Boundary Value Problems for Fractional Diﬀerential Equationsprr.hec.gov.pk/jspui/bitstream/123456789/2267/1/2508S.pdf · Boundary Value Problems for Fractional Diﬀerential Equations:

Boundary Value Problems for

Fractional Differential Equations:

Existence Theory and Numerical Solutions

Mujeeb ur Rehman

Centre for Advanced Mathematics and Physics

National University of Sciences and Technology

PhD Thesis

2011

Boundary Value Problems for

Fractional Differential Equations:

Existence Theory and Numerical Solutions

by

Mujeeb ur Rehman

Supervised by

Dr. Rahmat Ali Khan

Centre for Advanced Mathematics and Physics,

National University of Sciences and Technology, Islamabad

A thesis submitted for the degree of

Doctor of Philosophy

c⃝ Mujeeb ur Rehman 2011

Abstract

Fractional calculus can be considered as supper set of conventional calculus in the sense that it extends

the concepts of integer order differentiation and integration to an arbitrary (real or complex) order. This

thesis aims at existence theory and numerical solutions to fractional differential equations. Particular focus

of interest are the boundary value problems for fractional order differential equations. This thesis begins

with the introduction to some basic concepts, notations and definitions from fractional calculus, functional

analysis and the theory of wavelets. Existence and uniqueness results are established for boundary value

problems that include, two–point, three–point and multi–point problems. Sufficient conditions for the

existence of positive solutions and multiple positive solutions to scalar and systems of fractional differential

equations are established using the Guo–Krasnoselskii cone expansion and compression theorems.

Owning to the increasing use of fractional differential equations in basic sciences and engineering,

there exists strong motivation to develop efficient, reliable numerical methods. In this work wavelets are

used to develop a numerical scheme for solution of the boundary value problems for fractional ordinary

and partial differential equations. Some new operational matrices are developed and used to reduce the

boundary value problems to system of algebraic equations. Matlab programmes are developed to compute

the operational matrices. The simplicity and efficiency of the wavelet method is demonstrated by aid of

several examples and comparisons are made between exact and numerical solutions.

i

Acknowledgements

First and foremost, I would like to thank National University of Sciences and Technology (NUST) for its

financial support. I would like to express my gratitude to my Ph.D supervisor Dr. Rehmat Ali Khan for

his guidance and encouragements during the whole course of my studies leading to this thesis. I would like

to thank Prof. Asghar Qadir whose cardial, rigorous and elegant lectures on special functions sparkled

my interest in the field of fractional calculus. During my research phase, I was in constant need of getting

some books on the subject of fractional calculus and fractional differential equations. In this context, I am

also very thankful to Prof. Faiz Ahmad, who managed to provide me a number of good books despite the

limited availability of funds. I want to express my thanks to my colleagues for their encouragements and

many useful discussions. I am thankful to Dr. Tyab Kamran, Dr. M. Rafique, Dr. Rashid Farooq and

Dr. Jamil Raza for their moral support and encouragements. I am grateful to Principle CAMP, Professor

Azad Akhtar Siddiqui for providing an impressive research environment at the centre. I am also thankful

to the University of Malakand for its hospitality during my short stay there.

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.

Dedicated to My Mother

Contents

1 Introduction 1

2 Preliminaries 7

2.1 Some special functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.1 Euler’s gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.2 Mittag–Leffler function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Fractional calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 The Riemann–Liouville fractional integration . . . . . . . . . . . . . . . . . . . . . 11

2.2.2 The Riemann–Liouville fractional derivative . . . . . . . . . . . . . . . . . . . . . . 13

2.2.3 The Caputo fractional derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Fixed Point Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 Wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4.1 The Haar scaling function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.4.2 Multiresolution Analysis (MRA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4.3 The Haar wavelet function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.4.4 Orthogonality of the Haar wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.4.5 Function approximation by the Haar wavelets . . . . . . . . . . . . . . . . . . . . . 28

2.4.6 Error analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Existence and uniqueness of solutions 31

3.1 Two–point boundary value problems (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 Two–point boundary value problems (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3 Three–point boundary value problems (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.3.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3.2 Uniqueness of solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4 Three–point boundary value problems (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . 43



3.5 Multi–point boundary value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48



iv

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3.6 Boundary value problems with integral boundary conditions . . . . . . . . . . . . . . . . . 53



4 Existence and multiplicity of positive solutions 59

4.1 Positive solutions for three–point boundary value problems (I) . . . . . . . . . . . . . . . . 60

4.1.1 Green’s function and its properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

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

4.1.3 Existence of at least two positive solutions . . . . . . . . . . . . . . . . . . . . . . . 64

4.1.4 Existence of at least three positive solutions . . . . . . . . . . . . . . . . . . . . . . 65

4.2 Positive solutions to three–point boundary value problems (II) . . . . . . . . . . . . . . . 67


4.2.2 Existence of positive solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.2.3 Uniqueness of positive solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5 Existence and multiplicity of positive solutions for systems of fractional differential

equations 74

5.1 Positive solutions for a coupled system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74


5.1.2 Existence of at least one positive solution . . . . . . . . . . . . . . . . . . . . . . 76


5.2 Positive solutions to a system of fractional differential equations with three–point boundary

conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.2.1 Greens’s function and its properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.2.2 Existence of at least one positive solution . . . . . . . . . . . . . . . . . . . . . . . 88


6 Numerical solutions to fractional differential equations by the Haar wavelets 93

6.1 Numerical solutions to fractional ordinary differential equations . . . . . . . . . . . . . . . 93

6.1.1 Linear fractional differential equations with constant coefficients . . . . . . . . . . . 99

6.1.2 Linear fractional differential equations with variable coefficients . . . . . . . . . . . 108

6.2 Numerical solutions to fractional partial differential equations . . . . . . . . . . . . . . . . 114

6.2.1 Fractional partial differential equations with constant coefficients . . . . . . . . . . 114

6.2.2 Fractional partial differential equations with variable coefficients . . . . . . . . . . 118

7 Numerical solutions to fractional differential equations by the Legendre wavelets 125

7.0.3 The Legendre wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.0.4 Function approximations by the Legendre wavelets . . . . . . . . . . . . . . . . . . 126

7.1 An operational matrices of fractional order integration . . . . . . . . . . . . . . . . . . . . 127

7.2 Numerical solutions of fractional differential equations . . . . . . . . . . . . . . . . . . . . 128

vi

8 Conclusions 135

A Matlab and Mathematica programs 137

A.1 Computations of some operational matrices by Matlab . . . . . . . . . . . . . . . . . . . . 137

A.2 Computations by Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

B Useful results from Analysis 144

References 146

List of Tables

6.1 Absolute error for different values of m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.2 Maximum absolute error for α = 2, β = 0, 1 and different values of m, ω. . . . . . . . . . . 102

6.3 Absolute error for m = 32 and α = 1.2, 1.4, 1.6, 1.8, 2. . . . . . . . . . . . . . . . . . . . . . 105

6.4 Absolute error for α = 32 and different values of m. . . . . . . . . . . . . . . . . . . . . . . 106

6.5 Absolute error for different values of m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.6 Absolute error for different values of m and α = 114 . . . . . . . . . . . . . . . . . . . . . . 110

6.7 The maximum absolute error for m = 32, and different values of α and β. . . . . . . . . . 111

6.8 For problem (6.1.85), absolute error for m = 32 and different values of α. . . . . . . . . . 114

6.9 Maximum absolute error for different values of J and α. . . . . . . . . . . . . . . . . . . . 118

6.10 The Haar wavelet solutions and solutions obtained in [105], using ADM and VIM. . . . . 121

7.1 Absolute error for M = 3 and different values of k. . . . . . . . . . . . . . . . . . . . . . . 129

7.2 Absolute error for M = 3 and different values of k. . . . . . . . . . . . . . . . . . . . . . . 132

7.3 Numerical results with comparison to Ref. [104] and [12]. . . . . . . . . . . . . . . . . . . . 133

7.4 Maximum absolute error for the Haar wavelet and the Legendre wavelets. . . . . . . . . . 133

7.5 The absolute error for M = 3, k = 3 and different values of α. . . . . . . . . . . . . . . . 134

vii

List of Figures

2.1 Gamma function and its reciprocal. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 The Mittag–Leffler function Eα,β(−(3x)2), for β = 1/2 and different values of α. . . . . . . . . . 10

2.3 Fractional order integrals and derivatives of some elementary functions. . . . . . . . . . . . 15

2.4 The Maxican hat ψ(t) = (1− t2)e−12 t

2

and its dilated shifts. . . . . . . . . . . . . . . . . . . . 26

2.5 Haar wavelets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.6 Approximating f(t) = sin(9t) + 2 cos(11t) + 12 sin(50t) by the Haar wavelets. . . . . . . . . . . . 29

6.1 Exact and numerical solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.2 Solutions y(t) of problem (6.1.29), for 1 ≤ α ≤ 2, y0 = cosωπ − 1. . . . . . . . . . . . . . . 103

6.3 Numerical solutions of problem (6.1.38) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

6.4 Exact and Numerical solutions for problem (6.1.5) . . . . . . . . . . . . . . . . . . . . . . 105

6.5 Exact and Numerical solutions for problem (6.1.48) . . . . . . . . . . . . . . . . . . . . . . 107

6.6 Exact and Numerical solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

6.7 Exact and Numerical solutions for problem (6.1.73), (6.1.74) . . . . . . . . . . . . . . . . . 110

6.8 Numerical and exact solutions for the boundary value problems (6.1.73), (6.1.74) and

(6.1.75), (6.1.76) ((c)-(f)). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.9 Exact and Numerical solutions for problem (6.1.85). . . . . . . . . . . . . . . . . . . . . . 113

6.10 Numerical and exact solutions for telegraph equation (6.2.12). . . . . . . . . . . . . . . . . 116

6.11 Solutions for (6.2.13) and (6.2.15) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.12 Numerical and exact solutions for the equation (6.2.14) for α = 0.3, J = 5.0 . . . . . . . . 117

6.13 The absolute error between Haar wavelet solution and analytic solution. . . . . . . . . . . 120

6.14 Exact and numerical solutions for different values of J , α, β and γ. . . . . . . . . . . . . 123

6.15 The absolute error between exact and numerical solutions for different values of α, β and γ. 124

7.1 The Legendre wavelets for M = 3, k = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

7.2 Numerical and exact solutions of problem (7.2.1), (7.2.2) for 1 ≤ α ≤ 2. . . . . . . . . . . 128

7.3 Solutions y(t) for Example 7.2.2 for ω = 11, y0 = 0, y1 = 1 and different values of α, β. . . 131

viii

List of publications from thesis

1. Mujeeb ur Rehman and R.A. Khan, Existence and uniqueness of solutions for multi–point boundary

value problem for fractional differential equations, Appl. Math. Lett., 23 (2010) 1038–1044.

2. Mujeeb ur Rehman, R.A. Khan and Naseer Ahmad Asif, Three point boundary value problems for

nonlinear fractional differential equations, Acta Math. Sci., 31 (2011).

3. Mujeeb ur Rehman and R.A. Khan, Positive Solutions to Coupled system of fractional differential

equations, Int. J. Nonlin. Sci., 10 (2010) 96–104.

4. Mujeeb ur Rehman and R.A. Khan, Positive solutions to nonlinear higher-order nonlocal boundary

value problems for fractional differential equations, Abs. Appl. Anal., Vol. 2010, Article ID 501230,

15 pages doi:10.1155/2010/501230

5. R.A. Khan, Mujeeb ur Rehman and J. Hendersom, Existence and uniqueness of solutions for non-

linear fractional differential equations with integral boundary conditions, Fract. Diff. Cal., 1 (2011)

29–43.

6. R. A. Khan, Mujeeb ur Rehman, Existence of multiple positive solutions for a general system of

fractional differential equations, Commun. Appl. Nonlin. Anal. 18 (2011) 25–35.

7. Mujeeb ur Rehman and R.A. Khan, The Legendre wavelet method for solving fractional differential

equations, Commun. Nonlin. Sci. Numer. Simulat., 16 (2011) 4163–4173.

8. Mujeeb ur Rehman and R.A. Khan, A numerical method for solving boundary value problems for

fractional differential equations, Appl. Math. Mod., (2011), doi: 10.1016/j.apm.2011.07.045

9. Mujeeb ur Rehman and R.A. Khan, Existence and uniqueness of solutions for fractional order dif-

ferential equations with nonlocal boundary conditions, Int. J. Math. Anal., (Accepted).

10. Mujeeb ur Rehman, R.A. Khan and Paul W. Eloe, Positive solutions to three-point boundary value

problem for higher order fractional differential system, Dyn. Syst. Appl., (Accepted)

11. Mujeeb ur Rehman and R.A. Khan, Numerical solutions to a class of partial fractional differential

equations (submitted)

12. Mujeeb ur Rehman, Numerical solutions to initial and boundary value problems for fractional partial

differential equations with variable coefficients (submitted)

ix

Chapter 1

Introduction

The discovery of differential calculus is attributed to Isaac Newton (1642-1727) and Gottfried Leibniz

(1646-1716) who independently developed the foundations of the subject in the seventeenth century. The

first mention of the fractional calculus can be traced back to a letter exchange between Leibniz and a

French mathematician Marquis de L’Hôpital. Leibniz introduced the notation dnydxn (still used today) for

nth order derivative with the assumption that n ∈ N and repotted this to L’Hôpital. In his letter L’Hôpital

posed the question to Leibniz “What would be the result if n = 12?” Leibniz in his replay, dated 30th

September 1695, writes “... this is an apparent paradox from which, one day, useful consequences will

be drawn. Since there are little paradoxes without usefulness. ... ”. S. F. de Lacroix (1819) for the

first time introduced the fractional derivatives in published text. Subsequent contributions to fractional

calculus were made by many great mathematicians of the time such as J.P.J. Fourier (1822), N.H. Abel

(1823-1826), J. Liouville (1832), B. Riemann (1847), A.K. Grunwald (1867), A.V. Letnikov (1868), J.

Hadamard (1892), O. Heaviside (1892), H. Weyl (1917), A. Erdélyi (1939), H. Kober (1940) and M. Riesz

(1949). Excellent summary of key milestones in the history of fractional calculus can be found in [110,134].

Recently, in a survey report [134], J. T. Machado, V. Kiryakova and F. Mainardi have comprehensively

listed the major documents and key events in this area of mathematics since 1974 up to April 2010.

Fractional calculus has long and rich history, but due to lack of suitable physical and geometrical inter-

pretations, it remained unfamiliar to applied scientists up to recent years and was considered mathematical

curiosities, not useful for solving problems arising from applied sciences. Several attempts have been made

to provide physical and geometric interpretations for fractional operators. However, these interpretations

are limited to only a small collection of selected applications of fractional derivatives and integrals in the

context of hereditary effects and self-similarity. In 2002, I. Podlubny [115] proposed a convincing physical

and geometric interpretation of fractional derivatives and integrals.

There are several competing definitions of fractional derivatives and integrals. Some of them include,

the Riemann, the Liouville, the Riemann–Liouville, the Caputo, the Weyl, the Hadamard, the Marchaud,

the Gränwald-Letnikov, the Erdélyi-Kober and the Riesz-Feller fractional derivatives and integrals. In

general, these definitions are not equivalent except for some special cases. Probably the most frequently

used definition of fractional derivative and integral is due to B. Riemann and J. Liouville, commonly

known as the Riemann-Liouville fractional derivative (integral). But in some situations, this approach is

1

2

not useful due to lack of physical interpretation of initial and boundary conditions involving fractional

derivatives, and also the Riemann-Liouville approach may yield derivative of a constant different from zero.

A useful alternative to Riemann-Liouville derivative is the Caputo fractional derivative, introduced by M.

Caputo in 1967 and adopted by Caputo and Mainardi 1969 in the context of the theory of viscoelasticity.

Fractional derivatives are non–local in nature. Local fractional derivatives have been proposed in [13, 28]

to study properties of irregular functions.

As a first application of fractional calculus, consider the problem of determination of shape of a

frictionless wire lying in a vertical plane such that the time required for a bead placed on the wire to slide

to the lowest point of the wire is the same regardless of its starting position (tautochronous problem). This

problem can be formulated by the integral equation√2gT =

∫ y

η=0(y − n)−

12ds,

where g is gravitational acceleration, (x, y) is initial position. The arc length s may be expressed as a

function of the y, say s = H(η). Thus

T√

2g =

∫ η=y

η=0(y − η)−

12H ′(η)dη, or T

√2g = Γ

(12

)I

120 H

′(y).

In 1823, N.H. Abel solved this problem by applying operator D120 on both sides of the above equation and

obtained

T√

2gD120 1 =

√πH ′(y).

By computing the fractional derivative of constant,

H ′(y) =ds

dy=T√2g

π√y.

The fact that the constant functions may have nonzero fractional derivative, is not always a drawback.

Because the Abel’s solution of the tatochronous problem rests on this fact.

For almost three centuries fractional calculus had been treated as an interesting, but abstract, mathe-

matical concept. It had significantly been developed with in pure mathematics. However the applications

of the fractional calculus just emerged in last few decades in several diverse areas of sciences, such as

physics, bio-sciences, chemistry and engineering. It is realized widely that in many situations fractional

derivative based models are much better than integer order models. Being nonlocal in nature, the frac-

tional derivatives provide an excellent tool for the understanding of memory and hereditary properties

of various materials and processes. This is the main advantage of fractional derivatives in comparison

with classical integer order derivatives. A new application field for fractional calculus is psychological and

life sciences, to characterize the time variation of emotions of people [6, 132]. In addition to the above

mentioned applications, there are several applications of fractional calculus within different fields of math-

ematics itself. For example, the fractional operators are useful for the analytic investigation of various

spacial functions [70, 71]. There are several collections of articles such as [25, 54, 119], which exhibit wide

variety of applications of fractional calculus and present many of the key developments of the theory. The

3

list of applications of fractional calculus is still growing; perhaps “the fractional calculus is the calculus of

twenty–first century”.

There are a number of books and monographs dealing with theory and applications of fractional

calculus. For the first book, the merit is ascribed to K.B. Oldham and J. Spanier [110] which gives a

historical survey and comprehensive overview on the topic of fractional calculus. The text by K.S Miller and

B. Ross [103] provides an easy introduction to the theory of fractional derivatives and fractional differential

equations. The book of I. Podlubny [114] deals particularly with fractional differential equations and their

applications. S. Samko, A. Kilbas and O. Marichev [124] published an encyclopedic type monograph in

Russian in 1987 and in English in 1993. In addition numerous other works have also been appeared. These

include A.A. Kilbas, H.M. Srivastava, J.J. Trujillo [69], A. Carpinteri and F. Mainardi [25], F. Mainardi

[101], R. Hilfer [54], J. Sabatier, O.P. Agrawal and J.A. Tenreiro Machado [119], V. Lakshmikantham, S.

Leela, J. Vasundhara Devi [77] and K. Diethelm [39].

We make a remark about notations used for fractional differentiation and integration. As pointed

out by K.S Miller and B. Ross [103] “...some of the power and elegance of fractional calculus rests in its

simplified notation.” R. Hilfer [57] have summarized the notations used by various authors for fractional

derivatives and integrals. In this work we will denote the Riemann–Liouville fractional integral by Iαa ,

Riemann–Liouville derivative by Dαa and the Caputo derivative by cDα

a .

The thesis is organized as follows: In Chapter 2 we recall some basic results from spacial functions,

fractional calculus, fixed point theory and wavelet analysis that form basis for our further investigations.

The Chapter begins with a brief introduction to the gamma function and the Mittage-Lefller function.

Analytic results of fractional calculus, frequently needed for our intended investigations in the succeeding

chapters, are briefly discussed. For most of the results the concise and rigorous proofs are established using

fundamental properties of fractional operators. Comparisons of Riemann–Liouville and caputo approaches

are occasionally provided. We also state some commonly used fixed point theorems needed to establish

the existence results in the next three chapters, corresponding to boundary value problems for fractional

differential equations.

Due to frequent applications of fractional differential equations in many standard models, there has

been significant interest for obtaining exact analytical and numerical solutions for them. The exact so-

lutions of initial value problems for fractional differential equations have been investigated by classical

integral transform methods, such as the Laplace transform method, the Fourier transform method and the

Mellin transform method [103,114]. Undoubtedly, “boundary value problems” for classical as well as frac-

tional differential equations is one of the fundamental topic and an active area of research. In general there

exist no method to find an exact analytic solution for boundary value problems for fractional differential

equations. Several numerical methods for solving integer order differential equations have been generalized

to solve initial value problems for fractional differential equations. In contrast, boundary value problems

have received much less attention and few results can be found concerning their numerical solutions. So

there exists a strong motivation to develop an efficient numerical technique for the treatment of boundary

value problems for fractional differential equation.

The concept of approximating complicated functions, which are known implicitly via differential or

4

integral equations, with simpler functions plays a decisive role in many areas of modern mathematics and

its applications. In particular, if solution y(x) to some differential equation belongs to certain class of

functions, then we are interested to find basis functions ϕ0(x), ϕ1(x), ϕ2(x), . . . such that y has represen-

tation

y(x) =

∞∑j=0

cjϕj(x), (1.0.1)

for some coefficients cj . If (1.0.1) holds, one might hope that the finite partial sum

y(x) ≈n∑j=0

cjϕj(x), for some n ∈ N, (1.0.2)

approximates y well. This idea is similar to that of power series (or the Fourier series) with ϕj being

polynomials ( or trigonometric functions). The functions ϕ0(x), ϕ1(x), ϕ2(x), . . . are required to have

simple and convenient structure. Power series and the Fourier series fulfill these requirements, but working

with them have certain disadvantages. One of the disadvantage is that they are available for a limited

classes of functions. In Section 2.4, we will introduce wavelets, a comprehensive mathematical tool leading

to representation of type (1.0.1) for relatively a large class of functions. Our focus of interest will be the

Haar wavelets and the Legendre wavelets. Some of the basic properties of these wavelets are discussed

which will be used in Chapter 6 and Chapter 7 dealing with numerical solutions for boundary value

problems of fractional ordinary and partial differential equations.

In Chapter 3, we will establish existence and uniqueness of solutions for different types of fractional

differential equations. In Section 3.1, we will study existence and uniqueness of solutions to a nonlinear

class of fractional differential equations involving Caputo fractional derivative

cDαy(t) = g(t, y(t)), y(0) = y0, y(1) = y1, t ∈ [0, 1],

where, 1 < α ≤ 2, y0, y1 ∈ R. In Section 3.2, we study the existence and uniqueness of solutions to a more

general problem

cDαy(t) = g(t, y(t), cDβy(t)), y(0) = y0, y(1) = y1, 1 < α ≤ 2, 0 ≤ β ≤ 1.

Section 3.3, is concerned with the existence and uniqueness of solutions to the following three–point

boundary value problem for nonlinear fractional differential equations

cDαy(t) = g(t, y(t), cDβy(t)), y(0) = µy(η), y(a) = νy(η),

where 1 < α < 2, 0 < β < 1, µ, ν ∈ R, η ∈ (0, a), µη(1− ν)+ (1−µ)(a− νη) = 0. In Section 3.4 we study

the existence and uniqueness of solution to following three–point boundary value problem

Dαy(t) = g(t, y(t),Dβy(t)), y(0) = 0, Dβy(1) = γDβy(η),

where, 1 < α ≤ 2, 0 < β < 1, α − β > 1, ∆α,β := (1 − γηα−β−1) > 0. We also give some new results

for the uniqueness of solutions. Section 3.5 deals with the existence and uniqueness of solutions to the

following class of multi–point boundary value problem for nonlinear fractional differential equations

Dαy(t) = g(t, y(t),Dβy(t)), t ∈ (0, 1), y(0) = 0, Dβy(1) =m−2∑i=1

aiDβy(ξi) + y0,

5

where 1 < α ≤ 2, 0 < β < 1, 0 < ξi < 1 (i = 1, 2, · · · ,m − 2), with, Λα,β :=∑m−2

i=1 aiξα−β−1i < 1. In

Section, 3.6 we study existence and uniqueness of solutions to nonlinear fractional differential equations

cDαy(t) = f(t, y(t), cDβy(t)), for t ∈ [0, l],

satisfying integral boundary conditions

py(0)− qy′(0) =

∫ l

0g(s, y(s))ds, γy(1) + δy′(1) =

∫ l

0h(s, y(s))ds,

where 0 < β < 1, 1 < α ≤ 2, p, δ > 0, q, γ ≥ 0 (or p, δ ≥ 0, q, γ > 0). The functions f, g and h

are assumed to be continuous. For the existence of solutions, we employ the nonlinear alternative of the

Leray–Schauder and a uniqueness result is established using the Banach fixed point theorem.

In many situations, only positive solutions of boundary value problems, that model important physical

phenomena, are meaningful. This is what we will pursue in Chapter 4 and Chapter 5. In Chapter 4, we

study the existence and multiplicity of positive solutions to three–point boundary value problems. The

tool that will be used to achieve this goal is Gau-Krasnosel’skii’s fixed point theorem on cone expansion

and compression. In Section 4.1, we investigate the existence and multiplicity results for the following

class of nonlinear three–point boundary value problems for fractional differential equations of type

cDαy(t) + a(t)g(t, y(t)) = 0, t ∈ (0, 1), n− 1 < α ≤ n,

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = ξy(η),

where ξ, η ∈ (0, 1). Existence theorems for positive and multiple positive solutions are established by

assuming that f satisfies certain growth conditions. An existence theorem for triple positive solutions is

proved using the Leggett-Wiliam fixed point theorem.

In Section 4.2, we study the existence of positive solutions to a nonlinear higher order three–point

boundary value problem

cDδy(t) + f(t, y(t)) = 0, t ∈ (0, 1), 0 < t < 1, n− 1 < δ < n, n(≥ 3) ∈ N,

y(1) = βy(η) + λ2, y′(0) = αy′(η)− λ1, y

′′(0) = 0, y′′′(0) = 0 · · · y(n−1)(0) = 0,

where, 0 < η, α, β < 1. The boundary parameters are assumed to be nonnegative. Sufficient conditions

for the existence and uniqueness of positive solutions are obtained by employing Gau-Krasnosel’skii’s fixed

point theorem. Also, a result for the existence of a unique positive solution is established. Applicability

of the proposed results is demonstrated by including some examples.

In Chapter 5, some existence theory for positive and multiple positive solutions for nonlinear systems

of fractional differential equations is developed. Section 5.1, is concerned with existence results for positive

solutions to following system of fractional differential equationscDαx(t) + λφ(t)f(y(t)) = 0, n− 1 < α ≤ n,

cDβy(t) + λψ(t)g(x(t)) = 0, n− 1 < β ≤ n,(1.0.3)

6

satisfying the two point boundary conditionsx(1) = 0, x′(0) = 0, x′′(0) = 0, · · · , x(n−2)(0) = 0, x(n−1)(0) = 0,

y(1) = 0, y′(0) = 0, y′′(0) = 0, · · · , y(n−2)(0) = 0, y(n−1)(0) = 0,(1.0.4)

where t ∈ [0, 1], λ > 0. The nonlinear functions f, g : [0,∞) → [0,∞) are continuous and trφ(t), tsψ(t) :

[0, 1] → [0,∞) are also assumed to be continuous for r, s ∈ [0, 1) and do not vanish identically on any

subinterval. Several results on the existence of positive solutions are obtained for the above two point

boundary value problem. In section 5.2, we study existence and multiplicity results for a coupled system

of nonlinear three–point boundary value problems for higher order fractional differential equations of the

type cDαx(t) = λφ(t)f(x(t), y(t)), n− 1 < α ≤ n, n ∈ N,cDβy(t) = µψ(t)g(x(t), y(t)), n− 1 < α, β ≤ n

(1.0.5)

satisfying the boundary conditionsx′(0) = x′′(0) = x′′′(0) = · · · = x(n−1)(0) = 0, x(1) = θ1x(µ1),

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = θ2x(µ2),(1.0.6)

where λ, µ > 0, for n ∈ N; θi, µi ∈ (0, 1) for i = 1, 2. We will also derive explicit intervals for the

parameters λ, µ, for with the above system has positive solutions.

In Chapter 6 we focus on providing a numerical scheme based on the Haar wavelets for solving different

types of boundary value problems for fractional differential equations. We drive a useful operational matrix,

and use it together with some other operational matrices developed in [27, 73, 80] for solving boundary

value problems. Various types of examples are presented to demonstrate the accuracy and simplicity of

our proposed numerical scheme.

In Chapter 7, an operational matrix of fractional order integration for Legendre wavelets is developed.

A numerical scheme based on this operational matrix is used to solve fractional differential equations. The

numerical solutions obtained by using Legendre wavelet method are compared with exact solutions and

with solutions obtained by some other numerical methods to demonstrate the accuracy, simplicity and

validity of the method.

The thesis includes two Appendices. In Appendix A we developed some Matlab programs to compute

various operational matrices that are used in the numerical solutions to fractional boundary value problems.

In Appendix B, some basic results from functional analysis are reviewed.

Chapter 2

Preliminaries

We review some elements of fractional calculus, fixed point theory and wavelet analysis that will be used

throughout this work. We begin by introducing Euler’s gamma function and using it to give a general

introduction to the idea of differentiation and integration of arbitrary order. The analytic results of

fractional calculus presented in section 2.2 are well known in literature and can be found in [69,103,110,114].

Some fixed point theorems are outlined in section 2.3 which are needed for the analysis of fractional

differential equations. The last section of the chapter is a brief introduction to the theory of wavelets. For

the most part, we use the notations and symbols that are commonly used in the current literature.

2.1 Some special functions

The generalization of integer order derivatives and multiple integrals to the derivatives and integrals of

arbitrary order is associated with the generalization of factorial function to gamma function. Also, the

Mittag–Leffler function, which is generalization of exponential function, plays an important role in the

theory of fractional differential equations and is connected with gamma function. In what follows, we

define and briefly discuss some properties of these special functions that will be frequently used in this

work.

2.1.1 Euler’s gamma function

In 1729, Euler discovered the gamma function while investigating the interpolation problem for the factorial

function. There are several approaches leading to the definition of gamma function. The most preferred

way of defining it, is the use of Euler’s integral

ϕ(x) =

∫ ∞

0txe−tdt, x ∈ N, (2.1.1)

as the starting point. Integration of (2.1.1) by parts, yields

ϕ(x) = [−txe−t]∞0 + x

∫ ∞

0tx−1e−tdt

= xϕ(x− 1), x = 1, 2, 3, . . . .

(2.1.2)

7

8

Since, ϕ(1) = 1, therefore repeated application of (2.1.2) gives ϕ(x) = x!. Thus we have integral represen-

tation of the factorial function x! as

x! =

∫ ∞

0txe−tdt, x ∈ N. (2.1.3)

Legendre introduced the notation Γ(x) for the function (x − 1)!. The integral Γ(x) =∫∞0 tx−1e−tdt

converges for x ∈ R+. From (2.1.2) we have the following fundamental equation

Γ(x+ 1) = xΓ(x). (2.1.4)

For n ∈ N, if some value of the function Γ(x) is known on the interval (n − 1, n], then with the help of

(2.1.4), we can find its value on the interval (n, n+1]. Now, one can extend the domain of Γ(x) to include

the negative real numbers. The repeated application of equation (2.1.4) gives

Γ(x) =Γ(x+ n)

x(x+ 1) . . . (x+ n− 2)(x+ n− 1), x ∈ R\0,−1,−2, . . . . (2.1.5)

The equations (2.1.4), (2.1.5) are valid even for complex values of x, provided x ∈ C\0,−1,−2, . . . . At

this point, we have following definition of gamma function.

Definition 2.1.1. The Euler’s gamma function is defined as

Γ(x) =

∫ ∞

0tx−1e−tdt if R(x) > 0,Γ(x) = Γ(x+ 1)/x. (2.1.6)

Theorem 2.1.2. (Weierstrass infinite product) For any x ∈ C,

1

Γ(x)= xeγx

∞∏n=1

(1 +

x

n

)e−

xn , (2.1.7)

where γ is the Euler’s constant given by

γ = limn→∞

( n∑k=1

1

k− log n

).

From (2.1.5), it follows that the gamma function has poles at x = 0,−1,−2, . . . , but 1Γ(x) is entire

function with zeroes at these points.

Another special function, closely related to the gamma function is the beta function. It has a simple

and useful integral representation.

Definition 2.1.3. The beta function is defined by Euler’s integral of first kind:

B(x, y) =

∫ 1

0sx−1(1− s)y−1ds, ℜ(x) > 0,ℜ(y) > 0.

This function is related to Euler’s gamma function as

B(x, y) =Γ(x)Γ(y)

Γ(x+ y), x, y ∈ C\0,−1,−2, . . . .

Thus the beta function is analytically continued to entire complex plane. Instead of using combinations

of gamma function, it is convenient to use beta function.

9

−5 −4 −3 −2 −1 0 1 2 3 4−20

−15

−10

−5

0

5

x

Γ(x)

1/Γ(x)

Figure 2.1: Gamma function and its reciprocal.

2.1.2 Mittag–Leffler function

In 1902, a Swedish mathematician Gosta Mittag–Leffler introduced the Mittag–Leffler function. It is a

straightforward generalization of exponential function. In recent years, the Mittag–Leffler function have

received much attention from researchers due to its role played in the investigation of fractional differential

equations. It arises frequently in the solutions of fractional differential and integral equations. The one–

parameter Mittag–Leffler function Eα(x) is defined by

Eα(x) =

∞∑k=0

xk

Γ(αk + 1), x ∈ C,ℜ(α) > 0. (2.1.8)

For 0 < α < 1, the one–parameter Mittag–Leffler function interpolates between exponential function ex

and hypergeometric function 1x−1 . Some special cases of the Mittag–Leffler function are [101]

(i) E0(x) =1

1− x, (ii) E1(x) = ex,

(iii) E2(−x2) = cos(x), (iv) E2(x2) = cosh(x),

(v) E3(x) =1

2

[ex

13 + 2e−

12x13 cos

(√3

2x

13

)], (vi) E4(x) =

1

2

[cos(x

14 ) + cosh(x

14 )].

The two–parameter Mittag–Leffler function Eα,β(x), which is defined as

Eα,β(x) =

∞∑k=0

xk

Γ(αk + β), x ∈ C,ℜ(α) > 0, ℜ(β) > 0, (2.1.9)

is a generalization of the one–parameter Mittag–Leffler function, to which it reduces to for β = 1. This

function was originally introduced by Wiman in 1905 and later investigated by Agrawal and Humbert in

1953. The two–parameter Mittag–Leffler function is related to the generalized hyperbolic function of order

n as

Hr(n, x) =

∞∑k=0

xnk+r−1

(nk + r − 1)!= xr−1En,r(x

n), r ∈ N. (2.1.10)

10

0 2 4 6 8 10 12 14 16 18 20−4

−3

−2

−1

0

1

2

3

4

5

x

α=1.3

α=1.5

α=1.7

α=1.9

α=1.95

α=1.99

2.0

Figure 2.2: The Mittag–Leffler function Eα,β(−(3x)2), for β = 1/2 and different values of α.

Also, Eα,β(x) is related to the generalized trigonometric function as

Kr(n, x) =∞∑k=0

(−1)kxnk+r−1

(nk + r − 1)!= xr−1En,r(−xn), r ∈ N. (2.1.11)

Another generalization of the Mittag–Leffler function is discussed by Prabhakar in 1971. He introduced

the function

Eδα,β(x) =

∞∑k=0

(δ)kxk

Γ(αk + β)k!, x, δ ∈ C,ℜ(α) > 0, ℜ(β) > 0 ℜ(δ) > 0. (2.1.12)

were (δ)0 = 0, (δ)k = δ(δ+1) · · · (δ+k−1) (k ∈ N) is the Pochhammer symbol. Many other generalizations

of the Mittag–Leffler functions have been appeared recently [49,68,72].

The Mittag–Leffler function E1α,1(x) = Eα(x) have no zero for α ∈ (0, 1] and for α ∈ (1, 2), it has

odd number of zeros on negative real axis. Moreover, for α ≥ 2, Eα(x) have infinite number of zeros on

negative real axis. Note that Eα(x) have no positive zero. For detailed information describing the zeroes

of the Mittag–Leffler functions we refer to [53,143].

2.2 Fractional calculus

Fractional calculus deals with derivatives and integrals of arbitrary order that are joined under the name

of differintegration. There are several approaches to fractional differentiation and integration that are not

equivalent, except for some special classes of functions. B. Ross [103] provided the following set of criteria

for fractional differintegration.

• Zero property: The operation of zero order must leave the function unchanged.

• Compatibility: The fractional differintegrals must produce the same results as ordinary differentia-

tion and integration when the order of differintegrals is an integer.

• Linearity: The fractional operators must be linear.

11

• Law of exponents: The fractional integrals must satisfy the law of exponents.

The most commonly used definitions of fractional integration and differentiation that fulfill these demands

are due to Riemann, Liouville and Caputo. In what follows, we define fractional derivatives and integral and

discuss some of their most useful properties. We begin with the Riemann–Liouville fractional integration.

2.2.1 The Riemann–Liouville fractional integration

One of the possible approach to define non-integer order differentiation and integration is through the use

of well known Cauchy’s integral formula for n-fold integral

Ina f(t) =∫ t

a

∫ tn−1

a· · ·∫ t1

af(t0)dt0 · · · dtn−2dtn−1

=1

(n− 1)!

∫ t

a(t− s)n−1f(s)ds, n ∈ N,

(2.2.1)

where f ∈ L2[a, b], a, b ∈ R. The generalization of factorial function to gamma function allows us to

replace n in (2.2.1) with an arbitrary real number α (or, even complex number) provided that the integral

on right side converges. Hence, it is natural to define the fractional integral as follows:

Definition 2.2.1. [114] Let f ∈ L1[a, b], α ∈ R+, the Riemann–Liouville fractional integral operator of

order α is defined as

Iαa f(t) =1

Γ(α)

∫ t

a(t− s)α−1f(s)ds (2.2.2)

for all t ∈ [a, b]. In particular, when α = 0, we set I0a := I; the identity operator.

For arbitrary lower limit (2.2.2) is the Riemann version and for infinite lower limit, i.e., for a = −∞,

(2.2.2) is the Liouville version of fractional integral. The case when a = 0, namely Iα0 is called the

Riemann–Liouville fractional integral and is quite convenient for further manipulations. On the other

hand if we keep lower limit arbitrary, take upper limit ∞ and replace kernel in (2.2.2) with (s− t)α−1 then

the resulting integral operator, for a reasonable class of functions, is called the Weyl fractional integral of

order α and is usually denoted by xWα∞. Historically, Abel (1823) used the fractional integral operator Iα0 ,

for α = 1/2 to solve his celebrated integral equation. Later in (1826) he generalized it to order α ∈ (0, 1).

Therefor, some authors prefer to name Iα0 , as the “Abel–Riemann fractional integral”.

Lemma 2.2.2. If α ≥ 0, β > −1, then the Riemann–Liouville fractional integral of the function (x− a)β

is given by

(Iαa (s− a)β)(t) =Γ(β + 1)

Γ(β + α+ 1)(t− a)β+α.

Theorem 2.2.3. [39] Let f ∈ L1[a, b] and α ∈ R+. Then the integral Iαa exists almost every where on

[a, b] and also Iαa is an element of L1[a, b].

Proof. Define a function φ : [a, b]× [a, b] → R by

φ(t, s) =

(t− s)α−1, if s ≤ t,

0, if t < s.

12

Case 1. If α ≥ 1, then φ(t, s) is continuous on [a, b]. Since f ∈ L1[a, b], therefore the product φ(t, s)f(s)

is integrable on [a, b]. Thus Iαa ∈ L1[a, b] in this case.

Case 2. If 0 ≤ α ≤ 1 then∫ ba φ(t, s)dt =

1α(b− s)α and∫ b

a

(∫ b

aφ(t, s)|f(s)|dt

)ds =

∫ b

a|f(s)|

( ∫ b

aφ(t, s)dt

)ds

=

∫ b

a|f(s)|(b− s)

α

α

ds

≤ (b− a)α

α∥f(s)∥1 <∞.

Thus, the product φ(t, s)f(s) is integrable over [a, b] × [a, b]. Therefore by Fubini’s Theorem the func-

tion g(t) =∫ ba φ(t, s)f(s)ds is integrable over [a, b]. Hence the fractional integral Iαa f(t) exists almost

everywhere on [a, b].

Theorem 2.2.4. [39] Let α ≥ 1 and f ∈ L1[a, b]. Then Iαa f ∈ C[a, b].

Before establishing the composition of fractional integral with the Mittag–Leffler function, let us point

out that one of the fundamental property of integer order integral, namely the semi group property, caries

over the Riemann–Liouville fractional order integral.

Lemma 2.2.5. [69] Let α, β ∈ R+ ∪ 0 and f be an element of L1[a, b]. Then

Iαa Iβa f(t) = Iα+βa f(t) = Iβa Iαa f(t) (2.2.3)

is valid almost everywhere on [a, b]. In addition to this, if f ∈ C[a, b] or α + β ≥ 1, then (2.2.3) is

identically true for all t ∈ [a, b].

Proof. Since I0a = I is defined as identity operator. Therefore for α = 0 or β = 0 or α = 0 = β, the

statement of the theorem holds trivially. By the definition of fractional integral, we have

Iαa Iβa f(t) =1

Γ(α)Γ(β)

∫ t

a(t− s)α−1

(∫ s

af(τ)(s− τ)β−1dτ

)ds.

The integral exists and Fubini’s Theorem allows us to interchange the order of integration, that is,

Iαa Iβa f(t) =1

Γ(α)Γ(β)

∫ t

af(τ)

(∫ t

τ(t− s)α−1(s− τ)β−1ds

)dτ.

The substitution s = τ + x(t− τ) yields

Iαa Iβa f(t) =1

Γ(α)Γ(β)

∫ t

af(τ)(t− τ)α+β−1

∫ 1

0xβ−1(1− x)α−1dxdτ

=1

Γ(α+ β)

∫ t

a(t− τ)α+β−1f(τ)dτ = Iα+βa f(t).

Therefore (2.2.3) holds almost everywhere on [a, b].

Now, if f ∈ C[a, b], then by Theorem 2.2.4 Iα+βa f(t) ∈ C[a, b] for α + β ≥ 1 and any t ∈ [a, b]. An

application of Fubini’s Theorem yields Iαa Iβa f(t) ∈ C[a, b] and is equal to Iα+βa f(t) ∈ C[a, b] for every

t ∈ [a, b].

13

The composition relation between fractional integral and the Mittag–Leffler functions are useful for

evaluation of fractional integrals of some of the elementary functions and in solutions of differential equa-

tions of fractional order.

Theorem 2.2.6. [125] For α, β, γ, δ ∈ R+ and λ ∈ R, the following holds:

(Iαa [(s− a)γ−1Eδβ,γ(λ(s− a)β)])(t) = (t− a)α+γ−1Eδβ,α+γ(λ(t− a)β). (2.2.4)

Proof. By definition of the Riemann–Liouville fractional integral and the generalized Mittag–Leffler func-

tion Eδβ,γ , we have

(Iαa [(s− a)γ−1Eδβ,γ(λ(s− a)β)])(t) =

∞∑k=0

(δ)kλk

Γ(βk + γ)k!Iαa (t− a)kβ+γ−1

= (t− s)γ+α−1∞∑k=0

(δ)k(λ(t− a)β)k

Γ(α+ βk + γ)k!

= (t− a)α+γ−1Eδβ,α+γ(λ(t− a)β).

The convergence of series in the definition of Eδβ,γ allows us to interchange the order of integration and

summation. The Lemma 2.2.2 is used to evaluate the fractional integral involved.

2.2.2 The Riemann–Liouville fractional derivative

Having defined the concept of fractional integrals, we intend to develop the notion of fractional order

derivatives and some of their important basic properties.

From now and onwards, Dn will denote the ntn order differential operator with D1 := D. In accordance

with these notations, the fundamental theorem of integer order calculus becomes

DIaf = f. (2.2.5)

A repeated application of above equation yields the following relation

f = DnIna f, n ∈ N. (2.2.6)

Replacing n in (2.2.6) by m− n with n < m , m ∈ N and applying Dn on both sides, we have

Dnf = DnDm−nIm−na f = DmIm−n

a f. (2.2.7)

This relation is still valid and meaningful, for some reasonable class of functions, if n is replaced by α ∈ Rprovided that m − α > 0 (m ≥ ⌈α⌉). Using the semigroup property of fractional integrals together with

the index law of classical derivative D and the fact that D is inverse of I, we have

DnIn−αa f = DαDn−⌈α⌉In−⌈α⌉a I⌈α⌉−αa f = D⌈α⌉I⌈α⌉−αa f = Dα

a f.

It is worthmentioning that the operator defined in this way depends on the choice of lower limit a of

fractional integral operator involved. One can define a fractional order derivative of a function as follows.

14

Definition 2.2.7. [69] Let α ∈ R+, m = ⌈α⌉ and f ∈ ACm[a, b]. The Riemann–Liouville fractional

derivative of order α is defined by

Dαa f := Dm

a Im−αa f, D0

a := I (the identity operator). (2.2.8)

The above definition is valid for arbitrary integer m provided m > α. There is no loss of generality

while considering narrow condition m = ⌈α⌉ or m − 1 ≤ α < m. In view of (2.2.7), the operator Dαa

coincides with classical nth order derivative operator when α is replaced with a positive integer.

Lemma 2.2.8. If α ≥ 0, β > −1, then the Riemann–Liouville fractional derivative of function (x − a)β

is given by

(Dαa (s− a)β)(t) =

Γ(β + 1)

Γ(β − α+ 1)(t− a)β−α.

In particular, when β = α− j, (j = 1, 2, . . . , ⌈α⌉+ 1), we have (Dαa (t− a)α−j = 0.

The composition relation between the Riemann–Liouville fractional derivative and the generalized

Mittag–Leffler function Eδβ,γ is of significant importance for evaluating fractional derivatives of various

functions.

Theorem 2.2.9. [125] For α, β, γ, δ ∈ R+ and λ ∈ R, the following holds:

(Dαa [(s− a)γ−1Eδ

β,γ(λ(s− a)β)])(t) = (t− a)γ−α−1Eδβ,γ−α(λ(t− a)β). (2.2.9)

Proof. By definition of the Riemann–Liouville fractional integral, the generalized Mittag–Leffler functions

Eδβ,γ and Lemma 2.2.8 we have

(Dαa [(s− a)γ−1Eδβ,γ(λ(s− a)β)])(t) =

∞∑k=0

(δ)kλk

Γ(βk + γ)k!Dαa (t− a)kβ+r−1

= (t− s)γ−α−1∞∑k=0


Γ(βk + γ − α)k!

= (t− a)γ−α−1Eδβ,γ−α(λ(t− a)β).

The convergence of series in the definition of Eδβ,γ allows us to interchange the order of integration and

summation.

Remark 2.2.10. The fractional derivatives and integrals of exponential function et plotted in Figures 2.3

(a)-(c) are computed by applying Theorem 2.2.6 and Theorem 2.2.9. The fraction derivatives and integrals

of f(t) = t3 are evaluated by the application of Lemma 2.2.2 and Theorem 2.2.8. The fractional derivatives

and integrals of trigonometric and hyperbolic functions can be evaluated using the relation between the

Mittag–Leffler function and generalized trigonometric functions (2.1.11), generalized hyperbolic functions

(2.1.10). But, the numerical evaluation of the Mittag–Leffler functions is itself difficult. We have used

a much simpler method based on the Haar wavelets, (which will be discussed later in this chapter), to

evaluate the fractional integrals of some functions plotted in Figures 2.3(f)-(h). For the classical cases

i.e. α = 1, 2, the obtained results by the Haar wavelets are in good agreement with the exact values. In

particular, for f(t) = e−t cos(7t) and α = 1, 2, (m=32) the maximum absolute error is 5.15 × 10−4 and

8.52× 10−5 respectively.

15

0 0.5 1 1.5 2 2.50

2

4

6

8

10

t

α=0

α=1/3

α=7/10

α=1

(a) Iα0 e

t, α = 0, 13, 73, 1.

0 0.2 0.4 0.6 0.8 10

0.5

1−0.5

0

0.5

1

1.5

2

2.5

3

α

t

(b) Iα0 e

t, 0 ≤ α ≤ 1.

0 0.2 0.4 0.6 0.8 10

0.5

10.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

tα

(c) Dα0 e

t, 0 ≤ α ≤ 1.

0 0.4

0.61.2

1.62

00.5

11.5

22.5

30

2

4

6

8

10

12

14

tα

D0α t2

(d) Dα0 t

3, 0 ≤ α ≤ 3.

−2

−1.2

−0.4

−3−2−10123−20

−10

0

10

20

tα

−2

(e) Differintegration Dα−2t

3,−3 ≤ α ≤ 3.

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

t

α=1α=1(Exact)α=1.3α=1.5α=1.7α=2α=2(Exact)

(f) Iα0 sin(7t), α = 1, 1.3, 1.5, 1.7, 2

00.2

0.40.6

0.81

1

1.2

1.4

1.6

1.8

20

0.05

0.1

0.15

0.2

0.25

0.3

0.35

x

α

(g) Iα0 sin(7t), 1 ≤ α ≤ 2.

0 0.2 0.4 0.6 0.8 1−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

t

α=1α=1(Exact)α=1.3α=1.5α=1.7α=2α=2

(h) Iα0 e

−t cos(7t), 1 ≤ α ≤ 2.

Figure 2.3: Fractional order integrals and derivatives of some elementary functions.

16

The following theorem characterizes the conditions for the existence of the Riemann–Liouville fractional

derivative Dαa defined above.

Theorem 2.2.11. [69, 114] Let f ∈ ACm[a, b] and m − 1 ≤ α < m. Then, the Riemann–Liouville

fractional derivative Dαa exists almost everywhere on [a, b]. Furthermore Dα

a f ∈ Lp[a, b] for 1 ≤ p < 1α and

Dαa f(t) =

m−1∑j=0

Djf(a)

Γ(j − α+ 1)(t− a)j−α + Im−α

a Dmf(t). (2.2.10)

Proof. Since f ∈ ACm[a, b], therefore by equations (2.2.6), (2.2.8), Lemma 2.2.5, Lemma 2.2.8 and Theo-

rem B.0.3, we have

Dαa f(t) = Dα

a

[m−1∑j=0

Djf(a)

Γ(j + 1)(t− a)j + Ima Dmf(t)

]

=m−1∑j=0

Djf(a)

Γ(j − α+ 1)(t− a)j−α +DmIm−αIma Dmf(t)

=

m−1∑j=0

Djf(a)

Γ(j − α+ 1)(t− a)j−α + Im−α

a Dmf(t).

