Fractional-Order Equations and Inclusions 9783110522075, 9783110521382

Table of contents :
Acknowledgment
Preface
Contents
Introduction
1. Fractional Difference Equations
2. Fractional Integral Equations
3. Fractional Differential Equations
4. Fractional Evolution Equations: Continued
5. Fractional Differential Inclusions
A Appendix
A. Appendix
Bibliography
Index

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Michal Fečkan, JinRong Wang, Michal Pospíšil Fractional-Order Equations and Inclusions

Fractional Calculus in Applied Sciences and Engineering

| Editor in Chief Changpin Li Editorial Board Virginia Kiryakova Francesco Mainardi Dragan Spasic Bruce Ian Henry YangQuan Chen

Volume 3

Michal Fečkan, JinRong Wang, Michal Pospíšil

Fractional-Order Equations and Inclusions |

Mathematics Subject Classification 2010 26A33, 34A08, 34A33, 34K37, 35R11, 35Q93, 45G05, 47H08, 47H10, 47J35 Authors Prof. Dr. Michal Fečkan Department of Mathematical Analysis and Numerical Mathematics Comenius University in Bratislava 842 48 Bratislava Slovak Republic [email protected] Prof. Dr. JinRong Wang Department of Mathematics, Guizhou University Guizhou 550025 P. R. China [email protected] Dr. Michal Pospíšil Mathematical Institute Slovak Academy of Sciences 814 73 Bratislava Slovak Republic [email protected]

ISBN 978-3-11-052138-2 e-ISBN (PDF) 978-3-11-052207-5 e-ISBN (EPUB) 978-3-11-052155-9 Set-ISBN 978-3-11-052208-2 ISSN 2509-7210 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2017 Walter de Gruyter GmbH, Berlin/Boston Cover image: naddi/iStock/thinkstock Typesetting: Dimler & Albroscheit, Müncheberg Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

| To our beloved families.

Acknowledgment Michal Fečkan and Michal Pospíšil acknowledge the support by the Slovak Grant Agency VEGA through grants No. 2/0153/16 and No. 1/0078/17 and by the Slovak Research and Development Agency under the contract No. APVV-14-0378. JinRong Wang acknowledges the support by the National Natural Science Foundation of China (No. 11661016), Training Object of High Level and Innovative Talents of Guizhou Province [(2016)4006], Graduate Course of Guizhou University (ZDKC[2015]003) and Unite Foundation of Guizhou Province [(2015)7640].

DOI 10.1515/9783110522075-202

Preface It is clear that differential calculus is an important part of mathematics with many applications. Nowadays its extension, fractional differential calculus, is rapidly developing partially due to its challenging attraction for mathematicians and partially because more and more new applications are appearing. So it was a pleasure for us when Professor Changpin Li invited us to write a monograph in the series Fractional Calculus in Applied Sciences and Engineering covered by De Gruyter. The aim of this book is to present a broad variety of problems that appear in using fractional calculus as well as to show that the mathematical theory behind this is non-trivial but beautiful. This monograph is intended for post-graduate students, mathematicians, and researchers interested in fractional differences and differential equations and applying fractional calculus in physics, mechanics, engineering, biology, economics and other branches of sciences. Michal Fečkan, JinRong Wang and Michal Pospíšil Bratislava, Slovakia and Guiyang, China April 2017

DOI 10.1515/9783110522075-203

Contents Acknowledgment | VII Preface | IX Introduction | 1 1 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.2 1.2.1 1.2.2 1.2.3 1.2.4

Fractional Difference Equations | 5 Fractional difference Gronwall inequalities | 5 Introduction | 5 Caputo like fractional difference | 5 Linear fractional difference equation | 7 Fractional difference inequalities | 17 S-asymptotically periodic solutions | 20 Introduction | 20 Preliminaries | 21 Non-existence of periodic solutions | 22 Existence and uniqueness results | 27

2 2.1 2.1.1 2.1.2 2.1.3

Fractional Integral Equations | 37 Abel-type nonlinear integral equations | 37 Introduction | 37 Preliminaries | 38 Existence and uniqueness of non-trivial solution in an order interval | 40 General solutions of Erdélyi–Kober type integral equations | 48 Illustrative examples | 48 Quadratic Erdélyi–Kober type integral equations of fractional order | 51 Introduction | 51 Preliminaries | 52 Existence and limit property of solutions | 53 Uniqueness and another existence results | 60 Applications | 63 Fully nonlinear Erdélyi–Kober fractional integral equations | 67 Introduction | 67 Main result | 67 Example | 76 Quadratic Weyl fractional integral equations | 77 Introduction | 77

2.1.4 2.1.5 2.2 2.2.1 2.2.2 2.2.3 2.2.4 2.2.5 2.3 2.3.1 2.3.2 2.3.3 2.4 2.4.1

DOI 10.1515/9783110522075-204

XII | Contents

2.4.2 2.4.3 2.4.4 2.4.5 3 3.1 3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.4 3.4.1 3.4.2 3.4.3 3.5 3.5.1 3.5.2 3.5.3 3.5.4 3.6 3.6.1 3.6.2 3.6.3 3.6.4

Preliminaries | 78 Some basic properties of Weyl kernel | 79 Existence and uniform local attractivity of 2π-periodic solutions | 82 Example | 89 Fractional Differential Equations | 91 Asymptotically periodic solutions | 91 Introduction | 91 Preliminaries | 92 Non-existence results for periodic solutions | 92 Existence results for asymptotically periodic solutions | 94 Further extensions | 98 Modified fractional iterative functional differential equations | 103 Introduction | 103 Notation, definitions and auxiliary facts | 105 Existence | 106 Data dependence | 111 Examples | 113 Ulam–Hyers–Rassias stability for semilinear equations | 116 Introduction | 116 Ulam–Hyers–Rassias stability for surjective linear equations on Banach spaces | 116 Ulam–Hyers–Rassias stability for linear equations on Banach spaces with closed ranges | 123 Ulam–Hyers–Rassias stability for surjective semilinear equations between Banach spaces | 124 Practical Ulam–Hyers–Rassias stability for nonlinear equations | 126 Introduction | 126 Main results | 127 Examples | 131 Ulam–Hyers–Mittag-Leffler stability of fractional delay differential equations | 132 Introduction | 132 Preliminaries | 132 Main results | 133 Examples | 138 Nonlinear impulsive fractional differential equations | 139 Introduction | 139 Preliminaries | 140 Existence results for impulsive Cauchy problems | 141 Ulam stability results for impulsive fractional differential equations | 148

Contents

3.6.5 3.6.6 3.7 3.7.1 3.7.2 3.7.3 3.7.4 3.7.5 3.7.6 3.7.7 3.8 3.8.1 3.8.2 3.8.3 3.8.4 3.9 3.9.1 3.9.2 3.9.3 3.9.4 3.10 3.10.1 3.10.2 3.10.3 3.10.4 4 4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.1.5 4.2 4.2.1 4.2.2 4.2.3 4.2.4

| XIII

Existence results for impulsive boundary value problems | 151 Applications | 156 Fractional differential switched systems with coupled nonlocal initial and impulsive conditions | 158 Introduction | 158 Preliminaries | 159 Existence and uniqueness result via Banach fixed point theorem | 160 Existence result via Krasnoselskii fixed point theorem | 162 Existence result via Leray–Schauder fixed point theorem | 164 Existence result for the resonant case: Landesman–Lazer conditions | 166 Ulam type stability results | 169 Not instantaneous impulsive fractional differential equations | 172 Introduction | 172 Framework of linear impulsive fractional Cauchy problem | 173 Generalized Ulam–Hyers–Rassias stability concept | 175 Main results via fixed point methods | 176 Center stable manifold result for planar fractional damped equations | 182 Introduction | 182 Asymptotic behavior of Mittag-Leffler functions E α,β | 183 Planar fractional Cauchy problems | 184 Center stable manifold result | 185 Periodic fractional differential equations with impulses | 197 Introduction | 197 FDE with Caputo derivatives with varying lower limits | 198 FDE with Caputo derivatives with fixed lower limits | 209 Conclusions | 213 Fractional Evolution Equations: Continued | 215 Fractional evolution equations with periodic boundary conditions | 215 Introduction | 215 Homogeneous periodic boundary value problem | 216 Non-homogeneous periodic boundary value problem | 217 Parameter perturbation methods for robustness | 221 Example | 223 Abstract Cauchy problems for fractional evolution equations | 224 Introduction | 224 Preliminaries | 225 Existence and uniqueness theorems of solutions for Problem (I) | 226 Existence and uniqueness theorems of solutions for Problem (II) | 231

XIV | Contents

4.2.5 4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.4 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.5 4.5.1 4.5.2 4.5.3 4.5.4 4.5.5 5 5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.1.5 5.2 5.2.1 5.2.2 5.2.3 5.2.4 5.3 5.3.1 5.3.2 5.3.3 5.3.4

Existence and uniqueness theorems of solutions for Problem (III) | 233 Nonlocal Cauchy problems for Volterra–Fredholm type fractional evolution equations | 243 Introduction | 243 Preliminaries | 244 Existence of mild solutions | 245 Example | 257 Controllability of Sobolev type fractional functional evolution equations | 259 Introduction | 259 Preliminaries | 259 Characteristic solution operators and their properties | 260 Main results | 263 Example | 268 Relaxed controls for nonlinear impulsive fractional evolution equations | 271 Introduction | 271 Problem statement | 272 Original and relaxed fractional impulsive control systems | 273 Properties of relaxed trajectories for fractional impulsive control systems | 280 Example | 284 Fractional Differential Inclusions | 287 Fractional differential inclusions with anti-periodic conditions | 287 Introduction | 287 Preliminaries | 288 Existence results for (5.1) | 293 Existence results for (5.2) | 295 Applications to fractional lattice inclusions | 302 Nonlocal Cauchy problems for semilinear fractional differential inclusions | 304 Introduction | 304 Preliminaries and notation | 305 Existence results | 306 Examples | 315 Nonlocal impulsive fractional differential inclusions | 318 Introduction | 318 Preliminaries and notation | 320 Existence results for convex case | 322 Existence results for non-convex case | 334

Contents |

A A.1 A.1.1 A.1.2 A.1.3 A.1.4 A.1.5 A.2 A.3 A.4 A.5

Appendix | 337 Functional analysis | 337 Basic notation and results | 337 Banach functional spaces | 337 Linear operators | 339 Semigroup of linear operators | 339 Metric spaces | 339 Fractional differential calculus | 340 Henry–Gronwall’s inequality | 342 Measures of noncompactness | 343 Multifunctions | 346

Bibliography | 349 Index | 365

XV

Introduction Fractional calculus was introduced at the end of the nineteenth century by Liouville and Riemann, but the concept of non-integer calculus, as a generalization of the traditional integer-order calculus was mentioned already in 1695 by Leibniz and L’Hospital. The subject of fractional calculus has become a rapidly growing area and has found applications in diverse fields ranging from physical sciences, engineering to biological sciences and economics. It draws applications in nonlinear oscillations of earthquakes, many physical phenomena such as seepage flow in porous media and in fluid dynamic traffic model. Fractional calculus provides a powerful tool for the description of hereditary properties of various materials and memory processes. In particular, integral equations involving fractional integral operators (which can be regarded as an extension of Abel integral equations) appear naturally in the fields of biophysics, viscoelasticity, electrical circuits, etc. For some of these applications, one can see [120, 138] and the references therein. There are remarkable monographs of Baleanu et al. [36], Diethelm [91], Kilbas et al. [162], Lakshmikantham et al. [170], Miller and Ross [195], Michalski [194], Podlubny [223], Tarasov [263] and Zhou et al. [337] that provide main theoretical tools for the qualitative analysis of fractional-order differential equations, and at the same time, show the interconnection as well as the contrast between integer-order and fractionalorder differential models. Furthermore, studies of Cauchy problems, boundary value problems, nonlocal problems, impulsive problems for fractional differential equations and related optimal controls, feedback control and controllability have been investigated. For instance, Ahmad and Nieto [9] applied Leray–Schauder degree theory to derive the existence of solutions for fractional anti-periodic boundary value problems, Bai [28] used the tools of the contraction map principle and fixed point index theory to show the existence of positive solutions of a nonlocal fractional boundary value problem; Benchohra et al. [48] used the Banach fixed point theorem and the nonlinear alternative of Leray–Schauder type to prove the existence of solutions for fractional functional and neutral functional differential equations with infinite delay; Chang and Nieto [68] studied the existence of solutions for fractional differential inclusions with boundary conditions by means of the Bohnenblust–Karlin fixed point theorem when the multi-valued nonlinearity term has sub-linear or linear growth in the second variable; Zhou et al. [335, 336] studied fractional neutral differential equations with infinite delay and p-type fractional neutral differential equations, while some interesting new concepts and existence results were obtained. Next, Jiao and Zhou [149] succeeded to apply the critical point theory to prove existence results for a class of fractional boundary value problems involving left- and right-sided fractional derivatives. Further, Zhou and Jiao [334] introduced a suitable concept of mild solutions involving two characteristic solution operators, and derived the properties of boundedness, strong continuity and compactness, to prove existence results for the DOI 10.1515/9783110522075-205

2 | Introduction

nonlocal Cauchy problem for fractional evolution equations. We also mention some similar works: Hernández et al. [136] gave another possible method to introduce the suitable mild solutions for abstract differential equations with fractional derivatives by virtue of resolvent operators; Li [178] was concerned with Cauchy problems for Riemann–Liouville fractional differential equations in infinite-dimensional Banach spaces, introduced the so-called fractional resolvent and derived its properties, and presented sufficient conditions to guarantee the existence and uniqueness of weak and strong solutions of fractional Cauchy problems; meanwhile, Wang [316] introduced two characteristic solution operators for mild solutions of abstract fractional evolution equations by virtue of the theory of almost sectorial operators whose resolvent satisfies the estimate of growth in a sector of the complex plane; next, Kaslik and Sivasundaram [157] proved non-existence of periodic solutions of fractional-order differential equations in the interval [0, ∞) by using the Mellin transform approach. Numerical methods are treated in the books [174, 182, 260]. Of course, there are many other interesting achievements which we cannot list here. Let us mention that there was not a really substantial parallel development of a discrete fractional calculus until very recently, since the used method is much more difficult and technical than that for fractional differential equations. But nowadays, there are many interesting results in that direction, we refer to the monograph of Goodrich and Peterson [122]. In the past seventy years, Ulam type stability problems have been taken up by a large number of mathematicians and the study of this area has grown to be one of the most important subjects in the mathematical analysis area, since it is quite useful in many applications such as numerical analysis, optimization, biology and economics, where finding the exact solution is quite difficult. The stability of functional equations was originally raised by Ulam in 1940 in a talk given at Wisconsin University. The problem posed by Ulam was the following: “Under what conditions does there exist an additive mapping near an approximately additive mapping?” (for more details see [271]). The first answer to the question of Ulam was given by Hyers in 1941 in the case of Banach spaces (see [142]): Let E1 , E2 be two real Banach spaces and ϵ > 0. Then for every mapping f : E1 → E2 satisfying ‖f(x + y) − f(x) − f(y)‖ ≤ ϵ

for all x, y ∈ E1

there exists a unique additive mapping g : E1 → E2 with the property ‖f(x) − g(x)‖ ≤ ϵ

for all x ∈ E1 .

Thereafter, this type of stability is called the Ulam–Hyers stability. In 1978, Rassias [233] provided a remarkable generalization of the Ulam–Hyers stability of mappings by considering variables. The stability properties of all kinds of equations have attracted the attention of many mathematicians. For more details, one can see the monographs of Cădariu [78], Hyers [143] and Jung [150].

Introduction

| 3

In recent years, the theory of impulsive differential equations has been an object of increasing interest because of its wide applicability in biology, medicine and many other fields. The reason for this applicability arises from the fact that impulsive differential problems are an appropriate model for describing processes which at certain moments change their state rapidly and which cannot be described using the classical differential problems. For a wide bibliography and exposition on this object see for instance the monographs [29, 47, 167, 327]. After the above brief description of the main topics of this book, we now outline the contents of this book in more details. Chapter 1 is devoted to fractional difference equations. In Section 1.1, we derive inequalities of Gronwall type for fractional difference inequalities by applying them to asymptotic properties of a solution of a linear homogeneous fractional difference equation with constant coefficient. In Section 1.2, we show that a fractional difference equation with a periodic right-hand side can not possess a periodic solution but it can have an S-asymptotically periodic solution. This is related to [157]. Chapter 2 investigates fractional integral equations of several kinds. Section 2.1 studies Abel-type nonlinear integral equations with weakly singular kernels. Quadratic Erdélyi–Kober type integral equations of fractional order are considered in Section 2.2. In Section 2.3, a generalized nonlinear functional equation involving Erdélyi– Kober fractional integrals on an unbounded interval is studied. Periodic solutions of quadratic Weyl fractional integral equations are shown in Section 2.4. Chapter 3 studies fractional differential equations from several points of view. It is shown in Section 3.1 that there are no non-constant periodic solutions for fractional differential equations but there may exist asymptotically periodic ones. The weakly Picard operator technique is used in Section 3.2 to obtain some existence, uniqueness and data dependence results for modified fractional iterative functional differential equations. Section 3.3 is devoted to Ulam–Hyers–Rassias stability of linear and semilinear equations on Banach spaces from functional analysis theory. A practical Ulam–Hyers–Rassias stability for nonlinear equations is introduced and studied in Section 3.4. The existence, uniqueness and Ulam–Hyers–Mittag-Leffler stability of solutions of fractional-order differential equations are investigated in Section 3.5 with modified arguments. Nonlinear impulsive fractional differential equations are investigated in Section 3.6. Section 3.7 studies nonlocal impulsive fractional-order differential switched systems. Fractional differential equations with not instantaneous impulses are introduced and investigated in Section 3.8. A center stable manifold theorem is obtained in Section 3.9 for planar fractional damped equations involving two differential Caputo derivatives. Section 3.10 deals with the existence of periodic solutions for fractional differential equations with periodic impulses. Chapter 4 deals with fractional evolution equations. Section 4.1 discusses periodic boundary value problems for fractional evolution equations. Abstract Cauchy problems for several types of fractional evolution equations are considered in Section 4.2. Section 4.3 investigates nonlocal Cauchy problems for fractional evolution equations involving

4 | Introduction

Volterra–Fredholm type integral operators. Section 4.4 studies the controllability of a class of fractional functional evolution equations of Sobolev type via the theory of semigroups. In Section 4.5, relaxation of fractional impulsive control systems is studied. Chapter 5 is devoted to the study of fractional inclusions. Anti-periodic solutions are shown in Section 5.1 for fractional differential equations and inclusions. Section 5.2 presents existence results for nonlocal semilinear differential inclusions of fractional order. Existence of mild solutions is shown in Section 5.3 for nonlocal impulsive fractional differential inclusions. In the final part, Appendix A, some well-known results used in this book are summarized for the reader’s convenience. The motivation for writing this monograph is twofold for us. Firstly, here we are presenting a broad variety of problems that appear in using fractional calculus which justify the usefulness of developing the theory of fractional difference and differential equations. Secondly, we are showing that the mathematical theory behind this is sophisticated but nice and beautiful. So most results of this book are based on our published papers, but we have improved and modified the original papers to simplify and clarify final results. This monograph is intended for anyone who is interested in the theory of fractional difference and differential equations and their applications in science and practice as well.

1 Fractional Difference Equations 1.1 Fractional difference Gronwall inequalities 1.1.1 Introduction The main aim of this section is to prove new inequalities of Gronwall type for fractional difference inequalities with linear right-hand side and constant or variable coefficients. Asymptotic properties of a solution of a linear homogeneous fractional difference equation with constant coefficient are also studied. In our results, we use a convenient Green function. This section is based on [105]. Throughout the present section, ∇ denotes the backward difference operator defined as ∇u(n) = u(n) − u(n − 1). It possesses the following properties. Lemma 1.1. Let ∇k F(k, j) := F(k, j) − F(k − 1, j). The following holds true: k

k

∇[ ∑ F(k, j)] = F(k − 1, k) + ∑ ∇k F(k, j), j=a

j=a k−1

k

∑ A(k − j)∇ f(j) = ∑ A(j)∇k f(k − j), j=1

j=0 b−1

b

∑ f(j)∇g(j) = [f(j)g(j)]bj=a−1 − ∑ ∇ f(j + 1)g(j). j=a

j=a−1

So we will consider fractional differences corresponding to the backward difference operator. Of course, some related achievements different from those mentioned above are already done: we refer the reader to similar interesting results in [1, 24, 102]. Here we only note that in [1], the inequality of the form μ

∇∗ u(k) ≤ a(k)u(k),

k ∈ ℕ0

was investigated for k ∈ {q n | n ∈ ℤ} ∪ {0} for 0 < q < 1; Atici and Eloe [24] considered the Riemann–Liouville type fractional difference; and Ferreira [102] focused on the fractional difference corresponding to the forward difference operator. Hence our results are not covered by these papers. We denote by ℕa = {a, a + 1, . . . } the shifted set of positive integers, for simplicity ℕ = ℕ1 , and −ℕa = {. . . , a − 1, a}. We assume the property of empty sum and empty product, i.e., ∑bj=a f(j) = 0, ∏bj=a f(j) = 1 whenever a, b ∈ ℤ are such that a > b.

1.1.2 Caputo like fractional difference In this subsection, we recall some definitions of ∇-based fractional operators, and we show that fractional difference considered in [86] is of Caputo type. DOI 10.1515/9783110522075-001

6 | 1 Fractional Difference Equations Definition 1.2 (see [128]). Let α ∈ ℂ, p = max{0, p0 }, and p0 ∈ ℤ be such that 0 < Re(α + p0 ) ≤ 1, and let function f be defined on ℕa−p . We define the α-th fractional sum as k ∇p Σ α f(k) := (1.1) ∑ (k − ρ(j))p+α−1 f(j) Γ(p + α) j=a for k ∈ ℕa , where ρ(j) = j − 1 and Γ(α+β) if α, α + β ∉ {. . . , −1, 0}, { Γ(α) { { { { {1 if α = β = 0, (α)β = { {0 if α = 0, β ∉ {. . . , −1, 0}, { { { { {undefined otherwise

for α, β ∈ ℂ. If p ∈ ℕ, α ∉ −ℕ0 , k ∈ ℕa and f is defined on ℕa−p , formula (1.1) can be simplified to the case p = 0 (cf. [128, (2.2)]) using k ∇p 1 k ∑ (k − ρ(j))p+α−1 f(j) = ∑ (k − ρ(j))α−1 f(j), Γ(p + α) j=a Γ(α) j=a

(1.2)

i.e., Σ α f(k) =

1 k ∑ (k − ρ(j))α−1 f(j) Γ(α) j=a

(1.3)

for k ∈ ℕa , f defined on ℕa−p . We note that the latter equality is valid whenever α ∈ ℝ \ −ℕ0 . Sometimes Σ α is denoted as ∇−α . Fractional sum (1.3) is a discrete analogue to the Riemann–Liouville fractional integral (cf. [223]). Analogously to the forward difference based Riemann–Liouville [22] and the Caputo [17, 18] like fractional difference, Deekshitulu and Mohan [86] proposed the next definition of a ∇-based fractional difference. Definition 1.3. Let μ ∈ (0, 1) and let f be defined on ℕ0 . Then we define the μ-th fractional difference of a function f as k−1

∇μ f(k) = ∑ ( j=0

j−μ )∇k f(k − j) j

(1.4)

for k ∈ ℕ, where (bn) with b ∈ ℝ, n ∈ ℤ is a generalized binomial coefficient given by Γ(b+1)

{ { { Γ(b−n+1)Γ(n+1) b ( ) = {1 { n { {0

if n > 0, if n = 0, if n < 0.

1.1 Fractional difference Gronwall inequalities | 7

Here we used the lower index k to denote the variable affected by operator ∇. From now on, we assume μ ∈ (0, 1). Example 1.4. Consider u(k) = b k for k ∈ ℕ0 and b > 0. Then k−1

[∇μ u(k)]k=1 = [ ∑ ( j=0

j−μ −μ = ( )[∇u(k)]k=1 = b − 1. )∇k u(k − j)] j 0 k=1

On the other side, 0 < 1 − μ < 1. Thus p = 1 in (1.1). Moreover, since u is defined on ℕ0 , we have a = 1. Therefore, [Σ−μ u(k)]k=1 = [

k ∇ ∑ (k − ρ(j))−μ u(j)] Γ(1 − μ) j=1 k=1

=[

k 1 ∑ (k − ρ(j))−μ−1 u(j)] Γ(−μ) j=1 k=1

=[

k 1 Γ(k − j − μ) u(j)] = b. ∑ Γ(−μ) j=1 Γ(k − j + 1) k=1

Here we applied the identity (1.2). Σ−μ is closely related to the Riemann–Liouville like ∇-based fractional difference discussed in [21], while the following lemma states that ∇μ is of Caputo type. Lemma 1.5. Let μ ∈ (0, 1) and ν = 1 − μ. Let f be a function defined on ℕ0 . Then ∇μ f(k) = Σ ν (∇ f(k)) on ℕ. Proof. For k ∈ ℕ we expand the left-hand side by (1.4), and apply Lemma 1.1 to get k−1

∇μ f(k) = ∑ ( j=0

=

j−μ 1 k−1 Γ(j + ν) ∇k f(k − j) )∇k f(k − j) = ∑ j Γ(ν) j=0 Γ(j + 1)

1 k Γ(k − j + ν) 1 k ∇ f(j) = ∑ ∑ (k − ρ(j))ν−1 ∇ f(j), Γ(ν) j=1 Γ(k − j + 1) Γ(ν) j=1

which, by (1.1) with a = 1, p = 0, is exactly what was to be proved. In the sense of the above lemma, we add the lower index ∗ as done in [17, 18] to denote μ the Caputo nature of the difference, i.e., ∇∗ := ∇μ in the rest of Section 1.1.

1.1.3 Linear fractional difference equation In this subsection, we derive a solution of a non-homogeneous linear fractional difference equation in terms of a Green function. A particular case of a constant coefficient at linear term is investigated in details.

8 | 1 Fractional Difference Equations

First, we provide a lemma transforming a fractional difference equation to a corresponding fractional sum equation and a direct corollary. Lemma 1.6. Let μ ∈ (0, 1), u and let f be real functions defined on ℕ0 and ℕ0 × ℝ, respectively. For any n ∈ ℕ, if μ

∇∗ u(k + 1) = f(k, u(k)) for all k = 0, 1, . . . , n − 1,

(1.5)

then k−1

u(k) = u(0) + ∑ A μ (k − 1, j)f(j, u(j))

for all k = 0, 1, . . . , n

(1.6)

j=0

with A μ (k, j) = (k−j+μ−1 ) for 0 ≤ j ≤ k. k−j Proof. We show that applying Σ μ to equation (1.5) results in equation (1.6). Let k ∈ {0, 1, . . . , n − 1} be arbitrary and fixed. By Lemma 1.5, μ

Σ μ (∇∗ u(k + 1)) = Σ μ (Σ ν (∇u(k + 1))) for k ∈ ℕ0 , where ν = 1 − μ. Property 2 (ii) of [128] says that if μ ∈ ℂ and ν ∉ ℕ, then Σ μ Σ ν = Σ μ+ν . Hence k

μ

Σ μ (∇∗ u(k + 1)) = Σ1 (∇u(k + 1)) = ∑ ∇u(j + 1) = u(k + 1) − u(0) j=0

= Σ μ f(k, u(k)) = k

= ∑( j=0

1 k ∑ (k − ρ(j))μ−1 f(j, u(j)) Γ(μ) j=0

k k−j+μ−1 )f(j, u(j)) = ∑ A μ (k, j)f(j, u(j)). k−j j=0

This completes the proof. Corollary 1.7. Let μ ∈ (0, 1). Let u and f be real functions defined on ℕ0 and ℕ0 × ℝ, respectively. For any n ∈ ℕ, if μ

∇∗ u(k + 1) ≤ f(k, u(k)) for all k = 0, 1, . . . , n − 1, then k−1

u(k) ≤ u(0) + ∑ A μ (k − 1, j)f(j, u(j))

for all k = 0, 1, . . . , n.

j=0

Proof. Since A μ (k, j) > 0 for each 0 ≤ j ≤ k < n, the statement immediately follows from the definition of Σ μ . Note that the above corollary holds true with ≥ instead of ≤.

1.1 Fractional difference Gronwall inequalities | 9

Next, we derive a solution of a linear initial value problem. Denote by h(k) the solution of the problem μ

∇∗ h(k + 1) = a(k)h(k), { h(0) = 1,

k ∈ ℕ0 ,

(1.7)

and define a Green function {g j (k)}k∈ℕ0 , j ∈ ℕ0 as μ

{

∇∗ g j (k + 1) = a(k)g j (k) + δ j (k),

k ∈ ℕ0 ,

g j (0) = 0,

(1.8)

where δ j (k) = 0 for j ≠ k and δ j (j) = 1. Then v(k) = u0 h(k), k ∈ ℕ0 solves μ

∇∗ v(k + 1) = a(k)v(k),

{

k ∈ ℕ0 ,

v(0) = u0 ,

(1.9)

and k−1

u(k) = ∑ g j (k)b(j),

k ∈ ℕ0

(1.10)

j=0

solves the equation μ

{

∇∗ u(k + 1) = a(k)u(k) + b(k),

k ∈ ℕ0 ,

u(0) = 0.

(1.11)

Here we note that g j (k) = 0 for 0 ≤ k ≤ j and g j (j + 1) = 1. So if we set g̃ j (k) := g j (k + 1),

k = −1, 0, . . . ,

formula (1.10) becomes k−1

u(k) = ∑ g̃ j (k − 1)b(j),

k ∈ ℕ0 ,

(1.12)

j=0

while (1.8) gives μ

∇∗ g̃ j (k + 1) = a(k + 1)g̃ j (k),

{

g̃ j (j) = 1,

g̃ j (k) = 0,

k ≥ j, −1≤ k < j

μ

for j ∈ ℕ0 . Using ∇∗ g̃ j (j) = 1, we can directly verify that (1.12) solves (1.8). Thus there are two different ways, (1.10) and (1.12), how to define a solution u of (1.11). The following lemma uses (1.10), and concludes the above arguments. Lemma 1.8. The initial value problem μ

∇∗ u(k + 1) = a(k)u(k) + b(k), { u(0) = u0

k ∈ ℕ0 ,

has a solution k−1

u(k) = u0 h(k) + ∑ g j (k)b(j), j=0

k ∈ ℕ0 .

10 | 1 Fractional Difference Equations

Proof. The considered problem is decomposed to a homogeneous equation with a non-trivial initial condition, of the form (1.9), and a non-homogeneous equation with a zero initial condition, of the form (1.11). Consequently, the superposition principle is applied. The rest of this subsection is devoted to the case of a constant function a(k). Proposition 1.9. Let μ ∈ (0, 1), a, u0 ∈ ℝ and u fulfill μ

∇∗ u(k + 1) = au(k),

{

k ∈ ℕ0 ,

(1.13)

u(0) = u0 .

Then u has the form j

k

u(k) = u0 [1 + ∑

∏

∑

j=1 i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l ≤k−j

aΓ(i l + μ) ], Γ(μ)Γ(i l + 1)

k ∈ ℕ0 .

(1.14)

Proof. First, we apply Lemma 1.6 to get a corresponding fractional sum equation k−1

u(k) = u0 + ∑ A μ (k − 1, j)au(j),

k ∈ ℕ0 .

(1.15)

j=0

Note that A μ (k, j) = A μ (k + α, j + α) =

Γ(k − j + μ) Γ(μ)Γ(k − j + 1)

for any α ∈ [−j, ∞), 0 ≤ j ≤ k. For simplicity, we denote B μ (k − j) = aA μ (k, j) for 0 ≤ j ≤ k, i.e., aΓ(n + μ) , n ∈ ℕ0 . B μ (n) = Γ(μ)Γ(n + 1) Consequently, k−1

u(k) = u0 + ∑ B μ (k − 1 − j)u(j),

k ∈ ℕ0 .

(1.16)

j=0

We claim that then j

k

u(k) = u0 [1 + ∑

∑

∏ B μ (i l )],

k ∈ ℕ0 ,

(1.17)

j=1 i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l ≤k−j

which is the statement of the proposition. We prove the claim by induction with respect to k. If k = 0, then due to the empty sum property we have u(0) = u0 in both (1.16) and (1.17). Now, let us assume that (1.17) holds for 0, 1, . . . , k. We show that it is true also for k + 1. Using (1.16) and the

1.1 Fractional difference Gronwall inequalities | 11

inductive hypothesis, we have k

u(k + 1) = u0 + ∑ B μ (k − j)u(j) j=0 j

k

q

= u0 + ∑ B μ (k − j)u0 [1 + ∑

∏ B μ (i l )]

∑

q=1 i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l ≤j−q

j=0 k

j

k

q

= u0 [1 + ∑ B μ (k − j) + ∑ B μ (k − j) ∑ ∑ ∏ B μ (i l )]. q=1 i1 ,i2 ,...,i q ≥0 l=1 j=0 j=0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ q ∑l=1 i l ≤j−q =:S1 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ =:S2

Changing k − j + 1 → j, we get k+1

1

1

j

1

S1 = ∑ B μ (j − 1) = ∑ ∑ ∏ B μ (i j ) = ∑ j=1

j=1 0≤i j ≤k l=1

∑

∏ B μ (i l ).

j=1 i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l ≤k+1−j

Similarly in S2 : k+1−j

k+1

q

S2 = ∑ B μ (j − 1) ∑ q=1

j=1

∏ B μ (i l )

∑

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l ≤k+1−j−q

k+1 k+1−j

q

= ∑ ∑ B μ (j − 1) j=1 q=1

∏ B μ (i l ).

∑

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l +j−1≤k−q

k+1−j

k+1−q

k+1 Now we switch the sums ∑k+1 j=1 ∑q=1 = ∑q=1 ∑j=1 for q = k + 1. Hence

. Note that the second sum is empty q

k k+1−q

S2 = ∑ ∑ B μ (j − 1) q=1 j=1

∑

∏ B μ (i l ).

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l +j−1≤k−q

Note that j − 1 takes values 0, 1, . . . , k − q. So we can denote i q+1 = j − 1 and merge the last two sums to obtain q+1

k

S2 = ∑

∑

∏ B μ (i l ).

q=1 i1 ,i2 ,...,i q+1 ≥0 l=1 q+1 ∑l=1 i l ≤k−q

Finally, we change q + 1 → q: q

k+1

S2 = ∑

∑

∏ B μ (i l ).

q=2 i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l ≤k+1−q

After summing S1 and S2 , equation (1.17) is obtained for k + 1.

12 | 1 Fractional Difference Equations

Now we present an alternative proof without using the induction principle. Alternative proof of Proposition 1.9. If a = 0, then by (1.15), the statement is proved. From now on, we assume that a ≠ 0. By using the formula lim

n→∞

Γ(n + μ) 1−μ n =1 Γ(n + 1)

(1.18)

(see [172, Problem 7, p. 15]), it is clear that 1

= 1.

n

lim supn→∞ √|B μ (n)| So the power series ∞

∆ μ (x) = ∑ B μ (n)x n n=0

has the radius of convergence 1. Furthermore, (1.18) gives that 1 ∞ ∑ B μ (n) = +∞. a n=0 Using

B μ (n) a

> 0, we see that 1 lim ∆ μ (x) = +∞. a x→1−

(1.19)

We know that the sequence {|B μ (n)|}∞ n=0 is decreasing (see the proof of Lemma 1.17) with limn→∞ B μ (n) = 0 (see (1.18)). So the Leibniz criterion implies the convergence of the series ∞

∑ B μ (n)(−1)n , n=0

and the Abel theorem [269, p. 9] implies ∞

∞

n=0

n=0

lim + ∆ μ (x) = lim− ∑ B μ (n)(−1)n x n = ∑ B μ (n)(−1)n = ∆ μ (−1).

x→−1

x→1

Next, Lemma 1.17 implies |u(k)| ≤ |u0 |(1 + |a|)k , Hence

1 n

lim supn→∞ √|u(n)|

k ∈ ℕ0 . ≥

1 , 1 + |a|

thus the power series ∞

U(x) = ∑ u(n)x n n=0

(1.20)

1.1 Fractional difference Gronwall inequalities | 13

has the radius of convergence R U greater than or equal to 1 with 0 ≠ |x| < 1+|a| . Then using (1.16), we derive ∞

∞

i=0

j=0

∞

U(x)∆ μ (x) = ( ∑ u(i)x i )( ∑ B μ (j)x j ) = ∑

1 1+|a| .

∑

Consequently, we start

u(i)B μ (j)x k−1

k=1 i+j=k−1

∞ k−1

∞

k=1 j=0

k=1

= ∑ ∑ u(j)B μ (k − 1 − j)x k−1 = ∑ (u(k) − u0 )x k−1 =

∑∞ k=0

− u(0)

u(k)x k x

−

u0 U(x) u0 = − . 1−x x x(1 − x)

(1.21)

Solving (1.21), we obtain u0 . (1 − x)(1 − x∆ μ (x))

U(x) =

(1.22)

From (1.22) we obtain ∞

∞

∞

∞

n=0

j=0

i=0

k=0

i

∑ u(n)x n = u0 ( ∑ x j )( ∑ x i ( ∑ B μ (k)x k ) ).

(1.23)

By expanding the right-hand side of (1.23) and comparing the powers of x, we immediately get (1.17). If u0 = 0 then U(x) = 0. So we suppose that u0 ≠ 0. Next, using (1.18), we see that a as n → ∞. B μ (n) ∼ Γ(μ)n1−μ So by results of [269, pp. 224–225], we have ∞

(1 − x)∆ μ (x) ∼ (1 − x) ∑ n=1

a x n ∼ a(1 − x)1−μ Γ(μ)n1−μ

as x → 1− ,

so lim (1 − x)∆ μ (x) = 0,

x→1−

while we recall (1.19). Then clearly lim |U(x)| = +∞

x→1−

due to (1.22), thus the radius of convergence R U of U(x) is less than or equal to 1, and we get 1 ≤ R U ≤ 1. (1.24) 1 + |a| On the other hand, we know |∆ μ (x)| ≤ |a|[1 + ≤ |a|[1 +

1 ∞ Γ(n + μ) n |x| ] ∑ Γ(μ) n=1 Γ(n + 1) 1 ∞ |x|n 1 Li1−μ (|x|)] ] = |a|[1 + ∑ Γ(μ) n=1 n1−μ Γ(μ)

14 | 1 Fractional Difference Equations

where we applied the estimation of the ratio of gamma functions 1 Γ(n + μ) ≤ , Γ(n + 1) n1−μ

n ∈ ℕ, μ ∈ (0, 1)

from [118], and used the notation ∞

Liν (x) = ∑ n=1

∞

xn 1 xt ν−1 dt, = ∫ t ν n Γ(ν) e − x

x, ν ∈ (0, 1)

0

for the polylogarithm function [173, Section 7.12]. Consequently, |1 − x∆ μ (x)| ≥ 1 − |x| |∆ μ (x)| ≥ 1 − F(|x|) for F(x) = x|a|[1 +

(1.25)

1 Li1−μ (x)]. Γ(μ)

We note that F is increasing on (0, 1). The right-hand side of (1.25) is positive if and 1 only if |x| < F −1 (1). Note that F −1 (1) > 1+|a| . To see this, we estimate Li1−μ ( i.e.,

∞ n 1 1 Γ(μ) 1 < , ) < ∑( ) = 1 + |a| 1 + |a| |a| |a| n=1

1 |a| 1 1 Li1−μ ( [1 + )] = F( ) < 1. 1 + |a| Γ(μ) 1 + |a| 1 + |a|

This improves (1.24) to F −1 (1) ≤ R U ≤ 1.

(1.26)

Summarizing the above arguments, we obtain the next result. Proposition 1.10. The solution u(n) of initial value problem (1.13) with u0 ≠ 0 fulfills n

lim sup √|u(n)| = n→∞

1 RU

(1.27)

for R U satisfying (1.26). To get a better result, we note that if a > 0 then x∆ μ (x) is increasing on [0, 1) from 0 to +∞. So for any a > 0, there is a unique 1 > r a,μ > F −1 (1) solving equation 1 = r a,μ ∆ μ (r a,μ ). Then |1 − x∆ μ (x)| > 0 for any |x| < r|a|,μ . Hence (1.26) is improved to r|a|,μ ≤ R U ≤ 1.

(1.28)

Note that estimation (1.26) as well as (1.28) gives a better estimate on asymptotic property (1.27) than (1.20) derived from Lemma 1.17. Moreover, Proposition 1.10 yields the next corollary.

1.1 Fractional difference Gronwall inequalities | 15

Corollary 1.11. If the solution u(n) of (1.13) with u0 ≠ 0 satisfies u(n) → 0 as n → ∞, then the rate of convergence is slower than any exponential one, i.e., there are no constants c1 > 0 and ϖ ∈ (0, 1) so that |u(n)| ≤ c1 ϖ n for any n ∈ ℕ0 . Proof. On the contrary, if c1 > 0 and ϖ ∈ (0, 1) are such that |u(n)| ≤ c1 ϖ n for each n ∈ ℕ0 , then 1 1 RU = >1 ≥ n ϖ √ lim sup |u(n)| n→∞

which contradicts (1.28). On the other hand, if a > 0 and u0 ≠ 0, then 1 lim U(x) = +∞, u0 x→r−a,μ so R U = r a,μ , and thus {u(n)}n∈ℕ0 is unbounded satisfying (1.27). A related result is derived in [23, Theorem 5.1]. Remark 1.12. Taking limit μ → 1− in (1.13) gives ∇u(k + 1) = au(k),

{

k ∈ ℕ0

u(0) = u0

which has a solution u(k) = u0 (1 + a)k . On the other hand, from (1.14), we get k

lim− u(k) = u0 [1 + ∑ a j

μ→1

j=1

∑

1]

i1 ,i2 ,...,i j ≥0 j

∑l=1 i l ≤k−j k

k−j

j=1

k−j

k

= u0 [1 + ∑ a j ∑

∑

1] = u0 [1 + ∑ a j ∑ (

m=0 i1 ,i2 ,...,i j ≥0

j=1

m=0

m+j−1 )] j−1

j

∑l=1 i l =m k k = u0 [1 + ∑ ( )a j ] = u0 (1 + a)k . j j=1

So we call the bracket in (1.14) the generalized binomial. Note that linear Riemann– Liouville fractional difference equations are solved in [23] leading to discrete MittagLeffler functions. Next, we derive a formula for a Green function satisfying (1.8) with constant a. Proposition 1.13. Let μ ∈ (0, 1) and a ∈ ℝ. Then for any i ∈ ℕ0 , the Green function {g ic (k)}k∈ℕ0 satisfying μ

∇∗ g ic (k + 1) = ag ic (k) + δ i (k), { g ic (0) = 0

k ∈ ℕ0 ,

16 | 1 Fractional Difference Equations

has the form j

k−i

g ic (k) = ∑ a j−1

∑

j=1

∏

i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l =k−i−j

Γ(i l + μ) , Γ(μ)Γ(i l + 1)

k ∈ ℕ0 .

(1.29)

Proof. Let i ∈ ℕ0 be arbitrary and fixed. As in the proof of Proposition 1.9, applying Lemma 1.6, one can see that k−1

g ic (k) = ∑ (B μ (k − 1 − j)g ic (j) + A μ (k − 1, j)δ i (j)),

k ∈ ℕ0 .

(1.30)

j=0

In particular, g ic (0) = 0 which agrees with (1.29). Let (1.29) be valid at 0, 1, . . . , k. Then by (1.30), j−i

k

q

g ic (k + 1) = ∑ B μ (k − j) ∑ a q−1

k

∏ C μ (i l ) + ∑ C μ (k − j)δ i (j),

∑

q=1

j=0

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l =j−i−q ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

(1.31)

j=0

=:S

where C μ (k − j) = A μ (k, j). There are three possible cases. If i ∉ {0, 1, . . . , k}, then g ic (k + 1) = 0 due to the empty sum property. The same is true for (1.29). If i = k, then S = 0 and (1.31) implies k

g ic (k + 1) = ∑ C μ (k − j)δ i (j) = C μ (0) = 1. j=0

The same holds for (1.29). Finally, if i ∈ {0, 1, . . . , k − 1}, then k

j−i

q

S = ∑ ∑ a q C μ (k − j) j=i+1 q=1 k−i

∏ C μ (i l )

∑

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l =j−i−q q

k

= ∑ ∑ a q C μ (k − j) q=1 j=q+i

∏ C μ (i l ).

∑

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l =j−i−q

Note that k − j takes the values 0, 1, . . . , k − q − i. So we denote i q+1 = k − j and write q+1

k−i

S = ∑ aq q=1

∏ C μ (i l ).

∑

i1 ,i2 ,...,i q+1 ≥0 l=1 q+1 ∑l=1 i l =k−i−q

Substituting q + 1 → q, we get q

k+1−i

S = ∑ a q−1 q=2

and (1.31) becomes (1.29).

∑

∏ C μ (i l )

i1 ,i2 ,...,i q ≥0 l=1 q ∑l=1 i l =k+1−i−q

1.1 Fractional difference Gronwall inequalities | 17

1.1.4 Fractional difference inequalities First, we recall a discrete Gronwall lemma (see e.g. [236, Lemma 1.4.2]). Lemma 1.14. Let u, p, q be real functions defined on ℕ0 , and let p be nonnegative. For any n ∈ ℕ0 , if k−1

u(k) ≤ u(0) + ∑ (p(j)u(j) + q(j)) for all k = 0, 1, . . . , n, j=0

then k−1

k−1

k−1

u(k) ≤ u(0) ∏(1 + p(j)) + ∑ q(j) ∏ (1 + p(l)) for all k = 0, 1, . . . , n. j=0

j=0

l=j+1

We provide the following fractional difference inequalities of Gronwall type. Lemma 1.15. Let μ ∈ (0, 1), let u, a, b be real functions defined on ℕ0 , and let a be nonnegative. For any n ∈ ℕ, if μ

∇∗ u(k + 1) ≤ a(k)u(k) + b(k) for all k = 0, 1, . . . , n − 1, then k−1

u(k) ≤ u(0)h(k) + ∑ g j (k)b(j) for all k = 0, 1, . . . , n, j=0

where h solves (1.7) and g j is a solution of (1.8). Proof. Denote by v(k) a solution of the initial value problem μ

∇∗ v(k + 1) = a(k)v(k) + b(k),

{

k ∈ ℕ0 ,

v(0) = u(0).

Then by Lemma 1.8, k−1

v(k) = u(0)h(k) + ∑ g j (k)b(j) for k = 0, 1, . . . , n. j=0

So it is sufficient to prove u(k) ≤ v(k) for each k = 0, 1, . . . , n. Clearly, it is true for k = 0. Let it be valid for 1, 2, . . . , k. Then by Corollary 1.7 and Lemma 1.6, k

u(k + 1) ≤ u(0) + ∑ A μ (k, j)(a(j)u(j) + b(j)) j=0 k

≤ u(0) + ∑ A μ (k, j)(a(j)v(j) + b(j)) = v(k + 1). j=0

Hence the lemma is proved.

18 | 1 Fractional Difference Equations Lemma 1.16. Let μ ∈ (0, 1), a > 0, and let u, b be real functions defined on ℕ0 . For any n ∈ ℕ, if μ ∇∗ u(k + 1) ≤ au(k) + b(k) for all k = 0, 1, . . . , n − 1, then j

k

u(k) ≤ u(0)[1 + ∑

∑

∏

j=1 i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l ≤k−j k−1

j

k−i

+ ∑ b(i) ∑ a j−1 i=0

aΓ(i l + μ) ] Γ(μ)Γ(i l + 1)

j=1

∏

∑

i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l =k−i−j

Γ(i l + μ) , Γ(μ)Γ(i l + 1)

k = 0, 1, . . . , n.

Proof. The lemma can be proved as Lemma 1.15 using Propositions 1.9 and 1.13, and Lemma 1.8. Next, we state an estimation independent of μ. Lemma 1.17. If the assumptions of Lemma 1.15 are satisfied, then k−1

k−1

k−1

u(k) ≤ u(0) ∏(1 + a(j)) + ∑ b(j) ∏ (1 + a(l)) j=0

j=0

for all k = 0, 1, . . . , n.

l=j+1

Proof. Corollary 1.7 is applied to obtain the sum inequality k−1

u(k) ≤ u(0) + ∑ A μ (k − 1, j)(a(j)u(j) + b(j))

for all k = 0, 1, . . . , n.

(1.32)

j=0

Let us define a function g as g(x) =

Γ(x + μ) Γ(x + 1)

for x ≥ 0.

We claim that g is decreasing on [0, ∞). Clearly, g is positive on the whole domain. Next, g󸀠 (x) = (Ψ(x + μ) − Ψ(x + 1))g(x) where Ψ is a logarithmic derivative of the gamma function (psi function or digamma function, see e.g. [172]) which satisfies ∞

1 > 0, (x + j)2 j=0

Ψ󸀠 (x) = ∑

x > 0;

see [2, formula 6.4.10]. Thus indeed, g 󸀠 (x) < 0 for x ≥ 0. Consequently, since A μ (k − 1, j) =

g(k − 1 − j) , Γ(μ)

function A μ (⋅, j) is decreasing on [j, ∞), and A μ (k − 1, ⋅) is increasing on [1, k − 1]. So we get two estimates A μ (k − 1, j) ≤ A μ (j, j),

A μ (k − 1, j) ≤ A μ (k − 1, k − 1)

1.1 Fractional difference Gronwall inequalities | 19

Fig. 1.1. Solution of (1.33) (black with points) and estimate (1.34) (red with solid boxes).

for 1 ≤ j ≤ k − 1. It does not matter which one we use, as both give the same result A μ (j, j) = A μ (k − 1, k − 1) = 1. Hence by (1.32), k−1

u(k) ≤ u(0) + ∑ (a(j)u(j) + b(j)) for all k = 0, 1, . . . , n. j=0

Finally, the statement follows by Lemma 1.14. Remark 1.18. Lemmas 1.15, 1.16 and 1.17 remain valid when ≤ is replaced with ≥. Example 1.19. Consider the following initial value problem: 1

{∇∗2 u(k + 1) = u(k), { u(0) = 1. {

k ∈ ℕ0 ,

(1.33)

By Lemma 1.6, k−1

u(k) = u(0) + ∑ A 12 (k − 1, j)u(j),

k ∈ ℕ0 .

j=0

In Figure 1.1, one can compare values of the solution of (1.33) with its estimate u(k) ≤ 2k ,

k ∈ ℕ0

(1.34)

obtained from Lemma 1.17. Because of the simple form of (1.33), its solution can be explicitly calculated using Proposition 1.9 to get j

k

u(k) = 1 + ∑

∑

∏

j=1 i1 ,i2 ,...,i j ≥0 l=1 j ∑l=1 i l ≤k−j

Γ(i l + 12 ) , √πΓ(i l + 1)

k ∈ ℕ0 .

20 | 1 Fractional Difference Equations

Remark 1.20. We recall that Gronwall type inequalities are derived in [1, 24, 102] for fractional difference inequalities different from those discussed in Section 1.1. More general inequalities are presented in [18].

1.2 S-asymptotically periodic solutions In this section, it is shown that a fractional difference equation with a periodic righthand side can not possess a periodic solution but it can have an S-asymptotically periodic solution. Sufficient conditions are proved for the existence of a unique Sasymptotically periodic solution. An example is also given illustrating the obtained results. This section is based on [90] (see also Section 3.1).

1.2.1 Introduction In the present subsection, we consider the following initial value problem: μ

{

∆∗ u(k) = f(k),

k ∈ ℕ1−μ ,

u(0) = u0

(1.35)

μ

with Caputo like fractional difference operator ∆∗ of order 0 < μ < 1. Here and after, ℕa with a ∈ ℝ denotes the shifted set of positive integers, i.e., ℕa = {a, a + 1, a + 2, . . . }. We shortly denote ℕ := ℕ1 . We answer the question if there are any N-periodic solutions u : ℕ0 → X (see [20, 22] for the domain of u) of (1.35) for a given N-periodic function f : ℕ1−μ → X, where X is a Banach space endowed with a norm | ⋅ |. Later, we study the possible S-asymptotic property of the solution of (1.35). In Section 1.2.4, we prove sufficient conditions for the existence of an S-asymptotically N-periodic solution u : ℕ0 → X of a nonlinear initial value problem μ

{

∆∗ u(k) = f(k, u(k − 1 + μ)),

k ∈ ℕ1−μ ,

u(0) = u0

(1.36)

for f : ℕ1−μ × X → X such that f(k, ⋅) ∈ C(X, X) for all k ∈ ℕ1−μ . Finally, for X = ℝn , we give a result on the existence of a unique S-asymptotically N-periodic solution of (1.36) with an estimation of this solution. Throughout the whole Section 1.2, we assume the property of empty sum and empty product, i.e., b

b

∑ f(k) = 0,

∏ f(k) = 1

k=a

k=a

if a > b.

(1.37)

1.2 S-asymptotically periodic solutions | 21

1.2.2 Preliminaries In this subsection, we recall some definitions (see e.g. [17, 20, 55]), a known result from the theory of fractional difference equations and some important lemmas on gamma function. Definition 1.21. Let ν ∈ ℝ. The factorial function is defined as {0 t(ν) = { Γ(t+1) { Γ(t+1−ν)

if t + 1 − ν ∈ {. . . , −2, −1, 0}, otherwise.

Definition 1.22. Let a ∈ ℝ, ν > 0. The ν-th fractional sum of function f defined on ℕa is given by 1 k−ν ∆−ν f(k) = ∑ (k − σ(j))(ν−1) f(j) Γ(ν) j=a for any k ∈ ℕa+ν , where σ(j) = j + 1. Definition 1.23. Let a ∈ ℝ, μ > 0, m − 1 < μ < m for some m ∈ ℕ, ν := m − μ and let function f be defined on ℕa . The μ-th fractional Caputo like difference of f is defined as 1 k−ν μ ∆∗ f(k) = ∆−ν (∆ m f(k)) = ∑ (k − σ(j))(ν−1) (∆ m f)(j) Γ(ν) j=a for any k ∈ ℕa+ν . Here ∆ m is the m-th forward difference operator m m (∆ m f)(k) = ∑ ( )(−1)m−j f(k + j). j j=0

Definition 1.24. The R a -transform (Laplace transform on the time scale of integers) of a function f is defined as ∞

R a (f(k))(z) = ∑ ( k=a

1 k+1 ) f(k). z+1

Lemma 1.25 (see [20]). Let μ ∈ ℝ \ {. . . , −1, 0} and let f be defined on ℕa . Then R a+μ (∆−μ f(k))(z) =

1 R a (f(k))(z). zμ

Remark 1.26. (i) The above defined Caputo like difference operator is different from the Riemann–Liouville like difference given by ∆ μ f(k) = ∆ m (∆−ν f(k)) (cf. [17, 22]). (ii) The R-transform is closely related to a more common Z-transform (see e.g. [153, 229]) given by ∞ f(k) Z(f(k))(z) = ∑ k z k=0

22 | 1 Fractional Difference Equations

as can be seen from 1 Z(f(k))(z + 1), z+1 Z(f(k))(z) = zR0 (f(k))(z − 1).

R0 (f(k))(z) =

Definition 1.27. A function f : ℕa → X is called S-asymptotically N-periodic if there exists N ∈ ℕ such that limk→∞ (f(k + N) − f(k)) = 0. In this case, N is called asymptotic period of f . The set of all bounded and S-asymptotically N-periodic functions u : ℕa → X is denoted by SAPN,ℕa (X). It is a Banach space with a norm ‖u‖∞ := supk∈ℕa |u(k)|. For simplicity, we set SAPN (X) = SAPN,ℕ0 (X). The following lemmas play an important role in estimating difficult sums in this section. Lemma 1.28 (see [231]). For 0 < r < 1, n ∈ ℕ it holds Γ(n + r) n r − (n − 1)r ≤ . Γ(n + 1) r Lemma 1.29 (see [59]). For n ∈ ℕ it holds √ (n +

n + 12 1 1 Γ(n + 1) < + . )< 4 32(n + 1) Γ(n + 21 ) √ n + 3 + 1 4 32n+48

1.2.3 Non-existence of periodic solutions In this subsection, we prove the non-existence of periodic orbits in non-homogeneous initial value problem (1.35). This part is motivated by [284] where an analogous problem is studied for fractional differential equations (see also Section 3.1). First of all, we split the initial value problem (1.35) to a non-homogeneous equation with a homogeneous condition and a homogeneous equation with a non-homogeneous condition. More precisely, u = v + w and we have two initial value problems: μ

{

∆∗ v(k) = f(k), v(0) = 0, μ

{

k ∈ ℕ1−μ ,

∆∗ w(k) = 0,

k ∈ ℕ1−μ ,

w(0) = u0 .

(1.38)

(1.39)

Clearly, (1.39) has a constant solution w(k) = u0 for each k ∈ ℕ0 . Now we prove the first main result of this section. Theorem 1.30. For any N ∈ ℕ, there is no non-constant N-periodic solution of problem (1.35) with an N-periodic function f .

1.2 S-asymptotically periodic solutions | 23

Proof. From [69] we know that problem (1.38) is equivalent to the next summation equation v(k) = ∆−μ f(k), k ∈ ℕ0 , (1.40) i.e., k−μ

v(k) =

1 ∑ (k − σ(j))(μ−1) f(j), Γ(μ) j=1−μ

k ∈ ℕ0

assuming property (1.37) of empty sum. Taking the R0 -transform of (1.40) yields ∞

R0 (v(k))(z) = R0 (∆−μ f(k))(z) = ∑ ( k=0

1 k+1 −μ ) ∆ f(k). z+1

Since k−μ

1 󵄨 =0 ∆−μ f(k)󵄨󵄨󵄨k=0 = [ ∑ (k − σ(j))(μ−1) f(j)] Γ(μ) j=1−μ k=0 due to the empty sum property, we get R0 (v(k))(z) = R1 (∆−μ f(k))(z). By Lemma 1.25, R1 (∆−μ f(k))(z) =

1 R1−μ (f(k))(z), zμ

i.e., ∞

∑( k=0

1 ∞ 1 k+1 1 k+1 ) v(k) = μ ∑ ( ) f(k), z+1 z k=1−μ z + 1

z > 0.

Now, let N ∈ ℕ be fixed. We apply the assumption of N-periodicity of function u on ℕ0 and f on ℕ1−μ . Then of course, v is N-periodic. Therefore, N−1

∞

∑ v(k) ∑ ( j=0

k=0

N−μ

∞ 1 k+1+jN 1 k+1+jN 1 = μ ∑ f(k) ∑ ( , ) ) z+1 z k=1−μ z+1 j=0

so

N−μ

1 k+1 (z + 1)N N−1 1 k+1 1 (z + 1)N v(k) = ) ( ) f(k) ( ∑ ∑ z μ (z + 1)N − 1 k=1−μ z + 1 (z + 1)N − 1 k=0 z + 1 and, consequently, N−1

zμ ∑ ( k=0

N−μ

1 k+1 1 k+1 ) v(k) = ∑ ( ) f(k), z+1 z+1 k=1−μ

z > 0.

(1.41)

Putting N−1

P(z) = ∑ ( k=0

1 k+1 ) v(k), z+1

N−μ

Q(z) = ∑ ( k=1−μ

1 k+1 ) f(k), z+1

equation (1.41) reads z μ P(z) = Q(z),

z > 0.

(1.42)

24 | 1 Fractional Difference Equations

Using ∞

ln(1 + z) = ∑ (−1)j+1 j=1

(

zj , j

|z| < 1,

1 k+1 = e−(k+1) ln(1+z) , ) z+1

functions P k (z) and Q k (z) can be analytically extended to the open unit disc in ℂ, so there are their Taylor series ∞

P(z) = ∑ p j z j ,

∞

Q(z) = ∑ q j z j ,

j=0

p j , q j ∈ X.

j=0

Then (1.42) gives ∞

∞

j=0

j=0

zμ ∑ pj zj = ∑ qj zj ,

0 < z < 1.

(1.43)

Letting z → 0+ , we get q0 = 0. But then (1.43) implies ∞

∞

j=0

j=1

∑ p j z j = ∑ q j z j−μ ,

0 < z < 1.

Letting z → 0+ , we get p0 = 0. So following this argument step by step, we derive pj = qj = 0

for all j ∈ ℕ0 .

Hence P(z) = Q(z) = 0,

0 < z < 1,

which is equivalent to N−1

∑( k=0

1 k+1 ) v(k) = 0, z+1

N−μ

∑ ( k=1−μ

1 k+1 ) f(k) = 0, z+1

Then clearly v(k) = 0,

k = 0, 1, . . . , N − 1,

f(k) = 0,

k = 1 − μ, 2 − μ, . . . , N − μ.

Consequently, u(k) = u0 for each k ∈ ℕ0 .

0 < z < 1.

1.2 S-asymptotically periodic solutions | 25

Now, we investigate the potentially S-asymptotically periodic solutions of initial value problem (1.35). Theorem 1.31. Let N ∈ ℕ. If f is N-periodic on ℕ1−μ , then a solution u of (1.35) is S-asymptotically N-periodic on ℕ0 . Moreover, for f ̄ :=

N−μ

1 ∑ f(j) N j=1−μ

the following holds: (i) If f ̄ = 0, then ‖u‖∞ ≤ |u0 | + N‖f ‖∞ with ‖f ‖∞ = supr∈ℕ1−μ |f(r)|. (ii) If f ̄ ≠ 0, then |u(k)| → ∞ as k → ∞. Proof. Using the summation equation k−μ

u(k) = u0 +

1 ∑ (k − σ(j))(μ−1) f(j), Γ(μ) j=1−μ

k ∈ ℕ0 ,

we get k−μ+N

u(k + N) − u(k) =

k−μ

1 ( ∑ (k + N − σ(j))(μ−1) f(j) − ∑ (k − σ(j))(μ−1) f(j)) Γ(μ) j=1−μ j=1−μ k−μ

k−μ

=

1 ( ∑ (k − σ(j))(μ−1) f(j + N) − ∑ (k − σ(j))(μ−1) f(j)) Γ(μ) j=1−μ−N j=1−μ

=

1 Γ(μ)

−μ

∑

(k − σ(j))(μ−1) f(j + N)

j=1−μ−N

for any k ∈ ℕ0 . Hence, |u(k + N) − u(k)| ≤ =

‖f ‖∞ Γ(μ) ‖f ‖∞ Γ(μ)

−μ

∑ j=1−μ−N −μ

∑ j=1−μ−N

󵄨 󵄨󵄨 󵄨󵄨(k − σ(j))(μ−1) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 Γ(k − j) 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 Γ(k − j + 1 − μ) 󵄨󵄨

as k − j + 1 − μ ∈ ℕ for j = 1 − μ − N, 2 − μ − N, . . . , −μ. Since 0 < μ ≤ k + μ ≤ k − j ≤ k + μ + N − 1, k − j − μ + 1 > 0 and 1 − μ > 0, we have 󵄨󵄨 󵄨󵄨 Γ(k − j) B(k − j, 1 − μ) B(k + μ, 1 − μ) Γ(k − j) 󵄨󵄨 󵄨󵄨 = = ≤ , 󵄨󵄨 󵄨 Γ(1 − μ) Γ(1 − μ) 󵄨 Γ(k − j − μ + 1) 󵄨󵄨 Γ(k − j − μ + 1) where B(x, y) is the beta function [172]. Here we used that B(x, y) is decreasing in its variables. So, we obtain B(k + μ, 1 − μ) Γ(μ)Γ(1 − μ) N‖f ‖∞ sin(πμ) = B(k + μ, 1 − μ) , π

|u(k + N) − u(k)| ≤ N‖f ‖∞

k ∈ ℕ0 ,

26 | 1 Fractional Difference Equations

by the Euler reflection formula. Next, we have 1

B(k + μ, 1 − μ) = ∫ t k+μ−1 (1 − t)−μ dt 0 1

q

1

p

q

≤ √ ∫ t q(k+μ−1) dt √ ∫ t−pμ dt = √ 0

0 p p−1 .

for any k ∈ ℕ, p ∈ (1, 1μ ) and q = and then |u(k + N) − u(k)| ≤

p 1 1 √ q(k + μ − 1) + 1 1 − pμ

For instance, taking p =

N‖f ‖ sin(πμ) π

1+μ 2μ

√

2 1−μ

1+μ 1−μ

√

1+μ 2μ ,

we have q =

1−μ k + (−1 + k)μ + μ2

1+μ 1−μ

(1.44)

for any k ∈ ℕ. The proof of asymptotic N-periodicity of u is finished by (1.44). Next, from the above considerations, for k ∈ ℕ, we know k−μ

u(k) = u0 +

1 ∑ (k − σ(j))(μ−1) f(j) Γ(μ) j=1−μ k−μ

= u0 +

sin(πμ) ∑ B(k − j, 1 − μ)f(j) π j=1−μ

= u0 +

sin(πμ) sin(πμ) k ∑ a jk b jk ∑ B(k − j, 1 − μ)f ̄ + π π j=1 j=1−μ

k−μ

(1.45)

as k − j + 1 − μ > k − j > 0 for j = 1 − μ, 2 − μ, . . . , k − μ and where a jk := f(k + 1 − j − μ) − f ̄ ,

b jk := B(μ − 1 + j, 1 − μ).

We also have b1k > b2k > ⋅ ⋅ ⋅ > b kk > 0. Then the Abel lemma [269, p. 6] gives k

m k b1k ≤ ∑ a jk b jk ≤ M k b1k

(1.46)

j=1

for

r

r

m k := min ∑ a jk , r=1,...,k

j=1

M k := max ∑ a jk . r=1,...,k

j=1

Note that b1k = B(μ, 1 − μ) is independent of k and max{|m k |, |M k |} ≤ N‖f ‖∞ , since a j+Nk = a jk , |a jk | ≤ ‖f ‖∞ with ∑Nj=1 a jk = 0. Using (1.45) and (1.46), we get ‖u‖∞ ≤ |u0 | +

sin(πμ) NB(μ, 1 − μ)‖f ‖∞ . π

1.2 S-asymptotically periodic solutions | 27

Rewriting the beta function via gamma functions and using the reflection formula completes the proof of statement (i). If f ̄ ≠ 0, then we use B(k − j, 1 − μ) ≥ (k−j)c̄ 1−μ for a constant c̄ > 0. So we have k−μ

k−μ

∑ B(k − j, 1 − μ) ≥ ∑ j=1−μ

j=1−μ

k−1 c̄ c̄ = →∞ ∑ (k − j)1−μ j=0 (μ + j)1−μ

(1.47)

as k → ∞, since 0 < 1 − μ < 1. Using (1.45), (1.46) and (1.47) completes the proof of statement (ii). Remark 1.32. As p → 1μ , we have q → q

√

1 1−μ

and

1 1 − μ 1−μ →( ) . q(k + μ − 1) + 1 k

On the other hand, we know that B(k + μ, 1 − μ) ∼ Γ(1−μ) as k → ∞. This rate of conk1−μ vergence can be obtained by using Lemma 1.28 (see (1.55)).

1.2.4 Existence and uniqueness results Here we prove sufficient conditions for the existence of a unique S-asymptotically N-periodic solution u : ℕ0 → X for (1.36), i.e., u ∈ SAPN (X). Theorem 1.33. Let a, b : ℕ1−μ → ℝ+ be functions such that |f(k, u1 ) − f(k, u2 )| ≤ a(k)|u1 − u2 | { |f(k + N, u) − f(k, u)| ≤ b(k)(|u| + 1)

for all k ∈ ℕ1−μ , u1 , u2 ∈ X, for all k ∈ ℕ1−μ , u ∈ X.

(1.48)

If 󵄨󵄨 k−μ 󵄨󵄨 m := sup 󵄨󵄨󵄨󵄨 ∑ (k − σ(j))(μ−1) f(j, 0)󵄨󵄨󵄨󵄨 < ∞, 󵄨 󵄨 k∈ℕ j=1−μ k−μ

ρ := sup ∑ (k − σ(j))(μ−1) a(j) < Γ(μ), k∈ℕ j=1−μ

and k−μ

lim ∑ (k − σ(j))(μ−1) b(j) = 0,

k→∞

(1.49)

j=1−μ

then there exists a unique S-asymptotically N-periodic solution of initial value problem (1.36). Proof. Let us define operator F : SAPN (X) → {u : ℕ0 → X | ‖u‖∞ < ∞} as k−μ

(Fu)(k) := u0 +

1 ∑ (k − σ(j))(μ−1) f(j, u(j − 1 + μ)). Γ(μ) j=1−μ

(1.50)

28 | 1 Fractional Difference Equations

Clearly, u is a fixed point of F if and only if it is an S-asymptotically N-periodic solution of (1.36). First, we show that F(SAPN (X)) ⊂ SAPN (X). Let u ∈ SAPN (X). By Definition 1.27, for any ε > 0 there is N ε ∈ ℕ0 such that sup |u(k + N) − u(k)| < ε. k∈ℕN ε

Let ε > 0 be fixed and k ∈ ℕN ε . Then (Fu)(k + N) − (Fu)(k) k−μ+N

=

1 ∑ (k + N − σ(j))(μ−1) f(j, u(j − 1 + μ)) Γ(μ) j=1−μ k−μ

− =

1 ∑ (k − σ(j))(μ−1) f(j, u(j − 1 + μ)) Γ(μ) j=1−μ k−μ

1 Γ(μ)

(k − σ(j))(μ−1) f(j + N, u(j + N − 1 + μ))

∑ j=1−μ−N

k−μ

−

1 ∑ (k − σ(j))(μ−1) f(j, u(j − 1 + μ)) Γ(μ) j=1−μ k−μ

=

1 ∑ (k − σ(j))(μ−1) (f(j + N, u(j + N − 1 + μ)) − f(j, u(j + N − 1 + μ))) Γ(μ) j=1−μ k−μ

+

1 ∑ (k − σ(j))(μ−1) (f(j, u(j + N − 1 + μ)) − f(j, u(j − 1 + μ))) Γ(μ) j=1−μ

+

1 Γ(μ)

−μ

(k − σ(j))(μ−1) f(j + N, u(j + N − 1 + μ)).

∑ j=1−μ−N

Using (1.48), we estimate 󵄨󵄨 k−μ 󵄨 󵄨󵄨 ∑ (k − σ(j))(μ−1) (f(j + N, u(j + N − 1 + μ)) − f(j, u(j + N − 1 + μ)))󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 j=1−μ

k−μ

≤ (‖u‖∞ + 1) ∑ (k − σ(j))(μ−1) b(j) j=1−μ

and 󵄨󵄨 k−μ 󵄨 󵄨󵄨 ∑ (k − σ(j))(μ−1) (f(j, u(j + N − 1 + μ)) − f(j, u(j − 1 + μ)))󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 j=1−μ

k−μ

≤ ∑ (k − σ(j))(μ−1) a(j)|u(j + N − 1 + μ) − u(j − 1 + μ)| j=1−μ

(1.51)

1.2 S-asymptotically periodic solutions | 29

N ε −μ

= ∑ (k − σ(j))(μ−1) a(j)|u(j + N − 1 + μ) − u(j − 1 + μ)| j=1−μ k−μ

+

(k − σ(j))(μ−1) a(j)|u(j + N − 1 + μ) − u(j − 1 + μ)|

∑ j=N ε +1−μ N ε −μ

k−μ

j=1−μ

j=N ε +1−μ

< 2‖u‖∞ ∑ (k − σ(j))(μ−1) a(j) + ε

∑

(k − σ(j))(μ−1) a(j).

If N ε = 0, then the first sum is empty. Now suppose that N ε ∈ ℕ. Since k − j + 1 − μ ∈ ℕ for j = 1 − μ, 2 − μ, . . . , N ε − μ, one can write (k − σ(j))(μ−1) =

Γ(k − j) Γ(k − j + 1 − μ)

by Definition 1.21. Using formula Γ(x + 1) = xΓ(x) for x > 0, we expand Γ(N ε − j) ∏k−1 Γ(k − j) l=N ε (l − j) = . Γ(k − j + 1 − μ) Γ(N ε − j + 1 − μ) ∏k−1 (l − j + 1 − μ) l=N ε If k = N ε , the products are empty. In the other case, we set a = l − j ≥ μ > 0 and b = l − 1 + μ in the relation a≤b ⇒

a b ≤ , a+1−μ b+1−μ

and estimate k−1

∏ l=N ε

k−1 l−j l−1+μ ≤ ∏ =: P k−1 Nε . l − j + 1 − μ l=N l ε

Hence, (μ−1) (k − σ(j))(μ−1) ≤ P k−1 N ε (N ε − σ(j))

for j = 1 − μ, 2 − μ, . . . , N ε − μ. Moreover, k−μ

∑

k−μ

(k − σ(j))(μ−1) a(j) ≤ ∑ (k − σ(j))(μ−1) a(j) ≤ ρ.

j=N ε +1−μ

j=1−μ

Consequently, 󵄨󵄨 k−μ 󵄨 󵄨󵄨 ∑ (k − σ(j))(μ−1) (f(j, u(j + N − 1 + μ)) − f(j, u(j − 1 + μ)))󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 j=1−μ

N ε −μ

< 2‖u‖∞ P k−1 ∑ (N ε − σ(j))(μ−1) a(j) + ερ Nε j=1−μ

≤

(2‖u‖∞ P k−1 Nε

+ ε)ρ.

(1.52)

30 | 1 Fractional Difference Equations

We note that k

P kN ε = exp { ∑ ln(1 − l=N ε

k 1−μ 1−μ )} ≤ exp{− ∑ } → 0+ l l l=N ε

as k → ∞, since ln(1 + x) ≤ x for x > −1, and 0 < 1−μ l < 1 for each l ∈ ℕ. Recall that estimation (1.52) remains true for N ε = 0 when we define P0k := 0 for any k ∈ ℕ0 . Next, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨

−μ 󵄨󵄨 (k − σ(j))(μ−1) f(j + N, u(j + N − 1 + μ))󵄨󵄨󵄨󵄨 ≤ F ∗u ∑ (k − σ(j))(μ−1) 󵄨 j=1−μ−N j=1−μ−N −μ

∑

with F ∗u :=

max

j=1−μ,...,N−μ |v|≤‖u‖∞

|f(j, v)|.

(1.53)

Since k − j + 1 − μ ∈ ℕ for j = 1 − μ − N, 2 − μ − N, . . . , −μ, we get −μ

(k − σ(j))(μ−1) =

∑ j=1−μ−N

−μ

∑ j=1−μ−N

Γ(k − j) =: M k . Γ(k − j + 1 − μ)

(1.54)

Now, for k ∈ ℕ2 we apply Lemma 1.28 and a mean value theorem of differential calculus to obtain N−1

Mk = ∑ j=0

Γ(k + j + μ) N−1 (k + j)μ − (k + j − 1)μ ≤ ∑ Γ(k + j + 1) j=0 μ

(k + N − 1)μ − (k − 1)μ N = ≤ →0 μ (k − 1)1−μ

(1.55)

as k → ∞. So we obtain 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨

󵄨󵄨 (k − σ(j))(μ−1) f(j + N, u(j + N − 1 + μ))󵄨󵄨󵄨󵄨 ≤ F ∗u M k . 󵄨 j=1−μ−N −μ

∑

Combining (1.51), (1.52) and (1.56), we arrive at k−μ

|(Fu)(k + N) − (Fu)(k)|
0 and 0 < q+ ≤ q− < p+ ≤ p− , the function K(t, s) is nonnegative and it has the form either K(t, s) = k1 (t β − s β ) or K(t, s) = k2 (t, s) for some functions k1 , k2 specified later, and Γ is the Euler gamma function. Of course, we suppose a− M p− −p+ ≤ a+ , DOI 10.1515/9783110522075-002

b− M q− −q+ ≤ b+ .

(2.4)

38 | 2 Fractional Integral Equations

It is obvious that equation (2.1) or t

K(t, s)s γ−1 q β1−α x (t) = ]x (s)ds, ∫[ β Γ(α) (t − s β )1−α

t ∈ [0, T],

p

(2.5)

0

are special cases of equation (2.2); which of course all have trivial solutions. As an extension, we find general solutions in closed form of some Erdélyi–Kober type fractional integral equations (the special case of equation (2.5) if b = 0): φ m (x) = ax

β(m−N) N

α (EK I 0+;σ,η φ N )(x) + bx

βm N

,

x > 0,

(2.6)

α φ N denotes the Erdélyi– where α, b, σ ≥ 0, N ≠ 0 and η ∈ ℝ, and the symbol EK I0+;σ,η N Kober type fractional integrals [162] of the function φ , given by x

α (EK I0+;σ,η φ N )(x)

σx−σ(α+η) t ση+σ−1 φ N (t)dt , := ∫ Γ(α) (x σ − t σ )1−α

x > 0.

0

This section is based on [296].

2.1.2 Preliminaries Let M be the set M := {f ∈ C[0, T] | f(0) = 0} with the supremum norm ‖f ‖M := sup {|f(t)|}. 0 0 for all t > 0 and let Mq be defined as |f(t)| Mq := {f ∈ M | sup < ∞} 0 0 for all t ∈ (0, T] and k2 (t, s) ≥ 0 for all 0 ≤ s ≤ t ≤ T. (1) (n−1) (n) (n) (iii) k1 (0) = k1 (0) = ⋅ ⋅ ⋅ = k1 (0) = 0 and k1 (t) ≥ k1 (0) > 0 for all t ∈ (0, T]. For K(t, s) = k1 (t β − s β ) and n ≥ 1, we set (n)

𝕂low = k1 (0),

(n)

𝕂up = max {k1 (t)}. t∈[0,T]

(2.7)

For K(t, s) = k2 (t, s), we set n = 0 and 𝕂low = Kmin := min {k2 (t, s)}, 0≤s≤t≤T

𝕂up = Kmax := max {k2 (t, s)}. 0≤s≤t≤T

(2.8)

Similarly for K(t, s) = k1 (t β − s β ) and n = 0. The following lemma is an important estimate on the function K which will be used in the sequel. Lemma 2.2. The function K(⋅, ⋅) has the following estimate: 1 β 1 (t − s β )n 𝕂low ≤ K(t, s) ≤ (t β − s β )n 𝕂up , n! n!

0 ≤ s ≤ t ≤ T.

Proof. We only check the case of K(t, s) = k1 (t β − s β ) with n ≥ 1, since the other cases are trivial. Integrating n times step-by-step all sides of the inequality (n)

(n)

k1 (0) ≤ k1 (t) ≤ 𝕂up (1)

(n−1)

from 0 to t and using k1 (0) = k1 (0) = ⋅ ⋅ ⋅ = k1

(0) = 0, we immediately derive

t n (n) tn k1 (0) ≤ k1 (t) ≤ 𝕂up . n! n! Replacing t by (t β − s β ), we obtain the desired result. To end this subsection, we mention the following basic fact which will be used several times in the next subsection.

40 | 2 Fractional Integral Equations Lemma 2.3. Let λ, γ, μ and ν be constants such that λ > 0, Re(γ) > 0, Re(μ) > 0 and Re(ν) > 0. Then t

∫(t λ − s λ )ν−1 s μ−1 ds =

t λ(ν−1)+μ μ B( , ν), λ λ

t ∈ [0, +∞),

0

and t

γ

∫(t λ − s λ )ν−1 s γ−1 (s λ − a λ )μ ds ≥ a

where

(t λ − a λ )ν+ λ +μ−1 γ B( + μ, ν), λ λ

t ∈ [a, +∞), a ≥ 0,

1

B(ξ, η) = ∫ s ξ −1 (1 − s)η−1 ds,

Re(ξ) > 0, Re(η) > 0

0

is the well-known Euler beta function. Proof. The first result has been reported in [235] and [126, Formula 3.251]. We only verify the second inequality. In fact, for any t ∈ [a, +∞), a ≥ 0, we derive t

∫(t λ − s λ )ν−1 s γ−1 (s λ − a λ )μ ds a tλ

γ−1 1 1 = ∫ (t λ − u)ν−1 u λ (u − a λ )μ u λ −1 du λ

(set u = s λ )

aλ tλ

γ 1 = ∫ (t λ − u)ν−1 u λ −1 (u − a λ )μ du λ

aλ t λ −a λ

γ 1 = ∫ (t λ − a λ − z)ν−1 (a λ + z) λ −1 z μ du λ

(set u = a λ + z)

0 t λ −a λ

≥

γ 1 ∫ (t λ − a λ − z)ν−1 z λ +μ−1 du λ

0 γ

(t λ − a λ )ν+ λ +μ−1 γ = B( + μ, ν). λ λ

2.1.3 Existence and uniqueness of non-trivial solution in an order interval In this subsection, we will use the fixed point method to prove the existence and uniqueness of a non-trivial solution for equation (2.2) in an order interval.

2.1 Abel-type nonlinear integral equations | 41

For all t ∈ [0, T], we introduce the following functions: F(t) = At τ− ,

G(t) = Bt τ+ , 1

A=(

p+ −q− b− β−α 𝕂low β(n + α − 1)q− + γp+ B( , n + α)) , a+ Γ(α)n! β(p+ − q− ) 1

p− −q+ b+ β−α 𝕂up β(n + α − 1)q+ + γp− B( , n + α)) , a− Γ(α)n! β(p− − q+ ) β(n + α − 1) + γ β(n + α − 1) + γ τ− = , τ+ = , p+ − q− p− − q+

B=(

where 𝕂low and 𝕂up are defined in (2.7) or (2.8). Remark 2.4. Note β(n + α − 1)q∓ + γp± ≥ β(α − 1)q∓ + βp± = β(p± − q∓ ) + αβq∓ > 0, β(n + α − 1) + γ ≥ β(α − 1) + β = αβ > 0. Moreover, τ− ≥ τ+ . The following result is clear. Lemma 2.5. If AT τ− −τ+ ≤ B < MT −τ+ ,

(2.9)

then F(t) ≤ G(t) < M for all t ∈ [0, T]. Consequently, the order interval [F, G] ⊂ PM is well defined. Remark 2.6. If p+ = p− , q+ = q− and M = +∞, then (2.9) reads as a+ b+ 𝕂low ≥ a− b− 𝕂up which is satisfied, since (2.3) implies a+ ≥ a− > 0, b+ ≥ b− > 0 (see (2.4)) and, clearly, ̃ and g(x) = x q g(x) ̃ with 𝕂low /𝕂up ≤ 1. This case occurs for instance when h(x) = x p h(x) p > q and 0 < inf ℝ h(x) ≤ supℝ h(x) < ∞, 0 < inf ℝ g(x) ≤ supℝ g(x) < ∞. From now on, we suppose that all the above assumptions, (2.3), (2.4), (i)–(iii) and (2.9) hold. Lemma 2.7. Any solution x ∈ PM of (2.2), with M > x(t) > 0 for all t ∈ (0, T], satisfies x ∈ [F, G]. Proof. Step 1: We prove that x ≤ G for a solution x of (2.2). Set x+ (t) = maxs∈[0,t] x(s) = x(s t ). Then we obtain t

a− x p− (t) ≤ h(x(t)) = q

≤ b+ x++ (t)

β1−α K(t, s)s γ−1 g(x(s))ds ∫ β Γ(α) (t − s β )1−α

β1−α Γ(α)

0 st

∫ 0

K(s t , s)s γ−1 β

(s t − s β )1−α

ds

42 | 2 Fractional Integral Equations st

q

≤ b+ x++ (t)

β1−α 𝕂up β ∫(s t − s β )n+α−1 s γ−1 ds Γ(α)n! 0

β q b+ x++ (t)

−α 𝕂 up

γ β(n+α−1)+γ B( , n + α)s t Γ(α)n! β β−α 𝕂up γ q ≤ b+ x++ (t) B( , n + α)t β(n+α−1)+γ , Γ(α)n! β =

which implies that 1

x(t) ≤ x+ (t) ≤ ( Next we set

p− −q+ β(n+α−1)+γ b+ β−α 𝕂up γ B( , n + α)) t p− −q+ . a− Γ(α)n! β

(2.10)

1

Ξ := sup t∈(0,T]

p− −q+ b+ β−α 𝕂up x(t) γ ≤ B( , n + α)) . ( τ + t a− Γ(α)n! β

Then it holds t

K(t, s)s γ−1 β1−α a− x (t) ≤ h(x(t)) = g(x(s))ds ∫ β Γ(α) (t − s β )1−α p−

0

t

≤ b + Ξ q+

β1−α 𝕂up ∫(t β − s β )n+α−1 s γ−1 s q+ τ+ ds Γ(α)n! 0

β−α 𝕂up γ + q+ τ+ = b+ Ξ B( , n + α)t β(n+α−1)+γ+q+ τ+ Γ(α)n! β β−α 𝕂up β(n + α − 1)q+ + γp− ≤ b + Ξ q+ B( , n + α)t p− τ+ , Γ(α)n! β(p− − q+ ) q+

and so

1

q+ p− b+ β−α 𝕂up β(n + α − 1)q+ + γp− x(t) p− ( ≤ Ξ B( , n + α)) . t τ+ a− Γ(α)n! β(p− − q+ )

Hence q+

Ξ ≤ Ξ p− ( thus

1

p− b+ β−α 𝕂up β(n + α − 1)q+ + γp− B( , n + α)) , a− Γ(α)n! β(p− − q+ ) 1

Ξ≤(

p− −q+ b+ β−α 𝕂up β(n + α − 1)q+ + γp− B( , n + α)) . a− Γ(α)n! β(p− − q+ )

Consequently, 1

x(t) ≤ ( Since

p− −q+ b+ β−α 𝕂up β(n + α − 1)q+ + γp− B( , n + α)) t τ+ = G(t). a− Γ(α)n! β(p− − q+ )

β(n + α − 1)q+ + γp− γ > β(p− − q+ ) β

(2.11)

2.1 Abel-type nonlinear integral equations | 43

implies B(

γ β(n + α − 1)q+ + γp− , n + α) < B( , n + α), β(p− − q+ ) β

estimate (2.11) is an improvement of (2.10). Step 2: We prove that x ≥ F. Fix a ∈ (0, T) and set Υa := inf

t∈(a,T]

x(t) >0 (t β − a β )Θ

for Θ :=

β(n + α − 1) + γ > 0. β(p+ − q− )

Then like above, for t ∈ (a, T], we get t

a+ x p+ (t) ≥ h(x(t)) =

K(t, s)s γ−1 β1−α g(x(s))ds ∫ β Γ(α) (t − s β )1−α 0

t

q

≥ b − Υ a−

β1−α K(t, s)s γ−1 β (s − a β )q− Θ ds ∫ β Γ(α) (t − s β )1−α a

t

q

≥ b − Υ a−

β1−α 𝕂low ∫(t β − s β )n+α−1 s γ−1 (s β − a β )q− Θ ds Γ(α)n!

≥

a −α γ q β 𝕂low b − Υ a− B(

=

q b − Υ a−

+ q− Θ, n + α)(t β − a β )

n+α−1+ βγ +q− Θ

Γ(α)n! β β−α 𝕂low β(n + α − 1)q− + γp+ B( , n + α)(t β − a β )p+ Θ , Γ(α)n! β(p+ − q− )

which implies (

(t β

−α p+ β(n + α − 1)q− + γp+ x(t) q b − β 𝕂low B( , n + α). ) ≥ Υ a− β Θ a+ Γ(α)n! β(p+ − q− ) −a )

Hence p

q

Υ a + ≥ Υ a− and so

b− β−α 𝕂low β(n + α − 1)q− + γp+ B( , n + α), a+ Γ(α)n! β(p+ − q− ) 1

Υa ≥ (

p+ −q− b− β−α 𝕂low β(n + α − 1)q− + γp+ . B( , n + α)) a+ Γ(α)n! β(p+ − q− )

Consequently, we arrive at x(t) ≥ Υa (t β − a β )Θ 1

≥(

p+ −q− b− β−α 𝕂low β(n + α − 1)q− + γp+ B( , n + α)) (t β − a β )Θ . a+ Γ(α)n! β(p+ − q− )

Since a ∈ (0, T) is arbitrary, we have 1

x(t) ≥ (

p+ −q− b− β−α 𝕂low β(n + α − 1)q− + γp+ B( , n + α)) t βΘ = F(t). a+ Γ(α)n! β(p+ − q− )

Hence the proof is complete.

44 | 2 Fractional Integral Equations To solve (2.2), we introduce an operator S h,g : [F, G] ⊂ PM → C[0, T] by t

K(t, s)s γ−1 β1−α S h,g (x)(t) = h ( ]g(x(s))ds), ∫[ β Γ(α) (t − s β )1−α −1

t ∈ [0, T].

0

Lemma 2.8. The operator S h,g maps the order interval [F, G] into itself. Proof. To achieve our aim, we only need to verify that SF ≥ F and SG ≤ G: t

h(F(t)) ≤

β1−α K(t, s)s γ−1 ]g(F(s))ds, ∫[ β Γ(α) (t − s β )1−α

t ∈ [0, T],

(2.12)

t ∈ [0, T].

(2.13)

0

h(G(t)) ≥

β1−α Γ(α)

t

∫[ 0

K(t, s)s γ−1 ]g(G(s))ds, (t β − s β )1−α

First we show (2.12): t

t

β1−α K(t, s)s γ−1 b− β1−α A q− 𝕂low ≥ ]g(F(s))ds ∫[ β ∫(t β − s β )n+α−1 s q− τ− +γ−1 ds Γ(α) Γ(α)n! (t − s β )1−α 0

0

b− β−α A q− 𝕂low q− τ− + γ B( , n + α)t β(n+α−1)+q− τ− +γ = Γ(α)n! β = a+ A p+ t p+ τ− ≥ h(F(t)). Second we derive (2.13): t

t

b+ β1−α B q+ 𝕂up K(t, s)s γ−1 β1−α ]g(G(s))ds ≤ ∫[ β ∫(t β − s β )n+α−1 s q+ τ+ +γ−1 ds β 1−α Γ(α) Γ(α)n! (t − s ) 0

0

b+ β−α B q+ 𝕂up q+ b + γ B( , n + α)t β(n+α−1)+q+ τ+ +γ = Γ(α)n! β = a− B p− t p− τ+ ≤ h(G(t)). Obviously, the operator S is strictly increasing in [F, G], and if x ∈ [F, G], then F(t) ≤ x(t) ≤ G(t) < M for t ∈ [0, T]. Hence t

h(F(t)) ≤

β1−α K(t, s)s γ−1 ]g(F(s))ds ∫[ β Γ(α) (t − s β )1−α 0

≤

β1−α Γ(α)

t

∫[ 0 t

≤

K(t, s)s γ−1 ]g(x(s))ds (t β − s β )1−α

β1−α K(t, s)s γ−1 ]g(G(s))ds ≤ h(G(t)), ∫[ β Γ(α) (t − s β )1−α 0

2.1 Abel-type nonlinear integral equations | 45

so

t

K(t, s)s γ−1 β1−α ]g(x(s))ds ∈ [0, h(G(t)] = h([0, G(t)]. ∫[ β Γ(α) (t − s β )1−α 0

Consequently, S h,g is well defined and S h,g ([F, G]) ⊂ [F, G]. From the Arzelà–Ascoli theorem and since S h,g : [F, G] → [F, G] is nondecreasing, it follows that S h,g is compact. So the Schauder fixed point theorem implies the following existence result [330]. Theorem 2.9. Equation (2.2) has a solution in [F, G]. Moreover, lim S nh,g (F) = x−

n→∞

and

lim S nh,g (G) = x+

n→∞

are fixed points of S h,g with F ≤ x− ≤ x+ ≤ G. Now we are ready to state the following uniqueness result. But first we note that the above considerations can be repeated for any 0 < T1 ≤ T, so we get 𝕂low (T1 ), 𝕂up (T1 ), A(T1 ), B(T1 ), F T1 and G T1 as continuous functions of T1 . Note that 𝕂low (T1 ) is nonincreasing, 𝕂up (T1 ) is nondecreasing and 𝕂low (T1 ), 𝕂up (T1 ) can be continuously extended to T1 = 0. Then 𝕂low (0) = 𝕂up (0). We still keep notations 𝕂low = 𝕂low (T), 𝕂up = 𝕂up (T), F = F T and G = G T . Theorem 2.10. If there are constants ψ, χ and continuous functions a g (t) > 0 and a h (t) > 0 on [0, T] such that a h (T1 )t ψ |x(t) − y(t)| ≤ |h(x(t)) − h(y(t))|, |g(x(t)) − g(y(t))| ≤ a g (T1 )t χ |x(t) − y(t)| for all T1 ∈ (0, T], t ∈ (0, T1 ], x, y ∈ [F T1 , G T1 ], then equation (2.2) has a unique solution in [F, G] provided it holds β(n + α − 1) + χ + γ ≥ ψ (2.14) and Λ :=

a g (0)β−α 𝕂up (0) χ + γ + τ+ B( , n + α)0β(n+α−1)+χ+γ−ψ < 1, a h (0)Γ(α)n! β

(2.15)

where we set 00 = 1. Proof. For any x, y ∈ [F, G] we set x1 = S h,g (x) and y1 = S h,g (y). Clearly, it holds ‖x − y‖q := sup

t∈(0,T]

|x(t) − y(t)| ≤ 2B t τ+ (1 + ιt ϖ )

for q(t) = t τ+ (1 + ιt ϖ ) with ϖ > 0 and ι > 0 specified below. So [F, G] ⊂ Mq . Then for

46 | 2 Fractional Integral Equations any t ∈ (0, T], we derive a h (t)t ψ |x1 (t) − y1 (t)| ≤ |h(x1 (t)) − h(y1 (t))| t

K(t, s)s γ−1 β1−α ≤ ]|g(x(s)) − g(y(s))|ds ∫[ β Γ(α) (t − s β )1−α 0

t

≤

a g (t)β1−α 𝕂up (t) ∫(t β − s β )n+α−1 s χ+γ−1 |x(s) − y(s)|ds Γ(α)n! 0

≤

ag

(t)β1−α 𝕂 Γ(α)n!

up (t)

t

[ ∫(t β − s β )n+α−1 s χ+γ+τ+ −1 ds 0

t

+ ι ∫(t β − s β )n+α−1 s χ+γ+τ+ +ϖ−1 ds]‖x − y‖q 0

a g (t)β1−α 𝕂up (t) χ + γ + τ+ = , n + α)t β(n+α−1)+χ+γ+τ+ [B( Γ(α)n! β + ιB(

χ + γ + τ+ + ϖ , n + α)t β(n+α−1)+χ+γ+τ+ +ϖ ]‖x − y‖q , β

which implies a g (t)β1−α 𝕂up (t) |x1 (t) − y1 (t)| χ + γ + τ+ ≤ , n + α)t β(n+α−1)+χ+γ−ψ [B( τ ϖ ϖ + t (1 + ιt ) a h (t)Γ(α)n!(1 + ιt ) β + ιB(

χ + γ + τ+ + ϖ , n + α)t β(n+α−1)+χ+γ+ϖ−ψ ]‖x − y‖q . β

Consequently, we obtain ‖S h,g (x) − S h,g (y)‖q ≤ L‖x − y‖q

for all x, y ∈ [F, G]

with L := sup L(t), t∈(0,T]

L(t) := L1 (t) + L2 (t)

L1 (t) :=

a g (t)β1−α 𝕂up (t) χ + γ + τ+ B( , n + α)t β(n+α−1)+χ+γ−ψ , a h (t)Γ(α)n!(1 + ιt ϖ ) β

L2 (t) :=

ιa g (t)β1−α 𝕂up (t) χ + γ + τ+ + ϖ B( , n + α)t β(n+α−1)+χ+γ+ϖ−ψ . a h (t)Γ(α)n!(1 + ιt ϖ ) β

Since (note (2.14)) L2 (t) ≤ ≤

ιt ϖ a g (t)β1−α 𝕂up χ + γ + τ+ + ϖ B( , n + α)T β(n+α−1)+χ+γ−ψ a h (t)Γ(α)n!(1 + ιt ϖ ) β a g (t)β1−α 𝕂up χ + γ + τ+ + ϖ B( , n + α)T β(n+α−1)+χ+γ−ψ a h (t)Γ(α)n! β

2.1 Abel-type nonlinear integral equations | 47

and B( χ+γ+τβ + +ϖ , n + α) → 0 as ϖ → +∞, we see that sup L2 (t)
0 sufficiently large uniformly for any ι > 0. So we take and fix such ϖ. Next, by (2.15) there is t0 ∈ (0, T] so that L1 (t) ≤

a g (t)β1−α 𝕂up (t) χ + γ + τ+ 1+Λ B( , n + α)t β(n+α−1)+χ+γ−ψ < 0. Consequently, we get sup L1 (t) ≤

t∈(0,T]

1+Λ . 2

Summarizing, we see that there is ϖ > 0 and ι > 0 so that L≤

1+Λ 1−Λ 3+Λ + = < 1. 2 4 4

This yields that S h,g : [F, G] → [F, G] is a contraction with respect to the norm ‖⋅‖q with a constant L. By the contraction mapping principle, one obtains the statement immediately. Remark 2.11. Consider (2.5). Of course, we can suppose p > 1 = q. Then p± = p, q± = 1, a± = b± = 1 and τ± = τ := β(n+α−1)+γ . Moreover, Remark 2.6 can be applied to p−1 get an existence result. If in addition 𝕂low > 0, then B ≥ A > 0, and it is not difficult to see that ψ = (p − 1)τ, χ = 0, a g (T1 ) = 1 and a h (T1 ) = pA p−1 (T1 ) = p

β−α 𝕂low (T1 ) β(n + α − 1) + γp B( , n + α). Γ(α)n! β(p − 1)

Then β(n + α − 1) + χ + γ = ψ, so (2.14) holds. Next, we derive Λ=

B( γ+τ 𝕂up (0) 𝕂up (0) 1 β , n + α) = = < 1. β(n+α−1)+γp p𝕂low (0) B( , n + α) p𝕂low (0) p β(p−1)

Hence condition (2.15) is satisfied and then we get a uniqueness result by Theorem 2.10.

