝑛 − 𝛼)Γ(𝜇 − 𝑛 + 1)𝑥𝜇−𝑛

Γ(𝜇 − 𝑛 + 1)Γ(𝑛 − 𝛼)

Γ(𝜇 − 𝛼 + 1)

=Γ(𝜇 + 1)

Γ(𝜇 − 𝛼 + 1)𝑥𝜇−𝑛

35

Secondly, we can prove it by the relation between the Caputo and

Riemann-Liouville derivatives:

𝐷𝑎𝛼𝑐 𝑥𝜇 = 𝐷𝛼𝑥𝜇 −∑

𝑥𝑘−𝛼

Γ(𝑘 + 1 − 𝛼)

𝑛−1

𝑘=0

𝐷𝑘[(𝑥)𝜇]𝑥=0

Now, the 𝐷𝑘[(𝑥)𝜇]𝑥=0 = 0, for 𝑘 ≤ 𝑛 − 1 ≤ 𝜇

Then 𝐷𝑎𝛼𝑐 𝑥𝜇 =

Γ(𝜇+1)

Γ(μ−α+1)𝑥𝜇−𝛼 ∎

3) To prove it, we need to use the relation between Caputo and Riemann-Liouville

fractional derivative as in (1.93) and use the exponential case of Riemann-

Liouville 𝛼 −derivative in (1.80), then we have:

𝐷𝑎𝛼𝑒𝑐𝑥 = 𝐷𝛼𝑒𝑐𝑥 −∑

𝑥𝑘−𝛼

Γ(𝑘 + 1 − 𝛼)

𝑛−1

𝑘=0

𝐷𝑘𝑒𝑐𝑥|

𝑥=0

= 𝑥−𝛼𝐸1,1−𝛼(𝑐𝑥) −∑𝑥𝑘−𝛼𝑐𝑘

Γ(𝑘 + 1 − 𝛼)

𝑛−1

𝑘=0

= ∑(𝑐𝑥)𝑘𝑥−𝛼

Γ(k + 1 − α)

∞

𝑘=0

−∑𝑥𝑘−𝛼𝑐𝑘

Γ(𝑘 + 1 − 𝛼)

𝑛−1

𝑘=0

= ∑𝑥𝑘−𝛼𝑐𝑘

Γ(𝑘 + 1 − 𝛼)

∞

𝑘=𝑛

=∑𝑥𝑘+𝑛−𝛼𝑐𝑘+𝑛

Γ(𝑘 + 𝑛 − 𝛼 + 1)

∞

𝑘=0

= 𝑐𝑛𝑥𝑛−𝛼𝐸1,𝑛−𝛼+1(𝑐𝑥) ∎

4) Use (sin 𝑥 =𝑒𝑖𝑥−𝑒−𝑖𝑥

2𝑖), then by using (1.105)

𝐷𝑎𝛼𝑐 sin(𝑐𝑥) = 𝐷𝑎

𝛼𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖𝑐 =

1

2𝑖( 𝐷𝑎

𝛼𝑐 𝑒𝑖𝑐𝑥 − 𝐷𝑎𝛼𝑐 𝑒−𝑖𝑐𝑥)

36

=1

2𝑖[(𝑖𝑐)𝑛𝑥𝑛−𝛼𝐸1,𝑛−𝛼+1(𝑖𝑐𝑥) − (−𝑖𝑐)

𝑛𝑥𝑛−𝛼𝐸1,𝑛−𝛼+1(−𝑖𝑐𝑥)]

= −1

2𝑖(𝑖𝑐)𝑛𝑥𝑛−𝛼[𝐸1,𝑛−𝛼+1(𝑖𝑐𝑥) − (−1)

𝑛𝐸1,𝑛−𝛼+1(−𝑖𝑐𝑥)] ∎

5) Follows by using (cos𝑥 =𝑒𝑖𝑥+𝑒−𝑖𝑥

2) and the same as (4).

6) Follows by using (sinh 𝑥 =𝑒𝑥−𝑒−𝑥

2) and the same as (4).

7) Follows by using (cosh𝑥 =𝑒𝑥+𝑒−𝑥

2) and the same as (4).

1.6 Comparison between Riemann-Liouville and Caputo Fractional Derivative

Operators

Our goal in this section is to make a comparison between the definitions of

fractional derivative of Riemann-Liouville and Caputo, because the definition of

fractional integral is the same for both Riemann-Liouville and Caputo definitions

Remark 1.6.1. [11] If 𝑓(𝑐) = 𝑓′(𝑐) = ⋯ = 𝑓(𝑛−1)(𝑐) = 0, then

𝐷𝑎𝛼𝑓(𝑥) = 𝐷𝑎

𝛼𝑐 𝑓(𝑥)

Remark 1.6.2. [11] The difference between Caputo and Riemann-Liouville formulas

for the fractional derivatives leads to the following differences:

Caputo fractional derivative of a constant equals zero while (Riemann-Liouville)

fractional derivative of a constant does not equal zero.

The non-commutation, in Caputo fractional derivative we have:

37

𝐷𝑎𝛼𝑐 ( 𝐷𝑎

𝑚𝑐 𝑓(𝑥)) = 𝐷𝑎𝛼+𝑚𝑐 𝑓(𝑥) ≠ 𝐷𝑎

𝑚 ( 𝐷𝑎𝛼𝑐 𝑓(𝑥)) ,𝑐 (1.111)

where 𝛼 ∈ (𝑛 − 1, 𝑛), 𝑛 ∈ ℕ,𝑚 = 1,2,…

While for Riemann-Liouville derivative

𝐷𝑎𝑚(𝐷𝑎

𝛼𝑓(𝑥)) = 𝐷𝑎𝛼+𝑚𝑓(𝑥) ≠ 𝐷𝑎

𝛼(𝐷𝑎𝑚𝑓(𝑥)) , (1.112)

where 𝛼 ∈ (𝑛 − 1, 𝑛), 𝑛 ∈ ℕ,𝑚 = 1,2,…

Note that the formulas as in (1.111) and (1.112) become equalities under the following

additional conditions:

𝑓(𝑠)(𝑎) = 0 , 𝑠 = 𝑛, 𝑛 + 1,… ,𝑚 − 1 for 𝐷𝑐 𝛼

𝑓(𝑠)(𝑎) = 0 , 𝑠 = 0,1,2,… ,𝑚 − 1 for 𝐷𝛼.

38

Table 1: Comparison between Riemann-Liouville and Caputo [11]

𝑓(𝑡 )

=𝑐=

consta

nt

Non-c

om

muta

tion

Lin

earity

Inte

rpola

tion

Repre

senta

tion

Pro

perty

𝐷𝛼𝑐=

𝑐

Γ(1−𝛼)𝑡−𝛼≠0 ,𝑐=𝑐𝑜𝑛𝑠𝑡

𝐷𝑚𝐷𝛼𝑓(𝑡 )

=𝐷𝛼+𝑚𝑓(𝑡)

≠𝐷𝛼𝐷𝑚𝑓(𝑡)

𝐷𝛼(𝜆𝑓(𝑡 )+𝑔(𝑡) )

=𝜆𝐷𝛼𝑓(𝑡 )

+𝐷𝛼𝑔(𝑡)

lim𝛼→𝑛𝐷𝛼𝑓(𝑡 )

=𝑓(𝑛)(𝑡)

lim𝛼→𝑛−1𝐷𝛼𝑓(𝑡)

=𝑓(𝑛−1)(𝑡)

𝐷𝛼𝑓(𝑡 )

=𝐷𝛼(𝐷

−(𝑛−𝛼)𝑓(𝑥))

Rie

ma

nn

-Lio

uv

ille

𝐷𝑎 𝛼𝑐=0 ,𝑐=𝑐𝑜𝑛𝑠𝑡

𝐷𝑎 𝛼

𝑐𝐷𝑚𝑓(𝑡 )

=𝐷𝑎 𝛼+𝑚𝑓(𝑡)

≠𝐷𝑚𝐷𝑎 𝛼𝑓(𝑡)

𝐷𝑎 𝛼

𝑐(𝜆𝑓(𝑡 )+𝑔(𝑡) )

=𝜆𝐷𝑎 𝛼𝑓(𝑡 )

+𝐷𝑎 𝛼𝑔(𝑡)

lim𝛼→𝑛𝐷𝑎 𝛼

𝑐𝑓(𝑡 )

=𝑓(𝑛)(𝑡)

lim𝛼→𝑛−1𝐷𝑎 𝛼

𝑐𝑓(𝑡 )

=𝑓(𝑛−1)(𝑡 )

−𝑓(𝑛−1)(0)

𝐷 𝑐𝑎 𝛼𝑓(𝑡 )

=𝐷−(𝑛−𝛼)𝐷𝑎 𝑛

𝑐𝑓(𝑥)

Ca

pu

to

39

1.7 Ordinary Differential Equations [2] :

This section shows some basic information about ordinary differential equation

which is needed in this thesis.

