Boundary Value Problems for Fractional Differential Equations and Systems 9811224455, 9789811224454

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TRENDS IN ABSTRACT AND APPLIED ANALYSIS ISSN: 2424-8746

Series Editor: John R. Graef The University of Tennessee at Chattanooga, USA This series will provide state of the art results and applications on current topics in the broad area of Mathematical Analysis. Of a more focused nature than what is usually found in standard textbooks, these volumes will provide researchers and graduate students a path to the research frontiers in an easily accessible manner. In addition to being useful for individual study, they will also be appropriate for use in graduate and advanced undergraduate courses and research seminars. The volumes in this series will not only be of interest to mathematicians but also to scientists in other areas. For more information, please go to http://www.worldscientific.com/series/taaa Published Vol. 9 Boundary Value Problems for Fractional Differential Equations and Systems by Bashir Ahmad, Johnny Henderson & Rodica Luca Vol. 8

Ordinary Differential Equations and Boundary Value Problems Volume II: Boundary Value Problems by John R. Graef, Johnny Henderson, Lingju Kong & Xueyan Sherry Liu

Vol. 7

Ordinary Differential Equations and Boundary Value Problems Volume I: Advanced Ordinary Differential Equations by John R. Graef, Johnny Henderson, Lingju Kong & Xueyan Sherry Liu

Vol. 6

The Strong Nonlinear Limit-Point/Limit-Circle Problem by Miroslav Bartušek & John R. Graef

Vol. 5

Higher Order Boundary Value Problems on Unbounded Domains: Types of Solutions, Functional Problems and Applications by Feliz Manuel Minhós & Hugo Carrasco

Vol. 4

Quantum Calculus: New Concepts, Impulsive IVPs and BVPs, Inequalities by Bashir Ahmad, Sotiris Ntouyas & Jessada Tariboon

More information on this series can be found at https://www.worldscientific.com/series/taaa

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Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Names: Ahmad, Bashir, author. | Henderson, Johnny, author. | Luca, Rodica, author. Title: Boundary value problems for fractional differential equations and systems / Bashir Ahmad, King Abdulaziz University, Saudi Arabia; Johnny Henderson, Baylor University, USA; Rodica Luca, “Gheorghe Asachi” Technical University of Iasi, Romania. Description: USA : World Scientific, 2021. | Series: Trends in abstract and applied analysis, 2424-8746 ; vol 9 | Includes bibliographical references and index. Identifiers: LCCN 2020041568 (print) | LCCN 2020041569 (ebook) | ISBN 9789811224454 (hardcover) | ISBN 9789811224461 (ebook) | ISBN 9789811224478 (ebook other) Subjects: LCSH: Boundary value problems. | Functional differential equations. Classification: LCC QA379 .A35 2021 (print) | LCC QA379 (ebook) | DDC 515/.35--dc23 LC record available at https://lccn.loc.gov/2020041568 LC ebook record available at https://lccn.loc.gov/2020041569

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2021 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher. For any available supplementary material, please visit https://www.worldscientific.com/worldscibooks/10.1142/11942#t=suppl Desk Editors: George Vasu/Lai Fun Typeset by Stallion Press Email: [email protected] Printed in Singapore

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Dedication

Bashir Ahmad dedicates this book to the memory of his parents. Johnny Henderson dedicates this book to the memory of Kathryn Madora Strunk (February 5, 1991–March 1, 2007). Rodica Luca dedicates this book to the memory of her parents.

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b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Preface

Fractional-order differential and integral operators, and fractional differential equations have extensive applications in the mathematical modeling of real-world phenomena occurring in scientific and engineering disciplines, such as physics, biophysics, chemistry, biology, medical sciences, ecology, financial economics, control theory, signal and image processing, transport dynamics, aerodynamics, thermodynamics, hydrology, viscoelasticity, electromagnetics, statistical mechanics, astrophysics, cosmology, bioengineering and rheology (see [21, 26, 28, 30, 34, 55, 65, 66, 70, 86, 87, 92, 95, 96, 113]). For some recent developments on the topic, we mention [1, 3, 4, 13, 18–20, 22, 24, 31–33, 36–39, 54, 56–59, 61, 64, 73, 76, 88–91, 94, 97, 100, 101, 106, 111, 115] and the references cited therein. For example, Ding et al. [31] and Arafa et al. [18] developed a model for the primary infection with HIV which is a virus that targets the white blood cells-CD4+ T lymphocytes. This model can be described as a system with three fractional differential equations of different orders (α, β, γ > 0) in the unknown functions T (the concentration of uninfected CD4+ T cells), I (infected CD4+ T cells) and V (the free HIV virus particles in the blood). The fractional differential equations are also regarded as a better tool for the description of hereditary properties of various materials and processes than the corresponding integer order differential equations. For some recent results on the existence, nonexistence and multiplicity of solutions or positive solutions for fractional differential equations, inclusions and inequalities, and systems of fractional differential equations supplemented with various boundary conditions we refer the reader to [6, 11, 14–16, 35, 42, 43, 62, 63, 74, 75, 93, 98, 103, 105, 109, 112, 117–119].

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Preface

In this monograph, we present our results obtained in the last years related to the existence of solutions or positive solutions for various classes of Riemann–Liouville and Caputo fractional differential equations, and systems of fractional differential equations subject to nonlocal boundary conditions. Chapter 1 contains some preliminaries related to the fractional integral and fractional derivatives, and the fixed point theorems that we will use in the subsequent chapters. Chapter 2 deals with the existence of positive solutions or nonnegative solutions for some Riemann–Liouville fractional differential equations with parameters or without parameters, supplemented with multi-point or Riemann–Stieltjes integral boundary conditions, which contain fractional derivatives. The nonlinearities of equations are nonsingular or singular functions, with nonnegative values or real values. Chapter 3 is focused on the existence and multiplicity of positive solutions for systems of two Riemann–Liouville fractional differential equations, subject to uncoupled or coupled multi-point boundary conditions, which contain fractional derivatives, and the nonlinearities of systems are nonsingular or singular functions, with nonnegative values. Chapter 4 is devoted to the existence and nonexistence of positive solutions for systems of two Riemann–Liouville fractional differential equations with p-Laplacian operators and positive parameters, supplemented with uncoupled or coupled multi-point boundary conditions, which contain fractional derivatives, and the nonlinearities are nonnegative functions. In Chapter 5, we investigate the existence and nonexistence of positive solutions for systems of three Riemann–Liouville fractional differential equations with positive parameters, subject to uncoupled multi-point boundary conditions which contain fractional derivatives, and the nonlinearities of systems are nonnegative functions. Chapter 6 is concerned with the existence of solutions for Riemann–Liouville fractional differential equations and systems of Riemann–Liouville fractional differential equations with integral terms, supplemented with nonlocal boundary conditions which contain fractional derivatives and Riemann–Stieltjes integrals. In Chapter 7, we study the existence of solutions for some Caputo fractional differential equations and inclusions, and systems of Caputo fractional differential equations subject to nonlocal boundary conditions which contain Riemann–Stieltjes integrals. In each chapter, various examples are presented, which support the main results. The methods used in the proof of our theorems include results from the fixed point theory and the fixed point index theory. This book complements the existing literature in fractional differential equations, inclusions and systems. The monograph can serve as a good

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resource for mathematical and scientific researchers, and for graduate students in mathematics and science interested in the existence of solutions for fractional differential equations and systems. We would like to express our warm thanks to the editors at World Scientific Publishing: Trends in Abstract and Applied Analysis Series Editor John R. Graef, Executive Editor Rochelle Kronzek, Desk Editor Lai Fun Kwong, and Book Editor George Vasu for their support and work on our book. Bashir Ahmad Johnny Henderson Rodica Luca

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Contents

Preface Notations

vii xv

Chapter 1 1.1 1.2

2.2

2.3

1

Fractional Integral and Fractional Derivatives . . . . . . . Fixed Point Theorems . . . . . . . . . . . . . . . . . . . .

Chapter 2

2.1

Preliminaries

Riemann–Liouville Fractional Differential Equations with Nonlocal Boundary Conditions

Singular Fractional Differential Equations with Parameters and Multi-Point Boundary Conditions . . . . 2.1.1 Auxiliary results . . . . . . . . . . . . . . . . . . . 2.1.2 Existence of positive solutions . . . . . . . . . . . 2.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . A Fractional Differential Equation with Integral Terms and Multi-Point Boundary Conditions . . . . . . . . . . . . . . 2.2.1 Existence of nonnegative solutions . . . . . . . . . 2.2.2 An example . . . . . . . . . . . . . . . . . . . . . Semipositone Singular Fractional Boundary Value Problems with Integral Boundary Conditions . . . . . . . 2.3.1 Preliminary results . . . . . . . . . . . . . . . . .

xi

1 3

9 9 10 16 23 25 26 38 39 40

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Contents

2.3.2

2.4

2.5

2.6

Existence and multiplicity of positive solutions . . . . . . . . . . . . . . . . . . . 2.3.3 An example . . . . . . . . . . . . . . . . . Singular Fractional Differential Equations with General Integral Boundary Conditions . . . . . . . 2.4.1 Auxiliary results . . . . . . . . . . . . . . . 2.4.2 Existence of multiple positive solutions . . 2.4.3 An example . . . . . . . . . . . . . . . . . On a Singular Fractional Boundary Value Problem with Parameters . . . . . . . . . . . . . . . . . . . 2.5.1 Existence of positive solutions . . . . . . . 2.5.2 Some remarks on a related semipositone problem . . . . . . . . . . . . . . . . . . . . 2.5.3 Examples . . . . . . . . . . . . . . . . . . . A Singular Fractional Differential Equation with Integral Boundary Conditions . . . . . . . . . . . . 2.6.1 Preliminary results . . . . . . . . . . . . . 2.6.2 Existence and multiplicity of positive solutions . . . . . . . . . . . . . . . . . . . 2.6.3 An example . . . . . . . . . . . . . . . . .

Chapter 3

3.1

3.2

. . . . . . . .

43 50

. . . .

. . . .

53 54 58 64

. . . . . . . .

68 69

. . . . . . . .

77 78

. . . . . . . .

81 82

. . . . . . . .

83 92

. . . .

. . . .

Systems of Two Riemann–Liouville Fractional Differential Equations with Multi-Point Boundary Conditions

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions . . 3.1.1 Auxiliary results . . . . . . . . . . . . . . 3.1.2 Existence and multiplicity of positive solutions . . . . . . . . . . . . . . . . . . Systems of Fractional Differential Equations with Coupled Multi-Point Boundary Conditions . . . . 3.2.1 Preliminary results . . . . . . . . . . . . 3.2.2 Nonsingular nonlinearities . . . . . . . . . 3.2.3 Singular nonlinearities . . . . . . . . . . . 3.2.4 Examples . . . . . . . . . . . . . . . . . .

97 . . . . . . . . . .

97 98

. . . . . 101 . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

117 117 128 138 151

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Contents

Chapter 4

4.1

4.2

5.1

155 . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

155 156 159 181 187 189

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

191 192 196 218 227

Systems of Three Riemann–Liouville Fractional Differential Equations with Parameters and Multi-Point Boundary Conditions

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions . . 5.1.1 Auxiliary results . . . . . . . . . . . . . . 5.1.2 Existence of positive solutions . . . . . . 5.1.3 Nonexistence of positive solutions . . . . 5.1.4 Examples . . . . . . . . . . . . . . . . . .

Chapter 6

6.1

Systems of Two Riemann–Liouville Fractional Differential Equations with p-Laplacian Operators, Parameters and Multi-Point Boundary Conditions

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions . . 4.1.1 Auxiliary results . . . . . . . . . . . . . . 4.1.2 Existence of positive solutions . . . . . . 4.1.3 Nonexistence of positive solutions . . . . 4.1.4 An example . . . . . . . . . . . . . . . . 4.1.5 A relation between two supremum limits Systems of Fractional Differential Equations with Coupled Multi-Point Boundary Conditions . . . . 4.2.1 Preliminary results . . . . . . . . . . . . 4.2.2 Existence of positive solutions . . . . . . 4.2.3 Nonexistence of positive solutions . . . . 4.2.4 An example . . . . . . . . . . . . . . . .

Chapter 5

xiii

Existence of Solutions for Riemann–Liouville Fractional Boundary Value Problems

233 . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

233 234 238 266 272

279

Riemann–Liouville Fractional Differential Equations with Nonlocal Boundary Conditions . . . . . . . . . . . . 279

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Contents

6.2

6.3

6.1.1 Preliminary results . . . . . . . . . . . . . . 6.1.2 Existence of solutions . . . . . . . . . . . . . 6.1.3 Examples . . . . . . . . . . . . . . . . . . . . Systems of Riemann–Liouville Fractional Differential Equations with Uncoupled Boundary Conditions . . 6.2.1 Auxiliary results . . . . . . . . . . . . . . . . 6.2.2 Existence of solutions . . . . . . . . . . . . . 6.2.3 Examples . . . . . . . . . . . . . . . . . . . . Systems of Riemann–Liouville Fractional Differential Equations with Coupled Boundary Conditions . . . . 6.3.1 Preliminary results . . . . . . . . . . . . . . 6.3.2 Existence of solutions . . . . . . . . . . . . . 6.3.3 Examples . . . . . . . . . . . . . . . . . . . .

Chapter 7 7.1

7.2

7.3

. . . 280 . . . 281 . . . 291 . . . .

. . . .

. . . .

293 294 297 313

. . . .

. . . .

. . . .

316 317 324 350

Existence of Solutions for Caputo Fractional Boundary Value Problems

Sequential Caputo Fractional Differential Equations and Inclusions with Nonlocal Boundary Conditions . . . . 7.1.1 Auxiliary results . . . . . . . . . . . . . . . . . . . 7.1.2 Existence of solutions for problem (7.1), (7.3) . . . 7.1.3 Existence of solutions for problem (7.2), (7.3) . . . 7.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . Sequential Caputo Fractional Integro-Differential Systems with Coupled Integral Boundary Conditions . . . . . . . . 7.2.1 Preliminary results . . . . . . . . . . . . . . . . . 7.2.2 Existence of solutions . . . . . . . . . . . . . . . . 7.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . Caputo Fractional Differential Systems with Coupled Nonlocal Boundary Conditions . . . . . . . . . . 7.3.1 Auxiliary results . . . . . . . . . . . . . . . . . . . 7.3.2 Existence of solutions . . . . . . . . . . . . . . . . 7.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . .

355 355 357 360 374 385 387 388 395 416 418 419 422 436

Bibliography

439

Index

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Notations

N = {1, 2, . . .} R R+ = {x ∈ R, x ≥ 0} P(C) Br Br ∂Br Ω ∂Ω C[0, 1] C(A, B) CnA AC n D

L1 (0, 1)

the set of natural numbers the set of real numbers the set of nonnegative real numbers the family of all nonempty subsets of C the open ball centered at θ of radius r, (θ is the zero element) the closure of Br the boundary of Br the closure of the set Ω the boundary of Ω the space of continuous real functions defined on the interval [0, 1] the space of continuous functions defined on the set A with values in the set B the space of n-times continuously differentiable real functions defined on A the space of (n − 1)-times absolutely continuously differentiable real functions defined on D the space of Lebesgue integrable and measurable real functions (actually, classes of equivalent functions which are equal almost everywhere) on (0, 1)

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Notations

Lγ (0, 1), γ ∈ (1, ∞) the space of measurable real functions f (actually, classes of equivalent functions which are equal almost everywhere) on (0, 1), with |f |γ ∈ L1 (0, 1) ∞ the space of measurable almost everywhere L (0, 1) bounded real functions (actually, classes of equivalent functions which are equal almost everywhere) on (0, 1) i(f, U, X) the fixed point index of f over U with respect to X, (X is a retract of a real Banach space E, and U ⊂ X is an open set) α f the (left-sided) Riemann–Liouville fractional I0+ integral of order α of function f α f the (left-sided) Riemann–Liouville fractional D0+ derivative of order α of function f c α D0+ f the (left-sided) Caputo fractional derivative of order α of function f .

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Chapter 1

Preliminaries

In this chapter, we present the definitions and some properties of the fractional integral and the Riemann–Liouville and Caputo fractional derivatives, and the fixed point theorems that we will use in the following chapters.

1.1

Fractional Integral and Fractional Derivatives

In this section, we present the definitions of the Riemann–Liouville fractional integral and the Riemann–Liouville and Caputo fractional derivatives and some of their properties. Definition 1.1.1. The (left-sided) Riemann–Liouville fractional integral of order α > 0 of a function f : (0, ∞) → R is given by  t 1 α f )(t) = (t − s)α−1 f (s) ds, t > 0, (I0+ Γ(α) 0 provided the right-hand side is pointwise defined (0, ∞), where Γ(α) is  ∞ on α−1 −t e dt, α > 0. the Euler gamma function defined by Γ(α) = 0 t Definition 1.1.2. The (left-sided) Riemann–Liouville fractional derivative of order α ≥ 0 for a function f : (0, ∞) → R is given by  n  n−α  d α (D0+ f )(t) = I0+ f (t) dt  n  t 1 d f (s) = ds, t > 0, Γ(n − α) dt (t − s)α−n+1 0

1

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Boundary Value Problems for Fractional Differential Equations and Systems

where n = α + 1, provided that the right-hand side is pointwise defined on (0, ∞). The notation α stands for the largest integer not greater than α. For m f (t) = f (m) (t) for t > 0, and if f ∈ C m [0, ∞), if α = m ∈ N, then D0+ 0 α = 0, then D0+ f (t) = f (t) for t > 0. Definition 1.1.3. For u ∈ AC n [0, ∞), the (left-sided) Caputo fractional derivative of order α is defined as  t 1 c α D0+ u(t) = (t − s)n−α−1 u(n) (s) ds, Γ(n − α) 0 t > 0, n − 1 < α < n, n ∈ N. α u(t) = u(m) (t) for t > 0, and For u ∈ C m [0, ∞), if α = m ∈ N then cD0+ c 0 if α = 0 then D0+ u(t) = u(t) for t > 0.

Lemma 1.1.1 ([65]). (a) If α > 0, β > 0 and f ∈ Lp (0, 1), (1 ≤ p ≤ ∞), then the relation α+β α β I0+ f )(t) = (I0+ f )(t) is satisfied at almost every point t ∈ (0, 1). (I0+ If α + β > 1, then the above relation holds at any point of [0, 1]. (b) If α > 0 and f ∈ Lp (0, 1), (1 ≤ p ≤ ∞), then the relation α α I0+ f )(t) = f (t) holds almost everywhere on (0, 1). (D0+ (c) If α > β > 0 and f ∈ Lp (0, 1), (1 ≤ p ≤ ∞), then the relation β α−β α I0+ f )(t) = (I0+ f )(t) holds almost everywhere on (0, 1). (D0+ By using the above definitions and a direct computation, we obtain the following lemma. Lemma 1.1.2. Γ(β) α+β−1 α β−1 (a) If α > 0 and β > 0, then I0+ t = Γ(α+β) t . (b) If α ∈ N, β ∈ R, β = 1, 2, . . . , α, then α β−1 t = (β − 1)(β − 2) · · · (β − α)tβ−α−1 ; D0+ α β−1 t = 0; If α ∈ N and β = 1, 2, . . . , α, then D0+ (c) If α > 0, α ∈ N, n − 1 < α < n, β > 0 and β = α, α − 1, . . . , α − n + 1, then Γ(β) α β−1 t = Γ(β+n−α) (β + n − α − 1)(β + n − α − 2) · · · (β − α)tβ−α−1 ; D0+ If α > 0, α ∈ N, n − 1 < α < n and β = α, α − 1, . . . , α − n + 1, then α β−1 t = 0; D0+ Γ(β) β−α−1 α β−1 t = Γ(β−α) t . (d) If α ≥ 0 and β > α, then D0+

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3

Preliminaries

Lemma 1.1.3 ([65]). Let α > 0 and n = α + 1 for α ∈ N and n = α for α ∈ N; that is, n is the smallest integer greater than or equal to α (namely, α u(t) ∈ C(0, 1) ∩ L1 (0, 1), then n − 1 < α ≤ n). If u, D0+ α α D0+ u(t) = u(t) + I0+

n 

ci tα−i ,

0 < t < 1,

i=1

where c1 , . . . , cn ∈ R. Lemma 1.1.4 ([65]). Let α > 0 and n = α + 1 for α ∈ N and n = α for α ∈ N, (namely, n − 1 < α ≤ n). If u ∈ AC n [0, 1], then α c α D0+ u(t) = u(t) + I0+

n−1 

c i ti ,

0 < t < 1,

i=0

where c0 , . . . , cn−1 ∈ R. For other properties of the fractional integrals and fractional derivatives, we refer the reader to [65, 96].

1.2

Fixed Point Theorems

In this section, we present the fixed point theorems and the fixed point index theorems that will be used in the following chapters. The Banach contraction mapping principle Theorem 1.2.1 (see [29]). If Y is a nonempty complete metric space with the metric d, and T : Y → Y is a contraction mapping, then T has a unique fixed point x∗ ∈ Y, (T x∗ = x∗ ). The Guo–Krasnosel’skii fixed point theorem Let X be a real Banach space with the norm  · . Definition 1.2.1. A nonempty convex closed set C ⊂ X is a cone if it satisfies the conditions: (a) if u ∈ C and α ≥ 0, then αu ∈ C; (b) if u ∈ C and −u ∈ C, then u = θ, where θ is the zero element of X.

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Boundary Value Problems for Fractional Differential Equations and Systems

A cone C ⊂ X defines a partial ordering in X given by x ≤ y if and only if y − x ∈ C. Definition 1.2.2. An operator A : D(A) ⊂ X → X is compact if it maps bounded sets into relatively compact sets. An operator A : D(A) ⊂ X → X is completely continuous if it is continuous and compact. Theorem 1.2.2 (Fixed point theorem of cone expansion and compression of norm type — see [41]). Let X be a real Banach space with the norm ·, and let C ⊂ X be a cone in X. Assume Ω1 and Ω2 are bounded open subsets of X with θ ∈ Ω1 , Ω1 ⊂ Ω2 and let A : C ∩ (Ω2 \ Ω1 ) → C be a completely continuous operator such that, either (i) Au ≤ u, ∀ u ∈ C ∩ ∂Ω1 , and Au ≥ u, ∀ u ∈ C ∩ ∂Ω2 , or (ii) Au ≥ u, ∀ u ∈ C ∩ ∂Ω1 , and Au ≤ u, ∀ u ∈ C ∩ ∂Ω2 . Then A has at least one fixed point in C ∩ (Ω2 \ Ω1 ). The Krasnosel’skii fixed point theorem for the sum of two operators Theorem 1.2.3 (see [23, 67, 99]). Let M be a closed, convex, bounded and nonempty subset of a Banach space X. Let A1 and A2 be two operators such that (a) A1 x + A2 y ∈ M for all x, y ∈ M ; (b) A1 is a completely continuous operator; (c) A2 is a contraction mapping. Then there exists z ∈ M such that z = A1 z + A2 z. The Schauder fixed point theorem Theorem 1.2.4. Let X be a Banach space and Y ⊂ X a nonempty, bounded, convex and closed subset. If operator A : Y → Y is completely continuous, then A has at least one fixed point. The Leray–Schauder alternative Theorem 1.2.5. Let E be a Banach space and T : E → E be a completely continuous operator. Let F = {x ∈ E, x = νT (x) for some 0 < ν < 1}. Then either the set F is unbounded or T has at least one fixed point.

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5

Preliminaries

The nonlinear alternative of Leray–Schauder type Theorem 1.2.6. Let X be a Banach space and U be an open and bounded subset of X, θ ∈ U and A : U → X be a completely continuous operator. Then either A has a fixed point in U , or there exist u ∈ ∂U and ν ∈ (0, 1) such that u = νAu. Theorems from the fixed point index theory Let X be a real Banach space with the norm  · , C ⊂ X a cone, “≤” the partial ordering defined by C and θ the zero element of X. For r > 0, let Br = {u ∈ X, u < r} be the open ball of radius r centered at θ, its closure B r = {u ∈ X, u ≤ r} and its boundary ∂Br = {u ∈ X, u = r}. Theorem 1.2.7 (see [17]). Let A : B r ∩ C → C be a completely continuous operator. If Au = μu for all u ∈ ∂Br ∩ C and μ ≥ 1, then i(A, Br ∩ C, C) = 1. Theorem 1.2.8 (see [17]). Let A : B r ∩ C → C be a completely continuous operator which has no fixed points on ∂Br ∩ C. If Au ≤ u for all u ∈ ∂Br ∩ C, then i(A, Br ∩ C, C) = 1. Theorem 1.2.9 (see [17]). Let A : B r ∩ C → C be a completely continuous operator. If there exists u0 ∈ C \ {θ} such that u − Au = λu0 for all λ ≥ 0 and u ∈ ∂Br ∩ C, then i(A, Br ∩ C, C) = 0. Theorem 1.2.10 ([120]). Let A : B r ∩C → C be a completely continuous operator which has no fixed point on ∂Br ∩C. If there exists a linear operator L : C → C and u0 ∈ C \ {θ} such that (i) u0 ≤ Lu0 ,

(ii) Lu ≤ Au,

∀ u ∈ ∂Br ∩ C,

then i(A, Br ∩ C, C) = 0. Theorem 1.2.11 (see [17, 68]). Let Ω ⊂ X be a bounded open set with θ ∈ Ω. Assume that A : Ω ∩ C → C is a completely continuous operator. (a) If u ≤ Au for all u ∈ ∂Ω ∩ C, then the fixed point index i(A, Ω ∩ C, C) = 1. (b) If Au ≤ u for all u ∈ ∂Ω ∩ C, then the fixed point index i(A, Ω ∩ C, C) = 0. The Leggett–Williams theorem Let X be a real Banach space with the norm  · , and let P be a cone in X.

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Definition 1.2.3. A map ξ is a nonnegative concave functional on the cone P if ξ : P → [0, ∞) satisfies the condition ξ(tx + (1 − t)y) ≥ tξ(x) + (1 − t)ξ(y) for all x, y ∈ P and t ∈ [0, 1]. Let ξ be a nonnegative concave functional on P , and 0 < c < d. We define the convex sets Pc , P c , P (ξ, c, d) and S(ξ, c, d) by Pc = {u ∈ P, u < c}, P c = {u ∈ P, u ≤ c}, P (ξ, c, d) = {u ∈ P, ξ(u) ≥ c, u ≤ d}, S(ξ, c, d) = {u ∈ P, ξ(u) > c, u ≤ d}. Theorem 1.2.12 ([72]). Suppose that A : P c → P is completely continuous, and there exists a concave positive functional ξ with ξ(u) ≤ u for all u ∈ P, and positive constants 0 < a < b ≤ c satisfying the following conditions: (1) S(ξ, a, b) = ∅, and ξ(Au) > a if u ∈ P (ξ, a, b); (2) Au ∈ P c if u ∈ P (ξ, a, c); (3) ξ(Au) > a for all u ∈ P (ξ, a, c) with Au > b. Then the fixed point index i(A, S(ξ, a, c), P c ) = 1. The Krasnosel’skii fixed point index theorem With the same notations as that used in Theorem 1.2.12, we present the next theorem. Theorem 1.2.13 (see [41, 118]). Let P be a cone in the real Banach space X, and A : P → P be a completely continuous operator. Let a, b, c be three positive constants with 0 < a < b < c. (a) If Au > u for all u ∈ ∂Pa , and Au ≤ u for all u ∈ ∂Pb , then i(A, P b \ P a , P b ) = 1. (b) If Au > u for all u ∈ ∂Pa , and Au < u for all u ∈ ∂Pb , then i(A, Pb \ P a , P c ) = 1.

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7

The nonlinear alternative of Leray–Schauder type for Kakutani maps Theorem 1.2.14 ([40]). Let C a closed convex subset of a Banach space E and U be an open subset of C with θ ∈ U . Suppose that F : U → Pcp,c (C) is an upper semicontinuous compact map. Then either (i) F has a fixed point in U or (ii) there exists u ∈ ∂U and λ ∈ (0, 1) such that u ∈ λF (u). The Covitz–Nadler fixed point theorem Theorem 1.2.15 ([27]). Let (X, d) be a complete metric space. If N : X → Pcl (X) is a contraction, then Fix N = ∅. In Theorems 1.2.14 and 1.2.15, Pcp,c (C) = {Y ∈ P(C), Y is compact and convex}, Pcl (X) = {Y ∈ P(X), Y is closed}, and Fix N is the set of the fixed points of operator N .

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Chapter 2

Riemann–Liouville Fractional Differential Equations with Nonlocal Boundary Conditions

In this chapter, we investigate the existence of positive solutions or nonnegative solutions for some Riemann-Liouville fractional differential equations with parameters or without parameters, subject to various multi-point or Riemann–Stieltjes integral boundary conditions which contain fractional derivatives. The nonlinearities of equations are nonsingular or singular functions, with nonnegative values or real values. 2.1

Singular Fractional Differential Equations with Parameters and Multi-Point Boundary Conditions

We consider the nonlinear fractional differential equation α u(t) + λf (t, u(t)) = 0, D0+

t ∈ (0, 1),

(2.1)

with the multi-point fractional boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0,

p D0+ u(1) =

m 

q ai D0+ u(ξi ),

(2.2)

i=1

where λ is a positive parameter, α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, ξi ∈ R for all i = 1, . . . , m, (m ∈ N), 0 < ξ1 < · · · < ξm ≤ 1, p, q ∈ R, k denotes the Riemann-Liouville derivative of p ∈ [1, α − 1), q ∈ [0, p], D0+ order k (for k = α, p, q), and the nonlinearity f may change sign and may be singular at t = 0 and/or t = 1. 9

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We present intervals for the parameter λ such that problem (2.1), (2.2) has at least one positive solution. Because f is a sign-changing function, our problem (2.1), (2.2) is a semipositone boundary value problem. In the proof of the main existence results, we use the Guo-Krasnosel’skii fixed point theorem (Theorem 1.2.2). By a positive solution of (2.1), (2.2), we mean a function u ∈ C[0, 1] satisfying (2.1) and (2.2), with u(t) > 0 for all t ∈ (0, 1]. This problem is a generalization of the problem studied in [60], where p ∈ N and q = 1. Other particular cases of problem (2.1), (2.2) and of problem from [60] were investigated in [105] (where n = 3, p = 0 and ai = 0 for all i = 1, . . . , m) and in [110] (where p = 0 and ai = 0 for all i = 1, . . . , m and n is an arbitrary natural number, n ≥ 3). 2.1.1

Auxiliary results

In this section, we present some auxiliary results that will be used to prove our main results. We consider the fractional differential equation α u(t) + x (t) = 0, D0+

0 < t < 1,

(2.3)

with the boundary conditions (2.2), where x  ∈ C(0, 1)∩L1 (0, 1). We denote Γ(α) Γ(α) m α−q−1 . by Δ = Γ(α−p) − Γ(α−q) i=1 ai ξi Lemma 2.1.1. If Δ = 0, then the unique solution u ∈ C[0, 1] of problem (2.3), (2.2) is given by u(t) = −

1 Γ(α)

 0

α−1

t

(t − s)α−1 x (s) ds +

t − ΔΓ(α − q)

m  i=1

 ai

0

ξi

tα−1 ΔΓ(α − p) α−q−1

(ξi − s)

 0

1

(1 − s)α−p−1 x (s) ds 

x (s) ds ,

t ∈ [0, 1]. (2.4)

Proof. By Lemma 1.1.3, we deduce that the solutions u ∈ C(0, 1)∩L1 (0, 1) of the fractional differential equation (2.3) are given by α x (t) + c1 tα−1 + · · · + cn tα−n u(t) = −I0+  t 1 =− (t − s)α−1 x (s) ds + c1 tα−1 + · · · + cn tα−n , Γ(α) 0

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where c1 , c2 , . . . , cn ∈ R. By using the conditions u(0) = u (0) = · · · = u(n−2) (0) = 0, we obtain c2 = · · · = cn = 0. Then we conclude  t 1 u(t) = c1 tα−1 − (t − s)α−1 x (s) ds, t ∈ [0, 1]. (2.5) Γ(α) 0 For the obtained function (2.5), we find p u(t) = c1 D0+

Γ(α) α−p−1 α−p t − I0+ x (t), Γ(α − p)

Γ(α) α−q−1 α−q t − I0+ x (t). Γ(α − q) m p q u(1) = i=1 ai D0+ u(ξi ) gives us Then the condition D0+  1 1 Γ(α) − (1 − s)α−p−1 x (s) ds c1 Γ(α − p) Γ(α − p) 0    ξi m  Γ(α) α−q−1 1 α−q−1 = ξ ai c1 − (ξi − s) x (s) ds . Γ(α − q) i Γ(α − q) 0 i=1 q D0+ u(t) = c1

So, we deduce 1 c1 = ΔΓ(α − p)



1

0

(1 − s)α−p−1 x (s) ds

m

 1 − ai ΔΓ(α − q) i=1

 0

ξi

 α−q−1

(ξi − s)

x (s) ds .

Replacing the above constant c1 in (2.5), we obtain the expression (2.4) for the solution u ∈ C[0, 1] of problem (2.3), (2.2). Conversely, one easily verifies that u ∈ C[0, 1] given by (2.4) satisfies Eq. (2.3) and the boundary conditions (2.2).  Lemma 2.1.2. If Δ = 0, then the solution u of problem (2.3), (2.2) given by (2.4) can be written as  1 G(t, s) x(s) ds, t ∈ [0, 1], (2.6) u(t) = 0

where the Green function G is G(t, s) = g1 (t, s) +

m tα−1  ai g2 (ξi , s), Δ i=1

(2.7)

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and



tα−1 (1 − s)α−p−1 − (t − s)α−1 , 0 ≤ s ≤ t ≤ 1, tα−1 (1 − s)α−p−1 , 0 ≤ t ≤ s ≤ 1,  α−q−1 1 (1 − s)α−p−1 − (t − s)α−q−1 , 0 ≤ s ≤ t ≤ 1, t g2 (t, s) = α−q−1 (1 − s)α−p−1 , 0 ≤ t ≤ s ≤ 1, Γ(α − q) t

1 g1 (t, s) = Γ(α)

(2.8) for all (t, s) ∈ [0, 1] × [0, 1]. Proof. By Lemma 2.1.1 and relation (2.4), we deduce  t 1 u(t) = [tα−1 (1 − s)α−p−1 − (t − s)α−1 ] x(s) ds Γ(α) 0  1 α−1 α−p−1 t (1 − s) x (s) ds + t

1 − Γ(α)

 0

1

tα−1 (1 − s)α−p−1 x (s) ds

tα−1 + ΔΓ(α − p)



1

0

(1 − s)α−p−1 x (s) ds

  m ξi  tα−1 α−q−1 − ai (ξi − s) x (s) ds ΔΓ(α − q) i=1 0  t 1 [tα−1 (1 − s)α−p−1 − (t − s)α−1 ] x(s) ds = Γ(α) 0  1 α−1 α−p−1 + t (1 − s) x (s) ds t

−

1 ΔΓ(α − p)

1 + ΔΓ(α − q) +

tα−1 ΔΓ(α − p)



1

0





tα−1 (1 − s)α−p−1 x (s) ds

m  i=1 1

0 m

ai ξiα−q−1

 0

1

tα−1 (1 − s)α−p−1 x (s) ds

(1 − s)α−p−1 x (s) ds

 tα−1 − ai ΔΓ(α − q) i=1

 0

ξi

 (ξi − s)α−q−1 x (s) ds

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13

 t 1 = [tα−1 (1 − s)α−p−1 − (t − s)α−1 ] x(s) ds Γ(α) 0  1 α−1 α−p−1 + t (1 − s) x (s) ds t



m

 tα−1 + ai ΔΓ(α − q) i=1 

− =

ξi

0

1

0

ξiα−q−1 (1 − s)α−p−1 x (s) ds 

(ξi − s)α−q−1 x (s) ds

 t 1 [tα−1 (1 − s)α−p−1 − (t − s)α−1 ] x(s) ds Γ(α) 0  1 + tα−1 (1 − s)α−p−1 x (s) ds t

m

 tα−1 + ai ΔΓ(α − q) i=1 

1

+ ξi



1

= 0



1

= 0

ξiα−q−1 (1

 0

ξi

[ξiα−q−1 (1 − s)α−p−1 − (ξi − s)α−q−1 ] x(s) ds

α−p−1

− s)

g1 (t, s) x(s) ds +

x (s) ds

 1 m tα−1  ai g2 (ξi , s) x(s) ds Δ i=1 0

G(t, s) x(s) ds,

where G and g1 , g2 are given in (2.7) and (2.8). Therefore, we obtain the expression (2.6) for the solution u of problem (2.3), (2.2) given by (2.4).  Lemma 2.1.3. The functions g1 and g2 given by (2.8) have the following properties: (a) g1 (t, s) ≤ h1 (s) for all t, s ∈ [0, 1], where h1 (s) =

1 (1 − s)α−p−1 (1 − (1 − s)p ), s ∈ [0, 1]; Γ(α)

(b) g1 (t, s) ≥ tα−1 h1 (s) for all t, s ∈ [0, 1]; α−1 (c) g1 (t, s) ≤ tΓ(α) , for all t, s ∈ [0, 1];

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(d) g2 (t, s) ≥ tα−q−1 h2 (s) for all t, s ∈ [0, 1], where h2 (s) =

1 (1 − s)α−p−1 (1 − (1 − s)p−q ), s ∈ [0, 1]; Γ(α − q)

1 (e) g2 (t, s) ≤ Γ(α−q) tα−q−1 for all t, s ∈ [0, 1]; (f) The functions g1 and g2 are continuous on [0, 1] × [0, 1]; g1 (t, s) ≥ 0, g2 (t, s) ≥ 0 for all t, s ∈ [0, 1]; g1 (t, s) > 0, g2 (t, s) > 0 for all t, s ∈ (0, 1).

Proof. (a) The function g1 is nondecreasing in the first variable. Indeed, for s ≤ t, we have 1 ∂g1 (t, s) = [(α − 1)tα−2 (1 − s)α−p−1 − (α − 1)(t − s)α−2 ] ∂t Γ(α) ≥

1 [tα−2 (1 − s)α−2 − (t − s)α−2 ] Γ(α − 1)

=

1 [(t − ts)α−2 − (t − s)α−2 ] ≥ 0. Γ(α − 1)

Then, g1 (t, s) ≤ g1 (1, s) for all (t, s) ∈ [0, 1] × [0, 1] with s ≤ t. For s ≥ t, we obtain 1 ∂g1 (t, s) = tα−2 (1 − s)α−p−1 ≥ 0. ∂t Γ(α − 1) Hence, g1 (t, s) ≤ g1 (s, s) for all (t, s) ∈ [0, 1] × [0, 1] with s ≥ t. Therefore, we deduce that g1 (t, s) ≤ h1 (s) for all (t, s) ∈ [0, 1] × [0, 1], 1 (1 − s)α−p−1 (1 − (1 − s)p ) for all s ∈ [0, 1]. where h1 (s) = g1 (1, s) = Γ(α) (b) For s ≤ t, we have g1 (t, s) =

1 [tα−1 (1 − s)α−p−1 − (t − s)α−1 ] Γ(α)

≥

1 [tα−1 (1 − s)α−p−1 − (t − ts)α−1 ] Γ(α)

=

1 α−1 t [(1 − s)α−p−1 − (1 − s)α−1 ] Γ(α)

=

1 α−1 t (1 − s)α−p−1 (1 − (1 − s)p ) = tα−1 h1 (s). Γ(α)

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Fractional Differential Equations with Nonlocal BCs

For s ≥ t, we obtain g1 (t, s) =

1 α−1 1 α−1 t t (1 − s)α−p−1 ≥ (1 − s)α−p−1 (1 − (1 − s)p ) Γ(α) Γ(α)

= tα−1 h1 (s). Hence, we conclude that g1 (t, s) ≥ tα−1 h1 (s) for all (t, s) ∈ [0, 1] × [0, 1]. (c) For all (t, s) ∈ [0, 1] × [0, 1] we have g1 (t, s) ≤

1 α−1 tα−1 t . (1 − s)α−p−1 ≤ Γ(α) Γ(α)

(d) From (2.8), if s ≤ t, we obtain g2 (t, s) =

1 [tα−q−1 (1 − s)α−p−1 − (t − s)α−q−1 ] Γ(α − q)

≥

1 [tα−q−1 (1 − s)α−p−1 − (t − ts)α−q−1 ] Γ(α − q)

=

1 tα−q−1 [(1 − s)α−p−1 − (1 − s)α−q−1 ] Γ(α − q)

=

1 tα−q−1 (1 − s)α−p−1 (1 − (1 − s)p−q ) = tα−q−1 h2 (s), Γ(α − q)

1 (1 − s)α−p−1 (1 − (1 − s)p−q ), s ∈ [0, 1]. where h2 (s) = Γ(α−q) If t ≤ s, we deduce

g2 (t, s) =

1 tα−q−1 (1 − s)α−p−1 ≥ tα−q−1 h2 (s). Γ(α − q)

Hence, we conclude g2 (t, s) ≥ tα−q−1 h2 (s) for all t, s ∈ [0, 1]. (e) We have g2 (t, s) ≤

1 tα−q−1 tα−q−1 (1 − s)α−p−1 ≤ , Γ(α − q) Γ(α − q)

∀ t, s ∈ [0, 1].

(f) This property follows from the definitions of g1 and g2 and from the properties (b) and (d) above.  Lemma 2.1.4. Assume that ai ≥ 0 for all i = 1, . . . , m and Δ > 0. Then the Green function G given by (2.7) is a continuous function on [0, 1]×[0, 1] and satisfies the inequalities: (a) G(t, s) m 1 Δ

≤ J(s) for all t, s a i=1 i g2 (ξi , s), s ∈ [0, 1];

∈

[0, 1], where J(s)

=

h1 (s) +

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(b) G(t, s) ≥ tα−1 J(s) for all t, s ∈ [0, 1]; (c) G(t, s) ≤ σtα−1 , for all t, s ∈ m α−q−1 1 . i=1 ai ξi ΔΓ(α−q)

[0, 1], where σ

=

1 Γ(α)

+

Proof. By definition of the function G, we deduce that G is a continuous function. In addition, by using Lemma 2.1.3, we obtain for all t, s ∈ [0, 1] 1 m (a) G(t, s) ≤ h1 (s) + Δ i=1 ai g2 (ξi , s) = J(s); m tα−1 α−1 α−1 h1 (s) + Δ J(s); (b) G(t, s) ≥ t i=1 ai g2 (ξi , s) = t  α−1 α−1 m α−q−1 t t α−1 (c) G(t, s) ≤ Γ(α) + ΔΓ(α−q) i=1 ai ξi = σt .  Lemma 2.1.5. Assume that ai ≥ 0 for all i = 1, . . . , m, Δ > 0, x  ∈ (t) > 0 for all t ∈ (0, 1). Then the solution u of C(0, 1) ∩ L1 (0, 1) and x problem (2.3), (2.2) given by (2.4) satisfies the inequality u(t) ≥ tα−1 u(t ) for all t, t ∈ [0, 1]. Proof. By using Lemma 2.1.4, we obtain  1  1 u(t) = G(t, s) x(s) ds ≥ tα−1 J(s) x(s) ds 0

0

≥ tα−1

 0

1

G(t , s) x(s)ds = tα−1 u(t ),



for all t, t ∈ [0, 1]. 2.1.2



Existence of positive solutions

In this section, we investigate the existence of positive solutions for our problem (2.1), (2.2). First, we present the assumptions that we shall use in the sequel. (H1) α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, ξi ∈ R for all i = 1, . . . , m, (m ∈ N), 0 < ξ1 < · · · < ξm ≤ 1, p, q ∈ R, p ∈ [1, α − 1), q ∈ [0, p], ai ≥ 0 Γ(α) Γ(α) m α−q−1 − Γ(α−q) > 0. for all i = 1, . . . , m, λ > 0, Δ = Γ(α−p) i=1 ai ξi (H2) The function f ∈ C((0, 1) × [0, ∞), R) may be singular at t = 0 and/or t = 1, and there exist the functions r, z ∈ C((0, 1), [0, ∞)), g ∈ C([0, 1] × [0, ∞), [0, ∞)) such that −r(t) ≤ f (t, x) ≤ z(t)g(t, x) 1 for all t ∈ (0, 1) and x ∈ [0, ∞), with 0 < 0 r(t) dt < ∞, 0 < 1 z(t) dt < ∞. 0

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(H3) There exists c ∈ (0, 1/2) such that f∞ = lim

min

u→∞ t∈[c,1−c]

f (t, u) = ∞. u

(H4) There exists c ∈ (0, 1/2) such that lim inf u→∞ mint∈[c,1−c] f (t, u) > L0 , with   1  −1  1−c L0 = 2σ r(s) ds cα−1 J(s) ds , and 0

c

g∞ = lim max

u→∞ t∈[0,1]

g(t, u) = 0, u

where J and σ are given in Lemma 2.1.4. We consider the fractional differential equation α D0+ x(t) + λ(f (t, [x(t) − λw(t)]∗ ) + r(t)) = 0,

0 < t < 1,

(2.9)

with the multi-point fractional boundary conditions x(0) = x (0) = · · · = x(n−2) (0) = 0,

p D0+ x(1) =

m 

q ai D0+ x(ξi ), (2.10)

i=1 ∗

∗

where λ > 0 and ζ(t) = ζ(t) if ζ(t) ≥ 0, and ζ(t) = 0 if ζ(t) < 0. Here 1 w(t) = 0 G(t, s)r(s) ds, t ∈ [0, 1] is solution of problem α D0+ w(t) + r(t) = 0,

0 < t < 1,

w(0) = w (0) = · · · = w(n−2) (0) = 0,

p D0+ w(1) =

m 

q ai D0+ w(ξi ).

i=1

Under assumptions (H1), (H2), we have w(t) ≥ 0 for all t ∈ [0, 1]. We shall prove that there exists a solution x of the problem (2.9), (2.10) with x(t) ≥ λw(t) on [0, 1] and x(t) > λw(t) on (0, 1). In this case u = x − λw represents a positive solution of the problem (2.1), (2.2). Therefore, in what follows we shall investigate the problem (2.9), (2.10). By using Lemma 2.1.2, x is a solution of the integral equation  1 G(t, s)(f (s, [x(s) − λw(s)]∗ ) + r(s)) ds, t ∈ [0, 1], x(t) = λ 0

if and only if x is a solution for problem (2.9), (2.10). We consider the Banach space X = C[0, 1] with the supremum norm u = supt∈[0,1] |u(t)|, and we define the cone P = {x ∈ X, x(t) ≥ tα−1 x , ∀ t ∈ [0, 1]}.

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For λ > 0 we introduce the operator T : X → X defined by  T x(t) = λ

0

1

G(t, s)(f (s, [x(s) − λw(s)]∗ ) + r(s)) ds,

t ∈ [0, 1], x ∈ X.

It is clear that x is a solution of problem (2.9), (2.10) if and only if x is a fixed point of operator T . Lemma 2.1.6. If (H1) and (H2) hold, then operator T : P → P is a completely continuous operator. Proof. Let x ∈ P be fixed. By using (H1) and (H2), we deduce that T x(t) < ∞ for all t ∈ [0, 1]. Besides, by Lemma 2.1.4, we obtain  T x(t) ≤ λ

0

1

J(s)(f (s, [x(s) − λw(s)]∗ ) + r(s)) ds,

∀ t ∈ [0, 1],

and  T x(t) ≥ λ

0

1

tα−1 J(s)(f (s, [x(s) − λw(s)]∗ ) + r(s))ds

≥ tα−1 T x(t ),

∀ t, t ∈ [0, 1].

Therefore, T x(t) ≥ tα−1 T x for all t ∈ [0, 1]. We deduce that T x ∈ P , and hence T (P ) ⊂ P . By using standard arguments, we conclude that operator T : P → P is a completely continuous operator.  Theorem 2.1.1. Assume that (H1)–(H3) hold. Then there exists λ∗ > 0 such that, for any λ ∈ (0, λ∗ ], the boundary value problem (2.1), (2.2) has at least one positive solution. Proof. We choose a positive number R1 > σ the set Ω1 = {x ∈ X, x < R1 }(= BR1 ). We introduce   1

∗

λ = min 1, R1 M1

0

1 0

r(s) ds > 0, and we define

−1 J(s)(z(s) + r(s)) ds ,

  with M1 = max maxt∈[0,1], u∈[0,R1 ] g(t, u), 1 .

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Fractional Differential Equations with Nonlocal BCs

Let λ ∈ (0, λ∗ ]. Because w(t) ≤ σtα−1 deduce for any x ∈ P ∩ ∂Ω1 and t ∈ [0, 1]

1 0

r(s) ds for all t ∈ [0, 1], we

[x(t) − λw(t)]∗ ≤ x(t) ≤ x ≤ R1 , and x(t) − λw(t) ≥ t

α−1

≥t

α−1

≥t

α−1

x − λσt

α−1

  ∗ R1 − λ σ   R1 − σ

0

 0 1

0

1

1

r(s) ds = t

α−1

  R1 − λσ

0



1

 r(s) ds

r(s) ds 

r(s) ds

≥ 0.

Then for any x ∈ P ∩ ∂Ω1 and t ∈ [0, 1], we obtain  T x(t) ≤ λ

1

0

∗

J(s) (z(s)g(s, [x(s) − λw(s)]∗ ) + r(s)) ds

≤ λ M1

 0

1

J(s)(z(s) + r(s)) ds ≤ R1 = x .

Therefore, we conclude T x ≤ x ,

∀ x ∈ P ∩ ∂Ω1 .

(2.11)

On the other hand, for c from (H3), we choose a positive constant L > 0 such that   L ≥ 2 λc2(α−1)

c

1−c

−1 J(s) ds

.

By (H3), we deduce that there exists a constant M0 > 0 such that f (t, u) ≥ Lu,

∀ t ∈ [c, 1 − c],

u ≥ M0 .

(2.12)

Now, we define R2 = max{2R1 , 2M0 /cα−1 } and let Ω2 = {x ∈ X, x < R2 }(= BR2 ).

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Then for any x ∈ P ∩ ∂Ω2 , we obtain  1 α−1 α−1 x(t) − λw(t) ≥ t x − λσt r(s) ds   α−1 ≥t R2 − σ

0

1

0

≥ 0,



r(s) ds

≥t

α−1

  R1 − σ



1

r(s) ds

0

∀ t ∈ [0, 1].

Therefore, we conclude   [x(t) − λw(t)]∗ = x(t) − λw(t) ≥ tα−1 R2 − σ

0

c

≥

α−1

2

R2

1

 r(s) ds

≥ M0 , ∀ t ∈ [c, 1 − c].

(2.13)

Then for any x ∈ P ∩ ∂Ω2 and t ∈ [c, 1 − c], by (2.12) and (2.13), we deduce  1−c T x(t) ≥ λ G(t, s)(f (x, [x(s) − λw(s)]∗ ) + r(s)) ds c

 ≥λ ≥

1−c

c

λLc

 G(t, s)L[x(s) − λw(s)] ds ≥ λL

2(α−1)

2

R2



1−c

c

1−c

c

tα−1 J(s)

cα−1 R2 ds 2

J(s) ds ≥ R2 = x .

