Fractional Differential Equations: A Coincidence Degree Approach 9783111334349, 9783111334387, 9783111334462

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Mouffak Benchohra, Soufyane Bouriah, Abdelkrim Salim, Yong Zhou Fractional Differential Equations

Fractional Calculus in Applied Sciences and Engineering

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

Volume 12

Mouffak Benchohra, Soufyane Bouriah, Abdelkrim Salim, Yong Zhou

Fractional Differential Equations �

A Coincidence Degree Approach

Mathematics Subject Classification 2020 34A08, 34B10, 34B40, 26A33 Authors Prof. Dr. Mouffak Benchohra Laboratory of Mathematics Djillali Liabes University of Sidi Bel-Abbes P.O. Box 89 Sidi Bel-Abbes 22000, Algeria [email protected] Dr. Soufyane Bouriah Department of Mathematics Faculty of Exact Sciences and Informatics University Hassiba Benbouali of Chlef, Algeria [email protected]

Prof. Dr. Abdelkrim Salim Faculty of Technology Hassiba Benbouali University of Chlef P.O. Box 151 Chlef 02000, Algeria Laboratory of Mathematics Djillali Liabes University of Sidi Bel-Abbes P.O. Box 89 Sidi Bel-Abbes 22000, Algeria [email protected] Prof. Dr. Yong Zhou School of Computer Science and Engineering Macau University of Science and Technology Macau 999078, P.R. China Faculty of Mathematics and Computational Science Xiangtan University Hunan 411105, P.R. China [email protected]

ISBN 978-3-11-133434-9 e-ISBN (PDF) 978-3-11-133438-7 e-ISBN (EPUB) 978-3-11-133446-2 ISSN 2509-7210 Library of Congress Control Number: 2023942984 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. © 2024 Walter de Gruyter GmbH, Berlin/Boston Cover image: naddi/iStock/thinkstock Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

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We dedicate this book to our family members. In particular, Mouffak Benchohra makes his dedication to the memory of his wife K. Bencherif; Soufyane Bouriah dedicates it to his mother, father, wife, sons, and friends, particularly “Belkacemi Amin”; Abdelkrim Salim makes his dedication to his family; and Yong Zhou makes his dedication to his family.

Preface Fractional calculus is a branch of mathematics which deals with the derivatives and integrals of arbitrary (noninteger) order. Though the subject is as old as conventional calculus, the applications are rather recent. The researchers Grunwald, Letnikov, L’Hopital, Leibniz, Hardy, Caputo, Mainardi, and others did pioneering work in this field. The key feature of the models involving fractional derivatives is the flexibility in the choice of fractional order. Such models have been proved appropriate in modeling processes showing intermediate behavior. The fractional-order operators are nonlocal, in contrast to the classical integer-order operators. This nonlocality plays a vital role in modeling the memory and hereditary properties in natural systems. This book is devoted to the existence and uniqueness results for various classes of problems with periodic conditions. All of the problems in this book deal with fractional differential equations and some fractional derivatives such as the Riemann–Liouville, Caputo, and Hilfer fractional derivatives. We made certain that each chapter contains results that may be regarded as a generalization or a partial continuation of the prior chapter’s results. Classical fixed point theorems as well as Mawhin’s coincidence degree theory are employed as tools. Each chapter ends with a section devoted to remarks and bibliographical suggestions, and all abstract results are substantiated with illustrations. This monograph adds to the current literature on fractional calculus by providing original content. All of the chapters include some of the authors’ most current research work. This book is appropriate for use in advanced graduate courses, seminars, and research projects in numerous applied sciences. We are grateful to S. Abbas, D. Foukrach, J. Graef, J. Henderson, and J.J. Nieto for their contributions to the research on the problems covered in this book. Sidi Bel-Abbès and Chlef, Algeria, Macau, China, September 2023

https://doi.org/10.1515/9783111334387-201

Mouffak Benchohra Soufyane Bouriah Abdelkrim Salim Yong Zhou

Contents Preface � VII 1

Introduction � 1

2 2.1 2.1.1 2.1.2 2.1.3 2.2 2.2.1 2.2.2 2.3 2.3.1 2.3.2 2.3.3 2.4 2.5

Preliminary background � 9 Notations and functional spaces � 9 Space of continuous functions � 9 Spaces of integrable functions � 9 Spaces of weighted continuous functions � 10 Special functions of fractional calculus � 11 Gamma function � 11 k-Gamma and k-beta functions � 12 Elements from fractional calculus theory � 13 Fractional integrals � 13 Fractional derivatives � 14 Necessary lemmas, theorems, and properties � 17 Fixed point theorems � 23 Coincidence degree theory � 24

3 3.1 3.2 3.2.1 3.2.2 3.3 3.3.1 3.3.2 3.4 3.4.1 3.5

Caputo-type fractional differential equations � 26 Introduction and motivations � 26 Nonlinear implicit Caputo fractional differential equations � 27 Existence results � 27 Examples � 34 Existence results for nonlinear Caputo fractional differential systems � 35 Existence results � 35 An example � 50 Nonlinear periodic problems with Caputo exponential fractional derivative � 51 Existence results � 51 Notes and remarks � 65

4 4.1 4.2 4.2.1 4.2.2 4.3 4.3.1 4.3.2

ψ-Caputo fractional integro-differential equations � 66 Introduction and motivations � 66 Periodic solutions for nonlinear fractional differential equations � 67 Existence results � 68 Examples � 78 Nonlinear Volterra–Fredholm integro-differential equations � 80 Main results � 80 Examples � 92

X � Contents 4.4 5

Notes and remarks � 94

5.2.1 5.2.2 5.3 5.3.1 5.3.2 5.4 5.4.1 5.4.2 5.5

Nonlinear fractional pantograph differential equations with ψ-Caputo derivative � 95 Introduction and motivations � 95 Periodic solutions for some nonlinear fractional pantograph differential equations � 96 Existence results � 96 Examples � 106 Wide class of fractional pantograph differential equations � 107 Existence results � 107 An example � 119 Nonlinear fractional pantograph integro-differential equations � 120 Existence results � 121 An example � 136 Notes and remarks � 137

6 6.1 6.2 6.2.1 6.2.2 6.2.3 6.3

Nonlinear ψ-Caputo fractional pantograph coupled systems � 138 Introduction and motivations � 138 Periodic solutions for pantograph coupled systems � 140 Introduction � 140 Existence results � 140 An example � 155 Notes and remarks � 156

7

Nonlinear fractional differential equations with ψ-Hilfer fractional derivative � 157 Introduction and motivations � 157 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces � 159 Existence results � 159 Applications � 169 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations � 171 Existence results � 171 An example � 184 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville integral � 186 Main results � 186 An example � 199 Notes and remarks � 200

5.1 5.2

7.1 7.2 7.2.1 7.2.2 7.3 7.3.1 7.3.2 7.4 7.4.1 7.4.2 7.5

Contents �

8 8.1 8.2 8.2.1 8.2.2 8.3 8.3.1 8.3.2 8.4 9 9.1 9.2 9.2.1 9.2.2 9.3 10 10.1 10.2 10.2.1 10.2.2 10.3 11 11.1 11.2 11.2.1 11.2.2 11.3

XI

ψ-Hilfer fractional pantograph-type differential equations � 201 Introduction and motivations � 201 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer derivative � 202 Main results � 202 An example � 215 Fractional pantograph-type differential equations depending on the ψ-Riemann–Liouville integral � 216 Existence results � 216 An example � 231 Notes and remarks � 232 Nonlinear ψ-Hilfer fractional coupled systems � 233 Introduction and motivations � 233 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems � 233 Existence results � 233 Examples � 256 Notes and remarks � 259 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions � 260 Introduction and motivations � 260 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions � 261 Existence results � 261 An example � 274 Notes and remarks � 275 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems � 276 Introduction and motivations � 276 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions � 277 Existence results � 278 Examples � 304 Notes and remarks � 309

Bibliography � 311 Index � 323

1 Introduction Fractional calculus is a branch of mathematics that encompasses real- and complexorder derivatives and integrals. It is an extension of integer differential calculus [10, 11, 13–15, 32, 36, 58, 75, 77, 79, 217, 219–222]. The notion of fractional differential calculus has a lengthy history. One may question what meaning the derivative of a fractional dn y order might have, that is, dx n , where n is a fraction. Apparently, L’Hopital discussed this idea in a letter with Leibniz. L’Hopital wrote to Leibniz in 1695 asking, “What if n be 21 ?” The study of fractional calculus was born from this question. Leibniz responded to the 1 question, “d 2 x will be equal to x √dx : x. This is an apparent paradox from which, one day, useful consequences will be drawn.” “Fractional calculus” was born on September 30, 1695. Subsequently, many wellknown mathematicians have contributed to the development of this theory throughout the years. As a result, fractional calculus has its origins in the work of Leibniz, L’Hopital (1695), Bernoulli (1697), Euler (1730), and Lagrange (1772). Some years later, Laplace (1812), Fourier (1822), Abel (1823), Liouville (1832), Riemann (1847), Grünwald (1867), Letnikov (1868), Nekrasov (1888), Hadamard (1892), Heaviside (1892), Hardy (1915), Weyl (1917), Riesz (1922), P. Levy (1923), Davis (1924), Kober (1940), Zygmund (1945), Kuttner (1953), J. L. Lions (1959), and Liverman (1964) and several more contributed to the fundamental principles of fractional calculus. Ross organized the first fractional calculus conference at the University of New Haven in June 1974 and edited the proceedings [197]. Thereafter, Spanier published the first monograph devoted to “Fractional Calculus” in 1974 [181]. The integrals and derivatives of noninteger order, as well as fractional integro-differential equations, have seen various applications in studies in theoretical physics, mechanics, and applied mathematics. The exceptionally extensive encyclopedic-type monograph by Samko, Kilbas, and Marichev was published in Russian in 1987 and in English in 1993 [227] (for more details see [165]). The works devoted to fractional differential equations include the books of Miller and Ross (1993) [172], Podlubny (1999) [188], Kilbas et al. (2006) [143], Diethelm (2010) [114], Ortigueira (2011) [184], Abbas et al. (2012) [14], and Baleanu et al. (2012) [57]. As is generally known, the origins of fixed point theory may be traced back to the system of successive approximations (or the Picard iterative approach) used to resolve some differential equations. Banach derived the fixed point theorem through a series of repeated approximations. Over the last few decades, fixed point theory has grown enormously and independently of differential equations. However, the results of fixed points have recently been discovered to be the instruments for the solutions of differential equations. Differential fractional-order equations have recently been demonstrated to be an excellent tool for exploring several phenomena in various domains of science and engineering, including electrochemistry, electromagnetics, viscoelasticity, and economics. It is common in the literature to propose a solution to fractional differential equations by combining several types of fractional derivatives; see, e. g., [7–9, 14, 15, 18, https://doi.org/10.1515/9783111334387-001

2 � 1 Introduction 22, 26, 27, 32, 36, 48, 58, 62, 140, 141, 253]. On the other hand, there are more findings about the issue of boundary values for fractional differential equations [32, 53, 67, 69, 253]. Mawhin’s coincidence degree theory [126, 169] has been widely used in the study of many classes of nonlinear differential equations. Coincidence degree theory is useful in the study of nonlinear fractional differential equations (NFDEs). We can use this technique when other techniques do not work, such as the fixed point principle. For example, in the works [68, 73], the authors obtained the results by using coincidence degree theory. The pantograph equations have been widely used in the fields of quantum mechanics and dynamical systems [190, 198]. Actually, several researchers have investigated some new existence and uniqueness results for NFDE pantograph problems by applying fixed point theorems or the coincidence degree theorem [19, 19, 65, 68, 70, 73, 180, 190]. Many articles and monographs have been written recently in which the authors investigate numerous results for systems with different types of fractional differential equations and inclusions and various conditions. One may see [16, 39, 42, 130, 160, 162, 226] and the references therein. In this monograph, a new generalization of the well-known Hilfer fractional derivative is given. We took the publications of Diaz et al. [113] into account, where the authors presented the k-gamma and k-beta functions and demonstrated a number of their properties, many of which can also be found in [101, 175, 177, 178]. In addition, we were inspired by Sousa’s numerous publications [102–108], in which the authors established another sort of fractional operator known as the ψ-Hilfer fractional derivative with respect to a particular function and provided several essential properties about this type of fractional operator. Our work in this monograph may be viewed as a continuation and generalization of the preceding studies, i. e., several results in the fractional calculus literature. In the following we give an outline of the organization of this monograph, which consists of 11 chapters. Chapter 2 provides the notation and preliminary results, descriptions, theorems, and other auxiliary results that will be needed for this study. In the first section we give some notations and definitions of the functional spaces used in this book. In the second section, we give the definitions of the elements from fractional calculus theory, and then we present some necessary lemmas, theorems, and properties. In the third section, we give some properties of the measure of noncompactness. We finish the chapter by giving all the fixed point theorems that are used throughout the book and some definitions and necessary results of coincidence degree theory in the last section. Chapter 3 deals with some existence results for a class of problems for nonlinear implicit fractional differential equations (IFDEs) with Caputo-type fractional derivatives and periodic conditions by using coincidence degree theory. The chapter is divided into five sections. Section 3.1 provides an introduction and some motivations. Section 3.5 contains some remarks and suggestions. The main results of the chapter begin in Section 3.2, where we provide the existence of periodic solutions for the following nonlinear IFDE:

1 Introduction

c

{

D α y(t) = f (t, y(t), c D α y(t)),

� 3

for every t ∈ T := [0, b], b > 0, 0 < α ≤ 1,

y(0) = y(b),

where c D α is the fractional derivative of Caputo and f : T × ℝ × ℝ 󳨀→ ℝ is a continuous function. The results are based on coincidence degree theory. Next, some examples are included to show the applicability of our results. In Section 3.3, we establish some existence and uniqueness results for the following functional fractional differential problem: β

β

c α D + y(t) − A c D0+ y(t) = f (t, y(t), c D0+ y(t), c D0α+ y(t)); { 0 y(0) = y(b), y′ (0) = y′ (b),

t ∈ T := [0, b],

where b > 0, c D0α is the Caputo fractional derivative of order α, 0 < β ≤ 1 < α ≤ 2, 1 + β < α, f : T × ℝn × ℝn × ℝn → ℝn is a given continuous function, and A is an ℝn×n invertible matrix. In addition, an example is given to justify our results. Section 3.4 deals with the existence of solutions for the following nonlinear fractional differential problem: μ

e α D + y(t) = g(t, y(t), ec Da+ y(t)), t ∈ T := [a, b], 1 < α ≤ 2 and 0 < μ ≤ 1, {c a y(a) = y(b) and e Da1+ y(a) = e Da1+ y(b), μ

where ec Daα+ and ec Da+ are the Caputo exponential fractional derivatives of order α > 0 and μ > 0, respectively, and g : T × ℝ × ℝ 󳨀→ ℝ is a continuous function. The aim of Chapter 4 is to prove the existence and uniqueness of periodic solutions to a large class of problems for NFDEs with ψ-Caputo fractional derivative. As always, we base our arguments on Mawhin’s coincidence degree theory. Examples are included to show the applicability of our results in each section. The first result is provided in Section 4.2, where we establish the existence and uniqueness results for solutions to the NFDE of the following form: c

{

α;ψ

t

Da+ y(t) = f (t, y(t), ∫a κ(t, s, y(s))ds),

t ∈ J,̄

y(a) = y(b),

α;ψ

where c Da+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, J ̄ := [a, b] (−∞ < a < b < +∞), and f : J ̄ × ℝ × ℝ → ℝ,

κ :Δ×ℝ→ℝ

are continuous functions, where Δ = {(t, s) : a ⩽ s ⩽ t ⩽ b}. In Section 4.3, we consider the following nonlinear Volterra–Fredholm integro-differential problem:

4 � 1 Introduction t

α;ψ

b

c D + y(t) = f (t, y(t), ∫a κ1 (t, s, y(s))ds, ∫a κ2 (t, s, y(s))ds), { a y(a) = y(b),

t ∈ J,̄

α;ψ

where J ̄ := [a, b] (−∞ < a < b < +∞), c Da+ denotes the ψ-Caputo derivative of fractional order 0 < α ⩽ 1, and f : J ̄ × ℝ × ℝ × ℝ → ℝ,

κ1 : Δ × ℝ → ℝ,

and κ2 : Δ0 × ℝ → ℝ

are continuous functions, where Δ = {(t, s) : a ⩽ s ⩽ t ⩽ b} and Δ0 = J ̄ × J.̄ Examples are included to show the applicability of our results for each case. Chapter 5 deals with some existence and uniqueness of periodic solutions for a certain type of nonlinear fractional pantograph differential and integro-differential equations with ψ-Caputo derivative. The proofs are based on Mawhin’s coincidence degree theory. We provide illustrations to demonstrate the applicability of our results in each section. After the introduction section, in Section 5.2, we study a class of problems of these NFDEs: the nonlinear pantograph fractional equations with ψ-Caputo fractional derivative α;ψ

c D + y(t) = f (t, y(t), y(εt)), { 0 y(0) = y(b),

t ∈ T := [0, b],

α;ψ

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, ε ∈ (0, 1), and f : T × ℝ × ℝ → ℝ is a continuous function. In Section 5.3, we investigate the existence and uniqueness results for the nonlinear fractional pantograph differential equations involving the ψ-Caputo derivative operator supplemented with periodic conditions of the form c

α;ψ

D0+ y(t) = f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)),

t ∈ T := [0, b],

y(0) = y(b),

(1.1) (1.2)

α;ψ

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, p ∈ (0, 1), and f , g : T × ℝ × ℝ → ℝ are given continuous functions. Examples are included to show the applicability of our results. Section 5.4 presents the following nonlinear wide class of fractional integro-differential problems of pantograph type: α;ψ

qt

c D + y(t) = f (t, y(t), y(pt)) + ∫0 g(t, s, y(s))ds { { { 0 t + ∫0 h(t, s, y(s))ds, t ∈ T, { { { {y(0) = y(b), α;ψ

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, T := [0, b], and p, q ∈ (0, 1). Moreover, f : T ×ℝ×ℝ → ℝ and g, h : T ×T ×ℝ → ℝ are given continuous functions. Further, for the justification of our results we provide an example.

1 Introduction

� 5

Chapter 6 is devoted to proving some existence and uniqueness results of periodic solutions for a nonlinear fractional pantograph coupled system with ψ-Caputo derivative. We employ Mawhin’s coincidence degree theory to establish our proofs. Further, examples are provided in each section to illustrate our results. Section 6.2 deals with the following problem: α ;ψ

c D +1 y1 (t) = f1 (t, y1 (t), y2 (t)), { { {c 0α ;ψ D +2 y2 (t) = f2 (t, y1 (t), y2 (t)), { { { 0 {y1 (0) = y1 (b) and y2 (0) = y2 (b), α ;ψ

where t ∈ T := [0, b], c D0+i denotes the ψ-Caputo fractional derivative of order 0 < αi < 1, i ∈ {1, 2}, and f1 , f2 : T × ℝ2 → ℝ are continuous functions. In Chapter 7, we establish existence and uniqueness results for a class of problems for nonlinear fractional differential and integro-differential equations (NFDEs) with ψ-Hilfer fractional operator. Further, examples are given to illustrate the viability of our results in each section. In Section 7.2, we consider the following fractional integro-differential problem with ψ-Hilfer fractional derivative: H

{

α,β;ψ

t

D a+ y(t) = f (t, y(t), ∫a κ(t, s)y(s)ds), 1−γ;ψ

Ja+

1−γ;ψ

y(a) = Ja+

t ∈ (a, b],

y(b),

α,β;ψ

where H D a+ is the generalized Hilfer fractional derivative of order α ∈ (0, 1) and type 1−γ;ψ β ∈ [0, 1], Ja+ is the generalized fractional integral in the sense of Riemann–Liouville of order 1 − γ (γ = α + β − αβ), E is a Banach space, f : (a, b] × ℝ × ℝ → ℝ and κ : t [a, b] × [a, b] → ℝ are continuous functions, and By(t) = ∫a κ(t, s)y(s)ds is a linear integral operator. Further, examples are given to illustrate the viability of our results. In Section 7.3, we consider the following nonlinear class of Volterra–Fredholm integrodifferential fractional problems: H

{

α,β;ψ

D a+ y(t) = f (t, y(t), G y(t), H y(t)),

1−γ;ψ Ja+ y(a)

=

t ∈ (a, b],

1−γ;ψ Ja+ y(b),

where t

G y(t) = ∫ g(t, s, y(s))ds

b

and H y(t) = ∫ h(t, s, y(s))ds

a

a

and f : (a, b] × ℝ × ℝ × ℝ → ℝ,

g : Δ × ℝ → ℝ,

and

h : Δ0 × ℝ → ℝ

6 � 1 Introduction are continuous functions with J ̄ := [a, b] (−∞ < a < b < +∞), Δ0 = J ̄ × J,̄ and Δ = {(t, s) : α,β;ψ

a ⩽ s ⩽ t ⩽ b}; H D a+ denotes the generalized ψ-Hilfer fractional derivative of order 1−γ;ψ 0 < α ⩽ 1 and type β ∈ [0, 1]; Ja+ is the generalized fractional integral in the sense of Riemann–Liouville of order 1 − γ (γ = α + β − αβ). Further, illustrative examples are provided in support of the obtained results. In Section 7.4, we establish the existence and uniqueness results for a class of nonlinear ψ-Hilfer fractional differential equations depending on the ψ-Riemann–Liouville fractional problem: H

{

α,β;ψ

Da+

α,ψ

y(t) = f (t, y(t), Ja+ y(t)),

1−γ,ψ Ja+ y(a)

α,β;ψ

where H Da+

=

t ∈ (a, b],

1−γ,ψ Ja+ y(b),

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and type 1−γ,ψ

β ∈ [0, 1] and Ja+ is the ψ-Riemann–Liouville fractional integral of order 1 − γ (γ = α + β − αβ). Moreover, f : (a, b] × ℝ2 → ℝ is a given continuous function. Chapter 8 deals with the existence and uniqueness of periodic solutions for some class of nonlinear fractional pantograph systems with ψ-Hilfer derivative associated with periodic-type fractional integral boundary conditions in a weighted space of continuous functions. The proofs are based on Mawhin’s coincidence degree theory. Suitable illustrative examples are provided in each section. In Section 8.2, we study the classes of interest of these NFDEs: the nonlinear pantograph fractional equations with ψ-Hilfer fractional derivative H

{

α,β;ψ

Da +

y(t) = f (t, y(t), y(εt)),

1−γ,ψ J0+ y(0)

α,β;ψ

where H D0+

=

1−γ,ψ J0+ y(b),

t ∈ (0, b],

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and type 1−γ,ψ

β ∈ [0, 1], 0 < ε < 1, and J0+ is the ψ-Riemann–Liouville fractional integral of order 1 − γ (γ = α + β − αβ). Moreover, f : (0, b] × ℝ2 → ℝ is a given continuous function. In Section 8.3, we study some existence and uniqueness results of solutions for the following nonlinear ψ-Hilfer fractional differential equations of pantograph type depending on the ψ-Riemann–Liouville fractional integral in a weighted space of continuous functions: H

{

α,β;ψ

D0+

1−γ,ψ J0+ y(0)

α,β;ψ

where H D0+

1−γ,ψ [0, 1], J0+

α,ψ

y(t) = f (t, y(θt), J0+ y(εt)), =

1−γ,ψ J0+ y(b),

t ∈ (0, b],

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and type β ∈

is the ψ-Riemann–Liouville fractional integral of order 1−γ (γ = α +β −αβ), and θ, ε ∈ (0, 1]. Moreover, f : T × ℝ2 → ℝ is a given continuous function. Suitable illustrative examples are provided.

1 Introduction

� 7

The main goal of Chapter 9 is to study the existence and uniqueness of periodic solutions for some class of nonlinear fractional coupled systems with ψ-Hilfer derivative. In Section 9.2, we study the following problem: α ,β ;ψ

H D +1 1 y1 (t) = f1 (t, y1 (t), y2 (t)), { { {H aα ,β ;ψ Da+2 2 y2 (t) = f2 (t, y1 (t), y2 (t)), t ∈ (a, b], { { { 1−γ 1−γ ,ψ 1−γ ,ψ 1−γ ,ψ 1 ,ψ y1 (a) = Ia+ 1 y1 (b) and Ia+ 2 y2 (a) = Ia+ 2 y2 (b), {I a + α ,β ;ψ

1−γ ,ψ

where we denote by H Da+i i and Ia+ i the ψ-Hilfer fractional derivative of order 0 < αi ≤ 1 and type βi ∈ [0, 1] and the ψ-Riemann–Liouville fractional integral of order 1 − γi (γi = αi + βi − αi βi ), i ∈ {1, 2}, respectively. Moreover, f1 , f2 : J ̄ × ℝ2 → ℝ are continuous functions. Chapter 10 deals with some existence and uniqueness results for a class of problems with nonlinear k-generalized ψ-Hilfer fractional differential equations with periodic conditions. The arguments are based on Mawhin’s coincidence degree theory. Furthermore, an illustration is presented to demonstrate the plausibility of our results. In Section 10.2, we study the existence and uniqueness results for a class of nonlinear k-generalized ψ-Hilfer-type fractional differential equations with periodic condition: α,β;ψ

α,β;ψ

(H Da+ y)(t) = f (t, y(t), (H k Da+ y)(t)), { k k(1−ξ),k;ψ k(1−ξ),k;ψ + (Ja+ y)(a ) = (Ja+ y)(b), α,β;ψ

t ∈ (a, b],

k(1−ξ),k;ψ

are the k-generalized ψ-Hilfer fractional derivative of orwhere H and Ja+ k Da+ der α ∈ (0, k) and type β ∈ [0, 1] and the k-generalized ψ-fractional integral of order k(1 − ξ), respectively, where ξ = k1 (β(k − α) + α), k > 0, and f : [a, b] × ℝ × ℝ 󳨀→ ℝ are given functions. In the last part, we present some illustrations to demonstrate the practicability of our results. Chapter 11 deals with some existence and uniqueness results for a class of nonlinear fractional coupled systems with k-generalized ψ-Hilfer fractional differential equations and periodic conditions. The arguments are based on Mawhin’s coincidence degree theory. We demonstrate several results by changing the required conditions of the theorems. Furthermore, illustrative examples are presented to demonstrate the plausibility of our results. In Section 11.2, we study the existence and uniqueness results for a class of nonlinear k-generalized ψ-Hilfer-type fractional differential coupled systems with periodic condition: α ,β ;ψ

α ,β ;ψ

α ,β ;ψ

1 1 1 1 2 2 (H y1 )(t) = f1 (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)), { k Da + k Da + k Da + { {H α2 ,β2 ;ψ H α1 ,β1 ;ψ H α2 ,β2 ;ψ (k Da+ y2 )(t) = f2 (t, y1 (t), y2 (t), (k Da+ y1 )(t), (k Da+ y2 )(t)), { { { k(1−ξ k(1−ξ1 ),k;ψ k(1−ξ2 ),k;ψ k(1−ξ ),k;ψ 1 ),k;ψ y1 (a) = Ja+ y1 (b) and Ja+ y2 (a) = Ja+ 2 y2 (b), {Ja+

8 � 1 Introduction α ,β ;ψ

k(1−ξ ),k;ψ

i i are the k-generalized ψ-Hilfer fractional where t ∈ (a, b], H and Ja+ i k Da + derivative of order 0 < αi ≤ k and type βi ∈ [0, 1] and the k-generalized ψ-fractional integral of order k(1 − ξi ), ξi = k1 (αi + kβi − αi βi ), i ∈ {1, 2}, respectively. Moreover f1 , f2 : Θ × ℝ4 → ℝ are continuous functions. In the last part, we present some illustrations to demonstrate the practicability of our results.

2 Preliminary background This chapter covers the mathematical tools, notations, and concepts that will be required in subsequent chapters. We will look at some of the most important properties of fractional differential operators. We also go over some of the fundamental aspects of coincidence degree theory and fixed point theorems, which are important in our results for fractional differential equations.

2.1 Notations and functional spaces In this section, we will present all of the notations and definitions of functional spaces that are regarded as essential and remain constant throughout the book. In fact, these are only given once in this section. Let 0 < a < b, J = (a, b], J ̄ = [a, b], T = (0, b], and T = [0, b]. Consider the parameters α, β, γ satisfying γ = α + β − αβ and 0 < α, β, γ < 1. Let ξ = k1 (β(k − α) + α), where k > 0. Let ψ be an increasing and positive function on J ̄ such that ψ′ is continuous on J.̄

2.1.1 Space of continuous functions By C(J,̄ ℝ) we denote the Banach space of all continuous functions from J ̄ into ℝ with the norm 󵄨 󵄨 ̄ ‖u‖∞ = sup{󵄨󵄨󵄨u(t)󵄨󵄨󵄨 : t ∈ J}. Let AC n (J, ℝ) and C n (J, ℝ) be the spaces of n-times absolutely continuous functions and n-times continuously differentiable functions on J, respectively. We denote by ACen (T) the space defined by AC ne (T) := {y : T → ℝ : e D n−1 y(t) ∈ AC(T), e D = e−t

d }, dt

where n = [α] + 1, where [α] is the integer part of α, and e D is the exponential derivative defined in the sequel. In particular, if 0 < α ≤ 1, then n = 1 and ACe1 (T) := ACe (T). 2.1.2 Spaces of integrable functions p

Consider the space Xc (a, b) (c ∈ ℝ, 1 ≤ p ≤ ∞) of those real-valued Lebesgue measurable functions f on J ̄ for which ‖f ‖Xcp < ∞, where the norm is defined by

https://doi.org/10.1515/9783111334387-002

10 � 2 Preliminary background 1 p

b

‖f ‖Xcp

󵄨p dt 󵄨 ) . = (∫󵄨󵄨󵄨t c f (t)󵄨󵄨󵄨 t a

p

p

In particular, when c = p1 , the space Xc (a, b) coincides with the Lp (a, b)-space: X 1 (a, b) = p

Lp (a, b). By L1 (J), we denote the space of Bochner-integrable functions f : J 󳨀→ E with the norm b

󵄩 󵄩 ‖f ‖1 = ∫󵄩󵄩󵄩f (t)󵄩󵄩󵄩dt. a

p

Consider the space Xψ (a, b) (1 ≤ p ≤ ∞) of those real-valued Lebesgue measurable functions g on J ̄ for which ‖g‖ p < ∞, where the norm is defined by Xψ

b

‖g‖X p ψ

1 p

󵄨 󵄨p = (∫ ψ′ (t)󵄨󵄨󵄨g(t)󵄨󵄨󵄨 dt) , a

where ψ is an increasing and positive function on [a, b] such that ψ′ is continuous on J.̄ p In particular, when ψ(x) = x, the space Xψ (a, b) coincides with the Lp (a, b)-space. 2.1.3 Spaces of weighted continuous functions Consider the weighted Banach space C1−γ;ψ (J) = {u : J → ℝ : t → (ψ(t) − ψ(a))

1−γ

u(t) ∈ C(J,̄ ℝ)},

with the norm 1−γ 󵄨 󵄨 ‖u‖C1−γ;ψ = sup󵄨󵄨󵄨(ψ(t) − ψ(a)) u(t)󵄨󵄨󵄨, t∈J ̄

and n Cγ;ψ (J) = {u ∈ C n−1 (J) : u(n) ∈ C1−γ;ψ (J)}, 0 Cγ;ψ (J)

= C1−γ;ψ (J),

with the norm n−1

󵄩 󵄩 󵄩 󵄩 ‖u‖C n = ∑ 󵄩󵄩󵄩u(i) 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩u(n) 󵄩󵄩󵄩C . 1−γ;ψ γ;ψ i=0

n ∈ ℕ,

2.2 Special functions of fractional calculus

� 11

α,β

The weighted space C1−γ;ψ (J) is defined by α,β

C1−γ;ψ (J) = {u ∈ C1−γ;ψ (J),

H

α,β;ψ

Da +

u ∈ C1−γ;ψ (J)},

α,β;ψ

where H Da+ is a fractional derivative defined in the following sections. Consider the weighted Banach space Cξ;ψ (J) = {x : J → E : t → (ψ(t) − ψ(a))

1−ξ

x(t) ∈ C(J,̄ E)},

with the norm 1−ξ 󵄩 󵄩 ‖x‖Cξ;ψ = sup󵄩󵄩󵄩(ψ(t) − ψ(a)) x(t)󵄩󵄩󵄩, t∈J ̄

and n Cξ;ψ (J) = {x ∈ C n−1 (J) : x (n) ∈ Cξ;ψ (J)}, 0 Cξ;ψ (J)

n ∈ ℕ,

= Cξ;ψ (J),

with the norm n−1

󵄩 󵄩 󵄩 󵄩 ‖x‖C n = ∑ 󵄩󵄩󵄩x (i) 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩x (n) 󵄩󵄩󵄩C . ξ;ψ

ξ;ψ

i=0

α,β

The weighted space Cξ;ψ (J) is defined by α,β

Cξ;ψ (J) = {x ∈ Cξ;ψ (J), α,β;ψ

where H k Da+

H α,β;ψ k Da+ x

∈ Cξ;ψ (J)},

is defined in the sequel.

2.2 Special functions of fractional calculus 2.2.1 Gamma function Euler’s gamma function Γ(z), which extends the factorial n! and enables n to assume noninteger and even complex values, is definitely one of the fundamental functions of fractional calculus. Leonhard Euler was a Swiss mathematician, physicist, astronomer, geographer, logician, and engineer who pioneered and inspired breakthroughs in analytic number theory, complex analysis, and infinitesimal calculus. He developed a lot of today’s mathematical language and notation, such as the concept of a mathematical function. He is also recognized for his contributions to mechanics, fluid dynamics, optics, astronomy, and music theory.

12 � 2 Preliminary background The gamma function for complex numbers with a positive real part was established by Daniel Bernoulli. Daniel Bernoulli, who was a member of the famous Bernoulli family from Basel, was a mathematician and physicist. He is most recognized for his mathematical contributions to mechanics, particularly fluid mechanics, as well as his seminal work in probability and statistics. Definition 2.1 ([188]). The gamma function is defined via a convergent improper integral: +∞

Γ(z) = ∫ t z−1 e−t dt, 0

where Re(z) > 0. One of the basic properties of the gamma function is that it satisfies the functional equation Γ(z + 1) = zΓ(z), so for positive integer values n, the gamma function becomes Γ(n) = (n − 1)! and thus can be seen as an extension of the factorial function to real values. A useful particular value of the function, Γ( 21 ) = √π, is used throughout many examples in this monograph. 2.2.2 k-Gamma and k-beta functions In 2005, Diaz and Petruel [113] defined new functions, called k-gamma and k-beta functions: ∞

tk

Γk (α) = ∫ t α−1 e− k dt,

α > 0,

0

and 1

β α 1 Bk (α, β) = ∫ t k −1 (1 − t) k −1 dt. k

0

It is noteworthy that if k → 1, then Γk (α) → Γ(α) and Bk (α, β) → B(α, β). We also have the following useful relations: α α Γk (α) = k k −1 Γ( ), Γk (α + k) = αΓk (α), Γk (k) = Γ(1) = 1, k Γ (α)Γk (β) 1 α β Bk (α, β) = B( , ), Bk (α, β) = k . k k k Γk (α + β)

2.3 Elements from fractional calculus theory

� 13

2.3 Elements from fractional calculus theory In this section, we will review some definitions of fractional integral and fractional differential operators, which are used throughout this book. We finish it with some required lemmas, theorems, and properties.

2.3.1 Fractional integrals Definition 2.2 ([188]). The fractional (arbitrary)-order integral of the function h ∈ L1 ([0, b], ℝ+ ) of order α ∈ ℝ+ is defined by t

α

I h(t) =

1 ∫(t − s)α−1 h(s)ds. Γ(α) 0

Definition 2.3 ([179]). The exponential-type fractional integral of order α > 0 of a function g ∈ L1 (T) is defined by e α Ja+ g(t)

t

=

1 α−1 ∫(et − eϱ ) g(ϱ)eϱ dϱ, Γ(α)

for each t ∈ T.

a

Definition 2.4 (Generalized fractional integral [143]). Let α ∈ ℝ+ and g ∈ L1 (J). The generalized fractional integral of order α is defined by t

(ρ Jaα+ g)(t) = ∫ sρ−1 ( a

α−1

t ρ − sρ ) ρ

g(s) ds, Γ(α)

t > a, ρ > 0.

Definition 2.5 (ψ-Riemann–Liouville fractional integral [143]). Let (a, b) (−∞ ≤ a < b ≤ p ∞) be a finite or infinite interval of the real line ℝ, α > 0, c ∈ ℝ, and h ∈ Xc (a, b). Also let ψ(t) be an increasing and positive monotone function on J having a continuous derivative ψ′ (t) on (a, b). The left- and right-sided fractional integrals of a function h of order α with respect to another function ψ on J are defined by α;ψ (Ja+ h)(t)

t

α−1 h(τ)

= ∫ ψ′ (τ)(ψ(t) − ψ(τ)) a

Γ(α)

dτ

and α;ψ (Jb− h)(t)

b

α−1 h(τ)

= ∫ ψ′ (τ)(ψ(τ) − ψ(t)) t

Γ(α)

dτ.

14 � 2 Preliminary background p

Definition 2.6 (k-Generalized ψ-fractional integral [193]). Let g ∈ Xψ (a, b), ψ(t) > 0, be

an increasing function on J and let ψ′ (t) > 0 be continuous on (a, b) and α > 0. The generalized k-fractional integral operators of a function g (left-sided and right-sided) of order α are defined by α,k;ψ Ja+ g(t)

α,k;ψ

Jb−

t

ψ′ (s)g(s)ds 1 = ∫ α , kΓk (α) (ψ(t) − ψ(s))1− k a

b

g(t) =

ψ′ (s)g(s)ds 1 ∫ α , kΓk (α) (ψ(s) − ψ(t))1− k t

with k > 0. Also in [177], Nápoles Valdés gave a more generalized fractional integral operators defined by t

α,k;ψ

JG,a+ g(t) = α,k;ψ JG,b− g(t)

ψ′ (s)g(s)ds 1 , ∫ kΓk (α) G(ψ(t) − ψ(s), kα ) a

b

ψ′ (s)g(s)ds 1 = , ∫ kΓk (α) G(ψ(s) − ψ(t), kα ) t

where G(⋅, α) ∈ AC(J). 2.3.2 Fractional derivatives Definition 2.7 ([143]). For a function h given on the interval [0, b], the Caputo fractional derivative of order α is defined by (c D α h)(t) =

t

1 ∫(t − s)n−α−1 h(n) (s)ds, Γ(n − α) 0

where n = [α] + 1 and [α] denotes the integer part of the real number α. Definition 2.8 (Generalized fractional derivative [143]). Let α ∈ ℝ+ \ ℕ and ρ > 0. The generalized fractional derivative ρ Daα+ of order α is defined by (ρ Daα+ g)(t) = δρn (ρ Jan−α + g)(t) = (t 1−ρ

n t

n−α−1

d t ρ − sρ ) ∫ sρ−1 ( ) dt ρ a

d n where n = [α] + 1 and δρn = (t 1−ρ dt ) .

g(s) ds, Γ(n − α)

t > a, ρ > 0,

2.3 Elements from fractional calculus theory

� 15

Definition 2.9 ([179]). Let α ≥ 0 and g ∈ ACen (T). The exponential derivative of Riemann– Liouville type of order α > 0 is defined by (e Daα+ g)(t) :=

n t

d 1 n−α−1 (e−t ) ∫(et − eϱ ) g(ϱ)eϱ dϱ, Γ(n − α) dt a

t ∈ T,

where n = [α] + 1 and [α] denotes the integer part of the real number α. Definition 2.10 ([179]). Let α ≥ 0 and g ∈ AC ne (T). The Caputo exponential-type fractional derivative of order α is defined by (ec Daα+ g)(t)

t

n

dϱ 1 d n−α−1 = (e−t ) g(ϱ) −ϱ , ∫(et − eϱ ) Γ(n − α) dt e a

for each t ∈ T. Definition 2.11 (Generalized Hilfer-type fractional derivative [182]). Let order α and type β satisfy n − 1 < α < n, 0 ≤ β ≤ 1 and ρ > 0, with n ∈ ℕ. The generalized Hilfer-type fractional derivative of a function g is defined by α,β

β(n−α)

(ρ Da+ g)(t) = (ρ Ja+

(t ρ−1

n

d ) dt

ρ

(1−β)(n−α)

Ja+

g)(t)

β(n−α) n ρ (1−β)(n−α) g)(t). δρ Ja+

= (ρ Ja+

Definition 2.12 (ψ-Riemann–Liouville fractional derivative [143]). Let ψ′ (t) ≠ 0 (−∞ ≤ a < t < b ≤ ∞), α > 0, and n ∈ ℕ. The Riemann–Liouville derivatives of a function h of order α with respect to another function ψ on J ̄ are defined by α;ψ

n−α;ψ

(Da+ h)(t) = δn (Ja+

h)(t)

t

n−α−1

= δn ∫ ψ′ (τ)(ψ(t) − ψ(τ)) a

h(τ) dτ Γ(n − α)

and α;ψ

n−α;ψ

(Db− h)(t) = (−1)n δn (Ja+ b

h)(t) n−α−1

= (−1)n δn ∫ ψ′ (τ)(ψ(τ) − ψ(t)) t

d n where n = [α] + 1 and δn = ( ψ′1(t) dt ) .

h(τ) dτ, Γ(n − α)

16 � 2 Preliminary background Definition 2.13 ([43]). Let n − 1 < α < n with n ∈ ℕ and let y, ψ ∈ C n (J,̄ ℝ) be two functions such that ψ is increasing and positive where ψ′ (t) ≠ 0, for any t ∈ J.̄ The left ψ-Caputo fractional derivative of y of order α is given by c

n−α;ψ

α;ψ

Da+ y(t) := Ja+

n

(

1 d ) y(t), ′ ψ (t) dt

t ∈ J.̄

In particular, when 0 < α < 1, we have c

α;ψ Da+ y(t)

t

1 −α = ∫(ψ(t) − ψ(s)) u′ (s)ds, Γ(1 − α) a

t ∈ J.̄

Definition 2.14 (ψ-Hilfer fractional derivative [102]). Let order α and type β satisfy n − 1 < α < n and 0 ≤ β ≤ 1, with n ∈ ℕ, and let h, ψ ∈ C n (J,̄ ℝ) be two functions such that ψ is increasing and ψ′ (t) ≠ 0. The ψ-Hilfer fractional derivatives of a function h to t are defined by α,β;ψ

(H Da+

β(n−α);ψ

h)(t) = (Ja+

n

(

1 d (1−β)(n−α);ψ ) Ja + h)(t) ψ′ (t) dt

and α,β;ψ

(H Db−

β(n−α);ψ

h)(t) = (Jb−

n

(−

1 d (1−β)(n−α);ψ h)(t). ) Jb− ψ′ (t) dt

In this monograph we consider the case n = 1 only, because 0 < α < 1. Definition 2.15 (k-Generalized ψ-Hilfer derivative [214, 223]). Let n − 1 < kα ≤ n with n ∈ ℕ, let −∞ ≤ a < b ≤ ∞, and let g, ψ ∈ C n (J,̄ ℝ) be two functions such that ψ is increasing and ψ′ (t) ≠ 0 for all t ∈ J. The k-generalized ψ-Hilfer fractional derivatives (left-sided α,β;ψ H α,β;ψ and right-sided) H k Da+ (⋅) and k Db− (⋅) of a function g of order α and type 0 ≤ β ≤ 1 with k > 0 are defined by H α,β;ψ k Da+ g(t)

β(kn−α),k;ψ

= (Ja+

n

(

1 d (1−β)(kn−α),k;ψ ) (k n Ja+ g))(t) ′ ψ (t) dt

β(kn−α),k;ψ n n (1−β)(kn−α),k;ψ δψ (k Ja+ g))(t)

= (Ja+ and H α,β;ψ k Db− g(t)

β(kn−α),k;ψ

= (Jb−

β(kn−α),k;ψ

= (Jb− d n where δψn = ( ψ′1(t) dt ) .

n

(−

1 d (1−β)(kn−α),k;ψ ) (k n Jb− g))(t) ψ′ (t) dt

(−1)n δψn (k n Jb−

(1−β)(kn−α),k;ψ

g))(t),

2.3 Elements from fractional calculus theory

� 17

Property 2.1 ([214, 223]). It is worth noting that the k-generalized ψ-Hilfer fractional derivative is thought to be an expansion to many fractional operators defined over the years; indeed, in the following part, we will give a list of some of the most commonly used fractional derivatives that are considered to be a particular case of our operator. α,β;ψ The fractional derivative H interpolates the following fractional derivatives: k Da+ – the ψ-Hilfer fractional derivative (k = 1); – the ψ-Riemann–Liouville fractional derivative (k = 1, β = 0); – the ψ-Caputo fractional derivative (k = 1, β = 1); – the Hilfer fractional derivative (k = 1, ψ(t) = t); – the Riemann–Liouville fractional derivative (k = 1, ψ(t) = t, β = 0); – the Caputo fractional derivative (k = 1, ψ(t) = t, β = 1); – the Hilfer–Hadamard fractional derivative (k = 1, ψ(t) = ln(t)); – the Caputo–Hadamard fractional derivative (k = 1, ψ(t) = ln(t), β = 1); – the Hadamard fractional derivative (k = 1, ψ(t) = ln(t), β = 0); – the Hilfer-generalized fractional derivative (k = 1, ψ(t) = t ρ ); – the Caputo-generalized fractional derivative (k = 1, ψ(t) = t ρ , β = 1); – the generalized fractional derivative (k = 1, ψ(t) = t ρ , β = 0); – the Weyl fractional derivative (k = 1, ψ(t) = t ρ , β = 0, a = −∞).

2.3.3 Necessary lemmas, theorems, and properties Lemma 2.1 ([143]). From the definition of fractional integrals and Caputo derivatives, we have α c

n−1

h(k) (0) k t , k! k=0

α

I0+ ( D0+ h(t)) = h(t) − ∑

t > 0, n − 1 < α < n.

Especially, when 1 < α < 2, we have α c

α

′

I0+ ( D0+ h(t)) = h(t) − h(0) − th (0).

Theorem 2.2 ([143]). Let α > 0, β > 0, 1 ≤ p ≤ ∞, 0 < a < b < ∞. Then for g ∈ L1 (J) we have β

α+β

(ρ Jaα+ ρ Ja+ g)(t) = (ρ Ja+ g)(t). Property 2.3 ([179]). If γ, β > 0, then for each t ∈ T we have: γ Γ(β+1) 1. e Ja+ (et − ea )β = Γ(γ+β+1) (et − ea )γ+β ; 2.

3.

t a γ ) e γ Ja+ (c) = (e −e ; γ Γ(β+1) e γ t a β Da+ (e − e ) = Γ(β−γ++1) (et

− ea )β−γ .

18 � 2 Preliminary background Lemma 2.2 ([179]). Let γ > 0 and h ∈ AC ne (T). Then n−1

e γ e γ Ja+ (c Da+ h(t))

= h(t) − ∑

i=0

(et − ea )i e i D h(a), i!

where n = [γ] + 1. Lemma 2.3 ([179]). Let γ > 0. Then the differential equation e γ c Da+ h(t)

=0

has a solution: 2

h(t) = ϖ0 + ϖ1 (et − ea ) + ϖ2 (et − ea ) + ⋅ ⋅ ⋅ + ϖn−1 (et − ea )

n−1

,

where ϖi ∈ ℝ, i = 1, . . . , n, are constants and n = [γ] + 1. Theorem 2.4 ([43]). If y ∈ C n (J,̄ ℝ) and n − 1 < α < n, then α;ψ c

k

n−1

(ψ(t) − ψ(a))k 1 d ( ′ ) y(a). k! ψ (t) dt k=0

α;ψ

Ja+ Da+ y(t) = y(t) − ∑

In particular, when 0 < α < 1, we have α;ψ c

α;ψ

Ja+ Da+ y(t) = y(t) − y(a).

Theorem 2.5 ([43]). Let y ∈ C 1 (J,̄ ℝ) and α > 0. Then we have c

α;ψ

α;ψ

Da+ Ja+ y(t) = y(t).

Theorem 2.6 ([43]). Let y, υ ∈ C n (J,̄ ℝ) and α > 0. Then c

α;ψ

c

n−1

α;ψ

k

Da+ y(t) = Da+ υ(t) ⇐⇒ y(t) = υ(t) + ∑ ck (ψ(t) − ψ(a)) , k=0

where ck =

1 ( 1 d )k (y k! ψ′ (t) dt

− υ)(a).

Remark 2.1. Let y ∈ C n (J,̄ ℝ) and α > 0. Then c

α,β;ψ

Da +

n−1

k

y(t) = 0 ⇐⇒ y(t) = ∑ ck (ψ(t) − ψ(a)) . k=0

Theorem 2.7 ([43]). Let y, υ ∈ C n (J,̄ ℝ), 0 ⩽ β ⩽ 1, and α > 0. Then H

α,β;ψ

H

α,β;ψ

n

D a+ y(t) = D a+ υ(t) ⇐⇒ y(t) = υ(t) + ∑ ck (ψ(t) − ψ(a)) k=1

γ−k

,

2.3 Elements from fractional calculus theory

� 19

where ck =

n−k

1 d 1 ( ) Γ(γ + 1 − k) ψ′ (t) dt

(1−β)(n−α);ψ

Ja+

y(a)

and γ = α + β − αβ. Remark 2.2. Let y ∈ C n (J,̄ ℝ), 0 ⩽ β ⩽ 1, and α > 0. Then H

n

α,β;ψ

γ−k

D a+ y(t) = 0 ⇐⇒ y(t) = ∑ ck (ψ(t) − ψ(a)) k=1

.

α;ψ

Lemma 2.4 ([103]). Let α > 0 and 0 ≤ γ < 1. Then Ja+ is bounded from C1−γ;ψ (J) into α;ψ C1−γ;ψ (J). In addition, if γ ≤ α, then J + is bounded from C1−γ;ψ (J) into C(J,̄ ℝ). a

Theorem 2.8 ([177]). Let g : J ̄ → ℝ be an integrable function and take α > 0 and k > 0. α,k;ψ Then JG,a+ g exists for all t ∈ J.̄ p

α,k;ψ

Theorem 2.9 ([177]). Let g ∈ Xψ (a, b) and take α > 0 and k > 0. Then JG,a+ g ∈ C(J,̄ ℝ). Lemma 2.5 ([214, 223]). Let α > 0, β > 0, and k > 0. Then we have the following semigroup property: α,k;ψ

Ja+

β,k;ψ

Ja+

α+β,k;ψ

f (t) = Ja+

β,k;ψ

f (t) = Ja+

α,k;ψ

Ja+

f (t)

and α,k;ψ

Jb−

β,k;ψ

Jb−

α+β,k;ψ

f (t) = Jb−

β,k;ψ

f (t) = Jb−

α,k;ψ

Jb−

f (t).

Lemma 2.6 ([102, 143]). Let t > a. Then for α ≥ 0 and β > 0, we have β−1

α;ψ

[Ja+ (ψ(τ) − ψ(a))

](t) =

Γ(β) α+β−1 (ψ(t) − ψ(a)) . Γ(α + β)

Lemma 2.7 ([214, 223]). Let α, β > 0 and k > 0. Then we have β

α,k;ψ

[ψ(t) − ψ(a)] k

α,k;ψ

β

Ja+

−1

=

α+β Γk (β) −1 (ψ(t) − ψ(a)) k Γk (α + β)

=

α+β Γk (β) −1 (ψ(b) − ψ(t)) k . Γk (α + β)

and Jb−

[ψ(b) − ψ(t)] k

−1

α,β

Property 2.10 ([182]). The operator ρ Da+ can be written as ρ

α,β

Da + =

ρ

β(1−α)

Ja+

1−γ

β(1−α) ρ

δρ ρ Ja+ = ρ Ja+

γ

Da + ,

γ = α + β − αβ.

20 � 2 Preliminary background Lemma 2.8 ([143, 182]). Let α > 0 and 0 ≤ γ < 1. Then ρ Jaα+ is bounded from Cγ,ρ (J) into α,β

β(1−α) ρ

Cγ,ρ (J). Since ρ Da+ u = ρ Ja+

γ

Da+ u, we have α,β

γ

C1−γ,ρ (J) ⊂ C1−γ,ρ (J) ⊂ C1−γ,ρ (J). Lemma 2.9 ([182]). Let 0 < a < b < ∞, α > 0, 0 ≤ γ < 1, and u ∈ Cγ,ρ (J). If α > 1 − γ, then ρ Jaα+ u is continuous on J and (ρ Jaα+ u)(a) = lim+ (ρ Jaα+ u)(t) = 0. t→a

Lemma 2.10 ([102]). Let 0 < a < b < ∞, α > 0, 0 ≤ γ < 1, and u ∈ Cγ;ψ (J). If α > 1 − γ, then α;ψ J + u ∈ C(J,̄ ℝ) and a

α;ψ

α;ψ

(Ja+ u)(a) = lim+ (Ja+ u)(t) = 0. t→a

Theorem 2.11 ([214, 223]). Let 0 < a < b < ∞, α > 0, 0 ≤ ξ < 1, k > 0, and u ∈ Cξ;ψ (J). If α > 1 − ξ, then k α,k;ψ

(Ja+

α,k;ψ

u)(a) = lim+ (Ja+ t→a

u)(t) = 0.

Lemma 2.11 ([182]). Let α > 0, 0 ≤ γ < 1, and g ∈ Cγ,ρ (J). Then (ρ Daα+ ρ Jaα+ g)(t) = g(t)

for all t ∈ J.

1 Lemma 2.12 ([102]). Let α > 0, 0 ≤ β ≤ 1, and h ∈ Cγ;ψ (J). Then α,β;ψ

(H Da+

α;ψ

Ja+ h)(t) = h(t)

for all t ∈ J.

1 (J), where k > 0. Then for t ∈ J, Lemma 2.13 ([214, 223]). Let α > 0, 0 ≤ β ≤ 1, and u ∈ Cξ;ψ we have α,β;ψ

(H k Da+

α,k;ψ

Ja+

u)(t) = u(t).

Lemma 2.14 ([102, 234]). Let t > a, α > 0, and 0 ≤ β ≤ 1. Then for 0 < γ < 1, γ = α + β − αβ, we have γ;ψ

γ−1

[Da+ (ψ(τ) − ψ(a))

](t) = 0

and α,β;ψ

[H Da+

γ−1

(ψ(τ) − ψ(a))

](t) = 0.

2.3 Elements from fractional calculus theory

� 21

Lemma 2.15 ([214, 223]). Let t > a, α > 0, 0 ≤ β ≤ 1, and k > 0. Then for 0 < ξ < 1, ξ = k1 (β(k − α) + α), we have α,β;ψ

[H k Da+ (ψ(s) − ψ(a))

ξ−1

](t) = 0.

Lemma 2.16 ([143]). Let α > 0 and n = [α] + 1. Then α c

n−1

f (k) (0) k t . k! k=0

α

I ( D f (t)) = f (t) − ∑

Lemma 2.17 ([188]). Let α > 0 so that the homogenous differential equation of fractional order c

α

D h(t) = 0

has a solution: h(t) = c0 + c1 t + c2 t 2 + ⋅ ⋅ ⋅ + cn−1 t n−1 , where ci , i = 1, . . . , n, are constants and n = [α] + 1. 1 Lemma 2.18 ([182]). Let 0 < α < 1 and 0 ≤ γ < 1. If g ∈ Cγ,ρ (J) and ρ Ja1−α + g ∈ Cγ,ρ (J), then α−1 (ρ Ja1−α + g)(a) t ρ − aρ ( ) Γ(α) ρ

(ρ Jaα+ ρ Daα+ g)(t) = g(t) −

for all t ∈ J. γ

Lemma 2.19 ([182]). Let 0 < α < 1, 0 ≤ β ≤ 1, and γ = α + β − αβ. If u ∈ Cγ,ρ (J), then ρ

γ ρ

γ

Ja+ Da+ u =

ρ

α ρ

α,β

Ja+ Da+ u

and ρ

γ ρ

α

Da + J a + u =

ρ

β(1−α)

Da+

u.

1 Lemma 2.20 ([102, 234]). Let α > 0, 0 ≤ β ≤ 1, and h ∈ Cγ;ψ (J). Then α,β;ψ α;ψ (Ja+ H Da+ h)(t)

= h(t) −

1−γ;ψ

(Ja+

h)(a)

Γ(γ)

γ−1

(ψ(t) − ψ(a))

for all t ∈ J.

n Theorem 2.12 ([214, 223]). If f ∈ Cξ;ψ [a, b], n − 1 < α < n, 0 ≤ β ≤ 1, where n ∈ ℕ and k > 0, then α,k;ψ H α,β;ψ k Da+ f )(t)

(Ja+

n

=∑ i=1

−(ψ(t) − ψ(a))ξ−i k(n−ξ),k;ψ {δn−i (Ja+ f (a))} k i−n Γk (k(ξ − i + 1)) ψ

+ f (t),

22 � 2 Preliminary background where 1 (β(kn − α) + α). k

ξ= In particular, if n = 1, we have α,k;ψ H α,β;ψ k Da+ f )(t)

(Ja+

= f (t) −

(ψ(t) − ψ(a))ξ−1 (1−β)(k−α),k;ψ J f (a). Γk (β(k − α) + α) a+

α,β;ψ

Property 2.13 ([102]). The operator H Da+ H

α,β;ψ

Da+

β(1−α);ψ

= Ja +

can be written as γ;ψ

Da + ,

γ = α + β − αβ. p

Lemma 2.21 ([102, 143]). Let α > 0, β > 0, and 0 < a < b < ∞. Then for h ∈ Xc (a, b) the semigroup property is valid, i. e., α;ψ

β;ψ

α+β;ψ

(Ja+ Ja+ h)(t) = (Ja+

h)(t). γ

Lemma 2.22 ([182]). Let f be a function such that f ∈ Cγ,ρ (J). Then u ∈ Cγ,ρ (J) is a solution of the differential equation α,β

(ρ Da+ u)(t) = f (t) for each t ∈ J, 0 < α < 1, 0 ≤ β ≤ 1 if and only if u satisfies the following Volterra integral equation: t

1−γ

α−1 (ρ Ja+ u)(a+ ) t ρ − aρ γ−1 1 t ρ − sρ ( ) + ) sρ−1 f (s)ds, u(t) = ∫( Γ(γ) ρ Γ(α) ρ a

where γ = α + β − αβ. Lemma 2.23 ([214, 223]). Let α, β > 0 and k > 0. Then we have α,k;ψ

Ja+

α

α

𝔼αk ((ψ(t) − ψ(a)) k ) = 𝔼αk ((ψ(t) − ψ(a)) k ) − 1.

1 Lemma 2.24. If y, γ ∈ Cξ;ψ [a, b], 0
0, then

H α,β;ψ k Da+ y(t)

α,β;ψ

=H k Da+ γ(t)

if and only if y(t) = γ(t) + where ξ = k1 (β(k − α) + α).

(ψ(t) − ψ(a))ξ−1 (1−β)(k−α),k;ψ J (y − γ)(a), Γk (β(k − α) + α) a+

2.4 Fixed point theorems

� 23

Proof. Let us assume that H α,β;ψ k Da+ y(t)

α,β;ψ

=H k Da+ γ(t).

This implies that H α,β;ψ k Da+ (y(t) α,k;ψ

By applying the left integral operator Ja+

− γ(t)) = 0.

(⋅) on both sides of equation (2.1), we obtain

α,k;ψ H α,β;ψ k Da+ (y(t)

Ja+

(2.1)

− γ(t)) = 0.

Thus, by Theorem 2.12 we get y(t) = γ(t) +

(ψ(t) − ψ(a))ξ−1 (1−β)(k−α),k;ψ J (y − γ)(a). Γk (β(k − α) + α) a+

(2.2)

α,β;ψ

For the converse, we apply the derivative operator H k Da+ (⋅) on both sides of equation (2.2). Then we have H α,β;ψ k Da+ y(t)

α,β;ψ

=H k Da+ γ(t) (1−β)(k−α),k;ψ

+

(y − γ)(a) H α,β;ψ ψ −1 [k Da+ (Ψξ (s, a)) ](t). Γk (β(k − α) + α)

Ja+

By Lemma 2.15, we conclude that H α,β;ψ k Da+ y(t)

α,β;ψ

=H k Da+ γ(t).

2.4 Fixed point theorems In this part, we will go through all of the fixed point theorems that are employed in the various studies throughout the monograph. Is it commonly recognized that fixed points are useful instruments for the solution of differential equations. Fixed point theory has been one of the most intensely studied research subjects in recent decades. The fixed point notion dates back to the middle of the eighteenth century. While fixed point theory seems to be a separate academic field nowadays, it first emerged in articles dealing with the solution of certain differential equations; see, e. g., Liouville (1837) [163], Picard (1890) [187], and Poincaré (1886) [189]. One of the first independent fixed point results was obtained by Banach [61] by abstracting the successive approximation method of Picard.

24 � 2 Preliminary background Theorem 2.14 (Banach’s fixed point theorem [129]). Let D be a nonempty closed subset of a Banach space E. Then any contraction mapping N of D into itself has a unique fixed point. Banach’s fixed point theorem differs from previous fixed point theorems in that it ensures not only the existence but also the uniqueness of the fixed point. More crucially, it not only gives you the existence and uniqueness of a fixed point, but also how to obtain it. In what follows, we list some other fixed point theorems that have turned out to be useful instruments for the solution of differential equations. Theorem 2.15 (Schaefer’s fixed point theorem [129]). Let E be a Banach space and let N : E → E be a completely continuous operator. If the set D = {u ∈ E : u = λNu for some λ ∈ (0, 1)} is bounded, then N has a fixed point.

2.5 Coincidence degree theory The following definitions and coincidence degree theory are fundamental in the proof of our main result. We refer the reader to [126, 169]. Definition 2.16 ([126, 169]). Let X and Y be normed spaces. A linear operator L : Dom(L) ⊂ X → Y is said to be a Fredholm operator of index zero provided that: 1. Img L is a closed subset of Y ; 2. dim ker L = codim Img L < +∞. It follows from Definition 2.16 that there exist continuous projectors P : X → X and Q : Y → Y such that Img P = ker L,

ker Q = Img L,

X = ker L ⊕ ker P,

Y = Img L ⊕ Img Q,

which implies that the restriction of L to Dom L ∩ ker P, which we will denote by LP , is an isomorphism onto its image. Definition 2.17 ([126, 169]). Let L be a Fredholm operator of index zero, let Ω ⊆ X be a bounded subset, and let Dom L ∩ Ω ≠ 0. Then the operator N : Ω → Y is said to be L-compact in Ω if: 1. the mapping QN : Ω → Y is continuous and QN(Ω) ⊆ Y is bounded; 2. the mapping (LP )−1 (I − Q)N : Ω → X is completely continuous. Lemma 2.25 ([183]). Let X, Y be Banach spaces and let Ω ⊂ X be a bounded open set symmetric with 0 ∈ Ω. Suppose L : Dom L ⊂ X → Y is a Fredholm operator of index zero

2.5 Coincidence degree theory

� 25

with Dom L ∩ Ω ≠ 0 and N : X → Y is an L-compact operator on Ω. Assume, moreover, that Lx − Nx ≠ −λ(Lx + N(−x)), for all x ∈ Dom L ∩ 𝜕Ω and all λ ∈ (0, 1], where 𝜕Ω is the boundary of Ω with respect to X. Under these conditions, the equation Lx = Nx has at least one solution on Dom L ∩ Ω.

3 Caputo-type fractional differential equations 3.1 Introduction and motivations This chapter deals with some existence results for a class of problems of nonlinear IFDEs with Caputo-type fractional derivatives and periodic conditions by using coincidence degree theory. At the end of each section, examples are included to show the applicability of our results. The results obtained in this chapter are studied and presented as a consequence of the following: – We used the monographs of Abbas et al. [7, 14] and Benchohra et al. [75] and the papers of Ahmad et al. [32], Benchohra et al. [67, 69, 191, 192, 202, 204], and Zhou et al. [253], which are focused on linear and nonlinear initial and boundary value problems for fractional differential equations involving different kinds of fractional derivatives. – We used the monographs of Abbas et al. [7, 15], Kilbas et al. [143], and Zhou et al. [253] and the papers of Abbas et al. [10] and Benchohra et al. [77, 79]. – Much attention has been paid to the existence and uniqueness of the solutions of fractional dynamic systems [65, 72, 73, 131] on account of the fact that existence is the fundamental problem and a necessary condition for considering some other properties of fractional dynamic systems, such as controllability and stability. – Our results are essentially based on the works [25, 68, 73, 94, 123, 127] and are obtained using coincidence degree theory. – Sheng and Jiang [232] provided some existence results for the dynamical system c

{

β

D0α+ y(t) − A c D0+ y(t) = f (t, y(t)),

y(0) = y0 ,

–

y (0) = ′

Benchohra et al. [68] obtained some existence results for periodic solutions to the following nonlinear IFDEs with Caputo fractional derivatives: c α

{

–

t ∈ T := [0, b],

y′0 .

D y(t) = Ψ(t, y(t), c Dα y(t)),

t ∈ T := [0, b], b > 0, 0 < α ≤ 1,

y(0) = y(b),

where Ψ : T × ℝ × ℝ → ℝ is a continuous function. They based their arguments on coincidence degree theory. Bouriah et al. [94] considered the following nonlinear pantograph fractional equation with ψ-Caputo fractional derivative: c

{

α;ψ

D0+ y(t) = Ψ(t, y(t), y(εt)),

y(0) = y(b),

https://doi.org/10.1515/9783111334387-003

t ∈ T := [0, b],

3.2 Nonlinear implicit Caputo fractional differential equations

� 27

α;ψ

–

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, ε ∈ (0, 1), and Ψ : T × ℝ × ℝ → ℝ is a continuous function. The authors of [123] studied the following nonlinear problem: H

{ α,β;ψ

where H D0+

α,β;ψ

D0+

1−v,ψ

I0+

α,ψ

y(ξ) = F (ξ, y(θξ), I0+ y(εξ)), 1−v,ψ

y(0) = I0+

ξ ∈ (0, b],

y(b),

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and 1−v,ψ

type β ∈ [0, 1], I0+ is the ψ-Riemann–Liouville fractional integral of order 1 − v (v = α + β − αβ), and θ, ε ∈ (0, 1]. Moreover, F : J × R2 → R is a given continuous function.

3.2 Nonlinear implicit Caputo fractional differential equations In this section, we are concerned with the existence of periodic solutions for the following nonlinear IFDE: c

α

c

α

D y(t) = f (t, y(t), D y(t)),

for every t ∈ T := [0, b], b > 0, 0 < α ≤ 1,

y(0) = y(b),

(3.1) (3.2)

where c D α is the Caputo fractional derivative and f : T × ℝ × ℝ 󳨀→ ℝ is a continuous function. 3.2.1 Existence results We denote X = {y ∈ C(T, ℝ) : y(t) = I α u(t) : u ∈ C(T, ℝ), t ∈ T} with the norm 󵄩 󵄩 ‖y‖X = max{‖y‖∞ , 󵄩󵄩󵄩c D α y󵄩󵄩󵄩∞ } and Y = C(T, ℝ) with the norm 󵄨 󵄨 ‖u‖Y = sup{󵄨󵄨󵄨u(t)󵄨󵄨󵄨 : t ∈ T}. Now, we define the linear operator L : Dom L ⊆ X → Y by Ly := c D α y,

(3.3)

where Dom L = {y ∈ X : c D α y ∈ Y : y(0) = y(b)}. We define N : X → Y by Ny(t) := f (t, y(t), c D α y(t)),

t ∈ T.

Then the problem (3.1)–(3.2) can be equivalently rewritten as Ly = Ny.

28 � 3 Caputo-type fractional differential equations Lemma 3.1. Let L be defined by (3.3). Then ker L = y(0), b

Img L = {y ∈ Y : ∫(b − s)α−1 y(s)ds = 0}. 0

Proof. By Lemma 2.17, for t ∈ T, Ly(t) = c D α y(t) = 0 has solution y(t) = c,

where (c ∈ ℝ).

Then ker L = y(0). For u ∈ Img L, there exists y ∈ Dom L such that u = Ly ∈ Y . By Lemma 2.16, we have for each t ∈ T t

1 y(t) = y(0) + ∫(t − s)α−1 u(s)ds. Γ(α) 0

Since y ∈ Dom L, then u can satisfy b

1 ∫(b − s)α−1 u(s)ds = 0. Γ(α) 0

On the other hand, suppose u ∈ Y and satisfies b

∫(b − s)α−1 u(s)ds = 0. 0

Let y(t) = I α u(t). Then u(t) = c D α y(t) and y ∈ Dom L, so that u ∈ Img L. Then we have b

Img L = {y ∈ Y : ∫(b − s)α−1 y(s)ds = 0}. 0

The proof is complete. Lemma 3.2. Let L be defined by (3.3). Then L is a Fredholm operator of index zero, and the linear continuous projector operators P : X → X and Q : Y → Y can be defined as Py = y(0),

3.2 Nonlinear implicit Caputo fractional differential equations

� 29

b

α Qu(t) = α ∫(b − s)α−1 u(s)ds. b 0

Furthermore, the operator L−1 P : Img L → X ∩ ker P can be written as α L−1 P (u)(t) = I u(t).

Proof. Obviously, Img P = ker L and P2 = P. It follows for each y ∈ X, y = (y − Py) + Py that X = ker P + ker L. By a simple calculation, we can get ker P ∩ ker L = 0. Then we get X = ker P ⊕ ker L. A similar proof can show that for each u ∈ Y , Q2 u = Qu and u = (u − Q(u)) + Q(u), where (u − Q(u)) ∈ ker Q = Img L. It follows from Img L = ker Q and Q2 = Q that Img Q ∩ Img L = 0. Then we have Y = Img L ⊕ Img Q. Thus, dim ker L = dim Img Q = codim Img L. This means that L is a Fredholm operator of index zero. Now, we will prove that L−1 P is the inverse of L|Dom L∩ker P . In fact, for u ∈ Img L, we have c α α LL−1 P (u) = D (I u) = u.

(3.4)

Moreover, for y ∈ Dom L ∩ ker P we obtain α c α L−1 P (L(y(t))) = I ( D y(t)) = y(t) − y(0).

In view of y ∈ Dom L ∩ ker P, we know that y(0) = 0. Therefore, L−1 P (L(y(t))) = y(t).

(3.5)

−1 Combining (3.4) with (3.5), we know that L−1 P = (L|Dom L∩ker P ) . The proof is complete.

Lemma 3.3. Assume that the following hypothesis is satisfied. (3.3.1) There exist constants K, K > 0 with 21 ≤ K < 1 and K + K < 1 such that 󵄨󵄨 󵄨 󵄨󵄨f (t, u, v) − f (t, u,̄ v)̄ 󵄨󵄨󵄨 ≤ K|u − u|̄ + K|v − v|̄

for t ∈ T and u, u,̄ v, v̄ ∈ ℝ.

Then the operator N is L-compact on any bounded open set Ω ⊂ X.

30 � 3 Caputo-type fractional differential equations Proof. Define the bounded open set Ω = {y ∈ X : ‖y‖X < M}, where M is a positive constant. The proof will be given in several claims. Claim 1. We prove that QN is continuous. The continuity of QN follows from the hypothesis on f and the Lebesgue dominated convergence theorem. Claim 2. We show that QN(Ω) is bounded. For each y ∈ Ω and t ∈ T, we have b

󵄨 󵄨󵄨 󵄨 α α−1 󵄨 c α 󵄨󵄨QN(y)(t)󵄨󵄨󵄨 ≤ α ∫(b − s) 󵄨󵄨󵄨f (s, y(s), D y(s))󵄨󵄨󵄨ds b 0

b

≤

α 󵄨 󵄨 ∫(b − s)α−1 󵄨󵄨󵄨f (s, y(s), c D α y(s)) − f (s, 0, 0)󵄨󵄨󵄨ds bα 0

b

+

α 󵄨 󵄨 ∫(b − s)α−1 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds bα 0

b

α 󵄨 󵄨 󵄨 󵄨 ≤ f + α ∫(b − s)α−1 (K 󵄨󵄨󵄨y(s)󵄨󵄨󵄨 + K 󵄨󵄨󵄨c D α y(s)󵄨󵄨󵄨)ds b ∗

∗

0

≤ f + M(K + K), where f ∗ = supt∈T |f (t, 0, 0)|. Thus, 󵄩󵄩 󵄩 ∗ 󵄩󵄩QN(y)󵄩󵄩󵄩Y ≤ f + M(K + K) := R. This shows that QN(Ω) ⊆ Y is bounded. Claim 3. We show that L−1 P (I − Q)N : Ω → X is completely continuous. In view of the Arzelà–Ascoli theorem, we need to prove that L−1 P (I − Q)N(Ω) ⊂ X is equicontinuous and bounded. First, for each y ∈ Ω and t ∈ T, we have L−1 P (I − Q)Ny(t) = L−1 P (Ny(t) − QNy(t)) = I α [f (t, y(t), c D α y(t)) − t

=

b

α ∫(b − s)α−1 f (s, y(s), c D α y(s))] bα 0

1 ∫(t − s)α−1 f (s, y(s), c D α y(s))ds Γ(α) 0

3.2 Nonlinear implicit Caputo fractional differential equations

� 31

b

tα − α ∫(b − s)α−1 f (s, y(s), c D α y(s))ds. b Γ(α) 0

On the one hand, for each y ∈ Ω and t ∈ T we have b

2 󵄨 󵄨󵄨 −1 󵄨 󵄨 ∫(b − s)α−1 󵄨󵄨󵄨f (s, y(s), c D α y(s)) − f (t, 0, 0)󵄨󵄨󵄨ds 󵄨󵄨LP (I − Q)Ny(t)󵄨󵄨󵄨 ≤ Γ(α) 0

b

2 󵄨 󵄨 + ∫(b − s)α−1 󵄨󵄨󵄨f (t, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

≤ [f ∗ + M(K + K)]

2T α := B1 . Γ(α + 1)

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (I − Q)Ny󵄩󵄩󵄩∞ ≤ B1 .

(3.6)

On the other hand, we have c

α

c

−1

α

D (LP (I − Q)Ny(t)) = f (t, y(t), D y(t)) b

α − α ∫(b − s)α−1 f (s, y(s), c D α y(s)), b 0

which implies that for each y ∈ Ω and t ∈ T, we have 󵄨󵄨c α −1 󵄨 ∗ 󵄨󵄨 D (LP (I − Q)Ny(t))󵄨󵄨󵄨 ≤ 2f + 2M(K + K) := B2 . Thus, 󵄩󵄩c α −1 󵄩 󵄩󵄩 D (LP (I − Q)Ny)󵄩󵄩󵄩∞ ≤ B2 .

(3.7)

By inequalities (3.6) and (3.7), we have 󵄩󵄩 −1 󵄩 󵄩󵄩LP (I − Q)Ny󵄩󵄩󵄩X ≤ max{B1 , B2 }, which shows that L−1 P (I − Q)N(Ω) is uniformly bounded in X. Now, we need to prove that L−1 P (I − Q)N(Ω) is equicontinuous. For 0 ≤ t1 ≤ t2 ≤ b, y ∈ Ω, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (I − Q)Ny(t2 ) − LP (I − Q)Ny(t1 )󵄨󵄨󵄨 ≤

t2

t1

t1

0

f ∗ + M(K + K) 󵄨 󵄨 [∫(t2 − s)α−1 ds + ∫󵄨󵄨󵄨(t2 − s)α−1 − (t1 − s)α−1 󵄨󵄨󵄨ds] Γ(α)

32 � 3 Caputo-type fractional differential equations

+[

M(K + K) + f ∗ α α ](t2 − t1 ). Γ(α + 1)

As t1 → t2 the right-hand side of the above inequality tends to zero. Then L−1 P (I−Q)N(Ω) is equicontinuous in X. By the Arzelà–Ascoli theorem, L−1 (I −Q)N(Ω) is relatively compact. P As a consequence of Claims 1 to 3, we can conclude that the operator N is L-compact in Ω. The proof is complete. Lemma 3.4. If condition (3.3.1) holds, then there exists a positive number A which does not depend on λ such that if L(y) − N(y) = −λ[L(y) + N(−y)],

λ ∈ (0, 1],

(3.8)

then ‖y‖X ≤ A. Proof. Assume that (3.3.1) holds and that y ∈ X satisfies (3.8). Then L(y) − N(y) = −λL(y) − λN(−y), L(y) =

λ 1 N(y) − N(−y). 1+λ 1+λ

Thus, by using the definition of operators L and N, we obtain for each t ∈ T 󵄨󵄨 󵄨 󵄨c α 󵄨 󵄨󵄨Ly(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨 D y(t)󵄨󵄨󵄨 1 󵄨󵄨 λ 󵄨󵄨 󵄨 󵄨 c α c α ≤ 󵄨f (t, y(t), D y(t))󵄨󵄨󵄨 + 󵄨f (t, −y(t), − D y(t))󵄨󵄨󵄨 1 + λ󵄨 1 + λ󵄨 1 󵄨󵄨 󵄨 ≤ [󵄨f (t, y(t), c D α y(t)) − f (t, 0, 0)󵄨󵄨󵄨 + f ∗ ] 1+λ 󵄨 λ 󵄨󵄨 󵄨 + [󵄨f (t, −y(t), −c D α y(t)) − f (t, 0, 0)󵄨󵄨󵄨 + f ∗ ] 1+λ 󵄨 1 󵄨 󵄨 󵄨 󵄨 [K 󵄨󵄨y(t)󵄨󵄨󵄨 + K 󵄨󵄨󵄨c D α y(t)󵄨󵄨󵄨] ≤ f∗ + 1+λ 󵄨 λ 󵄨 󵄨 󵄨 󵄨 [K 󵄨󵄨−y(t)󵄨󵄨󵄨 + K 󵄨󵄨󵄨−c D α y(t)󵄨󵄨󵄨] + 1+λ 󵄨 󵄩 󵄩 ≤ f ∗ + K‖y‖∞ + K 󵄩󵄩󵄩c D α y󵄩󵄩󵄩∞ , which implies that 󵄩󵄩c α 󵄩󵄩 󵄩c α 󵄩 ∗ 󵄩󵄩 D y󵄩󵄩∞ ≤ f + K‖y‖∞ + K 󵄩󵄩󵄩 D y󵄩󵄩󵄩∞ .

(3.9)

We can also easily show that 󵄩 󵄩 ‖y‖∞ ≤ f ∗ + K‖y‖∞ + K 󵄩󵄩󵄩c D α y󵄩󵄩󵄩∞ . By inequalities (3.9) and (3.10), we deduce that

(3.10)

3.2 Nonlinear implicit Caputo fractional differential equations

� 33

󵄩 󵄩 ‖y‖X ≤ f ∗ + K‖y‖∞ + K 󵄩󵄩󵄩c D α y󵄩󵄩󵄩∞ ≤ f ∗ + (K + K)‖y‖X , which implies that ‖y‖X ≤

f∗

1 − (K + K)

:= A.

This completes the proof. Lemma 3.5. If condition (3.3.1) is satisfied, then there is a bounded open set Ω ⊂ X such that

L(y) − N(y) ≠ −λ[L(y) + N(−y)]

(3.11)

for all y ∈ 𝜕Ω and all λ ∈ (0, 1]. Proof. By (3.3.1) and Lemma 3.4, there exists a positive constant A which does not de-

pend on λ such that if y satisfies the equality L(y) − N(y) = −λ[L(y) + N(−y)], λ ∈ (0, 1],

then ‖y‖X ≤ A. Thus, if

Ω = {y ∈ X; ‖y‖X < B},

(3.12)

where B > A, we conclude that L(y) − N(y) ≠ −λ[L(y) − N(−y)] for every y ∈ 𝜕Ω = {y ∈ X; ‖y‖X = B} and λ ∈ (0, 1]. Theorem 3.1. If (3.3.1) holds, then the problem (3.1)–(3.2) has at least one solution. Proof. By (3.3.1), clearly, the set Ω defined in (3.12) is symmetric, 0 ∈ Ω, and X ∩Ω = Ω ≠ 0. Furthermore, it follows from Lemma 3.5 that if condition (3.3.1) is fulfilled, then L(y) − N(y) ≠ −λ[L(y) − N(−y)] for all y ∈ X ∩ 𝜕Ω = 𝜕Ω and all λ ∈ (0, 1]. This together with Lemma 2.25 implies that the

problem (3.1)–(3.2) has at least one solution, which completes the proof of the theorem.

34 � 3 Caputo-type fractional differential equations 3.2.2 Examples Example 1. Consider the following problem with nonlinear IFDEs: 1

1

D 2 y(t) =

y(t) D 2 y(t) 8e−t e−t ] [ ] − [ t t (11 + e ) 1 + y(t) (11 + e ) 1 + D 21 y(t)

for each t ∈ [0, 1],

(3.13) (3.14)

y(0) = y(1). Set f (t, u, v) =

8e−t u e−t v ( ) − ( ), (11 + et ) 1 + u (11 + et ) 1 + v

t ∈ [0, 1], and u, v ∈ ℝ.

Clearly, the function f is jointly continuous. For each u, u,̄ v, v̄ ∈ ℝ and t ∈ [0, 1], e−t 8e−t |u − u|̄ + |v − v|̄ t (11 + e ) (11 + et ) 8 1 ̄ ≤ |u − u|̄ + |v − v|. 12 12

󵄨󵄨 󵄨 󵄨󵄨f (t, u, v) − f (t, u,̄ v)̄ 󵄨󵄨󵄨 ≤

Hence, condition (3.3.1) is satisfied with 21 ≤ K = 128 < 1, K = 121 , and K + K = 129 < 1. It follows from Theorem 3.1 that the problem (3.13)–(3.14) has at least one solution. Example 2. Consider the following problem with nonlinear IFDEs: 1

1 1 1 1 (t sin y(t) − y(t) cos(t)) + sin D 2 y(t) + 3 100 2 y(0) = y(1).

D 2 y(t) =

for each t ∈ [0, 1],

(3.15) (3.16)

Set 1 1 1 f (t, u, v) = (t sin u − u cos(t)) + sin v + , 3 100 2

t ∈ [0, 1], u, v ∈ ℝ.

Clearly, the function f is jointly continuous. For any u, u,̄ v, v̄ ∈ ℝ and t ∈ [0, 1], 󵄨󵄨 󵄨 󵄨󵄨f (t, u, v) − f (t, u,̄ v)̄ 󵄨󵄨󵄨 1 1 1 | sin v − sin v|̄ ≤ |t|| sin u − sin u|̄ + | cos t||u − u|̄ + 3 3 100 1 1 1 ≤ |u − u|̄ + |u − u|̄ + |v − v|̄ 3 3 100 2 1 ̄ = |u − u|̄ + |v − v|. 3 100 Hence, condition (3.3.1) is satisfied with

1 2

≤K =

2 3

< 1, K =

1 , 100

and K + K =

203 300

< 1.

3.3 Existence results for nonlinear Caputo fractional differential systems

� 35

It follows from Theorem 3.1 that the problem (3.15)–(3.16) has at least one solution on T.

3.3 Existence results for nonlinear Caputo fractional differential systems The purpose of this section is to establish some existence and uniqueness results for the following functional fractional differential equation: c

α

c

β

c

β

c

α

D0+ y(t) − A D0+ y(t) = f (t, y(t), D0+ y(t), D0+ y(t)),

t ∈ T := [0, b],

(3.17)

with the periodic conditions y(0) = y(b),

y′ (0) = y′ (b),

(3.18)

where b > 0, c D0α is the Caputo fractional derivative of order α, 0 < β ≤ 1 < α ≤ 2, 1 + β < α, f : T × ℝn × ℝn × ℝn → ℝn is a given continuous function, and A is an ℝn×n invertible matrix.

3.3.1 Existence results Lemma 3.6. For any y ∈ C 1 (T, ℝn ) and 0 < β ≤ 1, we have b1−β 󵄩󵄩 ′ 󵄩󵄩 󵄩󵄩c β 󵄩󵄩 󵄩󵄩 D0+ y󵄩󵄩∞ ≤ 󵄩y 󵄩 , Γ(2 − β) 󵄩 󵄩∞

so

b1−β 󵄩󵄩c β 󵄩󵄩 ‖y‖ . 󵄩󵄩 D0+ y󵄩󵄩∞ ≤ Γ(2 − β) 1

Proof. Obviously, when β = 1, the conclusions are true. So, we only consider the case 0 < β < 1. In fact, by Definition 2.7, for any y ∈ C 1 (T, ℝn ) and t ∈ T, we have 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨 1 󵄨󵄨c β 󵄨 󵄨󵄨∫(t − s)−β y′ (s)ds󵄨󵄨󵄨 󵄨󵄨 D0+ y(t)󵄨󵄨󵄨 = 󵄨󵄨 󵄨 Γ(1 − β) 󵄨󵄨󵄨 󵄨󵄨 0

󵄩 󵄩 ≤ 󵄩󵄩󵄩y′ 󵄩󵄩󵄩∞

t 1−β 󵄩󵄩 ′ 󵄩󵄩 = 󵄩y 󵄩 Γ(2 − β) 󵄩 󵄩∞ ≤ ≤

t

1 ∫(t − s)−β ds Γ(1 − β)

b1−β 󵄩󵄩 ′ 󵄩󵄩 󵄩y 󵄩 Γ(2 − β) 󵄩 󵄩∞ b1−β ‖y‖ . Γ(2 − β) 1

0

36 � 3 Caputo-type fractional differential equations Lemma 3.7. Let h ∈ C(T, ℝn ). The function y ∈ C 1 (T, ℝn ) is a periodic solution of the fractional differential problem c

α

β

c

D0+ y(t) − A D0+ y(t) = h(t),

y(0) = y(b),

t ∈ T,

(3.19)

′

′

y (0) = y (b)

(3.20)

if and only if y is a solution of the fractional integral equation t

A At α−β y(t) = (1 − )y(0) + ty′ (0) + ∫(t − s)α−β−1 y(s)ds Γ(α − β + 1) Γ(α − β) 0

t

+

1 ∫(t − s)α−1 h(s)ds, Γ(α)

(3.21)

0

with b

y(0) = b1+β−α (α − β − 1) ∫(b − s)α−β−2 y(s)ds 0 b

1+β−α

+

b

Γ(α − β)A−1 ∫(b − s)α−2 h(s)ds Γ(α − 1)

(3.22)

0

and y′ (0) =

b

A ∫(b − s)α−β−2 y(s)ds (α − β)Γ(α − β − 1) 0

b

+

1 ∫(b − s)α−2 h(s)ds (α − β)Γ(α − 1) 0

b

−

A ∫(b − s)α−β−1 y(s)ds bΓ(α − β) 0

b

−

1 ∫(b − s)α−1 h(s)ds. bΓ(α)

(3.23)

0

Proof. Let y ∈ C 1 (T, ℝn ) be a solution of (3.19)–(3.20). Then we have α c

α

c

β

α

I0+ ( D0+ y(t) − A D0+ y(t)) = (I0+ h)(t).