1.7.1. Bernoulli Differential Equation

Let us take a look at differential equation on the form

𝑦′ + 𝑝(𝑥)𝑦 = 𝑞(𝑥)𝑦𝑛 (1.113)

where 𝑝(𝑥) and 𝑞(𝑥) both are continous, 𝑛 ∈ ℝ.

Differential equation above is called Bernoulli equation

Now we solve (1.113) by dividing both sides by 𝑦𝑛.

𝑦−𝑛𝑦′ + 𝑝(𝑥)𝑦1−𝑛 = 𝑞(𝑥) (1.114)

Let 𝑣 = 𝑦1−𝑛, then

𝑣′ = (1 − 𝑛)𝑦−𝑛𝑦′

Multiply (1.113) by (1 − 𝑛)𝑦−𝑛 , we get:

1

1 − 𝑛𝑣′ + 𝑝(𝑥)𝑣 = 𝑞(𝑥) (1.115)

This is a linear differential equation.

1.7.2 Second-Order Linear Differential Equations [2]:

A second-order linear differential equation has the form

40

𝐴(𝑥)𝑦′′ + 𝑃(𝑥)𝑦 + 𝑄(𝑥) = 𝐺(𝑥( (1.116)

where A,P,Q and G are continuous functions, when 𝐺(𝑥) = 0, for all 𝑥, in equation

(1.116). Such equations are called homogenous linear equations. Thus, the form of a

second-order linear homogenous differential equation is:

𝐴(𝑥)𝑑2𝑦

𝑑𝑥2+ 𝑃(𝑥)

𝑑𝑦

𝑑𝑥+ 𝑄(𝑥) = 0 (1.117)

( if 𝐺(𝑥) ≠ 0 for some 𝑥, equation (1.116) is called nonhomogeneous equation)

41

Chapter Two: Conformable Fractional Definition

2.1 Conformable Fractional Derivative

When we study the previous definitions of derivative, we can illustrate that those

definitions have some inconveniences. The following are some of these shortcomings:

i) The Riemann-liouville derivative does not satisfy 𝐷𝑎𝛼(1) = 0

(𝐷𝑎𝛼(1) = 0 for the Caupto derivative) , if α is not a natural number.

ii) All fractional derivatives do not satisfy the Known product rule:

𝐷𝑎𝛼(𝑓𝑔) = 𝑓𝐷𝑎

𝛼(𝑓) + 𝑔𝐷𝑎𝛼(𝑓)

iii) All fractional derivatives do not satisfy the known quotient rule:

𝐷𝑎𝛼(𝑓 𝑔) =

𝑔𝐷𝑎𝛼(𝑓)−𝑓𝐷𝑎

𝛼(𝑔)

𝑔2⁄

iv) All fractional derivatives do not satisfy the chain rule:

𝐷𝑎𝛼(𝑓 ∘ 𝑔)(𝑡) = 𝑓(𝛼)(𝑔(𝑡))𝑔(𝛼)(𝑡)

v) All fractional derivatives don't satisfy: 𝐷𝛼𝐷𝛽𝑓 = 𝐷𝛼+𝛽𝑓 in general

vi) The Caputo definition assumes that the function f is differentiable.

Let us write Tα to denote the operator which is called the "Conformable

fractional derivative of order α ".

Khalil, et al. [14] introduced a completely new definition of fractional calculus which

is more natural and effective than previous definitions of order 𝛼 ∈ (0, 1]. Also, this

definition can be generalized to include any α. However, the case 𝛼 ∈ (0, 1] is the most

important one, and the other cases become easy when it is established..

42

Definition 2.1.1. [14] Given a function ∶ 𝑓: [0,∞) → ℝ . Then the (conformable fractional

derivative) of 𝑓 of order 𝛼 is defined by

𝑇𝛼(𝑓)(𝑡) = 𝑙𝑖𝑚𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀

For all 𝑡 ˃0, 𝛼 ∈ (0.1), if ƒ is α-differentiable in some (0, 𝛼). 𝛼 ˃ 0 and, lim𝑡→0+ 𝑓(𝛼)(𝑡)

exists, then define𝑓(𝛼)(0) = lim𝑡→0+ 𝑓(𝛼)(𝑡)

We sometimes, write 𝑓(𝛼)(𝑡) for 𝑇𝛼(𝑓 )(𝑡), to denote the conformable fractional

derivatives of 𝑓 of order 𝛼. In addition, if the conformable fractional derivative of f

of order α exists, then we say 𝑓 is α-differentiable.

We should take into consideration that 𝑇𝛼(𝑡𝑝) = 𝑝𝑡𝑝−𝛼. Further, this definition

coincides happen with the same of traditional definition of Riemann–Liouville and

of Caputo on polynomials (up to a constant multiple).

Theorem 2.1.1. [14] if a function 𝑓: [0,∞) → ℝ is 𝛼-differentiable at 𝑡0 ˃ 0. 𝛼 ∈

(0.1] then 𝑓 is continuous at 𝑡0

Proof:

Because 𝑓(𝛼)is differentiable at 𝑥 = 𝑡0, we know that

𝑓(𝛼)(𝑡0) = 𝑙𝑖𝑚𝜀→0𝑓(𝑡0+𝜀𝑡0

1−𝛼)−𝑓(𝑡0)

𝜀 exists.

If we next assume that 𝑥 ≠ 𝑡0 we can write the following

𝑓(𝑡0 + 𝜀𝑡01−𝛼) − 𝑓(𝑡0) =

𝑓(𝑡0 + 𝜀𝑡01−𝛼) − 𝑓(𝑡0)

𝜀𝜀

43

Then some basic properties of limits give us

𝑙𝑖𝑚𝜀→0( 𝑓(𝑡0 + 𝜀𝑡0

1−𝛼) − 𝑓(𝑡0)) = 𝑙𝑖𝑚𝜀→0

𝑓(𝑡0 + 𝜀𝑡01−𝛼) − 𝑓(𝑡0)

𝜀. 𝑙𝑖𝑚𝜀→0

𝜀

𝑙𝑖𝑚𝜀→0( 𝑓(𝑡0 + 𝜀𝑡0

1−𝛼) − 𝑓(𝑡0)) = 𝑓′(𝑡0). 0

Let ℎ = 𝜀𝑡01−𝛼. Then,

𝑙𝑖𝑚ℎ→0 𝑓(𝑡0 + ℎ) = 𝑓(𝑡0) . Hence, f is continuous at 𝑡0

It can be easily shown that 𝑇𝛼 satisfies all properties in the following theorem

Theorem 2.1.2. [14] Let 𝛼 ∈ (0.1] and 𝑓, 𝑔 be α-differentiable at a point 𝑡 ˃ 0 .Then:

(1) 𝑇𝛼(𝑓)(𝑡) = 𝑡1−𝛼 𝑑𝑓

𝑑𝑡(𝑡), where f is differentiable

(2.1)

(2) 𝑇𝛼(af + bg) = a 𝑇𝛼 (f ) + b 𝑇𝛼 (g), for all 𝑎, 𝑏 ∈ ℝ (2.2)

(3) 𝑇𝛼 (𝑡𝑝) = 𝑝 𝑡𝑝−𝛼 for all 𝑝 ∈ ℝ (2.3)

(4) 𝑇𝛼 (λ)=0 , for all constant functions 𝑓 (𝑡) = 𝜆 (2.4)

(5) 𝑇𝛼 (fg) = f 𝑇𝛼 (g) + g 𝑇𝛼 (f ) (2.5)

(6) 𝑇𝛼 ( ƒ

𝑔 ) =

𝑔 𝑇𝛼ƒ − ƒ 𝑇𝛼(𝑔)

𝑔2 (2.6)

Proof:

(1) Let ℎ = 𝜀𝑡1−𝛼 in definition (2.1.1). Then 𝜀 = ℎ𝑡𝛼−1

44

𝑇𝛼(𝑓)(𝑡) = lim𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀= lim𝜀→0

𝑓(𝑡 + ℎ) − 𝑓(𝑡)