Then T x ≥ x ,

∀ x ∈ P ∩ ∂Ω2 .

(2.14)

By (2.11), (2.14) and Theorem 1.2.2 (i), we conclude that T has a fixed point x1 ∈ P ∩ (Ω2 \ Ω1 ), that is, R1 ≤ x1 ≤ R2 . Since x1 ≥ R1 , we deduce    1 r(s) ds x1 (t) − λw(t) ≥ tα−1 x1 − σλ   ≥ tα−1 R1 − σ

0

α−1

0

1

r(s) ds



= Λ1 tα−1 ,

and for all t ∈ [0, 1], where Λ1 = R1 −  1 so x1 (t) ≥ λw(t) + Λ1 t σ 0 r(s) ds > 0. Let u1 (t) = x1 (t) − λw(t) for all t ∈ [0, 1]. Then u1 is a positive solution of problem (2.1), (2.2) with u1 (t) ≥ Λ1 tα−1 for all t ∈ [0, 1]. This completes the proof of Theorem 2.1.1. 

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Fractional Differential Equations with Nonlocal BCs

Theorem 2.1.2. Assume that (H1), (H2) and (H4) hold. Then there exists λ∗ > 0 such that, for any λ ≥ λ∗ , the boundary value problem (2.1), (2.2) has at least one positive solution. Proof. By (H4) there exists M2 > 0 such that   2σ

f (t, u) ≥

1

0

r(s) ds

  cα−1

1−c

c

−1 J(s) ds

,

∀ t ∈ [c, 1 − c], u ≥ M2 . We define   λ∗ = M2 cα−1 σ

0

1

−1 r(s) ds

.

1 We assume now λ ≥ λ∗ . Let R3 = 2λσ 0 r(s) ds and Ω3 = {x ∈ X, x < R3 }(= BR3 ). Then for any x ∈ P ∩ ∂Ω3 , we deduce    1  1 α−1 α−1 α−1 x(t) − λw(t) ≥ t x − λσt r(s) ds = t r(s) ds R3 − λσ = tα−1 λσ =

0



1

0

r(s) ds ≥ tα−1 λ∗ σ

M2 tα−1 ≥ 0, cα−1

0



1

0

r(s) ds

∀ t ∈ [0, 1].

Therefore, for any x ∈ P ∩ ∂Ω3 and t ∈ [c, 1 − c], we have [x(t) − λw(t)]∗ = x(t) − λw(t) ≥

M2 tα−1 ≥ M2 . cα−1

Hence, for any x ∈ P ∩ ∂Ω3 and t ∈ [c, 1 − c], we conclude  1−c T x(t) ≥ λ G(t, s)f (s, [x(s) − λw(s)]∗ ) ds  ≥λ  ≥λ =

c

1−c c 1−c c

tα−1 J(s)f (s, u(s) − λw(s)) ds t

α−1

 J(s) 2σ

R3 tα−1 ≥ R3 . cα−1

 0

1

  α−1 r(s) ds c

c

1−c

−1 J(s)

ds

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Boundary Value Problems for Fractional Differential Equations and Systems

Therefore, we obtain ∀ x ∈ P ∩ ∂Ω3 .

T x ≥ x ,

(2.15)

On  1 the other hand, we consider the positive number ε = (2λ 0 J(s)z(s) ds)−1 . Then by (H4), we deduce that there exists M3 > 0 such that g(t, u) ≤ εu for all t ∈ [0, 1], u ≥ M3 . Therefore, we obtain g(t, u) ≤ M4 + εu for all t ∈ [0, 1], u ≥ 0, where M4 = maxt∈[0,1], u∈[0,M3 ] g(t, u). We define now   1 J(s)(z(s) + r(s)) ds R4 > max R3 , 2λ max{M4 , 1} 0

and Ω4 = {x ∈ X, x < R4 }(= BR4 ). Then for any x ∈ P ∩ ∂Ω4 we have  1 x(t) − λw(t) ≥ tα−1 x − λσtα−1 r(s) ds   = tα−1 R4 − λσ = tα−1 λσ =

0

1

0



1

0



r(s) ds

  ≥ tα−1 R3 − λσ

r(s) ds ≥ tα−1 λ∗ σ

M2 tα−1 ≥ 0, cα−1

 0

0

1

1

 r(s) ds

r(s) ds

∀ t ∈ [0, 1].

Then for any x ∈ P ∩ ∂Ω4 , we obtain  1 T x(t) ≤ λ J(s)[z(s)g(s, [x(s) − λw(s)]∗ + r(s)] ds  ≤λ

0

1 0

J(s)[z(s)(M4 + ε(x(s) − λw(s))) + r(s)] ds 

≤ λ max{M4 , 1} ≤

0

1

 J(s)(z(s) + r(s)) ds + λεR4

R4 R4 + = R4 = x , 2 2

0

1

J(s)z(s) ds

∀ t ∈ [0, 1].

Therefore, T x ≤ x ,

∀ x ∈ P ∩ ∂Ω4 .

(2.16)

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Fractional Differential Equations with Nonlocal BCs

By (2.15), (2.16) and Theorem 1.2.2 (ii), we conclude that T has a fixed point x1 ∈ P ∩ (Ω4 \ Ω3 ), so R3 ≤ x1 ≤ R4 . In addition, we deduce that for all t ∈ [0, 1]  1 α−1 r(s) ds x1 (t) − λw(t) ≥ x1 (t) − λσt 0

≥t

α−1

≥t

α−1

x1 − λσt R3 − λσt

= λσtα−1

 0

1

α−1

α−1



1

0



1

0

r(s) ds

r(s) ds

r(s) ds ≥ λ∗ σtα−1



1

0

r(s) ds =

M2 tα−1 . cα−1

 1 tα−1 for all Let u1 (t) = x1 (t) − λw(t) for all t ∈ [0, 1]. Then u1 (t) ≥ Λ M 2  1 = α−1 . Hence, we conclude that u1 is a positive solution t ∈ [0, 1], where Λ c of problem (2.1), (2.2), which completes the proof of Theorem 2.1.2.  In a similar manner to that used in the proof of Theorem 2.1.2, we deduce the following result. Theorem 2.1.3. Assume (H1), (H2) and  There exists c ∈ (0, 1/2) such that (H4) fˆ∞ = lim min f (t, u) = ∞ u→∞ t∈[c,1−c]

g∞ = lim max

u→∞ t∈[0,1]

and

g(t, u) = 0, u

∗ > 0 such that, for any λ ≥ λ ∗ , the boundary hold. Then there exists λ value problem (2.1), (2.2) has at least one positive solution. 2.1.3

Examples

Let α = 10/3 (n = 4), p = 3/2, q = 4/3, m = 3, ξ1 = 1/4, ξ2 = 1/2, ξ3 = 3/4, a1 = 1, a2 = 1/2, a3 = 1/3. We consider the fractional differential equation 10/3

D0+ u(t) + λf (t, u(t)) = 0,

t ∈ (0, 1),

with the boundary conditions ⎧ ⎪ u(0) = u (0) = u (0) = 0, ⎨       1 1 3 1 4/3 1 4/3 3/2 4/3 ⎪ D D u(1) = D u u u D + + . ⎩ 0+ 0+ 0+ 0+ 4 2 2 3 4

(2.17)

(2.18)

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Then we obtain Δ = Γ(10/3)(1/Γ(11/6) − 3/4) ≈ 0.86980822. So assumption (H1) is satisfied. Besides, we deduce t7/3 (1 − s)5/6 − (t − s)7/3 , 0 ≤ s ≤ t ≤ 1, 1 g1 (t, s) = Γ(10/3) t7/3 (1 − s)5/6 , 0 ≤ t ≤ s ≤ 1, t(1 − s)5/6 − t + s, 0 ≤ s ≤ t ≤ 1, g2 (t, s) = t(1 − s)5/6 , 0 ≤ t ≤ s ≤ 1,        1 1 3 1 1 t7/3 G(t, s) = g1 (t, s) + , s + g2 , s + g2 ,s , g2 Δ 4 2 2 3 4 ∀ t, s ∈ [0, 1]. 1 (1 − s)5/6 (1 − (1 − s)3/2 ) for all s ∈ [0, 1], We also obtain h1 (s) = Γ(10/3) σ = 1/Γ(10/3) + 3/(4Δ) ≈ 1.22220971 and ⎧   1 3 h1 (s) + Δ , 0 ≤ s < 14 , (1 − s)5/6 + 22s−9 ⎪ ⎪ 4 12 ⎪ ⎪   ⎪ ⎨h1 (s) + 1 3 (1 − s)5/6 + 5s−3 , 1 ≤ s < 1 , Δ 4 6 4 2 J(s) =   1 3 4s−3 1 5/6 ⎪ h (s) + Δ 4 (1 − s) + 12 , 2 ≤ s < 34 , ⎪ ⎪ ⎪ 1 ⎪ ⎩ 3 h1 (s) + 4Δ (1 − s)5/6 , 34 ≤ s ≤ 1.

Example 1. We consider the function u2 + u + 1 f (t, u) =  + ln t, 3 t(1 − t)2

t ∈ (0, 1),

u ≥ 0.

1 We have r(t) = − ln t and z(t) = √ for all t ∈ (0, 1), g(t, u) = u2 + 3 t(1−t)2 1 1 u+1 for all t ∈ [0, 1] and u ≥ 0, 0 r(t) dt = 1, 0 z(t) dt = Γ( 23 )Γ( 13 ) ≈ 3.63. Therefore, assumption (H2) is satisfied. In addition, for c ∈ (0, 1/2) fixed, assumption (H3) is also satisfied (f∞ = ∞). 1 After some computations, we deduce that 0 J(s)(z(s) + r(s)) ds ≈ 1 1.12036124. We choose R1 = 2, (R1 > σ 0 r(s) ds), and then we obtain M1 = 7 and λ∗ ≈ 0.255. By Theorem 2.1.1, we conclude that problem (2.17), (2.18) has at least one positive solution for any λ ∈ (0, λ∗ ].

Example 2. We consider the function √ 1 u+1 −√ , f (t, u) =  4 t t3 (1 − t)

t ∈ (0, 1),

u ≥ 0.

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Fractional Differential Equations with Nonlocal BCs

1 Here, we have r(t) = √1t and z(t) = √ for all t ∈ (0, 1), 4 3 t (1−t) √ g(t, u) = u + 1 for all t ∈ [0, 1] and u ≥ 0. For c ∈ (0, 1/2) fixed, 1 1 assumptions (H2) and (H4) are satisfied ( 0 r(t) dt = 2, 0 z(t) dt ≈ 4.44, limu→∞ mint∈[c,1−c] f (t, u) = ∞ and g∞ = 0).

 3/4 For c = 1/4, we obtain 1/4 J(s) ds ≈ 0.23472325 and L0 ≈ 529.001. From the proof of Theorem 2.1.2, we deduce that M2 ≈ 91277.5 and λ∗ ≈ 948407. Then by Theorem 2.1.2, we conclude that, for any λ ≥ λ∗ , our problem (2.17), (2.18) has at least one positive solution. Remark 2.1.1. The results presented in this section, under the assumption p ∈ [1, n − 2] instead of p ∈ [1, α − 1), were published in [46]. The properties (a)–(c) of Lemma 2.1.3 are from [45].

2.2

A Fractional Differential Equation with Integral Terms and Multi-Point Boundary Conditions

In this section, we consider the nonlinear fractional integro-differential equation α u(t) + f (t, u(t), T u(t), Su(t)) = 0, D0+

t ∈ (0, 1),

(2.19)

with the multi-point boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0,

p D0+ u(1) =

m  i=1

q ai D0+ u(ξi ), (2.20)

where α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, ai , ξi ∈ R for all i = 1, . . . , m, (m ∈ N), 0 < ξ1 < · · · < ξm ≤ 1, p, q ∈ R, p ∈ [1, α − 1), q ∈ [0, p], k denotes the Riemann-Liouville derivative of order k (for k = α, p, q), D0+ t 1 T u(t) = 0 K(t, s)u(s) ds, Su(t) = 0 H(t, s)u(s) ds for all t ∈ [0, 1], and f is a nonnegative and nonsingular function. We study the existence and uniqueness of nonnegative solutions for problem (2.19), (2.20), by using the Banach contraction mapping principle (Theorem 1.2.1) and the Krasnosel’skii fixed point theorem for the sum of two operators (Theorem 1.2.3). By a nonnegative solution of (2.19), (2.20), we mean a function u ∈ C([0, 1], R+ ) satisfying (2.19) and (2.20). In [104], the authors investigated the existence of nonnegative solutions for

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the Caputo fractional differential equation c

α D0+ u(t) + f (t, u(t), T u(t), Su(t)) = 0,

t ∈ (0, 1),

with the boundary conditions ⎧ ⎪ u(0) = b0 , u (0) = b1 , . . . , u(n−3) (0) = bn−3 , ⎨  1 ⎪ u(s) ds, ⎩ u(n−1) (0) = bn−1 , u(1) = μ 0

where n− 1 < α ≤ n, 0 ≤ μ < n− 1, n ≥ 3, bi ≥ 0 (i = 1, 2, . . . , n− 3, n− 1), α D0+ is the Caputo fractional derivative, and the operators T and S are defined as the operators from our problem, given above. We will use in what follows the notations (Δ, G and σ) and the auxiliary results (Lemmas 2.1.1–2.1.4) from Section 2.1. c

2.2.1

Existence of nonnegative solutions

In this section, we prove the existence of nonnegative solutions for our problem (2.19), (2.20). First, we present the assumptions that we shall use in the sequel. (I1) α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, ξi ∈ R for all i = 1, . . . , m, (m ∈ N), 0 < ξ1 < · · · < ξm ≤ 1, p, q ∈ R, p ∈ [1, α − 1), q ∈ [0, p], Γ(α) Γ(α) m α−q−1 − Γ(α−q) > 0. ai ≥ 0 for all i = 1, . . . , m, Δ = Γ(α−p) i=1 ai ξi 3 (I2) f : [0, 1] × R+ → R+ is measurable with respect to t on [0, 1]. (I3) There exists β ∈ [1, +∞] and the functions a, b, c ∈ Lβ ((0, 1), R+ ) such that |f (t, u, v, w) − f (t, u ¯, v¯, w)| ¯ ≤ a(t)|u − u ¯| + b(t)|v − v¯| + c(t)|w − w|, ¯ a.e. t ∈ (0, 1) and for all u, v, w, u ¯, v¯, w ¯ ∈ R+ . (I4) There exists γ ∈ [1, +∞] and the function h ∈ Lγ ((0, 1), R+ ) such that |f (t, u, v, w)| ≤ h(t),

a.e. t ∈ (0, 1), ∀ u, v, w ∈ R+ .

(I5) K ∈ C(D, R+ ), D = {(t, s) ∈ [0, 1] × [0, 1], t ≥ s}, and H ∈ C([0, 1] × [0, 1], R+ ). Remark 2.2.1. In the classical case when α = n ∈ N, n ≥ 3 and p, q ∈ N, p ∈ [1, α − 1), q ∈ [0, p], the condition from assumption (I1) becomes Δ = (n−1)! (n−1)! m n−q−1 > 0. i=1 ai ξi (n−p−1)! − (n−q−1)!

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Fractional Differential Equations with Nonlocal BCs

1

We denote by k ∗ = supt∈[0,1]

t 0

page 27

K(t, s) ds and h∗ = supt∈[0,1]

H(t, s) ds. We also denote by · γ (for γ ∈ R, γ ≥ 1) and · ∞ 0 the of the Lebesgue spaces Lγ (0, 1) and L∞ (0, 1), that is g γ =  1 norms ( 0 |g(s)|γ ds)1/γ for g ∈ Lγ (0, 1), and g ∞ = esssups∈(0,1) |g(s)| for g ∈ L∞ (0, 1). Theorem 2.2.1. Assume that (I1)–(I5) hold. If σω0 < 1, where ω0 = a + k ∗ b + h∗ c 1 , then problem (2.19), (2.20) has a unique nonnegative solution on [0, 1]. Proof. For u ∈ C([0, 1], R+ ) and t ∈ [0, 1], by (I4), we have (a) for γ = 1:



t

0

|(t − s)α−1 f (s, u(s), T u(s), Su(s))| ds 

≤  ≤

t

|f (s, u(s), T u(s), Su(s))| ds

0 1 0

h(s) ds = h 1 ;

(b) for γ ∈ (1, ∞):  t |(t − s)α−1 f (s, u(s), T u(s), Su(s))| ds 0

 ≤

t

0

(t − s)α−1 h(s) ds

 ≤ ≤t (c) for γ = ∞:

t

0

(t − s)

αγ−1 γ

 0

t



(α−1)γ γ−1

γ−1 αγ − 1

ds

 γ−1  γ

 γ−1 γ

t

0



h γ ≤

γ

1/γ

h (s) ds γ−1 αγ − 1

 γ−1 γ

|(t − s)α−1 f (s, u(s), T u(s), Su(s))| ds 

≤  ≤

t 0 t 0

(t − s)α−1 h(s) ds (t − s)α−1 ds h ∞ ≤

1 h ∞ . α

h γ ;

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We deduce that the function s → (t − s)α−1 f (s, u(s), T u(s), Su(s)) is Lebesgue integrable on [0, t] for all t ∈ [0, 1] and u ∈ C([0, 1], R+ ). In a similar manner, we show that the function s → (1 − s)α−p−1 f (s, u(s), T u(s), Su(s)) is Lebesgue integrable on [0, 1] for all u ∈ C([0, 1], R+ ), and the function s → (ξi − s)α−q−1 f (s, u(s), T u(s), Su(s)) is Lebesgue integrable on [0, ξi ] for all u ∈ C([0, 1], R+ ). We will consider in what follows the integral equation  t 1 u(t) = − (t − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) 0  1 tα−1 + (1 − s)α−p−1 f (s, u(s), T u(s), Su(s)) ds ΔΓ(α − p) 0   m ξi  tα−1 α−q−1 − ai (ξi − s) f (s, u(s), T u(s), Su(s)) ds , ΔΓ(α − q) i=1 0 (2.21) or equivalently

 u(t) =

1

0

G(t, s)f (s, u(s), T u(s), Su(s)) ds.

(2.22)

By Lemmas 2.1.1 and 2.1.2, we know that u is a solution of problem (2.19), (2.20) if and only if u is a solution of equation (2.21) (or equivalently (2.22)). For any γ ∈ [1, +∞], we deduce that h ∈ L1 (0, 1). So, let r = σ h 1 . We define the operator A on Ξ1 = {u ∈ C([0, 1], R+ ), u ≤ r}, where u = supt∈[0,1] |u(t)|, by  t 1 Au(t) = − (t − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) 0  1 tα−1 (1 − s)α−p−1 f (s, u(s), T u(s), Su(s)) ds + ΔΓ(α − p) 0   m ξi  tα−1 α−q−1 − ai (ξi − s) f (s, u(s), T u(s), Su(s)) ds , ΔΓ(α − q) i=1 0 or equivalently Au(t) =

 0

1

G(t, s)f (s, u(s), T u(s), Su(s)) ds.

The function u is a solution of equation (2.21) (or (2.22)) if and only if u is a fixed point of operator A.

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29

We will investigate the existence and uniqueness of fixed points of operator A by using the Banach contraction mapping principle. First, we show that for u ∈ Ξ1 , then Au ∈ Ξ1 . Indeed, we have  1    |Au(t + δ) − Au(t)| =  [G(t + δ, s) − G(t, s)]f (s, u(s), T u(s), Su(s)) ds  ≤

0

1

|G(t + δ, s) − G(t, s)|h(s) ds

0

⎧ maxs∈[0,1] |G(t + δ, s) − G(t, s)| · h 1 → 0 ⎪ ⎪ ⎪ ⎪ ⎪ for δ → 0, if γ = 1; ⎪ ⎪ ⎪  γ−1 ⎪ ⎪ 1 γ ⎪ γ ⎪ ⎨ |G(t + δ, s) − G(t, s)| γ−1 ds h γ → 0

≤

0

⎪ for δ → 0, if γ ∈ (1, ∞); ⎪ ⎪ ⎪  1 ⎪ ⎪ ⎪ ⎪ ⎪ |G(t + δ, s) − G(t, s)| ds · h ∞ → 0 ⎪ ⎪ ⎪ ⎩ 0 for δ → 0, if γ = ∞.

So, Au is a continuous function. By (I1), (I2) and Lemma 2.1.4, we obtain Au(t) ≥ 0 for all t ∈ [0, 1], so Au ∈ C([0, 1], R+ ). Moreover, for any u ∈ Ξ1 and all t ∈ [0, 1], we deduce 

1

(Au)(t) = 0

 ≤

0

≤ σt

1

G(t, s)f (s, u(s), T u(s), Su(s)) ds σtα−1 f (s, u(s), T u(s), Su(s)) ds

α−1

 0

1

h(s) ds ≤ σ h 1 = r,

and then Au ≤ r for all u ∈ Ξ1 , so A : Ξ1 → Ξ1 . Then we show that A is a contraction mapping on Ξ1 . For u1 , u2 ∈ Ξ1 , and any t ∈ [0, 1], by using (I3), we obtain |(Au1 )(t) − (Au2 )(t)|  1  =  G(t, s)f (s, u1 (s), T u1 (s), Su1 (s)) ds 0



−

0

1

  G(t, s)f (s, u2 (s), T u2 (s), Su2 (s)) ds

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 ≤

1 0

G(t, s)|f (s, u1 (s), T u1 (s), Su1 (s))

− f (s, u2 (s), T u2 (s), Su2 (s))| ds  1 ≤ G(t, s) [a(s)|u1 (s) − u2 (s)| + b(s)|T u1 (s) 0

− T u2(s)| + c(s)|Su1 (s) − Su2 (s)|] ds  1 ≤ u1 − u2 G(t, s) [a(s) + k ∗ b(s) + h∗ c(s)] ds 0

≤ u1 − u2 σt

α−1



1

0

[a(s) + k ∗ b(s) + h∗ c(s)] ds

≤ σ u1 − u2 · a + k ∗ b + h∗ c 1 = σω0 u1 − u2 , because  T u1 (s) − T u2 (s) =

0

s

K(s, τ )[u1 (τ ) − u2 (τ )] dτ 

≤ sup |u1 (τ ) − u2 (τ )| τ ∈[0,1]

s 0

K(s, τ ) dτ

≤ k ∗ u1 − u2 , ∀ s ∈ [0, 1],  1 H(s, τ )[u1 (τ ) − u2 (τ )] dτ Su1 (s) − Su2 (s) = 0

≤ sup |u1 (τ ) − u2 (τ )| τ ∈[0,1]



1 0

H(s, τ ) dτ

≤ h∗ u1 − u2 , ∀ s ∈ [0, 1], and a, b, c ∈ Lβ (0, 1) ⊂ L1 (0, 1). Then we deduce Au1 − Au2 ≤ σω0 u1 − u2 . Because σω0 < 1, we conclude that A is a contraction mapping. By Theorem 1.2.1, we deduce that A has a unique fixed point, which is the unique nonnegative solution of problem (2.19), (2.20). 

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31

Next, we denote by  1 1 ω1 = (1 − s)α−1 (a(s) + k ∗ b(s) + h∗ c(s)) ds, Γ(α) 0   ⎧ m  ⎪ 1 1 ⎪ ∗ ∗ ⎪ a + k b + h c 1 + ai , if β = 1, ⎪ ⎪ ΔΓ(α − p) ΔΓ(α − q) i=1 ⎪ ⎪ ⎪

⎪   β−1 ⎪ β ⎪ β − 1 1 ⎪ ∗ ∗ ⎪ a + k b + h c β ⎪ ⎪ ⎪ ΔΓ(α − p) αβ − pβ − 1 ⎪ ⎪ m   ⎪  β−1 ⎨ αβ−qβ−1 β  1 β−1 β ω2 = + ai ξi , ⎪ ΔΓ(α − q) i=1 αβ − qβ − 1 ⎪ ⎪ ⎪ ⎪ ⎪ if β ∈ (1, ∞), ⎪ ⎪ ⎪   ⎪ m ⎪  ⎪ 1 1 ⎪ α−q ⎪ a + k ∗ b + h∗ c ∞ + ai ξi , ⎪ ⎪ ⎪ ΔΓ(α − p + 1) ΔΓ(α − q + 1) i=1 ⎪ ⎪ ⎩ if β = ∞. (2.23) Theorem 2.2.2. Assume that assumptions (I1), (I2) f : [0, 1] × R3+ → R+ is a continuous function, and (I3)–(I5) hold. If max{ω1 , ω2 } < 1, then problem (2.19), (2.20) has at least one nonnegative solution on [0, 1]. Proof. We choose R ≥ R0 , where

  ⎧ m  ⎪ 1 ψ0 1 ⎪ −1 ⎪ (1 − ω1 ) + h 1 + ai , ⎪ ⎪ Γ(α + 1) ΔΓ(α − p) ΔΓ(α − q) i=1 ⎪ ⎪ ⎪ ⎪ ⎪ if γ = 1, ⎪ ⎪ ⎪ ⎪

 ⎪   γ−1 ⎪ ⎪ γ ⎪ γ − 1 ψ 1 0 ⎪ −1 ⎪ (1 − ω1 ) + h γ ⎪ ⎪ Γ(α + 1) ΔΓ(α − p) αγ − pγ − 1 ⎪ ⎪ ⎪  ⎪   γ−1 m ⎪ αγ−qγ−1 γ  ⎨ γ−1 1 γ + ai ξi , R0 = ΔΓ(α − q) αγ − qγ − 1 ⎪ i=1 ⎪ ⎪ ⎪ ⎪ if γ ∈ (1, ∞), ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ψ0 ⎪ −1 ⎪ + h ∞ (1 − ω1 ) ⎪ ⎪ ⎪ Γ(α + 1) ⎪   ⎪ m ⎪  ⎪ 1 1 ⎪ α−q ⎪ + × ai ξi , ⎪ ⎪ ⎪ ΔΓ(α − p + 1) ΔΓ(α − q + 1) i=1 ⎪ ⎪ ⎩ if γ = ∞,

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ψ0 = max{f (t, 0, 0, 0), t ∈ [0, 1]}, and ω1 is given by (2.23). We consider the set Ξ2 = {u ∈ C([0, 1], R+ ), u ≤ R}. Then Ξ2 is a closed, convex, bounded and nonempty subset of C([0, 1], R+ ). We define the operators B and C on Ξ2 by  t 1 (Bu)(t) = − (t − s)α−1 f (s, u(s), T u(s), Su(s)) ds, t ∈ [0, 1], Γ(α) 0  1 tα−1 (1 − s)α−p−1 f (s, u(s), T u(s), Su(s)) ds (Cu)(t) = ΔΓ(α − p) 0 m

− ×

 tα−1 ai ΔΓ(α − q) i=1  ξi

0



α−q−1

(ξi − s)

f (s, u(s), T u(s), Su(s)) ds ,

t ∈ [0, 1].

By (I1), (I2) and Lemma 2.1.4, we have (Bu)(t) + (Cu)(t) ≥ 0 for all t ∈ [0, 1]. For any u ∈ Ξ2 , by using (I3), we deduce |f (t, u(t), T u(t), Su(t))| ≤ |f (t, u(t), T u(t), Su(t)) − f (t, 0, 0, 0)| + |f (t, 0, 0, 0)| ≤ a(t)|u(t)| + b(t)|T u(t)| + c(t)|Su(t)| + ψ0 ,

∀ t ∈ [0, 1].

Then for any u ∈ Ξ2 and all t ∈ [0, 1], we obtain by using the above inequality    t   1 |(Bu)(t)| = − (t − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) 0  t 1 (t − s)α−1 (a(s)|u(s)| + b(s)|T u(s)| ≤ Γ(α) 0 +c(s)|Su(s)| + ψ0 ) ds  1 u ≤ (1 − s)α−1 (a(s) + k ∗ b(s) + h∗ c(s)) ds Γ(α) 0  t 1 (t − s)α−1 ψ0 ds + Γ(α) 0 ≤ ω1 R +

tα ψ0 ψ0 ≤ ω1 R + , Γ(α + 1) Γ(α + 1)

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t because (T u)(t) ≤ supt∈[0,1] u(t) 0 K(t, s) ds ≤ k ∗ u and (Su)(t) ≤ 1 supt∈[0,1] u(t) 0 H(t, s) ds ≤ h∗ u for all t ∈ [0, 1]. Then for any v ∈ Ξ2 and all t ∈ [0, 1], we deduce, by (I4) (a) for γ = 1: tα−1 |Cv(t)| ≤ ΔΓ(α − p)



1 0

(1 − s)α−p−1 f (s, v(s), T v(s), Sv(s)) ds

m

+

 tα−1 ai ΔΓ(α − q) i=1  ξi

× ≤

0

(ξi − s)

tα−1 ΔΓ(α − p)



1 0

≤

 tα−1 ai ΔΓ(α − q) i=1 

f (s, v(s), T v(s), Sv(s)) ds

(1 − s)α−p−1 h(s) ds

m

+



α−q−1

 0

ξi

(ξi − s)α−q−1 h(s) ds m

 1 ai ΔΓ(α − q) i=1 0  m  1 1 + ai ; ΔΓ(α − p) ΔΓ(α − q) i=1

1 ΔΓ(α − p) 

= h 1

1

h(s) ds +



1

h(s) ds

0

(b) for γ ∈ (1, ∞): |Cv(t)| ≤

tα−1 ΔΓ(α − p)



1

0

(1 − s)α−p−1 h(s) ds

m

 t ai + ΔΓ(α − q) i=1 α−1

1 ≤ ΔΓ(α − p)



1

0 m



ξi

0

(1 − s)

 1 + ai ΔΓ(α − q) i=1

(ξi − s)α−q−1 h(s) ds γ(α−p−1) γ−1

 0

ξi

ds

(ξi − s)

 γ−1  γ

γ(α−q−1) γ−1

1

0

γ

h (s) ds

 γ1

 γ−1  γ

ds

0

1

γ

h (s) ds

 γ1

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 = h γ

1 ΔΓ(α − p)

1 + ΔΓ(α − q)





γ−1 αγ − pγ − 1

γ−1 αγ − qγ − 1

 γ−1 γ

 γ−1 m γ  i=1

 αγ−qγ−1 γ

ai ξi

;

(c) for γ = ∞: tα−1 |Cv(t)| ≤ ΔΓ(α − p)



1 0

(1 − s)α−p−1 h(s) ds

m

 tα−1 + ai ΔΓ(α − q) i=1 ≤

1 h ∞ ΔΓ(α − p)



 0 1

0

ξi

 α−q−1

(ξi − s)

h(s) ds

(1 − s)α−p−1 ds

 ξi m  1 h ∞ ai (ξi − s)α−q−1 ds ΔΓ(α − q) 0 i=1   m  1 1 α−q + ai ξi . = h ∞ ΔΓ(α − p + 1) ΔΓ(α − q + 1) i=1 +

Then, for u, v ∈ Ξ2 , and t ∈ [0, 1], we have (a) for γ = 1: |Bu(t) + Cv(t)| ≤ |Bu(t)| + |Cv(t)| ≤ ω1 R +  + h 1

ψ0 Γ(α + 1) m

 1 1 + ai ΔΓ(α − p) ΔΓ(α − q) i=1

 ≤ R;

(b) for γ ∈ (1, ∞): ψ0 Γ(α + 1)    γ−1 γ γ−1 1 + h γ ΔΓ(α − p) αγ − pγ − 1

|Bu(t) + Cv(t)| ≤ |Bu(t)| + |Cv(t)| ≤ ω1 R +

1 + ΔΓ(α − q)



γ−1 αγ − qγ − 1

 γ−1 m γ  i=1

 αγ−qγ−1 γ

ai ξi

≤ R;

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Fractional Differential Equations with Nonlocal BCs

(c) for γ = ∞: |Bu(t) + Cv(t)| ≤ |Bu(t)| + |Cv(t)| ≤ ω1 R +  + h ∞

ψ0 Γ(α + 1) m

 1 1 + ai ξiα−q ΔΓ(α − p + 1) ΔΓ(α − q + 1) i=1

 ≤ R.

Therefore, for v1 , v2 ∈ Ξ2 and t ∈ [0, 1], by using (I3), we deduce |Cv1 (t) − Cv2 (t)| ≤

tα−1 ΔΓ(α − p)



1

0

(1 − s)α−p−1 |f (s, v1 (s), T v1 (s), Sv1 (s))

−f (s, v2 (s), T v2 (s), Sv2 (s))| ds  m ξi  tα−1 ai (ξi − s)α−q−1 |f (s, v1 (s), + ΔΓ(α − q) i=1 0  T v1 (s), Sv1 (s)) − f (s, v2 (s), T v2 (s), Sv2 (s))| ds ≤

1 ΔΓ(α − p)



1

0

(1 − s)α−p−1 (a(s)|v1 (s) − v2 (s)|

+ b(s)|T v1 (s) − T v2 (s)| + c(s)|Sv1 (s) − Sv2 (s)|) ds

 m ξi  1 + ai (ξi − s)α−q−1 (a(s)|v1 (s) ΔΓ(α − q) i=1 0 − v2 (s)| + b(s)|T v1 (s) − T v2 (s)|  + c(s)|Sv1 (s) − Sv2 (s)|) ds ≤

v1 − v2 ΔΓ(α − p)

 0

1

(1 − s)α−p−1 (a(s) + k ∗ b(s) + h∗ c(s)) ds

m

v1 − v2  + ai ΔΓ(α − q) i=1

 0



+ k ∗ b(s) + h∗ c(s)) ds . So, we obtain

ξi

(ξi − s)α−q−1 (a(s)

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( a) for β = 1: |Cv1 (t) − Cv2 (t)| ≤

v1 − v2 ΔΓ(α − p)

 0

1

(a(s) + k ∗ b(s) + h∗ c(s)) ds

m

+

v1 − v2  ai ΔΓ(α − q) i=1

 0

1

(a(s) + k ∗ b(s) + h∗ c(s)) ds

= v1 − v2 · a + k ∗ b + h∗ c 1   m  1 1 + × ai = ω2 v1 − v2 ; ΔΓ(α − p) ΔΓ(α − q) i=1 (b) for β ∈ (1, ∞):  β−1  1 β β(α−p−1) v1 − v2 β−1 (1 − s) ds |Cv1 (t) − Cv2 (t)| ≤ ΔΓ(α − p) 0  1  β1 ∗ ∗ β × (a(s) + k b(s) + h c(s)) ds 0

m

v1 − v2  + ai ΔΓ(α − q) i=1  ×

0

1

∗



ξi 0

(ξi − s) ∗

β(α−q−1) β−1

β

(a(s) + k b(s) + h c(s)) ds

 β−1 β ds

 β1

= v1 − v2 · a + k ∗ b + h∗ c β

  β−1 β 1 β−1 × ΔΓ(α − p) αβ − pβ − 1 

m

αβ−qβ−1  1 + ai ξi β ΔΓ(α − q) i=1

β−1 αβ − qβ − 1

  β−1 β

= ω2 v1 − v2 ; ( c) for β = ∞: v1 − v2 a + k ∗ b + h∗ c ∞ |Cv1 (t) − Cv2 (t)| ≤ ΔΓ(α − p)

 0

1

(1 − s)α−p−1 ds

 ξi m v1 − v2  ∗ ∗ ai a + k b + h c ∞ (ξi − s)α−q−1 ds + ΔΓ(α − q) i=1 0

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37

= v1 − v2 · a + k ∗ b + h∗ c ∞   m  1 1 α−q + × ai ξi ΔΓ(α − p + 1) ΔΓ(α − q + 1) i=1 = ω2 v1 − v2 , where ω2 is given by (2.23). Because ω2 < 1, we deduce that C is a contraction mapping. By using assumptions (I2) and (I5), we conclude that B is a continuous mapping. In addition, B is uniformly bounded on Ξ2 , because for any u ∈ Ξ2 , we have    t  1  α−1  |Bu(t)| =  (t − s) f (s, u(s), T u(s), Su(s)) ds Γ(α) 0  t  1 1 1 ≤ (t − s)α−1 h(s) ds ≤ h(s) ds Γ(α) 0 Γ(α) 0 =

1 h 1 , Γ(α)

∀ t ∈ [0, 1],

1 h 1 for all u ∈ Ξ2 . so Bu ≤ Γ(α) The operator B is also equicontinuous on Ξ2 . Indeed, let u ∈ Ξ2 , t1 , t2 ∈ [0, 1], with t1 < t2 . We obtain   t2  1  |Bu(t2 ) − Bu(t1 )| =  (t2 − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) 0   t1  1 − (t1 − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) 0  t1   1 ≤ (t2 − s)α−1 − (t1 − s)α−1 Γ(α) 0

×f (s, u(s), T u(s), Su(s)) ds  t2 1 + (t2 − s)α−1 f (s, u(s), T u(s), Su(s)) ds Γ(α) t1   t1   1 (t2 − s)α−1 − (t1 − s)α−1 ds ≤ ψ1 Γ(α) 0  t2 1 α−1 + (t2 − s) ds Γ(α) t1 =

ψ1 ψ1 (t2 − t1 ) (tα − tα , 1) ≤ Γ(α + 1) 2 Γ(α)

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where ψ1 = sup{f (t, u, v, w), t ∈ [0, 1], |u| ≤ R, |v| ≤ k ∗ R, |w| ≤ h∗ R}. So, it is clear that |Bu(t2 ) − Bu(t1 )| → 0 as t2 − t1 → 0. By using the Arzela–Ascoli theorem, we deduce that B(Ξ2 ) is relatively compact, and so B is completely continuous. By Theorem 1.2.3, we conclude that operator B + C has at least one fixed point, and then problem (2.19), (2.20) has at least one nonnegative solution.  2.2.2

An example

Let α = 10/3 (n = 4), p = 3/2, q = 4/3, m = 3, ξ1 = 1/4, ξ2 = 1/2, ξ3 = 3/4, a1 = 1, a2 = 1/2, a3 = 1/3. We consider the fractional differential equation 10/3

D0+ u(t) + f (t, u(t), T u(t), Su(t)) = 0,

t ∈ (0, 1),

(2.24)

with the boundary conditions ⎧ ⎪ u(0) = u (0) = u (0) = 0, ⎨       (2.25) 1 1 3 1 4/3 1 4/3 3/2 4/3 ⎪ + D0+ u + D0+ u , ⎩ D0+ u(1) = D0+ u 4 2 2 3 4 1 t where T u(t) = 0 K(t, s)u(s) ds and Su(t) = 0 H(t, s)u(s) ds for all t ∈ [0, 1], with K(t, s) = e−t s for all t, s ∈ [0, 1] with s ≤ t, and H(t, s) = e−2t (s + 1) for all t, s ∈ [0, 1]. Then we obtain Δ = Γ(10/3)(1/Γ(11/6) − 3/4) ≈ 0.86980822, and σ = 1/Γ(10/3) + 3/(4Δ) ≈ 1.22220971. So, assumptions (I1) and (I5) are satisfied. We define the function f (t, u, v, w) =

e−3t u e−2t v e−t w t + + + , 2 (1 + k)(1 + u) (1 + k 2 )(1 + v) (1 + k 3 )(1 + w)

for all t ∈ [0, 1] and u, v, w ∈ R+ , with k ≥ 1. We deduce that k ∗ = t 1 1 and h∗ = supt∈[0,1] 0 H(t, s) ds = 32 . In addisupt∈[0,1] 0 K(t, s) ds = 2e tion, we obtain the inequalities |f (t, u, v, w) − f (t, u ¯, v¯, w)| ¯ ≤

e−2t e−t e−3t |u − u¯| + |v − v ¯ | + |w − w|, ¯ 1+k 1 + k2 1 + k3

for all t ∈ [0, 1], u, v, w, u ¯, v¯, w ¯ ∈ R+ , and |f (t, u, v, w)| ≤

e−3t e−2t e−t t + + + , 2 1 + k 1 + k2 1 + k3

∀ t ∈ [0, 1], u, v, w ∈ R+ .

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We define a(t) = −2t

e 1+k2

−t

e 1+k3 , γ

e−3t 1+k ,

b(t) =

e−2t 1+k2 ,

c(t) =

e−t 1+k3 ,

and h(t) =

t 2

+

e−3t 1+k β

+

+ for all t ∈ [0, 1]. For β, γ ∈ [1, ∞], we have a, b, c ∈ L (0, 1) and h ∈ L (0, 1). So, assumptions (I2)–(I4) are also satisfied. 3 2 3(e−1) −1 −1 e3 −1 e2 −1 + 4e3e(1+k Besides, we obtain ω0 = 3ee3 (1+k) 2 ) + 2e(1+k3 ) ≤ 6e3 + 8e3 + 3(e−1) 4e

≈ 0.67222, and so ω0 < 1/σ ≈ 0.81819. Therefore, by Theorem 2.2.1, we deduce that problem (2.24), (2.25) has a unique nonnegative and nontrivial solution. Remark 2.2.2. The results presented in this section were published in [47].

2.3

Semipositone Singular Fractional Boundary Value Problems with Integral Boundary Conditions

In this section, we consider the nonlinear fractional differential equation α u(t) + f (t, u(t)) = 0, D0+

t ∈ (0, 1),

(2.26)

with the integral boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0,

p D0+ u(1) =

 0

1

q D0+ u(t) dH0 (t),

(2.27) where α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, p, q ∈ R, p ∈ [1, α − 1), q ∈ [0, p], k denotes the Riemann–Liouville derivative of order k (for k = α, p, q), D0+ the integral from the boundary condition (2.27) is the Riemann–Stieltjes integral with H0 a function of bounded variation, the nonlinearity f (t, u) may change sign and may be singular at the points t = 0, t = 1 and/or u = 0. We present conditions on the functions f and H0 such that problem (2.26), (2.27) has at least one positive solution or at least m positive solutions (m ∈ N, n ≥ 2). In the proof of our main results, we use an approximation method and the Guo-Krasnosel’skii fixed point theorem (Theorem 1.2.2). By a positive solution of (2.26), (2.27) we mean a function u ∈ C[0, 1] satisfying (2.26) and (2.27) with u(t) > 0 for all t ∈ (0, 1]. This problem is a generalization of the problem studied in [93], where the function H0 is a step function, that is, the boundary conditions are of the

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form u(0) = u (0) = · · · = u(n−2) (0) = 0,

p D0+ u(1) =

M  i=1

q ai D0+ u(ξi ),

and ai , ξi ∈ R for all i = 1, . . . , M , (M ∈ N), 0 < ξ1 < · · · < ξM < 1. 2.3.1

Preliminary results

We consider the fractional differential equation α u(t) + x (t) = 0, D0+

t ∈ (0, 1),

(2.28)

with the integral boundary conditions (2.27), where x  ∈ C(0, 1) ∩ L1 (0, 1). 1 α−q−1 Γ(α) Γ(α) dH0 (s). We denote by Δ1 = Γ(α−p) − Γ(α−q) 0 s Lemma 2.3.1. If Δ1 = 0, then the unique solution u ∈ C[0, 1] of problem (2.28), (2.27) is given by  t  1 1 tα−1 u(t) = − (t − s)α−1 x (s) ds + (1 − s)α−p−1 x (s) ds Γ(α) 0 Δ1 Γ(α − p) 0   1  s tα−1 α−q−1 − (s − τ ) x (τ ) dτ dH0 (s), t ∈ [0, 1]. Δ1 Γ(α − q) 0 0 (2.29) Proof. By Lemma 1.1.3, we deduce that the solutions u ∈ C(0, 1)∩L1 (0, 1) of the fractional differential equation (2.28) are given by α x (t) + c1 tα−1 + · · · + cn tα−n u(t) = −I0+  t 1 =− (t − s)α−1 x (s) ds + c1 tα−1 + · · · + cn tα−n , Γ(α) 0

where c1 , c2 , . . . , cn ∈ R. By using the conditions u(0) = u (0) = · · · = u(n−2) (0) = 0, we obtain c2 = · · · = cn = 0. Then we conclude  t 1 α−1 u(t) = c1 t − (t − s)α−1 x (s) ds, t ∈ [0, 1]. (2.30) Γ(α) 0 For the obtained function (2.30), we find p D0+ u(t) = c1

Γ(α) α−p−1 α−p − I0+ x (t), t Γ(α − p)

q D0+ u(t) = c1

Γ(α) α−q−1 α−q t − I0+ x (t). Γ(α − q)

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1 q p Then the condition D0+ u(1) = 0 D0+ u(t) dH0 (t) gives us    1 Γ(α) Γ(α) α−q−1 α−p α−q − I0+ x t c1 (1) = − I0+ x (t) dH0 (t). c1 Γ(α − p) Γ(α − q) 0 So, we find c1 =

1 Δ1 Γ(α − p) −



1

0

1 Δ1 Γ(α − q)

(1 − s)α−p−1 x (s) ds



1

0

 0

s

(s − τ )α−q−1 x (τ ) dτ

 dH0 (s).

Replacing the above constant c1 in (2.30), we obtain the expression (2.29) for the solution u ∈ C[0, 1] of problem (2.28), (2.27). Conversely, one easily verifies that u ∈ C[0, 1] given by (2.29) satisfies equation (2.28) and the boundary conditions (2.27).  Lemma 2.3.2. If Δ1 = 0, then the solution u of problem (2.28), (2.27) given by (2.29) can be written as  1 G1 (t, s) x(s) ds, t ∈ [0, 1], (2.31) u(t) = 0

where the Green function G1 is G1 (t, s) = g1 (t, s) +

tα−1 Δ1



1

0

g2 (τ, s) dH0 (τ ),

and g1 , g2 are given by (2.8). Proof. By Lemma 2.3.1 and relation (2.29), we deduce  t  α−1  1 u(t) = t (s) ds (1 − s)α−p−1 − (t − s)α−1 x Γ(α) 0  1 α−1 α−p−1 + t (1 − s) x (s) ds t

1 − Γ(α)

 0

1

tα−1 (1 − s)α−p−1 x (s) ds

tα−1 + Δ1 Γ(α − p) tα−1 − Δ1 Γ(α − q)

 

1

0

0

1

(1 − s)α−p−1 x (s) ds  0

s

α−q−1

(s − τ )

 x (τ ) dτ

dH0 (s)

(2.32)

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 t  α−1  1 t (s) ds = (1 − s)α−p−1 − (t − s)α−1 x Γ(α) 0  1 1 α−1 α−p−1 + t (1 − s) x (s) ds − Δ Γ(α − p) 1 t   1  1 1 α−1 α−p−1 α−q−1 t (1 − s) x (s) ds + s dH0 (s) × Δ1 Γ(α − q) 0 0  1  1 tα−1 α−1 α−p−1 × t (1 − s) x (s) ds + (1 − s)α−p−1 x (s) ds Δ1 Γ(α − p) 0 0   1  s tα−1 α−q−1 (s − τ ) x (τ ) dτ dH0 (s) − Δ1 Γ(α − q) 0 0  t  α−1  1 = t (s) ds (1 − s)α−p−1 − (t − s)α−1 x Γ(α) 0  1 + tα−1 (1 − s)α−p−1 x (s) ds t

 1  1 tα−1 τ α−q−1 dH0 (τ ) (1 − s)α−p−1 x (s) ds Δ1 Γ(α − q) 0 0    1  s − (s − τ )α−q−1 x (τ ) dτ dH0 (s)

+

0

=



0

t  1 tα−1 (1 − s)α−p−1 − (t − s)α−1 x (s) ds Γ(α) 0  1 + tα−1 (1 − s)α−p−1 x (s) ds t

  1  1 tα−1 τ α−q−1 (1 − s)α−p−1 dH0 (τ ) x (s) ds Δ1 Γ(α − q) 0 0    1  1 − (s − τ )α−q−1 dH0 (s) x (τ ) dτ +



0

1

= 0

 =

0

1

τ

g1 (t, s) x(s) ds +

tα−1 Δ1

 0

1

 0

1

 g2 (τ, s) dH0 (τ ) x (s) ds

G1 (t, s) x(s) ds.

Hence, we obtain the expression (2.31) for the solution u of problem (2.28), (2.27) given by (2.29). 

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43

Lemma 2.3.3. Assume that H0 : [0, 1] → R is a nondecreasing function and Δ1 > 0. Then the function G1 given by (2.32)is a continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G1 (t, s) ≤ J1 (s) for all t, s ∈ [0, 1], where J1 (s) = h1 (s) + Δ11 1 1 × 0 g2 (τ, s) dH0 (τ ), s ∈ [0, 1], and h1 (s) = Γ(α) (1 − s)α−p−1 (1 − (1 − s)p ), s ∈ [0, 1]; (b) G1 (t, s) ≥ tα−1 J1 (s) for all t, s ∈ [0, 1]; 1 1 + Δ1 Γ(α−q) (c) G1 (t, s) ≤ σ1 tα−1 , for all t, s ∈ [0, 1], where σ1 = Γ(α)  1 α−q−1 × 0 τ dH0 (τ ). Proof. By the definition of function G1 we deduce that G1 is a continuous function. In addition, by using Lemma 2.1.3, we obtain for all t, s ∈ [0, 1] 1 (a) G1 (t, s) ≤ h1 (s) + Δ11 0 g2 (τ, s) dH0 (τ ) = J1 (s); α−1  1 (b) G1 (t, s) ≥ tα−1 h1 (s) + tΔ1 0 g2 (τ, s) dH0 (τ ) = tα−1 J1 (s); 1 α−1 α−1 (c) G1 (t, s) ≤ tΓ(α) + Δ1tΓ(α−q) 0 τ α−q−1 dH0 (τ ) = σ1 tα−1 .  Lemma 2.3.4. Assume that H0 : [0, 1] → R is a nondecreasing function,  ∈ C(0, 1) ∩ L1 (0, 1) and x (t) ≥ 0 for all t ∈ (0, 1). Then the Δ1 > 0, x solution u of problem (2.28), (2.27) given by (2.29) satisfies the inequality u(t) ≥ tα−1 u(t ) for all t, t ∈ [0, 1]. Proof. By using Lemma 2.3.3, we obtain for all t, t ∈ [0, 1]  1  1 G1 (t, s) x(s) ds ≥ tα−1 J1 (s) x(s) ds u(t) = 0

≥ tα−1

2.3.2

 0

0

1

G1 (t , s) x(s) ds = tα−1 u(t ).



Existence and multiplicity of positive solutions

In this section, we prove the existence of at least one or at least m (m ∈ N, m ≥ 2) positive solutions for our problem (2.26), (2.27). We introduce the assumptions that we will use in the sequel. (L1) α ∈ R, α ∈ (n − 1, n], n ∈ N, n ≥ 3, p, q ∈ R, p ∈ [1, α − 1), q ∈ [0, p], H0 : [0, 1] → R is a nondecreasing function, and Δ1 = 1 α−q−1 Γ(α) Γ(α) dH0 (s) > 0. Γ(α−p) − Γ(α−q) 0 s

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(L2) The function f ∈ C((0, 1) × (0, ∞), R) and there exists a function r ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) such that f (t,  1x) ≥ −r(t) for all t ∈ (0, 1) and x ∈ (0, ∞), and r 1 > 0, ( r 1 = 0 |r(t)| dt is the norm of r in L1 (0, 1)). (L3) For any positive numbers θ1 , θ2 , θ1 < θ2 , there exists a function hθ1 ,θ2 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) such that f (t, x) ≤ hθ1 ,θ2 (t),

θ1 tα−1 ≤ x ≤ θ2 .