From Lemma 2.1, we get

3.3 Existence results for nonlinear Caputo fractional differential systems

� 37

t

At α−β A y(t) = y(0) + ty (0) − y(0) + ∫(t − s)α−β−1 y(s)ds Γ(α − β + 1) Γ(α − β) ′

0

t

+

1 ∫(t − s)α−1 h(s)ds. Γ(α) 0

Applying conditions (3.20), we obtain (3.22) and (3.23). Thus, y is solution of the integral equation (3.21). Conversely, assume that y satisfies the fractional integral equation (3.21). Using the fact that c D0α+ is the left inverse of I0α+ and the fact that c D0α+ C = 0, where C is a constant, we get c

α

c

β

D0+ y(t) − A D0+ y(t) = h(t)

for each t ∈ T.

Also, we can easily show that y(0) = y(b) and

y′ (0) = y′ (b).

We are now in a position to state and prove our existence result for the problem (3.17)–(3.18) based on Banach’s fixed point theorem. Theorem 3.2. Assume that the following hypothesis holds. (3.9.1) There exist constants L1 , L2 > 0 and 0 < L3 < 1 such that 󵄩󵄩 ̄ 󵄩󵄩󵄩󵄩 ≤ L1 ‖u − u‖̄ + L2 ‖v − v‖̄ + L3 ‖w − w‖ ̄ 󵄩󵄩f (t, u, v, w) − f (t, u,̄ v,̄ w) for any u, v, w, u,̄ v,̄ w̄ ∈ ℝn and t ∈ T. Set R1 = (

‖A‖bα−β ‖A‖bα−β + 1)M1 + Γ(α − β + 1) Γ(α − β + 1)

+

bα L1 Γ(2 − β) + b1−β+α (L3 ‖A‖ + L2 ) + TM2 , Γ(α + 1)Γ(2 − β)(1 − L3 )

R2 = M2 + +

(α − β)‖A‖bα−β−1 ‖A‖bα−β−1 M1 + Γ(α − β + 1) Γ(α − β)

bα−1 L1 Γ(2 − β) + bα−β (L3 ‖A‖ + L2 ) , Γ(α)Γ(2 − β)(1 − L3 )

where M1 = 1 +

bβ Γ(α − β)‖A−1 ‖[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] Γ(α)Γ(2 − β)(1 − L3 )

38 � 3 Caputo-type fractional differential equations and M2 =

bα−1 (2α − β)[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] 2T α−β−1 ‖A‖ + . (α − β)Γ(α + 1)Γ(2 − β)(1 − L3 ) Γ(α − β + 1)

If max{R1 , R2 } < 1,

(3.24)

then the problem (3.17)–(3.18) has a unique solution on T. Proof. Transform the problem (3.17)–(3.18) into a fixed point problem. Consider the operator N : C 1 (T, ℝn ) → C 1 (T, ℝn ) defined by t

(Ny)(t) = (1 −

At α−β A )B + tB2 + ∫(t − s)α−β−1 y(s)ds Γ(α − β + 1) 1 Γ(α − β) 0

t

+

1 ∫(t − s)α−1 g(s)ds, Γ(α)

(3.25)

0

with B1 = b

1+β−α

b

(α − β − 1) ∫(b − s)α−β−2 y(s)ds 0 b

+

b1+β−α Γ(α − β)A−1 ∫(b − s)α−2 g(s)ds Γ(α − 1) 0

and b

A B2 = ∫(b − s)α−β−2 y(s)ds (α − β)Γ(α − β − 1) 0

b

+

1 ∫(b − s)α−2 g(s)ds (α − β)Γ(α − 1) 0

b

−

A ∫(b − s)α−β−1 y(s)ds bΓ(α − β) 0

b

−

1 ∫(b − s)α−1 g(s)ds, bΓ(α) 0

where g ∈ C(T, ℝn ) is such that

3.3 Existence results for nonlinear Caputo fractional differential systems

� 39

β

β

g(t) = f (t, y(t), c D0+ y(t), g(t) + Ac D0+ y(t)). For every y ∈ C 1 (T, ℝn ) and each t ∈ T, we have (Ny)′ (t) = B2 −

t

(α − β)At α−β−1 A B + ∫(t − s)α−β−2 y(s)ds Γ(α − β + 1) 1 Γ(α − β − 1) 0

t

+

1 ∫(t − s)α−2 g(s). Γ(α − 1)

(3.26)

0

It is clear that (Ny)′ ∈ C(T, ℝn ); consequently, N is well defined. Clearly, the fixed points of operator N are solutions of problem (3.17)–(3.18). Let y, v ∈ C 1 (T, ℝn ). Then for t ∈ T, we have ‖A‖bα−β 󵄩󵄩 󵄩 )‖B1 − B1̃ ‖ + b‖B2 − B2̃ ‖ 󵄩󵄩(Ny)(t) − (Nv)(t)󵄩󵄩󵄩 ≤ (1 + Γ(α − β + 1) b

‖A‖ 󵄩 󵄩 + ∫(b − s)α−β−1 󵄩󵄩󵄩y(s) − v(s)󵄩󵄩󵄩ds Γ(α − β) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−1 󵄩󵄩󵄩g(s) − h(s)󵄩󵄩󵄩ds, Γ(α) 0

with b

B1̃ = b1+β−α (α − β − 1) ∫(b − s)α−β−2 v(s)ds 0 b

b1+β−α Γ(α − β)A−1 + ∫(b − s)α−2 h(s)ds Γ(α − 1) 0

and b

B2̃ =

A ∫(b − s)α−β−2 v(s)ds (α − β)Γ(α − β − 1) 0

b

+

1 ∫(b − s)α−2 h(s)ds (α − β)Γ(α − 1) 0

b

−

A ∫(b − s)α−β−1 v(s)ds bΓ(α − β) 0

40 � 3 Caputo-type fractional differential equations b

1 − ∫(b − s)α−1 h(s)ds, bΓ(α) 0

where h ∈ C(T, ℝn ) is such that β

β

h(t) = f (t, v(t), c D0+ v(t), h(t) + Ac D0+ v(t)). From (3.9.1), for each t ∈ T we have 󵄩 󵄩c β 󵄩 󵄩󵄩 󵄩 󵄩 c β 󵄩󵄩g(t) − h(t)󵄩󵄩󵄩 ≤ L1 󵄩󵄩󵄩y(t) − v(t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩 D0+ y(t) − D0+ v(t)󵄩󵄩󵄩 β β 󵄩 󵄩 + L3 󵄩󵄩󵄩g(t) + Ac D0+ y(t) − h(t) − Ac D0+ v(t)󵄩󵄩󵄩 β 󵄩 󵄩 󵄩 β 󵄩 ≤ L1 󵄩󵄩󵄩y(t) − v(t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩c D0+ y(t) − c D0+ v(t)󵄩󵄩󵄩 β 󵄩 󵄩 󵄩 β 󵄩 + L3 󵄩󵄩󵄩g(t) − h(t)󵄩󵄩󵄩 + L3 ‖A‖󵄩󵄩󵄩c D0+ y(t) − c D0+ v(t)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ L1 󵄩󵄩󵄩y(t) − v(t)󵄩󵄩󵄩 + L3 󵄩󵄩󵄩g(t) − h(t)󵄩󵄩󵄩 󵄩 󵄩 β + (L3 ‖A‖ + L2 )󵄩󵄩󵄩c D0+ (y(t) − v(t))󵄩󵄩󵄩. Thus, L1 󵄩󵄩 󵄩󵄩 󵄩 󵄩 L ‖A‖ + L2 󵄩󵄩c β 󵄩 󵄩󵄩g(t) − h(t)󵄩󵄩󵄩 ≤ 󵄩y(t) − v(t)󵄩󵄩󵄩 + 3 󵄩 D + (y(t) − v(t))󵄩󵄩󵄩 1 − L3 󵄩 1 − L3 󵄩 0 L ‖A‖ + L2 󵄩󵄩c β L1 󵄩 ≤ ‖y − v‖∞ + 3 󵄩 D + (y − v)󵄩󵄩󵄩∞ . 1 − L3 1 − L3 󵄩 0 Then by using Lemma 3.6, we get b1−β (L3 ‖A‖ + L2 ) L1 󵄩󵄩 󵄩 ‖y − v‖1 + ‖y − v‖1 󵄩󵄩g(t) − h(t)󵄩󵄩󵄩 ≤ 1 − L3 Γ(2 − β)(1 − L3 ) L Γ(2 − β) + b1−β (L3 ‖A‖ + L2 ) = 1 ‖y − v‖1 . Γ(2 − β)(1 − L3 )

Using (3.27) we obtain bβ Γ(α − β)‖A−1 ‖[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] ‖B1 − B1̃ ‖ ≤ [ + 1] Γ(α)Γ(2 − β)(1 − L3 ) × ‖y − v‖1

= M1 ‖y − v‖1 and ‖B2 − B2̃ ‖ ≤ [

bα−1 (2α − β)[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] (α − β)Γ(α + 1)Γ(2 − β)(1 − L3 )

(3.27)

3.3 Existence results for nonlinear Caputo fractional differential systems

� 41

2T α−β−1 ‖A‖ ]‖y − v‖1 Γ(α − β + 1) = M2 ‖y − v‖1 . +

Therefore, for each t ∈ T we have ‖A‖bα−β ‖A‖bα−β 󵄩 󵄩󵄩 + 1)M1 + TM2 + 󵄩󵄩(Ny)(t) − (Nv)(t)󵄩󵄩󵄩 ≤ [( Γ(α − β + 1) Γ(α − β + 1)

bα L1 Γ(2 − β) + b1−β+α (L3 ‖A‖ + L2 ) ]‖y − v‖1 Γ(α + 1)Γ(2 − β)(1 − L3 ) = R1 ‖y − v‖1 . +

On the other hand, for each t ∈ T we have (α − β)‖A‖bα−β−1 󵄩󵄩 ′ ′ 󵄩 ‖B1 − B1̃ ‖ 󵄩󵄩(Ny) (t) − (Nv) (t)󵄩󵄩󵄩 ≤ ‖B2 − B2̃ ‖ + Γ(α − β + 1) b

‖A‖ 󵄩 󵄩 + ∫(b − s)α−β−2 󵄩󵄩󵄩y(s) − v(s)󵄩󵄩󵄩ds Γ(α − β − 1) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s) − h(s)󵄩󵄩󵄩ds. Γ(α − 1) 0

Using (3.27) we get (α − β)‖A‖bα−β−1 ‖A‖bα−β−1 󵄩󵄩 ′ ′ 󵄩 M1 + 󵄩󵄩(Ny) (t) − (Nv) (t)󵄩󵄩󵄩 ≤ [M2 + Γ(α − β + 1) Γ(α − β)

bα−1 L1 Γ(2 − β) + bα−β (L3 ‖A‖ + L2 ) ]‖y − v‖1 Γ(α)Γ(2 − β)(1 − L2 ) = R2 ‖y − v‖1 . +

Thus, ‖N(y) − N(v)‖1 ≤ max{R1 , R2 }‖y − v‖1 . By (3.24), the operator N is a contraction. Hence, by the Banach contraction principle, the operator N has a unique fixed point, which is the unique solution of (3.17)–(3.18). Our second result is based on Schaefer’s fixed point theorem. Set R=1+

2‖A‖bα−β−1 2‖A‖(2T α−β + bα−β−1 ) + Γ(α − β) Γ(α − β + 1)

+ D(

(2α − β)(bα + bα−1 ) bβ Γ(α − β)‖A−1 ‖ + 2T α−1 + (α − β)Γ(α + 1) Γ(α)

42 � 3 Caputo-type fractional differential equations

+

bα (2α − β) ), (α − β)Γ(α + 1)

where D=

(L ‖A‖ + L2 )b1−β L1 + 3 . 1 − L3 (1 − L3 )Γ(2 − β)

Theorem 3.3. Assume that (3.9.1) holds. If R < 1, then the problem (3.17)–(3.18) has at least one solution on T. Proof. Let N be the operator defined in (3.25). The proof will be given in several steps. Step 1. The operator N is continuous. Let {yn } be a sequence such that yn → y in C 1 (T, ℝn ). Then for each t ∈ T, ‖A‖bα−β 󵄩󵄩 󵄩 + 1)‖B1 − Bñ 1 ‖ + b‖B2 − Bñ 2 ‖ 󵄩󵄩(Ny)(t) − (Nyn )(t)󵄩󵄩󵄩 ≤ ( Γ(α − β + 1) b

‖A‖ 󵄩 󵄩 ∫(b − s)α−β−1 󵄩󵄩󵄩y(s) − yn (s)󵄩󵄩󵄩ds Γ(α − β)

+

0

b

1 󵄩 󵄩 ∫(b − s)α−1 󵄩󵄩󵄩g(s) − gn (s)󵄩󵄩󵄩ds, Γ(α)

+

0

where Bñ 1 , Bñ 2 ∈ ℝn , with b

Bñ 1 = b1+β−α (α − β − 1) ∫(b − s)α−β−2 yn (s)ds 0

b

b1+β−α Γ(α − β)A−1 + ∫(b − s)α−2 gn (s)ds, Γ(α − 1) 0

b

Bñ 2 =

A ∫(b − s)α−β−2 yn (s)ds (α − β)Γ(α − β − 1) 0

b

+

1 ∫(b − s)α−2 gn (s)ds (α − β)Γ(α − 1) 0

b

−

A ∫(b − s)α−β−1 yn (s)ds bΓ(α − β) 0

b

−

1 ∫(b − s)α−1 gn (s)ds, bΓ(α) 0

3.3 Existence results for nonlinear Caputo fractional differential systems

� 43

and gn ∈ C(T, ℝn ) such that β

β

gn (t) = f (t, yn (t), c D0+ yn (t), gn (t) + Ac D0+ yn (t)). From (3.9.1) for each t ∈ T we have 󵄩 󵄩 󵄩󵄩 󵄩 󵄩󵄩g(t) − gn (t)󵄩󵄩󵄩 ≤ L1 󵄩󵄩󵄩y(t) − yn (t)󵄩󵄩󵄩 β β 󵄩 󵄩 + L2 󵄩󵄩󵄩g(t) + Ac D0+ y(t) − gn (t) − Ac D0+ yn (t)󵄩󵄩󵄩 β 󵄩 β 󵄩 + L3 󵄩󵄩󵄩c D0+ y(t) − c D0+ yn (t)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ L1 󵄩󵄩󵄩y(t) − yn (t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩g(t) − gn (t)󵄩󵄩󵄩 β 󵄩 β 󵄩 + L2 ‖A‖󵄩󵄩󵄩c D0+ y(t) − c D0+ yn (t)󵄩󵄩󵄩 β 󵄩 󵄩 β + L3 󵄩󵄩󵄩c D0+ y(t) − c D0+ yn (t)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ L1 󵄩󵄩󵄩y(t) − yn (t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩g(t) − gn (t)󵄩󵄩󵄩 󵄩 β 󵄩 + (L2 ‖A‖ + L3 )󵄩󵄩󵄩c D0+ (y(t) − yn (t))󵄩󵄩󵄩. Thus, L1 󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩g(t) − gn (t)󵄩󵄩󵄩 ≤ 󵄩y(t) − yn (t)󵄩󵄩󵄩 1 − L3 󵄩 L ‖A‖ + L2 󵄩󵄩c β 󵄩 + 3 󵄩 D + (y(t) − yn (t))󵄩󵄩󵄩 1 − L3 󵄩 0 L ‖A‖ + L2 󵄩󵄩c β L1 󵄩 ‖y − yn ‖∞ + 3 ≤ 󵄩 D + (y − yn )󵄩󵄩󵄩∞ . 1 − L3 1 − L3 󵄩 0 Then by using Lemma 3.6, we obtain b1−β (L3 ‖A‖ + L2 ) L1 󵄩󵄩 󵄩 ‖y − yn ‖1 + ‖y − yn ‖1 󵄩󵄩g(t) − gn (t)󵄩󵄩󵄩 ≤ 1 − L3 Γ(2 − β)(1 − L3 ) L Γ(2 − β) + b1−β (L3 ‖A‖ + L2 ) = 1 ‖y − yn ‖1 . Γ(2 − β)(1 − L3 )

Hence, bβ Γ(α − β)‖A−1 ‖[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] ] Γ(α)Γ(2 − β)(1 − L3 ) × ‖y − yn ‖1

‖B1 − Bñ 1 ‖ ≤ [1 +

= M1 ‖y − yn ‖1 and ‖B2 − Bñ 2 ‖ ≤ [

bα−1 (2α − β)[L1 Γ(2 − β) + b1−β (L3 ‖A‖ + L2 )] (α − β)Γ(α + 1)Γ(2 − β)(1 − L3 )

(3.28)

44 � 3 Caputo-type fractional differential equations 2T α−β−1 ‖A‖ ]‖y − yn ‖1 Γ(α − β + 1) = M2 ‖y − yn ‖1 . +

Therefore, for each t ∈ T we get ‖A‖bα−β ‖A‖bα−β 󵄩 󵄩󵄩 + 1)M1 + TM2 + 󵄩󵄩(Ny)(t) − (Nyn )(t)󵄩󵄩󵄩 ≤ [( Γ(α − β + 1) Γ(α − β + 1)

bα L1 Γ(2 − β) + b1−β+α (L3 ‖A‖ + L2 ) ]‖y − yn ‖1 Γ(α + 1)Γ(2 − β)(1 − L3 ) = R1 ‖y − yn ‖1 . +

On the other hand, for each t ∈ T we have (α − β)‖A‖bα−β−1 󵄩󵄩 ′ ′ 󵄩 ‖B1 − Bñ 1 ‖ 󵄩󵄩(Ny) (t) − (Nyn ) (t)󵄩󵄩󵄩 ≤ ‖B2 − Bñ 2 ‖ + Γ(α − β + 1) b

+

‖A‖ 󵄩 󵄩 ∫(b − s)α−β−2 󵄩󵄩󵄩y(s) − yn (s)󵄩󵄩󵄩ds Γ(α − β − 1) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s) − gn (s)󵄩󵄩󵄩ds. Γ(α − 1) 0

Using (3.28) we get (α − β)‖A‖bα−β−1 ‖A‖bα−β−1 󵄩󵄩 ′ ′ 󵄩 M1 + 󵄩󵄩(Ny) (t) − (Nyn ) (t)󵄩󵄩󵄩 ≤ [M2 + Γ(α − β + 1) Γ(α − β)

bα−1 L1 Γ(2 − β) + bα−β (L3 ‖A‖ + L2 ) ]‖y − yn ‖1 Γ(α)Γ(2 − β)(1 − L3 ) = R2 ‖y − yn ‖1 . +

Thus, ‖Ny − Nyn ‖1 → 0 as n → ∞, which implies that the operator N is continuous. Step 2. The operator N maps bounded sets into bounded sets in C 1 (T, ℝn ). It is enough to show that for any η∗ > 0, there exists a positive constant ℓ such that for each y ∈ Bη∗ = {y ∈ C 1 (T, ℝn ) : ‖y‖1 ≤ η∗ }, we have ‖N(y)‖1 ≤ ℓ. We have for each t∈T 󵄩󵄩 󵄩 󵄩 󵄩 c β c β 󵄩󵄩g(t)󵄩󵄩󵄩 = 󵄩󵄩󵄩f (t, y(t), g(t) + A D0+ y(t), D0+ y(t)) − f (t, 0, 0, 0) + f (t, 0, 0, 0)󵄩󵄩󵄩 β 󵄩 󵄩 󵄩 󵄩 󵄩 β 󵄩 󵄩 󵄩 ≤ L1 󵄩󵄩󵄩y(t)󵄩󵄩󵄩 + L3 󵄩󵄩󵄩g(t) + Ac D0+ y(t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩c D0+ y(t)󵄩󵄩󵄩 + 󵄩󵄩󵄩f (t, 0, 0, 0)󵄩󵄩󵄩 󵄩 󵄩 󵄩 β 󵄩 ≤ L1 ‖y‖∞ + L3 󵄩󵄩󵄩g(t)󵄩󵄩󵄩 + (L3 ‖A‖ + L2 )󵄩󵄩󵄩D0+ y󵄩󵄩󵄩∞ + f ∗ , where

3.3 Existence results for nonlinear Caputo fractional differential systems

� 45

󵄩 󵄩 f ∗ = sup󵄩󵄩󵄩f (t, 0, 0, 0)󵄩󵄩󵄩. t∈T

Thus, L ‖A‖ + L2 󵄩󵄩 β 󵄩󵄩 L1 f∗ 󵄩 󵄩󵄩 ‖y‖∞ + 3 . 󵄩󵄩g(t)󵄩󵄩󵄩 ≤ 󵄩󵄩D0+ y󵄩󵄩∞ + 1 − L3 1 − L3 1 − L3 Then by using Lemma 3.6, we have (L ‖A‖ + L2 )b1−β 󵄩󵄩 ′ 󵄩󵄩 L1 f∗ 󵄩󵄩 󵄩 ‖y‖∞ + 3 󵄩󵄩y 󵄩󵄩∞ + 󵄩󵄩g(t)󵄩󵄩󵄩 ≤ 1 − L3 (1 − L3 )Γ(2 − β) 1 − L3 ≤ ≤

(L ‖A‖ + L2 )b1−β L1 f∗ ‖y‖1 + 3 ‖y‖1 + 1 − L3 (1 − L3 )Γ(2 − β) 1 − L3

L1 ∗ (L3 ‖A‖ + L2 )b1−β ∗ f∗ η + η + := M1 , 1 − L3 (1 − L3 )Γ(2 − β) 1 − L3

which implies that b

󵄩 󵄩 ‖B1 ‖ ≤ b1+β−α (α − β − 1) ∫(b − s)α−β−2 󵄩󵄩󵄩y(s)󵄩󵄩󵄩ds 0 b

+

b1+β−α Γ(α − β)‖A−1 ‖ 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds Γ(α − 1) 0

bβ Γ(α − β)‖A−1 ‖ ≤ η∗ + M1 := M2 Γ(α) and b

‖B2 ‖ ≤

‖A‖ 󵄩 󵄩 ∫(b − s)α−β−2 󵄩󵄩󵄩y(s)󵄩󵄩󵄩ds (α − β)Γ(α − β − 1) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds (α − β)Γ(α − 1) 0

b

+

‖A‖ 󵄩 󵄩 ∫(b − s)α−β−1 󵄩󵄩󵄩y(s)󵄩󵄩󵄩ds bΓ(α − β) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−1 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds bΓ(α) 0 α−β−1

≤

(2α − β)bα−1 2‖A‖b η∗ + M := M3 . Γ(α − β + 1) (α − β)Γ(α + 1) 1

(3.29)

46 � 3 Caputo-type fractional differential equations Thus, (3.25) implies ‖A‖bα−β ‖A‖bα−β ∗ 󵄩 󵄩󵄩 M2 + η 󵄩󵄩(Ny)(t)󵄩󵄩󵄩 ≤ M2 + TM3 + Γ(α − β + 1) Γ(α − β + 1) bα + M Γ(α + 1) 1 := ℓ1 . On the other hand, we have for each t ∈ T (α − β)‖A‖bα−β−1 󵄩󵄩 ′ 󵄩 M2 󵄩󵄩(N y)(t)󵄩󵄩󵄩 ≤ M3 + Γ(α − β + 1) +

‖A‖bα−β−1 ∗ bα−1 η + M Γ(α − β) Γ(α) 1

:= ℓ2 . Thus, ‖Ny‖1 ≤ max{ℓ1 , ℓ2 } := ℓ. Step 3. The operator N maps bounded sets into equicontinuous sets of C 1 (T, ℝn ). Let t1 , t2 ∈ T and t1 < t2 , let Bη∗ be a bounded set of C 1 (T, ℝn ) as in Step 2, and let y ∈ Bη∗ . Then 󵄩󵄩 󵄩 󵄩󵄩(Ny)(t2 ) − (Ny)(t1 )󵄩󵄩󵄩 ≤ M3 (t2 − t1 ) + t2

+

‖A‖M2 α−β α−β (t − t1 ) Γ(α − β + 1) 2

‖A‖η∗ ∫(t2 − s)α−β−1 ds Γ(α − β) t1

t1

‖A‖η∗ + ∫[(t2 − s)α−β−1 − (t1 − s)α−β−1 ]ds Γ(α − β) t2

+

0

t1

M1 [∫(t2 − s)α−1 ds + ∫[(t2 − s)α−1 − (t1 − s)α−1 ]ds] Γ(α) t1

0

‖A‖M2 α−β α−β ≤ M3 (t2 − t1 ) + (t − t1 ) Γ(α − β + 1) 2 M1 ‖A‖η∗ α−β α−β + (t − t1 ) + (t α − t1α ). Γ(α − β + 1) 2 Γ(α + 1) 2 As t1 → t2 , the right-hand side of the above inequality tends to zero. Now from (3.26), we have

3.3 Existence results for nonlinear Caputo fractional differential systems

� 47

󵄩 󵄩󵄩 ′ ′ 󵄩󵄩(N y)(t2 ) − (N y)(t1 )󵄩󵄩󵄩 (α − β)‖A‖M2 α−β−1 α−β−1 ≤ (t − t1 ) Γ(α − β + 1) 2 t2

+

‖A‖η∗ ∫(t2 − s)α−β−2 ds Γ(α − β − 1) t1

t1

‖A‖η∗ + ∫[(t2 − s)α−β−2 − (t1 − s)α−β−2 ]ds Γ(α − β − 1) t2

+

0

t1

M1 [∫(t2 − s)α−2 ds + ∫[(t2 − s)α−2 − (t1 − s)α−2 ]ds] Γ(α − 1) t1

0

(α − β)‖A‖M2 α−β−1 α−β−1 (t − t1 ) ≤ Γ(α − β + 1) 2 M ‖A‖η∗ α−β−1 α−β−1 + (t − t1 ) + 1 (t2α−1 − t1α−1 ). Γ(α − β) 2 Γ(α) As t1 → t2 , the right-hand side of the above inequality tends to zero. As a consequence of Steps 1 to 3 together with the Arzelà–Ascoli theorem, we can conclude that N : C 1 (T, ℝn ) → C 1 (T, ℝn ) is completely continuous. Step 4. A priori bounds. Now it remains to show that the set E = {y ∈ C 1 (T, ℝn ) : y = λN(y) for some 0 < λ < 1} is bounded. Let y ∈ E. Then y = λN(y) for some 0 < λ < 1. Thus, for each t ∈ T we have t

λA λAt α−β B1 + y(t) = λB1 + λtB2 − ∫(t − s)α−β−1 y(s)ds Γ(α − β + 1) Γ(α − β) t

+

0

λ ∫(t − s)α−1 g(s)ds. Γ(α) 0

From (3.9.1), for each t ∈ T we have 󵄩󵄩 󵄩 󵄩 󵄩 c β c β 󵄩󵄩g(t)󵄩󵄩󵄩 = 󵄩󵄩󵄩f (t, y(t), D0+ y(t), g(t) + A D0+ y(t)) − f (t, 0, 0, 0) + f (t, 0, 0, 0)󵄩󵄩󵄩 β 󵄩 󵄩 󵄩 󵄩 󵄩 β 󵄩 󵄩 󵄩 ≤ L1 󵄩󵄩󵄩y(t)󵄩󵄩󵄩 + L3 󵄩󵄩󵄩g(t) + Ac D0+ y(t)󵄩󵄩󵄩 + L2 󵄩󵄩󵄩c D0+ y(t)󵄩󵄩󵄩 + 󵄩󵄩󵄩f (t, 0, 0, 0)󵄩󵄩󵄩 󵄩 󵄩 󵄩 β 󵄩 ≤ L1 ‖y‖∞ + L3 󵄩󵄩󵄩g(t)󵄩󵄩󵄩 + (L3 ‖A‖ + L2 )󵄩󵄩󵄩D0+ y󵄩󵄩󵄩∞ + f ∗ . Thus,

48 � 3 Caputo-type fractional differential equations L ‖A‖ + L2 󵄩󵄩 β 󵄩󵄩 L1 f∗ 󵄩 󵄩󵄩 ‖y‖∞ + 3 . 󵄩󵄩D0+ y󵄩󵄩∞ + 󵄩󵄩g(t)󵄩󵄩󵄩 ≤ 1 − L3 1 − L3 1 − L3 Then by using Lemma 3.6, we have (L ‖A‖ + L2 )b1−β 󵄩󵄩 ′ 󵄩󵄩 L1 f∗ 󵄩 󵄩󵄩 ‖y‖∞ + 3 󵄩󵄩y 󵄩󵄩∞ + 󵄩󵄩g(t)󵄩󵄩󵄩 ≤ 1 − L3 (1 − L3 )Γ(2 − β) 1 − L3 (L ‖A‖ + L2 )b1−β L1 f∗ + 3 ]‖y‖1 + 1 − L3 (1 − L3 )Γ(2 − β) 1 − L3 ∗ f , ≤ D‖y‖1 + 1 − L3

≤[

which implies that ‖B1 ‖ ≤ b

1+β−α

b

󵄩 󵄩 (α − β − 1) ∫(b − s)α−β−2 󵄩󵄩󵄩y(s)󵄩󵄩󵄩ds 0 b

+

b1+β−α Γ(α − β)‖A−1 ‖ 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds Γ(α − 1) 0

(L ‖A‖ + L2 )b1−β L bβ Γ(α − β)‖A−1 ‖ ≤ [1 + ( )( 1 + 3 )]‖y‖1 Γ(α) 1 − L3 (1 − L3 )Γ(2 − β) +

f ∗ bβ Γ(α − β)‖A−1 ‖ (1 − L3 )Γ(α)

≤ [1 +

f ∗ bβ Γ(α − β)‖A−1 ‖ bβ Γ(α − β)‖A−1 ‖ D]‖y‖1 + Γ(α) (1 − L3 )Γ(α)

‖B2 ‖ ≤

‖A‖ 󵄨 󵄨 ∫(b − s)α−β−2 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds (α − β)Γ(α − β − 1)

and b

0

b

+

1 󵄩 󵄩 ∫(b − s)α−2 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds (α − β)Γ(α − 1) 0

b

+

‖A‖ 󵄩 󵄩 ∫(b − s)α−β−1 󵄩󵄩󵄩y(s)󵄩󵄩󵄩ds bΓ(α − β) 0

b

+

1 󵄩 󵄩 ∫(b − s)α−1 󵄩󵄩󵄩g(s)󵄩󵄩󵄩ds bΓ(α) 0

≤ [(

(L ‖A‖ + L2 )b1−β L (2α − β)bα−1 )( 1 + 3 ) (α − β)Γ(α + 1) 1 − L3 (1 − L3 )Γ(2 − β)

3.3 Existence results for nonlinear Caputo fractional differential systems

+ ≤[

� 49

(2α − β)bα−1 f ∗ 2‖A‖bα−β−1 ]‖y‖1 + [ ] Γ(α − β + 1) (1 − L3 )(α − β)Γ(α + 1)

(2α − β)bα−1 2‖A‖bα−β−1 + D]‖y‖1 Γ(α − β + 1) (α − β)Γ(α + 1)

+[

(2α − β)bα−1 f ∗ ]. (1 − L3 )(α − β)Γ(α + 1)

Thus, for each t ∈ T, we have 󵄩󵄩 󵄩 󵄩󵄩y(t)󵄩󵄩󵄩 ≤ ‖B1 ‖ + b‖B2 ‖ + +

‖A‖bα−β ‖A‖bα−β |B1 | + ‖y‖ Γ(α − β + 1) Γ(α − β + 1) 1

(L ‖A‖ + L2 )b1−β L f ∗ bα bα [ 1 + 3 ]‖y‖1 + Γ(α + 1) 1 − L3 (1 − L3 )Γ(2 − β) (1 − L3 )Γ(α + 1)

≤ [(

bβ Γ(α − β)‖A−1 ‖ (2α − β)bα bα bα + + + )D Γ(α) (α − β)Γ(α + 1) (α − β)Γ(α) Γ(α + 1)

bβ Γ(α − β)‖A−1 ‖f ∗ 4‖A‖bα−β ]‖y‖1 + Γ(α − β + 1) (1 − L3 )Γ(α) (2α − β)bα f ∗ + (1 − L3 )(α − β)Γ(α + 1) bα f ∗ bα f ∗ + + (1 − L3 )(α − β)Γ(α) (1 − L3 )Γ(α + 1)

+1+

(2α − β)f ∗ (bα−1 + 2T α ) 2T α−1 f ∗ + bβ Γ(α − β)‖A−1 ‖f ∗ + . (1 − L3 )(α − β)Γ(α + 1) (1 − L3 )Γ(α)

≤ R‖y‖1 + Thus,

‖y‖∞ ≤ R‖y‖1 + +

(2α − β)f ∗ (bα−1 + 2T α ) (1 − L3 )(α − β)Γ(α + 1)

2T α−1 f ∗ + bβ Γ(α − β)‖A−1 ‖f ∗ . (1 − L3 )Γ(α)

On the other hand, for each t ∈ T we have (α − β)‖A‖bα−β−1 ‖A‖bα−β−1 󵄩󵄩 ′ 󵄩󵄩 |B1 | + ‖y‖1 󵄩󵄩y (t)󵄩󵄩 ≤ |B2 | + Γ(α − β + 1) Γ(α − β) +

(L ‖A‖ + L2 )b1−β L f ∗ bα−1 bα−1 [ 1 + 3 ]‖y‖1 + Γ(α) 1 − L3 (1 − L3 )Γ(2 − β) (1 − L3 )Γ(α)

≤ [( +

(2α − β)bα−1 2‖A‖bα−β−1 (1 + α − β) 2T α−1 + )D + ]‖y‖1 (α − β)Γ(α + 1) Γ(α) Γ(α − β + 1)

(2α − β)bα−1 f ∗ 2T α−1 f ∗ + . (1 − L3 )(α − β)Γ(α + 1) (1 − L3 )Γ(α)

(3.30)

50 � 3 Caputo-type fractional differential equations Hence, (2α − β)f ∗ (bα−1 + 2T α ) 󵄩󵄩 ′ 󵄩󵄩 󵄩󵄩y 󵄩󵄩∞ ≤ R‖y‖1 + (1 − L3 )(α − β)Γ(α + 1) +

2T α−1 f ∗ + bβ Γ(α − β)‖A−1 ‖f ∗ . (1 − L3 )Γ(α)

(3.31)

From (3.30) and (3.31), we get ‖y‖1 ≤ R‖y‖1 +

(2α − β)f ∗ (bα−1 + 2T α ) 2T α−1 f ∗ + bβ Γ(α − β)‖A−1 ‖f ∗ + . (1 − L3 )(α − β)Γ(α + 1) (1 − L3 )Γ(α)

Since R < 1, we have ‖y‖1 ≤

(2α − β)f ∗ (bα−1 + 2T α ) 2T α−1 f ∗ + bβ Γ(α − β)‖A−1 ‖f ∗ + (1 − R)(1 − L3 )(α − β)Γ(α + 1) (1 − R)(1 − L3 )Γ(α)

= ψ.

This shows that the set E is bounded. As a consequence of Schaefer’s fixed point theorem, we deduce that N has a fixed point which is a solution of problem (3.17)–(3.18).

3.3.2 An example Consider the fractional differential equation c

α

c

β

c

β

c

α

D0+ y(t) − A D0+ y(t) = f (t, y(t), D0+ y(t), D0+ y(t)),

t ∈ [0, 1],

(3.32)

with the periodic conditions y(0) = y(1),

y′ (0) = y′ (1),

(3.33)

where f : [0, 1] × ℝ2 × ℝ2 × ℝ2 → ℝ2 such that f = (f1 , f2 ) with fi (t, y, v, z) =

ci t 2 , 1 + ‖y‖ + ‖v‖ + ‖z‖

i = 1, 2,

ci > 0, y, v, z ∈ ℝ2 , A = ( 02 21 ), β = 41 , and α = 32 . It is clear that the hypothesis (3.9.1) is satisfied. A simple computation shows that all conditions of Theorem 3.3 are satisfied for an appropriate choice of the constants ci . It follows that the problem (3.32)–(3.33) has at least one solution defined on [0, 1].

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

� 51

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative This section deals with the existence of solutions for the nonlinear fractional differential equation e α c Da+ y(t)

μ

= g(t, y(t), ec Da+ y(t)),

t ∈ T := [a, b], 1 < α ≤ 2, and 0 < μ ≤ 1,

(3.34)

with the periodic conditions y(a) = y(b) and

e

1

e

1

Da+ y(a) = Da+ y(b),

(3.35)

μ

where ec Daα+ and ec Da+ are the Caputo exponential fractional derivatives of orders α > 0 and μ > 0 and g : T × ℝ × ℝ 󳨀→ ℝ is a continuous function.

3.4.1 Existence results We consider the Banach space Ω = {y ∈ C(T, ℝ) : y(t) = e Jaα+ η(t) : η ∈ C(T, ℝ), t ∈ T} with the norm 󵄩 󵄩 󵄩 μ 󵄩 ‖y‖Ω = max{‖y‖∞ , 󵄩󵄩󵄩ec Daα+ y󵄩󵄩󵄩∞ , 󵄩󵄩󵄩ec Da+ y󵄩󵄩󵄩∞ } and Θ = C(T, ℝ) with the norm 󵄨 󵄨 ‖η‖Θ = sup{󵄨󵄨󵄨η(t)󵄨󵄨󵄨 : t ∈ T}. Consider the linear operator K : Dom K ⊆ Ω → Θ defined by e

α

K y := c Da+ y,

1 < α < 2,

(3.36)

where Dom K = {y ∈ Ω : ec Daα+ y ∈ Θ; y(a) = y(b) and e Da1+ y(a) = e Da1+ y(b)}. Define Φ : Ω → Θ by μ

Φy(t) := g(t, y(t), ec Da+ y(t)),

t ∈ T and 0 < μ < 1.

Then the problem (3.34)–(3.35) can be rewritten as K y = Φy.

(3.37)

52 � 3 Caputo-type fractional differential equations Lemma 3.8. Let K be defined by (3.36). Then ker K = {y ∈ Ω : y(t) = y(a), t ∈ T} and b

Img K = {ω ∈ Θ :

1 α−2 ∫(eb − eϱ ) ω(ϱ)eϱ dϱ = 0}. Γ(α − 1) a

Proof. By Lemma 2.3, for t ∈ T, K y(t) = ec Daα+ y(t) = 0 has the following solution: y(t) = ϖ0 + ϖ1 (et − ea ),

t ∈ T,

where ϖ0 = y(a) and ϖ1 = e Da1+ y(a). Since y ∈ Dom K , we can get y(t) = y(a),

t ∈ T.

Then ker K = {y ∈ Ω : y(t) = y(a), t ∈ T}. For ω ∈ Img K , there exists y ∈ Dom K such that ω = K y ∈ Θ. By Lemma 2.2, we have for each t ∈ T y(t) = y(a) +

e

1 t Da+ y(a)(e

t

1 α−1 −e )+ ∫(et − eϱ ) ω(ϱ)eϱ dϱ. Γ(α) a

a

Then e

1 Da+ y(t)

=

e

1 Da+ y(a)

t

1 α−2 + ∫(et − eϱ ) ω(ϱ)eϱ dϱ. Γ(α − 1) a

Since y ∈ Dom K , then ω can satisfy b

1 α−2 ∫(eb − eϱ ) ω(ϱ)eϱ dϱ = 0. Γ(α − 1) a

On the other hand, suppose ω ∈ Θ and satisfies b

α−2

∫(eb − eϱ ) a

ω(ϱ)eϱ dϱ = 0.

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

� 53

Then we get ω(t) = ec Daα+ y(t) and e

e α−1

1

Da+ y(t) = Ja+ ω(t).

Therefore, y(a) = y(b) and e

e

1

1

Da+ y(a) = Da+ y(b),

which implies that y ∈ Dom K . Thus, ω ∈ Img K . Then we have b

α−2

Img K = {ω ∈ Θ : ∫(eb − eϱ )

ω(ϱ)eϱ dϱ = 0}.

a

Lemma 3.9. Let K be defined by (3.36). Then K is a Fredholm operator of index zero, and the linear continuous projector operators ψ1 : Ω → Ω and ψ2 : Θ → Θ can be defined as ψ1 η(t) = η(a) + e Da1+ η(a)(et − ea ) and b

α−1 α−2 ψ2 v(t) = b ∫(eb − eϱ ) v(ϱ)eϱ dϱ. a α−1 (e − e ) a

Furthermore, the operator Kψ−1 : Img K → Ω ∩ ker ψ1 can be given by 1 −1

e α

Kψ1 (v)(t) = T a+ v(t).

Proof. Clearly, Img ψ1 = ker K and ψ21 = ψ1 . It follows for each η ∈ Ω, η = (η − ψ1 η) + ψ1 η that Ω = ker ψ1 + ker K . Also, we have ker ψ1 ∩ ker K = 0. So Ω = ker ψ1 ⊕ ker K .

54 � 3 Caputo-type fractional differential equations Similarly, for each v ∈ Θ, ψ22 v = ψ2 v and v = (v − ψ2 (v)) + ψ2 (v), where (v − ψ2 (v)) ∈ ker ψ2 = Img K . Since Img K = ker ψ2 and ψ22 = ψ2 , we have Img ψ2 ∩ Img K = 0. Thus, Θ = Img K ⊕ Img ψ2 . Thus, dim ker K = dim Img ψ2 = codim Img K , which implies that K is a Fredholm operator of index zero. We will demonstrate that Kψ−1 = e Jaα+ is the inverse of K |Dom K ∩ker ψ1 . For v ∈ 1 Img K , we have −1

e

α e α

K Kψ1 (v) = c Da+ ( Ja+ v) = v.

(3.38)

Furthermore, for η ∈ Dom K ∩ ker ψ1 we get e α

−1

e

α

e

1

t

a

Kψ1 (K (η(t))) = Ja+ (c Da+ η(t)) = η(t) − η(a) − Da+ η(a)(e − e ).

Since η ∈ Dom K ∩ ker ψ1 , we have η(a) = 0 and e Da1+ η(a) = 0. Therefore, −1

Kψ1 (K (η(t))) = η(t).

(3.39)

Combining (3.38) with (3.39), we know that Kψ−1 = (K |Dom K ∩ker ψ1 )−1 . 1 Lemma 3.10. Assume that the following condition (3.13.1) is satisfied. (3.13.1) There exist constants ϖ, ϖ > 0 such that 󵄨󵄨 󵄨 󵄨󵄨g(t, p, q) − g(t, p,̄ q)̄ 󵄨󵄨󵄨 ≤ ϖ|p − p|̄ + ϖ|q − q|̄ for t ∈ T and p, p,̄ q, q̄ ∈ ℝ. Then the operator Φ is K -compact on any bounded open set S ⊂ Ω. Proof. Let S = {η ∈ Ω : ‖η‖Ω < K} be a bounded open set where K > 0. Claim 1. We prove that ψ2 Φ is continuous. Let (ηn )n∈ℕ be a sequence such that ηn 󳨀→ η in Ω. Then for each ϱ ∈ T we have 󵄨󵄨 󵄨 󵄨󵄨ψ2 Φ(ηn )(ϱ) − ψ2 Φ(η)(ϱ)󵄨󵄨󵄨 b

α−1 α−2 󵄨 󵄨 ≤ b ∫(eb − eϱ ) 󵄨󵄨󵄨Φ(ηn )(ϱ) − Φ(η)(ϱ)󵄨󵄨󵄨eϱ dϱ. a α−1 (e − e ) a

By (T1), we have

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

󵄨 󵄨󵄨 󵄨󵄨ψ2 Φ(ηn )(ϱ) − ψ2 Φ(η)(ϱ)󵄨󵄨󵄨 b

ϖ(α − 1) α−2 󵄨 󵄨 ≤ b ∫(eb − eϱ ) 󵄨󵄨󵄨(ηn )(ϱ) − (η)(ϱ)󵄨󵄨󵄨eϱ dϱ a α−1 (e − e ) a

b

+

ϖ(α − 1) α−2 󵄨 󵄨 ∫(eb − eϱ ) 󵄨󵄨󵄨ec Daμ (ηn )(ϱ) − ec Daμ (η)(ϱ)󵄨󵄨󵄨eϱ dϱ (eb − ea )α−1 a

b

(α − 1)(ϖ + ϖ)‖ηn − η‖Ω α−2 ≤ ∫(eb − eϱ ) eϱ dϱ b a α−1 (e − e ) a

≤ (ϖ + ϖ)‖ηn − η‖Ω . Thus, for each ϱ ∈ T, we get 󵄨󵄨 󵄨 󵄨󵄨ψ2 Φ(ηn )(ϱ) − y2 Φ(η)(ϱ)󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞,

and hence 󵄩󵄩 󵄩 󵄩󵄩ψ2 Φ(ηn )(ϱ) − ψ2 Φ(η)(ϱ)󵄩󵄩󵄩Θ 󳨀→ 0

as n 󳨀→ +∞.

Thus, ψ2 Φ is continuous. Claim 2. We show that ψ2 Φ(S) is bounded. For each η ∈ S and t ∈ T, we have 󵄨󵄨 󵄨 󵄨󵄨ψ2 Φ(η)(t)󵄨󵄨󵄨 b

≤

α−1 α−2 󵄨 󵄨 ∫(eb − eϱ ) eϱ 󵄨󵄨󵄨g(ϱ, η(ϱ), ec Daμ η(ϱ))󵄨󵄨󵄨dϱ (eb − ea )α−1 a

b

α−1 α−2 󵄨 󵄨 ≤ b ∫(eb − eϱ ) eϱ 󵄨󵄨󵄨g(ϱ, η(ϱ), ec Daμ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨dϱ (e − ea )α−1 a

b

+

α−1 α−2 󵄨 󵄨 ∫(eb − eϱ ) eϱ 󵄨󵄨󵄨g(ϱ, 0, 0))󵄨󵄨󵄨dϱ (eb − ea )α−1 a

b

α−1 α−2 󵄨 󵄨 󵄨 󵄨 ≤g + b ∫(eb − eϱ ) eϱ (ϖ 󵄨󵄨󵄨η(ϱ)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨ec Daμ η(ϱ)󵄨󵄨󵄨)dϱ (e − ea )α−1 ∗

a

≤ g ∗ + K(ϖ + ϖ), where g ∗ = supt∈J |g(t, 0, 0)|. Thus,

� 55

56 � 3 Caputo-type fractional differential equations 󵄩 󵄩󵄩 ∗ 󵄩󵄩ψ2 Φ(η)󵄩󵄩󵄩Θ ≤ g + K(ϖ + ϖ) := R. This shows that ψ2 Φ(S) is a bounded set in Θ. Claim 3. We show that Kψ−1 (id − ψ2 )Φ : S → Ω is completely continuous. 1 In view of the Arzelà–Ascoli theorem, we need to prove that Kψ−1 (id − ψ2 )Φ(S) ⊂ Ω 1

is equicontinuous and bounded. First, for each η ∈ S and t ∈ T, we have −1

Kψ1 (id − ψ2 )Φη(t)

= Kψ−1 (Φη(t) − ψ2 Φη(t)) 1 μ

= e Jaα+ [g(t, η(t), ec Da+ η(t)) b

−

α−1 α−2 μ ∫(eb − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ] (eb − ea )α−1 a

t

=

1 α−1 μ ∫(et − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ Γ(α) a

b

(α − 1)(et − ea )α α−2 μ − ∫(eb − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ. b a α−1 Γ(α + 1)(e − e ) a

On the one hand, for each η ∈ S and t ∈ T, we have 󵄨󵄨 −1 󵄨 󵄨󵄨Kψ1 (id − ψ2 )Φη(t)󵄨󵄨󵄨 b

1 α−1 󵄨 μ 󵄨 ≤ ∫(eb − eϱ ) 󵄨󵄨󵄨g(ϱ, η(ϱ), ec Da+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ Γ(α) a

b

+

1 α−1 󵄨 󵄨 ∫(eb − eϱ ) 󵄨󵄨󵄨g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ Γ(α) a

b

e

μ

ϱ

|g(ϱ, η(ϱ), c Da+ η(ϱ)) − g(ϱ, 0, 0)|e (α − 1)(eb − ea )α + dϱ ∫ b a α−1 Γ(α + 1)(e − e ) (eb − eϱ )2−α a

b

+

(α − 1)(eb − ea )α α−2 󵄨 󵄨 ∫(eb − eϱ ) 󵄨󵄨󵄨g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ Γ(α + 1)(eb − ea )α−1 a

2(eb − ea )α := β1 . ≤ [g ∗ + K(ϖ + ϖ)] Γ(α + 1) Therefore,

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

󵄩󵄩 −1 󵄩 󵄩󵄩Kψ1 (id − ψ2 )Φη󵄩󵄩󵄩∞ ≤ β1 .

� 57

(3.40)

On the other hand, we have e α −1 c Da+ (Kψ1 (id

− ψ2 )Φη(t)) μ

= g(t, η(t), ec Da+ η(t)) b

−

(α − 1) α−2 μ ∫(eb − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ, (eb − ea )(α−1) a

which implies that for each η ∈ S and t ∈ T, we have 󵄨󵄨e α 󵄨 −1 ∗ 󵄨󵄨c Da+ (Kψ1 (id − ψ2 )Φη(t))󵄨󵄨󵄨 ≤ 2g + 2K(ϖ + ϖ) := β2 . Thus, 󵄩󵄩e α 󵄩 −1 󵄩󵄩c Da+ (Kψ1 (id − ψ2 )Φη)󵄩󵄩󵄩∞ ≤ β2 . Also, we have e μ −1 c Da+ (Kψ1 (id

= e Ja+

(α−μ)

− ψ2 )Φη(t)) μ

[g(t, η(t), ec Da+ η(t)) b

−

(α − 1) α−2 μ ∫(eb − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ] (eb − ea )(α−1) a

t

=

1 (α−μ−1) μ g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ ∫(et − eϱ ) Γ(α − μ) a

b

t

(e − ea )(α−μ) (α − 1) α−2 μ − ∫(eb − eϱ ) g(ϱ, η(ϱ), ec Da+ η(ϱ))eϱ dϱ, Γ(α − μ + 1)(eb − ea )(α−1) a

which implies that for each η ∈ S and t ∈ T, we have 󵄨󵄨e μ 󵄨 −1 󵄨󵄨c Da+ (Kψ1 (id − ψ2 )Φη(t))󵄨󵄨󵄨 b

≤

1 (α−μ−1) 󵄨 󵄨󵄨g(ϱ, η(ϱ), e D μ+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ ∫(et − eϱ ) c a 󵄨 󵄨 Γ(α − μ) a

b

1 (α−μ−1) 󵄨 󵄨󵄨g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ + ∫(et − eϱ ) 󵄨 󵄨 Γ(α − μ) a

(3.41)

58 � 3 Caputo-type fractional differential equations

+

(eb − ea )(α−μ) (α − 1) Γ(α − μ + 1)(eb − ea )(α−1) b

α−2 󵄨

× ∫(eb − eϱ ) a

󵄨󵄨g(ϱ, η(ϱ), e D μ+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ c a 󵄨 󵄨 b

+

(eb − ea )(α−μ) (α − 1) α−2 󵄨 󵄨 ∫(eb − eϱ ) 󵄨󵄨󵄨g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ Γ(α − μ + 1)(eb − ea )(α−1) a

2(eb − ea )(α−μ) := β3 . ≤ [g ∗ + K(ϖ + ϖ)] Γ(α − μ + 1) Thus, 󵄩󵄩e μ 󵄩 −1 󵄩󵄩c Da+ (Kψ1 (id − ψ2 )Φη)󵄩󵄩󵄩∞ ≤ β3 .

(3.42)

By inequalities (3.40), (3.41), and (3.42), we have 󵄩󵄩 −1 󵄩 󵄩󵄩Kψ1 (id − ψ2 )Φη󵄩󵄩󵄩Ω ≤ max{β1 , β2 , β3 }, which shows that Kψ−1 (id − ψ2 )Φ(S) is uniformly bounded in Ω. 1

Now, we need to prove that Kψ−1 (id − ψ2 )Φ(S) is equicontinuous. Furthermore, for 1

a < t1 < t2 ≤ b and η ∈ S, firstly we have

󵄨󵄨 −1 󵄨 −1 󵄨󵄨Kψ1 (id − ψ2 )Φη(t2 ) − Kψ1 (id − ψ2 )Φη(t1 )󵄨󵄨󵄨 g ∗ + K(ϖ + ϖ) ≤ Γ(α) t2

t2

ϱ α−1 ϱ

× [∫(e − e ) t1

t1

α−1 α−1 󵄨 󵄨 e dϱ + ∫󵄨󵄨󵄨(et2 − eϱ ) − (et1 − eϱ ) 󵄨󵄨󵄨eϱ dϱ] a

K(ϖ + ϖ) + g α α [(et2 − ea ) − (et1 − ea ) ]. Γ(α + 1) ∗

+ Secondly, we have

󵄨󵄨e α 󵄨 −1 e α −1 󵄨󵄨c Da+ (Kψ1 (id − ψ2 )Φη(t2 )) − c Da+ (Kψ1 (id − ψ2 )Φη(t1 ))󵄨󵄨󵄨 μ μ 󵄨 󵄨 ≤ 󵄨󵄨󵄨g(t2 , η(t2 ), ec Da+ η(t2 )) − g(t1 , η(t1 ), ec Da+ η(t1 ))󵄨󵄨󵄨. Finally, we have e μ −1 c Da+ (Kψ1 (id

=

− ψ2 )Φη(t))

μ e (α−μ) 󵄨󵄨 Ja+ 󵄨󵄨g(t2 , η(t2 ), ec Da+ η(t2 ))

μ 󵄨 − g(t1 , η(t1 ), ec Da+ η(t1 ))󵄨󵄨󵄨.

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

� 59

We conclude that as t1 → t2 the right-hand side of the above three inequalities tends to zero. Thus, Kψ−1 (id − ψ2 )Φ(S) is equicontinuous in Ω. By the Arzelà–Ascoli theorem, 1 Kψ−1 (id − ψ2 )Φ(S) is relatively compact. Consequently, Φ is K -compact in S. 1

Lemma 3.11. Suppose that (3.13.1) is satisfied. If ϖ + ϖ ≤ min{1,

Γ(α + 1) Γ(α − μ + 1) , }, (eb − ea )α (eb − ea )α−μ

(3.43)

then there exists H > 0 independent of ε, where K (η) − Φ(η) = −ε[K (η) + Φ(−η)] 󳨐⇒ ‖η‖Ω ≤ H ,

ε ∈ (0, 1].

(3.44)

Proof. Let η ∈ Ω satisfy (3.44). Then K (η) − Φ(η) = −εK (η) − εΦ(−η),

and thus K (η) =

ε 1 Φ(η) − Φ(−η). 1+ε 1+ε

(3.45)

By employing the definition of K and Φ, we get for each t ∈ T 󵄨󵄨 󵄨 󵄨e α 󵄨 󵄨󵄨K η(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨c Da+ η(t)󵄨󵄨󵄨 1 󵄨󵄨 ε 󵄨󵄨 󵄨 󵄨 e μ e μ ≤ 󵄨g(t, η(t), c Da+ η(t))󵄨󵄨󵄨 + 󵄨g(t, −η(t), −c Da+ η(t))󵄨󵄨󵄨 1 + ε󵄨 1 + ε󵄨 1 󵄨󵄨 μ 󵄨 [󵄨g(t, η(t), ec Da+ η(t)) − g(t, 0, 0)󵄨󵄨󵄨 + g ∗ ] ≤ 1+ε 󵄨 ε 󵄨󵄨 μ 󵄨 [󵄨g(t, −η(t), −ec Da+ η(t)) − g(t, 0, 0)󵄨󵄨󵄨 + g ∗ ] + 1+ε 󵄨 ε 1 1 󵄨 󵄨 μ 󵄨 󵄨 + )g ∗ + [ϖ 󵄨󵄨η(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨ec Da+ η(t)󵄨󵄨󵄨] ≤( 1+ε 1+ε 1+ε 󵄨 ε μ 󵄨 󵄨 󵄨 󵄨 [ϖ 󵄨󵄨−η(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨−ec Da+ η(t)󵄨󵄨󵄨] + 1+ε 󵄨 ε 1 󵄨 󵄨 󵄨 μ 󵄨 + )[ϖ 󵄨󵄨󵄨η(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨ec Da+ η(t)󵄨󵄨󵄨] ≤ g∗ + ( 1+ε 1+ε 󵄨 μ 󵄨 󵄨 󵄨 ≤ g ∗ + [ϖ 󵄨󵄨󵄨η(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨ec Da+ η(t)󵄨󵄨󵄨] 󵄩 μ 󵄩 ≤ g ∗ + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ , which implies that 󵄩e μ 󵄩 󵄩󵄩e α 󵄩󵄩 ∗ 󵄩󵄩c Da+ η󵄩󵄩∞ ≤ g + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩c Da+ η󵄩󵄩󵄩∞ . By (3.45), for each t ∈ T, we have

(3.46)

60 � 3 Caputo-type fractional differential equations 1 ε −1 −1 Kψ1 Φ(η)(t) − K Φ(−η)(t). 1+ε 1 + ε ψ1

η(t) = Then t

1 α−1 󵄨 μ 󵄨 󵄨 󵄨󵄨 ∫(et − eϱ ) 󵄨󵄨󵄨g(ϱ, η(ϱ), ec Da+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ 󵄨󵄨η(t)󵄨󵄨󵄨 ≤ (1 + ε)Γ(α) a

ε α−1 󵄨 μ 󵄨 ∫(et − eϱ ) 󵄨󵄨󵄨g(ϱ, −η(ϱ), −ec Da+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ (1 + ε)Γ(α)

+

εg ∗ (eb − ea )α g (e − e ) + (1 + ε)Γ(α + 1) (1 + ε)Γ(α + 1) ∗

≤(

a a α

b

1 (eb − ea )α ε 󵄩 μ 󵄩 + ) (ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ) 1 + ε 1 + ε Γ(α + 1)

+( =

t

+

g ∗ (eb − ea )α ε 1 + ) 1+ε 1+ε Γ(α + 1)

g ∗ (eb − ea )α (eb − ea )α 󵄩 μ 󵄩 (ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ) + . Γ(α + 1) Γ(α + 1)

Hence, b a α 󵄩 μ 󵄩 (e − e ) ‖η‖∞ ≤ [g ∗ + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ] . Γ(α + 1)

On the one hand, for each t ∈ T, we have 1 ε −1 −1 K Φ(η)(t) − K Φ(−η)(t). 1 + ε ψ1 1 + ε ψ1

η(t) = Then e μ c Da+ η(t)

1 e (α−μ) ε e (α−μ) J + Φ(η)(t) − J + Φ(−η)(t). 1+ε a 1+ε a

=

Hence, 󵄨󵄨e μ 󵄨 󵄨󵄨c Da+ η(t)󵄨󵄨󵄨

t

1 (α−μ−1) 󵄨 󵄨󵄨g(ϱ, η(ϱ), e D μ+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ ≤ ∫(et − eϱ ) c a 󵄨 󵄨 (1 + ε)Γ(α − μ) a

ε + (1 + ε)Γ(α − μ) t

× ∫(et − eϱ ) a

󵄨󵄨g(ϱ, −η(ϱ), −e D μ+ η(ϱ)) − g(ϱ, 0, 0)󵄨󵄨󵄨eϱ dϱ c a 󵄨 󵄨

(α−μ−1) 󵄨

(3.47)

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

+ ≤(

εg ∗ (eb − ea )(α−μ) g ∗ (eb − ea )(α−μ) + (1 + ε)Γ(α − μ + 1) (1 + ε)Γ(α − μ + 1)

ε 1 (eb − ea )(α−μ) 󵄩 μ 󵄩 + ) (ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ) 1 + ε 1 + ε Γ(α − μ + 1)

+( =

� 61

g ∗ (eb − ea )(α−μ) ε 1 + ) 1+ε 1+ε Γ(α − μ + 1)

(eb − ea )(α−μ) ∗ 󵄩 μ 󵄩 (g + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ). Γ(α − μ + 1)

Hence, b a α−μ 󵄩󵄩e μ 󵄩 󵄩e μ 󵄩 (e − e ) ∗ . 󵄩󵄩c Da+ η(t)󵄩󵄩󵄩∞ ≤ [g + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩c Da+ η󵄩󵄩󵄩∞ ] Γ(α − μ + 1)

(3.48)

Using the definition of the norm ‖ ⋅ ‖Ω and the inequalities (3.46), we see that if ‖η‖Ω = ‖ec Daα+ η‖∞ , then 󵄩 μ 󵄩 ‖η‖Ω ≤ g ∗ + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ≤ g ∗ + (ϖ + ϖ)‖η‖Ω ,

which implies that ‖η‖Ω ≤

g∗ := H1 . 1 − (ϖ + ϖ)

On the other hand, if ‖η‖Ω = ‖η‖∞ , then (3.47) implies b a α 󵄩 μ 󵄩 (e − e ) ‖η‖Ω ≤ [g ∗ + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ] Γ(α + 1)

≤ [g ∗ + ϖ‖η‖Ω + ϖ‖η‖Ω ] ≤ [g ∗ + (ϖ + ϖ)‖η‖Ω ]

(eb − ea )α Γ(α + 1)

(eb − ea )α , Γ(α + 1)

so ‖η‖Ω ≤

Γ(α+1) (eb −ea )α

g∗

− (ϖ + ϖ)

:= H2 .

μ

If ‖η‖Ω = ‖ec Da+ η‖∞ , then by inequalities (3.48) we have b a α−μ 󵄩 μ 󵄩 (e − e ) ‖η‖X ≤ [g ∗ + ϖ‖η‖∞ + ϖ 󵄩󵄩󵄩ec Da+ η󵄩󵄩󵄩∞ ] Γ(α − μ + 1)

≤ [g ∗ + ϖ‖η‖X + ϖ‖η‖X ]

(eb − ea )α−μ Γ(α − μ + 1)

62 � 3 Caputo-type fractional differential equations

≤ [g ∗ + (ϖ + ϖ)‖η‖X ]

(eb − ea )α−μ , Γ(α − μ + 1)

so ‖η‖Ω ≤

Γ(α−μ+1) (eb −ea )α−μ

g∗ − (ϖ + ϖ)

:= H3 .

Therefore, ‖η‖Ω ≤ max{H1 , H2 , H3 } := H . Lemma 3.12. If the conditions (3.13.1) and (3.43) are satisfied, then there is a bounded open set S ⊂ Ω such that K (η) − Φ(η) ≠ −ε[K (η) + Φ(−η)]

for all η ∈ 𝜕S and all ε ∈ (0, 1]. Proof. By Lemma 3.11, there exists H > 0 independent of ε, where if η satisfies K (η) − Φ(η) = −ε[K (η) + Φ(−η)], ε ∈ (0, 1], then ‖η‖Ω ≤ H . Thus, if S = {η ∈ Ω; ‖η‖Ω < F },

(3.49)

where F > H , we conclude that K (η) − Φ(η) ≠ −ε[K (η) − Φ(−η)]

for every η ∈ 𝜕S = {η ∈ Ω; ‖η‖Ω = F } and ε ∈ (0, 1]. Theorem 3.4. If the conditions (3.13.1) and (3.43) hold, then the problem (3.34)–(3.35) has at least one solution. Proof. The set S given in (3.49) is symmetric, 0 ∈ S, and Ω ∩ S = S ≠ 0. Moreover, by Lemma 3.12, if (3.13.1) is satisfied, then K (η) − Φ(η) ≠ −ε[K (η) − Φ(−η)]

for all η ∈ Ω ∩ 𝜕S = 𝜕S and all ε ∈ (0, 1]. We conclude that (3.34)–(3.35) has at least one solution. Theorem 3.5. Assume that the hypothesis (3.13.1) is satisfied. Moreover, we assume that the following hypothesis holds.

3.4 Nonlinear periodic problems with Caputo exponential fractional derivative

� 63

(3.17.1) There exist constants γ > 0 and γ ≥ 0 such that 󵄨 󵄨󵄨 󵄨󵄨g(t, η1 , η2 ) − g(t, η̄ 1 , η̄ 2 )󵄨󵄨󵄨 ≥ γ|η1 − η̄ 1 | − γ|η2 − η̄ 2 | for t ∈ T and η1 , η̄ 1 , η2 , η̄ 2 ∈ ℝ. If max{(

γ 2(ϖ + ϖ)(eb − ea )α ϖ(eb − ea )α ϖ(eb − ea )α−μ + ); ( + )} < 1, Γ(α + 1) Γ(α − μ + 1) γ Γ(α + 1)

(3.50)

then the problem (3.34)–(3.35) has a unique solution in Dom K ∩ S. Proof. Note that condition (3.50) is stronger than condition (3.43). By Theorem 3.4, the problem (3.34)–(3.35) has at least one solution in Dom K ∩ S. Now, we prove the uniqueness result. Suppose that the problem (3.34)–(3.35) has two different solutions η1 , η2 ∈ Dom K ∩ S. Then we have for each t ∈ T μ

e α c Da+ η1 (t)

= g(t, η1 (t), ec Da+ η1 (t))

e α c Da+ η2 (t)

= g(t, η2 (t), ec Da+ η2 (t)),

and μ

where η1 (a) = η1 (b),

η2 (a) = η2 (b),

and e 1 c Da+ η1 (t)(a)

= ec Da1+ η1 (t)(b),

e 1 c Da+ η2 (t)(a)

= ec Da1+ η2 (t)(b).

Let U(t) = η1 (t) − η2 (t) for all t ∈ T. Then LU(t) = ec Daα+ U(t)

= ec Daα+ η1 (t) − ec Daα+ η2 (t) μ

μ

= g(t, η1 (t), ec Da+ η1 (t)) − g(t, η2 (t), ec Da+ η2 (t)).

(3.51)

Using the fact that Img K = ker ψ2 , we have b

μ

μ

(α − 1) ∫a (eb − eϱ )α−2 [g(ϱ, η1 (ϱ), ec Da+ η1 (ϱ)) − g(ϱ, η2 (ϱ), ec Da+ η2 (ϱ))]eϱ dϱ (eb − ea )(α−1)

Since g is a continuous function, there exists t0 ∈ T such that

= 0.

64 � 3 Caputo-type fractional differential equations μ

μ

g(t0 , η1 (t0 ), ec Da+ η1 (t0 )) − g(t0 , η2 (t0 ), ec Da+ η2 (t0 )) = 0. In view of (3.17.1), we have 󵄨 γ 󵄨 γ 󵄨e μ 󵄨󵄨 e μ 󵄨󵄨η1 (t0 ) − η2 (t0 )󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨c Da+ η1 (t0 ) − c Da+ η2 (t0 )󵄨󵄨󵄨 ≤ ‖η1 − η2 ‖Ω . γ γ Then 󵄨󵄨 󵄨 γ 󵄨󵄨U(t0 )󵄨󵄨󵄨 ≤ ‖U‖Ω . γ

(3.52)

On the other hand, by Lemma 2.2, we have e α e α Ja+ (c Da+ U(t))

= U(t) − U(a) − ec Da1+ U(a)(et − ea ),

which implies that U(a) + ec Da1+ U(a)(et − ea ) = U(t0 ) − e Jaα+ (ec Daα+ U(t0 )), and therefore U(t) = e Jaα+ (ec Daα+ U(t)) − U(t0 ) + e Jaα+ (ec Daα+ U(t0 )). Using (3.52), for every t ∈ T, we obtain 󵄨󵄨 󵄨 󵄨e α e α 󵄨 󵄨 󵄨 󵄨e α e α 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨 Ja+ (c Da+ U(t))󵄨󵄨󵄨 + 󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨 Ja+ (c Da+ U(t0 ))󵄨󵄨󵄨 γ 2(eb − ea )α 󵄩󵄩e α 󵄩󵄩 ≤ ‖U‖Ω + 󵄩 D +U󵄩 . γ Γ(α + 1) 󵄩c a 󵄩∞ Then γ 2(eb − ea )α 󵄩󵄩e α 󵄩󵄩 ‖U‖∞ ≤ ‖U‖Ω + 󵄩 D +U󵄩 . γ Γ(α + 1) 󵄩c a 󵄩∞

(3.53)

By (3.51) and (3.13.1), we find that 󵄨󵄨e α 󵄨 󵄨e α 󵄨 e α 󵄨󵄨c Da+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨c Da+ η1 (t) − c Da+ η2 (t)󵄨󵄨󵄨 μ μ 󵄨 󵄨 = 󵄨󵄨󵄨g(t, η1 (t), ec Da+ η1 (t)) − g(t, η2 (t), ec Da+ η2 (t)))󵄨󵄨󵄨 ≤ (ϖ + ϖ)‖W ‖Ω .

Then 󵄩󵄩e α 󵄩󵄩 󵄩󵄩c Da+ U 󵄩󵄩∞ ≤ (ϖ + ϖ)‖U‖Ω . By (3.54) and (3.53) we get

(3.54)

3.5 Notes and remarks �

65

γ 2(ϖ + ϖ)(eb − ea )α ‖U‖∞ ≤ ( + )‖U‖Ω . γ Γ(α + 1) On the other hand, we have 󵄨 󵄨e α e α e μ 󵄨 󵄨󵄨e μ 󵄨󵄨c Da+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨c Da+ [ Ja+ (c Da+ U(t))]󵄨󵄨󵄨 μ μ 󵄨 󵄨 = 󵄨󵄨󵄨ec Daα+ [e Jaα+ (ec Da+ η1 (t))] − ec Daα+ [e Jaα+ (ec Da+ η2 (t))]󵄨󵄨󵄨 μ α−μ μ 󵄨 = 󵄨󵄨󵄨g(t, e Jaα+ (ec Da+ η1 (t)), e Ja+ (ec Da+ η1 (t))) α−μ μ μ 󵄨 − g(t, e Jaα+ (ec Da+ η2 (t)), e Ja+ (ec Da+ η2 (t)))󵄨󵄨󵄨 ≤( ≤(

ϖ(eb − ea )α ϖ(eb − ea )α−μ 󵄩󵄩e μ 󵄩󵄩 + )󵄩 D + U 󵄩 Γ(α + 1) Γ(α − μ + 1) 󵄩c a 󵄩∞

ϖ(eb − ea )α ϖ(eb − ea )α−μ + )‖U‖Ω . Γ(α + 1) Γ(α − μ + 1)

Then ϖ(eb − ea )α ϖ(eb − ea )α−μ 󵄩󵄩e μ 󵄩󵄩 + )‖U‖Ω . 󵄩󵄩c Da+ U 󵄩󵄩∞ ≤ ( Γ(α + 1) Γ(α − μ + 1)

(3.55)

Hence, by (3.50), we conclude that ‖U‖Ω = 0. As a result, for any t ∈ T we get U(t) = 0 󳨐⇒ η1 (t) = η2 (t).

3.5 Notes and remarks Using the articles [5, 66, 100], we presented and proved several results. For more relevant results and studies, one can see the monographs [7, 14, 29, 35, 49, 75, 98, 115, 143, 171] and the papers [19, 43–45, 94, 94, 110, 241, 242].

4 ψ-Caputo fractional integro-differential equations 4.1 Introduction and motivations The aim of this chapter is to prove the existence and uniqueness of periodic solutions to a large class of problems with NFDEs with ψ-Caputo fractional derivative. As always, we base our arguments on Mawhin’s coincidence degree theory. Examples are included to show the applicability of our results in each section. The outcome of our study in this chapter can be considered as a partial continuation of the problems raised recently in the following: – The monographs of Abbas et al. [7, 8, 14], Baleanu et al. [58], and Rassias et al. [194] and the papers of Afshari et al. [26, 27], Benchohra et al. [67, 69, 84–86, 97, 225], Karapinar et al. [18, 22, 48, 140, 141], and Zhou et al. [253], which deal with various linear and nonlinear initial and boundary value problems with fractional differential equations involving different types of fractional derivatives. – The monographs of Benchohra et al. [75], Graef et al. [128], and Samoilenko et al. [228] and the papers of Abbas et al. [9] and Benchohra et al. [69]. – The monograph of Abbas et al. [15] and the papers of Abbas et al. [10, 13], Benchohra et al. [77–79], and Kucche et al. [153, 166, 235]. – Beginning with the classical operators of Riemman–Liouville and Caputo, a large number of their extensions have been presented by many authors; see [43, 45, 227]. Many researchers studied the existence and uniqueness of solutions by using various methods such as fixed point theory and coincidence degree theory [65, 68, 70, 73]. Existence results of such problems with different kernels and fractional derivatives can be found in [44, 110, 231]. – In [231], the authors established the existence and uniqueness results for fractional integro-differential equations with ψ-Caputo fractional derivative using the contraction mapping principle and Schaefer’s fixed point theorem of the problem (4.1) with t ∈ [0, b] (0 < b < +∞) and the condition ny(0) + my(b) = c,

–

where n, m, c are real constants, where n + m ≠ 0. The reasoning is mainly based upon different types of classical fixed point theory. However, if n+m = 0, this method is invalid. In [243], Tidke examined the following nonlinear mixed Volterra–Fredholm integrodifferential problem: t

b

y′ (t) = f (y(t), ∫0 κ(t, s, y(s))ds, ∫0 f (t, s, y(s))ds), { y(0) + g(y) = y0 .

https://doi.org/10.1515/9783111334387-004

t ∈ [0, b],

4.2 Periodic solutions for nonlinear fractional differential equations

–

He proved the existence of solutions by means of the Leray–Schauder nonlinear alternative theorem. Anguraj et al. [50] proved the existence and uniqueness of solutions to the following fractional mixed integro-differential equations with nonlocal integral initial conditions associated with the Caputo fractional derivative of order α ∈ (0, 1]: t

α

1

d y(t) = f (t, y(t), ∫0 κ(t, s, y(s))ds, ∫0 f (t, s, y(s))ds), α { dt 1 y(0) = ∫0 g(s)y(s)ds.

–

� 67

t ∈ [0, 1],

In [155], Laadjal and Ma studied the existence and the uniqueness of solutions for the following nonlinear Volterra–Fredholm integro-differential equation of Caputo fractional order with boundary value conditions: c

{

t

t

D α y(t) = g(t) + λ1 ∫0 κ1 (t, s)f1 (s, y(s))ds + λ2 ∫0 κ2 (t, s)f2 (s, y(s))ds,

ay(0) + by(1) = 0,

where t ∈ [0, 1], α ∈ (0, 1], and a, b are real constants with a + b ≠ 0. The reasoning is mainly based upon the contraction mapping principle and Krasnoselkii’s fixed point theorem. However, if a + b = 0, this method is invalid; one cannot show the existence of periodic solutions by this method.

4.2 Periodic solutions for nonlinear fractional differential equations In this section we establish the existence and uniqueness results to the NFDE of the form c

t

α;ψ

Da+ y(t) = f (t, y(t), ∫ κ(t, s, y(s))ds),

t ∈ J,̄

(4.1)

a

y(a) = y(b),

(4.2)

α;ψ

where c Da+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, J ̄ := [a, b] (−∞ < a < b < +∞), and f : J ̄ × ℝ × ℝ → ℝ and

κ :Δ×ℝ→ℝ

are continuous functions, where Δ = {(t, s) : a ⩽ s ⩽ t ⩽ b}. For the sake of brevity, we take t

By(t) = ∫ κ(t, s, y(s))ds. a

68 � 4 ψ-Caputo fractional integro-differential equations 4.2.1 Existence results Let the spaces α;ψ

X = {y ∈ C(J,̄ ℝ) : y(t) = Ja+ υ(t) : υ ∈ C(J,̄ ℝ)}

and Y = C(J,̄ ℝ)

be endowed with the norms ‖y‖X = ‖y‖Y = ‖y‖∞ . We consider the operator L : Dom L ⊆ X → Y defined by α;ψ

Ly := c Da+ y,

(4.3)

where α;ψ

Dom L = {y ∈ X : c Da+ y ∈ Y : y(a) = y(b)}. Lemma 4.1. Using the definition of L given in (4.3), we have ker L = {y ∈ X : y(t) = y(a), t ∈ J}̄ and b

α−1

Img L = {υ ∈ Y : ∫ ψ′ (s)(ψ(b) − ψ(s))

υ(s)ds = 0}.

a α;ψ

Proof. By Remark 2.1, if for all y ∈ X the equation Ly = c Da+ y = 0 has a solution of the form y(t) = c0 = y(a),

t ∈ J,̄

then ̄ ker L = {y ∈ X : y(t) = y(a), t ∈ J}. For υ ∈ Img L, there exists y ∈ Dom L such that υ = Ly ∈ Y . Using Lemma 2.4, we obtain for every t ∈ J ̄ α;ψ

y(t) = y(a) + Ja+ υ(t)

4.2 Periodic solutions for nonlinear fractional differential equations

� 69

t

1 α−1 = y(a) + ∫ ψ′ (s)(ψ(t) − ψ(s)) υ(s)ds. Γ(α) a

Since y ∈ Dom L, we have y(a) = y(b). Thus, b

∫ ψ′ (s)(ψ(b) − ψ(s))

α−1

υ(s)ds = 0.

α−1

υ(s)ds = 0,

a

Furthermore, if υ ∈ Y and satisfies b

∫ ψ′ (s)(ψ(b) − ψ(s)) a

α;ψ

α;ψ

then for any y(t) = Ja+ υ(t), using Lemma 2.5, we get υ(t) = c Da+ y(t). Therefore, y(b) = y(a), which implies that y ∈ Dom L. So υ ∈ Img L. So b

α−1

Img L = {υ ∈ Y : ∫ ψ′ (s)(ψ(b) − ψ(s))

υ(s)ds = 0},

a

which completes the proof. Lemma 4.2. Let L be the operator defined by (4.3). Then L is a Fredholm operator of index zero and the linear continuous projector operators Q : Y → Y and P : X → X can be written as b

α α−1 Q (υ) = ∫ ψ′ (s)(ψ(b) − ψ(s)) υ(s)ds (ψ(b) − ψ(a))α a

and P (y) = y(a).

Furthermore, the operator L−1 P : Img L → X ∩ ker P can be written as α;ψ

L−1 P (υ)(t) = Ja+ υ(t),

t ∈ J.̄

Proof. Obviously, for each υ ∈ Y , Q 2 υ = Q υ and υ = Q (υ)+(υ−Q (υ)), where (υ−Q (υ)) ∈ ker Q = Img L.

70 � 4 ψ-Caputo fractional integro-differential equations Using the fact that Img L = ker Q and Q 2 = Q , we have Img Q ∩ Img L = 0. So, Y = Img L ⊕ Img Q .

In the same way we get Img P = ker L and P 2 = P . It follows for each y ∈ X that if y = (y − P (y)) + P (y), then X = ker P + ker L. Clearly, we have ker P ∩ ker L = 0. So, X = ker P ⊕ ker L.

Therefore, dim ker L = dim Img Q = codim Img L. Consequently, L is a Fredholm operator of index zero. Now, we will show that the inverse of L|Dom L∩ker P is L−1 P . Effectively, for υ ∈ Img L, by Lemma 2.5, we have α;ψ

α;ψ

c LL−1 P (υ) = Da+ (Ja+ υ) = υ.

(4.4)

Furthermore, for y ∈ Dom L ∩ ker P we get α;ψ

α;ψ

c L−1 P (L(y(t))) = Ja+ ( Da+ y(t)) = y(t) − y(a),

t ∈ J.̄

Using the fact that y ∈ Dom L ∩ ker P , we have y(a) = 0. Thus, L−1 P L(y) = y.

(4.5)

−1 Using (4.4) and (4.5) together, we get L−1 P = (L|Dom L∩ker P ) , which completes the demonstration.

Define N : X → Y by t

N y(t) := f (t, y(t), ∫ κ(t, s, y(s))ds),

t ∈ J.̄

a

It follows that the operator N is well defined, because f and κ are continuous functions. We observe that the problem (4.1)–(4.2) is equivalent to the problem Ly = N y.

4.2 Periodic solutions for nonlinear fractional differential equations

� 71

Lemma 4.3. Suppose that the following hypotheses (4.3.1) and (4.3.2) are satisfied. (4.3.1) There exist positive constants γ, η with 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ γ|y − y|̄ + η|By − By|̄ 󵄨󵄨f (t, y, B(y)) − f (t, y,̄ B(y)) for every t ∈ J ̄ and y, ȳ ∈ ℝ. (4.3.2) There exists a constant ρ > 0 such that 󵄨󵄨 󵄨 󵄨󵄨κ(t, s, υ) − κ(t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ|υ − υ|̄ for every (t, s) ∈ Δ and υ, ῡ ∈ ℝ. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps. Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ J,̄ we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

⩽

α α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(a))α a

By (4.3.1), we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

αγ α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds α (ψ(b) − ψ(a)) a

b

+

αη α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(yn )(s) − B(y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

⩽

αγ‖yn − y‖Y α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) ds α (ψ(b) − ψ(a)) a

b

+

αη α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(yn )(s) − B(y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(a))α a

Using (H2), we get

72 � 4 ψ-Caputo fractional integro-differential equations 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

αγ‖yn − y‖Y α−1 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) ds α (ψ(b) − ψ(a)) a

b

+

αηρ(b − a)‖yn − y‖Y α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(a))α a

⩽ [γ + ηρ(b − a)]‖yn − y‖Y . Thus, for each t ∈ J,̄ we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞,

and hence 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN (Ω) is bounded. For t ∈ J ̄ and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α α ⩽ (ψ(b) − ψ(a))α

a

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

α−1 󵄨

󵄨󵄨f (s, y(s), B(y)(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

b

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

αγ α−1 󵄨 󵄨 ⩽f + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α ∗

a

b

+

αη α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(y)(s) − B(0)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

αη α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(0)(s)󵄨󵄨󵄨ds α (ψ(b) − ψ(a)) a

4.2 Periodic solutions for nonlinear fractional differential equations

� 73

⩽ f ∗ + γM + η(b − a)[κ∗ + ρM ], where f ∗ = ‖f (⋅, 0, 0)‖∞ and κ∗ = sup(t,s)∈Δ |κ(t, s, 0)|. Thus, 󵄩 󵄩󵄩 ∗ ∗ 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ f + γM + η(b − a)(κ + ρM ). So, QN (Ω) is a bounded set in Y . Step 3. We prove that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ J,̄ we get L−1 P (N y(t) − QN y(t)) α;ψ

= Ja+ [f (t, y(t), By(t)) b

−

α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), By(s))ds] (ψ(b) − ψ(a))α a

t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), By(s))ds Γ(α) a

b

−

(ψ(t) − ψ(a))α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), By(s))ds. Γ(α)(ψ(b) − ψ(a))α a

For all y ∈ Ω and t ∈ J,̄ we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

1 α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

b

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

74 � 4 ψ-Caputo fractional integro-differential equations t

2f ∗ γ α α−1 󵄨 󵄨 ⩽ (ψ(b) − ψ(a)) + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds αΓ(α) Γ(α) a

t

+

η α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B(y)(s) − B(0)(s)󵄨󵄨󵄨ds Γ(α) a

t

+

η α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B(0)(s)󵄨󵄨󵄨ds Γ(α) a

b

γ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) a

b

+

η α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(y)(s) − B(0)(s)󵄨󵄨󵄨ds Γ(α) a

b

η α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B(0)(s)󵄨󵄨󵄨ds Γ(α) a

2(ψ(b) − ψ(a))α ∗ ⩽ [f + κ∗ (b − a)η + γM + ηρ(b − a)M ]. Γ(α + 1) Therefore, 2(ψ(b) − ψ(a))α 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X ⩽ Γ(α + 1)

× [f ∗ + κ∗ (b − a)η + γM + ηρ(b − a)M ].