ℎ𝑡𝛼−1

= 𝑡1−𝛼 limℎ→0

𝑓(𝑡 + ℎ) − 𝑓(𝑡)

ℎ= 𝑡1−𝛼𝑓′(𝑡) ∎

(2) 𝑇𝛼(𝑎𝑓 + 𝑏𝑔) = lim𝜀→0(𝑎𝑓+𝑏𝑔)(𝑡+𝜀𝑡1−𝛼)−(𝑎𝑓+𝑏𝑔)(𝑡)

𝜀

= lim𝜀→0

𝑎𝑓(𝑡 + 𝜀𝑡1−𝛼) + 𝑏𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑎𝑓(𝑡) − 𝑏𝑔(𝑡)

𝜀

= lim𝜀→0

𝑎 (𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀) + lim

𝜀→0𝑏 (𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑔(𝑡)

𝜀)

= 𝑎𝑇𝛼(𝑓) + 𝑏𝑇𝛼(𝑔) ∎

(3) Recall (𝑎 + 𝑏)𝑛 = ∑(𝑛

𝑘) 𝑎𝑛−𝑘

𝑛

𝑘=0

𝑏𝑘

Thus,

(𝑡 + 𝜀𝑡1−𝛼)𝑝 =∑(𝑝

𝑘) 𝑡𝑝−𝑘

𝑝

𝑘=0

(𝜀𝑡1−𝛼)𝑘

(𝑡 + 𝜀𝑡1−𝛼)𝑝 = (𝑝

0) 𝑡𝑝 + (

𝑝

1) 𝑡𝑝−1(𝜀𝑡1−𝛼)1 +⋯+ (

𝑝

𝑝) 𝑡0(𝜀𝑡1−𝛼)𝑝

To proof that 𝑇𝛼(𝑡𝑝) = 𝑝𝑡𝑝−𝛼

𝑙𝑖𝑚𝜀→0

(𝑡 + 𝜀𝑡1−𝛼)𝑝 − 𝑡𝑝

𝜀= 𝑙𝑖𝑚

𝜀→0

𝑡𝑝 + (𝑝1)𝑡𝑝−1(𝜀𝑡1−𝛼) + ⋯+ (𝑝

𝑝) 𝜀𝑝(𝑡1−𝛼)𝑝 − 𝑡𝑝

𝜀

= 𝑙𝑖𝑚 𝜀→0

𝜀𝑝𝑡𝑝−1𝑡1−𝛼 +⋯+ (𝑝

𝑝) 𝜀𝑝−1(𝑡1−𝛼)𝑝

𝜀

= 𝑝𝑡𝑝−1𝑡1−𝛼 = 𝑝𝑡𝑝−𝛼

45

(4) 𝑇𝛼(𝜆) = lim𝜀→0𝑓(𝑡+𝜀𝑡1−𝛼)−𝑓(𝑡)

𝜀

= lim𝜀→0

𝜆 − 𝜆

𝜀= 0 ∎

(5)

= lim𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼)𝑔(𝑡 + 𝜀𝑡1−𝛼) + 𝑓(𝑡)𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)𝑔(𝑡)

𝜀

= lim𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀𝑔(𝑡 + 𝜀𝑡1−𝛼) + 𝑓(𝑡) lim

𝜀→0

𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑔(𝑡)

𝜀

= 𝑇𝛼(𝑓(𝑡)) lim𝜀→0

𝑔(𝑡 + 𝜀𝑡1−𝛼) + 𝑓(𝑡)𝑇𝛼(𝑔(𝑡))

= 𝑔(𝑡)𝑇𝛼(𝑓(𝑡)) + 𝑓(𝑡)𝑇𝛼(𝑔(𝑡)) ∎

(6)

= lim𝜀→0

(𝑓(𝑡 + 𝜀𝑡1−𝛼)

𝑔(𝑡 + 𝜀𝑡1−𝛼)−

𝑓(𝑡)

𝑔(𝑡 + 𝜀𝑡1−𝛼)+

𝑓(𝑡)

𝑔(𝑡 + 𝜀𝑡1−𝛼)−𝑓(𝑡)

𝑔(𝑡)) .1

𝜀

= lim𝜀→0

(𝑓(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)

𝜀. 𝑔(𝑡 + 𝜀𝑡1−𝛼)) + 𝑓(𝑡). lim

𝜀→0(

1

𝜀𝑔(𝑡 + 𝜀𝑡1−𝛼)−

1

𝜀𝑔(𝑡))

= 𝑇𝛼(𝑓) lim𝜀→0

1

𝑔(𝑡 + 𝜀𝑡1−𝛼)+ 𝑓(𝑡) lim

𝜀→0(−(

𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑔(𝑡)

𝜀𝑔(𝑡)𝑔(𝑡 + 𝜀𝑡1−𝛼)))

= 𝑇𝛼(𝑓)1

𝑔(𝑡)− 𝑓(𝑡) lim

𝜀→0(𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑔(𝑡)

𝜀) lim𝜀→0

(1

𝑔(𝑡)𝑔(𝑡 + 𝜀𝑡1−𝛼))

𝑇𝛼(𝑓𝑔) = lim𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼)𝑔(𝑡 + 𝜀𝑡1−𝛼) − 𝑓(𝑡)𝑔(𝑡)

𝜀

𝑇𝛼 (𝑓

𝑔) = lim

𝜀→0

𝑓(𝑡 + 𝜀𝑡1−𝛼)𝑔(𝑡 + 𝜀𝑡1−𝛼)

−𝑓(𝑡)𝑔(𝑡)

𝜀

46

=𝑇𝛼(𝑓)

𝑔(𝑡)− 𝑓(𝑡)𝑇𝛼(𝑔(𝑡)).

1

𝑔2(𝑡)

=𝑔(𝑡)𝑇𝛼(𝑓) − 𝑓(𝑡)𝑇𝛼(𝑔(𝑡))

𝑔2(𝑡) ∎

Theorem 2.1.3. [14] (Conformable fractional derivative of Known functions)

1) 𝑇𝛼(𝑒𝑐𝑡) = 𝑐𝑡1−𝛼𝑒𝑐𝑡 (2.7)

2) 𝑇𝛼(𝑠𝑖𝑛 (𝑎𝑡)) = 𝑎𝑡1−𝛼 𝑐𝑜𝑠 (𝑎𝑡) , 𝑎 ∈ ℝ (2.8)

3) 𝑇𝛼 (𝑐𝑜𝑠 (𝑎𝑡)) = −𝑎𝑡1−𝛼 𝑠𝑖𝑛 (𝑎𝑡), 𝑎 ∈ ℝ (2.9)

4) 𝑇𝛼(tan(𝑎𝑡)) = 𝑎𝑡1−𝛼𝑠𝑒𝑐2(𝑎𝑡) , 𝑎 ∈ ℝ (2.10)

5) 𝑇𝛼(𝑐𝑜𝑡(𝑎𝑡)) = −𝑎𝑡1−𝛼𝑐𝑠𝑐2(𝑎𝑡) , 𝑎 ∈ ℝ (2.11)

6) 𝑇𝛼(𝑠𝑒𝑐(𝑎𝑡)) = 𝑎𝑡1−𝛼 𝑠𝑒𝑐(𝑎𝑡) 𝑡𝑎𝑛(𝑎𝑡) , 𝑎 ∈ ℝ (2.12)

7) 𝑇𝛼(𝑐𝑠𝑐(𝑎𝑡)) = −𝑎𝑡1−𝛼 𝑐𝑠𝑐(𝑎𝑡) 𝑐𝑜𝑡(𝑎𝑡) , 𝑎 ∈ ℝ (2.13)

8) 𝑇𝛼 (1

𝛼𝑡𝛼) = 1 (2.14)

Proof:

1. 𝑇𝛼(𝑒𝑐𝑥) = 𝑙𝑖𝑚𝜀→0

𝑒𝑐(𝑡+𝜀𝑡1−𝛼)−𝑒𝑐𝑡

𝜀= 𝑒𝑐𝑡 𝑙𝑖𝑚𝜀→0

𝑒𝑐𝜀𝑡1−𝛼

−1

𝜀

= 𝑒𝑐𝑡 𝑙𝑖𝑚𝜀→0

𝑡1−𝛼𝑒𝑐𝜀𝑡1−𝛼

− 𝑡1−𝛼

𝜀𝑡1−𝛼= 𝑒𝑐𝑡𝑡1−𝛼 𝑙𝑖𝑚

𝜀→0

𝑒𝑐𝜀𝑡1−𝛼

− 1

𝜀𝑡1−𝛼

Let ℎ = 𝜀𝑡1−𝛼 . Then by using L’Hopital‘s rule, we get

47

= 𝑡1−𝛼𝑒𝑐𝑡 𝑙𝑖𝑚ℎ→0

𝑒𝑐ℎ − 1

ℎ= 𝑐𝑡1−𝛼𝑒𝑐𝑡 𝑙𝑖𝑚

ℎ→0

𝑒𝑐ℎ

1

= 𝑐𝑡1−𝛼𝑒𝑐𝑡 ∎

(By L’Hopital Rule)