Since, by Theorem B.0.3, Dmf(t) ∈ L[a, b] , therefore the integral Im−αa exists. Also we observe that the

existence of Dαa , α > 0 implies the existence of Dβ

a , 0 < β < α.

Before presenting properties of the Riemann–Liouville fractional derivative, we define a useful function

space

IαaL1[a, b] =f : Iαa f = φ, φ ∈ L1[a, b], α > 0

.

In the following , we establish composition relations between fractional derivative and integrals [114].

Lemma 2.2.12. If α, β ∈ R+, α > β and f ∈ L1[a, b], then DβaIαa f = Iα−βa f holds almost everywhere

on [a, b].

Proof. Using definition of the Riemann–Liouville fractional derivative, and semigroup property of frac-

tional integrals, we have DβaIαa f = D⌈β⌉I⌈β⌉−β

a Iαa f = Iα−βa f.

When α = β it immediately follows from above Lemma that the Riemann–Liouville fractional derivative

is left inverse to fractional integral operator. For some restricted class of functions the Riemann–Liouville

fractional derivative is also right inverse of fractional integral, as shown in the following lemma.

Lemma 2.2.13. Let a > 0 and f ∈ IαaL1[a, b]. Then IαaDαa f = f .

Proof. By same arguments as in the proof of the Lemma 2.2.12, we have

IαaDαa f = IαaD⌈α⌉I⌈α⌉−α

a f = IαaD⌈α⌉I⌈α⌉−αa Iαa φ = IαaD⌈α⌉I⌈α⌉

a φ = f.

In general, Fractional derivatives and integrals do not commute. We have the following result, which is

crucial for the investigations of fractional differential equations involving the Riemann–Liouville derivative.

We use it to convert differential equations into corresponding integral equations.

17

Theorem 2.2.14. [114] Let m ∈ N, m − 1 ≤ α < m and f ∈ L1[a, b]. Then, for Im−αa f ∈ ACm[a, b],

following holds:

IαaDαa f(t) = f(t)−

m∑j=1

(t− a)α−j

Γ(α− j + 1)

[Dα−ja f(t)

]t=a

. (2.2.11)

Proof. By definition of the Riemann–Liouville fractional derivative it follows that

Iα+1a Dα

a f(t) = Iα+1a DmDm−α

a f(t).

Integrating by parts and making use of Lemma 2.2.5 and Lemma 2.2.12, then by induction, we have

Iα+1a Dα

a f(t) =−k∑j=1

(t− a)α−j+1

Γ(α− j + 2)

[Dm−ja Im−α

a f(t)]t=a

+ Iα−m+1a Im−α

a f(t)

=Iaf(t)−k∑j=1

(t− a)α−j+1

Γ(α− j + 2)

[Dα−ja f(t)

]t=a

.

Applying D on both sides, we arrive at (2.2.11).

Having established relation between the Riemann–Liouville fractional derivative and integral, we now

turn to discuss the composition relation between two Riemann–Liouville fractional derivatives, namely

Dαa , (m − 1 ≤ α < m) and Dβ

a , (n − 1 ≤ α < n). The classical derivatives satisfy an unconditional

semigroup property. This is not the case when we are dealing with the fractional differential operators.

More precisely, we state and prove following result.

Theorem 2.2.15. [39] Let m, n ∈ N, m − 1 ≤ α < m, n − 1 ≤ α < n, α + β < m and f ∈ L1[a, b].

Then, for In−αa f ∈ ACn[a, b], following holds:

DαaDβ

af(t) = Dα+βa f(t)−

n∑j=1

Dβ−ja f(a)

Γ(1− α− j)(t− a)−α−j . (2.2.12)

Proof. By definition of the Riemann–Liouville fractional derivative and Lemma 2.2.5, we have

DαaDβ

af(t) = DmIm−αa Dβ

af(t) = DmIm−α−βa

[IβaDβ

af(t)]= Dα+β

a

[IβaDβ

af(t)].

Therefore, by Lemma 2.2.8 and Theorem 2.2.14 we obtain

DαaDβ

af(t) = Dα+βa

[f(t)−

n∑j=1

Dβ−ja f(a)

Γ(β − j + 1)(t− a)β−j

]= Dα+β

a f(t)−n∑

j=1

Dβ−ja f(a)

Γ(1− α− j)(t− a)−α−j .

In order to have a semigroup property for the Riemann–Liouville fractional derivative when α = β, we

must have DαaD

βaf(t) = Dβ

af(t)DβaDα

a f(t), which is possible only if Dα−ja f(a) and Dβ−j

a f(a) vanish simul-

taneously. For this, the function f must have a certain degree of smoothness. The following smoothness

assumptions on f are explored in [114, p. 215] by I. Podlubny.

Theorem 2.2.16. Assume that conditions of the Theorem 2.2.15 are satisfied. Then Dα−ja f(a), (j =

1, 2, . . . ,m) and Dβ−ja f(a), (j = 1, 2, . . . , n) vanish simultaneously, if and only if Djf(a) = 0 for j =

1, 2, . . . , q − 1, where q = maxm,n.

18

At this point, we introduce a useful function space for our purposes.

Lemma 2.2.17. [129] The space Bδ =y : y ∈ C[a, b] and Dδ

ay ∈ C[a, b], 0 < δ < 1

supplied with the

norm ∥y∥δ = max0≤t≤1

|y|+ max0≤t≤1

|Dδay| is a Banach space.

2.2.3 The Caputo fractional derivatives

The Riemann–Liouville fractional differential operators have played a significant role in the development

of the theory of differentiation and integration of arbitrary order. However, there are certain disadvan-

tages of using the Riemann–Liouville fractional derivatives for modeling the real world phenomena. In

Lemma 2.2.8 when β = 0 and f(t) = C(t − a)β , we observe that the fractional derivative of constant,

(Dαa (C))(t) =

C(t−a)−α

Γ(1−α) is a function of t and is never zero except for a = −∞. But for most of the physical

applications the lower limit is required to be a finite number. We also note that the initial value problems

for fractional differential equations with the Riemann–Liouville approach leads to the initial conditions

involving fractional derivative at lower limit. Mathematically, such problems can successfully be solved.

Being familiar with interpretation of real world problems with classical derivatives, we at least at the

present do not have any known physical interpretation of initial conditions involving fractional deriva-

tives. Applied problems, modeled using fractional operators, require an approach to fractional derivatives

which can utilize physically meaningful initial conditions involving classical derivatives. To cop with these

situations, M. Caputo in 1967 introduced another definition of fractional derivative and later in 1969

Caputo and Mainardi used it in the framework of viscoelasticity theory.

In what follows, we give a formal definition of the Caputo derivative and discuss its relations with the

Riemann–Liouville fractional integral and derivative.

Definition 2.2.18. [69] Let α ∈ R+ and f ∈ ACm[a, b], m = ⌈α⌉. Then the Caputo fractional derivative

of order α is defined bycDα

a f(t) = Im−αDmf(t). (2.2.13)

For α = m, the equation (2.2.13) yields cDαa f(t) = Dmf(t). Thus for integer values of α, the Caputo

fractional derivative becomes the conventional derivative. Historically, this concept was known to many

authors, including Rabotnov (1969), Dzherbashyan and Nersesian (1968), Gerasimov (1948), Gross (1947)

and even can traced back to Liouville (1832). Following the most common convention, we will name it

the Caputo derivative.

The definitions of fractional derivative, both in the sense of Riemann–Liouville and Caputo , utilize

the definition of Riemann–Liouville fractional integral but the order of fractional integration with integer

differentiation is interchanged. Also, note that both in the definition of the Riemann–Liouville fractional

derivative and the Caputo fractional derivative we require m = ⌈α⌉. This condition is not strict in the

case of the Riemann–Liouville definition of fractional derivative. We may chose any integer m such that

m ≥ α. However in the case of the Cputo fractional derivative, we may not use the condition m > ⌈α⌉.Another difference between Riemann–Liouville and caputo fractional derivatives is that the Riemann–

Liouville fractional derivative exists for a class of integrable function while the existence of the Caputo

19

fractional derivative requires the integrability of m times differentiable functions. This can be seen from

following Lemma.

Lemma 2.2.19. [39] If α ≥ 0 and f(t) = (t− a)β, m = ⌈α⌉, then

cDαa f(t) =

0, if β ∈ 0, 1, 2, . . . ,m− 1,Γ(β+1)

Γ(β−α+1)(t− a)β−α, if β ∈ N, and β ≥ m or β /∈ N, and β > m− 1.

Using this lemma, we can compute the Caputo fractional derivative of Mittag–Leffler function. More

precisely, we have following result.

Theorem 2.2.20. For α, β, γ, δ ∈ R+ and λ ∈ R, the following holds:

(cDαa [(s− a)γ−1Eδβ,γ(λ(s− a)β)])(t) = (t− a)γ−α−1Eδβ,γ−α(λ(t− a)β). (2.2.14)

Proof. By definition of the generalized Mittag–Leffler function and Lemma 2.2.19, we have

(cDαa [(s− a)γ−1Eδβ,γ(λ(s− a)β)])(t) =

∞∑k=0

(δ)kλk

Γ(βk + γ)k!Dαa (t− a)kβ+r−1

= (t− s)γ−α−1∞∑k=0


Γ(βk + γ − α)k!

= (t− a)γ−α−1Eδβ,γ−α(λ(t− a)β).

The convergence of series in the definition of Eδβ,γ allow us to interchange the order of integration.

The following result shows an important connection between the Riemann–Liouville and the Caputo

fractional derivatives and is key to the construction of the Caputo differential operator.

Theorem 2.2.21. For α ≥ 0, assume that f ∈ ACm[a, b], m = ⌈α⌉. Then

cDαa f(t) = Dα

a

[f(t)−

m−1∑j=0

Djf(a)

j!(t− a)j

](2.2.15)

Proof. In view of definition of the Riemann–Liouville fractional derivative, we have

Dαa

[f(t)−

m−1∑j=0

Djf(a)

j!(t− a)j

]= DmIm−α

a

[f(t)−

m−1∑j=0

Djf(a)

j!(t− a)j

].

Repeatedly integrating by parts, we obtain

Im−αa f(t) =

−1

Γ(m− α+ 1)

[(f(t)−

m−1∑j=0

Djf(a)

j!(t− a)j

)(t− a)m−α

]ta+ Im−α+1

a Df(t)

= Im−α+1a Df(t) = Im−α+2

a D2f(t) = · · · = I2m−αa Dmf(t).

Thus, we have DmI2m−αa Dmf(t) = Im−α

a Dmf(t) = cDαa f(t).

Using the above Theorem and Lemma 2.2.8 another relation between the Riemann–Liouville and

Caputo fractional derivatives can be given as.

20

Theorem 2.2.22. For α ≥ 0, m = ⌈α⌉ assume that for some f , both Dαa f(t) and cDα

a f(t) exist. Then

cDαa f(t) = Dα

a f(t)−m−1∑j=0

Djf(a)

Γ(j − α+ 1)(t− a)j−α. (2.2.16)

The above relation between the Riemann–Liouville and the Caputo derivatives says that they are

not equal in general. However, the equality cDαa f(t) = Dα

a f(t) holds if and only if Djf(a) = 0, j =

0, 1, 2, . . . ,m− 1. Having established relationship between the Riemann–Liouville and the Caputo deriva-

tives, we now turn to the composition relations between the Riemann–Liouville fractional integral and the

Caputo fractional derivatives.

Lemma 2.2.23. Let α, β ∈ R+, α ≥ β and f be continuous function. Then cDαaIαa f(t) = f(t).

Proof. First, we consider the case α = β. Since f is continuous on [a, b], therefore it is easy to see that

Iαa f(t) is bounded. Replacing f(t) by Iαa f(t) in (2.2.16), we get the required result.

In the case, α > β, using definition of the Caputo fractional derivative and semigroup property of

fractional integrals, we have

cDβaIαa f(t) = I⌈β⌉−β

a D⌈β⌉Iαa f(t) = I⌈β⌉−βa Iα−⌈β⌉

a f(t) = f(t).

Lemma 2.2.23 shows that the Caputo fractional derivative is left inverse of the Riemann–Liouville

fractional integral but on the other hand the following lemma shows the it is not right inverse of the

Riemann–Liouville fractional integral.

Lemma 2.2.24. Let α > 0, m = ⌈α⌉ and f ∈ ACm[a, b]. Then

Iαa [cDαa f(t)] = f(t)−

m−1∑j=0

Djf(a)

j!(t− a)j . (2.2.17)

Proof. Using (2.2.13), Lemma 2.2.5 and classical Taylor Theorem , we have

Iαa [cDαa f(t)] = Iαa Im−α

a Dmf(t) = Ima Dmf(t) = f(t)−m−1∑j=0

Djf(a)

j!(t− a)j .

The equation (2.2.17) also serves as the fractional Taylor series in term of the Caputo fractional

derivative. It has a significant importance in the theory of fractional differential equations involving the

Caputo fractional derivative for obtaining the corresponding integral representation.

We leave this section by introducing a function space which will be used to discuss matters.

Lemma 2.2.25. [152] The space Bα =y : y ∈ C[a, b] and cDα

a y ∈ C[a, b], α > 0

furnished with the

norm ∥y∥α = max0≤t≤1

|y|+ max0≤t≤1

|cDαa y| is a Banach space.

Remark 2.2.26. The conclusion of above Lemma holds if cDαa is replaced with Dα

a .

21

2.3 Fixed Point Theorems

One of the objective of this work is to provide a numerical technique for the treatment of boundary value

problems for fractional differential equations. While solving equations, or in particular solving differential

equations, the important thing to know is whether a particular equation has a solution or not. The

presence of solutions is guaranteed by so called fixed point theorems. So for in this chapter we have

been mainly concerned with the basic theory of fractional calculus. Now we also provide the terminology,

basic concepts and notations from analysis and fixed point theory. We state some important fixed point

theorems needed to establish existence of solutions for fractional differential equations.

Many of the most important problems of applied mathematical sciences can be described in a unitary

fission as: Find an object which belongs to a given class O of objects and is related to a given object φ

by a certain relation R. More precisely x ∈ O : x R φ

. (2.3.1)

It will be useful to explain this problem by a specific example. Consider the initial value problem for

fractional differential equationscDαa y(t) = f(t, y), α > 0, t ∈ [a, b]

Dky(a) = bk, k = 0, 1, 2, . . . ,m− 1, m = ⌈α⌉.(2.3.2)

Indeed, here we have O = C[a, b], φ = (f, a, bk), where y is a continuous function [a, b] :→ R and R is the

system (2.3.2). If f [a, b] : R is continuous, then by Lemma 2.2.23, the problem (2.3.2) can be transformed

into following Volterra integral equation

y(t) = ga(t) + Iαa f(t, y) (2.3.3)

where ga(t) =∑m−1

j=0 bj(t − a)j , bj = Djy(a)j! for j = 0, 1, 2, . . . ,m − 1. Since f is continuous, therefore

(2.3.2) can be recovered from (2.3.3). Thus (2.3.2) and (2.3.3) are equivalent.

Define an operator A : C[0, 1] → C[0, 1] as

Ay(t) = ga(t) + Iαa f(t, y). (2.3.4)

The continuity of f implies that the operator A is continuous. In view of (2.3.4) the equation (2.3.3) takes

the form

y = Ay(t). (2.3.5)

The first question in the study of this operator equation involves the existence of solutions. The solutions

of equation (2.3.5) are called the fixed points of A, which in turn are the solutions of problem (2.3.2). The

second questions involves the uniqueness of the fixed points of operator A. Perhaps the most simplest and

well known elementary result in fixed point theory employed to answer these questions is due to Banach

(1922)and is commonly known as the Banach Fixed Point Theorem or the Contraction Mapping Principle.

Being based on iterative process, it provides a constructive way of finding fixed points.

22

Theorem 2.3.1. Let M be a closed subset of the Banach space B and A : M → M. Then A has a

unique fixed point y in M and for any initial guess y0 ∈ M, the successive approximations ym+1 = Aymconverges to y, provided that ∥Ay −Ay∥ ≤ q∥Ay −Ay∥ for all y, y ∈ M and q < 1.

The Banach fixed point theorem yields a great deal of information about the solutions of certain

equations, but the class of equations to which it is applicable is very limited. Thus various alternative

have been developed.

Theorem 2.3.2. (The Brouwer Fixed Point Theorem) Suppose that M is a nonempty, convex, compact

subset of a finite dimensional normed vector space and A : M → M is a continuous mapping. Then Ahas a fixed point in M.

Now the question arises wether the condition of finite dimensionality, in the Brouwer fixed point

theorem, can be removed. The first powerful result, which avoids the restriction of this type, is the Schauder

fixed point theorem. This result uses the idea of compactness of operators to bridge the gape between

finite and infinite dimensional normed spaces. The boundary value problems for fractional differential

equations on a finite domain can often be posed in term of nonlinear compact operators using Green’s

functions. For this reason the Schauder fixed point theorem occupies a central position in the existence

theory of differential equations.

Theorem 2.3.3. (The Schauder Fixed Point Theorem) Suppose that M is a nonempty, convex, compact

subset of the Banach space B and A : M → B is a compact operator that maps M into itself. Then A has

a fixed point in M.

One of the possible way of solving a given operator equations is to embed it into continuum of equations

and then starting from one of the solution of simpler equation to obtain the solution of given equation.

More explicitly, for given two nonempty sets Σ and Ω consider a mapping Υ1 : Σ → Ω and U : a proper

subset of Ω. Our goal is to investigate the solvability of the problem

Υ1(x) ∈ U, x ∈ Σ. (2.3.6)

The idea behind the continuation methods is combining the problem (2.3.6) with a simpler one

Υ0(x) ∈ U, x ∈ Σ, (2.3.7)

using homotopy µ : Σ× [0, 1] → Ω such that

µ(., 0) = Υ0, µ(., 1) = Υ1.

Key role is played by conditions in continuation theorem which assure that the solvability of (2.3.7) implies

the solvability of (2.3.6). This important method is called Leray–Schauder continuation principle and is

key to establish the existence results for nonlinear boundary value problems for differential equations.

Historically, the continuation method was introduced by Poincaré (1884) and Bernstein (1912) used this

method to establish existence results by introducing technique of priori estimates. However, the first

rigorous study of continuation principle was carried out by Leray and Schauder (1934). In the following

we recall the statement of a Leray–Schauder continuation principle.

23

Theorem 2.3.4. [144, Theorem 6.A](Leray–Schauder Continuation Principle). Assume that

(i) the operator A : B → B is compact in the Banach space B;

(ii) there is a constant ρ > 0 and y = λAy for λ ∈ (0, 1) implies ∥y∥ ≤ ρ.

Then the equation (2.3.5) has a solution.

Another important fixed point theorem, derived from the Schauder fixed point theorem, commonly

used in the existence theory of differential equations is the nonlinear alternative of Leray–Schauder type.

Theorem 2.3.5. [48, Theorem 2.3] (Nonlinear alternative of Leray–Schauder type) Let B be a Banach

space and M be a nonempty convex subset of B and U be open in M with 0 ∈ U . Let A : U → M be a

compact operator. Then either

(i) A has a fixed point, or

(ii) there exists y ∈ ∂U and λ ∈ (0, 1) such that y = λAy.

In the case when the solutions of the operator equation (2.3.5) are not unique then the theorems

of non-uniqueness, that is the existence theorems for two or more solutions, are of basic interest. The

existence of solutions ensured by pure mathematical reasoning often requires some additional conditions

be satisfied. Most of the mathematical models in biomathematics, economics, physics and engineering

require the nonnegativity of solutions. Usually the existence theorems for positive solutions are established

in specially constricted cones defined as:

Definition 2.3.6. A closed convex subset P of a Banach space B is called cone, if λP ⊂ P, for all λ ≥ 0

and −P ∩ P = 0.

The most frequently used result to prove the existence of multiple positive solutions for differential

equations is the Guo-Krasnosel’skii Fixed Point Theorem, which is stated as:

Theorem 2.3.7. [75] Let B be Banach space and P ⊂ B be a cone. Assume that Ω1,Ω2 are open disks

in B such that 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2. Let A : P ∩ (Ω2\Ω1) → P be completely continuous such that either

(i) ∥Ay∥ ≤ ∥y∥, for y ∈ P ∩ ∂Ω1 and ∥Ay∥ ≥ ∥y∥, for y ∈ P ∩ ∂Ω2 or

(ii) ∥Ay∥ ≥ ∥y∥, for y ∈ P ∩ ∂Ω1 and ∥Ay∥ ≤ ∥y∥, for y ∈ P ∩ ∂Ω2.

Then, A has a fixed point in P ∩ (Ω2\Ω1).

In the following, we state another fixed point theorems which is commonly used to ensure the existence

of three fixed points.

Theorem 2.3.8. [83] (Leggett-Wiliam’s Fixed Point Theorem) Let θ : B → R+ be a continuous concave

functional and θ(y) ≤ ∥y∥, for all y ∈ Ec, where Ec = y ∈ E : ∥y∥ < c. Let 0 < a < b < d ≤ c be given

and define Eθ(b, d) = y ∈ E : b ≤ θ(y), ∥y∥ ≤ d. Assume that A : Ec → Ec is completely continuous and

satisfies

24

(i) y ∈ Eθ(b, d) : θ(y) > b = ∅ and θ(Ay) > b, for y ∈ Eθ(b, d),

(ii) ∥Ay∥ < a, for ∥y∥ ≤ a and

(iii) θ(Ay) > b for y ∈ Eθ(b, c) with ∥Ay∥ > d.

Then, A has at least three fixed points y1, y2 and y3 such that ∥y1∥ < a, b < θ(y2) and ∥y3∥ > a with

θ(y3) < b.

Finally, we make a remark about notation. In the case when lower limit a = 0, in fractional operators,

we shall use the notation Iα for Riemann–Liouville fractional integral and Dα, cDα will be used for

Riemann–Liouville and Caputo fractional derivatives. The symbol B will denote the Banach space C[a, b]

equipped with the norm ∥y∥ = maxt∈[a,b]

|y(t)|.

2.4 Wavelets

The origan of wavelets goes back to the beginning of 20th century, when Hungarian mathematician Alfred

Haar in 1910 constructed an orthonormal system of functions on the unit interval [0, 1] even though he did

not named it. Historically, the concept of wavelets was formally introduced at the beginning of eighties by

J. Morlet, a French geophysical engineer, as a family of functions constructed by translation and dilation

of single function, called the mother wavelet, for the analysis of seismic signals. The reason behinds the

discovery of wavelets is that Fourier series represents frequency of a signal well, but it does not model its

localized features appropriately. This is because the building blocks of Fourier series, the sine and cosine

functions, are periodic waves which continue forever.

The wavelet theory have drawn great deal of attention from scientists working in various disciplines

because of its comprehensive mathematical power and wide range applications in science and engineering.

Particularly, wavelets are very useful in signal processing, image processing, edge extraction, computer

graphics, approximation theory, biomedical engineering, differential equations, numerical analysis, etc. As

pointed out by Stephane Mallat,

“Wavelet theory is the result of a multidisciplinary effort that brought together mathematicians

physicists and engineers...”.

Wavelets are special kind of functions which exhibit oscillatory behavior for a short period of time and

then become zero. Wavelets are constructed from dilation and translation of single function ψ(t), called

mother wavelet and thus generating a two parameter family of functions ψa,b. It is convenient to define

ψa,b as follows.

ψa,b(t) =1√|a|ψ( t− b

a

), a, b ∈ R, a = 0,

where a is dilation parameter and b is translation parameter. If |a| < 1, then ψa,b is compressed form of

mother wavelet and corresponds to higher frequencies. On the other hand, for |a| > 1 the wavelet ψa,bcorresponds to lower frequencies. More precisely, we have the following definition of wavelets:

25

Definition 2.4.1. [43] A function ψ ∈ L2(R) is admissible as a wavelet if and only if

Cψ =

∫ ∞

−∞

|ψ(ω)|2

|ω|dω <∞,

where ψ is the Fourier transform of ψ, i.e.,

ψ(ω) =1

2√π

∫ ∞

−∞e−iωxψ(x)dx.

The admissibility condition requires that Cψ is finite. This implies that ψ(0) = 0 i.e. the mean value

of ψ should vanish:∫∞−∞ ψ(s)ds = 0. As an example of wavelets, the Maxican hat and its dilated shifts

are shown in Figure 2.4.

The continuous wavelets are not useful for many practical purposes. In particular they do not form

basis. For that reason, discretization is performed by fixing the positive constants a0 > 1, b0 > 0 and

setting a = a−j0 , b = kb0a−j0 , where n, k ∈ N. Thus, we define the following family of discrete wavelets as

ψj,k(t) = (a0)j2ψ(aj0t− kb0

).

Usually a0 is chosen to be 2 and b = 1. Ingrid Daubechies gave solid foundations for wavelet theory. In

1988 she provided major break through by constructing a system of orthonormal wavelets with compact

support. The Haar wavelet is the simplest example of orthogonal wavelets compactly supported on the

interval [0, 1] and was constructed by Haar in 1910 in his Ph.D. dissertation.

2.4.1 The Haar scaling function

In discrete wavelet transform we consider two sets of functions, scaling functions and wavelet functions.

The Haar scaling function φ(t) is defined on [0, 1] as

φ(t) = χ[0,1)(t) =

1, 0 ≤ t < 1,

0, elsewhere,(2.4.1)

which is a characteristics function of the interval [0, 1). The translates of Haar scaling function φ(t−k)k∈Zform an orthonormal set of functions. That is

∫R φ(t−m)φ(t−n) = δmn. The subspace of L2(R) spanned

by translates of the Haar scaling functions is denoted by V0. Scaling translates of φ(t) by 2j , we get the

functions φj,k = 2j2φ(2jt − k) supported on the dyadic subintervals Ij,k = [k2−j , (k + 1)2−j), j, k ∈ Z,

where φ0,0 is abbreviated as φ. For fixed j, the functions φj,k are orthonormal among themselves and

the space spanned by φj,k is denoted by Vj . The Haar scaling function φ satisfies the dilation equation

φ(t) =√2

∞∑k=−∞

ckφ(2t− k) (2.4.2)

where ck are given by

ck =√2

∫ ∞

−∞φ(s)φ(2s− k)ds. (2.4.3)

Evaluating (2.4.3), we have c0 = c1 = 1√2

and ck = 0 for k > 1. Therefore the dilation equation (2.4.2)

becomes

φ(t) = φ(2t) + φ(2t− 1). (2.4.4)

26

−10 −5 0 5 10−0.5

0

0.5

1

t

ψψ

−1,0

ψ−1,−1

ψ−1,1

Figure 2.4: The Maxican hat ψ(t) = (1− t2)e−12 t

2

and its dilated shifts.

The spaces spanned by the scaling function, that we briefly mentioned above, define a multiresolution

representation in L2(R). The idea of multiresolution is to express functions in L2(R) as limit of successive

approximations. These successive approximations use different levels of resolutions. Before proceeding

further, in the following subsection, we define multiresolution analysis.

2.4.2 Multiresolution Analysis (MRA)

The central idea of multiresolution analysis for the constructions of wavelet basis was formulated by S.

Mallat and Y. Meyer in 1986. The MRA is a formal approach for the construction of orthogonal wavelet

basis.

Definition 2.4.2. A multiresolution analysis is a sequence Vjj∈Z of closed subspaces of L2(R), such

that following conditions are satisfied:

(i) The sequence Vjj∈Z is nested: · · · ⊂ V−1 ⊂ V0 ⊂ V1 · · ·Vm ⊂ Vm+1 · · · .

(ii)∪j∈Z

Vj = L2(R) i.e.∪j∈Z

Vj is dense in L2(R).

(iii)∩j∈Z

Vj = 0.

(iv) f(t) ∈ Vm if and only if f(2t) ∈ Vm+1.

(v) There exists a function φ ∈ V0 such that φ0,k = φ(t− k), k ∈ Z is a Riez basis for V0, that is, for

every f ∈ V0, there exists a unique sequence ckk∈Z ∈ l2(Z) such that f(t) =∑

k∈Z ckφ(t− k) with

convergence in L2(R) and there exist two positive constants A, B independent of f ∈ V0 such that

A∑

k∈Z |ck|2 ≤ ∥f∥2 ≤ B∑

k∈Z |ck|2 where 0 < A < B <∞.

The subspaces Vjj∈Z of L2(R) spanned by the sets of functions φj,k is a MRA. Thus the Haar

scaling function generates the MRA.

27

0 1/2 1

(a) ψ0,0

000 1/4 1/21

(b) ψ1,0

0 1/2 13/4

(c) ψ1,1

Figure 2.5: Haar wavelets.

2.4.3 The Haar wavelet function

Based on the relation Vj ⊂ Vj+1 we are interested in the decomposition of Vj+1 as an orthonormal sum of

Vj and its orthogonal complement. For j = 0, the space V0 is spanned by the integer translates of scaling

function φ. This observation motivates for the construction of a function ψ(t) whose translate form the

basis for the orthogonal complement W0 of V0. The function ψ should be member of V1 and orthogonal to

V0. The simplest ψ that fulfills these requirements is ψ(t) = χ[0, 12) −χ[ 1

2,1) and is referred as Haar wavelet

function. The translates of ψ(t) i.e. ψ(t− k)k∈Z form an orthonormal set. Furthermore it is easy to see

that the system ψ(t− k)k∈Z forms basis for the space W0. For each pair of integers j, k ∈ Z, the dilated

translates of ψ are given as

ψj,k(t) = 2j2 (χIlj,k

(t)− χIrj,k(t)) = 2j2ψ(2jt− k). (2.4.5)

The functions ψj,k are compactly supported on the dyadic intervals Ij,k, j, k ∈ Z. For fixed j ∈ Z, we

define Wj as the orthogonal complement of Vj in Vj+1. The system ψj,kk∈Z, forms an orthonormal basis

of the complementary space Wj . In what follows, we show that the Haar wavelet functions are orthogonal

among themselves.

2.4.4 Orthogonality of the Haar wavelets

Theorem 2.4.3. The Haar wavelet system ψj,kj,k∈Z is orthonormal in L2(R).

Proof. Consider the inner product

⟨ψj,k, ψl,m⟩ =∫ ∞

−∞2

j2ψ(2jt− k)2

l2ψ(2lt−m)dt. (2.4.6)

Using change of variables 2jt− k = x, we find

⟨ψj,k, ψl,m⟩ = 212(l−j)

∫ ∞

−∞ψ(x)ψ(2l−j(x+ k)−m)dx. (2.4.7)

We consider the following cases:

Case 1. j = l. If k = m then ⟨ψj,k, ψj,k⟩ = ∥ψ∥2 = 1. On the other hand, if k = m then from (2.4.7) we

have ⟨ψj,k, ψj,k⟩ = ⟨ψ,ψ0,m−k⟩. Since m− k = 0, therefore the supports of ψ, ψ0,m−k are not overlapping,

i.e., I = [0, 1] ∩ I0,m−k = ∅ and the integral in (2.4.7) vanishes. Hence ⟨ψj,k, ψj,k⟩ = δk,m.

28

Case 2. j = l. It is sufficient to consider l > j. By symmetry, the result will hold for l < j. Substituting

r = l − j and n = m− 2rk in (2.4.7), we get.

⟨ψj,k, ψl,m⟩ = 2r2

∫ ∞

−∞ψ(x)ψ(2rx− n)dx. (2.4.8)

In the case of non overlapping diadic intervals, i.e., I ∩Ir,m = ∅, the integral (2.4.8) becomes 0. The diadic

intervals either do not overlap or one is contained in other. If one diadic interval is contained in other,

then it is contained in the left half or in the right half of it. So, if I ∩ Ir,m = ∅, then ψ(t) is constant over

Ir,m. Therefore the integral (2.4.7) vanishes.

In view of orthogonality of the system ψi,kj,k∈Z, the subspace Wj , j ∈ Z of L2(R) have following

important property:

· · ·W−1⊥W0⊥W1⊥W2⊥ · · ·Wj⊥Wj+1 · · · .

Furthermore the space Wj is an orthogonal compliment of Vj

Vj+1 =Wj ⊕ Vj , Wj⊥Vj . (2.4.9)

The repeated application of (2.4.9) yields the following relation

V0 ⊕W0 ⊕ · · · ⊕Wj = Vj+1. (2.4.10)

The nested structure of Vj together with properties (ii) and (iv) of MRA provide a decomposition of the

space L2(R). When m→ ∞, we have the decomposition.

L2(R) = V0 ⊕(⊕∞j=0Wj

). (2.4.11)

Thus MRA enables us to construct an orthonormal basis for L2(R).

2.4.5 Function approximation by the Haar wavelets

The decomposition of L2(R) into scaling space Vj and wavelet space Wj allows us decomposing any

function f(t) ∈ L2(R) such that components of f(t) lie in both subspaces. For each j ∈ Z, the successive

approximations of given function f(t) are defined as orthogonal projections Pj of function f ∈ L2(R) on

the space Vj given by

Pjf =∑k∈Z

⟨f, φj,k⟩φj,k(t), j ∈ Z, (2.4.12)

where ⟨f, φj,k⟩ are scaling coefficients. Since φj,k = 2j2χIj,k(t), therefore

Pjf = 2j∑k∈Z

(∫Ij,k

f(s)ds)χIj,k(t), j ∈ Z. (2.4.13)

Thus for each k ∈ Z, the projection Pjf is the average value of f(t) on the interval Ij,k. Therefore, in

relation to previous approximation Pj−1f , the function approximations determined at resolution level j

29

0 0.2 0.4 0.6 0.8 1-2

-1

0

1

2

Figure 2.6: Approximating f(t) = sin(9t) + 2 cos(11t) + 12 sin(50t) by the Haar wavelets.

lost any details about variation of f of scale smaller then 2−j on Ij,k. The lost details can be recovered

by orthogonal projection Qjf on the orthogonal complement Wj of Vj in Vj+1 so that

Qjf =∑k∈Z

⟨f, ψj,k⟩ψj,k(t), j ∈ Z, (2.4.14)

where ⟨f, ψj,k⟩ are the wavelet coefficients. The projection operator Qjf contains the details of f smaller

the 2−j but larger then 2−j−1. In view of (2.4.9),we have

Pj+1f = Pjf +Qjf. (2.4.15)

Therefore, in accordance with relation (2.4.11), we have multiresolution decomposition of the function

f ∈ L2(R) as

f(t) =∑k∈Z

⟨f, φ0,k⟩φ0,k(t) +∞∑j=0

∑k∈Z

⟨f, ψj,k⟩ψj,k(t) (2.4.16)

For the function f : [0, 1] → R, the Haar wavelet series representation is

f(t) = ⟨f, χ[0,1)⟩χ[0,1)(t) +

∞∑j=0

2j−1∑k=0

⟨f, ψj,k⟩ψj,k(t), (2.4.17)

The infinite series (2.4.17) is a powerful mathematical tool for the representation of large class of functions.

2.4.6 Error analysis

When dealing with series representation in practice, we are only able to deal with finite sums. We are

required to choose some fixed J such that the partial sum

SN (t) = ⟨f, χ[0,1]⟩χ[0,1](t) +

J−1∑j=0

2j−1∑k=0

⟨f, ψj,k⟩ψj,k(t), N = 2J , (2.4.18)

30

approximates f sufficiently well. The Haar wavelet approximation of the function f with bounded first

order derivative on (0, 1), gives the following estimate for error [15].

∥∥∥f(t)− SN (t)∥∥∥2 = ∥∥∥f − ⟨f, χ[0,1]⟩χ[0,1](t)−

J∑j=0

2j−1∑k=0

⟨f, ψj,k⟩ψj,k(t)∥∥∥2

=∥∥∥ ∞∑j=J

2j−1∑k=0

⟨f, ψj,k⟩ψj,k(t)∥∥∥2 = ∞∑

j=J

2j−1∑k=0

|⟨f, ψj,k⟩|2.

Now, by mean value theorem of integral calculus, there exist t1 ∈ I lj,k, t2 ∈ Irj,k, such that

⟨f, ψj,k⟩ = 2j2

(∫Ilj,k

f(t)dt−∫Irj,k

f(t)dt

)= 2−

j2−1(f(t1)− f(t2)

).

By mean value theorem of differential calculus, there exists ξ ∈ (t1, t2), such that

|⟨f, ψj,k⟩|2 = 2−j−2(f(t1)− f(t2)

)2= 2−j−2(t2 − t2)

2(f ′(ξ))2

≤ 2−3j−2L2, for some L > 0.

Thus, we have

∥∥∥f(t)− SN (t)∥∥∥2 ≤ L2

∞∑j=J

2j−1∑k=0

2−3j−2 = L2∞∑j=J

2−2j−2 =L2

N2= O

( 1N

).

Thus, with increasing J the error of approximation decreases.

Chapter 3

Existence and uniqueness of solutions

The objective of this chapter is to study some existence results for solutions of certain classes of nonlin-

ear two–point, three–point and multi–point boundary value problems for differential equations involving

fractional derivatives. The existence theory for initial value problems for fractional differential equations

have received considerable attention in the last two decades. However the existence theory for linear

and nonlinear boundary value problems for differential equations of fractional order is in its initial stages

of development. The theory of boundary value problems for fractional or even integer order differential

equations is much more complicated as compared with that of initial value problems.

This chapter is organized as follows: In section 3.1 and section 3.2, we are concerned with the existence

and uniqueness of solutions to two–point boundary value problems for nonlinear differential equations of

fractional order. In section 3.3, we establish sufficient conditions for the existence and uniqueness of

solutions to a general class of three–point boundary value problems by imposing some growth conditions

on the nonlinear functions involved. Section 3.4 deals with a class of three–point boundary value problems

for fractional order differential equations where the nonlinear term as well as the boundary conditions

involve the derivatives of unknown functions. Section 3.5 concerns with existence and uniqueness results

for a class of multi–point boundary value problems. Finally, in section 3.6, existence and uniqueness results

are established for nonlinear boundary value problems involving integral boundary conditions. Examples

are included to show the applicability of our results.

3.1 Two–point boundary value problems (I)

The question of existence and uniqueness of solution for two–point boundary value problems for fractional

differential equations have been studied in [3,21,126,128] and the references therein. Here we consider the

following nonlinear boundary value problem with the Caputo fractional derivative

cDαy(t) = g(t, y(t)), t ∈ [0, 1], (3.1.1)

y(0) = y0, y(1) = y1, (3.1.2)

where, 1 < α ≤ 2, y0, y1 ∈ R. The nonlinear function g : [0, 1] × R → R is assumed to be continuous.

Existence theory for the problem (3.1.1), (3.1.2) under much stronger hypothesis is developed in [128].

31

32

Different from the technique presented in [128], we prove the existence results not only in much simpler

way but also under weaker hypothesis on the nonlinear function g. Moreover, we also provide sufficient

conditions for the existence of a unique solution to the boundary value problem.

Lemma 3.1.1. A function y ∈ C[0, 1] is a solution of the boundary value problem (3.1.1), (3.1.2) if and

only if it is solution of the integral equation

y(t) = Iαg(t, y(t))− tIαg(1, y(1)) + (y1 − y0)t+ y0 =

∫ 1

0G(t, s)ds+ (y1 − y0)t+ y0, (3.1.3)

where

G(t, s) =

(t−s)α−1−t(1−s)α−1

Γ(α) , if s ≤ t,

− t(1−s)α−1

Γ(α) , if t ≤ s .(3.1.4)

Proof. Assume that y ∈ C[0, 1] is a solution of the boundary value problem (3.1.1), (3.1.2). Let w(t) :=

g(t, y(t)) = cDαy(t) = I2−αD2y(t). Therefore I2−α0 D2y(t) is continuous. Hence y(t) ∈ AC2[0, 1]. By

Lemma 2.2.24, we have

y(t) = Iαw(t) + c0 + c1t, c0, c1 ∈ R. (3.1.5)

Using the boundary conditions (3.1.2), we obtain c0 = y0, c1 = y1 − y0 − Iαw(1). Thus

y(t) = Iαg(t, y(t))− tIαg(1, y(1)) + (y1 − y0)t+ y0 =

∫ 1

0G(t, s)ds+ (y1 − y0)t+ y0.

Conversely suppose that y(t) is a solution of the integral equation (3.1.3). The application of the operatorcDα on both the sides of (3.1.3) and the use of Lemma 2.2.19 and Lemma 2.2.23, implies that y(t) satisfies

the differential equation (3.1.1) and the boundary conditions.

Lemma 3.1.2. The Green function G(t, s) defined in (3.1.4) has the following properties:

(i) |G(t, s)| ≤ 1Γ(α)(1− s)α−1, for all t, s ∈ [0, 1],

(ii)∣∣∣ ∂∂tG(t, s)∣∣∣ ≤ γ

Γ(α−1)(1− s)α−2, where γ := min1, Γ(α−1)Γ(α) .

Proof. For s ≤ t, using (3.1.4) we have G(t, s) = 1Γ(α)(t − s)α−1 − t(1 − s)α−1 ≤ (1 − t)(1 − s)α−1. Thus

G(t, s) ≤ 1Γ(α)(1− s)α−1. For s < t, obviously |G(t, s)| ≤ 1

Γ(α)(1− s)α−1.

Now, differentiating (3.1.4) with respect to t, we get

∂

∂tG(t, s) =

(t−s)α−2

Γ(α−1) − (1−s)α−1

Γ(α) , if s ≤ t,

− (1−s)α−1

Γ(α) , if t ≤ s .(3.1.6)

We observe that (1− s)α−2 ≥ (1− s)α−1 for s ∈ [0, 1] and 1 < α ≤ 2. Now, for s ≤ t we have

∂

∂tG(t, s) ≤ (t− s)α−2

Γ(α− 1)

and for t < s, we have ∣∣∣ ∂∂tG(t, s)

∣∣∣ ≤ (t− s)α−1

Γ(α)≤ (t− s)α−2

Γ(α).

Therefore∣∣∣ ∂∂tG(t, s)∣∣∣ ≤ γ

Γ(α−1)(1− s)α−2, where γ = min1, Γ(α−1)Γ(α) .

33

For the establishment of sufficient conditions for the existence of at least one solution to the fractional

boundary value problem (3.1.1), (3.1.2), we choose the Banach space of all continuous functions B :=

C[0, 1], equipped with Chebyshev norm.

Theorem 3.1.3. Assume that there exists continuous function φ : [0, 1] → R+ such that

|g(t, y)| ≤ φ(t) + c|y|θ, c, θ ∈ R+, θ = 1, (3.1.7)

then the boundary value problem (3.1.1), (3.1.2) has at least one solution.

Proof. Let R ≥ max2(Iαφ(1) + 2|y0| + |y1|,

(2c

Γ(α+1)

) 11−θ

and define a set U := y ∈ B : ∥y∥ ≤ R.Obviously U is nonempty closed convex subset of the Banach space B. Define an operator T : B → B as

T y(t) =∫ 1

0G(t, s)g(t, y(t))ds+ (y1 − y0)t+ y0. (3.1.8)

From (3.1.3) and (3.1.8), we have T y(t) = y(t). Thus in order to prove the existence of solutions to the

boundary value problem (3.1.1), (3.1.2), we have to prove that the operator T has a fixed points.

For the continuity of the operator T , consider a sequence of functions ynn∈N in B such that yn →y ∈ B as n→ ∞. For each t ∈ B, we have

|T yn(t)− T y(t)| ≤∫ 1

0|G(t, s)||g(s, yn(s))− g(s, y(s))|ds.

From the continuity of g, it follows that ∥T yn(t)− T y(t)∥ → 0 as n→ ∞. Hence, T is continuous.

Now to show that T maps U into itself, take y ∈ U and consider

|T y(t)| ≤∫ 1

0|G(t, s)||g(t, y(t))|ds+ |y1 − y0|t+ |y0|

≤∫ 1

0|G(t, s)|φ(s)ds+ cRθ

∫ 1

0|G(t, s)|ds+ 2|y0|+ |y1|

≤ Iαφ(1) + 2|y0|+ |y1|+cRθ

Γ(α+ 1)≤ R

2+R

2= R.

Hence, ∥T y(t)∥ ≤ R. Therefore, the set T (U) :=T y : y ∈ U

is bounded. In order to show that T (U)

is relatively compact, it remains to show that T (U) is equicontinuous.

Let L = maxt∈[0,1]

g(t, y(t)) : y ∈ U

. In view of (3.1.8) and Lemma 3.1.7, we obtain

∣∣∣ ddt(T y(t))

∣∣∣ ≤ ∫ 1

0

∣∣∣ ∂∂tG(t, s)

∣∣∣|g(s, y(s))|ds+ |y1|

≤ γL

Γ(α− 1)

∫ 1

0(1− s)α−2 + |y1| =

γL

Γ(α)+ |y1|.

For t1, t2 ∈ [0, 1] with t1 ≤ t2, using mean value theorem, we have∣∣∣T y(t2)− T y(t1)∣∣∣ = ∣∣∣[ d

dt(T y(t))

]t=ξ

(t2 − t1)∣∣∣ ≤ ( γL

Γ(α)+ |y1|

)(t2 − t1).

where ξ ∈ [t1, t2]. Hence, if for some δ > 0, (t2 − t1) < δ, then ∥T y(t2) − T y(t1)∥ < ε, where ε :=(γLΓ(α) + |y1|

)δ is independent of y, t1, t2. Therefore the set T (U) is equicontinuous. By the Arzela–Ascoli

theorem, T (U) is relatively compact. Hence, by the Schauder fixed point theorem, the boundary value

problem (3.1.1), (3.1.2) has at least one solution.

34

Theorem 3.1.4. Assume that there exists a continuous function φ : [0, 1] → R+, such that

|g(t, y)| ≤ φ(t) + c|y|, c ≤ Γ(α+ 1)

2, (3.1.9)

then the boundary value problem (3.1.1), (3.1.2) has at least one solution.

Proof. Let R ≥ min 2(Iαφ(1) + 2|y0|+ |y1|) and define the set U := y ∈ B : ∥y∥ ≤ R. Then, for y ∈ U ,

we have∣∣∣T y(t)∣∣∣ ≤ Iαφ(1) + Γ(α+1)

2 ≤ R. Thus T : U → U . The rest of the proof is similar to the proof

of the Theorem 3.1.3.

Remark 3.1.5. If φ(t) = 0 in (3.1.7) and (3.1.9), then we can choose R ≤(Γ(α+1)

2c

) 1θ−1 , θ = 1 in the

proof of the Theorem 3.1.3 and R > 0 in Theorem 3.1.4.

Theorem 3.1.6. Assume that there exists a constant k < Γ(α+ 1) such that

|g(t, y)− g(t, z)| ≤ k|y − z|, for each t ∈ [0, 1] and all y, z ∈ R. (3.1.10)

Then the boundary value problem (3.1.1), (3.1.2) has a unique solution.

Proof. Define U :=y ∈ B : ∥y∥ ≤ K

and using arguments similar to used in the Theorem 3.1.4, we

conclude that the operator T defined by (3.1.8) maps the bounded set U into itself. It remains to prove

that T is contraction. For y, z ∈ U , we have

∣∣T y(t)− T z(t)∣∣ ≤∫ 1

0|G(t, s)||g(s, y(s))− g(s, z(s))|ds

≤k∥y − z∥∫ 1

0|G(t, s)|ds ≤ k

Γ(α+ 1)∥y − z∥ ≤ q∥y − z∥,

where q = kΓ(α+1) < 1. Hence by contraction mapping principle the boundary value problem (3.1.1), (3.1.2)

has a unique solution.

3.2 Two–point boundary value problems (II)

In this section, we study the existence and uniqueness of solutions to the following fractional differential

equation

cDαy(t) = g(t, y(t), cDβy(t)), 1 < α ≤ 2, 0 ≤ β ≤ 1. (3.2.1)

subject to the boundary conditions

y(0) = y0, y(1) = y1. (3.2.2)

This problem generalizes the results studied in section 3.1 in the sense that the nonlinear function depends

on fractional derivative of y as well.

35

Lemma 3.2.1. Assume that g : [0, 1]×R×R → R is continuous. Then, a function y ∈ C[0, 1] is a solution

of the boundary value problem (3.2.1), (3.2.2) if and only if y(t) is a solution of the integral equation

y(t) = Iαg(t, y(t), cDβy(t))− tIαg(1, y(1), cDβy(1)) + (y1 − y0)t+ y0

=

∫ 1

0Gα(t, s)g(s, y(s),

cDβy(s))ds+ (y1 − y0)t+ y0.(3.2.3)

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

Gα(t, s) =

(t−s)α−1−t(1−s)α−1

Γ(α) , 0 ≤ s ≤ t,

− t(1−s)α−1

Γ(α) , t < s ≤ 1.(3.2.4)

The proof of the Lemma 3.2.1 is similar to the proof of Lemma 3.1.1. Following will be assumed to

establish sufficient conditions for the existence and uniqueness of solutions.

(A1) g : [0, 1]× R× R → R is continuous;

(A2) there exists a nonnegative function φ ∈ C[0, 1] such that |g(t, y, z)| ≤ φ(t) + µ1|y|ν1 + µ2|z|ν2 , where

µ1, µ2 ∈ R+ and 0 ≤ ν1, ν2 < 1;

(A3) |g(t, y, z)| ≤ µ1|y|ν1 + µ2|z|ν2 , where µ1, µ2 ∈ R+ and ν1, ν2 > 1;

(A4) there exists a constant k > 0 such that

|g(t, y, z)− g(t, y, z))| ≤ k(|y − y|+ |z − z|), for each t ∈ [0, 1] and all y, z, y, z ∈ R.

Consider the Banach space Bβ defined in Lemma 2.2.25. Define an operator A : Bβ → Bβ by

T y(t) =∫ 1

0Gα(t, s)g(s, y(s),

cDβy(s))ds+ (y1 − y0)t+ y0 (3.2.5)

In order to show that the boundary value problem (3.2.1), (3.2.2) has a solution, it is sufficient to prove

that the operator T has a fixed point.