This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. For a ⩽ t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (id − Q )N y(t2 ) − LP (id − Q )N y(t1 )󵄨󵄨󵄨 t1

⩽

1 α−1 α−1 󵄨 󵄨 ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨 Γ(α) a

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨]ds t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds Γ(α) t1

+

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] Γ(α)(ψ(b) − ψ(a))α b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds 󵄨 󵄨

4.2 Periodic solutions for nonlinear fractional differential equations

� 75

t1

1 α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

−

t1

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

t1

1 α−1 α−1 󵄨 󵄨 + ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ]󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] + Γ(α)(ψ(b) − ψ(a))α b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

+

α−1 󵄨

󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] Γ(α)(ψ(b) − ψ(a))α b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

α

α−1 󵄨

󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨 α

α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) + Λ[(ψ(t1 ) − ψ(a)) − (ψ(t2 ) − ψ(a)) ] α

α

+ Λ[(ψ(t2 ) − ψ(a)) − (ψ(t1 ) − ψ(a)) ], where Λ=

1 (f ∗ κ∗ + γM + η(b − a)(ρM + κ∗ )). Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 4.4. Assume that (4.3.1) and (4.3.2) hold. If the condition γ + ηρ(b − a) 1 α (ψ(b) − ψ(a)) < Γ(α + 1) 2

(4.6)

76 � 4 ψ-Caputo fractional integro-differential equations is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζL(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ J ̄ α;ψ

Ly(t) = c Da+ y(t) =

1 ζ f (t, y(t), By(t)) − f (t, −y(t), B(−y)(t)). 1+ζ 1+ζ

By Lemma 2.4 we get y(t) = y(a) +

1 α;ψ [J + (f (s, y(s), By(s)))(t) ζ +1 a

α;ψ

− ζ Ja+ (f (s, −y(s), B(−y)(s)))(t)]. Thus, for every t ∈ J ̄ we obtain 󵄨 󵄨 |y(t)| ⩽ 󵄨󵄨󵄨y(a)󵄨󵄨󵄨 +

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds (ζ + 1)Γ(α) t

+

a

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, −y(s), B(−y)(s))󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

1 󵄨 󵄨 ⩽ 󵄨󵄨󵄨y(a)󵄨󵄨󵄨 + (ζ + 1)Γ(α) t

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) a

+

󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α)

1 + (ζ + 1)Γ(α) t

a

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) a

󵄨󵄨f (s, −y(s), B(−y)(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

4.2 Periodic solutions for nonlinear fractional differential equations

� 77

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

∗ ∗ α 󵄨 2(f + η(b − a)κ ) 󵄨 ⩽ 󵄨󵄨󵄨y(a)󵄨󵄨󵄨 + (ψ(b) − ψ(a)) Γ(α + 1)

+

2(γ + ηρ(b − a)) α (ψ(b) − ψ(a)) ‖y‖X . Γ(α + 1)

Thus, ∗ ∗ α 󵄨 󵄨 2(f + η(b − a)κ ) ‖y‖X ⩽ 󵄨󵄨󵄨y(a)󵄨󵄨󵄨 + (ψ(b) − ψ(a)) Γ(α + 1) 2(γ + ηρ(b − a)) α + (ψ(b) − ψ(a)) ‖y‖X . Γ(α + 1)

We deduce that ‖y‖X ⩽

2(f ∗ +η(b−a)κ∗ ) (ψ(b) − ψ(a))α Γ(α+1) 2(γ+ηρ(b−a)) (ψ(b) − ψ(a))α ] Γ(α+1)

|y(a)| + [1 −

:= A .

The demonstration is completed. Lemma 4.5. If conditions (4.3.1), (4.3.2), and (4.6) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(4.7)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 4.4, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(4.8)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 4.1. Assume that (4.3.1), (4.3.2), and (4.6) hold. Then there exists at least one solution for the problem (4.1)–(4.2).

78 � 4 ψ-Caputo fractional integro-differential equations Proof. It is clear that the set Ω defined in (4.8) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 4.5, assume that (4.3.1), (4.3.2), and (4.6) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (4.1)–(4.2) has at least one solution on Dom L ∩ Ω, which completes the demonstration.

4.2.2 Examples Example 1. Consider the following problem: c

1

;2t

D02+ y(t) = e

t+2

+

1 (t sin(y(t)) − y(t) cos(t)) 4√π

t

+

t 2 y(s) 1 ds, ∫ −s+2 17 3e (1 + |y(s)|)

t ∈ [0, 1],

(4.9)

0

y(0) = y(1).

(4.10)

Set f (t, y(t), By(t)) = et+2 +

1 1 (t sin(y(t)) − y(t) cos(t)) + By(t), 17 4√π

t ∈ [0, 1],

with t

t

By(t) = ∫ κ(t, s, y(s))ds = ∫ 0

0

t 2 y(s) ds, + |y(s)|)

3e−s+2 (1

t ∈ [0, 1].

We have α=

1 2

and ψ(t) = 2t .

It is clear that the function f ∈ C([0, 1], ℝ). Then for 0 ≤ s ≤ t ≤ 1 and y, υ : [0, 1] 󳨀→ ℝ, we obtain 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨f (t, y(t), By(t)) − f (t, υ(t), Bυ(t))󵄨󵄨󵄨 ⩽ 󵄨y(t) − υ(t)󵄨󵄨󵄨 2√π 󵄨 1󵄨 󵄨 + 󵄨󵄨󵄨By(t) − Bυ(t)󵄨󵄨󵄨 17 and 󵄨󵄨 󵄨 1󵄨 󵄨 󵄨󵄨κ(t, s, y(s)) − κ(t, s, υ(s))󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨y(s) − υ(s)󵄨󵄨󵄨, 3e

4.2 Periodic solutions for nonlinear fractional differential equations

� 79

which implies that (4.3.1) and (4.3.2) are satisfied with γ=

1 , 2√π

1 , 17

η=

ρ=

1 . 3e

Further, by some simple calculations, we see that γ + ηρ(b − a) 1 α (ψ(b) − ψ(a)) ≈ 0.3264 < . Γ(α + 1) 2 Theorem 4.1 implies that problem (4.9)–(4.10) has at least one solution. Example 2. Consider the following problem with NFDE: c

1

;ln(t)

D12+

t

y(t) =

2 e−2−t 1 + ∫ t 3 e−6−s sin(y(s))ds, 10(1 + y(t)) 3√π

t ∈ [1, e],

(4.11)

1

y(1) = y(e).

(4.12)

Set t

f (t, y(t), By(t)) =

2 e−2−t 1 + ∫ t 3 e−6−s sin(y(s))ds, 10(1 + y(t)) 3√π

t ∈ [1, e],

1

with t

t

1

1

2

By(t) = ∫ κ(t, s, y(s))ds = ∫ t 3 e−6−s sin(y(s))ds,

t ∈ [1, e].

We have α=

1 2

and

ψ(t) = ln t.

It is clear that the function f ∈ C([1, e], ℝ). Then for 1 ≤ s ≤ t ≤ e and y, υ : [1, e] 󳨀→ ℝ, we obtain 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨f (t, y(t), By(t)) − f (t, υ(t), Bυ(t))󵄨󵄨󵄨 ⩽ 󵄨y(t) − υ(t)󵄨󵄨󵄨 10e3 󵄨 1 󵄨󵄨 󵄨 + 󵄨By(t) − Bυ(t)󵄨󵄨󵄨 3√π 󵄨 and 󵄨 󵄨 󵄨󵄨 −7 󵄨 󵄨󵄨κ(t, s, y(s)) − κ(t, s, υ(s))󵄨󵄨󵄨 ⩽ e 󵄨󵄨󵄨y(s) − υ(s)󵄨󵄨󵄨, which implies that (4.3.1) and (4.3.2) are satisfied with

80 � 4 ψ-Caputo fractional integro-differential equations γ=

1 , 10e3

η=

1 , 3√π

ρ = e−7 .

Further, by some simple calculations, we see that γ + ηρ(b − a) 1 α (ψ(b) − ψ(a)) ≈ 0.1386 < . Γ(α + 1) 2 With the use of Theorem 4.1, the problem (4.11)–(4.12) has at least one solution.

4.3 Nonlinear Volterra–Fredholm integro-differential equations In this section we consider the following nonlinear Volterra–Fredholm integrodifferential equation: c

α;ψ Da+ y(t)

t

b

= f (t, y(t), ∫ κ1 (t, s, y(s))ds, ∫ κ2 (t, s, y(s))ds), a

t ∈ J,̄

(4.13)

a

with the periodic conditions y(a) = y(b),

(4.14)

α;ψ

where J ̄ := [a, b] (−∞ < a < b < +∞), c Da+ denotes the ψ-Caputo derivative of fractional order 0 < α ⩽ 1, and f : J ̄ × ℝ × ℝ × ℝ → ℝ,

κ1 : Δ × ℝ → ℝ,

and κ2 : Δ0 × ℝ → ℝ

are continuous functions with Δ = {(t, s) : a ⩽ s ⩽ t ⩽ b} and Δ0 = J ̄ × J.̄ For the sake of brevity, we take b

t

B1 y(t) = ∫ κ1 (t, s, y(s))ds a

and

B2 y(t) = ∫ κ2 (t, s, y(s))ds.

(4.15)

a

4.3.1 Main results In this chapter, we will adopt the spaces, lemmas, and any auxiliary results given in Section 4.2.1. We will be using specifically Lemma 4.1 and Lemma 4.2. Let the spaces α;ψ

X = {y ∈ C(J,̄ ℝ) : y(t) = Ja+ υ(t) : υ ∈ C(J,̄ ℝ)}

and

4.3 Nonlinear Volterra–Fredholm integro-differential equations

� 81

Y = C(J,̄ ℝ)

be endowed with the norms ‖y‖X = ‖y‖Y = ‖y‖∞ . We consider the operator L : Dom L ⊆ X → Y defined by α;ψ

Ly := c Da+ y,

(4.16)

where α;ψ

Dom L = {y ∈ X : c Da+ y ∈ Y : y(a) = y(b)}. Define N : X → Y by t

b

N y(t) := f (t, y(t), ∫ κ1 (t, s, y(s))ds, ∫ κ2 (t, s, y(s))ds), a

t ∈ J.̄

a

The operator N is well defined, because f , κ1 , and κ2 are continuous functions. We remark that the problem (4.13)–(4.14) is equivalent to the problem Ly = N y. Lemma 4.6. Suppose that the following hypotheses (4.7.1), (4.7.2), and (4.7.3) are satisfied. (4.7.1) There exist positive constants γ, η1 , η2 with 󵄨󵄨 ̄ B2 (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, B1 (y), B2 (y)) − f (t, y,̄ B1 (y), ⩽ γ|y − y|̄ + η1 |B1 y − B1 y|̄ + η2 |B2 y − B2 y|̄ for every t ∈ J ̄ and y, ȳ ∈ ℝ. (4.7.2) There exists a constant ρ1 > 0 such that 󵄨󵄨 󵄨 󵄨󵄨κ1 (t, s, υ) − κ1 (t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ1 |υ − υ|̄ for every (t, s) ∈ Δ and υ, ῡ ∈ ℝ. (4.7.3) There exists a constant ρ2 > 0 such that 󵄨󵄨 󵄨 󵄨󵄨κ2 (t, s, υ) − κ2 (t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ2 |υ − υ|̄ for every (t, s) ∈ Δ0 and υ, ῡ ∈ ℝ. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps:

82 � 4 ψ-Caputo fractional integro-differential equations Step 1. We show that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ J,̄ we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(a))α a

By (4.7.1), we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

αγ α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

+

αη1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (yn )(s) − B1 (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

αη2 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (yn )(s) − B2 (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

⩽

αγ‖yn − y‖Y α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(a))α a

b

αη1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (yn )(s) − B1 (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

+

αη2 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (yn )(s) − B2 (y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(a))α a

Using (4.7.2) and (4.7.3), we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

⩽

αγ‖yn − y‖Y α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(a))α a

b

α(b − a)(η1 ρ1 + η2 ρ2 )‖yn − y‖Y α−1 + ∫ ψ′ (s)(ψ(b) − ψ(s)) ds α (ψ(b) − ψ(a)) a

⩽ [γ + (b − a)(η1 ρ1 + η2 ρ2 )]‖yn − y‖Y . Thus, for each t ∈ J,̄ we obtain

4.3 Nonlinear Volterra–Fredholm integro-differential equations

󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 󳨀→ 0

� 83

as n 󳨀→ +∞.

Therefore, 󵄩 󵄩󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We demonstrate that QN (Ω) is bounded. For t ∈ J ̄ and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α α ⩽ (ψ(b) − ψ(a))α

a

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

α−1 󵄨

󵄨󵄨f (s, y(s), B1 (y)(s), B2 (y)(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

b

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

⩽ f∗ +

b

αγ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

+

αη1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (y)(s) − B1 (0)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

+

αη1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (0)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

αη2 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (y)(s) − B2 (0)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(a))α a

b

αη2 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (0)(s)󵄨󵄨󵄨ds α (ψ(b) − ψ(a)) a

∗

⩽ f + γM + (b − a)[(ρ1 η1 + ρ2 η2 )M + κ1∗ η1 + κ2∗ η2 ], where 󵄩 󵄩 f ∗ = 󵄩󵄩󵄩f (⋅, 0, 0, 0)󵄩󵄩󵄩∞ ,

󵄨 󵄨 κ1∗ = sup 󵄨󵄨󵄨κ(t, s, 0, 0)󵄨󵄨󵄨, (t,s)∈Δ

84 � 4 ψ-Caputo fractional integro-differential equations and 󵄨 󵄨 κ2∗ = sup 󵄨󵄨󵄨κ(t, s, 0, 0)󵄨󵄨󵄨. (t,s)∈Δ0

Thus, 󵄩 󵄩󵄩 ∗ ∗ ∗ 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ f + γM + (b − a)[(ρ1 η1 + ρ2 η2 )M + κ1 η1 + κ2 η2 ]. So, QN (Ω) is a bounded set in Y . Step 3. We prove that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ ̄ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ J, we get L−1 P (N y(t) − QN y(t)) α;ψ

= Ja+ [f (t, y(t), B1 y(t), B2 y(t)) b

α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), B1 y(s), B2 y(s))ds] α (ψ(b) − ψ(a)) a

t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), B1 y(s), B2 y(s))ds Γ(α) a

−

(ψ(t) − ψ(a))α Γ(α)(ψ(b) − ψ(a))α b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

α−1

a

f (s, y(s), B1 y(s), B2 y(s))ds.

For all y ∈ Ω and t ∈ J,̄ we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

a

+

+

󵄨󵄨f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨 ds 󵄨 󵄨 Γ(α)

α−1 󵄨

⩽ ∫ ψ′ (s)(ψ(t) − ψ(s)) t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds Γ(α) a b ′ ∫a ψ (s)(ψ(b)

− ψ(s))α−1 |f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)|ds Γ(α)

4.3 Nonlinear Volterra–Fredholm integro-differential equations b

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) a

� 85

󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨 ds 󵄨 Γ(α) 󵄨

α−1 󵄨

t

2f ∗ γ α α−1 󵄨 󵄨 ⩽ (ψ(b) − ψ(a)) + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds αΓ(α) Γ(α) t

+

a

η1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B1 (y)(s) − B1 (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

η α−1 󵄨 󵄨 + 1 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B1 (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

η α−1 󵄨 󵄨 + 2 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B2 (y)(s) − B2 (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

+

η2 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨B2 (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

γ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) a

b

+

η1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (y)(s) − B1 (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

η α−1 󵄨 󵄨 + 1 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B1 (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

+

η2 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (y)(s) − B2 (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

η α−1 󵄨 󵄨 + 2 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨B2 (0)(s)󵄨󵄨󵄨ds Γ(α) a

2(ψ(b) − ψ(a))α ∗ ⩽ [f + γM + (b − a)[κ1∗ η1 + κ2∗ η2 + (η1 ρ1 + η2 ρ2 )M ]]. Γ(α + 1) Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X 2(ψ(b) − ψ(a))α ∗ ⩽ [f + γM + (b − a)[κ1∗ η1 + κ2∗ η2 + (η1 ρ1 + η2 ρ2 )M ]]. Γ(α + 1) This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous.

86 � 4 ψ-Caputo fractional integro-differential equations For a ⩽ t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (id − Q )N y(t2 ) − LP (id − Q )N y(t1 )󵄨󵄨󵄨 t1

1 α−1 α−1 󵄨 󵄨 ⩽ ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨 Γ(α) a

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), B1 y(s), B2 y(s))󵄨󵄨󵄨]ds t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), B1 y(s), B2 y(s))󵄨󵄨󵄨ds Γ(α) t1

+

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] Γ(α)(ψ(b) − ψ(a))α b

α−1 󵄨

󵄨󵄨f (s, y(s), B1 y(s), B2 y(s))󵄨󵄨󵄨ds 󵄨 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

⩽

a t1 ′ ∫a ψ (s)(ψ(t1 )

− +

− ψ(s))α−1 |f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)|ds

t ∫a1

ψ (s)(ψ(t2 ) − ψ(s))

t ∫a1

ψ (s)[(ψ(t1 ) − ψ(s))

α−1

′

|f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)|ds

α−1

′

Γ(α)

− (ψ(t2 ) − ψ(s))α−1 ]|f (s, 0, 0, 0)|ds Γ(α)

t2

+

Γ(α)

|f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)| 1 ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α t1

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] + Γ(α)(ψ(b) − ψ(a))α b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨 b

[(ψ(t2 ) − ψ(a))α − (ψ(t1 ) − ψ(a))α ] |f (s, 0, 0, 0)| + ds ∫ ψ′ (s) Γ(α)(ψ(b) − ψ(a))α (ψ(b) − ψ(s))1−α a α

α

α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) + Λ[(ψ(t1 ) − ψ(a)) − (ψ(t2 ) − ψ(a)) ] α

α

+ Λ[(ψ(t2 ) − ψ(a)) − (ψ(t1 ) − ψ(a)) ] α

= 2Λ(ψ(t2 ) − ψ(t1 )) ,

4.3 Nonlinear Volterra–Fredholm integro-differential equations

� 87

where Λ :=

1 [f ∗ + γM + (b − a)[κ1∗ η1 + κ2∗ η2 + (η1 ρ1 + η2 ρ2 )M ]]. Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 4.7. Assume that (4.7.1), (4.7.2), and (4.7.3) hold. If the condition γ + (η1 ρ1 + η2 ρ2 )(b − a) 1 α (ψ(b) − ψ(a)) < Γ(α + 1) 2

(4.17)

is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζL(y) − ζ N (−y). Then L(y) =

ζ 1 N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ J ̄ α;ψ

Ly(t) = c Da+ y(t) =

1 f (t, y(t), B1 y(t), B2 y(t)) 1+ζ ζ f (t, −y(t), B1 (−y)(t), B2 (−y)(t)). − 1+ζ

By Theorem 2.4 we get y(t) = c0 +

1 α;ψ [J + (f (s, y(s), B1 y(s), B2 y(s)))(t) ζ +1 a α;ψ

− ζ Ja+ (f (s, −y(s), B1 (−y)(s), B2 (−y)(s)))(t)], where c0 = y(a). Thus, for every t ∈ J ̄ we obtain |y(t)| ⩽ |c0 | +

1 (ζ + 1)Γ(α)

88 � 4 ψ-Caputo fractional integro-differential equations t

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) a

+

1 (ζ + 1)Γ(α) t

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) a

󵄨󵄨f (s, −y(s), B1 (−y)(s), B2 (−y)(s))󵄨󵄨󵄨ds 󵄨 󵄨

1 (ζ + 1)Γ(α)

⩽ |c0 | + t

× ∫ ψ′ (s) a

+

󵄨󵄨f (s, y(s), B1 y(s), B2 y(s))󵄨󵄨󵄨ds 󵄨 󵄨

|f (s, y(s), B1 y(s), B2 y(s)) − f (s, 0, 0, 0)| ds (ψ(t) − ψ(s))1−α t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

t

+

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) (ζ + 1)Γ(α) a

󵄨 󵄨 × 󵄨󵄨󵄨f (s, −y(s), B1 (−y)(s), B2 (−y)(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

2(f ∗ + (η1 κ1∗ + η2 κ2∗ )(b − a)) α (ψ(b) − ψ(a)) ⩽ |c0 | + Γ(α + 1) 2(γ + (η1 ρ1 + η2 ρ2 )(b − a)) α + (ψ(b) − ψ(a)) ‖y‖X . Γ(α + 1) Thus, 2(f ∗ + (η1 κ1∗ + η2 κ2∗ )(b − a)) α (ψ(b) − ψ(a)) Γ(α + 1) 2(γ + (η1 ρ1 + η2 ρ2 )(b − a)) α (ψ(b) − ψ(a)) ‖y‖X . + Γ(α + 1)

‖y‖X ⩽ |c0 | +

We deduce that ‖y‖X ⩽

|c0 | + [1 −

2(f ∗ +(η1 κ1∗ +η2 κ2∗ )(b−a)) (ψ(b) − ψ(a))α Γ(α+1) 2(γ+(η1 ρ1 +η2 ρ2 )(b−a)) (ψ(b) − ψ(a))α ] Γ(α+1)

:= A .

The demonstration is completed. Lemma 4.8. If conditions (4.7.1), (4.7.2), (4.7.3), and (4.17) are satisfied, then there exists a bounded open set Ω ⊂ X with

4.3 Nonlinear Volterra–Fredholm integro-differential equations

� 89

L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(4.18)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 4.7, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(4.19)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 4.2. Assume that (4.7.1), (4.7.2), (4.7.3), and (4.17) hold. Then there exists at least one solution for the problem (4.13)–(4.14). Proof. It is clear that the set Ω defined in (4.19) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 4.8, if (4.7.1), (4.7.2), (4.7.3), and (4.17) hold, then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (4.13)–(4.14) has at least one solution on Dom L ∩ Ω, which completes the demonstration. Now, we investigate the existence and uniqueness of periodic solutions for our problem (4.13)–(4.14). Theorem 4.3. Let (4.7.1), (4.7.2), and (4.7.3) be satisfied. Moreover, we assume that: (4.11.1) There exist constants γ > 0 and η1 , η2 ⩾ 0 such that 󵄨󵄨 ̄ B2 (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, B1 (y), B2 (y)) − f (t, y,̄ B1 (y),

⩾ γ|y − y|̄ − η1 |B1 y − B1 y|̄ − η2 |B2 y − B2 y|̄

for every t ∈ J ̄ and y, ȳ ∈ ℝ. If one has (η1 ρ1 + η2 ρ2 )(b − a) 2[γ + (η1 ρ1 + η2 ρ2 )(b − a)] α + (ψ(b) − ψ(a)) < 1, γ Γ(α + 1) then the problem (4.13)–(4.14) has a unique solution in Dom L ∩ Ω.

(4.20)

90 � 4 ψ-Caputo fractional integro-differential equations Proof. Note that condition (4.20) is stronger than condition (4.17). By Theorem 4.2, the problem (4.13)–(4.14) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (4.13)–(4.14) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ J ̄ c c

α;ψ

Da+ y1 (t) = f (t, y1 (t), B1 (y1 )(t), B2 (y1 )(t)), α;ψ

Da+ y2 (t) = f (t, y2 (t), B1 (y2 )(t), B2 (y2 )(t)),

where B1 , B2 are defined as in (4.15) and y1 (a) = y1 (b),

y2 (a) = y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ J.̄ Then α;ψ

LU(t) = c Da+ U(t) α;ψ

α;ψ

= c Da+ y1 (t) − c Da+ y2 (t)

= f (t, y1 (t), B1 (y1 )(t), B2 (y1 )(t)) − f (t, y2 (t), B1 (y2 )(t), B2 (y2 )(t)).

(4.21)

Using the fact that Img L = ker Q , we have b

∫ a

ψ′ (s)[f (s, y1 (s), B1 (y1 )(s), B2 (y1 )(s)) − f (s, y2 (s), B1 (y2 )(s), B2 (y2 )(s))]ds = 0. (ψ(b) − ψ(s))1−α

Since f is a continuous function, there exists t0 ∈ [a, b] such that f (t0 , y1 (t0 ), B1 (y1 )(t0 ), B2 (y1 )(t0 )) − f (t, y2 (t0 ), B1 (y2 )(t0 ), B2 (y2 )(t0 )) = 0. In view of (4.11.1) we have 󵄨󵄨 󵄨 (b − a)(η1 ρ1 + η2 ρ2 ) ‖y1 − y2 ‖X . 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ γ Then 󵄨󵄨 󵄨 (b − a)(η1 ρ1 + η2 ρ2 ) ‖U‖X . 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ γ On the other hand, by Theorem 2.4, we have α;ψ c

α;ψ

Ja+ Da+ U(t) = U(t) − U(a),

which implies that

(4.22)

4.3 Nonlinear Volterra–Fredholm integro-differential equations

α;ψ

� 91

α;ψ

U(a) = U(t0 ) − Ja+ c Da+ U(t0 ), and therefore α;ψ

α;ψ

α;ψ

α;ψ

U(t) = Ja+ c Da+ U(t) + U(t0 ) − Ja+ c Da+ U(t0 ). Using (4.22) we obtain for every t ∈ J ̄ 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨Ja+ Da+ U(t)󵄨󵄨󵄨 + 󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨Ja+ Da+ U(t0 )󵄨󵄨󵄨 (b − a)(η1 ρ1 + η2 ρ2 ) 2(ψ(b) − ψ(a))α 󵄩󵄩c α;ψ 󵄩󵄩 ⩽ ‖U‖X + 󵄩󵄩 Da+ U 󵄩󵄩X . γ Γ(α + 1)

(4.23)

By the hypotheses (4.7.1), (4.7.2), and (4.7.3) and the condition (4.21), we find 󵄨󵄨c α;ψ 󵄨 󵄨 󵄨󵄨 Da+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (t), B1 (y1 )(t), B2 (y1 )(t)) 󵄨 − f (t, y2 (t), B1 (y2 )(t), B2 (y2 )(t))󵄨󵄨󵄨

⩽ [γ + (b − a)(η1 ρ1 + η2 ρ2 )]‖U‖X .

Then 󵄩󵄩c α;ψ 󵄩󵄩 󵄩󵄩 Da+ U 󵄩󵄩X ⩽ [γ + (b − a)(η1 ρ1 + η2 ρ2 )]‖U‖X . Substituting (4.24) in the right side of (4.23), we get for every t ∈ J ̄ |U(t)| ⩽ [

(η1 ρ1 + η2 ρ2 )(b − a) γ

2[γ + (η1 ρ1 + η2 ρ2 )(b − a)] α (ψ(b) − ψ(a)) ]‖U‖X . Γ(α + 1)

+ Therefore, ‖U‖X ⩽ [

(η1 ρ1 + η2 ρ2 )(b − a) γ

+

2[γ + (η1 ρ1 + η2 ρ2 )(b − a)] α (ψ(b) − ψ(a)) ]‖U‖X . Γ(α + 1)

Hence, by (4.20), we conclude that ‖U‖X = 0. As a result, for any t ∈ J,̄ we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof.

(4.24)

92 � 4 ψ-Caputo fractional integro-differential equations 4.3.2 Examples In this section, we illustrate the applicability of Theorems 4.2 and 4.3 through two examples. Example 4.1. Consider the following problem with nonlinear Volterra–Fredholm integro-differential fractional equations: c

1

;2t

D02+ y(t) = f (t, y(t), B1 y(t), B2 y(t)),

t ∈ [0, 1],

y(0) = y(1),

(4.25) (4.26)

where f (t, y(t), B1 y(t), B2 y(t)) = et+1 +

t 1 1 󵄨 󵄨 ln(󵄨󵄨󵄨y(t)󵄨󵄨󵄨 + 2) + B1 y(t) + B2 y(t), 17 27 3√π

with t

t

B1 y(t) = ∫ κ1 (t, s, y(s))ds = ∫ 0

0

1 sin y(s)ds, t 2 + 5e2

t ∈ [0, 1],

y(s) ds, 3et+3 (1 + y(s))

t ∈ [0, 1].

and 1

1

B2 y(t) = ∫ κ2 (t, s, y(s))ds = ∫ 0

0

We have α=

1 2

and ψ(t) = 2t .

It is clear that the function f ∈ C([0, 1] × ℝ3 , ℝ). Then for all t ∈ [0, 1] and y, ȳ ∈ ℝ, we obtain 󵄨󵄨 ̄ B2 (y)) ̄ 󵄨󵄨󵄨󵄨 ⩽ γ|y − y|̄ + η1 |B1 y − B1 y|̄ 󵄨󵄨f (t, y, B1 (y), B2 (y)) − f (t, y,̄ B1 (y), + η2 |B2 y − B2 y|̄ and 󵄨󵄨 󵄨 ̄ 󵄨󵄨κ1 (t, s, y) − κ1 (t, s, y)̄ 󵄨󵄨󵄨 ⩽ ρ1 |y − y|, 󵄨󵄨 󵄨 󵄨 ̄ 󵄨󵄨κ2 (t, s, y) − κ2 (t, s, y)̄ 󵄨󵄨 ⩽ ρ2 |y − y|,

(t, s) ∈ Δ, (t, s) ∈ Δ0 ,

with Δ = {(t, s) : 0 ⩽ s ⩽ t ⩽ 1} and Δ0 = [0, 1] × [0, 1], which implies that (4.7.1), (4.7.2), and (4.7.3) are satisfied with

4.3 Nonlinear Volterra–Fredholm integro-differential equations

γ=

1 , 6√π

η1 =

1 , 17

η2 =

1 , 27

ρ1 =

1 , 5e2

and ρ2 =

� 93

1 . 3e3

Further, by some simple calculations, we see that γ + (η1 ρ1 + η2 ρ2 )(b − a) 1 α (ψ(b) − ψ(a)) ≈ 0.10859 < . Γ(α + 1) 2 Theorem 4.2 implies that problem (4.25)–(4.26) has at least one solution. Example 4.2. Consider the following problem with nonlinear Volterra–Fredholm integro-differential fractional equations: c

1

;ln t

D12+

y(t) = f (t, y(t), B1 y(t), B2 y(t)),

t ∈ J ̄ := [1, e],

y(1) = y(e),

where f (t, y(t), B1 y(t), B2 y(t)) =

1 1 3 1 + 2 (cos y(t) + y(t)) + B1 y(t) 2 + 2 3e 7√π 1 + B2 y(t), 23 t2

with t

t

1

1

2

B1 y(t) = ∫ κ1 (t, s, y(s))ds = ∫ t 3 e−6−t sin(y(s))ds,

t ∈ [1, e],

and e

e

B2 y(t) = ∫ κ2 (t, s, y(s))ds = ∫ 1

1

e−11−t ds, 37(1 + y(s))

t ∈ [1, e].

Here α = 21 and ψ(t) = ln t. It is easy to see that f ∈ C([1, e] × ℝ3 , ℝ). Then for all t ∈ [1, e] and y, ȳ ∈ ℝ, we obtain 󵄨󵄨 ̄ B2 (y)) ̄ 󵄨󵄨󵄨󵄨 ⩽ γ|y − y|̄ + η1 |B1 y − B1 y|̄ 󵄨󵄨f (t, y, B1 (y), B2 (y)) − f (t, y,̄ B1 (y), ̄ + η2 |B2 y − B2 y|, 󵄨󵄨󵄨κ1 (t, s, y) − κ1 (t, s, y)̄ 󵄨󵄨󵄨 ⩽ ρ1 |y − y|, ̄ (t, s) ∈ Δ, 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 ̄ (t, s) ∈ Δ0 , 󵄨󵄨κ2 (t, s, y) − κ2 (t, s, y)̄ 󵄨󵄨 ⩽ ρ2 |y − y|, and 󵄨󵄨 ̄ B2 (y)) ̄ 󵄨󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η1 |B1 y − B1 y|̄ 󵄨󵄨f (t, y, B1 (y), B2 (y)) − f (t, y,̄ B1 (y),

94 � 4 ψ-Caputo fractional integro-differential equations ̄ − η2 |B2 y − B2 y|, with Δ = {(t, s) : 1 ⩽ s ⩽ t ⩽ e} and Δ0 = [1, e] × [1, e], which implies that (4.7.1), (4.7.2), (4.7.3), and (4.11.1) are satisfied with γ=

5 , 6e2

γ=

1 , 6e2

η1 = η1 =

1 , 7√π

η2 = η2 =

1 , 23

ρ1 =

1 , e7

and

ρ2 =

1 . 37e12

By simple calculations, we see that (η1 ρ1 + η2 ρ2 )(b − a) 2[γ + (η1 ρ1 + η2 ρ2 )(b − a)] α + (ψ(b) − ψ(a)) ≈ 0.2604 γ Γ(α + 1) < 1. So, by Theorem 4.3, our problem has a unique solution.

4.4 Notes and remarks This chapter’s contents are based on the articles [122, 124]. The monographs [7, 8, 14, 35, 58, 138, 170, 194, 238, 251, 252] and the papers [90, 94, 100, 117, 118, 120, 122, 124, 147, 149– 152, 157, 173, 176] provide more important conclusions and analyses.

5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative 5.1 Introduction and motivations The present chapter deals with some existence and uniqueness results for periodic solutions for a certain type of nonlinear fractional pantograph differential and integrodifferential equations with ψ-Caputo derivative. The proofs are based on Mawhin’s coincidence degree theory. We provide illustrations to demonstrate the applicability of our results in each section. The outcome of our study in this chapter can be considered as a partial continuation of the problems raised recently in the following: – The monographs of Abbas et al. [7, 8, 14], Ahmed et al. [32], and Baleanu et al. [58] and the papers of Ahmed et al. [36], which deal with various linear and nonlinear initial and boundary value problems for fractional differential equations involving different kinds of fractional derivatives. – The monographs of Abbas et al. [7], Agarwal et al. [30], and Benchohra et al. [75], the papers of Abbas et al. [1–4, 6, 8, 17], Bai et al. [52], Hernández et al. [134], Kong et al. [146], and Wang et al. [247, 248], where the authors investigated the class of problems with fractional differential equations with impulsive conditions, and the books [116, 246], where the qualitative properties of solutions are considered. – The works [14, 15, 31, 227], where many definitions and results about fractional derivative and integral operators were generalized. Ricardo Almeida [43, 45] introduced a new generalized fractional derivative (the ψ-Caputo fractional derivative), and therefore some recent work on the existence and uniqueness of solutions for NFDEs with ψ-Caputo fractional derivative can be found in [44, 110]. – The pantograph equations have been widely used in the fields of quantum mechanics and dynamical systems [190, 198]. Actually, several researchers have investigated some new existence and uniqueness results for NFDE pantograph problems by applying fixed point theorems or the coincidence degree theorem [19, 56, 65, 68, 70, 73]. – In 2017, Jalilian and Ghasmi [137] studied an initial value problem with pantographtype nonlinear fractional integro-differential equations of the following form: qt

c α D u(t) = f (t, u(t), u(pt)) + ∫0 g1 (t, s, u(s))ds { { { t + ∫0 g2 (t, s, u(s))ds, t ∈ [0, T], { { { {u(0) = u0 ,

where 0 < p, q < 1. The symbol c Dα denotes the Caputo fractional derivative of order α ∈ (0, 1]. They obtained their results by using the fixed point approaches.

https://doi.org/10.1515/9783111334387-005

96 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative –

In [231], the authors studied the following nonlinear fractional pantograph equations with nonlocal boundary conditions by applying fixed point theory: c w,ψ

{

–

D

u(t) = f (t, u(t), u(ηt)),

t ∈ [0, T], η ∈ (0, 1),

au(0) + bu(T) = c,

where c Dw,ψ is the ψ-Caputo factional derivative of order 0 < w < 1 and a, b, c are real constants, where a + b ≠ 0. However, if a + b = 0, this technique is invalid. In [51], the authors discussed the existence and uniqueness of solutions for the following equation with nonlocal conditions: α;ρ c D y(t) = f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)), { y(0) − I β y(t),

–

t ∈ [0, T], 0 < t < T,

where c Dα;ρ denotes the Katugampola fractional derivative in Caputo sense of order α ∈ (0, 1], p ∈ (0, 1), ρ > 0, and I β is the integral operator of order β > 0. The famous delay differential equation is the pantograph-type equation that has been introduced in 1971 by Ockendon and Taylor [180]: u′ (t) = au(t) + bu(εt),

{

0 ≤ t ≤ T, 0 < ε < 1,

u(0) = u0 .

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations In the current section, we study a class of problems, namely, the nonlinear pantograph fractional equations with ψ-Caputo fractional derivative: c

α;ψ

D0+ y(t) = f (t, y(t), y(εt)),

y(0) = y(b),

t ∈ T := [0, b],

(5.1) (5.2)

α;ψ

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, ε ∈ (0, 1), and f : T × ℝ × ℝ → ℝ is a continuous function.

5.2.1 Existence results We will adopt the spaces, lemmas, and any auxiliary results given in the previous chapters. We will be using specifically Lemma 4.1 and Lemma 4.2.

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations

� 97

Let the spaces α;ψ

X = {y ∈ C(T, ℝ) : y(t) = J0+ υ(t) : υ ∈ C(T, ℝ)}

and Y = C(T, ℝ)

be endowed with the norms ‖y‖X = ‖y‖Y = ‖y‖∞ = sup |y(t)|. t∈T

We will now give the definition of the operator L : Dom L ⊆ X → Y : α;ψ

Ly := c D0+ y,

(5.3)

where α;ψ

Dom L = {y ∈ X : c D0+ y ∈ Y : y(0) = y(b)}. Define N : X → Y by N y(t) := f (t, y(t), y(εt)),

t ∈ T and ε ∈ (0, 1).

Then the problem (5.1)–(5.2) is equivalent to the problem Ly = N y. Lemma 5.1. Suppose that the following hypothesis (5.1.1) is satisfied. (5.1.1) There exist positive constants γ, η with 󵄨󵄨 󵄨 ̄ 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ γ|y − y|̄ + η|υ − υ|, for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps: Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ T, we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(0))α 0

98 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative By (5.1.1), we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

αγ α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds α (ψ(b) − ψ(0)) 0

b

αη α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (εs) − y(εs)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

α(γ + η)‖yn − y‖Y α−1 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(0))α 0

⩽ (γ + η)‖yn − y‖Y . Thus, for each t ∈ T, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞,

and hence 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN (Ω) is bounded. For t ∈ T and y ∈ Ω, we have b

α α−1 󵄨 󵄨󵄨 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (y)(s)󵄨󵄨󵄨ds 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 ⩽ (ψ(b) − ψ(0))α α ⩽ (ψ(b) − ψ(0))α

0

b

∫ ψ′ (s)(ψ(b) − ψ(s))

α−1 󵄨

󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds α (ψ(b) − ψ(0)) 0

b

αγ α−1 󵄨 󵄨 ⩽f + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α ∗

0

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations

� 99

b

αη α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(εs)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α ⩽ f ∗ + (γ + η)M ,

0

where f ∗ = ‖f (⋅, 0, 0)‖∞ . Thus, ‖QN (y)‖Y ⩽ f ∗ + (γ + η)M . So, QN (Ω) is a bounded set in Y . Step 3. We demonstrate that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ T, we get L−1 P (N y(t) − QN y(t)) α;ψ

= J0+ [f (t, y(t), y(εt)) b

−

α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(εs))ds] (ψ(b) − ψ(0))α 0

t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), y(εs))ds Γ(α) 0

b

(ψ(t) − ψ(0))α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(εs))ds. Γ(α)(ψ(b) − ψ(0))α 0

For all y ∈ Ω and t ∈ T, we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

100 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

⩽

t

2f ∗ γ α α−1 󵄨 󵄨 (ψ(b) − ψ(0)) + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds αΓ(α) Γ(α) 0

t

+

η α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(εs)󵄨󵄨󵄨ds Γ(α) 0

b

γ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) 0

b

+

η α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(εs)󵄨󵄨󵄨ds Γ(α) 0

2(ψ(b) − ψ(0))α ∗ ⩽ [f + (γ + η)M ]. Γ(α + 1) Therefore, 2(ψ(b) − ψ(0))α ∗ 󵄩󵄩 −1 󵄩 [f + (γ + η)M ]. 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X ⩽ Γ(α + 1) This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X .

It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. For 0 < t1 < t2 ⩽ b, y ∈ Ω, we have

󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (id − Q )N y(t2 ) − LP (id − Q )N y(t1 )󵄨󵄨󵄨 t

⩽

∫01 [ψ′ (s)|(ψ(t2 ) − ψ(s))α−1 − (ψ(t1 ) − ψ(s))α−1 ||f (s, y(s), y(εs))|]ds Γ(α)

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs))󵄨󵄨󵄨ds Γ(α) t1

+

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] Γ(α)(ψ(b) − ψ(0))α b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

t1

α−1 󵄨

󵄨󵄨f (s, y(s), y(εs))󵄨󵄨󵄨ds 󵄨 󵄨

1 α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations

� 101

t1

1 α−1 󵄨 󵄨 − ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t1

1 α−1 α−1 󵄨 󵄨 + ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ]󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] + Γ(α)(ψ(b) − ψ(0))α b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

α−1 󵄨

󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] |f (s, 0, 0)| + ds ∫ ψ′ (s) α Γ(α)(ψ(b) − ψ(0)) (ψ(b) − ψ(s))1−α 0 α

α

α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) + Λ[(ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) ] α

α

+ Λ[(ψ(t2 ) − ψ(0)) − (ψ(t1 ) − ψ(0)) ] α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) , where Λ=

f ∗ + (γ + η)M . Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 5.2. Assume (5.1.1) holds. If the condition (γ + η) 1 α (ψ(b) − ψ(0)) < Γ(α + 1) 2

(5.4)

is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

102 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative Proof. Let y ∈ X satisfy L(y) − N (y) = −ζL(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ T α;ψ

Ly(t) = c D0+ y(t) =

1 ζ f (t, y(t), y(εt)) − f (t, −y(t), −y(εt)). 1+ζ 1+ζ

By Lemma 2.4 we get y(t) = y(0) +

1 α;ψ α;ψ [J + (f (s, y(s), y(εs)))(t) − ζ J0+ (f (s, −y(s), −y(εs)))(t)]. ζ +1 0

Thus, for every t ∈ T we obtain t

1 α−1 󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs))󵄨󵄨󵄨ds 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨y(0)󵄨󵄨󵄨 + (ζ + 1)Γ(α) t

+

0

ζ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, −y(s), −y(εs))󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

t

|f (s, y(s), y(εs)) − f (s, 0, 0)| 1 󵄨 󵄨 ds ⩽ 󵄨󵄨󵄨y(0)󵄨󵄨󵄨 + ∫ ψ′ (s) (ζ + 1)Γ(α) (ψ(t) − ψ(s))1−α t

+

0

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

t

+

|f (s, −y(s), −y(εs)) − f (s, 0, 0)| ζ ds ∫ ψ′ (s) (ζ + 1)Γ(α) (ψ(t) − ψ(s))1−α 0

t

ζ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

∗ α 2(γ + η) α 󵄨 󵄨 2f (ψ(b) − ψ(0)) + (ψ(b) − ψ(0)) ‖y‖X . ⩽ 󵄨󵄨󵄨y(0)󵄨󵄨󵄨 + Γ(α + 1) Γ(α + 1)

Thus,

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations

� 103

∗ α 2(γ + η) α 󵄨 2f (ψ(b) − ψ(0)) 󵄨 ‖y‖X ⩽ 󵄨󵄨󵄨y(0)󵄨󵄨󵄨 + + (ψ(b) − ψ(0)) ‖y‖X . Γ(α + 1) Γ(α + 1)

We deduce that |y(0)| +

‖y‖X ⩽

[1 −

2f ∗ (ψ(b)−ψ(0))α Γ(α+1)

2(γ+η) (ψ(b) Γ(α+1)

− ψ(0))α ]

:= A .

The demonstration is completed. Lemma 5.3. If conditions (5.1.1) and (5.4) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(5.5)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 5.2, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(5.6)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 5.1. Assume (5.1.1) and (5.4) hold. Then there exists at least one solution for the problem (5.1)–(5.2) in Dom L ∩ Ω. Proof. It is clear that the set Ω defined in (5.6) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 5.3, assume (5.1.1) and (5.4) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (5.1)–(5.2) has at least one solution in Dom L ∩ Ω, which completes the demonstration. Theorem 5.2. Let (5.1.1) be satisfied. Moreover, we assume that the following hypothesis holds.

104 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative (5.5.1) There exist constants γ > 0 and η ⩾ 0 such that 󵄨 󵄨󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η|υ − υ|̄ for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. If one has η 2(γ + η) α + (ψ(b) − ψ(0)) < 1, γ Γ(α + 1)

(5.7)

then the problem (5.1)–(5.2) has a unique solution in Dom L ∩ Ω. Proof. Note that condition (5.7) is stronger than condition (5.4). By Theorem 5.1, the problem (5.1)–(5.2) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (5.1)–(5.2) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ T c c

α;ψ

D0+ y1 (t) = f (t, y1 (t), y1 (εt)), α;ψ

D0+ y2 (t) = f (t, y2 (t), y2 (εt)),

and y1 (0) = y1 (b),

y2 (0) = y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ T. Then α;ψ

LU(t) = c D0+ U(t) α;ψ

α;ψ

= c D0+ y1 (t) − c D0+ y2 (t)

= f (t, y1 (t), y1 (εt)) − f (t, y2 (t), y2 (εt)). Using the fact that Img L = ker Q , we have b

α−1

∫ ψ′ (s)(ψ(b) − ψ(s))

[f (s, y1 (s), y1 (εs)) − f (s, y2 (s), y2 (εs))]ds = 0.

0

Since f is a continuous function, there exists t0 ∈ [0, b] such that f (t0 , y1 (t0 ), y1 (εt0 )) − f (t0 , y2 (t0 ), y2 (εt0 )) = 0. In view of (5.5.1), we have

(5.8)

5.2 Periodic solutions for some nonlinear fractional pantograph differential equations

� 105

󵄨 η󵄨 󵄨 η 󵄨󵄨 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨y1 (εt0 ) − y2 (εt0 )󵄨󵄨󵄨 ⩽ ‖y1 − y2 ‖X . γ γ Then 󵄨 η 󵄨󵄨 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ ‖U‖X . γ

(5.9)

On the other hand, by Theorem 2.4, we have α;ψ c

J0+

α;ψ

D0+ U(t) = U(t) − U(0),

which implies that α;ψ

α;ψ

U(0) = U(t0 ) − J0+ c D0+ U(t0 ), and therefore α;ψ

α;ψ

α;ψ

α;ψ

U(t) = J0+ c D0+ U(t) + U(t0 ) − J0+ c D0+ U(t0 ). Using (5.9) we obtain for every t ∈ T 󵄨󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨J0+ D0+ U(t)󵄨󵄨󵄨 + 󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ U(t0 )󵄨󵄨󵄨 ⩽

η 2(ψ(b) − ψ(0))α 󵄩󵄩c α;ψ 󵄩󵄩 ‖U‖X + 󵄩󵄩 D0+ U 󵄩󵄩X . γ Γ(α + 1)

(5.10)

By (5.8) and (5.1.1), we find that 󵄨󵄨c α;ψ 󵄨 󵄨 󵄨 󵄨󵄨 D0+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (t), y1 (εt)) − f (t, y2 (t), y2 (εt))󵄨󵄨󵄨 ⩽ (γ + η)‖U‖X . Then 󵄩󵄩c α;ψ 󵄩󵄩 󵄩󵄩 D0+ U 󵄩󵄩X ⩽ (γ + η)‖U‖X . Substituting (5.11) in the right side of (5.10), we get for every t ∈ T η 2(γ + η)(ψ(b) − ψ(0))α 󵄨󵄨 󵄨 ]‖U‖X . 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ + γ Γ(α + 1) Therefore, η 2(γ + η)(ψ(b) − ψ(0))α ‖U‖X ⩽ [ + ]‖U‖X . γ Γ(α + 1) Hence, by (5.7), we conclude that

(5.11)

106 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative ‖U‖X = 0. As a result, for any t ∈ T, we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof. 5.2.2 Examples Example 1. Consider the following problem with NFDE: c

1

;2t

D03+ y(t) = f (t, y(t), y(εt)),

t ∈ T := [0, 1],

y(0) = y(1),

where f (t, y(t), y(εt)) = Here α = 31 , ψ(t) = 2t , and ε =

y(t) 1 t et + sin y( ) + . √2 5(1 + t) 3√π 3

1 . √2

It is clear that the function f ∈ C([0, 1], ℝ). Let y, y, υ, υ ∈ ℝ and t ∈ T. Then 1 󵄨󵄨 󵄨 1 ̄ |υ − υ|, 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ |y − y|̄ + 5 3√π

1 which implies that (5.1.1) is satisfied with γ = 51 and η = 3√π . Furthermore, by some simple calculations, we see that

(γ + η) 1 α (ψ(b) − ψ(0)) ≈ 0.4346 < . Γ(α + 1) 2 With the use of Theorem 5.1 our problem has at least one solution. Example 2. Consider the following problem with NFDE: c

1

;et

D02+ y(t) = f (t, y(t), y(εt)),

t ∈ T := [0, 1],

y(0) = y(1),

where f (t, y(t), y(εt)) = ln (t + 2) + Here α = 21 , ψ(t) = et , and ε =

1 . √π

1 3 e−11−t (sin y(t) + y(t)) + . t 2 23√π 37(1 + y( √π ))

� 107

5.3 Wide class of fractional pantograph differential equations

It is easy to see that f ∈ C([0, 1], ℝ). Let y, y, υ, υ ∈ ℝ and t ∈ T. Then 󵄨 󵄨󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽

5 1 |y − y|̄ + |υ − υ|̄ 37e11 46√π

󵄨 󵄨󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾

1 1 ̄ |y − y|̄ − |υ − υ|. 37e11 46√π

and

Hence, the assumptions (5.1.1) and (5.5.1) are satisfied with γ = 1 . 37e11

η=η= By simple calculations, we see that

5 , 46√π

γ =

1 , 46√π

and

η 2(γ + η) α + (ψ(b) − ψ(0)) ≈ 0.1814 < 1. γ Γ(α + 1) So, by Theorem 5.2, our problem has a unique solution.

5.3 Wide class of fractional pantograph differential equations Motivated by the works mentioned in the introduction, in this section, we investigate the existence and uniqueness of the solutions for the nonlinear fractional pantograph differential equations involving a ψ-Caputo derivative operator supplemented with periodic conditions of the form c

α;ψ

D0+ y(t) = f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)),

t ∈ T := [0, b],

y(0) = y(b),

α;ψ

(5.12) (5.13)

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, p ∈ (0, 1), and f , g : T × ℝ × ℝ → ℝ are given continuous functions.

5.3.1 Existence results Define N : X → Y by N y(t) := f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)),

t ∈ T and p ∈ (0, 1).

Then the problem (5.12)–(5.13) is equivalent to the problem Ly = N y.

108 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative Lemma 5.4. Suppose that the following hypothesis (5.6.1) is satisfied. (5.6.1) There exist nonnegative functions γ1 , γ2 , η1 , η2 ∈ C(T, ℝ+ ) such that 󵄨 󵄨󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ γ1 (t)|y − y|̄ + η1 (t)|υ − υ|̄ and 󵄨 󵄨󵄨 󵄨󵄨g(t, y, υ) − g(t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ γ2 (t)|y − y|̄ + η2 (t)|υ − υ|̄ for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps: Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ T, we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. α (ψ(b) − ψ(0)) 0

By (5.6.1), we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

αγ1∗ α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

αη∗1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (ps) − y(ps)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

αγ2∗ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

αη∗2 |y ((1 − p)s) − y((1 − p)s)| + ds ∫ ψ′ (s) n (ψ(b) − ψ(0))α (ψ(b) − ψ(s))1−α 0

b

α(γ1∗ + γ2∗ + η∗1 + η∗2 )‖yn − y‖Y α−1 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(0))α 0

5.3 Wide class of fractional pantograph differential equations

⩽ (γ1∗ + γ2∗ + η∗1 + η∗2 )‖yn − y‖Y , where γ1∗ = ‖γ1 ‖∞ , γ2∗ = ‖γ2 ‖∞ , η∗1 = ‖η1 ‖∞ , and η∗2 = ‖η2 ‖∞ . Thus, for each t ∈ T, we get 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞,

and hence 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We prove that QN (Ω) is bounded. For t ∈ T and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

⩽

α α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α

α ⩽ (ψ(b) − ψ(0))α

0

b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α α + (ψ(b) − ψ(0))α

0

b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

+

α α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

+ γ2∗ ) α−1 󵄨 󵄨 ⩽f +g + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds α (ψ(b) − ψ(0)) ∗

∗

α(γ1∗

0

b

+

αη∗1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(ps)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

� 109

110 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative b

αη∗2 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y((1 − p)s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α ⩽f +g + ∗

∗

(γ1∗

+

0 ∗ γ2 +

η∗1 + η∗2 )M ,

where f ∗ = ‖f (⋅, 0, 0)‖∞ and g ∗ = ‖g(⋅, 0, 0)‖∞ . Thus, ‖QN (y)‖Y ⩽ f ∗ + g ∗ + (γ1∗ + γ2∗ + η∗1 + η∗2 )M . So, QN (Ω) is a bounded set in Y . Step 3. We show that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ T, we get L−1 P (N y(t) − QN y(t)) α;ψ

= J0+ [f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)) b

−

α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(ps))ds (ψ(b) − ψ(0))α 0

b

α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) g(s, y(s), y((1 − p)s))ds] α (ψ(b) − ψ(0)) 0

t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), y(ps))ds Γ(α) 0

+

t

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) g(s, y(s), y((1 − p)s))ds Γ(α) 0

b

(ψ(t) − ψ(0))α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(ps))ds α Γ(α)(ψ(b) − ψ(0)) 0

b

(ψ(t) − ψ(0))α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) g(s, y(s), y((1 − p)s))ds. Γ(α)(ψ(b) − ψ(0))α 0

For all y ∈ Ω and t ∈ T, we get

5.3 Wide class of fractional pantograph differential equations

󵄨 󵄨󵄨 −1 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

1 α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

+

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

⩽

t

γ∗ 2(f ∗ + g ∗ ) α α−1 󵄨 󵄨 (ψ(b) − ψ(0)) + 1 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds αΓ(α) Γ(α) +

η∗1

Γ(α)

t

0

α−1 󵄨

∫ ψ′ (s)(ψ(t) − ψ(s))

󵄨󵄨y(ps)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

γ∗ α−1 󵄨 󵄨 + 1 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) 0

b

+

η∗1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(ps)󵄨󵄨󵄨ds Γ(α) 0

+

γ2∗

Γ(α)

t

α−1 󵄨

∫ ψ′ (s)(ψ(t) − ψ(s)) 0

󵄨󵄨y(s)󵄨󵄨󵄨ds 󵄨 󵄨

� 111

112 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative t

η∗ α−1 󵄨 󵄨 + 2 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y((1 − p)s)󵄨󵄨󵄨ds Γ(α) 0

b

+

γ2∗ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) 0

+

η∗2

Γ(α)

b

α−1 󵄨

∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨y((1 − p)s)󵄨󵄨󵄨ds 󵄨 󵄨

0

2(ψ(b) − ψ(0))α ∗ ⩽ [f + g ∗ + (γ1∗ + γ2∗ + η∗1 + η∗2 )M ]. Γ(α + 1) Therefore, 2(ψ(b) − ψ(0))α ∗ 󵄩 󵄩󵄩 −1 [f + g ∗ + (γ1∗ + γ2∗ + η∗1 + η∗2 )M ]. 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X ⩽ Γ(α + 1) This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. For 0 < t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (id − Q )N y(t2 ) − LP (id − Q )N y(t1 )󵄨󵄨󵄨 t1

1 α−1 α−1 󵄨 󵄨 ⩽ ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨 Γ(α) 0

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨]ds t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds Γ(α) t1

t1

1 α−1 α−1 󵄨 󵄨 + ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨 Γ(α) 0

󵄨 󵄨 × 󵄨󵄨󵄨g(s, y(s), y((1 − p)s))󵄨󵄨󵄨]ds t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s))󵄨󵄨󵄨ds Γ(α) t1

+

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] Γ(α)(ψ(b) − ψ(0))α b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

+

󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds 󵄨 󵄨

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] Γ(α)(ψ(b) − ψ(0))α

5.3 Wide class of fractional pantograph differential equations b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

⩽

� 113

󵄨󵄨g(s, y(s), y((1 − p)s))󵄨󵄨󵄨ds 󵄨 󵄨

t1

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

−

t1

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t1

1 α−1 α−1 󵄨 󵄨 + ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ]󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t1

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t1

1 α−1 󵄨 󵄨 − ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t1

1 α−1 α−1 󵄨 󵄨 + ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ]󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] + Γ(α)(ψ(b) − ψ(0))α b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] |f (s, 0, 0)| + ds ∫ ψ′ (s) α Γ(α)(ψ(b) − ψ(0)) (ψ(b) − ψ(s))1−α 0

114 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative

+

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] Γ(α)(ψ(b) − ψ(0))α b

α−1 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨g(s, y(s), y((1 − p)s)) − g(s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

+

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] |g(s, 0, 0)| ds ∫ ψ′ (s) Γ(α)(ψ(b) − ψ(0))α (ψ(b) − ψ(s))1−α 0 α

α

α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) + Λ[(ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) ] α

α

+ Λ[(ψ(t2 ) − ψ(0)) − (ψ(t1 ) − ψ(0)) ] α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) , where Λ=

f ∗ + g ∗ + (γ1∗ + γ2∗ + η∗1 + η∗2 )M . Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 5.5. Assume (5.6.1) holds. If the condition (γ1∗ + γ2∗ + η∗1 + η∗2 ) 1 α (ψ(b) − ψ(0)) < Γ(α + 1) 2

(5.14)

is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A , Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ T α;ψ

Ly(t) = c D0+ y(t)

ζ ∈ (0, 1].

5.3 Wide class of fractional pantograph differential equations

=

1 [f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t))] 1+ζ ζ − [f (t, −y(t), −y(pt)) + g(t, −y(t), −y((1 − p)t))]. 1+ζ

By Theorem 2.4 we get y(t) = c0 +

1 α;ψ [J + (f (s, y(s), y(ps) + g(s, y(s), y((1 − p)s)))(t) ζ +1 0 α;ψ

− ζ J0+ (f (s, −y(s), −y(ps)) + g(s, −y(s), −y((1 − p)s)))(t)], where c0 = y(0). Thus, for every t ∈ T we obtain t

1 α−1 󵄨 󵄨󵄨 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ |c0 | + (ζ + 1)Γ(α) t

+

0

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨g(s, y(s), y((1 − p)s))󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

t

+

ζ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, −y(s), −y(ps))󵄨󵄨󵄨ds (ζ + 1)Γ(α) t

+∫ 0

0

ζψ (s)(ψ(t) − ψ(s))α−1 |g(s, −y(s), −y((1 − p)s))|ds (ζ + 1)Γ(α) ′

t

⩽ |c0 | + ∫ 0

+

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) t

+∫ 0

+

ψ′ (s)(ψ(t) − ψ(s))α−1 |f (s, y(s), y(ps)) − f (s, 0, 0)| ds (ζ + 1)Γ(α)

0

ψ (s)(ψ(t) − ψ(s))α−1 |g(s, y(s), y((1 − p)s)) − g(s, 0, 0)| ds (ζ + 1)Γ(α) ′

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨g(s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) t

+∫ 0

0

ζψ (s)(ψ(t) − ψ(s))α−1 |f (s, −y(s), −y(ps)) − f (s, 0, 0)|ds (ζ + 1)Γ(α) ′

t

ζ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

� 115

116 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative t

+∫ 0

t

+∫ 0

ζψ′ (s)(ψ(t) − ψ(s))α−1 |g(s, −y(s), −y((1 − p)s)) − g(s, 0, 0)|ds (ζ + 1)Γ(α) ζψ′ (s)(ψ(t) − ψ(s))α−1 |g(s, 0, 0)|ds (ζ + 1)Γ(α)

2(f ∗ + g ∗ )(ψ(b) − ψ(0))α Γ(α + 1) 2(γ1∗ + γ2∗ + η∗1 + η∗2 ) α + (ψ(b) − ψ(0)) ‖y‖X . Γ(α + 1)

⩽ |c0 | +

Thus, 2(f ∗ + g ∗ )(ψ(b) − ψ(0))α Γ(α + 1) ∗ ∗ 2(γ1 + γ2 + η∗1 + η∗2 ) α + (ψ(b) − ψ(0)) ‖y‖X . Γ(α + 1)

‖y‖X ⩽ |c0 | +

We deduce that ‖y‖X ⩽

2(f ∗ +g ∗ )(ψ(b)−ψ(0))α Γ(α+1) 2(γ1∗ +γ2∗ +η∗1 +η∗2 ) (ψ(b) − ψ(0))α ] Γ(α+1)

|c0 | +

[1 −

:= A .

The demonstration is completed. Lemma 5.6. If conditions (5.6.1) and (5.14) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)],

(5.15)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 5.5, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ} such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1].

(5.16)

5.3 Wide class of fractional pantograph differential equations

� 117

Theorem 5.3. Assume that (5.6.1) and (5.14) hold. Then there exists at least one solution for the problem (5.12)–(5.13) in Dom L ∩ Ω. Proof. It is clear that the set Ω defined in (5.16) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 5.6, we assume that (5.6.1) and (5.14) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (5.12)–(5.13) has at least one solution in Dom L ∩ Ω, which completes the demonstration. Theorem 5.4. Let (5.6.1) be satisfied. Moreover, we assume that the following hypothesis holds. (5.10.1) There exist constants γ > 0 and η ⩾ 0 such that 󵄨󵄨 󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η|υ − υ|̄ for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. If one has η + γ2∗ + η∗2 2(γ1∗ + γ2∗ + η∗1 + η∗2 )(ψ(b) − ψ(0))α + < 1, γ Γ(α + 1)

(5.17)

then the problem (5.12)–(5.13) has a unique solution in Dom L ∩ Ω. Proof. Note that condition (5.17) is stronger than condition (5.14). By Theorem 5.3, the problem (5.12)–(5.13) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (5.12)–(5.13) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ T c c

α;ψ

D0+ y1 (t) = f (t, y1 (t), y1 (pt)) + g(t, y1 (t), y1 ((1 − p)t)), α;ψ

D0+ y2 (t) = f (t, y2 (t), y2 (pt)) + g(t, y2 (t), y2 ((1 − p)t)),

and y1 (0) = y1 (b), y2 (0) = y2 (b). Let U(t) = y1 (t) − y2 (t) for all t ∈ T. Then α;ψ

LU(t) = c D0+ U(t) α;ψ

α;ψ

= c D0+ y1 (t) − c D0+ y2 (t)

= f (t, y1 (t), y1 (pt)) + g(t, y1 (t), y1 ((1 − p)t))

118 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative − f (t, y2 (t), y2 (pt)) − g(t, y2 (t), y2 ((1 − p)t)).

(5.18)

Using the fact that ImgL = ker Q , we have b

∫ ψ′ (s)(ψ(b) − ψ(s))

α−1

[f (s, y1 (s), y1 (ps)) + g(s, y1 (s), y1 ((1 − p)s))

0

− f (s, y2 (s), y2 (ps)) − g(s, y2 (s), y2 ((1 − p)s))]ds = 0. Since f is a continuous function, there exists t0 ∈ [0, b] such that f (t0 , y1 (t0 ), y1 (pt0 )) + g(t0 , y1 (t0 ), y1 ((1 − p)t0 )) − f (t0 , y2 (t0 ), y2 (pt0 )) − g(t0 , y2 (t0 ), y2 ((1 − p)t0 )) = 0.

In view of (5.6.1) and (5.10.1), we have 󵄨󵄨 󵄨 1 󵄨 󵄨 󵄨 ∗󵄨 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ [η󵄨󵄨󵄨y1 (pt0 ) − y2 (pt0 )󵄨󵄨󵄨 + γ2 󵄨󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 γ 󵄨 󵄨 + η∗2 󵄨󵄨󵄨y1 ((1 − p)t0 ) − y2 ((1 − p)t0 )󵄨󵄨󵄨] η + γ2∗ + η∗2 ⩽ ‖y1 − y2 ‖X . γ Then ∗ ∗ 󵄨󵄨 󵄨 η + γ2 + η2 ‖U‖X . 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ γ

(5.19)

On the other hand, by Theorem 2.4, we have α;ψ c

J0+

α;ψ

D0+ U(t) = U(t) − U(0),

which implies that α;ψ

α;ψ

U(0) = U(t0 ) − J0+ c D0+ U(t0 ), and therefore α;ψ

α;ψ

α;ψ

α;ψ

U(t) = J0+ c D0+ U(t) + U(t0 ) − J0+ c D0+ U(t0 ). Using (5.19) we obtain for every t ∈ T 󵄨󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨J0+ D0+ U(t)󵄨󵄨󵄨 + 󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ U(t0 )󵄨󵄨󵄨 ⩽

η + γ2∗ + η∗2 2(ψ(b) − ψ(0))α 󵄩󵄩c α;ψ 󵄩󵄩 ‖U‖X + 󵄩󵄩 D0+ U󵄩󵄩X . γ Γ(α + 1)

(5.20)

5.3 Wide class of fractional pantograph differential equations

� 119

By (5.18) and (5.6.1), we find that 󵄨 󵄨 󵄨󵄨c α;ψ 󵄨󵄨 D0+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨(f (t, y1 (t), y1 (pt)) − f (t, y2 (t), y2 (pt))) 󵄨 + (g(t, y1 (t), y1 ((1 − p)t)) − g(t, y2 (t), y2 ((1 − p)t)))󵄨󵄨󵄨 ⩽ (γ1∗ + γ2∗ + η∗1 + η∗2 )‖U‖X .

Then 󵄩󵄩c α;ψ 󵄩󵄩 ∗ ∗ ∗ ∗ 󵄩󵄩 D0+ U󵄩󵄩X ⩽ (γ1 + γ2 + η1 + η2 )‖U‖X .

(5.21)

Substituting (5.21) in the right side of (5.20), we get for every t ∈ T η + γ2∗ + η∗2 2(γ1∗ + γ2∗ + η∗1 + η∗2 )(ψ(b) − ψ(0))α 󵄨󵄨 󵄨 + ]‖U‖X . 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ γ Γ(α + 1) Therefore, ‖U‖X ⩽ [

η + γ2∗ + η∗2 2(γ1∗ + γ2∗ + η∗1 + η∗2 )(ψ(b) − ψ(0))α + ]‖U‖X . γ Γ(α + 1)

Hence, by (5.17), we conclude that ‖U‖X = 0. As a result, for any t ∈ T, we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof.

5.3.2 An example We present an example for a nonlinear fractional differential equation of pantograph type with ψ-Caputo derivative operator to test our main findings: c

1

;2t

D05+ y(t) = f (t, y(t), y(pt)) + g(t, y(t), y((1 − p)t)),

t ∈ T,

y(0) = y(1),

where for any t ∈ T f (t, y(t), y(pt)) =

ln (e + √t) e−t t t + y(t) + cos y( ) 2 √ 3(1 + t) 3 95√π e + t+1

120 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative and et t+7 e−11−t . + sin y(t) + 13 17e7 √π 113(1 + y( 32 t))

g(t, y(t), y((1 − p)t)) =

Here T := [0, 1], α = 51 , ψ(t) = 2t , and p = 31 . It is clear that the functions f , g ∈ C(T × ℝ × ℝ, ℝ). Furthermore, for all t ∈ T and y, y, υ, υ ∈ ℝ, we get t e−t 󵄨󵄨 󵄨 ̄ |y − y|̄ + |υ − υ|, 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ 3(1 + t) 95√π

e−11−t t+7 󵄨󵄨 󵄨 ̄ |y − y|̄ + |υ − υ|, 󵄨󵄨g(t, y, υ) − g(t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ 7 113 17e √π and 󵄨󵄨 󵄨 ̄ 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η|υ − υ|, which implies that (5.6.1) and (5.10.1) are satisfied with γ1 (t) = η2 (t) =

e−t , 3(1 + t) e−11−t , 113

η1 (t) = γ=

e−1 , 6

By simple calculations, we get γ1∗ = 31 , η∗1 =

t , 95√π η= 1 , 95√π

γ2 (t) =

1 . 95√π

γ2∗ =

t+7 , 17e7 √π

8 , 17e7 √π

η∗2 =

1 , 113e11

and

η + γ2∗ + η∗2 2(γ1∗ + γ2∗ + η∗1 + η∗2 )(ψ(b) − ψ(0))α + ≈ 0.840359 < 1. γ Γ(α + 1) So, by Theorem 5.4, our problem has a unique solution.

5.4 Nonlinear fractional pantograph integro-differential equations Following the work of the previous section, in this section, we study the following nonlinear wide class of fractional integro-differential equations of pantograph type: c

α;ψ D0+ y(t)

qt

t

= f (t, y(t), y(pt)) + ∫ g(t, s, y(s))ds + ∫ h(t, s, y(s))ds,

under periodic conditions

0

0

t ∈ T,

(5.22)

5.4 Nonlinear fractional pantograph integro-differential equations

y(0) = y(b),

� 121

(5.23)

α;ψ

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, T := [0, b], and p, q ∈ (0, 1). Moreover, f : T × ℝ × ℝ → ℝ and g, h : T × T × ℝ → ℝ are given continuous functions.

5.4.1 Existence results Define N : X → Y by qt

N y(t) := f (t, y(t), y(pt)) + ∫ g(t, s, y(s))ds t

0

+ ∫ h(t, s, y(s))ds,

t ∈ T, p, q ∈ (0, 1).

0

Then the problem (5.22)–(5.23) is equivalent to the problem Ly = N y. Lemma 5.7. Suppose that the following hypotheses (5.11.1) and (5.11.2) are satisfied. (5.11.1) There exist nonnegative functions γ, η ∈ C(T, ℝ) such that 󵄨󵄨 󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ γ(t)|y − y|̄ + η(t)|υ − υ|̄ for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. (5.11.2) There exist nonnegative functions ρ1 , ρ2 ∈ C(T, ℝ) such that 󵄨󵄨 󵄨 󵄨󵄨g(t, s, υ) − g(t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ1 (t)|υ − υ|̄ and 󵄨󵄨 󵄨 󵄨󵄨h(t, s, υ) − h(t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ2 (t)|υ − υ|̄ for every (t, s) ∈ T × T and υ, ῡ ∈ ℝ. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps:

122 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative Step 1. We demonstrate that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ T, we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

α α−1 󵄨 󵄨 ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. (ψ(b) − ψ(0))α 0

By (5.11.1) and (5.11.2) we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

⩽

αγ∗ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

αη α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨yn (ps) − y(ps)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α ∗

0

b

qs

0

0

b

s

αρ∗1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨yn (t) − y(t)󵄨󵄨󵄨dt) ds (ψ(b) − ψ(0))α αρ∗2 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨yn (t) − y(t)󵄨󵄨󵄨dt) ds (ψ(b) − ψ(0))α 0

0

b

⩽

α(γ∗ + η∗ + b(qρ∗1 + ρ∗2 ))‖yn − y‖Y α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(0))α 0

⩽ (γ∗ + η∗ + b(qρ∗1 + ρ∗2 ))‖yn − y‖Y , where γ∗ = ‖γ‖∞ , η∗ = ‖η‖∞ , ρ∗1 = ‖ρ1 ‖∞ , and ρ∗2 = ‖ρ2 ‖∞ . Thus, for each t ∈ T, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞,

and hence 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0 We deduce that QN is continuous.

as n 󳨀→ +∞.

5.4 Nonlinear fractional pantograph integro-differential equations

Step 2. We prove that QN (Ω) is bounded. For t ∈ T and y ∈ Ω, we have 󵄨 󵄨󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨

b

⩽

α α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨N (y)(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α

⩽

α (ψ(b) − ψ(0))α

0

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

α−1 󵄨

󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

b

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α α + (ψ(b) − ψ(0))α

0

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

α−1

qs

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds 0

b

qs

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds (ψ(b) − ψ(0))α α + (ψ(b) − ψ(0))α

0

0

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

α−1

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds 0

b

s

α α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds (ψ(b) − ψ(0))α 0

⩽ f ∗ + b(qg∗ + h∗ ) +

+

+

+

0

b

αγ∗ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α 0

b

αη α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨y(ps)󵄨󵄨󵄨ds (ψ(b) − ψ(0))α ∗

αρ∗1

(ψ(b) − ψ(0))α αρ∗2

(ψ(b) − ψ(0))α

0 b

α−1

∫ ψ (s)(ψ(b) − ψ(s)) ′

0 b

α−1

∫ ψ′ (s)(ψ(b) − ψ(s)) 0

qs

󵄨 󵄨 (∫󵄨󵄨󵄨y(t)󵄨󵄨󵄨dt) ds 0 s

󵄨 󵄨 (∫󵄨󵄨󵄨y(t)󵄨󵄨󵄨dt) ds 0

� 123

124 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative ⩽ f ∗ + M (γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) + b(qg∗ + h∗ ), where f ∗ = ‖f (⋅, 0, 0)‖∞ . Thus, 󵄩 󵄩󵄩 ∗ ∗ ∗ ∗ ∗ ∗ ∗ 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ f + M (γ + η + b(qρ1 + ρ2 )) + b(qg + h ). So, QN (Ω) is a bounded set in Y . Step 3. We show that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ T, we get L−1 P (N y(t) − QN y(t)) =

α;ψ J0+ [f (t, y(t), y(pt))

qt

t

+ ∫ g(t, s, y(s))ds + ∫ h(t, s, y(s))ds 0

0

b

−

α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(ps))ds (ψ(b) − ψ(0))α 0

b

qs

0

0

b

s

0

0

α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫ g(s, t, y(t))dt)ds (ψ(b) − ψ(0))α −

α α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫ h(s, t, y(t))dt)ds] (ψ(b) − ψ(0))α t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), y(ps))ds Γ(α) 0

t

qs

0

0

1 α−1 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫ g(s, t, y(t))dt)ds Γ(α) t

+

s

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫ h(s, t, y(t))dt)ds Γ(α) 0

0

α

−

b

(ψ(t) − ψ(0)) α−1 ∫ ψ′ (s)(ψ(b) − ψ(s)) f (s, y(s), y(ps))ds α Γ(α)(ψ(b) − ψ(0)) 0

b

qs

0

0

(ψ(t) − ψ(0))α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫ g(s, t, y(t))dt)ds Γ(α)(ψ(b) − ψ(0))α

5.4 Nonlinear fractional pantograph integro-differential equations b

s

0

0

� 125

(ψ(t) − ψ(0))α α−1 − ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫ h(s, t, y(t))dt)ds. Γ(α)(ψ(b) − ψ(0))α For all y ∈ Ω and t ∈ T, we get 󵄨 󵄨󵄨 −1 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

+

+

t

qs

0

t

0 qs

0

0

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) t

+

0

t

+

s

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 0

s

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 0

0

b

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

b

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

+

+

b

qs

0

b

0 qs

0

0

b

s

0

0

b

s

0

0

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α)

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(b) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α)

126 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative

⩽

2(ψ(b) − ψ(0))α ∗ [f + M (γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) + b(qg∗ + h∗ )]. Γ(α + 1)

Therefore, 2(ψ(b) − ψ(0))α 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X ⩽ Γ(α + 1) × [f ∗ + M (γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) + b(qg∗ + h∗ )]. This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. For 0 < t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨LP (id − Q )N y(t2 ) − LP (id − Q )N y(t1 )󵄨󵄨󵄨 t

⩽

∫01 [ψ′ (s)|(ψ(t2 ) − ψ(s))α−1 − (ψ(t1 ) − ψ(s))α−1 ||f (s, y(s), y(ps))|]ds Γ(α)

t2

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds Γ(α) t1

t1

1 α−1 α−1 󵄨 󵄨 + ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨] Γ(α) qs

0

󵄨 󵄨 × (∫󵄨󵄨󵄨g(s, t, y(t))󵄨󵄨󵄨dt)ds 0

t2

qs

t1

0

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, y(t))󵄨󵄨󵄨dt)ds Γ(α) t1

+

1 α−1 α−1 󵄨 󵄨 ∫[ψ′ (s)󵄨󵄨󵄨(ψ(t2 ) − ψ(s)) − (ψ(t1 ) − ψ(s)) 󵄨󵄨󵄨] Γ(α) s

0

󵄨 󵄨 × (∫󵄨󵄨󵄨h(s, t, y(t))󵄨󵄨󵄨dt)ds 0

t2

s

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, y(t))󵄨󵄨󵄨dt)ds Γ(α) t1

+

0

[(ψ(t2 ) − ψ(0)) − (ψ(t1 ) − ψ(0))α ] Γ(α)(ψ(b) − ψ(0))α b

α

α−1 󵄨

× [∫ ψ′ (s)(ψ(b) − ψ(s)) 0

󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds 󵄨 󵄨

5.4 Nonlinear fractional pantograph integro-differential equations b

α−1

+ ∫ ψ (s)(ψ(b) − ψ(s)) ′

0

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, y(t))󵄨󵄨󵄨dt)ds 0

b

α−1

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

⩽

qs

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, y(t))󵄨󵄨󵄨dt)ds] 0

t1

1 α−1 α−1 ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ] Γ(α) 0

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds t1

+

1 α−1 α−1 󵄨 󵄨 ∫ ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ]󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t2

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) t1

t1

+

1 α−1 α−1 ∫[ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ] Γ(α) qs

0

󵄨 󵄨 × (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)]ds 0

t1

1 α−1 α−1 + ∫[ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ] Γ(α) qs

0

󵄨 󵄨 × (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)]ds 0

t2

qs

t1

0

t2

qs

t1

0

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) t1

+

1 α−1 α−1 ∫[ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ] Γ(α) 0

� 127

128 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative s

󵄨 󵄨 × (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)]ds 0

t1

1 α−1 α−1 + ∫[ψ′ (s)[(ψ(t1 ) − ψ(s)) − (ψ(t2 ) − ψ(s)) ] Γ(α) 0

s

󵄨 󵄨 × (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)]ds 0

+

+

t2

s

t1

0

t2

s

t1

0

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α) 1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds Γ(α)

[(ψ(t2 ) − ψ(0))α − (ψ(t1 ) − ψ(0))α ] + Γ(α)(ψ(b) − ψ(0))α b

α−1 󵄨

󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

× [∫ ψ′ (s)(ψ(b) − ψ(s)) 0 b

α−1 󵄨

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) 0 b

α−1

+ ∫ ψ (s)(ψ(b) − ψ(s)) ′

0 b

α−1

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

qs

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds 0 qs

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds 0

b

α−1

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds 0

b

α−1

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds] 0

α

α

α

⩽ Λ[(ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) + 2(ψ(t2 ) − ψ(t1 )) ] α

α

+ Λ[(ψ(t2 ) − ψ(0)) − (ψ(t1 ) − ψ(0)) ] α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) ,

5.4 Nonlinear fractional pantograph integro-differential equations

� 129

where Λ=

1 [f ∗ + M (γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) + b(qg∗ + h∗ )]. Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 5.8. Assume that (5.11.1) and (5.11.2) hold. If the condition γ∗ + η∗ + b(qρ∗1 + ρ∗2 ) 1 α (ψ(b) − ψ(0)) < Γ(α + 1) 2

(5.24)

is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζL(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ T α;ψ

Ly(t) = c D0+ y(t) =

qt

0

−

qt

0

ζ [f (t, −y(t), −y(pt)) + ∫ g(t, s, −y(s))ds 1+ζ t

+ ∫ h(t, s, −y(s))ds]. 0

By Theorem 2.4 we get y(t) = c0 +

t

1 [f (t, y(t), y(pt)) + ∫ g(t, s, y(s))ds + ∫ h(t, s, y(s))ds] 1+ζ

1 ζ +1

0

130 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative

×

α;ψ [J0+ (f (s, y(s), y(ps))

qs

s

+ ∫ g(s, t, y(t))dt + ∫ h(s, t, y(t))dt)(t) 0

0

qs

α;ψ

− ζ J0+ (f (s, −y(s), −y(ps)) + ∫ g(s, t, −y(t))dt 0

s

+ ∫ h(s, t, −y(t))dt)(t)], 0

where c0 = y(0). Thus, for every t ∈ T we obtain t

1 α−1 󵄨 󵄨󵄨 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(ps))󵄨󵄨󵄨ds 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ |c0 | + (ζ + 1)Γ(α) t

0

qs

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, y(t))󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) 0

0

s

t

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, y(t))󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) 0

0

t

+

ζ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, −y(s), −y(ps))󵄨󵄨󵄨ds (ζ + 1)Γ(α) 0

t

qs

0

0

ζ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, −y(t))󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) t

+

s

ζ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, −y(t))󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) 0

0

1 ⩽ |c0 | + (ζ + 1)Γ(α) t

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) 0

󵄨󵄨f (s, y(s), y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) 1 + (ζ + 1)Γ(α) t

0

α−1

× ∫ ψ (s)(ψ(t) − ψ(s)) ′

0

qs

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds 0

5.4 Nonlinear fractional pantograph integro-differential equations t

qs

0

0

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) 1 + (ζ + 1)Γ(α) t

α−1

× ∫ ψ′ (s)(ψ(t) − ψ(s)) 0

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds 0

t

s

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) ζ + (ζ + 1)Γ(α)

0

t

0

α−1 󵄨

× ∫ ψ′ (s)(ψ(t) − ψ(s)) 0

󵄨󵄨f (s, −y(s), −y(ps)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

t

ζ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds (ζ + 1)Γ(α) ζ + (ζ + 1)Γ(α)

0

t

α−1

× ∫ ψ (s)(ψ(t) − ψ(s)) ′

0

+

qs

󵄨 󵄨 (∫󵄨󵄨󵄨g(s, t, −y(t)) − g(s, t, 0)󵄨󵄨󵄨dt)ds 0

t

qs

ζ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨g(s, t, 0)󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α)

ζ + (ζ + 1)Γ(α)

0

t

0

α−1

× ∫ ψ′ (s)(ψ(t) − ψ(s)) 0

t

s

󵄨 󵄨 (∫󵄨󵄨󵄨h(s, t, −y(t)) − h(s, t, 0)󵄨󵄨󵄨dt)ds 0

s

ζ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) (∫󵄨󵄨󵄨h(s, t, 0)󵄨󵄨󵄨dt)ds (ζ + 1)Γ(α) 0

0

2(f ∗ + b(qg∗ + h∗ )) α (ψ(b) − ψ(0)) ⩽ |c0 | + Γ(α + 1) 2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α + (ψ(b) − ψ(0)) ‖y‖X . Γ(α + 1) Thus, ‖y‖X ⩽ |c0 | +

2(f ∗ + b(qg∗ + h∗ )) α (ψ(b) − ψ(0)) Γ(α + 1)

� 131

132 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative

+

2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α (ψ(b) − ψ(0)) ‖y‖X . Γ(α + 1)

We deduce that ‖y‖X ⩽

2(f ∗ +b(qg∗ +h∗ )) (ψ(b) − ψ(0))α Γ(α+1) 2(γ∗ +η∗ +b(qρ∗1 +ρ∗2 )) (ψ(b) − ψ(0))α ] Γ(α+1)

|c0 | + [1 −

:= A .

The demonstration is completed. Lemma 5.9. If conditions (5.11.1), (5.11.2), and (5.24) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(5.25)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 5.8, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(5.26)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 5.5. Assume that (5.11.1), (5.11.2), and (5.24) hold. Then there exists at least one solution for the problem (5.22)–(5.23) in Dom L ∩ Ω. Proof. It is clear that the set Ω defined in (5.26) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 5.9, assume that (5.11.1), (5.11.2), and (5.24) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (5.22)–(5.23) has at least one solution in Dom L ∩ Ω, which completes the demonstration. Theorem 5.6. Let (5.11.1) and (5.11.2) be satisfied. Moreover, we assume that the following hypothesis holds.

5.4 Nonlinear fractional pantograph integro-differential equations

� 133

(5.15.1) There exist constants γ > 0 and η ⩾ 0 such that 󵄨 󵄨󵄨 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η|υ − υ|̄ for every t ∈ T and y, y,̄ υ, ῡ ∈ ℝ. If one has η + b(qρ∗1 + ρ∗2 ) 2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α + (ψ(b) − ψ(0)) < 1, γ Γ(α + 1)

(5.27)

then the problem (5.22)–(5.23) has a unique solution in Dom L ∩ Ω. Proof. Note that condition (5.27) is stronger than condition (5.24). Then by Theorem 5.5 we obtain that the problem (5.22)–(5.23) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (5.22)–(5.23) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ T c

qt

α;ψ

t

D0+ y1 (t) = f (t, y1 (t), y1 (pt)) + ∫ g(t, s, y1 (s))ds + ∫ h(t, s, y1 (s))ds, 0

0

c

qt

α;ψ

t

D0+ y2 (t) = f (t, y2 (t), y2 (pt)) + ∫ g(t, s, y2 (s))ds + ∫ h(t, s, y2 (s))ds, 0

0

and y1 (0) = y1 (b),

y2 (0) = y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ T. Then α;ψ

LU(t) = c D0+ U(t) α;ψ

α;ψ

= c D0+ y1 (t) − c D0+ y2 (t)

= f (t, y1 (t), y1 (pt)) − f (t, y2 (t), y2 (pt)) qt

+ ∫[g(t, s, y1 (s)) − g(t, s, y2 (s))]ds 0

t

+ ∫[h(t, s, y1 (s)) − h(t, s, y2 (s))]ds. 0

Using the fact that Img L = ker Q , we have

(5.28)

134 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative b

α−1

∫ ψ′ (s)(ψ(b) − ψ(s))

[f (s, y1 (s), y1 (ps)) − f (s, y2 (s), y2 (ps))]ds

0 b

+ ∫ ψ′ (s)(ψ(b) − ψ(s))

α−1

qs

(∫[g(s, t, y1 (t)) − g(s, t, y2 (t))]dt)ds

0

0

b

+ ∫ ψ′ (s)(ψ(b) − ψ(s))

α−1

s

(∫[h(s, t, y1 (t)) − h(s, t, y2 (t))]dt)ds = 0.