2. 𝑇𝛼(𝑠𝑖𝑛(𝑎𝑡)) = 𝑙𝑖𝑚𝜀→0

𝑠𝑖𝑛 𝑎(𝑡 + 𝜀𝑡1−𝛼) − 𝑠𝑖𝑛(𝑎𝑡)

𝜀

= 𝑙𝑖𝑚𝜀→0

𝑠𝑖𝑛(𝑎𝑡) [𝑐𝑜𝑠(𝑎𝜀𝑡1−𝛼) − 1

𝜀] + 𝑙𝑖𝑚

𝜀→0

𝑐𝑜𝑠(𝑎𝑡) 𝑠𝑖𝑛(𝑎𝜀𝑡1−𝛼)

𝜀

= 𝑡1−𝛼 𝑠𝑖𝑛(𝑎𝑡) 𝑙𝑖𝑚𝜀→0

[𝑐𝑜𝑠(𝑎𝜀𝑡1−𝛼) − 1

𝜀𝑡1−𝛼]

+ 𝑡1−𝛼 𝑐𝑜𝑠(𝑎𝑡) 𝑙𝑖𝑚𝜀→0

𝑠𝑖𝑛(𝑎𝜀𝑡1−𝛼)

𝜀𝑡1−𝛼

Let ℎ = 𝜀𝑡1−𝛼 then we get

= 𝑡1−𝛼 𝑠𝑖𝑛(𝑎𝑡) 𝑙𝑖𝑚ℎ→0

[𝑐𝑜𝑠(𝑎ℎ) − 1

ℎ] + 𝑡1−𝛼 𝑐𝑜𝑠(𝑎𝑡) 𝑙𝑖𝑚

ℎ→0

𝑠𝑖𝑛(𝑎ℎ)

ℎ

By using L’Hoputal Rule, we get

= 𝑡1−𝛼sin (𝑎𝑡) 𝑙𝑖𝑚ℎ→0

−𝑎 𝑠𝑖𝑛(𝑎ℎ)

1+ 𝑡1−𝛼 𝑐𝑜𝑠(𝑎𝑡) . 𝑎

= 𝑎𝑡1−𝛼 𝑐𝑜𝑠(𝑎𝑡) ∎

3. Similar to (2)

= 𝑙𝑖𝑚𝜀→0

𝑠𝑖𝑛(𝑎𝑡) 𝑐𝑜𝑠(𝑎𝜀𝑡1−𝛼) + 𝑐𝑜𝑠(𝑎𝑡) 𝑠𝑖𝑛(𝑎𝜀𝑡1−𝛼) − 𝑠𝑖𝑛(𝑎𝑡)

𝜀

48

4. 𝑇𝛼(𝑡𝑎𝑛(𝑎𝑡)) = 𝑇𝛼 (𝑠𝑖𝑛(𝑎𝑡)

𝑐𝑜𝑠(𝑎𝑡))

=𝑐𝑜𝑠(𝑎𝑡) 𝑇𝛼(𝑠𝑖𝑛(𝑎𝑡)) − 𝑠𝑖𝑛(𝑎𝑡) 𝑇𝛼(𝑐𝑜𝑠 𝑎𝑡)

𝑐𝑜𝑠2(𝑎𝑡)

=𝑐𝑜𝑠(𝑎𝑡) (𝑎𝑡1−𝛼 𝑐𝑜𝑠(𝑎𝑡)) − 𝑠𝑖𝑛(𝑎𝑡)(−𝑎𝑡1−𝛼 𝑠𝑖𝑛(𝑎𝑡))

𝑐𝑜𝑠2(𝑎𝑡)

=𝑎𝑡1−𝛼 𝑐𝑜𝑠2(𝑎𝑡) + 𝑎𝑡1−𝛼 𝑠𝑖𝑛2(𝑎𝑡)

𝑐𝑜𝑠2(𝑎𝑡)

= 𝑎𝑡1−𝛼(1 + 𝑡𝑎𝑛2(𝑎𝑡))

= 𝑎𝑡1−𝛼 𝑠𝑒𝑐2(𝑎𝑡) ∎

5. Similar to (4)

6. 𝑇𝛼(𝑠𝑒𝑐(𝑎𝑡)) = 𝑇𝛼 (1

𝑐𝑜𝑠(𝑎𝑡)) =

(−1)(𝑇𝛼(𝑐𝑜𝑠(𝑎𝑡)))

𝑐𝑜𝑠2(𝑎𝑡)

=(−1)(−𝑎𝑡1−𝛼 𝑠𝑖𝑛(𝑎𝑡))

𝑐𝑜𝑠2(𝑎𝑡)= 𝑎𝑡1−𝛼

𝑠𝑖𝑛(𝑎𝑡)

𝑐𝑜𝑠(𝑎𝑡).

1

𝑐𝑜𝑠(𝑎𝑡)

= 𝑎𝑡1−𝛼 𝑡𝑎𝑛(𝑎𝑡) 𝑠𝑒𝑐(𝑎𝑡) ∎

7. Similar to (7)

8. 𝑇𝛼 (1

𝛼𝑡𝛼) = 𝑙𝑖𝑚

𝜀→0

1𝛼(𝑡 + 𝜀𝑡1−𝛼)𝛼 −

1𝛼 𝑡

𝛼

𝜀

=1

𝛼𝑙𝑖𝑚𝜀→0

(𝑡 + 𝜀𝑡1−𝛼)𝛼 − 𝑡𝛼

𝜀

=1

𝛼𝑙𝑖𝑚𝜀→0

𝑡𝛼 + (𝛼1) 𝑡𝛼−1𝜀𝑡1−𝛼 +⋯+ (

𝛼𝛼 − 1

)𝜀𝛼−1𝑡𝛼−1 + (𝛼𝛼)𝜀𝛼(𝑡1−𝛼)𝛼 − 𝑡𝛼

𝜀

49

=1

𝛼𝑙𝑖𝑚𝜀→0

𝜀 ((𝛼1) + (

𝛼𝛼 − 1

) 𝑡𝛼𝜀𝛼−2 +⋯+ (𝛼𝛼)𝑡𝛼−1(𝑡1−𝛼)𝛼)

𝜀

=1

𝛼. 𝛼 = 1 ∎

Corollary 2.1.1. (Conformable fractional derivative of certain functions)

i) 𝑇𝛼 (𝑠𝑖𝑛1

𝛼𝑡𝛼) = 𝑐𝑜𝑠

1

𝛼𝑡𝛼 (2.15)

ii) 𝑇𝛼 (𝑠𝑖𝑛1

𝛼𝑡𝛼) = 𝑐𝑜𝑠

1

𝛼𝑡𝛼 (2.16)

iii) 𝑇𝛼 (𝑒1𝛼𝑡𝛼) = 𝑒

1𝛼𝑡𝛼

(2.17)

Note: The function could be α-differentiable at a point but not differentiable. For

example, let 𝑓(𝑡) = 2√𝑡.

Then, 𝑇12

(𝑓)(0) = 𝑙𝑖𝑚𝑡→0+ 𝑇12

(𝑓)(𝑡) = 1 , when 𝑇12

(𝑓)(𝑡) = 1 , for all t>0 , but

𝑇1(𝑓)(0) does not exist.

The most important case for the range of 𝛼 ∈ (0,1), when 𝛼 ∈ (𝑛, 𝑛 + 1] the

definition would be as the following

Definition 2.1.2. [14] Let 𝛼 ∈ (𝑛, 𝑛 + 1], and f be an n-differentiable at t ,

where t > 0, then the conformable fractional derivative of f of order α is defined as:

𝑇𝛼(𝑓)(𝑡) = limε→0

ƒ(⌈α⌉−1) ( 𝑡 + 𝜀𝑡(⌈α⌉−α)) − ƒ(⌈α⌉−1)(t)

𝜀

where [α] is the smallest integer greater than or equal to α.

