SOLVING FRACTIONAL DIFFERENTIAL EQUATIONS BY USING ...SOLVING FRACTIONAL DIFFERENTIAL EQUATIONS BY...
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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 Masterโs Degree of Science in Mathematics
Faculty of Graduate Studies
Zarqa University
May, 2016
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COMMITTEE DECISION
This Thesis/Dissertation (Solving Fractional Differential Equations by Using Conformable
Fractional Derivatives Definition) was Successfully Defended and Approved on
โฆโฆโฆโฆโฆโฆโฆโฆโฆ..
Examination Committee
Signature
Dr. Khaled Jaber (Supervisor)
Assoc. Prof. of Mathematics ------------------------------
------------------------------
Dr. (Member)
------------------------------
Dr. (Member)
------------------------------
Dr. (Member)
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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 ALLAHโs 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.
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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
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2.4. Abelโs Formula and Wronskain for Conformable Fractional Differential Equation ...... 55
2.4.1. The Wronskain ............................................................................................................. 56
2.4.2. Abelโs 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
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List of Symbols
Symbol Denoted
โ The set of Natural Numbers
โ The set of Real Numbers
๐พ(๐ , ๐ฅ) The Lower Incomplete Gamma Function
๐(๐ฅ) The Digamma Function
๐ต๐ฅ(๐, ๐) The Incomplete Beta Function
๐ธ๐ผ,๐ฝ(๐ง) The Two-Parameters Mittage-Leffler Function
๐ธ๐ผ,๐ฝ(๐)(๐ฅ) The k-th Derivative of Mittage-Leffler Function
๐ท๐โ๐๐(๐ฅ) The Riemann-Liouville Fractional Integral
๐ท๐๐๐(๐ฅ) The Riemann-Liouville Fractional Derivative
๐ท๐ ๐โ๐๐(๐ฅ) The Caputo Fractional Derivative
๐๐ผ๐(๐ฅ) The Conformable Fractional Derivative
๐ฝ๐ผ๐(๐ฅ) The Conformable Fractional Integral
ฮ(๐ฅ) Gamma Function
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List of Abbreviations
Abbreviation Denoted
R-L Riemann-Liouville
FDEs Fractional Differential Equations
FRDE Fractional Riccati Differential Equation
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List of Figures and Tables
Figure/Table Page
Figure 1 : Graph of Gamma Function ฮ(๐ฅ) 5
Table 1: Comparison between Riemann-Liouville and Caputo
38
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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 โconformable fractional derivativeโ.
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.
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INTRODUCTION
The sense of differentiation operator ๐ท = ๐ ๐๐ฅโ is known to all who went through
ordinary calculus. And for proper function ๐, the ๐โth derivative of ๐, namely ๐ท๐๐(๐ฅ) =
๐๐๐(๐ฅ)๐๐ฅ๐โ is well defined where ๐ is positive integer.
The beginning of derivative theory of non-integers order dates back to leibnizโs note in
his letter to LโHopital, 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) donโt 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], Adomainโs decomposition method [32],
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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 โconformable fractional derivativeโ.
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 โconformable fractional derivativesโ 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.
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Chapter one: Basic Concepts and Preliminaries
This chapter shows popular fractional derivatives presented by Riemann-liouville
(R-L) and Caputoโs 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 โfractional derivativeโ started in 1695 by LโHopetal, when he
questioned Leibniz what would it mean ๐ท๐๐ฅ
๐ท๐ฅ๐ if ๐ =
1
2 in his letter, Leibniz answered that
would be a paradox. This was the beginning of โfractional derivativeโ 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 ๐(๐ฅ) =
๐ฅ๐, where m is a positive integer, and differentiated it n times where ๐ โฅ ๐ to get
๐ท๐
๐ท๐ฅ=
๐!
(๐โ๐)! ๐ฅ๐โ๐ ,then he used Legendres symbol ฮ to have
๐ท๐๐
๐ท๐ฅ๐=
ฮ(๐+1)
ฮ(๐โ๐+1)๐ฅ๐โ๐ .
Using this formula when ๐ = 1 ๐๐๐ ๐ =1
2 he obtained ๐ท
1
2๐ =2โ๐ฅ
โ๐.
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.
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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 ๐ฅ = 0,โ1 ,โ2 ,โฆ, also
ฮ(๐ฅ) has an integral representation for complex number ๐, 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 ๐ฅ > 0 The Gamma function is defined by
ฮ(๐ฅ) = โซ(โ log(๐ก))๐ฅโ1๐๐ก
1
0
, (1.1)
by elementary changes of variables these historical definitions take the more usual
forms:
Theorem 1.2.1. [17, 23] For ๐ฅ > 0,
ฮ(๐ฅ) = โซ ๐ก๐ฅโ1๐โ๐ก๐๐ก ,
โ
0
(1.2)
or sometimes
ฮ(๐ฅ) = 2โซ ๐ก2๐ฅโ1๐โ๐ก2๐๐ก
โ
0
. (1.3)
Proof: Use respectively the changes of variable ๐ข = โlog (๐ก) and ๐ข2 = โ log(๐ก) in
(1.1)
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Figure 1 : Graph of Gamma Function ฮ(๐ฅ)
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 ๐ฅ > 0 , ๐ โ ๐, define:
ฮ๐(๐ฅ) =
๐!. ๐๐ฅ
๐ฅ(๐ฅ + 1)โฆ (๐ฅ + ๐)
= ๐๐ฅ
๐ฅ(1 + ๐ฅ 1โ )โฆ(1 + ๐ฅ ๐โ )
(1.4)
Theorem 1.2.2. [23] (Weierstrass) For any real number, except the non-positive
integers {0,-1 โฆ} we have the infinite product
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1
ฮ(๐ฅ)= ๐ฅ๐๐พ๐ฅโ (1 +
๐ฅ
๐) ๐โ๐ฅ ๐โโ
๐=1 . (1.5)
where ฮณ is the Eulerโs constant ฮณ =0.5772156649015328606065120900824024310421...
which is defined by: ๐พ = ๐๐๐๐โโ (1 +1
2+โฏ+
1
๐โ ๐๐๐ (๐)) .
Below are two important properties of Gamma function.
Theorem 1.2.3. [16, 17, 23] let ๐ฅ โ 0, ๐ โ โ, then:
1. ฮ(n + 1) = n! (1.6)
2. ฮ(๐ฅ) = ฮ(๐ฅ+1)
๐ฅ, for negative value of x . (1.7)
3. ฮ(๐ฅ)ฮ(1 โ ๐ฅ) = ๐
๐ ๐๐(๐๐ฅ) . (1.8)
4. ๐๐
๐๐ฅ๐ฮ(๐ฅ) = โซ ๐ก๐ฅโ1๐โ๐ก(๐๐ ๐ก)๐๐๐ก
โ
0
, ๐ฅ > 0 . (1.9)
5. ฮ(๐ฅ) = ๐ฅโ1โ (1 +1
๐)๐ฅ
(1 +๐ฅ
๐)โ1
โ๐=1 . (1.10)
6. ฮ (1
2+ ๐ง)ฮ (
1
2โ ๐ง) = ๐ sec ๐๐ง. (1.11)
7. 1
ฮ(๐ง)= ๐ง lim
๐โโ{๐โ๐งโ (1 +
๐ง
๐)
๐
๐=1
} (1.12)
From the above we can get:
(a) ฮ (1
2) = โ๐
(b) ฮ (5
2) =
3
2ฮ (3
2) =
3
2.1
2ฮ (1
2) =
3
4โ๐
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(c) ฮ (โ3
2) =
ฮ(โ3 2โ + 1)
โ32โ
=ฮ(โ12 )
โ32โ=
ฮ(12)
โ32 โ
โ12
=4
3โ๐
Definition 1.2.3. The lower incomplete Gamma function is defined by [17, 19]:
๐พ(๐ , ๐ฅ) = โซ๐ก๐ โ1๐โ๐ก๐ฅ
0
. ๐๐ก (1.13)
and the upper incomplete Gamma function
๐ค(๐ , ๐ฅ) = โซ ๐ก๐ โ1๐โ๐ก. ๐๐ก
โ
๐ฅ
(1.14)
The Relation between Gamma function and incomplete Gamma function is given by
[17].
(a) ๐พ(๐ , ๐ฅ) =โ
๐ฅ๐ ๐โ๐ฅ๐ฅ๐
๐ (๐ + 1)โฆ(๐ + ๐)
โ
๐=0
= ๐ฅ๐ ๐ค(๐ )๐โ๐ฅโ๐ฅ๐
๐ค(๐ + ๐ + 1)
โ
๐=0
(1.15)
(b) ๐๐๐๐ฅโโ
๐พ(๐ , ๐ฅ) = ฮ(๐ ) (1.16)
(c) ๐พ(๐ , ๐ฅ) + ฮ(๐ , ๐ฅ) = ฮ(๐ ) (1.17)
Definition 1.2.4. The Digamma function ฯ(x) is defined by [17]
๐(๐ฅ) =
๐
๐๐ฅ๐๐ ฮ(๐ฅ) =
๐คโฒ(๐ฅ)
๐ค(๐ฅ) (1.18)
Here are some properties of Digamma functions:
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1) ๐(๐ง + ๐) = ๐(๐ง) +1
๐ง+
1
๐ง + 1+โฏ+
1
๐ง + ๐ โ 1 (1.19)
2) ๐(๐ง) โ ๐(1 โ ๐ง) =โ๐
tan(๐๐ง) (1.20)
1.2.2. The Beta Function
The Beta function is useful function related to the Gamma functions. It is defined
for ๐ฅ > 0 and ๐ฆ > 0 by the two equivalent identities:
Definition 1.2.5. [23] The Beta function (or Eulerian integral of the first kind) is given
by
๐ต(๐ฅ, ๐ฆ) = โซ ๐ก๐ฅโ1(1 โ ๐ก)๐ฆโ1๐๐ก1
0; 0 โค ๐ก โค 1 (1.21)
= 2 โซ ๐ ๐๐(๐ก)2๐ฅโ1 ๐๐๐ (๐ก)2๐ฆโ1 ๐๐ก
๐ 2โ
0
; 0 โค ๐ก โค๐
2
This definition is also applicable for complex numbers ๐ฅ and ๐ฆ such as ๐ ๐(๐ฅ) > 0
and ๐ ๐(๐ฆ) > 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 ๐ ๐(๐ฅ) > 0 and ๐ ๐(๐ฆ) > 0 , Then
๐ต(๐ฅ, ๐ฆ) =
ฮ(๐ฅ)ฮ(๐ฆ)
ฮ(๐ฅ + ๐ฆ)= ๐ต(๐ฆ, ๐ฅ) (1.22)
Proof: by using the definite integral (1.3)
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ฮ(๐ฅ)ฮ(๐ฆ) = 4โซ ๐ข2๐ฅโ1๐โ๐ข2๐๐ขโซ ๐ฃ2๐ฆโ1๐โ๐ฃ
2๐๐ฃ
โ
0
โ
0
= 4โซ โซ ๐โ(๐ข2+๐ฃ2) ๐ข2๐ฅโ1๐ฃ2๐ฆโ1
โ
0
๐๐ข๐๐ฃ
โ
0
Now by using the polar variables ๐ข = ๐ cos๐ and ๐ฃ = ๐ sin ๐ so that,
ฮ(๐ฅ)ฮ(๐ฆ) = 4โซ โซ ๐โ๐2
๐ 2โ
0
๐2(๐ฅ+๐ฆ)โ1 cos2xโ1 ๐ sin2yโ1 ๐ ๐๐๐๐
โ
0
= 2โซ ๐โ๐2
โ
0
๐2(๐ฅ+๐ฆ)โ1๐๐. 2 โซ cos2xโ1 ๐ sin2yโ1 ๐๐๐
๐ 2โ
0
= ฮ(x + y)B(x, y) โ
From relation (1.23) follows
๐ต(๐ฅ + 1, ๐ฆ) =ฮ(๐ฅ + 1)ฮ(๐ฆ)
ฮ(๐ฅ + ๐ฆ + 1)=
xฮ(๐ฅ)ฮ(๐ฆ)
(x + y)ฮ(๐ฅ + ๐ฆ)=
๐ฅ
๐ฅ + ๐ฆ๐ต(๐ฅ, ๐ฆ)
This is the beta function functional equation
๐ต(๐ฅ + 1, ๐ฆ) =๐ฅ
๐ฅ + ๐ฆ๐ต(๐ฅ, ๐ฆ) (1.23)
Definition 1.2.6. The incomplete Beta function ๐ต๐(๐ฅ, ๐ฆ)is defined by:
๐ต๐(๐ฅ, ๐ฆ) = โซ ๐ก๐ฅโ1(1 โ ๐ก)๐ฆโ1. ๐๐ก ,
๐
0
0 < ๐ < 1 (1.24)
Note that from the above:
๐ต (1
2,1
2) = ๐
10
๐ต (1
3,2
3) =
2 โ3
3๐
๐ต (1
4,3
4) = ๐ โ2
๐ต(๐ฅ, 1 โ ๐ฅ) = ๐
sin (๐๐ฅ)
๐ต(๐ฅ, 1) = 1
๐ฅ
๐ต(๐ฅ, ๐) = (๐ โ 1)!
๐ฅ. (๐ฅ + 1)โฆ (๐ฅ + ๐ โ 1) ๐ โฅ 1
๐ต(๐, ๐) = (๐ โ 1)! (๐ โ 1)!
(๐ + ๐ โ 1)! ๐ โฅ 1 , ๐ โฅ 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:
๐ธ๐(๐ฅ) = โ
๐ฅ๐
ฮ(๐๐ + 1) , ๐ > 0
โ
๐=0
(1.25)
๐ธ๐,๐(๐ฅ) = โ
๐ฅ๐
ฮ(๐๐ + ๐) , ๐ > 0, ๐ > 0
โ
๐=0
(1.26)
If ๐ = 1 and ๐ โ โ
11
๐ธ1,1(๐ฅ) = โ
๐ฅ๐
ฮ(๐ + 1)=
โ
๐=0
โ๐ฅ๐
n!
โ
๐=0
= ๐๐ฅ (1.27)
๐ธ1,2(๐ฅ) = โ
๐ฅ๐
ฮ(๐ + 2)
โ
๐=0
=โ๐ฅ๐
(๐ + 1)!
โ
๐=0
=1
๐ฅโ
๐ฅ๐+1
(n + 1)!
โ
๐=0
=๐๐ฅ โ 1
๐ฅ
(1.28)
๐ธ1,3(๐ฅ) = โ
๐ฅ๐
ฮ(๐ + 3)
โ
๐=0
=โ๐ฅ๐
(๐ + 2)!
โ
๐=0
=1
๐ฅ2โ
๐ฅ๐+2
(n + 2)!
โ
๐=0
=๐๐ฅ โ 1โ ๐ฅ
๐ฅ2
(1.29)
In general,
๐ธ1,๐ =1
๐ฅ๐โ1{๐๐ฅ โ โ
๐ฅ๐
n!
๐โ2
๐=0
} (1.30)
Easily, we can obtain the following:
(a) ๐ธ2,1(๐ฅ2) = โ
๐ฅ2๐
๐ค(2๐ + 1)
โ
๐=0
=โ๐ฅ2๐
(2๐)!
โ
๐=0
= ๐๐๐ โ(๐ฅ) (1.31)
(b) ๐ธ2,2(๐ฅ2) = โ
๐ฅ2๐
๐ค(2๐ + 2)
โ
๐=0
=โ๐ฅ2๐+1
๐ฅ(2๐ + 1)!
โ
๐=0
=๐ ๐๐โ(๐ฅ)
๐ฅ (1.32)
(c) ๐ธ2,1(โ๐ฅ2) = โ
(โ๐ฅ2)๐
ฮ(2๐ + 1)
โ
๐=0
=โ(โ1)๐๐ฅ2๐
(2๐)!
