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MULTI POINT BOUNDARY VALUE PROBLEMS FOR SYSTEMS

OF FRACTIONAL DIFFERENTIAL EQUATIONS: EXISTENCE

THEORY AND NUMERICAL SIMULATIONS

By

Kamal Shah

SUBMITTED IN PARTIAL FULFILLMENT OF THE

REQUIREMENTS FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

AT

UNIVERSITY OF MALAKAND

KPK, LOWER DIR

DECEMBER 2015

c© Copyright by Kamal Shah, 2015


DEPARTMENT OF

MATHEMATICS

The undersigned hereby certify that they have read and recommend

to the Faculty of Graduate Studies for acceptance a thesis entitled

“MULTI POINT BOUNDARY VALUE PROBLEMS FOR

SYSTEMS OF FRACTIONAL DIFFERENTIAL EQUATIONS:

EXISTENCE THEORY AND NUMERICAL SIMULATIONS”

by Kamal Shah in partial fulfillment of the requirements for the degree of

Doctor of Philosophy.

Dated: December 2015

External Examiner:

Research Supervisor:Rahmat Ali Khan

Examing Committee:

ii


Date: December 2015

Author: Kamal Shah

Title: MULTI POINT BOUNDARY VALUE PROBLEMS

FOR SYSTEMS OF FRACTIONAL DIFFERENTIAL

EQUATIONS: EXISTENCE THEORY AND

NUMERICAL SIMULATIONS

Department: Mathematics

Degree: Ph.D. Convocation: May Year: 2016

Permission is herewith granted to University of Malakand to circulate and to

have copied for non-commercial purposes, at its discretion, the above title upon the

request of individuals or institutions.

Signature of Author

THE AUTHOR RESERVES OTHER PUBLICATION RIGHTS, AND NEITHERTHE THESIS NOR EXTENSIVE EXTRACTS FROM IT MAY BE PRINTED OROTHERWISE REPRODUCED WITHOUT THE AUTHOR’S WRITTEN PERMISSION.

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iii

Dedicated: To my Late Grand father and dear parents.

iv

Table of Contents

Table of Contents v

Abstract ix

Acknowledgements xi

List of Publication from thesis xii

1 Introduction 1

2 Preliminaries 6

2.1 Special functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.2 The Beta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.3 The Mittag-Leffler function . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.4 The Mellin-Ross function . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Fractional order derivatives and integrals . . . . . . . . . . . . . . . . . . . 10

2.2.1 Riemann-Liouville fractional order integrals . . . . . . . . . . . . . . 11

2.2.2 Fractional order derivatives . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.3 The Riemann-Liouville’s fractional derivatives . . . . . . . . . . . . . 12

2.2.4 The Caputo’s fractional derivative . . . . . . . . . . . . . . . . . . . 14

2.3 Fixed point theorems and some results from analysis . . . . . . . . . . . . . 16

2.3.1 Fixed point theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Degree theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 Operational matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.5.1 Legendre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.5.2 Bernstein polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.5.3 The shifted Jacobi polynomials and its fundamental properties . . . 32

3 Multi-point boundary value problems 35

3.1 Multi-point boundary value problems . . . . . . . . . . . . . . . . . . . . . 36

v

3.1.1 Properties of Green’s function . . . . . . . . . . . . . . . . . . . . . . 38

3.1.2 Existence of at least one solution . . . . . . . . . . . . . . . . . . . . 39

3.1.3 Uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.1.4 Multiplicity: Existence of at least three positive solutions . . . . . . 43

3.2 Multi-point boundary value problems with nonlocal boundary conditions . . 44



3.3 General Class of multi-point boundary value problems . . . . . . . . . . . . 50

3.3.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 53


4 Coupled systems of fractional order multi-point boundary value problems 58

4.1 Coupled system with m-point boundary conditions . . . . . . . . . . . . . . 61

4.1.1 Existence of at least one positive solution . . . . . . . . . . . . . . . 63

4.1.2 Uniqueness of positive solutions . . . . . . . . . . . . . . . . . . . . . 65

4.2 Coupled system with movable type integral boundary conditions . . . . . . 67



4.3 Coupled system with integral boundary conditions . . . . . . . . . . . . . . 73



4.4 Coupled system with impulsive boundary conditions . . . . . . . . . . . . . 83



4.5 Existence of solutions of more general systems:

A topological degree theory approach . . . . . . . . . . . . . . . . . . . . . . 93

4.5.1 Coupled system with multi-point nonlocal boundary conditions(I) . 93



4.5.4 Coupled system with nonlinear boundary conditions(II) . . . . . . . 102



5 Multiplicity of positive solutions for coupled system of multi-point bound-

ary value problems 110

5.1 Multiplicity of positive solutions to a coupled system of higher order four

point BVPs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.1.1 Existence and uniqueness of solution . . . . . . . . . . . . . . . . . 114


5.1.3 Non-existence of positive solutions . . . . . . . . . . . . . . . . . . . 123

5.1.4 Existence of at least two positive solutions . . . . . . . . . . . . . . . 124

vi

5.1.5 Existence of Multiple positive solutions . . . . . . . . . . . . . . . . 126

6 Iterative techniques for solutions of Coupled systems of multi-point bound-

ary value problems 131

6.1 Iterative techniques for solutions of coupled systems of boundary value prob-

lem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.1.1 Existence of upper and lower solutions . . . . . . . . . . . . . . . . . 134

6.1.2 Uniqueness of upper and lower solutions . . . . . . . . . . . . . . . . 138

6.2 Iterative techniques for solutions of coupled systems with three point bound-

ary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140


6.3 Iterative techniques for solutions of coupled systems with m-point coupled

boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

6.3.1 Existence of Green’s functions . . . . . . . . . . . . . . . . . . . . . 144


7 Numerical solutions of systems of boundary value problem of fractional

differential equations 151

7.1 Numerical solutions of system of fractional order differential equations by

Legendre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

7.1.1 Numerical solutions of boundary value problems . . . . . . . . . . . 157

7.1.2 Numerical solutions of system of boundary value problems . . . . . 159

7.1.3 Illustrative examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

7.1.4 Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

7.2 Numerical solutions of boundary value problems by Bernstein polynomials . 166

7.2.1 Numerical solutions of boundary value problems . . . . . . . . . . . 174

7.2.2 Numerical solutions of system of boundary value problem . . . . . . 176


7.3 Numerical solutions boundary value problem by Shifted Jacobi polunomials 184

7.3.1 Applications of Shifted Jacobi’s polynomials operational matrices . . 188

7.3.2 Numerical solutions of system of boundary value problems . . . . . 190


7.4 Numerical solutions of system of partial differential equations . . . . . . . . 202

7.4.1 Numerical solutions of the coupled systems fractional partial differ-

ential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

7.5 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

7.6 Comparison between Bernstein, Shifted Legendre and Shifted Jacobi poly-

nomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

8 Matlab codes for numerical analysis 215

8.1 Computations of operational matrices using Legendre polynomials . . . . . 215

vii

8.2 Computations of operational matrices using Bernstein polynomials . . . . . 218

8.3 Computation of operational matrices using Shifted Jacobi polynomials . . . 220

9 Summary and conclusion 224

Bibliography 228

viii

Abstract

In recent years, it has been proved that differential equations of fractional order and their

systems are the best tools for modeling various phenomena of chemical and physical as well

as biological sciences. Besides from these, it is also proved that the aforementioned area

has also important applications in various field of engineering and technology. Therefore,

considerable attentions were given to study the subject of fractional order differential equa-

tions in last few decades. This thesis is concerned with a detailed study of the existence

theory and numerical solutions of multi-point boundary value problems of fractional order

differential equations. For such study, first we review some useful notations, definitions, re-

sults from fractional calculus, functional analysis and fixed point theory. Also for numerical

solutions, we use some orthogonal polynomials like Legendre, Bernstein and Jacobi polyno-

mials. We begin our work with the study of existence and uniqueness of positive solutions

for simple multi point boundary value problem. Then, we obtain necessary and sufficient

conditions for existence of at least three positive solutions for the considered problems in

chapter 3. Then another class with nonlocal boundary conditions is studied by topological

degree method for existence and uniqueness of positive solutions. While a class of fractional

order differential equations, where nonlinear function involved in it depending on fractional

derivative involve in it with multi point boundary conditions is also studied for existence

of solutions. These conditions are developed by using some classical fixed point theorems

and results of functional analysis. Existence and uniqueness of positive solution for multi

point boundary value problems for coupled systems are studied in chapter 4. Sufficient

conditions for existence and uniqueness results of multi point boundary value problems for

coupled systems are established with the help of fixed point theorems such as Banach, Gue-

Krasnoselskii’s fixed point theorem of cone expansion and compression, Schauder’s fixed

ix

x

point theorem and Perov’s fixed point theorem etc. Some multiplicity results for existence

of solutions to the nonlocal boundary value problem are discussed in chapter 5. For every

differential equation or system of differential equations of classical or arbitrary order it is

nor compulsory that it has a solution. Therefore conditions for nonexistence are developed

in a part of the same chapter 5. Moreover, for multiplicity of positive solutions, the nec-

essary and sufficient conditions are developed by means of monotone iterative technique

together with the method of upper and lower solutions in chapter 6. It is very difficult

to find exact solution for each and every problem of fractional order differential equations

due to the complexity of fractional order differential and integral operator involved in the

system. Therefore, there exists a strong motivation to develop numerical schemes which are

easily understandable and easily computable as well as efficient and reliable. By means of

some orthogonal polynomials such as Shifted Legendre, Bernstein and Shifted Jacobi poly-

nomials, we are developed some operational matrices of integrations and differentiations

for numerical solutions of boundary value problems for both ordinary and partial fractional

order differential equations in chapter 7. With the help of these operational matrices, we

convert the problems under consideration to algebraic equations which are easily soluble for

unknown coefficient matrices. The obtain coefficient matrices are used to fined numerical

solutions for the concerned problems. The method is extended to solve coupled systems of

boundary value problems of fractional order differential equations. To perform the com-

putations, we use Matlab and Maple software. The efficiency of the numerical methods is

checked by solving several examples, and the comparison of exact and numerical solutions

will also be demonstrated.

Acknowledgements

All glory be to Allah, Who gave me strength and ability to complete this work easily, Who

remove every obstacle from my way along this work and along my life.

I wish to express my sincere gratitude to my Ph.D supervisor, Prof. Dr. Rahmat Ali

Khan for his supervisory work and his availability, whose encouragement, guidance and

support from the initial to the final level enabled me to develop an understanding of the

subject. I am heartily thankful to the Chairman and all my teachers and colleagues in the

departments of mathematics university of Malakand for their encouragements and many

useful discussions and suggestions in this work.

My special prayer goes to my late Grandfather who provided me every facility during my

education. Also my thank goes to my parents, who prayer for me all the time and without

them this work would never come into being. I am thankful to my brothers and sister who

help me and always expressed noble wishes for my success in life. I am also thankful to

my wife and my dear children, Muhammad Anus, Rida Kamal , Hafsa Kamal and Asma

Kamal, who prayer for me all the time, which give me courage and comfort. Lastly, I

offer my regards and blessings to all of those who supported me in any respect during the

completion of this thesis.

Department of Mathematics University of Malakand Kamal Shah

April 7, 2015

xi

List of Publication from thesis

(1) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Investigation of positive solution

to a coupled system of impulsive boundary value problems for nonlinear fractional

order differential equations, Chaos Soliton and Fractals.,77, (2015) 240-246.

(2) Kamal Shah, Amjad Ali and Rahmat Ali Khan, Degree theory and existence of posi-

tive solutions to coupled systems of multi-point boundary value problems, Boundary

Value Problems., (2016) 2016:43.

(3) Rahmat Ali Khan and Kamal Shah, Existence and uniqueness of solutions to fractional

order multi-point boundary value problems, Communication in Applied Analysis.,

19,(2015) 515-526.

(4) Hammad Khalil, Rahmat Ali Khan and Kamal Shah, Corrigendum to Investigation

of positive solution to a coupled system of impulsive boundary value problems for

nonlinear fractional order differential equations, 78, (2015), 329-330.

(5) Kamal Shah, Rahmat Ali Khan, Existence and uniqueness of positive solutions to a

coupled system of nonlinear fractional order differential equations with anti-periodic

boundary conditions, Differential Equations and Applications., 7(2), (2015) 245-262.

(6) Kamal Shah, Hammad Khalil and Rahmat Ali Khan,Upper and lower solutions to

a coupled system of nonlinear fractional differential equations,Progress in Fractional

Differential equations and Applications., 1(1),(2016) 1-10 .

(7) Kamal Shah and Rahmat Ali Khan, The Applications of Natural Transform to the

Analytical Solutions of Some Fractional Order Ordinary Differential Equations,Sindh

University Research Journal., 47 (4),(2015) 1-4.

xii

xiii

(8) Kamal Shah and Rahmat Ali Khan, Iterative solutions to a coupled system of non-

linear fractional differential equation, Journal of Fractional Calculus and Applica-

tions., 7(2), (2016) 40-50.

(9) Kamal Shah, Salman Zeb and Rahmat Ali Khan, Existence of positive solutions for m-

point boundary value problem of nonlinear fractional differential equation, Fractional

Differential Calculus., 5 (2), (2015) 171-18.

(10) Kamal Shah, Amjad Ali and Rahmat Ali Khan, Numerical solutions of fractional

order system of Baglay-Torvik Equation using operational matrices, Sindh University

Research Journal., 47 (4),(2015) 1-8.

(11) Kamal Shah, Salman Zeb and Rahmat Ali Khan, Multiplicity of positive solutions for

boundary value problem of nonlinear fractional differential equation, (submitted).

(12) Kamal Shah, Muhammad Sarwar and Rahmat Ali Khan, Multiplicity of positive solu-

tions to a coupled systems of nonlinear fractional order differential equations,(submitted).

(13) Kamal Shah and Rahmat Ali Khan, Iterative scheme for a coupled system of fractional

order differential equations with three point boundary conditions, (submitted).

(14) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Existence and computational

results for coupled system of fractional order boundary value problems, (submitted).

(15) Kamal Shah and Rahmat Ali Khan, Existence and uniqueness results to a coupled

system of fractional order boundary value problems by topological Degree theory,

(submitted).

(16) Kamal Shah and Rahmat Ali Khan, Existence of solutions for nonlinear coupled sys-

tem of fractional differential equation with integral boundary conditions, (submitted).

(17) Kamal Shah and Rahmat Ali Khan, Existence of positive solutions to a coupled system

of nonlinear fractional order differential equations with m-point boundary conditions,

(submitted).

(18) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Numerical solutions of boundary

value problem of coupled system of linear fractional order differential equations by the

use of Legendre polynomials, (submitted).

xiv

(19) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Numerical solutions of coupled

systems of fractional order partial differential equations, (submitted).

(20) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Numerical Solutions of Bound-

ary Value Problems of Fractional order differential Equations by Using Berenstein

Polynomials, (submitted).

(21) Kamal Shah, Hammad Khalil and Rahmat Ali Khan, Numerical solutions of Coupled

system of boundary value problems of fractional order differential equations by shifted

Jacobi polynomials, (submitted).

Chapter 1

Introduction

The foundation of fractional calculus was laid by Isaac Newton (1642 − 1727) and L. G.

Leibniz (1646− 1716). It is the branch of Mathematics which investigates the properties of

derivatives and integrals of non-integer orders known as differintegrals. Basically fractional

calculus owes its origin to a question of wether or not the meaning of derivative of an integer

order n could be extended to non-integer. This question was first raised by L’ Hospital on

30the September, 1695. He wrote in a letter to Leibniz that whether the derivative of semi

order is possible or not? In reply Leibniz said

“It will lead to a paradox, from which one day useful consequence will be drawn.”

After this, fractional calculus was built on formal foundations by several famous mathe-

maticians such as Riemann, Liouville, Grunwald, Euler, Lagrange, Heaviside, Fourier and

Abel etc. However, a rich summary of contributions can be found in [1]. In 1819, S. F.

De Lacroix introduced fractional derivative for the first time in his paper [2]. For a posi-

tive integer m using Legendre’s symbol Γ, Lacroix found that the nth order derivative of a

function u = tm is given by

dnu

dtn=

Γ(m+ 1)

Γ(m− n+ 1)tm−n.

By choosing m = 1 and n = 12 , he obtained the result as

d12u

dt12

=2√t√π.

1

2

At the same time, N.H. Abel [3], in 1823 also applied fractional calculus for the solutions of

an integral equation that happened in the mathematical models of Tautochrone problems,

that is, problems which determine the shape of the curve such that the time of descent of

an object sliding down the curve under uniform gravity is independent of the object’s initial

position. In recent years, the major work and key events developed by J. T. Machado, V.

Kiryakove, F. Mainardi and M. Benchohra, etc, in the area of fractional calculus since 1974

to 2010 was reported in [4]. Just like the classical calculus, fractional calculus has long

history but for many centuries it was unfamiliar among the researchers of applied science

because of unavailability of proper physical and geometrical meaning of fractional order

derivatives and integrals. Many attempts were made to define proper physical and geomet-

rical interpretations but these attempts remaind limited to certain problems. In 2002, I.

Podlubny [5], gave a convincing physical and geometrical interpretation of fractional deriva-

tives and integrals.

There are various definitions of fractional order derivatives and integrals such as Riemann,

Liouville, Caputo, Riemann Liouville, Weyl, Hadamard, Marchaud, and the Granwald-

Letnikov definitions etc. All these definitions are not equivalent except for some special

cases. However, the most notable definitions of fractional order derivatives and integrals

are those introduced by B. Riemann , J. Liouville, Riemann- Liouville, and Cuputo. The

uses of Riemann- Liouville definitions are less as compared to Caputo’s definitions, because

in most situations it fails in describing the physical and geometrical interpretations of ini-

tial and boundary conditions involving fractional derivatives. Solving fractional differential

equations with initial conditions involving Riemann-Liouvilli fractional order derivatives by

any integral transform such as Laplace Sumudu and Natural transform causes problems

in describing their physical and geometrical interpretations. But using Caputo’s definition

3

one can easily describe the physical and geometrical interpretation of a differential equa-

tion solved by any integral transform. Another difference between Caputo’s and Riemann-

Liouvilli’s fractional order derivative is that the Caputo’s derivative of a constant is zero

while that of Riemann-Liouvilli fractional order derivative of a constant is not zero and

given by

0Dαt [C] =

C.t−α

Γ(1− α), where C = constant.

However, this is not a demerit because the N. H. Abel ’s solution for Tautochrone problem

was based on this fact.

One of the most important areas of fractional calculus is the area of fractional differential

equations to which this thesis is devoted. Recently it has been found that fractional dif-

ferential equations have a lot of applications in many engineering and scientific disciplines

such as physics, chemistry, biology, economics, control theory, signal and image processing

phenomenon etc, see[6, 7, 8, 9, 10]. These can also be used as best tools for the description

of hereditary properties of various materials and processes. As memory terms are present

in the mathematical models of the concerned process, which guarantee the history and its

impact to the present and future, for detail, we refer to [11]. It has been studied that most

of the physical and biological models that involve integer order derivative are less realistic

as compared to those in which fractional order derivatives are involved [12, 13, 14]. More

recently, it is investigated that differential and integral equations of arbitrary order have

also valuable and interesting applications in the field of electric networks, control theory

of dynamical systems, probability and statistic, electro chemistry of corrosion, chemical

physics and optics, see [15, 16, 17, 18, 19, 20, 21, 22, 23, 24]. Due to these applications,

fractional differential equations have attracted the attention of many researchers for further

developments in the theory. One of the most important areas in the theory of fractional

order differential equations is the theory of existence and uniqueness of positive solutions

of boundary value problems. In most of the applied problems, there has been significant

4

interest for obtaining positive solutions as in these situations only positive solutions are

meaningful. Existence theory for positive solutions of boundary value problems with ordi-

nary derivatives have been widely explored for detail, see[25, 26, 27, 28, 29], but boundary

value problems with fractional order derivatives is still in its initial stages and its different

aspects need to be properly explored. Recently, the theory has attracted the attention of

many researchers, we refer to [30, 31, 32, 33] and the references therein for some of the recent

development in the theory. The study of existence and uniqueness of positive solutions to

coupled systems of boundary value problems for fractional order differential equations have

also attracted some attention, we refer to [34, 35, 36, 37, 38] and the references there in for

sufficient conditions on existence and uniqueness results. This is due to the fact that most

of mathematical models of physical and biological phenomena are in the form of systems of

fractional order differential equations.

Organization of this thesis as follow: In chapter 2, we provide some fundamentals results,

lemmas, theorems which are required in this study. We also recall definitions of some special

functions, theorems of fractional calculus and fixed point theory and functional analysis.

We also provide definitions of some orthogonal polynomials that are required for numerical

solutions of fractional order differential equations. Moreover the basic concept about the

degree theory of topology is also discussed. Chapter 3 is devoted to the study on existence

of positive solutions to three classes of different type of multi-point boundary conditions and

nonlinear functions involved in the class. In Chapter 4, we solve different types of systems

of boundary value problems corresponding to fractional order differential equations. We

establish necessary and sufficient conditions for at least one solution and check its unique-

ness. In chapter 5, we discuss existence of multiple solutions to a coupled system of highly

nonlinear fractional order differential equations. For this purposes, some growth conditions

are imposed on nonlinear functions involved in system to develop necessary and sufficient

5

conditions for multiplicity of solutions. Chapter 6 is devoted to the study of iterative so-

lutions to some coupled system of multi-point boundary value problems of fractional order

differential equations. The concerned study is carried out by means of monotone iterative

coupled with the method of upper and lower solutions. Chapter 7 is concerned with numeri-

cal solutions of system of boundary value problems of fractional order differential equations.

The numerical codes based on Matlab programming will be provided in chapter 8. Brief

conclusions and summary is given in chapter 9.

Chapter 2

Preliminaries

In this chapter, we recall some basic results from fractional calculus, fixed point theory

and functional analysis. Moreover, for numerical analysis, we review some basic orthogonal

polynomials and provide convergence analysis.

We organize this chapter as: In section 2.1, we provide definitions of special functions.

Section 2.2 is related to the definitions and some results of fractional calculus. In section

2.3, some fixed point theorems and other required results are given. Section 2.4 is devoted

to the introduction and properties of topological degree theory, while the last sections 2.5

is related to the properties of some orthogonal polynomials.

2.1 Special functions

We first recall some important mathematical definitions that are inherently linked to frac-

tional calculus and commonly used. These functions are, The Gamma function, the Beta

function, the Mittage-Lefler function and the Mellin-Rose function.

6

7

2.1.1 The Gamma function

The most basic interpretation of the Gamma function introduced by Euler in 1729 is the

generalization of the factorial for all real numbers. Its definition is given by

Γ(z) =

∫ ∞

0e−ttz−1dt, z ∈ R+. (2.1.1)

Some unique properties of Gamma functions are given by

Γ(z + 1) = zΓ(z), z ∈ R+. (2.1.2)

Γ(z) = (z − 1)!, z ∈ N, (2.1.3)

and

Γ(1

2) =

∫ ∞

0e−tt−

12dt =

√π. (2.1.4)

−6 −4 −2 0 2 4 6−4

−3

−2

−1

0

1

2

3

4

5

x

Plot of Gamma(x)

Plot of reciprocal ofGamma(x)

Figure 2.1: Plot of Gamma(x) and its reciprocal function.

2.1.2 The Beta function

Another important special function is the Beta functions and defined by the definite integral.

Its definition is given by

β(x, y) =

∫ 1

0tx−1(1− t)y−1dt, x, y ∈ R+. (2.1.5)

8

The relation between Gamma and Beta function is given by

β(x, y) =Γ(x)Γ(y)

Γ(x+ y), x, y ∈ R+. (2.1.6)

−6−4

−20

24

6

−6−4

−20

24

6−600

−400

−200

0

200

400

xy

B(x

,y)

Figure 2.2: Plot of Beta function.

2.1.3 The Mittag-Leffler function

The Mittage-Leffler function is named after a Swedish mathematician who defined and

studied for detail we refer [39]. This function is the direct generalization of the exponential

function ex, it plays a major role in fractional differential equations and fractional order

integral equations. The one parameter representation of Mattige-Leffler function in terms

of power series is given by

Eα(t) =

∞∑

k=0

tk

Γ(kα+ 1), t ∈ C, Re(α) > 0. (2.1.7)

The generalization of (2.1.7) in terms of two parameter was given by R.P. Agarwal and

Humbert [53], in 1953 as

Eα,β(t) =

∞∑

k=0

tk

Γ(kα+ β), t ∈ C, Re(α) > 0, Re(β) > 0. (2.1.8)

9

Figure 2.3: Plot of two parameter Mittage-Leffler function for various values of parameterα and β = 1.

For α ∈ (0, 1), the one parameter Mittage-Leffler function interpolates between exponential

function et and hypergeometric function 11−t as given below

(a) E0(t) =1

1− t, (b) E1(x) = et.

Some special cases of two parameter Mittage-Leffler function are given by

(a) E1,1(t) = et, (b) E1,2(t) =et − 1

t, (c) E2,1(t

2) = cosh(t),

(d) E2,2(t2) =

sinh(t)

t.

For further detail properties of Mittage-Leffler function, we refer to [54].

2.1.4 The Mellin-Ross function

The Mellin-Ross function, Et(q, a), arises when finding the fractional derivatives and in-

tegrals of an exponential function ex and it is defined in term of Mittage-Leffler function

as

Et(q, a) = tq∞∑

k=0

(at)k

Γ(k + q + 1)= tqE1,q+1(at). (2.1.9)

10

2.2 Fractional order derivatives and integrals

The generalization of the classical derivatives and integrations to arbitrary order is called

fractional order derivative and integrations jointly called differintegrals. In literature, there

are various approaches available to define fractional order derivatives and integrations. But

these are not equivalent except for some special functions. I. Podlubny and K. S. Miller

[42, 43], gave the following properties for arbitrary order derivatives and integrations:

(P1) The zeroth order differentiation and integration of a function leave the function un-

change.

(P2) The fractional order differentiation and integration of a function coincide with ordi-

nary derivative and integration when the order of differintegral is an integer.

(P3) Similar to classical order differentiation and integration, fractional order differentia-

tion and integration are linear.

Dq (αf(t) + βg(t)) = αDqf(t) + βDqg(t), (2.2.1)

where Dq is any form of fractional order differentiation.

(P4) For α, β ∈ R and f an element of L1[a, b], the following

Iαa Iβaf(t) = Iα+βa f(t) = IβaI

αaf(t), (2.2.2)

holds everywhere on [a, b].

There are numerous definitions available for fractional order integral and derivative, but

some of them causes difficulties in computations as well as other technical problems which

some times waste times and memory. Therefore, Caputo’s, Riemann and Riemann-Liouville

approaches attracted more attentions. On ward in this chapter, we review some definitions

and results need for further study in this thesis. For detail see [44, 45].

11

2.2.1 Riemann-Liouville fractional order integrals

The frequently used definition of fractional order integral is Riemann-Liouville’s fractional

integral whose derivation is based on Cauchy integral formula for n-fold integrals. For any

f(t) ∈ Ln[a, b], a, b ∈ R, the Cauchy’s integral formula is given by

Inaf(t) =

∫ t

a

∫ tn−1

a· · ·∫ t1

af(t)dtdt1 · · · dtn−1 =

1

Γ(n)

∫ t

a(t− s)n−1f(s)ds, n ∈ N. (2.2.3)

The Riemann-Liouville’s fractional order integral of order q ∈ R+ of a function f(t) ∈

(L1[a, b],R) on interval [a, b] ⊂ R, is defined by

Iqa+f(t) =

1

Γ(q)

∫ t

a(t− s)q−1f(s)ds, q > 0 (2.2.4)

provided that the integral on right hand side exists. The formula (2.2.4) represents the

integral of arbitrary order q > 0, but does not allow q = 0 which formally corresponds to

the identity operator. For this purposes we consider q → 0 and extend the aforementioned

definition by setting Iq0f(t) = f(t). For illustration purposes, we provide fractional integral

of some simple functions as

Iqa+t

p =1

Γ(q)

∫ t

a(t− s)q−1spds =

1

Γ(q)

∫ t

a(1− s

t)q−1tq−1spds

=1

Γ(q)tq+p

∫ 1

0yp(1− y)q−1dy,

(2.2.5)

using substitution y = st , we obtain I

qa+t

p = 1Γ(q) t

q+pβ(p + 1, q) = Γ(p+1)Γ(p+q+1) t

p+q. Similarly,

for the function f(t) = eλt, where λ is constant, we obtain

Iqa+e

λt =1

Γ(q)

∫ t

0(t− s)q−1eλsds =

eλt

Γ(q)

∫ t

0yq−1e−λydy, q > 0, (2.2.6)

where t− s = y. In view of (2.1.9), we can write (2.2.6) as

Iqa+e

λt = Et(q, λ) = tqE1,q+1(λt). (2.2.7)

Similarly

Iqa+ cos(λt) =

1

Γ(q)

∫ t

0yq−1 cos[λ(t− s)]ds = Ct(q, λ), Re(q) > 0. (2.2.8)

12

Remark 2.2.1. The fractional integral exist almost every where on [a, b] and belonging to

L1[a, b] for any f ∈ L1[a, b] and α ∈ R+.

We need the following theorem in the sequel.

Theorem 2.2.1. For q, β, ν ∈ R+ and λ ∈ R, the following result holds:

(Iq0[(s− a)γ−1Eβ,γ(λ(s− a)β)])(t) = (t− a)γ+q−1Eβ,γ+q(λ(t− a)β). (2.2.9)

Proof. By definitions of the Riemann-liouville’s fractional order integral and the generalized

Mittage-Leffler’s function, we have

(Iq0[(s− a)γ−1Eβ,γ(λ(s− a)β)])(t) =

∞∑

k=0

λk

Γ(βk + γ)Γ(k + 1)Iqa(t− a)kβ+γ−1

= (t− s)γ+q−1∞∑

k=0

λk(t− a)kβ

Γ(kβ + γ + q)Γ(k + 1)= (t− a)γ+q−1Eβ,γ−q(λ(t− a)β).

(2.2.10)

The convergence of the series in Eβ,γ gives permission to interchange the order of integration

and summation.

2.2.2 Fractional order derivatives

In this subsection, we provide in detail the definitions of fractional order derivatives in

Riemann-Liouville and Caputo’s sense. We use the notation D throughout the thesis for

Riemann-Liouville’s fractional order derivatives and the notation I for fractional integral.

2.2.3 The Riemann-Liouville’s fractional derivatives

Definition 2.2.1. For a given function f(t) ∈ Cn[a, b], the Riemann-Liouville’s fractional

order derivative of order q is defined as

Dqaf(t) =

1

Γ(n− q)

(d

dt

)n ∫ t

af(s)(t− s)n−q+1ds, n− 1 ≤ q < n , n ∈ N, (2.2.11)

provided that the right side is point wise defined on (a,∞), where n = [q] + 1 in case q not

an integer and n = q in case q is an integer, in such a situation the fractional derivative and

classical derivative coincide.

13

Example 2.2.1. The fractional order derivative of order 0 < q < 1 of f(t) = tp, p ≥ 0 is

given by

Dq0f(t) = D1[D1(1−q)f(t)] = D1[D1(1−q)tp] = D1[

Γ(p+ 1)

Γ(q − p+ 1) + 1tp−q+1]

= (p− q + 1)Γ(q + 1)

(p− q + 1)Γ(p− q + 1)tp−q =

Γ(p+ 1)

Γ(p− q + 1)tp−q,

(2.2.12)

similarly

Dq0eλt = t−qE1,−q+1(λt). (2.2.13)

Theorem 2.2.2. For q, β, ν ∈ R+ and λ ∈ R, the following result holds:

(Dq0[(s− a)γ−1Eβ,γ(λ(s− a)β)])(t) = (t− a)γ−q−1Eβ,γ−q(λ(t− a)β). (2.2.14)

Proof. By definitions of the Riemann-liouville’s fractional derivative and the generalized

Mittage-Leffler function, it follows that

(Dq0[(s− a)γ−1Eβ,γ(λ(s− a)β)])(t) =

∞∑

k=0

λk

Γ(βk + γ)Γ(k + 1)Dqa(t− a)kβ+γ−1

= (t− s)γ−q−1∞∑

k=0

λk(t− a)kβ

Γ(kβ + γq)Γ(k + 1)= (t− a)γ−q−1Eβ,γ−q(λ(t− a)β).

(2.2.15)

The convergence of the series in Eβ,γ gives permission to interchange the order of differen-

tiation and summation.

Theorem 2.2.3. Let f ∈ ACn[a, b] and n−1 ≤ q < n. Then Riemann-Liouville’s fractional

derivative Dqa exist almost everywhere on [a, b]. Moreover D

qaf ∈ Lp[a, b] for 1 ≤ p < 1

q and

Dqaf(t) =

n−1∑

i=0

Dif(a)

Γ(i− q + 1)(t− a)i−q + In−qa Dnf(t). (2.2.16)

Some further properties of Riemann-Liouville’s fractional derivative are given in the

following sequel.

Lemma 2.2.4. If p, q ∈ R+, p > q and f ∈ L1[a, b] then

DqaIpaf(t) = Ip−qa f(t), (2.2.17)

14

holds almost everywhere on [a, b]. If p = q then Riemann-Liouville’s fractional derivative

is left inverse of fractional integral. In some special cases it is also right inverse as in the

next lemma.

Lemma 2.2.5. For a > 0 and f(t) ∈ IqaL1[a, b]. Then I

qaD

qaf(t) = f(t).

However in general fractional derivative and fractional integrals do not commute as in

the following theorem.

Theorem 2.2.6. Let n ∈ N, n− 1 ≤ q < n and f ∈ L1[a, b]. Then , for In−qq f ∈ ACn[a, b],

the following holds

IqaDqaf(t) = f(t)−

n∑

i=1

(t− a)q−i

Γ(q − i+ 1)[Dq−i

a f(t)]t=a. (2.2.18)

2.2.4 The Caputo’s fractional derivative

It is clear from history of fractional calculus that Riemann-Liouville’s fractional derivative

have played important role in the development of the theory of fractional calculus. But due

to some demerits of the Riemann-Liouville’s fractional derivative, it is not applicable for

modeling of real world problems. Generally, derivative of a constant is zero but in case of

the Riemann-Liouville fractional derivative, it is (Dqa[C])(t) = Ct−q

Γ(1−q) is a function of t and

never zero except for a = −∞. However, for most of the real world problems, a must be

finite. It also causes problem in solving boundary value problems and for most of the initial

value problems its results are not interpretable. Therefore, in 1967, M. Caputo’s introduced

another concept of fractional order derivative, which is later on applied in the framework

of viscoelasticity in 1969 by Caputo and Mainardi. We use the notation cD in this thesis

for Caputo’s fractional derivative.

Definition 2.2.2. For a given function f(t) ∈ Cn[a, b], the Caputo’s fractional order deriva-

tive of order q is defined as

cDqaf(t) =

1

Γ(n− q)

∫ t

a

f (n)(s)

(t− s)q+1−n ds, n− 1 ≤ q < n , n ∈ N, (2.2.19)

15

provided the right side is point wise defined on (a,∞), where n = [q] + 1 in case q not

an integer and n = q in case q is an integer. When q = n then Caputo’s fractional

derivative and classical derivative coincides. The definition of fractional order derivative

either Riemann-Liouville’s fractional derivative or Caputo’s utilizes the Riemann-Liouville

fractional integral. For Riemann-Liouville’s fractional derivative and Caputo’s both we need

n = [q]. In case of Caputo’s, it is must while in case of the Riemann-Liouville fractional

derivative it is not compulsory and we may use n ≥ q. For a class of integrable functions, the

Riemann-Liouville’s fractional derivative exist while for Caputo’s derivative the integrability

of n times differentiable function is required. For this, we provide the following lemma

Lemma 2.2.7. If q ≥ 0 and f(t) = (t− a)p, n = [q], then

cDqaf(t) =

0, if p ∈ 0, 1, 2, ..., n − 1,Γ(p+ 1)(t− a)p−q

Γ(p− q + 1), if p ∈ N and p ≥ n or p ∈ N and p > n− 1.

(2.2.20)

In the following theorem, we provide relation between the Caputo’s and Riemann-

Liouville’s fractional order derivative.

Theorem 2.2.8. For q ≥ 0 and f(t) ∈ ACn[a, b], n = [q]. Then

cDqaf(t) = Da

[f(t)−

n−1∑

i=0

Dif(a)

Γ(i+ 1)(t− a)i

]. (2.2.21)

Another relation is given by

Theorem 2.2.9. For q ≥ 0 and f(t) ∈ ACn[a, b], n = [q] such that Dq and cDqa exist for

f(t). Then

cDqaf(t) = Dq

a[f(t)]−n−1∑

i=0

Dif(a)

Γ(i− q + 1)(t− a)i. (2.2.22)

From these (2.2.8) and (2.2.9) we conclude that Caputo’s and Riemann-Liouville’s frac-

tional order derivative are not equal. The equality holds if Dif(a) = 0. The following

theorem is more important in this thesis.

16

Theorem 2.2.10. For q ≥ 0 and f(t) ∈ ACn[a, b], n = [q]. Then

Iqa[cDq

af(t)] = f(t)−n−1∑

i=0

Dif(a)

Γ(i− q + 1)(t− a)i. (2.2.23)

Remark 2.2.2. For simplicity, we usually apply the following relation during the solutions

of fractional order differential equations.

Iα[Dαu(t)] = u(t) + C1tα−1 + C2t

α−2 + · · ·+ Cntα−n, (2.2.24)

where Ci ∈ R, for i = 1, 2, . . . , n. The above relation (2.2.24) also hold if we use cD instead

of D.

2.3 Fixed point theorems and some results from analysis

In this section, we give brief introduction of fixed point theorems and some useful results

from functional analysis which can be found in [46, 47, 48], which are used in this thesis.

The following result is known

Lemma 2.3.1. The space X = u : u ∈ C[a, b] such that Dqu ∈ Cn[a, b], 0 < p < 1

equipped with the norm ||u|| = maxt∈[a,b] |u(t)|+maxt∈[a,b] |Dpu(t)| is a Banach space. The

norms in the product space X ×X are defined by

||(u, v)|| = max||u||, ||v|| or ||(u, v)|| = ||u||+ ||v||.

Remark 2.3.1. The conclusion of the above lemma is also valid if we replace D by cD.

Definition 2.3.1. Let X ,Y be Banach spaces and T : C ⊂ X → Y be an operator. Then

T is called compact operator if

(i) T is continuous;

(ii) T maps bounded sets B ⊂ C into relatively compact sets, that is T (B) is compact.

17

Definition 2.3.2. Let X be normed space and T : X → X be an operator. Then T is

called Lipschitzian if there exist a constant k ≥ 0 such that

‖T (x) − T (y)‖ ≤ k‖x− y‖, for all x, y ∈ X . (2.3.1)

If k < 1 in (2.3.1), then T is called a contraction. If k = 1 then T is called non-expansive.

Definition 2.3.3. Let D ⊂ Rn. Then a set M ⊂ C(D) is called equi-continuous set if and

only if for every positive ε, we can find δ > 0 such that

‖T (x)− T (y)‖ ≤ ε, with ‖x− y‖ < δ, for all x, y ∈ D, T ∈M. (2.3.2)

.

Theorem 2.3.2. (The Arzela-Ascoli Theorem) Let D be a bounded subset of Rn and let

M be a subset of C(D). Then M is relatively compact if and only if it is bounded and

equi-continuous.

Definition 2.3.4. Let X be a Banach space and C ⊂ X , then C is called cone if and only if

(i) C is closed, nonempty and C 6= 0;

(ii) α, β ∈ R+ ∪ 0 and x, y ∈ C implies that αx+ βy ∈ C;

(iii) x ∈ C, −x ∈ mathcalC implies x = 0.

Remark 2.3.2. The Cone in product space X × Y is defined by

P = (u, v) ∈ X × Y : mint∈I

[u(t) + v(t)] ≥ 0,

where I = [a, b].

Definition 2.3.5. A cone C ⊂ X of an ordered Banach space X is partial ordered i.e u v

if and only if v − u ∈ C. Similarly a nonempty closed convex subset C ⊂ X is a cone if it

satisfies:

18

(i) u ∈ C, κ ≥ 0 ⇒ κu ∈ C;

(ii) u ∈ C,−u ∈ C ⇒ u− v = 0X ∈ C, where 0X is additive identity of Banach space X .

A cone C is normal cone if and only if for all u, v ∈ X , 0X u v ⇒ ||u|| ≤ κ||v||, κ > 0.

For all u, v ∈ X such that u ∼ v means that there exist a, b > 0 such that au v bv.

Obviously the relations denoted by ∼ is an equivalence relation, for which we define a set

Dh = u ∈ X : u ∼ h. Easily, we can show that Dh ⊂ C for h 0X .

Definition 2.3.6. An operator T : C → C is said to be λ-concave if and only if T (λu)

θλTu, for all θ ∈ (0, 1), u ∈ C, with λ ∈ (0, 1).

Definition 2.3.7. An operator T : C → C is said to be increasing if u, v ∈ C, u v ⇒ Tu

Tv.

Lemma 2.3.3. Let T : C → C be an increasing λ−concave operator where C be a normal

cone in real Banach space X . Let there exist u 0X such that Tu ∈ Dh. Then, T has a

fixed point u ∈ Dh.

Remark : For a positive integer n ≥ 1, then the set Rn is partial order set with 0Rn is

additive identity and the partial order defined by u = (u1, u2, ..., un), v = (v1, v2, ..., vn) ∈

Rn, u v ⇔ ui ≤ vi,∀i = 1, 2, ..., n.

Definition 2.3.8. For a nonempty set X , the mapping d : X × X → Rn is called a

generalized metric on X if for all x, y, and u, v ∈ X with v 6= x, v 6= u, v 6= y it satisfies:

(M1) d(u, v) = 0Rn ⇔ u = v, ∀u, v ∈ X ;

(M2) d(u, v) = d(v, u), ∀u, v ∈ X ; (symmetric property)

(M3) d(x, y) n d(x, v) + d(v, u) + d(u, y),∀x, y, u, v,∈ X (tetrahedral inequality).

Note: It is to be noted that the properties of convergent sequence, Cauchy sequence,

open/closed subset are same for generalized metric spaces as hold for usual metric spaces.

19

Definition 2.3.9. Let M = Mn,n ∈ Rn×n+ ,the system of all n× n matrices with positive

element. For any matrix A the spectral radius is defined by ρ(A) = sup|λi|, i = 1, 2, ..., n,

where λi, (i = 1, 2, ..., n) is the eigen values of the matrix A.

Definition 2.3.10. A function f(t, x) : [0, 1]×R → R is said to satisfies the Carathodeory

conditions, if the following hold:

(i) f(t, x) is Lebesgue measurable with respect to t for x ∈ R,

(ii) f(t, x) is continuous with respect to x for t ∈ [0, 1].

Definition 2.3.11. A function µ ∈ D is said to a lower solution of the operator equation

(I − T )z = 0 if (I − T )µ ≤ 0 and ν ∈ D is said to an upper solution of (I − T )z = 0 if

(I − T )ν ≥ 0.

Lemma 2.3.4. [49](Comparison theorem) Let h(t) ∈ X and r(t) ∈ Y. If h(t) satisfies the

inequality

−Dαh(t) ≤ −r(t)h(t), t ∈ (0, 1),

h(0) ≤ 0, h(1) ≤ 0,

then h(t) ≤ 0, for all t ∈ [0, 1].

(2.3.3)

Remark 2.3.3. Set of upper and lower solutions is denoted in this thesis by U = [u0, u∗0] ⊂ X .

2.3.1 Fixed point theorems

Fixed point theory provide important information on the existence and uniqueness of solu-

tions of equations including differential and integral equations. It has been emerged as an

effective and powerful tool for studying a wide range of problems arises in economics, fi-

nance, image reconstruction, ecology, transportation and network etc. Fixed point problems

include many nonlinear problems such as variational inequality problems, complementary

20

problems and saddle point problems as special cases. Fixed point theory has been got con-

siderable attentions from researchers in last century and there is vast literature since 1922

with Banach fixed point theorem.

Definition 2.3.12. For a given mapping T : X → Y, where X ,Y are Banach or simply

normed spaces, every point x which satisfy the following equation

T (x) = x (2.3.4)

is said to be a fixed point of T .

The famous and simplest result in fixed point theory was given by Banach in 1922,

which is usually called the Banach fixed point theorem or contraction principle.

Theorem 2.3.5. [50](The Banach contraction theorem) Let D be a closed subset of

a Banach space X and T : D → D. Then T has a unique fixed point u in D and for any

initial value x0 ∈ D, the successive approximation converges to u if

‖Tu− T u‖ ≤ k‖u− u‖, for all u, u ∈ D with k < 1.

Although Theorem (2.3.5) provide great information about solutions of some equations but

it is applicable to a limited classes of problems.

Theorem 2.3.6. [50](The Brouwer fixed point theorem) Let D be a nonempty, convex

and compact subset of a finite dimensional normed space and T : D → D is continuous

function. Then T has a fixed point in D.

But due to finite dimensionality, the direct applications of this theorem are limited,

however the condition of finite dimensionality is removed by the idea of compact operators.

Theorem 2.3.7. [50] (The Schauder fixed point theorem) Let D be a nonempty,

convex and compact subset of a Banach space X and T : D → X is compact operator such

that D maps into itself. Then, T has a fixed point in D.

21

Utilizing homotopy and embedding the given operator equation into continuum equa-

tion and then starting from one of the solutions of simpler equation to get the required

solution of the given problem. This important method is called Leray-Schauder continuum

principle which provides the fundamental concept about the existence of solutions of nonlin-

ear boundary value problems for differential equations. The first detail study of continuum

principle was introduced by Leray and Schauder (1934).

Theorem 2.3.8. [51](The Leray- Schauder continuum principle) Let

(i) the operator T : X → X is compact operator;

(ii) there exist κ > 0 and λ ∈ (0, 1) such that u = λTu implies that ‖u‖ ≤ κ.

Then the corresponding equation has a solution.

In the existence theory of differential and integral equations, the next theorem is widely

used and was derived from the Schauder fixed point theorem.

Theorem 2.3.9. [51](Nonlinear alternative of Leray-Schauder type) Let X be a

Banach space and C a nonempty convex subset of X and D be open in C with 0 ∈ D and let

T : D → C be a continuous and compact mapping. Then either,

(i) T has a fixed point in D or,

(ii) There exist u ∈ ∂D with u = λTu, where 0 < λ < 1.

Next, we provide a well known fixed point theorem given by Guo-Krasnoselskii’s and

mostly applicable in the existence theory of differential equations.

Theorem 2.3.10. [52](Guo-Krasnoselskii’s fixed point theorem) Let D be a closed

convex and nonempty subset of Banach space X . Let F, G be the operators such that

(i) Fx+Gy ∈ D whenever x, y ∈ D;

22

(ii) F is compact and continuous;

(iii) G is contraction mapping;

then there exist z ∈ D such that z = Fz +Gz.

For multiplicity of solutions of differential equations, the above results will not work

and hence we need other results. One of them is the fixed point theorem due to Guo-

Krasnoselskii’s and is given below

Theorem 2.3.11. [52] (Fixed point theorem of cone expansion and compression

of norm type) Let C be a cone of real Banach space X and let Ω1 and Ω2 be two bounded

open sets in X such that 0 ∈ Ω1, Ω1 ⊂ Ω2. Let the operator T : C ∩ (Ω2 \ Ω1) → C be

completely continuous such that either the following conditions hold

(i) ||Tu|| ≤ ||u||, for all u ∈ C ∩ ∂Ω1 and ||Tu|| ≥ ||u||, for all u ∈ C ∩ ∂Ω2;

(ii) ||Tu|| ≥ ||u||, for all u ∈ C ∩ ∂Ω1 and ||Tu|| ≤ ||u||, for all u ∈ C ∩ ∂Ω2,

then T has at least one fixed point in C ∩ (Ω2 \ Ω1).

For existence of triple positive solutions for differential equations, the following fixed

point theorem is useful.

Theorem 2.3.12. [53] (Leggett - Williams ’s fixed point theorem) Let C be a cone in a

real Banach space X , B = u ∈ C : ‖u‖ ≤ c, θ a nonnegative continuous concave function

on C such that θ(u) ≤ ‖u‖ for all u ∈ B, and C(θ, b, d) = u ∈ C : b ≤ θ(u), ‖u‖ ≤ d.

Suppose T : B → B is completely continuous and there exist constants 0 < a < b < d ≤ c

such that

(i) u ∈ C(θ, b, d) | θ(u) > b 6= ∅, and θ(Tu) > b, for u ∈ C(θ, b, d)

(ii) ‖Tu‖< a, for u ≤ a

23

(iii) θ(Tu) > b, for u ∈ C(θ, b, c) with ‖Tu‖ > d,

then T has at least three fixed points u1, u2, u3 with ‖u1‖ < a, b < θ(u2), a < ‖u3‖ with

θ(u3) < b.

The following fixed point theorems and result can be found in [54, 55], which are im-

portant for our work.

Theorem 2.3.13. (Perov’s fixed point theorem) Let (X , d) be a complete generalized

metric space and let T : X → X be an operator such that there exist a matrix A ∈Mm,m(R)

with d(Tu, Tv) ≤ Ad(u, v), for all u, v ∈ X . If ρ(A) < 1, then T has a fixed point x∗ ∈ X ,

further for any x0 the iterative sequence xn+1 = Tzn converges to x0.

Lemma 2.3.14. Let T : C → C be an increasing λ−concave operator where C be a normal

cone in real Banach space X . Let there exist u 0X such that Tu ∈ Dh. Then, T has a

fixed point u ∈ Dh.

2.4 Degree theory

Classical fixed point theorems such as Banach contraction principle and Schauder fixed point

theorem are commonly used for existence of solutions to boundary value problems. However,

the use of these results require stronger conditions on the nonlinear functions involved which

restricts their applicability to limited classes of problems and to a very specialized systems

of boundary value problems. In order to enlarge the class of boundary value problems and

to impose less restrictive conditions, one need to search for other sophisticated tools of

functional analysis. One such tool is topological degree theory which was first defined by

Brouwer, who showed that the degree is homotopy invariant (invariant among homotopies)

and used it to prove the Brouwer’s fixed point theorem. The degree of a continuous mapping

arises in many contexts and theories.

24

Definition 2.4.1. For any continuous mapping f : M → M, the degree is just the winding

number. For Ω ⊂ Rn is any open set, F : Ω → Rn, u ∈ f(∂Ω), degree is given by

Deg(f,Ω, u) =∑

x;f(x)=u

sgn(det(f ′(x))

).

In this thesis we use topological degree only. Large number of articles and monographs

are available in literature which uses degree theory arguments for existence of solutions

to boundary value problems of differential and integral equations. The study of nonlocal

Cauchy problems by means of topological degree theory using condensing maps is an active

area of research today. Some fundamental results and definitions which are required in this

work are follow.

Definition 2.4.2. [56] The Kuratowski’s measure of non-compactness µ : S → R+ is defined

as

µ(S) = infd > 0 : S admits a finite cover by sets of diameter ≤ d,

where for a Banach space X , S ∈ S and S ∈ P (X ) .

Proposition 2.4.1. [56] The Kuratowski measure µ satisfy the following properties:

(i) µ(S) = 0 if and only if S is relatively compact;

(ii) µ is a semi-norm, i.e, µ(λS) = |λ|µ(S), λ ∈ R and µ(S1 + S2) ≤ µ(S1) + µ(S2);

(iii) S1 ⊂ S2 implies µ(S1) ≤ µ(S2);µ(S1 ∪ S2) = maxµ(S1), µ(S2);

(iv) µ(convS) = µ(S);

(v) µ(S) = µ(S).

Definition 2.4.3. [56] Let F : Ω → X be continuous bounded map, where Ω ⊂ X . Then

F is µ-Lipschitz if there exists K ≥ 0, such that

µ(F (S)) ≤ Kµ(S), for all S ⊂ Ω bounded.

25

Further, F will be strict µ-contraction if k < 1.

Definition 2.4.4. [57] A function F is µ-condensing if

µ(F (S)) < µ(S), for all S ⊂ Ω bounded with µ(S) > 0.

In other words, µ(F (S)) ≥ µ(S) implies µ(S) = 0.

Here, we denote the class of all strict µ-contractions F : Ω → X by ϑCµ(Ω) and denoting

the class of all µ-condensing maps F : Ω → X by Cµ(Ω).

Remark 2.4.1. ϑCµ(Ω) ⊂ Cµ(Ω) and every F ∈ Cµ(Ω) is µ-Lipschitz with constant K = 1.

For the existence of solution via degree theory, the following results are important.

Proposition 2.4.2. If F,G : Ω → X are µ-Lipschitz with constants K and K′

respectively,

then F +G : Ω → X is µ-Lipschitz with constant K +K′

.

Proposition 2.4.3. If F : Ω → X is compact, then F is µ-Lipschitz with constant K = 0.

Proposition 2.4.4. If F : Ω → X is Lipschitz with constant K, then F is µ-Lipschitz with

the same constant K.

The following theorem due to F.Isaia [57], play important role for our results.

Theorem 2.4.5. Let F : X → X be µ-condensing mapping and

Ψ = x ∈ X : ∃ λ ∈ [0, 1] such that x = λFx.

If Ψ is a bounded set in X , so there exists r > 0 such that Ψ ⊂ Sr(0), then the degree

D(I − λF, Sr(0), 0) = 1, ∀ λ ∈ [0, 1].

Consequently, F ha at least one fixed point and the set of the fixed points of F lies in Sr(0).

26

2.5 Operational matrices

In many cases, exact analytical solutions of differential or integral equations are not avail-

able, therefore it is strongly felt by researchers of applied sciences to introduced some best

numerical scheme for approximate solutions of differential and integral equations. For clas-

sical differential equations this area has been explored by many mathematicians, while for

fractional calculus it is in initial stages. In recent years, A. Saadatmandi and M.Dehgan

[58], E. H. Doha et al. [59], introduced some operational matrices by using some orthogonal

polynomials together with some spectral methods such as Collocation and Tau method.

These operational matrices convert differential equations and integral equations to system

of algebraic equations which are easily solvable by computational softwares. The reason

of using polynomials is that they are incredibly useful mathematical tools, because they

are simply defined and can be quickly computed on computer system and represent large

numbers of functions. Their integration and differentiation are easy. With the help of

polynomials, spline curves are formed by piecing them together and any functions can be

approximated to the desired accuracy. The area devoted to find approximate solutions

of classical differential equations has been explored and plenty of papers are available on

it. For this purposes the commonly used polynomials are Legendre polynomials, Bernstein

polynomials and Jacobi’s polynomials etc. However in the field of fractional calculus it

is in initial stages and very few papers are available on it. It is mainly founded that the

aforementioned polynomials provided a best tools for numerical solutions of fractional order

differential equations as compare to the Wavelet analysis and other methods like Adomian

decomposition method, homotopy analysis method and homotopy perturbation method,

variational iteration method etc. Because these methods are applicable mostly for initial

value problems and for boundary value problems, more computational work is required. On

the other hand operational matrices methods are easy to use and implement for both initial

27

and boundary value problems of fractional order differential equations.

2.5.1 Legendre polynomials

The most popular orthogonal polynomials are Legendre polynomials, which are defined on

[−1, 1]. They are determined from the following recurrence relation:

Li+1(x) =2i+ 1

i+ 1xLi(x)−

i

i+ 1Li−1(x), i = 0, 1, 2, 3, · · · , (2.5.1)

where L0(x) = 1 and L1(x) = x. By using the suitable substitution t = x+12 , t ∈ [0, 1], the

Shifted Legendre polynomials are obtained

Pi+1(t) =2i+ 1)(2t − 1)

i+ 1Pi(t)−

i

i+ 1Pi−1, i = 1, 2, · · · , (2.5.2)

where P0(t) = 1 and P1(t) = 2t−1. The analytical form of the aforesaid shifted polynomial

is

Pi(t) =

i∑

k=0

(−1)i+kΓ(i+ k + 1)tk

Γ(i− k + 1)(Γ(k + 1))2(2.5.3)

and the orthogonality condition is given by

∫ 1

0Pi(t)Pj(t)dt =

1

2i+ 1, for i = j,

0, for i 6= j.

(2.5.4)

Any function f(t) ∈ C[0, 1] can be approximated by Shifted Legendre polynomials as

f(t) ≈m∑

j=0

cjPj(x), where cj = (2j + 1)

∫ 1

0f(t)Pj(t)dt. (2.5.5)

In vector notation (2.5.5) can be written as

f(t) = KTM PM (t), (2.5.6)

whereM = m+1, KTM is the coefficient vector and PM (t) isM terms function vector. The

corresponding error analysis is computed by the following results available in [59].

28

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1

2

3

4

5

6

7

8

tf(t

)

Exact Function

Approximate solution

Figure 2.4: Approximation of f(t) = e2t + sin(t2) by the Legendre polynomials.

Lemma 2.5.1. (Error analysis) (i) Let f(t) ∈ CM+1[0, T ], then KTM PM (t) is the best

approximations of f(t) on [0, 1], then the error approximation is given by

‖f(t)−KTM PM (t)‖2 ≤ 1

4MM+1maxt∈[0,1]

|f (M+1)(t)|.

(ii) Let f(t) ∈ Y , where Y is the spanning set of first M Legendre polynomials and

f(t) =m∑

i=0

CiPTi (t), then |Ci| '

C

(δi)m‖fm‖,

and ‖f(t)−m∑

i=0

CiPTi (t)‖2 =

∞∑

i=m

εiC2i ,

where εi = i(i+ 1) and Ci =2j + 1

T

∫ T

0y(t)P Ti (t)dt,

where C is constant and m is taken in such away that y2m ∈ Y.

The two dimensional Legendre polynomials of order M are defined by

Pn(x, y) = Pa(x)Pb(y), n =Ma+ b+ 1, a, b = 0, 1, 2, · · · ,m (2.5.7)

and the orthognathy is given by

∫ 1

0

∫ 1

0Pa(x)Pb(y)Pc(x)Pd(y)dxdy =

1

(2a+ 1)(2b + 1), if a = c, b = d,

0, otherwise.