50

Remark 2.1.1. Let 𝛼 ∈ (𝑛, 𝑛 + 1], and f is (𝑛 + 1)-differentiable at 𝑡 > 0. Then:

𝑇𝛼(𝑓)(𝑡) = 𝑡(⌈α⌉−α) 𝑓

⌈α⌉(𝑡) (2.18)

Theorem 2.1.4 [14]

(Rolle’s Theorem for Conformable Fractional Differentiable Functions).

Let 𝑎 > 0 and 𝑓 ∶ [𝑎, 𝑏] → ℝ be a given function that satisfies

i. 𝑓 is continuous on [𝑎, 𝑏],

ii. 𝑓 is α-differentiable for some 𝛼 ∈ (0,1),

iii. 𝑓(𝑎) = 𝑓(𝑏).

Then, there exists 𝑐 ∈ (𝑎, 𝑏), such that 𝑓(𝛼)(𝑐) = 0.

Proof:

Since 𝑓 is continuous on [𝑎, 𝑏], and 𝑓(𝑎) = 𝑓(𝑏), there is 𝑐 ∈ (𝑎, 𝑏), which is

a point of local extrema. With no loss of generality, assume c is a point of local

minimum. So, 𝑓(𝛼)(𝑐) = lim𝜀→ 0+𝑓(𝑐+𝜀𝑐1−𝛼)−𝑓(𝑐)

𝜀= lim𝜀→ 0−

𝑓(𝑐+𝜀𝑐1−𝛼)−𝑓(𝑐)

𝜀, but the

first limit is non – negative, and the second limit is non-positive. Hence, 𝑓(𝛼)(𝑐) = 0.

Theorem 2.1.5. [14] (Mean Value Theorem for Conformable Fractional

Differentiable Functions). Let a > 0 and 𝑓 : [𝑎, 𝑏] → ℝ be a given function that

satisfies:

51

i) 𝑓 is continuous on [𝑎, 𝑏].

ii) 𝑓 is α-differentiable for some 𝛼 ∈ (0, 1).

Then, there exists 𝑐 ∈ (𝑎, 𝑏), such that

Proof:

The equation of the secant through (𝑎, 𝑓(𝑎)) and (𝑏, 𝑓(𝑏)) is

𝑦 − 𝑓(𝑎) =𝑓(𝑏) − 𝑓(𝑎)

1𝛼 𝑏

𝛼 −1𝛼 𝑎

𝛼(1

𝛼𝑥𝛼 −

1

𝛼𝑎𝛼)

which we can write as

𝑦 =𝑓(𝑏) − 𝑓(𝑎)

1𝛼 𝑏

𝛼 −1𝛼 𝑎

𝛼(1

𝛼𝑥𝛼 −

1

𝛼𝑎𝛼) + 𝑓(𝑎)

Let 𝑔(𝑥) = 𝑓(𝑥) − [𝑓(𝑏)−𝑓(𝑎)1

𝛼𝑏𝛼−

1

𝛼𝑎𝛼(1

𝛼𝑥𝛼 −

1

𝛼𝑎𝛼) + 𝑓(𝑎)].

Note that 𝑔(𝑎) = 𝑔(𝑏) = 0 , 𝑔 is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏). So

by Roll’s theorem there are 𝑐 in (𝑎, 𝑏) such that 𝑔(𝛼)(𝑐) = 0.

But

𝑓(𝛼)(𝑐) =ƒ(𝑏) − ƒ(𝑎)

1𝛼 𝑏

𝛼 − 1𝛼 𝑎

𝛼

𝑔(𝛼)(𝑥) = 𝑓(𝛼)(𝑥) − [𝑓(𝑏) − 𝑓(𝑎)

1𝛼 𝑏

𝛼 −1𝛼 𝑎

𝛼]

52

So

𝑓(𝛼)(𝑐) =𝑓(𝑏) − 𝑓(𝑎)

1𝛼 𝑏

𝛼 −1𝛼 𝑎

𝛼 ∎

2.2. Conformable Fractional Integrals

Suppose that the function is continuous

Let 𝛼 ∈ (0,∞). Define 𝐽𝛼(𝑡𝑝) =

𝑡𝑝+𝛼

𝑝+𝛼 , for any 𝑝 ∈ 𝑅 , 𝛼 ≠ −𝑝.

If 𝑓(𝑡) = ∑ 𝑏𝑘𝑡𝑘𝑛

𝑘=0 , then we define 𝐽𝛼(𝑓) = ∑ 𝑏𝑘𝐽𝛼(𝑡𝑘)𝑛

𝑘=0 = ∑ 𝑏𝑘𝑡𝑘+𝛼

𝑘+𝛼

𝑛𝑘=0

Cleary, 𝐽𝛼is linear in its domain. Further, if 𝛼 = 1, then 𝐽𝛼 the usual integral.

Now according to conformable fractional definition, if 𝛼 = 1 2⁄ ,then

sin 𝑡 = ∑(−1)𝑛

(2𝑘+1)!∞𝑛=0 𝑡2𝑛+1 then 𝐽𝛼(sin 𝑡) = ∑

(−1)𝑛𝑡2𝑛+

32

(2𝑛+3

2)(2𝑛+1)!

∞𝑛=0 .

Also, if 𝛼 =1

2

cos(𝑡) = ∑(−1)𝑛𝑡2𝑛

(2𝑛)!∞𝑛=0 then 𝐽𝛼(cos(𝑡)) = ∑

(−1)𝑛𝑡2𝑛+

12

(2𝑛+1

2)(2𝑛)!

∞𝑛=0

𝑒𝑡 = ∑𝑡𝑛

𝑛!∞𝑛=0 then 𝐽𝛼(𝑒

𝑡) = ∑𝑡𝑛+

12

(𝑛+1

2)(𝑛)!

∞𝑛=0

𝑔(𝛼)(𝑐) = 𝑓(𝛼)(𝑐) − [𝑓(𝑏) − 𝑓(𝑎)

1𝛼 𝑏

𝛼 −1𝛼 𝑎

𝛼] = 0

53

sinh(𝑡) = ∑𝑡2𝑛+1

(2𝑛+1)!∞𝑛=0 then 𝐽𝛼(sinh(𝑡)) = ∑

𝑡2𝑛+

32

(2𝑛+3

2)(2𝑛+1)!

∞𝑛=0

cosh (𝑡) = ∑𝑡2𝑛

(2𝑛)!∞𝑛=0 then 𝐽𝛼(cosh(𝑡)) = ∑

𝑡2𝑛+

12

(2𝑛+1

2)(2𝑛)!

∞𝑛=0 .

Definition 2.2.1 [14]

Let f be a continuous function. Then 𝛼-fractional integral of f is defined by:

𝐼𝛼𝑎𝑓(𝑡) = 𝐼1

𝑎(𝑡𝛼−1𝑓(𝑡)) = ∫𝑓(𝑥)

𝑥1−𝛼𝑑𝑥

𝑡

𝑎

(2.19)

where 𝑎 > 0,𝛼 ∈ (0,1) and the integral is the usual Riemann improper integral.

Examples:

1) 𝐼12

0(√𝑡 cos(𝑡)) = ∫cos(𝑥) . 𝑑𝑥

𝑡

0

= sin(𝑡)

2) 𝐼12

0(cos(2√𝑡)) = ∫cos(2√𝑥)

√𝑥. 𝑑𝑥

𝑡

0

= sin(2√𝑡)

Theorem 2.2.1 [14]

Let 𝑓 be any continuous function in the domain of 𝐼𝛼. Then

(𝑇𝛼𝐼𝛼𝑎(𝑓(𝑡)) = 𝑓(𝑡), for 𝑡 ≥ 𝑎) . (2.20)

54

Proof: since f is continues, then 𝐼𝛼𝑎 (𝑓)(𝑡) is differentiable. So

𝑇𝛼 (𝐼𝛼𝑎(𝑓(𝑡))) = 𝑡1−𝛼

𝑑

𝑑𝑡𝐼𝛼𝑎𝑓(𝑡)

= 𝑡1−𝛼𝑑

𝑑𝑡∫𝑓(𝑥)

𝑥1−𝛼

𝑡

𝑎

= 𝑡1−𝛼𝑓(𝑡)

𝑡1−𝛼 = 𝑓(𝑡) ∎

2.3 Applications [14]:

Now in this section we will solve fractional differential equations according to

conformable definitions:

Example (2.3.1):

𝑦(12⁄ ) + 𝑦 = 𝑥3 + 3𝑥5 2⁄ , 𝑦(0) = 0 (2.21)

To find

𝑦ℎ of 𝑦1 2⁄ + 𝑦 = 0

we use

𝑦ℎ = 𝑒𝑟√𝑥

Now

𝑦(1 2)⁄ + 𝑦 = 0

55

𝑟

2𝑒𝑟√𝑥 + 𝑒𝑟√𝑥 = 0

𝑒𝑟√𝑥 (𝑟

2+ 1) = 0

𝑟

2+ 1 = 0

𝑟 = −2

𝑦ℎ = 𝑒−2√𝑥

And simply the particular solution is 𝑦𝑝 = 𝑥3

And by plugging the initial condition 𝑦𝑝 = 𝑥3 then A = 0

∴ 𝑦 = 𝑦ℎ + 𝑦𝑝 = 𝑒−2√𝑥 + 𝑥3

For more examples see [14].