โ
๐=0
= ๐๐๐ (๐ฅ) (1.33)
(d) ๐ธ2,2(โ๐ฅ2) = โ
(โ๐ฅ2)๐
ฮ(2๐ + 2)
โ
๐=0
=โ(โ1)๐๐ฅ2๐+1
๐ฅ(2๐ + 1)!
โ
๐=0
=๐ ๐๐(๐ฅ)
๐ฅ (1.34)
The Mittage-Leffler function has the following relations :
12
๐ธ๐,๐(๐ฅ) = ๐ฅ ๐ธ๐,๐+๐(๐ฅ) +
1
๐ค(๐) (1.35)
๐ธ๐,๐(๐ฅ) = ๐๐ธ๐,๐+1(๐ฅ) + ๐๐ฅ
๐
๐๐ฅ ๐ธ๐,๐+1(๐ฅ) (1.36)
Obviously, from (1.36) we have
๐
๐๐ฅ ๐ธ๐,๐(๐ฅ) =
1
๐๐ฅ [๐ธ๐,๐โ1(๐ฅ) โ (๐ โ 1) ๐ธ๐,๐(๐ฅ) ] (1.37)
The ๐-th derivative of Mittage-Leffler function is given as follows:
๐๐
๐๐ฅ๐[๐ฅ๐โ1 ๐ธ๐,๐(๐ฅ
๐)] = ๐ฅ๐โ๐โ1๐ธ๐,๐โ๐(๐ฅ๐) , ๐ โ ๐ > 0 , ๐ = 0, 1 ,โฏ (1.38)
The integration of the Mittage-Leffler function is given as follows:
โซ ๐ธ๐,๐(๐ ๐ก
๐)๐ก๐โ1๐ฅ
0
๐๐ก = ๐ฅ๐๐ธ๐,๐+1(๐ ๐ฅ๐) (1.39)
The relation (1.39) is a special case and the following relation is more general:
1
ฮ(๐ฃ)โซ (๐ฅ โ ๐ก)๐ฃโ1๐ฅ
0
๐ธ๐,๐(๐ ๐ก๐)๐ก๐โ1๐๐ก = ๐ฅ๐+๐ฃโ1 ๐ธ๐,๐+๐ฃ(๐ ๐ฅ
๐) , ๐ฃ > 0 (1.40)
From (1.40) we obtain the following important formulas:
1
ฮ(๐)โซ (๐ฅ โ ๐ก)๐โ1๐๐๐ก ๐๐ก = ๐ฅ๐ ๐ธ1,๐+1(๐๐ฅ) , ๐ > 0 ๐ฅ
0
(1.41)
1
ฮ(๐)โซ (๐ฅ โ ๐ก)๐โ1 cosh(๐๐ก) ๐๐ก = ๐ฅ๐ ๐ธ2,๐+1((๐๐ฅ)
2) , ๐ > 0 ๐ฅ
0
(1.42)
1
๐ค(๐)โซ (๐ฅ โ ๐ก)๐โ1 ๐ ๐๐โ(๐๐ก) ๐๐ก = ๐ ๐ฅ๐+1 ๐ธ2,๐+2((๐๐ฅ)
2) , ๐ > 0 ๐ฅ
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]:
๐ท๐ก๐ผ
๐ ๐(๐ก) =1
๐ค(๐ โ ๐ผ)(๐
๐๐ก)๐
โซ๐(๐ฅ)
(๐ก โ ๐ฅ)๐ผโ๐+1๐๐ฅ
๐ก
๐
(1.44)
(๐ โ 1) โค ๐ผ < ๐ ,where ๐ผ is a real number, ๐ is integer.
1.3.2. M.Caputo (1967) [3,4]:
๐ท๐ก๐ผ
๐๐ ๐(๐ก) =
1
๐ค(๐ โ ๐ผ)โซ
๐(๐)(๐ฅ)
(๐ก โ ๐ฅ)๐ผ+1โ๐๐๐ฅ
๐ก
๐
(1.45)
(๐ โ 1) โค ๐ผ < ๐ , where ๐ผ is a real number and ๐ is integer
1.3.3. Oldham and Spainer (1974) [4]:
The scaling property for fractional derivatives
๐๐๐(๐ฝ๐ฅ)
๐๐ฅ๐= ๐ฝ๐
๐๐๐(๐ฝ๐ฅ)
๐(๐ฝ๐ฅ)๐ (1.46)
1.3.4. Kolwanker and Gangel (1994) [4]:
Kolwanker and Gangel (KG) defined a local fractional derivative to explain the
behavior of โcontinuous but nowhere differentiableโ function for 0 < ๐ < 1 , the local
fractional derivative at point ๐ฅ = ๐ฆ , for ๐: [0,1] โ โ is:
๐ท๐๐(๐ฆ) =
๐๐(๐(๐ฅ) โ ๐(๐ฆ))
๐(๐ฅ โ ๐ฆ)๐ (1.47)
1.3.5. Conformable Fractional Derivative (2014) [4]:
let ๐: [0,โ) โ ๐ , ๐ก > 0 , then the Conformable fractional derivative of ๐ of order ๐ผ is
defined by
14
๐๐ผ(๐)(๐ก) = ๐๐๐
๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐ (1.48)
for all ๐ก > 0 , ๐ผ โ (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 ๐ is an integrable function on[๐, ๐], then for ๐ โ โ and for ๐ฅ โ [๐, ๐], we have
๐ท๐ผโ๐๐(๐ฅ) =
1
(๐ โ 1)!โซ(๐ฅ โ ๐ก)๐โ1๐(๐ก)
๐ฅ
๐
๐๐ก. (1.49)
By using (๐ โ 1)! = ฮ(๐) and if we replace the order (๐) by the order(๐),
where ๐ โ ๐ , then we get the following definition:-
Definition 1.4.1. [3, 4, 19, 22] let ๐(๐ฅ) be a piecewise continuous on ๐ = (0,โ) and
intergrable on any finite subinterval of ๐โฒ = [0,โ) and for ๐ > 0 , ๐ฅ > 0 we call
๐ทโ๐๐(๐ฅ) =1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐(๐ก) ๐๐ก
๐ฅ
0
(1.50)
The Riemann-Liouville fractional integral of order ๐ of ๐
Remark 1.4.1. [19, 22] The Riemannโs definition is given by:
15
๐ท๐โ๐๐(๐ฅ) =
1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐(๐ก) ๐๐ก.
๐ฅ
๐
(1.51)
where ๐ > 0 , ๐ฅ > ๐
The Liouvilleโs definition is given by:
๐ทโโโ๐ ๐(๐ฅ) =
1
ฮ(๐) โซ(๐ฅ โ ๐ก)๐โ1๐(๐ก)๐๐ก.
๐ฅ
โโ
where p > 0
Note that we use the symbol ๐ทโ๐๐(๐ฅ) instead of ๐ท0โ๐๐(๐ฅ) when the lower limit of
the integral equals zero.
Properties 1.4.1. [19,22] If ๐(๐ฅ) and โ(๐ฅ) are continuous functions a, ๐ โ ๐ , and
๐,๐ > 0 , then:
๐ท๐โ๐(๐ท๐
โ๐๐(๐ฅ)) = ๐ท๐โ๐(๐ท๐
โ๐๐(๐ฅ)) = ๐ท๐โ(๐+๐)๐(๐ฅ) (1.52)
๐ท๐ผโ๐(๐๐(๐ฅ) + ๐โ(๐ฅ)) = ๐๐ท๐ผ
โ๐๐(๐ฅ) + ๐๐ท๐ผโ๐โ(๐ฅ) (1.53)
Theorem 1.4.1. [19,22] (Basic rules of Riemann-Liouville fractional integral)
Let ๐ > 0 , ๐ฅ > 0, then
1. ๐ทโ๐๐ฅ๐ =ฮ(๐ + 1)
ฮ(๐ + ๐ + 1)๐ฅ๐+๐ , ๐ > โ1 (1.54)
2. ๐ทโ๐๐ =๐
ฮ(๐ + 1)๐ฅ๐ , ๐ is a constant (1.55)
16
3. ๐ทโ๐๐๐๐ฅ =
๐๐๐ฅ
๐๐๐ค(๐)๐พ(๐, ๐๐ฅ) , ๐ > 0
where ฮณ(p,cx) is the lower incomplete Gamma functions
(1.56)
4. ๐ทโ๐(sin ๐๐ฅ) = ๐๐ฅ๐+1๐ธ2,๐+2 (โ(๐๐ฅ )2) (1.57)
5. ๐ทโ๐(cos ๐๐ฅ) = ๐ฅ๐+1๐ธ2,๐+1 (โ(๐๐ฅ )2) (1.58)
6. ๐ทโ๐(cosh ๐๐ฅ) = ๐ฅ๐๐ธ2,๐+1 ((๐๐ฅ )2) (1.59)
7. ๐ทโ๐(sinh ๐๐ฅ) = ๐๐ฅ๐+1๐ธ2,๐+2 ((๐๐ฅ )2) (1.60)
8. ๐ทโ๐ ๐๐ ๐ฅ =๐ฅ๐
ฮ(๐ + 1)[๐๐ ๐ฅ โ ๐พ โ ๐(๐ + 1)] (1.61)
where ๐ is the digamma function and ๐พ = โ๐(1) = โฮโฒ(1) โ
0.5772157 is Euler constant.
Proof:
(1) ๐ทโ๐๐ฅ๐ = 1
ฮ(๐)โซ (๐ฅ โ ๐ก)๐โ1๐ก๐๐ฅ
0๐๐ก
=1
ฮ(๐)โซ(1 โ
๐ก
๐ฅ)๐โ1
๐ฅ๐โ1๐ก๐๐ฅ
0
๐๐ก
By substituting ๐ข = ๐ก ๐ฅโ
= 1
ฮ(๐)โซ(1 โ ๐ข)๐โ1๐ฅ๐โ1(๐ข๐ฅ)๐1
0
๐ฅ๐๐ข
=1
ฮ(๐)โซ(1 โ ๐ข)๐โ1๐ข๐๐ฅ๐+๐1
0
๐๐ข
17
= 1
ฮ(๐)๐ฅ๐+๐๐ฃ(๐ + 1, ๐)
=ฮ(๐ + 1)
ฮ(๐ + ๐ + 1)๐ฅ๐+๐ โ
(2) If we set ๐ = 0 in (1.54), then the proof is complete
(3)
๐ทโ๐๐๐๐ฅ =1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐ฅ
0
๐๐๐ก๐๐ก
=1
ฮ(๐)โซ(
๐(๐ฅ โ ๐ก)
๐)
๐โ1๐ฅ
0
๐๐๐ก๐๐ก
=1
ฮ(๐)โซ๐ข๐โ1
๐๐โ1
๐๐ฅ
0
๐๐๐ฅโ๐ข๐๐ข
๐=
๐๐๐ฅ
๐๐๐ค(๐)โซ ๐ข๐โ1๐๐ฅ
0
๐โ๐ข๐๐ข ,
By substituting ๐ข = ๐(๐ฅ โ ๐ก)
=๐๐๐ฅ
๐๐ฮ(๐)๐พ(๐, ๐๐ฅ) โ
(4) ๐ทโ๐(๐ ๐๐ ๐๐ฅ) =1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐ฅ
0
๐ ๐๐(๐๐ก) ๐๐ก
=1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐ฅ
0
๐ ๐๐(๐๐ก)
๐๐ก๐๐ก ๐๐ก
Simply by using (1.34) and (1.40),
๐ทโ๐(๐ ๐๐ ๐๐ฅ) = ๐๐ฅ๐+1๐ธ2,๐+2(โ(๐๐ฅ)2) โ
18
(5) Follows by using (1.33) and (1.40) and the same as (4).
(6) By using (1.42), we get:
๐ทโ๐(๐๐๐ โ(๐๐ฅ)) =1
๐ค(๐)โซ(๐ฅ โ ๐ก)๐โ1 ๐๐๐ โ(๐๐ก) . ๐๐ก
๐ฅ
0
= ๐ฅ๐๐ธ2,๐+1((๐๐ฅ)2) โ
(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 ๐๐๐ (๐๐ฅ) , ๐ ๐๐(๐๐ฅ) can be expressed in
generalized ๐ ๐๐ and ๐๐๐ functions as:
๐ทโ๐ cos(๐๐ฅ) =1
ฮ(๐)โซ(๐ฅ โ ๐ก)๐โ1๐ฅ
0
cos(๐๐ก) ๐๐ก = ๐ถ๐ฅ(๐, ๐) (1.62)
๐ทโ๐ sin(๐๐ฅ) =1
ฮ(๐)โซ(๐ฅ โ ๐ก)
๐ฅ
0
sin(๐๐ก) ๐๐ก = ๐๐ฅ(๐, ๐) (1.63)
Remark 1.4.3 [19, 22] We can express the fractional integral function ๐๐๐ฅ by using
Mittage-Leffler function as
๐ทโ๐๐๐๐ฅ = ๐ฅ๐๐ธ1,๐+1(๐๐ฅ) (1.61)
Proof:
By using (1.15) and (1.27), then
Dโp๐๐๐ฅ =๐๐๐ฅ
๐๐ฮ(๐)๐พ(๐, ๐๐ฅ) =
๐๐๐ฅ
๐๐ฮ(๐)(๐๐ฅ)๐ฮ(๐)๐โ๐๐ฅ๐ธ1,๐+1(๐๐ฅ)
19
= ๐ฅ๐๐ธ1,๐+1(๐๐ฅ)
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:
๐ท๐ผ๐๐ = ๐ท๐ผ
๐(๐ท๐โ๐๐) , ๐, ๐ โ โ, ๐ > ๐
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 ๐(๐ฅ) of order ๐ผ, ๐ โ 1 < ๐ผ < ๐,
๐ โ โ is defined by:
๐ท๐๐ผ๐(๐ฅ) = ๐ท๐ (๐ท๐
โ(๐โ๐ผ)๐(๐ฅ))
=๐๐
๐๐ฅ๐1
ฮ(๐ โ ๐ผ)โซ(๐ฅ โ ๐ก)๐โ๐ผ+1๐ฅ
๐
๐(๐ก). ๐๐ก
(1.62)
Definition 1.4.3. [3, 5, 19, 22] Let ๐(๐ฅ) be a function defined on the closed interval
[๐, ๐] and let ๐ผ โ [0,1), then the left Riemann-Liouville ๐ผ derivative of ๐(๐ฅ) is:
๐ท๐๐ผ๐(๐ฅ) =
1
ฮ(1 โ ๐ผ)
๐
๐๐ฅโซ
๐(๐ก)
(๐ฅ โ ๐ก)๐ผ
๐ฅ
๐
. ๐๐ก (1.63)
The right Riemann-Liouville ๐ผ derivative of ๐(๐ฅ) is
๐ท๐๐ผ๐(๐ฅ) =
โ1
ฮ(1 โ ๐ผ)
๐
๐๐ฅโซ
๐(๐ก)
(๐ก โ ๐ฅ)๐ผ
๐
๐ฅ
. ๐๐ก (1.64)
20
But when ๐ผ is any number greater than 1. Then the definition will be as the
following
Definition 1.4.4. [3, 5, 19, 22]
Let ๐(๐ฅ) be a function defined on the closed interval [๐, ๐] and let ๐ผ โ
[๐ โ 1, ๐), ๐ โ โ. Then the left Riemann-Liouville ๐ผ derivative of ๐(๐ฅ) is:
๐ท๐๐ผ๐(๐ฅ) =
1
ฮ(๐ โ ๐ผ)
๐๐
๐๐ฅ๐โซ
๐(๐ก)
(๐ฅ โ ๐ก)๐ผโ๐+1
๐ฅ
๐
. ๐๐ก (1.65)
and the right Riemann-Liouville ๐ผ derivative of ๐(๐ฅ) is
๐ท๐๐ผ๐(๐ฅ) =
(โ1)๐
ฮ(๐ โ ๐ผ)
๐๐
๐๐ฅ๐โซ
๐(๐ก)
(๐ก โ ๐ฅ)๐ผโ๐+1
๐
๐ฅ
. ๐๐ก (1.66)
Note that the required condition required in the definitions is to be ๐-times
continuously differentiable.