(2.5.8)

29

Any function U(x, y) ∈ C([0, 1] × [0, 1]) can be approximated in terms of Pn(x, y) as

U(x, y) ≈m∑

a=0

m∑

b=0

CabPa(x)Pb(y), (2.5.9)

where

Cab = (2a+ 1)(2b + 1)

∫ 1

0

∫ 1

0u(x, y)Pa(x)Pb(y)dxdy. (2.5.10)

In vector notation, (2.5.9) can be written as

U(x, y) ≈M2∑

n=1

CabPn(x, y) = KTM2Φ(x, y), (2.5.11)

where n =Ma+b+1 and KM2 isM2×1 coefficient column vector and ΦM2(x, y) isM 2×1

column vector of functions defined by

ΦM2(x, y)

=(ψ11(x, y) · · · ψ1M (x, y) ψ21(x, y) · · · ψ2M (x, y) · · · ψMM (x, y)

)T,

(2.5.12)

where ψi+1,j+1(x, y) = (Pi(x)), (Pj(y)) , i, j = 0, 1, 2, · · · ,m.

Lemma 2.5.2. [60](Error analysis of approximations) For sufficiently smooth func-

tion U(x, y) on [0, 1] × [0, 1], the error of the approximation is given by

‖U(x, y)− PM,M (x, y)|2 ≤ (C1 + C2 + C31

MM+1)

1

MM+1,

where

C1 =1

4max

(x,y)∈[0,1]×[0,1]| ∂

M+1

∂xM+1g(x, y)|, C2 =

1

4max

(x,y)∈[0,1]×[0,1]| ∂

M+1

∂yM+1g(x, y)|

C3 =1

16max

(x,y)∈[0,1]×[0,1]| ∂2M+2

∂xM+1∂yM+1g(x, y)|.

The inequality is also satisfied if PM,M (x, y) is interpolating polynomial at point (xi, yj).

30

2.5.2 Bernstein polynomials

The Bernstein polynomials Bi,m(t) [61, 62, 63], are defined on [0, 1] by

Bi,m(t) =

m

i

ti(1− t)m−i, for i = 0, 1, 2, · · · ,m, (2.5.13)

where m

i

=

m!

(m− i)!i!, (2.5.14)

which on furthers simplification can be written in most simplified form as

Bi,m(t) =

m−i∑

k=0

Θ(i,k,m)tk+i, i = 0, 1, 2...m, (2.5.15)

where

Θ(i,k,m) = (−1)k

m

i

m− i

k

. (2.5.16)

In Hilbert space H = L2[0, 1], the inner product is defined by < f, g >=∫ 10 f(x).g(x)dx.

If Y = spanB0,m, B1,m, · · · , Bm,m is finite dimensional closed subspace, and f ∈ H is an

arbitrary element, then its best approximation is unique in Y . This terminology can be

achieved by using y0 ∈ Y and for all y ∈ Y we have ‖ f−y0 ‖≤‖ f−y ‖. Further any function

f(t) can be approximated in terms of Bernstein polynomials as f(t) ≈ ∑mj=0CjBi,m(t), i =

0, 1, 2, . . . ,m. The phenomenon is shown in Figure 2.5.

Lemma 2.5.3. Convergence analysis: Assume that the function g ∈ Cm+1[0, 1], that

is, m+1 times continuously differentiable function and let X = 〈B0,m, B1,m, . . . , Bm,m〉. If

CTΨ(x) is the best approximations of g out of X, then the error bound is presented as

‖g −CTΨ‖2 ≤√2MS

2m+32

Γ(m+ 2)√2m+ 3

,

where M = maxt∈[0,1] |g(m+1)(t)|, S = max1− t0, t0.

31

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−3

−2

−1

0

1

2

3

t

f(t)

Exact function

Approximate function

Figure 2.5: Approximation of f(t) = 2 cos(t)−t3 sin(t)−t2 cos(t)+e−2t in terms of Bernsteinpolynomials.

Proof. In view of Taylor’s polynomials, we have

F (t) = g(t0) + (t− t0)g(1)(t0) +

(t− t0)2

Γ3g(2) + . . .+

(t− t0)m

Γ(m+ 1)g(m),

from which we know that

|g − F (t)| = |g(m+1)(η)|(t− t0)m+1

Γ(m+ 2), where η ∈ (0, 1).

Due the best approximation CTΨ(t) of g, so we have

‖g −CTΨ(t)‖22 ≤ ‖g − F‖22 =∫ 1

0(g(t) − F (t))2dt

=

∫ 1

0

[|g(m+1)(η)|(t − t0)

m+1

Γ(m+ 2)

]2dt

≤ M2

Γ2(m+ 2)

∫ 1

0(t− t0)

2m+2dt ≤ 2M2S2m+3

Γ2(m+ 2)(2m+ 3)

⇒ ‖g −CTΨ(t)‖2 ≤√2MS

2m+32

Γ(m+ 2)√

(2m+ 3).

Remark 2.5.1. For mathematical convenience, we usually set Bi,n(t) = 0 for i < 0 and

i > n.

32

2.5.3 The shifted Jacobi polynomials and its fundamental properties

In this subsections, we provide basic properties of Shifted Jacobi polynomials [64, 65, 66].

The famous Jacobi polynomials of two parameters P(a,b)i (x) is defined over the interval

[−1, 1] are given by

P(a,b)ξ,i (x) =

(a+ b+ 2i− 1)[a2 − b2 + x(a+ b+ 2i− 2)(a+ b+ 2i− 2)]

2i(a + b+ i)(a+ b+ 2i− 2)P

(a,b)i−1 (x)

− (a+ i− 1)(b + i− 1)(a+ b+ 2i)

i(a+ b+ i)(a+ b+ 2i− 2)P

(a,b)i−2 (x), i = 2, 3, · · · ,

where P(a,b)0 (x) = 1, P

(a,b)1 (x) =

a+ b+ 2

2x+

a− b

2.

(2.5.17)

By the substitution x = 2tξ−1 , a new version of the above polynomials known as analytical

form of the Shifted Jacobi’s polynomials over the interval [0, T ] is obtained. The analytical

form of Shifted Jacobi polynomials is given by the relations

P(a,b)ξ,i (t) =

i∑

k=0

(−1)i−kΓ(i+ b+ 1)Γ(i+ k + a+ b+ 1)

Γ(k + b+ 1)Γ(i+ a+ b+ 1)Γ(i − k + 1)Γ(k + 1)ξktk, (2.5.18)

where

P(a,b)ξ,i (0) = (−1)i

Γ(i+ a+ 1)

Γ(b+ 1)Γ(i + 1), P

(a,b)ξ,i (t) =

Γ(i+ a+ 1)

Γ(a+ 1)Γ(i+ 1).

Some other special orthogonal polynomials which are interrelated with the Shifted Jacobi’s

polynomials are

(i) Pξ,i(t) = P(0,0)ξ,i (t) is the shifted Legendre polynomials by putting a = b = 0 in (2.5.18);

(ii) Tξ,i(t) =Γ(i+1)Γ( 1

2)

Γ(i+ 12)P

(−12,−1

2)

ξ,i (t), is said to be shifted Chebyshev polynomials by assign-

ing a = b = − 12 in (2.5.18);

(iii) Uξ,i(t) =Γ(i+2)Γ( 1

2)

Γ(i+ 32)P

( 12, 12)

ξ,i (t), is called shifted Chebyshev polynomials of second kind

when a = b = 12 in (2.5.18);

(iv) Caξ,i(t) =

Γ(i+1)Γ(a+ 12)

Γ(i+a+ 12)P

(a− 12,b− 1

2)

ξ,i (t) shifted Gegenbauer (Ultraspherical) polynomials

by putting a = b, in (2.5.18);

33

(v) Vξ,i(t) = (Γ(2i+1))(Γ(2i−1))P

( 12,−1

2)

ξ,i (t), is called shifted Chebyshev polynomials of third kinds,

when we put a = 12 , b =

−12 in (2.5.18);

(vi) Wξ,i(t) = (Γ(2i+1))(Γ(2i−1))P

(−12, 12)

ξ,i (t), is said to be Shifted Chebyshev polynomials of fourth

kinds if a = −12 , b =

12 in (2.5.18).

The orthogonality condition of the Shifted Jacobi polynomials is given by

∫ ξ

0P

(a,b)ξ,j (t)P

(a,b)ξ,i (t)W

(a,b)ξ (t)dt = Ω

(a,b)ξ,j δji,

where δji = 1, if i = j, other wise δji = 0,

(2.5.19)

the weight function is given by W(a,b)ξ (t) = (ξ − t)atb, and

Ω(a,b)ξ,j (t) =

ξa+b+1Γ(j + a+ 1)Γ(j + b+ 1)

(2j + a+ b+ 1)Γ(j + 1)Γ(j + a+ b+ 1). (2.5.20)

Any squared integrable function f(t) over [0, ξ], can be approximated in terms of Shifted

Jacobi polynomials as

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−0.5

0

0.5

1

1.5

2

t

f(t)

Exact function

Approximate function

Figure 2.6: Approximation of f(t) = e−sin(2t) + t6(1 − t3) + cos(8t) by the Shifted Jacobipolynomials at a = b = 0, ξ = 1.

f(t) 'm∑

k=0

BjP(a,b)ξ,k (t) = KT

MΦM (t), (2.5.21)

where the Shifted Jacobi coefficient vector is denoted byKM andΦM(t) isM terms function

vector. The approximation of a function f(t) is provided in Figure 2.6. Also M = m +

34

1, when m → ∞ the approximations converges to the exact value of the function. The

coefficient Bj can be calculated by using (2.5.18)−(2.5.21) as

Bj =1

Ω(a,b)ξ,j

∫ ξ

0W

(a,b)ξ (t)f(t)P

(a,b)ξ,i (t)dt, i = 0, 1, · · · . (2.5.22)

Convergence analysis : For convergence analysis consider a sufficiently smooth func-

tion f(t) ∈ C[0, ξ]. Consider∏M (t) is the space of M -terms Jacobi polynomials. Assume

that FM (t) is the best approximation in∏M (t). Then for any polynomials PM (t) of degree

≤M in variable t gives

||f(t)− FM (t)||2 ≤ ||f(t)− PM (t)||2, (2.5.23)

the above inequality (2.5.23) also holds for any interpolating polynomials PM (t) of the

function f(t) at points (ti) for ti = 0 + i(ξ−0)M . Then by same procedure as applied in [66],

we can obtain

||f(t)− PM (t)||2 ≤(

D1

MM+1

), (2.5.24)

whereD1 =14 maxt∈[0,ξ]

∣∣∣∣ ∂M+1

∂tM+1 f(t)

∣∣∣∣. The accuracy of this procedure is given in the following

lemma.

Lemma 2.5.4. Let f(t) ∈ ∏M (t) and f(t) =∑m

k=0CkP(a,b)ξ,k (t), then

|Ck| 'C

(λk)m||Fm||

and ||f(t)−m∑

k=0

CkP(a,b)ξ,k ||2 =

∞∑

k=m

γkC2k

where λk = k(k + a+ b+ 1) and Ck =1

D(a,b)ξ,j

∫ 1

0y(t)P

(a,b)ξ,k (t)W

(a,b)ξ (t)dt.

(2.5.25)

C is constant and m be chosen in any way such that y(2m) ∈∏M (t). Also we have FM =

1

W(a,b)(t)ξ

LF(M−1)(t) =

(L

W(a,b)T (t)

)mF (t), where L is the Strum-Liouville operator and g(0) =

g(t).

Chapter 3

Multi-point boundary valueproblems

This chapter is devoted to the study of existence and uniqueness of solutions to some classes

of nonlinear multi point boundary value problems for fractional order differential equations.

Existence and uniqueness of positive solutions to initial value problems for both ordinary

and fractional order differential equations are well studied and a plenty of research articles

are available in literature. However, the theory of existence of solutions to boundary value

problems corresponding to fractional differential equations have attracted the attention of

researchers in last few decades and some useful results have been developed, we refer to

[67, 68, 69, 70, 71, 72, 73, 74, 75] and the references therein. In these cited reference, ex-

istence of at least one solution is studied with the tools of classical fixed point theory. On

the other hand, multi-point boundary value problems for fractional differential equations

is in the initial stages and few results are available in the literature. The existence and

uniqueness of multi-point boundary value problems has also attracted some attention. The

coincidence degree theory approach to study existence of solutions of fractional order differ-

ential equations is quite recent and only few results can be found in the literature dealing

with boundary value problems of fractional order, see[76, 77, 78]. We use the coincidence

degree theory approach for condensing maps to handel such problems. Therefore, we study

35

36

a multi point boundary value problem with nonlinear nonlocal boundary conditions. Suf-

ficient conditions for existence and uniqueness of solution are developed with the help of

topological degree theory. Furthermore, with the help of Schauder fixed point theorem, we

study existence and uniqueness to a highly nonlinear multi-point boundary value problems

of fractional differential equations. Such type of problems are also studied by mean of other

fixed point theorem for two point, three point and anti periodic boundary conditions. For

detail, we refer few of them in [79, 80, 81].

Organization of the chapter: This chapter has three sections. Section 3.1 is devoted to

the study of existence, uniqueness and multiplicity of positive solutions to a class of rela-

tively simple nonlinear multi-point boundary value problems for fractional order differential

equations. In section 3.2 is devoted to highly nonlinear class of multi-point boundary value

problem [82]. We develop necessary and sufficient conditions for existence and uniqueness

of positive solutions to a more general and complicated class of multi-point boundary value

problems with nonlinear boundary conditions. We use topological degree theory approach

to study the concerned problem. The results of this section are published . In section 3.3,

we establish necessary and sufficient conditions for existence and uniqueness of solutions

to a more general class [83], of multi-point boundary value problems where the nonlinear

function explicitly depends on Caputo’s derivative.

3.1 Multi-point boundary value problems

In this section, we consider the following class of multi-point boundary value problems of

nonlinear fractional order differential equations of the form

−Dq0+u(t) = f(t, u(t)), 1 < q ≤ 2, 0 < t < 1

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi),(3.1.1)

37

where δi, ηi ∈ (0, 1) withm−2∑i=1

δiηq−1i < 1, and f : [0, 1] × [0,∞) → [0,∞) is continuous.

We obtain sufficient conditions for existence, uniqueness and multiplicity results of positive

solutions to the boundary value problem (3.1.1) and provide some examples.

Lemma 3.1.1. For y(t) ∈ C[0, 1], the linear boundary value problem

Dq0+u(t) + y(t) = 0; 0 < t < 1, 1 < q ≤ 2,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi),(3.1.2)

has a unique solution of the form u(t) =∫ 10 G(t, s)y(s) ds, where the Green function G(t, s)

is given by

G(t, s) =1

Γ(q)

tq−1

1−λ[(1− s)q−1 −

m−2∑j=i

δj(ηj − s)q−1]− (t− s)q−1; s ≤ t, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1,

tq−1

1−λ[(1− s)q−1 −

m−2∑j=i

δj(ηj − s)q−1]; t ≤ s, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1.(3.1.3)

Proof. In view of Lemma (2.2.6), we obtain

u(t) = −Iq0+y(t) + C1t

q−1 + C2tq−2, C1, C2 ∈ R (3.1.4)

The boundary condition u(0) = 0 implies C2 = 0 and the condition u(1) =m−2∑i=1

δiu(ηi),

yields C1 =1

1−λ[Iq0+y(1)−

m−2∑i=1

δiIq0+y(ηi)

], where λ =

m−2∑i=1

δiηq−1i < 1. Hence, (3.1.3) takes

the form

u(t) = −Iq0+y(t) +tq−1

1− λ

[Iq0+y(1)−

m−2∑

i=1

δiIq0+y(ηi)

]. (3.1.5)

We discuss several cases: for 0 ≤ t ≤ η1, we write (3.1.5) as

u(t) =

∫ t

0

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)]y(s) ds

+tq−1

Γ(q)(1− λ)

∫ η1

t

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)y(s) ds

+m−2∑

i=2

∫ ηi

ηi−1

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)y(s) ds+

∫ 1

ηm−2

(1− s)q−1y(s) ds.

38

For ηl−1 ≤ t ≤ ηl, 2 ≤ l ≤ m− 2 , we write (3.1.5) as

u(t) =

∫ η1

0

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=1


+

m−2∑

i=2

∫ ηi

ηi−1

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1 − λ)

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1)+ (1− s)q−1

]y(s) ds

+

∫ t

ηl−1

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=l


+tq−1

Γ(q)(1 − λ)

[ ∫ ηl

t

((1− s)q−1 −

m−2∑

j=l


+m−2∑

i=l+1

∫ ηi

ηi−1

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1)y(s) ds+

∫ 1

ηm−2

(1− s)q−1y(s) ds.

For ηm−2 ≤ t ≤ 1, we write (3.1.5) as

u(t) =

∫ η1

0

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=1


+

m−2∑

i=2

∫ ηi

ηi−1

[− (t− s)q−1

Γ(q)+

tq−1

Γ(q)(1 − λ)

((1− s)q−1 −

m−2∑

j=i


+tq−1

Γ(q)(1 − λ)

[ ∫ t

ηm−2

(1− λ)(t− s)q−1 + (1− s)q−1y(s) ds+

∫ 1

t(1− s)q−1y(s) ds

].

Therefore, the unique solution of the boundary value problem (3.1.1) is given by u(t) =∫ 10 G(t, s)y(s) ds, where G(t, s) is defined by (3.1.3).

We discuss some properties of the Green’s function G(t, s) in the form of the following

lemma.

3.1.1 Properties of Green’s function

Lemma 3.1.2. The Green’s function defined by (3.1.3), satisfies the following conditions:

(i) G(t, s) > 0, for t, s ∈ (0, 1);

(ii) There exists a positive function γ ∈ C(0, 1) such that

minηi−1≤t≤ηi

G(t, s) ≥ γ(s) max0≤t≤1

G(t, s) = γ(s)G(s, s), for 0 < s < 1. (3.1.6)

39

Proof. (i) The expression forG(t, s) in (3.1.3), clearly shows that G(t, s) > 0, for s, t ∈ (0, 1).(ii) Let us define

g1(t, s) =tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1)− (t− s)q−1

Γ(q),

g2(t, s) =tq−1

Γ(q)(1− λ)

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1),

then, ∂g1(t,s)∂t < 0 and ∂g2(t,s)

∂t > 0 for s ≤ t, which implies that g1(t, s) is decreasing andg2(t, s) is an increasing function for s ≤ t. It follows that G(t, s) is decreasing with respectto t for s ≤ t and increasing with respect to t for t ≤ s. Consequently,


G(t, s) =

g1(ηi, s), s ∈ (0, ηi−1],ming1(ηi, s), g2(ηi−1, s), s ∈ [ηi−1, ηi],g2(ηi−1, s), s ∈ [ηi, 1),

which implies that


G(t, s) =

g1(ηi, s), s ∈ (0, r],g2(ηi−1, s), s ∈ [r, 1),

where ηi−1 < r < ηi. Further, we note that

max0≤t≤1

G(t, s) = G(s, s) =sq−1

Γ(q)(1− λ)

[(1− s)q−1 −

m−2∑

j=i

δi(ηi − s)q−1],

for s ∈ (0, 1). As a result

γ(s) =

(ηis

)q−1 − (ηi−s)q−1

sq−1

(1−λ)

[(1−s)q−1−

m−2∑j=i

δj(ηj−s)q−1] , s ∈ (0, r],

(ηi−1

s

)q−1, s ∈ [r, 1).

3.1.2 Existence of at least one solution

Now, we study existence of at least one solution of the boundary value problem (3.1.1) .

In view of Lemma 3.1.1, the boundary value problem (3.1.1) is equivalent to the integral

equation

u(t) =

∫ 1

0G(t, s)f(s, u(s))ds (3.1.7)

40

and by a solution of the Boundary value problem (3.1.1), we mean a solution of the integral

equation (3.1.7), that is, a fixed point of the operator T : C → C defined by

Tu(t) =

∫ 1

0G(t, s)f(s, u(s))ds. (3.1.8)

Lemma 3.1.3. For nonnegative real-valued functions m, n ∈ L[0, 1] such that

f(t, u) ≤ m(t) + n(t)u, for almost every t ∈ [0, 1], and all u ∈ [0,∞), (3.1.9)

the operator T defined by (3.1.8) is completely continuous.

Proof. Due to nonnegativity and continuity of G(t, s) and f(t, s), the operator T is contin-uous. For each u ∈ Ω = u ∈ C : ‖u‖ ≤M, M > 0, we have

|Tu(t)| =∣∣∫ 1

0G(t, s)f(s, u(s))ds

∣∣ ≤∫ 1

0G(t, s)(m(s) + n(s)u(s))ds

≤∫ 1

0G(s, s)m(s)ds+M

∫ 1

0G(s, s)n(s)ds = l,

which implies that T (Ω) is bounded.

For equi-continuity of T : C → C, take t1, t2 ∈ [0, 1], such that t1 < t2 with t2 − t1 < δ

and taking ε > 0 and 1 < q ≤ 2, such that δ = 12

(εqA

) 1q−1

. Then for u ∈ Ω, we claim that

|Tu(t2)− Tu(t1)| < ε.. Then for u ∈ Ω, we have

|Tu(t2)− Tu(t1)| =∣∣∫ 1

0

[G(t2, s)f(s, u(s))−G(t1, s)f(s, u(s))

]ds∣∣

≤∫ 1

0|G(t2, s)−G(t1, s)|(m(s) + n(s)M)ds,

which in view of the continuity of G(t, s) and K = maxm(s) + n(s)M : s ∈ [0, 1] impliesthat

|Tu(t2)− Tu(t1)| ≤ 1

Γ(q)

∣∣tq−12 − tq−1

1

∣∣1∫

0

1

1− λ

[(1− s)q−1 −

m−2∑

j=1

δi(ηi − s)q−1]Kds

≤ 1

Γ(q)

∣∣tq−12 − tq−1

1

∣∣ K

q(1− λ).

Further, we explain the above process as in the following cases:Case I. δ ≤ t1 < t2 < 1 and using Mean value theorem on |tq−1

2 − tq−11 |, we have

|Tu(t2)− Tu(t1)| ≤ K

Γ(q)q(1− λ)(tq−1

2 − tq−11 ) <

K(q − 1)

Γ(q)q(1− λ)δq−2(t2 − t1)

< qAδq−1 = ε.

41

Case II. 0 ≤ t1 < δ, t2 < 2δ, we have

|Tu(t2)− Tu(t1)| ≤ K

Γ(q)q(1 − λ)(tq−1

2 − tq−11 ) <

K(q − 1)

Γ(q)q(1− λ)tq−22

< qA(2δ)q−1 = ε.

Hence T : C → C is equi-continuous. By Arzela-Ascoli theorem, we conclude that theoperator T : C → C is completely continuous.

Choose M =(∫ 1

0 G(s, s)ds)−1

and N =(∫ ηi

ηi−1γ(s)G(s, s)ds

)−1.

Theorem 3.1.4. Assume that the condition (3.1.9) of Lemma 3.1.1 holds and there existtwo positive constants r2 > r1 > 0 such that

f(t, u) ≤Mr2, for (t, u) ∈ [0, 1] × [0, r2] and f(t, u) ≤ Nr1, for (t, u) ∈ [0, 1] × [0, r1].(3.1.10)

Then the Boundary value problem (3.1.1) has at least one positive solution u such thatr1 ≤ ‖u‖ ≤ r2.

Proof. In view of (3.1.8) and Lemma 3.1.3, T is completely continuous. By Schauder fixedpoint theorem, T has a fixed point, Tu = u.

Define Ω2 = u ∈ C : ‖u‖ < r2. For u ∈ ∂Ω2, we have 0 ≤ u(t) ≤ r2 for all t ∈ [0, 1]and using (3.1.10) and Lemma 3.1.1, it follows that

‖Tu‖ = max0≤t≤1

∫ 1

0G(t, s)f(s, u(s))ds ≤Mr2

∫ 1

0G(s, s)ds = r2 = ‖u‖.

Similarly, define Ω1 = u ∈ C : ‖u‖ < r1. For u ∈ ∂Ω1, we have 0 ≤ u(t) ≤ r1 for allt ∈ [0, 1] and for t ∈ [ηi−i, ηi], we have

‖Tu‖ =

∫ 1

0G(t, s)f(s, u(s))ds ≥

∫ 1

0γ(s)G(s, s)f(s, u(s))ds

≥ Nr1

∫ ηi

ηi−1

γ(s)G(s, s)ds = r1 = ‖u‖.

Now, we provide an example to show the applicability of the result.

Example 3.1.1. For the boundary value problem

D320+u(t) + u2 +

cos t

4+

1

5= 0, 0 < t < 1,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi) =1

5,

(3.1.11)

42

We have M =(∫ 1

0 G(s, s)ds)−1

≈ 1.088 and N =

(∫ 3414

γ(s)G(s, s) ds

)−1

= 4.914. Choos-

ing r1 =125 , r2 =

34 , we obtain

f(t, u) = u2 +cos t

4+

1

5≤ 1.45 ≤Mr2, for (t, u) ∈ [0, 1] ×

[0,

3

4

],

f(t, u) = u2 +cos t

4+

1

5≥ 1

5≥ Nr1, for (t, u) ∈ [0, 1] ×

[0,

1

25

].

By Theorem 3.1.4, the Boundary value problem (3.1.11) has at least one solution u suchthat 1

25 ≤ ‖u‖ ≤ 34 .

3.1.3 Uniqueness of solutions

Theorem 3.1.5. Let there exists h(t) ∈ L[0, 1] such that1∫0

G(s, s)h(s)ds < 1 and

|f(t, u)− f(t, v)| ≤ h(t)|u − v|,

for almost t ∈ L[0, 1], and u, v ∈ [0,∞), then the Boundary value problem (3.1.1) has aunique positive solution.

Proof. For u, v ∈ C, we have

|Tu(t)− Tv(t)| =∣∣

1∫

0

G(t, s)(f(s, u(s)− f(s, v(s))ds∣∣ ≤

1∫

0

G(t, s)|(f(s, u(s)− f(s, v(s))|ds,

≤1∫

0

G(s, s)h(s)|u(s) − v(s)|ds ≤1∫

0

G(s, s)h(s)ds‖u − v‖ = α‖u− v‖,

where α =1∫0

G(s, s)h(s)ds < 1. Therefore, by Banach contraction principle, The Boundary

value problem (3.1.1) has a unique solution.

Example 3.1.2. Consider the boundary value problem

D320+u(t) +

e2tu

4(1 + e2t)(1 + u)+ cos2 t+ 1 = 0, 0 < t < 1,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi) =1

5.

(3.1.12)

Here, f(t, u) = e2tu4(1+e2t)(1+u)

+ cos2 t+ 1 and taking h(t) = e2t

4(1+e2t), then

∫ 1

0G(s, s)h(s)ds ≤

∫ 1

0

√s

5+

2√s(1− s)√π

e2s

4(1 + e2s)ds ≤

∫ 1

0

√s

5+

2√s(1− s)√π

ds = 0.2990

43

and|f(t, u)− f(t, v)| ≤ h(t)|u− v|, for (t, u), (t, v) ∈ [0, 1] × [0,∞).

By Theorem 3.1.5, the Boundary value problem (3.1.1) has a unique positive solution.

3.1.4 Multiplicity: Existence of at least three positive solutions

Theorem 3.1.6. Assume that the condition (3.1.9) of Lemma 3.1.1 holds and there existspositive constants 0 < a < b < c such that

(i) f(t, u) < Ma, for (t, u) ∈ [0, 1] × [0, a];

(ii) f(t, u) ≥ Nb, for (t, u) ∈ [ηi−1, ηi]× [b, c];

(iii) f(t, u) ≤Mc, for (t, u) ∈ [0, 1] × [0, c];

then, the Boundary value problem (3.1.1) has at least three positive solutions u1, u2, andu3 such that

max0≤t≤1

|u1(t)| < a, b < minηi−1≤t≤ηi

|u2(t)| ≤ c, a < max0≤t≤1

|u3(t)| < c, minηi−1≤t≤ηi

|u3(t)| < b.

(3.1.13)

Proof. For u ∈ Cc, the relation

‖Tu‖ = max0≤t≤1

∣∣∫ 1

0G(t, s)f(s, u(s))ds

∣∣ ≤∫ 1

0G(s, s)f(s, u)s))ds ≤

∫ 1

0G(s, s)Mcds ≤ c

follows from (iii) and completely continuity of T : Cc → Cc follows from Lemma 3.1.1.Define nonnegative continuous concave functional

θ(u) = minηi−2≤t≤ηi

|u(t)|, (3.1.14)

choose u(t) = b+c2 , 0 ≤ t ≤ 1. Then using (3.1.14), we have u(t) = b+c

2 ∈ C(θ, b, c), θ(u) =θ( (b+c)2 ) > b implies that u ∈ C(θ, b, c) |θ(u) > b 6= ∅. Hence, if u ∈ C(θ, b, c), thenb ≤ u(t) ≤ c for ηi−1 ≤ t ≤ ηi. Also, from assumption (ii), we have f(t, u(t)) ≥ Nb, forηi−1 ≤ t ≤ ηi and

θ(Tu) = minηi−1≤t≤ηi

|(T (u))| ≥∫ 1

0γ(s)G(s, s)f(s, u(s))ds >

∫ ηi

ηi−1

γ(s)G(s, s)Nbds = b,

which implies thatθ(Tu) > b, for all u ∈ C(θ, b, c).

Hence by Lemma 2.3.12, the Boundary value problem (3.1.1) has at least three positivesolutions u1, u2 and u3 satisfying (3.1.13).

44

Example 3.1.3. For the following boundary value problem

D320+u(t) + f(t, u) = 0, 0 < t < 1,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi) =1

5,

(3.1.15)

where

f(t, u) =

cos t100 + u2; u ≤ 1 ,3 + cos t

100 + u; u > 1 ,

we find that M ≈ 1.088 and N = 4.914. Choosing a = 125 , b =

34 and c = 4, we have

f(t, u) =cos t

100+ u2 ≤ 0.011 < Ma ≈ 0.04352, for (t, u) ∈ [0, 1] ×

[0,

1

25

],

f(t, u) = 3 +cos t

100+ u ≤ 4.01 ≥ Nb ≈ 3.7, for (t, u) ∈

[1

4,3

4

]×[3

4, 4

],

f(t, u) = 3 +cos t

100+ u ≤ 4.01 ≤Mc ≈ 4.352, for (t, u) ∈ [0, 1] × [0, 4].

Hence, by Theorem 3.1.6, the Boundary value problem (3.1.15) has at least three positivesolutions u1, u2, and u3 with

max0≤t≤1

|u1(t)| <1

25, 1 < min

14≤t≤ 3

4

|u2(t)| ≤ 4,1

25< max

0≤t≤1|u3(t)| < 4, min

14≤t≤ 3

4

|u3(t)| <3

4.

3.2 Multi-point boundary value problems with nonlocal bound-

ary conditions

In this section, we study sufficient conditions for existence and uniqueness of solutions to

the following more general class of nonlocal and multi-point boundary value problems with

nonlinear boundary conditions

−c Dqu(t) = f(t, u(t)); t ∈ J, 0 < q ≤ 1,

u(0) = g(u), u(1)−m−2∑

i=1

λiu(ηi) = h(u),(3.2.1)

where λi, ηi ∈ (0, 1) withm−2∑i=1

λiηi < 1, g, h : C(J,R) and f : J ×R×R → R is continuous.

We use the notations A = 1−∑m−2i=1 λi, ∆ = 1−∑m−2

i=1 λiηi, and in view of the definition

Iqf(t, u(t)) =1

Γ(q)

∫ t

0(t− s)q−1f(s, u(s)) ds (3.2.2)

45

we use

Iqf(1, u(1)) =1

Γ(q)

∫ 1

0(1− s)q−1f(s, u(s)) ds,

Iqf(ηi, u(ηi)) =1

Γ(q)

∫ ηi

0(ηi − s)q−1f(s, u(s)) ds.

(3.2.3)

Lemma 3.2.1. The Boundary value problem for fractional differential equation (3.2.1) hasa solution of the form

u(t) =(1− tA

∆)g(u(0)) +

t

∆h(u(1)) +

t

∆[Iqf(1, u(1)) − Σm−2

i=1 λiIqf(ηi, u(ηi))]− Iqf(t, u(t))

= (1− tA

∆)g(u(0)) +

t

∆h(u(1)) +

∫ 1

0G(t, s)f(s, u(s)) ds,

(3.2.4)

where G(t, s) is the Green function and is given by

G(t, s) =

s ≤ t

t∆Γq [(1 − s)q−1 −

m−2∑i=1

λi(ηi − s)q−1]− (t−s)q−1

Γq , s ∈ [0, η1]

t∆Γq [(1 − s)q−1 −

m−2∑i=j

λi(ηi − s)q−1]− (t−s)q−1

Γq , s ∈ [ηj−1, ηj ],

j = 2, 3, ...,m − 2,t

∆Γq (1− s)q−1 − (t−s)q−1

Γq , s ∈ [ηm−2, 1]

t ≤ s

t∆Γq [(1 − s)q−1 −

m−2∑i=1

λi(ηi − s)q−1], s ∈ [0, η1]

t∆Γq [(1 − s)q−1 −

m−2∑i=j

λi(ηi − s)q−1], s ∈ [ηj−1, ηj ], j = 2, 3, ...,m − 2,

t∆Γq (1− s)q−1, s ∈ [ηm−2, 1]

(3.2.5)

Proof. Applying Iq on the differential equation in (3.2.1) and using Lemma 2.2.9, we obtain

u(t) = −Iqf(t, u(t)) + C0 +C1t, C0, C1 ∈ R (3.2.6)

The boundary conditions u(0) = g(u(0)) and u(1)−m−2∑i=1

λiu(ηi) = 0 yield

c0 = g(u(0)), c1 =1

∆

[h(u(1)) + Iqf(1, u(1)) +Ag(u(0)) −

m−2∑

j=i

λiIqf(ηi, u(ηi))

].

46

Hence, it follows that

u(t) = −Iqf(t, u(t)) + g(u(0)) +t

∆

[h(u(1)) + Iqf(1, u(1)) +Ag(u(0)) −

m−2∑

i=1

λiIqf(ηi, u(ηi))

]

= −Iqf(t, u(t)) + (1 +tA

∆)g(u(0)) +

t

∆h(u(1)) +

t

∆

[Iqf(1, u(1)) −

m−2∑

i=1

λiIqf(ηi, u(ηi))

]

= (1 +tA

∆)g(u(0)) +

t

∆h(u(1)) + U1(t),

(3.2.7)

where

U1(t) =t

∆

[Iqf(1, u(1)) −

m−2∑

i=1

λiIqf(ηi, u(ηi))

]

=t

∆Γq

[ ∫ 1

0(1− s)q−1f(s, u(s))ds−

m−2∑

i=1

λi

∫ ηi

0(ηi − s)q−1f(s, u(s))ds

]

− 1

Γq

∫ t

0(t− s)q−1f(s, u(s))ds =

∫ 1

0G(t, s)f(s, u(s))ds.

Define the following operators F, G, T : C(J,R) → C(J,R) by

(Fu)(t) = g(u)− tA

∆g(u) +

t

∆h(u) = (1− tA

∆)g(u) +

t

∆h(u),

(Gu)(t) =t

Γ(q)∆)

∫ 1

0(1− s)q−1f(s, u(s)) ds−

m−2∑

j=i

λi

∫ ηi

0(ηi − s)q−1f(s, u(s)) ds

− 1

Γ(q)

∫ t

0(t− s)q−1f(s, u(s)) ds, and Tu = Fu+Gu.

The operator T is well defined as g, h, f are continuous functions. We can write (3.2.7) as

an operator equation

u = Tu = Fu+Gu, (3.2.8)

and solutions of the Boundary value problem (3.2.1) means solutions of the operator equa-

tion, that is, fixed points of T . Assume that the following hold

(A1) There exist constants Kg, Cg, mg, q1 ∈ [0, 1) such that

|g(u) − g(v)| ≤ Kg‖u− v‖, |g(u)| ≤ Cg‖u‖q1 +mg, for u, v ∈ C(J,R).

47

(A2) There exist constants Kh, Ch, q1, mh ∈ [0, 1) such that

|h(u)− h(v)| ≤ Kh‖u− v‖, |h(u)| ≤ Ch‖u‖q1 +mh, for u, v ∈ C(J,R).

(A3) There exist constants Cf , Mf , q2 ∈ [0, 1) such that

|f(t, u(s)| ≤ Cf‖u‖q2 +Mf , for (t, u) ∈ J × C(J,R).

Lemma 3.2.2. Under the assumptions (A1), (A2), the operator F : C(J,R) → C(J,R)satisfies the Lipschitz’s conditions with constant K and the growth condition

‖Fu‖ ≤ Cg‖u‖q1 + Ch‖u‖q1 , for every u ∈ C(J,R). (3.2.9)

Proof. Using the assumptions (A1) and (A2), we obtain

|Fu(t)− Fv(t)| =∣∣∣∣(1−

tA

∆)(g(u) − g(v)) +

t

∆

(h(u)− h(v)

)∣∣∣∣≤ Kg‖u− v‖+Kh‖u− v‖ = K‖u− v‖, where K = Kg +Kh,

By proposition 2.4.4, F is also µ-Lipschitz with constant K. For Growth condition, we get

‖Fu‖ = Cg‖u‖q1 + Ch‖u‖q1 .

Lemma 3.2.3. The operator G : C(J,R) → C(J,R) is continuous and under the assump-tion (A3) satisfies the growth condition

‖Gu‖ ≤ 2(Cf‖u‖q2 +Mf )

Γ∆(q + 1), u ∈ C(J,R). (3.2.10)

Proof. Let un be a sequence of a bounded set Bk = ‖u‖ ≤ r: u ∈ C(J,R) such thatun → u in Bk. Since f is continuous and un → u, it follows that f(s, un(s)) → f(s, u(s)) asn→ 0. Now consider

|(Gun)(t)− (Gu)(t)| ≤ 1

Γq

∫ t

0(t− s)q−1|f(s, un(s))− f(s, u(s))|ds+

t

Γq∆)

m−2∑

i=1

λi

∫ ηi

0(ηi − s)q−1|f(s, un(s))− f(s, u(s))|ds

+1

Γ(q)

∫ t

0(t− s)q−1|f(s, un(s))− f(s, u(s))|ds,

which in view of the Lebesgue Dominated Convergence theorem implies that

‖(Gun)(t)− (Gu)(t)‖ → 0 as n→ ∞.

48

For Growth conditions on G, using (A3), we obtain

|(Gu)(t)| =

∣∣∣∣t

Γ(q∆)

[∫ 1

0(1− s)q−1f(s, u(s)) ds−

m−2∑

i=1

λi

∫ ηi

0(ηi − s)q−1f(s, u(s)) ds

]

− 1

Γ(q)

∫ t

0(t− s)q−1f(s, u(s)) ds

∣∣∣∣ ≤ 2(Cf ||u(s)||q2 +Mf )

∆Γ(q + 1),

Lemma 3.2.4. The operator G : C(J,R) → C(J,R) is compact. Consequently, G is µ-Lipschitz with zero constant.

Proof. Take a bounded set D ⊂ Bk ⊆ C(J,R) and a sequence un in D, then using (3.2.10),we have

‖Gun‖ ≤ Cf ||u||q2 +Mf

∆Γ(q + 1),

which implies that G(D) is bounded. Now, for 0 ≤ t1 < t2 ≤ 1, consider

|Gun(t1)−Gun(t2)| = | 1Γq

∫ t1

0((t2 − s)q−1 − (t1 − s)q−1)f(s, un(s))ds

+(t1 − t2)

∆Γq

∫ 1

0(1− s)q−1f(s, un(s))ds−

(t1 − t2)

∆Γq

m−2∑

i=1

λi

∫ ηi

0(ηi − s)q−1f(s, un(s))ds|

which implies that

|(Gun)(t1)− (Gun)(t2)| ≤2

∆Γ(q + 1(Cf‖|u||q2 +Mf ) [(t

q2 − tq1) + 2(t1 − t2)− (t2 − t1)

q] .

The right side of the above inequality tends to zero as t2 → t1. Hence, Gun is equi-continuous. Therefore, G(D) is relatively compact in C(J,R) by Arzela-Ascoli theorem.Furthermore, by proposition 2.4.4, G is µ-Lipschitz with constant zero.


Theorem 3.2.5. Under the assumptions (A1)− (A3), the boundary value problem (3.2.1)has at least one solution u ∈ C(J,R). Moreover, the set of solutions of (3.2.1) is boundedin C(J,R).

Proof. From proposition 2.4.3, the operator T is a strict µ-contraction with constant K.Now setting

S = u ∈ C(J,R) : ∃ λ ∈ [0, 1] such that u = λTu.We need to prove that S is bounded in C(J,R). For this, consider

||u|| = ||λTu|| = λ‖Tu‖ ≤ λ(‖Fu‖+ ‖Gu‖

)

49

which in view of (3.2.9) and (3.2.10) together with q1 < 1, q2 < 1, implies that S is boundedin C(J,R). Therefore, by Theorem 2.4.5, T has at least one fixed point and the set of fixedpoints is bounded in C(J,R).

Assume that the following hold

(A4) There exist a constant Lf > 0 such that

|f(t, u)− f(t, v)| ≤ Lf |u− v|, for each t ∈ J, and for all u, v ∈ R.


Theorem 3.2.6. In addition to the assumption (A1) − (A4), assume that there exists aconstant M > 0 such that

M = (Kg +Kh +3Lf

∆Γ(q + 1)) < 1, (3.2.11)

then the Boundary value problem (3.2.1) has a unique solution.

Proof. We use Banach contraction principle, for u, v ∈ C(J,R), consider

|(Tu)(t)− (Tv)(t)| =∣∣(1− tA

∆)(g(u) − g(v)

)+

t

∆

(h(u) − h(v)

)

+t

∆Γ(q)

∫ 1

0(1 − s)q−1(f(s, u(s))− f(s, v(s)))ds

+

tm−2∑i=1

λi

∆Γ(q)

∫ ηi

0(ηi − s)q−1(f(s, u(s))− f(s, v(s)))ds

+1

Γ(q)

∫ t

0(t− s)q−1(f(s, u(s))− f(s, v(s)))ds

∣∣,

which in view of the assumptions (A1)− (A4) yields

|(Tu)(t)− (Tv)(t)| ≤ Kg‖u− v‖+Kh‖u− v‖+ 3Lf∆Γ(q + 1)

‖u− v‖,

which implies that

|(Tu)(t) − (Tv)(t)| ≤ (Kg +Kh +3Lf

∆Γ(q + 1))‖u− v‖ =M‖u− v‖.

Hence, the Boundary value problem (3.2.1) has a unique solution.

50

Example 3.2.1. Consider the following multi-point boundary value problem

cD23u(t) =

|u(t)| 12(1 + 12e2t)(1 + |u(t)| 12 )

t ∈ [0, 1],

u(0) = g(u), u(1) =m−2∑

i=1

λiu(ηi) + h(u),m−2∑

i=1

λiu(ηi) =1

5.

(3.2.12)

Take q = 23 , q1 = 1, q2 = λ = 1

2 , r = 2 ∈ (1, 3), λ = 12 ∈ [0, 12 ], Lf = Cf = 1

10 , mf = 0,Kg = Kh = 1

4 , the assumptions (A1) − (A4) satisfied. The solution of the boundary valueproblem (3.2.12) is given by

u(t) = g(u)[1 − t] + t(2

5+ h(u)) − I

23 y(t) + I

23 y(1).

Here

Fu(t) = (1− t)g(u) + t(2

5+ h(u)) and Gu(t) = −I

23 y(t) + I

23 y(1).

Since F, G are continuous and bounded so also T = F +G. Further

||Fu− Fv|| ≤ 1

2||u− v||,

that is F is µ Lipschitz and G is µ Lipschitz with zero constant implies that T is strict-µ-contraction with constant 1

2 . let

S = u ∈ C(J,R)∃λ ∈ [0, 1] : u =1

2Tu

||u|| = 1

2||Tu|| ≤ 1,

implies S is bounded and by Theorem 3.2.5 the Boundary value problem(3.2.1) has a solationu in C(J,R).

3.3 General Class of multi-point boundary value problems

In this section, we focus our attention to investigate existence and uniqueness of solutions to

more general class of multi-point boundary value problems for fractional order differential

equations of the form

−c Dqu(t) = f(t, u(t),cDq−1u(t)); 0 < t < 1, 1 < q ≤ 2,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi),(3.3.1)

51

where δi, ηi ∈ (0, 1) withm−2∑i=1

δiηi < 1, and f : J × R × R → R explicitly depends on the

fractional order derivative.

Lemma 3.3.1. For h ∈ L1(J,R), the boundary value problem for fractional differential

equation

cDqu(t) + h(t) = 0; 0 < t < 1, 1 < q ≤ 2,

u(0) = 0, u(1) =m−2∑

i=1

δiu(ηi),(3.3.2)

has a solution u of the form u(t) =∫ 10 G(t, s)h(s) ds, where G(t, s) is the Green function

and is given by

G(t, s) =1

Γ(q)

t1−∆

[(1− s)q−1 −

m−2∑j=i

δj(ηj − s)q−1

]− (t− s)q−1; s ≤ t, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1,

t1−∆

[(1− s)q−1 −

m−2∑j=i

δj(ηj − s)q−1

]; t ≤ s, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1.

,

(3.3.3)

where ∆ =m−2∑i=1

δiηi < 1.

Proof. Applying Iq on −cDqu(t) = h(t) and using Lemma 2.2.9, we have

u(t) = −Iqh(t)− C0 − C1t, C0, C1 ∈ R (3.3.4)

The boundary condition u(0) = 0 implies C0 = 0 and the boundary condition u(1) =m−2∑i=1

δiu(ηi) yields C1 =1

1−∆

[− Iqh(1) +

m−2∑i=1

δiIqh(ηi)

], It follows that

u(t) = −Iqh(t) + t

1−∆

[Iqh(1) −

m−2∑

i=1

δiIqh(ηi)

]. (3.3.5)

For 0 ≤ t ≤ η1, (3.3.5) implies

u(t) =

∫ t

0

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1 −∆)

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)]h(s) ds

52

+t

Γ(q)(1−∆)

∫ η1

t

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)h(s) ds

+m−2∑

i=2

∫ ηi

ηi−1

((1− s)q−1 −

m−2∑

j=1

δj(ηj − s)q−1)h(s) ds+

∫ 1

ηm−2

(1− s)q−1h(s) ds,

and for ηl−1 ≤ t ≤ ηl, 2 ≤ l ≤ m− 2, we obtain

u(t) =

∫ η1

0

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1 −∆)

((1− s)q−1 −

m−2∑

j=1


+

m−2∑

i=2

∫ ηi

ηi−1

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1−∆)

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1)+

(1− s)q−1]h(s) ds

+

∫ t

ηl−1

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1−∆)

((1− s)q−1 −

m−2∑

j=l


+t

Γ(q)(1−∆)

[ ∫ ηl

t

((1− s)q−1 −

m−2∑

j=l


+

m−2∑

i=l+1

∫ ηi

ηi−1

((1− s)q−1 −

m−2∑

j=i

δj(ηj − s)q−1)h(s) ds+

∫ 1

ηm−2

(1− s)q−1h(s) ds.

Further, for ηm−2 ≤ t ≤ 1, (3.3.5) yields

u(t) =

∫ η1

0

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1−∆)

((1− s)q−1 −

m−2∑

j=1


+

m−2∑

i=2

∫ ηi

ηi−1

[− (t− s)q−1

Γ(q)+

t

Γ(q)(1−∆)

((1− s)q−1 −

m−2∑

j=i


+t

Γ(q)(1−∆)

[ ∫ t

ηm−2

((1−∆)(t− s)q−1 + (1− s)q−1

)h(s) ds+

∫ 1

t(1− s)q−1h(s) ds

],

and the desired result follows.

53

3.3.1 Existence of solutions

Here the Banach space is denoted by X = C(J,R). In view of Lemma 3.3.1, we write the

Boundary value problem (3.3.1) is an equivalent integral equation

u(t) =

∫ 1

0G(t, s)f(s, u(s),cDq−1u(s))ds, t ∈ J (3.3.6)

and define operator T : X → X by

Tu(t) =

∫ 1

0G(t, s)f(s, u(s),cDq−1u(s)) ds. (3.3.7)

Solutions of the Boundary value problem (3.3.1) means fixed points of T . We also have

cDq−1Tu(t) =

∫ t

0f(s, u(s),cDq−1u(s))ds+

t2−q

(1−∆)Γ(q)Γ(3 − q)

×[ ∫ 1

0(1− s)q−1u(s)ds−

m−2∑

j=1

δj

∫ ηj

0(ηj − s)q−1u(s)ds

]. (3.3.8)

Theorem 3.3.2. Assume that f : J × R× R → R is continuous and the following hold

(a) There exist p ∈ C(J,R+) and a continuous nondecreasing function ψ : [0,∞) → (0,∞)

such that

|f(t, u, z)| ≤ p(t)ψ(|z|) for t ∈ J and each u, z ∈ R;

(b) There exists a constant ρ > 0 such that

ρ ≥ maxt∈J

ψ(ρ)p∗G∗, bψ(ρ)p∗Γ(3− q)Γ(q + 1) + 1

Γ(3− q)Γ(q + 1), (3.3.9)

where p∗ = supp(s), s ∈ J, G∗ = supt∈J∫ 10 |G(t, s)|ds,

then, the Boundary value problem (3.3.1) has at least one solution such that |u(t)| ≤ ρ for

each t ∈ J .

54

Proof. First we show that the operator T defined by (3.3.7) is continuous. Let ur be a

sequence such that ur → u ∈ X , and let σ > 0 be such that ‖ur‖C ≤ σ, ‖u‖C ≤ σ, then, for

each t ∈ J , using (3.3.7) and (3.3.8), we obtain

|(Tur)(t)− (Tu)(t)| ≤∫ 1

0|G(t, s)f(s, ur(s),cDq−1ur(s))− f(s, u(s),cDq−1u(s))|ds,

and

|(cDq−1Tur)(t)− (cDq−1Tu)(t)| ≤∫ t

0|f(s, ur(s),cDq−1ur(s))− f(s, u(s),cDq−1(s))|ds

+t2−q

(1−∆)Γ(q)Γ(3− q)

[( ∫ 1

0(1− s)q−1|f(s, ur(s),cDq−1ur(s))− f(s, u(s),cDq−1(s))|

)ds

+

m−2∑

j=1

δj

(∫ ηj

0(ηj − s)q−1u(s)|f(s, ur(s),cDq−1ur(s))− f(s, u(s),cDq−1(s))|

)]ds.

In view of the continuity of f and Lebesgue Dominated Convergence theorem, we have

‖Tur − Tu‖∞ → 0 as r → ∞,

and

‖(cDq−1Tur)(t)− (cDq−1Tu)(t)‖∞ → 0, as r → ∞,

implies that T is continuous.

Define D = u ∈ C(J,R), ‖u‖C ≤ ρ a closed and convex subset of X . We show that

T (D) ⊂ D. For u ∈ D and t ∈ J , using (a), (b), (3.3.7) and (3.3.8), we obtain

(Tu)(t)) ≤∫ 1

0|G(t, s)||f(s, u(s),cDq−1u(s)| ≤ ψ(‖u‖

CP ∗G∗, (3.3.10)

and

|(cDq−1Tu)(t) ≤||f(s, u(s),cDq−1u(s)|ds+ t2−q

(1−∆)Γ(q)Γ(3− q)×

(m−2∑

j=1

δj

∫ ηj

0(ηj − s)q−1|u(s)|ds+

∫ 1

0(1− s)q−1|u(s)|

)

≤ bψ(ρ)P ∗(Γ(3− q)Γ(q + 1) + 1

Γ(3− q)Γ(q + 1)

).

(3.3.11)

55

From (3.3.10) and (3.3.11), it follows that ‖Tu‖C ≤ ρ which implies that T (D) ⊂ D.

Finally, we show that T maps D into an equi-continuous set of X . Take t1, t2 ∈ J with

t1 < t2 and u ∈ D, we have

|(Tu)(t2)− (Tu)(t1)| ≤∫ 1

0|G(t2, s)−G(t1, s)||f(s, u(s),cDq−1u(s))|ds

≤ p∗ψ(ρ)∫ 1

0|G(t2, s)−G(t1, s)|ds

and from the continuity of G(t, s), it follows that |(Tu)(t2) − (Tu)(t1)| → 0 as t2 → t1.

Also,

|cDq−1Tu(t2)−c Dq−1Tu(t1)| ≤ |∫ t2

0f(s, u(s),cDq−1u(s)ds−

∫ t1

0f(s, u(s),cDq−1u(s)ds|

+t2−q2 − t2−q1

(1−∆)Γ(q)Γ(3− q)

(m−2∑

j=1

δj

∫ ηj

0(ηj − s)q−1|u(s)|ds+

∫ 1

0(1− s)q−1|u(s)|ds

)

≤ p∗ψ‖cDq−1u(s)‖t2 − t1 +t2−q2 − tq−1

1

(1−∆)Γ(q)Γ(3 − q)(1 +

m−2∑

j=1

δjηqi ),

which implies

‖(cDq−1Tu)(t2)− (cDq−1Tu)(t1)‖ → 0 as t2 → t1.

Consequently by Arzela- Ascoli Theorem, T is completely continuous. By Schauder’s fixed

point theorem, T has a fixed point u ∈ D for t ∈ J .


Theorem 3.3.3. Assuming that the following hold

(i) f : J × R× R → R is continuous.

(ii) There exists constant k > 0 such that for each t ∈ J and all x, y, x, y ∈ R,

|f(t, x, y)− f(t, x, y)| ≤ k(|x− x|+ |y − y|).

56

Further, if [max2G∗k,

2kΓ(3− q)Γ(q + 1) + 1

Γ(3− q)Γ(q + 1)]< 1, (3.3.12)

then the Boundary value problem (3.3.1) has a unique solution on J .

Proof. For the uniqueness of solutions of (3.3.1), we use Banach contraction principle. The

continuity of Tu(t) and cDq−1Tu(t) follow from the continuity of f and G. Let u, u ∈ X ,

then for each t ∈ J , we have

|(Tu)(t) − (T u)(t)| ≤∫ 1

0G(t, s)|f(s, u(s),cDq−1u(s))− f(s, u(s),cDq−1u(s))|ds

≤ G∗k(‖u− u‖∞ + ‖cDq−1u−c Dq−1u‖∞) ≤ 2G∗k‖u− u‖C ,

which implies that

‖Tu− T u‖ ≤ 2G∗k‖u− u‖C. (3.3.13)

Also,

|cDq−1(Tu)(t)−c Dq−1(T u)(t)| ≤ |f(s, u(s),cDq−1u(s))− f(s, u(s),cDq−1u(s))|ds

+t2−q

(1−∆)Γ(q)Γ(3− q)

[( ∫ 1

0(1− s)q−1|f(s, u(s),cDq−1u(s))− f(s, u(s),cDq−1u(s))|

)ds

−m−2∑

j=1

δj( ∫ ηj

0(ηj − s)q−1|f(s, u(s),cDq−1u(s))− f(s, u(s),cDq−1u(s))|

)]ds,

which implies that

|cDq−1(Tu)(t)−c Dq−1(T u)(t)| ≤ 2k‖u− u‖∞ +2k

Γ(3− q)Γ(q + 1)‖u− u‖∞

≤ 2k

(Γ(3− q)Γ(q + 1) + 1

Γ(3− q)Γ(q + 1)

)‖u− u‖

C.

(3.3.14)

From (3.3.13) and (3.3.14), it follows that

‖Tu− T u‖ ≤ max2G∗k,2kΓ(3 − q)Γ(q + 1) + 1

Γ(3− q)Γ(q + 1)‖u− u‖

C.

Hence, by Banach contraction principle, T has a unique fixed point.

57

Example 3.3.1. Consider the following m-points boundary value problem

cD32u(t) +

1

20e2t + 7

(1

1 + 2|u(t)|+ 3|cD 12u(t)|

), t ∈ J := [0, 1], 1 < q ≤ 2,

u(0) = 0, u(1) =

m−2∑

i=1

δiu(ηi) =1

5.

(3.3.15)

Here

f(t, u(t), v(t)) =1

20e2t + 7

(1

1 + 2|u(t)| + 3|cD 12u(t)|

)

and for u, v, u, v ∈ R and t ∈ J , we obtain

|f(t, u, v)− f(t, u, v| ≤ 1

9[|u− u|+ |v − v|] ,

which is condition (ii) of Theorem 3.3.3 with k = 19 .

Further, we have

G(t, s) =1

Γ(32)

5t4

[(1− s)

12 −

m−2∑j=i

δj(ηj − s)12

]− (t− s)

12 ; s ≤ t, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1,

5t4

[(1− s)

12 −

m−2∑j=i

δj(ηj − s)12

]; t ≤ s, ηi−1 < s ≤ ηi,

i = 1, 2, ...,m − 1,

(3.3.16)

G∗ = supt∈J

∫ 1

0|G(t, s)|ds < 5

3Γ(π)

and clearly

max29G∗,

2Γ(3− q)Γ(q + 1) + 1

9Γ(3 − q)Γ(q + 1) < 1. (3.3.17)

Hence by Theorem 3.3.3, the Boundary value problem(3.3.1) has a unique solution on [0, 1]

for each value of q ∈ (1, 2].

Chapter 4

Coupled systems of fractionalorder multi-point boundary valueproblems

In this chapter, we study existence of positive solutions to coupled systems of nonlinear

fractional order differential equations with various kinds of boundary conditions. Many

problems in applied sciences can be modeled as coupled system of differential equations

with different type of boundary conditions. Boundary values problems for coupled sys-

tems with ordinary derivatives are well studied, however, coupled systems with fractional

derivatives have attracted the attention quite recently. For example, Su [84], developed suf-

ficient conditions for existence of solutions for the coupled system with two point boundary

conditions of the form

Dαu(t) = f(t, v(t),Dµv(t)), Dβv(t) = g(t, u(t),Dvv(t)), 0 < t < 1,

u(0) = u(1) = v(0) = v(1) = 0,

where 1 < α, β < 2, µ, v > 0, α − v ≥ 1, β − µ ≥ 1, f, g : [0, 1] × R× R → R. B. Ahmad

and Neito [85], extended the results of Su to a three-point boundary value problem for the

following coupled system of fractional differential equations

Dαu(t) = f(t, v(t),Dµv(t)), Dβv(t) = g(t, u(t),Dνv(t)), 0 < t < 1,

u(0) = 0, u(1) = γu(η), v(0) = 0, v(1) = γv(η),

58

59

where 1 < α, β < 2, µ, ν, γ > 0, 0 < η < 1, α− ν ≥ 1, β − µ ≥ 1, γηα−1 < 1,

γηβ−1 < 1, f, g : [0, 1]×R×R → R are continuous. Wang et al. [86], studied existence and

uniqueness of positive solutions to a three-point boundary value problems for the coupled

system

Dαu(t) = f(t, v(t)), Dβv(t) = g(t, u(t)), 0 < t < 1,

u(0) = 0 = v(0), u(1) = au(ξ), v(1) = bv(ξ),

where 1 < α, β < 2, 0 ≤ a, b ≤ 1, 0 < ξ < 1, f, g : I × R× R → R are continuous.