48 | 2 Fractional Integral Equations

2.1.4 General solutions of Erdélyi–Kober type integral equations This subsection is devoted to derive explicit solutions of some Erdélyi–Kober type integral equations. In order to establish this, we introduce the following useful result. Lemma 2.12. Let ση + β > −σ and α, σ > 0. Then α (EK I0+;σ,η t β )(x) =

Γ(η + 1 + σβ ) Γ(η + 1 + α + σβ )

xβ .

Proof. Set t = xy. By using Lemma 2.3, we have x

σx−σ(α+η) t ση+σ−1 t β dt ∫ σ Γ(α) (x − t σ )1−α

α (EK I0+;σ,η t β )(x) =

0

Γ(η + 1 + σβ ) β 1 ση + σ + β B( , α) = x . Γ(α) σ σ Γ(η + 1 + α + σβ ) σx β

= This completes the proof.

Now we are ready to present our main result of this subsection. Theorem 2.13. Let α > 0, σ > 0, σβ + η + 1 > 0, m, b, β ∈ ℝ and a, N, m ≠ 0. Then equation (2.6) is solvable and its solution φ(x) can be written as 1

β

φ(x) = C N x N ,

(2.16)

where the constant C satisfies the equation C

m N

= aC

Γ(η + 1 + σβ ) Γ(η + 1 + α + σβ )

+ b.

(2.17)

Proof. With the help of Lemma 2.12, substituting (2.16) into (2.6), we find that C satisfies (2.17) which completes the proof.

2.1.5 Illustrative examples In this subsection, we present three numerical performance results. Example 2.14. We consider the problem: 1

x (t) = 2

( 14 ) 2 Γ( 12 )

t

∫[ 0

1

t2 s− 2 1

1

1

(t 4 − s 4 ) 2

]x(s)ds,

t ∈ [0, 1].

(2.18)

2.1 Abel-type nonlinear integral equations | 49

0.6 0.5 0.4 0.3 0.2 0.1

0.2

0.4

0.6

0.8

1.0

Fig. 2.1. Solution of (2.18) and the boundaries F and G for Example 2.14 coincide with the unique solution.

First, Theorem 2.13 gives an exact solution x(t) = (2.18). Next, by changing x(t) = z(t)t, we get 1

z (t) = 2

( 14 ) 2 Γ( 12 )

t

∫[ 0

88179√πt19/8 262144

≐ 0.596211t2.375 of

1

s2 1

1

1

(t 4 − s 4 ) 2

]z(s)ds,

t ∈ [0, 1].

(2.19)

11/8

Of course, we get a solution z(t) = 88179√πt . In equation (2.5) for (2.19), we set 262144 1 3 1 K(t, s) = 1, α = 2 , γ = 2 , β = 4 , n = 0, T = 1, p = 2 and q = 1. After some computation, we find that 11 2 23 1 11 88179√πt 8 . F(t) = G(t) = 1 B( , )t 8 = 2 2 262144 Γ( 2 ) Obviously, all the assumptions in Theorem 2.10 are satisfied. Numerical results are given in Figure 2.1. Example 2.15. In equation (2.5), we set K(t, s) = e4t , α = 43 , β = γ = 12 , n = 0, T = 1, p = 2 and q = 1. Now, we turn to consider the following homogeneous Abel type integral equation with weakly singular kernel and power law nonlinearity: 1

x (t) = 2

( 12 ) 4 Γ( 34 )

t

∫[ 0

1

e4t s− 2 1 2

1 2

1

(t − s ) 4

]x(s)ds,

t ∈ [0, 1].

After some computation, we find that 3

3

F(t) =

7 7 3 3 42 4 Γ( 4 ) 3 8 = B( t 8 ≐ 1.16274t0.375 , , )t 4 4 3√π Γ( 34 )

( 12 )− 4

3

G(t) =

e4 ( 21 )− 4 Γ( 34 )

3

B(1,

3 3 42 4 e4 3 t 8 ≐ 99.9092t0.375 . )t 8 = 4 3Γ( 34 )

(2.20)

50 | 2 Fractional Integral Equations 100 90 80 70

x

60 50 40 30 20 10 0

0

0.2

0.4

0.6

t

0.8

1

Fig. 2.2. Solution (solid black) of (2.20) and the boundaries F (dashed red) and G (dot-dashed blue) for Example 2.15.

Obviously, all the assumptions in Theorem 2.10 are satisfied. Then, the problem (2.15) has a unique solution in [F, G]. Numerical results are given in Figure 2.2. Example 2.16. In equation (2.2), we set K(t, s) = e4t , α = 34 , β = γ = 21 , n = 0, T = 1, 1 p± = 2, q± = 1, h(x) = x2 (1 − 300 x), g(x) = x and M = 200. Now, we turn to consider the following homogeneous Abel type integral equation with weakly singular kernel and polynomial law nonlinearity: 1

x2 (t)(1 −

t

(1)4 1 e4t s− 2 x(t)) = 2 3 ∫[ 1 ]x(s)ds, 1 1 300 Γ( 4 ) (t 2 − s 2 ) 4 0

It is clear that now a+ = 1, a− = we find that

1 3

1

t ∈ [0, 1].

(2.21)

and b± = 1, so (2.4) holds. After some computation, 3

3

7 7 3 3 42 4 Γ( 4 ) 3 8 = F(t) = B( , t 8 ≐ 1.16274t0.375 , )t 4 4 3√π Γ( 34 )

( 12 )− 4

3

G(t) =

3e4 ( 21 )− 4 Γ( 34 )

3

7 7 3 3 42 4 e4 Γ( 4 ) 3 t 8 ≐ 190.45t0.375 . B( , )t 8 = 4 4 √π

Since now A ≐ 1.16274 < 190.45 ≐ B < M = 200, obviously, all the assumptions of Theorem 2.9 are satisfied. Then, problem (2.16) has a solution in [F, G]. Numerical results are given in Figure 2.3.

2.2 Quadratic Erdélyi–Kober type integral equations | 51

200 180 160 140

x

120 100 80 60 40 20 0

0

0.2

0.4

0.6

t

0.8

1

Fig. 2.3. Solution (solid black) of (2.21) and the boundaries F (dashed red) and G (dot-dashed blue) for Example 2.16.

2.2 Quadratic Erdélyi–Kober type integral equations of fractional order 2.2.1 Introduction In 2009, Banaś and Zaja¸c [44] studied the solvability of a functional integral equation of fractional order: t

x(t) = f1 (t, x(t))+

f2 (t, x(t)) ∫(t − s)α−1 u(t, s, x(s))ds, Γ(α)

t ∈ [0, 1], α ∈ (0, 1), (2.22)

0

where Γ(⋅) is the gamma function. Here, the term (t − s)α−1 can be named as Riemann– Liouville singular kernel since it appears in the standard Riemann–Liouville fractional integral of order α of a continuous function y defined by t

I α y(t) =

1 ∫(t − s)α−1 y(s)ds, Γ(α)

t > 0, α > 0.

0

Moreover, Erdélyi–Kober fractional integrals, Hadamard fractional integrals and Riesz fractional integrals are also widely used to describe the medium with non-integer mass dimension. One can also find more details of such fractional integrals in physics, viscoelasticity, electrochemistry and porous media [93, 117, 120, 138, 190, 193, 263].

52 | 2 Fractional Integral Equations

The Erdélyi–Kober fractional integral [163] of a continuous function y is defined by t

γ,α I β y(t)

t−β(γ+α) = ∫(t β − s β )α−1 s βγ y(s)d(s β ), Γ(α) 0

(t β

where α, β and γ > 0. The term − s β )α−1 can be named as Erdélyi–Kober singular kernel. Obviously, the Riemann–Liouville singular kernel is a special case of the Erdélyi–Kober singular kernel which implies that the Erdélyi–Kober fractional integral can better describe the memory property than the Riemann–Liouville fractional integral. Thus, quadratic integral equations involving Erdélyi–Kober singular kernels may be better applicable in the theory of kinetic theory of gases [140] and in the theory of neutron transport [160]. Let us pay attention to the fact that only a few papers investigated the existence and local stability of solutions of Erdélyi–Kober type integral equations of fractional order on an unbounded interval [275]. Motivated by the above fact, we are going to study the following quadratic Erdélyi– Kober type integral equations of fractional order: t

f2 (t, x(t)) x(t) = f1 (t, x(t)) + ∫(t β − s β )α−1 s γ u(t, s, x(s))ds, Γ(α)

(2.23)

0

where t ∈ ℝ+ := [0, ∞), 1 > α > 0,

β > 0,

γ > β(1 − α) − 1,

(2.24)

f1 , f2 and u are three functions that will be defined later. Obviously, equation (2.22) is a particular case of equation (2.23) when β = 1 and γ = 0. We will use certain measure of noncompactness from [44] and some of the methods of [275] to derive a new existence and limit property of solutions to equation (2.23) with a restriction. Moreover, we also prove the uniqueness and other existence results of the solutions to equation (2.23). This section is based on [295].

2.2.2 Preliminaries In this subsection, we collect some definitions and results which will be needed in our investigation. We shall use a measure of noncompactness in the space BC(ℝ+ ) (see [40, 270]). In order to define this measure let us fix a nonempty bounded subset X of the space BC(ℝ+ ) and a positive number T. For x ∈ X and ε > 0 denote ω T (x, ε) := sup{|x(t) − x(s)| | t, s ∈ [0, T], |t − s| ≤ ε}. Further, let us set ω T (X, ε) := sup{ω T (x, ε) | x ∈ X},

ω0T (X) := lim ω T (X, ε), ε→0

ω0 (X) := lim ω0T (X). T→∞

2.2 Quadratic Erdélyi–Kober type integral equations | 53

Next, for a fixed T > 0 and for a function x ∈ X, we define β T (X) := sup{β T (x) | x ∈ X},

β T (x) := sup{|x(s) − x(t)| | s ≥ T, t ≥ T}.

Let β(X) := limT→∞ β T (X). Finally, we define the measure of noncompactness in the space BC(ℝ+ ) by μ(X) := ω0 (X) + β(X). The kernel ker μ of this measure consists of all sets X ∈ MBC(ℝ+ ) such that functions belonging to X are locally equicontinuous on ℝ+ and tend to their limits at infinity uniformly with respect to the set X. The following basic equality [235], which is a modification of Lemma 2.3, will be used in the sequel. Lemma 2.17. Let α, β, γ and p be constants such that α > 0, p(γ − 1) + 1 > 0 and p(β − 1) + 1 > 0. Then t

∫(t α − s α )p(β−1) s p(γ−1) ds =

tθ p(γ − 1) + 1 B( , p(β − 1) + 1) α α

0

for t ∈ ℝ+ := [0, +∞) and θ = p[α(β − 1) + γ − 1] + 1.

2.2.3 Existence and limit property of solutions In this subsection, we will investigate the existence and limit property of solutions to equation (2.23) by using the fixed point theorem of Darbo type (see Theorem A.21 in Appendix A) via the measure of noncompactness defined in the preceding part. We introduce the following assumptions (see [44]): (H1 ) f i ∈ C(ℝ+ × ℝ, ℝ), i = 1, 2 and there are nondecreasing functions k i : ℝ+ → ℝ+ such that |f i (t, x) − f i (t, y)| ≤ k i (r)|x − y|, i = 1, 2 for any t ∈ ℝ+ , r ≥ 0 and x, y ∈ [−r, r]. Moreover, f i (t, 0) ∈ BC(ℝ+ ), i = 1, 2. (H2 ) u ∈ C(∆ × ℝ, ℝ) and there exist n ∈ C(∆, ℝ+ ) and a nondecreasing function ϕ ∈ C(ℝ+ , ℝ+ ) with ϕ(0) = 0 such that |u(t, s, x) − u(t, s, y)| ≤ n(t, s)ϕ(|x − y|) for all (t, s) ∈ ∆ and x, y ∈ ℝ. Here ∆ := {(t, s) ∈ ℝ+ × ℝ+ | t ≥ s}. It is clear that (2.23) is equivalent to the fixed point problem x = V(x),

(2.25)

where the operator V is defined by (Vx)(t) := (F1 x)(t) + (F2 x)(t)(Ux)(t),

t ∈ ℝ+

54 | 2 Fractional Integral Equations with (F1 x)(t) := f1 (t, x(t)), (F2 x)(t) := f2 (t, x(t)) and t

(Ux)(t) :=

1 ∫(t β − s β )α−1 s γ u(t, s, x(s))ds. Γ(α) 0

Our aim is to solve (2.25) on BC(ℝ+ ) by following a method of [44]. For the sake of convenience, we shall split our main result into several lemmas. Obviously by (H1 ), we get the first one: Lemma 2.18. Assume (H1 ). Then F1 , F2 ∈ C(BC(ℝ+ ), BC(ℝ+ )). To show that U ∈ C(BC(ℝ+ ), BC(ℝ+ )) and V ∈ C(BC(ℝ+ ), BC(ℝ+ )), we introduce functions n(t) and u(t) on ℝ+ as follows: t

t

n(t) = ∫ n(t, s)(t β − s β )α−1 s γ ds,

u(t) = ∫ |u(t, s, 0)|(t β − s β )α−1 s γ ds.

0

0

Since n t := max{n(t, s) | s ∈ [0, t]} < ∞ for any t > 0, by Lemma 2.17 and (2.24), we have t

t

∫ n(t, s)(t β − s β )α−1 s γ ds ≤ n t ∫(t β − s β )α−1 s γ ds 0

0

n t t β(α−1)+γ+1 γ + 1 = B( , α) < ∞. β β

(2.26)

Consequently, n(⋅) and u(⋅) are well defined. Lemma 2.19. Assume (H2 ) and (2.24). Then n(t) and u(t) are continuous functions on ℝ+ . Proof. It is enough to verify the continuity of n(t), since the same method can be applied for u(t). First, by (2.26), we have a continuity at t = 0. Next, fix arbitrary T > 0, ε > 0 and t1 , t2 ∈ [0, T] such that |t2 − t1 | ≤ ε with t1 < t2 . We denote ω1T (n, ε) := sup{|n(s2 , s3 ) − n(s1 , s3 )| | s1 , s2 , s3 ∈ [0, T], s3 ≤ s1 , s3 ≤ s2 , |s1 − s2 | ≤ ε}. Since γ > β(1 − α) − 1 and α > 0, we can take ζ > 1 such that ζγ > ζβ(1 − α) − 1 and ζ ζ(α − 1) + 1 > 0. Set ζ ∗ := ζ −1 . By Lemma 2.17 and Hölder’s inequality, we have t

t

1 󵄨󵄨 2 󵄨󵄨 β β |n(t2 ) − n(t1 )| ≤ 󵄨󵄨󵄨󵄨 ∫ n(t2 , s)(t2 − s β )α−1 s γ ds − ∫ n(t1 , s)(t2 − s β )α−1 s γ ds󵄨󵄨󵄨󵄨 󵄨 󵄨

0

0

t1

t

0

0

1 󵄨󵄨 󵄨󵄨 β β β α−1 γ 󵄨 󵄨 + 󵄨󵄨 ∫ n(t1 , s)(t2 − s ) s ds − ∫ n(t1 , s)(t1 − s β )α−1 s γ ds󵄨󵄨󵄨󵄨 󵄨 󵄨

2.2 Quadratic Erdélyi–Kober type integral equations | 55

t2 β

≤ ∫ n(t2 , s)(t2 − s β )α−1 s γ ds t1 t

t

0

0

t1

t

1 󵄨󵄨 󵄨󵄨 1 β β + 󵄨󵄨󵄨󵄨 ∫ n(t2 , s)(t2 − s β )α−1 s γ ds − ∫ n(t1 , s)(t2 − s β )α−1 s γ ds󵄨󵄨󵄨󵄨 󵄨 󵄨 1 󵄨󵄨 󵄨󵄨 β β β α−1 γ 󵄨 + 󵄨󵄨󵄨 ∫ n(t1 , s)(t2 − s ) s ds − ∫ n(t1 , s)(t1 − s β )α−1 s γ ds󵄨󵄨󵄨󵄨 󵄨 󵄨

0

0

t1

t2

β

β

≤ n t2 ∫(t2 − s β )α−1 s γ ds + ∫ |n(t2 , s) − n(t1 , s)|(t2 − s β )α−1 s γ ds t1

0 t1

β 󵄨 β 󵄨 + ∫ n(t1 , s)󵄨󵄨󵄨(t2 − s β )α−1 − (t1 − s β )α−1 󵄨󵄨󵄨s γ ds 0

≤ n t2

ζ∗

ζ

t2

√ t2 − t1 √ ∫(t2β − s β )ζ(α−1) s ζγ ds t1 t1 β

+ ∫ ω1T (n, ε)(t2 − s β )α−1 s γ ds 0 t1

t2 β

β

+ n t1 ( ∫(t1 − s β )1−α s γ ds − ∫(t2 − s β )1−α s γ ds) 0

0

t2 β

+ n t1 ∫(t2 − s β )α−1 s γ ds t1 ζ

ζ∗

≤ (n t1 + n t2 ) √ε √ + ω1T (n, ε)

T ζβ(α−1)+ζγ+1 B( ζγ+1 β , ζ(α − 1) + 1) β

T β(α−1)+γ+1 β

β(α−1)+γ+1

+ n t1

t2

ζ∗

ζ

≤ (n t1 + n t2 ) √ε √ +

B( γ+1 β , β

α)

B(

γ+1 , α) β

β(α−1)+γ+1

− t1 β

B(

γ+1 , α) β

T ζβ(α−1)+ζγ+1 B( ζγ+1 β , ζ(α − 1) + 1) β

(ω1T (n, ε)T β(α−1)+γ+1 + ω T (h, ε)n t1 ),

(2.27)

56 | 2 Fractional Integral Equations where h(t) = t β(α−1)+γ+1 is continuous on [0, T]. By (H2 ), we have limε→0 ω1T (n, ε) = 0. So, using (2.27), we derive that n is continuous on [0, T]. Since T is arbitrary, the proof is complete. Now we are ready to prove the following result. Lemma 2.20. Assume (H1 ), (H2 ) and (2.24). Then for any x ∈ BC(ℝ+ ), functions Ux and Vx are continuous on ℝ+ . Proof. We need to verify only the continuity of U. Let x ∈ BC(ℝ+ ). First, by (2.26), we have continuity at t = 0. Next, we take T > 0, ε > 0 and set u T := sup{|u(t, s, 0)| | t, s ∈ [0, T], s ≤ t} and ω1T (u, ε; ‖x‖) := sup{|u(t2 , s, x) − u(t1 , s, x)| | t2 , t1 , s ∈ [0, T], s ≤ t1 , s ≤ t2 , |t2 − t1 | ≤ ε, |x| ≤ ‖x‖}. Now, assume that t1 , t2 ∈ [0, T] are such that |t1 − t2 | ≤ ε with t1 < t2 . Then we get |(Ux)(t2 ) − (Ux)(t1 )| t1

t2

0

t1

1 󵄨󵄨󵄨 󵄨󵄨 ∫(t2β − s β )α−1 s γ u(t2 , s, x(s))ds + ∫(t2β − s β )α−1 s γ u(t2 , s, x(s))ds = Γ(α) 󵄨󵄨 t1

󵄨󵄨 β − ∫(t1 − s β )α−1 s γ u(t1 , s, x(s))ds󵄨󵄨󵄨󵄨 󵄨 0

t1

1 β ≤ ∫(t2 − s β )α−1 s γ |u(t2 , s, x(s)) − u(t1 , s, x(s))|ds Γ(α) 0 t1

1 + ∫[|u(t1 , s, x(s)) − u(t1 , s, 0)| + |u(t1 , s, 0)|] Γ(α) 0 β

β

× |(t2 − s β )α−1 − (t1 − s β )α−1 |s γ ds t2

1 β + ∫[|u(t2 , s, x(s)) − u(t2 , s, 0)| + |u(t2 , s, 0)|](t2 − s β )α−1 s γ ds Γ(α) t1

≤

ω1T (u,

ε; ‖x‖) T β(α−1)+γ+1 γ + 1 B( , α) Γ(α) β β t1

1 β β + ∫[n(t1 , s)ϕ(‖x‖) + |u(t1 , s, 0)|]|(t2 − s β )α−1 − (t1 − s β )α−1 |s γ ds Γ(α) 0 t2

+

1 β ∫[n(t2 , s)ϕ(‖x‖) + |u(t2 , s, 0)|](t2 − s β )α−1 s γ ds Γ(α) t1

2.2 Quadratic Erdélyi–Kober type integral equations | 57

≤

ω1T (u, ε; ‖x‖)T β(α−1)+γ+1 γ + 1 B( , α) βΓ(α) β t1

+

n T ϕ(‖x‖) + u T β β ∫[(t1 − s β )α−1 − (t2 − s β )α−1 ]s γ ds Γ(α) 0 t2

n T ϕ(‖x‖) + u T β + ∫(t2 − s β )α−1 s γ ds Γ(α) t1

≤

ω1T (u,

ε; ‖x‖)T β(α−1)+γ+1 βΓ(α)

B(

γ+1 , α) β

ζ T n T ϕ(‖x‖) + u T ζ ∗ √ √ε +2 Γ(α)

+

[n T ϕ(‖x‖) +

u T ]B( γ+1 β , βΓ(α)

ζβ(α−1)+ζγ+1 B( ζγ+1 , β

α)ω T (h, ε)

ζ(α − 1) + 1)

β →0

as ε → 0,

(2.28)

where ζ , ζ ∗ and h are defined in the proof of Lemma 2.19. This implies the continuity of Ux on [0, T], and so on ℝ+ as well. Now by Lemma 2.18, the continuity of Vx on ℝ+ is clear. To proceed, we suppose the next condition on n(⋅) and u(⋅): (H3 ) limt→∞ n(t) = limt→∞ u(t) = 0. Lemma 2.21. Assume (H1 )–(H3 ) and (2.24). Then V : BC(ℝ+ ) → BC(ℝ+ ). Proof. Due to (H1 ), Lemma 2.18 and (H3 ), we can set F i := sup{|f i (t, 0)| | t ∈ ℝ+ } < ∞,

i = 1, 2,

N := sup {n(t) | t ∈ ℝ+ } < ∞, U := sup{u(t) | t ∈ ℝ+ } < ∞. Then for any x ∈ BC(ℝ+ ) and t ∈ ℝ+ we calculate |(Vx)(t)| ≤ |f1 (t, x(t)) − f1 (t, 0)| + |f1 (t, 0)| 1 + [|f2 (t, x(t)) − f2 (t, 0)| + |f2 (t, 0)|] Γ(α) t

× ∫(t β − s β )α−1 s γ [|u(t, s, x(s)) − u(t, s, 0)| + |u(t, s, 0)|]ds 0

≤ k1 (‖x‖)‖x‖ + F 1 +

k2 (‖x‖)‖x‖ + F 2 Γ(α)

t

× ∫(t β − s β )α−1 s γ [n(t, s)ϕ(‖x‖) + |u(t, s, 0)|]ds 0

58 | 2 Fractional Integral Equations

= ‖x‖k1 (‖x‖) + F 1 +

k2 (‖x‖)‖x‖ + F 2 (ϕ(‖x‖)n(t) + u(t)) Γ(α)

≤ ‖x‖k1 (‖x‖) + F 1 +

k2 (‖x‖)‖x‖ + F 2 (ϕ(‖x‖)N + U), Γ(α)

which implies ‖Vx‖ ≤ ‖x‖k1 (‖x‖) + F 1 +

k2 (‖x‖)‖x‖ + F 2 (ϕ(‖x‖)N + U). Γ(α)

(2.29)

From Lemma 2.20 and (2.29) we obtain Vx ∈ BC(ℝ+ ). Lemma 2.22. Assume (H1 )–(H3 ) and (2.24). Then V ∈ C(BC(ℝ+ ), BC(ℝ+ )). Proof. By Lemmas 2.18 and 2.21, and construction of V, we only need to show that U ∈ C(BC(ℝ+ ), BC(ℝ+ )). But for any ε > 0, x, y ∈ BC(ℝ+ ) with ‖x − y‖ ≤ ε and t ∈ ℝ+ we obtain t

|(Ux)(t) − (Uy)(t)| ≤

1 ∫(t β − s β )α−1 s γ |u(t, s, x(s)) − u(t, s, y(s))|ds Γ(α) 0 t

ϕ(ε)N ϕ(ε) . ≤ ∫(t β − s β )α−1 s γ n(t, s)ds ≤ Γ(α) Γ(α) 0

That is, ‖Ux − Uy‖ ≤

ϕ(ε)N , Γ(α)

which implies the desired continuity of the operator U on BC(ℝ+ ). Next, we introduce the final assumption: (H4 ) There exists an r0 > 0 such that 0 )r 0 +F 2 (ϕ(r0 )N + U) ≤ r0 , (i) r0 k1 (r0 ) + F 1 + k2 (rΓ(α) (ii) κ := k1 (r0 ) +

1 Γ(α) [Nϕ(r 0 )

+ U]k2 (r0 ) < 1,

(iii) limT→∞ sup{|f1 (t, x) − f1 (s, x)| | t, s ≥ T, |x| ≤ r0 } = 0. Clearly (2.29) and (H4 ) (i) give V : B r0 → B r0 . Now, we are ready to state the main result of this section. Theorem 2.23. Under the assumptions (H1 )–(H4 ) and (2.24), equation (2.23) has at least one solution x ∈ BC(ℝ+ ) and it tends to a limit at infinity. Proof. We intend to apply Theorem A.21 with the standard measure of noncompactness μ introduced in Section 2.2.2, and E = BC(ℝ+ ) with T = V and Q = B r0 . First, Lemma 2.22 implies V ∈ C(B r0 , B r0 ). Now, take a nonempty subset X of the ball B r0 . For any ε > 0, T > 0, x ∈ X and t1 , t2 ∈ [0, T] such that |t1 − t2 | ≤ ε, by setting ω1T (f i , ε; r0 ) := sup{|f i (s, x) − f i (t, x)| | t, s ∈ [0, T], |t − s| ≤ ε, |x| ≤ r0 },

i = 1, 2,

2.2 Quadratic Erdélyi–Kober type integral equations | 59

and using (2.28), we obtain |(Vx)(t2 ) − (Vx)(t1 )| ≤ |f1 (t2 , x(t2 )) − f1 (t2 , x(t1 ))| + |f1 (t2 , x(t1 )) − f1 (t1 , x(t1 ))| + |(Ux)(t2 )|[|f2 (t2 , x(t2 )) − f2 (t2 , x(t1 ))| + |f2 (t2 , x(t1 )) − f2 (t1 , x(t1 ))|] + [|f2 (t1 , x(t1 )) − f2 (t1 , 0)| + |f2 (t1 , 0)|]|(Ux)(t2 ) − (Ux)(t1 )| ≤ k1 (r0 )|x(t2 ) − x(t1 )| + ω1T (f1 , ε; r0 ) +

k2 (r0 )|x(t2 ) − x(t1 )| + ω1T (f2 , ε; r0 ) Γ(α)

t2 β

× ∫(t2 − s β )α−1 s γ [|u(t2 , s, x(s)) − u(t2 , s, 0)| + |u(t2 , s, 0)|]ds 0

+

r0 k2 (r0 ) + F 2 |(Ux)(t2 ) − (Ux)(t1 )| αβΓ(α)

k2 (r0 )ω T (x, ε) + ω1T (f2 , ε; r0 ) (ϕ(r0 )N + U) Γ(α) r0 k2 (r0 ) + F 2 ω1T (u, ε; r0 )T β(α−1)+γ+1 γ + 1 + B( , α) { αβΓ(α) βΓ(α) β

≤ k1 (r0 )ω T (x, ε) + ω1T (f1 , ε; r0 ) +

+2 +

ζ T n T ϕ(r0 ) + u T ζ ∗ √ √ε Γ(α)

[n T ϕ(r0 ) +

u T ]B( γ+1 β ,

ζβ(α−1)+ζγ+1 B( ζγ+1 , β

ζ(α − 1) + 1)

β

α)ω T (h,

ε)

βΓ(α)

},

(2.30)

where ζ , ζ ∗ and h are defined in the proof of Lemma 2.19. So (2.30) gives ω T (VX, ε) ≤ k1 (r0 )ω T (X, ε) + ω1T (f1 , ε; r0 ) k2 (r0 )ω T (X, ε) + ω1T (f2 , ε; r0 ) (ϕ(r0 )N + U) Γ(α) r0 k2 (r0 ) + F 2 ω1T (u, ε; r0 )T β(α−1)+γ+1 γ + 1 + B( , α) { αβΓ(α) βΓ(α) β

+

ζ T n T ϕ(r0 ) + u T ζ ∗ √ √ε +2 Γ(α)

+

[n T ϕ(r0 ) +

u T ]B( γ+1 β ,

ζβ(α−1)+ζγ+1 B( ζγ+1 , β

α)ω T (h,

βΓ(α)

ζ(α − 1) + 1)

β ε)

}.

(2.31)

For fixed T, according to our assumption, we get ω1T (f i , ε; r0 ) → 0 (i = 1, 2),

ω1T (u, ε; r0 ) → 0,

ω T (h, ε) → 0

as ε → 0. From (2.31) we derive ω0T (VX) ≤ k1 (r0 )ω0T (X) +

k2 (r0 )(ϕ(r0 )N + U) T ω0 (X), Γ(α)

60 | 2 Fractional Integral Equations

which yields that ω0 (VX) ≤ [k1 (r0 ) +

k2 (r0 )(ϕ(r0 )N + U) ]ω0 (X). Γ(α)

(2.32)

Similarly, for any t1 ≥ T, t2 ≥ T, we have |(Vx)(t2 ) − (Vx)(t1 )| ≤ |f1 (t2 , x(t2 )) − f1 (t1 , x(t1 ))| t2

|f2 (t2 , x(t2 ))| β + ∫(t2 − s β )α−1 s γ |u(t2 , s, x(s))|ds Γ(α) 0 t1

+

|f2 (t1 , x(t1 ))| β ∫(t1 − s β )α−1 s γ |u(t1 , s, x(s))|ds Γ(α) 0

≤ k1 (r0 )|x(t2 ) − x(t1 )| + |f1 (t2 , x(t1 )) − f1 (t1 , x(t1 ))| +

r0 k2 (r0 ) + F 2 {ϕ(r0 )[n(t2 ) + n(t1 )] + u(t2 ) + u(t1 )}, Γ(α)

which implies β T (Vx) ≤ k1 (r0 )β T (x) + β1T (f1 ; r0 ) + 2

r0 k2 (r0 ) + F 2 {ϕ(r0 )n T + u T } Γ(α)

(2.33)

for β1T (f1 ; r0 ) = sup{|f1 (t2 , x) − f1 (t1 , x)| | t1 ≥ T, t2 ≥ T, |x| ≤ r0 }, n T = sup{n(t) | t ≥ T},

u T = sup{u(t) | t ≥ T}.

Now, by (H3 ), (H4 ) (iii) and (2.33), we derive β(VX) ≤ k1 (r0 )β(X).

(2.34)

Consequently, (2.32), (2.34) and (H4 ) give μ(VX) ≤ [k1 (r0 ) +

1 (ϕ(r0 )N + U)k2 (r0 )]μ(X) = κμ(X) Γ(α)

with κ < 1. Therefore by Theorem A.21, the operator V has a fixed point in B r0 , which is a solution of equation (2.23). Moreover, we obtain that all solutions of equation (2.23) have finite limits at infinity.

2.2.4 Uniqueness and another existence results In this subsection, we establish some sufficient conditions for the uniqueness and another existence results for solutions to (2.23).

2.2 Quadratic Erdélyi–Kober type integral equations | 61

Theorem 2.24. Suppose that the assumptions of Theorem 2.23 hold with Lipschitz ϕ, i.e., |u(t, x) − u(t, y)| ≤ n(t, s)|x − y|, (2.35) for any t ∈ ℝ+ and all x, y ∈ ℝ. Then (2.23) has a unique solution. Proof. Suppose that y is another solution of (2.23). Then ‖y‖ ≤ r0 and like above, we have |x(t) − y(t)| ≤ |f1 (t, x(t)) − f1 (t, y(t))| t

+

|f2 (t, x(t)) − f2 (t, s, y(t))| ∫(t β − s β )α−1 s γ |u(t, s, x(s))|ds Γ(α) 0 t

+

|f2 (t, s, y(t))| ∫(t β − s β )α−1 s γ |u(t, s, x(s)) − u(t, s, y(s))|ds Γ(α) 0

≤ k1 (r0 )|x(t) − y(t)| t

k2 (r0 )|x(t) − y(t)| + ∫(t β − s β )α−1 s γ [|u(t, s, x(s)) − u(t, s, 0)| + |u(t, s, 0)|]ds Γ(α) 0 t

+

|f2 (t, s, y(t)) − f2 (t, s, 0)| + |f2 (t, s, 0)| ∫(t β − s β )α−1 s γ n(t, s)|x(s) − y(s)|ds Γ(α) 0

k2 (r0 )(ϕ(r0 )N + U) ≤ [k1 (r0 ) + ]|x(t) − y(t)| Γ(α) t

n t [r0 k2 (r0 ) + F 2 ] + ∫(t β − s β )α−1 s γ |x(s) − y(s)|ds. Γ(α) 0

This yields that k2 (r0 )(ϕ(r)N + U) ]|x(t) − y(t)| Γ(α)

[1 − k1 (r0 ) −

t

≤

n t [r0 k2 (r0 ) + F 2 ] ∫(t β − s β )α−1 s γ |x(s) − y(s)|ds Γ(α) 0 t

t

0

0

ζ∗ n t [r0 k2 (r0 ) + F 2 ] ζ √ ∫(t β − s β )ζ(α−1) s ζγ ds √ ∫ |x(s) − y(s)|ζ ∗ ds ≤ Γ(α)

≤

n t [r0 k2 (r0 ) + F 2 ] ζ t ζβ(α−1)+ζγ+1 ζγ + 1 √ B( , ζ(α − 1) + 1) Γ(α) β β ζ∗

t

∗ × √ ∫ |x(s) − y(s)|ζ ds,

0

62 | 2 Fractional Integral Equations where ζ and ζ ∗ are defined in the proof of Lemma 2.19. In view of (H4 ), we can rewrite the above inequality to t

t

̂ ∫ z(s)ds ≤ (1 + c(t)) ̂ z(t) ≤ c(t) ∫ z(s)ds, 0

(2.36)

0

∗

where z(t) := |x(t) − y(t)|ζ and ζ∗

̂ := c(t)

n t [k2 (r0 )r0 + F 2 ]ζ [1 − k1 (r0 ) −

∗

∗ k2 (r0 )(ϕ(r0 )N+U) ζ ∗ ] Γ(α)ζ Γ(α)

t ζβ(α−1)+ζγ+1 ζγ + 1 ×( B( , ζ(α − 1) + 1)) β β

ζ∗ ζ

.

From (2.36), we get t

z(s) z(t) ̂ ≤ ∫(1 + c(s))[ ]ds, ̂ ̂ 1 + c(t) 1 + c(s) 0

and Gronwall’s inequality implies

z(t) ̂ 1+c(t)

= 0, so z(t) = 0.

Now suppose Hölder continuity of ϕ, i.e., ϕ(|x − y|) = |x − y|ν with 0 < ν < 1 for any x, y ∈ ℝ. Then like above, we derive |x(t) − y(t)| ≤

r0 k2 (r0 ) + F 2 (1 − k1 (r0 ))Γ(α) − k2 (r0 )(ϕ(r0 )N + U) t

× ∫(t β − s β )α−1 s γ n(t, s)|x(s) − y(s)|ν ds 0

≤

(r0 k2 (r0 ) + F 2 )N (1 − k1 (r0 ))Γ(α) − k2 (r0 )(ϕ(r0 )N + U)

‖x − y‖ν ,

which gives ‖x − y‖ ≤

1−ν

√

(r0 k2 (r0 ) + F 2 )N (1 − k1 (r0 ))Γ(α) − k2 (r0 )(ϕ(r0 )N + U)

.

So we have some localization of all possible solutions. Finally, we weaken (H3 ) and (H4 ) as follows: (H󸀠3 ) There exist lim supt→∞ n(t) = N ∞ < ∞ and lim supt→∞ u(t) = U ∞ < ∞. (H󸀠4 ) There exists an r0 > 0 satisfying (H4 ) (i), (ii). Then (2.32) remains and instead of (2.34) we get β(VX) ≤ k1 (r0 )β(X) + β∞ 1 (f 1 ; r 0 ) + 2

r0 k2 (r0 ) + F 2 {ϕ(r0 )N ∞ + U ∞ }, Γ(α)

(2.37)

2.2 Quadratic Erdélyi–Kober type integral equations | 63

∞ T where β∞ 1 (f 1 ; r 0 ) := limT→∞ β 1 (f 1 ; r 0 ). From (H1 ) we deduce β 1 (f 1 ; r 0 ) < ∞. Consequently, there is a solution of (2.23), and the set fix V of all solutions satisfies

β(fix V) ≤

r0 k2 (r0 )+F 2 β∞ {ϕ(r0 )N ∞ + U ∞ } 1 (f 1 ; r 0 ) + 2 Γ(α)

1 − k1 (r0 )

.

(2.38)

Estimate (2.38) measures uniform asymptotic oscillation bounds of all solutions of equation (2.23). Remark 2.25. It is clear that (H4󸀠 ) (and so (H4 ) (i), (ii)) holds for an r0 satisfying r0 k1 (r0 ) + F 1 +

k2 (r0 )r0 + F 2 (ϕ(r0 )N + U) < r0 . Γ(α)

(2.39)

Hence it is enough to verify (2.39). Note that N ∞ ≤ N and U ∞ ≤ U.

2.2.5 Applications In this subsection, we present two examples illustrating the main results contained in Theorems 2.23 and 2.24 which are partly motivated by an example in [44]. Example 2.26. Consider the following quadratic Erdélyi–Kober type integral equation of fractional order: t

1

1

1

qx2 (t) (t 2 − s 2 ) 4 x(s) + es−t ds, x(t) = p arctan(t + cos t + x(t)) + ∫ 1 1 1 3 2 (1 + t )Γ( 4 ) (1 + t2 + s2 )(t 2 − s 2 ) 4

(2.40)

0

where t ∈ ℝ+ , p > 0 and q > 0. Clearly, (2.40) has a form of (2.23) with α = β = 12 > 0, γ = 0 ∈ (− 78 , ∞). Thus (2.24) holds, and

3 4

∈ (0, 1),

f1 (t, x) := p arctan(t + cos t + x), f2 (t, x) :=

qx2 , 1 + t2

1

u(t, s, x) :=

1

1

(t 2 − s 2 ) 4 x + es−t . 1 + t2 + s2

Now we verify the assumptions of Theorem 2.23 for (2.40). (i) For any r > 0, x, y ∈ [−r, r] and t ∈ ℝ+ , we get |f1 (t, x) − f1 (t, y)| ≤ p|t + cos t + x − (t + cos t + y)| = p|x − y|, y2 󵄨󵄨󵄨 󵄨󵄨 x2 󵄨󵄨 ≤ q|x + y| |x − y| ≤ 2rq|x − y|. |f2 (t, x) − f2 (t, y)| = q󵄨󵄨󵄨󵄨 − 󵄨 1 + t2 1 + t2 󵄨󵄨 So f i (t, x) satisfies the Lipschitz condition from (H1 ) with k1 (r) = p, k2 (r) = 2rq and F 1 = 2π p, F 2 = 0.

64 | 2 Fractional Integral Equations

(ii) Using

π 2

− arctan y ≤

for any y > 0, we obtain

1 y

|f1 (t, x) − f1 (s, x)| = p| arctan(t + cos t + x) − arctan(s + cos t + x)| p p 2p ≤ + ≤ t + cos t + x s + cos s + x T − r for any t, s ≥ T > r > 0 and x ∈ ℝ, |x| ≤ r. Thus, f1 satisfies (H4 ) (iii). (iii) For (t, s) ∈ ∆ and x, y ∈ ℝ, we get 1

1

1

(t 2 − s 2 ) 4 |x − y| . |u(t, s, x) − u(t, s, y)| = 1 + t2 + s2 Thus u(t, s, x) satisfies (H2 ) with n(t, s) = (iv) Since t

1

1 2

1 2

− 41

n(t) = ∫(t − s ) 0

1

1

(t1/2 −s1/2 )1/4 1+t2 +s2

and ϕ(r) = r, and (2.35) holds.

t

(t 2 − s 2 ) 4 1 t 1 arctan ds = ∫ ds = 1 + t2 + s2 1 + t2 + s2 √1 + t2 √1 + t2 0

and by Lemma 2.17, we have t

t

1 2

1 2

u(t) = ∫(t − s )

− 41

0

=

1 1 1 es−t 1 ds ≤ ∫(t 2 − s 2 )− 4 ds 1 + t2 + s2 1 + t2

0

7 8

2t B(2,

3 4)

1 + t2

=

7 8

32t . 21(1 + t2 )

We get limt→∞ n(t) = 0 and limt→∞ u(t) = 0. So (H3 ) holds. Further, numerically by Mathematica, we have N = sup{n(t) | t ∈ ℝ+ } ≐ 0.4383 and 7

32( √37 ) 8 32t 8 38 = U = sup{u(t) | t ∈ ℝ+ } ≤ sup = 2 9 ≐ 0.7679. 2 7 21(1 + 9 ) t∈ℝ+ 21(1 + t ) 7 16 7

1

(v) Finally, (2.39) now possesses a form r0 > pr0 +

2qr2 π p + 3 0 [0.4383r0 + 0.7679] 2 Γ( 4 )

= 0.7154qr30 + 1.2533qr20 + pr0 + 1.5708p, which is equivalent to F p (r0 ) :=

(1 − p)r0 − 1.5708p 0.7154r30 + 1.2533r20

> q.

(2.41)

2.2 Quadratic Erdélyi–Kober type integral equations | 65

r

m

25 20 15 10 5

0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0

p

0.2 0.4 0.6 0.8 1.0

p

Fig. 2.4. The graphs of functions r(m) and m(p) on (0, 1), and the range of q: 0 < q < m(p).

Since we require p > 0 and q > 0, we need p ∈ (0, 1) and look for m(p) := maxr>0 F p (r). Solving F 󸀠p (r) = 0, r > 0, we get r = r(p) :=

0.5(1.2533 − 4.6246p − 1.0714√9.6743 − p√0.1415 + p) . 1.4308(p − 1)

Hence m(p) = F p (r(p)) =

2.9291(p − 1)3 2

(1.2533 − 4.6246p − 1.0714√9.6743 − p√0.1415 + p) ×

−0.438 + 0.0453p + 0.3744√9.6743 − p√0.1415 + p −0.3362 + 0.0348p − 0.0958√9.6743 − p√0.1415 + p

.

So a necessary and sufficient condition for solvability of (2.41) for some r0 > 0 is m(p) > q. Consequently, we can take r0 = r(p). Note that limp→0 r(p) = 0, limp→0 m(r) = ∞, limp→1 r(p) = ∞ and limp→1 m(r) = 0 (see Figure 2.4). In summary, by Theorems 2.23 and 2.24, we arrive at the following result. Theorem 2.27. If p ∈ (0, 1) and q > 0 satisfy m(p) > q, then (2.40) has a unique solution. For instance, for p = 0.2 we have r(0.2) ≐ 0.6983 and m(0.2) ≐ 0.286 > q > 0. Example 2.28. Consider another quadratic integral equation involving the Erdélyi– Kober singular kernel t

x(t) = f1 (t, x(t)) +

f2 (t, x(t)) ∫(t β − s β )α−1 s γ u(t, s, x(s))ds, Γ(α) 0

where α, β, γ, f i , i = 1, 2 are the same as in the Example 2.26 and 1

u(t, s, x) :=

1

1

[(t 2 − s 2 )x(s) + es−t ] 4 . 1 + t2 + s2

t ∈ ℝ+ ,

(2.42)

66 | 2 Fractional Integral Equations Clearly, (t, s) ∈ ∆ for all x, y ∈ ℝ. Then by Lemma 2.17, we derive 1 1 1 1 1 1 1 1 󵄨 󵄨󵄨 12 󵄨󵄨[(t − s 2 ) 4 x + es−t ] 4 − [(t 2 − s 2 ) 4 y + es−t ] 4 󵄨󵄨󵄨 2 2 1+t +s 1 1 1 1 (t 2 − s 2 ) 4 |x − y| 4 ≤ . 1 + t2 + s2 Thus the function u(t, s, x) satisfies (H2 ) with the same n(t, s) as in the Example 2.26, but now ϕ(r) = r1/4 , and (2.35) is satisfied for ν = 14 . Meanwhile, n(t) is the same as in Example 2.26, but now

|u(t, s, x) − u(t, s, y)| =

t

t

1

1

1

1

u(t) = ∫(t 2 − s 2 )− 4 0

1

1

1

7

(es−t ) 4 (t 2 − s 2 )− 4 32t 8 ds ≤ ∫ ds = . 2 2 2 2 1+t +s 1+t +s 21(1 + t2 ) 0

Hence, (2.39) now possesses the form 9

r0 > 0.7154qr04 + 1.2533qr20 + pr0 + 1.5708p,

(2.43)

which is equivalent to F p (r0 ) :=

(1 − p)r0 − 1.5708p 9

0.7154r04 + 1.2533r20

> q.

Since again we require p > 0 and q > 0, we need p ∈ (0, 1) and look for m(p) := 󸀠 maxr>0 F p (r). Solving F p (r) = 0, r > 0 is equivalent to 1

5

7.6932p + 4.9403pr 4 − 2.4488(1 − p)r − 1.7473(1 − p)r 4 = 0,

(2.44)

which is a polynomial of order 5 in r1/4 . So we cannot expect its explicit solution. On the other hand, by the Descartes sign rule [192], we know that (2.44) has precisely ̄ one positive solution r(p). But still we cannot find it explicitly. For this reason, we use ̄ numerical tools of Mathematica for plotting the graphs of r(p) and m(p) in Figure 2.5. Note F p (r) is decreasing in p, so m(p) is decreasing in p and again, limp→0 m(r) = ∞ ̄ from (2.44): and limp→1 m(r) = 0. Furthermore, we solve explicitly the inverse of r−1 1

̄ (r) = r−1

1.7473(1.4015 + r 4 )r

. 1 5 7.6932 + 4.9403r 4 + 2.4488r + 1.7473r 4 ̄ is increasing with r−1 ̄ (0) = 0 and limr→∞ r−1 ̄ (r) = 1. It is elementary to check that r−1 ̄ ̄ ̄ So r(p) is also increasing in p, limp→0 r(p) = 0 and limp→1 r(p) = ∞. So a necessary and sufficient condition for solvability of (2.43) for some r0 > 0 is ̄ m(p) > q. Then we can take r0 = r(p). In summary, by Theorem 2.23, we arrive at the following result. Theorem 2.29. If p ∈ (0, 1) and q > 0 satisfy m(p) > q, then (2.42) has a solution. ̄ For instance, for p = 0.2 we have r(0.2) ≐ 0.754 and m(0.2) ≐ 0.2648 > q > 0. Moreover, using (2.37), a distance between any two solutions is less than or equal to 3

0.593168(

4 q ) < 0.362516. 0.980333 − 1.77396q

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations | 67

r

m

30 25 20 15 10 5

0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0

p

0.2 0.4 0.6 0.8 1.0

p

Fig. 2.5. The range of q: 0 < q < m(p).

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations 2.3.1 Introduction Recently, Olszowy [215] and Banaś and Cabrera [39] presented interesting existence theorems for solutions of the functional integral equation t

t

x(t) = f (t, ∫ x(s)ds, ∫ x(h(s, x(s)))ds), 0

t ∈ ℝ+ := [0, ∞)

0

in an appropriate Banach or Fréchet space. This section is an extension and continuation of the papers [42, 44, 215, 275, 295]. We will study the following generalized nonlinear functional equation involving Erdélyi–Kober fractional integrals on an unbounded interval: t

t

0

0

β α−1 β α−1 x(t) = f (t, ∫(t β − s β )α−1 s γ−1 x(s)ds, ∫(t β − s β )α−1 s γ−1 x(h(s, x(s)))ds), Γ(α) Γ(α) (2.45) where t ∈ ℝ+ , 0 < α < 1, β and γ > 0. This section is based on [297].

2.3.2 Main result Suppose that E is a real Banach space with the norm ‖⋅‖. Denote by B(r) the closed ball centered at zero element and with radius r. In this section, we will study existence of solutions of equation (2.45) under the following assumptions: (H1) f : ℝ+ × E × E → E is uniformly continuous on [0, T] × B(R) × B(R) for any T, R > 0. (H2) h : ℝ+ × E → ℝ+ is uniformly continuous on [0, T] × B(R) for any T, R > 0, and h(t, x) ≤ t for all t ∈ ℝ+ and x ∈ E. (H3) There exist continuous functions K i : ℝ+ → ℝ+ , i = 1, 2 such that μ(f(t, U, V)) ≤ K1 (t)μ(U) + K2 (t)μ(V)

68 | 2 Fractional Integral Equations for any bounded sets U, V ⊂ E, where μ is the Kuratowski measure of noncompactness in E. (H4) There exist continuous functions L i : ℝ+ → ℝ+ , i = 0, 1, 2 such that ‖f(t, x, y)‖ ≤ L0 (t) + L1 (t)‖x‖ + L2 (t)‖y‖ holds for each t ∈ ℝ+ and x, y ∈ E. Now, we formulate the following theorem which is the main result of this section. Theorem 2.30. Assume that (H1)–(H4) are satisfied and β(α − 1) + γ > 0. Then equation (2.45) has at least one solution in C(ℝ+ , E). Proof. For t ≥ 0, we define F on the space C(ℝ+ , E) as t

t

0

0

β α−1 β α−1 (Fx)(t) = f (t, ∫(t β − s β )α−1 s γ−1 x(s)ds, ∫(t β − s β )α−1 s γ−1 x(h(s, x(s)))ds). Γ(α) Γ(α) Since β(α − 1) + γ > 0 (for more details see Step 3 below, or [295]), we see that F is well defined and transforms the space C(ℝ+ , E) into C(ℝ+ , E). Step 1. We take ∆ = {x ∈ C(ℝ+ , E) | ‖x(t)‖ ≤ ψ(t), t ≥ 0} where ψ(⋅) is a continuous, nondecreasing and positive function. Clearly ∆ is convex and closed. We look for ψ(⋅) satisfying t

β α−1 L1 (t) L0 (t) + ∫(t β − s β )α−1 s γ−1 ψ(s)ds Γ(α) 0 t

β α−1 L2 (t) + ∫(t β − s β )α−1 s γ−1 ψ(s)ds ≤ ψ(t) on ℝ+ . Γ(α)

(2.46)

0

Let us begin with a construction of a positive, nondecreasing ψ n (⋅) satisfying t

β α−1 (L1 (n) + L2 (n)) L0 (n) + ∫(t β − s β )α−1 s γ−1 ψ n (s)ds ≤ ψ n (t), Γ(α)

t ∈ [0, n]

0

for any n ∈ ℕ, where L i (t) = sups≤t L i (s), i = 0, 1, 2. For simplicity we set A n := L0 (n),

B n :=

β α−1 (L1 (n) + L2 (n)) , Γ(α)

and we look for ψ n (t) = 2A n + d n t ω n with some d n > 0, ω n > 0 specified below. We compute t

A n + B n ∫(t β − s β )α−1 s γ−1 (2A n + d n s ω n )ds 0

= A n + 2A n B n

t β(α−1)+γ γ t β(α−1)+γ+ω n γ + ωn B( , α) + B n d n B( , α). β β β β

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations | 69

n Since B( γ+ω β , α) → 0 as ω n → ∞, we take and fix ω n > 0 such large that

Bn

γ + ωn 1 n β(α−1)+γ B( , α) ≤ . β β 2

Now we take t n ∈ (0, n] such that β(α−1)+γ

Bn

tn

β

γ 1 B( , α) ≤ . β 2

Then for any t ∈ [0, t n ], we have t

A n + B n ∫(t β − s β )α−1 s γ−1 ψ n (s)ds 0

= A n + 2A n B n

t β(α−1)+γ t β(α−1)+γ+ω n γ γ + ωn B( , α) + B n d n B( , α) β β β β β(α−1)+γ

γ n β(α−1)+γ t ω n γ + ωn B( , α) + B n d n B( , α) β β β

≤ A n + 2A n B n

tn

≤ 2A n + d n t

= ψ n (t).

ωn

β

Next, taking d n = max{2(2A n B n

γ n β(α−1)+γ −ω B( , α) − A n )t n n , 1}, β β

for any t ∈ [t n , n] we obtain t

A n + B n ∫(t β − s β )α−1 s γ−1 ψ n (s)ds 0

= A n + 2A n B n

t β(α−1)+γ γ t β(α−1)+γ+ω n γ + ωn B( , α) + B n d n B( , α) β β β β

≤ A n + 2A n B n

n β(α−1)+γ γ n β(α−1)+γ t ω n γ + ωn B( , α) + B n d n B( , α) β β β β

≤ A n + 2A n B n

n β(α−1)+γ γ d n ωn B( , α) + t β β 2

= A n + 2A n B n

n β(α−1)+γ γ d n ωn B( , α) − t + d n t ωn β β 2

≤ A n + 2A n B n

γ d n ωn n β(α−1)+γ B( , α) − t n + d n t ωn β β 2

≤ 2A n + d n t n := ψ n (t).

70 | 2 Fractional Integral Equations

Clearly, we have t

L0 (t) +

β α−1 L1 (t) ∫(t β − s β )α−1 s γ−1 ψ n (s)ds Γ(α) 0

+

β α−1 L

2 (t)

Γ(α)

t

∫(t β − s β )α−1 s γ−1 ψ n (s)ds ≤ ψ n (t) 0

for any t ∈ [0, n]. Next, consider the operator t

A(ψ)(t) := L0 (t) +

β α−1 L1 (t) ∫(t β − s β )α−1 s γ−1 ψ(s)ds Γ(α) 0

+

β α−1 L

t

2 (t)

∫(t β − s β )α−1 s γ−1 ψ(s)ds.

Γ(α)

0

Then A : [0, ψ n ] → [0, ψ n ] where [0, ψ n ] := {ψ ∈ C([0, n], ℝ+ ) | ψ(t) ≤ ψ n (t)}. Since L i are nondecreasing, like in [283, Theorem 6.2], we see that if ψ ∈ [0, ψ n ] is nondecreasing in t, then A(ψ)(t) is nondecreasing in t as well. So we take [0, ψ n ]m := {ψ ∈ [0, ψ n ] | ψ is nondecreasing} and A : [0, ψ n ]m → [0, ψ n ]m . Clearly A is nondecreasing and, like below in the present subsection, we can verify that A is compact. So it has a fixed point ψ n , i.e., t

ψ n (t) = L0 (t) +

β α−1 L1 (t) ∫(t β − s β )α−1 s γ−1 ψ n (s)ds Γ(α) 0

+

β α−1 L

2 (t)

Γ(α)

t

∫(t β − s β )α−1 s γ−1 ψ n (s)ds 0

for any t ∈ [0, n]. ̃ n . Then since Assume that the above equation has another solution ψ βq(α − 1) + q(γ − 1) + 1 > 0, for q > 1 sufficiently near to 1, for

1 p

+

q(γ − 1) + 1 > 0, 1 q

q(α − 1) + 1 > 0

= 1, we have

̃ n (t)| |ψ n (t) − ψ t

β α−1 (L1 (n) + L2 (n)) ̃ n (s)|ds ≤ ∫(t β − s β )α−1 s γ−1 |ψ n (s) − ψ Γ(α) 0

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations | 71

≤

≤

t

t

0

0

p β α−1 (L1 (n) + L2 (n)) q ̃ n (s)|p ds √ ∫(t β − s β )q(α−1) s q(γ−1) ds √ ∫ |ψ n (s) − ψ Γ(α)

β α−1 (L

1 (n)

+ L2 (n)) √ Γ(α)

q t βq(α−1)+q(γ−1)+1

β

B(

q(γ − 1) + 1 , q(α − 1) + 1) β

t

p ̃ n (s)|p ds, × √ ∫ |ψ n (s) − ψ 0

which reduces to ̃ n (t)|p |ψ n (t) − ψ ≤(

p β α−1 (L1 (n) + L1 (n) q t βq(α−1)+q(γ−1)+1 q(γ − 1) + 1 √ B( , q(α − 1) + 1)) Γ(α) β β t

̃ n (s)|p ds. × ∫ |ψ n (s) − ψ 0

̃ n (t) on [0, n]. By the standard Gronwall’s inequality we see that ψ n (t) = ψ Summarizing, we see that there is a unique ψ(⋅) ≥ 0 satisfying t

β α−1 L1 (t) ψ(t) = L0 (t) + ∫(t β − s β )α−1 s γ−1 ψ(s)ds Γ(α) 0 t

+

β α−1 L2 (t) ∫(t β − s β )α−1 s γ−1 ψ(s)ds Γ(α) 0

for any t ≥ 0, which is also nondecreasing. Then clearly (2.46) holds on ℝ+ . Step 2. We prove F(∆) ⊂ ∆. For any x ∈ ∆, it follows from (H4), monotonicity of ψ and (2.46) that t

‖(Fx)(t)‖ ≤ L0 (t) +

β α−1 L1 (t) ∫(t β − s β )α−1 s γ−1 ‖x(s)‖ds Γ(α) 0

+

β α−1 L

2 (t)

Γ(α)

t

∫(t β − s β )α−1 s γ−1 ‖x(h(s, x(s)))‖ds 0 t

≤ L0 (t) +

β α−1 L1 (t) ∫(t β − s β )α−1 s γ−1 ψ(s)ds Γ(α) 0

+

β α−1 L

2 (t)

Γ(α)

t

∫(t β − s β )α−1 s γ−1 ψ(s)ds ≤ ψ(t), 0

which immediately implies that F(∆) ⊂ ∆.

72 | 2 Fractional Integral Equations Step 3. We show that F(∆) is equicontinuous on compact intervals of ℝ+ . Let us fix x ∈ ∆, T > 0, ε > 0 and take t1 , t2 ∈ [0, T], t1 < t2 such that |t1 − t2 | ≤ ε. Taking p and q like above, by Hölder’s inequality, we have t

t

2 󵄨󵄨 1 β 󵄨 󵄨󵄨 ∫(t − s β )α−1 s γ−1 x(s)ds − ∫(t β − s β )α−1 s γ−1 x(s)ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 1 2 󵄨 󵄨

0

0

t2

t1

≤

β ψ(T) ∫((t1

−s )

β α−1

−

β (t2

−s )

β α−1

)s

γ−1

β

ds + ψ(T) ∫(t2 − s β )α−1 s γ−1 ds t1

0 t1

t2 β

β

= ψ(T)( ∫(t1 − s β )α−1 s γ−1 ds − ∫(t2 − s β )α−1 s γ−1 ds) 0

0 t2 β

+ 2ψ(T) ∫(t2 − s β )α−1 s γ−1 ds t1

≤ ψ(T)

β(α−1)+γ t1

β(α−1)+γ

− t2 β

p

q

γ B( , α) β

t2

β + 2ψ(T) √ t2 − t1 √ ∫(t2 − s β )q(α−1) s q(γ−1) ds t1 q

p

≤ 2ψ(T) √ε √

T βq(α−1)+q(γ−1)+1 q(γ − 1) + 1 B( , q(α − 1) + 1), β β

and similarly t

t

0

0

2 󵄨󵄨 1 β 󵄨 󵄨󵄨 ∫(t − s β )α−1 s γ−1 x(h(s, x(s)))ds − ∫(t β − s β )α−1 s γ−1 x(h(s, x(s)))ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1 󵄨 󵄨

p

q

≤ 2ψ(T) √ε √

T βq(α−1)+q(γ−1)+1 q(γ − 1) + 1 B( , q(α − 1) + 1). β β

Hence clearly, ‖(Fx)(t1 ) − (Fx)(t2 )‖ t1 t1 󵄩󵄩 β α−1 β α−1 󵄩 β β = 󵄩󵄩󵄩󵄩f (t1 , ∫(t1 − s β )α−1 s γ−1 x(s)ds, ∫(t1 − s β )α−1 s γ−1 x(h(s, x(s)))ds) Γ(α) Γ(α) 󵄩󵄩 0 0 t2 t2 󵄩󵄩 β α−1 β α−1 󵄩 β β β α−1 γ−1 − f (t2 , ∫(t2 − s ) s x(s)ds, ∫(t2 − s β )α−1 s γ−1 x(h(s, x(s)))ds)󵄩󵄩󵄩󵄩 Γ(α) Γ(α) 󵄩󵄩 0

≤ ν (f, ε), T

0

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations | 73

where ν T (f, ε) = sup{‖f(t1 , x1 , y1 ) − f(t2 , x2 , y2 )‖ | t1 , t2 ∈ [0, T], |t1 − t2 | ≤ ε}, and x1 , x2 , y1 , y2 ∈ B(

β α−2 T β(α−1)+γ B( βγ , α)ψ(T) Γ(α)

)

and max{‖y1 − y2 ‖, ‖x1 − x2 ‖} 2ψ(T)β α−1 q T βq(α−1)+q(γ−1)+1 q(γ − 1) + 1 √ B( , q(α − 1) + 1). Γ(α) β β

p

≤ √ε

From (H1) follows the uniform continuity of the function f = f(t, x, y) on the set [0, T] × B(

β α−2 T β(α−1)+γ B( βγ , α)ψ(T) Γ(α)

) × B(

β α−2 T β(α−1)+γ B( βγ , α)ψ(T) Γ(α)

).

Thus, we derive that ν T (f, ε) → 0

as ε → 0,

which yields that F(∆) is equicontinuous on compact intervals [0, T]. Step 4. We show that F is continuous on ∆. We take x, x n ∈ ∆ such that x n → x in C(ℝ+ , E). Then, x n → x uniformly on [0, T] for any T > 0. We only need to prove that Fx n → Fx uniformly on [0, T]. Fix T > 0 and take t ∈ [0, T]. We have ‖(Fx)(t) − (Fx n )(t)‖ t t 󵄩󵄩 β α−1 β α−1 󵄩 ≤ 󵄩󵄩󵄩󵄩f (t, ∫(t β − s β )α−1 s γ−1 x(s)ds, ∫(t β − s β )α−1 s γ−1 x(h(s, x(s)))ds) Γ(α) Γ(α) 󵄩󵄩 0 0 t t 󵄩󵄩 β α−1 β α−1 󵄩 β β α−1 γ−1 − f (t, ∫(t − s ) s x(s)ds, ∫(t β − s β )α−1 s γ−1 x(h(s, x n (s)))ds)󵄩󵄩󵄩󵄩 Γ(α) Γ(α) 󵄩󵄩 0

0

󵄩󵄩 β α−1 β α−1 󵄩 + 󵄩󵄩󵄩󵄩f (t, ∫(t β − s β )α−1 s γ−1 x(s)ds, ∫(t β − s β )α−1 s γ−1 x(h(s, x n (s)))ds) Γ(α) Γ(α) 󵄩󵄩 0 0 t

t

t t 󵄩󵄩 β α−1 β α−1 󵄩 β β α−1 γ−1 − f (t, ∫(t − s ) s x n (s)ds, ∫(t β − s β )α−1 s γ−1 x n (h(s, x n (s)))ds)󵄩󵄩󵄩󵄩 Γ(α) Γ(α) 󵄩󵄩

≤ ν T (f,

0 α−2 β(α−1)+γ β T B( βγ ,

+ ν T (f,

0

α)

Γ(α) β α−2 T β(α−1)+γ B( βγ , Γ(α)

sup ν1T (h, x, ‖x(s) − x n (s)‖)) s≤T

α)

sup‖x(s) − x n (s)‖), s≤T

where ν T (f, δ) = sup{‖f(t, x1 , y1 ) − f(t, x2 , y2 )‖ | t ∈ [0, T]}

74 | 2 Fractional Integral Equations

with the supremum taken over all x1 , x2 , y1 , y2 ∈ B(

β α−2 T β(α−1)+γ B( βγ , α)ψ(T) Γ(α)

), ‖x1 − x2 ‖ ≤ δ, ‖y1 − y2 ‖ ≤ δ,

and ν1T (h, x, δ) = sup{‖x(h(s, y1 )) − x(h(s, y2 ))‖ | s ∈ [0, T], y1 , y2 ∈ B(

β α−2 T β(α−1)+γ B( βγ , α)ψ(T) Γ(α)

), ‖y1 − y2 ‖ ≤ δ}.

Now using (H1) and (H2), i.e., the uniform continuity of f and h on bounded sets, we have Fx n → Fx uniformly on [0, T]. Denote Ω = conv F(∆). By Lemma A.2 we derive that Ω is equicontinuous on compact intervals of ℝ+ . Step 5. We check that the continuous mapping F : Ω → Ω satisfies all the assumptions of Lemma A.28. Fix T > 0 and define H : C(ℝ+ , E) → C(ℝ+ , E) by H(x)(t) = x(h(t, x(t))). For a fixed X ⊂ Ω, we note that conv X(t) ⊂ conv(X(t)). (2.47) Further, it follows [215] that μ(H(X)(t)) ≤ sup{μ(X(s)) | s ≤ t}.

(2.48)

Denote A := sup{μ(X(s)) | s ≤ T} and M := sup{K1 (s) + K2 (s) | s ≤ T}. Note that A < ∞. By F(X) ⊂ f (t,

t

t

0

0

β α−1 β α−1 ∫(t β − s β )α−1 s γ−1 X(s)ds, ∫(t β − s β )α−1 s γ−1 H(X)(s)ds), Γ(α) Γ(α)

assumption (H3), Lemma A.27 and (2.48), we get for any t ≤ T, ̂ 1 (X)(t)) μ(F t

t

0

0

β α−1 β α−1 ≤ μ(f (t, ∫(t β − s β )α−1 s γ−1 X(s)ds, ∫(t β − s β )α−1 s γ−1 H(X)(s)ds)) Γ(α) Γ(α) t

≤

β α−1 K1 (t) ∫(t β − s β )α−1 s γ−1 μ(X(s))ds Γ(α) 0 t

β α−1 K2 (t) + ∫(t β − s β )α−1 s γ−1 μ(H(X)(s))ds Γ(α) 0 t

AMβ α−2 B( βγ , α)t β(α−1)+γ AMβ α−1 β β α−1 γ−1 ≤ . ∫(t − s ) s ds ≤ Γ(α) Γ(α) 0

2.3 Fully nonlinear Erdélyi–Kober fractional integral equations | 75

Using (2.47), we have ̂ 1 (X)(t)) ≤ μ(conv F

AMβ α−2 B( βγ , α)t β(α−1)+γ Γ(α)

.

Repeating the same process as above, we get t

α−1 ̂ 1 (X)(s)ds, ̂ 2 (X)(t)) ≤ μ(f (t, β μ(F ∫(t β − s β )α−1 s γ−1 conv F Γ(α) 0 t

β α−1 ̂ 1 (X)(s))ds)) ∫(t β − s β )α−1 s γ−1 H(conv F Γ(α) 0

≤

β α−1 K

1 (t)

Γ(α)

t

̂ 1 (X)(s))ds ∫(t β − s β )α−1 s γ−1 μ(conv F 0 t

+

β α−1 K2 (t) ̂ 1 (X)(s)))ds ∫(t β − s β )α−1 s γ−1 μ(H(conv F Γ(α) 0

≤

≤

AM 2 β2α−3 B( βγ , (Γ(α))2 AB( βγ ,

α)

t

∫(t β − s β )α−1 s β(α−1)+2γ−1 ds

0 β(α−1)+2γ α)B( , β

α)(β α−2 Mt β(α−1)+γ )2

(Γ(α))2

.

Using (2.47) again, we have ̂ 2 (X)(t)) ≤ μ(conv F

AB( βγ , α)B( β(α−1)+2γ , α)(β α−2 Mt β(α−1)+γ )2 β (Γ(α))2

.

By the method of mathematical induction, we can prove ̂ n (X)(t)) ≤ μ(F =

, α) . . . B( (n−1)β(α−1)+nγ , α)(β α−2 Mt β(α−1)+γ )n AB( βγ , α)B( β(α−1)+2γ β β (Γ(α))n (n−1)β(α−1)+nγ Γ( βγ )Γ( β(α−1)+2γ ) . . . Γ( )A(β α−2 Mt β(α−1)+γ )n β β Γ( βγ + α)Γ( β(α−1)+2γ + α) . . . Γ( (n−1)β(α−1)+nγ + α) β β

,

which yields that ̂ n (X)(t)) sup μ(F t≤T

≤

(n−1)β(α−1)+nγ Γ( βγ )Γ( β(α−1)+2γ ) . . . Γ( )A(β α−2 MT β(α−1)+γ )n β β

Γ( βγ + α)Γ( β(α−1)+2γ + α) . . . Γ( (n−1)β(α−1)+nγ + α) β β

=: G T (n).

76 | 2 Fractional Integral Equations

In view of the formula [126, p. 895, formula 8.328.2], lim

x→∞

Γ(x + α) −α x = 1, Γ(x)

there is a constant K > 0 such that K Γ(x) ≤ Γ(x + α) x α

for all x ≥ c > 0

for arbitrary fixed c > 0. Using β(α − 1) + γ > 0, and

1 > α > 0,

β(α − 1) + γ β(α − 1) + γ (i − 1)β(α − 1) + iγ =i +1−α > i β β β

for i ∈ ℕ, we derive G T (n) = ≤ ≤ =

(n−1)β(α−1)+nγ Γ( βγ )Γ( β(α−1)+2γ ) . . . Γ( )A(β α−2 MT β(α−1)+γ )n β β

Γ( βγ + α)Γ( β(α−1)+2γ + α) . . . Γ( (n−1)β(α−1)+nγ + α) β β A(Kβ α−2 MT β(α−1)+γ )n α β(α−1)+2γ α ) β

( βγ ) (

. . . ( (n−1)β(α−1)+nγ ) β

α

A(Kβ α−2 MT β(α−1)+γ )n (n!)α ( β(α−1)+γ ) β

nα

1 A(Kβ2(α−1) MT β(α−1)+γ )n (n!)α (β(α − 1) + γ)nα α

2(α−1) MT β(α−1)+γ ) n α 1 √ A(Kβ =( ( ) ) . n! (β(α − 1) + γ)

Now it is clear that limn→∞ G T (n) = 0 for any T > 0. As a result, we have ̂ n (X)(t)) ≤ G T (n) → 0 sup μ(F t≤T

as n → ∞.

Finally, one can apply Lemma A.28 to show that F has at least one fixed point in Ω.

2.3.3 Example In this subsection, we present an example illustrating the result contained in Theorem 2.30.

2.4 Quadratic Weyl fractional integral equations | 77

Consider the following quadratic Erdélyi–Kober fractional integral equation: t

1

( 1 )− 2 e−2−t 1 1 1 1 2 3 x(t) = |sin t|(t+1) + 4 1 ∫(t 4 − s 4 )− 2 s− 4 x(s)ds 5 Γ( 2 ) 0

+

− 12

( 41 )

Γ( 21 )

t

1

1

8 3

10t + 11

1

1

1

∫(t 4 − s 4 )− 2 s− 4 x( 0

2s2 )ds, 1 + s2 + sin2 x(s)

(2.49)

for t ∈ ℝ+ . Observe that the above equation is a special case of equation (2.45) when α = 12 , β = 41 , γ = 34 , and f(t, x, y) =

2 3 1 y, |sin t|(t+1) + e−2−t x + 8 5 10t 3 + 11

h(t, x) =

2t2 . 1 + t2 + sin2 x(s)

Clearly, we have β(α − 1) + γ = 0.625 > 0 and 2 3 1 |sin t|(t+1) + e−2−t ‖x‖ + ‖y‖ 8 5 10t 3 + 11 =: L0 (t) + L1 (t)‖x‖ + L2 (t)‖y‖,

‖f(t, x, y)‖ ≤

and h(t, x) ≤

2t2 ≤ t, 1 + t2

2

1 where L0 (t) = 35 |sin t|(t+1) , L1 (t) = e−2−t and L2 (t) = 10t8/3 , t ∈ ℝ+ . +11 1 −2−t Now taking K1 (t) = e , K2 (t) = 10t8/3 +11 , t ∈ ℝ+ , all the assumptions of Theorem 2.30 are satisfied. Hence our result can be applied to equation (2.49).

2.4 Quadratic Weyl fractional integral equations 2.4.1 Introduction The aim of this section is to study the following quadratic Weyl fractional integral equation on an unbounded interval: x(t) = p(t) + f(t, x(t))(W D−α 0,2π u(t, s, x(s))),

(2.50)

where t ∈ ℝ+ := [0, +∞), α ∈ (0, ∞), p, f and u are 2π-periodic functions of the time variable t and satisfy some additional conditions. The symbol W D−α 0,2π g denotes the Weyl fractional integral of a continuous function g of order α: 2π

−α W D 0,2π g(t)

:=

1 ∫ ψ α (t − s)g(s)ds. 2π 0

78 | 2 Fractional Integral Equations The kernel ψ α (⋅) can be called the Weyl kernel (see more details in [320]), and it is defined as ∞ cos(nt − απ ) 2 ψ α (t) := 2 ∑ . (2.51) α n n=1 The Weyl kernel ψ α (⋅) defined in (2.51) is 2π-periodic and seems to be much different from the Riemann–Liouville kernel t α−1 , which appears in the definition of Riemann– Liouville fractional integral of the continuous function g of order α: t

−α RL D 0,t g(t)

:=

1 ∫(t − s)α−1 g(s)ds, Γ(α) 0

where Γ(⋅) is the Euler gamma function. However, Weyl showed that the Weyl fractional integral of a nice 2π-periodic function g can be coincided with the properly interpreted Riemann–Liouville fractional integral of g in the theory of one-dimensional fractional integrals (see [251, Lemma 19.3]). In order to study 2π-periodic solutions of equation (2.50), we have to study some basic properties of the Weyl kernel such as convergence, periodicity, continuity and boundedness (see Lemma 2.32). Under some modifications of the assumptions imposed in [198], we adopt the method of [41] and prove the existence of 2π-periodic solutions of equation (2.50) in the space of real 2π-periodic functions defined, continuous and bounded on an unbounded interval ℝ+ by using a technique of measure of noncompactness from [40]. Meanwhile, we obtain an important property, uniform local attractivity, of the 2π-periodic solutions of equation (2.50). Let us mention that there are other fractional integrals such as Erdélyi–Kober fractional integrals, Hadamard fractional integrals, Riesz fractional integro-differentiation, etc. This section is based on [70].

2.4.2 Preliminaries In this subsection, we collect some definitions and results which will be needed later. Let E be a Banach space with the standard norm ‖⋅‖. Denote by B(x, r) the closed ball centered at x and with radius r. The symbol B r stands for the ball B(θ, r) where θ is the zero element. We introduce the following 2π-periodic function space: BC2π (ℝ+ ) := {x | x ∈ BC(ℝ+ ), x(t + 2π) = x(t), t ∈ ℝ+ }. Obviously, (BC2π (ℝ+ ), ‖⋅‖2π ) is a Banach space with ‖x‖2π = sup{|x(t)| | t ∈ [0, 2π]}. We will use a measure of noncompactness in the space BC2π (ℝ+ ). We need the following preparations inspired by [41]. Take a nonempty bounded subset X ⊂ BC2π (ℝ+ ) and T := 2iπ > 0, i ∈ ℕ. For x ∈ X and ε ≥ 0, denote ω T (x, ε) := sup{|x(t) − x(s)| | t, s ∈ [0, T], |t − s| ≤ ε}.

2.4 Quadratic Weyl fractional integral equations | 79

Set ω T (X, ε) = sup{ω T (x, ε) | x ∈ X},

ω0T (X) = lim ω T (X, ε), ε→0

ω0 (X) = lim ω0T (X). i→∞

For t ∈ ℝ+ , denote X(t) = {x(t) | x ∈ X}

and

diam X(t) = sup{|x(t) − y(t)| | x, y ∈ X}.

Now, consider the function μ defined on the family MBC(ℝ+ ) by the formula μ(X) = ω0 (X) + lim sup diam X(t).

(2.52)

t→∞

Then the function μ is a measure of noncompactness in the space BC2π (ℝ+ ) (see [41]). The kernel ker μ of this measure consists of nonempty and bounded sets X such that functions from X are locally equicontinuous on ℝ+ and the thickness of the bundle formed by functions from X tends to zero at infinity. This property can help us to characterize solutions of equation (2.50) and other kinds of integral equations. In order to introduce some basic concepts, we choose Ω ⊂ BC2π (ℝ+ ), Ω ≠ 0, define Q : Ω → BC2π (ℝ+ ) and consider the operator equation x(t) = (Qx)(t),

t ∈ ℝ+ .

(2.53)

Analogously to the concept of [41], we can give the following locally attractive concept for the 2π-periodic solutions of the above operator equation (2.53). Definition 2.31. A 2π-periodic solution of equation (2.53) is said to be locally attractive if there exists a ball B(x0 , r) in the space BC2π (ℝ+ ) such that two arbitrary solutions x, y ∈ B(x0 , r) ∩ Ω of equation (2.53) satisfy lim (x(t) − y(t)) = 0.

t→∞

(2.54)

In the case when the limit (2.54) is uniform with respect to the set B(x0 , r) ∩ Ω, i.e., for each ε > 0 there exists T := 2iπ > 0 for some i ∈ ℕ such that |x(t) − y(t)| ≤ ε for all x, y ∈ B(x0 , r) ∩ Ω and for t ≥ T, we will say that 2π-periodic solutions of equation (2.53) are uniformly locally attractive.

2.4.3 Some basic properties of Weyl kernel We give the following basic properties of the Weyl kernel ψ α (⋅) defined by (2.51). Lemma 2.32. The operator ψ α (⋅) has the following properties: απ (i) For α ∈ (0, ∞), ψ α (⋅) is convergent for t ≠ 2iπ + 2n for n = 1, 2, . . . and i ∈ ℤ+ := ∞ cos nt {0, 1, 2, . . . } if and only if ∑n=1 n α is convergent for t ≠ 2iπ for i ∈ ℤ+ . Furthermore, the set Π := {t | t = 2iπ, i ∈ ℤ+ } is at most countable, and mes(Π) = 0 where mes(Π) denotes the measure of the set Π.

80 | 2 Fractional Integral Equations (ii) ψ α (⋅) is 2π-periodic. (iii) ψ α (⋅) is continuous on ℝ+ for α ∈ (1, ∞). (iv) ψ α (⋅) is continuous in (2(i − 1)π, 2iπ), i ∈ ℕ for α ∈ (0, 1]. Moreover, the limits limt→±2iπ ψ1 (t) exist but are different. (v) There exists a constant λ α such that |ψ α (t)| ≤ |λ α |,

(2.55)

where 󵄨 󵄨 ∞ 󵄨󵄨 ∞ cos nt 󵄨󵄨 󵄨󵄨 + 2󵄨󵄨󵄨 ∑ sin nt 󵄨󵄨󵄨 { 2󵄨󵄨󵄨󵄨 ∑ { 󵄨 󵄨 󵄨 { α α 󵄨 󵄨 { 󵄨 n=1 n 󵄨 󵄨 n=1 n 󵄨󵄨 { { { { { |π − t| |λ α | := { { 2 { { { ∞ { { 1 { {4 ∑ α n { n=1

if α ∈ (0, ∞), if α = 1, t ∈ (2(i − 1)π, 2iπ), i ∈ ℕ, if α ∈ (1, ∞).

(vi) For α ∈ (0, ∞), t ∈ [2(i − 1)π, 2π], i ∈ ℕ, one has 2π

󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨 ∫ ψ α (t − s)ds󵄨󵄨󵄨 ≤ |λ α |. 󵄨 󵄨󵄨 2π 󵄨 0

(vii)For α ∈ (0, ∞), 2(i − 1)π ≤ t1 < t2 ≤ 2π, i ∈ ℕ, one has 2π

1 󵄨󵄨 α 󵄨 ∫ 󵄨ψ (t2 − s) − ψ α (t1 − s)󵄨󵄨󵄨ds = O(|t2 − t1 |). 2π 󵄨 0

Proof. (i) Using the formula for trigonometric functions, cos(nt −

απ απ απ cos nt + 2 sin sin nt, ) = cos 2 2 2

we have ψ α (t) = 2 cos

απ ∞ cos nt απ ∞ sin nt + 2 sin . ∑ ∑ α 2 n=1 n 2 n=1 n α

(2.56)

sin nt But the series ∑∞ n=1 n α is convergent for any t ∈ ℝ and α ∈ (0, ∞), and the series ∞ cos nt ∑n=1 n α is convergent for α ∈ (0, ∞) and t ≠ 2iπ with i ∈ ℤ+ . Thus, the desired equivalent result in (i) is obvious. Furthermore, since Π ∼ ℤ+ and ℤ+ is a countable set, the set Π is at most countable and mes(Π) = 0. (ii) For 0 < t < 2π, it follows

cos(n(t + 2π) − nα n=1 ∞

ψ α (t + 2π) = 2 ∑

i.e., ψ α (⋅) is a 2π-periodic function.

απ 2 )

cos(nt − nα n=1 ∞

=2∑

απ 2 )

= ψ α (t),

2.4 Quadratic Weyl fractional integral equations | 81

(iii) Let α > 1. Without loss of generality, for an arbitrary ε > 0, any t1 , t2 ∈ (0, 2π) and t1 < t2 with 0 < |t2 − t1 | ≤ nε < πn , we have 󵄨󵄨 ∞ cos(nt2 − |ψ α (t2 ) − ψ α (t1 )| = 2󵄨󵄨󵄨󵄨 ∑ 󵄨 n=1 ∞ 󵄨 sin( 󵄨 ≤ 4 ∑ 󵄨󵄨󵄨󵄨 n=1󵄨

n(t2 +t1 ) 2

∞ n|t2 −t1 | 2 α n n=1

≤4∑

απ 2 )

−

− cos(nt1 − nα

απ 󵄨 2 ) 󵄨󵄨󵄨

n(t2 −t1 ) 󵄨 απ ) 󵄨󵄨 2 ) sin( 2 󵄨󵄨 α 󵄨󵄨 n ∞

≤ 2ε ∑ n=1

󵄨󵄨 󵄨 ∞ 󵄨󵄨sin( n(t2 −t1 ) )󵄨󵄨 󵄨 󵄨󵄨 2 ≤4∑ 󵄨 α n n=1

1 = O(|t2 − t1 |) → 0 nα

as t2 → t1 ,

|sin nt| where we used sin t < t for t ∈ (0, 2π ). The series ∑∞ is convergent for any n=1 n α 1 t ∈ (0, π) and α > 1, and ∑∞ n=1 n α is convergent for α > 1. (iv) Case 1: Let α ∈ (0, 1). From Zygmund’s book [338, p. 42, formula (2.6)], we see that ψ α (t) is continuous on (0, 2π). Then using the periodicity, we can deduce that ψ α (t) is continuous in (2(i − 1)π, 2iπ) for each i ∈ ℕ. Case 2: Let α = 1. From Zygmund’s book [338, p. 5, formula (2.8)], we have sin nt 1 = π−t ψ1 (t) = ∑∞ n=1 n 2 for t ∈ (0, 2π). So it is continuous in (0, 2π). But ψ (0) = 0. Thus it is discontinuous at t = 0. Now using the periodicity again, we obtain the desired result. (v) By (2.56) we can deduce 󵄨 󵄨󵄨 ∞ cos nt 󵄨󵄨 󵄨 ∞ 󵄨󵄨 + 2󵄨󵄨󵄨 ∑ sin nt 󵄨󵄨󵄨. |ψ α (t)| ≤ 2󵄨󵄨󵄨󵄨 ∑ 󵄨󵄨 󵄨 󵄨󵄨 α α n 󵄨 n 󵄨󵄨 󵄨 󵄨 n=1

Clearly,

n=1

󵄨󵄨 ∞ cos nt 󵄨󵄨 ∞ 1 󵄨󵄨 ≤ ∑ 󵄨󵄨 ∑ , 󵄨 󵄨󵄨 󵄨 n=1 n α 󵄨󵄨 n=1 n α

󵄨󵄨 ∞ sin nt 󵄨󵄨 ∞ 1 󵄨󵄨 ≤ ∑ 󵄨󵄨 ∑ . 󵄨 󵄨󵄨 󵄨 n=1 n α 󵄨󵄨 n=1 n α

In view of the above inequalities and the definition of |λ α | in (v), we get the estimation (2.55) for α ∈ (0, ∞) and α ∈ (1, ∞). The estimation (2.55) for α = 1 can be directly obtained from (iv). (vi) By (ii) we only need to check the result for t ∈ [0, 2π]. Note that 2π

∫ ψ α (t − s)ds = 0

∫

ψ α (t − s)ds + ∫ ψ α (t − s)ds.

[0,2π]\Π

Π

∫

|ψ α (t − s)|ds + ∫ |ψ α (t − s)|ds

Then by (iii) and (iv), we have 2π

∫ |ψ α (t − s)|ds ≤ 0

Π

[0,2π]\Π

≤

∫ [0,2π]\Π

|λ α |ds + ∫ |λ α |ds Π

≤ 2π|λ α | + mes(Π)|λ α | = 2π|λ α |. (vii) Using (i)–(iv) and the same procedure as in (vi), one can prove (vii).

82 | 2 Fractional Integral Equations For more results on ψ1 (⋅), we refer the reader to Zygmund’s book [338].

2.4.4 Existence and uniform local attractivity of 2π-periodic solutions In this subsection, we will study the existence and uniform local attractivity of the 2π-periodic solutions of equation (2.50). We introduce the following assumptions. [H1] p : ℝ+ → ℝ is 2π-periodic, continuous and bounded on ℝ+ . [H2] f : ℝ+ × ℝ → ℝ is 2π-periodic in t and continuous, and there exists a function m : ℝ+ → ℝ+ continuous on ℝ+ such that |f(t, x) − f(t, y)| ≤ m(t)|x − y| for all x, y ∈ ℝ. [H3] u : ℝ+ × ℝ+ × ℝ → ℝ is 2π-periodic in t and continuous. Moreover, there exist a function n : ℝ+ → ℝ+ continuous on ℝ+ and a function Φ : ℝ+ → ℝ+ continuous and nondecreasing on ℝ+ with Φ(0) = 0 and such that |u(t, s, x) − u(t, s, y)| ≤ n(t)Φ(|x − y|) for all x, y ∈ ℝ. Define u1 : ℝ+ → ℝ+ as u1 = max{|u(t, s, 0)| | 0 ≤ s ≤ t}. Clearly, u1 is 2π-periodic in t and continuous on ℝ+ . In what follows we shall consider equation (2.50) under the following assumptions. [H4] The functions a, b, c, d : ℝ+ → ℝ+ defined by a(t) = m(t)n(t)|λ α |,

b(t) = m(t)u1 (t)|λ α |,

c(t) = n(t)|f(t, 0)| |λ α |,

d(t) = u1 (t)|f(t, 0)| |λ α |

are bounded on ℝ+ , and a(⋅), c(⋅) satisfy limt→∞ a(t) = limt→∞ c(t) = 0. For brevity, define A = sup{a(t) | t ∈ ℝ+ },

B = sup{b(t) | t ∈ ℝ+ },

C = sup{c(t) | t ∈ ℝ+ },

D = sup{d(t) | t ∈ ℝ+ }.

[H5] There exists a number r0 > 0 satisfying the inequality ‖p‖2π + Ar0 Φ(r0 ) + Br0 + CΦ(r0 ) + D ≤ r0 . The inequality AΦ(r0 ) + B < 1 also holds. In order to use the technique of the fixed point theorem, we introduce the operator V : BC2π (ℝ+ ) → BC(ℝ+ ) given by (Vx)(t) = p(t) + (Fx)(t)(Ux)(t),

2.4 Quadratic Weyl fractional integral equations | 83

where F, U : BC2π (ℝ+ ) → BC(ℝ+ ) are defined by (Fx)(t) = f(t, x(t)), 2π

(Ux)(t) =

1 ∫ ψ α (t − s)u(t, s, x(s))ds, 2π

α > 1.

0

Obviously, F, U and V are well defined. Define B2π r0 = {x | x ∈ BC 2π (ℝ+ ), ‖x‖2π ≤ r 0 } ⊂ BC 2π (ℝ+ ), where r0 satisfies assumption [H5]. 2π Lemma 2.33. V : B2π r0 → B r0 .

Proof. We subdivide the proof into three steps. Step 1. We verify that V is continuous. To achieve our aim, we only need to verify that F, U are continuous operators. In fact, for any function x ∈ BC2π (ℝ+ ), it is clear that the function Fx is continuous on ℝ+ . We only need to show that the function Ux is continuous on ℝ+ . Let x ∈ BC2π (ℝ+ ) be arbitrary and fix T := 2iπ > 0 and ε := πi > 0, i ∈ ℕ. Without loss of generality we can assume that 0 ≤ t1 < t2 ≤ T with |t2 − t1 | ≤ ε. After some standard computation via Lemma 2.32, we obtain |(Ux)(t2 ) − (Ux)(t1 )| 2π

2π

0

0

󵄨󵄨 󵄨󵄨 1 1 = 󵄨󵄨󵄨󵄨 ∫ ψ α (t2 − s)u(t2 , s, x(s))ds − ∫ ψ α (t1 − s)u(t1 , s, x(s))ds󵄨󵄨󵄨󵄨 2π 󵄨 2π 󵄨 2π

≤

󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨 ∫ ψ α (t2 − s)[u(t2 , s, x(s)) − u(t1 , s, x(s))]ds󵄨󵄨󵄨 󵄨 󵄨󵄨 2π 󵄨 0

2π

󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨 ∫ [ψ α (t2 − s) − ψ α (t1 − s)]u(t1 , s, x(s))ds󵄨󵄨󵄨 + 󵄨󵄨 2π 󵄨󵄨 0

1 ≤ 2π

󵄨 󵄨 |ψ α (t2 − s)| 󵄨󵄨󵄨u(t2 , s, x(s)) − u(t1 , s, x(s))󵄨󵄨󵄨ds

∫ [0,2π]\Π

+

1 󵄨 󵄨 ∫ |ψ α (t2 − s)| 󵄨󵄨󵄨u(t2 , s, x(s)) − u(t1 , s, x(s))󵄨󵄨󵄨ds 2π Π

1 + 2π

∫

󵄨󵄨 α 󵄨 󵄨󵄨ψ (t2 − s) − ψ α (t1 − s)󵄨󵄨󵄨 |u(t1 , s, x(s))|ds

[0,2π]\Π

1 󵄨󵄨 α 󵄨 + ∫󵄨ψ (t2 − s) − ψ α (t1 − s)󵄨󵄨󵄨 |u(t1 , s, x(s))|ds 2π 󵄨 Π

84 | 2 Fractional Integral Equations 2π

≤

|λ α | T |λ α | ω (u, ε; ‖x‖2π ) mes(Π) ∫ ω1T (u, ε; ‖x‖2π )ds + 2π 2π 1 0 2π

1 + ∫ O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )]ds 2π 0

1 + O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )] mes(Π) 2π = |λ α |ω1T (u, ε; ‖x‖2π ) + O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )], where ω1T (u, ε; ‖x‖2π ) = sup{|u(t2 , s, y) − u(t1 , s, y)| | s, t1 , t2 ∈ [0, T], s ≤ t1 , s ≤ t2 , |t2 − t1 | ≤ ε, |y| ≤ ‖x‖2π }. Obviously, we have that ω1T (u, ε; ‖x‖2π ) → 0 as ε → 0 due to the uniform continuity of the function u(t, s, y) on the set [0, T] × [0, T] × [−‖x‖2π , ‖x‖2π ]. Denote n(T) := max{n(t) | [0, T]},

u1 (T) = max{u1 (t) | [0, T]}.