The relationship between integration and differentiation of Riemann-Liouville
operators for the arbitrary order ๐ are shown as follows:
The Derivative of fractional integral could be shown as:
๐ท๐ผ๐ (๐ท๐ผ
โ๐๐(๐ฅ)) = ๐ท๐โ๐๐(๐ฅ), (1.67)
where ๐(๐ฅ) is continuous also ๐ โฅ ๐ โฅ 0
precisely, when ๐ โฅ 0 then ๐ท๐ผ๐ (๐ท๐ผ
โ๐๐(๐ฅ)) = ๐(๐ฅ) (1.68)
21
Preposition 1.4.2. [19,22] Let ๐1(๐ฅ), ๐2(๐ฅ) be two functions defined on[๐, ๐], and let
๐ผ โ [๐ โ 1, ๐), ๐ โ โ, ๐, ๐ฝ โ โ and ๐ท๐๐ผ๐1(๐ฅ),๐ท๐
๐ผ๐2(๐ฅ) exist, then
๐ท๐๐ผ[๐๐1(๐ฅ) + ๐ฝ๐2(๐ฅ)] = ๐๐ท๐
๐ผ๐1(๐ฅ) + ๐ฝ๐ท๐๐ผ๐2(๐ฅ) (1.69)
Proof:
๐ท๐๐ผ[๐๐1(๐ฅ) + ๐ฝ๐2(๐ฅ)] =
1
ฮ(๐ โ ๐ผ)
๐๐
๐๐ฅ๐โซ[๐๐1(๐ฅ) + ๐ฝ๐2(๐ฅ)]
(๐ฅ โ ๐ก)๐ผโ๐+1
๐ฅ
๐
๐๐ก
=๐
ฮ(๐ โ ๐ผ)
๐๐
๐๐ฅ๐โซ
๐1(๐ฅ)
(๐ฅ โ ๐ก)๐ผโ๐+1
๐ฅ
๐
๐๐ก +๐ฝ
ฮ(๐ โ ๐ผ)
๐๐
๐๐ฅ๐โซ
๐2(๐ฅ)
(๐ฅ โ ๐ก)๐ผโ๐+1
๐ฅ
๐
๐๐ก
= ๐๐ท๐๐ผ๐1(๐ฅ) + ๐ฝ๐ท๐
๐ผ๐2(๐ฅ) โ
Preposition 1.4.3. (Interpolation Property)
Let ๐(๐ฅ) be a function defined on [๐, ๐] and let ๐ผ โ [0,1). Let ๐(๐ฅ) have a
continuous derivative of sufficient order and ๐ท๐๐ผ๐(๐ฅ) exists, then
lim๐ผโ1
๐ท๐๐ผ๐(๐ฅ) = ๐โฒ(๐ฅ) (1.70)
and lim๐ผโ0
๐ท๐๐ผ๐ (๐ฅ) = ๐(๐ฅ) (1.71)
Proof: see [22]
We can generalize the above equalities in preposition 1.4.3 for any positive
number ๐ผ to be
lim๐ผโ๐
๐ท๐๐ผ๐(๐ฅ) = ๐(๐)(๐ฅ) (1.72)
22
and lim๐ผโ๐โ1
๐ท๐๐ผ๐ (๐ฅ) = ๐(๐โ1)(๐ฅ) (1.73)
where ๐ผ โ [๐ โ 1, ๐) , ๐ โ โ 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
๐ท๐โ๐(๐ท๐
๐๐(๐ฅ)) = ๐ท๐
๐โ๐๐(๐ฅ) โโ[๐ท๐ผ
๐โ๐โ1๐(๐ฅ)]
๐ฅ=๐
(๐ฅ โ ๐)๐โ๐โ1
ฮ(๐ โ ๐)
๐โ1
๐=0
(1.74)
where ๐ โ 1 < ๐ < ๐ , ๐ โ โ
2) ๐ท๐โ๐ผ(๐ท๐
๐ผ๐(๐ฅ)) = ๐(๐ฅ) โโ[๐ท๐ผ๐ผโ๐โ1๐(๐ฅ)]๐ฅ=๐
(๐ฅ โ ๐)๐ผโ๐โ1
ฮ(๐ผ โ ๐)
๐โ1
๐=0
(1.75)
3) The fractional derivative of fractional derivative is shown as:-
๐ท๐๐(๐ท๐
๐ผ๐(๐ฅ)) = ๐ท๐+๐ผ๐(๐ฅ) โ โ[๐ท๐ผ๐ผโ๐โ1๐(๐ฅ)]๐ฅ=๐
(๐ฅ โ ๐)โ๐โ๐โ1
ฮ(โp โ k)
๐โ1
๐=0
where ๐ โ 1 < ๐ < ๐ , ๐ โ 1 < ๐ผ < ๐, ๐,๐ โ โ
(1.76)
Remark 1.4.4.
๐ท๐๐ท๐๐(๐ฅ) = ๐ท๐+๐๐(๐ฅ) = ๐ท๐๐ท๐๐(๐ฅ) (1.77)
if and only if
๐(๐)(0) = 0 , ๐ = 0,1,โฆ , ๐ where ๐ = max(๐,๐),
where ๐โ 1 โค ๐ < ๐ and ๐ โ 1 โค ๐ < ๐
23
Theorem 1.4.2. [19,22] The Riemann-Liouville ๐ derivative does not satisfy the
following
1) ๐ท๐ผ๐(๐โ) = ๐๐ท๐
๐(โ) + โ๐ท๐๐(๐)
2) ๐ท๐๐(๐ โ โ) = ๐(๐)(โ(๐ฅ))โ(๐ผ)(๐ฅ)
3) ๐ท๐ผ๐ (๐โโ ) =
โ๐ท๐๐(๐) โ ๐๐ท๐
๐(โ)
โ2
Theorem 1.4.2. [19,22] The Riemann-Liouville ๐ derivative of known functions:
Let ๐ > 0 , ๐ฅ > 0 , ๐, ๐ โ โ , then
1) ๐ท๐๐ฅ๐ =ฮ(๐ + 1)
ฮ(๐ โ ๐ + 1)๐ฅ๐โ๐ , ๐ > โ1 (1.78)
2) ๐ท๐๐ =๐
ฮ(1 โ ๐)๐ฅโ๐ (1.79)
3) ๐ท๐๐๐๐ฅ = ๐ฅโ๐๐ธ1,1โ๐(๐๐ฅ) (1.80)
4) ๐ท๐ cos(๐๐ฅ) = ๐ฅโ๐๐ธ2,1โ๐(โ(๐๐ฅ)2) (1.81)
5) ๐ท๐ sin(๐๐ฅ) = ๐๐ฅ1โ๐๐ธ2,2โ๐((๐๐ฅ)2) (1.82)
6) ๐ท๐ cosh(๐๐ฅ) = ๐ฅโ๐๐ธ2,1โ๐((๐๐ฅ)2) (1.83)
7) ๐ท๐ sinh(๐๐ฅ) = ๐๐ฅ1โ๐๐ธ2,2โ๐((๐๐ฅ)2) (1.84)
8) ๐ท๐ ln(๐ฅ) =๐ฅโ๐
ฮ(1 โ ๐)[ln(๐ฅ) โ ๐พ โ ๐(1 โ ๐)] (1.85)
Proof:
(1) Let ๐ โ 1 < ๐ < ๐ , ๐ โ โ, then
๐ท๐๐ฅ๐ = ๐ท๐[๐ทโ(๐โ๐)๐ฅ๐]
24
= ๐ท๐ [ฮ(๐ + 1)
ฮ(๐ + ๐ โ ๐ + 1)๐ฅ๐+๐โ๐]
=ฮ(๐ + 1)
ฮ(๐ + ๐ โ ๐ + 1).ฮ(๐ + ๐ โ ๐ + 1)
ฮ(๐ + ๐ โ ๐ โ ๐ + 1)๐ฅ๐+๐โ๐โ๐
=ฮ(ฮผ + 1)
ฮ(ฮผ โ p + 1)๐ฅ๐โ๐ โ
(2) It follows by substituting ๐ = 0 in (1.78)
(3) Using ๐ทโ๐๐๐๐ฅ = ๐ฅ๐๐ธ1,๐+1(๐๐ฅ)
= ๐ฅ๐โ(๐๐ฅ)๐
ฮ(๐ + ๐ + 1)
โ
๐=0
=โ๐๐๐ฅ๐+๐
ฮ(๐ + ๐ + 1)
โ
๐=0
=โ๐๐
ฮ(๐ + 1)
โ
๐=0
.ฮ(๐ + 1)
ฮ(๐ + ๐ + 1)๐ฅ๐+๐
=โ๐๐
ฮ(๐ + 1)
โ
๐=0
๐ทโ๐๐ฅ๐
(1.86)
Now, by using (1.86) we have
๐ท๐๐๐๐ฅ = ๐ท๐[๐ทโ(๐โ๐)๐๐๐ฅ] = ๐ท๐ [โ๐๐
ฮ(k + 1)
โ
๐=0
๐ทโ(๐โ๐)๐ฅ๐]
=โ๐๐
ฮ(k + 1)
โ
๐=0
๐ท๐๐ฅ๐ =โ๐๐
ฮ(k + 1).ฮ(๐ + 1)๐ฅ๐โ๐
ฮ(๐ โ ๐ + 1)
โ
๐=0
= ๐ฅโ๐โ(๐๐ฅ)๐
ฮ(๐ โ ๐ + 1)
โ
๐=0
= ๐ฅโ๐๐ธ1,1โ๐(๐๐ฅ) โ
(4) ๐ท๐ cos(๐๐ฅ) = ๐ท๐[๐ทโ(๐โ๐) cos(๐๐ฅ)]
25
= ๐ท๐[๐ฅ๐โ๐๐ธ2,๐โ๐+1(โ(๐๐ฅ)2)]
= ๐ฅ๐โ๐+1โ๐+1๐ธ2,๐โ๐+1โ๐(โ(๐๐ฅ)2)
= ๐ฅโ๐๐ธ2,1โ๐(โ(๐๐ฅ)2) โ
(5) Similarly of (4).
(6) ๐ท๐ cosh(๐๐ฅ) = ๐ท๐[๐ทโ(๐โ๐) cosh(๐๐ฅ)]
= ๐ท๐[๐ฅ๐โ๐๐ธ2,๐โ๐+1((๐๐ฅ)2)]
= ๐ฅ๐โ๐+1โ๐โ1๐ธ2,๐โ๐+1โ๐(โ(๐๐ฅ)2)
= ๐ฅโ๐๐ธ2,1โ๐((๐๐ฅ)2) โ
(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 ๐ be ๐ โtimes differentiable function,
๐ฅ, ๐ โ โ , ๐ฅ > ๐ and ๐ผ โ [0,1). Then the Caputo fractional differential operator of
order ๐ผ of ๐ is defined by:
๐ท๐๐ผ๐ ๐(๐ฅ) =
1
ฮ(1 โ ๐ผ)โซ
๐โฒ(๐ก)
(๐ฅ โ ๐ก)๐ผ
๐ฅ
๐
๐๐ก
26
Definition 1.5.2. [3, 4, 5, 11, 19,22] let ๐ be ๐-times differentiable function, ๐ฅ, ๐ โ
โ , ๐ฅ > ๐ and ๐ผ โ (๐, ๐ โ 1). Then the caputo fractional differential operator of ๐ผ is
defined as:
๐ท๐๐ผ๐(๐ฅ)๐ =
1
ฮ(๐ โ ๐ผ)โซ
๐(๐)(๐ก)
(๐ฅ โ ๐ก)๐ผโ๐+1
๐ฅ
๐
๐๐ก
Remark 1.5.1.
Because of similarity between (R-L) and Caputo fractional integration, the symbol
๐ท๐โ๐ผ๐(๐ฅ) will be indicated to (R-L) and Caputo fractional integral.
Remark 1.5.2
The symbol ๐ท๐ ๐๐ผ๐(๐ฅ)is used to denote Caputo fractional derivative of order ๐ผ
with lower limit ๐ and the symbol ๐ท๐ผ๐(๐ฅ)๐ is used to denote caputo fractional
derivative of order ๐ผ with lower limit 0.
Preposition 1.5.1. [11, 19,22] let ๐(๐ฅ), ๐(๐ฅ) be two functions such that both
๐ท๐ ๐๐ผ๐(๐ฅ), ๐ท๐
๐ผ๐ ๐(๐ฅ) exist for ๐ผ โ [0,1) and let ๐, ๐ โ โ.
Then
๐ท๐๐ผ(๐๐(๐ฅ) + ๐๐(๐ฅ)) = ๐ ๐ท๐
๐ผ๐ ๐(๐ฅ) +๐ ๐ ๐ท๐๐ผ๐ ๐(๐ฅ) (1.87)
Proof: using the definition of Caputo fractional ๐ผ derivative
๐ท๐๐ผ๐ (๐๐(๐ฅ) + ๐๐(๐ฅ)) =
1
ฮ(1 โ ๐ผ)โซ(๐๐(๐ก) + ๐๐(๐ก))โฒ
(๐ฅ โ ๐ก)๐ผ
๐ฅ
๐
๐๐ก
27
=1
ฮ(1 โ ๐ผ)[๐โซ
๐โฒ(๐ฅ)
(๐ฅ โ ๐ก)๐ผ
๐ฅ
๐
๐๐ก + ๐โซ๐โฒ(๐ฅ)
(๐ฅ โ ๐ก)๐ผ๐๐ก
๐ฅ
๐
]
=1
ฮ(1โ๐ผ)๐ โซ
๐โฒ(๐ฅ)
(๐ฅโ๐ก)๐ผ
๐ฅ
๐๐๐ก +
1
ฮ(1โ๐ผ)๐ โซ
๐โฒ(๐ฅ)
(๐ฅโ๐ก)๐ผ๐๐ก
๐ฅ
๐
= ๐ ๐ท๐๐ผ๐ ๐(๐ฅ) + ๐ ๐ท๐
๐ผ๐ ๐(๐ฅ) โ
We can generalize the previous result for any ๐ผ โ [๐ โ 1, ๐)
The Relation between integration and differentiation of Caputo operator of order
๐ผ are given as shown:
The Caputo derivative of fractional integral is
๐ท๐๐ผ๐ (๐ท๐
โ๐ผ๐(๐ฅ)) = ๐(๐ฅ) (1.88)
The fractional integral of Caputo derivative is
๐ท๐โ๐ผ( ๐ท๐ ๐
๐ผ๐(๐ฅ)) = ๐(๐ฅ) โ โ(๐ฅ โ ๐)๐
๐!
๐โ1
๐=0
๐(๐)(๐) (1.89)
From (1.88) and (1.89) we have
๐ท๐๐ผ๐ (๐ท๐
โ๐ผ๐(๐ฅ)) โ ๐ท๐โ๐ผ( ๐ท๐ ๐
๐ผ๐(๐ฅ)) (1.90)
Generally, we can conclude:
๐ท๐[๐ทโ(๐โ๐ผ)๐(๐ฅ)] โ ๐ทโ(๐โ๐ผ)[๐ท๐๐(๐ฅ)]
28
Thus
๐ท๐๐ผ๐(๐ฅ) โ ๐ท๐
๐ผ๐ ๐(๐ฅ) (1.91)
which implies that the Caputo derivative is not equivalent with (Riemann-
Liouville) derivative.