Moreover, Coupled system with integral boundary conditions was also studied by W.Yang

[87]. He obtained sufficient conditions for existence of positive solutions to the following

coupled system of boundary values problems

Dαu(t) + f(t, v(t)) = 0, Dβv(t) + g(t, u(t)), 0 < t < 1

u(0) = v(0) = 0, u(1) =

∫ 1

0φ(t)u(t)dt, v(1) =

∫ 1

0ψ(t)v(t)dt,

where 1 < α, β ≤ 2, φ, ψ ∈ L1[0, 1] under the conditions∫ 10 φ(t)t

α−1dt 6= 1,∫ 10 ψ(t)t

β−1 6= 1.

We generalize the system studied by W. Yang, to the case, where the nonlinear functions

explicitly depend on the fractional derivatives. On the other hand, the problem with inte-

gral boundary conditions which we study include several problems as special cases like the

aforesaid problem studied by Su, Yang and Neito etc.

Another important class of differential equations is known as impulsive differential equa-

tions. This class plays the role of an effective mathematical tool for those evolution processes

that are subject to abrupt changes in their states. There are many physical systems that

exhibit impulsive behavior such as action of a pendulum clock, mechanical systems sub-

ject to impacts, the maintenance of a species through periodic stocking or harvesting, the

thrust impulse maneuver of a spacecraft, and functions of the heart, we refer to [88], for an

introduction to the theory of impulsive differential equations. It is well known that in the

evolution processes, the impulsive phenomena can be found in many situations. For exam-

ple, disturbances in cellular neural networks [89], operation of a damper subjected to the

60

percussive effects [90], change of the valve shutter speed in its transition from open to closed

state [91], fluctuations of pendulum systems in the case of external impulsive effects [92],

Percussive systems with vibrations [93], Relaxational oscillations of the electromechanical

systems [94], Dynamic of system with automatic regulation, Control of the satellite orbit,

using the radial acceleration [95], and so on. The theory of impulsive differential equations

is well studied and large number of research articles are available in the literature on im-

pulsive differential equations, we refer to [96, 97, 98, 99, 100, 101, 102] and the references

therein for some of the recent development in the theory. In very recent years Feng et al.

[103], Wang et al. [104], Y.Tian[105] and Zhange et al. [106], etc studied various classes

and system of fractional order impulsive differential equations.

In this chapter, we study solutions of system of fractional order differential equation by

using the technique of topological degree theory also. The study of positive solutions of

boundary value problems for fractional order differential equations using the techniques of

topological degree theory is rarely available in the literature and therefore this area of re-

search need further explorations. To the best of our knowledge, existence theory of positive

solutions to a coupled systems of fractional differential equations with topological degree

theory has not been studied previously.

We organize this chapter as: In section 4.1, we develop sufficient conditions for existence

and uniqueness of positive solutions for multi-point boundary value problems of nonlin-

ear fractional order differential equations via using standard fixed point theorem of Leray-

Schauder type and Banach contraction principle. In section 4.2, we study a coupled system

involving movable type integral boundary conditions. Sufficient conditions for the existence

of at least one positive solution are obtained by means of Krasnoselskii’s fixed theorem.

Moreover, by the use of Banach Contraction principle, conditions for uniqueness are also

developed. In section 4.3, existence and uniqueness of solutions to a coupled systems with

integral boundary conditions is studied, where the nonlinear functions depend on fractional

61

derivative. The corresponding problem is more general form of section 4.2. With the help

of Krasnoselskii’s fixed point theorem, we develop necessary and sufficient conditions for

the existence of at least one positive solutions for a general dynamical system of coupled

system of impulsive boundary conditions in section 4.4. In section 4.5, we provide some

existence and uniqueness results of positive solutions for nonlinear coupled system of two

different classes with nonlinear more general and nonlocal four-point boundary conditions

by means of topological degree theory.

4.1 Coupled system with m-point boundary conditions

In this section, we study the following system of m-point boundary value problems

Dα0+u(t) = f(t, u(t), v(t)), D

β0+v(t) = g(t, u(t), v(t)), 0 ≤ t ≤ 1,

u(0) = 0, u(1) =n−2∑

j=1

λju(ηj), v(0) = 0, v(1) =n−2∑

j=1

µjv(ξj),(4.1.1)

where λj , µj ∈ (0,∞) and 0 < η1 < η2 < .... < ηn−2 < 1, 0 < ξ1 < ξ2 < .... < ξn−2 < 1,

such thatn−2∑j=1

λjηαj < 1,

n−2∑j=1

µjξβj < 1 and 1 < α, β ≤ 2. The nonlinear functions f, g :

[0, 1] × R × R → R are assumed to be continuous. We develop necessary and sufficient

conditions for the existence and uniqueness of positive solutions by using Leray-Schauder

fixed point theorem and Banach contraction principle.

Lemma 4.1.1. Let ∆1 = 1 −m−2∑j=1

λjηj , ∆2 = 1 −m−2∑j=1

µjξj and u, v ∈ C(I,R). Then

Boundary value problem (4.1.1) has a unique positive solution given by

(u, v) =

( 1∫

0

G1(t, s)f(s, u(s), v(s))ds,

1∫

0

G2(t, s)g(s, u(s), v(s))ds

),

62

where G1(t, s), G2(t, s) are the Green’s functions and are given by

G1(t, s) =

tα−1

∆1Γ(α)

(1− s)α−1 +

m−2∑

i=j

λj(ηj − s)α−1

− 1

Γ(α)(t− s)α−1; s ≤ t,

ηj−1 < s ≤ ηj , j = 1, 2, . . . ,m− 1,

tα−1

∆1Γ(α)

(1− s)α−1 +

m−2∑

i=j

λj(ηj − s)α−1

; t ≤ s,

ηj−1 < s ≤ ηj , j = 1, 2, . . . ,m− 1.

(4.1.2)

G2(t, s) =

tβ−1

∆2Γ(β)

(1− s)β−1 +

m−2∑

i=j

µj(ξj − s)β−1

− 1

Γ(β)(t− s)β−1; s ≤ t,

ξj−1 < s ≤ ξj, j = 1, 2, . . . ,m− 1,

tβ−1

∆2Γ(β)

(1− s)β−1 +

m−2∑

i=j

µj(ξj − s)β−1

; t ≤ s,

ξj−1 < s ≤ ξj; j = 1, 2, . . . ,m− 1.

(4.1.3)

Proof. Proof of Lemma 4.1.1 is same as proof of Lemma 3.1.1.

Lemma 4.1.2. If 0 < ∆i < 1(i = 1, 2) then each Gi(t, s)(i = 1, 2) satisfies the followingproperties:

(P1) Gi(t, s) ≥ 0(i = 1, 2) is continuous ∀t, s ∈ [0, 1],Gi(t, s) > 0(i = 1, 2),∀t, s ∈ (0, 1);

(P2) Gi(t, s) ≤ Ki(s)(i = 1, 2), for each t, s ∈ [0, 1],and mint∈[θ,1−θ]Gi(t, s) ≥ γiGi(s)(i =

1, 2), where 0 < θ < 12 .

Proof. The proof of properties of Green’s function is same as given by(3.1.2).

Writing the system(4.1.1), as equivalent system of integral equations given by

u(t) =

∫ 1

0G1(t, s)f(s, u(s), v(s))ds, v(t) =

∫ 1

0G2(t, s)g(s, u(s), v(s))ds. (4.1.4)

Define the operator T : X × Y → X ×Y by

T (u, v)(t) =

(∫ 1

0G1(t, s)f(s, u(s), v(s))ds,

∫ 1

0G2(t, s)g(s, u(s), v(s))ds

)

= (T1(x, y)(t), T2(x, y)(t)).

(4.1.5)

In view of Lemma 4.1.1, solutions of the system of Boundary value problem (4.1.1) are fixed

points of T .

63

4.1.1 Existence of at least one positive solution

Now, we study existence of solutions of the system of boundary value problem (4.1.1).

Theorem 4.1.3. Assume that f(t, u, v), g(t, u, v) are continuous on [0, 1] × R2 → R, thenthe operator defined in (4.1.5) T : P → P is completely continuous.

Proof. We first prove that T : P → P defined in (4.1.5) is completely continuous. From thecontinuity and nonnegativity of Gi(t, s) for (i = 1, 2), f and g, the operator T is continuousfor all (u, v) ∈ P. Let D ⊂ P be bounded then there exist a positive constants K and Lsuch that |f(t, u, v)| ≤ K and |g(t, u, v)| ≤ L for all (u, v) ∈ U . Then for every (u, v) ∈ U ,using |t| ≤ 1, we have

|T1(u, v)(t)| =∣∣∣∣tα−1

∆1Γ(α)

m−2∑

j=1

λj

∫ ηj

0(ηj − s)α−1f(s, x(s), y(s))ds

− tα−1

∆1Γ(α)

∫ 1

0(1− s)α−1f(s, u(s), v(s))ds+

tα−1

Γ(α)

∫ t

0(t− s)α−1f(s, u(s), v(s))ds

∣∣∣∣

≤∣∣∣∣tα−1

∆1Γ(α)

m−2∑

j=1

λj

∫ ηj

0(ηj − s)α−1f(s, u(s), v(s))ds

− tα−1

∆1Γ(α)

∫ 1

0(1− s)α−1f(s, u(s), v(s))ds

∣∣∣∣+∣∣∣∣tα−1

Γ(α)

∫ t

0(t− s)α−1f(s, u(s), v(s))ds

∣∣∣∣

≤ |tα−1|∆1Γ(α)

m−2∑

j=1

λj

∫ ηj

0(ηj − s)α−1|f(s, u(s), v(s))|ds + 1

Γ(α)

∫ t

0(t− s)α−1|f(s, u(s), v(s))|ds,

which implies that |T1(u, v)(t)| ≤ 2K∆1Γ(α+1) . Similarly, |T2(u, v)(t)| ≤ 2L

∆2Γ(β+1) . Hence, theoperator T is uniformly bounded.Next we show that T is equi-continuous. Choose 0 ≤ t ≤ τ ≤ 1, we have

|T1(u, v)(τ) − T1(u, v)(t)| =∣∣∣∣τα−1

∆1Γ(α)

m−2∑

j=1

λj

∫ ηj


+τα−1

Γ(α)

∫ τ

0(τ − s)α−1f(s, u(s), v(s))ds − tα−1

∆1Γ(α)

m−2∑

j=1

∫ ηj


− tα−1

Γ(α)

∫ t

0(t− s)α−1f(s, u(s), v(s))ds

∣∣∣∣ ≤K

∆1Γ(α)|τα−1 − tα−1|

m−2∑

j=1

λj

∣∣∣∣∫ ηj

0(ηj − s)α−1

∣∣∣∣ds

+K

Γ(α)

(∫ t

0[(τ − s)α−1 − (t− s)α−1]ds−

∫ t

0(t− s)α−1ds

).

64

Thus after simplification, we get

|T1(u, v)(τ) − T1(u, v)(t)| ≤K

∆1Γ(α+ 1)

([τα−1 − tα−1]

m−2∑

j=1

λjηαj + [τα − tα]

)

≤ K

∆1Γ(α+ 1)

(τα−1 − tα−1 + τα − tα

).

(4.1.6)

Similarly, |T2(u, v)(τ)−T2(u, v)(t)| ≤ L∆2Γ(β+1)

(τβ−1− tβ−1+ τβ− tβ

). From the uniformly

continuity on [0, 1] of the right hand sides of the above inequalities, it follows that theoperator T is equi-continuous. Thus T is completely continuous.

Theorem 4.1.4. Assume that f(t, u, v), g(t, u, v) are continuous on [0, 1] × R2 → R andsatisfy

(A1) |f(t, u(t), v(t))| ≤ a0(t) + a(t)[|u(t)| + |v(t)|];

(A2) |g(t, u(t), v(t))| ≤ b0(t) + b(t)[|u(t)| + |v(t)|];

(A3) 0 < Υ1 =∫ 10 G1(t, s)a0(s)ds <∞, Υ2 =

∫ 10 G1(t, s)a(s)ds < 1;

(A4) 0 < Υ3 =∫ 10 G2(t, s)b0(s)ds <∞, Υ4 =

∫ 10 G2(t, s)b(s)ds < 1.

Then, the system (4.1.1) has at least one positive solution in

U =(u, v) ∈ P : ‖(u, v)‖ < min

( Υ1

1−Υ2,

Υ3

1−Υ4

). (4.1.7)

Proof. Let U = (u, v) ∈ X × Y : ‖(u, v)‖ < ε, where ε = min(

Υ11−Υ2

, Υ31−Υ4

).

For (u, v) ∈ U , we obtain

‖T1(u, v)‖ = maxt∈[0,1]

|∫ 1

0G1(t, s)f(s, u(s), v(s))ds|

≤∫ 1

0G1(t, s)a0(s)ds+

∫ 1

0G1(t, s)a(s)[|u(s)| + |v(s)|]ds

≤∫ 1

0G1(t, s)a0(s)ds+

∫ 1

0G1(t, s)a(s)‖(u, v)‖ds

= Υ1 +Υ2‖(u, v)‖ ≤ ε.

Similarly, ‖T2(u, v)‖ ≤ ε,

which implies that ‖T (u, v)‖ ≤ ε, that is, T (U) ⊂ U , which implies that T : U → U iscompletely continuous. Now let us consider the eigenvalue problem

(u, v) = ρT (u, v), ρ ∈ (0, 1). (4.1.8)

65

Then, we obtain

‖u‖ = ‖ρT1(u, v)‖ = ρ maxt∈[0,1]

∥∥∥∥∫ 1

0G1(t, s)f(s, u(s), v(s)

∥∥∥∥

<

∫ 1

0G1(s, s)[a0(s) + a(s)(‖u‖ + ‖v‖)ds

=

∫ 1

0G1(s, s)a0(s)ds+

∫ 1

0G1(s, s)a(s)‖(u, v)‖ds

= Υ1 +Υ2‖(u, v)‖ ≤ ε.

Similarly ‖v‖ = ‖ρT2(u, v)‖ < ε.

Hence, ‖(u, v)‖ < ε which show that (u, v) is not in ∂U . Hence by Lemma 2.3.9, T hasat least one fixed point in U . Consequently, the coupled system (4.1.1) has at least onepositive solution.

4.1.2 Uniqueness of positive solutions

Theorem 4.1.5. Assume that f, g : I × R × R → R are continuous functions and thereexist two positive functions φ(t), ψ(t) such that for all t ∈ [0, 1] and ui, vi ∈ R, i = 1, 2,

(A5) |f(t, u1, v1)− f(t, u2, v2)| ≤ φ(t)[|u1 − u2|+ |v1 − v2|],

(A6) |g(t, u1, v1)− g(t, u2, v2)| ≤ ψ(t)[|u1 − u2|+ |v1 − v2|].Then the system(4.1.1) has a unique positive solutions if

λ =

∫ 1

0G1(s, s)φ(s)ds < 1, µ =

∫ 1

0G2(s, s)φ(s)ds < 1. (4.1.9)

Proof. Since Gi(t, s), i = 1, 2 and f(t, u, v), g(t, u, v), are nonnegative so for any (u, v) ∈ Pwe have T (u, v) ≥ 0 also T (P) ⊂ P.

‖T1(u2, v2)− T1(u1, v1)‖ = maxt∈[0,1]

|T1(u2, v2)− T1(u1, v1)|

= maxt∈[0,1]

∣∣∣∣∫ 1

0G1(t, s)[f(s, u2(s), v2(s))− f(s, u1(s), v1(s))]ds

∣∣∣∣

≤∫ 1

0G1(s, s)φ(s)ds[‖u2 − u1‖+ ‖v2 − v1‖]

≤ λ[‖u2 − u1‖+ ‖v2 − v1‖].

(4.1.10)

Similarly we can show that

‖T2(u2, v2)− T2(u1, v1)‖ ≤ µ[‖u2 − v1‖+ ‖v2 − v1‖]. (4.1.11)

From (4.1.10) and (4.1.11), we have

‖T (u2, v2)− T (u1, v1)‖ ≤ max(λ, µ)‖(u2, v2)− (u1, v1)‖, (4.1.12)

66

which under the assumption (4.1.9) implies that T is a contraction operator. Also T is com-pletely continuous operator by Lemma 4.1.3, hence, by Banach’s fixed-point theorem, theoperator T has a unique fixed point, which is the the unique positive solution of Boundaryvalue problem (4.1.1).

Example 4.1.1. Consider the following system

D74u(t) = 1 +

t2

4+

sin(u(t))

4+t ln(1 + v(t))

t2 + 3, t ∈ (0, 1),

D32 v(t) = 10 + e−2πt +

cos(2u(t))

16+

v(t)

15 + t, t ∈ (0, 1),

u(0) = 0, u(1) =10∑

j=1

1

3ju(

1

2j), v(0) = 0, v(1) =

10∑

j=1

1

2jv(

1

3j).

(4.1.13)

Here, we have

|f(t, u(t), v(t))| =

∣∣∣∣1 +t2

4+

sin(u(t))

4+t ln(1 + v(t))

t2 + 3

∣∣∣∣

≤ (1 +t2

4) +

|u(t)|4

+t

t2 + 3|v(t)|,

|g(t, u(t), v(t))| =

∣∣∣∣10 + e−2πt +cos(2u(t))

16+

v(t)

15 + t

∣∣∣∣

≤ (10 + e−2πt) +|u(t)|16

+1

15 + t|v(t)|.

Hence

Υ1 =

∫ 1

0G1(s, s)a0(s)ds =

∫ 1

0G1(s, s)(1 +

s2

4)ds <∞,

Υ2 =

∫ 1

0G1(s, s)a(s)ds ≤

∫ 1

0G1(s, s)ds ≈ 0.99984 < 1,

Υ3 =

∫ 1

0G2(s, s)b0(s)ds =

∫ 1

0G2(s, s)(10 + e−2πs)ds <∞,

Υ4 =

∫ 1

0G2(s, s)b(s)ds ≤

∫ 1

0G2(s, s)ds ≈ 0.76889 < 1.

All the conditions of Theorem 4.1.4 are satisfied and hence By Theorem 4.1.4, the system(4.1.13) has at least one positive solution in

U = (u, v) ∈ P : ‖(u, v)‖ < min( Υ2

1−Υ1,

Υ4

1−Υ3

).

67

Example 4.1.2. Consider the following system of multi-point Boundary value problems

D32u(t) =

1

t2 + 1+

cos2(u(t))

4+

sin2(v(t))

8, t ∈ [0, 1],

D32 v(t) = e−t

2+

cos 2(u(t))

16+

1

15 + v(t), t ∈ [0, 1],

u(0) = 0, u(1) =

100∑

j=1

1

2ju

(1

4j

), v(0) = 0, v(1) =

100∑

j=1

1

3jv

(1

2j

).

(4.1.14)

Here, we have

α =3

2, β =

3

2, λj =

1

2j, µj =

1

3j, ηj =

1

4j, ξj =

1

2j, j = 1, 2, 3, . . . , 100

f(t, u, v) =1

t2 + 1+

cos2(u(t))

4+

sin2(v(t))

8,

g(t, u, v) = e−t2+

cos 2u(t)

16+

1

15 + v(t).

Set ui(t), vi(t), (i = 1, 2) ∈ [0,∞) and t ∈ [0, 1], then we have

|f(t, u2, v2)− f(t, u1, v1)| ≤1

2[|u2 − u1|+ |v2 − v1|],

|g(t, u2, v2)− g(t, u1, v1)| ≤1

8[|u2 − u1|+ |v2 − v1|],

φ(s) =1

2, ψ(s) =

1

8.

(4.1.15)

λ =

∫ 1

0G1(s, s)φ(s)ds ≤

∫ 1

0G1(s, s)ds = 0.99984 < 1,

µ =

∫ 1

0G2(s, s)ψ(s)ds ≤

∫ 1

0G1(s, s)ds = 0.76889 < 1.

(4.1.16)

By Theorem 4.1.4, the system of Boundary value problem 4.1.14 has at least one positivesolution. Further, conditions of Theorem 4.1.5 are also satisfied, hence, the system ofBoundary value problem (4.1.14) has a unique positive solution.

4.2 Coupled system with movable type integral boundaryconditions

In this section, we study existence and uniqueness of positive solutions for the following cou-

pled system of nonlinear boundary values problems for fractional order differential equations

cDαu(t) = f(t, v(t)), cDβv(t) = g(t, v(t)), 0 < t < 1,

u(0) = 0, v(0) = 0, u(1) =

∫ η

0u(s)ds, v(1) =

∫ ξ

0v(s)ds,

(4.2.1)

68

where 1 < α, β ≤ 2 and 0 < η, ξ < 1 and f, g : I × [0,∞) → [0,∞) are continuous. The

boundary conditions in (4.2.1) are not fixed but of moving type and represent area under

the curve of solutions u(t) and v(t) from t = 0 to t = η and t = ξ respectively.

Lemma 4.2.1. Integral representation of the system of boundary value problems (4.2.1) isgiven by

u(t) =1

Γ(α)

∫ t

0(t− s)α−1f(s, v(s))ds− 2t

(2− η2)Γ(α)

∫ 1

0(1− s)α−1f(s, v(s))ds

−∫ η

0(

∫ s

0(s− τ)α−1)f(τ, v(τ))dτ)ds

,

v(t) =1

Γ(β)

∫ t

0(t− s)β−1g(s, u(s))ds− 2t

(2− ξ2)Γ(β)

∫ 1

0(1− s)β−1g(s, u(s))ds

−∫ ξ

0(

∫ s

0(s− τ)β−1)g(τ, u(τ))dτ)ds

.

(4.2.2)

Proof. Applying the operator Iα on the first equation of (4.2.1) and using Lemma 2.2.23,we obtain

u(t) = Iαf(t, v(t)) − C0 − C1t, C0, C1 ∈ R, (4.2.3)

which implies that∫ η

0u(s)ds =

1

Γ(α)

∫ η

0(

∫ s

0(s− τ)α−1f(τ, v(τ))dτ)ds − c0η − c1

η2

2. (4.2.4)

The boundary condition u(0) = 0 yields C0 = 0 and the boundary condition u(1) =∫ η0 u(s)ds yields

C1 =2

(2− η2)Γ(α)

∫ 1

0(1− s)α−1f(s, v(s))ds−

∫ η

0(

∫ s

0(s− τ)α−1f(τ, v(τ))dτ)ds

.

Substituting the values of C0 and C1 in (4.2.3), we obtain the first part of (4.2.1).Similarly repeating the above process with the second equation of the system, we obtain

the second part of (4.2.1).

Define T1 : Y → Y and T2 : X → X by

T1v(t) =1

Γ(α)

∫ t

0(t− s)α−1f(s, v(s))ds− 2t

(2− η2)Γ(α)

∫ 1

0(1− s)α−1f(s, v(s))ds

−∫ η

0(

∫ s

0(s− τ)α−1)f(τ, v(τ))dτ)ds

,

T2u(t) =1

Γ(β)

∫ t

0(t− s)β−1g(s, u(s))ds− 2t

(2− ξ2)Γ(β)

∫ 1

0(1− s)β−1g(s, u(s))ds

−∫ ξ

0(

∫ s

0(s− τ)β−1)g(τ, u(τ))dτ)ds

.

(4.2.5)

69

In view of Lemma 4.2.1, the system (4.2.2) of integral equations can be written as a coupled

system of operators equations of the form

u(t) = T1v(t), v(t) = T2u(t) (4.2.6)

Define T : X×Y → X×Y by T (u, v) = (T1v, T2u), then the problem of existence of solutions

of the operator equations (4.2.6) reduces to the problem of existence of fixed points of T .

For simplicity, we use the notations

A =1

Γ(α+ 1)(1 +

2[α + 1 + ηα+1]

(α+ 1)|2 − η2| ) and B =1

Γ(β + 1)(1 +

2[β + 1 + ξβ+1]

(β + 1)|2 − ξ2| ).

Assume the following hypothesis

(A7) f, g : I × [0,∞) → [0,∞) are continuous.

(A8) There exist L > 0 such that |f(t, u)− f(t, u)| ≤ L|u− u|, t ∈ [0, 1], u, v, u, v ∈ R.

(A9) There exist M > 0 such that |g(t, v) − g(t, v)| ≤M |v − v|, t ∈ [0, 1], u, v, u, v ∈ R.


Now, introducing the operators F1, G1 : Y → Y and F2, G2 : X → X by

F1v(t) =1

Γ(α)

∫ t

0(t− s)α−1f(s, v(s))ds,

G1v(t) = − 2t

Γ(α)(2− η2)[

∫ η

0(

∫ s

0((s− τ)α−1f(τ, v(τ))dτ)ds −

∫ 1

0(1− s)α−1f(s, v(s))ds],

F2u(t) =1

Γ(β)

∫ t

0(t− s)β−1g(s, u(s))ds,

G2u(t) = − 2t

Γ(β)(2 − ξ2)[

∫ η

0(

∫ s

0((s− τ)β−1g(τ, u(τ))dτ)ds −

∫ 1

0(1− s)β−1g(s, u(s))ds].

(4.2.7)

Then, we write T1 = F1+G1, T2 = F2+G2 and consequently, the operator T can be written

as

T = F +G, where F (u, v) = (F1v, F2u), G(u, v) = (G1v,G2u).


70

(A10) There exist constants Cf ,Mf > 0 such that |f(t, v(t))| ≤ Cf‖v‖Y +Mf on I × Y.

(A11) There exist constants Cg,Mg > 0 such that |g(t, u(t))| ≤ Cg‖u‖X +Mg on I ×X .

Theorem 4.2.2. Assume that (A7), (A10) and (A11) hold. Further assume that 2L[(α+1+ηα+1)]Γ(α+2)|2−η2 | <

1 and 2M [(β+1+ξβ+1)]Γ(β+2)|2−ξ2| < 1, then the System of boundary value problem (4.2.1) has at least

one solution.

Proof. The continuity of F on X × Y follows from the continuity of f and g. Consider abounded subset Ω ⊆ X × Y. In view of (A10), for every (u, v) ∈ Ω, we have

|(F1v)(t)| = | 1

Γ(α)

∫ t

0(t− s)α−1f(s, v(s))ds| ≤ 1

Γ(α)

∫ t

0(t− s)α−1(Cf‖v‖Y +Mf )ds

which implies that ‖F1v‖Y ≤ (Cf ‖v‖Y+Mf )Γ(α+1) . Similarly in view of A11, we can show that

‖F1u‖X ≤ (Cg‖u‖X+Mg)Γ(β+1) . Hence F (Ω) is bounded.

For equi-continuity of F , choose t1, t2 ∈ [0, 1] with t1 < t2 and any (u, v) ∈ X × Y, wehave

|F1v(t1)− F1v(t2)| =1

Γ(α)|∫ t1

0[(t2 − s)α−1 − (t1 − s)α−1]f(s, v(s))ds

+

∫ t2

t1

(t2 − s)α−1f(s, v(s))ds|,

which implies that ‖F1v(t1)− F1v(t2)‖Y ≤ Cf‖v‖Y+Mf

Γ(α+1) |2(t2 − t1)α + t1

α − t2α|. Similarly,

‖F2u(t1)− F2u(t2)‖X ≤ Cg‖u‖X +Mg

Γ(β + 1)|2(t2 − t1)

β + t1β − t2

β|.

The right hand sides of these inequalities tends to zero as t1 → t2. Hence, F is equi-continuous and by Arzela-Ascoli’s theorem F is compact. Now, we show that G is contrac-tion. For v, v ∈ Y, we have,

|G1v(t)−G1v(t)| ≤| 2t

(2− η2)Γ(α)|∫ 1

0(1− s)α−1|f(s, v(s))− f(s, v(s))|ds

+ | 2t

(2− η2)Γ(α)|∫ η

0(

∫ 1

0(s− τ)α−1|f(s, v(s))− f(s, v(s))|dτ)ds

≤ 2L|v − v|(α+ 1 + ηα+1)

|2− η2|Γ(α + 2)

which implies that ‖G1v −G1v‖Y ≤ 2L(α+1+ηα+1)|2−η2|Γ(α+2)

‖v − v‖Y . Similarly, we have

‖G2u−G2u‖X ≤ 2L(β + 1 + ξβ+1)

|2− ξ2|Γ(β + 2)‖u− u‖X .

Hence, G is a contraction and by Lemma 2.3.10 it follow that T has a fixed point.

71


Theorem 4.2.3. Under the assumptions (A7), (A8) and (A9) with AL < 1 and BM < 1,the Boundary value problem (4.2.1) has a unique solution.

Proof. DefineK = maxt∈I |f(t, 0)| and N = maxt∈I |g(t, 0)| and choose r ≥ max AK1−AL ,

BN1−MB.

Consider a closed bounded and convex subset U = (u, v) ∈ X × Y : ||(u, v)||X×Y ≤ r ofX × Y. For (u, v) ∈ U , using the definition of T1 and the assumptions (A8) and (A9), weobtain

|T1v(t)| ≤1

Γ(α)

∫ t

0(t− s)α−1|f(s, v(s))|ds+ 2t

(2− η2)Γ(α)

∫ 1

0(1− s)α−1|f(s, v(s))|ds

+2t

(2− η2)Γ(α)

∫ η

0(

∫ s

0(s− τ)α−1)|f(τ, v(τ))|dτ)ds

≤ 1

Γ(α)

∫ t

0(t− s)α−1(|f(s, v(s))− f(s, 0)|+ |f(s, 0)|)ds

+2t

(2− η2)Γ(α)

∫ 1

0(1− s)α−1(|f(s, v(s)) − f(s, 0)|+ |f(s, 0)|)ds

+2t

(2− η2)Γ(α)

∫ η

0(

∫ s

0(s− τ)α−1)(|f(τ, v(τ)) − f(τ, 0)|+ |f(τ, 0)|)dτ)ds

≤ 1

Γ(α)

∫ t

0(t− s)α−1(Lr +K)ds+ | 2t

(2− η2)Γ(α)|∫ 1

0(1− s)α−1(Lr +K)ds

| 2t

(2− η2)Γ(α)|∫ η

0(

∫ s

0(s− τ)α−1(Lr +K)dτ)ds

which implies that

|T1v(t)| ≤Lr +K

Γ(α+ 1)(tα + | 2t

(2− η2)|+ | 2t

(2− η2)(α+ 1)|ηα+1).


‖T1v‖Y ≤ Lr +K

Γ(α+ 1)(1 +

2(α+ 1 + ηα+1)

|(2− η2)|(α + 1)≤ (Lr +K)A ≤ r. (4.2.8)

Similarly, we can show that

‖T2u‖X ≤ (Mr +N)B ≤ r. (4.2.9)

From (4.2.8) and (4.2.9), it follows that ‖T (u, v)‖X×Y ≤ r.

72

Further, for u, v ∈ U and t ∈ [0, 1], we obtain

|T1v(t)− T1v(t)| ≤1

Γ(α)

∫ t

0(t− s)α−1|f(s, v(s))− f(s, v(s))|ds

+ | 2t

(2− η2)Γ(α)|∫ 1

0(1− s)α−1|f(s, v(s))− f(s, v(s))|ds

+ | 2t

(2− η2)Γ(α)|∫ η

0(

∫ 1

0(s− τ)α−1|f(s, v(s))− f(s, v(s))|dτ)ds

≤ 1

Γ(α)

∫ t

0(t− s)α−1L|v(s)− v(s)|ds+ | 2t

(2− η2)Γ(α)|∫ 1

0(1− s)α−1L|v(s)− v(s)|ds

+ | 2t

(2− η2)Γ(α)|∫ η

0(

∫ 1

0(1− s)α−1L|v(s)− v(s)|dτ)ds,

which implies that

‖T1v − T1v‖Y ≤ L

Γ(α+ 1)[‖v − v‖Y +

2‖v − v‖Y|2− η2| +

2ηα+1‖v − v‖Y|2− η2|(α + 1)

] ≤ AL‖v − v‖Y .


‖T2u− T2u‖Y ≤ BM‖u− u‖X (4.2.10)

which in view of the assumption AL < 1 and BM < 1 implies that

‖T (u, v) − T (u, v)‖X×Y < ‖(u, v) − (u, v)‖X×Y ,

that is, T is a contraction. Hence, by Banach fixed point theorem, the Boundary valueproblem (4.2.1) has a unique solution.

Example 4.2.1. Consider the following coupled system of boundary value problem

cD74u(t) =

||v||(t+ 3)3(1 + ||v||) ,

cD32 v(t) =

9||u||32π(1 + 4||u||) , t ∈ [0, 1]

u(0) = v(0) = 0, u(1) =

∫ 34

0u(s)ds, and v(1) =

∫ 12

0v(s)ds.

Here f(t, v) = ||v||(t+3)3(1+||v||) , g(t, u) =

9||u||32π(1+4||u||) , α = 7

4 , β = 32 , η = 3

4 , ξ =12 . Since

‖f(t, v) − f(t, v)‖ ≤ 1

81‖v − v‖, ‖g(t, u) − g(t, u)‖ ≤ 9

32π‖u− u‖,

it follows that

AL =L

Γ(α+ 1)(1 +

2[α + 1 + ηα+1]

(α+ 1)|2 − η2| ) < 1, BM =M

Γ(β + 1)(1 +

2[β + 1 + ξβ+1]

(β + 1)|2 − ξ2| ) < 1,

where L = 181 , M = 9

32π . Hence, by Theorem 4.2.3, the system of boundary value problemshas a unique solution.

73

4.3 Coupled system with integral boundary conditions

We study sufficient conditions for existence and uniqueness of positive solutions to the

following more general coupled systems of nonlinear boundary value problems

Dαu(t) = f(t, v(t),Dpv(t)), Dβv(t) = g(t, u(t),Dqu(t)), t ∈ I = [0, 1]

u(0) = u′(0) = 0, u(1) =

∫ 1

0φ(t)u(t)dt,

v(0) = v′(0) = 0, v(1) =

∫ 1

0ψ(t)v(t)dt,

(4.3.1)

where 2 < α, β ≤ 3, 0 < p, q < 1 such that α − q ≥ 2, β − p ≥ 2 and φ, ψ ∈ L1[0, 1] are

non-negative, and f, g : I × R× R → R are continuous.

Assume that

(A12) f, g : I × R× R → R are continues.

(A13) φ, ψ ∈ L1[0, 1] such that∫ 10 φ(t)t

α−1dt 6= 1 and∫ 10 ψ(t)t

β−1dt 6= 1.

Lemma 4.3.1. Under the assumptions (A12) and (A13), the system of boundary valueproblems (4.3.1) has a unique solution given by

u(t) =

∫ 1

0Gα(t, s)f(s, v(s),D

pv(s))ds, v(t) =

∫ 1

0Gβ(t, s)g(s, u(s),D

qu(s))ds, (4.3.2)

where Gα(t, s) = G1α(t, s) +G2α(t, s), Gβ(t, s) = G1β(t, s) +G2β(t, s),

G1α(t, s) =1

Γ(α)

(t− s)α−1 − (t(1− s))α−1 ; 0 ≤ s ≤ t ≤ 1,

− (t(1− s))α−1 ; 0 ≤ t ≤ s ≤ 1,(4.3.3)

G1β(t, s) =1

Γ(β)

(t− s)β−1 − (t(1− s))β−1 , 0 ≤ s ≤ t ≤ 1,

− (t(1− s))β−1 , 0 ≤ t ≤ s ≤ 1,(4.3.4)

and

G2α(t, s) =tα−1

∫ 10 φ(t)G1α(t, s)dt

1−∫ 10 φ(t)t

α−1dt, G2β(t, s) =

tβ−1∫ 10 ψ(t)G1β(t, s)dt

1−∫ 10 ψ(t)t

β−1dt. (4.3.5)

Proof. Applying the operator Iα on both the sides of the first equation of the system (4.3.1),we obtain u(t) = Iαf(t, v(t),Dpv(t)) + C1t

α−1 + C2tα−2 + C3t

α−3, Ci(i = 1, 2, 3) ∈ R.

74

The boundary condition u(0) = u′(0) = 0 give C3 = C2 = 0 and the condition u(1) =∫ 10 φ(t)u(t)dt yields

C1 =

∫ 1

0φ(t)u(t)dt − Iαf(1, v(1),Dpv(1)).


u(t) = Iαf(t, v(t),Dpv(t)) + tα−1

∫ 1

0φ(t)u(t)dt− tα−1Iαf(1, v(1),Dpv(1)). (4.3.6)

Now, using (4.3.6), we have

∫ 1

0φ(t)u(t)dt =

∫ 1

0φ(t)(Iαf(t, v(t),Dpv(t))dt +

∫ 1

0φ(t)tα−1

( ∫ 1

0φ(t)u(t)dt

)dt

−∫ 1

0φ(t)tα−1Iαf(1, v(1),Dpv(1))dt

which implies that∫ 10 φ(t)u(t)dt =

∫ 10 φ(t)[I

αf(t,v(t),Dpv(t))−tα−1Iαf(1,v(1),Dpv(1))]dt

(1−∫ 10 φ(t)t

α−1dt). Conse-

quently, (4.3.6) takes the form

u(t) = Iαf(t, v(t),Dpv(t))− tα−1Iαf(1, v(1),Dpv(1))

+

∫ 10 φ(t)[I

αf(t, v(t),Dpv(t))− tα−1Iαf(1, v(1),Dpv(1))]dt

(1−∫ 10 φ(t)t

α−1dt)

=

∫ 1

0G1α(t, s)f(s, v(s),D

pv(s))ds +

∫ 1

0G2α(t, s)f(s, v(s),D

pv(s))ds

=

∫ 1


pv(s))ds.

Further, we have

Dqu(t) = Iα−qf(t, v(t),Dpv(t)) − tα−q−1

(1− ρ)Γ(α− q − 2)Iαf(1, v(1),Dpv(1))

+tα−q−1

∫ 10 φ(t)[I

αf(t, v(t),Dpv(t))− tα−1Iαf(1, v(1),Dpv(1))]dt

(1− ρ)Γ(α − q − 2).

(4.3.7)

Similarly, using the second equation of the system, we can show that

v(t) =

∫ 1

0G1β(t, s)g(s, u(s),D

qu(s))ds+

∫ 1

0G2β(t, s)g(s, u(s),D

qu(s))ds

=

∫ 1


qu(s))ds.

75

Thus in view of Lemma 4.3.1, the Coupled system (4.3.1) in equivalent integral equations

forms is

u(t) =

∫ 1


pv(s))ds, t ∈ [0, 1],

v(t) =

∫ 1


qu(s))ds, t ∈ [0, 1].

(4.3.8)

Lemma 4.3.2. Under the assumptions of Lemma (4.3.1), (u, v) ∈ X × Y is a solutionof the system (4.3.1) if and only if (u, v) ∈ X × Y is a solution of the system of integralequations (4.3.8).

Proof. If (u, v) ∈ X × Y is a solution of system (4.3.1), then by Lemma (4.3.1), (u, v) is asolution of the system of integral equations (4.3.8).

Conversely, let (u, v) satisfies (4.3.8), using Dαtα−k = 0, k = 1, 2, ...N where N is thesmallest integer greater than or equal to α, we have

Dαu(t) =Dα[Iαf(t, v(t),Dpv(t))

+tα−1

(1− ρ)Γ(α)

[ ∫ 1

0

( ∫ 1

sφ(w)(w − s)α−1dw − (1− s)α−1

)f(s, v(s),Dpv(s))ds

]

which on simplification implies that Dαu(t) = f(t, v(t),Dpv(t)) and by similar procedure,we have Dβv(t) = f(t, u(t),Dqu(t)). Further, it is easy to verify that

u(0) = u′(0) = 0, v(0) = v′(0) = 0, u(1) =

∫ 1

0φ(t)u(t)dt, y(1) =

∫ 1

0ψ(t)v(t)dt.

We list the following hypothesis:

(A14) There exist ci, di, ∈ R+ ∪ 0 and 0 ≤ µi, ρi < 1(i = 1, 2) such that

|f(t, u, v)| ≤ a(t) + c1|u|ρ1 + c2|v|ρ2 and |g(t, u, v)| ≤ b(t) + d1|u|µ1 + d2|v|µ2 ,

where a(t), b(t) ∈ C[0, 1] are non-negative.

(A15) There exist ci, di, ∈ R+ and µi, ρi > 1(i = 1, 2) such that

|f(t, u, v)| ≤ c1|u|ρ1 + c2|v|ρ2 and |g(t, u, v)| ≤ d1|u|µ1 + d2|v|µ2 .

(A16) Since φ(t), ψ(t) ∈ L1[0, 1], there exist positive real numbers h, l such that maxt∈[0,1] |φ(t)| ≤

h,maxt∈[0,1] |ψ(t)| ≤ l.

76

Define T1 : X → X and T2 : Y → Y by

T1u(t),=

∫ 1

0Gα(t, s)g(s, u(s),D

qu(s))ds, T2v(t) =

∫ 1

0Gβ(t, s)f(s, v(s),D

pv(s))ds

and T : X × Y → X × Y by T (x, y)(t) = (T2v(t), T1u(t)). Then by Lemma 4.3.1, solutions

of the system (4.3.1) are fixed points of the operator T . In view of (A16), it should be noted

that

∫ 1

0Gα(t, s)ds ≤

∫ 1

0

(1− s)α−1

Γ(α)ds+

∫ 1

0

[ tα−1

(1− ρ)Γ(α)(

∫ 1

sφ(w)(w − s)α−1dw − (1− s)α−1)

]ds

=1

Γ(α+ 1)

(1 +

h(α+ 1)

α(1− ρ)

)

(4.3.9)

and similarly, ∫ 1

0Gβ(t, s)ds ≤

1

Γ(β + 1)

(1 +

l(β + 1)

β(1 − θ)

). (4.3.10)

Let us use the following notations

A =1

Γ(α+ 1)(1 +

h(α+ 1)

α(1− ρ)) +

1

Γ(α− q + 1)+

1

α(1− ρ)Γ(α − q)+

h

α(α+ 1)(1 − ρ)Γ(α− q),

B =1

Γ(β + 1)(1 +

l(β + 1)

β(1− θ)) +

1

Γ(β − p+ 1)+

1

β(1− θ)Γ(β − p)+

l

β(β + 1)(1 − θ)Γ(β − p),

M = maxt∈[0,1]

∫ 1

0|Gα(t, s)a(s)|ds+

1

Γ(α− q)[

∫ 1

0(1− s)α−q−1a(s)ds+

1

1− ρ

∫ 1

0(1− s)α−1a(s)ds

+h

α(1 − ρ)

∫ 1

0(1− s)α−1a(s)ds],

N = maxt∈[0,1]

∫ 1

0|Gβ(t, s)b(s)|ds+

1

Γ(β − p)[

∫ 1

0(1− s)β−p−1b(s)ds+

1

1− θ

∫ 1

0(1− s)β−1b(s)ds

+l

β(1− θ)

∫ 1

0(1− s)β−1b(s)ds]


Theorem 4.3.3. Under the assumptions (A12), (A13), (A14) and (A15), the operator T :X × Y → X × Y is completely continuous.

Proof. Choose R ≥ max(3Ac1)1

1−ρ1 , (3Ac2)1

1−ρ2 , (3Bd1)1

1−µ1 , (3Bd2)1

1−µ2 , 3M, 3N and de-fine U = (u, v) ∈ X × Y : ||(u, v)||X×Y ≤ R. For (u, v) ∈ U , using (A14) and (4.3.1), we

77

have

|T1u(t)| = |∫ 1


pv(s))ds| ≤∫ 1

0|Gα(t, s)[a(s) + c1|R|ρ1 + c2|R|ρ2 ]ds

≤∫ 1

0|Gα(t, s)a(s)|ds+ (c1|R|ρ1 + c2|R|ρ2)

∫ 1

0|Gα(t, s)|ds

≤∫ 1

0|Gα(t, s)a(s)|ds+

(c1|R|ρ1 + c2|R|ρ2)Γ(α+ 1)

(1 +

h(α+ 1)

α(1− ρ)

),

which implies that

|T1u(t)| ≤∫ 1

0|Gα(t, s)a(s)|ds +

(c1|R|ρ1 + c2|R|ρ2)Γ(α+ 1)

(1 +

h(α + 1)

α(1− ρ)

). (4.3.11)

Now using (4.3.7), (A13) and (A15), we obtain

|T1Dqu(t)| ≤∣∣∣∣Iα−qf(t, v(t),Dpv(t))

+tα−q−1

(1− ρ)Γ(α− q)

[ ∫ 1

0

( ∫ 1

sφ(w)(w − s)α−1dw − (1− s)α−1


]∣∣∣∣

≤∣∣∣∣

1

Γ(α− q)[

∫ t

0(t− s)α−q−1f(s, v(s),Dpv(s))ds]

∣∣∣∣+∣∣∣∣

tα−q−1

(1− ρ)Γ(α− q)

[ ∫ 1

0

( ∫ 1

sφ(w)(w − s)α−1dw − (1− s)α−1


]∣∣∣∣

≤ 1

Γ(α− q)

[ ∫ t

0(t− s)α−q−1a(s)ds+ (c1|R|ρ1 + c2|R|ρ2)

∫ t

0(t− s)α−q−1ds

]

+1

(1− ρ)Γ(α− q)[

∫ 1

0

∫ 1

s|φ(w)|(w − s)α−1(a(s) + (c1|R|ρ1 + c2|R|ρ2)dwds

+

∫ 1

0(1− s)α−1(a(s) + c1|R|ρ1 + c2|R|ρ2)ds]

≤ 1

Γ(α− q)

∫ 1

0(1− s)α−q−1a(s)ds+

c1|R|ρ1 + c2|R|ρ2Γ(α− q + 1)

+

h

(1− ρ)αΓ(α − q)

∫ 1

0(1− s)αa(s)ds+

1

(1− ρ)Γ(α− q)

∫ 1

0(1− s)α−1a(s)ds

+h

(1− ρ)α(α + 1)Γ(α− q)(c1|R|ρ1 + c2|R|ρ2 +

1

(1− ρ)Γ(α− q)(c1|R|ρ1 + c2|R|ρ2

⇒ |T1Dqu(t)| ≤ 1

Γ(α− q)

∫ 1

0(1− s)α−q−1a(s)ds+

h

(1− ρ)αΓ(α− q)

∫ 1

0(1− s)αa(s)ds

+1

(1− ρ)Γ(α− q)

∫ 1

0(1− s)α−1a(s)ds+ (c1|R|ρ1 + c2|R|ρ2)

[ 1

Γ(α− q + 1)

+1

α(1 − ρ)Γ(α− q)+

h

α(α + 1)(1 − ρ)Γ(α− q)

].

(4.3.12)

78

It follows from (4.3.11) and (4.3.12) that

‖T1(u)‖X = maxt∈[0,1]

|T1(u)| + maxt∈[0,1]

|DqT1(u)| ≤M + (c1|R|ρ1 + c2|R|ρ2)A.

Hence

‖T1(u)‖X ≤ R

3+R

3+R

3= R. (4.3.13)

Similarly, we obtain

‖T2(v)‖Y ≤ R. (4.3.14)

Thus, we have ‖T (u, v)‖X×Y ≤ R which implies that T : U → U and is uniformly bounded.The continuity of the operator T follows from the continuity of Gα, Gβ , f and g.

In order to show that T is completely continues operator, choose (u, v) ∈ U and t, τ ∈ Isuch that t < τ and use the mean value theorem, we have

|Gα(t, s)−Gα(τ, s)| ≤ Hc,d,e(s)(t− τ),

|Gβ(t, s)−Gβ(τ, s)| ≤ Kc,d,e(s)(t− τ),(4.3.15)

where c, d, e ∈ (t, τ),

Hc,d,e(s) =α− 1

Γ(α)

[(c− s)α−2 − 1

1− ρ(dα−2 − eα−2)

( ∫ φ

s(w)(w − s)α−1dw − (1− s)α−1

)].

Further,

|T1u(t)− T1x(τ)| = |∫ 1

0(Gα(t, s)−Gα(τ, s))f(s, v(s),D

pv(s))|ds

≤∫ 1

0Hc,d,e(s)|f(s, v(s),Dpv(s))|(t− τ)ds

≤∫ 1

0Hc,d,e(s)[a(s) + c1|R|ρ1 + c2|R|ρ2 ](t− τ)ds

≤ (t− τ)

∫ 1

0Hc,d,e(s)[a(s) + c1|R|ρ1 + c2|R|ρ2 ]ds,

(4.3.16)

|DqT1u(t)−DqT1u(τ)| ≤ |Iα−qf(t, v(t),Dpv(t))− Iα−qf(τ, v(τ),Dpv(τ))|

+1

(1− ρ)Γ(α− q)[

∫ 1

0

∫ 1

s|φ(w)|(w − s)α−1|f(s, v(s),Dpv(s))|dwds

+

∫ 1

0(1− s)α−1|f(s, v(s),Dpv(s))|ds](tα−q−1 − τα−q−1) ≤ Hc,d(s)(t− τ)

(4.3.17)

79

where c, d ∈ (t, τ),

Hc,d(s) =

∫ 1

0

[α− q − 1

Γ(α− q)(c− s)α−q−2 +

(1− s)α

α(1− ρ)+α− q − 1

Γ(α− q)(1− s)α−1dα−q−2

]×

|f(s, v(s),Dpv(s))|ds.


|T2v(t)− T2y(τ)| ≤ (t− τ)

∫ 1

0Kc,d,e(s)[b(s) + d1|R|µ1 + d2|R|µ2 ]ds,

|DqT2v(t)−DqT2v(τ)| ≤ (t− τ)

∫ 1

0Kc,d(s)ds

(4.3.18)

where

Kc,d(s) =

∫ 1

0

[β − p− 1

Γ(β − p)(c− s)β−p−2 +

(1− s)β

β(1− θ)+β − p− 1

Γ(β − p)(1− s)n−1dβ−p−2

]×

|g(s, u(s),Dqu(s))|ds.

If t→ τ , then from (4.3.16), (4.3.17) and (4.3.18) it follows that

|T1u(t)− T1u(τ)| → 0, |DqT1u(t)−DqT1u(τ)| → 0,

|T2v(t)− T2v(τ)| → 0, |DpT2v(t)−DpT2v(τ)| → 0.

By Arzela- Ascoli’s theorem, it follows that T : X × Y → X × Y is completely continuous.Hence, by the Schauder fixed point theorem, T has a fixed point.

Theorem 4.3.4. Under the assumptions (A12), (A13), (A15), the system of the Boundaryvalue problems (4.3.1) has at least one solution.

Proof. The proof is similar to that of Theorem (4.3.3).

Assume that there exist constants K1, K2 > 0 such that

(A17) |f(t, u, v)−f(t, u, v)| ≤ K1(|u− u|+ |v− v|), for each t ∈ [0, 1] and for all u, u, v, v ∈ R.

(A18) |g(t, u, v)−g(t, x, y)| ≤ K2(|x− x|+ |y− y|), for each t ∈ [0, 1] and for all x, x, y, y ∈ R.

(A19) there exist continuous and non-decreasing functions ψ if , ψig : [0,∞) → [0,∞) and

φif , φig : [0, 1] → (0,∞) for i = 1, 2 such that

|f(t, u, v)| ≤ φ1f (t)ψ1f (|u|) + φ2f (t)ψ

2f (|v|) for t ∈ [0, 1], u, v ∈ R

|g(t, u, v)| ≤ φ1g(t)ψ1g(|u|) + φ2g(t)ψ

2g(|v|) for t ∈ [0, 1], u, v ∈ R.

80

(A20) There exist γ > 0 and δ such that

γ[

1Γ(α+1)

(1 + h+α

α(1−ρ)

)+ 1

Γ(α−q)(

1α−q +

h+α+1(1−ρ)α(α+1)

)]Q

> 1,

δ[

1Γ(β+1)

(1 + l+β

β(1−θ)

)+ 1

Γ(β−p)(

1β−p +

l+β+1(1−θ)β(β+1)

)]S

> 1,

where Q = φ1f (η)ψ1f (|γ|) + |φ2f (η)|ψ2

f (|γ|) and S = φ1g(ξ)ψ1g(|γ|) + φ2g(ξ)ψ

2g(|δ|) where

ξ, η ∈ R+.

Theorem 4.3.5. Assume that (A12)− (A15), (A19) and (A20) hold. Define

Ur = (u, v) ∈ X × Y : ‖(u, v)‖X×Y ≤ r

and assume that there exist (u, v) ∈ ∂Ur such that (u, v) = λT (u, v) and λ ∈ (0, 1), then thesystem of the Boundary value problem (4.3.1) has at least one solution.

Proof. In view of (4.3.2) and (A19), we have

|u(t)| ≤∫ 1

0|Gα(t, s)||f(s, v(s),Dpv(s))|ds

≤ 1

Γ(α+ 1)

(φ1f (s)ψ

1f (‖v‖Y ) + φ2f (s)ψ

2f (‖y‖Y)

)

+1

Γ(α+ 1)(1 − ρ)

(1 +

h

α

)[φ1f (s)ψ

1f (‖v‖Y ) + |φ2f (s)|ψ2

f (‖v‖Y )]

≤ 1

Γ(α+ 1)

(1 +

h+ α

α(1− ρ)

)(φ1f (s)ψ

1f (‖v‖Y ) + φ2f (s)ψ

2f (‖v‖Y )

),

|Dpu(t)| ≤ 1

Γ(α− q)

( 1

α− q+

h+ α+ 1

(1− ρ)α(α + 1)

)[φ1f (s)ψ

1f (‖v‖Y ) + φ2f (s)ψ

2f (‖v‖Y )

].


|u(t)| + |Dpu(t)| ≤[ 1

Γ(α+ 1)

(1 +

h+ α

α(1 − ρ)

)+

1

Γ(α− q)

( 1

α− q+

h+ α+ 1

(1− ρ)α(α + 1)

)]×

(φ1f (s)ψ

1f (‖v‖Y ) + φ2f (s)ψ

2f (‖v‖Y )

),

which implies that

‖u‖X ≤[ 1

Γ(α+ 1)

(1 +

h+ α

α(1 − ρ)

)+

1

Γ(α− q)

( 1

α− q+

h+ α+ 1

(1− ρ)α(α+ 1)

)]

(φ1f (s)ψ

1f (‖v‖Y ) + φ2f (s)ψ

2f (‖v‖Y )

).

81


‖v‖Y ≤[ 1

Γ(β + 1)

(1 +

l + β

β(1 − θ)

)+

1

Γ(β − p)

( 1

β − p+

l + β + 1

(1− θ)β(β + 1)

)]

(φ1g(s)ψ

1g(‖u‖X ) + φ2g(s)ψ

2g(‖u‖X )

).


‖u‖X[1

Γ(α+1)

(1 + h+α

α(1−ρ))+ 1

Γ(α−q)(

1α−q +

h+α+1(1−ρ)α(α+1)

)]Q< 1,

‖v‖Y[1

Γ(β+1)

(1 + l+β

β(1−θ))+ 1

Γ(β−p)(

1β−p +

l+β+1(1−θ)β(β+1)

)]S< 1,

a contradiction to (A20), hence it follows that (u, v) 6= λT (u, v) for (u, v) ∈ ∂Ur, λ ∈(0, 1). By Theorem 2.3.9, the system of Boundary value problem (4.3.1) has at least onesolution.


Theorem 4.3.6. Assume that (A12), (A17), (A18) hold such that K1A < 1 and K2B < 1,then the system of Boundary value problems (4.3.1) has a unique solution.

Proof. We use Banach Contraction principle. Let (u, v), (u, v) ∈ X × Y, the in view of(A17) and (A18), we obtain

|T1u(t)− T1u(t)| ≤∫ 1

0

(1− s)α−1

Γ(α)K1‖v − v‖Yds

+1

(1− ρ)Γ(α)

∫ 1

0

(∫ 1

s|φ(w)|(w − s)α−1dw + (1− s)α−1

)dsK1‖v − v‖Y

≤ K1‖v − v‖Y(

1

Γ(α+ 1)+

h+ 1

(1− ρ)αΓ(α+ 1)

)

≤ K1

Γ(α+ 1)

(1 +

h+ 1

α(1 − ρ)

)‖v − v‖Y

and

|DqT1u(t)−DqT1u(t)| ≤1

Γ(α− q)

∫ t

0(t− s)α−q−1 max

t∈[0,1]

∣∣∣∣f(t, v(t),Dpv(t))− f(t, v(t),Dpv(t))

∣∣∣∣ds

+1

(1− ρ)Γ(α− q)

[ ∫ 1

0

(|φ(w)|(w − s)α−1dw + (1− s)α−1

)×

maxt∈[0,1]

∣∣∣∣f(t, v(t),Dpv(t)) − f(t, v(t),Dpv(t))

∣∣∣∣ds]

≤ K1

Γ(α− q)

(1

α− q+

h+ 1 + α

α(α + 1)(1− ρ))

)‖v − v‖Y .

82

Now

‖T1u(t)− T1u(t)‖X

≤ K1

[(1 + h+1α(1−ρ) )

Γ(α+ 1)+

1

Γ(α− q)

(1

α− q+

h+ 1 + α

α(α+ 1)(1 − ρ)

)]‖v − v‖Y

≤ K1A‖v − v‖Y < ‖v − v‖Y .

(4.3.19)

Similarly,

‖T2v(t)− T2v(t) Y | ≤ K2B‖u− u‖X < ‖u− u‖X . (4.3.20)

Hence, using (4.3.19) and (4.3.20), we get ‖T (u, v) − T (u, v)‖ < ‖(u, v) − (u, v)‖X×Y .By Banach contraction principle, the operator T has a unique fixed point.

Example 4.3.1. Consider the system of Boundary value problem

D73u(t) =

1

1 + t3+

1

πv(t)

13 +

t

π(D

12 v(t))

12 ; t ∈ [0, 1],

D73 v(t) =

1

(20π + t2)+

1

π(sin

13 (u(t)) + sin

12

(D

13u(t)

); t ∈ [0, 1],

subject to the boundary conditions

u(0) = u′(0) = 0, v(0) = v′(0) = 0, u(1) =

∫ 1

0tu(t)dt, v(1) =

∫ 1

0tv(t)dt.