2.4. Abel’s Formula and Wronskain for Conformable Fractional Differential

Equation

In this section we will discuss the differential equation

𝑦′′ + 𝑃(𝑥)𝑦′ +𝑄(𝑥)𝑦 = 0 (2.22)

In the sense of conformable fractional derivative, Abu Hammad, et al. [7] replaced

the derivative by conformable fractional derivative. They studied the form of

Wronskain for conformable fractional linear differential equation with variable

56

coefficients. Finally, they study the Abel's formula. The result is similar to the case of

ordinary differential equation.

2.4.1. The Wronskain

For 𝛼 ∈ (0,1], Abu Hammad, et al. discussed the equation [7].

𝑇𝛼𝑇𝛼𝑦 + 𝑃(𝑥)𝑇𝛼𝑦 + 𝑄(𝑥)𝑦 = 0 (2.23)

They discussed also the fractional Wronskain of two functions.

Definition 2.4.1. [7] For two functions 𝑦1 and 𝑦2 satisfying (2.24) and 𝛼 ∈ (0,1] we set

𝑊𝛼[𝑦1, 𝑦2] = |𝑦1 𝑦2𝑇𝛼𝑦1 𝑇𝛼𝑦2

|

Theorem 2.4.1. [7] assume that 𝑦1, 𝑦2 satisfy equation (2.23), Then

𝑊𝛼[𝑦1, 𝑦2] = 𝑒−𝐼𝛼(𝑃)

Proof: applying the operator 𝑇𝛼 on 𝑊𝛼[𝑦1, 𝑦2] to get

𝑇𝛼(𝑊𝛼[𝑦1, 𝑦2]) = 𝑇𝛼(𝑦1𝑇𝛼𝑦2 − 𝑦2𝑇𝛼𝑦1)

= 𝑇𝛼𝑦1𝑇𝛼𝑦2 + 𝑦1𝑇𝛼𝑇𝛼𝑦2 − 𝑇𝛼𝑦2𝑇𝛼𝑦1 − 𝑦2𝑇𝛼𝑇𝛼𝑦1

But, 𝑦1 and 𝑦2 satisfy (2.24). So

𝑇𝛼𝑇𝛼𝑦1 = −𝑃(𝑥)𝑇𝛼𝑦1 − 𝑄(𝑥)𝑦1,

and

𝑇𝛼𝑇𝛼𝑦2 = −𝑃(𝑥)𝑇𝛼𝑦2 − 𝑄(𝑥)𝑦2,

57

therefore,

𝑇𝛼(𝑊𝛼[𝑦1, 𝑦2]) = −𝑃(𝑥)(𝑦1𝑇𝛼𝑦2 − 𝑦2𝑇𝛼𝑦1)

= −𝑃(𝑥)𝑊𝛼[𝑦1, 𝑦2] ,

thus,

𝑇𝛼(𝑊𝛼[𝑦1, 𝑦2])

𝑊𝛼[𝑦1, 𝑦2]= −𝑃(𝑥)

Consequently,

𝑊𝛼[𝑦1, 𝑦2] = 𝑒−𝐼𝛼(𝑃) (2.24)

2.4.2. Abel’s Formula

First of all, it is important to discuss linear fractional differential equation

𝑇𝛼𝑦 + 𝑎(𝑥)𝑦 = 𝑏(𝑥), 𝛼 ∈ [0, 1] (2.25)

Multiply (2.26) by 𝑒𝐼𝛼(𝑎(𝑥)) to get

𝑒𝐼𝛼(𝑎(𝑥))𝑇𝛼𝑦 + 𝑒𝐼𝛼(𝑎(𝑥))𝑎(𝑥)𝑦 = 𝑒𝐼𝛼(𝑎(𝑥))𝑏(𝑥)

𝑇𝛼(𝑒𝐼𝛼(𝑎(𝑥))𝑦) = 𝑒𝐼𝛼(𝑎(𝑥))𝑏(𝑥).

Hence

𝑦 = 𝑒−𝐼𝛼(𝑎(𝑥))𝐼𝛼(𝑒𝐼𝛼(𝑎(𝑥))𝑏(𝑥)) (2.26)

Is a solution of (2.26).

Now, let 𝑦1 be a solution of (2.24). To find a second solution 𝑦2 for equation (2.24).

58

We have 𝑊𝛼[𝑦1, 𝑦2] = 𝑒−𝐼𝛼(𝑃), from which we get:

𝑦1𝑇𝛼𝑦2 − 𝑦2𝑇𝛼𝑦1 = 𝑒−𝐼𝛼(𝑃),

And so

𝑇𝛼𝑦2 − 𝑦2

𝑇𝛼𝑦1𝑦1

=𝑒−𝐼𝛼(𝑃)

𝑦1 (2.27)

Equation (2.28) is a fractional linear equation, with 𝑎(𝑥) =𝑇𝛼𝑦1

𝑦1, and 𝑏(𝑥) =

𝐼𝛼(−𝑃(𝑥))

𝑦1.

Hence, using the fact:

𝐼𝛼 (𝑇𝛼𝑦1𝑦1

) = ln𝑦1,

And formula (2.27) to get:

𝑦2 = 𝑦1𝐼𝛼 (

𝑒−𝐼𝛼(𝑃)

𝑦12 ). (2.28)

59

Chapter 3: Exact Solution of Riccati Fractional Differential Equation

Riccati differential equation refers to the Italian Nobleman Count Jacopo Francesco

Riccati (1676-1754).

The fractional Riccati equation was studied by many researchers by using different

numerical methods [6, 9, 12, 13, 15, 20, 21, 24- 34]. Our interest in solving fractional

differential equations began when Prof. Khalil, et al.[14], presented the new and simple

conformable definition of fractional derivative.

In the rest of this chapter, we will find an exact solution to the fractional Riccati

differential equation (FRDE) precisely, we consider the following Problem:-

𝑦(𝛼) = 𝐴(𝑥)𝑦2 + 𝐵(𝑥)𝑦 + 𝐶(𝑥) (3.1)

𝑦(0) = 𝑘 , 𝑘: constant (3.2)

where 𝑦(𝛼) is the conformable fractional derivative of order 𝛼 ∈ (0,1] , we should remark

that the method can be generalized to include any 𝛼 .

3.1 Fractional Riccati Differential Equation (FRDE)

Riccati equation is studied by many researchers [8]. In this section, we found the exact

solution of fractional Riccati equation with known particular solution.

Theorem 3.1.1. (Reduction to second order equation)

The non-linear fractional Riccati equation can be reduced to a second order linear

ordinary differential equation of the form:

60

𝑢′′ − (

𝛼 − 1

𝑥+ 𝑅(𝑥)) 𝑢′ + 𝑥𝛼−1𝑆(𝑥)𝑢 = 0 (3.3)

When 𝐴(𝑥) is non-zero and differentiable, such that 𝛼 ∈ (0,1] ,also the solution of this

equation leads us to the solution.