For convenience, we define the following constants:

kα = Iα|φ(1)| + 2|y0| + |y1|, kα,β = σα,βIα−βφ(s) + |y1−y0|Γ(2−β) , where σα,β =

(1

Γ(α−β) +1

Γ(α)Γ(2−β)

)and

qα,β = 1Γ(α+1) +

σα,β

Γ(α−β+1) .

Theorem 3.2.2. Assume that (A1) and (A2) hold. Then the fractional boundary value problem (3.2.1),

(3.2.2) has at least one solution.

Proof. Continuity of T follows form the continuity of g. Choose

R ≥ max3(kα + kα,β), (3qα,βµ1)

11−ν1 , (3qα,βµ2)

11−ν2

and define a set M =

y ∈ Bβ : ∥y∥β ≤ R

. Then for an arbitrary y ∈ M, we have the following estimate.

|T y(t)| ≤∫ 1

0|Gα(t, s)|φ(s)ds+ (µ1R

ν1 + µ2Rν2)

∫ 1

0|Gα(t, s)|ds+ 2|y0|+ |y1|

≤ Iαφ(1) + 2|y0|+ |y1|+ (µ1Rν1 + µ2R

ν2)

∫ 1

0

(t− s)α−1

Γ(α)ds

≤ kα +1

Γ(α+ 1)(µ1R

ν1 + µ2Rν2).

36

In view of Lemma 2.2.2 and Lemma 2.2.5, we have

cDβ(T y)(t) = I1−β(cD(T y))(t)

= I1−β(Iα−1g(t, y(t), cDβy(t))− Iαg(1, y(1), cDβy(1)) + y1 − y0

)= Iα−βg(t, y(t), cDβy(t))−

(Iαg(1, y(1), cDβy(1))− y1 + y0

) t1−β

Γ(2− β)

=

∫ 1

0Gα,β(t, s)g(s, y(s),

cDβy(s))ds+(y1 − y0)

Γ(2− β)t1−β

where Gα,β is defined by

Gα,β =

(t−s)α−β−1

Γ(α−β) − t1−β(1−s)α−1

Γ(α)Γ(2−β) , 0 ≤ s ≤ t,

− t1−β(1−s)α−1

Γ(α)Γ(2−β) , t < s ≤ 1.(3.2.6)

For, s ≤ t, we have |Gα,β | ≤ (1−s)α−β−1

Γ(α−β) + (1−s)α−1

Γ(α)Γ(2−β) ≤(

1Γ(α−β) +

1Γ(α)Γ(2−β)

)(1− s)α−β−1 and for t < s,

we have |Gα,β | ≤ (1−s)α−1

Γ(α)Γ(2−β) ≤(1−s)α−β−1

Γ(α)Γ(2−β) . Thus, |Gα,β | ≤σα,β(1−s)α−β−1

Γ(α−β) . Hence, it follows that∣∣cDβ(T y)(t)∣∣ ≤ ∫ 1

0|Gα,β(t, s)||g(s, y(s), cDβy(s))|ds+ |y1 − y0|

Γ(2− β)

≤∫ 1

0|Gα,β(t, s)|φ(s)ds+

|y1 − y0|Γ(2− β)

+ (µ1Rν1 + µ2R

ν2)

∫ 1

0|Gα,β(t, s)|ds

≤ σα,βIα−βφ(s) +|y1 − y0|Γ(2− β)

+ (µ1Rν1 + µ2R

ν2)σα,βt

α−β

Γ(α− β + 1)

≤ kα,β + (µ1Rν1 + µ2R

ν2)σα,β

Γ(α− β + 1).

Therefore,

∥T y)(t)∥β ≤ kα + kα,β + qα,β(µ1Rν1 + µ2R

ν2) ≤ R

3+R

3+R

3= R,

which imply that T maps the bounded set M of the Banach space Bβ into itself.Now, we prove that the set T (M) is equicontinuous. Let Λ = max

t∈[0,1]

∣∣g(t, y(t), cDβy(t))∣∣ : y ∈ M

,

then for 0 ≤ t ≤ τ ≤ 1, we have

|T y(τ)− T y(t)|

≤∣∣Iαg(τ, y(τ), cDβy(τ))− Iαg(t, y(t), cDβy(t))

∣∣+ (∣∣Iαg(1, y(1), cDβy(1))∣∣+ |y1 − y0|

)(τ − t)

≤ Λ

(∫ t

0

(τ − s)α−1 − (t− s)α−1

Γ(α)ds+

∫ τ

t

(τ − s)α−1

Γ(α)ds

)+

(Λ

∫ 1

0

(1− s)α−1

Γ(α)ds+ |y1 − y0|

)(τ − t)

≤ Λ

(−(τ − t)α + τα − tα

αΓ(α)+

(τ − t)α

αΓ(α)

)+

(Λ

αΓ(α)+ |y1 − y0|

)(τ − t)

=Λ(τα − tα)

Γ(α+ 1)+

(Λ

Γ(α+ 1)+ |y1 − y0|

)(τ − t),

∣∣cDβ(T y)(τ)− cDβ(T y)(t)∣∣ ≤ ∣∣Iα−βg(τ, y(τ), cDβy(τ))− Iα−βg(t, y(t), cDβy(t))

∣∣+(∣∣Iαg(1, y(1), cDβy(1))

∣∣+ |y1 − y0|) (τ1−β − t1−β)

Γ(2− β)

≤Λ(τα−β − tα−β

)Γ(α− β + 1)

+

(Λ

Γ(2− β)Γ(α− β + 1)+

|y1 − y0|Γ(2− β)

)(τ1−β − t1−β)

Γ(2− β).

37

Hence, it follows that ∥T y(τ)−T y(t)∥ → 0, as τ − t→ 0. By the Arzela–Ascoli theorem, T : M → M is

compact operator. Thus by Theorem 2.3.3, the boundary value problem (3.2.1), (3.2.2) has a solution.

Theorem 3.2.3. Assume that (A1) and (A3) hold. Then the boundary value problem (3.2.1), (3.2.2) has

at least one solution.

Remark 3.2.4. The proof of the Theorem 3.2.3 is similar to the proof of the Theorem 3.2.2. The bounded

convex subset of Bβ can be defined by chosing

0 < R ≤ min( 1

2qα,βµ1

) 1ν1−1

,( 1

2qα,βµ2

) 1ν2−1

.

Remark 3.2.5. Theorem 3.2.2 and Theorem 3.2.3 do not cover the cases ν1 = ν2 = 1, ν1 = 1, ν2 = 1, (or

ν1 = 1, ν2 = 1). We consider these cases separately.

Theorem 3.2.6. Assume that A1 holds. Furthermore, there exists a nonnegative function φ(t) and

µ1, µ2 ≥ 0 such that

|g(t, y, z)| ≤ φ(t) + µ1|y|+ µ2|z|, µ1 + µ2 ≤1

2qα,β, or

|g(t, y, z)| ≤ φ(t) + µ1|y|+ µ2|z|ν , 0 < ν < 1, µ1 ≤1

3qα,β.

Then, the boundary value problem (3.2.1), (3.2.2) has a solution.

The proof is similar to the proof of Theorem 3.2.2. Therefore, it is omitted.

Theorem 3.2.7. Assume that A1 is satisfied and there exists a constant k <(

1Γ(α−β+1) +

1Γ(2−β)Γ(α+1)

)such that

|g(t, y, z)− g(t, y, z)| ≤ k(|y − y|+ |z − z|), for each t ∈ [0, 1] and all y, y, z, z ∈ R, (3.2.7)

then there exists a unique solution of the boundary value problem (3.2.1), (3.2.2).

Proof. From (3.2.5), we have,∣∣T y(t)− T y(t)∣∣ ≤ ∫ 1

0|G(t, s)||g(s, y(s), cDαy(t))− g(s, y(s), cDαy(t))|ds

≤ ∥y − y∥kΓ(α)

(∫ t

0(t− s)α−1ds+ t

∫ 1

0(1− s)α−1ds

)=

∥y − y∥kΓ(α+ 1)

(tα + t) ≤ 2k

Γ(α+ 1)||y − y||,

∣∣cDα(T y)(t)− cDα(T y)(t)∣∣ ≤ ∫ 1

0|G(t, s)||g(s, y(s), cDαy(t))− g(s, y(s), cDαy(t))|ds

≤ ∥y − y∥k(∫ t

0

(t− s)α−β−1

Γ(α− β)ds+

t1−β

Γ(2− β)

∫ 1

0

(1− s)α−1

Γ(α)ds

)= ∥y − y∥k

(tα−β

Γ(α− β + 1)+

t1−β

Γ(2− β)Γ(α+ 1)

)≤ ∥y − y∥k

(1

Γ(α− β + 1)+

1

Γ(2− β)Γ(α+ 1)

).

38

Thus, we have ∥∥T y(t)− T y(t)∥∥ ≤ ρ∥y − y∥,

where, ρ = k(

1Γ(α−β+1) +

1Γ(2−β)Γ(α+1)

)< 1. Therefore, by the contraction mapping principle, the bound-

ary value problem (3.2.1), (3.2.2) has a unique solution.

3.3 Three–point boundary value problems (I)

Multi–point boundary value problems arise in different areas of physics and mathematics. The most

commonly quoted example in this respect is modeling the vibration of a guy-wire with n parts of different

densities, but having uniform cross-section. Bitsadze and Samarski [22] initiated the study of multi–point

boundary value problems for integer order differential equations. Later II’in and Moreover [58,59] played

the leading role for the development of the existence theory of such problems. Since then, the multi–

point boundary value problems have been investigated by several researchers including [51, 97, 98, 100].

In contrast, the multi–point boundary value problems for fractional differential equations have received

attention quite recently. For details, we refer to [4,19,44,123,153]. In this section, we study existence and

uniqueness of solutions to a class of nonlinear fractional differential equations of the type [87]

cDαy(t) = g(t, y(t), cDβy(t)), t ∈ [0, a], a > 0, (3.3.1)

y(0) = µy(η), y(a) = νy(η), (3.3.2)

where 1 < α < 2, 0 < β ≤ 1, µ, ν ∈ R, η ∈ (0, a), µη(1− ν) + (1− µ)(a− νη) = 0.

We need the following fundamental lemma for our main result.

Lemma 3.3.1. Let h ∈ C[0, a] and 1 < α ≤ 2. Then the linear problem

cDαy(t) = h(t), 0 < t < 1, t ∈ [0, a], (3.3.3)

y(0) = µy(η), y(a) = νy(η), (3.3.4)

has solution

y(t) = Iαh(t) + 1

∆(µa+ (ν − µ)t)Iαh(η)− (µη + (1− µ)t)Iαh(a) =

∫ a

0G(t, s)h(s)ds, (3.3.5)

where ∆ = µη(1− ν) + (1− µ)(a− νη) = 0 and G(t, s) is the Green function

G(t, s) =

(t−s)α−1

Γ(α) + (µa+(ν−µ)t)(η−s)α−1

∆Γ(α) − (µη+(1−µ)t)(a−s)α−1

∆Γ(α) , if 0 ≤ s < t, η ≥ s,

(µa+(ν−µ)t)(η−s)α−1

∆Γ(α) − (µη+(1−µ)t)(a−s)α−1

∆Γ(α) , if 0 ≤ t ≤ s ≤ η,

(t−s)α−1

Γ(α) − (µη+(1−µ)t)(a−s)α−1

∆Γ(α) , if η ≤ s ≤ t ≤ 1,

− (µη+(1−µ)t)(a−s)α−1

∆Γ(α) , if 0 ≤ t ≤ s, s ≥ η .

(3.3.6)

Proof. In view of Theorem 2.2.24, the general solution of (3.3.3), (3.3.4) is

y(t) = Iαh(t) + c0 + c1t, (3.3.7)

39

where c0, c1 ∈ R. Using (3.3.4) and (3.3.7), we have

(1− µ)c0 − µηc1 = µIαh(η), (1− ν)c0 + (a− νη)c1 = νIαh(η)− Iαh(a),

which implies that

c0 =µ

∆(aIαh(η)− ηIαh(a)), c1 =

1

∆((ν − µ)Iαh(η) + (µ− 1)Iαh(a).

Therefore, the solution of (3.3.3), (3.3.4) is

y(t) = Iαh(t) + 1

∆(µa+ (ν − µ)t)Iαh(η)− (µη + (1− µ)t)Iαh(a). (3.3.8)

Now, for t ≤ η, equation (3.3.1) can be written as

y(t) =

∫ t

0

((t− s)α−1

Γ(α)+

(µa+ (ν − µ)t)(η − s)α−1

∆Γ(α)− (µη + (1− µ)t)(a− s)α−1

∆Γ(α)

)h(s)ds

+

∫ η

t

((t− s)α−1

Γ(α)+

(µa+ (ν − µ)t)(η − s)α−1

∆Γ(α)− (µη + (1− µ)t)(a− s)α−1

∆Γ(α)

)h(s)ds

−∫ a

η

(µη + (1− µ)t)(a− s)α−1

∆Γ(α)h(s)ds =

∫ a

0

G(t, s)h(s)ds.

For t > η, we have

y(t) =

∫ η

0

((t− s)α−1

Γ(α)+

(µa+ (ν − µ)t)(η − s)α−1

∆Γ(α)− (µη + (1− µ)t)(a− s)α−1

∆Γ(α)

)h(s)ds

+

∫ t

η

((t− s)α−1

Γ(α)− (µη + (1− µ)t)(a− s)α−1

∆Γ(α)

)h(s)ds

−∫ a

t

(µη + (1− µ)t)(a− s)α−1

∆Γ(α)h(s)ds =

∫ a

0

G(t, s)h(s)ds.

Define an operator T : Bβ → Bβ by

T y(t) =∫ a

0G(t, s)g(s, y(s), cDβy(s))ds. (3.3.9)

In order to find the solutions of the boundary value problem (3.3.1), (3.3.2), we need to find the fixed

points of the operator T .

For convenience, define the constants:

l∗ = max(∫ a

0 | ∂∂tG(t, s)φ(s)|ds), l = max

(∫ a0 |G(t, s)φ(s)|ds

), Nα = |ν − µ|ηα + |1− µ|aα,

Mα,β =[(∆ + |µ|η)aα + |µ|aηα + aβ−1

Γ(2−β)(∆aα−1 +Nα)

], kαµ,ν = 2|µ|+|ν|

∆αΓ(α) , kαµ = 1+|µ|

∆αΓ(α) and kα = aα

∆αΓ(α) .

3.3.1 Existence of solutions

Theorem 3.3.2. Assume that (A1), (A2) or (A1), (A3) are satisfied. Then the boundary value problem

(3.3.1), (3.3.2) has at least one solution.

Proof. Suppose (A2) holds. Choose a constant R ≥ max3(l + l∗a1−β

Γ(2−β)), (3Mα,βµ1)1

1−ν1 , (3Mα,βµ2)1

1−ν2

and define W = y ∈ Bβ : ∥y∥β ≤ R. W is bounded convex subset of the Banach space Bβ . For an

40

arbitrary y ∈ W , using assumption (A2), we have

|T y(t)| =∣∣∣∣∫ a

0

G(t, s)g(s, y(s), cDβy(s))ds

∣∣∣∣ ≤ ∫ a

0

|G(t, s)φ(s)|ds+ (µ1Rν1 + µ2R

ν2)

[∫ t

0

(t− s)α−1

Γ(α)ds

+1

∆(|µ|a+ |ν − µ|t)

∫ η

0

(η − s)α−1

Γ(α)ds+

1

∆(|µ|η + |1− µ|t)

∫ a

0

(a− s)α−1

Γ(α)ds

]≤l + (µ1R

ν1 + µ2Rν2)

(tα

αΓ(α)+

1

∆(|µ|a+ |ν − µ|t) ηα

αΓ(α)+

1

∆(|µ|η + |1− µ|t) aα

Γ(α)

),

which implies that |T y(t)| ≤ l + (µ1Rν1+µ2R

ν2 )α∆Γ(α) (∆aα + |µ|(aηα + ηaα) +NαT ). Also,

|(T y)′(t)| ≤∫ a

0

| ∂∂tG(t, s)||g(s, y(s), cDβy(s))|ds ≤

∫ a

0

| ∂∂tG(t, s)φ(s)|ds+ (µ1R

ν1 + µ2Rν2)

(∫ t

0

(t− s)α−2

Γ(α− 1)ds

+|ν − µ|

∆

∫ η

0

(η − s)α−1

Γ(α)ds+

|1− µ|∆

∫ a

0

(a− s)α−1

Γ(α)ds

)≤l∗ + (µ1R

ν1 + µ2Rν2)

(tα−1

(α− 1)Γ(α− 1)+

Nα

α∆Γ(α)

).

Hence,

|DβT (y)| =∣∣∣∣ 1

Γ(1− β)

∫ t

0

(t− s)−β(T y)′(s)ds∣∣∣∣ ≤ 1

Γ(1− β)

∫ t

0

(t− s)−β |(T u)′(s)|ds

≤ l∗t1−β

(1− β)Γ(1− β)+

(µ1Rν1 + µ2R

ν2)t1−β

(1− β)Γ(1− β)

(tα−1

Γ(α)+

|ν − µ|ηα

∆αΓ(α)+

|1− µ|aα

∆αΓ(α)

),

which gives |DβT (y)| ≤ l∗a1−β

Γ(2−β) +(µ1R

ν1+µ2Rν2 )

α∆Γ(α)Γ(2−β)

(∆aα−1 +Nα

)a1−β . Finally, we have,

∥T y(t)∥β ≤l + l∗a1−β

Γ(2− β)+µ1R

ν1 + µ2Rν2

∆αΓα

[(∆ + |µ|η)aα + |µ|aηα +

aβ−1

Γ(2− β)(∆aα−1 +Nα)

]≤R

3+ (µ1R

ν1 + µ2Rν2)Mα,β ≤ R

3+R

3+R

3= R.

Thus, T : W → W. The continuity of T follows from the continuity of f and G.Now, if (A3) holds, we choose 0 < R ≤ min

3(l + l∗a1−β

Γ(2−β) ), (1

3Mα,βµ1)

11−ν1 , ( 1

3Mα,βµ1)

11−ν1

and by the same

process as above, we obtain ∥T y(t)∥β ≤ R3 + (µ1R

ν1 + µ2Rν2)Mα,β ≤ R, which implies that T : W → W.

Now we show that T is completely continuous operator. Let K = max|g(t, y(t), cDβy(t))| : t ∈ [0, a], y ∈ W. Fort1, t2 ∈ [0, a] such that t1 < t2, we have

|T y(t1)− T y(t2)| =∣∣∣∣∫ a

0

(G(t1, s)−G(t2, s))g(s, y(s),cDβy(s))ds

∣∣∣∣≤K

[∫ t1

0

|G(t1, s)−G(t2, s)|ds+∫ t2

t1

|G(t1, s)−G(t2, s)|ds +

∫ a

t2

|G(t1, s)−G(t2, s)|ds],

which in view of (3.3.6) implies that

|T y(t1)− T y(t2)| ≤ K[ ∫ t2

0

(t2 − s)α−1

Γ(α)ds−

∫ t1

0

(t1 − s)α−1

Γ(α)ds

+|ν − µ|(t2 − t1)

∆Γ(α)

∫ η

0

(η − s)α−1ds+|1− µ|(t2 − t1)

∆Γ(α)

∫ a

0

(η − s)α−1ds]

≤ K

α∆Γ(α)[∆(tα2 − tα1 ) + (|ν − µ|ηα + |1− µ|aα)(t2 − t1)]

=K

αpΓ(α)

[∆(tα2 − tα1 ) +Nα(t2 − t1)

].

41

Also,

|cDβT (y)(t1)− cDβT (y)(t2)| =1

Γ(1− β)

∣∣∣∣∫ t2

0

(t2 − s)−β(T y)′(s)ds−∫ t1

0

(t1 − s)−β(T y)′(s)ds∣∣∣∣

≤ 1

Γ(1− β)

∣∣∣∣∫ t2

0

(t2 − s)−β(T y)′(s)ds−∫ t1

0

(t2 − s)−β(T y)′(s)ds∣∣∣∣

+1

Γ(1− β)

∣∣∣∣∫ t1

0

(t2 − s)−β(T y)′(s)ds−∫ t1

0

(t1 − s)−β(T y)′(s)ds∣∣∣∣

≤ 1

Γ(1− β)

(∫ t2

t1

(t2 − s)−β |(T y)′(s)|ds+∫ t1

0

((t2 − s)−β − (t1 − s)−β)|(T y)′(s)|ds)

≤ 1

Γ(1− β)

[∫ t2

t1

(t2 − s)−β

(∫ a

0

| ∂∂sG(s, t)||g(z, y(t), cDβy(t))|dt

)ds

+

∫ t1

0

((t2 − s)−β − (t1 − s)−β)

(∫ a

0

| ∂∂sG(s, t)||g(t, y(t), cDβy(t))|dt

)ds

]≤ K

α∆Γ(α)Γ(1− β)(∆aα−1 +Nα)

[∫ t2

t1

(t2 − s)−βds+

∫ t1

0

((t2 − s)−β − (t1 − s)−β)ds

]≤K(∆aα−1 +Nα)

α∆Γ(α)Γ(2− β)(t1−β

2 − t1−β1 ).

Hence, ∥T y(t1) − T y(t2)∥β ≤ Kα∆Γ(α)

[∆(tα2 − tα1 ) +Nα(t2 − t1) +

∆aα−1+Nα

Γ(2−β) (t1−β2 − t1−β

1 )], which implies that

∥T y(t1) − T y(t2)∥β → 0 as t1 → t2. Thus, by Arzela–Ascoli theorem, it follows that T is compact operator.Therefore, using the Schauder fixed point theorem, we conclude that T has at least one fixed point which is solutionof the boundary value problem (3.3.1), (3.3.2).

Now, if (A3) holds then, we can choose 0 < R ≤ max

(1

3Mα,βµ1

) 1ν1−1

,(

13Mα,βµ2

) 1ν2−1

. Using the same arguments

as above, we conclude that the problem (3.3.1), (3.3.2) has at least one solution.

Example 3.3.3. Consider the boundary value problem,

cDαy(t) =Γ(α+ 1)

64√π

et +Γ(α+ 1)

120e−κt(y(t))ν1 +

cos te−πt

187(cDβy(t))ν2 , t ∈ [0, 2], (3.3.10)

y(0) =3

7y(32

), y(1) =

9

11y(32

), (3.3.11)

where 1 < α < 2, 0 < β < 1 and κ > 0. Choose µ = 37 , ν = 9

11 , η = 32 , φ(t) =

Γ(α+1)64

√πet, µ1 = Γ(α+1)

120 ,

µ2 =1

187 and g(t, y, z) = Γ(α+1)64

√πet + Γ(α+1)

120 e−κt(y(t))ν1 + e−πt cos t187 (z(t))ν2 , then ∆ = 5847

1078 and for t ∈ [0, 2],

we have |g(t, y, z)| ≤ φ(t) + µ1|y|ν1 + µ1|z|ν2 .For 0 < ν1, ν2 < 1, condition (A2) of the Theorem 3.3.2 is satisfied and for ν1, ν2 > 1, condition (A3) of

the Theorem 3.3.2 is satisfied. Therefore the boundary value problem (3.3.10), (3.3.11) has a solution.

3.3.2 Uniqueness of solution

Theorem 3.3.4. Assume (A1), (A4) are satisfied: Furthermore, if there exists a constant

k <

(kαµ,ν

(aα +

ηαa1−β

Γ(2− β)

)+ (kαµa

α+1 + kα)

(1 +

a−β

Γ(2− β)

))−1

,

then the boundary value problem (3.3.1), (3.3.2) has a unique solution.

42

Proof. From (3.3.9), we have following estimates

|T (y)(t)− T (y)(t)| ≤∫ a

0

|G(t, s)||g(s, y(s), cDβy(s))− g(s, y(s), cDβy(s))|ds

≤k∥y − y∥β∆Γ(α)

[∫ t

0

(t− s)α−1ds+ (|µ|a+ |ν − µ|t)∫ η

0

(η − s)α−1ds+ (|µ|η + |1− µ|t)∫ a

0

(a− s)α−1ds

]≤k∥y − y∥β

∆αΓ(α)(tα + (|µ|a+ |ν − µ|t)ηα + (|µ|η + |1− µ|t)aα)

≤k∥y − y∥β(

aα

∆αΓ(α)+

(2|µ|+ |ν|)ηaα

∆αΓ(α)+

(1 + |µ|)aα+1

∆αΓ(α)

)= k∥y − y∥β

(kα + kαµ,νηa

α + kαµaα+1

),

|cDβ(T u)(t)− cDβ(T y)(t)| =∣∣∣∣ 1

Γ(1− β)

∫ t

0

(t− s)−β((T y)′(s)− (Ay)′(s))ds

∣∣∣∣≤ 1

Γ(1− β)

∫ t

0

(t− s)−β

(∫ 1

0

∣∣∣∣ ∂∂sG(s, z)∣∣∣∣ |g(z, y(z), cDβy(z)− g(z, y(z), cDβy(z)|dz

)ds

≤k∥y − y∥βΓ(1− β)

∫ t

0

(t− s)−β

(∫ 1

0

∣∣∣∣ ∂∂sG(s, z)∣∣∣∣ dz) ds.

Using equation (3.3.6), we obtain∫ 1

0

| ∂∂tG(t, s)|ds ≤

∫ t

0

(t− s)α−2

Γ(α− 1)ds+

|ν − µ|∆Γ(α)

∫ η

0

(η − s)α−1ds+|1− µ|∆Γ(α)

∫ a

0

(a− s)α−1ds

≤ tα−1

αΓ(α)+

|ν − µ|ηα + |1− µ|aα

∆αΓ(α)≤ aα−1

αΓ(α)+

|ν − µ|ηα + |1− µ|aα

∆αΓ(α).

Consequently,

|cDβ(T y)(t)− cDβ(T y)(t)| ≤ k∥y − y∥βt1−β

(1− β)Γ(1− β)

(aα−1

αΓ(α)+

|ν − µ|ηα + |1− µ|aα

∆αΓ(α)

)≤ k∥y − y∥βa1−β

Γ(2− β)

(aα−1

∆αΓ(α)+

(2|µ|+ |ν|)ηα

∆αΓ(α)+

(1 + |µ|)aα

∆αΓ(α)

)=k∥y − y∥βΓ(2− β)

(kαa−β + kαµ,νη

αa1−β + kαµaα−β+1

).

Hence, it follows that ∥T y − T y∥β ≤ q∥y − y∥β , where q = k(kαµ,ν(a

α + ηαa1−β

Γ(2−β) ) + (kαµaα+1 + kα)(1 + a−β

Γ(2−β) )).

Obviously, q < 1. Thus, by the contraction mapping principle guarantees the unique solution of the boundary valueproblem (3.3.1), (3.3.2) has a unique solution.

Example 3.3.5. Consider the following fractional differential equation,

cD32 y(t)

(24

√π + e−πt)(1 + |y|+ |cD

12 y(t)|

)= e−πt(|y|+ |cD

12 y(t)|), t ∈ [0, 1], (3.3.12)

y(0) =5

7y(

1

4), y(1) =

9

7y(

1

4). (3.3.13)

Set g(t, y, z) = e−πt(y(t)+z(t))(24

√π+e−πt)(1+y(t)+z(t))

, t ∈ [0, 1], y, z ∈ [0,∞). For t ∈ [0, 1] and y, y, v, v ∈ [0,∞), we have

|g(t, y, z)− g(t, y, v)| = e−πt

(24√π + e−πt)

∣∣∣∣ y(t) + z(t)

1 + y(t) + z(t)− y(t) + z(t)

1 + y(t) + z(t)

∣∣∣∣≤ e−πt(|y(t)− y(t)|+ |z(t)− z(t)|)

24√π + e−ct

≤ 1

24√π(|y(t)− y(t)|+ |z(t)− z(t)|).

For µ = 57 , ν = 9

7 , η = 14 , α = 3

2 and β = 12 , we have ∆ = 3, kαµ, ν = 76

63√π, kαµ = 16

3√π

and kα = 49√π.

Therefore q = 102261π < 1. Hence by Theorem 3.3.4 the fractional boundary value problem (3.3.12), (3.3.13)


43

3.4 Three–point boundary value problems (II)

In this section we study existence and uniqueness of solutions to a class of nonlinear boundary value

problems for fractional differential equations with three–point boundary conditions involving standard

Riemann–Liouville fractional derivative. A particular focus concerns the nonlinear term satisfying the

Caratheodory conditions in Lp[0, 1]. The studies dealing with existence of solutions for boundary value

problems using Caratheodory conditions in Lp[0, 1] space are well studied in [51,74,98,100] and references

therein.

N. Kosmatov [74] established sufficient conditions for the existence of solutions for the following bound-

ary value problem,

Dαy(t) = f(t, y(t), y′(t)), t ∈ [0, 1], y(0) = 0, y(1) = 0, (3.4.1)

where Dα is the Riemann–Liouville fractional derivative. We consider a more general boundary value

problem [90],

Dαy(t) = g(t, y(t),Dβy(t)), t ∈ [0, 1], (3.4.2)

y(0) = 0, Dβy(1) = γDβy(η), (3.4.3)

where, 1 < α ≤ 2, 0 < β ≤ 1, α − β > 1, ∆α,β := (1 − γηα−β−1) > 0. The nonlinear function g and the

boundary condition involves the fractional derivative of an unknown function. We study the existence as

well as uniqueness of the solutions by imposing some growth conditions on g.

Lemma 3.4.1. Let h ∈ C[0, 1], 1 < α ≤ 2, 0 < β ≤ 1, α− β > 1 and ∆α,β > 0. Then the linear problem

Dαy(t) = h(t), t ∈ [0, 1], (3.4.4)

y(0) = 0, Dβy(1) = γDβy(η), (3.4.5)

has solution

y(t) =Iαh(t) + Γ(α− β)tα−1

∆α,βΓ(α)

[γIα−βh(η)− Iα−βh(1)

]=

∫ 1

0G(t, s)h(s)ds, (3.4.6)

where

G(t, s) =

(t−s)α−1

Γ(α) + [γ(η−s)α−β−1−(1−s)α−β−1]tα−1

∆α,βΓ(α), s ≤ t, η ≥ s,

[γ(η−s)α−β−1−(1−s)α−β−1]tα−1

∆α,βΓ(α), t ≤ s ≤ η ≤ 1,

(t−s)α−1

Γ(α) − tα−1(1−s)α−β−1

∆α,βΓ(α), η ≤ s ≤ t ≤ 1,

− tα−1(1−s)α−β−1

∆α,βΓ(α), s ≥ t, η ≤ s,

(3.4.7)

is the Green’s function associated with the problem (3.4.4), (3.4.5).

Proof. From Theorem 2.2.14 and equation (3.4.4), we have

y(t) = Iαh(t) + c0tα−1 + c1t

α−2. (3.4.8)

44

Using boundary condition y(0) = 0 in (3.4.8) we obtain c1 = 0 and in view of Lemma 2.2.8 and Lemma

2.2.12, we have, c0 =Γ(α−β)

Γ(α)(1−γηα−β−1)

[γIα−βh(η)−Iα−βh(1)

]. Therefore the unique solution of the bound-

ary value problem (3.4.3), (3.4.4) is given by

y(t) = Iαh(t) + Γ(α− β)tα−1

∆α,βΓ(α)


]. (3.4.9)

For t ≤ η, using (3.4.9) we have the following estimates

y(t) =

∫ 1

0

(t− s)α−1

Γ(α)h(s)ds+

∫ t

0

γtα−1(η − s)α−β−1

∆α,βΓ(α)h(s)ds+

∫ η

t


∆α,βΓ(α)h(s)ds

−∫ t

0

tα−1(1− s)α−β−1

∆α,βΓ(α)h(s)ds−

∫ η

t

tα−1(1− s)α−β−1


∫ 1

η

tα−1(1− s)α−β−1


=

∫ t

0

[ (t− s)α−1

Γ(α)+

[γ(η − s)α−β−1 − (1− s)α−β−1]tα−1

∆α,βΓ(α)

]h(s)ds

+

∫ η

t

[γ(η − s)α−β−1 − (1− s)α−β−1]tα−1


∫ 1

η

tα−1(1− s)α−β−1

∆α,βΓ(α)h(s)ds =

∫ 1

0

G(t, s)h(s)ds.

For η < t, we have

y(t) =

∫ η

0

(t− s)α−1

Γ(α)h(s)ds+

∫ t

η

(t− s)α−1

Γ(α)h(s)ds+

∫ η

0



−∫ η

0

tα−1(1− s)α−β−1


∫ t

η

tα−1(1− s)α−β−1


∫ 1

t

tα−1(1− s)α−β−1


=

∫ η

0

[ (t− s)α−1

Γ(α)+

[γ(η − s)α−β−1 − (1− s)α−β−1]tα−1

∆α,βΓ(α)

]h(s)ds

+

∫ t

η

[ (t− s)α−1

Γ(α)− (1− s)α−β−1tα−1

∆α,βΓ(α)

]h(s)ds−

∫ 1

t

tα−1(1− s)α−β−1

∆α,βΓ(α)h(s)ds =

∫ 1

0

G(t, s)h(s)ds.

Consider the Banach space Bβ and define an operator T : Bβ → Bβ by

T y(t) =Iαg(s, y(s),Dβy(t)) +Γ(α− β)tα−1

∆α,βΓ(α)

[γIα−βg(η, y(η),Dβy(η))− Iα−βg(1, y(1),Dβy(1))

]=

∫ 1

0

G(t, s)g(s, y(s),Dβy(s))ds.

(3.4.10)

The fixed points of T are the solutions of the boundary value problem (3.4.2), (3.4.3).

Define Ω1,α,β,γ,η,q =1

((α−1)q+1)1q Γ(α)

+ γηα−β−1+1

q +1

((α−β−1)q+1)1q ∆α,βΓ(α)

and Ω2,α,β,γ,η,q =2−γηα−β−1(1−η

1q )

((α−β−1)q+1)1q ∆α,βΓ(α−β)

.

In the following, we prove that the operator T is completely continuous.

Lemma 3.4.2. Assume that g satisfies the Carathéodory conditions. Then T : Bβ → Bβ is completely

continuous.

Proof. Let M ⊂ Bβ be bounded i.e. there exists a positive number r > 0 such that ∥y∥β ≤ r for ally ∈ M . Since g is Carathéodory function. Therefore, there exists a nonnegative function ψr such that|g(t, y(t),Dβy(t))| ≤ ψr(t), for all y ∈ M and a.e. t ∈ [0, 1]. First we prove that T (M) ⊂ Bα is bounded.

45

For t ∈ [0, 1], y ∈M using the Hölder’s inequality, we have

|y(t)| ≤∫ t

0

(t− s)α−1

Γ(α)|ψr(s)|ds+

tα−1γ

∆α,βΓ(α)

[ ∫ η

0

(η − s)α−β−1|ψr(s)|ds+∫ 1

0

(1− s)α−β−1|ψr(s)|ds

≤ 1

Γ(α)

(∫ t

0

(t− s)(α−1)qds

) 1q

∥ψ∥p]+

1

∆α,βΓ(α)

[γ

(∫ η

0

(η − s)(α−β−1)qds

) 1q

+

(∫ 1

0

(1− s)(β−α−1)qds

) 1q

]∥ψr∥p ≤ tα−1+ 1

q

((α− 1)q + 1)1q Γ(α)

∥ψ∥p +γηα−β−1+ 1

q + 1

((α− β − 1)q + 1)1q∆α,βΓ(α)

∥ψr∥p

≤Ω1,α,β,γ,η,q∥ψr∥p,

Therefore

∥y(t)∥ ≤ Ω1,α,β,γ,η,q∥ψr∥p. (3.4.11)

Now, in view of Lemma 2.2.8 and Lemma 2.2.12, we have

|Dβy(t)| =∣∣∣Iα−βg(s, y(s),Dβy(t)) +

tα−β−1

∆α,β

[γIα−βg(η, y(η),Dβy(η))− Iα−βg(1, y(1),Dβy(1))

]∣∣∣≤∫ t

0

(t− s)α−β−1

Γ(α− β)|ψr(s)|ds+

tα−β−1

∆α,βΓ(α− β)

[γ

∫ η

0

(η − s)α−β−1|ψr(s)|ds+∫ 1

0

(1− s)α−β−1|ψr(s)|ds]

≤(∫ t

0

(t− s)(α−β−1)q

Γ(α− β)ds

) 1q

∥ψ∥p +1

∆α,βγ

(∫ t

0

((η − s)(α−β−1)

Γ(α− β)

)q

ds

) 1q

∥ψr∥p

+1

∆α,β

(∫ t

0

((1− s)(α−β−1)

Γ(α− β)

)q

ds

) 1q

∥ψr∥p

≤ tα−β−1+ 1q

((α− β − 1)q + 1)1q Γ(α− β)

∥ψ∥p +γηα−β−1+ 1

q + 1

((α− β − 1)q + 1)1q ∆α,βΓ(α− β)

∥ψr∥p

≤ 2− γηα−β−1(1− η1q )

((α− β − 1)q + 1)1q ∆α,βΓ(α− β)

∥ψr∥p = Ω2,α,β,γ,η,q∥ψr∥p,

Therefore

∥Dβy∥β ≤ Ω2,α,β,γ,η,q∥ψr∥p. (3.4.12)

Hence, from (3.4.11) and (3.4.12), we have

∥y∥β ≤ (Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q)∥ψr∥p. (3.4.13)

Next we prove that the operator T is equicontinuous. Let τ, t ∈ [0, 1], τ < t. For any y ∈M ,

|T y(t)− T y(τ)| =∣∣∣Iαh(t)− Iαh(τ) +

Γ(α− β)(tα−1 − τα−1)

∆α,βΓ(α)


]≤ 1

Γ(α)

[∫ τ

0

[(t− s)α−1 − (τ − s)α−1]|ψr(t)|ds+∫ t

τ

(t− s)α−1|ψr(t)|ds

+tα−1 − τα−1

1− γηα−β−1

∣∣∣γ ∫ η

0

(η − s)α−β−1|ψr(t)|ds−∫ 1

0

(1− s)α−β−1|ψr(t)|ds∣∣∣]

≤( (t− τ)(α−1)+ 1

q + tα−1+ 1q − τα−1+ 1

q

((α− 1)q + 1)1qΓ(α)

+(tα−1 − τα−1)γηα−β−1+ 1

q

((α− β − 1)q + 1)1q∆α,βΓ(α)

)∥ψr∥p

<(t− τ)(α−1)+ 1

q + tα−1+ 1q − τα−1+ 1

q + (tα−1 − τα−1)γηα−β−1+ 1q

((α− β − 1)q + 1)1q∆α,βΓ(α)

∥ψr∥p

46

|Dβ(T u)(t)−Dβ(T y)(τ)|(α− β)[ ∫ τ

0

[(t− s)α−β−1 − (τ − s)α−β−1]|ψr(t)|ds+∫ t

τ

(t− s)α−β−1|ψr(t)|ds

+tα−β−1 − τα−β−1

1− γηα−β−1

∣∣∣γ ∫ η

0

(η − s)α−β−1|ψr(t)|ds−∫ 1

0

(1− s)α−β−1|ψr(t)|ds∣∣∣]

≤( (t− τ)α−β−1+ 1

q + tα−β−1+ 1q − τα−β−1+ 1

q

((α− β − 1)q + 1)1qΓ(α− β)

+(tα−β−1 − τα−β−1)γηα−β−1+ 1

q

((α− β − 1)q + 1)1q ∆α,βΓ(α− β)

)∥ψr∥p

<(t− τ)α−β−1+ 1

q + tα−β−1+ 1q − τα−β−1+ 1

q + (tα−β−1 − τα−β−1)γηα−β−1+ 1q

((α− β − 1)q + 1)1q∆α,βΓ(α− β)

∥ψr∥p

Now, it is obvious that ∥T y(t) − T y(τ)∥β → 0 as τ → t. We conclude that T (M) is an equicontinuous

set. Obviously it is uniformly bounded since T (M) ⊂ Bβ . Thus, by Arzela–Ascoli theorem we conclude

that T : Bβ → Bβ is completely continuous.

Corollary 3.4.3. Assume that h ∈ Lp[0, 1], then the solution y(t) of the boundary value problem (3.4.4),

(3.4.5) satisfies

∥y∥β ≤(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)∥h∥p.


Theorem 3.4.4. Assume that the function g : [0, 1]×R2 → R satisfies the Carathéodory conditions B.0.8

and there exist functions a(t), b(t), φ(t) ∈ Lp[0, 1] such that

|g(t, u, v)| ≤ a(t)|y|+ b(t)|z|+ φ(t), (3.4.14)

for a.e. t ∈ [0, 1] and y, z ∈ R.Then boundary value problem (3.4.2), (3.4.3) has at least one solution

y ∈ Bβ provided that (Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)(∥a∥p + ∥b∥p∥

)< 1. (3.4.15)

Proof. By Lemma 3.4.2, the operator T is completely continuous. We apply Theorem 2.3.4 to obtain the

existence of at least one solution for boundary value problem (3.4.2), (3.4.3) in B. Now we prove that for

λ ∈ [0, 1] the possible solution set of family of boundary value problems

Dαy(t) = λg(t, y(t),Dβy(t)), t ∈ (0, 1), (3.4.16)

y(0) = 0, Dβy(1) = γDβy(η), (3.4.17)

is a priori bound in Lp[0, 1]. The domain of Riemann–Liouville linear operator

z = Dα(.) : Bβ → Lp[0, 1] is given by

D(L) :=y ∈ AC[0, 1], Dβy ∈ AC[0, 1], y(0) = 0, Dβy(1) = γDβy(η)

.

For y ∈ D(L), in view of Corollary 3.4.3, we have

∥y∥β ≤(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)∥g(t, y,Dβy(t))∥p

≤(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)(∥a∥p∥y∥+ ∥b∥p∥Dβy∥+ ∥φ∥p

)≤(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)((∥a∥p + ∥b∥p∥)∥y∥β + ∥φ∥p

).

47

Hence,

∥y∥β ≤

(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)∥φ∥p

1−(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)(∥a∥p + ∥b∥p∥

) .Therefore solution set of (3.4.16) and (3.4.17) is bounded in Bβ . All conditions of the Theorem 2.3.4 are

satisfied. Thus the problem (3.4.2), (3.4.3) has at least one solution in Bβ .


Theorem 3.4.5. Let g(t, y, z) satisfies Carathéodory conditions and there exists ϕ(t) ∈ Lp[0, 1] such that

|g(t, y, z)− g(t, y, z)| ≤ ϕ(t)(|y − y|+ |z − z|), for all y, z, y, z ∈ R. (3.4.18)

If 0 <∫ 10 φ(s)ds < ln 2, where φ(s) = max

0≤t≤1|G(t, s)||ϕ(s)|, then the boundary value problem (3.4.2), (3.4.3)


Proof. Let ω(t) =∫ t0 φ(s)ds, then ω′(t) = φ(t) for a.e. t ∈ [0, 1]. Define norm

∥y∥ω = max0≤t≤1

e−ω(t)|y(t)|+ max0≤t≤1

e−ω(t)|Dβy(t)|.

This norm is equivalent to the norm ∥.∥, because

e−∥ϱ∥1∥y∥ ≤ ∥y∥ω ≤ ∥y∥, where ∥ϱ∥1 =∫ 1

0|φ(t)|dt.

Consequently (Bω, ∥.∥ω) is a Banach space. Next, we show that T is a contraction on (Bω, ∥.∥ω). Let u,

y ∈ B, then

e−ω(t)|T y(t)− T y(t)| ≤e−ω(t)∫ 1

0|G(t, s)||g(s, y(s),Dβy(s))− g(s, y(s),Dβ y(s))|ds

≤e−ω(t)∫ 1

0|G(t, s)||ϕ(s)|(|y(s)− y(s)|+ |Dβy(s)−Dβ y(s))|)ds

≤e−ω(t)∫ 1

0φ(s)eω(s)

e−ω(s)|y(s)− y(s)|+ e−ω(s)|Dβy(s)−Dβ y(s))|)

ds

≤e−ω(t)(eω(1) − 1)∥y − y∥ω.

Therefore

∥T y(t)− T y(t)∥ω ≤ e−ω(t)(eω(1) − 1)∥y − y∥ω.

Since e−ω(t)(eω(1) − 1) < 1, the Banach contraction principle guarantees the existence of unique solution

for the boundary value problem (3.4.2), (3.4.3).

Theorem 3.4.6. Let g(t, y, z) satisfies Carathéodory conditions and there exists ϕ(t) ∈ Lp[0, 1] such that

(3.4.18) is satisfied. Furthermore, if(Ω1,α,β,γ,η,q + Ω2,α,β,γ,η,q

)∥ϕ∥p < 1 then the boundary value problem

(3.4.2), (3.4.3) has a unique solution.

48

Proof. From (3.4.18) we have

|g(t, y, z)| ≤ |g(t, 0, 0)|+ ϕ(t)(|y|+ |z|). (3.4.19)

By similar arguments used in the Theorem 3.4.4, we conclude that there exists at least one solution for the

boundary value problem (3.4.2), (3.4.3). In order to prove uniqueness of solution, we assume that there

exist two solutions y, y for the the boundary value problem (3.4.2), (3.4.3). Let z = y − y, then we have

Dαz(t) = g(t, y(t),Dβy(t))− g(t, y(t),Dβ y(t)), t ∈ [0, 1], (3.4.20)

z(0) = 0, Dβz(1) = γDβz(η), (3.4.21)

Now, in view of (3.4.18) and Corollary 3.4.3, we have following estimates

∥z∥β ≤(Ω1,α,β,γ,η,q +Ω2,α,β,γ,η,q

)(∥g(t, y(t),Dβy(t))− g(t, y(t),Dβ y(t))∥p


)∥ϕ∥p∥y − y∥β


)∥ϕ∥p∥z∥β

Which implies that z(t) = 0 for all t ∈ [0, 1]. Hence y(t) = y(t).

3.5 Multi–point boundary value problems

Multi–point boundary value problems for fractional differential equations have received some attention

from researchers quite recently, for example, Moustafa El-Shahed and Juan J. Nieto [44], Hussein A.H.

Salem [123], Z. Bai and Y. Zhang [19], B. Ahmad and S. Sivasundaram [7], W. Zhong, and W. Lin, [153]. In

this section, we establish sufficient conditions for the existence and uniqueness of solutions to the following

class of multi–point boundary value problem for nonlinear fractional differential equations [86]

Dαy(t) = g(t, y(t),Dβy(t)), t ∈ [0, 1] (3.5.1)

y(0) = 0, Dβy(1) =

m−2∑i=1

aiDβy(ξi) + y0, (3.5.2)

where 1 < α ≤ 2, 0 < β ≤ 1, 0 < ξi < 1 (i = 1, 2, · · · ,m − 2), ai ≥ 0 with, Λα,β =∑m−2

i=1 aiξα−β−1i < 1.

The nonlinear function f and boundary conditions involve fractional derivatives of unknown functions. For

the degenerate case (i.e. β = 1), the above problem is general compared to the problem studied in [153]

in the sense that the nonlinearity still depends on the first derivative. Moreover, the boundary conditions

are general too.

Lemma 3.5.1. Let 1 < α ≤ 2 and 0 < β ≤ 1. Then a function y ∈ Bβ is a solution of the boundary valueproblem (3.5.1), (3.5.2) if and only if it is solution of the integral equation

y(t) = Iαg(t, y(t),Dβy(t))+Γ(α− β)

Γ(α) (1− Λα,β)

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβy(ξi))−Iα−βg(1, y(1),Dβy(1))+ y0

)tα−1.

(3.5.3)

49

Proof. Assume that y ∈ Bβ is a solution of the boundary value problem (3.5.1), (3.5.2). By Theorem

2.2.15 and equation (3.5.1), we have

y(t) = Iαg(t, y(t),Dβy(t)) + c0tα−1 + c1t

α−2, for c0, c1 ∈ R. (3.5.4)

The boundary condition y(0) = 0 implies that c1 = 0. Using Lemma (2.2.8), equation (3.5.4) reduces to

Dβy(t) = Iα−βg(t, y(t),Dβy(t)) +Γ(α)tα−β−1

Γ(α− β)c0. (3.5.5)

Using the second boundary condition, from equation (3.5.5), we obtain

c0 =Γ(α− β)

Γ(α)(1− Λα,β)

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβy(ξi))− Iα−βg(1, y(1),Dβy(1)) + y0

).

Thus we have

y(t) = Iαg(t, y(t),Dβy(t))+Γ(α− β)

Γ(α) (1− Λα,β)

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβy(ξi))−Iα−βg(1, y(1),Dβy(1))+ y0

)tα−1.

Conversely suppose that y ∈ Bβ is a solution of the integral equation (3.5.3). Denoting the right handside of (3.5.3) by w(t), that is,

w(t) =Iαg(t, y(t),Dβy(t)) +Γ(α− β)

Γ(α) (1− Λα,β)

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβξiy(ξi))− Iα−βg(1, y(1),Dβ

1 y(1)) + y0

)tα−1.

Using the Lemma 2.2.12 we have Dαw(t) = g(t, y(t),Dβy(t)), that is Dαy(t) = g(t, y(t),Dβy(t)). On theother hand it is quite easy to verify that y(0) = 0. Now, in view of Lemma 2.2.8 , we obtain

Dβy(1) =1

1− Λα,β

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβξiy(ξi))− Iβ−αg(1, y(1),Dβ

1 y(1))Λα,β + y0

),

m−2∑i=1

aiDβy(ξi) =1

1− Λα,β

(m−2∑i=1


1 y(1))− y0Λα,β

)= Dβy(1) + y0.

Thus, we conclude that y ∈ Bβ is a solution of the boundary value problem (3.5.1), (3.5.2).