0

0

Since f , g, and h are continuous functions, there exists t0 ∈ [0, b] such that f (t0 , y1 (t0 ), y1 (pt0 )) − f (t0 , y2 (t0 ), y2 (pt0 )) qt0

+ ∫ [g(t0 , s, y1 (s)) − g(t0 , s, y2 (s))]ds 0

t0

+ ∫[h(t0 , s, y1 (s)) − h(t0 , s, y2 (s))]ds = 0. 0

In view of (5.11.2) and (5.15.1), we have qt0

󵄨 󵄨 󵄨󵄨 󵄨 󵄨 1 󵄨 ∗ 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ [η󵄨󵄨󵄨y1 (pt0 ) − y2 (pt0 )󵄨󵄨󵄨 + ρ1 ∫ 󵄨󵄨󵄨y1 (s) − y2 (s)󵄨󵄨󵄨ds γ 0

t0

󵄨 󵄨 + ρ∗2 ∫󵄨󵄨󵄨y1 (s) − y2 (s)󵄨󵄨󵄨ds] 0

η + b(qρ∗1 + ρ∗2 ) ‖y1 − y2 ‖X . ⩽ γ Then ∗ ∗ 󵄨󵄨 󵄨 η + b(qρ1 + ρ2 ) ‖U‖X . 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ γ

On the other hand, by Theorem 2.4, we have α;ψ c

J0+

α;ψ

D0+ U(t) = U(t) − U(0),

which implies that α;ψ

α;ψ

U(0) = U(t0 ) − J0+ c D0+ U(t0 ), and therefore

(5.29)

5.4 Nonlinear fractional pantograph integro-differential equations

α;ψ

α;ψ

α;ψ

� 135

α;ψ

U(t) = J0+ c D0+ U(t) + U(t0 ) − J0+ c D0+ U(t0 ). Using (5.29), we obtain for every t ∈ T 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨 󵄨 󵄨 󵄨 α;ψ c α;ψ 󵄨󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨J0+ D0+ U(t)󵄨󵄨󵄨 + 󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ U(t0 )󵄨󵄨󵄨 η + b(qρ∗1 + ρ∗2 ) 2(ψ(b) − ψ(0))α 󵄩󵄩c α;ψ 󵄩󵄩 ⩽ ‖U‖X + 󵄩󵄩 D0+ U 󵄩󵄩X . γ Γ(α + 1)

(5.30)

By (5.11.1), (5.11.2), and (5.28), we find that 󵄨󵄨c α;ψ 󵄨 󵄨 󵄨 󵄨󵄨 D0+ U(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨f (t, y1 (t), y1 (pt)) − f (t, y2 (t), y2 (pt))󵄨󵄨󵄨 qt

󵄨 󵄨 + ∫ 󵄨󵄨󵄨g(t, s, y1 (s)) − g(t, s, y2 (s))󵄨󵄨󵄨ds 0

t

󵄨 󵄨 + ∫ 󵄨󵄨󵄨h(t, s, y1 (s)) − h(t, s, y2 (s))󵄨󵄨󵄨ds 0

⩽ (γ∗ + η∗ + b(qρ∗1 + ρ∗2 ))‖U‖X . Then 󵄩󵄩c α;ψ 󵄩󵄩 ∗ ∗ ∗ ∗ 󵄩󵄩 D0+ U 󵄩󵄩X ⩽ (γ + η + b(qρ1 + ρ2 ))‖U‖X . Substituting (5.31) in the right side of (5.30) we get for every t ∈ T η + b(qρ∗1 + ρ∗2 ) 2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α 󵄨󵄨 󵄨 + (ψ(b) − ψ(0)) ]‖U‖X . 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ γ Γ(α + 1) Therefore, ‖U‖X ⩽ [

η + b(qρ∗1 + ρ∗2 ) 2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α + (ψ(b) − ψ(0)) ]‖U‖X . γ Γ(α + 1)

Hence, by (5.27), we conclude that ‖U‖X = 0. As a result, for any t ∈ T we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof.

(5.31)

136 � 5 Nonlinear fractional pantograph differential equations with ψ-Caputo derivative 5.4.2 An example Example 5.1. We present an example of nonlinear fractional integro-differential equations of pantograph type with ψ-Caputo derivative operator to test our main findings: c

1 t ;2 3 0+

D

t 2

t

t y(t) = f (t, y(t), y( )) + ∫ g(t, s, y(s))ds + ∫ h(t, s, y(s))ds, √3 0

t ∈ T,

0

y(0) = y(1), where for any t ∈ T f (t, y(t), y(pt)) =

2et e−t t t + y(t) + cos y( ), √3 5 7(1 + t) 55√π

and for all (t, s, y) ∈ T × T × ℝ we have g(t, s, y) =

2st 3

e13+t

Here T := [0, 1], α = 31 , ψ(t) = 2t , p =

2

2

sin(y), 1 , √3

2s2 e−7−t . 73(1 + y)

h(t, s, y) =

and q = 21 .

It is clear that the function f ∈ C(T × ℝ × ℝ, ℝ). Furthermore, for all (t, s) ∈ T × T and y, y, υ, υ ∈ ℝ, we get e−t t 󵄨󵄨 󵄨 ̄ |y − y|̄ + |υ − υ|, 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩽ 7(1 + t) 55√π 2t 3 󵄨󵄨 󵄨 ̄ 󵄨󵄨g(t, s, y) − g(t, s, y)̄ 󵄨󵄨󵄨 ⩽ 13+t2 |y − y|, e 2

−7−t 󵄨󵄨 󵄨 2e ̄ |y − y|, 󵄨󵄨h(t, s, y) − h(t, s, y)̄ 󵄨󵄨󵄨 ⩽ 73

and 󵄨󵄨 󵄨 ̄ 󵄨󵄨f (t, y, υ) − f (t, y,̄ υ)̄ 󵄨󵄨󵄨 ⩾ γ|y − y|̄ − η|υ − υ|, which implies that (5.11.1), (5.11.2), and (5.15.1) are satisfied with e−t t γ(t) = , η(t) = , 7(1 + t) 55√π 1 1 γ= , η= . 14 e 55√π By simple calculations, we get γ∗ = 71 , η∗ =

ρ1 (t) =

1 , 55√π

2t 3

e13+t

ρ∗1 =

2

2

2 , e14

,

2e−7−t ρ2 (t) = , 73

ρ∗2 =

2e−7 , 73

and

5.5 Notes and remarks �

137

η + b(qρ∗1 + ρ∗2 ) 2(γ∗ + η∗ + b(qρ∗1 + ρ∗2 )) α + (ψ(b) − ψ(0)) ≈ 0.73438 < 1. γ Γ(α + 1) So, by Theorem 5.6, our problem has a unique solution.

5.5 Notes and remarks The findings in this chapter are based on the papers [90, 94, 120]. We suggest to the reader the monographs [7, 30, 35, 58, 99, 115, 125, 144, 167, 238] and the papers [19, 38, 44, 46, 54– 56, 62, 69, 90, 94, 94, 120, 121, 123, 190, 241, 242] for more information.

6 Nonlinear ψ-Caputo fractional pantograph coupled systems 6.1 Introduction and motivations This chapter is devoted to proving some existence and uniqueness results of periodic solutions for a nonlinear fractional pantograph coupled system with ψ-Caputo derivative. We employ Mawhin’s coincidence degree theory to establish our proofs. Further, examples are provided in each section to illustrate our results. We took as motivation the following works: – The papers [14, 15, 31, 33, 95, 111, 112, 133, 135, 156, 201, 206, 216, 223, 227], where the authors studied several problems with ordinary and fractional differential equations and inclusions with many new defined fractional operators; for more details see [43–45, 110] and the references therein. – In [231], Shah et al. investigated the ψ-Caputo fractional pantograph problem c α;ψ

D

y(t) = H(t, y(t), y(λt)) for each t ∈ I := [0, T],

with the nonlocal boundary condition ξ1 y(0) + ξ2 y(b) = c,

–

where c Dα;ψ is the ψ-Caputo fractional derivative of order α, 0 < α < 1, and λ ∈ (0, 1). Also, H : I × ℝ × ℝ → ℝ and ψ : I → ℝ are continuous functions and ξ1 , ξ2 , and c are real constants, where ξ1 + ξ2 ≠ 0. Some existence and uniqueness results are obtained by using the Banach contraction theorem and Schaefer’s fixed point theorem. If ξ1 + ξ2 = 0, this method is invalid. The paper of Wongcharoen et al. [250], where the authors considered the following coupled system with Hilfer fractional derivative and boundary conditions: H α,β D x(t) = f (t, x(t), y(t)), t ∈ [a, b], { { { { H { { Dα1 ,β1 y(t) = g(t, x(t), y(t)), t ∈ [a, b], { {x(a) = 0, x(b) = ∑m θi I φi y(ξi ), { i=1 { { { n ψj y(a) = 0, y(b) = ∑ j=1 ζj I x(zj ), {

where H Dα,β and H Dα1 ,β1 are the Hilfer fractional derivatives of order α and α1 , 1 < α, α1 < 2, and with parameter β and β1 , respectively, 0 ≤ β, β1 ≤ 1, I φi and I ψj are the Riemann–Liouville fractional integrals of order φi > 0 and ψj > 0, respectively, the points ξi , zj ∈ [a, b], where a ≥ 0, f , g : [a, b] × ℝ × ℝ 󳨀→ ℝ are continuous functions, and θi , ζj ∈ ℝ, i = 1, 2, . . . , m, j = 1, 2, . . . , n, are given real constants. They base their arguments on the Leray–Schauder and Krasnoselskii’s fixed point theorems and Banach’s contraction mapping principle. https://doi.org/10.1515/9783111334387-006

6.1 Introduction and motivations

–

� 139

In [21], Abdo et al. investigated the existence, uniqueness, and Ulam–Hyers stability of solutions of the following coupled system with generalized Hilfer fractional derivative: θ ,ξ ;ψ

Da1+ 1 y(t) = f1 (t, x(t)), { { { { { {Dθa2+,ξ2 ;ψ x(t) = f2 (t, y(t)), { {y(T) = w1 ∈ ℝ, { { { { {x(T) = w2 ∈ ℝ,

a < t ≤ T, a > 0, a < t ≤ T, a > 0,

θ ,ξ ψ

–

where 0 < θi < 1, 0 ≤ ξi ≤ 1, Dai+ i (i = 1, 2) is the Hilfer fractional derivative of order θi and type ξi with respect to ψ, and f : (a, T] × ℝ → ℝ is a given function. The tools employed for their demonstrations are the fixed point techniques of Banach and Krasnoselskii. Using the Banach contraction principle and Schauder’s fixed point theorem, in [37], Ahmed et al. studied the existence and uniqueness of solutions of the following switched coupled implicit ψ-Hilfer fractional differential system: p,q;ψ

p,q;ψ

t ∈ J = (a, b], { H D + u(t) = f (t, u(t), H Da+ v(t)), { { ap,q;ψ p,q;ψ γ = p + q − pq, H Da+ v(t) = g(t, H Da+ u(t), v(t)), { { { 1−γ;ψ 1−γ;ψ {Ia+ u(t)|t=a = ua , Ia+ v(t)|t=a = va , ua , va ∈ ℝ, p,q

where H Da+ represents the ψ-Hilfer fractional derivative of order p and type q with 1−γ;ψ

–

p ∈ (0, 1) and q ∈ (0, 1] and Ia+ denotes the ψ-Hilfer fractional integral of order 1 − γ. Moreover, f , g : J × X × X → X are continuous and nonlinear functions on a Banach space X . The linear function ψ : J → ℝ satisfies ψ′ (t) ≠ 0, t ∈ J . In [12], Abbas et al. considered the coupled system with Hilfer–Hadamard fractional derivative α ,β

(H D1 1 1 u)(t) = g1 (t, u(t), v(t)),

{

α ,β

(H D1 2 2 v)(t) = g2 (t, u(t), v(t)),

t ∈ [1, T],

with initial conditions 1−γ1

(H I1

{

u)(1) = ψ1 ,

1−γ (H I1 2 v)(1)

= ψ2 ,

where T > 1, αi ∈ (0, 1), βi ∈ [0, 1], γi = αi + βi − αi βi , ψi ∈ E, gi : [1, T] × E × E → E 1−γ (i = 1, 2) are given functions, H I1 i is the left-sided mixed Hadamard integral of α ,β

order 1 − γi , and H D1 i i is the Hilfer–Hadamard fractional derivative of order αi and type βi (i = 1, 2). They employed a technique that relies on the concept of the measure of noncompactness and fixed point theory.

140 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems

6.2 Periodic solutions for pantograph coupled systems 6.2.1 Introduction In this section, we consider the following problem: c

{c

α ;ψ

D0+1 y1 (t) = f1 (t, y1 (t), y2 (t)), α ;ψ

D0+2 y2 (t) = f2 (t, y1 (t), y2 (t)),

y1 (0) = y1 (b) and

t ∈ T := [0, b],

y2 (0) = y2 (b),

(6.1) (6.2)

α ;ψ

where c D0+i denotes the ψ-Caputo fractional derivative of order 0 < αi < 1, i ∈ {1, 2}, and f1 , f2 : T × ℝ2 → ℝ are continuous functions. 6.2.2 Existence results Consider the spaces α ;ψ

α ;ψ

ℏ1 = {y = (y1 , y2 ) ∈ C(T, ℝ2 ) : (y1 , y2 ) = (J0+1 ϰ1 , J0+2 ϰ2 ) where ϰ = (ϰ1 , ϰ2 ) ∈ C(T, ℝ2 )}

and ℏ2 = C(T, ℝ2 ), with the norms 󵄨 󵄨 ‖y‖ℏ1 = ‖y‖ℏ2 = max{sup󵄨󵄨󵄨yj (t)󵄨󵄨󵄨}. 1≤j≤2

t∈T

Define the operator ℘ : Dom ℘ ⊆ ℏ1 → ℏ2 by α ;ψ

α ;ψ

℘y = (℘1 y1 , ℘2 y2 ) := (c D0+1 y1 , c D0+2 y2 ), where Dom ℘ = {y ∈ ℏ1 : ℘y ∈ ℏ2 : y1 (0) = y1 (b) and y2 (0) = y2 (b)}. Lemma 6.1. By (6.3), we have ker ℘ = {y ∈ ℏ1 : y(t) = (y1 (t), y2 (t)) = (y1 (0), y2 (0)), t ∈ T}

(6.3)

6.2 Periodic solutions for pantograph coupled systems

� 141

and Img ℘ b

αi −1

= {ϰ = (ϰ1 , ϰ2 ) ∈ ℏ2 : ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

ϰi (ϱ)dϱ = 0, i ∈ {1, 2}}.

0 α ;ψ

α ;ψ

Proof. By Remark 2.1, for y ∈ ℏ1 , if ℘y = (c D0+1 y1 , c D0+2 y2 ) = (0, 0) has a solution of the form y(t) = (y1 (t), y2 (t)) = (c1 , c2 ) = (y1 (0), y2 (0)),

t ∈ T,

then ker ℘ = {y ∈ ℏ1 : y(t) = (y1 (t), y2 (t)) = (y1 (0), y2 (0)), t ∈ T}. For ϰ = (ϰ1 , ϰ2 ) ∈ Img ℘, there exists y = (y1 , y2 ) ∈ Dom ℘ such that (ϰ1 , ϰ2 ) = (℘1 y1 , ℘2 y2 ) ∈ ℏ2 . Using Theorem 2.4, we obtain for every t ∈ T and i ∈ {1, 2} α ;ψ

yi (t) = yi (0) + J0+i ϰi (t) t

= yi (0) +

1 α −1 ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) i ϰi (ϱ)dϱ. Γ(αi ) 0

Since y ∈ Dom ℘, we have yi (0) = yi (b). Thus, b

αi −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

ϰi (ϱ)dϱ = 0.

0

If ϰ = (ϰ1 , ϰ2 ) ∈ ℏ2 and satisfies b

αi −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

ϰi (ϱ)dϱ = 0,

i ∈ {1, 2},

0 α ;ψ

α ;ψ

then for y(t) = (y1 (t), y2 (t)) = (J0+1 ϰ1 (t), J0+2 ϰ2 (t)), using Theorem 2.5, we get (ϰ1 (t), ϰ2 (t)) =

α ;ψ α ;ψ (c D0+1 y1 (t), c D0+2 y2 (t))

= (℘1 y1 , ℘2 y2 ). Therefore,

yi (b) = yi (0), which gives us y ∈ Dom ℘. So, ϰ ∈ Img ℘.

i ∈ {1, 2},

142 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems So b

αi −1

Img ℘ = {ϰ = (ϰ1 , ϰ2 ) ∈ ℏ2 : ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

ϰi (ϱ)dϱ = 0,

0

i ∈ {1, 2}}. Lemma 6.2. The operator ℘ is a Fredholm operator of index zero, and the linear continuous projector operators Φ2 : ℏ2 → ℏ2 and Φ1 : ℏ1 → ℏ1 can be written as Φ2 (ϰ) = (Φ21 ϰ1 , Φ22 ϰ2 ), such that b

αi α −1 Φ2i ϰi , = ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) i ϰi (ϱ)dϱ, (ψ(b) − ψ(0))αi

i ∈ {1, 2},

0

and Φ1 (y) = (Φ11 y1 , Φ12 y2 ) = (y1 (0), y2 (0)). Furthermore, the operator ℘−1 Φ1 : Img ℘ → ℏ1 ∩ ker Φ1 can be written as α ;ψ

α ;ψ

−1 −1 1 2 ℘−1 Φ1 (ϰ)(t) = (℘Φ11 ϰ1 (t), ℘Φ12 ϰ2 (t)) = (J0+ ϰ1 (t), J0+ ϰ2 (t)),

t ∈ T.

Proof. For ϰ ∈ ℏ2 , Φ2 2 ϰ = Φ2 ϰ and ϰ = Φ2 (ϰ) + (ϰ − Φ2 (ϰ)), where (ϰ − Φ2 (ϰ)) ∈ ker Φ2 = Img ℘. Using Img ℘ = ker Φ2 and Φ2 2 = Φ2 , we have Img Φ2 ∩ Img ℘ = {0}. So, ℏ2 = Img ℘ ⊕ Img Φ2 . Also, we may find that Img Φ1 = ker ℘ and Φ1 2 = Φ1 . It follows for each y ∈ ℏ1 that if y = (y − Φ1 (y)) + Φ1 (y), then ℏ1 = ker Φ1 + ker ℘. Clearly, we have ker Φ1 ∩ ker ℘ = {0}. So ℏ1 = ker Φ1 ⊕ ker ℘. Using the rank–nullity theorem, we get codim Img L = dim ℏ2 − dim Img L

= [dim ker Φ2 + dim Img Φ2 ] − dim Img L,

and since Img L = ker Φ2 , we have

6.2 Periodic solutions for pantograph coupled systems

codim Img L = dim Img Φ2 .

� 143

(6.4)

Using again the rank–nullity theorem, we obtain dim ker ℘ = dim ℏ1 − dim Img L = codim Img L, which implies that dim ker ℘ = codim Img L.

(6.5)

By (6.4) and (6.5) we have dim ker ℘ = codim Img L = dim Img Φ2 , and since dim Img Φ2 < ∞, we have dim ker ℘ = codim Img L < ∞. Since Img ℘ is a closed subset of ℏ2 , ℘ is a Fredholm operator of index zero. Now, we will demonstrate that ℘|Dom ℘∩ker Φ1 is ℘−1 Φ1 . Effectively, for ϰ ∈ Img ℘, by Theorem 2.5, we have α ;ψ

α ;ψ

α ;ψ

α ;ψ

c c 1 1 2 2 ℘℘−1 Φ1 (ϰ) = ( D0+ (J0+ ϰ1 ), D0+ (J0+ ϰ2 )) = (ϰ1 , ϰ2 ) = ϰ.

(6.6)

Furthermore, for y ∈ Dom ℘ ∩ ker Φ1 and t ∈ T we get α ;ψ

α ;ψ

α ;ψ c

α ;ψ

c 1 1 2 2 ℘−1 Φ1 (℘y)(t) = (J0+ ( D0+ y1 )(t), J0+ ( D0+ y2 ))

= (y1 (t) − y1 (0), y2 (t) − y2 (0)).

Using the fact that y ∈ Dom ℘ ∩ ker Φ1 , we have (y1 (0), y2 (0)) = (0, 0). Thus, ℘−1 Φ1 ℘(y) = y.

(6.7)

−1 Using (6.6) and (6.7) together, we get ℘−1 Φ1 = (℘|Dom ℘∩ker Φ1 ) .

We define ℵ : ℏ1 → ℏ2 by ℵy(t) = (ℵ1 y(t), ℵ2 y(t)) := (f1 (t, y1 (t), y2 (t)), f2 (t, y1 (t), y2 (t))), Then (6.1)–(6.2) is equivalent to ℘y = ℵy.

t ∈ T.

144 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems Lemma 6.3. Assume that the following condition is satisfied. (6.3.1) For each j ∈ {1, 2}, there exist positive constants βj , λj with 󵄨 󵄨󵄨 󵄨󵄨fj (t, y, ϰ) − fj (t, y, ϰ)󵄨󵄨󵄨 ⩽ βj |y − y| + λj |ϰ − ϰ| for every t ∈ T and y, y, ϰ, ϰ ∈ ℝ. For any bounded open set Ψ ⊂ ℏ1 , ℵ is ℘-compact. Proof. Let ϖ > 0 and Ψ = {y ∈ ℏ1 : ‖y‖ℏ1 < ϖ}. Step 1. We prove that Φ2 ℵ is continuous. Let (yκ )κ∈ℕ be a sequence where yκ 󳨀→ y in ℏ2 . Then for j ∈ {1, 2} and t ∈ T we have 󵄨 󵄨󵄨 󵄨󵄨Φ2j ℵj (yκ )(t) − Φ2j ℵj (y)(t)󵄨󵄨󵄨 ⩽

αj

b

(ψ(b) − ψ(0))αj

αj −1 󵄨

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

󵄨󵄨ℵj (yκ )(ϱ) − ℵj (y)(ϱ)󵄨󵄨󵄨dϱ. 󵄨 󵄨

0

By (6.3.1), we have 󵄨 󵄨󵄨 󵄨󵄨Φ21 ℵ1 (yκ )(t) − Φ21 ℵ1 (y)(t)󵄨󵄨󵄨 b

⩽

α1 β1 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 󵄨󵄨󵄨y1κ (ϱ) − y1 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α1 0

b

α1 λ1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 󵄨󵄨󵄨y2κ (ϱ) − y2 (ϱ)󵄨󵄨󵄨dϱ α 1 (ψ(b) − ψ(0)) 0

⩽

α1 β1 supt∈T |y1κ (t) − y1 (t)| (ψ(b) − ψ(0))α1

+

b

dϱ

0

α1 λ1 supt∈T |y2κ (t) − y2 (t)| (ψ(b) − ψ(0))α1

α1 −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) b

α1 −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

dϱ

0

⩽ 2(β∗ + λ∗ )‖yκ − y‖ℏ2 and 󵄨󵄨 󵄨 󵄨󵄨Φ22 ℵ2 (yκ )(t) − Φ22 ℵ2 (y)(t)󵄨󵄨󵄨 b

α2 β2 α −1 󵄨 󵄨 ⩽ ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨y1κ (ϱ) − y1 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2 0

6.2 Periodic solutions for pantograph coupled systems

� 145

b

α2 λ2 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨y2κ (ϱ) − y2 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2 0

⩽

α2 β2 supt∈T |y1κ (t) − y1 (t)| (ψ(b) − ψ(0))α2

+

b

α2 −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

dϱ

0

α2 λ2 supt∈T |y2κ (t) − y2 (t)| (ψ(b) − ψ(0))α2

b

α2 −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

dϱ

0

⩽ 2(β∗ + λ∗ )‖yκ − y‖ℏ2 , where β∗ = max1≤j≤2 βj and λ∗ = max1≤j≤2 λj . Thus, for each j ∈ {1, 2}, we get 󵄨 󵄨 sup󵄨󵄨󵄨Φ2j ℵj (yκ )(t) − Φ2j ℵj (y)(t)󵄨󵄨󵄨 ⩽ 2(β∗ + λ∗ )‖yκ − y‖ℏ2 , t∈T

and hence 󵄩󵄩 󵄩 󵄩󵄩Φ2 ℵ(yκ ) − Φ2 ℵ(y)󵄩󵄩󵄩ℏ2 󳨀→ 0

as κ 󳨀→ +∞.

We conclude that Φ2 ℵ is continuous. Step 2. We show that Φ2 ℵ(Ψ) is bounded. For each t ∈ T and y ∈ Ψ, we have b

α1 α −1 󵄨 󵄨 󵄨󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 󵄨󵄨󵄨ℵ1 (y)(ϱ)󵄨󵄨󵄨dϱ 󵄨󵄨Φ21 ℵ1 (y)(t)󵄨󵄨󵄨 ⩽ (ψ(b) − ψ(0))α1 α1 ⩽ (ψ(b) − ψ(0))α1

0

b

α1 −1 󵄨

󵄨󵄨f1 (ϱ, y1 (ϱ), y2 (ϱ)) − f1 (ϱ, 0, 0)󵄨󵄨󵄨dϱ 󵄨 󵄨

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0 b

α1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 󵄨󵄨󵄨f1 (ϱ, 0, 0)󵄨󵄨󵄨dϱ α 1 (ψ(b) − ψ(0)) 0

b

⩽

f1∗

α1 β1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 󵄨󵄨󵄨y1 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α1 0

b

+

α1 λ1 α −1 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 |y2 (ϱ)|dϱ (ψ(b) − ψ(0))α1

⩽ f ∗∗ + (β∗ + λ∗ )ϖ and

0

146 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems

󵄨 󵄨󵄨 󵄨󵄨Φ22 ℵ2 (y)(t)󵄨󵄨󵄨 ⩽

b

α2 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨ℵ2 (y)(ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2

α2 ⩽ (ψ(b) − ψ(0))α2

0

b

α2 −1 󵄨

󵄨󵄨f2 (ϱ, y1 (ϱ), y2 (ϱ)) − f2 (ϱ, 0, 0)󵄨󵄨󵄨dϱ 󵄨 󵄨

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0 b

+

α2 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨f2 (ϱ, 0, 0)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2 0

b

⩽ f2∗ +

α2 β2 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨y1 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2 0

b

+ ⩽f

α2 λ2 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 󵄨󵄨󵄨y2 (ϱ)󵄨󵄨󵄨dϱ (ψ(b) − ψ(0))α2

∗∗

+ (β + λ )ϖ, ∗

0

∗

where fj∗ = supt∈T |fj (t, 0, 0)|, j ∈ {1, 2}, and f ∗∗ = max1≤j≤2 fj∗ . Thus, for each j ∈ {1, 2} 󵄨 󵄨 sup󵄨󵄨󵄨Φ2j ℵj (y)(t)󵄨󵄨󵄨 ⩽ f ∗∗ + (β∗ + λ∗ )ϖ, t∈T

which implies that 󵄩󵄩 󵄩 ∗∗ ∗ ∗ 󵄩󵄩Φ2 ℵ(y)󵄩󵄩󵄩ℏ2 ⩽ f + (β + λ )ϖ. So, Φ2 ℵ(Ψ) is a bounded set in ℏ2 . Step 3. We demonstrate that ℘−1 Φ1 (id − Φ2 )ℵ : Ψ → ℏ1 is completely continuous. Firstly, for any y ∈ Ψ and t ∈ T, we get −1 −1 ℘−1 Φ1 (ℵy(t) − Φ2 ℵy(t)) = (℘Φ11 (ℵ1 y(t) − Φ21 ℵ1 y(t)), ℘Φ12 (ℵ2 y(t) − Φ22 ℵ2 y(t))),

where ℘−1 Φ11 (ℵ1 y(t) − Φ21 ℵ1 y(t)) α ;ψ

= J0+1 [f1 (t, y1 (t), y2 (t)) b

α1 α −1 − ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 f1 (ϱ, y1 (ϱ), y2 (ϱ))dϱ] (ψ(b) − ψ(0))α1 0

6.2 Periodic solutions for pantograph coupled systems

� 147

t

1 α −1 = ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) 1 f1 (ϱ, y1 (ϱ), y2 (ϱ))dϱ Γ(α1 ) 0

b

−

(ψ(t) − ψ(0))α1 α −1 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 1 f1 (ϱ, y1 (ϱ), y2 (ϱ))dϱ Γ(α1 )(ψ(b) − ψ(0))α1 0

and ℘−1 Φ12 (ℵ2 y(t) − Φ22 ℵ2 y(t)) α ;ψ

= J0+2 [f2 (t, y1 (t), y2 (t)) b

α2 α −1 − ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 f2 (ϱ, y1 (ϱ), y2 (ϱ))dϱ] α 2 (ψ(b) − ψ(0)) 0

t

=

1 α −1 ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) 2 f2 (ϱ, y1 (ϱ), y2 (ϱ))dϱ Γ(α2 ) 0

b

−

(ψ(t) − ψ(0))α2 α −1 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 2 f2 (ϱ, y1 (ϱ), y2 (ϱ))dϱ. Γ(α2 )(ψ(b) − ψ(0))α2 0

For all y ∈ Ψ, t ∈ T and j ∈ {1, 2}, we get 󵄨󵄨 −1 󵄨 󵄨󵄨℘Φ1j (id − Φ2j )ℵj y(t)󵄨󵄨󵄨 t

⩽

1 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) 0

t

1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) 0

b

1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) 0

b

+

1 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) 0

⩽

2fj∗

αj Γ(αj ) +

λj

Γ(αj )

α

(ψ(b) − ψ(0)) j + t

βj

Γ(αj )

0

αj −1 󵄨

∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) 0

t

αj −1 󵄨

∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) 󵄨󵄨y2 (ϱ)󵄨󵄨󵄨dϱ 󵄨 󵄨

󵄨󵄨y1 (ϱ)󵄨󵄨󵄨dϱ 󵄨 󵄨

148 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems

+

+

b

βj

Γ(αj )

αj −1 󵄨

󵄨󵄨y1 (ϱ)󵄨󵄨󵄨dϱ 󵄨 󵄨

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0 b

λj

Γ(αj )

αj −1 󵄨

󵄨󵄨y2 (ϱ)󵄨󵄨󵄨dϱ 󵄨 󵄨

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0

2(ψ(b) − ψ(0))αj ∗∗ [f + (β∗ + λ∗ )ϖ] ⩽ Γ(αj + 1) ⩽ 2 max{ 1≤j≤2

(ψ(b) − ψ(0))αj }[f ∗∗ + (β∗ + λ∗ )ϖ], Γ(αj + 1)

which implies that for each j ∈ {1, 2} we get α

(ψ(b) − ψ(0)) j 󵄨 󵄨󵄨 }[f ∗∗ + (β∗ + λ∗ )ϖ]. sup󵄨󵄨󵄨℘−1 Φ1j (id − Φ2j )ℵj y(t)󵄨󵄨 ⩽ 2 max{ 1≤j≤2 Γ(α + 1) j t∈T Therefore, α

(ψ(b) − ψ(0)) j 󵄩 󵄩󵄩 −1 }[f ∗∗ + (β∗ + λ∗ )ϖ]. 󵄩󵄩℘Φ1 (id − Φ2 )ℵy󵄩󵄩󵄩ℏ1 ⩽ 2 max{ 1≤j≤2 Γ(α + 1) j

This means that ℘−1 Φ1 (id − Φ2 )ℵ(Ψ) is uniformly bounded in ℏ1 .

It remains to show that ℘−1 Φ1 (id − Φ2 )ℵ(Ψ) is equicontinuous. For 0 < t1 < t2 ⩽ b, y ∈ Ψ, and j ∈ {1, 2}, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨℘Φ1j (id − Φ2j )ℵj y(t2 ) − ℘Φ1j (id − Φ2j )ℵj y(t1 )󵄨󵄨󵄨 t1

1 α −1 α −1 󵄨 󵄨 ⩽ ∫[ψ′ (ϱ)󵄨󵄨󵄨(ψ(t2 ) − ψ(ϱ)) j − (ψ(t1 ) − ψ(ϱ)) j 󵄨󵄨󵄨 Γ(αj ) 0

󵄨 󵄨 × 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ))󵄨󵄨󵄨]dϱ t2

1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(t2 ) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ))󵄨󵄨󵄨dϱ Γ(αj ) t1

+

[(ψ(t2 ) − ψ(0))αj − (ψ(t1 ) − ψ(0))αj ] Γ(αj )(ψ(b) − ψ(0))αj b

αj −1 󵄨

× ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0

t1

󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ))󵄨󵄨󵄨dϱ 󵄨 󵄨

1 α −1 󵄨 󵄨 ⩽ ∫ ψ′ (ϱ)(ψ(t1 ) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) 0

6.2 Periodic solutions for pantograph coupled systems

� 149

t1

1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(t2 ) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) +

t ∫01

0

ψ (ϱ)[(ψ(t1 ) − ψ(ϱ))αj −1 − (ψ(t2 ) − ψ(ϱ))αj −1 ]|fj (ϱ, 0, 0)|dϱ ′

Γ(αj )

t2

+

1 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(t2 ) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) t1

t2

1 α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(t2 ) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ Γ(αj ) t1

+

[(ψ(t2 ) − ψ(0))αj − (ψ(t1 ) − ψ(0))αj ] Γ(αj )(ψ(b) − ψ(0))αj b

αj −1 󵄨

× ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

󵄨󵄨fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ 󵄨 󵄨

0

+

[(ψ(t2 ) − ψ(0))αj − (ψ(t1 ) − ψ(0))αj ] Γ(αj )(ψ(b) − ψ(0))αj b

αj −1 󵄨

× ∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ)) 0

󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ 󵄨 󵄨

α

α

α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) j + Λ[(ψ(t1 ) − ψ(0)) j − (ψ(t2 ) − ψ(0)) j ] α

α

+ Λ[(ψ(t2 ) − ψ(0)) j − (ψ(t1 ) − ψ(0)) j ] α

⩽ 2Λ(ψ(t2 ) − ψ(t1 )) j , where Λ=

f ∗∗ + (β∗ + λ∗ )ϖ . Γ(αj + 1)

The operator ℘−1 Φ1 (id−Φ2 )ℵ(Ψ) is equicontinuous in ℏ1 because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà– Ascoli theorem implies that ℘−1 Φ1 (id − Φ2 )ℵ(Ψ) is relatively compact in ℏ1 . Consequently, ℵ is ℘-compact in Ψ. Lemma 6.4. Assume that (6.3.1) is satisfied. If max{ 1≤j≤2

(ψ(b) − ψ(0))αj 1 }(β∗ + λ∗ ) < , Γ(αj + 1) 2

(6.8)

150 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems then there exists ℑ > 0 which is independent of ω, where for y ∈ ℏ1 we get ℘(y) − ℵ(y) = −ω[℘(y) + ℵ(−y)] 󳨐⇒ ‖y‖ℏ1 ⩽ ℑ,

ω ∈ (0, 1].

Proof. Let y ∈ ℏ1 satisfy ℘(y) − ℵ(y) = −ω℘(y) − ωℵ(−y). Then ℘(y) =

1 ω ℵ(y) − ℵ(−y). 1+ω 1+ω

So, we obtain for any j ∈ {1, 2} and t ∈ T α ;ψ

℘j yj (t) = c D0+j yj (t) =

1 ω f (t, y1 (t), y2 (t)) − f (t, −y1 (t), −y2 (t)). 1+ω j 1+ω j

By Lemma 2.4 we get yj (t) = yj (0) +

1 α ;ψ [J0+j (fj (ϱ, y1 (ϱ), y2 (ϱ)))(t) ω+1

α ;ψ

− ωJ0+j (fj (ϱ, −y1 (ϱ), −y2 (ϱ)))(t)]. Thus, for every j ∈ {1, 2} and t ∈ T we obtain t

′ αj −1 󵄨 󵄨 ∫ ψ (ϱ)(ψ(t) − ψ(ϱ)) |fj (ϱ, y1 (ϱ), y2 (ϱ))|dϱ |yj (t)| ⩽ 󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 + 0 (ω + 1)Γ(αj ) t

ω α −1 󵄨 󵄨 + ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, −y1 (ϱ), −y2 (ϱ))󵄨󵄨󵄨dϱ (ω + 1)Γ(αj ) 0

󵄨 󵄨 ⩽ 󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 t

+

∫0 ψ′ (ϱ)(ψ(t) − ψ(ϱ))αj −1 |fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, 0, 0)|dϱ (ω + 1)Γ(αj )

t

+

+

1 α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ (ω + 1)Γ(αj ) t ω ∫0

0

ψ (ϱ)(ψ(t) − ψ(ϱ))αj −1 |fj (ϱ, −y1 (ϱ), −y2 (ϱ)) − fj (ϱ, 0, 0)|dϱ ′

(ω + 1)Γ(αj )

t

+

ω α −1 󵄨 󵄨 ∫ ψ′ (ϱ)(ψ(t) − ψ(ϱ)) j 󵄨󵄨󵄨fj (ϱ, 0, 0)󵄨󵄨󵄨dϱ (ω + 1)Γ(αj )

󵄨 󵄨 ⩽ 󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 +

0 ∗ 2fj (ψ(b)

− ψ(0))αj

Γ(αj + 1)

+

2(βj + λj ) Γ(αj + 1)

α

(ψ(b) − ψ(0)) j ‖y‖ℏ1

6.2 Periodic solutions for pantograph coupled systems

� 151

∗∗ α 󵄨 2f (ψ(b) − ψ(0)) j 󵄨 ⩽ max󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 + 1≤j≤2 Γ(αj + 1)

+

2(β∗ + λ∗ ) α (ψ(b) − ψ(0)) j ‖y‖ℏ1 Γ(αj + 1)

α

(ψ(b) − ψ(0)) j 󵄨 󵄨 ⩽ max󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 + 2 max{ }[f ∗∗ + (β∗ + λ∗ )‖y‖ℏ1 ]. 1≤j≤2 1≤j≤2 Γ(αj + 1) Thus, for every j ∈ {1, 2} we have 󵄨 󵄨 󵄨 󵄨 sup󵄨󵄨󵄨yj (t)󵄨󵄨󵄨 ⩽ max󵄨󵄨󵄨yj (0)󵄨󵄨󵄨 1≤j≤2 t∈T

+ 2 max{ 1≤j≤2

(ψ(b) − ψ(0))αj }[f ∗∗ + (β∗ + λ∗ )‖y‖ℏ1 ]. Γ(αj + 1)

We deduce that αj

‖y‖ℏ1 ⩽

max1≤j≤2 |yj (0)| + 2f ∗∗ max1≤j≤2 { (ψ(b)−ψ(0)) } Γ(α +1) j

αj

[1 − 2 max1≤j≤2 { (ψ(b)−ψ(0)) }(β∗ + λ∗ )] Γ(α +1)

:= ℑ.

j

Lemma 6.5. If (6.3.1) and (6.8) are met, then there exists a bounded open set Ψ ⊂ ℏ1 with ℘(y) − ℵ(y) ≠ −ω[℘(y) + ℵ(−y)]

(6.9)

for any y ∈ 𝜕Ψ and any ω ∈ (0, 1]. Proof. By Lemma 6.4, we have the existence of ℑ > 0 which is independent of ω, where if y satisfies ℘(y) − ℵ(y) = −ω[℘(y) + ℵ(−y)],

ω ∈ (0, 1],

then ‖y‖ℏ1 ⩽ ℑ. So, if Ψ = {y ∈ ℏ1 ; ‖y‖ℏ1 < ϑ},

(6.10)

where ϑ > ℑ, we obtain ℘(y) − ℵ(y) ≠ −ω[℘(y) − ℵ(−y)] for all y ∈ 𝜕Ψ = {y ∈ ℏ1 ; ‖y‖ℏ1 = ϑ} and ω ∈ (0, 1]. Theorem 6.1. Assume that (6.3.1) and (6.8) are met. Then there exists at least one solution for (6.1)–(6.2) in Dom ℘ ∩ Ψ. Proof. Clearly, Ψ given in (6.10) is symmetric, 0 ∈ Ψ, and ℏ1 ∩ Ψ = Ψ ≠ 0. In addition, by Lemma 6.5, assume that (6.3.1) and (6.8) hold. Then

152 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems

℘(y) − ℵ(y) ≠ −ω[℘(y) − ℵ(−y)] for each y ∈ ℏ1 ∩ 𝜕Ψ = 𝜕Ψ and each ω ∈ (0, 1]. Thus, problem (6.1)–(6.2) has at least one solution in Dom ℘ ∩ Ψ. Theorem 6.2. Let (6.3.1) be satisfied. Further, we assume that the following hypothesis holds. (6.7.1) There exist constants βj > 0 and λj ⩾ 0, j ∈ {1, 2}, where 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y, ϰ) − f1 (t, y,̄ ϰ)̄ 󵄨󵄨󵄨 ⩾ β1 |y − y|̄ − λ1 |ϰ − ϰ|̄ and 󵄨󵄨 󵄨 󵄨󵄨f2 (t, y, ϰ) − f2 (t, y,̄ ϰ)̄ 󵄨󵄨󵄨 ⩾ β2 |ϰ − ϰ|̄ − λ2 |y − y|̄ for every t ∈ T and y, y,̄ ϰ, ϰ̄ ∈ ℝ. If one has max{

λ1 λ2 (ψ(b) − ψ(0))αj , } + 2 max{ }(β∗ + λ∗ ) < 1, 1≤j≤2 Γ(αj + 1) β1 β2

(6.11)

then the problem (6.1)–(6.2) has a unique solution in Dom ℘ ∩ Ψ. Proof. Note that condition (6.11) is stronger than condition (6.8). By Theorem 6.1, the problem (6.1)–(6.2) has at least one solution in Dom ℘ ∩ Ψ. Now, we prove the uniqueness result. Suppose that the problem (6.1)–(6.2) has two different solutions y, y ∈ Dom ℘ ∩ Ψ. Then we have for each t ∈ T and j ∈ {1, 2} c c

α ;ψ

D0+j yj (t) = fj (t, y1 (t), y2 (t)), α ;ψ

D0+j yj (t) = fj (t, y1 (t), y2 (t)),

and yj (0) = yj (b),

yj (0) = yj (b).

Let S(t) = y(t) − y(t) for all t ∈ T, which means that S(t) = (S1 (t), S2 (t)) = (y1 (t) − y1 (t), y2 (t) − y2 (t)) Then ℘S(t) = (℘1 S1 (t), ℘2 S2 (t)) α ;ψ

α ;ψ

= (c D0+1 S1 (t), c D0+2 S2 (t))

for all t ∈ T.

6.2 Periodic solutions for pantograph coupled systems

α ;ψ

α ;ψ

α ;ψ

� 153

α ;ψ

= (c D0+1 y1 (t) − c D0+1 y1 (t), c D0+2 y2 (t) − c D0+2 y2 (t))

= (f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t)),

f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))).

(6.12)

Using Img ℘ = ker Φ2 , for j ∈ {1, 2} we have b

αj −1

∫ ψ′ (ϱ)(ψ(b) − ψ(ϱ))

[fj (ϱ, y1 (ϱ), y2 (ϱ)) − fj (ϱ, y1 (ϱ), y2 (ϱ))]dϱ = 0.

0

Since for each j ∈ {1, 2} the functions fj are continuous, there exists t0 ∈ [0, b] such that fj (t0 , y1 (t0 ), y2 (t0 )) − fj (t0 , y1 (t0 ), y2 (t0 )) = 0. In view of (6.7.1), we have 󵄨 λ 󵄨 󵄨 λ 󵄨 󵄨 󵄨󵄨 󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 ⩽ 1 󵄨󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ 1 sup󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 β1 β1 t∈T and 󵄨 󵄨󵄨 󵄨 λ 󵄨 󵄨 λ 󵄨 󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ 2 󵄨󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 ⩽ 2 sup󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨, β2 β2 t∈T which implies that λ λ 󵄨󵄨 󵄨 󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 ⩽ max{ 1 , 2 }‖y − y‖ℏ1 β1 β2 and λ λ 󵄨󵄨 󵄨 󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ max{ 1 , 2 }‖y − y‖ℏ1 . β1 β2 Then for each j ∈ {1, 2} we have λ λ 󵄨󵄨 󵄨 󵄨󵄨Sj (t0 )󵄨󵄨󵄨 ⩽ max{ 1 , 2 }‖S‖ℏ1 . β1 β2 By Theorem 2.4, for any t ∈ [0, b] and j ∈ {1, 2} we have α ;ψ c

J0+j

which implies that

α ;ψ

D0+j Sj (t) = Sj (t) − Sj (0),

(6.13)

154 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems α ;ψ c

Sj (0) = Sj (t0 ) − J0+j

α ;ψ

D0+j Sj (t0 ),

and therefore α ;ψ c

Sj (t) = J0+j

α ;ψ

α ;ψ c

D0+j Sj (t) + Sj (t0 ) − J0+j

α ;ψ

D0+j Sj (t0 ).

Using (6.13), we obtain for every t ∈ T and for each j ∈ {1, 2} 󵄨󵄨 󵄨 󵄨 αj ;ψ c αj ;ψ 󵄨 󵄨 󵄨 󵄨 αj ;ψ c αj ;ψ 󵄨 󵄨󵄨Sj (t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨J0+ D0+ Sj (t)󵄨󵄨󵄨 + 󵄨󵄨󵄨Sj (t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ Sj (t0 )󵄨󵄨󵄨 λ λ 2(ψ(b) − ψ(0))αj 󵄨 α ;ψ 󵄨 ⩽ max{ 1 , 2 }‖S‖ℏ1 + sup󵄨󵄨󵄨c D0+j Sj (t)󵄨󵄨󵄨 Γ(αj + 1) β1 β2 t∈T ⩽ max{

λ1 λ2 (ψ(b) − ψ(0))αj 󵄨 α ;ψ 󵄨 , }‖S‖ℏ1 + 2 max{ } sup󵄨󵄨󵄨c D0+j Sj (t)󵄨󵄨󵄨. 1≤j≤2 Γ(α + 1) β1 β2 j t∈T

Then λ λ (ψ(b) − ψ(0))αj 󵄩󵄩c α;ψ 󵄩󵄩 󵄨󵄨 󵄨 }󵄩󵄩 D0+ S󵄩󵄩ℏ . 󵄨󵄨Sj (t)󵄨󵄨󵄨 ⩽ max{ 1 , 2 }‖S‖ℏ1 + 2 max{ 1 1≤j≤2 Γ(αj + 1) β1 β2

(6.14)

By (6.12) and (6.3.1), for each j ∈ {1, 2} we find that 󵄨 󵄨 󵄨 󵄨󵄨c αj ;ψ 󵄨󵄨 D0+ Sj (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨fj (t, y1 (t), y2 (t)) − fj (t, y1 (t), y2 (t))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ βj sup 󵄨󵄨󵄨S1 (t)󵄨󵄨󵄨 + λj sup 󵄨󵄨󵄨S2 (t)󵄨󵄨󵄨 t∈T

⩽ (βj + λj )‖S‖ℏ1

t∈T

⩽ (β∗ + λ∗ )‖S‖ℏ1 .

Then 󵄩󵄩c α;ψ 󵄩󵄩 ∗ ∗ 󵄩󵄩 D0+ S󵄩󵄩ℏ1 ⩽ (β + λ )‖S‖ℏ1 .

(6.15)

Substituting (6.15) in the right side of (6.14), we get for every t ∈ T and for each j ∈ {1, 2} λ λ (ψ(b) − ψ(0))αj 󵄨󵄨 󵄨 }(β∗ + λ∗ )]‖S‖ℏ1 . 󵄨󵄨Sj (t)󵄨󵄨󵄨 ⩽ [max{ 1 , 2 } + 2 max{ 1≤j≤2 Γ(αj + 1) β1 β2 Therefore, ‖S‖ℏ1 ⩽ [max{

λ1 λ2 (ψ(b) − ψ(0))αj , } + 2 max{ }(β∗ + λ∗ )]‖S‖ℏ1 . 1≤j≤2 Γ(αj + 1) β1 β2

Hence, by (6.11) we conclude that ‖S‖ℏ1 = 0.

6.2 Periodic solutions for pantograph coupled systems

� 155

As a result, for any t ∈ T we get S(t) = 0 󳨐⇒ y(t) = y(t).

6.2.3 An example Example 6.1. Consider the following system: 1

;et

c 2 { D + y1 (t) = f1 (t, y1 (t), y2 (t)), { { { 01 ;et c 3 D0+ y2 (t) = f2 (t, y1 (t), y2 (t)), { { { { {y1 (0) = y1 (1) and y2 (0) = y2 (1),

t ∈ T := [0, 1],

where f1 (t, y1 (t), y2 (t)) = ln (t + 2) +

2et e9 t cos y1 (t). + y2 (t) + t 5 55√π 7(1 + 3 ) −t

f2 (t, y1 (t), y2 (t)) =

3 1 e−11−t (sin y1 (t) + y1 (t)) + , 2 37(1 + y2 (t)) 23√π

Here α1 = 21 , α2 = 31 , and ψ(t) = et . It is easy to see that f1 , f2 ∈ C([0, 1] × ℝ2 , ℝ). Let y, y, ϰ, ϰ ∈ ℝ and t ∈ T. Then ̄ ⩽ {|f1 (t, y, ϰ) − f1 (t, y,̄ ϰ)| { ̄ ⩽ |f (t, y, ϰ) − f2 (t, y,̄ ϰ)| { 2

5 |y 46√π 1 |y 55√π

Hence, the assumption (6.3.1) is satisfied with β1 = 5 which implies that β∗ = 46√π and λ∗ = 71 . By simple calculations, we see that

max{ 1≤j≤2

− y|̄ + − y|̄ +

1 ̄ |ϰ − ϰ|, 37e11 1 ̄ |ϰ − ϰ|. 7

5 , β2 46√π

=

1 ,λ 55√π 1

(ψ(b) − ψ(0))αj 1 }(β∗ + λ∗ ) ≈ 0.302 < . Γ(αj + 1) 2

With the use of Theorem 6.1, our problem has at least one solution. Otherwise, for each y, y, ϰ, ϰ ∈ ℝ and t ∈ T, we have ̄ ⩾ {|f1 (t, y, ϰ) − f1 (t, y,̄ ϰ)| { ̄ ̄ {|f2 (t, y, ϰ) − f2 (t, y, ϰ)| ⩾

1 ̄ |y − y|̄ − 37e1 11 |ϰ − ϰ|, 46√π −1 3e 9 1 ̄ |ϰ − ϰ|̄ − 55√π |y − y|. 28

=

1 , and λ2 37e11

= 71 ,

156 � 6 Nonlinear ψ-Caputo fractional pantograph coupled systems Then (6.7.1) is satisfied with β1 = that max{ λ1 ,

λ2 } β1 β2

We have

≈ 0.035.

max{

1 , β2 46√π

−1

=

3e 9 28

, λ1 =

1 , and λ2 37e11

=

1 , which implies 55√π

λ1 λ2 (ψ(b) − ψ(0))αj }(β∗ + λ∗ ) ≈ 0.64 < 1. , } + 2 max{ 1≤j≤2 Γ(αj + 1) β1 β2

So, by Theorem 6.2, our problem has a unique solution.

6.3 Notes and remarks This chapter’s results are based on the article [87]. The monographs [99, 109, 115, 125, 138, 144, 159, 233] and the papers [95, 123, 158, 240] include additional pertinent results and investigations.

7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative 7.1 Introduction and motivations In this chapter, we establish the existence and uniqueness results for a class of problems with nonlinear fractional differential and integro-differential equations (NFDEs) with ψ-Hilfer fractional operator. Further, examples are given to illustrate the viability of our results in each section. The results of our analysis in this chapter can be viewed as a conditional extension of the problems discussed fairly recently in the following: – The monographs of Abbas et al. [7, 14, 15] and Benchohra et al. [75] and the papers of Abbas et al. [10, 13, 148, 157], Benchohra et al. [71, 77, 79, 80], Liu et al. [164], Kharade et al. [142], and Sousa et al. [103, 104], where considerable attention has been given to the study of the existence results and Ulam stability of a class of initial and boundary value problems for fractional differential equations. – Several researchers have investigated different extensions of some classical fractional operators. In 2018, Vanterler et al. discussed the so-called ψ-Hilfer fractional derivative [43]. For some new research related to the study of some class of fractional differential equations involving the generalized Hilfer fractional derivative, see [231] and the references therein. – The paper of Krim et al. [148], where the authors discussed the following terminal value problem for fractional differential equations with Katugampola fractional derivative: (ρ D0r+ x)(τ) = κ(τ, x(τ), (ρ D0r+ x)(τ)),

{

–

τ ∈ I := [0, T],

x(T) = xT ∈ ℝ,

where T > 0 and the function κ : I × ℝ × ℝ → ℝ is continuous. Here, ρ D0r+ is the Katugampola fractional derivative of order r ∈ (0, 1]. Their reasoning is mainly based upon on α-ϕ-Geraghty-type contraction and fixed point theory. The paper of Almalahi et al. [41], where, using the Krasnoselskii fixed point theorem, the Picard operator method, and Gronwall’s inequality lemma, the authors proved some existence and stability results for the following nonlocal initial value problem for differential equations involving the ψ-Hilfer fractional derivative: α,β;ψ

H D0+ x(t) = f (t, x(t), x(g(t))), t ∈ (0, b], { { { 1−γ;ψ J x(0+ ) = ∑ki=1 ci x(τi ), τi ∈ (0, b), { { 0+ { {x(t) = φ(t), t ∈ [−r, 0], https://doi.org/10.1515/9783111334387-007

158 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative α,β;ψ

where H D0+

–

(⋅) is the ψ-Hilfer fractional derivative operator of order α ∈ (0, 1) and 1−γ,ψ

type β ∈ [0, 1], J0+ (⋅) is the ψ-fractional integral in Riemann–Liouville sense of order 1 − γ, where γ = α + β(1 − α), 0 < γ < 1, τi , i = 1, 2, . . . , k, are prefixed points satisfying 0 < τ1 ≤ τ2 ≤ ⋅ ⋅ ⋅ ≤ τi < b, ci ∈ ℝ, φ ∈ C[−r, 0], f : (0, b] × ℝ × ℝ → ℝ is a given continuous function, and g ∈ C(0, b] → [−r, b] with g(t) ≤ t, r > 0. Some interesting results can be found in the paper [40]. In the work [231], the authors presented existence and uniqueness results for fractional integro-differential equations with ψ-Hilfer fractional derivative: H

α,β;ψ D a+ y(t)

t

= f (t, y(t), ∫ κ(t, s)y(s)ds), a 1−γ;ψ + Ja+ [ny(a )

t ∈ (a, b], a < b < +∞,

+ my(b− )] = c,

α,β;ψ

–

where H D a+ is the generalized Hilfer fractional derivative of order α ∈ (0, 1) and 1−γ;ψ type β ∈ [0, 1], Ja+ is the generalized fractional integral in the sense of Riemann– Liouville of order 1−γ (γ = α +β −αβ), n, m ∈ ℝ, where n+m ≠ 0, c ∈ E (E is a Banach space), f : (a, b] × ℝ × ℝ → ℝ and κ : [a, b] × [a, b] → ℝ are continuous functions, t and By(t) = ∫a κ(t, s)y(s)ds is a linear integral operator. The reasoning is mainly based upon different types of classical fixed point theory. However, if n + m = 0, this method is invalid. In [243], using fixed point theory, Tidke studied the existence and uniqueness of solutions for the mixed Volterra–Fredholm integro-differential problem t

b

y′ (t) = f (y(t), ∫0 κ(t, s, y(s))ds, ∫0 h(t, s, y(s))ds), t ∈ [0, b], { y(0) + g(y) = y0 . –

In [50], Anguraj et al. studied the existence and uniqueness of solutions for the mixed Volterra–Fredholm integro-differential problem with Caputo fractional derivative of order 0 < α ≤ 1, associated with integral boundary conditions, of the form d α y(t) α { dt

t

y(0) =

–

1

= f (t, y(t), ∫0 κ(t, s, y(s))ds, ∫0 h(t, s, y(s))ds),

t ∈ [0, 1],

1 ∫0 g(s)y(s)ds.

By using fixed point theory for the boundary value problems of ψ-Hilfer fractional integro-differential equations, Sudsutad et al. [236] studied the existence, uniqueness, and stability of solutions for the following problem with mixed nonlocal boundary conditions: H

{

α,ρ;ψ

D0+

ϕ,ψ

x(t) = f (t, x(t), I0+ x(t)),

x(0) = 0,

∑m i=1 δi x(ηi )

+

t ∈ (0, T],

β ,ψ ∑nj=1 ωj I0+j x(θj )

μ ,ρ;ψ

+ ∑rk=1 λk H D0+k

x(tk ) = κ.

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces

� 159

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces Motivated by the above papers, we investigate some existence and uniqueness results for solutions to fractional integro-differential equations with ψ-Hilfer fractional derivative: H

t

α,β;ψ

D a+ y(t) = f (t, y(t), ∫ κ(t, s)y(s)ds), 1−γ;ψ

Ja+

1−γ;ψ

y(a) = Ja+

for each t ∈ (a, b], a < b < +∞,

(7.1)

a

y(b),

(7.2)

α,β;ψ

where H D a+ is the generalized Hilfer fractional derivative of order α ∈ (0, 1) and type 1−γ;ψ β ∈ [0, 1], Ja+ is the generalized fractional integral in the sense of Riemann–Liouville of order 1 − γ (γ = α + β − αβ), E is a Banach space, f : (a, b] × ℝ × ℝ → ℝ and κ : t [a, b] × [a, b] → ℝ are continuous functions, and By(t) = ∫a κ(t, s)y(s)ds is a linear integral operator. 7.2.1 Existence results Let α;ψ

X = {y ∈ C1−γ;ψ (J,̄ ℝ) : y(t) = Ja+ υ(t) : υ ∈ C1−γ;ψ (J,̄ ℝ), t ∈ (a, b]}

and let Y = C1−γ;ψ (J,̄ ℝ) with the norm ‖y‖X = ‖y‖Y = ‖y‖C1−γ;ψ . We now give the definition of the operator L : Dom L ⊆ X → Y : α,β;ψ

Ly := H D a+ y,

(7.3)

where α,β;ψ

1−γ;ψ

Dom L = {y ∈ X : H D a+ y ∈ Y : Ja+

1−γ;ψ

y(a) = Ja+

y(b)}.

Lemma 7.1. Using the definition of L given in (7.3), we have ker L = {y ∈ X : y(t) =

1−γ;ψ

Ja+

y(a)

Γ(γ)

γ−1

(ψ(t) − ψ(a))

and 1+β(α−1);ψ

Img L = {υ ∈ Y : Ja+

υ(b) = 0}.

, t ∈ (a, b]}

160 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative H

Proof. By Remark 2.2, for all y ∈ X the equation Ly = solution given by y(t) =

1−γ;ψ

Ja+

y(a)

Γ(γ)

γ−1

(ψ(t) − ψ(a))

α,β;ψ

D a+ y = 0 in (a, b] has a

t ∈ (a, b],

,

which implies that ker L = {y ∈ X : y(t) =

1−γ;ψ

Ja+

y(a)

Γ(γ)

γ−1

(ψ(t) − ψ(a))

, t ∈ (a, b]}.

For υ ∈ Img L, there exists y ∈ Dom L such that υ = Ly ∈ Y . Using Lemma 2.20, we obtain for each t ∈ (a, b] y(t) =

1−γ;ψ

Ja +

y(a)

Γ(γ)

(ψ(t) − ψ(a))

γ−1

α;ψ

+ Ja+ υ(t).

Using Lemma 2.6, we obtain 1−γ;ψ

Ja+

1−γ;ψ

1+β(α−1);ψ

y(t) = Ja+

y(a) + Ja+

1−γ;ψ

1−γ;ψ

Since y ∈ Dom L, we have Ja+

y(a) = Ja+

y(b). Thus,

1+β(α−1);ψ

υ(b) = 0.

1+β(α−1);ψ

υ(b) = 0,

Ja+

υ(t).

Furthermore, if υ ∈ Y and satisfies Ja+

α,β;ψ

α;ψ

then for any y(t) = Ja+ υ(t), we get υ(t) = H D a+ y(t). Therefore, 1−γ;ψ

Ja+

1−γ;ψ

y(b) = Ja+

y(a),

which implies that y ∈ Dom L, so that υ ∈ Img L. So 1+β(α−1);ψ

Img L = {υ ∈ Y : Ja+

υ(b) = 0},

which completes the proof. Lemma 7.2. Let L be defined by (7.3). Then L is a Fredholm operator of index zero, and the linear continuous projector operators Q : Y → Y and P : X → X can be written as

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces

Q υ(t) =

� 161

Γ(2 + β(α − 1)) 1+β(α−1);ψ J + υ(b) (ψ(b) − ψ(a))1+β(α−1) a

and 1−γ;ψ

Ja+

P (y)(t) =

y(a)

Γ(γ)

(ψ(t) − ψ(a))

γ−1

.

Furthermore, the operator L−1 P : Img L → X ∩ ker P can be written as α;ψ

L−1 P (υ)(t) = Ja+ υ(t). Proof. Obviously, for each υ ∈ Y , Q 2 υ = Q υ and υ = (υ−Q (υ))+Q (υ), where (υ−Q (υ)) ∈ ker Q = Img L. Using the fact that Img L = ker Q and Q 2 = Q , we have Img Q ∩ Img L = 0. So, Y = Img L ⊕ Img Q .

In the same way, we get Img P = ker L and P 2 = P . It follows for each y ∈ X that if y = (y − P (y)) + P (y), then X = ker P + ker L. Clearly, we have ker P ∩ ker L = 0. Thus, X = ker P ⊕ ker L.

Therefore, dim ker L = dim Img Q = codim Img L. Consequently, L is a Fredholm operator of index zero. Now, we will show that the inverse of L|Dom L∩ker P is L−1 P . Effectively, for υ ∈ Img L, by Lemma 2.12 we have α,β;ψ

1−γ;ψ

H LL−1 P (υ) = D a+ (Ja+

υ) = υ.

(7.4)

Furthermore, for y ∈ Dom L ∩ ker P we get 1−γ;ψ H

L−1 P (L(y(t))) = Ja+

α,β;ψ

( D a+ y(t)) = y(t) −

1−γ;ψ

Ja +

y(a)

Γ(γ)

Using the fact that y ∈ Dom L ∩ ker P , we have 1−γ;ψ

Ja+

y(a)

Γ(γ)

Thus,

γ−1

(ψ(t) − ψ(a))

= 0.

(ψ(t) − ψ(a))

γ−1

.

162 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative L−1 P L(y) = y.

(7.5)

−1 Using (7.4) and (7.5) together, we get L−1 P = (L|Dom L∩ker P ) , which completes the demonstration.

Lemma 7.3. For all y, ȳ ∈ C1−γ;ψ (J,̄ ℝ) and t ∈ J ̄ we get 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ η‖y − y‖̄ C1−γ;ψ , 󵄨󵄨By(t) − By(t) where η=

(ψ(b) − ψ(a))γ 󵄨 󵄨 max󵄨󵄨κ(t, s)󵄨󵄨󵄨. γ mint∈J ̄ ψ′ (t) t,s∈J ̄ 󵄨

Proof. For any t ∈ J,̄ we have t

󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ ∫󵄨󵄨󵄨󵄨κ(t, s)󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨y(s) − y(s) ̄ 󵄨󵄨󵄨󵄨ds 󵄨󵄨By(t) − By(t) a

t

γ−1 󵄨 󵄨 ⩽ max󵄨󵄨󵄨κ(t, s)󵄨󵄨󵄨‖y − y‖̄ C1−γ ;ψ ∫(ψ(s) − ψ(a)) ds ̄ t,s∈J

a

t

γ−1 󵄨 󵄨 ⩽ max󵄨󵄨󵄨κ(t, s)󵄨󵄨󵄨‖y − y‖̄ C1−γ ;ψ ∫ ψ′ (s)(ψ(s) − ψ(a)) t,s∈J ̄

⩽

a

‖y − y‖̄ C1−γ ;ψ maxt,s∈J ̄ |κ(t, s)| mint∈[a,b] ψ′ (t)

t

1 ds ψ′ (s) γ−1

∫ ψ′ (s)(ψ(s) − ψ(a)) a

ds

(ψ(b) − ψ(a))γ 󵄨 󵄨 ⩽ max󵄨󵄨κ(t, s)󵄨󵄨󵄨‖y − y‖̄ C1−γ ;ψ := η‖y − y‖̄ C1−γ;ψ . γ mint∈[a,b] ψ′ (t) t,s∈J ̄ 󵄨 Define N : X → Y by t

N y(t) := f (t, y(t), ∫ κ(t, s)y(s)ds),

t ∈ (a, b].

a

Then the problem (7.1)–(7.2) is equivalent to the problem Ly = N y. Lemma 7.4. Suppose that the following assumptions are satisfied. (7.4.1) The function f : (a, b] × ℝ × ℝ → ℝ is such that f (⋅, y(⋅), B(y)(⋅)) ∈ C1−γ;ψ (J,̄ ℝ) for all y ∈ C1−γ;ψ (J,̄ ℝ), t

where (By)(t) := ∫a κ(t, s)y(s)ds.

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces

� 163

(7.4.2) There exist positive constants R, R such that 󵄨󵄨 󵄨󵄨 󵄨 ̄ 󵄨󵄨󵄨󵄨 ̄ B(y)(t)) ̄ ̄ 󵄨󵄨󵄨󵄨 + R󵄨󵄨󵄨󵄨By(t) − By(t) 󵄨󵄨f (t, y(t), B(y(t))) − f (t, y(t), 󵄨󵄨 ⩽ R󵄨󵄨󵄨y(t) − y(t) for t ∈ (a, b] and y, ȳ ∈ C1−γ;ψ (J,̄ ℝ). Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps: Step 1. We prove that QN is continuous. By the hypotheses on f and the Lebesgue dominated convergence theorem we get the continuity of QN . Step 2. We show that QN (Ω) is bounded. For t ∈ J ̄ and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 ⩽

Γ(2 + β(α − 1)) (ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1)) b

β(α−1) 󵄨

󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds 󵄨 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

⩽

Γ(2 + β(α − 1)) (ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1)) b

β(α−1) 󵄨 󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds × [∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 [a b

β(α−1) 󵄨

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) a

⩽

󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds] 󵄨 󵄨 ]

Γ(2 + β(α − 1)) (ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1)) b

β(α−1) 󵄨 󵄨 󵄨 󵄨 × [∫ ψ′ (s)(ψ(b) − ψ(s)) [R󵄨󵄨󵄨y(s)󵄨󵄨󵄨 + R󵄨󵄨󵄨By(s)󵄨󵄨󵄨]ds [a b

β(α−1) 󵄨

+ ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds] 󵄨 󵄨 ]

(RM + f ∗ )Γ(2 + β(α − 1)) 1+β(α−1);ψ γ−1 ⩽ Ja+ (ψ(s) − ψ(a)) (b) + ηRM (ψ(b) − ψ(a))1+β(α−1)

164 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative

⩽

Γ(γ)Γ(2 + β(α − 1)) γ−1 (RM + f ∗ )(ψ(b) − ψ(a)) + ηRM , Γ(α + 1)

where f ∗ = ‖f (⋅, 0, 0)‖C1−γ;ψ . Thus, Γ(γ)Γ(2 + β(α − 1)) 1−γ 󵄩 󵄩󵄩 (RM + f ∗ ) + ηRM (ψ(b) − ψ(a)) . 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ Γ(α + 1) Consonantly, QN (Ω) is a bounded set in Y . Step 3. We prove that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ J,̄ we get L−1 P (N y(t) − QN y(t)) α;ψ

= Ja+ [f (t, y(t), By(t)) Γ(2 + β(α − 1)) 1+β(α−1);ψ J + f (s, y(s), By(s))(b)] (ψ(b) − ψ(a))1+β(α−1) a

−

t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), By(s))ds Γ(α) a

Γ(2 + β(α − 1)) α (ψ(t) − ψ(a)) 1+β(α−1) Γ(α + 1)(ψ(b) − ψ(a))

−

1+β(α−1);ψ

× Ja+

f (s, y(s), By(s))(b).

For all y ∈ Ω and t ∈ J,̄ we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

+

t

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

Γ(2 + β(α − 1))(ψ(b) − ψ(a))γ−1 + Γ(α + 1)Γ(1 + β(α − 1)) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

+

󵄨󵄨f (s, y(s), By(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

Γ(2 + β(α − 1))(ψ(b) − ψ(a))γ−1 Γ(α + 1)Γ(1 + β(α − 1))

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces b

β(α−1) 󵄨

󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

Γ(γ) γ+α−1 ⩽ (RM + f ∗ )(ψ(b) − ψ(a)) Γ(γ + α) Γ(γ)Γ(2 + β(α − 1)) γ+α−1 + (RM + f ∗ )(ψ(b) − ψ(a)) 2 Γ (α + 1)

2ηRM (ψ(b) − ψ(a))α Γ(α + 1) Γ(2 + β(α − 1)) 1 γ+α−1 ⩽[ ]Γ(γ)(RM + f ∗ )(ψ(b) − ψ(a)) + Γ(γ + α) Γ2 (α + 1) +

2ηRM (ψ(b) − ψ(a))α . Γ(α + 1)

+ Therefore,

󵄩 󵄩󵄩 −1 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X Γ(2 + β(α − 1)) 1 α ]Γ(γ)(RM + f ∗ )(ψ(b) − ψ(a)) ⩽[ + Γ(γ + α) Γ2 (α + 1) +

2ηRM (ψ(b) − ψ(a))α+1−γ . Γ(α + 1)

This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to prove that L−1 P (id − Q )N (Ω) is equicontinuous. For a ⩽ t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 L−1 (id − Q )N y(t ) L−1 (id − Q )N y(t ) 󵄨󵄨 󵄨󵄨 P 1 󵄨󵄨 2 − P 󵄨󵄨 󵄨 (ψ(t1 ) − ψ(a))γ−1 󵄨󵄨󵄨 󵄨󵄨 (ψ(t2 ) − ψ(a))γ−1 t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 1 󵄨 󵄨󵄨 2 1 − ⩽ ∫[ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(γ1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

a

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨]ds t2

α−1

+ ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) t1

+

(ψ(t2 ) − ψ(a))

1−γ |f (s, y(s), By(s))|

Γ(α)

Γ(2 + β(α − 1))[(ψ(t2 ) − ψ(a))1+α−γ − (ψ(t1 ) − ψ(a))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1)) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds 󵄨 󵄨

ds

� 165

166 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 (RM + f ∗ ) 󵄨 󵄨󵄨 2 1 ⩽ − ∫[ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

a

× (ψ(s) − ψ(a))

γ−1

]ds

t2

(ψ(t2 ) − ψ(a))1−γ (RM + f ∗ ) γ−1 + (ψ(s) − ψ(a)) ds ∫ ψ′ (s) 1−α Γ(α) (ψ(t2 ) − ψ(s)) t1

+

(RM + f ∗ )Γ(2 + β(α − 1)) Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1))

× [(ψ(t2 ) − ψ(a))

1+α−γ

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

1+α−γ

− (ψ(t1 ) − ψ(a)) β(α−1)

]

γ−1

(ψ(s) − ψ(a))

ds

t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 ηRM 󵄨 󵄨󵄨 1 2 − + ∫ ψ′ (s)󵄨󵄨󵄨 󵄨ds 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

a

t2

+

ηRM α−1 1−γ ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) (ψ(t2 ) − ψ(a)) ds Γ(α) t1

+

ηRM Γ(2 + β(α − 1))[(ψ(t2 ) − ψ(a))1+α−γ − (ψ(t1 ) − ψ(a))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) Γ(1 + β(α − 1)) b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

β(α−1)

ds.

a

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, we can conclude that the operator N is L-compact in Ω, which completes the demonstration. Lemma 7.5. Assume that (7.4.1) and (7.4.2) are satisfied. If the condition 2RΓ(γ) 2ηR α α+1−γ (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A , Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y).

ζ ∈ (0, 1].

(7.6)

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces

Then L(y) =

ζ 1 N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ (a, b] α,β;ψ

Ly(t) = H D a+ y(t) =

1 ζ f (t, y(t), By(t)) − f (t, −y(t), −By(t)). 1+ζ 1+ζ

Using Lemma 2.20, we obtain y(t) =

c1 (ψ(t) − ψ(a))γ−1 1 + Γ(γ) ζ +1 α;ψ

α;ψ

× [Ja+ (f (s, y(s), By(s)))(t) − ζ Ja+ (f (s, −y(s), −By(s)))(t)], 1−γ;ψ

where c1 = Ja+

y(a). Thus, for each t ∈ (a, b] we have

γ−1 󵄨󵄨 󵄨 |c |(ψ(t) − ψ(a)) 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ 1 Γ(γ) t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), By(s))󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, −y(s), −By(s))󵄨󵄨󵄨ds (ζ + 1)Γ(α) a

⩽

|c1 |(ψ(t) − ψ(a))γ−1 Γ(γ) +

t

+

t

2(R‖y‖X + f ∗ ) α−1 γ−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(a)) ds (ζ + 1)Γ(α) a

2ηR‖y‖X α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) ds (ζ + 1)Γ(α) a

|c |(ψ(t) − ψ(a))γ−1 2(R‖y‖X + f ∗ )Γ(γ) α+γ−1 ⩽ 1 + (ψ(t) − ψ(a)) Γ(γ) Γ(α + γ) +

2ηR‖y‖X α (ψ(t) − ψ(a)) . Γ(α + 1)

Thus, 2RΓ(γ) 2ηR α α+1−γ (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) ]‖y‖X Γ(α + γ) Γ(α + 1) |c | 2f ∗ Γ(γ) α + 1 + (ψ(b) − ψ(a)) . Γ(γ) Γ(α + γ)

‖y‖X ⩽ [

� 167

168 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative We deduce that ‖y‖X ⩽

|c1 | Γ(γ)

+

2f ∗ Γ(γ)(ψ(b)−ψ(a))α Γ(α+γ)

2RΓ(γ) 1 − [ Γ(α+γ) (ψ(b) − ψ(a))α +

2ηR (ψ(b) Γ(α+1)

− ψ(a))α+1−γ ]

:= A .

The demonstration is completed. Lemma 7.6. If conditions (7.4.1), (7.4.2), and (7.6) are satisfied, then there exists a bounded open set Ω ⊂ X such that L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(7.7)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 7.5, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(7.8)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 7.1. Assume that (7.4.1), (7.4.2), and (7.6) hold. Then there exists at least one solution for the problem (7.1)–(7.2). Proof. It is clear that the set Ω defined in (7.8) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 7.6, if conditions (7.4.1), (7.4.2), and (7.6) are satisfied, then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (7.1)–(7.2) has at least one solution on Dom L ∩ Ω. The demonstration is completed.

7.2 Periodic solutions for nonlinear fractional integro-differential equations in Banach spaces

� 169

7.2.2 Applications Application 1. Consider the following NFDE problem: H

D

1 1 t , ;e 2 3 0+ 1

;et

t

1 t3 y(t) = sin y(t) + + ∫ e1−s y(s)ds, 1 67 (et − 1) 3 67 0 cos(2t) 1

t ∈ (0, 1],

;et

J03+ y(0) = J03+ y(1).

(7.10)

Set t

t3 1 f (t, y(t), By(t)) = + sin y(t) + ∫ e1−s y(s)ds, 1 t 67 67 (e − 1) 3 0 cos(2t)

t ∈ (0, 1],

with t

By(t) = ∫ e1−s y(s)ds,

t ∈ (0, 1].

0

We have 1

C 1 ;et ([0, 1], ℝ) = {f : (0, 1] → ℝ : (et − 1) 3 f ∈ C([0, 1], ℝ)}, 3

with 2 γ= , 3

1 α= , 2

1 β= , 3

(7.9)

ψ(t) = et ,

and

η=

2 3e (e − 1) 3 ≈ 5.85. 2

Clearly, the function f ∈ C 1 ;et ([0, 1], ℝ). Hence, condition (7.4.1) is satisfied. 3 For each y, ȳ ∈ C 1 ;et ([0, 1], ℝ) and t ∈ (0, 1], 3

t3 󵄨󵄨 ̄ By(t)) ̄ 󵄨󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨󵄨sin y(t) − sin y(t) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y(t), By(t)) − f (t, y(t), 67 t 󵄨t 󵄨󵄨 󵄨󵄨 1 󵄨󵄨󵄨󵄨 1−s 1−s ̄ + 󵄨󵄨∫ e y(s)ds − ∫ e y(s)ds󵄨󵄨󵄨 󵄨󵄨 67 󵄨󵄨󵄨 󵄨 0 0 ⩽

1 󵄨󵄨 1 ̄ 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨By(t) − By(t) ̄ 󵄨󵄨󵄨󵄨. 󵄨y(t) − y(t) 67 󵄨 67

Hence, condition (7.4.2) is satisfied with R = R =

1 . 67

170 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative And we have 2RΓ(γ) 2ηR α α+1−γ (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) ≈ 0.366 < 1. Γ(α + γ) Γ(α + 1) Thus, by Theorem 7.1 the problem (7.9)–(7.10) has at least one solution. Application 2. Consider the following NFDE problem: H

D

1 1 , ;ln(t) 2 3 0+ 1

;ln(t)

J03+

t

−1 y(s) 1 1 ds + (ln t) 3 , y(t) = + ∫ 10e−t+2 (1 + y(t)) 3√π 2 + s 1

t ∈ (1, e],

y(0) = J03+

;ln(t)

(7.12)

y(1).

Set t

f (t, y(t), By(t)) =

−1 y(s) 1 1 + ds + (ln t) 3 , ∫ −t+2 10e (1 + y(t)) 3√π 2 + s

t ∈ (1, e],

0

with t

By(t) = ∫ 0

y(s) ds, 2+s

t ∈ (1, e].

We have 1

C 1 ;ln t ([1, e], ℝ+ ) = {f : (1, e] → ℝ+ : (ln t) 3 f ∈ C([1, e], ℝ+ )}, 3

with 2 γ= , 3

(7.11)

0

1 α= , 2

1 β= , 3

ψ(t) = ln t,

and η =

Clearly, the function f ∈ C 1 ;ln t ([1, e], ℝ+ ). 3 Hence, condition (7.4.1) is satisfied. For all y, ȳ ∈ C 1 ;ln t ([1, e], ℝ+ ) and t ∈ (1, e], 3

󵄨󵄨 ̄ By(t)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y(t), By(t)) − f (t, y(t), 1 󵄨󵄨󵄨󵄨 1 1 󵄨󵄨󵄨󵄨 ⩽ − 󵄨󵄨 󵄨 −t+2 󵄨 ̄ 󵄨󵄨󵄨 10e 󵄨 1 + y(t) 1 + y(t) t 󵄨t 󵄨󵄨 󵄨󵄨 ̄ y(s) 1 󵄨󵄨󵄨󵄨 y(s) + ds − ∫ ds󵄨󵄨󵄨 󵄨󵄨∫ 2 + s 󵄨󵄨󵄨 3√π 󵄨󵄨󵄨 2 + s 0 0 ̄ − y(t)| |y(t) 1 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ ee−2 + 󵄨By(t) − By(t) ̄ |1 + y(t)||1 + y(t)| 3√π 󵄨

3 . 4

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

󵄨 ̄ 󵄨󵄨󵄨󵄨 + ⩽ ee−2 󵄨󵄨󵄨y(t) − y(t)

� 171

1 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨. 󵄨By(t) − By(t) 3√π 󵄨 1 . 3√π

Hence, condition (7.4.2) is satisfied with R = ee−2 and R = And we have

2RΓ(γ) 2ηR α α+1−γ (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) ≈ 0.47 < 1. Γ(α + γ) Γ(α + 1) Hence, all assumptions of Theorem 7.1 are satisfied and therefore the problem (7.11)–(7.12) has at least one solution.

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations In this section, we consider the following nonlinear class of Volterra–Fredholm integrodifferential fractional equations: H

α,β;ψ

D a+ y(t) = f (t, y(t), G y(t), H y(t)),

t ∈ (a, b],

(7.13)

with the fractional integral conditions 1−γ;ψ

Ja+

1−γ;ψ

y(a) = Ja+

y(b),

(7.14)

where t

G y(t) = ∫ g(t, s, y(s))ds

b

and H y(t) = ∫ h(t, s, y(s))ds,

a

(7.15)

a

and f : (a, b] × ℝ × ℝ × ℝ → ℝ,

g : Δ × ℝ → ℝ,

and

h : Δ0 × ℝ → ℝ

are continuous functions with J ̄ := [a, b] (−∞ < a < b < +∞), Δ0 = J ̄ × J,̄ and Δ = {(t, s) : α,β;ψ

a ⩽ s ⩽ t ⩽ b}; H D a+ denotes the generalized ψ-Hilfer fractional derivative of order 1−γ;ψ 0 < α ⩽ 1 and type β ∈ [0, 1], and Ja+ is the generalized fractional integral in the sense of Riemann–Liouville of order 1 − γ (γ = α + β − αβ). 7.3.1 Existence results Let α;ψ

X = {y ∈ C1−γ;ψ (J,̄ ℝ) : y(t) = Ja+ υ(t) : υ ∈ C1−γ;ψ (J,̄ ℝ), t ∈ (a, b]}

172 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative and let Y = C1−γ;ψ (J,̄ ℝ) with the norm ‖y‖X = ‖y‖Y = ‖y‖C1−γ;ψ . To prove the main findings, we need to employ the lemmas of the previous sections, notably, Lemma 7.1 and Lemma 7.2. Lemma 7.7. For all y, ȳ ∈ C1−γ;ψ (J,̄ ℝ) and t ∈ (a, b], we get 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ λ1 ‖y − y‖̄ X , 󵄨󵄨G y(t) − G y(t) 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ λ2 ‖y − y‖̄ X , 󵄨󵄨H y(t) − H y(t) where λ1 =

(ψ(b) − ψ(a))γ ρ γ mint∈[a,b] ψ′ (t) 1

and

λ2 =

(ψ(b) − ψ(a))γ ρ. γ mint∈[a,b] ψ′ (t) 2

Before establishing the proof, we will need to introduce the following hypotheses. (7.8.1) The function f : (a, b] × ℝ × ℝ × ℝ → ℝ is such that f (⋅, y(⋅), G (y)(⋅), H (y)(⋅)) ∈ C1−γ;ψ (J,̄ ℝ) for all y ∈ C1−γ;ψ (J,̄ ℝ). (7.8.2) There exist positive constants ϱ, η1 , η2 with 󵄨󵄨 ̄ H (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, G (y), H (y)) − f (t, y,̄ G (y),

⩽ ϱ|y − y|̄ + η1 |G y − G y|̄ + η2 |H y − H y|̄

for every t ∈ (a, b] and y, ȳ ∈ C1−γ;ψ (J,̄ ℝ). (7.8.3) There exists a constant ρ1 > 0 such that 󵄨󵄨 󵄨 󵄨󵄨g(t, s, υ) − g(t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ1 |υ − υ|̄ for every (t, s) ∈ Δ and υ, ῡ ∈ C1−γ;ψ (J,̄ ℝ). (7.8.4) There exists a constant ρ2 > 0 such that 󵄨󵄨 󵄨 󵄨󵄨h(t, s, υ) − h(t, s, υ)̄ 󵄨󵄨󵄨 ⩽ ρ2 |υ − υ|̄ for every (t, s) ∈ Δ0 and υ, ῡ ∈ C1−γ;ψ (J,̄ ℝ). Proof. Using (7.8.3), we have for any t ∈ (a, b] t

󵄨󵄨 󵄨󵄨 ̄ ̄ 󵄨󵄨󵄨󵄨 ⩽ ∫󵄨󵄨󵄨󵄨g(t, s, y(s)) − g(t, s, y(s)) 󵄨󵄨ds 󵄨󵄨G y(t) − G y(t) a

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations t

⩽ ρ1 ‖y − y‖̄ X ∫(ψ(s) − ψ(a))

γ−1

� 173

ds

a

t

⩽ ρ1 ‖y − y‖̄ X ∫ ψ′ (s)(ψ(s) − ψ(a)) a

γ−1

1

ψ′ (s)

ds

b

⩽

ρ1 ‖y − y‖̄ X γ−1 ∫ ψ′ (s)(ψ(s) − ψ(a)) ds mint∈[a,b] ψ′ (t) γ

⩽

a

(ψ(b) − ψ(a)) ρ ‖y − y‖̄ X := λ1 ‖y − y‖̄ X . γ mint∈[a,b] ψ′ (t) 1

By using a similar argument and (7.8.4), we get 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 ⩽ λ2 ‖y − y‖̄ X . 󵄨󵄨H y(t) − H y(t) Now, we define N : X → Y by N y(t) := f (t, y(t), G y(t), H y(t)),

t ∈ (a, b].

The operator N is well defined, because f , g, and h are continuous functions. We can remark that the problem (7.13)–(7.14) is equivalent to the problem Ly = N y. Lemma 7.8. Suppose that (7.8.1), (7.8.2), (7.8.3), and (7.8.4) are satisfied. Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps: Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ J ̄ we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 (1 + β(α − 1)) ⩽ (ψ(b) − ψ(a))1+β(α−1) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨N (yn )(s) − N (y)(s)󵄨󵄨󵄨ds. 󵄨 󵄨

By (7.8.2), we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

ϱ(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(a)) a

174 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative b

|G (yn )(s) − G (y)(s)| η1 (1 + β(α − 1)) ds + ∫ ψ′ (s) 1+β(α−1) (ψ(b) − ψ(a)) (ψ(b) − ψ(s))β(1−α) a

b

+

η2 (1 + β(α − 1)) |H (yn )(s) − H (y)(s)| ds ∫ ψ′ (s) (ψ(b) − ψ(a))1+β(α−1) (ψ(b) − ψ(s))β(1−α) a

b

ϱ(1 + β(α − 1)) α−1 ⩽ ‖yn − y‖Y ∫ ψ′ (s)(ψ(b) − ψ(s)) ds (ψ(b) − ψ(a))1+β(α−1) a

b

+

η1 (1 + β(α − 1)) |G (yn )(s) − G (y)(s)| ds ∫ ψ′ (s) (ψ(b) − ψ(a))1+β(α−1) (ψ(b) − ψ(s))β(1−α) a

b

|H (yn )(s) − H (y)(s)| η2 (1 + β(α − 1)) ds. + ∫ ψ′ (s) (ψ(b) − ψ(a))1+β(α−1) (ψ(b) − ψ(s))β(1−α) a

Using Lemma 7.7, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 ϱ γ−1 ⩽ (1 + β(α − 1))(ψ(b) − ψ(a)) ‖yn − y‖Y + (η1 λ1 + η2 λ2 )‖yn − y‖Y α ϱ γ−1 ⩽ [ (1 + β(α − 1))(ψ(b) − ψ(a)) + (λ1 η1 + λ2 η2 )]‖yn − y‖Y . α Thus, for each t ∈ J ̄ we obtain 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(b) − ψ(t)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 ϱ 1−γ ⩽ [ (1 + β(α − 1)) + (λ1 η1 + λ2 η2 )(ψ(b) − ψ(a)) ]‖yn − y‖Y . α

Then for all t ∈ J ̄ we get 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(b) − ψ(t)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞.