Preposition 1.5.2. [11, 19,22] let ๐ โ โ , ๐ผ โ [๐ โ 1, ๐). Let the function ๐(๐ฅ) be an n-
times differentiable function. Then the representation of the Caputo ๐ผ derivative:
๐ท๐๐ผ๐ ๐(๐ฅ) = ๐ท๐
โ(๐โ๐ผ) ๐ท๐๐๐ ๐(๐ฅ) (1.92)
where ๐ท๐โ๐ผ๐(๐ฅ) =
1
ฮ(๐ผ)โซ
๐(๐ก)
(๐ฅโ๐ก)1โ๐ผ
๐ผ
๐๐๐ก
is the Riemann-Liouville ๐ผ integral
Theorem1.5.1. [11, 19,22] (Relation between Caputo ๐ผ derivative and Riemann-
Liouville ๐ผ derivative).
Let ๐ โ โ,๐ผ โ [๐ โ 1, ๐). And let ๐(๐ฅ) be a function such that ๐ท๐๐ผ๐ ๐(๐ฅ) and
๐ท๐๐ผ๐(๐ฅ) exist. Then the relation between the (R-L) and the Caputo derivatives is given
by:
๐ท๐๐ผ๐ ๐(๐ฅ) = ๐ท๐
๐ผ๐(๐ฅ) โโ(๐ฅ โ ๐)๐โ๐ผ
ฮ(๐ + 1 โ ๐ผ)
๐โ1
๐=0
๐(๐)(๐) (1.93)
Proof: The well-known Taylor Series expansion of ๐ about ๐ฅ = 0 is
29
๐(๐ฅ) = ๐(0) + ๐ฅ๐โฒ(0) +๐ฅ2
๐ฅ!๐โฒโฒ(0) +โฏ+
๐ฅ๐โ1
(๐ โ 1)!๐(๐โ1)(0) + ๐ ๐โ1
=โ๐ฅ๐
ฮ(๐ + 1)
๐โ1
๐=0
๐(๐)(0) + ๐ ๐โ1
(1.94)
where, considering the following
๐ทโ๐๐(๐ก) = โซโซ โฆ
๐ก1
๐
๐ก
๐
โซ ๐(๐)
๐ก๐โ1
๐
๐๐โฆ๐๐2 ๐๐1
=1
(๐ โ 1)!โซ๐(๐)
๐ก
๐
(๐ก โ ๐)(๐โ1)๐๐
(1.95)
The previous formula is called cauchyโs formula for repeated integration.
๐ ๐โ1 = โซ๐(๐)(๐ก)(๐ฅ โ ๐ก)๐โ1
(๐ โ 1)!
๐ฅ
0
๐๐ก =1
ฮ(๐)โซ๐(๐)(๐ก)
๐ฅ
0
(๐ฅ โ 1)๐โ1๐๐ก (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.
๐ท๐๐ผ๐(๐ก) = ๐ท๐
๐ผ [โ๐ฅ๐
ฮ(๐ + 1)๐(๐)(0) + ๐ ๐โ1
๐โ1
๐=0
] = โ๐ท๐๐ผ
๐โ1
๐=0
๐ฅ๐
ฮ(๐ + 1)๐(๐)(0) + ๐ท๐
๐ผ๐ ๐โ1
=โฮ(๐ + 1)
ฮ(๐ โ ๐ผ + 1)
๐โ1
๐=0
๐ฅ๐โ๐ผ
ฮ(๐ + 1)๐(๐)(0) + ๐ท๐
๐ผ๐ทโ๐๐(๐)(๐ฅ)
30
=โ๐ฅ๐โ๐ผ
ฮ(๐ โ ๐ผ + 1)
๐โ1
๐=0
๐(๐)(0) + ๐ทโ(๐โ๐ผ)๐(๐)(๐ฅ)
= โ๐ฅ๐โ๐ผ
ฮ(๐ โ ๐ผ + 1)
๐โ1
๐=0
๐(๐)(0) + ๐ท๐๐ผ๐ ๐(๐ฅ)
โด ๐ท๐๐ผ๐ ๐(๐ฅ) = ๐ท๐
๐ผ๐(๐ก) โโ๐ฅ๐โ๐ผ
ฮ(๐ โ ๐ผ + 1)
๐โ1
๐=0
๐(๐)(0)
Preposition 1.5.3. [11, 19, 22] Let ๐ผ โ [0,1], let ๐(๐ฅ) be a function with second
continuous bounded derivative in [๐, ๐] for every ๐ > ๐ and ๐ท๐๐ผ๐ ๐(๐ฅ) exist, then:
1) lim๐โ1
๐ท๐๐ผ๐ ๐(๐ฅ) = ๐โฒ(๐ฅ) (1.97)
2) lim๐โ0
๐ท๐ ๐๐ผ๐(๐ฅ) = ๐(๐ฅ) โ ๐(๐) (1.98)
To proof see [11].
We can generalize the above equations in preposition 1.5.3 for any positive ๐ผ to be:
lim๐โ๐
๐ท๐๐ผ๐ ๐(๐ฅ) = ๐(๐)(๐ฅ) (1.99)
and lim๐โ๐โ1
๐ท๐ ๐๐ผ๐(๐ฅ) = ๐(๐โ1)(๐ฅ) โ ๐(๐โ1)(0) (1.100)
where ๐ผ โ [๐ โ 1, ๐), ๐ โ โ 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) ๐ท๐๐ผ(๐โ)๐ = ๐ ๐ท๐
๐ผ๐ (โ) + โ ๐ท๐๐ผ๐ (๐)
2) ๐ท๐๐ผ๐
โ๐ =
โ ๐ท๐๐ผ๐ (๐) + ๐ ๐ท๐
๐ผ๐ (โ)
โ2
3) ๐ท๐๐ผ(๐ โ โ)๐ = ๐(๐ผ)(โ(๐ฅ))โ(๐ผ)(๐ฅ)
where ๐(๐ผ)(๐ฅ), โ(๐ผ)(๐ฅ) are the Caputo ๐ผ derivative.
Now, I will give counter example to show that the above rule does not satisfy for
Caputo Operator considers that:
๐ท๐ผ๐ (๐ก) =1
ฮ(2 โ ๐ผ)๐ก1โ๐ผ โด ๐ท
13๐ (๐ก) = 1.1077๐ก2 3โ
๐ท๐ผ๐ (๐ก2) =2
ฮ(3 โ ๐ผ)๐ก2โ๐ผ โด ๐ท
13๐ (๐ก2) = 1.3293๐ก5 3โ
๐ท๐ผ๐ (๐ก3) =6
ฮ(4 โ ๐ผ)๐ก3โ๐ผ โด ๐ท
13๐ (๐ก3) = 1.4954๐ก8 3โ
Let ๐(๐ฅ) = ๐ก , ๐(๐ฅ) = ๐ก2 , โ(๐ก) = ๐ก3
๐ท13๐ (๐๐) = ๐ท
13๐ (๐ก3) = 1.4954๐ก8 3โ
๐ ๐ท13๐ (๐) + ๐ ๐ท
13๐ (๐) = ๐ก(1.3293๐ก5 3โ ) + ๐ก2(1.1077๐ก2 3โ )
= 1.3293๐ก8 3โ + 1.1077๐ก8 3โ = 2.4370 ๐ก8 3โ
Obviously
๐ท13(๐๐)๐ โ ๐ ๐ท
13๐ (๐) + ๐ ๐ท
13๐ (๐)
Also
๐ท13๐ (โ(๐ก)
๐(๐ก)) = ๐ท
13๐ (๐ก3
๐ก) = ๐ท
13๐ (๐ก2) = 1.3293๐ก5 3โ
But
32
๐(๐ก) ๐ท13๐ โ(๐ก) โ โ(๐ก) ๐ท
13๐ ๐(๐ก)
๐2(๐ก)=๐ก(1.4954๐ก5 3โ ) โ ๐ก3(1.1077๐ก2 3โ )
๐ก2
= 1.4954๐ก2 3โ โ 1.1077๐ก5 3โ
Thus
๐ท13๐ (โ
๐) โ
๐ ๐ท13๐ (โ) โ โ ๐ท
13๐ (๐)
โ2
It is easy to show that the composition Rule does not satisfy.
Preposition 1.5.5. [11, 19, 22] suppose that ๐ผ โ [๐ โ 1, ๐),๐, ๐ โ โ, and ๐ท๐๐ผ๐(๐ฅ)๐
exist. Then
๐ท๐๐ผ๐ ๐ท๐๐ ๐(๐ฅ) = ๐ท๐ผ+๐๐ ๐(๐ฅ) โ ๐ท๐๐ ๐ท๐
๐ผ๐ ๐(๐ฅ) (1.101)
Now we give counter example to show that the Caputo derivative is not commute
Example:
๐ท๐๐ผ๐ ๐ฅ๐ = {
ฮ(๐ + 1)
ฮ(๐ โ ๐ผ + 1)๐ฅ๐โ๐ผ ๐๐ ๐ โ 1 โค ๐ผ < ๐ , ๐ > ๐ โ 1, ๐ โ โ
0 ๐๐ ๐ โ 1 โค ๐ผ < ๐ , ๐ โค ๐ โ 1, ๐, ๐ โ โ
Now if we take ๐ผ =1
2 , ๐ = 3 , ๐ = 2. Then
๐ท1 2โ ๐ท3๐๐ [๐ฅ2] = 0
But ๐ท3๐ ๐ท1 2โ๐ [๐ฅ2] = ๐ท3๐ [ฮ(3)
ฮ(5
2)๐ฅ3
2]
= โ38 ฮ(3)ฮ(52)๐ฅโ
32
33
Corollary 1.5.1. let ๐ โ โ,๐ผ โ [๐ โ 1, ๐), ๐ = ๐ผ โ (๐ โ 1). Let ๐(๐ฅ) be a function
such that ๐ท๐๐ผ๐(๐ฅ)๐ exist, then
๐ท๐๐ผ๐ ๐(๐ฅ) = ๐ท๐๐ ๐ท๐โ1๐ ๐(๐ฅ) (1.102)
Theorem 1.5.2. [11, 19,22] (Some basic rules of Caputo fractional derivative):
Let ๐ผ โ [๐ โ 1, ๐), ๐ โ โ,
1) ๐ท๐๐ผ๐ ๐ = 0 , ๐ is constant (1.103)
2) ๐ท๐๐ผ๐ ๐ฅ๐ = {
ฮ(๐ + 1)
ฮ(๐ โ ๐ผ + 1)๐ฅ๐โ๐ผ ๐๐ ๐ > ๐ โ 1, ๐ฅ > 0, ๐ โ โ
0 ๐๐ ๐ โค ๐ โ 1, ๐ฅ > 0, ๐ โ โ
(1.104)
3) ๐ท๐๐ผ๐ ๐๐๐ฅ = ๐๐๐ฅ๐โ๐ผ๐ธ1,๐โ๐ผ+1(๐๐ฅ) (1.105)
4) ๐ท๐๐ผ๐ sin(๐๐ฅ) = โ
1
2๐(๐๐)๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐๐ฅ) โ (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐๐ฅ)] (1.106)
5) ๐ท๐๐ผ๐ cos(๐๐ฅ) =
1
2(๐๐)๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐๐ฅ) + (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐๐ฅ)] (1.107)
6) ๐ท๐๐ผ๐ cos(๐๐ฅ) =
1
2(๐๐)๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐๐ฅ) + (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐๐ฅ)] (1.108)
7) ๐ท๐๐ผ๐ sinh(๐๐ฅ) = โ
1
2๐๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐ฅ) โ (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐ฅ)] (1.109)
8) ๐ท๐๐ผ๐ cosh(๐๐ฅ) =
1
2๐๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐ฅ) + (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐ฅ)] (1.110)
Proof:
34
1) By applying the Caputo definition and because of the nโth derivative ๐(๐), (๐ โ
โ , ๐ โฅ 1) of constant equals 0, then
๐ท๐๐ผ๐ ๐ =
1
ฮ(๐ โ ๐ผ)โซ
๐(๐)
(๐ฅ โ ๐)๐ผโ๐+1
๐ฅ
๐
๐๐ก = 0
2) The second case has an easy proof
( ๐ท๐๐ผ๐ ๐ก๐ = 0, ๐ผ โ (๐ โ 1, ๐), ๐ โค ๐ โ 1, ๐ โ โ)
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 ฮฑ โ (๐ โ 1, ๐), ๐ > ๐ โ 1, ๐ โ โ
๐ท๐๐ผ๐ฅ๐ =
1
ฮ(๐ โ ๐ผ)๐ โซ
๐ท๐๐ก๐
(๐ฅ โ ๐ก)๐ผ+1โ๐
๐ฅ
0
๐๐ก
=1
ฮ(๐ โ ๐ผ)โซ
ฮ(๐ + 1) ๐ก๐โ๐
(๐ฅ โ ๐ก)๐ผ+1โ๐ฮ(๐ โ ๐ + 1)
๐ฅ
0
๐๐ก
Now by plugging t= ๐ฅ๐ข ; 0 โค ๐ข โค 1
=ฮ(๐ + 1)
ฮ(๐ โ ๐ผ)ฮ(๐ โ ๐ + 1)โซ(๐ฅ๐ข)๐โ๐((1 โ ๐ข)๐ฅ)๐โ๐ผโ11
0
๐ฅ๐๐ข
=ฮ(๐ + 1)
ฮ(๐ โ ๐ผ)ฮ(๐ โ ๐ + 1)๐ฅ๐โ๐๐ฝ(๐ โ ๐ + 1, ๐ โ ๐ผ)
=ฮ(๐ + 1)
ฮ(๐ โ ๐ผ)ฮ(๐ โ ๐ + 1)๐ฅ๐โ๐
ฮ(๐ โ ๐ + 1)ฮ(๐ โ ๐ผ)
ฮ(๐ โ ๐ผ + 1)
=ฮ(๐ + 1)
ฮ(๐ โ ๐ผ + 1)๐ฅ๐โ๐
35
Secondly, we can prove it by the relation between the Caputo and
Riemann-Liouville derivatives:
๐ท๐๐ผ๐ ๐ฅ๐ = ๐ท๐ผ๐ฅ๐ โโ
๐ฅ๐โ๐ผ
ฮ(๐ + 1 โ ๐ผ)
๐โ1
๐=0
๐ท๐[(๐ฅ)๐]๐ฅ=0
Now, the ๐ท๐[(๐ฅ)๐]๐ฅ=0 = 0, for ๐ โค ๐ โ 1 โค ๐
Then ๐ท๐๐ผ๐ ๐ฅ๐ =
ฮ(๐+1)
ฮ(ฮผโฮฑ+1)๐ฅ๐โ๐ผ โ
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 ๐ผ โderivative in (1.80), then we have:
๐ท๐๐ผ๐๐๐ฅ = ๐ท๐ผ๐๐๐ฅ โโ
๐ฅ๐โ๐ผ
ฮ(๐ + 1 โ ๐ผ)
๐โ1
๐=0
๐ท๐๐๐๐ฅ|
๐ฅ=0
= ๐ฅโ๐ผ๐ธ1,1โ๐ผ(๐๐ฅ) โโ๐ฅ๐โ๐ผ๐๐
ฮ(๐ + 1 โ ๐ผ)
๐โ1
๐=0
= โ(๐๐ฅ)๐๐ฅโ๐ผ
ฮ(k + 1 โ ฮฑ)
โ
๐=0
โโ๐ฅ๐โ๐ผ๐๐
ฮ(๐ + 1 โ ๐ผ)
๐โ1
๐=0
= โ๐ฅ๐โ๐ผ๐๐
ฮ(๐ + 1 โ ๐ผ)
โ
๐=๐
=โ๐ฅ๐+๐โ๐ผ๐๐+๐
ฮ(๐ + ๐ โ ๐ผ + 1)
โ
๐=0
= ๐๐๐ฅ๐โ๐ผ๐ธ1,๐โ๐ผ+1(๐๐ฅ) โ
4) Use (sin ๐ฅ =๐๐๐ฅโ๐โ๐๐ฅ
2๐), then by using (1.105)
๐ท๐๐ผ๐ sin(๐๐ฅ) = ๐ท๐
๐ผ๐๐๐ฅ โ ๐โ๐๐ฅ
2๐๐ =
1
2๐( ๐ท๐
๐ผ๐ ๐๐๐๐ฅ โ ๐ท๐๐ผ๐ ๐โ๐๐๐ฅ)
36
=1
2๐[(๐๐)๐๐ฅ๐โ๐ผ๐ธ1,๐โ๐ผ+1(๐๐๐ฅ) โ (โ๐๐)
๐๐ฅ๐โ๐ผ๐ธ1,๐โ๐ผ+1(โ๐๐๐ฅ)]
= โ1
2๐(๐๐)๐๐ฅ๐โ๐ผ[๐ธ1,๐โ๐ผ+1(๐๐๐ฅ) โ (โ1)
๐๐ธ1,๐โ๐ผ+1(โ๐๐๐ฅ)] โ
5) Follows by using (cos๐ฅ =๐๐๐ฅ+๐โ๐๐ฅ
2) and the same as (4).