For convenience, we define:

ρα,β =(Γ(α− β) + Γ(α)

)∑m−2i=1 aiξ

α−βi + (2− Λα,β)Γ(α) + Γ(α− β),

Q1α,β = max

t∈[0,1]Iα|φ(t)|+ Γ(α−β)

Γ(α)(1−Λα,β)

(∣∣∣∑m−2i=1 aiIα−βφ(ξi)− Iα−βφ(1)

∣∣∣+ |y0|

),

Q2α,β = max

t∈[0,1]Iα−β |ϕ(t)|+ 1

1−Λα,β

(∣∣∣∑m−2i=1 aiIα−β |ϕ(ξi)| − Iα−β |ϕ(1)|

∣∣∣+ |y0|

),

Q3α,β = ρα,β +

((Γ(α− β))2 + Γ(α)

)|y0|, and

Kα,β = 1Γ(α+1) +

1(1−Λα,β)

(|∑m−2

i=1 aiξα−βi −1|

Γ(α)(α−β) +|∑m−2

i=1 aiξα−βi −Λα,β |

Γ(α−β+1)

).

50


a solution.

Proof. Define an operator T : Bβ → Bβ by

T y(t) = Iαg(t, y(t),Dβy(t)) +Γ(α− β)tα−1

Γ(α) (1− Λα,β)

(m−2∑i=1


1 y(1)) + y0

).

(3.5.6)

By Lemma 3.5.1, fixed points of the operator T are solutions of the boundary value problem (3.5.1), (3.5.2).

In view of the continuity of f , the operator T is continuous. Define U =y ∈ Bβ : ∥y∥Bβ

≤ R, t ∈ [0, 1].

Now we show that T : U → U . Choose

R ≥ max3(Q1

α,β +Q2α,β), (3µ1Kα,β)

11−ν1 , (3µ2Kα,β)

11−ν2

then for every y ∈ U , we have

|T y(t)| ≤ Iα|g(t, y(t),Dβy(t))|

+Γ(α− β)

Γ(α) (1− Λα,β)

(∣∣∣m−2∑i=1


1 y(1))∣∣∣+ |y0|

)

≤Iα|φ(t)|+ Γ(α− β)

Γ(α) (1− Λα,β)

∣∣∣m−2∑i=1

aiIα−βφ(ξi)− Iα−βφ(1)∣∣∣+ Γ(α− β)|y0|

Γ(α) (1− Λα,β)

+ (µ1Rν1 + µ2R

ν2)

(∫ t

0

(t− s)α−1

Γ(α)ds+

Γ(α− β)

Γ(α) (1− Λα,β)

∣∣∣m−2∑i=1

ai

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds−

∫ 1

0

(1− s)α−β−1

Γ(α− β)ds∣∣∣)

≤Q1α,β + (µ1R

ν1 + µ2Rν2)

(1

Γ(α+ 1)+

|∑m−2

i=1 aiξα−βi − 1|

Γ(α)(α− β)(1− Λα,β)

).

In view of Lemma 2.2.8 and equation (3.5.6), we have the following estimate:

|Dβ(T y)(t)| =∣∣∣Iα−βg(t, y(t),Dβy(t)) +

Γ(α− β)

Γ(α) (1− Λα,β)

×

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβy(ξi))− Iα−βg(1, y(1),Dβy(1)) + y0

)Dβtα−1

∣∣∣≤Iα−β |ϕ(t)|+ 1

1− Λα,β

(∣∣∣m−2∑i=1

aiIα−β |ϕ(ξi)| − Iα−β |ϕ(1)|∣∣∣+ |y0|

)

+ (µ1Rν1 + µ2R

ν2)∣∣∣ ∫ t

0

(t− s)α−β−1

Γ(α− β)ds+

1

1− Λα,β

(m−2∑i=1

ai

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds−

∫ 1

0

(1− s)α−β−11

Γ(α− β)

)∣∣∣≤Q2

α,β + (µ1Rν1 + µ2R

ν2)|∑m−2

i=1 aiξα−βi − Λα,β |

(1− Λα,β)Γ(α− β + 1).

Hence, ∥T y(t)∥Bβ≤ Q1

α,β +Q2α,β + (µ1R

ν1 + µ2Rν2)Kα,β ≤ R

3 + R3 + R

3 = R. Consequently, T : U → U .In the following, we show that the operator T is completely continuous.

Let N = maxt∈[0,1]

|g(t, y(t),Dβy(t))|, for y ∈ U . Taking t, τ ∈ [0, 1] such that t < τ . Then, for given ε > 0,

51

choose δ = εN

(αηα−1

Γ(α) +Q3

α,β(α−1)ηα−2

(1−Λα,β)Γ(α)Γ(α−β−1)

)−1

such that τ − t < δ.

|T y(τ)− T y(t)| ≤ |Iαg(τ, y(τ),Dβy(τ))− Iαg(t, y(t),Dβy(t))|

+Γ(α− β)

Γ(α) (1− Λα,β)

(∣∣∣m−2∑i=1


1 y(1))∣∣∣+ |y0|

)(τα−1 − tα−1)

≤N

∣∣∣∣∣∫ τ

0

(τ − s)α−1 − (t− s)α−1

Γ(α)ds

∣∣∣∣∣+

Γ(α− β)

Γ(α)(1− Λα,β)

(N(m−2∑

i=1

ai

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds+

∫ 1

0

(1− s)α−β−1

Γ(α− β)ds)+ |y0|

)(τα−1 − tα−1)

=N

Γ(α+ 1)(τα − tα) +

N(∑m−2

i=1 aiξα−βi + Γ(α− β)|y0|+ 1

)Γ(α)(1− Λα,β)(α− β)

(τα−1 − tα−1),

|(DβT y)(τ)− (DβT y)(t)| ≤∣∣Iα−βg(τ, y(τ),Dβy(τ))− Iα−βg(t, y(t),Dβ

t y(t))∣∣

+N

1− Λα,β

(m−2∑i=1

aiIα−βg(ξi, y(ξi),Dβy(ξi)) + Iα−βg(1, y(1),Dβy(1)) + |y0|)(τα−β − tα−β)

≤N

∣∣∣∣∣∫ τ

0

(τ − s)α−β−1 − (t− s)α−β−1

Γ(α− β)ds

∣∣∣∣∣+

N1− Λα,β

(m−2∑i=1

ai

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds+

∫ 1

0

(1− s)α−β−1

Γ(α− β)ds)+ |y0|

)(τα−β − tα−β)

≤ NΓ(α− β + 1)

(τα−β − tα−β) +N(∑m−2

i=1 aiξα−βi + |y0|+ 1

)(1− Λα,β)Γ(α− β + 1)

(τα−β−1 − tα−β−1)

≤N(2− Λα,β +

∑m−2i=1 aiξ

α−βi + |y0|

)(1− Λα,β)Γ(α− β + 1)

(τα−β−1 − tα−β−1).

By Mean Value Theorem, there exists η ∈ (0, 1) such that

τα − tα = αηα−1(τ − t), τα−1 − tα−1 = (α− 1)ηα−2(τ − t) and

τα−β−1 − tα−β−1 = (α − 1)ηα−β−2(τ − t) ≤ (α − β − 1)ηα−β−2(τ − t). Obviously ηα−β−2 ≥ ηα−2. This

gives τα−1 − tα−1 ≤ (α− 1)ηα−β−2(τ − t). Therefore, we have

∥T y(τ)− T y(t)∥ ≤ N

(αηα−1

Γ(α)+

Q3α,β(α− 1)ηα−β−2

(1− Λα,β)Γ(α)Γ(α− β − 1)

)(τ − t) < ε.

Thus, we conclude that T U is equicontinuous set. Also T U is a uniformly bounded set. We have T U ⊂ U .By the Arzela–Ascoli theorem, T : U → U is completely continuous. Hence by the Schauder fixed point

theorem the boundary value problem (3.5.1), (3.5.2) has at least one solution.


a solution.

Example 3.5.4. Consider the fractional boundary value problem

Dαy(t) =ω1t

νeγt

1 + t2+ω2 sinπt√π + |y|

|y|ν1 + ω3e−υt√

2 + |Dαy||Dβy|ν2 , t ∈ (0, 1), (3.5.7)

y(0) = 0, Dβy(1) =1

2Dβy

(14

)+

1

4Dβy

(12

)+

1

4Dβy

(34) + eπ, (3.5.8)

52

where 1 < α < 2, 0 < β < 1, α−β > 1 and ν, γ, ωi ∈ R+ (i = 1, 2, 3), υ ≥ 0. Choose ai =(i−1)!2i

and ξi = i4

(i = 1, 2, 3). Then Λα,β =∑3

i=1(i−1)!2i

( i4)α−β−1 < (34)

α−β−1 < 1. For g(t, y, z) = ω1tνeγt

1+t2+ ω2 sinπt√

π+|y||y|ν1 +

ω3e−υt√2+|Dαy|

|Dβy|ν2 , t ∈ (0, 1). Thus, we have |g(t, y, z)| < φ(t) + µ1|y|ν1 + µ2|z|ν2 , where φ(t) = ω1tνeγt

1+t2,

µ1 =ω2√π, µ2 = ω3√

2. For 0 < ν1, ν2 < 1, the assumption (A2) is satisfied and for ν1, ν2 > 1, the assumption

(A3) holds. Therefore, by Theorems 3.5.2 and 3.5.5, the boundary value problem (3.5.1), (3.5.2) has a

solution.


Theorem 3.5.5. Assume that (A1) and (A4) hold. If there exists k ∈ R+, such that

k <( 1

Γ(α+ 1)+

ρα,β(1− Λα,β)Γ(α)Γ(α− β + 1)

)−1.

Then the boundary value problem (3.5.1), (3.5.2) has a unique solution.

Proof. By assumption (A4), we have

|T y(t)− T y(t)| = |Iα(g(t, y(t),Dβy(t))− g(t, y(t),Dβy(t)))|+ Γ(α− β)

Γ(α) (1− Λα,β)

×

(∣∣∣m−2∑i=1

aiIα−β(g(ξi, y(ξi),Dβξiy(ξi))− g(ξi, y(ξi),Dβ

ξiy(ξi)))− Iα−β(g(1, y(1),Dβy(1))− g(1, y((1),Dβy((1)))

∣∣∣)

≤k∥y − y∥β

(∫ t

0

(t− s)α−1

Γ(α)ds+

Γ(α− β)

Γ(α)(1− Λα,β)

(m−2∑i=1

aiIα−β(1)

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds+

∫ 1

0

(1− s)α−β−1

Γ(α− β)ds))

<k( 1

Γ(α+ 1)+

(∑m−2i=1 aiξ

α−βi + 1

)Γ(α)(1− Λα,β)(α− β)

)∥y − y∥β ,

|(DβT y(t)− (Dβ(T y)(t)| ≤

(Iα−β |g(t, y(t),Dβy(t))− g(t, y(t),Dβy(t))|

+1

1− Λα,β

(m−2∑i=1

aiIα−β |g(ξi, y(ξi),Dβy(ξi))− g(ξi, y(ξi),Dβy(ξi))|

+ Iα−β |g(1, y(1),Dβ1 y(1))− g(1, y(1),Dβy(1))|

))tα−β−1

≤k∥y − y∥β

(∫ t

0

(t− s)α−β−1

Γ(α− β)ds+

1

1− Λα,β

(m−2∑i=1

ai

∫ ξi

0

(ξi − s)α−β−1

Γ(α− β)ds+

∫ 1

0

(1− s)α−β−1

Γ(α− β)ds

))

≤k∥y − y∥β

(tα−β

Γ(α− β + 1)+

∑m−2i=1 aiξ

α−βi + 1

(1− Λα,β)Γ(α− β + 1)

)<k(2 +

∑m−2i=1 aiξ

α−βi − Λα,β

)(1− Λα,β)Γ(α− β + 1)

∥y − y∥β .

Thus, we have ∥T y(t)− (T y)(t)∥β < L∥y−y∥β, where L = k(

1Γ(α+1) +

ρα,β

(1−Λα,β)Γ(α)Γ(α−β+1)

)< 1. Hence,

by the Banach fixed point theorem, the multi–point boundary value problem problem (3.5.1), (3.5.2) has

a unique solution.

53

Example 3.5.6. Consider the five–point fractional boundary value problem

D32 y(t)(|D

12 y|+ |y|+ 1) =

(|y|+ |D12 y(t)|)e−κt

(42√π + 85e−κt)

, t ∈ (0, 1), (3.5.9)

y(0) = 0, D12 y(1) =

5

7D

12 y(25

)+

2

3D

12 y(35

)+

8

21D

12 y(45

), (3.5.10)

where κ > 0. Let g(t, y, z) = e−γt(y+z)(42

√π+85e−κt)(1+y+z)

, t ∈ (0, 1), y, z ∈ [0,∞), α = 32 , β = 1

2 , a1 = 57 , a2 = 2

3 ,

a3 = 821 , and ξk = k+1

5 (k = 1, 2, 3). Also Λα,β = 3435 < 1. For y, y, z, z we have |g(t, y, z) − g(t, y, z)| <

142

√π+85

(|y + z| + |y + z|

). Hence the condition (A4) is satisfied. By calculations, L = 146

105√π< 1. Thus

by Theorem 3.5.2 the boundary value problem (3.5.1), (3.5.2) has a unique solution.

3.6 Boundary value problems with integral boundary conditions

In [24], Cannon initiated the study of nonlocal boundary value problems with integral boundary conditions.

Since then the subject has been addressed by many authors [5,34,64,148]. Boundary value problems with

integral boundary conditions arises in thermoplasticity, papulation dynamics, underground water flow and

blood flow problems. For details,we refer to [29, 32, 47, 131]. This interesting class of boundary value

problem includes two-point, three–point and multi–point boundary value problems as special cases.

In this section, we study existence and uniqueness of solutions to a nonlinear fractional differential

equations with integral boundary conditions in an ordered Banach space. We use the Caputo fractional

differential operator and the nonlinearity depends on the fractional derivative of an unknown function. For

the existence of solutions, we employ the nonlinear alternative of the Leray–Schauder and a uniqueness

result is established using the Banach fixed point theorem. We are concerned with existence and uniqueness

of solutions to nonlinear fractional differential equations of the type [89]

cDαy(t) = f(t, y(t), cDβy(t)), for t ∈ [0, l], (3.6.1)

subject to the integral boundary conditions

py(0)− qy′(0) =

∫ l

0g(s, y(s))ds, γy(1) + δy′(1) =

∫ l

0h(s, y(s))ds, (3.6.2)

where 0 < β ≤ 1, 1 < α ≤ 2, p, δ > 0, q, γ ≥ 0 (or p, δ ≥ 0, q, γ > 0). The functions f, g and h are assumed

to be continuous. For particular case, when β = 0, p = q = γ = δ = 1, the existence of solutions for the

boundary value problem (3.6.1), (3.6.2) is studied in [3]. The boundary value problem (3.6.1), (3.6.2) is

more general than that considered in [3] in the sense that the nonlinear function g depends on fractional

derivative of y as well.

Lemma 3.6.1. A function y ∈ Bβ is solution of fractional boundary value problem (3.6.1), (3.6.2) if and

only if y ∈ Bβ is solution of the fractional integral equation

y(t) =

∫ l

0G(t, s)f(s, y(s), cDβy(s))ds+ φ(t), (3.6.3)

54

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

G(t, s) =

(t−s)α−1

Γ(α) − (q+pt)γ(l−s)α−1

∆p,qΓ(α)− (q+pt)δ(α−1)(l−s)α−2

∆p,qΓ(α), 0 ≤ s ≤ t,

− (q+pt)γ(l−s)α−1

∆p,qΓ(α)− (q+pt)δ(α−1)(l−s)α−2

∆p,qΓ(α), 0 ≤ t ≤ s,

(3.6.4)

φ(t) =δ + γ(l − t)

∆p,q

∫ l

0g(s, y(s))ds+

q(1 + t)

∆p,q

∫ l

0h(s, y(s))ds and ∆p,q = p(δ + γl) + qγ.

Proof. Taking into account the equation (3.6.1) and Lemma 2.2.24, we infer that the general solution is

y(t) = Iαf(t, y(t), cDβy(t)) + c0 + c1t, c0, c1 ∈ R. (3.6.5)

Using boundary condition (3.6.2), we obtain

c0 = − q

∆p,q(γIαf(l, y(l), cDβy(l))+ δIα−1f(l, y(l), cDβy(l)))+

1

∆p,q

∫ l

0((δ+γl)g(s, y(s))+ qh(s, y(s)))ds,

(3.6.6)

c1 = − p

∆p,q(γIαy(l) + δIα−1y(l)) +

1

∆p,q

∫ l

0(ph(s, y(s))− γg(s, y(s)))ds. (3.6.7)

Therefore, we have the integral equation

y(t) =Iαf(s, y(s), cDβy(t))− 1

∆p,q(γIαf(l, y(l), cDβy(l)) + δIα−1y(l))(q + pt)

+δ + γ(l − t)

∆p,q

∫ l

0

g(s, y(s))ds+(q + pt)

∆p,q

∫ l

0

h(s, y(s))ds,

which can be written as

y(t) =

∫ l

0G(t, s)f(s, y(s), cDβy(s))ds+ φ(t).

Conversely, let y ∈ Bβ satisfy (3.6.3) and denote the right hand side of equation (3.6.3) by w(t). Then, byLemmas 2.2.5 and Lemma 2.2.23, we obtain

w(t) =

∫ l

0

G(t, s)f(s, y(s), cDβy(s))ds+ φ(t)

=Iαf(t, y(t), cDβy(t))− γ

∆p,qIαf(l, y(l), cDβy(l))(q + pt)− δ

∆p,qIα−1f(l, y(l), cDβy(l))(q + pt) + φ(t),

which implies that

cDαw(t) =cDαIαf(t, y(t), cDβy(t))− γ

∆p,q

cDαIαf(l, y(l), cDβy(l))(q + pt)

− δ

∆p,q

cDαIα−1f(l, y(l), cDβy(l))(q + pt) = f(t, y(t), cDβy(t)).

Hence, y(t) is a solution of the fractional differential equation cDαy(t) = f(t, y(t), cDβy(t)). Also, one can

easily verify that py(0)− qy′(0) =∫ l0 g(s, y(s))ds, γy(t) + δy′(l) =

∫ l0 h(s, y(s))ds.

Remark 3.6.2. The Green function (3.6.4) satisfies the following properties:

(i)∫ 10 |G(t, s)| ≤ lα

Γ(α+1) +q+plα+1

∆p,qΓ(α)

(γα + δ

l

), t ∈ (0, 1).

(ii)∫ l0 |

∂∂tG(t, s)|ds ≤

lα−1

Γ(α) +q+plα

∆p,qΓ(α−1)

(γ

α−1 + δl

).

55

Proof. (i) By definition of G(t, s),∫ l

0

|G(t, s)|ds ≤∫ t

0

(t− s)α−1

Γ(α)ds+

∫ l

0

(q + pt)

(γ(l − s)α−1

∆p,qΓ(α)+δ(α− 1)(l − s)α−2

∆p,qΓ(α)

)ds

≤ tα

αΓ(α)+

q + pt

∆p,qαΓ(α)(γlα + δlα−1) ≤ lα

Γ(α+ 1)+q + plα+1

∆p,qΓ(α)

(γα+δ

l

).

(ii) Also,∫ l

0

| ∂∂tG(t, s)|ds ≤

∫ t

0

(t− s)α−2

Γ(α− 1)ds+

p

∆p,qΓ(α)

∫ l

0

(γ(l − s)α−1 + δ(α− 1)(l − s)α−2)ds

≤ tα−1

Γ(α)+

γplα

∆p,q(α− 1)Γ(α− 1)+

δplα−1

∆p,qΓ(α− 1)≤ lα−1

Γ(α)+

q + plα

∆p,qΓ(α− 1)

( γ

α− 1+δ

l

).

Define an operator T : Bβ → Bβ by

T y(t) =∫ l

0G(t, s)f(s, y(s), cDαy(s))ds+ φ(t). (3.6.8)

Then, the boundary value problem (3.6.1), (3.6.2) is equivalent to the fixed point problem T y = y.

In what follows, we establish an existence result using the nonlinear alternative of Leray–Schauder type

by imposing some growth conditions on f , g and h.

(A5) There exist continuous functions ψif : [0,∞) → (0,∞), ϕif ∈ L1[0, l], (i = 1, 2) such that |f(t, y, z)| ≤ϕ1f (t)ψ

1f (|y|) + ϕ2f (t)ψ

2f (|z|).

(A6) There exist continuous and nondecreasing function ψg : [0,∞) → (0,∞) and a function ϕg ∈ L1[0, l]

such that |g(t, y)| ≤ ϕg(t)ψg(|y|), for t ∈ [0, l],

(A7) There exist continuous and nondecreasing function ψh : [0,∞) → (0,∞) and a function ϕh ∈ L1[0, l]

such that |h(t, y)| ≤ ϕh(t)ψh(|y|), for t ∈ [0, l].

Define a =∫ l0 ϕg(s)ds, b =

∫ l0 ϕh(s)ds, Nα := lα

Γ(α+1) + q+plα+1

∆p,qΓ(α)

(γα + δ

l

), kη,α :=

(ψ1f (η)∥ϕ1f∥L1 +

ψ2f (η)∥ϕ2f∥L1

)Nα and Mβ,η =

l1−β

Γ(2−β)

(kη,α−1 +

aγψg(η)∆p,q

+ bpψh(η)∆p,q

)kη,α.

(A8) There exists r > 0 such that rMβ,r

> 1,

(A9) There exists constants k1, k2 > 0 such that

|g(t, y)− g(t, y)| ≤ k1|y − y| and |h(t, y)− h(t, y)| ≤ k2|y − y|, for each t ∈ [0, l].


Theorem 3.6.3. Under the assumptions (A5) − (A8), the fractional boundary value problem (3.6.1),

(3.6.2) has at least one solution on [0, l].

Proof. In view of the continuity of f , g and h, the operator T is continuous. To show that T mapsbounded sets into bounded sets in Bβ , choose η > 0 (fixed).Let ρ ≥ maxη,Mβ,η. Define U = y ∈ B : ∥y∥β < η and V = y ∈ B : ∥y∥β < ρ. For t ∈ [0, l] andy ∈ U , we have

|T y(t)| ≤∫ l

0

|G(t, s)||f(s, y(s), cDβy(s))|ds+ δ + γ(l − t)

∆p,q

∫ l

0

|g(s, y)|ds+ q + pt

∆p,q

∫ l

0

|h(s, y)|ds.

56

Using (A5)− (A7) and Remark 3.6.2, we obtain

|T y(t)| ≤

(lα

Γ(α+ 1)+q + plα+1

∆p,qΓ(α)

( γα+δ

l

))(|ϕ1f (s)|ψ1

f (∥y∥β) + |ϕ2f (s)|ψ2f (∥y∥β))ds

+δ + γ(l − t)

∆p,q

∫ l

0

|ϕg(s)|ψg(∥y∥β)ds+q + pt

∆p,q

∫ l

0

|ϕh(s)|ψh(∥y∥β)ds

≤(ψ1f (η)∥ϕ1f∥L1 + ψ2

f (η)∥ϕ2f∥L1

)Nα

+(δ + 2γl)ψg(η)

∆p,q

∫ l

0

ϕg(s)ds+(q + pl)ψh(η)

∆p,q

∫ l

0

ϕh(s)ds

≤kη, α +a

∆p,q(δ + 2γl)ψg(η) +

b

∆p,q(q + pl)ψh(η) = kη, α +Mη,

|(T y)′(t)| ≤∫ l

0

| ∂∂tG(t, s)||f(s, y(s), cDβy(s))|ds+ γ

∆p,q

∫ l

0

|g(s, y)|ds+ p

∆p,q

∫ l

0

|h(s, y)|ds

≤

(lα−1

Γ(α)+

q + plα

∆p,qΓ(α− 1)

( γ

α− 1+δ

l

))(|ϕ1f (s)|ψ1

f (∥y∥β) + |ϕ2f (s)|ψ2f (∥y∥β)

)ds

+γ

∆p,q

∫ l

0

|ϕg(s)|ψg(∥y∥β)ds+p

∆p,q

∫ l

0


≤(ψ1f (η)∥ϕ1f∥L1 + ψ2

f (η)∥ϕ2f∥L1

)Nα−1 +

γψg(η)

∆p,q

∫ l

0

ϕg(s)d+pψh(η)

∆p,q

∫ l

0

ϕh(s)ds

≤kη, α−1 +aγψg(η)

∆p,q+bpψh(η)

∆p,q.

Hence, it follows that

|cDβT (y)(t)| ≤ 1

Γ(1− β)

∫ t

0

(t− s)−β |(T y)′(s)|ds ≤ l1−β

Γ(2− β)

(kη,α−1 +

aγψg(η)

∆p,q+bpψh(η)

∆p,q

).

Therefore, ∥T y∥β ≤ ρ, which implies that T y ∈ V . Hence, T maps bounded sets into bounded setsin Bβ . Now, we show that T maps bounded sets into equicontinuous sets of Bβ . For this, we takeK = max|f(t, y(t), cDβy(t))| : y ∈ U, t ∈ [0, l], L1 = max|g(t, y(t)| : y ∈ U, t ∈ [0, l] and L2 =

max|h(t, y(t)| : y ∈ U, t ∈ [0, l]. Taking t, τ ∈ (0, l] such that t < τ and y ∈ U . Then,

|T y(τ)− T y(t)| ≤∫ l

0

|G(τ, s)−G(t, s)||f(s, y(s), cDβy(s))|ds+ |φ(τ − t)|

≤K

[∫ t

0

|G(τ, s)−G(t, s)|ds+∫ τ

t

|G(τ, s)−G(t, s)|ds

+

∫ l

τ

|G(τ, s)−G(t, s)|ds

]+

l

∆p,q(γL1 + pL2)(τ − t)

≤K[∫ t

0

((τ − s)α−1 − (t− s)α−1

Γ(α)+

(pγ(l − s)α−1

Γ(α)+pδ(α− 1)(l − s)α−2)

∆p,qΓ(α)

)(τ − t)

)ds

+

∫ τ

t

((τ − s)α−1

Γ(α)+

(pγ(l − s)α−1

Γ(α)+pδ(α− 1)(l − s)α−2)

∆p,qΓ(α)

)(τ − t)

)ds

+

∫ l

τ

(pγ(l − s)α−1

Γ(α)+pδ(α− 1)(l − s)α−2)

∆p,qΓ(α)

)(τ − t)ds

]+

l

∆p,q(γL1 + pL2)(τ − t)

≤K

[∫ l

0

(pγ(l − s)α−1

Γ(α)+pδ(α− 1)(l − s)α−2)

∆p,qΓ(α)

)(τ − t)ds

]+

l

∆p,q(γL1 + pL2)(τ − t)

=Kp

∆p,qΓ(α)

(γlα

α+ δlα−1

)(τ − t) +

K(τα − tα)

qΓ(α)+

l

∆p,q(γL1 + pL2)(τ − t),

57

|cDβT y(τ)− cDβT y(t)| = 1

Γ(1− β)

∣∣∣∣∫ τ

0

(τ − s)−β(T y)′(s)ds−∫ t

0

(t− s)−β(T y)′(s)ds∣∣∣∣

≤ 1

Γ(1− β)

(∫ τ

t

(τ − s)−β |(T y)′(s)|ds+∫ t

0

((τ − s)−β − (t− s)−β)|(T y)′(s)|ds)

≤ 1

Γ(1− β)

[∫ τ

t

(τ − s)−β

(∫ l

0

| ∂∂sG(s, z)||f(z, y(z), cDβy(z))|dz + φ′(s)

)ds

+

∫ t

0

((τ − s)−β − (t− s)−β)

(∫ l

0

| ∂∂sG(s, z)||f(z, y(z), cDβy(z))|dz + φ′(s)

)ds

]

≤ K

∆p,qΓ(α)Γ(1− β)((pδ + p)lα−1 + pγlα)

[∫ τ

t

(τ − s)−βds+

∫ t

0

((τ − s)−β − (t− s)−β)ds

]+

(γL1 + pL2)l

∆p,qΓ(1− β)

[∫ τ

t

(τ − s)−βds+

∫ t

0

((τ − s)−β − (t− s)−β)ds

]≤ K

∆p,qΓ(α)Γ(2− β)((pδ + p)lα−1 + pγlα)(τ1−β − t1−β) +

(γL1 + pL2)l

∆p,qΓ(2− β)(τ1−β − t1−β)

≤(

K

∆p,qΓ(α)Γ(2− β)((pδ + p)lα−1 + pγlα) +

(γL1 + pL2)l

∆p,qΓ(2− β)

)(τ1−β − t1−β).

Obviously, |T y(τ)−T y(t)| → 0 and |cDβT y(τ)−cDβT y(t)| → 0 as t→ τ . Therefore, ∥T y(τ)−T y(t)∥ →0, as t→ τ . By the Arzela–Ascoli theorem it follows that T : Bβ → Bβ is completely continuous.Define U1 = y ∈ Bβ : ∥y∥β < r and assume that there exists y ∈ ∂U1 such that y = λT y for someλ ∈ (0, 1). Therefore, y(t) = λ

(∫ l0 G(t, s)f(s, y(s),

cDβy(s))ds+ φ(t)).

In view (A5)− (A7) and Remark 3.6.2, we have

|y(t)| <∫ 1

0

|G(t, s)|ds+ |φ(t)|

≤Nα(|ϕ1f (s)|ψ1f (∥y∥β) + |ϕ2f (s)|ψ2

f (∥y∥β))

+δ + γ(l − t)

∆p,q

∫ l

0

|ϕg(s)|ψg(∥y∥β)ds+q + pt

∆p,q

∫ l

0


≤k∥y∥β , α +a

∆p,q(δ + 2γl)ψg(∥y∥β) +

b

∆p,q(q + pl)ψh(∥y∥β).

Similarly,

|y′(t)| < k∥y∥β ,α−1 +aγψg(∥y∥β)

∆p,q+bpψh(∥y∥β)

∆p,q.

Hence,

|cDβy(t)| < 1

Γ(1− β)

∫ t

0

(t− s)−β |y′(s)|ds ≤ l1−β

Γ(2− β)

(k∥y∥β , α−1 +

aγψg(∥y∥β)∆p,q

+bpψh(∥y∥β)

∆p,q

).

Therefore, ∥y∥βMβ,∥y∥β

< 1, a contradiction to (A8). Hence, y = λT y for y ∈ ∂U1, λ ∈ [0, 1]. By Theorem

2.3.5, the boundary value problem (3.6.1), (3.6.2) has at least one solution.


The uniqueness result is based on the Banach contraction principal.

Theorem 3.6.4. Assume that (A4), (A9) hold. Furthermore, if k < 13

(Nα + l1−β

Γ(2−β)Nα−1

)−1, k1 <

∆p,q

3l

(δ + 2γl + pl2−β

Γ(2−β)

)−1and k2 <

∆p,q

3l

(q + pl + pl1−β

Γ(2−β)

)−1. Then the boundary value problem (3.6.1),

(3.6.2) has a unique solution.

58

Proof. Let y, y ∈ Bβ , then for each t ∈ [0, l] we have

|T y(t)− T y(t)| ≤∫ l

0

G(t, s)|f(s, y(s), cDβy(s))− f(s, y(s), cDβy(s))|ds

+δ + γ(l − t)

∆p,q

∫ l

0

|g(s, y(s))− g(s, y(s))|ds+ (q + pt)

∆p,q

∫ l

0

|g(s, y(s))− g(s, y(s))|ds

≤∥y − y∥β

(k

(lα

Γ(α+ 1)+q + plα+1

∆p,qΓ(α)

(γα+δ

l

))+

l

∆p,q(k1(δ + γ(l − t)) + k2(q + pt)

)

≤∥y − y∥β(kNα +

l

∆p,q(k1(δ + 2γl) + k2(q + pl))

).

Also,

|cDβT y(t)− cDβT y(t)| =∣∣∣∣ 1

Γ(1− β)

∫ t

0

(t− s)−β((T y)′(s)− (T y)′(s))ds∣∣∣∣

≤ 1

Γ(1− β)

∫ t

0

(t− s)−β

(∫ l

0

∣∣∣∣ ∂∂sG(s, z)∣∣∣∣ |f(z, y(z), cDβy(z)− f(z, y(z), cDβy(z)|dz

+γ

∆p,q

∫ l

0

|g(z, y(z))|dz + p

∆p,q

∫ l

0

|h(z, y(z))|dz

)ds

≤ l1−β

Γ(2− β)

(kNα−1 +

(γk1 + pk2)l

∆p,q

)∥y − y∥β .

Therefore, ∥T y − T y∥β ≤ L∥y − y∥β , where

L =

(Nα +

l1−β

Γ(2− β)Nα−1

)k +

lk1∆p,q

(δ + 2γl +

pl2−β

Γ(2− β)

)+

lk2∆p,q

(q + pl +

pl1−β

Γ(2− β)

)< 1.

By the contraction mapping principle, the boundary value problem (3.6.1), (3.6.2) has a unique solution.

Example 3.6.5. Consider the following fractional differential equation,

cD 32 y(t)(e−λt + 124

√π)(|y|+ |cD 1

2 y(t)|+ 1) = e−λt(|y|+ |cD 12 y(t)|), t ∈ [0, 1], (3.6.9)

y(0)− y′(0) =1

9

∫ 1

0

|y(s)|e−s

(1 + |y(s)|)ds, y(1) + y′(1) =

1

36

∫ 1

0

|y(s)| sin sds. (3.6.10)

Set f(t, y, z) = e−λt(y(t)+z(t))(24

√π+e−λt)(1+y(t)+z(t))

, g(t, y) = y(t)e−t

9(1+y(t)) and h(t, y) = y(t) sin t36 for t ∈ [0, 1] and y, z ∈

[0,∞). Let y, y, z, z ∈ [0,∞), then we have

|f(t, y, z)− f(t, y, z)| = e−λt

(124√π + e−λt)

∣∣∣∣ y(t) + z(t)

1 + y(t) + z(t)− y(t) + z(t)

1 + y(t) + z(t)

∣∣∣∣≤ e−λt(|y(t)− y(t)|+ |z(t)− z(t)|)

(124√π + e−λt)(1 + y(t) + z(t))(1 + y(t) + z(t))

≤ e−λt(|y(t)− y(t)|+ |z(t)− z(t)|)124

√π + e−λt

≤ 1

124√π(|y(t)− y(t)|+ |z(t)− z(t)|).

By some calculations, ∆p,q = 3, Nα = 143√π, 13Nα

(1 + l−β

Γ(2−β)

)−1= π

14(√π+2)

. Here

k = 1124

√π< π

14(√π+2)

. Furthermore, |g(t, y)− g(t, y)| ≤ e−t|y(t)−y(t)|9(1+y(t))(1+y(t)) ≤

19 |y(t)− y(t)|, where k1 = 1

9 <∆p,q

3l(δ+2γl) = 13 . Also, note that |h(t, y) − h(t, y)| = sin t

36 |y(t) − y(t)| ≤ 136 |y(t) − y(t)|, where k2 = 1

36 <∆p,q

3l(q+pl) = 13 . All conditions of the Theorem 3.6.4 are satisfied. Therefore the boundary value problem

(3.6.9), (3.6.10) has a unique solution.

Chapter 4

Existence and multiplicity of positive

solutions

In many situations, we require positive solutions of boundary value problems that arise in the modeling of

problems from natural and social sciences such as papulation dynamics, engineering physics and finance.

One of the most useful tool which have been used effectively for proving existence of positive solutions for

boundary value problems, is Krasnosel’skii’s fixed point theorem on cone expansion and compression and

its norm type version due to Gau [50]. Most early work related to the application of Krasnosel’skii’s fixed

point theorem to eigne value problems to establish the interval of parameter for which there exists at least

one positive solution, was carried out by Wang [137]. Since this pioneering work, notable contributions

to the existence theory of positive solutions have been carried out [63, 99, 139]. The attention drawn to

the theory of existence and multiplicity of positive solutions for fractional differential equations is quite

evident from the increasing number of recent publications. To identify few, we refer to [20,81,145,146] and

references cited therein. However, few results can be found in the literature concerning existence of positive

solutions to nonlinear three–point boundary value problems for fractional differential equations [20,81].

In this chapter, we are concerned with the existence, multiplicity and uniqueness of positive solutions

to boundary value problems of fractional differential equations. A particular focus concerns boundary

value problems with three–point boundary conditions. In section 4.1, we investigate sufficient conditions

for the existence and multiplicity of positive solutions to nonlinear three–point boundary value problems

for fractional differential equations of type

cDαy(t) + a(t)g(t, y(t)) = 0, t ∈ [0, 1], n− 1 < α ≤ n,

satisfying boundary conditions

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = ξy(η),

We apply superlinear, sublinear type growth conditions on nonlinearity which will enable us to apply

the Guo-Krasnoselskii and the Leggett-William fixed point theorems to establish several existence and

multiplicity results for positive solutions. In section 4.2, we study the existence and uniqueness of positive

solutions for three–point boundary value problems for fractional differential equations of the type [85]:

59

60

cDαy(t) + g(t, y(t)) = 0, t ∈ [0, 1], n − 1 < α ≤ n, n ∈ N, satisfying the boundary conditions y′(0) =

µy′(η) − θ1, y′′(0) = 0, y′′′(0) = 0, . . . , y(n−1)(0) = 0, y(1) = νy(η) + θ2, where, 0 < η, µ, ν < 1, n > 2,

the boundary parameters θ1, θ2 ∈ R+. Examples are included to show the applicability of some of our

results.

4.1 Positive solutions for three–point boundary value problems (I)

In this section, we study existence and multiplicity results for a class of nonlinear three–point boundary

value problems for fractional differential equations of the type

cDαy(t) + a(t)g(t, y(t)) = 0, t ∈ [0, 1], n− 1 < α ≤ n, (4.1.1)

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = ξy(η), (4.1.2)

where ξ, η ∈ (0, 1).

We require the following assumptions:

(C1) (i) g : [0, 1]× [0,∞) → [0,∞) is continuous; (ii) a : [0, 1] → (0,∞) is continuous;

(C2) there exist 0 < µ1, µ2 ≤ 1 such that limy→0

g(t,y(t))yµ1 = ∞, lim

y→∞g(t,y(t))yµ2 = 0, for all t ∈ (0, 1);

(C3) there exist ν1, ν2 ≥ 1 such that limy→0

g(t,y(t))yν1 = 0, lim

y→∞g(t,y(t))yν2 = ∞, for all t ∈ (0, 1);

(C4) there exist positive constants r∗1 < r∗2 and κ satisfying κ∫ 10 Φα(s)a(s)ds = 1, such that

(i) g(t, y(t)) ≤ κ−1r∗1 , for y ∈ [0, r∗1],

(ii) g(t, y(t)) ≥ κ−1r∗2 , for y ∈ [γαr∗2, r

∗2];

(C5) there exist 0 < µ1 ≤ 1, µ2 ≥ 1, such that limy→0

g(t,y(t))yµ1 = lim

y→∞g(t,y(t))yµ2 = ∞, for all t ∈ (0, 1);

(C6) there exist ψ1(y) ∈ C([0,∞), [0,∞)) and φ1(t) ∈ C([0, 1], [0,∞)) such that

g(t, y(t)) ≤ φ1(t)ψ1(y) and there exists ρ > 0 such that ψ1(y) ≤ ερ, for y ∈ [0, ρ] and 0 < ε <

(∫ 10 Φα(s)a(s)φ1(t)ds)

−1;

(C7) there exist µ1 ≥ 1, 0 < µ2 ≤ 1 such that limy→0

g(t,y(t))yµ1 = lim

y→∞g(t,y(t))yµ2 = 0, for all t ∈ (0, 1);

(C8) there exists ψ2(y) ∈ C([0,∞), [0,∞)), φ2(t) ∈ C([0, 1], [0,∞)) such that g(t, y(t)) ≥ φ2(t)ψ2(y) and

there exists ρ∗ > 0 such that ψ2(y) ≥ ε∗ρ∗, for y ∈ [γαρ∗, ρ∗].

4.1.1 Green’s function and its properties

In this subsection, we derive Green’s function for the boundary value problem (4.1.1), (4.1.1) and list some

of its properties useful for the proof of our existence and multiplicity results.

61

Lemma 4.1.1. [93] Let h ∈ C(0, 1]. Then the linear three–point boundary value problem

cDαy(t) + h(t) = 0, t ∈ (0, 1), n− 1 < α ≤ n, (4.1.3)

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = ξy(η), (4.1.4)

has solution given by y(t) =∫ 10 G(t, s)h(s)ds, where

G(t, s) =1

(1− ξ)Γ(α)

(1− s)α−1 − (1− ξ)(t− s)α−1 − ξ(η − s)α−1, s ≤ t, η ≥ s,

(1− s)α−1 − (1− ξ)(t− s)α−1, η ≤ s ≤ t ≤ 1,

(1− s)α−1 − ξ(η − s)α−1, 0 ≤ t ≤ s ≤ η,

(1− s)α−1, t ≤ s, s ≥ η.

(4.1.5)

Proof. In view of Lemma 2.2.24, the general solution of differential equation (4.1.3) is given by

y(t) = −Iαh(t) +n∑i=1

citi−1, ci ∈ R, i = 1, 2, . . . , n. (4.1.6)

By Lemma 2.2.23 and Lemma 2.2.19, we obtain

y(m)(t) = −Iα−mh(t) +n∑

i=m+1

(i− 1)!ci(i−m− 1)!

ti−m−1, (4.1.7)

where m = 1, 2, . . . , n− 1. Using the boundary conditions we have

ci = 0, i = 2, 3, . . . , n and c1 =

∫ 1

0

(1− s)α−1

Γ(α)(1− ξ)h(s)ds−

∫ η

0

ξ(η − s)α−1

Γ(α)(1− ξ)h(s)ds.

Substituting values of ci in (4.1.18), we get

y(t) =

∫ 1

0

(1− s)α−1


∫ η

0

ξ(η − s)α−1


∫ t

0

(t− s)α−1

Γ(α)h(s)ds. (4.1.8)

Thus, the solution of the boundary value problem (4.1.3),(4.1.4) is given by y(t) =∫ 10 G(t, s)h(s)ds.

Lemma 4.1.2. [93] The Green’s function G(t, s) defined by (4.1.5) satisfies the following properties:

(P1) G(t, s) > 0 for all t, s ∈ (0, 1);

(P2) For each s ∈ [0, 1], G(t, s) is nonincreasing in t;

(P3) For ℓ ∈ (0, 1), Φα(s) ≥ G(t, s) ≥ minℓ≤t≤1

G(t, s) ≥ γαΦα(s) where

γα = ξ(1− ηα−1), Φα(s) =(1− s)α−1

(1− ξ)Γ(α).

Proof. (P1): For η ≥ s,

G(t, s) =(1− s)α−1 − (1− ξ)(t− s)α−1 − ξ(η − s)α−1

(1− ξ)Γ(α).

For t < η,

G(t, s) >(1− s)α−1 − (η − s)α−1

(1− ξ)Γ(α)> 0

62

and for t ≥ η,

G(t, s) ≥ (1− s)α−1 − (t− s)α−1

(1− ξ)Γ(α)> 0.

For η ≤ s ≤ t ≤ 1,

G(t, s) =(1− s)α−1 − (1− ξ)(t− s)α−1

(1− ξ)Γ(α)≥ (1− s)α−1 − (t− s)α−1

(1− ξ)Γ(α)> 0.

Thus, for each case, η ≤ s ≤ t ≤ 1 and 0 ≤ t ≤ s, s ≥ η, G(t, s) > 0.

(P2) : From (4.1.5), if t > s,

∂

∂tG(t, s) = −(α− 1)

Γ(α)(t− s)α−2 = −(t− s)α−2

Γ(α− 1)≤ 0.

If t ≤ s, then∂

∂tG(t, s) ≡ 0.

Thus, G(t, s) is nonincreasing in t.

(P3): Clearly, G(t, s) ≤ (1−s)α−1

Γ(α)(1−ξ) = Φα(s).

Case(i): s ≤ η, using (P2), we have

G(t, s) ≥ G(1, s) =ξ(1− ηα−1)(1− s)α−1

(1− ξ)Γ(α)= γαΦα(s), for t ≥ s.

For t ≤ s, we have

G(t, s) ≥ (1− s)α−1 − ξηα−1(1− s)α−1

(1− ξ)Γ(α)≥ ξ(1− ηα−1)Φα(s) = γαΦα(s).

Case(ii): s > η: for t ≥ s,

G(t, s) ≥ G(1, s) =ξ(1− s)α−1

(1− ξ)Γ(α)≥ γαΦα(s).

For t < s, we have G(t, s) = (1−s)α−1

(1−ξ)Γ(α) = Φα(s) ≥ γαΦα(s).

Hence, minℓ≤t≤1

G(t, s) ≥ γαΦα(s) where ℓ ∈ (0, 1), γα = ξ(1− ηα−1) and Φα(s) =(1−s)α−1

(1−ξ)Γ(α) .

We write the three–point boundary value problem (4.1.1),(4.1.2) as an equivalent integral equation

y(t) =

∫ 1

0G(t, s)a(s)g(s, y(s))ds. (4.1.9)

By a solution of the boundary value problem (4.1.1),(4.1.2), we mean a solution of the integral equation

(4.1.9). Define a cone U ⊂ B = (C[0, 1], ∥.∥) by

U = y ∈ B : y(t) ≥ 0, minℓ≤t≤1

y(t) ≥ γα∥y∥

and an operator T : U → U by

T y(t) =∫ 1

0G(t, s)a(s)g(s, y(s))ds, 0 ≤ t ≤ 1. (4.1.10)

By a solution of the integral equation (4.1.9), we mean a fixed point of the operator T .

63

Lemma 4.1.3. Under the assumption (C1), the operator T : U → U is completely continuous.

Proof. The operator T is continuous and T y(t) ≥ 0. Also, for all y ∈ U ,

minℓ≤t≤1

T y(t) = minℓ≤t≤1

(∫ 1

0G(t, s)a(s)g(s, y(s))ds

)≥ γα

(∫ 1

0Φα(s)a(s)g(s, y(s))ds

)≥ γα∥T y(t)∥,

which implies that T (U) ⊂ U . For fixed R > 0, consider a bounded subset M of U defined by

M = y ∈ U : ∥y∥ < R.

Define L = max0≤y≤R

g(t, y(t)) + 1 and K = max0≤t≤1

a(t) + 1, then for y ∈ M, we have

|T y(t)| =∣∣∣∣∫ 1

0G(t, s)a(s)g(s, y(s))ds

∣∣∣∣ ≤ KL

∫ 1

0Φα(s)ds =

KL

αΓ(α)(1− ξ).

Hence ∥T y(t)∥ ≤ KLαΓ(α)(1−ξ) , which implies that T (M) is uniformly bounded. Since g is continuous, it

follows that T is completely continuous.

4.1.2 Existence of at least one positive solution

Theorem 4.1.4. Assume that (C1)-(C3) hold. Then the boundary value problem (4.1.1), (4.1.2) has at

least one positive solution.

Proof. The proof is similar to the proof of the Theorem 3.6 in [127].

Now, we study existence of at least one positive solution under a weaker hypothesis on g, that is,

limy→0

g(t,y(t))yµ , lim

y→∞g(t,y(t))yν ∈ 0,∞ for µ, ν > 0.

Theorem 4.1.5. Assume that (C1) and (C4) are satisfied, then, the boundary-value problem (4.1.1),

(4.1.2) has at least one positive solution y∗ such that r∗1 < ∥y∗∥ < r∗2.

Proof. Define Er∗1 = y ∈ B : ∥y∥ < r∗1 and Er∗2 = y ∈ B : ∥y∥ < r∗2. For y ∈ U ∩ ∂Er∗1 , using Lemma

4.1.2, we obtain

|T y(t)| ≤ κ−1r∗1

∫ 1

0Φα(s)a(s)ds < r∗1 = ∥y∥,

which implies that

∥T y(t)∥ ≤ ∥y∥, for y ∈ U ∩ ∂Er∗1 . (4.1.11)

For y ∈ U ∩ ∂Er∗2 , we have, y ≥ minℓ≤t≤1

y(t) ≥ γαr∗2. Using Lemma 4.1.2, we get

|T y(t)| ≥ κ−1r∗2

∫ 1

0Φα(s)a(s)ds > r2 = ∥y∥,

which yields

∥T y(t)∥ ≥ ∥y∥, for y ∈ U ∩ ∂Er∗2 . (4.1.12)

Hence, in view of (4.1.11), (4.1.12) and Theorem 2.3.7, it follows that T has at least one fixed point y∗ in

U ∩ (Er∗2\Er∗1 ) satisfying r∗1 < ∥y∗∥ < r∗2.

64

4.1.3 Existence of at least two positive solutions

Theorem 4.1.6. Assume that (C1), (C5) and (C6) are satisfied, then the boundary value problem (4.1.1),

(4.1.2) has at least two positive solution y1, y2 such that 0 < ∥y1∥ < ρ < ∥y2∥.

Proof. In view of (C5), there exists m∗1 ∈ (0, ρ) such that for all ε∗1 > γ−2

α κ,

g(t, y(t)) ≥ ε∗1yµ1 , for y ≥ m∗

1.

Define Er1 = y ∈ B : ∥y∥ < r1, where 0 < r1 < m∗1 and using Lemma 4.1.2, we have

|T y(t)| ≥ ε∗1γ2α

∫ 1

0Φα(s)a(s)∥y∥µ1ds > r1 = ∥y∥ for y ∈ U ∩ ∂Er1 .

Hence,

∥T y∥ ≥ ∥y∥, for y ∈ U ∩ ∂Er1 . (4.1.13)

Now, for µ2 ≥ 1, using limu→∞g(t,y(t))yµ2 = ∞, it follows that there exists m∗

2 > ργα

such that for all

0 < ε∗2 < κγ−(µ2+1)α , we have

g(t, y(t)) ≥ ε∗2yµ2 , for y ≥ γαm

∗2.

Let Er2 = y ∈ B : ∥y∥ < r2, where r2 ≥ m∗2. Since y ≥ min

ℓ≤t≤1y(t) ≥ γα∥y∥ for y ∈ U ∩ ∂Er2 , it follows

that

|T y(t)| ≥ ε∗2γ1+µ2α

∫ 1

0Φα(s)a(s)ds > r2 = ∥y∥,

which implies that

∥T y(t)∥ ≥ ∥y∥, for y ∈ U ∩ ∂Er2 . (4.1.14)

Using (C6), we obtain

g(t, y(t)) ≤ ερφ1(t), for 0 ≤ y ≤ ρ.