𝑦 =

−𝑈′(𝑥) 𝑥1−𝛼

𝐴(𝑥)𝑈(𝑥) (3.4)

Proof:

Let 𝑣 = 𝑦𝐴(𝑥)

𝑣(𝛼) = (𝑦𝐴(𝑥))(𝛼) = 𝑦(𝛼)𝐴(𝑥) + 𝑦𝑥1−𝛼𝐴′(𝑥)

𝑦(𝛼) satisfies the FRDE also by substituting 𝑦 =𝑣

𝐴 and some algebraic steps, then:

𝑥1−𝛼𝑣′(𝑥) = 𝑣2 + 𝐵𝑣 + 𝐶𝐴 + 𝑣𝑥1−𝛼𝐴′

𝐴

Divided both sides by 𝑥1−𝛼, then:

𝑣′(𝑥) = 𝑥𝛼−1𝑣2 + 𝑥𝛼−1𝐵𝑣 + 𝑥𝛼−1𝐶𝐴 + 𝑣𝐴′

𝐴

Combining like terms, to get:

𝑣′(𝑥) = 𝑥𝛼−1𝑣2 + (𝑥𝛼−1𝐵 +

𝐴′

𝐴) 𝑣 + 𝑥𝛼−1𝐶𝐴 (3.5)

Assume: 𝑅(𝑥) = 𝑥𝛼−1𝐵 +𝐴′

𝐴 and 𝑆(𝑥) = 𝑥𝛼−1𝐶𝐴 , to get:

𝑣′(𝑥) = 𝑥𝛼−1𝑣2 + 𝑅(𝑥)𝑣 + 𝑆(𝑥)

61

Let 𝑥𝛼−1𝑣 = −

𝑢′

𝑢 (3.6)

(𝛼 − 1)𝑥𝛼−2𝑣 + 𝑥𝛼−1𝑣′ =−𝑢𝑢′′ + (𝑢′)2

𝑢2

(𝛼 − 1)𝑥𝛼−2𝑣 + 𝑥𝛼−1𝑣′ =−𝑢′′

𝑢+ 𝑣2(𝑥𝛼−1)2

Divide both sides by 𝑥𝛼−1

(𝛼 − 1)𝑥−1𝑣 + 𝑣′ = −𝑥1−𝛼𝑢′′

𝑢+ 𝑥𝛼−1𝑣2

𝛼 − 1

𝑥𝑣 + 𝑥1−𝛼

𝑢′′

𝑢= 𝑥𝛼−1𝑣2 − 𝑣′

From equation (3.5)

𝛼 − 1

𝑥𝑣 + 𝑥1−𝛼

𝑢′′

𝑢= −(𝑥𝛼−1𝐵 +

𝐴′

𝐴) 𝑣 − 𝑥𝛼−1𝐶𝐴

𝛼 − 1

𝑥𝑣 + 𝑥1−𝛼

𝑢′′

𝑢= −𝑅(𝑥)𝑣 − 𝑆(𝑥)

combining like terms to get:

𝑥1−𝛼𝑢′′

𝑢+ (𝛼 − 1

𝑥+ 𝑅(𝑥)) 𝑣 + 𝑆(𝑥) = 0

divide both sides by 𝑥1−𝛼 after substitute 𝑣 = −𝑢′

𝑢𝑥1−𝛼

𝑥1−𝛼𝑢′′

𝑢+ (𝛼 − 1

𝑥+ 𝑅(𝑥)) (−

𝑢′

𝑢𝑥1−𝛼) + 𝑆(𝑥) = 0

𝑢′′

𝑢− (𝛼 − 1

𝑥+ 𝑅(𝑥))

𝑢′

𝑢+ 𝑥𝛼−1𝑆(𝑥) = 0

∴ 𝑢′′ − (𝛼 − 1

𝑥+ 𝑅(𝑥))𝑢′ + 𝑥𝛼−1𝑆(𝑥)𝑢 = 0

62

An answer of this equation will lead us to

𝑦 =𝑣

𝐴=−𝑢′𝑥1−𝛼

𝑢𝐴 ∎

Theorem 3.1.2. (Transform FRDE to the Bernoulli equation)

For non-linear fractional Riccati equation the substitution 𝑣(𝑥) = 𝑦(𝑥) − 𝑦1(𝑥) will

transform the (FRDE) into Bernoulli equation (ordinary differential equation of the first

order), when 𝑦1 is a known particular solution,

Proof:

Since 𝑣(𝑥) = 𝑦(𝑥) − 𝑦1(𝑥)

∴ 𝑦(𝑥) = 𝑣(𝑥) + 𝑦1(𝑥)

And 𝑦(𝛼)(𝑥) = 𝑣(𝛼)(𝑥) + 𝑦1(𝛼)(𝑥)

Since 𝑦1(𝑥) solves the (FRDE), it must be that

𝑦1(𝛼) = 𝐴(𝑥)𝑦1

2 + 𝐵(𝑥)𝑦1 + 𝐶(𝑥)

Substitute in (3.1)

𝑣(𝛼)(𝑥) + 𝑦1(𝛼)(𝑥)⏟

𝑦(𝛼)(𝑥)

= 𝐴(𝑥) [𝑣 + 𝑦1]⏟ 𝑦(𝑥)

2+ 𝐵(𝑥) [𝑣 + 𝑦1]⏟

𝑦(𝑥)

+ 𝐶(𝑥)

𝑥1−𝛼𝑣 ′(𝑥)⏟ 𝑣(𝛼)(𝑥)

+ 𝐴𝑦12 + 𝐵𝑦1 + 𝐶 = 𝐴𝑣

2 + 2𝐴𝑣𝑦1 + 𝐴𝑦12 + 𝐵𝑣 + 𝐵𝑦1 + 𝐶

𝑥1−𝛼𝑣 ′(𝑥) = 𝐴𝑣2(𝑥) + 2𝐴𝑦1𝑣(𝑥) + 𝐵𝑣(𝑥)

𝑣 ′(𝑥) = 𝐴𝑥𝛼−1𝑣2(𝑥) + 2𝐴𝑥𝛼−1𝑦1𝑣(𝑥) + 𝐵𝑥𝛼−1𝑣(𝑥)

𝑣 ′(𝑥) + [−2𝑥𝛼−1𝐴(𝑥)𝑦1 − 𝑥𝛼−1𝐵(𝑥)]⏟

𝜑(𝑥)

𝑣 = 𝐴𝑥𝛼−1⏟ 𝑞(𝑥)

𝑣2(𝑥) (3.7)

This equation is of the form of Bernoulli equation with n=2 ∎

63

which could be transformed to first order linear differential equation.

Let 𝑢 = 𝑣−1(𝑥).

𝑑𝑢

𝑑𝑥= −𝑣−2(𝑥)

𝑑𝑣

𝑑𝑥

Multiply (3.7) by – 𝑣(𝑥)−2

−𝑣−2𝑣 ′ + [2𝑥𝛼−1𝐴𝑦1 + 𝑥𝛼−1𝐵]𝑣−2𝑣 = −𝐴𝑥𝛼−1

𝑣′+ [2𝑥𝛼−1𝐴𝑦1 + 𝑥𝛼−1𝐵]𝑣 = 𝐴 𝑥𝛼−1⏟

𝑞(𝑥)

(3.8)

The general solution is given by

𝑣 =

∫𝜇(𝑥)𝑞(𝑥). 𝑑𝑥 + 𝑐(𝑥)

𝜇(𝑥) (3.9)

where 𝜇(𝑥) = 𝑒(∫[2𝑥𝛼−1𝐴𝑦1+𝑥

𝛼−1𝐵]𝑑𝑥) (3.10)

Theorem 3.1.3. (Obtaining solution of FRDE by Abel’s formula)

Let 𝑦1 be a solution of (3.1), and assume that 𝑧 = 1

𝑦− 𝑦1, then the solution of FRDE is

𝑧 = 𝑒−𝐼(2𝐴𝑦1+𝐵)𝐼𝛼(𝑒𝐼(2𝐴𝑦1+𝐵)(−𝐴(𝑥))) (3.11)

Proof: suppose that 𝑦1 is a solution of FRDE, and let =1

𝑦−𝑦1 , then

𝑧(𝑦 − 𝑦1) = 1

𝑦 =

1

𝑧+ 𝑦1 (3.12)

64

Apply 𝛼-derivative definition to both sides of (3.12)

𝑇𝛼𝑦 = 𝑇𝛼 (1

𝑧) + 𝑇𝛼𝑦1

𝑇𝛼𝑦 = −𝑧−1−𝛼𝑧′ + 𝑇𝛼𝑦1

Substituting in the original FRDE

−𝑧−1−𝛼𝑧′ + 𝑇𝛼𝑦1 = 𝐴 [1

𝑧+ 𝑦1]

2

+ 𝐵 [1

𝑧+ 𝑦1] + 𝐶

−𝑧−1−𝛼𝑧′ = 𝐴 [1

𝑧2+2𝑦1𝑧+ 𝑦1

2] + 𝐵 [1

𝑧+ 𝑦1] + 𝐶 − 𝑇𝛼𝑦1

𝑇𝛼𝑦1 satisfies the FRDE

−𝑧−1−𝛼𝑧′ =𝐴

𝑧2+2𝑦1𝐴

𝑧+ 𝐴𝑦1

2 +𝐵

𝑧+ 𝐵𝑦1 + 𝐶 − 𝐴𝑦

2 − 𝐵𝑦1 − 𝐶

Combining like terms and divide both sides by −𝑧−1−𝛼

𝑧′ = −(2𝐴𝑦1 + 𝐵)𝑧𝛼 − 𝐴𝑧𝛼−1, then

𝑧′ + (2𝐴𝑦1 + 𝐵)𝑧𝛼 = −𝐴𝑧𝛼−1 (3.13)

Multiply both sides of equation (3.13) by 𝑧1−𝛼

𝑧1−𝛼𝑧′ + (2𝐴𝑦1 + 𝐵)𝑧 = −𝐴

𝑧(𝛼) + (2𝐴𝑦1 + 𝐵)𝑧 = −𝐴 (3.14)

which is Abel’s formula as we mentioned in the previous chapter.