6) Follows by using (sinh ๐ฅ =๐๐ฅโ๐โ๐ฅ
2) and the same as (4).
7) Follows by using (cosh๐ฅ =๐๐ฅ+๐โ๐ฅ
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 ๐(๐) = ๐โฒ(๐) = โฏ = ๐(๐โ1)(๐) = 0, then
๐ท๐๐ผ๐(๐ฅ) = ๐ท๐
๐ผ๐ ๐(๐ฅ)
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
๐ท๐๐ผ๐ ( ๐ท๐
๐๐ ๐(๐ฅ)) = ๐ท๐๐ผ+๐๐ ๐(๐ฅ) โ ๐ท๐
๐ ( ๐ท๐๐ผ๐ ๐(๐ฅ)) ,๐ (1.111)
where ๐ผ โ (๐ โ 1, ๐), ๐ โ โ,๐ = 1,2,โฆ
While for Riemann-Liouville derivative
๐ท๐๐(๐ท๐
๐ผ๐(๐ฅ)) = ๐ท๐๐ผ+๐๐(๐ฅ) โ ๐ท๐
๐ผ(๐ท๐๐๐(๐ฅ)) , (1.112)
where ๐ผ โ (๐ โ 1, ๐), ๐ โ โ,๐ = 1,2,โฆ
Note that the formulas as in (1.111) and (1.112) become equalities under the following
additional conditions:
๐(๐ )(๐) = 0 , ๐ = ๐, ๐ + 1,โฆ ,๐ โ 1 for ๐ท๐ ๐ผ
๐(๐ )(๐) = 0 , ๐ = 0,1,2,โฆ ,๐ โ 1 for ๐ท๐ผ.
38
Table 1: Comparison between Riemann-Liouville and Caputo [11]
๐(๐ก )
=๐=
consta
nt
Non-c
om
muta
tion
Lin
earity
Inte
rpola
tion
Repre
senta
tion
Pro
perty
๐ท๐ผ๐=
๐
ฮ(1โ๐ผ)๐กโ๐ผโ 0 ,๐=๐๐๐๐ ๐ก
๐ท๐๐ท๐ผ๐(๐ก )
=๐ท๐ผ+๐๐(๐ก)
โ ๐ท๐ผ๐ท๐๐(๐ก)
๐ท๐ผ(๐๐(๐ก )+๐(๐ก) )
=๐๐ท๐ผ๐(๐ก )
+๐ท๐ผ๐(๐ก)
lim๐ผโ๐๐ท๐ผ๐(๐ก )
=๐(๐)(๐ก)
lim๐ผโ๐โ1๐ท๐ผ๐(๐ก)
=๐(๐โ1)(๐ก)
๐ท๐ผ๐(๐ก )
=๐ท๐ผ(๐ท
โ(๐โ๐ผ)๐(๐ฅ))
Rie
ma
nn
-Lio
uv
ille
๐ท๐ ๐ผ๐=0 ,๐=๐๐๐๐ ๐ก
๐ท๐ ๐ผ
๐๐ท๐๐(๐ก )
=๐ท๐ ๐ผ+๐๐(๐ก)
โ ๐ท๐๐ท๐ ๐ผ๐(๐ก)
๐ท๐ ๐ผ
๐(๐๐(๐ก )+๐(๐ก) )
=๐๐ท๐ ๐ผ๐(๐ก )
+๐ท๐ ๐ผ๐(๐ก)
lim๐ผโ๐๐ท๐ ๐ผ
๐๐(๐ก )
=๐(๐)(๐ก)
lim๐ผโ๐โ1๐ท๐ ๐ผ
๐๐(๐ก )
=๐(๐โ1)(๐ก )
โ๐(๐โ1)(0)
๐ท ๐๐ ๐ผ๐(๐ก )
=๐ทโ(๐โ๐ผ)๐ท๐ ๐
๐๐(๐ฅ)
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
๐ฆโฒ + ๐(๐ฅ)๐ฆ = ๐(๐ฅ)๐ฆ๐ (1.113)
where ๐(๐ฅ) and ๐(๐ฅ) both are continous, ๐ โ โ.
Differential equation above is called Bernoulli equation
Now we solve (1.113) by dividing both sides by ๐ฆ๐.
๐ฆโ๐๐ฆโฒ + ๐(๐ฅ)๐ฆ1โ๐ = ๐(๐ฅ) (1.114)
Let ๐ฃ = ๐ฆ1โ๐, then
๐ฃโฒ = (1 โ ๐)๐ฆโ๐๐ฆโฒ
Multiply (1.113) by (1 โ ๐)๐ฆโ๐ , we get:
1
1 โ ๐๐ฃโฒ + ๐(๐ฅ)๐ฃ = ๐(๐ฅ) (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
๐ด(๐ฅ)๐ฆโฒโฒ + ๐(๐ฅ)๐ฆ + ๐(๐ฅ) = ๐บ(๐ฅ( (1.116)
where A,P,Q and G are continuous functions, when ๐บ(๐ฅ) = 0, for all ๐ฅ, in equation
(1.116). Such equations are called homogenous linear equations. Thus, the form of a
second-order linear homogenous differential equation is:
๐ด(๐ฅ)๐2๐ฆ
๐๐ฅ2+ ๐(๐ฅ)
๐๐ฆ
๐๐ฅ+ ๐(๐ฅ) = 0 (1.117)
( if ๐บ(๐ฅ) โ 0 for some ๐ฅ, 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 ๐ท๐๐ผ(1) = 0
(๐ท๐๐ผ(1) = 0 for the Caupto derivative) , if ฮฑ is not a natural number.
ii) All fractional derivatives do not satisfy the Known product rule:
๐ท๐๐ผ(๐๐) = ๐๐ท๐
๐ผ(๐) + ๐๐ท๐๐ผ(๐)
iii) All fractional derivatives do not satisfy the known quotient rule:
๐ท๐๐ผ(๐ ๐) =
๐๐ท๐๐ผ(๐)โ๐๐ท๐
๐ผ(๐)
๐2โ
iv) All fractional derivatives do not satisfy the chain rule:
๐ท๐๐ผ(๐ โ ๐)(๐ก) = ๐(๐ผ)(๐(๐ก))๐(๐ผ)(๐ก)
v) All fractional derivatives don't satisfy: ๐ท๐ผ๐ท๐ฝ๐ = ๐ท๐ผ+๐ฝ๐ in general
vi) The Caputo definition assumes that the function f is differentiable.
Let us write Tฮฑ to denote the operator which is called the "Conformable
fractional derivative of order ฮฑ ".
Khalil, et al. [14] introduced a completely new definition of fractional calculus which
is more natural and effective than previous definitions of order ๐ผ โ (0, 1]. Also, this
definition can be generalized to include any ฮฑ. However, the case ๐ผ โ (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 โถ ๐: [0,โ) โ โ . Then the (conformable fractional
derivative) of ๐ of order ๐ผ is defined by
๐๐ผ(๐)(๐ก) = ๐๐๐๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐
For all ๐ก ห0, ๐ผ โ (0.1), if ฦ is ฮฑ-differentiable in some (0, ๐ผ). ๐ผ ห 0 and, lim๐กโ0+ ๐(๐ผ)(๐ก)
exists, then define๐(๐ผ)(0) = lim๐กโ0+ ๐(๐ผ)(๐ก)
We sometimes, write ๐(๐ผ)(๐ก) for ๐๐ผ(๐ )(๐ก), to denote the conformable fractional
derivatives of ๐ of order ๐ผ. In addition, if the conformable fractional derivative of f
of order ฮฑ exists, then we say ๐ is ฮฑ-differentiable.
We should take into consideration that ๐๐ผ(๐ก๐) = ๐๐ก๐โ๐ผ. Further, this definition
coincides happen with the same of traditional definition of RiemannโLiouville and
of Caputo on polynomials (up to a constant multiple).
Theorem 2.1.1. [14] if a function ๐: [0,โ) โ โ is ๐ผ-differentiable at ๐ก0 ห 0. ๐ผ โ
(0.1] then ๐ is continuous at ๐ก0
Proof:
Because ๐(๐ผ)is differentiable at ๐ฅ = ๐ก0, we know that
๐(๐ผ)(๐ก0) = ๐๐๐๐โ0๐(๐ก0+๐๐ก0
1โ๐ผ)โ๐(๐ก0)
๐ exists.
If we next assume that ๐ฅ โ ๐ก0 we can write the following
๐(๐ก0 + ๐๐ก01โ๐ผ) โ ๐(๐ก0) =
๐(๐ก0 + ๐๐ก01โ๐ผ) โ ๐(๐ก0)
๐๐
43
Then some basic properties of limits give us
๐๐๐๐โ0( ๐(๐ก0 + ๐๐ก0
1โ๐ผ) โ ๐(๐ก0)) = ๐๐๐๐โ0
๐(๐ก0 + ๐๐ก01โ๐ผ) โ ๐(๐ก0)
๐. ๐๐๐๐โ0
๐
๐๐๐๐โ0( ๐(๐ก0 + ๐๐ก0
1โ๐ผ) โ ๐(๐ก0)) = ๐โฒ(๐ก0). 0
Let โ = ๐๐ก01โ๐ผ. Then,
๐๐๐โโ0 ๐(๐ก0 + โ) = ๐(๐ก0) . Hence, f is continuous at ๐ก0
It can be easily shown that ๐๐ผ satisfies all properties in the following theorem
Theorem 2.1.2. [14] Let ๐ผ โ (0.1] and ๐, ๐ be ฮฑ-differentiable at a point ๐ก ห 0 .Then:
(1) ๐๐ผ(๐)(๐ก) = ๐ก1โ๐ผ ๐๐
๐๐ก(๐ก), where f is differentiable
(2.1)
(2) ๐๐ผ(af + bg) = a ๐๐ผ (f ) + b ๐๐ผ (g), for all ๐, ๐ โ โ (2.2)
(3) ๐๐ผ (๐ก๐) = ๐ ๐ก๐โ๐ผ for all ๐ โ โ (2.3)
(4) ๐๐ผ (ฮป)=0 , for all constant functions ๐ (๐ก) = ๐ (2.4)
(5) ๐๐ผ (fg) = f ๐๐ผ (g) + g ๐๐ผ (f ) (2.5)
(6) ๐๐ผ ( ฦ
๐ ) =
๐ ๐๐ผฦ โ ฦ ๐๐ผ(๐)
๐2 (2.6)
Proof:
(1) Let โ = ๐๐ก1โ๐ผ in definition (2.1.1). Then ๐ = โ๐ก๐ผโ1
44
๐๐ผ(๐)(๐ก) = lim๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐= lim๐โ0
๐(๐ก + โ) โ ๐(๐ก)
โ๐ก๐ผโ1
= ๐ก1โ๐ผ limโโ0
๐(๐ก + โ) โ ๐(๐ก)
โ= ๐ก1โ๐ผ๐โฒ(๐ก) โ
(2) ๐๐ผ(๐๐ + ๐๐) = lim๐โ0(๐๐+๐๐)(๐ก+๐๐ก1โ๐ผ)โ(๐๐+๐๐)(๐ก)
๐
= lim๐โ0
๐๐(๐ก + ๐๐ก1โ๐ผ) + ๐๐(๐ก + ๐๐ก1โ๐ผ) โ ๐๐(๐ก) โ ๐๐(๐ก)
๐
= lim๐โ0
๐ (๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐) + lim
๐โ0๐ (๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐)
= ๐๐๐ผ(๐) + ๐๐๐ผ(๐) โ
(3) Recall (๐ + ๐)๐ = โ(๐
๐) ๐๐โ๐
๐
๐=0
๐๐
Thus,
(๐ก + ๐๐ก1โ๐ผ)๐ =โ(๐
๐) ๐ก๐โ๐
๐
๐=0
(๐๐ก1โ๐ผ)๐
(๐ก + ๐๐ก1โ๐ผ)๐ = (๐
0) ๐ก๐ + (
๐
1) ๐ก๐โ1(๐๐ก1โ๐ผ)1 +โฏ+ (
๐
๐) ๐ก0(๐๐ก1โ๐ผ)๐
To proof that ๐๐ผ(๐ก๐) = ๐๐ก๐โ๐ผ
๐๐๐๐โ0
(๐ก + ๐๐ก1โ๐ผ)๐ โ ๐ก๐
๐= ๐๐๐
๐โ0
๐ก๐ + (๐1)๐ก๐โ1(๐๐ก1โ๐ผ) + โฏ+ (๐
๐) ๐๐(๐ก1โ๐ผ)๐ โ ๐ก๐
๐
= ๐๐๐ ๐โ0
๐๐๐ก๐โ1๐ก1โ๐ผ +โฏ+ (๐
๐) ๐๐โ1(๐ก1โ๐ผ)๐
๐
= ๐๐ก๐โ1๐ก1โ๐ผ = ๐๐ก๐โ๐ผ
45
(4) ๐๐ผ(๐) = lim๐โ0๐(๐ก+๐๐ก1โ๐ผ)โ๐(๐ก)
๐
= lim๐โ0
๐ โ ๐
๐= 0 โ
(5)
= lim๐โ0
๐(๐ก + ๐๐ก1โ๐ผ)๐(๐ก + ๐๐ก1โ๐ผ) + ๐(๐ก)๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)๐(๐ก)
๐
= lim๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐๐(๐ก + ๐๐ก1โ๐ผ) + ๐(๐ก) lim
๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐
= ๐๐ผ(๐(๐ก)) lim๐โ0
๐(๐ก + ๐๐ก1โ๐ผ) + ๐(๐ก)๐๐ผ(๐(๐ก))
= ๐(๐ก)๐๐ผ(๐(๐ก)) + ๐(๐ก)๐๐ผ(๐(๐ก)) โ
(6)
= lim๐โ0
(๐(๐ก + ๐๐ก1โ๐ผ)
๐(๐ก + ๐๐ก1โ๐ผ)โ
๐(๐ก)
๐(๐ก + ๐๐ก1โ๐ผ)+
๐(๐ก)
๐(๐ก + ๐๐ก1โ๐ผ)โ๐(๐ก)
๐(๐ก)) .1
๐
= lim๐โ0
(๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐. ๐(๐ก + ๐๐ก1โ๐ผ)) + ๐(๐ก). lim
๐โ0(
1
๐๐(๐ก + ๐๐ก1โ๐ผ)โ
1
๐๐(๐ก))
= ๐๐ผ(๐) lim๐โ0
1
๐(๐ก + ๐๐ก1โ๐ผ)+ ๐(๐ก) lim
๐โ0(โ(
๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐๐(๐ก)๐(๐ก + ๐๐ก1โ๐ผ)))
= ๐๐ผ(๐)1
๐(๐ก)โ ๐(๐ก) lim
๐โ0(๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)
๐) lim๐โ0
(1
๐(๐ก)๐(๐ก + ๐๐ก1โ๐ผ))
๐๐ผ(๐๐) = lim๐โ0
๐(๐ก + ๐๐ก1โ๐ผ)๐(๐ก + ๐๐ก1โ๐ผ) โ ๐(๐ก)๐(๐ก)
๐
๐๐ผ (๐
๐) = lim
๐โ0
๐(๐ก + ๐๐ก1โ๐ผ)๐(๐ก + ๐๐ก1โ๐ผ)
โ๐(๐ก)๐(๐ก)
๐
46
=๐๐ผ(๐)
๐(๐ก)โ ๐(๐ก)๐๐ผ(๐(๐ก)).