0 < t < 1,

(L4) There exists a positive constant γ1 > σ1 r 1 such that ϕ(s, γ1 ) ds ≤ γ1 , where

1 0

J1 (s)

ϕ(t, γ1 ) = max{f (t, x), (γ1 − σ1 r 1 )tα−1 ≤ x ≤ γ1 } + r(t),

∀ t ∈ (0, 1).

(L5) There exists a positive constant γ2 > γ1 such that ψ(s, γ2 ) ds ≥ γ2 , where

1 0

J1 (s)

ψ(t, γ2 ) = min{f (t, x), (γ2 − σ1 r 1 )tα−1 ≤ x ≤ γ2 } + r(t),

∀ t ∈ (0, 1).

We approximate the problem (2.26), (2.27) by the approximating fractional differential equation α D0+ x(t) + f (t, χn (x(t) − w(t))) + r(t) = 0,

0 < t < 1,

(2.33)

with the boundary conditions 

(n−2)

x(0) = x (0) = · · · = x

(0) = 0,

p D0+ x(1)

 = 0

1

q D0+ x(t) dH0 (t),

(2.34)   where χn (x) = x, x ≥ n1 ; n1 , x < n1 , , n ∈ N. 1 Here, w(t) = 0 G1 (t, s)r(s) ds, t ∈ [0, 1] is solution of problem α D0+ w(t) + r(t) = 0,

0 < t < 1,

w(0) = w (0) = · · · = w(n−2) (0) = 0,

p D0+ w(1) =

1 0

q D0+ w(t) dH0 (t).

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45

By (L1), (L2), Lemmas 2.3.3 and 2.3.4, we obtain w(t) ≥ 0 and w(t) ≥ tα−1 maxs∈[0,1] w(s) for all t ∈ [0, 1], and  1  1 w(t) = G1 (t, s)r(s) ds ≤ σ1 tα−1 r(s) ds = σ1 tα−1 r 1 , ∀ t ∈ [0, 1]. 0

0

By using Lemma 2.3.2, x is solution of the integral equation  1 G1 (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds, t ∈ [0, 1], x(t) =

(2.35)

0

if and only if x is solution for problem (2.33), (2.34). We consider the Banach space X = C[0, 1] with the supremum norm · , and we define the cone P = {x ∈ X, x(t) ≥ tα−1 x , ∀ t ∈ [0, 1]}. For r > 0, we denote by Br = {x ∈ X, x < r} and B r = {x ∈ X, x ≤ r}. We introduce now the operators An : X → X, n ∈ N defined by  1 G1 (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds, (An x)(t) = 0

∀ t ∈ [0, 1],

x ∈ X.

It is clear that xn is a solution of problem (2.33), (2.34) if and only if xn is a fixed point of operator An . Lemma 2.3.5. If (L1)–(L3) hold, then for any θ1 , θ2 with σ r 1 < θ1 < θ2 , the operator An : P ∩ (B θ2 \ Bθ1 ) → P for n ≥ [1/θ1 ] + 1 fixed, is completely continuous. Proof. Let n ≥ [1/θ1 ] + 1 be fixed and let x ∈ P ∩ (B θ2 \ Bθ1 ). Then for all t ∈ [0, 1] we obtain x(t) − w(t) ≥ tα−1 x − σ1 tα−1 r 1 ≥ tα−1 (θ1 − σ1 r 1 ) ≥ 0, and

 1 tα−1 (θ1 − σ1 r 1 ) ≤ max x(t) − w(t), ≤ θ2 , n

that is tα−1 (θ1 − σ1 r 1 ) ≤ χn (x(t) − w(t)) ≤ θ2 .

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Then by (L1)–(L3) and Lemma 2.3.3, we deduce  1 1 1 J1 (s) ≤ + τ α−q−1 dH0 (τ ) Γ(α) Δ1 Γ(α − q) 0 1 1 + (H0 (1) − H0 (0)) =: M0 , ∀ s ∈ [0, 1], Γ(α) Δ1 Γ(α − q)

≤ and



0 ≤ (An x)(t) ≤

0

 ≤

1

0

1

J1 (s)(f (x, χn (x(s) − w(s))) + r(s)) ds J1 (s)(hθ1 −σ1 r1 ,θ2 (s) + r(s)) ds

≤ M0 ( hθ1 −σ1 r1 ,θ2 1 + r 1 ) < ∞,

∀ t ∈ [0, 1]. (2.36)

We conclude that An : P ∩ (B θ2 \ Bθ1 ) → X is well defined. Besides, by Lemma 2.3.3, we obtain for all t, t ∈ [0, 1] that  1 (An x)(t) ≤ J1 (s)(f (s, χn (x(s) − w(s)) + r(s)) ds, 0

 (An x)(t) ≥

0

1

tα−1 J1 (s)(f (x, χn (x(s) − w(s)) + r(s)) ds ≥ tα−1 (An x)(t ).

Therefore, (An x)(t) ≥ tα−1 An x for all t ∈ [0, 1]. We deduce that An x ∈ P , and hence An (P ∩ (B θ2 \ Bθ1 )) ⊂ P . In what follows, we will prove that An is a completely continuous operator. By assumptions (L2)–(L3) and the Lebesgue’s dominated convergence theorem, we obtain that An is continuous in the space X. Besides, An is compact, that is, for any bounded set E ⊂ P ∩ (B θ2 \ Bθ1 ), the set An (E) is relatively compact. Indeed, let E be a bounded set in P ∩ (B θ2 \ Bθ1 ). By (2.36), we deduce that An (E) is uniformly bounded. We will prove next that An (E) is equicontinuous. For this, we estimate (An x) with x ∈ E, and we obtain  1    ∂G1   (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds |(An x) (t)| =  ∂t 0  1     ∂g1 (α − 1)tα−2 1  (t, s) + ≤ g2 (τ, s) dH0 (τ ) ∂t Δ1 0 0   × (f (s, χn (x(s) − w(s))) + r(s)) ds

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 ≤

0

1

page 47

 (α − 1) tα−1 (1 − s)α−p−1 α−2

+ (t − s)

tα−2 + Δ1



1

0

 g2 (τ, s) dH0 (τ )

× (f (s, χn (x(s) − w(s))) + r(s)) ds  1 ≤ (α − 1)[(1 − s)α−p−1 + (1 − s)α−2 0

1 + Δ1

 0

1

g2 (τ, s) dH0 (τ )](hθ1 −σ1 r1 ,θ2 (s) + r(s)) ds

 ≤ (α − 1) 2 +

 1 (H0 (1) − H0 (0)) Δ1 Γ(α − q)

× ( hθ1 −σ1 r1 ,θ2 1 + r 1 ) =: M1 , ∀ t ∈ (0, 1). Then     t    |(An x)(t ) − (An x)(t)| =  (An x) (s) ds  t  ≤ M1 |t − t|,

∀ t, t ∈ [0, 1],

x ∈ E.

Hence, we conclude that An (E) is equicontinuous. By using Arzela– Ascoli theorem, we deduce that An (E) is relatively compact, and then An : P ∩ (B θ2 \ Bθ1 ) → P is a compact operator, and so it is completely continuous.  Lemma 2.3.6. If (L1)–(L5) hold and n ≥ n0 is fixed, where n0 = [1/γ1 ] + 1, then the operator An : P ∩ (B γ2 \ Bγ1 ) → P has a fixed point in P ∩ (B γ2 \ Bγ1 ). Proof. Let x ∈ P ∩ ∂Bγ1 . Then we have x(t) ≥ tα−1 x = γ1 tα−1 for all t ∈ [0, 1], and tα−1 (γ1 − σ1 r 1 ) ≤ χn (x(t) − w(t)) ≤ γ1 ,

∀ t ∈ [0, 1].

From the definition of ϕ(t, γ1 ), we deduce f (t, χn (x(t) − w(t))) + r(t) ≤ ϕ(t, γ1 ),

∀ t ∈ (0, 1).

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By using (L4) and Lemma 2.3.3, we obtain  1 (An x)(t) = G1 (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds 0

 ≤

1

0

J1 (s)ϕ(s, γ1 ) ds ≤ γ1 ,

∀ t ∈ [0, 1].

Then An x ≤ γ1 = x ,

∀ x ∈ P ∩ ∂Bγ1 .

(2.37)

Next, let x ∈ P ∩ ∂Bγ2 . Then we have x(t) ≥ tα−1 x = γ2 tα−1 for all t ∈ [0, 1], and tα−1 (γ2 − σ1 r 1 ) ≤ χn (x(t) − w(t)) ≤ γ2 . From the definition of ψ(t, γ2 ), we deduce f (t, χn (x(t) − w(t))) + r(t) ≥ ψ(t, γ2 ),

∀ t ∈ (0, 1).

By using (L5) and Lemma 2.3.3, we obtain  1 α−1 (An x)(t) ≥ t J1 (s)(f (s, χn (x(s) − w(s))) + r(s)) ds 0

≥ tα−1



0

1

J1 (s)ψ(s, γ2 ) ds ≥ tα−1 γ2 ,

∀ t ∈ [0, 1].

Then An x ≥ sup tα−1 γ2 = γ2 = x , ∀ x ∈ P ∩ ∂Bγ2 .

(2.38)

t∈[0,1]

By (2.37), (2.38), Lemma 2.3.5 and Theorem 1.2.2, we conclude that the operator An (with n ≥ n0 ) has a fixed point xn ∈ P ∩ (B γ2 \ Bγ1 ).  Theorem 2.3.1. Assume that the assumptions (L1)–(L5) hold. Then the problem (2.26), (2.27) has at least one positive solution. Proof. By Lemma 2.3.6 we know that the operators An , n ≥ n0 have the fixed points xn ∈ P ∩ (B γ2 \ Bγ1 ), that is, xn (t) = (An xn )(t) for all t ∈ [0, 1] or  1 G1 (t, s)(f (s, χn (xn (s) − w(s))) + r(s)) ds, ∀ t ∈ [0, 1]. xn (t) = 0

(2.39)

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In addition, the functions xn , n ≥ n0 satisfy the inequalities xn (t) − w(t) ≥ tα−1 ( xn − σ1 r 1 ) ≥ tα−1 (γ1 − σ1 r 1 ),

∀ t ∈ [0, 1]. (2.40)

By the proof of Lemma 2.3.5 we deduce that the set {xn , n ≥ n0 } is relatively compact in the space X. Then (xn )n≥n0 contains a subsequence (xnk )k≥n0 which converges in X for k → ∞ to a function x ∈ X. By taking k → ∞ in relations (2.40) and (2.39) (with n = nk ), we obtain x(t) − w(t) ≥ tα−1 (γ1 − σ1 r 1 ),

∀ t ∈ [0, 1],

(2.41)

and  x(t) =

1 0

G1 (t, s)(f (s, x(s) − w(s)) + r(s)) ds,

∀ t ∈ [0, 1].

(2.42)

By (2.41) and (2.42), we conclude that the function u(t) = x(t) − w(t), t ∈ [0, 1] satisfies the relations u(t) ≥ tα−1 (γ1 − σ1 r 1 ) for all t ∈ [0, 1] and  1 u(t) = G1 (t, s)f (s, u(s)) ds, t ∈ [0, 1], 0

that is, u is a positive solution of problem (2.26), (2.27).



In a similar manner as that used in the proof of Theorem 2.3.1, we obtain the existence of m (m ∈ N, m ≥ 2) positive solutions for problem (2.26), (2.27) if we replace the assumptions (L4) and (L5) by the following ones:   There exist γi > σ1 r 1 , i = 1, . . . , m such that 1 J1 (s)ϕ(s, γi ) ds ≤ (L4) 0 γi , i = 1, . . . , m, where ϕ(t, γi ) = max{f (t, x), (γi − σ1 r 1 )tα−1 ≤ x ≤ γi } + r(t), ∀ t ∈ (0, 1),

i = 1, . . . , m;

 There exist Γi , i = 1, . . . , m with 0 < γ1 < Γ1 < γ2 < Γ2 < · · · < (L5) 1 γm < Γm such that 0 J1 (s)ψ(s, Γi ) ds ≥ Γi , i = 1, . . . , m, where ψ(t, Γi ) = min{f (t, x), (Γi − σ1 r 1 )tα−1 ≤ x ≤ Γi } + r(t), ∀ t ∈ (0, 1),

i = 1, . . . , m.

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 (L5)  hold. Theorem 2.3.2. Assume that the assumptions (L1)–(L3), (L4), Then the boundary value problem (2.26), (2.27) has at least m positive solutions.

2.3.3

An example

Let α = 10/3 (n = 4), p = 5/4, q = 1/2, and H0 (t) = {1, if t ∈ [0, 1/4); 4/3, if t ∈ [1/4, 1/2); 2, if t ∈ [1/2, 3/4]; 4t − 1, if t ∈ (3/4, 1]}. We consider the fractional differential equation 10/3 D0+ u(t)

1 +  4 8 t(1 − t)3



1 u3 (t) +  5 u(t)

 −

1 √ = 0, 10 t

t ∈ (0, 1), (2.43)

with the boundary conditions ⎧ ⎪ ⎨

u(0) = u (0) = u (0) = 0,      1 (2.44) 1 1 1 1/2 2 1/2 5/4 1/2 ⎪ D0+ u(t) dt. + D0+ u +4 ⎩ D0+ u(1) = D0+ u 3 4 3 2 3/4 1 (x3 t(1−t)3

Here, f (t, x) = √ 4 8

+

1 √ 5 x)

−

1√ 10 t

for all t ∈ (0, 1), x ∈ (0, ∞),

and r(t) = 101√t for all t ∈ (0, 1). We obtain Δ1 ≈ 1.06472292 > 0, σ1 ≈ 0.90470354, and r 1 = 0.2. Besides, f satisfies the inequality f (t, x) ≤ hθ1 ,θ2 (t) for 0 < t < 1 and θ1 tα−1 ≤ x ≤ θ2 , where 1 hθ1 ,θ2 (t) =  (θ23 + (θ1 t7/3 )−1/5 ), 4 8 t(1 − t)3

∀ t ∈ (0, 1),

and hθ1 ,θ2 ∈ L1 (0, 1). In addition, we deduce 1 (1 − s)13/12 (1 − (1 − s)5/4 ), s ∈ [0, 1], Γ(10/3)  11/6 1 (1 − s)13/12 − (t − s)11/6 , 0 ≤ s ≤ t ≤ 1, t g2 (t, s) = 11/6 (1 − s)13/12 , 0 ≤ t ≤ s ≤ 1, Γ(17/6) t h1 (s) =

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J1 (s) = h1 (s) +

1 Δ1

 0

1

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51

g2 (τ, s) dH0 (τ ),



     1 1 1 2 1 1 g2 , s + g2 ,s + 4 = h1 (s) + g2 (τ, s) dτ Δ1 3 4 3 2 3/4 ⎧ 1 13/12 ⎪ (1 − (1 − s)5/4 ) ⎪ Γ(10/3) (1 − s) ⎪ ⎪ ⎪   ! ⎪ !11/6 " ⎪ 1 1 1 11/6 1 ⎪ 13/12 ⎪ + (1 − s) − − s ⎪ Δ1 Γ(17/6) 3 4 4 ⎪ ⎪ ⎪  ! " ⎪ ! ⎪ 11/6 11/6 ⎪ ⎪ (1 − s)13/12 − 12 − s + 23 12 ⎪ ⎪ ⎪ ⎪  ⎪ ! ⎪ ⎪ 6 6 3 17/6 13/12 ⎪ − 17 (1 − s)13/12 + 4 ⎪ 17 (1 − s) 4 ⎪ ⎪ ⎪ ⎪ !17/6 "# ⎪ ⎪ 6 6 3 ⎪ , 0 ≤ s < 14 , − 17 (1 − s)17/6 + 17 ⎪ 4 −s ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ (1 − s)13/12 (1 − (1 − s)5/4 ) ⎪ ⎪ Γ(10/3) ⎪ ⎪  ⎪ ! ⎪ 1 1 1 11/6 ⎪ ⎪ (1 − s)13/12 + ⎪ Δ Γ(17/6) 3 4 1 ⎪ ⎪ ⎪  ! ⎪ !11/6 " ⎪ 2 1 11/6 1 ⎪ 13/12 ⎪ + (1 − s) − − s ⎪ 3 2 2 ⎪ ⎪ ⎪  ⎪ ! ⎪ ⎪ 6 6 3 17/6 ⎪ + 4 17 (1 − s)13/12 − 17 (1 − s)13/12 ⎪ 4 ⎪ ⎪ ⎨ !17/6 "# 1 6 6 3 , 4 ≤ s < 12 , − 17 (1 − s)17/6 + 17 − s = 4 ⎪ ⎪ ⎪ 1 13/12 ⎪ (1 − (1 − s)5/4 ) ⎪ Γ(10/3) (1 − s) ⎪ ⎪ ⎪ ⎪  ⎪ ! ⎪ 1 1 1 11/6 ⎪ + (1 − s)13/12 ⎪ ⎪ Δ Γ(17/6) 3 4 1 ⎪ ⎪ ⎪ !11/6 ⎪ ⎪ ⎪ + 23 12 (1 − s)13/12 ⎪ ⎪ ⎪ ⎪  ⎪ ! ⎪ 6 6 3 17/6 6 ⎪ + 4 17 (1 − s)13/12 − 17 (1 − s)13/12 − 17 (1 − s)17/6 ⎪ ⎪ 4 ⎪ "# ⎪ !17/6 ⎪ ⎪ 6 3 ⎪ , 12 ≤ s < 34 , + 17 ⎪ 4 −s ⎪ ⎪ ⎪ ⎪ ⎪ 1 (1 − s)13/12 (1 − (1 − s)5/4 ) ⎪ Γ(10/3) ⎪ ⎪  ⎪ ! ⎪ ⎪ 1 1 1 11/6 ⎪ + (1 − s)13/12 ⎪ Δ1 Γ(17/6) 3 4 ⎪ ⎪ ⎪ !11/6 ⎪ ⎪ ⎪ + 23 12 (1 − s)13/12 ⎪ ⎪ ⎪ ⎪  ⎪ ! ⎪ 6 6 3 17/6 ⎪ + 4 (1 − s)13/12 − 17 (1 − s)13/12 ⎪ ⎪ 17 4 ⎪ ⎪ "# ⎪ ⎪ ⎩ − 6 (1 − s)17/6 , 3 ≤ s ≤ 1. 17 4

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For γ1 = 1, we obtain   1 1 3 7/3 max x + √ ≤x≤1 ϕ(t, 1) =  , (1 − σ1 r 1 )t 5 x 8 4 t(1 − t)3   1 1 ≤  1+  , t ∈ (0, 1), 5 8 4 t(1 − t)3 (1 − σ1 r 1 )t7/3 and for γ2 = 15, we get   1 1 3 7/3 √ min x + r )t ≤ x ≤ 15 ψ(t, 15) =  , (15 − σ 1 1 5 x 8 4 t(1 − t)3   1 1 7/3 3 √ ≥  r )t ) + ((15 − σ , t ∈ (0, 1). 1 1 5 15 8 4 t(1 − t)3 Then after some computations, we deduce  0

1

 J1 (s)ϕ(s, 1) ds ≤

1

1 J1 (s)  4 8 s(1 − s)3 0   1 × 1+  ds ≈ 0.1315 ≤ 1, 5 (1 − σ1 r 1 )s7/3

and  0

1

 J1 (s)ψ(s, 15) ds ≥

1

1 J1 (s)  4 8 s(1 − s)3 0   1 × ((15 − σ1 r 1 )s7/3 )3 + √ ds ≈ 19.2383 ≥ 15. 5 15

Therefore, all assumptions (L1)–(L5) are satisfied, and so by Theorem 2.3.1, we conclude that the boundary value problem (2.43), (2.44) has at least one positive solution u(t), t ∈ [0, 1] which satisfies the inequality u(t) ≥ (1 − 0.2σ1 )t7/3 for all t ∈ [0, 1]. Remark 2.3.1. The results presented in this section were published in [78].

page 52

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2.4

Singular Fractional Differential Equations with General Integral Boundary Conditions

In this section, we investigate the nonlinear fractional differential equation α u(t) + f (t, u(t)) = 0, D0+

t ∈ (0, 1),

(2.45)

with the general integral boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0,

β0 D0+ u(1) =

m   i=1

1 0

βi D0+ u(t) dHi (t),

(2.46) where α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, k denotes the Riemann– 0 ≤ β1 < β2 < · · · < βm ≤ β0 < α − 1, β0 ≥ 1, D0+ Liouville derivative of order k (for k = α, β0 , β1 , . . . , βm ), the integrals from the boundary condition (2.46) are Riemann–Stieltjes integrals with Hi , i = 1, . . . , m, functions of bounded variation, the nonlinearity f (t, u) may change sign and may be singular at the points t = 0, t = 1 and/or u = 0. We present conditions on the functions f and Hi , i = 1, . . . , m, such that problem (2.45), (2.46) has multiple positive solutions u ∈ C([0, 1], R+ ), u(t) > 0 for all t ∈ (0, 1]. The boundary conditions (2.46) cover various cases, such as multi-point boundary conditions when the functions Hi are step functions, or classical integral boundary conditions, or a combination of them. In our main results, we use various height functions of the nonlinearity of equation defined on special bounded sets, an approximation method, and two theorems from the fixed point index theory (Theorems 1.2.12 and 1.2.13). In [118], the authors proved the existence of at least three positive solutions for equation (2.45) with the boundary conditions  η β β u(1) = λ h(t)D0+ u(t) dt, u(0) = u (0) = · · · = u(n−2) (0) = 0, D0+ 0

(2.47) η where β ≥ 1, α − β − 1 > 0, 0 < η ≤ 1, 0 ≤ λ 0 h(t)tα−β−1 dt < 1, h ∈ L1 (0, 1) is nonnegative and may be singular at t = 0 and t = 1, and the function f is nonnegative and may be singular at the points t = 0, t = 1 and u = 0. Our boundary conditions (2.46) are more general than the above boundary conditions (2.47). Indeed, the last rela1 β β u(1) = 0 D0+ u(t) dH(t), with tion from (2.47) can be written as D0+

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  # η t H(t) = λ 0 h(s) ds, t ∈ [0, η]; λ 0 h(s) ds, t ∈ (η, 1] , and in the righthand side of the last condition in (2.46) we have a sum of Riemann-Stieltjes integrals from Riemann-Liouville derivatives of various orders. In [118], the authors used different height functions of the nonlinear term on special bounded sets and the Krasnosel’skii and the Leggett-Williams fixed point index theorems. 2.4.1

Auxiliary results

We consider the fractional differential equation α u(t) + x (t) = 0, D0+

t ∈ (0, 1),

(2.48)

with the boundary conditions (2.46), where x  ∈ C(0, 1)∩L1 (0, 1). We denote m Γ(α)  1 α−βi −1 Γ(α) dHi (s). by Δ2 = Γ(α−β0 ) − i=1 Γ(α−βi ) 0 s Lemma 2.4.1. If Δ2 = 0, then the unique solution u ∈ C[0, 1] of problem (2.48), (2.46) is given by  t tα−1 1 (t − s)α−1 x (s) ds + u(t) = − Γ(α) 0 Δ2 Γ(α − β0 )  1 m 1 tα−1  × (1 − s)α−β0 −1 x (s) ds − Δ2 i=1 Γ(α − βi ) 0   1  s α−βi −1 (s − τ ) x (τ ) dτ dHi (s), t ∈ [0, 1]. (2.49) × 0

0

Proof. By Lemma 1.1.3, we deduce that the solutions u ∈ C(0, 1)∩L1 (0, 1) of the fractional differential equation (2.48) are given by α x (t) + c1 tα−1 + · · · + cn tα−n u(t) = −I0+  t 1 =− (t − s)α−1 x (s) ds + c1 tα−1 + · · · + cn tα−n , Γ(α) 0

where c1 , c2 , . . . , cn ∈ R. By using the conditions u(0) = u (0) = · · · = u(n−2) (0) = 0, we obtain c2 = · · · = cn = 0. Then we conclude  t 1 α−1 − (t − s)α−1 x (s) ds, t ∈ [0, 1]. (2.50) u(t) = c1 t Γ(α) 0 For the obtained function (2.50), we find βi u(t) = c1 D0+

Γ(α) α−βi tα−βi −1 − I0+ x (t), Γ(α − βi )

i = 0, 1, . . . , m.

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m  1

β0 Then the condition D0+ u(1) =

c1

i=1 0

βi D0+ u(t) dHi (t) gives us

Γ(α) α−β0 − I0+ x (1) Γ(α − β0 )  m  1  Γ(α) α−βi α−βi −1 t − I0+ x (t) dHi (t), = c1 Γ(α − βi ) i=1 0

or  c1 =

m

 Γ(α) Γ(α) − Γ(α − β0 ) i=1 Γ(α − βi ) 1 Γ(α − β0 ) −

m  i=1



0

1

 0

1

 s

α−βi −1

dHi (s)

(1 − s)α−β0 −1 x (s) ds

1 Γ(α − βi )



1

0

 0

s

(s − τ )α−βi −1 x (τ ) dτ

 dHi (s).

So, we deduce  1 1 c1 = (1 − s)α−β0 −1 x (s) ds Δ2 Γ(α − β0 ) 0   s m  1  1 1 α−βi −1 (s − τ ) x (τ ) dτ dHi (s). − Δ2 i=1 0 Γ(α − βi ) 0 Replacing the above constant c1 in (2.50), we obtain the expression (2.49) for the solution u ∈ C[0, 1] of problem (2.48), (2.46). Conversely, one easily verifies that u ∈ C[0, 1] given by (2.49) satisfies equation (2.48) and the boundary conditions (2.46).  Lemma 2.4.2. If Δ2 = 0, then the solution u of problem (2.48), (2.46) given by (2.49) can be written as  u(t) =

0

1

G2 (t, s) x(s) ds,

t ∈ [0, 1],

(2.51)

where the Green function G2 is  m  1 tα−1  G2 (t, s) = g0 (t, s) + g1i (τ, s) dHi (τ ) , Δ2 i=1 0

(2.52)

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and 1 g0 (t, s) = Γ(α)



tα−1 (1 − s)α−β0 −1 − (t − s)α−1 , t

α−1

0 ≤ s ≤ t ≤ 1,

α−β0 −1

(1 − s) , 0 ≤ t ≤ s ≤ 1, ⎧ α−βi −1 (1 − s)α−β0 −1 − (t − s)α−βi −1 , ⎪ ⎨t 1 0 ≤ s ≤ t ≤ 1, g1i (t, s) = Γ(α − βi ) ⎪ ⎩ α−βi −1 t (1 − s)α−β0 −1 , 0 ≤ t ≤ s ≤ 1,

(2.53)

for all (t, s) ∈ [0, 1] × [0, 1], i = 1, . . . , m. Proof. By using Lemma 2.4.1 and relation (2.49), we have u(t) =

 t  α−1  1 t (s) ds (1 − s)α−β0 −1 − (t − s)α−1 x Γ(α) 0  1 + tα−1 (1 − s)α−β0 −1 x (s) ds t

−

1 Γ(α)



1

0

tα−1 (1 − s)α−β0 −1 x (s) ds

α−1

+

t Δ2 Γ(α − β0 ) α−1

t − Δ2  = 0

1

m  i=1

 0

1

(1 − s)α−β0 −1 x (s) ds

1 Γ(α − βi )

g0 (t, s) x(s) ds −

 0

1

 0

s

α−βi −1

(s − τ )

1 Δ2 Γ(α − β0 )

 0

1

 x (τ ) dτ

dHi (s)

tα−1 (1 − s)α−β0 −1 x (s) ds

 1 tα−1 (1 − s)α−β0 −1 x (s) ds Δ2 Γ(α) 0 m   Γ(α)  1 α−βi −1 × t dHi (t) Γ(α − βi ) 0 i=1  1 tα−1 + (1 − s)α−β0 −1 x (s) ds Δ2 Γ(α − β0 ) 0   1  s m 1 tα−1  α−βi −1 (s − τ ) x (τ ) dτ dHi (s) − Δ2 i=1 Γ(α − βi ) 0 0 +



1

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57

  1 tα−1 α−β0 −1 = g0 (t, s) x(s) ds + (1 − s) x (s) ds Δ2 0 0 m   1  1 × τ α−βi −1 dHi (τ ) Γ(α − β ) i 0 i=1    s  m 1  1 α−βi −1 − (s − τ ) x (τ ) dτ dHi (s) Γ(α − βi ) 0 0 i=1   1  m  tα−1 1  1 = g0 (t, s) x(s) ds + g2i (τ, s) dHi (τ ) x (s) ds Δ2 0 0 i=1 0  1 = G2 (t, s) x(s) ds. 

1

0

Therefore, we obtain the expression (2.51) for the solution u of problem (2.48), (2.46) given by (2.49).  Based on Lemma 2.1.3, we obtain some properties of the functions g0 and g1i , i = 1, . . . , m in the following lemma. Lemma 2.4.3. The functions g0 and g1i , i = 1, . . . , m given by (2.53) have the following properties: (a) g0 (t, s) ≤ h0 (s) for all t, s ∈ [0, 1], where h0 (s) =

1 (1 − s)α−β0 −1 (1 − (1 − s)β0 ), s ∈ [0, 1]; Γ(α)

(b) g0 (t, s) ≥ tα−1 h0 (s) for all t, s ∈ [0, 1]; α−1 (c) g0 (t, s) ≤ tΓ(α) , for all t, s ∈ [0, 1]; (d) g1i (t, s) ≥ tα−βi −1 h1i (s) for all t, s ∈ [0, 1], where h1i (s) =

1 (1−s)α−β0 −1 (1−(1−s)β0 −βi ), s ∈ [0, 1], i = 1, . . . , m; Γ(α − βi )

1 tα−βi −1 for all t, s ∈ [0, 1], i = 1, . . . , m; (e) g2i (t, s) ≤ Γ(α−β i) (f) The functions g0 and g1i are continuous on [0, 1] × [0, 1]; g0 (t, s) ≥ 0, g1i (t, s) ≥ 0 for all t, s ∈ [0, 1]; g0 (t, s) > 0, g1i (t, s) > 0 for all t, s ∈ (0, 1), i = 1, . . . , m.

By the definitions of the functions G2 , g0 , g1i , i = 1, . . . , m, we obtain, in a similar manner as we proved Lemmas 2.3.3 and 2.3.4, the following results.

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Lemma 2.4.4. Assume that Hi : [0, 1] → R, i = 1, . . . , m are nondecreasing functions and Δ2 > 0. Then the function G2 given by (2.52) is a continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G2 (t, s) ≤ J2 (s) for all t, s ∈ [0, 1], where J2 (s) = h0 (s) + 1 0 g1i (τ, s) dHi (τ ), s ∈ [0, 1]; (b) G2 (t, s) ≥ tα−1 J2 (s) for all t, s ∈ [0, 1]; 1 + (c) G2 (t, s) ≤ σ2 tα−1 , for all t, s ∈ [0, 1], where σ2 = Γ(α)  1 α−βi −1 1 dHi (τ ). Γ(α−βi ) 0 τ

1 Δ2

1 Δ2

m

i=1

m

i=1

Lemma 2.4.5. Assume that Hi : [0, 1] → R, i = 1, . . . , m are nondecreas ∈ C(0, 1) ∩ L1 (0, 1) and x (t) ≥ 0 for all t ∈ (0, 1). ing functions, Δ2 > 0, x Then the solution u of problem (2.48), (2.46) given by (2.51) satisfies the inequality u(t) ≥ tα−1 u(t ) for all t, t ∈ [0, 1]. 2.4.2

Existence of multiple positive solutions

In this section, we prove the existence of at least two positive solutions for problem (2.45), (2.46). We introduce the first assumptions that we will use in the sequel. (D1) α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, 0 ≤ β1 < β2 < · · · < βm ≤ β0 < α − 1, β0 ≥ 1, Hi : [0, 1] → Γ(α) − R, i = 1, . . . , m are nondecreasing functions, and Δ2 = Γ(α−β 0)  m 1 Γ(α) α−βi −1 dHi (s) > 0. i=1 Γ(α−βi ) 0 s (D2) The function f ∈ C((0, 1) × (0, ∞), R) and there exists a function r ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) such that f (t,  1x) ≥ −r(t) for all t ∈ (0, 1) and x ∈ (0, ∞), and r 1 > 0, ( r 1 = 0 |r(t)| dt). (D3) For any positive numbers θ1 , θ2 with θ1 < θ2 , there exists a function hθ1 ,θ2 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) such that f (t, x) ≤ hθ1 ,θ2 (t),

0 < t < 1,

θ1 tα−1 ≤ x ≤ θ2 .

For γ, γ1 , γ2 > 0 with σ2 r 1 < γ, σ2 r 1 < γ1 < γ2 we introduce the height functions: ϕ(t, γ) = max{f (t, x), (γ − σ2 r 1 )tα−1 ≤ x ≤ γ} + r(t), ψ(t, γ) = min{f (t, x), (γ − σ2 r 1 )t

α−1

≤ x ≤ γ} + r(t),

∀ t ∈ (0, 1), ∀ t ∈ (0, 1),

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59

Φ(t, γ1 , γ2 ) = max{f (t, x), (γ1 − σ2 r 1 )tα−1 ≤ x ≤ γ2 } + r(t), ∀ t ∈ (0, 1), Ψ(t, γ1 , γ2 ) = min{f (t, x), (γ1 − σ2 r 1 )tα−1 ≤ x ≤ γ2 } + r(t), ∀ t ∈ (0, 1). We approximate the problem (2.45), (2.46) by an approximating fractional differential equation where the nonlinearity is nonnegative and in the second variable of f we have a nonnegative function, namely the equation α x(t) + f (t, χn (x(t) − w(t))) + r(t) = 0, D0+

0 < t < 1,

(2.54)

with the boundary conditions x(0) = x (0) = · · · = x(n−2) (0) = 0,

β0 D0+ x(1) =

m   i=1

1

0

βi D0+ x(t) dHi (t),

(2.55)   where χn (x) = x, x ≥ n1 ; n1 , x < n1 , n ∈ N. 1 Here, w(t) = 0 G2 (t, s)r(s) ds, t ∈ [0, 1] is solution of problem ⎧ α ⎪ ⎨D0+ w(t) + r(t) = 0, 0 < t < 1, m  1  β0 βi  (n−2) ⎪ (0) = 0, D0+ w(1) = D0+ w(t) dHi (t). ⎩w(0) = w (0) = · · · = w i=1 0

By (D1), (D2), Lemmas 2.4.4 and 2.4.5, we deduce that w(t) ≥ 0, w(t) ≥ tα−1 maxs∈[0,1] w(s) for all t ∈ [0, 1], and  1  1 G2 (t, s)r(s) ds ≤ σ2 tα−1 r(s) ds = σ2 tα−1 r 1 , ∀ t ∈ [0, 1]. w(t) = 0

0

By Lemma 2.4.2, the function x ∈ C[0, 1] is solution of the integral equation  1 G2 (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds, t ∈ [0, 1], (2.56) x(t) = 0

if and only if x ∈ C[0, 1] is solution for problem (2.54), (2.55). We consider the Banach space X = C[0, 1] with the supremum norm · , and we define the cone P = {x ∈ X, x(t) ≥ tα−1 x , ∀ t ∈ [0, 1]}.

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We introduce the operators An : X → X, n ∈ N defined by  1 (An x)(t) = G2 (t, s)(f (s, χn (x(s)−w(s)))+r(s)) ds, ∀ t ∈ [0, 1], x ∈ X. 0

We can easily see that if (D1)–(D3) hold, then xn is a fixed point of operator An if and only if xn is a solution of problem (2.54), (2.55). Remark 2.4.1. By using standard arguments, we can show that under assumptions (D1)–(D3), for any θ1 , θ2 with σ2 r 1 < θ1 < θ2 , the operator An : P θ2 \ Pθ1 → P for n ≥ [1/θ1 ] + 1 fixed, is completely continuous, where Pθ1 = {u ∈ P, u < θ1 }, P θ2 = {u ∈ P, u ≤ θ2 }. Theorem 2.4.1. Assume that (D1)–(D3) hold, and c, d ∈ R, 0 < c < d ≤ 1. In addition, there exist five positive constants δi , i = 1, . . . , 5 such that σ2 r 1 < δ1 < δ2 < δ3 < δ4 ≤ δ5 with δ3 c1−α ≤ δ4 and 1 (I1 ) 0 J2 (s)ϕ(s, δ2 ) ds < δ2 ; 1 (I2 ) 0 J2 (s)ψ(s, δ1 ) ds > δ1 ; 1 (I3 ) 0 J2 (s)Φ(s, δ3 , δ5 ) ds ≤ δ5 ; d (I4 ) c J2 (s)Ψ(s, δ3 , δ4 ) ds > δ3 c1−α . Let n ≥ n0 be fixed, where n0 = [1/δ1 ] + 1. Then the operator An has at least three fixed points in P δ5 . Proof. By Remark 2.4.1, for θ1 = δ1 and θ2 = δ5 , we know that the operator An : P δ5 \ Pδ1 → P for n ≥ [1/δ1 ] + 1 fixed, is completely continuous. By the extension theorem, An has a completely continuous extension (also denoted by An ) from P into P . We define the functional ξ(u) = mint∈[c,d] u(t) for any u ∈ P . We will verify the conditions of the Leggett–Williams theorem (Theorem 1.2.12). First, we show that for n ≥ n0 we have i(An , S(ξ, δ3 , δ5 ), P δ5 ) = 1. We set x0 (t) = (δ3 +δ4 )/2, t ∈ [0, 1]. Then x0 ∈ P (that is, x0 (t) ≥ tα−1 x0 for all t ∈ [0, 1]), ξ(x0 ) = (δ3 + δ4 )/2 > δ3 , and x0 = (δ3 + δ4 )/2 ≤ δ4 . So, x0 ∈ S(ξ, δ3 , δ4 ) and then S(ξ, δ3 , δ4 ) = ∅. In addition, if x ∈ P (ξ, δ3 , δ4 ), that is x ∈ P, ξ(x) ≥ δ3 and x ≤ δ4 , we obtain x = max x(t) ≥ min x(t) ≥ δ3 , t∈[0,1]

t∈[c,d]

x(t) ≥ tα−1 x ≥ tα−1 δ3 , ∀ t ∈ [0, 1],  1 α−1 t (δ3 − σ2 r 1 ) ≤ x(t) − w(t) ≤ max x(t) − w(t), n = χn (x(t) − w(t)) ≤ δ4 ,

∀ t ∈ [0, 1].

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From the definition of Ψ(t, δ3 , δ4 ) we deduce that f (t, χn (x(t) − w(t))) + r(t) ≥ Ψ(t, δ3 , δ4 ),

∀ t ∈ [c, d].

By using Lemma 2.4.4 and assumption (I4 ), we conclude that  1 (An x)(t) = G2 (t, s)(f (s, χn (x(s) − w(s))) + r(s)) ds 0

 ≥

c

d

tα−1 J2 (s)Ψ(s, δ3 , δ4 ) ds

≥ cα−1

 c

d

J2 (s)Ψ(s, δ3 , δ4 ) ds > cα−1 c1−α δ3 = δ3 ,

∀ t ∈ [0, 1],

and then ξ(An x) = min (An x)(t) > δ3 .

(2.57)

t∈[c,d]

So, we have the condition (1) from Theorem 1.2.12. Next, we prove that if x ∈ P (ξ, δ3 , δ5 ), then An x ∈ P δ5 . So, let x ∈ X, x(t) ≥ tα−1 x , mint∈[c,d] x(t) ≥ δ3 and x ≤ δ5 . We obtain δ3 ≤ min x(t) ≤ max x(t) = x ≤ δ5 , t∈[c,d]

t∈[0,1]

and as above, tα−1 (δ3 − σ2 r 1 ) ≤ χn (x(t) − w(t)) ≤ δ5 ,

∀ t ∈ [0, 1].

(2.58)

By Lemma 2.4.4 and assumption (I3 ),  1 (An x)(t) ≤ J2 (s)(f (s, χn (x(s) − w(s))) + r(s)) ds 0

 ≤

0

1

J2 (s)Φ(s, δ3 , δ5 ) ds ≤ δ5 ,

∀ t ∈ [0, 1],

(2.59)

and so An x ≤ δ5 , which means that An x ∈ P δ5 . Therefore, we obtain condition (2) from Theorem 1.2.12. Finally, we consider x ∈ P (ξ, δ3 , δ5 ) with An x > δ4 . By using Lemma 2.4.5, ξ(An x) = min (An x)(t) ≥ min tα−1 An x t∈[c,d]

t∈[c,d]

= cα−1 An x > cα−1 δ4 ≥ δ3 , that is, we obtain condition (3) of Theorem 1.2.12.

(2.60)

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Therefore, by Theorem 1.2.12 with A = An , a = δ3 , b = δ4 and c = δ5 , we conclude that i(An , S(ξ, δ3 , δ5 ), P δ5 ) = 1.

(2.61) α−1

α−1

Next, if x ∈ ∂Pδ5 , then x ∈ P with x = δ5 and δ3 t ≤ δ5 t ≤ x(t) ≤ δ5 for all t ∈ [0, 1]. In addition, the relation (2.58) holds. Hence, by (2.58), assumption (I3 ), and Lemma 2.4.4, in a similar manner as we proved relation (2.59), we obtain An x ≤ δ5 ,

∀ x ∈ ∂Pδ5 .

(2.62)

If u ∈ ∂Pδ2 , then x ∈ P , x = δ2 and δ2 tα−1 ≤ x(t) ≤ δ2 , for all t ∈ [0, 1]. Thus, (δ2 − σ2 r 1 )tα−1 ≤ χn (x(t) − w(t)) ≤ δ2 ,

∀ t ∈ [0, 1].

By assumption (I1 ) and Lemma 2.4.4,  1 (An x)(t) ≤ J2 (s)(f (x, χn (x(s) − w(s))) + r(s)) ds 0

 ≤

1

0

J2 (s)ϕ(s, δ2 ) ds < δ2 ,

∀ t ∈ [0, 1],

and then An x < δ2 ,

∀ x ∈ ∂Pδ2 .

(2.63)

If u ∈ ∂Pδ1 , then u ∈ P , u = δ1 and δ1 tα−1 ≤ x(t) ≤ δ1 for all t ∈ [0, 1]. Therefore, (δ1 − σ2 r 1 )tα−1 ≤ χn (x(t) − w(t)) ≤ δ1 . Then by assumption (I2 ) and Lemma 2.4.4,  1 (An x)(t) ≥ tα−1 J2 (s)f (s, χn (x(s) − w(s))) + r(s)) ds 0

≥ tα−1



1

0

J2 (s)ψ(t, δ1 ) ds,

and therefore An x ≥ max tα−1 t∈[0,1]

 =

0

1

 0

1

∀ t ∈ [0, 1],

J2 (s)ψ(t, δ1 ) ds

J2 (s)ψ(t, δ1 ) ds > δ1 , ∀ x ∈ ∂Pδ1 .

(2.64)

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Then by using (2.62)–(2.64) and Theorem 1.2.13, we have i(An , P δ5 \ P δ1 , P δ5 ) = 1,

(2.65)

i(An , Pδ2 \ P δ1 , P δ5 ) = 1.

(2.66)

By relation (2.63), we deduce that the operator An has no fixed point on ∂Pδ2 . Besides, for x ∈ P (ξ, δ3 , δ4 ) we have ξ(An x) > δ3 (see relation (2.57)), and for x ∈ P (ξ, δ3 , δ5 ) with An x > δ4 we obtain ξ(An x) > δ3 (see relation (2.60)). Then An has no fixed point on P (ξ, δ3 , δ5 )\S(ξ, δ3 , δ5 ). Therefore, by (2.61), (2.65), (2.66), and the additivity property of the fixed point index, we conclude i(An , P δ5 \ (P (ξ, δ3 , δ5 ) ∪ P δ2 ), P δ5 ) = i(An , P δ5 \ P δ1 , P δ5 ) − i(An , Pδ2 \ P δ1 , P δ5 ) − i(An , S(ξ, δ3 , δ5 ), P δ5 ) = −1.

(2.67)

Hence, by (2.61), (2.66) and (2.67), we deduce that An has at least three fixed points x1n ∈ Pδ2 \P δ1 , x2n ∈ S(ξ, δ3 , δ5 ), x3n ∈ P δ5 \(P (ξ, δ3 , δ5 )∪P δ2 ) that satisfy the conditions δ1 ≤ x1n < δ2 , x2n ≤ δ5 , δ2 < x3n ≤ δ5 , min x2n (t) > δ3 , min x3n (t) < δ3 .

t∈[c,d]

t∈[c,d]



Theorem 2.4.2. Assume that (D1)–(D3) hold, and c, d ∈ R, 0 < c < d ≤ 1. In addition, assume there exist five positive constants δi , i = 1, . . . , 5 such that σ2 r 1 < δ1 < δ2 < δ3 < δ4 ≤ δ5 with δ3 c1−α ≤ δ4 , and the assumptions (I1 ) − (I4 ) are satisfied. Then the boundary value problem (2.45), (2.46) has at least two positive solutions u1 , u2 satisfying δ1 − σ2 r 1 ≤ u1 ≤ δ2 , δ3 − σ2 r 1 ≤ u2 ≤ δ5 and mint∈[c,d] u2 (t) ≥ δ3 − σ2 r 1 . Proof. By Theorem 2.4.1, we know that the operators An , n ≥ n0 have the fixed points xin ∈ P , i = 1, 2, 3, that is xin (t) = (An xin )(t) for all t ∈ [0, 1] or  1 xin (t) = G2 (t, s)(f (s, χn (xin (s) − w(s))) + r(s)) ds, ∀ t ∈ [0, 1], 0

i = 1, 2, 3.

(2.68)

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In addition, the functions xin , n ≥ n0 satisfy the inequalities xin (t) − w(t) ≥ tα−1 ( xin − σ2 r 1 ) ≥ tα−1 (δ1 − σ2 r 1 ) ≥ 0, ∀ t ∈ [0, 1], i = 1, 2, 3.

(2.69)

By Remark 2.4.1, the sets {xin , n ≥ n0 }, i = 1, 2, 3, are relatively compact in the space X. Then (xin )n≥n0 contains a subsequence (xink )k≥n0 which converges in X for k → ∞ to some function xi ∈ X, i = 1, 2, 3. By taking k → ∞ in the relations (2.69) and (2.68) (with n = nk ), we obtain xi (t) − w(t) ≥ tα−1 (δ1 − σ2 r 1 ), ∀ t ∈ [0, 1],  1 G2 (t, s)(f (s, xi (s) − w(s)) + r(s)) ds, ∀ t ∈ [0, 1], xi (t) =

(2.70) (2.71)

0

and δ1 ≤ x1 ≤ δ2 , δ2 ≤ x3 ≤ δ5 ,

x2 ≤ δ5 ,

min x2 (t) ≥ δ3 ,

t∈[c,d]

min x3 (t) ≤ δ3 .

t∈[c,d]

Here, we remark that x2 may be equal to x3 . By the above relations, we conclude that the functions ui (t) = xi (t) − w(t), t ∈ [0, 1], i = 1, 2, satisfy the relations ui (t) ≥ tα−1 (δ1 − σ2 r 1 ) for all t ∈ [0, 1], and  ui (t) =

0

1

G2 (t, s)f (s, ui (s)) ds,

∀ t ∈ [0, 1], i = 1, 2,

that is, ui , i = 1, 2, are positive solutions of problem (2.45), (2.46). Besides, u1 , u2 satisfy the relations δ1 −σ2 r 1 ≤ u1 ≤ δ2 , δ3 −σ2 r 1 ≤ u2 ≤ δ5 and mint∈[c,d] u2 (t) ≥ δ3 − σ2 r 1 .  2.4.3

An example

Let us choose α = 7/2 (n = 4), β0 = 5/3, m = 2, β1 = 1/4, β2 = 6/5, H1 (t) = t for all t ∈ [0, 1], H2 (t) = {0, for t ∈ [0, 1/2); 2, for t ∈ [1/2, 1]}, for all t ∈ (0, 1). and r(t) = 1041√ 4 t

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We consider the fractional differential equation 7/2

D0+ u(t) + f (t, u(t)) = 0,

0 < t < 1,

(2.72)

with the boundary conditions 



u(0) = u (0) = u (0) = 0,

5/3 D0+ u(1)

 = 0

1

1/4 D0+ u(t) dt

+

6/5 2D0+ u

  1 , 2 (2.73)

where 1 1 (x) − 4 √ , f (t, x) =  5 2 3 10 4 t 9 t (1 − t) with

0 < t < 1, x > 0,

⎧√ 1 √ ⎪ ⎨ x + 7 x , 0 < x ≤ 1, (x) = x7 + 1, 1 < x ≤ 3, ⎪ √ ⎩√ 3 x + 2188 − 3 3, x > 3.