Therefore, 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0 We deduce that QN is continuous.

as n 󳨀→ +∞.

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

� 175

Step 2. We demonstrate that QN (Ω) is bounded. For t ∈ J ̄ and y ∈ Ω, we have 󵄨 󵄨󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨N (y)(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1)

(1 + β(α − 1)) ⩽ (ψ(b) − ψ(a))1+β(α−1)

a

b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

β(α−1) 󵄨

󵄨󵄨f (s, y(s), G (y)(s), H (y)(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

a b

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

(1 + β(α − 1))f ∗ γ−1 (ψ(b) − ψ(a)) ⩽ α b

+

ϱ(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨y(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

b

η1 (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨G (y)(s) − G (0)(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(a)) a

b

+

η1 (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨G (0)(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

b

η2 (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨H (y)(s) − H (0)(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(a)) a

b

+

η2 (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨H (0)(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

(1 + β(α − 1)) ∗ γ−1 [f + M (ϱ + λ1 η1 + λ2 η2 )](ψ(b) − ψ(a)) ⩽ α + (g∗ η1 + h∗ η2 )(b − a), where 󵄩 󵄩 f ∗ = 󵄩󵄩󵄩f (⋅, 0, 0, 0)󵄩󵄩󵄩C , 1−γ;ψ and

󵄨 󵄨 g∗ = sup 󵄨󵄨󵄨g(t, s, 0, 0)󵄨󵄨󵄨, (t,s)∈Δ

176 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative 󵄨 󵄨 h∗ = sup 󵄨󵄨󵄨h(t, s, 0, 0)󵄨󵄨󵄨. (t,s)∈Δ0

Thus, (1 + β(α − 1)) ∗ 󵄩 󵄩󵄩 [f + M (ϱ + λ1 η1 + λ2 η2 )] 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ α 1−γ

+ (g∗ η1 + h∗ η2 )(b − a)(ψ(b) − ψ(a))

.

So, QN (Ω) is a bounded set in Y . Step 3. We show that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ (a, b], we get L−1 P (N y(t) − QN y(t)) α;ψ

= Ja+ [f (t, y(t), G y(t), H y(t)) −

Γ(2 + β(α − 1)) 1+β(α−1);ψ J + f (s, y(s), G y(s), H y(s))(b)] (ψ(b) − ψ(a))1+β(α−1) a t

=

1 α−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), G y(s), H y(s))ds Γ(α) a

−

Γ(2 + β(α − 1))(ψ(t) − ψ(a))α 1+β(α−1);ψ J + f (s, y(s), G y(s), H y(s))(b). Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) a

For all y ∈ Ω and t ∈ (a, b], we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), G y(s), H y(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds Γ(α) a

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0, 0)󵄨󵄨󵄨ds Γ(α) a

1 γ−1 + (1 + β(α − 1))(ψ(b) − ψ(a)) Γ(α + 1) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, y(s), G y(s), H y(s)) − f (s, 0, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨 b

(1 + β(α − 1))(ψ(b) − ψ(a))γ−1 |f (s, 0, 0, 0)| + ds ∫ ψ′ (s) Γ(α + 1) (ψ(b) − ψ(s))β(1−α) a

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

⩽

f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 (ψ(t) − ψ(a)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ t

+

ϱ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) a

t

η α−1 󵄨 󵄨 + 1 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨G (y)(s) − G (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

η α−1 󵄨 󵄨 + 1 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨G (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

+

η2 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨H (y)(s) − H (0)(s)󵄨󵄨󵄨ds Γ(α) a

t

η α−1 󵄨 󵄨 + 2 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨H (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

ϱ(1 + β(α − 1))(ψ(b) − ψ(a))γ−1 ψ′ (s)|y(s)| + ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) a

b

+

η1 (1 + β(α − 1))(ψ(b) − ψ(a))γ−1 ψ′ (s)|G (y)(s) − G (0)(s)| ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) a

γ−1

η (1 + β(α − 1))(ψ(b) − ψ(a)) + 1 Γ(α + 1)

b

∫ a

ψ′ (s)|G (0)(s)| ds (ψ(b) − ψ(s))β(1−α)

b

+

η2 (1 + β(α − 1))(ψ(b) − ψ(a))γ−1 ψ′ (s)|H (y)(s) − H (0)(s)| ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) a

γ−1

η (1 + β(α − 1))(ψ(b) − ψ(a)) + 2 Γ(α + 1)

b

∫ a

ψ′ (s)|H (0)(s)| ds. (ψ(b) − ψ(s))β(1−α)

Using Lemma 7.7, we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(a)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ ϱM Γ(γ)Γ(2 + β(α − 1)) ϱM Γ(γ) α+γ−1 + (ψ(t) − ψ(a)) + 2 Γ(γ + α) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ 2M α + (λ1 η1 + λ2 η2 )(ψ(b) − ψ(a)) Γ(α + 1) 2(b − a) ∗ α + (g η1 + h∗ η2 )(ψ(b) − ψ(a)) . Γ(α + 1)

� 177

178 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative So ∗ α+γ−1 󵄨󵄨 −1 󵄨 (f + ϱM )Γ(γ) (ψ(t) − ψ(a)) 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 ⩽ Γ(α + γ) (f ∗ + ϱM )Γ(γ)Γ(2 + β(α − 1)) α+γ−1 + (ψ(b) − ψ(a)) Γ2 (α + 1) 2(ψ(b) − ψ(a))α + Γ(α + 1)

× [(λ1 η1 + λ2 η2 )M + (g∗ η1 + h∗ η2 )(b − a)].

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X (f ∗ + ϱM )Γ(γ) (f ∗ + ϱM )Γ(γ)Γ(2 + β(α − 1)) α + ⩽[ ](ψ(b) − ψ(a)) 2 Γ(α + γ) Γ (α + 1) 2[(λ1 η1 + λ2 η2 )M + (g∗ η1 + h∗ η2 )(b − a)] . + Γ(α + 1)(ψ(b) − ψ(a))γ−α−1 This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. For a < t1 < t2 ⩽ b, y ∈ Ω, we have 󵄨󵄨 L−1 (id − Q )N y(t ) L−1 (id − Q )N y(t ) 󵄨󵄨 󵄨󵄨 P 1 󵄨󵄨 2 − P 󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(a))γ−1 (ψ(t1 ) − ψ(a))γ−1 󵄨󵄨󵄨 t

⩽

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 1 󵄨 󵄨󵄨 2 1 − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

a

󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), G y(s), H y(s))󵄨󵄨󵄨ds t2

(ψ(t2 ) − ψ(a))1−γ 󵄨󵄨 1 󵄨 + ∫ ψ′ (s) 󵄨f (s, y(s), G y(s), H y(s))󵄨󵄨󵄨ds Γ(α) (ψ(t2 ) − ψ(s))1−α 󵄨 t1

+

(1 + β(α − 1))[(ψ(t2 ) − ψ(a))1+α−γ − (ψ(t1 ) − ψ(a))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f (s, y(s), G y(s), H y(s))󵄨󵄨󵄨ds 󵄨 󵄨

t1

(f ∗ + ϱM ) ⩽ ∫ ψ′ (s) Γ(α) a

󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 γ−1 󵄨 󵄨󵄨 2 1 × 󵄨󵄨󵄨 − 󵄨(ψ(s) − ψ(a)) ds 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

� 179

t1

1 [(λ η + λ2 η2 )M + (η1 g∗ + η2 h∗ )(b − a)] ∫ ψ′ (s) + Γ(α) 1 1 a

󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 󵄨 󵄨󵄨 2 1 × 󵄨󵄨󵄨 − 󵄨ds 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 t2

+

(ψ(t2 ) − ψ(a))1−γ (f ∗ + ϱM ) γ−1 (ψ(s) − ψ(a)) ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α t1

1 + [(λ η + λ2 η2 )M + (η1 g∗ + η2 h∗ )(b − a)] Γ(α) 1 1 t2

α−1

× ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) t1

+[

1−γ

(ψ(t2 ) − ψ(a))

ds

(f ∗ + ϱM )Γ(2 + β(α − 1))Γ(γ) Γ(α + 1)(ψ(b) − ψ(a))1−γ

+ (λ1 η1 + λ2 η2 )M + (η1 g∗ + η2 h∗ )(b − a)] ×

1 1+α−γ 1+α−γ [(ψ(t2 ) − ψ(a)) − (ψ(t1 ) − ψ(a)) ]. Γ(α + 1)

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 7.9. Assume that (7.8.1), (7.8.2), (7.8.3), and (7.8.4) are satisfied. If the condition (λ η + λ2 η2 ) ϱΓ(γ) 1 α 1+α−γ (ψ(b) − ψ(a)) + 1 1 (ψ(b) − ψ(a)) < Γ(γ + α) Γ(α + 1) 2 is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A , Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ (a, b]

ζ ∈ (0, 1].

(7.16)

180 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative α,β;ψ

Ly(t) = H D a+ y(t) =

1 f (t, y(t), G y(t), H y(t)) 1+ζ ζ − f (t, −y(t), G (−y)(t), H (−y)(t)). 1+ζ

By Theorem 2.20 we get y(t) =

c1 (ψ(t) − ψ(a))γ−1 1 α;ψ + [J + (f (s, y(s), G y(s), H y(s))(t)) Γ(γ) ζ +1 a α;ψ

− ζ Ja+ (f (s, −y(s), G (−y)(s), H (−y)(s))(t))], 1−γ;ψ

where c1 = Ja+

y(a). Thus, for each t ∈ (a, b] we have

γ−1 2f ∗ Γ(γ) α+γ−1 󵄨󵄨 󵄨 |c |(ψ(t) − ψ(a)) + (ψ(t) − ψ(a)) 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ 1 Γ(γ) (ζ + 1)Γ(γ + α) 2(g∗ η1 + h∗ η2 )(b − a) α + (ψ(t) − ψ(a)) (ζ + 1)Γ(α + 1) 2‖y‖X (λ1 η1 + λ2 η2 ) ϱΓ(γ) + [ ]. + (ζ + 1) Γ(γ + α)(ψ(t) − ψ(a))1−α−γ Γ(α + 1)(ψ(t) − ψ(a))−α

Thus, ‖y‖X ⩽

|c1 | 2(g∗ η1 + h∗ η2 )(b − a) 2f ∗ Γ(γ) α + (ψ(b) − ψ(a)) + Γ(γ) Γ(γ + α) Γ(α + 1)(ψ(b) − ψ(a))γ−1−α + 2[

(λ1 η1 + λ2 η2 ) ϱΓ(γ) + ]‖y‖X . Γ(γ + α)(ψ(b) − ψ(a))−α Γ(α + 1)(ψ(b) − ψ(a))γ−1−α

We deduce that ‖y‖X ⩽

|c1 | Γ(γ)

1

∗ +h∗ η2 )(b−a) 2f ∗ Γ(γ) (ψ(b) − ψ(a))α + 2(g η1Γ(α+1) (ψ(b) − ψ(a))1+α−γ Γ(γ+α) η1 +λ2 η2 ) ϱΓ(γ) − 2[ Γ(γ+α) (ψ(b) − ψ(a))α + (λ1Γ(α+1) (ψ(b) − ψ(a))1+α−γ ]

+

:= A . The demonstration is completed. Lemma 7.10. If conditions (7.8.1)–(7.8.4) and (7.16) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)]

(7.17)

for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 7.9, there exists a positive constant A which is independent of ζ such that if y satisfies

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

L(y) − N (y) = −ζ [L(y) + N (−y)],

� 181

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(7.18)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. To prove the main result in this subsection, we need the following lemma. Lemma 7.11. Assume that 0 < δ < 1 and 0 < μ ≤ 1. Then the following inequality holds: Γ(μ) 1 ≤ . Γ(δ + 1) Γ(δ + μ) Proof. Using Lemma 2.6, we have for t ∈ (a, b] 1 δ (ψ(t) − ψ(a)) Γ(δ + 1) t

1 δ−1 = ∫ ψ′ (s)(ψ(t) − ψ(s)) ds Γ(δ) a

t

1 δ−1 μ−1 1−μ = ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(a)) (ψ(s) − ψ(a)) ds Γ(δ) a

≤ (ψ(t) − ψ(a))

1−μ

δ;ψ

μ−1

Ja+ (ψ(s) − ψ(a))

(t)

Γ(μ) δ+μ−1 1−μ (ψ(t) − ψ(a)) (ψ(t) − ψ(a)) Γ(δ + μ) Γ(μ) δ ≤ (ψ(t) − ψ(a)) , Γ(δ + μ) ≤

which is the desired result. Theorem 7.2. Assume that (7.8.1)–(7.8.4) and (7.16) hold. Then there exists at least one solution for the problem (7.13)–(7.14). Proof. It is clear that the set Ω defined in (7.18) is symmetric, 0 ∈ Ω, and X ∩Ω = Ω ≠ 0. In addition, by Lemma 7.10, assume that (7.8.1), (7.8.2), (7.8.3), (7.8.4), and (7.16) are satisfied. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)]

182 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (7.13)–(7.14) has at least one solution on Dom L ∩ Ω, which completes the demonstration. Now, we investigate the existence and uniqueness of periodic solutions for our problem (7.13)–(7.14). Theorem 7.3. Let (7.8.1), (7.8.2), (7.8.3), and (7.8.4) be satisfied. Moreover, we assume that the following hypothesis holds. (7.14.1) There exist constants ϱ > 0 and η1 , η2 ⩾ 0 such that 󵄨󵄨 ̄ H (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, G (y), H (y)) − f (t, y,̄ G (y), ⩾ ϱ|y − y|̄ − η1 |G y − G y|̄ − η2 |H y − H y|̄ for every t ∈ (a, b] and y, ȳ ∈ C1−γ;ψ (J,̄ ℝ). If one has (η1 λ1 + η2 λ2 ) 2Γ(γ)(η1 λ1 + η2 λ2 ) 2ϱΓ(γ) + + < 1, −α γ−1−α Γ(α + γ)(ψ(b) − ψ(a)) Γ(α + γ)(ψ(b) − ψ(a)) ϱ(ψ(b) − ψ(a))γ−1

(7.19)

then the problem (7.13)–(7.14) has a unique solution in Dom L ∩ Ω. Proof. By Lemma 7.11 we can see that the condition (7.19) is stronger than condition (7.16). By Theorem 7.2, the problem (7.13)–(7.14) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (7.13)–(7.14) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ (a, b] α,β;ψ

H

D a+ y1 (t) = f (t, y1 (t), G (y1 )(t), H (y1 )(t)),

H

D a+ y2 (t) = f (t, y2 (t), G (y2 )(t), H (y2 )(t)),

α,β;ψ

where G and H are defined as in (7.15) and y1 (a) = y1 (b),

y2 (a) = y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ (a, b]. Then α,β;ψ

LU(t) = H D a+ U(t) α,β;ψ

α,β;ψ

= H D a+ y1 (t) − H D a+ y2 (t)

= f (t, y1 (t), G (y1 )(t), H (y1 )(t)) − f (t, y2 (t), G (y2 )(t), H (y2 )(t)). Using the fact that Img L = ker Q , we have

(7.20)

� 183

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations b

β(α−1)

∫ ψ′ (s)(ψ(b) − ψ(s)) a

[f (s, y1 (s), G (y1 )(s), H (y1 )(s)) − f (s, y2 (s), G (y2 )(s), H (y2 )(s))]ds = 0. Since f ∈ C1−γ;ψ (J,̄ ℝ), there exists t0 ∈ (a, b] such that f (t0 , y1 (t0 ), G (y1 )(t0 ), H (y1 )(t0 )) − f (t, y2 (t0 ), G (y2 )(t0 ), H (y2 )(t0 )) = 0. In view of (7.14.1) we have 󵄨󵄨 󵄨 (η λ + η2 λ2 ) ‖y1 − y2 ‖X . 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ 1 1 ϱ Then 󵄨󵄨 󵄨 (η λ + η2 λ2 ) ‖U‖X . 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ 1 1 ϱ

(7.21)

On the other hand, by Theorem 2.20, we have α;ψ H

Ja+

α,β;ψ

D a+ y(t) = y(t) −

c1 (ψ(t) − ψ(a))γ−1 , Γ(γ)

which implies that α;ψ H

c1 = [y(t0 ) − Ja+

α,β;ψ

D a+ y(t0 )]Γ(γ)(ψ(t0 ) − ψ(a))

1−γ

,

and therefore α;ψ H

U(t) = Ja+

α,β;ψ

D a+ U(t) α;ψ H

+ [U(t0 ) − Ja+

α,β;ψ

1−γ

D a+ U(t0 )](ψ(t0 ) − ψ(a))

γ−1

(ψ(t) − ψ(a))

.

Using (7.21), we obtain for every t ∈ (a, b] 1−γ γ−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 α;ψ H α,β;ψ 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨Ja+ D a+ U(t0 )󵄨󵄨󵄨](ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) α,β;ψ 󵄨 α;ψ 󵄨 + 󵄨󵄨󵄨Ja+ H D a+ U(t)󵄨󵄨󵄨 ‖U‖X (η1 λ1 + η2 λ2 ) 1−γ γ−1 (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) ⩽ ϱ Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 α γ−1 + 󵄩 D + U󵄩󵄩X (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) Γ(γ + α) 󵄩 a

+

Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 γ+α−1 . 󵄩 D + U󵄩󵄩X (ψ(t) − ψ(a)) Γ(γ + α) 󵄩 a

By (7.8.2), (7.8.3), (7.8.4), and (7.20), we find that

(7.22)

184 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative 󵄨 󵄨 󵄨 󵄨󵄨H α,β;ψ 󵄨󵄨 D a+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (t), G (y1 )(t), H (y1 )(t)) − f (t, y2 (t), G (y2 )(t), H (y2 )(t))󵄨󵄨󵄨 ⩽ [ϱ(ψ(t) − ψ(a))

γ−1

+ η1 λ1 + η2 λ2 ]‖U‖X .

Then 1−γ 󵄩󵄩H α,β;ψ 󵄩󵄩 󵄩󵄩 D a+ U󵄩󵄩X ⩽ [ϱ + (η1 λ1 + η2 λ2 )(ψ(b) − ψ(a)) ]‖U‖X .

(7.23)

Substituting (7.23) in the right side of (7.22), we get for every t ∈ (a, b] (η λ + η2 λ2 ) 1−γ γ−1 󵄨󵄨 󵄨 (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ 1 1 ϱ Γ(γ) 1−γ (ϱ + (η1 λ1 + η2 λ2 )(ψ(b) − ψ(a)) ) Γ(γ + α) Γ(γ) α γ−1 × (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) + Γ(γ + α) +

1−γ

× (ϱ + (η1 λ1 + η2 λ2 )(ψ(b) − ψ(a))

)(ψ(t) − ψ(a))

γ+α−1

]‖U‖X .

Therefore, 2Γ(γ)(η1 λ1 + η2 λ2 ) 2ϱΓ(γ) α (ψ(b) − ψ(a)) + Γ(α + γ) Γ(α + γ)(ψ(b) − ψ(a))γ−1−α (η λ + η2 λ2 ) 1−γ + 1 1 (ψ(b) − ψ(a)) ]‖U‖X . ϱ

‖U‖X ⩽ [

Hence, by (7.19), we conclude that ‖U‖X = 0. As a result, for any t ∈ (a, b] we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof.

7.3.2 An example We present an example of a Volterra–Fredholm integro-differential equation to test our main results. We have H

1 1

, ;ln t

D 12+ 3

y(t) = f (t, y(t), G y(t), H y(t)), y(1) = y(e),

t ∈ (1, e],

7.3 The periodic solutions for nonlinear Volterra–Fredholm integro-differential equations

� 185

where for any t ∈ (1, e], we have −1

3 1 ln 3 (t) (sin y(t) + y(t)) + f (t, y(t), G y(t), H y(t)) = 2t 2 (e + 3) 17√π 1 1 + G y(t) + H y(t), 19 13e3 with t

t

5 −7−t 2

G y(t) = ∫ g(t, s, y(s))ds = ∫ t e 1

cos(y(s))ds,

t ∈ J,̄

1

and e

e

3

e−9−t ds, H y(t) = ∫ h(t, s, y(s))ds = ∫ 19(1 + y(s)) 1

t ∈ J.̄

1

Here J ̄ := [1, e], α = 21 , β = 31 . and ψ(t) = ln t. It is easy to see that f ∈ C 1 ;ψ (J,̄ ℝ). Hence, condition (7.8.1) is satisfied. 3 Furthermore, for all t ∈ (1, e] and y, ȳ ∈ C 1 (J,̄ ℝ), we obtain 3

;ψ

󵄨󵄨 ̄ H (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, G (y), H (y)) − f (t, y,̄ G (y), ̄ ⩽ ϱ|y − y|̄ + η1 |G y − G y|̄ + η2 |H y − H y|, 󵄨 󵄨󵄨 󵄨 ̄ (t, s) ∈ Δ, 󵄨󵄨g(t, s, y) − g(t, s, y)̄ 󵄨󵄨 ⩽ ρ1 |y − y|, 󵄨󵄨 󵄨 󵄨 ̄ ̄ 󵄨󵄨h(t, s, y) − h(t, s, y)󵄨󵄨 ⩽ ρ2 |y − y|, (t, s) ∈ Δ0 , and 󵄨󵄨 ̄ H (y)) ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y, G (y), H (y)) − f (t, y,̄ G (y), ̄ ⩾ ϱ|y − y|̄ − η1 |G y − G y|̄ − η2 |H y − H y|, with Δ = {(t, s) : 1 ⩽ s ⩽ t ⩽ e} and Δ0 = J ̄ × J,̄ which implies that (7.8.2), (7.8.3), (7.8.4), and (7.14.1) are satisfied with 5 1 1 , ϱ= , η1 = η1 = , 13e3 34√π 34√π 1 1 1 η2 = η2 = , ρ1 = 6 , and ρ2 = . 19 e 19e8 ϱ=

By simple calculations, we get λ1 =

3 ,λ e5 2

=

3 , 19e7

and

186 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative

[

2Γ(γ) 2ϱΓ(γ) α 1+α−γ (ψ(b) − ψ(a)) + (η λ + η2 λ2 )(ψ(b) − ψ(a)) Γ(α + γ) Γ(α + γ) 1 1 (η λ + η2 λ2 ) 1−γ + 1 1 (ψ(b) − ψ(a)) ] ≈ 0.39934 < 1. ϱ

So, by Theorem 7.3, our problem has a unique solution.

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville integral Inspired by the previous studies and employing Mawhin’s coincidence degree theory [126, 169, 183], we establish the existence and uniqueness results for a class of nonlinear ψ-Hilfer fractional differential equations depending on the ψ-Riemann–Liouville fractional integral: H

α,β;ψ

Da +

α,ψ

y(t) = f (t, y(t), Ja+ y(t)),

t ∈ (a, b],

(7.24)

under periodic-type integral boundary conditions 1−γ,ψ

Ja+ α,β;ψ

where H Da+

1−γ,ψ

y(a) = Ja+

y(b),

(7.25)

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and type 1−γ,ψ

β ∈ [0, 1] and Ja+ is the ψ-Riemann–Liouville fractional integral of order 1 − γ (γ = α + β − αβ). Moreover, f : (a, b] × ℝ2 → ℝ is a given continuous function.

7.4.1 Main results Let α;ψ

X = {y ∈ C1−γ;ψ (J,̄ ℝ) : y(t) = Ja+ υ(t) : υ ∈ C1−γ;ψ (J,̄ ℝ), t ∈ (a, b]}

and let Y = C1−γ;ψ (J,̄ ℝ) with the norm ‖y‖X = ‖y‖Y = ‖y‖C1−γ;ψ . To prove the main findings, we need the lemmas mentioned in Section 7.2.1. Now, we need to prove the following result. Lemma 7.12. For all y, ȳ ∈ C1−γ;ψ (J,̄ ℝ) and t ∈ (a, b], we get Γ(γ) α+γ−1 α,ψ 󵄨󵄨 α,ψ ̄ 󵄨󵄨󵄨󵄨 ⩽ (ψ(t) − ψ(a)) ‖y − y‖̄ C1−γ;ψ . 󵄨󵄨Ja+ y(t) − Ja+ y(t) Γ(α + γ)

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

� 187

Proof. Using Lemma 2.6, we have for any t ∈ (a, b] α,ψ 󵄨󵄨 α,ψ ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨Ja+ y(t) − Ja+ y(t) t

1 α−1 󵄨 ̄ 󵄨󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s) − y(s) Γ(α) a

⩽

‖y − y‖̄ C1−γ ;ψ Γ(α)

t

∫ ψ′ (s)(ψ(t) − ψ(s)) a

α−1

γ−1

α,ψ

⩽ ‖y − y‖̄ C1−γ ;ψ Ja+ ((ψ(s) − ψ(a)) ⩽

γ−1

(ψ(s) − ψ(a))

ds

)(t)

Γ(γ) α+γ−1 (ψ(t) − ψ(a)) ‖y − y‖̄ C1−γ ;ψ . Γ(α + γ)

Now, we define N : X → Y by α,ψ

N y(t) := f (t, y(t), Ja+ y(t)),

t ∈ (a, b].

The operator N is well defined, because f is a continuous function. We can remark that the problem (7.24)–(7.25) is equivalent to the problem Ly = N y. Lemma 7.13. Suppose that the following hypotheses are satisfied. (7.16.1) The function f : (a, b] × ℝ × ℝ → ℝ is such that α,ψ f (⋅, y(⋅), Ja+ y(⋅)) ∈ C1−γ;ψ (J,̄ ℝ) for all y ∈ C1−γ;ψ (J,̄ ℝ).

(7.16.2) There exist nonnegative functions ϱ, η ∈ C(J,̄ ℝ+ ) such that α,ψ 󵄨󵄨 ̄ Jaα,ψ ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (t, y(t), Ja+ y(t)) − f (t, y(t), + y(t)) α,ψ 󵄨 ̄ 󵄨󵄨󵄨󵄨 + η(t)󵄨󵄨󵄨󵄨Jaα,ψ ̄ 󵄨󵄨󵄨󵄨 ⩽ ϱ(t)󵄨󵄨󵄨y(t) − y(t) + y(t) − Ja+ y(t)

for every t ∈ (a, b] and y, ȳ ∈ C1−γ;ψ (J,̄ ℝ). Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps:

188 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ (a, b] we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

|N (yn )(s) − N (y)(s)| (1 + β(α − 1)) ⩽ ds. ∫ ψ′ (s) 1+β(α−1) (ψ(b) − ψ(a)) (ψ(b) − ψ(s))β(1−α) a

By (7.16.2) and Lemma 7.12, we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

ϱ∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(a)) a

b

+

α,ψ

α,ψ

|Ja+ yn (s) − Ja+ y(s)| η∗ (1 + β(α − 1)) ′ ψ (s) ds ∫ (ψ(b) − ψ(a))1+β(α−1) (ψ(b) − ψ(s))β(1−α) a

ϱ∗ Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ ⩽ ‖yn − y‖Y Ja+ (ψ(s) − ψ(a)) (b) (ψ(b) − ψ(a))1+β(α−1) η∗ Γ(γ)Γ(2 + β(α − 1))‖yn − y‖Y α+γ−1 1+β(α−1);ψ + Ja+ (ψ(s) − ψ(a)) (b), Γ(α + γ)(ψ(b) − ψ(a))1+β(α−1) where ϱ∗ = ‖ϱ‖∞ and η∗ = ‖η‖∞ . Using Lemma 2.6, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 ⩽ Γ(2 + β(α − 1))‖yn − y‖Y ×[

η∗ Γ(γ) ϱ∗ Γ(γ) γ−1 α+γ−1 (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) ]. Γ(α + 1) Γ(2α + 1)

Thus, for each t ∈ J ̄ we obtain 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 ⩽ Γ(2 + β(α − 1))‖yn − y‖Y

×[

η∗ Γ(γ) ϱ∗ Γ(γ) α + (ψ(b) − ψ(a)) ]. Γ(α + 1) Γ(2α + 1)

Then for all t ∈ J ̄ we get 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 󳨀→ 0

Therefore,

as n 󳨀→ +∞.

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

󵄩 󵄩󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN (Ω) is bounded. For t ∈ (a, b] and y ∈ Ω, we have 󵄨 󵄨󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨N (y)(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1)

(1 + β(α − 1)) ⩽ (ψ(b) − ψ(a))1+β(α−1)

a

b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

β(α−1) 󵄨

󵄨 󵄨󵄨f (s, y(s), J α,ψ y(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 a+

a b

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(a))1+β(α−1) a

b

ϱ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨y(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(a)) ∗

η∗ (1 + β(α − 1)) + (ψ(b) − ψ(a))1+β(α−1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

a

β(α−1) 󵄨

a

α,ψ 󵄨󵄨 󵄨󵄨J α,ψ 󵄨 a+ y(s) − Ja+ (0)(s)󵄨󵄨ds.

Using Lemma 2.6 and Lemma 7.12, we get ∗ ∗ γ−1 󵄨󵄨 󵄨 (ϱ M + f )Γ(γ)Γ(2 + β(α − 1)) (ψ(b) − ψ(a)) 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 ⩽ Γ(α + 1)

+

η∗ Γ(2 + β(α − 1))Γ(γ)M α+γ−1 (ψ(b) − ψ(a)) , Γ(2α + 1)

where f ∗ = ‖f (⋅, 0, 0)‖C1−γ;ψ . Thus, (ϱ∗ M + f ∗ )Γ(γ)Γ(2 + β(α − 1)) 󵄩󵄩 󵄩 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ Γ(α + 1)

� 189

190 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative η∗ Γ(2 + β(α − 1))Γ(γ)M α (ψ(b) − ψ(a)) . Γ(2α + 1)

+

So, QN (Ω) is a bounded set in Y . Step 3. We demonstrate that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ (a, b], we get L−1 P (N y(t) − QN y(t)) α;ψ

α,ψ

= Ja+ [f (t, y(t), Ja+ y(t)) Γ(2 + β(α − 1)) 1+β(α−1);ψ α,ψ J + f (s, y(s), Ja+ y(s))(b)] (ψ(b) − ψ(a))1+β(α−1) a

−

t

1 α−1 α,ψ = ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(s), Ja+ y(s))ds Γ(α) a

Γ(2 + β(α − 1))(ψ(t) − ψ(a))α 1+β(α−1);ψ α,ψ J + f (s, y(s), Ja+ y(s))(b). − Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) a For all y ∈ Ω and t ∈ (a, b], we get 󵄨 󵄨󵄨 −1 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 α,ψ 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), Ja+ y(s)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) a

1 γ−1 (1 + β(α − 1))(ψ(b) − ψ(a)) + Γ(α + 1) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨 󵄨󵄨f (s, y(s), J α,ψ y(s)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 a+ b

(1 + β(α − 1))(ψ(b) − ψ(a))γ−1 ψ′ (s)|f (s, 0, 0)| + ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) a

f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(a)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ t

ϱ∗ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) a

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

� 191

t

η∗ α−1 󵄨 α,ψ α,ψ 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨Ja+ y(s) − Ja+ (0)(s)󵄨󵄨󵄨ds Γ(α) a

b

ϱ (1 + β(α − 1))(ψ(b) − ψ(a))γ−1 β(α−1) 󵄨 󵄨󵄨y(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 Γ(α + 1) ∗

+

γ−1

η (1 + β(α − 1))(ψ(b) − ψ(a)) Γ(α + 1) ∗

b

a

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

α,ψ 󵄨󵄨 󵄨󵄨J α,ψ 󵄨 a+ y(s) − Ja+ (0)(s)󵄨󵄨ds.

Using Lemma 2.6 and Lemma 7.12, we get 󵄨 󵄨󵄨 −1 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(a)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ ∗ ϱ∗ M Γ(γ)Γ(2 + β(α − 1)) ϱ M Γ(γ) α+γ−1 + (ψ(t) − ψ(a)) + 2 Γ(γ + α) Γ (α + 1)(ψ(b) − ψ(a))1−α−γ ∗ η M Γ(γ) 2α+γ−1 + (ψ(t) − ψ(a)) Γ(2α + γ) η∗ M Γ(γ)Γ(2 + β(α − 1)) + . Γ(α + 1)Γ(2α + 1)(ψ(b) − ψ(a))1−2α−γ So 1−γ −1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) LP (id − Q )N y(t)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α α ⩽ (ψ(t) − ψ(a)) + (ψ(b) − ψ(a)) Γ(α + γ) Γ2 (α + 1) ϱ∗ M Γ(γ)Γ(2 + β(α − 1)) ϱ∗ M Γ(γ) α α (ψ(t) − ψ(a)) + + (ψ(b) − ψ(a)) 2 Γ(α + γ) Γ (α + 1) η∗ M Γ(γ)Γ(2 + β(α − 1)) η∗ M Γ(γ) 2α + (ψ(t) − ψ(a)) + . Γ(2α + γ) Γ(α + 1)Γ(2α + 1)(ψ(b) − ψ(a))−2α

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X Γ(γ)Γ(2 + β(α − 1)) Γ(γ) α ⩽( + )(f ∗ + ϱ∗ M )(ψ(b) − ψ(a)) Γ(α + γ) Γ2 (α + 1) +(

Γ(2 + β(α − 1)) 1 2α + )η∗ M Γ(γ)(ψ(b) − ψ(a)) . Γ(2α + γ) Γ(α + 1)Γ(2α + 1)

This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X .

192 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous.

Using Lemma 2.6 and Lemma 7.12, we have for a < t1 < t2 ⩽ b and y ∈ Ω 󵄨󵄨 L−1 (id − Q )N y(t ) L−1 (id − Q )N y(t ) 󵄨󵄨 󵄨󵄨 P 2 1 󵄨󵄨 − P 󵄨󵄨 󵄨 (ψ(t1 ) − ψ(a))γ−1 󵄨󵄨󵄨 󵄨󵄨 (ψ(t2 ) − ψ(a))γ−1 t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γ (ψ(t ) − ψ(a))1−γ 󵄨󵄨 1 󵄨󵄨 󵄨 1 2 ⩽ − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

a

α,ψ 󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(s), Ja+ y(s))󵄨󵄨󵄨ds t2

α,ψ

|f (s, y(s), Ja+ y(s))| (ψ(t2 ) − ψ(a))1−γ + ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α t1

+

(1 + β(α − 1))[(ψ(t2 ) − ψ(a))1+α−γ − (ψ(t1 ) − ψ(a))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(a))1+β(α−1) b

β(α−1) 󵄨

󵄨 󵄨󵄨f (s, y(s), J α,ψ y(s))󵄨󵄨󵄨ds 󵄨 a+

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

⩽

t1

(f ∗ + ϱ∗ M )(ψ(t1 ) − ψ(a))1−γ (ψ(s) − ψ(a))γ−1 ds ∫ ψ′ (s) Γ(α) (ψ(t1 ) − ψ(s))1−α a

1−γ

(f + ϱ M )(ψ(t2 ) − ψ(a)) − Γ(α) ∗

∗

t1

+

t1

∫ ψ′ (s) a

(ψ(s) − ψ(a))γ−1 ds (ψ(t2 ) − ψ(s))1−α

η∗ M Γ(γ)(ψ(t1 ) − ψ(a))1−γ (ψ(s) − ψ(a))α+γ−1 ds ∫ ψ′ (s) Γ(α)Γ(α + γ) (ψ(t1 ) − ψ(s))1−α a

1−γ

η M Γ(γ)(ψ(t2 ) − ψ(a)) − Γ(α)Γ(α + γ) ∗

t1

∫ ψ′ (s) a

(ψ(s) − ψ(a))α+γ−1 ds (ψ(t2 ) − ψ(s))1−α

t2

+

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(a))1−γ (ψ(s) − ψ(a))γ−1 ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α

+

η M Γ(γ)(ψ(t2 ) − ψ(a)) Γ(α)Γ(α + γ) t2

× ∫ ψ′ (s) t1

+

1−γ

∗

t1

(ψ(s) − ψ(a))α+γ−1 ds (ψ(t2 ) − ψ(s))1−α

Γ(γ)Γ(2 + β(α − 1)) (f ∗ + ϱ∗ M ) η∗ M α [ + (ψ(b) − ψ(a)) ] 1−γ Γ(α + 1) Γ(2α + 1) Γ(α + 1)(ψ(b) − ψ(a))

× [(ψ(t2 ) − ψ(a))

1+α−γ

1+α−γ

− (ψ(t1 ) − ψ(a))

]

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

⩽

(f ∗ + ϱ∗ M )Γ(γ) α (ψ(t1 ) − ψ(a)) Γ(α + γ) −

� 193

t2

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(a))1−γ (ψ(s) − ψ(a))γ−1 ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α a

1−γ

(f + ϱ M )(ψ(t2 ) − ψ(a)) + Γ(α) ∗

+

∗

η M Γ(γ) 2α (ψ(t1 ) − ψ(a)) Γ(2α + γ)

t2

∫ ψ′ (s) t1

(ψ(s) − ψ(a))γ−1 ds (ψ(t2 ) − ψ(s))1−α

∗

t2

η∗ M Γ(γ)(ψ(t2 ) − ψ(a))1−γ (ψ(s) − ψ(a))α+γ−1 − ds ∫ ψ′ (s) Γ(α)Γ(α + γ) (ψ(t2 ) − ψ(s))1−α a

1−γ

η M Γ(γ)(ψ(t2 ) − ψ(a)) + Γ(α)Γ(α + γ) ∗

t2

∫ ψ′ (s) t1

(ψ(s) − ψ(a))α+γ−1 ds (ψ(t2 ) − ψ(s))1−α

t2

(f ∗ + ϱ∗ M )(ψ(t1 ) − ψ(a))γ−1 α−1 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) ds Γ(α)(ψ(t2 ) − ψ(a))γ−1 t1

t2

+

η∗ M Γ(γ)(ψ(t2 ) − ψ(a))α+1−γ α−1 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) ds Γ(α)Γ(α + γ)(ψ(t1 ) − ψ(a))1−γ t1

Γ(γ)Γ(2 + β(α − 1)) η∗ M (f ∗ + ϱ∗ M ) α + + (ψ(b) − ψ(a)) ] [ 1−γ Γ(α + 1) Γ(2α + 1) Γ(α + 1)(ψ(b) − ψ(a)) 1+α−γ

× [(ψ(t2 ) − ψ(a)) ⩽

1+α−γ

− (ψ(t1 ) − ψ(a))

]

(f + ϱ M )Γ(γ) (f + ϱ M )Γ(γ) α α (ψ(t1 ) − ψ(a)) − (ψ(t2 ) − ψ(a)) Γ(α + γ) Γ(α + γ) ∗

∗

∗

∗

2(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(a))1−γ (ψ(t1 ) − ψ(a))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1) η∗ M Γ(γ) η∗ M Γ(γ) 2α 2α + (ψ(t1 ) − ψ(a)) − (ψ(t2 ) − ψ(a)) Γ(2α + γ) Γ(2α + γ) +

+ +

2η∗ M Γ(γ)(ψ(t2 ) − ψ(a))α+1−γ (ψ(t1 ) − ψ(a))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)Γ(α + γ)

Γ(γ)Γ(2 + β(α − 1)) (f ∗ + ϱ∗ M ) η∗ M α [ + (ψ(b) − ψ(a)) ] Γ(α + 1) Γ(2α + 1) Γ(α + 1)(ψ(b) − ψ(a))1−γ 1+α−γ

× [(ψ(t2 ) − ψ(a))

1+α−γ

− (ψ(t1 ) − ψ(a))

].

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side

of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The

Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As

194 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 7.14. Assume that (7.16.1) and (7.16.2) hold. If the condition ϱ∗ Γ(γ) η∗ Γ(γ) 1 α 2α (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) < Γ(γ + α) Γ(2α + γ) 2 is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y). Then L(y) =

ζ 1 N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ (a, b] α,β;ψ

Ly(t) = H Da+

y(t) =

1 α,ψ f (t, y(t), Ja+ y(t)) 1+ζ ζ α,ψ − f (t, −y(t), Ja+ (−y)(t)). 1+ζ

By Theorem 2.20 we get y(t) =

c1 (ψ(t) − ψ(a))γ−1 1 α;ψ α,ψ + [J + (f (s, y(s), Ja+ y(s)))(t) Γ(γ) ζ +1 a α;ψ

α,ψ

− ζ Ja+ (f (s, −y(s), Ja+ (−y)(s)))(t)], 1−γ;ψ

where c1 = Ja+

y(a). Thus, for each t ∈ (a, b] we have

γ−1 2f ∗ Γ(γ) α+γ−1 󵄨󵄨 󵄨 |c |(ψ(t) − ψ(a)) + (ψ(t) − ψ(a)) 󵄨󵄨y(t)󵄨󵄨󵄨 ⩽ 1 Γ(γ) (ζ + 1)Γ(γ + α)

+ +

2‖y‖X ϱ∗ Γ(γ) α+γ−1 [ (ψ(t) − ψ(a)) (ζ + 1) Γ(γ + α)

η∗ Γ(γ) 2α+γ−1 (ψ(t) − ψ(a)) ]. Γ(2α + γ)

Thus, ‖y‖X ⩽

|c1 | 2f ∗ Γ(γ) α + (ψ(b) − ψ(a)) Γ(γ) Γ(γ + α)

(7.26)

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

+ 2[

� 195

η∗ Γ(γ) ϱ∗ Γ(γ) α 2α (ψ(b) − ψ(a)) + (ψ(b) − ψ(a)) ]‖y‖X . Γ(γ + α) Γ(2α + γ)

We deduce that |c1 | Γ(γ)

‖y‖X ⩽

ϱ Γ(γ) 1 − 2[ Γ(γ+α) (ψ(b) ∗

2f ∗ Γ(γ) (ψ(b) − ψ(a))α Γ(γ+α) η∗ Γ(γ) − ψ(a))α + Γ(2α+γ) (ψ(b)

+

− ψ(a))2α ]

:= A . The demonstration is completed. Lemma 7.15. If conditions (7.16.1), (7.16.2), and (7.26) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)], for any y ∈ 𝜕Ω and any ζ ∈ (0, 1]. Proof. Using Lemma 7.14, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(7.27)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. To prove the main result in this subsection, we need the following lemma. Lemma 7.16. Assume that 0 < δ < 1 and 0 < μ ≤ 1. Then the following inequality holds: Γ(δ + μ) Γ(μ) ⩽ . Γ(2δ + μ) Γ(δ + μ) Proof. We have for t ∈ (a, b] δ,ψ

Ja+ (ψ(s) − ψ(a)) t

=

δ+μ−1

(t)

1 δ−1 δ+μ−1 ds ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(a)) Γ(δ) a

196 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative t

1 δ−1 δ μ−1 = ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(a)) (ψ(s) − ψ(a)) ds Γ(δ) a

t

(ψ(t) − ψ(a))δ δ−1 μ−1 ⩽ ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(a)) ds Γ(δ) δ

a

δ,ψ

⩽ (ψ(t) − ψ(a)) Ja+ (ψ(s) − ψ(a))

μ−1

(t).

Using Lemma 2.6, we get for t ∈ (a, b] Γ(δ + μ) Γ(μ) 2δ+μ−1 2δ+μ−1 (ψ(t) − ψ(a)) ⩽ (ψ(t) − ψ(a)) , Γ(2δ + μ) Γ(δ + μ) which is the desired result. Theorem 7.4. Assume that (7.16.1), (7.16.2), and (7.26) hold. Then there exists at least one solution for the problem (7.24)–(7.25). Proof. It is clear that the set Ω defined in (7.27) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 7.15, assume that (7.16.1), (7.16.2), and (7.26) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (7.24)–(7.25) has at least one solution on Dom L ∩ Ω, which completes the demonstration. Now, we investigate the existence and uniqueness of periodic solutions for our problem (7.24)–(7.25). Theorem 7.5. Let (7.16.1) and (7.16.2) be satisfied. Moreover, we assume that the following hypothesis holds. (7.21.1) There exist constants ϱ > 0 and η ⩾ 0 such that α,ψ α,ψ 󵄨󵄨 󵄨 󵄨󵄨f (t, y(t), Ja+ y(t)) − f (t, y(t), Ja+ y(t))󵄨󵄨󵄨 α,ψ 󵄨 󵄨󵄨 ̄ 󵄨󵄨󵄨󵄨 − η󵄨󵄨󵄨󵄨Jaα,ψ ⩾ ϱ󵄨󵄨󵄨y(t) − y(t) + y(t) − Ja+ y(t)󵄨󵄨

for every t ∈ (a, b] and y, ȳ ∈ C1−γ;ψ (J,̄ ℝ). If one has [

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(a))α α (ψ(b) − ψ(a)) + ϱ Γ(α + γ) Γ(α + γ) +

2η∗ Γ2 (γ) 2α (ψ(b) − ψ(a)) ] < 1, Γ2 (α + γ)

then the problem (7.24)–(7.25) has a unique solution in Dom L ∩ Ω.

(7.28)

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

� 197

Proof. By Lemma 7.16 we can see that the condition (7.28) is stronger than condition (7.26). By Theorem 7.4, the problem (7.24)–(7.25) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (7.24)–(7.25) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ (a, b] H H

α,β;ψ

Da +

α,β;ψ

Da +

α,ψ

y1 (t) = f (t, y1 (t), Ja+ y1 (t)), α,ψ

y2 (t) = f (t, y2 (t), Ja+ y2 (t)),

and 1−γ,ψ

Ja+

1−γ,ψ

y1 (a) = Ja+

y1 (b),

1−γ,ψ

Ja+

1−γ,ψ

y2 (a) = Ja+

y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ (a, b]. Then α,β;ψ

LU(t) = H Da+ = =

H

U(t)

α,β;ψ Da+ y1 (t)

α,β;ψ

− H Da +

α,ψ f (t, y1 (t), Ja+ y1 (t))

y2 (t)

α,ψ

− f (t, y2 (t), Ja+ y2 (t)).

(7.29)

Using the fact that Img L = ker Q , we have b

β(α−1)

∫ ψ′ (s)(ψ(b) − ψ(s)) a

α,ψ

α,ψ

[f (s, y1 (s), Ja+ y1 (s)) − f (s, y2 (s), Ja+ y2 (s))]ds = 0. Since f ∈ C1−γ;ψ (J,̄ ℝ), there exists t0 ∈ (a, b] such that α,ψ

α,ψ

f (t0 , y1 (t0 ), Ja+ y1 (t0 )) − f (t, y2 (t0 ), Ja+ y2 (t0 )) = 0. In view of (7.21.1), we have α+γ−1 󵄨󵄨 󵄨 η Γ(γ)(ψ(t0 ) − ψ(a)) × ‖y1 − y2 ‖X . 󵄨󵄨y1 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ ϱ Γ(α + γ)

Then α+γ−1 󵄨󵄨 󵄨 η Γ(γ)(ψ(t0 ) − ψ(a)) × ‖U‖X . 󵄨󵄨U(t0 )󵄨󵄨󵄨 ⩽ ϱ Γ(α + γ)

On the other hand, by Theorem 2.20, we have

(7.30)

198 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative α;ψ H

Ja+

α,β;ψ

Da +

y(t) = y(t) −

c1 (ψ(t) − ψ(a))γ−1 , Γ(γ)

which implies that α;ψ H

c1 = [y(t0 ) − Ja+

α,β;ψ

Da +

y(t0 )]Γ(γ)(ψ(t0 ) − ψ(a))

1−γ

,

and therefore α;ψ H

U(t) = Ja+

α,β;ψ

Da+

U(t) α;ψ H

+ [U(t0 ) − Ja+

α,β;ψ

Da +

1−γ

U(t0 )](ψ(t0 ) − ψ(a))

γ−1

(ψ(t) − ψ(a))

.

Using (7.30), we obtain for every t ∈ (a, b] 1−γ γ−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 α;ψ H α,β;ψ 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [󵄨󵄨󵄨U(t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨Ja+ Da+ U(t0 )󵄨󵄨󵄨](ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) α,β;ψ 󵄨 α;ψ 󵄨 + 󵄨󵄨󵄨Ja+ H Da+ U(t)󵄨󵄨󵄨 η Γ(γ)(ψ(t0 ) − ψ(a))α γ−1 ⩽ × ‖U‖X (ψ(t) − ψ(a)) ϱ Γ(α + γ) Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 α γ−1 + 󵄩 D + U󵄩󵄩X (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) Γ(γ + α) 󵄩 a Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 γ+α−1 + . 󵄩󵄩 Da+ U󵄩󵄩X (ψ(t) − ψ(a)) Γ(γ + α)

(7.31)

By (7.16.2), (7.29), and Lemma 7.12, we find α,ψ α,ψ 󵄨󵄨H α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 Da+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (t), Ja+ y1 (t)) − f (t, y2 (t), Ja+ y2 (t))󵄨󵄨󵄨 η∗ Γ(γ) α+γ−1 γ−1 (ψ(t) − ψ(a)) ]‖U‖X . ⩽ [ϱ∗ (ψ(t) − ψ(a)) + Γ(α + γ)

Then η∗ Γ(γ) α 󵄩󵄩H α,β;ψ 󵄩󵄩 ∗ (ψ(t) − ψ(a)) ]‖U‖X . 󵄩󵄩 Da+ U󵄩󵄩X ⩽ [ϱ + Γ(α + γ) Substituting (7.32) in the right side of (7.31), we get for every t ∈ (a, b] η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(a))α α 󵄨󵄨 󵄨 (ψ(b) − ψ(a)) + 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ ϱ Γ(α + γ) Γ(α + γ) +

2η∗ Γ2 (γ) 2α γ−1 (ψ(b) − ψ(a)) ](ψ(t) − ψ(a)) ‖U‖X . Γ2 (α + γ)

Therefore, ‖U‖X ⩽ [

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(a))α α (ψ(b) − ψ(a)) + ϱ Γ(α + γ) Γ(α + γ)

(7.32)

7.4 Nonlinear fractional differential equations depending on the Ψ-Riemann–Liouville…

+

� 199

2η∗ Γ2 (γ) 2α (ψ(b) − ψ(a)) ]‖U‖X . Γ2 (α + γ)

Hence, by (7.28) we conclude that ‖U‖X = 0. As a result, for any t ∈ (a, b] we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof. 7.4.2 An example Example 7.1. We present an example of a Volterra–Fredholm integro-differential equation to test our main results. We have 1 1

, ;ln t

D12+ 7 J

3 ,ψ 7 1+

α,ψ

y(t) = f (t, y(t), Ja+ y(t)), 3 ,ψ 7 1+

y(1) = J

t ∈ (1, e],

y(e),

where for any t ∈ (1, e] we have α,ψ

α,ψ

2

Ja+ y(t) t 3 ln 7 (t) e−7−t + 3 y(t) + ( ). 5t 11 α,ψ e +2 33e e √π 1 + Ja+ y(t) −3

f (t, y(t), Ja+ y(t)) =

Here J ̄ := [1, e], α = 21 , β = 71 , and ψ(t) = ln t. It is easy to see that f ∈ C 3 ;ψ (J,̄ ℝ). Hence, condition (7.16.1) is satisfied. 7 Furthermore, for all t ∈ (1, e] and y, y ∈ C 3 (J,̄ ℝ) we obtain 7

;ψ

α,ψ α,ψ 󵄨󵄨 󵄨 󵄨󵄨f (t, y(t), Ja+ y(t)) − f (t, y(t), Ja+ y(t))󵄨󵄨󵄨 α,ψ 󵄨 󵄨 󵄨 α,ψ 󵄨 ⩽ ϱ(t)󵄨󵄨󵄨y(t) − y(t)󵄨󵄨󵄨 + η(t)󵄨󵄨󵄨Ja+ y(t) − Ja+ y(t)󵄨󵄨󵄨

and α,ψ α,ψ 󵄨󵄨 󵄨 󵄨󵄨f (t, y(t), Ja+ y(t)) − f (t, y(t), Ja+ y(t))󵄨󵄨󵄨 α,ψ 󵄨 󵄨 󵄨 α,ψ 󵄨 ⩾ ϱ󵄨󵄨󵄨y(t) − y(t)󵄨󵄨󵄨 − η󵄨󵄨󵄨Ja+ y(t) − Ja+ y(t)󵄨󵄨󵄨,

which implies that (7.16.2) and (7.21.1) are satisfied with ϱ(t) =

t , e3 √π

2

η(t) =

e−7−t , 33e11

ϱ=

1 , e3 √π

and

η=

1 . 33e19

200 � 7 Nonlinear fractional differential equations with ψ-Hilfer fractional derivative By simple calculations, we get ϱ∗ = [

1 , e2 √π

η∗ =

1 , 33e19

and

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(a))α α (ψ(b) − ψ(a)) + ϱ Γ(α + γ) Γ(α + γ) +

2η∗ Γ2 (γ) 2α (ψ(b) − ψ(a)) ] ≈ 0.247024 < 1. 2 Γ (α + γ)

So, by Theorem 7.5, our problem has a unique solution.

7.5 Notes and remarks The conclusions of the present chapter are based on the papers [91–93]. One can see the monographs [7, 14, 15, 29, 47, 49, 75, 154, 237, 239] and the papers [102–108, 153, 164, 166, 176, 178, 182, 203, 205–207, 210, 213, 215, 234, 235, 250] for additional details and results.

8 ψ-Hilfer fractional pantograph-type differential equations 8.1 Introduction and motivations This chapter deals with the existence and uniqueness of periodic solutions for some class of nonlinear fractional pantograph systems with ψ-Hilfer derivative associated with periodic-type fractional integral boundary conditions in a weighted space of continuous functions. The proofs are based upon Mawhin’s coincidence degree theory. Suitable illustrative examples are provided in each section. We explore and demonstrate the results obtained in this chapter by taking into consideration the publications stated in the preceding chapters and the works that follow: – The papers [38, 54, 117–119, 139, 168, 173, 174, 195, 196, 229, 230, 241, 242], where the authors provided many interesting and insightful results for diverse fractional differential problems with varied conditions and based on multiple approaches. – The works [19, 56, 65, 68, 70, 73, 190, 198], where several researchers investigated some new existence and uniqueness results for NFDE pantograph problems by applying fixed point theorems or the coincidence degree theorem. – As some of the significant and particular problems in fractional calculus, there were some studies on the fractional differential equations of pantograph type. Some scholars established the existence and uniqueness of the solutions to some nonlinear classes of pantograph-type fractional differential equations with a variety of boundary conditions by using the fixed point approaches [137, 180, 231]. – The paper of Wang and Zhang [249], where they proved some existence results using the Krasnoselskii, Schaefer, and Schauder fixed point theorems for the following nonlocal initial value problem for differential equations involving the Hilfer fractional derivative: α,β

Da+ u(t) = f (t, u(t)),

{ –

1−γ (Ja+ u)(a+ )

=

t ∈ (a, b],

∑m i=1 λi u(τi ).

Jalilian and Ghasmi [137] studied the following initial value problem of nonlinear fractional integro-differential equations of pantograph type: qt

c α D u(t) = f (t, u(t), u(pt)) + ∫0 g1 (t, s, u(s))ds { { { t + ∫0 g2 (t, s, u(s))ds, t ∈ [0, T], { { { {u(0) = u0 ,

–

where c Dα is the Caputo fractional derivative of order α ∈ (0, 1] and 0 < p, q < 1. Sudsutad et al. [236] employed Banach’s contraction principle, Krasnoselskii’s fixed point theorem, and the Leray–Schauder nonlinear alternative to establish the exis-

https://doi.org/10.1515/9783111334387-008

202 � 8 ψ-Hilfer fractional pantograph-type differential equations tence, uniqueness, and stability results for the following problem with mixed nonlocal boundary conditions: H

{

α,ρ;ψ

D0+

ϕ,ψ

x(t) = f (t, x(t), I0+ x(t)),

x(0) = 0,

∑m i=1 δi x(ηi )

+

t ∈ (0, T],

β ,ψ ∑nj=1 ωj I0+j x(θj )

μ ,ρ;ψ

+ ∑rk=1 λk H D0+k

x(tk ) = κ.

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer derivative In the current section, we study a class of interest of these NFDEs, the nonlinear pantograph fractional equations with ψ-Hilfer fractional derivative: H

α,β;ψ

Da +

1−γ,ψ

J0+ α,β;ψ

where H D0+

y(t) = f (t, y(t), y(εt)), 1−γ,ψ

y(0) = J0+

t ∈ (0, b],

y(b),

(8.1) (8.2)

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1, 0 < ε < 1, 1−γ,ψ

and type β ∈ [0, 1] and J0+ is the ψ-Riemann–Liouville fractional integral of order 1− γ (γ = α + β − αβ). Moreover, f : (0, b] × ℝ2 → ℝ is a given continuous function.

8.2.1 Main results Let α;ψ

X = {y ∈ C1−γ;ψ (T, ℝ) : y(ξ) = J0+ υ(ξ) : υ ∈ C1−γ;ψ (T, ℝ), ξ ∈ (0, b]}

and let Y = C1−γ;ψ (T, ℝ) with the norm ‖y‖X = ‖y‖Y = ‖y‖C1−γ;ψ . To prove the main findings, we need the lemmas mentioned in Section 7.2.1. Lemma 8.1. Let ε ∈ (0, 1) and α, γ ∈ (0, 1). Then for all ξ ∈ (a, b] we have α+γ−1

(ψ(εξ) − ψ(0))

⩽

′ 1 maxξ∈T ψ (ξ) α+γ−1 (ψ(ξ) − ψ(0)) ε minξ∈T ψ′ (ξ)

and γ−1

(ψ(εξ) − ψ(0))

1−γ

(ψ(ξ) − ψ(0))

⩽[

′ 1−γ 1 maxξ∈T ψ (ξ) ] . ε minξ∈T ψ′ (ξ)

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer…

� 203

Proof. Let ε ∈ (0, 1). Then for all ξ ∈ (a, b] we have α+γ−1

(ψ(εξ) − ψ(0))

α+γ

= (ψ(εξ) − ψ(0))

(ψ(εξ) − ψ(0)) α+γ−1 (ψ(ξ) − ψ(0)) ⩽ (ψ(ξ) − ψ(0)) (ψ(εξ) − ψ(0)) α+γ−1

⩽ (ψ(ξ) − ψ(0))

ξ

∫0 ψ′ (t)dt εξ

∫0 ψ′ (t)dt

α+γ−1

⩽ (ψ(ξ) − ψ(0)) ⩽

−1

ξ maxξ∈T ψ′ (ξ)

εξ minξ∈T ψ′ (ξ)

′ 1 maxξ∈T ψ (ξ) α+γ−1 (ψ(ξ) − ψ(0)) . ε minξ∈T ψ′ (ξ)

On the other hand, we have γ−1

(ψ(εξ) − ψ(0))

1−γ

(ψ(ξ) − ψ(0))

1−γ

=[

(ψ(ξ) − ψ(0)) ] ψ(εξ) − ψ(0)) ξ

⩽[

⩽[

∫0 ψ′ (t)dt εξ

∫0 ψ′ (t)dt

1−γ

]

′ 1−γ 1 maxξ∈T ψ (ξ) ] . ε minξ∈T ψ′ (ξ)

Lemma 8.2. Suppose that the following hypotheses are satisfied. (8.2.1) The function f : (0, b] × ℝ × ℝ → ℝ is such that f (⋅, y(⋅), y(ε⋅)) ∈ C1−γ;ψ (T, ℝ) for all y ∈ C1−γ;ψ (T, ℝ). (8.2.2) There exist nonnegative functions ϱ, η ∈ C(T, ℝ+ ) such that 󵄨󵄨 󵄨 󵄨󵄨f (ξ, y(ξ), y(εξ)) − f (ξ, y(ξ), y(εξ))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ ϱ(ξ)󵄨󵄨󵄨y(ξ) − y(ξ)󵄨󵄨󵄨 + η(ξ)󵄨󵄨󵄨y(εξ) − y(εξ)󵄨󵄨󵄨 for every ξ ∈ (0, b] and y, y ∈ C1−γ;ψ (T, ℝ). Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps:

204 � 8 ψ-Hilfer fractional pantograph-type differential equations Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each ξ ∈ (0, b] we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(ξ) − QN (y)(ξ)󵄨󵄨󵄨 b

|N (yn )(s) − N (y)(s)| (1 + β(α − 1)) ⩽ ds. ∫ ψ′ (s) 1+β(α−1) (ψ(b) − ψ(0)) (ψ(b) − ψ(s))β(1−α) 0

By (8.2.2) we have 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(ξ) − QN (y)(ξ)󵄨󵄨󵄨 b

ϱ∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

b

+

η∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨yn (εs) − y(εs)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

ϱ∗ Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ ‖yn − y‖Y J0+ (ψ(s) − ψ(0)) (b) ⩽ (ψ(b) − ψ(0))1+β(α−1) η∗ Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ + ‖yn − y‖Y J0+ (ψ(s) − ψ(0)) (εb), (ψ(b) − ψ(0))1+β(α−1) where ϱ∗ = ‖ϱ‖∞ and η∗ = ‖η‖∞ . Using Lemma 2.6, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(ξ) − QN (y)(ξ)󵄨󵄨󵄨 Γ(2 + β(α − 1))ϱ∗ Γ(γ) γ−1 (ψ(b) − ψ(0)) ‖yn − y‖Y ⩽ Γ(α + 1) Γ(2 + β(α − 1))η∗ Γ(γ) γ−1 + (ψ(εb) − ψ(0)) ‖yn − y‖Y Γ(α + 1) Γ(2 + β(α − 1))(ϱ∗ + η∗ )Γ(γ) γ−1 ⩽ (ψ(εb) − ψ(0)) ‖yn − y‖Y . Γ(α + 1) Thus, for each ξ ∈ T we obtain 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(ξ) − ψ(0)) (QN (yn )(ξ) − QN (y)(ξ))󵄨󵄨󵄨 Γ(2 + β(α − 1))(ϱ∗ + η∗ )Γ(γ) γ−1 ⩽ (ψ(εb) − ψ(0)) Γ(α + 1) 1−γ

× (ψ(b) − ψ(0))

‖yn − y‖Y .

Then for all ξ ∈ T we get 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(ξ) − ψ(0)) (QN (yn )(ξ) − QN (y)(ξ))󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞.

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer…

Therefore, 󵄩 󵄩󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN (Ω) is bounded. For ξ ∈ (0, b] and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(ξ)󵄨󵄨󵄨 b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨N (y)(s)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1)

(1 + β(α − 1)) ⩽ (ψ(b) − ψ(0))1+β(α−1)

0

b

β(α−1) 󵄨

󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

b

+

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

b

ϱ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨y(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(0)) ∗

η∗ (1 + β(α − 1)) + (ψ(b) − ψ(0))1+β(α−1) b

0

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s))

󵄨󵄨y(εs)󵄨󵄨󵄨ds. 󵄨 󵄨

0

Using Lemma 2.6, we get ∗ ∗ γ−1 󵄨󵄨 󵄨 [ϱ M + f ]Γ(γ)Γ(2 + β(α − 1)) (ψ(b) − ψ(0)) 󵄨󵄨QN (y)(ξ)󵄨󵄨󵄨 ⩽ Γ(α + 1)

η∗ M Γ(γ)Γ(2 + β(α − 1)) γ−1 (ψ(εb) − ψ(0)) Γ(α + 1) [(ϱ∗ + η∗ )M + f ∗ ]Γ(γ)Γ(2 + β(α − 1)) γ−1 ⩽ (ψ(εb) − ψ(0)) , Γ(α + 1) +

where f ∗ = ‖f (⋅, 0, 0)‖C1−γ;ψ .

� 205

206 � 8 ψ-Hilfer fractional pantograph-type differential equations Thus, [(ϱ∗ + η∗ )M + f ∗ ]Γ(γ)Γ(2 + β(α − 1)) 󵄩 󵄩󵄩 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ Γ(α + 1) × (ψ(εb) − ψ(0))

γ−1

1−γ

(ψ(b) − ψ(0))

.

So, QN (Ω) is a bounded set in Y . Step 3. We demonstrate that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and ξ ∈ (0, b], we get L−1 P (N y(ξ) − QN y(ξ)) α;ψ

= J0+ [f (ξ, y(ξ), y(εξ)) −

Γ(2 + β(α − 1)) 1+β(α−1);ψ J+ f (s, y(s), y(εs))(b)] (ψ(b) − ψ(0))1+β(α−1) 0 ξ

1 α−1 = ∫ ψ′ (s)(ψ(ξ) − ψ(s)) f (s, y(s), y(εs))ds Γ(α) 0

−

Γ(2 + β(α − 1))(ψ(ξ) − ψ(0))α 1+β(α−1);ψ J+ f (s, y(s), y(εs))(b). Γ(α + 1)(ψ(b) − ψ(0))1+β(α−1) 0

For all y ∈ Ω and ξ ∈ (0, b], we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(ξ)󵄨󵄨󵄨 ξ

⩽

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(ξ) − ψ(s)) 󵄨󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

ξ

+

1 α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(ξ) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

1 γ−1 (1 + β(α − 1))(ψ(b) − ψ(0)) + Γ(α + 1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

β(α−1) 󵄨

󵄨󵄨f (s, y(s), y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

0 b

(1 + β(α − 1))(ψ(b) − ψ(0))γ−1 ψ′ (s)|f (s, 0, 0)| ds + ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) 0

⩽

f ∗ Γ(γ)Γ(2 + β(α − 1)) f Γ(γ) α+γ−1 (ψ(ξ) − ψ(0)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0))1−α−γ ∗

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer… � 207 ξ

ϱ∗ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(ξ) − ψ(s)) 󵄨󵄨󵄨y(s)󵄨󵄨󵄨ds Γ(α) 0

ξ

η∗ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(ξ) − ψ(s)) 󵄨󵄨󵄨y(εs)󵄨󵄨󵄨ds Γ(α) 0

b

ϱ∗ (1 + β(α − 1))(ψ(b) − ψ(0))γ−1 β(α−1) 󵄨 󵄨󵄨y(s)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 Γ(α + 1) η∗ (1 + β(α − 1))(ψ(b) − ψ(0))γ−1 + Γ(α + 1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

0

β(α−1) 󵄨

󵄨󵄨y(εs)󵄨󵄨󵄨ds. 󵄨 󵄨

0

Using Lemma 2.6 and Lemma 8.1, we get f ∗ Γ(γ) α+γ−1 󵄨󵄨 −1 󵄨 (ψ(ξ) − ψ(0)) 󵄨󵄨LP (id − Q )N y(ξ)󵄨󵄨󵄨 ⩽ Γ(α + γ) f ∗ Γ(γ)Γ(2 + β(α − 1)) α+γ−1 + (ψ(b) − ψ(0)) Γ2 (α + 1) ϱ∗ M Γ(γ) α+γ−1 + (ψ(ξ) − ψ(0)) Γ(γ + α) η∗ M Γ(γ) α+γ−1 (ψ(εξ) − ψ(0)) + Γ(γ + α) M Γ(γ)Γ(2 + β(α − 1)) + Γ2 (α + 1) × [ϱ∗ (ψ(b) − ψ(0)) ⩽

α+γ−1

+ η∗ (ψ(εb) − ψ(0))

]

f Γ(γ) α+γ−1 (ψ(ξ) − ψ(0)) Γ(α + γ) f ∗ Γ(γ)Γ(2 + β(α − 1)) α+γ−1 + (ψ(b) − ψ(0)) Γ2 (α + 1) ϱ∗ M Γ(γ) α+γ−1 + (ψ(ξ) − ψ(0)) Γ(γ + α) ∗

+ +

′ η∗ M Γ(γ) maxξ∈T ψ (ξ) α+γ−1 ( )(ψ(ξ) − ψ(0)) εΓ(γ + α) minξ∈T ψ′ (ξ)

M Γ(γ)Γ(2 + β(α − 1))

Γ2 (α + 1)

× [ϱ∗ (ψ(b) − ψ(0)) So

α+γ−1

α+γ−1

α+γ−1

+ η∗ (ψ(εb) − ψ(0))

].

208 � 8 ψ-Hilfer fractional pantograph-type differential equations 1−γ −1 󵄨 󵄨󵄨 󵄨󵄨(ψ(ξ) − ψ(0)) LP (id − Q )N y(ξ)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α α ⩽ (ψ(b) − ψ(0)) (ψ(b) − ψ(0)) + Γ(α + γ) Γ2 (α + 1)

+ +

′ ϱ∗ M Γ(γ) η∗ M Γ(γ) maxξ∈T ψ (ξ) α α )(ψ(ξ) − ψ(0)) (ψ(b) − ψ(0)) + ( ′ Γ(γ + α) εΓ(γ + α) minξ∈T ψ (ξ)

M Γ(γ)Γ(2 + β(α − 1))

Γ2 (α

+ 1)

1−γ

× (ψ(b) − ψ(0))

α

[ϱ∗ (ψ(b) − ψ(0)) + η∗ (ψ(εb) − ψ(0))

α+γ−1

].

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α α ⩽ (ψ(b) − ψ(0)) + (ψ(b) − ψ(0)) Γ(α + γ) Γ2 (α + 1) + +

′ ϱ∗ M Γ(γ) η∗ M Γ(γ) maxξ∈T ψ (ξ) α α (ψ(b) − ψ(0)) + ( )(ψ(b) − ψ(0)) Γ(γ + α) εΓ(γ + α) minξ∈T ψ′ (ξ)

M Γ(γ)Γ(2 + β(α − 1))

Γ2 (α + 1)

1−γ

× (ψ(b) − ψ(0))

α

α+γ−1

[ϱ∗ (ψ(b) − ψ(0)) + η∗ (ψ(εb) − ψ(0))

].

This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. By using Lemma 2.6, we have for 0 < ξ1 < ξ2 ⩽ b and y ∈ Ω 󵄨󵄨 L−1 (id − Q )N y(ξ ) L−1 (id − Q )N y(ξ ) 󵄨󵄨 󵄨󵄨 P 1 󵄨󵄨 2 − P 󵄨󵄨 󵄨 󵄨󵄨 (ψ(ξ2 ) − ψ(0))γ−1 (ψ(ξ1 ) − ψ(0))γ−1 󵄨󵄨󵄨 ξ

⩽

1 󵄨󵄨 (ψ(ξ ) − ψ(0))1−γ (ψ(ξ ) − ψ(0))1−γ 󵄨󵄨 1 󵄨 󵄨󵄨 2 1 − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(ξ2 ) − ψ(s))1−α (ψ(ξ1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

0

ξ2

(ψ(ξ2 ) − ψ(0))1−γ ψ′ (s)|f (s, y(s), y(εs))| 󵄨 󵄨 ds × 󵄨󵄨󵄨f (s, y(s), y(εs))󵄨󵄨󵄨ds + ∫ Γ(α) (ψ(ξ2 ) − ψ(s))1−α 1+α−γ

+

ξ1

(1 + β(α − 1))[(ψ(ξ2 ) − ψ(0)) − (ψ(ξ1 ) − ψ(0))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(0))1+β(α−1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s))

β(α−1) 󵄨

󵄨󵄨f (s, y(s), y(εs))󵄨󵄨󵄨ds 󵄨 󵄨

0 ξ

1 󵄨󵄨 (ψ(ξ ) − ψ(0))1−γ (ψ(ξ ) − ψ(0))1−γ 󵄨󵄨 (f ∗ + ϱ∗ M ) 󵄨 󵄨󵄨 2 1 ⩽ − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(ξ2 ) − ψ(s))1−α (ψ(ξ1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

0

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer…

γ−1

× (ψ(s) − ψ(0))

� 209

ξ1

η∗ M ds + ∫ ψ′ (s) Γ(α) 0

󵄨󵄨 (ψ(ξ ) − ψ(0))1−γ (ψ(ξ ) − ψ(0))1−γ 󵄨󵄨 γ−1 󵄨 󵄨󵄨 2 1 × 󵄨󵄨󵄨 − 󵄨(ψ(εs) − ψ(0)) ds 󵄨󵄨 (ψ(ξ2 ) − ψ(s))1−α (ψ(ξ1 ) − ψ(s))1−α 󵄨󵄨󵄨 ξ2

(f ∗ + ϱ∗ M )(ψ(ξ2 ) − ψ(0))1−γ ψ′ (s)(ψ(s) − ψ(0))γ−1 + ds ∫ Γ(α) (ψ(ξ2 ) − ψ(s))1−α ξ1

ξ2

+

η∗ M (ψ(ξ2 ) − ψ(0))1−γ (ψ(ξ2 ) − ψ(s))α−1 ds ∫ ψ′ (s) Γ(α) (ψ(εs) − ψ(0))1−γ ξ1

+

Γ(γ)Γ(2 + β(α − 1)) η∗ M f ∗ + ϱ∗ M + ] [ 1−γ 2 (ψ(b) − ψ(0)) (ψ(εb) − ψ(0))1−γ Γ (α + 1)

× [(ψ(ξ2 ) − ψ(0))

1+α−γ

1+α−γ

− (ψ(ξ1 ) − ψ(0))

].

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as ξ1 → ξ2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 8.3. Assume that (8.2.1) and (8.2.2) hold. If the condition maxξ∈T ψ′ (ξ) Γ(γ) 1 α ∗ ∗ 1 [ϱ + η ( )](ψ(b) − ψ(0)) < Γ(γ + α) ε minξ∈T ψ′ (ξ) 2

(8.3)

is satisfied, then there exists A > 0 which is independent of ζ , where for y ∈ X we get L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any ξ ∈ (0, b] α,β;ψ

Ly(ξ) = H D0+

y(ξ) =

1 f (ξ, y(ξ), y(εξ)) 1+ζ ζ − f (ξ, −y(ξ), −y(εξ)). 1+ζ

210 � 8 ψ-Hilfer fractional pantograph-type differential equations By Theorem 2.20 we get y(ξ) =

c1 (ψ(ξ) − ψ(0))γ−1 1 α;ψ + [J + (f (s, y(s), y(εs)))(ξ), Γ(γ) ζ +1 0 α;ψ

− ζ J0+ (f (s, −y(s), −y(εs)))(ξ)], 1−γ;ψ

where c1 = J0+

y(0). Thus, by using Lemma 8.1, for each ξ ∈ (0, b] we have

γ−1 2f ∗ Γ(γ) α+γ−1 󵄨󵄨 󵄨 |c |(ψ(ξ) − ψ(0)) + (ψ(ξ) − ψ(0)) 󵄨󵄨y(ξ)󵄨󵄨󵄨 ⩽ 1 Γ(γ) (ζ + 1)Γ(γ + α) 2ϱ∗ Γ(γ)‖y‖X α+γ−1 + (ψ(ξ) − ψ(0)) (ζ + 1)Γ(γ + α) 2η∗ Γ(γ)‖y‖X α+γ−1 (ψ(εξ) − ψ(0)) + (ζ + 1)Γ(γ + α)

⩽

|c1 |(ψ(ξ) − ψ(0))γ−1 2f ∗ Γ(γ) α+γ−1 + (ψ(ξ) − ψ(0)) Γ(γ) (ζ + 1)Γ(γ + α) 2ϱ∗ Γ(γ)‖y‖X α+γ−1 + (ψ(ξ) − ψ(0)) Γ(γ + α) +

′ 2η∗ Γ(γ)‖y‖X maxξ∈T ψ (ξ) α+γ−1 ( )(ψ(ξ) − ψ(0)) . εΓ(γ + α) minξ∈T ψ′ (ξ)

Thus, ‖y‖X ⩽

|c1 | 2f ∗ Γ(γ) α + (ψ(b) − ψ(0)) Γ(γ) Γ(γ + α) 2ϱ∗ Γ(γ)‖y‖X α + (ψ(b) − ψ(0)) Γ(γ + α) +

′ 2η∗ Γ(γ)‖y‖X maxξ∈T ψ (ξ) α ( )(ψ(b) − ψ(0)) . εΓ(γ + α) minξ∈T ψ′ (ξ)

We deduce that ‖y‖X ⩽

1−

|c1 | Γ(γ)

+

2Γ(γ) [ϱ∗ Γ(γ+α)

+

2f ∗ Γ(γ) (ψ(b) Γ(γ+α)

− ψ(0))α

′ η∗ maxξ∈T ψ (ξ) ( )](ψ(b) ε minξ∈T ψ′ (ξ)

− ψ(0))α

:= A .

The demonstration is completed. Lemma 8.4. If conditions (8.2.1), (8.2.2), and (8.3) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)] for any y ∈ 𝜕Ω and any ζ ∈ (0, 1].

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer… � 211

Proof. Using Lemma 8.3, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(8.4)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. To prove the main result in this subsection, we need the following lemma. Lemma 8.5. Assume that 0 < δ < 1 and 0 < μ ≤ 1. Then the following inequality holds: Γ(δ + μ) Γ(μ) ⩽ . Γ(2δ + μ) Γ(δ + μ) Proof. We have for ξ ∈ (0, b] δ+μ−1

δ,ψ

J0+ (ψ(s) − ψ(0))

(ξ)

ξ

1 δ−1 δ+μ−1 ds = ∫ ψ′ (s)(ψ(ξ) − ψ(s)) (ψ(s) − ψ(0)) Γ(δ) 0

ξ

=

1 δ−1 δ μ−1 ∫ ψ′ (s)(ψ(ξ) − ψ(s)) (ψ(s) − ψ(0)) (ψ(s) − ψ(0)) ds Γ(δ) 0

ξ

⩽

(ψ(ξ) − ψ(0))δ δ−1 μ−1 ∫ ψ′ (s)(ψ(ξ) − ψ(s)) (ψ(s) − ψ(0)) ds Γ(δ) δ

0

δ,ψ

⩽ (ψ(ξ) − ψ(0)) J0+ (ψ(s) − ψ(0))

μ−1

(ξ).

By using Lemma 2.6, we get for ξ ∈ (0, b] Γ(δ + μ) Γ(μ) 2δ+μ−1 2δ+μ−1 (ψ(ξ) − ψ(0)) ⩽ (ψ(ξ) − ψ(0)) , Γ(2δ + μ) Γ(δ + μ) which is the desired result. Theorem 8.1. Assume that (8.2.1), (8.2.2), and (8.3) hold. Then there exists at least one solution for the problem (8.1)–(8.2).

212 � 8 ψ-Hilfer fractional pantograph-type differential equations Proof. It is clear that the set Ω defined in (8.4) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 8.4, assume that (8.2.1), (8.2.2), and (8.3) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Thus, problem (8.1)–(8.2) has at least one solution on Dom L ∩ Ω, which completes the demonstration. Now, we investigate the existence and uniqueness of periodic solutions for our problem (8.1)–(8.2). Theorem 8.2. Let (8.2.1) and (8.2.2) be satisfied. Moreover, we assume that the following hypothesis holds. (8.7.1) There exist constants ϱ > 0 and η ⩾ 0 such that 󵄨󵄨 󵄨 󵄨 ̄ 󵄨󵄨󵄨󵄨 󵄨󵄨f (ξ, y(ξ), y(εξ)) − f (ξ, y(ξ), y(εξ))󵄨󵄨󵄨 ⩾ ϱ󵄨󵄨󵄨y(ξ) − y(ξ) 󵄨 󵄨 − η󵄨󵄨󵄨y(εξ) − y(εξ)󵄨󵄨󵄨 for every ξ ∈ (0, b] and y, ȳ ∈ C1−γ;ψ (T, ℝ). If one has ′ 1−γ maxξ∈T ψ′ (ξ) 1−γ η 1 maxξ∈T ψ (ξ) 2Γ(γ) α ∗ ∗ 1 ( [ϱ + η ( ) + ) ](ψ(b) − ψ(0)) ϱ ε minξ∈T ψ′ (ξ) Γ(α + γ) ε minξ∈T ψ′ (ξ)

< 1,

(8.5)

then the problem (8.1)–(8.2) has a unique solution in Dom L ∩ Ω. Proof. Since 0 < 1 − v < 1, we can see that the condition (8.5) is stronger than condition (8.3). By Theorem 8.1, the problem (8.1)–(8.2) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (8.1)–(8.2) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each ξ ∈ (0, b] H H

α,β;ψ

D0+

α,β;ψ

D0+

y1 (ξ) = f (ξ, y1 (ξ), y1 (εξ)), y2 (ξ) = f (ξ, y2 (ξ), y2 (εξ)),

and 1−γ,ψ

J0+

1−γ,ψ

y1 (0) = J0+

y1 (b),

Let U(ξ) = y1 (ξ) − y2 (ξ) for all ξ ∈ (0, b]. Then α,β;ψ

LU(ξ) = H D0+

U(ξ)

1−γ,ψ

J0+

1−γ,ψ

y2 (0) = J0+

y2 (b).

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer…

α,β;ψ

= H D0+

α,β;ψ

y1 (ξ) − H D0+

� 213

y2 (ξ)

= f (ξ, y1 (ξ), y1 (εξ)) − f (ξ, y2 (ξ), y2 (εξ)).

(8.6)

Using the fact that Img L = ker Q , we have b

β(α−1)

∫ ψ′ (s)(ψ(b) − ψ(s)) 0

[f (s, y1 (s), y1 (εs)) − f (s, y2 (s), y2 (εs))]ds = 0. Since f ∈ C1−γ;ψ (T, ℝ), there exists ξ0 ∈ (0, b] such that f (ξ0 , y1 (ξ0 ), y1 (εξ0 )) − f (ξ, y2 (ξ0 ), y2 (εξ0 )) = 0. In view of (8.7.1), we have γ−1 󵄨󵄨 󵄨 η(ψ(εξ0 ) − ψ(0)) × ‖y1 − y2 ‖X . 󵄨󵄨y1 (ξ0 ) − y2 (ξ0 )󵄨󵄨󵄨 ⩽ ϱ

Then γ−1 󵄨󵄨 󵄨 η(ψ(εξ0 ) − ψ(0)) × ‖U‖X . 󵄨󵄨U(ξ0 )󵄨󵄨󵄨 ⩽ ϱ

(8.7)

On the other hand, by Theorem 2.20, we have α;ψ H

J0+

α,β;ψ

D0+

U(ξ) = U(ξ) −

c1 (ψ(ξ) − ψ(0))γ−1 , Γ(γ)

which implies that α;ψ H

c1 = [U(ξ0 ) − J0+

α,β;ψ

D0+

U(ξ0 )]Γ(γ)(ψ(ξ0 ) − ψ(0))

1−γ

.

We deduce that α;ψ H

U(ξ) = J0+

α,β;ψ

D0+

U(ξ) α;ψ H

+ [U(ξ0 ) − J0+

α,β;ψ

D0+

1−γ

U(ξ0 )](ψ(ξ0 ) − ψ(0))

(ψ(ξ) − ψ(0))

γ−1

.

Using (8.7) and Lemma 8.1, we obtain for every ξ ∈ (0, b] 1−γ γ−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 α;ψ H α,β;ψ 󵄨 󵄨󵄨U(ξ)󵄨󵄨󵄨 ⩽ [󵄨󵄨󵄨U(ξ0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ U(ξ0 )󵄨󵄨󵄨](ψ(ξ0 ) − ψ(0)) (ψ(ξ) − ψ(0)) α,β;ψ 󵄨 α;ψ 󵄨 + 󵄨󵄨󵄨J0+ H D0+ U(ξ)󵄨󵄨󵄨 η γ−1 1−γ γ−1 ⩽ ‖U‖X (ψ(εξ0 ) − ψ(0)) (ψ(ξ0 ) − ψ(0)) (ψ(ξ) − ψ(0)) ϱ

214 � 8 ψ-Hilfer fractional pantograph-type differential equations Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 α γ−1 󵄩󵄩 D0+ U󵄩󵄩X (ψ(ξ0 ) − ψ(0)) (ψ(ξ) − ψ(0)) Γ(γ + α) Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 γ+α−1 + 󵄩 D + U󵄩󵄩X (ψ(ξ) − ψ(0)) Γ(γ + α) 󵄩 0 +

⩽

′ 1−γ η 1 maxξ∈T ψ (ξ) γ−1 ‖U‖X [ ] (ψ(ξ) − ψ(0)) ′ ϱ ε minξ∈T ψ (ξ)

Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 α γ−1 󵄩󵄩 D0+ U󵄩󵄩X (ψ(ξ0 ) − ψ(0)) (ψ(ξ) − ψ(0)) Γ(γ + α) Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 γ+α−1 + . 󵄩 D + U󵄩󵄩X (ψ(ξ) − ψ(0)) Γ(γ + α) 󵄩 0 +

(8.8)

By (8.2.2) and (8.6), we find 󵄨󵄨H α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 D0+ U(ξ)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (ξ, y1 (ξ), y1 (εξ)) − f (ξ, y2 (ξ), y2 (εξ))󵄨󵄨󵄨 γ−1

⩽ [ϱ∗ (ψ(ξ) − ψ(0))

γ−1

+ η∗ (ψ(εξ) − ψ(0))

]‖U‖X .

Using Lemma 8.1, we have 1−γ H α,β;ψ 󵄨󵄨 󵄨 D0+ U(ξ)󵄨󵄨󵄨 󵄨󵄨(ψ(ξ) − ψ(0))

γ−1

⩽ [ϱ∗ + η∗ (ψ(εξ) − ψ(0)) ⩽ [ϱ∗ + η∗ (

1−γ

(ψ(ξ) − ψ(0))

]‖U‖X

1−γ

1 maxξ∈T ψ (ξ) ) ε minξ∈T ψ′ (ξ) ′

]‖U‖X .