Hence, for y ∈ U ∩ ∂Eρ, where Eρ = y ∈ B : ∥y∥ < ρ, we have

|T y(t)| ≤ ερ

∫ 1

0Φα(s)a(s)φ1(t)ds < ρ = ∥y∥,

which implies that

∥T y(t)∥ ≤ ∥y∥, for y ∈ U ∩ ∂Eρ. (4.1.15)

Since r1 < ρ < r2, by (4.1.13), (4.1.14), (4.1.15) and Theorem 2.3.7, it follows that T has at least two

fixed points y1 ∈ U ∩ (∂Eρ\∂Er1) and y2 ∈ U ∩ (∂Er2\∂Eρ) such that 0 < ∥y1∥ < ρ < ∥y2∥.

Theorem 4.1.7. Assume that (C1), (C7), (C8) hold, then the boundary value problem (4.1.1), (4.1.2)

has at least two positive solution y∗1 and y∗2 satisfying 0 < ∥y∗1∥ < ρ∗ < ∥y∗2∥.

65

Proof. In view of limy→0g(t,y(t))yµ1 = 0, for any ε ∈ (0, κ] there exists m ∈ [0, ρ∗] such that

g(t, y(t)) ≤ εyµ1 , for u ∈ (0,m].

Let Er = y ∈ B : ∥y∥ < r, for 0 < r ≤ m. For any y ∈ U ∩ ∂Er we have

|T y(t)| ≤ εr

∫ 1

0Φα(s)a(s)ds < r = ∥y∥,

which implies

∥T y(t)∥ ≤ ∥y∥, for y ∈ U ∩ ∂Er. (4.1.16)

Also, in view of limu→∞

g(t,y(t))yµ2 = 0, for any ε∗ ∈ (0, κ2 ], there exists m∗ > ρ∗ such that

g(t, y(t)) ≤ ε∗yµ2 , for y ≥ m∗.

Hence, g(t, y(t)) ≤ ε∗yµ2 + L, for y ∈ [0,∞) where L = max

0≤t≤1,0≤y≤m∗|g(t, y(t))| + 1. Choose r∗ >

max2m∗, 2Lκ−1 and define Er∗ = y ∈ B : ∥y∥ < r∗. For y ∈ U ∩ ∂Er∗ , using Lemma 4.1.2, we obtain

|T y(t)| ≤ ε∗rµ2∗

∫ 1

0Φα(s)a(s)ds+ L

∫ 1

0Φα(s)a(s)ds <

r∗2

+r∗2

= r∗ = ∥y∥,

which implies that

∥T y(t)∥ ≤ ∥y∥ for y ∈ U ∩ ∂Er∗ . (4.1.17)

Finally, for y ∈ U ∩ ∂Eρ∗ , where Eρ∗ = y ∈ B : ∥y∥ < ρ∗, using lemma ?? and (C8), we obtain

|T y(t)| ≥ ε∗ρ∗∫ 1

0Φα(s)a(s)φ2(t)ds > ρ∗ = ∥y∥,

which implies

∥T y(t)∥ ≥ ∥y∥, for y ∈ U ∩ ∂Eρ∗ . (4.1.18)

Since r < ρ∗ < r∗, by (4.1.16), (4.1.17), (4.1.18) and Theorem 2.3.7, it follows that T has at least two

fixed points, y∗1 ∈ U ∩ (∂Eρ∗\∂Er) and y∗2 ∈ U ∩ (∂Er∗\∂Eρ∗) such that 0 < ∥y∗1∥ < ρ < ∥y∗2∥.

4.1.4 Existence of at least three positive solutions

Theorem 4.1.8. Suppose there exist constants 0 < a < b < b/γα ≤ c such that

(i) g(t, y) < κa, for 0 ≤ t ≤ 1, 0 ≤ y ≤ a,

(ii) g(t, y) ≥ κ bγα

, for 0 6 t ≤ 1, b ≤ y ≤ b/γα,

(iii) g(t, y) ≤ κc, for 0 ≤ t ≤ 1, 0 6 y ≤ c.

Then the boundary value problem (4.1.16), (4.1.16) has at least three positive solutions y1, y2 and y3 such

that ∥y1∥ < a, b < θ(y2) and ∥y3∥ > a with θ(y3) < b.

66

Proof. Let y ∈ Uc, then ∥y∥ 6 c and by (iii), we have

∥T y(t)∥ = max0≤t≤1

∣∣∣∣∫ 1

0G(t, s)a(s)g(t, s)ds

∣∣∣∣ ≤ ∫ 1

0Φα(s)a(s)κcds = c,

which implies that T : Uc → Uc.Similarly if y ∈ Ua then in view of (i), we have ∥T y(t)∥ < a . Furthermore, T : Uc → Uc is completely

continuous operator. Hence, condition (ii) of Theorem 2.3.8 is satisfied.

Now choose y(t) = b/γα, 0 ≤ t ≤ 1. Obviously y(t) = bγα

∈ Uθ(b, bγα), which implies that y ∈

Uθ(b, bγα) : θ(y) > b = ∅. Hence, if y ∈ Uθ(b, c), then b ≤ y(t) ≤ c for 0 ≤ t ≤ 1. Also, in view of (ii) and

Lemma 4.1.2, we have

θ(T y(t)) = min0≤t≤1

|T y(t)| ≥∫ 1

0γαΦα(s)a(s)g(t, s)ds >

∫ 1

0Φα(s)a(s)κcds = b

Therefore ∥T y(t)∥ > b, for y ∈ Uθ(b, c) which is condition (i) of Theorem 2.3.8.

Finally to check condition (iii) of Theorem 2.3.8 choose y ∈ Uθ(b, c) with ∥T y∥ > bγα

then,

θ(T y(t)) = min0≤t≤1

|T y(t)| ≥ γα∥T y∥ > γαb

γα= b.

Therefore condition (iii) of Theorem 2.3.8 is satisfied. Thus, the boundary value problem (4.1.2), (4.1.2)

has at least three positive solutions y1, y2 and y3 such that ∥y1∥ < a, b < θ(y2) and ∥y3∥ > a with

θ(y3) < b.

Example 4.1.9. Consider the problem

cD5/3y(t) + a(t)g(t, y) =0, 0 < t < 1, (4.1.19)

y′(0) = 0,1

2y(

1

8) =y(1), (4.1.20)

where

a(t) = e−t, g(t, y) =

Γ(53)u

23 + e−t

12 , 0 ≤ y ≤ 10,

Γ(53)u23 + y2

√u−10

5√3

+ e−t

12 , 10 < y ≤ 30,

2367375000 + e−t

12 , y ≥ 30.

Where κ = 1.59923, a = 10, b = 15, c = 350; consequently we have

g(t, y) ≤ 4.27351 < κa ≈ 15.92228, for 0 ≤ y ≤ 10,

g(t, y) ≥ 63.616 > κb

γα≈ 47.768, for 15 ≤ y ≤ 30,

g(t, y) ≤ 473.557 < κc ≈ 557.28, for 0 ≤ y ≤ 350.

Thus all the assumptions of Theorem 4.1.8 are satisfied. Therefore the problem (4.1.19), (4.1.20) has at

least three positive solutions y1, y2 and y3 such that ∥y1∥ < 10, 15 < θ(y2) and ∥y3∥ > 10 with θ(y3) < 15.

67

4.2 Positive solutions to three–point boundary value problems (II)

In this section, we study existence of positive solutions to nonlinear higher order nonlocal boundary value

problems corresponding to fractional differential equations of the type [85]

cDαy(t) + g(t, y(t)) = 0, t ∈ [0, 1], n− 1 < α < n, n ∈ N, (4.2.1)

y′(0) = µy′(η)− θ1, y′′(0) = 0, y′′′(0) = 0, . . . , y(n−1)(0) = 0, y(1) = νy(η) + θ2, (4.2.2)

where, 0 < η, µ, ν < 1, n > 2, the boundary parameters θ1, θ2 ∈ R+. Sufficient conditions are obtained

for the existence of positive solutions using the Guo–Krasnosel’skii fixed point theorem. The conditions

for the existence of unique positive solution are also obtained.


Lemma 4.2.1. Let h ∈ C[0, 1], then, the linear problem

cDαy(t) + ψ(t) = 0, t ∈ (0, 1), n− 1 < α < n, (4.2.3)

y′(0) = µy′(η)− θ1, y′′(0) = 0, y′′′(0) = 0, . . . , y(n−1)(0) = 0, y(1) = νy(η) + θ2, (4.2.4)

has solution

y(t) =

∫ 1

0Gα(t, s)ψ(s)ds+

∫ 1

0Gα,β(t; η, s)ψ(s)ds+ φ(t), (4.2.5)

where,

Gα(t, s) =

(1−s)α−1−(t−s)α−1

Γ(α) , s ≤ t,

(1−s)α−1

Γ(α) , t ≤ s,Gα,β(t; η, s) =

ν[(1−s)α−1−(η−s)α−1]

(1−ν)Γ(α)

+µ[1−νη−(1−ν)t](η−s)α−2

(1−µ)(1−ν)Γ(α−1) , s ≤ η,

ν(1−s)α(1−ν)Γ(α−1) , η ≤ s

and φ(t) =(1−νη−(1−ν)t(1−µ)(1−ν)

)θ1 +

θ21−ν .

Proof. In view of Lemma 2.2.24, the equation (4.2.3) is equivalent to the integral equation

y(t) = −Iαψ(t) + c1 + c2t+ c3t2+, . . . ,+cnt

n−1.

Using Lemma 2.2.23, we obtain

y′(t) = −Iα−1ψ(t) + c2 + 2c3t+, . . . ,+(n− 1)cntn−2,

y′′(t) = −Iα−2ψ(t) + 2c3+, . . . ,+(n− 1)(n− 2)cntn−3,

. . . ,

y(n−1)(t) = −Iα−(n−1)ψ(t) + (n− 1)!cn,

where c1, c2, . . . , cn ∈ R. Using the boundary conditions, y′′(0) = 0, y′′′(0) = 0, . . . , y(n−1)(0) = 0, we get

c3 = 0, c4 = 0,· · · , cn = 0. By the boundary conditions y′(0) = µy′(η)− θ1, y(1) = νy(η) + θ2, we have

(1− µ)c2 = −µIα−1ψ(η)− θ1, (1− ν)c1 + (1− νη)c2 = Iαψ(1)− νIαψ(η) + θ2,

68

which implies that c2 = − µ1−µI

α−1ψ(η)− θ11−µ and

c1 =1

1− νIαψ(1)− ν

1− νIαψ(η) + µ(1− νη)

(1− µ)(1− ν)Iα−1ψ(η) +

(1− νη)θ1(1− µ)(1− ν)

+θ2

(1− ν).

Therefore, the solution of the linear boundary value problem (4.2.1), (4.2.2) is given by

y(t) =− Iαψ(t) +1

1− νIαψ(1)− ν

1− νIαψ(η) +

µ(1− νη)− µ(1− ν)t

(1− µ)(1− ν)Iα−1ψ(η) + φ(t)

=− Iαψ(t) + Iαψ(1) +ν

1− νIαψ(1)− ν

1− νIαψ(η) +

µ(1− νη)− µ(1− ν)t

(1− µ)(1− ν)Iα−1ψ(η) + φ(t)

=

∫ t

0

(1− s)α−1 − (t− s)α−1

Γ(α)ψ(s)ds+

∫ 1

t

(1− s)α−1

Γ(α)ψ(s)ds+

∫ η

0

[ν(1− s)α−1 − ν(η − s)α−1

(1− ν)Γ(α)

+((µ(1− νη)− µ(1− ν)t)(η − s)α−2

(1− µ)(1− ν)Γ(α− 1)

]ψ(s)ds+

∫ 1

η

ν(1− s)α−1

(1− ν)Γ(α)ψ(s)ds+ φ(t)

=

∫ 1

0

Gα(t, s)ψ(s)ds+

∫ 1

0

Gα,β(t; η, s)ψ(s)ds+ φ(t).

Lemma 4.2.2. The functions Gα(t, s) and Gα,β(t; η, s) satisfy the following properties:

(i) Gα(t, s) ≥ 0 , Gα,β(t; η, s) ≥ 0 and Gα(t, s) ≤ Gα(s, s) for all 0 ≤ s, t ≤ 1;

(ii) minξ≤t≤τ

Gα(t, s) ≥ (1− τα−1) max0≤t≤1

Gα(t, s) = (1− τα−1)Gα(s, s), for s ∈ (0, 1), 0 < ξ < τ < 1,

(iii) ν(1− ηα−1)(1− s)α−1 ≤ (1− ν)Γ(α)Gα,β(t; η, s) < 2(α− 1)(1− s)α−2.

(iv) minξ≤t≤τ

Gα,β(t; η, s) ≥ (1− τ) max0≤t≤1

Gα,β(t; η, s), for s ∈ (0, 1), 0 < ξ < τ < 1.

Proof. (i) : For α > 1, by definition of Gα(t, s), it follows that Gα(t, s) ≥ 0 and Gα(t, s) ≤ Gα(s, s) for all

0 ≤ s, t ≤ 1.

For s ≤ η and 0 < ν < 1, we have 1− νη − (1− ν)t > (1− νη)(1− t). Hence, Gα,β(t, η, s) > 0. Also we

note that Gα,β(t, η, s) > 0 for s ≥ η.

(ii) : Obviously, maxt∈[0,1]Gα(t, s) = (1−s)α−1

Γ(α) . Now, for s ≤ t, t ∈ [ξ, τ ] we have following estimate for

Gα(t, s):

Gα(t, s) =1

Γ(α)[(1− s)α−1 − (t− s)α−1] = max

t∈[0,1]Gα(t, s)−

(t− s)α−1

Γ(α)≥ max

t∈[0,1]Gα(t, s)− τα−1 (1−

st )α−1

Γ(α).

Since (1− st )α−1 ≤ (1− s)α−1. Which implies that

mint∈[ξ,τ ]

Gα(t, s) ≥ (1− τα−1)maxt∈[0,1]

Gα(t, s), for s ∈ (0, 1).

Now for s ≤ η, we have

Gα,β(t; η, s) =ν[(1− s)α−1 − (η − s)α−1]

(1− ν)Γ(α)+µ[1− νη − (1− ν)t](η − s)α−2

(1− µ)(1− ν)Γ(α− 1)

≤ν(1− s)α−1

(1− ν)Γ(α)+

µ(η − s)α−2

(1− µ)(1− ν)Γ(α)<

2(α− 1)(1− s)α−2

(1− µ)(1− ν)Γ(α).

69

For s ≥ η, obviously Gα,β(η, s) <2(α−1)(1−s)α−2

(1−µ)(1−ν)Γ(α) . From the definition of Gα,β(η, s), it clearly follows that

Gα,β(t; η, s) ≥ν(1− ηα−1)(1− s)α−1

(1− ν)Γ(α).

(iii) : From the definition of Gα,β(t; η, s), we have

∂

∂t(Gα,β(t; η, s)) =

−(η − s)α−2

(1− µ)Γ(α− 1)≤ 0.

Therefore Gα,β(t; η, s) is non–increasing in t, so its minimum value occurs at t = τ for t ∈ [ξ, τ ] and its

maximum value occurs at t = 0 for t ∈ [0, 1]. That is

minξ≤t≤τ

Gα,β(t; η, s) =ν[(1− s)α−1 − (η − s)α−1]

(1− ν)Γ(α)+µ[1− νη − (1− ν)τ ](η − s)α−2

(1− ν)Γ(α)(4.2.6)

and

max0≤t≤1

Gα,β(t; η, s) =ν[(1− s)α−1 − (η − s)α−1]

(1− ν)Γ(α)+µ(1− νη)(η − s)α−2

(1− ν)Γ(α). (4.2.7)

Since (1− s)α−1 − (η − s)α−1 ≥ 0 and 1− τ < 1, therefore

(1− s)α−1 − (η − s)α−1 ≥ (1− τ)((1− s)α−1 − (η − s)α−1). (4.2.8)

Also, as 1− ν ≤ 1− νη. Thus,

1− νη − (1− ν)τ ≥1− νη − (1− νη)τ

≥(1− τ)(1− νη).(4.2.9)

Subletting (4.2.8) and (4.2.9) in (4.2.6) and using (4.2.7), we have

minξ≤t≤τ

Gα,β(t; η, s) ≥ (1− τ)

ν[(1− s)α−1 − (η − s)α−1]

(1− ν)Γ(α)+µ(1− νη)(η − s)α−2

(1− ν)Γ(α)

= (1− τ) max

0≤t≤1Gα,β(t; η, s).

Remark 4.2.3. For θ1, θ2 > 0, minξ≤t≤τ

φ(t) ≥ (1− τ) max0≤t≤1

φ(t).

Proof. As ddtφ(t) = − θ1

(1−µ) < 0, therefor φ(t) is a decreasing function. Hence

max0≤t≤1

φ(t) = φ(0) =(1− νη)θ1

(1− µ)(1− ν)+

θ2(1− ν)

and

minξ≤t≤τ

φ(t) = φ(τ) =[1− νη − (1− ν)τ ]θ1

(1− µ)(1− ν)+

θ2(1− ν)

≥ (1− τ)(1− νη)θ1(1− µ)(1− ν)

+(1− τ)θ2(1− ν)

= (1− τ) max0≤t≤1

φ(t).

Thus, we have minξ≤t≤τ

φ(t) ≥ (1− τ) max0≤t≤1

φ(t).

70

Define a cone V in B by V = y ∈ B : min y(t)ξ≤t≤τ

≥ (1− τ)∥y∥. Also, define T : B → B as

T y(t) =∫ 1

0Gα(t, s)g(s, y(s))ds+

∫ 1

0Gα,β(t; η, s)g(s, y(s))ds+ φ(t). (4.2.10)

The boundary value problem (4.2.1), (4.2.2) has a solution if and only if T has a fixed point.

Lemma 4.2.4. Assume that (C1)(i) is satisfied. Then operator T : V → V is completely continuous.

Proof. First we prove that T (V) ⊂ V. From (4.2.10), Lemma 4.2.2 and Remark 4.2.3, we have

minξ≤t≤τ

(T y(t)) ≥ (1− τα−1)

∫ 1

0

Gα(s, s)g(s, y(s))ds+ (1− τ)

∫ 1

0

maxt∈[0,1]

Gα,β(t; η, s)g(s, y(s))ds+ (1− τ) maxt∈[0,1]

φ(t)

≥ (1− τ)

∫ 1

0

Gα(s, s)g(s, y(s))ds+

∫ 1

0

maxt∈[0,1]

Gα,β(t; η, s)g(s, y(s))ds+ maxt∈[0,1]

φ(t)

≥ (1− τ)∥T y∥.

Hence, T (V) ⊂ V.

Consider a bounded subset M of V defined by M = y ∈ V : ∥y∥ ≤ ℓ, ℓ ∈ R+. Let Λ = max0≤y≤ℓ

g(t, y(t))+1.

Then, for y ∈ M, we have

|T y(t)| ≤∫ 1

0Gα(t, s)

∣∣g(s, y(s))∣∣ds+ ∫ 1

0Gα,β(t; η, s)

∣∣g(s, y(s))∣∣ds+ maxt∈[0,1]

φ(t)

≤ Λ

Γ(α)

∫ 1

0(1− s)α−1ds+

2Λ(α− 1)

(1− µ)(1− ν)Γ(α)

∫ 1

0(1− s)α−2ds+

(1− νη)θ1 + (1− µ)θ2(1− µ)(1− ν)

≤ Λ

(1− µ)(1− ν)

(Iα(1) + 2Iα(1)

)+

(1− νη)θ1 + (1− µ)θ2(1− µ)(1− ν)

=(3α− 1)Λ

α(1− µ)(1− ν)(α− 1)+

(1− νη)θ1 + (1− µ)θ2(1− µ)(1− ν)

.

Hence T (M) is a bounded set.Finally we show that T is compact operator. Define δ = α(1−µ)Γ(α)ε

Λ[α(1−µ)+µηµ−1+θ1]and choose t > τ such that

t− τ < δ. Then, for all ε > 0 and y ∈ M, we have

|T y(t)− T y(τ)| =∣∣∣∣∫ 1

0

(Gα(t, s)−Gα(τ, s))g(s, y(s))ds+

∫ 1

0

(Gα,β(t; η, s)−Gα,β(t; η, s))g(s, y(s))ds+ φ(t− τ)

∣∣∣∣≤ Λ

[∫ 1

0

∣∣Gα(t, s)−Gα(τ, s)∣∣ds+ ∫ 1

0

∣∣Gα,β(t; η, s)−Gα,β(t; η, s)∣∣ds+ θ1

1− µ(t− τ)

]= Λ

[1

Γ(α)

∫ t

0

((t− s)α−1 − (τ − s)α−1)ds+µ(t− τ)

(1− µ)Γ(α− 1)

∫ η

0

(η − s)α−2ds+θ1

1− µ(t− τ)

]=

Λ

αΓ(α)

[tα − τα +

µηα−1 + θ11− µ

(t− τ)].

An application of the Mean Value theorem yields, tα − τα ≤ α(t− τ) < αδ. Hence, it follows that

|T y(t)− T y(τ)| < Λδ[α(1− µ) + µηµ−1 + θ1]

α(1− µ)Γ(α)< ε.

The Arzela–Ascoli theorem guarantees that the operator T : V → V is completely continuous.

For simplicity, we introduce following notations:

Kµ,ν =(3Iαψ1(1) +

6Iα−1ψ1(1)

(1− µ)(1− ν)

)−1, Kν =

([(1− τ) +

ν(1− η)

1− ν

]Iαξ+ψ2(τ)

)−1

where, ψ1, ψ2 ∈ C([0, 1] → R+), and

cµ =1

3(1− µ), cν =

1

3(1− ν), γα = 1− τα−1.

71

4.2.2 Existence of positive solutions

In the following theorem, we establish sufficient conditions for the existence of at least one positive solution

for the boundary value problem (4.2.1), (4.2.2).

Theorem 4.2.5. Suppose that (C1)(i) is satisfied. Furthermore, there exist constants r1, r2 ∈ R+ such

that r2 > r1 and functions ψ1, ψ2 ∈ C([0, 1] → R+) such that

(i) g(t, u) ≥ r2Kνψ2(t), for (t, y) ∈ [0, 1]× [γαr2, r2],

(ii) g(t, u) ≤ r1K1µ,νψ1(t), for (t, y) ∈ [0, 1]× [0, r1],

then, the boundary value problem (4.2.1), (4.2.2) has at least one positive solution for θ1, θ2 small enough

and has no positive solution for θ1, θ2 such that θ1 ≤ cµr1 and θ2 ≤ cνr1.

Proof. By Lemma 4.2.4, the operator T is completely continuous. We show that the problem (4.2.1),

(4.2.2) has at least one positive solution in V. Define an open subset of the Banach space B as Er1 = y ∈B : ∥y∥ < r1. Then, for any y ∈ V ∩ ∂Er1 , we have ∥y∥ = r1 and in view of Lemma 4.2.2 and equation

(4.2.10), it follows that

|T y(t)| =∣∣∣ ∫ 1


∫ 1

0Gα,β(t; η, s)g(s, y(s))ds+

(1− νη − (1− ν)t

(1− µ)(1− ν)

)θ1 +

θ21− ν

∣∣∣≤ Iαg(1, y(1)) + 2

(1− µ)(1− ν)Iα−1g(1, y(1)) +

(1− νη)θ1 + (1− µ)θ2(1− µ)(1− ν)

≤ Kµ,νr1

[Iαψ1(1) +

2Iα−1ψ1(1)

(1− µ)(1− ν)

]+

θ11− µ

+θ2

1− ν≤ r1

3+r13

+r13

= ∥y∥.

Therefore, we have ∥T y∥ ≤ ∥y∥, for y ∈ V ∩ ∂Er1 .Define Er2 = y ∈ V : ∥y∥ < r2. For any t ∈ [ξ, τ ] and y ∈ V ∩ ∂Er2 , using Lemma 4.2.1, we have

minξ≤t<τ

y(t) ≥ (1− τ)∥y∥. Therefore, by Remark 4.2.3 and equation (4.2.10), we have the following estimate

|T y(t)| =∣∣∣ ∫ 1


∫ 1

0Gα,β(t; η, s)g(s, y(s))ds+ φ(t)

∣∣∣≥[(1− τα−1) +

ν(1− ηα−1)

1− ν

]Iαξ+g(τ, y(τ)) + φ(t)

≥ r2Kν

[(1− τ) +

ν(1− η)

1− ν

]Iαξ+ψ2(τ) = r2 = ∥y∥.

Thus ∥T y∥ ≥ ∥y∥ for y ∈ V ∩ ∂Er2 .Hence, by Theorem 2.3.7, it follows that T has a fixed point y in V ∩ (Er2\Er1), which is a solution to the

boundary value problem (4.2.1), (4.2.2).

Example 4.2.6. Consider the boundary value problem

cDαy(t) =t2e

32t + 1

8(1 + t2)+t2y2(1 + sin t)

16π, t ∈ [0, 1], (4.2.11)

y(1) = νy(η) + θ2, y′(0) = µy′(η)− θ1, y

′′(0) = 0, y′′′(0) = 0, (4.2.12)

72

where, α = 72 , η = 1

2 , µ = 23 and ν = 1

3 . Let g(t, y) = t2e32 t+1

8(1+t2)+ t2y2(1+sin t)

16π , (t, y) ∈ [0, 1] × [0,∞]. For

r1 = 1, r2 = 550, we observe that

g(t, y) ≤ 1

8

(e+

1

π

)t2, for (t, y) ∈ [0, 1]× [0, 1]

and

g(t, y) ≥(

1√e

)t2

1 + t2for (t, y) ∈ [0, 1]× [283, 550].

Choose ψ1(t) = t2 and ψ2(t) =t2

1+t2and ξ = 1

4 , τ = 34 . By computations we obtain cµ = 1

9 , cν = 29 ,

Iαξ+ψ2(τ) =1

60√π

(49

√7 tan−1

(√27

)− tan−1

(√2)− 1132

√2

21

)≈ 0.001291, also Iαψ1(1) =

12810395

√π. There-

fore Kµ,ν ≈ 0.583566 and Kν ≈ 0.000993874. Hence g(t, y) ≤ 0.583566t2, for (t, y) ∈ [0, 1] × [0, 1], and

g(t, y) ≥ 0.546631(

t2

1+t2

), for (t, y) ∈ [0, 1] × [283, 550]. The assumptions (i) and (ii) of the Theorem

4.2.5 are satisfied. Therefore the boundary value problem (4.2.11), (4.2.12) has a positive solution for

θ1 ∈ [0, 19 ], θ2 ∈ [0, 29 ].

4.2.3 Uniqueness of positive solution

Theorem 4.2.7. Assume that there exists ω(t) ∈ C([0, 1] → R+) such that

|g(t, y)− g(t, z)| ≤ ω(t)|y − z|, for t ∈ [0, 1], y, z ∈ [0,∞),

and

λ := Iα|ω(1)|+ 2

(1− µ)(1− ν)Iα−1|ω(1)| < 1,

then, the boundary value problem (4.2.1), (4.2.2) has a unique positive solution.

Proof. For y, z ∈ V, using (4.2.10) and Lemma 4.2.2, we obtain

|T y(t)− T z(t)| ≤∫ 1

0Gα(t, s)|g(s, y(s))− g(s, z(s))|ds+

∫ 1

0Gα,β(t; η, s)|g(s, y(s))− g(s, z(s))|ds

< ∥y − z∥[∫ 1

0

(1− s)α−1

Γ(α)|ω(s)|ds+

∫ 1

0

2(α− 1)(1− s)α−2

(1− µ)(1− ν)Γ(α)|ω(s)|ds

]= ∥y − z∥

[Iα|ω(1)|+ 2

(1− µ)(1− ν)Iα−1|ω(1)|

]= λ∥y − z∥.

Therefore, ∥T y(t) − T z(t)∥ ≤ λ∥y − z∥. Hence, it follows by Banach Contraction principle that the

boundary value problem (4.2.1), (4.2.2) has a unique positive solution.


(14√π + 425e

18t)cDαy(t)(1 + y(t)) = e−33t(11

√πe33t cos(25t) + 500 sin2 t)y2, (4.2.13)

y(1) = νy(η)+θ2 , y′(0) = µy′(η)− θ1, y

′′(0) = 0, y′′′(0) = 0, (4.2.14)

73

where α = 72 , ν = 3

5 , µ = 23 , η = 1

2 and θ1, θ2 ∈ R.

Let g(t, y) = (11√πe33t cos(25t)+500 sin2 t)y2e−33t

(14√π+425e

18 t)(1+y)

, (t, y) ∈ [0, 1]× [0,∞]. For u, v ∈ V, t ∈ [0, 1], we have

|g(t, y)− g(t, z)| ≤ e−33t(11√πe33t + 500)

(14√π + 425e

18t)

(∣∣∣∣ y2

1 + y+

v2

1 + z

∣∣∣∣)≤ e−33t(11

√πe33t + 500)

(14√π + 425e

18t)

(|y − z|(y + z + yz)

(1 + y)(1 + z)

)≤ e−33t(11

√πe33t + 500)

(14√π + 425e

18t)

(|y − z|) .

Here, ω(t) = e−33t(11√πe33t+500)

(14√π+425e

18 t)

. After some calculations, we have Iα−1ω(1) ≈ 0.013469, λ ≈ 0.215504 < 1.

All the conditions of Theorem 4.2.7 are satisfied. Therefore, by Theorem 4.2.7, the boundary value problem

(4.2.13), (4.2.14) has a unique positive solution.

Chapter 5

Existence and multiplicity of positive

solutions for non linear systems of

fractional differential equations

In this chapter we are concerned with the existence and multiplicity of positive solutions for systems of

fractional order differential equations with two-point and three–point boundary conditions. Systems of

fractional differential equations are used to model the problems in physics, biosciences and engineering.

For example, in [11] Atanackovic and Stankovic have described the motion of an elastic column fixed at

one end and loaded at other by modeling it as a system of fractional differential equations. The existence

theory for coupled systems of fractional differential equations is well established in [8,95,129,150]. There

are some recent works dealing with the existence and multiplicity of positive solutions for systems of

fractional differential equations,for example, [14, 18,142].

In section 5.1, we study a coupled system of fractional differential equation, which is an extension

of scalar fractional differential equation considered in [145]. Some growth conditions are imposed on the

nonlinear functions (f and g) involved to obtain the conditions for the existence of positive solutions,

and also the explicit interval are obtained for the parameter for which there exists positive or multiple

positive solutions for the problem. In section 5.2, we present some existence results for positive and

multiple positive solutions for more general systems differential equations satisfying three–point boundary

conditions.

5.1 Positive solutions for a coupled system

In this section, we study existence results for positive solutions to a coupled system of fractional differential

equations [88] cDαx(t) + λφ(t)f(y(t)) = 0, n− 1 < α ≤ n,

cDβy(t) + λψ(t)g(x(t)) = 0, n− 1 < β ≤ n,(5.1.1)

74

75

satisfying the two point boundary conditionsx(1) = 0, x′(0) = 0, x′′(0) = 0, · · · , x(n−2)(0) = 0, x(n−1)(0) = 0,

y(1) = 0, y′(0) = 0, y′′(0) = 0, · · · , y(n−2)(0) = 0, y(n−1)(0) = 0,(5.1.2)

where t ∈ [0, 1], λ > 0. It is assumed that f, g : [0,∞) → [0,∞) are continuous and trφ(t), tsψ(t) : [0, 1] →[0,∞) are also assumed to be continuous for r, s ∈ [0, 1) and do not vanish identically on any subinterval.

We introduce following notations:

f0 = limx→0

f(x)

x, f∞ = lim

x→∞

f(x)

x, g0 = lim

x→0

g(x)

x, g∞ = lim

x→∞

g(x)

x.

The following assumptions will be used in the proof of main results.

(H1) f0 = 0, g0 = 0 and f∞ = ∞ , g∞ = ∞;

(H2) f0 = ∞, g0 = ∞ and f∞ = 0 , g∞ = 0;

(H3) f0 = ∞, g0 = ∞, f∞ = ∞, g∞ = ∞;

(H4) f0 = 0, g0 = 0 and f∞ = 0, g∞ = 0;

(H5) g(0) = 0 and f is an increasing function.


Lemma 5.1.1. [145] Let h ∈ C[0, 1], h ∈ C[0, 1], 0 ≤ r < 1, then the boundary value problem

Dαx(t) + h(t) = 0, t ∈ [0, 1], n− 1 < α ≤ n, (5.1.3)

x(1) = x′(0) =x′′(0) = · · · = x(n−2)(0) = x(n−1)(0) = 0, (5.1.4)

has a unique solution x(t) =∫ 10 Gα(t, s)h(s)ds, where

Gα(t, s) =

(1−s)α−1−(t−s)α−1

Γ(α) , 0 ≤ s ≤ t ≤ 1,

(1−s)α−1

Γ(α) , 0 ≤ t ≤ s ≤ 1.(5.1.5)

Green’s function Gα(t, s) defined by (5.1.5) has the following properties:

(i) Gα(t, s) ≥ 0 for all t, s∈ [0, 1] and Gα(t, s) > 0 for all t, s∈ (0, 1) .

(ii) max0≤t≤1

Gα(t, s) = Gα(s, s), s ∈ [0, 1].

(iii) min1/4≤t≤3/4

Gα(t, s) ≥ γαGα(s, s), where γα :=(1−

(34

)α−1).

The system of boundary value problems (5.1.1), (5.1.2) is equivalent to the following system of integral

equations x(t) = λ∫ 10 Gα(t, s)φ(s)f(y(s))ds, 0 ≤ t ≤ 1,

y(t) = λ∫ 10 Gβ(t, s)ψ(s)g(x(s))ds, 0 ≤ t ≤ 1.

(5.1.6)

76

A pair (x, y) ∈ C[0, 1]×C[0, 1] is a solution of boundary value problem (5.1.1), (5.1.2) if and only if (x, y)

is a solution of the system of integral equations (5.1.6)

Define Υ : B → B by

Υx(t) = λ

∫ 1

0Gβ(t, s)ψ(s)g(x(s))ds, (5.1.7)

then the system of integral equations (5.1.1) takes the formx(t) = λ∫ 10 Gα(t, s)φ(s)f(Υx(s))ds, 0 ≤ t ≤ 1,

y(t) = Υx(t), 0 ≤ t ≤ 1,(5.1.8)

Define T : B → B by

T x(t) = λ

∫ 1

0Gα(t, s)φ(s)f (Υx(s)) ds. (5.1.9)

If x is a fixed point of T , then (x(t),Υx(t)) is a solution of the system (5.1.8).

We use the following notations:

µ := max

∫ 1

0Gα(s, s)φ(s)ds,

∫ 1

0Gβ(s, s)ψ(s)ds

,

ν := min

∫ 3/4

1/4γαGα(s, s)φ(s)ds,

∫ 3/4

1/4γβGβ(s, s)ψ(s)ds

.

Define a cone V ⊂ C[0, 1] by V =

x ∈ C[0, 1] : x(t) ≥ 0, min

1/4≤t≤3/4x(t) ≥ γ∥x∥

, where γ := minγα, γβ.

The cone V ⊂ C[0, 1] induces a partial ordering in the Banach space C[0, 1], i.e, x ≤ y if x− y ∈ V.

Lemma 5.1.2. The operator T : V → V is completely continuous.

Proof. Since f , g, Gα and Gβ are positive functions. Therefore, for any x ∈ V, Υx(t) ≥ 0 and T x(t) ≥ 0.

Using (5.1.7), (5.1.9) and property (iii) of Green’s function, we obtain

min1/4≤t≤3/4

Υx(t) ≥ γβλ

∫ 1

0Gβ(s, s)ψ(s)g(x(s))ds

≥ γ max0≤t≤1

λ

∫ 1

0Gβ(t, s)ψ(s)g(x(s))ds = γ∥Υx∥,

which implies that Υ(V) ⊂ P. Further, for x ∈ V, we have

min1/4≤t≤3/4

T x(t) ≥ γαλ

∫ 1

0Gα(s, s)φ(s)f(Υ(x(s)))ds

≥ γ max0≤t≤1

λ

∫ 1

0Gα(t, s)φ(s)f(Υ(x(s)))ds = γ∥T x∥.

It follows T (V) ⊂ V. By Arzela–Ascoli’s theorem, T : V → V is completely continuous operator.


In this section, we study existence of at least one positive solution of the system (5.1.1), (5.1.2).

77

Theorem 5.1.3. Assume that µmaxf0, g0 < νminf∞, g∞ then for

λ ∈((νminf∞, g∞)−1, (µmaxf0, g0)−1

), (5.1.10)

the boundary value problem (5.1.1), (5.1.2) has at least one positive solution. Moreover, if (H1) holds,

then for any λ ∈ (0,∞), the boundary value problem (5.1.1), (5.1.2) has at least one positive solution.

Proof. In view of (5.1.10), choose sufficiently small ε > 0 such that

(νminf∞ − ε, g∞ − ε)−1 ≤ λ ≤ (µmaxf0 + ε, g0 + ε)−1. (5.1.11)

Furthermore, by the definitions of f0 and g0, there exists a constant r > 0 such that

f(x) ≤ (f0 + ε)x, g(x) ≤ (g0 + ε)x, for x ≤ r. (5.1.12)

Define Er = x ∈ C[0, 1] : ∥x∥ < r. For any x ∈ V ∩ ∂Er, by (5.1.7), (5.1.11) and (5.1.12), we have

Υx(t) = λ

∫ 1

0Gβ(t, z)ψ(z)g(x(z))dz ≤ λ

∫ 1

0Gβ(z, z)ψ(z)(g0 + ε)rdz ≤ λµ(g0 + ε)r ≤ r.

Using (5.1.12), we get

f(Υx(t)) ≤ (f0 + ε)Υx(t), for any x ∈ V ∩ ∂Er. (5.1.13)

In view of (5.1.9) and (5.1.13), we obtain

T x(t) = λ

∫ 1

0Gα(t, s)φ(s)f (Υx(s)) ds

≤ λ

∫ 1

0Gα(s, s)φ(s)(f0 + ε)Υx(t)ds ≤ λµ(f0 + ε)r ≤ r = ∥x∥.

Hence,

∥T x∥ ≤ ∥x∥, for all x ∈ V ∩ ∂Er. (5.1.14)

Next, we consider two cases:

Case 1. f∞, g∞ are finite. Let ε∗ > 0 such that

0 < (νmin(f∞ − ε∗, g∞ − ε∗)−1 ≤ λ. (5.1.15)

It follows from the definitions of f∞, g∞ that there exists constant r > r such that

f(x) ≥ (f∞ − ε∗)x, g(x) ≥ (g∞ − ε∗)x, for x ≥ γr. (5.1.16)

For any x ∈ V ∩ ∂Er, where Er = x ∈ C[0, 1] : ∥x∥ < r, we have

x(t) ≥ min1/4≤t≤3/4

x(t) ≥ ∥x∥ ≥ γr, t ∈ [1/4, 3/4].

Using, (5.1.7) and (5.1.16), we have

Υx(t) ≥λ∫ 3/4

1/4γβGβ(z, z)ψ(z)(g∞ − ε∗)x(z)dz ≥ λr

∫ 3/4

1/4γ22Gβ(z, z)ψ(z)(g∞ − ε∗)dz

≥ rλν(g∞ − ε∗) ≥ γr.

78

Which in view of (5.1.16) gives

f(Υx(t)) ≤ (f∞ − ε∗)Υx(t), for any x ∈ V ∩ ∂Er. (5.1.17)

Using (5.1.9) and (5.1.17), we get

T x(t) ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)f(Υx(s))ds ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f∞ − ε∗)A(x(s))ds

≥ λ

∫ 3/4

1/4γ2Gα(s, s)φ(s)(f∞ − ε∗)rds ≥ λν(f∞ − ε∗)r ≥ r = ∥x∥.

Hence

∥T x∥ ≥ ∥x∥, for all x ∈ V ∩ ∂Er. (5.1.18)

By the Theorem 2.3.7 and inequalities (5.1.14), (5.1.18) it follows that the operator T has a fixed point

x ∈ V ∩ E r\Er such that r < ∥x∥ < r.

Case 2. f∞ = ∞ and g∞ = ∞. Choose a constant N ∈ R+ satisfying N ≥ (λν)−1. There exists r > r

such that

f(x) ≥ Nx, g(x) ≥ Nx, for x ≥ γr. (5.1.19)

For any x ∈ V ∩ ∂Er, we have

x(t) ≥ min1/4≤t≤3/4

x(t) ≥ ∥x∥ ≥ γr, t ∈ [1/4, 3/4].

Using (5.1.19), we obtain

g(x) ≥ Nx, for any x ∈ V ∩ ∂Er. (5.1.20)

In view of (5.1.7), (5.1.19) and (5.1.20), it follows that

Υx(t) ≥ λ

∫ 3/4

1/4γβGβ(z, z)ψ(z)g(x(z))dz ≥ λ

∫ 3/4

1/4γβGβ(z, z)ψ(z)Nx(z)dz ≥ λνNγr ≥ γr,

This gives

f(Υx(t)) ≥ Nx(t), x ∈ V ∩ ∂Er.

Thus, by (5.1.9), we have

T x(t) ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)f(Υx(s))ds ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)NA(x(s))ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)Nrds ≥ λνNr ≥ r = ∥x∥.

Hence,

∥T x∥ ≥ ∥x∥, for all x ∈ V ∩ ∂Er. (5.1.21)

By Theorem 2.3.7, and the inequalities (5.1.14), (5.1.21) T has at least one fixed point x ∈ V ∩E r\Er such

that r < ∥x∥ < r.

79

Theorem 5.1.4. Assume that (H5) is satisfied and µmaxf∞, g∞ < νminf0, g0, then for

λ ∈((νminf0, g0)−1, (µmaxf∞, g∞)−1

), (5.1.22)

the boundary value problem (5.1.1), (5.1.2) has at least one positive solution. Moreover, if (H5) holds,

then for all λ ∈ (0,∞), the boundary value problem (5.1.1), (5.1.2) has at least one positive solution.

Proof. We consider two cases:

Case 1. f0, g0 are finite. Choose ε > 0 such that 0 < (νminf0 − ε, g0 − ε)−1 ≤ λ. There exists r > 0

such that

f(x) ≥ (f0 − ε)x, g(x) ≥ (g0 − ε)x, for x ∈ (0, r). (5.1.23)

By (H5), we choose 0 < ρ < rmin1, (λµ(g0 − ε))−1 such that

g(x) ≤ (λµ)−1r for x ∈ [0, ρ].

Using (5.1.7), we have

Υx(t) ≤ λ

∫ 1

0Gβ(z, z)ψ(z)g(x(z))dz ≤ λ

∫ 1

0Gβ(z, z)ψ(z)(λµ)

−1rdz ≤ r.

Which in view of (5.1.23) implies

f(Υx(t)) ≥ (f0 − ε)Υx(t) for x ∈ [0, ρ].

For any x ∈ V ∩ ∂Eρ, where Eρ = x ∈ C[0, 1] : ∥x∥ < ρ, we have x(t) ≥ min1/4≤t≤3/4

x(t) ≥ γ∥x∥ = γρ, t ∈

[1/4, 3/4]. Since, νλ(g0 − ε) ≥ 1. Therefore, using (5.1.9) we obtain

T x(t) ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f0 − ε)Υx(s)ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f0 − ε)λ

∫ 3/4

1/4γGβ(z, z)ψ(z)g(x(z))dzds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f0 − ε)λ

∫ 3/4

1/4γ2Gβ(z, z)ψ(z)(g0 − ε)∥x∥dzds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f0 − ε)νλ(g0 − ε)∥x∥ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(f0 − ε)∥x∥ds ≥ λν(f0 − ε)∥x∥ ≥ ∥x∥.

Which yields

∥T x∥ ≥ ∥x∥, for all x ∈ V ∩ ∂Eρ. (5.1.24)

Case 2. f0 = ∞ and g0 = ∞. Choose a constant N > 0 satisfying λνN ≥ 1. There exists r > 0 such that

f(x) ≥ Nx, g(x) ≥ Nx, for x ∈ (0, r). (5.1.25)

80

By (H5), there exists 0 < ρ < r such that g(x) ≤ (λµ)−1r, for x ∈ [0, ρ]. Therefore,

Υx(t) ≤ λ

∫ 1

0γGβ(s, s)ψ(s)g(x(s))ds ≤ λ

∫ 1

0γGβ(s, s)ψ(s)(λµ)

−1rds ≤ r, x ∈ [0, r1].

Which in view of (5.1.25) implies that

f(Υx(t)) ≤ NΥx(t), x ∈ [0, r1].

For any x ∈ V ∩ ∂Eρ, where Eρ = x ∈ C[0, 1] : ∥x∥ < ρ, using (5.1.9), (5.1.25), we have

T x(t) ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)NΥx(s)ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)Nλ

∫ 3/4


≥ λ

∫ 3/4


∫ 3/4

1/4γ2Gβ(z, z)ψ(z)N∥x∥dzds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)NνλN∥x∥ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)N∥x∥ds ≥ λνN∥x∥ ≥ ∥x∥.

Thus

∥T x∥ ≥ ∥x∥, for all x ∈ V ∩ ∂Eρ. (5.1.26)

Now, let λ be as given in (5.1.22) and choose ε∗ > 0 small enough such that

(νminf0 − ε∗, g0 − ε∗)−1 ≤ λ ≤ (µmaxf∞ + ε∗, g∞ + ε∗)−1. (5.1.27)

The right hand side of (5.1.27), gives µλ(f0 − ε∗) ≤ 1. Now, by the definitions of f∞, g∞, there exists a

constant r > ρ such that

f(x) ≤ (f∞ + ε∗)x, g(x) ≤ (g∞ + ε∗)x for x ≥ r. (5.1.28)

We, again consider two cases:

Case 1. g is bounded. There exists a constant l > 0, such that

g(x) ≤ l for x ∈ (0,∞).

Let l∗ = maxf(x) : x ∈ [0, λµl] and choose r2 ≥ max2r, λl∗µ. Then for any x ∈ V ∩ ∂Er2 , where

Er2 = x ∈ C[0, 1] : ∥x∥ < r2, using (5.1.7), we get

Υx(t) ≤∫ 1

0λGβ(z, z)ψ(z)ldz ≤ λµl.

Which implies that f(Υx(t)) ≤ l∗. Thus, we have,

T x(t) ≤ λ

∫ 1

0Gα(s, s)φ(s)l

∗ds ≤ λµl∗ ≤ ∥x∥.

81

Hence,

∥T x∥ ≤ ∥x∥, for all x ∈ V ∩ ∂Er2 . (5.1.29)

Case 2. g is unbounded. There exists a constant r∗0 ≥ max2ρ, r such that

g(x) ≤ g(r∗0) for x ∈ (0, r∗0].

Since the function f is increasing, therefore there exists a constant ρ∗ ≥ maxr∗0, µg(r∗0) such that

f(x) ≤ f(ρ∗), for x ∈ (0, ρ∗].

Let r2 ≥ max2r∗0, λµf(ρ∗) and define Er2 = x ∈ C[0, 1] : ∥x∥ < r2. Then for x ∈ V ∩ ∂Er2 , we obtain

Υx(t) ≤ λ

∫ 1

0Gβ(s, s)ψ(s)g(r

∗0)ds ≤ λµg(r∗0) ≤ ρ∗.

Hence,

T x(t) ≤ λ

∫ 1

0Gα(s, s)φ(s)f(Υx(s))ds ≤ λµf(ρ∗) ≤ ∥x∥.

Which implies that

∥T x∥ ≤ ∥x∥, for all x ∈ V ∩ ∂Er2 . (5.1.30)

Therefore, by Theorem 2.3.7, the inequalities (5.1.24) and (5.1.29) or (5.1.24) and (5.1.30) or (5.1.26) and

(5.1.29) or (5.1.26) and (5.1.30) implies that T has a fixed point x ∈ V ∩ Er2\Eρ.

Example 5.1.5. Consider the following system of nonlinear fractional differential equationscD2.5x(t) + λ (56754y2+160)yt72

1+y2= 0, 0 ≤ t ≤ 1,

cD2.7y(t) + λx(x+1)(23471e−x+120)t92

2+x = 0,(5.1.31)

satisfying the boundary conditions x(1) = 0, x′(0) = 0, x′′(0) = 0,

y(1) = 0, y′(0) = 0, y′′(0) = 0,(5.1.32)

Here, α = 2.5, β = 2.7, φ(t) = t72 , ψ(t) = t

92 , f(y) = (56754y2+160)

1+y2and g(x) = (x+1)(23471e−x+120)

2+x . By

direct calculations γ = 0.512860, µ = 0.016155 and ν = 0.001530. By calculations f0 = 160, f∞ = 56754,

g0 = 39295 and g∞ = 120. All conditions of the Theorem 5.1.3 are satisfied. Therefore, the boundary

value problem (5.1.31) and (5.1.32) has a positive solution for each λ ∈ (0.01151, 0.38687).


Theorem 5.1.6. Assume that (H3), (H5) holds and for λ > 0, there exists r ∈ R+ such that

f(x) < (λµ)−1r, g(x) < (λµ)−1r, for x ∈ [0, r], (5.1.33)

then, the boundary value problem (5.1.1), (5.1.2) has at least two positive solutions.

82

Proof. Set Er = x ∈ C[0, 1] : ∥x∥ < r. For any x ∈ V ∩ ∂Er, using (5.1.33), we obtain

Υ(x(t)) ≤ λ

∫ 1

0Gβ(s, s)ψ(s)g(x(s))ds < λ

∫ 1

0Gβ(s, s)ψ(s)(λµ)

−1rds ≤ r.

Which gives f(Υx(t)) ≤ (λµ)−1r, x ∈ V ∩ ∂Er. Hence,

T (x(t)) ≤ λ

∫ 1

0Gα(s, s)φ(s)f(Υx(s))ds < λ

∫ 1

0Gα(s, s)φ(s)(λµ)

−1rds ≤ r, x ∈ V ∩ ∂Er.