(4.3.21)

Here, we have

|f(t, u, v))| = | 1

1 + t3+

1

π|v(t)| 13 +

t

π|D 1

2 v(t)| 12 | ≤ | 1

1 + t3|+ 1

π|v(t)| 13 +

1

π|D 1

2 v(t)| 12 ,

|g(t, u, v)| = | 1

(20π + t2)+

1

π

(sin

13 (u(t)) + sin

12

(D

13u(t)

))|

≤ | 1

(20π + t2)|+ 1

π|u(t)| 13 +

1

π|D 1

3u(t))| 12 .

Taking

a(t) = | 1

1 + t3|, b(t) = | 1

(20π + t2)|, c1 = c2 = d1 = d2 =

1

π, α = β =

7

3, p =

1

2, q =

1

3

and

φ(t) = t = ψ(t),

∫ 1

0t·tα−1dt =

3

10,

∫ 1

0t·tβ−1dt =

3

10, h = 1, l = 1, ρ1 =

1

3, ρ2 =

1

2, µ1 =

1

3

and µ2 = 13 , we see that the assumptions (A12) − (A14), (A16) are satisfied. Hence, by

Theorem 4.3.3, the system of Boundary value problem(4.3.1) has at least one solution.

83

Example 4.3.2. Consider the following system

D52u(t) =

1

20 + π2cos2(v(t)) +

t

20 + π2sin3

(D

12 v(t)

), t ∈ [0, 1],

D52 v(t) =

1

40u(t)2 +

1

eπ(D

13u(t))2, t ∈ [0, 1],

subject to the conditions

u(0) = u′(0) = 0, v(0) = v′(0) = 0, u(1) =

∫ 1

0tu(t)dt, v(1) =

∫ 1

0tv(t)dt.

(4.3.22)

Here,

|f(t, u, v))| = | 1

20 + π2cos2(v(t)) +

t

20 + π2sin3

(D

12 v(t)

)

≤ 1

20 + π2|v(t)|2 + 1

20 + π2|D 1

2 v(t)|3

and

|g(t, u, v)| = | 140u(t)2 +

1

eπ(D

13u(t))2| ≤ 1

40||u(t)|2 + 1

eπ|D 1

3u(t)|2.

Further,

α = β =5

2, p =

1

2, q =

1

3and φ(t) = t = ψ(t).

We have, |φ(t)| ≤ 1 and |ψ(t)| ≤ 1,∫ 1

0t · tα−1dt =

∫ 1

0t52 dt =

2

7,

∫ 1

0t · tβ−1dt =

∫ 1

0t52 dt =

2

7,

where h = 1, l = 1, ρ1 = 2, ρ2 = 3, µ1 = 2, µ2 = 2. Thus assumptions (A12), (A13), (A15)are satisfied, by Theorem 4.3.3, the Boundary value problem(4.3.2) has at least one solution.

4.4 Coupled system with impulsive boundary conditions

In this section, we develop sufficient conditions for existence and uniqueness of solutions to

the following complex dynamical network [107], in the form of a coupled system of p+q+2-

point boundary conditions for impulsive fractional differential equations given by

cDαu(t) = f(t, u(t), v(t)), t ∈ [0, 1], t 6= tj, j = 1, 2, ...,m,

cDβv(t) = g(t, u(t), v(t)), t ∈ [0, 1], t 6= ti, i = 1, 2, . . . , n,

u(0) = h(u), u(1) = g(u) and v(0) = κ(v), v(1) = f(v),

∆u(tj) = Ij(u(tj)), ∆u′(tj) = Ij(u(tj)), j = 1, 2, ...,m,

∆v(ti) = Ii(v(ti)), ∆v′(ti) = Ii(v(ti)), i = 1, 2, . . . , n,

(4.4.1)

84

where 1 < α, β ≤ 2, f, g : [0, 1]×R2 → R are continuous functions and g, h : X → R, f , κ :

Y → R are continuous functionals define by

g(u) =

p∑

j=1

λju(ξj), h(u) =

p∑

j=1

λju(ηj), f(v) =

q∑

i=1

δiv(ξi), κ(v) =

q∑

i=1

δiv(ηi),

ξi, ηi, ξj, ηj ∈ (0, 1) for i = 1, 2, . . . , p, j = 1, 2, . . . , q, and

∆u(tj) = u(t+j )− u(t−j ), ∆u′(tj) = u′(t+j )− u′(t−j )

∆v(ti) = v(t+i )− v(t−i ), ∆v′(ti) = v′(t+i )− v′(t−i ).

In the above, the notations u(t+j ), v(t+i ) are right and u(t

−j ), v(t

−i ) are left limits respectively.

Moreover, Ir, Ir(r = j, i) : R → R are continuous functions. We assume thatp∑r=1

λjηα−1r <

1,q∑r=1

δrξα−1r < 1, r = i, j. Here, we remark that the system (4.4.1) includes as a special

case the following well studied Hopfield neural networks and design model in the presence

of impulses[109]

duidt

= −aiui(t) +m∑

j=1

bijfj(uj(t)) + ci, t 6= tk, i = 1, 2, . . . ,m, k = 1, 2, . . . ,m,

∆ui(tk) = Ii(ui(tk)).

We obtain necessary and sufficient conditions for the existence and uniqueness of positive

solution via Krasnoselskii’s fixed point theorem and Banach contraction principle. For

tj ∈ (0, 1) such that t1 < t2 < · · · < tm, and I ′ = I − t1, t2, . . . , tm define the Banach

space

X = u : [0, 1] → R : u ∈ C(I ′) and left u(t+j ) and right Limit u(t−j ) exist and

u(t−j )− u(tj , ), 1 ≤ j ≤ m

and

Y = v : [0, 1] → R : v ∈ C(I ′) and left v(t+i ) and right Limit v(t−i ) exist and

85

v(t−i )− v(ti, ), 1 ≤ i ≤ n.


(A21) for any u, v ∈ C([0, 1],R),∃ Kg,Kh,Kf ,Kκ such that

‖g(u)− g(v)‖ ≤ Kg‖u− v‖, ‖f(u)− f(v)‖ ≤ Kf‖u− v‖,

‖h(u) − h(v)‖ ≤ Kh‖u− v‖, ‖κ(u)− κ(v)‖ ≤ Kκ‖u− v‖;

(A22) for all u, u ∈ R for all t ∈ [0, 1] there exist Lf , such that

|f(t, u, v) − f(t, u, v)| ≤ Lf [|u− u|+ |v − v|];

(A23) for all v, v ∈ R for all t ∈ [0, 1] there exist Lg, such that

|g(t, u, v) − g(t, u, v)| ≤ Lg[|u− u|+ |v − v|].

(A24) There exist constants Al, Bl > 0 (l = 1, 2) such that for u, u, v, v ∈ R,

|Ij(u)− Ij(u)| ≤ A1|u− u|, |Ij(u)− Ij(u| ≤ A2|u− u|, j = 1, 2, . . . ,m

|Ii(v)− Ii(v)| ≤ B1|v − v|, |Ii(v) − Ii(v| ≤ B2|v − v|, i = 1, 2, . . . , n.

Define ∆1 =2Lf

Γ(α+1) +m(A1 +A2) +Kg +Kh and ∆2 =2Lg

Γ(β+1) + n(B1 +B2) +Kf +Kk.

Theorem 4.4.1. For w ∈ C([0, 1],R), the following impulsive boundary value problem

cDαu(t) = w(t), t ∈ [0, 1], t 6= tj , j = 1, 2, . . . ,m, 1 < α ≤ 2,

∆u(tj) = Ij(u(tj)), ∆u′(tj) = Ij(u(tj)), j = 1, 2, . . . ,m,

u(0) = h(u), u(1) = g(u)

(4.4.2)

has a solution of the form

u(t) =1

Γ(α)

t∫

0

(t− s)α−1w(s)ds− t

Γ(α)

1∫

0

(1− s)α−1w(s)ds

+ tg(u) + (1− t)h(u) +m∑

j=1

Gj(t, uj),

86

where

Gj(t, uj) =

− t[Ij(u(tj)) + (1− tj)Iju(tj)], 0 ≤ t ≤ tj;

(1− t)[Ij(u(tj))− tj Iju(tj)], tj < t ≤ 1,

Proof. Using lemma 2.2.9 and the differential equation in (4.4.1), there exist real constantsC0, C1 such that

u(t) = Iαw(t)− C0 − C1t =1

Γα

∫ t

0(t− s)α−1w(s)ds−C0 − C1t, t ∈ [0, t1], (4.4.3)

from which it follows that

u′(t) = Iα−1w(t)− C1 =1

Γ(α− 1)

∫ t

0(t− s)α−2w(s)ds− C1, t ∈ [0, t1]. (4.4.4)

Similarly, for t ∈ (t1, t2], there exist real numbers d0, d1 such that

u(t) =1

Γα

∫ t

t1

(t− s)α−1w(s)ds− d0 − d1(t− t1),

u′(t) =1

Γ(α− 1)

∫ t

t1

(t− s)α−2w(s)ds− d1.

(4.4.5)


u(t−1 ) =1

Γα

∫ t1

0(t1 − s)αw(s)ds− C0 − C1t1,

u(t+1 ) = −d0,

u′(t−1 ) =1

Γ(α− 1)

∫ t1

0(t1 − s)α−2w(s)ds− C1,

u′(t+1 ) = −d1.

(4.4.6)

Using the impulsive conditions ∆u(t1) = u(t+1 )− u(t−1 ) = I1(u(t1)) and ∆u′(t1) = u′(t+1 )−u′(t−1 ) = I1(u(t1)), we obtain

− d0 =1

Γα

∫ t1

0(t1 − s)αw(s)ds− C0 − C1t1 + I1(u(t1)),

− d1 =1

Γ(α− 1)

∫ t1

0(t1 − s)α−2w(s)ds− C1 + I1(u(t1)).

(4.4.7)

Hence we have

u(t) =1

Γα

∫ t

t1

(t− s)α−1w(s)ds+1

Γα

∫ t1

0(t1 − s)αw(s)ds

+t− t1

Γ(α− 1)

∫ t1

0(t1 − s)α−2w(s)ds+ (t− t1)I1(u(t1)) + I1(u(t1))− C0 − C1t, t ∈ (t1, t2].

(4.4.8)

87

Repeating in the same fashion, we obtain expression for the solution u(t) for t ∈ (ti, ti+1]as follows

u(t) =1

Γα

∫ t

tj

(t− s)α−1w(s)ds+1

Γα

k∑

j=1

∫ tj

tj−1

(tj − s)α−1w(s)ds

+1

Γ(α− 1)

k∑

j=1

(t− tj)

∫ tj

tj−1

(tj − s)α−2w(s)ds+k∑

j=1

(t− tj)Ij(u(tj))

+

k∑

j=1

Ij(u(tj))− C0 − C1t, t ∈ (tj , tj+1], k = 1, 2, ...,m.

(4.4.9)

Now, using the boundary conditions u(0) = g(u), u(1) = h(u) on (4.4.9), calculating thevalues of C0, C1, the equation (4.4.3) can be rewritten as follows

u(t) =1

Γ(α)

t∫

0

(t− s)α−1w(s)ds− t

Γ(α)

1∫

0

(1− s)α−1w(s)ds

+ tg(u) + (1− t)h(u) +m∑

j=1

Gj(t, u(tj)),

(4.4.10)

where

Gj(t, u(tj)) =

− t[Ij(v(tj)) + (1− tj)Ij(v(tj))], 0 ≤ t ≤ tj;

(1− t)[Ij(v(tj))− tj Ijv(tj)], tj < t ≤ 1.

Similarly, the impulsive boundary value problem

cDβv(t) = w(t), t ∈ [0, 1], t 6= ti, i = 1, 2, . . . , n, 1 < β ≤ 2,

∆v(ti) = Ii(v(ti)), ∆v′(ti) = Ii(v(ti)), i = 1, 2, . . . , n,

v(0) = κ(v), v(1) = f(v)

(4.4.11)

has a solution of the form

v(t) =1

Γ(β)

t∫

0

(t− s)β−1w(s)ds− t

Γ(β)

1∫

0

(1− s)β−1w(s)ds

+ tf(v) + (1− t)κ(v) +

n∑

i=1

Gi(t, vi),

where

Gi(t, v(ti)) =

− t[Ii(v(ti)) + (1− ti)Iiv(ti)], 0 ≤ t ≤ ti,

(1− t)[Ii(v(ti))− tiIiv(ti)], ti < t ≤ 1.

88

In view of Theorem (4.5.2), if (u, v) is a solution of the coupled system (4.4.1), then, wehave

u(t) =1

Γ(α)

t∫

0

(t− s)α−1f(s, u(s), v(s))ds − t

Γ(α)

1∫

0

(1− s)α−1f(s, u(s), v(s))ds

+ tg(u) + (1− t)h(u) +

m∑

j=1

Gj(t, u(tj)),

v(t) =1

Γ(β)

t∫

0

(t− s)β−1g(s, u(s), v(s))ds − t

Γ(β)

1∫

0

(1− s)β−1g(s, u(s), v(s))ds

+ tf(v) + (1− t)κ(v) +

n∑

i=1

Gi(t, v(ti)).

(4.4.12)

Theorem 4.4.2. Let f, g are continuous, then (u, v) ∈ X×Y is a solution of the Boundaryvalue problem(4.4.1) if and only if (u, v) is the solution of the coupled system of integralequations

u(t) =1

Γ(α)

t∫

0

(t− s)α−1f(s, u, v)ds− t

Γ(α)

1∫

0

(1− s)α−1f(s, u, v)ds

+ tg(u) + (1− t)h(u) +

m∑

j=1

Gj(t, u(tj))

v(t) =1

Γ(β)

t∫

0

(t− s)β−1g(s, u, v)ds − t

Γ(β)

1∫

0

(1− s)β−1g(s, u, v)ds

+ tf(v) + (1− t)κ(v) +

n∑

i=1

Gi(t, v(ti)).

(4.4.13)

Proof. If (u, v) is a solution of the Boundary value problem(4.4.1), then it is a solution ofthe system of integral equation (4.4.13). Conversely Let (u, v) is solution of the system ofintegral equations (4.4.13), then

cDαu(t) = f(t, u(t), v(t)),cDβv(t) = g(t, u(t), v(t)),

and u(0) = h(u), u(1) = g(u) and v(0) = κ(v), v(1) = f(v)where t ∈ [0, 1], t 6= tj, j = 1, 2, ...,m and t 6= ti, i = 1, 2, . . . , n.It is also clear that

Gj(0, u(tj)) = Gj(1, u(tj)) = 0, j = 1, 2, . . . ,m

89

Gi(0, v(ti)) = Gi(1, v(ti)) = 0, i = 1, 2, . . . , n,

and∆u(tj) = Ij(u(tj)), ∆u′(tj) = Ij(u(tj)), j = 1, 2, . . . ,m

∆v(ti) = Ii(v(ti)), ∆v′(ti) = Ii(v(ti)), i = 1, 2, . . . , n.

Thus (u, v) is the solution of the Boundary value problem(4.4.1).

Define T1, T2 : X × Y → X × Y by

T1(u, v) =1

Γ(α)

t∫

0

(t− s)α−1f(s, u(s), v(s))ds− t

Γ(α)

1∫

0

(1− s)α−1f(s, u(s), v(s))ds

+ tg(u) + (1− t)h(u) +

m∑

j=1

Gj(t, uj),

T2(u, v) =1

Γ(β)

t∫

0

(t− s)β−1g(s, u(s), v(s))ds − t

Γ(β)

1∫

0

(1− s)β−1g(s, u(s), v(s))ds

+ tf(v) + (1− t)κ(v) +

n∑

i=1

Gi(t, vi),

and T (u, v) = (T1(u, v), T2(u, v)). Then, solutions of the coupled system of integral equa-

tions (4.4.13) are fixed points of T .


Theorem 4.4.3. In addition to the assumption (A21) − (A24), assume that the followinghold

(A25) there exist constants Cl, Dl > 0 for l = 1, 2 such that j=1,2,. . . ,m, i=1,2,. . . ,n,

|Ij(uj)| ≤ C1, |Ij(uj)| ≤ C2, |Ii(ui)| ≤ D1, |Ii(vi)| ≤ D2.

(A26) there exists constants ρl, µl (l = 1, 2) such that

|g(u)| ≤ ρ1, |h(u)| ≤ ρ2 ∀u ∈ X , |f(v)| ≤ µ1, |κ(v)| ≤ µ2 ∀v ∈ Y.

If m(A1 +A2) +Kg +Kh < 1 and n(B1 +B2) +Kf +Kκ < 1, then the system(4.4.1) hasat least one positive solution.

90

Proof. Choose ρ(t) = f(t, 0, 0), σ(t) = g(t, 0, 0), ‖ρ‖L1 = max1∫0

|f(s, 0, 0)|ds, ‖σ‖L1 =

max1∫0

|g(s, 0, 0)|ds andR ≥ max

2‖ρ‖L1

Γ(α) +m(C1 + C2) + ρ1 + ρ2,2‖σ‖L1

Γ(β) + n(D1 +D2) + µ1 + µ2

.

Construct a closed ball P ⊂ X × Y such that

P = (u, v) ∈ X × Y : ‖(u, v)‖ ≤ R.

Split the operator T in two parts as T = F +G with F = (F1, F2) and G = (G1, G2) where

F1(u, v) =1

Γ(α)

t∫

0

(t− s)α−1f(s, u(s), v(s))ds − t

Γ(α)

1∫

0

(1− s)α−1f(s, u(s), v(s))ds

F2(u, v) =1

Γ(β)

t∫

0

(t− s)β−1g(s, u, v)ds − t

Γ(β)

1∫

0

(1− s)β−1g(s, u, v)ds

G1u(t) = tg(u) + (1− t)h(u) +

m∑

j=1

Gj(t, u(tj))

G2v(t) = tf(v) + (1− t)κ(v) +n∑

i=1

Gi(t, v(ti)).

Clearly T1 = F1 + G1, T2 = F2 + G2. Now we show that T (u, v) = F (u, v) + G(u, v) ∈P, for all (u, v) ∈ P. For any (u, v) ∈ P, we have

|T1(u, v)| ≤ 1Γ(α)

t∫0

(t− s)α−1|f(s, u, v)|ds+ 1Γ(α)

1∫0

(1− s)α−1|f(s, u, v)|ds

+ |tg(u)|(1 − t)h(u)|+m∑j=1

|Ij(u(tj))|+m∑j=1

|Ij(u(tj))|

≤ 2‖ρ‖L1

Γ(α) + ρ1 + ρ2 +m(C1 + C2) ≤ R.

Similarly, we have |T2(u, v)| ≤ 2‖σ‖L1

Γ(β) + µ1 + µ2 + n(D1 + D2) ≤ R. Hence, we have

|T (u, v) ≤ R which implies that T (D) ⊆ D.Now, we show that G is contraction. Using (A21) to (A23) for any (u, v), (u, v) ∈ P, we

have

|G1(u)−G1(u)| ≤ |g(u)− g(u)|+ |h(u)− h(u)|+m∑

j=1

|Gj(t, u(tj))−Gj(t, u(tj))|

≤ Kg‖u− u‖+Kh‖u− u‖+m(A1 +A2)‖u− u‖= (Kg +Kh +m(A1 +A2))‖u− u‖.

Similarly, |G2v − G2v| ≤ (Kf + Kκ + n(B1 + B2))‖v − v‖. The contraction of G follows

from the assumption that (Kg +Kh +m(A1 +A2)) < 1 and (Kf +Kκ + n(B1 +B2)) < 1.

91

Next F = (F1 +F2) is compact. The continuity of F follows from the continuity of f, g.For (u, v) ∈ P, we have

|F1(u, v)| =1

Γ(α)

t∫

0

(t− s)α−1|f(s, u(s), v(s))|ds + 1

Γ(α)

1∫

0

(1− s)α−1|f(s, u(s), v(s))|ds

≤ 2

Γ(α)

1∫

0

(1− s)α−1|f(s, u(s), v(s))|ds ≤ 2‖ρ‖L1

Γ(α).

Similarly |F2v(t)| ≤ 2‖σ‖L1

Γ(β).

Hence, ‖F (u, v)‖ ≤ 2( ‖ρ‖

L1

Γ(α) +‖σ‖

L1

Γ(β)

)which implies that F is uniformly bounded on P.

Take a bounded subset D of P and (u, v) ∈ D. Then for t, τ ∈ J with 0 ≤ t ≤ τ ≤ 1, using(A23) and (A25), we obtain

‖F1(u(t), v(t))−F1(u(τ), v(τ))‖ =∣∣∣ 1

Γ(α)

t∫

0

((t− s)α−1 − (τ − s)α−1

)f(s, u(s), v(s))ds

+1

Γ(α)

τ∫

t

(τ − s)α−1f(s, u(s), v(s))ds +(t− τ)

Γ(α)

1∫

0

(1− s)α−1f(s, u(s), v(s))ds∣∣∣

≤ 1

Γ(α+ 1)

(2(τ − t)α + tα − τα + t− τ

)→ 0 as t→ τ.

Similarly, we have

‖F2(u(t), v(t)) − F2(u(τ), v(τ))‖ ≤ 1

Γ(β + 1)

(2(τ − t)β + tβ − τβ + t− τ

)→ 0 as t→ τ.

Hence, F is equi-continuous and by Arzela Ascoli theorem, F is compact. By Lemma 2.3.10,the system (4.4.1) has at least one positive solution.


Theorem 4.4.4. If max(∆1,∆2) < 1, where ∆1 =2Lf

Γ(α+1)Kg +Kh +m(A1 +A2) and

∆2 =2Lg

Γ(β+1)Kf+Kκ+n(B1+B2), then under the assumptions (A21)−(A24), the Impulsive

coupled system (4.4.1) has a unique positive solution.

92

Proof. Let u, u ∈ X , v, v ∈ Y and each t ∈ J , consider

|T1(u, v)− T1(u, v)| =∣∣∣ 1

Γ(α)

t∫

0

(t− s)α−1(f(s, u, v)− f(s, u, v)

)ds

+t

Γ(α)

1∫

0

(1− s)α−1(f(s, u, v)− f(s, u, v)

)ds

+ t(g(u)− g(u)

)+ (1− t)

(h(u) − h(u)

)+

m∑

j=1

(Gj(t, u(tj))−Gj(t, u(tj))

)∣∣∣

≤ 2LfΓ(α+ 1)

[‖u− u‖+ ‖v − v‖] +Kg‖u− u‖+Kh‖u− u‖

+

m∑

j=1

|Ij(u(tj))− Ij(u(tj))|+m∑

j=1

|Ij(u(tj))− Ij(u(tj))|

≤ 2LfΓ(α+ 1)

[‖u− u‖+ ‖v − v‖] +Kg‖u− u‖+Kh‖u− u‖

+ m(A1 +A2)‖u− u‖,which implies that

|T1(u, v)−T1(u, v)| ≤[ 2LfΓ(α+ 1)

Kg+Kh+m(A1+A2)](‖u−u‖+‖v−v‖

)= ∆1

(‖u−u‖+‖v−v‖

).

Similarly, we can show that |T2(u, v)− T2(u, v)| ≤ ∆2

(‖u− u‖+ ‖v− v‖

). Hence, it follows

that|T (u, v) − T (u, v)| ≤ max(∆1,∆2)(‖u− u‖+ ‖v − v‖),

which implies that T is Contractions hence it has a unique fixed point.

Example 4.4.1. Consider the following impulsive coupled system of fractional differentialequations

cD32u(t) =

1 + |u(t)| + |v(t)| sin(t)(2 + t)2(2 + |u|+ |v|) , t 6=

1

3, cD

32 v(t) =

2|u(t)| + 2| cos(v(t))|(4 + t2)

e−t, t 6= 1

4,

u(0) = g(u) =60∑

j=1

u(ξj)

ξ2j + 100u(1) = h(u) =

60∑

j=1

u(ηj)

ηj + 50,

v(0) = f(v) =100∑

i=1

v(ξi)

ξ4i + 100, v(1) = κ(v) =

100∑

i=1

v(ηi)

2η2i + 60,

∆u

(1

3

)= Iu

(1

3

)=

1

6 + ‖u‖ , ∆u′(1

3) = Iu

(1

3

)=

1

6 + ‖u‖

∆v

(1

4

)= Iv

(1

4

)=

1

50 + ‖v‖ , ∆v′(1

4

)= Iv

(1

4

)=

1

100 + ‖v‖(4.4.14)

93

where tj 6= 13 for j = 1, 2, 3, . . . , 60, and ti 6= 1

4 for i = 1, 2, 3, . . . , 100.For any u, u, v, v ∈ R, t ∈ [0, 1], we obtain

|f(t, u, v)− f(t, u, v)| ≤ 1

8[‖u− u‖+ ‖v − v‖],

|g(t, u, v) − g(t, u, v)| ≤ 1

2[‖u− u‖+ ‖v − v‖],

and from impulsive conditions, we have

|Iu(tj)− Iu(tj)| ≤1

8|u− u|, |Iu(tj)− I u(tj)| ≤

1

16|u− u|,

|Iv(ti)− Iv(ti)| ≤1

50|v − v|, |Iv(ti)− Iv(ti)| ≤

1

100|v − v|,

|g(u)− g(u)| ≤ Kg‖u− u‖, |h(u) − h(u)| ≤ Kh‖u− u‖,|f(v) − f(v)| ≤ Kf‖v − v‖, ‖Kk(v) −Kf (v)| ≤ Kk‖v − v‖,

where Kg =1

100 ,Kh = 150 ,Kf = 1

100 ,Kκ = 160 , A1 =

16 , A2 =

16 , B1 =

150 , B2 =

1100 and m =

1, n = 1. Further, we obtain Lf = 18 , Lg =

12 ,

∆1 =2Lf

Γ(α+ 1)+m(A1 +A2) +Kg +Kh

=2× 1

8

Γ(32 + 1)+ 1(

1

6+

1

6) +

1

100+

1

50=

1

3√π+

1

3+

3

100< 1 and ∆2 = 0.8089194447 < 1.

Hence by Theorem 4.4.4, coupled system (4.4.1) has a unique positive solution.

4.5 Existence of solutions of more general systems:

A topological degree theory approach

This section is devoted to the study of more general systems with nonlocal multi point

boundary conditions. The concerned study is carried out via topological degree method. It

has two subsection, where two different systems of problems are studied.

4.5.1 Coupled system with multi-point nonlocal boundary conditions(I)

In this subsection, we use coincidence degree theory approach for condensing maps to ob-

tain sufficient conditions for existence and uniqueness of solutions to more general coupled

systems[108], of nonlinear multi-point boundary value problems. The boundary conditions

94

are also nonlocal. The system is of the form

cDαu(t) = f(t, u(t), v(t)), t ∈ I = [0, 1],

cDβv(t) = g(t, u(t), v(t)), t ∈ I = [0, 1],

u(0) = φ(u), u(1) = δu(η), 0 < η < 1,

v(0) = ψ(v), v(1) = γv(ξ), 0 < ξ < 1,

(4.5.1)

where α, β ∈ (1, 2], and 0 < δ, γ < 1 such that δηα < 1, γξβ < 1, φ, ψ ∈ C(I,R) are nonlocal

boundary functions and f, g : I × R× R → R are continuous.

The following definitions and assumptions are required for further study in this chapter.

(A27) There exist constants Kφ, Kψ ∈ [0, 1) such that for u1, u2, v1, v2,∈ C(I,R),

|φ(u2)− φ(u1)| ≤ Kφ‖u2 − u1‖, |ψ(v2)− ψ(v1)| ≤ Kψ‖v2 − v1‖.

(A28) There exist constants Cφ, Cψ, Mφ,Mψ > 0 such that for u, v ∈ C(I,R),

|φ(u)| ≤ Cφ‖u‖q1 +Mφ, |ψ(v)| ≤ Cψ‖v‖q1 +Mψ.

(A29) There exist constants C(i)f , C

(i)g (i = 1, 2), Mf ,Mg such that for t ∈ I, u, v ∈ C(I,R),

|f(t, u(s), v(s))| ≤ C1f‖u‖q2 + C2

f‖v‖q2 +Mf ,

|g(t, v(s), v(s))| ≤ C1g‖u‖q2 + C2

g‖v‖q2 +Mg.

(A30) For arbitrary t ∈ I, u1, u2, v1, v2 ∈ C(I,R), there exist constants Lf , Lg > 0 such that

|f(t, u2, v2)− f(t, u1, v2)| ≤ Lf (|u2 − u1|+ |v2 − v1|),

|g(t, u2, v2)− g(t, u1, v2)| ≤ Lg(|u2 − u1|+ |v2 − v1|).

Lemma 4.5.1. If h : I → R is α times integrable function, then the solution of the linearboundary value problem of fractional order

cDαu(t) = h(t), t ∈ I = [0, 1],

u(0) = φ(u), u(1) = δu(η), 0 < η < 1(4.5.2)

95

is a solution of the following Fredholm integral equation

u(t) =

(1− t(1− δ)

1− δη

)φ(u) +

∫ 1

0Gα(t, s)h(s)ds, t ∈ [0, 1], (4.5.3)

where Gα(t, s) is defined by

Gα(t, s) =1

Γ(α)

(t− s)α−1 +tδ(η − s)α−1

1− δη− t(1− s)α−1

1− δη, 0 ≤ s ≤ t ≤ η ≤ 1,

(t− s)α−1 − t(1− s)α−1

1− δη, 0 ≤ η ≤ s ≤ t ≤ 1,

tδ(η − s)α−1

1− δη− t(1− s)α−1

1− δη, 0 ≤ t ≤ s ≤ η ≤ 1,

− t(1− s)α−1

1− δη, 0 ≤ η ≤ t ≤ s ≤ 1.

(4.5.4)

Proof. Applying Iα on cDαu(t) = h(t) and using Lemma 2.2.10, we have

u(t) = Iαh(t) +C0 + C1t, (4.5.5)

for some C0, C1 ∈ R. The conditions u(0) = φ(u) and u(1) = δu(η) implies C0 = φ(u) andC1 =

δ1−δη I

αh(η) − 1−δ1−δηφ(u)− 1

1−δη Iαh(1). Hence, we obtain

u(t) = Iαh(t) + φ(u) + t

[δ

1− δηIαh(η)− 1− δ

1− δηφ(u)− 1

1− δηIαh(1)

], (4.5.6)

which after rearranging can be written as

u(t) =

(1− t(1− δ)

1− δη

)φ(u) +

∫ 1

0Gα(t, s)h(s)ds.

In view of Lemma 4.5.1, solutions of the coupled systems of Boundary value problems (4.5.1)

are solutions of the following coupled systems of Fredholm integral equations

u(t) =

(1− t(1− δ)

1− δη

)φ(u) +

∫ 1

0Gα(t, s)f(t, u(s), v(s))ds, t ∈ [0, 1],

v(t) =

(1− t(1− γ)

1− γξ

)ψ(v) +

∫ 1

0Gβ(t, s)g(t, u(s), v(s))ds, t ∈ [0, 1],

(4.5.7)

96

where Gβ(t, s) is defined by

Gβ(t, s) =1

Γ(β)

(t− s)β−1 +tγ(ξ − s)β−1

1− γξ− t(1− s)β−1

1− γξ, 0 ≤ s ≤ t ≤ ξ ≤ 1,

(t− s)β−1 − t(1− s)β−1

1− γξ, 0 ≤ ξ ≤ s ≤ t ≤ 1,

tγ(ξ − s)β−1

1− γξ− t(1− s)β−1

1− γξ, 0 ≤ t ≤ s ≤ ξ ≤ 1,

− t(1− s)β−1

1− γξ, 0 ≤ ξ ≤ t ≤ s ≤ 1,

(4.5.8)

from Green’s functions, for further computation, clealy we write as

maxt∈[0,1]

| Gα(t, s) |=(1− s)α−1

(1− δη)Γ(α), maxt∈[0,1]

| Gβ(t, s) |=(1− s)β−1

(1− γξ)Γ(β), s ∈ [0, 1]. (4.5.9)


Define the operators F1 : X → X , F2 : Y → Y by

F1(u)(t) =

(1− t(1− δ)

1− δη

)φ(u), F2(v)(t) =

(1− t(1− γ)

1− γξ

)ψ(v)

and the operators G1, G2 : X × Y → X ×Y by

G1(u, v)(t) =

∫ 1

0Gα(t, s)f(s, u(s), v(s))ds, G2(u, v)(t) =

∫ 1

0Gβ(t, s)g(t, u(s), v(s))ds.

Further, we define F = (F1, F2), G = (G1, G2) and T = F +G. Then, the system of integral

equations (4.5.7) can be written as an operator equation of the form

(u, v) = T (u, v) = F (u, v) +G(u, v), (4.5.10)

and solutions of the System (4.5.7) are fixed points of T .

Lemma 4.5.2. Under the assumptions (A27) and (A28), the operator F satisfies Lipschitzcondition and the following growth condition

‖F (u, v)‖ ≤ C‖(u, v)‖ +M, for every (u, v) ∈ X × Y. (4.5.11)

97

Proof. Using the assumptions (A27), we obtain

∣∣∣∣F (x, y)(t) − F (u, v)(t)

∣∣∣∣ =∣∣∣∣(1− t(1− δ)

1− δη

)(φ(x)− φ(u)) +

(1− t(1− γ)

1− γξ

)(ψ(y)− ψ(v))

∣∣∣∣≤ Kφ‖x− u‖+Kψ‖y − v‖ ≤ K‖(x, y) − (u, v)‖,

(4.5.12)

where K = maxKφ,Kψ. By proposition 2.4.2, F is also µ-Lipschitz with constant K.Now for the Growth condition using (A28), we get

‖F (u, v)‖ = ‖(F1(u), F2(v))‖ = ‖(F1(u)‖+‖F2(v))‖ ≤ C‖(u, v)‖ +M,

where C = maxCφ, Cψ, M = maxMφ,Mψ.

Lemma 4.5.3. The operator G is continuous and under the assumption (A29) satisfies thegrowth condition

‖G(u, v)‖ ≤ ∆‖(u, v)‖ +Λ, for every (u, v) ∈ X × Y, (4.5.13)

where ∆ = θ(c+d), θ = max 1(1−δη)Γ(α) ,

1(1−γξ)Γ(β), c = maxc1, c2, d = maxd1, d2, Λ =

θ(Mf +Mg).

Proof. Let (un, vn) be a sequence of a bounded set UR = ‖(u, v)‖ ≤ R: (u, v) ∈ X ×Ysuch that (un, vn) → (u, v) in UR. We need to prove that ‖G(un, vn) − G(u, v)‖ → 0.Consider∣∣∣∣G1(un, vn)(t)−G1(u, v)(t)

∣∣∣∣ ≤1

Γ(α)

∫ t

0(t− s)α−1

∣∣∣∣f(s, un(s), vn(s))− f(s, u(s), v(s))

∣∣∣∣ds

+δ

(1− δη)Γ(α)

∫ η

0(η − s)α−1

∣∣∣∣f(s, un(s), vn(s))− f(s, u(s), v(s))

∣∣∣∣ds

+1

Γ(α)

1

1− δη

∫ 1

0(1− s)α−1

∣∣∣∣f(s, un(s), vn(s))− f(s, u(s), v(s))

∣∣∣∣ds.

From the continuity of f and g, it follows that f(s, un(s), vn(s)) → f(s, u(s), v(s)) asn → ∞. For each t ∈ I, using (A29), we obtain (t − s)α−1 |f(s, un, vn)− f(s, u, v)| ≤(t − s)α−12 ((c1 + c2)r +Mf ) which implies the integrability for s, t ∈ I and by using

the Lebesgue Dominated convergence theorem, we obtain∫ t0 (t − s)α−1Γ(α)|f(s, un, vn) −

f(s, u, v)Γ(α)| ds → 0 as n → ∞. Similarly, the other terms approaches 0 as n → ∞. Itfollows that

‖G1(un, vn)(t)−G1(u, v)(t)‖ → 0 as n→ ∞,

and similarly we can show that

‖G2(un, vn)(t)−G2(u, v)(t)‖ → 0 as n→ ∞.

98

Now for growth condition on G, using (A29) and (4.5.9), we obtain

|G1(u, v)(t)| =∣∣∣∣∫ 1

0Gα(t, s)f(s, u(s), v(s)) ds

∣∣∣∣ ≤1

(1− δη)Γ(α)(c1‖u‖ + c2‖v‖ +Mf )

and

|G2(u, v)(t)| =∣∣∣∣∫ 1

0Gβ(s, t)g(s, u(s), v(s)) ds

∣∣∣∣ ≤1

(1− γξ)Γ(β)(d1‖u‖+ d2‖v‖ +Mg) .


‖G(u, v)‖ =‖G1(u, v)‖ + ‖G2(u, v)‖ ≤ θ(c1‖u‖+ c2‖v‖+Mf ) + θ(d1‖u‖ + d2‖v‖ +Mg)

≤ θ(c+ d)(‖u‖ + ‖v‖) + θ(Mf +Mg) = ∆‖(u, v)‖ +Λ.

Which completes the proof.

Lemma 4.5.4. The operator G : X×Y → X×Y is compact. Consequently, G is µ-Lipschitzwith zero constant.

Proof. Take a bounded set D ⊂ UR ⊆ X × Y and a sequence (un, vn) in D, then using(4.5.13), we have

‖G(un, vn)‖ ≤ ∆r +Λ, for every (u, v) ∈ X × Y,

which implies that G(D) is bounded. Now, for equi-continuity, choose 0 ≤ t < τ ≤ 1, forgiven ε > 0, taking

δ = min

δ1 =

1

2

(εΓ(α+ 1)

6 ([c1 + c2]R+Mf )

) 1α

, δ2 =1

2

(εΓ(β + 1)

6 ([d1 + d2]R+Mg)

) 1β

.

For each (un, vn) ∈ D, we claim that if t, τ ∈ [0, 1] and 0 < τ − t < δ1, there holds

|G1(un, vn)(t)−G1(un, vn)(τ)| <ε

2.

Now consider

|G1(un, vn)(t)−G1(un, vn)(τ)| =∣∣∣∣

1

Γ(α)

∫ t

0

[(t− s)α−1 − (τ − s)α−1

]f(s, un(s), vn(s))ds

+1

Γ(α)

∫ τ

t(τ − s)α−1f(s, un(s), vn(s))ds+

δ(t− τ)

(1− δη)Γ(α)

∫ η

0(η − s)α−1f(s, un(s), vn(s))ds

+(τ − t)

(1− δη)Γ(α)

∫ 1

0(1− s)α−1f(s, un(s), vn(s))ds

∣∣∣∣

≤ (c1|un|+ c2|vn|+Mf )

Γ(α+ 1)[(tα − τα) + 2(τ − t)α] ≤ (c1 + c2)R+Mf )

Γ(α+ 1)[(tα − τα) + 2(τ − t)α] .

99

Similarly

|(G2(un, vn)(t)− (G2(un, vn)(τ))| ≤((d1 + d2)R+Mg)

Γ(β + 1)

[(tβ − τβ) + 2(τ − t)β

].

Onward we continue the proof with several cases.Case 1. δ1 ≤ t < τ < 1.

|G1(un, vn)(t)−G1(un, vn)(τ)| <(c1 + c2)r +Mf )

Γ(α+ 1)(2 + α)δα−1

1 (τ − t)

<(c1 + c2)R+Mf )

Γ(α+ 1)(2 + α)δα1 <

ε

2.

Onward, we discuss the above results in the following cases.Case 2. 0 ≤ t < δ1, τ < 2δ1.

|G1(un, vn)(t)−G1(un, vn)(τ)| <(c1 + c2)R +Mf )

Γ(α+ 1)[3τα]

<(c1 + c2)R +Mf )

Γ(α+ 1)3(2δ1)

α =ε

2.

Similarly for second part, for each (un, vn) ∈ D, we claim that if t, τ ∈ [0, 1] and 0 < τ − t <δ2, there holds |G2(un, vn)(t)−G2(un, vn)(τ)| < ε

2 . ThenCase 1. δ2 ≤ t < τ < 1.

|G2(un, vn)(t)−G2(un, vn)(τ)| <(d1 + d2)R+Mg)

Γ(β + 1)(2 + β)δβ−1

2 (τ − t)

<(d1 + d2)R+Mg)

Γ(β + 1)(2 + β)δβ2 <

ε

2.

Case 2. 0 ≤ t < δ2, τ < 2δ2.

|G2(un, vn)(t) −G2(un, vn)(τ)| <(d1 + d2)R +Mg)

Γ(β + 1)[3τβ ]

<(d1 + d2)R +Mg)

Γ(β + 1)3(2δ2)

β =ε

2.

Hence, we have

‖G(xn, yn)−G(xn, yn)‖ < ε. (4.5.14)

Thus G(D) is equi-continuous. Hence, G(un, vn) is equi-continuous and by Arzela-Ascolitheorem G is compact. Furthermore, by proposition 2.4.2, G is µ-Lipschitz with constantzero.

Theorem 4.5.5. Under the assumptions (A27)− (A29), the System (4.5.1) has at least onesolution (u, v) ∈ X × Y provided C + ∆ < 1. Moreover, the set of solutions of (4.5.1) isbounded in X × Y.

100

Proof. By Lemma 4.5.2 F is µ- Lipschitz with constant K ∈ [0, 1) and by Lemma 4.5.4 G isµ- Lipschitz with constant 0. It follows by proposition 2.4.3 that T is strict µ- contractionwith constant K. Define

U = (u, v) ∈ X × Y : ∃ λ ∈ [0, 1] such that (u, v) = λT (u, v).We have to prove that U is bounded in X × Y. For this, choose (u, v) ∈ U, then in view ofthe growth conditions as in Lemmas 4.5.2 and 6.2.1, we have

‖(u, v)‖ = ‖λT (u, v)‖ = λ‖T (u, v)‖ ≤ λ(‖F (u, v)‖ + ‖G(u, v)‖

)

≤ λ [C‖(u, v)‖ +M +∆‖(u, v)‖ +Λ]

= λ(C +∆)‖(u, v)‖ + λ(M +Λ),

which implies that U is bounded in X ×Y. Therefore, by Theorem 2.4.5, T has at least onefixed point and the set of fixed points is bounded in X × Y.


Theorem 4.5.6. In addition to the assumption (A27)−(A30), assume that K+θ(Lφ+Lψ) <1, then the system (4.5.1) has a unique solution.

Proof. We use Banach contraction theorem, for (u, v), (u, v) ∈ R×R, we have from (4.5.12)that

|F (u, v)− F (u, v)| ≤ K‖(u, v)− (u, v)‖. (4.5.15)

Using A30 and (4.5.9), we obtain

|G1(u, v)−G1(u, v)| ≤∫ 10 |Gα(t, s)||f(s, u(s), v(s))− f(s, u(s), v(s))|ds

≤ θLf (|u− u|+ |v − v|)which implies that

‖G1(u, v)−G1(u, v)‖ ≤ θLf (‖u− u‖+ ‖v − v‖)= θLf‖(u− u, v − v)‖ = θLf‖(u, v)− (u, v)‖. (4.5.16)

Similarly, we have‖G2(u, v)−G2(u, v)‖ ≤ θLg‖(u, v)− (u, v)‖. (4.5.17)


‖G(u, v)−G(u, v)‖ = ‖G1(u, v) −G1(u, v)‖+‖G2(u, v)−G2(u, v)‖≤ θ(Lg + Lg)‖(u, v)− (u, v)‖. (4.5.18)

Hence, in view of (4.5.15) and (4.5.18), we obtain

|T (u, v)− T (u, v)| ≤ ‖F (u, v)− F (u, v)‖ + ‖G(u, v)−G(u, v)‖≤ (K + θ(Lg + Lg))‖(u, v)− (u, v)‖

which implies that T is a contraction. By Banach contraction principle, the system (4.5.1)has a unique solution.

101

Example 4.5.1. Consider the following multi-point Boundary value problem

cD32u(t) =

1

50 + t2(1 + |u(t)|+ |v(t)|), t ∈ [0, 1],

cD32 v(t) =

1 + |u(t)|+ |v(t)|50 + |cos u(t)|+ |sin v(t) |, t ∈ [0, 1],

u(0) =φ(u)

2, u(1) =

1

2u

(1

2

),

v(0) =ψ(v)

2, v(1) =

1

3v

(1

3

).

(4.5.19)

The solution of the Boundary value problem (4.5.19) is given by

u(t) =φ(u)

2

(1− 2t

3

)+

∫ 1

0Gα(t, s)f(s, u(s), v(s))ds,

v(t) =ψ(v)

2

(1− 3t

4

)+

∫ 1

0Gβ(t, s)g(s, u(s), v(s))ds,

where Gα, Gβ are the Green’s functions and can be obtained easily as obtained generally in(4.5.4) and (4.5.8) respectively. From the system (4.5.19) we take

α = β =3

2, δ = η =

1

2, γ = ξ =

1

3,

and r = 2 ∈ (1, 3), and let us take λ = 12 ∈ [0, 1]. Then by the use of Theorem 4.5.6, we

have

Lf = Lg =1

50, Mf =Mg =

1

50= ci = di, forı = 1, 2

and taking Kφ = 12 , Kψ = 1

2 , then assumptions (A27)− (A30) are satisfied. Since

F1u(t) =φ(u)

2

(1− 2t

3

), G1u(t) =

∫ 1

0Gα(t, s)f(s, u(s), v(s))ds,

F2v(t) =ψ(v)

2

(1− 3t

4

), G2v(t) =

∫ 1

0Gβ(t, s)g(s, u(s), v(s))ds.

Since F1, F2, G1, G2 are continuous and bounded so also F = (F1, F2), G = (G1, G2),whichfurther implies that T = F +G is continuous and bounded. Further

‖F (u, v)− F (u, v)‖ ≤ 1

2‖(u, v)− (u, v)‖,

that is F is µ- Lipschitz with Lipschitz constant 12 and G is µ- Lipschitz with zero constant

implies that T is strict-µ- contraction with constant 12 . Further it is easy to calculate θ =

1.5045. As

U = (u, v) ∈ C(×R× R,R), there exists λ ∈ [0, 1] : (u, v) =1

2T (u, v).

102

Then the solution

‖(u, v)‖ ≤ 1

2‖T (u, v)‖ ≤ 1,

implies that U is bounded and by Theorem 4.5.5 the Boundary value problem (4.5.19) hasat least a solution (u, v) in C(I ×R×R,R). Further more K + θ(Lf + Lg) = 0.56018 < 1.Hence by Theorem 4.5.6 the Boundary value problem (4.5.19) has a unique solution.

4.5.4 Coupled system with nonlinear boundary conditions(II)

In this subsection, we study existence and uniqueness results for a coupled system of non-

linear fractional order differential subject to nonlinear more general four -point boundary

condition of the following type

cDαu(t) = f(t, u(t), v(t)), cDβv(t) = g(t, u(t), v(t)), t ∈ I = [0, 1],

λ1u(0)− γ1u(η)− µ1u(1) = φ(u), λ2v(0)− γ2v(ξ)− µ2v(1) = ψ(v),

(4.5.20)

where 0 < α, β ≤ 1 and f, g ∈ C([0, 1] × R2,R) are continuous and the nonlocal functions

φ, ψ : (I,R) → R are also continuous. Also η, ξ satisfy 0 < η < 1, 0 < ξ < 1 and the

parameters λi, γi, µi for (i = 1, 2) are real numbers satisfy λi 6= γi + µi for (i = 1, 2).

Lemma 4.5.7. For u , v ∈ C[0, 1], 0 < α, β ≤ 1, λi 6= µi + γi(i = 1, 2), and λi, µi, γi ∈ R

and the non local functions φ(u), ψ(v) : C([0, 1],R) → R, the coupled system of non-local

boundary value problem(4.5.20) has a solution of the form

u(t) =φ(u)

λ1 − (µ1 + γ1)+

µ1λ1 − (µ1 + γ1)

1

Γα

∫ 1

0(1− s)α−1f(s, u(s), v(s))ds

+γ1

λ1 − (µ1 + γ1)

1

Γα

∫ η

0(η − s)α−1f(s, u(s), v(s))ds +

∫ t

0(t− s)α−1f(s, u(s), v(s))ds,

v(t) =ψ(v)

λ2 − (µ2 + γ2)+

µ2λ2 − (µ2 + γ2)

1

Γβ

∫ 1

0(1− s)β−1g(s, u(s), v(s))ds

+γ1

λ2 − (µ2 + γ2)

1

Γβ

∫ ξ

0(ξ − s)β−1g(s, u(s), v(s))ds +

∫ t

0(t− s)β−1g(s, u(s), v(s))ds.

(4.5.21)

Proof. The proof of this Lemma is similar to the proof of Lemma 3.2.1.

103


For the existence of solution of the Coupled system (4.5.20), it is enough to show that the

integral equation (4.5.21) has at least one solution (u, v) ∈ X × Y. Define the following

operators F, G, T : X × Y → X × Y by

F (u, v)(t) = (F1u(t), F2v(t)), G(u, v)(t) = (G1(u, v)(t), G2(u, v)(t))

and T (u, v) = F (u, v) +G(u, v),

where,

F1u(t) =φ(u)

λ1 − (µ1 + γ1), F2v(t) =

ψ(v)

λ2 − (µ2 + γ2)

G1(u, v)(t) =µ1

λ1 − (µ1 + γ1)

1

Γα

∫ 1

0(1− s)α−1f(s, u(s), v(s))ds

+γ1

λ1 − (µ1 + γ1)

1

Γα

∫ η

0(η − s)α−1f(s, u(s), v(s))ds+

∫ t

0(t− s)α−1f(s, u(s), v(s))ds,

G2(u, v)(t) =µ2

λ2 − (µ2 + γ2)

1

Γβ

∫ 1

0(1− s)β−1g(s, u(s), v(s))ds

+γ1

λ2 − (µ2 + γ2)

1

Γβ

∫ ξ

0(ξ − s)β−1g(s, u(s), v(s))ds +

∫ t

0(t− s)β−1g(s, u(s), v(s))ds.

The continuity of f, g implies that the operator T is well defined. The Integral equation

(4.5.21) can be written as an operator equation

(u, v) = T (u, v) = F (u, v) +G(u, v) (4.5.22)

and fixed points of the operator equation (4.5.22) are solutions of the Integral equation

(4.5.21).

Lemma 4.5.8. Under the assumption A27 and A28, the operator F : X × Y → X × Y is

Lipschitz with constant K and satisfies the following growth condition

‖F (u, v)‖≤ C‖(u, v)‖q1+M. (4.5.23)

104

Proof. For (u, v), (u, v) ∈ X × Y, using (A27) and (4.5.22), we obtain

‖F1u− F1u‖≤1

|λ1 − (µ1 + γ1)||φ(u)− φ(u)|≤ K1

|λ1 − (µ1 + γ1)|‖u− u‖

and ‖F2v − F2v‖≤K2

|λ2 − (µ2 + γ2)|‖v − v‖.

which implies that

‖F (u, v)−F (u, v)‖≤ max

(K1

|λ1 − (µ1 + γ1)|,

K2

|λ2 − (µ2 + γ2)|

)‖(u, v)−(u, v)‖ = K‖(u, v)−(u, v)‖,

where K = max(

K1|λ1−(µ1+γ1)| ,

K2|λ2−(µ2+γ2)|

). Hence by proposition (2.4.2), F is µ-Lipschitz

with constant K. For growth condition, using the assumption (A28), we obtain

|F1u(t)|=| φ(u)

λ1 − (µ1 + γ1)| ≤ Cφ‖u‖q1

|λ1 − (µ1 + γ1)|+Mφ,

|F2v(t)|=| ψ(v)

λ2 − (µ2 + γ2)| ≤ Cψ‖v‖q1

|λ2 − (µ2 + γ2)|+Mψ.

Hence, it follows that ‖F (u, v)‖≤ C‖(u, v)‖q1+M , where C = max(

Cφ|λ1−(µ1+γ1)| ,

Cψ|λ2−(µ2+γ2)|

),

and M = max(Mφ,Mψ).

Lemma 4.5.9. Under the assumption A29, the operator G : X ×Y → X ×Y is continuous

and satisfies the following growth condition

‖G(u, v)‖≤ ∆(‖(u, v)‖q2+M∗), (u, v) ∈ X × Y (4.5.24)

where ∆ = max

(Cf [|λ1|+|γ1|+|µ1|]

|λ1−(µ1+γ1)|Γ(α+1) ,Cg [|λ2|+|γ2|+|µ2|]

|λ2−(µ2+γ2)|Γ(β+1)

), M∗ = max(Mf ,Mg),

Cf = maxC1f , C

2f and Cg = maxC1

g , C2g.

Proof. Choose a bounded subset

UR = (u, v) ∈ X × Y :‖(u, v)‖≤ R ⊂ X × Y

and consider a sequence zn = (un, vn) ∈ UR such that zn → z = (u, v) as n → ∞ in

UR. We need to show that ‖Gzn − Gz‖→ 0, n → ∞. From the continuity of f(t, u, v), it

105

follows that f(s, un, vn) → f(s, u, v), as n→ ∞. In view of (A29), we obtain the following

relations

(1− s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|≤ (1− s)α−1[C1fR+ C2

fR+Mf ]

(η − s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|≤ (η − s)α−1[C1fR+ C2

fR+Mf ],

(t− s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|≤ (1− s)α−1[C1fR+C2

fR+Mf ]

which implies that each term on the left is integrable. By Lebesgue Dominated Convergent

theorem, we obtain

∫ 1

0(1− s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|ds → 0, as n→ ∞

∫ η

0(η − s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|ds → 0, as n→ ∞

∫ t

0(t− s)α−1|f(s, un(s), vn(s))− f(s, u(s), v(s))|ds → 0, as n→ ∞.

Hence, |G1(un, vn)−G1(u, v)|→ 0, as n→ ∞. Similarly, we obtain |G2(un, vn)−G2(u, v)|→

0, as n → ∞. It follows that |G(un, vn) − G(u, v)|→ 0, as n → ∞ which implies the

continuity of the operator G. For the growth conditions, we note that

|G1(u, v)(t)| ≤1

|λ1 − (µ1 + γ1)|Γα(|µ1|

∫ 1

0(1− s)α−1|f(s, u(s), v(s))|+

|γ1|∫ η

0(η − s)α−1|f(s, u(s), v(s))|

)+

|γ1||λ1 − (µ1 + γ1)|Γα

∫ t

0(t− s)α−1|f(s, u(s), v(s))|

≤[ |λ1|+ |µ1|+ |γ1||λ1 − (µ1 + γ1)|Γ(α+ 1)

](C1

f‖u‖q2+C2f‖v‖q2+Mf ).

(4.5.25)


‖G2(u, v)(t)‖≤[ |λ2|+ |µ2|+ |γ2||λ2 − (µ2 + γ2)|Γ(β + 1)

](C1

g‖u‖q2+C2g‖v‖q2+Mg). (4.5.26)

The growth condition (4.5.24) follows from (4.5.25) and (4.5.26).

Lemma 4.5.10. The operator G : X × Y → X × Y is compact and µ- Lipschitz with

constant zero.

106

Proof. Consider a bounded set D ⊂ UR ⊂ X ×Y. We need to prove that G(D) is relatively

compact in X × Y. For any zn = (un, vn) ∈ D ⊂ UR, the growth condition (4.5.24) implies

that

‖G(un, vn)‖≤ ∆(Rq2 +M∗),

that is, G(D) is uniformly bounded. For equi-continuity of G, choose 0 ≤ t ≤ τ ≤ 1, then

we have

‖G1(un, vn)(t)−G1(un, vn)(τ)‖≤1

Γ(α)

∫ t

0[(t− s)α−1 − (τ − s)α−1]|f(s, un(s), vn(s))|ds

+

∫ τ

t(τ − s)α−1|f(s, un(s), vn(s))|ds

≤ 1

Γ(α+ 1)[tα − τα + (τ − t)α + (τ − t)α] (C1

f‖u‖q2+C2f‖v‖q2+Mf )

≤((C1

f + C2f )R

q2 +Mf

Γ(α+ 1)

)[tα − τα + (τ − t)α + (t− τ)α] ,

(4.5.27)


‖G2(un, vn)(t)−G2(un, vn)(τ)‖≤((C1

g +C2g )R

q2 +Mg

Γ(β + 1)

)[tβ − τβ + (τ − t)β + (t− τ)β

].

(4.5.28)


‖G1(un, vn)(t)−G1(un, vn)(τ)‖→ 0, ‖G2(un, vn)(t)−G2(un, vn)(τ)‖→ 0 as t→ τ,

which implies that G(u, v) is equi-continuous. By Arzela-Ascoli theorem G(u, v) is compact

and hence by proposition 2.4.2, G is µ- Lipschitz with constant 0.

Theorem 4.5.11. Assume that (A27) − (A29) hold, then the Boundary value problem

(4.5.20) has at least one solution (u, v) ∈ X × Y and the set of the solutions is bounded

in X × Y.

107

Proof. By Lemma 4.5.8, F is µ Lipschitz with constant K and by Lemma 4.5.10, G is µ-

Lipschitz with constant 0. Consequently, T is µ- Lipschitz with constant K. Now consider

the following set

U = (u, v) ∈ X × Y : there exist ρ ∈ [0, 1] such that (u, v) = ρT (u, v).

We prove that U is bounded. For (u, v) ∈ U, we have

(u, v) = λT (u, v) = λ(F (u, v) +G(u, v)),

which implies that

‖u‖≤ λ[‖F1u‖+‖G1u‖]

≤ λ

[Cφ||u||q1

|λ1 − (µ1 + γ1)|+Mφ +

|λ1|+ |µ1|+ |γ1||λ1 − (µ1 + γ1)|Γ(α + 1)

(C1f‖u‖q2+C2

f‖v‖q2+Mf

)].

(4.5.29)


‖v‖≤ λ

[Cψ||v||q1

|λ2 − (µ2 + γ2)|+Mψ +

|λ2|+ |µ2|+ |γ2||λ2 − (µ2 + γ2)|Γ(β + 1)

(C1g‖u‖q2+C2

g‖v‖q2+Mg

)].

(4.5.30)

The above inequalities (4.5.29) and (4.5.30) together with q1, q2 ∈ [0, 1) show that U is

bounded in X × Y . If the assertion is not true, then dividing (4.5.29) by ‖u‖ and letting

‖u‖→ ∞, we arrive at

1 ≤ limλ→∞

ρ[

Cψλq1

|λ1−(µ1+γ1)| +Mφ +|λ1|+|µ1|+|γ1|

|λ1−(µ1+γ1)|Γ(α+1) [(C1f +C2

f ]λq2 +Mf

]

λ= 0,

a contradiction. Similar contradiction occur when we divide (4.5.30) by ‖v‖ and letting

‖v‖→ ∞. Thus T has at least one fixed point which corresponds to a solution of (4.5.20).