Then, we have ω T (Ux, ε) ≤ |λ α |ω1T (u, ε; ‖x‖2π ) + O(|t2 − t1 |)[n(T)Φ(‖x‖2π ) + u1 (T)], which implies that the function Ux is continuous on the interval [0, T] for any T > 0. Furthermore, we obtain the continuity of Ux on ℝ+ . As a result, the continuity of Vx on ℝ+ can be derived. Step 2. We show that the function Vx is bounded on ℝ+ . For an arbitrary x ∈ B2π r0 and a fixed t ∈ ℝ+ we have |(Vx)(t)| ≤ |p(t)| + ×[

1 [|f(t, x(t)) − f(t, 0)| + f(t, 0)] 2π ∫

|ψ α (t − s)|[|u(t, s, x(s)) − u(t, s, 0)| + |u(t, s, 0)|]ds

[0,2π]\Π

+ ∫ |ψ α (t − s)|[|u(t, s, x(s)) − u(t, s, 0)| + |u(t, s, 0)|]ds] Π

≤ ‖p‖2π +

m(t)|x(t)| + |f(t, 0)| 2π

∫

|ψ α (t − s)|[n(t)Φ(|x(s)|) + u1 (t)]ds

[0,2π]\Π

m(t)|x(t)| + |f(t, 0)| + ∫ |ψ α (t − s)|[n(t)Φ(|x(s)|) + u1 (t)]ds 2π Π

2.4 Quadratic Weyl fractional integral equations | 85

≤ ‖p‖2π +

m(t)‖x‖2π + |f(t, 0)| [n(t)Φ(‖x‖2π ) + u1 (t)] 2π

∫

|λ α |ds

[0,2π]\Π

+

m(t)‖x‖2π + |f(t, 0)| [n(t)Φ(‖x‖2π ) + u1 (t)] ∫ |λ α |ds 2π Π

≤ ‖p‖2π + a(t)‖x‖2π Φ(‖x‖2π ) + b(t)‖x‖2π + c(t)Φ(‖x‖2π ) + d(t). By assumption [H4], we can show that the function Vx is bounded on ℝ+ . Combining with the continuity of Vx on ℝ+ , we obtain Vx ∈ BC(ℝ+ ). Moreover, we get the inequality ‖Vx‖ ≤ ‖p‖2π + A‖x‖2π Φ(‖x‖2π ) + B‖x‖2π + CΦ(‖x‖2π ) + D. Combining the above estimate with assumption [H5], we deduce that there exists r0 > 0 such that V : B2π r0 → BC(ℝ+ ). Step 3. We show that the function Vx is 2π-periodic. To achieve our aim, we need to show that Vx ∈ BC2π (ℝ+ ) for any x ∈ BC2π (ℝ+ ). It is obvious that the function p is 2π-periodic. Meanwhile, for any x ∈ BC2π (ℝ+ ), (Fx)(t + 2π) = f(t + 2π, x(t + 2π)) = (Fx)(t). Thus, the operator F is 2π-periodic. Next, we check that Ux is also 2π-periodic. In fact, for any x ∈ BC2π (ℝ+ ), 2π

(Ux)(t + 2π) =

1 ∫ ψ α (t + 2π − s)u(t + 2π, s, x(s))ds 2π 0 2π

=

1 ∫ ψ α (t − s)u(t, s, x(s))ds = (Ux)(t), 2π 0

which implies the 2π-periodicity of Ux. Thus, we can deduce the 2π-periodicity of Vx. 2π Finally, we can deduce V : B2π r0 → B r0 due to Steps 1–3. The following assertion is an immediate consequence of Lemma 2.33. Lemma 2.34. The fixed points of the operator V are just the 2π-periodic solutions of equation (2.50). Now, we are ready to state and prove one of the main results of this section. Theorem 2.35. Let the assumptions [H1]–[H5] be satisfied. Then: (i) Equation (2.50) has at least one 2π-periodic solution x ∈ BC2π (ℝ+ ). (ii) The 2π-periodic solutions of equation (2.50) are uniformly locally attractive. 2π Proof. (i) Take X ∈ B2π r0 and X ≠ 0, where B r0 is the ball described in Lemma 2.34. Then, for x, y ∈ X and for an arbitrarily fixed t ∈ ℝ+ , using our assumptions [H2]–[H4] and Lemma 2.32, we obtain

|(Vx)(t) − (Vy)(t)| 2π

2π

0

0

󵄨󵄨 f(t, x(t)) 󵄨󵄨 f(t, y(t)) = 󵄨󵄨󵄨󵄨 ∫ ψ α (t − s)u(t, s, x(s))ds − ∫ ψ α (t − s)u(t, s, y(s))ds󵄨󵄨󵄨󵄨 2π 󵄨 2π 󵄨

86 | 2 Fractional Integral Equations 2π

≤

1 󵄨󵄨 󵄨 󵄨f(t, x(t)) − f(t, y(t))󵄨󵄨󵄨 ∫ |ψ α (t − s)| |u(t, s, x(s))|ds 2π 󵄨 0 2π

+

|f(t, y(t))| 󵄨 󵄨 ∫ |ψ α (t − s)| 󵄨󵄨󵄨u(t, s, x(s)) − u(t, s, y(s))󵄨󵄨󵄨ds 2π 0 2π

≤

1 m(t)|x(t) − y(t)| ∫ |ψ α (t − s)|[|u(t, s, x(s)) − u(t, s, 0)| + |u(t, s, 0)|]ds 2π 0 2π

1 + [|f(t, y(t)) − f(t, 0)| + |f(t, 0)|] ∫ |ψ α (t − s)|n(t)Φ(|x(s) − y(s)|)ds 2π 0 2π

≤

m(t)|x(t) − y(t)| ∫ |λ α |[n(t)Φ(|x(s)|) + u1 (t)]ds 2π 0 2π

[m(t)|y(t)| + |f(t, 0)|]n(t) + ∫ |λ α |Φ(|x(s) − y(s)|)ds 2π 0 2π

≤

|λ α |m(t)n(t)(|x(t)| + |y(t)|) ∫ Φ(|x(s)|)ds 2π 0 2π

|λ α |m(t)u1 (t) + |x(t) − y(t)| ∫ 1ds 2π 0 2π

+

|λ α |m(t)n(t)|y(t)| ∫ Φ(|x(s)| + |y(s)|)ds 2π 0 2π

+

|λ α |n(t)|f(t, 0)| ∫ Φ(|x(s)| + |y(s)|)ds 2π 0

≤ 2|λ α |m(t)n(t)r0 Φ(r0 ) + |λ α |m(t)u1 (t) diam X(t) + |λ α |m(t)n(t)r0 Φ(2r0 ) + |λ α |n(t)|f(t, 0)|Φ(2r0 ) = 2a(t)r0 Φ(r0 ) + a(t)r0 Φ(2r0 ) + c(t)Φ(2r0 ) + b(t) diam X(t). From the above estimation we derive the inequality diam(Vx)(t) ≤ 2a(t)r0 Φ(r0 ) + a(t)r0 Φ(2r0 ) + c(t)Φ(2r0 ) + b(t) diam X(t). Hence, by assumption [H4] we get lim sup diam(Vx)(t) ≤ k̄ lim sup diam X(t), t→∞

where due to assumption [H5].

t→∞

k̄ = B ≤ AΦ(r0 ) + B < 1

(2.57)

2.4 Quadratic Weyl fractional integral equations | 87

Further, take T := 2iπ > 0 and ε := πi > 0, i ∈ ℕ. Let x ∈ X be arbitrary and t1 , t2 ∈ [0, T] with |t2 − t1 | ≤ ε. Without loss of generality we assume that t1 < t2 . Then, using our assumptions, Lemma 2.32 and the previously obtained estimate, we get 󵄨 󵄨 |(Vx)(t2 ) − (Vx)(t1 )| ≤ |p(t2 ) − p(t1 )| + 󵄨󵄨󵄨(Fx)(t2 )(Ux)(t2 ) − (Fx)(t1 )(Ux)(t2 )󵄨󵄨󵄨 󵄨 󵄨 + 󵄨󵄨󵄨(Fx)(t1 )(Ux)(t2 ) − (Fx)(t1 )(Ux)(t1 )󵄨󵄨󵄨 2π

󵄨 󵄨 1 ≤ ω (p, ε) + 󵄨󵄨󵄨f(t2 , x(t2 )) − f(t1 , x(t1 ))󵄨󵄨󵄨 ∫ |ψ α (t2 − s)| |u(t2 , s, x(s))|ds 2π T

0

+

|f(t1 , x(t1 ))|{|λ α |ω1T (u,

ε; ‖x‖2π ) + O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )]}

|f(t2 , x(t2 )) − f(t2 , x(t1 ))| + |f(t2 , x(t1 )) − f(t1 , x(t1 ))| ≤ ω T (p, ε) + 2π ×

∫

|λ α |[|u(t2 , s, x(s)) − u(t2 , s, 0)| + |u(t2 , s, 0)|]ds

[0,2π]\Π

+ [|f(t1 , x(t1 )) − f(t1 , 0)| + |f(t1 , 0)|] × {|λ α |ω1T (u, ε; ‖x‖2π ) + O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )]} 2π

m(t2 )|x(t2 ) − x(t1 )| + ω1T (f, ε) ≤ ω (p, ε) + ∫ |λ α |[n(t2 )Φ(‖x‖2π ) + u1 (t2 )]ds 2π T

0

+ [m(t1 )|x(t1 )| + |f(t1 , 0)|] × {|λ α |ω1T (u, ε; ‖x‖2π ) + O(|t2 − t1 |)[n(t1 )Φ(‖x‖2π ) + u1 (t1 )]} ≤ ω T (p, ε) + |λ α |[m(t2 )ω T (x, ε) + ω1T (f, ε)][n(t2 )Φ(r0 ) + u1 (t2 )] + [m(T)r0 + f (T)]{|λ α |ω1T (u, ε; r0 ) + O(|t2 − t1 |)[n(t1 )Φ(r0 ) + u1 (t1 )]} ≤ ω T (p, ε) + |λ α |[m(t2 )n(t2 )Φ(r0 ) + m(t2 )u1 (t2 )]ω T (x, ε) + ω1T (f, ε)|λ α |[n(T)Φ(r0 ) + u1 (T)] + [m(T)r0 + f (T)]{|λ α |ω1T (u, ε; r0 ) + O(|t2 − t1 |)[n(T)Φ(r0 ) + u1 (T)]} ≤ ω T (p, ε) + [AΦ(r0 ) + B]ω T (x, ε) + ω1T (f, ε)|λ α |[n(T)Φ(r0 ) + u1 (T)] + [m(T)r0 + f (T)]{|λ α |ω1T (u, ε; r0 ) + O(|t2 − t1 |)[n(T)Φ(r0 ) + u1 (T)]}, where ω1T (f, ε) = sup{|f(t2 , x) − f(t1 , x)| | t1 , t2 ∈ [0, T], |t2 − t1 | ≤ ε, x ∈ [−r0 , r0 ]}, m(T) = max{m(t) | t ∈ [0, T]},

f (T) = max{|f(t, 0)| | t ∈ [0, T]}.

Noting the uniform continuity of the function u = u(t, s, x) on [0, T] × [0, T] × [−r0 , r0 ] and the uniform continuity of the function f = f(t, x) on [0, T] × [−r0 , r0 ], from the estimate above we obtain ω0T (Vx) ≤ kω0T (X). Consequently, ω0 (Vx) ≤ kω0 (X).

(2.58)

88 | 2 Fractional Integral Equations

Combining (2.57) and (2.58) and noting the definition of the measure of noncompactness μ given by the formula (2.52), we get μ(Vx) ≤ kμ(X).

(2.59)

Now, put B2π,1 = conv V(B2π r0 r0 ),

B2π,2 = conv(B2π,1 r0 r0 )

and so on. Clearly, from Lemma 2.34 we have B2π,1 ⊂ B2π r0 r0 . 2π,n 2π,n Further, B2π,n+1 ⊂ B r0 for n = 1, 2, . . . . The sets B r0 are closed, convex and r0 nonempty. Moreover, we get n 2π μ(B2π,n r0 ) ≤ k μ(B r0 )

(2.60)

for any n = 1, 2, . . . due to (2.59). Combining μ(B2π r0 ) = 4r 0 with (2.60) and 0 < k < 1, we obtain lim μ(B2π,n r0 ) = 0. n→∞

2π,n Thus, one can see that the set Y = ⋂∞ is nonempty, bounded, closed and convex. n=1 B r0 Also the set Y is a member of the kernel ker μ of the measure of noncompactness μ. Particularly, lim sup diam Y(t) = lim diam Y(t) = 0. (2.61) t→∞

t→∞

Therefore, V : Y → Y. Next, we prove that V is continuous on the set Y. To show this fact, fix ε > 0 and take x, y ∈ Y such that ‖x − y‖2π ≤ ε. From (2.61) and the fact that V(Y) ⊂ Y, it follows that there exists T := 2iπ > 0, i ∈ ℕ such that for an arbitrary t ≥ T, |(Vx)(t) − (Vy)(t)| ≤ ε.

(2.62)

Now we only need to investigate the case t ∈ [0, T]. Applying the assumptions, after some standard computation, we obtain |(Vx)(t) − (Vy)(t)| ≤

m(t)|x(t) − y(t)| 2π

∫

|λ α |[n(t)Φ(|x(s)|) + u1 (t)]ds

[0,2π]\Π

[m(t)|y(t)| + |f(t, 0)|]n(t) + 2π

∫

|λ α |Φ(|x(s) − y(s)|)ds

[0,2π]\Π

≤ [m(t)n(t)Φ(r0 ) + m(t)u1 (t)]|λ α |ε + [m(t)n(t)r0 + |f(t, 0)|n(t)]λ α |Φ(ε) = [a(t)Φ(r0 ) + b(t)]ε + [a(t)r0 + c(t)]Φ(ε) ≤ [AΦ(r0 ) + B]ε + [Ar0 + C]Φ(ε).

(2.63)

2.4 Quadratic Weyl fractional integral equations | 89

In view of (2.62), (2.63) and assumption [H4], the operator V is continuous on the set Y. By means of the Schauder fixed point theorem, the operator V has at least one fixed point x ∈ Y. Using Lemma 2.34, we see that this fixed point x must be a 2π-periodic solution of equation (2.50). Noting that Y ∈ ker μ, the characterization of sets belonging to ker μ and Definition 2.31, we deduce the following property. Corollary 2.36. Let all the assumptions of Theorem 2.35 hold. Then all 2π-solutions of equation (2.50) are uniformly locally attractive.

2.4.5 Example Here we present an example illustrating the main results contained in Theorem 2.35. Example 2.37. Consider the following quadratic Weyl fractional integral equation: 1

x(t) = ρ1 sin t +

1

(sin t) 3 + (sin t) 3 x(t) 2π

2π

3

× ∫ ψ2 (t − s)[ρ2 (sin t) 3 √ x2 (t) + ρ3 sin t]ds, 1

t ∈ ℝ+ ,

(2.64)

0

where ρ i ∈ ℝ+ , i = 1, 2, 3, and ψ2 (⋅) is the Weyl type kernel defined as cos(nt − π) . n2 n=1 ∞

ψ2 (t) = 2 ∑

Observe that (2.64) is a special case of equation (2.50). Set α = 2,

󵄨󵄨 ∞ cos nt ∞ sin nt 󵄨󵄨 󵄨󵄨, λ2 = 2󵄨󵄨󵄨󵄨 ∑ +∑ 2 󵄨󵄨󵄨 n 󵄨 n=1 n2 n=1 1

1

p(t) = ρ1 sin t, 1

f(t, x) = (sin t) 3 + (sin t) 3 x(t), |f(t, 0)| = (sin t) 3 , 3 1 ρ3 u(t, s, x) = ρ2 √ x2 (t) + , m(t) = (sin t) 3 , 10 sin t + 1 2 Φ(r) = r 3 , u(t, s, 0) = u1 (t) = ρ3 sin t.

1

n(t) = ρ2 (sin t) 3 ,

Clearly, p, f and u are all 2π-periodic functions. The functions a, b, c, d take the form 4

a(t) = c(t) = ρ2 (sin t) 3 λ2 ,

4

b(t) = d(t) = ρ3 (sin t) 3 λ2 .

It is easily seen that a(t) → 0 as t → ∞, and A = ρ2 λ2 . Further, we have that the function b(t) is bounded on ℝ+ , and B = ρ3 λ2 . It is also easy to check that c(t) → 0 as t → ∞.

90 | 2 Fractional Integral Equations

0.1

x

0.05

0

−0.05

−0.1 0

2

4

6

t

8

10

12

Fig. 2.6. The 2π-periodic solution of equation (2.64).

Moreover, we have C = ρ2 λ2 . Also we see that d(t) → 0 as t → ∞, and D = ρ3 λ2 . Next, the inequality 5 2 L(r) = ρ1 + ρ2 λ2 r 3 + ρ3 λ2 r + ρ2 λ2 r 3 + ρ3 λ2 ≤ r has a solution r0 = 1, if we choose ρ1 =

1 10

AΦ(1) + B =

and ρ2 = ρ3 =

1 10λ2 .

Moreover,

1 1 + < 1. 10 10

Then all assumptions given in Theorem 2.35 and Corollary 2.36 are satisfied, our results can be applied to equation (2.64). The 2π-periodic solution of equation (2.64) is displayed in Figure 2.6.

3 Fractional Differential Equations 3.1 Asymptotically periodic solutions 3.1.1 Introduction A standard approach to derive T-periodic solutions is to define the Poincaré operator which maps an initial value along the unique solution by T-units. Then the key periodicity and compactness conditions are given such that some fixed point theorems can be applied to get fixed points for the Poincaré operator, which give rise to T-periodic solutions. However, it seems that it is not possible to study the existence of periodic solutions for the fractional differential equations by constructing a suitable Poincaré operator directly. Since the fractional derivatives provide the description of memory property, it does not guarantee the property of semigroup computation for the Poincaré operator which holds back the periodicity of the Poincaré operator in some sense. We remark that Cuevas et al. [79–81] study the asymptotically periodic mild solutions for fractional functional integro-differential equations with infinite delay and fractional evolution equations of order q ∈ (1, 2) by using a characterization of generators of solution operators, which are nice and interesting results. However, there are few works on the same problem for fractional differential equations of order q ∈ (0, 1), which can be used for the analysis of periodic phenomena in nonlinear oscillation circuit [154]. Anticipating a wide interest in the subject, this section contributes in filling this important gap. We firstly consider the non-homogeneous fractional Cauchy problem c q D0,t u(t) = f(t), t ∈ J := [0, ∞), (3.1) { u(0) = u0 , q

where c D0,t is the Caputo fractional derivative of order q ∈ (0, 1) with the lower limit zero and u, f are T-periodic L∞ -functions where T > 0. We derive that problem (3.1) does not have non-constant periodic L∞ -solutions on J. Secondly, we study asymptotically periodic solutions to the semilinear fractional differential equation c q D0,t u(t) = f(t, u(t)), t ∈ J, (3.2) { u(0) = u0 , where f : J × ℝ → ℝ satisfies some additional assumptions. We give existence and uniqueness results for an S-asymptotically T-periodic solution and an Sv-asymptotically T-periodic solution for problem (3.2) in Banach space SAPT (ℝ). Furthermore, we extend the existence and uniqueness results to B ψ,χ , which is a closed nonempty convex subset of a Fréchet space. Some examples are given to illustrate the results. This section is based on [284].

DOI 10.1515/9783110522075-003

92 | 3 Fractional Differential Equations

3.1.2 Preliminaries Here we recall some basic definitions of asymptotically periodic functions. Definition 3.1 ([134, Definition 3.1]). A function f ∈ C b (J, X) is called S-asymptotically T-periodic if there exists T > 0 such that limt→∞ (f(t + T) − f(t)) = 0. In this case, we say that T is an asymptotic period of f . Let SAPT (X) represent the space formed by all the X-valued S-asymptotically T-periodic functions endowed with the uniform convergence norm denoted ‖⋅‖∞ . Then SAPT (X) is a Banach space (see [134, Proposition 3.5]). Definition 3.2 ([221, Definition 2.5]). A function f ∈ C b (J, X) is called weighted Sv= 0, where v ∈ C b (J, (0, ∞)). asymptotically T-periodic if limt→∞ f(t+T)−f(t) v(t) Let SAPv,T (X) represent the space formed by all the Sv-asymptotically T-periodic functions endowed with the norm ‖f ‖SAPv,T (X) = sup‖f(t)‖X + sup t≥0

t≥0

‖f(t + T) − f(t)‖X . v(t)

Then SAPv,T (X) are Banach spaces (see [221, Proposition 2.3]).

3.1.3 Non-existence results for periodic solutions It is well known that problem (3.1) is equivalent to the integral equation u(t) = u0 + v(t) for

t

1 v(t) = ∫(t − s)q−1 f(s)ds. Γ(q)

(3.3)

0

Assume u, f are T-periodic functions and u, f ∈ L∞ (0, T). Then taking the Laplace transform of (3.3), we derive 1 ̂ f (λ), λq

̂ v(λ) =

λ > 0.

But since v and f are T-periodic, we have T

̂ v(λ) =

∫0 v(t)e−λt dt 1 − e−λT

Then

T q

f ̂(λ) =

∫0 f(t)e−λt dt 1 − e−λT

T

λ ∫ v(t)e

−λt

0

T

,

dt = ∫ f(t)e−λt dt, 0

.

3.1 Asymptotically periodic solutions | 93

i.e., ∞

∑ λ i+q i=0

Letting λ → 0+ , we derive ∞

∑ λi i=0

T

T

0

0

∞ (−1)i (−1)i ∫ v(t)t i dt = ∑ λ i ∫ f(t)t i dt. i! i! i=0

T ∫0

f(t)dt = 0. Then T

T

0

0

∞ (−1)i (−1)i ∫ v(t)t i dt = ∑ λ i−q ∫ f(t)t i dt. i! i! i=1 T

Letting λ → 0+ , we derive ∫0 v(t)dt = 0. Then ∞

∑λ

i−1+q (−1)

i

i!

i=1

T

∞

∫ v(t)t dt = ∑ λ i

i=1

0

i−1 (−1)

i!

i

T

∫ f(t)t i dt. 0

T

Letting λ → 0+ , we derive ∫0 f(t)tdt = 0. By mathematical induction we derive T

T

∫ v(t)t i dt = 0,

∫ f(t)t i dt = 0

0

0

for any i = 0, 1, 2, . . . . But then f = v = 0. So there are no non-constant T-periodic L∞ -solutions on J of problem (3.1). On the other hand, we have t+T

t

1 1 u(t + T) − u(t) = ∫ (t + T − s)q−1 f(s)ds − ∫(t − s)q−1 f(s)ds Γ(q) Γ(q) =

0

0

t

t

−T

0

1 1 ∫ (t − s)q−1 f(s + T)ds − ∫(t − s)q−1 f(s)ds Γ(q) Γ(q) 0

=

1 ∫ (t − s)q−1 f(s)ds. Γ(q) −T

So we obtain 0

󵄨󵄨 󵄨󵄨 1 |u(t + T) − u(t)| ≤ 󵄨󵄨󵄨󵄨 ∫ (t − s)q−1 f(s)ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 −T

0

‖f ‖∞ ‖f ‖∞ T ≤ , ∫ (t − s)q−1 ds ≤ Γ(q) Γ(q)t1−q −T

which implies limt→∞ (u(t + T) − u(t)) = 0. Thus, u is asymptotically T-periodic on J.

94 | 3 Fractional Differential Equations

3.1.4 Existence results for asymptotically periodic solutions In this subsection, we discuss the existence of weighted asymptotically periodic solutions for problem (3.2). To begin, we recall the definition of a solution for problem (3.2). Definition 3.3. A function u ∈ C(J, ℝ) is said to be a solution of problem (3.2) if u q satisfies the equation c D0,t u(t) = f(t, u(t)) a.e. on J, and the condition u(0) = u0 . It is not hard to see that problem (3.2) is equivalent to the following integral equation: t

1 u(t) = u0 + ∫(t − s)q−1 f(s, u(s))ds. Γ(q) 0

We are ready to state the existence and uniqueness results for asymptotically periodic solutions. Theorem 3.4. Assume in addition to the continuity of f that there exist two functions a, b ∈ C(J, ℝ) such that |f(t, u1 ) − f(t, u2 )| ≤ a(t)|u1 − u2 |

{

|f(t + T, u) − f(t, u)| ≤ b(t)(|u| + 1)

for all t ∈ J, u1 , u2 ∈ ℝ, for all t ∈ J, u ∈ ℝ.

(3.4)

If t

󵄨󵄨 󵄨󵄨 m∗ = sup 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 f(s, 0)ds󵄨󵄨󵄨󵄨 < ∞, 󵄨 t∈J 󵄨

(3.5)

0

t

lim ∫(t − s)q−1 b(s)ds = 0,

(3.6)

t→∞

0 t

ρ := sup ∫(t − s)q−1 a(s)ds < Γ(q), t≥0

(3.7)

0

then problem (3.2) has a unique S-asymptotically T-periodic solution on J. Proof. Let F : SAPT (ℝ) → C b (J, ℝ) be the operator defined by t

(Fu)(t) = u0 +

1 ∫(t − s)q−1 f(s, u(s))ds. Γ(q)

(3.8)

0

Step 1. We show that F(SAPT (ℝ)) ⊂ SAPT (ℝ). If u ∈ SAPT (ℝ) then for any ϵ > 0 there is a positive number t ϵ > 0 such that sups≥t ϵ |u(s + T) − u(s)| ≤ ϵ. Consequently, for t > t ϵ , we observe that (Fu)(t + T) − (Fu)(t) t+T

t

0

0

1 1 = ∫ (t + T − s)q−1 f(s, u(s))ds − ∫(t − s)q−1 f(s, u(s))ds Γ(q) Γ(q)

3.1 Asymptotically periodic solutions | 95

=

t

t

−T

0

1 1 ∫ (t − s)q−1 f(s + T, u(s + T))ds − ∫(t − s)q−1 f(s, u(s))ds Γ(q) Γ(q) t

=

1 ∫(t − s)q−1 (f(s + T, u(s + T)) − f(s, u(s + T)))ds Γ(q) 0 t

1 + ∫(t − s)q−1 (f(s, u(s + T)) − f(s, u(s)))ds Γ(q) 0 0

1 ∫ (t − s)q−1 f(s + T, u(s + T))ds. Γ(q)

+

(3.9)

−T

Next we derive t

󵄨󵄨 󵄨 󵄨󵄨 ∫(t − s)q−1 (f(s + T, u(s + T)) − f(s, u(s + T)))ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 0

t

≤ (‖u‖∞ + 1) ∫(t − s)q−1 b(s)ds,

(3.10)

0

and t

󵄨󵄨 󵄨 󵄨󵄨 ∫(t − s)q−1 (f(s, u(s + T)) − f(s, u(s)))ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 0

t

󵄨󵄨 ϵ 󵄨󵄨 ≤ 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 (f(s, u(s + T)) − f(s, u(s)))ds󵄨󵄨󵄨󵄨 󵄨 󵄨 0

t

󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 (f(s, u(s + T)) − f(s, u(s)))ds󵄨󵄨󵄨󵄨 󵄨 󵄨 tϵ

tϵ

t

≤ ∫(t − s)q−1 a(s)ds + ϵ ∫(t − s)q−1 a(s)ds ≤ ρ(( tϵ

0

t ϵ 1−q + ϵ), ) t

since t − s ≥ ttϵ (t ϵ − s) for any s ∈ [0, t ϵ ]. Furthermore, since f is continuous, it holds max

s∈[0,T], |v(s)|≤‖u‖∞

and we derive

|f(s, v(s))| = M ∗u < ∞,

0

∗ 󵄨󵄨 󵄨 󵄨󵄨 ∫ (t − s)q−1 f(s + T, u(s + T))ds󵄨󵄨󵄨 ≤ M u T . 󵄨󵄨 󵄨󵄨 1−q 󵄨 󵄨 t −T

(3.11)

96 | 3 Fractional Differential Equations

Therefore, we obtain t

ρ t ϵ 1−q ‖u‖∞ + 1 + ϵ), |(Fu)(t + T) − (Fu)(t)| ≤ (( ) ∫(t − s)q−1 b(s)ds + Γ(q) Γ(q) t 0

which implies lim ((Fu)(t + T) − (Fu)(t)) = 0.

t→∞

Moreover, t

t

0

0

󵄨󵄨 ‖u‖∞ 1 󵄨󵄨󵄨 󵄨󵄨 ∫(t − s)q−1 f(s, 0)ds󵄨󵄨󵄨 |(Fu)(t)| ≤ |u0 | + ∫(t − s)q−1 a(s)ds + 󵄨 󵄨󵄨 Γ(q) Γ(q) 󵄨 ‖u‖∞ ρ + m∗ . ≤ |u0 | + Γ(q) As a result, F(SAPT (ℝ)) ⊂ SAPT (ℝ). Step 2. We prove that F is a contraction mapping. For all u, v ∈ SAPT (ℝ), one has t

1 |(Fu)(t) − (Fv)(t)| ≤ ∫(t − s)q−1 |u(s) − v(s)|ds Γ(q) 0 t

≤

1 ( ∫(t − s)q−1 a(s)ds)‖u − v‖∞ , Γ(q) 0

which implies ‖Fu − Fv‖∞ ≤

ρ ‖u − v‖∞ . Γ(q)

ρ It comes from condition (3.7) that Γ(q) < 1. Then F is a contraction mapping and there exists a unique S-asymptotically T-periodic solution on J.

Remark 3.5. Instead of (3.11), we can use 0

∗ 󵄨󵄨 󵄨 󵄨󵄨 ∫ (t − s)q−1 f(s + T, u(s + T))ds󵄨󵄨󵄨 ≤ M u ((t + T)q − t q ) for all t ∈ J. 󵄨󵄨 󵄨󵄨 q 󵄨 󵄨 −T

Note t 󳨃→ (t + T)q − t q is a positive decreasing function on J satisfying (t + T)q − t q ≤

Tq t1−q

for all t ∈ J, q

q

T which gives (3.11). So we can replace t1−q with (t+T)q −t in the above considerations. This term is better, since it is regular at t = 0. But for studying asymptotic properties as T t → ∞, the term t1−q is sufficient.

3.1 Asymptotically periodic solutions | 97

Example 3.6. Motivated by [154], consider the following electrical circuit like fractional differential equation: 1 sin u(t) c 2 { + cos 2πt, { D0,t u(t) = α √t + 1 { { u(0) = 0. {

t ∈ J, α ∈ ℝ,

(3.12)

Then set T = 1, a(t) = √|α| , b(t) = √|α| and f(t, 0) = cos 2πt. We already know − √|α| t+1 t+1 t+2 from arguments of Section 3.1.3 that m∗ < ∞. As a matter of fact, we have t

∫ 0 s

cos 2πs ds = C(2√t) cos 2πt + S(2√t) sin 2πt, √t − s s

2

2

where C(s) = ∫0 cos( πs2 )ds and S(s) = ∫0 sin( πs2 )ds are Fresnel integrals. Hence we have m∗ ≤ 2. Next, we compute t

lim ∫(t − s)− 2 b(s)ds = lim 2|α|(arctan √t − arctan 1

t→∞

t→∞

√2t )=0 2

0

and

t

∫(t − s)− 2 a(s)ds = 2|α| arctan √t. 1

0

So ρ = |α|π, and since Γ( 12 ) 1 |α| < √π .

= √π, the assumptions of Theorem 3.4 hold for (3.12) when

Theorem 3.7. Assume the assumptions of Theorem 3.4 hold. If a function t

a(t)M ∗ T a(t)(‖u‖∞ + 1) μ(t) = 1−q u + ∫(t − s)q−1 b(s)ds Γ(q) t Γ(q) 0

is nondecreasing on J and lim

t→∞

μ(t) E q (a(t)t q ) = 0, a(t)v(t)

(3.13)

then problem (3.2) has a unique weighted Sv-asymptotically T-periodic solution on J. Proof. Let F : SAPT (ℝ) → C b (J, ℝ) be the operator defined by (3.8). Due to Theorem 3.4, F has a unique fixed point u which is the solution of problem (3.2). We only need to check that this u is a weighted Sv-asymptotically T-periodic function.

98 | 3 Fractional Differential Equations In fact, for t ∈ J we observe that ω(t) :=

u(t + T) − u(t) v(t) 0

1 = ∫ (t − s)q−1 f(s + T, u(s + T))ds v(t)Γ(q) −T t

1 + ∫(t − s)q−1 (f(s + T, u(s + T)) − f(s, u(s + T)))ds v(t)Γ(q) 0 t

+

1 ∫(t − s)q−1 (f(s, u(s + T)) − f(s, u(s)))ds. v(t)Γ(q) 0

By (3.4), (3.10) and (3.11), we arrive at t

|ω(t)| ≤

1 μ(t) + ∫(t − s)q−1 a(s)v(s)|ω(s)|ds. a(t)v(t) v(t)Γ(q) 0

Setting θ(t) = a(t)v(t)|ω(t)|, we obtain t

θ(t) ≤ μ(t) +

a(t) ∫(t − s)q−1 θ(s)ds. Γ(q) 0

Applying Lemma A.18, we obtain θ(t) ≤ μ(t)E q (a(t)t q ) which implies |ω(t)| ≤

μ(t) E q (a(t)t q ). a(t)v(t)

Due to condition (3.13), we obtain lim ω(t) = 0.

t→∞

This completes the proof.

3.1.5 Further extensions Consider the space X = C(J, ℝ) with the topology of uniform convergence on finite subintervals. Then X is a Fréchet space. Let ψ, χ ∈ C(J, [0, ∞)). Take a subset B ψ,χ = {u ∈ X | |u(t)| ≤ ψ(t), |u(t + T) − u(t)| ≤ χ(t) for all t ∈ J}.

3.1 Asymptotically periodic solutions | 99

Then B ψ,χ is a closed nonempty convex subset of X. Next, consider F given by (3.8). Then F : X → X is continuous. Indeed, if {u n }n≥1 ⊂ X and u n → u in X, then for any k > 0, u n (t) → u(t) uniformly on J k = [0, k]. Since f is continuous, we have q q t f(t, u n (t)) → f(t, u(t)) uniformly on J k . Using ∫0 (t − s)q−1 ds = tq ≤ kq , we see that F(u n )(t) → F(u)(t) uniformly on J k . So the continuity of F is verified. Now we prove the pre-compactness of F(B ψ,χ ). If u ∈ B ψ,χ then t

󵄨󵄨 󵄨󵄨 kq mk 1 |F(u)(t)| = 󵄨󵄨󵄨󵄨u0 + ∫(t − s)q−1 f(s, u(s))ds󵄨󵄨󵄨󵄨 ≤ |u0 | + Γ(q) Γ(q + 1) 󵄨 󵄨

for all t ∈ J k ,

0

where mk =

max

t∈J k |v(t)|≤maxt∈J k ψ(t)

|f(t, v(t))|.

Next, let t1 , t2 ∈ J k , t1 < t2 . Then |F(u)(t2 ) − F(u)(t1 )| t

t

0

0

1 󵄨󵄨 󵄨󵄨 1 2 1 = 󵄨󵄨󵄨󵄨 ∫(t2 − s)q−1 f(s, u(s))ds − ∫(t1 − s)q−1 f(s, u(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 Γ(q)

t2

t

1 󵄨󵄨 󵄨󵄨 1 󵄨󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨 ∫(t2 − s)q−1 f(s, u(s))ds󵄨󵄨󵄨 + 󵄨󵄨 ∫((t2 − s)q−1 − (t1 − s)q−1 )f(s, u(s))ds󵄨󵄨󵄨 ≤ 󵄨 󵄨 󵄨 󵄨󵄨 Γ(q) 󵄨 󵄨 Γ(q) 󵄨

t1

≤

0

)q

q (t2

− (2(t2 − t1 + Γ(q + 1)

q t1 ))m k

.

We see that the restriction of F(B ψ,χ ) onto J k is uniformly bounded and equicontinuous. So by the Arzelà–Ascoli theorem, it is pre-compact in C(J k , ℝ). Since this holds for any k > 0, F(B ψ,χ ) is pre-compact in X. Now we find conditions for F(B ψ,χ ) ⊂ B ψ,χ .

(3.14)

By the assumptions of Theorem 3.4, we derive t

󵄨󵄨 󵄨󵄨 1 |F(u)(t)| = 󵄨󵄨󵄨󵄨u0 + ∫(t − s)q−1 f(s, u(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 0

t

t

0

0

󵄨󵄨 1 1 󵄨󵄨󵄨 󵄨󵄨 ∫(t − s)q−1 f(s, 0)ds󵄨󵄨󵄨 ≤ |u0 | + ∫(t − s)q−1 a(s)ψ(s)ds + 󵄨 󵄨󵄨 Γ(q) Γ(q) 󵄨 for all t ∈ J. So we need to suppose t

t

0

0

󵄨󵄨 1 󵄨󵄨󵄨 1 󵄨󵄨 ∫(t − s)q−1 f(s, 0)ds󵄨󵄨󵄨 ≤ ψ(t) |u0 | + ∫(t − s)q−1 a(s)ψ(s)ds + 󵄨󵄨 Γ(q) Γ(q) 󵄨󵄨

(3.15)

100 | 3 Fractional Differential Equations for all t ∈ J. Next using (3.4) and (3.9), we arrive at t

1 |(Fu)(t + T) − (Fu)(t)| ≤ ∫(t − s)q−1 (b(s)(ψ(s + T) + 1) + a(s)χ(s))ds Γ(q) 0

(a∗ ψ∗ + M0∗ )T + , Γ(q)t1−q where a∗ = maxs∈[0,T] |a(s)| and ψ∗ = maxs∈[0,T] |ψ(s)|. Hence we need to suppose t

(a∗ ψ∗ + M0∗ )T 1 ≤ χ(t) ∫(t − s)q−1 (b(s)(ψ(s + T) + 1) + a(s)χ(s))ds + Γ(q) Γ(q)t1−q

(3.16)

0

for all t ∈ J. Consequently, (3.15) and (3.16) imply (3.14). Summarizing, we have the next result. Theorem 3.8. Assume condition (3.4) holds. If (3.15) and (3.16) hold, then problem (3.2) has a unique solution in B ψ,χ . Proof. The existence follows from the Tikhonov fixed point theorem [165, 166, 204], and the uniqueness from condition (3.4) and Lemma A.18. Remark 3.9. Assume in (3.15) that ‖a‖∞ < ∞ and ‖f(⋅, 0)‖∞ < ∞. Then instead of (3.15), we consider the following: t

|u0 | +

‖a‖∞ ‖f(⋅, 0)‖∞ q t ≤ ψ(t) ∫(t − s)q−1 ψ(s)ds + Γ(q) Γ(q + 1)

(3.17)

0

for all t ∈ J. Take ψ(t) =

Υeϖt

for some positive Υ > 0 and ϖ > 0. Then using t

∫(t − s)q−1 eϖs ds ≤

eϖt , ϖq

0

from (3.17) we derive |u0 | +

Υ‖a‖∞ ϖt ‖f(⋅, 0)‖∞ q e + t ≤ Υeϖt Γ(q)ϖ q Γ(q + 1)

for all t ∈ J,

|u0 | +

‖f(⋅, 0)‖∞ q ‖a‖∞ t ≤ Υ(1 − )eϖt Γ(q + 1) Γ(q)ϖ q

for all t ∈ J.

i.e.,

Putting 1

ϖ=( we obtain 2|u0 | +

2‖a‖∞ q ) , Γ(q)

2‖f(⋅, 0)‖∞ q t ≤ Υeϖt Γ(q + 1)

for all t ∈ J.

(3.18)

3.1 Asymptotically periodic solutions | 101

q q ϖt ) e for all t ∈ J, we compute Since t q ≤ ( eϖ

2|u0 | +

2‖f(⋅, 0)‖∞ q q ϖt 2‖f(⋅, 0)‖∞ q t ≤ (2|u0 | + ( ) )e . Γ(q + 1) Γ(q + 1) eϖ

So we take Υ = 2|u0 | +

2‖f(⋅, 0)‖∞ q q ( ) Γ(q + 1) eϖ

to satisfy (3.18). In summary, inequality (3.17) holds for ψ(t) := 2(|u0 | +

1 ‖f(⋅, 0)‖∞ q )e(2‖a‖∞ /Γ(q)) t . q 1−q 2e q ‖a‖∞

Next, instead of (3.16), we consider t

1 ∫(t − s)q−1 (b(s)(ΥeϖT + 1)eϖs + ‖a‖∞ χ(s))ds Γ(q) 0

+

(‖a‖∞ ψ∗ + M0∗ )((t + T)q − t q ) = χ(t) Γ(q + 1)

(3.19)

for all t ∈ J. Equation (3.19) has a solution χ on J by the above considerations. Moreover, an upper growth of χ on J can be estimated by Lemma A.18. This remark is also applicable to electrical circuits modeled by fractional differential equations [154]. Furthermore, our approach can be directly extended to the system of fractional differential equations. Next suppose condition (3.7) holds. Then we can take ψ(t) = and

󵄨 t 󵄨 Γ(q)|u0 | + 󵄨󵄨󵄨∫0 (t − s)q−1 f(s, 0)ds󵄨󵄨󵄨

(3.20)

Γ(q) − ρ

t

χ(t) =

∫0 (t − s)q−1 b(s)(ψ(s + T) + 1)ds + (a∗ ψ∗ + M0∗ )T/t1−q Γ(q) − ρ

.

(3.21)

So we have the next result. Theorem 3.10. Assume conditions (3.4) and (3.7) hold. Then problem (3.2) has a unique solution in B ψ,χ with (3.20) and (3.21). Let us assume also conditions (3.5) and (3.6). Then ψ(t) ≤

Γ(q)|u0 | + m∗ Γ(q) − ρ

and t

χ(t) ≤

(m∗ + 1) ∫0 (t − s)q−1 b(s)ds + (a∗ ψ∗ + M0∗ )T/t1−q Γ(q) − ρ

→0

as t → ∞.

102 | 3 Fractional Differential Equations

This gives an alternative proof of Theorem 3.4, but in addition, now we get the above growth estimates (3.20) and (3.21) for a unique S-asymptotically T-periodic solution on J. Example 3.11. Again, motivated by [154], consider the following electrical circuit like fractional differential equation: 1 sin u(t) c 2 { + 1 + cos 2πt, { D0,t u(t) = α √t + 1 { { u(0) = 0. {

t ∈ J, α ∈ ℝ,

(3.22)

Set T = 1,

|α| , √t + 1

a(t) =

b(t) =

|α| |α| − , √t + 1 √t + 2

f(t, 0) = 1 + cos 2πt.

We already know from Example 3.6 that ρ = |α|π, and assumption (3.7) holds for 1 |α| < √π . Next, from (3.20) we derive ψ(t) =

t 1+cos 2πs ds √t−s

∫0

√π − |α|π

2√t + C(2√t) cos 2πt + S(2√t) sin 2πt 2√t + 2 ≤ . √π − |α|π √π − |α|π

=

By (3.21), this implies t

χ(t) ≤

Since a∗ = |α|, ψ∗ ≤ (

1

− ∫0 (t − s)− 2 ( √|α| s+1

1 √s + 1

+ 1)ds +

√π − |α|π

(a∗ ψ∗ +M0∗ ) √t

.

M0∗ = 2 and

4 , √π−|α|π

−

√s+1+2 |α| )( 2√π−|α|π √s+2

1 √s + 2

)(

2√s + 1 + 2 5 + 1) ≤ , √π − |α|π 2(s + 2)(√π − |α|π)

we compute χ(t) ≤ =

t 5|α| ∫ (t 2(√π−|α|π) 0

1

− s)− 2 (s + 2)−1 ds + √π − |α|π

5|α| 2 2(√π−|α|π) √2+t

arcsinh( √√t ) + 2

√π − |α|π

1 ( 4|α| √t √π−|α|π

1 ( 4|α| √t √π−|α|π

+ 2)

+ 2)

→0

as t → ∞.

Furthermore, (3.22) gives t

u(t) =

1 1 sin u(s) + 1 + cos 2πs)ds, ∫(t − s)− 2 (α √π √s + 1

0

which implies t

−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨u(t) − 2√t 󵄨󵄨󵄨 ≤ |α| ∫ (t − s) 2 ds + 󵄨󵄨󵄨 cos 2πtC(2√t) + sin 2πtS(2√t) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 √π 󵄨 √π √s + 1 √π 󵄨 󵄨 󵄨 0

≤ √π|α| +

2 . √π

3.2 Modified fractional iterative functional differential equations | 103

1 So when |α| < √π , all the assumptions of Theorem 3.10 hold for (3.22), and thus (3.22) √t has a unique solution on J which in addition has the properties u(t) = 2√π + O(1) and limt→∞ (u(t + 1) − u(t)) = 0.

3.2 Modified fractional iterative functional differential equations 3.2.1 Introduction Picard and weakly Picard operator methods are a powerful tool to study the nonlinear differential equations. For more details on this novel methods to discuss existence and uniqueness and the data dependence on data of solutions for some differential equations and integral equations, we refer the reader to Rus et al. [239–243, 247, 248], Şerban et al. [256], Muresan [202, 203] and Olaru [213]. Wang et al. [312] apply these interesting methods to study nonlocal Cauchy problems and impulsive Cauchy problems for nonlinear differential equations. Motivated by [72, 103, 247, 312], we study boundary value problems for the following modified fractional iterative functional differential equation: q

c D a,t x(t) = f(t, x(t), x(x v (t))), t ∈ [a, b], v ∈ ℝ \ {0}, q ∈ (1, 2), { { { x(t) = φ(t), t ∈ [a1 , a], { { { x(t) = ψ(t), t ∈ [b, b1 ], {

(3.23)

q

where c D a,t is the Caputo fractional derivative of order q with the lower limit a and (C1 ) a, b, a1 , b1 ∈ ℝ, a1 ≤ a < b ≤ b1 , and the function Υ(z) = z v satisfies Υ ∈ C([a1 , b1 ], [a1 , b1 ]); (C2 ) f ∈ C([a, b] × [a1 , b1 ]2 , ℝ); (C3 ) φ ∈ C([a1 , a], [a1 , b1 ]) and ψ ∈ C([b, b1 ], [a1 , b1 ]); (C4 ) there exists L f > 0 such that |f(t, u1 , w1 ) − f(t, u2 , w2 )| ≤ L f (|u1 − u2 | + |w1 − w2 |) for all t ∈ [a, b], u i , w i ∈ [a1 , b1 ], i = 1, 2. A function x ∈ C([a1 , b1 ], [a1 , b1 ]) is said to be a solution of problem (3.23) if q x satisfies the equation c D a,t x(t) = f(t, x(t), x(x v (t))) on [a, b], and the conditions x(t) = φ(t) for t ∈ [a1 , a], and x(t) = ψ(t) for t ∈ [b, b1 ]. It is easy to verify that problem (3.23) is equivalent to the following fixed point equation: φ(t) { { { x(t) = {w(φ, ψ)(t) + { { {ψ(t)

if t ∈ [a1 , a], b 1 Γ(q) ∫a

G(t, s)f(s, x(s), x(x v (s)))ds

if t ∈ [a, b], if t ∈ [b, b1 ],

(3.24)

104 | 3 Fractional Differential Equations and x ∈ C([a1 , b1 ], [a1 , b1 ]), where w(φ, ψ)(t) := φ(a) +

ψ(b) − φ(a) (t − a), b−a

G is the Green function defined by t−a {(t − s)q−1 − b−a (b − s)q−1 G(t, s) := { t−a − (b − s)q−1 { b−a

if a ≤ s ≤ t ≤ b, if a ≤ t ≤ s ≤ b,

and b

b

∫ G(t, s)f(s, x(s), x(x v (s)))ds = −

t−a ∫(b − s)q−1 f(s, x(s), x(x v (s)))ds b−a a

a

t

+ ∫(t − s)q−1 f(s, x(s), x(x v (s)))ds. a t−a Remark 3.12. Note G(t, s) ≤ 0 for a ≤ t ≤ s ≤ b, while G(t, t) = − b−a (b − t)q−1 < 0 and q−2 q−2 G(t, a) = (t − a)((t − a) − (b − a) ) > 0 for a < t < b. Next

∂ (b − s)q−2 (t − a) − (t − s)q−2 (b − a) G(t, s) = (q − 1) < 0 for a < s < t < b. ∂s b−a So for any t ∈ (a, b) there is a unique s(t) ∈ (a, t) such that G(t, s(t)) = 0, G(t, s) < 0 for s(t) < s < b, and G(t, s) > 0 for s(t) > s ≥ a. Note that G(t, b) = 0 and 1

s(t) = b +

(b − a) q−1 (b − t)

. 1 1 (t − a) q−1 − (b − a) q−1 Furthermore, s(t) = a for q = 2, which is a great difference for the Green function in comparison to q ∈ (1, 2), since we cannot expect monotonicity of the integral operator B f defined in (3.25) below. On the other hand, we derive b

∫ G(t, s)ds = a b ∫a

t−a ((t − a)q−1 − (b − a)q−1 ). q b

Hence G(t, s)ds ≤ 0 for t ∈ [a, b], and ∫a G(t, s)ds < 0 for t ∈ (a, b], which holds also for q = 2. So G(t, ⋅) is nonpositive in average on [a, b]. On the other hand, the first equation of problem (3.23) is equivalent to x(t) if t ∈ [a1 , a], { { { b 1 v x(t) := {w(x|[a1 ,a] , x|[b,b1 ] )(t) + Γ(q) ∫ G(t, s)f(s, x(s), x(x (s)))ds if t ∈ [a, b], a { { x(t) if t ∈ [b, b1 ], { and x ∈ C([a1 , b1 ], [a1 , b1 ]). We use the weakly Picard operator technique to obtain some existence, uniqueness and data dependence results for the solution of problem (3.23). This section is based on [286].

3.2 Modified fractional iterative functional differential equations | 105

3.2.2 Notation, definitions and auxiliary facts We need some notions and results from the weakly Picard operator theory; for more details see Rus [242, 243]. Let (X, d) be a metric space and A : X → X an operator. We shall use the following notation: ∙ F A = {x ∈ X | A(x) = x} denotes the fixed point set of A; ∙ I(A) = {Y ∈ P(X) | A(Y) ⊆ Y, Y ≠ 0}; ∙ A n+1 = A n ∘ A, A1 = A, A0 = I, n ∈ ℕ; ∙ P(X) = {Y ⊆ X | Y ≠ 0}; ∙ O A (x) = {x, A(x), A2 (x), . . . , A n (x), . . . } is the A-orbit of x ∈ X; ∙ H : P(X) × P(X) → ℝ+ ∪ {+∞}; ∙ H(Y, Z) = max{supy∈Y inf z∈Z d(y, z), supz∈Z inf y∈Y d(y, z)} is the Pompeiu–Hausdorff functional on P(X) × P(X). Definition 3.13. Let (X, d) be a metric space. An operator A : X → X is a Picard operator if there exists x∗ ∈ X such that F A = {x∗ } and the sequence (A n (x0 ))n∈ℕ converges to x∗ for all x0 ∈ X. Theorem 3.14 (Contraction principle). Let (X, d) be a complete metric space and let A : X → X be a γ-contraction. Then (i) F A = {x∗ }; (ii) (A n (x0 ))n∈ℕ converges to x∗ for all x0 ∈ X; γn (iii) d(x∗ , A n (x0 )) ≤ 1−γ d(x0 , A(x0 )) for all n ∈ ℕ. Remark 3.15. According to Definition 3.13, the contraction principle insures that, if A : X → X is a γ-contraction on the complete metric space X, then it is a Picard operator. Theorem 3.16. Let (X, d) be a complete metric space and A, B : X → X two operators. We suppose the following: (i) A is a contraction with contraction constant γ and F A = {x∗A }. (ii) B has fixed points and x∗B ∈ F B . (iii) There exists η > 0 such that d(A(x), B(x)) ≤ η for all x ∈ X. Then η . d(x∗A , x∗B ) ≤ 1−γ Definition 3.17. Let (X, d) be a metric space. An operator A : X → X is a weakly Picard operator if the sequence (A n (x0 ))n∈ℕ converges for all x0 ∈ X and its limit (which may depend on x0 ) is a fixed point of A. Theorem 3.18. Let (X, d) be a metric space. Then A : X → X is a weakly Picard operator if and only if there exists a partition X = ⋃λ∈Λ X λ of X such that (i) X λ ∈ I(A) for all λ ∈ Λ; (ii) A |X λ : X λ → X λ is a Picard operator for all λ ∈ Λ.

106 | 3 Fractional Differential Equations Definition 3.19. If A is a weakly Picard operator, then we consider the operator A∞ defined by A∞ : X → X, A∞ (x) = lim A n (x). n→∞

It is clear that A∞ (X) = F A and ω A (x) = {A∞ (x)}, where ω A (x) is the ω-limit point set of mapping A for point x. Definition 3.20. Let A be a weakly Picard operator and c > 0. The operator A is a c-weakly Picard operator if d(x, A∞ (x)) ≤ cd(x, A(x))

for all x ∈ X.

Remark 3.21. Let (X, d) be a complete metric space and A : X → X a continuous operator. We suppose that there exists γ ∈ [0, 1) such that d(A2 (x), A(x)) ≤ γd(x, Ax) for all x ∈ X. Then A is a c-weakly Picard operator with c =

1 1−γ .

Theorem 3.22. Let (X, d) be a complete metric space and A i : X → X for i = 1, 2. We suppose that (i) A i is a c i -weakly Picard operator, i = 1, 2; (ii) there exists α > 0 such that d(A1 (x), A2 (x)) ≤ α for all x ∈ X. Then H(F A1 , F A2 ) ≤ α max(c1 , c2 ).

3.2.3 Existence In what follows we consider the fixed point equation (3.24). Consider the operator B f : C([a1 , b1 ], [a1 , b1 ]) → C([a1 , b1 ], [a1 , b1 ]) where φ(t) { { { b B f (x)(t) := {w(φ, ψ)(t) + ∫ G(t, s)f(s, x(s), x(x v (s)))ds a { { {ψ(t)

if t ∈ [a1 , a], if t ∈ [a, b],

(3.25)

if t ∈ [b, b1 ].

It is clear that x is a solution of problem (3.23) if and only if x is a fixed point of the operator B f . So, the problem is to study the fixed point equation x = B f (x). Let L > 0 and introduce the following notation: C L ([a1 , b1 ], [a1 , b1 ]) = {x ∈ C([a1 , b1 ], [a1 , b1 ]) | |x(t1 ) − x(t2 )| ≤ L|t1 − t2 |} for all t1 , t2 ∈ [a1 , b1 ]. Remark that C L ([a1 , b1 ], [a1 , b1 ]) ⊆ C([a1 , b1 ], ℝ) is also a complete metric space with respect to the metric d(x1 , x2 ) := maxa1 ≤t≤b1 |x1 (t) − x2 (t)|.

3.2 Modified fractional iterative functional differential equations | 107

Theorem 3.23. We suppose that (i) the conditions (C1 )–(C4 ) are satisfied, but in addition v ≥ 1; (ii) φ ∈ C L ([a1 , a], [a1 , b1 ]), ψ ∈ C L ([b, b1 ], [a1 , b1 ]); (iii) there are m f , M f ∈ ℝ such that m f ≤ f(t, u, w) ≤ M f

for all t ∈ [a, b], u, w ∈ [a1 , b1 ],

and moreover, a1 ≤ min(φ(a), ψ(b)) − max(0,

m f (b − a)q M f (b − a)q ) + min(0, ) Γ(q + 1) Γ(q + 1)

and max(φ(a), ψ(b)) − min(0, (iv) it holds

m f (b − a)q M f (b − a)q ) + max(0, ) ≤ b1 ; Γ(q + 1) Γ(q + 1)

|ψ(b) − φ(a)| (b − a)q−1 (1 + q) max{|m f |, |M f |} + < L; b−a Γ(q + 1)

(v) it holds

2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2) < 1. Γ(q + 1)

Then problem (3.23) has in C L ([a1 , b1 ], [a1 , b1 ]) a unique solution. Moreover, the operator B f : C L ([a1 , b1 ], [a1 , b1 ]) → C L ([a, b], [a1 , b1 ]) is a c-Picard operator with c :=

Γ(q + 1) . Γ(q + 1) − 2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2)

Proof. First of all we remark that (iii) and (iv) imply that C L ([a1 , b1 ], [a1 , b1 ]) is an invariant subset for B f . Indeed, for t ∈ [a1 , a] ∪ [b, b1 ], we have B f (x)(t) ∈ [a1 , b1 ]. Furthermore, we obtain a1 ≤ B f (x)(t) ≤ b1

for all t ∈ [a, b],

if and only if a1 ≤ min B f (x)(t)

(3.26)

max B f (x)(t) ≤ b1

(3.27)

t∈[a,b]

and t∈[a,b]

hold. Since min B f (x)(t) ≥ min(φ(a), ψ(b)) − max(0,

t∈[a,b]

M f (b − a)q m f (b − a)q ) + min(0, ) Γ(q + 1) Γ(q + 1)

108 | 3 Fractional Differential Equations

and M f (b − a)q m f (b − a)q ) + max(0, ), Γ(q + 1) Γ(q + 1)

max B f (x)(t) ≤ max(φ(a), ψ(b)) − min(0,

t∈[a,b]

the requirements (3.26) and (3.27) are equivalent to the conditions appearing in (iii). Now, consider a1 ≤ t1 < t2 ≤ a. Then, |B f (x)(t2 ) − B f (x)(t1 )| = |φ(t2 ) − φ(t1 )| ≤ L|t1 − t2 | as φ ∈ C L ([a1 , a], [a1 , b1 ]), due to (ii). Similarly, for b ≤ t1 < t2 ≤ b1 , |B f (x)(t2 ) − B f (x)(t1 )| = |ψ(t2 ) − ψ(t1 )| ≤ L|t1 − t2 | which also follows from (ii). On the other hand, for a ≤ t1 < t2 ≤ b, |B f (x)(t2 ) − B f (x)(t1 )| ≤ |w(φ, ψ)(t2 ) − w(φ, ψ)(t1 )| b

+

1 󵄨 󵄨 ∫ |G(t2 , s) − G(t1 , s)| 󵄨󵄨󵄨f(s, x(s), x(x v (s)))󵄨󵄨󵄨ds Γ(q) a

b

|ψ(b) − φ(a)| |t2 − t1 | 󵄨 󵄨 ≤ |t2 − t1 | + ∫(b − s)q−1 󵄨󵄨󵄨f(s, x(s), x(x v (s)))󵄨󵄨󵄨ds b−a (b − a)Γ(q) a

t1

+

1 󵄨 󵄨 ∫[(t2 − s)q−1 − (t1 − s)q−1 ]󵄨󵄨󵄨f(s, x(s), x(x v (s)))󵄨󵄨󵄨ds Γ(q) a

t2

+

1 󵄨 󵄨 ∫(t2 − s)q−1 󵄨󵄨󵄨f(s, x(s), x(x v (s)))󵄨󵄨󵄨ds Γ(q) t1

(b − a)q−1 max{|m f |, |M f |}|t2 − t1 | |ψ(b) − φ(a)| ≤ |t2 − t1 | + b−a Γ(q + 1) max{|m f |, |M f |} q + ((t2 − a) − (t1 − a)q − (t2 − t1 )q ) Γ(q + 1) max{|m f |, |M f |} + (t2 − t1 )q Γ(q + 1) |ψ(b) − φ(a)| (b − a)q−1 (1 + q) max{|m f |, |M f |} ≤( + )|t1 − t2 |, b−a Γ(q + 1) where we used the inequality r q − s q ≤ qr q−1 (r − s) for all r ≥ s ≥ 0. Therefore, due to (iv), the function B f (x) is L-Lipschitz in t. Thus, according to the above, we have C L ([a1 , a], [a1 , b1 ]) ∈ I(B f ).

3.2 Modified fractional iterative functional differential equations | 109

From condition (v) it follows that B f is an L B f -contraction with L B f :=

2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2) . Γ(q + 1)

Indeed, for all t ∈ [a1 , a] ∪ [b, b1 ], we have |B f (x1 )(t) − B f (x2 )(t)| = 0. Moreover, for t ∈ [a, b] we get b

1 |B f (x1 )(t) − B f (x2 )(t)| ≤ ∫ |G(t, s)f(s, x1 (s)) − G(t, s)f(s, x2 (s))|ds Γ(q) a

b

≤

t−a 󵄨 󵄨 ∫(b − s)q−1 󵄨󵄨󵄨f(s, x1 (s), x1 (x1v (s))) − f(s, x2 (s), x2 (x2v (s)))󵄨󵄨󵄨ds (b − a)Γ(q) a

t

+

1 󵄨 󵄨 ∫(t − s)q−1 󵄨󵄨󵄨f(s, x1 (s), x1 (x1v (s))) − f(s, x2 (s), x2 (x2v (s)))󵄨󵄨󵄨ds Γ(q) a

b

≤

Lf 󵄨 󵄨 ∫(b − s)q−1 [|x1 (s) − x2 (s)| + 󵄨󵄨󵄨x1 (x1v (s)) − x2 (x2v (s))󵄨󵄨󵄨]ds Γ(q) a

t

+

Lf 󵄨 󵄨 ∫(t − s)q−1 [|x1 (s) − x2 (s)| + 󵄨󵄨󵄨x1 (x1v (s)) − x2 (x2v (s))󵄨󵄨󵄨]ds Γ(q) a

b

≤

Lf 󵄨 󵄨 ∫(b − s)q−1 [|x1 (s) − x2 (s)| + 󵄨󵄨󵄨x1 (x1v (s)) − x1 (x2v (s))󵄨󵄨󵄨 Γ(q) a

󵄨 󵄨 + 󵄨󵄨󵄨x1 (x2v (s)) − x2 (x2v (s))󵄨󵄨󵄨]ds t

+

Lf 󵄨 󵄨 ∫(t − s)q−1 [|x1 (s) − x2 (s)| + 󵄨󵄨󵄨x1 (x1v (s)) − x1 (x2v (s))󵄨󵄨󵄨 Γ(q) a

󵄨 󵄨 + 󵄨󵄨󵄨x1 (x2v (s)) − x2 (x2v (s))󵄨󵄨󵄨]ds b

Lf 󵄨 󵄨 ≤ ∫(b − s)q−1 [2‖x1 − x2 ‖C + L󵄨󵄨󵄨x1v (s) − x2v (s)󵄨󵄨󵄨]ds Γ(q) a

t

+

Lf 󵄨 󵄨 ∫(t − s)q−1 [2‖x1 − x2 ‖C + L󵄨󵄨󵄨x1v (s) − x2v (s)󵄨󵄨󵄨]ds Γ(q) a

b

≤

Lf ∫(b − s)q−1 [2‖x1 − x2 ‖C + Lv max{|a1 |, |b1 |}v−1 |x1 (s) − x2 (s)|]ds Γ(q) a

t

Lf + ∫(t − s)q−1 [2‖x1 − x2 ‖C + Lv max{|a1 |, |b1 |}v−1 |x1 (s) − x2 (s)|]ds Γ(q) a

110 | 3 Fractional Differential Equations b

≤

Lf ∫(b − s)q−1 [2‖x1 − x2 ‖C + Lv max{|a1 |, |b1 |}v−1 ‖x1 − x2 ‖C ]ds Γ(q) a

t

Lf + ∫(t − s)q−1 [2‖x1 − x2 ‖C + Lv max{|a1 |, |b1 |}v−1 ‖x1 − x2 ‖C ]ds Γ(q) a

2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2) ≤ ‖x1 − x2 ‖C . Γ(q + 1) So, B f is a c-Picard operator, with c = (1 −

2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2) −1 ) . Γ(q + 1)

This completes the proof. In what follows, consider the operator E f : C L ([a1 , b1 ], [a1 , b1 ]) → C L ([a1 , b1 ], [a1 , b1 ]) where x(t) { { { { { {w(x|[a1 ,a] , x|[b,b1 ] )(t) (E f x)(t) := { b 1 { + Γ(q) ∫a G(t, s)f(s, x(s), x(x v (s)))ds { { { { {x(t)

if t ∈ [a1 , a], if t ∈ [a, b], if t ∈ [b, b1 ].

Theorem 3.24. Under the conditions of Theorem 3.23, the operator E f : C L ([a1 , b1 ], [a1 , b1 ]) → C L ([a1 , b1 ], [a1 , b1 ]) is a weakly Picard operator. Proof. The operator E f is continuous but it is not a contraction operator. Let us take the following notation: X φ,ψ := {x ∈ C L ([a1 , b1 ], [a1 , b1 ]) | x|[a1 ,a] = φ, x|[b,b1 ] = ψ}. Then we can write C L ([a1 , b1 ], [a1 , b1 ]) =

⋃

X φ,ψ .

φ,ψ∈C L ([a1 ,a],[a1 ,b1 ])

We have that X φ,ψ ∈ I(E f ) and E f |X φ,ψ is a Picard operator, because it is the operator which appears in the proof of Theorem 3.23. By applying Theorem 3.18, we obtain that E f is a weakly Picard operator. Finally, in general, from the proof of Theorem 3.23 and the Schauder fixed point theorem we immediately obtain the following existence result.

3.2 Modified fractional iterative functional differential equations | 111

Theorem 3.25. Suppose conditions (C1 )–(C4 ) are satisfied together with assumptions (ii)–(iv) of Theorem 3.23. Then problem (3.23) has a solution in C L ([a1 , b1 ], [a1 , b1 ]). We do not know about uniqueness. But this is not so surprising, since B f is not Lipschitzian in general. So we cannot apply metric fixed point theorems, only topological ones. This can be simply illustrated on the problems { and

x󸀠 (t) = Ax(x2 (t)),

(3.28)

x(0) = 0, 4

{x󸀠 (t) = Ax( √ x(t)), { { x(0) = 0,

(3.29)

for A > 0. Rewriting (3.28) as t

x(t) = B1 (x)(t) = A ∫ x(x2 (s))ds, 0

and applying the above procedure to B1 , it follows that B1 : C Ab1 ([0, b1 ], [0, b1 ]) → C Ab1 ([0, b1 ], [0, b1 ]),

Ab1 ≤ 1, 0 < b1 ≤ 1

is Ab1 (1 + 2Ab21 )-Lipschitzian, so its only solution is x(t) = 0 in that space when Ab1 (1 + 2Ab21 ) < 1. On the other hand, rewriting (3.29) as t

4

x(t) = B2 (x)(t) = A ∫ x( √ x(s))ds, 0

it follows that B2 : C Ab1 ([0, b1 ], [0, b1 ]) → C Ab1 ([0, b1 ], [0, b1 ]), satisfies

b1 ≥ 1, Ab1 ≤ 1

4

‖B2 (x1 ) − B2 (x2 )‖C ≤ Ab1 (‖x1 − x2 ‖C + Ab1 √‖x1 − x2 ‖C ), so it is not Lipschitzian. Hence (3.29) should have a nonzero solution, and it does have x(t) = A4 t2 .

3.2.4 Data dependence In this subsection, we consider problem (3.23) and suppose that the conditions of Theorem 3.23 are satisfied. Denote by x(⋅; φ, ψ, f) the solution of this problem.

112 | 3 Fractional Differential Equations Theorem 3.26. Let φ i , ψ i , f i for i = 1, 2 be as in Theorem 3.23. Furthermore, we suppose that (i) there exists η1 > 0 such that |φ1 (t) − φ2 (t)| ≤ η1 ,

t ∈ [a1 , a],

|ψ1 (t) − ψ2 (t)| ≤ η1 ,

t ∈ [b, b1 ];

and (ii) there exists η2 > 0 such that |f1 (t, u, w) − f2 (t, u, w)| ≤ η2

for all t ∈ [a, b], u, w ∈ [a1 , b1 ].

Then 󵄨󵄨 󵄨 󵄨󵄨x(t; φ1 , ψ1 , f1 ) − x(t; φ2 , ψ2 , f2 )󵄨󵄨󵄨 ≤

2(b−a)q Γ(q+1) η 2 2(b−a)q L f (Lv max{|a1 |,|b1 |}v−1 +2) Γ(q+1)

3η1 +

1−

,

where L f = min{L f1 , L f2 }. Proof. Consider the operators B φ i ,ψ i ,f i , i = 1, 2. From Theorem 3.23 these operators are contractions. Additionally, 󵄨 󵄨 ‖B φ1 ,ψ1 ,f1 (x) − B φ2 ,ψ2 ,f2 (x)‖C ≤ 󵄨󵄨󵄨w(φ1 , ψ1 )(t) − w(φ2 , ψ2 )(t)󵄨󵄨󵄨 b

1 󵄨 󵄨 + ∫ G(t, s)󵄨󵄨󵄨f1 (s, x(s), x(x v (s))) − f2 (s, x(s), x(x v (s)))󵄨󵄨󵄨ds Γ(q) a

≤ 3η1 +

2(b − a)q η2 . Γ(q + 1)

Now, the proof follows from Theorem 3.16, with A := B φ1 ,ψ1 ,f1 , and γ := L A =

B := B φ2 ,ψ2 ,f2 ,

η := 3η1 +

2(b − a)q η2 , Γ(q + 1)

2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2) , Γ(q + 1)

where we can suppose that L f = L f1 = min{L f1 , L f2 }. Remark 3.27. Let φ i , ψ i , f i , i ∈ ℕ, and φ, ψ, f be as in Theorem 3.23. We suppose that unif.

φ i 󳨀󳨀󳨀→ φ, Then

unif.

ψ i 󳨀󳨀󳨀→ ψ, unif.

x(⋅; φ i , ψ i , f i ) 󳨀󳨀󳨀→ x(⋅; φ, ψ, f)

unif.

f i 󳨀󳨀󳨀→ f. as i → ∞.

3.2 Modified fractional iterative functional differential equations | 113

Theorem 3.28. Let f1 and f2 be as in Theorem 3.23. Let F E fi be the solution set of the first equation of problem (3.23) corresponding to f i , i = 1, 2. Suppose that there exists η > 0 such that 󵄨󵄨󵄨f1 (t, u1 , w1 ) − f2 (t, u2 , w2 )󵄨󵄨󵄨 ≤ η for all t ∈ [a, b], u i , w i ∈ [a1 , b1 ], i = 1, 2. 󵄨 󵄨 Then H‖⋅‖C (F E f1 , F E f2 ) ≤

Γ(q + 1) − 2(b −

η2(b − a)q , v−1 + 2) f (Lv max{|a 1 |, |b 1 |}

a)q L

where L f = max{L f1 , L f2 } and H‖⋅‖C denotes the Pompeiu–Hausdorff functional with respect to ‖⋅‖C on C L ([a1 , b1 ], [a1 , b1 ]). Proof. We will look for those c i , for which in the condition of Theorem 3.23 the operators E f i , i = 1, 2 are c i -weakly Picard. Set X φ,ψ := {x ∈ C L ([a1 , b1 ], [a1 , b1 ]) | x|[a1 ,a] = φ, x|[b,b1 ] = ψ}. It is clear that E f i |X φ,ψ = B f i . So, from Theorems 3.18 and 3.23 we have ‖E2f i (x) − E f i (x)‖C ≤

2(b − a)q L f i (Lv max{|a1 |, |b1 |}v−1 + 2) ‖E f i (x) − x‖C Γ(q + 1)

for all x ∈ C L ([a1 , b1 ], [a1 , b1 ]) and i = 1, 2. Now, choosing λi =

2(b − a)q L f i (Lv max{|a1 |, |b1 |}v−1 + 2) , Γ(q + 1)

1 , i = 1, 2. we get that E f i are c i -weakly Picard operators with c i = 1−λ i Next, we obtain 2(b − a)q ‖E f1 (x) − E f2 (x)‖C ≤ η Γ(q + 1) for all x ∈ C L ([a1 , b1 ], [a1 , b1 ]). Applying Theorem 3.22, we have that

H‖⋅‖C (F E f1 , F E f2 ) ≤

η2(b − a)q . Γ(q + 1) − 2(b − a)q L f (Lv max{|a1 |, |b1 |}v−1 + 2)

3.2.5 Examples Consider the following problem: 3

c 2 { D 2 ,t x(t) = { { 5 { { { { x(t) = { { { { { { { x(t) = {

where x ∈ C L ([ 51 , 45 ], [ 15 , 45 ]).

μx(x(t)), 1 , 2 1 , 2

2 t∈[ , 5 1 t∈[ , 5 3 t∈[ , 5

3 ], μ > 0, 5 2 ], 5 4 ], 5

(3.30)

114 | 3 Fractional Differential Equations

Proposition 3.29. Consider problem (3.30). We suppose that { 3L√5π μ < { 15√8 5π { 8(L+2)

if 0 < L ≤ √6 − 1, if √6 − 1 ≤ L.

(3.31)

Then problem (3.30) has in C L ([ 15 , 45 ], [ 51 , 45 ]) a unique solution. Proof. First of all notice that according to Theorem 3.23 we have v = 1, q = 23 , a = 25 , b = 53 , ψ( 35 ) = 12 , φ( 52 ) = 12 , a1 = 15 , b1 = 45 . Observe that the Lipschitz constant for the function f(t, u1 , u2 ) = μu2 is L f = μ and |f(t, u1 , u2 ) − f(t, w1 , w2 )| ≤ μ|u2 − w2 |, u i , w i ∈ [ 15 , 45 ]. So we choose m f = 5μ and M f = 4μ 5 . By a common check in the conditions of Theorem 3.23 we can make sure that μ≤

45√5π 32

M f (b − a)q m f (b − a)q { a ≤ min(φ(a), ψ(b)) − max(0, ) + min(0, ), { { 1 Γ(q + 1) Γ(q + 1) ⇔ { q q { { max(φ(a), ψ(b)) − min(0, m f (b − a) ) + max(0, M f (b − a) ) ≤ b 1 { Γ(q + 1) Γ(q + 1) and 8 |ψ(b) − φ(a)| (b − a)q−1 (1 + q) max{|m f |, |M f |} μ 0, 5 2 ], 5 4 ], 5

(3.32)

3.2 Modified fractional iterative functional differential equations | 115

and

3

c 2 { D 2 ,t x(t) = μ2 x(x(t)), { { 5 { { { { x(t) = φ2 , { { { { { { { x(t) = ψ2 , {

2 t∈[ , 5 1 t∈[ , 5 3 t∈[ , 5

3 ], μ2 > 0, 5 2 ], 5 4 ]. 5

(3.33)

Suppose that we have satisfied the following assumptions: (H1 ) φ i ∈ C L ([ 51 , 25 ], [ 15 , 45 ]), ψ i ∈ C L ([ 53 , 45 ], [ 51 , 45 ]) such that φ i ( 52 ) = 12 , ψ i ( 35 ) = 12 , i = 1, 2; (H2 ) conditions (3.31) of Proposition 3.29 are fulfilled for both problems (3.32) and (3.33). Let x∗1 be the unique solution of problem (3.32) and x∗2 be the unique solution of problem (3.33). We are looking for an estimation for ‖x∗1 − x∗2 ‖C . Then, using Theorems 3.26 and 3.28 and a common substitution, we can deduce the following result. Proposition 3.30. Consider the problems (3.32), (3.33) and suppose the requirements (H1 ), (H2 ) hold. Additionally, we suppose that (i) there exists η1 > 0 such that |φ1 (t) − φ2 (t)| ≤ η1

1 2 for all t ∈ [ , ], 5 5

|ψ1 (t) − ψ2 (t)| ≤ η1

3 4 for all t ∈ [ , ]; 5 5

and

(ii) there exists η2 > 0 such that |μ1 − μ2 | ≤

5 η2 . 4

Then 󵄨󵄨 ∗ 󵄨 󵄨󵄨x1 (⋅; ψ1 , ψ1 , f1 ) − x∗2 (⋅; ψ2 , ψ2 , f2 )󵄨󵄨󵄨 ≤

45√5πη1 + 8η2 15√5π − 8(L + 2) min{μ1 , μ2 }

.

Furthermore, if F E f1 is the solution set for the first equation of problem (3.32) and F E f2 is the solution set for the first equation of problem (3.33), then H‖⋅‖C (F E f1 , F E f2 ) ≤

15√5πη2 15√5π − 8(L + 2) max{μ1 , μ2 }

,

where H‖⋅‖C denotes the Pompeiu–Hausdorff functional with respect to C L ([ 15 , 45 ], [ 15 , 54 ]).

116 | 3 Fractional Differential Equations

3.3 Ulam–Hyers–Rassias stability for semilinear equations 3.3.1 Introduction The theory of Ulam–Hyers–Rassias stability has been an important subject of mathematical analysis, functional analysis and differential equations. The reader is referred to monographs mentioned in the Introduction of this book and to other recent contributions [62, 75, 94, 132, 151, 152, 187, 196, 197, 225, 232, 245, 246, 326]. In particular, Rezaei et al. [238] succeeded to use the Laplace transform method to derive Hyers–Ulam stability of linear differential equations. Popa and Raşa [226] derived an interesting result on the stability of Bernstein–Schnabl operators by applying a new way to the Fréchet functional equation. Moreover, Popa and Raşa [227] made some progress in the study of best constant in Hyers–Ulam stability of Stancu, Bernstein and Kantorovich operators. Meanwhile, Popa and Raşa [228] studied Hyers–Ulam stability of Bernstein, Szász–Mirakjan, Stancu and Bleimann–Butzer–Hahn discrete operators, and Kantorovich and Beta integral operators from approximation theory. Indeed, the above results are related to Takagi et al. [261, Theorem 2] from a functional analysis point of view. The goal of this section is to obtain results on Ulam–Hyers–Rassias stability for surjective linear equations on Banach spaces, linear equations on Banach spaces with closed ranges and surjective semilinear equations between Banach spaces from functional analysis theory with several illustrative examples. This section is based on [278].

3.3.2 Ulam–Hyers–Rassias stability for surjective linear equations on Banach spaces Let X, Y be Banach spaces and T ∈ L(X, Y). Since ker T, the kernel of T, is a closed subset of X, the factor space Z := X/ ker T is a Banach space with a norm [268] ‖[x]‖ = inf{‖z‖ | z ∈ [x]}, where [x] := {z ∈ X | z − x ∈ ker T}. Then we extend T to [T] : Z → Y as [T][x] := Tx. It is well defined, [T] ∈ L(Z, Y) and ‖[T]‖ ≤ ‖T‖. The next theorem seems to be known [228], but here we state it and give its proof. Theorem 3.31. Let T ∈ L(X, Y) be surjective. Then for any x ∈ X, y ∈ Y, δ > 0 and ε > 0 such that ‖Tx − y‖ ≤ ε there is x y ∈ X such that Tx y = y and ‖x − x y ‖ ≤ (‖[T]−1 ‖ + δ)ε. Proof. Clearly [T] : Z → Y is injective and surjective, so by the Banach inverse theorem, we have [T]−1 ∈ L(Y, Z). Set y1 := Tx − y. Then ‖y1 ‖ ≤ ε, so [x1 ] := [T]−1 y1 satisfies ‖[x1 ]‖ ≤ ‖[T]−1 ‖ε. We take x2 ∈ [x1 ] with ‖x2 ‖ ≤ cε for c := ‖[T]−1 ‖ + δ. Then

3.3 Ulam–Hyers–Rassias stability for semilinear equations | 117

Tx2 = Tx1 = y1 . Setting x y := x − x2 , we get Tx y = Tx − Tx2 = Tx − y1 = Tx − (Tx − y) = y and ‖x y − x‖ = ‖x2 ‖ ≤ cε. Remark 3.32. The number c = ‖[T]−1 ‖ + δ is called the Hyers–Ulam stability constant (briefly HUS-constant) [228, 261], and the infimum of all HUS-constants is ‖[T]−1 ‖. Next, setting d(x, ker T) = inf{‖y − z‖ | z ∈ ker T}, clearly ‖[x]‖ = d(x, ker T). Let Q : X → ker T be the metric projection defined as (see [237, p. 300]) Qx = {w ∈ ker T | ‖x − w‖ = d(x, ker T)}. It is known [177, 237] that Qx is nonempty when ker T is reflexive, moreover Qx is a singleton when ker T is in addition strictly convex. Then by the above proof we see that, if ker T is reflexive then the best HUS-constant is ‖[T]−1 ‖. In particular, this is true when dim ker T < ∞. Example 3.33. Let A ∈ L(ℝn ) and for γ > max{Re λ | λ ∈ σ(A)}. Take −γt ̇ < ∞}, X := {x ∈ C1 (ℝ+ , ℝn ) | sup |x(t)|e−γt < ∞, sup |x(t)|e t∈ℝ+

t∈ℝ+

Y := {y ∈ C(ℝ+ , ℝn ) | sup |y(t)|e−γt < ∞}, t∈ℝ+

Tx := ẋ − Ax,

where ℝ+ := [0, ∞). Then X and Y are Banach spaces with norm −γt ̇ ‖x‖ := max{sup |x(t)|e−γt , sup |x(t)|e }, t∈ℝ+

t∈ℝ+

‖y‖ := sup |y(t)|e−γt , t∈ℝ+

respectively. Clearly T ∈ L(X, Y) with ‖T‖ ≤ (1 + ‖A‖). Lemma 3.34. T is surjective. Proof. Solve Tx = y ∈ Y, i.e., ẋ − Ax = y. t ∫0

It has a solution x(t) = eA(t−s) y(s)ds. Take γ > γ1 > r(A) = max{Re λ | λ ∈ σ(A)}, i.e., r(A) is the spectral radius of A. It is well known that there is a constant k > 0 such that ‖eAt ‖ ≤ keγ1 t ,

t ∈ ℝ+ .

(3.34)

Then we derive t

|x(t)| ≤ k ∫ e 0

so ‖x‖ ≤

k γ−γ1 ‖y‖.

t γ1 (t−s)

|y(s)|ds ≤ k‖y‖ ∫ eγ1 (t−s)+γs ds ≤ 0

k ‖y‖eγt , γ − γ1

118 | 3 Fractional Differential Equations Applying Theorem 3.31, we see that there is a constant c > 0 such that for any x ∈ X, ̇ − Ax(t) − y(t)| ≤ εeγt , t ∈ ℝ+ there is x y ∈ X such that y ∈ Y and ε > 0 such that |x(t) ẋ y − Ax y = y and |x(t) − x y (t)| ≤ cεeγt , t ∈ ℝ+ . Example 3.35. We can extend the above approach in Example 3.33 to A ∈ C(ℝ+ , L(ℝn )) such that supt∈ℝ+ ‖A(t)‖ < ∞ and the fundamental solution X(t) of ̇ = A(t)x(t) x(t) ‖X(t)X −1 (s)‖

satisfies ≤ t ≥ s ≥ 0 for some k > 0 and γ > γ1 . These assumptions hold if limt→∞ A(t) = A and γ1 > r(A). keγ1 (t−s) ,

Example 3.36. We continue with Example 3.35 by supposing exponential splitting, i.e., there are k > 0, θ− < θ+ and a projection P ∈ L(ℝn ) such that ‖X(t)PX −1 (s)‖ ≤ keθ− (t−s) , ‖X(t)(I − P)X (s)‖ ≤ ke −1

θ+ (t−s)

,

t ≥ s ≥ 0, 0 ≤ t ≤ s.

Then we take γ ∈ (θ− , θ+ ), and a solution of ẋ − A(t)x = y is given by t

∞

x(t) = ∫ X(t)PX −1 (s)y(s)ds − ∫ X(t)(I − P)X −1 (s)y(s)ds, t

0

which satisfies t

|x(t)| ≤ k‖y‖ ∫ e

∞ θ− (t−s)+γs

ds + k‖y‖ ∫ eθ+ (t−s)+γs ds = ( t

0

k k + )‖y‖eγt , γ − θ− θ+ − γ

which gives k k + )‖y‖. γ − θ− θ+ − γ These assumptions hold if limt→∞ A(t) = A and [θ− , θ+ ] ∩ σ(A) = 0. ‖x‖ ≤ (

Remark 3.37. The above examples are related to [238] where the Laplace transform method is applied to obtain the results. We remark that the Laplace transform method can be used only for constant linear differential systems [238]. Even for the case of semilinear equations, it is not easy to establish a uniform framework to solve the problem. We will give more details in Section 3.3.4. Concerning the approximation theory to discuss Ulam–Hyers stability, we refer the reader to the recent contributions of Popa and Raşa [227]. Next we show that our method can also be applied to study impulsive systems. Example 3.38. Let γ > r(A), 0 < θ, A, B ∈ L(ℝn ), let {t j }j∈ℕ ⊂ ℝ+ be an increasing sequence with θ ≤ t j+1 − t j , and take −γt ̇ X PC := {x ∈ PC1 (ℝ+ , ℝn ) | sup |x(t)|e−γt < ∞, sup |x(t)|e < ∞}, t∈ℝ+

Y

PC

t∈ℝ+

:= {y ∈ PC(ℝ+ , ℝ ) | sup |y(t)|e n

−γt

t∈ℝ+

< ∞},

3.3 Ulam–Hyers–Rassias stability for semilinear equations | 119

ℓγ := {c = {c j }j∈ℕ ⊂ ℝn | sup |c j |e−γt j < ∞},

Z := Y PC × ℓγ ,

j∈ℕ

T PC x := (ẋ − Ax, t ∈ ℝ+ \ {t j }j∈ℕ , {∆x(t j ) − Bx(t j )}j∈ℕ ). Then X PC , Y PC , ℓγ and Z are Banach spaces with norm −γt ̇ ‖x‖ := max{sup |x(t)|e−γt , sup |x(t)|e }, t∈ℝ+

t∈ℝ+

‖y‖ := sup |y(t)|e

−γt

‖c‖ := sup |c j |e−γt j ,

,

t∈ℝ+

‖z‖ := max{‖y‖, ‖c‖},

k∈ℕ

respectively. Clearly T PC ∈ L(X PC , Z) with ‖T PC ‖ ≤ (3 + ‖A‖ + ‖B‖). Lemma 3.39. T PC is surjective. Proof. Solve T PC x = z = (y, c) ∈ Z, i.e., {

ẋ − Ax = y

on ℝ+ \ {t j }j∈ℕ ,

∆x(t j ) = Bx(t j ) + c j ,

j ∈ ℕ,

which has a solution t

j−1

x(t) = ∫ eA(t−s) y(s)ds + ∑ eA(t−t i ) (Bx(t i ) + c i ),

t ∈ (t j−1 , t j ], j ∈ ℕ,

i=1

0

where we set t0 := 0. By (3.34), for t ∈ (t j−1 , t j ], j ∈ ℕ, γ > γ1 > r(A), it holds t

|x(t)|e

−γt

j−1

≤ k ∫ e(γ1 −γ)(t−s) ‖y‖ds + k ∑ eγ1 (t−t i )−γt (‖B‖|x(t i )| + |c i |) i=1

0 j−1

≤k

j−1

‖y‖ + k‖c‖ ∑ e(γ1 −γ)(t−t i ) + k‖B‖ ∑ e(γ1 −γ)(t−t i ) |x(t i )|e−γt i . γ − γ1 i=1 i=1

(3.35)

Setting Γ := supj∈ℕ |x(t j )|e−γt j (note that x ∈ X PC implies Γ < ‖x‖ < ∞), we have j−1

|x(t j )|e−γt j ≤ k

j−1

‖y‖ + k‖c‖ ∑ e(γ1 −γ)(t j −t i ) + k‖B‖ ∑ e(γ1 −γ)(t j −t i ) |x(t i )|e−γt i γ − γ1 i=1 i=1 j−1

j−1

≤k

‖y‖ + k‖c‖ ∑ eθ(γ1 −γ)(j−i) + k‖B‖Γ ∑ eθ(γ1 −γ)(j−i) γ − γ1 i=1 i=1

≤k

eθ(γ1 −γ) eθ(γ1 −γ) ‖y‖ + k‖c‖ + k‖B‖Γ . θ(γ −γ) γ − γ1 1−e 1 1 − eθ(γ1 −γ)

So Γ≤k

‖y‖ eθ(γ1 −γ) eθ(γ1 −γ) + k‖c‖ + k‖B‖Γ γ − γ1 1 − eθ(γ1 −γ) 1 − eθ(γ1 −γ)

120 | 3 Fractional Differential Equations

and (1 − k‖B‖

eθ(γ1 −γ) eθ(γ1 −γ) ‖y‖ + k‖c‖ ≤ k . )Γ γ − γ1 1 − eθ(γ1 −γ) 1 − eθ(γ1 −γ)

θ(γ1 −γ)

e Hence, if 1 − k‖B‖ 1−e θ(γ1 −γ) > 0, i.e.,

γ > γ1 +

ln(1 + k‖B‖) , θ

(3.36)

then Γ ≤ C1 ‖z‖ for a constant C1 > 0 given by C1 := k

eθ(γ1 −γ) 1−eθ(γ1 −γ) eθ(γ1 −γ) k‖B‖ 1−e θ(γ1 −γ)

1 γ−γ1

1−

+

.

Then (3.35) for t ∈ (t j−1 , t j ] implies j−1

|x(t)|e−γt ≤ k ≤(

j−1

‖y‖ + k‖c‖ ∑ e(γ1 −γ)(t j−1 −t i ) + k‖B‖ ∑ e(γ1 −γ)(t j−1 −t i ) Γ γ − γ1 i=1 i=1 k‖B‖C1 k k + + )‖z‖. γ − γ1 1 − eθ(γ1 −γ) 1 − eθ(γ1 −γ)

Consequently, we arrive at sup |x(t)|e−γt ≤ C2 ‖z‖

t∈ℝ+

for C2 :=

k k‖B‖C1 k + . + γ − γ1 1 − eθ(γ1 −γ) 1 − eθ(γ1 −γ)

Summarizing, we get ‖x‖ ≤ max{1 + ‖A‖C2 , C2 }‖z‖. Hence we can take γ fulfilling (3.36). Note that then γ > γ1 . Applying Theorem 3.31, we see that fixing γ > γ1 > r(A) satisfying (3.36), there is a constant c̃ > 0 such that for any x ∈ X PC , y ∈ Y PC , c ∈ ℓγ and ε > 0 such that ̇ − Ax(t) − y(t)| ≤ εeγt , t ∈ ℝ+ \ {t k }k∈ℕ with |∆x(t k ) − Bx(t k ) − c k | ≤ εeγt k , k ∈ ℕ, |x(t) there is x y,c ∈ X PC such that ẋ y,c − Ax y = y with ∆x y,c (t k ) = Bx y (t k ) + c k , k ∈ ℕ, and ̇ − ẋ y,c (t)|} ≤ cεe ̃ γt , max{|x(t) − x y,c (t)|, |x(t)

t ∈ ℝ+ \ {t k }k∈ℕ .

Now we continue with fractional differential equations. Example 3.40. Let λ, γ > 0 and take α X α := {x ∈ C(α) (ℝ+ , ℝ) | sup |x(t)|e−γt < ∞, sup |c D0,t x(t)|e−γt < ∞}, t∈ℝ+

t∈ℝ+

Y := {y ∈ C(ℝ+ , ℝ) | sup |y(t)|e

−γt

t∈ℝ+

< ∞},

α T α x := c D0,t x + λx,

α ∈ (0, 1).

3.3 Ulam–Hyers–Rassias stability for semilinear equations | 121

Then X α and Y are Banach spaces with norm α ‖x‖α := max{sup |x(t)|e−γt , sup |c D0,t x(t)|e−γt }, t∈ℝ+

t∈ℝ+

‖y‖ := sup |y(t)|e−γt , t∈ℝ+

respectively. Clearly T α ∈ L(X α , Y) with ‖T α ‖ ≤ (1 + λ). We recall that the generalized Caputo derivative [162] is defined as t

c

α D0,t x(t) =

d x(s) − x(0) 1 ds ∫ Γ(1 − α) dt (t − s)α 0

and

t

C(α) (ℝ+ , ℝ) = {x ∈ C(ℝ+ , ℝ) | ∫ 0

x(s) − x(0) ds ∈ C1 (ℝ+ , ℝ)}. (t − s)α

Lemma 3.41. T α is surjective. Proof. Solve T α x = y ∈ Y, i.e., α D0,t x + λx = y.

(3.37)

This equation has a solution [162, pp. 140–141, (3.1.32)–(3.1.34)] t

x(t) = ∫(t − s)α−1 Eα,α (−λ(t − s)α )y(s)ds, 0

where Eα,α (z) is the generalized Mittag-Leffler function [162] defined as ∞

Eα,β (z) := ∑ k=0

Note that Eα,α (−t α λ) ≤

1 Γ(α)

zk . Γ(αk + β)

for t ∈ ℝ+ . Then we derive

t

t

0

0

1 ‖y‖ ‖y‖eγt |x(t)| ≤ . ∫(t − s)α−1 |y(s)|ds ≤ ∫(t − s)α−1 eγs ds < Γ(α) Γ(α) γα So ‖x‖ ≤

1 γ α ‖y‖,

and (3.37) gives ‖x‖α ≤ (1 +

λ γ α )‖y‖.

Applying Theorem 3.31, we see that there is a constant c > 0 such that for any x ∈ X α , α y ∈ Y and ε > 0 such that |c D0,t x(t) + λx(t) − y(t)| ≤ εeγt , t ∈ ℝ+ , there is x y ∈ X α such α c that D0,t x y + λx y = y and |x(t) − x y (t)| ≤ cεeγt , t ∈ ℝ+ . Remark 3.42. Certainly, more complicated fractional differential equations can be studied, but the above one is enough to demonstrate our method. Finally, motivated by the above examples, we have the following result related to Theorem 3.31.