Thus,

∥T x∥ ≤ ∥x∥, for x ∈ V ∩ ∂Er. (5.1.34)

Now, choose N ∈ R+ satisfying N ≥ (λν)−1. From H3, f0 = ∞, g0 = ∞. Therefore, there exists

0 < ρ0 < r such that

f(x) ≥ Nx, g(x) ≥ Nx for x ∈ (0, ρ0). (5.1.35)

By assumption (H5) and (5.1.33), there exists 0 < r∗ < ρ0 such that

g(x) < (λµ)−1ρ0, for x ∈ [0, r∗].

For any x ∈ V ∩ ∂Er∗ , where Er∗ = x ∈ C[0, 1] : ∥x∥ < r∗, we have

Υx(t) ≤ λ

∫ 1

0Gβ(z, z)ψ(z)g(x(z))dz ≤ λ

∫ 1

0Gβ(z, z)ψ(z)(λµ)

−1ρ0dz ≤ ρ0.

Which in view of (5.1.35) implies that f(Υx) ≥ Nx, x ∈ V ∩ ∂Er∗ . Hence,

T x(t) ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)NΥx(s)ds

≥ λ

∫ 3/4


∫ 3/4


≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)N(λνN)∥x∥ds

≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)N∥x∥ds ≥ λνN∥x∥ ≥ ∥x∥, x ∈ V ∩ ∂Er∗ .

Consequently,

∥T x∥ ≥ ∥x∥ for x ∈ V ∩ ∂Er∗ . (5.1.36)

Again, from H3, we have f∞ = ∞, g∞ = ∞. Let M ∈ R+, such that M ≥ (λν)−1. Then there exists

ρ∗0 > r such that

f(x(t)) ≥Mx, g(x(t)) ≥Mx, for x ≥ ρ∗0. (5.1.37)

Let r = max2r∗,ρ∗0γ and define Er = x ∈ C[0, 1] : ∥x∥ < r. Then for x ∈ V ∩ ∂Er, we have

min1/4≤t≤3/4

x(t) ≥ γ∥x∥ = γr ≥ ρ∗0. Therefore,

Υx(t) ≥ λ

∫ 3/4

1/4γGβ(s, s)ψ(s)g(x(s))ds ≥ λ

∫ 3/4

1/4γGβ(s, s)ψ(s)Mx(s)ds ≥ λνMρ∗0 ≥ ρ∗0.

83

Thus, in view of (5.1.37) we have f(Υx) ≥Mu, x ∈ V ∩ ∂Er. Hence,

T (x) ≥ λM

∫ 3/4

1/4γGα(s, s)φ(s)A(x(s))ds

≥ λM

∫ 3/4

1/4γGα(s, s)φ(s)

(λ

∫ 3/4

1/4γ2Gβ(z, z)ψ(z)M∥x∥dz

)ds

≥ λM

∫ 3/4

1/4γGα(s, s)φ(s)(λνM)rds ≥ λνMr ≥ r = ∥x∥, x ∈ V ∩ ∂Er.

This implies that,

∥T x∥ ≥ ∥x∥ for x ∈ V ∩ ∂Er. (5.1.38)

From (5.1.34), (5.1.36) and (5.1.38), using Theorem 2.3.7, it follows that the operator T has two fixed

points x1 ∈ V ∩ (Er\Er∗) and x2 ∈ V ∩ (E r\Er) such that 0 < ∥x1∥ < r < ∥x2∥.

Example 5.1.7. Consider the system of nonlinear fractional differential equationscD3.8x(t) + λ√1 + t2(

√y + y3) = 0, 0 < t < 1,

cD3.5y(t) + λ√t(x

34 + x

72 ) = 0, 0 < t < 1,

(5.1.39)

satisfying the boundary conditionsx(1) = 0, x′(0) = 0, · · · , xiii(0) = 0,

y(1) = 0, y′(0) = 0, · · · , yiii(0) = 0.(5.1.40)

Here, α = 3.8, β = 3.5, φ(t) =√1 + t2, ψ(t) =

√t, f(y) =

√y + y3 and g(x) = x

34 + x7. Obviously,

f0 = ∞, g0 = ∞ and f∞ = ∞, g∞ = ∞. By some calculations we have γ = 0.634645, µ = 0.057976 and

ν = 0.012333. Also, for 0 < λ < 0.07657 and x ∈ (0, 15], we have f(y) < 3378 and g(x) < 3378. Hence,

by Theorem 5.1.6, the system of fractional differential equations (5.1.39), (5.1.40) has at least two positive

solutions.

Theorem 5.1.8. Assume that (H4), (H5) are satisfied and for λ > 0, there is r ∈ R+ such that

f(x) ≥ (λν)−1r, g(x) ≥ (λν)−1r, for x ∈ [γr, r], (5.1.41)

then, the boundary value problem (5.1.1), (5.1.2) has at least two positive solutions.

Proof. Choose N ∈ R+ satisfying N ≤ (λµ)−1. Since, from H4, we have f0 = 0, g0 = 0. Thus, we may

choose 0 < ρ < r such that

f(x) < Nx and g(x) < Nx for x ∈ (0, ρ), (5.1.42)

For any x ∈ V ∩ ∂Eρ, where Eρ = x ∈ C[0, 1] : ∥x∥ < ρ, we have

Υ(x(t)) ≤ λ

∫ 1

0Gβ(s, s)ψ(s)g(x(s))ds < λ

∫ 1

0Gβ(s, s)ψ(s)N∥x∥ds ≤ λµN∥x∥ ≤ ρ. (5.1.43)

84

Using (5.1.42) and (5.1.43), we have

T (x(t)) ≤ λ

∫ 1

0Gα(s, s)φ(s)f(Υx(s))ds

≤ λ

∫ 1

0Gα(s, s)φ(s)NΥx(s)ds

< λ

∫ 1

0Gα(s, s)φ(s)N∥x∥ds ≤ λµN∥x∥ ≤ ∥x∥.

Which gives,

∥T x∥ ≤ ∥x∥, for x ∈ V ∩ ∂Eρ. (5.1.44)

Also, from H4, we have f∞ = 0, g∞ = 0. Thus, here exists constant r∗2 > r, such that for M satisfying

M ≤ (λµ)−1, we have

f(x) < Mx, g(x) < Mx for x ∈ [r∗2,∞). (5.1.45)

We consider following two cases:

Case 1. g is bounded. There exists a constant ω > 0 such that g(x) ≤ ω for all x ∈ (0,∞). Therefore, we

have

Υx(t) ≤ λω

∫ 1

0Gβ(s, s)ψ(s)ds ≤ ωλµ.

Let τ = maxf(x) : 0 ≤ u ≤ ωλµ. For r2 ≥ max2r, τλµ, define Er2 = x ∈ C[0, 1] : ∥x∥ < r2. For

any x ∈ V ∩ ∂Er2 , we get

T x(t) ≤∫ 1

0λGα(s, s)φ(s)f(A(x(s)))ds ≤

∫ 1

0λGα(s, s)φ(s)τds ≤ τλµ ≤ ∥x∥.

Hence,

∥T x∥ ≤ ∥x∥ for all x ∈ V ∩ ∂Er2 .

Case 2. g is unbounded. There exists r2 > r∗2 such that g(x) ≤ g(r2) ≤ Mr2, for x ∈ (0, r2]. Therefore,

Υx(t) ≤ λ

∫ 1

0Gβ(s, s)ψ(s)Mr2ds ≤ λMµr2 ≤ r2.

Since f is increasing function, therefore f(x) ≤ f(r2) ≤ Mr2, for x ∈ (0, r2]. It follows that for x ∈ V∩∂Er2 ,we have

T x(t) ≤ λ

∫ 1

0Gα(s, s)φ(s)f(Υx(s))ds ≤ (λMµ)r2 ≤ ∥x∥.

Therefore,

∥T x∥ ≤ ∥x∥, for all x ∈ V ∩ ∂Er2 .

Consequently, in both the cases, we have

∥T x∥ ≤ ∥x∥, for all x ∈ V ∩ ∂Er2 . (5.1.46)

Now, for any x ∈ V ∩ ∂Er, where Er = x ∈ C[0, 1] : ∥x∥ < r, we have

x(t) ≥ min1/4≤t≤3/4

x(t) ≥ ∥x∥ ≥ γr, t ∈ [1

4,3

4],

85

which in view of (5.1.41) implies that

Υx(t) ≥λ∫ 3/4

1/4γGβ(s, s)ψ(s)(g(x(s))ds ≥ γr.

Also, using (5.1.43), we obtain Υx(t) ≤ ρ for any x ∈ V ∩ ∂Eρ. Since, ρ < r, we have γr ≤ Υx(t) ≤ r, for

x ∈ V ∩ ∂Er. Consequently, f(Υx) ≥ (λν)−1r. Hence,

T x(t) ≥ λ

∫ 3/4

1/4γGα(t, s)φ(s)f(Υx)ds ≥ λ

∫ 3/4

1/4γGα(s, s)φ(s)(λν)

−1rds ≥ ∥x∥.

Thus,

∥T x∥ ≥ ∥x∥ for x ∈ V ∩ ∂Er. (5.1.47)

From (5.1.44), (5.1.46) and (5.1.47), using Theorem 2.3.7, we conclude that T has two fixed points x1 ∈Er\Eρ and x2 ∈ Er2\Er.

Example 5.1.9. cD5.4x(t) + λ(1 + t2)y2e−πy = 0, 0 < t < 1,

cD5.8y(t) + λt2x52 e−πx = 0, 0 < t < 1,

(5.1.48)

satisfying the boundary conditionsx(1) = 0, x′(0) = 0, · · · , x5(0) = 0,

y(1) = 0, y′(0) = 0, · · · , y5(0) = 0.(5.1.49)

Set α = 5.4, β = 5.8, φ(t) = (1+ t2), ψ(t) = t2, f(y) = y2e−πy and g(x) = x52 e−πx. By direct calculations,

we get γ = 0.78849004, µ = 0.00432759 and ν = 0.00007918.

Furthermore, for λ > 190716, we have f(y) > 0.05221499, g(x) > 0.05221499, for x ∈ (0.78849004, 1).

Also, f0 = 0, f∞ = 0, g0 = 0 and g∞ = 0. All the conditions of Theorem 5.1.8 are satisfied. Hence, the

system of boundary value problems (5.1.48), (5.1.49) has at least two positive solution.

5.2 Positive solutions to a system of fractional differential equations

with three–point boundary conditions

In this section, we study existence and multiplicity results for a coupled system of nonlinear three–point

boundary value problems for higher order fractional differential equations of the type [93]cDαx(t) = λφ(t)f(x(t), y(t)), n− 1 < α ≤ n, n ∈ N,cDβy(t) = µψ(t)g(x(t), y(t)), n− 1 < α, β ≤ n

(5.2.1)

satisfying the boundary conditionsx′(0) = x′′(0) = x′′′(0) = · · · = x(n−1)(0) = 0, x(1) = θ1x(µ1),

y′(0) = y′′(0) = y′′′(0) = · · · = y(n−1)(0) = 0, y(1) = θ2x(µ2),(5.2.2)

where λ, µ > 0, for n ∈ N; θi, µi ∈ (0, 1) for i = 1, 2. We use Guo-Krasnosel’skii fixed point theorem

to establish existence and multiplicity results for positive solutions. We drive explicit intervals for the

parameters λ and µ for which the system possess the positive solutions or multiple positive solutions.

86

5.2.1 Greens’s function and its properties

Lemma 5.2.1. Let h ∈ C[0, 1], then the linear three–point boundary value problem

cDαx(t) + h(t) = 0, t ∈ (0, 1), n− 1 < α ≤ n,

x′(0) = x′′(0) = x′′′(0) = · · · = x(n−1)(0) = 0, x(1) = θ1x(µ1),(5.2.3)

has a solution given by

x(t) =

∫ 1

0Hα(t, s)h(s)ds,

where,

Hα(t, s) =

(1−s)α−1−(1−θ1)(t−s)α−1−θ1(µ1−s)α−1

(1−θ1)Γ(α) , s ≤ t, µ1 ≥ s,

(1−s)α−1−(1−θ1)(t−s)α−1

(1−θ1)Γ(α) , µ1 ≤ s ≤ t ≤ 1,

(1−s)α−1−θ1(µ1−s)α−1

(1−θ1)Γ(α) , 0 ≤ t ≤ s ≤ µ1,

(1−s)α−1

(1−θ1)Γ(α) , t ≤ s, s ≥ µ1.

(5.2.4)

Proof. The proof follows by slight modifications in the proof of Lemma 4.1.1.

Lemma 5.2.2. The Green’s function Hα(t, s) defined by (5.2.4) satisfies the following properties:

(i) Hα(t, s) > 0 for all t, s ∈ (0, 1),

(ii) For t ∈ [0, 1], Hα(t, s) is decreasing.

(iii) For ℓ ∈ (0, 1), minℓ≤t≤1

Hα(t, s) ≥ ξαΦα(s) where ξα = θ1(1 − µα−11 ), Hα(t, s) ≤ Φα(s) and Φα(s) =

(1−s)α−1

(1−θ1)Γ(α) .

Proof. The proof follows by slight modifications in the proof of Lemma 4.1.2.

Now we write the system of boundary value problem (5.2.1), (5.2.2) as an equivalent system of integral

equations x(t) = λ∫ 10 Hα(t, s)φ(s)f(x(s), y(s))ds

y(t) = µ∫ 10 Hβ(t, s)ψ(s)g(x(s), y(s))ds.

(5.2.5)

Starting from now, we will work in the Banach space B2 = C[0, 1]× C[0, 1] furnished with the norm

∥(x, y)∥2 = ∥x∥+ ∥v∥ = max0≤t≤1

|x(t)|+ max0≤t≤1

|y(t)|.

We define an operators Tλ, Tµ : B2 → B2 by Tλ(x, y)(t) = λ∫ 10 Hα(t, s)φ(s)f(x(s), y(s))ds and

Tµ(x, y)(t) = µ∫ 10 Hβ(t, s)ψ(s)g(x(s), y(s))ds. Define an other operator T : B2 → B2 by

T (x, y) = (Tλ(x, y), Tµ(x, y)). The fixed points of T are the solutions of (5.2.1), (5.2.2).

Let ξ = minξα, ξβ, and define a cone in B2 by

W =(x, y) ∈ B2 : x(t) ≥ 0, y(t) ≥ 0, min

l≤t≤1(x(t) + y(t)) ≥ ξ∥(x, y)∥2

, where ξ := min(ξα, ξβ).

In this section for the system (5.2.1), (5.2.2) we admit following assumptions:

87

(H6) f, g ∈ C(R+ × R+,R), and the limits

f0 = limx+y→0

f(x, y)

x+ y, f∞ = lim

x+y→∞

f(x, y)

x+ y,

g0 = limx+y→0

g(x, y)

x+ y, g∞ = lim

x+y→∞

g(x, y)

x+ y

exist and f0, f∞, g0, g∞ ∈ [0,∞);

(H7) φ,ψ ∈ C([0, 1], (0,∞)) such that Iαφ(1) and Iβψ(1) exist and are finite;

(H8) (1− θ1)Iαφ(1)f0 < ξαξIαl φ(1)f∞, (1− θ2)Iβψ(1)g0 < (1− ξ)ξβIβl ψ(1)g∞;

(H9) (1− θ1)Iαφ(1)f∞ < ξξαIαl φ(1)f0, (1− θ2)Iβψ(1)g∞ < (1− ξ)ξβIβl ψ(1)g0;

(H10) there exist constants r, γ, ζ with

(1− θ1)ζIαφ(1) < ξ2γIαl φ(1) and (1− θ2)ζIβψ(1) < ξ(1− ξ)γIβl ψ(1) such that

(i) f0 = 0, g0 = 0, f∞ = 0, g∞ = 0;

(ii) f(x, y) ≥ γr, g(x, y) ≥ γr, for ∥(x, y)∥ ∈ [ξr, r];

(H11) there exist constants r, γ, ζ with

(1− θ1)γIαφ(1) < ξ2ζIαl φ(1) and (1− θ2)γIβψ(1) < ξ(1− ξ)ζIβl ψ(1) such that

(i) f0 = ∞, g0 = ∞, f∞ = ∞, g∞ = ∞;

(ii) f(x, y) ≤ γr, g(x, y) ≤ γr, for ∥(x, y)∥ ∈ [0, r].

(H12) (i) f0 = 0, g0 = 0, f∞ = ∞, g∞ = ∞; (ii) f0 = ∞, g0 = ∞, f∞ = 0, g∞ = 0.

Lemma 5.2.3. Assume that H6, holds. Then, the operator T : W → W is completely continuous.

Proof. First, we prove that T (W) ⊂ W. For any (t, s) ∈ [l, 1]× [0, 1], by Lemma 5.2.2, we have

minl≤t≤1

(Tλ(x, v)(t) + Tµ(x, v)(t))

= minl≤t≤1

(λ

∫ 1

0Hα(t, s)φ(s)f(x(s), y(s))ds,+µ

∫ 1

0Hβ(t, s)ψ(s)g(x(s), y(s))ds

)≥λξα

∫ 1

0Φα(s)φ(s)f(x(s), y(s))ds+ µξβ

∫ 1

0Φβ(s)ψ(s)g(x(s), y(s))ds

≥max0≤t≤1

(λξα

∫ 1

0Hα(t, s)φ(s)f(x(s), y(s))ds+ µξβ

∫ 1


)=ξα∥Tλ(x, y)∥2 + ξβ∥Tµ(x, y)∥2 ≥ ξ∥T (x, y)∥2.

Therefore minl≤t≤1

(Tλ(x, v)(t) + Tµ(x, v)(t)) ≥ ξ∥T (x, y)∥2. Hence T (W) ⊂ W.

Now we prove that the operator T maps bonded sets into uniformly bonded sets. For fixed M ∈ R+,

consider a bounded subset M of W defined by M =(x, y) ∈ W : ∥(x, y)∥2 ≤ M

. Also, define

88

L1 = maxf(x(t), y(t)) : (x, y) ∈ M

, L2 = max

g(x(t), y(t)) : (x, y) ∈ M

, then for (x, y) ∈ M, using

(H7) and Lemma 5.2.2, we have

|Tλ(x, y)| =∣∣∣λ ∫ 1

0Hα(t, s)φ(s)f(x(s), y(s))ds

∣∣∣≤ λL1

∫ 1

0

(1− s)α−1φ(s)

Γ(α)(1− θ1)ds =

L1λ

1− θ1Iαφ(1) < +∞.

Similarly, one can show that |Tµ(x, y)| < +∞. Therefore |Tλ(x, y)|+ |Tµ(x, y)| < +∞. Which implies that

T (M) is bounded.

Finally, it remains to show that T (M) is equicontinuous. By (ii) of lemma 5.2.2, we have

∣∣ ddt

(Tλ(x, y)(t))∣∣ ≤λ ∫ 1

0

(t− s)α−2

Γ(α− 1)φ(s)|f(x(s), y(s))|ds

≤L1λ

∫ 1

0

(1− s)α−2

Γ(α− 1)φ(s)ds = L1λIα−1φ(1) < +∞.

Define δ =(λL1Iα−1φ(1) + µL2Iβ−1ψ(1)

)−1, and chose t, τ ∈ [0, 1] such that t < τ and τ − t < δ. Then

for all ε > 0 and (x, y) ∈ M, we obtain

∣∣Tλ(x, y)(τ)− Tλ(x, y)(t)∣∣ =∣∣∣ ∫ τ

t

d

ds(Tλ(x, y)(s)) ds

∣∣∣≤L1λIα−1φ(1)(τ − t).

Similarly, ∣∣Tµ(x, y)(τ)− Tµ(x, y)(t)∣∣ = ∣∣∣ ∫ τ

t

d

ds(Tµ(x, y)(s)) ds

∣∣∣ ≤ L2µIβ−1ψ(1)(τ − t).

Hence, it follows that∥∥T (x, y)(τ)− T (x, y)(t)∥∥2≤(L1λIα−1φ(1) + L2µIβ−1ψ(1)

)(τ − t) < ε.

Therefore, by the Arzela–Ascoli Theorem, T : W → W is completely continuous.


Theorem 5.2.4. Assume that (H6)− (H8) hold, then for every

λ ∈(

1−θ1ξαIα

l φ(1)f∞, ξ

Iαφ(1)f0

), and µ ∈

(0, 1−ξ

Iβψ(1)g0

)or λ ∈

(0, ξ

Iαφ(1)f0

)and µ ∈

(1−θ2

ξβIβl ψ(1)g∞

, 1−ξIβψ(1)g0

)the boundary value problem (5.2.1), (5.2.2) has at least one positive solution. Moreover, if (H12)(i) holds

then for each λ, µ ∈ (0,∞) the boundary value problem (5.2.1), (5.2.2) has at least one positive solution.

Proof. Choose ε > 0 such that following hold:

1− θ1ξαIαl φ(1)(f∞ − ε)

≤ λ ≤ ξ

Iαφ(1)(f0 + ε), 0 < µ ≤ 1− ξ

Iβψ(1)(g0 + ε).

By definitions of f0 and g0 there exists a constant r > 0 such that

f(x, y) ≤ (f0 + ε)(x+ y), g(x, y) ≤ (g0 + ε)(x+ y), for x+ y ∈ [0, r].

89

Define Er = (x, y) ∈ W : ∥(x, y)∥2 ≤ r . For any (x, y) ∈ W ∩ ∂Er, by Lemma 5.2.2, we have

Tλ(x, y)(t) =λ∫ 1


≤λ∫ 1

0

(1− s)α−1

Γ(α)φ(s)(f0 + ε)(x+ y)ds

≤λ(f0 + ε)Iαφ(1)∥(x, y)∥2 ≤ ξ∥(x, y)∥2,

Tµ(x, y)(t) =µ∫ 1

0Hβ(t, s)ψ(s)f(x(s), y(s))ds

≤µ∫ 1

0

(1− s)β−1

Γ(β)ψ(s)(g0 + ε)(x+ y)ds

≤µ(g0 + ε)Iβψ(1)∥(x, y)∥2 ≤ (1− ξ)∥(x, y)∥2.

Hence,

∥T (x, y)∥2 ≤ξ∥(x, y)∥2 + (1− ξ)∥(x, y)∥2

=∥(x, y)∥2, for all (x, y) ∈ W ∪ ∂Er.(5.2.6)

Next, we consider two cases:

Case 1. f∞, g∞ are finite.

Let ε1 > 0 such that 0 < 1−θ1ξαIα

l φ(1)(f∞−ε1) ≤ λ. By the definition of f∞, g∞ there exists a constant r∗ > r,

such that

f(x, y) ≥ (f∞ − ε1)(x+ y), g(x, y) ≥ (g∞ − ε1)(x+ y), for x+ y ≥ ξr∗.

Define Er∗ = (x, y) ∈ W : ∥(x, y)∥2 ≤ r∗ . Then for (x, y) ∈ W ∩ ∂Er∗ , we have

x(t) + y(t) ≥ minl≤t≤0

(x(t) + y(t)) ≥ ξ∥(x, y)∥2 = ξr∗.

Therefore, by Lemma 5.2.2, we obtain

Tλ(x, y)(t) =λ∫ 1


≥λξα∫ 1

l

(1− s)α−1

(1− θ1)Γ(α)φ(s)(f∞ − ε)(x+ y)ds

≥ λξα1− θ1

Iαl φ(1)(f∞ − ε)∥(x, y)∥2 ≥ ∥(x, y)∥2.

Thus

∥T (x, y)∥2 ≥ ∥Tλ(x, y)∥2 ≥ ∥(x, y)∥2, for all (x, y) ∈ W ∩ ∂Er∗ . (5.2.7)

By Theorem 2.3.7 and inequalities (5.2.6), (5.2.7), the operator T has a fixed point (x, y) ∈ W ∩ Er∗\Ersuch that r ≤ ∥(x, y)∥2 ≤ r∗.

Case 2. f∞ = ∞, g∞ = ∞.

For λ > 0, choose a constant ζ > 0 such that ζ ≥ min

1−θ1λξαIα

l φ(1), 1−θ2µξβIβ

l ψ(1)

. By (H12)(i), f∞ = ∞,

g∞ = ∞. Therefore, there exists r∗ > r, such that

f(x, y) ≥ ζ(x+ y), for x+ y ≥ ξr∗.

90

For (x, y) ∈ W ∩ ∂Er∗ , we have x(t) + y(t) ≥ minl≤t≤0

(x(t) + y(t)) ≥ ξ∥(x, y)∥2 = ξr∗. Hence,

f(x, y) ≥ ζ(x+ y), for any (x, y) ∈ W ∩ ∂Er∗ .

By Lemma 5.2.2, we have

Tλ(x, y)(t) =λ∫ 1


≥λξαζ∫ 1

l

(1− s)α−1

(1− θ1)Γ(α)φ(s)(x+ y)ds

≥ λξαζ

1− θ1Iαl φ(1)∥(x, y)∥2 ≥ ∥(x, y)∥2.

Thus, we have

∥T (x, y)∥2 ≥ ∥Tλ(x, y)∥2 ≥ ∥(x, y)∥2, for all (x, y) ∈ W ∩ ∂Er∗ . (5.2.8)

Hence, by Theorem 2.3.7 and inequalities (5.2.6), (5.2.8), the operator T has a fixed point (x, y) ∈W ∩ Er∗\Er such that r ≤ ∥(x, y)∥2 ≤ r∗.

Theorem 5.2.5. Assume that (H6), (H7) and (H9) hold, then for every

λ ∈(

1−θ1ξαIα

l φ(1)f0, ξ

Iαφ(1)f∞

)and µ ∈

(0, 1−ξ

Iβψ(1)g∞

)or λ ∈

(0, ξ

Iαφ(1)f∞

)and µ ∈

(1−θ2

ξβIβψ(1)g0, 1−ξIβl ψ(1)g∞

),

the boundary value problem (5.2.1), (5.2.2) has at least one positive solution. Moreover, if (H12)(ii) holds,

then for each λ, µ ∈ (0,∞) the boundary value problem (5.2.1), (5.2.2) has at least one positive solution.

Proof. The proof is similar to the Theorem 5.2.4. Therefore it is omitted.

Example 5.2.6. Consider the system of fractional differential equationscD

52x(t) = λ

(π − 6(x+ y + π

2 )− 3

2

)(243578 + eπt),

cD2710 y(t) = µ

((1 + 23√

x+y

)− 52 + π

257

)(sin t+ 256eπt),

(5.2.9)

satisfying the boundary conditionsx′(0) = x′′(0) = 0, x(1) = 12x(

23),

y′(0) = y′′(0) = 0, y(1) = 34x(

35),

(5.2.10)

Set f(x, y) = π − 6(x + y + π2 )

− 32 , g(x, y) =

(1 + 23√

x+y

)− 52+ π

257 , θ1 = 12 , θ2 = 3

4 , µ1 = 23 , µ2 = 3

5 .

Clearly f0 = π − 6( 2π )32 , f∞ = π, g0 = π

257 and g∞ = 1 + π257 . Let l = 1

4 . By calculations ξα ≈ 0.227834,

ξβ ≈ 0.4352, ξ ≈ 0.2278; Iαφ(1) ≈ 1.0794, Iαl φ(1) ≈ 0.7865, Iβψ(1) ≈ 182.2013 and Iβl ψ(1) ≈ 133.3916.

Since, (1 − θ1)Iαφ(1)f0 ≈ 0.050678 < ξαξIαl φ(1)f∞ ≈ 0.128268 and (1 − θ2)Iβψ(1)g0 ≈ 0.556812 <

(1 − ξ)ξβIβl ψ(1)g∞ ≈ 45.382788. We conclude that for λ, µ satisfying 0.888114 < λ < 2.247865 and

0 < µ < 0.346692 or 0 < λ < 2.247865 and 0.004254µ < 0.346692, the system of fractional differential

equations (5.2.9), (5.2.10) has at least one positive solution.

91


Theorem 5.2.7. Assume that (H6), (H7), (H10) hold, then for any

λ ∈[

1−θ1ξγIα

l φ(1), ξζIαφ(1)

]and µ ∈

(0, 1−ξ

ζIβψ(1)

], or λ ∈

(0, ξ

ζIαφ(1)

]and µ ∈

[1−θ2

ξγIβl ψ(1)

, 1−ξζIβψ(1)

], the boundary

value problem (5.2.1), (5.2.2) has at least two positive solutions (x1, y1), (x2, y2) such that 0 < ∥(x1, y1)∥2 <r < ∥(x2, y2)∥2 for some r > 0.

Proof. By (H10)(i), f0 = 0, g0 = 0, there exists r1 ∈ (0, r) such that

f(x, y) ≤ ζ(x+ y), g(x, y) ≤ ζ(x+ y), for x+ y ∈ (0, r1).

Define Er1 = (x, y) ∈ W : ∥(x, y)∥2 < r1. For any (x, y) ∈ W ∩ ∂Er1 ,by Lemma 5.2.2 we have

Tλ(x, y)(t) =λ∫ 1


≤λζ∫ 1

0

(1− s)α−1

Γ(α)φ(s)(x+ y)ds

≤λζIαφ(1)∥(x, y)∥2 ≤ θ∥(x, y)∥2,

Tµ(x, y)(t) =λ∫ 1


≤µζ∫ 1

0

(1− s)β−1

Γ(β)ψ(s)(x+ y)ds

≤µζIβψ(1)∥(x, y)∥2 ≤ (1− θ)∥(x, y)∥2.

Which implies that

∥T (x, y)∥2 ≤ ∥(x, y)∥2, for all (x, y) ∈ W ∩ ∂Er1 . (5.2.11)

Also, from (H10)(i), we have f∞ = 0, g∞ = 0. Therefore, there exists r2 > r such that for some positive

constant ζ, we have

f(x, y) ≤ ζ(x+ y), g(x, y) ≤ ζ(x+ y), for x+ y ≥ r2. (5.2.12)

Set Er2 = (x, y) ∈ W : ∥(x, y)∥2 < r2. For any (x, y) ∈ W ∩ ∂Er2 , by Lemma 5.2.2 we have

∥T (x, y)∥2 ≤ ∥(x, y)∥2, for all (x, y) ∈ W ∩ ∂Er2 . (5.2.13)

Next define Er = (x, y) ∈ W : ∥(x, y)∥2 < r. For any (x, y) ∈ W ∩ ∂Er, by Lemma 5.2.2 we have

x(t) + y(t) ≥ ∥(x, y)∥2 ≥ ξr, for all t ∈ [l, 1].

Therefore

Tλ(x, y)(t) =λ∫ 1


≥λξ∫ 1

l

(1− s)α−1

(1− θ1)Γ(α)φ(s)γrds

=λξIαl φ(1)γr = r = ∥(x, y)∥2.

92

Finally, we have

∥T (x, y)∥2 ≥ ∥(x, y)∥2, for all (x, y) ∈ W ∩ ∂Er. (5.2.14)

Hence, by (5.2.11), (5.2.13), (5.2.11) and by Theorem 2.3.7, it follows that the operator T has two fixed

points x1 ∈ Er\Er1 and x2 ∈ Er2\Er.

Theorem 5.2.8. Assume that (H6), (H7) and (H11) hold, then for any

λ ∈[

1−θ1ξζIα

l φ(1), ξγIαφ(1)

]and µ ∈

(0, 1−ξ

γIβψ(1)

]or λ ∈

(0, ξ

γIαφ(1)

], µ ∈

[1−θ2

ξζIβl ψ(1)

, 1−ξγIβψ(1)

], the boundary

value problem (5.2.1), (5.2.2) has at least two positive solutions (x1, y1), (x2, y2) such that 0 < ∥(x1, y1)∥2 <r < ∥(x2, y2)∥2.

Proof. The proof is similar to the Theorem 5.2.7, so we omit it.

Example 5.2.9. Consider the system of fractional differential equationscD

145 x(t) = λ

(194t2

1+t

)(12π

x+y+12π − e−π(x+y)),

cD125 y(t) = µ

(78125t2

1+t2

)(1√x+y

− e−π(x+y)),

(5.2.15)

satisfying the boundary conditionsx′(0) = x′′(0) = 0, x(1) = 3750x(

1125),

y′(0) = y′′(0) = 0, y(1) = 1120x(

1950).

(5.2.16)

Set f(x, y) = 194(

12πx+y+12π − e−π(x+y)

), g(x, y) = 78125

(12π

x+y+12π − e−π(x+y)). One can easily verify that

f0 = 0, f∞ = 0, g0 = 0 and g∞ = 0. Choose constants l = 12 , r = 5, ζ = 12, γ = 2449. By computations

Iαφ(1) ≈ 0.015652, Iαl φ(1) ≈ 0.007547, Iβψ(1) ≈ 0.034126, Iβl ψ(1) ≈ 0.018676,

(1− θ1)ζIαφ(1) ≈ 0.326817 < ξ2γIαl φ(1) ≈ 0.959355

and (1 − θ2)ζIβψ(1) ≈ 0.184282 < ξ(1 − ξ)γIβl ψ(1) ≈ 8.046569. Also f(x, y) > 12245, for ∥(x, y)∥2 ∈(1.1397, 5) and g(x, y) > 12245, for ∥(x, y)∥2 ∈ (1.1397, 5). Therefore, for any

λ ∈ [0.0617, 1.2130] and µ ∈ (0, 1.8855] or λ ∈ (0, 1.2130] and µ ∈ [0.0431, 1.8855], by Theorem 5.2.7 the

boundary value problem (5.2.15), (5.2.16) has at least two positive solutions.

Chapter 6

Numerical solutions to fractional

differential equations by the Haar wavelets

Fractional differential equations is rapidly growing field of mathematics both in theory and in applica-

tions to real word problems. Numerous problems in physics, engineering and other applied sciences can

be modeled as differential equations of fractional order. However, in spite of a large number of recent

applications of fractional differential equations in applied problems, the state of art is far less developed

for boundary value problems. Exact analytic solutions to most of the differential equations of fractional

order are not available in general. Therefore various numerical schemes for integer order differential equa-

tions are generalized to approximate solutions of fractional differential equations (differential equations

of arbitrary order). Some of these are listed as Adomian decomposition method [42], homotype analy-

sis method [52], homotype perturbation method [1], finite difference method [149], variational iteration

method [140], fractional linear multi-step method [84], generalized differential transform method [109], ex-

trapolation method [36] and predictor-corrector method [38]. Most of these numerical schemes are applied

to deal with initial value problems and so far there has been no considerable advancement in extending

these methods for the boundary value problems of fractional order differential equations. In the literature,

one can find only few papers [9, 33, 60] that deal with numerical solutions of boundary value problems

for fractional ordinary and partial differential equations. Therefore, it appears to be very important to

develop efficient numerical techniques to solve the boundary value problems for differential equations of

fractional order. In this chapter and in chapter 7, we will focus on providing numerical techniques based

on wavelets for the solutions of boundary value problems for fractional differential equations.

6.1 Numerical solutions to fractional ordinary differential equations

In 1997 Chen and Hsiao [27] first proposed the idea of Haar operational matrix for the integration of Haar

function vectors and used it for solving differential equations. This remarkable idea deals with a general

formalism of construction of an operational matrix of integration for Haar wavelets. Since then numerous

operational matrices based on different orthogonal functions, such as the Legendre [10, 67, 117, 147], the

Chebyscev [16,133], the Fourier [112], sine–cosine [62,116] and block–pulse [102] have been established and

93

94

used to solve various problems in engineering. J. Wu, C. Chen and Chih-Fan Chen [141] have developed a

unified approach for deriving the operational matrices of several orthogonal functions including the square

wave group and sinusoidal group of orthogonal functions. In the following we review some operational

matrices of integration and derive a new operational matrix that will be used together existing operational

matrices to solve the boundary value problems for fractional differential equations. In the literature, as

for as we know, the boundary value problems are not solved by operational matrices approach even for

integer order differential equations.

Definition 6.1.1. [102] For l > 0, a set of block-pulse functions is defined on [0, l) as

bi(t) = χIi(t), where Ii :=[(i− 1)l

m,il

m

), i = 1, 2, . . . ,m.

The block-pulse functions are disjoint. That is, for t ∈ [0, 1) and i, j = 1, 2, . . . ,m, we have

bi(t)bj(t) = biδij .

The block-pulse functions are orthogonal among themselves. That is ⟨bi, bj⟩ = lmδij . The orthogonality

property of block-pulse function is obtained from the disjointness property. Any function f ∈ L2[0, l) can

be approximated by block-pulse series as follows

f(t) ≈m−1∑i=0

kibi(t) = kTmBm(t),

where the coefficients ki are given by ki = ml ⟨f, bi⟩. Furthermore, km = [k0, k1, . . . , km−1]

T , is the block-

pulse coefficient vector and Bm(t) = [b0(t), b1(t), . . . , bm−1(t)]T is the block-pulse function vector. For

ti =2i−1m , i = 1, 2, . . . ,m− 1, the block-pulse matrix is simply m×m identity matrix. By Integrating the

block-pulse function vector and expanding it into block-pulse series, we have∫ t

0Bm(s)ds ≈ Bm×mBm(t), (6.1.1)

where Bm×m is the operational matrix of integration for the block-pulse functions [102] and it is given by

Bm×m =1

m

12 1 1 · · · 1

0 12 1 · · · 1

0 0 12 · · · 1

......

.... . .

...0 0 0 · · · 1

2

.

Our aim is to develop a numerical scheme to solve two–point and multi–point boundary value problems

for fractional order differential equations. We derive a matrix and use it together with the existing Haar

wavelet operational matrices and the operational matrix of fractional order integration to solve boundary

value problems for fractional order differential equations. The work is accepted for publication in [94].

Usually the Haar scaling function φ and the mother wavelet ψ are defined on unit interval [0, 1). In

many situations, we need the Haar wavelet system defined on an interval [0, η), η > 0. In particular,

whenever we are interested to solve three–point (or multi–point) boundary value problems on [0, 1], we

95

choose η ∈ (0, 1]. Let I = I00 = [0, η) and Ij,k =[2−jkη, 2−j(k + 1)η

), then the Haar scaling and wavelet

functions on [0, η) are defined as follows:

φ(t) =1√ηχI(t), ψj,k =

2j/2√η

(χIlj,k

(t)− χIrj,k(t)), J ≥ 0, j ≤ J − 1, k ≤ 2j − 1,

where ψ0,0(t) = ψ1(t) = 1√η

(χ[0, η

2)(t) − χ[ η

2, η)(t)

)is the mother wavelet function for the Haar system

ψj,k = 2j/2ψ1(2jt− k. An arbitrary function y ∈ L2[0, η] can be expanded into the Haar wavelet series

as follows:

y(t) = ⟨y, φ⟩φ(t) +J−1∑j=0

2j−1∑k=0

⟨y, ψj,k⟩cj,kψj,k(t)

= cφ(t) +

J−1∑j=0

2j−1∑k=0

cj,kψj,k(t) = CTmΨm(t),

(6.1.2)

where m = 2J , for some fixed J ∈ N. The Haar coefficient vector Cm and the Haar wavelet vectors Ψm(t)

are given as

Cm =[c, c0,0, c1,0, c1,1, c2,0, c2,1, c2,2, c2,3, . . . , cJ−1,0, cJ−1,1, cJ−1,2, . . . , cJ−1,2J−1]T ,

Ψm(t) =[φ(t), ψ0,0(t), ψ1,0(t), ψ1,1(t), ψ2,0(t), ψ2,1(t), ψ2,2(t), ψ2,3(t), . . . ,

ψJ−1,0(t), ψJ−1,1(t), ψJ−1,2(t), . . . , ψJ−1,2J−1(t)]T .

For ti =(2i−1)η

2m , i = 1, 2, . . . ,m, the m×m Haar matrix is defined as

Ψm×m =[Ψm

( η

2m

)Ψm

( 3η

2m

)· · · Ψm

((2m− 1)η

2m

)]. (6.1.3)

For instance, when m = 8, the Haar matrix is given by

Ψ8×8 =1√η

1 1 1 1 1 1 1 1

1 1 1 1 −1 −1 −1 −1√2

√2 −

√2 −

√2 0 0 0 0

0 0 0 0√2

√2 −

√2 −

√2

2 −2 0 0 0 0 0 0

0 0 2 −2 0 0 0 0

0 0 0 0 2 −2 0 0

0 0 0 0 0 0 2 −2

.

The fractional order integral of the Haar wavelets can be approximated by the Haar wavelet series

Iαψj′,k′(t) =⟨Iαψj′,k′ , φ⟩φ(t) +J−1∑j=0

2j−1∑k=0

⟨Iαψj′,k′ , ψj,k⟩ψj,k(t).

Therefore,

IαΨm(t) = Pα,ηm×mΨm(t), (6.1.4)

where Pα,ηm×m = [pα,ηil ]m×m, pα,ηil = ⟨Iαψj′,k′ , ψj,k⟩, i, l ∈ N. The indices, i, l are determined as i = 2j

′+ k′,

l = 2j + k (j, j′ ≤ J, J − 1 ∈ N; k ≤ 2j − 1, k′ ≤ 2j′ − 1). In particular pα,ηi0 := ⟨Iαψj′,k′ , φ⟩ and

pα,η00 := ⟨Iαφ,φ⟩.

96

The Haar wavelets can be approximated by the block-pulse series as

ψj,k(t) =

m−1∑i=0

⟨ψj,k, bi⟩bi(t).

Consequently,

Ψm(t) = Ψm×mBm(t). (6.1.5)

Applying fractional integral of order α > 0 on both side of (6.1.5), we have

IαΨm(t) = Ψm×mIαBm(t)

= Ψm×mFα,ηm×mBm(t),

(6.1.6)

where Fα,ηm×m is the operational matrix of integration for the block-pulse functions [73] and is given by

Fα,ηm×m =( ηm

)α 1

Γ(α+ 2)

1 ζ1 ζ2 · · · ζm−1

0 1 ζ1 · · · ζm−2

0 0 1 · · · ζm−3

0 0 0. . .

...

0 0 0 · · · 1

,

with ζ0 = 1, ζj = (j + 1)α+1 − 2jα+1 + (j − 1)α+1 j = 1, 2, . . . ,m − i + 1 for i = 0, 1, 2, . . . ,m + 1.

Substituting (6.1.5), (6.1.8), into (6.1.4) we have

Pα,ηm×mΨm×m = Ψm×mFα,ηm×m,

Since Ψm×m is non-singular, therefore

Pα,ηm×m =

1

mΨm×mFα,ηm×mΨ

Tm×m, (6.1.7)

where Ψ−1m×m = 1

mΨTm×m. The matrix Pα,η

m×m is called the operational matrix of fractional order integration

for the Haar wavelets supported on [0, η]. In particular when η = 1 then Pα,ηm×m = Pα

m×m is operational

matrix of fractional order integration for the Haar wavelets supported on unit interval [80]. One can easily

verify that

Pα,ηm×m =

(1η

)1/2Pαm×m. (6.1.8)

For the special case, when α = 1, η = 1, the matrix Pα,ηm×m reduces to the well known operational matrix

Pm×m given in [27] and expressed by

Pm×m =1

2m

2mPm2×m

2−1

2Ψm2×m

2

12Ψ

−1m2×m

20

, (6.1.9)

with P1×1 =12 , Ψ

−1m×m = η

mΨTm×m.

Next, we derive a useful matrix for the Haar wavelets which is crucial for developing a numerical scheme

for solutions of boundary value problems to fractional ordinary and partial differential equations.

97

Consider functions Jn ∈ L2[0, η], n = 1, 2, . . . , l. Then for m-dimensional vector Jnm = [Jn1 , Jn2 , . . . , J

nm]

(Jni = Jn(2i−12m ), i = 1, 2, . . . ,m) and for any m×m matrix Am×m = [Ai,j ]m×m, define the product

Jnm ⊗Am×m = [Jnj Ai,j ]m×m.

In general Jnm⊗Am×m = Am×m⊗Jnm and also Jnm⊗(Am×mBm×m

)=(Jnm⊗Am×m

)Bm×m. By equation

(6.1.7), we have

Jnm ⊗(Pα,ηm×mΨm×m

)= Jnm ⊗

(Ψm×mF

α,ηm×m

). (6.1.10)

A function of two variables K(t, s) ∈ (L2[0, η))2 can be approximated by the Haar wavelets as

K(t, s) = ⟨φ(t), ⟨K(t, s), φ(s)⟩⟩φ(t)φ(s) +J−1∑j=0

2j−1∑k=0

⟨ψj,k(t), ⟨K(t, s), φ(s)⟩⟩ψj,k(t)φ(s)

+J−1∑j′=0

2j′−1∑

k′=0

⟨φ(t), ⟨K(t, s), ψj′,k′(s)⟩⟩φ(t)ψj′,k′(s)

+

J−1∑j=0

2j−1∑k=0

J−1∑j′=0

2j′−1∑

k′=0

⟨ψj,k(t), ⟨K(t, s), ψj′,k′(s)⟩⟩ψj,k(t)ψj′,k′(s)

= ΨTm×m(t)Km×mΨm×m(s).

Consider continuous functions ϕn : [0, η] → R, n = 1, 2, . . . , then

ϕn(t)Iαψj1,k1(η) =∫ η

0

(η − s)α−1

Γ(α)ϕn(t)ψj1,k1(s)ds =

∫ η

0Qnα(t, s)ψj1,k1(s)ds

=(⟨φ(t), ⟨Qnα(t, s), φ(s)⟩⟩φ(t) +

J−1∑j=0

2j−1∑k=0

⟨ψj,k(t), ⟨Qnα(t, s), φ(s)⟩⟩ψj,k(t))⟨φ(s), ψj1,k1(s)⟩

+

(J−1∑j′=0

2j′−1∑

k′=0

⟨φ(t), ⟨Qnα(t, s), ψj′,k′(s)⟩⟩φ(t)

+J−1∑j=0

2j−1∑k=0

J−1∑j′=0

2j′−1∑

k′=0

⟨ψj,k(t), ⟨Qnα(t, s), ψj′,k′(s)⟩⟩ψj,k(t)

)⟨ψj′,k′(s), ψj1,k1(s)⟩,

=(Cφ(t) +

J−1∑j=0

2j−1∑k=0

Cj,kψj,k(t))⟨φ(s), ψj1,k1(s)⟩

+

(J−1∑j′=0

2j′−1∑

k′=0

Cj′,k′φ(t) +

J−1∑j=0

2j−1∑k=0

J−1∑j′=0

2j′−1∑

k′=0

Cj,kj′,k′ψj,k(t)

)⟨ψj′,k′(s), ψj1,k1(s)⟩,

where Qnα(t, s) =(η−s)α−1

Γ(α) ϕn(t). Since the Haar wavelets are orthonormal on [0, η], i.e.

⟨ψj′,k′(s), ψj1,k1(s)⟩ = δj′,k1δk′,k1 .

98

Therefore,

ϕn(t)IαΨm(η) = ϕn(t)[Iαφ(t), Iαψ0,0(t), Iαψ1,0(t), Iαψ1,1(t), Iαψ2,0(t), Iαψ2,1(t),

Iαψ2,2(t), Iαψ2,3(t), . . . , IαψJ,0(t), IαψJ−1,1(t), IαψJ−1,2(t), . . . , IαψJ−1,2J−1(t)]T

=

Cφ(t) +∑J−1

j=0

∑2j−1j=0 Cj,kψj,k(t)

C0,0φ(t) +∑J−1

j=0

∑2j−1j=0 Cj,k0,0ψj,k(t)

C1,0φ(t) +∑J−1

j=0

∑2j−1j=0 Cj,k1,0ψj,k(t)

C1,1φ(t) +∑J−1

j=0

∑2j−1j=0 Cj,k1,1ψj,k(t)

...

CJ−1,0φ(t) +∑J−1

j=0

∑2j−1j=0 Cj,kJ−1,0ψj,k(t)

CJ−1,1φ(t) +∑J−1

j=0

∑2j−1j=0 Cj,kJ−1,1ψj,k(t)

...

CJ−1,2Jφ(t) +∑J−1

j=0

∑2j−1j=0 Cj,k

J−1,2Jψj,k(t)

.

Hence,

ϕn(t)IαΨm(η) = Qα,η,nm×mΨm(t). (6.1.11)

Since the Haar wavelets ψj,k, j = 0, 1, . . . , J , 0 ≤ k < 2j − 1 are supported on dyadic subintervals

Ij,k =[ηk2j, η(k+1)

2j

)of [0, η]. Therefore, we have,∫ η

0

(η − s)α−1

Γ(α)ψj,k(s)ds =

∫Ij,k

(η − s)α−1

Γ(α)ψj,k(s)ds

=

∫Ilj,k

(η − s)α−1

Γ(α)ψj,k(s)ds−

∫Irj,k

(η − s)α−1

Γ(α)ψj,k(s)ds,

where, I lj,k =[ηk2j, η(2k+1)

2j+1

)and Irj,k =

(η(2k+1)2j+1 , η(k+1)

2j

]are the left and right halves of the interval Ij,k.

Thus, we have

Λα,ηφ :=

∫ η

0

(η − s)α−1

Γ(α)φ(s)ds =

ηα

Γ(α+ 1)

and

Λα,ηj,k :=

∫ η

0

(η − s)α−1

Γ(α)ψj,k(s)ds =

ηα

Γ(α+ 1)

[(1− k

2j

)α+(1− k + 1

2j

)α− 2(1− 2k + 1

2j+1

)α].

Hence,

IαΨm(η) = [Λα,ηφ ,Λα,η0,0 ,Λα,η1,0 ,Λ

α,η1,1 ,Λ

α,η2,0 , . . . ,Λ

α,η2,3 , . . . ,Λ

α,ηJ−1,0, . . .Λ

α,ηJ−1,2J−1

]T

= [Λα,η0 , Λα,η1 , . . . ,Λα,ηm−1]T . (Λα,η0 := Λα,ηφ , Λα,ηi := Λα,ηj,k , i = 2j + k)

For a continuous function ϕn : [0, η] → R, we define a matrix at collocation points ti = 2i+12J+1 , i =

0, 1, . . . ,m− 1 as follows

Λα,nm×m =

ϕn(t0)Λ

α,η0 ϕn(t1)Λ

α,η0 . . . ϕn(tm−1)Λ

α,η0

ϕn(t0)Λα,η1 ϕn(t1)Λ

α,η1 . . . ϕn(tm−1)Λ

α,η1

......

. . ....