1
๐2(๐ก)
=๐(๐ก)๐๐ผ(๐) โ ๐(๐ก)๐๐ผ(๐(๐ก))
๐2(๐ก) โ
Theorem 2.1.3. [14] (Conformable fractional derivative of Known functions)
1) ๐๐ผ(๐๐๐ก) = ๐๐ก1โ๐ผ๐๐๐ก (2.7)
2) ๐๐ผ(๐ ๐๐ (๐๐ก)) = ๐๐ก1โ๐ผ ๐๐๐ (๐๐ก) , ๐ โ โ (2.8)
3) ๐๐ผ (๐๐๐ (๐๐ก)) = โ๐๐ก1โ๐ผ ๐ ๐๐ (๐๐ก), ๐ โ โ (2.9)
4) ๐๐ผ(tan(๐๐ก)) = ๐๐ก1โ๐ผ๐ ๐๐2(๐๐ก) , ๐ โ โ (2.10)
5) ๐๐ผ(๐๐๐ก(๐๐ก)) = โ๐๐ก1โ๐ผ๐๐ ๐2(๐๐ก) , ๐ โ โ (2.11)
6) ๐๐ผ(๐ ๐๐(๐๐ก)) = ๐๐ก1โ๐ผ ๐ ๐๐(๐๐ก) ๐ก๐๐(๐๐ก) , ๐ โ โ (2.12)
7) ๐๐ผ(๐๐ ๐(๐๐ก)) = โ๐๐ก1โ๐ผ ๐๐ ๐(๐๐ก) ๐๐๐ก(๐๐ก) , ๐ โ โ (2.13)
8) ๐๐ผ (1
๐ผ๐ก๐ผ) = 1 (2.14)
Proof:
1. ๐๐ผ(๐๐๐ฅ) = ๐๐๐๐โ0
๐๐(๐ก+๐๐ก1โ๐ผ)โ๐๐๐ก
๐= ๐๐๐ก ๐๐๐๐โ0
๐๐๐๐ก1โ๐ผ
โ1
๐
= ๐๐๐ก ๐๐๐๐โ0
๐ก1โ๐ผ๐๐๐๐ก1โ๐ผ
โ ๐ก1โ๐ผ
๐๐ก1โ๐ผ= ๐๐๐ก๐ก1โ๐ผ ๐๐๐
๐โ0
๐๐๐๐ก1โ๐ผ
โ 1
๐๐ก1โ๐ผ
Let โ = ๐๐ก1โ๐ผ . Then by using LโHopitalโs rule, we get
47
= ๐ก1โ๐ผ๐๐๐ก ๐๐๐โโ0
๐๐โ โ 1
โ= ๐๐ก1โ๐ผ๐๐๐ก ๐๐๐
โโ0
๐๐โ
1
= ๐๐ก1โ๐ผ๐๐๐ก โ
(By LโHopital Rule)
2. ๐๐ผ(๐ ๐๐(๐๐ก)) = ๐๐๐๐โ0
๐ ๐๐ ๐(๐ก + ๐๐ก1โ๐ผ) โ ๐ ๐๐(๐๐ก)
๐
= ๐๐๐๐โ0
๐ ๐๐(๐๐ก) [๐๐๐ (๐๐๐ก1โ๐ผ) โ 1
๐] + ๐๐๐
๐โ0
๐๐๐ (๐๐ก) ๐ ๐๐(๐๐๐ก1โ๐ผ)
๐
= ๐ก1โ๐ผ ๐ ๐๐(๐๐ก) ๐๐๐๐โ0
[๐๐๐ (๐๐๐ก1โ๐ผ) โ 1
๐๐ก1โ๐ผ]
+ ๐ก1โ๐ผ ๐๐๐ (๐๐ก) ๐๐๐๐โ0
๐ ๐๐(๐๐๐ก1โ๐ผ)
๐๐ก1โ๐ผ
Let โ = ๐๐ก1โ๐ผ then we get
= ๐ก1โ๐ผ ๐ ๐๐(๐๐ก) ๐๐๐โโ0
[๐๐๐ (๐โ) โ 1
โ] + ๐ก1โ๐ผ ๐๐๐ (๐๐ก) ๐๐๐
โโ0
๐ ๐๐(๐โ)
โ
By using LโHoputal Rule, we get
= ๐ก1โ๐ผsin (๐๐ก) ๐๐๐โโ0
โ๐ ๐ ๐๐(๐โ)
1+ ๐ก1โ๐ผ ๐๐๐ (๐๐ก) . ๐
= ๐๐ก1โ๐ผ ๐๐๐ (๐๐ก) โ
3. Similar to (2)
= ๐๐๐๐โ0
๐ ๐๐(๐๐ก) ๐๐๐ (๐๐๐ก1โ๐ผ) + ๐๐๐ (๐๐ก) ๐ ๐๐(๐๐๐ก1โ๐ผ) โ ๐ ๐๐(๐๐ก)
๐
48
4. ๐๐ผ(๐ก๐๐(๐๐ก)) = ๐๐ผ (๐ ๐๐(๐๐ก)
๐๐๐ (๐๐ก))
=๐๐๐ (๐๐ก) ๐๐ผ(๐ ๐๐(๐๐ก)) โ ๐ ๐๐(๐๐ก) ๐๐ผ(๐๐๐ ๐๐ก)
๐๐๐ 2(๐๐ก)
=๐๐๐ (๐๐ก) (๐๐ก1โ๐ผ ๐๐๐ (๐๐ก)) โ ๐ ๐๐(๐๐ก)(โ๐๐ก1โ๐ผ ๐ ๐๐(๐๐ก))
๐๐๐ 2(๐๐ก)
=๐๐ก1โ๐ผ ๐๐๐ 2(๐๐ก) + ๐๐ก1โ๐ผ ๐ ๐๐2(๐๐ก)
๐๐๐ 2(๐๐ก)
= ๐๐ก1โ๐ผ(1 + ๐ก๐๐2(๐๐ก))
= ๐๐ก1โ๐ผ ๐ ๐๐2(๐๐ก) โ
5. Similar to (4)
6. ๐๐ผ(๐ ๐๐(๐๐ก)) = ๐๐ผ (1
๐๐๐ (๐๐ก)) =
(โ1)(๐๐ผ(๐๐๐ (๐๐ก)))
๐๐๐ 2(๐๐ก)
=(โ1)(โ๐๐ก1โ๐ผ ๐ ๐๐(๐๐ก))
๐๐๐ 2(๐๐ก)= ๐๐ก1โ๐ผ
๐ ๐๐(๐๐ก)
๐๐๐ (๐๐ก).
1
๐๐๐ (๐๐ก)
= ๐๐ก1โ๐ผ ๐ก๐๐(๐๐ก) ๐ ๐๐(๐๐ก) โ
7. Similar to (7)
8. ๐๐ผ (1
๐ผ๐ก๐ผ) = ๐๐๐
๐โ0
1๐ผ(๐ก + ๐๐ก1โ๐ผ)๐ผ โ
1๐ผ ๐ก
๐ผ
๐
=1
๐ผ๐๐๐๐โ0
(๐ก + ๐๐ก1โ๐ผ)๐ผ โ ๐ก๐ผ
๐
=1
๐ผ๐๐๐๐โ0
๐ก๐ผ + (๐ผ1) ๐ก๐ผโ1๐๐ก1โ๐ผ +โฏ+ (
๐ผ๐ผ โ 1
)๐๐ผโ1๐ก๐ผโ1 + (๐ผ๐ผ)๐๐ผ(๐ก1โ๐ผ)๐ผ โ ๐ก๐ผ
๐
49
=1
๐ผ๐๐๐๐โ0
๐ ((๐ผ1) + (
๐ผ๐ผ โ 1
) ๐ก๐ผ๐๐ผโ2 +โฏ+ (๐ผ๐ผ)๐ก๐ผโ1(๐ก1โ๐ผ)๐ผ)
๐
=1
๐ผ. ๐ผ = 1 โ
Corollary 2.1.1. (Conformable fractional derivative of certain functions)
i) ๐๐ผ (๐ ๐๐1
๐ผ๐ก๐ผ) = ๐๐๐
1
๐ผ๐ก๐ผ (2.15)
ii) ๐๐ผ (๐ ๐๐1
๐ผ๐ก๐ผ) = ๐๐๐
1
๐ผ๐ก๐ผ (2.16)
iii) ๐๐ผ (๐1๐ผ๐ก๐ผ) = ๐
1๐ผ๐ก๐ผ
(2.17)
Note: The function could be ฮฑ-differentiable at a point but not differentiable. For
example, let ๐(๐ก) = 2โ๐ก.
Then, ๐12
(๐)(0) = ๐๐๐๐กโ0+ ๐12
(๐)(๐ก) = 1 , when ๐12
(๐)(๐ก) = 1 , for all t>0 , but
๐1(๐)(0) does not exist.
The most important case for the range of ๐ผ โ (0,1), when ๐ผ โ (๐, ๐ + 1] the
definition would be as the following
Definition 2.1.2. [14] Let ๐ผ โ (๐, ๐ + 1], and f be an n-differentiable at t ,
where t > 0, then the conformable fractional derivative of f of order ฮฑ is defined as:
๐๐ผ(๐)(๐ก) = limฮตโ0
ฦ(โฮฑโโ1) ( ๐ก + ๐๐ก(โฮฑโโฮฑ)) โ ฦ(โฮฑโโ1)(t)
๐
where [ฮฑ] is the smallest integer greater than or equal to ฮฑ.
50
Remark 2.1.1. Let ๐ผ โ (๐, ๐ + 1], and f is (๐ + 1)-differentiable at ๐ก > 0. Then:
๐๐ผ(๐)(๐ก) = ๐ก(โฮฑโโฮฑ) ๐
โฮฑโ(๐ก) (2.18)
Theorem 2.1.4 [14]
(Rolleโs Theorem for Conformable Fractional Differentiable Functions).
Let ๐ > 0 and ๐ โถ [๐, ๐] โ โ be a given function that satisfies
i. ๐ is continuous on [๐, ๐],
ii. ๐ is ฮฑ-differentiable for some ๐ผ โ (0,1),
iii. ๐(๐) = ๐(๐).
Then, there exists ๐ โ (๐, ๐), such that ๐(๐ผ)(๐) = 0.
Proof:
Since ๐ is continuous on [๐, ๐], and ๐(๐) = ๐(๐), there is ๐ โ (๐, ๐), which is
a point of local extrema. With no loss of generality, assume c is a point of local
minimum. So, ๐(๐ผ)(๐) = lim๐โ 0+๐(๐+๐๐1โ๐ผ)โ๐(๐)
๐= lim๐โ 0โ
๐(๐+๐๐1โ๐ผ)โ๐(๐)
๐, but the
first limit is non โ negative, and the second limit is non-positive. Hence, ๐(๐ผ)(๐) = 0.
Theorem 2.1.5. [14] (Mean Value Theorem for Conformable Fractional
Differentiable Functions). Let a > 0 and ๐ : [๐, ๐] โ โ be a given function that
satisfies:
51
i) ๐ is continuous on [๐, ๐].
ii) ๐ is ฮฑ-differentiable for some ๐ผ โ (0, 1).
Then, there exists ๐ โ (๐, ๐), such that
Proof:
The equation of the secant through (๐, ๐(๐)) and (๐, ๐(๐)) is
๐ฆ โ ๐(๐) =๐(๐) โ ๐(๐)
1๐ผ ๐
๐ผ โ1๐ผ ๐
๐ผ(1
๐ผ๐ฅ๐ผ โ
1
๐ผ๐๐ผ)
which we can write as
๐ฆ =๐(๐) โ ๐(๐)
1๐ผ ๐
๐ผ โ1๐ผ ๐
๐ผ(1
๐ผ๐ฅ๐ผ โ
1
๐ผ๐๐ผ) + ๐(๐)
Let ๐(๐ฅ) = ๐(๐ฅ) โ [๐(๐)โ๐(๐)1
๐ผ๐๐ผโ
1
๐ผ๐๐ผ(1
๐ผ๐ฅ๐ผ โ
1
๐ผ๐๐ผ) + ๐(๐)].
Note that ๐(๐) = ๐(๐) = 0 , ๐ is continuous on [๐, ๐] and differentiable on (๐, ๐). So
by Rollโs theorem there are ๐ in (๐, ๐) such that ๐(๐ผ)(๐) = 0.
But
๐(๐ผ)(๐) =ฦ(๐) โ ฦ(๐)
1๐ผ ๐
๐ผ โ 1๐ผ ๐
๐ผ
๐(๐ผ)(๐ฅ) = ๐(๐ผ)(๐ฅ) โ [๐(๐) โ ๐(๐)
1๐ผ ๐
๐ผ โ1๐ผ ๐
๐ผ]
52
So
๐(๐ผ)(๐) =๐(๐) โ ๐(๐)
1๐ผ ๐
๐ผ โ1๐ผ ๐
๐ผ โ
2.2. Conformable Fractional Integrals
Suppose that the function is continuous
Let ๐ผ โ (0,โ). Define ๐ฝ๐ผ(๐ก๐) =
๐ก๐+๐ผ
๐+๐ผ , for any ๐ โ ๐ , ๐ผ โ โ๐.
If ๐(๐ก) = โ ๐๐๐ก๐๐
๐=0 , then we define ๐ฝ๐ผ(๐) = โ ๐๐๐ฝ๐ผ(๐ก๐)๐
๐=0 = โ ๐๐๐ก๐+๐ผ
๐+๐ผ
๐๐=0
Cleary, ๐ฝ๐ผis linear in its domain. Further, if ๐ผ = 1, then ๐ฝ๐ผ the usual integral.
Now according to conformable fractional definition, if ๐ผ = 1 2โ ,then
sin ๐ก = โ(โ1)๐
(2๐+1)!โ๐=0 ๐ก2๐+1 then ๐ฝ๐ผ(sin ๐ก) = โ
(โ1)๐๐ก2๐+
32
(2๐+3
2)(2๐+1)!
โ๐=0 .
Also, if ๐ผ =1
2
cos(๐ก) = โ(โ1)๐๐ก2๐
(2๐)!โ๐=0 then ๐ฝ๐ผ(cos(๐ก)) = โ
(โ1)๐๐ก2๐+
12
(2๐+1
2)(2๐)!
โ๐=0
๐๐ก = โ๐ก๐
๐!โ๐=0 then ๐ฝ๐ผ(๐
๐ก) = โ๐ก๐+
12
(๐+1
2)(๐)!