122 | 3 Fractional Differential Equations Theorem 3.43. Let X and Y be linear spaces and T : X → Y be linear and surjective, possessing a right inverse T r : Y → X, i.e., TT r y = y for any y ∈ Y. Moreover, there are linear subspaces X1 ⊂ X and Y1 ⊂ Y which are normed so that T r : Y1 → X1 is continuous with them, i.e., ‖T r y‖ ≤ ‖T r ‖ ‖y‖ for any y ∈ Y1 . Then for any x ∈ X, y ∈ Y and ε > 0 such that ‖Tx − y‖ ≤ ε (so Tx − y ∈ Y1 ), there is x y ∈ X such that Tx y = y and ‖x − x y ‖ ≤ ‖T r ‖ε. Proof. We set x y = x − T r Tx + T r y. Then Tx y = Tx − TT r Tx + TT r y = Tx − Tx + y = y and x − x y = T r (Tx − y). So using Tx − y ∈ Y1 and x − x y ∈ X1 , we get ‖x − x y ‖ = ‖T r (Tx − y)‖ ≤ ‖T r ‖ ‖Tx − y‖ ≤ ‖T r ‖ε. The proof is finished. Remark 3.44. The above results are related to [261, Theorem 2]. But we do not need the boundedness of T. Moreover, only the existence of the right inverse T r is needed. Related results will be further discussed in Section 3.3.4. Now we present an example. Example 3.45. We continue with Example 3.40. Let J ⊂ ℝ+ be a closed interval not necessarily bounded, 0 ∈ J and let ϕ ∈ C(J, ℝ+ ) be such that ϕ(t) > 0 for t > 0. We set t

ψ(t) := ∫(t − s)α−1 ϕ(s)ds, 0

X := C(α) (J, ℝ),

X1α := {x ∈ C(α) (J, ℝ) | sup

α

Y := C(J, ℝ),

t∈J\{0}

Y1 := {y ∈ C(J, ℝ) | sup

t∈J\{0}

α T α x := c D0,t x + λx,

|x(t)| < ∞}, ψ(t)

|y(t)| < ∞}, ϕ(t)

α ∈ (0, 1).

We consider the norms |x(t)| , t∈J\{0} ψ(t)

‖x‖α := sup

|y(t)| t∈J\{0} ϕ(t)

‖y‖ := sup

on X1α and Y1 , respectively. We take t

T r y(t) = ∫(t − s)α−1 Eα,α (−(t − s)α λ)y(s)ds. 0

Then

t

|T r y(t)| ≤

1 1 ψ(t)‖y‖ ∫(t − s)α−1 ϕ(s)ds‖y‖ = Γ(α) Γ(α) 0

3.3 Ulam–Hyers–Rassias stability for semilinear equations | 123

α

1 for any t ∈ J. So ‖T r ‖ ≤ Γ(α) . For instance, if ϕ(t) = 1 then ψ(t) = tα . Thus applying Theorem 3.43, we see that there is a constant c > 0 such that for any x ∈ C(α) (J, ℝ), α y ∈ C(J, ℝ) and ε > 0 such that |c D0,t x(t) + λx(t) − y(t)| ≤ ε, t ∈ J, there is x y ∈ C(α) (J, ℝ) α such that c D0,t x y + λx y = y and |x(t) − x y (t)| ≤ cεt α , t ∈ J.

3.3.3 Ulam–Hyers–Rassias stability for linear equations on Banach spaces with closed ranges Now we suppose that T ∈ L(X, Y) has a closed range im T ⊂ Y. We set d(y, im T) := inf{‖y − z‖ | z ∈ im T} and consider [T] : Z → im T, i.e., [T]−1 : im T → Z. Now we extend Theorem 3.31. Theorem 3.46. Let T ∈ L(X, Y) have a closed range im T. Then for any x ∈ X, y ∈ Y, δ1 > 0, δ2 > 0 and ε > 0 such that ‖Tx − y‖ ≤ ε, there is x y ∈ X such that ‖y − Tx y ‖ ≤ d(y, im T) + δ1 ,

‖x − x y ‖ ≤ (‖[T]−1 ‖ + δ2 )(d(y, im T) + δ1 + ε).

Note that d(y, im T) ≤ ε. If ‖y − y∗ ‖ = d(y, im T) for some y∗ ∈ im T, then T x y = y∗ (so ‖y − Tx y ‖ = d(y, im T)) and ‖x − x y ‖ ≤ (‖[T]−1 ‖ + δ2 )(d(y, im T) + ε). Proof. Take y∗ ∈ im T so that ‖y − y∗ ‖ ≤ d(y, im T) + δ1 . Then ‖y∗ − Tx‖ ≤ ‖y∗ − y‖ + ‖y − Tx‖ ≤ d(y, im T) + δ1 + ε. Applying Theorem 3.31 to T : X → im T and y∗ completes the proof. When Y is a Hilbert space, we have the following consequence. Corollary 3.47. Let Y be a Hilbert space and let T ∈ L(X, Y) have a closed range im T. Let P : H → im T be the orthogonal projection. Then for any x ∈ X, y ∈ Y and ε > 0 such that ‖Tx − y‖ ≤ ε, there is x y ∈ X such that ‖y − Tx y ‖ = ‖y − Py‖,

‖x − x y ‖ ≤ ‖[T]−1 ‖(‖y − Py‖ + ε).

Note that ‖y − Py‖ ≤ ε. Proof. Take y∗ = Py and the proof is finished by Remark 3.32. Remark 3.48. The above results are related to [261, Theorem 2] by Takagi et al. We prove some new results from our point of view. We can consider in Corollary 3.47 that Y is reflexive and strictly convex [177, 268] with the metric projection P : Y → im T. We already know [177, 237] that Py is a singleton when Y is reflexive and strictly convex. But we only need Py to be nonempty, which is satisfied if im T is reflexive.

124 | 3 Fractional Differential Equations Example 3.49. Consider X = {x ∈ W 1,2 (0, 1) | x(0) = x(1)}, Y = L2 (0, 1) and Tx = ẋ 1 with the usual integral norms. Then Py = y − ∫0 y(s)ds. Applying Corollary 3.47, we see that there is a constant c > 0 such that for any x ∈ X, y ∈ Y and ε > 0 such that 1 ̇ − y(t))2 dt ≤ ε, there is x y ∈ X such that ∫0 (x(t) 1

1

ẋ y = y − ∫ y(s)ds, 0

1

̇ − ẋ y (t))2 + (x(t) − x y (t))2 )dt ≤ c(ε + ∫ y(s)ds). ∫((x(t) 0

0

3.3.4 Ulam–Hyers–Rassias stability for surjective semilinear equations between Banach spaces Now we consider a map F(x) = Tx + f(x) with T ∈ L(X, Y) surjective and that there exists L > 0 such that ‖f(x1 ) − f(x2 )‖ ≤ L‖x1 − x2 ‖, so f is globally Lipschitz continuous and F is semilinear. Moreover, we assume that there is a splitting X = ker T ⊕ W for a closed subspace W ⊂ X. Then by the Banach inverse theorem, T W := T : W → Y has a −1 continuous inverse T W ∈ L(Y, W). We set W := {W ⊂ X | X = ker T ⊕ W, W is closed} and put −1 Γ := inf ‖T W ‖. W∈W

−1 If [x] = [T]−1 y and x = x1 + w, x1 ∈ ker T, w ∈ W, then T W y = w and w ∈ [x]. So −1 −1 ‖[T]−1 y‖ = ‖[x]‖ ≤ ‖w‖ = ‖T W y‖ ≤ ‖T W ‖ ‖y‖ for all y ∈ Y,

which gives −1 ‖[T]−1 ‖ ≤ ‖T W ‖ for all W ∈ W,

i.e., ‖[T]−1 ‖ ≤ Γ.

(3.38)

We do not know whether there is an equality in (3.38) in general, but certainly it is there when X is a Hilbert space since then we take W = ker T ⊥ . Now we have the following extension of Theorem 3.31. Theorem 3.50. Suppose W ≠ 0. If LΓ < 1 then there is a constant c > 0 such that for any x ∈ X, y ∈ Y and ε > 0 such that ‖F(x) − y‖ ≤ ε, there is x y ∈ X such that F(x y ) = y and Γ ‖x − x y ‖ ≤ cε. Moreover, inf c ≤ 1−LΓ . −1 Proof. Since LΓ < 1, we take W ∈ W such that L‖T W ‖ < 1. Let x ∈ X, y ∈ Y and ε > 0 be such that ‖F(x) − y‖ ≤ ε. Setting e := F(x) − y, we get

y + e = F(x) = Tw1 + f(x1 + w1 )

(3.39)

3.3 Ulam–Hyers–Rassias stability for semilinear equations | 125

−1 for x = x1 + w1 , x1 ∈ ker T and w1 ∈ W. Consider the map F W (w) := T W (y − f(x1 + w)), −1 F W : W → W. It easy to verify that F W is a contraction with the constant L‖T W ‖ < 1. So the Banach fixed point theorem gives a unique fixed point w y ∈ W of F W , i.e.,

y = F(x y ) = Tw y + f(x1 + w y )

(3.40)

for x y = x1 + w y . Furthermore, subtracting (3.40) from (3.39), we arrive at e = F(x) − F(x y ) = T(w1 − w y ) + f(x1 + w1 ) − f(x1 + w y ), which gives 󵄩 −1 󵄩 ‖x − x y ‖ = ‖w1 − w y ‖ = 󵄩󵄩󵄩T W (e + f(x1 + w y ) − f(x1 + w1 ))󵄩󵄩󵄩 −1 −1 ≤ ‖T W ‖(‖e‖ + L‖w1 − w y ‖) = ‖T W ‖(‖e‖ + L‖x − x y ‖).

Hence ‖x − x y ‖ ≤ So c =

−1 ‖T W ‖ −1 . 1−L‖T W ‖

−1 ‖T W ‖ ‖e‖ −1 1 − L‖T W ‖

≤

−1 ‖T W ‖ −1 1 − L‖T W ‖

ε.

Clearly inf

W∈W

−1 ‖T W ‖

1−

−1 L‖T W ‖

=

Γ . 1 − LΓ

The proof is finished. Of course, F : X → Y is surjective. Note that the existence of W ∈ W is equivalent to the existence of a right inverse T r−1 ∈ L(Y, X), i.e., TT r−1 y = y for any y ∈ Y. The equivalent −1 correspondence is given by im T r−1 = W and T r−1 = T W . So Γ is the infimum of ‖T r−1 ‖ for continuous right inverses T r−1 of T. Here we use the following result. Lemma 3.51. T r−1 is continuous if and only if im T r−1 is closed. Proof. Assume T r−1 is continuous. Let x n = T r−1 y n → x as n → ∞. Then y n = Tx n → Tx = y, and thus x n = T r−1 y n → T r−1 y. So x = T r−1 y, i.e., x ∈ im T r−1 . Hence im T r−1 is closed. Next, suppose that im T r−1 is closed. Let y n → y and x n = T r−1 y n → x. Since im T r−1 is closed, we have x ∈ im T r−1 , i.e., there is y0 ∈ Y such that x = T r−1 y0 . Consequently, y n = Tx n → Tx = y0 . Thus y = y0 and x = T r−1 y. This means that the graph of T r−1 is closed, and by the closed graph theorem, T r−1 is continuous. As we have mentioned, Lemma 3.51 is applied in the above examples where we find T r−1 . For instance, we have the following extension of Example 3.33: Let h ∈ C(ℝn × ℝ+ , ℝn ) and let there be L > 0 such that |h(x1 , t) − h(x2 , t)| ≤ L|x1 − x2 | kL for any x1 , x2 ∈ ℝn and t ∈ ℝ+ . Assume γ−γ < 1. Then for any x ∈ X, y ∈ Y and 1 ̇ − Ax(t) − h(x(t), t) − y(t)| ≤ εeγt , t ∈ ℝ+ , there is x y ∈ X such that ε > 0 such that |x(t)

126 | 3 Fractional Differential Equations ẋ y − Ax y − h(x y , t) = y and |x(t) − x y (t)| ≤ γ−γ1k−kL εeγt , t ∈ ℝ+ . The other above examples can be extended analogously. Results of these examples are known as weighted shadowing results in dynamical systems [234]. Remark 3.52. Of course, the results of Section 3.3.3 can be extended to the corresponding semilinear case, like the results of Section 3.3.2 are extended in this subsection. But we do not go into details.

3.4 Practical Ulam–Hyers–Rassias stability for nonlinear equations 3.4.1 Introduction The notion of practical stability for nonlinear equations [171, 257] has attracted more and more attention by many practical problems: a travel by a space vehicle between two points, an aircraft or a missile which may oscillate around a mathematically unstable course yet whose performance may be acceptable, a chemical process of keeping the temperature within certain bounds, etc. However, there are a few papers concerning the Ulam–Hyers–Rassias stability for nonlinear equations on bounded subsets, which we call a practical Ulam–Hyers–Rassias stability for nonlinear equations. Now we are ready to formulate our problem as follows. Let X and Y be Banach spaces. Consider a mapping F ∈ C1 (X, Y). The problem of a practical Ulam–Hyers–Rassias stability (PUHRS for short) of F can be formulated as follows: PUHRS: There is a function φ ∈ C(ℝ2+ , ℝ+ ) with φ(r, 0) = 0 for any r ∈ ℝ+ , and nondecreasing in each variable such that for any x ∈ X and y ∈ F(X), there is x y ∈ X such that F(x y ) = y and |x − x y | ≤ φ(|x|, |y − F(x)|). There is no sense involving also |y| in the function φ, since |y| ≤ |y − F(x)| + |F(x)| and |F(x)| is controlled by |x| in many cases. If φ(r1 , r2 ) is independent of r1 , i.e., φ(r1 , r2 ) = φ(r2 ), then we get the Ulam–Hyers–Rassias stability (UHRS for short) of F. The meaning of PUHRS consists in a restriction of UHRS on bounded subsets: If F is PUHRS then it holds: For any M > 0 there is a nondecreasing function φ M ∈ C(ℝ+ , ℝ+ ) with φ M (0) = 0 such that for any x ∈ X, |x| ≤ M and y ∈ F(X), |y| ≤ M, there is x y ∈ X such that F(x y ) = y and |x − x y | ≤ φ M (|y − F(x)|). Indeed, we take φ M (r) = φ(M, r). Moreover, we may take φ M (r) = c M r for c M > 0 in many reasonable cases (see Corollary 3.55 below). In what follows, we give results answering these interesting questions from a nonlinear functional analysis point of view. This section is based on [279].

3.4 Ulam–Hyers–Rassias stability for nonlinear equations | 127

3.4.2 Main results First, we recall the following Bihari’s inequality [53]. Theorem 3.53. If w(t) is a nonnegative continuous function satisfying t

w(t) ≤ α + β ∫ g(w(s))ds 0

with constants α ≥ 0, β ≥ 0 and g : ℝ+ → (0, ∞) nondecreasing continuous, then w(t) ≤ G−1 (G(α) + βt) x

for all t ≥ 0 for which G(α) + βt belongs to the domain of G−1 and G(x) = ∫1

1 g(u) du.

Now we suppose the following: (i) There is a mapping R : X → L(Y, X) such that (1) R is locally Lipschitz, i.e., for any x ∈ X, there is an open neighborhood U x of x and a constant L x such that ‖R(x1 ) − R(x2 )‖ ≤ L x |x1 − x2 | for all x1 , x2 ∈ U x ; (2) R(x) is a right inverse of DF(x) for all x ∈ X, i.e., DF(x)R(x) = 𝕀Y for all x ∈ X, where 𝕀Y : Y → Y is the identity map on Y. (ii) There is a continuous nondecreasing function g : ℝ+ → (0, ∞) such that ‖R(x)‖ ≤ g(|x|) for any x ∈ X. Theorem 3.54. Assume (i) and (ii). If ∞

∫

1 du = ∞, g(u)

1

then F is PUHRS. Proof. Let x ∈ X and y ∈ Y. Setting e := F(x) − y, we get F(x) = y + e.

(3.41)

We plug (3.41) into the homotopy (see [14] for more complex homotopy theory) {

F(z(t)) = y + (1 − t)e,

t ∈ [0, 1],

z(0) = x.

(3.42)

Assuming that z ∈ C1 ([0, 1], X) and differentiating (3.42), we obtain {

DF(z(t))z󸀠 (t) = −e, z(0) = x.

t ∈ [0, 1],

(3.43)

128 | 3 Fractional Differential Equations

If the differential equation {

z󸀠 (t) = −R(z(t))e,

t ∈ [0, 1],

z(0) = x

(3.44)

has a solution z ∈ C1 ([0, 1], X), then it satisfies (3.43), which gives F(z(t)) + te = c,

t ∈ [0, 1]

(3.45)

for a constant c. But putting t = 0 into (3.45), we derive c = F(z(0)) = F(x) = y + e, which gives (3.42). So we need to solve (3.44). Since R(x) is locally Lipschitz, the Cauchy problem (3.44) has a unique local solution. To prolong it, we note t

|z(t)| ≤ |x| + |e| ∫‖R(z(s))‖ds.

(3.46)

0

So Theorem 3.53 gives |z(t)| ≤ G−1 (G(|x|) + |e|),

t ∈ [0, 1],

which by (3.44) implies 1

|x y − x| = |z(1) − z(0)| ≤ |e| ∫‖R(z(s))‖ds 0 1

≤ |e| ∫ g(|z(s)|)ds ≤ |e|g(G−1 (G(|x|) + |e|)). 0

So φ(r1 , r2 ) = r2 g(G−1 (G(r1 ) + r2 )). The proof is finished. Corollary 3.55. In addition to Theorem 3.54, if F is locally bounded, i.e., F M := sup |F(x)| < ∞ |x|≤M

for any M > 0, then we can take φ(r) = c M r with c M = g(G−1 (G(M) + M + F M )) whenever |x| ≤ M and |y| ≤ M. Remark 3.56. Taking x = 0 in Theorem 3.54, we see that F is surjective and moreover, for any y ∈ Y there is x y ∈ X so that F(x y ) = y with |x y | ≤ (|y| + |F(0)|)g(G−1 (G(0) + |y| + |F(0)|)). Hence Theorem 3.54 is an extension of global invertible mapping results to a global surjective mapping result, especially of one by Hadamard [52, 73].

3.4 Ulam–Hyers–Rassias stability for nonlinear equations | 129

Following the proof of Theorem 3.54, we get the next result. Theorem 3.57. Assume (i) and (ii). Then for any x ∈ X and y ∈ Y such that ∞

|F(x) − y| < ∫

1 du, g(u)

(3.47)

|x|

there is x y ∈ X such that F(x y ) = y and |x y − x| ≤ |F(x) − y|g(G−1 (G(|x|) + |F(x) − y|)). Of course, estimate (3.47) is useful when G(∞) < ∞ for ∞

G(∞) := ∫

1 du. g(u)

1

Then Theorem 3.57 gives an error estimate for PUHRS of F. The next simple example shows that (3.47) is optimal in some sense. Example 3.58. Take X = Y = ℝ and F(x) = arctan x. Then DF(x) = F 󸀠 (x) = R(x) = 1 + x2 . So g(u) = 1 + u2 and (3.47) has the form y − arctan x
0 and y > arctan x. But now (3.48) cannot be improved since the range of F is (− 2π , 2π ). Now we present a simple local result. Theorem 3.59. Assume there is a locally Lipschitz right inverse R : B r → L(Y, X) of DF(x) on the ball B r := {x ∈ X | |x| < r} such that ‖R(x)‖ ≤ M for any x ∈ B r and a constant M > 0. Then for any x ∈ B r and y ∈ Y such that |F(x) − y|
0. Since DF(x)v = ∇F(x)v∗ , we take R(x)y = |∇F(x)| 2 ∇F(x). Thus ‖R(x)‖ = |∇F(x)| . Setting 1 , M := max x∈B r |∇F(x)| we can apply Theorem 3.59. For m = 2, this could express a climbing on the hill, when the relief of the hill is given by F ∈ C2 (ℝ3 , (0, ∞)). So we are at the position (x1 , x2 , F(x1 , x2 )), (x1 , x2 ) ∈ B r , and we try to reach the given altitude y within the region B r . Inequality (3.49) is sufficient to reach the altitude y at the region B r and the location of this new position is given by (3.50).

3.5 Ulam–Hyers–Mittag-Leffler stability of fractional delay differential equations 3.5.1 Introduction Motivated by [145–147, 217], this section is devoted to investigating existence, uniqueness and Ulam–Hyers–Mittag-Leffler stability of the following fractional-order differential equation with modification of the argument: c

D αt x(t) = f(t, x(t), x(g(t))),

t ∈ I ⊂ ℝ, α ∈ (0, 1),

where c D αt is the Caputo fractional derivative of order α with the lower limit zero and we consider I = [0, b], b ∈ ℝ+ and the functions f ∈ C(I × ℝ2 , ℝ), g ∈ C(I, [−h, b]) with g(t) ≤ t and h > 0. Finally, two examples are also provided to illustrate our results. This section is based on [300].

3.5.2 Preliminaries Inspired by the well-known Ulam type stability concepts of ordinary differential equations [245], we recall the Ulam–Hyers–Mittag-Leffler stability concept [294] to describe our current problem. For f ∈ C(I × ℝ2 , ℝ) and ε > 0, we consider the following equations: c

D αt x(t) = f(t, x(t), x(g(t))), x(t) = ψ(t),

t ∈ [−h, 0],

t ∈ I,

(3.54) (3.55)

3.5 Ulam–Hyers–Mittag-Leffler stability | 133

associated with the inequality 󵄨 󵄨󵄨c α 󵄨󵄨 D t y(t) − f(t, y(t), y(g(t)))󵄨󵄨󵄨 ≤ εE α (t α ),

t ∈ I,

(3.56)

where E α is the Mittag-Leffler function. Definition 3.65. Equation (3.54) is Ulam–Hyers–Mittag-Leffler stable with respect to E α (t α ) if there exists c E α > 0 such that for each ε > 0 and for each solution y ∈ C([−h, b], ℝ) to (3.56), there exists a solution x ∈ C([−h, b], ℝ) to (3.54) with |y(t) − x(t)| ≤ c E α εE α (t α ),

t ∈ [−h, b].

Remark 3.66. A function y ∈ C(I, ℝ) is a solution of inequality (3.56) if and only if there exists a function h̃ ∈ C(I, ℝ) (depending on y) such that ̃ (i) |h(t)| ≤ εE α (t α ) for all t ∈ I; α ̃ for all t ∈ I. (ii) c D t y(t) = f(t, y(t), y(g(t))) + h(t) By the above remark, we have: Remark 3.67. Let y ∈ C(I, ℝ) be a solution of inequality (3.56). Then y is a solution of the following integral inequality: t

󵄨 󵄨󵄨 󵄨󵄨y(t) − y(0) − 1 ∫(t − s)α−1 f(s, y(s), y(g(s)))ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 Γ(α) 󵄨 󵄨 0

t

t

∞ s kα ε ε ≤ ds ∫(t − s)α−1 E α (s α )ds ≤ ∫(t − s)α−1 ∑ Γ(α) Γ(α) Γ(kα + 1) k=0 0

0

t

=

ε ∞ 1 ε ∞ t(k+1)α Γ(α)Γ(kα + 1) ∑ ∑ ∫(t − s)α−1 s kα ds = Γ(α) k=0 Γ(kα + 1) Γ(α) k=0 Γ(kα + 1) Γ((k + 1)α + 1) 0

∞

∞ t(k+1)α t nα =ε∑ ≤ε∑ ≤ εE α (t α ). Γ((k + 1)α + 1) Γ(nα + 1) n=0 k=0

3.5.3 Main results In this subsection, we will prove existence, uniqueness and Ulam–Hyers–Mittag-Leffler stability for equation (3.54) on the impact interval I. In the sequel we need to use the abstract Gronwall lemma from Rus [244]. Lemma 3.68. Let (X, d, ≤) be an ordered metric space and A : X → X be an increasing Picard operator (F A = {x∗A }). Then for x ∈ X, x ≤ A(x) implies x ≥ x∗A . We impose the following assumptions: (H1 ) f ∈ C(I × ℝ2 , ℝ), g ∈ C(I, [a − h, b]), g(t) ≤ t, h > 0.

134 | 3 Fractional Differential Equations (H2 ) There exists L f > 0 such that 2

|f(t, u1 , u2 ) − f(t, v1 , v2 )| ≤ L f ∑ |u i − v i | i=1

for all t ∈ I and u i , v i ∈ ℝ, i = 1, 2. (H3 ) The following inequality holds: 2L f b α < 1. Γ(α + 1) Now we are ready to state our first result. Theorem 3.69. Assume that (H1 ), (H2 ) and (H3 ) are satisfied. Then: (i) Problem (3.54), (3.55) has a unique solution in C([−h, b], ℝ) ∩ C(I, ℝ). (ii) Equation (3.54) is Ulam–Hyers–Mittag-Leffler stable. Proof. It is easy to see that equations (3.54) and (3.55) are equivalent to the singular integral system {ψ(t) x(t) = { ψ(0) + {

if t ∈ [−h, 0], t 1 Γ(α) ∫0 (t

−

s)α−1 f(s,

x(s), x(g(s)))ds

if t ∈ I.

(3.57)

In order to prove our first claim, we only prove the existence of a solution for system (3.57) which can be turned into a fixed point problem in X := C([−h, b], ℝ) for an operator B f defined as {ψ(t) B f (x)(t) = { ψ(0) + {

if t ∈ [−h, 0], t 1 Γ(α) ∫0 (t

−

s)α−1 f(s,

x(s), x(g(s)))ds

if t ∈ I.

(3.58)

Next, we need to show that B f is a contraction mapping on X := C([−h, b], ℝ) with respect to the Chebyshev norm ‖⋅‖C . Consider B f : X → X defined in (3.58). For t ∈ [−h, 0], we have |B f (x)(t) − B f (y)(t)| = 0,

x, y ∈ C([−h, b], ℝ).

For all t ∈ I, it follows by (H2 ) that |B f (x)(t) − B f (y)(t)| t

≤

1 󵄨 󵄨 ∫(t − s)α−1 󵄨󵄨󵄨f(s, x(s), x(g(s))) − f(s, y(s), y(g(s)))󵄨󵄨󵄨ds Γ(α) 0 t

Lf ≤ ∫(t − s)α−1 ( max |x(s) − y(s)| + max |x(g(s)) − y(g(s))|)ds Γ(α) −h≤s≤b −h≤s≤b 0

t

2L f 2L f b α ≤ ‖x − y‖C ∫(t − s)α−1 ds ≤ ‖x − y‖C , Γ(α) Γ(α + 1) 0

3.5 Ulam–Hyers–Mittag-Leffler stability | 135

which implies that ‖B f (x) − B f (y)‖C ≤

2L f b α ‖x − y‖C , Γ(α + 1)

x, y ∈ C([−h, b], ℝ).

Thus, B f is a contraction via the Chebyshev norm ‖⋅‖C on X. The rest of the proof follows from the Banach contraction principle. Now we prove our second claim. Let y ∈ C([−h, b], ℝ) ∩ C(I, ℝ) be a solution to (3.56). We denote by x ∈ C([−h, b], ℝ) ∩ C(I, ℝ) the unique solution to the problem c

{

D αt x(t) = f(t, x(t), x(g(t))), x(t) = y(t),

t ∈ I,

t ∈ [−h, 0].

By (H1 ) we have {y(t) x(t) = { y(0) + {

if t ∈ [−h, 0], t 1 Γ(α) ∫0 (t

−

s)α−1 f(s,

x(s), x(g(s)))ds

if t ∈ I.

Obviously, Remark 3.67 gives t

󵄨 󵄨󵄨 󵄨󵄨y(t) − y(0) − 1 ∫(t − s)α−1 f(s, y(s), y(g(s)))ds󵄨󵄨󵄨 ≤ εE (t α ) α 󵄨󵄨 󵄨󵄨 Γ(α) 󵄨 󵄨 0

for t ∈ I, and |y(t) − x(t)| = 0 for t ∈ [−h, 0]. For all t ∈ I, it follows by (H2 ) that |y(t) − x(t)| t

󵄨󵄨 󵄨󵄨 1 ≤ 󵄨󵄨󵄨󵄨y(t) − y(0) − ∫(t − s)α−1 f(s, y(s), y(g(s)))ds󵄨󵄨󵄨󵄨 Γ(α) 󵄨 󵄨 0

t

t

󵄨󵄨 󵄨󵄨 1 1 + 󵄨󵄨󵄨󵄨 ∫(t − s)α−1 f(s, y(s), y(g(s)))ds − ∫(t − s)α−1 f(s, x(s), x(g(s)))ds󵄨󵄨󵄨󵄨 Γ(α) 󵄨 󵄨 Γ(α) 0

0

t

≤ εE α (t α ) +

Lf ( ∫(t − s)α−1 [|y(s) − x(s)| + |y(g(s)) − x(g(s))|]ds). Γ(α)

(3.59)

0

For u ∈ C([−h, b], ℝ+ ) we consider the operator A : C([−h, b], ℝ+ ) → C([−h, b], ℝ+ ) defined by {0 A(u)(t) = { εE (t α ) + { α

if t ∈ [−h, 0], t Lf Γ(α) (∫0 (t

−

t s)α−1 u(s)ds + ∫0 (t

−

s)α−1 u(g(s))ds)

if t ∈ I.

136 | 3 Fractional Differential Equations

Next, we verify that A is a Picard operator. For all t ∈ I, it follows by (H2 ) that t

Lf |A(u)(t) − A(v)(t)| ≤ ( ∫(t − s)α−1 [|u(s) − v(s)|ds + |u(g(s)) − v(g(s))|]ds) Γ(α) 0

2L f b α ≤ ‖u − v‖C Γ(α + 1) for all u, v ∈ C([a − h, b], ℝ+ ). So, we obtain ‖A(u) − A(v)‖C ≤

2L f b α ‖u − v‖C Γ(α + 1)

for all u, v ∈ C([−h, b], ℝ+ ). Thus, A is a contraction via the Chebyshev norm ‖⋅‖C on C([−h, b], ℝ+ ) due to (H3 ). Applying the Banach contraction principle to A, we derive that A is a Picard operator and F A = {u∗ }. Then, for t ∈ I, t

t

0

0

Lf u (t) = εE α (t ) + ( ∫(t − s)α−1 u∗ (s)ds + ∫(t − s)α−1 u∗ (g(s))ds). Γ(α) ∗

α

It remains to verify that the solution u∗ is increasing. Indeed, for 0 ≤ t1 < t2 ≤ b and denoting m := mins∈I [u∗ (s) + u∗ (g(s))] ∈ ℝ+ , we have u∗ (t2 ) − u∗ (t1 ) = ε[E α (t2α ) − E α (t1α )] t1

+

Lf ∫[(t2 − s)α−1 − (t1 − s)α−1 ][u∗ (s) + u∗ (g(s))]ds Γ(α) 0 t2

Lf + ∫(t2 − s)α−1 [u∗ (s) + u∗ (g(s))]ds Γ(α) t1

t1

≥

ε[E α (t2α )

−

E α (t1α )]

mL f + ∫[(t2 − s)α−1 − (t1 − s)α−1 ]ds Γ(α) 0

t2

+

mL f ∫(t2 − s)α−1 ds Γ(α) t1

=

ε[E α (t2α )

− E α (t1α )] +

mL f (t α − t α ) > 0. Γ(α + 1) 2 1

Then we obtain that u∗ is increasing. So, u∗ (g(t)) ≤ u∗ (t) due to g(t) ≤ t and t

2L f u (t) ≤ εE α (t ) + ∫(t − s)α−1 u∗ (s)ds. Γ(α) ∗

α

0

3.5 Ulam–Hyers–Mittag-Leffler stability | 137

Using Lemma A.18 and Remark A.19, we obtain u∗ (t) ≤ c E α εE α (t α ),

t ∈ [−h, b],

where c E α := E α (2L f b α ). In particular, if u = |y − x|, from (3.59), u ≤ Au and applying Lemma 3.68, we obtain u ≤ u∗ , where A is an increasing Picard operator. As a result, we get |y(t) − x(t)| ≤ c E α εE α (t α ),

t ∈ [−h, b].

Thus, equation (3.54) is Ulam–Hyers–Mittag-Leffler stable. Next, we use the Bielecki norm ‖⋅‖B (‖x‖B := maxt∈J |x(t)|e−θt , θ > 0, J ⊂ ℝ+ ) to derive results for equation (3.54) similar to those above. We change (H3 ) to (H4 ): (H4 ) The following inequality holds for some α ∈ ( 12 , 1) and θ > 0: 2L f b α eθb Γ(α)√2(2α − 1)θ

< 1.

Theorem 3.70. Assume that (H1 ), (H2 ) and (H4 ) are satisfied. Then: (i) Problem (3.54), (3.55) has a unique solution in C([−h, b], ℝ) ∩ C(I, ℝ). (ii) Equation (3.54) is Ulam–Hyers–Mittag-Leffler stable. Proof. Just like in the discussion in Theorem 3.69, we only prove that B f defined in (3.58) is a contraction on X via the Bielecki norm ‖⋅‖B . Since the process is standard, we only give the main difference in the proof: For all t ∈ I, we have t

|B f (x)(t) − B f (y)(t)| ≤

1 󵄨 󵄨 ∫(t − s)α−1 󵄨󵄨󵄨f(s, x(s), x(g(s))) − f(s, y(s), y(g(s)))󵄨󵄨󵄨ds Γ(α) 0 t

≤

Lf ∫(t − s)α−1 eθs ( max |x(s) − y(s)|e−θs Γ(α) −h≤s≤b 0

+ max |x(g(s)) − y(g(s))|e−θs )ds −h≤s≤b

t

≤

2L f ‖x − y‖B ∫(t − s)α−1 eθs ds. Γ(α) 0

Note that, by the well-known Hölder’s inequality, for α ∈ ( 12 , 1) we have t

∫(t − s) 0

t

e ds ≤ ( ∫(t − s)

α−1 θs

0

2(α−1)

1 2

t 2θs

ds) ( ∫ e 0

1 2

ds) ≤

1 bα eθb . √2θ √2α − 1

138 | 3 Fractional Differential Equations

Then ‖B f (x) − B f (y)‖B ≤

2L f b α eθb Γ(α)√2(2α − 1)θ

‖x − y‖B .

So, for α ∈ ( 12 , 1) we have ‖B f (x) − B f (y)‖B ≤

2L f b α eθb Γ(α)√2(2α − 1)θ

‖x − y‖B ,

x, y ∈ C([−h, b], ℝ).

Thus, B f is a contraction via the Bielecki norm ‖⋅‖B on X. The rest of the proof follows from the Banach contraction principle due to (H4 ). In order to verify that A in Theorem 3.69 is a Picard operator, as defined in Definition 3.13, we prove that A is a contraction. For all t ∈ I and α ∈ ( 12 , 1) we have |A(u)(t) − A(v)(t)| ≤

2L f b α eθb Γ(α)√2(2α − 1)θ

‖u − v‖B

for all u, v ∈ C([−h, b], ℝ+ ). So, ‖A(u) − A(v)‖B ≤

2L f b α eθb Γ(α)√2(2α − 1)θ

‖u − v‖B

for all u, v ∈ C([−h, b], ℝ+ ). Thus, A is a contraction via the Bielecki norm ‖⋅‖B on C([−h, b], ℝ+ ). The proof of Ulam–Hyers–Mittag-Leffler stability is just like in Theorem 3.69, so we omit it here. Remark 3.71. In Theorem 3.70, we discussed the problem by using Hölder’s inequality and had to put the restriction on α ∈ ( 12 , 1). Here, we can also extend α ∈ ( 12 , 1) to α ∈ (0, 1). Let us give some additional computation to support our attitude. In fact, for all t ∈ I, t

|B f (x)(t) − B f (y)(t)| ≤

2L f b α eθb 2L f ‖x − y‖B ∫(t − s)α−1 eθs ds ≤ ‖x − y‖B . Γ(α) Γ(α + 1) 0

If we restrict 2L f b α eθb /Γ(α + 1) < 1 for some α ∈ (0, 1) and θ > 0, then we can prove that B f is a contraction with respect to the Bielecki norm ‖⋅‖B on X as well.

3.5.4 Examples In this subsection, we give two examples to illustrate our main results. Example 3.72. Consider the fractional-order system 1 1 x2 (t − 1) 1 { + sin(2x(t)), {c D t2 x(t) = 2 5 1 + x (t − 1) 5 { { x(0) = 0, t ∈ [−1, 0], {

t ∈ [0, 1],

(3.60)

3.6 Nonlinear impulsive fractional differential equations | 139

and the inequality 1 󵄨 󵄨󵄨c 21 󵄨󵄨 D t y(t) − f(t, y(t), y(t − 1))󵄨󵄨󵄨 ≤ εE 12 (t 2 ).

Set b = 1, α = 21 , L f = 25 , and g(⋅) = ⋅ − 1,

f(⋅, x(⋅), g(x(⋅))) =

1 x2 (⋅ − 1) 1 + sin(2x(⋅)). 2 5 1 + x (⋅ − 1) 5

Thus,

2L f b α 8 = ≈ 0.9 < 1. Γ(α + 1) 5√π Now all the assumptions of Theorem 3.69 are satisfied, problem (3.60) has a unique solution and the first equation in (3.60) is Ulam–Hyers–Mittag-Leffler stable with 1

|y(t) − x(t)| ≤ c E 1 εE 12 (t 2 ),

t ∈ [−1, 1],

2

where c E 1 = E 21 ( 45 ). 2

Example 3.73. Consider the fractional-order system 2 1 x2 (t) 1 { + cos(2x(t − h)), {c D t3 x(t) = 2 (t) 4 4 1 + x { { x(0) = 0, t ∈ [−1, 0], {

t ∈ [0,

1 ], 4

(3.61)

and the inequality 2 󵄨 󵄨󵄨c 23 󵄨󵄨 D t y(t) − f(t, y(t), y(t − h))󵄨󵄨󵄨 ≤ εE 23 (t 3 ).

Following Theorem 3.70, letting α = 23 , θ =

1 2

and L f = 21 , we derive 2

2L f b α eθb Γ(α)√2(2α − 1)θ

=

1

( 14 ) 3 e 8 Γ( 23 )√ 13

< 0.58 < 1.

Now all the assumptions of Theorem 3.70 are satisfied, problem (3.61) has a unique solution and the first equation in (3.61) is Ulam–Hyers–Mittag-Leffler stable with 2

|y(t) − x(t)| ≤ c E 2 εE 23 (t 3 ), 3

t ∈ [−1,

1 ], 4

2

where c E 2 = E 32 (( 14 ) 3 ). 3

3.6 Nonlinear impulsive fractional differential equations 3.6.1 Introduction The theory of impulsive differential equations of integer order has found extensive applications in realistic mathematical modeling of a wide variety of practical situations and has emerged as an important area of investigation in recent years. For the

140 | 3 Fractional Differential Equations

general theory and applications of impulsive differential equations, we refer the reader to the monographs mentioned in the Introduction of this book. However, impulsive differential equations of fractional order have not been extensively studied and many aspects of these equations are yet to be explored. For some recent works on impulsive fractional differential equations, see [5, 10, 33, 50, 274] and the references therein. The main purpose of this section is threefold. Firstly, we consider the Cauchy problems for nonlinear impulsive fractional differential equations q

c D t u(t) = f(t, u(t)), t ∈ J 󸀠 := J \ {t1 , . . . , t m }, J := [0, T], { { { ∆u(t k ) := u(t+k ) − u(t−k ) = I k (u(t−k )), k = 1, 2, . . . , m, { { { { u(0) = u0 ,

(3.62)

q

where c D t is the Caputo fractional derivative of order q ∈ (0, 1) with the lower limit zero, u0 ∈ ℝ, f : J × ℝ → ℝ is jointly continuous, I k : ℝ → ℝ, t k satisfy 0 = t0 < t1 < ⋅ ⋅ ⋅ < t m < t m+1 = T, and u(t+k ) = limϵ→0+ u(t k + ϵ) and u(t−k ) = limϵ→0− u(t k + ϵ) represent the right- and left-sided limits of u(t) at t = t k , respectively. Secondly, we discuss Ulam stability for impulsive fractional differential equations c

{

q

D t u(t) = f(t, u(t)),

t ∈ J󸀠 ,

∆u(t k ) = I k (u(t−k )),

k = 1, 2, . . . , m.

(3.63)

Lastly, we consider boundary problems for nonlinear impulsive fractional differential equations c q D t u(t) = f(t, u(t)), t ∈ J 󸀠 , { { { (3.64) ∆u(t k ) = I k (u(t−k )), k = 1, 2, . . . , m, { { { {au(0) + bu(T) = c, where a, b, c are real constants with a + b ≠ 0. This section is based on [307].

3.6.2 Preliminaries This subsection is devoted to certain impulsive fractional inequalities. Lemma 3.74. Let u ∈ PC(J, ℝ) satisfy the inequality t

|u(t)| ≤ c1 (t) + c2 ∫(t − s)β−1 |u(s)|ds + ∑ θ k |u(t−k )|, 0 0 and λ ∈ [0, 1) such that |f(t, u) − f(t, v)| ≤ L f |u − v|λ

for all t ∈ J, u, v ∈ ℝ.

[H4] I k : ℝ → ℝ and there exists a constant K Ik > 0 such that |I k (u) − I k (v)| ≤ K Ik |u − v|

for all u, v ∈ ℝ, k = 1, 2, . . . , m.

i Moreover, it holds K I = ∑m i=1 K I ∈ [0, 1). [H5] For arbitrary u ∈ ℝ, there exist C I , M I > 0 and q2 ∈ [0, 1) such that

|I k (u)| ≤ C I |u|q2 + M I ,

k = 1, 2, . . . , m.

Now, we define the following operators: Let F : PC(J, ℝ) → PC(J, ℝ) be given by k

(Fu)(t) = ∑ (F i u)(t),

t ∈ (t k , t k+1 ], k = 0, 1, 2, . . . , m,

i=0

where (F 0 u)(t) = u0 and (F i u)(t) = I i (u(t−i )) for i = 1, 2, . . . , m.

144 | 3 Fractional Differential Equations Let G : PC(J, ℝ) → PC(J, ℝ) be given by t

1 (Gu)(t) = ∫(t − s)q−1 f(s, u(s))ds, Γ(q)

t ∈ (t k , t k+1 ], k = 0, 1, 2, . . . , m.

0

Let 𝕋 : PC(J, ℝ) → PC(J, ℝ) be given by 𝕋u = Fu + Gu. Thus, the existence of a solution for problem (3.62) is equivalent to the existence of a fixed point of the operator 𝕋. Lemma 3.79. The operator F : PC(J, ℝ) → PC(J, ℝ) is Lipschitz with constant K I . Consequently, F is α-Lipschitz with the same constant K I . Moreover, F satisfies the growth condition q ‖Fu‖PC ≤ |u0 | + m(C I ‖u‖C2 + M I ) (3.72) for every u ∈ PC(J, ℝ). Proof. For t ∈ [0, t1 ], it is obvious that |(Fu)(t) − (Fv)(t)| = |(F 0 u)(t) − (F 0 v)(t)| = 0 for every u, v ∈ PC(J, ℝ). For t ∈ (t1 , t2 ], using [H4], |(Fu)(t) − (Fv)(t)| = |(F 1 u)(t) − (F 1 v)(t)| ≤ K 1I ‖u − v‖PC for every u, v ∈ PC(J, ℝ). In general, for t ∈ (t k , t k+1 ], k = 2, 3, . . . , m, using [H4] step by step, F i is α-Lipschitz with constant K Ii . Using Proposition A.30, F is α-Lipschitz i with constant ∑m i=1 K I . Relation (3.72) is a simple consequence of [H5]. Lemma 3.80. The operator G : PC(J, ℝ) → PC(J, ℝ) is continuous. Moreover, G satisfies the growth condition q T q (C f ‖u‖PC1 + M f ) ‖Gu‖PC ≤ (3.73) Γ(q + 1) for every u ∈ PC(J, ℝ). Proof. For our purpose, we need to check the continuity of the operators G on C((t k , t k+1 ], ℝ), k = 1, 2, . . . , m step by step. For that, let {u n } be a sequence of the bounded set BK1 := {‖u‖C([0,t1 ],ℝ) ≤ K1 | u ∈ C([0, t1 ], ℝ)} ⊆ C([0, t1 ], ℝ) such that u n → u in BK1 . We have to show that ‖Gu n − Gu‖C([0,t1 ],ℝ) → 0.

3.6 Nonlinear impulsive fractional differential equations | 145

It is easy to see that f n = maxs∈J |f(s, u n (s)) − f(s, u(s))| → 0 as n → ∞ due to the continuity of f . Then, for all t ∈ [0, t1 ], t

|(Gu n )(t) − (Gu)(t)| ≤

1 󵄨 󵄨 ∫(t − s)q−1 󵄨󵄨󵄨f(s, u n (s)) − f(s, u(s))󵄨󵄨󵄨ds Γ(q) 0 t

fn fn T q ≤ . ∫(t − s)q−1 ds ≤ Γ(q) Γ(q) 0

Therefore, Gu n → Gu as n → ∞ which implies that G is continuous on [0, t1 ]. Repeating the above process step by step, one obtains the continuity of the operator G on (t k , t k+1 ], k = 1, 2, . . . , m. Relation (3.73) is a simple consequence of [H2]. Lemma 3.81. The operator G : PC(J, ℝ) → PC(J, ℝ) is compact. Consequently, G is αLipschitz with zero constant. Proof. In order to prove the compactness of G, we consider a bounded set D ⊂ BK1 ⊆ C((t k , t k+1 ], ℝ), k = 1, 2, . . . , m and we will show that G(D) is relatively compact in C((t k , t k+1 ], ℝ) with the help of the Arzelà–Ascoli theorem. For t ∈ [0, t1 ], let {u n } be a sequence on D ⊂ BK1 ⊆ C([0, t1 ], ℝ). By (3.73), we have q1 T q (C f ‖u‖C([0,t + Mf ) 1 ],ℝ) ‖Gu‖C([0,t1 ],ℝ) ≤ Γ(q + 1) for every u n ∈ D. So, G(D) is bounded in C([0, t1 ], ℝ). Now we prove that {Gu n } is equicontinuous. For 0 ≤ τ1 < τ2 ≤ t1 , we get τ1

1 󵄨󵄨 󵄨 ∫((τ1 − s)q−1 − (τ2 − s)q−1 )|f(s, u n (s))|ds 󵄨󵄨(Gu n )(τ1 ) − (Gu n )(τ2 )󵄨󵄨󵄨 ≤ Γ(q) 0 τ2

1 + ∫(τ2 − s)q−1 |f(s, u n (s))|ds Γ(q) τ1

τ1

≤

1 ∫((τ1 − s)q−1 − (τ2 − s)q−1 )(C f |u n (s)|q1 + M f )ds Γ(q) 0 τ2

+

1 ∫(τ2 − s)q−1 (C f |u n (s)|q1 + M f )ds Γ(q) τ1

q

q

q

(C f K11 + M f ) τ1 τ2 (τ2 − τ1 )q (τ2 − τ1 )q ≤ + + [ − ] Γ(q) q q q q q

≤

2(C f K11 + M f )(τ2 − τ1 )q . Γ(q)

146 | 3 Fractional Differential Equations As τ2 → τ1 , the right-hand side of the above inequality tends to zero. Therefore {Gu n } is equicontinuous. Hence, G(D) ⊂ C([0, t1 ], ℝ) satisfies the hypotheses of the Arzelà– Ascoli theorem, yielding that G(D) is relatively compact in C([0, t1 ], ℝ). Repeating the above process for each (t k , t k+1 ], k = 1, 2, . . . , m, one obtains the compactness of the operator G on C((t k , t k+1 ], ℝ), k = 1, 2, . . . , m. By Proposition A.31, G is α-Lipschitz with zero constant. Now, we are ready to prove the main results of this section. Theorem 3.82. Assume that [H1], [H2], [H4] and [H5] hold. Then problem (3.62) has at least one solution u ∈ PC(J, ℝ), and the set of the solutions of problem (3.62) is bounded in PC(J, ℝ). Proof. Let F, G, 𝕋 : PC(J, ℝ) → PC(J, ℝ) be the operators defined in the beginning of this subsection. They are continuous and bounded. Moreover, F is α-Lipschitz with constant K I ∈ [0, 1) and G is α-Lipschitz with zero constant (see Lemmas 3.79, 3.80 and 3.81). Also, Proposition A.30 shows that 𝕋 is a strict α-contraction with constant K I . Set S = {u ∈ PC(J, ℝ) | ∃ λ ∈ [0, 1] such that u = λ𝕋u}. Next, we prove that S is bounded in PC(J, ℝ). Consider u ∈ S and λ ∈ [0, 1] such that u = λ𝕋u. It follows from (3.72) and (3.73) that ‖u‖PC ≤ λ(‖Fu‖PC + ‖Gu‖PC ) q

≤ λ[|u0 | +

T q (C f ‖u‖PC1 + M f ) q + m(C I ‖u‖PC2 + M I )]. Γ(q + 1)

(3.74)

This inequality, together with q1 < 1 and q2 < 1, shows that S is bounded in PC(J, ℝ). If not, we suppose in contrary, ρ := ‖u‖PC → ∞. Dividing both sides of (3.74) by ρ, and taking ρ → ∞, we have 1 ≤ lim

ρ→∞

|u0 | +

T q (C f ρ q1 +M f ) Γ(q+1)

ρ

+ m(C I ρ q2 + M I )

= 0.

This is a contradiction. Consequently, by Theorem A.33 we deduce that 𝕋 has at least one fixed point and the set of the fixed points of 𝕋 is bounded in PC(J, ℝ). Remark 3.83. (i) If the growth condition [H2] is formulated for q1 = 1, then the conT q Cf clusions of Theorem 3.82 remain valid provided that Γ(q+1) < 1. (it) If the growth condition [H5] is formulated for q2 = 1, then the conclusions of Theorem 3.82 remain valid provided that mC I < 1. (iii) If the growth conditions [H2] and [H5] are formulated for q2 = 1 and q3 = 1, T q Cf then the conclusions of Theorem 3.82 remain valid provided that Γ(q+1) + mC I < 1. (iv) For q1 = q2 = 1, one can also obtain the boundedness of the set S by virtue of Lemma A.18 and Remark A.19. (v) For q1 , q2 ∈ [0, 1), one can also obtain the boundedness of the set S by virtue of Lemma 3.75.

3.6 Nonlinear impulsive fractional differential equations | 147

Now, we give the following data dependence result. Theorem 3.84. Assume that [H1]–[H5] hold. Let u(⋅) and v(⋅) be the solutions of problem (3.62) with initial values u0 and v0 , respectively. Then it holds ‖u − v‖PC ≤ M ∗ , where M ∗ is the only positive solution of the equation M∗ =

L f T q M ∗λ |u0 − v0 | . + 1 − KI (1 − K I )Γ(q + 1)

Proof. Note that by [H3] and [H4], we obtain t

Lf |u(t) − v(t)| ≤ |u0 − v0 | + K I ‖u − v‖PC + ∫(t − s)q−1 |u(s) − v(s)|λ ds, Γ(q) 0

which implies that (1 − K I )|u(t) − v(t)| ≤ (1 − K I )‖u − v‖PC t

≤ |u0 − v0 | +

Lf ∫(t − s)q−1 |u(s) − v(s)|λ ds. Γ(q) 0

Thus, t

|u(t) − v(t)| ≤

Lf |u0 − v0 | + ∫(t − s)q−1 |u(s) − v(s)|λ ds. 1 − KI (1 − K I )Γ(q) 0

Lemma 3.75 completes the proof. In order to obtain the uniqueness of solutions, we revise [H3] as follows: [H3󸀠 ] There exists a constant L f > 0 such that |f(t, u) − f(t, v)| ≤ L f |u − v|

for all t ∈ J, u, v ∈ ℝ.

Finally, we give the following uniqueness result. Theorem 3.85. Assume that [H1], [H2], [H3󸀠 ], [H4] and [H5] hold. Then problem (3.62) has a unique solution u ∈ PC(J, ℝ). Proof. By Theorem 3.82, problem (3.62) has a solution u(⋅) in PC(J, ℝ). Let v(⋅) be another solution of problem (3.62) with initial value v0 . Note that by [H3󸀠 ] and [H4], repeating the same process as in the proof of Theorem 3.84, we obtain t

Lf 1 |u(t) − v(t)| ≤ |u0 − v0 | + ∫(t − s)q−1 |u(s) − v(s)|ds. 1 − KI (1 − K I )Γ(q) 0

This yields the uniqueness of u(⋅) due to Lemma A.18 and Remark A.19.

148 | 3 Fractional Differential Equations

3.6.4 Ulam stability results for impulsive fractional differential equations In this subsection, we investigate the Ulam stability of impulsive fractional differential equations (3.63). Let ϵ be a positive real number and φ : J → ℝ+ be a continuous function. We consider the following inequalities q

{

|c D t y(t) − f(t, y(t))| ≤ ϵ, |∆u(t k ) − I k (y(t−k ))| ≤ ϵ,

t ∈ J󸀠 , k = 1, 2, . . . , m,

q

{

|c D t y(t) − f(t, y(t))| ≤ φ(t), |∆u(t k ) − I k (y(t−k ))| ≤ φ(t), q

|c D t y(t) − f(t, y(t))| ≤ ϵφ(t), { |∆u(t k ) − I k (y(t−k ))| ≤ ϵφ(t),

t ∈ J󸀠 , k = 1, 2, . . . , m, t ∈ J󸀠 , k = 1, 2, . . . , m.

(3.75)

(3.76)

(3.77)

Definition 3.86. Equation (3.63) is Ulam–Hyers stable if there exists a real number c f,m > 0 such that for each ϵ > 0 and for each solution y ∈ PC(J, ℝ) of inequality (3.75) there exists a solution x ∈ PC(J, ℝ) of equation (3.63) with |y(t) − x(t)| ≤ c f,m ϵ,

t ∈ J.

Definition 3.87. Equation (3.63) is generalized Ulam–Hyers stable if there exists θ f,m ∈ PC(J, ℝ+ ), θ f,m (0) = 0 such that for each solution y ∈ PC(J, ℝ) of inequality (3.75) there exists a solution x ∈ PC(J, ℝ) of equation (3.63) with |y(t) − x(t)| ≤ θ f,m (ϵ),

t ∈ J.

Definition 3.88. Equation (3.63) is Ulam–Hyers–Rassias stable with respect to φ if there exists c f,m,φ > 0 such that for each ϵ > 0 and for each solution y ∈ PC(J, ℝ) of inequality (3.77) there exists a solution x ∈ PC(J, ℝ) of equation (3.63) with |y(t) − x(t)| ≤ c f,m,φ ϵφ(t),

t ∈ J.

Definition 3.89. Equation (3.63) is generalized Ulam–Hyers–Rassias stable with respect to φ if there exists c f,m,φ > 0 such that for each solution y ∈ PC(J, ℝ) of inequality (3.76) there exists a solution x ∈ PC(J, ℝ) of equation (3.63) with |y(t) − x(t)| ≤ c f,m,φ φ(t),

t ∈ J.

Remark 3.90. It is clear that: (i) Definition 3.86 ⇒ Definition 3.87; (ii) Definition 3.88 ⇒ Definition 3.89; (iii) Definition 3.88 for φ(t) = 1 ⇒ Definition 3.86. Remark 3.91. A function y ∈ PC(J, ℝ) is a solution of inequality (3.75) if and only if there exist a function g ∈ PC(J, ℝ) and a sequence g k , k = 1, 2, . . . , m (which depend on y) such that

3.6 Nonlinear impulsive fractional differential equations | 149

(i) |g(t)| ≤ ϵ, t ∈ J and |g k | ≤ ϵ, k = 1, 2, . . . , m; q (ii) c D t y(t) = f(t, y(t)) + g(t), t ∈ J 󸀠 ; (iii) ∆y(t k ) = I k (y(t−k )) + g k , k = 1, 2, . . . , m. Remarks analogous to Remark 3.91 hold for the inequalities (3.76) and (3.77). So, the Ulam stabilities of the impulsive fractional differential equations are some special types of data dependence of the solutions of impulsive fractional differential equations. Remark 3.92. Let 0 < q < 1. If y ∈ PC(J, ℝ) is a solution of inequality (3.75), then y is a solution of the integral inequality t

k 󵄨 󵄨󵄨 tq 󵄨󵄨y(t) − y(0) − ∑ I (y(t− )) − 1 ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨 ≤ (m + )ϵ i 󵄨󵄨 󵄨󵄨 i Γ(q) Γ(q + 1) 󵄨 󵄨 i=1 0

for t ∈ J. Indeed, by Remark 3.91 we have that c

{

D α y(t) = f(t, y(t)) + g(t),

t ∈ J󸀠 ,

∆y(t k ) = I k (y(t−k )) + g k ,

k = 1, 2, . . . , m.

Then k

k

y(t) = y(0) + ∑ I i (y(t−i )) + ∑ g i + i=1

i=1

t

1 ∫(t − s)q−1 f(s, y(s))ds Γ(q) 0

t

+

1 ∫(t − s)q−1 g(s)ds, Γ(q)

t ∈ (t k , t k+1 ].

0

From this it follows that t

k 󵄨󵄨 󵄨 󵄨󵄨y(t) − y(0) − ∑ I (y(t− )) − 1 ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨 i 󵄨󵄨 󵄨󵄨 i Γ(q) 󵄨 󵄨 i=1 0

t

󵄨󵄨 󵄨󵄨 k 1 = 󵄨󵄨󵄨󵄨 ∑ g i + ∫(t − s)q−1 g(s)ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 i=1 0

t

m

1 ≤ ∑ |g i | + ∫(t − s)q−1 |g(s)|ds Γ(q) i=1 0

t

≤ mϵ +

tq ϵ )ϵ. ∫(t − s)q−1 ds ≤ (m + Γ(q) Γ(q + 1) 0

Remarks analogous to the above remark hold for the inequalities (3.76) and (3.77). Now, we give the main results of this section on the generalized Ulam–Hyers– Rassias stability.

150 | 3 Fractional Differential Equations Theorem 3.93. Let the assumptions [H1], [H2], [H3󸀠 ], [H4] and [H5] hold. Suppose that there exists λ φ > 0 such that t

1 ∫(t − s)q−1 φ(s)ds ≤ λ φ φ(t) for all t ∈ J, Γ(q) 0

where φ ∈ C(J, ℝ+ ) is nondecreasing. Then equation (3.63) is generalized Ulam–Hyers– Rassias stable. Proof. Let y ∈ PC(J, ℝ) be a solution of inequality (3.76). By Theorem 3.85 there is a unique solution x of the impulsive Cauchy problem q

c D t x(t) = f(t, x(t)), { { { ∆x(t k ) = I k (x(t−k )), { { { { x(0) = y(0).

t ∈ J󸀠 , k = 1, 2, . . . , m,

Then t

1 y(0) + Γ(q) ∫0 (t − s)q−1 f(s, x(s))ds { { { { t { 1 { y(0) + I1 (x(t−1 )) + Γ(q) ∫0 (t − s)q−1 f(s, x(s))ds { { { { t − − 1 q−1 x(t) = {y(0) + I1 (x(t1 )) + I2 (x(t2 )) + Γ(q) ∫0 (t − s) f(s, x(s))ds {. { { { .. { { { { { t m − 1 q−1 {y(0) + ∑k=1 I k (x(t k )) + Γ(q) ∫0 (t − s) f(s, x(s))ds By differential inequality (3.76), for each t ∈ (t k , t k+1 ], we have

if t ∈ [0, t1 ], if t ∈ (t1 , t2 ], if t ∈ (t2 , t3 ], if t ∈ (t m , T].

t

k 󵄨󵄨 󵄨 󵄨󵄨y(t) − y(0) − ∑ I (y(t− )) − 1 ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨 k 󵄨󵄨 󵄨󵄨 i Γ(q) 󵄨 󵄨 i=1 0

t

m

≤ ∑ |g(t−k )| + k=1

1 ∫(t − s)q−1 φ(s)ds ≤ (m + λ φ )φ(t), Γ(q)

t ∈ J.

0

Hence for each t ∈ (t k , t k+1 ], it follows t

k 󵄨󵄨 󵄨󵄨 1 |y(t) − x(t)| ≤ 󵄨󵄨󵄨󵄨y(t) − y(0) − ∑ I i (x(t−i )) − ∫(t − s)q−1 f(s, x(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 i=1 0

t

k 󵄨󵄨 󵄨󵄨 1 ≤ 󵄨󵄨󵄨󵄨y(t) − y(0) − ∑ I i (y(t−i )) − ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 i=1 0

t

k

+

∑ |I i (x(t−i )) i=1

−

I i (y(t−i ))|

0 t

≤ (m + λ φ )φ(t) +

1 + ∫(t − s)q−1 |f(s, y(s)) − f(s, x(s))|ds Γ(q)

m Lf ∫(t − s)q−1 |y(s) − x(s)|ds + ∑ K Ik |y(t−k ) − x(t−k )|. Γ(q) k=1 0

3.6 Nonlinear impulsive fractional differential equations | 151

By Lemma 3.74, there exists a constant M ∗f,m > 0 independent of λ φ φ(t) such that |y(t) − x(t)| ≤ M ∗f,m (m + λ φ )φ(t) := c f,m,φ φ(t),

t ∈ J.

Thus, equation (3.63) is generalized Ulam–Hyers–Rassias stable. Remark 3.94. (i) Under the assumptions of Theorem 3.93, we consider equation (3.63) and inequality (3.77). One can repeat the same process to verify that equation (3.63) is Ulam–Hyers–Rassias stable. (ii) Under the assumptions of Theorem 3.93, we consider equation (3.63) and inequality (3.75). One can repeat the same process to verify that equation (3.63) is Ulam–Hyers stable. (iii) One can extend the above results to the case of equation (3.63) with T = +∞.

3.6.5 Existence results for impulsive boundary value problems This subsection deals with the existence and data dependence of solutions for problem (3.64). Definition 3.95. A function u ∈ PC(J, ℝ) is said to be a solution of problem (3.64) if u(t) = u k (t) for t ∈ (t k , t k+1 ), and u k ∈ C([0, t k+1 ], ℝ) for k = 0, 1, 2, . . . , m satisfies c D q u (t) = f(t, u (t)) a.e. on (0, t k k+1 ), the restriction of u k (t) onto [0, t k ) is just u k−1 (t), t k and u(t+k ) = u(t−k ) + I k (u(t−k )), k = 1, 2, . . . , m, along with au(0) + bu(T) = c. Lemma 3.96. Let q ∈ (0, 1) and h : J → ℝ be continuous. A function u is a solution of the fractional integral equation t

{ 1 { { ∫(t − s)q−1 h(s)ds − U { { { Γ(q) { { 0 { { { { t { { { {I (u(t− )) + 1 ∫(t − s)q−1 h(s)ds − U { { 1 1 { { Γ(q) { { 0 { { { { t u(t) = { 1 − − { I1 (u(t1 )) + I2 (u(t2 )) + ∫(t − s)q−1 h(s)ds − U { { Γ(q) { { { 0 { { { { . { { .. { { { { { { t { m { { 1 { − q−1 { { { ∑ I i (u(t i )) + Γ(q) ∫(t − s) h(s)ds − U 0 { i=1 with

T

if t ∈ [0, t1 ],

if t ∈ (t1 , t2 ],

if t ∈ (t2 , t3 ],

if t ∈ (t m , T],

m 1 b U= [ ∑ bI i (u(t−i )) + ∫(T − s)q−1 h(s)ds − c] a + b i=1 Γ(q) 0

(3.78)

152 | 3 Fractional Differential Equations

if and only if u is a solution of the following impulsive boundary value problem: q

c D t u(t) = h(t), t ∈ (0, T], { { { ∆u(t k ) = I k (u(t−k )), k = 1, 2, . . . , m, { { { {au(0) + bu(T) = c.

(3.79)

Proof. Assume that u satisfies (3.79). If t ∈ [0, t1 ] then c

q

D t u(t) = h(t),

t ∈ (0, t1 ].

(3.80)

Integrating expression (3.80) from 0 to t by the fractional integral definition, one obtains t

1 ∫(t − s)q−1 h(s)ds. Γ(q)

u(t) = u(0) +

0

If t ∈ (t1 , t2 ] then

c

{

q

D t u(t) = h(t), u(t+1 )

=

u(t−1 )

t ∈ (t1 , t2 ], + I1 (u(t−1 )).

By Lemma A.16, one obtains t1

u(t) =

u(t+1 )

t

1 1 − ∫(t − s)q−1 h(s)ds ∫(t1 − s)q−1 h(s)ds + Γ(q) Γ(q) 0

0

=

u(t−1 )

+

I1 (u(t−1 ))

t1

t

0

0

1 1 − ∫(t − s)q−1 h(s)ds ∫(t1 − s)q−1 h(s)ds + Γ(q) Γ(q) t

= u(0) + I1 (u(t−1 )) +

1 ∫(t − s)q−1 h(s)ds. Γ(q) 0

If t ∈ (t2 , t3 ] then again from Lemma A.16 we get t2

u(t) = u(t+2 ) −

t

1 1 ∫(t2 − s)q−1 h(s)ds + ∫(t − s)q−1 h(s)ds Γ(q) Γ(q) 0

0 t2

=

u(t−2 )

+

I2 (u(t−2 ))

t

1 1 − ∫(t2 − s)q−1 h(s)ds + ∫(t − s)q−1 h(s)ds Γ(q) Γ(q) 0

0 t

= u(0) + I1 (u(t−1 )) + I1 (u(t−2 )) +

1 ∫(t − s)q−1 h(s)ds. Γ(q) 0

If t ∈ (t m , T] then again from Lemma A.16 we derive t

m

u(t) = u(0) + ∑ i=1

I i (u(t−i ))

1 + ∫(t − s)q−1 h(s)ds. Γ(q) 0

3.6 Nonlinear impulsive fractional differential equations | 153

From the boundary value condition au(0) + bu(T) = c we get T

m b 1 u(0) = − [ ∑ bI i (u(t−i )) + ∫(T − s)q−1 h(s)ds − c]. a + b i=1 Γ(q) 0

This gives formula (3.78). q Conversely, assume that u satisfies (3.78). If t ∈ (0, t1 ] and using the fact that c D t q is the left inverse of I t , we get (3.80). If t ∈ (t k , t k+1 ], k = 1, 2, . . . , m and using the q fact that the Caputo derivative of a constant is equal to zero, we obtain c D t u(t) = h(t), t ∈ (t k−1 , t k ] and u(t+k ) − u(t−k ) = I k (u(t−k )), k = 1, 2, . . . , m. For t ∈ (t m , T], we have au(0) + bu(T) = c. Now, define operators F1 , G11 , G12 : PC(J, ℝ) → PC(J, ℝ) as (F1 u)(t) =

k m b a c , ∑ I i (u(t−i )) − ∑ I i (u(t−i )) + a + b i=1 a + b i=k+1 a+b t

(G11 u)(t) =

1 ∫(t − s)q−1 f(s, u(s))ds, Γ(q) 0 T

b (G12 u)(t) = − ∫(T − s)q−1 f(s, u(s))ds (a + b)Γ(q) 0

for t ∈ (t k , t k+1 ], k = 0, 1, 2, . . . , m. Let 𝕋1 : PC(J, ℝ) → PC(J, ℝ) be given by 𝕋1 u = F1 u + G11 u + G12 u. Thus, the existence of a solution for problem (3.64) is equivalent to the existence of a fixed point for operator 𝕋1 . Proofs of the following lemmas follow like above, so we omit them. Lemma 3.97. Under assumptions [H1], [H4] and [H5], the operator F1 : PC(J, ℝ) → PC(J, ℝ) is Lipschitz with constant KI =

max{|a|, |b|} m i ∑ KI . |a + b| i=1

Consequently, F1 is α-Lipschitz with the same constant K I . Moreover, F1 satisfies the growth condition ‖F1 u‖PC ≤ for every u ∈ PC(J, ℝ).

max{|a|, |b|} |c| q m(C I ‖u‖C2 + M I ) + |a + b| |a + b|

(3.81)

154 | 3 Fractional Differential Equations Lemma 3.98. Under assumptions [H1] and [H2], the operators G11 , G12 : PC(J, ℝ) → PC(J, ℝ) are continuous and compact, and satisfy the growth conditions q

‖G11 u‖PC ≤

T q (C f ‖u‖PC1 + M f ) , Γ(q + 1)

q

‖G12 u‖PC ≤

|b|T q (C f ‖u‖PC1 + M f ) |a + b|Γ(q + 1)

(3.82)

for every u ∈ PC(J, ℝ). Now, we are ready to prove the main result of this subsection. Theorem 3.99. Assume that [H1], [H2], [H4] and [H5] hold. Suppose that K I < 1. Then problem (3.64) has at least one solution u ∈ PC(J, ℝ) and the set of the solutions of problem (3.64) is bounded in PC(J, ℝ). Proof. Let F1 , G11 , G12 , 𝕋1 : PC(J, ℝ) → PC(J, ℝ) be the operators defined at the beginning of this subsection. They are continuous and bounded. Moreover, F1 is α-Lipschitz with the constant K I , and G11 + G12 is α-Lipschitz with zero constant. Also, Proposition A.30 shows that 𝕋1 is a strict α-contraction with the constant K I . Set S1 = {u ∈ PC(J, ℝ) | ∃ λ ∈ [0, 1] such that u = λ𝕋1 u}. Next, we prove that S1 is bounded in PC(J, ℝ). Consider u ∈ S1 and λ ∈ [0, 1] such that u = λ𝕋1 u. It follows from (3.81) and (3.82) that ‖u‖PC = λ‖𝕋1 u‖PC ≤ λ(‖F1 u‖PC + ‖G11 u‖PC + ‖G12 u‖PC ) q

≤

m max{|a|, |b|}(C I ‖u‖C2 + M I ) |c| + |a + b| |a + b| q T q (C f ‖u‖C1 + M f ) |b| + (1 + . ) |a + b| Γ(q + 1)

(3.83)

Inequality (3.83) together with q1 < 1, q2 < 1 shows that S1 is bounded in PC(J, ℝ). Consequently, by Theorem A.33 we deduce that 𝕋1 has at least one fixed point and the set of the fixed points of 𝕋1 is bounded in PC(J, ℝ). Remark 3.100. (i) If the growth condition [H2] is formulated for q1 = 1, then the conclusions of Theorem 3.99 remain valid provided that (1 +

T q Cf |b| < 1. ) |a + b| Γ(q + 1)

(ii) If the growth condition [H5] is formulated for q2 = 1, then the conclusions of Theorem 3.99 remain valid provided that m max{|a|, |b|}C I < 1. |a + b| (iii) If the growth conditions [H2] and [H5] are formulated for q1 = 1 and q2 = 1, then the conclusions of Theorem 3.99 remain valid provided that (1 +

T q Cf |b| m max{|a|, |b|}C I + < 1. ) |a + b| Γ(q + 1) |a + b|

3.6 Nonlinear impulsive fractional differential equations | 155

(iv) For q1 , q2 ∈ [0, 1), one can also obtain the boundedness of the set S1 by virtue of Lemma 3.75. (v) We do not obtain the boundedness of the set S1 without putting any restriction condition. To end this subsection, we give the following data dependence results. Theorem 3.101. Assume that [H1]–[H5] hold and K I < 1. Let u(⋅) and v(⋅) be solutions of problem (3.64) with boundary conditions au(0) + bu(T) = c and av(0) + bv(T) = c. Then it holds ‖u − v‖PC ≤ [

1

Lf T q (1 − K I )Γ(q + 1)

(1 +

max{|a|, |b|} 1−λ )] . |a| + |b|

Proof. Without loss of generality, for t ∈ (t k , t k+1 ], k = 0, 1, 2, . . . , m, using [H3] and [H4], we obtain |u(t) − v(t)| ≤

max{|a|, |b|} ∑ki=1 K Ii ‖u − v‖C((t k ,t k+1 ],ℝ) |a + b| t

Lf + ∫(t − s)q−1 |u(s) − v(s)|λ ds Γ(q) 0 T

|b|L f + ∫(T − s)q−1 |u(s) − v(s)|λ ds, |a + b|Γ(q) 0

which implies t

Lf (1 − K I )|u(t) − v(t)| ≤ ∫(t − s)q−1 |u(s) − v(s)|λ ds Γ(q) 0 T

|b|L f + ∫(T − s)q−1 |u(s) − v(s)|λ ds. |a + b|Γ(q) 0

By Lemma 3.75, we obtain ‖u − v‖PC ≤ M ∗ , where M ∗ is the only positive solution of the equation M∗ = Note that M∗ = [

L f T q M ∗λ (1 − K I )Γ(q + 1) Lf T q

(1 − K I )Γ(q + 1)

(1 +

max{|a|, |b|} ). |a| + |b| 1

(1 +

max{|a|, |b|} 1−λ )] . |a| + |b|

This completes the proof. Remark 3.102. Under the assumptions of Theorem 3.101, we do not obtain the uniqueness of the solutions.

156 | 3 Fractional Differential Equations

3.6.6 Applications In this subsection, we present two examples to indicate how our theorems can be applied to concrete problems. Example 3.103. Let us consider the Cauchy problem for nonlinear impulsive fractional differential equations 1

|u(t)| 2 { c 23 { D u(t) = , { 1 { { (1 + 99et )(1 + |u(t)| 2 ) { { { { − |u( 12 )| 1 { { ∆u( = ) { − 1 , { 2 { 100(1 + |u( 12 )| 2 ) { { { { u(0) = 0, {

1 t ∈ (0, 1] \ { }, 2 (3.84)

and the boundary value problem for nonlinear impulsive fractional differential equations 1 |u(t)| 2 1 { c 23 { u(t) = , t ∈ (0, 1] \ { }, D { 1 { t )(1 + |u(t)| 2 ) 2 { (1 + 99e { { { { − (3.85) |u( 12 )| 1 { { , { ∆u( ) = 1 − { 2 { 100(1 + |u( 12 )| 2 ) { { { { { 99u(0) = −u(1). Set

1

f(t, u) =

u2 1

(1 + 99et )(1 + u 2 )

,

(t, u) ∈ [0, 1] × [0, +∞).

Let u1 , u2 ∈ [0, ∞) and t ∈ [0, 1]. Then we have 1 󵄨󵄨 u 12 u22 1 󵄨󵄨 1 󵄨󵄨 |f(t, u1 ) − f(t, u2 )| = − 1 1 (1 + 99et ) 󵄨󵄨󵄨 1 + u12 1 + u22 1

=

󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨

1

|u12 − u22 | 1 2

1 2

(1 + 99et )(1 + u1 )(1 + u2 )

≤

1 1 |u1 − u2 | 2 . 100

Obviously, for all u ∈ [0, ∞) and each t ∈ [0, 1], |f(t, u)| = Set

1 󵄨󵄨 1 󵄨󵄨 u 2 󵄨 󵄨 1 + 99et 󵄨󵄨󵄨 1 + u 12

󵄨󵄨 1 1 󵄨󵄨 2 󵄨󵄨 ≤ 󵄨󵄨 100 |u| . 󵄨 −

I1 (u(

|u( 12 )| 1− )) = − 1 . 2 100(1 + |u( 21 )| 2 )

3.6 Nonlinear impulsive fractional differential equations | 157

Also, 1− 1 − 󵄨󵄨 − − 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨I (u ( 1 )) − I (u ( 1 ))󵄨󵄨󵄨 ≤ 󵄨󵄨u1 ( 2 ) − u2 ( 2 )󵄨󵄨 , 2 1 󵄨󵄨 1 1 󵄨󵄨 2 2 100 󵄨 󵄨 1− 2 − 󵄨󵄨 󵄨 󵄨󵄨I (u( 1 ))󵄨󵄨󵄨 ≤ |u( 2 )| . 1 󵄨󵄨 󵄨󵄨 2 100 󵄨 󵄨 1

Denote q = 23 , q1 = λ = 12 , q2 = 21 , m = 1, L f = C f = a = 99, b = 1, c = 0.

1 100 ,

KI = CI =

1 100 ,

M I = M f = 0,

Remark 3.104. (i) It is not difficult to see that all the assumptions of Theorem 3.82 are satisfied. Thus, problem (3.84) has a solution in PC([0, 1], ℝ). (ii) It is obvious that all the assumptions of Theorem 3.99 are satisfied. Thus, problem (3.85) has a solution in PC([0, 1], ℝ). Example 3.105. We consider the impulsive fractional differential equation 1 c q D t x(t) = 0, t ∈ (0, 1] \ { }, { { { 2 { − 1 1 { |x( 2 )| 2 1 { { { ∆x( ) = − 1 , 2 1 + |x( 1 )| 2 {

(3.86)

2

and the inequalities q

|c D t y(t)| ≤ ϵ, { { { { − 1 󵄨󵄨 { |y( 12 )| 2 󵄨󵄨󵄨󵄨 1 󵄨󵄨 { { 󵄨 {󵄨󵄨∆y( ) − − 1 󵄨 ≤ ϵ, 󵄨 2 1 + |y( 12 )| 2 󵄨󵄨󵄨 {󵄨󵄨

1 t ∈ (0, 1] \ { }, 2 (3.87) ϵ > 0.

Let y ∈ PC([0, 1], ℝ) be a solution of inequality (3.87). Then there exist a function g ∈ PC([0, 1], ℝ) and g1 ∈ ℝ such that |g(t)| ≤ ϵ, c

t ∈ [0, 1],

|g1 | ≤ ϵ, 1 t ∈ [0, 1] \ { }, 2

q

D t y(t) = g(t),

−

1

|y( 12 )| 2 1 ∆y( ) = − 1 + g1 . 2 1 + |y( 12 )| 2

(3.89)

Integrating (3.88) from 0 to t by virtue of Lemma A.16 via (3.89), we have −

(3.88)

1

|y( 12 )| 2

t

1 y(t) = y(0) + χ( 21 ,1] (t)( + g1 ) + ∫(t − s)q−1 g(s)ds 1 − 12 Γ(q) 1 + |y( 2 )| 0 for the characteristic function χ( 21 ,1] (t) of ( 12 , 1].

158 | 3 Fractional Differential Equations

Let us take the unique solution x(t) of (3.86) given by 1

x(t) = y(0) + χ( 12 ,1] (t)

|y(0)| 2 1

1 + |y(0)| 2

.

Then we have t

− 1 󵄨󵄨 󵄨󵄨 |y( 12 )| 2 1 |y(0)| 2 󵄨 󵄨 1 |y(t) − x(t)| = 󵄨󵄨χ( 2 ,1] (t)( − + g1 ) + ∫(t − s)q−1 g(s)ds󵄨󵄨󵄨󵄨 1 1 − Γ(q) 󵄨 󵄨 1 + |y( 12 )| 2 1 + |y(0)| 2 0 1

t

1 󵄨󵄨 1 − 1 1 − 󵄨󵄨 2 ≤ χ( 21 ,1] (t)󵄨󵄨󵄨󵄨y( ) − x( )󵄨󵄨󵄨󵄨 + |g1 | + ∫(t − s)q−1 |g(s)|ds 2 󵄨 Γ(q) 󵄨 2

0

t

󵄨󵄨 1 ϵ 1 󵄨󵄨 ≤ χ( 21 ,1] (t)󵄨󵄨󵄨󵄨y( ) − x( )󵄨󵄨󵄨󵄨 + ϵ + ∫(t − s)q−1 ds 2 󵄨 Γ(q) 󵄨 2 −

−

1 2

0

1 󵄨󵄨 1 − 1 − 󵄨󵄨 2 ϵ ≤ χ( 21 ,1] (t)󵄨󵄨󵄨󵄨y( ) − x( )󵄨󵄨󵄨󵄨 + ϵ + , 2 󵄨 Γ(q + 1) 󵄨 2

t ∈ [0, 1],

which gives |y(t) − x(t)| ≤ (1 +

1 1 √ϵ, )ϵ + √ 1 + Γ(q + 1) Γ(q + 1)

t ∈ [0, 1].

So, equation (3.86) is generalized Ulam–Hyers stable.

3.7 Fractional differential switched systems with coupled nonlocal initial and impulsive conditions 3.7.1 Introduction There are some works on existence results for initial problems for first-order nonlinear differential systems with different nonlocal conditions [57, 208–210, 232]. Impulsive switched systems [139, 180, 258, 272], one kind of hybrid systems, have been at the center of increasing attention in recent years due to their wide applications in modern systems. In general, impulsive switched systems consist of several modes of operation that undergo abrupt changes, meanwhile, a switching rule orchestrates the switching between them. Moreover, impulsive switched models have been used extensively to describe systems in a wide range of applications, including information science, electronics and automatic control systems. It should be noted that, although the classical impulsive switched systems have received many recent studies in the control engineering and mathematics literature [131, 186, 325, 331, 332], there are only a few works dealing with the basic theory of fractional-order impulsive switched

3.7 Fractional differential switched systems | 159

systems. Such systems are a special class of hybrid systems. It seems that there are a few papers discussing the existence and stability results in this new area. The aim of this section is to investigate a nonlocal impulsive fractional-order differential switched system of the form q

c D t ,t u(t) = f(t, u(t)), t ∈ (t i , t i+1 ), i = 0, 1, . . . , m, { { { i u(0) = α0 (u), { { { u(t+i ) = α i (u), i = 1, 2, . . . , m, {

(3.90)

q

where q ∈ (0, 1), c D t i ,t u(t) is the generalized Caputo fractional derivative with lower limit at t i , 0 = t0 < t1 < ⋅ ⋅ ⋅ < t m < t m+1 = 1 is a given sequence, f : J × ℝn → ℝn and nonlocal impulsive terms α i : PC(J, ℝn ) → ℝn , i = 0, 1, . . . , m. This section is based on [289].

3.7.2 Preliminaries The first convention is to identify the elements u = (u1 , u2 , . . . , u n ) of a space ℝn with column matrices. Next, the scalar norm of a normed space ℝn is denoted by | ⋅ |, while for an element u ∈ ℝn , u = (u1 , u2 , . . . , u n ), we use the symbol ‖⋅‖ to denote its vectorvalued norm ‖u‖ = maxi=1,2,...,n |u i | on the set ℝn . We suppose that α i ∈ L(PC(J, ℝn ), ℝn ), i = 0, 1, . . . , m, when PC(J, ℝn ) is endowed with the Chebyshev PC-norm ‖u‖∞ , i.e., we have |α i (u)| ≤ ‖α i ‖ ‖u‖∞ ,

u ∈ PC(J, ℝn ).

If we denote t

u f i,q (t)

1 := ∫(t − s)q−1 f(s, u(s))ds, Γ(q)

t ∈ (t i , t i+1 ], i = 0, 1, . . . , m,

ti

then a general solution of (3.90) on each (t i , t i+1 ), i = 0, . . . , m is u u(t) = c i + f i,q (t)

for constant vectors c i ∈ ℝn . Introducing the characteristic functions χ i (t) = 1 if t ∈ (t i , t i+1 ] and χ i (t) = 0 if t ∈ J \ (t i , t i+1 ], and setting u f qu (t) := f i,q (t),

we can write

t ∈ (t i , t i+1 ], i = 0, 1, . . . , m, m

u(t) = ∑ χ i (t)c i + f qu (t), i=0

t ∈ J,

(3.91)

160 | 3 Fractional Differential Equations where χ0 is extended to t0 = 0 as χ0 (t0 ) = 1. Thus setting n

c i = ∑ c ik e k k=1

for the standard basis {e k }nk=1 of ℝn , we get m

n

u(t+i ) = c i + f qu (t i ) = α i ( ∑ ∑ χ j c jk e k + f qu ) j=0 k=1 m

n

m

= ∑ ∑ α i (χ j e k )c jk + α i (f qu ) = ∑ A ij c j + α i (f qu ), j=0 k=1

where

j=0

n

A ij c j := ∑ α i (χ j e k )c jk k=1

ℝmn .

for matrices A ij ∈ Denote by M(k) the set of all square matrices of order k with elements in ℝ. Setting (m+1)n , 𝔸 = {A ij }m {c = (c0 , . . . , c m ) ∈ ℝ i,j=0 ∈ M((m + 1)n), { u u u u b (f ) = (α0 (f q ) − f q (t0 ), . . . , α m (f q ) − f qu (t m )) ∈ ℝ(m+1)n , { q

(3.92)

we get (𝕀 − 𝔸)c = b q (f u ),

(3.93)

which has a solution c = (𝕀 − 𝔸)−1 b q (f u ) if det(𝕀 − 𝔸) ≠ 0.

(3.94)

Hence by (3.91) and (3.94), we derive the solution m

u(t) = ∑ χ i (t)[(𝕀 − 𝔸)−1 b q (f u )]i + f qu (t),

t ∈ J,

(3.95)

i=0

where (𝕀 − 𝔸)−1 b q (f u ) = ([(𝕀 − 𝔸)−1 b q (f u )]0 , . . . , [(𝕀 − 𝔸)−1 b q (f u )]m ).

3.7.3 Existence and uniqueness result via Banach fixed point theorem Throughout this subsection, we consider system (3.90) under the following three assumptions:

3.7 Fractional differential switched systems | 161

[A0] det(𝕀 − 𝔸) ≠ 0. [A1] f ∈ C(J × ℝn , ℝn ) and there exists a constant L f > 0 such that |f(t, u) − f(t, v)| ≤ L f |u − v| for all t ∈ [0, 1] and u, v ∈ ℝn . [A2] α i ∈ L(PC(J, ℝn ), ℝn ), i = 0, 1, . . . , m. Theorem 3.106. Assume that [A0]–[A2] are satisfied. If ρ := ρ1 L f < 1, { { { { { {ρ1 := [‖(𝕀 − 𝔸)−1 ‖ max (‖α i ‖ + 1)(t i+1 − t i )q i=0,1,...,m { { { { 1 { { , + max (t i+1 − t i )q ] Γ(q + 1) i=0,1,...,m {

(3.96)

then system (3.90) has a unique solution. Proof. We transform equation (3.90) into a fixed point problem by considering the mapping T : PC(J, ℝn ) → PC(J, ℝn ) given by (Tu)(t) = (Pu)(t) + (Qu)(t), where

t ∈ J,

(3.97)

m

(Pu)(t) = ∑ χ i (t)[(𝕀 − 𝔸)−1 b q (f u )]i ,

t∈J

(3.98)

i=0

and (Qu)(t) = f qu (t),

t ∈ J.

(3.99)

Obviously, T is well defined, since P and Q are well defined due to our assumptions. In what follows, we check that T is contractive. On the one hand, for any u, v ∈ PC(J, ℝn ) and t ∈ (t i , t i+1 ], we have 󵄨 󵄨 |(Pu)(t) − (Pv)(t)| = 󵄨󵄨󵄨[(𝕀 − 𝔸)−1 (b q (f u ) − b q (f v ))]i 󵄨󵄨󵄨 󵄨 󵄨 ≤ ‖(𝕀 − 𝔸)−1 ‖ 󵄨󵄨󵄨(b q (f u ) − b q (f v ))i 󵄨󵄨󵄨 󵄩 󵄩 ≤ ‖(𝕀 − 𝔸)−1 ‖ 󵄩󵄩󵄩b q (f u ) − b q (f v )󵄩󵄩󵄩,

(3.100)

where we consider the norm ‖c‖ = maxi=0,1,...,m |c i | on the set ℝ(m+1)n . Note that 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨[b q (f u ) − b q (f v )]i 󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨α i (f qu ) − α i (f qv )󵄨󵄨󵄨 + 󵄨󵄨󵄨f qu (t i ) − f qv (t i )󵄨󵄨󵄨 ≤ (‖α i ‖ + 1)‖f qu − f qv ‖∞ . Meanwhile, for any t ∈ J i := (t i , t i+1 ], we have t

1 󵄨󵄨 u 󵄨 󵄨 󵄨 ∫(t − s)q−1 󵄨󵄨󵄨f(s, u(s)) − f(s, v(s))󵄨󵄨󵄨ds 󵄨󵄨f q (t) − f qv (t)󵄨󵄨󵄨 ≤ Γ(q) ti

(3.101)

162 | 3 Fractional Differential Equations t

≤

1 ∫(t − s)q−1 L f ds‖u − v‖∞ Γ(q) ti

L f (t i+1 − t i )q ≤ ‖u − v‖∞ . Γ(q + 1)

(3.102)

Linking (3.101) and (3.102), we have (‖α i ‖ + 1)(t i+1 − t i )q L f 󵄩 󵄩󵄩 ‖u − v‖∞ . 󵄩󵄩b q (f u ) − b q (f v )󵄩󵄩󵄩 ≤ max Γ(q + 1) i=0,1,...,m

(3.103)

Now linking (3.100) and (3.103), we have (‖α i ‖ + 1)(t i+1 − t i )q L f 󵄨󵄨 󵄨 ‖u − v‖∞ 󵄨󵄨(Pu)(t) − (Pv)(t)󵄨󵄨󵄨 ≤ ‖(𝕀 − 𝔸)−1 ‖ max Γ(q + 1) i=0,1,...,m for any t ∈ J. Consequently, we arrive at ‖Pu − Pv‖∞ ≤ ‖(𝕀 − 𝔸)−1 ‖

max

i=0,1,...,m

(‖α i ‖ + 1)(t i+1 − t i )q L f ‖u − v‖∞ . Γ(q + 1)

(3.104)

On the other hand, from (3.102) we derive ‖Qu − Qv‖∞ = ‖f qu − f qv ‖∞ ≤

maxi=0,1,...,m (t i+1 − t i )q L f ‖u − v‖∞ Γ(q + 1)

(3.105)

for any u, v ∈ PC(J, ℝn ). In summary, it follows by (3.104) and (3.105) that ‖Tu − Tv‖∞ ≤ ρ‖u − v‖∞ . By (3.96), T defined in (3.97) is contractive. So one can use the Banach fixed point theorem to derive the existence and uniqueness result.

3.7.4 Existence result via Krasnoselskii fixed point theorem We suppose that [A3] f ∈ C(J × ℝn , ℝn ) and there is a constant m f ≥ 0 such that |f(t, u)| ≤ m f

for all t ∈ J, u ∈ ℝn .

Theorem 3.107. Assume that [A0]–[A3] are satisfied. Then system (3.90) has at least one solution. Proof. Let us set B r = {u ∈ PC(J, ℝn ) | ‖u‖PCB ≤ r}, where r := ρ1 m f

(3.106)

3.7 Fractional differential switched systems | 163

and ρ1 is given in (3.96). Now we reconsider the operators P and Q defined in (3.98) and (3.99) on B r , respectively. For the sake of convenience, we divide the proof into several steps. Step 1. Pu + Qu ∈ B r for any u ∈ B r . Like for (3.104) and (3.105), we derive (‖α i ‖ + 1)(t i+1 − t i )q m f , Γ(q + 1) i=0,1,...,m (t i+1 − t i )q m f , Γ(q + 1)

‖Pu‖∞ ≤ ‖(𝕀 − 𝔸)−1 ‖ ‖Qu‖∞ ≤

max

i=0,1,...,m

max

which implies ‖Tu‖PCB ≤ ‖Pu‖PCB + ‖Qu‖PCB ≤ r. Thus, we obtain that Px + Qx ∈ B r for any x ∈ B r and λ > 0. Step 2. Q is a contraction on B r for λ > 0 sufficiently large. For any t ∈ J i , we have t

1 󵄨 󵄨 󵄨󵄨 u 󵄨 ∫(t − s)q−1 󵄨󵄨󵄨f(s, u(s)) − f(s, v(s))󵄨󵄨󵄨ds 󵄨󵄨f q (t) − f qv (t)󵄨󵄨󵄨 ≤ Γ(q) ti

t

Lf 1 ≤ ∫(t − s)q−1 L f eλs ds‖u − v‖PCB ≤ q eλt ‖u − v‖PCB , Γ(q) λ

(3.107)

0

where we used the fact

t

1 eλt ∫(t − s)q−1 eλs ds < q . Γ(q) λ 0

From (3.107) we derive that ‖Qu − Qv‖PCB = ‖f qu − f qv ‖PCB ≤

Lf ‖u − v‖PCB λq

for any u, v ∈ PC(J, ℝn ). Obviously Q is a contraction on B r for any sufficient large λ. Step 3. P is continuous. Let {u n } be a sequence of B r such that u n → u in PC(J, ℝn ). Like for (3.104), we derive ‖Pu − Pu n ‖∞ ≤ ‖(𝕀 − 𝔸)−1 ‖

max

i=0,1,...,m

(‖α i ‖ + 1)(t i+1 − t i )q 󵄩󵄩 󵄩 󵄩󵄩f(⋅, u(⋅)) − f(⋅, u n (⋅))󵄩󵄩󵄩∞ , Γ(q + 1)

from which we infer that ‖Pu n − Pu‖PCB → 0 as n → ∞. Step 4. P is compact. First, P is uniformly bounded on B r due to Step 1. Next, for any t i ≤ τ1 < τ2 ≤ t i+1 and any u ∈ B r , one has |(Pu)(τ2 ) − (Pu)(τ1 )| = 0,

164 | 3 Fractional Differential Equations

so P is equicontinuous. As a consequence of Steps 3–4 together with the Arzelà–Ascoli theorem, we can conclude that Q : B r → B r is continuous and completely continuous. By using the Krasnoselskii fixed point theorem, T = P + Q has a fixed point, which gives at least one solution.

3.7.5 Existence result via Leray–Schauder fixed point theorem We suppose that [A4] f ∈ C(J × ℝn , ℝn ) and there are constants M f ≥ 0 and m f ≥ 0 such that |f(t, u)| ≤ m f + M f |u|

for all t ∈ J, u ∈ ℝn .