ϕn(t0)Λα,ηm−1 ϕn(t1)Λ

α,ηm−1 . . . ϕn(tm−1)Λ

α,ηm−1

. (6.1.12)

99

Therefore

ϕn(t)IαΨm(η) =η

mΛα,nm×mΨ

Tm×mΨm(t) (6.1.13)

From equation (6.1.11) and (6.1.13), we have

Qα,η,nm×m =

η

mΛα,nm×mΨ

Tm×m. (6.1.14)

In particular, when η = 1, the matrix Qα,η,nm×m will be denoted by Qα,n

m×m. It is worth mentioning that Qα,η,nm×m,

is very useful for solving multi–point boundary value problems for differential equations of arbitrary order.In particular, for m = 8, α = 1.5, η = 0.98 and ϕ1(t) = tα−1eπt, the matrix Qα,η,n

m×m is given by

Qα,η,nm×m =

0.00214 0.0217 0.0788 0.2103 0.4835 1.0174 2.0185 3.8400

0.00151 0.0152 0.0553 0.1478 0.3398 0.7149 1.4183 2.6981

0.00073 0.0074 0.0269 0.0719 0.1654 0.3481 0.6907 1.3141

0.00031 0.0032 0.0116 0.0310 0.0714 0.1503 0.2981 0.5672

0.00029 0.0029 0.0107 0.0286 0.0657 0.1383 0.2745 0.5223

0.00022 0.0022 0.0083 0.0222 0.0510 0.1074 0.2132 0.4056

0.00015 0.0015 0.0056 0.0151 0.0347 0.0731 0.1452 0.2762

0.00006 0.0006 0.0024 0.0065 0.0150 0.0315 0.0626 0.1192

.

Before applying the Haar wavelets to solve the boundary value problems for fractional differential

equations, we need to give a note of caution. The direct use of the Haar wavelets for the solutions of

fractional order differential equations my not suitable in some situations, for example when derivative

involved are of integer order. This is because the Haar wavelets are discontinuous. The obvious way to

overcome this difficulty is to convert the underlying differential equation into an integral equation and

approximating the solution by truncated orthogonal series and using operational matrices of integration

to eliminate the integral operators. An other possibility is that, one can expand the term involving highest

derivative, appearing in the underlying differential equation. The main characteristics of the method is to

convert the fractional order differential equation into system of algebraic equations.

6.1.1 Linear fractional differential equations with constant coefficients

In this section the numerical solutions for boundary value problems for linear fractional differential equa-

tions with constant coefficients are discussed. The linear problems can be treated in two ways.

Let us consider the following boundary value problem:

aDαy(t)+bDβy(t) + cy(t) = g(t), (6.1.15)

y(0) = 0, y(1) = y0, (6.1.16)

where t ∈ [0, 1], 1 ≤ α ≤ 2, 0 ≤ β ≤ 1, α ≥ β + 1 and a, b, c, y0 ∈ R.

Method.1: We approximate the fractional derivative and integral as follows

Dαy(t) = CTmΨm(t), (6.1.17)

IαΨm(t) = CTmPα

m×mΨm(t). (6.1.18)

100

Using properties of fractional derivatives and integrals, together with boundary conditions, from equation

(6.1.17), we have

y(t) = CTmIαΨm(t)− CT

mϕ1(t)IαΨm(1) + y0t

α−1, (6.1.19)

Dβy(t) = CTmIα−βΨm(t)− CT

mϕ2(t)IαΨm(1) +

Γ(α)y0tα−1

Γ(α− β)(6.1.20)

where ϕ1(t) = tα−1 and ϕ2(t) = Γ(α)tα−β−1

Γ(α−β) . In view of (6.1.11) and (6.1.18) the equations (6.1.19) and

(6.1.20) become

y(t) = CTmPα

m×mΨm(t)− CTmQα,1

m×mΨm(t) + y0tα−1, (6.1.21)

Dβy(t) = CTmPα−β

m×mΨm(t)− CTmQα,2

m×mΨm(t) +Γ(α)y0t

α−1

Γ(α− β), (6.1.22)

Substituting (6.1.17) (6.1.21) and (6.1.22) into (6.1.15), we have following linear system of algebraic

equations

CTm

aIm×m + b

(Pα−βm×m − Qα,2

m×m)+ c(Pαm×m − Qα,1

m×m)

Ψm(t) = FTmΨm(t), (6.1.23)

where we have approximated the function f(t;α, β) = g(t)−(

bΓ(α)Γ(α−β) + c

)y0t

α−1 as

f(t;α, β) = FTmΨm(t).

The Bagley–Torvik equation

The Bagley–Torvik equation [17] arises in the modeling of the motion of a rigid plate in a Newtonian fluid

and a gas in a fluid. Several authors have investigated this equation. In [114], I. Podlubny solved this

equation numerically with homogenous initial conditions and he compared the numerical solutions with

exact solutions obtained with the help of fractional Green’s function. S. Saha Ray and R.K. Bera, [122]

solved the Bagley–Torvik equation using the Adomian decomposition method. K. Diethelm and N.J.

Ford [37, 38] have applied fractional linear multistep method and predictor-corrector method of Adams

type to solve the Bagley–Torvik equation. Z.H. Wang, X. Wang [136] provided a general solution of the

Bagley–Torvik equation. In the present work, we solve this equation with two–point boundary conditions

by using Haar wavelets.

Example 6.1.2. Consider the Bagley–Torvik equation with boundary conditions,

y′′(t) +(1817

)Dαy(t) +

(1351

)y(t) =

t−12

89250√π

(48p(t) + 7

√tq(t)

), t ∈ [0, 1],

y(0) = 0, y(1) = 0,

(6.1.24)

where p(t) = 16000t4−32480t3+21280t2−4746t+189 and q(t) = 3250t5−9425t4+264880t3−448107t2+

233262t−34578. The equation (6.1.24) is a prototype fractional differential equation with two derivatives.

Let

y′′(t) = CTmΨm(t). (6.1.25)

101

0 0.2 0.4 0.6 0.8 1

0

0.002

0.004

0.006

0.008

0.01

0.012

Figure 6.1: m = 8 −−−, m = 16 −−−, m = 64 −−−, Exact Solution . . .Then, by Theorem 2.2.14 and boundary conditions in (6.1.24), the equation (6.1.25) reduces to

y(t) = CTmP2

m×mΨm(t)− CTmQ

2,1m×mΨm(t), (6.1.26)

where ϕ1(t) = t. Also,

Dαy(t) = CTmP2−α

m×mΨm(t)− CTmQ

2,2m×mΨm(t), (6.1.27)

where ϕ2(t) = t1−α

Γ(2−α) . Substituting equation (6.1.25), (6.1.26) and (6.1.27) in equation (6.1.24), we have

CTm

Im×m +

(1817

)(P2−αm×m −Q2,2

m×m)+(1351

)(Pαm×m −Q2,1

m×m)

Ψm(t) = FTmΨm(t), (6.1.28)

The function f(t) = t−12

89250√π

(48p(t) + 7

√tq(t)

)is approximated as f(t) = FTmΨm(t). The exact solution

for the boundary value problem (6.1.24) is y(t) = t5 − 29t4

10 + 76t3

25 − 339t2

250 + 27t125 . The numerical and exact

solutions are plotted in Figure 6.1. The absolute error for different values of m is shown in the Table 6.1.

Obviously, the absolute error decreases with increasing m.

t m = 8 m = 16 m = 32 m = 64 m = 128 m = 256

0.1 3.597(−4) 9.128(−5) 9.128(−5) 3.157(−5) 1.107(−5) 3.899(−6)

0.2 1.582(−3) 1.328(−4) 1.328(−4) 4.072(−5) 1.286(−5) 4.171(−6)

0.3 1.786(−3) 1.215(−4) 1.390(−4) 4.110(−5) 1.253(−5) 3.942(−6)

0.4 1.634(−3) 8.992(−4) 1.215(−4) 3.560(−5) 1.079(−5) 3.373(−6)

0.5 1.157(−3) 1.766(−5) 8.992(−5) 2.677(−5) 8.238(−6) 2.609(−6)

0.6 5.835(−4) 5.361(−5) 5.361(−5) 1.683(−5) 5.425(−6) 1.788(−6)

0.7 1.271(−4) 2.153(−5) 2.153(−5) 7.930(−6) 2.888(−6) 1.040(−6)

0.8 1.196(−4) 1.648(−6) 1.648(−6) 2.108(−6) 1.128(−6) 4.920(−7)

0.9 5.540(−4) 1.420(−6) 1.420(−6) 1.327(−6) 6.274(−7) 2.610(−7)

Table 6.1: Absolute error for different values of m.


Dαy(t) +Dβy(t) + (ωπ)2y(t) = −ωπ(sin(ωπt) + ωπ

), y(0) = 0, y(1) = −2. (6.1.29)

102

When α = 2, β = 1 and ω = 1, 3, 5, . . . , the boundary value problem (6.1.29) has exact solution y(t) =

cos(ωπ)− 1. Also, when α = 2 and β = 0, the exact solution of (6.1.29) is given by

y(t) = a(t cos√π2ω2 + 1− 1)− ωπ sinπωt+ bt sin

√π2ω2 + 1, (6.1.30)

where a = π2ω2

π2ω2+1and b = 1

sin√π2ω2+1

a(1 − cos

√π2ω2 + 1) + πω sinπω + y0

. The solutions of the

problem (6.1.29) for different values of α, β and ω are plotted in Figure 6.2. Solutions obtained for α = 2,

β = 0 and α = 2, β = 1 are in good agreement with the exact solutions. As α = 2 and β approaches 0 or 1

the solutions of fractional differential equation approach the solution of integer order ordinary differential

equations. The maximum absolute error is tabulated in Table 6.2. The error grows as ω increases. The

results in the Table 6.2 and Figures 6.2(d), 6.2(e) and 6.2(f) show that increasing ω, while not increasing

m produces a solution whose error is large. For large ω, the maximum absolute error can be reduced by

increasing m.

m = 32 m = 128 m = 512

ω β = 0 β = 1 β = 0 β = 1 β = 0 β = 1

0.5 2.966(−4) 1.25(−4) 1.85(−5) 7.85(−5) 1.15(−6) 4.91(−7)

1.5 0.00491 0.00380 3.08(−4) 2.38(−4) 1.93(−5) 1.49(−5)

2.0 0.02108 0.01918 0.00133 0.001209 8.33(−5) 7.56(−5)

3.5 0.05795 0.05552 0.00371 0.003543 2.32(−4) 2.22(−4)

Table 6.2: Maximum absolute error for α = 2, β = 0, 1 and different values of m, ω.

Method 2: The corresponding integral equation for equation (6.1.17), (6.1.18) is

y(t) = −1

a

(bIα−βy(t) + cIαy(t)

)+tα−1

a

(bIα−βy(1) + cIαy(1)

)+ f(t), (6.1.31)

where f(t) = 1a

(Iαg(t)− tα−1Iαg(1)

)+ y0t

α−1,

Set

y(t) = CTmΨm(t), (6.1.32)

then

Iαy(t) = CTmPα

m×mΨm(t), (6.1.33)

Iα−βy(t) = CTmPα−β

m×mΨm(t), (6.1.34)

ϕ1(t)Iαy(1) = CTmQ

α,1m×mΨm(t) (6.1.35)

ϕ1(t)Iα−βy(1) = CTmQ

α−β,1m×mΨm(t), (6.1.36)

where ϕ1(t) = tα−1. Inserting (6.1.32)- (6.1.36), in equation (6.1.31), we have following linear algebraic

system

CTm

Im×m +

( ba

)(Pα−βm×m −Qα−β,1

m×m

)+( ca

)(Pαm×m −Qα,1

m×m

)Ψm(t) = FTmΨm(t). (6.1.37)

103

0 0.2 0.4 0.6 0.8 1−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

t

y

β=1

β=1 (Exact)

β=0

β=0 (Exact)

β=0.2

β=0.4

β=0.6

β=0.8

(a) ω = 1, J = 6

0 0.2 0.4 0.6 0.8 10

0.5

1

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

β

t

y

(b) ω = 1, J = 6

0 0.2 0.4 0.6 0.8 1−2.5

−2

−1.5

−1

−0.5

0

0.5

t

y

β=1 (Exact)

β=1

β=0 (Exact)

β=0

β=0.2

β=0.4

β=0.6

β=0.8

(c) ω = 1.7, J = 6, α = 2

0 0.2 0.4 0.6 0.8 1−8

−6

−4

−2

0

2

4

6

t

y

β=1(Exact)

β=1

β=0 (Exact)

β=0

β=0.2

β=0.4

β=0.6

β=0.8

(d) ω = 2, J = 6, α = 2

0 0.2 0.4 0.6 0.8 1−15

−10

−5

0

5

10

t

y

β=1 (Exact)

β=1

β=0 (Exact)

β=0

β=0.2

β=0.4

β=0.6

β=0.8

(e) ω = 3, α = 2, J = 6

0 0.2 0.4 0.6 0.8 1−15

−10

−5

0

5

10

t

y

β=1 (Exact)

β=1

β=0 (Exact)

β=0

β=0.2

β=0.4

β=0.6

β=0.8

(f) α = 2, ω = 3, J = 8

0

0.5

1

00.20.40.60.81−30

−20

−10

0

10

20

30

tβ

y

(g) α = 2, ω = 3, J = 8, 0 ≤ β ≤ 1.

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

−4

−2

0

2

4

6

β

t

y

(h) α = 1.5, w = 1, j = 7, 0 ≤ β ≤ 1

Figure 6.2: Solutions y(t) of problem (6.1.29), for 1 ≤ α ≤ 2, y0 = cosωπ − 1.

104

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

Figure 6.3: Numerical solutions of problem (6.1.38)

(α = 1.2 —, α = 1.4 —, α = 1.6 —, α = 1.8 —, α = 2 —).

Example 6.1.4. Consider the boundary value problem for inhomogeneous linear fractional differential

equation

Dαy(t) +( 3

57

)y(t) = t+

3tα+1

57Γ(α+ 2), t ∈ [0, 1], y(0) = 0, y(1) =

1

Γ(α+ 2), (6.1.38)

where 1 < α ≤ 2, a ∈ R. The exact solution of boundary value problem is y(t) = tα+1

Γ(α+1) . The equivalent

integral equation for the boundary value problem (6.1.38) is given by

y(t) = − 3

57

Iαy(t)− tα−1Iαy(1)

+ f(t;α), (6.1.39)

where f(t;α) = Iαg(t)− tα−1Iαg(1)+ tα−1

Γ(α+2) . Inserting (6.1.32), (6.1.33) and (6.1.35) in equation (6.1.39)

we have following system of algebraic equations

CTm

Im×m +

3

57

(Pαm×m −Qα,1

m×m

)Ψm(t) = FTmΨm(t). (6.1.40)

where, f is approximated as f(t;α) = FTmΨm(t).

We solve (6.1.40) for Haar coefficient vector, for m = 32 and α = 1.2, 1.4, 1.6, 1.8, 2. The absolute error

is given in the Table 6.3. Also, numerical solutions are plotted in Figure 6.3.

Three–point boundary value problems

Usually the physical systems are modeled as initial (or terminal) value problems or, as two–point boundary

value problems. There are some applications in science and engineering where multi–point boundary value

problems arise. For example, multi–point boundary value problems appear in wave propagation and in

elastic stability. In the following we solve a three–point boundary value problem numerically using the

Haar wavelets.

Example 6.1.5. Consider boundary value problems for a class of fractional differential equations with

three–point boundary conditions,

Dαy(t) + ay(t) = g(t), 1 ≤ α < 2, y(0) = 0, y(1) = ξy(η), (6.1.41)

105

t α = 1.2 α = 1.4 α = 1.6 α = 1.8 α = 2,

0.1 1.53063(−6) 1.45713(−6) 1.56047(−7) 1.49112(−7) 3.50621(−8)

0.2 1.52699(−7) 6.98702(−9) 3.89064(−8) 8.90900(−7) 6.58227(−8)

0.3 8.07661(−7) 3.38714(−8) 5.73057(−7) 1.14669(−7) 8.79828(−8)

0.4 6.31371(−7) 3.13234(−7) 3.89575(−7) 1.86018(−7) 9.72476(−8)

0.5 5.19845(−7) 7.84210(−7) 3.64839(−7) 2.66526(−7) 8.93295(−8)

0.6 1.82879(−6) 1.58534(−6) 2.33905(−7) 6.05455(−7) 5.99495(−8)

0.7 2.43150(−6) 4.81312(−7) 3.65764(−8) 1.05672(−6) 4.84022(−9)

0.8 3.11752(−6) 7.98561(−7) 2.27054(−7) 1.85259(−7) 8.02523(−8)

0.9 3.96605(−6) 1.17157(−6) 5.58515(−7) 3.54806(−7) 1.99566(−7)

Table 6.3: Absolute error for m = 32 and α = 1.2, 1.4, 1.6, 1.8, 2.

0 0.2 0.4 0.6 0.8 1

-0.02

-0.01

0

0.01

0.02

0.03

0.04

Figure 6.4: Exact and Numerical solutions for problem (6.1.5)

(Exact solution —, Numerical solution . . . ).

where η ∈ (0, 1), a, ξ ∈ R and ∆ := 1− ξη = 0.

We transform the differential equation in (6.1.41), by incorporating the boundary conditions, to an

equivalent integral equation

y(t) = −aIαy(t)− atα−1

∆

(ξIαy(η)− Iαy(1)

)+ f(t;α), (6.1.42)

where f(t;α) := Iαg(t) + atα−1

∆

(ξIαg(η) − Iαg(1)

)may be expanded into the Haar series as f(t;α) =

FTmΨm(t). Let

y(t) = CTmΨm(t), (6.1.43)

then

Iαy(t) = CTmPα

m×mΨm(t), (6.1.44)

ϕ1(t)Iαy(η) = CTmQ

α,η,1m×mΨm(t), (6.1.45)

ϕ1(t)Iαy(1) = CTmQ

α,1m×mΨm(t), (6.1.46)

106

where ϕ1(t) = tα−1. Substituting (6.1.43)-(6.1.46) into equation (6.1.42), we obtain following system

CTm

Im×m + aPα

m×m +a

∆

(ξQα,η

m×m +Qαm×m

)Ψm(t) = FTmΨm(t). (6.1.47)

In particular, for α = 32 , a = e−3π

√π

, g(t) = e−3π

40√π

(t2(40t2 − 74t+ 33) + 4e3π

√t(128t2 − 148t+ 33)

), ξ =

−125196 , η = 2

5 and y(1) = − 140 the exact solution is y(t) = t2(t3 − 37t

20 + 3340). The numerical and exact

solutions for m = 32 are shown in Figure 6.4. The absolute error for different values of m is shown in the

Table 6.4

t m = 8 m = 16 m = 32 m = 64 m = 128, m = 256

0.1 6.973(−5) 2.871(−6) 4.426(−7) 1.076(−8) 1.737(−9) 2.417(−10)

0.2 4.592(−5) 7.082(−6) 1.786(−7) 2.729(−8) 8.407(−10) 3.595(−10)

0.3 4.592(−5) 7.082(−6) 1.800(−7) 2.734(−8) 4.338(−10) 8.063(−10)

0.4 1.133(−4) 2.869(−6) 4.422(−7) 1.089(−8) 3.148(−9) 6.640(−10)

0.5 1.373(−4) 8.583(−6) 5.358(−7) 3.289(−8) 2.368(−9) 6.698(−10)

0.6 1.133(−4) 2.870(−6) 4.418(−7) 9.466(−9) 1.439(−9) 1.415(−9)

0.7 5.800(−5) 7.080(−6) 1.782(−7) 2.677(−8) 5.683(−10) 7.945(−10)

0.8 6.679(−5) 7.079(−6) 1.781(−7) 2.629(−8) 7.103(−10) 8.230(−10)

0.9 6.275(−4) 4.172(−6) 4.408(−7) 1.045(−8) 1.469(−9) 5.191(−10)

Table 6.4: Absolute error for α = 32 and different values of m.

Fractionally damped mechanical oscillator

In [113] Attila Pálfalvi have applied the Adomian decomposition method on a fractionally damped mechan-

ical oscillator equation for a sine excitation. We solve this equation with two–point boundary conditions

using the Haar wavelets.

Example 6.1.6. Consider the fractionally damped mechanical oscillator equation with boundary condi-

tions

Dαy(t) + λDβy(t) + νy(t) = g(t), t ∈ [0, 1], y(0) = 0, y(1) = 0, (6.1.48)

where 1 < α ≤ 2, 0 < β ≤ 1, α − β > 1, λ, ν are prescribed constants and g(t) is the forcing function.

If α = 2, β = 1 then (6.1.48) reduces to the usual differential equation of harmonic oscillator. The

corresponding integral equation for equation (6.1.48) is

y(t) = −λIα−βy(t)− νIαy(t) + tα−1λIα−βy(1) + νIαy(1)

+ f(t;α), (6.1.49)

where f(t;α) = Iαg(t)− tα−1Iαg(1),Inserting (6.1.32)-(6.1.36) into equation (6.1.49), we have

CTm

Im×m + λ

(Pα−βm×m −Qα−β

m×m

)+ ν(Pαm×m −Qα

m×m

)Ψm(t) = FTmΨm(t). (6.1.50)

107

0 0.2 0.4 0.6 0.8 1-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

Figure 6.5: Exact and Numerical solutions for problem (6.1.48)


In particular, for α = 74 , β = 1

2 and λ = 1, ν = − 1√π

and g(t) = 1√π

(16t

32 p(t)

45045 + 24t14 q(t)

9945Γ( 54)− t2(5t− 3)2r(t)

),

where p(t), q(t) and r(t) are polynomials given by

p(t) = 28028000t3 − 14620320t2 − 21527571t− 270270,

q(t) = 6400000t5 − 15360000t4 + 13328000t3 − 5021120t2 + 757809t− 29835,

r(t) = 25t3 − 50t2 + 29t− 4.

One can easily verify that the exact solution is y(t) = 625t7 − 2000t6 + 2450t5 − 1420t4 + 381t3 − 36t2.

For m = 64 we solve the system (6.1.50) for the vector Cm. The numerical results for the approximate

solution for the problem (6.1.48) are shown in Figure 6.5. The absolute error for different values of m is

shown in the Table 6.5.

t m = 8 m = 16 m = 32 m = 64 m = 128, m = 256

0.1 0.040484 1.018(−3) 2.271(−4) 8.777(−6) 2.823(−6) 6.499(−7)

0.2 0.017310 1.815(−4) 7.112(−5) 1.361(−5) 3.352(−6) 6.356(−7)

0.3 0.010302 1.455(−3) 9.529(−5) 1.446(−5) 2.577(−6) 3.715(−7)

0.4 0.003190 5.008(−4) 4.983(−5) 2.315(−6) 2.387(−6) 9.482(−7)

0.5 0.026157 7.389(−4) 2.966(−5) 1.007(−5) 5.020(−6) 1.595(−6)

0.6 0.039802 5.940(−5) 2.196(−5) 1.663(−6) 2.822(−6) 1.054(−6)

0.7 0.020226 1.014(−4) 5.453(−5) 4.072(−6) 7.854(−7) 6.346(−7)

0.8 0.018284 3.448(−3) 8.472(−5) 1.099(−7) 5.572(−6) 1.886(−6)

0.9 0.295680 4.654(−3) 5.406(−4) 7.195(−6) 8.658(−6) 3.139(−6)

Table 6.5: Absolute error for different values of m.

108

6.1.2 Linear fractional differential equations with variable coefficients

Consider the following boundary value problem with variable coefficients

Dαy(t) +A(t)Dβy(t) +B(t)y(t) = g(t), (6.1.51)

y(0) = 0, y(1) = y0, 1 < α ≤ 2, 0 < β ≤ 1 (6.1.52)

Approximating the fractional derivative and integral as follows

Dαy(t) = CTmΨm(t), (6.1.53)

IαΨm(t) = CTmPα

m×mΨm(t). (6.1.54)

Using properties of fractional derivatives and integrals, together with boundary conditions, from equation

(6.1.53), we have

y(t) = CTm

(IαΨm(t)− tα−1IαΨm(1)

)+ y0t

α−1, (6.1.55)

Dβy(t) = CTm

(Iα−βΨm(t)−

Γ(α)tα−β−1

Γ(α− β)IαΨm(1)

)+

Γ(α)y0tα−1

Γ(α− β). (6.1.56)

Let ϕ1(t) = tα−1, ϕ2(t) = B(t)ϕ1(t), ϕ3(t) = A(t)Γ(α)tα−β−1

Γ(α−β) , then from (6.1.55) and (6.1.56) we have

y(t) = CTm

(IαΨm(t)− ϕ1(t)IαΨm(1)

)+ y0t

α−1, (6.1.57)

B(t)y(t) = CTm

(B(t)IαΨm(t)− ϕ2(t)IαΨm(1)

)+ y0B(t)tα−1, (6.1.58)

A(t)Dβy(t) = CTm

(A(t)Iα−βΨm(t)− ϕ3(t)IαΨm(1)

)+

Γ(α)y0A(t)tα−1

Γ(α− β). (6.1.59)

Substituting (6.1.4) and (6.1.11) in (6.1.57)-(6.1.59), we obtain

y(t) = CTm

(Pαm×mΨm(t)−Qα,1

m×mΨm(t))+ y0t

α−1, (6.1.60)

B(t)y(t) = CTm

(B(t)Pα

m×mΨm(t)−Qα,2m×mΨm(t)

)+ y0B(t)tα−1, (6.1.61)

A(t)Dβy(t) = CTm

(A(t)Pα−β

m×mΨm(t)−Qα,3m×mΨm(t)

)+

Γ(α)y0A(t)tα−1

Γ(α− β). (6.1.62)

Inserting (6.1.60)-(6.1.62) into equation (6.1.51) we have following system of algebraic equations

CTm

Im×m +B(t)Pα

m×m +A(t)Pα−βm×m −Qα,2

m×m −Qα,3m×m

Ψm(t) = FTmΨm(t), (6.1.63)

where the function f(t;α, β) = g(t)− Γ(α)Γ(α−β)y0B(t)tα−1 − y0A(t)t

α−1 is approximated by FTmΨm(t).

Let ti = 2i−12m (i = 1, 2, . . . ,m), then from (6.1.63) we have

CTm

Im×m +BT

m ⊗(Pαm×mΨm×m

)+AT

m ⊗(Pα−βm×mΨm×m

)−(Qα,2m×m +Qα,3

m×m)Ψm×m

. = FTmΨm×m,

(6.1.64)

Using (6.1.10) in (6.1.64), we get

CTm

Im×m +BT

m ⊗(Ψm×mFαm×m

)+AT

m ⊗(Ψm×mFα−βm×m

)−(Qα,2m×m +Qα,3

m×m)Ψm×m

= FTmΨm×m.

(6.1.65)

109

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

t

α=2.00

.

α=2.25

.

α=2.50

.

α=2.75

.

α=3.00

.

(a)

00.2

0.40.6

0.81

22.5

33.5

4−0.1

0

0.1

0.2

0.3

t

α

y

(b)

Figure 6.6: Numerical (solid circles) and exact (solid lines) solutions for problem (6.1.66), (6.1.67).

Example 6.1.7. Consider the boundary value problem for fractional differential equation with variable

coefficients

Dαy(t) + e−9πty(t) = e−9πt(tα−1 − tα

)− Γ(α+ 1), 2 ≤ α < 4, (6.1.66)

y(0) = 0, y′(0) = 0, y′′(0) = 0, y(1) = 0. (6.1.67)

Let B(t) = e−9πt, g(t;α) = e−9πt(tα−1 − tα

)− Γ(α + 1), ϕ1(t) = tα−1 and ϕ2(t) = B(t)tα−1. We

approximate the fractional derivative (of order α) of y(t) by the Haar wavelets as follows

Dαy(t) = CTmΨm(t). (6.1.68)

Together with boundary conditions, we have

y(t) = CTm

(Pαm −Qα,1

m×m)Ψm(t), (6.1.69)

B(t)y(t) = CTm

(B(t)Pα

m −Qα,2m×m

)Ψm(t). (6.1.70)

Substituting (6.1.68) and (6.1.70) into (6.1.66), we have

CTm

Im×m +B(t)Pα

m×m −Qα,1m×m

Ψm(t) = GT

mΨm(t), (6.1.71)

where we have used the approximation g(t;α) = GTmΨm(t). Let ti = 2i−1

2m (i = 1, 2, . . . ,m), then

(6.1.71) reduces to system of algebraic equations

CTm

(Im×m −Qα,1

m×m)Ψm×m +BT

m ⊗(Ψm×mFαm×m

= GT

mΨm×m. (6.1.72)

The numerical solution of the boundary value problem (6.1.66), (6.1.67) is obtained form (6.1.69) by

solving the linear system of algebraic equations (6.1.72) for the vector Cm. One can verify that the exact

solution of the boundary value problem (6.1.66), (6.1.67) is y(t) =(1t − 1

)tα. The numerical solutions for

m = 8 and different values of α are shown in Figure 6.6. The error analysis is given in Table 6.6.

110

t m = 8 m = 16 m = 32 m = 64 m = 128, m = 256

0.1 3.666(−4) 1.452(−4) 3.442(−5) 8.709(−6) 2.176(−6) 5.437(−7)

0.2 9.604(−4) 2.328(−4) 5.858(−5) 1.464(−5) 3.659(−6) 9.139(−7)

0.3 1.280(−3) 3.175(−4) 7.938(−5) 1.984(−5) 4.959(−6) 1.237(−6)

0.4 1.570(−3) 3.940(−4) 9.850(−5) 2.462(−5) 6.152(−6) 1.535(−6)

0.5 1.868(−3) 4.659(−4) 1.164(−4) 2.910(−5) 7.271(−6) 1.813(−6)

0.6 2.135(−3) 5.340(−4) 1.335(−4) 3.336(−5) 8.335(−6) 2.077(−6)

0.7 2.398(−3) 5.995(−4) 1.498(−4) 3.745(−5) 9.355(−6) 2.331(−6)

0.8 2.652(−3) 6.626(−4) 1.656(−4) 4.139(−5) 1.033(−5) 2.574(−6)

0.9 2.882(−3) 7.238(−4) 1.809(−4) 4.522(−5) 1.129(−5) 2.810(−6)

Table 6.6: Absolute error for different values of m and α = 114 .

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

Figure 6.7: Exact and Numerical solutions for problem (6.1.73), (6.1.74)


Example 6.1.8. Consider the fractional boundary value problem

Dαy(t) + sin(t)Dβy(t) + e−ty(t) =e−t

2tα(tβ − 1)(2t− 1) + q1(t;α, β) + q2(t;α, β), (6.1.73)

y(0) = 0, y(1) = 0, (6.1.74)

where, 1 < α ≤ 2, 0 < β ≤ 1, q1(t;α, β) = Γ(α + 1)(12 − t(1 + α)

)− Γ(α+β+1)tβ

Γ(β+1)

(12 − (α+β+1)t

β+1

)and

q2(t;α, β) = sin(t)tα−βΓ(α+1)Γ(α−β+1)

(12 − (α+1)t

α−β+1

)− tαΓ(α+β+1)

Γ(α+1)

(12 − (α+β+1)t

α+1

). Furthermore y0 = 0, ϕ1(t) =

tα−1, ϕ2(t) = sin(t)tα−1 and ϕ3(t) = e−t Γ(α)tα−β−1

Γ(α−β) . Also, f(t;α, β) = g(t;α, β) = e−t

2 tα(tβ − 1)(2t− 1) +

q1(t;α, β)+q2(t;α, β). The exact solution is y(t) = 12 tα(tβ−1)(2t−1). Solving the corresponding system of

algebraic equations (6.1.65) we obtain the numerical solution at points ti =(2i−1)2m (i = 1, 2, . . . ,m). Exact

and numerical solutions are plotted in Figures 6.8(a)-6.8(b). The maximum absolute error for different

values of α and β is displayed in Table 6.7.

Example 6.1.9. Now we consider following boundary value problem for the Caputo fractional order

111

β α = 1.3 α = 1.6 α = 1.8 α = 2.0

0.2 3.53(−4) 1.58(−4) 6.51(−5) 3.76(−5)

0.4 5.41(−4) 2.25(−4) 8.74(−5) 6.39(−5)

0.6 6.06(−4) 2.44(−4) 8.92(−5) 8.62(−5)

0.8 6.01(−4) 2.41(−4) 9.94(−5) 1.04(−4)

1.0 5.69(−4) 2.30(−4) 9.54(−5) 1.14(−4)

Table 6.7: The maximum absolute error for m = 32, and different values of α and β.

differential equations with variable coefficients

cDα0+y(t) +A(t)cDβ

0+y(t) +B(t)y(t) = g(t), t ∈ [0, 1], (6.1.75)

y(0) = y0 y(1) = y1, 1 < α ≤ 2, 0 < β ≤ 1. (6.1.76)

Taking ϕ1(t) = t, ϕ2(t) = A(t)ϕ1(t), ϕ3(t) = B(t)Γ(α)t−β

Γ(2−β) . Using the same method as used for bound-

ary value problem (6.1.51) involving the Riemann–Liouville fractional derivative, we arrive at system of

algebraic equations (6.1.65). In particular, for the choice α = 2, β = 1, A(t) = − 2t1+t2

, B(t) = 21+t2

and g(t) = (1 + t2)et, the exact solution of the boundary value problem (6.1.75), (6.1.76) is y(t) =

y1t+ (1− t)2((1− t)et − (1− y0)(t+ 1)

). The exact and numerical solutions for different values of α, β

are plotted in Figures 6.8(c)-6.8(f).

A nonlinear fractional differential equation

Since the method based on Haar wavelets converts the underlying nonlinear problem to a system of

nonlinear algebraic equations. A Nonlinear system having large dimension is itself difficult to solve due

to increased computational complexity. However, for some simple cases one can apply wavelet method to

solve the boundary value problems numerically.

Consider the following class of boundary value problem for nonlinear fractional order differential equa-

tion

Dαy(t) + a(y(t))n = g(t), (6.1.77)

y(0) = 0, y(1) = y0, (6.1.78)

where 1 < α ≤ 2, y0 ∈ R, n ∈ N and g(t) is a given function. The properties of fractional derivatives

and integrals allow us to reduce the the boundary value problem (6.1.77), (6.1.78) into following integral

equation

y(t) = −aIα(y(t))n − atα−1Iα(y(1))n + f(t;α), (6.1.79)

where,

f(t;α) = Iαg(t)− Iαg(1)tα−1 + y0tα−1. (6.1.80)

112

0 0.2 0.4 0.6 0.8 1−0.025

−0.02

−0.015

−0.01

−0.005

0

0.005

0.01

0.015

0.02

t

y

(a) α = 1.5, β = 0.5, m = 32.

00.2

0.40.6

0.81

0

0.5

1

−0.04

−0.02

0

0.02

t

β

y

(b) α = 2, 0 ≤ β ≤ 1.

0 0.2 0.4 0.6 0.8 10.9

1

1.1

1.2

1.3

1.4

1.5

t

y(t)

ExactNumerical

(c) α = 2, β = 1, m = 64

0 0.2 0.4 0.6 0.8 1 00.2

0.40.6

0.81

0.9

1

1.1

1.2

1.3

1.4

1.5

β

y(t;

β)

t

(d) α = 2, 0 < β ≤ 1, y0 = 1.5, y1 = 1.

0 0.2 0.4 0.6 0.8 1 1

1.5

20.2

0.4

0.6

0.8

1

1.2

1.4

1.6

α

y(t;α

)

t

(e) 1 < α ≤ 2, β = 0.5, y0 = 1.5, y1 = 1

00.2

0.40.6

0.81 1

1.2

1.4

1.6

1.8

2

1

1.5

2

2.5

y(t;α

)

αt

(f) 1 < α ≤ 2, β = 0.5, y0 = 2, y1 = 2.

Figure 6.8: Numerical and exact solutions for the boundary value problems (6.1.73), (6.1.74) and (6.1.75),

(6.1.76) ((c)-(f)).

113

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

y

ExactNumerical

Figure 6.9: Exact and Numerical solutions for problem (6.1.85).

We approximate y(t) as

y(t) = CTmΨm(t)

= CTmΨm×mBm(t),

(6.1.81)

Let CTmΨm×m = [d0, d1, . . . , dm−1], then using properties of block-pulse function, we have

(y(t))2 =([d0, d1, . . . , dm−1]Bm(t)

)2= d20b0(t) + d21b1(t) + · · ·+ d2m−1bm−1(t)

By induction, we have

(y(t))n = dn0b0(t) + dn1 b1(t) + · · ·+ dnm−1bm−1(t)

= [dn0 , dn1 , . . . , d

nm−1]Bm(t)

= DTmBm(t)

where Dm = [dn0 , dn1 , . . . , d

nm−1]

T . Now

Iα(y(t))n = DTmIαBm(t)

= DTmFαm×mBm(t).

(6.1.82)

Let φ(t) ∈ L2[0, 1] be a given function, then

ν(t)Iα(y(1))n =

∫ 1

0

ν(t)(1− s)α−1

Γ(α)(y(s))nds

=

∫ 1

0kα(s, t)(y(s))

nds

= DTm

∫ 1

0Bm(s)Ψ

T (s)Kαm×mΨm(t)ds

= DTm

(∫ 1

0Bm(s)BT

m(s)ds

)Ψm×mK

αm×mΨm(t)

=1

mDTmΨ

Tm×mK

αm×mΨm×mBm(t).

(6.1.83)

114

Defining ν := atα−1 and substituting (6.1.81)-(6.1.83) into equation (6.1.79), we have following system of

nonlinear algebraic equations.

CTmΨm(t) = −aDT

mFαm×mBm(t) +1

mDTmΨm×mK

αm×mΨm×mBm(t) + FTmΨm(t), (6.1.84)

where the function f(t) is approximated by f(t) = FTmΨm(t). This nonlinear system can be solved using

Matlab built in function fsolve.

Example 6.1.10. Consider the following boundary value problem for nonlinear fractional order differential

equation

Dαy(t) + e−5π(y(t))2 =Γ(α+ 3)

2t2 + e−5πt2α+4, y(0) = 0, y(1) = 1, (6.1.85)

where 1 < α ≤ 2. The exact solution is y(t) = tα+2. We compare the numerical solution with exact

solution for m = 32 and different values of α and the results are given below in the Table 6.8.

t α = 1.2 α = 1.5 α = 1.7 α = 1.9 α = 2

0.1 0.0056 0.0028 0.0017 0.0011 0.0009

0.3 0.0495 0.0345 0.0271 0.0213 0.0188

0.5 0.1088 0.0884 0.0769 0.0669 0.0625

0.7 0.1368 0.1229 0.1145 0.1066 0.1028

0.9 0.0793 0.0768 0.0752 0.0736 0.0728

Table 6.8: For problem (6.1.85), absolute error for m = 32 and different values of α.

6.2 Numerical solutions to fractional partial differential equations

In this section we apply the numerical scheme developed in previous section to linear fractional partial

differential equations. Our focus of interest are the two classes of boundary value problems for fractional

partial differential equations. Throughout this section, the fractional partial derivative will be considered

in the Caputo sense.

6.2.1 Fractional partial differential equations with constant coefficients

This subsection considers following class of linear fractional partial differential equations with variable

coefficients

∂αu(x, t)

∂tα+ λ

∂βu(x, t)

∂tβ+ µu(x, t) = η

∂γu(x, t)

∂xγ+ f(t, x), 0 < α ≤ 2, 0 ≤ β ≤ 1, 1 ≤ γ ≤ 2, (6.2.1)

satisfying non–homogenous initial and boundary conditions conditions

u(x, 0) = ρ(x),∂u(x, t)

∂t|t=0 = σ(x), u(0, t) = ξ(t), u(1, t) = ζ(t). (6.2.2)

115

In particular, for 1 < α ≤ 2, λ, µ, η > 0, the equation (6.2.1) reduces to the fractional telegraph

equation. It also includes heat, wave and the Poisson equations as special cases. Many researchers have

recently studied the equation using various numerical techniques. In [35] S. Das et al. have employed the

homotopy analysis method for numerical solutions of the equation satisfying initial conditions u(x, t)|t = 0,∂u(x,t)∂t |t=0 = 0. Some other contributions have been mad by authors, such as S. Momani [106], F. Huang

[55], R. C. Cascaval et al. [26], R. F. Camargo et al. [23], J. Chen et al. [30], and E. Orsingher, L.

Beghin [111]. We use the Haar wavelet method developed in Section 6.1 to provide numerical solutions.

Approximating ∂αu(x,t)∂tα by two dimensional Haar wavelet series as

∂αu(x, t)

∂tα= ΨT

m(x)Km×mΨm(t). (6.2.3)

Operating on both sides of (6.2.3) by fractional integral Iαt in variable t, and using Lemma 2.2.24 we have

u(x, t) = ΨTm(x)Km×m

(∫ t

0

(t− s)β

Γ(α)Ψm(s)ds

)+ p(x)t+ q(x). (6.2.4)

Using the initial conditions u(x, 0) = ρ(x), ∂u(x,t)∂t |t=0 = σ(x), from (6.2.4) we have q(x) = ρ(x) and

p(x) = σ(x). Thus, the equation (6.2.4) takes the form

u(x, t) = ΨTm(x)Km×mP

αm×mΨm(t) + σ(x)t+ ρ(x). (6.2.5)

Applying ∂β

∂tβon both sides of (6.2.5) and using Lemma 2.2.19, we have

∂βu(x, t)

∂tβ= ΨT

m(x)Km×mPα−βm×mΨm(t) + σ(x)

t1−β

Γ(2− β). (6.2.6)

Substituting (6.2.3), (6.2.5) and (6.2.6) in (6.2.1), we get

η∂γu(x, t)

∂xγ=ΨT

m(x)Km×mΨm(t) + λΨTm(x)Km×mP

α−βm×mΨm(t)

+ µΨTm(x)Km×mP

αm×mΨm(t) + g(x, t)

=ΨTm(x)

Km×m(I + λPα−β

m×m + µPαm×m) +Gm×m

Ψm(t),

(6.2.7)

where

g(x, t) = σ(x)( λt1−β

Γ(2− β)+ µt

)+ µρ(x)− f(x, t) = ΨT

m(x)Gm×mΨm(t).

Applying Iγx on both sides of (6.2.7), we have

ηu(x, t) = IγxΨTm(x)



Ψm(t) + xφ1(t) + φ2(t). (6.2.8)

Now, applying the boundary conditions u(0, t) = ξ(t), we have φ2(t) = ξ(t), and u(1, t) = ζ(t) implies

φ1(t) = −IγxΨTm(1)



Ψm(t) + ζ(t)− ξ(t). (6.2.9)

Substituting (6.2.9) in (6.2.8) we have

ηu(x, t) =ΨTm(x)

((Pγ

m×m)T − (Qγ

m×m)T)

Km×m(I + λPα−βm×m + µPα

m×m)

+Gm×mΨm(t) + x(ζ(t)− ξ(t)) + ξ(t).

(6.2.10)

116

0 0.2 0.4 0.6 0.8 10

0.5

1−1

−0.5

0

0.5

1

t

x

u

HaarExact

Figure 6.10: Numerical and exact solutions for telegraph equation (6.2.12).

where, IγxΨm(x) = Pγm×mΨm(x) = ΨT

m(x)(Pγm×m)

T and xIγxΨTm(1) = Qγ

m×mΨm(x). From (6.2.5) and

(6.2.10) we get Sylvester matrix equation

((Pγm×m)

T − (Qγm×m)

T )Km×m(I + λPα−βm×m + µPα

m×m)− ηKm×mPαm×m

= Sm×m − ((Pγm×m)

T − (Qγm×m)

T )Gm×m,(6.2.11)

where, s(x, t) := x(ζ(t)− ξ(t)) + ξ(t)− η(σ(x)t+ ϱ(t)) = ΨTm(x)Sm×mΨm(t). Solving (6.2.11) for Km×m

and using (6.2.5) or (6.2.10) we can get the solution of the problem (6.2.1).

Example 6.2.1. We consider the time–fractional telegraph equation

∂αu(x, t)

∂tα+∂α−1u(x, t)

∂tα−1+ u(x, t) =

∂2u(x, t)

∂x2+

(Γ(2α+ 1)

Γ(α+ 1)(1 +

t

α+ 1)− 49tα

)t2α cos (7x), (6.2.12)

satisfying initial and boundary conditions u(x, 0) = 0, ∂u(x,t)∂t |t=0 = 0, u(0, t) = t2α, u(1, t) = 0.7539022t2α.

One can easily verify that the exact solution for the problem is u(x, t) = t2α cos (7x). The numerical so-

lutions are obtained for the points (xi, ti) = (2i−12n , 2i−1

2n ), i = 1, 2, . . . , n, n = 2J . Computer plots for

numerical and exact solutions for J = 4, α = 1.3 are shown in Figure 6.10.

Example 6.2.2. Consider the time–fractional heat equation

∂αu(x, t)

∂tα= λ

∂2u(x, t)

∂x2, 0 < α ≤ 1, (6.2.13)

satisfying the initial condition, u(x, 0) = xα(1− x) and the boundary conditions u(0, t) = 0, u(1, t) = 0.

For integer case, i.e., α = 1 and λ = 0.2 the problem is solved numerically in [79]. The numerical

solutions for fractional case, α = 0.15, 0.35, 0.50, 1.0 are obtained by the method discussed above. The

numerical solutions are plotted in Figure 6.11a. It is observed that the solutions for the fractional cases

are approaching to the solution of classical heat equation.

117

00.2

0.40.6

0.81

00.2

0.40.6

0.81

0

0.05

0.1

0.15

0.2

0.25

tx

u

α=0.15

α=0.35

α=0.50

α=1.00

(a) Numerical solutions of heat equation for J = 5, α = 0.15, α =

0.35, α = 0.50 and α = 1.00.

0

0.5

1

00.20.40.60.81

−0.04

−0.02

0

0.02

0.04

t

x

u(x,

t)

HaarExact

(b) Numerical and exact solutions ( J = 5.0,α = 0.3).

Figure 6.11: Solutions for (6.2.13) and (6.2.15)

Example 6.2.3. Consider the fractional partial differential equation

∂αu(x, t)

∂xα+ λ

∂2u(x, t)

∂t2= xα−1et, 1 ≤ α < 2, x, t ∈ [0, 1], (6.2.14)

with boundary conditions u(0, t) = 0, u(1, t) = et, u(x, 0) = xα−1, u(x, 1) = exα−1.

For α = 2, the equation reduces to Poisson equation having exact solution u(x, t) = xey. Numerical

solutions obtained for α = 1.2, α = 2 are plotted in Figure 6.12.

0 0.2 0.4 0.6 0.8 1 0

0.5

1−1

−0.5

0

0.5

1

1.5

2

2.5

3

xt

u

α=2.0

α=1.2

Figure 6.12: Numerical and exact solutions for the equation (6.2.14) for α = 0.3, J = 5.0

118

Example 6.2.4. Consider partial fractional differential equation

∂αu(x, t)

∂tα−λ∂

2u(x, t)

∂x2=

(t1−α

Γ(2− α)− Γ(3α+ 1)t2α

Γ(2α+ 1)

)+144λt(1−t3α−1) sin(12x), 0 ≤ α ≤ 1, (6.2.15)

satisfying initial and boundary conditions u(x, 0) = 0, u(0, t) = 0, u(1, t) = −0.536573t(1 − t3α−1). The

exact solution of the problem is u(x, t) = t(1−t3α−1) sin(12x). In order to show the accuracy and simplicity

of the method, we compare numerical solutions for different values of α and J by computing the maximum

absolute differences between solutions at collocation points. The numerical results are shown in Table 6.9.

Also, for α = 0.3 and J = 5, the plots for solutions are shown in Figure 6.11b.

J α = 0.1 α = 0.3 α = 0.5 α = 0.7 α = 0.9 α = 1.0

4 5.2008(−4) 7.2442(−5) 3.5490(−4) 7.6899(−4) 1.1825(−3) 1.3946(−3)

5 5.2278(−5) 7.3221(−6) 3.6103(−5) 7.8885(−5) 1.2262(−4) 1.4573(−4)

6 5.4590(−6) 7.6600(−7) 3.7860(−6) 8.3071(−6) 1.2998(−5) 1.5500(−5)

7 6.0733(−7) 8.5326(−8) 4.2210(−7) 9.2925(−7) 1.4640(−6) 1.7403(−6)

8 7.0972(−8) 9.9905(−9) 4.9472(−8) 1.0927(−7) 1.7236(−7) 2.0439(−7)

Table 6.9: Maximum absolute error for different values of J and α.

6.2.2 Fractional partial differential equations with variable coefficients

This subsection concerns with the numerical solutions for the for the class of fractional partial differential

equations∂γu(x, t)

∂tγ− a(x)

∂αu(x, t)

∂xα+ b(x)

∂βu(x, t)

∂xβ+ d(x)u(x, t) = f(x, t), (6.2.16)

( 0 < γ ≤ 2, 1 < α ≤ 2, 0 < β ≤ 1) with initial and boundary conditions,

(i) u(x, 0) = ϕ1(x),∂u(x, t)

∂t|t=0 = ψ1(x), or (ii) u(x, 0) = ϕ1(x), u(x, 1) = ψ2(x), (6.2.17)

and

u(0, t) = µ(t), u(1, t) = ν(t), (6.2.18)

For a(x) > 0, b(x) > 0, d(x) ≥ 0 and 0 < α < 1, 1 < β < 2, the problem reduces to fractional convection–

diffusion equations. In [151] Y. Zhang have used stable difference scheme for the numerical solutions of the

convection–diffusion equation supplemented with initial condition u(x, 0) = 0 and boundary conditions

u(0) = u(1) = 0. For a, b > 0, d = 0 and f(x, t) = 0, the problem reduces to space–time fractional

advection–dispersion equation. Many authors have discussed numerical methods for convection–diffusion,

including [31, 40, 118, 130, 138]. Motivated by the work mentioned above, we give a numerical algorithm

based on operational matrices of integration and the matrix derived in Section 6.1. We approximate∂αu(x,t)∂xα by the Haar wavelets as

∂αu(x, t)

∂xα= ΨT

m(x)Km×mΨm(t). (6.2.19)

119

Applying the fractional integral operator Iαx on both sides of (6.2.19), we have

u(x, t) = IαxΨTm(x)Km×mΨm(t) + p(t)x+ q(t). (6.2.20)

Using the boundary condition (6.2.18), from (6.2.20), we have

q(t) = µ(t), p(t) = −IαxΨTm(1)Km×mΨm(t) + ν(t)− µ(t).