For θ1 , θ2 > 0, θ1 < θ2 , we have f (t, x) ≤ hθ1 ,θ2 (t) for 0 < t < 1 and θ1 t5/2 ≤ x ≤ θ2 , where ⎧√ θ2 + (θ1 t5/2 )−1/7 , for θ2 ≤ 1, ⎪ ⎪ ⎪ ⎪ ⎨max{1 + (θ t5/2 )−1/7 , θ7 + 1}, for 1 < θ ≤ 3, 1 1 2 2 √ hθ1 ,θ2 (t) =  √ 3 5/2 −1/7 3 t ) , θ + 2188 − 3}, max{1 + (θ 9 5 t2 (1 − t)3 ⎪ ⎪ 1 2 ⎪ ⎪ ⎩ for θ2 > 3. This function hθ1 ,θ2 ∈ L1 (0, 1), and in addition we obtain Δ2 ≈ 4 0.81820862 > 0 and r 1 = 3·10 4 > 0. Thus the assumptions (D1)–(D3) are satisfied. We also deduce σ2 ≈ 1.29928727. We take c = 3/4, d = 1, δ1 = 0.01, δ2 = 1, δ3 = 5, δ4 = 15, δ5 = 1000. The conditions σ2 r 1 < δ1 and δ3 c1−α ≤ δ4 are satisfied because σ2 r 1 ≈ 0.00017 < δ1 and δ3 c1−α ≈ 10.264 < δ4 . In addition, we obtain 1 (1 − s)5/6 (1 − (1 − s)5/3 ), Γ(7/2) t9/4 (1 − s)5/6 − (t − s)9/4 , 0 ≤ s ≤ t ≤ 1, 1 g11 (t, s) = Γ(13/4) t9/4 (1 − s)5/6 , 0 ≤ t ≤ s ≤ 1, t13/10 (1 − s)5/6 − (t − s)13/10 , 0 ≤ s ≤ t ≤ 1, 1 g12 (t, s) = Γ(23/10) t13/10 (1 − s)5/6 , 0 ≤ t ≤ s ≤ 1, h0 (s) =

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⎧ 1 (1 − s)5/6 (1 − (1 − s)5/3 ) ⎪ ⎪ ⎪ Γ(7/2) ⎪ 4  ⎪ ⎪ 1 4 5/6 ⎪ + − 13 (1 − s)13/4 ⎪ Δ2 Γ(13/4) 13 (1 − s) ⎪ ⎪ ⎪  ! ⎪ !13/10 " ⎪ 2 1 13/10 ⎪ ⎪ + Δ2 Γ(23/10) (1 − s)5/6 − 12 − s , ⎪ 2 ⎪ ⎨ 0 ≤ s < 12 , J2 (s) = ⎪ ⎪ ⎪ ⎪ 1 (1 − s)5/6 (1 − (1 − s)5/3 ) ⎪ ⎪ Γ(7/2) ⎪ ⎪ ⎪ 4  ⎪ 1 4 ⎪ ⎪ (1 − s)5/6 − 13 (1 − s)13/4 + Δ2 Γ(13/4) ⎪ 13 ⎪ ⎪ ⎪ ! ⎪ ⎩ + 2 1 13/10 (1 − s)5/6 , 12 ≤ s ≤ 1. Δ2 Γ(23/10) 2 Because ϕ(t, 1) = max{f (t, x), (1 − σ2 r 1 )t5/2 ≤ x ≤ 1} + r(t)

  √ 1 1  x+ √ = max , (1 − σ2 r 1 )t5/2 ≤ x ≤ 1 7 x 9 5 t2 (1 − t)3   1 1 ≤  1 + , ∀ t ∈ (0, 1), ((1 − σ2 r 1 )t5/2 )1/7 9 5 t2 (1 − t)3 we deduce by using a computer program    1  1 1 1 J2 (s)ϕ(s, 1) ds ≤ J2 (s)  1+ ds ((1 − σ2 r 1 )s5/2 )1/7 9 5 s2 (1 − s)3 0 0 ≈ 0.25127612 < 1, so the assumption (I1 ) is satisfied. Because      1 1 1 5/2 − σ2 r 1 t ≤x≤ ψ t, = min f (t, x), + r(t) 100 100 100   √ 1 1  √ x + = min , 7 x 9 5 t2 (1 − t)3

  1 1 − σ2 r 1 t5/2 ≤ x ≤ 100 100  $  √ 1 1 7 5/2 ≥  − σ2 r 1 t + 100 , 100 9 5 t2 (1 − t)3 ∀ t ∈ (0, 1),

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we conclude    1  1 1 1 J2 (s)ψ s, J2 (s)  ds ≥ 5 2 100 9 s (1 − s)3 0 0 $   √ 1 7 − σ2 r 1 s5/2 + 100 ds × 100 ≈ 0.20792415 > 0.01, and so the assumption (I2 ) is satisfied. Next, we obtain Φ(t, 5, 1000) = max{f (t, x), (5 − σ2 r 1 )t5/2 ≤ x ≤ 1000} + r(t) 1 , 0 < t ≤ t1 , 1+ √ 1 7 (5−σ2 r1 )t5/2 ≤  √ 5 9 t2 (1 − t)3 2198 − 3 3, t1 < t < 1, √ where t1 = ((5 − σ2 r 1 )(2197 − 3 3)7 )−2/5 ≈ 2.31256 × 10−10. We find that  0

1

 J2 (s)Φ(s, 5, 1000) ds =

t1

0



J2 (s)Φ(s, 5, 1000) ds 1

+ t1

 ≤

t1

0



J2 (s)Φ(s, 5, 1000) ds

1 J2 (s)  5 2 9 s (1 − s)3 1



× 1+  ds 7 (5 − σ2 r 1 )s5/2  1 √ 1 3 + J2 (s)  (2198 − 3) ds 5 2 3 9 s (1 − s) t1 ≈ 231.13 < 1000, so the assumption (I3 ) is satisfied. We also deduce for t ∈ [3/4, 1) that Ψ(t, 5, 15) = min{f (t, x), (5 − σ2 r 1 )t5/2 ≤ x ≤ 15} + r(t) [(5 − σ2 r 1 )t5/2 ]7 + 1, 3/4 ≤ t ≤ t2 , 1 ≥   √ 9 5 t2 (1 − t)3 3 (5 − σ2 r 1 )t5/2 + 2188 − 3 3, t2 < t < 1,

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where t2 = (3/(5 − σ2 r 1 ))2/5 ≈ 0.815204. Then we get  t2  1  1 J2 (s)Ψ(s, 5, 15) ds = J2 (s)Ψ(s, 5, 15) ds + J2 (s)Ψ(s, 5, 15) ds 3/4

≥



t2

3/4

t2

1 J2 (s)  ([(5 − σ2 r 1 )s5/2 ]7 + 1) ds 5 2 9 s (1 − s)3 3/4  %  1 √ 1 3 5/2 + 2188 − 3 3 + J2 (s)  (5 − σ r )s ds 2 1 9 5 s2 (1 − s)3 t2

≈ 35.0712 > 5 · (3/4)−5/2 ≈ 10.264. Hence, the assumption (I4 ) is satisfied. Therefore, by Theorem 2.4.2, we conclude that the boundary value problem (2.72), (2.73) has at least two positive solutions u1 (t), u2 (t), t ∈ [0, 1] that satisfy the conditions 0.009826 ≤ u1 ≤ 1, 4.999826 ≤ u2 ≤ 1000 and mint∈[3/4,1] u2 (t) ≥ 4.999826. Remark 2.4.2. The results presented in this section were published in [2]. The existence of nonnegative solutions for the fractional differential equation (2.19) subject to the boundary conditions (2.46) was studied in [85].

2.5

On a Singular Fractional Boundary Value Problem with Parameters

We consider the nonlinear fractional differential equation α u(t) + λh(t)f (t, u(t)) = 0, D0+

t ∈ (0, 1),

(2.74)

with the nonlocal boundary conditions 

u(0) = u (0) = · · · = u

(n−2)

(0) = 0,

β0 D0+ u(1)

=

m   i=1

1 0

βi D0+ u(t) dHi (t),

(2.75) where α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, 0 ≤ β1 < β2 < · · · < βm ≤ β0 < α − 1, β0 ≥ 1, λ is a positive k denotes the Riemann-Liouville derivative of order k (for parameter, D0+ k = α, β0 , β1 , . . . , βm ), the integrals from the boundary conditions (2.75) are Riemann–Stieltjes integrals with Hi , i = 1, . . . , m, functions of bounded

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variation, the nonnegative function f (t, u) may have singularity at u = 0 and the nonnegative function h(t) may be singular at t = 0 and/or t = 1. Under some assumptions for the functions h and f , we establish intervals for the parameter λ which are dependent on the principal characteristic value of an associated linear operator, for which the problem (2.74), (2.75) has positive solutions u ∈ C([0, 1], R+ ), u(t) > 0 for all t ∈ (0, 1]. In the proof of the main theorems, we use the fixed point index theory and an application of the Krein–Rutman theorem in the space of continuous functions defined on [0, 1] (see Theorem 2.5.1). In the case in which h ≡ 1 and f is a function which changes sign and has singularities at t = 0 and/or t = 1, we present two existence results for the positive solutions of this problem, in the proof of which we apply the Guo–Krasnosel’skii fixed point theorem. In [116] the authors prove the existence of positive solutions of fractional differential equation (2.74) supplemented with the boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0,

β D0+ u(1) =

∞  i=1

γ αi D0+ u(ξi ),

(2.76)

where β ∈ [1, n − 2], γ ∈ [0, β], αi ≥ 0, i − 1, 2, . . ., 0 < ξ1 < ξ2 < · · · < ∞ ξi−1 < ξi < · · · < 1, and Γ(α − γ) > Γ(α − β) i=1 αi ξiα−γ−1 . The last β u(1) = condition of the boundary conditions (2.76) can be written as D0+ 1 γ D u(t)dH(t), where H is the step function defined by H(t) = 0+ 0 n {0, t ∈ [0, ξ1 ]; α1 , t ∈ (ξ1 , ξ2 ]; α1 + α2 , t ∈ (ξ2 , ξ3 ]; . . . ; i=1 αi , t ∈ (ξn , ξn+1 ]; . . .}, so this condition is a particular case of our condition from (2.75). In this section, we will use the notations (Δ2 , G2 , g0 , g1i , J2 , h0 , σ2 ) and the auxiliary results (Lemmas 2.4.1–2.4.5) from Section 2.4. 2.5.1

Existence of positive solutions

In this section, we present intervals for the parameter λ such that our problem (2.74), (2.75) has at least one positive solution. We consider the Banach space X = C[0, 1] with the supremum norm u = supt∈[0,1] |u(t)|, and we define the cones P0 = {u ∈ X, u(t) ≥ 0, ∀ t ∈ [0, 1]}, P = {u ∈ X, u(t) ≥ tα−1 u , ∀ t ∈ [0, 1]} ⊂ P0 .

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We define the operator D : P0 → P0 and the linear operator L : X → X by  Du(t) = λ  Lu(t) =

0

1

0 1

G2 (t, s)h(s)f (s, u(s)) ds,

G2 (t, s)h(s)u(s) ds,

t ∈ [0, 1], u ∈ P0 ,

t ∈ [0, 1], u ∈ X.

We see that u is a solution of problem (2.74), (2.75) if and only if u is a fixed point of operator D. As in Section 2.4, for r > 0, we denote by Pr = {x ∈ P, x < r} and P r = {x ∈ P, x ≤ r}. We introduce now the assumptions that we will use in what follows. (K1) α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, 0 ≤ β1 < β2 < · · · < βm ≤ β0 < α − 1, β0 ≥ 1, Hi : [0, 1] → R, Γ(α) − i = 1, . . . , m, are nondecreasing functions, λ > 0, and Δ2 = Γ(α−β 0)  m 1 α−βi −1 Γ(α) dHi (s) > 0. i=1 Γ(α−βi ) 0 s 1 (K2) The function h ∈ C((0, 1), [0, ∞)) and 0 J2 (s)h(s) ds < ∞. (K3) The function f ∈ C([0, 1] × (0, ∞), [0, ∞)) and for any 0 < r < R we have  sup

lim

n→∞

u∈P R \Pr

En

h(s)f (s, u(s)) ds = 0,

where En = [0, n1 ] ∪ [ n−1 n , 1]. Lemma 2.5.1. Assume that assumptions (K1)–(K3) hold. Then for any 0 < r < R, the operator D : P R \ Pr → P is completely continuous. Proof. By (K3), we deduce that there exists a natural number n1 ≥ 3 such that  sup u∈P R \Pr

En 1

h(s)f (s, u(s)) ds < 1.

For u ∈ P R \ Pr , there exists r1 ∈ [r, R] such that u = r1 , and then tα−1 r ≤ tα−1 r1 ≤ u(t) ≤ r1 ≤ R,

∀ t ∈ [0, 1].

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     1 Let L1 = max f (t, x), t ∈ n11 , n1n−1 r, R . By Lemma 2.4.4, , x ∈ nα−1 1 1

(K2) and (K3), we find  sup u∈P R \Pr

≤ ≤

λ

1

0

G2 (t, s)h(s)f (s, u(s)) ds 

sup u∈P R \Pr

sup u∈P R \Pr

λ  λ

1 0

J2 (s)h(s)f (s, u(s)) ds

En 1

 +

λ

sup u∈P R \Pr

 + λL1

1

0

n1 −1 n1 1 n1

 ≤ λJ0 + λL1

J2 (s)h(s)f (s, u(s)) ds

n1 −1 n1 1 n1

J2 (s)h(s)f (s, u(s)) ds

J2 (s)h(s) ds ≤ λJ0

J2 (s)h(s) ds < ∞,

where J0 = maxt∈[0,1] J2 (t). This implies that the operator D is well defined. We show next that D : P R \ Pr → P . Indeed, for any u ∈ P R \ Pr and t ∈ [0, 1], we have  (Du)(t) = λ

0

1

 G2 (t, s)h(s)f (s, u(s)) ds ≤ λ

0

1

J2 (s)h(s)f (s, u(s)) ds,

and then  Du ≤ λ

0

1

J2 (s)h(s)f (s, u(s)) ds.

On the other hand, by Lemma 2.4.4, we obtain  1 (Du)(t) ≥ λtα−1 J2 (s)h(s)f (s, u(s)) ds ≥ tα−1 Du , 0

∀ t ∈ [0, 1],

so Du ∈ P . Therefore, D(P R \ Pr ) ⊂ P . We prove now that D : P R \ Pr → P is completely continuous. We assume that S ⊂ P R \ Pr is an arbitrary bounded set. From the first part of the proof, we know that D(S) is uniformly bounded. Then we show that

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D(S) is equicontinuous. Indeed, for ε > 0, there exists a natural number n2 ≥ 3 such that  ε sup h(s)f (s, u(s)) ds < . 4λJ0 u∈P R \Pr En2 Since G2 (t, s) is uniformly continuous on [0, 1] × [0, 1], for the above ε > 0 there  exists δ> 0 such that for any t1 , t2 ∈ [0, 1], with |t1 − t2 | < δ, and , we have s ∈ n12 , n2n−1 2 ε |G2 (t1 , s) − G2 (t2 , s)| < ¯ , 2λhL2   1     , x ∈ nα−1 r, R and where L2 = max 1, max f (t, x), t ∈ n12 , n2n−1 2 2   1 n2 −1  ¯ = max 1, max{h(t), t ∈ h . , n2

n2

Then for any u ∈ S, t1 , t2 ∈ [0, 1], with |t1 − t2 | < δ, we deduce  1    |(Du)(t1 ) − (Du)(t2 )| = λ  (G2 (t1 , s) − G2 (t2 , s))h(s)f (s, u(s)) ds 0  ≤ 2λ J2 (s)h(s)f (s, u(s)) ds En 2

 +λ sup u∈S

n2 −1 n2 1 n2

|G2 (t1 , s) − G2 (t2 , s)|h(s)f (s, u(s)) ds

ε ελ + ¯ ≤ 2λJ0 4λJ0 2λhL2



n2 −1 n2 1 n2

 h(s) ds L2

ε ε + = ε. 2 2 This gives us that D(S) is equicontinuous. By the Arzela–Ascoli theorem, we conclude that D : P R \ Pr → P is compact. Finally we prove that D : P R \ Pr → P is continuous. We suppose that un , u0 ∈ P R \ Pr for all n ≥ 1 and un − u0 → 0 as n → ∞. Then r ≤ un ≤ R for all n ≥ 0. By (K3), for ε > 0 there exists a natural number n3 ≥ 3 such that  ε sup h(s)f (s, u(s)) ds < . (2.77) 4λJ0 u∈P R \Pr En3     1 Because f (t, x) is uniformly continuous in n13 , n3n−1 × nα−1 r, R , we 3 3 obtain   1 n3 − 1 lim |f (s, u(s)) − f (s, u0 (s))| = 0, uniformly for s ∈ , . n→∞ n3 n3 ≤

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Then the Lebesgue dominated convergence theorem gives us 

n3 −1 n3 1 n3

h(s)|f (s, un (s)) − f (s, u0 (s))| ds → 0, as n → ∞.

Thus, for the above ε > 0, there exists a natural number N such that for n > N , we have 

n3 −1 n3 1 n3

h(s)|f (s, un (s)) − f (s, u0 (s))| ds
0 such that cAv ≥ v, then the spectral radius r(A) = 0 and A has an eigenvector u0 ∈ P0 \ {θ} corresponding to its principal characteristic value λ1 = (r(A))−1 , that is λ1 Au0 = u0 or Au0 = r(A)u0 , and so r(A) > 0. Lemma 2.5.2. Assume that assumptions (K1) and (K2) hold. Then the spectral radius r(L) = 0 and L has an eigenfunction ψ1 ∈ P0 \ {θ} corresponding to the principal eigenvalue r(L), that is Lψ1 = r(L)ψ1 . So r(L) > 0.

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Proof. The operator L : X → X is a linear completely continuous operator. By Lemma 2.4.4 we know that G2 (t, s) > 0 for all t, s ∈ (0, 1). By (K2) we deduce that there exists an interval [c, d] ⊂ (0, 1) (0 < c < d < 1) such that h(t) > 0 for all t ∈ [c, d]. We consider a function ϕ ∈ C[0, 1] satisfying the conditions ϕ(t) > 0 for t ∈ (c, d) and ϕ(t) = 0 for t ∈ (c, d). Then for all t ∈ [c, d] we have  1 G2 (t, s)h(s)ϕ(s) ds (Lϕ)(t) = 0

 ≥

d

c

G2 (t, s)h(s)ϕ(s) ds > 0,

∀ t ∈ [c, d].

maxt∈[c,d] ϕ(t) ! Hence, there exists a constant a > 0 a = mint∈[c,d] (Lϕ)(t) which satisfies the inequality a(Lϕ)(t) ≥ ϕ(t), ∀ t ∈ [0, 1]. By Theorem 2.5.1 we conclude that the spectral radius r(L) = 0 and L has an eigenfunction ψ1 ∈ P0 \ {θ} corresponding to its principal characteristic value λ1 = (r(L))−1 such that  Lψ1 = r(L)ψ1 , and so r(L) > 0.

Using a similar argument as that used in the proof of Lemma 2.5.1 for operator L, we obtain that L(P ) ⊂ P . Theorem 2.5.2. Assume that assumptions (K1)–(K3) hold. If f (t, u) f (t, u) < f0i := lim inf min ≤ ∞, u→0+ t∈[0,1] u u & ' 1 1 then for any λ ∈ f i r(L) , the problem (2.74), (2.75) has at least , f s r(L) s 0 ≤ f∞ := lim sup max

u→∞ t∈[0,1]

∞

0

one positive solution u(t), t ∈ [0, 1], (with the conventions 01+ = ∞ and 1 ∞ = 0+ ). ' & 1 1 , f s r(L) . For f0i we have the cases: f0i ∈ Proof. We consider λ ∈ f i r(L) 0

∞

1 (0, ∞) with f0i > λr(L) , and f0i = ∞. In the first case, f0i ∈ (0, ∞) with 1 we obtain f0i > λr(L)

∀ ε > 0 ∃ δ(ε) > 0 s.t. By taking ε = f0i − f (t,u) u

1 λr(L)

f (t, u) ≥ f0i − ε, u

∀ t ∈ [0, 1], u ∈ (0, δ(ε)].

we deduce that there exists r1 > 0 such that

1 ≥ λr(L) for all t ∈ [0, 1] and u ∈ (0, r1 ], and so f (t, u) ≥ t ∈ [0, 1] and u ∈ [0, r1 ].

u λr(L)

for all

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In the case f0i = ∞, we have ∀ ε > 0 ∃ δ(ε) > 0 s.t.

f (t, u) ≥ ε, ∀ t ∈ [0, 1], u ∈ (0, δ(ε)]. u

1 u So for ε = λr(L) we deduce that there exists r1 > 0 such that f (t, u) ≥ λr(L)  for all t ∈ [0, 1] and u ∈ [0, r1 ]. Hence, in both the above cases, we conclude that there exists r1 > 0 u for all t ∈ [0, 1] and u ∈ [0, r1 ]. such that f (t, u) ≥ λr(L) Then for any u ∈ ∂Pr1 we find  1 Du(t) = λ G2 (t, s)h(s)f (s, u(s)) ds 0

≥

1 r(L)



1

0

G2 (t, s)h(s)u(s) ds =

1 Lu(t), r(L)

∀ t ∈ [0, 1].

We assume that D has no fixed point on ∂Pr1 , (otherwise the proof is finished). We will prove that u − Du = μψ1 ,

∀ u ∈ ∂Pr1 ,

μ ≥ 0,

(2.79)

where ψ1 is given in Lemma 2.5.2. If not, there exist u1 ∈ ∂Pr1 and μ1 ≥ 0 such that u1 − Du1 = μ1 ψ1 . Then μ1 > 0 and u1 = Du1 + μ1 ψ1 ≥ μ1 ψ1 . We denote by μ0 = sup{μ, u1 ≥ μψ1 }. Then μ0 ≥ μ1 , u1 ≥ μ0 ψ1 and Du1 ≥

1 1 Lu1 ≥ μ0 Lψ1 = μ0 ψ1 . r(L) r(L)

Hence, u1 = Du1 + μ1 ψ1 ≥ μ0 ψ1 + μ1 ψ1 = (μ0 + μ1 )ψ1 , which contradicts the definition of μ0 . So, relation (2.79) holds, and by Theorem 1.2.9, we deduce that i(D, Pr1 , P ) = 0.

(2.80)

1 s s s For f∞ , we also have two cases: f∞ ∈ (0, ∞) with f∞ < λr(L) , and 1 s s s = 0. In the first case, f∞ ∈ (0, ∞) with f∞ < λr(L) , we obtain f∞

∀ ε > 0 ∃ δ(ε) > 0 s.t. By taking ε = f (t, u) ≤ (0, 1).

1 2λr(L)

θ1 λr(L) u

−

s f∞ 2 ,

f (t, u) s ≤ f∞ + ε, u

∀ t ∈ [0, 1], u ≥ δ(ε).

we deduce that there exists r2 > r1 such that

for all t ∈ [0, 1] and u ∈ [r2 , ∞), where θ1 =

1 2

+

s f∞ λr(L) 2

∈

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Boundary Value Problems for Fractional Differential Equations and Systems s In the case f∞ = 0, we have

f (t, u) ≤ ε, ∀ t ∈ [0, 1], u ≥ δ(ε). u

∀ ε > 0 ∃ δ(ε) > 0 s.t.

1 So, for ε = 2λr(L) , we deduce that there exists r2 > r1 such that f (t, u) ≤ 1  2λr(L) u for all t ∈ [0, 1] and u ∈ [r2 , ∞). Therefore, in both the above cases, we conclude that there exist 1 u for all t ∈ [0, 1] and θ ∈ (0, 1) and r2 > r1 such that f (t, u) ≤ θ λr(L) u ∈ [r2 , ∞). We define now the operator L1 : X → X by  1 θ 1 L1 u = θ Lu = G2 (t, s)h(s)u(s) ds, t ∈ [0, 1], u ∈ X. r(L) r(L) 0

The operator L1 is linear and bounded, and L1 (P ) ⊂ P . Because θ ∈ (0, 1) we obtain r(L1 ) = θ < 1. We consider the set Z = {u ∈ P \ Br1 , μu = Du, with μ ≥ 1}. For u ∈ P , we denote by D(u) = {t ∈ [0, 1], u(t) ≥ r2 }. Then for u ∈ P , we have u(t) ≥ r2 for all t ∈ D(u), and so f (t, u(t)) ≤ θ

1 u(t), λr(L)

∀ t ∈ D(u).

(2.81)

By (2.81) and the definition of operator L, for any u ∈ Z, μ ≥ 1 and t ∈ [0, 1] we deduce  u(t) ≤ μu(t) = (Du)(t) = λ  =λ

D(u)

1 0

G2 (t, s)h(s)f (s, u(s)) ds

G2 (t, s)h(s)f (s, u(s)) ds



+λ [0,1]\D(u)

≤

θ r(L)

≤

θ r(L)

G2 (t, s)h(s)f (s, u(s)) ds



D(u)



0

1

G2 (t, s)h(s)u(s) ds + λ

 0

1

J2 (s)h(s)f (s, u (s)) ds

G2 (t, s)h(s)u(s) ds + λJ0 M1 = (L1 u)(t) + λJ0 M1 , (2.82)

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77

where u (t) = min{u(t), r2 } for all t ∈ [0, 1], (which satisfies the (t) ≤ r2 for all t ∈ [0, 1]), and M1 = inequalities r1 tα−1 ≤ u 1 supu∈P r \Pr 0 h(s)f (s, u(s)) ds, (as in the proof of Lemma 2.5.1 we obtain 2 1 that M1 < ∞). By the Gelfand formula we know that (I − L1 )−1 exists and ∞ (I −L1 )−1 = i=1 Li1 , which implies (I −L1 )−1 (P ) ⊂ P . This together with (2.82) gives us u(t) ≤ (I −L1 )−1 (λJ0 M1 ) and so u(t) ≤ λJ0 M1 (I −L1 )−1 for all t ∈ [0, 1], which means that Z is bounded. Now we choose R > max{r2 , sup{ u , u ∈ Z}}. Then we obtain that μu = Du for all u ∈ ∂PR and μ ≥ 1. By Theorem 1.2.7, we conclude that i(D, PR , P ) = 1.

(2.83)

By (2.80), (2.83) and the additivity property of the fixed point index, we deduce that i(D, PR \ P r1 , P ) = i(D, PR , P ) − i(D, Pr1 , P ) = 1. So, operator D has at least one fixed point on PR \ P r1 , which is a positive solution of problem (2.74),(2.75).  By using a similar approach as that used in the proof of Theorem 2.5.2, we obtain the following result. Theorem 2.5.3. Assume that assumptions (K1)–(K3) hold. If f (t, u) f (t, u) i < f∞ ≤ ∞, := lim inf min u→∞ t∈[0,1] u u ' 1 1 , s r(L) f r(L) , the problem (2.74), (2.75) has at least

0 ≤ f0s := lim sup max

u→0+ t∈[0,1]

then for any λ ∈

&

i f∞

0

one positive solution u(t), t ∈ [0, 1]. 2.5.2

Some remarks on a related semipositone problem

In this section, we present two existence results for a semipositone problem associated to problem (2.74), (2.75). More precisely, we consider the fractional differential equation α u(t) + λf(t, u(t)) = 0, D0+

t ∈ (0, 1),

(2.84)

subject to the boundary conditions (2.75). We suppose that assumption (K1) holds and f satisfies the following conditions: (K2) The function f ∈ C((0, 1) × [0, ∞), R) may be singular at t = 0 and/or t = 1, and there exist the functions p, q ∈ C((0, 1), [0, ∞)),

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g ∈ C([0, 1] × [0, ∞), [0, ∞)) such that −p(t) ≤ f(t, u) ≤ q(t)g(t, u) 1 for all t ∈ (0, 1) and u ∈ [0, ∞), with 0 < 0 p(t) dt < ∞, 0 < 1 q(t) dt < ∞. 0 (K3) There exists ζ ∈ (0, 1/2) such that limu→∞ mint∈[ζ,1−ζ]

f(t,u) u

= ∞.

By using the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2) and similar arguments as those used in the proofs of Theorem 2.1.1 and Theorem 2.1.2, we obtain the following results for the problem (2.84), (2.75). Theorem 2.5.4. Assume that (K1), (K2) and (K3) hold. Then there exists λ∗ > 0 such that, for any λ ∈ (0, λ∗ ], the boundary value problem (2.84), (2.75) has at least one positive solution. 1 In the proof of Theorem 2.5.4 we consider R1 > σ2 0 p(t) dt > 0,  1 and we define λ∗ = min{1, R1 (M2 0 J2 (s)(q(s) + p(s)) ds)−1 }, with M2 = max{maxt∈[0,1], u∈[0,R1 ] g(t, u), 1}. The solution u(t), t ∈ [0, 1] satisfies the 1 condition u(t) ≥ Λ1 tα−1 for all t ∈ [0, 1], where Λ1 = R1 −σ2 0 p(s) ds > 0. Theorem 2.5.5. Assume that (K1), (K2) and (K4) There exists ζ ∈ (0, 1/2) such that limu→∞ mint∈[ζ,1−ζ] f(t, u) = ∞ = 0, and limu→∞ maxt∈[0,1] g(t,u) u hold. Then there exists λ∗ > 0 such that, for any λ ≥ λ∗ , the boundary value problem (2.84), (2.75) has at least one positive solution. By (K4), we know that for ζ ∈ (0, 1/2) and for a fixed number L0 > 0 there exists M3 > 0 such that f(t, u) ≥ L0 for all t ∈ [ζ, 1 − ζ]and u ≥ M3 . 1 In the proof of Theorem 2.5.5, we define λ∗ = M3 (ζ α−1 σ2 0 p(s) ds)−1 .  1 tα−1 for all The solution u(t), t ∈ [0, 1] satisfies the condition u(t) ≥ Λ M3  t ∈ [0, 1], where Λ1 = ζ α−1 . 2.5.3

Examples

Let α = 10/3 (n = 4), β0 = 11/5, m = 2, β1 = 1/2, β2 = 5/4, H1 (t) = t for all t ∈ [0, 1], H2 (t) = {0, for t ∈ [0, 1/2); 1, for t ∈ [1/2, 1]}.

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We consider the fractional differential equations 10/3

D0+ u(t) + λh(t)f (t, u(t)) = 0,

t ∈ (0, 1),

(2.85)

D0+ u(t) + λf(t, u(t)) = 0,

t ∈ (0, 1),

(2.86)

10/3

subject to the boundary conditions u(0) = u (0) = u (0) = 0,

11/5



D0+ u(1) =

1

0

1/2

5/4

D0+ u(t) dt + D0+ u

  1 . 2 (2.87)

We have Δ2 ≈ 1.12792427 > 0 and σ2 ≈ 0.94443688. So assumption (K1) is satisfied. In addition we obtain t11/6 (1 − s)2/15 − (t − s)11/6 , 0 ≤ s ≤ t ≤ 1, 1 g11 (t, s) = Γ(17/6) t11/6 (1 − s)2/15 , 0 ≤ t ≤ s ≤ 1, t13/12 (1 − s)2/15 − (t − s)13/12 , 0 ≤ s ≤ t ≤ 1, 1 g12 (t, s) = Γ(25/12) t13/12 (1 − s)2/15 , 0 ≤ t ≤ s ≤ 1, 1 (1 − s)2/15 (1 − (1 − s)11/5 ), s ∈ [0, 1], Γ(10/3)  ⎧ 1 1 ⎪ h0 (s) + Δ12 Γ(23/6) (1 − s)2/15 − Γ(23/6) (1 − s)17/6 ⎪ ⎪ ⎪ ⎪ ⎪  1 13/12 # ⎪ 1 1 2/15 13/12 ⎪ + ) (1 − s) − ( − s) ( , ⎪ Γ(25/12) 2 2 ⎪ ⎪ ⎨ 0 ≤ s ≤ 1/2, J2 (s) = ⎪ ⎪  ⎪ ⎪ 1 1 ⎪ (1 − s)2/15 − Γ(23/6) (1 − s)17/6 h0 (s) + Δ12 Γ(23/6) ⎪ ⎪ ⎪ ⎪ # ⎪ ⎪ ⎩ + 1 1 13/12 (1 − s)2/15 , 1/2 < s ≤ 1. Γ(25/12) ( 2 )

h0 (s) =

Example 1. We consider the functions h(t) =  3

1 , t ∈ (0, 1); t(1 − t)2

f (t, u) =

√ 1 , u+t+ √ 4 u

t ∈ [0, 1], u > 0.

The cone P from Section 2.5.1 is here P = {u ∈ C[0, 1], u(t) ≥ t7/3 u , ∀ t ∈ [0, 1]}. For 0 < r < R and u ∈ P R \ Pr we deduce f (t, u(t)) ≤

√ 1 , R+1+ √ 4 t7/3 r

∀ t ∈ (0, 1].

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1 Besides, we obtain 0 J2 (s)h(s) ds ≤ J0 Γ(2/3)Γ(1/3) < ∞, where J0 = maxs∈[0,1] J2 (s) ≈ 0.781. Hence, assumption (K2) is satisfied. For u ∈ P R \ Pr and En = [0, n1 ] ∪ [ n−1 n , 1], we find      1 1  h(s)f (s, u(s)) ds = u(s) + s +  Cn = ds 3 4 s(1 − s)2 u(s) En En    √ 1 1  R+1+ √ ≤ ds 4 3 s(1 − s)2 s7/3 r En   √ 1 ds 1  + √ ds, = ( R + 1) 4 11/12 3 2 r En s (1 − s)2/3 s(1 − s) En and then limn→∞ supu∈P R \Pr Cn = 0, because f1 (s) = √ 3

1 s(1−s)2

∈ L1 (0, 1)

1 1 and f2 (s) = s11/12 (1−s) 2/3 ∈ L (0, 1). Hence, assumption (K3) is satisfied. s We also have f∞ = 0 and f0i = ∞. Then by using Theorem 2.5.2, we deduce that, for any λ ∈ (0, ∞), the problem (2.85), (2.87) has at least one positive solution u(t), t ∈ [0, 1], which satisfies the condition u(t) ≥ t7/3 u for all t ∈ [0, 1].

Example 2. We consider the function u3 + u + 1 f(t, u) =  + ln t, 4 t(1 − t)3

t ∈ (0, 1), u ≥ 0.

For this example, we have p(t) = − ln t and q(t) = √ 4

1 for all t t(1−t)3 1 0, 0 p(t) dt = 

∈

(0, 1), g(t, u) = u3 + u + 1 for all t ∈ [0, 1] and u ≥ 1, 1 3 1 q(t) dt = Γ( )Γ( ) ≈ 4.44288. Then assumption (K2) is satisfied. In 4 4 0 addition, for ζ ∈ (0, 1/2) fixed, assumption (K3) is also satisfied. By some 1 computations, we obtain 0 J2 (s)(q(s) + p(s)) ds ≈ 2.71742073. We choose 1 R1 = 2, which satisfies the condition R1 > σ2 0 p(t) dt ≈ 0.944, and then we deduce M2 = 11 and λ∗ ≈ 0.0669084. By Theorem 2.5.4, we conclude that, for any λ ∈ (0, λ∗ ], the problem (2.86), (2.87) has at least one positive solution u(t), t ∈ [0, 1], which satisfies the condition u(t) ≥ Λ1 t7/3 for all t ∈ [0, 1], where Λ1 ≈ 1.05556. Example 3. We consider the function  u + 1/3 1  f (t, u) =  − √ , 3 5 3 2 t t (1 − t)

t ∈ (0, 1), u ≥ 0.

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1 1 Here, we have p(t) = √ and q(t) = √ for all t ∈ (0, 1), 3 5 3 t t (1−t)2  1 g(t, u) = u + 1/3 for all t ∈ [0, 1] and u ≥ 0. Because 0 p(t) dt = 1 3/2, 0 q(t) dt ≈ 3.30327, the assumption (K2) is satisfied. In addition, for ζ ∈ (0, 1/2), we obtain limu→∞ mint∈[ζ,1−ζ] f(t, u) = ∞ and limu→∞ maxt∈[0,1] g(t, u)/u = 0, and then assumption (K4) is also satisfied. We choose ζ = 1/4 and L0 = 100, and then we find M3 = 5805 and λ∗ ≈ 104075. Then by Theorem 2.5.5, we deduce that, for any λ ≥ λ∗ , the problem (2.86), (2.87) has at least one positive solution u(t), t ∈ [0, 1],  1 t7/3 for all t ∈ [0, 1], where which satisfies the inequality u(t) ≥ Λ  1 ≈ 147438. Λ

Remark 2.5.1. The results presented in this section will be published in [102]. 2.6

A Singular Fractional Differential Equation with Integral Boundary Conditions

We consider the nonlinear fractional differential equation α D0+ u(t) + f (t, u(t)) = 0,

t ∈ (0, 1),

with the boundary conditions ⎧  (n−2) (0) = 0, ⎪ ⎨ u(0) = u (0) = · · · = u  m 1  β0 βi ⎪ ai (t)D0+ u(t) dHi (t), ⎩ D0+ u(1) = i=1

(2.88)

(2.89)

0

where α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, k denotes the Riemann– 0 ≤ β1 < β2 < · · · < βm < α−1, 1 ≤ β0 < α−1, D0+ Liouville derivative of order k (for k = α, β0 , β1 , . . . , βm ), the integrals from the boundary conditions (2.89) are Riemann-Stieltjes integrals with Hi , i = 1, . . . , m, functions of bounded variation, the functions ai ∈ C(0, 1) ∩ L1 (0, 1), i = 1, . . . , m, and the nonlinearity f (t, u) is nonnegative and it may be singular at the points t = 0, t = 1 and/or u = 0. We will present conditions for the data of problem (2.88), (2.89) connected to the spectral radii of some associated linear operators such that this problem has at least one or two positive solutions u ∈ C([0, 1], R+ ), u(t) > 0 for all t ∈ (0, 1]. In the proof of the main existence theorems, we use an application of the Krein-Rutman theorem in the space C[0, 1] (Theorem 2.5.1) and the fixed point index theory. In [103], the author

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investigates the fractional differential equation (2.88) supplemented with the boundary conditions  η β γ u(1) = a(t)D0+ u(t) dV (t), u(0) = u (0) = · · · = u(n−2) (0) = 0, D0+ 0

where β ∈ (0, 1), γ ∈ [0, α − 1), η ∈ (0, 1], a(t) ∈ L1 (0, 1) ∩ C(0, 1), and the function f (t, u) is nonnegative and it may be singular at t = 0, t = 1 and x = 0. The author proves in [103] some existence and multiplicity results which are closely associated with the relationship between 1 and the spectral radii corresponding to the relevant linear operators. 2.6.1

Preliminary results

We consider the fractional differential equation α u(t) + x (t) = 0, D0+

t ∈ (0, 1),

(2.90)

with the boundary conditions (2.89), where x  ∈ C(0, 1) ∩ L1 (0, 1). We m Γ(α)  1 α−βi −1 Γ(α) ai (s) dHi (s). In a simdenote by Δ3 = Γ(α−β0 ) − i=1 Γ(α−βi ) 0 s ilar manner as we proved Lemma 2.4.2, we obtain the following result. Lemma 2.6.1. If Δ3 = 0, then the unique solution u ∈ C[0, 1] of problem (2.90), (2.89) is given by  1 G3 (t, s) x(s) ds, t ∈ [0, 1], (2.91) u(t) = 0

where the Green function G3 is G3 (t, s) = g0 (t, s) + and 1 g0 (t, s) = Γ(α)

 m  1 tα−1  ai (τ )g1i (τ, s) dHi (τ ) , Δ3 i=1 0

(2.92)

tα−1 (1 − s)α−β0 −1 − (t − s)α−1 , 0 ≤ s ≤ t ≤ 1,

tα−1 (1 − s)α−β0 −1 , 0 ≤ t ≤ s ≤ 1, ⎧ α−βi −1 (1 − s)α−β0 −1 − (t − s)α−βi −1 , ⎨t 1 g1i (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(α − βi ) ⎩ α−βi −1 (1 − s)α−β0 −1 , 0 ≤ t ≤ s ≤ 1, t for all (t, s) ∈ [0, 1] × [0, 1], i = 1, . . . , m.

(2.93)

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Here, the functions g1i may have negative values, because we did not impose a relation between β0 and βi , i = 1, . . . , m. In Sections 2.4 and 2.5, we used the condition βm ≤ β0 which implies that g1i , i = 1, . . . , m are nonnegative functions. Based on some properties of the function g0 given by (2.93) from Lemma 2.4.3, we obtain the following lemma. m Lemma 2.6.2. We suppose that Δ3 = 0 and F (s) := Δ13 i=1 1 a (τ )g1i (τ, s) dHi (τ ) ≥ 0, for all s ∈ [0, 1]. Then the Green function 0 i G3 given by (2.92) is a continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G3 (t, s) ≤ J3 (s) for all t, s ∈ [0, 1], where J3 (s) = h0 (s) + F (s), s ∈ 1 (1 − s)α−β0 −1 (1 − (1 − s)β0 ), s ∈ [0, 1]; [0, 1], and h0 (s) = Γ(α) (b) G3 (t, s) ≥ tα−1 J3 (s) for all t, s ∈ [0, 1]; 1 (1 − (c) G3 (t, s) ≤ tα−1 K(s) for all t, s ∈ [0, 1], where K(s) = Γ(α) α−β0 −1 + F (s), s ∈ [0, 1]. s) In a similar manner as we proved Lemma 2.4.5, we deduce here the following result. Lemma 2.6.3. We suppose that Δ3 = 0, F (s) ≥ 0 for all s ∈ [0, 1], (t) ≥ 0 for all t ∈ (0, 1). Then the solution u of x  ∈ C(0, 1) ∩ L1 (0, 1) and x problem (2.90), (2.89) given by (2.91) satisfies the inequality u(t) ≥ tα−1 u for all t ∈ [0, 1], where u = supt∈[0,1] |u(t)|, and so u(t) ≥ 0 for all t ∈ [0, 1]. 2.6.2

Existence and multiplicity of positive solutions

We give in this section some theorems for the existence of at least one or two positive solutions for problem (2.88), (2.89). We present firstly the assumptions that we will use in the sequel. (V1) α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 3, βi ∈ R for all i = 0, . . . , m, 0 ≤ β1 < β2 < · · · < βm < α − 1, 1 ≤ β0 < α − 1. (V2) ai ∈ C(0, 1) ∩ L1 (0, 1) for all i = 1, . . . , m, and Hi : [0, 1] → R, i = 1, . . . , m are functions of bounded variation. m Γ(α)  1 α−βi −1 Γ(α) ai (s) dHi (s) = 0, and (V3) Δ3 = Γ(α−β0 ) − i=1 Γ(α−βi ) 0 s m  1 1 F (s) = Δ3 i=1 0 ai (τ )g2i (τ, s) dHi (τ ) ≥ 0 for all s ∈ [0, 1].

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(V4) The function f : (0, 1) × (0, ∞) → [0, ∞) is continuous. Besides, for any 0 < r < R, there exists φr,R ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) such that f (t, u) ≤ φr,R (t) for all t ∈ (0, 1) and u ∈ [rtα−1 , R]. (V5) There exist R1 > 0 and a function p1 ∈ C((0, 1), [0, ∞))∩L1 (0, 1) with 1 p (t) dt > 0 such that f (t, u) ≥ p1 (t)u for all (t, u) ∈ (0, 1)×(0, R1 ]. 0 1 (V6) There exist R2 > 0 and a function p2 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) 1 with 0 p2 (t) dt > 0 such that f (t, u) ≤ p2 (t)u for all (t, u) ∈ (0, 1) × [R2 , ∞). (V7) There exist R3 > 0 and a function p3 ∈ C((0, 1), [0, ∞))∩L1 (0, 1) with 1 0 p3 (t) dt > 0 such that f (t, u) ≤ p3 (t)u for all (t, u) ∈ (0, 1)×(0, R3 ]. (V8) There exist R4 > 0 and a function p4 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) 1 with 0 p4 (t) dt > 0 such that f (t, u) ≥ p4 (t)u for all (t, u) ∈ (0, 1) × [R4 , ∞). We consider X = C[0, 1] the space of continuous functions defined on [0, 1] with the supremum norm u = supt∈[0,1] |u(t)|, and the cone P = {u ∈ X, u(t) ≥ tα−1 u , ∀ t ∈ [0, 1]}. We define the operators  Gu(t) =

0

 Pi u(t) =

1

0

1

G3 (t, s)f (s, u(s)) ds, G3 (t, s)pi (s)u(s) ds,

i = 1, . . . , 4.

Lemma 2.6.4. Assume that (V1)−(V4) hold. Then for any r > 0, the operator G : P \ Br → P is completely continuous, (Br = {u ∈ X, u < r}). Proof. For any u ∈ P \ Br , we have rtα−1 ≤ u(t) ≤ u . Let R = u . By (V4) it follows that there exists a function φr,R ∈ C((0, 1), [0, ∞))∩L1 (0, 1) such that f (t, u(t)) ≤ φr,R (t) for all t ∈ (0, 1). Then by using Lemma 2.6.2, we find  Gu(t) =

0

 ≤

1

0

1

 G3 (t, s)f (s, u(s)) ds ≤ J3 (s)φr,R (s) ds ≤ J0

 0

1 0

1

J3 (s)f (s, u(s)) ds

φr,R (s) ds < ∞, ∀ t ∈ [0, 1],

where J0 = maxt∈[0,1] J3 (t), and then Gu is well defined.

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On the other hand, we obtain  1 Gu(t) = G3 (t, s)f (s, u(s)) ds 0

≥ tα−1



1

J3 (s)f (s, u(s)) ds ≥ tα−1 Gu(t1 ),

0

for all t, t1 ∈ [0, 1], and so Gu(t) ≥ tα−1 Gu for all t ∈ [0, 1]. Therefore, G(P \ Br ) ⊂ P . We will prove next that G is completely continuous. Firstly, we show that G is continuous. Let {un }n≥1 ⊂ P \ Br and un − u0 → 0 as n → ∞, with u0 ∈ P \ Br . Then there exists R > r such that r ≤ un ≤ R for all n = 0, 1, . . .. By (V4), for the above r, R we deduce (by the absolute continuity of the integral function) that for any  1ε > 0 there exists θ ∈ θ (0, 1/2) such that 0 φr,R (s) ds < ε/(6J0 ) and 1−θ φr,R (s) ds < ε/(6J0 ). Because f (t, u) is uniformly continuous on [θ, 1 − θ] × [θα−1 r, R], then there exists N > 0 such that for any n > N , we have |f (t, un (t)) − f (t, u0 (t))|
N , we find  1 Gun − Gu0 ≤ max G3 (t, s)|f (s, un (s)) − f (s, u0 (s))| ds t∈[0,1]



≤

1

0

J3 (s)|f (s, un (s)) − f (s, u0 (s))| ds



≤2

0



θ

J3 (s)φr,R (s) ds +

0



− f (s, u0 (s))| ds + 2 < 2J0

 0

θ

 φr,R (s) ds +

θ 1

1−θ

θ

− f (s, u0 (s))| ds + 2J0

1−θ

J3 (s)φr,R (s) ds

1−θ



J3 (s)|f (s, un (s))

J3 (s)|f (s, un (s))

1

1−θ

φr,R (s) ds < ε.

Hence, Gun − Gu0 → 0 as n → ∞, and so G is a continuous operator. Next, we will show that G is a compact operator, that is, it maps bounded sets into relatively compact sets. For this, let E ⊂ P \ Br be

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a bounded set. Then there exists R1 > r such that r ≤ u ≤ R1 for all u ∈ E. By the above proof, we obtain  1  1 J3 (s)φr,R1 (s) ds ≤ J0 φr,R1 (s) ds, ∀ t ∈ [0, 1], u ∈ E, Gu(t) ≤ 0

0

which implies that G(E) is uniformly bounded. The function G3 (t, s) is uniformly continuous on [0, 1]×[0, 1]. So, for any ε > 0, there exists ζ1 > 0 such that for any t1 , t2 ∈ [0, 1] with |t1 − t2 | < ζ1 , and for any s ∈ [0, 1] we have ε . |G3 (t1 , s) − G3 (t2 , s)| < 1 2 max{ 0 φr,R1 (s) ds, 1} Therefore, for any u ∈ E, we deduce  1 |G3 (t1 , s) − G3 (t2 , s)|f (s, u(s)) ds |Gu(t1 ) − Gu(t2 )| ≤ 0

 ≤

0

 ≤

1

0

|G3 (t1 , s) − G3 (t2 , s)|φr,R1 (s) ds

1

2 max{

1 0

ε φr,R1 (s) ds, 1}

φr,R1 (s) ds < ε.

This implies that G(E) is equicontinuous. By using the Arzela–Ascoli theorem, we conclude that G(E) is relatively compact, and then G : P \ Br → P is a compact operator.  Under assumptions (V1)–(V4), by the extension theorem, for any r > 0, the operator G has a completely continuous extension (also denoted by G) from P to P . By using Theorem 2.5.1 and similar arguments as those used in the proof of Lemma 2.5.2, we obtain the following result. Lemma 2.6.5.  1 Assume that (V1)−(V3) hold, and pi ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) with 0 pi (t) dt > 0, i = 1, . . . , 4. Then the operators Pi : P → P are linear and completely continuous. Besides, the spectral radius r(Pi ) > 0 and Pi has an eigenfunction ψi ∈ P \ {θ} corresponding to the eigenvalue r(Pi ), that is, Pi ψi = r(Pi )ψi , i = 1, . . . , 4. Theorem 2.6.1. We assume that (V1)–(V4) hold, and there exist R2 > R1 > 0 such that (V5) and (V6) are satisfied. Besides, we suppose that r(P1 ) ≥ 1 > r(P2 ) > 0. Then the boundary value problem (2.88), (2.89) has at least one positive solution.

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Proof. By (V5), for any u ∈ ∂BR1 ∩ P , we obtain  1  1 G3 (t, s)f (s, u(s)) ds ≥ G3 (t, s)p1 (s)u(s) ds Gu(t) = 0

0

= P1 u(t),

∀ t ∈ [0, 1].

We assume that G has no fixed points on ∂BR1 ∩ P (otherwise, the theorem is proved). We will show that u − Gu = μψ1 ,

∀ u ∈ ∂BR1 ∩ P, μ ≥ 0,

(2.94)

where ψ1 is given in Lemma 2.6.5. In fact, if not, there exist u1 ∈ ∂BR1 ∩ P and μ1 ≥ 0 such that u1 −Gu1 = μ1 ψ1 . Then μ1 > 0 and u1 = Gu1 +μ1 ψ1 ≥ μ1 ψ1 . We denote by μ0 = sup{μ, u1 ≥ μψ1 }. Then μ0 ≥ μ1 , u1 ≥ μ0 ψ1 and Gu1 ≥ P1 u1 ≥ μ0 P1 ψ1 = μ0 r(P1 )ψ1 ≥ μ0 ψ1 . Hence, u1 = Gu1 +μ1 ψ1 ≥ μ0 ψ1 +μ1 ψ1 = (μ0 +μ1 )ψ1 , which contradicts the definition of μ0 . We deduce that relation (2.94) holds, and by Theorem 1.2.9 we deduce i(G, BR1 ∩ P, P ) = 0.

(2.95)

Now, we consider the set V = {u ∈ P \ BR1 , Gu = μu with μ ≥ 1}. We will prove next that the set V is bounded. For any u ∈ V , we find  1 u(t) ≤ μu(t) = Gu(t) = G3 (t, s)f (s, u(s)) ds 0

 = D1

G3 (t, s)f (s, u(s)) ds +

 ≤

D1

 ≤

0

1

 

G3 (t, s)p2 (s)u(s) ds +  G3 (t, s)p2 (s)u(s) ds +

D2 1

0

0

1

G3 (t, s)f (s, u(s)) ds

G3 (t, s)f (s, u (s)) ds G3 (t, s)f (s, u (s)) ds

= P2 u(t) + G u(t) ≤ P2 u(t) + J0 M, (s) = min{u(s), R2 }, where D1 = {s, u(s) ≥ R2 }, D2 = {s, u(s) < R2 }, u 1 M = 0 φR1 ,R2 (s) ds < ∞ (by (V4)). Then we obtain (I − P2 )u(t) ≤ M for all t ∈ [0, 1]. Because r(P2 ) < 1, we deduce that the inverse operator

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∞ of (I − P2 ) exists and (I − P2 )−1 = i=1 P2i . Therefore we find u(t) ≤ (I − P2 )−1 (M ) and u(t) ≤ M (I − P2 )−1 for all t ∈ [0, 1], which means 2 > max{R2 , sup{ u , u ∈ V }}. Then that V is bounded. We choose R Gu = μu for all μ ≥ 1, u ∈ ∂BR2 ∩ P , and by Theorem 1.2.7, we obtain i(G, BR2 ∩ P, P ) = 1.

(2.96)

By (2.95) and (2.96), we conclude i(G, (BR2 \ B R1 ) ∩ P, P ) = i(G, BR2 ∩ P, P ) − i(G, BR1 ∩ P, P ) = 1. Then we deduce that G has at least one fixed point u  on (BR2 \ B R1 ) ∩ P , which is a positive solution of problem (2.88), (2.89). Taking into account the remark from the beginning of the proof (that G may have fixed points  of problem (2.88), (2.89) on ∂BR1 ∩ P ), we conclude that the solution u  u < R2 .  satisfies R1 ≤  Theorem 2.6.2. We assume that (V1)–(V4) hold and there exist R4 > R3 > 0 such that (V7) and (V8) are satisfied. Besides, we suppose that r(P4 ) > 1 ≥ r(P3 ) > 0. Then the boundary value problem (2.88), (2.89) has at least one positive solution. Proof. We suppose that G has no fixed points on ∂BR3 ∩ P (otherwise, the theorem is proved). We will show that ∀ u ∈ ∂BR3 ∩ P,

Gu = μu,

μ > 1.