Then maxξ∈T ψ′ (ξ) 1−γ 󵄩󵄩H α,β;ψ 󵄩󵄩 ∗ ∗ 1 ) ]‖U‖X . 󵄩󵄩 D0+ U󵄩󵄩X ⩽ [ϱ + η ( ε min ψ′ (ξ) ξ∈T

Substituting (8.9) in the right side of (8.8), we get for every ξ ∈ (0, b] ′ 1−γ η 1 maxξ∈T ψ (ξ) 󵄨󵄨 󵄨 ) 󵄨󵄨U(ξ)󵄨󵄨󵄨 ⩽ [ ( ϱ ε min ψ′ (ξ) ξ∈T

+

′ 1−γ 2Γ(γ) 1 maxξ∈T ψ (ξ) α ) )(ψ(b) − ψ(0)) ] (ϱ∗ + η∗ ( Γ(α + γ) ε minξ∈T ψ′ (ξ) γ−1

× (ψ(ξ) − ψ(0))

‖U‖X .

Therefore, ‖U‖X

′ 1−γ η 1 maxξ∈T ψ (ξ) ⩽[ ( ) ′ ϱ ε minξ∈T ψ (ξ)

(8.9)

8.2 Periodic solutions for some nonlinear fractional pantograph problems with Ψ-Hilfer…

+

� 215

′ 1−γ 2Γ(γ) 1 maxξ∈T ψ (ξ) α ) )(ψ(b) − ψ(0)) ]‖U‖X . (ϱ∗ + η∗ ( ′ Γ(α + γ) ε minξ∈T ψ (ξ)

Hence, by (8.5), we conclude that ‖U‖X = 0. As a result, for any ξ ∈ (0, b] we get U(ξ) = 0 󳨐⇒ y1 (ξ) = y2 (ξ). This completes the proof.

8.2.2 An example We present an example of a Volterra–Fredholm integro-differential equation to test our main results. We have 1 1

, ;eξ

D02+ 7

y(ξ) = f (ξ, y(ξ), y(εξ)),

3 ξ ,e 7 0+

J

3 ξ ,e 7 0+

y(1) = J

ξ ∈ (0, 1],

y(e),

where for any ξ ∈ (0, 1] we have y( ξ3 ) e−7−ξ 3 ln 7 (ξ) ξ + 1 + y(ξ) + ( ). 33e11 1 + y( ξ ) e3 √π e5ξ + 2 3 2

−3

f (ξ, y(ξ), y(εξ)) =

Here T := [0, 1], α = 21 , β = 71 , ε = 31 , and ψ(ξ) = eξ . It is easy to see that f ∈ C 3 ;ψ (T, ℝ). Hence, condition (8.2.1) is satisfied. 7

Furthermore, for all ξ ∈ (0, 1] and y, y ∈ C 3 ;ψ (T, ℝ), we obtain 7

󵄨󵄨 󵄨 󵄨󵄨f (ξ, y(ξ), y(εξ)) − f (ξ, y(ξ), y(εξ))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ ϱ(ξ)󵄨󵄨󵄨y(ξ) − y(ξ)󵄨󵄨󵄨 + η(ξ)󵄨󵄨󵄨y(εξ) − y(εξ)󵄨󵄨󵄨. Then (8.2.2) is satisfied with ϱ(ξ) =

ξ+1 e3 √π ∗

By simple calculations, we get ϱ = 1, and

2 , η∗ e3 √π

=

2

e−7−ξ . 33e11 1 , maxξ∈T 33e18

and η(ξ) =

ψ′ (ξ) = e, minξ∈T ψ′ (ξ) =

maxξ∈T ψ′ (ξ) Γ(γ) 1 α ∗ ∗ 1 [ϱ + η ( )](ψ(b) − ψ(0)) ≈ 0.119 < . Γ(γ + α) ε minξ∈T ψ′ (ξ) 2 With the use of Theorem 8.1 our problem has at least one solution.

216 � 8 ψ-Hilfer fractional pantograph-type differential equations Otherwise, for each ξ ∈ (0, 1] and y, y ∈ C 3 ;ψ (T, ℝ), we have 7

󵄨󵄨 󵄨 󵄨󵄨f (ξ, y(ξ), y(εξ)) − f (ξ, y(ξ), y(εξ))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ⩾ ϱ󵄨󵄨󵄨y(ξ) − y(ξ)󵄨󵄨󵄨 − η󵄨󵄨󵄨y(εξ) − y(εξ)󵄨󵄨󵄨. Then (8.7.1) is satisfied with ϱ =

1 e3 √π

and η =

1 , 33e18

which implies that

′ 1−γ maxξ∈T ψ′ (ξ) 1−γ η 1 maxξ∈T ψ (ξ) 2Γ(γ) α ∗ ∗ 1 ( ) + [ϱ + η ( ) ](ψ(b) − ψ(0)) ϱ ε minξ∈T ψ′ (ξ) Γ(α + γ) ε minξ∈T ψ′ (ξ)

≈ 0.238 < 1. So, by Theorem 8.2, our problem has a unique solution.

8.3 Fractional pantograph-type differential equations depending on the ψ-Riemann–Liouville integral Motivated by the previous works and the recent development in the field of pantographtype equations, this section is devoted to the study of existence and uniqueness results of solutions for the following nonlinear ψ-Hilfer fractional differential equations of pantograph type depending on the ψ-Riemann–Liouville fractional integral in a weighted space of continuous functions: H

α,β;ψ

D0+

α,ψ

y(t) = f (t, y(θt), J0+ y(εt)),

t ∈ (0, b],

(8.10)

associated with periodic-type fractional integral boundary conditions 1−γ,ψ

J0+ α,β;ψ

where H D0+

1−γ,ψ [0, 1], J0+

1−γ,ψ

y(0) = J0+

y(b),

(8.11)

denotes the ψ-Hilfer fractional derivative of order 0 < α ≤ 1 and type β ∈

is the ψ-Riemann–Liouville fractional integral of order 1−γ (γ = α +β −αβ), and θ, ε ∈ (0, 1]. Moreover, f : T × ℝ2 → ℝ is a given continuous function.

8.3.1 Existence results Let α;ψ

X = {y ∈ C1−γ;ψ (T, ℝ) : y(t) = J0+ υ(t) : υ ∈ C1−γ;ψ (T, ℝ), t ∈ (0, b]}

8.3 Fractional pantograph-type differential equations depending…

� 217

and let Y = C1−γ;ψ (T, ℝ) with the norm ‖y‖X = ‖y‖Y = ‖y‖C1−γ;ψ . We give the definition of the operator L : Dom L ⊆ X → Y : α,β;ψ

Ly := H D0+

y,

(8.12)

where α,β;ψ

Dom L = {y ∈ X : H D0+

1−γ;ψ

y ∈ Y : J0+

1−γ;ψ

y(0) = J0+

y(b)}.

Before stating and proving our findings, let us prove some lemmas that will be used later. Lemma 8.6. For any ε ∈ (0, 1], γ > 1, and t ∈ (0, b] we have γ−1

α,ψ

J0+ ((ψ(εs) − ψ(0)) min

)(t) ⩽

ϒΓ(γ) α+γ−1 (ψ(t) − ψ(0)) , Γ(α + γ)

ψ′ (t)

where ϒ = εγ−1 ( maxt∈T ψ′ (t) )γ−1 . t∈T

Proof. Using Lemma 2.6, we have for any t ∈ (0, b] α,ψ

J0+ ((ψ(εs) − ψ(0))

γ−1

)(t)

t

=

1 α−1 γ−1 ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(εs) − ψ(0)) ds Γ(α) 0

t

γ−1 1 α−1 γ−1 (ψ(εs) − ψ(0)) ds = ∫ ψ′ (s)(ψ(t) − ψ(s)) (ψ(s) − ψ(0)) Γ(α) (ψ(s) − ψ(0))γ−1 0

⩽ ⩽

α,ψ ϒJ0+ ((ψ(s)

− ψ(0))

γ−1

)(t)

ϒΓ(γ) α+γ−1 (ψ(t) − ψ(0)) . Γ(α + γ)

Thus, the proof is completed. Lemma 8.7. For all y, ȳ ∈ C1−γ;ψ (T, ℝ) and t ∈ (0, b], we get Γ(γ) α+γ−1 α,ψ 󵄨󵄨 α,ψ 󵄨󵄨 ̄ (ψ(εt) − ψ(0)) ‖y − y‖̄ C1−γ;ψ . 󵄨󵄨J0+ y(εt) − J0+ y(εt) 󵄨󵄨 ⩽ Γ(α + γ)

218 � 8 ψ-Hilfer fractional pantograph-type differential equations Proof. Using Lemma 2.6 we have for any t ∈ (0, b] α,ψ 󵄨󵄨 󵄨󵄨 α,ψ ̄ 󵄨󵄨J0+ y(εt) − J0+ y(εt) 󵄨󵄨 εt

1 α−1 󵄨 ̄ 󵄨󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(εt) − ψ(s)) 󵄨󵄨󵄨y(s) − y(s) Γ(α) 0

⩽

‖y − y‖̄ C1−γ ;ψ Γ(α)

εt

α−1

∫ ψ′ (s)(ψ(εt) − ψ(s)) 0

γ−1

α,ψ

⩽ ‖y − y‖̄ C1−γ ;ψ J0+ ((ψ(s) − ψ(0)) ⩽

γ−1

(ψ(s) − ψ(0))

ds

)(εt)

Γ(γ) α+γ−1 (ψ(εt) − ψ(0)) ‖y − y‖̄ C1−γ ;ψ , Γ(α + γ)

which is the desired result. Now, we define N : X → Y by α,ψ

N y(t) := f (t, y(θt), J0+ y(εt)),

t ∈ (0, b].

The operator N is well defined, because f is a continuous function. We can remark that the problem (8.10)–(8.11) is equivalent to the problem Ly = N y. Lemma 8.8. Suppose that the following hypotheses are satisfied. (8.10.1) The function f : (0, b] × ℝ × ℝ → ℝ is such that α,ψ

f (⋅, y(⋅), J0+ y(⋅)) ∈ C1−γ;ψ (T, ℝ) for all y ∈ C1−γ;ψ (T, ℝ). (8.10.2) There exist nonnegative functions ϱ, η ∈ C(T, ℝ+ ) such that α,ψ α,ψ 󵄨󵄨 󵄨󵄨 ̄ ̄ J0+ y(εt)) 󵄨󵄨 󵄨󵄨f (t, y(θt), J0+ y(εt)) − f (t, y(θt), α,ψ 󵄨 󵄨󵄨 󵄨󵄨 󵄨 α,ψ ̄ ̄ ⩽ ϱ(t)󵄨󵄨󵄨y(θt) − y(θt) 󵄨󵄨 󵄨󵄨 + η(t)󵄨󵄨󵄨J0+ y(εt) − J0+ y(εt)

for every t ∈ (0, b] and y, ȳ ∈ C1−γ;ψ (T, ℝ). Then for any bounded open set Ω ⊂ X , the operator N is L-compact. Proof. We consider for M > 0 the bounded open set Ω = {y ∈ X : ‖y‖X < M }. We split the proof into three steps:

8.3 Fractional pantograph-type differential equations depending…

� 219

Step 1. We show that QN is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Y . Then for each t ∈ (0, b] we have 󵄨 󵄨󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

|N (yn )(s) − N (y)(s)| (1 + β(α − 1)) ⩽ ds. ∫ ψ′ (s) 1+β(α−1) (ψ(b) − ψ(0)) (ψ(b) − ψ(s))β(1−α) 0

By (8.10.2) and Lemmas 8.6 and 8.7, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 b

⩽

ϱ∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨yn (θs) − y(θs)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

b

α,ψ

α,ψ

|J + yn (εs) − J0+ y(εs)| η (1 + β(α − 1)) ds + ∫ ψ′ (s) 0 1+β(α−1) (ψ(b) − ψ(0)) (ψ(b) − ψ(s))β(1−α) ∗

0

ϱ Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ ‖yn − y‖Y J0+ (ψ(s) − ψ(0)) (b) (ψ(b) − ψ(0))1+β(α−1) η∗ Γ(γ)Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ + ‖yn − y‖Y J0+ (ψ(εs) − ψ(0)) (b) 1−γ Γ(α + γ)(ψ(b) − ψ(0)) ϱ∗ Γ(2 + β(α − 1)) γ−1 1+β(α−1);ψ ⩽ ‖yn − y‖Y J0+ (ψ(s) − ψ(0)) (b) 1+β(α−1) (ψ(b) − ψ(0)) ∗

⩽

+

η∗ ϒΓ2 (γ)Γ(2 + β(α − 1)) α+γ−1 (ψ(b) − ψ(0)) ‖yn − y‖Y , Γ(α + γ)Γ(α + 1)

where ϱ∗ = ‖ϱ‖∞ and η∗ = ‖η‖∞ . Using Lemma 2.6, we get 󵄨󵄨 󵄨 󵄨󵄨QN (yn )(t) − QN (y)(t)󵄨󵄨󵄨 Γ(γ)Γ(2 + β(α − 1)) ‖yn − y‖Y ⩽ Γ(α + 1) η∗ ϒΓ(γ) γ−1 α+γ−1 × [ϱ∗ (ψ(b) − ψ(0)) + (ψ(b) − ψ(0)) ]. Γ(α + γ) Thus, for each t ∈ T, we obtain 1−γ 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(0)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 Γ(γ)Γ(2 + β(α − 1)) ∗ η∗ ϒΓ(γ) α ⩽ [ϱ + (ψ(b) − ψ(0)) ]‖yn − y‖Y . Γ(α + 1) Γ(α + γ)

220 � 8 ψ-Hilfer fractional pantograph-type differential equations Then for all t ∈ T we get 1−γ 󵄨 󵄨󵄨 󵄨󵄨(ψ(t) − ψ(0)) (QN (yn )(t) − QN (y)(t))󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞.

Therefore, 󵄩 󵄩󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We demonstrate that QN (Ω) is bounded. For t ∈ (0, b] and y ∈ Ω, we have 󵄨󵄨 󵄨 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 b

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨N (y)(s)󵄨󵄨󵄨ds ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(0)) (1 + β(α − 1)) ⩽ (ψ(b) − ψ(0))1+β(α−1)

0

b

β(α−1) 󵄨

󵄨 󵄨󵄨f (s, y(θs), J α,ψ y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 0+

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

b

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds + ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(0)) 0

b

⩽

(1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 1+β(α−1) (ψ(b) − ψ(0)) 0

b

+

ϱ∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨y(θs)󵄨󵄨󵄨ds ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 󵄨 (ψ(b) − ψ(0))1+β(α−1) 0

b

+

η∗ (1 + β(α − 1)) β(α−1) 󵄨 󵄨󵄨 󵄨󵄨J α,ψ ∫ ψ′ (s)(ψ(b) − ψ(s)) 󵄨 0+ y(εs)󵄨󵄨ds. (ψ(b) − ψ(0))1+β(α−1) 0

Using Lemma 2.6 and Lemma 8.7, we get ∗ ∗ γ−1 󵄨󵄨 󵄨 (ϱ M + f )Γ(γ)Γ(2 + β(α − 1)) (ψ(b) − ψ(0)) 󵄨󵄨QN (y)(t)󵄨󵄨󵄨 ⩽ Γ(α + 1)

η∗ ϒΓ2 (γ)Γ(2 + β(α − 1))M α+γ−1 (ψ(b) − ψ(0)) Γ(α + 1)Γ(α + γ) Γ(γ)Γ(2 + β(α − 1)) γ−1 ⩽ (ψ(b) − ψ(0)) Γ(α + 1) +

8.3 Fractional pantograph-type differential equations depending…

× [f ∗ + ϱ∗ M +

� 221

η∗ ϒΓ(γ)M α (ψ(b) − ψ(0)) ], Γ(α + γ)

where f ∗ = ‖f (⋅, 0, 0)‖C1−γ;ψ . Thus, Γ(γ)Γ(2 + β(α − 1)) 󵄩󵄩 󵄩 󵄩󵄩QN (y)󵄩󵄩󵄩Y ⩽ Γ(α + 1) η∗ ϒΓ(γ)M α ∗ (ψ(b) − ψ(0)) ]. × [f + ϱ∗ M + Γ(α + γ) So, QN (Ω) is a bounded set in Y . Step 3. We prove that L−1 P (id − Q )N : Ω → X is completely continuous. We will use the Arzelà–Ascoli theorem, so we have to show that L−1 P (id − Q )N (Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ (0, b], we get L−1 P (N y(t) − QN y(t)) α;ψ

α,ψ

= J0+ [f (t, y(θt), J0+ y(εt)) −

Γ(2 + β(α − 1)) 1+β(α−1);ψ α,ψ J+ f (s, y(θs), J0+ y(εs))(b)] (ψ(b) − ψ(0))1+β(α−1) 0 t

1 α−1 α,ψ = ∫ ψ′ (s)(ψ(t) − ψ(s)) f (s, y(θs), J0+ y(εs))ds Γ(α) 0

Γ(2 + β(α − 1))(ψ(t) − ψ(0))α 1+β(α−1);ψ α,ψ − J+ f (s, y(θs), J0+ y(εs))(b). Γ(α + 1)(ψ(b) − ψ(0))1+β(α−1) 0 For all y ∈ Ω, using Lemma 8.7, we can get for each t ∈ (0, b] 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 t

⩽

1 α−1 󵄨 α,ψ 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, y(θs), J0+ y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

t

1 α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨ds Γ(α) 0

1 γ−1 + (1 + β(α − 1))(ψ(b) − ψ(0)) Γ(α + 1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

β(α−1) 󵄨

󵄨 󵄨󵄨f (s, y(θs), J α,ψ y(εs)) − f (s, 0, 0)󵄨󵄨󵄨ds 󵄨 0+

222 � 8 ψ-Hilfer fractional pantograph-type differential equations b

(1 + β(α − 1))(ψ(b) − ψ(0))γ−1 ψ′ (s)|f (s, 0, 0)| ds + ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) 0

f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(0)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0))1−α−γ t

+

ϱ∗ α−1 󵄨 󵄨 ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(θs)󵄨󵄨󵄨ds Γ(α) 0

t

η∗ α−1 󵄨 α,ψ 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨J0+ y(εs)󵄨󵄨󵄨ds Γ(α) 0

b

+

ϱ∗ (1 + β(α − 1))(ψ(b) − ψ(0))γ−1 ψ′ (s)|y(θs)| ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) 0

b

+

α,ψ

|J0+ y(εs)| η∗ (1 + β(α − 1))(ψ(b) − ψ(0))γ−1 ds ∫ ψ′ (s) Γ(α + 1) (ψ(b) − ψ(s))β(1−α) 0

f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(0)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0))1−α−γ t

ϱ∗ α−1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 󵄨󵄨󵄨y(θs)󵄨󵄨󵄨ds Γ(α) 0

η∗ Γ(γ)M (ψ(t) − ψ(0))α α,ψ γ−1 + J0+ (ψ(εs) − ψ(0)) (t) Γ(α + γ) b

+

ϱ∗ (1 + β(α − 1))(ψ(b) − ψ(0))γ−1 ψ′ (s)|y(θs)| ds ∫ Γ(α + 1) (ψ(b) − ψ(s))β(1−α) η M Γ(γ)Γ(2 + β(α − ∗

+

0 1+β(α−1),ψ 1))J0+

(ψ(εs) − ψ(0))γ−1 (b)

Γ(α + γ)Γ(α + 1)(ψ(b) − ψ(0))1−α−γ

.

Using Lemma 2.6 and Lemma 8.6, we get 󵄨󵄨 −1 󵄨 󵄨󵄨LP (id − Q )N y(t)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α+γ−1 ⩽ (ψ(t) − ψ(0)) + 2 Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0))1−α−γ ∗ ϱ∗ M Γ(γ)Γ(2 + β(α − 1)) ϱ M Γ(γ) α+γ−1 + (ψ(t) − ψ(0)) + 2 Γ(γ + α) Γ (α + 1)(ψ(b) − ψ(0))1−α−γ + +

η∗ ϒM Γ2 (γ) 2α+γ−1 (ψ(t) − ψ(0)) Γ2 (α + γ)

η∗ ϒM Γ2 (γ)Γ(2 + β(α − 1)) 2α+γ−1 (ψ(b) − ψ(0)) . 2 Γ (α + 1)Γ(α + γ)

8.3 Fractional pantograph-type differential equations depending…

� 223

So 1−γ −1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(0)) LP (id − Q )N y(t)󵄨󵄨󵄨 f ∗ Γ(γ)Γ(2 + β(α − 1)) f ∗ Γ(γ) α α ⩽ (ψ(t) − ψ(0)) + (ψ(b) − ψ(0)) Γ(α + γ) Γ2 (α + 1) ϱ∗ M Γ(γ)Γ(2 + β(α − 1)) ϱ∗ M Γ(γ) α α + (ψ(t) − ψ(0)) + (ψ(b) − ψ(0)) Γ(α + γ) Γ2 (α + 1)

+ +

η∗ ϒM Γ2 (γ) 2α (ψ(t) − ψ(0)) Γ2 (α + γ)

η∗ ϒM Γ2 (γ)Γ(2 + β(α − 1)) 2α (ψ(b) − ψ(0)) . Γ2 (α + 1)Γ(α + γ)

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q )N y󵄩󵄩󵄩X Γ(γ)Γ(2 + β(α − 1)) Γ(γ) α )(f ∗ + ϱ∗ M )(ψ(b) − ψ(0)) ⩽( + 2 Γ(α + γ) Γ (α + 1) +(

Γ(2 + β(α − 1)) η∗ ϒM Γ2 (γ) 1 2α + ) (ψ(b) − ψ(0)) . 2 Γ(α + γ) Γ(α + γ) Γ (α + 1)

This means that L−1 P (id − Q )N (Ω) is uniformly bounded in X . It remains to show that L−1 P (id − Q )N (Ω) is equicontinuous. Using Lemma 2.6, Lemma 8.6, and Lemma 8.7, we have for 0 < t1 < t2 ⩽ b and y ∈ Ω 󵄨󵄨 L−1 (id − Q )N y(t ) L−1 (id − Q )N y(t ) 󵄨󵄨 󵄨󵄨 P 1 󵄨󵄨 2 − P 󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(0))γ−1 (ψ(t1 ) − ψ(0))γ−1 󵄨󵄨󵄨 t

⩽

1 󵄨󵄨 (ψ(t ) − ψ(0))1−γ (ψ(t ) − ψ(0))1−γ 󵄨󵄨 1 󵄨 󵄨󵄨 2 1 − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−α (ψ(t1 ) − ψ(s))1−α 󵄨󵄨󵄨 Γ(α)

0

α,ψ 󵄨 󵄨 × 󵄨󵄨󵄨f (s, y(θs), J0+ y(εs))󵄨󵄨󵄨ds t2

α,ψ

′ (ψ(t2 ) − ψ(0))1−γ ψ (s)|f (s, y(θs), J0+ y(εs))| + ds ∫ Γ(α) (ψ(t2 ) − ψ(s))1−α t1

+

(1 + β(α − 1))[(ψ(t2 ) − ψ(0))1+α−γ − (ψ(t1 ) − ψ(0))1+α−γ ] Γ(α + 1)(ψ(b) − ψ(0))1+β(α−1) b

β(α−1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) 0

󵄨 󵄨󵄨f (s, y(θs), J α,ψ y(εs))󵄨󵄨󵄨ds 󵄨 0+ t1

(f ∗ + ϱ∗ M )(ψ(t1 ) − ψ(0))1−γ (ψ(s) − ψ(0))γ−1 ⩽ ds ∫ ψ′ (s) Γ(α) (ψ(t1 ) − ψ(s))1−α 0

224 � 8 ψ-Hilfer fractional pantograph-type differential equations t1

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(s) − ψ(0))γ−1 − ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α 0

t1

+

η∗ M Γ(γ) 1−γ α−1 ∫ ψ′ (s)[(ψ(t1 ) − ψ(0)) (ψ(t1 ) − ψ(s)) Γ(α)Γ(α + γ) 0 1−γ

− (ψ(t2 ) − ψ(0))

α−1

(ψ(t2 ) − ψ(s))

α+γ−1

](ψ(εs) − ψ(0))

ds

t2

+

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(s) − ψ(0))γ−1 ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α

+

η M Γ(γ)(ψ(t2 ) − ψ(0)) Γ(α)Γ(α + γ)

1−γ

∗

t2

α−1

× ∫ ψ′ (s)(ψ(t2 ) − ψ(s))

t1

α+γ−1

(ψ(εs) − ψ(0))

ds

t1

+

Γ(γ)Γ(2 + β(α − 1)) η∗ ϒM Γ(γ) α ∗ ∗ [f + ϱ M + (ψ(b) − ψ(0)) ] Γ(α + γ) Γ2 (α + 1)(ψ(b) − ψ(0))1−γ 1+α−γ

× [(ψ(t2 ) − ψ(0)) ⩽

1+α−γ

− (ψ(t1 ) − ψ(0))

(f + ϱ M )Γ(γ) α (ψ(t1 ) − ψ(0)) Γ(α + γ) ∗

−

]

∗

t2

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(s) − ψ(0))γ−1 ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α 0

t2

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(s) − ψ(0))γ−1 + ds ∫ ψ′ (s) Γ(α) (ψ(t2 ) − ψ(s))1−α t1

t1

+

(ψ(t1 ) − ψ(s))α−1 η∗ M Γ(γ) ψ′ (s) [ ∫ 1−α−γ Γ(α)Γ(α + γ) (ψ(s) − ψ(0)) (ψ(t1 ) − ψ(0))γ−1 0

1−γ

− (ψ(t2 ) − ψ(0))

α−1

(ψ(t2 ) − ψ(s))

]

t2

(ψ(εs) − ψ(0))γ−1 ds (ψ(s) − ψ(0))γ−1

(f ∗ + ϱ∗ M )(ψ(t1 ) − ψ(0))γ−1 α−1 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) ds Γ(α)(ψ(t2 ) − ψ(0))γ−1 t1

t2

+

η∗ M Γ(γ)(ψ(εt1 ) − ψ(0))γ−1 α−1 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) ds Γ(α)Γ(α + γ)(ψ(t2 ) − ψ(0))γ−α−1 t1

Γ(γ)Γ(2 + β(α − 1)) η∗ ϒM Γ(γ) α ∗ ∗ + 2 [f + ϱ M + (ψ(b) − ψ(0)) ] 1−γ Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0)) 1+α−γ

× [(ψ(t2 ) − ψ(0))

1+α−γ

− (ψ(t1 ) − ψ(0))

]

8.3 Fractional pantograph-type differential equations depending…

⩽

� 225

(f ∗ + ϱ∗ M )Γ(γ) (f ∗ + ϱ∗ M )Γ(γ) α α (ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) Γ(α + γ) Γ(α + γ) t2

+

(f ∗ + ϱ∗ M )(ψ(t1 ) − ψ(0))γ−1 α−1 ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) ds Γ(α)(ψ(t2 ) − ψ(0))γ−1

+

η ϒM Γ(γ)(ψ(t1 ) − ψ(0)) Γ(α + γ)

−

η∗ ϒM Γ(γ)(ψ(t2 ) − ψ(0))1−γ (ψ(s) − ψ(0))α+γ−1 ds ∫ ψ′ (s) Γ(α)Γ(α + γ) (ψ(t2 ) − ψ(s))1−α

1−γ

∗

t1

α+γ−1

α,ψ

J0+ (ψ(s) − ψ(0))

(t1 )

t1

0

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(t1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α + Γ(α + 1) + +

η∗ M Γ(γ)(ψ(t2 ) − ψ(0))α+1−γ (ψ(εt1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)Γ(α + γ)

Γ(γ)Γ(2 + β(α − 1)) η∗ ϒM Γ(γ) α ∗ ∗ [f + ϱ M + (ψ(b) − ψ(0)) ] 2 1−γ Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0)) 1+α−γ

× [(ψ(t2 ) − ψ(0)) ⩽

1+α−γ

− (ψ(t1 ) − ψ(0))

]

(f + ϱ M )Γ(γ) (f + ϱ M )Γ(γ) α α (ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) Γ(α + γ) Γ(α + γ) ∗

+ + − +

∗

∗

∗

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(t1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1) η∗ ϒM Γ(γ)(ψ(t1 ) − ψ(0))1−γ α,ψ α+γ−1 J0+ (ψ(s) − ψ(0)) (t1 ) Γ(α + γ)

η∗ ϒM Γ(γ)(ψ(t2 ) − ψ(0))1−γ α,ψ α+γ−1 J0+ (ψ(s) − ψ(0)) (t2 ) Γ(α + γ) η∗ ϒM Γ(γ)(ψ(t2 ) − ψ(0))1−γ Γ(α)Γ(α + γ) t2

α−1

× ∫ ψ′ (s)(ψ(t2 ) − ψ(s))

α+γ−1

(ψ(s) − ψ(0))

ds

t1

+ + +

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(t1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)

η∗ M Γ(γ)(ψ(t2 ) − ψ(0))α+1−γ (ψ(εt1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)Γ(α + γ)

Γ(γ)Γ(2 + β(α − 1)) η∗ ϒM Γ(γ) α [f ∗ + ϱ∗ M + (ψ(b) − ψ(0)) ] 1−γ Γ(α + γ) + 1)(ψ(b) − ψ(0))

Γ2 (α

1+α−γ

× [(ψ(t2 ) − ψ(0)) ⩽

1+α−γ

− (ψ(t1 ) − ψ(0))

]

(f + ϱ M )Γ(γ) (f + ϱ M )Γ(γ) α α (ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) Γ(α + γ) Γ(α + γ) ∗

∗

∗

∗

226 � 8 ψ-Hilfer fractional pantograph-type differential equations (f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(t1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1) η∗ ϒM Γ(γ) 2α 2α + [(ψ(t1 ) − ψ(0)) − (ψ(t2 ) − ψ(0)) ] Γ(2α + γ) +

+ + + +

η∗ ϒM Γ(γ)(ψ(t2 ) − ψ(0))α+1−γ (ψ(t1 ) − ψ(0))γ−1 α (ψ(t2 ) − ψ(t1 )) Γ(α + 1)Γ(α + γ)

(f ∗ + ϱ∗ M )(ψ(t2 ) − ψ(0))1−γ (ψ(t1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)

η∗ M Γ(γ)(ψ(t2 ) − ψ(0))α+1−γ (ψ(εt1 ) − ψ(0))γ−1 (ψ(t2 ) − ψ(t1 ))α Γ(α + 1)Γ(α + γ)

Γ(γ)Γ(2 + β(α − 1)) η∗ ϒM Γ(γ) α ∗ ∗ [f + ϱ M + (ψ(b) − ψ(0)) ] 2 1−γ Γ(α + γ) Γ (α + 1)(ψ(b) − ψ(0)) 1+α−γ

× [(ψ(t2 ) − ψ(0))

1+α−γ

− (ψ(t1 ) − ψ(0))

].

The operator L−1 P (id − Q )N (Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y. The Arzelà–Ascoli theorem implies that L−1 P (id − Q )N (Ω) is relatively compact in X . As a consequence of steps 1 to 3, N is L-compact in Ω, which completes the demonstration. Lemma 8.9. Assume that (8.10.1) and (8.10.2) hold. If the condition ϱ∗ Γ(γ) η∗ ϒΓ2 (γ) 1 α 2α (ψ(b) − ψ(0)) + 2 (ψ(b) − ψ(0)) < Γ(α + γ) 2 Γ (α + γ) is satisfied, then there exists A > 0 which is independent of ζ such that L(y) − N (y) = −ζ [L(y) + N (−y)] 󳨐⇒ ‖y‖X ⩽ A ,

ζ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N (y) = −ζ L(y) − ζ N (−y). Then L(y) =

1 ζ N (y) − N (−y). 1+ζ 1+ζ

So, from the expression of L and N , we get for any t ∈ (0, b] α,β;ψ

Ly(t) = H D0+

y(t) =

1 α,ψ f (t, y(θt), J0+ y(εt)) 1+ζ ζ α,ψ − f (t, −y(θt), J0+ (−y)(εt)). 1+ζ

(8.13)

8.3 Fractional pantograph-type differential equations depending…

� 227

By Theorem 2.20 we get y(t) =

c1 (ψ(t) − ψ(0))γ−1 1 α;ψ α,ψ + [J + (f (s, y(θs), J0+ y(εs)))(t) Γ(γ) ζ +1 0 α;ψ

α,ψ

− ζ J0+ (f (s, −y(θs), J0+ (−y)(εs)))(t)], 1−γ;ψ

where c1 = J0+ t ∈ (0, b]

|y(t)| ⩽

y(0). Using Lemma 2.6, Lemma 8.6, and Lemma 8.7, we have for each |c1 |(ψ(t) − ψ(0))γ−1 2f ∗ Γ(γ) α+γ−1 + (ψ(t) − ψ(0)) Γ(γ) (ζ + 1)Γ(α + γ) + +

⩽

2‖y‖X ϱ∗ Γ(γ) α+γ−1 [ (ψ(t) − ψ(0)) (ζ + 1) Γ(α + γ)

η∗ Γ(γ) α+γ−1 α,ψ J0+ (ψ(εs) − ψ(0)) (t)] Γ(α + γ)

|c1 |(ψ(t) − ψ(0))γ−1 2f ∗ Γ(γ) α+γ−1 + (ψ(t) − ψ(0)) + 2‖y‖X Γ(γ) Γ(α + γ) ×[

ϱ∗ Γ(γ) η∗ ϒΓ2 (γ) α 2α (ψ(b) − ψ(0)) + 2 (ψ(b) − ψ(0)) ] Γ(α + γ) Γ (α + γ) γ−1

× (ψ(t) − ψ(0))

.

Thus, ‖y‖X ⩽

|c1 | 2f ∗ Γ(γ) α + (ψ(b) − ψ(0)) Γ(γ) Γ(α + γ) + 2[

ϱ∗ Γ(γ) η∗ ϒΓ2 (γ) α 2α (ψ(b) − ψ(0)) + 2 (ψ(b) − ψ(0)) ]‖y‖X . Γ(α + γ) Γ (α + γ)

We deduce that ‖y‖X ⩽

|c1 | Γ(γ)

+

2f ∗ Γ(γ) (ψ(b) Γ(α+γ)

ϱ Γ(γ) 1 − 2[ Γ(α+γ) (ψ(b) − ψ(0))α + ∗

− ψ(0))α

η∗ ϒΓ2 (γ) (ψ(b) Γ2 (α+γ)

− ψ(0))2α ]

:= A . The demonstration is completed. Lemma 8.10. If conditions (8.10.1), (8.10.2), and (8.13) are satisfied, then there exists a bounded open set Ω ⊂ X with L(y) − N (y) ≠ −ζ [L(y) + N (−y)] for any y ∈ 𝜕Ω and any ζ ∈ (0, 1].

228 � 8 ψ-Hilfer fractional pantograph-type differential equations Proof. Using Lemma 8.9, there exists a positive constant A which is independent of ζ such that if y satisfies L(y) − N (y) = −ζ [L(y) + N (−y)],

ζ ∈ (0, 1],

then ‖y‖X ⩽ A . So, if Ω = {y ∈ X ; ‖y‖X < ϑ}

(8.14)

such that ϑ > A , we deduce that L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for all y ∈ 𝜕Ω = {y ∈ X ; ‖y‖X = ϑ} and ζ ∈ (0, 1]. Theorem 8.3. Assume that (8.10.1), (8.10.2), and (8.13) hold. Then there exists at least one solution for the problem (8.10)–(8.11). Proof. It is clear that the set Ω defined in (8.14) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 8.10, assume that (8.10.1), (8.10.2), and (8.13) hold. Then L(y) − N (y) ≠ −ζ [L(y) − N (−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ζ ∈ (0, 1]. Then problem (8.10)–(8.11) has at least one solution on Dom L ∩ Ω, which completes the demonstration. Now, we investigate the existence and uniqueness result for our problem (8.10)–(8.11). Theorem 8.4. Let (8.10.1) and (8.10.2) be satisfied. Moreover, we assume that the following hypothesis holds. (8.14.1) There exist constants ϱ > 0 and η ⩾ 0 such that α,ψ α,ψ 󵄨󵄨 󵄨 󵄨󵄨f (t, y(θt), J0+ y(εt)) − f (t, y(θt), J0+ y(εt))󵄨󵄨󵄨 α,ψ 󵄨 󵄨󵄨 󵄨 α,ψ 󵄨 ̄ ⩾ ϱ󵄨󵄨󵄨y(θt) − y(θt) 󵄨󵄨 − η󵄨󵄨󵄨J0+ y(εt) − J0+ y(εt)󵄨󵄨󵄨

for every t ∈ (0, b] and y, ȳ ∈ C1−γ;ψ (T, ℝ). If one has [

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(0))α α (ψ(b) − ψ(0)) + ϱ Γ(α + γ) Γ(α + γ) +

2η∗ ϒΓ2 (γ) 2α (ψ(b) − ψ(0)) ] < 1, 2 Γ (α + γ)

then the problem (8.10)–(8.11) has a unique solution in Dom L ∩ Ω.

(8.15)

8.3 Fractional pantograph-type differential equations depending…

� 229

Proof. We can see that the condition (8.15) is stronger than condition (8.13). By Theorem 8.3, the problem (8.10)–(8.11) has at least one solution in Dom L ∩ Ω. Now, we prove the uniqueness result. Suppose that the problem (8.10)–(8.11) has two different solutions y1 , y2 ∈ Dom L ∩ Ω. Then we have for each t ∈ (0, b] H H

α,β;ψ

D0+

α,β;ψ

D0+

α,ψ

y1 (t) = f (t, y1 (θt), J0+ y1 (εt)), α,ψ

y2 (t) = f (t, y2 (θt), J0+ y2 (εt)),

and 1−γ,ψ

J0+

1−γ,ψ

y1 (0) = J0+

y1 (b),

1−γ,ψ

J0+

1−γ,ψ

y2 (0) = J0+

y2 (b).

Let U(t) = y1 (t) − y2 (t) for all t ∈ (0, b]. Then α,β;ψ

LU(t) = H D0+

α,β;ψ

= H D0+ =

U(t) α,β;ψ

y1 (t) − H D0+

y2 (t)

α,ψ f (t, y1 (θt), J0+ y1 (εt))

α,ψ

− f (t, y2 (θt), J0+ y2 (εt)).

(8.16)

Using the fact that Img L = ker Q , we have b

∫ ψ′ (s)(ψ(b) − ψ(s)) 0

β(α−1)

α,ψ

α,ψ

[f (s, y1 (θs), J0+ y1 (εs)) − f (s, y2 (θs), J0+ y2 (εs))]ds = 0. Since f ∈ C1−γ;ψ (T, ℝ), there exists t0 ∈ (0, b] such that α,ψ

α,ψ

f (t0 , y1 (θt0 ), J0+ y1 (εt0 )) − f (t0 , y2 (θt0 ), J0+ y2 (εt0 )) = 0. In view of (8.14.1), we have α+γ−1 󵄨󵄨 󵄨 η Γ(γ)(ψ(εt0 ) − ψ(0)) ‖y1 − y2 ‖X . 󵄨󵄨y1 (θt0 ) − y2 (θt0 )󵄨󵄨󵄨 ⩽ ϱ Γ(α + γ)

Then α+γ−1 󵄨󵄨 󵄨 η Γ(γ)(ψ(εt0 ) − ψ(0)) ‖U‖X . 󵄨󵄨U(θt0 )󵄨󵄨󵄨 ⩽ ϱ Γ(α + γ)

On the other hand, by Theorem 2.20, we have α;ψ H

J0+

α,β;ψ

D0+

y(t) = y(t) −

c1 (ψ(t) − ψ(0))γ−1 , Γ(γ)

(8.17)

230 � 8 ψ-Hilfer fractional pantograph-type differential equations which implies that α;ψ H

c1 = [y(θt0 ) − J0+

α,β;ψ

D0+

1−γ

y(θt0 )]Γ(γ)(ψ(θt0 ) − ψ(0))

,

and therefore α;ψ H

U(t) = J0+

α,β;ψ

D0+

U(t) α;ψ H

+ [U(θt0 ) − J0+

α,β;ψ

D0+

1−γ

U(θt0 )](ψ(θt0 ) − ψ(0))

γ−1

(ψ(t) − ψ(0))

.

Using (8.17), we obtain for every t ∈ (0, b] 1−γ γ−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 α;ψ H α,β;ψ 󵄨 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [󵄨󵄨󵄨U(θt0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨J0+ D0+ U(θt0 )󵄨󵄨󵄨](ψ(θt0 ) − ψ(0)) (ψ(t) − ψ(0)) α,β;ψ 󵄨 α;ψ 󵄨 + 󵄨󵄨󵄨J0+ H D0+ U(t)󵄨󵄨󵄨 η Γ(γ)(ψ(εt0 ) − ψ(0))α γ−1 × ‖U‖X (ψ(t) − ψ(0)) ⩽ ϱ Γ(α + γ) Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 α γ−1 + 󵄩 D + U󵄩󵄩X (ψ(εt0 ) − ψ(0)) (ψ(t) − ψ(0)) Γ(α + γ) 󵄩 0 Γ(γ) 󵄩󵄩H α,β;ψ 󵄩󵄩 γ+α−1 + . (8.18) 󵄩 D + U󵄩󵄩X (ψ(t) − ψ(0)) Γ(α + γ) 󵄩 0

By (8.10.2), (8.16), and Lemma 8.7, we find α,ψ α,ψ 󵄨󵄨H α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 D0+ U(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (θt), J0+ y1 (εt)) − f (t, y2 (θt), J0+ y2 (εt))󵄨󵄨󵄨 η∗ Γ(γ) γ−1 α+γ−1 ⩽ [ϱ∗ (ψ(t) − ψ(0)) + (ψ(εt) − ψ(0)) ]‖U‖X . Γ(α + γ)

Then η∗ Γ(γ) α+γ−1 1−γ 󵄩󵄩H α,β;ψ 󵄩󵄩 ∗ (ψ(εt) − ψ(0)) (ψ(t) − ψ(0)) ]‖U‖X . 󵄩󵄩 D0+ U󵄩󵄩X ⩽ [ϱ + Γ(α + γ) Substituting (8.19) in the right side of (8.18), we get for every t ∈ (0, b] η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(0))α α 󵄨󵄨 󵄨 (ψ(b) − ψ(0)) + 󵄨󵄨U(t)󵄨󵄨󵄨 ⩽ [ ϱ Γ(α + γ) Γ(α + γ) +

2η∗ ϒΓ2 (γ) 2α γ−1 (ψ(b) − ψ(0)) ](ψ(t) − ψ(0)) ‖U‖X . 2 Γ (α + γ)

‖U‖X ⩽ [

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(0))α α (ψ(b) − ψ(0)) + ϱ Γ(α + γ) Γ(α + γ)

Therefore,

+

2η∗ ϒΓ2 (γ) 2α (ψ(b) − ψ(0)) ]‖U‖X . Γ2 (α + γ)

(8.19)

8.3 Fractional pantograph-type differential equations depending…

� 231

Hence, by (8.15), we conclude that ‖U‖X = 0. As a result, for any t ∈ (0, b] we get U(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof.

8.3.2 An example Example 8.1. We present an example for a ψ-Hilfer fractional differential equation of pantograph type depending on the ψ-Riemann–Liouville integral operator to test our main results. We have 1 1

1

, ;2t

D12+ 3 y(t) = f (t, y(θt), J02+ y(εt)), 1 ,ψ 3 0+

I

1 ,ψ 3 0+

y(0) = I

,ψ

t ∈ (0, 1],

y(1),

where for any t ∈ (0, 1] we have α,ψ

f (t, y(θt), J0+ y(εt)) =

(2t − 1) 3 et t + 3 y( ) 5t e +2 e √π 3 −1

α,ψ

J0+ y( 2t ) e−7−t ( ). 33e11 1 + J α,ψ y( 2t ) 0+ 2

+

Here T := [0, 1], α = 21 , β = 31 , γ = 32 , θ = 31 , ε = 21 , and ψ(t) = 2t . It is easy to see that f ∈ C 1 ;ψ (T, ℝ). Hence, condition (8.10.1) is satisfied. 3

Furthermore, for all t ∈ (0, 1] and y, y ∈ C 1 ;ψ (T, ℝ), we obtain 3

󵄨󵄨 t t t 󵄨󵄨󵄨 t α,ψ α,ψ 󵄨󵄨 󵄨󵄨f (t, y( ), J0+ y( )) − f (t, y( ), J0+ y( ))󵄨󵄨󵄨 󵄨󵄨 3 2 3 2 󵄨󵄨 󵄨 󵄨󵄨 t 󵄨 󵄨󵄨 α,ψ t t 󵄨󵄨 t 󵄨󵄨󵄨 α,ψ 󵄨 ⩽ ϱ(t)󵄨󵄨󵄨y( ) − y( )󵄨󵄨󵄨 + η(t)󵄨󵄨󵄨J0+ y( ) − J0+ y( )󵄨󵄨󵄨 󵄨󵄨 3 󵄨󵄨 3 󵄨󵄨 2 2 󵄨󵄨 and 󵄨󵄨 t t t t 󵄨󵄨󵄨 α,ψ α,ψ 󵄨󵄨 󵄨󵄨f (t, y( ), J0+ y( )) − f (t, y( ), J0+ y( ))󵄨󵄨󵄨 3 2 3 2 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨󵄨 t 󵄨 󵄨 t 󵄨 t 󵄨󵄨󵄨 α,ψ 󵄨 α,ψ t ⩾ ϱ󵄨󵄨󵄨y( ) − y( )󵄨󵄨󵄨 − η󵄨󵄨󵄨J0+ y( ) − J0+ y( )󵄨󵄨󵄨, 󵄨󵄨 3 󵄨 󵄨 3 󵄨 2 2 󵄨󵄨 󵄨

232 � 8 ψ-Hilfer fractional pantograph-type differential equations which implies that (8.10.2) and (8.14.1) are satisfied with et ϱ(t) = 3 , e √π

By simple calculations, we get ϱ∗ = [

2

e−7−t η(t) = , 33e11

ϱ=

1 , e2 √π

1 , e3 √π

η∗ =

1 , 33e18

and

η=

1 . 33e18

ϒ = ( 21 ) 3 , and −2

η Γ(γ) 2ϱ∗ Γ(γ)(ψ(b) − ψ(0))α α (ψ(b) − ψ(0)) + ϱ Γ(α + γ) Γ(α + γ) +

2η∗ ϒΓ2 (γ) 2α (ψ(b) − ψ(0)) ] ≈ 0.222897 < 1. Γ2 (α + γ)

So, by Theorem 8.4, our problem has a unique solution.

8.4 Notes and remarks The results discussed in this chapter are taken from the articles [88, 121]. We refer the reader to the monographs [35, 47, 49, 59, 60, 125, 199] and the papers [37, 40–42, 71, 74, 89, 91–93, 96, 101, 121, 130, 132, 142] and the references therein for additional results.

9 Nonlinear ψ-Hilfer fractional coupled systems 9.1 Introduction and motivations The main goal of this chapter is to study the existence and uniqueness of periodic solutions for some class of nonlinear fractional coupled systems with ψ-Hilfer derivative. The arguments are based on Mawhin’s coincidence degree theory. Furthermore, an illustration is presented to demonstrate the plausibility of our results. We investigate and demonstrate the results in this chapter taking into account the previously stated papers in the preceding chapters.

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems Motivated by the aforementioned publications and with the objective of generalizing earlier results, in this section, we study the following problem: H

α ,β1 ;ψ

y1 (t) = f1 (t, y1 (t), y2 (t)), t ∈ (a, b], α2 ,β2 ;ψ Da + y2 (t) = f2 (t, y1 (t), y2 (t)), 1−γ ,ψ 1−γ ,ψ 1−γ ,ψ 1−γ ,ψ Ia+ 1 y1 (a) = Ia+ 1 y1 (b) and Ia+ 2 y2 (a) = Ia+ 2 y2 (b), {H

Da+1

α ,β ;ψ

(9.1) (9.2)

1−γ ,ψ

where we denote by H Da+i i and Ia+ i the ψ-Hilfer fractional derivative of order 0 < αi ≤ 1 and type βi ∈ [0, 1] and the ψ-Riemann–Liouville fractional integral of order 1 − γi (γi = αi + βi − αi βi ), i ∈ {1, 2}, respectively. Moreover f1 , f2 : J ̄ × ℝ2 → ℝ are continuous functions.

9.2.1 Existence results We denote by C1−γ;ψ (J,̄ ℝ2 ) = C1−γ1 ;ψ (J,̄ ℝ) × C1−γ2 ;ψ (J,̄ ℝ) the product weighted space with the norm ‖y‖C = max{‖yj ‖C1−γ ;ψ }. 1≤j≤2

j

Also, α ;ψ α ;ψ ℏ1 = {y = (y1 , y2 ) ∈ C1−γ;ψ (J,̄ ℝ2 ) : (y1 , y2 ) = (Ia+1 υ1 , Ia+2 υ2 ),

where υ = (υ1 , υ2 ) ∈ C1−γ;ψ (J,̄ ℝ2 )}

and https://doi.org/10.1515/9783111334387-009

234 � 9 Nonlinear ψ-Hilfer fractional coupled systems ℏ2 = C1−γ;ψ (J,̄ ℝ2 ), with the norms ‖y‖ℏ1 = ‖y‖ℏ2 = ‖y‖C . We define the operator μ : Dom μ ⊆ ℏ1 → ℏ2 by α ,β1 ;ψ

μy = (μ1 y1 , μ2 y2 ) := (H Da+1

α ,β2 ;ψ

y1 , H Da+2

y2 ),

(9.3)

where 1−γ1 ,ψ

Dom μ = {y ∈ ℏ1 : μy ∈ ℏ2 : Ia+ 1−γ ,ψ Ia+ 2 y2 (a)

and

=

1−γ1 ,ψ

y1 (a) = Ia+

1−γ ,ψ Ia+ 2 y2 (b)}.

y1 (b)

Lemma 9.1. Using the definition of μ given in (9.3), we have ker μ = {y = (y1 , y2 ) ∈ ℏ1 : yj (t) =

1−γj ;ψ

Ia+

yj (a)

Γ(γj )

(ψ(t) − ψ(a))

γj −1

,

t ∈ (a, b], j ∈ {1, 2}} and 1+βj (αj −1);ψ

Img μ = {v = (v1 , v2 ) ∈ ℏ2 : Ia+

vj (b) = 0, j ∈ {1, 2}}. α ,βj ;ψ

Proof. By Remark 2.1, for all y ∈ ℏ1 and j ∈ {1, 2} the equation μj yj = H D0+j (a, b] has a solution given by yj (t) =

1−γj ;ψ

Ia+

y(a)

Γ(γj )

(ψ(t) − ψ(a))

γj −1

yj = 0 in

t ∈ (a, b],

,

which implies that ker μ = {y = (y1 , y2 ) ∈ ℏ1 : yj (t) =

1−γj ;ψ

Ia+

yj (a)

Γ(γj )

(ψ(t) − ψ(a))

γj −1

,

t ∈ (a, b], j ∈ {1, 2}}. For v = (v1 , v2 ) ∈ Img μ, there exists y = (y1 , y2 ) ∈ Dom μ such that (v1 , v2 ) = (μ1 y1 , μ2 y2 ) ∈ ℏ2 . Using Theorem 2.4, we obtain for each t ∈ (a, b] and j ∈ {1, 2} yj (t) =

1−γj ;ψ

Ia+

yj (a)

Γ(γj )

(ψ(t) − ψ(a))

γj −1

α ;ψ

+ Ia+j vj (t).

� 235

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

Using Lemma 2.6 we obtain that for each j ∈ {1, 2} 1−γj ;ψ

Ia+

1−γj ;ψ

yj (t) = Ia+

1+βj (αj −1);ψ

yj (a) + Ia+ 1−γj ;ψ

Since y ∈ Dom μ, for any j ∈ {1, 2} we have Ia+ 1+βj (αj −1);ψ

Ia+

vj (b) = 0,

vj (t).

1−γj ;ψ

yj (a) = Ia+

yj (b). Thus,

j ∈ {1, 2}.

Moreover, if v = (υ1 , υ2 ) ∈ ℏ2 and satisfies 1+βj (αj −1);ψ

Ia+

vj (b) = 0,

j ∈ {1, 2},

α ;ψ

α ;ψ

then for any y(t) = (y1 (t), y2 (t)) = (Ia+1 v1 (t), Ia+2 v2 (t)), we get (v1 (t), v2 (t)) = α ,β ;ψ

α ,β2 ;ψ

(H Da+1 1 y1 (t), H Da+2 Therefore,

y2 (t)) = (μ1 y1 (t), μ2 y2 (t)). 1−γj ;ψ

Ia +

1−γj ;ψ

yj (b) = Ia+

yj (a),

j ∈ {1, 2}.

Then y ∈ Dom μ, so that v ∈ Img μ. So 1+βj (αj −1);ψ

Img μ = {v = (v1 , v2 ) ∈ ℏ2 : Ia+

vj (b) = 0, j ∈ {1, 2}}.

Lemma 9.2. Let μ be defined by (9.3). Then μ is a Fredholm operator of index zero, and the linear continuous projector operators Φ : ℏ2 → ℏ2 and Λ : ℏ1 → ℏ1 can be written as Φ(v) = (Φ1 v1 , Φ2 v2 ), such that Φj vj (t) =

Γ(2 + βj (αj − 1))

1+βj (αj −1)

(ψ(b) − ψ(a))

1+βj (αj −1);ψ

Ia +

vj (b),

t ∈ J,̄ j ∈ {1, 2}.

For all t ∈ J,̄ Λ(y)(t) = (Λ1 (y1 )(t), Λ2 (y2 )(t)) 1−γ1 ;ψ

=(

Ia+

y1 (a)

Γ(γ1 )

(ψ(t) − ψ(a))

γ1 −1

1−γ2 ;ψ

,

Ia+

y2 (a)

Γ(γ2 )

γ2 −1

(ψ(t) − ψ(a))

).

Furthermore, the operator μ−1 Λ : Img μ → ℏ1 ∩ ker Λ can be written as α ;ψ

α ;ψ

−1 −1 1 2 μ−1 Λ (v)(t) = (μΛ 1 (v1 )(t), μΛ 2 (v2 )(t)) = (Ia+ v1 (t), Ia+ v2 (t)),

t ∈ J.̄

236 � 9 Nonlinear ψ-Hilfer fractional coupled systems Proof. Obviously, for each v ∈ ℏ2 , Φ2 v = Φv and v = (v − Φ(v)) + Φ(v), where (v − Φ(v)) ∈ ker Φ = Img μ. Using the fact that Img μ = ker Φ and Φ2 = Φ, we have Img Φ ∩ Img μ = 0. So, ℏ2 = Img μ ⊕ Img Φ. In the same way, we get Img Λ = ker μ and Λ2 = Λ. It follows for each y ∈ ℏ1 that if y = (y − Λ(y)) + Λ(y), then ℏ1 = ker Λ + ker μ. Clearly, we have ker Λ ∩ ker μ = 0. Thus, ℏ1 = ker Λ ⊕ ker μ. Using the rank–nullity theorem, we get codim Img L = dim ℏ2 − dim Img L

= [dim ker Φ + dim Img Φ] − dim Img L,

and since Img L = ker Φ, we have codim Img L = dim Img Φ.

(9.4)

Using again the rank–nullity theorem, we obtain dim ker μ = dim ℏ1 − dim Img L = codim Img L, which implies that dim ker μ = codim Img L.

(9.5)

By (9.4) and (9.5) we have dim ker μ = codim Img L = dim Img Φ, and since dim Img Φ < ∞, we have dim ker μ = codim Img L < ∞. Since Img μ is a closed subset of ℏ2 , μ is a Fredholm operator of index zero. We will now demonstrate that μ|Dom μ∩ker Λ is μ−1 Λ . Effectively, for v ∈ Img μ, by Theorem 2.5 we have α ,β1 ;ψ

H 1 μμ−1 Λ (v) = ( Da+

1−γ1 ;ψ

Ia+

α ,β2 ;ψ

v1 , H Da+2

Furthermore, for y ∈ Dom μ ∩ ker Λ we get

1−γ2 ;ψ

Ia +

v2 ) = (v1 , v2 ) = v.

(9.6)

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

μ−1 Λ (μy)(t)

1−γ1 ;ψ H

= (Ia+

= (y1 (t) −

α ,β1 ;ψ

Da+1

1−γ2 ;ψ H

y1 (t), Ia+

1−γ1 ;ψ

Ia+

y1 (a)

Γ(γ1 )(ψ(t) −

ψ(a))1−γ1

α ,β2 ;ψ

Da+2

, y2 (t) −

y2 (t))

1−γ2 ;ψ

Ia +

y2 (a)

Γ(γ2 )(ψ(t) − ψ(a))1−γ2

� 237

).

Using the fact that y ∈ Dom μ ∩ ker Λ, we have 1−γj ;ψ

Ia +

yj (a)

Γ(γj )

γj −1

(ψ(t) − ψ(a))

= 0,

j ∈ {1, 2}.

Thus, −1 −1 μ−1 Λ μ(y) = (μΛ 1 μ1 (y1 ), μΛ 2 μ2 (y2 )) = (y1 , y2 ) = y.

(9.7)

−1 Using (9.6) and (9.7) together, we get μ−1 Λ = (μ|Dom μ∩ker Λ ) .

Now, we define ℵ : ℏ1 → ℏ2 by ℵy(t) = (ℵ1 y(t), ℵ2 y(t)) := (f1 (t, y1 (t), y2 (t)), f2 (t, y1 (t), y2 (t))),

t ∈ (a, b].

We can remark that the problem (9.1)–(9.2) is equivalent to the problem μy = ℵy. Lemma 9.3. Suppose that the hypotheses (9.3.1) and (9.3.2) are satisfied. (9.3.1) For each j ∈ {1, 2}, the function fj : (a, b] × ℝ2 → ℝ is such that fj (⋅, y1 (⋅), y2 (⋅)) ∈ C1−γj ;ψ (J,̄ ℝ) for all yj ∈ C1−γj ;ψ (J,̄ ℝ). (9.3.2) For each j ∈ {1, 2}, there exist nonnegative functions ϑj , ηj ∈ C(J,̄ ℝ+ ) such that 󵄨󵄨 󵄨 󵄨󵄨fj (t, y1 (t), y2 (t)) − fj (t, y1 (t), y2 (t))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ ϑj (t)󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 + ηj (t)󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 for every t ∈ J ̄ and yj , yj ∈ C1−γj ;ψ (J,̄ ℝ). Then for any bounded open set Ψ ⊂ ℏ1 the operator ℵ is μ-compact. Proof. Let Ψ = {y ∈ ℏ1 : ‖y‖ℏ1 < ℘} be a bounded open set where ℘ > 0. Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence where yn 󳨀→ y in ℏ2 ; thus, for j ∈ {1, 2} and t ∈ (a, b], we have

238 � 9 Nonlinear ψ-Hilfer fractional coupled systems 󵄨 󵄨󵄨 󵄨󵄨Φj ℵj (yn )(t) − Φj ℵj (y)(t)󵄨󵄨󵄨 ⩽

(1 + βj (αj − 1))

1+βj (αj −1)

(ψ(b) − ψ(a))

b

∫ ψ′ (s) a

|ℵj (yn )(s) − ℵj (y)(s)| (ψ(b) − ψ(s))βj (1−αj )

ds.

By (9.3.2), for each t ∈ J,̄ we have 1−γ1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (Φ1 ℵ1 (yn )(t) − Φ1 ℵ1 (y)(t))󵄨󵄨󵄨 b

(ψ(t) − ψ(a))1−γ1 ϑ1 (t)(1 + β1 (α1 − 1)) β (α −1) ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 1 1 1+β (α −1) 1 1 (ψ(b) − ψ(a)) a

󵄨 󵄨 × 󵄨󵄨󵄨yn1 (s) − y1 (s)󵄨󵄨󵄨ds

b

(ψ(t) − ψ(a))1−γ1 η1 (t)(1 + β1 (α1 − 1)) β (α −1) + ∫ ψ′ (s)(ψ(b) − ψ(s)) 1 1 (ψ(b) − ψ(a))1+β1 (α1 −1) a

󵄨 󵄨 × 󵄨󵄨󵄨yn2 (s) − y2 (s)󵄨󵄨󵄨ds ⩽

(ψ(b) − ψ(a))1−γ1 ϑ ∗ Γ(2 + β1 (α1 − 1)) ‖yn − y‖ℏ2 (ψ(b) − ψ(a))1+β1 (α1 −1) 1+β1 (α1 −1);ψ

× Ia + +

γ1 −1

(b)

1−γ1 ∗

(ψ(b) − ψ(a)) η Γ(2 + β1 (α1 − 1)) ‖yn − y‖ℏ2 (ψ(b) − ψ(a))1+β1 (α1 −1) 1+β1 (α1 −1);ψ

× Ia + ⩽[

(ψ(s) − ψ(a))

(ψ(s) − ψ(a))

γ2 −1

(b)

Γ(2 + β1 (α1 − 1))ϑ Γ(γ1 ) Γ(α1 + 1)

+

∗

Γ(2 + β1 (α1 − 1))η∗ Γ(γ2 ) γ −γ (ψ(b) − ψ(a)) 2 1 ]‖yn − y‖ℏ2 Γ(γ2 + β1 (α1 − 1) + 1)

⩽ ϒ1 ‖yn − y‖ℏ2 and

1−γ2 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (Φ2 ℵ2 (yn )(t) − Φ2 ℵ2 (y)(t))󵄨󵄨󵄨 b

(ψ(t) − ψ(a))1−γ2 ϑ2 (t)(1 + β2 (α2 − 1)) β (α −1) ⩽ ∫ ψ′ (s)(ψ(b) − ψ(s)) 2 2 1+β (α −1) 2 2 (ψ(b) − ψ(a)) 󵄨 󵄨 × 󵄨󵄨󵄨yn1 (s) − y1 (s)󵄨󵄨󵄨ds

a

b

+

(ψ(t) − ψ(a))1−γ2 η2 (t)(1 + β2 (α2 − 1)) β (α −1) ∫ ψ′ (s)(ψ(b) − ψ(s)) 2 2 (ψ(b) − ψ(a))1+β2 (α2 −1)

󵄨 󵄨 × 󵄨󵄨󵄨yn2 (s) − y2 (s)󵄨󵄨󵄨ds

a

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

⩽

� 239

(ψ(b) − ψ(a))1−γ2 ϑ ∗ Γ(2 + β2 (α2 − 1)) ‖yn − y‖ℏ2 (ψ(b) − ψ(a))1+β2 (α2 −1) 1+β2 (α2 −1);ψ

× Ia + +

(b)

1−γ2 ∗

(ψ(b) − ψ(a)) η Γ(2 + β2 (α2 − 1)) ‖yn − y‖ℏ2 (ψ(b) − ψ(a))1+β2 (α2 −1) 1+β2 (α2 −1);ψ

× Ia + ⩽[

γ1 −1

(ψ(s) − ψ(a))

γ2 −1

(ψ(s) − ψ(a))

(b)

Γ(2 + β2 (α2 − 1))ϑ Γ(γ1 ) γ −γ (ψ(b) − ψ(a)) 1 2 Γ(γ1 + β2 (α2 − 1) + 1)

+

∗

Γ(2 + β2 (α2 − 1))η∗ Γ(γ2 ) ]‖yn − y‖ℏ2 Γ(α2 + 1)

⩽ ϒ2 ‖yn − y‖ℏ2 ,

where ϑ ∗ = maxj∈{1,2} ‖ϑj ‖∞ , η∗ = maxj∈{1,2} ‖ηj ‖∞ , ϒ1 =

Γ(2 + β1 (α1 − 1))ϑ ∗ Γ(γ1 ) Γ(2 + β1 (α1 − 1))η∗ Γ(γ2 ) γ −γ + (ψ(b) − ψ(a)) 2 1 , Γ(α1 + 1) Γ(γ2 + β1 (α1 − 1) + 1)

ϒ2 =

Γ(2 + β2 (α2 − 1))ϑ ∗ Γ(γ1 ) Γ(2 + β2 (α2 − 1))η∗ Γ(γ2 ) γ −γ (ψ(b) − ψ(a)) 1 2 + . Γ(γ1 + β2 (α2 − 1) + 1) Γ(α2 + 1)

and

Thus, for each j ∈ {1, 2}, we get 󵄩󵄩 󵄩 󵄩󵄩Φj ℵj (yn ) − Φj ℵj (y)󵄩󵄩󵄩C

1−γj ;ψ

⩽ max{ϒ1 , ϒ2 }‖yn − y‖ℏ2 .

Therefore, 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩ℏ2 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN (Ψ) is bounded. Using Lemma 2.6, we have for any t ∈ (a, b] and y ∈ Ψ 1−γ1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) Φ1 ℵ1 (y)(t)󵄨󵄨󵄨 b

|ℵ1 (y)(s)| (ψ(t) − ψ(a))1−γ1 (1 + β1 (α1 − 1)) ds ⩽ ∫ ψ′ (s) 1+β (α −1) 1 1 (ψ(b) − ψ(a)) (ψ(b) − ψ(s))β1 (1−α1 ) (ψ(t) − ψ(a))1−γ1 (1 + β1 (α1 − 1)) ⩽ (ψ(b) − ψ(a))1+β1 (α1 −1)

a

240 � 9 Nonlinear ψ-Hilfer fractional coupled systems b

β1 (α1 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨f1 (s, y1 (s), y2 (s)) − f1 (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨 b

+

(ψ(t) − ψ(a))1−γ1 (1 + β1 (α1 − 1)) ψ′ (s)|f1 (s, 0, 0)| ds ∫ 1+β (α −1) (ψ(b) − ψ(a)) 1 1 (ψ(b) − ψ(s))β1 (1−α1 ) a

b

⩽

(ψ(t) − ψ(a))1−γ1 (1 + β1 (α1 − 1)) ψ′ (s)|f1 (s, 0, 0)| ds ∫ (ψ(b) − ψ(a))1+β1 (α1 −1) (ψ(b) − ψ(s))β1 (1−α1 ) a

+

+

b

1−γ1 ∗

ψ′ (s)|y1 (s)| (ψ(t) − ψ(a)) ϑ (1 + β1 (α1 − 1)) ds ∫ (ψ(b) − ψ(a))1+β1 (α1 −1) (ψ(b) − ψ(s))β1 (1−α1 ) 1−γ1 ∗

(ψ(t) − ψ(a)) η (1 + β1 (α1 − 1)) (ψ(b) − ψ(a))1+β1 (α1 −1) b

a

β1 (α1 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨y2 (s)󵄨󵄨󵄨ds 󵄨 󵄨

[ϑ ∗ ℘ + f ∗ ]Γ(γ1 )Γ(2 + β1 (α1 − 1)) ⩽ Γ(α1 + 1) ∗ η ℘Γ(γ2 )Γ(2 + β1 (α1 − 1)) γ −γ + (ψ(b) − ψ(a)) 2 1 Γ(γ2 + β1 (α1 − 1) + 1) ⩽ ϒ3

and 1−γ2 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) Φ2 ℵ2 (y)(t)󵄨󵄨󵄨 b

(ψ(t) − ψ(a))1−γ2 (1 + β2 (α2 − 1)) ψ′ (s)|ℵ2 (y)(s)| ⩽ ds ∫ (ψ(b) − ψ(a))1+β2 (α2 −1) (ψ(b) − ψ(s))β2 (1−α2 ) 1−γ2

⩽

(ψ(t) − ψ(a)) (1 + β2 (α2 − 1)) (ψ(b) − ψ(a))1+β2 (α2 −1) b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

a

β2 (α2 −1) 󵄨

󵄨󵄨f2 (s, y1 (s), y2 (s)) − f2 (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨 b

ψ′ (s)|f2 (s, 0, 0)| (ψ(t) − ψ(a))1−γ2 (1 + β2 (α2 − 1)) ds + ∫ (ψ(b) − ψ(a))1+β2 (α2 −1) (ψ(b) − ψ(s))β2 (1−α2 ) a

b

⩽

(ψ(t) − ψ(a))1−γ2 (1 + β2 (α2 − 1)) ψ′ (s)|f2 (s, 0, 0)| ds ∫ (ψ(b) − ψ(a))1+β2 (α2 −1) (ψ(b) − ψ(s))β2 (1−α2 ) a

1−γ2 ∗

b

(ψ(t) − ψ(a)) ϑ (1 + β2 (α2 − 1)) ψ′ (s)|y1 (s)| + ds ∫ (ψ(b) − ψ(a))1+β2 (α2 −1) (ψ(b) − ψ(s))β2 (1−α2 ) a

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

+

� 241

(ψ(t) − ψ(a))1−γ2 η∗ (1 + β2 (α2 − 1)) (ψ(b) − ψ(a))1+β2 (α2 −1) b

β2 (α2 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

󵄨󵄨y2 (s)󵄨󵄨󵄨ds 󵄨 󵄨

ϑ ∗ ℘Γ(γ1 )Γ(2 + β2 (α2 − 1)) γ −γ ⩽ (ψ(b) − ψ(a)) 1 2 Γ(γ1 + β2 (α2 − 1) + 1) [η∗ ℘ + f ∗ ]Γ(γ2 )Γ(2 + β2 (α2 − 1)) + Γ(α2 + 1)

⩽ ϒ4 ,

where f ∗ = maxj∈{1,2} ‖fj (., 0, 0)‖C1−γ ;ψ , j

ϒ3 =

[ϑ ∗ ℘ + f ∗ ]Γ(γ1 )Γ(2 + β1 (α1 − 1)) Γ(α1 + 1) ∗ η ℘Γ(γ2 )Γ(2 + β1 (α1 − 1)) γ −γ + (ψ(b) − ψ(a)) 2 1 , Γ(γ2 + β1 (α1 − 1) + 1)

and ϒ4 =

ϑ ∗ ℘Γ(γ1 )Γ(2 + β2 (α2 − 1)) γ −γ (ψ(b) − ψ(a)) 1 2 Γ(γ1 + β2 (α2 − 1) + 1) [η∗ ℘ + f ∗ ]Γ(γ2 )Γ(2 + β2 (α2 − 1)) + , Γ(α2 + 1)

which implies that ‖Φ1 ℵ1 y‖C1−γ ;ψ ≤ ϒ3 1

and ‖Φ2 ℵ2 y‖C1−γ ;ψ ≤ ϒ4 . 2

Thus, ‖QN (y)‖ℏ2 ⩽ max{ϒ3 , ϒ4 }. So, QN (Ψ) is a bounded set in ℏ2 . Step 3. We demonstrate that μ−1 Λ (id − Φ)ℵ : Ψ → ℏ1 is completely continuous. We will demonstrate that μ−1 Λ (id − Φ)ℵ(Ψ) ⊂ ℏ1 is equicontinuous and bounded. Firstly, for any y ∈ Ψ and t ∈ (a, b], we get −1 −1 μ−1 Λ (ℵy(t) − QN y(t)) = (μΛ 1 (ℵ1 y(t) − Φ1 ℵ1 y(t)), μΛ 2 (ℵ2 y(t) − Φ2 ℵ2 y(t))),

242 � 9 Nonlinear ψ-Hilfer fractional coupled systems where for j ∈ {1, 2} we have μ−1 Λ j (ℵj y(t) − Φj ℵj y(t)) α ;ψ

= Ia+j [fj (t, y1 (t), y2 (t)) −

Γ(2 + βj (αj − 1))

(ψ(b) − ψ(a))1+βj (αj −1)

1+βj (αj −1);ψ

Ia +

fj (s, y1 (s), y2 (s))(b)]

t

1 α −1 = ∫ ψ′ (s)(ψ(t) − ψ(s)) j fj (s, y1 (s), y2 (s))ds Γ(αj ) a

−

Γ(2 + βj (αj − 1))(ψ(t) − ψ(a))αj Γ(αj + 1)(ψ(b) − ψ(a))1+βj (αj −1)

1+βj (αj −1);ψ

Ia+

fj (s, y1 (s), y2 (s))(b).

Using Lemma 2.6, for all y ∈ Ψ and t ∈ (a, b] we have 1−γ1 −1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) μΛ 1 (id − Φ1 )ℵ1 y(t)󵄨󵄨󵄨 t

|f (s, y1 (s), y2 (s)) − f1 (s, 0, 0)| (ψ(t) − ψ(a))1−γ1 ds ⩽ ∫ ψ′ (s) 1 Γ(α1 ) (ψ(t) − ψ(s))1−α1 a

+ +

1−γ1

(ψ(t) − ψ(a)) Γ(α1 )

1−γ1

(ψ(t) − ψ(a)) Γ(α1 + 1)

t

α1 −1 󵄨

a

(1 + β1 (α1 − 1))(ψ(b) − ψ(a))

b

󵄨󵄨f1 (s, y1 (s), y2 (s)) − f1 (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

a

(1 + β1 (α1 − 1))(ψ(t) − ψ(a))1−γ1 (ψ(b) − ψ(a))γ1 −1 Γ(α1 + 1) b

β1 (α1 −1) 󵄨

󵄨󵄨f1 (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

⩽

γ1 −1

β1 (α1 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) +

󵄨󵄨f1 (s, 0, 0)󵄨󵄨󵄨ds 󵄨 󵄨

∫ ψ′ (s)(ψ(t) − ψ(s))

(ψ(t) − ψ(a))1−γ1 f ∗ Γ(γ1 ) α +γ −1 (ψ(t) − ψ(a)) 1 1 Γ(α1 + γ1 ) +

(ψ(t) − ψ(a))1−γ1 f ∗ Γ(γ1 )Γ(2 + β1 (α1 − 1)) α +γ −1 (ψ(b) − ψ(a)) 1 1 Γ2 (α1 + 1) t

(ψ(t) − ψ(a))1−γ1 ϑ ∗ α −1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t) − ψ(s)) 1 󵄨󵄨󵄨y1 (s)󵄨󵄨󵄨ds Γ(α1 ) a

1−γ1 ∗

(ψ(t) − ψ(a)) + Γ(α1 )

η

t

α1 −1 󵄨

∫ ψ′ (s)(ψ(t) − ψ(s)) a

󵄨󵄨y2 (s)󵄨󵄨󵄨ds 󵄨 󵄨

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

+

(ψ(t) − ψ(a))1−γ1 ϑ ∗ (1 + β1 (α1 − 1))(ψ(b) − ψ(a))γ1 −1 Γ(α1 + 1) b

β1 (α1 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

+

󵄨󵄨y1 (s)󵄨󵄨󵄨ds 󵄨 󵄨

(ψ(t) − ψ(a))1−γ1 η∗ (1 + β1 (α1 − 1))(ψ(b) − ψ(a))γ1 −1 Γ(α1 + 1) b

β1 (α1 −1) 󵄨

× ∫ ψ′ (s)(ψ(b) − ψ(s)) ⩽

� 243

a ∗

󵄨󵄨y2 (s)󵄨󵄨󵄨ds 󵄨 󵄨

f ∗ Γ(γ1 )Γ(2 + β1 (α1 − 1)) f Γ(γ1 ) α α (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 ) Γ2 (α1 + 1) η∗ ℘Γ(γ2 ) ϑ ∗ ℘Γ(γ1 ) α α +γ −γ (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 2 1 + Γ(γ1 + α1 ) Γ(γ2 + α1 ) ϑ ∗ ℘Γ(γ1 )Γ(2 + β1 (α1 − 1)) α + (ψ(b) − ψ(a)) 1 2 Γ (α1 + 1) η∗ ℘Γ(γ2 )Γ(2 + β1 (α1 − 1)) 1+γ −γ +β (α −1) + (ψ(b) − ψ(a)) 2 1 1 1 . Γ(α1 + 1)Γ(γ2 + β1 (α1 − 1) + 1)

So 󵄩󵄩 −1 󵄩 󵄩󵄩μΛ 1 (id − Φ1 )ℵ1 y󵄩󵄩󵄩C1−γ ;ψ 1 ⩽

f ∗ Γ(γ1 ) f ∗ Γ(γ1 )Γ(2 + β1 (α1 − 1)) α α (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 ) Γ2 (α1 + 1) η∗ ℘Γ(γ2 ) ϑ ∗ ℘Γ(γ1 ) α α +γ −γ (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 2 1 + Γ(γ1 + α1 ) Γ(γ2 + α1 ) ϑ ∗ ℘Γ(γ1 )Γ(2 + β1 (α1 − 1)) α + (ψ(b) − ψ(a)) 1 Γ2 (α1 + 1) η∗ ℘Γ(γ2 )Γ(2 + β1 (α1 − 1)) 1+γ −γ +β (α −1) + (ψ(b) − ψ(a)) 2 1 1 1 Γ(α1 + 1)Γ(γ2 + β1 (α1 − 1) + 1)

= ϒ5 .

Similarly, we get 󵄩󵄩 −1 󵄩 󵄩󵄩μΛ 2 (id − Φ2 )ℵ2 y󵄩󵄩󵄩C1−γ ;ψ 2 ⩽

f ∗ Γ(γ2 )Γ(2 + β2 (α2 − 1)) f ∗ Γ(γ2 ) α α (ψ(b) − ψ(a)) 2 + (ψ(b) − ψ(a)) 2 Γ(γ2 + α2 ) Γ2 (α2 + 1) η∗ ℘Γ(γ2 ) ϑ ∗ ℘Γ(γ1 ) α +γ −γ α + (ψ(b) − ψ(a)) 2 1 2 + (ψ(b) − ψ(a)) 2 Γ(γ1 + α2 ) Γ(γ2 + α2 ) ϑ ∗ ℘Γ(γ1 )Γ(2 + β2 (α2 − 1)) 1+γ −γ +β (α −1) + (ψ(b) − ψ(a)) 1 2 2 2 Γ(α2 + 1)Γ(γ1 + β2 (α2 − 1) + 1)

244 � 9 Nonlinear ψ-Hilfer fractional coupled systems η∗ ℘Γ(γ2 )Γ(2 + β2 (α2 − 1)) α (ψ(b) − ψ(a)) 2 Γ2 (α2 + 1)

+

= ϒ6 . We deduce that

󵄩󵄩 −1 󵄩 󵄩󵄩μΛ (id − Φ)ℵy󵄩󵄩󵄩ℏ1 ≤ max{ϒ5 , ϒ6 }. −1 Then μ−1 Λ (id − Φ)ℵ(Ψ) is uniformly bounded in ℏ1 . We prove now that μΛ (id − Φ)ℵ(Ψ) is equicontinuous. Using Lemma 2.6, we have for a < t1 < t2 ⩽ b, y ∈ Ψ, and j ∈ {1, 2} −1 󵄨󵄨 μ−1 󵄨 󵄨󵄨 Λ j (id − Φj )ℵj y(t2 ) μΛ j (id − Φj )ℵj y(t1 ) 󵄨󵄨󵄨 − 󵄨󵄨 󵄨󵄨 γ −1 γ −1 󵄨󵄨 (ψ(t2 ) − ψ(a)) j (ψ(t1 ) − ψ(a)) j 󵄨󵄨 t1

ψ′ (s) 󵄨󵄨󵄨󵄨 (ψ(t2 ) − ψ(a))1−γj (ψ(t1 ) − ψ(a))1−γj 󵄨󵄨󵄨󵄨 − 󵄨 󵄨 Γ(αj ) 󵄨󵄨󵄨 (ψ(t2 ) − ψ(s))1−αj (ψ(t1 ) − ψ(s))1−αj 󵄨󵄨󵄨 a 󵄨 󵄨 × 󵄨󵄨󵄨fj (s, y1 (s), y2 (s))󵄨󵄨󵄨ds

⩽∫

t2

(ψ(t2 ) − ψ(a))1−γj α −1 󵄨 󵄨 + ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) j 󵄨󵄨󵄨fj (s, y1 (s), y2 (s))󵄨󵄨󵄨ds Γ(αj ) t1

+

(1 + βj (αj − 1))[(ψ(t2 ) − ψ(a))1+αj −γj − (ψ(t1 ) − ψ(a))1+αj −γj ] Γ(αj + 1)(ψ(b) − ψ(a))1+βj (αj −1)

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

βj (αj −1) 󵄨

󵄨󵄨fj (s, y1 (s), y2 (s))󵄨󵄨󵄨ds 󵄨 󵄨

t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γj (ψ(t ) − ψ(a))1−γj 󵄨󵄨 (f ∗ + (ϑ ∗ + η∗ )℘) 󵄨 󵄨󵄨 2 1 ⩽ − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−αj (ψ(t1 ) − ψ(s))1−αj 󵄨󵄨󵄨 Γ(αj )

a

× max {(ψ(s) − ψ(a))

γj −1

j∈{1,2}

+

}ds

(f ∗ + (ϑ ∗ + η∗ )℘)(ψ(t2 ) − ψ(a))1−γj Γ(αj ) t2

× ∫ ψ′ (s)(ψ(t2 ) − ψ(s)) t1

+[

γj −1

max {(ψ(s) − ψ(a))

j∈{1,2}

f ∗ Γ(γj )Γ(2 + βj (αj − 1))

Γ2 (αj + 1)(ψ(b) − ψ(a))1−γj

b

× ∫ ψ′ (s)(ψ(b) − ψ(s)) a

αj −1

βj (αj −1)

+

}ds

(1 + βj (αj − 1))(ϑ ∗ + η∗ )℘

Γ(αj + 1)(ψ(b) − ψ(a))1+βj (αj −1)

max {(ψ(s) − ψ(a))

j∈{1,2}

γj −1

}ds]

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

1+αj −γj

× [(ψ(t2 ) − ψ(a))

1+αj −γj

− (ψ(t1 ) − ψ(a))

� 245

].

The operator μ−1 Λ (id − Φ)ℵ(Ψ) is equicontinuous in ℏ1 because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y for j ∈ {1, 2}. By the Arzelà–Ascoli theorem, μ−1 Λ (id − Φ)ℵ(Ψ) is relatively compact in ℏ1 . Then ℵ is μ-compact in Ψ. Let ϒ7 :=

η∗ Γ(γ2 ) ϑ ∗ Γ(γ1 ) α α +γ −γ (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 2 1 Γ(γ1 + α1 ) Γ(γ2 + α1 )

ϒ8 :=

η∗ Γ(γ2 ) ϑ ∗ Γ(γ1 ) α +γ −γ α (ψ(b) − ψ(a)) 2 1 2 + (ψ(b) − ψ(a)) 2 . Γ(γ1 + α2 ) Γ(γ2 + α2 )

and

Lemma 9.4. Assume that (9.3.1) and (9.3.2) hold. If the condition max{ϒ7 , ϒ8 }
0 which is independent of ϖ where for y ∈ ℏ1 we get μ(y) − ℵ(y) = −ϖ[μ(y) + ℵ(−y)] 󳨐⇒ ‖y‖ℏ1 ⩽ ϱ,

ϖ ∈ (0, 1].

Proof. Let y ∈ ℏ1 satisfy μ(y) − ℵ(y) = −ϖμ(y) − ϖℵ(−y). Then μ(y) =

1 ϖ ℵ(y) − ℵ(−y). 1+ϖ 1+ϖ

So, from the expression of μ and ℵ, we get for any t ∈ (a, b] and j ∈ {1, 2} α ,βj ;ψ

μj yj (t) = H Da+j

yj (t)

1 = f (t, y1 (t), y2 (t)) 1+ϖ j ϖ − f (t, −y1 (t), −y2 (t)). 1+ϖ j

By Theorem 2.4, we get yj (t) =

cj∗ (ψ(t) − ψ(a))γj −1 Γ(γj )

+

1 α ;ψ [I +j (fj (s, y1 (s), y2 (s)))(t) ϖ+1 a

246 � 9 Nonlinear ψ-Hilfer fractional coupled systems α ;ψ

− ϖ Ia+j (fj (s, −y1 (s), −y2 (s)))(t)], 1−γj ;ψ

where cj∗ = Ia+

yj (a). Thus, for each t ∈ (a, b], we have

∗ γ −1 2f ∗ Γ(γ1 ) α +γ −1 󵄨 |c |(ψ(t) − ψ(a)) 1 󵄨󵄨 + (ψ(t) − ψ(a)) 1 1 󵄨󵄨y1 (t)󵄨󵄨󵄨 ⩽ 1 Γ(γ1 ) (ϖ + 1)Γ(γ1 + α1 ) 2ϑ ∗ Γ(γ1 )‖y‖ℏ1 α +γ −1 + (ψ(t) − ψ(a)) 1 1 (ϖ + 1)Γ(γ1 + α1 ) 2η∗ Γ(γ2 )‖y‖ℏ1 α +γ −1 (ψ(t) − ψ(a)) 1 2 . + (ϖ + 1)Γ(γ2 + α1 )

Thus, ∗ 2f ∗ Γ(γ1 ) 1−γ1 α 󵄨󵄨 󵄨 |c | (ψ(t) − ψ(a)) 1 󵄨󵄨(ψ(t) − ψ(a)) y1 (t)󵄨󵄨󵄨 ⩽ 1 + Γ(γ1 ) Γ(γ1 + α1 ) 2ϑ ∗ Γ(γ1 )‖y‖ℏ1 α + (ψ(t) − ψ(a)) 1 Γ(γ1 + α1 ) 2η∗ Γ(γ2 )‖y‖ℏ1 α +γ −γ + (ψ(t) − ψ(a)) 1 2 1 Γ(γ2 + α1 ) |c∗ | 2f ∗ Γ(γ1 ) α (ψ(b) − ψ(a)) 1 ⩽ 1 + Γ(γ1 ) Γ(γ1 + α1 ) 2ϑ ∗ Γ(γ1 )‖y‖ℏ1 α + (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 ) 2η∗ Γ(γ2 )‖y‖ℏ1 α +γ −γ + (ψ(b) − ψ(a)) 1 2 1 . Γ(γ2 + α1 )

We deduce that |c1∗ | 2f ∗ Γ(γ1 ) α + (ψ(b) − ψ(a)) 1 Γ(γ1 ) Γ(γ1 + α1 ) 2ϑ ∗ Γ(γ1 )‖y‖ℏ1 α + (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 ) 2η∗ Γ(γ2 )‖y‖ℏ1 α +γ −γ + (ψ(b) − ψ(a)) 1 2 1 Γ(γ2 + α1 ) |c∗ | 2f ∗ Γ(γ1 ) α ⩽ 1 + (ψ(b) − ψ(a)) 1 Γ(γ1 ) Γ(γ1 + α1 )

‖y1 ‖C1−γ ;ψ ⩽ 1

+ 2[

ϑ ∗ Γ(γ1 ) α (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 )

η∗ Γ(γ2 ) α +γ −γ (ψ(b) − ψ(a)) 1 2 1 ]‖y‖ℏ1 Γ(γ2 + α1 ) |c∗ | 2f ∗ Γ(γ1 ) α ⩽ 1 + (ψ(b) − ψ(a)) 1 + 2ϒ7 ‖y‖ℏ1 . Γ(γ1 ) Γ(γ1 + α1 ) +

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

� 247

Similarly, we get ‖y2 ‖C1−γ ;ψ ⩽ 2

|c2∗ | 2f ∗ Γ(γ2 ) α + (ψ(b) − ψ(a)) 2 Γ(γ2 ) Γ(γ2 + α2 ) + 2[

ϑ ∗ Γ(γ1 ) α +γ −γ (ψ(b) − ψ(a)) 2 1 2 Γ(γ1 + α2 )

η∗ Γ(γ2 ) α (ψ(b) − ψ(a)) 2 ]‖y‖ℏ1 Γ(γ2 + α2 ) |c∗ | 2f ∗ Γ(γ2 ) α (ψ(b) − ψ(a)) 2 + 2ϒ8 ‖y‖ℏ1 , ⩽ 2 + Γ(γ2 ) Γ(γ2 + α2 ) +

which implies that ‖y‖ℏ1 ⩽ max { j∈{1,2}

|cj∗ |

Γ(γj )

+

2f ∗ Γ(γj )

Γ(γj + αj )

α

(ψ(b) − ψ(a)) j } + 2 max{ϒ7 , ϒ8 }‖y‖ℏ1 .