Thus, the solution is

65

𝑧 = 𝑒−𝐼(2𝐴𝑦1+𝐵)𝐼𝛼(𝑒𝐼(2𝐴𝑦1+𝐵)(−𝐴(𝑥)))

Theorem 3.1.4. Assume that the coefficients 𝐶(𝑥) + 𝐵(𝑥) + 𝐴(𝑥) = 0 of the fractional

Ricatii (3.1), if 𝐶(𝑥) satisfies the integral condition, which is

𝐶(𝑥) =𝑓1(𝑥) − {𝐵(𝑥) + 𝐴(𝑥) [∫

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)− 𝐴1

𝑥]}2

4𝐴

(3.15)

where 𝐴1 is an arbitrary constant of integration.

and 𝑓1 is the new generating function satisfying the differential condition (3.15) given by:

𝐵2(𝑥) + 4𝐴(𝑥)𝑥1−𝛼

𝑑𝑦𝑝

𝑑𝑥= 𝑓1(𝑥) (3.16)

Then the general solution is given by:

𝑦(𝑥) =1

𝑒−𝐼(2𝐴𝑦1+𝐵)𝐼𝛼(𝑒𝐼(2𝐴𝑦1+𝐵)(−𝐴(𝑥)))

+1

2[∫

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)𝑑𝜙

𝑥

− 𝐴1],

where 𝐴0 is an arbitrary constant of integration.

(3.17)

Proof.

Assume that the arbitrary function 𝐵(𝑥), 𝐴(𝑥) and 𝑓1(𝑥) satisfying (3.15) then the

particular solution

𝑦𝑝±(𝑥) =−𝐵 ± √𝑓1 − 4𝐴𝐶

2𝐴

66

=

−𝐵 ±√𝑓1 − 4𝐴

𝑓1(𝑥) − {𝐵(𝑥) + 𝐴(𝑥) [∫𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)𝑥

− 𝐴1]}2

4𝐴

2𝐴

=

−𝐵 ± √𝑓1 − 𝑓1(𝑥) − {𝐵(𝑥) + 𝐴(𝑥) [∫𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)− 𝐴1

𝑥]}2

2𝐴

=−𝐵 + 𝐵(𝑥) + 𝐴(𝑥) [∫

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)𝑥

− 𝐴1]

2𝐴

=𝐴(𝑥) [∫

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)𝑥

− 𝐴1]

2𝐴

=1

2[∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

− 𝐴1]

Thus

𝑦𝑝±(𝑥) =−𝐵 ± √𝑓1 − 4𝐴𝐶

2𝐴=1

2[∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

− 𝐴1] (3.18)

Differentiate equation (3.18)

𝑑𝑦

𝑑𝑥[−𝐵 ± √𝑓1 − 4𝐴𝐶

2𝐴] =

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙) (3.19)

Equation (3.19) can be integrated to get

−𝐵 ± √𝑓1 − 4𝐴𝐶

𝐴=∫𝑓1(𝜙) − 𝐵

2(𝜙)2𝐴(𝜙)

𝑥

1

67

−𝐵 ± √𝑓1 − 4𝐴𝐶 = 𝐴 [∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

]

√𝑓1 − 4𝐴𝐶 = 𝐵 + 𝐴 [∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

]

𝑓1 − 4𝐴𝐶 = {𝐵 + 𝐴 [∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

]}

2

−4𝐴𝐶 = −𝑓1 + {𝐵 + 𝐴 [∫𝑓1(𝜙) − 𝐵

2(𝜙)

2𝐴(𝜙)

𝑥

]}

2

𝐶(𝑥) =𝑓1(𝑥) − {𝐵(𝑥) + 𝐴(𝑥) [∫

𝑓1(𝜙) − 𝐵2(𝜙)

2𝐴(𝜙)− 𝐴1

𝑥]}2

4𝐴

3.2 Applications:

Example: - find the solution of

𝑦(1

2) = (𝑦 − 2√𝑥)

2+ 1 , 𝑦1(𝑥) = 2√𝑥 ; 𝑦(0) = 1 (3.20)

Solution: First we need to verify that 𝑦1 = 2√𝑥 is a solution to this equation by computing,

we find that 𝑦1 is a solution of (3.20).

Now we solve the equation.

Step1. Make the change of variables

Substituting 𝑦 = 𝑣 + 2√𝑥 and 𝑦(1

2) = 𝑣(

1

2) + 1 yields

68

𝑣(12)+ 1 = (𝑣 + 2√𝑥 − 2√𝑥)

2+ 1

Step 2. Simplify to a Bernoulli equation for 𝑣

𝑥12𝑣′ = 𝑣2

𝑣 ′ = 𝑥−

12𝑣2 (3.21)

This is a Bernoulli equation.

Step3. Solve the Bernoulli equation

Let 𝑢 = 𝑣−1

𝑢′ = −𝑣−2𝑣′

Multiply equation (3.21) by −𝑣−2

−𝑣−2𝑣 ′ = −𝑥−12𝑣−2𝑣2

𝑢′ = −𝑥−12 =

−1

√𝑥

𝑑𝑢

𝑑𝑥=−1

√𝑥 → 𝑑𝑢 =

−1

√𝑥𝑑𝑥

𝑢 = ∫−1

√𝑥. 𝑑𝑥 = −2√𝑥 + 𝑐

1

𝑣= −2√𝑥 + 𝑐

𝑣 =1

−2√𝑥 + 𝑐

Step 4. Reverse the substitution 𝑦 = 𝑣 + 2√𝑥

69

𝑦 =1

−2√𝑥 + 𝑐− 2√𝑥

Finally, we use the initial condition 𝑦(0) = 1

∴ 𝑐 = 1

∴ The general solution is

𝑦 =

3

−4√𝑥3 + 1−𝑥2

2 (3.22)

70

Future Work

The main aspect of the future work in the thesis is to take other conditions of fractional

Riccati Differential Equation (FRDE) and solve it.

71

Conclusions

The objective of the present thesis is to use conformable fractional derivative which is

simpler and more efficient. The new definition reflects a natural extension of normal

derivative to solve fractional differential equation specifically fractional Riccati differential

equation.

In this thesis we found an exact solution of fractional Riccati differential equation and

introduced some theorems which lead us to find a second solution when we have a given

particular solution.

72

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76

حل معادالت تفاضلية كسرية باستخدام تعريف المطابق للمشتقات الكسرية

إعداد

شادي أحمد الطراونة

المشرف

د. خالد جابر

الملخص

ع المكانيكية، واألحياء ، والفيزياء ، ئالمعادالت التفاضلية العادية والجزئية مهمة جداً في مجاالت عديدة؛ مثل الموا

ونظرية التحكم في األنظمة والبصريات، والكهروكيميائية ، والهندسة، واللزوجة المطاطية ، والشبكات الكهربائية ،

الديناميكية.

من قبل باحثين عدة، باستخدام طرق عديدة مختلفة. بدأ موضوع اهتمامنا وهو حل ريكاتيتّم دراسة معادلة

ومجموعة باحثين آخرين لتقديم تعريف جديد وبسيط وأكثر خليل رشديمعادالت تفاضلية كسرية، عندما قام الدكتور

رية. هذا التعريف الجديد هو امتداد للمشتقات العادية والذي يسّمى "تعريف المطابق للمشتقات كفاءةً للمشتقات الكس

الكسرية".

التفاضلية الكسرية، وقدّمنا بعض النظريات الذي تساعدنا في ريكاتيفي هذا البحث، أوجدنا حل دقيق لمعادلة

ة.التفاضلية الكسري د حل ثاني عندما يعطي حل لمعادلة ريكاتيإيجا