Equation (6.2.20) takes the form

u(x, t) = IαxΨTm(x)Km×mΨm(t)− xIαxΨ

Tm(1)Km×mΨm(t) + x(ν(t)− µ(t)) + µ(t), (6.2.21)

Since IαxΨm(x) = Pαm×mΨm(x) and ϕ1(x)IαxΨm(1) = Qα,1

m×mΨm(x), where ϕ1(x) = x. Therefore, (6.2.21)

takes the form

u(x, t) =ΨTm(x)(P

αm×m)

TKm×mΨm(t)−ΨTm(x)(Q

α,1m×m)

TKm×mΨm(t)

+ x(ν(t)− µ(t)) + µ(t).(6.2.22)

Applying the caputo operator ∂β

∂xβon (6.2.21), using the Lemma 2.2.19 and Lemma 2.2.23, we have

∂βu(x, t)

∂xβ= Iα−βx ΨT

m(x)Km×mΨm(t)−x1−β

Γ(2− β)IαΨT

m(1)Km×mΨm(t) +(ν(t)− µ(t))x1−β

Γ(2− β). (6.2.23)

For simplicity, we introduced some convenient notations.

ϕ2(x) =b(x)x1−β

Γ(2− β), ϕ3(x) = xd(x), r(x, t) =

(ν(t)− µ(t))b(x)

Γ(2− β)x1−β + xd(x)(ν(t)− µ(t)) + d(x)q(t),

s(x, t) = x(ν(t)− µ(t)) + µ(t) + ψ1(x)t+ ϕ1(x),

g(x, t) = −x(ν(t)− µ(t))− µ(t) + (ψ2(x)− ϕ1(x))t+ ϕ1(x)

d(x)IαxΨm(x) = Pαm×mΨm(x), b(x)IαxΨm(x) = Pα

m×mΨm(x).

Substituting (6.2.19), (6.2.21) and (6.2.23) in (6.2.16), we have

∂γu(x, t)

∂tγ=(a(x)ΨT

m(x)−ΨTm(x)(P

α−βm×m)

T +ΨTm(x)(Q

α,2m×m)

T −ΨTm(x)(P

αm×m)

T

+ΨTm(x)(Q

α,3m×m)

T)Km×mΨm(t) +ΨT

m(x)Rm×mΨm(t)

Applying Iγt on both sides of above equation

u(x, t) =(a(x)ΨT

m(x)−ΨTm(x)(P

α−βm×m)

T +ΨTm(x)(Q

α,2m×m)

T −ΨTm(x)(P

αm×m)

T

+ΨTm(x)(Q

α,3m×m)

T)Km×mΨm(t) +ΨT

m(x)Rm×mIγt Ψm(t) + w(x)t+ ω(x).

By initial conditions (6.2.17)(i) we have w(x) = ϕ1(x) and v(x) = ψ1(x). Therefore

u(x, t) =[(a(x)ΨT

m(x)−ΨTm(x)(P

α−βm×m)

T +ΨTm(x)(Q

α,2m×m)

T −ΨTm(x)(P

αm×m)

T

+ΨTm(x)(Q

α,3m×m)

T)Km×m +ΨT

m(x)Rm×m

]Pγm×mΨm(t) + ψ1(x)t+ ϕ1(x).

(6.2.24)

120

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

t

x

u(x,

t)

Haar

Exact

(a) Exact and numerical solutions for J = 5.

00.2

0.40.6

0.81

0

0.5

10

1

2

x 10−4

tx

Err

or

(b) For J = 6, the error between exact and numerical solutions.

Figure 6.13: The absolute error between Haar wavelet solution and analytic solution.

Now, taking into account the boundary conditions (6.2.17)(ii), we have w(x) = ϕ1(x) and

υ(x) =[(a(x)ΨT

m(x)−ΨTm(x)(P

α−βm×m)

T +ΨTm(x)(Q

α,2m×m)

T −ΨTm(x)(P

αm×m)

T

+ΨTm(x)(Q

α,3m×m)

T)Km×m +ΨT

m(x)Rm×m

]Iγt Ψm(1) + (ψ2(x)− ϕ1(x).

(6.2.25)

Therefore, (6.2.24) becomes

u(x, t) =[(a(x)ΨT

m(x)−ΨTm(x)(P

α−βm×m)

T +ΨTm(x)(Q

α,2m×m)

T

−ΨTm(x)(P

αm×m)

T +ΨTm(x)(Q

α,3m×m)

T)Km×m +ΨT

m(x)Rm×m

](Pγm×m −Qγ

m×m

)Ψm(t) + (ψ2(x)− ϕ1(x))t+ ϕ1(x).

(6.2.26)

From 6.2.22 and (6.2.24) we have the Sylvester equation

((Pαm×m)

T − (Qα,1m×m)

T )Km×m −( ηmΨm×mAm×mΨ

Tm×m

− (Pα−βm×m)

T − (Pαm×m)

T + (Qα,2m×m)

T + (Qα,3m×m)

T)Km×mP

γm×m = Rm×mP

γm×m + Sm×m,

(6.2.27)

where Am×m := diag[a(xi)], xi = 2i−12m , i = 1, 2, . . . ,m. Also, from (6.2.22) and (6.2.26), we get following

matrix equation

((Pαm×m)T − (Qα,1

m×m)T )Km×m −( ηmΨm×mAm×mΨT

m×m − (Pα−βm×m)T − (Pα

m×m)T + (Qα,2m×m)T

+ (Qα,3m×m)T

)Km×m

(Pγ

m×m −Qγm×m

)= Rm×m

(Pγ

m×m −Qγm×m

)+Gm×m.

(6.2.28)

Solving equations (6.2.32) and (6.2.28), for Gm×m and then substituting in (6.2.22), we get the approxi-

mate solution of the problem (6.2.16).

Now, we implement the Haar wavelet method to solve different types of fractional partial differential

equations. Also, we compare the results obtained with exact solutions and solutions obtained in literature

by other methods.

121

Example 6.2.5. Consider the partial differential equation

∂αu(x, t)

∂tα=

1

2x2∂2u(x, t)

∂x2, 1 < α ≤ 2, (6.2.29)

subject to initial and boundary conditions

u(x, 0) = x,∂u(x, t)

∂t|t=0 = 0, (6.2.30)

u(0, t) = 0, u(1, t) = 1 + tEα,2(tα). (6.2.31)

The series solution of the problem is [105] u(x, t) = x+x2∑∞

k=0tkα+1

Γ(kα+2) . The numerical solutions by Haar

wavelets for J = 7 and different values of α are shown in Table 6.10. Also we compare the results with

the solutions obtained in reference [105] using Adomian decomposition method and variational iteration

method. Solutions by the Haar wavelets agree well with the solutions by ADM and VIM.

α = 1.5 α = 1.75

t x uHarr uADM uV IM uHarr uADM uV IM

0.2 0.25 0.26284409 0.26284061 0.26269693 0.26267320 0.26266989 0.26248505

0.50 0.55136710 0.55136246 0.55078773 0.55068355 0.55067959 0.55059402

0.75 0.86557129 0.86556550 0.86427239 0.86403415 0.86402909 0.86383654

0.4 0.25 0.27697847 0.27697113 0.27642739 0.27616359 0.27567202 0.27567202

0.50 0.60792985 0.60788455 0.60570958 0.60463521 0.60268808 0.60268808

0.75 0.99272600 0.99274024 0.98784655 0.98542677 0.98104818 0.98104818

0.6 0.25 0.29310668 0.29309481 0.29198616 0.29110091 0.28979084 0.28979084

0.50 0.66272310 0.67237923 0.66794464 0.66437353 0.65916338 0.65916339

0.75 1.14235387 1.13785532 1.12787544 1.11793023 1.10811762 1.10811764

Table 6.10: The Haar wavelet solutions and solutions obtained in [105], using ADM and VIM.

Example 6.2.6. Consider the fractional diffusion equation

∂u(x, t)

∂t= a(x)

∂1.8u(x, t)

∂x1.8+ f(x, t), (6.2.32)

with initial conditions u(x, 0) = x2(1 − x), and Dirichlet boundary Condition u(0, t) = 0, u(1, t) = 0.

In particular, for a(x) = Γ(1.2)x1.8, f(x, t) = 3x2(2x− 1)e−t, the problem has been studied for numerical

solutions in [66] by a method based upon Chebyshev approximations. It can be easily verified that the

exact solution is u(x, t) = x2(1 − x)e−t. The exact and numerical solutions by the Haar wavelets are

plotted in Figure 6.13(a). Figure 6.13(b) show the maximum absolute error between exact and numerical

solutions.

Example 6.2.7. Finally, we consider the fractional convection–diffusion equation

∂γu(x, t)

∂tγ= −a(x)∂

αu(x, t)

∂xα+ b(x)

∂βu(x, t)

∂xβ+ f(t, x), (6.2.33)

122

where 0 < γ ≤ 2, 1 < α ≤ 2, 0 < β ≤ 1 and, with boundary conditions u(x, 0) = u(x, 1) = 0, u(0, t) =

u(1, t) = 0. We solve this problem for

a(x) =Γ(2 + β)Γ(5− α− β)xβ , b(x) = Γ(2β − α+ 2)Γ(5− 2α)xα,

f(x, t) =(2π)x2β+1 − x4−α)t1−γE2,2−γ(−(2πt)2) +Γ(2β + 2)(Γ(5− α− β)

− Γ(5− 2α))x2β+1 + Γ(5− α)(Γ(2β − α+ 2)− Γ(β + 2))x4−αsin(2πt).

The exact solution of the problem is u(x, t) = (x2β+1 − x4−α) sin(2πt). Numerical solutions are obtained

for different values of α, β and γ. The results are shown in Figure 6.14. The absolute error computed

between exact and numerical solutions by the Haar wavelets, for J = 7 and different values of α, β, γ is

shown in figure Figure 6.15.

123

00.2

0.40.6

0.81

0

0.5

1−0.06

−0.04

−0.02

0

0.02

0.04

0.06

t

x

u(x,

t)

HaarExact

(a) Exact and numerical solutions for J = 5, α = 2.0, β = 0.35,

γ = 0.50.

00.2

0.40.6

0.81

0

0.5

1−0.1

−0.05

0

0.05

0.1

tx

u(x,

t)

HaarExact

(b) Error for J = 5, α = 2.0, β = 0.35, γ = 0.50.

00.2

0.40.6

0.81

0

0.5

1−0.2

−0.1

0

0.1

0.2

t

x

u(x,

t)

HaarExact

(c) For J = 5, α = 2.0, β = 0.75, γ = 0.50.

Figure 6.14: Exact and numerical solutions for different values of J , α, β and γ.

124

00.2

0.40.6

0.810

0.5

10

2

4

6

8

x 10−4

xt

Err

or

(a) For J = 5, α = 2.0, β = 0.75, γ = 0.50.

00.2

0.40.6

0.810

0.20.4

0.60.8

10

0.2

0.4

0.6

0.8

1

x 10−3

x

t

Err

or

(b) For J = 5, α = 0.75, β = 0.50, γ = 0.75.

00.2

0.40.6

0.810

0.20.4

0.60.8

1

0

2

4

6

8

x 10−3

x

t

Err

or

(c) Error for J = 7, α = 0.75, β = 0.5, γ = 0.75.

Figure 6.15: The absolute error between exact and numerical solutions for different values of α, β and γ.

Chapter 7

Numerical solutions to fractional

differential equations by the Legendre

wavelets

Approximation by orthogonal families of functions have been playing a vital role in the development of

physical sciences engineering and technology. These functions have received considerable attention in

the evaluation of new numerical techniques to solve various problems of dynamic system and optimal

controls. The most commonly used classes of orthogonal functions include the sets of piecewise constant

basis functions such as block-pulse functions and the Walsh functions; the sets of orthogonal polynomials,

for example the Laguerre, the Chebshev, and the widely used sine-cosine functions in the Fourier series.

The application of the Legendre wavelets for solving differential and integral equations is thoroughly

considered in [10, 67, 117, 147] and the references therein. The main characteristics of this technique is

that it converts the underlying problems into equivalent algebraic systems. The objective of the present

chapter is to introduce a Legendre wavelet operational matrix of fractional order integration and use it to

solve different types of fractional differential equations. The derivation of the Legendre wavelet operational

matrix of integration is similar to the derivation of the Haar wavelet operational matrix of integration [80]

and CAS operational matrix of integration [120, 121]. The Legendre wavelet method efficiently works

for initial value and boundary value problems for fractional order differential equations. The illustrative

examples are provided to demonstrate the applicability of the numerical scheme based on the Legendre

wavelet operational matrix of integration. Throughout in this chapter, we consider the derivative in the

Caputo sense.

7.0.3 The Legendre wavelets

The Legendre polynomials denoted by Pm(t), m ∈ N, are defined on [−1, 1] and can be determined by

recurrence formulae:

P0 = 1, P1 = t, Pm+1(t) =2m+ 1

m+ 1tPm(t)−

m

m+ 1Pm−1(t), m = 1, 2, 3, . . . .

125

126

0 0.2 0.4 0.6 0.8 1−3

−2

−1

0

1

2

3

4

5

6

t

L1,0

L1,1

L1,2

L2,0

L2,1

L2,2

Figure 7.1: The Legendre wavelets for M = 3, k = 2.

The Legendre wavelets are defined on interval [0, 1) as [10]

Ln,m(t) =

2k2√2m+ 1Pm(2

kt− n), n−12k

≤ t < n+12k,

0, elsewhere,

where k = 2, 3, . . . , n = 2n − 1, n = 1, 2, 3, . . . , 2k−1 and, for some fixed positive integer M , m =

0, 1, 2, . . . ,M − 1 is the order of the Legendre polynomial.

7.0.4 Function approximations by the Legendre wavelets

Any function g(t) ∈ L2[0, 1) can be expanded into Legendre wavelet series as [10]:

g(t) =

∞∑n=1

∞∑m=0

cn,mLn,m(t), (7.0.1)

where cn,m = ⟨f(t), Ln,m(t)⟩. When working with the infinite series representation (7.0.1) in practice, we

can only deal with finite number of terms.

g(t) =

2k−1∑n=1

M−1∑m=0

cn,mLn,m(t), (7.0.2)

where Cm and Lm are m× 1 (m = 2k−1M) matrices, given by

Cm =[c1,0, c1,1, . . . , c1,M−1, c2,0, c2,1, . . . , c2,M−1, . . . , c2k−1,0, c2k−1,1, . . . , c2k−1,m−1]T ,

Lm(t) =[L1,0(t), L1,1(t), . . . , L1,M−1(t), L2,0(t), L2,1(t), . . . ,

L2,M−1(t), . . . , L2k−1,0(t), L2k−1,1(t), . . . , L2k−1,m−1(t)]T .

The convergence of the Legendre wavelet series (7.0.2) is established in [82]. A function of two variables

k(s, t) ∈ L2([0, 1)× [0, 1)) can be expanded into Legendre wavelet series as

k(s, t) =

2k−1∑n1=1

M−1∑m1=0

2k−1∑n2=1

M−1∑m2=0

⟨Ln1,m1(s), ⟨k(s, t), Ln2,m2(t)⟩⟩Ln1,m1(s)Ln2,m2(t). (7.0.3)

127

In order to simplify notations, we define i =M(n−1)+m+1, where n = 1, 2, 3, . . . , 2k−1, (k = 2, 3, . . . , n =

2n− 1, m = 0, 1, 2, . . . ,M − 1). Thus, we can write (7.0.2) as

g(t) ≈m∑i=1

ciLi(t) = CTmLm(t), (7.0.4)

where, Cm = [c1, c2, · · · , cm]T , Lm(t) = [L1(t), L2(t), · · · , Lm(t)]T . Similarly, (7.0.3) can be represented as

k(s, t) =

m∑i=1

m∑j=1

kijLi(s)Lj(t) = LTm(s)Km×mLm(t), (7.0.5)

where Km×m = [kij ]m×m and kij = ⟨Li(s), ⟨k(s, t), Lj(t)⟩⟩. The number of needed terms increases fast

with the desired accuracy. Therefore, these type of calculations are performed with computers.

7.1 An operational matrices of fractional order integration

The Legendre function vector Lm(t) can be represented in terms of block-pulse functions as

Lm(t) = Lm×mBm(t), (7.1.1)

We recall the fractional integral of block-pulse function vector

(IαBm)(t) = Fαm×m Bm(t), (7.1.2)

where Fαm×m is the block-pulse operational matrix of the fractional order integration [73].

On the other hand, since (IαLm)(t) ∈ L2[0, 1], there fore we have following Legendre wavelet series

(IαLm)(t) = Rαm×mLm(t), (7.1.3)

where Rαm×m is the Legendre wavelet operational matrix of fractional order integration. Substituting

(7.1.1), (7.1.2) into (7.1.3), we have

Rαm×m = Lm×mFαm×mL

−1m×m.

In particular, let α = 0.75, M = 3, k = 2, the operational matrix of fractional order integration Rαm×m

is given by

R0.756×6 =

0.3741 0.5006 0.1738 −0.0468 −0.0082 0.0066

0 0.3741 0 0.1738 0 −0.0082

−0.1873 0.0392 0.0721 −0.0196 0.0609 0.0046

0 −0.1873 0 0.0721 0 0.0609

0.1660 0.2773 −0.0004 −0.0323 0.0404 0.0057

0 0.1660 0 −0.0004 0 0.0404

.

Remark 7.1.1. The Legendre wavelet operational matrix can be obtained by computing inner products

directly. But this approach leads to higher computational complexity. Based on above arguments, we

have developed an efficient Matlab program, given in Appendix A, which can be utilized to compute the

Legendre operational matrix of fractional order integration easily.

128

0 0.2 0.4 0.6 0.8 1 1

1.5

2

0.2

0.4

0.6

0.8

1

1.2

α

t

y

(a)

0 0.2 0.4 0.6 0.8 1

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

t

y

α=1.0

α=1.0 (Exact)

α=1.2

α=1.4

α=1.6

α=1.6

α=2.0

α=2.0 (Exact)

(b)

Figure 7.2: Numerical and exact solutions of problem (7.2.1), (7.2.2) for 1 ≤ α ≤ 2.

7.2 Numerical solutions of fractional differential equations

The general formulism of the Legendre wavelet method is much similar to the Haar wavelet method for

solutions of different types of fractional differential equations discussed in previous chapter. Since there

is no general formulism for the comparison of the Legendre wavelet method to other existing methods

that can be applied to solve fractional differential equations. Therefore we will compare the method with

the Haar wavelet method and some other methods by considering some specific examples for which exact

solutions are available.

Example 7.2.1. Consider the fractional differential equation

cDαy(t) + y(t) = 0, 0 < α ≤ 2, (7.2.1)

satisfying

y(0) = 1, y′(0) = 0. (7.2.2)

This problem have been studied by P. Kumar and O.P. Agrawal [76], K. Diethelm, N.J. Ford and A.D.

Freed [38]. The second initial condition is taken into account only if 1 < α ≤ 2. Using the Laplace

transform, the exact solution of the problem (7.2.1), (7.2.2) is given by

y(t) = Ea(−tα), (7.2.3)

where Ea(t) is the Mittag–Leffler function of order α. The corresponding integral representation for equa-

tions (7.2.1), (7.2.2) is

y(t) = −Iαy(t) + y(0) + y′(0)t = −Iαy(t) + 1. (7.2.4)

Now, approximating the solution y(t) in term of the Legendre wavelets as

y(t) = CTmLm(t). (7.2.5)

Applying the fractional integral operator Iα on both sides of (7.2.5), we have

Iαy(t) = CTmIαLm(t),

= CTmRα

m×mLm(t).(7.2.6)

129

Inserting (7.2.5) and (7.2.6) into equation (7.2.4) we have following system of algebraic equations.

CTm

(Im×m + Rα

m×m)Lm(t) = [1, 0, 0, . . . , 0]T . (7.2.7)

Solving the linear system (7.2.7), for the vector Cm, and substituting it in (7.2.5) we get approximate

solution of the initial value problem (7.2.1), (7.2.2). First, we consider the case when α is an integer i.e.,

α = 1,or α = 2. For the case when α = 1, we observe that the exact solution is y(t) = e−t and for

α = 2 we have y = cos(t). The numerical and exact solutions plotted in the Figure 7.2 indicate that, as

α approaches to 1 or 2, the numerical solutions converge to the exact solution. The absolute error for

k = 3, 4, . . . , 8, α = 1.5 is shown in Table 7.1. It can be observed that the error decreases with increasing

k.

t k = 3 k = 4 k = 5 k = 6 k = 7 k = 8

0.1 5.094(−4) 1.051(−4) 2.698(−5) 6.731(−6) 1.682(−6) 4.207(−7)

0.2 1.915(−4) 4.990(−5) 1.244(−5) 3.109(−6) 7.776(−7) 1.944(−7)

0.3 5.720(−5) 1.455(−5) 3.649(−6) 9.124(−7) 2.282(−7) 5.705(−8)

0.4 4.650(−5) 1.178(−5) 2.948(−6) 7.368(−7) 1.842(−7) 4.605(−8)

0.5 1.314(−4) 3.282(−5) 8.205(−5) 2.051(−6) 5.128(−7) 1.282(−7)

0.6 1.987(−4) 4.974(−5) 1.244(−5) 3.111(−6) 7.777(−7) 1.944(−7)

0.7 2.525(−4) 6.324(−5) 1.581(−5) 3.954(−6) 9.886(−7) 2.471(−7)

0.8 2.942(−4) 7.365(−5) 1.842(−5) 4.605(−6) 1.151(−6) 2.878(−7)

0.9 3.235(−4) 8.126(−5) 2.032(−5) 5.081(−6) 1.270(−6) 3.176(−7)

Table 7.1: Absolute error for M = 3 and different values of k.

Example 7.2.2. Consider the problem [46],

cDαy(t) + ωα−βcDβy(t) = 0, y(0) = A, y′(0) = B, (7.2.8)

where α ∈ (1, 2], β ∈ (0, 1]. β = 0 the equations (7.2.8) describes fractional oscillator and has been studied

in details by B.N. Narahari Achar et.al. in [107].

The corresponding integral equation for fractional order differential equation (7.2.8) is

y(t) = −ωα−βIα−βy(t) + (1− ωα−β)A+Bt. (7.2.9)

We approximate y as

y = CTmLm(t). (7.2.10)

Applying Iα on both sides of (7.2.10), we have

IαcDβy(t) = Iα−βy(t)

= CTmIα−βLm(t)

= CTmRα−β

m×mLm(t).

(7.2.11)

130

Substituting (7.2.10) and (7.2.11) into equation (7.2.9) we have system of algebraic equations.

CTm = FTm

(Im×m + ωα−βCT

mRα−βm×m

)−1, (7.2.12)

where f(t;α, β) := (1− ωα−β)A+Bt = FTmLm(t). Thus the numerical solution of the problem (7.2.8) is

y(t) = FTm(Im×m + ωα−βCT

mRα−βm×m

)−1Lm(t). (7.2.13)

For the classical case α = 2, β = 0 with the constants A = 0 and B = 1, the exact solution is

y(t) = 1ω sin(ωt). Also, for α = 2, β = 1 (A = 0, B = 1), the exact solution is y(t) = − 1

ω (eωt − 1).

Numerical results for different values of α and β are shown in Figure 7.3. From Figure 7.3, we see that,

as α approaches to 1 or 2, the numerical solution converges to that of differential equations of order 1 or

2 respectively.

Example 7.2.3. Consider following linear non–homogenous boundary value problem with constant coef-

ficients

acDαy(t) + bcDβy(t) + cy(t) = g(t), y(0) = A, y′(0) = B, (7.2.14)

where α = 2, β ∈ (1, 2), a = 0 and b, c ∈ R. For α = 2, β = 3/2, equation (7.2.14) reduces to the

Bargely–Torvik equation studied in [17] for modeling of a rigid plate immersed in a Newtonian fluid.

We approximate cDαy(t) ascDαy(t) = CT

mLm(t). (7.2.15)

Then

cDβy(t) = Iα−β(cDαy)(t)

= CTmIα−βLm(t)

= CTmRα−β

m×mLm(t).

(7.2.16)

Applying Iα on both sides of (7.2.16) and using Lemma 2.2.24, we have

y(t) = CTmRα−β

m×mLm(t) +A+Bt. (7.2.17)

Substituting equation (7.2.15), (7.2.16) and (7.2.17) into equation (7.2.14), we have following system of

algebraic equations.

aCTmLm(t) + bCT

mRα−βm×mLm(t) + cCT

mRα−βm×mLm(t) = FTmLm(t), (7.2.18)

where the function f(t) := g(t)− c(A+Bt) = FTmLm(t). We choose α = 2, β = 32 , a = 12, b = 8

17 , c =59 ,

g(t) = 72t + 6417

√πt32 + 5

9 t3 and A = B = 0, one can verify that the exact solution of the boundary value

problem (7.2.14) is y = t3. We compare the numerical solution with exact solution for different values of

k and the results are given in the Table 7.2. For the comparison of the Legendre wavelet method with

some other numerical methods, we choose α = 2, β = 1/2, a = b = c = 1 and g(t) = 8. The numerical

solutions of this problem are discussed in Ref. [12] and [104] by some other numerical methods namely,

the Adomian decomposition method, variational iteration method, fractional finite difference method,

131

0

0.2

0.4

0.6

0.8

1 00.2

0.40.6

0.81

−0.1

0

0.1

0.2

0.3

0.4

βt

y

(a) α = 1.3, 0 ≤ β ≤ 1

0

0.5

1 00.2

0.40.6

0.81

−0.1

0

0.1

0.2

0.3

βt

y

(b) α = 1.6, 0 ≤ β ≤ 1

0

0.2

0.4

0.6

0.8

1 0 0.2 0.4 0.6 0.8 1

−0.1

−0.05

0

0.05

0.1

βt

y

(c) α = 2, 0 ≤ β ≤ 1

0

0.5

1 1

1.5

2

−0.1

−0.05

0

0.05

0.1

αt

y

(d) 1 ≤ α ≤ 2, β = 0

0

0.5

1 11.2

1.41.6

1.82

−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

αt

y

(e) 1 ≤ α ≤ 2, β = 1.5

0

0.5

1 1 1.2 1.4 1.6 1.8 2

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

αt

y

(f) 1 ≤ α ≤ 2, β = 1

Figure 7.3: Solutions y(t) for Example 7.2.2 for ω = 11, y0 = 0, y1 = 1 and different values of α, β.

132

t k = 4 k = 5 k = 6 k = 7 k = 8, k = 9

0.1 1.293(−4) 3.229(−5) 8.070(−6) 2.016(−6) 5.041(−7) 1.260(−7)

0.2 2.574(−4) 6.433(−5) 1.607(−5) 4.017(−6) 1.004(−6) 2.510(−7)

0.3 3.849(−4) 9.617(−5) 2.403(−5) 6.006(−6) 1.501(−6) 3.753(−7)

0.4 5.117(−4) 1.278(−4) 3.195(−5) 7.984(−6) 1.996(−6) 4.988(−7)

0.5 6.378(−4) 1.593(−4) 3.982(−5) 9.952(−6) 2.487(−6) 6.218(−7)

0.6 7.632(−4) 1.907(−4) 4.765(−5) 1.191(−5) 2.976(−6) 7.441(−7)

0.7 8.879(−4) 2.218(−4) 5.544(−5) 1.385(−5) 3.463(−6) 8.656(−7)

0.8 1.012(−3) 2.528(−4) 6.317(−5) 1.579(−5) 3.946(−6) 9.865(−7)

0.9 1.135(−3) 2.836(−4) 7.087(−5) 1.771(−5) 4.427(−6) 1.107(−6)

Table 7.2: Absolute error for M = 3 and different values of k.

fractional differential transform method. The numerical results in table 7.3 indicate that, at least for this

particular problem the Legendre wavelet method is in good agreement with above mentioned numerical

methods. It is worth mentioning that some of the numerical methods, such as Adomian decomposition

method require stronger differentiability assumptions on y(t). On the other hand the Haar wavelet and

the Legendre wavelet methods require y to be an element of L2(R). How ever the operational matrices

approach has some draw backs too. For example, it becomes cumbersome to deal the nonlinear problems

by this approach, since the method reduces nonlinear boundary value problems to nonlinear system of

algebraic equations. To solve these nonlinear algebraic systems, we have to look for some iterative methods,

such as Newton’s method, which requires some appropriate initial guess. Also, since we need to solve

large algebraic systems which may cause grater computational complexity. However for well–posed linear

problems the Haar and the Legendre wavelet methods work effectively. Also the Haar and the Legendre

wavelet methods are much simpler then other methods.

Example 7.2.4. In example (7.2.5), we take α = 2, 0 ≤ β ≤ 1, a = b = c = 1, A = 0, B = 0 and

g(t;α, β) = 6(

t−α

Γ(4−α) −t−β

Γ(4−β)

)t3. The exact solution in this case is y(t) = t3. The numerical results

obtained for the Haar wavelet and the Legendre wavelets are shown in Table 7.4. The Legendre wavelet

method seems to giving almost the same results as the Haar wavelet method. However the calculations

with the Legendre wavelets are much complicated as compared with the Haar wavelets. This causes

relatively greater computational complexity and large storage requirements.

Example 7.2.5. Consider the following boundary value problem for nonlinear fractional order differential

equation

cDαy(t) + a(y(t))n = g(t), y(0) = A, y(1) = B, (7.2.19)

where < α ≤ 2, A,B ∈ R, n ∈ N and g(t) is a given function. Using properties of fractional derivatives

and integrals the differential equation (7.2.19) can be reduced into following integral equation

y(t) = −aIα(y(t))n + atIα(y(1))n + f(t), (7.2.20)

133

t yFDM [12] yADM [104] yFDTM [12] yVIM [104] yLWM yexact

0.0 0.000000 0.000000 0.000000 0.000000 0.000000 00000000

0.1 0.039473 0.039874 0.039750 0.039874 0.039750 0.039750

0.2 0.157703 0.158512 0.157036 0.158512 0.157035 0.157036

0.3 0.352402 0.353625 0.347370 0.353625 0.347370 0.347370

0.4 0.620435 0.622083 0.604695 0.622083 0.604695 0.604695

0.5 0.957963 0.960047 0.921768 0.960047 0.921767 0.921768

0.6 1.360551 1.363093 1.290457 1.363093 1.290456 1.290457

0.7 1.823267 1.826257 1.702008 1.826257 1.702007 1.702008

0.8 2.340749 2.344224 2.147287 2.344224 2.147286 2.147287

0.9 2.907324 2.911278 2.617001 2.911278 2.617000 2.617001

1.0 3.517013 3.521462 3.101906 3.521462 3.101905 3.101906

Table 7.3: Numerical results with comparison to Ref. [104] and [12].

(k = 10)

Haar Legendre

β m = 8 m = 32 m = 128 m = 512 m = 6 m = 24 m = 96 m = 384

0.25 0.0076 4.80(−4) 3.00(−5) 1.87(−6) 0.0135 8.54(−4) 5.34(−5) 3.33(−6)

0.50 0.0071 4.47(−4) 2.80(−5) 1.75(−6) 0.0127 7.96(−4) 4.97(−5) 3.11(−6)

0.75 0.0066 4.16(−4) 2.60(−5) 1.62(−6) 0.0118 7.40(−4) 4.63(−5) 2.89(−6)

1.00 0.0062 3.90(−4) 2.44(−5) 1.52(−6) 0.0111 6.94(−4) 4.34(−5) 2.71(−6)

Table 7.4: Maximum absolute error for the Haar wavelet and the Legendre wavelets.

where,

f(t) = Iαg(t) + (y1 − y0)t− tIαg(1) + y0. (7.2.21)

We approximate y(t) as

y(t) = CTmLm(t) = CT

mLm×mBm(t), (7.2.22)

Let CTmLm×m = [s1, s2, . . . , sm], then using orthogonality of block-pulse function, we obtain

(y(t))n = [sn1 , sn2 , . . . , s

nm]Bm(t) = STmBm(t). (7.2.23)

where Sm = [sn1 , sn2 , . . . , s

nm]T . Now

Iα(y(t))n = STmIαBm(t) = STmFαm×mBm(t). (7.2.24)

134

Let µ(t) ∈ L2[0, 1], then

µ(t)Iα(y(1))n =

∫ 1

0

µ(t)(1− s)α−1

Γ(α)(y(s))nds

=

∫ 1

0Kα(s, t)(y(s))

nds

= STm

∫ 1

0Bm(s)L

Tm(s)K

αm×mLm(t)ds

= STm(∫ 1

0Bm(s)BT

m(s)ds)LTm×mKα

m×mLm(t)

=1

mSTmL

Tm×mKα

m×mLm×mBm(t).

(7.2.25)

We take µ(t) = at and substitute (7.2.22), (7.2.24) and (7.2.25) into equation (7.2.20), then

CTmLm(t) = −aSTmFαm×mBm(t) +

1

mSTmLm×mKα

m×mLm×mBm(t) + FTmLm(t), (7.2.26)

where the function f(t) is approximated by f(t) = FTmLm(t). For α = 1.5, a = e−2π, A = 0, B = 1, n = 2,

and g(t) = t2(e2πt5 + 10532 ), the exact solution is y(t) = t

72 . We compare the numerical solution with exact

solution for M = 3 and different values of k. The results are given below in the Table 7.5.

t α = 1.1 α = 1.3 α = 1.5 α = 1.7 α = 1.9, α = 2

0.1 2.9111(−4) 1.9756(−4) 9.6996(−5) 4.7793(−5) 2.8511(−5) 2.4492(−5)

0.2 5.4223(−3) 2.2168(−3) 9.3927(−4) 4.0536(−4) 1.7477(−4) 1.1436(−4)

0.3 6.0275(−3) 3.0160(−3) 1.5087(−3) 7.5162(−4) 3.7264(−4) 2.6165(−4)

0.4 1.3892(−3) 6.8755(−4) 3.3989(−4) 1.6667(−4) 7.8889(−5) 5.3659(−5)

0.5 8.4144(−3) 4.5061(−3) 2.4163(−3) 1.2916(−3) 6.8293(−4) 4.9443(−4)

0.6 1.2710(−3) 6.5427(−4) 3.1023(−4) 1.1972(−4) 1.3952(−5) 1.6752(−5)

0.7 5.7669(−3) 3.0298(−3) 1.4799(−3) 5.9606(−4) 9.2122(−5) 7.0255(−5)

0.8 5.1134(−3) 2.3439(−3) 6.3407(−4) 4.5289(−4) 1.1664(−3) 1.4317(−3)

0.9 3.4295(−3) 4.0232(−3) 4.6701(−3) 5.3913(−3) 6.1995(−3) 6.6398(−3)

Table 7.5: The absolute error for M = 3, k = 3 and different values of α.

Chapter 8

Summery and Conclusions

In the first sections of Chapter 2, we introduced some spacial functions that were required for the devel-

opment of our results. In the second section some fundamental concepts and definitions from fractional

calculus including some basic results about the Riemann–Liouville and the Caputo fractional operators

are provided. In the remaining two sections of the Chapter 2 some fixed point theorems are stated and

some basic properties of the Haar wavelets are reviewed.

With the help of classical tools from functional analysis, operator theory and fixed point theory,

the theory on existence and uniqueness of solutions to boundary value problems for nonlinear fractional

differential equations, with the Riemann–Liouville and the Caputo fractional derivatives, is developed. In

Chapter 3, we established sufficient conditions for existence and uniqueness results for different classes

of nonlinear boundary value problems involving fractional derivatives, subject to two–point three–point,

multi–point and integral boundary conditions.

In Chapter 4, several existence results for positive and multiple positive solutions to two different

types boundary value problems for fractional differential equations are obtained. For the boundary value

problem 4.1, the existence of at least one, at least two and at least three positive solutions is guaranteed in

a specially constructed cone in the Banach space B. It is observed that the Green functions Gα(t, s) and

Gα,β(t; η, s) for the boundary value problem (4.2.1), (4.2.2), satisfy some interesting and useful properties

and they are related to each other. This helps us to construct a cone in the Banach space B. Then we

established existence and uniqueness results for positive solutions in this cone. The existence of positive

solution is assured only if λ1 ≤ cαρ1 and λ2 ≤ cβρ1.

In chapter Chapter 5, we derived some existence results for positive and multiple positive solutions

for two different types of boundary value problems concerning systems of nonlinear fractional differential

equations. The existence and multiplicity results for the boundary value problem (5.1.1), (5.1.2) are

established by assuming that the nonlinear functions f and g satisfy superlinear–sublinear type conditions.

For some of the established results, we found the explicit intervals for the parameter λ for which the

boundary value problem (5.1.1), (5.1.2) has positive or multiple positive solutions. In section 5.2, we have

established some existence theorems for positive and multiple positive solutions for a higher order system

of nonlinear fractional differential equations with three–point boundary conditions. Explicate intervals for

the parameters λ, µ are derived, for which the system possesses positive solutions. Validity of proposed

135

136

results is shown by two examples.

Owing to the challenging difficulties of having an exact analytic solutions for fractional boundary value

problems, we have used the Haar and the Legendre wavelets to obtain numerical solutions. We obtained

interesting results demonstrating that these mathematical tools are useful to develop efficient numerical

schemes to treat the wider class of problems.

In Chapter 6, a numerical scheme, based on the Haar wavelets, is proposed to solve the boundary

value problems for fractional differential equations. A new operational matrix Qα,η,nm×m is obtained and is

used along with some existing Haar wavelet operational matrices of integration to solve different types of

boundary value problems for fractional differential equations. The method is successfully applied to linear

problems with constant and variable coefficients and also to some nonlinear problems. Also we noticed

that the matrix Qα,η,nm×m is very useful to deal with three–point (or multi–point) boundary value problems

as well. The wavelet based method reduces the problem to a system of algebraic equations. The numerical

results obtained are compared with exact solutions by tabulating their absolute error and by comparing

their respective graphs. It is worth mentioning that results obtained, agree well with exact solutions even

for small number of collocation points. The method seems to be more efficient and convenient for solving

linear boundary value problems for fractional order differential equations. It takes care of the boundary

conditions automatically. However much is needed to be done for nonlinear problems. The method, when

applied to nonlinear problems, reduces the problem to system of nonlinear algebraic equations leading to

extra difficulties.

Despite of effectiveness, efficiency, and simplicity of the wavelet method, there are some issues, related

to this approach, needed to be addressed. While solving the linear systems for wavelet coefficients ci, we

have to invert some matrix which of course involves the Haar matrix. The Haar wavelet basis leads to a

sparse matrix. Thus for large values of m, the matrix appears to be nearly singular. In this situation, the

calculation of the wavelet coefficients becomes impossible with required accuracy.

In Chapter 6, we have used the Haar basis to represent the solutions of boundary value problems.

However, it is possible to have a wavelet representation of solution for boundary value problems on an

infinitely many different wavelet bases. In Chapter 7 we used the Legendre polynomials as a basis for the

wavelet representation of solutions of fractional differential equations. An operational matrix of fractional

order integration is obtained. The procedure for obtaining solution remains the same as in the case of

the Haar wavelets. The results obtained by the numerical scheme based on the Legendre wavelets are

compared with exact solutions and also with solutions obtained by some other numerical methods. It has

been observed that, in some situations, the method gives good results as compared with other methods.

The method is convenient for solving linear initial value problems as well as boundary value problems,

science the boundary conditions are taken care of automatically.

Appendix A

Matlab and Mathematica programs

In this chapter, we present some of the most important Matlab and Mathematica codes that have been

developed for the numerical analysis of the fractional differential equations.

A.1 Computations of some operational matrices by Matlab

The following Matlab code generates the Haar matrix of order m by m. Here we point out that F. Chanal

have developed a Matlab code for computing Haar matrices, available at

http://www.mathworks.com/matlabcentral/fx_files/4619/1/haarmtx.m. The computation by the code

developed here are little bit faster then that of F. Chanal.

function H=Hmtx(J)

format long J=3; m=2.^J;

for l=1:m;

x=(2*l-1)/(2*m);

G(1,l)=1;

end

for l=1:m;

for j=0:J-1

for k=0:2.^j-1;

i=2.^j+k+1;

a=(k)/2^j; b=(2*k+1)/(2*2^j); c=(k+1)/2^j;

x=(2*l-1)/(2*m); cc=min(x,c);

if x >= a & x <= b;

G(i,l) = 1;

elseif x >= b & x <= c;

G(i,l) =-1;

else

G(i,l) =0;

end

137

138

end

end

end

H=G

The following programm is developed to compute the block–pulse operational matrix of fractional order

integration. The computations of most of the other operational matrices of fractional order integration

are based on this matrix.

function F=Falpha(J,alpha) m=2^J;

for j=1:m;

for i=1:m;

if i>j;

xi(i,j)=0;

elseif i<=j&i>=j;

xi(i,j)=1;

else

k=j-i;

xi(i,j)=(k+1).^(alpha+1)-2.*(k).^(alpha+1)+(k-1).^(alpha+1);

end

end

end

F=xi; F=(1./(m.^(alpha)*gamma(alpha+2)))*xi

Next, we present a Matlab code for the computation of the operational matrix of fractional order integration

Pαm×m.

function P=PFalpha(J,alpha)

m=2^J;

for j=1:m;

for i=1:m;

if i>j;

xi(i,j)=0;

elseif i<=j&i>=j;

xi(i,j)=1;

else

k=j-i;


end

end

end

R=xi;

139

F=(1./(m.^(alpha)*gamma(alpha+2)))*xi;

P=H*F*inv(H);

The following code is developed to compute an other important operational matrix, namely Qαm×m.

functionQ=Qmtx(J,alpha)

m=2.^J;

eta=1;

psi=@(s)s;

psi2=@(s)((eta-s).^(alpha-1))./gamma(alpha); U(1)=quad8(psi,0,1);

for j=1:J;

for k=1:2.^(j-1);

i=2.^(j-1)+k;

a=(k-1)./2.^(j-1); b=(2*k-1)./(2.^j); c=k./2^(j-1);

h0=@(s)2.^((j-1)./2)*psi;

U(i)=quad8(h0,a,b)-quad8(h0,b,c);

end

end

V(1)=quad8(psi2,0,eta);

for j=1:J;

for k=1:2.^(j-1);

i=2.^(j-1)+k;

a=(k-1).*eta./2.^(j-1); b=min(eta,(2*k-1).*eta./(2.^j));

c=min(eta,k.*eta./2^(j-1));

h0=@(s)2.^((j-1)./2)*((eta-s).^(alpha-1))./gamma(alpha);

V(i)=quad8(h0,a,b)-quad8(h0,b,c);

end

end for i=1:m;

for j=1:m;

Qe(j,i)=[U(i)*V(j)];

end

end

Q=Qe;

The Legendre matrix can be computed by using following code.

function L=Lmtx(k)

M=3; N=2.^(k-1); m=M*N; for i=1:m;

t=(2*i-1)./(2*m);

for n=1:N;

a=(2*n-2)./2.^k;

b=(2*n)./2.^k;

140

if t>=a & t<b;

FF(n,i)=2.^(k./2).*(1./2).^(1./2)+0*n;

FF(n+N,i)=2.^(k./2).*(3./2).^(1./2)*(2.^k*t-2*n+1);

FF(n+2*N,i)=2.^(k./2).*(5./2).^(1./2)*((5./2).*(2.^k*t-2*n+1).^2-1./2);

elseif t>b;

FF(n,i)=0;

end

end

end L=FF;

The following programme is developed to compute the Legendre wavelet operational matrix of fractional

order integration.

function P=PFalpha(J,alpha)

m=2^J; for j=1:m;

for i=1:m;

if i>j;

xi(i,j)=0;

elseif i<=j&i>=j;

xi(i,j)=1;

else

k=j-i;


end

end

end R=xi; F=(1./(m.^(alpha)*gamma(alpha+2)))*xi;

P=haarmtx(J)*F*inv(haarmtx(J));

In the following we give an other important code in the context of boundary value problems for fractional

order differential equations.

function KL=KLmtx(k,alpha)

syms t;

M=3; N=2.^(k-1); m=M*N; psi=t.^(alpha-1);

ptt=((1-t).^(alpha-1))./gamma(alpha);

for n=1:N;

a=(2*n-2)./2.^k;

b=(2*n)./2.^k;

L0=2.^(k./2).*(1./2).^(1./2)*psi;

L1=2.^(k./2).*(3./2).^(1./2)*(2.^k*t-2*n+1)*psi;;

L2=2.^(k./2).*(5./2).^(1./2)*((5./2).*(2.^k*t-2*n+1).^2-1./2)*psi;

U(n)=eval(int(L0,t,a,b));

141

U(n+N)=eval(int(L1,t,a,b));

U(n+2*N)=eval(int(L2,t,a,b));

end

for n=1:N;

aa=(2*n-2)./2.^k;

bb=(2*n)./2.^k;

L0=(2.^(k./2).*(1./2).^(1./2)+0.*n+0.*t)*ptt;

L1=2.^(k./2).*(3./2).^(1./2)*(2.^k*t-2*n+1)*ptt;

L2=2.^(k./2).*(5./2)...

.^(1./2)*((5./2).*(2.^k*t-2*n+1).^2-1./2)*ptt;

V(n)=eval(int(L0,t,aa,bb));

V(n+N)=eval(int(L1,t,aa,bb));

V(n+2*N)=eval(int(L2,t,aa,bb));

end

for i=1:m;

for j=1:m;

Ke(j,i)=[U(i)*V(j)];

end

end KL=Ke;

142

A.2 Computations by Mathematica

143

Appendix B

Useful results from Analysis

For convenience, in this chapter we review some useful result from analysis.

Definition B.0.1. A function f [a, b] → R is said to be absolutely continuous on [a, b], if for each ε > 0,

there is a δ > 0, such that given any collection (ak, bk) : 1 ≤ k ≤ n of pairwise disjoint open subintervals

of [a, b], we haven∑k=1

(bk − ak) < δ impliesn∑k=1

(f(bk)− f(ak)) < ε.

Absolutely continuous functions on [a, b] are uniformly continuous but converse is not true. The space

of absolutely continuous functions is denoted by AC[a, b].

We define the space

ACm[a, b] =f : [a, b] → R and Dm−1f(x) ∈ AC[a, b]

.

Theorem B.0.2. [68] Let f : [a, b] → R and m ∈ N. Then f ∈ ACm[a, b] if and only if f can be

represented as

f(x) =

m−1∑k=0

ck(x− a)k + Ima φ,

where k = 0, 1, . . . ,m− 1, φ ∈ L(a, b).

Theorem B.0.3. (Taylor’s Theorem) Let f ∈ ACm[a, b], m ∈ N. The for any x ∈ [a, b]

f(x) =

m−1∑j=0

(x− a)j

j!Djf(a) + Ima Dmf(x), (B.0.1)

where Dmf(x) ∈ L[a, b].

Definition B.0.4. [56] Let Ω ⊂ Rn. A set M ⊂ C(Ω) is said to be equicontinuous if and only if, for

every ε > 0 there exist a δ > 0 such that |f(x), f(y)| < ε with ∥x− y∥ < δ, for all x, y ∈ Ω and all f ∈ M.

Theorem B.0.5. [56] (The Arzela-Ascoli Theorem) Let Ω be a bounded subset of Rn, and let M be a

subset of C(Ω). Then M is relatively compact if and only if it is bounded and equicontinuous.

144

145

The space obtained by generalizing Rn or Cn with an infinite sequence f = (f1, f2, f3, . . . ) as its

element is called the sequence space and is denoted by ℓp. For 0 ≤ p ≤ ∞, the sequence space ℓp with the

norm defined by

∥f∥p =

∞∑n=1

|fn|p 1

p

(1 ≤ p <∞), ∥f∥∞ = sup ∥fn∥,

is a Banach space.

Now we define an other important space which is a continues analog of sequence space.

Definition B.0.6. Let f be measureable and set

∥f∥p =∫

|f |pdµ 1

p

(1 ≤ p <∞), ∥f∥∞ = ess sup |f |,

where µ is the Lebesgue measure defined on a σ-algebra of subsets of a set X in Rn. The space Lp(X) is

the space of all measurable functions with the ∥f∥p <∞.

The space Lp, (p ≥ 1) is a Banach space under the norm ∥.∥p. In particular, the space L2 is the Hilbert

space supplied with the inner product ⟨f, g⟩ =∫fgdµ.

We use the notation Lp[0, 1] for the space of measurable functions such that∫ 10 |f(s)|ds <∞, where the

integral is understood in the sense of Lebesgue and the norm on Lp[0, 1] is defined as ∥f∥p = (∫ 10 |f(s)|pds)

1p

where 1 < p <∞.

Theorem B.0.7. (Fubini’s Theorem) Suppose that f : R2 → R is measureable and one of the integrals∫ ∫|f(t, x)|dtdx,

∫dt∫|f(t, x)|dx,

∫dx∫|f(t, x)|dt is finite. Then the functions g(t) =

∫f(t, x)dx and

h(x) =∫f(t, x)dt are measureable and∫

dt

∫f(t, x)dx =

∫dx

∫f(t, x)dt =

∫ ∫f(x, y)dxdy.

Definition B.0.8. A function f : [0, 1]× Rn → R satisfies the Carathéodory conditions if

(i) f(t, x) is Lebesgue measureable on [0, 1] for each x ∈ Rn .

(ii) f(t, x) is continuous on Rn for almost every t ∈ [0, 1].

(iii) For each r > 0, there exists an ψr ∈ Lp([0, 1]) such that |x| ≤ r implies that |f(t, x)| ≤ ψr(t) for

a.e., t ∈ [0, 1].

Definition B.0.9. A sequence of elements φkk∈Z in a Hilbert space H is a Riesz basis if for every

f ∈ H there exists a unique sequence αkk∈Z ∈ ℓ2(Z) such that the sequence φkk∈Z is complete and

A∥f∥22 ≤∑k∈Z

|αk|2 ≤ B∥f∥22

with 0 < A ≤ B <∞ constants independent of f ∈ H.

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