We deduce that |c∗ |

‖y‖ℏ1 ⩽

maxj∈{1,2} { Γ(γj ) + j

2f ∗ Γ(γj ) (ψ(b) Γ(γj +αj )

− ψ(a))αj }

1 − 2 max{ϒ7 , ϒ8 }

:= ϱ.

Lemma 9.5. If conditions (9.3.1), (9.3.2), and (9.8) are satisfied, then there exists a bounded open set Ψ ⊂ ℏ1 with μ(y) − ℵ(y) ≠ −ϖ[μ(y) + ℵ(−y)] for any y ∈ 𝜕Ψ and any ϖ ∈ (0, 1]. Proof. Using Lemma 9.4, there exists a positive constant ϱ which is independent of ϖ such that if y satisfies μ(y) − ℵ(y) = −ϖ[μ(y) + ℵ(−y)],

ϖ ∈ (0, 1],

then ‖y‖ℏ1 ⩽ ϱ. So, if Ψ = {y ∈ ℏ1 ; ‖y‖ℏ1 < ς}

(9.9)

such that ς > ϱ, then μ(y) − ℵ(y) ≠ −ϖ[μ(y) − ℵ(−y)] for all y ∈ 𝜕Ψ = {y ∈ ℏ1 ; ‖y‖ℏ1 = ς} and ϖ ∈ (0, 1]. Theorem 9.1. Assume that (9.3.1), (9.3.2), and (9.8) hold. Then there exists at least one solution for the problem (9.1)–(9.2).

248 � 9 Nonlinear ψ-Hilfer fractional coupled systems Proof. It is clear that the set Ψ defined in (9.9) is symmetric, 0 ∈ Ψ, and ℏ1 ∩ Ψ = Ψ ≠ 0. In addition, by Lemma 9.5, assume that (9.3.1), (9.3.2), and (9.8) hold. Then μ(y) − ℵ(y) ≠ −ϖ[μ(y) − ℵ(−y)] for each y ∈ ℏ1 ∩ 𝜕Ψ = 𝜕Ψ and each ϖ ∈ (0, 1]. Thus, problem (9.1)–(9.2) has at least one solution on Dom μ ∩ Ψ. We have proved our precedent existence result by assuming some condition of the constants α1 , α2 , γ1 , and γ2 , which are often regarded as strong requirements. Now, we will employ different conditions on the functions fj , j ∈ {1, 2}. We will use them to demonstrate existence and uniqueness results. In all that follows, we will present and demonstrate our results using the same approach as before, which will allow us to skip certain steps. Lemma 9.6. Assume that the hypothesis (9.3.1) and the following condition are satisfied. (9.7.1) For each j ∈ {1, 2}, there exist nonnegative functions ϑ̂j , η̂j ∈ C(J,̄ ℝ+ ) such that 󵄨󵄨 󵄨 󵄨󵄨fj (t, y1 (t), y2 (t)) − fj (t, y1 (t), y2 (t))󵄨󵄨󵄨 1−γ 󵄨 󵄨 ⩽ ϑ̂j (t)(ψ(t) − ψ(a)) 1 󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 1−γ 󵄨 󵄨 + η̂j (t)(ψ(t) − ψ(a)) 2 󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 for every t ∈ J ̄ and yj , yj ∈ C1−γj ;ψ (J,̄ ℝ). ̂ ⊂ ℏ1 , the operator ℵ is μ-compact. Then for any bounded open set Ψ ̂ = {y ∈ ℏ1 : ‖y‖ℏ < B } be an bounded open set. Proof. For B > 0, let Ψ 1 Step 1. We show that QN is continuous. Let (yn )n∈ℕ be a sequence where yn 󳨀→ y in ℏ2 ; thus, for j ∈ {1, 2} and by hypothesis (9.7.1), we obtain for t ∈ J ̄ 1−γ1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (Φ1 ℵ1 (yn )(t) − Φ1 ℵ1 (y)(t))󵄨󵄨󵄨 ̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }‖y − y‖ ⩽ (ϑ n ℏ

2

j∈{1,2}

and 1−γ2 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) (Φ2 ℵ2 (yn )(t) − Φ2 ℵ2 (y)(t))󵄨󵄨󵄨 1−γj

̂∗ + η̂∗ ) max {(ψ(b) − ψ(a)) ⩽ (ϑ j∈{1,2}

}‖yn − y‖ℏ2 ,

̂∗ = max ̂ ̂∗ ̂j ‖∞ . Thus, for each j ∈ {1, 2}, we get where ϑ j∈{1,2} ‖ϑj ‖∞ and η = maxj∈{1,2} ‖η

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

� 249

󵄩 󵄩󵄩 ̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }‖y − y‖ , 󵄩󵄩Φj ℵj (yn ) − Φj ℵj (y)󵄩󵄩󵄩C1−γ ;ψ ⩽ (ϑ n ℏ2 j j∈{1,2} which implies 󵄩󵄩 󵄩 󵄩󵄩QN (yn ) − QN (y)󵄩󵄩󵄩ℏ2 󳨀→ 0

as n 󳨀→ +∞.

We may conclude then that QN is continuous. ̂ is bounded. Step 2. We demonstrate that QN (Ψ) ̂ Employing Lemma 2.6, we get for any t ∈ (a, b] and y ∈ Ψ 1−γ1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) Φ1 ℵ1 (y)(t)󵄨󵄨󵄨 Γ(γj )Γ(2 + βj (αj − 1)) } ⩽ f ∗ max { j∈{1,2} Γ(αj + 1)

1−γj

̂∗ + η̂∗ ) max {(ψ(b) − ψ(a)) + (ϑ

}B

j∈{1,2}

and 1−γ2 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) Φ2 ℵ2 (y)(t)󵄨󵄨󵄨 Γ(γj )Γ(2 + βj (αj − 1)) ⩽ f ∗ max { } j∈{1,2} Γ(αj + 1)

1−γj

̂∗ + η̂∗ ) max {(ψ(b) − ψ(a)) + (ϑ j∈{1,2}

}B .

Then for each j ∈ {1, 2}, we obtain ‖Φj ℵj y‖C1−γ ;ψ ⩽ f ∗ max { j∈{1,2}

j

Γ(γj )Γ(2 + βj (αj − 1)) Γ(αj + 1)

}

̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }B . + (ϑ j∈{1,2}

Thus, Γ(γj )Γ(2 + βj (αj − 1)) 󵄩󵄩 󵄩 ∗ } 󵄩󵄩QN (y)󵄩󵄩󵄩ℏ2 ⩽ f max { j∈{1,2} Γ(α + 1) j

̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }B . + (ϑ j∈{1,2}

̂ is a bounded set in ℏ2 . So, QN (Ψ)

250 � 9 Nonlinear ψ-Hilfer fractional coupled systems ̂ Step 3. We prove that μ−1 Λ (id − Φ)ℵ : Ψ → ℏ1 is completely continuous. ̂ and t ∈ (a, b], we have For any y ∈ Ψ μ−1 Λ (ℵy(t) − QN y(t)) −1 = (μ−1 Λ 1 (ℵ1 y(t) − Φ1 ℵ1 y(t)), μΛ 2 (ℵ2 y(t) − Φ2 ℵ2 y(t))),

̂ and t ∈ (a, b] where j ∈ {1, 2}. By Lemma 2.6, we have for all y ∈ Ψ 1−γ1 −1 󵄨󵄨 󵄨 󵄨󵄨(ψ(t) − ψ(a)) μΛ 1 (id − Φ1 )ℵ1 y(t)󵄨󵄨󵄨 f ∗ Γ(γ1 ) f ∗ Γ(γ1 )Γ(2 + β1 (α1 − 1)) α α ⩽ (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 2 Γ(γ1 + α1 ) Γ (α1 + 1) ̂∗ + η̂∗ )B (ϑ 1−γ +α 1+β (α −1) [(ψ(b) − ψ(a)) 1 1 + (ψ(b) − ψ(a)) 1 1 ]. + Γ(α1 + 1)

Thus, 󵄩󵄩 −1 󵄩 󵄩󵄩μΛ 1 (id − Φ1 )ℵ1 y󵄩󵄩󵄩C1−γ ;ψ 1 ⩽

f ∗ Γ(γ1 ) f ∗ Γ(γ1 )Γ(2 + β1 (α1 − 1)) α α (ψ(b) − ψ(a)) 1 + (ψ(b) − ψ(a)) 1 Γ(γ1 + α1 ) Γ2 (α1 + 1) ̂∗ + η̂∗ )B (ϑ 1−γ +α 1+β (α −1) + [(ψ(b) − ψ(a)) 1 1 + (ψ(b) − ψ(a)) 1 1 ] Γ(α1 + 1)

= ϒ5 .

Also, we may obtain 󵄩󵄩 −1 󵄩 󵄩󵄩μΛ 2 (id − Φ2 )ℵ2 y󵄩󵄩󵄩C1−γ ;ψ 2 ⩽

f ∗ Γ(γ2 ) f ∗ Γ(γ2 )Γ(2 + β2 (α2 − 1)) α α (ψ(b) − ψ(a)) 2 + (ψ(b) − ψ(a)) 2 2 Γ(γ2 + α2 ) Γ (α2 + 1) ∗ ∗ ̂ ̂ (ϑ + η )B 1−γ +α 1+β (α −1) [(ψ(b) − ψ(a)) 2 2 + (ψ(b) − ψ(a)) 2 2 ] + Γ(α2 + 1)

= ϒ6 .

We conclude that 󵄩󵄩 −1 󵄩 󵄩󵄩μΛ (id − Φ)ℵy󵄩󵄩󵄩ℏ ≤ max{ϒ5 , ϒ6 }. 1

̂ This means that μ−1 Λ (id − Φ)ℵ(Ψ) is uniformly bounded in ℏ1 . ̂ is equicontinuous. Let us now demonstrate μ−1 (id − Φ)ℵ(Ψ) Λ

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

� 251

̂ and j ∈ {1, 2}, we have For a < t1 < t2 ⩽ b, y ∈ Ψ, −1 󵄨󵄨 μ−1 󵄨 󵄨󵄨 Λ j (id − Φj )ℵj y(t2 ) μΛ j (id − Φj )ℵj y(t1 ) 󵄨󵄨󵄨 − 󵄨󵄨 󵄨󵄨 γ −1 γ −1 󵄨󵄨 (ψ(t2 ) − ψ(a)) j (ψ(t1 ) − ψ(a)) j 󵄨󵄨 t

1 󵄨󵄨 (ψ(t ) − ψ(a))1−γj (ψ(t ) − ψ(a))1−γj 󵄨󵄨 f∗ 󵄨 󵄨󵄨 2 1 ⩽ − ∫ ψ′ (s)󵄨󵄨󵄨 󵄨 󵄨󵄨 (ψ(t2 ) − ψ(s))1−αj (ψ(t1 ) − ψ(s))1−αj 󵄨󵄨󵄨 Γ(αj )

a

× max {(ψ(s) − ψ(a))

γj −1

j∈{1,2}

t1

}ds +

̂∗ + η̂∗ )B (ϑ ∫ ψ′ (s) Γ(αj ) a

󵄨󵄨 (ψ(t ) − ψ(a))1−γj (ψ(t ) − ψ(a))1−γj 󵄨󵄨 󵄨 󵄨󵄨 1 2 × 󵄨󵄨󵄨 − 󵄨ds 󵄨󵄨 (ψ(t2 ) − ψ(s))1−αj (ψ(t1 ) − ψ(s))1−αj 󵄨󵄨󵄨 +

f ∗ (ψ(t2 ) − ψ(a))1−γj Γ(αj ) t2

× ∫ ψ′ (s)(ψ(t2 ) − ψ(s))

αj −1

γj −1

max {(ψ(s) − ψ(a))

j∈{1,2}

t1

+

}ds

̂∗ + η̂∗ )B (ψ(t ) − ψ(a))1−γj (ϑ α 2 (ψ(t2 ) − ψ(t1 )) j Γ(αj + 1)

+[

f ∗ Γ(γj )Γ(2 + βj (αj − 1))

Γ2 (α

j

1−γj

+ 1)(ψ(b) − ψ(a))

× [(ψ(t2 ) − ψ(a))

1+αj −γj

+

̂∗ + η̂∗ )B (ϑ ] Γ(αj + 1) 1+αj −γj

− (ψ(t1 ) − ψ(a))

].

̂ As in the precedent result, we can deduce that ℵ is μ-compact in Ψ. Lemma 9.7. Suppose that (9.3.1) and (9.7.1) are satisfied. If ̂∗ + η̂∗ ) max { (ϑ j∈{1,2}

1 1 α +1−γj (ψ(b) − ψ(a)) j }< , Γ(αj + 1) 2

then there exists ϱ̂ > 0 which is independent of ϖ where for y ∈ ℏ1 we get μ(y) − ℵ(y) = −ϖ[μ(y) + ℵ(−y)] 󳨐⇒ ‖y‖ℏ1 ⩽ ϱ̂, Proof. Let y ∈ ℏ1 , where μ(y) − ℵ(y) = −ϖμ(y) − ϖℵ(−y). Then μ(y) =

1 ϖ ℵ(y) − ℵ(−y). 1+ϖ 1+ϖ

ϖ ∈ (0, 1].

(9.10)

252 � 9 Nonlinear ψ-Hilfer fractional coupled systems So, from the expression of μ and ℵ, we get for any t ∈ (a, b] and j ∈ {1, 2} α ,βj ;ψ

μj yj (t) = H Da+j

yj (t) =

1 f (t, y1 (t), y2 (t)) 1+ϖ j ϖ − f (t, −y1 (t), −y2 (t)). 1+ϖ j

By Theorem 2.4, we get ‖y‖ℏ1 ⩽ max { j∈{1,2}

|cj∗ |

Γ(γj )

+

2f ∗ Γ(γj )

Γ(γj + αj )

̂∗ + η̂∗ ) max { + 2(ϑ j∈{1,2}

α

(ψ(b) − ψ(a)) j }

1 α +1−γj (ψ(b) − ψ(a)) j }‖y‖ℏ1 . Γ(αj + 1)

We deduce that |c∗ |

‖y‖ℏ1 ⩽

maxj∈{1,2} { Γ(γj ) + j

2f ∗ Γ(γj ) (ψ(b) Γ(γj +αj )

− ψ(a))αj }

1 αj +1−γj ̂∗ + η̂∗ ) max } 1 − 2(ϑ j∈{1,2} { Γ(α +1) (ψ(b) − ψ(a))

:= ϱ̂.

j

Lemma 9.8. If conditions (9.3.1), (9.7.1), and (9.10) are satisfied, then there exists a ̂ ⊂ ℏ1 with bounded open set Ψ μ(y) − ℵ(y) ≠ −ϖ[μ(y) + ℵ(−y)] ̂ and any ϖ ∈ (0, 1]. for any y ∈ 𝜕Ψ Proof. The proof is the same as the proof of Lemma 9.5. Theorem 9.2. Suppose that the hypotheses (9.3.1) and (9.7.1) and the condition (9.10) are satisfied. Then there exists at least one solution for the problem (9.1)–(9.2). Proof. The proof is the same as the proof of Theorem 9.1. Lemma 9.9. Let 0 < λ1 < 1 and 0 < λ2 ≤ 1. Then the following inequality holds: Γ(λ2 ) 1 ≤ . Γ(λ1 + 1) Γ(λ1 + λ2 ) Proof. Using Lemma 2.6, we have for ε ∈ (a, b] 1 λ (ψ(ε) − ψ(a)) 1 Γ(λ1 + 1) ε

1 λ −1 = ∫ ψ′ (s)(ψ(ε) − ψ(s)) 1 ds Γ(λ1 ) a ε

=

1 λ −1 λ −1 1−λ ∫ ψ′ (s)(ψ(ε) − ψ(s)) 1 (ψ(s) − ψ(a)) 2 (ψ(s) − ψ(a)) 2 ds Γ(λ1 ) a

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

≤ (ψ(ε) − ψ(a))

1−λ2

λ ;ψ

Ia+1 (ψ(s) − ψ(a))

λ2 −1

� 253

(ε)

Γ(λ2 ) λ +λ −1 1−λ (ψ(ε) − ψ(a)) 1 2 (ψ(ε) − ψ(a)) 2 Γ(λ1 + λ2 ) Γ(λ2 ) λ ≤ (ψ(ε) − ψ(a)) 1 . Γ(λ1 + λ2 ) ≤

Theorem 9.3. Suppose that (9.3.1) and (9.7.1) are satisfied. Moreover, we suppose that the following hypothesis holds. (9.12.1) There exist ϑj > 0 and ηj ⩾ 0, j ∈ {1, 2}, where 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t))󵄨󵄨󵄨 1−γ 󵄨 󵄨 ⩾ ϑ1 (ψ(t) − ψ(a)) 1 󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 1−γ 󵄨 󵄨 − η1 (ψ(t) − ψ(a)) 2 󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 and 󵄨󵄨 󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))󵄨󵄨󵄨 1−γ 󵄨 󵄨 ⩾ ϑ2 (ψ(t) − ψ(a)) 2 󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 1−γ 󵄨 󵄨 − η2 (ψ(t) − ψ(a)) 1 󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 for every t ∈ (a, b], y1 , y1 ∈ C1−γ1 ;ψ (J,̄ ℝ), and y2 , y2 ∈ C1−γ2 ;ψ (J,̄ ℝ). If one has [max { j∈{1,2}

ηj ϑj

̂∗ + η̂∗ ) max { } + 2(ϑ j∈{1,2}

1−γj

× max {(ψ(b) − ψ(a)) j∈{1,2}

Γ(γj )

Γ(γj + αj )

α

(ψ(b) − ψ(a)) j }]

} < 1,

(9.11)

̂ then the problem (9.1)–(9.2) has a unique solution in Dom μ ∩ Ψ. Proof. Since the condition (9.11) is stronger than (9.10), by Theorem 9.2 we may get that ̂ (9.1)–(9.2) has at least one solution in Dom μ ∩ Ψ.

̂ Then we We assume that (9.1)–(9.2) has two different solutions y, y ∈ Dom μ ∩ Ψ. have for each t ∈ (a, b] and j ∈ {1, 2} H

α ,βj ;ψ

Da+j

yj (t) = fj (t, y1 (t), y2 (t)),

H αj ,βj ;ψ Da+ yj (t)

and

= fj (t, y1 (t), y2 (t)),

254 � 9 Nonlinear ψ-Hilfer fractional coupled systems 1−γj ,ψ

Ia+

1−γj ,ψ

yj (a) = Ia+

1−γj ,ψ

yj (b), Ia+

1−γj ,ψ

yj (a) = Ia+

yj (b).

Let U(t) = y(t) − y(t) for all t ∈ (a, b], which means that U(t) = (U1 (t), U2 (t)) = (y1 (t) − y1 (t), y2 (t) − y2 (t)). Then μU(t) = (μ1 U1 (t), μ2 U2 (t)) α ,β1 ;ψ

= (H Da+1

=(

H

α ,β2 ;ψ

U1 (t), H Da+2

α ,β ;ψ Da+1 1 y1 (t)

−

H

U2 (t))

α ,β ;ψ H α ,β ;ψ Da+1 1 y1 (t), Da+2 2 y2 (t)

= (f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t)),

α ,β2 ;ψ

− H Da+2

y2 (t))

f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))).

(9.12)

Using the fact that Img μ = ker Φ, for j ∈ {1, 2} we have b

βj (αj −1)

∫ ψ′ (s)(ψ(b) − ψ(s))

[fj (s, y1 (s), y2 (s)) − fj (s, y1 (s), y2 (s))]ds = 0.

b

Since for each j ∈ {1, 2} the functions fj ∈ C1−γj ;ψ (J,̄ ℝ), there exists t0 ∈ (a, b] such that fj (t0 , y1 (t0 ), y2 (t0 )) − fj (t, y1 (t0 ), y2 (t0 )) = 0. In view of (9.12.1), we have 󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 ⩽ η1 (ψ(t0 ) − ψ(a))1−γ2 󵄨󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ϑ1 η ⩽ 1 × ‖y2 − y2 ‖C1−γ ,ψ 2 ϑ1 ηj ⩽ max { }‖U‖ℏ1 j∈{1,2} ϑ j

1−γ1 󵄨

(ψ(t0 ) − ψ(a))

and 󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ⩽ η2 (ψ(t0 ) − ψ(a))1−γ1 󵄨󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ϑ2 η ⩽ 2 × ‖y1 − y1 ‖C1−γ ,ψ 1 ϑ2 ηj ⩽ max { }‖U‖ℏ1 . j∈{1,2} ϑ j

1−γ2 󵄨

(ψ(t0 ) − ψ(a))

Then for each j ∈ {1, 2},

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

|Uj (t0 )| ⩽ max { j∈{1,2}

ηj ϑj

� 255

(9.13)

}‖U‖ℏ1 .

By Theorem 2.4, for any t ∈ (a, b] and j ∈ {1, 2}, we have α ;ψ H

Ia+j

α ,βj ;ψ

Da+j

Uj (t) = Uj (t) −

cj (ψ(t) − ψ(a))γj −1 Γ(γj )

,

which implies that α ;ψ H

cj = [Uj (t0 ) − Ia+j

α ,βj ;ψ

Da+j

1−γj

Uj (t0 )]Γ(γj )(ψ(t0 ) − ψ(a))

,

and therefore α ;ψ H

Uj (t) = Ia+j

α ,βj ;ψ

Da+j

α ;ψ H

Uj (t) + [Uj (t0 ) − Ia+j 1−γj

× (ψ(t0 ) − ψ(a))

(ψ(t) − ψ(a))

γj −1

α ,βj ;ψ

Da+j

Uj (t0 )]

.

Using (9.13), we obtain for every t ∈ (a, b] and j ∈ {1, 2} 1−γ

󵄨󵄨 󵄨 󵄨 󵄨 󵄨 αj ;ψ H αj ,βj ;ψ 󵄨 (ψ(t0 ) − ψ(a)) j 󵄨󵄨Uj (t)󵄨󵄨󵄨 ⩽ [󵄨󵄨󵄨Uj (t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨Ia+ Da+ Uj (t0 )󵄨󵄨󵄨] (ψ(t) − ψ(a))1−γj α ;ψ H

+ |Ia+j ⩽ max { j∈{1,2}

+ +

ϑj

Uj (t)|

}‖U‖ℏ1 (ψ(b) − ψ(a))

1−γj

γj −1

(ψ(t) − ψ(a))

Γ(γj )

αj γj −1 󵄩󵄩H αj ,βj ;ψ 󵄩󵄩 󵄩 D + Uj 󵄩󵄩C1−γ ,ψ (ψ(t0 ) − ψ(a)) (ψ(t) − ψ(a)) j Γ(γj + αj ) 󵄩 a

Γ(γj ) 󵄩H αj ,βj ;ψ 󵄩 αj γj −1 󵄩󵄩 D + Uj 󵄩󵄩 󵄩C1−γj ,ψ (ψ(t) − ψ(a)) (ψ(t) − ψ(a)) Γ(γj + αj ) 󵄩 a

⩽ max { j∈{1,2}

+

ηj

α ,βj ;ψ

Da+j

ηj

ϑj

}‖U‖ℏ1 (ψ(b) − ψ(a))

1−γj

γj −1

(ψ(t) − ψ(a))

2Γ(γj ) 󵄩H αj ,βj ;ψ 󵄩 αj γj −1 󵄩󵄩 D + Uj 󵄩󵄩 . 󵄩 󵄩 C1−γj ,ψ (ψ(b) − ψ(a)) (ψ(t) − ψ(a)) a Γ(γj + αj )

By (9.7.1) and (9.12), for each j ∈ {1, 2} we find 1−γj H αj ,βj ;ψ 󵄨󵄨 󵄨 Da+ Uj (t)󵄨󵄨󵄨 󵄨󵄨(ψ(t) − ψ(a)) 1−γ 󵄨 󵄨 = (ψ(t) − ψ(a)) j 󵄨󵄨󵄨fj (t, y1 (t), y2 (t)) − fj (t, y1 (t), y2 (t))󵄨󵄨󵄨 ̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }‖U‖ . ⩽ (ϑ ℏ j∈{1,2}

Thus,

1

(9.14)

256 � 9 Nonlinear ψ-Hilfer fractional coupled systems 󵄩󵄩H αj ,βj ;ψ 󵄩󵄩 ̂∗ + η̂∗ ) max {(ψ(b) − ψ(a))1−γj }‖U‖ . 󵄩󵄩 Da+ Uj 󵄩󵄩C1−γ ,ψ ⩽ (ϑ ℏ1 j j∈{1,2} Substituting (9.15) in the right side of (9.14), for every t ∈ (a, b] and j ∈ {1, 2} we get Γ(γj ) ηj α 󵄨󵄨 󵄨 ̂∗ + η̂∗ ) max { (ψ(b) − ψ(a)) j }] 󵄨󵄨Uj (t)󵄨󵄨󵄨 ⩽ [max { } + 2(ϑ j∈{1,2} Γ(γj + αj ) j∈{1,2} ϑ j × max {(ψ(b) − ψ(a))

1−γj

j∈{1,2}

}(ψ(t) − ψ(a))

γj −1

‖U‖ℏ1 .

Therefore, ‖U‖ℏ1 ⩽ [max { j∈{1,2}

ηj ϑj

̂∗ + η̂∗ ) max { } + 2(ϑ j∈{1,2}

× max {(ψ(b) − ψ(a)) j∈{1,2}

1−γj

Γ(γj )

Γ(γj + αj )

α

(ψ(b) − ψ(a)) j }]

}‖U‖ℏ1 .

Hence, by (9.11), ‖U‖ℏ1 = 0. Consequently, for any t ∈ (a, b] we get U(t) = 0 󳨐⇒ y(t) = y(t).

9.2.2 Examples Example 9.1. Consider the following system: 1 1

, ;ln t

H 4 3 { D+ y1 (t) = f1 (t, y1 (t), y2 (t)), { { { 11 , 1 ;ln t H 5 7 D1+ y2 (t) = f2 (t, y1 (t), y2 (t)), { { { 24 24 1 { 21 ;ln t ;ln t ;ln t ;ln t 35 35 2 {I0+ y1 (1) = I0+ y1 (e) and I0+ y2 (1) = I0+ y2 (e),

t ∈ (1, e],

where y1 (t) ln 2 (t) t−1 f1 (t, y1 (t), y2 (t)) = + + 7 sin y2 (t), 3 53(1 + t) 3e √π −1

2

y (t) 3 ln 35 (t) t+1 e−7−t + 13 y1 (t) + ( 2 ). 5t te + 2 33e11 1 + y2 (t) e √π −24

f2 (t, y1 (t), y2 (t)) =

Here α1 = 41 , α2 = 51 , β1 = 31 , β2 = 71 , γ1 = 21 , γ2 =

11 , 35

and ψ(t) = ln(t).

(9.15)

9.2 Periodic solutions for some nonlinear ψ-Hilfer fractional coupled systems

� 257

It is easy to see that fj ∈ C1−γj ;ln(t) ([1, e], ℝ), j ∈ {1, 2}. Hence, condition (9.3.1) is satisfied. Furthermore, for all t ∈ (1, e] and yj , yj ∈ C1−γj ;ln(t) ([1, e], ℝ), j ∈ {1, 2}, we obtain 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨y (t) − y1 (t)󵄨󵄨󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩽ 53(1 + t) 󵄨 1 t − 1 󵄨󵄨 󵄨 + 7 󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 3e √π 󵄨 and t + 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩽ 13 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 e √π 󵄨 2

+

e−7−t 󵄨󵄨 󵄨 󵄨y2 (t) − y2 (t)󵄨󵄨󵄨. 33e11 󵄨

Then (9.3.2) is satisfied with ϑ1 (t) =

1 , 53(1 + t)

ϑ2 (t) =

t+1 e13 √π

and t−1 η1 (t) = 7 , 3e √π By simple calculations, we get ϑ ∗ = Also, we have

1 106

2

e−7−t η2 (t) = . 33e11

and η∗ =

t−1 . 3e7 √π

α1 + γ2 ≈ 0.564 ≥ γ1 = 0.5,

{

α2 + γ1 ≈ 0.7 ≥ γ2 ≈ 0.314

and {ϒ7 := { ϒ := { 8

ϑ ∗ Γ(γ1 ) η∗ Γ(γ2 ) (ψ(b) − ψ(a))α1 + Γ(γ (ψ(b) − ψ(a))α1 +γ2 −γ1 Γ(γ1 +α1 ) 2 +α1 ) ∗ η∗ Γ(γ2 ) ϑ Γ(γ1 ) (ψ(b) − ψ(a))α2 +γ1 −γ2 + Γ(γ (ψ(b) − ψ(a))α2 Γ(γ1 +α2 ) 2 +α2 )

≈ 0.0689, ≈ 0.0645,

which implies that 1 max{ϒ7 , ϒ8 } ≈ 0.0645 < . 2 With the use of Theorem 9.1 our problem has at least one solution.

258 � 9 Nonlinear ψ-Hilfer fractional coupled systems Example 9.2. Consider the following system: 1 1

, ;et

H 2 7 { D + y1 (t) = f1 (t, y1 (t), y2 (t)), { { { 01 , 1 ;et H 3 7 D0+ y2 (t) = f2 (t, y1 (t), y2 (t)), { { { 3 t 4 t 4 t { 73 ;et ;e ;e ;e 7 7 7 {I0+ y1 (0) = I0+ y1 (1) and I0+ y2 (0) = I0+ y2 (1),

t ∈ (0, 1],

where f1 (t, y1 (t), y2 (t)) =

ln (t + 2) 3

(et − 1) 7 +

3

+

(et − 1) 7 3 (sin y1 (t) + y1 (t)) 11 2 23e √π 4

e−11−t (et − 1) 7 37e9 (1 + y2 (t))

and 4

(et − 1) 7 e 9 + f2 (t, y1 (t), y2 (t)) = y2 (t) 4 7e7 (1 + 3t ) 5(et − 1) 7 2et

−t

3

t(et − 1) 7 + cos y1 (t). 55e13 √π

Here α1 = 21 , α2 = 31 , β1 = β2 = 71 , γ1 = 73 , γ2 = 47 , and ψ(t) = et . It is easy to see that fj ∈ C1−γj ;et ([0, 1], ℝ), j ∈ {1, 2}. Hence, condition (9.3.1) is satisfied. Furthermore, for all t ∈ (0, 1] and yj , yj ∈ C1−γj ;et ([0, 1], ℝ), j ∈ {1, 2}, we obtain 3

t 󵄨󵄨 󵄨 5(e − 1) 7 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩽ 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 46e11 √π 󵄨 4

+

e−11−t (et − 1) 7 󵄨󵄨 󵄨 󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 37e9

and 3

t 󵄨 t(e − 1) 7 󵄨󵄨 󵄨 󵄨󵄨 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩽ 55e13 √π 󵄨 4

e 9 (et − 1) 7 󵄨󵄨 󵄨 + y (t) − y2 (t)󵄨󵄨󵄨. t 󵄨󵄨 2 7 7e (1 + 3 ) −t

Then (9.7.1) is satisfied with ̂(t) = ϑ 1 and

5 , 46e11 √π

̂ (t) = ϑ 2

t 55e13 √π

9.3 Notes and remarks

e−11−t ̂1 (t) = η , 37e9 By simple calculations, we get ϑ̂ ∗ = ̂∗ + η̂∗ ) max { (ϑ j∈{1,2}

� 259

e9 η̂2 (t) = . 7 7e (1 + 3t ) −t

5 46e11 √π

1 , 7e7

and η̂∗ =

and

1 1 α +1−γj (ψ(b) − ψ(a)) j } ≈ 0.00244 < . Γ(αj + 1) 2

With the use of Theorem 9.2 our problem has at least one solution. Otherwise, for t ∈ (0, 1], y1 , y1 ∈ C1−γ1 ;et ([0, 1], ℝ), and y2 , y2 ∈ C1−γ2 ;et ([0, 1], ℝ), we have 3

t 󵄨󵄨 󵄨 (e − 1) 7 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t)) − f1 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩾ 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 46e11 √π 󵄨 4

−

(et − 1) 7 󵄨󵄨 󵄨 󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 37e20 󵄨

and 4

t 󵄨 3e 9 (e − 1) 7 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t)) − f2 (t, y1 (t), y2 (t))󵄨󵄨󵄨 ⩾ 28e7 −1

3

− Then (9.12.1) is satisfied with ϑ1 = We have [max { j∈{1,2}

ηj ϑj

1

46e11 √π

, ϑ2 =

̂∗ + η̂∗ ) max { } + 2(ϑ j∈{1,2}

1−γj

× max {(ψ(b) − ψ(a)) j∈{1,2}

−1

3e 9 28e7

(et − 1) 7 󵄨󵄨 󵄨 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨. 55e13 √π 󵄨

, η1 =

Γ(γj )

Γ(γj + αj )

1 , 37e20

and η2 =

1 . 55e13 √π

α

(ψ(b) − ψ(a)) j }]

} ≈ 0.00112 < 1.

So, by Theorem 9.3, our problem has a unique solution.

9.3 Notes and remarks The paper [96] contains the results of Chapter 9. More related results and investigations may be found in the monographs [49, 98, 115, 125, 144, 145, 161, 185, 186, 244, 245] and the articles [10–13, 16, 20, 21, 28, 34, 110].

10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions 10.1 Introduction and motivations This chapter deals with some existence and uniqueness results for a class of systems with nonlinear k-generalized ψ-Hilfer fractional differential equations with periodic conditions. The arguments are based on Mawhin’s coincidence degree theory. Furthermore, an illustration is presented to demonstrate the plausibility of our results. We investigate and demonstrate the results in this chapter by taking into account the previously mentioned publications in the prior chapters as well as the publications that follow: – The papers [16, 39, 42, 130, 162, 226], where the authors investigated numerous results for systems with different types of fractional differential equations and inclusions with various conditions. – The papers [71, 120, 215, 240], where the authors studied a wide range of results for various forms of fractional differential equations and inclusions with different types of conditions. – Several generalizations for various special functions and fractional operators that have been presented, opening the door to more extended results for a variety of problems; interest in these types of operators is growing each year. For some information, we refer the reader to the papers of Chu et al. [101], Diaz et al. [113], Kucche et al. [153], Mubeen et al. [175], Salim et al. [211, 212], and Sousa et al. [102]. – In [94], Bouriah et al. consider the following nonlinear pantograph fractional problem with ψ-Caputo fractional derivative: c

{

α;ψ

D0+ u(t) = h(t, u(t), u(εt)),

t ∈ J := [0, b],

u(0) = u(b),

α;ψ

–

where c D0+ denotes the ψ-Caputo fractional derivative of order 0 < α < 1, ε ∈ (0, 1), and h : J × ℝ × ℝ → ℝ is a continuous function. They base their arguments on Mawhin’s coincidence degree theory. Using the Banach contraction principle and Schauder’s fixed point theorem, in [211], Salim et al. studied the boundary value problem for the nonlinear implicit k-generalized ψ-Hilfer-type fractional differential equation involving both retarded and advanced arguments: ϑ,r;ψ

ϑ,r;ψ

H (H t ∈ (a, b], { k Da+ x)(t) = f (t, xt (⋅), (k Da+ x)(t)), { { { k(1−ϱ),k;ψ k(1−ϱ),k;ψ { {α1 (Ja+ x)(a+ ) + α2 (Ja+ x)(b) = α3 , { {x(t) = ϖ(t), t ∈ [a − α, a], α > 0, { { { { ̃ ̃ α̃ > 0, {x(t) = ϖ(t), t ∈ [b, b + α], https://doi.org/10.1515/9783111334387-010

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

ϑ,r;ψ

� 261

k(1−ϱ),κ,ψ

are, respectively, the k-generalized ψ-Hilfer fractional where H k Da+ and Ja+ derivative of order ϑ ∈ (0, k) and type r ∈ [0, 1] and the k-generalized ψ-fractional ̃ ℝ) × integral of order k(1 − ϱ), where ϱ = k1 (r(k − ϑ) + ϑ), k > 0, f : [a, b] × C([−α, α], ̃ ℝ 󳨀→ ℝ is a given function, α1 , α2 , α3 ∈ ℝ such that α1 + α2 ≠ 0, and ϖ(t) and ω(t) ̃ For each function are, respectively, continuous functions on [a − α, a] and [b, b + α]. x defined on [a − α, b + α]̃ and for any t ∈ (a, b], we denote by xt the element defined ̃ by xt (τ) = x(t + τ), τ ∈ [−α, α].

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions Motivated by the aforementioned publications and with the objective of generalizing earlier results, in this section, we study the existence and uniqueness results for a class of nonlinear k-generalized ψ-Hilfer-type fractional differential equations with periodic conditions: α,β;ψ

α,β;ψ

H (H k Da+ y)(t) = f (t, y(t), (k Da+ y)(t)),

k(1−ξ),k;ψ (Ja+ y)(a+ ) α,β;ψ

=

t ∈ (a, b],

k(1−ξ),k;ψ (Ja+ y)(b),

(10.1) (10.2)

k(1−ξ),k;ψ

where H and Ja+ are, respectively, the k-generalized ψ-Hilfer fractional k Da+ derivative of order α ∈ (0, k) and type β ∈ [0, 1] and the k-generalized ψ-fractional integral of order k(1 − ξ), where ξ = k1 (β(k − α) + α), k > 0, and f : [a, b] × ℝ × ℝ 󳨀→ ℝ is a given function.

10.2.1 Existence results For the sake of simplification, we will suppose the following: ψ

Ψξ (t, a) = (ψ(t) − ψ(a)) (ψ(t) − ψ(s)) Ψ̄ k,ψ α (t, s) = kΓk (α)

1−ξ

,

α −1 k

.

Consider the spaces α,k;ψ

Ω = {y ∈ Cξ;ψ (J)̄ : y(t) = Ja+

̄ t ∈ (a, b]} γ(t) : γ ∈ Cξ;ψ (J),

and Θ = Cξ;ψ (J)̄ with the norm ‖y‖Ω = ‖y‖Θ = ‖y‖Cξ;ψ .

262 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions The following lemmas are required to prove the main results. Before we proceed, we will define the operator K : Dom K ⊆ Ω → Θ by α,β;ψ

H

K y := k Da+ y,

(10.3)

where α,β;ψ

k(1−ξ),k;ψ

Dom K = {y ∈ Ω : H k Da+ y ∈ Θ : Ja+

k(1−ξ),k;ψ

y(a) = Ja+

y(b)}.

Lemma 10.1. We can assume from (10.3) that ker K = {y ∈ Ω : y(t) =

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

, t ∈ (a, b]}

and k(1−ξ)+α,k;ψ

Img K = {γ ∈ Θ : Ja+

γ(b) = 0}.

Proof. Using Lemma 2.24 and taking γ(t) = 0 for all t ∈ (a, b], for all y ∈ Ω the equation α,β;ψ

Ky=H k Da+ y = 0 in (a, b] has a solution given by

y(t) =

(ψ(t) − ψ(a))ξ−1 (1−β)(k−α),k;ψ y(a), J Γk (β(k − α) + α) a+

t ∈ (a, b],

which implies that ker K = {y ∈ Ω : y(t) =

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

, t ∈ (a, b]}.

For γ ∈ Img K , there exists y ∈ Dom K such that γ = K y ∈ Θ. Using Theorem 2.12, we obtain for each t ∈ (a, b] y(t) = k(1−ξ),k;ψ

Applying Ja+ ing t = b, we have

(ψ(t) − ψ(a))ξ−1 k(1−ξ),k;ψ α,k;ψ Ja+ y(a) + Ja+ γ(t). Γk (kξ)

(⋅) on both sides of (10.4), using Lemma 2.5 and Lemma 2.7, and takk(1−ξ),k;ψ

Ja+

k(1−ξ),k;ψ

k(1−ξ)+α,k;ψ

y(b) = Ja+

y(a) + Ja+

k(1−ξ),k;ψ

k(1−ξ),k;ψ

Since y ∈ Dom K , we have Ja+

Thus,

(10.4)

y(a) = Ja+

y(b).

γ(b).

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

k(1−ξ)+α,k;ψ

γ(b) = 0.

k(1−ξ)+α,k;ψ

γ(b) = 0,

Ja+

� 263

Furthermore, if γ ∈ Θ and satisfies Ja+ α,k;ψ

then for any y(t) = Ja+ Lemma 2.13, we get

γ(t), by applying

H α,β;ψ k Da+ (⋅)

on both sides and using

α,β;ψ

γ(t) = H k Da+ y(t). k(1−ξ)+α,k;ψ

Applying Ja+

(10.5)

(⋅) on both sides of (10.5), we obtain

k(1−ξ)+α,k;ψ

Ja+

k(1−ξ),k;ψ

γ(t) = Ja+

α,k;ψ H α,β;ψ k Da+ y(t).

Ja+

By employing Lemma 2.7 and Lemma 2.12, we get k(1−ξ)+α,k;ψ

Ja+

k(1−ξ),k;ψ

γ(t) = Ja+

k(1−ξ),k;ψ

y(t) − Ja+

y(a).

Then, by taking t = b, we obtain k(1−ξ),k;ψ

Ja+

k(1−ξ),k;ψ

y(b) = Ja+

y(a),

which implies that y ∈ Dom K , so that γ ∈ Img K . So k(1−ξ)+α,k;ψ

Img K = {γ ∈ Θ : Ja+

γ(b) = 0}.

Lemma 10.2. Let K be defined by (10.3). Then K is a Fredholm operator of index zero, and the linear continuous projector operators Ψ : Θ → Θ and ϒ : Ω → Ω can be written as 2 k,ψ k(1−ξ)+α,k;ψ Ψγ(t) = kΓk (β(k − α)) Ψ̄ β(k−α) (b, a)Ja+ γ(b)

and ϒ(y)(t) =

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

.

Furthermore, the operator Kϒ−1 : Img K → Ω ∩ ker ϒ can be written as −1

α,k;ψ

Kϒ (γ)(t) = Ja+

γ(t).

264 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions Proof. Obviously, for each γ ∈ Θ, Ψ2 γ = Ψγ and γ = (γ − Ψ(γ)) + Ψ(γ), where (γ − Ψ(γ)) ∈ ker Ψ = Img K . Using the fact that Img K = ker Ψ and Ψ2 = Ψ, we have Img Ψ ∩ Img K = 0. So, Θ = Img K ⊕ Img Ψ. In the same way, we get Img ϒ = ker K and ϒ2 = ϒ. It follows for each y ∈ Ω that if y = (y − ϒ(y)) + ϒ(y), then Ω = ker ϒ + ker K . Clearly, we have ker ϒ ∩ ker K = 0. Thus, Ω = ker ϒ ⊕ ker K . Therefore, dim ker K = dim Img Ψ = codim Img K . Consequently, K is a Fredholm operator of index zero. Now, we will show that the inverse of K |Dom K ∩ker ϒ is Kϒ−1 . Effectively, for γ ∈ Img K , by Lemma 2.13 we have H

−1

α,β;ψ

k(1−ξ),k;ψ

K Kϒ (γ) = k Da+ (Ja+

γ) = γ.

(10.6)

Furthermore, for y ∈ Dom K ∩ ker ϒ we get −1

k(1−ξ),k;ψ H α,β;ψ (k Da+ y(t))

Kϒ (K (y(t))) = Ja+

= y(t) −

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

.

Since y ∈ Dom K ∩ ker ϒ, we have k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

= 0.

Thus, −1

Kϒ K (y) = y.

(10.7)

Using (10.6) and (10.7), we get Kϒ−1 = (K |Dom K ∩ker ϒ )−1 . Now, we define S : Ω → Θ by H

α,β;ψ

S y(t) := f (t, y(t), (k Da+ y)(t)),

t ∈ (a, b].

The operator S is well defined, because f is a continuous function. We can remark that the problem (10.1)–(10.2) is equivalent to the problem K y = S y.

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

� 265

Theorem 10.1. Assume that the following hypotheses are satisfied. (10.3.1) The function f : J ̄ × ℝ × ℝ → ℝ is continuous. (10.3.2) There exist constants ζ > 0 and 1 > ϖ > 0 such that 󵄨 󵄨󵄨 󵄨󵄨f (t, y1 , γ1 ) − f (t, y2 , γ2 )󵄨󵄨󵄨 ≤ ζ |y1 − y2 | + ϖ|γ1 − γ2 | for any yi , γi ∈ ℝ and t ∈ (a, b], where i = 1, 2. (10.3.3) There exist constants ζ ̃ > 0 and ϖ̃ ≥ 0 such that 󵄨󵄨 󵄨 ̃ 1 − γ2 | 󵄨󵄨f (t, y1 , γ1 ) − f (t, y2 , γ2 )󵄨󵄨󵄨 ≥ ζ ̃ |y1 − y2 | − ϖ|γ for any yi , γi ∈ ℝ and t ∈ (a, b], where i = 1, 2. Then for any bounded open set G ⊂ Ω, the operator S is K -compact. Proof. Let G = {y ∈ Ω : ‖y‖Ω < M } be a bounded open set where M > 0. Claim 1. We demonstrate that ΨS is continuous. Let (yn )n∈ℕ be a sequence such that yn 󳨀→ y in Θ. Then for each t ∈ (a, b] we have 󵄨󵄨 󵄨 󵄨󵄨ΨS (yn )(t) − ΨS (y)(t)󵄨󵄨󵄨 2 k,ψ ⩽ kΓk (β(k − α)) Ψ̄

β(k−α)

k(1−ξ)+α,k;ψ 󵄨󵄨

(b, a)(Ja+

󵄨 󵄨󵄨S (yn )(s) − S (y)(s)󵄨󵄨󵄨)(b).

Thus, 󵄨󵄨 󵄨 󵄨󵄨ΨS (yn )(t) − ΨS (y)(t)󵄨󵄨󵄨 2 k,ψ ⩽ kΓk (β(k − α)) Ψ̄

β(k−α)

(b, a)

k(1−ξ)+α,k;ψ 󵄨󵄨 H α,β;ψ 󵄨󵄨(k Da+ yn )(s)

× (Ja+

α,β;ψ 󵄨󵄨 − (H k Da+ y)(s)󵄨󵄨)(b).

By (10.3.2) we have 󵄨󵄨 H α,β;ψ 󵄨 H α,β;ψ 󵄨󵄨(k Da+ yn )(t) − (k Da+ y)(t)󵄨󵄨󵄨 α,β;ψ 󵄨 󵄨󵄨 H α,β;ψ = 󵄨󵄨󵄨f (t, yn (t), (H k Da+ yn )(t)) − f (t, y(t), (k Da+ y)(t))󵄨󵄨 α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 H α,β;ψ ⩽ ζ 󵄨󵄨󵄨yn (t) − y(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨(H k Da+ yn )(t) − (k Da+ y)(t)󵄨󵄨. Then ζ 󵄨󵄨 󵄨󵄨 H α,β;ψ 󵄨 󵄨 H α,β;ψ 󵄨󵄨(k Da+ yn )(t) − (k Da+ y)(t)󵄨󵄨󵄨 ⩽ 󵄨y (t) − y(t)󵄨󵄨󵄨. 1−ϖ󵄨 n Consequently, we have

266 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions 󵄨 󵄨󵄨 󵄨󵄨ΨS (yn )(t) − ΨS (y)(t)󵄨󵄨󵄨 k,ψ ζkΓk (β(k − α))2 Ψ̄ β(k−α) (b, a) k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨 ⩽ (Ja+ 󵄨󵄨yn (s) − y(s)󵄨󵄨󵄨)(b) 1−ϖ k,ψ ζkΓk (β(k − α))2 Ψ̄ β(k−α) (b, a)‖yn − y‖Θ k(1−ξ)+α,k;ψ 󵄨󵄨 ψ 󵄨 ⩽ (Ja+ 󵄨󵄨Ψξ (s, a)󵄨󵄨󵄨)(b). 1−ϖ Using Lemma 2.7, we get 󵄨󵄨 󵄨 󵄨󵄨ΨS (yn )(t) − ΨS (y)(t)󵄨󵄨󵄨 k,ψ ζkΓk (β(k − α))2 Ψ̄ β(k−α) (b, a)‖yn − y‖Θ ⩽ 1−ϖ k(1−ξ)+α,k;ψ

× (Ja+

ψ

−1

(Ψξ (s, a)) )(b)

ζ Γ (β(k − α))(ψ(b) − ψ(a)) ⩽ k 1−ϖ k(1−ξ)+α,k;ψ

× (Ja+ ⩽

ψ

β(k−α) −1 k

‖yn − y‖Θ

−1

(Ψξ (s, a)) )(b)

ζ Γk (kξ − α)Γk (kξ) ψ

(1 − ϖ)Ψξ (b, a)Γk (k + α)

‖yn − y‖Θ .

Thus, for each t ∈ J,̄ we obtain 󵄨󵄨 ψ 󵄨 ζ Γ (kξ − α)Γk (kξ) ‖y − y‖Θ . 󵄨󵄨Ψξ (t, a)(ΨS (yn )(t) − ΨS (y)(t))󵄨󵄨󵄨 ⩽ k (1 − ϖ)Γk (k + α) n Then for all t ∈ J,̄ we get 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)(ΨS (yn )(t) − ΨS (y)(t))󵄨󵄨󵄨 󳨀→ 0

as n 󳨀→ +∞.

Therefore, 󵄩󵄩 󵄩 󵄩󵄩ΨS (yn ) − ΨS (y)󵄩󵄩󵄩Θ 󳨀→ 0

as n 󳨀→ +∞.

We conclude that ΨS is continuous. Claim 2. We prove that ΨS (G) is bounded. For t ∈ (a, b] and y ∈ G, we have 2 k,ψ k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨 S (y)(b). 󵄨󵄨ΨS (y)(t)󵄨󵄨󵄨 ⩽ kΓk (β(k − α)) Ψ̄ β(k−α) (b, a)Ja+

By (10.3.2) we have 󵄨 󵄨󵄨 H α,β;ψ 󵄨 󵄨 H α,β;ψ 󵄨󵄨(k Da+ y)(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y(t), (k Da+ y)(t)) − f (t, 0, 0) + f (t, 0, 0)󵄨󵄨󵄨

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

� 267

α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ ζ 󵄨󵄨󵄨y(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨(H k Da+ y)(t)󵄨󵄨 + 󵄨󵄨f (t, 0, 0)󵄨󵄨.

Thus, ζ 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 H α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨(k Da+ y)(t)󵄨󵄨󵄨 ⩽ 󵄨y(t)󵄨󵄨󵄨 + 󵄨f (t, 0, 0)󵄨󵄨󵄨. 1−ϖ󵄨 1−ϖ󵄨 Then for t ∈ (a, b] and y ∈ G we have 󵄨󵄨 󵄨 󵄨󵄨ΨS (y)(t)󵄨󵄨󵄨 ζ 2 k,ψ k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨 kΓ (β(k − α)) Ψ̄ β(k−α) (b, a)(Ja+ ⩽ 󵄨󵄨y(s)󵄨󵄨󵄨)(b) 1−ϖ k 1 2 k,ψ k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨 + kΓ (β(k − α)) Ψ̄ β(k−α) (b, a)(Ja+ 󵄨󵄨f (s, 0, 0)󵄨󵄨󵄨)(b) 1−ϖ k ζ ‖y‖Ω + f ∗ −1 2 k,ψ ψ k(1−ξ)+α,k;ψ (Ψξ (s, a)) )(b), ⩽ kΓk (β(k − α)) Ψ̄ β(k−α) (b, a)(Ja+ 1−ϖ ψ

where f ∗ = supt∈J ̄ |Ψξ (t, a)f (s, 0, 0)|. Using Lemma 2.7, we get ∗ 󵄨󵄨 󵄨 (ζ ‖y‖Ω + f )Γk (kξ − α)Γk (kξ) 󵄨󵄨ΨS (y)(t)󵄨󵄨󵄨 ⩽ ψ (1 − ϖ)Ψξ (b, a)Γk (k + α)

⩽

(ζ M + f ∗ )Γk (kξ − α)Γk (kξ) ψ

(1 − ϖ)Ψξ (b, a)Γk (k + α)

.

Thus, (ζ M + f ∗ )Γk (kξ − α)Γk (kξ) 󵄩󵄩 󵄩 . 󵄩󵄩ΨS (y)󵄩󵄩󵄩Θ ⩽ (1 − ϖ)Γk (k + α) Then ΨS (G) is a bounded set in Θ. Claim 3. We show that Kϒ−1 (id − Ψ)S : G → Ω is completely continuous. In order to employ the Arzelà–Ascoli theorem, we will demonstrate that Kϒ−1 (id − Ψ)S (G) ⊂ Ω is equicontinuous and bounded. For any y ∈ G and t ∈ (a, b], we have −1

Kϒ (S y(t) − ΨS y(t)) α,k;ψ

= Ja+

α,β;ψ

[f (t, y(t), (H k Da+ y)(t))

2 k,ψ k(1−ξ)+α,k;ψ − kΓk (β(k − α)) Ψ̄ β(k−α) (b, a)Ja+ S (y)(b)] α,k;ψ

= (Ja+

α,β;ψ

f (s, y(s), (H k Da+ y)(s)))(t)

2 k,ψ k,ψ k(1−ξ)+α,k;ψ − k 2 Γk (β(k − α)) Ψ̄ β(k−α) (b, a)Ψ̄ α+k (t, a)Ja+ S (y)(b).

268 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions For all y ∈ G, using the same approach as in Claim 2, we can get for each t ∈ (a, b] 󵄨 󵄨󵄨 −1 󵄨󵄨Kϒ (id − Ψ)S y(t)󵄨󵄨󵄨 ζ 1 α,k;ψ 󵄨󵄨 α,k;ψ 󵄨󵄨 󵄨 󵄨 ⩽ J J 󵄨y(t)󵄨󵄨󵄨 + 󵄨f (t, 0, 0)󵄨󵄨󵄨 1 − ϖ a+ 󵄨 1 − ϖ a+ 󵄨 ζk 2 2 k,ψ k,ψ k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨 + Γ (β(k − α)) Ψ̄ β(k−α) (b, a)Ψ̄ α+k (b, a)(Ja+ 󵄨󵄨y(s)󵄨󵄨󵄨)(b) 1−ϖ k k2 2 k,ψ k,ψ Γ (β(k − α)) Ψ̄ β(k−α) (b, a)Ψ̄ α+k (b, a) + 1−ϖ k k(1−ξ)+α,k;ψ 󵄨󵄨 󵄨󵄨 × (Ja+ 󵄨󵄨f (s, 0, 0)󵄨󵄨)(b) ζ ‖y‖Ω + f ∗ −1 α,k;ψ ψ ⩽ [(Ja+ (Ψξ (s, a)) )(t) 1−ϖ 2 k,ψ k,ψ + k 2 Γk (β(k − α)) Ψ̄ (b, a)Ψ̄ (b, a) ×

α+k β(k−α) −1 ψ k(1−ξ)+α,k;ψ (Ψξ (s, a)) )(b)]. (Ja+

Using Lemma 2.7, we get ∗ Γ (kξ) −1 ψ 󵄨󵄨 −1 󵄨 ζ ‖y‖Ω + f [ k (Ψ α (t, a)) 󵄨󵄨Kϒ (id − Ψ)S y(t)󵄨󵄨󵄨 ⩽ 1−ϖ Γk (α + kξ) ξ+ k k,ψ k Ψ̄ α+k (b, a)Γk (kξ − α)Γk (kξ) + ]. ψ Ψξ (b, a)Γk (k + α)

So ∗ α Γ (kξ) 󵄨󵄨 ψ 󵄨 ζ ‖y‖Ω + f −1 [ k (ψ(t) − ψ(a)) k 󵄨󵄨Ψξ (t, a)Kϒ (id − Ψ)S y(t)󵄨󵄨󵄨 ⩽ 1−ϖ Γk (α + kξ) k,ψ ψ ̄ k Ψα+k (b, a)Γk (kξ − α)Γk (kξ)Ψξ (t, a) ]. + ψ Ψξ (b, a)Γk (k + α)

Therefore, 󵄩󵄩 −1 󵄩 󵄩󵄩Kϒ (id − Ψ)S y󵄩󵄩󵄩Ω α Γ (kξ) ζ ‖y‖Ω + f ∗ [ k (ψ(b) − ψ(a)) k ⩽ 1−ϖ Γk (α + kξ) k,ψ ̄ k Ψα+k (b, a)Γk (kξ − α)Γk (kξ) + ]. Γk (k + α) This means that Kϒ−1 (id − Ψ)S (G) is uniformly bounded in Ω. Now, we demonstrate that Kϒ−1 (id − Ψ)S (G) is equicontinuous.

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

Let τ1 , τ2 ∈ (a, b], τ1 < τ2 , and y ∈ G. Then ψ 󵄨󵄨 ψ 󵄨 −1 −1 󵄨󵄨Ψξ (τ1 , a)Kϒ (id − Ψ)S y(τ1 ) − Ψξ (τ2 , a)Kϒ (id − Ψ)S y(τ2 )󵄨󵄨󵄨 α,β;ψ α,k;ψ 󵄨 󵄨 ψ 󵄨󵄨 ≤ 󵄨󵄨󵄨Ψξ (τ1 , a)(Ja+ 󵄨󵄨󵄨f (s, y(s), (H k Da+ y)(s))󵄨󵄨)(τ1 ) α,β;ψ ψ α,k;ψ 󵄨 󵄨󵄨 󵄨󵄨 − Ψξ (τ2 , a)(Ja+ 󵄨󵄨󵄨f (s, y(s), (H k Da+ y)(s))󵄨󵄨)(τ2 )󵄨󵄨 −1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β(k−α) (τ1 , a)) − (Ψ̄ β(k−α) (τ2 , a)) 󵄨󵄨󵄨 kΓ (β(k − α)) ̄ k,ψ k(1−ξ)+α,k;ψ S (y)(b) Ψβ(k−α) (b, a)Ja+ × k Γk (α + k) τ1

ψ 󵄨 ψ 󵄨󵄨 ̄ k,ψ ≤ ∫󵄨󵄨󵄨Ψξ (τ1 , a)Ψ̄ k,ψ α (τ1 , s) − Ψξ (τ2 , a)Ψα (τ2 , s)󵄨󵄨 a

α,β;ψ 󵄨 󵄨󵄨 × 󵄨󵄨󵄨ψ′ (s)f (s, y(s), (H k Da+ y)(s))󵄨󵄨ds α,β;ψ α,k;ψ 󵄨 󵄨 ψ 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨Ψξ (τ2 , a)(Jτ + 󵄨󵄨󵄨f (s, y(s), (H k Da+ y)(s))󵄨󵄨)(τ2 )󵄨󵄨 1 −1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β(k−α) (τ1 , a)) − (Ψ̄ β(k−α) (τ2 , a)) 󵄨󵄨󵄨 kΓ (β(k − α)) ̄ k,ψ k(1−ξ)+α,k;ψ Ψβ(k−α) (b, a)Ja+ × k S (y)(b). Γk (α + k)

By Lemma 2.7, we get ψ 󵄨󵄨 ψ 󵄨 −1 −1 󵄨󵄨Ψξ (τ1 , a)Kϒ (id − Ψ)S y(τ1 ) − Ψξ (τ2 , a)Kϒ (id − Ψ)S y(τ2 )󵄨󵄨󵄨 τ1

ζ M + f ∗ 󵄨󵄨 ψ ψ 󵄨󵄨 ̄ k,ψ ≤ ∫󵄨󵄨Ψξ (τ1 , a)Ψ̄ k,ψ α (τ1 , s) − Ψξ (τ2 , a)Ψα (τ2 , s)󵄨󵄨 1−ϖ a

−1 󵄨 ψ 󵄨 × 󵄨󵄨󵄨ψ′ (s)(Ψξ (s, a)) 󵄨󵄨󵄨ds ζ M + f ∗ 󵄨󵄨 ψ −1 α,k;ψ ψ 󵄨 + 󵄨Ψ (τ , a)(Jτ + (Ψξ (s, a)) )(τ2 )󵄨󵄨󵄨 1 1−ϖ 󵄨 ξ 2 −1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β(k−α) (τ1 , a)) − (Ψ̄ β(k−α) (τ2 , a)) 󵄨󵄨󵄨 kΓ (β(k − α)) ̄ k,ψ k(1−ξ)+α,k;ψ Ψβ(k−α) (b, a)Ja+ S (y)(b). × k Γk (α + k)

Thus, ψ 󵄨󵄨 ψ 󵄨 −1 −1 󵄨󵄨Ψξ (τ1 , a)Kϒ (id − Ψ)S y(τ1 ) − Ψξ (τ2 , a)Kϒ (id − Ψ)S y(τ2 )󵄨󵄨󵄨 τ1

≤

ζ M + f ∗ 󵄨󵄨 ψ ψ 󵄨󵄨 ̄ k,ψ ∫󵄨󵄨Ψξ (τ1 , a)Ψ̄ k,ψ α (τ1 , s) − Ψξ (τ2 , a)Ψα (τ2 , s)󵄨󵄨 1−ϖ a

−1 󵄨 ψ 󵄨 × 󵄨󵄨󵄨ψ′ (s)(Ψξ (s, a)) 󵄨󵄨󵄨ds α (ζ M + f ∗ )Γk (kξ) ψ + Ψξ (τ2 , a)(ψ(τ2 ) − ψ(τ1 )) k (1 − ϖ)Γk (α + kξ)

� 269

270 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions −1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β(k−α) (τ1 , a)) − (Ψ̄ β(k−α) (τ2 , a)) 󵄨󵄨󵄨

×

kΓk (β(k − α)) ̄ k,ψ k(1−ξ)+α,k;ψ Ψβ(k−α) (b, a)Ja+ S (y)(b). Γk (α + k)

As τ1 → τ2 , the right-hand side of the above inequality tends to zero. This exhibits the equicontinuity of the operator Kϒ−1 (id − Ψ)S (G) in Ω. By the Arzelà–Ascoli theorem, we deduce that Kϒ−1 (id − Ψ)S (G) is relatively compact in Ω. From Claim 1 to Claim 3, we conclude that S is K -compact in G. Lemma 10.3. Suppose that the hypotheses (10.3.1) and (10.3.2) are satisfied. If r :=

α Γ (kξ) 1 ζ [ k (ψ(b) − ψ(a)) k ] < , (1 − ϖ) Γk (α + kξ) 2

(10.8)

then there exists H > 0 independent of ε, where K (y) − S (y) = −ε[K (y) + S (−y)] 󳨐⇒ ‖y‖Ω ⩽ H ,

ε ∈ (0, 1].

Proof. Let y ∈ Ω satisfy K (y) − S (y) = −εK (y) − εS (−y).

Then K (y) =

1 ε S (y) − S (−y). 1+ε 1+ε

Then for any t ∈ (a, b] we get H

α,β;ψ

K y(t) = k Da+ y(t) =

1 α,β;ψ f (t, y(t), (H k Da+ y)(t)) 1+ε ε α,β;ψ − f (t, −y(t), (H − y)(t)). k Da+ 1+ε

By Theorem 2.12, we obtain y(t) =

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

−

+

1 α,β;ψ α,k;ψ (Ja+ f (s, y(s), (H k Da+ y)(s)))(t) 1+ε

ε α,β;ψ α,k;ψ (Ja+ f (s, −y(s), (H − y)(s)))(t). k Da+ 1+ε

Using Lemma 2.7, we have for each t ∈ (a, b] k(1−ξ),k;ψ

y(a)| 2ζ ‖y‖Ω + 2f ∗ −1 α,k;ψ ψ 󵄨󵄨 󵄨 |Ja+ + [(Ja+ (Ψξ (s, a)) )(t)]. 󵄨󵄨y(t)󵄨󵄨󵄨 ≤ ψ (1 − ϖ)(1 + ε) Ψξ (t, a)Γk (kξ) Thus,

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

� 271

k(1−ξ),k;ψ

‖y‖Ω ≤

α |Ja+ y(a)| 2ζ ‖y‖Ω + 2f ∗ Γ (kξ) + [ k (ψ(b) − ψ(a)) k ]. Γk (kξ) (1 − ϖ)(1 + ε) Γk (α + kξ)

We deduce that k(1−ξ),k;ψ

‖y‖Ω ⩽

|Ja+ y(a)| Γk (kξ)

1−

α Γ (kξ) 2f ∗ [ k (ψ(b) − ψ(a)) k ] (1−ϖ)(1+ε) Γk (α+kξ) α Γ (kξ) 2ζ [ k (ψ(b) − ψ(a)) k ] (1−ϖ)(1+ε) Γk (α+kξ)

+

:= H . Lemma 10.4. If (10.3.1), (10.3.2), and (10.8) are met, then there exists a bounded open set G ⊂ Ω with K (y) − S (y) ≠ −ε[K (y) + S (−y)]

for any y ∈ 𝜕G and any ε ∈ (0, 1]. Proof. By employing Lemma 10.3, we deduce the existence of H > 0 independent of ε, where if y satisfies K (y) − S (y) = −ε[K (y) + S (−y)],

ε ∈ (0, 1],

then ‖y‖Ω ⩽ H . So, if G = {y ∈ Ω; ‖y‖Ω < ϱ}

(10.9)

such that ϱ > H , we deduce that K (y) − S (y) ≠ −ε[K (y) − S (−y)]

for all y ∈ 𝜕G = {y ∈ Ω; ‖y‖Ω = ϱ} and ε ∈ (0, 1]. Theorem 10.2. Suppose that the hypotheses (10.3.1) and (10.3.2) and the condition (10.8) are met. Then there exists at least one solution for (10.1)–(10.2). Proof. Clearly, the set G defined in (10.9) is symmetric, 0 ∈ G, and Ω ∩ G = G ≠ 0. Also, Lemma 10.4 implies that K (y) − S (y) ≠ −ε[K (y) − S (−y)]

for each y ∈ Ω ∩ 𝜕G = 𝜕G and each ε ∈ (0, 1]. Thus, problem (10.1)–(10.2) has at least one solution on Dom K ∩ G. Let us now investigate the uniqueness result for our problem (10.1)–(10.2).

272 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions Theorem 10.3. Assume that the hypotheses (10.3.1), (10.3.2), and (10.3.3) are met. If the condition (10.8) is satisfied, then the problem (10.1)–(10.2) has a unique solution in Dom K ∩ G. Proof. Firstly, Theorem 10.2 implies that the problem (10.1)–(10.2) has at least one solution in Dom K ∩ G. Let us demonstrate the uniqueness result. Assume that the problem (10.1)–(10.2) has two different solutions y1 , y2 ∈ Dom K ∩ G. Then we have for each t ∈ (a, b] α,β;ψ

α,β;ψ

α,β;ψ

α,β;ψ

H (H k Da+ y1 )(t) = f (t, y1 (t), (k Da+ y1 )(t)),

{

H (H k Da+ y2 )(t) = f (t, y2 (t), (k Da+ y2 )(t))

and k(1−ξ),k;ψ

k(1−ξ),k;ψ

(J y1 )(a+ ) = (Ja+ y1 )(b), { a+ k(1−ξ),k;ψ k(1−ξ),k;ψ + (Ja+ y2 )(a ) = (Ja+ y2 )(b). Let Λ(t) = y1 (t) − y2 (t) for all t ∈ (a, b]. Then H

α,β;ψ

K Λ(t) = k Da+ Λ(t) α,β;ψ

α,β;ψ

H =H k Da+ y1 (t) − k Da+ y2 (t) α,β;ψ

α,β;ψ

H = f (t, y1 (t), (H k Da+ y1 )(t)) − f (t, y2 (t), (k Da+ y2 )(t)).

(10.10)

Since Img K = ker Ψ, we have k(1−ξ)+α,k;ψ

(Ja+

α,β;ψ

[f (s, y1 (s), (H k Da+ y1 )(s)) α,β;ψ

− f (s, y2 (s), (H k Da+ y2 )(s))])(b) = 0.

Since f is a continuous function, there exists τ0 ∈ (a, b] such that α,β;ψ

α,β;ψ

H f (τ0 , y1 (τ0 ), (H k Da+ y1 )(τ0 )) − f (τ0 , y2 (τ0 ), (k Da+ y2 )(τ0 )) = 0.

By (10.3.3), we have 󵄨󵄨 󵄨 󵄨󵄨y1 (τ0 ) − y2 (τ0 )󵄨󵄨󵄨 ⩽ 0. Then 󵄨󵄨 󵄨 󵄨󵄨Λ(τ0 )󵄨󵄨󵄨 = 0. On the other hand, by Theorem 2.12, we have

(10.11)

10.2 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions

α,k;ψ H α,β;ψ k Da+ y)(t)

(Ja+

= y(t) −

k(1−ξ),k;ψ

Ja+

y(a)

ψ Ψξ (t, a)Γk (kξ)

� 273

,

which implies that k(1−ξ),k;ψ

Ja+

α,k;ψ H α,β;ψ ψ k Da+ y)(τ0 )]Ψξ (τ0 , a)Γk (kξ),

y(a) = [y(τ0 ) − (Ja+

and therefore α,k;ψ H α,β;ψ k Da+ Λ)(t)

Λ(t) = (Ja+

−1 α,k;ψ H α,β;ψ ψ ψ k Da+ Λ)(τ0 )]Ψξ (τ0 , a)[Ψξ (t, a)] .

+ [y(τ0 ) − (Ja+

Using (10.11), we obtain for every t ∈ (a, b] α,k;ψ H α,β;ψ 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨Λ(t)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨(Ja+ k Da+ Λ)(t)󵄨󵄨󵄨 −1 α,β;ψ α,k;ψ ψ 󵄨 󵄨 󵄨 󵄨󵄨 ψ + [󵄨󵄨󵄨y(τ0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨(Ja+ H k Da+ Λ)(τ0 )󵄨󵄨]Ψξ (τ0 , a)[Ψξ (t, a)] α

⩽

Γk (kξ)(ψ(τ0 ) − ψ(a)) k ψ −1 󵄩 α,β;ψ 󵄩 󵄩 [Ψξ (t, a)] 󵄩󵄩󵄩H k Da+ Λ󵄩 󵄩Ω Γk (α + kξ) Γk (kξ) −1 󵄩 α,β;ψ 󵄩 ψ 󵄩 . (Ψξ+ α (t, a)) 󵄩󵄩󵄩H + k Da+ Λ󵄩 󵄩Ω Γk (α + kξ) k

(10.12)

By (10.3.2) and (10.10), we get 󵄨󵄨 H α,β;ψ 󵄨 󵄨 H α,β;ψ 󵄨󵄨(k Da+ Λ)(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t, y1 (t), (k Da+ y1 )(t))

α,β;ψ 󵄨󵄨 − f (t, y2 (t), (H k Da+ y2 )(t))󵄨󵄨 α,β;ψ 󵄨 󵄨 󵄨 󵄨󵄨 ⩽ ζ 󵄨󵄨󵄨Λ(t)󵄨󵄨󵄨 + ϖ 󵄨󵄨󵄨(H k Da+ Λ)(t)󵄨󵄨.

Thus, ζ 󵄨󵄨 󵄨 󵄨󵄨 H α,β;ψ 󵄨 󵄨Λ(t)󵄨󵄨󵄨. 󵄨󵄨(k Da+ Λ)(t)󵄨󵄨󵄨 ⩽ 1−ϖ󵄨 Then 󵄩󵄩 H α,β;ψ 󵄩 󵄩󵄩(k Da+ Λ)(t)󵄩󵄩󵄩Ω ⩽

ζ ‖Λ‖Ω . 1−ϖ

By substituting (10.13) in the right side of (10.12), we obtain for every t ∈ (a, b] α

Γ (kξ)(ψ(τ0 ) − ψ(a)) k ψ ζ −1 󵄨󵄨 󵄨 [ k [Ψξ (t, a)] 󵄨󵄨Λ(t)󵄨󵄨󵄨 ⩽ 1−ϖ Γk (α + kξ) +

Γk (kξ) −1 ψ (Ψξ+ α (t, a)) ]‖Λ‖Ω . Γk (α + kξ) k

(10.13)

274 � 10 k-Generalized ψ-Hilfer fractional differential equations with periodic conditions Therefore, ‖Λ‖Ω ⩽

α Γ (kξ) ζ [ k (ψ(b) − ψ(a)) k ]‖Λ‖Ω . (1 − ϖ) Γk (α + kξ)

Hence, by (10.8), we conclude that ‖Λ‖Ω ≤ 2r‖Λ‖Ω . Thus, ‖Λ‖Ω = 0. As a result, for any t ∈ (a, b] we get Λ(t) = 0 󳨐⇒ y1 (t) = y2 (t). This completes the proof. 10.2.2 An example In this section, we provide an illustration to our results. This example can be considered as a specific case of our problem (10.1)–(10.2). We take α → 21 , β = 21 , k = 1, ψ(t) = t, J ̄ = [1, 2], and ξ = 43 . We obtain the following periodic problem with Hilfer fractional derivative: 1 1

1 1

, ;ψ

1 1

,

,

2 2 {(H y)(t) = (H 𝔻12+ 2 y)(t) = f (t, y(t), (H 𝔻12+ 2 y)(t)), 1 D1+ 1 1 { ,1;ψ ,1;ψ + 2 2 {(J1+ y)(1 ) = (J1+ y)(2),

t ∈ (1, 2],

(10.14)

where f (t, y, γ) =

arctan(t) (156 + 22et+7 )(1 + |y| + |γ|)

and t ∈ J,̄ y, γ ∈ ℝ. We have 1

Cξ;ψ (J)̄ = C 3 ;ψ (J)̄ = {y : (1, 2] → ℝ : (t − 1) 4 y ∈ C(J,̄ ℝ)}. 4

Since it is clear that the function f is continuous, the condition (10.3.1) is satisfied. Further, for each y1 , γ1 , y2 , γ2 ∈ ℝ and t ∈ J,̄ we have 󵄨󵄨 󵄨 󵄨󵄨f (t, y1 , γ1 ) − f2 (t, y2 , γ2 )󵄨󵄨󵄨 ≤ and

π (|y − y2 | + |γ1 − γ2 |) 156 + 22et+7 1

10.3 Notes and remarks

� 275

π 󵄨 󵄨󵄨 (|y − y2 | − |γ1 − γ2 |). 󵄨󵄨f (t, y1 , γ1 ) − f2 (t, y2 , γ2 )󵄨󵄨󵄨 ≥ 156 + 22et+7 1 Thus, the conditions (10.3.2) and (10.3.3) are satisfied with ζ =ϖ=

1 156 + 22e8

ζ ̃ = ϖ̃ =

1 . 156 + 22e9

and

Also, r :=

Γ( 43 ) (155 +

22e8 )Γ( 45 )

1 ≈ 2.05664312241225 × 10−5 < . 2

As all the conditions of Theorem 10.3 are met, the problem (10.14) has a unique solution.

10.3 Notes and remarks The results of this chapter are taken from the paper of Salim et al. [218]. More results and investigations may be found in the monographs [49, 76, 98, 115, 125, 144, 145, 161, 185, 186, 244, 245] and the articles [158, 175, 178, 208, 211, 212, 214, 216, 218, 218, 223, 224].

11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems 11.1 Introduction and motivations This chapter deals with some existence and uniqueness results for a class of nonlinear fractional coupled systems with k-generalized ψ-Hilfer fractional differential equations and periodic conditions. The arguments are based on Mawhin’s coincidence degree theory. We demonstrate several results by changing the required conditions of the theorems. Furthermore, illustrative examples are presented to demonstrate the plausibility of our results. We studied and proved the results in this chapter by considering the previous chapters’ publications as well as the publications that follow: – The papers [16, 39, 42, 130, 162, 226], where the authors investigated numerous results for systems with different types of fractional differential equations and inclusions and various conditions. – The books of Herrmann [135], Hilfer [136], Kilbas et al. [143], and Samko et al. [227]. In [14, 15], Abbas et al. studied several problems with advanced fractional differential and integral equations and presented various applications. Agrawal [31] introduced some generalizations of fractional integrals and derivatives and present some of their properties. In [203, 206], Salim et al. demonstrated the existence results for some hybrid generalized Hilfer fractional differential equations. – In the papers of Adiguzel et al. [22–24] and Afshari et al. [26–28], the authors presented the existence and uniqueness of solutions for some nonlinear fractional differential equations. Almeida et al. [43–45] studied some problems with ψ-Caputo fractional derivative and gave some of their applications. Derbazi et al. [110, 111] presented some results on the fractional hybrid differential equations with hybrid conditions. In the papers of Bouriah et al. [68, 95] and Salim et al. [63, 64, 81–83, 133, 156, 200, 201, 223], the authors addressed the existence, stability, and uniqueness of solutions for diverse problems with differential equations using different types of conditions. – Several generalizations for various special functions and fractional operators have recently been proposed, paving the way for more extended solutions to a variety of problems. Interest in these sorts of operators is rising each year. For more information, we refer the reader to the papers of Chu et al. [101], Diaz et al. [113], Kucche et al. [153], Mubeen et al. [175], Salim et al. [211, 212], and Sousa et al. [102]. – In [12], Abbas et al. considered the following coupled system with Hilfer–Hadamard fractional derivative and initial integral conditions: α,β

(H D 1 u)(t) = g1 (t, u(t), v(t)), { H ϑ1 ,β ( D1 2 2 v)(t) = g2 (t, u(t), v(t)), https://doi.org/10.1515/9783111334387-011

t ∈ [1, T],

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 277

with the following initial conditions: 1−γ1

(H I1

{

u)(1) = ψ1 ,

1−γ (H I1 2 v)(1)

= ψ2 ,

where T > 1, ϑi ∈ (0, 1), βi ∈ [0, 1], γi = ϑi + βi − ϑi βi , ψi ∈ E, gi : [1, T] × E × E → E, 1−γ i = 1, 2, are given functions, H I1 i is the left-sided mixed Hadamard integral of order ϑ ,β

–

1 − γi , and H D1 i i is the Hilfer–Hadamard fractional derivative of order ϑi and type βi , i = 1, 2. They employed a technique that relies on the concept of the measure of noncompactness and fixed point theory. Using the Banach contraction principle, in [158], Lazreg et al. consider the initial value problem with k-generalized ψ-Hilfer-type fractional differential equation: ϑ,r;ψ

(H Da+ x)(t) = f (t, x(t)), { k k(1−ξ),k;ψ x)(a+ ) = x0 , (Ja+

t ∈ (a, b],

k(1−ξ),k;ψ

ϑ,r;ψ

where H and Ja+ are the k-generalized ψ-Hilfer fractional derivative k Da+ of order ϑ ∈ (0, 1) and type r ∈ [0, 1] and the k-generalized ψ-fractional integral of order k(1 − ξ) defined in [193], respectively, where ξ = k1 (r(k − ϑ) + ϑ), θ < k, x0 ∈ ℝ, k > 0, and f ∈ C([a, b] × ℝ, ℝ).

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions Motivated by the aforementioned articles and with the goal of generalizing previous results, in this chapter, we study the existence and uniqueness results for a coupled system of nonlinear k-generalized ψ-Hilfer-type IFDEs and periodic conditions as follows: α ,β ;ψ

α ,β ;ψ

α ,β ;ψ

1 1 1 1 2 2 (H y1 )(t) = f1 (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)), k Da + k Da+ k Da + {H α2 ,β2 ;ψ H α1 ,β1 ;ψ H α2 ,β2 ;ψ (k Da+ y2 )(t) = f2 (t, y1 (t), y2 (t), (k Da+ y1 )(t), (k Da+ y2 )(t)),

k(1−ξ1 ),k;ψ

Ja+

k(1−ξ1 ),k;ψ

y1 (a) = Ja+

α ,β ;ψ

y1 (b) and

k(1−ξ2 ),k;ψ

Ja+

k(1−ξ ),k;ψ

k(1−ξ2 ),k;ψ

y2 (a) = Ja+

(11.1) y2 (b), (11.2)

i i where t ∈ (a, b] and H and Ja+ i are the k-generalized ψ-Hilfer fractional k Da + derivative of order 0 < αi ≤ k and type βi ∈ [0, 1] and the k-generalized ψ-fractional integral of order k(1 − ξi ), ξi = k1 (αi + kβi − αi βi ), i ∈ {1, 2}, respectively. Moreover, f1 , f2 : J ̄ × ℝ4 → ℝ are continuous functions.

278 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems 11.2.1 Existence results As in the previous chapter, we will suppose the following: ψ

Ψξ (t, a) = (ψ(t) − ψ(a)) (ψ(t) − ψ(s)) Ψ̄ k,ψ α (t, s) = kΓk (α)

1−ξ

,

α −1 k

.

We denote by Cξ;ψ (J,̄ ℝ2 ) = Cξ1 ;ψ (J,̄ ℝ) × Cξ2 ;ψ (J,̄ ℝ) the product weighted space with the norm ‖y‖C = max{‖yj ‖Cξ ;ψ }. 1≤j≤2

j

Also, α ,k;ψ α ,k;ψ X = {y = (y1 , y2 ) ∈ Cξ;ψ (J,̄ ℝ2 ) : (y1 , y2 ) = (Ja+1 υ1 , Ja+2 υ2 ),

where υ = (υ1 , υ2 ) ∈ Cξ;ψ (J,̄ ℝ2 )}

and Y = Cξ;ψ (J,̄ ℝ2 ), with the norms ‖y‖X = ‖y‖Y = ‖y‖C . We define the operator L : Dom L ⊆ X → Y by α ,β1 ;ψ

1 Ly = (L1 y1 , L2 y2 ) := (H k Da +

α ,β2 ;ψ

2 y1 , H k Da +

y2 ),

(11.3)

where k(1−ξ1 ),k;ψ

Dom L = {y ∈ X : Ly ∈ Y : Ja+ k(1−ξ2 ),k;ψ

and Ja+

k(1−ξ1 ),k;ψ

y1 (a) = Ja+

k(1−ξ2 ),k;ψ

y2 (a) = Ja+

y1 (b)

y2 (b)}.

To prove the main findings, we need the following lemmas. Lemma 11.1. Using the definition of L given in (11.3), we have ker L = {y = (y1 , y2 ) ∈ X : yj (t) = t ∈ (a, b], j ∈ {1, 2}}

k(1−ξj ),k;ψ

Ja+

yj (a)

Γk (kξj )

(ψ(t) − ψ(a))

ξj −1

,

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 279

and k(1−ξj )+αj ,k;ψ

Img L = {ϕ = (ϕ1 , ϕ2 ) ∈ Y : Ja+

ϕj (b) = 0, j ∈ {1, 2}}.

Proof. Using Lemma 2.24 and taking γ(t) = 0 for all t ∈ (a, b], for all y ∈ X and j ∈ {1, 2} α ,βj ;ψ

j the equation Lj yj = H k Da +

yj (t) =

yj = 0 in (a, b] has a solution given by

k(1−ξj ),k;ψ

Ja+

yj (a)

Γk (kξj )

ξj −1

(ψ(t) − ψ(a))

,

t ∈ (a, b],

which implies that ker L = {y = (y1 , y2 ) ∈ X : yj (t) =

k(1−ξj ),k;ψ

Ja+

yj (a)

Γk (kξj )

(ψ(t) − ψ(a))

ξj −1

,

t ∈ (a, b], j ∈ {1, 2}}. For ϕ = (ϕ1 , ϕ2 ) ∈ Img L, there exists y = (y1 , y2 ) ∈ Dom L such that (ϕ1 , ϕ2 ) = (L1 y1 , L2 y2 ) ∈ Y . Using Theorem 2.12, we obtain for each t ∈ (a, b] and j ∈ {1, 2} yj (t) =

k(1−ξj ),k;ψ

Ja+

yj (a)

Γk (kξj )

ξj −1

(ψ(t) − ψ(a))

α ,k;ψ

+ Ja+j

ϕj (t).

k(1−ξ ),k;ψ

(11.4)

yj (⋅) on both sides of (11.4), using Lemmas 2.5 and 2.7, and taking Applying Ja+ j t = b, we have for each j ∈ {1, 2} k(1−ξj ),k;ψ

Ja+

k(1−ξj ),k;ψ

yj (b) = Ja+

k(1−ξj )+αj ,k;ψ

yj (a) + Ja+

ϕj (b).