Theorem 4.5.12. Under the assumption (A27)− (A28) and if maxΥ1,Υ2 < 1 where

Υ1 =K1Γ(α+ 1) + Lf (|λ1|+ |µ1|+ |γ1|)

|λ1 − (µ1 + γ1)|Γ(α+ 1), Υ2 =

K2Γ(β + 1) + Lg(|λ2|+ |µ2|+ |γ2|)|λ2 − (µ2 + γ2)|Γ(β + 1)

,

108

then operator T has a unique fixed point which corresponds to a unique solution of Boundary

value problem (4.5.20).

Proof. For (u, v), (u, v) ∈ X × Y, consider

‖(F1(u) +G1(u, v)) − (F1(u) +G1(u, v))‖

≤ K1‖u− u‖|λ1 − (µ1 + γ1)|

+|µ1|

|λ1 − (µ1 + γ1)|Γα

∫ 1

0(1− s)α−1|f(s, u(s), v(s)) − f(s, u(s), v(s))|ds

+|γ1|

|λ1 − (µ1 + γ1)|Γα

∫ η

0(η − s)α−1|f(s, u(s), v(s)) − f(s, u(s), v(s))|ds

+1

Γα

∫ t

0(t− s)α−1|f(s, u(s), v(s)) − f(s, u(s), v(s))|ds

≤[

K1

|λ1 − (µ1 + γ1)|+

LfΓ(α+ 1)

( |λ1|+ |µ1|+ |γ1||λ1 − (µ1 + γ1)|

)] (‖u− u‖+‖v − v‖

)

≤ Υ1

(‖u− u‖+‖v − v‖

).

Similarly, we obtain ‖(F2(v)+G2(u, v))−(F2(v)+G2(u, v))‖≤ Υ2

(‖u− u‖+‖v− v‖

). Hence

from the above two expressions, we have

‖T (u, v)− T (u, v)‖≤ max(Υ1,Υ2)

(‖(u, v)‖−‖(u, v)‖

)= Υ

(‖(u, v)‖−‖(u, v)‖

).

Hence T is contraction as Υ < 1 and by Banach contraction principle T has a unique fixed

point.

Remark The growth conditions in Theorem 4.5.11 are also true for q1, q2 = 1 provided

that

max

Cφ +

|λ1|+ |µ1|+ |γ1||λ1 − (µ1 + γ1)|Γ(α+ 1)

(C1f + C2

f ), Cψ +|λ2|+ |µ2|+ |γ2|

|λ2 − (µ2 + γ2)|Γ(β + 1)(C1

g + C2g )

< 1.

For the demonstration of the above developed theory we give the following example of

nonlocal more general boundary conditions of fractional order differential equations.

109

Example 4.5.2. Consider the following coupled system of boundary values problems

cD12u(t) =

1

4+e−πt

√|u(t)|

16 +√

|u(t)|+

cos√

|v(t)|16

, t ∈ [0, 1],

cD23 v(t) =

1

8+

sin√|u(t)|

24+

√|v(t)|24

, t ∈ [0, 1],

u(0) − 1

4u(1)− 1

2u

(1

2

)=

10∑

i=1

δi|u(ti)|,

v(0) +1

7v(1) − 1

8v

(1

2

)=

9∑

i=1

νi|v(ti)|,

where10∑

i=1

δi <1

10,

9∑

i=1

νi <1

9.

(4.5.31)

From Coupled system (4.5.31), we have

f(t, u, v) =1

4+e−πt

√|u(t)|

16 +√|u(t)|

+cos√|v(t)|

16, g(t, u, v) =

1

8+

sin√|u(t)|

24+

√|v(t)|24

,

α =1

2, β =

2

3, λ1 = 1, µ1 =

1

4, γ1 =

1

2, λ2 = 1, µ2 = −1

7, γ2 =

1

8, η =

1

2= ξ.

Let ρ =1

2. Since λi 6= µi + γi(i = 1, 2)

‖f(t, u, v) − f(t, u, v)‖ ≤ 1

16[‖u− u‖+ ‖v − v‖],

‖g(t, u, v) − g(t, u, v)‖ ≤ 1

24[‖u− u‖+ ‖v − v‖],

it follows that

K1 =1

10,K2 =

1

9,Mφ =

1

4,Mψ =

1

8, Lf =

1

16, Lg =

1

24, C1

f = C2f =

1

16, C1

g = C2g =

1

24,

and q1 =12 , q2 = 1. Then all the assumptions of Theorem 4.5.11 are satisfied and also

Υ1 =

[2

5+

7

8√π

]< 1, Υ2 =

[4

9+

71

656Γ(23 )

]< 1.

Hence, by Theorem 4.5.12, the system of Boundary value problems (4.5.31) has a unique

solution.

Chapter 5

Multiplicity of positive solutionsfor coupled system of multi-pointboundary value problems

In this chapter, we establish sufficient conditions for the existence of solutions to a general

class of multi-point boundary value problems for coupled system of fractional differential

equations. As in recent decades, fractional-order differential equations have been proved to

be valuable tools in the modeling of many phenomena in various fields of science and engi-

neering. The area devoted to the existence and uniqueness of positive solutions of coupled

system of fractional order differential equations has been explored and plenty of work is

available on this. The area devoted to study multiplicity of positive solutions to ordinary

differential equations has been studied, see for detail [110, 111, 112]. However, the area of

multiplicity of solutions to fractional order differential equations need further explorations,

as this area has very recently attracted the researchers toward itself, see [113, 114, 115]. By

means of some classical fixed point theorems, we develop necessary and sufficient conditions

under which the considered system has at least one , two or more positive solutions. Further

the criteria is extended for existence of multiple solutions. Uniqueness is derived by the

use of generalized metric space and with the help of Perov’s fixed point theorem.. Also, we

develop some conditions under which the considered system has no solution. For multiple

110

111

solutions, we impose some growth conditions on nonlinear functions involved in the system.

Based on these conditions sufficient conditions are derived for multiple solutions as well as

for no solution.

Organization of the chapter as follow: In section 5.1, we study existence, nonexistence and

uniqueness of positive solutions as well as multiplicity of positive solutions to a coupled

system with four point more general nonlinear boundary conditions.

5.1 Multiplicity of positive solutions to a coupled system of

higher order four point BVPs

In this section, we study existence, uniqueness, multiplicity as well as non-existence of

positive solutions to the following system of non-linear higher order fractional differential

equations with four point boundary conditions

Dαu(t) + f(t, u(t), v(t)) = 0, t ∈ (0, 1), n− 1 < α ≤ n,

Dβv(t) + g(t, u(t), v(t)) = 0, t ∈ (0, 1), n− 1 < β ≤ n,

u(0) = u′(0) = u′′(0) = 0 · · · = u(n−2)(0) = 0, Dδu(1) = λDδu(η),

v(0) = v′(0) = v′′(0) = 0 · · · = v(n−2)(0) = 0, Dγv(1) = µDγv(ξ),

(5.1.1)

where n > 3, α − δ ≥ 1, β − γ ≥ 1 and 0 < δ, γ < 1, λ, µ ∈ (0,∞), 0 < η, ξ < 1,

f, g : [0, 1]× [0,∞)× [0,∞) → [0,∞) are continuous functions. By means of some classical

fixed point theorems, we develop necessary and sufficient conditions under which the system

has at least one , two or more solutions. Also we develop conditions for nonexistence of

positive solutions of (5.1.1).

Lemma 5.1.1. Let y ∈ C(0, 1] then the boundary value problem

Dαu(t) + y(t) = 0, 0 < t < 1, α ∈ (n− 1, n],

u(0) = u′(0) = · · · = u(n−2)(0) = 0, Dδu(1) = λDδu(η),(5.1.2)

where 0 < η < 1, λ ∈ (0,∞) 0 < δ < 1 has a unique solution given by

u(t) =

1∫

0

Gα(t, s)y(s)ds.

112

Where

Gα(t, s) =1

Γ(α)

d1tα−1(1− s)α−δ−1 − (t− s)α−1 − d1λt

α−1(η − s)α−δ−1, 0 ≤ s ≤ t ≤ η ≤ 1,

d1tα−1(1− s)α−δ−1 − d1t

α−1λ(η − s)α−δ−1, 0 ≤ t ≤ η ≤ s ≤ 1,

d1tα−1(1− s)α−δ−1 − (t− s)α−1, 0 ≤ η ≤ s ≤ t ≤ 1,

d1tα−1(1− s)α−δ−1, 0 ≤ η ≤ t ≤ s ≤ 1,

(5.1.3)where d1 =

11−ληα−δ−1 .

Proof. By applying Iα and using Lemma 2.2.6, the general solution of (5.1.2) is given by

u(t) = −Iαy(t) + C1tα−1 + C2t

α−2 + · · · + Cntα−n. (5.1.4)

Using the boundary conditions of equation (5.1.2) and the result Dδ[tα−1] = Γ(α)Γ(α−δ) t

α−δ−1,

we obtain C2 = C3 = ... = Cn = 0 and using d1 =1

1−ληα−δ−1 , we have

C1 =d1

Γ(α)

[ ∫ 1

0(1− s)α−δ−1y(s)ds− λ

∫ η

0(η − s)α−δ−1y(s)ds

].

Hence, (5.1.4) becomes

u(t) = −Iαy(t) +d1t

α−1

Γ(α)

[ ∫ 1

0(1− s)α−δ−1y(s)ds− λ

∫ η


]

= − 1

Γ(α)

∫ t

0(t− s)α−1y(s)ds

+d1t

α−1

Γ(α)

[ ∫ 1

0(1− s)α−δ−1y(s)ds− λ

∫ η


]

=

∫ 1

0Gα(t, s)y(s)ds,

where Gα(t, s) is the Green’ function of (5.1.2).

Thus inview of Lemma 5.1.1, the coupled system (5.1.1) is equivalent to the following

coupled system of Fredholm integral equations

u(t) =

∫ 1

0Gα(t, s)f(s, u(s), v(s))ds, t ∈ [0, 1],

v(t) =

∫ 1

0Gβ(t, s)g(s, u(s), v(s))ds, t ∈ [0, 1],

(5.1.5)

113

where Gβ(t, s) is the Green’s function for the second equation of the system(5.1.1) and given

by

Gβ(t, s) =1

Γβ

d2tβ−1(1− s)β−γ−1 − (t− s)β−1 − d2µt

β−1(ξ − s)β−γ−1, 0 ≤ s ≤ t ≤ ξ ≤ 1,

d2tβ−1(1− s)β−γ−1 − d2t

β−1µ(ξ − s)β−γ−1, 0 ≤ t ≤ ξ ≤ s ≤ 1,

d2tβ−1(1− s)β−γ−1 − (t− s)β−1, 0 ≤ ξ ≤ s ≤ t ≤ 1,

d2tβ−1(1− s)β−γ−1, 0 ≤ ξ ≤ t ≤ s ≤ 1,

(5.1.6)

where d2 =1

1−µξβ−γ−1 .

Theorem 5.1.2. We call G(t, s) = (Gα(t, s), Gβ(t, s)) is the Green’s function of (5.1.1).This Green’s function G(t, s) has the following properties:

(P1) G(t, s) is continuous function on the unit square for all (t, s) ∈ [0, 1] × [0, 1];

(P2) G(t, s) ≥ 0 for all t, s ∈ [0, 1] and G(t, s) > 0 for all t, s ∈ (0, 1);

(P3) max0≤t≤1

G(t, s) = G(1, s),∀ s ∈ [0, 1];

(P4) mint∈[θ,1−θ]

G(t, s) ≥ γ(s)G(1, s) for each θ ∈(0, 12), s ∈ [0, 1],

where γ = minγα, γβ = θα−1, θβ−1.Proof. The proof of properties (P1) − (P3) can be obtained similarly by making slightmodification in the proof of Lemma 3.1.2.

Let J = [θ, 1− θ], then define the cone P ⊂ X × Y by

P = (u, v) ∈ X × Y : mint∈J

[u(t) + v(t)] ≥ γ||(u, v)||

and

Pr = (u, v) ∈ P : ‖(u, v)‖ ≤ r, ∂Pr = (u, v) ∈ P : ‖(u, v)‖ = r.

Define T : X × Y → X × Y be the operator defined as

T (u, v)(t) =

1∫

0

Gα(t, s)f(s, u(s), v(s))ds,

1∫

0

Gβ(t, s)g(s, u(s), v(s))ds

= (T1u(t), T2v(t)).

. (5.1.7)

Then the fixed points of operator T coincide with the solutions of System (5.1.1).

114

5.1.1 Existence and uniqueness of solution

In this subsection, we provide existence and uniqueness of solution to the coupled System

(5.1.1).

Theorem 5.1.3. Assume that f and g are continuous on [0, 1]× [0,∞)× (0,∞) → [0,∞),and there exist fi(t), gi(t), (i = 1, 2) : (0, 1) → [0,∞) that satisfy

(H1) |f(t, u, v) − f(t, u, v)| ≤ f1(t)|u− u|+ g1(t)|v − v|, t ∈ (0, 1), for u, v, u, v ≥ 0;

(H2) |g(t, u, v) − g(t, u, v)| ≤ f2(t)|u− u|+ g2(t)|v − v|, for t ∈ (0, 1), u, u, u, v ≥ 0;

(H3) ρ(A) < 1, where A ∈M2,2(R+) is the matrix given by

∫ 10 Gα(1, s)f1(s)ds

∫ 10 Gα(1, s)g1(s)ds

∫ 10 Gβ(1, s)f2(s)ds

∫ 10 Gβ(1, s)g2(s)ds

.

Then the System (5.1.1) has a unique positive solution (u, v) ∈ P.

Proof. Let us define a generalized metric d : X 2 × Y2 → R2 by

d((u, v), (u, v) =

[||u− u||||v − v||

], for all (u, v), (u, v) ∈ X × Y.

Obviously, (X ×Y, d) is a generalized complete metric space. For any (u, v), (u, v) ∈ X ×Y,using property (P3), we get

|T1(u, v)(t) − T1(u, v)(t)| ≤ maxt∈[0,1]

1∫

0

|Gα(t, s)|[|f(s, u(s), v(s)) − f(s, u(s), v(s))|]ds)

≤∫ 1

0Gα(1, s)[f1(s)‖u− u‖+ g1(s)‖v − v‖]ds

⇒ ‖T1(u, v) − T1(u, v)‖ ≤∫ 1

0f1(s)Gα(1, s)ds‖u − u‖+

∫ 1

0g1(s)Gα(1, s)ds‖v − v‖.


‖T2(u, v)− T2(u, v)‖ ≤∫ 1

0f2(s)Gβ(1, s)ds‖u − u‖+

∫ 1

0g2(s)Gβ(1, s)ds‖v − v‖.

Thus we have

‖T (u, v)− T (u, v)‖ ≤ Ad ((u, v), (u, v)) , for all (u, v), (u, v) ∈ X × Y,

115

where

A =

∫ 10 Gα(1, s)f1(s)ds

∫ 10 Gα(1, s)g1(s)ds

∫ 10 Gβ(1, s)f2(s)ds

∫ 10 Gβ(1, s)g2(s)ds

.

As ρ(A) < 1, hence, T is a contraction and by Theorem 2.3.13, the System (5.1.1) has aunique positive solution.

Next we define some assumptions and notations.

(C1) f, g : [0, 1] × [0,∞) × [0,∞) → [0,∞) are uniformly continuous with respect to t on

[0, 1].

(C2) Let fσ = lim(u,v)→(σ,σ) supt∈[0,1]f(t, u, v)

u+ v, gσ = lim(u,v)→(σ,σ) supt∈[0,1]

g(t, u, v)

u+ v,

fσ = lim(u,v)→(σ,σ) inft∈[0,1]f(t, u, v)

u+ v, gσ = lim(u,v)→(σ,σ) inft∈[0,1]

g(t, u, v)

u+ v,

where σ ∈ 0,∞;

(C3) σα = maxt∈[0,1]1∫0

Gα(t, s)ds, σβ = maxt∈[0,1]1∫0

Gβ(t, s)ds.


Theorem 5.1.4. Assume that (C1)− (C3) hold. Further assume that the following condi-tions are satisfied:

(H4) there exist two constants a, b > 0, with 1 ≤ a+ b <∞ such that

a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

< f0 <∞, b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

< g0 <∞;

(H5) there exist constants λ1, λ2 with λ1 + λ2 ≤ 1 such that

0 ≤ σαf∞ < λ1, 0 ≤ σβg

∞ < λ2.

Then, the boundary value problem (5.1.1) has at least one positive solution.

Proof. From (H4) we have

f0 − a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

> λ1 > 0, g0 − b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

> λ2 > 0.

116

Then, there exist real constant ε1 > 0, such that

f(t, u, v) ≥(f0 − r1)(u(t) + v(t)), t ∈ [0, 1], u, v ∈ [0, ε1], where r1 > 0,

g(t, u, v) ≥(g0 − r2)(u(t) + v(t)), t ∈ [0, 1], u, v ∈ [0, ε1], where r2 > 0.

We define an open set

U1 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε1.

Let τ ∈ [θ, 1 − θ]. From property (P4) of Green’s function, for any (u, v) ∈ P ∩ ∂U1, weobtain

‖T1(u, v)‖ ≥ T1(u, v)(τ) ≥1∫

0

Gα(τ, s)f(s, u(s), v(s))ds

≥ γα

1−θ∫

θ

Gα(1, s)(f0 − r1)[u(s) + v(s)]ds

≥ (f0 − r1)

(γ2

1−θ∫

θ

Gα(1, s)ds

)‖(u, v)‖ ≥ a‖(u, v)‖.

Similarly we get

‖T2(u, v)‖ ≥ T2(u, v)(τ) ≥1∫

0

Gβ(τ, s)g(s, u(s), v(s))ds

≥ γβ

1−θ∫

θ

Gβ(1, s)(g0 − r2)[u(s) + v(s)]ds

≥ (g0 − r2)

(γ2

1−θ∫

θ

Gβ(1, s)ds

)‖(u, v)‖ ≥ b‖(u, v)‖.

Therefore, it follows that

‖T (u, v)‖ = ‖T1(u, v)‖ + ‖T2(u, v)‖ ≥ (a+ b)‖(u, v)‖ ≥ ‖(u, v)‖.

Also from (H5), we have

λ1σ−1α − f∞ > r3 > 0, λ2σ

−1β − g∞ > r4 > 0.

There exist ε2 > 0 such that

f(t, u, v) ≤ (f∞ + r3)(u+ v), t ∈ [0, 1], u, v ≥ 0, u+ v > ε2,

117

g(t, u, v) ≤ (g∞ + r4)(u+ v), t ∈ [0, 1], u, v ≥ 0, u+ v > ε2.

From (C2), for ε2 > 0, we take

K = supt∈[0,1],u+v∈[0,ε2]

f(t, u, v), L = supt∈[0,1],u+v∈[0,ε2]

g(t, u, v),

then, we havef(t, u, v) ≤ (f∞ + r3)(u+ v) +K, t ∈ [0, 1], u, v ≥ 0,

g(t, u, v) ≤ (g∞ + r4)(u+ v) + L, t ∈ [0, 1], u, v ≥ 0.

DefineU2 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε2,

wheremaxε1,Kσα(λ1 − (f∞ + r3)σα)

−1, Lσβ(λ2 − (g∞ + r4)σβ)−1 < ε.

From (P3) and for any (u, v) ∈ P ∩ ∂U2, for t ∈ [0, 1], we have

T1(u, v)(t) =

1∫

0

Gα(t, s)f(s, u(s), v(s))ds ≤1∫

0

Gα(1, s)[(f∞ + r3)(u(s) + v(s) +K)]ds

≤ K

1∫

0

Gα(1, s)ds + (f∞ + r3)ε

1∫

0

Gα(1, s)ds ≤ Kσα + (f∞ + r3)εσα

≤ λ1ε− (f∞ + r3)εσα + (f∞ + r3)εσα = λ1ε

⇒ ‖T1(u, v) ≤ λ1‖(u, v)‖.

Similarly we can obtain for

T2(u, v)(t) =

1∫

0

Gβ(t, s)g(s, u(s), v(s))ds ≤1∫

0

Gβ(1, s)[(g∞ + r4)(u(s) + v(s) + L)]ds

≤ L

1∫

0

Gβ(1, s)ds+ (g∞ + r4)ε

1∫

0

Gβ(1, s)ds ≤ Lσα + (g∞ + r4)εσα

≤ λ2ε− (g∞ + r4)εσβ + (g∞ + r4)εσβ = λ2ε

⇒ ‖T2(u, v) ≤ λ2‖(u, v)‖.

Then, we have ‖T (u, v)‖ = ‖T1(u, v)‖ + ‖T2(u, v)‖ ≤ (λ1 + λ2)‖(u, v)‖ ≤ ‖(u, v)‖.Since T : P ∩ (U2 \ U1) → P is completely continuous operator. Thus by using Theorem2.3.11, T has at least one fixed point (u, v) ∈ P ∩ (U2 \ U1), which is the correspondingpositive solution to (5.1.1).

Theorem 5.1.5. Assume that (C1) holds and also the following conditions are satisfied:

118

(H6) f0 = g0 = ∞;

(H7) there exist a, b > 0 such that a+ b ≤ 1 with

a ≥ f∞σα ≥ 0, b ≥ g∞σβ ≥ 0.

Then (5.1.1) has at least one positive solutions.

Proof. Choose

2r1 >

(γ2

1∫

0

Gα(1, s)ds

)−1

, 2r2 >

(γ2

1∫

0

Gβ(1, s)ds

)−1

.

From (C2), there exist ε such that

f(t, u, v) ≥ r1(u+ v), t ∈ [0, 1], u, v, t ∈ [0, ε1],

g(t, u, v) ≥ r2(u+ v), t ∈ [0, 1], u, v, t ∈ [0, ε1].

DefineU1 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε1.

Let τ ∈ [θ, 1− θ], using (P3), for any (u, v) ∈ P ∩ ∂U1, we get

‖T (u, v)‖ ≥ T1(u, v)(τ) ≥1∫

0

Gα(τ, s)f(s, u(s), v(s))ds ≥ γα

1−θ∫

θ

Gα(1, s)[u(s) + v(s)]ds

≥ r1

(γ2

1−θ∫

θ

Gα(1, s)ds

)‖(u, v)‖ ≥ 1

2‖(u, v)‖.

Similarly

‖T2(u, v) ≥ T2(u, v)(τ) ≥1∫

0

Gβ(τ, s)g(s, u(s), v(s))ds ≥ γβ

1−θ∫

θ

Gβ(1, s)(g0 − r2)[u(s) + v(s)]ds

≥ r2

(γ2

1−θ∫

θ

Gβ(1, s)ds

)‖(u, v)‖ ≥ 1

2‖(u, v)‖.

Therefore‖T (u, v)‖ = ‖T1(u, v)‖ + ‖T2(u, v)‖ ≥ ‖(u, v)‖.

Also from (H7), and from Theorem 2.3.11, there exist ε2 < ε1 such that

‖T (u, v)‖ ≤ ‖(u, v)‖, (u, v) ∈ P ∩ ∂U2,

119

whereU2 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε2.

By Theorem2.3.11, T has at least one fixed point (u, v) ∈ P ∩ (U2 \ U1), which is thecorresponding positive solution of (5.1.1).


(H8) there exist two constants a, b > 0, with 1 ≤ a+ b <∞ such that

a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

< f∞ <∞, b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

< g∞ <∞;

(H9) there exist constants λ1, λ2 with λ1 + λ2 ≤ 1 such that

0 ≤ σαf0 < µ, 0 ≤ σβg

0 < ν.

Then, the System (5.1.1) has at least one positive solution.

Proof. From (H9) we have

λ1σ−1α − f0 > r1 > 0, λ2σ

−1β − g0 > r2 > 0.

Then, there exists ε > 0 such that

f(t, u, v) ≤ (f 0 + r1)(u+ v), t ∈ [0, 1], u, v ∈ [0, ε],

andg(t, u, v) ≤ (g0 + r2)(u+ v), t ∈ [0, 1], u, v ∈ [0, ε].

DefineU1 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε.

From (P3) and for any (u, v) ∈ P ∩ ∂U1, t ∈ [0, 1], we get

T1(u, v)(t) =

1∫

0


0

Gα(1, s)(f0 + r1)(u(s) + v(s))ds

≤ (f0 + r1)ε

1∫

0

Gα(1, s)ds ≤ (f 0 + r1)εσα ≤ λ1ε

⇒ ‖T1(u, v) ≤ λ1‖(u, v)‖.

Similarly, for t ∈ [0, 1], we obtain

‖T2(u, v)‖ ≤ λ2‖(u, v)‖.

120


‖T (u, v)‖ = ‖T1(u, v)‖ + ‖T2(u, v)‖ ≤ (λ1 + λ2)‖(u, v)‖ ≤ ‖(u, v)‖.

Further from (H8), we obtain

f∞ − a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

> r3 > 0, g∞ − b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

> r4 > 0.

Then, there exist ε > 0 so that

f(t, u, v) ≥ (f∞ − r3)(u+ v), t ∈ [0, 1], u, v ≥ 0, u+ v ≥ ε,

g(t, u, v) ≥ (g∞ − r4)(u+ v), t ∈ [0, 1], u, v ≥ 0, u+ v ≥ ε.

Define U2 = (u, v) ∈ X × Y : ‖(u, v)‖ < ε1, where maxε, εγ < ε1. Then, for (u, v) ∈P ∩ ∂U2 and for all s ∈ [0, 1], we get

u(s) + v(s) ≥ mint∈[0,1]

[u(t) + v(t)] ≥ γ‖(u, v)‖ ≥ γε1 ≥ ε.

Taking τ ∈ [θ, 1− θ] and using(P4) we obtain that

‖T1(u, v)‖ ≥ T1(u, v)(τ) ≥1−θ∫

θ

Gα(τ, s)f(s, u(s), v(s))ds

≥ γα

1−θ∫

θ

Gα(τ, s)(f∞ − r3)(u(s) + v(s))ds

≥ (f∞ − r3)

(γ2α

1−θ∫

θ

Gα(1, s)ds

)‖(u, v)‖ ≥ a‖(u, v)‖

⇒ ‖T1(u, v) ≥ a‖(u, v)‖.

Similarly,‖T2(u, v)‖ ≥ b‖(u, v)‖.

From these, we have‖T (u, v)‖ ≥ (a+ b)‖(u, v)‖ ≥ ‖(u, v)‖.

Thus by Theorem 2.3.11 , T has at least one fixed point (u, v) ∈ P ∩ (U2 \ U1), which is thecorresponding positive solutions of (5.1.1).


(H10) f∞ = g∞ = ∞;

121

(H11) there exist two constants λ1, λ2 > 0, with λ1 + λ2 ≤ 1 such that

λ1σ−1α > f0 > 0, λ2σ

−1β > g0 > 0.

Then the Boundary value problem (5.1.1) has at least one positive solution.

Proof. The proof is similar to that of Theorem 5.1.6 so we omit it.


(H12) for all t ∈ [0, 1], we have

f(t, 0, 0) 6= 0, f(t, 1, 1) 6= 1, g(t, 0, 0) 6= 0, g(t, 1, 1) 6= 0;

(H13) for all t ∈ [0, 1] such that

0 ≤ u ≤ u1, 0 ≤ v ≤ v1 ⇒ f(t, u, v) ≤ f(t, u1, v1), g(t, u, v) ≤ f(t, u1, v1);

(H14) there exist λ, µ ∈ (0, 1) such that

t ∈ [0, 1], ξ ∈ (0, 1), u, v ≥ 0 ⇒ f(t, ξu, ξv) ≥ ξλf(t, u, v),

t ∈ [0, 1], ξ ∈ (0, 1), u, v ≥ 0 ⇒ g(t, ξu, ξv) ≥ ξλg(t, u, v).

Then, the System (5.1.1) has at least one positive solution in Sh where h(t) = (tα−1, tβ−1).

Proof. Assume that κ = maxλ, µ and (u, v) ∈ P. For all t ∈ [0, 1], using (H14), we have

T1(ξu, ξv)(t) =

1∫

0

Gα(t, s)f(s, ξu(s), ξv(s))ds ≥ ξλ1∫

0

Gα(t, s)f(s, u(s), v(s))ds

= ξλT1(u, v)(t) ≥ ξκT1(u, v)(t)

similarly, we getT2(ξu, ξv)(t) ≥ ξκT2(u, v)(t).

Thus, we haveT (ξu, ξv) ξκT, ξ ∈ (0, 1), (u, v) ∈ P,

where is partial order in X × Y induced by the cone P. Which implies that T is ξ−concave operator. Also from assumption (H13), T is nondecreasing operator with respectto the partial order. Now let us consider h ∈ P defined by

h(t) = (tα−1, tβ−1) = (fα(t), fβ(t)), t ∈ [0, 1]

122

and choose

µ = max

1

Γ(α)

1∫

0

f(s, 1, 1)ds,1

Γβ

1∫

0

g(s, 1, 1)ds

ν = max

1

Γ(α)

1∫

0

Kα(s)f(s, 0, 0)ds,1

Γβ

1∫

0

Kβ(s)g(s, 0, 0)ds

where from Green’s functions we can obtain that

Kα(s) = (1− s)α−δ−1[d1 − (1− s)δ],Kβ(s) = (1− s)β−γ [d2 − (1− s)γ ]. (5.1.8)

Now from nondecreasing property of f, g from (H13) we get µ > 0, ν > 0. Thus from (5.1.8)and using (H14) we have

T1h(t) =

1∫

0

Gα(t, s)f(s, hα(s), hβ(s))ds =

1∫

0

Gα(t, s)f(s, sα−1, sβ−1)ds

≤1∫

0

Gα(t, s)f(s, 1, 1)ds ≤(

1

Γα

1∫

0

(1− s)α−δ−1f(s, 1, 1)ds

)tα−1 ≤ µhα(t)

similarly, we haveT2h(t) ≤ µhβ(t).

Then, we obtainTh µh. (5.1.9)

Further, using (5.1.8) and (H14), we have for all t ∈ [0, 1]

T1h(t) =

1∫

0

Gα(t, s)f(s, sα−1, sβ−1)ds ≥

1∫

0

Gα(t, s)f(s, 0, 0)ds

≥(

1

Γα

1∫

0

Kα(s)f(s, 0, 0)ds

)tα−1 ≥ νhα(t)

similarly,T2h(t) ≥ νhβ(t).

Thus, we haveTh(t) νh. (5.1.10)

From (5.1.9) and (5.1.10) it follows that

νh Th µh,

which implies that Th ∈ P. Hence by Lemma 2.3.14 the operator T has a at least one fixedpoint(u, v) ∈ P.

123

5.1.3 Non-existence of positive solutions

In this subsection, the following two theorems discuss the non-existence of positive solutions.

Theorem 5.1.9. Suppose (C1) − (C3) in previous section hold true and there exist twoconstants a, b > 0 with a+ b < 1, and

f(t, u, v) <a‖(u, v)‖

σαand g(t, u, v) <

b‖(u, v)‖σβ

, for all t ∈ [0, 1], u > 0, v > 0,

then the Boundary value problem (5.1.1) has no positive solution.

Proof. On contrary let (u, v) be the positive solution of Boundary value problem (5.1.1) .Then (u, v) ∈ P for 0 < t < 1 and

‖(u, v)‖ = maxt∈[0,1]

|T (u(t), v(t))|

= maxt∈[0,1]

|T1(u(t), v(t))| + maxt∈[0,1]

|T2(u(t), v(t))|

≤ maxt∈[0,1]

1∫

0

Gα(t, s)|f(s, u(s), v(s))|ds + maxt∈[0,1]

1∫

0

Gβ(t, s)|g(s, u(s), v(s)|ds

⇒ ‖(u, v)‖ <1∫

0

Gα(1, s)a‖(u, v)‖

σαds+

1∫

0

Gβ(1, s)b‖(u, v)‖

σβds

< a‖(u, v)‖ + b‖(u, v)‖ = (a+ b)‖(u, v)‖ < ‖(u, v)‖⇒ ‖(u, v)‖ < ‖(u, v)‖.

(5.1.11)

Which is contradiction so Boundary value problem (5.1.1) has no positive solution. Henceproof is completed.

Theorem 5.1.10. Let (C1)− (C3) hold and

f(t, u(t), v(t)) > λ‖(u, v)‖(γ2α

1−θ∫

θ

Gα(1, s)ds

)−1

,

g(t, u(t), v(t)) > µ‖(u, v)‖(γ2β

1−θ∫

θ

Gβ(1, s)ds

)−1

,

(5.1.12)

for all t ∈ [0, 1], u > 0 and v > 0, further there exist two constants λ, µ > 0 such thatλ+ µ > 1. Then the Boundary value problem (5.1.1) has no positive solution.

Proof. Proof is just like the proof of Theorem5.1.9, so we omit it.

124

5.1.4 Existence of at least two positive solutions

Theorem 5.1.11. Assume that (C1) − (C3) together with the following assumption aresatisfied:

(H16) there a, b > 0 with a+ b ≥ 1 such that

f0 > a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

, f∞ > a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

;

g0 > b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

, g∞ > b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

;

(H17) there exist r > 0, c, d > 0 with c+ d < 1 such that

maxt∈[0,1],(u,v)∈∂Pr

f(t, u, v) ≤ cr

σα, maxt∈[0,1],(u,v)∈∂Pr

g(t, u, v) ≤ dr

σβ.

Moreover f0 = f∞ = ∞, g0 = g∞ = ∞ are also hold. Then Boundary value problem (5.1.1)has at least two positive solutions (u, v) and (u, v), which satisfy

0 < ‖(u, v)‖ < r < ‖(u, v)‖.Proof. Let ε, R with 0 < ε < r < R. If

∞ ≥ f0 > a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

,∞ ≥ g0 > b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

hold, then similar to the proof of Theorem 5.1.5, we have

‖T (u, v)‖ ≥ (a+ b)‖(u, v)‖ ≥ ‖(u, v)‖, for all (u, v) ∈ ∂Pε.If

∞ ≥ f∞ > a

(γ2

1−θ∫

θ

Gα(1, s)ds

)−1

,∞ ≥ g∞ > b

(γ2

1−θ∫

θ

Gβ(1, s)ds

)−1

hold, then similar to the proof of Theorem 5.1.6, we have

‖T (u, v)‖ ≥ (a+ b)‖(u, v)‖ ≥ ‖(u, v)‖, for all (u, v) ∈ ∂PR.Also together with (H16), for (u, v) ∈ P ∩ ∂Pr, we get

T1(u, v) =

1∫

0


0

Gα(1, s)f(s, u(s), v(s))ds

≤ cr

σα

1∫

0

Gα(1, s)ds = cr,

125

which implies that

T1(u, v) ≤ cr, similarly we can get T2(u, v) ≤ dr. (5.1.13)

From (5.1.13), we have‖T (u, v)‖ ≤ (c+ d)r < r.

Which implies that

‖T (u, v)‖ < r = ‖(u, v)‖, for all (u, v) ∈ P ∩ ∂Pr.

Thus by means of Theorem 2.3.11 and Theorems 5.1.5 and (5.1.6) give that T has a fixedpoint (u, v) ∈ P∩(Pε \Pr). Applying Theorem (2.3.11) and Theorems 5.1.4 and 5.1.5 whichgive that T has a fixed point (u, v) ∈ P ∩ (PR \ Pε). Further from Theorem 5.1.5 we havethat ‖(u, v)‖ 6= r and ‖(u, v)‖ 6= r. Thus

0 < ‖(u, v)‖ < r < ‖(u, v)‖.

Hence we complete the proof.

Theorem 5.1.12. Let (C1)− (C3) hold and suppose that the following condition hold:

(H18) there exist constants a, b, c, d > 0 such that a+ b ≤ 1, c + d ≤ 1 such that

a

σα> f0 ≥ 0,

b

σβ> g0 ≥ 0, and

c

σα> f∞ ≥ 0,

d

σβ> g∞ ≥ 0;

(H19) there exist ρ, λ, µ > 0 constant with λ+ µ > 1 such that

mint∈[θ,1−θ],(u,v)∈[0,ρ]×[0,ρ]

f(t, u(t), v(t)) > ρλ

(γα

1−θ∫

θ

Gα(1, s)ds

)−1

and

mint∈[θ,1−θ],(u,v)∈[0,ρ]×[0,ρ]

g(t, u(t), v(t)) > ρµ

(γβ

1−θ∫

θ

Gβ(1, s)ds

)−1

.

Then the Boundary value problem(5.1.1) has at least two positive solutions (u1, v1) and(u2, v2) which obey

0 < ‖(u1, v1)‖ < ρ < ‖(u2, v2)‖.

Proof. Assume that 0 < ε < ρ < R and constructing two open balls as in the proof ofTheorem5.1.5, as

Uε = (u, v) ∈ X × Y : ‖(u, v)‖ < ε, UR = (u, v) ∈ X × Y : ‖(u, v)‖ < R.

126

Now from (H19), foraσα

> f0 ≥> 0, bσβ

> g0 ≥ 0 and from the proof of Theorem 5.1.4, weget

‖T (u, v)‖ ≤ ‖(u, v)‖, ∀ (u, v) ∈ P ∩ ∂Uε. (5.1.14)

Further for cσα

> f∞ ≥ 0, dσβ> g∞ ≥ 0, and from the proof of Theorem 5.1.4, we have

‖T (u, v)‖ ≤ ‖(u, v)‖, ∀ (u, v) ∈ P ∩ ∂UR. (5.1.15)

On the other hand define an open ball Uρ = (u, v) ∈ X × Y : ‖(u, v)‖ < ρ, and from theassumption H19, if (u, v) ∈ P ∩ ∂Uρ, and for any τ ∈ [0, 1], we have

‖T1(u, v) ≥ T1(u, v)(τ)

≥1∫

0

Gα(t, s)f(s, u(s), v(s))ds

≥ λρ

(γα

1−θ∫

θ

Gα(1, s)ds

)(γα

1−θ∫

θ

Gα(1, s)ds

)−1

= λρ.

Similarly we have ‖T2(u, v)‖ ≥ µρ. Which implies that

‖T (u, v)‖ = ‖T1(u, v)‖ + ‖T2(u, v)‖ ≥ ρ(λ+ µ) > ρ = ‖(u, v)‖.

Thus we have from these

‖T (u, v)‖ > ρ = ‖(u, v)‖, ∀ (u, v) ∈ P ∩ ∂Uρ. (5.1.16)

By the application of Theorem 2.3.11 to (5.1.14) and (5.1.16), which yields that T has afixed point (u1, v1) ∈ P ∩ (Uρ \ Uε and similarly applying Theorem2.3.11 to (5.1.15) and(5.1.16), we get that T has a fixed point (u2, v2) ∈ P ∩ (UR \Uρ. Also we know from (5.1.16)that ρ 6= ‖(u1, v1)‖ and ρ 6= ‖(u2, v2)‖. Therefore

0 < ‖(u1, v1)‖ < ρ < ‖(u2, v2)‖.

Which completes the proof.

5.1.5 Existence of Multiple positive solutions

Theorem 5.1.13. Let (C1) − (C3) hold. If there exist 2m positive numbers wi, wi, i =1, 2 . . . m with w1 < γαw1 < w1 < w2 < γαw2 < w2 . . .wm < γαwm < wm and w1 <γβw1 < w1 < w2 < γβw2 < w2 . . .wm < γβwm < wm such that

(H20) f(t, u(t), v(t)) ≥ wi

(γα

1∫0

Gα(1, s)ds

)−1

, for (t, u, v) ∈ [0, 1]× [γαwi,wi]× [γβwi,wi],

and

f(t, u(t), v(t)) ≤ σ−1α wi, for (t, u, v) ∈ [0, 1] × [γαwi, wi]× [γβwk,wi], i = 1, 2 . . . m,

127

(H21) g(t, u(t), v(t)) ≥ wi

(γβ

1∫0

Gβ(1, s)ds)−1

)−1

, for (t, u, v) ∈ [0, 1]×[γβwi,wi]×[γαwi,wi],

and

g(t, u(t), v(t)) ≤ σ−1β wi, for (t, u, v) ∈ [0, 1] × [γαwi,wi]× [γβwk, wi], i = 1, 2 . . . m.

Then the Boundary value problem (5.1.1) has at least m- positive solutions (ui, vi), obeying

wi ≤ ‖(ui, vi)‖ ≤ wi, i = 1, 2 . . . m.

Proof. Proof of Theorem 5.1.13 is like the proof of Theorems 5.1.11 and 5.1.12, so we omitit.

Theorem 5.1.14. Suppose that (C1) − (C3) hold. If there exist 2m- positive numberswi, wi, i = 1, 2 . . . m, with w1 < w1 < w2 < w2 . . . < wm < wm, such that

(H22) f and g are non-decreasing on [0, wm], for all t ∈ [0, 1];

(H23) f(t, u(t), v(t)) ≥ wi

(γα

1−θ∫θ

Gα(1, s)ds)

)−1

, f(t, u(t), v(t)) ≤ wi

σα, i = 1, 2 . . . m,

g(t, u(t), v(t)) ≥ wi

(γβ

1−θ∫

θ

Gβ(1, s)ds)

)−1

, g(t, u(t), v(t)) ≤ wi

σβ, i = 1, 2 . . . m.

Then the Boundary value problem (5.1.1) has at least m- positive solutions (ui, vi), satisfying

wi ≤ ‖(ui, vi)‖ ≤ wi, i = 1, 2 . . . m.

Proof. Proof of Theorem 5.1.14 is like the proof of Theorem 5.1.11 and Theorem 5.1.12, sowe omit it.

Example 5.1.1. Consider the coupled system as follow

D52u(t) +

(t+ 1

4

)[Γ

(5

2

)|u(t)| + cos |v(t)|

], t ∈ [0, 1], u, v ≥ 0,

D52 v(t) =

(t2 + 1

4

)[sin |u(t)|+ v(t)

], t ∈ [0, 1], u, v ≥ 0,

u(i)(0) = 0, D32u(1) =

1

2D

32u

(1

2

),

v(i)(0) = 0, D32 v(1) =

1

2D

32 v

(1

2

),

128

Then, from the coupled system, we have

f(t, u(t), v(t)) =t+ 1

4[Γ

(52

)|u(t)|+cos |v(t)|], g(t, u(t), v(t)) =

(t2 + 1

4

)[sin |u(t)|+v(t)].

|f(t, u2, v2)− f(t, u1, v1)| ≤ Γ

(5

2

)(t+ 1

4

)|u2 − u1|+

(t+ 1

4

)|v2 − v1|,

|g(t, u2, v2)− g(t, u1, v1)| ≤(t2 + 1

4

)|u2 − u1|+

t2 + 1

56|v2 − v1|.

where f1(t) =t+ 1

4= g1(t), f2(t) = g2(t) =

t2 + 1

4. Using these values, we have

A =

1∫0

Gα(1, s)f1(s)ds)1∫0

Gα(1, s)g1(s(ds)

1∫0

Gα(1, s)f2(s)ds)1∫0

Gα(1, s)g2(s)ds)

=

[1235

48105

√π

56135

√π

4135

√π

].

By calculation we can obtain that ρ(A) = 0.4746 < 1, hence by the use of Theorem 5.1.3Boundary value problem (5.1.1) has a unique positive solution.

Example 5.1.2. Consider the system of non-linear fractional differential equations.

D73u(t) + a(t)

√u(t) + v(t) = 0, t ∈ (0, 1),

D52 v(t) + b(t) 3

√u(t) + v(t) = 0, t ∈ (0, 1),

u(0) = u′(0) = 0, D13u(1) =

1

3D

13u

(1

2

),

v(0) = v′(0) = 0, D12 v(1) =

1

2D

12 v

(1

2

)

and where f(t, u, v) = a(t)√u(t) + v(t) and g(t, u, v) = b(t) 3

√u(t) + v(t),

and a(t), b(t) : [0, 1] → [0,∞) are continuous.

Now f0 = lim(u,v)→0

f(t, u, v)

u+ v= ∞, similarly g0 = ∞. Also by simple calculation, we can get

that f∞ = 0 = g∞. Thus by Theorem 5.1.4, Boundary value problem (5.1.2) has at leastone positive solution.

Example 5.1.3. Consider the following boundary value problem

D92u(t) + (1− t2)[u(t) + v(t)]2 = 0, t ∈ (0, 1),

D143 v(t) + [u(t) + v(t)]3 = 0, t ∈ (0, 1),

u(i)(0) = 0, D52u(1) =

1

4D

52u

(1

4

),

v(i) = 0,D72 v(1) =

1

3D

72 v

(1

3

),

i = 0, 1, 2, 3.

129

By simple calculation, we obtain that f 0 = g0 = 0 and f∞ = g∞ = ∞. Thus by Theorem5.1.5,the Boundary value problem (5.1.3) has a positive solution.


D72u(t) + (1− t2) + 3

√u(t)v(t) = 0, t ∈ (0, 1),

D94 v(t) + 1 + t+ 4

√u(t)v(t) = 0, t ∈ (0, 1),

u(i)(0) = 0, D32u(1) =

1

4D

32u

(1

4

),

v(i) = 0, D74 v(1) =

1

3D

74 v

(1

3

),

i = 0, 1, 2.

Clearly f(t, 0, 0) 6= 0, g(t, 0, 0) 6= 0, f(t, 1, 1) 6= 0, g(t, 1, 1) 6= 0. By simple calculation,one can show that f, g are non decreasing for all t ∈ (0, 1). Further for ξ ∈ (0, 1) and

t ∈ (0, 1), u, v ≥ 0, we have by taking max

14 ,

13

= 1

3 ,

f(t, ξu, ξv) ≥ ξ13 f(t, u, v), g(t, ξu, ξv) ≥ ξ

13 g(t, u, v).

Thus all the assumption of Theorem5.1.6 are satisfied , so the Boundary value problem(5.1.4) has a unique positive solution in Cf where f(t) = (t

52 , t

54 ).


D92u(t) + (t2 + 1)

[u2(t) + v(t)

(4t2 + 4)σα

]= 0, t ∈ (0, 1),

D143 v(t) + (t3 + 1)

[u(t) + v2(t)

(8t4 + 8)σβ

]= 0, t ∈ (0, 1),

u(i)(0) = 0, D52u(1) =

1

4D

52u

(1

4

),

v(i) = 0, D72 v(1) =

1

3D

72 v

(1

3

),

i = 0, 1, 2, 3.

By simple calculation, we obtain that f0 = g0 = ∞ and f∞ = g∞ = ∞.Further for all (t, u, v) ∈ [0, 1] × [0, 1] × [0, 1], we have

f(t, u, v) ≤ (t+ 1)2

4(t+ 1)σα=σ−1α

2, g(t, u, v) ≤ (t2 + 1)2

8(t+ 1)σβ=σ−1β

4.

Thus all the assumptions of Theorem 5.1.6 are satisfied and also a = 12 , b =

14 , r = 1. Hence

by Theorem 5.1.7, the Boundary value problem (5.1.5) has at least two positive solutions(u1, v1) and (u2, v2) which satisfy

0 < ‖(u1, v1)‖ < 1 < ‖(u2, v2)‖.

130

Example 5.1.6. Consider the system of non linear fractional differential equations:

D52u(t) +

[2u2(t) + u(t)][20 + cos v(t)]

u(t) + v(t) + 1= 0, t ∈ (0, 1),

D52 v(t) +

[2v2(t) + v(t)][20 + cos v(t)]

u(t) + v(t) + 1= 0, t ∈ (0, 1),

u(i)(0) = 0, D12u(1) =

1

10D

12u

(1

2

),

v(i)(0) = 0, D12 v(1) =

1

10D

12 v

(1

2

),

0, i = 0, 1.

Since (C1)− (C3) hold and also

f0 = g0 = 20, f∞ = g∞ = 43, 20‖(u, v)‖ < f(t, u(t), v(t)) < 43‖(u, v)‖

20‖(u, v)‖ < g(t, u(t), v(t)) < 43‖(u, v)‖ and f(t, u(t), v(t)) < 43‖(u, v)‖ < ‖(u, v)‖σα

where σα ≈ 0.876126 and σβ ≈ .475675

(i) Now f(t, u(t), v(t)) <‖(u, v)‖σα

≈ 1.1413‖(u, v)‖.From which it implies that f(t, u(t), v(t)) < 43‖(u, v)‖ ≈ 1.1413‖(u, v)‖and g(t, u(t), v(t)) < 43‖(u, v)‖ ≈ 2.1022‖(u, v)‖. Thus by Theorem 5.1.9, Boundary valueproblem (5.1.6) has no positive solution.(ii) Also , we have

f(t, u(t), v(t)) > 20‖(u, v)‖ > ‖(u, v)‖(γ2α

34∫

14

Gα(1, s)ds

)−1

≈ 15.8688‖(u, v)‖,

g(t, u(t), v(t)) > 2‖(u, v)‖ > ‖(u, v)‖(γ2β

34∫

14

Gβ(1, s)ds

)−1

≈ 12.678‖(u, v)‖.

Then by Theorem 5.1.10, Boundary value problem (5.1.6) has no solution.

Chapter 6

Iterative techniques for solutions ofCoupled systems of multi-pointboundary value problems

In this chapter, we study iterative solutions for some coupled system of multi point bound-

ary value problems of fractional order differential equations. We develop some iterative

schemes for approximate solutions for systems of nonlinear fractional order differential

equations subject to different types of multi-point boundary conditions. As in the last

few decades, existence and uniqueness of solutions for systems of nonlinear fractional order

differential equations have been studied by many mathematicians and the area is well ex-

plored [116, 117, 118, 119, 120]. It is well known that the upper and lower solution method

is a powerful tool for proving existence results for boundary value problems. In many cases

it is possible to find a minimal solution and a maximal solution between the lower solution

and the upper solution by the monotone iterative technique. Recently the area devoted to

the existence of approximate solutions and iterative techniques for systems of fractional dif-

ferential equations has attracted some attentions, for example [121, 122]. Moreover in this

chapter, we develop sufficient conditions for existence of multiple solutions by using mono-

tone iterative technique together with the method of upper and lower solutions as studied in

[123, 124, 125, 126]. Recently, some authors studied coupled system with coupled boundary

131

132

conditions. Such system are rarely studied and only few articles are available in literature,

for example (see[127]and the references therein). In this chapter, we investigate multiple

solutions for a coupled system with coupled m-point boundary conditions. The considered

system is more general and complicated one.

Organization of the chapter as: Section 6.1 is related to the existence of iterative solutions

for a system of nonlinear fractional order differential equations with two point boundary

conditions. The nonlinear terms explicitly depend on the fractional integral involved in the

nonlinear functions. Section 6.2 is devoted to the investigation of sufficient conditions for the

existence of iterative solutions for a coupled system with three point boundary conditions.

In section 6.3, a coupled system with coupled m-point boundary conditions is studied with

upper and lower solutions. Moreover iterative sequences are developed for approximating

the solutions for the system under consideration.

6.1 Iterative techniques for solutions of coupled systems of

boundary value problem

We study the following coupled systems [128], of nonlinear fractional order differential

equations via monotone iterative technique together with the method of upper and lower

solutions

Dαu(t) + f1(t, v(t), Iβv(t)) = 0, t ∈ [0, 1],

Dβv(t) + f2(t, u(t), Iαu(t)) = 0, t ∈ [0, 1],

u(0) = 0, u′(1) = 0, v(0) = 0, v′(1) = 0.

(6.1.1)

where 1 < α, β ≤ 2 and f1, f2 : I × R × R → R are satisfying Caratheodory conditions.

We use various tools of functional analysis to obtain sufficient conditions for existence and

uniqueness of maximal and minimal solutions. We need the following assumptions in sequel.

Assume that

(H1) f(t, u), g(t, u) are non decreasing in u for each t ∈ [0, 1].

133

Lemma 6.1.1. Iffi(i = 1, 2) : I×R×R → R are satisfying Caratheodory conditions, then(6.1.1) has integral representation of the form given by

u(t) =

∫ 1

0Gα(t, s)f1(s, v(s), I

βv(s))ds, t ∈ [0, 1],

v(t) =

∫ 1

0Gβ(t, s)f2(s, u(s), I

αu(s))ds, t ∈ [0, 1],

(6.1.2)

where Gα, Gβare Green’s functions given by

Gα(t, s) =1

Γ(α)

tα−1(1− s)α−2 − (t− s)α−1, 0 ≤ s ≤ t ≤ 1,

tα−1(1− s)α−2, 0 ≤ t ≤ s ≤ 1,(6.1.3)

Gβ(t, s) =1

Γ(β)

tβ−1(1− s)β−2 − (t− s)β−1, 0 ≤ s ≤ t ≤ 1,

tβ−1(1− s)β−2, 0 ≤ t ≤ s ≤ 1.(6.1.4)

Proof. Applying Iα on both sides of (6.1.1) , we obtain

u(t) = −Iαf1(t, v(t), Iαv(t))− C1t

α−1 − C2tα−2, C1, C2 ∈ R, (6.1.5)

Using the boundary conditions u(0) = 0 and u′(1) = 0, we have C2 = 0, andC1 = − 1

α−1Iα−1f1(1, v(s), I

βv(t)). Hence, (6.1.5), takes the form

u(t) =tα−1

α− 1Iαf1(1, v(1), I

βv(1)) − Iαf1(t, v(t), Iβv(t))

=1

(α− 1)Γ(α − 1)

∫ 1

0tα−1(1− s)α−2f1(s, v(s), I

βv(s))ds

+1

Γ(α)

∫ t

0(t− s)α−1f1(s, v(s), I

βv(s))ds =

∫ 1


βv(s))ds.

Similarly repeating the above process with the second equation of the system, we obtainthe second part of (6.1.1)

v(t) =∫ 10 Gβ(t, s)f2(s, u(s), I

αu(s))ds.

Lemma 6.1.2. Let G = (Gα, Gβ) be the Green’s functions of (6.1.1), then it satisfies thefollowing properties:

(P1) G(t, s) ≥ 0, for all t, s ∈ [0, 1];

(P2)∫ 10 G(t, s)ds ≤

(1

(α−1)Γ(α+1) ,1

(β−1)Γ(β+1)

), for all t ∈ [0, 1].

134

Proof. For all t, s ∈ [0, 1], we have

tα−1(1− s)α−2 − (t− s)α−1

Γ(α)≥ 0, as 0 ≤ s ≤ t ≤ 1,

thus Gα(t, s) ≥ 0 for all t, s ∈ [0, 1], Gβ(t, s) ≥ 0 for all t, s ∈ [0, 1]. Now∫ 1

0Gα(t, s)ds =

∫ 1

0

tα−1(1− s)α−2

Γ(α)ds−

∫ t

0

(t− s)α−1

Γ(α)ds ≤ 1

(α− 1)Γ(α + 1),

similarly∫ 10 Gβ(t, s)ds ≤ 1

(β−1)Γ(β−1) . Hence,∫ 10 G(s, t)ds ≤ ( 1

(α−1)Γ(α+1) ,1

(β−1)Γ(β+1) ).

Now we write (6.1.1) in integral equation form as

u(t) =

∫ 1


βv(s))ds

=

∫ 1

0Gα(t, s)

f1(s,

∫ 1

0Gβ(s, x)f2(x, u(x), I

αu(x))dx, Iβ[ ∫ 1


αu(x))dx])ds.

(6.1.6)

6.1.1 Existence of upper and lower solutions

For further investigation we give the following assumptions and notations:

(H2) fi(i = 1, 2) : [0, 1] × R× R → R satisfy Caratheodory conditions;

(H3) For any u1, u2, v1, v2 ∈ U with ui ≤ vi, i = 1, 2 there exist constants Πi > 0, for

i = 1, 2, 3, 4, such that

0 ≤ f1(t, v1, v2)− f1(t, u1, u2) ≤ Π1(v1 − u1) + Π2(v2 − u2), t ∈ [0, 1],

0 ≤ f2(t, v1, v1)− f2(t, u1, u2) ≤ Π3(v1 − u1) + Π4(v2 − u2), t ∈ [0, 1];

(H4) (u0, v0) ∈ X ×Y and (u0, v0) ∈ X ×Y are lower and upper solution respectively with

u0 ≤ u0, v0 ≤ v0.

We use

Υ =Π1Π3

(β − 1)Γ(β + 1)+

Π1Π4

(α− 1)(β − 1)Γ(α+ 1)Γ(β + 1)

+Π2Π3

(β − 1)2Γ(β + 1)+

Π2Π4

(α− 1)(β − 1)Γ(α + 1)Γ2(β + 1).

135

Lemma 6.1.3. Let assumptions (H1) − (H3) holds and Υ < 1, then there exist maximaland minimal solutions (u∗, v∗), (u∗, v∗) for (6.1.1), by using u∗0, u

∗0 as initial iterations and

get the following iterative sequences for each n ∈ Z+,

un(t) =

∫ 1

0Gα(t, s)f1

(s,

∫ 1

0Gβ(s, x)f2(x, un−1(x), I

αun−1(x))dx,

Iβ[∫ 1


αun−1(x))dx

])ds,

(6.1.7)

u∗n(t) =∫ 1

0Gα(t, s)f2

(s,

∫ 1

0Gβ(s, x)f2(x, u

∗n−1(x), I

αu∗n−1(x))dx,

Iβ[∫ 1

0Gβ(s, x)f2(x, u

∗n−1(x), I

αu∗n−1(x))dx

])ds,

(6.1.8)

and we haveu0 ≤ u1 ≤ u2 ≤ . . . ≤ un ≤ . . . ≤ u∗n ≤ . . . u∗1 ≤ u∗0,

u∗(t) = limn→∞

un(t), v∗(t) =∫ 1

0Gβ(t, s)f2(s, u

∗(s), Iαu∗(s))ds,


u∗n(t), v∗(t) =∫ 1

0Gβ(t, s)f2(s, u


further the error estimate of the minimal solutions is given by

‖u∗(t)− un(t)‖ ≤ Υn

1−Υ‖u1(t)− u0(t)‖

and maximal solutions is given by

‖u∗n(t)− u∗(t)‖ ≤ Υn

1−Υ‖u∗0(t)− u∗1(t)‖.

Proof. First define an operator T : U → X by

Tu(t) =

∫ 1

0Gα(t, s)

f1(s,

∫ 1


αu(x))dx, Iβ[ ∫ 1


αu(x))dx])ds.

(6.1.9)

Clearly T is continuous as Gα(t, s), Gβ(t, s), fi, i = 1, 2 are continuous and fixed points ofT are solutions of (6.1.1). Hence we need to prove the existence of fixed points of T . Let(u0, v0) ∈ X ×Y be a lower solution of (6.1.1), then

v0(t) ≤∫ 1

0Gβ(t, s)f2(s, u0(s), I

αu0(s))ds, u0(t) ≤∫ 1

0Gβ(t, s)f1(s, v0(s), I

βv0(s))ds.