Theorem 3.108. Assume that [A0], [A2] and [A4] are satisfied. If aM f Eq (M f ) < 1, where a := ‖(𝕀 − 𝔸)−1 ‖

max

i=0,1,...,m

and

∞

Eq (z) := ∑ k=0

(‖α i ‖ + 1)(t i+1 − t i )q , Γ(q + 1)

(3.108)

(3.109)

zk Γ(kq + 1)

is the Mittag-Leffler function, then system (3.90) has at least one solution. Proof. Now we consider T : PC(J, ℝn ) → PC(J, ℝn ) given in (3.97). Repeating the same process as in Steps 3–4 of the proof of Theorem 3.107, we can verify that P : PC(J, ℝn ) → PC(J, ℝn ) is continuous and completely continuous. Now we show that Q is continuous. Let {u n } be a sequence of B r such that u n → u in PC(J, ℝn ). For any t ∈ (t i , t i+1 ], t

1 󵄨󵄨 󵄨 󵄩 󵄩 ∫(t − s)q−1 eλs ds󵄩󵄩󵄩f(⋅, u n (⋅)) − f(⋅, u(⋅))󵄩󵄩󵄩PCB 󵄨󵄨(Qu n )(t) − (Qu)(t)󵄨󵄨󵄨 ≤ Γ(q) ti

eλt 󵄩 󵄩 ≤ q 󵄩󵄩󵄩f(⋅, u n (⋅)) − f(⋅, u(⋅))󵄩󵄩󵄩PCB , λ which implies that ‖Qu n − Qu‖PCB ≤

1 󵄩󵄩 󵄩 󵄩f(⋅, u n (⋅)) − f(⋅, u(⋅))󵄩󵄩󵄩PCB . λq 󵄩

Thus, we infer that ‖Qu n − Qu‖PCB → 0 as n → ∞. Next, we prove that Q is compact. First, Q is uniformly bounded on bounded sets due to Step 1 of the proof of Theorem 3.107. Now, we show that Q maps a bounded set

3.7 Fractional differential switched systems | 165

into an equicontinuous set. For any t ∈ (t i , t i+1 ], t i ≤ τ1 < τ2 ≤ t i+1 and any ‖u‖∞ ≤ r,̄ one has 󵄨󵄨 󵄨 󵄨󵄨(Qu)(τ2 ) − (Qu)(τ1 )󵄨󵄨󵄨 ≤

τ2

τ1

τ1

ti

m f + M f r̄ m f + M f r̄ ∫(τ2 − s)q−1 ds + ∫[(τ1 − s)q−1 − (τ2 − s)q−1 ]ds Γ(q) Γ(q)

m f + M f r̄ ≤ (2(τ2 − τ1 )q + (τ1 − t i )q − (τ2 − t i )q ), Γ(q + 1) which implies |(Qu)(τ2 ) − (Qu)(τ1 )| → 0 as τ2 → τ1 uniformly with respect to any ‖u‖∞ ≤ r.̄ So Q is equicontinuous. Therefore, T is continuous and completely continuous. In what follows, it remains to give a priori bounds, i.e., to show that the set ∆(T) := {u ∈ PC(J, ℝn ) | u = σT(u), σ ∈ [0, 1]} is a bounded subset of PC(J, ℝn ). Taking into account (3.104) and the linear growth condition, we derive t

1 |u(t)| ≤ a(m f + M f ‖u‖∞ ) + ∫(t − s)q−1 (m f + M f |u(s)|)ds Γ(q) ti

t

≤ (a +

Mf maxi=0,1,...,m (t i+1 − t i )q )m f + aM f ‖u‖∞ + ∫(t − s)q−1 |u(s)|ds Γ(q + 1) Γ(q) ti

t

≤ r + aM f ‖u‖∞ +

Mf ∫(t − s)q−1 |u(s)|ds Γ(q) 0

for t ∈ J i , where r is given by (3.106). Using singular Gronwall’s inequality [328, Collorary 2], we get |u(t)| ≤ (r + aM f ‖u‖∞ )Eq (M f ), which gives ‖u‖∞ ≤ (r + aM f ‖u‖∞ )Eq (M f ). So by (3.108), we have ‖u‖∞ ≤ M ∗ :=

rEq (M f ) . 1 − aM f Eq (M f )

Of course, this implies ‖u‖PCB ≤ M ∗ . This shows that the set ∆(T) is bounded. As a consequence of the Leray–Schauder fixed point theorem, we deduce that T has a fixed point. Certainly, Theorem 3.108 is stronger than Theorem 3.107, but a location of the solution in Theorem 3.107 is more precise than in Theorem 3.108: ‖u‖∞ ≤ r in Theorem 3.107, while ‖u‖∞ ≤ M ∗ in Theorem 3.108 and M ∗ > r. Next, the function 𝔼q (r) := rEq (r) is −1 increasing on ℝ+ , so (3.108) is equivalent to M f < 𝔼q ( 1a ).

166 | 3 Fractional Differential Equations

3.7.6 Existence result for the resonant case: Landesman–Lazer conditions In this subsection, we consider system (3.90) under the assumptions [A2], [A3] and [A5] dim ker(𝕀 − 𝔸) = codim im(𝕀 − 𝔸) = 1. So we consider the simplest case when [A0] does not hold. Then (3.93) has a solution c = r c̄ + (𝕀 − 𝔸)−1 Pb q (f u ), r ∈ ℝ whenever it holds (𝕀 − P)b q (f u ) = 0,

(3.110)

where ker(𝕀 − 𝔸) is spanned by c̄ = (c̄ 0 , . . . , c̄ m ), P : → is a projection −1 (m+1)n onto im(𝕀 − 𝔸) and (𝕀 − 𝔸) : im 𝔸 → ℝ is a right inverse of 𝕀 − 𝔸. Thus by (3.91), the solution of (3.90) is given by ℝ(m+1)n

̄ + F(u)(t), u(t) = r χ(t)

ℝ(m+1)n

t∈J

for m

̄ := ∑ c̄ i χ i (t), χ(t)

t ∈ J,

i=0 m

F(u)(t) := ∑ χ i (t)[(𝕀 − 𝔸)−1 Pb q (f u )]i + f qu (t),

t ∈ J,

i=0

where (𝕀 − 𝔸)−1 Pb q (f u ) = ([(𝕀 − 𝔸)−1 Pb q (f u )]0 , . . . , [(𝕀 − 𝔸)−1 Pb q (f u )]m ). Taking 0 ≠ ψ ∈ ker(𝕀 − 𝔸)∗ , (3.110) has the form B(u) := ψ∗ b q (f u ) = 0. Summarizing, we obtain that the solvability of (3.90) is reduced to u = r χ̄ + F(u),

B(r χ̄ + F(u)) = 0.

(3.111)

Now we plug (3.111) into the homotopy u = λ(r χ̄ + F(u)),

B(r χ̄ + λF(u)) = 0,

λ ∈ [0, 1].

(3.112)

Like in the proof of Theorem 3.107, there is a constant Θ > 0 so that ‖F(u)‖∞ ≤ Θ

for all u ∈ PC(J, ℝn ).

(3.113)

Moreover, we also know that F is compact. Next we derive m

̄ r χ+λF(u)

̄ B(r χ̄ + λF(u)) = ψ∗ b q (f r χ+λF(u) ) = ∑ ψ∗i (α i (f q i=0

̄ r χ+λF(u)

) − fq

(t i ))

3.7 Fractional differential switched systems | 167

with ψ = (ψ0 , . . . , ψ m ). Furthermore, for t ∈ (t i , t i+1 ], i = 0, 1, . . . , m we have t

̄ r χ+λF(u)

fq

̄ r χ+λF(u)

(t) = f i,q

(t) =

1 ∫(t − s)q−1 f(s, r c̄ i + λF(u)(s))ds. Γ(q)

(3.114)

ti

When c̄ i = 0, then (3.114) and [A3] give t

(t i+1 − t i )q m f 1 ̄ 󵄨 󵄨󵄨 r χ+λF(u) (t)󵄨󵄨󵄨 = ∫(t − s)q−1 |f(s, λF(u)(s))|ds ≤ 󵄨󵄨f q Γ(q) Γ(q + 1)

(3.115)

ti

for t ∈ (t i , t i+1 ], i = 0, 1, . . . , m. To proceed, we suppose the following: [A6] There is a function f∞ ∈ C(J × S1 , ℝn ) such that lim f(t, rw) = f∞ (t, w)

r→∞

uniformly with respect to w ∈ S1 := {w ∈ ℝn | |w| = 1} and t ∈ J. Certainly, [A6] implies [A3], so we consider mf =

sup (t,x)∈J×ℝn

|f(t, x)|

(3.116)

in this subsection. Setting t

∞ f i,q (t,

1 w) := ∫(t − s)q−1 f∞ (s, w)ds Γ(q) ti

for t ∈ (t i , t i+1 ], i = 0, 1, . . . , m, when r c̄ i ≠ 0, we deduce from (3.114) that 󵄨󵄨 r χ+λF(u) c̄ i 󵄨󵄨󵄨 ∞ 󵄨󵄨f ̄ (t) − f i,q (t, sgn r )󵄨󵄨 󵄨󵄨 q |c̄ i | 󵄨󵄨 󵄨 t

≤

󵄨󵄨 1 c̄ i 󵄨󵄨󵄨 )󵄨󵄨ds ∫(t − s)q−1 󵄨󵄨󵄨󵄨f(s, r c̄ i + λF(u)(s)) − f∞ (s, sgn r Γ(q) |c̄ i | 󵄨󵄨 󵄨

(3.117)

ti

for t ∈ (t i , t i+1 ], i = 0, 1, . . . , m. Since by (3.113), lim |r c̄ i + λF(u)(s)| = ∞,

r→±∞

lim

r→±∞

c̄ i r c̄ i + λF(u)(s) =± |r c̄ i + λF(u)(s)| |c̄ i |

uniformly with respect to s ∈ J and u ∈ PC(J, ℝn ), from (3.117) we obtain 󵄩󵄩 r χ+λF(u) c̄ i 󵄩󵄩󵄩 ̄ ∞ lim 󵄩󵄩󵄩󵄩f q − f i,q )󵄩󵄩 = 0 (⋅, ± r→±∞󵄩 |c̄ i | 󵄩󵄩∞ uniformly with respect u ∈ PC(J, ℝn ) and λ ∈ [0, 1]. Setting ∞ f±,q (t)

c̄ i ∞ { (t, ± ) {f i,q |c̄ i | := { { {0

if t ∈ (t i , t i+1 ], c i ≠ 0, if t ∈ (t i , t i+1 ], c i = 0

(3.118)

168 | 3 Fractional Differential Equations for i = 0, 1, . . . , m, from (3.115) and (3.118) we derive 󵄩 󵄩󵄩 mf ̄ r χ+λF(u) ∞ 󵄩 lim sup󵄩󵄩󵄩b q (f q ) − b q (f±,q )󵄩󵄩󵄩 ≤ 󵄩 Γ(q + 1) r→±∞ 󵄩

max

c i =0,i=0,...,m

(‖α i ‖ + 1)(t i+1 − t i )q

uniformly with respect to u ∈ PC(J, ℝn ) and λ ∈ [0, 1]. So we arrive at m f ‖ψ‖ 󵄨 󵄨 lim sup󵄨󵄨󵄨B(r χ̄ + λF(u)) − B± 󵄨󵄨󵄨 ≤ Γ(q + 1) r→±∞

max

c i =0,i=0,...,m

(‖α i ‖ + 1)(t i+1 − t i )q

uniformly with respect to u ∈ PC(J, ℝn ) and λ ∈ [0, 1], where m ∞ B± := ∑ ψ∗i [b q (f±,q )]i ∈ ℝ.

(3.119)

i=0

Now we can prove the following result. Theorem 3.109. Suppose [A2], [A5] and [A6]. If m f ‖ψ‖ Γ(q + 1)

max

c i =0,i=0,...,m

(‖α i ‖ + 1)(t i+1 − t i )q < min{|B− |, |B+ |}

(3.120)

and B− B+ < 0,

(3.121)

where B± , m f are given by (3.119), (3.116), respectively, and 0 ≠ ψ ∈ ker(𝕀 − 𝔸)∗ . Then (3.90) has a solution. Proof. By (3.120), it holds B(r χ̄ + λF(u))Bsgn r > 0

(3.122)

for any |r| > 0 large. So we fix such ±r0 , r0 > 0. Then we consider (3.112) on the set Ω := {(u, r) ∈ PC(J, ℝn ) × ℝ | ‖u‖∞ ≤ r0 ‖c‖̄ + Θ + 1, |r| ≤ r0 }. If (u, r) ∈ ∂Ω is a solution of (3.112), then either ‖u‖∞ = r0 ‖c‖̄ + Θ + 1, and the first equation of (3.112) implies that r0 ‖c‖̄ + Θ + 1 = ‖u‖∞ ≤ r0 ‖c‖̄ + Θ, which is a contradiction; or r = ±r0 , and the second equation of (3.112) gives a contradiction to (3.122). Hence (3.112) has no solution on the boundary ∂Ω of Ω for any λ ∈ [0, 1]. Hence for the Schauder topological degree we get ̄ 0) deg(Ω, (u − r χ̄ − F(u), B(r χ̄ + F(u))), 0) = deg(Ω, (u, B(r χ)), ̄ 0) = ±1, = deg((−r0 , r0 ), B(r χ), where we used (3.122). Remark 3.110. If c̄ i ≠ 0 for any i = 0, . . . , m, then we just need (3.121), which is a Landesman–Lazer condition [52] for this resonant problem.

3.7 Fractional differential switched systems | 169

3.7.7 Ulam type stability results In this subsection, we introduce the concept of Ulam type stability for equation (3.90). Set PC(J, ℝ+ ) := {x ∈ PC(J, ℝ) | x(t) ≥ 0} and t

PC(q) (J, ℝn ) = {x ∈ PC(J, ℝn ) | ∫ ti

x(s) − x(t i ) ds ∈ C1 ((t i , t i+1 ), ℝn ), i = 0, 1, . . . , m}, (t − s)q

see [162, pp. 140–141, (3.1.32)–(3.1.34)]. Let ε > 0, ψ ≥ 0 and φ ∈ PC(J, ℝ+ ). We consider the inequalities q

|c D t i ,t y(t) − f(t, y(t))| ≤ ε, { { { |y(0) − α0 (y)| ≤ ε, { { { |y(t+i ) − α i (y)| ≤ ε, {

t ∈ (t i , t i+1 ), i = 0, 1, . . . , m, (3.123) i = 1, 2, . . . , m,

and q

|c D t i ,t y(t) − f(t, y(t))| ≤ φ(t), t ∈ (t i , t i+1 ), i = 0, 1, . . . , m, { { { |y(0) − α0 (y)| ≤ ψ, { { { |y(t+i ) − α i (y)| ≤ ψ, i = 1, 2, . . . , m, {

(3.124)

and q

|c D t i ,t y(t) − f(t, y(t))| ≤ εφ(t), t ∈ (t i , t i+1 ), i = 0, 1, . . . , m, { { { |y(0) − α0 (y)| ≤ εψ, { { { |y(t+i ) − α i (y)| ≤ εψ, i = 1, 2, . . . , m. {

(3.125)

Definition 3.111. Equation (3.90) is Ulam–Hyers stable if there exists a real number c f,m > 0 such that for each ε > 0 and for each solution y ∈ PC(q) (J, ℝn ) of inequality (3.123) there exists a solution u ∈ PC(q) (J, ℝn ) of equation (3.90) with ‖y − u‖∞ ≤ c f,m ε. Definition 3.112. Equation (3.90) is generalized Ulam–Hyers stable if there exists θ f,m ∈ C(ℝ+ , ℝ+ ), θ f,m (0) = 0 such that for each solution y ∈ PC(q) (J, ℝn ) of inequality (3.123) there exists a solution u ∈ PC(q) (J, ℝn ) of equation (3.90) with ‖y − u‖∞ ≤ θ f,m (ε). The following definitions will extend the original (generalized) Ulam–Hyers–Rassias stability concepts for the equations without impulses to equations with impulses.

170 | 3 Fractional Differential Equations Definition 3.113. Equation (3.90) is Ulam–Hyers–Rassias stable with respect to (φ, ψ) if there exists c f,m,φ > 0 such that for each ε > 0 and for each solution y ∈ PC(q) (J, ℝn ) of inequality (3.125) there exists a solution u ∈ PC(q) (J, ℝn ) of equation (3.90) with ‖y − u‖∞ ≤ c f,m,φ ε(‖φ‖∞ + ψ). Definition 3.114. Equation (3.90) is generalized Ulam–Hyers–Rassias stable with respect to (φ, ψ) if there exists c f,m,φ > 0 such that for each solution y ∈ PC(q) (J, ℝn ) of inequality (3.124) there exists a solution u ∈ PC(q) (J, ℝn ) of equation (3.90) with ‖y − u‖∞ ≤ c f,m,φ (‖φ‖∞ + ψ),

t ∈ J.

Remark 3.115. It follows directly from inequality (3.125) that a function y ∈ PC(q) (J, ℝn ) is a solution of inequality (3.125) if and only if there are g ∈ C(J 󸀠 , ℝn ), ψ ≥ 0, J 󸀠 := J \ {t1 , . . . , t m } and a sequence h i , i = 0, 1, 2, . . . , m (which depend on y) such that (i) |g(t)| ≤ εφ(t) and |h i | ≤ εψ, t ∈ J 󸀠 , i = 0, 1, 2, . . . , m; q (ii) c D t i ,t y(t) − f(t, y(t)) = g(t), t ∈ (t i , t i+1 ), i = 1, 2, . . . , m; (iii) y(0) − α0 (y) = h0 ; (iv) y(t+i ) − α i (y) = h i , i = 1, 2, . . . , m. Remark 3.116. If y ∈ PC(q) (J, ℝn ) is a solution of inequality (3.125), then y satisfies m 󵄩󵄩 󵄩 󵄩󵄩y − ∑ χ [(𝕀 − 𝔸)−1 b (f y )] − f y 󵄩󵄩󵄩 ≤ ε[ρ ‖φ‖ + ‖(𝕀 − 𝔸)−1 ‖ψ], i q 1 ∞ 󵄩󵄩 q󵄩 i 󵄩󵄩∞ 󵄩

(3.126)

i=0

where b q (f y ) is defined by (3.92) and ρ1 by (3.96). Proof. It follows from Remark 3.115 that we have q

c D t ,t y(t) = f(t, y(t)) + g(t), t ∈ (t i , t i+1 ), i = 1, 2, . . . , m, { { { i y(0) = α0 (y) + h0 , { { { y(t+i ) = α i (y) + h i , i = 1, 2, . . . , m. {

Thus the solution can be written as m

y

y(t) = ∑ χ i (t)[(𝕀 − 𝔸)−1 b q,g,h (f y )]i + f q,g (t),

t ∈ J,

i=0

where y

y

f q,g (t) = f i,q,g (t) ∈ J i ,

i = 1, 2, . . . , m,

t

y

f i,q,g (t) = b q,g,h (f y ) =

1 ∫(t − s)q−1 (f(s, y(s)) + g(s))ds, Γ(q)

ti y (α0 (f q,g )

y

y

t ∈ J i , i = 1, 2, . . . , m, y

+ h0 − f q,g (t0 ), . . . , α m (f q,g ) + h m − f q,g (t m )) ∈ ℝ(m+1)n .

3.7 Fractional differential switched systems | 171

As a result, for t ∈ J i we find that y 󵄨 󵄨󵄨 󵄨󵄨y(t) − [(𝕀 − 𝔸)−1 b q (f y )]i − f q (t)󵄨󵄨󵄨 y 󵄨 󵄨 󵄨 󵄨 y ≤ 󵄨󵄨󵄨(𝕀 − 𝔸)−1 [b q,g,h (f y ) − b q (f y )]i 󵄨󵄨󵄨 + 󵄨󵄨󵄨f q,g (t) − f q (t)󵄨󵄨󵄨 t

1 󵄨 󵄨 ≤ ‖(𝕀 − 𝔸) ‖ 󵄨󵄨󵄨[b q,g,h (f y ) − b q (f y )]i 󵄨󵄨󵄨 + ∫(t − s)q−1 |g(s)|ds Γ(q) −1

ti

≤ ‖(𝕀 − 𝔸)−1 ‖

max

i=0,1,...,m

+ ‖(𝕀 − 𝔸)−1 ‖

(‖α i ‖ + 1)(t i+1 − t i Γ(q + 1)

max

i=0,1,...,m

)q

‖g‖∞ +

(t i+1 − t i )q ‖g‖∞ Γ(q + 1)

|h i |

≤ ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ]. The proof is complete. Analogously to Remark 3.116, we have the following results. Remark 3.117. If y ∈ PC(q) (J, ℝn ) is a solution of inequality (3.123), then y satisfies m 󵄩󵄩 󵄩 󵄩󵄩y − ∑ χ [(𝕀 − 𝔸)−1 b (f y )] − f y 󵄩󵄩󵄩 ≤ ε[ρ + ‖(𝕀 − 𝔸)−1 ‖]. i q 1 󵄩󵄩 q󵄩 i 󵄩󵄩∞ 󵄩 i=0

Remark 3.118. If y ∈ PC(q) (J, ℝn ) is a solution of inequality (3.124), then y satisfies m 󵄩󵄩 󵄩 󵄩󵄩y − ∑ χ [(𝕀 − 𝔸)−1 b (f y )] − f y 󵄩󵄩󵄩 ≤ [ρ ‖φ‖ + ‖(𝕀 − 𝔸)−1 ‖ψ]. i q 1 ∞ 󵄩󵄩 q󵄩 i 󵄩󵄩∞ 󵄩 i=0

Now we can prove the following result. Theorem 3.119. Assume [A0]–[A2] are satisfied. If either a𝔼q (L f ) < 1,

(3.127)

where a is given by (3.109), or (3.108) holds, then system (3.90) is Ulam–Hyers–Rassias stable with respect to (φ, ψ). Moreover, (3.90) has a unique solution. Proof. Let y ∈ PC(q) (J, ℝn ) be a solution of inequality (3.125). Certainly, [A2] implies [A3] with m f = maxt∈J |f(t, 0)| and M f = L f . So (3.127) implies (3.108). Let u be a solution of (3.90) following from Theorem 3.108. Then by (3.95), (3.102), (3.103) and (3.126), for t ∈ J i , we have y 󵄨 󵄨 |y(t) − u(t)| ≤ 󵄨󵄨󵄨y(t) − [(𝕀 − 𝔸)−1 b q (f y )]i − f q (t)󵄨󵄨󵄨 y 󵄨 󵄨 + 󵄨󵄨󵄨[(𝕀 − 𝔸)−1 b q (f y )]i + f q (t) − [(𝕀 − 𝔸)−1 b q (f u )]i − f qu (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 y 󵄨 ≤ ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ] + ‖(𝕀 − 𝔸)−1 ‖ 󵄨󵄨󵄨[b q (f y ) − b q (f u )]i 󵄨󵄨󵄨 + 󵄨󵄨󵄨f q (t) − f qu (t)󵄨󵄨󵄨 t

Lf ≤ ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸) ‖ψ] + aL f ‖y − u‖∞ + ∫(t − s)q−1 |y(s) − u(s)|ds. Γ(q) −1

0

172 | 3 Fractional Differential Equations

Using singular Gronwall’s inequality [328, Theorem 1], we derive ‖y − u‖∞ ≤ (ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ] + aL f ‖y − u‖∞ )Eq (L f ) ≤ c f,m,φ ε(φ(t) + ψ),

t ∈ (t i , t i+1 ].

Firstly, assuming (3.127), we get ‖y − u‖∞ ≤ ε

[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ]Eq (L f ) ≤ c f,m,φ ε(‖φ‖∞ + ψ), 1 − aL f Eq (L f )

where c f,m,φ :=

[ρ1 + ‖(𝕀 − 𝔸)−1 ‖]Eq (L f ) > 0. 1 − aL f Eq (L f )

On the other hand, like above, we get |y(t) − u(t)| ≤ ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ] + aL f ‖y − u‖∞ t

Lf + ∫(t − s)q−1 |y(s) − u(s)|ds Γ(q) 0

≤ ε[ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ] + ρ‖y − u‖∞ . Secondly, assuming (3.108), we derive ‖y − u‖∞ ≤ ε

ρ1 ‖φ‖∞ + ‖(𝕀 − 𝔸)−1 ‖ψ ≤ c f,m,φ ε(‖φ‖∞ + ψ), 1−ρ

where c f,m,φ :=

ρ1 + ‖(𝕀 − 𝔸)−1 ‖ > 0. 1−ρ

Thus, system (3.90) is Ulam–Hyers–Rassias stable with respect to (φ, ψ) in both cases. The uniqueness of solutions follows from the above inequalities. One can proceed as in the proof of Theorem 3.119 to prove the following results. Corollary 3.120. Under assumptions [A0]–[A2], system (3.90) is generalized Ulam– Hyers–Rassias stable with respect to (φ, ψ), if either (3.108) or (3.127) holds. Corollary 3.121. Under assumptions [A0]–[A2], system (3.90) is Ulam–Hyers stable, if either (3.108) or (3.127) holds.

3.8 Not instantaneous impulsive fractional differential equations 3.8.1 Introduction Assume that the hemodynamic equilibrium of a person, the introduction of drugs in the bloodstream and the consequent absorption for the body are gradual and continuous

3.8 Not instantaneous impulsive fractional differential equations | 173

processes. In fact, this situation should be characterized by a new case of impulsive action, which starts at an arbitrary fixed point and stays active on a finite time interval. Motivated by [135, 222], we consider a class of new fractional differential equations with not instantaneous impulses of the form c

{

q

D0,t x(t) = f(t, x(t)), x(t) = g k (t, x(t)),

t ∈ (t k , s k ], k = 0, 1, . . . , m, 0 < q < 1, t ∈ (s k−1 , t k ], k = 1, . . . , m,

(3.128)

q

where c D0,t is a generalization of the classical Caputo derivative [162] of order q with the lower limit 0, 0 = t0 < s0 < t1 < s1 < ⋅ ⋅ ⋅ < t m < s m = T, T is a pre-fixed number, f : [0, T] × ℝ → ℝ is continuous and g k : [s k−1 , t k ] × ℝ → ℝ is continuous for each k = 1, 2, . . . , m and represents not instantaneous impulses. In the present section, we first establish a standard framework to derive a suitable formula for solutions of a fractional Cauchy problem with not instantaneous impulses which will inspire researchers to study existence and stability results on such new fields. Secondly, we introduce a new concept of generalized Ulam–Hyers–Rassias stability for equation (3.128). This section is based on [310].

3.8.2 Framework of linear impulsive fractional Cauchy problem In this subsection, we establish a standard framework to derive a suitable formula for solutions of an impulsive fractional Cauchy problem of the form q

c D0,t x(t) = f(t), t ∈ (t k , s k ], k = 0, 1, . . . , m, 0 < q < 1, { { { x(t) = g k (t), t ∈ (s k−1 , t k ], k = 1, . . . , m, { { { x(0) = x0 . {

(3.129)

Lemma 3.122. A function x is a solution of the fractional integral equation t

1 { { { ∫(t − s)q−1 f(s)ds + x0 , t ∈ (0, s0 ], { { Γ(q) { { { 0 { { { { t { { { { 1 ∫(t − s)q−1 f(s)ds { x(t) = { Γ(q) { 0 { { tk { { { 1 { { { + g (t ) − (t k − s)q−1 f(s)ds, t ∈ (t k , s k ], k = 1, . . . , m, ∫ k k { { Γ(q) { { { 0 { { g (t), t ∈ (s , t k−1 k ], k = 1, . . . , m, { k if and only if x is a solution of problem (3.129). Proof. Assume that x solves problem (3.129).

174 | 3 Fractional Differential Equations When t ∈ [0, s0 ], we consider c

{

q

D0,t x(t) = f(t),

t ∈ (0, s0 ],

x(0) = x0 .

(3.130)

Integrating the expression (3.130) from 0 to t by virtue of Lemma A.16, one obtains t

1 x(t) = ∫(t − s)q−1 f(s)ds + c. Γ(q) 0

Clearly, x(0) = x0 implies c = x0 . Thus, t

x(t) =

1 ∫(t − s)q−1 f(s)ds + x0 , Γ(q)

t ∈ [0, s0 ].

0

When t ∈ (s0 , t1 ], x(t) = g1 (t). When t ∈ (t1 , s1 ], we consider c

{

q

D0,t x(t) = f(t),

t ∈ (t1 , s1 ],

x(t1 ) = g1 (t1 ).

By Lemma A.16, we have t

t1

0

0

1 1 x(t) = ∫(t − s)q−1 f(s)ds + g1 (t1 ) − ∫(t1 − s)q−1 f(s)ds, Γ(q) Γ(q)

t ∈ (t1 , s1 ].

When t ∈ (s1 , t2 ], x(t) = g2 (t). When t ∈ (t2 , s2 ], we consider c

{

q

D0,t x(t) = f(t),

t ∈ (t2 , s2 ],

x(t2 ) = g2 (t2 ).

Again by Lemma A.16, we have t2

t

x(t) =

1 1 ∫(t − s)q−1 f(s)ds + g2 (t2 ) − ∫(t2 − s)q−1 f(s)ds, Γ(q) Γ(q) 0

t ∈ (t2 , s2 ].

0

Generally speaking, we consider c

{

q

D0,t x(t) = f(t),

t ∈ (t k , s k ],

x(t k ) = g k (t k ).

By Lemma A.16, we have t

tk

0

0

1 1 x(t) = ∫(t − s)q−1 f(s)ds + g k (t k ) − ∫(t k − s)q−1 f(s)ds, Γ(q) Γ(q)

t ∈ (t k , s k ].

One can verify the converse by performing the standard steps to complete the rest of the proof.

3.8 Not instantaneous impulsive fractional differential equations | 175

3.8.3 Generalized Ulam–Hyers–Rassias stability concept Motivated by the concepts of stability in [245, 282], we can introduce a generalized Ulam–Hyers–Rassias stability concept for equation (3.128). Let ϵ > 0, ψ ≥ 0 and φ ∈ PC(J, ℝ+ ) be nondecreasing. Consider q

|c D0,t y(t) − f(t, y(t))| ≤ φ(t),

{

|y(t) − g k (t, y(t))| ≤ ψ,

t ∈ (t k , s k ], k = 0, 1, 2, . . . , m, 0 < q < 1, t ∈ (s k−1 , t k ], k = 1, 2, . . . , m. (3.131)

Definition 3.123. Equation (3.128) is generalized Ulam–Hyers–Rassias stable with respect to (φ, ψ) if there exists c f,q,g i ,φ > 0 such that for each solution y ∈ PC(J, ℝ) of inequality (3.131) there exists a solution x ∈ PC(J, ℝ) of equation (3.128) with |y(t) − x(t)| ≤ c f,q,g i ,φ (φ(t) + ψ),

t ∈ J.

Remark 3.124. A function y ∈ PC(J, ℝ) is a solution of inequality (3.131) if and only if there are G ∈ PC(J, ℝ) and a sequence G k , k = 1, 2, . . . , m (which depend on y) such that (i) |G(t)| ≤ φ(t), t ∈ J and |G k | ≤ ψ, k = 1, 2, . . . , m; q (ii) c D0,t y(t) = f(t, y(t)) + G(t), t ∈ (t k , s k ], k = 0, 1, 2, . . . , m; (iii) y(t) = g k (t, y(t)) + G k , t ∈ (s k−1 , t k ], k = 1, 2, . . . , m. Remark 3.125. If y ∈ PC(J, ℝ) is a solution of inequality (3.131), then y is a solution of the integral inequality |y(t) − g k (t, y(t))| ≤ ψ, t ∈ (s k−1 , t k ], k = 1, 2, . . . , m, { { { { t { {󵄨 󵄨󵄨 { 1 󵄨󵄨 { q−1 󵄨󵄨 { 󵄨 { y(t) − y(0) − (t − s) f(s, y(s))ds ∫ 󵄨 󵄨󵄨 { { Γ(q) 󵄨󵄨 󵄨 { { 0 { { { { t { { { 1 { { (t − s)q−1 φ(s)ds, t ∈ (0, s0 ], ≤ ∫ { { { Γ(q) { { 0 { { { { t { { {󵄨󵄨󵄨 1 { q−1 { { {󵄨󵄨󵄨󵄨y(t) − Γ(q) ∫(t − s) f(s, y(s))ds − g k (t k , y(t k )) 0 { { { { tk { { 󵄨󵄨 { 1 { { + (t k − s)q−1 f(s, y(s))ds󵄨󵄨󵄨󵄨 ∫ { { { Γ(q) 󵄨 { { 0 { { { { t { { { 1 { { ≤ ∫(t − s)q−1 φ(s)ds { { { Γ(q) { { 0 { { { { tk { { { 1 { { + ∫(t k − s)q−1 φ(s)ds + ψ, t ∈ (t k , s k ], k = 1, 2, . . . , m. { { Γ(q) 0 {

(3.132)

176 | 3 Fractional Differential Equations

In fact, by Remark 3.124 we get c

{

q

D0,t y(t) = f(t, y(t)) + G(t),

t ∈ (t k , s k ], k = 0, 1, 2, . . . , m,

y(t) = g k (t, y(t)) + G k ,

t ∈ (s k−1 , t k ], k = 1, 2, . . . , m.

(3.133)

Clearly, the solution of equation (3.133) is given by g k (t, y(t)) + G k , t ∈ (s k−1 , t k ], k = 1, 2, . . . , m, { { { { t { { { 1 { { ∫(t − s)q−1 [f(s, y(s)) + G(s)]ds + y0 , t ∈ (0, s0 ], { { { Γ(q) { { 0 { { { { t y(t) = { 1 { (t − s)q−1 [f(s, y(s)) + G(s)]ds + g k (t k , y(t k )) + G k ∫ { { { Γ(q) { { 0 { { { { tk { { { 1 { q−1 { { { − Γ(q) ∫(t k − s) [f(s, y(s)) + G(s)]ds, t ∈ (t k , s k ], k = 1, 2, . . . , m. 0 { For t ∈ (t k , s k ], k = 1, 2, . . . , m, we get t

t

k 󵄨󵄨 󵄨 󵄨󵄨y(t) − 1 ∫(t − s)q−1 f(s, y(s))ds − g (t , y(t )) + 1 ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨 k k k k 󵄨󵄨 󵄨󵄨 Γ(q) Γ(q) 󵄨 󵄨

0

0

tk

t

󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 ≤ 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 G(s)ds󵄨󵄨󵄨󵄨 + |G k | + 󵄨󵄨󵄨󵄨 ∫(t k − s)q−1 G(s)ds󵄨󵄨󵄨󵄨 󵄨 󵄨 Γ(q) 󵄨 󵄨 Γ(q) 0

≤

0

t

tk

0

0

1 1 ∫(t − s)q−1 φ(s)ds + ∫(t k − s)q−1 φ(s)ds + ψ. Γ(q) Γ(q)

Proceeding as above, we derive |y(t) − g k (t, y(t))| ≤ |G k | ≤ ψ,

t ∈ (s k−1 , t k ], k = 1, 2, . . . , m

and t

t

0

0

󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨y(t) − y(0) − 1 ∫(t − s)q−1 f(s, y(s))ds󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨 1 ∫(t − s)q−1 G(s)ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 Γ(q) 󵄨 󵄨 󵄨 Γ(q) 󵄨 t

1 ≤ ∫(t − s)q−1 φ(s)ds, Γ(q) 0

3.8.4 Main results via fixed point methods We introduce the following assumptions:

t ∈ (0, s0 ].

3.8 Not instantaneous impulsive fractional differential equations | 177

(H1 ) f ∈ C(J × ℝ, ℝ). (H2 ) There exists a positive constant L f such that 󵄨 󵄨󵄨 󵄨󵄨f(t, u1 ) − f(t, u2 )󵄨󵄨󵄨 ≤ L f |u1 − u2 | for all t ∈ J, u1 , u2 ∈ ℝ. (H3 ) g k ∈ C([s k−1 , t k ] × ℝ, ℝ) and there are positive constants L g k , k = 1, 2, . . . , m such that 󵄨󵄨 󵄨 󵄨󵄨g k (t, u1 ) − g k (t, u2 )󵄨󵄨󵄨 ≤ L g k |u1 − u2 |

for all t ∈ (s k−1 , t k ], u1 , u2 ∈ ℝ.

(H4 ) Let φ ∈ C(J, ℝ+ ) be a nondecreasing function and there exists c φ > 0 such that t 1

p

( ∫(φ(s)) p ds) ≤ c φ φ(t) for all t ∈ J. 0

Now, we discuss stability of equation (3.128) by using the concept of generalized Ulam–Hyers–Rassias stability from the latter subsection. Theorem 3.126. Assume that (H1 ), (H2 ), (H3 ), (H4 ) are satisfied and that a function y ∈ PC(J, ℝ) fulfills (3.131). Then there exists a unique solution y0 of equation (3.128) such that t

1 { { { if t ∈ [0, s0 ], ∫(t − s)q−1 f(s, y0 (s))ds + x0 { { Γ(q) { { { 0 { { { { { g k (t, y0 (t)) if t ∈ (s k−1 , t k ], k = 1, 2, . . . , m, { { { { t y0 (t) = { 1 { { ∫(t − s)q−1 f(s, y0 (s))ds + g k (t k , y0 (t k )) { { { Γ(q) { { 0 { { { tk { { 1 { { { − if t ∈ (t k , s k ], k = 1, 2, . . . , m, ∫(t k − s)q−1 f(s, y0 (s))ds { Γ(q) { 0 (3.134) and 2c 1−p q−p T + 1](φ(t) + ψ) [ Γ(q)φ ( 1−p q−p ) (3.135) |y(t) − y0 (t)| ≤ 1−M for all t ∈ J provided that 0 < p < q < 1 and M = max{M1 , M2 } < 1, where L f c φ 1 − p 1−p q−p q−p ( ) (s k + t k ) + L g k | k = 0, 1, 2, . . . , m}, Γ(q) q − p Lf q q M2 = max{ (s + t k ) + L g k | k = 1, 2, . . . , m}. Γ(q + 1) k M1 = max{

(3.136)

178 | 3 Fractional Differential Equations

Proof. Consider the space of piecewise continuous functions, X = {g : J → ℝ | g ∈ PC(J, ℝ)}, endowed with the generalized metric on X defined by d(g, h) = inf{C1 + C2 ∈ [0, +∞] | |g(t) − h(t)| ≤ (C1 + C2 )(φ(t) + ψ) for all t ∈ J}, (3.137) where C1 ∈ {C ∈ [0, +∞] | |g(t) − h(t)| ≤ Cφ(t) for all t ∈ (t k , s k ], k = 0, 1, 2, . . . , m}, C2 ∈ {C ∈ [0, +∞] | |g(t) − h(t)| ≤ Cψ for all t ∈ (s k−1 , t k ], k = 1, 2, . . . , m}. It is easy to verify that (X, d) is a complete generalized metric space. Define an operator Λ : X → X by t

1 { { { if t ∈ [0, s0 ], ∫(t − s)q−1 f(s, x(s))ds + x0 { { Γ(q) { { { 0 { { { { { g (t, x(t)) if t ∈ (s k−1 , t k ], k = 1, 2, . . . , m, k { { { { (Λx)(t) = { 1 t { { ∫(t − s)q−1 f(s, x(s))ds + g k (t k , x(t k )) { { Γ(q) { { { 0 { { { tk { { 1 { { { (t k − s)q−1 f(s, x(s))ds if t ∈ (t k , s k ], k = 1, 2, . . . , m − ∫ { Γ(q) { 0 for all x ∈ X and t ∈ [0, T]. Clearly, Λ is a well-defined operator according to (H1 ). Next, we shall verify that Λ is strictly contractive on X. Note that by the definition of (X, d), for any g, h ∈ X, it is possible to find C1 , C2 ∈ [0, ∞] such that {C1 φ(t) if t ∈ (t k , s k ], k = 0, 1, 2, . . . , m, |g(t) − h(t)| ≤ { (3.138) C2 ψ if t ∈ (s k−1 , t k ], k = 1, 2, . . . , m. { From the above definition of Λ, (H2 ), (H3 ), and (3.138), we obtain the following: Case 1: For t ∈ [0, s0 ], t

t

0

0

󵄨󵄨 1 󵄨󵄨 󵄨 󵄨󵄨 1 ∫(t − s)q−1 f(s, g(s))ds − ∫(t − s)q−1 f(s, h(s))ds󵄨󵄨󵄨󵄨 󵄨󵄨(Λg)(t) − (Λh)(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨 Γ(q) Γ(q) 󵄨 󵄨 t

≤

1 󵄨 󵄨 ∫(t − s)q−1 󵄨󵄨󵄨f(s, g(s)) − f(s, h(s))󵄨󵄨󵄨ds Γ(q) 0 t

Lf ≤ ∫(t − s)q−1 |g(s) − h(s)|ds Γ(q) 0 t

L f C1 ≤ ∫(t − s)q−1 |φ(s)|ds Γ(q) 0

3.8 Not instantaneous impulsive fractional differential equations | 179

≤

t

t

0

0

1−p p q−1 1 L f C1 ( ∫(t − s) 1−p ds) ( ∫(φ(s)) p ds) Γ(q)

L f C1 c φ 1 − p 1−p q−p φ(t)( ≤ ) t Γ(q) q−p L f c φ 1 − p 1−p q−p ≤ ( ) s0 C1 φ(t). Γ(q) q − p Case 2: For t ∈ (s k−1 , t k ], |(Λg)(t) − (Λh)(t)| = |g k (t, g(t)) − g k (t, h(t))| ≤ L g k |g(t) − h(t)| ≤ L g k C2 ψ. Case 3: For t ∈ (t k , s k ] and s ∈ (t k , s k ], 󵄨󵄨 󵄨 󵄨󵄨(Λg)(t) − (Λh)(t)󵄨󵄨󵄨 tk t 󵄨󵄨󵄨 1 1 q−1 󵄨 󵄨 = 󵄨󵄨 ∫(t − s) f(s, g(s))ds + g k (t k , g(t k )) − ∫(t k − s)q−1 f(s, g(s))ds Γ(q) 󵄨󵄨 Γ(q) 0 0 −[

tk t 󵄨󵄨 1 1 󵄨 ∫(t − s)q−1 f(s, h(s))ds + g k (t k , h(t k )) − ∫(t k − s)q−1 f(s, h(s))ds]󵄨󵄨󵄨󵄨 Γ(q) Γ(q) 󵄨󵄨 0

0

t

t

0

0

󵄨󵄨 󵄨󵄨 1 1 ≤ 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 f(s, g(s))ds − ∫(t − s)q−1 f(s, h(s))ds󵄨󵄨󵄨󵄨 Γ(q) Γ(q) 󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨g k (t k , g(t k )) − g k (t k , h(t k ))󵄨󵄨󵄨 t

t

0

0

k 󵄨󵄨 󵄨󵄨 1 k 1 q−1 󵄨 󵄨 + 󵄨󵄨 ∫(t k − s) f(s, h(s))ds − ∫(t k − s)q−1 f(s, g(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 Γ(q)

L f c φ 1 − p 1−p q−p L f c φ 1 − p 1−p q−p ≤ ( ) s k C1 φ(t) + L g k C2 ψ + ( ) t k C1 φ(t) Γ(q) q − p Γ(q) q − p L f c φ 1 − p 1−p q−p q−p = ( ) (s k + t k )C1 φ(t) + L g k C2 ψ Γ(q) q − p L f c φ 1 − p 1−p q−p q−p ≤[ ( ) (s k + t k ) + L g k ](C1 + C2 )(φ(t) + ψ) Γ(q) q − p and for t ∈ (t k , s k ] and s ∈ (s k−1 , t k ], 󵄨󵄨 󵄨 󵄨󵄨(Λg)(t) − (Λh)(t)󵄨󵄨󵄨 t

t

0

0

󵄨󵄨 󵄨󵄨 1 1 ≤ 󵄨󵄨󵄨󵄨 ∫(t − s)q−1 f(s, g(s))ds − ∫(t − s)q−1 f(s, h(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 Γ(q) 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨g k (t k , g(t k )) − g k (t k , h(t k ))󵄨󵄨󵄨 t

t

0

0

k 󵄨󵄨 1 k 󵄨󵄨 1 q−1 󵄨 󵄨 + 󵄨󵄨 ∫(t k − s) f(s, h(s))ds − ∫(t k − s)q−1 f(s, g(s))ds󵄨󵄨󵄨󵄨 Γ(q) 󵄨 󵄨 Γ(q)

180 | 3 Fractional Differential Equations tk

t

≤

Lf Lf ∫(t − s)q−1 |g(s) − h(s)|ds + L g k C2 ψ + ∫(t k − s)q−1 |h(s) − g(s)|ds Γ(q) Γ(q) 0

≤

0 t

tk

0

0

L f C2 ψ L f C2 ψ ∫(t − s)q−1 ds + L g k C2 ψ + ∫(t k − s)q−1 ds Γ(q) Γ(q)

L f C2 ψ q L f C2 ψ q ≤ t + L gk C2 ψ + t qΓ(q) qΓ(q) k Lf q Lf q ≤[ s + L gk + t ]C2 ψ qΓ(q) k qΓ(q) k Lf q q =[ (s + t k ) + L g k ]C2 ψ Γ(q + 1) k Lf q q ≤[ (s + t k ) + L g k ](C1 + C2 )(φ(t) + ψ). Γ(q + 1) k From above, we get 󵄨󵄨 󵄨 󵄨󵄨(Λg)(t) − (Λh)(t)󵄨󵄨󵄨 ≤ M(C1 + C2 )(φ(t) + ψ),

t ∈ J.

That is d(Λg, Λh) ≤ M(C1 + C2 )(φ(t) + ψ). Hence, we can conclude that d(Λg, Λh) ≤ Md(g, h) for any g, h ∈ X. Using condition (3.136) proves the property of strict continuity. Let us take g0 ∈ X. From the property of piecewise continuity of g0 and Λg0 , it follows that there exists a constant 0 < G1 < ∞ such that t

󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 1 ∫(t − s)q−1 f(s, g0 (s))ds + x0 − g0 (t)󵄨󵄨󵄨󵄨 󵄨󵄨(Λg0 )(t) − g0 (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨 󵄨 Γ(q) 󵄨 0

≤ G1 φ(t) ≤ G1 (φ(t) + ψ), There exists a constant 0 < G2 < ∞ such that 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨(Λg0 )(t) − g0 (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨g k (t, g0 (t)) − g0 (t)󵄨󵄨󵄨 ≤ G2 ψ ≤ G2 (φ(t) + ψ),

t ∈ [0, s0 ].

t ∈ (s k−1 , t k ], k = 1, 2, . . . , m.

There exists a constant 0 < G3 < ∞ such that t

󵄨󵄨 󵄨 󵄨󵄨 1 ∫(t − s)q−1 f(s, g0 (s))ds + g k (t k , g0 (t k )) 󵄨󵄨(Λg0 )(t) − g0 (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨 󵄨 Γ(q) 0

t

k 󵄨󵄨 1 − ∫(t k − s)q−1 f(s, g0 (s))ds − g0 (t)󵄨󵄨󵄨󵄨 Γ(q) 󵄨

0

≤ G3 (φ(t) + ψ),

t ∈ (t k , s k ], k = 1, 2, . . . , m,

3.8 Not instantaneous impulsive fractional differential equations | 181

since f , g k and g0 are bounded on J and φ(⋅) + ψ > 0. Thus, (3.137) implies that d(Λg0 , g0 ) < ∞. By using the Banach fixed point theorem, there exists a continuous function y0 : J → ℝ such that Λ n g0 → y0 in (X, d) as n → ∞ and Λy0 = y0 . That is, y0 satisfies equation (3.134) for every t ∈ J. Next, we check that {g ∈ X | d(g0 , g) < ∞} = X. For any g ∈ X, since g and g0 are bounded on J and mint∈J (φ(t) + ψ) > 0, there exists a constant 0 < C g < ∞ such that |g0 (t) − g(t)| ≤ C g (φ(t) + ψ) for any t ∈ J. Hence, we have d(g0 , g) < ∞ for all g ∈ X, that is, {g ∈ X | d(g0 , g) < ∞} = X. Hence, we conclude that y0 is the unique continuous function with the property (3.134). On the other hand, from (3.132) and (H4 ) it follows that 2c φ 1 − p 1−p q−p d(y, Λy) ≤ + 1. ( ) T Γ(q) q − p Summarizing, we have d(y, y0 ) ≤

d(Λy, y) ≤ 1−M

2c φ 1−p 1−p q−p T Γ(q) ( q−p )

1−M

+1

,

which means that (3.135) is true for t ∈ J. Finally, we give an example to illustrate the above results. Consider 1 |x(t)| 2 {c D0,t x(t) = , t ∈ (0, 1] ∪ (2, 3], { { 8 + et + t 2 { |x(t)| { { x(t) = , t ∈ (1, 2], 2 (3 + t )(1 + |x(t)|) { and

󵄨󵄨 1 󵄨 󵄨󵄨c D 2 y(t) − |y(t)| 󵄨󵄨󵄨 ≤ et , t ∈ [0, 1] ∪ (2, 3], { 󵄨 { t 2 { 󵄨󵄨󵄨 0,t 8 + e + t 󵄨󵄨 { 󵄨󵄨 |y(t)| {󵄨󵄨󵄨 󵄨󵄨 ≤ 1, t ∈ (1, 2]. { 󵄨󵄨y(t) − 󵄨 2 󵄨 (3 + t )(1 + |y(t)|) 󵄨󵄨 {󵄨 Let J = [0, 3], q = 12 , p =

1 3

and 0 = t0 < s0 = 1 < t1 = 2 < s1 = 3. Denote f(t, x(t)) =

with L f =

1 9

for t ∈ (0, 1] ∪ (2, 3], and g1 (t, x(t)) =

with L g1 =

|x(t)| 8 + et + t 2

1 4

|x(t)| (3 + t2 )(1 + |x(t)|)

for t ∈ (1, 2]. Putting φ(t) = et , ψ = 1 and c φ = 1, we have t

1 3

( ∫(es )3 ds) ≤ et . 0

182 | 3 Fractional Differential Equations

Let 2 2 1 1 1 1 1 43 , 4 3 (3 6 + 2 6 ) + } ≈ 0.6171, 4 9√π 9√π 1 1 1 1 M2 = (3 2 + 2 2 ) + ≈ 0.5599, 3 4 9Γ( 2 )

M1 = max{

so M = 0.6171 < 1. By Theorem 3.126, there exists a unique solution y0 : [0, 3] → ℝ such that t

{ 1 1 |y0 (t)| { { { ds + x0 ∫(t − s)− 2 { 1 { 8 + et + t 2 { Γ( 2 ) { { 0 { { { { |y0 (t)| { { { { 2 { (3 + t )(1 + |y0 (t)|) { { t y0 (t) = { { 1 1 |y0 (t)| |y0 (2)| { { (t − s)− 2 ds + ∫ { 1 t 2 { 7(1 + |y0 (2)|) 8+e +t { { Γ( 2 ) 0 { { { { { 2 { { { { − 1 ∫(2 − s)− 12 |y0 (t)| ds { { { 8 + et + t 2 Γ( 12 ) 0 { and |y(t) − y0 (t)| ≤

2 √π

2

if t ∈ (1, 2],

if t ∈ (2, 3],

1

× 43 × 36 + 1 1 − 0.7

if t ∈ [0, 1],

(et + 1) ≈ 14.7157(et + 1)

for all t ∈ [0, 3].

3.9 Center stable manifold result for planar fractional damped equations 3.9.1 Introduction Very recently, Cong et al. [76] obtained a local stable manifold theorem near a hyperbolic equilibrium point for planar fractional differential equations of order α ∈ (0, 1) by defining a related Lyapunov–Perron operator for two-dimensional fractional systems and dealing with asymptotic behavior of Mittag-Leffler function E α,α . Then, the fixed points of the Lyapunov–Perron operator describe the set of all solutions tending to zero near the fixed point, which is called the stable manifold of the hyperbolic fixed point. In this section, we prove a center stable manifold theorem for planar fractional damped equations involving two Caputo derivatives. We develop and extend the idea and methods in [76] by using asymptotic behavior of the Mittag-Leffler function with two different parameters, E α,β , α ∈ (1, 2), β ∈ (0, 1), which will be used to obtain a center stable manifold result. This section is based on [288].

3.9 Planar fractional damped equations | 183

3.9.2 Asymptotic behavior of Mittag-Leffler functions E α,β To give some results on the asymptotic behavior of Mittag-Leffler functions E α−β,2 and E α−β,α for α ∈ (1, 2) and β ∈ (0, 1), we recall the following. ̄ the Lemma 3.127 ([124]). Let ᾱ ∈ (0, 2) and β̄ ∈ ℝ be arbitrary. Then for p̄ = [β/̄ α], following asymptotic expansions hold: p̄

1 z−k 1 1−̄ β̄ { { z α exp(z ᾱ ) − ∑ + O(z−1−p̄ ) as z → ∞, { { ̄ { ̄ α ̄ Γ( β − αk) { k=1 E α,̄ β̄ (z) = { ̄ p { −k { z { { + O(|z|−1−p̄ ) as z → −∞. {− ∑ ̄ ̄ { k=1 Γ(β − αk)

Putting ᾱ = α − β, β̄ = 2 and z = t α−β λ, we get the first asymptotic expansions for MittagLeffler functions which extend [76, Lemma 3] to our case. Lemma 3.128. For any λ > 0, α ∈ (1, 2), β ∈ (0, 1) and p = [2/(α − β)], it holds 1

tE α−β,2 (λt α−β ) =

exp(λ α−β t) λ

1 α−β

(α − β)

p

tE α−β,2 (−λt α−β ) = − ∑ k=1

p

−∑ k=1

t1−k(α−β) + O(t1−(α−β)(1+p) ), − (α − β)k)

λ k Γ(2

(3.139)

t1−k(α−β) + O(t1−(α−β)(1+p) ) − (α − β)k)

(−λ)k Γ(2

as t → ∞. Putting ᾱ = α − β, β̄ = α and z = t α−β λ, we get asymptotic expansions for Mittag-Leffler functions E α−β,α . Lemma 3.129. For any λ > 0, α ∈ (1, 2) and β ∈ (0, 1) with p = 1, it holds 1

t α−1 E α−β,α (λt α−β ) =

exp(λ α−β t) λ

α−1 α−β

(α − β)

−

t β−1 + O(t2β−α−1 ), λΓ(β)

t β−1 t α−1 E α−β,α (−λt α−β ) = + O(t2β−α−1 ) λΓ(β) as t → ∞. Putting ᾱ = α − β, β̄ = α + 1 and z = t α−β λ, we get asymptotic expansions for MittagLeffler functions E α−β,α+1 . Lemma 3.130. For any λ > 0, α ∈ (1, 2), β ∈ (0, 1) with α − β ≥ 1, it holds 1

t E α−β,α+1 (λt α

α−β

)=

t α E α−β,α+1 (−λt α−β ) = as t → ∞.

exp(λ α−β t) λ

α α−β

(α − β)

−

tβ + O(t2β−α ), λΓ(β + 1)

tβ + O(t2β−α ) λΓ(β + 1)

184 | 3 Fractional Differential Equations

3.9.3 Planar fractional Cauchy problems Consider the following nonautonomous fractional Cauchy problems: β

c α D t x(t) + A c D t x(t) = f(x(t), t), { { { x(0) = x = (x1 , x2 )T , { { { x󸀠 (0) = x = (x3 , x4 )T , {

α ∈ (1, 2), β ∈ (0, 1), t ≥ 0, (3.140)

where A=(

λ1 0

0 ), λ2

f(x, t) = (

f1 (x, t) ) f2 (x, t)

and λ1 > 0, λ2 < 0, and f ∈ C(ℝ2 × ℝ+ , ℝ2 ) is a locally Lipschitz function, ‖f(x, t) − f(y, t)‖ ≤ l f (r)κ(t)‖x − y‖ for ‖x‖, ‖y‖ ≤ r with f(0, t) = 0, limr→0 l f (r) = 0 and κ ∈ C∞ (ℝ+ , (0, ∞)) satisfying t

M κ := sup ∫(t − s)β−1 κ(s)ds < ∞. t>0

0

Remark 3.131. The conditions f(0, t) = 0 and limr→0 l f (r) = 0 are reasonable. We study properties of our FDEs near (0, 0). But since we want to find all bounded solutions (not just possible) near (0, 0), and our perturbation is nonautonomous, we need to control the growth of f(x, t) by the function κ(t). We want the center stable manifold to really contain starting points bounded on [0, ∞) near (0, 0), which could not be possible without κ. We consider the maximum norm ‖(x1 , x2 )‖ = max{|x1 |, |x2 |} on ℝ2 . Definition 3.132. By a local center stable manifold of (3.140) we mean the set of all small x and x̄ for which the solution of (3.140) is bounded on ℝ+ . By [32, formula (5)], taking the Laplace transform and the inverse transform, substituting the Laplace transform of the Mittag-Leffler function and the Laplace convolution operator, the solution φ(⋅, x, x) of (3.140) is given by φ(t, x, x) = E α−β (−t α−β A)x + E α−β,α−β+1 (−t α−β A)At α−β x + E α−β,2 (−t α−β A)tx t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β A)f(φ(s, x, x), s)ds. 0

On the other hand, using [123, formula (4.2.3)], we have 1 = E α−β (z) − zE α−β,α−β+1 (z), which implies E α−β (−t α−β A) + E α−β,α−β+1 (−t α−β A)At α−β = I.

(3.141)

3.9 Planar fractional damped equations | 185

Consequently, formula (3.141) is simplified to φ(t, x, x) = x + E α−β,2 (−t α−β A)tx t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β A)f(φ(s, x, x), s)ds.

(3.142)

0

Thus the homogeneous linear equation has the solution x + E α−β,2 (−t α−β A)tx. By the results of Section 3.9.2, we obtain lim |E α−β,2 (−t α−β λ2 )|t = ∞,

t→∞

0 { { { lim |E α−β,2 (−t α−β λ1 )|t = { λ1 t→∞ { { 1 {∞

if 1 < α − β, if 1 = α − β, if 1 > α − β.

Hence the linear equation has an asymptotically stable part only if α − β > 1. Furthermore, for nonzero constant f = c ∈ ℝ2 , we get t

x + E α−β,2 (−t

α−β

A)tx + ∫(t − s)α−1 E α−β,α (−(t − s)α−β A)cds 0

= x + E α−β,2 (−t

α−β

A)tx + t α E α−β,α+1 (−t α−β A)c,

where we used t

∫(t − s)α−1 E α−β,α (−(t − s)α−β A)ds = t α E α−β,α+1 (−t α−β A). 0

By applying Lemma 3.130 with α − β > 1, we see 󵄨 󵄨 lim 󵄨󵄨t α E α−β,α+1 (−t α−β A)c󵄨󵄨󵄨 = ∞.

t→∞󵄨

So we cannot get a center stable manifold in general, but we need to control it with κ.

3.9.4 Center stable manifold result Based on the analysis of Section 3.9.3, we suppose α − β > 1. Define an operator T = (T1 , T2 ) : C∞ (ℝ+ , ℝ2 ) → C∞ (ℝ+ , ℝ2 ) as follows: T1 (ξ)(t) = x1 + E α−β,2 (−t α−β λ1 )tx3 t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ1 )f1 (ξ(s), s)ds, 0

186 | 3 Fractional Differential Equations

and T2 (ξ)(t) = x2 − (−λ2 )

2−α α−β

∞

E α−β,2 (−t

α−β

1

λ2 )t ∫ exp ( − (−λ2 ) α−β s)f2 (ξ(s), s)ds 0

t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ2 )f2 (ξ(s), s)ds. 0

To verify that T is well defined, we need the next estimates. Lemma 3.133. For λ > 0 we define 1−α E α−β,α (λ) + λ α−β E α−β,2 (λ)}. α Then for any function g ∈ C∞ (ℝ+ , ℝ) the following statements hold for all t ∈ [0, 1]: 1−α

K(α, β, λ) = max{E α−β,α+1 (λ), λ α−β E α−β,2 (λ),

t

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫(t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 ≤ K(α, β, λ)‖g‖∞ , α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨

(3.143)

0

∞

󵄨󵄨 2−α 1 󵄨󵄨 α−β 󵄨󵄨λ α−β E λ)t ∫ exp(−λ α−β s)g(s)ds󵄨󵄨󵄨󵄨 ≤ K(α, β, λ)‖g‖∞ , α−β,2 (t 󵄨󵄨 󵄨 󵄨

(3.144)

t

󵄨󵄨 󵄨󵄨 t 1 2−α 󵄨 󵄨󵄨 󵄨󵄨 ∫[(t − s)α−1 E α−β,α ((t − s)α−β λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨0 ≤ K(α, β, λ)‖g‖∞ .

(3.145)

Proof. Using t

∫(t − s)α−1 E α−β,α ((t − s)α−β λ)ds = t α E α−β,α+1 (λt α−β ) 0

and the fact that Mittag-Leffler functions are increasing on [0, ∞), we get (3.143): t

t

0

0 α

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫(t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 ≤ ∫(t − s)α−1 E α−β,α ((t − s)α−β λ)|g(s)|ds α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨 ≤ t E α−β,α+1 (λt α−β )‖g‖∞ ≤ E α−β,α+1 (λ)‖g‖∞ ≤ K(α, β, λ)‖g‖∞ . In the same way, we obtain (3.144): ∞

∞

󵄨󵄨 󵄨󵄨 2−α 1 2−α 1 α−β 󵄨󵄨λ α−β E λ)t ∫ exp(−λ α−β s)g(s)ds󵄨󵄨󵄨󵄨 ≤ λ α−β E α−β,2 (λ) ∫ exp(−λ α−β s)ds‖g‖∞ α−β,2 (t 󵄨󵄨 󵄨 󵄨 t

0

≤λ

1−α α−β

E α−β,2 (λ)‖g‖∞

≤ K(α, β, λ)‖g‖∞ .

3.9 Planar fractional damped equations | 187

Similarly, we derive (3.145): t

󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨 ∫[(t − s)α−1 E λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨󵄨 α−β,α ((t − s) 󵄨󵄨 󵄨 󵄨 0

t 2−α

1

≤ ∫[(t − s)α−1 E α−β,α (λ) + λ α−β E α−β,2 (λ) exp(−λ α−β s)]ds‖g‖∞ 0 −1 1 −1 2−α 1 = [ t α E α−β,α (λ) + λ α−β E α−β,2 (λ)(−λ α−β exp(−λ α−β t) + λ α−β )]‖g‖∞ α 1−α E α−β,α (λ) ≤[ + λ α−β E α−β,2 (λ)]‖g‖∞ α ≤ K(α, β, λ)‖g‖∞ .

The proof of the lemma is complete. By Lemmas 3.128 and 3.129, we can define finite constants 󵄨 exp(λ α−β t) 󵄨󵄨󵄨 α−β 󵄨󵄨, m1 (α, β, λ) := sup t 󵄨󵄨tE α−β,2 (λt ) − 1 󵄨 󵄨 t≥1 λ α−β (α − β) 󵄨 󵄨 󵄨 m2 (α, β, λ) := sup t α−β−1 󵄨󵄨󵄨tE α−β,2 (−λt α−β )󵄨󵄨󵄨, 1

α−β−1 󵄨󵄨󵄨

t≥1

m3 (α, β, λ) := sup t

󵄨 󵄨󵄨t 󵄨

α+1−2β 󵄨󵄨󵄨 α−1

1

E α−β,α (λt

α−β

)−

exp(λ α−β t) α−1 α−β

λ (α − β) 󵄨󵄨 t β−1 󵄨󵄨󵄨 󵄨󵄨. m4 (α, β, λ) := sup t α+1−2β 󵄨󵄨󵄨󵄨t α−1 E α−β,α (−λt α−β ) − λΓ(β) 󵄨󵄨 󵄨 t≥1 t≥1

+

t β−1 󵄨󵄨󵄨 󵄨󵄨, λΓ(β) 󵄨󵄨

Now we can prove the next result. Lemma 3.134. For λ > 0 we define M(α, β, λ, M κ ) = max{E α−β,α+1 (λ) + +

Mκ max{m3 (α, β, λ), m4 (α, β, λ)} + α − 2β λΓ(β)

−α 1 1−α 1 λ α−β exp(λ α−β ) + 2m1 (α, β, λ)λ α−β }. α−β

󵄨 󵄨󵄨 Then for any function g ∈ C∞ (ℝ+ , ℝ) with ‖g‖κ := supt≥0 󵄨󵄨󵄨 g(t) κ(t) 󵄨󵄨 < ∞, the following statements hold for all t > 1: t

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫(t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 ≤ M(α, β, λ, M κ )(‖g‖∞ + ‖g‖κ ), α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨

(3.146)

0

∞

󵄨󵄨 2−α 󵄨󵄨 1 α−β 󵄨󵄨λ α−β E λ)t ∫ exp(−λ α−β s)g(s)ds󵄨󵄨󵄨󵄨 ≤ M(α, β, λ, M κ )‖g‖∞ , α−β,2 (t 󵄨󵄨 󵄨 󵄨 t

(3.147)

188 | 3 Fractional Differential Equations 󵄨󵄨 t 󵄨󵄨 2−α 1 󵄨󵄨 󵄨 󵄨󵄨 ∫[(t − s)α−1 E α−β,α ((t − s)α−β λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨0 󵄨 ≤ M(α, β, λ, M κ )(‖g‖∞ + ‖g‖κ ).

(3.148)

Proof. Note that t

∫ (t − s)α−1 E α−β,α ((t − s)α−β λ)ds = E α−β,α+1 (λ). t−1

So t

t

t−1

t−1

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫ (t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 ≤ ∫ (t − s)α−1 E α−β,α ((t − s)α−β λ)|g(s)|ds α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨 ≤ E α−β,α+1 (λ)‖g‖∞ . On the other hand, we get t−1

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫ (t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨 0

t−1

t−1

|g(s)| 1 |g(s)| ≤ m4 (α, β, λ) ∫ ds + ds ∫ α−2β+1 λΓ(β) (t − s) (t − s)−β+1 0

0

t−1

≤

m4 (α, β, λ)‖g‖∞ 1 |g(s)| + ds ∫ (t − s)β−1 κ(s) α − 2β λΓ(β) κ(s) 0

m4 (α, β, λ)‖g‖∞ M κ ‖g‖κ + . ≤ α − 2β λΓ(β) Consequently, we get (3.146): t

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫(t − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 α−β,α (−(t − s) 󵄨󵄨 󵄨 󵄨 0

≤ (E α−β,α+1 (λ) +

Mκ m4 (α, β, λ) ‖g‖κ )‖g‖∞ + α − 2β λΓ(β)

≤ M(α, β, λ, M κ )(‖g‖∞ + ‖g‖κ ). Similarly, we get (3.147): ∞

󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨λ α−β E λ)t ∫ exp(−λ α−β s)g(s)ds󵄨󵄨󵄨󵄨 α−β,2 (t 󵄨󵄨 󵄨 󵄨 t

∞

∞

t

t

1−α 1 2−α 1 1 ≤ λ α−β ∫ exp(λ α−β (t − s))|g(s)|ds + λ α−β m1 (α, β, λ) ∫ exp(−λ α−β s)g(s)ds α−β

3.9 Planar fractional damped equations | 189

−α 1−α 1 λ α−β + λ α−β m1 (α, β, λ))‖g‖∞ α−β

≤(

≤ M(α, β, λ, M κ )‖g‖∞ . Like above, we obtain 󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨(t − s)α−1 E λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)󵄨󵄨󵄨󵄨 α−β,α ((t − s) 󵄨󵄨 󵄨 󵄨 2−α 1 m3 (α, β, λ) 1 ≤ + + m1 (α, β, λ)λ α−β exp(−λ α−β s) α−2β+1 −β+1 (t − s) λΓ(β)(t − s) for t − s ≥ 1 and t ≥ 1. Since s ≥ 0, it is enough to suppose t − s ≥ 1, because then t ≥ 1 + s ≥ 1. So we derive t−1

󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨 ∫ [(t − s)α−1 E λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨󵄨 α−β,α ((t − s) 󵄨󵄨 󵄨 󵄨 0

t−1

≤ ∫ 0

t−1

|g(s)| m3 (α, β, λ)|g(s)| ds + ∫ ds (t − s)α−2β+1 λΓ(β)(t − s)−β+1 0

t−1 2−α

1

+ m1 (α, β, λ)λ α−β ∫ exp(−λ α−β s)ds‖g‖∞ 0 (1−α) m3 (α, β, λ) Mκ ≤[ + m1 (α, β, λ)λ α−β ds]‖g‖∞ + ‖g‖κ . α − 2β λΓ(β)

On the other hand, t

󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨 ∫ [(t − s)α−1 E λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨󵄨 α−β,α ((t − s) 󵄨󵄨 󵄨 󵄨 t−1

t 2−α

1

≤ [E α−β,α+1 (λ) + ∫ λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)ds]‖g‖∞ . t−1

Furthermore, we obtain t

󵄨󵄨 󵄨󵄨 1 α−β 󵄨󵄨 ∫ λ 2−α α−β E λ)t exp(−λ α−β s)ds󵄨󵄨󵄨󵄨 α−β,2 (t 󵄨󵄨 󵄨 󵄨 t−1

≤

t

t

t−1

t−1

(2−α) 1 1 1−α 1 λ α−β ∫ exp(λ α−β (t − s))ds + m1 (α, β, λ)λ α−β ∫ exp(−λ α−β s)ds α−β

1 1−α −α 1 ≤ λ α−β exp(λ α−β ) + m1 (α, β, λ)λ α−β . α−β

190 | 3 Fractional Differential Equations

Consequently, we get (3.148): t

󵄨󵄨 󵄨󵄨 2−α 1 α−β 󵄨󵄨 ∫[(t − s)α−1 E λ) − λ α−β E α−β,2 (t α−β λ)t exp(−λ α−β s)]g(s)ds󵄨󵄨󵄨󵄨 α−β,α ((t − s) 󵄨󵄨 󵄨 󵄨 0

≤[

−α 1 1−α m3 (α, β, λ) 1 + E α−β,α+1 (λ) + λ α−β exp(λ α−β ) + 2m1 (α, β, λ)λ α−β ]‖g‖∞ α − 2β α−β Mκ ‖g‖κ + λΓ(β)

≤ M(α, β, λ, M κ )(‖g‖∞ + ‖g‖κ ). Proposition 3.135. Define C(α, β, λ1 , λ2 , κ) = (1 + ‖κ‖∞ ) max{K(α, β, λ1 ), M(α, β, λ1 , M κ ), 2K(α, β, −λ2 ), 2M(α, β, −λ2 , M κ )}, where K and M are the functions defined in Lemmas 3.133 and 3.134. For any ξ, ξ ̂ ∈ C∞ (ℝ+ , ℝ2 ), it holds ‖T(ξ) − T(ξ ̂ )‖∞ ≤ C(α, β, λ1 , λ2 , κ)l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))‖ξ − ξ ̂ ‖∞ and ‖T(ξ)‖∞ ≤ ‖x‖ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 | + C(α, β, λ1 , λ2 , κ)l f (‖ξ‖∞ )‖ξ‖∞ , where ‖ξ‖∞ = max(‖ξ1 ‖∞ , ‖ξ2 ‖∞ ). Proof. Note that t

󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨T1 (ξ)(t) − T1 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨󵄨 ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ1 )(f1 (ξ(s), s) − f1 (ξ ̂ (s), s))ds󵄨󵄨󵄨󵄨. 󵄨 󵄨 0

Using Lemmas 3.133 and 3.134, we have 󵄨 󵄨 sup 󵄨󵄨󵄨T1 (ξ)(t) − T1 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ K(α, β, λ1 )l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))‖κ‖∞ ‖ξ − ξ ̂ ‖∞

t∈[0,1]

and 󵄨 󵄨 sup󵄨󵄨󵄨T1 (ξ)(t) − T1 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ M(α, β, λ1 , M κ )l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))(1 + ‖κ‖∞ )‖ξ − ξ ̂ ‖∞ . t>1

So, 󵄨 󵄨 sup󵄨󵄨󵄨T1 (ξ)(t) − T1 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ C(α, β, λ1 , λ2 , κ)l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))‖ξ − ξ ̂ ‖∞ . t≥0

3.9 Planar fractional damped equations | 191

On the other hand, 󵄨 󵄨󵄨 󵄨󵄨T2 (ξ)(t) − T2 (ξ ̂ )(t)󵄨󵄨󵄨 ∞ 󵄨󵄨󵄨 2−α 1 = 󵄨󵄨󵄨󵄨−(−λ2 ) α−β E α−β,2 (−λ2 t α−β )t ∫ exp(−(−λ2 ) α−β s)(f2 (ξ(s), s) − f2 (ξ ̂ (s), s))ds 󵄨󵄨 0 󵄨󵄨 󵄨 + ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ2 )(f2 (ξ(s), s) − f2 (ξ ̂ (s), s))ds󵄨󵄨󵄨󵄨 󵄨󵄨 0 t

󵄨󵄨 t 󵄨 = 󵄨󵄨󵄨󵄨 ∫((t − s)α−1 E α−β,α (−(t − s)α−β λ2 ) 󵄨󵄨 0 − (−λ2 ) α−β E α−β,2 (−λ2 t α−β )t exp(−(−λ2 ) α−β s))(f2 (ξ(s), s) − f2 (ξ ̂ (s), s))ds 2−α

− (−λ2 )

2−α α−β

1

󵄨󵄨 1 󵄨 )t ∫ exp(−(−λ2 ) α−β s)(f2 (ξ(s), s) − f2 (ξ ̂ (s), s))ds󵄨󵄨󵄨󵄨. 󵄨󵄨 t ∞

E α−β,2 (−λ2 t

α−β

By Lemmas 3.133 and 3.134, we have 󵄨 󵄨 sup 󵄨󵄨󵄨T2 (ξ)(t) − T2 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ 2K(α, β, −λ2 )l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))‖κ‖∞ ‖ξ − ξ ̂ ‖∞

t∈[0,1]

and 󵄨 󵄨 sup󵄨󵄨󵄨T2 (ξ)(t) − T2 (ξ ̂ )(t)󵄨󵄨󵄨 t>1

≤ 2M(α, β, −λ2 , M κ )l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))(1 + ‖κ‖∞ )‖ξ − ξ ̂ ‖∞ . So, 󵄨 󵄨 sup󵄨󵄨󵄨T2 (ξ)(t) − T2 (ξ ̂ )(t)󵄨󵄨󵄨 ≤ C(α, β, λ1 , λ2 , κ)l f (max(‖ξ‖∞ , ‖ξ ̂ ‖∞ ))‖ξ − ξ ̂ ‖∞ . t≥0

Consequently, we get the first inequality of the proposition. Next, we derive 󵄨 󵄨 |T1 (0)(t)| ≤ |x1 | + 󵄨󵄨󵄨tE α−β,2 (−t α−β λ1 )󵄨󵄨󵄨 |x3 | ≤ ‖x‖ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 |, |T2 (0)(t)| ≤ ‖x‖. Hence we get the second inequality. Lemma 3.136. For any function g ∈ C∞ (ℝ+ , ℝ) and λ > 0, it holds u

lim

u→∞

∫0 (u − s)α−1 E α−β,α ((u − s)α−β λ)g(s)ds uE α−β,2

(λu α−β )

2−α

∞

0

Proof. According to (3.139), we obtain 1

1

= λ α−β ∫ exp(−λ α−β s)g(s)ds.

1 exp(λ α−β u) lim = (α − β)λ α−β . α−β ) u→∞ uE (λu α−β,2

192 | 3 Fractional Differential Equations Next, since for u > 1, u

󵄨󵄨 󵄨󵄨 α−β 󵄨󵄨 ∫ (u − s)α−1 E λ)g(s)ds󵄨󵄨󵄨󵄨 ≤ E α−β,α+1 (λ)‖g‖∞ , α−β,α ((u − s) 󵄨󵄨 󵄨 󵄨 u−1

we have

u

(u − s)α−1 E α−β,α ((u − s)α−β λ)

lim ∫

1

u→∞

g(s)ds = 0.

exp(λ α−β u)

u−1

Next we have 1−α

1

󵄨󵄨 u−1 (u − s)α−1 E α−β,α ((u − s)α−β λ) − 1 λ α−β exp(λ α−β (u − s)) 󵄨󵄨 󵄨 󵄨 α−β lim 󵄨󵄨󵄨󵄨 ∫ g(s)ds󵄨󵄨󵄨󵄨 1 u→∞ 󵄨 󵄨󵄨 󵄨 0 exp(λ α−β u) u−1

≤ lim ∫ u→∞

0

1 exp(λ

1 α−β

(u − s)β−1 m3 (α, β, λ) + )ds‖g‖∞ = 0, λΓ(β) (u − s)α−2β+1 u) (

since α − 2β > 0. So we get u−1

lim ∫

(u − s)α−1 E α−β,α ((u − s)α−β λ) 1

u→∞

g(s)ds

exp(λ α−β u)

0 u−1

= lim ∫ u→∞

0

1−α

1

λ α−β exp(λ α−β (u − s))

1−α

∞

1 λ α−β g(s)ds = ∫ exp(−λ α−β s)g(s)ds. 1 α − β (α − β) exp(λ α−β u) 0

The proof is complete. Let V ⊂ U ⊂ ℝ2 and W ⊂ ℝ2 be open neighborhoods of 0. Define a local center stable manifold W cs (V × W, U) = {(x, x) | x ∈ V, x ∈ W, φ(t, x, x) ∈ U for all t ≥ 0}. Proposition 3.137. (x, x) ∈ W cs (V × W, U) if and only if φ(⋅, x, x) is a fixed point of T along with φ(t, x, x) ∈ U for all t ≥ 0. Proof. If (x, x) ∈ W cs (V × W, U), then by (3.141), we get φ1 (t, x, x) = x1 + E α−β,2 (−t α−β λ1 )tx3 t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ1 )f1 (φ(s, x, x), s)ds, 0

φ2 (t, x, x) = x2 + E α−β,2 (−t α−β λ2 )tx4 t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ2 )f2 (φ(s, x, x), s)ds. 0

Since α − β > 1, by the above results we know that φ1 (⋅, x, x) = T1 (φ(⋅, x, x)).

3.9 Planar fractional damped equations | 193

Furthermore, limt→∞ E α−β,2 (−t α−β λ2 )t = ∞. Then using φ2 ∈ C∞ (ℝ+ , ℝ), we arrive at t

x4 = − lim

∫0 (t − s)α−1 E α−β,α (−(t − s)α−β λ2 )f2 (φ(s, x, x), s)ds E α−β,2 (−t α−β λ2 )t

t→∞ 2−α

∞

1

= −(−λ2 ) α−β ∫ exp(−(−λ2 ) α−β s)f2 (φ(s, x, x), s)ds 0

due to Lemma 3.136. Hence φ2 (t, x, x) = x2 − (−λ2 )

2−α α−β

∞

E α−β,2 (−t

α−β

1

λ2 )t ∫ exp(−(−λ2 ) α−β s)f2 (φ(s, x, x), s)ds 0

t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ2 )f2 (φ(s, x, x), s)ds 0

= T2 (φ2 (⋅, x, x)). Hence, φ(t, x, x) is a fixed point of T. Clearly, φ(t, x, x) ∈ U for all t ≥ 0. On the other hand, let ξ ∈ C∞ (ℝ+ , ℝ2 ) ∩ C1 (ℝ+ , ℝ2 ), (ξ(0), ξ 󸀠 (0)) ∈ V × W be a fixed point of T such that ξ(t) ∈ U for all t ≥ 0. Then t

ξ1 (t) = x1 + E α−β,2 (−t

α−β

λ1 )tx3 + ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ1 )f1 (ξ(s), s)ds, 0

ξ2 (t) = x2 − (−λ2 )

2−α α−β

∞

E α−β,2 (−t

α−β

1

λ2 )t ∫ exp ( − (−λ2 ) α−β s)f2 (ξ(s), s)ds 0

t

+ ∫(t − s)α−1 E α−β,α (−(t − s)α−β λ2 )f2 (ξ(s), s)ds. 0

Defining x4 = −(−λ2 )

2−α α−β

∞

1

∫ exp(−(−λ2 ) α−β s)f2 (ξ(s), s)ds, 0

we get t

ξ(t) = x + E α−β,2 (−t

α−β

A)tx + ∫(t − s)α−1 E α−β,α (−(t − s)α−β A)f(φ(s, x, x))ds 0

with ξ(0) = (x1 , x2 )T and ξ 󸀠 (0) = (x3 , x4 )T , which is a bounded solution of (3.140). Now we are ready to present the main result of this section.

194 | 3 Fractional Differential Equations Theorem 3.138. Take r∗ > 0 such that q = l f (r∗ )C(α, β, λ1 , λ2 , κ) < 1 and set r= r

∗∗

(1 − q)r∗ , 1 + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))

= (−λ2 )

2−α α−β

∞

1

l f (r )r ∫ exp(−(−λ2 ) α−β s)κ(s)ds. ∗

∗

0

Then for any x̃ = (x1 , x2 , x3 ) ∈ (−r, r)3 , there exists a unique s(x)̃ ∈ (−r∗∗ , r∗∗ ) such that ̃ ∈ W cs (V × W, U) with V = (−r, r)2 , W = (−r, r) × (−r∗∗ , r∗∗ ) and U = (−r∗ , r∗ )2 . (x,̃ s(x)) Furthermore, s : (−r, r)3 → (−r∗∗ , r∗∗ ) satisfies the following properties: (i) s(0) = 0; (ii) s is Lipschitz continuous: ̃ ≤ |s(x)̃ − s(y)|

r∗∗ (‖x − y‖∞ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 − y3 |) − q)

r∗ (1

for any x̃ = (x, x3 ), ỹ = (y, y3 ) ∈ (−r, r)3 . Proof. Define B r∗ (0) = {ξ ∈ C∞ (ℝ+ , ℝ2 ) | ‖ξ‖∞ ≤ r∗ }. By Proposition 3.135, we have ‖T(ξ) − T(ξ ̂ )‖∞ ≤ C(α, β, λ1 , λ2 , κ)l f (r∗ )‖ξ − ξ ̂ ‖∞ = q‖ξ − ξ ̂ ‖∞ , ‖T(ξ)‖∞ ≤ (1 + (E α−β,2 (λ1 ) + m2 (α, β, λ1 )))r + C(α, β, λ1 , λ2 , κ)l f (r∗ )r∗ = r∗ for any x̃ ∈ (−r, r)3 and ξ, ξ ̂ ∈ B r∗ (0). Since r∗ (1 − q) = r, we find ‖T(ξ)‖∞ ≤ r∗ , i.e., T : B r∗ (0) → B r∗ (0). The Banach fixed point theorem uniquely determines ξ x̃ by ξ x̃ = T(ξ x̃ ) with ‖ξ x̃ ‖∞ < r∗ . We set 2−α

∞

1

s(x)̃ = −(−λ2 ) α−β ∫ exp(−(−λ2 ) α−β s)f2 (ξ x̃ (s), s)ds.

(3.149)

0

Then 2−α

∞

1

̃ ≤ (−λ2 ) α−β l f (r∗ )r∗ ∫ exp(−(−λ2 ) α−β s)κ(s)ds = r∗∗ . |s(x)| 0

̃ ∈ W cs (V × W, U). Clearly, (i) holds. Furthermore, Proposition 3.137 implies (x,̃ s(x)) Next we have T = T x̃ , in particular ξ x̃ = T x̃ (ξ x̃ ). Then for any x̃ = (x, x3 ), ỹ = (y, y3 ) ∈ (−r, r)3 , it follows from Proposition 3.135 and the definition of T x̃ that ‖ξ x̃ − ξ ỹ ‖∞ = ‖T x̃ (ξ x̃ ) − T ỹ (ξ ỹ )‖∞ ≤ ‖T x̃ (ξ x̃ ) − T x̃ (ξ ỹ )‖∞ + ‖T x̃ (ξ ỹ ) − T ỹ (ξ ỹ )‖∞ ≤ q‖ξ x̃ − ξ ỹ )‖∞ + ‖T x−̃ ỹ (0)‖∞ ≤ q‖ξ x̃ − ξ ỹ ‖∞ + ‖x − y‖∞ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 − y3 |,

3.9 Planar fractional damped equations | 195

since T x̃ (ξ) − T ỹ (ξ) = T x−̃ ỹ (0) for any ξ ∈ C∞ (ℝ+ , ℝ2 ). This yields that ‖ξ x̃ − ξ ỹ ‖∞ ≤

1 (‖x − y‖∞ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 − y3 |). 1−q

Hence ∞

2−α 1 ̃ ≤ (−λ2 ) α−β ∫ exp(−(−λ2 ) α−β s)󵄨󵄨󵄨󵄨f2 (ξ x̃ (s), s) − f2 (ξ ỹ (s), s)󵄨󵄨󵄨󵄨ds |s(x)̃ − s(y)|

0

≤ (−λ2 )

2−α α−β

∞

1

l f (r ) ∫ exp(−(−λ2 ) α−β s)κ(s)ds‖ξ x̃ − ξ ỹ ‖∞ ∗

0

r∗∗ ≤ ∗ (‖x − y‖∞ + (E α−β,2 (λ1 ) + m2 (α, β, λ1 ))|x3 − y3 |), r (1 − q) which gives (ii). Remark 3.139. In [76, Theorem 10], Cong et al. extended the classical stable manifold results to planar fractional differential equations. In the above Theorem 3.138, we extend [76, Theorem 10] to the local center stable manifold case for planar fractional damped equations. Remark 3.140. In general, the local center manifold is defined for autonomous ODEs, and it contains all possible bounded solutions on ℝ near the equilibrium. For instance, the simple system of ODEs ẋ 1 = x1 , ẋ 2 = x32 has a local center manifold as a graph of a function x1 = 0. But this system has the only zero solution staying near (0, 0) on ℝ, even on [0, ∞). So the center manifold is not the set of bounded solutions, but possible bounded solutions near (0, 0). For FDEs the situation is much different, since a system of FDEs is not a dynamical system and solutions are defined only for t ≥ 0. For this reason, we are looking for all solutions of FDEs which are bounded on [0, ∞). In the classical ODEs, it is a center stable manifold. So we extend this notion to FDEs. Finally, an example is given to illustrate our theoretical result. Example 3.141. Consider the fractional damped equations motivated by [76], 3

3

c 2 { D t x1 (t) + c D t10 x1 (t) = 0, { { { { 3 { 3 x1 (t)2 c 2 D t x2 (t) − c D t10 x2 (t) = , { 2 { { 5 (1 + t) { { { T 󸀠 T {x(0) = x = (x1 , x2 ) , x (0) = x = (x3 , x4 ) .

Set α = 23 , β =

3 10

and κ(t) =

t

∫(t − s) 0

1 . (1+t)3/5

Clearly, α − β =

6 5

> 1. Next, we have

t β−1

7

2

1

κ(s)ds ≤ ∫(t − s)− 10 s− 5 ds = t− 10 B( 0

3 3 3 3 , ) ≤ B( , ) 10 5 10 5

(3.150)

196 | 3 Fractional Differential Equations for t ≥ 1, where B(⋅, ⋅) is the beta function, while t

t

∫(t − s)

β−1

7

κ(s)ds ≤ ∫(t − s)− 10 ds =

0

10 3 10 t 10 ≤ 3 3

0

for t ∈ (0, 1). Hence M κ ≤ max{

10 3 3 , B( , )} ≐ 4.16891. 3 10 5

Using (3.142), the solution φ(t, x, x) is given by 6

φ1 (t, x, x) = x1 + E 6 ,2 (−t 5 )tx3 , 5

t 1

6 5

6

φ2 (t, x, x) = x2 + E 6 ,2 (t )tx4 + ∫(t − s) 2 E 6 , 3 ((t − s) 5 ) 5

φ1 (s, x, x)2 2

(1 + s) 5

5 2

0

ds.

Hence, the Lyapunov–Perron operator T has the form 6

T1 (ξ)(t) = x1 + E 6 ,2 (−t 5 )tx3 , 5

∞

6

T2 (ξ)(t) = x2 − E 6 ,2 (t 5 )t ∫ exp(−s)

φ1 (s, x, x)2 2

(1 + s) 5

5

0 t 6

1

+ ∫(t − s) 2 E 6 , 3 (t − s) 5

ds

φ1 (s, x, x)2 2

(1 + s) 5

5 2

0

ds.

By (3.149) we derive ∞

s(x)̃ = − ∫ exp(−s)

φ1 (s, x, x)2 (1 + s)

0

2 5

∞

ds = − ∫ 0

exp(−s) (1 + s)

2 5

6

2

(x1 + E 6 ,2 (−s 5 )sx3 ) ds. 5

Consequently, Proposition 3.137 and Theorem 3.138 imply that the local center stable manifold around the origin is given by ∞

{(x1 , x2 , x3 , − ∫ 0

e−s (1 + s)

6

2 5

2

(x1 + E 6 ,2 (−s 5 )sx3 ) ds) | x1 , x2 , x3 are small} 5

= {(x1 , x2 , x3 , −0.7985x21 − 2.735x1 x3 − 1.2271x23 ) | x1 , x2 , x3 are small}, where we computed ∞

∫ 0

e−s ∞

= x21 ∫ 0

≐

6

2

(1 + s) 5

2

(x1 + E 6 ,2 (−s 5 )sx3 ) ds 5

∞

e−s (1 + s)

0.7985x21

2 5

ds + 2x1 x3 ∫

+ 2.735x1 x3 +

0

e−s (1 + s)

6

2 5

∞

E 6 ,2 (−s 5 )sds + x23 ∫ 5

0

e−s (1 + s)

6

2 5

E 6 ,2 (−s 5 )2 s2 ds 5

1.2271x23 .

The graph of a local center stable manifold around the origin is plotted in Figure 3.1.

3.10 Periodic fractional differential equations with impulses | 197

Fig. 3.1. Local center stable manifold of (3.150) around the origin for x1 , x3 ∈ [−1, 1], x4 = −0.7985x12 − 2.735x1 x3 − 1.2271x32 .

Remark 3.142. Consider again the simple system of ODEs: ẋ 1 = x1 , ẋ 2 = x32 , t ≥ 0. By taking ẋ 1 = x1 , ẋ 2 = κ(t)x32 , for “good” κ(t) > 0 the set x1 = 0 will really be the only set of starting bounded solutions: the solution is x1 (t) = 0,

x2 (t) =

x0 √1 −

t 2x20 ∫0

. κ(s)ds

∞

So we need ∫0 κ(s)ds < ∞, and the local (global) center stable manifold is {{0} × (−

1 ∞ √2 ∫0

κ(s)ds

,

1 ∞ √2 ∫0

)}.

κ(s)ds

Obviously, we can see that limt→∞ (x1 (t), x2 (t)) ≠ (0, 0) for x0 ≠ 0. Unfortunately, it is not easy to extend explicitly this example to FDEs.

3.10 Periodic fractional differential equations with impulses 3.10.1 Introduction It is well known that fractional differential equations (FDE) have no nonconstant periodic solutions [157]. Nevertheless, the existence of asymptotically periodic solutions was proved [80, 284], see also Section 3.1. This section is devoted to this problem. First we study impulsive FDEs (IFDEs) with Caputo derivatives when the lower limits are

198 | 3 Fractional Differential Equations

periodically varying at each impulse. We show that such kind of IFDE may possess periodic solutions. We present several particular cases of IFDEs along with non-existence results. For completeness of the section, we also discuss IFDEs when the lower limit of the Caputo derivative is fixed, which is also studied in Section 3.6. Then there is also a problem on the definition of solutions for such kind of IFDE, this is mentioned in [287]. Here we just study periodic boundary value conditions on finite intervals. Several concrete examples are given to illustrate the theoretical results. At the end, we briefly outline further possible ways for studies. This section is based on [111].

3.10.2 FDE with Caputo derivatives with varying lower limits 3.10.2.1 General impulses Let ℕ0 = {0, 1, 2, . . . }. We consider q

c D t ,t x(t) = f(t, x(t)), t ∈ (t k , t k+1 ), k ∈ ℕ0 , { { { k + x(t k ) = x(t−k ) + ∆ k (x(t−k )), k ∈ ℕ, { { { x(0) = x0 , {

(3.151)

q

where q ∈ (0, 1), and c D t k ,t x(t) is the generalized Caputo fractional derivative with lower limit at t k . We suppose the following: (i) f : ℝ × ℝm → ℝm is continuous and T-periodic in t. (ii) There are constants K > 0 and L k > 0 such that ‖f(t, x) − f(t, y)‖ ≤ K‖x − y‖ and ‖x + ∆ k (x) − y − ∆ k (y)‖ ≤ L k ‖x − y‖ for any t ∈ ℝ, x, y ∈ ℝm and each k ∈ ℕ. (iii) There is N ∈ ℕ such that T = t N+1 , t k+N+1 = t k + T and ∆ k+N+1 = ∆ k for any k ∈ ℕ. It is well known [162] that under assumptions (i) and (ii), (3.151) has a unique solution on ℝ+ . So we can consider the Poincaré mapping P(x0 ) = x(T − ) + ∆ N+1 (x(T − )). Clearly, fixed points of P determine T-periodic solutions of (3.151) (see [109, Lemma 2.2]). Lemma 3.143. Under assumptions (i) and (ii), it holds ‖P(x) − P(y)‖ ≤ Θ‖x − y‖

for all x, y ∈ ℝm

for Θ = ∏Nk=0 L k+1 E q (K(t k+1 − t k )q ), where E q is the Mittag-Leffler function. Proof. On each of the intervals (t k , t k+1 ), k = 0, 1, . . . , N, equation (3.151) is equivalent to t

x(t) =

x(t+k )

1 + ∫(t − s)q−1 f(s, x(s))ds. Γ(q) tk

(3.152)

3.10 Periodic fractional differential equations with impulses | 199

Consider two solutions x and y of (3.151) with x(0) = x0 and y(0) = y0 , respectively. Then by (3.152), we get t

‖x(t) − y(t)‖ ≤ ‖x(t+k ) − y(t+k )‖ +

1 ∫(t − s)q−1 ‖f(s, x(s)) − f(s, y(s))‖ds Γ(q) tk

t

≤

‖x(t+k )

−

y(t+k )‖

K + ∫(t − s)q−1 ‖x(s) − y(s)‖ds. Γ(q)

(3.153)

tk

Applying Gronwall’s fractional inequality of Lemma A.18 to (3.153), we obtain ‖x(t) − y(t)‖ ≤ ‖x(t+k ) − y(t+k )‖E q (K(t − t k )q ),

t ∈ (t k , t k+1 ).

(3.154)

Then using (3.154) with (ii), we get ‖x(t+k+1 ) − y(t+k+1 )‖ ≤ L k+1 E q (K(t k+1 − t k )q )‖x(t+k ) − y(t+k )‖,

k = 0, 1, . . . , N,

which implies N

‖P(x0 ) − P(y0 )‖ ≤ ∏ L k+1 E q (K(t k+1 − t k )q )‖x0 − y0 ‖.