(2.97)

If not, there exist u1 ∈ ∂BR3 ∩ P and μ1 > 1 such that Gu1 = μ1 u1 . By using (V7) we obtain  1 G3 (t, s)f (s, u1 (s)) ds μ1 u1 (t) = Gu1 (t) =  ≤

0

0

1

G3 (t, s)p3 (s)u1 (s) ds = P3 u1 (t), ∀ t ∈ [0, 1].

Because P3 is a nondecreasing operator, we deduce μ21 u1 (t) ≤ μ1 P3 u1 (t) = P3 (μ1 u1 )(t) ≤ P3 (P3 u1 (t)) = P32 u1 (t), ∀ t ∈ [0, 1]. Repeating the process, we find μn1 u1 (t) ≤ P3n u1 (t),

∀ t ∈ [0, 1], n ≥ 1,

and so μn1 u1 = μn1 u1 ≤ P3n u1 ≤ P3n u1 ,

∀ n ≥ 1.

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89

We conclude that P3n ≥ μn1 for all n ≥ 1, and then r(P3 ) =  n n limn→∞ P3 ≥ μ1 > 1, which is a contradiction, because r(P3 ) ≤ 1. Therefore, the relation (2.97) is satisfied, and by Theorem 1.2.7, we deduce that i(G, BR3 ∩ P, P ) = 1.

(2.98)

Now, we consider a decreasing sequence (cn )n≥1 , with 0 < cn < 1, for all n ≥ 1, convergent to 0, and we define the operators  1 Fn u(t) = G3 (t, s)p4 (s)u(s) ds. cn

By Theorem 3.7 from [107], the sequence of spectral radii (r(Fn ))n is increasing and converges to r(P4 ). Then we can choose n0 sufficiently large such that r(Fn0 ) > 1. We define Rn0 = R4 c1−α n0 . Then for any u ∈ ∂BRn0 ∩ P , we have u(t) ≥ tα−1 u = tα−1 Rn0 ≥ cα−1 n0 Rn0 = R4 ,

∀ t ∈ [cn0 , 1].

(2.99)

In a similar manner as we obtained Lemma 2.6.5, we deduce that Fn0 has an eigenfunction ψ0 ∈ P \ {θ} corresponding to the eigenvalue r(Fn0 ), that is Fn0 ψ0 = r(Fn0 )ψ0 . Let u ∈ ∂BRn0 ∩ P . By (V8) and (2.99), we find  Gu(t) =

0

 ≥

1

 G3 (t, s)f (s, u(s)) ds ≥

1

cn0

1

cn0

G3 (t, s)f (s, u(s)) ds

G3 (t, s)p4 (s)u(s) ds = Fn0 u(t),

∀ t ∈ [0, 1].

We assume that G has no fixed points on ∂BRn0 ∩ P (otherwise, the theorem is proved). We will show that u − Gu = μψ0 ,

∀ u ∈ ∂BRn0 ∩ P, μ > 0.

(2.100)

In fact, if not, there exist u2 ∈ ∂BRn0 ∩ P and μ2 > 0 such that u2 − Gu2 = μ2 ψ0 . We denote by μ0 = sup{μ, u2 ≥ μψ0 }. Then μ0 ≥ μ2 and u2 ≥ μ0 ψ0 . In addition, we have u2 (t) = Gu2 (t) + μ2 ψ0 (t) ≥ Fn0 u2 (t) + μ2 ψ0 (t) ≥ μ0 Fn0 ψ0 (t) + μ2 ψ0 (t) = μ0 r(Fn0 )ψ0 (t) + μ2 ψ0 (t) ≥ μ0 ψ0 (t) + μ2 ψ0 (t) = (μ0 + μ2 )ψ0 (t),

∀ t ∈ [0, 1].

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The obtained inequality contradicts the definition of μ0 . Therefore, the relation (2.100) is satisfied, and by Theorem 1.2.9, we deduce i(G, BRn0 ∩ P, P ) = 0.

(2.101)

Then by (2.98) and (2.101), we conclude that i(G, (BRn0 \ B R3 ) ∩ P, P ) = i(G, BRn0 ∩ P, P ) − i(G, BR3 ∩ P, P ) = −1. This means that G has at least one fixed point u  on (BRn0 \ B R3 )∩P , which is a positive solution of problem (2.88), (2.89). Taking into account that G  satisfies may have fixed points on (∂BRn0 ∪ ∂BR3 ) ∩ P , then the solution u u ≤ Rn0 .  R3 ≤  We can also obtain existence results for multiple positive solutions by imposing various conditions similar to (V5)–(V8). Theorem 2.6.3. We assume that (V1)–(V4) hold and there exist R4 > R5 > R1 > 0 such that (V5), (V8) and (V9) There exists a function p5 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) with 1 α−1 , R5 ] 0 p5 (t) dt > 0 such that f (t, u) ≤ p5 (t)R5 for all u ∈ [R1 t and t ∈ (0, 1), hold. Besides,  1 we suppose that r(P1 ) ≥ 1, r(P4 ) > 1 and P5 < 1, where P5 u(t) = 0 G3 (t, s)p5 (s)u(s) ds, for t ∈ [0, 1] and u ∈ P . Then the boundary value problem (2.88), (2.89) has at least two positive solutions u1 and u2 with R1 ≤ u1 < R5 < u2 . Proof. We will show that for any u ∈ ∂BR5 ∩ P , we have Gu = λu for all λ ≥ 1. If not, there exist u0 ∈ ∂BR5 ∩ P and λ0 ≥ 1 such that Gu0 = λ0 u0 . Then we obtain  1 G3 (t, s)f (s, u0 (s)) ds u0 (t) ≤ λ0 u0 (t) = Gu0 (t) =  ≤

0

0

1

G3 (t, s)p5 (s)R5 ds = R5 (P5 Id )(t) ≤ R5 P5 < R5 , ∀ t ∈ [0, 1],

where Id (t) = t for all t ∈ [0, 1]. So we deduce u0 < R5 , which is a contradiction, because u0 ∈ ∂BR5 ∩ P . Therefore, by Theorem 1.2.7, we conclude i(G, BR5 ∩ P, P ) = 1.

(2.102)

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By the proof of Theorem 2.6.1, we obtain that i(G, BR1 ∩ P, P ) = 0,

(2.103)

or G has a fixed point on ∂BR1 ∩ P. By the proof of Theorem 2.6.2, we deduce that there exists Rn0 > R4 such that i(G, BRn0 ∩ P, P ) = 0,

(2.104)

or G has a fixed point on ∂BRn0 ∩ P. If G has no fixed points on (∂BR1 ∪ ∂BRn0 ) ∩ P , then by the relations (2.102)–(2.104) we conclude i(G, (BRn0 \ B R5 ) ∩ P, P ) = −1, and i(G, (BR5 \ B R1 ) ∩ P, P ) = 1. Therefore, the operator G has at least two fixed points u1 ∈ (BRn0 \B R5 )∩P and u2 ∈ (BR5 \ B R1 ) ∩ P , which are positive solutions for problem (2.88), (2.89). Because G may have fixed points on (∂BRn0 ∪ ∂BR1 ) ∩ P , we deduce that the solutions u1 , u2 of problem (2.88), (2.89) satisfy R1 ≤ u1 <  R5 < u2 ≤ Rn0 . Theorem 2.6.4. We assume that (V1)–(V4) hold and there exist R2 > R6 > R3 > 0 such that (V6), (V7) and (V10) There exist c ∈ (0, 1) and a function p6 ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1) 1 with 0 p6 (t) dt > 0 such that f (t, u) ≥ p6 (t)R6 for all t ∈ [c, 1) and u ∈ [cα−1 R6 , R6 ], 1 hold. Besides, we suppose that r(P2 ) < 1, r(P3 ) ≤ 1 and c J3 (s) p6 (s) ds > 1. Then the boundary value problem (2.88), (2.89) has at least two positive solutions u1 and u2 with R3 ≤ u1 < R6 < u2 . Proof. For any u ∈ ∂BR6 ∩P , we have u(t) ≥ tα−1 u = tα−1 R6 ≥ cα−1 R6 for all t ∈ [c, 1]. Then we deduce  1  1 G3 (t, s)f (s, u(s)) ds ≥ max tα−1 J3 (s)f (s, u(s)) ds Gu = max t∈[0,1]

0

≥ max tα−1 t∈[0,1]



≥

1 c

 c

t∈[0,1]

1



J3 (s)f (s, u(s)) ds =

J3 (s)p6 (s)R6 ds > R6 = u .

c

1

0

J3 (s)f (s, u(s)) ds

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Hence, Gu > u for all u ∈ ∂BR6 ∩ P . This last inequality implies that Gu ≤ u for all u ∈ ∂BR6 ∩ P , and then we obtain (see [17]) i(G, BR6 ∩ P, P ) = 0.

(2.105)

2 > R2 By the proof of Theorem 2.6.1, we deduce that there exists R such that i(G, BR2 ∩ P, P ) = 1.

(2.106)

By the proof of Theorem 2.6.2, we conclude that i(G, BR3 ∩ P, P ) = 1,

(2.107)

or G has a fixed point on ∂BR3 ∩ P. If G has no fixed points on ∂BR3 ∩ P , then by the relations (2.105)– (2.107), we obtain i(G, (BR2 \ B R6 ) ∩ P, P ) = 1, and i(G, (BR6 \ B R3 ) ∩ P, P ) = −1. Hence, the operator G has at least two fixed points u1 ∈ (BR2 \B R6 )∩P and u2 ∈ (BR6 \B R3 )∩P , which are positive solutions for problem (2.88), (2.89). Because G may have fixed points on ∂BR3 ∩P , we deduce that the solutions 2 .  u1 , u2 of problem (2.88), (2.89) satisfy R3 ≤ u1 < R6 < u2 < R 2.6.3

An example

Let α = 5/2 (n = 3), m = 2, β0 = 5/4, β1 = 1/2, β2 = 4/3, H1 (t) = {0, t ∈ [0, 1/3); 1, t ∈ [1/3, 1]}, H2 (t) = t for all t ∈ [0, 1], a1 = 1, a2 = 1/2. We consider the fractional differential equation 5/2

D0+ u(t) + f (t, u(t)) = 0,

t ∈ (0, 1),

(2.108)

with the boundary conditions u(0) = u (0) = 0,

5/4

1/2

D0+ u(1) = D0+ u

   1 1 1 4/3 D u(t) dt. + 3 2 0 0+ (2.109)

We obtain Δ3 ≈ 0.40939289 = 0. We also deduce t3/2 (1 − s)1/4 − (t − s)3/2 , 0 ≤ s ≤ t ≤ 1, 1 g0 (t, s) = Γ(5/2) t3/2 (1 − s)1/4 , 0 ≤ t ≤ s ≤ 1,

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g11 (t, s) =

page 93

93

t(1 − s)1/4 − (t − s), 0 ≤ s ≤ t ≤ 1,

t(1 − s)1/4 , 0 ≤ t ≤ s ≤ 1, t1/6 (1 − s)1/4 − (t − s)1/6 , 0 ≤ s ≤ t ≤ 1, 1 g12 (t, s) = Γ(7/6) t1/6 (1 − s)1/4 , 0 ≤ t ≤ s ≤ 1, ⎧1 ! 1 (1 − s)1/4 − 13 − s + 2Γ(13/6) [(1 − s)1/4 − (1 − s)7/6 ], ⎪ ⎪ 3 ⎪ ⎪ 1 1 ⎨ 0 ≤ s < 3, F (s) = 1 1 1/4 Δ3 ⎪ + 2Γ(13/6) [(1 − s)1/4 − (1 − s)7/6 ], ⎪ 3 (1 − s) ⎪ ⎪ ⎩ 1 3 ≤ s ≤ 1, G3 (t, s) = g0 (t, s) + t3/2 F (s), t, s ∈ [0, 1], J3 (s) =

1 (1 − s)1/4 (1 − (1 − s)5/4 ) + F (s), s ∈ [0, 1]. Γ(5/2)

We have F (s) ≥ 0, for all s ∈ [0, 1]. We mention here that the function g12 has negative values in the vicinity of t = 1. We consider the function ⎧ ⎪ 10(t − 14 )2 t−1/3 u−1/5 + (t − 13 )2 (1 − t)−1/4 u, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (t, u) ∈ (0, 1) × (0, 1], ⎪ ⎪ ⎪  ⎪ ⎨ 10(t − 14 )2 t−1/3 + (t − 13 )2 (1 − t)−1/4 cos2 π(u−1) , 7 f (t, u) = ⎪ (t, u) ∈ (0, 1) × (1, 64], ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 5 1 2 −1/3 1/3 ⎪ u + 18 (t − 13 )2 (1 − t)−1/4 u1/2 , ⎪ 2 (t − 4 ) t ⎪ ⎪ ⎩ (t, u) ∈ (0, 1) × (64, ∞). For any 0 < r < R, we obtain the inequality f (t, u) ≤ φr,R (t) for all (t, u) ∈ (0, 1) × [rt3/2 , R], where φr,R (t), t ∈ (0, 1) is defined by

φr,R (t) =

⎧ 10(t − 14 )2 t−1/3 (rt3/2 )−1/5 + (t − 13 )2 (1 − t)−1/4 R, 0 < R ≤ 1, ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ max 10(t − 14 )2 t−1/3 + (t − 13 )2 (1 − t)−1/4 ; ⎪ ⎪  ⎪ ⎪ ⎨ 10(t − 1 )2 t−1/3 (rt3/2 )−1/5 + (t − 1 )2 (1 − t)−1/4 , 4

3

1 < R ≤ 64, ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ max 52 (t − 14 )2 t−1/3 R1/3 + 18 (t − 13 )2 (1 − t)−1/4 R1/2 , ⎪ ⎪ ⎪  ⎩ 10(t − 14 )2 t−1/3 (rt3/2 )−1/5 + (t − 13 )2 (1 − t)−1/4 , R > 64.

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The above function φr,R ∈ C((0, 1), [0, ∞)) ∩ L1 (0, 1), and so the assumptions (V1)–(V4) are satisfied. We choose R1 = 1, R2 = 64 and the functions   2 2 1 1 t−1/3 + t − (1 − t)−1/4 , t ∈ (0, 1), p1 (t) = 10 t − 4 3 p1 (t) , 3

p2 (t) =

t ∈ (0, 1).

We have the inequalities f (t, u) ≥ p1 (t)u for all (t, u) ∈ (0, 1) × (0, 1], and f (t, u) ≤ p2 (t)u for all (t, u) ∈ (0, 1) × [64, ∞). Evidently,  1 p1 , p2 ∈ 1 C((0, 1), [0, ∞)) ∩ L1 (0, 1), 0 p1 (t) dt ≈ 1.88182 > 0, 0 p2 (t) dt ≈ 0.62727 > 0, and so assumptions (V5) and (V6) are also satisfied. We define the linear operators P1 , P2 : P → P , where P = {u ∈ C[0, 1], u(t) ≥ t3/2 u , ∀ t ∈ [0, 1]}, by  1 G3 (t, s)p1 (s)u(s) ds, P1 u(t) =  P2 u(t) =

0

1 0

1 P1 x(t), 3

G3 (t, s)p2 (s)u(s) ds =

t ∈ [0, 1], u ∈ P.

We will show that r(P1 ) ≥ 1 and r(P2 ) < 1. We denote by Id (t) = t, ∀ t ∈ [0, 1], and ζ(t) = t3/2 , ∀ t ∈ [0, 1]. Then we find  1  1 P1 ζ(t) = G3 (t, s)p1 (s)ζ(s) ds = G3 (t, s)p1 (s)s3/2 ds 0

 ≥

0

1

0

t

3/2

J3 (s)p1 (s)s

3/2

and therefore P1n ζ(t) = P1 (P1n−1 ζ)(t) ≥

 ds = 

1

0

0

1

J3 (s)p1 (s)s

J3 (s)p1 (s)s3/2 ds



3/2

ds ζ(t),

n ζ(t).

The last inequality gives us % r(P1 ) = lim n P1n ≥ lim n→∞

 = 0

1

n→∞

$ n

0

1

J3 (s)p1 (s)s3/2 ds

J3 (s)p1 (s)s3/2 ds ≈ 1.70534,

and then r(P1 ) > 1.

n max ζ(t)

t∈[0,1]

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On the other hand, we obtain  1  G3 (t, s)p1 (s) ds ≤ (P1 Id )(t) = 0

0

1

page 95

95

J3 (s)p1 (s) ds ≈ 2.41727, ∀ t ∈ [0, 1],

and so P1 Id < 2.418. Therefore, we find r(P2 ) =

1 1 1 r(P1 ) ≤ P1 = P1 Id < 1. 3 3 3

Then 0 < 13 < r(P2 ) < 1 < r(P1 ). By Theorem 2.6.1, we conclude that problem (2.108), (2.109) has at least one positive solution u(t), t ∈ [0, 1], which satisfies the inequality u(t) ≥ t3/2 u for all t ∈ [0, 1]. Remark 2.6.1. The results presented in this section will be published in [83].

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Chapter 3

Systems of Two Riemann–Liouville Fractional Differential Equations with Multi-Point Boundary Conditions

In this chapter, we study the existence and multiplicity of positive solutions for systems of two Riemann-Liouville fractional differential equations, subject to uncoupled or coupled multi-point boundary conditions which contain fractional derivatives, and the nonlinearities of systems are nonsingular or singular functions, with nonnegative values.

3.1

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions

We consider the system of nonlinear fractional differential equations  α D0+ u(t) + f (t, u(t), v(t)) = 0, t ∈ (0, 1), (3.1) β v(t) + g(t, u(t), v(t)) = 0, t ∈ (0, 1), D0+ with the uncoupled multi-point boundary conditions ⎧ N  ⎪ ⎪ p1 q1 (j) ⎪ ai D0+ u(ξi ), ⎪ ⎨ u (0) = 0, j = 0, . . . , n − 2, D0+ u(1) = ⎪ ⎪ (k) ⎪ ⎪ ⎩ v (0) = 0,

k = 0, . . . , m − 2,

p2 D0+ v(1)

=

i=1 M 

(3.2) q2 bi D0+ v(ηi ),

i=1

where α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], 97

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ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi , bi ∈ R ζ denotes the for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, and D0+ Riemann-Liouville derivative of order ζ (for ζ = α, β, p1 , q1 , p2 , q2 ). Under sufficient conditions on the functions f and g, we study the existence and multiplicity of positive solutions of problem (3.1), (3.2) by using some theorems from the fixed point index theory. By a positive solution of problem (3.1), (3.2), we mean a pair of functions (u, v) ∈ (C([0, 1], R+ ))2 satisfying (3.1) and (3.2) with u(t) > 0 for all t ∈ (0, 1] or v(t) > 0 for all t ∈ (0, 1]. The system (3.1) with two positive parameters λ1 and λ2 , and with the boundary conditions  u(i) (0) = v (i) (0) = 0, i = 0, . . . , n − 2, (3.3) μ μ ν ν u(1) = η1 D0+ u(ξ1 ), D0+ v(1) = η2 D0+ v(ξ2 ), D0+ where α, β ∈ (n − 1, n], n ∈ N, n ≥ 3, 1 ≤ μ, ν ≤ n − 2, ξ1 , ξ2 ∈ (0, 1), 0 < η1 ξ1α−μ−1 < 1, 0 < η2 ξ2β−ν−1 < 1, was investigated in [108]. 3.1.1

Auxiliary results

In this section, we present some auxiliary results that can be proved in a similar manner as the auxiliary results from Section 2.1.1. We consider the fractional differential equation α u(t) + x(t) = 0, D0+

0 < t < 1,

(3.4)

with the multi-point boundary conditions u(j) (0) = 0,

j = 0, . . . , n − 2;

p1 D0+ u(1) =

N 

q1 ai D0+ u(ξi ),

(3.5)

i=1

where α ∈ (n − 1, n], n ∈ N, n ≥ 3, ai , ξi ∈ R, i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, p1 , q1 ∈ R, p1 ∈ [1, α − 1), q1 ∈ [0, p1 ], and x ∈ Γ(α) Γ(α) N α−q1 −1 − Γ(α−q . C(0, 1) ∩ L1 (0, 1). We denote by Δ1 = Γ(α−p i=1 ai ξi 1) 1) Lemma 3.1.1. If Δ1 = 0, then the unique solution u ∈ C[0, 1] of problem (3.4), (3.5) is given by

1 G1 (t, s)x(s) ds, t ∈ [0, 1], (3.6) u(t) = 0

where the Green function G1 is N tα−1  ai g2 (ξi , s), G1 (t, s) = g1 (t, s) + Δ1 i=1

∀ (t, s) ∈ [0, 1] × [0, 1],

(3.7)

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and 1 g1 (t, s) = Γ(α)



tα−1 (1 − s)α−p1 −1 − (t − s)α−1 , t

α−1

α−p1 −1

(1 − s)

,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1,

⎧ α−q1 −1 (1 − s)α−p1 −1 − (t − s)α−q1 −1 , ⎪ ⎨t 1 0 ≤ s ≤ t ≤ 1, g2 (t, s) = Γ(α − q1 ) ⎪ ⎩ α−q1 −1 (1 − s)α−p1 −1 , 0 ≤ t ≤ s ≤ 1. t

(3.8)

Lemma 3.1.2. The functions g1 and g2 given by (3.8) have the properties: (a) g1 (t, s) ≤ h1 (s) for all t, s ∈ [0, 1], where h1 (s) =

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 ), s ∈ [0, 1]; Γ(α)

(b) g1 (t, s) ≥ tα−1 h1 (s) for all t, s ∈ [0, 1]; α−1 (c) g1 (t, s) ≤ tΓ(α) , for all t, s ∈ [0, 1]; (d) g2 (t, s) ≥ tα−q1 −1 h2 (s) for all t, s ∈ [0, 1], where h2 (s) =

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 −q1 ), s ∈ [0, 1]; Γ(α − q1 )

1 (e) g2 (t, s) ≤ Γ(α−q tα−q1 −1 for all t, s ∈ [0, 1]; 1) (f) The functions g1 and g2 are continuous on [0, 1] × [0, 1]; g1 (t, s) ≥ 0, g2 (t, s) ≥ 0 for all t, s ∈ [0, 1]; g1 (t, s) > 0, g2 (t, s) > 0 for all t, s ∈ (0, 1).

Lemma 3.1.3. Assume that ai ≥ 0 for all i = 1, . . . , N and Δ1 > 0. Then the function G1 given by (3.7) is a nonnegative continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G1 (t, s) ≤ J1 (s) for all t, s ∈ [0, 1], where J1 (s) = h1 (s) + N 1 i=1 ai g2 (ξi , s), s ∈ [0, 1]; Δ1 (b) G1 (t, s) ≥ tα−1 J1 (s) for all t, s ∈ [0, 1]; 1 + (c) G1 (t, s) ≤ σ1 tα−1 , for all t, s ∈ [0, 1], where σ1 = Γ(α) N α−q1 −1 1 . i=1 ai ξi Δ1 Γ(α−q1 ) Lemma 3.1.4. Assume that ai ≥ 0 for all i = 1, . . . , N , Δ1 > 0, x ∈ C(0, 1) ∩ L1 (0, 1) and x(t) ≥ 0 for all t ∈ (0, 1). Then the solution u of problem (3.4), (3.5) given by (3.6) satisfies the inequality u(t) ≥ tα−1 u(t ) for all t, t ∈ [0, 1].

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We consider now the fractional differential equation β v(t) + y(t) = 0, D0+

0 < t < 1,

(3.9)

with the multi-point boundary conditions v

(j)

(0) = 0,

j = 0, . . . , m − 2;

p2 D0+ v(1)

=

M 

q2 bi D0+ v(ηi ),

(3.10)

i=1

where β ∈ (m − 1, m], m ∈ N, m ≥ 3, bi , ηi ∈ R, i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, p2 , q2 ∈ R, p2 ∈ [1, β − 1), q2 ∈ [0, p2 ], and Γ(β) Γ(β) M β−q2 −1 − Γ(β−q . y ∈ C(0, 1)∩L1 (0, 1). We denote by Δ2 = Γ(β−p i=1 bi ηi 2) 2) Lemma 3.1.5. If Δ2 = 0, then the unique solution of problem (3.9), (3.10) is given by

1 v(t) = G2 (t, s)y(s) ds, t ∈ [0, 1], (3.11) 0

where the Green function G2 is G2 (t, s) = g3 (t, s) +

M tβ−1  bi g4 (ηi , s), Δ2

∀ (t, s) ∈ [0, 1] × [0, 1], (3.12)

i=1

and 1 g3 (t, s) = Γ(β)



tβ−1 (1 − s)β−p2 −1 − (t − s)β−1 , t

β−1

β−p2 −1

(1 − s)

,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1,

⎧ β−q2 −1 ⎪ (1 − s)β−p2 −1 − (t − s)β−q2 −1 , ⎨t 1 0 ≤ s ≤ t ≤ 1, g4 (t, s) = Γ(β − q2 ) ⎪ ⎩ β−q2 −1 (1 − s)β−p2 −1 , 0 ≤ t ≤ s ≤ 1. t

(3.13)

Lemma 3.1.6. The functions g3 and g4 given by (3.13) have the properties: (a) g3 (t, s) ≤ h3 (s) for all t, s ∈ [0, 1], where h3 (s) =

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 ), s ∈ [0, 1]; Γ(β)

(b) g3 (t, s) ≥ tβ−1 h3 (s) for all t, s ∈ [0, 1]; β−1 (c) g3 (t, s) ≤ tΓ(β) , for all t, s ∈ [0, 1]; (d) g4 (t, s) ≥ tβ−q2 −1 h4 (s) for all t, s ∈ [0, 1], where h4 (s) =

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 −q2 ), s ∈ [0, 1]; Γ(β − q2 )

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101

1 (e) g4 (t, s) ≤ Γ(β−q tβ−q2 −1 for all t, s ∈ [0, 1]; 2) (f) The functions g3 and g4 are continuous on [0, 1] × [0, 1]; g3 (t, s) ≥ 0, g4 (t, s) ≥ 0 for all t, s ∈ [0, 1]; g3 (t, s) > 0, g4 (t, s) > 0 for all t, s ∈ (0, 1).

Lemma 3.1.7. Assume that bi ≥ 0 for all i = 1, . . . , M and Δ2 > 0. Then the function G2 given by (3.12) is a nonnegative continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G2 (t, s) ≤ J2 (s) for all t, s ∈ [0, 1], where J2 (s) = h3 (s) + M 1 i=1 bi g4 (ηi , s), s ∈ [0, 1]; Δ2 (b) G2 (t, s) ≥ tβ−1 J2 (s) for all t, s ∈ [0, 1]; 1 + (c) G2 (t, s) ≤ σ2 tβ−1 , for all t, s ∈ [0, 1], where σ2 = Γ(β) M β−q2 −1 1 . i=1 bi ηi Δ2 Γ(β−q2 ) Lemma 3.1.8. Assume that bi ≥ 0 for all i = 1, . . . , M, Δ2 > 0, y ∈ C(0, 1) ∩ L1 (0, 1) and y(t) ≥ 0 for all t ∈ (0, 1). Then the solution v of problem (3.9), (3.10) given by (3.11) satisfies the inequality v(t) ≥ tβ−1 v(t ) for all t, t ∈ [0, 1]. 3.1.2

Existence and multiplicity of positive solutions

In this section, we give sufficient conditions on f and g such that positive solutions with respect to a cone for our problem (3.1), (3.2) exist. We present the assumptions that we shall use in the sequel. (H1) α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N; n, m ≥ 3; ξi ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi ∈ R for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], ai ≥ 0 for all i = 1, . . . , N , bi ≥ 0 Γ(α) Γ(α) N α−q1 −1 − Γ(α−q > 0, for all i = 1, . . . , M , Δ1 = Γ(α−p i=1 ai ξi 1) 1) Γ(β) Γ(β) M β−q2 −1 > 0. Δ2 = Γ(β−p2 ) − Γ(β−q2 ) i=1 bi ηi (H2) The functions f, g : [0, 1] × R+ × R+ → R+ are continuous. (H3) There exist the functions a, b ∈ C(R+ , R+ ) such that (a) a(·) is concave and strictly increasing on R+ with a(0) = 0; (b) There exists σ ∈ (0, 1) such that f0i = lim inf v→0+

f (t, u, v) ∈ (0, ∞], uniformly with respect to a(v)

(t, u) ∈ [σ, 1] × R+ ,

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g0i = lim inf u→0+

g(t, u, v) ∈ (0, ∞], uniformly with respect to b(u)

(t, v) ∈ [σ, 1] × R+ ; a(Cb(u)) = ∞ exists for any constant C > 0. u (H4) There exist α1 , α2 > 0 with α1 α2 ≤ 1 such that (c) limu→0+

s f∞ = lim sup v→∞

f (t, u, v) ∈ [0, ∞), uniformly with respect to v α1

(t, u) ∈ [0, 1] × R+ , g(t, u, v) = 0 exists uniformly with respect to uα2 (t, v) ∈ [0, 1] × R+ .

s = lim g∞

u→∞

(H5) There exist the functions c, d ∈ C(R+ , R+ ) such that (a) c(·) is concave and strictly increasing on R+ ; (b) There exists σ ∈ (0, 1) such that i = lim inf f∞ v→∞

f (t, u, v) ∈ (0, ∞], uniformly with respect to c(v)

(t, u) ∈ [σ, 1] × R+ , i = lim inf g∞ u→∞

g(t, u, v) ∈ (0, ∞], uniformly with respect to d(u)

(t, v) ∈ [σ, 1] × R+ ; c(Cd(u)) = ∞ exists for any constant C > 0. u (H6) There exist β1 , β2 > 0 with β1 β2 ≥ 1 such that (c) lim

u→∞

f0s = lim sup v→0+

f (t, u, v) ∈ [0, ∞), uniformly with respect to v β1

(t, u) ∈ [0, 1] × R+ , g0s = lim

u→0+

g(t, u, v) = 0 exists uniformly with respect to uβ2

(t, v) ∈ [0, 1] × R+ .

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103

(H7) The functions f (t, u, v) and g(t, u, v) are nondecreasing with respect to u and v, and there exists N0 > 0 such that f (t, N0 , N0 ) < where m0 = max{ Section 3.1.1.

1 0

N0 , 2m0

g(t, N0 , N0 )
0, and the assumptions (H5) and (H6) with β1 = r, β2 = 1/r , r > 0 were used in Theorems 3.1 and 3.2 from [108] for problem (3.1) with two positive parameters λ1 and λ2 , and with the boundary conditions (3.3). By using the Green functions G1 and G2 from Section 3.1.1, we consider the following nonlinear system of integral equations: ⎧

1 ⎪ ⎪ G1 (t, s)f (s, u(s), v(s)) ds, 0 ≤ t ≤ 1, ⎨ u(t) =

01 (3.14) ⎪ ⎪ ⎩ v(t) = G2 (t, s)g(s, u(s), v(s)) ds, 0 ≤ t ≤ 1. 0

By Lemmas 3.1.1 and 3.1.5, (u, v) is solution of problem (3.1), (3.2) if and only if (u, v) is solution of the system (3.14). We consider the Banach space X = C[0, 1] with supremum norm ·

and the Banach space Y = X × X with the norm (u, v) Y = u + v . We define the cones P1 = {x ∈ X, x(t) ≥ tα−1 x , ∀ t ∈ [0, 1]} ⊂ X, P2 = {y ∈ X, y(t) ≥ tβ−1 y , ∀ t ∈ [0, 1]} ⊂ X, and P = P1 × P2 ⊂ Y . We introduce the operator Q : P → Y by Q(u, v) = (Q1 (u, v), Q2 (u, v)), (u, v) ∈ P , with Q1 , Q2 : P → X defined by

1 G1 (t, s)f (s, u(s), v(s)) ds, t ∈ [0, 1], (u, v) ∈ P, Q1 (u, v)(t) = 0

Q2 (u, v)(t) =

0

1

G2 (t, s)g(s, u(s), v(s)) ds,

t ∈ [0, 1],

(u, v) ∈ P.

Lemma 3.1.9. If (H1)−(H2) hold, then Q : P → P is a completely continuous operator.

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Proof. Let (u, v) ∈ P be an arbitrary element. Because Q1 (u, v) and Q2 (u, v) satisfy the problem (3.4), (3.5) for x(t) = f (t, u(t), v(t)), t ∈ [0, 1], and the problem (3.9), (3.10) for y(t) = g(t, u(t), v(t)), t ∈ [0, 1], respectively, then by Lemmas 3.1.4 and 3.1.8, we obtain Q1 (u, v)(t) ≥ tα−1 Q1 (u, v)(t ), ∀ t, t ∈ [0, 1],

Q2 (u, v)(t) ≥ tβ−1 Q2 (u, v)(t ),

(u, v) ∈ P,

and so Q1 (u, v)(t) ≥ tα−1 max Q1 (u, v)(t ) = tα−1 Q1 (u, v) , t ∈[0,1]

∀ t ∈ [0, 1], Q2 (u, v)(t) ≥ t

β−1

(u, v) ∈ P,



max Q2 (u, v)(t ) = tβ−1 Q2 (u, v) ,

t ∈[0,1]

∀ t ∈ [0, 1],

(u, v) ∈ P.

Therefore, Q(u, v) = (Q1 (u, v), Q2 (u, v)) ∈ P and then Q(P ) ⊂ P . By using standard arguments, we can easily show that Q1 and Q2 are completely continuous, and then Q is a completely continuous operator.  The pair of functions (u, v) is a solution of problem (3.14) if and only if (u, v) ∈ P is a fixed point of operator Q. So, we will investigate the existence of fixed points of operator Q. Theorem 3.1.1. Assume that (H1)–(H4) hold. Then the problem (3.1), (3.2) has at least one positive solution. Proof. By (H3), there exist C1 > 0, C2 > 0 and a sufficiently small r1 > 0 such that f (t, u, v) ≥ C1 a(v),

∀ (t, u) ∈ [σ, 1] × R+ ,

v ∈ [0, r1 ],

g(t, u, v) ≥ C2 b(u),

∀ (t, v) ∈ [σ, 1] × R+ ,

u ∈ [0, r1 ],

(3.15)

and a(C3 b(u)) ≥

2C3 u , C1 C2 m1 m2 γσ α+β−2

∀ u ∈ [0, r1 ],

(3.16)

where C3 = max{C2 σ β−1 J2 (s), s ∈ [0, 1]} > 0, γ = min{σ α−1 , σ β−1 }, 1 1 m1 = σ J1 (s) ds, m2 = σ J2 (s) ds. We will show that (Q1 (u, v), Q2 (u, v)) ≤ (u, v) for all (u, v) ∈ ∂Br1 ∩ P . We suppose that there exists (u, v) ∈ ∂Br1 ∩P , that is, (u, v) Y = r1 , such

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that (Q1 (u, v), Q2 (u, v)) ≤ (u, v). Then u ≥ Q1 (u, v) and v ≥ Q2 (u, v). By using the monotonicity and concavity of a(·), the Jensen inequality, Lemma 3.1.3, the relations (3.15) and (3.16), we obtain

1 G1 (t, s)f (s, u(s), v(s)) ds u(t) ≥ Q1 (u, v)(t) =

≥

0

1

σ



G1 (t, s)f (s, u(s), v(s)) ds ≥ C1

≥ C1 σ

α−1

≥ C1 σ

α−1

≥ C1 σ

α−1

≥ C1 σ

α−1

≥ C1 σ

α−1



1

σ



1

σ



1

σ



1

σ





J1 (s)a J1 (s)a

1

σ

a

≥

C1 C2 C3−1 σ α+β−2 m1

=

2 m2 γ

σ

ds

β−1

1

σ

ds

J2 (τ )b(u(τ )) dτ

ds

  a C2 σ β−1 J2 (τ )b(u(τ )) dτ

 −1   C3 C2 σ β−1 J2 (τ ) C3 b(u(τ )) dτ

≥

1

C2 σ

σ

J1 (s) ds

C1 C2 C3−1 σ α+β−2 m1



1



1



C2 G2 (s, τ )b(u(τ )) dτ

σ

J1 (s)a

ds

1





G2 (s, τ )g(τ, u(τ ), v(τ )) dτ

σ

J1 (s)a

tα−1 J1 (s)a(v(s)) ds

1



σ

G2 (s, τ )g(τ, u(τ ), v(τ )) dτ

0



σ

= C1 σ α−1 m1

1

1



1

J2 (τ )a (C3 b(u(τ ))) dτ

σ



1

J2 (τ )

σ

J2 (τ )u(τ ) dτ ≥

2C3 u(τ ) dτ C1 C2 m1 m2 γσ α+β−2

2 α−1 σ m2 u ≥ 2 u , m2 γ

∀ t ∈ [σ, 1].

So, u ≥ maxt∈[σ,1] |u(t)| ≥ 2 u , and then

u = 0. In a similar manner, by Lemma 3.1.7, we deduce 1

a(v(t)) ≥ a(Q2 (u, v)(t)) = a G2 (t, τ )g(τ, u(τ ), v(τ )) dτ

≥

0

0

1

a (G2 (t, τ )g(τ, u(τ ), v(τ ))) dτ

(3.17)

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≥

σ

≥

1

1

σ



1

= σ

a (G2 (t, τ )g(τ, u(τ ), v(τ ))) dτ   a σ β−1 J2 (τ )C2 b(u(τ )) dτ   a (C3−1 C2 σ β−1 J2 (τ ))C3 b(u(τ )) dτ

≥ C2 C3−1 σ β−1 ≥ C2 C3−1 σ β−1



1

σ



1

σ

2 = C1 m1 m2 γσ α−1 ≥

2 C1 m1 m2 γσ α−1

J2 (τ )a(C3 b(u(τ ))) dτ J2 (τ )

1

σ



1

σ

2C3 u(τ ) dτ C1 C2 m1 m2 γσ α+β−2

J2 (τ )u(τ ) dτ

J2 (τ )

1

0

G1 (τ, z)f (z, u(z), v(z)) dz dτ

1 2 α−1 J (τ ) σ J (z)C a(v(z)) dz dτ 2 1 1 C1 m1 m2 γσ α−1 σ σ

1

1 2 = J2 (τ ) dτ J1 (z)a(v(z)) dz m1 m2 γ σ σ

1 2 2 ≥ J1 (z)a(σ β−1 v ) dz ≥ σ β−1 a( v ) ≥ 2a( v ), m1 γ σ γ

≥

1



∀ t ∈ [σ, 1]. Then we conclude that a( v ) = a(supt∈[0,1] v(t)) ≥ a(v(σ)) ≥ 2a( v ), and hence a( v ) = 0. By (H3)(a), we obtain

v = 0.

(3.18)

Therefore, by (3.17) and (3.18), we deduce that (u, v) Y = 0, which is a contradiction. Hence, (Q1 (u, v), Q2 (u, v)) ≤ (u, v) for all (u, v) ∈ ∂Br1 ∩ P . By Theorem 1.2.11 (b), we conclude that the fixed point index i(Q, Br1 ∩ P, P ) = 0.

(3.19)

On the other hand, by (H4), we deduce that there exist C4 > 0, C5 > 0 and C6 > 0 such that f (t, u, v) ≤ C4 v α1 + C5 ,

∀ (t, u, v) ∈ [0, 1] × R+ × R+ ,

g(t, u, v) ≤ ε1 uα2 + C6 ,

∀ (t, u, v) ∈ [0, 1] × R+ × R+ ,

(3.20)

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  1 with ε1 = min M2 (8C14 M1 )α2 , 8M2 (C14 M1 )α2 > 0, M1 = 0 J1 (s) ds, and 1 M2 = 0 J2 (s) ds. Then, by (3.20), we have

Q1 (u, v)(t) =

1

0

G1 (t, s)f (s, u(s), v(s)) ds ≤



= C4

Q2 (u, v)(t) =

0

1

0

0

J1 (s)(v(s))α1 ds + C5 M1 ,

1

0

J1 (s)(C4 (v(s))α1 + C5 ) ds

∀ t ∈ [0, 1],



G2 (t, s)g(s, u(s), v(s)) ds ≤



= ε1

1

1

1

0

J2 (s)(u(s))α2 ds + C6 M2 ,

J2 (s)(ε1 (u(s))α2 + C6 ) ds

∀ t ∈ [0, 1].

(3.21)

We consider the functions p, q : R+ → R+ given by  p(w) = C4 M1 q(w) =

w 8C4 M1

α2

α1 + C6 M2

+ C5 M1 ,

1 (C4 M1 wα1 + C5 M1 )α2 + C6 M2 , 8(C4 M1 )α2

w ∈ R+ , w ∈ R+ .

Because p(w) q(w) = lim = w→∞ w w→∞ w lim



0, if α1 α2 < 1, 1/8, if α1 α2 = 1,

we conclude that there exists R1 > r1 such that p(w) ≤

1 1 w, q(w) ≤ w, ∀ w ≥ R1 . 4 4

(3.22)

We will show that (u, v) ≤ (Q1 (u, v), Q2 (u, v)) for all (u, v) ∈ ∂BR1 ∩ P . We suppose that there exists (u, v) ∈ ∂BR1 ∩ P , that is, (u, v) Y = R1 , such that (u, v) ≤ (Q1 (u, v), Q2 (u, v)). So, by (3.21), we obtain

u(t) ≤ Q1 (u, v)(t) ≤ C4

v(t) ≤ Q2 (u, v)(t) ≤ ε1

1

0

0

1

J1 (s)(v(s))α1 ds + C5 M1 ,

∀ t ∈ [0, 1],

J2 (s)(u(s))α2 ds + C6 M2 ,

∀ t ∈ [0, 1].

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Then, we deduce

1 J1 (s) ε1 u(t) ≤ C4 0

0



= C4 M1 ε1

1



≤ C4 M1 ε1

1

0

0

1

α2

J2 (τ )(u(τ )) α2

J2 (τ )(u(τ )) J2 (τ ) u

α2

α1 dτ + C6 M2

ds + C5 M1

α1 dτ + C6 M2

+ C5 M1

α1 dτ + C6 M2

+ C5 M1

α

= C4 M1 (ε1 M2 u α2 + C6 M2 ) 1 + C5 M1 

α2 α1

u

≤ C4 M1 + C6 M2 + C5 M1 8C4 M1 

α2 α1

(u, v) Y ≤ C4 M1 + C6 M2 + C5 M1 , 8C4 M1

∀ t ∈ [0, 1] (3.23)

and

v(t) ≤ ε1

0

≤ ε1

1

0

1

J2 (s)(u(s))α2 ds + C6 M2

J2 (s) C4

0

1

α1

J1 (τ )(v(τ ))

α2 dτ + C5 M1

ds + C6 M2

≤ ε1 M2 (C4 M1 v α1 + C5 M1 )α2 + C6 M2 ≤

1 α (C4 M1 v α1 + C5 M1 ) 2 + C6 M2 8(C4 M1 )α2

≤

1 α2 1 (C4 M1 (u, v) α + C6 M2 , ∀ t ∈ [0, 1]. Y + C5 M1 ) 8(C4 M1 )α2 (3.24)

By using (3.23), (3.24) and (3.22), we conclude that u(t) ≤ 14 (u, v) Y and v(t) ≤ 14 (u, v) Y for all t ∈ [0, 1]. Therefore, we obtain that

(u, v) Y ≤ 12 (u, v) Y , and so (u, v) Y = 0, which is a contradiction, because (u, v) Y = R1 > 0. So, (u, v) ≤ (Q1 (u, v), Q2 (u, v)) for all (u, v) ∈ ∂BR1 ∩ P . By Theorem 1.2.11 (a), we deduce that the fixed point index i(Q, BR1 ∩ P, P ) = 1.

(3.25)

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Because Q has no fixed points on ∂Br1 ∪ ∂BR1 , by (3.19) and (3.25), we conclude that i(Q, (BR1 \ B r1 ) ∩ P, P ) = i(Q, BR1 ∩ P, P ) − i(Q, Br1 ∩ P, P ) = 1. Therefore, the operator Q has at least one fixed point (u1 , v1 ) ∈ (BR1 \ B r1 )∩P , with r1 < (u1 , v1 ) Y < R1 , that is u1 > 0 or v1 > 0. Because u1 ∈ P1 and v1 ∈ P2 , we obtain u1 (t) > 0 for all t ∈ (0, 1] or v1 (t) > 0 for all t ∈ (0, 1].  Theorem 3.1.2. Assume that (H1), (H2), (H5) and (H6) hold. Then the problem (3.1), (3.2) has at least one positive solution. Proof. By (H5) there exist Ci > 0, i = 7, . . . , 11 such that f (t, u, v) ≥ C7 c(v) − C8 , g(t, u, v) ≥ C9 d(u) − C10 , ∀ (t, u, v) ∈ [σ, 1] × R+ × R+ ,

(3.26)

and c(C12 d(u)) ≥

2C12 u − C11 , C7 C9 m1 m2 γσ α+β−2

∀ u ∈ R+ ,

(3.27)

where C12 = max{C9 σ β−1 J2 (s), s ∈ [0, 1]} > 0. Then we have

1 G1 (t, s)f (s, u(s), v(s)) ds Q1 (u, v)(t) = 0

≥

1

σ

≥σ

α−1

Q2 (u, v)(t) =



1

σ 1

0

≥

G1 (t, s)(C7 c(v(s)) − C8 ) ds

1

σ

J1 (s)(C7 c(v(s)) − C8 ) ds,

∀ t ∈ [σ, 1],

G2 (t, s)g(s, u(s), v(s)) ds G2 (t, s)(C9 d(u(s)) − C10 ) ds

≥ σ β−1



1

σ

J2 (s)(C9 d(u(s)) − C10 ) ds,

∀ t ∈ [σ, 1].

(3.28)

We will prove that the set U = {(u, v) ∈ P, (u, v) = Q(u, v) + λ(ϕ1 , ϕ2 ), λ ≥ 0} is bounded, where (ϕ1 , ϕ2 ) ∈ P \ {(0, 0)}. Indeed,

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(u, v) ∈ U implies that u ≥ Q1 (u, v), v ≥ Q2 (u, v) for some ϕ1 , ϕ2 ≥ 0. By (3.28), we obtain u(t) ≥ Q1 (u, v)(t) ≥ σ α−1 C7 = σ α−1 C7



1 σ

=σ

C9

1

σ

1

J1 (s)c(v(s)) ds − σ α−1 C8

σ



1

J1 (s) ds

σ

J1 (s)c(v(s)) ds − C13 , ∀ t ∈ [σ, 1],

v(t) ≥ Q2 (u, v)(t) ≥ σ β−1



β−1

C9

1

J2 (s)d(u(s)) ds − σ

σ

β−1

(3.29)

C10

J2 (s)d(u(s)) ds − C14 , ∀ t ∈ [σ, 1],

1

σ

J2 (s) ds (3.30)

where C13 = m1 σ α−1 C8 , C14 = m2 σ β−1 C10 . By the monotonicity and concavity of c(·) and the Jensen inequality, the inequality (3.30) implies that

β−1 c(v(t) + C14 ) ≥ c σ C9

≥



1

J2 (s)d(u(s)) ds

σ

1 σ 1

= σ

  c σ β−1 C9 J2 (s)d(u(s)) ds   −1 c (σ β−1 C12 C9 J2 (s))C12 d(u(s)) ds

−1 ≥ σ β−1 C12 C9



1 σ

J2 (s)c(C12 d(u(s))) ds,

∀ t ∈ [σ, 1]. (3.31)

Since c(v(t)) ≥ c(v(t) + C14 ) − c(C14 ), by the relations (3.27), (3.29), (3.31), we deduce u(t) ≥ σ α−1 C7 ≥σ

α−1

=σ

α−1



σ

C7

1

σ

C7

1

1

σ

J1 (s)c(v(s)) ds − C13 J1 (s)[c(v(s) + C14 ) − c(C14 )] ds − C13 J1 (s)c(v(s) + C14 ) ds − C15

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≥σ

α−1

=σ

α+β−2

=σ

α+β−2

≥σ

α+β−2

C7

1

σ

 J1 (s) σ

−1 C7 C9 C12

β−1







1

1

σ



1 σ

 J2 (τ )c(C12 d(u(τ ))) dτ ds − C15

J1 (s) ds

σ

−1 C7 C9 C12 m1





1

σ

−1 C7 C9 C12 m1

−1 C12 C9

1

σ

J2 (τ )c(C12 d(u(τ ))) dτ

− C15

J2 (τ )c(C12 d(u(τ ))) dτ − C15 J2 (τ )

2C12 u(τ ) − C 11 dτ − C15 C7 C9 m1 m2 γσ α+β−2

1

2 2σ α−1 1 = J2 (τ )u(τ ) dτ − C16 ≥ J2 (τ ) u dτ − C16 m2 γ σ m2 γ σ ×

≥ 2 u − C16 ,

∀ t ∈ [σ, 1],

−1 where C15 = σ α−1 C7 c(C14 )m1 + C13 , C16 = σ α+β−2 C7 C9 C12 m1 m2 C11 + C15 . Therefore, u ≥ u(σ) ≥ 2 u − C16 , and then

u ≤ C16 .

(3.32)

Since c(v(t)) ≥ c(σ β−1 v ) ≥ σ β−1 c( v ) for all t ∈ [σ, 1] and v ∈ P2 , then by the relations (3.27), (3.29)–(3.31), we obtain c(v(t)) ≥ c(v(t) + C14 ) − c(C14 )

1 −1 ≥ σ β−1 C12 C9 J2 (s)c(C12 d(u(s))) ds − c(C14 ) σ

≥σ

β−1

−1 C12 C9



1

σ

2 = C7 m1 m2 γσ α−1

J2 (s)

1

σ

2C12 u(s) − C11 ds − c(C14 ) C7 C9 m1 m2 γσ α+β−2

J2 (s)u(s) ds − C17

1 2 ≥ J2 (s) C7 m1 m2 γσ α−1 σ

1 α−1 × σ C7 J1 (τ )c(v(τ )) dτ − C13 ds − C17 σ

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2 = m1 γ 2 ≥ m1 γ ≥



σ



1

σ

β−1

2σ γ

1

J1 (τ )c(v(τ )) dτ − C18 J1 (τ )c(σ β−1 v ) dτ − C18

c( v ) − C18 ≥ 2c( v ) − C18 ,

∀ t ∈ [σ, 1],

−1 13 where C17 = σ β−1 C12 C9 C11 m2 + c(C14 ), C18 = C7 m2C α−1 + C17 . 1 γσ Then c( v ) ≥ c(v(σ)) ≥ 2c( v ) − C18 , and so c( v ) ≤ C18 . By (H5) (a) and (c), we deduce that limv→∞ c(v) = ∞, thus there exists C19 > 0, such that

v ≤ C19 .