136

Since Iβv0(t) ≤ Iβ∫ 10 Gβ(t, s)f2(s, u0(s), I

αu0(s))ds. In view of (H1), we get

f1(t, v0(t), Iβv0(t))

≤ f1(t,

∫ 1

0Gβ(t, s)f2(s, u0(s), I

αu0(s))ds, Iβ

∫ 1

0Gβ(t, s)f2(s, u0(s), I

αu0(s))ds).

Thus,

u0(t) ≤∫ 1

0Gα(t, s)f1(s, v0(s), I

βv0(s))ds ≤∫ 1

0Gα(t, s)

f1(t,

∫ 1

0Gβ(t, s)f2(s, u0(s), I

αu0(s)ds, Iβ

∫ 1

0Gβ(t, s)f2(s, u0(s), I

αu0(s)ds)ds,

which implies that u0(t) ≤ Tu0(t) and hence u0 is lower solution of T . Now in view of (H1)for any u, v ∈ U with u ≤ v ⇒ Tu ≤ Tv which implies that T is increasing operator, so Tcan be written from the iterative sequence (6.1.7) as un = Tun−1 for all n ∈ Z+. Due to(H1) we have

u0 ≤ u1 ≤ u2 ≤ . . . ≤ un. (6.1.10)

For any u, v ∈ U with u ≤ v ⇒ Iαu ≤ Iαv, we have f2(t, u, Iαu) ≤ f2(t, v, I

αv) andGβ(t, s) ≥ 0 we obtain

∫ 1


αu(x))dx ≤∫ 1

0Gβ(s, x)f2(s, v(s), I

αv(s))dx

⇒ Iβ[ ∫ 1


αu(x))dx]≤ Iβ

[ ∫ 1

0Gβ(s, x)f2(s, v(s), I

αv(s))dx].

In view of (H1), we have

0 ≤ f1(t,

∫ 1

0Gβ(t, x)f2(x, v(x), I

αv(x))dx, Iβ∫ 1


αv(x))dx)

− f1(t,

∫ 1

0Gβ(t, x)f2(x, u(x), I

αu(x))dx, Iβ∫ 1


αu(x))dx)

≤ Π1

( ∫ 1


αv(x))dx, Iβ∫ 1


αv(x))dx)

−Π1

( ∫ 1


αu(x))dx, Iβ∫ 1


αu(x))dx)

≤ Π1

∫ 1

0Gβ(s, x)[Π3(v − u) + Π4(I

αv − Iαu)]dx

+Π2

Γ(β)

∫ t

0(t− s)β−1

∫ 1

0Gβ(s, x)[Π3(v − u) + Π4(I

αv − Iαu)]dxds

137

which implies that

∥∥f1(t,

∫ 1


αv(x))dx, Iβ∫ 1


αv(x))dx)

− f1(t,

∫ 1

0Gβ(t, x)f2(t, u(x), I

αu(x))dx, Iβ∫ 1

0Gβ(t, x)f2(t, u(x), I

αu(x))dx)∥∥

≤[

Π1Π3

(β − 1)Γ(β + 1)+

Π1Π4

(β − 1)Γ(α+ 1)Γ(β + 1)

]‖v − u‖

+

[Π2Π3

(β − 1)Γ2(β + 1)+

Π2Π4

(β − 1)Γ2(β + 1)Γ(α + 1)

]‖v − u‖.

(6.1.11)

Using (6.1.11), we have

‖Tv − Tu‖ ≤ Υ‖v − u‖. (6.1.12)

Thus from (6.1.10) and (6.1.12), we have

‖u2 − u1‖ = ‖Tu1 − Tu0‖ ≤ Υ‖u1 − u0‖,‖u3 − u2‖ = ‖Tu2 − Tu1‖ ≤ Υ2‖u1 − u0‖,...

‖un+1 − un‖ = ‖Tun − Tun−1‖ ≤ Υn‖u1 − u0‖,

therefore for each n, m ∈ Z+, we have

‖un+m − un‖ ≤ ‖un+m − un+m−1‖+ ‖un+m−1 − un+m−2‖+ . . . ‖un+1 − un‖,

which implies that

‖un+m − un‖ ≤ Υn(1−Υm)

1−Υ‖u1 − u0‖, (6.1.13)

for 0 < Υ < 1, we have ‖un+m−un‖ → 0 as n→ ∞. Thus un is a cauchy sequence in S. Letu∗(t) = limn→∞ un(t). So Tu

∗ = u∗. Hence (6.1.1) has a pairs of solutions (u∗(t), v∗(t)) andv∗(t) =

∫ 10 Gβ(t, s)f2(s, u

∗(s), Iαu∗(s))ds. Let m→ ∞ in (6.1.13), then we have ‖u∗−un‖ ≤Υn

1−Υ‖u1−u0‖. Similar if we use u∗0 as initial iteration for upper solution, then we have (u∗, v∗)

of (6.1.1) and u∗(t) = limn→∞ u∗n(t), v∗(t) =

∫ 10 Gβ(t, s)f2(s, u

∗(s), Iαu∗(s))ds. Hence, it isclear to observe that

u0 ≤ u1 ≤ . . . ≤ un ≤ . . . u∗n ≤ . . . ≤ u∗2 ≤ u∗1 ≤ u∗0.

Further to prove maximal and minimal solution of (6.1.1), take (u∗, v∗) as maximal and(u∗, v∗) as minimal solutions, then for any w(t) ∈ S with Tw = w, we have un ≤ w ≤ u∗n. AsT is increasing operator so, Tun ≤ Tw ≤ Tu∗n as n→ ∞ we get u∗(t) ≤ w(t) ≤ u∗(t). Thusu∗(t), u∗(t) are the minimal and maximal fixed points of T respectively. Thus (u∗(t), v∗(t))and (u∗(t), v∗(t)) are the minimal and maximal solutions of (6.1.1) respectively.

138

6.1.2 Uniqueness of upper and lower solutions

Theorem 6.1.4. Under the assumptions (H2), (H3) with Υ < 1, the extremal (maximaland minimal) solutions of BVP (6.1.1) are unique.

Proof. Let x0, y0 ∈ X be minimal and maximal solution of operator equation Tw = w suchthat x0 ≤ Tx0, y0 ≥ Ty0, t ∈ [0, 1].Assume that x0(t), y0(t) be the initial iterations respectively such that xn → x∗, yn →y∗ as n → ∞, also Tx∗ = x∗, T y∗ = y∗. We need to prove that x∗ = u∗ and y∗ = v∗.As x0 ≤ x∗ and T is increasing operator so we have xn = T nx0 ≤ T nx∗, for each n ∈ Z+.Thus u0 ≤ u1 ≤ u2 ≤ . . . ≤ un ≤ . . . ≤ x∗. Now by mathematical inductions and (6.1.12),we obtain

‖x∗ − un‖ = ‖T nx∗ − T nu0‖ ≤ Υn‖x∗ − u0‖ → 0 as n→ ∞.

Thuslimn→∞

‖x∗ − un‖ = 0 ⇒ x∗ = limn→∞

un = u∗ ⇒ u∗ = x∗.

Similarly, we can show that v∗ = y∗.

Let (u, v), (u∗, v∗) are unique minimal and maximal solutions then error estimate can

be calculated as

un(t) =

∫ 1

0Gα(t, s)f1

(s,

∫ 1


αun−1(x))dx

)ds, n = 1, 2, 3, ...

u∗n(t) =∫ 1

0Gα(t, s)f1

(s,

∫ 1

0Gβ(s, x)f2(x, u∗n−1(x), I

αu∗n−1(x))dx

)ds, n = 1, 2, 3, ...

as u0 ≤ u1 ≤ ... ≤ un ≤ ...u∗n ≤ ... ≤ u∗2 ≤ u∗1 ≤ u∗0,


un(t), v∗(t) =

∫ 1

0Gβ(t, s)f2(s, u



un(t), v∗(t) =

∫ 1

0K2(t, s)f2(s, u


we obtain the following error estimates for lower and upper solutions as

‖u− un‖ ≤ Υn

1−Υ|u1 − u0‖, |u∗n − u∗‖ ≤ Υn

1−Υ|u∗0 − u∗1‖.

Example 6.1.1. Consider the following coupled system of boundary values problem

D32u(t) +

(1− t

4

)2

v(t) +

(1− t3

6

)I

32 v(t) = 0, t ∈ [0, 1],

D32 v(t) +

(t2 − 1

3

)2

u(t) +

(t− 1

2

)3

I32u(t) = 0, , t ∈ [0, 1],

u(0) = u′(1) = 0, v(0) = v′(1) = 0.

139

Since

f1(t, v, I32 v) =

(1− t

4

)2

v(t) +

(1− t3

6

)I

32 v(t),

f2(t, u, I32u) =

(t2 − 1

3

)2

u(t) +

(t− 1

2

)3

I32u(t),

where α = 32 , β = 3

2 . For any u(t) ≤ v(t), we have

0 ≤ f1(t, v, I32 v)− f1(t, u, I

32u) ≤ 1

16(v − u) +

1

6I

32 (v − u),

0 ≤ f2(t, v, I32 v)− f2(t, u, I

32u) ≤ 1

9(v − u) +

1

9I

32 (v − u),

which implies that

Π1 =1

16, Π2 =

1

6, Π3 =

1

9, Π4 =

1

8

and Υ = 0.638522 < 1 is satisfied. Clearly (0, 0) is the unique solutions of (6.1.1). Letus take (u0, v0) = (−2,−2) and (u∗0, v

∗0) = (2, 2) as initial iteration for lower and upper

solutions respectively and the iterative sequences by taking n is large enough as

u∗(t) = u9(t), v∗(t) =∫ 1

0Gβ(t, s)

[(s2 − 1

3

)2u9(s) +

(s− 1

2

)3I

32u9(s)

]ds,

u∗(t) = u∗9(t), v∗(t) =

∫ 1

0Gβ(t, s)

[(s2 − 1

3

)2u∗9(s) +

(s− 1

2

)3I

32u∗9(s)

]ds.

The error estimates are

‖u(t) − u9(t)‖ ≤ Υ9

1−Υ‖u1(t)− u0(t)‖ ≤ (.638522)9

1− 0.638522maxt∈[0,1]

|u1(t) + 2| ' 5.8× 10−2,

similarly

‖u(t) − u∗9(t)‖ ≤ Υ9

1−Υ‖u∗1(t)− u∗0(t)‖ ≤ (0.638522)9

1− 0.638522)maxt∈[0,1]

|u∗1(t)− 2‖ ' 4.8× 10−2.

140

6.2 Iterative techniques for solutions of coupled systems withthree point boundary conditions

We study the following coupled system of nonlinear three point boundary value problems

for fractional order differential equations

Dαu(t) + f(t, v(t)) = 0, t ∈ [0, 1],

Dβv(t) + g(t, u(t)) = 0, t ∈ [0, 1],

u(0) = Dα−2u(0) = 0, u(1) = δu(η),

v(0) = Dβ−2v(0) = 0, v(1) = δv(η),

(6.2.1)

where 0 < η < 1, δηα < 1, δηβ < 1 and 2 < α, β ≤ 3, u, v ∈ C[0, 1] and f, g : I ×

R → R satisfy the Caratheodory conditions. We develop the method of upper and lower

solutions and monotone iterative technique to obtain sufficient conditions for existence and

approximation of solutions of the system.

Lemma 6.2.1. Under the assumption (H2), the integral representation of the system ofboundary value problems (6.2.1) is given by

u(t) =

∫ 1

0G1(t, s)f(t, v(t))ds, t ∈ [0, 1],

v(t) =

∫ 1

0G2(t, s)g(s, u(s))ds, t ∈ [0, 1],

(6.2.2)

where Gi(t, s), i = 1, 2 are given by

G1(t, s) =1

Γ(α)

tα−1

∆1

[δ(η − s)α−1 + (1− s)α−1

]− (t− s)α−1, 0 ≤ s ≤ t ≤ 1, t ≤ η,

tα−1

∆1

[δ(η − s)α−1 + (1− s)α−1

], 0 ≤ t ≤ s ≤ 1, t ≤ η,

tα−1

∆1(1− s)α−1 − (t− s)α−1, 0 ≤ s ≤ t ≤ 1, η ≤ t,

(t(1− s))α−1

∆1, 0 ≤ t ≤ s ≤ 1, η ≤ t,

141

G2(t, s) =1

Γ(β)

tβ−1

∆2

[δ(η − s)β−1 + (1− s)β−1

]− (t− s)β−1, 0 ≤ s ≤ t ≤ 1, t ≤ η,

tβ−1

∆2

[δ(η − s)β−1 + (1− s)β−1

], 0 ≤ t ≤ s ≤ 1, t ≤ η,

tβ−1

∆2(1− s)β−1 − (t− s)β−1, 0 ≤ s ≤ t ≤ 1, η ≤ t,

(t(1 − s))β−1

∆2, 0 ≤ t ≤ s ≤ 1, η ≤ t.

Proof. Using Iα on the first equation of the system (6.2.1) and Lemma 2.2.6, we obtain

u(t) = −Iαf(t, v(t)) + C1tα−1 + C2t

α−2 + C3tα−3, C1, C2, C3 ∈ R. (6.2.3)

The boundary conditions u(0) = 0, Dα−2u(0) = 0 and u(1) = δu(η) imply that

C2 = 0 = C3 and C1 =1

∆1[Iαf(η, u(η)) + Iαf(1, v(s))].

Hence, (6.2.3) can be rewritten as

u(t) = −Iαf(t, v(t)) +tα−1

∆1[δIαf(η, v(η)) + Iαf(1, v(1)) =

∫ 1

0G1(t, s)f(s, v(s)ds.

Similarly, using Iβ on the second equation of the System (6.2.1) and Lemma 2.2.6, we obtainthe second part of the System (6.2.2).

Lemma 6.2.2. The functions G1, G2 satisfies the following property

(P1) The Green’s functions G1(t, s) ≥ 0, G2(t, s) ≥ 0 for all t, s ∈ [0, 1].

(P2)∫ 10 G1(t, s)ds ≤ 2

∆1Γ(α+1) ,∫ 10 G2(t, s)ds ≤ 2

∆2Γ(β+1) ), for all t ∈ [0, 1].

Proof. (P1) is obvious.

(P2) Since G1(t, s) ≤ tα−1

Γ(α)∆1

(δ(η − s)α−1 + (1− s)α−1

), it follows that

∫ 1

0G1(t, s)ds ≤

tα−1

∆1Γα[δ

∫ η

0(η − s)α−1ds+

∫ 1

0(1− s)α−1ds]

≤ tα−1

∆1Γ(α+ 1)[δηα + 1] ≤ 2

∆1Γ(α+ 1).

Similarly, we have∫ 10 G2(t, s)ds ≤ 2

∆2Γ(β+1) .

142


We write the system of integral equations (6.2.2) in the following equivalent form

u(t) =

∫ 1

0G1(t, s)f(t, v(t))ds =

∫ 1

0G1(t, s)f

(s,

∫ 1

0G2(s, τ)g(τ, u(τ))dτ

)ds. (6.2.4)

and define an operator T : X → X by

Tu(t) =

∫ 1

0G1(t, s)f

(s,

∫ 1


)ds. (6.2.5)

Then, the integral equation (6.2.4) can be written as an operator equation

(I − Tu)(t) = 0, t ∈ [0, 1] (6.2.6)

and solutions of the integral equations (6.2.4) are solutions of the operator equations (6.2.6),

that is, fixed points of T . In view of (H1), for u, v ∈ U with u ≤ v, we have

Tu(t) =

∫ 1

0G1(t, s)f

(s,

∫ 1


)ds

≤∫ 1

0G1(t, s)f

(s,

∫ 1

0G2(s, τ)g(τ, v(τ))dτ

)ds = Tv(t)

(6.2.7)

that is, T is nondecreasing. Use the notation ∆ = AB∆1∆2Γ(α+1)Γ(β+1) and assume that

(H4) α0, β0 ∈ E are lower and upper solutions of (6.2.6) such that α0 ≤ β0 on [0, 1].

Lemma 6.2.3. Under the assumptions H1, H2 and H4, there exists a monotone sequenceof solutions of linear problems converging uniformly to a solution of the nonlinear integralequation (6.2.4).

Proof. In view of H2 and H4, the operator T is continuous and nondecreasing. Chooseu0 = p, using the definition of lower and upper solutions and the nondecreasing property ofT , we obtain

u0 ≤ Tu0 ≤ Tβ ≤ β0, that is, u0 ≤ u1 ≤ β0 on [0, 1],

where u1 is a solution of the linear problem u1 = Tu0. Again the nondecreasing propertyof T implies that

Tu0 ≤ Tu1 ≤ Tβ0 ≤ β0, that is, u1 ≤ u2 ≤ β0 on [0, 1],

143

where u2 is a solution of the linear problem u2 = Tu1. Continuing in the above fashion, weobtain a bounded monotone sequence un of solutions satisfying

u0 ≤ u1 ≤ u2 ≤ ...un−1 ≤ un ≤ β0 on [0, 1],

where un is a solution of the linear problem un = Tun−1. The monotonicity and bounded-ness of the sequence implies the existence of u ∈ U such that un → u as n → ∞. Hence,passing to the limit n → ∞, the equation un = Tun−1 implies that u = Tu, that is, u is asolution of the integral equation (6.2.4)

u(t) =

∫ 1

0G1(t, s)f

(s,

∫ 1


)ds, t ∈ [0, 1].

Now, define the error en = x− un, then from the monotonicity of the sequence un, itfollows that en−1 ≥ en, that is, the sequence en is monotonically decreasing and convergesto 0.

Remark 6.2.1. If we choose u0 = q, we obtain a bounded monotone decreasing sequenceun such that

u0 ≥ u1 ≥ u2 ≥ ...un−1 ≥ un ≥ p on [0, 1],

which converges to a solution of (6.2.4).

Theorem 6.2.4. Under the assumptions H2, H3 and H4 the system of of Boundary valueproblem (6.2.1) has a unique upper and lower solutions.

Proof. The uniqueness of solutions follows from the monotonicity and uniform boundednessof the sequence.

6.3 Iterative techniques for solutions of coupled systems with

m-point coupled boundary conditions

In this section, we study the following coupled system of nonlinear fractional order differ-

ential equations [129], by means of monotone iterative technique together with the method

of upper and lower solutions

cDαu(t) + f(t, u(t), v(t)) = 0, cDβv(t) + g(t, u(t), v(t)) = 0, 0 < t < 1,

u(0) = v(0) = 0, u(1) =

m−2∑

i=1

δiv(ηi), v(1) =

m−2∑

i=1

λiu(ξi),(6.3.1)

where 1 < α, β ≤ 2, ηi, ξi(i = 1, 2, ...m − 2) ∈ (0, 1),∑m−2

i=1 δiηi < 1,∑m−2

i=1 γiξi < 1 and

f, g : (0, 1) × R2 → R2 are given continuous functions. We use various tools of functional

144

analysis to obtain sufficient conditions for existence and uniqueness and multiplicity of

solutions. We assume that (x0, y0), (µ0, ν0) ∈ X ×X are ordered lower and upper solutions

respectively of Coupled boundary conditions (6.3.1). Further we define the ordered sector

as

Θ = (u, v) ∈ X × X : (x0, y0) ≤ (u, v) ≤ (µ0, ν0), t ∈ [0, 1]. (6.3.2)

For further study we make the following assumptions which will be use through out in this

paper.

(H5) 0 <∑m−2

i=1 δi∑m−2

i=1 λi < 1;

(H6) f, g : [0, 1] × R2 → R;

(H7) f(t, u, v) is non-decreasing in v and there exist U(t) > 0 such that

f(t, u1, v)− f(t, u2, v) ≤ −U(t)(u1 − u2);

(H8) g(t, u, v) is non-decreasing in u and there exist V (t) > 0 such that

g(t, u, v1)− g(t, u, v2) ≤ −V (t)(v1 − v2).

6.3.1 Existence of Green’s functions

Lemma 6.3.1. Let y(t), w(t) ∈ C[0, 1],then the corresponding Hammerstein integral repre-sentation of the Coupled boundary conditions(6.3.1), is given by

u(t) =

∫ 1

0K11(t, s)y(s)ds+

∫ 1

0K12(t, s)w(s)ds, t ∈ (0, 1),

v(t) =

∫ 1

0K2(t, s)w(s)ds +

∫ 1

0K22(t, s)y(s)ds, t ∈ (0, 1),

(6.3.3)

where

K11(t, s) = Gα(t, s) +t∑m−2

i=1 δiηi∑m−2

i=1 λi∆

Gα(ξi, s), K12 =t∑m−2

i=1 δi∆

Gβ(ηi, s),

K2(t, s) = Gβ(t, s) +t∑m−2

i=1 λiξi∑m−2

i=1 δi∆

Gβ(ηi, s), K22 =t∑m−2

i=1 λi∆

Gα(ξi, s),

∆ = 1−m−2∑

i=1

λiξi

m−2∑

i=1

δiηi,

(6.3.4)

145

also Gα(t, s), Gβ(t, s) are given by

Gα(t, s) =

t(1− s)α−1 − (t− s)α−1

Γ(α), , 0 ≤ s ≤ t ≤ 1,

t(1− s)α−1

Γ(α), 0 ≤ t ≤ s ≤ 1

(6.3.5)

Gβ(t, s) =

t(1− s)β−1 − (t− s)β−1

Γ(β), 0 ≤ s ≤ t ≤ 1,

t(1− s)β−1

Γ(β), 0 ≤ t ≤ s ≤ 1.

(6.3.6)

Proof. Now by applying Iα, Iβ on both sides of Coupled boundary value problem (6.3.1)corresponding to homogenous boundary conditions u(0) = v(0) = u(1) = v(1) = 0, weobtain

uc(t) = tIαy(1)− Iαy(t) =

∫ 1

0Gα(t, s)y(s)ds,

vc(t) = tIβw(1) − Iβw(t) =

∫ 1

0Gβ(t, s)w(s)ds,

(6.3.7)

as (6.3.1) is equivalent to the following system of integral equations

u(t) = u(1)t+

∫ 1

0Gα(t, s)y(s)ds, v(t) = v(1)t+

∫ 1

0Gβ(t, s)w(s)ds, t ∈ [0, 1]. (6.3.8)

By coupled m-point boundary conditions of (6.3.1) on (6.3.8), we obtain

v(1) =m−2∑

i=1

λiu(ξi) =m−2∑

i=1

λiξiu(1) +m−2∑

i=1

λi

∫ 1

0Gα(ξi, s)y(s)ds, (6.3.9)

u(1) =

m−2∑

i=1

δiv(ηi) =

m−2∑

i=1

δiηiv(1) +

m−2∑

i=1

δi

∫ 1

0Gβ(ηi, s)w(s)ds. (6.3.10)

By simple calculation using ∆ = 1−∑m−2i=1 λiξi

∑m−2i=1 δiηi, we get

u(1) =

1

∆

[t(

m−2∑

i=1

δiηi)

m−2∑

i=1

λi

∫ 1

0Gα(ξi, s)y(s)ds+ t

m−2∑

i=1

δi

∫ 1

0Gβ(ηi, s)w(s)ds

],

(6.3.11)

v(1) =

1

∆

[t(m−2∑

i=1

λiξi)m−2∑

i=1

δi

∫ 1

0Gβ(ηi, s)w(s)ds+ t

m−2∑

i=1

λi

∫ 1

0Gα(ξi, s)y(s)ds

],

(6.3.12)

putting (6.3.11) in (6.3.9) and (6.3.12) in (6.3.10) then substituting the values of u(1), v(1)in (6.3.8) respectively we get the desire results. Hence the proof.

146


Now we will study the following existence and uniqueness results for the system

cDαu(t)− U(t)u(t) + y(t) = 0, t ∈ (0, 1), 1 < α ≤ 2,

cDβv(t)− V (t)v(t) + w(t) = 0, t ∈ (0, 1), 1 < β ≤ 2,

u(0) = v(0) = 0, u(1) =m−2∑

i=1

δiv(ξi), v(1) =m−2∑

i=1

λiu(ηi),

(6.3.13)

where U(t), V (t) > 0.

Now let us define an operator T : C[0, 1] × C[0, 1] → C[0, 1] × C[0, 1] by

T (u, v) =

(−∫ 1

0K11(t, s)U(s)u(s)ds−

∫ 1

0K12(t, s)V (s)v(s)ds,

−∫ 1

0K2(t, s)V (s)v(s)ds−

∫ 1

0K22(t, s)U(s)u(s)ds

)+ (µ, ν),

where

µ =

∫ 1

0K11(t, s)y(s)ds+

∫ 1

0K12(t, s)w(s)ds,

ν =

∫ 1

0K2(t, s)w(s)ds +

∫ 1

0K22(t, s)y(s)ds.

(6.3.14)

Lemma 6.3.2. Let U(t), V (t) > 0 ,then the operator T has a unique fixed point which isthe corresponding unique solutions of (6.3.13).

Proof. As the operator defined in (6.3.14) is linear completely continuous operator by ap-plying the Fredholm theorem also the operator equations T (u, v) = (u, v) has only the zerosolution. Hence the operator equations defined in (6.3.14) ,for (µ, ν) ∈ C[0, 1]×C[0, 1] hasa unique solutions. Thus by the Fredholm theorem the operator T has a unique solution inC[0, 1]× C[0, 1]. Hence the proof.

Lemma 6.3.3. Assume that (H5) − (H8) hold. Then there exist a monotone sequences(un, vn) and (µn, νn) for ordered lower and upper solutions respectively of (6.3.1) insector Θ such that

(un, vn) → (u0, v0), (µn, νn) → (µ0, ν0) as n→ ∞,

where (u0, v0) is initial approximation of ordered lower system of solutions and (µ0, ν0)is initial approximation of ordered upper system of solutions of Coupled boundary valueproblem (6.3.1) respectively.

147

Proof. Let us define a sequence by

cDαun+1(t) + f(t, un(t), vn(t)) = U(t)[un+1(t)− un(t)], t ∈ (0, 1)cDβvn+1(t) + g(t, un(t), vn(t)) = V (t)[vn+1(t)− vn(t)], t ∈ (0, 1)

un+1(0) = 0, vn+1(0) = 0, un+1(1) =

m−2∑

i=1

δivn+1(ξi), vn+1(1) =

m−2∑

i=1

λiun+1(ηi).

(6.3.15)

Then the above system has a unique solutions (un+1, vn+1) by Lemma 2.3.4, putting n = 0in (6.3.15), we obtain

cDαu1(t) + f(t, u0(t), v0(t)) = U(t)[u1(t)− u0(t)], t ∈ (0, 1)cDβv1(t) + g(t, u0(t), v0(t)) = V (t)[v1(t)− v0(t)], t ∈ (0, 1)

u1(0) = 0, v1(0) = 0, u1(1) =

m−2∑

i=1

δiv1(ξi), v1(1) =

m−2∑

i=1

λiu1(ηi).

(6.3.16)

To show that (u0, v0) ≤ (u1, v1) ≤ (µ0, ν0) setting φ(t) = u0(t)−u1(t), ψ(t) = v0(t)− v1(t)in (7.2.47). We have

cDαφ(t) =c Dα[u0(t)− u1(t)] ≤ −U(t)φ(t), t ∈ (0, 1),cDβψ(t) =c Dβ[v0(t)− v1(t)] ≤ −V (t)ψ(t), t ∈ (0, 1),

φ(0) ≤ 0, ψ(0) ≤ 0, φ(1) ≤m−2∑

i=1

δiψ(ξi), ψ(1) ≤m−2∑

i=1

λiφ(ηi).

(6.3.17)

Hence by Lemma 2.3.4,cDαφ(t) ≤ −U(t)φ(t), t ∈ (0, 1),

cDβψ(t) ≤ −V (t)ψ(t), t ∈ (0, 1),(6.3.18)

which give φ(t) ≤ 0, ψ(t) ≤ 0.Hence u0(t) ≤ u1(t), v0(t) ≤ v1(t). Thus (u0, v0) ≤ (u1, v1), t ∈ [0, 1]. Similarly settingφ(t) = u1(t) − µ1(t), ψ(t) = v1(t) − ν(t) in (6.3.16), we get (u1, v1) ≤ (µ1, ν1), t ∈ [0, 1].Next to prove that (µ1, ν1) ≤ (µ0, ν0). Setting φ(t) = µ1(t) − µ0(t), ψ(t) = ν1(t) − ν0(t),we easily obtain that µ1(t) ≤ µ0(t), ν1(t) ≤ ν0(t) we have (µ1, ν1) ≤ (µ0, ν0). Thus aftercollecting these values, we have

(u0, v0) ≤ (u1, v1) ≤ (µ1, ν1) ≤ (µ0, ν0).

Now assume that by inductions k > 1 set φ(t) = uk − uk+1, ψ(t) = vk − vk+1 in System(6.3.15), we have

cDαφ(t) =c Dα[uk(t)− uk+1(t)] ≤ −U(t)φ(t), t ∈ (0, 1),cDβψ(t) =c Dβ[vk(t)− vk+1(t)] ≤ −V (t)ψ(t), t ∈ (0, 1),

φ(0) ≤ 0, ψ(0) ≤ 0, φ(1) ≤m−2∑

i=1

δiψ(ξi), ψ(1) ≤m−2∑

i=1

λiφ(ηi).

(6.3.19)

148

Thus by Lemma2.3.4, we have φ(t) ≤ 0, ψ(t) ≤ 0. Hence uk ≤ uk+1, vk ≤ vk+1 ⇒(uk, vk) ≤ (uk+1, vk+1), similarly we can prove that ((uk+1, vk+1) ≤ (µk+1, νk+1). Now usingthe corresponding system of Hammerstein integral equations

un+1 =

∫ 1

0K11(t, s)f(s, un, vn)ds+

∫ 1

0K12(t, s)g(s, un, vn)ds, t ∈ (0, 1),

vn+1 =

∫ 1

0K2(t, s)g(s, un, vn)ds+

∫ 1

0K22(t, s)f(s, un, vn)ds, t ∈ (0, 1),

(6.3.20)

it follows that (u, v) is solution of the system (6.3.20). Next we prove that (u0, v0), (µ0, ν0)are ordered minimal and maximal solutions respectively of the Coupled boundary valueproblem (6.3.1). Let (x, y) be another solution different from (u0, v0), (µ0, ν0) such thatfor some n ∈ Z+ we have (uk, vk) ≤ (xk, yk) ≤ (µk, νk), t ∈ [0, 1]. We set φ(t) = uk+1 −xk, ψ(t) = vk+1 − yk, then by comparison result we have φ(t) ≤ 0, ψ(t) ≤ 0, hence uk+1 ≤xk, vk+1 ≤ yk,on [0, 1]. Similarly we can show that xk ≤ µk+1, yk ≤ νk+1,∀n ∈ Z+. Thus(uk, vk) ≤ (xk, yk) ≤ (µk, νk), applying limit k → ∞ we have (u0, v0) ≤ (x0, y0) ≤ (µ0, ν0),hence maximal and minimal solutions are follow in the sector Θ.


cD32u(t) + cos(t)− 1− 8u(t) +

t2

4v2(t) = 0, t ∈ (0, 1)

cD53 v(t) + e

t2 +

u2(t)

2− v2(t) = 0, t ∈ (0, 1),

u(0) = v(0) = 0, u(1) =m−2∑

i=1

δiv(ηi), v(1) =m−2∑

i=1

λiu(ξi),

where

m−2∑

i=1

λi <1

100,

m−2∑

i=1

δi <1

50.

(6.3.21)

Since

f(t, u, v) = cos(t)− 1− 8u(t) +t2

4v2(t), g(t, u, v) = e

t2 +

u2(t)

2− v2(t),

taking (−1,−1) = (u0, v0) and (2, 2) = (µ0, ν0) be initial approximation of ordered lowerand upper solutions respectively,then

cD32u0(t) + cos(t)− 1− 8u0(t) +

t2

4v20(t) = cos t− 1 + 8 +

t2

4≥ 0, t ∈ (0, 1),

cD53 v0(t) + e

t2 + 8u0(t) +

u2(t)

2− v20(t) =

2et2 − 1

2≥ 0, t ∈ (0, 1),

u0(0) ≤ 0, v0(0) ≤ 0, u0(1) ≤m−2∑

i=1

δiv0(ηi), v0(1) ≤m−2∑

i=1

λiu0(ξi),

wherem−2∑

i=1

λi <1

100,m−2∑

i=1

δi <1

50.

149

Similarly by taking (2, 2) = (µ0, ν0),

cD32µ0(t) + cos(t)− 1− 8µ0(t) +

t2

4ν20(t) = cos t− 17 + t2 ≤ 0, t ∈ (0, 1),

cD53 ν0(t) + e

t2 + 8µ0(t) +

µ2(t)

2− ν20(t) = e

t2 − 2 ≤ 0, t ∈ (0, 1),

µ0(0) ≥ 0, ν0(0) ≥ 0, µ0(1) ≥m−2∑

i=1

δiν0(ηi), ν0(1) ≥m−2∑

i=1

λiµ0(ξi),

wherem−2∑

i=1

λi <1

100,m−2∑

i=1

δi <1

50.

Hence it follows that (−1,−1), (2, 2) are ordered lower and upper solutions respectively ofthe Coupled boundary value problem (6.3.21). Also assumptions (H7) and (H8) holds forU(t) = 8, V (t) = 1

4 .


D53u(t) + 4t3[t− u(t)]3 − 4t3v2(t) = 0, t ∈ (0, 1)

D74 v(t) + 6t3[t− v(t)]3 − 6t3u2(t) = 0, t ∈ (0, 1),

u(0) = v(0) = 0, u(1) =50∑

i=1

δiv(ηi), v(1) =50∑

i=1

λiu(ξi),

where m = 52,50∑

i=1

λi <1

10,

50∑

i=1

δi <1

20.

(6.3.22)

Since

f(t, u, v) = 4t3[t− u(t)]3 − 4t3v2(t), g(t, u, v) = 6t3[t− v(t)]3 − 6t3u2(t),

taking (−1,−1) = (u0, v0) and (1, 1) = (µ0, ν0) be initial approximation of ordered lowerand upper solutions respectively,then

D53u0(t) + 4t3[t− u0(t)]

3 − 4t3v20(t) = 4t3[(t+ 1)3 − 1] ≥ 0, t ∈ (0, 1),

D74 v0(t) + 6t3[t+ v0(t)]

3 − 6t3u20(t) = 6t3[(t+ 1)3 − 1] ≥ 0, t ∈ (0, 1),

u0(0) ≤ 0, v0(0) ≤ 0, u0(1) ≤50∑

i=1

δiv0(ηi), v0(1) ≤50∑

i=1

λiu0(ξi),

where

m−2∑

i=1

λi <1

10,

m−2∑

i=1

δi <1

20.

150

Figure 6.1: Plot of Upper and Lower solutions of Example (6.3.2).

Similarly by taking (1, 1) = (µ0, ν0),

D53µ0(t) + 4t3[t− µ0(t)]

3 − 4t3ν20(t) = 4t3[(t− 1)3 − 1] ≤ 0, t ∈ (0, 1),

D74 ν0(t) + 6t3[t− ν0(t)]

3 − 6t3µ20(t) = 6t3[(t− 1)3 − 1] ≤ 0, t ∈ (0, 1),

µ0(0) ≥ 0, ν0(0) ≥ 0, µ0(1) ≥50∑

i=1

δiν0(ηi), ν0(1) ≥50∑

i=1

λiµ0(ξi),

where

m−2∑

i=1

λi <1

10,

m−2∑

i=1

δi <1

20.

Hence it follows that (−1,−1), (1, 1) are ordered lower and upper solutions respectively ofthe Coupled boundary value problem (6.3.22). Also assumptions (H7) and (H8) holds forU(t) = 4t3, V (t) = 6t3.

Chapter 7

Numerical solutions of systems ofboundary value problem offractional differential equations

In the last few decades, fractional order differential equations has gain much attentions

from the researchers of mathematics , physics, computer science and engineers. This is

due to the large numbers of applications of fractional order differential equations in var-

ious fields of science and applied nature. The applications of fractional order differential

equations are found in physics, mechanics, viscoelasticity, photography, biology, chemistry,

fluid mechanics, image and signal processing phenomenons etc. The fractional order op-

erator both for integral and derivative is a global operators and provide greater degree of

freedom in the model as compare to classical order which is local operator and does not

allow greater degree of freedom for modeling. To find exact solutions(FDEs), some time

it is very difficult to solve these differential equations due to the complexities of fractional

calculus involving in these equations. However the solution of these equations can be easily

approximated with a large variety of methods. For example in [130, 131, 132], the authors

developed Adomain decomposition methods, Homotopy analysis methods and Variational

iterational methods to obtained good approximate solutions for certain class of fractional

order differential equations.

151

152

In recent years spectral methods have been given much attention as they were applied to

solve some real word problems of various fields of science and engineering. High accuracy

were obtained for solving such problems [133]. Spectral method need an operational ma-

trices for the numerical solutions, which have been constructed by using some polynomials,

for example in [134], the authors developed an operational matrix for Shifted Legendre

polynomials corresponding to fractional order derivative. The authors developed opera-

tional matrices for coupled system of initial value problem of fractional order derivative

and integration. These operational matrices were obtained by using the Shifted Legendre

polynomials. We review some already existing operational matrices and also develop a new

operational matrices corresponding to boundary condition of the system of differential equa-

tions. The obtained matrix plays important roles in the approximate solutions of fractional

order differential equations. In [135], the authors introduced operational matrix by using

Shifted Legendre polynomials corresponding to fractional order integrations. Similarly in

[136, 137], authors constructed operational matrices for fractional order derivative by using

Chebyshev and Jacobi Polynomials. In all these cases these matrices were applied to solve

multi-terms fractional order differential equations together with Tau-collocation method. In

[138], Singh et al. derived an operational matrices for Berenstein polynomials correspond-

ing to fractional order integrations. Haar wavelet operational matrices were also developed

and some problems of fractional order differential equations were solved. Moreover the

operational matrices were developed and applied together with the method of collocations

to obtained numerical solutions, (for detail see [139, 140, 141, 142]). In this chapter, we

develop an efficient numerical scheme for system of boundary value problems of fractional

order differential equations. The aforesaid scheme is developed by using polynomials such

as Shifted Legendre polynomials and Jacobi Polynomials and Berenstein polynomials.

153

Moreover, in this chapter, we also established a numerical scheme by using two dimen-

sional Legendre polynomials for a general coupled system of fractional order partial dif-

ferential equations. We developed some operational matrices, based on these matrices,

without collocations method, we developed necessary and sufficient conditions. For detail

see [143, 144, 145, 146].

We organize this chapter as: In section 7.1, we develop operational matrices for Legendre

polynomials and solve coupled system of boundary value problem with the help of these

matrices. Section 7.2 is devoted to the numerical solutions of coupled system by Berenstein

polynomials. In section 7.3, we approximate system of boundary value problems by means

of Shifted Jacobi polynomials. In last section 7.4, we develop operational matrices for the

numerical solutions of coupled system of partial differential equations by means of Shifted

Legendre polynomials. In last section 7.4, we give comparison between the aforementioned

polynomials for numerical solutions.

7.1 Numerical solutions of system of fractional order differ-ential equations by Legendre polynomials

Any function f(t) ∈ C[0, 1] can be approximated by Shifted Legendre polynomials as

f(t) ≈m∑

j=0

cjPj(t), where cj = (2j + 1)

∫ 1

0f(t)Pj(t)dt. (7.1.1)

In vector notation (7.1.1) can be written as

f(t) = KTM PM (t), (7.1.2)

where M = m+1, KTM is the coefficient vector and PM (t) is M terms function vector. For

further procedure we give the following lemmas.

Lemma 7.1.1. Let PM (t) be the function vector as defined in (7.1.2) then the integra-tions of order σ of PM (t)(t) is defined by Iσ(φ(t)) ≈ Pσ

M×M PM (t), where PσM×M is the

154

operational matrix of integration of order σ and is given by

PσM×M =

∑0k=0Ω0,0,k

∑0k=0Ω0,1,k · · · ∑0

k=0Ω0,j,k · · · ∑0k=0Ω0,m,j∑1

k=0Ω1,0,k

∑1k=0Ω1,1,k · · · ∑1

k=0Ω1,j,k · · · ∑1k=0Ω1,m,j

......

......

......∑i

k=0Ωi,0,k∑i

k=0Ωi,1,k · · · ∑ik=0Ωi,j,k · · · ∑i

k=0Ωi,m,k...

......

......

...∑mk=0Ωm,0,k

∑mk=0Ωm,1,k · · · ∑m

k=0Ωm,j,k · · · ∑mk=0Ωm,m,k

,

(7.1.3)where

Ωi,j,k = (2j + 1)

j∑

l=0

(−1)i+j+k+lΓ(i+ k + 1)Γ(l + j + 1)

Γ(i− k + 1)Γ(k + 1)Γ(k + σ + 1)Γ(j − l + 1)Γ2(l + 1)(k + l + σ + 1).

(7.1.4)

Proof. In view of (2.5.3) and linearity of fractional integral operator, we have

IσPi(t) =i∑

k=0

(−1)i+kΓ(i+ k + 1)

Γ(i− k + 1)(Γ(k + 1))2Iσtk

=

i∑

k=0

(−1)i+kΓ(i+ k + 1)

Γ(i− k + 1)(Γ(k + 1))Γ(k + σ + 1)tk+σ, i = 1, 2, · · · ,m.

(7.1.5)

Now approximating tk+σ in terms of Shifted Legendre polynomials as

tk+σ 'm∑

j=0

Ck,jPj(t), (7.1.6)

where, by the use of (2.5.4), from (7.1.6), we have

Cj,k = (2j + 1)

∫ 1

0tk+σPj(t)dt

= (2j + 1)

j∑

l=0

(−1)j+lΓ(j + l + 1)

Γ(j − l + 1)(Γ(l + 1))2

∫ 1

0tk+σ+ldt

= (2j + 1)

j∑

l=0

(−1)j+lΓ(j + l + 1)

Γ(j − l + 1)(Γ(l + 1))2(k + l + σ + 1).

(7.1.7)

Using (7.1.7) in (7.1.5), we obtain

IσPi(t) 'i∑

k=0

m∑

j=0

Ωi,j,k, (7.1.8)

155

where

Ωi,j,k = (2j+1)

j∑

l=0

(−1)i+j+k+lΓ(k + i+ 1)Γ(l + j + 1)

Γ(k + σ + 1)Γ(k + 1)Γ(i− k + 1)Γ(j − l + 1)(Γ(l + 1))2(k + l + σ + 1).

Evaluating the right hand sides of (7.1.8), we get the required operational matrix PσM×M .

To evaluate (7.1.3), let us take M = 6, σ = 75 , we have

PσM×M =

0.3354 0.4144 0.0628 −0.0098 0.0031 −0.0013−0.1381 −0.0915 0.0767 0.0220 −0.0044 0.00160.0126 −0.0460 −0.0381 0.0377 0.0121 −0.00260.0014 0.0094 −0.0270 −0.0229 0.0237 0.00800.0003 0.0015 0.0067 −0.0185 −0.0158 0.01680.0001 0.0004 0.0012 0.0051 −0.0137 −0.0119

. (7.1.9)

Lemma 7.1.2. Let PM (t) be the function vector as defined in (7.1.2) then the differentia-tion of order σ of PM (t) w.r.t t is defined by DσPM (t) ≈ Hσ

M×MPM (t), where HσM×M is

the operational matrix of differentiation of order σ and is given by

HσM×M =

∑1k=0Φ1,1,k

∑1k=0Φ1,2,k · · · ∑1

k=0Φ1,j,k · · · ∑1k=0Φ1,m,j∑1

k=0Φ2,1,k

∑2k=0Φ2,2,k · · · ∑2

k=0Φ2,j,k · · · ∑2k=0Φ2,m,j

......

......

......∑i

k=0Φi,1,k∑i

k=0Φi,2,k · · · ∑ik=0Φi,j,k · · · ∑i

k=0Φi,m,k...

......

......

...∑mk=0Φm,1,k

∑mk=0Φm,2,k · · · ∑m

k=0Φm,j,k · · · ∑mk=0Φm,m,k

,

(7.1.10)where

Φi,j,k = (2j + 1)

j∑

k=[σ]

(−1)i+j+k+lΓ(i+ k + 1)Γ(l + j + 1)

Γ(i− k + 1)Γ(k + 1)Γ(k − σ + 1)Γ(j − l + 1)Γ2(l + 1)(k + l − σ + 1),

j = 0, 1, 2...m, where Φi,j,k = 0, if j < σ.

(7.1.11)

Proof. The proof of this lemma is similar as of Lemma 7.1.1.

In example point of view to calculate operational matrices (7.1.10) of fractional deriva-

tive as taking σ = 75 ,M = 6, then the operational matrix of fractional derivative is given

by

156

HσM×M =

0 0 0 0 0 0

1.5045 0.9027 −0.2149 0.1003 −0.0586 0.0386

−0.9027 1.9344 1.5045 −0.4103 0.2083 −0.1290

1.2896 −0.6018 2.2274 1.9906 −0.5862 0.3132

−1.0030 1.5240 −0.3858 2.4619 2.4099 −0.7466

1.2310 −0.8101 1.6999 −0.2110 2.6625 2.7844

. (7.1.12)

Lemma 7.1.3. let φ(t) be any function defined in [0, 1], and u(t) = KTM PM (t) then

φ(t)0Iσ1u(t) = KT

MQσ,φM×M PM (t), (7.1.13)

where Qσ,φM×M is the operational matrix and is defined as

Qσ,φM×M =

Φ0,0 Φ0,1 · · · Φ0,j · · · Φ0,m

Φ1,0 Φ1,1 · · · Φ1,j · · · Φ1,m...

......

......

...Φi,0 Φi,1 · · · Φi,j · · · Φi,m...

......

......

...Φm,0 Φm,1 · · · Φm,j · · · Φm,m

, (7.1.14)

where

Φi,j = (2j + 1)4i

∫ 1

0φ(t)Pj(t)dt

and

4i =i∑

l=0

(−1)i+lΓ(i+ l + 1)

Γ(i− l + 1)Γ(l + 1)Γ(σ + l)

Proof. Consider the general term of PM (t),

0Iσ1Pi(t) =

1

Γ(σ)

∫ 1

0(1− s)σ−1Pi(s)ds, (7.1.15)

Using the definition of Shifted Legendre polynomials we get

0Iσ1Pi(t) =

i∑

l=0

(−1)i+lΓ(i+ l + 1)

Γ(i− l + 1)Γ2(l + 1)

1

Γ(σ)

∫ 1

0(1− s)σ−1slds. (7.1.16)

To evaluate the integral we use the well known Laplace transform, that is

£

(∫ 1

0(1− s)σ−1slds

)=

Γ(σ)Γ(l + 1)

s(σ+l+1). (7.1.17)

157

Using the inverse laplace transform we get∫ 1

0(1− s)σ−1slds =

Γ(σ)Γ(l + 1)

Γ(σ + l). (7.1.18)

In view of (7.1.18) and (7.1.16) we may write

0Iσ1Pi(t) =

i∑

l=0

(−1)i+lΓ(i+ l + 1)

Γ(i− l + 1)Γ(l + 1)Γ(σ + l)= 4i. (7.1.19)

Using (7.1.19) we may writeφ(t) 0I

σ1Pi(t) = 4iφ(t). (7.1.20)

Which can further be written in terms of Shifted Legendre polynomials as

4iφ(t) =

m∑

j=0

Φi,jPj(t), (7.1.21)

where Φi,j can be calculated using relation

Φi,j = (2j + 1)

∫ 1

04iφ(t)Pj(t)dt (7.1.22)

Now evaluating the above relation for i = 0, 1, · · ·m and j = 0, 1 · · ·m we can get the desiredresult. For example let M = 6, σ = 7

5 , φ(t) =∑10

k=0tk

Γ(k+1) , then

Qσ,φM×M =

0.9808 0.2779 0.1126 0.0112 0.0008 0.0000−0.1635 −0.0463 −0.0188 −0.0019 −0.0001 −0.0000−0.0288 −0.0082 −0.0033 −0.0003 −0.0000 −0.0000−0.0105 −0.0030 −0.0012 −0.0001 −0.0000 −0.0000−0.0051 −0.0014 −0.0006 −0.0001 −0.0000 −0.0000−0.0028 −0.0008 −0.0003 −0.0000 −0.0000 −0.0000

(7.1.23)

7.1.1 Numerical solutions of boundary value problems

The operational matrices developed in the previous section play a very important role in the

approximate solution of boundary value problem. As a first application we solve fractional

differential equations with boundary conditions. Consider the following generalized class of

fractional differential equations

cDν2u(t) = λ1Dν1u(t) + λ2u(t) + f(t), 0 ≤ t ≤ 1,

u(0) = u0, u(1) = u1, u0, u1 ∈ R,

(7.1.24)

158

We seek the solution of this problem such that the following holds

cDν2u(t) = KTM PM (t). (7.1.25)

Then by the application of fractional integration of order ν2 we get

u(t) = KTMPν2

M×M PM (t) + C1 + C2t. (7.1.26)

Using the initial condition we can easily get C1 = u0, to calculate the value of C2 we use

the boundary condition. So we get

C2 = u1 − u0 −KTMP

ν2M×M PM (1). (7.1.27)

Using the values of C1 and C2 in (7.1.26) we may write

u(t) = KTMPν2

M×M PM (t) + u0 + (u1 − u0)t− tKTMPν2

M×M PM (1). (7.1.28)

Making use of Lemma 3.0.3 and after simplification we may write

u(t) = KTM

(Pν2M×M −Q

ν2,φM×M

)PM (t) + FTM PM (t), (7.1.29)

where FTM PM (t) = u0 + (u1 − u0)t, φ = t. Now by the application of Lemma 7.1.1 we may

write

cDν1u(t) = KTM

(Pν2M×M −Q

ν2,φM×M

)Hν1M×M PM (t) + FTMHν1

M×MPM (t), (7.1.30)

Using (7.1.30),(7.1.29) and (7.1.25) in (7.1.24) we may write

KTM PM (t) = λ1K

TM

(Pν2M×M −Q

ν2,φM×M

)Hν1M×M PM (t) + λ2K

TM

(Pν2M×M −Q

ν2,φM×M

)PM (t)+

λ1FTMH

ν1M×M PM (t) + λ2F

TM PM (t) + GT

M PM (t).

(7.1.31)

159

Where GTM PM (t) = f(t). After simplification and rearranging we may write

(KTM −KT

M

(Pν2M×M −Q

ν2,φM×M

) (λ1H

ν1M×M + λ2IM×M

)−CM×M

)PM (t) = 0 (7.1.32)

On canceling out the common term we may write

KTM −KT

M

(Pν2M×M −Q

ν2,φM×M

) (λ1H

ν1M×M + λ2IM×M

)−CM×M = 0 (7.1.33)

It is to be noted that (7.1.33) is a generalized algebraic type matrix equation. The solution

of this matrix equation will lead us to the approximate solution of the problem.

7.1.2 Numerical solutions of system of boundary value problems

Consider the following generalized class of coupled system of fractional differential equations

cDαu = λc1Dν1u+ λc2D

ν2v + λ3u+ λ4v + f(t), 0 ≤ t ≤ 1

cDβv = τ c1Dω1u+ τ c2D

ω2v + τ3u+ τ4v + g(t), 0 ≤ t ≤ 1.

(7.1.34)

Subject to the boundary conditions

u(0) = u0, u(1) = u1, v(0) = v0, v(1) = v1.

As in the previous part assume the solution of the coupled System in Shifted Legendre

polynomials such that the following hold.

cDαu(t) = KTM PM (t), cDβv(t) = LTM PM (t). (7.1.35)

Then by the application of fractional integration of order α and β on the corresponding

equations we get

u(t) = KTMPα

M×M PM (t)+C1+C2t, v(t) = LTMPβM×M PM (t)+ C1+ C2t. (7.1.36)

160

Using the initial condition we can easily get C1 = u0 and C1 = v0, to calculate the value of

C2 and C2 we use the boundary condition. So we get

C2 = u1 − u0 −KTMPα

M×M PM (1), C2 = v1 − v0 − LTMPβM×M PM (1) (7.1.37)

Using the values of C1, C2, C1 and C2 in (7.1.36) we may write

u(t) = KTMPα

M×M PM (t) + u0 + (u1 − u0)t− tKTMPα

M×M PM (1),

V (t) = LTMPβM×M PM (t) + v0 + (v1 − v0)t− tLTMP

βM×M PM (1).

(7.1.38)

Making use of Lemma 7.1.2 and after simplification we may write

u(t) = KTM

(PαM×M −Q

α,φM×M

)PM (t) + FTM PM (t),

v(t) = LTM

(PβM×M −Q

β,φM×M

)PM (t) + FTM PM (t).

(7.1.39)

where FTM PM (t) = u0+(u1−u0)t, FTM PM (t) = v0+(v1−v0)t, φ = t. Now by the application

of Lemma 7.1.1 we may write

cDν1u(t) = KTM

(PαM×M −Q

α,φM×M

)Hν1M×MPM (t) + FTMHν1

M×M PM (t),

cDν2v(t) = LTM

(PβM×M −Q

β,φM×M

)Hν2M×MPM (t) + FTMH

ν2M×M PM (t),

cDω1u(t) = KTM

(PαM×M −Q

α,φM×M

)Hω1M×MPM (t) + FTMHω1

M×M PM (t),

cDω2v(t) = LTM

(PβM×M −Q

β,φM×M

)Hω2M×MPM (t) + FTMHω2

M×M PM (t).

(7.1.40)

Using (7.1.40),(7.1.39) and (7.1.35) in (7.1.34), we may write

KT

M PM (t)

LTM PM (t)

=

λ1K

TMDα,ν1PM (t)

τ2LTMDβ,ω2PM (t)

+

λ2L

TMDα,ν2PM (t)

τ1KTMDβ,ω1PM (t)

+

λ3K

TMDαPM (t)

τ4LTMDβPM (t)

+

λ4L

TMDαPM (t)

τ3KTMDβPM (t)

+

︷︸︸︷F1 PM (t)︷︸︸︷F2 PM (t)

(7.1.41)

Where Dα,νi =(PαM×M −Q

α,φM×M

)HνiM×M ,

︷︸︸︷F1 = λ1F

TMHν1

M×M+λ2FTMHν2

M×M+λ3FTM+

λ4FTM+W1,

︷︸︸︷F2 = τ1F

TMH

ω1M×M+τ2F

TMH

ω2M×M+τ3F

TM+τ4F

TM+W2.Where W1PM (t) =

161

f(t) and W2PM (t) = g(t) Taking the transpose of the above matrix equation and rear-

ranging the above equation we may write

[KTM LTM

] PM (t) OM

OM PM (t)

=

[KTM LTM

] λ1D

α,ν1 OM×M

OM×M τ2Dβ,ω2

PM (t) OM

OM PM (t)

+[KTM LTM

] OM×M τ1D

β,ω1

λ2Dα,ν2 OM×M

PM (t) OM

OM PM (t)

+[KTM LTM

] λ3D

α OM×M

OM×M τ4Dβ

PM (t) OM

OM PM (t)

[KTM LTM

] OM×M τ3D

β

λ4Dα OM×M

PM (t) OM

OM PM (t)

+

[ ︷︸︸︷F1

︷︸︸︷F2

] PM (t) OM

OM PM (t)

(7.1.42)

Where OM×M is M order zero matrix, OM is zero vector of order M . After simplification

and canceling out the common terms we can write the above equations as

[KTM LTM

]−[KTM LTM

] λ1D

α,ν1 + λ3Dα τ1D

β,ω1 + τ3Dβ

λ2Dα,ν2 + λ4D

α τ2Dβ,ω2 + τ4D

β

−

[ ︷︸︸︷F1

︷︸︸︷F2

]= 0

(7.1.43)

The above equation is generalized matrix equation, which can be easily solved for the

unknown matrix KTM ,L

TM .We can get the approximate solution using (7.1.39)

7.1.3 Illustrative examples

In this section we solve some problems to show the efficiency of the proposed method.

Example 7.1.1. As a first example consider the following fractional order boundary valueproblem.

cDσu(t) = cc1Dβ1u(t) + c2u(t) + f(t), (7.1.44)

where the order of derivative 1 < σ ≤ 2, 1 < β1, c1 and c2 are generalized constants.We approximate solution of the above problem under the following sets of parameters.

S1 = σ = 1.8, β1 = 0.8, c1 = 4, c2 = 8,

162

0 0.2 0.4 0.6 0.8 1−9

−8

−7

−6

−5

−4

−3

−2

−1

u(t

)

0 0.2 0.4 0.6 0.8 10

1

2

3

4

5

6x 10

−3

|u−uapp|

Absolute error atM = 6



Approximate u(t)atM = 6

Exact u(t)

(a) (b)

Figure 7.1: (a) Comparison of exact and approximate solution with parameter set S1. (b)Absolute error of Example(7.1.1) at different value of M .

0 0.2 0.4 0.6 0.8 1

−8

−6

−4

−2

(b)

u(t

)

0.2 0.4 0.6 0.8 1

10−6

10−4

(a)

|u(t

)−uapp|

Approximate u(t)

Exact u(t)

M = 8

M = 10M = 6

Figure 7.2: (a) Comparison of exact and approximate solution with parameter set S2. (b)Absolute error of Example(7.1.1) at different value of M .

S2 = σ = 1.4, β1 = 0.5, c1 = 4.9, c2 = 2.7,By taking the source term for S1 and S2 as

f1(t) = 8 t (t− 3)3 − 8 (t− 1)4 − 39239842694707435 t15 (25 t− 42)

36028797018963968

− 3567258426791585 t15

(125 t2 − 385 t+ 253

)

9007199254740992

and

f2(t) =27 t (t− 3)3

10− 27 (t− 1)4

10− 15120947718337785 t

35 (25 t− 56)

18014398509481984

− 249006631826735069√t(8 t2 − 28 t + 23

)

45035996273704960

respectively, the exact and unique solution of the problem is u(t) = t(t−3)3+(1− t)4 underthe following conditions.

u(0) = 1 u(1) = −8 (7.1.45)

163

We solve this example with the new technique developed in this thesis and observe that thesolutions obtained with the new method matches very well with the exact solutions. We solvethe problem with the parameter set S1, the results are displayed in Figure (7.1). In Figure(7.1), (a) the exact solution is compared with the approximate solution at scale level M = 6,and in (b) the absolute error between exact solution and approximate solution is displayed.