Since y ∈ Dom L, for any j ∈ {1, 2} we have k(1−ξj ),k;ψ

Ja+

k(1−ξj ),k;ψ

yj (a) = Ja+

yj (b).

Thus, k(1−ξj )+αj ,k;ψ

Ja+

ϕj (b) = 0,

j ∈ {1, 2}.

Moreover, if ϕ = (υ1 , υ2 ) ∈ Y and satisfies k(1−ξj )+αj ,k;ψ

Ja+

ϕj (b) = 0,

j ∈ {1, 2},

then for any α ,k;ψ

y(t) = (y1 (t), y2 (t)) = (Ja+1

α ,k;ψ

ϕ1 (t), Ja+2

ϕ2 (t)),

280 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems α ,βj ;ψ

j by applying H k Da +

(⋅) on both sides and using Lemma 2.13, we get α ,β1 ;ψ

1 (ϕ1 (t), ϕ2 (t)) = (H k Da +

k(1−ξj )+αj ,k;ψ

Applying Ja+ b, we obtain

α ,β2 ;ψ

2 y1 (t), H k Da+

y2 (t)) = (L1 y1 (t), L2 y2 (t)).

(⋅) on both sides, employing Lemmas 2.7 and 2.12, and taking t = k(1−ξj ),k;ψ

Ja +

k(1−ξj ),k;ψ

yj (b) = Ja+

yj (a),

j ∈ {1, 2}.

Then y ∈ Dom L, so that ϕ ∈ Img L. So k(1−ξj )+αj ,k;ψ

Img L = {ϕ = (ϕ1 , ϕ2 ) ∈ Y : Ja+

ϕj (b) = 0, j ∈ {1, 2}}.

Lemma 11.2. Let L be defined by (11.3). Then L is a Fredholm operator of index zero, and the linear continuous projector operators Q : Y → Y and P : X → X can be written as Q(ϕ) = (Q1 ϕ1 , Q2 ϕ2 ), such that k(1−ξ )+α ,k;ψ 2 k,ψ Qj ϕj (t) = kΓk (βj (k − αj )) Ψ̄ β (k−α ) (b, a)Ja+ j j ϕj (b), j

t ∈ J,̄ j ∈ {1, 2}.

j

For all t ∈ J,̄ we have P(y)(t) = (P1 (y1 )(t), P2 (y2 )(t)) k(1−ξ1 ),k;ψ

=(

Ja+

y1 (a)

Γk (kξ1 )

ψ Ψξ (t, a), 1

k(1−ξ2 ),k;ψ

Ja+

y2 (a)

Γk (kξ2 )

ψ

Ψξ (t, a)). 2

Furthermore, the operator L−1 P : Img L → X ∩ ker P can be written by α ,k;ψ

−1 −1 1 L−1 P (ϕ)(t) = (LP1 (ϕ1 )(t), LP2 (ϕ2 )(t)) = (Ja+

α ,k;ψ

ϕ1 (t), Ja+2

ϕ2 (t)),

t ∈ J.̄

Proof. Obviously, for each ϕ ∈ Y , Q2 ϕ = Qϕ and ϕ = (ϕ−Q(ϕ))+Q(ϕ), where (ϕ−Q(ϕ)) ∈ ker Q = Img L. Using the fact that Img L = ker Q and Q2 = Q, we have Img Q ∩ Img L = {0}. So, Y = Img L ⊕ Img Q. In the same way, we get Img P = ker L and P2 = P. It follows for each y ∈ X that if y = (y − P(y)) + P(y), then X = ker P + ker L. Clearly, we have ker P ∩ ker L = {0}. Thus, X = ker P ⊕ ker L.

� 281

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

Using the rank–nullity theorem, we get codim Img L = dim Y − dim Img L = [dim ker Q + dim Img Q] − dim Img L, and since Img L = ker Q, we have codim Img L = dim Img Q.

(11.5)

Using again the rank–nullity theorem, we obtain dim ker L = dim X − dim Img L = codim Img L, which implies that dim ker L = codim Img L.

(11.6)

By (11.5) and (11.6) we have dim ker L = codim Img L = dim Img Q, and since dim Img Q < ∞, we have dim ker L = codim Img L < ∞. Since Img L is a closed subset of Y , L is a Fredholm operator of index zero. We will now demonstrate that L|Dom L∩ker P is L−1 P . Effectively, for ϕ ∈ Img L, by Lemma 2.13 we have α ,β1 ;ψ

H 1 LL−1 P (ϕ) = (k Da+

k(1−ξ1 ),k;ψ

Ja +

α ,β2 ;ψ

2 ϕ1 , H k Da +

k(1−ξ2 ),k;ψ

Ja+

ϕ2 ) = (ϕ1 , ϕ2 ) = ϕ.

Furthermore, for y ∈ Dom L ∩ ker P we get L−1 P (Ly)(t)

k(1−ξ1 ),k;ψ H α1 ,β1 ;ψ k(1−ξ ),k;ψ H α2 ,β2 ;ψ y1 (t), Ja+ 2 y2 (t)) k Da + k Da + k(1−ξ1 ),k;ψ k(1−ξ2 ),k;ψ Ja+ y1 (a) Ja+ y2 (a) (y1 (t) − , y (t) − 2 Γk (kξ1 )(ψ(t) − ψ(a))1−ξ1 Γk (kξ2 )(ψ(t) − ψ(a))1−ξ2

= (Ja+ =

Using the fact that y ∈ Dom L ∩ ker P, we have k(1−ξj ),k;ψ

Ja+

Thus,

yj (a)

Γk (kξj )

ξj −1

(ψ(t) − ψ(a))

= 0,

j ∈ {1, 2}.

).

(11.7)

282 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems −1 −1 L−1 P L(y) = (LP1 L1 (y1 ), LP2 L2 (y2 )) = (y1 , y2 ) = y.

(11.8)

−1 Using (11.7) and (11.8) together, we get L−1 P = (L|Dom L∩ker P ) .

Now, we define N : X → Y by Ny(t) = (N1 y(t), N2 y(t))

:= (f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)), f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t))),

where t ∈ (a, b], and for j ∈ {1, 2}, ℧j ∈ Cξj ;ψ (J,̄ ℝ) satisfy the following system of functional equations: ℧1 (t) = f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)),

{

℧2 (t) = f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)).

The operator N is well defined, because f1 and f2 are continuous functions. We can remark that the problem (11.1)–(11.2) is equivalent to the problem Ly = Ny. Lemma 11.3. Suppose that the following hypotheses are satisfied. (11.3.1) For each j ∈ {1, 2}, the functions fj : (a, b] × ℝ4 → ℝ are such that fj (⋅, y1 (⋅), y2 (⋅), ŷ1 (⋅), ŷ2 (⋅)) ∈ Cξj ;ψ (Θ, ℝ) for all yj , ŷj ∈ Cξj ;ψ (Θ, ℝ). (11.3.2) For each j ∈ {1, 2}, there exist nonnegative functions γj , ηj , θj , ℑj ∈ C(Θ, ℝ+ ) such that 󵄨 󵄨󵄨 󵄨󵄨fj (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − fj (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ γj (t)󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 + ηj (t)󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 + θj (t)󵄨󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + ℑj (t)󵄨󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 for every t ∈ Θ, yj , yj , σj , σj ∈ Cξj ;ψ (Θ, ℝ), such that γ∗ = max ‖γj ‖∞ ,

η∗ = max ‖ηj ‖∞ ,

θ = max ‖θj ‖∞ ,

ℑ = max ‖ℑj ‖∞ ,

j∈{1,2}

∗

j∈{1,2}

j∈{1,2}

∗

j∈{1,2}

and 0 < ℑ∗ + θ∗ < 1. Then for any bounded open set Ω ⊂ X, the operator N is L-compact.

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 283

Proof. Let Ω = {y ∈ X : ‖y‖X < ℘} be a bounded open set where ℘ > 0. Step 1. We prove that QN is continuous. Let (yn )n∈ℕ be a sequence where yn 󳨀→ y in Y . Then for j ∈ {1, 2} and t ∈ (a, b], we have 󵄨 󵄨󵄨 󵄨󵄨Qj Nj (yn )(t) − Qj Nj (y)(t)󵄨󵄨󵄨 2 k,ψ ≤ kΓk (βj (k − αj )) Ψ̄

βj (k−αj )

k(1−ξj )+αj ,k;ψ 󵄨

(b, a)(Ja+

󵄨󵄨Nj (yn )(s) − Nj (y)(s)󵄨󵄨󵄨)(b). 󵄨 󵄨

Thus, 2 k,ψ 󵄨󵄨 󵄨 󵄨󵄨Qj Nj (yn )(t) − Qj Nj (y)(t)󵄨󵄨󵄨 ≤ kΓk (βj (k − αj )) Ψ̄ β (k−α ) (b, a) j j k(1−ξj )+αj ,k;ψ 󵄨 󵄨 × (Ja+ 󵄨󵄨󵄨℧j,n (s) − ℧j (s)󵄨󵄨󵄨)(b),

where for j ∈ {1, 2}, ℧j , ℧j,n ∈ Cξj ;ψ (J,̄ ℝ) satisfy the following systems of functional equations: ℧1 (t) = f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)), { ℧2 (t) = f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) and ℧1,n (t) = f1 (t, y1n (t), y2n (t), ℧1,n (t), ℧2,n (t)), { ℧2,n (t) = f2 (t, y1n (t), y2n (t), ℧1,n (t), ℧2,n (t)). By (11.3.2), for each t ∈ J,̄ we have 󵄨 󵄨󵄨 󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨f1 (t, y1n (t), y2n (t), ℧1,n (t), ℧2,n (t)) − f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t))󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ γ1 (t)󵄨󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + η1 (t)󵄨󵄨󵄨y2n (t) − y2 (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 + θ1 (t)󵄨󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 + ℑ1 (t)󵄨󵄨󵄨℧2,n (t) − ℧2 (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ γ∗ 󵄨󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2n (t) − y2 (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 + θ∗ 󵄨󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 + ℑ∗ 󵄨󵄨󵄨℧2,n (t) − ℧2 (t)󵄨󵄨󵄨, which implies that η∗ 󵄨󵄨 γ∗ 󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + 󵄨y (t) − y2 (t)󵄨󵄨󵄨 󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 ≤ ∗ 1−θ 1 − θ∗ 󵄨 2n ℑ∗ 󵄨󵄨 󵄨 + 󵄨℧ (t) − ℧2 (t)󵄨󵄨󵄨. 1 − θ∗ 󵄨 2,n Similarly, one can find that

284 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems γ∗ 󵄨󵄨 η∗ 󵄨󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨 + y (t) − y (t) 󵄨 󵄨󵄨℧2,n (t) − ℧2 (t)󵄨󵄨󵄨 ≤ 󵄨 󵄨y (t) − y2 (t)󵄨󵄨󵄨 1 󵄨 1 − ℑ∗ 󵄨 1n 1 − ℑ∗ 󵄨 2n θ∗ 󵄨󵄨 󵄨 + 󵄨℧ (t) − ℧1 (t)󵄨󵄨󵄨. 1 − ℑ∗ 󵄨 1,n Therefore, γ∗ 󵄨󵄨 η∗ 󵄨󵄨 󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 ≤ 󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + 󵄨y (t) − y2 (t)󵄨󵄨󵄨 ∗ 1−θ 1 − θ∗ 󵄨 2n ℑ∗ γ∗ 󵄨 󵄨󵄨 + 󵄨y (t) − y1 (t)󵄨󵄨󵄨 (1 − ℑ∗ )(1 − θ∗ ) 󵄨 1n ℑ∗ η∗ 󵄨 󵄨󵄨 + 󵄨y (t) − y2 (t)󵄨󵄨󵄨 (1 − ℑ∗ )(1 − θ∗ ) 󵄨 2n ℑ∗ θ∗ 󵄨󵄨 󵄨 + 󵄨℧ (t) − ℧1 (t)󵄨󵄨󵄨. (1 − ℑ∗ )(1 − θ∗ ) 󵄨 1,n Then η∗ γ∗ 󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + 󵄨y (t) − y2 (t)󵄨󵄨󵄨. 󵄨󵄨℧1,n (t) − ℧1 (t)󵄨󵄨󵄨 ≤ ∗ ∗ 1−ℑ −θ 1 − ℑ∗ − θ∗ 󵄨 2n By following the same approach, we can also obtain the following: η∗ γ∗ 󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨℧2,n (t) − ℧2 (t)󵄨󵄨󵄨 ≤ 󵄨󵄨y1n (t) − y1 (t)󵄨󵄨󵄨 + 󵄨y (t) − y2 (t)󵄨󵄨󵄨. ∗ ∗ 1−ℑ −θ 1 − ℑ∗ − θ∗ 󵄨 2n Consequently, we have 󵄨󵄨 󵄨 󵄨󵄨Q1 N1 (yn )(t) − Q1 N1 (y)(t)󵄨󵄨󵄨 ≤

k,ψ kΓk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a) 1

1

1 − ℑ∗ − θ∗ k(1−ξ1 )+α1 ,k;ψ ∗ 󵄨󵄨 󵄨 󵄨 󵄨 × (Ja+ γ 󵄨󵄨y1n (s) − y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2n (s) − y2 (s)󵄨󵄨󵄨)(b).

Using Lemma 2.7, we get 󵄨󵄨 󵄨 󵄨󵄨Q1 N1 (yn )(t) − Q1 N1 (y)(t)󵄨󵄨󵄨 k,ψ kΓk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a)‖yn − y‖Y 1 1 ≤ 1 − ℑ∗ − θ∗ k(1−ξ1 )+α1 ,k;ψ ∗

× (Ja+ ≤

ψ

γ (Ψξ (s, a)) 1

−1

β1 (k−α1 ) −1 k

Γk (β1 (k − α1 ))(ψ(b) − ψ(a)) 1 − ℑ∗ − θ∗ k(1−ξ1 )+α1 ,k;ψ ∗

× (Ja+ ≤[

ψ

γ (Ψξ (s, a)) 1

γ∗ Γk (kξ1 − α1 )Γk (kξ1 ) ψ

−1

−1

2

‖yn − y‖Y ψ

+ η∗ (Ψξ (s, b)) )(b)

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k + α1 ) 1

ψ

+ η∗ (Ψξ (s, b)) )(b)

2

−1

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

+

η∗ Γk (kξ1 − α1 )Γk (kξ2 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ2 − ξ1 ) + α1 )

� 285

]‖yn − y‖Y .

2

Thus, for each t ∈ J, we obtain 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)(Q1 N1 (yn )(t) − Q1 N1 (y)(t))󵄨󵄨󵄨 1 ≤[

γ∗ Γk (kξ1 − α1 )Γk (kξ1 ) (1 − ℑ∗ − θ∗ )Γk (k + α1 )

+

η∗ Γk (kξ1 − α1 )Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 ]‖yn − y‖Y (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ2 − ξ1 ) + α1 )

≤ ϒ1 ‖yn − y‖Y , where ϒ1 =

γ∗ Γk (kξ1 − α1 )Γk (kξ1 ) η∗ Γk (kξ1 − α1 )Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + . (1 − ℑ∗ − θ∗ )Γk (k + α1 ) (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ2 − ξ1 ) + α1 )

Similarly, we may obtain 󵄨󵄨 󵄨 󵄨󵄨Q2 N2 (yn )(t) − Q2 N2 (y)(t)󵄨󵄨󵄨 k,ψ kΓk (β2 (k − α2 ))2 Ψ̄ β (k−α ) (b, a) 2 2 ≤ 1 − ℑ∗ − θ∗ k(1−ξ )+α ,k;ψ 󵄨 󵄨 󵄨 󵄨 × (Ja+ 2 2 γ∗ 󵄨󵄨󵄨y1n (s) − y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2n (s) − y2 (s)󵄨󵄨󵄨)(b) k,ψ kΓk (β2 (k − α2 ))2 Ψ̄ β (k−α ) (b, a)‖yn − y‖Y 2 2 ≤ 1 − ℑ∗ − θ∗ k(1−ξ2 )+α2 ,k;ψ ∗

≤[

ψ

γ (Ψξ (s, a))

× (Ja+

1

η∗ Γk (kξ2 − α2 )Γk (kξ2 )

−1

ψ

+ η∗ (Ψξ (s, b)) )(b) −1

2

ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k + α2 ) 2

+

γ∗ Γk (kξ2 − α1 )Γk (kξ1 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ1 − ξ2 ) + α2 )

]‖yn − y‖Y .

1

Thus, for each t ∈ J, we obtain 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)(Q2 N2 (yn )(t) − Q2 N2 (y)(t))󵄨󵄨󵄨 2 ≤[

η∗ Γk (kξ2 − α2 )Γk (kξ2 ) (1 − ℑ∗ − θ∗ )Γk (k + α2 )

+

γ∗ Γk (kξ2 − α2 )Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 ]‖yn − y‖Y (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ1 − ξ2 ) + α2 )

≤ ϒ2 ‖yn − y‖Y ,

286 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems where ϒ2 =

η∗ Γk (kξ2 − α2 )Γk (kξ2 ) γ∗ Γk (kξ2 − α2 )Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 + . (1 − ℑ∗ − θ∗ )Γk (k + α2 ) (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ1 − ξ2 ) + α2 )

Thus, for each j ∈ {1, 2}, we get 󵄩 󵄩󵄩 󵄩󵄩Qj Nj (yn ) − Qj Nj (y)󵄩󵄩󵄩Cξ ;ψ ≤ max{ϒ1 , ϒ2 }‖yn − y‖Y . j Therefore, 󵄩󵄩 󵄩 󵄩󵄩QN(yn ) − QN(y)󵄩󵄩󵄩Y 󳨀→ 0

as n 󳨀→ +∞.

We deduce that QN is continuous. Step 2. We show that QN(Ω) is bounded. For t ∈ (a, b], j ∈ {1, 2}, and y ∈ Ω, we have k(1−ξj )+αj ,k;ψ 󵄨 2 k,ψ 󵄨󵄨 󵄨 󵄨󵄨Nj (y)(s)󵄨󵄨󵄨)(b). 󵄨󵄨Qj Nj (y)(t)󵄨󵄨󵄨 ≤ kΓk (βj (k − αj )) Ψ̄ β (k−α ) (b, a)(Ja+ 󵄨 󵄨 j j

By (11.3.2), for each t ∈ J,̄ we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨℧j (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨fj (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) − fj (t, 0, 0, 0, 0)󵄨󵄨󵄨 + 󵄨󵄨󵄨fj (t, 0, 0, 0, 0)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ γj (t)󵄨󵄨󵄨y1 (t)󵄨󵄨󵄨 + ηj (t)󵄨󵄨󵄨y2 (t)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + θj (t)󵄨󵄨󵄨℧1 (t)󵄨󵄨󵄨 + ℑj (t)󵄨󵄨󵄨℧2 (t)󵄨󵄨󵄨 + 󵄨󵄨󵄨fj (t, 0, 0, 0, 0)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ γ∗ 󵄨󵄨󵄨y1 (t)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (t)󵄨󵄨󵄨 + θ∗ 󵄨󵄨󵄨℧1 (t)󵄨󵄨󵄨 + ℑ∗ 󵄨󵄨󵄨℧2 (t)󵄨󵄨󵄨 + f ∗ , which implies that γ∗ η∗ f∗ 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 y (t) + y (t) + , 󵄨󵄨℧j (t)󵄨󵄨󵄨 ≤ 󵄨 󵄨 󵄨 󵄨 1 2 󵄨 󵄨 󵄨 󵄨 1 − ℑ∗ − θ∗ 1 − ℑ∗ − θ∗ 1 − ℑ∗ − θ∗ where 󵄨 󵄨 󵄨 󵄨 f ∗ = max{sup󵄨󵄨󵄨f1 (t, 0, 0, 0, 0)󵄨󵄨󵄨, sup󵄨󵄨󵄨f2 (t, 0, 0, 0, 0)󵄨󵄨󵄨}. t∈J ̄

t∈J ̄

Thus, for any t ∈ (a, b] and y ∈ Ω, we have 2 ̄ k,ψ 󵄨󵄨 󵄨󵄨 kΓk (β1 (k − α1 )) Ψβ1 (k−α1 ) (b, a) 󵄨󵄨Q1 N1 (y)(t)󵄨󵄨 ≤ 1 − ℑ∗ − θ∗ k(1−ξ1 )+α1 ,k;ψ ∗ 󵄨󵄨 󵄨 󵄨 󵄨 × (Ja+ γ 󵄨󵄨y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (s)󵄨󵄨󵄨 + f ∗ )(b) k,ψ kΓk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a) 1 1 ≤ 1 − ℑ∗ − θ∗

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

k(1−ξ1 )+α1 ,k;ψ ∗

ψ

γ (Ψξ (s, a))

× [‖y‖Y (Ja+

−1

1

k(1−ξ1 )+α1 ,k;ψ ∗

ψ

+ η∗ (Ψξ (s, b)) )(b) −1

2

f )(b)].

+ (Ja+ Using Lemma 2.7, we get 󵄨󵄨 󵄨 󵄨󵄨Q1 N1 (y)(t)󵄨󵄨󵄨 ≤ [

γ∗ Γk (kξ1 − α1 )Γk (kξ1 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k + α1 ) 1

+

η∗ Γk (kξ1 − α1 )Γk (kξ2 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ2 − ξ1 ) + α1 )

]‖y‖Y

2

+ ≤

f ∗ Γk (β1 (k − α1 )) (1 − ℑ∗ − θ∗ )Γk (k(1 − ξ1 ) + α1 + k) ℘γ∗ Γk (kξ1 − α1 )Γk (kξ1 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k + α1 ) 1

+

℘η∗ Γk (kξ1 − α1 )Γk (kξ2 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ2 − ξ1 ) + α1 ) 2

f ∗ Γk (β1 (k − α1 )) + . (1 − ℑ∗ − θ∗ )Γk (k(1 − ξ1 ) + α1 + k) Thus, ∗ 󵄨󵄨 ψ 󵄨 ℘γ Γk (kξ1 − α1 )Γk (kξ1 ) 󵄨󵄨Ψξ (t, a)Q1 N1 (y)(t)󵄨󵄨󵄨 ≤ 1 (1 − ℑ∗ − θ∗ )Γk (k + α1 )

+

℘η∗ Γk (kξ1 − α1 )Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ2 − ξ1 ) + α1 ) ψ

+

f ∗ Γk (β1 (k − α1 ))Ψξ (b, a) 1

(1 − ℑ∗ − θ∗ )Γk (k(1 − ξ1 ) + α1 + k)

:= ϒ3 . Similarly, we find that

2 ̄ k,ψ 󵄨󵄨 󵄨󵄨 kΓk (β2 (k − α2 )) Ψβ2 (k−α2 ) (b, a) 󵄨󵄨Q2 N2 (y)(t)󵄨󵄨 ≤ 1 − ℑ∗ − θ∗ k(1−ξ2 )+α2 ,k;ψ ∗ 󵄨󵄨 󵄨 󵄨 󵄨 × (Ja+ γ 󵄨󵄨y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (s)󵄨󵄨󵄨 + f ∗ )(b) η∗ Γk (kξ2 − α2 )Γk (kξ2 ) ≤[ ψ (1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k + α2 ) 2

+

γ∗ Γk (kξ2 − α1 )Γk (kξ1 ) ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ1 − ξ2 ) + α2 ) 1

� 287

f ∗ Γk (β2 (k − α2 )) + . (1 − ℑ∗ − θ∗ )Γk (k(1 − ξ2 ) + α2 + k)

]‖yn − y‖Y

288 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems Thus, ∗ 󵄨󵄨 ψ 󵄨 ℘η Γk (kξ2 − α2 )Γk (kξ2 ) 󵄨󵄨Ψξ (t, a)Q2 N2 (y)(t)󵄨󵄨󵄨 ≤ 2 (1 − ℑ∗ − θ∗ )Γk (k + α2 )

+

℘γ∗ Γk (kξ2 − α2 )Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 (1 − ℑ∗ − θ∗ )Γk (k(1 + ξ1 − ξ2 ) + α2 ) ψ

+

f ∗ Γk (β2 (k − α2 ))Ψξ (b, a) 2

(1 − ℑ∗ − θ∗ )Γk (k(1 − ξ2 ) + α2 + k)

:= ϒ4 . This implies that

‖Q1 N1 y‖Cξ ;ψ ≤ ϒ3 1

and ‖Q2 N2 y‖Cξ ;ψ ≤ ϒ4 . 2

Thus, 󵄩󵄩 󵄩 󵄩󵄩QN(y)󵄩󵄩󵄩Y ≤ max{ϒ3 , ϒ4 }. So, QN(Ω) is a bounded set in Y . Step 3. We demonstrate that L−1 P (id − Q)N : Ω → X is completely continuous. We will demonstrate that L−1 P (id − Q)N(Ω) ⊂ X is equicontinuous and bounded. Firstly, for any y ∈ Ω and t ∈ (a, b], we get −1 −1 L−1 P (Ny(t) − QNy(t)) = (LP1 (N1 y(t) − Q1 N1 y(t)), LP2 (N2 y(t) − Q2 N2 y(t))),

where for j ∈ {1, 2}, we have L−1 P j (Nj y(t) − Qj Nj y(t)) α ;ψ

α ,β1 ;ψ

1 = Ja+j [fj (t, y1 (t), y2 (t), (H k Da +

− kΓk (βj (k − αj )) α,k;ψ

= (Ja+

2

y2 )(t))

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b)] j j α ,β1 ;ψ

1 fj (s, y1 (s), y2 (s), (H k Da +

− k 2 Γk (βj (k − αj ))

α ,β2 ;ψ

2 y1 )(t), (H k Da +

2

α ,β2 ;ψ

2 y1 )(s), (H k Da +

y2 )(s)))(t)

k(1−ξ )+α ,k;ψ k,ψ k,ψ Ψ̄ β (k−α ) (b, a)Ψ̄ α +k (t, a)Ja+ j j Nj (y)(b). j j j

For all y ∈ Ω and t ∈ (a, b], we have

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

󵄨 󵄨󵄨 −1 󵄨󵄨LP1 (id − Q1 )N1 y(t)󵄨󵄨󵄨 1 α ,k;ψ 󵄨 󵄨 󵄨 󵄨 ≤ (Ja+1 γ∗ 󵄨󵄨󵄨y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (s)󵄨󵄨󵄨 + f ∗ )(b) 1 − ℑ∗ − θ∗ k,ψ k,ψ k 2 Γk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a)Ψ̄ α +k (t, a) 1 1 1 + 1 − ℑ∗ − θ∗ k(1−ξ )+α ,k;ψ 󵄨 󵄨 󵄨 󵄨 × (Ja+ 1 1 γ∗ 󵄨󵄨󵄨y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (s)󵄨󵄨󵄨 + f ∗ )(b) k,ψ k,ψ k 2 Γk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a)Ψ̄ α +k (t, a) 1 1 1 ≤ 1 − ℑ∗ − θ∗ k(1−ξ1 )+α1 ,k;ψ ∗

ψ

γ (Ψξ (s, a))

× [‖y‖Y (Ja+

−1

1

k(1−ξ1 )+α1 ,k;ψ ∗

ψ

+ η∗ (Ψξ (s, b)) )(b) −1

2

f )(b)]

+ (Ja+

1 −1 −1 α ,k;ψ ψ ψ [‖y‖Y (Ja+1 γ∗ (Ψξ (s, a)) + η∗ (Ψξ (s, b)) )(b) 1 2 1 − ℑ∗ − θ∗ α ,k;ψ + (Ja+1 f ∗ )(b)]

+

≤

k,ψ ℘γ∗ kΓk (kξ1 − α1 )Γk (kξ1 )Ψ̄ α +k (b, a) 1

(1 −

ℑ∗

−

ψ θ∗ )Ψξ (b, a)Γk (k 1

+ α1 )

k,ψ

℘η kΓk (kξ1 − α1 )Γk (kξ2 )Ψ̄ α +k (b, a) ∗

+

1

ψ

(1 − ℑ∗ − θ∗ )Ψξ (b, a)Γk (k(1 + ξ2 − ξ1 ) + α1 ) 2

k,ψ f kΓk (β1 (k − α1 ))Ψ̄ α +k (b, a) ∗

+

1

(1 − ℑ∗ − θ∗ )Γk (k(1 − ξ1 ) + α1 + k) ψ α ξ1 + k1

℘γ∗ Γk (kξ1 )(Ψ + +

(t, a))−1

(1 − ℑ∗ − θ∗ )Γk (α1 + kξ1 )

℘η∗ Γk (kξ2 )(Ψ

+

α1 k

ψ α ξ2 + k1

(t, a))−1

(1 − ℑ∗ − θ∗ )Γk (α1 + kξ2 )

f ∗ (ψ(b) − ψ(a)) . (1 − ℑ∗ − θ∗ )Γk (α1 + k)

So 󵄩󵄩 −1 󵄩 󵄩󵄩LP1 (id − Q1 )N1 y󵄩󵄩󵄩Cξ ;ψ 1 ≤

k,ψ ℘γ∗ kΓk (kξ1 − α1 )Γk (kξ1 )Ψ̄ α +k (b, a) 1

+ +

(1 − ℑ∗ − θ∗ )Γk (k + α1 ) k,ψ ℘η∗ kΓk (kξ1 − α1 )Γk (kξ2 )Ψ̄ α +k (b, a)(ψ(b) − ψ(a))ξ2 −ξ1 1

(1 − ℑ∗ − θ∗ )Γk (k(1 + ξ2 − ξ1 ) + α1 ) k,ψ ψ f ∗ kΓk (β1 (k − α1 ))Ψ̄ α +k (b, a)Ψξ (b, a) 1

1

(1 − ℑ∗ − θ∗ )Γk (k(1 − ξ1 ) + α1 + k) α1

α1

℘γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k ℘η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + + (1 − ℑ∗ − θ∗ )Γk (α1 + kξ1 ) (1 − ℑ∗ − θ∗ )Γk (α1 + kξ2 )

� 289

290 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems f ∗Ψ +

ψ α ξ1 − k1

(b, a)

(1 − ℑ∗ − θ∗ )Γk (α1 + k)

:= ϒ5 . Similarly, we get

󵄩 󵄩󵄩 −1 󵄩󵄩LP2 (id − Q2 )N2 y󵄩󵄩󵄩Cξ ;ψ 2 ≤

k,ψ ℘η∗ kΓk (kξ2 − α2 )Γk (kξ2 )Ψ̄ α +k (b, a) 2

+ +

(1 − ℑ∗ − θ∗ )Γk (k + α2 ) k,ψ ℘γ∗ kΓk (kξ2 − α2 )Γk (kξ1 )Ψ̄ α +k (b, a)(ψ(b) − ψ(a))ξ1 −ξ2 2

(1 − ℑ∗ − θ∗ )Γk (k(1 + ξ1 − ξ2 ) + α2 ) k,ψ ψ f ∗ kΓk (β2 (k − α2 ))Ψ̄ (b, a)Ψ (b, a) α2 +k

ξ2

(1 − ℑ∗ − θ∗ )Γk (k(1 − ξ2 ) + α2 + k) α2

℘η∗ Γk (kξ2 )(ψ(b) − ψ(a)) k ℘γ∗ Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 + + + (1 − ℑ∗ − θ∗ )Γk (α2 + kξ2 ) (1 − ℑ∗ − θ∗ )Γk (α2 + kξ1 ) ψ α ξ2 − k2

f ∗Ψ

+

(b, a)

(1 − ℑ∗ − θ∗ )Γk (α2 + k)

:= ϒ6 . We deduce that

󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q)Ny󵄩󵄩󵄩X ≤ max{ϒ5 , ϒ6 }. Then L−1 P (id − Q)N(Ω) is uniformly bounded in X. We prove now that L−1 P (id − Q)N(Ω) is equicontinuous. For a < t1 < t2 ≤ b, y ∈ Ω, and j ∈ {1, 2}, we have ψ 󵄨󵄨 ψ 󵄨 −1 −1 󵄨󵄨Ψξ (t1 , a)LPj (id − Qj )Nj y(t1 ) − Ψξ (t2 , a)LPj (id − Qj )Nj y(t2 )󵄨󵄨󵄨 j j α ,k;ψ 󵄨 󵄨 ψ 󵄨 ≤ 󵄨󵄨󵄨Ψξ (t1 , a)(Ja+j 󵄨󵄨󵄨fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨)(t1 ) j

ψ

α ,k;ψ 󵄨

󵄨󵄨fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨)(t2 )󵄨󵄨󵄨 󵄨 󵄨 󵄨 −1 −1 󵄨 k,ψ 󵄨󵄨 ̄ k,ψ + 󵄨󵄨(Ψβ (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j j j j − Ψξ (t2 , a)(Ja+j j

× t1

kΓk (βj (k − αj )) Γk (αj + k)

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b) j

j

ψ 󵄨󵄨 󵄨 ψ ̄ k,ψ ≤ ∫󵄨󵄨󵄨Ψξ (t1 , a)Ψ̄ k,ψ αj (t1 , s) − Ψξ (t2 , a)Ψαj (t2 , s)󵄨󵄨 a

j

j

󵄨 󵄨 × 󵄨󵄨󵄨ψ′ (s)fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨ds

α12 k

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 291

α ,k;ψ 󵄨 󵄨 󵄨 ψ 󵄨 + 󵄨󵄨󵄨Ψξ (t2 , a)(Jt+j 󵄨󵄨󵄨fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨)(t2 )󵄨󵄨󵄨 j

1

−1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j j j j

×

kΓk (βj (k − αj )) Γk (αj + k)

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b). j

j

By Lemma 2.7, we get ψ 󵄨 󵄨󵄨 ψ −1 −1 󵄨󵄨Ψξ (t1 , a)LPj (id − Qj )Nj y(t1 ) − Ψξ (t2 , a)LPj (id − Qj )Nj y(t2 )󵄨󵄨󵄨 j j ψ

≤

(γ∗ + η∗ )℘ + f ∗ Ψξ (b, a) 1 − ℑ∗ − θ∗

j

t1

ψ 󵄨 ψ 󵄨󵄨 ̄ k,ψ ∫󵄨󵄨󵄨Ψξ (t1 , a)Ψ̄ k,ψ αj (t1 , s) − Ψξj (t2 , a)Ψαj (t2 , s)󵄨󵄨 j a

−1 󵄨 ψ 󵄨 × 󵄨󵄨󵄨ψ′ (s)(Ψξ (s, a)) 󵄨󵄨󵄨ds j ψ

(γ∗ + η∗ )℘ + f ∗ Ψξ (b, a)

αj ,k;ψ −1 ψ 󵄨 󵄨󵄨 ψ 󵄨󵄨Ψξ (t2 , a)(Jt+ (Ψξ (s, a)) )(t2 )󵄨󵄨󵄨 j j 1 −1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j j j j

+

×

1 − ℑ∗ − θ∗

kΓk (βj (k − αj )) Γk (αj + k)

j

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b). j

j

Thus, ψ 󵄨 󵄨󵄨 ψ −1 −1 󵄨󵄨Ψξ (t1 , a)LPj (id − Qj )Nj y(t1 ) − Ψξ (t2 , a)LPj (id − Qj )Nj y(t2 )󵄨󵄨󵄨 j j ψ

≤

(γ∗ + η∗ )℘ + f ∗ Ψξ (b, a) 1 − ℑ∗ − θ∗

j

t1

ψ 󵄨 ψ 󵄨󵄨 ̄ k,ψ ∫󵄨󵄨󵄨Ψξ (t1 , a)Ψ̄ k,ψ αj (t1 , s) − Ψξj (t2 , a)Ψαj (t2 , s)󵄨󵄨 j a

−1 󵄨 ψ 󵄨 × 󵄨󵄨󵄨ψ′ (s)(Ψξ (s, a)) 󵄨󵄨󵄨ds j ψ

+

[(γ∗ + η∗ )℘ + f ∗ Ψξ (b, a)]Γk (kξj ) j

Γk (αj + kξj )(1 − ℑ∗ − θ∗ )

ψ

αj

Ψξ (t2 , a)(ψ(t2 ) − ψ(t1 )) k j

−1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j

×

j

kΓk (βj (k − αj )) Γk (αj + k)

j

j

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b). j

j

The operator L−1 P (id − Q)N(Ω) is equicontinuous in X because the right-hand side of the above inequality tends to zero as t1 → t2 and the limit is independent of y for j ∈ {1, 2}. By the Arzelà–Ascoli theorem, L−1 P (id − Q)N(Ω) is relatively compact in X. Then N is L-compact in Ω.

292 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems Let α1

α1

α2

α2

ϒ7 :=

γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + ∗ ∗ (1 − ℑ − θ )Γk (α1 + kξ1 ) (1 − ℑ∗ − θ∗ )Γk (α1 + kξ2 )

ϒ8 :=

η∗ Γk (kξ2 )(ψ(b) − ψ(a)) k γ∗ Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 + k + . ∗ ∗ (1 − ℑ − θ )Γk (α2 + kξ2 ) (1 − ℑ∗ − θ∗ )Γk (α2 + kξ1 )

and

Lemma 11.4. Assume that (11.3.1) and (11.3.2) hold. If the condition max{ϒ7 , ϒ8 }
0, which is independent of ϖ, where for y ∈ X we get L(y) − N(y) = −ϖ[L(y) + N(−y)] 󳨐⇒ ‖y‖X ≤ ϱ,

ϖ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N(y) = −ϖL(y) − ϖN(−y). Then L(y) =

1 ϖ N(y) − N(−y). 1+ϖ 1+ϖ

So, from the expression of L and N, we get for any t ∈ (a, b] and j ∈ {1, 2} α ,βj ;ψ

j Lj yj (t) = H k Da+

yj (t)

1 = f (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) 1+ϖ j ϖ − f (t, −y1 (t), −y2 (t), ℧̃ 1 (t), ℧̃ 2 (t)), 1+ϖ j

where α1 ,β1 ;ψ ℧̃ 1 (t) = (H − y1 )(t) k Da +

and α2 ,β2 ;ψ ℧̃ 2 (t) = (H − y2 )(t). k Da +

By Theorem 2.12, we get

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

yj (t) =

� 293

cj∗

ψ

Ψξ (t, a)Γk (kξj ) j

1 α ,k;ψ [J +j (fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s)))(t) ϖ+1 a α ,k;ψ − ϖ J +j (fj (s, −y1 (s), −y2 (s), ℧̃ 1 (s), ℧̃ 2 (s)))(t)],

+

a

k(1−ξj ),k;ψ

where cj∗ = (Ja+ 󵄨󵄨 󵄨 󵄨󵄨y1 (t)󵄨󵄨󵄨 ≤

yj )(a). Thus, for each t ∈ (a, b] we have

|c1∗ |

ψ

Ψξ (t, a)Γk (kξ1 ) 1

2 α ,k;ψ 󵄨 󵄨 󵄨 󵄨 (Ja+1 γ∗ 󵄨󵄨󵄨y1 (s)󵄨󵄨󵄨 + η∗ 󵄨󵄨󵄨y2 (s)󵄨󵄨󵄨 + f ∗ )(b) (1 − ℑ∗ − θ∗ )(ϖ + 1) |c1∗ | 2 + ≤ ψ ∗ − θ∗ )(ϖ + 1) (1 − ℑ Ψ (t, a)Γ (kξ ) +

k

ξ1

×

1

−1 α ,k;ψ ψ [‖y‖X (Ja+1 γ∗ (Ψξ (s, a)) 1 α ,k;ψ ∗

+ (Ja+1

+

−1

2

f )(b)] ψ α ξ1 + k1

‖y‖X γ∗ Γk (kξ1 )(Ψ ≤[

ψ

+ η∗ (Ψξ (s, b)) )(b)

(t, a))−1

ψ α ξ2 + k1

‖y‖X η∗ Γk (kξ2 )(Ψ +

Γk (α1 + kξ1 )

Γk (α1 + kξ2 )

α1 k

f (ψ(b) − ψ(a)) 2 ]× . Γk (α1 + k) (1 − ℑ∗ − θ∗ )(ϖ + 1) ∗

Then |c1∗ | 2 󵄨󵄨 ψ 󵄨 + 󵄨󵄨Ψξ (t, a)y1 (t)󵄨󵄨󵄨 ≤ ∗ 1 Γk (kξ1 ) (1 − ℑ − θ∗ )(ϖ + 1)

α1

‖y‖ γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k ×[ X Γk (α1 + kξ1 )

α1

‖y‖X η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + Γk (α1 + kξ2 ) f ∗Ψ

+

ψ α ξ1 − k1

(b, a)

Γk (α1 + k)

].

We deduce that ‖y1 ‖C1−ξ ;ψ ≤ 1

|c1∗ | 2 + Γk (kξ1 ) (1 − ℑ∗ − θ∗ )(ϖ + 1)

α1

‖y‖ γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k ×[ X Γk (α1 + kξ1 )

(t, a))−1

294 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems α1

‖y‖X η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + Γk (α1 + kξ2 ) f ∗Ψ

+

ψ α ξ1 − k1

(b, a)

Γk (α1 + k)

] ψ

2f ∗ Ψ α1 (b, a) ξ1 − k |c1∗ | ≤ + Γk (kξ1 ) (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α1 + k) α1

γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k + 2[ (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α1 + kξ1 ) α1

η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + ]‖y‖X (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α1 + kξ2 ) ψ

2f ∗ Ψ α1 (b, a) ξ1 − k |c1∗ | ≤ + + 2ϒ7 ‖y‖X . Γk (kξ1 ) (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α1 + k) Similarly, we get ψ

‖y2 ‖Cξ ;ψ 2

2f ∗ Ψ α2 (b, a) ξ2 − k |c2∗ | + ≤ Γk (kξ2 ) (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α2 + k) + 2[

α2

η∗ Γk (kξ2 )(ψ(b) − ψ(a)) k (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α2 + kξ2 ) α2

γ∗ Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 + k ]‖y‖X + (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α2 + kξ1 ) ψ

2f ∗ Ψ α2 (b, a) ξ2 − k |c2∗ | ≤ + + 2ϒ8 ‖y‖X . Γk (kξ2 ) (1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (α2 + k) This implies that ‖y‖X ≤ max { j∈{1,2}

|cj∗ |

ψ

2f ∗ Ψ

Γk (kξj )

+

αj

ξj − k

(b, a)

(1 − ℑ∗ − θ∗ )(ϖ + 1)Γk (αj + k)

}

+ 2 max{ϒ7 , ϒ8 }‖y‖X . We deduce that ψ

‖y‖X ≤

|cj∗ | maxj∈{1,2} { Γ (kξ k j)

2f ∗ Ψ

+

ξj −

αj k

(b,a)

(1−ℑ∗ −θ∗ )(ϖ+1)Γk (αj +k)

1 − 2 max{ϒ7 , ϒ8 }

}

:= ϱ.

Lemma 11.5. If conditions (11.3.1), (11.3.2), and (11.9) are satisfied, then there exists a bounded open set Ω ⊂ X with

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 295

L(y) − N(y) ≠ −ϖ[L(y) + N(−y)] for any y ∈ 𝜕Ω and any ϖ ∈ (0, 1]. Proof. Using Lemma 11.4, there exists a positive constant ϱ which is independent of ϖ such that if y satisfies L(y) − N(y) = −ϖ[L(y) + N(−y)],

ϖ ∈ (0, 1],

then ‖y‖X ≤ ϱ. So, if Ω = {y ∈ X; ‖y‖X < ς}

(11.10)

such that ς > ϱ, then L(y) − N(y) ≠ −ϖ[L(y) − N(−y)] for all y ∈ 𝜕Ω = {y ∈ X; ‖y‖X = ς} and ϖ ∈ (0, 1]. Theorem 11.1. Assume that (11.3.1), (11.3.2), and (11.9) hold. Then there exists at least one solution for the problem (11.1)–(11.2). Proof. It is clear that the set Ω defined in (11.10) is symmetric, 0 ∈ Ω, and X ∩ Ω = Ω ≠ 0. In addition, by Lemma 11.5, assume that (11.3.1), (11.3.2), and (11.9) hold. Then L(y) − N(y) ≠ −ϖ[L(y) − N(−y)] for each y ∈ X ∩ 𝜕Ω = 𝜕Ω and each ϖ ∈ (0, 1]. Thus, problem (11.1)–(11.2) has at least one solution on Dom L ∩ Ω. We have proved our precedent existence result by assuming some condition of the constants α1 , α2 , ξ1 , and ξ2 , which are often regarded as strong requirements. Now, we will employ different conditions on the functions fj , j ∈ {1, 2}. We will use them to demonstrate existence and uniqueness results. In all that follows, we will present and demonstrate our results using the same approach as before, which will allow us to skip certain steps. Lemma 11.6. If (11.3.1) and the following hypothesis are satisfied: ̂j ∈ C(J,̄ ℝ+ ) such (11.7.1) For each j ∈ {1, 2}, there exist nonnegative functions γ̂j , η̂j , θ̂j , ℑ that 󵄨󵄨 󵄨 󵄨󵄨fj (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − fj (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 ψ ψ 󵄨 󵄨 󵄨 󵄨 ≤ γ̂j (t)Ψξ (t, a)󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 + η̂j (t)Ψξ (t, a)󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 1

2

󵄨 󵄨 ̂ 󵄨󵄨 󵄨󵄨 + θ̂j (t)󵄨󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + ℑ j (t)󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨

296 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems for every t ∈ J,̄ yj , yj , σj , σj ∈ Cξj ;ψ (J,̄ ℝ), such that γ̂∗ = max ‖γj ‖∞ ,

η̂∗ = max ‖ηj ‖∞ ,

θ̂∗ = max ‖θj ‖∞ ,

̂∗ = max ‖ℑj ‖∞ , ℑ

j∈{1,2}

j∈{1,2}

j∈{1,2}

j∈{1,2}

̂∗ + θ̂∗ < 1, and 0 < ℑ ̂ ⊂ X, the operator N is L-compact. then for any bounded open set Ω ̂ = {y ∈ X : ‖y‖X < B } be a bounded open set. Proof. For B > 0, let Ω Step 1. We show that QN is continuous. Let (yn )n∈ℕ be a sequence where yn 󳨀→ y in Y . Then for j ∈ {1, 2}, by hypothesis (11.7.1), we obtain for t ∈ J ̄ 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)(Q1 N1 (yn )(t) − Q1 N1 (y)(t))󵄨󵄨󵄨 1 k,ψ (γ̂∗ + η̂∗ )kΓk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a) 1 1 ≤ ̂∗ − θ̂∗ 1−ℑ ψ

k(1−ξ1 )+α1 ,k;ψ

× max {Ψξ (b, a)}‖yn − y‖Y (Ja+ j∈{1,2}

≤

j

(1))(b)

(γ̂∗ + η̂∗ )Γk (β1 (k − α1 )) ψ × max {Ψ (b, a)}‖yn − y‖Y ̂∗ − θ̂∗ )Γk (k(1 − ξ1 ) + α1 + k) j∈{1,2} ξj (1 − ℑ

and 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)(Q2 N2 (yn )(t) − Q2 N2 (y)(t))󵄨󵄨󵄨 2 (γ̂∗ + η̂∗ )Γk (β2 (k − α2 )) ψ × max {Ψ (b, a)}‖yn − y‖Y . ≤ ̂∗ − θ̂∗ )Γk (k(1 − ξ2 ) + α2 + k) j∈{1,2} ξj (1 − ℑ Thus, for each j ∈ {1, 2}, we get 󵄩󵄩 󵄩 󵄩󵄩Qj Nj (yn ) − Qj Nj (y)󵄩󵄩󵄩Cξ ;ψ ≤ j

(γ̂∗ + η̂∗ )Γk (βj (k − αj )) ̂∗ − θ̂∗ )Γk (k(1 − ξj ) + αj + k) (1 − ℑ ψ

× max {Ψξ (b, a)}‖yn − y‖Y , j∈{1,2}

j

which implies that 󵄩󵄩 󵄩 󵄩󵄩QN(yn ) − QN(y)󵄩󵄩󵄩Y 󳨀→ 0 We may conclude then that QN is continuous.

as n 󳨀→ +∞.

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 297

̂ is bounded. Step 2. We demonstrate that QN(Ω) ̂ We get for any t ∈ (a, b] and y ∈ Ω 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)Q1 N1 (y)(t)󵄨󵄨󵄨 1 [(γ̂∗ + η̂∗ )B + f ∗ ]Γk (βj (k − αj )) ψ ≤ max { } × max {Ψξ (b, a)} j ̂∗ − θ̂∗ )Γk (k(1 − ξj ) + αj + k) j∈{1,2} (1 − ℑ j∈{1,2} and 󵄨󵄨 ψ 󵄨 󵄨󵄨Ψξ (t, a)Q2 N2 (y)(t)󵄨󵄨󵄨 2

≤ max { j∈{1,2}

[(γ̂∗ + η̂∗ )B + f ∗ ]Γk (βj (k − αj )) ψ } × max {Ψξ (b, a)}. j ̂∗ − θ̂∗ )Γ (k(1 − ξ ) + α + k) j∈{1,2} (1 − ℑ k

j

j

Then for each j ∈ {1, 2} we obtain ‖Qj Nj y‖Cξ ;ψ ≤ max { j∈{1,2}

j

[(γ̂∗ + η̂∗ )B + f ∗ ]Γk (βj (k − αj )) ψ } × max {Ψξ (b, a)}. j ̂∗ − θ̂∗ )Γ (k(1 − ξ ) + α + k) j∈{1,2} (1 − ℑ k

j

j

Thus, [(γ̂∗ + η̂∗ )B + f ∗ ]Γk (βj (k − αj )) ψ 󵄩󵄩 󵄩 } × max {Ψξ (b, a)}. 󵄩󵄩QN(y)󵄩󵄩󵄩Y ≤ max { j ̂∗ − θ̂∗ )Γk (k(1 − ξj ) + αj + k) j∈{1,2} (1 − ℑ j∈{1,2} ̂ is a bounded set in Y . So, QN(Ω) ̂ Step 3. We prove that L−1 P (id − Q)N : Ω → X is completely continuous. ̂ and t ∈ (a, b], we have For any y ∈ Ω L−1 P (Ny(t) − QNy(t)) −1 = (L−1 P1 (N1 y(t) − Q1 N1 y(t)), LP2 (N2 y(t) − Q2 N2 y(t))),

̂ and t ∈ (a, b] where j ∈ {1, 2}. We have for all y ∈ Ω 󵄨󵄨 ψ 󵄨 −1 󵄨󵄨Ψξ (t, a)LP1 (id − Q1 )N1 y(t)󵄨󵄨󵄨 1 (γ̂∗ + η̂∗ )B + f ∗ α ,k;ψ ψ ≤ × max {Ψξ (b, a)}(Ja+1 (1))(b) j ̂∗ − θ̂∗ j∈{1,2} 1−ℑ +

k,ψ k,ψ [(γ̂∗ + η̂∗ )B + f ∗ ]k 2 Γk (β1 (k − α1 ))2 Ψ̄ β (k−α ) (b, a)Ψ̄ α +k (t, a) ψ

k(1−ξ1 )+α1 ,k;ψ

× max {Ψξ (b, a)}(Ja+ j∈{1,2}

j

̂∗ − θ̂∗ 1−ℑ

(1))(b).

1

1

1

298 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems Thus, 󵄩󵄩 −1 󵄩 󵄩󵄩LP1 (id − Q1 )N1 y󵄩󵄩󵄩Cξ ;ψ 1

k,ψ

[(γ̂∗ + η̂∗ )B + f ∗ ]kΓk (β1 (k − α1 ))Ψ̄ α +k (b, a) 1

≤

̂∗ − θ̂∗ )Γk (k(1 − ξ1 ) + α1 + k) (1 − ℑ

+

[(γ̂∗ + η̂∗ )B + f ∗ ](ψ(b) − ψ(a)) ̂∗ − θ̂∗ )Γ (α + k) (1 − ℑ k

= ϒ5 .

α1 k

ψ

× max {Ψξ (b, a)} j∈{1,2}

j

ψ

× max {Ψξ (b, a)} j∈{1,2}

1

j

Also, we may obtain 󵄩󵄩 −1 󵄩 󵄩󵄩LP2 (id − Q2 )N2 y󵄩󵄩󵄩C

ξ2 ;ψ

k,ψ [(γ̂∗ + η̂∗ )B + f ∗ ]kΓk (β2 (k − α2 ))Ψ̄ α +k (b, a) 2

≤

̂∗ − θ̂∗ )Γk (k(1 − ξ2 ) + α2 + k) (1 − ℑ

+

[(γ̂∗ + η̂∗ )B + f ∗ ](ψ(b) − ψ(a)) ̂∗ − θ̂∗ )Γ (α + k) (1 − ℑ k

= ϒ6 .

α2 k

2

ψ

× max {Ψξ (b, a)} j∈{1,2}

ψ

× max {Ψξ (b, a)} j∈{1,2}

j

We conclude then that 󵄩󵄩 −1 󵄩 󵄩󵄩LP (id − Q)Ny󵄩󵄩󵄩X ≤ max{ϒ5 , ϒ6 }. ̂ This means that L−1 P (id − Q)N(Ω) is uniformly bounded in X. ̂ is equicontinuous. Let us now demonstrate L−1 (id − Q)N(Ω) P

̂ and j ∈ {1, 2}, we have For a < t1 < t2 ≤ b, y ∈ Ω,

ψ 󵄨󵄨 ψ 󵄨 −1 −1 󵄨󵄨Ψξ (t1 , a)LPj (id − Qj )Nj y(t1 ) − Ψξ (t2 , a)LPj (id − Qj )Nj y(t2 )󵄨󵄨󵄨 j j t1

ψ 󵄨 ψ 󵄨󵄨 ̄ k,ψ ≤ ∫󵄨󵄨󵄨Ψξ (t1 , a)Ψ̄ k,ψ αj (t1 , s) − Ψξ (t2 , a)Ψαj (t2 , s)󵄨󵄨 a

j

j

󵄨 󵄨 × 󵄨󵄨󵄨ψ′ (s)fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨ds α ,k;ψ 󵄨 󵄨 󵄨 󵄨 ψ + 󵄨󵄨󵄨Ψξ (t2 , a)(Jt+j 󵄨󵄨󵄨fj (s, y1 (s), y2 (s), ℧1 (s), ℧2 (s))󵄨󵄨󵄨)(t2 )󵄨󵄨󵄨 j 1 −1 󵄨 −1 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j j j j ×

kΓk (βj (k − αj )) Γk (αj + k)

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b) j

j

j

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 299

t1

[(γ̂∗ + η̂∗ )B + f ∗ ] 󵄨󵄨 ψ ψ 󵄨󵄨 󵄨󵄨 ′ 󵄨󵄨 ̄ k,ψ ≤ ∫󵄨󵄨Ψξ (t1 , a)Ψ̄ k,ψ αj (t1 , s) − Ψξj (t2 , a)Ψαj (t2 , s)󵄨󵄨 × 󵄨󵄨ψ (s)󵄨󵄨ds j ∗ ∗ ̂ ̂ (1 − ℑ − θ ) a

+

[(γ̂∗

αj

+ f ](ψ(t2 ) − ψ(t1 )) k ψ × Ψξ (t2 , a) j ∗ ̂ (1 − ℑ − θ̂∗ )Γ (α + k)

+

η̂∗ )B

∗

k

j

−1 −1 󵄨 k,ψ 󵄨 k,ψ + 󵄨󵄨󵄨(Ψ̄ β (k−α ) (t1 , a)) − (Ψ̄ β (k−α ) (t2 , a)) 󵄨󵄨󵄨 j

×

j

kΓk (βj (k − αj )) Γk (αj + k)

j

j

k(1−ξ )+α ,k;ψ k,ψ Ψ̄ β (k−α ) (b, a)Ja+ j j Nj (y)(b). j

j

̂ As in the precedent result, we can deduce that N is L-compact in Ω. Lemma 11.7. Suppose that (11.3.1) and (11.7.1) are satisfied. If ψ

max{

2(γ̂∗ + η̂∗ )Ψ

αj

ξj − k

(b, a)

̂∗ − θ̂∗ )Γk (αj + k) (1 − ℑ

1 }< , 2

(11.11)

then there exists ϱ̂ > 0 which is independent of ϖ where for y ∈ X we get L(y) − N(y) = −ϖ[L(y) + N(−y)] 󳨐⇒ ‖y‖X ≤ ϱ̂,

ϖ ∈ (0, 1].

Proof. Let y ∈ X satisfy L(y) − N(y) = −ϖL(y) − ϖN(−y). Then L(y) =

ϖ 1 N(y) − N(−y). 1+ϖ 1+ϖ

So, from the expression of L and N, we get for any t ∈ (a, b] and j ∈ {1, 2} α ,βj ;ψ

j Lj yj (t) = H k Da +

=

yj (t)

1 f (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) 1+ϖ j ϖ f (t, −y1 (t), −y2 (t), ℧̃ 1 (t), ℧̃ 2 (t)). − 1+ϖ j

By Theorem 2.12, we get 󵄨󵄨 󵄨 󵄨󵄨yj (t)󵄨󵄨󵄨 ≤

ψ

|cj∗ |

Ψξ (t, a)Γk (kξj ) j

+

2[(γ̂∗ + η̂∗ )‖y‖X + f ∗ ] α ,k;ψ (Ja+j (1))(b) ∗ ∗ ̂ ̂ − θ )(ϖ + 1) (1 − ℑ αj

2[(γ̂∗ + η̂∗ )‖y‖X + f ∗ ](ψ(b) − ψ(a)) k + . ≤ ψ ̂∗ − θ̂∗ )(ϖ + 1)Γk (αj + k) (1 − ℑ Ψξ (t, a)Γk (kξj ) |cj∗ |

j

300 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems We deduce that

‖yj ‖Cξ ;ψ ≤ j

|cj∗ |

Γk (kξj )

ψ 2[(γ̂∗ + η̂∗ )‖y‖X + f ∗ ]Ψ

+

αj

ξj − k

(b, a)

̂∗ − θ̂∗ )(ϖ + 1)Γk (αj + k) (1 − ℑ

.

We deduce that ψ

‖y‖X ≤

|cj∗ | maxj∈{1,2} { Γ (kξ k j)

1 − 2 max{

2f ∗ Ψ

+

αj (b,a) ξj − k ∗ ∗ (1−ℑ −θ )(ϖ+1)Γk (αj +k)

2(γ̂∗ +η̂∗ )Ψ

ψ

ξj −

αj k

(b,a)

}

:= ϱ.

̂∗ −θ̂∗ )(ϖ+1)Γk (αj +k) } (1−ℑ

Lemma 11.8. If conditions (11.3.1), (11.7.1), and (11.11) are satisfied, then there exists a ̂ ⊂ X with bounded open set Ω L(y) − N(y) ≠ −ϖ[L(y) + N(−y)] ̂ and any ϖ ∈ (0, 1]. for any y ∈ 𝜕Ω Proof. The proof is the same as the proof of Lemma 11.5. Theorem 11.2. Suppose that the hypotheses (11.3.1) and (11.7.1) and the condition (11.11) are satisfied. Then there exists at least one solution for the problem (11.1)–(11.2). Proof. The proof is the same as the proof of Theorem 11.1. Theorem 11.3. Suppose that (11.3.1) and (11.7.1) are satisfied. Moreover, suppose that the following hypothesis holds. (11.11.1)There exist γj , θj , ℑj > 0 and ηj ≥ 0, j ∈ {1, 2}, where 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 ψ ψ 󵄨 󵄨 󵄨 󵄨 ≥ γ1 Ψξ (t, a)󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 − η1 Ψξ (t, a)󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 1 2 ψ ψ 󵄨 󵄨 󵄨 󵄨 + θ1 Ψξ (t, a)󵄨󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 − ℑ1 Ψξ (t, a)󵄨󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 1 2 and 󵄨󵄨 󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 ψ ψ 󵄨 󵄨 󵄨 󵄨 ≥ γ2 Ψξ (t, a)󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 − η2 Ψξ (t, a)󵄨󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 2 1 ψ ψ 󵄨 󵄨 󵄨 󵄨 − θ2 Ψξ (t, a)󵄨󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + ℑ2 Ψξ (t, a)󵄨󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 1 2 for every t ∈ J ̄ and yj , yj , σj , σj ∈ Cξj ;ψ (J,̄ ℝ).

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 301

If one has αj

2(γ̂∗ + η̂∗ )Γk (kξj )(ψ(b) − ψ(a)) k max { } + max { } ̂∗ )Γ (α + kξ ) j∈{1,2} j∈{1,2} γj (1 − θ̂∗ − ℑ ηj

×

ψ max {Ψξ (b, a)} j j∈{1,2}

k

j

j

< 1,

(11.12)

̂ then the problem (11.1)–(11.2) has a unique solution in Dom L ∩ Ω. Proof. Since the condition (11.12) is stronger than (11.11), by Theorem 11.2 we may get ̂ that (11.1)–(11.2) has at least one solution in Dom L ∩ Ω. ̂ Then we We assume that (11.1)–(11.2) has two different solutions y, y ∈ Dom L ∩ Ω. have for each t ∈ (a, b] and j ∈ {1, 2} H αj ,βj ;ψ yj (t) k Da + H αj ,βj ;ψ yj (t) k Da +

α ,β1 ;ψ

1 = fj (t, y1 (t), y2 (t), (H k Da +

=

α ,β2 ;ψ

2 y1 )(t), (H k Da +

y2 )(t)),

α1 ,β1 ;ψ α2 ,β2 ;ψ fj (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)), k Da + k Da +

and k(1−ξj ),k;ψ

Ja+

k(1−ξj ),k;ψ

yj (a) = Ja+

yj (b),

k(1−ξj ),k;ψ

Ja+

k(1−ξj ),k;ψ

yj (a) = Ja+

yj (b).

Let U(t) = y(t) − y(t) for all t ∈ (a, b], which means that U(t) = (U1 (t), U2 (t)) = (y1 (t) − y1 (t), y2 (t) − y2 (t)). Then LU(t) = (L1 U1 (t), L2 U2 (t)) α ,β1 ;ψ

1 = (H k Da +

=

α ,β2 ;ψ

2 U1 (t), H k Da+

α1 ,β1 ;ψ (H y1 (t) k Da +

−

U2 (t))

α2 ,β2 ;ψ H α1 ,β1 ;ψ y1 (t), H y2 (t) k Da + k Da +

= (℧1 (t) − ℧1 (t), ℧2 (t) − ℧2 (t)),

α ,β2 ;ψ

2 −H k Da +

y2 (t)) (11.13)

where for j ∈ {1, 2}, ℧j , ℧j ∈ Cξj ;ψ (J,̄ ℝ) satisfy the following systems of functional equations: ℧1 (t) = f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)),

{

℧2 (t) = f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t))

and ℧1 (t) = f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)), { ℧2 (t) = f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)).

302 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems Using the fact that Img L = ker Q, for j ∈ {1, 2} we have k(1−ξj )+αj ,k;ψ

(Ja+

[℧j (s) − ℧j (s)])(b) = 0.

Since for each j ∈ {1, 2}, the functions fj ∈ Cξj ;ψ (J,̄ ℝ), there exists t0 ∈ (a, b] such that ℧j (t0 ) − ℧j (t0 ) = 0. In view of (11.11.1), we have ψ 󵄨 󵄨 󵄨 η ψ 󵄨 Ψξ (t0 , a)󵄨󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 ≤ 1 Ψξ (t0 , a)󵄨󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 1 γ1 2 η ≤ 1 × ‖y2 − y2 ‖Cξ ,ψ 2 γ1 ηj ≤ max { }‖U‖X j∈{1,2} γj

and ψ 󵄨 󵄨 η ψ 󵄨 󵄨 Ψξ (t0 , a)󵄨󵄨󵄨y2 (t0 ) − y2 (t0 )󵄨󵄨󵄨 ≤ 2 Ψξ (t0 , a)󵄨󵄨󵄨y1 (t0 ) − y1 (t0 )󵄨󵄨󵄨 2 γ2 1 η ≤ 2 × ‖y1 − y1 ‖Cξ ,ψ 1 γ2 ηj ≤ max { }‖U‖X . j∈{1,2} γj

Then for each j ∈ {1, 2}, we have ηj ψ 󵄨 󵄨 Ψξ (t0 , a)󵄨󵄨󵄨Uj (t0 )󵄨󵄨󵄨 ≤ max { }‖U‖X . j j∈{1,2} γj By Theorem 2.12, for any t ∈ (a, b] and j ∈ {1, 2} we have α ,k;ψ H αj ,βj ;ψ Uj )(t) k Da+

(Ja+j

k(1−ξj ),k;ψ

= Uj (t) −

Ja+

Uj (a) , ψ Ψξ (t, a)Γk (kξj ) j

which implies that k(1−ξj ),k;ψ

Ja+

α ,k;ψ H αj ,βj ;ψ ψ Uj )(t0 )]Ψξ (t0 , a)Γk (kξj ), k Da+ j

Uj (a) = [Uj (t0 ) − (Ja+j

and therefore α ,k;ψ H αj ,βj ;ψ Uj (t) k Da +

Uj (t) = Ja+j

(11.14)

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 303

−1 α ;ψ H αj ,βj ;ψ ψ ψ Uj (t0 )]Ψξ (t0 , a)(Ψξ (t, a)) . k Da + j j

+ [Uj (t0 ) − Ja+j

Using (11.14), we obtain for every t ∈ (a, b] and j ∈ {1, 2} −1 ψ 󵄨 ψ 󵄨 󵄨 αj ,k;ψ H αj ,βj ;ψ 󵄨 󵄨 󵄨󵄨 󵄨󵄨Uj (t)󵄨󵄨󵄨 ≤ [󵄨󵄨󵄨Uj (t0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨Ja+ k Da+ Uj (t0 )󵄨󵄨󵄨]Ψξ (t0 , a)(Ψξ (t, a)) j j 󵄨 󵄨 α ,k;ψ H αj ,βj ;ψ Uj (t)󵄨󵄨󵄨 + 󵄨󵄨󵄨Ja+j k Da + ηj −1 ψ ≤ max { }‖U‖X (Ψξ (t, a)) j j∈{1,2} γj αj

+ +

Γk (kξj )(ψ(t0 ) − ψ(a)) k Γk (αj + kξj )

Γk (kξj )

Γk (αj + kξj )

≤ max { j∈{1,2}

+

ηj γj

ψ

(Ψ

αj

ξj + k

αj ,βj ;ψ 󵄩 −1 󵄩 ψ [Ψξ (t, a)] 󵄩󵄩󵄩H Uj 󵄩󵄩󵄩C k Da + ξj ,ψ j

αj ,βj ;ψ 󵄩 −1 󵄩 (t, a)) 󵄩󵄩󵄩H Uj 󵄩󵄩󵄩C k Da + ξj ,ψ

ψ

−1

}‖U‖X (Ψξ (t, a)) j

αj

2Γk (kξj )(ψ(b) − ψ(a)) k Γk (αj + kξj )

αj ,βj ;ψ 󵄩 −1 󵄩 ψ [Ψξ (t, a)] 󵄩󵄩󵄩H Uj 󵄩󵄩󵄩C . k Da + ξj ,ψ j

(11.15)

By (11.7.1) and (11.13), for each j ∈ {1, 2} we find 󵄨󵄨H α1 ,β1 ;ψ 󵄨 U1 (t)󵄨󵄨󵄨 󵄨󵄨k Da+ 󵄨 󵄨 = 󵄨󵄨󵄨f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) − f1 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t))󵄨󵄨󵄨 󵄨 α1 ,β1 ;ψ 󵄨 ̂∗ 󵄨󵄨H α2 ,β2 ;ψ 󵄨 ≤ (γ̂∗ + η̂∗ )‖U‖X + θ̂∗ 󵄨󵄨󵄨H U1 (t)󵄨󵄨󵄨 + ℑ U2 (t)󵄨󵄨󵄨 󵄨󵄨k Da+ k Da + and 󵄨󵄨H α2 ,β2 ;ψ 󵄨 U2 (t)󵄨󵄨󵄨 󵄨󵄨k Da+ 󵄨 󵄨 = 󵄨󵄨󵄨f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t)) − f2 (t, y1 (t), y2 (t), ℧1 (t), ℧2 (t))󵄨󵄨󵄨 󵄨 α1 ,β1 ;ψ 󵄨 ̂∗ 󵄨󵄨H α2 ,β2 ;ψ 󵄨 ≤ (γ̂∗ + η̂∗ )‖U‖X + θ̂∗ 󵄨󵄨󵄨H U1 (t)󵄨󵄨󵄨 + ℑ U2 (t)󵄨󵄨󵄨. 󵄨󵄨k Da+ k Da + Thus, 󵄨󵄨H αj ,βj ;ψ 󵄨 󵄨󵄨k Da+ Uj (t)󵄨󵄨󵄨 ≤

γ̂∗ + η̂∗ ‖U‖X . ̂∗ 1 − θ̂∗ − ℑ

Then 󵄩󵄩H αj ,βj ;ψ 󵄩󵄩 󵄩󵄩k Da+ Uj 󵄩󵄩Cξ ,ψ ≤ j

γ̂∗ + η̂∗ ψ max {Ψ (b, a)}‖U‖X . ̂∗ j∈{1,2} ξj 1 − θ̂∗ − ℑ

(11.16)

Substituting (11.16) in the right side of (11.15), for every t ∈ (a, b] and j ∈ {1, 2} we get

304 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems αj

ηj 2Γk (kξj )(ψ(b) − ψ(a)) k −1 −1 ψ ψ 󵄨 󵄨󵄨 [Ψξ (t, a)] 󵄨󵄨Uj (t)󵄨󵄨󵄨 ≤ [max { }(Ψξ (t, a)) + j j j∈{1,2} γj Γk (αj + kξj ) γ̂∗ + η̂∗ ψ max {Ψ (b, a)}]‖U‖X . ̂∗ j∈{1,2} ξj 1 − θ̂∗ − ℑ

× Therefore,

αj

2(γ̂∗ + η̂∗ )Γk (kξj )(ψ(b) − ψ(a)) k } ‖U‖X ≤ [max { } + max { ̂∗ )Γ (α + kξ ) j∈{1,2} j∈{1,2} γj (1 − θ̂∗ − ℑ ηj

k

j

j

ψ

× max {Ψξ (b, a)}]‖U‖X . j∈{1,2}

j

Hence, by (11.12), ‖U‖X = 0. Consequently, for any t ∈ (a, b] we get U(t) = 0 󳨐⇒ y(t) = y(t).

11.2.2 Examples Example 11.1. Consider the following system: 1 1

1 1

1 1

, ;ln t

, ;ln t

, ;ln t

3 5 2 4 2 4 (H y1 )(t) = f1 (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)), { k D1+ k D1+ k D1+ { { 1 1 1 1 1 1 { { H 3 , 5 ;ln t H 2 , 4 ;ln t H 3 , 5 ;ln t { {(k D1+ y2 )(t) = f2 (t, y1 (t), y2 (t), (k D1+ y1 )(t), (k D1+ y2 )(t)), 3 3 { ,1;ln t ,1;ln t {J 8 { y1 (1) = J1+8 y1 (π), { 1+ { { 8 { 158 ,1;ln t ,1;ln t 15 y2 (1) = J1+ y2 (π), {J1+

where t ∈ (1, π], 1 1

, ;ln t

2 4 f1 (t, y1 (t), y2 (t), (H k D1+

=

1 1

, ;ln t

3 5 y1 )(t), (H k D1+

y2 )(t))

1 1 , ;ln t 2 4 1+

H 1 1 y1 )(t) t 2 y1 (t) + (k D t+1 , ;ln t 3 5 + + [sin(y2 (t)) + (H y2 )(t)], k D1+ 5 11 152(1 + t) 13e √π

and 1 1

, ;ln t

2 4 f2 (t, y1 (t), y2 (t), (H k D1+

=

1 1

, ;ln t

3 5 y1 )(t), (H k D1+

1 t+3 + [y1 (t) + (H kD te10t + 2 e7 √π

1 1 , ;ln t 2 4 1+

y2 )(t))

y1 )(t)]

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 305

1 1

, ;ln t

3 5 y2 )(t) y2 (t) + (H e−3−t k D1+ + ). ( 1 1 11 , ;ln t 111e 3 5 1 + y2 (t) + (H D y )(t) + 2 k 1 3

Let k = 1, α1 = 21 , α2 = 31 , β1 = 41 , β2 = 51 , ξ1 = 85 , ξ2 = 157 , and ψ(t) = ln(t). It is easy to see that fj ∈ Cξj ;ln(t) ([1, π], ℝ), j ∈ {1, 2}. Hence, condition (11.3.1) is satisfied. Furthermore, for all t ∈ (1, π] and yj , yj , σj , σj , ∈ Cξj ;ln(t) ([1, π], ℝ), j ∈ {1, 2}, we obtain 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 1 󵄨󵄨󵄨y (t) − y1 (t)󵄨󵄨󵄨 + t + 1 󵄨󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 ≤ 󵄨 13e5 √π 󵄨 󵄨 152(1 + t) 󵄨 1 t + 1 󵄨󵄨 1 󵄨󵄨 󵄨 󵄨 + 󵄨σ (t) − σ1 (t)󵄨󵄨󵄨 + 󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 152(1 + t) 󵄨 1 13e5 √π 󵄨 and 󵄨 󵄨󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 3

−3−t t + 3 󵄨󵄨 󵄨󵄨 󵄨 e 󵄨 ≤ 7 󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 󵄨󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 + 111e11 󵄨 e √π 3

+

−3−t t + 3 󵄨󵄨 󵄨 e 󵄨󵄨 󵄨 󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨. 󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + 7 111e11 󵄨 e √π

Then (11.3.2) is satisfied with γ1 (t) = θ1 (t) =

1 , 152(1 + t)

γ2 (t) = θ2 (t) =

t+3 e7 √π

and t+1 η1 (t) = ℑ1 (t) = , 13e5 √π

3

e−3−t η2 (t) = ℑ2 (t) = . 111e11

By simple calculations, we get γ∗ = θ ∗ =

1 , 304

η∗ = ℑ∗ =

π+1 . 13e5 √π

We also have α1

{ αk

2

k

+ ξ2 ≈ 0.9666 ≥ ξ1 ≈ 0.625,

+ ξ1 ≈ 0.9583 ≥ ξ2 ≈ 0.466,

ℑ∗ + θ∗ =

304 + 304π + 13√πe5 3952√πe5

306 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems ≈ 0.004500564423988 < 1, ϒ7 := =

α1

α1

γ∗ Γk (kξ1 )(ψ(b) − ψ(a)) k η∗ Γk (kξ2 )(ψ(b) − ψ(a))ξ2 −ξ1 + k + ∗ ∗ (1 − ℑ − θ )Γk (α1 + kξ1 ) (1 − ℑ∗ − θ∗ )Γk (α1 + kξ2 ) 304(1 −

41

Γ( 85 )√ln(π)

+

304+304π+13√πe5 )Γ( 89 ) 3952√πe5

(π + 1)Γ( 157 )(ln(π)) 120

13e5 √π(1 −

304+304π+13√πe5 29 ) )Γ( 30 3952√πe5

≈ 0.00775495076133521, and α2

α2

η∗ Γk (kξ2 )(ψ(b) − ψ(a)) k γ∗ Γk (kξ1 )(ψ(b) − ψ(a))ξ1 −ξ2 + k ϒ8 := + ∗ ∗ (1 − ℑ − θ )Γk (α2 + kξ2 ) (1 − ℑ∗ − θ∗ )Γk (α2 + kξ1 ) 1

=

59

(π + 1)Γ( 157 )(ln(π)) 3

13e5 √π(1 −

304+304π+13√πe5 12 )Γ( 15 ) 3952√πe5

+

304(1 −

Γ( 85 )(ln(π)) 120

304+304π+13√πe5 23 )Γ( 24 ) 3952√πe5

≈ 0.00701266368328416, which implies that 1 max{ϒ7 , ϒ8 } ≈ 0.0645 < . 2 Consequently, with the use of Theorem 11.1 our problem has at least one solution. Example 11.2. Consider the following system: 1 1

1 1

1 1

, ;et

, ;et

, ;et

3 7 2 7 2 7 (H y2 )(t)), y1 )(t) = f1 (t, y1 (t), y2 (t), (H y1 )(t), (H { k D0+ k D0+ k D0+ { { 1 1 1 1 t 1 1 t t { { , ;e , ;e , ;e { 3 7 3 7 2 7 {(H y2 )(t) = f2 (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)), k D0+ k D0+ k D0+ 3 3 t t { ,1;e ,1;e { 7 7 { J0+ y1 (0) = J0+ y1 (1), { { { 4 { 47 ,1;et ,1;et 7 {J0+ y2 (0) = J0+ y2 (1),

where t ∈ (0, 1], 1 1

1 1

, ;et

, ;et

3 7 2 7 f1 (t, y1 (t), y2 (t), (H y1 )(t), (H y2 )(t)) k D0+ k D0+

=

ln (t + 2) 3

(et − 1) 7 +

and

3

+

1 1 t 5(et − 1) 7 , ;e 2 7 (sin y1 (t) + (H y1 )(t)) k D0+ 11 46e √π 4

e−11−t (et − 1) 7

1 1

, ;et

37e9 (1 + y2 (t) + (H D 3 7 y2 )(t)) k 0+

,

11.2 k-Generalized ψ-Hilfer fractional differential coupled systems with periodic conditions

� 307

1 1

1 1

, ;et

, ;et

3 7 2 7 y1 )(t), (H y2 )(t)) f2 (t, y1 (t), y2 (t), (H k D0+ k D0+

=

4

2et

4

5(et − 1) 7 +

+ 3

1 1 t (et − 1) 7 e 9 , ;e 3 7 [y2 (t) + (H y2 )(t)] k D0+ t 7 7e (1 + 3 ) −t

1 1 t t(et − 1) 7 , ;e 2 7 [cos y1 (t) + (H y1 )(t)]. k D0+ 13 55e √π

Let k = 1, α1 = 21 , α2 = 31 , β1 = β2 = 71 , ξ1 = 73 , ξ2 = 47 , and ψ(t) = et . It can easily be seen that fj ∈ Cξj ;et ([0, 1], ℝ), j ∈ {1, 2}. Hence, condition (11.3.1) is satisfied. Furthermore, for all t ∈ (0, 1] and yj , yj , σj , σj ∈ Cξj ;et ([0, 1], ℝ), j ∈ {1, 2}, we obtain 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 3

4

−11−t t 5(et − 1) 7 󵄨󵄨 (e − 1) 7 󵄨󵄨 󵄨󵄨 e 󵄨 ≤ y (t) + y (t) − 󵄨 󵄨 󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 1 1 󵄨 󵄨 37e9 46e11 √π 4

3

+

−11−t t (e − 1) 7 󵄨󵄨 5(et − 1) 7 󵄨󵄨 󵄨 e 󵄨 󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + 󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 11 37e9 46e √π

and 󵄨󵄨 󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 3

≤

4

t t(et − 1) 7 󵄨󵄨 󵄨󵄨 e 9 (e − 1) 7 󵄨󵄨 󵄨 y (t) y (t) − + 󵄨 󵄨 󵄨󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 1 󵄨 1 55e13 √π 󵄨 7e7 (1 + 3t ) −t

3

+

4

t t(et − 1) 7 󵄨󵄨 󵄨 e 9 (e − 1) 7 󵄨󵄨 󵄨 󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨. 󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 + 13 7 55e √π 7e (1 + 3t ) −t

Then (11.7.1) is satisfied with γ̂1 (t) = ̂1 (t) = η

5

46e11 √π

γ̂2 (t) =

,

e−11−t , 37e9

t

55e13 √π

e9 , 7 7e (1 + 3t )

,

−t

η̂2 (t) = 3

5(et − 1) 7 θ̂1 (t) = , 46e11 √π

3

t(et − 1) 7 θ̂2 (t) = , 55e13 √π

and 4

−11−t t (e − 1) 7 ̂1 (t) = e ℑ , 37e9

By simple calculations, we get

4

t 9 7 ̂2 (t) = e (e − 1) . ℑ t 7 7e (1 + 3 ) −t

308 � 11 Nonlinear implicit k-generalized ψ-Hilfer fractional coupled systems 5

γ̂∗ =

46e11 √π

η̂∗ =

,

3

4

5(e − 1) 7 θ̂∗ = , 46e11 √π

1 , 7e7

7 ̂ ∗ = (e − 1) , ℑ 7 7e

and max{

ψ 2(γ̂∗ + η̂∗ )Ψ

αj

ξj − k

(b, a)

̂∗ − θ̂∗ )Γk (αj + k) (1 − ℑ

2( 46e115 √π +

} = max{

3

(1 −

5(e−1) 7 46e11 √π

1 )(e 7e7

−

19

− 1) 25 4

(e−1) 7 7e7

)Γ( 43 )

}

≈ 0.00044379227911538 1 < . 2

Consequently, with the use of Theorem 11.2 our problem has at least one solution. Otherwise, for each t ∈ (0, 1], yj , yj , σj , σj ∈ Cξj ;et ([0, 1], ℝ), j ∈ {1, 2}, we have 󵄨󵄨 󵄨 󵄨󵄨f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f1 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 4

3

≥

t (et − 1) 7 󵄨󵄨 󵄨󵄨 (e − 1) 7 󵄨󵄨 󵄨 y (t) − y (t) − 󵄨 󵄨y2 (t) − y2 (t)󵄨󵄨󵄨 󵄨 1 󵄨 1 37e20 󵄨 46e11 √π 󵄨 3

+

4

t (et − 1) 7 󵄨󵄨 󵄨 (e − 1) 7 󵄨󵄨 󵄨 󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨 󵄨󵄨σ1 (t) − σ1 (t)󵄨󵄨󵄨 − 11 37e20 󵄨 46e √π

and 󵄨󵄨 󵄨 󵄨󵄨f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t)) − f2 (t, y1 (t), y2 (t), σ1 (t), σ2 (t))󵄨󵄨󵄨 4

3

t 3e 9 (et − 1) 7 󵄨󵄨 󵄨󵄨 (e − 1) 7 󵄨󵄨 󵄨 y (t) − y (t) − 󵄨 󵄨 󵄨y1 (t) − y1 (t)󵄨󵄨󵄨 2 󵄨 󵄨 2 28e7 55e13 √π 󵄨 −1

≥

3

4

t (et − 1) 7 󵄨󵄨 󵄨󵄨 3e 9 (e − 1) 7 󵄨󵄨 󵄨 σ (t) + − σ (t) − 󵄨 󵄨 󵄨󵄨σ2 (t) − σ2 (t)󵄨󵄨󵄨. 1 1 󵄨 󵄨 28e7 55e13 √π −1

Then (11.11.1) is satisfied with γ1 = θ1 =

1

46e11 √π

3e 9 28e7 −1

,

γ2 = ℑ2 =

and η1 = ℑ1 =

1 , 37e20

η2 = θ2 =

1

55e13 √π

.

We have αj

ηj

2(γ̂∗ + η̂∗ )Γk (kξj )(ψ(b) − ψ(a)) k ψ max { } + max { } max {Ψξ (b, a)} j ̂∗ )Γ (α + kξ ) j∈{1,2} γj j∈{1,2} j∈{1,2} (1 − θ̂∗ − ℑ k

j

j

11.3 Notes and remarks

=

� 309

3 5 1 4 46e11 √π 2( 46e11 √π + 7e7 )Γ( 7 )√e − 1 + × (e − 1) 7 4 3 20 37e (e−1) 7 5(e−1) 7 13 (1 − 46e )Γ( 14 ) 11 √π − 7e7

≈ 0.00119853344106452 < 1.

So, by Theorem 11.3, our problem has a unique solution.

11.3 Notes and remarks The results of this chapter are taken from the paper of Salim et al. [209]. For more information, we recommend the books [49, 98, 144, 145, 185, 186, 244] and the articles [158, 175, 178, 208, 211, 212, 214, 218, 223].

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Index absolutely continuous functions 9

Gronwall lemma 157

Banach space 9–11, 24, 51, 139, 158, 159 boundary value problem 2, 26, 66, 95, 157

Hilfer 2, 15, 16, 157, 158, 201

Caputo 14–17, 26, 27, 35, 51, 66, 67, 80, 96, 107, 119, 121, 136, 138, 140, 158, 201, 260 coincidence degree 9, 24, 26, 66, 95, 138, 186, 201, 233, 260 completely continuous 24, 30, 47, 56, 73, 84, 146, 164, 176, 190, 206, 221, 241, 250, 267 contraction 24, 41, 66, 67, 138, 139, 157, 201, 260, 277 coupled systems 138, 139, 233 equicontinuous 30–32, 46, 56, 58, 59, 73–75, 84, 85, 87, 99–101, 110, 112, 114, 124, 126, 129, 148, 149, 164–166, 176, 178, 179, 190, 192, 193, 206, 208, 209, 221, 223, 226, 241, 244, 245, 250, 267, 268 existence 24, 35, 37, 51, 67, 89, 95, 96, 107, 151, 158, 159, 182, 186, 196, 202, 212, 216, 228, 233, 248, 261, 271 fixed point 2, 24, 37–39, 41, 50, 66, 67, 95, 96, 138, 139, 157, 158, 201 fractional integral 13, 14, 158

https://doi.org/10.1515/9783111334387-013

implicit equations 26, 27, 34, 139, 260 initial value problem 157, 201 measure of noncompactness 2 nonlocal 157, 201 periodic solutions 26, 27, 35, 36, 50, 51, 66, 67, 80, 89, 95, 107, 120, 138, 182, 186, 196, 201, 212, 216, 233, 261, 274 Riemann–Liouville 15, 158 terminal value problem 157 Ulam stability 157 uniqueness 3, 24, 26, 35, 63, 66, 67, 90, 96, 104, 107, 117, 133, 138, 139, 152, 186, 196, 197, 201, 202, 212, 216, 228, 229, 248, 261, 271, 272

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