(3.33)

By (3.32) and (3.33), we conclude that (u, v) Y ≤ C16 + C19 for all (u, v) ∈ U , that is, the set U is bounded. Then there exists a sufficiently large R2 > 0 such that (u, v) = Q(u, v) + λ(ϕ1 , ϕ2 ) for all (u, v) ∈ ∂BR2 ∩ P and λ ≥ 0. By Theorem 1.2.9, we deduce that i(Q, BR2 ∩ P, P ) = 0.

(3.34)

On the other hand, by (H6), there exist C20 > 0 and a sufficiently small r2 > 0, (r2 < R2 , r2 ≤ 1) such that f (t, u, v) ≤ C20 v β1 , β2

g(t, u, v) ≤ ε2 u ,

∀ (t, u) ∈ [0, 1] × R+ , ∀ (t, v) ∈ [0, 1] × R+ ,

v ∈ [0, r2 ], u ∈ [0, r2 ],

(3.35)

where ε2 = (2C20 M1 M2β1 )−1/β1 > 0. We will show that (u, v) ≤ Q(u, v) for all (u, v) ∈ ∂Br2 ∩ P . We suppose that there exists (u, v) ∈ ∂Br2 ∩ P , that is (u, v) Y = r2 ≤ 1, such that (u, v) ≤ (Q1 (u, v), Q2 (u, v)), or u ≤ Q1 (u, v) and v ≤ Q2 (u, v). Then by (3.35), we obtain

1

1 G1 (t, s)f (s, u(s), v(s))ds ≤ C20 J1 (s)(v(s))β1 ds u(t) ≤ Q1 (u, v)(t) =

≤ C20

0

≤ C20

0

1

0

1



J1 (s)

J1 (s)

1 0 1 0

β1 G2 (s, τ )g(τ, u(τ ), v(τ )) dτ β2

J2 (τ )ε2 (u(τ ))

β1 dτ

ds

0

ds

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= C20 M1 εβ2 1 ≤

C20 M1 εβ2 1





1

J2 (τ )(u(τ ))β2 dτ

0

β1

1

J2 (τ ) dτ

0

β1

u β1 β2

= C20 M1 εβ2 1 M2β1 u β1 β2 ≤ C20 εβ2 1 M1 M2β1 u =

1

u , 2

∀ t ∈ [0, 1].

Therefore, u ≤ 12 u , so

u = 0. In addition,



v(t) ≤ Q2 (u, v)(t) = β2

≤ ε2 M2 u ,

1

0

(3.36)

G2 (t, s)g(s, u(s), v(s)) ds ≤

0

1

J2 (s)ε2 (u(s))β2 ds

∀ t ∈ [0, 1].

(3.37)

By (3.36) and (3.37), we deduce that v = 0, and then (u, v) Y = 0, which is a contradiction, because (u, v) Y = r2 > 0. Then (u, v) ≤ Q(u, v) for all (u, v) ∈ ∂Br2 ∩ P . By Theorem 1.2.11 (a), we conclude that i(Q, Br2 ∩ P, P ) = 1.

(3.38)

Because Q has no fixed points on ∂Br2 ∪ ∂BR2 , by (3.34) and (3.38), we deduce that i(Q, (BR2 \ B r2 ) ∩ P, P ) = i(Q, BR2 ∩ P, P ) − i(Q, Br2 ∩ P, P ) = −1. So, the operator Q has at least one fixed point (u1 , v1 ) ∈ (BR2 \B r2 )∩P , with r2 < (u1 , v1 ) Y < R2 , which is a positive solution for our problem (3.1), (3.2).  Remark 3.1.2. In a similar manner, we can prove that Theorem 3.1.1 remains valid if assumption (H3) is replaced by (H3) There exist the functions a, b ∈ C(R+ , R+ ) such that (a) b(·) is concave and strictly increasing on R+ with b(0) = 0; (b) There exists σ ∈ (0, 1) such that f0i = lim inf v→0+

f (t, u, v) ∈ (0, ∞], uniformly with respect to a(v)

(t, u) ∈ [σ, 1] × R+ ,

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g0i = lim inf u→0+

g(t, u, v) ∈ (0, ∞], uniformly with respect to b(u)

(t, v) ∈ [σ, 1] × R+ ; (c) lim

v→0+

b(Ca(v)) = ∞ exists for any constant C > 0. v

Remark 3.1.3. Theorem 3.1.2 remains valid if assumption (H5) is replaced by (H5) There exist the functions c, d ∈ C(R+ , R+ ) such that (a) d(·) is concave and strictly increasing on R+ ; (b) There exists σ ∈ (0, 1) such that i = lim inf f∞ v→∞

f (t, u, v) ∈ (0, ∞], uniformly with respect to c(v)

(t, u) ∈ [σ, 1] × R+ , i = lim inf g∞ u→∞

g(t, u, v) ∈ (0, ∞], uniformly with respect to d(u)

(t, v) ∈ [σ, 1] × R+ ; (c) lim

v→∞

d(Cc(v)) = ∞ exists for any constant C > 0. v

Remark 3.1.4. Theorem 3.1.1 remains valid if assumption (H4) is replaced by (H4) There exist α1 , α2 > 0 with α1 α2 ≤ 1 such that f (t, u, v) = 0, exists uniformly with respect to v→∞ v α1 (t, u) ∈ [0, 1] × R+ ,

s f∞ = lim

s = lim sup g∞ u→∞

g(t, u, v) ∈ [0, ∞), uniformly with respect to uα2

(t, v) ∈ [0, 1] × R+ . Remark 3.1.5. Theorem 3.1.2 remains valid if assumption (H6) is replaced by

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(H6) There exist β1 , β2 > 0 with β1 β2 ≥ 1 such that f0s = lim

v→0+

f (t, u, v) = 0, exists uniformly with respect to v β1

(t, u) ∈ [0, 1] × R+ , g0s = lim sup u→0+

g(t, u, v) ∈ [0, ∞), uniformly with respect to uβ2

(t, v) ∈ [0, 1] × R+ . Theorem 3.1.3. Assume that assumptions (H1)–(H3), (H5) and (H7) hold. Then the problem (3.1), (3.2) has at least two positive solutions. Proof. By using (H7), for any (u, v) ∈ ∂BN0 ∩ P , we obtain

1

1 N0 J1 (s)f (s, N0 , N0 ) ds < J1 (s) ds Q1 (u, v)(t) ≤ 2m0 0 0 N0 M 1 N0 , ∀ t ∈ [0, 1], ≤ 2m0 2

1

1 N0 Q2 (u, v)(t) ≤ J2 (s)g(s, N0 , N0 ) ds < J2 (s) ds 2m0 0 0 =

=

N0 M 2 N0 , ∀ t ∈ [0, 1]. ≤ 2m0 2

Then we deduce

Q(u, v) Y = Q1 (u, v) + Q2 (u, v) < N0 = (u, v) Y ,

∀ (u, v) ∈ ∂BN0 ∩ P.

Because Q has no fixed points on ∂BN0 , by Theorem 1.2.8 we conclude that i(Q, BN0 ∩ P, P ) = 1.

(3.39)

On the other hand, from (H3) and (H5), and the proofs of Theorems 3.1.1 and 3.1.2, we know that there exist a sufficiently small r1 > 0 (r1 < N0 ) and a sufficiently large R2 > N0 such that i(Q, Br1 ∩ P, P ) = 0,

i(Q, BR2 ∩ P, P ) = 0.

(3.40)

Because Q has no fixed points on ∂Br1 ∪ ∂BR2 ∪ ∂BN0 , by the relations (3.39) and (3.40), we obtain i(Q, (BR2 \ B N0 ) ∩ P, P ) = i(Q, BR2 ∩ P, P ) − i(Q, BN0 ∩ P, P ) = −1, i(Q, (BN0 \ B r1 ) ∩ P, P ) = i(Q, BN0 ∩ P, P ) − i(Q, Br1 ∩ P, P ) = 1.

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Then Q has at least one fixed point (u1 , v1 ) ∈ (BR2 \ B N0 ) ∩ P and has at least one fixed point (u2 , v2 ) ∈ (BN0 \ B r1 ) ∩ P . Therefore, problem  (3.1), (3.2) has two distinct positive solutions (u1 , v1 ), (u2 , v2 ). Remark 3.1.6. Theorem 3.1.3 remains valid if (H3) is replaced by (H3) or (H5) is replaced by (H5) . Remark 3.1.7. In (H3), if a(v) = v p with p ≤ 1, and b(u) = uq with q > 0, the condition from (H3)(c) is satisfied if pq < 1. In (H5), if c(v) = v p with p ≤ 1, and d(u) = uq with q > 0, the condition from (H5) (c) is satisfied if pq > 1. Examples (1) We consider f (t, u, v) = et (1 + e−(u+v)) and g(t, u, v) = (1 + e−t)uθ , for (t, u, v) ∈ [0, 1] × R+ × R+ . For a(v) = v p with p ≤ 1, and b(u) = uq for q > 0 and pq < 1, then the assumptions (H3) and (H4) are satisfied if q > θ and α2 > θ. For example, if θ = 32 , p = 13 , q = 2, α1 = 14 and α2 = 3, we can apply Theorem 3.1.1, and we deduce that the problem (3.1), (3.2) has at least one positive solution. (2) We consider f (t, u, v) = (1 + e−u )v θ1 and g(t, u, v) = (1 + e−v )uθ2 , for (t, u, v) ∈ [0, 1] × R+ × R+ . For c(v) = v p with p ≤ 1, and d(u) = uq for q > 0 and pq > 1, then the assumptions (H5) and (H6) are satisfied if p < θ1 , q < θ2 , β1 < θ1 , and β2 < θ2 . For example, if θ1 = 3, θ2 = 23 , 7 , β1 = 52 and β2 = 12 , we can apply Theorem 3.1.2, p = 2, q = 12 and we conclude that the problem (3.1), (3.2) has at least one positive solution. Remark 3.1.8. The results presented in this section under the assumptions p1 ∈ [1, n− 2] and p2 ∈ [1, m− 2] instead of p1 ∈ [1, α− 1) and p2 ∈ [1, β − 1) were published in [50]. Remark 3.1.9. Under different assumptions on the nonlinearities f and g as those used in this section, the existence and multiplicity of positive solutions of the system (3.1) with f (t, u, v) = f(t, v) and g(t, u, v) =  g(t, u) which can be nonsingular or singular functions at the points t = 0 and/or t = 1, subject to the boundary conditions (3.2) were investigated in [49]. The existence and nonexistence of positive solutions of the system (3.1) with two positive parameters supplemented with the boundary conditions (3.2) were studied in the paper [52].

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3.2

Systems of Fractional Differential Equations with Coupled Multi-Point Boundary Conditions

We consider the system of nonlinear ordinary fractional differential equations  α u(t) + f (t, u(t), v(t)) = 0, t ∈ (0, 1), D0+ (3.41) β D0+ v(t) + g(t, u(t), v(t)) = 0, t ∈ (0, 1), with the coupled multi-point boundary conditions ⎧ N  ⎪ ⎪ p1 q1  (n−2) ⎪ u(0) = u (0) = · · · = u (0) = 0, D u(1) = ai D0+ v(ξi ), ⎪ 0+ ⎨ ⎪ ⎪  (m−2) ⎪ ⎪ (0) = 0, ⎩ v(0) = v (0) = · · · = v

i=1

p2 D0+ v(1) =

M  i=1

q2 bi D0+ u(ηi ),

(3.42) where α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), q1 ∈ [0, p2 ], q2 ∈ [0, p1 ], ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi , bi ∈ R for all ζ denotes the i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, and D0+ Riemann-Liouville derivative of order ζ (for ζ = α, β, p1 , q1 , p2 , q2 ). Under sufficient conditions on the functions f and g, which can be nonsingular or singular at the points t = 0 and/or t = 1, we study the existence and multiplicity of positive solutions of problem (3.41), (3.42). We use some theorems from the fixed point index theory and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2). By a positive solution of problem (3.41), (3.42) we mean a pair of functions (u, v) ∈ (C([0, 1], R+ ))2 satisfying (3.41) and (3.42) with u(t) > 0 for all t ∈ (0, 1] or v(t) > 0 for all t ∈ (0, 1]. 3.2.1

Preliminary results

We consider the fractional differential system  α u(t) + x(t) = 0, t ∈ (0, 1), D0+ β v(t) + y(t) = 0, D0+

t ∈ (0, 1),

(3.43)

with the coupled multi-point boundary conditions (3.42), where x, y ∈ C(0, 1) ∩ L1 (0, 1).

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We denote by Δ the constant

Δ=

Γ(α)Γ(β) Γ(α)Γ(β) − Γ(α − p1 )Γ(β − p2 ) Γ(α − q2 )Γ(β − q1 )  M  N   α−q −1 β−q1 −1 2 . × ai ξi bi ηi i=1

i=1

Lemma 3.2.1. If Δ = 0 and x, y ∈ C(0, 1) ∩ L1 (0, 1), then the unique solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (3.43), (3.42) is given by 1 u(t) = − Γ(α)



t

0

(t − s)α−1 x(s) ds



1 tα−1 Γ(β) (1 − s)α−p1 −1 x(s) ds + Δ Γ(α − p1 )Γ(β − p2 ) 0  

N ξi  Γ(β) β−q1 −1 ai (ξi − s) y(s) ds − Γ(β − q1 )Γ(β − p2 ) i=1 0 

N 1  Γ(β) β−q1 −1 + ai ξi (1 − s)β−p2 −1 y(s) ds Γ(β − q1 )Γ(β − p2 ) i=1 0 N   Γ(β) β−q1 −1 ai ξi − Γ(α − q2 )Γ(β − q1 ) i=1

 ηi M  α−q2 −1 × bi (ηi − s) x(s) ds , t ∈ [0, 1], i=1

v(t) = −

1 Γ(β)

0

0

t

(t − s)β−1 y(s) ds



1 Γ(α) tβ−1 (1 − s)β−p2 −1 y(s) ds + Δ Γ(α − p1 )Γ(β − p2 ) 0

ηi M  Γ(α) − bi (ηi − s)α−q2 −1 x(s) ds Γ(α − p1 )Γ(α − q2 ) i=1 0 M   α−q −1 1 Γ(α) 2 + bi η (1 − s)α−p1 −1 x(s) ds Γ(α − p1 )Γ(α − q2 ) i=1 i 0

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 M  α−q −1 Γ(α) 2 − bi ηi Γ(α − q2 )Γ(β − q1 ) i=1  

N ξi  β−q1 −1 × ai (ξi − s) y(s) ds , 0

i=1

t ∈ [0, 1].

page 119

119

(3.44)

Proof. By Lemma 1.1.3, we deduce that the solutions (u, v) ∈ (C(0, 1) ∩ L1 (0, 1))2 of the fractional differential system (3.43) are given by α x(t) + c1 tα−1 + · · · + cn tα−n u(t) = −I0+

t 1 =− (t − s)α−1 x(s) ds + c1 tα−1 + · · · + cn tα−n , Γ(α) 0 β v(t) = −I0+ y(t) + d1 tβ−1 + · · · + dn tβ−m

t 1 =− (t − s)β−1 y(s) ds + d1 tβ−1 + · · · + dm tβ−m , Γ(β) 0

where c1 , c2 , . . . , cn , d1 , d2 , . . . , dm ∈ R. By using the conditions u(0) = u (0) = · · · = u(n−2) (0) = 0 and v(0) = v  (0) = · · · = v (m−2) (0) = 0, we obtain c2 = · · · = cn = 0 and d2 = · · · = dm = 0. Then we conclude (t) = c1 tα−1 − v(t) = d1 t

β−1

1 Γ(α)

1 − Γ(β)



t

0



0

t

(t − s)α−1 x(s) ds, t ∈ [0, 1], (3.45) β−1

(t − s)

y(s) ds, t ∈ [0, 1].

By using some properties of the fractional integrals and fractional derivatives from Lemmas 1.1.1 and 1.1.2, we obtain p1 u(t) = c1 D0+

Γ(α) α−p1 tα−p1 −1 − I0+ x(t), Γ(α − p1 )

q2 D0+ u(t) = c1

Γ(α) α−q2 tα−q2 −1 − I0+ x(t), Γ(α − q2 )

p2 v(t) = d1 D0+

Γ(β) β−p2 y(t), tβ−p2 −1 − I0+ Γ(β − p2 )

q1 D0+ v(t) = d1

Γ(β) β−q1 tβ−q1 −1 − I0+ y(t). Γ(β − q1 )

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N

p1 Then the conditions D0+ u(1) = M q2 i=1 bi D0+ u(ηi ) give us

c1

i=1

1 Γ(α) − Γ(α − p1 ) Γ(α − p1 ) =

N 

ai d1

i=1



ξi

=

i=1

0

(1 − s)α−p1 −1 x(s) ds

0

 β−q1 −1

(ξi − s)

1 Γ(β) − d1 Γ(β − p2 ) Γ(β − p2 ) b i c1

1

Γ(β) ξ β−q1 −1 Γ(β − q1 ) i

1 − Γ(β − q1 )

M 



q1 p2 ai D0+ v(ξi ) and D0+ v(1) =



1

0

y(s) ds ,

(1 − s)β−p2 −1 y(s) ds

Γ(α) η α−q2 −1 Γ(α − q2 ) i

1 − Γ(α − q2 )



ηi

0

α−q2 −1

(ηi − s)

x(s) ds .

(3.46)

The above system in c1 and d1 has the determinant Δ, which by the assumption of this lemma is different from 0. So, the system (3.46) has the unique solution 

1 1 Γ(β) (1 − s)α−p1 −1 x(s) ds c1 = Δ Γ(α − p1 )Γ(β − p2 ) 0 N

 Γ(β) − ai Γ(β − q1 )Γ(β − p2 ) i=1 N 

Γ(β) + Γ(β − q1 )Γ(β − p2 )

i=1

Γ(β) − Γ(α − q2 )Γ(β − q1 ) ×

M  i=1



bi

0

ηi



N  i=1



0

 β−q1 −1

(ξi − s)

ai ξiβ−q1 −1



0

 ai ξiβ−q1 −1

α−q2 −1

(ηi − s)

ξi

 x(s) ds ,

1

y(s) ds

(1 − s)β−p2 −1 y(s) ds

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121



1 Γ(α) 1 d1 = (1 − s)β−p2 −1 y(s) ds Δ Γ(α − p1 )Γ(β − p2 ) 0

ηi M  Γ(α) − bi (ηi − s)α−q2 −1 x(s) ds Γ(α − p1 )Γ(α − q2 ) i=1 0 M   α−q −1 1 Γ(α) 2 + bi ηi (1 − s)α−p1 −1 x(s) ds Γ(α − p1 )Γ(α − q2 ) i=1 0 M   α−q −1 Γ(α) 2 − bi ηi Γ(α − q2 )Γ(β − q1 ) i=1  

N ξi  × ai (ξi − s)β−q1 −1 y(s) ds . 0

i=1

Replacing the above constants c1 and d1 in (3.45), we obtain the expression (3.44) for the solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (3.43), (3.42). Conversely, one easily verifies that (u, v) ∈ C[0, 1] × C[0, 1] given by (3.44) satisfies the system (3.43) and the boundary conditions (3.42).  Lemma 3.2.2. If Δ = 0 and x, y ∈ C(0, 1) ∩ L1 (0, 1), the solution (u, v) of problem (3.43), (3.42) given by (3.44) can be written as ⎧

1

1 ⎪   2 (t, s)y(s) ds, t ∈ [0, 1], ⎪ G1 (t, s)x(s) ds + G ⎨u(t) = 0 0

1 (3.47)

1 ⎪ ⎪ 3 (t, s)y(s) ds +  4 (t, s)x(s) ds, t ∈ [0, 1], ⎩v(t) = G G 0

0

 i , i = 1, . . . , 4 are where the Green functions G N   M ⎧ α−1   ⎪ Γ(β) t β−q −1 ⎪ 1 1 (t, s) =  ⎪ g1 (t, s) + ai ξi bi g2 (ηi , s) , ⎪G ⎪ ΔΓ(β − q1 ) i=1 ⎪ ⎪ i=1 ⎪ N  ⎪ ⎪  ⎪ tα−1 Γ(β) ⎪ ⎪ ai g3 (ξi , s) , ⎪ ⎨G2 (t, s) = ΔΓ(β − p2 ) i=1 M   N β−1  α−q −1  ⎪ Γ(α) t ⎪ ⎪ 3 (t, s) =  ⎪ G g4 (t, s) + bi ηi 2 ai  g3 (ξi , s) , ⎪ ⎪ ΔΓ(α − q ) 2 ⎪ ⎪ i=1  i=1  ⎪ ⎪ M ⎪  ⎪ tβ−1 Γ(α) ⎪  ⎪ G (t, s) = bi  g2 (ηi , s) , ∀ t, s ∈ [0, 1], ⎩ 4 ΔΓ(α − p1 ) i=1 (3.48)

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and  ⎧ tα−1 (1 − s)α−p1 −1 − (t − s)α−1 , 0 ≤ s ≤ t ≤ 1, 1 ⎪ ⎪ ⎪g1 (t, s) = ⎪ ⎪ Γ(α) tα−1 (1 − s)α−p1 −1 , 0 ≤ t ≤ s ≤ 1, ⎪ ⎪ ⎧ α−q −1 ⎪ ⎪ 2 ⎪ t (1 − s)α−p1 −1 − (t − s)α−q2 −1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0 ≤ s ≤ t ≤ 1, ⎪ ⎪ 1 ⎪ ⎪ g  (t, s) = 2 ⎪ ⎪ Γ(α − q2 ) ⎪ ⎪ tα−q2 −1 (1 − s)α−p1 −1 , ⎪ ⎨ ⎪ ⎪ ⎩ 0 ≤ t ≤ s ≤ 1. ⎪ ⎧ β−q1 −1 ⎪ ⎪ (1 − s)β−p2 −1 − (t − s)β−q1 −1 , t ⎪ ⎪ ⎪ ⎨ ⎪ 1 ⎪ ⎪ ⎪ g3 (t, s) = 0 ≤ s ≤ t ≤ 1, ⎪ ⎪ Γ(β − q1 ) ⎪ ⎪ ⎩ β−q1 −1 ⎪ ⎪ t (1 − s)β−p2 −1 , 0 ≤ t ≤ s ≤ 1. ⎪  ⎪ ⎪ ⎪ tβ−1 (1 − s)β−p2 −1 − (t − s)β−1 , 0 ≤ s ≤ t ≤ 1, ⎪ 1 ⎪ ⎪ ⎩g4 (t, s) = Γ(β) tβ−1 (1 − s)β−p2 −1 , 0 ≤ t ≤ s ≤ 1. (3.49) Proof. By Lemma 3.2.1 and relation (3.44), we conclude  t 1 u(t) = [tα−1 (1 − s)α−p1 −1 − (t − s)α−1 ]x(s) ds Γ(α) 0 

1 α−1 α−p1 −1 t (1 − s) x(s) ds + t

1 − Γ(α)



1

0

tα−1 (1 − s)α−p1 −1 x(s) ds

tα−1 Γ(β) + ΔΓ(α − p1 )Γ(β − p2 ) tα−1 Γ(β) − ΔΓ(α − q2 )Γ(β − q1 ) ×

M  i=1



bi

0

ηi



1

0



(1 − s)α−p1 −1 x(s) ds

N  i=1

α−q2 −1

(ηi − s)

 ai ξiβ−q1 −1

x(s) ds

 

N ξi  tα−1 Γ(β) β−q1 −1 − ai (ξi − s) y(s) ds ΔΓ(β − q1 )Γ(β − p2 ) i=1 0 N 

1  tα−1 Γ(β) β−q1 −1 + ai ξi (1 − s)β−p2 −1 y(s) ds ΔΓ(β − p2 )Γ(β − q1 ) i=1 0

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page 123

123

 t 1 = [tα−1 (1 − s)α−p1 −1 − (t − s)α−1 ]x(s) ds Γ(α) 0 

1 tα−1 (1 − s)α−p1 −1 x(s) ds + t

−



Γ(β) ΔΓ(α − p1 )Γ(β − p2 )

0

Γ(β) + ΔΓ(α − q2 )Γ(β − q1 )  ×

M  i=1

bi ηiα−q2 −1



1

0

1



tα−1 (1 − s)α−p1 −1 x(s) ds

N  i=1

 ai ξiβ−q1 −1

tα−1 (1 − s)α−p1 −1 x(s) ds

1 tα−1 Γ(β) (1 − s)α−p1 −1 x(s) ds ΔΓ(α − p1 )Γ(β − p2 ) 0  N  tα−1 Γ(β) β−q1 −1 ai ξi − ΔΓ(α − q2 )Γ(β − q1 ) i=1

+

×

M 



bi

ηi

0

i=1

α−q2 −1

(ηi − s)

x(s) ds

tα−1 Γ(β) ΔΓ(β − q1 )Γ(β − p2 ) 

N 1  ai ξiβ−q1 −1 (1 − s)β−p2 −1 y(s) ds − ×

+

0

i=1

=

ξi

0

 β−q1 −1

(ξi − s)

y(s) ds

 t 1 [tα−1 (1 − s)α−p1 −1 − (t − s)α−1 ]x(s) ds Γ(α) 0 

1 α−1 α−p1 −1 t (1 − s) x(s) ds + t

tα−1 Γ(β) + ΔΓ(α − q2 )Γ(β − q1 ) ×

M  i=1



bi

0

1



N  i=1

 ai ξiβ−q1 −1

ηiα−q2 −1 (1 − s)α−p1 −1 x(s) ds −

0

ηi

(ηi − s)α−q2 −1 x(s) ds



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tα−1 Γ(β) ΔΓ(β − q1 )Γ(β − p2 ) 

N 1  × ai ξiβ−q1 −1 (1 − s)β−p2 −1 y(s) ds −

+

0

i=1

ξi

0

 β−q1 −1

(ξi − s)

y(s) ds .

Therefore, we obtain  t 1 u(t) = [tα−1 (1 − s)α−p1 −1 − (t − s)α−1 ]x(s) ds Γ(α) 0 

1 + tα−1 (1 − s)α−p1 −1 x(s) ds t

tα−1 Γ(β) + ΔΓ(α − q2 )Γ(β − q1 ) ×

M 



bi

ηi 0

i=1



N  i=1

 ai ξiβ−q1 −1

 ηiα−q2 −1 (1 − s)α−p1 −1

 − (ηi − s)α−q2 −1 x(s) ds +



1

ηi

ηiα−q2 −1 (1 − s)α−p1 −1 x(s) ds

α−1



Γ(β) t ΔΓ(β − q1 )Γ(β − p2 ) 

N ξi    ξiβ−q1 −1 (1 − s)β−p2 −1 − (ξi − s)β−q1 −1 y(s) ds × ai +

0

i=1



1

+ ξi



1

= 0

×



ξiβ−q1 −1 (1

− s)

i=1

bi

0

1

N

= 0



N  i=1

 ai ξiβ−q1 −1

g2 (ηi , s)x(s) ds

tα−1 Γ(β)  ai + ΔΓ(β − p2 ) i=1 1

y(s) ds

tα−1 Γ(β) g1 (t, s)x(s) ds +  ΔΓ(β − q1 )

M 



β−p2 −1

 1 (t, s)x(s) ds + G



1

0

0

1

g3 (ξi , s)y(s) ds  2 (t, s)y(s) ds. G

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In a similar manner, we deduce

v(t) =

1

0

×

tβ−1 Γ(α) g4 (t, s)y(s) ds + ΔΓ(α − q2 )

N  i=1

ai

0

1



M  i=1

 bi ηiα−q2 −1

g3 (ξi , s)y(s) ds

1 M tβ−1 Γ(α)  + bi g2 (ηi , s)x(s) ds ΔΓ(α − p1 ) i=1 0

1

1  4 (t, s)x(s) ds, G3 (t, s)y(s) ds + G = 0

0

i ,  where G gi , i = 1, . . . , 4, are given in (3.48) and (3.49). Therefore, we obtain the expression (3.47) for the solution (u, v) of problem (3.43), (3.42) given by (3.44).  Using similar arguments as those used in the proof of Lemma 2.1.3 from Section 2.1, we obtain the following properties of the functions gi , i = 1, . . . , 4. Lemma 3.2.3. The functions  gi , i = 1, . . . , 4 given by (3.49) have the properties: (a1 ) g1 (t, s) ≤  h1 (s) for all t, s ∈ [0, 1], where  h1 (s) =

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 ), s ∈ [0, 1]; Γ(α)

(a2 ) g1 (t, s) ≥ tα−1  h1 (s) for all t, s ∈ [0, 1]; 1 tα−1 , for all t, s ∈ [0, 1]; (a3 ) g1 (t, s) ≤ Γ(α) (b1 ) g2 (t, s) ≥ tα−q2 −1  h2 (s) for all t, s ∈ [0, 1], where  h2 (s) = (b2 ) g2 (t, s) ≤ (b3 ) g2 (t, s) ≤ (c1 ) g3 (t, s) ≥

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 −q2 ), s ∈ [0, 1]; Γ(α − q2 )

1 α−q2 −1 (1 − s)α−p1 −1 for all t, Γ(α−q2 ) t 1 α−q2 −1 for all t, s ∈ [0, 1]; Γ(α−q2 ) t tβ−q1 −1  h3 (s) for all t, s ∈ [0, 1], where

 h3 (s) =

s ∈ [0, 1];

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 −q1 ), s ∈ [0, 1]; Γ(β − q1 )

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(c2 ) g3 (t, s) ≤ (c3 ) g3 (t, s) ≤

1 β−q1 −1 (1 − s)β−p2 −1 for all Γ(β−q1 ) t 1 β−q1 −1 for all t, s ∈ [0, 1]; Γ(β−q1 ) t

t, s ∈ [0, 1];

(d1 ) g4 (t, s) ≤  h4 (s) for all t, s ∈ [0, 1], where  h4 (s) =

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 ), s ∈ [0, 1]; Γ(β)

h4 (s) for all t, s ∈ [0, 1]; (d2 ) g4 (t, s) ≥ tβ−1 1 tβ−1 , for all t, s ∈ [0, 1]; (d3 ) g4 (t, s) ≤ Γ(β) (e) The functions gi , i = 1, . . . , 4, are continuous on [0, 1] × [0, 1]; gi (t, s) ≥ 0 for all t, s ∈ [0, 1]; gi (t, s) > 0 for all t, s ∈ (0, 1), i = 1, . . . , 4.  i , i = 1, . . . , 4, and Now, from the definitions of the Green functions G the properties of functions gi , i = 1, . . . , 4, we obtain the following lemma. Lemma 3.2.4. If Δ > 0, ai ≥ 0 for all i = 1, . . . , N , and bi ≥ 0 for all  i , i = 1, . . . , 4, have the properties i = 1, . . . , M , then the functions G 1 (t, s) ≤ J1 (s), ∀ (t, s) ∈ [0, 1] × [0, 1], where (a1 ) G N   M   Γ(β) β−q1 −1   J1 (s) = h1 (s) + ai ξi bi g2 (ηi , s) , ΔΓ(β − q1 ) i=1 i=1 ∀ s ∈ [0, 1]; 1 (t, s) ≥ tα−1 J1 (s), ∀ (t, s) ∈ [0, 1] × [0, 1]; (a2 ) G 1 (t, s) ≤ δ1 tα−1 , ∀ (t, s) ∈ [0, 1] × [0, 1], where (a3 ) G N  M    Γ(β) 1 + ai ξiβ−q1 −1 bi ηiα−q2 −1 ; δ1 = Γ(α) ΔΓ(β − q1 )Γ(α − q2 ) i=1 i=1 2 (t, s) ≤ J2 (s), ∀ (t, s) ∈ [0, 1] × [0, 1], where (b1 ) G J2 (s) =

N

 Γ(β) ai  g3 (ξi , s), ΔΓ(β − p2 ) i=1

∀ s ∈ [0, 1];

2 (t, s) = tα−1 J2 (s), ∀ (t, s) ∈ [0, 1] × [0, 1]; (b2 ) G 2 (t, s) ≤ δ2 tα−1 , ∀ (t, s) ∈ [0, 1] × [0, 1], where (b3 ) G N

 Γ(β) δ2 = ai ξiβ−q1 −1 ; ΔΓ(β − p2 )Γ(β − q1 ) i=1

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3 (t, s) ≤ J3 (s), ∀ (t, s) ∈ [0, 1] × [0, 1], where (c1 ) G Γ(α) J3 (s) =  h4 (s) + ΔΓ(α − q2 )



M  i=1

 bi ηiα−q2 −1

N 

 ai g3 (ξi , s) ,

i=1

∀ s ∈ [0, 1]; 3 (t, s) ≥ tβ−1 J3 (s), ∀ (t, s) ∈ [0, 1] × [0, 1]; (c2 ) G 3 (t, s) ≤ δ3 tβ−1 , ∀ (t, s) ∈ [0, 1] × [0, 1], where (c3 ) G Γ(α) 1 + δ3 = Γ(β) ΔΓ(α − q2 )Γ(β − q1 )



M  i=1

 bi ηiα−q2 −1

N  i=1

 ai ξiβ−q1 −1

;

4 (t, s) ≤ J4 (s), ∀ (t, s) ∈ [0, 1] × [0, 1], where (d1 ) G J4 (s) =

M

 Γ(α) bi g2 (ηi , s), ΔΓ(α − p1 ) i=1

∀ s ∈ [0, 1];

4 (t, s) = tβ−1 J4 (s), ∀ (t, s) ∈ [0, 1] × [0, 1]; (d2 ) G 4 (t, s) ≤ δ4 tβ−1 , ∀ (t, s) ∈ [0, 1] × [0, 1], where (d3 ) G M

 α−q −1 Γ(α) δ4 = bi ηi 2 . ΔΓ(α − p1 )Γ(α − q2 ) i=1 i , i = 1, . . . , 4, are continuous on [0, 1] × [0, 1], and (e) The functions G i (t, s) ≥ 0 for all t, s ∈ [0, 1], i = 1, . . . , 4. G Lemma 3.2.5. If Δ > 0, ai ≥ 0 for all i = 1, . . . , N, bi ≥ 0 for all i = 1, . . . , M, and x, y ∈ C(0, 1) ∩ L1 (0, 1) with x(t) ≥ 0, y(t) ≥ 0 for all t ∈ (0, 1), then the solution (u, v) of problem (3.43), (3.42) given by (3.44) satisfies the inequalities u(t) ≥ 0, v(t) ≥ 0 for all t ∈ [0, 1]. Moreover, we have the inequalities u(t) ≥ tα−1 u(t ) and v(t) ≥ tβ−1 v(t ) for all t, t ∈ [0, 1]. Proof. Under the assumptions of this lemma, by using relations (3.47) and Lemma 3.2.4, we deduce that u(t) ≥ 0 and v(t) ≥ 0 for all t ∈ [0, 1]. Besides,

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for all t, t ∈ [0, 1], we obtain the following inequalities:

1

1  1 (t, s)x(s) ds + 2 (t, s)y(s) ds G G u(t) = 0

≥ tα−1 ≥ tα−1

v(t) =

0

1



0

1

0



1

0

J1 (s)x(s) ds +

≥ tβ−1





1 0 1 0

1

0

1 (t , s)x(s) ds + G

 3 (t, s)y(s) ds + G

≥ tβ−1





1

0

J3 (s)y(s) ds +

J2 (s)y(s) ds



1

0

 2 (t , s)y(s) ds G

= tα−1 u(t ),

4 (t, s)x(s) ds G

1

0

 3 (t , s)y(s) ds + G

J4 (s)x(s) ds

0

1

4 (t , s)x(s) ds G

= tβ−1 v(t ).



Remark 3.2.1. Under the assumptions of Lemma 3.2.5, for c ∈ (0, 1), the solution (u, v) of problem (3.43), (3.42) given by (3.44) satisfies the inequalities mint∈[c,1] u(t) ≥ cα−1 maxt ∈[0,1] u(t ) and mint∈[c,1] v(t) ≥ cβ−1 maxt ∈[0,1] v(t ). 3.2.2

Nonsingular nonlinearities

In this section, we investigate the existence and multiplicity of positive solutions for our problem (3.41), (3.42) under various assumptions on the nonsingular functions f and g. We present the basic assumptions that we shall use in the sequel. (I1) α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), q1 ∈ [0, p2 ], q2 ∈ [0, p1 ], ξi ∈ R, N ai ≥ 0 for all i = 1, . . . , N (N ∈ N), i=1 ai > 0, 0 < ξ1 < · · · < ξN ≤ 1, ηi ∈ R, bi ≥ 0 for all i = 1, . . . , M (M ∈ N), M Γ(α)Γ(β) i=1 bi > 0, 0 < η1 < · · · < ηM ≤ 1, and Δ = Γ(α−p1 )Γ(β−p2 ) −     N M Γ(α)Γ(β) β−q1 −1 α−q2 −1 > 0. i=1 ai ξi i=1 bi ηi Γ(α−q2 )Γ(β−q1 ) (I2) The functions f, g : [0, 1] × [0, ∞) × [0, ∞) → [0, ∞) are continuous. By using Lemma 3.2.2, (u, v) is solution of problem (3.41), (3.42) if and only if (u, v) is solution of the following nonlinear system of integral

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equations ⎧

1

1 ⎪ 1 (t, s)f (s, u(s), v(s)) ds + 2 (t, s)g(s, u(s), v(s)) ds, ⎪u(t) = G G ⎪ ⎪ ⎪ 0 0 ⎪ ⎨ t ∈ [0, 1],

1

1 ⎪ ⎪  4 (t, s)f (s, u(s), v(s)) ds, ⎪ G G v(t) = (t, s)g(s, u(s), v(s)) ds + 3 ⎪ ⎪ ⎪ 0 0 ⎩ t ∈ [0, 1]. (3.50) We consider the Banach space X = C[0, 1] with supremum norm ·

and the Banach space Y = X ×X with the norm (u, v) Y = u + v . We define the cone P ⊂ Y by P = {(u, v) ∈ Y, u(t) ≥ 0, v(t) ≥ 0 for all t ∈ [0, 1]}. We introduce the operators Q1 , Q2 : Y → X and Q : Y → Y defined by

1

1 1 (t, s)f (s, u(s), v(s)) ds + 2 (t, s)g(s, u(s), v(s)) ds, G G Q1 (u, v)(t) = 0

Q2 (u, v)(t) =

0

0

0 ≤ t ≤ 1, 1

3 (t, s)g(s, u(s), v(s)) ds + G

0

1

4 (t, s)f (s, u(s), v(s)) ds, G

0 ≤ t ≤ 1, and Q(u, v) = (Q1 (u, v), Q2 (u, v)), (u, v) ∈ Y . Under the assumptions (I1) and (I2), it is easy to see that operator Q : P → P is completely continuous. It is clear that (u, v) is a solution of the system (3.50) if and only if (u, v) is a fixed point of operator Q. Therefore, we will investigate the existence and multiplicity of fixed points of operator Q. Theorem 3.2.1. Assume that (I1) and (I2) hold. If the functions f and g also satisfy the conditions (I3) There exist p ≥ 1 and q ≥ 1 such that f (t, u, v) =0 p u+v→0 t∈[0,1] (u + v)

f0s = lim

sup

u, v≥0

g(t, u, v) = 0; q u+v→0 t∈[0,1] (u + v)

g0s = lim

u, v≥0

sup

and

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(I4) There exists c ∈ (0, 1) such that i = lim f∞

u+v→∞

u, v≥0

i g∞ = lim

u+v→∞

u, v≥0

f (t, u, v) =∞ t∈[c,1] u + v inf

or

g(t, u, v) = ∞, t∈[c,1] u + v inf

then problem (3.41), (3.42) has at least one positive solution (u(t), v(t)), t ∈ [0, 1]. Proof. For c given in (I4), we define the cone P0 = {(u, v) ∈ P , min u(t) ≥ cα−1 u , min v(t) ≥ cβ−1 v }. t∈[c,1]

t∈[c,1]

From our assumptions and Remark 3.2.1, for any (u, v) ∈ P , we deduce that Q(u, v) = (Q1 (u, v), Q2 (u, v)) ∈ P0 , that is Q(P ) ⊂ P0 . We consider the functions u0 , v0 : [0, 1] → R defined by ⎧

1

1 ⎪ ⎪  1 (t, s) ds + 2 (t, s) ds, 0 ≤ t ≤ 1, G G ⎨ u0 (t) = 0 0

1

1 ⎪ ⎪ 3 (t, s) ds +  4 (t, s) ds, 0 ≤ t ≤ 1, ⎩ v0 (t) = G G 0

0

that is, (u0 , v0 ) is solution of problem (3.43), (3.42) with x(t) = x0 (t), y(t) = y0 (t), x0 (t) = 1, y0 (t) = 1 for all t ∈ [0, 1]. Hence, (u0 , v0 ) = Q(x0 , y0 ) ∈ P0 . We define the set M = {(u, v) ∈ P , there exists λ ≥ 0 such that (u, v) = Q(u, v) + λ(u0 , v0 )}. We will show that M ⊂ P0 and M is a bounded set of Y . If (u, v) ∈ M , then there exists λ ≥ 0 such that (u, v) = Q(u, v)+ λ(u0 , v0 ) or equivalently ⎧

1 ⎪  1 (t, s)(f (s, u(s), v(s)) + λ) ds ⎪ u(t) = G ⎪ ⎪ ⎪ 0 ⎪

1 ⎪ ⎪ ⎪ ⎪  2 (t, s)(g(s, u(s), v(s)) + λ) ds, 0 ≤ t ≤ 1, G + ⎨

10 ⎪ ⎪  3 (t, s)(g(s, u(s), v(s)) + λ) ds ⎪ G v(t) = ⎪ ⎪ ⎪ 0

⎪ 1 ⎪ ⎪ ⎪  4 (t, s)(f (s, u(s), v(s)) + λ) ds, 0 ≤ t ≤ 1. ⎩ G + 0

By Remark 3.2.1, we obtain (u, v) ∈ P0 , so M ⊂ P0 , and 1 1

u ≤ α−1 min u(t), v ≤ β−1 min v(t), ∀ (u, v) ∈ M . c c t∈[c,1] t∈[c,1]

(3.51)

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131

i From (I4), we suppose that f∞ = ∞, (in a similar manner, we can study i the case g∞ = ∞). Then for ε1 = max{ cα−12 m1 , cβ−12 m4 } > 0, there exists C1 > 0 such that

f (t, u, v) ≥ ε1 (u + v) − C1 , ∀ (t, u, v) ∈ [c, 1] × [0, ∞) × [0, ∞), (3.52) 1 where mi = c Ji (s) ds and Ji , i = 1, 4 are defined in Lemma 3.2.4. For (u, v) ∈ M and t ∈ [c, 1], by using Lemma 3.2.4 and relation (3.52), it follows that u(t) = Q1 (u, v)(t) + λu0 (t) ≥ Q1 (u, v)(t)

1

1   2 (t, s)g(s, u(s), v(s)) ds G1 (t, s)f (s, u(s), v(s)) ds + G = 0

≥

0

1

0

 1 (t, s)f (s, u(s), v(s)) ds ≥ G

≥ cα−1



1

c

≥ cα−1 ε1



1

c

tα−1 J1 (s)f (s, u(s), v(s)) ds

J1 (s)[ε1 (u(s) + v(s)) − C1 ] ds



1

c

J1 (s)u(s) ds − cα−1 m1 C1

≥ cα−1 ε1 m1 min u(s) − cα−1 m1 C1 s∈[c,1]

≥ 2 min u(s) − C2 , C2 = cα−1 m1 C1 , s∈[c,1]

v(t) = Q2 (u, v)(t) + λv0 (t) ≥ Q2 (u, v)(t)

1

1  3 (t, s)g(s, u(s), v(s)) ds +  4 (t, s)f (s, u(s), v(s)) ds G G = 0

≥

0

1

0

 4 (t, s)f (s, u(s), v(s)) ds ≥ G

≥c

β−1

≥c

β−1

c

ε1

1



c

1

tβ−1 J4 (s)f (s, u(s), v(s)) ds

J4 (s)[ε1 (u(s) + v(s)) − C1 ] ds

c

1

J4 (s)v(s) ds − cβ−1 m4 C1

≥ cβ−1 ε1 m4 min v(s) − cβ−1 m4 C1 s∈[c,1]

≥ 2 min v(s) − C3 , s∈[c,1]

C3 = cβ−1 m4 C1 .

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Therefore, we deduce min u(t) ≤ C2 ,

t∈[c,1]

min v(t) ≤ C3 ,

∀ (u, v) ∈ M .

t∈[c,1]

(3.53)

Now from relations (3.51) and (3.53), we obtain C2 , cα−1

C3 , and cβ−1 C2 C3

(u, v) Y = u + v ≤ α−1 + β−1 = C4 , c c

u ≤

v ≤

for all (u, v) ∈ M , that is M is a bounded set of Y . Besides, there exists a sufficiently large R1 > 1 such that (u, v) = Q(u, v) + λ(u0 , v0 ),

∀ (u, v) ∈ ∂BR1 ∩ P , ∀ λ ≥ 0.

From Theorem 1.2.9, we deduce that the fixed point index of operator Q over BR1 ∩ P with respect to P is i(Q, BR1 ∩ P , P ) = 0. Next, from assumption (I3), we conclude that for ε2 = min   1 and ε3 = min 8M , 1 , there exists r1 ∈ (0, 1] such that 2 8M3



1 8M1 ,

(3.54)  1 8M4

f (t, u, v) ≤ ε2 (u + v)p , g(t, u, v) ≤ ε3 (u + v)q , (3.55) ∀ t ∈ [0, 1], u, v ≥ 0, u + v ≤ r1 , 1 where Mi = 0 Ji (s) ds, i = 1, . . . , 4. By using (3.55), we deduce that for all (u, v) ∈ B r1 ∩ P and t ∈ [0, 1]

1

1 p  J1 (s)ε2 (u(s) + v(s)) ds + J2 (s)ε3 (u(s) + v(s))q ds Q1 (u, v)(t) ≤ 0

0

≤ ε2 M1 (u, v) pY + ε3 M2 (u, v) qY ≤ 1

(u, v) Y , 4

1

J3 (s)ε3 (u(s) + v(s))q ds + Q2 (u, v)(t) ≤

1 1

(u, v) Y + (u, v) Y 8 8

=

0

0

1

J4 (s)ε2 (u(s) + v(s))p ds

≤ ε3 M3 (u, v) qY + ε2 M4 (u, v) pY ≤ =

1

(u, v) Y . 4

1 1

(u, v) Y + (u, v) Y 8 8

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These imply that 1

(u, v) Y , 4 1

Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤ (u, v) Y , 2  ∀ (u, v) ∈ ∂Br1 ∩ P .

Q1 (u, v) ≤

1

(u, v) Y , 4

Q2 (u, v) ≤

From Theorem 1.2.8, we conclude that the fixed point index of operator Q over Br1 ∩ P with respect to P is i(Q, Br1 ∩ P, P ) = 1.

(3.56)

Combining (3.54) and (3.56), we obtain i(Q, (BR1 \ B r1 ) ∩ P , P) = i(Q, BR1 ∩ P, P) − i(Q, Br1 ∩ P , P ) = −1. We deduce that Q has at least one fixed point (u, v) ∈ (BR1 \ B r1 ) ∩ P, that is r1 < (u, v) Y < R1 or r1 < u + v < R1 . By Lemma 3.2.5, we obtain that u(t) > 0 for all t ∈ (0, 1] or v(t) > 0 for all t ∈ (0, 1]. The proof of the theorem is completed.  Theorem 3.2.2. Assume that (I1) and (I2) hold. If the functions f and g also satisfy the conditions s = lim (I5) f∞

f (t, u, v) g(t, u, v) s = 0 and g∞ = 0; = lim sup u+v→∞ u + v t∈[0,1] t∈[0,1] u + v sup

u+v→∞

u, v≥0

u, v≥0

(I6) There exist c ∈ (0, 1), pˆ ∈ (0, 1] and qˆ ∈ (0, 1] such that f0i = lim

u+v→0

u, v≥0

inf

t∈[c,1]

f (t, u, v) =∞ (u + v)pˆ

or

g0i = lim

u+v→0

u, v≥0

inf

t∈[c,1]

g(t, u, v) = ∞, (u + v)qˆ

then problem (3.41), (3.42) has at least one positive solution (u(t), v(t)), t ∈ [0, 1]. Proof.  From the assumption (I5), we  deduce that for ε4 = 1 1 1 1 min 8M1 , 8M4 and ε5 = min 8M2 , 8M3 there exist C5 , C6 > 0 such

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that f (t, u, v) ≤ ε4 (u + v) + C5 ,

g(t, u, v) ≤ ε5 (u + v) + C6 ,

∀ (t, u, v) ∈ [0, 1] × [0, ∞) × [0, ∞).

(3.57)

Hence, for (u, v) ∈ P, by using (3.57), we obtain

1 J1 (s)(ε4 (u(s) + v(s)) + C5 ) ds Q1 (u, v)(t) ≤ 0



1

+ 0

J2 (s)(ε5 (u(s) + v(s)) + C6 ) ds

≤ ε4 ( u + v )

×

1

0

1

0

J1 (s) ds + C5

J2 (s) ds + C6



1

0



1

0

J1 (s) ds + ε5 ( u + v )

J2 (s) ds

= ε4 (u, v) Y M1 + C5 M1 + ε5 (u, v) Y M2 + C6 M2 1 1 1

(u, v) Y + (u, v) Y + C7 = (u, v) Y + C7 , 8 8 4 ∀ t ∈ [0, 1], C7 = C5 M1 + C6 M2 ,

1 J3 (s)(ε5 (u(s) + v(s)) + C6 ) ds Q2 (u, v)(t) ≤ ≤

0



1

+ 0

J4 (s)(ε4 (u(s) + v(s)) + C5 ) ds

≤ ε5 ( u + v )

×

0

1

0

1

J3 (s) ds + C6

J4 (s) ds + C5



1

0



1

0

J3 (s) ds + ε4 ( u + v )

J4 (s) ds

= ε5 (u, v) Y M3 + C6 M3 + ε4 (u, v) Y M4 + C5 M4 ≤

1 1 1

(u, v) Y + (u, v) Y + C8 = (u, v) Y + C8 , 8 8 4 ∀ t ∈ [0, 1], C8 = C6 M3 + C5 M4 ,

and so

Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤ C9 = C7 + C8 .

1

(u, v) Y + C9 , 2

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Then there exists a sufficiently large R2 ≥ max{4C9 , 1} such that

Q(u, v) Y ≤

3

(u, v) Y , 4

∀ (u, v) ∈ P ,

(u, v) Y ≥ R2 .

Hence, Q(u, v) Y < (u, v) Y for all (u, v) ∈ ∂BR2 ∩ P and from Theorem 1.2.8 we have i(Q, BR2 ∩ P , P ) = 1.

(3.58)

On the other hand, from (I6), we suppose that f0i = ∞, (in a similar manner we can study the case g0i = ∞). We conclude that for ε6 = max{ cα−11 m1 , cβ−11 m4 }, there exists r2 ∈ (0, 1) such that f (t, u, v) ≥ ε6 (u + v)pˆ,

∀ t ∈ [c, 1], u, v ≥ 0, u + v ≤ r2 .