(3.155)

k=0

The proof is finished. Theorem 3.144. Suppose that (i)–(iii) are satisfied. If Θ < 1, then (3.151) has a unique T-periodic solution, which is in addition asymptotically stable. Proof. Inequality (3.155) implies that P : ℝm → ℝm is a contraction. So, applying the Banach fixed point theorem shows that P has a unique fixed point x0 . Moreover, since ‖P n (x0 ) − P n (y0 )‖ ≤ Θ n ‖x0 − y0 ‖ for any y0 ∈ ℝm , we see that the corresponding periodic solution is asymptotically stable. 3.10.2.2 Systems with weak nonlinearities Now we consider (3.151) with small nonlinearities of the form q

c D t ,t x(t) = εf(t, x(t)), t ∈ (t k , t k+1 ), k ∈ ℕ0 , { { { k + x(t k ) = x(t−k ) + ε∆ k (x(t−k )), k ∈ ℕ, { { { x(0) = x0 , {

(3.156)

where ε is a small parameter. Then (3.156) has a unique solution x(ε, t) and the Poincaré mapping is given by P(ε, x0 ) = x(ε, T − ) + ε∆ N+1 (x(ε, T − )). We suppose that (C1) f and ∆ k are C2 -smooth.

200 | 3 Fractional Differential Equations Then P(ε, x0 ) is C2 -smooth as well. Moreover, we have x(ε, t) = x0 + εv(t) + O(ε2 ), and then {

P(ε, x0 ) = x0 + εM(x0 ) + O(ε2 ), M(x0 ) = v(T − ) + ∆ N+1 (x0 ).

(3.157)

Note that v(t) solves q

c D t ,t v(t) = f(t, x0 ), t ∈ (t k , t k+1 ), k = 0, 1, . . . , N, { { { k + v(t k ) = v(t−k ) + ∆ k (x0 ), k = 1, 2, . . . , N + 1, { { { v(0) = 0. {

So we easily derive t k+1

1 N v(T ) = ∑ ∆ k (x0 ) + ∑ ∫ (t k+1 − s)q−1 f(s, x0 )ds. Γ(q) k=0 k=1 N

−

tk

Consequently, (3.157) gives N+1

M(x0 ) = ∑ ∆ k (x0 ) + k=1

t k+1

1 N ∑ ∫ (t k+1 − s)q−1 f(s, x0 )ds. Γ(q) k=0 tk

Now we are ready to prove the following theorem. Theorem 3.145. Suppose assumptions (i) and (C1). If there is a simple zero x0 ∈ ℝm of M, i.e., M(x0 ) = 0 and det DM(x0 ) ≠ 0, then (3.156) has a unique T-periodic solution located near x0 for any ε ≠ 0 small. Moreover, if ℜσ(DM(x0 )) < 0, then it is asymptotically stable, and if ℜσ(DM(x0 )) ∩ (0, ∞) ≠ 0, then it is unstable. If M(x0 ) ≠ 0 for any x0 ∈ ℝm , then (3.156) has no T-periodic solution for ε ≠ 0 small. Proof. To find a T-periodic solution of (3.156), we need to solve P(ε, x0 ) = x0 , which by (3.157) is equivalent to M(x0 ) + O(ε) = 0. (3.158) If there is a simple zero x0 of M, then (3.158) can be solved by the implicit function theorem to get its solution x0 (ε) with x0 (0) = x0 . Moreover, DP(ε, x0 (ε)) = 1 + εDM(x0 ) + O(ε2 ). Hence, the stability and instability results follow directly by the known arguments (see [201]), so we omit the details.

3.10 Periodic fractional differential equations with impulses | 201

3.10.2.3 Systems with equidistant and periodically shifted impulses We consider q

c D x(t) = f(t, x(t)), t ∈ (kh, (k + 1)h), k ∈ ℕ0 , { { { kh + x(kh ) = x(kh− ) + ∆ k , k ∈ ℕ, { { { x(0) = x0 , {

(3.159)

where h > 0, q ∈ (0, 1), and the norm ‖x‖ = maxi=1,...,m |x i | for x = (x1 , . . . , x m ) ∈ ℝm . We suppose the following: (I) f : ℝ × ℝm → ℝm is continuous, locally Lipschitz in x and T-periodic in t, where T = (N + 1)h for some N ∈ ℕ. (II) There is a constant M > 0 such that ‖f(t, x)‖ ≤ M for any t, x ∈ ℝ. (III) ∆ k ∈ ℝm satisfies ∆ k+N+1 = ∆ k for any k ∈ ℕ. We are looking for T-periodic solutions of (3.159) on ℝ+ . It is well known [162] that under the above assumptions, (3.159) has a unique solution x(x0 , t) on ℝ+ , which is continuous in x0 ∈ ℝm , t ∈ ℝ+ \ {kh | k ∈ ℕ} and left-continuous in t at impulse points {kh | k ∈ ℕ}. So we can consider the Poincaré mapping P(x0 ) = x(x0 , T + ). Clearly, fixed points of P determine T-periodic solutions of (3.159). Lemma 3.146. Under assumptions (I)–(III), it holds (k+1)h

r(N+1)

P r (x0 ) = x0 + ∑ ∆ k + k=1

1 rN ∑ ∫ ((k + 1)h − s)q−1 f(s, x(x0 , s))ds Γ(q) k=0

(3.160)

kh

and

N

‖x(x0 , t) − x0 ‖ ≤ ∑ ‖∆ k ‖ + k=1

M(N + 1)h q Γ(q + 1)

(3.161)

for t ∈ I = [0, T] and r ∈ ℕ, where P r is the r-th iteration of P. Proof. On each of the intervals (kh, (k + 1)h), k ∈ ℕ0 , equation (3.159) is equivalent to t

x(x0 , t) = x(kh+ ) +

1 ∫ (t − s)q−1 f(s, x(x0 , s))ds, Γ(q) kh

which implies (k+1)h

n

x(x0 , t) = x0 + ∑ ∆ k + k=1

kh

t

+

1 n−1 ∑ ∫ ((k + 1)h − s)q−1 f(s, x(x0 , s))ds Γ(q) k=0

1 ∫ (t − s)q−1 f(s, x(x0 , s))ds Γ(q) nh

202 | 3 Fractional Differential Equations for t ∈ (nh, (n + 1)h). This implies (3.160), since P r (x0 ) = x(x0 , rT + ). Moreover, we have (k+1)h

N

‖x(x0 , t) − x0 ‖ ≤ ∑ ‖∆ k ‖ + k=1

M N ∑ ∫ ((k + 1)h − s)q−1 ds Γ(q) k=0 kh

N

= ∑ ‖∆ k ‖ + k=1

M(N + 1)h q Γ(q + 1)

for t ∈ I. This implies (3.161). Now we can prove the following. Theorem 3.147. Suppose assumptions (I)–(III). If 󵄩󵄩 N+1 󵄩󵄩 󵄩󵄩∑k=1 ∆ k 󵄩󵄩 Mh q > , N+1 Γ(q + 1)

(3.162)

then (3.159) has no rT-periodic solution for any r ∈ ℕ. Proof. We need to solve P r (x0 ) = x0 , which is equivalent to (k+1)h

r(N+1)

1 rN − ∑ ∆k = ∑ ∫ ((k + 1)h − s)q−1 f(s, x(x0 , s))ds. Γ(q) k=1 k=0

(3.163)

kh

This gives (k+1)h

󵄩󵄩N+1 󵄩󵄩 Mr(N + 1)h q 1 rN r󵄩󵄩󵄩󵄩 ∑ ∆ k 󵄩󵄩󵄩󵄩 ≤ , ∑ ∫ ((k + 1)h − s)q−1 |f(s, x(x0 , s))|ds ≤ Γ(q + 1) 󵄩 k=1 󵄩 Γ(q) k=0 kh

which contradicts (3.162). Example 3.148. We consider the Lorenz system f(x1 , x2 , x3 ) = (a(x2 − x1 ), x1 (b − x3 ) − x2 , x1 x2 − cx3 )

(3.164)

which for q = 0.995, a = 10, b = 120, c = 83 and without impulses presents a chaotic attractor (see Figure 3.2). The utilized numerical scheme to integrate the Lorenz system is the predictor-corrector Adams–Bashforth–Moulton method for FDEs [92]. Take N = 1 and t k = kh, k ∈ ℕ0 . Let C = {(x1 , x2 , x3 ) ∈ ℝ3 | −50 ≤ x1 ≤ 50, −100 ≤ x2 ≤ 100, 50 ≤ x3 ≤ 200} which embeds the Lorenz attractor. To apply Theorem 3.147, we need to find M = max‖f(x)‖ x∈C

= max max{|a(x2 − x1 )|, |x1 (b − x3 ) − x2 |, |x1 x2 − cx3 |} ≐ 533.333. x∈C

3.10 Periodic fractional differential equations with impulses | 203

200

x3

150

100

100

50 −50

0

0 50

−100

x2 x1

Fig. 3.2. Chaotic attractor of (3.164).

Let ∆ k = (∆ k1 , ∆ k2 , ∆ k3 ). For ∆11 = ∆21 = ∆1 , ∆12 = 0, ∆22 = ∆2 and ∆31 = ∆32 = ∆3 , condition (3.162) has the form max{|∆1 |,

|∆2 | Mh q , ∆3 |} > ≐ 2.19767 2 Γ(q + 1)

for h = 0.004. Note that T = (N + 1)h = 0.008. Summarizing, we obtain the following result. Proposition 3.149. Consider (3.159) for N = 1, h = 0.004, q = 0.995 and with (3.164) for a = 10, b = 120 and c = 38 . Let ∆11 = ∆21 = ∆1 , ∆12 = 0, ∆22 = ∆2 and ∆31 = ∆32 = ∆3 . If either |∆1 | > 2.19767 or |∆2 | > 4.39535 or |∆3 | > 2.19767, then our system has no 0.008r periodic solution in C for any r ∈ ℕ. Now we present existence results. Let f(t, x) = (f1 (t, x), . . . , f m (t, x)),

∆ k = (∆ k1 , . . . , ∆ km ),

and for each i = 1, . . . , m set x̂ i = (x1 , . . . , x i−1 , x i+1 , . . . , x m ) ∈ ℝm−1 . Theorem 3.150. Suppose (I)–(III) and in addition that there are constants f±i ∈ ℝ, i = 1, . . . , m such that either (C1) there holds lim inf f i (t, x) ≥ f+i > f−i ≥ lim sup f i (t, x) x i →∞

uniformly with respect to t ∈ I and x̂ i ∈

x i →−∞

ℝm−1

for all i = 1, . . . , m, or

204 | 3 Fractional Differential Equations

(C2) there holds lim sup f i (t, x) ≤ f+i < f−i ≤ lim inf f i (t, x) x i →−∞

x i →∞

uniformly with respect to t ∈ I and x̂ i ∈ ℝm−1 for all i = 1, . . . , m. If −

N+1 ∆ ki ∑k=1 hq ∈ (f i , f i ) or N+1 Γ(q + 1) − +

−

∑N+1 hq k=1 ∆ ki ∈ (f i , f i ), N+1 Γ(q + 1) + −

(3.165)

for all i = 1, . . . , m, then (3.159) has a T-periodic solution. Proof. We need to solve (3.163). For simplicity, we set (k+1)h

1 N F(x0 ) = ∑ ∫ ((k + 1)h − s)q−1 f(s, x(x0 , s))ds. Γ(q) k=0 kh

So we need to solve

N+1

− ∑ ∆ k = F(x0 ).

(3.166)

k=1

Let F(x0 ) = (F1 (x0 ), . . . , F m (x0 )). By (3.161), we have lim x i (x0 , t) = ±∞

x0i →±∞

uniformly with respect to t ∈ I and x̂ 0i ∈ ℝm−1 , where we denote x0 = (x01 , . . . , x0m ) and x(x0 , t) = (x1 (x0 , t), . . . , x m (x0 , t)). So, by the first possibility (C1), for any ε > 0 there is m ε > 0 such that f i (s, x(x0 , s)) > f+i − ε

for all x0i > m ε ,

f i (s, x(x0 , s))
0 sufficiently small, (3.165) implies N+1

H i (λ, x0 ) ≠ − ∑ ∆ ki

(3.167)

k=1

for any x0i = ±m ε , x̂ 0i ∈ ℝm−1 , ‖x̂ 0i ‖ ≤ m ε . Similarly we verify that (3.167) holds also in the second possibility (C2). Consequently, we derive N+1

H i (λ, x0 ) ≠ − ∑ ∆ ki k=1

for any λ ∈ [0, 1] and x0 ∈ ∂Ω, where ∂Ω is the border of Ω = {x0 ∈ ℝm | ‖x0 ‖ ≤ m ε }. This gives N+1

N+1

deg(F, Ω, − ∑ ∆ k ) = deg(G, Ω, − ∑ ∆ k ), k=1

k=1

where deg is the Brouwer topological degree. On the other hand, a linear equation ̄ G(x0 ) = − ∑N+1 k=1 ∆ k has the only solution x 0 , which by (3.165) is located in Ω. Moreover, ̄ the linearization DG(x0 ) is a nonsingular matrix, so deg(G, Ω, − ∑N+1 k=1 ∆ k ) = ±1 and thus deg(F, Ω, − ∑N+1 ∆ ) = ±1 = ̸ 0. Summarizing, we see that (3.166) has a solution k k=1 in Ω, i.e., (3.163) is solvable. Following the proof of Theorem 3.150, we have a modified result. Theorem 3.151. Suppose (I)–(III) and in addition that there are constants f±i ∈ ℝ, i = 1, . . . , m and a permutation σ : {1, . . . , m} → {1, . . . , m} such that (C3) either σ(i) σ(i) lim inf f σ(i) (t, x) > f+ > f− > lim sup f σ(i) (t, x) x i →∞

x i →∞

or σ(i)

lim sup f σ(i) (t, x) < f+ x i →∞

σ(i)

< f−

< lim inf f σ(i) (t, x) x i →∞

uniformly with respect to t ∈ I and x̂ i ∈ ℝm−1 for all i = 1, . . . , m. If (3.165) holds for all i = 1, . . . , m, then (3.159) has a T-periodic solution. Clearly, either (C1) or (C2) implies (C3). Of course, assumptions (3.162) and (3.165) are complementary, since (f−i , f+i ) ⊂ (−M, M) or

(f+i , f−i ) ⊂ (−M, M)

for all i = 1, . . . , m. Now we present concrete examples. Example 3.152. Consider the system m

f i (t, x) = A i tanh x i + B i cos t cos x i sin ( ∑ x k ) k=1

206 | 3 Fractional Differential Equations for A i > B i > 0 and i = 1, . . . , m. Take f+i = A i − B i ,

f−i = −A i + B i ,

M = max {A i + B i }, i=1,...,m

T = 2π,

h=

2π . N+1

Hence, assumption (3.162) has the form 󵄩󵄩 N+1 󵄩󵄩 󵄩󵄩∑k=1 ∆ k 󵄩󵄩 maxi=1,...,m {A i + B i }h q > , N+1 Γ(q + 1) and assumption (3.165) has the form −

∑N+1 hq k=1 ∆ ki ∈ (−A i + B i , A i − B i ) N+1 Γ(q + 1)

for all i = 1, . . . , m. Consequently, Theorem 3.150 can be applied. 3.10.2.4 Systems with small equidistant and shifted impulses We consider c q D x(t) = f(x(t)), t ∈ (kh, (k + 1)h), k ∈ ℕ0 , { { { kh + x(kh ) = x(kh− ) + ∆h q , k ∈ ℕ, { { { x(0) = x0 , {

(3.168)

where h > 0, q ∈ (0, 1), ∆ ∈ ℝm and f : ℝm → ℝm is locally Lipschitz. Under the above assumptions [162], (3.168) has a unique solution x(x0 , t) on ℝ+ , which is continuous in x0 ∈ ℝm , t ∈ ℝ+ \ {kh | k ∈ ℕ} and left-continuous in t at impulse points {kh | k ∈ ℕ}. So we can consider the Poincaré mapping P h (x0 ) = x(x0 , h+ ). On each of the intervals (kh, (k + 1)h), k ∈ ℕ0 , equation (3.168) is equivalent to t

1 x(x0 , t) = x(kh ) + ∫ (t − s)q−1 f(x(x0 , s))ds Γ(q) +

kh t−kh

1 = x(kh ) + ∫ (t − kh − s)q−1 f(x(x(kh+ ), s))ds. Γ(q) +

(3.169)

0

Hence x((k + 1)h+ ) = P h (x(kh+ ))

(3.170)

and h

P h (x0 ) = x(x0 , h+ ) = x0 + ∆h q +

1 ∫(h − s)q−1 f(x(x0 , s))ds. Γ(q) 0

(3.171)

3.10 Periodic fractional differential equations with impulses | 207

Inserting x(x0 , t) = x0 + h q y(x0 , t),

t ∈ [0, h]

into (3.169), we get t

y(x0 , t) =

1 ∫(t − s)q−1 f(x0 + h q y(x0 , s))ds Γ(q)h q 0 t

1 1 = f(x0 ) + ∫(t − s)q−1 (f(x0 + h q y(x0 , s)) − f(x0 ))ds Γ(q + 1) Γ(q)h q 0

1 = f(x0 ) + O(h q ), Γ(q + 1) since t

t

0

0

󵄩󵄩 󵄩 󵄩󵄩∫(t − s)q−1 (f(x + h q y(x , s)) − f(x ))ds󵄩󵄩󵄩 ≤ ∫(t − s)q−1 󵄩󵄩󵄩f(x + h q y(x , s)) − f(x )󵄩󵄩󵄩ds 0 0 0 0 0 󵄩 󵄩󵄩 󵄩󵄩 󵄩 0 󵄩 󵄩 ≤ h q max ‖y(x0 , t)‖ Lloc t∈[0,h]

≤ h2q max ‖y(x0 , t)‖ t∈[0,h]

tq q

Lloc , q

where Lloc is a local Lipschitz constant of f . Then x(x0 , t) = x0 +

hq f(x0 ) + O(h2q ), Γ(q + 1)

t ∈ [0, h],

and (3.171) gives P h (x0 ) = x0 + ∆h q +

hq f(x0 ) + O(h2q ). Γ(q + 1)

Hence (3.170) becomes x((k + 1)h+ ) = x(kh+ ) + h q (∆ +

1 f(x(kh+ ))) + O(h2q ). Γ(q + 1)

(3.172)

We note that (3.169) implies ‖x(x0 , t) − x(kh+ )‖ = O(h q )

(3.173)

for t ∈ [kh, (k + 1)h]. We see that (3.172) leads to its approximation z((k + 1)h+ ) = z(kh+ ) + h q (∆ +

1 f(z(kh+ ))), Γ(q + 1)

which is the Euler numerical approximation of z󸀠 (t) = ∆ +

1 f(z(t)) Γ(q + 1)

(3.174)

with the step size h q . Using (3.173) and known results on the Euler approximation method [115], we arrive at the following result.

208 | 3 Fractional Differential Equations Theorem 3.153. Let z(t) be a solution of (3.174) with z(0) = x0 on [0, T]. Then the solution x(x0 , t) of (3.168) exists on [0, T] and satisfies x(x0 , t) = z(th q−1 ) + O(h q ) for t ∈ [0, Th1−q ]. If z(t) is a stable (hyperbolic) periodic solution of (3.174), then there is a stable (hyperbolic) invariant curve of Poincaré mapping P h of (3.168) in an O(h q ) neighborhood of z(t). Of course, h is sufficiently small. Now we extend (3.168) to periodic impulses of the form q

c D x(t) = f(x(t)), t ∈ (kh, (k + 1)h), k ∈ ℕ0 , { { { kh + x(kh ) = x(kh− ) + ∆ k h q , k ∈ ℕ, { { { x(0) = x0 , {

(3.175)

where ∆ k ∈ ℝm satisfy ∆ k+N+1 = ∆ k for each k ∈ ℕ and some N ∈ ℕ. So we can consider the Poincaré mapping P h (x0 ) = x(x0 , (N + 1)h+ ), which has the form P h = P N+1,h ∘ ⋅ ⋅ ⋅ ∘ P1,h

(3.176)

for P k,h (x0 ) = ∆ k h q + x(x0 , h). We already know that P k,h (x0 ) = ∆ k h q + x(x0 , h) = x0 + ∆ k h q +

hq f(x0 ) + O(h2q ). Γ(q + 1)

Then (3.176) implies N+1

P h (x0 ) = x0 + h q ∑ ∆ k + k=1

(N + 1)h q f(x0 ) + O(h2q ). Γ(q + 1)

On the other hand, the ODE z󸀠 (t) =

∑N+1 1 k=1 ∆ k + f(z(t)) N+1 Γ(q + 1)

(3.177)

has the Euler numerical approximation x0 + h q (

∑N+1 1 k=1 ∆ k + f(x0 )) N+1 Γ(q + 1)

with step size h q , and its (N + 1)-st iteration has the form N+1

x0 + h q ∑ ∆ k + k=1

(N + 1)h q f(x0 ) + O(h2q ), Γ(q + 1)

the same one as of P h . Hence instead of (3.174) we have (3.177), and Theorem 3.153 is directly extended to (3.175) with (3.177).

3.10 Periodic fractional differential equations with impulses | 209

3.10.3 FDE with Caputo derivatives with fixed lower limits Let T = (N + 1)h for an N ∈ ℕ and h > 0. We consider q

c D0 x(t) = f(t, x(t)), t ∈ [0, T], { { { x(kh+ ) = x(kh− ) + ∆ k , k ∈ {1, . . . , N}, { { { { x(0) = x0 ,

(3.178)

q

where q ∈ (0, 1), c D0 x(t) is the generalized Caputo fractional derivative with lower limit 0. We suppose the following: (ci) f : [0, T] × ℝm → ℝm is continuous and locally Lipschitz in x uniformly for t ∈ [0, T]. (cii) There is a constant M > 0 such that |f(t, x)| ≤ M for any t ∈ [0, T] and x ∈ ℝm . We are looking for a solution of (3.178) satisfying x(0) = x(T). Following [284] (see also Section 3.6), by a solution of (3.178) we mean a function x(t) = x(x0 , t) which is continuous in x0 ∈ ℝm , t ∈ [0, T] \ {kh | k = 1, . . . , N} and left-continuous in t at impulse points kh, and satisfies t

k

1 x(t) = x0 + ∑ ∆ i + ∫(t − s)q−1 f(s, x(s))ds, Γ(q) i=1

t ∈ (t k , t k+1 ].

(3.179)

0

It is known by Section 3.6 that under the above assumptions, (3.179) has a unique solution x(x0 , t) on [0, T]. Theorem 3.154. Suppose assumptions (ci)–(cii). If 󵄩󵄩 N 󵄩 󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩 Mh q > , q (N + 1) Γ(q + 1)

(3.180)

then (3.178) has no solution satisfying x(0) = x(T). Note that T = (N + 1)h for some N ∈ ℕ. Proof. We need to solve N

T

1 x(0) = x(T) = x0 + ∑ ∆ i + ∫(T − s)q−1 f(s, x(s))ds, Γ(q) i=1 0

which is equivalent to N

1 − ∑ ∆k = Γ(q) k=1

(N+1)h

∫ ((N + 1)h − s)q−1 f(s, x(x0 , s))ds. 0

This gives 󵄩󵄩 N 󵄩 󵄩󵄩 ∑ ∆ 󵄩󵄩󵄩 ≤ 1 k󵄩 󵄩󵄩 󵄩k=1 󵄩󵄩 Γ(q)

(N+1)h

∫ ((N + 1)h − s)q−1 |f(s, x(x0 , s))|ds ≤ 0

which contradicts (3.180).

M(N + 1)q h q , Γ(q + 1)

210 | 3 Fractional Differential Equations

Remark 3.155. Note that (3.180) is equivalent to 󵄩 󵄩󵄩 N 󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩 M > . Tq Γ(q + 1)

(3.181)

Now we suppose that (ciii) ∆ k+p = ∆ k for all k ∈ ℕ and some p ∈ ℕ. Corollary 3.156. Suppose assumptions (ci)–(ciii). Then: (a) If 󵄩󵄩 p 󵄩󵄩 q 󵄩󵄩∑k=1 ∆ k 󵄩󵄩 > Mh , q (p + 1) Γ(q + 1)

(3.182)

then (3.178) has no solution satisfying x(0) = x(T) with T = (rp + 1)h for any r ∈ ℕ. (b) If p

∑ ∆ k ≠ 0,

(3.183)

k=1

then there is no solution of (3.178) satisfying x(0) = x(T) with T = (rp + 1)h for any r ∈ ℕ such that M(p + 1)q h q 1−q r > √󵄩 p . (3.184) 󵄩 󵄩󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩Γ(q + 1) Proof. We need to verify (3.180) for N = rp. First, we note that 󵄩 󵄩󵄩 p 󵄩󵄩 󵄩󵄩 N 󵄩 󵄩 rp 󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩 󵄩󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩 󵄩∑k=1 ∆ k 󵄩󵄩 , 1−q 󵄩 = ≥ r (N + 1)q (rp + 1)q (p + 1)q

(3.185)

since r ≥ 1. If (3.182) holds, then (3.185) implies (3.180). This proves (a). If (3.183) and (3.184) hold, then (3.185) implies (3.180). This proves (b). Of course, (b) implies (a). Next, supposing that h > 0 is a small variable, we consider q

c D0 x(t) = f(x(t)), t ∈ (kh, (k + 1)h), k ∈ ℕ0 , { { { x(kh+ ) = x(kh− ) + ∆ k h q , k ∈ ℕ, { { { { x(0) = x0 ,

where q ∈ (0, 1), f : ℝm → ℝm satisfies (ci), (cii) and ∆ k ∈ ℝm satisfy (civ) ∆ k+p = ∆ k for all k ∈ ℕ and some p ∈ ℕ. Then condition (3.182) becomes 󵄩󵄩 p 󵄩 󵄩󵄩∑k=1 ∆ k 󵄩󵄩󵄩 M > , (p + 1)q Γ(q + 1) condition (3.183)

p

∑ ∆ k ≠ 0, k=1

(3.186)

3.10 Periodic fractional differential equations with impulses | 211

and (3.184) r>

M(p + 1)q . √󵄩 p 󵄩󵄩 󵄩󵄩∑ 󵄩 k=1 ∆ k 󵄩󵄩Γ(q + 1)

1−q

(3.187)

Summarizing, we have the following result. Corollary 3.157. Under assumptions (ci), (cii) and (civ), there is no solution of (3.186) satisfying x(0) = x(T) for any T = (rp + 1)h with r ∈ ℕ satisfying (3.187). Corollary 3.157 states that under assumptions (ci), (cii) and (civ), (3.186) has no “periodic” solutions with large periods. Example 3.158. We consider system (3.164) for a = 10, b = 120 and c = 83 . Motivated by [175, Theorem 2], for the sphere S = {(x1 , x2 , x3 ) ∈ ℝ3 | x21 + x22 + (x3 − a − b)2 ≤ R2 } with R=

(a + b)c = 134.263, √4(c − 1)

we find M = max‖f(x)‖ x∈S

= max max{|a(x2 − x1 )|, |x1 (b − x3 ) − x2 |, |x1 x2 − cx3 |} ≐ 8077.26. x∈S2

By Corollary 3.156, for x1 , ∆1 applied at 3kh, for x2 , ∆2 applied at kh, for x3 , ∆3 applied at 2kh, condition (3.182) has the form max{|∆1 |, 3|∆2 |, |∆3 |} > (p + 1)q

Mh q ≐ 132.214 Γ(q + 1)

for h = 0.004 and q = 0.995. For the cuboid C = {(x1 , x2 , x3 ) ∈ ℝ3 | −50 ≤ x1 ≤ 50, −100 ≤ x2 ≤ 100, 50 ≤ x3 ≤ 200} from Example 3.148, we have M = max‖f(x)‖ ≐ 533.333. x∈C

Now condition (3.182) has the form max{|∆1 |, 3|∆2 |, |∆3 |} > (p + 1)q again for h = 0.004 and q = 0.995.

Mh q ≐ 8.72997 Γ(q + 1)

212 | 3 Fractional Differential Equations Motivated by Section 3.10.2.4, we consider (3.178) with ∆ k = h∆ and c

∆ t1−q + f(t, y(t)), Γ(2 − q)

q

D0 y(t) =

y(0) = x0 .

(3.188)

Then we have t

1 ∫(t − s)q−1 f(s, x(s))ds, Γ(q)

x(t) = x0 + kh∆ +

t ∈ (kh, (k + 1)h),

0 t

y(t) = x0 +

1 ∆ s1−q + f(s, y(s)))ds ∫(t − s)q−1 ( Γ(q) Γ(2 − q) 0

t

B(q, 2 − q)∆ 1 = x0 + t+ ∫(t − s)q−1 f(s, y(s))ds Γ(q)Γ(2 − q) Γ(q) 0

t

= x0 + ∆t +

1 ∫(t − s)q−1 f(s, y(s))ds, Γ(q) 0

which implies t

1 󵄩 󵄩 ‖x(t) − y(t)‖ ≤ (t − kh)∆ + ∫(t − s)q−1 󵄩󵄩󵄩f(s, x(s)) − f(s, y(s))󵄩󵄩󵄩ds Γ(q) 0 t

≤ h∆ +

L ∫(t − s)q−1 ‖x(s) − y(s)‖ds, Γ(q)

t ∈ (kh, (k + 1)h),

(3.189)

0

where L is the Lipschitz constant from assumption (ci). Applying the generalized Gronwall’s inequality (see Remark A.19) to (3.189), we get ‖x(t) − y(t)‖ ≤ h∆E q (Lt q ),

t ∈ I.

Consequently, we obtain the following statement. Theorem 3.159. The solution x(t) of the impulsive FDE (3.178) with ∆ k = h∆ is approximated by the solution y(t) of the FDE (3.188) without impulses by the order O(h∆). Finally, for the completeness of the section, we present the following existence result, whose proof is analogous to the proof of Theorem 3.151, so we omit it. Theorem 3.160. Suppose (ci), (cii) and (C3). If −

∑Nk=1 ∆ ki 1 ∈ (f i , f i ) or q T Γ(q + 1) − +

−

∑Nk=1 ∆ ki 1 ∈ (f i , f i ) q T Γ(q + 1) + −

(3.190)

holds for all i = 1, . . . , m, then (3.186) has a solution satisfying x(0) = x(T). Again, assumptions (3.181) and (3.190) are complementary. More related achievements are given in [287] and the references therein.

3.10 Periodic fractional differential equations with impulses | 213

3.10.4 Conclusions To conclude this section, we roughly outline a possible further progress on this topic. Firstly, we can consider a non-homogeneous linear case q

c D t ,t x(t) = Ax(t) + f(t), t ∈ (t k , t k+1 ), k = 0, 1, . . . , N, { { { k + x(t k ) = (𝕀 + B k )x(t−k ) + ∆ k , k = 1, 2, . . . , N + 1, { { { x(0) = x0 , {

(3.191)

where A, B k : ℝm → ℝm are matrices, ∆ k ∈ ℝ, and then consider a semilinear case q

c D t ,t x(t) = Ax(t) + f(t, x(t)), t ∈ (t k , t k+1 ), k = 0, 1, . . . , N, { { { k + x(t k ) = (𝕀 + B k )x(t−k ) + ∆ k (x(t−k )), k = 1, 2, . . . , N + 1, { { { x(0) = x0 {

(3.192)

to achieve more specific results similar to the ones for impulsive ODEs [252] for the finitedimensional case. Of course, we can extend (3.191) and (3.192) to infinite-dimensional space. That is, we consider the associated impulsive fractional evolution equation by setting A as a generator of a C0 -semigroup on an infinite-dimensional Banach space, and then using the theory of semigroups with nonlinear functional analysis. In addition, the investigation of the existence of almost periodic solutions for (3.191) and (3.192) may be more interesting. Secondly, we can also extend our recent results of [280] to consider linear and semilinear differential equations with periodic not instantaneous impulses such as x󸀠 (t) = Ax(t), t ∈ (s k , t k+1 ], k = 0, 1, 2, . . . , { { { { { { x(t+i ) = (𝕀 + B k )x(t−k ) + B k x(t−k ), k = 1, 2, . . . , { { x(t) = (𝕀 + B k )x(t− ) + B k x(t− ), t ∈ (t k , s k ], k = 1, 2, . . . , { k k { { { + − x(s ) = x(s ), k = 1, 2, . . . , k { k

(3.193)

x󸀠 (t) = Ax(t) + f(t, x(t)), t ∈ (s k , t k+1 ], k = 0, 1, 2, . . . , { { { { { { x(t+i ) = (𝕀 + B k )x(t−i ) + B k x(t−k ), k = 1, 2, . . . , { { x(t) = (𝕀 + B k )x(t−k ) + B k x(t−k ), t ∈ (t k , s k ], k = 1, 2, . . . , { { { { + − {x(s k ) = x(s k ), k = 1, 2, . . . ,

(3.194)

and

where t k act as impulse points and s k as junction points satisfying t0 = s0 < t1 < s1 < t2 ⋅ ⋅ ⋅ < t k < s k < t k+1 ⋅ ⋅ ⋅, t k → ∞ and the periodicity conditions t k+N+1 = t k + T and s k+N+1 = s k + T. In addition, one can study the associated fractional-order and infinitedimensional cases for (3.193) and (3.194). The issues on the existence and stability of almost periodic solutions would be another interesting branch. Finally, we can consider controllability and iterative learning control for the above equations with impulsive periodic controls arising from some real-world problems in engineering.

4 Fractional Evolution Equations: Continued 4.1 Fractional evolution equations with periodic boundary conditions 4.1.1 Introduction Periodic motions play an important role in the dynamics of climate, food supplement, insecticide population, sustainable development, etc., see for instance [16, 104, 184, 206, 254, 321, 322]. Periodic boundary problems for fractional differential equations serve as a class of important models to study the dynamics of processes that are subject to periodic changes in their initial and final state. There are some papers discussing periodic (or anti-periodic) boundary problems for fractional differential equations in finitedimensional spaces [5, 9]. In this section, we discuss periodic boundary value problems (BVP for short) for fractional evolution equations such as c

{

D q x(t) = Ax(t) + f(t),

t ∈ J = [0, T], q ∈ (0, 1),

x(0) = x(T),

(4.1)

where c D q denotes the Caputo fractional derivative of order q, the unbounded operator A is the generator of a strongly continuous semigroup {S(t) | t ≥ 0} on a Banach space X, the function f : J → X will be specified later and T > 0. For that, we have to first discuss the homogeneous linear periodic BVP c

{

D q x(t) = Ax(t),

t ∈ J, q ∈ (0, 1),

x(0) = x(T),

(4.2)

and the associated Cauchy problem c

{

D q x(t) = Ax(t), x(0) = x,̄

t ∈ J, q ∈ (0, 1),

x̄ ∈ X,

(4.3)

and introduce a suitable definition of mild solution for BVP (4.2). To study BVP (4.2), we construct the evolution operator {T (⋅)} associated with the unbounded operator A, the order q, and a probability density function defined on (0, ∞), which is very important in the sequel. It can be deduced from the discussion on BVP (4.1) that the invertibility of [I − T (T)] is the key for the existence of mild solutions of BVP (4.2). For the invertibility of [I − T (T)], compactness or other additional conditions on {S(t) | t ≥ 0} generated by the unbounded operator A are needed. We remark that the constructed evolution operators {T (⋅)} can be used to reduce the existence of mild solutions for BVP (4.2) to the existence of roots for an operator DOI 10.1515/9783110522075-004

216 | 4 Fractional Evolution Equations: Continued equation T (T)x̄ = x,̄ where x̄ can be taken as the initial value in Cauchy problem (4.3). Applying the well-known Fredholm alternative theorem, we can present alternative results on mild solutions for BVP (4.1) and (4.2). To evaluate the probability of BVP (4.1), we study the following periodic BVP with parameter perturbations: c

{

D q x(t) = Ax(t) + f(t) + p(t, x(t), ξ),

t ∈ J, q ∈ (0, 1),

x(0) = x(T),

(4.4)

where p is a given function and ξ ∈ Λ = (−ξ ̃ , ξ ̃ ) is a small perturbation parameter that may be caused by some adaptive control algorithms or parameter drift. For a given periodic motion of BVP (4.1), periodic motion controllers that are robust to parameter drift are designed by virtue of BVP with parameter perturbations (4.4). This section is based on [306].

4.1.2 Homogeneous periodic boundary value problem In this subsection, we consider the homogeneous linear periodic BVP (4.2). We make the following assumption. [HA] A : D(A) → X is the infinitesimal generator of a strongly continuous semigroup {S(t) | t ≥ 0}. We introduce the following definition of a mild solution for our problem. Definition 4.1. By the mild solution of the periodic BVP (4.2), we mean the function x ∈ C(J, X) satisfying x(t) = T (t)x(0), t ∈ (0, T], { x(0) = x(T), where ∞

T (t) = ∫ ξ q (θ)S(t q θ)dθ,

ξ q (θ) =

1 1 −1− 1q θ ϖ q (θ− q ) ≥ 0, q

0

ϖ q (θ) =

1 ∞ Γ(nq + 1) sin(nπq), ∑ (−1)n−1 θ−qn−1 π n=1 n!

θ ∈ (0, ∞),

and ξ q is a probability density function defined on (0, ∞), that is, ∞

ξ q (θ) ≥ 0,

θ ∈ (0, ∞) and

∫ ξ q (θ)dθ = 1. 0

Let us put M = supt∈J ‖S(t)‖L b (X,X) , which is a finite number. The following results will be used throughout this section.

4.1 Fractional evolution equations with periodic boundary conditions | 217

Lemma 4.2 ([301, Lemma 2.9]). The operator T has the following properties: (i) For any fixed t ≥ 0, T (t) is a linear and bounded operator, i.e., for any x ∈ X, ‖T (t)x‖ ≤ M‖x‖. (ii) {T (t) | t ≥ 0} is also strongly continuous. (iii) For every t > 0, T (t) is also a compact operator provided that S(t) is compact. Theorem 4.3. Assume that [HA] holds and {S(t) | t ≥ 0} is a compact semigroup in X. Then either the homogeneous linear periodic BVP (4.2) has a unique trivial mild solution, or it has finitely many linearly independent non-trivial mild solutions in C(J, X). Proof. The periodic BVP (4.2) has a mild solution x if and only if T (T) has a fixed point. In fact, if the periodic BVP (4.2) has a mild solution x, then we have x(T) = T (T)x(0) = x(0). Thus, x̄ := x(0) is a fixed point of T (T). On the other hand, if x̄ = x(0) is a fixed point of T (T), the mild solution of the Cauchy problem c

{

D q x(t) = Ax(t),

t ∈ J, q ∈ (0, 1),

x(0) = x,̄

is given by x(t) = T (t)x,̄ which implies that x(T) = T (T)x̄ = x(0). This yields that x is just the mild solution of the periodic BVP (4.2). By (iii) of Lemma 4.2, T (T) is a compact operator. By the Fredholm alternative theorem, either (i) T (T)x(0) = x(0) has only a trivial mild solution and [I − T (T)]−1 exists, or (ii) T (T)x(0) = x(0) has non-trivial mild solutions which form a finitedimensional subspace of X. This implies that every mild solution of the periodic BVP i (4.2) can be written as x(t) = ∑m i=1 α i T (t)x(0) , where m is finite and α 1 , α 2 , . . . , α m are constants. Remark 4.4. (i) If ‖S(t)‖ < 1 for t ∈ (0, T], then T (nT) → 0 as n → ∞, and the operator I − T (T) is invertible and [I − T (T)]−1 ∈ L b (X). (ii) If ‖T (T)‖ < 1, then the operator I − T (T) is invertible and [I − T (T)]−1 ∈ L b (X).

4.1.3 Non-homogeneous periodic boundary value problem In this subsection, we consider the non-homogeneous linear periodic BVP (4.1). For that, we need the following assumption. [HF] Input f : J → X is measurable for t ∈ J, and there exist a constant q1 ∈ (0, q) and a real-valued function h(⋅) ∈ L1/q1 (J, ℝ) such that ‖f(t)‖ ≤ h(t) for all t ∈ J. We use the following definition of a mild solution for our problem.

218 | 4 Fractional Evolution Equations: Continued

Definition 4.5. By the mild solution of the periodic BVP (4.1) we mean a function x ∈ C(J, X) satisfying t

{ { { { x(t) = T (t)x(0) + ∫(t − s)q−1 S (t − s)f(s)ds, { 0 { { { {x(0) = x(T), where

t ∈ (0, T],

∞

S (t) = q ∫ θξ q (θ)S(t q θ)dθ, 0

while T (t), ξ q and ϖ q (θ) are as in Definition 4.1. Lemma 4.6 ([301, Lemma 2.9]). The operator S has the following properties: (i) For any fixed t ≥ 0, S (t) is a linear and bounded operator, i.e., for any x ∈ X, ‖S (t)x‖ ≤

M ‖x‖. Γ(q)

(ii) {S (t) | t ≥ 0} is also strongly continuous. (iii) For every t > 0, S (t) is a compact operator if S(t) is compact. Now we are ready to state and prove the main result of this section. Theorem 4.7. Let [HA], [HF] hold and {S(t) | t ≥ 0} be a compact semigroup in X. If the periodic BVP (4.2) has no non-trivial mild solutions, then the periodic BVP (4.1) has a unique mild solution given by t

x T (t) = T (t)[I − T (T)] z + ∫(t − s)q−1 S (t − s)f(s)ds, −1

0

where

T

z = ∫(T − s)q−1 S (T − s)f(s)ds. 0

Further, we have the estimate ‖x T (t)‖ ≤

M(ML1 + 1) T q−q1 H , 1−q Γ(q) ( q−q1 ) 1 1−q1

where L1 = ‖[I − T (T)]−1 ‖ and H = ‖h‖

1

L q1 (J,ℝ)

.

Proof. By the Fredholm alternative theorem, [I − T (T)]−1 exists and is bounded. Since the periodic BVP (4.2) has no non-trivial mild solutions, the operator equation [I − T (T)]x(0) = z has a unique solution x(0) = [I − T (T)]−1 z =: x.̄ Now we consider

4.1 Fractional evolution equations with periodic boundary conditions | 219

the Cauchy problem c

{

D q x(t) = Ax(t) + f(t),

t ∈ J, q ∈ (0, 1),

(4.5)

x(0) = [I − T (T)] z =: x.̄ −1

One can verify that the mild solution x T (⋅) of (4.5) corresponding to initial value x̄ is just the mild solution of the periodic BVP (4.1). Moreover, the uniqueness of the mild solution follows from the Fredholm alternative theorem again. For the estimation of x T (⋅), by Lemmas 4.2 and 4.6, t

󵄩 󵄩 ‖x T (t)‖ ≤ 󵄩󵄩󵄩T (t)[I − T (T)]−1 z󵄩󵄩󵄩 + ∫(t − s)q−1 ‖S (t − s)f(s)‖ds 0 t

M 󵄩 󵄩 ≤ M 󵄩󵄩󵄩[I − T (T)]−1 󵄩󵄩󵄩 ‖z‖ + ∫(t − s)q−1 h(s)ds Γ(q) 0 T

t

M 󵄩 󵄩 ≤ M 󵄩󵄩󵄩[I − T (T)]−1 󵄩󵄩󵄩 ∫(T − s)q−1 ‖S (T − s)f(s)‖ds + ∫(t − s)q−1 h(s)ds Γ(q) 0

0 T

≤

t

M 2 ‖[I − T (T)]−1 ‖ M ∫(T − s)q−1 h(s)ds + ∫(t − s)q−1 h(s)ds Γ(q) Γ(q) 0

0 T

≤

T

1−q1 q1 q−1 1 M 2 ‖[I − T (T)]−1 ‖ ( ∫(T − s) 1−q1 ds) ( ∫ h(s) q1 ds) Γ(q) 0

0

t

+

1−q1

q−1 M ( ∫(t − s) 1−q1 ds) Γ(q)

0

t 1

( ∫ h(s) q1 ds)

q1

0

M(ML1 + 1) T q−q1 H ≤ , 1−q Γ(q) ( q−q1 ) 1 1−q1

where L1 = ‖[I − T (T)]−1 ‖ and H = ‖h‖

1

L q1 (J,ℝ)

.

Remark 4.8. By Remark 4.4, we can replace the assumption of {S(t) | t ≥ 0} being compact by ‖S(t)‖ < 1 for t ∈ (0, T] or ‖T (T)‖ < 1 directly. It is obvious that all the results of Theorem 4.7 also hold. Let X be a Hilbert space. In the case when [I − T (T)]−1 does not exist, we need to consider the adjoint problem to BVP (4.1): c

{

D q y(t) = −A∗ y(t), y(T) = y(0) ∈ X , ∗

t ∈ J,

(4.6)

where A∗ is the adjoint operator of A. Due to reflexivity of X and Theorem A.8, A∗ is the infinitesimal generator of a strongly semigroup {S∗ (t) | t ≥ 0} in X ∗ .

220 | 4 Fractional Evolution Equations: Continued

Before we prove the main theorems of this subsection, we need the following results. Lemma 4.9 ([302, Lemma 2.11]). Let ∞

∞

T (t) = ∫ ξ q (θ)S (t θ)dθ, ∗

∗

S (t) = q ∫ θξ q (θ)S∗ (t q θ)dθ.

q

∗

0

0

In the Hilbert space X, the following holds: (i) For any fixed t ≥ 0, T ∗ (t) and S ∗ (t) are linear and bounded operators. (ii) {T ∗ (t) | t ≥ 0} and {S ∗ (t) | t ≥ 0} are strongly continuous. (iii) For every t > 0, T ∗ (t) and S ∗ (t) are compact operators provided that S(t) is compact. Now, we are ready to prove our main results of this subsection. Theorem 4.10. Let [HA], [HF] hold and let {S(t) | t ≥ 0} be a compact semigroup in a Hilbert space X. Suppose that [I − T (T)]−1 does not exist. Then the adjoint equation (4.6) has m linearly independent mild solutions y1 , y2 , . . . , y m . Proof. Since the operators T (T) and T ∗ (T) are compact, T (T)y(0) = y(0) has m nontrivial mild solutions y i (0), i = 1, 2, . . . , m, which form a finite-dimensional subspace of X. It comes from dim ker[I − T (T)] = dim ker[I − T ∗ (T)] = m < +∞ that [I − T ∗ (T)]y(0) = 0 has m non-trivial mild solutions y i (0), i = 1, 2, . . . , m, i.e., T ∗ (T)y i (0) = y i (0), i = 1, 2, . . . , m. Consider the Cauchy problem c

{

D q y(t) = −A∗ y(t),

t ∈ J,

y(T) = y (0). i

(4.7)

It is easy to see that the mild solutions of Cauchy problem (4.7) are just the mild solutions of the periodic BVP (4.6). This implies that every mild solution of the periodic BVP i (4.6) can be written as y(t) = ∑m i=1 α i T (t)y(0) where m is finite and α 1 , α 2 , . . . , α m are constants. Theorem 4.11. Let [HA], [HF] hold and let {S(t) | t ≥ 0} be a compact semigroup in a Hilbert space X. Suppose that [I − T (T)]−1 does not exist. Then the periodic BVP (4.1) has a mild solution if and only if ⟨z, y i (0)⟩X∗ ,X = 0,

i = 1, 2, . . . , m,

which is equivalent to T

∫ ⟨(T − θ)q−1 S (T − θ)f(θ), y i (0)⟩X,X∗ dθ = 0. 0

Otherwise, the periodic BVP (4.1) has no mild solution.

4.1 Fractional evolution equations with periodic boundary conditions | 221

Furthermore, every mild solution of the periodic BVP (4.1) can be given by m

x(t) = x T (t) + ∑ α i x i (t), i=1

where t ∈ J, x T is a mild solution of the periodic BVP (4.1), x1 , x2 , . . . , x m are m linearly independent mild solutions of the periodic BVP (4.2), and α1 , . . . , α m are constants. Proof. From compactness of T (T), T ∗ (T) is compact and dim ker[I − T (T)] = dim ker[I − T (T)] = m < +∞. The operator equation [I − T (T)]y(0) = 0 has m non-trivial linearly independent solui i tions {y i (0)}m i=1 . Let y be the solution of the periodic BVP (4.6) corresponding to y (0), i = 1, 2, . . . , m which is just a mild solution of the Cauchy problem (4.7). It is well known that the operator equation [I − T (T)]x̄ = z has a solution if and only if ⟨z, y i (0)⟩X∗ ,X = 0,

i = 1, 2, . . . , m,

which is equivalent to T

0 = ⟨z, y (0)⟩X,X∗ = ∫ ⟨(T − θ)q−1 S (T − θ)f(θ), y i (0)⟩X,X∗ dθ. i

0

This completes the proof.

4.1.4 Parameter perturbation methods for robustness Define C T (J, X) = {x ∈ C(J, X) | x(0) = x(T)} with ‖x‖C T = sup{‖x(t)‖ | t ∈ J} for x ∈ C T (J, X). It can be seen that endowed with the norm ‖⋅‖C T , C T (J, X) is a Banach space. Denote Sρ = {x ∈ C T (J, X) | ‖x‖C T < ρ}, B(x, ρ1 ) = {x ∈ C T (J, X) | ‖x − x T ‖C T ≤ ρ1 }, where ρ=

M(ML1 + 1) T q−q1 H Tq sup χ(ξ)], [ q−q 1−q + Γ(q) q |ξ|≤ξ ̃ ( 1) 1 1−q1

and χ is a nonnegative function. We introduce assumption [HP]:

ρ1 =

M(ML1 + 1)T q sup χ(ξ), Γ(1 + q) |ξ|≤ξ ̃

222 | 4 Fractional Evolution Equations: Continued [HP1] p : J × Sρ × Λ → X is measurable in t. [HP2] There exists a nonnegative function ϖ such that limξ →0 ϖ(ξ) = ϖ(0) = 0, and for any t ∈ J, x, y ∈ Sρ and ξ ∈ Λ we have ‖p(t, x, ξ) − p(t, y, ξ)‖ ≤ ϖ(ξ)‖x − y‖. [HP3] There exists a nonnegative function χ such that limξ →0 χ(ξ) = χ(0) = 0, and for any t ∈ J, x ∈ Sρ and ξ ∈ Λ we have ‖p(t, x, ξ)‖ ≤ χ(ξ). Now we introduce the mild solution of the periodic BVP (4.4). Definition 4.12. By the mild solution of the periodic BVP (4.4), we mean the function x ∈ C(J, X) satisfying t

{ { { { x(t) = T (t)x(0) + ∫(t − s)q−1 S (t − s)[f(s) + p(s, x(s), ξ)]ds, { 0 { { { x(0) = x(T). {

t ∈ (0, T],

The following result shows that given a periodic motion we can design periodic motion controllers that are robust with respect to parameter drift. Theorem 4.13. Let [HA], [HF] and [HP] hold, let {S(t) | t ≥ 0} be a compact semigroup in X, and let the periodic BVP (4.2) have no trivial mild solution. Then there is ξ0 ∈ (0, ξ ̃ ) ξ such that for |ξ| ≤ ξ0 the periodic BVP (4.4) has a unique mild solution x T satisfying ξ

‖x T − x T ‖C T ≤ ρ1

and

ξ

lim x T (t) = x T (t)

ξ →0

uniformly for t ∈ J, where x T is the mild solution of the periodic BVP (4.1). Proof. Let T

x0 = [I − T (T)] [z + ∫(T − s)q−1 S (T − s)p(s, x(s), ξ)ds] ∈ X −1

0

be fixed. Define the map O on B(x T , ρ1 ) by t

(Ox)(t) = T (t)x0 + ∫(t − s)q−1 S (t − s)[f(s) + p(s, x(s), ξ)]ds. 0

It is not difficult to show that Ox ∈ C T (J, X). By assumption [HP], we can choose ξ0 ∈ (0, ξ ̃ ) such that M(ML1 + 1)T q sup χ(ξ) ≤ ρ1 Γ(1 + q) |ξ|≤ξ0

4.1 Fractional evolution equations with periodic boundary conditions | 223

and η=

M(ML1 + 1)T q sup ϖ(ξ) < 1. Γ(1 + q) |ξ|≤ξ0

For ξ ∈ (−ξ0 , ξ0 ) and x, y ∈ B(x T , ρ1 ), one can verify that ‖Ox − x T ‖C T ≤

M(ML1 + 1)T q sup χ(ξ) ≤ ρ1 Γ(1 + q) |ξ|≤ξ0

(4.8)

and ‖Ox − Oy‖C T ≤ η‖x − y‖C T . This implies that O is a contraction mapping on B(x T , ρ1 ). Then O has a unique fixed ξ point x T ∈ B(x T , ρ1 ) given by t ξ

ξ

x T (t) = T (t)x0 + ∫(t − s)q−1 S (t − s)[f(s) + p(s, x T (s), ξ)]ds,

(4.9)

0

which is just the unique mild solution of the periodic BVP (4.4). ξ From (4.8) and (4.9), one can get ‖x T − x T ‖C T ≤ ρ1 . It is easy to see that ξ

lim x T (t) = x T (t)

ξ →0

uniformly for t ∈ J.

4.1.5 Example In this subsection, an example is given to illustrate our theory. Consider the problem ∂2 c q { D x(t, y) = x(t, y) + |sin(t, y)| + ξ|cos x(t, y)|, { { { ∂y2 { { x(0, y) = x(2π, y), y ∈ Ω, { { { { x(t, y) = 0, y ∈ ∂Ω, t > 0, {

y ∈ Ω, 0 < t ≤ 2π, (4.10)

where q ∈ (0, 1), Ω = (0, π) and ξ ∈ (−1, 1). Let X = Y = L2 (0, π) and let A : D(A) → X be defined by Ax = x yy , x ∈ D(A), where D(A) = {x ∈ X | x, x y are absolutely continuous, x(0) = x(π) = 0 and x yy ∈ X}. 2 Then Ax = ∑∞ n=1 n (x, x n )x n for x ∈ X, where x n (s) = √2/π sin(ns), n = 1, 2, 3, . . . is the orthogonal set of eigenfunctions of A. It can be easily shown that A is the infinitesimal generator of a compact analytic semigroup {S(t) | t ≥ 0} in X and it is given by −n2 t (x, x )x . So there exists a constant M ≥ 1 such that ‖S(t)‖ ≤ M. As S(t)x = ∑∞ n n n=1 e a matter of fact, ‖S(t)‖ ≤ e−t for any t ≥ 0, i.e., M = 1.

224 | 4 Fractional Evolution Equations: Continued Let us set J = [0, 2π] and x(⋅)(y) = x(⋅, y),

f(⋅)(y) = |sin(⋅, y)|,

p(⋅, x(⋅, y), ξ) = ξ|cos x(⋅, y)|.

Consequently, problem (4.10) can be rewritten as c

{

D q x(t) = Ax(t) + f(t) + p(t, x(t), ξ),

t ∈ J, q ∈ (0, 1),

x(0) = x(2π).

It satisfies all the assumptions given in Remark 4.4 (i) and Theorem 4.13, so our results can be applied to problem (4.10).

4.2 Abstract Cauchy problems for fractional evolution equations 4.2.1 Introduction In this section we consider Cauchy problems, impulsive problems and nonlocal problems in a Banach space X by applying a generalized Darbo fixed point theorem associated with the Hausdorff measure of noncompactness, which is inspired by Liu et al. [185]. Problem (I): Cauchy problem c

{

D q u(t) = f(t, u(t)),

t ∈ J = [0, T],

u(0) = 0,

(4.11)

q

where the symbol c D q ≡ c D0+ denotes the Caputo fractional derivative of order q ∈ (0, 1) with the lower limit zero, and f : J × X → X satisfies Carathéodory type conditions. Problem (II): impulsive problem c q D u(t) = f(t, u(t)), t ∈ J 󸀠 = J \ ̃ D, ̃ D = {t1 , . . . , t σ }, { { { + − u(t k ) = u(t k ) + y k , k = 1, 2, . . . , σ, y k ∈ X, { { { { u(0) = 0,

(4.12)

where f is the same as in Problem (I), impulse times t k satisfy 0 = t0 < t1 < ⋅ ⋅ ⋅ < t σ < t σ+1 = T, and u(t+k ) = limϵ→0+ u(t k + ϵ) and u(t−k ) = limϵ→0− u(t k + ϵ) represent the right- and left-sided limits of u(t) at t = t k . Problem (III): nonlocal problem c

{

D q u(t) = −Au(t) + f ̄(t, u(t)), u(0) = g(u),

t ∈ J,

(4.13)

where −A : D(A) → X is the infinitesimal generator of an analytic semigroup of uniformly bounded linear operators {S(t) | t ≥ 0}, f ̄ : J × X β → X or X β is a Carathéodory

4.2 Abstract Cauchy problems for fractional evolution equations | 225

type function, g : C(J, X β ) → X β contributes nonlocal conditions, where X β = D(A β ) (0 < β < 1) is a Banach space with the norm ‖u‖β = ‖A β u‖ for u ∈ X β . We recall that existence results for solutions of Problem (I) with u(0) = u0 and Problem (II) with nonlinear impulsive conditions u(t+k ) = u(t−k ) + I k (y k ) in ℝ were reported by Agarwal et al. [5] by utilizing the contraction mapping principle, the Schaefer fixed point theorem, and nonlinear alternative results of Leray–Schauder type when f and I k satisfy Lipschitz continuity conditions. Moreover, existence results for mild solutions for Problem (III) with g(u) = u0 were investigated by Wang and Zhou [301] by virtue of the Leray–Schauder fixed point theorem when {S(t) | t ≥ 0} is compact. This section is based on [308].

4.2.2 Preliminaries Let C(J, X β ) be the Banach space endowed with the supremum norm ‖u‖∞ := supt∈J ‖u‖β for x ∈ C(J, X β ). Suppose that −A : D(A) → X is the infinitesimal generator of an analytic semigroup of uniformly bounded linear operators {S(t) | t ≥ 0}. This means that there exists M ≥ 1 such that ‖S(t)‖ ≤ M. We assume without loss of generality that 0 ∈ ρ(A). This allows us to define the fractional power A β for 0 < β < 1, as a closed linear operator on its domain D(A β ) with inverse A−β (see Pazy [220]). Theorem 4.14. Suppose that 0 < ε < 1, 0 < z < 1, h > 0 and let n

SΓ = ∑ C in ε n−i h i i=0

Γ i (z + 1) , Γ(iz + 1)

n ∈ ℕ+ .

Then SΓ = o( n1s ) as n → +∞, where s > 0 is an arbitrary real number. Proof. Take ε̂ =

1−ε 2 n

for 0 < ε < 1. It is easy to see that ε̂ > 0 and ε + ε̂ =

1+ε 2

< 1. Then,

+ 1) h S Γ = ∑ C in ε n−i ε̂ i ( ) ̂ Γ(iz + 1) ε i=0 i

Γ i (z

n n h i Γ i (z + 1) hΓ(z + 1) ≤ ( ∑ C in ε n−i ε̂ i )( ∑ ( ) ) = (ε + ε)̂ n E z ( ). ̂ Γ(iz + 1) ε ε̂ i=0 i=0

Since limn→∞ (ε + ε)̂ n n s = 0 for any s > 0, we obtain the result immediately. To prove our main results, we need the following fixed point theorem associated with the Hausdorff measure of noncompactness. We point out that the corresponding results associated with the Kuratowski measure of noncompactness were reported by Liu et al. [185]. Theorem 4.15. Let B0 be a closed and convex subset of a Banach space X. Suppose that F : B0 → B0 is a continuous operator and F(B0 ) is bounded. For each bounded subset B ⊂ B0 , set F 1 (B) = F(B),

F n (B) = F(conv(F n−1 (B))),

n = 2, 3, . . . .

226 | 4 Fractional Evolution Equations: Continued If there exist a constant 0 ≤ k < 1 and a positive integer n0 such that for each bounded subset B ⊂ B0 , χ(F n0 (B)) ≤ kχ(B), then F has a fixed point in B0 . Proof. Keeping in mind the relationship between the Hausdorff measure of noncompactness χ and the Kuratowski measure of noncompactness α, i.e., χ(B) ≤ α(B) ≤ 2χ(B), one can repeat the same procedure as in the proof of [185, Lemma 2.4] and obtain the statement immediately.

4.2.3 Existence and uniqueness theorems of solutions for Problem (I) This subsection deals with existence and uniqueness of solutions for Problem (I). We make the following assumptions: (H1) The function f : J × X → X satisfies the Carathéodory type conditions, i.e., f(⋅, u) is strongly measurable for all u ∈ X, and f(t, ⋅) is continuous for a.e. t ∈ J. (H2) There exist a function m ∈ L1/q1 (J, ℝ+ ), q1 ∈ [0, q) and a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that ‖f(t, u)‖ ≤ m(t)Ω(‖u‖) for all x ∈ X and a.e. t ∈ J. (H3) There exists a constant L > 0 such that for any bounded D ⊂ X, χ(f(t, D)) ≤ Lχ(D) for a.e. t ∈ J. (H4) The inequality T

Ω(r) ∫(t − s)q−1 m(s)ds ≤ r, Γ(q)

t∈J

0

has at least one positive solution. Let us recall a definition of a solution for Problem (I) and a lemma on this solution. See Agarwal et al. [5] for more details, and also Lemma A.16 in Appendix A. Definition 4.16. A function u ∈ C(J, X) is said to be a solution of equation (4.11) if u satisfies the equation c D q u(t) = f(t, u(t)) a.e. on J, and the condition u(0) = 0. Lemma 4.17. A function u ∈ C(J, X) is a solution of the fractional integral equation t

u(t) =

1 ∫(t − s)q−1 f(s, u(s))ds Γ(q) 0

if and only if u is a solution of equation (4.11).

(4.14)

4.2 Abstract Cauchy problems for fractional evolution equations | 227

Under the assumptions (H1)–(H4), we will show that fractional integral equation (4.14) has at least one solution u ∈ C(J, X). Define an operator F1 : C(J, X) → C(J, X) as follows: t

(F1 u)(t) =

1 ∫(t − s)q−1 f(s, u(s))ds, Γ(q)

t ∈ J.

0

It is obvious that F1 is well defined due to (H1) and (H2). Then fractional integral equation (4.14) can be written as the operator equation u = F1 u.

(4.15)

Thus, the existence of a solution for equation (4.11) is equivalent to the existence of a fixed point of operator F1 , which satisfies operator equation (4.15). Lemma 4.18. Let K1 > 0 satisfy (H4). Set W1 = {u ∈ C(J, X) | ‖u‖C ≤ K1 } ⊆ C(J, X). Then B0 = conv(F1 W1 ) is equicontinuous. Proof. By Proposition A.3, it suffices to show that F1 (W1 ) ⊂ C(J, X) is equicontinuous. In fact, for any u ∈ W1 , 0 ≤ t1 < t2 ≤ T, we get 󵄩󵄩 󵄩 󵄩󵄩(F1 u)(t1 ) − (F1 u)(t2 )󵄩󵄩󵄩 t1

1 ≤ ∫((t1 − s)q−1 − (t2 − s)q−1 )‖f(s, u(s))‖ds Γ(q) 0 t2

1 + ∫(t2 − s)q−1 ‖f(s, u(s))‖ds Γ(q) t1

t1

≤

t2

Ω(K1 ) Ω(K1 ) ∫((t1 − s)q−1 − (t2 − s)q−1 )m(s)ds + ∫(t2 − s)q−1 m(s)ds Γ(q) Γ(q) t1

0

≤ ≤

Ω(K1 )‖m‖

1 q1

q−q1 1−q1

L (J,ℝ+ ) {[t1 q−q1 1−q1 Γ(q)( 1−q1 )

2Ω(K1 )‖m‖

q−q1 1−q1

− t2

+ (t2 − t1 )

q−q1 1−q1

1−q1

]

+ (t2 − t1 )q−q1 }

1

L q1 (J,ℝ+ ) (t2 q−q1 1−q1 Γ(q)( 1−q1 )

− t1 )q−q1 .

As t2 → t1 , the right-hand side of the above inequality tends to zero. Therefore F1 (W1 ) is equicontinuous. Lemma 4.19. The operator F1 : C(J, X) → C(J, X) is continuous. Proof. Let {u n } be a sequence of a bounded set W1 , such that u n → u in W1 . We have to show that ‖F1 u n − F1 u‖C → 0.

228 | 4 Fractional Evolution Equations: Continued It is easy to see that f(s, u n (s)) → f(s, u(s)) as n → ∞ due to the Carathéodory continuity of f . On the other hand, using (H2), we get for each t ∈ J, 󵄩 󵄩 (t − s)q−1 󵄩󵄩󵄩f(s, u n (s)) − f(s, u(s))󵄩󵄩󵄩 ≤ (t − s)q−1 2m(s)Ω(K1 ). Using the fact that the function s 󳨃→ (t − s)q−1 2m(s)Ω(K1 ) is integrable for s ∈ [0, t] and t ∈ J, by the Lebesgue dominated convergence theorem we get t

󵄩 󵄩 ∫(t − s)q−1 󵄩󵄩󵄩f(s, u n (s)) − f(s, u(s))󵄩󵄩󵄩ds → 0

as n → ∞.

0

Consequently, for all t ∈ J, t

1 󵄩󵄩 󵄩 󵄩 󵄩 ∫(t − s)q−1 󵄩󵄩󵄩f(s, u n (s)) − f(s, u(s))󵄩󵄩󵄩ds → 0. 󵄩󵄩(F1 u n )(t) − (F1 u)(t)󵄩󵄩󵄩 ≤ Γ(q) 0

Therefore, F1 u n → F1 u pointwisely on J as n → ∞. But using Lemma 4.18, it follows that F1 u n → F1 u uniformly on J as n → ∞, which implies that F1 is continuous. Lemma 4.20. The operator F1 : W1 → W1 is bounded. Proof. For any u ∈ W1 , we have t

T

0

0

󵄩󵄩 Ω(K1 ) 1 󵄩󵄩󵄩 󵄩󵄩∫(t − s)q−1 f(s, u(s))ds󵄩󵄩󵄩 ≤ ‖(F1 u)(t)‖ = ∫(t − s)q−1 m(s)ds. 󵄩 󵄩󵄩 Γ(q) 󵄩 Γ(q) Since K1 satisfies (H4), we have T

Ω(K1 ) ∫(t − s)q−1 m(s)ds ≤ K1 , Γ(q) 0

which implies that F1 : W1 → W1 is bounded. Now, we have the possibility to prove the main results for equation (4.11). Theorem 4.21. Assume that (H1)–(H4) hold. Then equation (4.11) has at least one solution u ∈ C(J, X) and the set of solutions of equation (4.11) is bounded in C(J, X). Proof. Let F1 : C(J, X) → C(J, X) be the operator defined in the beginning of this subsection. For any B ⊂ B0 , by the definition of operator F1 and Lemmas 4.18–4.20, we get that F1 (B) is bounded and equicontinuous on J. For any B ⊂ B0 , we know from Propositions A.22–A.24 that for any ε > 0 there is a sequence {u n }∞ n=1 ⊂ B such that χ(F11 B(t)) = χ(F1 B(t)) t

1 ≤ 2χ({ ∫(t − s)q−1 f(s, {u n (s)}∞ n=1 )ds}) + ε Γ(q) 0

4.2 Abstract Cauchy problems for fractional evolution equations | 229

t

≤

4 ∫(t − s)q−1 χ(f(s, {u n (s)}∞ n=1 ))ds + ε Γ(q) 0 t

4Lχ({u n }∞ n=1 ) ≤ ∫(t − s)q−1 ds + ε Γ(q) 0 t

≤

4Lt q 4Lχ(B) χ(B) + ε. ∫(t − s)q−1 ds + ε ≤ Γ(q) Γ(q + 1) 0

Since ε > 0 is arbitrary, we have χ(F11 B(t)) ≤

4Lt q χ(B). Γ(q + 1)

1 From Proposition A.24, for any ε > 0 there is a sequence {y n }∞ n=1 ⊂ conv(F 1 B) such that

χ(F12 B(t)) = χ(F1 (conv(F11 B(t)))) t

≤ 2χ({

1 ∫(t − s)q−1 f(s, {y n (s)}∞ n=1 )ds}) + ε Γ(q) 0 t

≤

4 ∫(t − s)q−1 χ(f(s, {y n (s)}∞ n=1 ))ds + ε Γ(q) 0 t

≤

4 ∫(t − s)q−1 Lχ({y n (s)}∞ n=1 )ds + ε Γ(q) 0 t

4L ≤ ∫(t − s)q−1 χ(F11 B(s))ds + ε Γ(q) 0 t

≤

4L (4Lt q )2 χ(B) + ε. ∫(t − s)q−1 (cs q )χ(B)ds + ε ≤ Γ(q) Γ(1 + 2q) 0

Thanks to the identity t

∫(t − s)q−1 s nq̄ ds = 0

̄ Γ(q)Γ(1 + nq) t q+nq̄ , ̄ Γ(1 + q + nq)

it can be shown, by mathematical induction, that χ(F1n̄ B(t)) ≤

(4Lt q )n̄ χ(B). ̄ Γ(1 + nq)

Using the well-known Stirling’s formula we get ̄ = √2π nq ̄ ( Γ(1 + nq)

θ ̄ nq̄ nq ) e 12nq̄ e

n̄ ∈ ℕ,

230 | 4 Fractional Evolution Equations: Continued for some 0 < θ < 1. Hence, lim

̄ n→+∞

(4LT q )n̄ (4LT q )n̄ = lim = 0. θ ̄ ̄ nq ̄ ̄ Γ(1 + nq) n→+∞ ̄ 12nq √2π nq( ̄ nq e ) e

Consequently, by Theorem 4.15, F1 has a fixed point, which is a solution of our equation. q

4LT Remark 4.22. If we assume Γ(q+1) < 1 directly, one can apply the Darbo fixed point theorem to obtain the result. However, from the above arguments, one can see that we do not need this redundant condition by virtue of the generalized Darbo fixed point theorem.

Theorem 4.23. Assume that (H1)–(H3) hold and T

∫(t − s)q−1 m(s)ds < lim inf r→∞

rΓ(q) , Ω(r)

t ∈ J.

0

Then equation (4.11) has at least one solution u ∈ C(J, X) and the set of solutions of equation (4.11) is bounded in C(J, X). Proof. We know that there exists a constant K1 > 0 such that T

Ω(K1 ) ∫(t − s)q−1 m(s)ds ≤ K1 , Γ(q)

t ∈ J.

0

The rest of the proof follows the very same lines of the proof of Theorem 4.21, so we do not give the details here. In order to obtain the uniqueness of solutions, we add the following assumption: (H2+ ) There exists a constant a f > 0 such that ‖f(t, u) − f(t, v)‖ ≤ a f ‖u − v‖

for all t ∈ J, u, v ∈ X.

Theorem 4.24. Let the assumptions of Theorem 4.21 (or Theorem 4.23) and (H2+ ) hold. Then equation (4.11) has a unique solution u ∈ C(J, X). Proof. By Theorem 4.21, equation (4.11) has a solution u(⋅) in C(J, X). Let v(⋅) be another solution of equation (4.11). Note that by (H2+ ), we have t

‖u(t) − v(t)‖ ≤

af ∫(t − s)q−1 ‖u(s) − v(s)‖ds. Γ(q) 0

By the standard singular Gronwall’s inequality, ‖u(t) − v(t)‖ = 0, which yields the uniqueness of u(⋅).

t ∈ J,

4.2 Abstract Cauchy problems for fractional evolution equations | 231

4.2.4 Existence and uniqueness theorems of solutions for Problem (II) This subsection deals with existence and uniqueness of solutions for Problem (II). In addition to the assumptions (H1)–(H3), we replace (H4) by the following condition: (H4󸀠 ) The inequality T

σ

Ω(r) ∑ ‖y i ‖ + ∫(t − s)q−1 m(s)ds ≤ r, Γ(q) i=1

t∈J

0

has at least one positive solution. Let us introduce a definition of a solution for Problem (II) which is inspired by Section 3.6. Definition 4.25. A function u ∈ PC(J, X) is said to be a solution of equation (4.12) if u(t) = u k (t) for t ∈ (t k , t k+1 ), and u k ∈ C([0, t k+1 ], ℝ), k = 0, 1, 2, . . . , σ, satisfies c D q u (t) = f(t, u (t)) a.e. on (0, t k k+1 ), the restriction of u k+1 (t) onto [0, t k ) is just u k (t), t k and u(t+k ) = u(t−k ) + y k , k = 0, 1, 2, . . . , σ, along with u(0) = 0. As a consequence of Lemma A.16 we have the following result, which is useful in what follows. Lemma 4.26 ([113, Lemma 2.6]). Let h ∈ PC(J, X). A function u ∈ PC(J, X) given by t

1 q−1 h(s)ds { { Γ(q) ∫0 (t − s) { { t { 1 { y1 + Γ(q) ∫0 (t − s)q−1 h(s)ds { { { { t 1 q−1 u(t) = {y1 + y2 + Γ(q) ∫0 (t − s) h(s)ds { { . { { .. { { { { { σ t 1 q−1 {∑i=1 y i + Γ(q) ∫0 (t − s) h(s)ds

if t ∈ [0, t1 ), if t ∈ (t1 , t2 ), if t ∈ (t2 , t3 ), if t ∈ (t σ , T]

is a unique solution of the impulsive problem c q D u(t) = h(t), t ∈ (0, T], { { { u(t+k ) = u(t−k ) + y k , k = 1, 2, . . . , σ, { { { { u(0) = 0.

Next, we define an operator F2 : PC(J, X) → PC(J, X) as follows: t

1 q−1 f(s, u(s))ds { { Γ(q) ∫0 (t − s) { { t { 1 { {y1 + Γ(q) ∫0 (t − s)q−1 f(s, u(s))ds { { { t 1 q−1 (F2 u)(t) = {y1 + y2 + Γ(q) ∫0 (t − s) f(s, u(s))ds { {. { { . { {. { { { σ t 1 q−1 {∑i=1 y i + Γ(q) ∫0 (t − s) f(s, u(s))ds

if t ∈ [0, t1 ), if t ∈ (t1 , t2 ), if t ∈ (t2 , t3 ), if t ∈ (t σ , T].

232 | 4 Fractional Evolution Equations: Continued

It is obvious that F2 is well defined. Then the fractional integral equation (4.14) can be written as the operator equation u = F2 u.

(4.16)

Thus, the existence of a solution for equation (4.12) is equivalent to the existence of a fixed point of operator F2 , which satisfies operator equation (4.16). Lemma 4.27. Let K2 > 0 satisfy (H4󸀠 ). Set W2 = {u ∈ PC(J, X) | ‖u‖PC ≤ K2 } ⊆ PC(J, X). Then B0 = conv(F2 W2 ) is equicontinuous. Proof. Repeating the process of the proof of Lemma 4.18 for each interval [t k , t k+1 ), k = 1, 2, . . . , σ, we can show that F2 (W2 ) is equicontinuous. Lemma 4.28. The operator F2 : PC(J, X) → PC(J, X) is continuous. Proof. Let {u n } be a sequence of a bounded set W2 such that u n → u in W2 . Repeating the process of the proof of Lemma 4.19 for each interval [t k , t k+1 ), k = 1, 2, . . . , σ, we can show that ‖F2 u n − F2 u‖PC → 0. Lemma 4.29. The operator F2 : W2 → W2 is bounded. Proof. For any u ∈ W2 , we have t

󵄩󵄩 1 󵄩󵄩󵄩 󵄩󵄩∫(t − s)q−1 f(s, u(s))ds󵄩󵄩󵄩 ‖(F2 u)(t)‖ ≤ ∑ ‖y i ‖ + 󵄩󵄩 󵄩󵄩 Γ(q) i=1 σ

0

σ

≤ ∑ ‖y i ‖ + i=1

T

Ω(K2 ) ∫(t − s)q−1 m(s)ds. Γ(q) 0

Since K2 satisfies (H4󸀠 ), repeating the process of the proof of Lemma 4.20 for each interval [t k , t k+1 ), k = 1, 2, . . . , σ, we have σ

‖F2 u‖PC

T

Ω(K2 ) ≤ ∑ ‖y i ‖ + ∫(t − s)q−1 m(s)ds ≤ K2 , Γ(q) i=1 0

which implies that F2 : W2 → W2 is bounded. Now, we have the possibility to prove the main results for equation (4.12). Theorem 4.30. Assume that (H1)–(H3) and (H4󸀠 ) hold. Then equation (4.12) has at least one solution u ∈ PC(J, X), and the set of solutions of equation (4.12) is bounded in PC(J, X). Proof. Let F2 : PC(J, X) → PC(J, X) be the operator defined in the beginning of this subsection. For any B ⊂ B0 , by the definition of operator F2 and Lemmas 4.27–4.29, we get that F2 (B) is bounded and equicontinuous on J.

4.2 Abstract Cauchy problems for fractional evolution equations | 233

Applying the same methods as in the proof of Theorem 4.30, we also have χ(F2n̄ B(t)) ≤

(4Lt q )n̄ χ(B) ̄ Γ(1 + nq)

for each n̄ ∈ ℕ. Using the same discussion as in the proof of Theorem 4.21, the value (4Lt q )n̄ of the term Γ(1+ ̄ can be made less than 1 and so, by Theorem 4.15, F 2 has a fixed nq) point, which is a solution to our equation. Theorem 4.31. Assume that (H1)–(H3) hold and T

∫(t − s)q−1 m(s)ds < lim inf r→∞

(r − ∑σi=1 ‖y i ‖)Γ(q) , Ω(r)

t ∈ J.

(4.17)

0

Then equation (4.12) has at least one solution u ∈ PC(J, X) and the set of solutions of equation (4.12) is bounded in PC(J, X). Proof. By (4.17), we know that there exists a constant K2 > 0 such that σ

∑ ‖y i ‖ + i=1

T

Ω(K2 ) ∫(t − s)q−1 m(s)ds ≤ K2 , Γ(q)

t ∈ J.

0

The rest of the proof follows the lines of the proof of Theorem 4.30, so we do not give the details here. The proof of the following result on uniqueness of solutions is standard and trivial, so we omit it. Theorem 4.32. Let the assumptions of Theorem 4.30 (or Theorem 4.31) and (H2+ ) hold. Then equation (4.12) has a unique solution u ∈ PC(J, X).

4.2.5 Existence and uniqueness theorems of solutions for Problem (III) This subsection deals with existence and uniqueness of solutions for Problem (III). Our results will cover the cases of the nonlinearity term f ̄ taking values in X β or X. 4.2.5.1 Case 1: f ̄ takes values in X β We make the following assumptions: (H5) f ̄ : J × X β → X β satisfies the Carathéodory type conditions. (H6) There exist a function m ∈ L1/q2 (J, ℝ+ ), q2 ∈ [0, q) and a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that ‖f ̄(t, u)‖β ≤ m(t)Ω(‖u‖β ) for all u ∈ X β and a.e. t ∈ J.

234 | 4 Fractional Evolution Equations: Continued (H7) There exists L ∈ L1/q3 (J, ℝ+ ), q3 ∈ [0, q) such that for any bounded D ⊂ X β , χ(f ̄(t, D)) ≤ L(t)χ(D) for a.e. t ∈ J. (H8) g : C(J, X β ) → X β is continuous, compact, and there exists a constant l g > 0 such that ‖g(u)‖β ≤ l g ‖u‖∞ for arbitrary u ∈ C(J, X β ). (H9) The inequality T

Ml g r +

MΩ(r) ∫(t − s)q−1 m(s)ds ≤ r Γ(q) 0

has at least one positive solution. Let us recall the definition and lemma of mild solutions for Problem (III). For more details, we refer the reader to Wang and Zhou [301] and Zhou and Jiao [334]. Definition 4.33. The mild solution of (4.13) is a function u ∈ C(J, X β ) satisfying t

u(t) = T (t)g(u) + ∫(t − s)q−1 S (t − s)f ̄(s, u(s))ds,

t∈J

0

(see Definition 4.5). The following boundedness properties of T and S are widely used in this subsection. Lemma 4.34 ([301, Lemma 2.9 (5)]). For fixed t ≥ 0 and any x ∈ X β , ‖T (t)x‖β ≤ M‖x‖β ,

‖S (t)x‖β ≤

M ‖x‖β . Γ(q)

We remark that we do not need the compactness of the semigroup {S(t) | t ≥ 0} in this subsection, we only need the continuity of T (⋅) and S (⋅). Lemma 4.35. Mappings T : (0, ∞) → L(X β ) and S : (0, ∞) → L(X β ) are continuous. Proof. For 0 ≤ t1 < t2 ≤ T, x ∈ D with the unit closed ball D of X β , we have ∞

q 󵄩󵄩 󵄩 󵄩 q 󵄩 󵄩󵄩T (t2 )x − T (t1 )x󵄩󵄩󵄩β ≤ ∫ ξ q (θ)󵄩󵄩󵄩S(t2 θ)x − S(t1 θ)x󵄩󵄩󵄩β dθ, 0 ∞

q 󵄩󵄩 󵄩 󵄩 q 󵄩 󵄩󵄩S (t2 )x − S (t1 )x󵄩󵄩󵄩β ≤ ∫ θξ q (θ)󵄩󵄩󵄩S(t2 θ)x − S(t1 θ)x󵄩󵄩󵄩β dθ. 0

Since −A is the generator of an analytic semigroup S(t), then S(t) is an equicontinuq q ous C0 -semigroup on (0, ∞). Thus, ‖S(t2 θ)x − S(t1 θ)x‖β → 0 with respect to x ∈ D as

4.2 Abstract Cauchy problems for fractional evolution equations | 235

t2 → t1 ∈ (0, ∞) for each fixed θ ∈ (0, ∞). Since ∞

∞

∫ ξ q (θ)dθ = 1

and

∫ θξ q (θ)dθ = 0

0

Γ(2) , Γ(1 + q)

one can observe the results immediately. Next, we define an operator F3 : C(J, X β ) → C(J, X β ) as follows: t

(F3 u)(t) = T (t)g(u) + ∫(t − s)q−1 S (t − s)f ̄(s, u(s))ds,

t ∈ J.

0

Lemma 4.36. Let K3 > 0 satisfy (H9). Set W3 = {u ∈ C(J, X β ) | ‖u‖∞ ≤ K3 } ⊆ C(J, X β ). Then B0 = conv(F3 W3 ) is equicontinuous. Proof. By Proposition A.3, it suffices to show that F3 (W3 ) ⊂ C(J, X β ) is equicontinuous. In fact, for any u ∈ W3 , 0 ≤ t1 < t2 ≤ T, we get 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩(F3 u)(t1 ) − (F3 u)(t2 )󵄩󵄩󵄩β ≤ 󵄩󵄩󵄩T (t1 )g(u) − T (t2 )g(u)󵄩󵄩󵄩β t

󵄩󵄩 1 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 [S (t1 − s) − S (t2 − s)]f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 󵄩 󵄩β 0 t

󵄩󵄩 󵄩󵄩 1 + 󵄩󵄩󵄩󵄩∫[(t1 − s)q−1 − (t2 − s)q−1 ]S (t2 − s)f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 󵄩 󵄩β 0 t

󵄩󵄩 2 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(t2 − s)q−1 S (t2 − s)f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 . 󵄩 󵄩β t1

Denote 󵄩 󵄩 I1 = 󵄩󵄩󵄩T (t1 )g(u) − T (t2 )g(u)󵄩󵄩󵄩β , t

󵄩󵄩 1 󵄩󵄩 I2 = 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 [S (t1 − s) − S (t2 − s)]f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 , 󵄩 󵄩β 0 t

󵄩󵄩 1 󵄩󵄩 I3 = 󵄩󵄩󵄩󵄩∫[(t1 − s)q−1 − (t2 − s)q−1 ]S (t2 − s)f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 , 󵄩 󵄩β 0 t

󵄩󵄩 󵄩󵄩 2 I4 = 󵄩󵄩󵄩󵄩∫(t2 − s)q−1 S (t2 − s)f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 . 󵄩 󵄩β t1

Next, we need to check that I1 , I2 , I3 , I4 tend to 0 independently of u ∈ W3 when t2 → t1 .

236 | 4 Fractional Evolution Equations: Continued For I1 , from T : (0, ∞) → L(X β ) we get the uniform continuity of (t, x) 󳨃→ T (t)x on J × D1 for any compact subset D1 ⊂ X β . So from (H8) we obtain that limt2 →t1 I1 = 0 uniformly on J. For I2 , we use for ϵ > 0 small and t1 > ϵ that I2 ≤ I2ε + I2,ε , where t −ϵ

I2ε

󵄩󵄩 1 󵄩󵄩 = 󵄩󵄩󵄩󵄩 ∫ (t1 − s)q−1 [S (t1 − s) − S (t2 − s)]f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 , 󵄩 󵄩β 0 t

󵄩󵄩 1 󵄩󵄩 I2,ε = 󵄩󵄩󵄩󵄩 ∫ (t1 − s)q−1 [S (t1 − s) − S (t2 − s)]f ̄(s, u(s))ds󵄩󵄩󵄩󵄩 . 󵄩 󵄩β t1 −ϵ

Next, for s ∈ [0, t1 − ϵ] we have T ≥ t2 − s ≥ t1 − s ≥ ϵ, and then t1 −ϵ

I2ε

≤ Ω(K3 )

≤ Ω(K3 )

max

s1 ,s2 ∈[ϵ,T] s1 ≤s2 ≤s1 +δ

max

s1 ,s2 ∈[ϵ,T] s1 ≤s2 ≤s1 +δ

‖S (s1 ) − S (s2 )‖β ∫ (t1 − s)q−1 m(s)ds 0

‖S (s1 ) − S (s2 )‖β

Moreover, we have

‖m‖

1

L q2 (J,ℝ+ ) q−q2 T , q−q2 1−q2 ( 1−q2 )

t2 − t1 < δ.

M ‖m‖L q2 (J,ℝ+ ) q−q2 ϵ . Γ(q) ( q−q2 )1−q2 1

I2,ε ≤ 2Ω(K3 )

1−q2

Keeping in mind lim

max

δ→0 s1 ,s2 ∈[ϵ,T] s1 ≤s2 ≤s1 +δ

‖S (s1 ) − S (s2 )‖β = 0,

for 0 ≤ t1 ≤ ϵ we derive M ‖m‖L q2 (J,ℝ+ ) q−q2 ϵ . Γ(q) ( q−q2 )1−q2 1

I2 ≤ 2Ω(K3 )

1−q2

Summarizing, we see that for a given ε > 0 there are ϵ > 0 and δ > 0 so that for any 0 ≤ t1 ≤ t2 ≤ T with t2 − t1 < δ, it holds I2 < ε, which implies that limt2 →t1 I2 = 0 uniformly on J. For I3 , we use t1

∫((t1 − s)q−1 − (t2 − s)q−1 )m(s)ds ≤ 0

This gives limt2 →t1 I3 = 0 uniformly on J.

‖m‖

1

L q2 (J,ℝ+ ) (t2 q−q2 1−q2 ( 1−q2 )

− t1 )q−q2 .

4.2 Abstract Cauchy problems for fractional evolution equations | 237

For I4 , we use t2

∫(t2 − s)q−1 m(s)ds ≤ t1

‖m‖

1

L q2 (J,ℝ+ ) (t2 q−q2 1−q2 ( 1−q2 )

− t1 )q−q2 .

This also gives limt2 →t1 I4 = 0 uniformly on J. Therefore, F3 (W3 ) is equicontinuous. Lemma 4.37. The operator F3 : C(J, X β ) → C(J, X β ) is continuous. Proof. Let {u n } be a sequence of a bounded set W3 such that u n → u in W3 . We have to show that ‖F3 u n − F3 u‖∞ → 0. Keeping in mind (H8), we have g(u n ) → g(u) as n → ∞. It is easy to see that f ̄(s, u n (s)) → f ̄(s, u(s)) as n → ∞ due to the Carathéodory continuity of f ̄. Using (H6), we get for each t ∈ J, 󵄩 󵄩 (t − s)q−1 󵄩󵄩󵄩f ̄(s, u n (s)) − f ̄(s, u(s))󵄩󵄩󵄩β ≤ (t − s)q−1 2m(s)Ω(K3 ). On the other hand, using the fact that the function s 󳨃→ (t − s)q−1 2m(s)Ω(K3 ) is integrable for s ∈ [0, t], t ∈ J, we deduce from the Lebesgue dominated convergence theorem that t

󵄩 󵄩 ∫(t − s)q−1 󵄩󵄩󵄩f ̄(s, u n (s)) − f ̄(s, u(s))󵄩󵄩󵄩β ds → 0 as n → ∞. 0

Then, for all t ∈ J, 󵄩󵄩 󵄩 󵄩󵄩(F3 u n )(t) − (F3 u)(t)󵄩󵄩󵄩β ≤ M‖g(u n ) − g(u)‖β t

M 󵄩 󵄩 + ∫(t − s)q−1 󵄩󵄩󵄩f ̄(s, u n (s)) − f ̄(s, u(s))󵄩󵄩󵄩β ds → 0 Γ(q)

as n → ∞.

0

Therefore, F3 u n → F3 u pointwisely on J as n → ∞. So, Lemma 4.36 implies that F3 u n → F3 u uniformly on J as n → ∞, and F3 is continuous. Lemma 4.38. The operator F3 : W3 → W3 is bounded. Proof. For any u ∈ W3 , we have t

M ‖(F3 u)(t)‖β ≤ Ml g K3 + ∫(t − s)q−1 ‖f ̄(s, u(s))‖β ds Γ(q) 0 t

≤ Ml g K3 +

MΩ(K3 ) ∫(t − s)q−1 m(s)ds ≤ K3 , Γ(q) 0

due to (H9), which implies that F3 : W3 → W3 is bounded.

238 | 4 Fractional Evolution Equations: Continued

Now, we are ready to prove the main results of this subsection. Theorem 4.39. Assume that (H5)–(H9) hold. Then equation (4.13) has at least one solution u ∈ C(J, X β ) and the set of solutions of equation (4.13) is bounded in C(J, X β ). Proof. Let F3 : C(J, X β ) → C(J, X β ) be the operator defined in the beginning of this subsection. For any B ⊂ B0 , by the definition of operator F and Lemmas 4.36–4.38, we get that F3 (B) is bounded and equicontinuous on J. For any B ⊂ B0 , we know from Propositions A.22–A.24 and (H8), that for any ε > 0 there is a sequence {u n }∞ n=1 ⊂ B such that χ(F31 B(t)) = χ(F3 B(t)) t

≤ 2χ({ ∫(t − s)q−1 S (t − s)f(s, {u n (s)}∞ n=1 )ds}) + ε 0 t

4M ≤ ∫(t − s)q−1 χ(f(s, {u n (s)}∞ n=1 ))ds + ε Γ(q) 0 t

≤

4Mχ({u n }∞ n=1 ) ∫(t − s)q−1 L(s)ds + ε Γ(q) 0 t

≤

4Mχ(B) ∫(t − s)q−1 L(s)ds + ε. Γ(q) 0

There is a continuous function ϕ : J → ℝ+ and M = max{|ϕ(t)| | t ∈ J} such that for arbitrary fixed 0 < γ̄ < Γ(q) 4M , T

∫(t − s)q−1 |L(s) − ϕ(s)|ds < γ.̄ 0

Then t

χ(F31 B(t))

t

4Mχ(B) ≤ [ ∫(t − s)q−1 |L(s) − ϕ(s)|ds + ∫(t − s)q−1 |ϕ(s)|ds] + ε Γ(q) 0

≤

0

Mt q

4Mχ(B) [γ̄ + ] + ε. Γ(q) q

Since ε > 0 is arbitrary, we have χ(F31 B(t)) ≤ (ā + bt̄ q )χ(B), where ā =

4M γ̄ Γ(q)

and b̄ =

̄ aM q γ̄ .

4.2 Abstract Cauchy problems for fractional evolution equations | 239

1 From Proposition A.24, for any ε > 0 there exists a sequence {y n }∞ n=1 ⊂ conv(F 3 B) such that

χ(F32 B(t)) = χ(F3 (conv(F31 B(t)))) t

≤ 2χ({ ∫(t − s)q−1 S (t − s)f(s, {y n (s)}∞ n=1 )ds}) + ε 0 t

≤

4M ∫(t − s)q−1 χ(f(s, {y n (s)}∞ n=1 ))ds + ε Γ(q) 0 t

≤

ā ∫(t − s)q−1 L(s)χ({y n (s)}∞ n=1 )ds + ε γ̄ 0

t

≤

ā ∫(t − s)q−1 L(s)χ(F31 B(s))ds + ε γ̄ 0

t

ā ≤ ∫(t − s)q−1 |L(s) − ϕ(s)|ds(ā + bt̄ q )χ(B) γ̄ 0

t

+

ā ̄ q )dsχ(B) + ε ∫(t − s)q−1 |ϕ(s)|(ā + bs γ̄ 0

t

̄ aM ̄ q )dsχ(B) + ε ≤ a(̄ ā + bt̄ q )χ(B) + ∫(t − s)q−1 (ā + bs γ̄ 0

≤ [ā 2 + 2ā b̄

Γ(q + 1) q Γ2 (q + 1) 2q t + b̄ 2 t ]χ(B) + ε. Γ(q + 1) Γ(2q + 1)

It can be shown, by mathematical induction, that for every n̄ ∈ ℕ, n̄

Γ (q + 1) iq ̄ χ(F3n̄ B(t)) ≤ [ ∑ C in ā n−i b̄ i t ]χ(B). Γ(iq + 1) i=0 i

Thanks to Theorem 4.14, there exists a positive integer n̂ such that n̂

Γ (q + 1) iq ̂ b̄ i t = ρ < 1. ∑ C in ā n−i Γ(iq + 1) i=0 i

Hence, χ(F3n̂ B(t)) ≤ ρχ(B). It follows from Theorem 4.15 that F3 has a fixed point, which is a mild solution to our equation.

240 | 4 Fractional Evolution Equations: Continued

Theorem 4.40. If (H5)–(H8) hold, then equation (4.13) has at least one solution u ∈ C(J, X β ) provided that T

∫(t − s)q−1 m(s)ds < lim inf

r − Ml g r

r→∞

MΩ(r) Γ(q)

0

,

t ∈ J.

(4.18)

Moreover, the set of solutions of equation (4.13) is bounded in C(J, X β ). Proof. By (4.18), we know that there exists a constant K4 > 0 such that T

∫(t − s)q−1 m(s)ds ≤ 0

K4 − Ml g K4 MΩ(K4 ) Γ(q)

,

t ∈ J.

The rest of the proof follows the very same lines of the proof of Theorem 4.39, so we do not give the details. In order to obtain the uniqueness of solutions, we add the following assumption: (H6+ ) There exists a constant ā f > 0 such that ‖f(t, u) − f(t, v)‖β ≤ ā f ‖u − v‖β

for all t ∈ J, u, v ∈ X β .

(H8+ ) There exists a constant l ḡ > 0 such that ‖g(u) − g(v)‖β ≤ l ḡ ‖u − v‖∞

for all u, v ∈ C(J, X β ).