Example 7.1.2. Consider the following coupled system of fractional order differential equa-tions.

cD1.8u(t) =c D0.6u(t) +c D0.5v(t) + 3u(t)− 2v(t) + f(t),cD1.6v(t) = 4cD0.6u(t) + 3cD0.5v(t) + 2u(t) + v(t) + g(t),

(7.1.46)

where

f =2229536516744740625 t

165 (10 t − 7)

1008806316530991104− 5081767996463981 t

72 (10 t− 9)

22166154415964160

− 1982745796388234375 t225 (10 t− 9)

6632113401256476672− 2 t4 + 6 t5 − 4 t6 − 6

g =1015165847750775625 t

125 (25 t− 17)

2143713422628356096− 1982745796388234375 t

225 (10 t− 9)

1658028350314119168

− 5081767996463981 t72 (10 t− 9)

7388718138654720+ t4 + t5 − 2 t6 − 5.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.94

1.95

1.96

1.97

1.98

1.99

2

2.01

(a)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.92

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

1.01

(b)

vApprox at,M = 5

vApprox at,M = 7

Exact v(x)

uApprox at,M = 5

uApprox at,M = 7

Exact U(x)

Figure 7.3: (a) Comparison of exact u(t) of Example (7.1.2) with the approximate solutionat different scale level. (b)Comparison of exact v(t) of Example (7.1.2) with the approximatesolution at different scale level.

The exact solution of the problems is

u(t) = t6 − t5 + 2 v(t) = t5 − t4 + 1.

We approximate the solution of this problem with our proposed method. And observe thatthe solution matches very well with the exact solution.

164

0 0.2 0.4 0.6 0.8

10−8

10−6

10−4

10−2

(a)

0.2 0.4 0.6 0.8 1

10−7

10−6

10−4

10−2

10−2

(b)

|u(t)− ︷︸︸︷u | atM = 5

|u(x)− ︷︸︸︷u | atM = 6

|u(t)− ︷︸︸︷u | atM = 7

|v(t)− ︷︸︸︷v | atM = 5

|v(t)− ︷︸︸︷v | atM = 6

|v(t)− ︷︸︸︷v | atM = 7

Figure 7.4: (a) Absolute error in u(t) of Example (7.1.2) at different scale level. (b)Absoluteerror in v(t) of Example (7.1.2) at different scale level.

In Figure (7.3), the comparison of the exact solution with an approximate solution is made.One can see that the exact solution matches very well with the exact solution. The absolutedifference between the exact and approximate solution is displayed in Figure (7.4). One caneasily note that the absolute difference falls below 10−3.

Example 7.1.3. Consider the following fractional order coupled system of differential equa-tions

cD1.7u(t) =c Du(t) + 9cD0.7v(t) + 2u(t) − v(t) + f(t),cD1.8v(t) = −6cDu(t) +c D0.6v(t)− u(t) + 0v(t) + g(t),

(7.1.47)

where

f(t) = t−1.7E1,−0.7(2t)− 2e2t − 9t−0.7E1,0.3(πt)− 2e2t + eπt

= t−1.7E1,−0.7(2t)− 4e2t − 9t−0.7E1,0.3(πt) + eπt

g(t) = t−1.8E1,−0.8(πt) + 12e2t − t−0.6E1,0.4(πt) + e2t

= t−1.8E1,−0.8(πt) + 13e2t − t−0.6E1,0.4(πt).

The exact solution of the problems is

u(t) = e2t v(t) = eπt.

We solve this problem with the proposed method and as expected we get a high accuracy ofapproximate solution. The results are displayed in Figure(7.5) and Figure(7.6).

165

0 0.2 0.4 0.6 0.8 1

1

2

3

4

5

6

7

8

(a)

0 0.2 0.4 0.6 0.8 10

5

10

15

20

(b)

uapp at M=6

uexact

vapp at M=6

vexact

Figure 7.5: (a) Comparison of exact u(t) of Example (7.1.3) with the approximate solutionat level M=6. (b)Comparison of exact v(t) of Example (7.1.3) with the approximate solutionat scale level M=6.

0 0.2 0.4 0.6 0.8 110−10

10−9

10−8

10−7

10−6

10−5

10−4

10−3

10−2

t

0.20.40.60.81

10−8

10−7

10−6

10−5

10−4

10−3

10−2

t

|u− uapp| at M = 6




|v− vapp| at M = 6




Figure 7.6: (a) Absolute error in u(t) of Example (7.1.3) at different scale level. (b)Absoluteerror in v(t) of Example (7.1.3) at different scale level.

166

7.1.4 Observations

We solve many problem with the proposed method. We observe that the method works wellfor all problems having smooth solution. The basic property of the spectral methods of suchtype is that it converges to the exact solution as we increase the scale level. So we solve theabove three problems with a wide range of scale level, i.e. M = 5 : 25. And measure thequantity ‖U‖e at each scale level. Where ‖U‖e is defined by the relation

‖U‖e = maxt∈[0,1]

∫ 1

0|(u(t)− uM (t)|dt,

where uM (t) is the approximate solution at scale level M.

(7.1.48)

The results obtained are displayed in Table 5.1. One can easily note that the solution of theproblem converges to the exact solution.

Example 1 Example 2 Example 3

M ‖Y ‖e ‖U‖e ‖V ‖e ‖U‖e ‖V ‖e5 1.9001(−3) 3.9111(−3) 1.7123(−3) 1.3113(−2) 2.0000(−3)

6 1.1001(−3) 1.9004(−3) 1.2009(−3) 6.7112(−3) 1.5300(−4)

7 7.5001(−4) 6.5001(−4) 9.7001(−4) 6.7001(−3) 7.0001(−4)

8 2.9002(−4) 1.0002(−4) 2.8001(−4) 7.3001(−3) 9.2712(−5)

9 2.0500(−4) 1.9001(−4) 1.1011(−4) 6.1710(−5) 6.2323(−5)

10 5.1003(−5) 6.5001(−5) 1.7001(−5) 3.6700(−6) 8.2889(−6)

15 1.0001(−7) 4.9101(−7) 9.5001(−8) 1.7123(−8) 6.3001(−7)

20 2.6004(−9) 1.0501(−9) 5.1123(−9) 1.7001(−10) 2.0701(−10)

25 7.3983(−12) 8.9983(−12) 1.3113(−13) 1.7101(−12) 8.9701(−13)

Table 7.1: Norm of errors of illustrative examples at different value of M .

7.2 Numerical solutions of boundary value problems by Bern-stein polynomials

In this section, we develop operational matrices for boundary value problems by using

Bernstein polynomials. Thus any function can be approximated in the form of Bernstein

polynomials defined in (2.5.13) as

f(t) =m∑

i=0

CiB(i,m), (7.2.1)

167

where coefficient Ci can easily be calculated by multiplying (7.2.1) by B(j,m)(t), j =

0, 1, 2, . . . ,m and integrating from 0 to 1 by using inner product and

di =

∫ 1

0B(i,m)(t)f(t)dt, θ(i,j) =

∫ 1

0B(i,m)(t)B(j,m)(t)dt, i, j = 0, 1, 2 . . . ,m.

We have

∫ 1

0f(t)B(j,m)(t)dt =

∫ 1

0

m∑

i=0

ciB(i,m)(t).B(j,m)(t)dt, j = 0, 1, 2, . . . ,m.

⇒∫ 1

0f(t)B(j,m)(t)dt =

m∑

i=0

ci

∫ 1

0B(i,m)(t).B(j,m)(t)dt, j = 0, 1, 2...m.

⇒ [d0 d1 . . . dm] = [C0 C1....Cm]

θ(0,0) θ(0,1) · · · θ(0,r) · · · θ(0,m)

θ(1,0) θ(1,1) · · · θ(1,r) · · · θ(1,m)

......

......

......

θ(r,0) θ(r,1) · · · θ(r,r) · · · θ(r,m)

......

......

......

θ(m,0) θ(m,1) · · · θ(m,r) · · · θ(m,m)

.

(7.2.2)

Let XM = [d0 d1 . . . dm], CM = [C0 C1, . . . , Cm], where M = m+1 whereM is scale level

and

ΦM×M =

θ(0,0) θ(0,1) · · · θ(0,r) · · · θ(0,m)

θ(1,0) θ(1,1) · · · θ(1,r) · · · θ(1,m)

......

......

......

θ(r,0) θ(r,1) · · · θ(r,r) · · · θ(r,m)

......

......

......

θ(m,0) θ(m,1) · · · θ(m,r) · · · θ(m,m)

, (7.2.3)

so

XM = CMΦM×M ⇒ CM = XMΦ−1M×M . (7.2.4)

Where ΦM×M is called dual matrix of the Bernstein polynomials. After calculating Ci

(7.2.1) can be written as

f(t) = CMBTM (t), (7.2.5)

168

C is the coefficient matrix. where

BTM (t) = [B(0,m), B(1,m), . . . , B(m,m)]

T . (7.2.6)

Lemma 7.2.1. Let BTM (t) be the function vector defined in (2.5.15), then the fractional

order integration of BTM (t) is given by

IαBTM (t) = PαM×MBT

M (t), (7.2.7)

where PM×M is fractional integration’s operational matrix defined as

PM×M = PαM×MΦ−1

M×M

and Φ−1M×M is given in (7.2.3) and Pα

M×M is given by

PαM×M =

Ψ(0,0) Ψ(0,1) · · · Ψ(0,r) · · · Ψ(0,m)

Ψ(1,0) Ψ(1,1) · · · Ψ(1,r) · · · Ψ(1,m)...

......

......

...Ψ(r,0) Ψ(r,1) · · · Ψ(r,r) · · · Ψ(r,m)

......

......

......

Ψ(m,0) Ψ(m,1) · · · Ψ(m,r) · · · Ψ(m,m)

, (7.2.8)

where

Ψi,j =

m−i∑

k=0

m−j∑

l=0

θ(i,k,m)θ(j,l,m)Γ(k + i+ 1)

(i+ j + k + l + α+ 1)Γ(k + i+ α+ 1). (7.2.9)

Proof. Consider

Bi,m(t) =

m−i∑

k=0

θ(i,k,m)tk+i (7.2.10)

taking fractional integration of both sides we have

IαBi,m(t) =

m−i∑

k=0

θ(i,k,m)Iαtk+i =

m−i∑

k=0

θ(i,k,m)Γ(k + i+ α)

Γ(k + i+ α+ 1)tk+i+α. (7.2.11)

Now to approximate right sides of above

m−i∑

k=0


Γ(k + i+ α+ 1)tk+i+α = C

(i)MBT

M (t) (7.2.12)

where C(i)M can be approximated by using (7.2.3) as

C(i)M = X

(i)MΦ−1

M×M , (7.2.13)

169

where entries of the vector X(i)M can be calculated in generalized form as

X(j)M =

∫ 1

0

m−i∑

k=0


Γ(k + i+ α+ 1)tk+i+αBj,m(t)dt, j = 0, 1, 2....m

⇒ X(j)M =

∫ 1

0

m−i∑

k=0


Γ(k + i+ α+ 1)

m−j∑

l=0

θ(j,l,m)tk+i+α.tl+jdt, j = 0, 1, 2, . . . ,m

=m−i∑

k=0

θ(i,k,m)

m−j∑

l=0

θ(j,l,m)Γ(k + i+ α)

Γ(k + i+ α+ 1)

1

(k + l + j + i+ α+ 1), j = 0, 1, 2, ....m

(7.2.14)

evaluating this result for i=0,1,2,. . . ,m we have

IαB0,m(t)IαB1,m(t)

...

...

...IαBm,m(t)

=

X(0)M Φ−1

M×MBTM (t)

X(1)M Φ−1

M×MBTM (t)

...

...

...

X(m)M Φ−1

M×MBTM (t))

(7.2.15)

further witting

Ψi,j =m−i∑

k=0

m−j∑

l=0

θ(i,k,m).θ(j,l,m)Γ(k + i+ α)

Γ(k + i+ α+ 1)

1

(k + l + j + i+ α+ 1)

we getIαBT

M (t) = PαM×M .Φ

−1M×M .B

TM (t). (7.2.16)

Let representPαM×M .Φ

−1M×M = PM×M

thusIαBT

M (t) = PM×M .BTM (t). (7.2.17)

Lemma 7.2.2. Let BTM (t) be the function vector as defined in (2.5.15) then fractional order

derivative is defined ascDαBT

M (t) = HM×M .BTM (t) (7.2.18)

where HM×M is the operational matrix of fractional order derivative given by

HM×M = GαM×MΦ−1

M×M , (7.2.19)

170

whereΦM×M is the dual matrix given in (7.2.3) and

GαM×M =

Ψ(0,0) Ψ(0,1) · · · Ψ(0,r) · · · Ψ(0,m)

Ψ(1,0) Ψ(1,1) · · · Ψ(1,r) · · · Ψ(1,m)...

......

......

...Ψ(r,0) Ψ(r,1) · · · Ψ(r,r) · · · Ψ(r,m)

......

......

......

Ψ(m,0) Ψ(m,1) · · · Ψ(m,r) · · · Ψ(m,m)

, (7.2.20)

where Ψ(i,j) is defined for two different cases asCaseI:(i < [α])

Ψi,j =

m−i∑

k=[α]

m−j∑

l=0

θ(i,k,m).θ(j,l,m)Γ(k + i− α)

Γ(k + i− α+ 1)

1

(k + l + j + i− α+ 1)(7.2.21)

. CaseII:(i ≥ [α])

Ψi,j =

m−i∑

k=0

m−j∑

l=0

θ(i,k,m).θ(j,l,m)Γ(k + i− α)

Γ(k + i− α+ 1)

1

(k + l + j + i− α+ 1). (7.2.22)

Proof. Consider the general element as

cDαBi,m(t) =c Dα(

m−i∑

k=0

θ(i,k,m).tk+i) =

m−i∑

k=0

θc(i,k,m)Dαtk+i. (7.2.23)

It is to be noted in the polynomial function Bi,mthe power of the variable′t′ is an ascending

order and the lowest power is ′i′ therefore the first [α−1] terms becomes zero when we takederivative of order α.CaseI:(i < [α])By the use of definition of fractional derivative

cDαBi,m(t) =

m−i∑

k=[α]

θ(i,k,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)tk+i−α. (7.2.24)

Now approximating R.H.S of (7.2.24) as

m−i∑

k=[α]

θ(i,k,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)tk+i−α = C

(i)MBT

M (t) (7.2.25)

171

which further implies that

X(j)M =

∫ 1

0

m−i∑

k=[α]

θ(i,k,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)tk+i−αBj,m(t)dt, j = 0, 1, 2....m

⇒ X(j)M =

m−i∑

k=[α]

θ(i,k,m)

m−j∑

l=0

θ(j,l,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)(k + i+ l + j − α+ 1), j = 0, 1, 2, . . . ,m

(7.2.26)

CaseII:(i ≥ [α]) if i ≤ [α] then

X(j)M =

m−i∑

k=0

θ(i,k,m)

m−j∑

l=0

θ(j,l,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)(k + i+ l + j − α+ 1), j = 0, 1, 2, . . . ,m.

(7.2.27)After careful simplification we get

cDαB0,m(t)cDαB1,m(t)

...

...

...cDαBm,m(t)

=

X(0)M Φ−1

M×MBTM (t)

X(1)M Φ−1

M×MBTM (t)

...

...

...

X(m)M Φ−1

M×MBTM (t))

. (7.2.28)

On further simplification we have

Ψi,j =m−i∑

k=[α]

m−j∑

l=0

θ(i,k,m).θ(j,l,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)

1

(k + l + j + i− α+ 1)(i < [α])

Ψi,j =

m−i∑

k=0

m−j∑

l=0

θ(i,k,m).θ(j,l,m)Γ(k + i+ 1)

Γ(k + i− α+ 1)

1

(k + l + j + i− α+ 1)

we get cDαBTM (t) = Gα

M×MΦ−1M×M .B

TM (t).

(7.2.29)

LetGαM×MΦ−1

M×M = HαM×M

socDαBT

M (t) = HαM×MBT

M (t)

which is the desired result.

172

Lemma 7.2.3. An operational matrix corresponding to the boundary condition by takingBTM (t) is function vector and KM is coefficient vector by taking the approximation

u(t) = KMBTM (t)

is given by

Qα,φM×M =

Ω(0,0) Ω(0,1) · · · Ω(0,r) · · · Ω(0,m)

Ω(1,0) Ω(1,1) · · · Ω(1,r) · · · Ω(1,m)...

......

......

...Ω(r,0) Ω(r,1) · · · Ω(r,r) · · · Ω(r,m)...

......

......

...Ω(m,0) Ω(m,1) · · · Ω(m,r) · · · Ω(m,m)

, (7.2.30)

where

Ωi,j =

∫ 1

0∆i,mφ(t)Bj(t)dt, i, j = 0, 1, 2, . . . ,m.

Proof. Let us take u(t) = KMBTM (t) then

0Iα1KMBT

M (t) = KM0Iα1B

TM (t) = KM

0Iα1B0(t)

0Iα1B1(t).........

0Iα1Bm(t)

.

Let us evaluate a general terms

0Iα1Bi(t)dt =

1

Γα

∫ 1

0(1− s)α−1Bi,m(s)ds

=1

Γα

m−i∑

k=0

Θi,k,m

∫ 1

0(1− s)α−1sk+ids.

(7.2.31)

Since by

L(

∫ 1

0(1− s)α−1.sk+ids) =

Γα.Γ(k + i+ 1)

τk+α+i

taking inverse Laplace of both sides we get∫ 1

0(1− s)α−1.sk+ids = L−1

[Γα.Γ(k + i+ 1)

τk+α+i

]=

Γα.Γ(k + i+ 1)

Γ(k + i+ α+ 1)

now equation (7.2.31) implies that

0Iα1Bi(t)dt =

m−i∑

k=0

Θi,k,mΓ(k + i+ 1)

Γ(k + i+ α+ 1)= ∆i,m. (7.2.32)

173

Now using the approximation ∆i,mφ(t) =∑m

i=0 ciBi(t) = CiMBT

M (t), using equation (7.2.3)

we get CiM = Ki

MΦ−1M×MBT

M (t) and also using cj =∫ 10 φ(t)Bj(t)dt,

φ(t)KMIαBTM (t) = KM

∆0,mφ(t)∆1,mφ(t)

...

...

...∆m,mφ(t)

= KM

C0MΦ−1

M×MBTM (t)

C1MΦ−1

M×MBTM (t)

...

...

...

CmMΦ−1

M×MBTM (t)

(7.2.33)

= KM

c00 c01 · · · c0r · · · c0mc10 c11 · · · c1r · · · c1m...

......

......

...cr0 cr1 · · · crr · · · crm...

......

......

...cm0 cm1 · · · cmr · · · cmm

Φ−1M×MBT

M (t)

Φ−1M×MBT

M (t).........

Φ−1M×MBT

M (t)

. (7.2.34)

On further simplification we get

φ(t)KM IαBTM (t) = KM

Ω(0,0) Ω(0,1) · · · Ω(0,r) · · · Ω(0,m)

Ω(1,0) Ω(1,1) · · · Ω(1,r) · · · Ω(1,m)...

......

......

...Ω(r,0) Ω(r,1) · · · Ω(r,r) · · · Ω(r,m)

......

......

......

Ω(m,0) Ω(m,1) · · · Ω(m,r) · · · Ω(m,m)

B0(t)B1(t)

...

...

...Bm(t)

.

(7.2.35)

Soφ(t)0I

α1u(t) = KMQ

α,φM×MBT

M (t),

and

Ωi,j =

∫ 1

0∆i,mφ(t)Bj(t)dt, i, j = 0, 1, 2, . . . ,m. (7.2.36)

which is the required result. To evaluate Qα,φM×M , let us take α = 9

5 ,M = 6, φ(t) = t, then

Qα,φM×M =

0 0.0229 0.0459 0.0688 0.0918 0.11470 0.0202 0.0405 0.0607 0.0810 0.10120 0.0175 0.0349 0.0524 0.0698 0.08730 0.0145 0.0291 0.0436 0.0582 0.07270 0.0115 0.0230 0.0344 0.0459 0.05740 0.0082 0.0164 0.0246 0.0328 0.04100 0.0046 0.0091 0.0137 0.0182 0.0228

. (7.2.37)

174

In next subsection, we are going to approximate a boundary value problem of fractional

order differential equation as well as a coupled system of fractional order boundary value

problem. The application of obtained operational matrices are shown in following procedure.

7.2.1 Numerical solutions of boundary value problems

Consider the following problem in generalized form of fractional order differential equation

cDαu(t) +AcDµu(t) +Bu(t) = f(t), 1 < α ≤ 2, 0 < µ ≤ 1, (7.2.38)


u(0) = a, u(1) = b.

Where f(t) is a source term,A,B are any real constants and u(t) is a known solution which

we want to determined. To obtain numerical solution of the above problem in terms of

Bernstein polynomials we proceed as

Let cDαu(t) = KMBTM (t). (7.2.39)

Applying fractional integral of order α we have

u(t) = KMPM×MBTM (t)− C0 + C1t

using boundary conditions we have

C0 = a, C1 = b− a−KMPM×MBTM (t)|t=1.

Using the approximation and Lemma (6.3.8)

a+ t(b− a) = F(1)M BT

M (t), tPM×MBTM (t)|t=1 = Q

α,φM×MBT

M (t).

175

Hence

u(t) = KMPM×MBTM (t) + a+ t(b− a)− tKMPM×MBT

M (t)|t=1

u(t) = KMPαM×MBT

M (t) + F(1)M BT

M (t)−Qα,φM×MBT

M (t).

Which on implies that

= KM (PαM×M −Q

α,φM×M )BT

M (t) + F(1)M BT

M (t). (7.2.40)

Now

cDµu(t) =c Dµ[KM (Pα

M×M −Qα,φM×M)BT

M (t) + F(1)M BT

M (t)]

= KM (PαM×M −Q

α,φM×M )Hµ

M×MBTM (t) + F

(1)M H

µM×MBT

M (t).

(7.2.41)

And

f(t) = F(2)M BT

M (t). (7.2.42)

Putting equation(7.2.39),(7.2.40),(7.2.41) and (7.2.42) in equation(7.2.38) we get

KMBTM (t) +AKM (Pα

M×M −Qα,φM×M)Gµ

M×MBTM (t) +AF

(1)M G

µM×MBT

M (t)

+BKM (PαM×M −Q

α,φM×M )BT

M (t) +BF(1)M BT

M (t) = F(2)M BT

M (t)KMBTM (t)

+AKM (PαM×M −Q

α,φM×M)Hµ

M×MBTM (t) +AF

(1)M H

µM×MBT

M (t)

+BKM (PαM×M −Q

α,φM×M )BT

M (t) +BF(1)M BT

M (t)− F(2)M BT

M (t) = 0

KMBTM (t) +KM (PαM×MBT

M (t) + QαM×MBT

M (t) = 0,

where

PαM×M = A(Pα

M×M −Qα,φM×M)Hµ

M×M +BPαM×M ,

QαM×M = AF

(1)M H

µM×M +BF

(1)M −BQα

M×M − F(2)M ,

(7.2.43)

which is an algebraic equation and easily soluble for unknown coefficient vector KM . Putting

this in equation(7.2.40) we get required approximate solution of the problem.

176

7.2.2 Numerical solutions of system of boundary value problem

Consider a generalized form of a fractional order boundary value problem

cDαu(t) +Ac1Dµ1u(t) +Bc

1Dν1v(t) + C1u(t) +D1v(t) = f(t),

cDβv(t) +Ac2Dµ2u(t) +Bc

2Dν2v(t) + C2u(t) +D2v(t) = g(t),

(7.2.44)


u(0) = a, u(1) = b, v(0) = c, v(1) = d. (7.2.45)

Where Ai, Bi, Ci, Di(i = 1, 2) are any real constant,f(t), g(t) are given source terms, while

1 < α ≤ 2, 0 < µ1, ν1 < 1 and 1 < β ≤ 2, 0 < µ2, ν2 < 1 . We approximate the solution of

above system in Bernstein polynomials such as

cDαu(t) = KMBTM (t), cDβv(t) = LMBT

M (t)

u(t) = KMPαM×MBT

M (t) + C0 + C1t, v(t) = LM (PβM×MBT

M (t) + C0 + C1t

applying boundary conditions we have

u(t) = KM(PαM×MBT

M (t) + a+ t(b− a)− tKMPαM×MBT

M (t)|t=1,

v(t) = KM(PβM×MBT

M (t) + c+ t(d− c)− tKMPβM×MBT

M (t)|t=1.

Let us approximate

a+ t(b− a) = F1MBT

M (t), c+ t(d− c) = F2MBT

M (t)

tPαM×MBT

M (t)|t=1 = Qα,φM×MBT

M (t), tPβM×MBT

M (t)|t=1 = Qβ,φM×MBT

M (t),

then

u(t) = KMPαM×MBT

M (t) + F(1)M BT

M (t)−KMQα,φM×MBT

M (t)

v(t) = LMPβM×MBT

M (t) + F(2)M BT

M (t)− LMQβ,φM×MBT

M (t)

cDµ1u(t) =[KMPα

M×MBTM (t) + F

(1)M BT


M (t)]

= KM (PαM×M −Q

α,φM×M)Hµ1

M×M + F(1)M H

µ1M×MBT

M (t),

cDν1v(t) =c Dν1[LMP

βM×MBT

M (t) + F(2)M BT


M (t)]

= LM (PβM×M −Q

β,φM×M)Hν1

M×M + F(2)M H

ν1M×MBT

M (t),

177

cDµ2u(t) =c Dµ2[KMPα

M×MBTM (t) + F

(1)M BT


M (t)]

= KM (PαM×M −Q

α,φM×M )Hµ2

M×M + F(1)M H

µ2M×MBT

M (t)

cDν2v(t) =c Dν2[KMP

βM×MBT

M (t) + F(2)M BT

M (t)−KMQβ,φM×MBT

M (t)]

= LM (PβM×M −Q

β,φM×M)Hν2

M×M + F(2)M H

ν2M×MBT

M (t)

f(t) = F(3)BTM (t), g(t) = F(4)BT

M (t).

Thus equation(7.2.44) implies that

KMBTM (t) +A1KM (Pα

M×M −Qα,φM×M)Hµ1

M×M +A1F(1)M H

µ1M×MBT

M (t)

+B1LM (PβM×M −Q

β,φM×M)Hν1

M×M +B1F(2)M H

ν1M×MBT

M (t) + C1KMPαM×MBT

M (t)

+ C1F(1)M BT

M (t)−C1KMQα,φM×MBT

M (t) +D1LMPβM×MBT

M (t) +D1F(2)M BT

M (t)

−D1LMQβ,φM×MBT

M (t) = F(3)BTM (t)

LMBTM (t) +A2KM (Pα

M×M −Qα,φM×M)Hµ2

M×M +A2F(1)M H

µ2M×MBT

M (t)

+B2LM (PβM×M −Q

β,φM×M)Hν2

M×M +B2F(2)M Hν2

M×MBTM (t) + C2KMPα

M×MBTM (t)

+ C2F(1)M BT

M (t)−C2KMQα,φM×MBT

M (t) +D2LMPβM×MBT

M (t) +D2F(2)M BT

M (t)

−D2LMQβ,φM×MBT

M (t) = F(4)BTM (t).

(7.2.46)

Rearranging the terms in above system and using the following notation for simplicity in

equation(7.3.1)

HαM×M = A1(P

αM×M −Q

α,φM×M)Hµ1

M×M + C1(PαM×M −Q

α,φM×M )

HβM×M = B1(P

βM×M −Q

β,φM×M)Hν1

M×M +D1(PβM×M −Q

β,φM×M )

SαM×M = A2(PαM×M −Q

α,φM×M)Hµ2

M×M + C2(PαM×M −Q

α,φM×M )

SβM×M = B2(P

βM×M −Q

β,φM×M)Hν2

M×M +D2(PβM×M −Q

β,φM×M )

FM = A1F(1)M H

µ1M×M +B1F

(2)M Hν1

M×M + C1F(1)M +D1F

(2)M − F

(3)M

GM = A2F(1)M H

µ2M×M +B2F

(2)M H

ν2M×M + C2F

(1)M +D2F

(2)M − F

(4)M ,

178

then the above system (7.3.1) becomes

KMBTM (t) +KMHα

M×MBTM (t) + LMH

βM×MBT

M (t) + FMBTM (t) = 0

LMBTM (t) +KMSαM×MBT

M (t) + LMSβM×MBT

M (t) +GMBTM (t) = 0.

(7.2.47)

Now (7.2.47) is a system of algebraic equations which ca be written in matrix form as

[ KM LM ]

BT

M (t) 0

0 BTM (t)

+ [ KM LM ]

Hα

M×M 0

0 SβM×M

BT

M (t) 0

0 BTM (t)

+ [ KM LM ]

0 SαM×M

HβM×M 0

BT

M (t) 0

0 BTM (t)

+ [ FM GM ]

BT

M (t) 0

0 BTM (t)

= 0

[ KM LM ] + [ KM LM ]

Hα

M×M SαM×M

HβM×M S

βM×M

+ [ FM GM ] = 0,

(7.2.48)

which is an algebraic equations can be easily solved by using Matlab functional solver or

mathematica for unknown matrix [KM LM ]. Calculating the coefficient matrix KM , LM

and putting in equation equations

u(t) = KMPαM×MBT

M (t) + F(1)M BT


M (t)

v(t) = LMPβM×MBT

M (t) + F(2)M BT


M (t),

we get the required approximate solution.


Example 7.2.1. Consider

cDαu(t) + cc1Dνu(t) + c2u(t) = f(t), 1 < α < 2 (7.2.49)


u(0) = 0, u(1) = 0.

Solution: We solve this problem under the following parameters sets defined as S1=α =1.8, ν = 0.8, c1 = 10, c2 = 100, S2=α = 1.5, ν = 0.5, c1 = 1/10, c2 = 1/100, and selectsource term for S1 as

179

0 0.2 0.4 0.6 0.8 1

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

x 10−3

(a)

0 0.2 0.4 0.6 0.8 110−12

10−10

10−8

10−6

10−4

10−2

(b)

U5

U6

U7

U8

U9

U10

u5

u6

u7

u8

u

Figure 7.7: Comparison of exact and approximate solution of Example (7.2.1), under pa-rameters set S1.

f1(t) =11147682583723703125 t

215

(1750 t3 − 4200 t2 + 3255 t − 806

)

406548945561989414912

+278692064593092578125 t

265

(5250 t3 − 14350 t2 + 12915 t − 3813

)

25002760152062349017088+ 100 t6 (t− 1)3

(7.2.50)

f2(t) =5081767996463981 t

92

(1344 t3 − 3360 t2 + 2730 t − 715

)

264146673456906240

+5081767996463981 t

112

(1344 t3 − 3808 t2 + 3570 t − 1105

)

22452467243837030400+t6 (t− 1)3

100

(7.2.51)

The exact solution of the above problem is

u(t) = t6 (t− 1)3.

We solve this problem with the proposed method under different sets of parameters as definedin S1, S2. The observation and simulation demonstrates that the solution obtained with theproposed method is highly accurate. The comparison of exact solution with approximatesolution obtained using the parameters set S1 is displayed in subplot (a), while in subplot(b) we plot absolute difference between the exact and approximate solution obtained usingdifferent scale levels of Figure (7.7) respectively. One can easily observe that the absoluteerror is much more less than 10−12.

By solving the problem under parameters set S2, we observe the same phenomena. Theapproximate solution matches very well with the exact solution. See Figure (7.8).

180

0 0.2 0.4 0.6 0.8 1

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

x 10−3

(a)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 110−11

10−10

10−9

10−8

10−7

10−6

10−5

10−4

10−3

(b)

u5(t)

u6(t)

u7(t)

u8(t)

Exact u(t)

U5

U6

U7

U8

U9

U10

Figure 7.8: Comparison of exact and approximate solution of Example (7.2.1), under pa-rameters set S2.

Example 7.2.2. Consider the following coupled system of fractional differential equations

cD1.8u(t) +c Du(t) + 9cD0.8v(t) + 2u(t)− v(t) = f(t)cD1.8v(t)− 6cD0.8u(t) +c Dv(t)− u(t) = g(t)

(7.2.52)


u(0) = 1, u(1) = 2 and v(0) = 2, v(1) = 2.

In Figure(7.10), we have visualized the absolute errorU, V of u(t), v(t). Solution: Theexact solution are

u(t) = t5(1− t), v(t) = t4(1− t).

We approximate the solution of this problem with this new method. The source terms aregiven by

f(t) = 2 t5 (x− 1)− t4 (t− 1) + t4 (6 t− 5)− 2229536516744740625 t165 (10 t− 7)

1008806316530991104

+1337721910046844375 t

165 (25 t − 21)

1008806316530991104(7.2.53)

g(t) = t3 (5 t− 4)− t5 (t− 1)− 11147682583723703125 t215 (15 t− 13)

6557241057451442176

− 89181460669789625 t115 (25 t− 16)

144115188075855872.

(7.2.54)

181

0 0.2 0.4 0.6 0.8 1−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

(a)

0 0.2 0.4 0.6 0.8 1

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

(b)

v3(t)

v5(t)

v7(t)

Exact v(t)

u3(t)

u5(t)

u7(t)

Exact u(t)

Figure 7.9: (a) Comparison of exact solution of Example (7.2.2) of u(t) with the approximatesolutions at different scale level. (b) Comparison of exact solution of Example (7.2.2) ofv(t) with the approximate solutions at different scale level.

0 0.2 0.4 0.6 0.8 110−10

10−8

10−6

10−4

10−2

100

(a)

0 0.2 0.4 0.6 0.8 110−10

10−8

10−6

10−4

10−2

100

(b)

U3

U4

U5

U6

U7

V3

V4

V5

V6

V7

Figure 7.10: (a) Absolute error in u(t) for different scale levelM = 3 : 7 of Example (7.2.2).(b) Absolute error in v(t) for different scale level M = 3 : 7 of Example (7.2.2).

182

As expected the method provides a very good approximation to the solution of the prob-lem.At first we approximate the solutions of the problem at α = 2, because the exact solutionat α = 2 is known. We observe that at much small scale level the method provides a verygood to the solution. We approximate the absolute error by the formula

U = maxt∈[0,1]

|uexact − uapp|.

andV = max

t∈[0,1]|vexact − vapp|.

We approximate the absolute error at different scale level of M . And observe that theabsolute error is much more less than 10−10 at scale level M = 7 , see Figure(7.10). Wealso approximate the solution at some fractional value of α and observe that as α → 2 theapproximate solutions approaches to the exact solutions . Which guarantees the accuracyof the solution at fractional value of α. Figure shows this phenomena. In Figure(7.9), thesubplot (a) represents the approximation of u(t) and subplot (b) represents the approximationofv(t).

Example 7.2.3. Consider the following coupled system

cD0.8u(t)− u(t) + 3v(t) = f(t)cD0.8v(t) + 4u(t)− 2v(t) = g(t), 0 < α, β ≤ 1

(7.2.55)


u(0) = −1, u(1) = −1 and v(0) = −1, v(1) = −1.

Solution:The exact solution for α = β = 1 is

u(t) = t5 − t4 − 1, and v(t) = t4 − t3 − 2. (7.2.56)

The source terms are given by

f(t) =445907303348948125t3.2(25t− 21)

3026418949592973312+ t3 − 4t4 + 3t5 − 2

g(t) =89181460669789625t2.5(5t− 4)

14411518807585872− 4t3 + 6t4 − 2t5 − 2

Approximating the solution with this method, we observe that our scheme gives high ac-curacy of approximate solutions. In the next Figure, we plot the exact solutions togetherwith the approximate solutions in sub Figure(a) and (b) for u(t) and v(t) in Figure(9) re-spectively. We see from the subplots (a) and (b) of Figure(9) that our approximations haveclose agreement to that of exact solutions as shown. This accuracy me be make better byincreasing scale level and observe that absolute error is below 10−10 in Figure (10) which isvery high accuracy for such type practical problems. In Figure(7.12), the subplot (a) repre-sents absolute error for u(t) while subplot (b) represents the same quantity for v(t). From

183

0 0.2 0.4 0.6 0.8 1

1.93

1.94

1.95

1.96

1.97

1.98

1.99

2

(a)

0 0.2 0.4 0.6 0.8 10.91

0.92

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

(b)

u3(t)

u7(t)

Exact u(t)

v3(t)

v7(t)

Exact v(t)

Figure 7.11: (a) Comparison of exact u(t) with its approximate solutions at scale levelM = 3 : 7 for Example (7.2.3). (b) Comparison of exact v(t) with its approximate solutionsat scale level M = 3 : 7 for Example (7.2.3).

00.20.40.60.8110−10

10−8

10−6

10−4

10−2

100

(a)

00.20.40.60.8110−10

10−8

10−6

10−4

10−2

100

(b)

U3

U4

U5

U6

U7

V3

V4

V5

V6

V7

Figure 7.12: (a) Absolute error in u(t) for different scale levelM = 3 : 7 of Example (7.2.3).(b) Absolute error in v(t) for different scale level M = 3 : 7 of Example (7.2.3).

184

Scale Our proposed method Proposed method in [134]

M u(t) v(t) u(t) v(t)

3 1.0034(−3) 1.0018(−3) 2.0017(−2) 1.0017(−2)

4 1.0038(−5) 2.0029(−4) 2.5017(−3) 4.5017(−3)

5 4.0004(−6) 1.4017(−6) 1.1017(−5) 1.0018(−5)

6 1.0367(−8) 3.0012(−8) 3.2017(−6) 2.0317(−6)

7 1.0017(−10) 2.0015(−10) 2.5017(−7) 1.0683(−7)

8 2.8000(−11) 5.1011(−11) 3.6018(−8) 1.0065(−7)

9 1.4017(−12) 1.0019(−12) 4.1017(−9) 1.5696(−8)

10 1.2017(−13) 1.0016(−13) 2.1017(−10) 1.0019(−9)

12 2.0017(−16) 2.0017(−16) 1.0116(−12) 1.0017(−12)

Table 7.2: Maximum absolute error for Bernstein and Shifted Jacobi polynomial operationalmatrices for Example (7.2.3) at different scale levels.

the sub plots we see that maximum absolute error for our proposed method for the givenproblem (7.2.3) is below 10−10. Which is very small amount and justifies the efficiency ofour constructed method.In[134], this problem has been solved by using Jacobian polynomials operational matricesand the comparison between the mentioned scheme and our proposed method in their abso-lute error are given in the following table at different scale levels. From the Tabel(7.2), weobserve that the operational matrices of Bernstein polynomials give best approximations ascompare to Jacobi polynomials.

7.3 Numerical solutions boundary value problem by ShiftedJacobi polunomials

In this section we provide some operational matrices for fractional order differentiation as

well as of fractional order integration for Shifted Jacobi polynomials. Further in this section

we give construction of new matrix for the boundary conditions.

Theorem 7.3.1. Let ΦM(t) be the function vector then the operational matrix of fractionalorder derivative α is given by

cDα[ΦM (t)] ' G(α)M×MΦM(t), (7.3.1)

185

where G(α)M×M is the operational matrix of fractional order derivative α given by

G(α)M×M =

Υ0,0,k Υ0,1,k · · · Υ0,j,k · · · Υ0,m,k

Υ2,1,k Υ2,2,k · · · Υ2,r,k · · · Υ1,m,k...

......

......

...Υi,0,k Υi,1,k · · · Υi,j,k · · · Υi,m,k

......

......

......

Υm,0,k Υm,1,k · · · Υm,j,k · · · Υm,m,k

, (7.3.2)

where

Υi,j,k =

i∑

k=0

(−1)i−kΓ(i+ b+ 1)Γ(i + k + a+ b+ 1)Γ(k + 1)

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i− k + 1)Γ(k + 1)Γ(k + 1− α)ξk

×j∑

l=0

(−1)j−l(2j + a+ b+ 1)Γ(j + 1)Γ(j + l + a+ b+ 1)Γ(k − alpha+ l + b+ 1)Γ(a+ 1)ξα

Γ(j + a+ 1)Γ(l + b+ 1)Γ(j − l + 1)Γ(l + 1)Γ(k − α+ l + b+ a+ 2).

If Υi,j,k = 0, for i < α.

(7.3.3)

Proof. The proof of this theorem is same as proof of Lemma 7.1.1.

Theorem 7.3.2. Let ΦM (t) be the function vector corresponding to the Shifted Jacobipolynomials and α > 0 then the operational matrix corresponding to the fractional orderintegration is given by

Iα[ΦM (t)] ' H(α)M×MΦM (t), (7.3.4)

where H(α)M×M is the operational matrix of fractional order integration α and is given by

H(α)M×M =

Υ0,0,k Υ0,1,k · · · Υ0,j,k · · · Υ0,m,k

Υ2,1,k Υ2,2,k · · · Υ2,r,k · · · Υ1,m,k...

......

......

...Υi,0,k Υi,1,k · · · Υi,j,k · · · Υi,m,k

......

......

......

Υm,0,k Υm,1,k · · · Υm,j,k · · · Υm,m,k

, (7.3.5)

where

Υi,j,k =

i∑

k=0

(−1)i−kΓ(i+ b+ 1)Γ(i + k + a+ b+ 1)Γ(k + 1)

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i − k + 1)Γ(k + 1)Γ(k + 1 + α)ξk

×j∑

l=0

(−1)j−l(2j + a+ b+ 1)Γ(j + 1)Γ(j + l + a+ b+ 1)Γ(k + α+ l + b+ 1)Γ(a + 1)ξα

Γ(j + a+ 1)Γ(l + b+ 1)Γ(j − l + 1)Γ(l + 1)Γ(k + α+ l + b+ a+ 2).

(7.3.6)

186

Proof. The proof of this theorem is same as Lemma 7.1.1 . Therefore we omit the proof.

Theorem 7.3.3. Let ΦM (t) be a function vector and let φ(t) be any function defined asφ(t) = ctn, n = 0, 1, 2, . . . , c ∈ R and u(t) = KT

MΦM(t) . Then

φ(t)0Iαξ u(t) = KT

MQ(c,φ,α)M×MΦM(t), (7.3.7)

where Q(c,φ,α)M×M is an operational matrix corresponding to some boundary value and is given

by

Q(c,φ,α)M×M =

Υ0,0,k Υ0,1,k · · · Υ0,j,k · · · Υ0,m,k

Υ2,1,k Υ2,2,k · · · Υ2,r,k · · · Υ1,m,k...

......

......

...Υi,0,k Υi,1,k · · · Υi,j,k · · · Υi,m,k

......

......

......

Υm,0,k Υm,1,k · · · Υm,j,k · · · Υm,m,k

, (7.3.8)

where

Υi,j,k =

M∑

j=0

∆i,k,αBjP(a,b)ξ,j (t), (7.3.9)

where

∆i,k,α =

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i+ a+ b+ 1)cξα

Γ(k + b+ 1)Γ(i+ a+ b+ 1)Γ(i− k + 1)Γ(k + α)(7.3.10)

and

Bj =

j∑

l=0

(−1)j−l(2j + a+ b+ 1)Γ(j + 1)Γ(j + l + a+ b+ 1)Γ(n+ b+ 1)Γ(a + 1)ξn−l

Γ(j + a+ 1)Γ(l + b+ 1)Γ(j − l + 1)Γ(l + 1)Γ(n+ a+ b+ 1).

(7.3.11)

Proof. Consider the general term of ΦM (t) as

0Iαξ P

(a,b)ξ,i (t) =

1

Γα

∫ ξ

0(ξ − s)α−1P

(a,b)ξ,i (s)ds

=1

Γα

∫ ξ

0(ξ − s)α−1

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i + k + a+ b+ 1)sk

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i − k + 1)Γ(k + 1)ξkds

=1

Γα

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i + k + a+ b+ 1)

Γ(k + b+ 1)Γ(i+ a+ b+ 1)Γ(i − k + 1)Γ(k + 1)ξk

∫ ξ

0(ξ − s)α−1skds.

(7.3.12)

Now by applying Laplace transform we

L(∫ ξ

0(ξ − s)α−1skds

)=

Γ(α)Γ(k + 1)

sk+α+1

⇒∫ ξ

0(ξ − s)α−1skds =

Γ(k + 1)Γα ξk+α

Γ(α+ 1).

(7.3.13)

187

Now putting (7.3.13) in (7.3.12) we get

0Iαξ P

(a,b)ξ,i (t) =

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i + a+ b+ k + 1)ξα

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i − k + 1)Γ(k + α). (7.3.14)

Now if φ(t) = tn, then

cφ(t)0Iαξ P

(a,b)ξ,i (t) =

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i + a+ b+ k + 1)ξαctn

Γ(k + b+ 1)Γ(i+ a+ b+ 1)Γ(i− k + 1)Γ(k + α). (7.3.15)

Representing tn in terms of Shifted Jacobi polynomials as

tn =M∑

j=0

BjP(a,b)ξ,j (t). (7.3.16)

By the use of Orthogonality relation (2.5.19), we have

Bj =1

Ω(a,b)ξ,j

∫ ξ

0tnP

(a,b)ξ,j (t)W

(a,b)ξ (ξ)dt

=1

Ω(a,b)ξ,j

j∑

l=0

(−1)(j−l)Γ(j + b+ 1)Γ(j + l + a+ b+ 1)

Γ(l + b+ 1)Γ(j + a+ b+ 1)Γ(j − l + 1)Γ(l + 1)ξ l

∫ ξ

0tn+b(ξ − t)adt.

(7.3.17)

Now by mean of Laplace transform we have

L(∫ ξ

0tn+b(ξ − t)adt

)=

Γ(n+ b+ 1)Γ(a+ 1)

ξn+b+a+2

⇒∫ ξ

0tn+b(ξ − t)adt =

Γ(n+ b+ 1)Γ(a+ 1)

Γ(n+ a+ b+ 1)ξn+a+b+1.

(7.3.18)

Thus putting (7.3.18) in (7.3.17) and using(2.5.20), we get

Bj =

j∑

l=0

(−1)j−l(2j + a+ b+ 1)Γ(j + 1)Γ(j + l + a+ b+ 1)Γ(n+ b+ 1)Γ(a + 1)ξn−l

Γ(j + a+ 1)Γ(l + b+ 1)Γ(j − l + 1)Γ(l + 1)Γ(n+ a+ b+ 1).

(7.3.19)

Now (7.3.15) implies that

φ(t)0Iαξ P

(a,b)ξ,i (t) =

M∑

j=0

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i+ a+ b+ k + 1)ξαc

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i − k + 1)Γ(k + α)BjP

(a,b)ξ,j (t)

=

M∑

j=0

∆i,k,αBjP(a,b)ξ,j (t) =

M∑

j=0

Υi,j,kP(a,b)ξ,j (t),

(7.3.20)

188

where

Υi,j,k =

M∑

j=0

∆i,k,αBj ,

∆i,k,α =

i∑

k=0

(−1)i−k(i+ b+ 1)Γ(i+ a+ b+ k + 1)ξαc

Γ(k + b+ 1)Γ(i + a+ b+ 1)Γ(i+ k + 1)Γ(k + α).

Evaluating the result (7.3.20) for different values of j we get the required result. For examplelet M = 6, φ(t) = ctn, n = 6, c = 1, α = 1.9, then

Q(c,φ,α)M×M =

0.0782 0.1759 0.1629 0.0912 0.0320 0.0065−0.0243 −0.0546 −0.0505 −0.0283 −0.0099 −0.0020−0.0006 −0.0014 −0.0013 −0.0007 −0.0003 −0.0001−0.0001 −0.0003 −0.0002 −0.0001 −0.0000 −0.0000−0.0005 −0.0011 −0.0010 −0.0006 −0.0002 −0.0000−0.0004 −0.0009 −0.0009 −0.0005 −0.0002 −0.0000

. (7.3.21)

7.3.1 Applications of Shifted Jacobi’s polynomials operational matrices

In this subsection we show fundamental importance of operational matrices of fractional

order derivative and integration. We apply them to solve some multi terms fractional order

boundary value problems of fractional differential equations. Consider the following general

fractional differential equation with constant coefficient and given boundary conditions

cDαu(t) +AcDνu(t) +BcDωu(t) + Cu(t) = f(t), t ∈ [0, ξ], A,B,C ∈ R, (7.3.22)

subject the boundary conditions

u(0) = u0, u(ξ) = u1,

where f(t) ∈ C[0, ξ] is a source terms and 1 < α ≤ 2, 0 < ν, ω < 1.

To obtain the solutions of the (7.3.22) in term of Shifted Jacobi polynomials we proceed as

cDαu(t) = KTMΦM (t). (7.3.23)

189

Applying Iα, definition of fractional derivative and Remark 2.2.24 of Theorem 2.2.10, then

we get

u(t) = KTMH

(α)M×MΦM(t) + C0 + C1t. (7.3.24)

By means of boundary conditions we get

u(t) = KTMH

(α)M×MΦM (t) + x0 + t

(x1 − x0)

ξ− t

ξKTMH

(α)M×MΦM(t)|t=ξ (7.3.25)

using the approximation x0 + t (x1−x0)ξ ' F1MΦM (t), t

ξ = cφ(t), c = 1ξ also using

cφ(t)KTMH

(α)M×MΦM (t)|t=ξ = KT

MQ(α,φ)M×MΦM (t),

then (7.3.25) implies that

u(t) = KTMH

(α)M×MΦM(t) + F1

MΦ(t)− cφ(t)KTMH

(α)M×MΦM (t)|t=ξ

= KTMH

(α)M×MΦM (t)−KT

MQ(α,φ)M×MΦM(t) + F1

MΦM(t)

= KTM

(HαM×M −Q

(α,φ)M×M

)ΦM(t) + F1

MΦM(t).

(7.3.26)

Now from(7.3.26) we have

cDνu(t) = KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ν)M×MΦM (t) + F1

MG(ν)M×MΦM(t)

cDωu(t) = KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ω)M×MΦM (t) +F1

MG(ω)M×MΦM (t),

(7.3.27)

and approximating the source term as

f(t) ' F2MΦM (t). (7.3.28)

190

Putting (7.3.23), (7.3.24),(7.3.25),(7.3.26), (7.3.28) and(7.3.27) in (7.3.22) which yields

KTMΦM (t) +A

[KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ν)M×MΦM(t) + F1

MG(ν)M×MΦM (t)

]

+B[KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ω)M×MΦM(t) + F1

MG(ω)M×MΦM(t)

]

+ C[KTM

(H

(α)M×M −Q

(α,φ)M×M

)ΦM(t) + F1

MΦM(t)]− F2

MΦM(t) = 0

KTMΦM (t) +KT

M

(H

(α)M×M −Q

(α,φ)M×M

) [AG

(ν)M×M +BG

(ω)M×M + CI

]ΦM(t)

+[F1M

(AG

(ν)M×M +BG

(ω)M×M

)+ CF1

M − F2M

]ΦM (t) = 0

KTM +KT

M

(HαM×M −Q

(α,φ)M×M

) [AG

(ν)M×M +BG

(ω)M×M + CI

]

+ F1M

(AG

(ν)M×M +BG

(ω)M×M + CI

)− F2

M = 0.

(7.3.29)

Which is a simple algebraic equation. Solving this equation forKM and putting it in (7.3.26)

we get the required approximate solution of the corresponding boundary value problem.

7.3.2 Numerical solutions of system of boundary value problems

In this subsection, we use operational matrices to derive procedure for the numerical so-

lutions of coupled system. We consider the following general coupled system of fractional

differential equations as

cDαu(t) +Ac1Dν1u(t) +Bc

1Dν2v(t) + C1u(t) +D1v(t) = f(t), t ∈ [0, ξ],

cDβv(t) +Ac2Dω1u(t) +Bc

2Dω2v(t) + C2u(t) +D2v(t) = g(t), t ∈ [0, ξ],

(7.3.30)

subject to the boundary conditions given by

u(0) = u0, u(ξ) = u1,

v(0) = v0, v(ξ) = v1,

(7.3.31)

where 1 < α, β ≤ 2, 0 < νi, ωi < 1(i = 1, 2). Further f(t), g(t) ∈ C[0, ξ] are the source term

of the system (7.3.30),(7.3.31) and u0, v0, u1, v1 are real constants, Ai, Bi, Ci, Di(i = 1, 2)

are any constants. To approximate the above system in terms of Shifted Jacobi polynomials

we take

cDαu(t) = KTMΦM (t) (7.3.32)

191

and

cDβv(t) = LTMΦM(t) (7.3.33)

Applying Iα, Iβ and Remark 2.2.24 of Theorem 2.2.10 and boundary conditions, equations

(7.3.32),(7.3.33) implies that

u(t) = KTM

(H

(α)M×M −Q

(α,φ)M×M

)ΦM(t) + F1

MΦM(t) (7.3.34)

v(t) = LTM

(H

(β)M×M −Q

(β,φ)M×M

)ΦM(t) + F2

MΦM(t). (7.3.35)

Now taking fractional order derivative of (7.3.34),(7.3.35), we have

cDν1u(t) = KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ν1)M×MΦM(t) + F1

MG(ν1)M×MΦM(t)

cDω1u(t) = KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ω1)M×MΦM (t) + F1

MG(ω1)M×MΦM(t)

(7.3.36)

and

cDν2v(t) = LTM

(H

(β)M×M −Q

(β,φ)M×M

)G


MG(ν2)M×MΦM (t)

cD(ω2)y(t) = LTM

(H

(β)M×M −Q

(β,φ)M×M

)G

(ω2)M×MΦM (t) + F2

MG(ω2)M×MΦM (t).

(7.3.37)

Approximating the source functions as:

f(t) ' F3MΦM (t), g(t) ' F4

MΦM (t),(7.3.38)

Putting (7.3.33), (7.3.34),(7.3.35),(7.3.36),(7.3.37) and (7.3.38) in (7.3.30), we obtain

KTMΦM(t) +A1

[KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ν1)M×MΦM (t) +F1

MG(ν1)M×MΦM (t)

]

+B1

[LTM

(H

(β)M×M −Q

(β,φ)M×M

)G


MG(ν2)M×MΦM(t)

]

+ C1

[KTM

(H

(α)M×M −Q

(α,φ)M×M

)ΦM (t) + F1

MΦM (t)]

+D1

[LTM

(H

(β)M×M −Q

(β,φ)M×M

)ΦM (t) + F2

MΦM (t)]

− F3MΦM(t) = 0,

(7.3.39)

192

LTMΦM (t) +A2

[KTM

(H

(α)M×M −Q

(α,φ)M×M

)G

(ω1)M×MΦM(t) + F1

MG(ω1)M×MΦM(t)

]

+B2

[LTM

(H

(β)M×M −Q

(β,φ)M×M

)G

(ω2)M×MΦM(t) + F2

MG(ω2)M×MΦM (t)

]

+C2

[KTM

(H

(α)M×M −Q

(α,φ)M×M

)ΦM(t) + F1

MΦM(t)]

+D2

[LTM

(H

(β)M×M −Q

(β,φ)M×M

)ΦM(t) + F2

MΦM(t)]

−F4MΦM (t) = 0.

(7.3.40)

Re arranging the terms in system (7.3.39), (7.3.40), and using the following notations for

simplicity

S = F1M (A1G

(ν1) + C1I) + F2M (B1G

(ν2) +D1I)− F3M

R = F1M (A2G

(ω1) + C2I) + F2M (B2G

(ω2) +D2I)−F4M

H(α) −Q(α,φ) =︷︸︸︷P , H(β) −Q(β,φ) =

︷︸︸︷Q .

Then witting in matrix form we have

[KTM LTM

] ΦM (t) 0

0 ΦM(t)

+[KTM LTM

]︷︸︸︷P (A1G

ν1 + C1I)︷︸︸︷Q (B1G

ν2 +D1I)︷︸︸︷P (A2G

ω1 + C2I)︷︸︸︷Q (B2G

ω2 + C2I)

ΦM (t) 0

0 ΦM(t)

+[S R

] ΦM (t) 0

0 ΦM (t)

= 0.

Which implies that

[KTM LTM

]+[KTM LTM

]︷︸︸︷P (A1G

ν1 + C1I)︷︸︸︷Q (B1G

ν2 +D1I)︷︸︸︷P (A2G

ω1 + C2I)︷︸︸︷Q (B2G

ω2 + C2I)

+[S R

]= 0

(7.3.41)

which is an algebraic equation and can easily be solved for unknown matrices KM ,LM ,

then putting it in(7.3.25) and(7.3.26), we get required approximations for u(t), v(t).


In this subsection, we will apply our proposed scheme to some practical problems.

193

Example 7.3.1. Consider the problems of fractionally damped mechanical oscillator, whosegoverning equation with boundary conditions is given by

cDαu(t) + λc1Dνu(t) + λ2u(t) = f(t), t ∈ [0, 1], 1 < α ≤ 2, (7.3.42)


u(0) = 0, u(1) = 0,

where 0 < ν < 1 are λi(i = 1, 2) are prescribed constants and f(t) is source of force function.As at α = 2, ν = 1, (7.3.42) is classical differential equations of harmonic oscillator. It iseasy to verify that

u(t) = 625t7 − 2000t6 + 2549t5 − 1420t4 + 381t3 − 36t2

is exact solution of (7.3.42). We solve (7.3.42) corresponding to α = 1.5, ν = 0.5, ω = 0,and A = λ1 = 1, B = 0, C = λ2 =

−1√π, where source term is

f(t) =1√π

(16t1.5A(t)

45045+

24t0.25B(t)9945Γ( 54 )

− t2(5t− 3)2C(t)),

where

A(t) = 28028000t3 − 14620320t2 − 21527571t − 270270

B(t) = 6400000t5 − 15360000t4 + 13328000t3 − 5021120t2 + 757809t − 29835

C(t) = 25t3 − 50t2 + 29t− 4.

We apply our technique introduced in the current section for various choices of a, b,M andα, ν, ωM . The observations are given in Table(7.3.1) for the absolute error |uapp − uexact|.

M a b |uapp − uexact| a b |uapp − uexact|8 0 0 1.200(−2) 1 1 4.0001(−2)

10 0 0 4.0017(−3) 1 1 3.0017(−3)

8 0.5 0 1.8000(−2) 2 2 2.0007(−2)

12 0.5 0.5 8.0018(−4) 1.5 1.5 2.0018(−4)

12 0 0 1.0001(−5) −0.5 −0.5 1.0317(−4)

16 2 0 3.5000(−3) 0 2 4.0003(−4)

Table 7.3: Maximum absolute error for ξ = 1, α = 1.5, ν = 0.5, ω = 0 for Example (7.3.1).