โ๐=0
๐(๐ผ)(๐) = ๐(๐ผ)(๐) โ [๐(๐) โ ๐(๐)
1๐ผ ๐
๐ผ โ1๐ผ ๐
๐ผ] = 0
53
sinh(๐ก) = โ๐ก2๐+1
(2๐+1)!โ๐=0 then ๐ฝ๐ผ(sinh(๐ก)) = โ
๐ก2๐+
32
(2๐+3
2)(2๐+1)!
โ๐=0
cosh (๐ก) = โ๐ก2๐
(2๐)!โ๐=0 then ๐ฝ๐ผ(cosh(๐ก)) = โ
๐ก2๐+
12
(2๐+1
2)(2๐)!
โ๐=0 .
Definition 2.2.1 [14]
Let f be a continuous function. Then ๐ผ-fractional integral of f is defined by:
๐ผ๐ผ๐๐(๐ก) = ๐ผ1
๐(๐ก๐ผโ1๐(๐ก)) = โซ๐(๐ฅ)
๐ฅ1โ๐ผ๐๐ฅ
๐ก
๐
(2.19)
where ๐ > 0,๐ผ โ (0,1) and the integral is the usual Riemann improper integral.
Examples:
1) ๐ผ12
0(โ๐ก cos(๐ก)) = โซcos(๐ฅ) . ๐๐ฅ
๐ก
0
= sin(๐ก)
2) ๐ผ12
0(cos(2โ๐ก)) = โซcos(2โ๐ฅ)
โ๐ฅ. ๐๐ฅ
๐ก
0
= sin(2โ๐ก)
Theorem 2.2.1 [14]
Let ๐ be any continuous function in the domain of ๐ผ๐ผ. Then
(๐๐ผ๐ผ๐ผ๐(๐(๐ก)) = ๐(๐ก), for ๐ก โฅ ๐) . (2.20)
54
Proof: since f is continues, then ๐ผ๐ผ๐ (๐)(๐ก) is differentiable. So
๐๐ผ (๐ผ๐ผ๐(๐(๐ก))) = ๐ก1โ๐ผ
๐
๐๐ก๐ผ๐ผ๐๐(๐ก)
= ๐ก1โ๐ผ๐
๐๐กโซ๐(๐ฅ)
๐ฅ1โ๐ผ
๐ก
๐
= ๐ก1โ๐ผ๐(๐ก)
๐ก1โ๐ผ = ๐(๐ก) โ
2.3 Applications [14]:
Now in this section we will solve fractional differential equations according to
conformable definitions:
Example (2.3.1):
๐ฆ(12โ ) + ๐ฆ = ๐ฅ3 + 3๐ฅ5 2โ , ๐ฆ(0) = 0 (2.21)
To find
๐ฆโ of ๐ฆ1 2โ + ๐ฆ = 0
we use
๐ฆโ = ๐๐โ๐ฅ
Now
๐ฆ(1 2)โ + ๐ฆ = 0
55
๐
2๐๐โ๐ฅ + ๐๐โ๐ฅ = 0
๐๐โ๐ฅ (๐
2+ 1) = 0
๐
2+ 1 = 0
๐ = โ2
๐ฆโ = ๐โ2โ๐ฅ
And simply the particular solution is ๐ฆ๐ = ๐ฅ3
And by plugging the initial condition ๐ฆ๐ = ๐ฅ3 then A = 0
โด ๐ฆ = ๐ฆโ + ๐ฆ๐ = ๐โ2โ๐ฅ + ๐ฅ3
For more examples see [14].
2.4. Abelโs Formula and Wronskain for Conformable Fractional Differential
Equation
In this section we will discuss the differential equation
๐ฆโฒโฒ + ๐(๐ฅ)๐ฆโฒ +๐(๐ฅ)๐ฆ = 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 ๐ผ โ (0,1], Abu Hammad, et al. discussed the equation [7].
๐๐ผ๐๐ผ๐ฆ + ๐(๐ฅ)๐๐ผ๐ฆ + ๐(๐ฅ)๐ฆ = 0 (2.23)
They discussed also the fractional Wronskain of two functions.
Definition 2.4.1. [7] For two functions ๐ฆ1 and ๐ฆ2 satisfying (2.24) and ๐ผ โ (0,1] we set
๐๐ผ[๐ฆ1, ๐ฆ2] = |๐ฆ1 ๐ฆ2๐๐ผ๐ฆ1 ๐๐ผ๐ฆ2
|
Theorem 2.4.1. [7] assume that ๐ฆ1, ๐ฆ2 satisfy equation (2.23), Then
๐๐ผ[๐ฆ1, ๐ฆ2] = ๐โ๐ผ๐ผ(๐)
Proof: applying the operator ๐๐ผ on ๐๐ผ[๐ฆ1, ๐ฆ2] to get
๐๐ผ(๐๐ผ[๐ฆ1, ๐ฆ2]) = ๐๐ผ(๐ฆ1๐๐ผ๐ฆ2 โ ๐ฆ2๐๐ผ๐ฆ1)
= ๐๐ผ๐ฆ1๐๐ผ๐ฆ2 + ๐ฆ1๐๐ผ๐๐ผ๐ฆ2 โ ๐๐ผ๐ฆ2๐๐ผ๐ฆ1 โ ๐ฆ2๐๐ผ๐๐ผ๐ฆ1
But, ๐ฆ1 and ๐ฆ2 satisfy (2.24). So
๐๐ผ๐๐ผ๐ฆ1 = โ๐(๐ฅ)๐๐ผ๐ฆ1 โ ๐(๐ฅ)๐ฆ1,
and
๐๐ผ๐๐ผ๐ฆ2 = โ๐(๐ฅ)๐๐ผ๐ฆ2 โ ๐(๐ฅ)๐ฆ2,
57
therefore,
๐๐ผ(๐๐ผ[๐ฆ1, ๐ฆ2]) = โ๐(๐ฅ)(๐ฆ1๐๐ผ๐ฆ2 โ ๐ฆ2๐๐ผ๐ฆ1)
= โ๐(๐ฅ)๐๐ผ[๐ฆ1, ๐ฆ2] ,
thus,
๐๐ผ(๐๐ผ[๐ฆ1, ๐ฆ2])
๐๐ผ[๐ฆ1, ๐ฆ2]= โ๐(๐ฅ)
Consequently,
๐๐ผ[๐ฆ1, ๐ฆ2] = ๐โ๐ผ๐ผ(๐) (2.24)
2.4.2. Abelโs Formula
First of all, it is important to discuss linear fractional differential equation
๐๐ผ๐ฆ + ๐(๐ฅ)๐ฆ = ๐(๐ฅ), ๐ผ โ [0, 1] (2.25)
Multiply (2.26) by ๐๐ผ๐ผ(๐(๐ฅ)) to get
๐๐ผ๐ผ(๐(๐ฅ))๐๐ผ๐ฆ + ๐๐ผ๐ผ(๐(๐ฅ))๐(๐ฅ)๐ฆ = ๐๐ผ๐ผ(๐(๐ฅ))๐(๐ฅ)
๐๐ผ(๐๐ผ๐ผ(๐(๐ฅ))๐ฆ) = ๐๐ผ๐ผ(๐(๐ฅ))๐(๐ฅ).
Hence
๐ฆ = ๐โ๐ผ๐ผ(๐(๐ฅ))๐ผ๐ผ(๐๐ผ๐ผ(๐(๐ฅ))๐(๐ฅ)) (2.26)
Is a solution of (2.26).
Now, let ๐ฆ1 be a solution of (2.24). To find a second solution ๐ฆ2 for equation (2.24).
58
We have ๐๐ผ[๐ฆ1, ๐ฆ2] = ๐โ๐ผ๐ผ(๐), from which we get:
๐ฆ1๐๐ผ๐ฆ2 โ ๐ฆ2๐๐ผ๐ฆ1 = ๐โ๐ผ๐ผ(๐),
And so
๐๐ผ๐ฆ2 โ ๐ฆ2
๐๐ผ๐ฆ1๐ฆ1
=๐โ๐ผ๐ผ(๐)
๐ฆ1 (2.27)
Equation (2.28) is a fractional linear equation, with ๐(๐ฅ) =๐๐ผ๐ฆ1
๐ฆ1, and ๐(๐ฅ) =
๐ผ๐ผ(โ๐(๐ฅ))
๐ฆ1.
Hence, using the fact:
๐ผ๐ผ (๐๐ผ๐ฆ1๐ฆ1
) = ln๐ฆ1,
And formula (2.27) to get:
๐ฆ2 = ๐ฆ1๐ผ๐ผ (
๐โ๐ผ๐ผ(๐)
๐ฆ12 ). (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:-
๐ฆ(๐ผ) = ๐ด(๐ฅ)๐ฆ2 + ๐ต(๐ฅ)๐ฆ + ๐ถ(๐ฅ) (3.1)
๐ฆ(0) = ๐ , ๐: constant (3.2)
where ๐ฆ(๐ผ) is the conformable fractional derivative of order ๐ผ โ (0,1] , we should remark
that the method can be generalized to include any ๐ผ .
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
๐ขโฒโฒ โ (
๐ผ โ 1
๐ฅ+ ๐ (๐ฅ)) ๐ขโฒ + ๐ฅ๐ผโ1๐(๐ฅ)๐ข = 0 (3.3)
When ๐ด(๐ฅ) is non-zero and differentiable, such that ๐ผ โ (0,1] ,also the solution of this
equation leads us to the solution.
๐ฆ =
โ๐โฒ(๐ฅ) ๐ฅ1โ๐ผ
๐ด(๐ฅ)๐(๐ฅ) (3.4)
Proof:
Let ๐ฃ = ๐ฆ๐ด(๐ฅ)
๐ฃ(๐ผ) = (๐ฆ๐ด(๐ฅ))(๐ผ) = ๐ฆ(๐ผ)๐ด(๐ฅ) + ๐ฆ๐ฅ1โ๐ผ๐ดโฒ(๐ฅ)
๐ฆ(๐ผ) satisfies the FRDE also by substituting ๐ฆ =๐ฃ
๐ด and some algebraic steps, then:
๐ฅ1โ๐ผ๐ฃโฒ(๐ฅ) = ๐ฃ2 + ๐ต๐ฃ + ๐ถ๐ด + ๐ฃ๐ฅ1โ๐ผ๐ดโฒ
๐ด
Divided both sides by ๐ฅ1โ๐ผ, then:
๐ฃโฒ(๐ฅ) = ๐ฅ๐ผโ1๐ฃ2 + ๐ฅ๐ผโ1๐ต๐ฃ + ๐ฅ๐ผโ1๐ถ๐ด + ๐ฃ๐ดโฒ
๐ด
Combining like terms, to get:
๐ฃโฒ(๐ฅ) = ๐ฅ๐ผโ1๐ฃ2 + (๐ฅ๐ผโ1๐ต +
๐ดโฒ
๐ด) ๐ฃ + ๐ฅ๐ผโ1๐ถ๐ด (3.5)
Assume: ๐ (๐ฅ) = ๐ฅ๐ผโ1๐ต +๐ดโฒ
๐ด and ๐(๐ฅ) = ๐ฅ๐ผโ1๐ถ๐ด , to get:
๐ฃโฒ(๐ฅ) = ๐ฅ๐ผโ1๐ฃ2 + ๐ (๐ฅ)๐ฃ + ๐(๐ฅ)
61
Let ๐ฅ๐ผโ1๐ฃ = โ
๐ขโฒ
๐ข (3.6)
(๐ผ โ 1)๐ฅ๐ผโ2๐ฃ + ๐ฅ๐ผโ1๐ฃโฒ =โ๐ข๐ขโฒโฒ + (๐ขโฒ)2
๐ข2
(๐ผ โ 1)๐ฅ๐ผโ2๐ฃ + ๐ฅ๐ผโ1๐ฃโฒ =โ๐ขโฒโฒ
๐ข+ ๐ฃ2(๐ฅ๐ผโ1)2
Divide both sides by ๐ฅ๐ผโ1
(๐ผ โ 1)๐ฅโ1๐ฃ + ๐ฃโฒ = โ๐ฅ1โ๐ผ๐ขโฒโฒ
๐ข+ ๐ฅ๐ผโ1๐ฃ2
๐ผ โ 1
๐ฅ๐ฃ + ๐ฅ1โ๐ผ
๐ขโฒโฒ
๐ข= ๐ฅ๐ผโ1๐ฃ2 โ ๐ฃโฒ
From equation (3.5)
๐ผ โ 1
๐ฅ๐ฃ + ๐ฅ1โ๐ผ
๐ขโฒโฒ
๐ข= โ(๐ฅ๐ผโ1๐ต +
๐ดโฒ
๐ด) ๐ฃ โ ๐ฅ๐ผโ1๐ถ๐ด
๐ผ โ 1
๐ฅ๐ฃ + ๐ฅ1โ๐ผ
๐ขโฒโฒ
๐ข= โ๐ (๐ฅ)๐ฃ โ ๐(๐ฅ)
combining like terms to get:
๐ฅ1โ๐ผ๐ขโฒโฒ
๐ข+ (๐ผ โ 1
๐ฅ+ ๐ (๐ฅ)) ๐ฃ + ๐(๐ฅ) = 0
divide both sides by ๐ฅ1โ๐ผ after substitute ๐ฃ = โ๐ขโฒ
๐ข๐ฅ1โ๐ผ
๐ฅ1โ๐ผ๐ขโฒโฒ
๐ข+ (๐ผ โ 1
๐ฅ+ ๐ (๐ฅ)) (โ
๐ขโฒ
๐ข๐ฅ1โ๐ผ) + ๐(๐ฅ) = 0
๐ขโฒโฒ
๐ขโ (๐ผ โ 1
๐ฅ+ ๐ (๐ฅ))
๐ขโฒ
๐ข+ ๐ฅ๐ผโ1๐(๐ฅ) = 0
โด ๐ขโฒโฒ โ (๐ผ โ 1
๐ฅ+ ๐ (๐ฅ))๐ขโฒ + ๐ฅ๐ผโ1๐(๐ฅ)๐ข = 0
62
An answer of this equation will lead us to
๐ฆ =๐ฃ
๐ด=โ๐ขโฒ๐ฅ1โ๐ผ
๐ข๐ด โ
Theorem 3.1.2. (Transform FRDE to the Bernoulli equation)
For non-linear fractional Riccati equation the substitution ๐ฃ(๐ฅ) = ๐ฆ(๐ฅ) โ ๐ฆ1(๐ฅ) will
transform the (FRDE) into Bernoulli equation (ordinary differential equation of the first
order), when ๐ฆ1 is a known particular solution,
Proof:
Since ๐ฃ(๐ฅ) = ๐ฆ(๐ฅ) โ ๐ฆ1(๐ฅ)
โด ๐ฆ(๐ฅ) = ๐ฃ(๐ฅ) + ๐ฆ1(๐ฅ)
And ๐ฆ(๐ผ)(๐ฅ) = ๐ฃ(๐ผ)(๐ฅ) + ๐ฆ1(๐ผ)(๐ฅ)
Since ๐ฆ1(๐ฅ) solves the (FRDE), it must be that
๐ฆ1(๐ผ) = ๐ด(๐ฅ)๐ฆ1
2 + ๐ต(๐ฅ)๐ฆ1 + ๐ถ(๐ฅ)
Substitute in (3.1)
๐ฃ(๐ผ)(๐ฅ) + ๐ฆ1(๐ผ)(๐ฅ)โ
๐ฆ(๐ผ)(๐ฅ)
= ๐ด(๐ฅ) [๐ฃ + ๐ฆ1]โ ๐ฆ(๐ฅ)
2+ ๐ต(๐ฅ) [๐ฃ + ๐ฆ1]โ
๐ฆ(๐ฅ)
+ ๐ถ(๐ฅ)
๐ฅ1โ๐ผ๐ฃ โฒ(๐ฅ)โ ๐ฃ(๐ผ)(๐ฅ)
+ ๐ด๐ฆ12 + ๐ต๐ฆ1 + ๐ถ = ๐ด๐ฃ
2 + 2๐ด๐ฃ๐ฆ1 + ๐ด๐ฆ12 + ๐ต๐ฃ + ๐ต๐ฆ1 + ๐ถ
๐ฅ1โ๐ผ๐ฃ โฒ(๐ฅ) = ๐ด๐ฃ2(๐ฅ) + 2๐ด๐ฆ1๐ฃ(๐ฅ) + ๐ต๐ฃ(๐ฅ)
๐ฃ โฒ(๐ฅ) = ๐ด๐ฅ๐ผโ1๐ฃ2(๐ฅ) + 2๐ด๐ฅ๐ผโ1๐ฆ1๐ฃ(๐ฅ) + ๐ต๐ฅ๐ผโ1๐ฃ(๐ฅ)
๐ฃ โฒ(๐ฅ) + [โ2๐ฅ๐ผโ1๐ด(๐ฅ)๐ฆ1 โ ๐ฅ๐ผโ1๐ต(๐ฅ)]โ
๐(๐ฅ)
๐ฃ = ๐ด๐ฅ๐ผโ1โ ๐(๐ฅ)
๐ฃ2(๐ฅ) (3.7)
This equation is of the form of Bernoulli equation with n=2 โ
63
which could be transformed to first order linear differential equation.