Lastly, we state the following uniqueness result. Theorem 4.41. Let the assumptions of Theorem 4.39 (or Theorem 4.40) and (H6+ ), (H8+ ) hold. Then equation (4.13) has a unique solution u ∈ C(J, X β ). Proof. By Theorem 4.39, equation (4.13) has a solution u(⋅) in C(J, X β ). Let v(⋅) be another solution of equation (4.13) with initial value v0 . Note that by (H6+ ) and (H8+ ), we obtain t

‖u(t) − v(t)‖β ≤ M l ḡ ‖u0 − v0 ‖β +

ā f M ∫(t − s)q−1 ‖u(s) − v(s)‖β ds. Γ(q) 0

This yields the uniqueness of u(⋅) due to the standard singular Gronwall’s inequality.

4.2.5.2 Case 2: f ̄ takes values in X In addition to (H8), we replace (H5), (H6), (H7) and (H9) in Section 4.2.5.1 by the following.

4.2 Abstract Cauchy problems for fractional evolution equations | 241

(H5󸀠 ) f ̄ : J × X β → X satisfies the Carathéodory type conditions. (H6󸀠 ) There exist a function m ∈ L1/q2 (J, ℝ+ ), q2 ∈ [0, q(1 − β)) and a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that ‖f ̄(t, u)‖ ≤ m(t)Ω(‖u‖β )

(H7󸀠 )

for all u ∈ X β and a.e. t ∈ J. There exists L ∈ L1/q3 (J, ℝ+ ), q3 ∈ [0, q(1 − β)) such that for any bounded D ⊂ Xβ , χ(f ̄(t, D)) ≤ L(t)χ(D)

for a.e. t ∈ J. (H9󸀠 ) The inequality T

Ml g r +

M β Γ(2 − β)Ω(r) ∫(t − s)q(1−β)−1 m(s)ds ≤ r Γ(q(1 − β)) 0

has at least one positive solution, where M β is given in Lemma 4.42. Lemma 4.42 ([301, Lemma 2.9 (4)]). For any x ∈ X, η ∈ (0, 1) and β ∈ (0, 1), we have AS (t)x = A1−η S (t)A η x, ‖A β S (t)‖ ≤

t ∈ J,

M β Γ(2 − β) −βq t , Γ(q(1 − β))

0 < t ≤ T.

Let F3 : C(J, X β ) → C(J, X β ) be the operator defined in Section 4.2.5.1. Repeating the proofs of Lemmas 4.36–4.38, one can verify that F3 is continuous, bounded, and that B0 is also equicontinuous after some computation. We remark that there are two main differences: the singular kernel (t − s)q−1 is changed to (t − s)q(1−β)−1 , and the term M β Γ(2−β) M Γ(q) is changed to Γ(q(1−β)) in the whole computation. Now, we can state the main results of this subsection. Theorem 4.43. Assume that (H5󸀠 ), (H6󸀠 ), (H7󸀠 ), (H8) and (H9󸀠 ) hold. Then equation (4.13) has at least one solution u ∈ C(J, X β ) and the set of solutions of equation (4.13) is bounded in C(J, X β ). Proof. For any B ⊂ B0 , since F3 is continuous, bounded, and B0 is equicontinuous, one obtains that F3 (B) is bounded and equicontinuous on J. For any B ⊂ B0 , we know from Propositions A.22–A.24, (H8) and Lemma 4.42, that for any ε > 0 there is a sequence {u n }∞ n=1 ⊂ B such that t

χ(F31 B(t)) = χ(F3 B(t)) ≤

4M β Γ(2 − β)χ(B) ∫(t − s)q(1−β)−1 L(s)ds + ε. Γ(q(1 − β)) 0

There is a continuous function ϕ : J → ℝ+ and M = max{|ϕ(t)| | t ∈ J} such that for

242 | 4 Fractional Evolution Equations: Continued arbitrary fixed 0 < γ̄
0 is arbitrary, we have χ(F31 B(t)) ≤ (ā + bt̄ q(1−β) )χ(B), where ā =

and b̄ =

4M β Γ(2−β)γ̄ Γ(q(1−β))

̄ aM q(1−β)γ̄ .

1 From Proposition A.24, for any ε > 0 there is a sequence {y n }∞ n=1 ⊂ conv(F 3 B) such

that χ(F32 B(t)) = χ(F3 (conv(F31 B(t)))) t

4M β Γ(2 − β) ≤ ∫(t − s)q(1−β)−1 χ(f(s, {y n (s)}∞ n=1 ))ds + ε Γ(q(1 − β)) 0

t

≤

ā ∫(t − s)q(1−β)−1 L(s)χ({y n (s)}∞ n=1 )ds + ε γ̄ 0

t

ā ≤ ∫(t − s)q(1−β)−1 L(s)χ(F31 B(s))ds + ε γ̄ 0

t

≤

ā ∫(t − s)q(1−β)−1 |L(s) − ϕ(s)|ds(ā + bt̄ q(1−β) )χ(B) γ̄ 0

t

+

ā ̄ q(1−β) )dsχ(B) + ε ∫(t − s)q(1−β)−1 |ϕ(s)|(ā + bs γ̄ 0

t

̄ aM ̄ q(1−β) )dsχ(B) + ε ≤ a(̄ ā + bt̄ q(1−β) )χ(B) + ∫(t − s)q(1−β)−1 (ā + bs γ̄ 0

Γ(q(1 − β) + 1) q(1−β) Γ2 (q(1 − β) + 1) 2q(1−β) t + b̄ 2 t ≤ [ā 2 + 2ā b̄ ]χ(B) + ε. Γ(q(1 − β) + 1) Γ(2q(1 − β) + 1)

4.3 Volterra–Fredholm type fractional evolution equations | 243

It can be shown, by mathematical induction, that for every n̄ ∈ ℕ, n̄

Γ (q(1 − β) + 1) iq(1−β) ̄ t χ(F3n̄ B(t)) ≤ [ ∑ C in ā n−i b̄ i ]χ(B). Γ(iq(1 − β) + 1) i=0 i

Due to Theorem 4.14, there exists a positive integer n̂ such that n̂

̂ b̄ i ∑ C in ā n−i i=0

Γ i (q(1 − β) + 1) iq(1−β) t = ρ < 1. Γ(iq(1 − β) + 1)

Hence, χ(F3n̂ B(t)) ≤ ρχ(B). It follows from Theorem 4.15 that F3 has a fixed point, which is a mild solution to our equation. Theorem 4.44. If (H5󸀠 ), (H6󸀠 ), (H7󸀠 ) and (H8) hold, then equation (4.13) has at least one solution u ∈ C(J, X β ) provided that T

∫(t − s)q(1−β)−1 m(s)ds < lim inf r→∞

0

r − Ml g r M β Γ(2−β)Ω(r) Γ(q(1−β))

,

t ∈ J.

(4.19)

Moreover, the set of solutions of equation (4.13) is bounded in C(J, X β ). Proof. By (4.19), we know that there exists a constant K4 > 0 such that T

∫(t − s)q(1−β)−1 m(s)ds ≤ 0

K4 − Ml g K4 M β Γ(2−β)Ω(K4 ) Γ(q(1−β))

,

t ∈ J.

The rest of the proof follows the very same lines of the proof of Theorem 4.43, so we do not give the details here. To end this subsection, we also give the following uniqueness result. The proof is similar to the proof of Theorem 4.41, so we omit it. (H6󸀠+ ) There exists a constant ā f > 0 such that ‖f(t, u) − f(t, v)‖ ≤ ā f ‖u − v‖β

for all t ∈ J, u, v ∈ X β .

Theorem 4.45. Let the assumptions of Theorem 4.43 (or Theorem 4.44), (H6󸀠+ ) and (H8+ ) hold. Then equation (4.13) has a unique solution u ∈ C(J, X β ).

4.3 Nonlocal Cauchy problems for Volterra–Fredholm type fractional evolution equations 4.3.1 Introduction Motivated by [67, 179, 183, 200, 299, 302, 318, 333], the main purpose of this section is to consider nonlocal Cauchy problems for fractional evolution equations involving

244 | 4 Fractional Evolution Equations: Continued

Volterra–Fredholm type integral operators such as c

{

q

D t x(t) = −Ax(t) + t n f (t, x(t), (Kx)(t), (Hx)(t)), t ∈ J = [0, T], n ∈ ℤ+ , q ∈ (0, 1), x(0) = g(x) + x0 ,

(4.20) q where the fractional derivative c D t is understood in the Caputo sense, −A : D(A) → X is the infinitesimal generator of a compact analytic semigroup of uniformly bounded linear operators {S(t) | t ≥ 0}, the Volterra type integral operator K and the Fredholm type integral operator H are defined by t

T

(Kx)(t) = ∫ k(t, s, x(s))ds, 0

(Hx)(t) = ∫ h(t, s, x(s))ds. 0

The function f : J × X α × X α × X α → X (or X α , or X μ ) is continuous where X α = D(A α ), 0 ≤ μ ≤ α ≤ 1, is a Banach space with the norm ‖x‖α = ‖A α x‖ for x ∈ X α . Functions f , k, h and g are specified later. We remark that the term t n will help us to overcome essential difficulties arising from the singular term (t − s)q−1 in the formula for solutions due to the well-known beta function. In this section, we discuss the existence and uniqueness of mild solutions for system (4.20). Our results cover the cases for the nonlinear term f taking values in the spaces X, X α , X μ , 0 ≤ μ ≤ α ≤ 1, where the nonlocal term g is compact and continuous or satisfies the Lipschitz continuity condition. This section is based on [298].

4.3.2 Preliminaries We denote by X a Banach space with the norm ‖⋅‖, and −A : D(A) → X is the infinitesimal generator of a compact analytic semigroup of uniformly bounded linear operators {S(t) | t ≥ 0}. This means that there exists M > 1 such that ‖S(t)‖ ≤ M. We assume without loss of generality that 0 ∈ ρ(A). This allows us to define the fractional power A α for 0 ≤ α ≤ 1, as a closed linear operator on its domain D(A α ) with inverse A−α (see [220]). We denote by C α the Banach space C(J, X α ) endowed with supremum norm, ‖x‖∞ = supt∈J ‖x(t)‖α for x ∈ C α . Motivated by [333, Definition 3.1], we adopt the following concept of mild solution for our problem. Definition 4.46. By the mild solution of system (4.20), we mean a function x : J → X α satisfying t

x(t) = T (t)[x0 + g(x)] + ∫(t − s)q−1 s n S (t − s)f (s, x(s), (Kx)(s), (Hx)(s))ds 0

for any t ∈ J (see Definition 4.5).

4.3 Volterra–Fredholm type fractional evolution equations | 245

Remark 4.47. It is not difficult to verify that for v ∈ [0, 1], ∞

∞

∫ θ v ξ q (θ)dθ = ∫ θ−qv ϖ q (θ)dθ = 0

0

Γ(1 + v) . Γ(1 + qv)

The following results, extensions of Lemmas 4.2 and 4.6, are very useful and will be used throughout this section. Lemma 4.48 ([302, Lemma 2.9]). The operators T and S have the following properties: (1) For fixed t ≥ 0, T (t) and S (t) are linear and bounded operators, i.e., for any x ∈ X, ‖T (t)x‖ ≤ M‖x‖,

‖S (t)x‖ ≤

M ‖x‖. Γ(q)

(2) {T (t) | t ≥ 0} and {S (t) | t ≥ 0} are strongly continuous. (3) For every t > 0, T (t) and S (t) are compact operators. (4) For any x ∈ X, β ∈ [0, 1] and α ∈ [0, 1], we have AS (t)x = A1−β S (t)A β x, ‖A α S (t)‖ ≤

t ∈ J,

M α qΓ(2 − α) −αq t , Γ(1 + q(1 − α))

0 < t ≤ T.

(5) For fixed t ≥ 0 and any x ∈ X α , we have ‖T (t)x‖α ≤ M‖x‖α ,

‖S (t)x‖α ≤

M ‖x‖α . Γ(q)

(6) For a positive number μ with 0 ≤ μ ≤ α ≤ 1, fixed t ≥ 0 and any x ∈ X μ , we have ‖T (t)x‖α ≤ ‖A α−μ ‖ M ‖x‖μ ,

‖S (t)x‖α ≤ ‖A α−μ ‖

M ‖x‖μ . Γ(q)

(7) Tα (t) and Sα (t), t > 0 are uniformly continuous, i.e., for any fixed t > 0 and ϵ > 0 there exists h > 0 such that

where

‖Tα (t + ϵ) − Tα (t)‖α < ε,

t + ϵ ≥ 0, |ϵ| < h,

‖Sα (t + ϵ) − Sα (t)‖α < ε,

t + ϵ ≥ 0, |ϵ| < h,

∞

Tα (t) = ∫ ξ q (θ)S α (t q θ)dθ, 0

∞

Sα (t) = q ∫ θξ q (θ)S α (t q θ)dθ. 0

4.3.3 Existence of mild solutions In this subsection, we prove the existence and uniqueness of mild solutions for system (4.20). First, we make the following assumptions:

246 | 4 Fractional Evolution Equations: Continued [Hf1] f : J × X α × X α × X α → X is continuous and there exist m1 , m2 , m3 > 0 such that ‖f(t, x1 , x2 , x3 ) − f(t, y1 , y2 , y3 )‖ ≤ m1 ‖x1 − y1 ‖α + m2 ‖x2 − y2 ‖α + m3 ‖x3 − y3 ‖α for all x i , y i ∈ X α , i = 1, 2, 3 and t ∈ J. [Hk1] Let D k = {(t, s) ∈ ℝ2 | 0 ≤ s ≤ t ≤ T}. The function k : D k × X α → X α is continuous and there exists m k (t, s) ∈ C(D k , ℝ+ ) such that ‖k(t, s, x) − k(t, s, y)‖α ≤ m k (t, s)‖x − y‖α for each (t, s) ∈ D k and x, y ∈ X α . We set t

K = max ∫ m k (t, s)ds. ∗

t∈J

0

[Hh1] Let D h = {(t, s) ∈ ℝ2 | 0 ≤ s, t ≤ T}. The function h : D h × X α → X α is continuous and there exists m h (t, s) ∈ C(D h , ℝ+ ) such that ‖h(t, s, x) − h(t, s, y)‖α ≤ m h (t, s)‖x − y‖α for each (t, s) ∈ D h and x, y ∈ X α . We set T

H = max ∫ m h (t, s)ds. ∗

t∈J

0

[Hg1] g : C α → X α and there exists a constant l g > 0 such that ‖g(x) − g(y)‖α ≤ l g ‖x − y‖∞

for all x, y ∈ C α .

[HΩ] A constant Ω n,α,q,T defined by Ω n,α,q,T := Ml g +

M α qΓ(2 − α)B(q, n + 1) n+(1−α)q T (m1 + K ∗ m2 + H ∗ m3 ) Γ(1 + q(1 − α))

satisfies Ω n,α,q,T < 1, where B(⋅, ⋅) denotes the beta function. Now we are ready to give our first result based on the Banach contraction mapping principle. Theorem 4.49. Assume that [Hf1], [Hk1], [Hh1], [Hg1] and [HΩ] are satisfied. If x0 ∈ X α then system (4.20) has a unique mild solution x ∈ C α . Proof. Define the function Γ : C α → C α by (Γx)(t) = T (t)[x0 + g(x)] t

+ ∫(t − s)q−1 s n S (t − s)f (s, x(s), (Kx)(s), (Hx)(s))ds, 0

t ∈ J.

(4.21)

4.3 Volterra–Fredholm type fractional evolution equations | 247

Note that Γ is well defined on C α . Now, take t ∈ J and x, y ∈ C α . Then we have ‖(Γx)(t) − (Γy)(t)‖α ≤ ‖T (t)(g(x) − g(y))‖α t

󵄩 + ∫(t − s)q−1 s n 󵄩󵄩󵄩S (t − s)[f(s, x(s), (Kx)(t), (Hx)(t)) 0

󵄩 − f(s, y(s), (Ky)(s), (Hy)(s))]󵄩󵄩󵄩α ds ≤ M‖g(x) − g(y)‖α t

󵄩 + ∫(t − s)q−1 s n ‖A α S (t − s)‖ 󵄩󵄩󵄩f(s, x(s), (Kx)(s), (Hx)(s)) 0

󵄩 − f(s, y(s), (Ky)(s), (Hy)(s))󵄩󵄩󵄩ds, from which, according to [Hf1], [Hk1], [Hh1], [Hg1], (4)–(5) of Lemma 4.48 and Hölder’s inequality, we obtain ‖(Γx)(t) − (Γy)(t)‖α t

≤ Ml g ‖x − y‖∞ + M α qt−αq

Γ(2 − α) m1 ∫(t − s)q−1 s n ‖x(s) − y(s)‖α ds Γ(1 + q(1 − α)) 0

t

+ M α qt−αq

Γ(2 − α) m2 ∫(t − s)q−1 s n ‖(Kx)(s) − (Ky)(s)‖α ds Γ(1 + q(1 − α)) 0

t

+ M α qt

−αq

Γ(2 − α) m3 ∫(t − s)q−1 s n ‖(Hx)(s) − (Hy)(s)‖α ds Γ(1 + q(1 − α)) 0

Γ(2 − α) ≤ Ml g ‖x − y‖∞ + M α qt−αq (m1 + K ∗ m2 + H ∗ m3 ) Γ(1 + q(1 − α)) t

× ‖x − y‖∞ ∫(t − s)q−1 s n ds 0

M α qΓ(2 − α)B(q, n + 1) n+(1−α)q t (m1 + K ∗ m2 + H ∗ m3 )}‖x − y‖∞ , ≤ {Ml g + Γ(1 + q(1 − α)) due to t

∫(t − s)q−1 s n ds = B(q, n + 1)t n+q , 0

‖(Kx)(s) − (Ky)(s)‖α ≤ K ∗ ‖x − y‖∞ ,

‖(Hx)(s) − (Hy)(s)‖α ≤ H ∗ ‖x − y‖∞ .

248 | 4 Fractional Evolution Equations: Continued

Therefore, we can deduce that ‖Γx − Γy‖∞ ≤ {Ml g +

M α qΓ(2 − α)B(q, n + 1) n+(1−α)q t (m1 + K ∗ m2 + H ∗ m3 )}‖x − y‖∞ Γ(1 + q(1 − α))

≤ Ω n,α,q,T ‖x − y‖∞ . Hence, in view of the contraction mapping principle, [HΩ] allows us to conclude that Γ has a unique fixed point x ∈ C α , and t

x(t) = T (t)[x0 + g(x)] + ∫(t − s)q−1 s n S (t − s)f(s, x(s), (Kx)(s), (Hx)(s))ds 0

which is the unique mild solution of system (4.20). Our second result uses the Schauder fixed point theorem. We assume the following conditions. [Hf2] The function f : J × X α × X α × X α → X α is continuous and there exists a positive function ρ ∈ L p (J, ℝ+ ) for some p ∈ ( 1q , ∞) such that ‖f(t, x, y, z)‖α ≤ ρ(t) for all x, y, z ∈ X α and t ∈ J. [Hk2] The function k : D k × X α → X α is continuous and there exists L1 > 0 such that ‖k(t, s, x) − k(t, s, y)‖α ≤ L1 ‖x − y‖α for each (t, s) ∈ D k and x, y ∈ X α . [Hh2] The function h : D h × X α → X α is continuous and there exists L2 > 0 such that ‖h(t, s, x) − h(t, s, y)‖α ≤ L2 ‖x − y‖α for each (t, s) ∈ D h and x, y ∈ X α . [Hg2] The function g : C α → X α is compact, continuous and there exist β1 ≥ 0, β2 ≥ 0 such that ‖g(x)‖α ≤ β1 ‖x‖∞ + β2 . Now we are ready to state and prove the following existence result. Theorem 4.50. Assume that the conditions [Hf2], [Hk2], [Hh2], [Hg2] are satisfied. If x0 ∈ X α then system (4.20) has at least one mild solution on J provided that Mβ1 < 1. Proof. Define the function F : C α → C α by t

(Fx)(t) = T (t)[x0 + g(x)] + ∫(t − s)q−1 s n S (t − s)f(s, x(s), (Kx)(s), (Hx)(s))ds, 0

4.3 Volterra–Fredholm type fractional evolution equations | 249

and for n ∈ ℤ+ , we choose r such that r≥

M pq − 1 (n + 1)p − 1 1 B( , [M(‖x0 ‖α + β2 ) + ) 1 − Mβ1 Γ(q) p−1 p−1

p−1 p

T

pq+np−1 p

‖ρ‖L p (J,ℝ+ ) ].

Let B r = {x ∈ C α | ‖x‖∞ ≤ r}. Then we proceed in three steps. Step 1. We show that FB r ⊂ B r . Let x ∈ B r . Then for t ∈ J, using (5) of Lemma 4.48 and Hölder’s inequality, we have ‖(Fx)(t)‖α ≤ ‖T (t)(x0 + g(x))‖α t

󵄩 󵄩 + ∫(t − s)q−1 s n 󵄩󵄩󵄩S (t − s)f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds 0

≤ M(‖x0 ‖α + ‖g(x)‖α ) t

+

M 󵄩 󵄩 ∫(t − s)q−1 s n 󵄩󵄩󵄩f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds, Γ(q) 0

which according to [Hf2], [Hg2] and pq > 1 (⇔

(q−1)p p−1

> −1) gives

t

M ‖(Fx)(t)‖α ≤ M(‖x0 ‖α + β1 ‖x‖∞ + β2 ) + ( ∫(t − s)q−1 s n ρ(s)ds) Γ(q) 0 t

≤ M(‖x0 ‖α + β1 ‖x‖∞ + β2 ) +

(q−1)p np M ( ∫(t − s) p−1 s p−1 ds) Γ(q)

p−1 p

t

( ∫ ρ(s)p ds)

0

M pq − 1 (n + 1)p − 1 ≤ M(‖x0 ‖α + β1 ‖x‖∞ + β2 ) + B( , ) Γ(q) p−1 p−1 ×t

pq+np−1 p

‖ρ‖L p (J,ℝ+ ) ≤ r,

1 p

0 p−1 p

t ∈ J.

Hence, we deduce that ‖Fx‖∞ ≤ r. Step 2. We prove that F is continuous. Let {x m } be a sequence in B r such that x m → x in B r . It comes from the continuity of k, h and assumptions [Hk2], [Hh2] that s

s

∫ k(s, τ, x m (τ))dτ → ∫ k(s, τ, x(τ))dτ, 0

0

T

T

∫ h(s, τ, x m (τ))dτ → ∫ h(s, τ, x(τ))dτ 0

0

as m → ∞ uniformly in s ∈ J on C α . Consequently, f(s, x m (s), (Kx m )(s), (Hx m )(s)) → f(s, x(s), (Kx)(s), (Hx)(s)) as m → ∞, since the function f is continuous on J × X α × X α .

(4.22)

250 | 4 Fractional Evolution Equations: Continued

Further, one has g(x m ) → g(x) as m → ∞

(4.23)

because g is continuous on C α . Now for t ∈ J, according to [Hf2], [Hg2], (5) of Lemma 4.48 and Hölder’s inequality, we have 󵄩󵄩 󵄩 󵄩󵄩(Fx m )(t) − (Fx)(t)󵄩󵄩󵄩α t

󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩T (t)(g(x m ) − g(x))󵄩󵄩󵄩α + ∫(t − s)q−1 s n 󵄩󵄩󵄩S (t − s)[f(s, x m (s), (Kx m )(s), (Hx m )(s)) 0

󵄩 − f(s, x(s), (Kx)(s), (Hx)(s))]󵄩󵄩󵄩α ds t

M 󵄩 ≤ M‖g(x m ) − g(x)‖α + ∫(t − s)q−1 s n 󵄩󵄩󵄩f(s, x m (s), (Kx m )(s), (Hx m )(s)) Γ(q) 0

󵄩 − f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds t

≤ M‖g(x m ) − g(x)‖α +

(q−1)p np M ( ∫(t − s) p−1 s p−1 ds) Γ(q)

p−1 p

0 t

󵄩 󵄩p × ( ∫󵄩󵄩󵄩f(s, x m (s), (Kx m )(s), (Hx m )(s)) − f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩 ds)

1 p

0

M pq − 1 (n + 1)p − 1 ≤ M‖g(x m ) − g(x)‖α + B( , ) Γ(q) p−1 p−1

p−1 p

t

pq+np−1 p

t

1

p 󵄩 󵄩p × ( ∫󵄩󵄩󵄩f(s, x m (s), (Kx m )(s), (Hx m )(s)) − f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩 ds)

0

≤ M‖g(x m ) − g(x)‖α +

M pq − 1 (n + 1)p − 1 B( , ) Γ(q) p−1 p−1

p−1 p

T

pq+np−1 p

T

1

p 󵄩 󵄩p × ( ∫󵄩󵄩󵄩f(s, x m (s), (Kx m )(s), (Hx m )(s)) − f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩 ds) .

0

Therefore, using (4.22), (4.23), [Hf2] and the Lebesgue dominated convergence theorem, it can be easily shown that lim ‖Fx m − Fx‖∞ = 0 as m → ∞.

m→∞

That is, F is continuous. Step 3. We show that F is compact by using the Arzelà–Ascoli theorem. First we prove that {(Fx)(t) | x ∈ B r } is relatively compact in X α , for all t ∈ J. Obviously, {(Fx)(0) | x ∈ B r } is compact. Let t ∈ (0, T]. For each h ∈ (0, t), arbitrary δ > 0 and

4.3 Volterra–Fredholm type fractional evolution equations | 251

x ∈ B r , we define the operator F h by t−h

(F h,δ x)(t) = T (t)[x0 + g(x)] + S(h δ) ∫ (t − s)q−1 s n q

0 ∞

× (q ∫ θξ q (θ)S((t − s)q θ − h q δ)dθ)f(s, x(s), (Kx)(s), (Hx)(s))ds δ t−h ∞

= T (t)[x0 + g(x)] + q ∫ ∫ θ(t − s)q−1 s n ξ q (θ)S((t − s)q θ) 0 δ

× f(s, x(s), (Kx)(s), (Hx)(s))dθds. From the above expression, we can see that the sets {(F h,δ x)(t) | x ∈ B r } are also relatively compact in X α , since the operator S α (h q δ) is compact in X α for h q δ > 0. Moreover, using [Hf2] and Hölder’s inequality, we have 󵄩󵄩 󵄩 󵄩󵄩(Fx)(t) − (F h,δ x)(t)󵄩󵄩󵄩α t δ

󵄩󵄩 󵄩󵄩 ≤ q󵄩󵄩󵄩󵄩∫ ∫ θ(t − s)q−1 s n ξ q (θ)S((t − s)q θ)f(s, x(s), (Kx)(s), (Hx)(s))dθds󵄩󵄩󵄩󵄩 󵄩 󵄩α 0 0

t ∞

󵄩󵄩 + q󵄩󵄩󵄩󵄩∫ ∫ θ(t − s)q−1 s n ξ q (θ)S((t − s)q θ)f(s, x(s), (Kx)(s), (Hx)(s))dθds 󵄩 0 δ

t−h ∞

󵄩󵄩 − ∫ ∫ θ(t − s)q−1 s n ξ q (θ)S((t − s)q θ)f(s, x(s), (Kx)(s), (Hx)(s))dθds󵄩󵄩󵄩󵄩 󵄩α 0 δ

t δ

󵄩 󵄩 ≤ q ∫ ∫ θ(t − s)q−1 s n ξ q (θ)󵄩󵄩󵄩S((t − s)q θ)f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α dθds 0 0 t ∞

󵄩 󵄩 + q ∫ ∫ θ(t − s)q−1 s n ξ q (θ)󵄩󵄩󵄩S((t − s)q θ)f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α dθds t−h δ t δ

t ∞

≤ qM ∫ ∫ θ(t − s)

s ξ q (θ)ρ(s)dθds + qM ∫ ∫ θ(t − s)q−1 s n ξ q (θ)ρ(s)dθds

q−1 n

0 0

t−h δ

t

δ

≤ qM( ∫(t − s)q−1 s n ρ(s)ds) ∫ θξ q (θ)dθ 0

0 t

∞

+ qM( ∫ (t − s)q−1 s n ρ(s)ds) ∫ θξ q (θ)dθ. t−h

0

252 | 4 Fractional Evolution Equations: Continued

It comes from ∞

∫ θξ q (θ)dθ = 0

1 , Γ(1 + q)

t

t

∫(t − s)

s ρ(s)ds ≤ ( ∫(t − s)

q−1 n

0

(q−1)p p−1

s

np p−1

t

( ∫ ρ(s)p ds)

0

≤ B( and, for a fixed χ ∈ (1,

1 p

0

pq − 1 (n + 1)p − 1 , ) p−1 p−1

p−1 p

t

pq+np−1 p

‖ρ‖L p (J,ℝ+ ) ,

p−1 (1−q)p ),

t

t

∫ (t − s)

ds)

p−1 p

s ρ(s)ds ≤ ( ∫ (t − s)

q−1 n

t−h

(q−1)p p−1

s

np p−1

ds)

p−1 p

t

≤ ( ∫ ds)

χ(p−1) (χ−1)p

t−h

≤h

( ∫ ρ(s)p ds)

1 p

t−h

t−h

χ(p−1) (χ−1)p

t

t

( ∫ (t − s)

χ(q−1)p p−1

s

χnp p−1

ds)

p−1 χp

‖ρ‖L p (J,ℝ+ )

t−h

B(

χ(q − 1)p χnp + 1, + 1) p−1 p−1

p−1 χp

t

χp((q−1)p+n) +1 p−1

‖ρ‖L p (J,ℝ+ ) ,

that 󵄩󵄩 󵄩 󵄩󵄩(Fx)(t) − (F h,δ x)(t)󵄩󵄩󵄩α ≤ qMB( +

pq − 1 (n + 1)p − 1 , ) p−1 p−1

p−1 p

δ

t

pq+np−1 p

M χ(q − 1)p χnp B( + 1, + 1) Γ(q) p−1 p−1

‖ρ‖L p (J,ℝ+ ) ∫ θξ q (θ)dθ 0 p−1 χp

t

χp((q−1)p+n) +1 p−1

χ(p−1)

‖ρ‖L p (J,ℝ+ ) h (χ−1)p → 0

as δ → 0+ and h → 0+ . Therefore, {(Fx)(t) | x ∈ B r } is relatively compact in X α for all t ∈ (0, T]. Since it is compact at t = 0, we have the relative compactness in X α for all t ∈ J. Next, let us prove that F(B r ) is equicontinuous. For 0 ≤ t2 < t1 ≤ T, we have 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩(Fx)(t1 ) − (Fx)(t2 )󵄩󵄩󵄩α ≤ 󵄩󵄩󵄩(T (t1 ) − T (t2 ))[x0 − g(x)]󵄩󵄩󵄩α t

󵄩󵄩 2 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 s n [S (t1 − s) − S (t2 − s)]f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 󵄩 󵄩α 0 t

󵄩󵄩 󵄩󵄩 2 + 󵄩󵄩󵄩󵄩∫[(t1 − s)q−1 − (t2 − s)q−1 ]s n S (t2 − s)f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 󵄩 󵄩α 0 t

󵄩󵄩 1 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 s n S (t1 − s)f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 . 󵄩 󵄩α t2

4.3 Volterra–Fredholm type fractional evolution equations | 253

Denote 󵄩 󵄩 I1 := 󵄩󵄩󵄩(T (t1 ) − T (t2 ))[x0 − g(x)]󵄩󵄩󵄩α , t

󵄩󵄩 󵄩󵄩 2 I2 := 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 s n [S (t1 − s) − S (t2 − s)]f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 , 󵄩α 󵄩 0 t

󵄩󵄩 󵄩󵄩 2 I3 := 󵄩󵄩󵄩󵄩∫[(t1 − s)q−1 − (t2 − s)q−1 ]s n S (t2 − s)f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 , 󵄩α 󵄩 0 t

󵄩󵄩 1 󵄩󵄩 I4 := 󵄩󵄩󵄩󵄩∫(t1 − s)q−1 s n S (t1 − s)f(s, x(s), (Kx)(s), (Hx)(s))ds󵄩󵄩󵄩󵄩 . 󵄩 󵄩α t2

Now, we need to check that I1 , I2 , I3 , I4 tend to 0 independently of x ∈ B r when t1 → t2 . In fact, by the compactness of the set g(B r ) in view of (2) of Lemma 4.48, one can deduce that limt1 →t2 I1 = 0 uniformly. Next for 0 < h < t2 , when t2 > 0, we similarly derive t2

󵄩 󵄩 󵄩 󵄩 I2 ≤ ∫(t1 − s)q−1 s n 󵄩󵄩󵄩Sα (t1 − s) − Sα (t2 − s)󵄩󵄩󵄩α 󵄩󵄩󵄩f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds 0 t2 −h

󵄩 󵄩 ≤ ∫ (t2 − s)q−1 s n ρ(s)󵄩󵄩󵄩Sα (t1 − s) − Sα (t2 − s)󵄩󵄩󵄩α ds 0 t2

󵄩 󵄩 + ∫ (t2 − s)q−1 s n ρ(s)󵄩󵄩󵄩Sα (t1 − s) − Sα (t2 − s)󵄩󵄩󵄩α ds t2 −h t2 −h

≤

󵄩 󵄩 max 󵄩󵄩󵄩Sα (t1 − s) − Sα (t2 − s)󵄩󵄩󵄩α ∫ (t2 − s)q−1 s n ρ(s)ds s∈[0,t −h] 2

0

t2

+

2M ∫ (t2 − s)q−1 s n ρ(s)ds Γ(q) t2 −h

󵄩 󵄩 ≤ max 󵄩󵄩󵄩Sα (t1 − s) − Sα (t2 − s)󵄩󵄩󵄩α s∈[0,t −h] 2

× B(

pq − 1 (n + 1)p − 1 , ) p−1 p−1

p−1 p

pq+np−1 p

t2

‖ρ‖L p (J,ℝ+ )

χ(p−1) χ(q − 1)p 2M (χ−1)p χnp h B( + 1, + 1) + Γ(q) p−1 p−1

p−1 χp

χp((q−1)p+n) +1 p−1

t2

‖ρ‖L p (J,ℝ+ ) ,

from which we deduce that lim(h,t2 )→(0,t1 ) I2 = 0 uniformly, due to (7) of Lemma 4.48.

254 | 4 Fractional Evolution Equations: Continued Using the inequality (x − y)a ≤ x a − y a for any x ≥ y ≥ 0 and a > 1, analogously we derive t2

I3 ≤

M 󵄩 󵄨 󵄩 󵄨 ∫󵄨󵄨(t1 − s)q−1 − (t2 − s)q−1 󵄨󵄨󵄨s n 󵄩󵄩󵄩f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds Γ(q) 󵄨 0 t2

M 󵄨 󵄨 ≤ ∫󵄨󵄨(t1 − s)q−1 − (t2 − s)q−1 󵄨󵄨󵄨s n ρ(s)ds Γ(q) 󵄨 0 t2

≤

1

t2

np p M 󵄨 󵄨 p ( ∫ ρ(s)p ds) ( ∫ s p−1 󵄨󵄨󵄨(t1 − s)q−1 − (t2 − s)q−1 󵄨󵄨󵄨 p−1 ds) Γ(q)

0

0

t2

≤

p−1 p

np (q−1)p (q−1)p M ( ∫ s p−1 ((t2 − s) p−1 − (t1 − s) p−1 )ds) Γ(q)

p−1 p

‖ρ‖L p (J,ℝ+ )

0

≤

pq+np−1 pq+np−1 M pq − 1 (n + 1)p − 1 , [B( )(t2 p − t1 p ) Γ(q) p−1 p−1

t1 np

+ ∫ s p−1 (t1 − s)

(q−1)p p−1

ds]

p−1 p

‖ρ‖L p (J,ℝ+ )

t2

≤

χ(p−1) M χ(q − 1)p χnp ‖ρ‖L p (J,ℝ+ ) (t1 − t2 ) (χ−1)p B( + 1, + 1) Γ(q) p−1 p−1

p−1 χp

χp((q−1)p+n) +1 p−1

t1

.

Thus, limt1 →t2 I3 = 0 uniformly. Finally, t1

󵄩 󵄩 I4 ≤ ∫(t1 − s)q−1 s n 󵄩󵄩󵄩S (t1 − s)f(s, x(s), (Kx)(s), (Hx)(s))󵄩󵄩󵄩α ds t2 t1

M ≤ ∫(t1 − s)q−1 s n ρ(s)ds Γ(q) t2

t1

≤

(q−1)p np M ( ∫ s p−1 (t1 − s) p−1 ds) Γ(q)

p−1 p

‖ρ‖L p (J,ℝ+ )

t2

≤

χ(p−1) M χ(q − 1)p χnp (t1 − t2 ) (χ−1)p B( + 1, + 1) Γ(q) p−1 p−1

p−1 χp

χp((q−1)p+n) +1 p−1

t1

‖ρ‖L p (J,ℝ+ ) ,

from which we deduce that limt1 →t2 I4 = 0 uniformly. In summary, we have proven that F(B r ) is relatively compact for t ∈ J, {Fx | x ∈ B r } is a family of equicontinuous functions. Hence by the Arzelà–Ascoli theorem, F is compact. By the Schauder fixed point theorem F has a fixed point x ∈ B r . Consequently, system (4.20) has at least one mild solution on J. Our next result is based on the following well-known fixed point theorem.

4.3 Volterra–Fredholm type fractional evolution equations | 255

Lemma 4.51 ([168]). Let Γ be a condensing operator on a Banach space X. If Γ(B) ⊂ B for a convex, closed and bounded set B of X, then Γ has a fixed point in B. Now, we assume the following conditions and apply the above fixed point theorem. [Hf3] (1) There exists μ with 0 ≤ μ ≤ α ≤ 1 such that f : J × X α × X α × X α → X μ is con(1) (2) (3) tinuous and there exist L f , L f , L f > 0 such that 󵄩󵄩 󵄩 󵄩󵄩f(t, x1 , x2 , x3 ) − f(t, y1 , y2 , y3 )󵄩󵄩󵄩μ (1)

(2)

(3)

≤ L f ‖x1 − y1 ‖α + L f ‖x2 − y2 ‖α + L f ‖x3 − y3 ‖α for all x i , y i ∈ X α , i = 1, 2, 3 and t ∈ J. (2) There exist two positive constants c(1) , d(1) such that for each (t, x, y, z) ∈ J × Xα × Xα × Xα , ‖f(t, x, y, z)‖μ ≤ c(1) (‖x‖α + ‖y‖α + ‖z‖α ) + d(1) . (1)

[Hk3] (1) The function k : D k × X α → X α is continuous and there exists L k > 0 such that for (t, s) ∈ D k and x, y ∈ X α , t

󵄩󵄩 󵄩 󵄩󵄩∫[k(t, s, x) − k(t, s, y)]ds󵄩󵄩󵄩 ≤ L(1) ‖x − y‖ . α 󵄩󵄩 󵄩󵄩 k 󵄩 󵄩α 0

(2)

(2) There exists a constant L k > 0 such that for (t, s) ∈ D k and x, y ∈ X α , t

󵄩󵄩 󵄩 󵄩󵄩∫ k(t, s, x)ds󵄩󵄩󵄩 ≤ L(2) (1 + ‖x‖ ). α 󵄩󵄩 󵄩󵄩 k 󵄩 󵄩α 0

(1)

[Hh3] (1) The function h : D h × X α → X α is continuous and there exists L h > 0 such that for (t, s) ∈ D h and x, y ∈ X α , T

󵄩󵄩 󵄩 󵄩󵄩∫[h(t, s, x) − h(t, s, y)]ds󵄩󵄩󵄩 ≤ L(1) ‖x − y‖ . α 󵄩󵄩 󵄩󵄩 h 󵄩 󵄩α 0

(2)

(2) There exists a constant L h > 0 such that for (t, s) ∈ D h and x, y ∈ X α , T

󵄩󵄩 󵄩 󵄩󵄩∫ h(t, s, x)ds󵄩󵄩󵄩 ≤ L(2) (1 + ‖x‖ ). α 󵄩󵄩 󵄩󵄩 h 󵄩 󵄩α 0

[Hg3] The function g : C α → X α is compact, continuous and there exists a nondecreasing function ϕ : ℝ+ → ℝ+ such that for all x ∈ C α , ‖g(x)‖α ≤ ϕ(‖x‖∞ ) and

lim inf l→+∞

ϕ(l) = δ < ∞. l

256 | 4 Fractional Evolution Equations: Continued

Now we are ready to state and prove the mentioned existence result. Theorem 4.52. Assume that the conditions [Hf3], [Hk3], [Hh3], [Hg3] are satisfied. If x0 ∈ X α then system (4.20) admits at least one mild solution on J provided that M{δ + and

‖A α−μ ‖Γ(n) (1) (2) (2) [c (1 + L k + L h )]T n+q } < 1 Γ(n + q + 1)

M‖A α−μ ‖Γ(n) n+q (1) (2) (1) (3) (1) T (L f + L f L k + L f L h ) < 1. Γ(n + q + 1)

(4.24)

(4.25)

Proof. Define the operator Γ : C α → C α by (4.21). For each positive number l, let B l = {x ∈ C α | ‖x‖∞ ≤ l}. Then for each l, B l is obviously a bounded closed convex set in C α . Firstly, we claim that Γ(B l ) ⊂ B l for some l > 0. If it is not true, then for each l > 0 there would exist x l ∈ B l and t l ∈ J such that ‖(Γx l )(t l )‖α > l. However, by [Hf3], [Hk3] and [Hh3], l ≤ ‖(Γx l )(t l )‖α tl

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩T (t)(x0 + g(x l ))󵄩󵄩󵄩α + ∫(t l − s)q−1 s n 󵄩󵄩󵄩S (t l − s)f(s, x l (s), (Kx l )(t), (Hx l )(t))󵄩󵄩󵄩α ds 0

≤ M(‖x0 ‖α + ‖g(x l )‖α ) tl

M‖A α−μ ‖ (1) (2) (2) + [c (l + L k (1 + l) + L h (1 + l)) + d(1) ] ∫(t l − s)q−1 s n ds Γ(q) 0

≤ M(‖x0 ‖α + ϕ(l)) +

‖A α−μ ‖M (1) n+q (2) (2) [c (l + L k (1 + l) + L h (1 + l)) + d(1) ]B(q, n + 1)t l Γ(q)

≤ M(‖x0 ‖α + ϕ(l)) +

M‖A α−μ ‖Γ(n) (1) (2) (2) [c (l + L k (1 + l) + L h (1 + l)) + d(1) ]T n+q . Γ(n + q + 1)

Dividing both sides by l and taking the lower limit as l → +∞, we obtain 1 ≤ M{δ +

‖A α−μ ‖Γ(n) (1) (2) (2) [c (1 + L k + L h )]T n+q }, Γ(n + q + 1)

which contradicts condition (4.24). Thus, Γ(B l ) ⊂ B l for some positive number l. We decompose Γ = Γ1 + Γ2 as (Γ1 x)(t) = T (t)[x0 + g(x)], t

(Γ2 x)(t) = ∫(t − s)q−1 s n S (t − s)f(s, x(s), (Kx)(s), (Hx)(s))ds. 0

4.3 Volterra–Fredholm type fractional evolution equations | 257

Secondly, we show that Γ1 is compact, continuous, and Γ2 is a contraction. By [Hg3], we can infer that Γ1 is compact and continuous on X α . Next, we prove that Γ2 is a contraction on B l . In fact, for each t ∈ J, x, y ∈ B l , by [Hg3] we have 󵄩󵄩 󵄩 󵄩󵄩(Γ2 x)(t) − (Γ2 y)(t)󵄩󵄩󵄩α t

≤ ‖A

α−μ

M 󵄩 ‖ ∫(t − s)q−1 s n 󵄩󵄩󵄩f(s, x(s), (Kx)(s), (Hx)(s)) Γ(q) 0

󵄩 − f(s, y(s), (Ky)(s), (Hy)(s))󵄩󵄩󵄩μ ds t

≤ ‖A α−μ ‖

M (1) ∫(t − s)q−1 s n [L f ‖x(s) − y(s)‖α Γ(q) 0

+

(2) 󵄩 L f 󵄩󵄩󵄩(Kx)(s)

(3) 󵄩 󵄩 󵄩 − (Ky)(s)󵄩󵄩󵄩α + L f 󵄩󵄩󵄩(Hx)(s) − (Hy)(s)󵄩󵄩󵄩α ]ds t

≤ ‖A

α−μ

M (1) (2) (1) (3) (1) ‖ (L + L f L k + L f L h ) ∫(t − s)q−1 s n ds‖x − y‖∞ Γ(q) f 0

MΓ(n) (1) (2) (1) (3) (1) ≤ ‖A α−μ ‖ T n+q (L f + L f L k + L f L h )‖x − y‖∞ . Γ(n + q + 1) Thus, ‖Γ2 x − Γ2 y‖∞ ≤

M‖A α−μ ‖Γ(n) n+q (1) (2) (1) (3) (1) T (L f + L f L k + L f L h )‖x − y‖∞ , Γ(n + q + 1)

which implies that Γ2 is a contraction by (4.25). At last, we can conclude that Γ = Γ1 + Γ2 is a condensing map on B l . By Lemma 4.51, system (4.20) admits at least one mild solution on J.

4.3.4 Example In this subsection, we present the following example illustrating how our theorems can be applied to concrete problems: −t

e c q { + e−t ) D t x(t, y) − ∆x(t, y) = ( t { { { e + e−t { { { { t T { { { { { × cos[x(t, y) + ∫ sin(t + s)x(s, y)ds + ∫ cos(ts)x(s, y)ds], { { { { { 0 0 { { 19 { , y ∈ Ω, t, s ∈ (0, T], q = { { 20 { { { { { x(t, y)|y∈∂Ω = 0, t > 0, { { { { { T { { { 1 { { { {x(0, y) = ∫ ∫ h(t, y) log(1 + |x(t, ξ)| 2 )dtdξ, { Ω 0

(4.26)

258 | 4 Fractional Evolution Equations: Continued where ∆ is the Laplace operator in ℝ3 , Ω ⊂ ℝ3 is a bounded domain, ∂Ω ∈ C3 , and h(t, y) ∈ C(J × Ω). We apply Theorem 4.50 by taking X = L2 (Ω), D(A) = H 2 (Ω) ∩ H01 (Ω), Ax = −∆x for x ∈ D(A), and set α = 0. Define x(t)(y) = x(t, y), t

T

(Kx)(t)(y) = ∫ sin(t + s)x(s, y)ds,

(Hx)(t)(y) = ∫ cos(ts)x(s, y)ds,

0

0

f(t, x(t), (Kx)(t), (Hx)(t))(y) t

T

0

0

e−t =( t + e−t ) cos[x(t) + ∫ sin(t + s)x(s)ds + ∫ cos(ts)x(s)ds](y), e + e−t and

T 1

g(x)(y) = ∫ ∫ h(t, y) log(1 + |x(t, ξ)| 2 )dtdξ,

y ∈ Ω, x ∈ C(J, X).

Ω 0

Then, A generates a compact analytic semigroup in X with M = 1, and 1 e−t 󵄩󵄩 󵄩 + e−t )(mes(Ω)) 2 󵄩󵄩f(t, x(t), (Kx)(t), (Hx)(t))󵄩󵄩󵄩 ≤ ρ(t) = ( t −t e +e

with ρ ∈ L p (J, ℝ+ ), p = 10. Moreover, g is compact (see [179]). Next, using log(1 + a) ≤ a for any a ≥ 0, we derive 3

1

1

3

1

∫ log(1 + |x(ξ)| 2 )dξ ≤ ∫ |x(ξ)| 2 dξ ≤ (mes(Ω)) 4 ‖x‖ 2 ≤ Ω

(mes(Ω)) 4 (1 + ‖x‖) 2

Ω

for any x ∈ X. Hence, using ‖x(t, ⋅)‖ ≤ ‖x‖∞ for any x ∈ C(J, X), we obtain ‖g(x)‖ ≤

7 T (mes(Ω)) 4 max |h(t, y)|(1 + ‖x‖∞ ), 2 t∈J,y∈Ω

x ∈ C(J, X).

Thus problem (4.26) can be rewritten as c

{

q

D t x(t) = −Ax(t) + t n f(t, x(t), (Kx)(t), (Hx)(t)),

t ∈ J, n ∈ ℤ+ , q ∈ (0, 1),

x(0) = g(x) + x0 .

Obviously, q =

19 20

>

1 10

= 1p . Furthermore, if T and h(t, y) satisfy 7

T(mes(Ω)) 4 max |h(t, y)| < 2, t∈J,y∈Ω

then all the assumptions given in Theorem 4.50 are satisfied. Therefore, problem (4.26) has at least one mild solution.

4.4 Controllability of Sobolev evolution equations | 259

4.4 Controllability of Sobolev type fractional functional evolution equations 4.4.1 Introduction Recently, existence of mild solutions, Ulam–Hyers–Mittag-Leffler stability, controllability and optimal controls for all kinds of fractional semilinear evolution systems in Banach spaces were reported by many researchers. We refer the reader to [34, 85, 97, 98, 136, 164, 178, 250, 277, 301–305, 311, 314–316, 333, 334] and the references therein. Balachandra and Dauer [31] provided some sufficient conditions for controllability of integer functional evolution equations of Sobolev type by virtue of the theory of semigroups via the technique of fixed point theorems. In 2012, Li et al. [176] obtained the existence results for fractional evolution equations of Sobolev type by virtue of the theory of propagation family via techniques of a measure of noncompactness and condensing maps. However, controllability of fractional functional evolution equations of Sobolev type has not been extensively studied via the theory of semigroups or propagation family. Motivated by [31, 176, 333], we study the controllability of a class of fractional functional evolution equations of Sobolev type via the theory of semigroups. Our aim in this section is to provide suitable sufficient conditions for the controllability results corresponding to two admissible control sets without assuming the semigroup to be compact. This section is based on [108].

4.4.2 Preliminaries Let X and Y be two real Banach spaces. We consider the following fractional functional evolution equations of Sobolev type: c q 0 D t (Ex(t))

{ q

+ Ax(t) = f(t, x t ) + Bu(t), x(t) = ϕ(t),

t ∈ J := [0, a],

−r ≤ t ≤ 0,

(4.27)

where 0c D t is the Caputo fractional derivative of order 0 < q < 1 with the lower limit zero, A : D(A) ⊂ X → Y and E : D(E) ⊂ X → Y are operators, the state x(⋅) takes values in X and the control function u(⋅) is given in U . For the Banach space of admissible control functions U we denote by U either a Banach space U := L2 (J, U) for 12 < q < 1, or U := L∞ (J, U) for 0 < q < 1. Moreover, the operator B is a bounded linear operator from U into Y. The nonlinear term f : J × C → Y with C := C([−r, 0], X) will be specified later. For a continuous function x : J ∗ := [−r, a] → X, x t is the element of C defined by x t (s) := x(t + s), −r ≤ s ≤ 0. The domain D(E) of E becomes a Banach space with the norm ‖x‖D(E) := ‖Ex‖Y , x ∈ D(E) and ϕ ∈ C(E) := C([−r, 0], D(E)).

260 | 4 Fractional Evolution Equations: Continued

4.4.3 Characteristic solution operators and their properties In this subsection, we consider the following fractional functional evolution equations: c q 0 D t (Ex(t))

{

+ Ax(t) = f(t, x t ),

t ∈ J,

x(t) = ϕ(t),

−r ≤ t ≤ 0.

(4.28)

We introduce the following assumptions on the operators A and E. [H1 ] A and E are linear operators, and A is closed. [H2 ] D(E) ⊂ D(A) and E is bijective. [H3 ] The linear operator E−1 : Y → D(E) ⊂ X is compact (which implies that E−1 is bounded). Note that [H3 ] implies that E is closed due to the following fact: if E−1 is closed and injective, then its inverse is also closed. From [H1 ]–[H3 ] and the closed graph theorem, we obtain the boundedness of the linear operator −AE−1 : Y → Y. Conse−1 quently, −AE−1 generates a semigroup {T(t) | t ≥ 0}, where T(t) := e−AE t . We denote M1 := supt≥0 ‖T(t)‖ < ∞. According to Lemma A.16, it is suitable to rewrite system (4.28) in the equivalent fractional integral equation t

{ 1 { { ∫(t − s)q−1 [−Ax(s) + f(s, x s )]ds, {Ex(t) = Eϕ(0) + Γ(q) { 0 { { { x(t) = ϕ(t), −r ≤ t ≤ 0, {

t ∈ J,

(4.29)

provided that the integral in (4.29) exists. Lemma 4.53. We have t

{ { { {x(t) = TE (t)Eϕ(0) + ∫(t − s)q−1 SE (t − s)f(s, x s )ds, { 0 { { { {x(t) = ϕ(t), −r ≤ t ≤ 0.

t ∈ J,

Here TE (⋅) and SE (⋅) are called characteristic solution operators and are given by ∞

∞

TE (t) := ∫ E ξ q (θ)T(t θ)dθ, q

−1

SE (t) := q ∫ E−1 θξ q (θ)T(t q θ)dθ

0

0

(see Definitions 4.1 and 4.5). Proof. Formally applying the Laplace transforms ∞

ν(λ) := ∫ e 0

∞ −λs

x(s)ds,

ω(λ) := ∫ e−λs f(s, x s )ds, 0

(4.30)

4.4 Controllability of Sobolev evolution equations | 261

for λ > 0 to the first equation of (4.29), we obtain Eν(λ) =

1 1 1 Eϕ(0) − q (AE−1 )Eν(λ) + q ω(λ), λ λ λ

which implies Eν(λ) = λ q−1 (λ q I + AE−1 )−1 Eϕ(0) + (λ q I + AE−1 )−1 ω(λ) ∞

∞

q

q

= λ q−1 ∫ e−λ s T(s)Eϕ(0)ds + ∫ e−λ s T(s)ω(λ)ds, 0

0

where I is the identity operator defined on Y. Thus, ∞

∞ q

q

ν(λ) = λ q−1 ∫ E−1 e−λ s T(s)Eϕ(0)ds + ∫ E−1 e−λ s T(s)ω(λ)ds. 0

0

Now, we consider the one-sided stable probability density ϖ q (θ) :=

1 ∞ Γ(nq + 1) sin(nπq), ∑ (−1)n−1 θ−qn−1 π n=1 n!

θ ∈ (0, ∞),

whose Laplace transform is given by ∞ q

∫ e−λθ ϖ q (θ)dθ = e−λ ,

q ∈ (0, 1).

0

Using the identity (4.31) again and again, we have ∞

∞

ν(λ) = ∫ e−λt [ ∫ E−1 ϖ q (θ)T( 0

tq )Eϕ(0)dθ θq

0 t ∞

+ q ∫ ∫ E−1 ϖ q (θ)T(

(t − s)q (t − s)q−1 dθds]dt. )f(s, x s ) q θ θq

0 0

Now we can invert the last Laplace transform to get ∞

x(t) = ∫ E−1 ϖ q (θ)T(

tq )Eϕ(0)dθ θq

0 t

+ q ∫(t − s)

∞ q−1

0

∫ E−1 ϖ q (θ)T(

1 (t − s)q )f(s, x s ) q dθds θq θ

0

∞

= ∫ E−1 ξ q (θ)T(t q θ)Eϕ(0)dθ 0 t

+ q ∫(t − s) 0

∞ q−1

∫ E−1 θξ q (θ)T((t − s)q θ)f(s, x s )dθds. 0

Note that by (4.30), the proof is complete.

(4.31)

262 | 4 Fractional Evolution Equations: Continued Definition 4.54. For each u ∈ U and ϕ ∈ C(E), by a mild solution of system (4.27) we mean a function x ∈ C(J ∗ , X) satisfying t

{ { { x(t) = TE (t)Eϕ(0) + ∫(t − s)q−1 SE (t − s)f(s, x s )ds { { { { { 0 { { t { { { + ∫(t − s)q−1 SE (t − s)Bu(s)ds, t ∈ J, { { { { { 0 { { x(t) = ϕ(t), −r ≤ t ≤ 0. { The following results for TE (⋅) and SE (⋅) will be used throughout this section. Lemma 4.55. Assume that [H1 ]–[H3 ] hold. The following properties hold: (i) For any fixed t ≥ 0, TE (t) and SE (t) are linear and bounded operators, i.e., for any x ∈ X, M1 ‖E−1 ‖ ‖TE (t)x‖ ≤ M1 ‖E−1 ‖ ‖x‖, ‖SE (t)x‖ ≤ ‖x‖. Γ(q) (ii) {TE (t) | t ≥ 0} and {SE (t) | t ≥ 0} are compact. Proof. (i) For any fixed t ≥ 0, since E−1 and T(t) are linear operators, it is easy to see that TE (t) and SE (t) are also linear operators. For any x ∈ X, according to (4.30), we have ∞

‖TE (t)x‖ ≤ ∫ ξ q (θ)‖E−1 T(t q θ)x‖dθ ≤ M1 ‖E−1 ‖ ‖x‖, 0 ∞

‖SE (t)x‖ ≤ q ∫ θξ q (θ)‖E−1 T(t q θ)x‖dθ ≤

M1 ‖E−1 ‖ ‖x‖, Γ(q)

0

where we used

∞

∞

∫ θξ q (θ)dθ = ∫ 0

0

ϖ q (θ) 1 dθ = . θq Γ(1 + q)

(ii) For each positive constant k, set Y k := {x ∈ Y | ‖x‖ ≤ k}. Then Y k is clearly a bounded subset of Y. We need to prove that for any positive constant k and t ≥ 0, the sets ∞

V1 (t) := { ∫ E−1 ξ q (θ)T(t q θ)xdθ | x ∈ Y k }, 0 ∞

V2 (t) := {q ∫ E−1 θξ q (θ)T(t q θ)xdθ | x ∈ Y k } 0

are relatively compact in Y. By assertion (i), we know that the operators TI (t) : Y → Y and SI (t) : Y → Y are also linear and bounded. So they map Y k into bounded sub-

4.4 Controllability of Sobolev evolution equations | 263

sets of Y. Consequently, V1 (t) = E−1 TI (t)(Y k ) and V2 (t) = qE−1 SI (t)(Y k ) are relatively compact in Y for any k > 0 and t ≥ 0 due to the compactness of E−1 : Y → X.

4.4.4 Main results In this subsection, we study the controllability of system (4.27) via the well-known Schauder fixed point theorem. Definition 4.56. The fractional system (4.27) is said to be controllable on the interval J if for every continuous initial function ϕ ∈ C(E) and every x1 ∈ D(E) there exists a control u ∈ U such that the mild solution x of system (4.27) satisfies x(a) = x1 . In addition to [H1 ]–[H3 ], we need the following assumption: [H4 ] B : U → Y is a bounded linear operator and the linear operator W : U → D(E) defined by a

Wu := ∫(a − s)q−1 SE (a − s)Bu(s)ds 0

has a bounded right inverse W −1 : D(E) → U , i.e., WW −1 = I D(E) , and thus there exist two constants M2 , M3 > 0 such that ‖B‖ ≤ M2 and ‖W −1 ‖ ≤ M3 , where we consider the norm ‖⋅‖D(E) on D(E) for determining M3 . It is obvious that Wu ∈ D(E), and W is well defined. In fact, it holds a

󵄩󵄩 󵄩󵄩 ‖EWu‖ = 󵄩󵄩󵄩󵄩∫(a − s)q−1 SI (a − s)Bu(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩 0

a

M1 ‖B‖ ≤ ∫(a − s)q−1 ‖u(s)‖ds Γ(q) 0

≤

M1 ‖B‖ a2q−1 M1 ‖B‖ a2q−1 √ √ ‖u‖L2 (J,U) =: ‖u‖U Γ(q) 2q − 1 Γ(q) 2q − 1

for q ∈ ( 21 , 1) and u ∈ L2 (J, U), whereas M1 ‖B‖a q M1 ‖B‖a q ‖u‖L∞ (J,U) =: ‖u‖U Γ(q + 1) Γ(q + 1)

‖EWu‖ ≤

for q ∈ (0, 1) and u ∈ L∞ (J, U). We also see that t

∫(t − s)q−1 ‖u(s)‖ds ≤ K q ‖u‖U 0

for any t ∈ J, where {√ a K q := { q 2q−1 a {q 2q−1

if q ∈ ( 21 , 1), u ∈ L2 (J, U), if q ∈ (0, 1), u ∈ L∞ (J, U).

264 | 4 Fractional Evolution Equations: Continued

Next we assume: [H5 ] f satisfies the following two conditions: (i) For each x ∈ C the function f(⋅, x) : J → Y is strongly measurable, and for each t ∈ J the function f(t, ⋅) : C → Y is continuous. (ii) For each k > 0 there is a measurable function g k such that sup ‖f(t, x)‖ ≤ g k (t) ‖x‖≤k

with ‖g k ‖∞ := sup g k (s) < ∞, s∈J

t

sup ∫(t − s)q−1 g k (s)ds ≤ γk t∈J

0

for any k > 0 sufficiently large and some γ. Note that we may take any γ > lim sup k→∞

a q ‖g k ‖∞ . qk

For the sake of simplicity, we set the standard framework to deal with controllability problems as follows. Based on our assumptions, for an arbitrary function x(⋅) it is suitable to define the control function a

u(t) := W −1 [x1 − TE (a)Eϕ(0) − ∫(a − s)q−1 SE (a − s)f(s, x s )ds].

(4.32)

0

In what follows, it is necessary to show that when using the control u in (4.32), the operator P defined by t

{ { { TE (t)Eϕ(0) − ∫(t − s)q−1 SE (t − s)f(s, x s )ds { { { { { 0 { { t (Px)(t) := { { { + ∫(t − s)q−1 SE (t − s)Bu(s)ds { { { { { 0 { { ϕ(t) {

if t ∈ J, if − r ≤ t ≤ 0,

from C(J ∗ , X) into C(J ∗ , X) for each x ∈ C(J ∗ , X), has a fixed point. Clearly, this fixed point is just a solution of system (4.27). Furthermore, one can check that a

(Px)(a) = TE (a)Eϕ(0) + ∫(a − s)q−1 SE (a − s)f(s, x s )ds 0 a

+ ∫(a − s)q−1 SE (a − s)Bu(s)ds 0

4.4 Controllability of Sobolev evolution equations | 265

a

a

= TE (a)Eϕ(0) + ∫(a − s)q−1 SE (a − s)f(s, x s )ds + ∫(a − s)q−1 SE (a − s)BW −1 0

0 a

× [x1 − TE (a)Eϕ(0) − ∫(a − τ)q−1 SE (a − τ)f(τ, x τ )dτ]ds = x1 , 0

which means that u steers the fractional system (4.27) from ϕ(0) to x1 in a finite time a. Consequently, we can claim that system (4.27) is controllable on J. For each number k > 0, define Bk := {x ∈ C(J ∗ , X) | ‖x(t)‖ ≤ k, t ∈ J ∗ }. Of course, Bk is a bounded closed convex subset of C(J ∗ , X). Under the assumptions [H1 ]–[H5 ], we will establish three important results. Lemma 4.57. There exists K ≥ max{maxt∈[−r,0] ‖ϕ(t)‖, √aM1 ‖B‖K q ‖W −1 ‖ γ‖E−1 ‖M1 { { (1 + ) 12 , we take U := L2 (J1 , U), and so K 43 := √2. Next, let u(s, y) := x(y) ∈ U. 1

Wu = b ∫(1 − s)

− 14

∞

3 3 ∫ E−1 θξ 43 (θ)T((1 − s) 4 θ)dθxds 4

0

0 1 1

= b ∫(1 − s)− 4 0

∞

∞ 3 2 3 − n (1−s) 4 θ ⟨x, x n ⟩x n dθds ∫ E−1 θξ 43 (θ) ∑ e 1+n2 4 n=1 0

∞ 1

∞

= b ∫ E−1 ξ 34 (θ) ∑ ∫ n=1 0

0

∞ 1

∞

= b ∫ ξ (θ) ∑ ∫ 3 4

0 ∞

n=1 0 ∞ 1

= b ∫ ξ 34 (θ) ∑ ∫ 0

n=1 0

3 2 1 − n 3 (1−s) 4 θ ds⟨x, x n ⟩x n dθ θ(1 − s)− 4 e 1+n2 4

3 n2 3 − 41 − 1+n2 (1−s) 4 θ θ(1 − s) e ds⟨x, x n ⟩x n dθ 4(1 + n2 )

1 d − n22 (1−s) 43 θ [e 1+n ]ds⟨x, x n ⟩x n dθ n2 ds

270 | 4 Fractional Evolution Equations: Continued ∞

∞

= b ∫ ξ 34 (θ) ∑

n=1

0 ∞

=b∑ n=1

2 1 − n θ [1 − e 1+n2 ]⟨x, x n ⟩x n dθ 2 n

1 n2 [1 − 𝔼 34 (− )]⟨x, x n ⟩x n , 2 n 1 + n2

where ∞

∞

0

0

2 2 4 n2 − n 2θ − n θ 4 −(1+ 43 ) 1+n 3 (θ)dθ = ∫ e 1+n2 𝔼 43 (− := e ξ θ ϖ 34 (θ− 3 )dθ ) ∫ 4 3 1 + n2

is the Mittag-Leffler function (for more details see [230, (24)–(27)]). Note that 2 − n θ 0 < 1 − e 1+n2 < 1 − e−θ < 1 for any θ > 0. So we have

1 n2 1 − 𝔼 34 (− ) ≤ 1 − 𝔼 34 (− ) ≤ 1 − 𝔼 34 (−1). 2 1 + n2 From the above computations we know that W is surjective. Hence, we define the inverse W −1 : D(E) → U by (W −1 x)(t, y) :=

n2 ⟨x, x n ⟩x n 1 ∞ ∑ b n=1 [1 − 𝔼 3 (− n2 2 )] 1+n 4

for x = ∑∞ n=1 ⟨x, x n ⟩x n . Since ∞

‖x‖D(E) := ‖Ex‖ = √ ∑ (1 + n2 )2 ⟨x, x n ⟩2 n=1

for x ∈ D(E), we derive ‖(W −1 x)(t, ⋅)‖ =

1 ∞ n4 ⟨x, x n ⟩2 √∑ b n=1 [1 − 𝔼 3 (− n2 2 )]2 1+n

4

≤

∞ ‖x‖D(E) 1 √ ∑ (1 + n2 )2 ⟨x, x n ⟩2 = . 1 b[1 − 𝔼 43 (− 2 )] n=1 b[1 − 𝔼 34 (− 21 )]

Note that W −1 x is independent of t ∈ J1 . Consequently, we obtain ‖W −1 ‖ ≤

1 . b[1 − 𝔼 34 (− 12 )]

Next, we suppose [C2 ] f : J1 × ℝ → ℝ is such that f(⋅, x) is measurable for each x ∈ ℝ, and f(t, ⋅) is continuous for each t ∈ J1 . Moreover, lim sup k→∞

1 k

sup t∈J1 ,|x|≤k

|f(t, x)| =: γ < ∞.

4.5 Relaxed controls for impulsive equations | 271

Define F : J1 × C([−1, 0], X) → Y by F(t, z)(y) = f(t, z(−r)(y)). Now, system (4.35) can be abstracted as 3

{0c D t4 (Ex(t)) = −Ax(t) + F(t, x t ) + Bu(t), t ∈ J1 , { x(t) = ϕ(t), −1 ≤ t ≤ 0. { Clearly, all the assumptions of Theorem 4.60 are satisfied if (4.33) holds, i.e., √2 γ [1 + 3 ] < 1. Γ( 34 ) Γ( 4 )[1 − 𝔼 34 (− 21 )]

(4.36)

Therefore, system (4.35) is controllable on J1 . To compute 𝔼 34 (− 12 ) numerically, we use the integral representation formula [51, formula (34)], ∞

𝔼q (−z) :=

1 s q−1 sin(qπ) q e−z s ds ∫ q 2q π 1 + 2s cos(qπ) + s

0

for q =

3 4

and z =

1 2,

to get ∞

−

s

1 1 e 24/3 𝔼 34 (− ) = ds. ∫ 1 2 √2π s 4 (1 − √2s 43 + s 23 ) 0 Let h(s) = 1 − √2s 4 + s 2 . Since h has a unique stationary point s∗ = (√2/2) 3 , and 󸀠󸀠 h (s) > 0 for s ≥ 0, we have h(s) ≥ h(s∗ ) = 21 for s ≥ 0. Thus, we obtain 3

∞

4

3

∞

s

− η √2 1 e 24/3 24/3 √2 − 4/3 − s ds ≤ e 2 ≤ 0.000004 ∫ 1 ∫ e 24/3 ds = 1 1 √2π s 4 h(s) πη 4 η πη 4 η

for η ≥ 30. On the other hand, a numerical computation in Mathematica shows 30

−

s

1 e 24/3 ds ≐ 0.60379. ∫ 1 √2π s 4 h(s) 0 So we get 𝔼 34 (− 12 ) ≐ 0.60379. Finally, using √2 ≐ 1.41421, 1 − 𝔼 34 (− 12 ) ≐ 0.39621

and Γ( 34 ) ≐ 1.22542, one obtains that (4.36) holds if γ < 0.17508.

4.5 Relaxed controls for nonlinear impulsive fractional evolution equations 4.5.1 Introduction In general, we need a convexity hypothesis on a certain orientor field to derive existence of optimal state-control pairs. Once this key convexity hypothesis is removed, we turn to discuss a relaxed control system, of which the orientor field has been convexified.

272 | 4 Fractional Evolution Equations: Continued

Relaxed control systems are considered in [11, 219, 224, 323, 324]. In this section, relaxation of fractional impulsive control systems is studied. This section is based on [285].

4.5.2 Problem statement In this section, we will consider a control system described by the following fractional impulsive evolution equation: c α α ∈ (0, 1], t ∈ J := [0, b], t ≠ t k , u ∈ Uad , { 0 D t x(t) = Ax(t) + f(t, x(t), u(t)), { { x(0) = x0 ∈ X, { { { { ∆x(t k ) = I k (x(t k )), k = 1, 2, . . . , m, (4.37) where 0c D αt denotes the Caputo fractional derivative of order α with the lower limit zero, A : D(A) ⊆ X → X is the generator of a C0 -semigroup {S(t) | t ≥ 0} on a Banach space X, the functions f and I k are continuous nonlinear operators from X to X, the state x takes values in X, and the control function u is given in a suitable admissible control set Uad . Next, ∆x(t k ) := x(t+k ) − x(t−k ) where x(t+k ) := limϵ→0+ x(t k + ϵ) and x(t−k ) = x(t k ) represent the right- and left-sided limits of x(t) at t = t k with 0 =: t0 < t1 < ⋅ ⋅ ⋅ < t m < t m+1 := b. It is clear that system (4.37) contains the jump in the state-control pairs (x, u) at time t k with I k determining the size of the jump at t k . For the original system (4.37), we consider Problem (P0 ): find a control policy u0 ∈ Uad such that it imparts a minimum to the cost functional

J(u) := J(x u , u) := ∫ l(t, x u (t), u(t))dt, J

where x u is the solution of the original system (4.37) corresponding to the control u ∈ Uad . First, we will prove the existence of state-control pairs of system (4.37). Moreover, by defining a cost functional J(u), we can give some sufficient conditions to guarantee the existence of optimal state-control pairs when the convexity conditions on a certain orientor field are not assumed, that is, to consider the relaxation problem. To achieve our aim, we need to convexify the original fractional control system (4.37) and construct the corresponding relaxed fractional impulsive control system c α α ∈ (0, 1], t ∈ J, t ≠ t k , μ ∈ U r , { 0 D t x(t) = Ax(t) + F(t, x(t))μ(t), { { x(0) = x0 , { { { { ∆x(t k ) = I k (x(t k )), k = 1, 2, . . . , m,

(4.38)

4.5 Relaxed controls for impulsive equations | 273

where U r is a given relaxed controls set. For a regular countably additive measures set Σrca (Ξ), the function F : J × X × Σrca (Ξ) → X is defined by F(t, x)μ := ∫ f(t, x, σ)μ(dσ), Ξ

where Ξ is a compact Polish space. For the relaxed control system (4.38), we consider Problem (Pr ): find a control policy μ0 ∈ U r such that it imparts a minimum to the cost functional J(μ) := J(x μ , μ) := ∫ ∫ l(t, x μ (t), σ)μ(t)dt, J Ξ

where x μ is a solution of system (4.38) corresponding to the control μ ∈ U r . For system (4.38), we will study the existence and uniqueness of the piecewise continuous mild solution and discuss the properties of relaxed trajectories. Further, we will investigate the existence of optimal relaxed controls for the problem (Pr ) and present the relation between problems (P0 ) and (Pr ). In the case of α = 1, the related results were reported in [224]. To achieve our results, we are partly motivated by the approach of [224]. However, we emphasize that we have to establish some necessary fundamental results such as solvability, data dependence and properties of relaxed trajectories for the current original and relaxed fractional impulsive control systems, which are much different from integer impulsive control systems.

4.5.3 Original and relaxed fractional impulsive control systems In this subsection, first we consider the original fractional impulsive evolution equations c α D t x(t) = Ax(t) + f(t, x(t)), α ∈ (0, 1], t ∈ J, t ≠ t k , { { { (4.39) x(0) = x0 , { { { { ∆x(t k ) = I k (x(t k )), k = 1, 2, . . . , m. Now, following Section 3.6 along with Definitions 4.1 and 4.5 (see also [281, Definition 3.1]), we can introduce the following definition. Definition 4.61. A PC-mild solution of system (4.39) is a function x ∈ PC(J, X) satisfying the integral equation t

x(t) = T (t)x0 + ∑ T (t − t i )I i (x(t i )) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds, 0 0 such that ‖f(t, x1 ) − f(t, x2 )‖ ≤ L1 (ρ)‖x1 − x2 ‖,

x1 , x2 ∈ B ρ := {x ∈ X | ‖x‖ ≤ ρ};

(ii) there exists a constant K > 0 such that ‖f(t, x)‖ ≤ K(1 + ‖x‖), [I]

x ∈ X.

The functions I k : X → X, k = 1, 2, . . . , m, are such that (i) I k maps a bounded set to a bounded set; (ii) there exist constants h k > 0 such that ‖I k (x) − I k (y)‖ ≤ h k ‖x − y‖,

x, y ∈ X.

Set M := supt≥0 ‖S(t)‖L b (X) . Now we are ready to prove the following result. Theorem 4.62. Suppose that the hypotheses [A], [F] and [I] are satisfied. Then it holds: (i) For every x0 ∈ X, system (4.39) has a unique mild solution x ∈ PC(J, X). (ii) There exists a constant C∗ > 0 such that for any x0 , y0 ∈ X, the corresponding mild solutions x(t), y(t) of (4.39) with x(0) = x0 , y(0) = y0 satisfy ‖x − y‖PC(J,X) ≤ C∗ ‖x0 − y0 ‖. Proof. We start by considering the following semilinear fractional evolution equation: c α 0 D t x(t)

{

= Ax(t) + f(t, x(t)),

α ∈ (0, 1], t ∈ J,

x(0) = x0

(4.40)

on C(J, X) with the weighted norm ‖x‖r := maxt∈J ‖x(t)‖e−rt , where r > 0 is specified later. Define a map P : C(J, X) → C(J, X) by t

(Px)(t) := T (t)x0 + ∫(t − s)α−1 S (t − s)f(s, x(s))ds. 0

Using [F] and Lemmas 4.2 and 4.6, we derive t

‖(Px)(t)‖ ≤ ‖T (t)x0 ‖ + ∫(t − s)α−1 ‖S (t − s)‖ ‖f(s, x(s))‖ds 0 t

MK MK ≤ M‖x0 ‖ + bα + ‖x‖r ∫(t − s)α−1 ers ds Γ(α + 1) Γ(α) 0

4.5 Relaxed controls for impulsive equations | 275

MK MK bα + ‖x‖r r−α ert γ(α, rt) Γ(α + 1) Γ(α) MK b α + MKert r−α ‖x‖r , ≤ M‖x0 ‖ + Γ(α + 1)

= M‖x0 ‖ +

x

where γ(α, x) := ∫0 z α−1 e−z dz is the lower incomplete gamma function. Note that t

∫(t − s)α−1 ers ds = r−α ert γ(α, rt) 0

by the substitution r(t − s) = z. So we arrive at ‖Px‖r ≤ M‖x0 ‖ +

MK b α + MKr−α ‖x‖r . Γ(α + 1)

From the above consideration we also derive that any possible mild solution of equation (4.40) satisfies t

‖x(t)‖ ≤ M‖x0 ‖ +

MK MK bα + ∫(t − s)α−1 ‖x(s)‖ds. Γ(α + 1) Γ(α) 0

By Remark A.19 we obtain ‖x‖C ≤ ρ̃ := (M‖x0 ‖ +

MK MK b α )E α ( Γ(α)b α ). Γ(α + 1) Γ(α)

Hence, any possible mild solution x of equation (4.40) satisfies ‖x‖C ≤ ρ.̃ Let x1 , x2 ∈ C(J, X) with ‖x1 ‖C ≤ ρ,̃ ‖x2 ‖C ≤ ρ.̃ By [F] (i), we have t

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩(Px1 )(t) − (Px2 )(t)󵄩󵄩󵄩 ≤ ∫(t − s)α−1 ‖S (t − s)‖ 󵄩󵄩󵄩f(s, x1 (s)) − f(s, x2 (s))󵄩󵄩󵄩ds 0 t

ML1 (ρ)̃ ‖x1 − x2 ‖r ∫(t − s)α−1 ers ds ≤ Γ(α) 0

̃ −α ‖x1 − x2 ‖r ert . ≤ ML1 (ρ)r On setting

1

̃ α, r0 := (2M max{K, L1 (ρ)}) we obtain ‖Px‖r0 ≤ M‖x0 ‖ +

MK 1 1 󵄩 󵄩 b α + ‖x‖r0 󵄩󵄩󵄩(Px1 )(t) − (Px2 )(t)󵄩󵄩󵄩r0 ≤ ‖x1 − x2 ‖r0 . Γ(α + 1) 2 2

Taking ρ0 := 2(M‖x0 ‖ +

MK b α ), Γ(α + 1)

276 | 4 Fractional Evolution Equations: Continued

we see that P is a contraction map on B r0 (ρ0 ) := {x ∈ C(J, X) | ‖x‖r0 ≤ ρ0 },

ρ0 ≤ ρ.̃

This implies that system (4.40) has a unique mild solution on J. Now, we are ready to construct a mild solution for the impulsive system (4.39). For t ∈ [0, t1 ), the above result implies that t

x(t) = T (t)x0 + ∫(t − s)α−1 S (t − s)f(s, x(s))ds 0

is the unique mild solution of system (4.39) on [0, t1 ]. The jump is uniquely determined by the expression x(t1 + 0) = x(t1 − 0) + I1 (x(t1 − 0)) = x(t1 ) + I1 (x(t1 )) = x1 . For the time t ∈ (t1 , t2 ), we have t

x(t) = T (t)x0 + T (t − t1 )I1 (x(t1 )) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds 0 t1

= T (t)x0 + T (t − t1 )I1 (x(t1 )) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds 0 t

+ ∫(t − s)α−1 S (t − s)f(s, x(s))ds t1 t

= f1 (t) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds, t1

where t1

f1 (t) := T (t)x0 + T (t − t1 )I1 (x(t1 )) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds. 0

The above equation can be solved on C([t1 , t2 ], X) like above. By the previous result, x is a mild solution of system (4.39) on (t1 , t2 ]. The above procedures can be repeated on t ∈ (t2 , t3 ], (t3 , t4 ], . . . , (t m , b]. So we obtain a unique mild solution of system (4.39) on J and it is given by t

x(t) = T (t)x0 + ∑ T (t − t i )I i (x(t i )) + ∫(t − s)α−1 S (t − s)f(s, x(s))ds, 0 0. Γ(α + 1)

The proof is complete. Before we study the original control system (4.37), we have to introduce an admissible controls set Uad := {u : J → Ξ | u is strongly measurable}, where Ξ is a compact Polish space. Our original control system (4.37) will be considered under the following assumptions: [F+ ] The function f : J × X × Ξ → X is such that (i) t 󳨃→ f(t, ξ, η) is measurable, and (ξ, η) 󳨃→ f(t, ξ, η) is continuous on X × Ξ;

278 | 4 Fractional Evolution Equations: Continued (ii) for any finite number ρ > 0 there exists a constant L(ρ) > 0 such that 󵄩 󵄩󵄩 󵄩󵄩f(t, x1 , σ) − f(t, x2 , σ)󵄩󵄩󵄩 ≤ L(ρ)‖x1 − x2 ‖, for all ‖x1 ‖, ‖x2 ‖ < ρ, t ∈ J and σ ∈ Ξ; (iii) there exists a constant K F > 0 such that ‖f(t, x, σ)‖ ≤ K F (1 + ‖x‖),

t ∈ J, σ ∈ Ξ.

Theorem 4.63. Under the assumptions [A], [I] and [F+ ], system (4.37) has a unique mild solution x ∈ PC(J, X) given by t

x(t) = T (t)x0 + ∑ T (t − t i )I i (x(t i )) + ∫(t − s)α−1 S (t − s)f(s, x(s), u(s))ds, 0 R∗ , S1F(.,x(.)) .

m α α M 2 ‖B‖Ω(R) ∑ |a k |I0,t φ(t k ) + MΩ(R) sup I0,t φ(t) < R. k t∈J

k=1

Denote U = {x ∈ C(J, E) | ‖x‖∞ < R0 }

and

D = {x ∈ C(J, E) | ‖x‖∞ ≤ R0 },

(5.46)

308 | 5 Fractional Differential Inclusions where R0 = R∗ + 1. Our aim is to prove that the multivalued operator 𝕋 : D → 2C(J,E) satisfies the assumptions of Lemma A.40. Since the values of F are convex, it is clear that the values of 𝕋 are convex. Step 1. Any possible solution of the inclusion x ∈ λ𝕋(x),

x ∈ D, λ ∈ (0, 1)

(5.47)

satisfies x ∈ D − U. Let x be a solution of (5.47). Then, recalling the definition of 𝕋, condition (5.42) and Lemma 5.18, for any t ∈ J we get ‖x(t)‖ ≤

tk

t

0

0

M 2 ‖B‖Ω(‖x‖∞ ) m MΩ(‖x‖∞ ) ∑ |a k | ∫(t k − s)α−1 φ(s)ds + ∫(t − s)α−1 φ(s)ds Γ(α) Γ(α) k=1 m

α α ≤ M 2 ‖B‖Ω(R0 ) ∑ |a k |I0,t φ(t k ) + MΩ(R0 ) sup I0,t φ(t). k t∈J

k=1

This inequality with (5.46) implies ‖x‖∞ < R0 , i.e., x ∈ D − U. Step 2. 𝕋 maps bounded sets into equicontinuous sets in C(J, E). Set B r = {x ∈ C(J, E) | ‖x‖C(J,E) ≤ r} for r ∈ ℝ. Let y ∈ 𝕋(B r ) and τ1 , τ2 ∈ J, τ1 < τ2 . Then there is x ∈ B r with y ∈ 𝕋(x). By (5.45), there is f ∈ S1F(.,x(.)) such that m

‖y(τ2 ) − y(τ1 )‖ ≤ ‖K1 (τ2 ) − K1 (τ1 )‖ ‖B‖ ∑ |a k | ‖g(t k )‖ k=1 τ1

τ

0 τ1

0 τ1

0 τ2

0

1 󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(τ2 − s)α−1 K2 (τ2 − s)f(s)ds − ∫(τ1 − s)α−1 K2 (τ2 − s)f(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩

󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(τ1 − s)α−1 K2 (τ2 − s)f(s)ds − ∫(τ1 − s)α−1 K2 (τ1 − s)f(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫(τ2 − s)α−1 K2 (τ2 − s)f(s)ds󵄩󵄩󵄩󵄩. 󵄩 󵄩 τ1

Then m

‖y(τ2 ) − y(τ1 )‖ ≤ ‖K1 (τ2 ) − K1 (τ1 )‖ ‖B‖ ∑ |a k | ‖g(t k )‖ k=1 τ1

+

MΩ(r) ∫((τ1 − s)α−1 − (τ2 − s)α−1 )φ(s)ds Γ(α) 0 τ1

+ Ω(r) ∫(τ1 − s)α−1 ‖K2 (τ2 − s) − K2 (τ1 − s)‖φ(s)ds 0 τ2

MΩ(r) + ∫(τ2 − s)α−1 φ(s)ds = S1 + S2 + S3 + S4 , Γ(α) τ1

5.2 Nonlocal Cauchy problems for inclusions | 309

where m

S1 := ‖K1 (τ2 ) − K1 (τ1 )‖ ‖B‖ ∑ |a k | ‖g(t k )‖, k=1 α α S2 := MΩ(r)(I0,τ φ(τ1 ) − I0,τ φ(τ2 )), 1 2 τ1

S3 := Ω(r) ∫(τ1 − s)α−1 ‖K2 (τ2 − s) − K2 (τ1 − s)‖φ(s)ds, 0

S4 := 2MΩ(r)I τα1 ,τ2 φ(τ2 ). Remark 5.17 and (H0 ) yield limτ2 →τ1 S1 = 0 uniformly on J. Next, (H2 ) implies limτ2 →τ1 S2 = 0 and limτ2 →τ1 S4 = 0 uniformly on J. Finally, lim ‖K2 (τ2 − s) − K2 (τ1 − s)‖ = 0

τ2 →τ1

uniformly for 0 ≤ s ≤ τ1 < τ2 ≤ 1, and τ1 α φ(t). ∫(τ1 − s)α−1 φ(s) ≤ Γ(α) sup I0,t t∈J

0

So we obtain limτ2 →τ1 S3 = 0 uniformly on J. Note that all S i , i = 1, 2, 3, 4 are independent of x ∈ B r . Step 3. Implication (A.1) in Lemma A.40 holds with x0 = 0. Let Z = conv({x0 } ∪ 𝕋(Z)) ⊆ D, Z = C with C ⊆ Z countable. We claim that Z is relatively compact. Since C is countable and C ⊆ Z = conv({x0 } ∪ 𝕋(Z)), we can find a countable set ℍ = {y n | n ≥ 1} ⊆ 𝕋(Z) with C ⊆ conv({x0 } ∪ ℍ). Then there exists x n ∈ Z with y n ∈ 𝕋(x n ). This means that there is f n ∈ S1F(.,x n (.)) such that for t ∈ J, m

y n (t) = K1 (t) ∑ a k B(g n (t k )) + g n (t),

t ∈ J,

(5.48)

k=0

where

t

g n (t) = ∫(t − s)α−1 K2 (t − s)f n (s)ds. 0

From Z ⊆ C ⊆ conv({x0 } ∪ ℍ) we find that for t ∈ J, χ(Z(t)) ≤ χ(C(t)) ≤ χ(ℍ(t)) m

tk

≤ χ{K1 (t) ∑ a k B ∫(t k − s)α−1 K2 (t k − s)f n (s)ds | n ≥ 1} k=0

0

t

+ χ{ ∫(t − s)α−1 K2 (t − s)f n (s) | n ≥ 1}ds. 0

(5.49)

310 | 5 Fractional Differential Inclusions Note that for every n ≥ 1, 󵄩 󵄩󵄩 󵄩󵄩(t k − s)α−1 K2 (t k − s)f n (s)󵄩󵄩󵄩 ≤ MΩ(R0 )(t k − s)α−1 φ(s) for a.e. s ∈ [0, t k ] and each k = 1, 2, . . . , m. Analogously, 󵄩 󵄩󵄩 󵄩󵄩(t − s)α−1 K2 (t − s)f n (s)󵄩󵄩󵄩 ≤ MΩ(R0 )(t − s)α−1 φ(s) for a.e. t ∈ J, s ∈ [0, t] and each k = 1, 2, . . . , m. Then, using Proposition A.23, Lemma 5.18, (H3 ) and the properties of the measure of noncompactness, we see that equalities (5.48) and (5.49) imply tk

M 2 ‖B‖ m χ(Z(t)) ≤ ∑ |a k | ∫(t k − s)α−1 χ{f n (s) | n ≥ 1}ds Γ(α) k=1 0

t

+

M ∫(t − s)α−1 χ{f n (s) | n ≥ 1}ds Γ(α) 0

≤

tk

M 2 ‖B‖ m Γ(α)

∑ |a k | ∫(t k − s)α−1 β(s)χ{x n (s) | n ≥ 1}ds

k=1

0

t

+

M ∫(t − s)α−1 β(s)χ{f n (s) | n ≥ 1}ds Γ(α) 0 tk

m

M 2 ‖B‖ ≤ ∑ |a k | ∫(t k − s)α−1 β(s)χ(Z(s))ds Γ(α) k=1 0

t

+

M ∫(t − s)α−1 β(s)χ(Z(s))ds. Γ(α)

(5.50)

0

Recalling Z = conv({x0 } ∪ 𝕋(Z)), we deduce from Step 2 that Z is equicontinuous. Hence, we find from Proposition A.22 and (5.50) that χ C(J,E) (Z) = max χ(Z(s)) t∈J

m α α ≤ [M 2 ‖B‖ ∑ |a k |I0,t β(t k ) + M sup I0,t β(t)]χ C(J,E) (Z). k k=1

t∈J

This inequality with (5.43) implies χ C(J,E) (Z) = 0, i.e., Z is relatively compact. Step 4. 𝕋 maps compact sets into relatively compact sets. Let G be a compact subset of D. From Step 2, 𝕋(G) is equicontinuous. Let t ∈ J. Recalling the definition of 𝕋, for any x ∈ G and y ∈ 𝕋(x) there is f y,x ∈ S1F(.,x(.)) such that m

y(t) = K1 (t) ∑ a k B(g y (t k )) + g y (t), k=1

t ∈ J,

5.2 Nonlocal Cauchy problems for inclusions | 311

where

t

g y (t) = ∫(t − s)α−1 K2 (t − s)f y,x (s)ds. 0

Therefore, χ(𝕋(G)(t)) ≤ χ{y(t) | y ∈ 𝕋(x), x ∈ G} tk

m

≤ χ{K1 (t) ∑ a k B ∫(t k − s)α−1 K2 (t k − s)f y,x (s)ds | x ∈ G} k=1

0

t

+ χ{ ∫(t − s)α−1 K2 (t − s)f y,x (s) | x ∈ G}ds. 0

Arguing as in (5.50), we obtain tk

M 2 ‖B‖ m χ(𝕋(G)(t)) ≤ ∑ |a k | ∫(t k − s)α−1 χ{f y,x (s) | x ∈ G}ds Γ(α) k=1 0

t

+

M ∫(t − s)α−1 χ{f y,x (s) | x ∈ G}ds Γ(α) 0 tk

m

≤

M 2 ‖B‖ ∑ |a k | ∫(t k − s)α−1 β(s)χ{x(s) | x ∈ G}ds Γ(α) k=1 0

t

+

M ∫(t − s)α−1 β(s)χ{x(s) | x ∈ G}ds = 0. Γ(α) 0

Consequently, Proposition A.22 implies χ C(J,E) (𝕋(G)) = supt∈J χ(𝕋(G)(t)) = 0, i.e., the set 𝕋(G) is relatively compact. Step 5. The graph of 𝕋 is closed. Let {x n } be a sequence in D with x n → x, and let y n ∈ 𝕋(x n ) with y n → y by the convergence in C(J, E). Clearly, x ∈ D. We have to show that y ∈ 𝕋(x). Then for any n ≥ 1, there is f n ∈ S1F(.,x n (.)) such that m

y n (t) = K1 (t) ∑ a k B(g n (t k )) + g n (t),

t ∈ J,

(5.51)

k=1

where

t

g n (t) = ∫(t − s)α−1 K2 (t − s)f n (s)ds. 0

Observe that for every n ≥ 1 and for a.e. t ∈ J, ‖f n (t)‖ ≤ φ(t)Ω(‖x n ‖C(J,E) ) ≤ φ(t)Ω(R0 ).

(5.52)

312 | 5 Fractional Differential Inclusions This shows that the set {f n | n ≥ 1} is integrably bounded. In addition, the set {f n (t) | n ≥ 1} is relatively compact for a.e. t ∈ J, since assumption (H3 ) along with the convergence of {x n } implies that χ{f n (t) | n ≥ 1} ≤ χ(F(t, {x n (t) | n ≥ 1})) ≤ β(t)χ{x n (t) | n ≥ 1} = 0. So, the sequence {f n }n≥1 is semicompact. Hence, by Lemma A.5, it is weakly compact in L1 (J, E). So, without loss of generality we can assume that f n converges weakly to a function f ∈ L1 (J, ℝ+ ). By the Mazur lemma, for every natural number j there are a natural number k0 (j) > j and a sequence of nonnegative real numbers λ j,k , k = j, . . . , k0 (j) k0 (j) k0 (j) such that ∑k=j λ j,k = 1 and the sequence of convex combinations z j = ∑k=j λ j,k f k , j ≥ 1 1 converges strongly to f in L (J, ℝ+ ) as j → ∞, i.e., we may suppose that z j (t) → f(t) for a.e. t ∈ J. Then we get for a.e. t ∈ J, f(t) ∈ ⋂ {z k (t) | k ≥ j} ⊆ ⋂ conv{⋃ F(t, x k (t))}. j≥1

j≥1

k≥j

Since F is upper semicontinuous with compact and convex values, by Lemma A.1, we conclude that f(t) ∈ F(t, x(t)) for a.e. t ∈ J. Next, taking k0 (n)

y n (t) = ∑ λ n,k y k , k=n

we have, by (5.51) and (5.52), m

y n (t) = K1 (t) ∑ a k B(g n (t k )) + g n (t),

t ∈ J,

(5.53)

k=1

where

t

g n (t) = ∫(t − s)α−1 K2 (t − s)z n (s)ds.