In the Table (7.4), we have shown the maximum absolute error for various choices ofa, b,M . Also in Figure(7.13) and (7.14), we have shown the images of their comparisonbetween exact and approximate values. Further we observed that as the α → 2 the errorreducing and at α = 2 the absolute error is 3.5 × 10−14 at integral values of α, ν,M = 6which shows that the method provide a very accurate results at a small scale level.

194

0 0.2 0.4 0.6 0.8 1−20

0

20

40

60

80

100

(a)0 0.2 0.4 0.6 0.8 1

0

0.002

0.004

0.006

0.008

0.01

0.012

(b)

|uecact − uapprox|

Exact u(t)Approximate u(t)

Figure 7.13: Subplot (a) represents comparison of approximate and exact u(t) of Example(7.3.1). Setting α = 1.5, ν = 0.5, ω = 0, M = 8, ξ = 1, a = b = 0. Stars dots represent exactsolution while the red curve represents approximate solution.

0 0.2 0.4 0.6 0.8 1−20

0

20

40

60

80

100

(a)

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10−3

(b)

|uapprox − uexact|approximate u(t)

exact u(t)

Figure 7.14: Subplot (a) represents comparison of approximate and exact u(t) of Example(7.3.1) at α = 1.9, ν = 0.9, a = b = 0.5,M = 8. The subplot(b) represents their absoluteerror at α = 1.9, ν = 0.9, a = b = 0.5,M = 8.

195

M α ν ω |uapprox − uexact|6 1.5 0.5 0 4.5000(−1)

6 1.7 0.7 1.0000(−1)

8 1.5 0.5 1.2000(−2)

6 1.9 0.9 4.5000(−3)

6 2 1 3.5000(−14)

Table 7.4: Maximum absolute error at various values of α, ν,M and ω = 0 for exam-ple(7.3.1).

Example 7.3.2. Consider the following general problem

cDαu(t) +AcDν1u(t) +BcDν2u(t) + Cu(t) = f(t), 1 < α ≤ 2, 0 < ν1, ν2 ≤ 1,

subject to the boundary conditions u(0) = 1, u(1) = 5.(7.3.43)

Let the exact solution u(t) = t4 + t3 + t2 + t+1, and A = 1, B = 2, C = 3, the source terms

f(t) = 3t4 + 3t3 + 15t2 + 9t+ 5 +24

Γ215

t165 +

6

Γ165

t115 +

2

Γ115

t65 +

1

Γ65

t15

+48

Γ235

t185 +

12

Γ185

t135 +

4

Γ135

t85 +

2

Γ85

t35 .

We approximate the solutions of this problem with proposed method by setting α = 2, ν1 =0.8, ν2 = 0.4, a = b = 1,M = 10, ξ = 1. and observed that the scheme provide much moreaccurate approximations for scale level M = 10 maximum absolute error is 3.5 × 10−5 asshown in Figure (7.15). In Figure (7.16), comparison of approximate and exact u(t) ofExample(7.3.2)is given for various value of order of differential operator involved in theproblem at a = b = 0,M = 6.

The absolute error for different choices of a, b and M and for α = 2, ν1 = 0.8, ν2 = 0.4is given in table(7.5)

M a b |uapp − uexact| a b |uapp − uexact|4 0 0 1.000(−3) 1 0 6.0001(−2)

4 0 1 4.0017(−2) 0.5 0.5 5.0000(−3)

6 0.5 0 1.8000(−3) 2 2 2.0007(−2)

8 0 0 8.0018(−4) 1.5 1.5 2.0018(−3)

10 0 0 3.5000(−5) −0.5 −0.5 1.0317(−4)

12 2 2 3.5000(−6) 0 0 4.0003(−8)

Table 7.5: Maximum absolute error at α = 2, ν1 = 0.8, ν2 = 0.4 for example(7.3.2).

From the Table (7.6), we observed that as the orders of derivatives tend to their corre-sponding integer values the approximate solutions tend to its exact value which demonstratethe accuracy of numerical solutions obtain at our proposed method.

196

0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

4.5

5

(a)

(a)

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

3.5x 10−5 (b)

approximate u

exact u

|uapprox− uexact|

Figure 7.15: Subplot (a) represents comparison of approximate and exact u(t) of Exam-ple(7.3.2). The subplot (b) represents their absolute error at M = 10.

M α ν1 ν2 |uapp − uexact| α ν1 ν2 |uapp − uexact|4 2 1 1 2.5000(−2) 1.8 0.7 0.5 3.0008(−2)

6 2 1 1 7.0000(−16) 1.8 0.9 0.7 1.0008(−4)

8 1.8 0.9 0.8 2.0000(−4) 1.9 0.9 1 2.0009(−5)

8 1.7 1 0.8 2.0004(−5) 2 0.8 0.4 1.0001(−5)

10 1.9 0.999 0.999 2.0001(−8) 1.9 1 1 6.0001(−13)

Table 7.6: Maximum absolute error at a = b = 0, and for different α, ν1, ν2 for exam-ple(7.3.2).

In the following example, we solve a coupled system under the given conditions by the

operational matrices of Shifted Jacobi polynomials.

Example 7.3.3. Consider the the coupled system given by

cDαu(t) +Ac1Dν1u(t) +Bc

1Dν2v(t) + C1u(t) +D1v(t) = f(t), 1 < α ≤ 2, 0 < ν1, ν2 ≤ 1,

cDβv(t) +Ac2Dω1u(t) +Bc

2Dω2v(t) +C2u(t) +D2v(t) = g(t), 1 < β ≤ 2, 0 < ω1, ω2 ≤ 1,


u(0) = 0.5, u(1) = 0.5

v(0) = 0.6, v(1) = 0.6.

We solve this problems under the given parameters

A1 = 2, A2 = 2, B1 = 2, B2 = 2, C1 = 3, C2 = 4, D1 = 5, D2 = 5,

α = β = 1.8, ν1 = 0.9, ω1 = 0.8, ν2 = ω2 = 1, a = b = 0.

197

0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

4.5

5

(a)00.20.40.60.81

0

1

2

x 10−4

(b)

|uexact − uapprox|

Exact u(t)Approximate u(t)

Figure 7.16: Subplot (a) represents comparison of approximate and exact u(t) of Exam-ple(7.3.2) at α = 1.8, ν1 = 0.9, ν2 = 0.7,M = 6. The subplot (b) represents their absoluteerror at α = 1.8, ν1 = 0.9, ν2 = 0.7,M = 6.

Let the exact solutions at α = β = 2, νi = ωi = 1(i = 1, 2) is given by

u(t) = t2(ξ − t)2 +1

2, v(t) = t(ξ − t)3 +

3

5,

the source terms are given by

f(t) = 4t(2t − 2)− 6t(t− 1)2 − 5t(t− 1)3 + 2(t− 1)2 − 2(t− 1)3 + 3t2(t− 1)2

+(3172393274221175t(

75)(50t2 − 85t+ 34))

33495522228568064+ 2t2 +

9

2g(t) = 2t2(2t− 2)− 2t(t− 1)2 − 5t(t− 1)3 − 6(t− 1)2 − 2(t− 1)3 − 3t(2t− 2)

+ 3t2(t− 1)2 +9

2

We approximate the coupled system at scale level M = 6 for the given parametersa = b = 0. Absolute error for various a, b,M are given in next Table (7.7),

M a b |uapp − uexact| |vapp − vexact|4 0 0 1.0000(−3) 3.5000(−3)

6 1.0009(−4) 2.5000(−4)

6 1 1 1.0000(−4) 4.5000(−4)

6 0.5 0.5 3.0000(−4) 3.0000(−4)

8 0.5 0.5 1.4000(−4) 1.2000(−4)

8 2 2 4.0004(−4) 3.0002(−4)

Table 7.7: Maximum absolute error at various values of a, b and M for Example(7.3.3).

198

0 0.2 0.4 0.6 0.8 10.6

0.605

0.61

0.615

0.62

0.625

0.63

0.635

0.64

(a)0 0.2 0.4 0.6 0.8 1

0

0.5

1

1.5

2

2.5

3

3.5x 10−4

(b)

|uexact − uapprox|

exact u(t)approximate u(t)

Figure 7.17: Subplot (a) represents comparison of approximate and exact u(t) of Example(7.3.3). The subplot (b) represents their absolute error at M = 6.

0 0.2 0.4 0.6 0.8 1

0.5

0.51

0.52

0.53

0.54

0.55

0.56

(a)

00.20.40.60.810

0.5

1

1.5

2

2.5

3

3.5x 10−4

(b)

|vexact − vapprox|

Exact v(t)Approximate v(t)

Figure 7.18: Subplot (a) represents comparison of approximate and exact v(t) of Example(7.3.3). The subplot (b) represents their absolute error at M = 6.

199

Absolute error for specific a = b = 0 and for various α, β, ν1, ν2, ω1, ω2 are given in Table(7.8).

M α β ν1 ν2 ω1 ω2 |uapp − uexact| |vapp − vexact|4 2 2 1 1 1 1 2.5000(−2) 1.2000(−2)

6 1.9 1.9 1 1 1 1 2.0001(−4) 1.9000(−4)

6 1.9 1.9 0.9 1 0.9 1 3.5000(−4) 3.5000(−4)

8 1.8 1.8 0.8 0.8 0.8 0.8 4.5000(−4) 1.7765(−4)

8 1.9 1.9 1 1 1 1 9.0000(−5) 2.0000(−5)

10 2 2 0.9 0.7 0.5 0.5 4.5000(−8) 6.0000(−8)

Table 7.8: Maximum absolute error at specific values of a = b = 0 for Example(7.3.3).

From Table (7.7), we see the effect of a, b at various scale level M . While Table (7.8),indicates the effect of various choices of α, β, ν1, ν2, ω1, ω2 at different scale level . As thescale level increases and values of α, β, ν1, ν2, ω1, ω2 are tend to their integer values. Theabsolute error reducing and the solutions are tending to their exact value which demonstratethe applicability of the proposed method.

Example 7.3.4. Consider the mathematical model of fractionally damped coupled systemof spring masses system whose governing equation is given by

cDαu+λ

m1

c

Dν1u− µ

m1

cDν2v +

κ1m1

u− κ2m1

v = f(t), 1 < α ≤ 2, 0 < ν1, ν2 ≤ 1,

cDβv − λ

m2

c

Dω1u+λ+ µ

m2

c

Dω2v − κ1m2

u+κ1 + κ2m2

v = g(t), 1 < β ≤ 2, 0 < ω1, ω2 ≤ 1,


u(0) = 0, u(1) = 0

v(0) = 1, v(1) = −1.

Where the source terms are given by

f(t) = cos(πt)− sin(πt)− π cos(πt)− π sin(πt)− π2 sin(πt)

g(t) = sin(πt)− 2 cos(πt) + π cos(πt) + 2π sin(πt)− π2 cos(πt).

Where λ, µ are damping parameters and κ1, κ2 are spring constants and m1,m2 are massesof objects and springs are hanged from both ends. We solve this problems under the followingvalues

λ = 2, µ = 2, κ1 = 2, κ2 = 2,m1 = m2 = 2,

then calculating the coefficient of the problems as

A1 = 1, A2 = −1, B1 = −1, B2 = 2, C1 = 1, C2 = −1, D1 = −1, D2 = 1.

200

0 0.2 0.4 0.6 0.8 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(a)

0 0.2 0.4 0.6 0.8 1−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

(b)

approximate u(t) at M=8exact u(t)

approximate v(t) at M=8exact v(t)

Figure 7.19: Subplot (a) represents comparison of approximate and exact u(t) at M = 7, 8.The subplot (b)represents comparison of approximate and exact v(t) atM = 7, 8 for Example(7.3.4).

0 0.2 0.4 0.6 0.8 1

0

1

2

3

4

5

6

7

8

9

x 10−6

(a)

0 0.2 0.4 0.6 0.8 1

0

1

2

3

4

5

6

7

x 10−5

(b)

absolute error in u(t) at M=7absolute error in u(t) at M=8

absolute error in v(t) at M=7absolute error in v(t) at M=8

Figure 7.20: Subplot (a) represents absolute error |uapp−uexact| at M = 7, 8. (b) representstheir absolute error |vapp − vexact| at M = 7, 8 for Example (7.3.4) for a = b = 0.

201

M a b |uapp − uexact| |vapp − vexact| a b |uapp − uexact| |vapp − vexact|4 0 0 7.0000(−3) 6.0000(−3) 1 1 3.0000(−2) 4.0001(−2)

5 0 0 1.0000(−3) 5.0000(−3) 1 1 5.9000(−3) 6.8000(−3)

6 0 0 9.1243(−4) 9.6754(−4) 0.5 0.5 1.9000(−3) 1.5600(−3)

7 0 0 9.0000(−6) 7.0000(−5) 2 2 5.7008(−4) 3.8000(−4)

8 0 0 8.0000(−6) 6.0000(−5) 1.5 1.5 2.2400(−5) 1.2300(−5)

8 1 0.5 3.5000(−5) 1.2000(−5) −0.5 −0.5 2.4500(−5) 1.9002(−5)

Table 7.9: Maximum absolute error at various values of a, b and M for Example(7.3.4).

M α β ν1 ν2 ω1 ω2 |uapp − uexact| |vapp − vexact|6 1.8 1.9 1 1 1 1 1.6000(−3) 5.0000(−3)

6 1.8 1.9 0.9 0.7 0.9 0.7 1.0000(−3) 5.5000(−3)

6 1.99 1.99 0.99 0.99 0.99 0.99 9.1243(−4) 9.6754(−4)

8 1.8 1.8 0.8 0.8 0.8 0.8 3.7001(−5) 1.9234(−5)

9 1.9 1.9 0.9 0.9 0.9 0.9 1.6000(−6) 1.7700(−6)

10 2 2 1 1 1 1 1.6000(−8) 1.8000(−8)

Table 7.10: Maximum absolute error at a = b = 0 and for various α, β, ν1, ν2, ω1, ω2 forExample(7.3.4).

Let the exact solutions at α = β = 2, νi = ωi = 1(i = 1, 2) is given byu(t) = sin(πt), v(t) = cos(πt).

The above model is approximated for the solutions by our proposed methods. We observedthat our method provides best approximate solutions to the problems for various scale levelM. We also find numerical solutions for fractional values of α, β, νi, ωi(i = 1, 2). Weobserved that as these order tends to their integer values the solutions are tending to theexact. in the following tables

Absolute error for various a, b,M and α = β = 2, νi = ωi = 1(i = 1, 2) are given in nextTable (7.9).

We observe from the Table (7.9), that values of a, b play important rule in the ap-proximate solution. By giving integral value to a, b, the aforementioned method gives bestapproximate solutions for the consider problem. Similarly for a = b = 0 and for variouschoices of α, β, ν1, ν2, ω1, ω2 the absolute errors are given in the Table (7.10). From Table(7.10), it is obvious that when the orders of the derivatives approach to their integral values,the error is reducing and the approximate solutions are converging to the exact solutions.While the same is observed for scale level M , as scale is increasing, the absolute errors arealso decreasing and vice versa.

202

7.4 Numerical solutions of system of partial differential equa-tions

In this section, we extend the notion to the case of two variables and develop operational

matrices of fractional order integrations and differentiations to solve a generalized class of

multi-term coupled systems of fractional order partial differential equations of the form

∂αu(x, y)

∂xα+ λ1

∂α1v(x, y)

∂xα12 ∂y

α12

+ λ2∂β1u(x, y)

∂yβ1= f(x, y),

∂αv(x, y)

∂xα+ µ1

∂α2u(x, y)

∂xα22 ∂y

α22

+ µ2∂β2v(x, y)

∂yβ2= g(x, y),

(7.4.1)

subject to the initial conditions

u(i)(0, y) = fi(y), v(i)(0, y) = gi(y), i = 1 · · · n, (7.4.2)

where n− 1 < α,αi, βi ≤ n, λi, µi(i = 1, 2)

and fi, gi are all real constants and u(x, y), v(x, y), f(x, y), g(x, y) ∈ C([0, 1] × [0, 1]).

Here, we remark that our considered system uses the coupled system studied in [145] as a

special case. In the absence of source terms and taking λ1 = 0, λ2 = 1 and µ1 = 0, µ2 = 1,

the considered system reduced to a famous coupled system of Laplace equations of fractional

order by taking 1 < α = β1 = β2 ≤ 2. Moreover in the presence of external or source terms

and 1 < α = β1 = β2 ≤ 2, λ1 = µ1 = 0, the considered system reduce to a famous

coupled system of Poisson’s equation [147], of arbitrary order, which has a large number

of applications in electrostatics, mechanical engineering and theoretical physics. It is used,

for instance, to describe the potential energy field caused by a given charge or mass density

distribution etc. We reduce the coupled system of fractional order differential equations to a

coupled system of algebraic equations without using Tau-collocations method. We directly

convert the system of differential equations to algebraic equations. Thus our technique is

simple and reduces the computational complexity of the resulting algebraic system.

203

Lemma 7.4.1. Let ΦM2(x, y) be as defined in (2.5.12), then the integration of order α ofΦM2(x, y) w.r.t x is given by

Iαx (ΦM2(x, y)) ' S(α,x)M2×M2ΦM2(x, y) (7.4.3)

where S(α,x)M2×M2 is the operational matrix of integration of order α and is defined by

S(α,x)M2×M2 =

Ψ1,1,k Ψ1,2,k · · · Ψ1,p,k · · · Ψ1,M2,k

Ψ2,1,k Ψ2,2,k · · · Ω2,p,k · · · Ψ2,M2,k...

......

......

...Ψq,1,k Ψq,2,k · · · Ψq,p,k · · · Ψq,M2,k

......

......

......

ΨM2,1,k ΨM2,2,k · · · ΨM2,p,k · · · ΨM2,M2,k

,

p =Mi+ j + 1, q =Ma+ b+ 1, Ψq,p,k = ∆i,j,a,b,k for i, j, a, b = 0, 1, 2, ...,m,

∆i,j,a,b,k =a∑

k=0

δi,b(2i+1)i∑

l=0

(−1)i+l+a+kΓ(i+ l + 1)Γ(a+ k + 1)

Γ(i− l + 1)Γ2(l + 1)Γ(a− k + 1)Γ(k + α+ 1)(k + l + α+ 1),

δi,b =

0, if i 6= b,

1, if i = b..

Proof. We take Pn(x, y) as defined by (2.5.7), then the fractional integral of order α ofPn(x, y) with respect to x is given by relation

IαxPn(x, y) = IαxPa(x)Pb(y) =

a∑

k=0

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ2(k + 1)Iαx x

kPb(y),

which implies that

IαxPa(x)Pb(y) =

a∑

k=0

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ2(k + 1)Γ(k + α+ 1)xk+αPb(y) (7.4.4)

Now approximating xk+αPb(y) by Shifted Legendre polynomials, we obtain

xk+αPb(y) ≈m∑

i=0

m∑

j=0

∆i,jPi(x)Pj(y), (7.4.5)

where ∆i,j = (2i+1)(2j +1)∫ 10

∫ 10 x

k+αPb(y)Pi(x)Pj(y)dxdy. Using orthogonality relation,we obtain

∆i,j =

0, if j 6= b,

(2i + 1)

j∑

l=0

(−1)i+lΓ(i+ l + 1)

Γ(i− l + 1)Γ2(l + 1)(k + l + α+ 1), if j = b

= δi,b(2i+ 1)

i∑

l=0

(−1)i+lΓ(i+ l + 1)

Γ(i− l + 1)Γ2(l + 1)(k + l + α+ 1).

(7.4.6)

204

Hence (7.4.4) takes the form

IαxPa(x)Pb(y) =

a∑

k=0

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ2(k + 1)Γ(k + α+ 1)

m∑

i=0

m∑

j=0

∆i,j,bPi(x)Pj(y)

=m∑

i=0

m∑

j=0

a∑

k=0

(−1)a+kΓ(a+ k + 1)∆i,j,b

Γ(a− k + 1)Γ(k + 1)Γ(k + α+ 1)Pi(x)Pj(y) =

m∑

i=0

m∑

j=0

∆i,j,a,b,kPi(x)Pj(y),

(7.4.7)

where ∆i,j,a,b,k =∑a

k=0(−1)a+kΓ(a+k+1)∆i,j,b

Γ(a−k+1)Γ(k+1)Γ(k+α+1) . Using the notations p = Mi + j + 1, q =Ma+ b+ 1, Ψq,p,k = ∆i,j,b,a,k for i, j, a, b = 0, 1, 2, 3, ...,m, we get the desired result

Lemma 7.4.2. The α- order fractional derivative of ΦM2(x, y) defined in (2.5.12), w.r.t yis given by

Dαy (ΦM2(x, y)) ' T

(α,y)M2×M2ΨM2(x, y) (7.4.8)

where T(α,y)M2×M2 is the operational matrix of α order w.r.t y and is defined by

T(α,y)M2×M2 =

Ψ1,1,k Ψ1,2,k · · · Ψ1,p,k · · · Ψ1,M2,k

Ψ2,1,k Ψ2,2,k · · · Ψ2,p,k · · · Ψ2,M2,k...

......

......

...Ψq,1,k Ψq,2,k · · · Ψq,p,k · · · Ψq,M2,k

......

......

......

ΨM2,1,k ΨM2,2,k · · · ΨM2,p,k · · · ΨM2,M2,k

, (7.4.9)

p =Mi+ j + 1, q =Ma+ b+ 1, Ψq,r,k = Ωi,j,a,b,k for i, j, a, b = 0, 1, 2, ...,m and

Ωi,j,a,b,k =b∑

l=dαeδi,a

j∑

l=0

(−1)j+l+b+kΓ(j + l + 1)Γ(b + k + 1)

Γ(j − l + 1)Γ2(l + 1)Γ(b− k + 1)Γ(k + 1)Γ(k − α+ 1)(k + l − α+ 1),

with Ωi,j,a,b,k = 0, if b < α.

(7.4.10)

Proof. The proof of this lemma is similar to that of Lemma 7.4.1.

Lemma 7.4.3. The α-order fractional derivative of ΦM2(x, y) as defined in (2.5.12), w.r.tx, y is given by

∂α

∂x(α/2)∂y(α/2)(ΦM2(x, y)) ' T

(α,x,y)M2×M2ΦM2(x, y) (7.4.11)

205

where T(α,x,y)M2×M2 is the operational matrix of differentiations w.r.t. x, y and is given by

T(α,x,y)M2×M2 =

Ψ1,1,k Ψ1,2,k · · · Ψ1,p,k · · · Ψ1,M2,k

Ψ2,1,k Ψ2,2,k · · · Ψ2,p,k · · · Ψ2,M2,k...

......

......

...Ψq,1,k Ψq,2,k · · · Ψq,p,k · · · Ψq,M2,k

......

......

......

ΨM2,1,k ΨM2,2,k · · · ΨM2,p,k · · · ΨM2,M2,k

, (7.4.12)

and p =Mi+ j + 1, q =Ma+ b+ 1, Ψq,p,k = Ωi,j,a,b,k for i, j, a, b = 0, 1, 2, ...,m,

Ωi,j,a,b,k =

a∑

k=0

∆a,b,k,l,αDi,j,a,b (7.4.13)

Di,j,a,b = δi,j,a,b(2i + 1)(2j + 1)

i∑

k=0

j∑

l=0

(−1)i+j+k+lΓ(j + l + 1)Γ(i + k + 1)

Γ(j − l + 1)Γ(i − k + 1)Γ2(k + 1)Γ2(l + 1)(k + l − α/2 + 1)2,

(7.4.14)

and

∆a,b,k,l,α =a∑

k=dα/2e

b∑

l=dα/2e

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ(k + 1)

(−1)b+lΓ(b+ l + 1)

Γ(b− l + 1)Γ(l + 1)(7.4.15)

Proof. Consider the general element Pn(x, y) as defined by (2.5.7), then the fractionalderivative of order α of Pn(x, y) with respect to y and x is given by relation

∂α

∂x(α/2)∂y(α/2)Pn(x, y) =

∂α

∂x(α/2)∂y(α/2)Pa(x)Pb(y)

=

a∑

k=0

b∑

l=0

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ(k + 1)

(−1)b+lΓ(b+ l + 1)

Γ(b− l + 1)Γ(l + 1)cD

α/2x xkcDα/2

y yl

=

a∑

k=dα/2e

b∑

l=dα/2e

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ(k + 1)

(−1)b+lΓ(b+ l + 1)

Γ(b− l + 1)Γ(l + 1)

xk−α/2yl−α/2

Γ(k − α/2 + 1)Γ(l − α/2 + 1).

(7.4.16)

Approximating x(k−α/2)y(l−α/2) by Shifted Legendre polynomials,

x(k−α/2)y(l−α/2) ≈m∑

i=0

m∑

j=0

D(i,j,k,l)Pi(x)Pj(y), (7.4.17)

206

where

D(i,j,k,l) = (2i+ 1)(2j + 1)

∫ 1

0

∫ 1

0Pa(x)Pb(y)Pi(x)Pj(y)x

k−α/2yl−α/2dxdy. (7.4.18)

Using the orthogonality conditions, expansion and evaluating the integrals by convolutiontheorem of Laplace transform, we obtain

D(i,j,k,l) = δi,j,a,b(2i+ 1)(2j + 1)×i∑

k=0

j∑

l=0

(−1)i+j+k+lΓ(j + l + 1)Γ(i+ k + 1)

Γ(j − l + 1)Γ(i− k + 1)Γ2(k + 1)Γ2(l + 1)(k + l − α/2 + 1)2(7.4.19)

where

δi,j,a,b =

1, if i = a, j = b,

0, if i 6= a, j 6= b.

Hence putting (7.4.17) in (7.4.16), we have

∂α

∂x(α/2)∂y(α/2)Pn(x, y)

=

a∑

k=dα/2e

b∑

l=dα/2e

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ(k + 1)

(−1)b+lΓ(b+ l + 1)

Γ(b− l + 1)Γ(l + 1)

m∑

i=0

m∑

j=0

D(i,j,k,l)Pi(x)Pj(y)

=m∑

i=0

m∑

j=0

a∑

k=dα/2e

b∑

l=dα/2e

(−1)a+kΓ(a+ k + 1)

Γ(a− k + 1)Γ(k + 1)

(−1)b+lΓ(b+ l + 1)

Γ(b− l + 1)Γ(l + 1)D(i,j,k,l)Pi(x)Pj(y).

(7.4.20)

Now using the notation p =Ma = b+ 1, q =Mi+ j + 1 and Ψq,p,k = Ωi,j,a,b,k,lfor i, j, a, b = 0, 1, 2, ...,m, we obtain the required results.

7.4.1 Numerical solutions of the coupled systems fractional partial dif-ferential equations

We use the operational matrices to find approximate solutions of the class of coupled system

of fractional order partial differential equations (7.4.1), (7.4.2).

Approximating the fractional order partial derivatives of u(x, y) and v(x, y) by Shifted

Lagendre polynomials as follows

∂αu(x, y)

∂xα= KM2ΦM2(x, y)

∂αv(x, y)

∂xα= LM2ΦM2(x, y) (7.4.21)

207

and applying integral of order α on the corresponding equations, we have

u(x, y)−n∑

i=0

cixi = KM2S

(α,x)M2×M2ΦM2(x, y)

v(x, y)−n∑

i=0

dixi = LM2S

(α,x)M2×M2ΦM2(x, y).

(7.4.22)

Using the initial conditions, we obtain

u(x, y) = KM2S(α,x)M2×M2ΦM2(x, y) +X1

M2ΦM2(x, y)

v(x, y) = LM2S(α,x)M2×M2ΦM2(x, y) +X2

M2ΦM2(x, y),

(7.4.23)

where X1M2ΦM2(x, y) =

∑ni=0 fi(y)x

i, X2M2ΦM2(x, y) =

∑ni=0 gi(y)x

i. For simplicity of

notation, we write

KM2S(α,x)M2×M2 +X1

M2 = KM2 LM2S(α,x)M2×M2 +X2

M2 = LM2 . (7.4.24)

Hence (7.4.23) can be rewritten as

u(x, y) = KM2ΦM2(x, y) v(x, y) = LM2ΦM2(x, y) (7.4.25)

Using (7.4.25) and operational matrices of derivatives, we obtain the following estimates

∂β1u(x, y)

∂yβ1= KM2T

(β1,y)M2×M2ΦM2(x, y),

∂β2v(x, y)

∂yβ2= LM2T

(β2,y)M2×M2ΦM2(x, y),

∂α1v(x, y)

∂xα12 ∂y

α12

= LM2T(α1,x,y)M2×M2ΦM2(x, y),

∂α2u(x, y)

∂xα22 ∂y

α22

= KM2T(α2,x,y)M2×M2

f(x, y) = X3M2ΦM2(x, y), g(x, y) = X4

M2ΦM2(x, y).

(7.4.26)

Substituting in (7.4.1), we obtain

KM2ΦM2(x, y) + λ1LM2T(α1,x,y)M2×M2ΦM2(x, y) + λ2KM2T

(β1,y)M2×M2ΦM2(x, y) = X3

M2ΦM2(x, y)

LM2ΦM2(x, y) + µ1KM2T(α2 ,x,y)M2×M2ΦM2(x, y) + µ2LM2T

(β2,y)M2×M2ΦM2(x, y) = X4

M2ΦM2(x, y).

(7.4.27)

208

Then witting system (7.4.27) in matrix form as KM2ΦM2(x, y)

LM2ΦM2(x, y)

+

λ1LM2T

(α1,x,y)M2×M2ΦM2(x, y)

µ1KM2T(α2 ,x,y)M2×M2ΦM2(x, y)

+

λ2KM2T

(γ2,y)M2×M2ΦM2(x, y)

µ2LM2T(ρ2,y)M2×M2ΦM2(x, y)

−

X3

M2ΦM2(x, y)

X4M2ΦM2(x, y)

= O,

(7.4.28)

which can be written as

[KM2 LM2

]D+

[ ︷︸︸︷KM2

︷︸︸︷LM2

] OM2×M2 µ1T

(α2,x,y)M2×M2

λ1T(α1,x,y)M2×M2 OM2×M2

D

+[KM2 LM2

] λ2T

(β1,y)M2×M2 OM2×M2

OM2×M2 µ2T(β2,y)M2×M2

D−

[X3M2 X4

M2

]D = O

[KM2 LM2

]+[ ︷︸︸︷KM2

︷︸︸︷LM2

] λ2T

(β1,y)M2×M2 µ1T

(α2 ,x,y)M2×M2

λ1T(α1,x,y)M2×M2 µ2T

(β2,y)M2×M2

−

[X3M2 X4

M2

]= O

(7.4.29)

where

D =

ΦM2(x, y) OM2

OM2 ΦM2(x, y)

.

OM2 is a column vector of zeros, and OM2×M2 and O are null matrices.

On simplification (7.4.29) implies that

[KM2 LM2

]−[KM2 LM2

] λ2T

(β1,y)M2×M2 µ1T

(α2,x,y)M2×M2

λ1T(α1,x,y)M2×M2 µ2T

(β2,y)M2×M2

− [X3

M2X4M2 ] = O.

(7.4.30)

Using the values of KM2 and LM2 , we can write the above equations as

[KM2 LM2

]−[KM2 LM2

]P− [X1

M2 X2M2 ]P− [F1 F2] = O, (7.4.31)

where

P =

λ2S

(α,x)M2×M2T

(β1,y)M2×M2 µ1S

(α,x)M2×M2T

(α2,x,y)M2×M2

λ2S(α,x)M2×M2T

(α1,x,y)M2×M2 µ2S

(α,x)M2×M2T

(β2,y)M2×M2

,

209

F1 = λ2X1M2T

(β1,y)M2×M2 + λ1X

2M2T

(α1,x,y)M2×M2 −X3

M2 ,

F2 = µ1X1M2T

(α2,x)M2×M2 + µ2X

2M2T

(β2,y)M2×M2 −X4

M2 .

As (7.4.31) is an algebraic equation which can easily be solved for the unknown coefficient

matrix [KM2 LM2 ]. After calculating coefficient matrix and putting its value in (7.4.25),

we will get the approximate solution of the problem.

7.5 Illustrative Examples

In this section, we justify the applicability of our proposed method by solving some exam-

ples. For all the simulations and plotting we use Matlab.

Example 7.5.1. Consider the following coupled system of fractional partial differentialequations

∂1.8u(x, y)

∂x1.8− ∂2v(x, y)

∂x∂y− 4

∂u(x, y)

∂y= f(x, y),

∂1.8v(x, y)

∂x1.8− 6

∂2u(x, y)

∂x∂y+ 3

∂v(x, y)

∂y= g(x, y),

subject to the initial conditions u(0, y) = u′(0, y) = 0, v(0, y) = v′(0, y) = 0. Where thesource terms f(x, y) and g(x, y) are defined by

f(x, y) = 27x2 y3 (2 y − 2) (x− 1)3 + 27x3 y3 (2 y − 2) (x− 1)2 + 81x2 y2 (x− 1)3 (y − 1)2

+ 81x3 y2 (x− 1)2 (y − 1)2 − 16x4 y3 (x− 1)2 (y − 1)3 − 12x4 y4 (x− 1)2 (y − 1)2

− 89181460669789625 x115 y4 (y − 1)3

(125x2 − 175x+ 56

)

504403158265495552,

g(x, y) = 3x3 y3 (2 y − 2) (x− 1)3 − 18x3 y3 (2 y − 2) (x− 1)2 − 18x2 y3 (2 y − 2) (x− 1)3

+ 36x4 y3 (2x− 2) (y − 1)3 + 27x4 y4 (2x− 2) (y − 1)2 − 54x2 y2 (x− 1)3 (y − 1)2

− 54x3 y2 (x− 1)2 (y − 1)2 + 9x3 y2 (x− 1)3 (y − 1)2 + 144x3 y3 (x− 1)2 (y − 1)3

− 17836292133957925 x65 y3 (y − 1)2

(1250x3 − 2625x2 + 1680x − 308

)

1008806316530991104

+ 108x3 y4 (x− 1)2 (y − 1)2.

The exact solution of the above system is

u(x, y) = (1− y)[(xy)2(1− x− y + xy)]2, v(x, y) = xy(1− y)[xy(1− x− y)]2.

We approximate this problem with our proposed technique and observe that the method worksvery well, the approximate solution obtained by our method is highly accurate. Comparison

210

00.2

0.40.6

0.81

0

0.5

1−5

0

5

10

15

x 10−5

xy

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−2

0

2

4

x 10−4

xy

Exact u(x, y)

Approximate u(x, y)

Exact v(x, y)

Approximate v(x, y)

Figure 7.21: Comparison of approximate and exact u(x, y), v(x, y) at M = 10 of Example(7.5.1).

0

0.5

1

0

0.5

10

0.5

1

x 10−5

xy 0

0.5

1

0

0.5

10

1

2

3

4

x 10−6

xy

|v− v10|

|u− u10|

Figure 7.22: Absolute error in u(x, y) and v(x, y) at M = 10 of Example (7.5.1)

of the exact and the approximate solutions is shown in Figure (7.21). The absolute errorat M = 10 is also shown for u(x, y) and v(x, y) in Figure (7.22). Also in Table (7.11), wehave given the maximum absolute error for different choices of x, y.

Example 7.5.2. Consider the following coupled system of fractional order partial differen-tial equations

∂1.8u

∂x1.8− ∂2v

∂x∂y+ 11

∂0.99u

∂y0.99= f(x, y)

∂1.8v

∂x1.8+

∂2u

∂x∂y− 11

∂0.99v

∂y0.99= g(x, y),

subject to the initial conditions u(0, y) = y, u′(0, y) = y, v(0, y) = 0, v′(0, y) = ey.

211

(x, y) |u− u8| |u− u10| |v − v8| |v − v10|(0.1, 0.1) 1.707(−4) 1.005(−6) 1.8250(−4) 4.006(−6)

(0.1, 0.5) 1.956(−4) 6.006(−6) 2.596(−4) 3.508(−6)

(0.1, 0.9) 2.236(−4) 5.803(−6) 3.773(−4) 3.009(−6)

(0.5, 0.1) 1.873(−4) 4.600(−6) 4.385(−4) 4.036(−6)

(0.5, 0.5) 1.887(−4) 3.008(−6) 5.858(−4) 3.005(−6)

(0.5, 0.9) 2.092(−4) 2.519(−6) 3.619(−4) 3.001(−6)

(0.9, 0.1) 1.973(−4) 2.363(−6) 4.363(−3) 2.506(−6)

(0.9, 0.5) 2.907(−4) 5.009(−6) 5.009(−3) 2.005(−6)

(0.9, 0.9) 2.306(−3 6.000(−6) 6.000(−3) 1.500(−6)

Table 7.11: Absolute error at M = 8, 10 in u(x, y) and v(x, y) at various x,y of Example(7.5.1).

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−1

0

1

2

xy 0

0.20.4

0.60.8

1

00.2

0.40.6

0.81

−1

0

1

2

xy

Exact u(x, y)

Approximate u(x, y) Exact v(x, y)

Approximate v(x, y)


Where the source terms are given by

f(x, y) =y

x95

E1,− 45(x)− ey +

11y110

Γ(199100),

g(x, y) =ey

x45Γ(15)

+ ex − 11x

y99100

E1, 110.

The exact solution of (7.5.2) is u(x, y) = yex, v(x, y) = xey. The comparison of approx-imate and numerical results for example (7.5.2) are shown in Figure (7.23), for variousvalues of x, y and at M = 10. We also plotted the absolute error in u(x, y) and v(x, y) atM = 10 in Figure (7.24). The maximum absolute error is bellow 10−6, which shows theefficiency of our method. Also we give the absolute error for various values of x, y in Table(7.12).

212

00.2

0.40.6

0.81

00.2

0.40.6

0.810

0.5

1

1.5

2

x 10−5

xy

00.2

0.40.6

0.81

00.2

0.40.6

0.810

2

4

6

8

x 10−5

xy

|u− u10||v − v10|

Figure 7.24: Absolute error in u(x, y) and v(x, y) at M = 10 of Example (7.5.2).

(x, y) |u− u8| |u− u10| |v − v8| |v − v10|(0.1, 0.1) 1.000(−3) 2.005(−5) 1.006(−3) 8.008(−5)

(0.1, 0.5) 2.000(−3) 1.596(−5) 2.000(−3) 6.009(−5)

(0.1, 0.9) 4.006(−3) 1.003(−5) 2.126(−3) 4.019(−5)

(0.5, 0.1) 1.003(−3) 5.000(−6) 2.451(−3) 4.008(−5)

(0.5, 0.5) 4.448(−3) 4.000(−6) 1.081(−3) 3.052(−6)

(0.5, 0.9) 1.500(−3) 5.619(−6) 5.003(−4) 2.051(−6)

(0.9, 0.1) 1.527(−3) 5.063(−6) 6.050(−4) 2.094(−6)

(0.9, 0.5) 1.5307(−4) 4.129(−6) 8.008(−4) 1.096(−6)

(0.9, 0.9) 4.706(−4) 5.002(−6) 9.004(−4) 5.009(−6)

Table 7.12: Absolute error at M = 8, 10 in u(x, y) and v(x, y) at various x,y of Exam-ple(7.5.2).

Example 7.5.3.

∂1.8u

∂x1.8+

∂2v

∂x∂y+ 4

∂0.8u

∂y0.8= f(x, y),

∂1.8v

∂y1.8+ 9

∂2u

∂x∂y+ 3

∂0.8v

∂y0.8= g(x, y),

subject to initial conditions u(0, y) = −y5, u′(0, y) = 0, v(0, y) = y5, v′(0, y) = 0.

Where

f(x, y) =6(x2 + y2)x

145

Γ(115 )+

2x15

Γ(15 )+ 6x2y − 4xy +

24

Γ(165 )(x2 + y2)x

115 +

8

Γ(115 )x

65 (x3 − y3),

g(x, y) = (y2 − x2)6x

65

Γ(115 )− (x3 + y3)

2x15

Γ(65)+ 54(x2y − xy2) +

18y65

Γ(115 )(x2 − y2)− (x3 + y3)

6y15

Γ(65)

Let the exact solution of the example (7.5.3) is given by

u(x, y) = (x2 + y2)(x3 − y3), v(x, y) = (y2 − x2)(x3 + y3).

213

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−1

0

1

xy

00.2

0.40.6

0.81

00.2

0.40.6

0.81

−1

−0.5

0

0.5

1

xy

Exact u(x, y)

Approximate u(x, y)

Exact v(x, y)

Approximate v(x, y)


00.2

0.40.6

0.81

0

0.5

10

1

2

3

4

x 10−3

xy 0

0.20.4

0.60.8

1

0

0.5

10

1

2

3

4

x 10−3

xy

|u− u8| |v − v8|

Figure 7.26: Absolute error in u(x, y) and v(x, y) at M = 8 for various choices of x,y ofExample (7.5.3).

(x, y) |u− u8| |u− u10| |v − v8| |v − v10|(0.1, 0.1) 5.707(−4) 1.325(−4) 1.826(−4) 1.000(−4)

(0.1, 0.5) 5.956(−4) 1.196(−4) 2.638(−4) 2.009(−4)

(0.1, 0.9) 7.236(−4) 1.703(−4) 2.126(−4) 2.900(−4)

(0.5, 0.1) 9.873(−4) 6.038(−4) 4.451(−4) 2.945(−4)

(0.5, 0.5) 1.000(−3) 4.058(−3) 3.781(−3) 1.520(−4)

(0.5, 0.9) 2.002(−3) 9.019(−3) 3.933(−3) 3.051(−4)

(0.9, 0.1) 5.027(−3) 2.063(−3) 1.65(−3) 3.194(−3)

(0.9, 0.5) 2.007(−3) 1.029(−3) 1.8(−3) 1.096(−3)

(0.9, 0.9) 1.006(−3) 1.008(−3) 1.746(−3) 3.209(−3)

Table 7.13: Absolute error in u(x, y) and v(x, y) of Example (7.5.3) atM = 8, 10 for variouschoices of x,y.

214

We approximate the solutions of Example(7.5.3) at scale level M = 8, 10 for variouschoices of x, y. Their comparative images with exact solutions are shown in Figure (7.25).We see that our proposed method works very well and the absolute error for u(x, y) andv(x, y) are shown in Figure (7.26). The maximum absolute error at M = 8 is below 10−3,which is very low value for such types of complicated problems. In Table (7.13) for scalelevel M = 8, 10, we have given maximum absolute error at different choices of x, y. From allthese observations, we see that the method provides best approximate solutions to a varietyof such types problems of applied nature.

7.6 Comparison between Bernstein, Shifted Legendre andShifted Jacobi polynomials

Many problems can be solved, and many difficulties can be removed by appropriate choice

of the basis of a polynomials during approximation of problems. The Shifted Legendre

polynomials are obtained from The Shifted Jacobi Polynomials. Shifted Legendre polyno-

mials are orthogonal, while Shifted Jacobi polynomials are orthogonal with respect to some

Weight function. However Bernstein polynomials are famous for their geometric properties,

but they are not orthogonal. Bernstein polynomials are symmetric, and their basis form

partition of unity. Also the basis form of Bernstein polynomials is optimally stable. These

properties and others make the Bernstein polynomials important for numerical solutions

of fractional order differential equations as compared to Shifted Legendre and Jacobi ’s

polynomials, because they need orthogonality condition. Moreover, the numerical results

clearly indicate that Bernstein polynomials provide best approximate solutions to fractional

order differential equations as compared to the other mentioned polynomials.

Chapter 8

Matlab codes for numericalanalysis

In this chapter, we give some basic codes for operational matrices used for computations in

this thesis. The concerned codes and commands are developed by us for numerical analysis

of the boundary value problems of fractional differential equations.

Organizations of this chapter as follow: In section 8.1, we provide the basic codes for

computional of operational matrices and computing fractional integral and derivative of

functions f(x) by using Shifted Legendre polynomials for numerical analysis. Similarly the

codes and commands for numerical analysis using Berstein polynomials’s operational ma-

trices are provided in section8.2. The fundamentals codes of computing Jacobi polynomials

operational matrices are given in section 8.3.

8.1 Computations of operational matrices using Legendrepolynomials

The operational matrices for fractional differentiation using Legendre polynomials is devel-

oped by using the following code.

Codes to find fractional derivative of a function f(x) is

function g=fracdiff(f,alpha,x)

syms s

215

216

n=ceil(alpha);

if n==alpha

g=diff(f,x,n)

else

fo=subs(diff(f,x,n),x,s)

ss = (x− s) ∧ (alpha+ 1− n)

g = (1/gamma(n − alpha)) ∗ int(fo/ss, s, 0, x)

end

With the help of above codes, we developed the following code to generate operational

matrices HαM×M of fractional derivative :

function H=leg der(alpha,M)

syms k l i j

b=ceil(alpha)

m=M-1;

for i=0:m;

for j=0:m;

if i¡b

p(i+1,j+1)=0

else

aa = (2 ∗ j + 1) ∗ symsum(((((−1) ∧ (i+ k + j + l)) ∗ gamma(l + j + 1) ∗ gamma(i + k +

1))/((gamma(i−k+1)) ∗ (gamma(k+1)) ∗ (gamma(k−alpha+1)) ∗ (gamma(j− l+1)) ∗

((gamma(l + 1)) ∧ 2) ∗ (k + l − alpha+ 1))), l, 0, j);

P (i+ 1, j + 1) = symsum(aa, k, b, i);

end

end

217

end

H= double(H)

furthermore, we present the following codes to compute fractional order integration of a

function f(x)

function g=fracint(f,alpha)

syms x t

ft=subs(f,x,t)

ss = (t− x) ∧ (alpha− 1)/(gamma(alpha))

g = int(ft ∗ ss, t, 0, x)

Next code is developed for computation of operational matrices of fractional order

integration PαM×M

function P=leg inta(alpha,M)

syms k l

m=M-1;

for i=0:m;

for j=0:m;

aa = (2 ∗ j + 1) ∗ symsum(((((−1) ∧ (i+ k + j + l)) ∗ gamma(l + j + 1) ∗ gamma(i + k +

1))/((gamma(i−k+1)) ∗ (gamma(k+1)) ∗ (gamma(k+alpha+1)) ∗ (gamma(j− l+1)) ∗

((gamma(l + 1))2) ∗ (k + l + alpha+ 1))), l, 0, j); P(i+1,j+1)=symsum(aa ,k,0,i);

end

end

P= double(P) The important operational matrix Qα,φM×M for boundary value problem is

calculated this code function Qc=Qmtx(phi,alpha,M)

syms x

Q=def int(alpha,M)

QP = Q ∗ phi ∗ x ∧ 0

218

for i=1:M

Qa(i, :) = coefvect(QP (i),M)

end

Qc=double(Qa)

8.2 Computations of operational matrices using Bernsteinpolynomials

The following codes are used to compute the fractional order derivative and integration as

well as their operational matrices using Bernstein polynomials

computation of HαM×M of fractional order derivative

function H=bern der(alpha,N)

syms k l

load INVMATRIX.mat

A=INVMATRIX;

B=INVMATRIX:,N-2;

Bi=inv(B);

for i=0:N

for j=0:N

if i¡ ceil(alpha)

H(i+ 1, j + 1) = symsum((symsum(((((−1) ∧ k) ∗ (gamma(N + 1)/(gamma(N − i+ 1) ∗

gamma(i+1)))∗(gamma(N −i+1)/(gamma(N −i−k+1)∗gamma(k+1))))∗(((−1)∧ l)∗

(gamma(N +1)/(gamma(N − j+1) ∗ gamma(j +1))) ∗ (gamma(N − j+1)/(gamma(N −

j − l+1) ∗ gamma(l+1)))) ∗ (gamma(i+ k+1)/(gamma(i+ k− alpha+1) ∗ (i+ j + k+

l − alpha+ 1)))), l, 0, N − j)), k, ceil(alpha), N − i)

else

H(i+ 1, j + 1) = symsum((symsum(((((−1) ∧ k) ∗ (gamma(N + 1)/(gamma(N − i+ 1) ∗

219



j − l+1) ∗ gamma(l+1)))) ∗ (gamma(i+ k+1)/(gamma(i+ k− alpha+1) ∗ (i+ j + k+

l − alpha+ 1)))), l, 0, N − j)), k, 0, N − i)

end

end

end

H=double(H)*Bi

Where the Inversive matrix is calculated by using this code

syms x

for M=3:15

F=bren funvec(M,x)

Fo = F ∗ transpose(F )

INVMATRIX :,M − 2 = int(Fo, 0, 1)

end Next, we give code for computation of fractional order integration PαM×M for Bernstein

polynomials function P=bern int(alpha,N)

syms k l

load INVMATRIX.mat

A=INVMATRIX;

B = INVMATRIX:, N − 2;

Bi=inv(B);

for i=0:N

for j=0:N

D(i+ 1, j + 1) = symsum((symsum(((((−1) ∧ k) ∗ (gamma(N + 1)/(gamma(N − i+ 1) ∗



220

j − l+1) ∗ gamma(l+1)))) ∗ (gamma(i+ k+1)/(gamma(i+ k+ alpha+1) ∗ (i+ j + k+

l + alpha+ 1)))), l, 0, N − j)), k, 0, N − i)

end

end

P=double(P)*Bi

Next code is used to Compute Qα,φM×M

function Q=bern Qmtx(alpha,phi,N)

syms x

C=bern constant(alpha,N)

fg=phi*transpose(C)

for i=1:N+1

ac=bern approx(fg(i,1),N,x)

for j=1:N+1

A(i,j)=ac(j)

end

end

Q=A

8.3 Computation of operational matrices using Shifted Ja-

cobi polynomials

In this section, we present the code to compute corresponding operational matrices of

fractional order derivative and integration by using Shifted Jacobi polynomials. Here the

following code is used to compute HαM×M

function H=jacob der(alpha,m,a,b,eta)

syms k l

for i=0:m-1;

221

for j=0:m-1;

aa = ((−1) ∧ (i− k)) ∗ gamma(i+ b+ 1) ∗ gamma(i+ k + a+ b+ 1);

bb = gamma(k+ b+1) ∗ gamma(i+ a+ b+1) ∗ gamma(i− k+1) ∗ gamma(k− alpha+1);

cc = ((−1)(j − l)) ∗ gamma(j + l+ a+ b+1) ∗ gamma(a+1) ∗ gamma(l+ k− alpha+ b+

1) ∗ (2 ∗ j + a+ b+ 1) ∗ gamma(j + 1) ∗ (eta( − alpha));

dd = gamma(j+ a+1) ∗ gamma(l+ b+1) ∗ gamma(j− l+1) ∗ gamma(l+1) ∗ gamma(l+

k + a+ b− alpha+ 2);

H(i+ 1, j + 1) = symsum(((aa/bb) ∗ symsum((cc/dd), l, 0, j)), k, 1, i);

end

end

The proceeding code is used to compute PαM×M

function P=jacob inta(alpha,m,a,b,eta)

syms k l

for i=0:m-1;

for j=0:m-1;

aa = ((−1) ∧ (i− k)) ∗ gamma(i+ b+ 1) ∗ gamma(i+ k + a+ b+ 1);

bb = gamma(k+ b+1) ∗ gamma(i+ a+ b+1) ∗ gamma(i− k+1) ∗ gamma(k+ alpha+1);

cc = ((−1) ∧ (j − l)) ∗ gamma(j + l+ a+ b+ 1) ∗ gamma(a+ 1) ∗ gamma(l + k + alpha+

b+ 1) ∗ (2 ∗ j + a+ b+ 1) ∗ gamma(j + 1) ∗ (etaalpha);

dd = gamma(j+ a+1) ∗ gamma(l+ b+1) ∗ gamma(j− l+1) ∗ gamma(l+1) ∗ gamma(l+

k + a+ b+ alpha+ 2);

P (i+ 1, j + 1) = symsum(((aa/bb) ∗ symsum((cc/dd), l, 0, j)), k, 0, i);

end

end

Next, we present code for computation of Q(α,c,φ)M×M

222

function Q=Qmtx manul(c,n,alpha,m,a,b,eta)

syms x

phi = c ∗ x ∧ n

B = jacobfunvector(m,x, a, b, eta);

P = jacobinta(alpha,m, a, b, eta);

PB = P ∗ B

PBeta = subs(PB, x, eta)

Pphi = PBeta ∗ phi

for i=1:m

Q(i,:)=jacob approx(Pphi(i),m,x,a,b,eta)

end

Q=double(Q)

To compute Q(α,c,φ)M×M , the following code is also used

function Q=Qmtx m(c,n,alpha,m,a,b,eta)

syms k l

for i=0:m-1;

for j=0:m-1;

aa = ((−1)(i− k)) ∗ gamma(i + b+ 1) ∗ gamma(i + k + a+ b+ 1) ∗ (eta(k + alpha)) ∗ c;

bb = gamma(k+ b+1) ∗ gamma(i+ a+ b+1) ∗ gamma(i− k+1) ∗ gamma(k+ alpha+1);

aabb = symsum(aa/bb, k, 0, i)

cc = ((−1) ∧ (j − l)) ∗ gamma(j + b+ 1) ∗ gamma(j + l + a+ b+ 1) ∗ gamma(n+ b+ 1) ∗

gamma(1 + a) ∗ (eta(n+ b+ a+ 1− l));

dd = gamma(j+a+b+1)∗gamma(l+b+1)∗gamma(j−l+1)∗gamma(l+1)∗gamma(n+

a+ b+ 1);

ccdd = symsum(cc/dd, l, 0, j)

or = ort(j + 1, a, b, eta);

223

Q(i+ 1, j + 1) = (1/or) ∗ aabb ∗ ccdd

end

end

Q=double(Q)

Where orthogonality is computed by using

function B=ort(m,a,b,eta)

n=m-1;

B = double(eta∧ (a+ b+1) ∗ gamma(n+ a+1) ∗ gamma(n+ b+1))/((2 ∗ n+ a+ b+1) ∗

gamma(n+ 1) ∗ gamma(n+ a+ b+ 1));

Chapter 9

Summary and conclusion

This thesis is consisted of nine chapters. In chapter 1, we provide a brief introduction in-

cluding short history of fractional calculus.

In chapter 2, some special functions are introduced which are needed in this study. In the

same chapter, definitions of fractional derivative both in Riemann-Liouville and Caputo’s

sense, are given and some properties of fractional integration and derivatives are also pro-

vided. In section 2.3, some results from analysis which are required in this thesis are given

and in the same section several fixed point theorems are stated on which the existence the-

ory of our study is based. In section 2.4, basic concepts about topological degree theory are

provided. Moreover, in the aforesaid chapter, some orthogonal polynomials are introduced

in the last section 2.5.

In chapter 3, with the help of classical fixed point theorems such as Schauder’s fixed point

theorem and Banach contraction principle, we established necessary and sufficient conditions

for existence and uniqueness of positive solution for multi-point boundary value problems

of two different kinds. Also conditions for existence of triple solutions and properties of

Green’s function are provided in the same chapter. Furthermore in the same chapter, we

have also developed some necessary and sufficient conditions for the existence and unique-

ness of solution for a multi -point nonlocal Cauchy problem of a class of nonlinear fractional

differential equations by means of topological degree theory.

224

225

In chapter 4, a detailed study of the existence and uniqueness of positive solutions for sev-

eral coupled systems of boundary value problems are provided. We studied various coupled

systems with different boundary conditions for their existence of solutions. For existence

and uniqueness, we have defined cone and Banach spaces for product spaces in which the

existence of solutions for system of fractional differential equations with multi-point bound-

ary conditions, movable type integral boundary conditions, integral boundary conditions,

impulsive boundary conditions. For this study, we impose some growth conditions on the

nonlinear functions involved in the systems of boundary value problems of fractional differ-

ential equations. Based on these conditions we derive sufficient conditions for existence of at

least one and uniqueness solution. Moreover in this chapter, we also studied two systems of

nonlinear fractional differential equations with different type nonlocal multi-point boundary

conditions with the help of topological degree theory.

Chapter 5 is devoted to the study of existence of multiple solutions for a coupled systems

of differential equations of fractional order. A coupled system with multi-point boundary

conditions is studied. Some growth conditions were imposed on nonlinear functions involved

in the system. Based on these conditions necessary and sufficient conditions for existence

of at least one solutions, two solutions and m-solutions are studied. Further, non-existence

of positive solution is also studied in the same chapter. For the validity of these established

results, several examples have been provided.

Monotone iterative technique have been studied in chapter 6. In this chapter, first a system

with two-point boundary conditions have been studied by using monotone iterative tech-

nique together with the method of upper and lower solutions. In the aforesaid chapter, we

use monotone iterative technique together with the method of upper and lower solutions

to establish necessary and sufficient conditions for the existence of iterative solutions for

coupled system with three-point boundary value problems. In the last section of chapter 6,

a coupled system with coupled m-point boundary conditions has been studied. Sufficient

226

conditions for existence of upper and lower solutions were studied by using comparison the-

orem on cone expansions. Two examples are provided to illustrate these results.

As always it is impossible for every system or problem of fractional differential equations to

obtain its exact solutions, due to the complexity of fractional differential operators involved

in these differential equations . So an efficient numerical scheme is needed to obtain the

best approximate solutions which have a close agreement with that of its exact solutions

of the concerned differential equation. In this thesis, we have used three well-known poly-

nomials, the Shifted Legendre’s polynomials, the Bernstein’s polynomials and the Shifted

Jacobi’s polynomials to establish operational matrices of fractional order derivative HσM×M

and integration PσM×M . Moreover, an operational matrix Q

(c,σ,φ)M×M is introduced correspond-

ing to the boundary conditions which is of great need for numerical solutions of boundary

value problems of fractional differential equations. Based on these operational matrices,

we reduced the system of differential equations to a system of simple algebraic equations

which are easily soluble for the un-known coefficient matrices required for the approxima-

tion of solutions for the concerned system of boundary value problems. We established the

mentioned numerical scheme without using the tau-collocation method. Also, comparison

of the results with that of exact solutions are done and their respective graphs are shown

by using Matlab. From the graphs, we observed that our scheme provided best accuracy

for the numerical solutions of system of boundary value problems of fractional differential

equations. Moreover, the scheme is extended to solve a highly generalized coupled system

of fractional order partial differential equations. From the observations, we examined that

the scheme provided very accurate approximations for the concerned coupled system of

fractional order partial differential equations. The efficiency and accuracy of the numerical

scheme was checked at various scale level, from which it is concluded that accuracy may

increase with the increase of scale level and vice versa. The mentioned numerical analysis

227

is given in chapter 7. A comparison between the polynomials approximation has been pro-

vided in last section of Chapter 7.

While chapter 8 is concerned with the Matlab codes and commands developed during this

research for numerical analysis of boundary value problems of fractional order differential

equations and their systems. The numerical analysis is carried out by mean of these codes.

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