Let ๐ข = ๐ฃโ1(๐ฅ).
๐๐ข
๐๐ฅ= โ๐ฃโ2(๐ฅ)
๐๐ฃ
๐๐ฅ
Multiply (3.7) by โ ๐ฃ(๐ฅ)โ2
โ๐ฃโ2๐ฃ โฒ + [2๐ฅ๐ผโ1๐ด๐ฆ1 + ๐ฅ๐ผโ1๐ต]๐ฃโ2๐ฃ = โ๐ด๐ฅ๐ผโ1
๐ฃโฒ+ [2๐ฅ๐ผโ1๐ด๐ฆ1 + ๐ฅ๐ผโ1๐ต]๐ฃ = ๐ด ๐ฅ๐ผโ1โ
๐(๐ฅ)
(3.8)
The general solution is given by
๐ฃ =
โซ๐(๐ฅ)๐(๐ฅ). ๐๐ฅ + ๐(๐ฅ)
๐(๐ฅ) (3.9)
where ๐(๐ฅ) = ๐(โซ[2๐ฅ๐ผโ1๐ด๐ฆ1+๐ฅ
๐ผโ1๐ต]๐๐ฅ) (3.10)
Theorem 3.1.3. (Obtaining solution of FRDE by Abelโs formula)
Let ๐ฆ1 be a solution of (3.1), and assume that ๐ง = 1
๐ฆโ ๐ฆ1, then the solution of FRDE is
๐ง = ๐โ๐ผ(2๐ด๐ฆ1+๐ต)๐ผ๐ผ(๐๐ผ(2๐ด๐ฆ1+๐ต)(โ๐ด(๐ฅ))) (3.11)
Proof: suppose that ๐ฆ1 is a solution of FRDE, and let =1
๐ฆโ๐ฆ1 , then
๐ง(๐ฆ โ ๐ฆ1) = 1
๐ฆ =
1
๐ง+ ๐ฆ1 (3.12)
64
Apply ๐ผ-derivative definition to both sides of (3.12)
๐๐ผ๐ฆ = ๐๐ผ (1
๐ง) + ๐๐ผ๐ฆ1
๐๐ผ๐ฆ = โ๐งโ1โ๐ผ๐งโฒ + ๐๐ผ๐ฆ1
Substituting in the original FRDE
โ๐งโ1โ๐ผ๐งโฒ + ๐๐ผ๐ฆ1 = ๐ด [1
๐ง+ ๐ฆ1]
2
+ ๐ต [1
๐ง+ ๐ฆ1] + ๐ถ
โ๐งโ1โ๐ผ๐งโฒ = ๐ด [1
๐ง2+2๐ฆ1๐ง+ ๐ฆ1
2] + ๐ต [1
๐ง+ ๐ฆ1] + ๐ถ โ ๐๐ผ๐ฆ1
๐๐ผ๐ฆ1 satisfies the FRDE
โ๐งโ1โ๐ผ๐งโฒ =๐ด
๐ง2+2๐ฆ1๐ด
๐ง+ ๐ด๐ฆ1
2 +๐ต
๐ง+ ๐ต๐ฆ1 + ๐ถ โ ๐ด๐ฆ
2 โ ๐ต๐ฆ1 โ ๐ถ
Combining like terms and divide both sides by โ๐งโ1โ๐ผ
๐งโฒ = โ(2๐ด๐ฆ1 + ๐ต)๐ง๐ผ โ ๐ด๐ง๐ผโ1, then
๐งโฒ + (2๐ด๐ฆ1 + ๐ต)๐ง๐ผ = โ๐ด๐ง๐ผโ1 (3.13)
Multiply both sides of equation (3.13) by ๐ง1โ๐ผ
๐ง1โ๐ผ๐งโฒ + (2๐ด๐ฆ1 + ๐ต)๐ง = โ๐ด
๐ง(๐ผ) + (2๐ด๐ฆ1 + ๐ต)๐ง = โ๐ด (3.14)
which is Abelโs formula as we mentioned in the previous chapter.
Thus, the solution is
65
๐ง = ๐โ๐ผ(2๐ด๐ฆ1+๐ต)๐ผ๐ผ(๐๐ผ(2๐ด๐ฆ1+๐ต)(โ๐ด(๐ฅ)))
Theorem 3.1.4. Assume that the coefficients ๐ถ(๐ฅ) + ๐ต(๐ฅ) + ๐ด(๐ฅ) = 0 of the fractional
Ricatii (3.1), if ๐ถ(๐ฅ) satisfies the integral condition, which is
๐ถ(๐ฅ) =๐1(๐ฅ) โ {๐ต(๐ฅ) + ๐ด(๐ฅ) [โซ
๐1(๐) โ ๐ต2(๐)
2๐ด(๐)โ ๐ด1
๐ฅ]}2
4๐ด
(3.15)
where ๐ด1 is an arbitrary constant of integration.
and ๐1 is the new generating function satisfying the differential condition (3.15) given by:
๐ต2(๐ฅ) + 4๐ด(๐ฅ)๐ฅ1โ๐ผ
๐๐ฆ๐
๐๐ฅ= ๐1(๐ฅ) (3.16)
Then the general solution is given by:
๐ฆ(๐ฅ) =1
๐โ๐ผ(2๐ด๐ฆ1+๐ต)๐ผ๐ผ(๐๐ผ(2๐ด๐ฆ1+๐ต)(โ๐ด(๐ฅ)))
+1
2[โซ
๐1(๐) โ ๐ต2(๐)
2๐ด(๐)๐๐
๐ฅ
โ ๐ด1],
where ๐ด0 is an arbitrary constant of integration.
(3.17)
Proof.
Assume that the arbitrary function ๐ต(๐ฅ), ๐ด(๐ฅ) and ๐1(๐ฅ) satisfying (3.15) then the
particular solution
๐ฆ๐ยฑ(๐ฅ) =โ๐ต ยฑ โ๐1 โ 4๐ด๐ถ
2๐ด
66
=
โ๐ต ยฑโ๐1 โ 4๐ด
๐1(๐ฅ) โ {๐ต(๐ฅ) + ๐ด(๐ฅ) [โซ๐1(๐) โ ๐ต2(๐)
2๐ด(๐)๐ฅ
โ ๐ด1]}2
4๐ด
2๐ด
=
โ๐ต ยฑ โ๐1 โ ๐1(๐ฅ) โ {๐ต(๐ฅ) + ๐ด(๐ฅ) [โซ๐1(๐) โ ๐ต2(๐)
2๐ด(๐)โ ๐ด1
๐ฅ]}2
2๐ด
=โ๐ต + ๐ต(๐ฅ) + ๐ด(๐ฅ) [โซ
๐1(๐) โ ๐ต2(๐)
2๐ด(๐)๐ฅ
โ ๐ด1]
2๐ด
=๐ด(๐ฅ) [โซ
๐1(๐) โ ๐ต2(๐)
2๐ด(๐)๐ฅ
โ ๐ด1]
2๐ด
=1
2[โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
โ ๐ด1]
Thus
๐ฆ๐ยฑ(๐ฅ) =โ๐ต ยฑ โ๐1 โ 4๐ด๐ถ
2๐ด=1
2[โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
โ ๐ด1] (3.18)
Differentiate equation (3.18)
๐๐ฆ
๐๐ฅ[โ๐ต ยฑ โ๐1 โ 4๐ด๐ถ
2๐ด] =
๐1(๐) โ ๐ต2(๐)
2๐ด(๐) (3.19)
Equation (3.19) can be integrated to get
โ๐ต ยฑ โ๐1 โ 4๐ด๐ถ
๐ด=โซ๐1(๐) โ ๐ต
2(๐)2๐ด(๐)
๐ฅ
1
67
โ๐ต ยฑ โ๐1 โ 4๐ด๐ถ = ๐ด [โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
]
โ๐1 โ 4๐ด๐ถ = ๐ต + ๐ด [โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
]
๐1 โ 4๐ด๐ถ = {๐ต + ๐ด [โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
]}
2
โ4๐ด๐ถ = โ๐1 + {๐ต + ๐ด [โซ๐1(๐) โ ๐ต
2(๐)
2๐ด(๐)
๐ฅ
]}
2
๐ถ(๐ฅ) =๐1(๐ฅ) โ {๐ต(๐ฅ) + ๐ด(๐ฅ) [โซ
๐1(๐) โ ๐ต2(๐)
2๐ด(๐)โ ๐ด1
๐ฅ]}2
4๐ด
3.2 Applications:
Example: - find the solution of
๐ฆ(1
2) = (๐ฆ โ 2โ๐ฅ)
2+ 1 , ๐ฆ1(๐ฅ) = 2โ๐ฅ ; ๐ฆ(0) = 1 (3.20)
Solution: First we need to verify that ๐ฆ1 = 2โ๐ฅ is a solution to this equation by computing,
we find that ๐ฆ1 is a solution of (3.20).
Now we solve the equation.
Step1. Make the change of variables
Substituting ๐ฆ = ๐ฃ + 2โ๐ฅ and ๐ฆ(1
2) = ๐ฃ(
1
2) + 1 yields
68
๐ฃ(12)+ 1 = (๐ฃ + 2โ๐ฅ โ 2โ๐ฅ)
2+ 1
Step 2. Simplify to a Bernoulli equation for ๐ฃ
๐ฅ12๐ฃโฒ = ๐ฃ2
๐ฃ โฒ = ๐ฅโ
12๐ฃ2 (3.21)
This is a Bernoulli equation.
Step3. Solve the Bernoulli equation
Let ๐ข = ๐ฃโ1
๐ขโฒ = โ๐ฃโ2๐ฃโฒ
Multiply equation (3.21) by โ๐ฃโ2
โ๐ฃโ2๐ฃ โฒ = โ๐ฅโ12๐ฃโ2๐ฃ2
๐ขโฒ = โ๐ฅโ12 =
โ1
โ๐ฅ
๐๐ข
๐๐ฅ=โ1
โ๐ฅ โ ๐๐ข =
โ1
โ๐ฅ๐๐ฅ
๐ข = โซโ1
โ๐ฅ. ๐๐ฅ = โ2โ๐ฅ + ๐
1
๐ฃ= โ2โ๐ฅ + ๐
๐ฃ =1
โ2โ๐ฅ + ๐
Step 4. Reverse the substitution ๐ฆ = ๐ฃ + 2โ๐ฅ
69
๐ฆ =1
โ2โ๐ฅ + ๐โ 2โ๐ฅ
Finally, we use the initial condition ๐ฆ(0) = 1
โด ๐ = 1
โด The general solution is
๐ฆ =
3
โ4โ๐ฅ3 + 1โ๐ฅ2
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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73
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76
ุญู ู ุนุงุฏุงูุช ุชูุงุถููุฉ ูุณุฑูุฉ ุจุงุณุชุฎุฏุงู ุชุนุฑูู ุงูู ุทุงุจู ููู ุดุชูุงุช ุงููุณุฑูุฉ
ุฅุนุฏุงุฏ
ุดุงุฏู ุฃุญู ุฏ ุงูุทุฑุงููุฉ
ุงูู ุดุฑู
ุฏ. ุฎุงูุฏ ุฌุงุจุฑ
ุงูู ูุฎุต
ุน ุงูู ูุงููููุฉุ ูุงุฃูุญูุงุก ุ ูุงูููุฒูุงุก ุ ุฆุงูู ุนุงุฏุงูุช ุงูุชูุงุถููุฉ ุงูุนุงุฏูุฉ ูุงูุฌุฒุฆูุฉ ู ูู ุฉ ุฌุฏุงู ูู ู ุฌุงุงูุช ุนุฏูุฏุฉุ ู ุซู ุงูู ูุง
ููุธุฑูุฉ ุงูุชุญูู ูู ุงุฃููุธู ุฉ ูุงูุจุตุฑูุงุชุ ูุงูููุฑูููู ูุงุฆูุฉ ุ ูุงูููุฏุณุฉุ ูุงููุฒูุฌุฉ ุงูู ุทุงุทูุฉ ุ ูุงูุดุจูุงุช ุงูููุฑุจุงุฆูุฉ ุ
ุงูุฏููุงู ูููุฉ.
ู ู ูุจู ุจุงุญุซูู ุนุฏุฉุ ุจุงุณุชุฎุฏุงู ุทุฑู ุนุฏูุฏุฉ ู ุฎุชููุฉ. ุจุฏุฃ ู ูุถูุน ุงูุชู ุงู ูุง ููู ุญู ุฑููุงุชูุชูู ุฏุฑุงุณุฉ ู ุนุงุฏูุฉ
ูู ุฌู ูุนุฉ ุจุงุญุซูู ุขุฎุฑูู ูุชูุฏูู ุชุนุฑูู ุฌุฏูุฏ ูุจุณูุท ูุฃูุซุฑ ุฎููู ุฑุดุฏูู ุนุงุฏุงูุช ุชูุงุถููุฉ ูุณุฑูุฉุ ุนูุฏู ุง ูุงู ุงูุฏูุชูุฑ
ุฑูุฉ. ูุฐุง ุงูุชุนุฑูู ุงูุฌุฏูุฏ ูู ุงู ุชุฏุงุฏ ููู ุดุชูุงุช ุงูุนุงุฏูุฉ ูุงูุฐู ูุณูู ู "ุชุนุฑูู ุงูู ุทุงุจู ููู ุดุชูุงุช ููุงุกุฉู ููู ุดุชูุงุช ุงููุณ
ุงููุณุฑูุฉ".
ุงูุชูุงุถููุฉ ุงููุณุฑูุฉุ ููุฏูู ูุง ุจุนุถ ุงููุธุฑูุงุช ุงูุฐู ุชุณุงุนุฏูุง ูู ุฑููุงุชููู ูุฐุง ุงูุจุญุซุ ุฃูุฌุฏูุง ุญู ุฏููู ูู ุนุงุฏูุฉ
ุฉ.ุงูุชูุงุถููุฉ ุงููุณุฑู ุฏ ุญู ุซุงูู ุนูุฏู ุง ูุนุทู ุญู ูู ุนุงุฏูุฉ ุฑููุงุชูุฅูุฌุง