(3.59)

From (3.59), we deduce that for any (u, v) ∈ B r2 ∩ P

Q1 (u, v)(t) ≥

c

≥

1

c

1

 1 (t, s)f (s, u(s), v(s)) ds + G



≥ ε6

c

≥

c

1

c

 2 (t, s)g(s, u(s), v(s)) ds G

c 1

 1 (t, s)(u(s) + v(s))pˆ ds G  1 (t, s)(u(s) + v(s)) ds =: L1 (u, v)(t), G

 3 (t, s)g(s, u(s), v(s)) ds + G

c

1

∀ t ∈ [0, 1],

 4 (t, s)f (s, u(s), v(s)) ds G

 4 (t, s)f (s, u(s), v(s)) ds G



≥ ε6

≥ ε6

1

c 1

1

 1 (t, s)f (s, u(s), v(s)) ds G

≥ ε6

Q2 (u, v)(t) ≥



1 c 1 c

 4 (t, s)(u(s) + v(s))pˆ ds G  4 (t, s)(u(s) + v(s)) ds =: L2 (u, v)(t), G

∀ t ∈ [0, 1],

Hence, Q(u, v) ≥ L(u, v),

∀ (u, v) ∈ ∂Br2 ∩ P,

(3.60)

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where the linear operator L : P → P is defined by L(u, v) = u0 , v0 ) ∈ P \ {(0, 0)} defined by (L1 (u, v), L2 (u, v)). For (

1

1  1 (t, s) ds, v0 (t) =  4 (t, s) ds, t ∈ [0, 1], G G u 0 (t) = c

c

v0 ) = (L1 ( u0 , v0 ), L2 ( u0 , v0 )) with we have L( u0 ,  1

1

  G1 (t, s) G1 (s, τ ) dτ + u0 , v0 )(t) = ε6 L1 ( c

≥ ε6

c

1

c

≥ ε6

1

c

= ε6 c

 1 (t, s) G  1 (t, s) G

α−1

m1

1

c



1

c



1

c

 1 (s, τ ) dτ G c

α−1



c

1

c

≥ ε6

1

c

≥ ε6

c

1

c

 4 (t, s) G  4 (t, s) G

= ε6 cβ−1 m4 = v0 (t),

c

1



1

c



1

c

 4 (s, τ ) dτ G c

β−1

c

ds

ds

 1 (t, s) ds G

1

1

c

 4 (s, τ ) dτ G

ds

ds

J4 (τ ) dτ

 4 (t, s) ds ≥ G



=u 0 (t), ∀ t ∈ [0, 1], 1

1   1 (s, τ ) dτ + G4 (t, s) G L2 ( u0 , v0 )(t) = ε6 c

 4 (s, τ ) dτ G

ds

J1 (τ ) dτ

1 (t, s) ds ≥ G

1

ds

4 (t, s) ds G

∀ t ∈ [0, 1],

So, u0 , v0 ). L( u0 , v0 ) ≥ (

(3.61)

We may suppose that Q has no fixed point on ∂Br2 ∩ P (otherwise the proof is finished). From (3.60), (3.61) and Theorem 1.2.10, we conclude that i(Q, Br2 ∩ P, P ) = 0.

(3.62)

Therefore, from (3.58) and (3.62), we have i(Q, (BR2 \ B r2 ) ∩ P , P ) = i(Q, BR2 ∩ P , P) − i(Q, Br2 ∩ P, P ) = 1.

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137

Then Q has at least one fixed point in (BR2 \ B r2 ) ∩ P, that is r2
0 such that f (t, N0 , N0 )
0 such that (u, v) Y ≤ M1 for all (u, v) ∈ D. By the continuity of q1 and q2 , there exists M2 > 0 such that ! " M2 = max supt∈[0,1], u,v∈[0,M 1 (t, u, v), supt∈[0,1], u,v∈[0,M 2 (t, u, v) . 1 ] q 1 ] q By using Lemma 3.2.4, for any (u, v) ∈ D and t ∈ [0, 1], we obtain

1 1 (u, v)(t) ≤ Q J1 (s)f (s, u(s), v(s)) ds 0



1

J2 (s)g(s, u(s), v(s)) ds

+

≤

0

0 1

J1 (s) p1 (s) q1 (s, u(s), v(s)) ds



1

J2 (s) p2 (s) q2 (s, u(s), v(s)) ds

+ 0

≤ M2

1

0

J1 (s) p1 (s) ds



+ M2

0

1

 J2 (s) p2 (s) ds = M2 ( α + β),

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2 (u, v)(t) ≤ Q



1

0

≤

1

0

J3 (s)g(s, u(s), v(s)) ds +

1

0

J4 (s)f (s, u(s), v(s)) ds

J3 (s) p2 (s) q2 (s, u(s), v(s)) ds



1

J4 (s) p1 (s) q1 (s, u(s), v(s)) ds

+ 0

≤ M2



page 141

1

0

J3 (s) p2 (s) ds



+ M2

1

0

 J4 (s) p1 (s) ds = M2 ( γ + δ).

 Q  for all  2 (u, v) ≤ M2 ( 1 (u, v) ≤ M2 ( α + β), γ + δ), Therefore, Q    (u, v) ∈ D, and so Q1 (D), Q2 (D) and Q(D) are bounded.  In what follows, we shall prove that Q(D) is equicontinuous. By using Lemma 3.2.2, we have for (u, v) ∈ D and t ∈ [0, 1] 1 (u, v)(t) = Q



1



0

tα−1 Γ(β) g1 (t, s) +  ΔΓ(β − q1 )



N  i=1

 ai ξiβ−q1 −1

M 

 bi g2 (ηi , s)

i=1

× f (s, u(s), v(s)) ds

1 α−1 N t Γ(β)  ai g3 (ξi , s)g(s, u(s), v(s)) ds + 0 ΔΓ(β − p2 ) i=1

t 1 = [tα−1 (1 − s)α−p1 −1 − (t − s)α−1 ]f (s, u(s), v(s)) ds 0 Γ(α)

1 1 α−1 + t (1 − s)α−p1 −1 f (s, u(s), v(s)) ds t Γ(α) N   tα−1 Γ(β) β−q1 −1 ai ξi + ΔΓ(β − q1 ) i=1 

1  M × bi  g2 (ηi , s) f (s, u(s), v(s)) ds 0

i=1

tα−1 Γ(β) + ΔΓ(β − p2 )

0

1



N  i=1

 ai g3 (ξi , s) g(s, u(s), v(s)) ds.

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Therefore, for any t ∈ (0, 1), we conclude 1 (u, v)) (t) = (Q

0

t

 1 # (α − 1)tα−2 (1 − s)α−p1 −1 − (α − 1)(t − s)α−2 Γ(α)

× f (s, u(s), v(s)) ds

1 1 + (α − 1)tα−2 (1 − s)α−p1 −1 f (s, u(s), v(s)) ds t Γ(α)  N (α − 1)tα−2 Γ(β)  β−q1 −1 ai ξi + ΔΓ(β − q1 ) i=1 

1  M × bi g2 (ηi , s) f (s, u(s), v(s)) ds 0

i=1

(α − 1)tα−2 Γ(β) + ΔΓ(β − p2 )



1



0

N 

 ai g3 (ξi , s)

i=1

× g(s, u(s), v(s)) ds. So, for any t ∈ (0, 1), we deduce 1 (u, v)) (t)| ≤ |(Q

1 Γ(α − 1)

0

t

[tα−2 (1 − s)α−p1 −1 + (t − s)α−2 ]

× p1 (s) q1 (s, u(s), v(s)) ds

1 1 + tα−2 (1 − s)α−p1 −1 Γ(α − 1) t × p1 (s) q1 (s, u(s), v(s)) ds N  (α − 1)tα−2 Γ(β)  β−q1 −1 + ai ξi ΔΓ(β − q1 ) i=1 

1  M × bi g2 (ηi , s) p1 (s) q1 (s, u(s), v(s)) ds 0

i=1

(α − 1)tα−2 Γ(β) + ΔΓ(β − p2 )

0

1



N  i=1

q2 (s, u(s), v(s)) ds. × p2 (s)

 ai g3 (ξi , s)

page 142

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Then we obtain 1 (u, v)) (t)| |(Q 

t 1 ≤ M2 [tα−2 (1 − s)α−p1 −1 + (t − s)α−2 ] p1 (s) ds Γ(α − 1) 0

1 1 + tα−2 (1 − s)α−p1 −1 p1 (s) ds Γ(α − 1) t    M N 1  (α − 1)tα−2 Γ(β)  β−q1 −1 ai ξi bi  g2 (ηi , s) p1 (s) ds + ΔΓ(β − q1 ) 0 i=1 i=1  

N (α − 1)tα−2 Γ(β) 1  + ai  g3 (ξi , s) p2 (s) ds . (3.66) ΔΓ(β − p2 ) 0 i=1 We denote h(t) =

1 Γ(α − 1) +



t

0

[tα−2 (1 − s)α−p1 −1 + (t − s)α−2 ] p1 (s) ds



1 Γ(α − 1)

1

t

tα−2 (1 − s)α−p1 −1 p1 (s) ds,

 N (α − 1)tα−2 Γ(β)  β−q1 −1 μ(t) = h(t) + ai ξi ΔΓ(β − q1 ) i=1 

1  M × bi g2 (ηi , s) p1 (s) ds 0

i=1

(α − 1)tα−2 Γ(β) + ΔΓ(β − p2 )



1



0

N 

 ai g3 (ξi , s) p2 (s) ds.

i=1

For the integral of the function h, by exchanging the order of integration, we obtain

0

1

1 h(t)dt = Γ(α − 1)



1



t

[t 0

1 + Γ(α − 1)

α−2

0



1 0



t

1

t

α−p1 −1

(1 − s)

α−2

+ (t − s)

α−p1 −1

(1 − s)

α−2

] p1 (s) ds dt

p1 (s) ds dt

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1 = Γ(α − 1)



1



1

[t 0

s

α−2

α−p1 −1

(1 − s)

α−2

+ (t − s)

]dt p1 (s)ds

1 s 1 α−2 α−p1 −1 + t (1 − s) dt p1 (s) ds Γ(α − 1) 0 0

1 (1 − s)α−p1 −1 sα−1 (1 − s)α−p1 −1 1 − = Γ(α − 1) 0 α−1 α−1  (1 − s)α−1 + p1 (s) ds α−1

1 α−1 s (1 − s)α−p1 −1 1 p1 (s) ds + Γ(α − 1) 0 α−1

1 1 (1 − s)α−p1 −1 (1 + (1 − s)p1 ) p1 (s) ds = Γ(α) 0

1 2 (1 − s)α−p1 −1 p1 (s) ds < ∞. ≤ Γ(α) 0

For the integral of the function μ, we have 

N

1

1 1 (α − 1)Γ(β)  β−q1 −1 μ(t) dt = h(t) dt + ai ξi tα−2 dt ΔΓ(β − q1 ) i=1 0 0 0   M   1  × bi g2 (ηi , s) p1 (s) ds 0

i=1

 

 1  1 N (α − 1)Γ(β) α−2 + t dt ai  g3 (ξi , s) p2 (s) ds ΔΓ(β − p2 ) 0 0 i=1

1 2 ≤ (1 − s)α−p1 −1 p1 (s) ds Γ(α) 0 N   Γ(β) β−q1 −1 ai ξi + ΔΓ(β − q1 ) i=1    M  1  bi ηiα−q2 −1 α−p1 −1 (1 − s) × p1 (s) ds Γ(α − q2 ) 0 i=1   N   1  Γ(β) ai ξiβ−q1 −1 β−p2 −1 + (1 − s) p2 (s) ds ΔΓ(β − p2 ) Γ(β − q1 ) 0 i=1

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2 = Γ(α)



1

0

page 145

(1 − s)α−p1 −1 p1 (s) ds

N   Γ(β) β−q1 −1 ai ξi + ΔΓ(β − q1 )Γ(α − q2 ) i=1 M   α−q −1 1 × bi ηi 2 (1 − s)α−p1 −1 p1 (s) ds 0

i=1

N   Γ(β) β−q1 −1 + ai ξi ΔΓ(β − p2 )Γ(β − q1 ) i=1

1 × (1 − s)β−p2 −1 p2 (s) ds. 0

Therefore, we obtain N  

1  Γ(β) 2 β−q1 −1 + μ(t) dt ≤ ai ξi Γ(α) ΔΓ(β − q1 )Γ(α − q2 ) i=1 0 M   α−q −1 2 × bi ηi i=1

× ×

0

(1 − s)α−p1 −1 p1 (s) ds +

N 

×

1

i=1

0

1



Γ(β) (3.67) ΔΓ(β − p2 )Γ(β − q1 )

ai ξiβ−q1 −1

(1 − s)β−p2 −1 p2 (s) ds < ∞.

We deduce that μ ∈ L1 (0, 1). Thus, for any t1 , t2 ∈ [0, 1] with t1 ≤ t2 and (u, v) ∈ D, by (3.66) and (3.67), we conclude $ t2 $

t2 $ $   $   (Q1 (u, v)) (t) dt$$ ≤ M2 μ(t) dt. |Q1 (u, v)(t1 ) − Q1 (u, v)(t2 )| = $ t1

t1

(3.68) From (3.67), (3.68) and the absolute continuity of the integral function, 1 (D) is equicontinuous. In a similar manner, we deduce we obtain that Q 2 (D) is also equicontinuous. By the Arzela–Ascoli theorem, we conthat Q 2 (D) are relatively compact sets, and so Q(D)  1 (D) and Q is clude that Q  also relatively compact. Therefore, Q is a compact operator. Besides, we can

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 is continuous on P, (see the proof of Lemma 1.4.1 from [43]). show that Q  : P → P is completely continuous. Hence Q  For c ∈ (0, 1) we define the cone   α−1 β−1 

u , min v(t) ≥ c

v . P0 = (u, v) ∈ P , min u(t) ≥ c t∈[c,1]

t∈[c,1]

 (I2)  and Remark 3.2.1, we have Q(  P ) ⊂ P0 and Under assumptions (I1),  is also a completely continuous  P0 : P0 → P0 (denoted again by Q) so Q| operator.  and (I2)  hold. If the functions q1 , q2 , f Theorem 3.2.4. Assume that (I1) and g also satisfy the conditions  There exist a ≥ 1 and b ≥ 1 such that (I3) q10 = lim

u+v→0

u,v≥0

q20 = lim

u+v→0

u,v≥0

q1 (t, u, v) =0 a t∈[0,1] (u + v) sup

and

q2 (t, u, v) = 0; b t∈[0,1] (u + v) sup

 There exists c ∈ (0, 1/2) such that (I4) i f∞ = lim

u+v→∞

u,v≥0

i g∞ = lim

u+v→∞

u,v≥0

inf

f (t, u, v) =∞ u+v

inf

g(t, u, v) = ∞, u+v

t∈[c,1−c]

t∈[c,1−c]

or

then problem (3.41), (3.42) has at least one positive solution (u(t), v(t)), t ∈ [0, 1].   Proof. We consider the above " P0 with c given " From (I3), we ! 1 1cone ! 1 in 1(I4). deduce that for ε7 = min 4  > 0, ε8 = min 4β  , 4 α , 4δ γ > 0 there exists r3 ∈ (0, 1) such that q1 (t, u, v) ≤ ε7 (u + v)a ,

q2 (t, u, v) ≤ ε8 (u + v)b ,

∀ t ∈ [0, 1], u, v ≥ 0, u + v ≤ r3 .

(3.69)

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147

Then by (3.69) and Lemma 3.2.4, for any (u, v) ∈ ∂Br3 ∩ P0 and t ∈ [0, 1], we obtain 1 (u, v)(t) ≤ Q



1 0

J1 (s) p1 (s) q1 (s, u(s), v(s)) ds



1

J2 (s) p2 (s) q2 (s, u(s), v(s)) ds

+

≤ ε7

0 1

J1 (s) p1 (s)(u(s) + v(s))a ds

0



+ ε8

1

0

J2 (s) p2 (s)(u(s) + v(s))b ds

   (u, v) aY + ε8 β (u,  (u, v) Y + ε8 β (u, v) bY ≤ ε7 α v) Y ≤ ε7 α 1 1 1

(u, v) Y + (u, v) Y = (u, v) Y , 4 4 2

1 2 (u, v)(t) ≤ J3 (s) Q p2 (s) q2 (s, u(s), v(s)) ds ≤

0



1

J4 (s) p1 (s) q1 (s, u(s), v(s)) ds

+

≤ ε8

0 1

0

J3 (s) p2 (s)(u(s) + v(s))b ds



+ ε7

0

1

J4 (s) p1 (s)(u(s) + v(s))a ds

  ≤ ε8 γ  (u, v) bY + ε7 δ (u, γ (u, v) Y + ε7 δ (u, v) aY ≤ ε8  v) Y ≤

1 1 1

(u, v) Y + (u, v) Y = (u, v) Y . 4 4 2

 2 (u, v) ≤ 1 (u, v) Y  1 (u, v) ≤ 1 (u, v) Y , Q Therefore, we deduce Q 2 2 for all (u, v) ∈ ∂Br3 ∩ P0 , and so  v) Y ≤ (u, v) Y ,

Q(u,

∀ (u, v) ∈ ∂Br3 ∩ P0 .

(3.70)

i  we suppose that f∞ From (I4), = ∞, (in a similar manner, we can i α−1 = ∞). So, for ε = 2(c m  1 min{cα−1 , cβ−1 })−1 , (m 1 = study the case g  9 ∞  1−c J (s) ds), there exists C > 0 such that 1 10 c

f (t, u, v) ≥ ε9 (u + v) − C10 ,

∀ t ∈ [c, 1 − c], u, v ≥ 0.

(3.71)

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Then by using (3.71), for any (u, v) ∈ P0 and t ∈ [c, 1], we have

1−c 1 (u, v)(t) ≥  1 (t, s)f (s, u(s), v(s)) ds Q G c



1−c

+ c

≥ cα−1 ≥c

α−1



 2 (t, s)g(s, u(s), v(s)) ds G

1−c

c



1−c

c

≥ cα−1 ε9 m 1

J1 (s)f (s, u(s), v(s)) ds J1 (s)(ε9 (u(s) + v(s)) − C10 ) ds min (u(s) + v(s)) − cα−1 m  1 C10

s∈[c,1−c]

≥ cα−1 ε9 m  1 min (u(s) + v(s)) − cα−1 m  1 C10 s∈[c,1]

≥ cα−1 ε9 m 1 ≥c

α−1



ε9 m  1 (c

 1 C10 min u(s) + min v(s) − cα−1 m

s∈[c,1] α−1

u + c

s∈[c,1]

β−1

v ) − cα−1 m  1 C10

≥ cα−1 ε9 m  1 min{cα−1 , cβ−1 } (u, v) Y − C11 = 2 (u, v) Y − C11 , C11 = cα−1 m  1 C10 .  1 (u, v) ≥ 2 (u, v) Y − C11 for all (u, v) ∈ P0 . We can Hence, we obtain Q choose R3 ≥ max{C11 , 1}, and then we deduce 1 (u, v) ≥ (u, v) Y ,  v) Y ≥ Q

Q(u,

∀ (u, v) ∈ ∂BR3 ∩ P0 .

(3.72)

By (3.70), (3.72) and the Guo–Krasnosel’skii fixed point theorem  has a fixed point (u, v) ∈ (B R3 \ (Theorem 1.2.2), we conclude that Q Br3 ) ∩ P0 , that is r3 ≤ (u, v) Y ≤ R3 . By Lemma 3.2.5, we obtain that u(t) > 0 for all t ∈ (0, 1] or v(t) > 0 for all t ∈ (0, 1]. The proof of the theorem is completed.   and (I2)  hold. If the functions q1 , q2 , f Theorem 3.2.5. Assume that (I1) and g also satisfy the conditions  (I5)

q1 (t, u, v) =0 u+v→∞ u+v t∈[0,1]

q1∞ = lim

sup

u,v≥0

q2 (t, u, v) = 0; u+v→∞ u+v t∈[0,1]

q2∞ = lim

u,v≥0

sup

and

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 There exist c ∈ (0, 1/2), a (I6) ˆ ∈ (0, 1] and ˆb ∈ (0, 1] such that f0i = lim

u+v→0

u,v≥0

g0i = lim

u+v→0

u,v≥0

inf

t∈[c,1−c]

inf

t∈[c,1−c]

f (t, u, v) =∞ (u + v)aˆ g(t, u, v) (u + v)ˆb

or

= ∞,

then problem (3.41), (3.42) has at least one positive solution (u(t), v(t)), t ∈ [0, 1].  From (I5)  in (I6). Proof. We consider againthe coneP0 with c given   1 1 and ε11 ∈ 0, 2(β+ we deduce that for ε10 ∈ 0, 2(   γ ) , there exist α+δ) C12 , C13 > 0 such that q1 (t, u, v) ≤ ε10 (u + v) + C12 , q2 (t, u, v) ≤ ε11 (u + v) + C13 ,

∀ t ∈ [0, 1], u, v ≥ 0.

(3.73)

 for any (u, v) ∈ P0 , we conclude By using (3.73) and (I2), 1 (u, v)(t) ≤ Q

0

1

J1 (s) p1 (s) q1 (s, u(s), v(s)) ds



1

+

≤

0

0 1

J2 (s) p2 (s) q2 (s, u(s), v(s)) ds

J1 (s) p1 (s)(ε10 (u(s) + v(s)) + C12 ) ds



1

+ 0

J2 (s) p2 (s)(ε11 (u(s) + v(s)) + C13 ) ds

α + (ε11 (u, v) Y + C13 )β ≤ (ε10 (u, v) Y + C12 )   = (ε10 α  + ε11 β) (u,  + C13 β, v) Y + C12 α

1 2 (u, v)(t) ≤ J3 (s) Q p2 (s) q2 (s, u(s), v(s)) ds 0



+ 0

1

J4 (s) p1 (s) q1 (s, u(s), v(s)) ds

∀ t ∈ [0, 1],

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≤

1

0

J3 (s) p2 (s)(ε11 (u(s) + v(s)) + C13 ) ds



1

+ 0

J4 (s) p1 (s)(ε10 (u(s) + v(s)) + C12 ) ds

γ + (ε10 (u, v) Y + C12 )δ ≤ (ε11 (u, v) Y + C13 )   ∀ t ∈ [0, 1]. = (ε11 γ v) Y + C13   + ε10 δ) (u, γ + C12 δ, Therefore,   1 (u, v) ≤ (ε10 α  + ε11 β) (u,  + C13 β,

Q v) Y + C12 α   2 (u, v) ≤ (ε11 γ

Q  + ε10 δ) (u, γ + C12 δ, v) Y + C13  and so  + ε11 (β + γ  + C13 (β + γ  v) Y ≤ [ε10 ( α + δ) )] (u, v) Y + C12 ( α + δ) )

Q(u,  + C13 (β + γ < (u, v) Y + C14 , C14 = C12 ( α + δ) ). We can choose large R4 > 1 such that  v) Y ≤ (u, v) Y , ∀ (u, v) ∈ ∂BR4 ∩ P0 .

Q(u,

(3.74)

 From (I6), we suppose that f0i = ∞, (in a similar manner we 1 can study the case g0i = ∞). So, we deduce that for ε12 = (cα−1 m 1−c min{cα−1 , cβ−1 })−1 > 0, (m  1 = c J1 (s) ds), there exists r4 ∈ (0, 1] such that f (t, u, v) ≥ ε12 (u + v)aˆ ,

∀ t ∈ [c, 1 − c], u, v ≥ 0, u + v ≤ r4 .

(3.75)

Then by using (3.75), for any (u, v) ∈ ∂Br4 ∩ P0 and t ∈ [c, 1], we have

1−c 1 (u, v)(t) ≥  1 (t, s)f (s, u(s), v(s)) ds Q G c



1−c

+ c

≥c

α−1

≥c

α−1



 2 (t, s)g(s, u(s), v(s)) ds G

1−c

c



c

1−c

J1 (s)ε12 (u(s) + v(s))aˆ ds J1 (s)ε12 (u(s) + v(s)) ds

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≥c

α−1

≥c

α−1

ε12

1−c

c

J1 (s)



min (u(s) + v(s)) ds

s∈[c,1−c]



ε12 min (u(s) + v(s)) s∈[c,1]

1 ≥ cα−1 ε12 m



1−c c

J1 (s) ds

min u(s) + min v(s)

s∈[c,1]

s∈[c,1]

 1 (cα−1 u + cβ−1 v ) ≥ cα−1 ε12 m ≥ cα−1 ε12 m  1 min{cα−1 , cβ−1 }( u + v ) = (u, v) Y .  1 (u, v) ≥ (u, v) Y for all (u, v) ∈ ∂Br4 ∩ P0 , and so Therefore, Q  1 (u, v) ≥ (u, v) Y ,  v) Y ≥ Q

Q(u,

∀ (u, v) ∈ ∂Br4 ∩ P0 .

(3.76)

By (3.74), (3.76) and the Guo–Krasnosel’skii fixed point theorem  has at least one fixed point (u, v) ∈ (Theorem 1.2.2), we deduce that Q (B R4 \ Br4 ) ∩ P0 , that is r4 ≤ (u, v) Y ≤ R4 . The proof of the theorem is complete.  3.2.4

Examples

Let α = 9/2 (n = 5), β = 8/3 (m = 3), p1 = 4/3, p2 = 1, q1 = 1/2, q2 = 2/3, N = 1, M = 2, ξ1 = 1/2, a1 = 2, η1 = 1/3, η2 = 2/3, b1 = 1 and b2 = 1/2. We consider the system of fractional differential equations  9/2 D0+ u(t) + f (t, u(t), v(t)) = 0, t ∈ (0, 1), (3.77) 8/3 D0+ v(t) + g(t, u(t), v(t)) = 0, t ∈ (0, 1), with the multi-point boundary conditions  4/3 1/2   u(0) = u (0) = u (0) = u (0) = 0, D0+ u(1) = 2D0+ v 12 , 2/3   2/3   v(0) = v  (0) = 0, v  (1) = D0+ u 13 + 12 D0+ u 23 . Then we obtain Δ = deduce g1 (t, s) = 

1 Γ(9/2)

5Γ(9/2) 3Γ(19/6)



−

Γ(9/2)Γ(8/3)(1+211/6 ) 21/6 317/6 Γ(13/6)Γ(23/6)

(3.78)

≈ 7.6683666. We also

t7/2 (1 − s)13/6 − (t − s)7/2 , 0 ≤ s ≤ t ≤ 1, t7/2 (1 − s)13/6 , 0 ≤ t ≤ s ≤ 1,

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1 g2 (t, s) = Γ(23/6)



t17/6 (1 − s)13/6 − (t − s)17/6 , 0 ≤ s ≤ t ≤ 1, t17/6 (1 − s)13/6 , 0 ≤ t ≤ s ≤ 1,

 7/6 1 t (1 − s)2/3 − (t − s)7/6 , 0 ≤ s ≤ t ≤ 1, g3 (t, s) = Γ(13/6) t7/6 (1 − s)2/3 , 0 ≤ t ≤ s ≤ 1,  5/3 1 t (1 − s)2/3 − (t − s)5/3 , 0 ≤ s ≤ t ≤ 1, g4 (t, s) = Γ(8/3) t5/3 (1 − s)2/3 , 0 ≤ t ≤ s ≤ 1,

1 1  (1 − s)13/6 (1 − (1 − s)4/3),  h2 (s) = Γ(23/6) (1 − s)13/6 (1 − (1 − h1 (s) = Γ(9/2) 1 h3 (s) = (1 − s)2/3 (1 − (1 − s)1/2 ),  h4 (s) = 1 s(1 − s)2/3 s)2/3 ),  Γ(13/6)

Γ(8/3)

for all s ∈ [0, 1]. For the functions Ji , i = 1, . . . , 4, we obtain ⎧ 1 ⎪ ⎪ (1 − s)13/6 (1 − (1 − s)4/3 ) ⎪ ⎪ Γ(9/2) ⎪ ⎪ ⎪ # Γ(8/3) ⎪ ⎪ ⎪ + 7/6 17/6 2(1 − s)13/6 − 2(1 − 3s)17/6 ⎪ ⎪ 2 3 ΔΓ(13/6)Γ(23/6) ⎪  ⎪ ⎪ ⎪ + 217/6 (1 − s)13/6 − (2 − 3s)17/6 , 0 ≤ s < 13 , ⎪ ⎪ ⎪ ⎪ ⎪ 1 (1 − s)13/6 (1 − (1 − s)4/3 ) ⎪ ⎪ ⎪ ⎨ Γ(9/2) # Γ(8/3) J1 (s) = 2(1 − s)13/6 + 7/6 17/6 ⎪ ⎪ 2 3 ΔΓ(13/6)Γ(23/6) ⎪  ⎪ ⎪ ⎪ + 217/6 (1 − s)13/6 − (2 − 3s)17/6 , 13 ≤ s < 23 , ⎪ ⎪ ⎪ ⎪ ⎪ 1 (1 − s)13/6 (1 − (1 − s)4/3 ) ⎪ ⎪ ⎪ Γ(9/2) ⎪ ⎪ ⎪ # Γ(8/3) ⎪ ⎪ (1 − s)13/6 + 1/6 17/6 ⎪ ⎪ ⎪ 2 3 ΔΓ(13/6)Γ(23/6) ⎪ ⎪  ⎩ +211/6 (1 − s)13/6 , 23 ≤ s ≤ 1, J2 (s) =

J3 (s) =

5 3 · 21/6 ΔΓ(13/6)



(1 − s)2/3 − (1 − 2s)7/6 , 0 ≤ s < 12 , (1 − s)2/3 , 12 ≤ s ≤ 1,

⎧ # 1 (1 + 211/6 )Γ(9/2) ⎪ 2/3 ⎪ s(1 − s) + (1 − s)2/3 ⎪ ⎪ 1/6 317/6 ΔΓ(13/6)Γ(23/6) Γ(8/3) ⎪ 2 ⎪  ⎪ ⎨ −(1 − 2s)7/6 , 0 ≤ s < 12 , ⎪ 1 (1 + 211/6 )Γ(9/2) ⎪ ⎪ ⎪ (1 − s)2/3 , s(1 − s)2/3 + 1/6 17/6 ⎪ ⎪ Γ(8/3) 2 3 ΔΓ(13/6)Γ(23/6) ⎪ ⎩ 1 2 ≤ s ≤ 1,

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153

Γ(9/2) 2 · 317/6 ΔΓ(19/6)Γ(23/6) ⎧ 2(1 − s)13/6 − 2(1 − 3s)17/6 + 217/6 (1 − s)13/6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −(2 − 3s)17/6 , 0 ≤ s < 13 , ⎪ ⎨ × 2(1 − s)13/6 + 217/6 (1 − s)13/6 ⎪ ⎪ ⎪ ⎪ −(2 − 3s)17/6 , 13 ≤ s < 23 , ⎪ ⎪ ⎪ ⎩ 2(1 − s)13/6 + 217/6 (1 − s)13/6 , 23 ≤ s ≤ 1. 1 1 We also deduce M1 = 0 J1 (s) ds ≈ 0.00912385, M2 = 0 J2 (s) ds ≈ 1 1 0.06605546, M3 = 0 J3 (s) ds ≈ 0.16869513 and M4 = 0 J4 (s) ds ≈ 0.00432433. J4 (s) =

Example 1. We consider the functions f (t, u, v) = a0 tγ0 (u + v)α0 , g(t, u, v) = b0 (t + 1)δ0 (u + v)β0 , t ∈ [0, 1],

u, v ≥ 0,

where α0 > 1, β0 ∈ (0, 1), a0 , b0 , γ0 , δ0 > 0. In addition, we have m0 = max{Mi , i = 1, . . . , 4} = M3 . The functions f (t, u, v) and g(t, u, v) are nondecreasing with respect to u and v for any t ∈ [0, 1], and for qˆ = 1 and c ∈ (0, 1) the assumptions (I4) and (I6) are satisfied; indeed we i = ∞ and g0i = ∞. We take N0 = 1 and then f (t, N0 , N0 ) ≤ obtain f∞ 1 α0 and b0 < 2 a0 , g(t, N0 , N0 ) ≤ 2β0 +δ0 b0 for all t ∈ [0, 1]. If a0 < 2α0 +2 m0 1 , then the assumption (I7) is satisfied. For example, for α0 = 2, 2β0 +δ0 +2 m0 β0 = 12 and δ0 = 1, if a0 ≤ 0.37 and b0 ≤ 0.52, then by Theorem 3.2.3, we deduce that problem (3.77), (3.78) has at least two positive solutions. Example 2. We consider the functions f t, u, v) =

(u + v)a0 , tζ1

g(t, u, v) =

(u + v)b0 , t ∈ (0, 1), u, v ≥ 0, (1 − t)ζ2

with a0 , b0 > 1 and ζ1 , ζ2 ∈ (0, 1). Here, f (t, u, v) = p1 (t) q1 (t, u, v), 1 q2 (t, u, v), where p1 (t) = tζ11 , p2 (t) = (1−t) for all g(t, u, v) = p2 (t) ζ2 t ∈ (0, 1), and q1 (t, u, v) = (u + v)a0 , q2 (t, u, v) = (u + v)b0 for all 1 t ∈ [0, 1], u, v ≥ 0. We have 0 < 0 (1 − s)13/6 p1 (s) ds < ∞, 0 < 1 2/3 p2 (s) ds < ∞. 0 (1 − s)  for a = b = 1, we obtain q10 = 0 and q20 = 0. In (I4),  for In (I3), i  c ∈ (0, 1/2), we have f = ∞. Then, by Theorem 3.2.4, we deduce that ∞

problem (3.77), (3.78) has at least one positive solution.

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Remark 3.2.2. The results presented in this section under the assumptions p1 ∈ [1, n− 2] and p2 ∈ [1, m− 2] instead of p1 ∈ [1, α− 1) and p2 ∈ [1, β − 1) were published in [48]. The existence and nonexistence of positive solutions for the system (3.41) with two positive parameters, subject to the boundary conditions (3.42) were investigated in [53].

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Chapter 4

Systems of Two Riemann–Liouville Fractional Differential Equations with p-Laplacian Operators, Parameters and Multi-Point Boundary Conditions

In this chapter, we investigate the existence and nonexistence of positive solutions for systems of two Riemann–Liouville fractional differential equations with p-Laplacian operators and positive parameters, subject to uncoupled or coupled multi-point boundary conditions which contain fractional derivatives, and the nonlinearities of systems are nonnegative functions. 4.1

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions

We consider the system of nonlinear ordinary fractional differential equations with r1 -Laplacian and r2 -Laplacian operators  α1 β1 u(t))) + λf (t, u(t), v(t)) = 0, t ∈ (0, 1), D0+ (ϕr1 (D0+ (4.1) β2 α2 D0+ (ϕr2 (D0+ v(t))) + μg(t, u(t), v(t)) = 0, t ∈ (0, 1),

155

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with the multi-point boundary conditions ⎧ β1 ⎪ u(j) (0) = 0, j = 0, . . . , n − 2; D0+ u(0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ N ⎪  ⎪ p1 q1 ⎪ ⎪ D u(1) = ai D0+ u(ξi ), ⎪ 0+ ⎨ i=1

⎪ v (j) (0) = 0, j = 0, . . . , m − 2; ⎪ ⎪ ⎪ ⎪ ⎪ M ⎪  ⎪ ⎪ q2 ⎪ Dp2 v(1) = bi D0+ v(ηi ), ⎪ 0+ ⎩

β2 D0+ v(0) = 0,

(4.2)

i=1

where α1 , α2 ∈ (0, 1], β1 ∈ (n − 1, n], β2 ∈ (m − 1, m], n, m ∈ N, n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, β1 − 1), p2 ∈ [1, β2 − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi , bi ∈ R for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, r1 , r2 > 1, 1 1 ϕri (s) = |s|ri −2 s, ϕ−1 ri = ϕi , ri + i = 1, i = 1, 2, λ, μ > 0, f, g ∈ k denotes the Riemann-Liouville C([0, 1] × [0, ∞) × [0, ∞), [0, ∞)), and D0+ derivative of order k (for k = α1 , β1 , α2 , β2 , p1 , q1 , p2 , q2 ). Under some assumptions on the functions f and g, we give intervals for the positive parameters λ and μ such that positive solutions of (4.1), (4.2) exist. By a positive solution of problem (4.1), (4.2), we mean a pair of functions (u, v) ∈ (C([0, 1], R+ ))2 , satisfying (4.1) and (4.2) with u(t) > 0 for all t ∈ (0, 1], or v(t) > 0 for all t ∈ (0, 1]. The nonexistence of positive solutions for the above problem is also studied. 4.1.1

Auxiliary results

We consider first the nonlinear fractional differential equation β1 α1 (ϕr1 (D0+ u(t))) + h(t) = 0, D0+

t ∈ (0, 1),

with the boundary conditions ⎧ ⎪ u(j) (0) = 0, j = 0, . . . , n − 2; ⎪ ⎨ N  β1 p1 q1 ⎪ D u(0) = 0, D u(1) = ai D0+ u(ξi ), ⎪ 0+ ⎩ 0+

(4.3)

(4.4)

i=1

where α1 ∈ (0, 1], β1 ∈ (n − 1, n], n ∈ N, n ≥ 3, p1 , q1 ∈ R, p1 ∈ [1, β1 − 1), q1 ∈ [0, p1 ], ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, and h ∈ C[0, 1].

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157

β1 If we denote by ϕr1 (D0+ u(t)) = x(t), then problem (4.3), (4.4) is equivalent to the following two boundary value problems: α1 x(t) + h(t) = 0, D0+

0 < t < 1,

(4.5)

with the boundary condition x(0) = 0,

(4.6)

and β1 u(t) = ϕ1 x(t), D0+

0 < t < 1,

(4.7)

with the boundary conditions u(j) (0) = 0, j = 0, . . . , n − 2;

p1 D0+ u(1) =

N 

q1 ai D0+ u(ξi ).

(4.8)

i=1

For the first problem (4.5), (4.6), the function  t 1 α1 (t − s)α1 −1 h(s) ds, x(t) = −I0+ h(t) = − Γ(α1 ) 0

t ∈ [0, 1]

(4.9)

is the unique solution x ∈ C[0, 1] of (4.5), (4.6). Γ(β1 ) Γ(β1 ) − Γ(β For the second problem (4.7), (4.8), if Δ1 = Γ(β 1 −p1 ) 1 −q1 ) N × i=1 ai ξiβ1 −q1 −1 = 0, then by Lemma 2.1.2 from Section 2.1, we deduce that the function  1 G1 (t, s)ϕ1 x(s) ds, t ∈ [0, 1], (4.10) u(t) = − 0

is the unique solution u ∈ C[0, 1] of (4.7), (4.8). Here, the Green function G1 is given by G1 (t, s) = g1 (t, s) + with 1 g1 (t, s) = Γ(β1 )



N tβ1 −1  ai g2 (ξi , s), Δ1 i=1

(t, s) ∈ [0, 1] × [0, 1],

tβ1 −1 (1 − s)β1 −p1 −1 − (t − s)β1 −1 , β1 −1

(4.11)

0 ≤ s ≤ t ≤ 1,

β1 −p1 −1

(1 − s) , 0 ≤ t ≤ s ≤ 1, ⎧ β1 −q1 −1 (1 − s)β1 −p1 −1 − (t − s)β1 −q1 −1 , ⎪ ⎨t 1 g2 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(β1 − q1 ) ⎪ ⎩ β1 −q1 −1 (1 − s)β1 −p1 −1 , 0 ≤ t ≤ s ≤ 1. t t

(4.12)

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Therefore by (4.9) and (4.10) we obtain the following lemma. Lemma 4.1.1. If Δ1 = 0, then the function  1 α1 u(t) = G1 (t, s)ϕ1 (I0+ h(s)) ds, 0

t ∈ [0, 1]

(4.13)

is the unique solution u ∈ C[0, 1] of problem (4.3), (4.4). Next, we consider the nonlinear fractional differential equation β2 α2 D0+ (ϕr2 (D0+ v(t))) + k(t) = 0,

t ∈ (0, 1),

with the boundary conditions ⎧ ⎪ v (j) (0) = 0, j = 0, . . . , m − 2; ⎪ ⎨ M  β2 p2 q2 ⎪ D v(0) = 0, D v(1) = bi D0+ v(ηi ), ⎪ 0+ 0+ ⎩

(4.14)

(4.15)

i=1

where α2 ∈ (0, 1], β2 ∈ (m−1, m], m ∈ N, m ≥ 3, p2 , q2 ∈ R, p2 ∈ [1, β2 −1), q2 ∈ [0, p2 ], ηi , bi ∈ R for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, and k ∈ C[0, 1]. Γ(β2 ) Γ(β2 ) M β2 −q2 −1 − Γ(β , and by We denote by Δ2 = Γ(β i=1 bi ηi 2 −p2 ) 2 −q2 ) G2 , g3 , g4 the following functions: G2 (t, s) = g3 (t, s) + 1 g3 (t, s) = Γ(β2 )



M tβ2 −1  bi g4 (ηi , s), Δ2 i=1

(t, s) ∈ [0, 1] × [0, 1],

tβ2 −1 (1 − s)β2 −p2 −1 − (t − s)β2 −1 , β2 −1

(4.16)

0 ≤ s ≤ t ≤ 1,

β2 −p2 −1

(1 − s) , 0 ≤ t ≤ s ≤ 1, ⎧ β2 −q2 −1 t (1 − s)β2 −p2 −1 − (t − s)β2 −q2 −1 , ⎪ ⎨ 1 g4 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(β2 − q2 ) ⎪ ⎩ β2 −q2 −1 t (1 − s)β2 −p2 −1 , 0 ≤ t ≤ s ≤ 1. t

(4.17) In a similar manner as above, we obtain the following result. Lemma 4.1.2. If Δ2 = 0, then the function  1 α2 v(t) = G2 (t, s)ϕ2 (I0+ k(s)) ds, 0

t ∈ [0, 1]

is the unique solution v ∈ C[0, 1] of problem (4.14), (4.15).

(4.18)

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Based on the properties of the functions gi , i = 1, . . . , 4 given by (4.12) and (4.17) (see Lemma 2.1.3 from Section 2.1), we obtain the following properties of the Green functions G1 and G2 that will be used in the following sections. Lemma 4.1.3. Assume that ai , bj ≥ 0 for all i = 1, . . . , N and j = 1, . . . , M, and Δ1 , Δ2 > 0. Then the functions G1 , G2 given by (4.11) and (4.16), respectively, have the properties (a) G1 , G2 : [0, 1] × [0, 1] → [0, ∞) are continuous functions; (b) G1 (t, s) ≤ J1 (s) for all t, s ∈ [0, 1], where J1 (s) = h1 (s) + N 1 1 β1 −p1 −1 (1 − i=1 ai g2 (ξi , s), s ∈ [0, 1] and h1 (s) = Γ(β1 ) (1 − s) Δ1 p1 (1 − s) ), s ∈ [0, 1]; (c) G1 (t, s) ≥ tβ1 −1 J1 (s) for all t, s ∈ [0, 1]; (d) G2 (t, s) ≤ J2 (s) for all t, s ∈ [0, 1], where J2 (s) = h3 (s) + M 1 1 β2 −p2 −1 (1 − i=1 bi g4 (ηi , s), s ∈ [0, 1] and h3 (s) = Γ(β2 ) (1 − s) Δ2 p2 (1 − s) ), s ∈ [0, 1]; (e) G2 (t, s) ≥ tβ2 −1 J2 (s) for all t, s ∈ [0, 1]. 4.1.2

Existence of positive solutions

In this section, we present sufficient conditions on the functions f, g, and intervals for the parameters λ, μ such that positive solutions with respect to a cone for our problem (4.1), (4.2) exist. We present now the assumptions that we will use in the sequel. (H1) α1 , α2 ∈ (0, 1], β1 ∈ (n − 1, n], β2 ∈ (m − 1, m], n, m ∈ N, n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, β1 − 1), p2 ∈ [1, β2 − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], ξi ∈ R, ai ≥ 0 for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi ∈ R, bi ≥ 0 for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < Γ(β1 ) Γ(β1 ) N β1 −q1 −1 − Γ(β > 0, ηM ≤ 1, λ, μ > 0, Δ1 = Γ(β i=1 ai ξi 1 −p1 ) 1 −q1 ) M Γ(β2 ) Γ(β2 ) β2 −q2 −1 Δ2 = Γ(β2 −p2 ) − Γ(β2 −q2 ) i=1 bi ηi > 0, ri > 1, ϕri (s) = ri = ϕ ,  = , i = 1, 2. |s|ri −2 s, ϕ−1 i i ri ri −1 (H2) The functions f, g : [0, 1] × [0, ∞) × [0, ∞) → [0, ∞) are continuous. For [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, we introduce the following extreme limits: f0s = lim sup max u+v→0+

u,v≥0

t∈[0,1]

f (t, u, v) , (u + v)r1 −1

g0s = lim sup max u+v→0+

u,v≥0

t∈[0,1]

g(t, u, v) , (u + v)r2 −1

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f (t, u, v) , (u + v)r1 −1

f0i = lim inf min u+v→0+

u,v≥0

t∈[c1 ,c2 ]

s f∞ = lim sup max u+v→∞

u,v≥0

t∈[0,1]

f (t, u, v) , (u + v)r1 −1

f (t, u, v) , u+v→∞ t∈[c1 ,c2 ] (u + v)r1 −1

g0i = lim inf min u+v→0+

u,v≥0

t∈[c1 ,c2 ]

s g∞ = lim sup max u+v→∞

u,v≥0

t∈[0,1]

g(t, u, v) , (u + v)r2 −1

g(t, u, v) , (u + v)r2 −1

g(t, u, v) . u+v→∞ t∈[c1 ,c2 ] (u + v)r2 −1

i = lim inf min f∞

i g∞ = lim inf min

u,v≥0

u,v≥0

By using Lemmas 4.1.1 and 4.1.2 (relations (4.13) and (4.18)), (u, v) is a solution of the following nonlinear system of integral equations ⎧  1 ⎪ α1 1 −1 ⎪ ⎪ u(t) = λ G1 (t, s)ϕ1 (I0+ f (s, u(s), v(s))) ds, t ∈ [0, 1], ⎨ 0  1 ⎪ ⎪ α2 ⎪ ⎩v(t) = μ2 −1 G2 (t, s)ϕ2 (I0+ g(s, u(s), v(s))) ds, t ∈ [0, 1], 0

if and only if (u, v) is solution of problem (4.1), (4.2). We consider the Banach space X = C[0, 1] with the supremum norm · , and the Banach space Y = X ×X with the norm (u, v) Y = u + v . We define the cones P1 = {u ∈ X, u(t) ≥ tβ1 −1 u , ∀t ∈ [0, 1]} ⊂ X, P2 = {v ∈ X, v(t) ≥ tβ2 −1 v , ∀t ∈ [0, 1]} ⊂ X, and P = P1 × P2 ⊂ Y . We define now the operators Q1 , Q2 : Y → X and Q : Y → Y by  1 α1 G1 (t, s)ϕ1 (I0+ f (s, u(s), v(s))) ds, t ∈ [0, 1], Q1 (u, v)(t) = λ1 −1 0

Q2 (u, v)(t) = μ2 −1



0

1

α2 G2 (t, s)ϕ2 (I0+ g(s, u(s), v(s))) ds,

t ∈ [0, 1],

and Q(u, v) = (Q1 (u, v), Q2 (u, v)), (u, v) ∈ Y . Then (u, v) is a solution of problem (4.1), (4.2) if and only if (u, v) is a fixed point of operator Q. Lemma 4.1.4. If (H1)–(H2) hold, then Q : P → P is a completely continuous operator. Proof. Let (u, v) ∈ P be an arbitrary element. Because Q1 (u, v) and Q2 (u, v) satisfy problem (4.3), (4.4) for h(t) = λf (t, u(t), v(t)), t ∈ [0, 1],

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and problem (4.14), (4.15) for k(t) = μg(t, u(t), v(t)), t ∈ [0, 1], respectively, then we obtain  1 α1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds, t ∈ [0, 1], Q1 (u, v)(t) ≤ λ 0

Q2 (u, v)(t) ≤ μ2 −1



1

0

α2 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds,

and so Q1 (u, v) ≤ λ1 −1 Q2 (u, v) ≤ μ2 −1



1

0



0

1

t ∈ [0, 1],

α1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds, α2 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds.

Therefore, we conclude that  1 α1 1 −1 Q1 (u, v)(t) ≥ λ tβ1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds 0

≥t

β1 −1

≥t

β2 −1

Q1 (u, v) , ∀t ∈ [0, 1],  1 α2 Q2 (u, v)(t) ≥ μ2 −1 tβ2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds 0

Q2 (u, v) ,

∀t ∈ [0, 1].

Hence, Q(u, v) = (Q1 (u, v), Q2 (u, v)) ∈ P , and then Q(P ) ⊂ P . By the continuity of the functions f, g, G1 , G2 and the Arzela–Ascoli theorem, we can show that Q1 and Q2 are completely continuous operators, and then Q is a completely continuous operator.  For [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, we denote by  c2 1 A= (s − c1 )α1 (1 −1) J1 (s) ds, (Γ(α1 + 1))1 −1 c1  1 1 B= sα1 (1 −1) J1 (s) ds, (Γ(α1 + 1))1 −1 0  c2 1 (s − c1 )α2 (2 −1) J2 (s) ds, C= (Γ(α2 + 1))2 −1 c1  1 1 D= sα2 (2 −1) J2 (s) ds, (Γ(α2 + 1))2 −1 0 where J1 and J2 are defined in Lemma 4.1.3.

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i i First, for f0s , g0s , f∞ , g∞ ∈ (0, ∞) and numbers α1 , α2 ≥ 0, α

1 , α

2 > 0  

1 + α

2 = 1, we define the numbers such that α1 + α2 = 1 and α L1 , L2 , L3 , L4 , L2 , L4 by

  r1 −1  r1 −1   r2 −1 α1

1 α2 1 1 α 1 , L2 = s , L3 = i , L1 = i f∞ γγ1 A f0 B g∞ γγ2 C  r2 −1

2 1 α 1 1 , L2 = s r1 −1 , L4 = s r2 −1 , L4 = s g0 D f0 B g0 D where γ1 = cβ1 1 −1 , γ2 = cβ1 2 −1 , γ = min{γ1 , γ2 }. Theorem 4.1.1. Assume that (H1) and (H2) hold, [c1 , c2 ] ⊂ [0, 1] with

1 , α

2 > 0 such that α1 + α2 = 1, α

1 + α

2 = 1. 0 < c1 < c2 ≤ 1, α1 , α2 ≥ 0, α i i (1) If f0s , g0s , f∞ , g∞ ∈ (0, ∞), L1 < L2 and L3 < L4 , then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ ∈ (0, ∞) and L3 < L4 , then for each λ ∈ (L1 , ∞) (2) If f0s = 0, g0s , f∞  and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ ∈ (0, ∞) and L1 < L2 , then for each λ ∈ (L1 , L2 ) (3) If g0s = 0, f0s , f∞ and μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ ∈ (0, ∞), then for each λ ∈ (L1 , ∞) and (4) If f0s = g0s = 0, f∞ μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ is ∞, then for each (5) If f0s , g0s ∈ (0, ∞) and at least one of f∞ λ ∈ (0, L2 ) and μ ∈ (0, L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ is ∞, then for each (6) If f0s = 0, g0s ∈ (0, ∞) and at least one of f∞  λ ∈ (0, ∞) and μ ∈ (0, L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ is ∞, then for each (7) If f0s ∈ (0, ∞), g0s = 0 and at least one of f∞  λ ∈ (0, L2 ) and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). i i , g∞ is ∞, then for each λ ∈ (0, ∞) (8) If f0s = g0s = 0 and at least one of f∞ and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2).