(5.54)

0

Note that for every t ∈ J, s ∈ (0, t] and n ≥ 1, k0 (n)

󵄩󵄩 󵄩 󵄩󵄩(t − s)α−1 K2 (t − s)z n (s)󵄩󵄩󵄩 ≤ (t − s)α−1 ‖K2 (t − s)‖ ∑ λ k,n ‖f k (s)‖ k=n

M ≤ M(t − s)α−1 φ(s)Ω(R0 ) ∈ L1 ((0, t], ℝ+ ). Γ(α) But z n (t) → f(t) for a.e. t ∈ J. Then the continuity of K2 (⋅) implies that for every t ∈ J, K2 (t − s)z n (s) → K2 (t − s)f(s) for a.e. s ∈ (0, t). Therefore, passing to the limit as n → ∞ in (5.53) and (5.54), we obtain from the Lebesgue dominated convergence theorem that y n (t) → y(t) for m

y(t) = K1 (t) ∑ a k B(g(t k )) + g(t), k=1

t ∈ J,

5.2 Nonlocal Cauchy problems for inclusions | 313

where

t

g(t) = ∫(t − s)α−1 K2 (t − s)f(s)ds. 0

Therefore, y ∈ 𝕋(x). This shows that the graph of 𝕋 is closed. As a result of the Steps 1–5 the multivalued function 𝕋 satisfies all the assumptions of Lemma A.40. Consequently, 𝕋 has a fixed point which is a solution of (5.39). Remark 5.22. In [290, 334] the following condition is assumed: (H∗2 ) There exist a function φ ∈ L1/q (J, ℝ+ ), 0 < q < α and a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that for any x ∈ E, ‖F(t, x)‖ ≤ φ(t)Ω(‖x‖) for a.e. t ∈ J. It is not difficult to verify that (H∗2 ) implies (H2 ). Indeed, by Hölder’s inequality, for t1 , t2 ∈ J, t1 > t2 we have t

t

0

0

2 1 α−1 󵄨󵄨 (t2 − s)α−1 󵄨󵄨 α 󵄨󵄨 󵄨󵄨󵄨󵄨 (t1 − s) α φ(s)ds − φ(s)ds󵄨󵄨󵄨󵄨 φ(t ) = ∫ ∫ 󵄨󵄨I0,t1 φ(t1 ) − I0,t 󵄨 2 󵄨 󵄨󵄨 2 Γ(α) Γ(α) 󵄨 󵄨

t2

t1

1 1 ≤ ∫((t2 − s)α−1 − (t1 − s)α−1 )φ(s)ds + ∫(t1 − s)α−1 φ(s)ds Γ(α) Γ(α) t2

0 t2

≤

t2

q 1−q 1 1 1 ( ∫((t2 − s)α−1 − (t1 − s)α−1 ) 1−q ds) ( ∫ φ(s) q ds) Γ(α) 0

+

0 t1

t1

t2

t2

1−q q 1 α−1 1 ( ∫(t1 − s) 1−q ds) ( ∫ φ(s) q ds) Γ(α) t2

≤

1

q 1−q 1 1 ( ∫((t2 − s)α−1 − (t1 − s)α−1 )ds) ( ∫ φ(s) q ds) Γ(α) 0

+

0

)α−q

(t1 − t2 Γ(α)

1

(

q 1 1 − q 1−q ) ( ∫ φ(s) q ds) α−1 0

1

=

q 1 t α − t α + (t1 − t2 )α 1−q 1 1 − q 1−q + (t1 − t2 )α−q ( ( ∫ φ(s) q ds) (( 2 1 ) ) ). Γ(α) α α−1 0

α Now, the continuity of I0,t φ(t) is clear. Similarly, for λ > 0 small and t ∈ J, we derive t+λ

󵄨󵄨 (t + λ − s)α−1 󵄨󵄨 α 󵄨 󵄨󵄨 φ(s)ds󵄨󵄨󵄨󵄨 󵄨󵄨I t,t+λ φ(t + λ)󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨 ∫ Γ(α) 󵄨 󵄨 t

314 | 5 Fractional Differential Inclusions t+λ

≤

t+λ

1−q q α−1 1 1 ( ∫ (t + λ − s) 1−q ds) ( ∫ φ(s) q ds) Γ(α) t

t

1

≤

q 1 λ α−q 1 − q 1−q ( ) ( ∫ φ(s) q ds) , Γ(α) α − 1 0

α φ(t + λ) = 0 uniformly for t ∈ J. which certainly implies limλ→0+ I t,t+λ

Next, we present our second existence result for (5.39). Theorem 5.23. Let F : J × E → P ck (E) be a multifunction. In addition to conditions (H0 ), (H1 ) and (C1 ), we assume the following conditions: α (H5 ) There exists a function φ ∈ L([0, t m ], ℝ+ ) such that the function t 󳨃→ I0,t φ(t) is α + continuous with limλ→0 I t,t+λ φ(t + λ) = 0 uniformly for t ∈ [0, t m ]. There exists a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that for any x ∈ E, ‖F(t, x)‖ ≤ φ(t)Ω(‖x‖) for a.e. t ∈ [0, t m ]. α There exists a function p ∈ L([t m , 1], ℝ+ ) such that the function t 󳨃→ I0,t p(t) is α continuous with limλ→0+ I t,t+λ p(t + λ) = 0 uniformly for t ∈ [t m , 1], and for any x ∈ E, ‖F(t, x)‖ ≤ p(t) for a.e. t ∈ [t m , 1].

(H6 ) The following inequality holds: m

−1

α lim sup ρ(MΩ(ρ)(M‖B‖ ∑ |a k | + 1) sup I0,t φ(t)) ρ→∞

k=0

t∈[0,t m ]

> 1.

(5.55)

Then problem (5.39) has a solution. Proof. Just like in the proof of Theorem 5.21, we need to consider the existence of fixed points of 𝕋 : C(J, E) → 2C(J,E) in (5.45). Here we only present the main differences in comparison to the proof of Theorem 5.21. First we check that any possible solution of the inclusion (5.47) satisfies x ∈ D − U, where U = {x ∈ C(J, E) | ‖x‖∞ < R0 }, D = {x ∈ C(J, E) | ‖x‖∞ ≤ R0 } with R0 = R∗ + 1 for some R∗ > 0. Let x be a solution of (5.47) and set R1 = max{‖x(t)‖ | t ∈ [0, t m ]}. Using (H5 ) and Lemma 5.18, for any t ∈ [0, t m ], we obtain tk

t

0

0

M 2 ‖B‖Ω(R1 ) m MΩ(R1 ) ‖x(t)‖ ≤ ∑ |a k | ∫(t k − s)α−1 φ(s)ds + ∫(t − s)α−1 φ(s)ds Γ(α) Γ(α) k=1 m α ≤ MΩ(R1 )(M‖B‖ ∑ |a k | + 1) sup I0,t φ(t), k=1

t∈[0,t m ]

5.2 Nonlocal Cauchy problems for inclusions | 315

which implies m α R1 ≤ MΩ(R1 )(M‖B‖ ∑ |a k | + 1) sup I0,t φ(t).

(5.56)

t∈[0,t m ]

k=1

Obviously, (5.55) with (5.56) implies that R1 < r1 for a constant r1 > 0 independent of x. For any t ∈ [t m , 1], using (H5 ) and Lemma 5.18, we obtain tk

M 2 ‖B‖Ω(r1 ) m ‖x(t)‖ ≤ ∑ |a k | ∫(t k − s)α−1 φ(s)ds Γ(α) k=1 0

tm

+

t

M MΩ(r1 ) ∫ (t − s)α−1 p(s)ds ∫ (t − s)α−1 φ(s)ds + Γ(α) Γ(α) tm

0 m

α α ≤ M[Ω(r1 )(M‖B‖ ∑ |a k | + 1) sup I0,t φ(t) + sup I0,t p(t)] =: r2 . t∈[0,t m ]

k=1

t∈[t m ,1]

On setting R∗ = max{r1 , r2 } + 1, we conclude that ‖x‖∞ < R∗ , i.e., x ∈ D − U. Next, taking the exchange φ ↔ φ + p, Ω ↔ Ω + 1, we see that assumption (H2 ) is satisfied. So we can follow the arguments of Steps 2–5 of the proof of Theorem 5.21 to show that 𝕋 satisfies all the assumptions of Lemma A.40. Thus, 𝕋 has a fixed point which gives the existence result.

5.2.4 Examples In this subsection, we first apply the above abstract results to a model of fractional differential inclusions onto lattices with global neighborhood interactions represented by (5.40). Set E = c0 = {x = {x n }n∈ℤ | x n ∈ ℝ, lim x n = 0} n→±∞

with the norm ‖x‖ = supn∈ℤ |x n |. Then E is a separable Banach space. Now, (Ax)n = ∑j∈ℤ α nj x j for α nj = λ j−n , n ≠ j and α nn = −2 ∑j∈ℕ λ j . Next, we compute ∑ |α nj | = ∑ λ|n−j| + 2 ∑ λ j = 4 ∑ λ j . j∈ℤ

n=j∈ℤ ̸

j∈ℕ

j∈ℕ

Then by [267, Theorem 6.3, p. 221] we know that A ∈ L(c0 ) and ‖A‖ = sup ∑ |α nj | = 4 ∑ λ j . n∈ℤ j∈ℤ

j∈ℕ

Furthermore, for any λ > 0, we have (λx − Ax)n = (λ + 2 ∑ λ j )x n − ∑ λ j−n x j , j∈ℕ

n=j∈ℤ ̸

316 | 5 Fractional Differential Inclusions i.e., (λ + 2 ∑j∈ℕ λ j )𝕀 − A for (Ax)n = ∑ λ j−n x j . n=j∈ℤ ̸

Then ‖A‖ = 2 ∑j∈ℕ λ j . Using 2 ∑j∈ℕ λ j < λ + 2 ∑j∈ℕ λ j , we see that the Neumann lemma implies (λ𝕀 − A)−1 ∈ L(c0 ) with ‖(λ𝕀 − A)−1 ‖ = 1λ . Consequently, by the Hille–Yosida theorem, T(⋅) = eA⋅ is a uniformly continuous semigroup of contractions, so M = 1 in (i) of Lemma 5.18. Let α = 12 , m = 1, a1 = 14 and t1 = 14 . Then x n (0) = 14 x n ( 41 ), n ∈ ℤ. Since ‖K1 ( 14 )‖ ≤ 1 1 1 1 −1 and M ∑m is k=0 |a k | = 4 < 1, by Lemma 5.19, the linear operator B = [𝕀 − 4 K 1 ( 4 )] defined with ‖B‖ ≤ 34 . The following two examples will illustrate the feasibility of our assumptions in the above theorems. Example 5.24. Let f n : J × ℝ → 2ℝ be multifunctions defined by f n (t, x n ) =

cos 2πt 1 1 + [− x n , xn ] 16(|n| + 1) 16 16

for any n ∈ ℤ. Then we define F : J × E → 2E as F(t, x) = {f n (t, x n )}n∈ℤ , and condition (H1 ) is easily verified. Next, for (t, x) ∈ J × E, we have ‖F(t, x)‖ = sup{‖y‖ | y ∈ F(t, x)} ≤

1 (‖x‖ + 1). 16

1 So, condition (H2 ) is satisfied with φ(t) = 16 and Ω(z) = z + 1 for t ∈ J and z ∈ ℝ+ . Now 1 cos 2πt + λ n x n , |λ n | ≤ | 16 |, n ∈ ℤ. Taking let t ∈ J, x, y ∈ E and u ∈ F(t, x). Then u n = 16(|n|+1)

v={ we get

cos 2πt + λn yn } ∈ F(t, y), 16(|n| + 1) n∈ℤ

1 󵄨 󵄨 ‖x − y‖. d(u, F(t, y)) ≤ ‖u − v‖ = sup󵄨󵄨󵄨λ n x n − λ n y n 󵄨󵄨󵄨 ≤ 16 n∈ℤ

This yields sup d(u, F(t, y)) ≤

1 ‖x − y‖. 16

sup d(v, F(t, x)) ≤

1 ‖x − y‖. 16

u∈F(t,x)

Similarly, we can show that v∈F(t,y)

Therefore,

1 ‖x − y‖, 16 where h denotes the standard Hausdorff distance. So by [155, Corollary 2.2.1, p. 47], 1 for every bounded subset D ⊆ E, χ(F(t, D)) ≤ β(t)χ(D) for a.e. t ∈ J, with β(t) = 16 for t ∈ J. Note that the relation (5.43) holds in (H3 ), since h(F(t, x), F(t, y)) ≤

t1α 1 M 35 + < 1, (M 2 ‖B‖ |a1 | )≤ 16 Γ(α + 1) Γ(α + 1) 256√π

5.2 Nonlocal Cauchy problems for inclusions | 317

while the relation (5.44) holds in (H4 ), since lim sup 16ρ(M 2 ‖B‖(ρ + 1)|a1 | ρ→∞

t1α M(ρ + 1) −1 256√π + > 1. ) ≥ Γ(α + 1) Γ(α + 1) 35

Finally, Theorem 5.21 shows that problem (5.40) has a solution. Example 5.25. Let f n : J × ℝ → 2ℝ be multifunctions defined as cos 2πt 1 cos 2πt { if t ∈ [0, ), { {[− 16 x n , 16 x n ] 4 f n (t, x n ) = { { {[− cos 2πt , cos 2πt ] if t ∈ [ 1 , 1] 4 { 16(|n| + 1) 16(|n| + 1) for any n ∈ ℤ. Then we define F : J × E → 2E as F(t, x) = {f n (t, x n )}n∈ℤ , and condition (H1 ) is easily verified. Like above we derive 1 1 { { { 16 ‖x‖ if (t, x) ∈ [0, 4 ) × E, ‖F(t, x)‖ ≤ { 1 { { 1 if (t, x) ∈ [ , 1] × E. 16 4 { Arguing as in the previous example, we can show that for every bounded subset 1 D ⊆ E, χ(F(t, D)) ≤ β(t)χ(D) for a.e. t ∈ [0, 14 ) with β(t) = 16 . On the other hand, 1 it is clear that F(t, D) is precompact for t ∈ [ 4 , 1], i.e., χ(F(t, D)) = 0. Therefore, 1 χ(F(t, D)) ≤ β(t)χ(D) for a.e. t ∈ J, with β(t) = 16 for t ∈ J. Hence, condition (H5 ) 1 is satisfied with φ(t) = p(t) = 16 and Ω(z) = z for t ∈ J and z ∈ ℝ+ . Moreover, t1α 19 1 M(M‖B‖ |a1 | + 1) ≤ < 1, 16 Γ(α + 1) 256√π so condition (H6 ) is satisfied as well. Finally, Theorem 5.23 shows that problem (5.40) has a solution. Motivated by [101, 106, 107, 255, 262], the above considerations can be extended to higher dimensional lattices. Example 5.26. For instance we can study fractional differential inclusions on twodimensional lattices ℤ2 of the form c α D0,t x n,m (t) ∈ ∑ λ j−n,k−m (x j,k (t) − x n,m (t)) { { { j,k∈ℤ { { { + f n,m (t, x n,m (t)), a.e. on J, n, m ∈ ℤ, { { m { { { {x n,m (0) = ∑ a k x n,m (t k ), n, m ∈ ℤ, { k=1

(5.57)

where λ j,k ≥ 0, j, k ∈ ℤ, λ0,0 = 0, f n,m : J × ℝ → 2ℝ is specified below, and x n,m ∈ ℝ with limn,m→±∞ x n,m = 0. Moreover, we suppose ̃ M := ∑j,k∈ℤ λ j,k < +∞. In the following, we study forced localized solutions of (5.57).

318 | 5 Fractional Differential Inclusions

We proceed like above by setting E = {x = {x n,m }n,m∈ℤ | x n,m ∈ ℝ,

lim

n,m→±∞

x n,m = 0}

with the norm ‖x‖ = supn,m∈ℤ |x n,m |. Now, (Ax)n,m = ∑ α nmjk x j,k j,k∈ℤ

for α nmjk = λ j−n,k−m , (n, m) ≠ (j, k) and α nmnm = − ∑j,k∈ℤ λ j,k . Next, we compute M. ∑ |α nmjk | = 2 ∑ λ j,k = 2̃ j,k∈ℤ

j,k∈ℤ

Then A ∈ L(E) with ‖A‖ = 2̃ M. Furthermore, for any λ > 0, we have (λx − Ax)n,m = (λ + ∑ λ j,k )x n,m −

∑

λ j−n,k−m x j,k ,

2 (n,m)=(j,k)∈ℤ ̸

j,k∈ℤ

i.e., (λ + ̃ M)𝕀 − A for (Ax)n,m =

∑

λ j−n,k−m x j,k .

2 (n,m)=(j,k)∈ℤ ̸

Then ‖A‖ = ̃ M and the Neumann lemma implies (λ𝕀 − A)−1 ∈ L(E) with ‖(λ𝕀 − A)−1 ‖ = 1 A⋅ λ . Consequently, by the Hille–Yosida theorem, T(⋅) = e is a uniformly continuous semigroup of contractions, so M = 1 in (i) of Lemma 5.18. Keeping α = 21 , m = 1, a1 = 14 , t1 = 41 , and taking either f n,m (t, x) =

cos 2πt 1 1 + [− x, x], 16(|n| + |m| + 1) 16 16

(t, x) ∈ J × ℝ,

or cos 2πt cos 2πt 1 { if (t, x) ∈ [0, ) × ℝ, { {[− 16 x, 16 x] 4 f n,m (t, x) = { 1 cos 2πt cos 2πt { {[− , ] if (t, x) ∈ [ , 1] × ℝ, 4 { 16(|n| + |m| + 1) 16(|n| + |m| + 1) for n, m ∈ ℤ, we deduce from Theorems 5.21 and 5.23 that problem (5.57) has a solution.

5.3 Nonlocal impulsive fractional differential inclusions 5.3.1 Introduction The theory of impulsive differential equations and impulsive differential inclusions has been an object of interest because of its numerous applications in physics, biology, engineering, medical fields, industry and technology. The reason for this applicability

5.3 Nonlocal impulsive fractional differential inclusions | 319

arises from the fact that impulsive differential problems are an appropriate model for describing processes which at certain moments change their state rapidly and which cannot be described using the classical differential problems. For some of these applications we refer to [7, 38]. During the last ten years, impulsive differential inclusions with different conditions have been intensely studied by many mathematicians. At present, the foundations of the general theory of impulsive differential equations and inclusions are already laid, and many of them are investigated in details in the book of Benchohra et al. [47]. Moreover, a strong motivation for investigating the nonlocal Cauchy problems, which are a generalization of the classical Cauchy problems with initial conditions, comes from physical problems. For example, the nonlocal Cauchy problem is used to determine the unknown physical parameters in some inverse heat condition problems. For the applications of nonlocal condition problems we refer to [37, 114]. In the past few years, several papers have studied the existence of solutions for differential equations or differential inclusions with nonlocal conditions [13]. For impulsive differential equations or inclusions with nonlocal conditions of order one we refer to [64, 99]. For impulsive differential equations or inclusions of fractional order we refer to [6, 84, 249, 290] and the references therein. In this section we are concerned with the existence of mild solutions to the following nonlocal impulsive fractional differential inclusions: c α D x(t) ∈ Ax(t) + F(t, x(t)), α ∈ (0, 1), a.e. on J − {t1 , t2 , . . . , t m }, { { { x(0) = x0 − g(x), { { { + { x(t i ) = x(t i ) + I i (x(t i )), i = 1, 2, . . . , m,

(5.58)

where J := [0, b] with fixed b > 0; c D α is the Caputo fractional derivative of order α ∈ (0, 1) with the lower limit zero; A is a fractional sectorial operator like in [30] defined on a separable Banach space E; F : J × E → 2E − {ϕ} is a multifunction; 0 = t0 < t1 < ⋅ ⋅ ⋅ < t m < t m+1 = b; I i : E → E for i = 1, 2, . . . , m are impulsive functions characterizing the jumps of the solutions at impulse points t i ; g : PC(J, E) → E is a nonlinear function related to the nonlocal condition at the origin; x(t+i ) and x(t−i ) are the right- and left-sided limits of x at the point t i , respectively; and PC(J, E) is defined in Appendix A. Concerning the main problem (5.58), we have to study the following impulsive fractional evolution equations with nonlocal conditions: c α D x(t) = Ax(t) + F(t, x(t)), α ∈ (0, 1), a.e. on J − {t1 , t2 , . . . , t m }, { { { x(0) = x0 − g(x), { { { + { x(t i ) = x(t i ) + I i (x(t i )), i = 1, 2, . . . , m.

This section is based on [292].

320 | 5 Fractional Differential Inclusions

5.3.2 Preliminaries and notation We recall some facts on a fractional Cauchy problem. Bajlekova [30] studied the following linear fractional Cauchy problem: c

{

D α x(t) = Ax(t),

t ≥ 0,

(5.59)

x(0) = x0 ∈ E,

where A is linear closed and D(A) is dense. Definition 5.27 ([30, Definition 2.3]). A family {S α (t) | t ≥ 0} ⊂ L(E) is called a solution operator for (5.59) if the following conditions are satisfied: (a) S α (t) is strongly continuous for t ≥ 0, and S α (0) = I; (b) S α (t)D(A) ⊂ D(A) and AS α (t)x = S α (t)Ax for all x ∈ D(A) and t ≥ 0; (c) S α (t)x is a solution of (5.59) for all x ∈ D(A) and t ≥ 0. Definition 5.28 ([30, Definition 2.4]). An operator A is said to belong to eα (M, ω) if the solution operator S α (⋅) of (5.59) satisfies ‖S α (t)‖L(E) ≤ Meωt ,

t≥0

for some constants M ≥ 1 and ω ≥ 0. Definition 5.29 ([30, Definition 2.13]). A solution operator S α (t) of (5.59) is called analytic if it admits an analytic extension to a sector Σ θ0 = {λ ∈ ℂ − {0} | |arg λ| < θ0 } for some θ0 ∈ (0, 2π ]. An analytic solution operator is said to be of analyticity type (θ0 , ω0 ) if for each θ < θ0 and ω > ω0 there is M = M(θ, ω) such that ‖S α (t)‖L(E) ≤ Meω Re t ,

t ∈ Σθ .

Set eα (ω) := ⋃{eα (M, ω) | M ≥ 1},

eα := ⋃{eα (ω) | ω ≥ 0},

A α (θ0 , ω0 ) := {A ∈ eα | A generates an analytic solution operator S α of type (θ0 , ω0 )}. Remark 5.30 ([30, Theorem 2.14]). Let α ∈ (0, 2). A linear closed densely defined operator A belongs to A α (θ0 , ω0 ) if and only if λ α ∈ ρ(A) for each λ ∈ Σ θ0 + 2π (w0 ) = {λ ∈ ℂ − {0} | |arg(λ − w0 )| < θ0 +

π }, 2

and for any ω > ω0 , θ < θ0 there is a constant C = C(θ, ω) such that ‖λ α−1 R(λ α , A)‖L(E) ≤ for λ ∈ Σ θ0 + 2π (w).

C |λ − ω|

5.3 Nonlocal impulsive fractional differential inclusions | 321

According to the proof of [30, Theorem 2.14], if A ∈ A α (θ0 , w0 ) for some θ0 ∈ (0, π) and w0 ∈ ℝ, the solution operator for equation (5.59) is given by S α (t) =

1 ∫ eλt λ α−1 R(λ α , A)dλ 2πi Γ

for a suitable path Γ. Next, following [4, 30, 259], a mild solution of the Cauchy problem c

{

D α x(t) = Ax(t) + f(t),

t ∈ J,

x(0) = x0 ∈ E

can be defined by t

u(t) = S α (t)x0 + ∫ T α (t − s)f(s)ds, 0

where

1 ∫ eλt R(λ α , A)dλ 2πi

T α (t) =

Γ

for a suitable path Γ and f : J → E is continuous. We need the following estimates. Lemma 5.31 (see [30, (2.26)], [259]). If A ∈ A α (θ0 , ω0 ) then ‖S α (t)‖L(E) ≤ Meωt

and ‖T α (t)‖L(E) ≤ Ceωt (1 + t α−1 )

for every t > 0, ω > ω0 . So putting M S := sup ‖S α (t)‖L(E) ,

M T := sup Ceωt (1 + t1−α ),

0≤t≤b

0≤t≤b

we get ‖S α (t)‖L(E) ≤ M S ,

‖T α (t)‖L(E) ≤ t α−1 M T .

(5.60)

Based on the above consideration (see also [281]), we introduce the definition of a mild solution for (5.58). Definition 5.32. Let A ∈ A α (θ0 , ω0 ) with θ0 ∈ (0, 2π ] and ω0 ∈ ℝ. A function x ∈ PC(J, E) is called a mild solution of (5.58) if t

i

x(t) = S α (t)(x0 − g(x)) + ∑ S α (t − k=1

t k )I k (x(t−k ))

+ ∫ T α (t − s)f(s)ds 0

for t ∈ J i , i = 0, 1, . . . , m, where f ∈ S1F(⋅,x(⋅)) . Here and subsequently, set ∑0k=1 ⋅ := 0. Remark 5.33. There are also other concepts of solutions for impulsive fractional differential equations, see [5, 10, 33, 50]. According to the paper [273], we are now aware that the definition of solutions of impulsive Caputo fractional equations (ICFE) is questionable. We tried to explain our attitude in [110]. It is now clear that the definition

322 | 5 Fractional Differential Inclusions

of solutions of ICFE is not so simple as for natural-order differential equations with impulses. Our approach is based on the fact that the lower limit in the Caputo derivative is given, so it is fixed. This means that a family of solutions is set at the beginning by the Cauchy initial conditions. Then at each impulse the solution is kicked by the impulse on one of these solutions. This was first demonstrated in the derivation of formula (10) in [112, Lemma 2.7] for non-homogeneous linear Caputo fractional differential equations. Later, this formula was adapted for determining a mild solution for linear Caputo fractional abstract differential equations with impulses in [281, formula (3.4)] (see also [288, Definition 4.1]). We follow this path also in this section by introducing Definition 5.32. Summarizing, we do not claim that our definition of solutions for ICFE is the best one, but we tried to follow a way similar to the one for ordinary differential equations with impulses, since there, for an ODE, a solution is kicked by an impulse on a solution with different Cauchy condition at the beginning.

5.3.3 Existence results for convex case In this subsection, we give some existence results for (5.58) when A is a sectorial operator. Theorem 5.34. Let A ∈ A α (θ0 , ω0 ) with θ0 ∈ (0, 2π ] and ω0 ∈ ℝ, let F : J × E → P ck (E) be a multifunction, g : PC(J, E) → E and I i : E → E for i = 1, 2, . . . , m. We assume the following conditions: (HF1) For every x ∈ E, t 󳨃→ F(t, x) is measurable; for a.e. t ∈ J, x 󳨃→ F(t, x) is upper semicontinuous. (HF2) There exist a function φ ∈ L1/q (J, ℝ+ ), q ∈ (0, α) and a nondecreasing continuous function Ω : ℝ+ → ℝ+ such that for any x ∈ E, ‖F(t, x)‖ ≤ φ(t)Ω(‖x‖)

for a.e. t ∈ J.

(HF3) There exists a function β ∈ L1/q (J, ℝ+ ), q ∈ (0, α) satisfying 2ηM T ‖β‖ where η =

b α−q ϖ1−q

and ϖ =

α−q 1−q ,

1

L q (J,ℝ+ )

< 1,

(5.61)

and for every bounded subset D ⊆ E, χ(F(t, D)) ≤ β(t)χ(D)

(Hg)

(HI)

for a.e. t ∈ J, where χ is the Hausdorff measure of noncompactness in E. The function g is continuous, compact and there are two positive constants a, d such that ‖g(x)‖ ≤ a‖x‖PC(J,E) + d for all x ∈ PC(J, E). For every i = 1, 2, . . . , m, I i is continuous and compact, and there exists a positive constant h i such that ‖I i (x)‖ ≤ h i ‖x‖, x ∈ E.

5.3 Nonlocal impulsive fractional differential inclusions | 323

Then problem (5.58) has a mild solution provided that there is r > 0 such that M S (‖x0 ‖ + ar + d + hr) + M T ηΩ(r)‖φ‖

1

L q (J,ℝ+ )

≤ r,

(5.62)

where h = ∑m i=1 h i . Proof. We turn problem (5.58) into a fixed point problem and define a multifunction R : PC(J, E) → 2PC(J,E) as follows: for x ∈ PC(J, E), R(x) is the set of all functions y ∈ R(x) such that t

i

y(t) = S α (t)(x0 − g(x)) + ∑ S α (t −

t k )I k (x(t−k ))

+ ∫ T α (t − s)f(s)ds

k=1

(5.63)

0

for t ∈ J i , i = 0, 1, . . . , m, where f ∈ S1F(⋅,x(⋅)) . In view of (HF1) the values of R are nonempty. It is easy to see that any fixed point of R is a mild solution of (5.58). So our goal is to prove, by using Lemma A.37, that R has a fixed point. The proof will be given in several steps. Step 1. The values of R are closed. Let x ∈ PC(J, E) and {y n | n ≥ 1} be a sequence in R(x) and converging to y in PC(J, E). Then, according to the definition of R, there is a sequence {f n | n ≥ 1} in S1F(⋅,x(⋅)) such that t

i

y n (t) = S α (t)(x0 − g(x)) + ∑ S α (t − t k )I k (x(t−k )) + ∫ T α (t − s)f n (s)ds k=1

(5.64)

0

for t ∈ J i , i = 0, 1, . . . , m. In view of (HF2) for every n ≥ 1 and for a.e. t ∈ J, ‖f n (t)‖ ≤ φ(t)Ω(‖x(t)‖) ≤ φ(t)Ω(‖x‖PC(J,E) ). This shows that {f n | n ≥ 1} is integrably bounded. Moreover, because {f n (t) | n ≥ 1} ⊂ F(t, x(t)) for a.e. t ∈ J, the set {f n (t) | n ≥ 1} is relatively compact in E for a.e. t ∈ J. Therefore, the set {f n | n ≥ 1} is semicompact and, by Lemma A.5, it is weakly compact in L1 (J, E). So, without loss of generality we can assume that f n converges weakly to a function f ∈ L1 (J, E). By the Mazur lemma, for every natural number j there is a natural number k0 (j) > j and a sequence of nonnegative real numbers λ j,k , k = j, . . . , k0 (j) such that k0 (j)

∑ λ j,k = 1,

k=j k (j)

0 and the sequence of convex combinations z j = ∑k=j λ j,k f k , j ≥ 1 converges strongly to f in L1 (J, E) as j → ∞. So, we may suppose that z j (t) → f(t) for a.e. t ∈ J. Since F takes convex and closed values, we obtain for a.e. t ∈ J,

f(t) ∈ ⋂ {z k (t) | k ≥ j} ⊆ ⋂ conv{f k | k ≥ j} ⊆ F(t, x(t)). j≥1

j≥1

324 | 5 Fractional Differential Inclusions Note that, by (5.60) for every t, s ∈ J, s ∈ (0, t] and every n ≥ 1, ‖T α (t − s)z n (s)‖ ≤ |t − s|α−1 M T φ(s)Ω(‖x‖PC(J,E) ) ∈ L1 ((0, t], ℝ+ ). k (j)

0 Next, taking ỹ n (t) = ∑k=j λ j,k y k , we see that (5.64) implies

t

i

ỹ n (t) = S α (t)(x0 − g(x)) + ∑ S α (t − t k )I k (x(t−k )) + ∫ T α (t − s)z n (s)ds k=1

(5.65)

0

for t ∈ J i , i = 0, 1, . . . , m. But ỹ n (t) → y(t) and z n (t) → f(t) for a.e. t ∈ J. Therefore, passing to the limit n → ∞ in (5.65) shows that the Lebesgue dominated convergence theorem yields t

i

y(t) = S α (t)(x0 − g(x)) + ∑ S α (t −

t k )I k (x(t−k ))

+ ∫ T α (t − s)f(s)ds

k=1

0

for t ∈ J i , i = 0, 1, . . . , m. This proves that R(x) is closed. Step 2. Set B0 = {x ∈ PC(J, E) | ‖x‖ ≤ r}. Obviously, B0 is a bounded, closed and convex subset of PC(J, E). We claim that R(B0 ) ⊆ B0 . To prove this relation, let x ∈ B0 and y ∈ R(x). Using (5.60), (5.62), (5.64), (HF2), (Hg) and Hölder’s inequality, we get for t ∈ J0 , t

‖y(t)‖ ≤ M S (‖x0 ‖ + ar + d) + M T Ω(r) ∫(t − s)α−1 φ(s)ds 0 t

≤ M S (‖x0 ‖ + ar + d) + M T Ω(r)‖φ‖

α−1

1 Lq

(J,ℝ+ )

1−q

( ∫(t − s) 1−q ds) 0

b α−q M T Ω(r) ≤ M S (‖x0 ‖ + ar + d) + ‖φ‖ 1q L (J,ℝ+ ) ϖ1−q = M S (‖x0 ‖ + ar + d) + M T ηΩ(r)‖φ‖ 1q ≤ r. L (J,ℝ+ )

Similarly, by using (HI) in addition, we get for t ∈ J i , i = 1, 2, . . . , m, ‖y(t)‖ ≤ M S (‖x0 ‖ + ar + d + hr) + M T ηΩ(r)‖φ‖

1

L q (J,ℝ+ )

≤ r.

Therefore, R(B0 ) ⊆ B0 . Step 3. Let Z = R(B0 ). We claim that the set Z|J i is equicontinuous for every i = 0, 1, 2, . . . , m, where Z|J i = {y∗ ∈ C(J i , E) | y∗ (t) = y(t), t ∈ J i , y∗ (t i ) = y(t+i ), y ∈ Z}. Let y ∈ Z. Then there is x ∈ B0 with y ∈ R(x). By recalling the definition of R, there is f ∈ S1F(⋅,x(⋅)) such that i

t

y(t) = S α (t)(x0 − g(x)) + ∑ S α (t − t k )I k (x(t−k )) + ∫ T α (t − s)f(s)ds k=1

for t ∈ J i , i = 0, 1, . . . , m.

0

5.3 Nonlocal impulsive fractional differential inclusions | 325

We consider the following cases: Case 1. When i = 0, let t, t + δ be two points in J0 . Then 󵄩 󵄩 󵄩 󵄩󵄩 ∗ 󵄩󵄩y (t + δ) − y∗ (t)󵄩󵄩󵄩 = 󵄩󵄩󵄩y(t + δ) − y(t)󵄩󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩S α (t + δ)(x0 − g(x)) − S α (t)(x0 − g(x))󵄩󵄩󵄩 t+δ

t

0

0

󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩 ∫ T α (t + δ − s)f(s)ds − ∫ T α (t − s)f(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩S α (t + δ)(x0 − g(x)) − S α (t)(x0 − g(x))󵄩󵄩󵄩 t+δ

󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩 ∫ T α (t + δ − s)f(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩 t

t

󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩∫[T α (t + δ − s) − T α (t − s)]f(s)ds󵄩󵄩󵄩󵄩 󵄩 󵄩 0

= G1 + G2 + G3 , where 󵄩 󵄩 G1 := 󵄩󵄩󵄩S α (t + δ)(x0 − g(x)) − S α (t)(x0 − g(x))󵄩󵄩󵄩, t+δ

󵄩󵄩 󵄩󵄩 G2 := 󵄩󵄩󵄩󵄩 ∫ T α (t + δ − s)f(s)ds󵄩󵄩󵄩󵄩, 󵄩 󵄩 t

t

󵄩󵄩 󵄩󵄩 G3 := 󵄩󵄩󵄩󵄩∫[T α (t + δ − s) − T α (t − s)]f(s)ds󵄩󵄩󵄩󵄩. 󵄩 󵄩 0

We only need to verify that G i → 0 as δ → 0 for every i = 1, 2, 3. For G1 we have 󵄩 󵄩 lim G1 = lim 󵄩󵄩󵄩S α (t + δ)(x0 − g(x)) − S α (t)(x0 − g(x))󵄩󵄩󵄩 δ→0 δ→0 󵄩 󵄩 ≤ lim 󵄩󵄩󵄩S α (t + δ) − S α (t)󵄩󵄩󵄩(‖x0 ‖ + ar + d) = 0 δ→0 uniformly for x ∈ B0 . For G2 , by Hölder’s inequality we have t+δ

󵄩󵄩 󵄩󵄩 lim G2 = lim 󵄩󵄩󵄩󵄩 ∫ T α (t + δ − s)f(s)ds󵄩󵄩󵄩󵄩 δ→0 δ→0󵄩 󵄩 t

t+δ

≤ M T lim ∫ (t + δ − s)α−1 ‖f(s)‖ds δ→0

t t+δ

≤ M T Ω(r) lim ∫ (t + δ − s)α−1 φ(s)ds δ→0

t

(5.66)

326 | 5 Fractional Differential Inclusions t+δ α−1

≤ M T Ω(r) lim [ ∫ (t + δ − s) 1−q ds]

1−q

δ→0

‖φ‖

t

1

L q (J,ℝ+ )

δ ϖ 1−q ≤ M T Ω(r) lim [ ] ‖φ‖ 1q =0 L (J,ℝ+ ) δ→0 ϖ uniformly for x ∈ B0 . For G3 , using the Lebesgue dominated convergence theorem and the definition of T α , we get t

󵄩󵄩 󵄩󵄩 lim G3 ≤ lim 󵄩󵄩󵄩󵄩∫ T α (t + δ − s)f(s) − T α (t − s)f(s)󵄩󵄩󵄩󵄩ds τ→0 δ→0󵄩 󵄩 0

t

󵄩 󵄩 ≤ ∫ lim 󵄩󵄩󵄩T α (t + δ − s)f(s) − T α (t − s)f(s)󵄩󵄩󵄩ds = 0 δ→0

0

independently of x. Case 2. When i ∈ {1, 2, . . . , m}, let t, t + δ be two points in J i . Invoking the definition of R, we have 󵄩󵄩 ∗ 󵄩 󵄩 󵄩 󵄩󵄩y (t + δ) − y∗ (t)󵄩󵄩󵄩 = 󵄩󵄩󵄩y(t + δ) − y(t)󵄩󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩S α (t + δ)(x0 − g(x)) − S α (t)(x0 − g(x))󵄩󵄩󵄩 i

󵄩 󵄩 + ∑ 󵄩󵄩󵄩S α (t + δ − t k )J k (x(t−k )) − S α (t − t k )I k (x(t−k ))󵄩󵄩󵄩 k=1

t+δ

t

0

0

󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩󵄩 ∫ T α (t + δ − s)f(s)ds − ∫ T α (t − s)f(s)ds󵄩󵄩󵄩󵄩. 󵄩 󵄩 Arguing as in Case 1, we get 󵄩 󵄩 lim 󵄩󵄩󵄩y(t + δ) − y(t)󵄩󵄩󵄩 = 0.

δ→0

Case 3. When t = t i , i = 1, . . . , m, let λ > 0 be such that t i + λ ∈ J i and σ > 0 such that t i < σ < t i + δ ≤ t i+1 . Then we have 󵄩󵄩 ∗ 󵄩 󵄩 󵄩 󵄩󵄩y (t i + δ) − y∗ (t i )󵄩󵄩󵄩 = lim+ 󵄩󵄩󵄩y(t i + δ) − y(σ)󵄩󵄩󵄩. σ→t i

According to the definition of R we get 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩y(t i + δ) − y(σ)󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩S α (t i + δ)(x0 − g(x)) − S α (σ)(x0 − g(x))󵄩󵄩󵄩 i

󵄩 󵄩 + ∑ 󵄩󵄩󵄩S α (t i + δ − t k )I k (x(t−k )) − S α (σ − t k )I k (x(t−k ))󵄩󵄩󵄩 k=1

t +δ

σ

0

0

󵄩󵄩 i 󵄩󵄩 + 󵄩󵄩󵄩󵄩 ∫ (t i + δ − s)α−1 T α (t i + δ − s)f(s)ds − ∫(σ − s)α−1 T α (σ − s)f(s)ds󵄩󵄩󵄩󵄩. 󵄩 󵄩

5.3 Nonlocal impulsive fractional differential inclusions | 327

Arguing as in Case 1, we can see that lim δ→0,

σ→t+i

󵄩 󵄩󵄩 󵄩󵄩y(t + δ) − y(σ)󵄩󵄩󵄩 = 0. 󵄩󵄩 󵄩󵄩 i 󵄩 󵄩

(5.67)

From (5.66) and (5.67) we conclude that Z|J i is equicontinuous for every i = 0, 1, 2, . . . , m. Now for every n ≥ 1, set B n = convR(B n−1 ). By Step 1, B n is a nonempty, closed and convex subset of PC(J, E). Moreover, B1 = convR(B0 ) ⊆ B0 . Also B2 = convR(B1 ) ⊆ convR(B0 ) ⊆ B1 . By induction, the sequence (B n ), n ≥ 1 is a decreasing sequence of nonempty, closed and bounded subsets of PC(J, E). Our goal is to show that the subset B = ⋂∞ n=1 B n is nonempty and compact in PC(J, E). By Lemma A.25, it is enough to show that lim χ PC (B n ) = 0,

n→∞

(5.68)

where χ PC is the Hausdorff measure of noncompactness on PC(J, E) defined in Section 5.3.2. In the following step we prove (5.68). Step 4. Let n ≥ 1 be a fixed natural number and ε > 0. In view of Proposition A.24, there exists a sequence (y k ), k ≥ 1 in R(B n−1 ) such that χ PC (B n ) = χ PC R(B n−1 ) ≤ 2χ PC {y k | k ≥ 1} + ε. From the definition of χ PC , the above inequality becomes χ PC (B n ) ≤ 2

max

i=0,1,...,m

χ i (S|J i ) + ε,

(5.69)

where S = {y k | k ≥ 1} and χ i is the Hausdorff measure of noncompactness on C(J i , E). Arguing as in the previous step, we can show that B n|J i , i = 0, 1, . . . , m is equicontinuous. Consequently, applying Proposition A.22, we obtain χ i (S|J i ) = sup χ(S(t)), t∈J i

where χ is the Hausdorff measure of noncompactness on E. Therefore, by using the nonsingularity of χ, inequality (5.69) becomes χ PC (B n ) ≤ 2

max [sup χ(S(t))] + ε

i=0,1,...,m

t∈J i

= 2 sup χ(S(t)) + ε = 2 sup χ{y k (t) | k ≥ 1} + ε. t∈J

(5.70)

t∈J

Now, since y k ∈ R(B n−1 ), k ≥ 1, there is x k ∈ B n−1 such that y k ∈ R(x k ), k ≥ 1. Recalling the definition of R, we have that for every k ≥ 1 there is f k ∈ S1F(⋅,x k (⋅)) such

328 | 5 Fractional Differential Inclusions

that χ{y k (t) | k ≥ 1} ≤ χ{S α (t)(x0 − g(x k )) | k ≥ 1} i

+ ∑ χ{S α (t − t p )I p (x k (t−p )) | k ≥ 1} p=1 t

+ χ{ ∫ T α (t − s)f k (s)ds | k ≥ 1}

(5.71)

0

for t ∈ J i , i = 0, 1, . . . , m. Since g is compact, the set {g(x k ) | k ≥ 1} is relatively compact. Hence, for every t ∈ J we have χ{S α (t)(x0 − g(x k )) | k ≥ 1} = 0. Furthermore, condition (HI) implies, for every p = 1, 2, . . . , m and every t ∈ J, χ{S α (t − t p )(I p (x k (t−p ))) | k ≥ 1} = 0.

(5.72)

In order to estimate t

χ{ ∫ T α (t − s)f k (s)ds | k ≥ 1}, 0

we observe that from (HF3) it holds χ{f k (t) | k ≥ 1} ≤ χ{F(t, x k (t)) | k ≥ 1} ≤ β(t)χ{x k (t) | k ≥ 1} ≤ β(t)χ(B n−1 (t)) ≤ β(t)χ PC (B n−1 ) = γ(t) for a.e. t ∈ J. Furthermore, for any k ≥ 1, by (HF2), ‖f k (t)‖ ≤ φ(t)Ω(r) for a.e. t ∈ J. Consequently, f k ∈ L1/q (J, E), k ≥ 1. Note that γ ∈ L1/q (J, ℝ+ ). Then, by virtue of Lemma A.26, there exists a compact set K ϵ ⊆ E, a measurable set J ϵ ⊂ J with measure less than ϵ, and a sequence of functions {g kϵ } ⊂ L1/q (J, E) such that for all s ∈ J, {g kϵ (s) | k ≥ 1} ⊆ K ϵ , and ‖f k (s) − g kϵ (s)‖ < 2γ(s) + ϵ

for all k ≥ 1, s ∈ J 󸀠ϵ = J − J ϵ .

Then using Minkowski’s inequality, we get q 󵄩󵄩 󵄩 󵄩󵄩∫ T (t − s)(f (s) − g ϵ (s))ds󵄩󵄩󵄩 ≤ M η[ ∫(2γ(s) + ϵ) 1q ds] α k T 󵄩󵄩 󵄩󵄩 k 󵄩 󵄩 J 󸀠ϵ

J 󸀠ϵ

≤ M T η‖2γ(s) + ϵ‖ ≤ M T η(‖2γ‖

1

1

L q (J,ℝ+ )

L q (J,ℝ+ )

+ ‖ϵ‖

1

L q (J,ℝ+ )

)

5.3 Nonlocal impulsive fractional differential inclusions | 329

≤ 2M T η(‖γ‖

1

L q (J,ℝ+ )

+ ϵb q )

= 2M T η(χ PC (B n−1 )‖β‖

1

L q (J,ℝ+ )

+ ϵb q ),

(5.73)

and by Hölder’s inequality, q 󵄩 󵄩󵄩 󵄩󵄩∫ T (t − s)f (s)ds󵄩󵄩󵄩 ≤ M ηΩ(r)[ ∫ φ(s) 1q ds] . α k T 󵄩󵄩 󵄩󵄩 󵄩 󵄩 Jϵ

(5.74)

Jϵ

So, by (5.73) and (5.74), we derive t

χ({ ∫ T α (t − s)f k (s)ds | k ≥ 1}) 0

≤ χ({ ∫ T α (t − s)f k (s)ds | k ≥ 1}) + χ({ ∫ T α (t − s)f k (s)ds | k ≥ 1}) J 󸀠ϵ

Jϵ

≤ χ({ ∫ T α (t − s)(f k (s) − g kϵ (s))ds | k ≥ 1}) J 󸀠ϵ

+ χ({ ∫ T α (t − s)g kϵ (s)ds | k ≥ 1}) + χ({ ∫ T α (t − s)f k (s)ds | k ≥ 1}) J 󸀠ϵ

Jϵ

≤ 2M T η(χ PC (B n−1 )‖β‖

q

1

1 Lq

(J,ℝ+ )

+ ϵb q ) + M T ηΩ(r)[ ∫ φ(s) q ds] . Jϵ

Taking into account that ϵ is arbitrary, we get for all t ∈ J, t

χ({ ∫ T α (t − s)f k (s)ds | k ≥ 1}) ≤ 2M T ηχ PC (B n−1 )‖β‖ 0

1

L q (J,ℝ+ )

.

Next, by (5.71) and (5.72), for every t ∈ J, χ{y k (t) | k ≥ 1} ≤ 2M T ηχ PC (B n−1 )‖β‖

1

L q (J,ℝ+ )

.

This inequality with (5.70) and the fact that ε is arbitrary imply χ PC (B n ) ≤ 2M T ηχ PC (B n−1 )‖β‖

1

L q (J,ℝ+ )

.

By means of a finite number of steps, we can write 0 ≤ χ PC (B n ) ≤ (2M T η‖β‖

1

L q (J,ℝ+ )

)

n−1

χ PC (B1 )

for all n ≥ 1.

Since this inequality is true for every n ∈ ℕ, by (5.61) and passing to the limit n → +∞, we obtain (5.68). Therefore, our aim in this step is achieved. Step 5. At this point, we are in a position to apply Lemma A.25. We claim that the set B = ⋂∞ n=1 B n is a nonempty and compact subset of PC(J, E). Moreover, every B n

330 | 5 Fractional Differential Inclusions

being bounded, closed and convex, B is also bounded closed and convex. Let us verify that R(B) ⊆ B. Indeed, R(B) ⊆ R(B n ) ⊆ convR(B n ) = B n+1 for every n ≥ 1. Therefore, R(B) ⊂ ⋂∞ n=2 B n . On the other hand, B n ⊂ B 1 for every n ≥ 1. So, ∞

∞

n=2

n=1

R(B) ⊂ ⋂ B n = ⋂ B n = B. Step 6. The graph of the multi-valued function R|B : B → 2B is closed. Consider a sequence {x n }n≥1 in B with x n → x in B, and let y n ∈ R(x n ) with y n → y in PC(J, E). We have to show that y ∈ R(x). Recalling the definition of R, for any n ≥ 1 there is f n ∈ S1F(⋅,x n (⋅)) such that t

i

y n (t) = S α (t)(x0 − g(x n )) + ∑ S α (t − t k )I k (x n (t−k )) + ∫ T α (t − s)f n (s)ds k=1

(5.75)

0

for t ∈ J i , i = 0, 1, . . . , m. Observe that for every n ≥ 1 and for a.e. t ∈ J, ‖f n (t)‖ ≤ φ(t)Ω(‖x n (t)‖) ≤ φ(t)Ω(r). This shows that the set {f n | n ≥ 1} is integrably bounded. In addition, {f n (t) | n ≥ 1} is relatively compact for a.e. t ∈ J because assumption (HF3) along with the convergence of {x n }n≥1 implies that χ{f n (t) | n ≥ 1} ≤ χ{F(t, {x n (t) | n ≥ 1})} ≤ β(t)χ{x n (t) | n ≥ 1} = 0. So, the sequence {f n }n≥1 is semicompact. Hence, by Lemma A.5 it is weakly compact in L1 (J, E). So, without loss of generality we can assume that f n converges weakly to a function f ∈ L1 (J, E). By the Mazur lemma, for every natural number j there is a natural number k0 (j) > j, and a sequence of nonnegative real numbers λ j,k , k = j, . . . , k0 (j) such that k0 (j)

∑ λ j,k = 1

k=j k (j)

0 and the sequence of convex combinations z j = ∑k=j λ j,k f k , j ≥ 1 converges strongly to f in L1 (J, E) as j → ∞. So we may suppose that z j (t) → f(t) for a.e. t ∈ J. Let t be such that F(t, ⋅) is upper semicontinuous. Then, for any neighborhood U of F(t, ⋅), there is a natural number n0 so that for any n ≥ n0 we have

F(t, x n (t)) ⊆ U. Because the values of F are convex and compact, Lemma A.1 says that ⋂ conv( ⋃ F(t, x n (t))) ⊆ F(t, x(t)). j≥1

n≥j

5.3 Nonlocal impulsive fractional differential inclusions | 331

As in Step 1, by the Mazur theorem, there is a sequence {z n | n ≥ 1} of convex combinations of f n such that for a.e. t ∈ J, f(t) ∈ ⋂ {z n (t) | n ≥ j} ⊆ ⋂ conv{f n (t) | n ≥ j} j≥1

j≥1

and z n converges strongly to f ∈

L1 (J,

E). Then for a.e. t ∈ J,

f(t) ∈ ⋂ {z n (t) | n ≥ j} ⊆ ⋂ conv{f n (t) | n ≥ j} j≥1

j≥1

⊆ ⋂ conv( ⋃ F(t, x n (t))) ⊆ F(t, x(t)). j≥1

n≥j

Hence, by the continuity of g, S α , T α , I k for k = 1, 2, . . . , m, and by the same arguments as used in Step 1, we conclude from relation (5.75) that t

i

y(t) = S α (t)(x0 − g(x)) + ∑ S α (t −

t k )I k (x(t−k ))

k=1

+ ∫(t − s)α−1 T α (t − s)f(s)ds 0

for t ∈ J i , i = 0, 1, . . . , m. Therefore, y ∈ R(x). This shows that the graph of R is closed. As a result of Steps 1–5 the multivalued function R|B : B → 2B is closed and χ PC condensing, with nonempty convex compact values. Applying the fixed point theorem, Lemma A.37, there is x ∈ B such that x ∈ R(x). Clearly, x is a PC-mild solution of problem (5.58). In the following theorem we prove that the set of mild solutions of (5.58) is compact. Theorem 5.35. If the function Ω in (HF2) is given in the form Ω(r) = r + 1, and the assumptions (HA), (HF1), (HF3), (Hg), (HI) of Theorem 5.34 are satisfied, then the set of mild solutions of (5.58) is compact in PC(J, E) provided that [M S (a + h) + M T η‖φ‖

1

L q (J,ℝ+ )

] < 1.

Proof. Note that by Theorem 5.34 the set of solutions of (5.58) is nonempty. Indeed, we take M S (‖x0 ‖ + d) + M T η‖φ‖ 1q L (J,ℝ+ ) r= 1 − (M S (a + h) + M T η‖φ‖ 1q ) L (J,ℝ+ )

in (5.62). So we have a solution in B0 . Now we show that any mild solution of (5.58) belongs to B0 . Let x be a mild solution of (5.58). Then i

t

x(t) = S α (t)(x0 − g(x)) + ∑ S α (t − t k )I k (x(t−k )) + ∫ T α (t − s)f(s)ds k=1

0

for t ∈ J i , i = 0, 1, . . . , m, where f is an integrable selection of F(⋅, x(⋅)). Arguing as in Step 2 of the proof of Theorem 5.34, we get ‖x‖PC(J,E) ≤ M S (‖x0 ‖ + a‖x‖PC(J,E) + d + h‖x‖PC(J,E) ) + M T η(‖x‖PC(J,E) + 1)‖φ‖

1

L q (J,E)

.

332 | 5 Fractional Differential Inclusions

Therefore, ‖x‖PC(J,E) ≤

M S (‖x0 ‖ + d) + M T η‖φ‖

1

L q (J,E)

1 − [M S (a + h) + M T η‖φ‖

1 Lq

(J,E)

]

= r.

Due to Lemma A.38, the proof is complete. Next, we present another existence result for a PC-mild solution of problem (5.58). Theorem 5.36. Let E be a separable Banach space, let F : J × E → P ck (E) be a multifunction, g : PC(J, E) → E and I i : E → E for i = 1, 2, . . . , m. We assume the following conditions: (HA) A ∈ A α (θ0 , ω0 ) with θ0 ∈ (0, 2π ] and ω0 ∈ ℝ. (H1 ) For every x ∈ E, t 󳨃→ F(t, x) is measurable. (H2 ) There is a function ς ∈ L1/q (J, ℝ+ ), q ∈ (0, α) such that (i) for every x, y ∈ E, h(F(t, x), F(t, y)) ≤ ς(t)‖x − y‖

for a.e. t ∈ J,

where h : P clb (E) × P clb (E) → ℝ+ is the Hausdorff distance; (ii) for every x ∈ E, sup{‖x‖ | x ∈ F(t, 0)} ≤ ς(t)

for a.e. t ∈ J.

(H3 ) There is a positive constant a such that ‖g(x1 ) − g(x2 )‖ ≤ a‖x1 − x2 ‖PC(J,E)

for all x1 , x2 ∈ PC(J, E).

(H4 ) For each i = 1, 2, . . . , m there is ξ i > 0 such that ‖I i (x) − I i (y)‖ ≤ ξ i ‖x − y‖

for all x, y ∈ E.

(H5 ) The following inequality holds: m

M S (a + ξ) + M T ‖ς‖

1 Lq

(J,ℝ+ )

η < 1,

ξ = ∑ ξi . i=1

Then problem (5.58) has a PC-mild solution. Proof. For x ∈ PC(J, E), set S1F(⋅,x(⋅)) = {f ∈ L1 (J, E) | f(t) ∈ F(t, x(t)) for a.e. t ∈ J}. By Lemma A.35, (H1 ), (H2 ) (i) and [155, Theorems 1.1.9, 1.3.1], F(⋅, x(⋅)) has a measurable selection which, by hypothesis (H1 ) (ii), belongs to L1 (J, E). Thus S1F(⋅,x(⋅)) is nonempty. Let us transform the problem to a fixed point problem. Consider the multifunction R : PC(J, E) → 2PC(J,E) defined as follows: for x ∈ PC(J, E), R(x) is the set of all functions y ∈ R(x) given by (5.63). It is easy to see that any fixed point for R is a mild

5.3 Nonlocal impulsive fractional differential inclusions | 333

solution of (5.58). So, we shall show that R satisfies the assumptions of Lemma A.39. The proof will be given in two steps. Step 1. The values of R are closed. By [155, Theorem 1.1.9], assumptions (H1 ) and (H2 ) (ii) imply assumption (HF1) of Theorem 5.34. Next, since by (H2 ), ‖F(t, x)‖ = h(F(t, x), {0}) ≤ h(F(t, x), F(t, 0)) + h(F(t, 0), {0}) ≤ ς(t)‖x‖ + ‖F(t, 0)‖ ≤ ς(t)(1 + ‖x‖), assumption (HF2) of Theorem 5.34 holds as well. Thus the statement follows from Step 1 of the proof of Theorem 5.34. Step 2. R is a contraction. Let x1 , x2 ∈ PC(J, E) and y1 ∈ R(x1 ). Then there is f ∈ S1F(⋅,x1 (⋅)) such that t

i

y1 (t) = S α (t)(x0 − g(x1 )) + ∑ S α (t − t k )I k (x1 (t−k )) + ∫ T α (t − s)f(s)ds k=1

(5.76)

0

for t ∈ J i , i = 0, 1, . . . , m. Consider the multifunction Z : J → 2E defined by Z(t) = {u ∈ E | ‖f(t) − u‖ ≤ ς(t)‖x2 (t) − x1 (t)‖}. For each t ∈ J, Z(t) is nonempty. Indeed, let t ∈ J. From (H2 ) (i), we have h(F(t, x2 (t)), F(t, x1 (t))) ≤ ς(t)‖x1 (t) − x2 (t)‖. Hence, there exists u⋅ ∈ F(⋅, x2 (⋅)) such that ‖u t − f(t)‖ ≤ ς(t)‖x1 (t) − x2 (t)‖. Since the functions f , ς, x1 , x2 are measurable, [65, Proposition 3.4] (or [155, Corollary 1.3.1 (a)]) says that the multifunction V : t 󳨃→ Z(t) ∩ F(t, x2 (t)) is measurable. Since its values are nonempty and compact, we see by [137, Theorem 1.3.1] that there is h ∈ S1F(⋅,x2 (⋅)) with ‖h(t) − f(t)‖ ≤ ς(t)‖x2 (t) − x1 (t)‖

for a.e. t ∈ J.

Let us define t

i

y2 (t) = S α (t)(x0 − g(x2 )) + ∑ S α (t − t k )I k (x2 (t−k )) + ∫ T α (t − s)h(s)ds k=1

(5.77)

0

for t ∈ J i , i = 0, 1, . . . , m. Obviously, y2 ∈ R(x2 ). If t ∈ J0 , we get from (5.76), (5.77), (H3 ) and (H4 ), t

‖y2 (t) − y1 (t)‖ ≤ M S ‖g(x1 ) − g(x2 )‖ + M T ∫(t − s)α−1 ‖h(s) − f(s)‖ds 0 t

≤ M S a‖x1 − x2 ‖PC(J,E) + M T ‖x1 − x2 ‖PC(J,E) ∫(t − s)α−1 ς(s)ds 0

334 | 5 Fractional Differential Inclusions ≤ M S a‖x1 − x2 ‖PC(J,E) + M T ‖x1 − x2 ‖PC(J,E) ‖ς‖ ≤ [M S a + M T ‖ς‖

1

L q (J,ℝ+ )

1

L q (J,ℝ+ )

η

η]‖x1 − x2 ‖PC(J,E) .

(5.78)

Similarly, if t ∈ J i , i = 1, . . . , m, we get from (5.76), (5.77), (H3 ) and (H4 ), ‖y2 (t) − y1 (t)‖ ≤ [M S (a + ξ) + M T ‖ς‖

1

L q (J,ℝ+ )

η]‖x1 − x2 ‖PC(J,E) .

(5.79)

Interchanging the role of y2 and y1 , we obtain from (5.78) and (5.79), h(R(x2 ), R(x1 )) ≤ [M S (a + ξ) + M T ‖ς‖

1

L q (J,ℝ+ )

η]‖x1 − x2 ‖PC(J,E) .

Therefore, R is a contraction due to (H5 ). Thus by Lemma A.39, R has a fixed point which is a mild solution of (5.58).

5.3.4 Existence results for non-convex case In this subsection, we give a non-convex version of Theorem 5.36 under the following hypothesis: (HF− ) F : J × E → P cl (E) is a multifunction such that (i) (t, x) 󳨃→ F(t, x) is graph measurable and x 󳨃→ F(t, x) is lower semicontinuous; (ii) there exists a function φ ∈ L1/q (J, ℝ+ ), 0 < q < α such that for any x ∈ E, ‖F(t, x)‖ ≤ φ(t)

for a.e. t ∈ J.

Theorem 5.37. Let A ∈ A α (θ0 , ω0 ) with θ0 ∈ (0, 2π ] and ω0 ∈ ℝ, let F : J × E → P ck (E) be a multifunction, g : PC(J, E) → E and I i : E → E for i = 1, 2, . . . , m. If conditions (HF3), (Hg), (HI) and (HF− ) hold, then problem (5.58) has a PC-mild solution provided that there is r > 0 such that M S (‖x0 ‖ + ar + d + hr) + M T η‖φ‖

1

L q (J,ℝ+ )

≤ r.

Proof. Consider the multivalued Nemytskii operator N : PC(J, E) → 2L

1

(J,E)

defined by

N(x) = S1F(⋅,x(⋅)) = {f ∈ L1 (J, E) | f(t) ∈ F(t, x(t)) for a.e. t ∈ J}. First we note that F is superpositionally measurable by [339]. Next, we shall prove that N has a nonempty closed decomposable value and that it is lower semicontinuous. Let x ∈ PC(J, E). Since F has closed values, S1F(⋅,x(⋅)) is closed [137]. Because F is integrably bounded, S1F(⋅,x(⋅)) is nonempty (see [137, Theorem 3.2]). It is readily verified that S1F(⋅,x(⋅)) is decomposable (see [137, Theorem 3.1]). To check the lower semicontinuity of N, we need to show that, for every u ∈ L1 (J, E), x 󳨃→ d(u, N(x)) is upper semicontinuous (see [141, Proposition 1.2.26]). Let u ∈ L1 (J, E) be fixed. From [137, Theorem 2.2]

5.3 Nonlocal impulsive fractional differential inclusions | 335

with ϕ(t, v) = ‖u(t) − v‖, we have b

d(u, N(x)) = inf ‖u − v‖L1 = v∈N(x)

inf

v(t)∈F(t,x(t))

∫‖u(t) − v(t)‖dt 0

b

=∫

b

‖u(t) − z(t)‖dt = ∫ d(u(t), F(t, x(t)))dt.

inf

z(t)∈F(t,x(t))

0

(5.80)

0

We shall show that for any λ ≥ 0, the set u λ = {x ∈ PC(J, E) | d(u, N(x)) ≥ λ} is closed. For this purpose, let {x n } ⊆ u λ and assume that x n → x in PC(J, E). Then for all t ∈ J, x n (t) → x(t) in E. By virtue of (HF− ) (i) the function z 󳨃→ d(u(t), F(t, z)) is upper semicontinuous. So, via the Fatou lemma and (5.80) we have b

λ ≤ lim sup d(u, N(x n )) = lim sup ∫ d(u(t), F(t, x n (t)))dt n→∞

n→∞

0 b

b

≤ ∫ lim sup d(u(t), F(t, x n (t)))dt ≤ ∫ d(u(t), F(t, x(t)))dt = d(u, N(x)). n→∞

0

0

Therefore, x ∈ u λ and this proves the lower semicontinuity of N. This allows us to apply [60, Theorem 3] and obtain a continuous map Z : PC(J, E) → L1 (J, E) such that Z(x) ∈ N(x) for every x ∈ PC(J, E). Then Z(x)(s) ∈ F(s, x(s)) for a.e. s ∈ J. Consider a map Φ : PC(J, E) → PC(J, E) defined by i

(Φx)(t) = S α (t)(t)(x0 − g(x1 )) + ∑ S α (t)(t − t k )I k (x1 (t−k )) k=1 t

+ ∫(t − s)α−1 T α (t)(t − s)Z(x)(s)ds 0

for t ∈ J i , i = 0, 1, . . . , m. Arguing as in the proof of Theorem 5.36, we can show that there is a convex compact subset B of E such that the function Φ|B : B → B satisfies the conditions of the Schauder fixed point theorem. Then there exists x ∈ PC(J, E) such that x(⋅) = (Φx)(⋅). This means that x is a PC-mild solution of the problem (5.58).

A Appendix For the reader’s convenience this chapter summarizes some well-known results used in this book.

A.1 Functional analysis A.1.1 Basic notation and results Let ℝ denote the set of real numbers, ℕ the set of positive integers, ℂ the set of complex numbers, ℝ+ = [0, ∞). Let E be a Banach space with a norm ‖⋅‖. Denote B r := {x ∈ X | ‖x‖ ≤ r}. For any X ⊂ E, the closure, convex hull and closed convex hull of X are denoted by X, conv X and conv(X), respectively. The two-parametric Mittag-Leffler function E α,β (z) is defined as ∞

E α,β (z) = ∑ i=0

zi , Γ(iα + β)

z ∈ ℂ and ℜα > 0, ℜβ > 0. Next, E α (z) = E α,1 (z). We need the following simple result related to [26, Theorem 1.1.4]. Lemma A.1. Let {Z n }n≥1 be a sequence of subsets of E. Suppose there is a compact and convex subset Z ⊂ X such that for any neighborhood U of Z there is n so that for any m ≥ n, Z m ⊂ U. Then ⋂N>0 conv(⋃n≥N Z n ) ⊂ Z. Let Ω ⊂ E. We also recall that F : Ω → E is Lipschitz if there exists κ > 0 such that ‖Fx − Fy‖ ≤ κ‖x − y‖ for all x, y ∈ Ω, and that F is a strict contraction if κ < 1.

A.1.2 Banach functional spaces BC(ℝ+ ) denotes the set of all real functions defined, continuous and bounded on ℝ+ , and endowed with the standard norm ‖x‖ := sup{|x(t)| | t ∈ ℝ+ }. Next, for any given continuous function f : ℝ+ × ℝ → ℝ, the Nemytskii operator F : BC(ℝ+ ) → BC(ℝ+ ) is defined by (Fx)(t) := f(t, x(t)), t ∈ ℝ for x ∈ BC(ℝ+ ). It is well known that F is continuous [19]. C(ℝ+ , E) denotes the space of continuous functions from ℝ+ to E. It is known that C(ℝ+ , E) furnished with the standard distance is a locally convex Fréchet space. It follows [214] that a sequence {x n } is convergent to x in C(ℝ+ , E) if and only if {x n } is uniformly convergent to x on any compact subset of ℝ+ . For any X ⊂ C(ℝ+ , E) and t ≥ 0 let X(t) = {x(t) | x ∈ X}. Next, we denote conv X(t) := (conv X)(t). DOI 10.1515/9783110522075-006

338 | A Appendix C b (J, E), J = ℝ+ denotes the Banach space consisting of all continuous and bounded functions from J into E with the norm of the uniform convergence, i.e., ‖u‖∞ = supt∈ℝ+ |u(t)|. For E = ℝ we use just ‖u‖ instead of ‖u‖∞ . Let J = [a, b] ⊂ ℝ and let C(J, E) be the Banach space of all continuous functions from J into E with the usual sup norm ‖u‖C = supt∈J ‖u(t)‖. We sometimes use the notations ‖u‖∞ or ‖u‖C(J,E) instead of ‖u‖C . Let a = t0 < t1 < ⋅ ⋅ ⋅ < t m < t m+1 = b. Denote PC(J, E) := {u : J → E | u ∈ C((t i , t i+1 ], E), i = 0, 1, . . . , m and there exist u(t−i ) and u(t+i ), i = 1, . . . , m with u(t−i ) = u(t i )} endowed either with the Chebyshev PC-norm ‖u‖∞ := sup{|u(t)| | t ∈ J} or with the Bielecki PCB-norm ‖u‖PCB := sup{|u(t)|e−λt | t ∈ J} for some λ > 0. We sometimes use the notations ‖u‖PC or ‖u‖ instead of ‖u‖∞ . C n (J, E) is the space of E-valued functions f on J, which have continuous derivatives up to the order n on J. AC n (J, E) is the space of E-valued functions f on J such that f ∈ C n−1 (J, E) and f n−1 is absolutely continuous on J for an interval J ⊂ ℝ. L p (J, E) is the space of all E-valued Bochner integrable functions on J with the norm ‖f‖L p (J,E) defined by 1

‖f ‖L p (J,E)

p

( ∫ |f(t)| p dt) { { { { ={ J { { { inf {sup |f(t)|} {μ(J)=0 t∈J−J ̄

if 1 < p < ∞, if p = ∞,

where μ(J) is the Lebesgue measure of J. Here J ⊂ ℝ is an interval. Now we state some useful results. Lemma A.2 ([130]). If X ⊂ C(ℝ+ , E) is equicontinuous on compact intervals of ℝ+ , then conv X is also equicontinuous on the compact intervals where X was equicontinuous. Proposition A.3. If W ⊂ C(J, E) is bounded and equicontinuous, then convW ⊂ C(J, E) is also bounded and equicontinuous. Definition A.4 ([26]). A sequence {f n | n ∈ ℕ} ⊂ L1 (J, E) is said to be semicompact if (i) it is integrably bounded, i.e., there is q ∈ L1 (J, ℝ+ ) such that ‖f n (t)‖ ≤ q(t) for a.e. t ∈ J; (ii) the set {f n (t) | n ∈ ℕ} is relatively compact in E for a.e. t ∈ J. We recall a fundamental result following the Dunford–Pettis theorem. Lemma A.5 ([155]). Every semicompact sequence in L1 (J, E) is weakly compact in L1 (J, E).

A.1 Functional analysis | 339

A.1.3 Linear operators As usual, for a linear operator A, we denote by D(A) the domain of A, by σ(A) its spectrum, while ρ(A) = ℂ − σ(A) is the resolvent set of A and R(z, A) = (zI − A)−1 , z ∈ ρ(A) is the resolvent operator of A. Moreover, we denote by L(Z, Y) (or L b (Z, Y)) the space of all bounded linear operators between two normed spaces Z and Y. When Y = Z we write L(Z) (or L b (Z)). L(Z, Y) is a Banach space with the standard norm ‖A‖ = supz∈Z,‖z‖≤1 ‖Az‖. The identity matrix is denoted by 𝕀 or just I. The adjoint operator to A is denoted by A∗ .

A.1.4 Semigroup of linear operators Definition A.6. A family {S(t) | t ≥ 0} in L(E) is a semigroup of linear operators on E, or simply semigroup if: (i) S(0) = 𝕀; (ii) S(t + s) = S(t)S(s) for each t, s ≥ 0. If, in addition, it satisfies the continuity condition limt→0+ S(t) = 𝕀 in the norm topology of L(E), the semigroup is called uniformly continuous. If, in addition, it satisfies the continuity condition limt→0+ S(t)x = x in the norm of E for each x ∈ E, the semigroup is called strongly continuous. If we have t ∈ ℝ instead of t ≥ 0 in the above definition, then we have a group. Definition A.7. The infinitesimal generator, or generator of the semigroup of linear operators {S(t) | t > 0} is the operator A : D(A) ⊂ E → E, defined by D(A) = {x ∈ E | ∃ lim+ t→0

S(t)x − x } t

and

S(t)x − x . t Equivalently, we say that A generates {S(t) | t ≥ 0}. Ax = lim+ t→0

We refer the reader to the basic monograph [220] for further results in this field. Here we just state a result used in Section 4.1.3. Theorem A.8 ([12]). Let X be a reflexive Banach space and let {S(t) | t ≥ 0} be a strongly continuous semigroup on X with A as its infinitesimal generator. Then the adjoint semigroup {S∗ (t) | t ≥ 0} is a strongly continuous semigroup on X ∗ and its generator is just A∗ .

A.1.5 Metric spaces For a nonempty set X, a function d : X × X → [0, ∞] is called a generalized metric on X if and only if d satisfies

340 | A Appendix (i) d(x, y) = 0 if and only if x = y; (ii) d(x, y) = d(y, x) for all x, y ∈ X; (iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X. Now we state the following generalized Banach fixed point theorem. Lemma A.9 ([89]). Let (X, d) be a generalized complete metric space. Assume that T : X → X is a strictly contractive operator with the Lipschitz constant L < 1. If there exists a nonnegative integer k such that d(T k+1 x, T k x) < ∞ for some x ∈ X, then the following statements are true: (i) The sequence {T n x} converges to a fixed point x∗ of T. (ii) x∗ is the unique fixed point of T in X ∗ = {y ∈ X | d(T k x, y) < ∞}. (iii) If y ∈ X ∗ , then d(y, x∗ ) ≤

1 d(Ty, y). 1−L

Of course, when d : X × X → ℝ+ then X is the standard metric space.

A.2 Fractional differential calculus Let J = ℝ+ and let E be a Banach space. Definition A.10 ([162]). The fractional integral of order γ > 0 with the lower limit zero for a function f : J → E is defined as t

γ

I t f(t) =

1 f(s) ds, ∫ Γ(γ) (t − s)1−γ

t > 0, γ > 0,

0

provided that the right side is pointwise defined on [0, ∞), where Γ(⋅) is the Euler ∞ gamma function defined by Γ(γ) := ∫0 t γ−1 e−t dt. We set I 0 f(t) = f(t). γ

γ

γ

It is also common to use the notation I γ or 0 I t with a lower limit 0, or I a,t or I a with a lower limit a ∈ ℝ. Clearly, it holds I γ f(t) = (g γ ∗ f)(t) for γ−1

{t g γ (t) = { Γ(γ) 0 {

if t > 0, if t ≤ 0,

where ∗ denotes the convolution of functions. It is known [162] that I q I β f(t) = I q+β f(t) for β, q ≥ 0. Moreover, by applying Young’s inequality, it follows that ‖I q f‖L p (J,E) = ‖g q ∗ f‖L p (J,E) ≤ ‖g q ‖L1 (J,ℝ) ‖f ‖L p (J,E) = g q+1 (T)‖f ‖L p (J,E) .

A.2 Fractional differential calculus | 341

Then I q maps L p (J, E) to L p (J, E). Let ⌈q⌉ be the least integer greater than or equal to the number q. Let q > 0, m = ⌈q⌉, then we set m−1

W m,1 (J, E) = {f(t) = ∑ c k k=0

tk + I m φ(t), t ∈ J | φ ∈ L1 (J, E), c k ∈ E}. k!

Note that in the above definition φ = f (m) and c k = f (k) (0), k = 0, 1, . . . , m − 1. Definition A.11 ([162]). The Riemann–Liouville derivative of order γ > 0 with the lower limit zero for a function f : J → E can be written as t

γ D t f(t)

1 dn f(s) = ds, ∫ n Γ(n − γ) dt (t − s)γ+1−n

t > 0, n − 1 < γ < n.

0

γ

γ

γ

It is also common to use the notation L D γ or L D t or 0L D t with a lower limit 0, L D a,t or γ γ D a,t or D a+ with a lower limit a ∈ ℝ. In the following lemma, we mention some elementary properties of the Riemann– Liouville fractional integrals and derivatives. Lemma A.12. Let q > 0 and m = ⌈q⌉. (i) If f ∈ L1 (J, E), then g m−q ∗ (I q f) ∈ W m,1 (J, E) and D q I q f(t) = f(t) a.e. (ii) If γ > q and f ∈ L1 (J, E), then D q I γ f(t) = I γ−q f(t) a.e. In particular, if γ > k, k ∈ ℕ, then D k I γ f(t) = I γ−k f(t) a.e. (iii) If p > 1q and f ∈ L p (J, E), then I q f is continuous on J. Proofs of these properties are exactly as in the scalar case [251, Chapter 1]. Definition A.13 ([162]). The Caputo derivative of order γ > 0 with a lower limit 0 for a function f : J → E can be written as c

γ

γ

n−1 k t

D t f(t) = D t [f(t) − ∑

k=0 γ

k!

f (k) (0)],

t > 0, n − 1 < γ < n.

γ

γ

γ

We also use the notation c D γ or 0c D t or c D g with a lower limit 0, c D a,t or c D a+ with a lower limit a ∈ ℝ. Remark A.14 ([162, Theorem 2.1]). Let E be a reflexive Banach space, m ∈ ℕ and q q ∈ (m − 1, m). If f ∈ AC m (J, E), then c D g f(t) exists a.e. and t

c

D αg f(t)

=I

m−q

(f

(m)

1 (t)) = ∫(t − s)m−q−1 f (m) (s)ds Γ(m − q)

for a.e. t ∈ J.

0

Moreover, if f ∈ C m (J, E), then this equation is valid for all t ∈ J. A simple consequence of Lemma A.12 and formula [251, (3.13)] is the following result.

342 | A Appendix Lemma A.15. Let m ∈ ℕ, q ∈ (m − 1, m) and f ∈ L1/γ (J, E), where γ ∈ (0, q − (m − 1)). Then: (i) If σ > γ, then the function t 󳨃→ I σ f(t) is continuous on J. (ii) For any t ∈ T and k ∈ {0, 1, 2, . . . , m − 1}, the function (I q f)(k) (t) is continuous satisfying (I q f)(k) (0) = 0. Moreover, c

q

D g (I q f(t)) = D q (I q f(t))

for a.e. t ∈ J,

q

and hence, c D g (I q f(t)) = f(t) a.e. on J. Note that the integrals in the three previous definitions are understood in Bochner sense. For further reading and details on fractional calculus, we refer to [30, 162, 223]. Lemma A.16 ([112, Lemma 2.6]). Let q ∈ (0, 1) and h : J → E be a continuous function. A function x ∈ C(J, ℝ) given by x(t) = x a −

a

t

0

0

1 1 ∫(a − s)q−1 h(s)ds + ∫(t − s)q−1 h(s)ds Γ(q) Γ(q)

is the only solution of the following fractional Cauchy problem c

q

D0,t x(t) = h(t), x(a) = x a ,

t ∈ J, a > 0.

Remark A.17. One can find that if the continuous function is weaken to the integrable function, the result of Lemma A.16 is still valid.

A.3 Henry–Gronwall’s inequality We state Henry–Gronwall’s inequality [143, Lemma 7.1.1], [328, Theorem 1] which can be used in fractional-order differential equations. ̃ is a nonnegative function locally integrable on Lemma A.18. Suppose that β > 0, a(t) ̃ is a nonnegative, nondecreasing continuous function such that g(t) ̃ ≤ M, [a, b), g(t) t ∈ [a, b), and y(t) is nonnegative and locally integrable on [a, b) with t

̃ + g(t) ̃ ∫(t − s)β−1 y(s)ds, y(t) ≤ a(t)

t ∈ [a, b).

0

Then

t

n ̃ (g(t)Γ(β)) ̃ (t − s)nβ−1 a(s)]ds, Γ(nβ) n=1 ∞

̃ + ∫[ ∑ y(t) ≤ a(t) 0

t ∈ [a, b).

A.4 Measures of noncompactness | 343

̃ be a nondecreasing funcRemark A.19. Under the hypothesis of Lemma A.18, let a(t) tion on [a, b). Then we have β ̃ ̃ y(t) ≤ a(t)E ), β ( g(t)Γ(β)t

where E β is the Mittag-Leffler function.

A.4 Measures of noncompactness Now we recall some results from the theory of measures of noncompactness [40, 270]. Let E be a real Banach space with a norm ‖⋅‖. Let ME denote the family of all nonempty and bounded subsets of E, and NE all relatively compact subsets. Definition A.20. A mapping μ : ME → ℝ+ is said to be a measure of noncompactness in E if it satisfies the following conditions: (i) The family ker μ = {X ∈ ME | μ(X) = 0} is nonempty and ker μ ⊂ NE . (ii) X ⊂ Y ⇒ μ(X) ≤ μ(Y). (iii) μ(X) = μ(X). (iv) μ(conv X) = μ(X). (v) μ(λX + (1 − λ)Y) ≤ λμ(X) + (1 − λ)μ(Y) for λ ∈ [0, 1]. (vi) If {X n }∞ n=1 is a sequence of closed sets from ME such that X n+1 ⊂ X n , n = 1, 2, . . . and if limn→∞ μ(X n ) = 0, then the intersection X∞ = ⋂∞ n=1 X n is nonempty. The family ker μ described in (i) is said to be the kernel of the measure of noncompactness μ. Let us observe that the intersection set X∞ from (vi) belongs to ker μ. In fact, since μ(X∞ ) ≤ μ(X n ) for every n, we have μ(X∞ ) = 0. Now we state a fixed point theorem of Darbo type [40]. Theorem A.21. Let Q ⊆ E be a nonempty, bounded, closed and convex set and let T : Q → Q be a continuous mapping. If there exists a constant k ∈ [0, 1) such that μ(TX) ≤ kμ(X) for any nonempty subset X ⊆ Q, then T has a fixed point in the set Q. Moreover, the set fix T of all fixed points of T belonging to Q is a member of the family ker μ. The measure of noncompactness μ is said to be (i) nonsingular if μ({a} ∪ B) = μ(B) for every a ∈ E and every nonempty subset B ⊆ E; (ii) regular if μ(B) = 0 if and only if B is relatively compact in E. Two of the most important examples of measure of noncompactness are the Hausdorff measure of noncompactness χ defined on each bounded subset B of E by χ(B) = inf{ε > 0 | B admits a finite cover by balls of radius ≤ ε}, and the Kuratowski measure of noncompactness α defined on each bounded subset B of E by α(B) = inf{ε > 0 | B is covered by a finite number of sets with diameter ≤ ε}.

344 | A Appendix

It is well known that the Hausdorff measure of noncompactness χ and the Kuratowski measure of noncompactness α enjoy the above properties (i) and (ii) and other properties (see Banas [40], Deimling [88], Heinz [133], Lakshmikantham and Leela [169]) such as (iii) μ(B1 + B2 ) ≤ μ(B1 ) + μ(B2 ); (iv) μ(B1 ∪ B2 ) ≤ max{μ(B1 ), μ(B2 )}; (v) μ(λB) = |λ|μ(B) for any λ ∈ ℝ. In particular, the relationship of the Hausdorff measure of noncompactness χ and the Kuratowski measure of noncompactness α is given by (vi) χ(B) ≤ α(B) ≤ 2χ(B). Let J = [0, T], T > 0. For any W ⊂ C(J, E), we define t

t

∫ W(s)ds = { ∫ u(s)ds | u ∈ W}, 0

t ∈ J.

0

Proposition A.22 ([130]). Let χ C(J,E) be the Hausdorff measure of noncompactness on C(J, E). If W ⊆ C(J, E) is bounded, then for every t ∈ J, χ(W(t)) ≤ χ C(J,E) (W). Furthermore, if W is equicontinuous on J, then the map t 󳨃→ χ{x(t) | x ∈ W} is continuous on J and χ C(J,E) (W) = sup χ{x(t) | x ∈ W}, t∈J

t

t

χ( ∫ W(s)ds) ≤ ∫ χ(W(s))ds, 0

t ∈ J.

0

Proposition A.23. Let {u n }∞ n=1 be a sequence of Bochner integrable functions from J into ̃ (t) for almost all t ∈ J and every n ≥ 1, where m ̃ ∈ L1 (J, ℝ+ ). Then E with ‖u n (t)‖ ≤ m t

t

̃ (s)ds. χ({∫ u n (s)ds | n ≥ 1}) ≤ 2 ∫ m 0

0

1 If E is separable, then the function ψ(t) = χ({u n (t)}∞ n=1 ) belongs to L (J, ℝ+ ) and satisfies t

t

χ({∫ u n (s)ds | n ≥ 1}) ≤ ∫ ψ(s)ds. 0

0

Proposition A.24. Let B be a bounded set in E. Then for every ε > 0 there is a sequence {x n }n≥1 in B such that χ(B) ≤ 2χ({x n | n ≥ 1}) + ε.

A.4 Measures of noncompactness | 345

Lemma A.25 ([64]). If {B n }n≥1 is a decreasing sequence of nonempty closed subsets of E and limn→∞ χ(B n ) = 0, then ⋂∞ n=1 B n is nonempty and compact. Lemma A.26 ([27, Lemma 4]). Let {f n | n ∈ ℕ} ⊂ L p (J, E), p ≥ 1 be an integrably bounded sequence such that χ({f n | n ≥ 1}) ≤ γ(t) for a.e. t ∈ J, where γ ∈ L1 (J, ℝ+ ). Then for each ϵ > 0 there exist a compact set K ϵ ⊆ E, a measurable set J ϵ ⊂ J with measure less than ϵ, and a sequence of functions {g nϵ } ⊂ L p (J, E) such that {g nϵ (t) | n ≥ 1} ⊆ K ϵ ,

t∈J

and ‖f n (t) − g nϵ (t)‖ < 2γ(t) + ϵ

for all n ≥ 1, t ∈ J − J ϵ .

In this book, we also need the following results. Lemma A.27 ([130]). Assume that X ⊂ C(ℝ+ , E) is equicontinuous on compact intervals of ℝ+ and X(t) is bounded for all t ≥ 0. Then t

t

α( ∫ X(s)ds) ≤ ∫ α(X(s))ds 0

for all t ≥ 0.

0

Lemma A.28 ([215, Lemma 2.1]). Let Ω be equicontinuous on compact intervals of ℝ+ , a closed and convex subset of C(ℝ+ , E) such that Ω(t) is a bounded set in E for each t ≥ 0, and let F : Ω → Ω be a continuous operator. For any X ⊂ Ω, set ̂ 1 (X) = F(X), F

̂ n (X) = F(conv F ̂ n−1 (X)), F

n = 2, 3, . . .

If for any X ⊂ Ω and T > 0, ̂ n (X)(t)) = 0, lim sup α(F

n→∞ t≤T

then F has a fixed point in Ω. For a detailed description of the following notions we refer the reader to Deimling [88]. Definition A.29. Consider Ω ⊂ E and a continuous bounded map F : Ω → E. We say that F is α-Lipschitz if there exists κ ≥ 0 such that α(F(B)) ≤ κα(B) for all B ⊂ Ω bounded. If, in addition, κ < 1, then we say that F is a strict α-contraction. We say that F is α-condensing if α(F(B)) < α(B) for all B ⊂ Ω bounded with α(B) > 0. In other words, α(F(B)) ≥ α(B) implies α(B) = 0. The class of all strict α-contractions F : Ω → E is denoted by SC α (Ω), and the class of all α-condensing maps F : Ω → E is denoted by C α (Ω). We remark that SC α (Ω) ⊂ C α (Ω) and every F ∈ C α (Ω) is α-Lipschitz with constant κ = 1. Next, we collect some properties of the notations defined above.

346 | A Appendix Proposition A.30. If F, G : Ω → E are α-Lipschitz maps with constants κ, κ󸀠 , respectively, then F + G : Ω → E is α-Lipschitz with constant κ + κ󸀠 . Proposition A.31. If F : Ω → E is compact, then F is α-Lipschitz with constant κ = 0. Proposition A.32. If F : Ω → E is Lipschitz with constant κ, then F is α-Lipschitz with the same constant κ. Now we state a fixed point theorem which will be used in proofs of some of our main existence results. For more details, see Isaia [148]. Theorem A.33 ([148, Theorem 2]). Let F : E → E be α-condensing and S = {x ∈ E | ∃ λ ∈ [0, 1] such that x = λF(x)}. If S is a bounded set in E, so there exists r > 0 such that S ⊂ B r (0), then F has at least one fixed point and the set of fixed points of F lies in B r (0).

A.5 Multifunctions Here, we recall some information on multifunctions, see [26, 65, 141, 155]. Let E be a Banach space. We set P(E) = {B ⊆ E | B is nonempty}, P c (E) = {B ⊆ E | B is nonempty and convex}, P b (E) = {B ⊆ E | B is nonempty and bounded}, P clb (E) = {B ⊆ E | B is nonempty, closed and bounded}, P cl (E) = {B ⊆ E | B is nonempty and closed}, P k (E) = {B ⊆ E | B is nonempty and compact}, P ck (E) = {B ⊆ E | B is nonempty, convex and compact}, P cl,cv (E) = {B ⊆ E | B is nonempty, closed and convex}. Definition A.34 ([26, 65, 137, 141]). Let X and Y be Banach spaces. A multifunction G : X → P(Y) is said to be upper semicontinuous if G−1 (V) = {x ∈ X | G(x) ⊆ V} is an open subset of X for every open V ⊆ Y. If its graph Γ G = {(x, y) ∈ X × Y | y ∈ G(x)} is a closed subset of the topological space X × Y, then G is called closed. If G(B) is relatively compact for every bounded subset B of X, then G is said to be completely continuous (or compact). We recall from [26] that if a multifunction G is completely continuous with nonempty compact values, then G is upper semicontinuous if and only if G is closed.

A.5 Multifunctions | 347

Let G : J → P(E) be a multifunction. By S1G we denote the set of all integrable selections of G, i.e., S1G = {f ∈ L1 (J, E) | f(t) ∈ G(t) a.e.}. This set may be empty. For a P cl (E)-valued measurable multifunction, S1G is nonempty and bounded in L1 (J, E) if and only if t 󳨃→ sup{‖x‖ | x ∈ G(t)} ∈ L1 (J, ℝ+ ) (such a multifunction is said to be integrably bounded); see [137, Theorem 3.2]. Note that S1G ⊆ L1 (J, E) is closed if the values of G are closed, and it is convex if and only if for almost all t ∈ J, G(t) is a convex set in E. Lemma A.35 ([155, Theorem 1.3.5]). Let X0 , X be (not necessarily separable) Banach spaces, and let F : J × X0 → P k (X) be such that (i) for every x ∈ X0 the multifunction F(⋅, x) has a strongly measurable selection; (ii) for a.e. t ∈ J the multifunction F(t, ⋅) is upper semicontinuous. Then for every strongly measurable function z : J → X0 there exists a strongly measurable function f : J → X such that f(t) ∈ F(t, z(t)) for a.e. t ∈ J. Remark A.36 ([155, Theorem 1.3.1]). For single-valued or compact-valued multifunctions acting on a separable Banach space the notions of measurability and strong measurability coincide. So, if X0 , X are separable Banach spaces, we can replace strongly measurable with measurable in the above lemma, and by [155, Theorem 1.3.4], for every measurable multifunction G : J → P k (X0 ) the multifunction Ω : J → P k (X), Ω(t) = F(t, G(t)) is measurable. The following fixed point theorems are crucial in proofs of some of our main results. Lemma A.37 ([155, Corollary 3.3.1]). Let W be a closed convex subset of a Banach space X and let R : W → P ck (X) be a closed multifunction which is χ-condensing, where χ is a nonsingular measure of noncompactness defined on subsets of W. Then R has a fixed point. Lemma A.38 ([155, Proposition 3.5.1]). Let W be a closed subset of a Banach space X and let R : W → P k (X) be a closed multifunction which is χ-condensing on every bounded subset of W, where χ is a monotone measure of noncompactness defined on X. If the set of fixed points of R is a bounded subset of X, then it is compact. Lemma A.39 ([77]). Let (X, d) be a complete metric space. If R : X → P clb (X) is a contraction, then R has a fixed point. To end this section, we state the following O’Regan–Precup fixed point theorem, which is a generalization of the Mönch fixed point theorem.

348 | A Appendix

Lemma A.40 ([216, Theorem 3.2]). Let D be a closed convex subset of E, U a relatively open subset of D, and 𝕋 : U → P c (D). Assume that the graph of 𝕋 is closed, 𝕋 maps compact sets into relatively compact sets and that, for some x0 ∈ U, one has Z ⊆ D,

Z = conv({x0 } ∪ 𝕋(Z)),

Z = C with C ⊆ Z countable

󳨐⇒ Z is relatively compact. If x ∉ (1 − λ)x0 + λ𝕋(x) then 𝕋 has a fixed point.

for all x ∈ U − U, λ ∈ (0, 1),

(A.1)

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Index ⌈⋅⌉, 341 A∗ , 339 AC n (J, E), 338 BC(ℝ+ ), 337 ℂ, 337 C(J, E), 338 C(ℝ+ , E), 337 C b (J, E), 338 C n (J, E), 338 conv X, 337 conv(X), 337 conv X(t), 337 D(A), 339 𝕀, 339 L(Z), 339 L(Z, Y), 339 L b (Z), 339 L b (Z, Y), 339 L p (J, E), 338 ℕ, 337 PC(J, E), 338 ℝ, 337 σ(A), 339 X, 337 Abel type integral equation, 37 abstract Gronwall lemma, 133 adjoint operator, 339 α-condensing map, 345 anti-periodic BVP, 287 backward difference operator, 5 Banach fixed point theorem, 340 Bihari’s inequality, 127 Caputo derivative, 341 Caputo fractional difference, 5 Carathéodory type conditions, 226, 233 center stable manifold, 184 characteristic solution operators, 260 continuous asymptotically periodic functions, 92 continuous Green function, 104 discrete asymptotically periodic function, 22 discrete Green function, 7 discrete Gronwall lemma, 17 DOI 10.1515/9783110522075-008

discrete Laplace transform, 21 discrete Mittag-Leffler function, 15 Erdélyi–Kober fractional integral equation, 48, 67 Euler numerical approximation, 207 fractional differential inclusions on lattices, 302, 305, 317 fractional discrete Gronwall lemma, 17 fractional integral, 340 fractional iterative functional differential equation, 103 fractional Lorenz system, 202 fractional-order differential switched system, 159 Fredholm alternative theorem, 217 generalized metric, 339 generalized Ulam–Hyers–Rassias stability, 149, 175 Hausdorff measure of noncompactness, 343 Hyers–Ulam stability constant, 117 identity matrix, 339 impulsive Cauchy problem, 141 impulsive fractional inequality, 140 infinitesimal generator of semigroup, 339 Kuratowski measure of noncompactness, 343 Lagrange optimal control, 283 Landesman–Lazer condition, 166 Lipschitz mapping, 337 Lorenz attractor, 202 measure of noncompactness, 343 Mittag-Leffler operator functions, 306 Moore–Penrose inverse, 130 multifunctions, 346 Nemytskii operator, 337 not instantaneous impulses, 173 PC-mild solution, 273 Picard operator, 105 Poincaré mapping, 198, 199, 201, 206, 208 Pompeiu–Hausdorff functional, 105 practical Ulam–Hyers–Rassias stability, 126 probability density function, 216

366 | Index

quadratic Erdélyi–Kober type integral equation, 52 quadratic Weyl fractional integral equation, 77

solution operator, 320 strict α-contraction, 345 strict contraction, 337

relaxed control systems, 272 Riemann–Liouville derivative, 341

Ulam–Hyers–Mittag-Leffler stability, 133 Ulam–Hyers–Rassias stability, 116

semigroup, 339 – strongly continuous, 339 – uniformly continuous, 339

Volterra–Fredholm type integral equations, 244 weakly Picard operator, 105

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