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Proof. We consider the above cone P ⊂ Y and the operators Q1 , Q2 and Q. Because the proofs of the above cases are similar, in what follows, we will prove some representative cases. i i Case (1). We have f0s , g0s , f∞ , g∞ ∈ (0, ∞), L1 < L2 and L3 < L4 . Let i , λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ). We consider ε > 0 such that ε < f∞ i ε < g∞ and

1 i −ε f∞ 1 i g∞ − ε

 

α1 γγ1 A α2 γγ2 C

r1 −1 r2 −1

1 ≤λ≤ s f0 + ε 1 ≤μ≤ s g0 + ε

 

α

1 B α

2 D

r1 −1 r2 −1

, .

By using (H2) and the definitions of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ (f0s + ε)(u + v)r1 −1 ,

g(t, u, v) ≤ (g0s + ε)(u + v)r2 −1 ,

for all t ∈ [0, 1] and u, v ≥ 0, u + v ≤ R1 . We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. Now, let (u, v) ∈ P ∩ ∂Ω1 , that is (u, v) ∈ P with (u, v) Y = R1 , or equivalently u + v = R1 . Then u(t) + v(t) ≤ R1 for all t ∈ [0, 1] and by Lemma 4.1.3, we obtain 



1

1 J1 (s)ϕ1 Q1 (u, v)(t) ≤ λ Γ(α 1) 0 × f (τ, u(τ ), v(τ )) dτ ds 1 −1



1



1 Γ(α1 ) 0 r1 −1 dτ ds × (u(τ ) + v(τ ))

≤ λ1 −1

1 −1

≤λ

(f0s

J1 (s)ϕ1

1 −1

+ ε)



1

0 r1 −1

× ( u + v )

dτ



s

0

 0

s

(s − τ )α1 −1

(s − τ )α1 −1 (f0s + ε)

 J1 (s)ϕ1

ds

1 Γ(α1 )

 0

s

(s − τ )α1 −1

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1 −1

=λ

(f0s 

× ϕ1

1 −1

+ ε)

1 Γ(α1 )



s

0

 ( u + v )

J1 (s)

0

(s − τ )α1 −1 dτ

= λ1 −1 (f0s + ε)1 −1 (u, v) Y

1



ds 

1

J1 (s)ϕ1

0

sα1 α1 Γ(α1 )

ds

= λ1 −1 (f0s + ε)1 −1 (u, v) Y  1 1 × J1 (s) sα1 (1 −1) ds (Γ(α + 1))1 −1 1 0 = λ1 −1 (f0s + ε)1 −1 B (u, v) Y ≤ α

1 (u, v) Y ,

∀t ∈ [0, 1].

Therefore, we have Q1 (u, v) ≤ α

1 (u, v) Y . In a similar manner, we conclude 



1

1 J2 (s)ϕ2 Q2 (u, v)(t) ≤ μ Γ(α2 ) 0 × g(τ, u(τ ), v(τ )) dτ ds 2 −1





1

1 ≤μ J2 (s)ϕ2 Γ(α 2) 0 × (u(τ ) + v(τ ))r2 −1 dτ ds 2 −1

≤ μ2 −1 (g0s + ε)2 −1



1

0

× ( u + v )r2 −1 dτ



s 0



s 0

× ϕ2 =μ

2 −1

1 Γ(α2 )

(g0s

 0

s





+ ε)

1 Γ(α2 )



s

0

(s − τ )α2 −1

ds

α2 −1

(s − τ )

2 −1

(s − τ )α2 −1 (g0s + ε)

J2 (s)ϕ2

= μ2 −1 (g0s + ε)2 −1 ( u + v ) 

(s − τ )α2 −1

1

0

J2 (s)



dτ

 (u, v) Y



0

1

ds 

J2 (s)ϕ2

sα2 α2 Γ(α2 )

ds

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165

= μ2 −1 (g0s + ε)2 −1 (u, v) Y  1 1 × J2 (s) sα2 (2 −1) ds (Γ(α2 + 1))2 −1 0 = μ2 −1 (g0s + ε)2 −1 D (u, v) Y ≤ α

2 (u, v) Y ,

∀t ∈ [0, 1].

Hence, we get Q2 (u, v) ≤ α

2 (u, v) Y . Therefore, for (u, v) ∈ P ∩ ∂Ω1 , we deduce Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤α

1 (u, v) Y + α

2 (u, v) Y = (u, v) Y .

(4.19)

i i Next, by the definitions of f∞ and g∞ there exists R2 > 0 such that i f (t, u, v) ≥ (f∞ − ε)(u + v)r1 −1 ,

i g(t, u, v) ≥ (g∞ − ε)(u + v)r2 −1 ,

for all t ∈ [c1 , c2 ] and u, v ≥ 0, u + v ≥ R2 . We consider R2 = max{2R1 , R2 /γ} and we define the set Ω2 = {(u, v) ∈ Y , (u, v) Y < R2 }. Then for (u, v) ∈ P ∩ ∂Ω2 , we obtain u(t) + v(t) ≥ min tβ1 −1 u + min tβ2 −1 v = cβ1 1 −1 u + cβ1 2 −1 v t∈[c1 ,c2 ]

t∈[c1 ,c2 ]

= γ1 u + γ2 v ≥ γ (u, v) Y = γR2 ≥ R2 ,

∀t ∈ [c1 , c2 ].

Then, by Lemma 4.1.3, we conclude 1 −1

Q1 (u, v)(c1 ) ≥ λ

 0

1

α1 cβ1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

≥ λ1 −1 cβ1 1 −1



c2 c1

 J1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ



1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

 s 1 J1 (s)ϕ1 (s − τ )α1 −1 ≥ Γ(α1 ) c1 c1 i r1 −1 × (f∞ − ε)(u(τ ) + v(τ )) dτ ds λ1 −1 cβ1 1 −1



c2

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 s 1 (s − τ )α1 −1 Γ(α ) 1 c1 c1 i × (f∞ − ε)(γ (u, v) Y )r1 −1 dτ ds

≥ λ1 −1 cβ1 1 −1



c2

J1 (s)ϕ1

i − ε)1 −1 γ (u, v) Y = λ1 −1 cβ1 1 −1 (f∞   s  c2 1 α1 −1 × J1 (s)ϕ1 (s − τ ) dτ ds Γ(α1 ) c1 c1 i = λ1 −1 cβ1 1 −1 (f∞ − ε)1 −1 γ (u, v) Y   c2 (s − c1 )α1 × J1 (s)ϕ1 ds α1 Γ(α1 ) c1 i − ε)1 −1 (u, v) Y = γγ1 λ1 −1 (f∞  c2 1 × J1 (s) (s − c1 )α1 (1 −1) ds (Γ(α + 1))1 −1 1 c1 i − ε)1 −1 A (u, v) Y ≥ α1 (u, v) Y . = γγ1 λ1 −1 (f∞

Therefore, we obtain Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ α1 (u, v) Y . In a similar manner, we deduce Q2 (u, v)(c1 ) ≥ μ2 −1 ≥

 0

1

α2 cβ1 2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

μ2 −1 cβ1 2 −1



c2

c1

 J2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ 

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds 

 s 1 (s − τ )α2 −1 Γ(α ) 2 c1 c1 i × (g∞ − ε)(u(τ ) + v(τ ))r2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1



c2

J2 (s)ϕ2



 s 1 (s − τ )α2 −1 Γ(α2 ) c1 c1 i r2 −1 × (g∞ − ε)(γ (u, v) Y ) dτ ds

≥ μ2 −1 cβ1 2 −1

c2

J2 (s)ϕ2

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167

i = μ2 −1 cβ1 2 −1 (g∞ − ε)2 −1 γ (u, v) Y   s  c2 1 × J2 (s)ϕ2 (s − τ )α2 −1 dτ ds Γ(α2 ) c1 c1 i = μ2 −1 cβ1 2 −1 (g∞ − ε)2 −1 γ (u, v) Y   c2 (s − c1 )α2 × J2 (s)ϕ2 ds α2 Γ(α2 ) c1 i − ε)2 −1 (u, v) Y = γγ2 μ2 −1 (g∞  c2 1 × J2 (s) (s − c1 )α2 (2 −1) ds (Γ(α2 + 1))2 −1 c1 i − ε)2 −1 C (u, v) Y ≥ α2 (u, v) Y . = γγ2 μ2 −1 (g∞

Hence, we get Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ α2 (u, v) Y . Then for (u, v) ∈ P ∩ ∂Ω2 , we obtain Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≥ (α1 + α2 ) (u, v) Y = (u, v) Y . (4.20) By using Lemma 4.1.4, (4.19), (4.20) and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2), we conclude that the operator Q has a fixed point (u, v) ∈ P ∩ (Ω2 \ Ω1 ), so u(t) ≥ tβ1 −1 u , v(t) ≥ tβ2 −1 v for all t ∈ [0, 1] and R1 ≤ u + v ≤ R2 . If u > 0 then u(t) > 0 for all t ∈ (0, 1], and if v > 0 then v(t) > 0 for all t ∈ (0, 1]. i i Case (3). We have g0s = 0, f0s , f∞ , g∞ ∈ (0, ∞) and L1 < L2 . Let λ ∈ 

1 and α

2 used in the (L1 , L2 ) and μ ∈ (L3 , ∞). Instead of the numbers α

1 ∈ (B(λf0s )1 −1 , 1) and α

2 = 1 − α

1 . first case, we choose α

1 such that α  s r1 −1 ). Then let ε > 0 The choice of the α

1 is possible because λ < 1/(f0 B i i , ε < g∞ and with ε < f∞   r1 −1   r1 −1 1 α1 α

1 1 ≤ λ ≤ , s i f∞ − ε γγ1 A f0 + ε B   r2 −1   r2 −1 α2

2 1 α 1 ≤ μ ≤ . i −ε g∞ γγ2 C ε D

By using (H2) and the definitions of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ (f0s + ε)(u + v)r1 −1 , for all t ∈ [0, 1] and u, v ≥ 0, u + v ≤ R1 .

g(t, u, v) ≤ ε(u + v)r2 −1 ,

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We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. In a similar manner as in the proof of Case (1), for any (u, v) ∈ P ∩ ∂Ω1 , we obtain Q1 (u, v)(t) ≤ λ1 −1 (f0s + ε)1 −1 B (u, v) Y ≤ α

1 (u, v) Y , Q2 (u, v)(t) ≤ μ2 −1 ε2 −1 D (u, v) Y ≤ α

2 (u, v) Y ,

∀t ∈ [0, 1],

∀t ∈ [0, 1],

and so Q(u, v) Y ≤ (

α1 + α

2 ) (u, v) Y = (u, v) Y . The second part of the proof is the same as the corresponding one from Case (1). For Ω2 defined in Case (1) and for any (u, v) ∈ P ∩ ∂Ω2 , we conclude i Q1 (u, v)(c1 ) ≥ γγ1 λ1 −1 (f∞ − ε)1 −1 A (u, v) Y ≥ α1 (u, v) Y , i Q2 (u, v)(c1 ) ≥ γγ2 μ2 −1 (g∞ − ε)2 −1 C (u, v) Y ≥ α2 (u, v) Y ,

and then Q(u, v) Y ≥ (α1 + α2 ) (u, v) Y = (u, v) Y . Therefore, we deduce the conclusion of the theorem. i Case (6). We consider here f0s = 0, g0s ∈ (0, ∞) and f∞ = ∞. Let λ ∈ 

1 and α

2 used in the first (0, ∞) and μ ∈ (0, L4 ). Instead of the numbers α case, we choose α

2 such that α

2 ∈ (D(μg0s )2 −1 , 1) and α

1 = 1 − α

2 . The  s r2 −1 ). Then let ε > 0 such choice of the α

2 is possible because μ < 1/(g0 D that  r1 −1   r1 −1   r2 −1 1

1 α

2 1 α 1 ε ≤λ≤ , μ≤ s . γγ1 A ε B g0 + ε D

By using (H2) and the definitions of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ ε(u + v)r1 −1 ,

g(t, u, v) ≤ (g0s + ε)(u + v)r2 −1 ,

for all t ∈ [0, 1] and u, v ≥ 0, u + v ≤ R1 . We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. In a similar manner as in the proof of Case (1), for any (u, v) ∈ P ∩ ∂Ω1 we obtain Q1 (u, v)(t) ≤ λ1 −1 ε1 −1 B (u, v) Y ≤ α

1 (u, v) Y ,

∀ t ∈ [0, 1],

Q2 (u, v)(t) ≤ μ2 −1 (g0s + ε)2 −1 D (u, v) Y ≤ α

2 (u, v) Y , and so Q(u, v) Y ≤ (

α1 + α

2 ) (u, v) Y = (u, v) Y .

∀t ∈ [0, 1],

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169

i For the second part of the proof, by the definition of f∞ , there exists R2 > 0 such that

f (t, u, v) ≥

1 (u + v)r1 −1 , ε

∀t ∈ [c1 , c2 ],

u, v ≥ 0,

u + v ≥ R2 .

We consider R2 = max{2R1 , R2 /γ} and we define Ω2 = {(u, v) ∈ Y , (u, v) Y < R2 }. Then for (u, v) ∈ P ∩ ∂Ω2 we deduce as in Case (1) that u(t) + v(t) ≥ γR2 ≥ R2 for all t ∈ [c1 , c2 ]. Then by Lemma 4.1.3, we have Q1 (u, v)(c1 ) ≥ λ1 −1 ≥

 0

1

α1 cβ1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

λ1 −1 cβ1 1 −1



c2 c1

 J1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ ≥ λ1 −1 cβ1 1 −1



c2 c1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

J1 (s)ϕ1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

1 × (u(τ ) + v(τ ))r1 −1 dτ ds ε   c2  s 1 1 −1 β1 −1 ≥λ c1 J1 (s)ϕ1 (s − τ )α1 −1 Γ(α1 ) c1 c1 1 r1 −1 × (γ (u, v) Y ) dτ ds ε  1 −1 1 1 −1 β1 −1 =λ c1 γ (u, v) Y ε   c2 (s − c1 )α1 × J1 (s)ϕ1 ds Γ(α1 + 1) c1  1 −1 1 γ (u, v) Y = λ1 −1 cβ1 1 −1 ε  c2 1 × J1 (s) (s − c1 )α1 (1 −1) ds (Γ(α + 1))1 −1 1 c1  1 −1 1 A (u, v) Y ≥ (u, v) Y . = γγ1 λ1 −1 ε

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So, we conclude that Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q1 (u, v) ≥ (u, v) Y . Therefore, we deduce the conclusion of the theorem. i = ∞. Let λ ∈ (0, ∞) and Case (8). We consider f0s = g0s = 0 and g∞ μ ∈ (0, ∞), and let ε > 0 such that  r2 −1 1 1 1 , ε ≤μ≤ . λ≤ ε(2B)r1 −1 γγ2 C ε(2D)r2 −1

By using (H2) and the definition of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ ε(u + v)r1 −1 ,

g(t, u, v) ≤ ε(u + v)r2 −1 ,

for all t ∈ [0, 1], u, v ≥ 0, u + v ≤ R1 . We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. In a similar manner as in the proof of Case (1), for any (u, v) ∈ P ∩ ∂Ω1 , we obtain 1 (u, v) Y , 2 1 Q2 (u, v)(t) ≤ μ2 −1 ε2 −1 D (u, v) Y ≤ (u, v) Y , 2

Q1 (u, v)(t) ≤ λ1 −1 ε1 −1 B (u, v) Y ≤

∀t ∈ [0, 1], ∀ t ∈ [0, 1],

and so Q(u, v) Y ≤ ( 12 + 12 ) (u, v) Y = (u, v) Y . i , there exists For the second part of the proof, by the definition of g∞ R2 > 0 such that g(t, u, v) ≥

1 (u + v)r2 −1 , ε

∀t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≥ R2 .

We consider R2 = max{2R1 , R2 /γ} and we define Ω2 = {(u, v) ∈ Y, (u, v) Y < R2 }. Then for (u, v) ∈ P ∩ ∂Ω2 we deduce as in Case (1) that u(t) + v(t) ≥ γR2 ≥ R2 for all t ∈ [c1 , c2 ]. Then by Lemma 4.1.3, we have  1 α2 cβ1 2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds Q2 (u, v)(c1 ) ≥ μ2 −1 0

≥

μ2 −1 cβ1 2 −1



c2

c1

 J2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ ≥ μ2 −1 cβ1 2 −1



c2

c1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds 

J2 (s)ϕ2

1 Γ(α2 )



s

c1

(s − τ )α2 −1

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171

1 (u(τ ) + v(τ ))r2 −1 dτ ds ε   s  c2 1 ≥ μ2 −1 cβ1 2 −1 J2 (s)ϕ2 (s − τ )α2 −1 Γ(α2 ) c1 c1 1 r2 −1 × (γ (u, v) Y ) dτ ds ε  2 −1 1 γ (u, v) Y = μ2 −1 cβ1 2 −1 ε   c2 (s − c1 )α2 × J2 (s)ϕ2 ds Γ(α2 + 1) c1  2 −1 1 γ (u, v) Y = μ2 −1 cβ1 2 −1 ε  c2 1 × J2 (s) (s − c1 )α2 (2 −1) ds (Γ(α2 + 1))2 −1 c1  2 −1 1 C (u, v) Y ≥ (u, v) Y . = γγ2 μ2 −1 ε ×

Then we conclude that Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q2 (u, v) ≥ (u, v) Y . Therefore, we deduce the conclusion of the theorem.  s s , g∞ ∈ (0, ∞) and numbers α1 , α2 ≥ 0, In what follows, for f0i , g0i , f∞  

2 > 0 such that α1 + α2 = 1 and α

1 + α

2 = 1, we define the numbers α

1 , α

1 , L

2, L

3 , L

4, L

 and L

 by L 2

4

 r1 −1   r2 −1 r1 −1 α1 α

1 α2

2 = 1

3 = 1 , L , L , s γγ1 A f∞ B g0i γγ2 C  r2 −1 α

2 1 1 1

 =

 = L4 = s , L , L . 2 4 s B r1 −1 s D r2 −1 g∞ D f∞ g∞

1 = 1 L f0i



Theorem 4.1.2. Assume that (H1) and (H2) hold, [c1 , c2 ] ⊂ [0, 1] with

1 , α

2 > 0 such that α1 + α2 = 1, α

1 + α

2 = 1. 0 < c1 < c2 ≤ 1, α1 , α2 ≥ 0, α s s

1 < L

2 and L

3 < L

4 , then for each λ ∈ , g∞ ∈ (0, ∞), L (1) If f0i , g0i , f∞





(L1 , L2 ) and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2).

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s s

1 < L

 , then for each λ ∈ (L

1, L

 ) (2) If f0i , g0i , f∞ ∈ (0, ∞), g∞ = 0 and L 2 2

3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] and μ ∈ (L for (4.1), (4.2). s s

3 < L

4 , then for each λ ∈ (L

1 , ∞) ∈ (0, ∞), f∞ = 0 and L (3) If f0i , g0i , g∞ 



and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). s s

1 , ∞) and = g∞ = 0, then for each λ ∈ (L (4) If f0i , g0i ∈ (0, ∞), f∞

μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2). s s , g∞ ∈ (0, ∞) and at least one of f0i , g0i is ∞, then for each (5) If f∞

4 ) there exists a positive solution (u(t), v(t)),

2 ) and μ ∈ (0, L λ ∈ (0, L t ∈ [0, 1] for (4.1), (4.2). s s ∈ (0, ∞), g∞ = 0 and at least one of f0i , g0i is ∞, then for each (6) If f∞ 

) and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), λ ∈ (0, L 2 t ∈ [0, 1] for (4.1), (4.2). s s = 0, g∞ ∈ (0, ∞) and at least one of f0i , g0i is ∞, then for each (7) If f∞

4 ) there exists a positive solution (u(t), v(t)), λ ∈ (0, ∞) and μ ∈ (0, L t ∈ [0, 1] for (4.1), (4.2). s s = g∞ = 0 and at least one of f0i , g0i is ∞, then for each λ ∈ (0, ∞) (8) If f∞ and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.1), (4.2).

Proof. We consider the cone P ⊂ Y and the operators Q1 , Q2 and Q defined at the beginning of this section. Because the proofs of the above cases are similar, in what follows, we will prove some representative cases. s s

1 < L

2 and L

3 < L

4 . Let , g∞ ∈ (0, ∞), L Case (1). We have f0i , g0i , f∞ i





λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ). We consider ε > 0 such that ε < f0 , ε < g0i and   r1 −1  r1 −1 1 1 α1 α

1 ≤λ≤ s , f0i − ε γγ1 A f∞ + ε B   r2 −1  r2 −1 α2 α

2 1 1 ≤μ≤ s . i γγ C g + ε D g0 − ε 2 ∞

By using (H2) and the definitions of f0i and g0i , we deduce that there exists R3 > 0 such that f (t, u, v) ≥ (f0i − ε)(u + v)r1 −1 ,

g(t, u, v) ≥ (g0i − ε)(u + v)r2 −1 ,

for all t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≤ R3 .

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173

We denote by Ω3 = {(u, v) ∈ Y, (u, v) Y < R3 }. Let (u, v) ∈ P with (u, v) Y = R3 , that is u + v = R3 . Because u(t)+v(t) ≤ u + v = R3 for all t ∈ [0, 1], then by Lemma 4.1.3, we obtain  1 α1 1 −1 Q1 (u, v)(c1 ) ≥ λ cβ1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds 0

≥ λ1 −1 cβ1 1 −1



c2 c1

 J1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ 

1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

 s 1 (s − τ )α1 −1 Γ(α ) 1 c1 c1 × (f0i − ε)(u(τ ) + v(τ ))r1 −1 dτ ds

≥ λ1 −1 cβ1 1 −1

c2



J1 (s)ϕ1



 s 1 ≥ J1 (s)ϕ1 (s − τ )α1 −1 Γ(α1 ) c1 c1 × (f0i − ε)(γ (u, v) Y )r1 −1 dτ ds λ1 −1 cβ1 1 −1

c2

= λ1 −1 cβ1 1 −1 (f0i − ε)1 −1 γ (u, v) Y   c2  s 1 × J1 (s)ϕ1 (s − τ )α1 −1 dτ ds Γ(α1 ) c1 c1 = γγ1 λ1 −1 (f0i − ε)1 −1 A (u, v) Y ≥ α1 (u, v) Y . Therefore, we conclude Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ α1 (u, v) Y . In a similar manner, we deduce  1 α2 cβ1 2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds Q2 (u, v)(c1 ) ≥ μ2 −1 0

≥

μ2 −1 cβ1 2 −1



c2

c1

 J2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ 



1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds 

 s 1 ≥ J2 (s)ϕ2 (s − τ )α2 −1 Γ(α2 ) c1 c1 × (g0i − ε)(u(τ ) + v(τ ))r2 −1 dτ ds μ2 −1 cβ1 2 −1

c2

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 s 1 (s − τ )α2 −1 Γ(α ) 2 c1 c1 × (g0i − ε)(γ (u, v) Y )r2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1



c2

J2 (s)ϕ2

= μ2 −1 cβ1 2 −1 (g0i − ε)2 −1 γ (u, v) Y   s  c2 1 × J2 (s)ϕ2 (s − τ )α2 −1 dτ ds Γ(α2 ) c1 c1 = γγ2 μ2 −1 (g0i − ε)2 −1 C (u, v) Y ≥ α2 (u, v) Y . Hence, we get Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ α2 (u, v) Y . Then for (u, v) ∈ P ∩ ∂Ω3 , we obtain Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≥ (α1 + α2 ) (u, v) Y = (u, v) Y . (4.21) Now, we define the functions f ∗ , g ∗ : [0, 1] × [0, ∞) → [0, ∞) by f ∗ (t, x) = max0≤u+v≤x f (t, u, v), g ∗ (t, x) = max0≤u+v≤x g(t, u, v), for all t ∈ [0, 1] and x ∈ [0, ∞). Then f (t, u, v) ≤ f ∗ (t, x),

g(t, u, v) ≤ g ∗ (t, x),

∀t ∈ [0, 1], u, v ≥ 0, u + v ≤ x.

The functions f ∗ (t, ·), g ∗ (t, ·) are nondecreasing for every t ∈ [0, 1] and they satisfy the conditions (see Section 4.1.5) f ∗ (t, x) s = f∞ , t∈[0,1] xr1 −1

lim sup max x→∞

g ∗ (t, x) s = g∞ . t∈[0,1] xr2 −1

lim sup max x→∞

Therefore, for ε > 0 there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1], we have f ∗ (t, x) f ∗ (t, x) s ≤ lim sup max + ε = f∞ + ε, r −1 xr1 −1 x→∞ t∈[0,1] x 1 g ∗ (t, x) g ∗ (t, x) s ≤ lim sup max + ε = g∞ + ε, r −1 xr2 −1 x→∞ t∈[0,1] x 2 s s and so f ∗ (t, x) ≤ (f∞ + ε)xr1 −1 and g ∗ (t, x) ≤ (g∞ + ε)xr2 −1 .

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175

We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y, (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definitions of f ∗ and g ∗ , we conclude f (t, u(t), v(t)) ≤ f ∗ (t, (u, v) Y ), g(t, u(t), v(t)) ≤ g ∗ (t, (u, v) Y ),

∀t ∈ [0, 1].

(4.22)

Then for all t ∈ [0, 1], we obtain 

1



1 Γ(α1 ) 0 × f (τ, u(τ ), v(τ )) dτ ds

Q1 (u, v)(t) ≤ λ1 −1





1

1 ≤λ J1 (s)ϕ1 Γ(α1 ) 0 ∗ × f (τ, (u, v) Y ) dτ ds 1 −1



1



J1 (s)ϕ1

s

0



s

0

(s − τ )α1 −1

(s − τ )α1 −1



1 s (f∞ + ε) (u, v) rY1 −1 Γ(α ) 1 0  s × (s − τ )α1 −1 dτ ds 1 −1

≤λ

J1 (s)ϕ1

0

1 −1

=λ

s (f∞ + ε)1 −1 B (u, v) Y ≤ α

1 (u, v) Y ,

1 (u, v) Y . and so Q1 (u, v) ≤ α In a similar manner, we deduce 

1



1 Γ(α2 ) 0 × g(τ, u(τ ), v(τ )) dτ ds

Q2 (u, v)(t) ≤ μ

2 −1



1

J2 (s)ϕ2



1 ≤μ J2 (s)ϕ2 Γ(α 2) 0 × g ∗ (τ, (u, v) Y ) dτ ds 2 −1



s

0

 0

s

(s − τ )α2 −1

(s − τ )α2 −1

∀t ∈ [0, 1],

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1



1 s (g∞ + ε) (u, v) rY2 −1 Γ(α 2) 0  s × (s − τ )α2 −1 dτ ds

≤μ

2 −1

=μ

2 −1

J2 (s)ϕ2

0

s (g∞ + ε)2 −1 D (u, v) Y ≤ α

2 (u, v) Y ,

∀ t ∈ [0, 1],

and then Q2 (u, v) ≤ α

2 (u, v) Y . Therefore, for (u, v) ∈ P ∩ ∂Ω4 , it follows that α1 + α

2 ) (u, v) Y = (u, v) Y . Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤ (

(4.23) By using Lemma 4.1.4, (4.21), (4.23) and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2), we conclude that Q has a fixed point (u, v) ∈ P ∩ (Ω4 \ Ω3 ). s s

3 < L

 . Let λ ∈ = 0, f0i , g0i , g∞ ∈ (0, ∞) and L Case (3). We have f∞ 4   s 2 −1





2 ∈ (D(μg∞ ) , 1) and α

1 = 1−

α2. (L1 , ∞) and μ ∈ (L3 , L4 ). We choose α i i Let ε > 0 with ε < f0 , ε < g0 and

1 i f0 − ε 1 g0i − ε

 

α1 γγ1 A α2 γγ2 C

r1 −1 r2 −1

1 ≤λ≤ ε ≤μ≤



α

1 B

r1 −1

1 s +ε g∞



α

2 D

, r2 −1

.

The first part of the proof is the same as the corresponding one from Case (1). For Ω3 defined in Case (1), for (u, v) ∈ P ∩ ∂Ω3 , we obtain Q1 (u, v)(c1 ) ≥ γγ1 λ1 −1 (f0i − ε)1 −1 A (u, v) Y ≥ α1 (u, v) Y , Q2 (u, v)(c1 ) ≥ γγ2 μ2 −1 (g0i − ε)2 −1 C (u, v) Y ≥ α2 (u, v) Y and so Q(u, v) Y ≥ (α1 + α2 ) (u, v) Y = (u, v) Y . For the second part, we use the same functions f ∗ and g ∗ from Case (1) which satisfy in this case the conditions f ∗ (t, x) = 0, x→∞ t∈[0,1] xr1 −1 lim max

g ∗ (t, x) s = g∞ . t∈[0,1] xr2 −1

lim sup max x→∞

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Therefore, for ε > 0, there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1], we have f ∗ (t, x) f ∗ (t, x) ≤ lim max + ε = ε, r −1 x→∞ t∈[0,1] xr1 −1 x1 g ∗ (t, x) g ∗ (t, x) s ≤ lim sup max + ε = g∞ + ε, r −1 xr2 −1 x→∞ t∈[0,1] x 2 s + ε)xr2 −1 . and so f ∗ (t, x) ≤ εxr1 −1 and g ∗ (t, x) ≤ (g∞ We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y, (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definitions of f ∗ and g ∗ , we obtain the relations (4.22). In addition, in a similar manner as in the proof of Case (1), we conclude

1 (u, v) Y , Q1 (u, v)(t) ≤ λ1 −1 ε1 −1 (u, v Y B ≤ α

∀t ∈ [0, 1],

s Q2 (u, v)(t) ≤ μ2 −1 (g∞ + ε)2 −1 (u, v) Y D ≤ α

2 (u, v) Y ,

∀t ∈ [0, 1],

α1 + α

2 ) (u, v Y = (u, v) Y . and so Q(u, v) Y ≤ (

Therefore, we deduce the conclusion of the theorem. s s = 0, f∞ ∈ (0, ∞) and g0i = ∞. Let λ ∈ Case (6). We consider here g∞   s 1 −1

1 ∈ (B(λf∞ ) , 1) and α

2 = 1 − α

1 , (0, L2 ) and μ ∈ (0, ∞). We choose α and let ε > 0 such that

1 λ≤ s f∞ + ε



α

1 B

r1 −1

 ,

ε

1 γγ2 C

r2 −1

1 ≤μ≤ ε



α

2 D

r2 −1

.

By (H2) and the definition of g0i , we deduce that there exists R3 > 0 such that g(t, u, v) ≥

1 (u + v)r2 −1 , ε

∀t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≤ R3 .

We denote by Ω3 = {(u, v) ∈ Y, (u, v) Y < R3 }. Let (u, v) ∈ P with (u, v) Y = R3 , that is u + v = R3 . Because u(t)+v(t) ≤ u + v = R3

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for all t ∈ [0, 1], then by Lemma 4.1.3, we obtain Q2 (u, v)(c1 ) ≥ μ ≥

2 −1

 0

1

α2 cβ1 2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

μ2 −1 cβ1 2 −1



c2

c1

 J2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ ≥ μ2 −1 cβ1 2 −1



c2

c1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds 

J2 (s)ϕ2

1 Γ(α2 )



s

c1

(s − τ )α2 −1

1 (u(τ ) + v(τ ))r2 −1 dτ ds ε   c2  s 1 ≥ μ2 −1 cβ1 2 −1 J2 (s)ϕ2 (s − τ )α2 −1 Γ(α2 ) c1 c1 1 × (γ (u, v) Y )r2 −1 dτ ds ε  2 −1 1 = μ2 −1 cβ1 2 −1 γ (u, v) Y ε  c2 1 × J2 (s) (s − c1 )α2 (2 −1) ds 2 −1 (Γ(α + 1)) 2 c1  2 −1 1 = γγ2 μ2 −1 C (u, v) Y ≥ (u, v) Y . ε ×

Hence, we get Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q2 (u, v) ≥ (u, v) Y . For the second part of the proof, we consider the functions f ∗ and g ∗ from Case (1), which satisfy in this case the conditions f ∗ (t, x) s = f∞ , t∈[0,1] xr1 −1

lim sup max x→∞

g ∗ (t, x) = 0. x→∞ t∈[0,1] xr2 −1 lim max

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Then for ε > 0, there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1] we have f ∗ (t, x) f ∗ (t, x) s ≤ lim sup max + ε = f∞ + ε, r −1 xr1 −1 x→∞ t∈[0,1] x 1 g ∗ (t, x) g ∗ (t, x) ≤ lim max + ε = ε, x→∞ t∈[0,1] xr2 −1 xr2 −1 s + ε)xr1 −1 and g ∗ (t, x) ≤ εxr2 −1 . and so f ∗ (t, x) ≤ (f∞ We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y, (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definitions of f ∗ and g ∗ , we deduce the relations (4.22). In addition, in a similar manner as in the proof of Case (1), we conclude s Q1 (u, v)(t) ≤ λ1 −1 (f∞ + ε)1 −1 (u, v) Y B ≤ α

1 (u, v) Y ,

Q2 (u, v)(t) ≤ μ2 −1 ε2 −1 (u, v) Y D ≤ α

2 (u, v) Y ,

∀t ∈ [0, 1],

∀t ∈ [0, 1],

α1 + α

2 ) (u, v Y = (u, v) Y . and so Q(u, v) Y ≤ (

Therefore, we obtain the conclusion of the theorem. s s = g∞ = 0 and f0i = ∞. Let λ ∈ (0, ∞) and Case (8). We consider f∞ μ ∈ (0, ∞). We choose ε > 0 such that

 ε

1 γγ1 A

r1 −1

≤λ≤

1 , ε(2B)r1 −1

μ≤

1 . ε(2D)r2 −1

By using (H2) and the definition of f0i , we deduce that there exists R3 > 0 such that f (t, u, v) ≥

1 (u + v)r1 −1 , ε

∀t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≤ R3 .

We denote by Ω3 = {(u, v) ∈ Y, (u, v) Y < R3 }. Let (u, v) ∈ P with (u, v) Y = R3 , that is u + v = R3 . Because u(t)+v(t) ≤ u + v = R3

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for all t ∈ [0, 1], then by Lemma 4.1.3, we obtain Q1 (u, v)(c1 ) ≥ λ1 −1 ≥

 0

1

α1 cβ1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

λ1 −1 cβ1 1 −1



c2 c1

 J1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ ≥ λ1 −1 cβ1 1 −1



c2 c1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

J1 (s)ϕ1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

1 (u(τ ) + v(τ ))r1 −1 dτ ds ε   c2  s 1 ≥ λ1 −1 cβ1 1 −1 J1 (s)ϕ1 (s − τ )α1 −1 Γ(α1 ) c1 c1 1 r1 −1 × (γ (u, v) Y ) dτ ds ε  1 −1 1 = λ1 −1 cβ1 1 −1 γ (u, v) Y ε  c2 1 × J1 (s) (s − c1 )α1 (1 −1) ds (Γ(α + 1))1 −1 1 c1  1 −1 1 A (u, v) Y ≥ (u, v) Y . = γγ1 λ1 −1 ε ×

Hence, we get Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q1 (u, v) ≥ (u, v) Y . For the second part of the proof, we consider the functions f ∗ and g ∗ from Case (1), which satisfy in this case the conditions f ∗ (t, x) = 0, x→∞ t∈[0,1] xr1 −1 lim max

g ∗ (t, x) = 0. x→∞ t∈[0,1] xr2 −1 lim max

Then for ε > 0 there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1] we have f ∗ (t, x) f ∗ (t, x) ≤ lim max r1 −1 + ε = ε, r −1 1 x→∞ t∈[0,1] x x

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g ∗ (t, x) g ∗ (t, x) ≤ lim max r2 −1 + ε = ε, r −1 2 x→∞ t∈[0,1] x x and so f ∗ (t, x) ≤ εxr1 −1 and g ∗ (t, x) ≤ εxr2 −1 . We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y, (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definitions of f ∗ and g ∗ , we obtain the relations (4.22). In addition, in a similar manner as in the proof of Case (1), we deduce 1 (u, v) Y , 2 1 Q2 (u, v)(t) ≤ μ2 −1 ε2 −1 (u, v) Y D ≤ (u, v) Y , 2

1 1 and so Q(u, v) Y ≤ 2 + 2 (u, v Y = (u, v) Y . Therefore, we obtain the conclusion of the theorem. Q1 (u, v)(t) ≤ λ1 −1 ε1 −1 (u, v) Y B ≤

4.1.3

∀t ∈ [0, 1], ∀t ∈ [0, 1],



Nonexistence of positive solutions

In this section, we present intervals for λ and μ for which there exist no positive solutions of problem (4.1), (4.2). Theorem 4.1.3. Assume that (H1) and (H2) hold. If there exist positive numbers M1 , M2 such that f (t, u, v) ≤ M1 (u + v)r1 −1 , g(t, u, v) ≤ M2 (u + v)r2 −1 ,

∀t ∈ [0, 1], u, v ≥ 0,

(4.24)

then there exist positive constants λ0 and μ0 such that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ) the boundary value problem (4.1), (4.2) has no positive solution. Proof. We define λ0 =

1 M1 (2B)r1 −1

and μ0 =

1 B= (Γ(α1 + 1))1 −1 1 D= (Γ(α2 + 1))2 −1



1

0



0

1

1 , M2 (2D)r2 −1

where

sα1 (1 −1) J1 (s) ds, sα2 (2 −1) J2 (s) ds.

We will prove that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ), the problem (4.1), (4.2) has no positive solution.

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Let λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ). We suppose that the problem (4.1), (4.2) has a positive solution (u(t), v(t)), t ∈ [0, 1]. Then we obtain u(t) = Q1 (u, v)(t) = λ1 −1 ≤ λ1 −1



1

0





 J1 (s)ϕ1 

1

0

α1 G1 (t, s)ϕ1 (I0+ f (s, u(s), v(s))) ds

1 Γ(α1 )



s

0

(s − τ )α1 −1 f (τ, u(τ ), v(τ )) dτ

ds

 s 1 (s − τ )α1 −1 Γ(α1 ) 0 0 r1 −1 × M1 (u(τ ) + v(τ )) dτ ds

≤ λ1 −1

1 −1

≤λ

1



1

0

J1 (s)ϕ1

 J1 (s)ϕ1 (M1 )ϕ1 r1 −1

× ( u + v )

dτ

1 Γ(α1 )



s

0

(s − τ )α1 −1

ds

= λ1 −1 M11 −1 B (u, v) Y , ∀t ∈ [0, 1] and v(t) = Q2 (u, v)(t) = μ ≤μ

2 −1



1

0

2 −1



 J2 (s)ϕ2 

1

0

α2 G2 (t, s)ϕ2 (I0+ g(s, u(s), v(s))) ds

1 Γ(α2 )



s

0

α2 −1

(s − τ )

g(τ, u(τ ), v(τ )) dτ

 s 1 ≤μ J2 (s)ϕ2 (s − τ )α2 −1 Γ(α2 ) 0 0 × M2 (u(τ ) + v(τ ))r2 −1 dτ ds 2 −1

≤μ

2 −1



1



1

0

 J2 (s)ϕ2 (M2 )ϕ2

× ( u + v )r2 −1 dτ



1 Γ(α2 )

 0

ds

= μ2 −1 M22 −1 D (u, v) Y ,

∀t ∈ [0, 1].

s

(s − τ )α2 −1

ds

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183

Then we deduce 1 (u, v) Y , 2 1 v ≤ μ2 −1 M22 −1 D (u, v) Y < μ02 −1 M22 −1 D (u, v) Y = (u, v) Y , 2 and so (u, v) Y = u + v < (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.1), (4.2) has no positive solution.  u ≤ λ1 −1 M11 −1 B (u, v) Y < λ01 −1 M11 −1 B (u, v) Y =

Remark 4.1.1. In the proof Theorem 4.1.3, we can also define λ0 =

α1 r1 −1

α2 ofr2 −1 1 1 and μ = with α1 , α2 > 0 and α1 + α2 = 1. 0 M1 B M2 D s s Remark 4.1.2. If f0s , g0s , f∞ , g∞ < ∞, then there exist positive constants M1 , M2 such that relation (4.24) holds, and then we obtain the conclusion of Theorem 4.1.3.

Theorem 4.1.4. Assume that (H1) and (H2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m1 > 0 such that f (t, u, v) ≥ m1 (u + v)r1 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.25)

0 such that for every λ > λ

0 and then there exists a positive constant λ μ > 0, the boundary value problem (4.1), (4.2) has no positive solution.  c2 1 1

0 = , where A = (Γ(α1 +1)) (s − Proof. We define λ 1 −1 c1 m1 (γγ1 A)r1 −1

c1 )α1 (1 −1) J1 (s) ds.

0 and μ > 0, the problem (4.1), (4.2) We will show that for every λ > λ has no positive solution.

0 and μ > 0. We suppose that the problem (4.1), (4.2) has a Let λ > λ positive solution (u(t), v(t)), t ∈ [0, 1]. Then we obtain  1 α1 1 −1 G1 (c1 , s)ϕ1 (I0+ f (s, u(s), v(s))) ds u(c1 ) = Q1 (u, v)(c1 ) = λ ≥ λ1 −1

 0

0

1

α1 cβ1 1 −1 J1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

≥ λ1 −1 cβ1 1 −1



c2

c1

 J1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ

ds

1 Γ(α1 )



s

c1

(s − τ )α1 −1

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c2



1 Γ(α 1) c1 × m1 (u(τ ) + v(τ ))r1 −1 dτ ds

≥ λ1 −1 cβ1 1 −1

≥



c2

J1 (s)ϕ1



1 Γ(α1 ) c1 × m1 γ r1 −1 (u, v) rY1 −1 dτ ds λ1 −1 cβ1 1 −1

J1 (s)ϕ1



s

c1



s

c1

(s − τ )α1 −1

(s − τ )α1 −1

= λ1 −1 γγ1 m11 −1 (u, v) Y  c2 1 × J1 (s) (s − c1 )α1 (1 −1) ds (Γ(α1 + 1))1 −1 c1 = γγ1 λ1 −1 m11 −1 A (u, v) Y . Then we conclude u ≥ u(c1 ) ≥ γγ1 A(λm1 )1 −1 (u, v) Y

0 m1 )1 −1 (u, v) Y = (u, v) Y , > γγ1 A(λ and so (u, v) Y = u + v ≥ u > (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.1), (4.2) has no positive solution.  Theorem 4.1.5. Assume that (H1) and (H2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m2 > 0 such that g(t, u, v) ≥ m2 (u + v)r2 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.26)

0 and then there exists a positive constant μ

0 such that for every μ > μ λ > 0, the boundary value problem (4.1), (4.2) has no positive solution. Proof. We define μ

0 =

1 , m2 (γγ2 C)r2 −1

where C =

1 (Γ(α2 +1))2 −1

 c2 c1

(s −

c1 )α2 (2 −1) J2 (s) ds. We will show that for every μ > μ

0 and λ > 0, the problem (4.1), (4.2) has no positive solution.

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185

Let μ > μ

0 and λ > 0. We suppose that the problem (4.1), (4.2) has a positive solution (u(t), v(t)), t ∈ [0, 1]. Then we obtain v(c1 ) = Q2 (u, v)(c1 ) = μ2 −1 ≥ μ2 −1 ≥

 0

1



1

0

α1 G2 (c1 , s)ϕ2 (I0+ g(s, u(s), v(s))) ds

α2 cβ1 2 −1 J2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

μ2 −1 cβ1 2 −1



c2

c1

 J2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ 

c2

1 Γ(α2 )





c2

J2 (s)ϕ2



1 Γ(α2 ) c1 × m2 γ r2 −1 (u, v) rY2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1

s

c1

(s − τ )α2 −1

ds

1 Γ(α 2) c1 × m2 (u(τ ) + v(τ ))r2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1



J2 (s)ϕ2



s

c1



s

c1

(s − τ )α2 −1

(s − τ )α2 −1

= μ2 −1 γγ2 m22 −1 (u, v) Y  c2 1 × J2 (s) (s − c1 )α2 (2 −1) ds 2 −1 (Γ(α + 1)) 2 c1 = γγ2 μ2 −1 m22 −1 C (u, v) Y . Then we deduce v ≥ v(c1 ) ≥ γγ2 C(μm2 )2 −1 (u, v) Y > γγ2 C(

μ0 m2 )2 −1 (u, v) Y = (u, v) Y , and so (u, v) Y = u + v ≥ v > (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.1), (4.2) has no positive solution. 

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Theorem 4.1.6. Assume that (H1) and (H2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m1 , m2 > 0 such that f (t, u, v) ≥ m1 (u + v)r1 −1 , g(t, u, v) ≥ m2 (u + v)r2 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.27)

ˆ 0 and ˆ0 and μ ˆ0 such that for every λ > λ then there exist positive constants λ μ>μ ˆ0 , the boundary value problem (4.1), (4.2) has no positive solution. 1 ˆ0 = Proof. We define λ and μ ˆ0 = m2 (2γγ12 C)r2 −1 . Then for m1 (2γγ1 A)r1 −1 ˆ0 and μ > μ ˆ0 , the problem (4.1), (4.2) has no positive solution. every λ > λ ˆ 0 and μ > μ ˆ0 . We suppose that the problem (4.1), (4.2) Indeed, let λ > λ

has a positive solution (u(t), v(t)), t ∈ [0, 1]. In a similar manner as that used in the proofs of Theorems 4.1.4 and 4.1.5 we obtain u ≥ u(c1 ) ≥ γγ1 A(λm1 )1 −1 (u, v) Y , v ≥ v(c1 ) ≥ γγ2 C(μm2 )2 −1 (u, v) Y , and so (u, v) Y = u + v ≥ γγ1 A(λm1 )1 −1 (u, v) Y +γγ2 C(μm2 )2 −1 (u, v) Y ˆ 0 m1 )1 −1 (u, v) Y + γγ2 C(ˆ > γγ1 A(λ μ0 m2 )2 −1 (u, v) Y =

1 1 (u, v) Y + (u, v) Y = (u, v) Y , 2 2

which is a contradiction. Therefore, the boundary value problem (4.1), (4.2) has no positive solution.  i >0 Remark 4.1.3. (i) If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have f0i , f∞ and f (t, u, v) > 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.25) holds and we obtain the conclusion of Theorem 4.1.4. i > 0 and g(t, u, v) > (ii) If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have g0i , g∞ 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.26) holds and we obtain the conclusion of Theorem 4.1.5. i i , g0i , g∞ > 0 (iii) If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have f0i , f∞ and f (t, u, v) > 0, g(t, u, v) > 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.27) holds and we obtain the conclusion of Theorem 4.1.6.

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4.1.4

187

An example

Let α1 = 1/2, α2 = 1/3, n = 3, β1 = 7/3, m = 4, β2 = 15/4, p1 = 1, q1 = 1/3, p2 = 3/2, q2 = 6/5, N = 2, M = 1, ξ1 = 1/4, ξ2 = 3/4, a1 = 3, a2 = 1/4, η1 = 1/3, b1 = 2, r1 = 4, 1 = 4/3, ϕr1 (s) = s3 , ϕ1 (s) = s1/3 , r2 = 3, 2 = 3/2, ϕr2 (s) = s|s|, ϕ2 (s) = s|s|−1/2 . We consider the system of fractional differential equations ⎧   ⎨D1/2 (ϕ4 (D7/3 u(t))) + λ(t + 1)a e(u(t)+v(t))3 − 1 = 0, 0+ 0+ ⎩D1/3 (ϕ (D15/4 v(t))) + μ(2 − t)b (u3 (t) + v 3 (t)) = 0, 3 0+ 0+

t ∈ (0, 1),

t ∈ (0, 1), (4.28)

with the multi-point boundary conditions ⎧   1 3 1 1/3 ⎪ 7/3 1/3   ⎪ ⎪ ⎨u(0) = u (0) = 0, D0+ u(0) = 0, u (1) = 3D0+ u 4 + 4 D0+ u 4 ,  ⎪ 1 ⎪ 15/4 3/2 6/5   ⎪ , ⎩v(0) = v (0) = v (0) = 0, D0+ v(0) = 0, D0+ v(1) = 2D0+ v 3 (4.29) where a, b > 0. 3 Here, we have f (t, u, v) = (t + 1)a (e(u+v) − 1), g(t, u, v) = (2 − t)b (u3 + v 3 ), ∀t ∈ [0, 1], u, v ≥ 0. Then we obtain Δ1 ≈ 0.21710894 > 0, Δ2 ≈ 2.73417069 > 0, and so the assumptions (H1) and (H2) are satisfied. In addition, we deduce 1 g1 (t, s) = Γ(7/3)  g2 (t, s) =



t4/3 (1 − s)1/3 − (t − s)4/3 , t4/3 (1 − s)1/3 ,

t(1 − s)1/3 − t + s,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1,

0 ≤ s ≤ t ≤ 1,

t(1 − s)1/3 , 0 ≤ t ≤ s ≤ 1,  11/4 (1 − s)5/4 − (t − s)11/4 , 0 ≤ s ≤ t ≤ 1, t 1 g3 (t, s) = Γ(15/4) t11/4 (1 − s)5/4 , 0 ≤ t ≤ s ≤ 1, 1 g4 (t, s) = Γ(51/20)



t31/20 (1 − s)5/4 − (t − s)31/20 ,

t31/20 (1 − s)5/4 ,   1 t4/3 ,s + G1 (t, s) = g1 (t, s) + 3g2 Δ1 4

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1,  3 1 g2 ,s , 4 4

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2t11/4 G2 (t, s) = g3 (t, s) + g4 Δ2 h1 (s) =



1 s(1 − s)1/3 , Γ(7/3)

1 ,s , 3 h3 (s) =

1 (1 − s)5/4 (1 − (1 − s)3/2 ). Γ(15/4)

For the functions J1 and J2 , we obtain   ⎧ 1 1 15 13s 15 ⎪ 1/3 1/3 ⎪ s(1 − s) (1 − s) − + + , ⎪ ⎪ Γ(7/3) Δ1 16 4 16 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ≤ s < 1, ⎪ ⎪ 4 ⎪ ⎪ ⎪   ⎨ 3 1 1 15 s J1 (s) = s(1 − s)1/3 + (1 − s)1/3 + − , ⎪ Γ(7/3) Δ1 16 4 16 ⎪ ⎪ ⎪ ⎪ 3 1 ⎪ ⎪ ⎪ ≤s< , ⎪ ⎪ 4 4 ⎪ ⎪ ⎪ ⎪ 15 3 1 ⎪ ⎪ ⎩ s(1 − s)1/3 + ≤ s ≤ 1, (1 − s)1/3 , Γ(7/3) 16Δ1 4 ⎧ 1 2 ⎪ ⎪ (1 − s)5/4 (1 − (1 − s)3/2 ) + 31/20 ⎪ ⎪ Γ(15/4) 3 Δ 2 Γ(51/20) ⎪ ⎪ ⎪ ⎪ 1 ⎪ 5/4 31/20 ⎪ ⎪ ], 0 ≤ s < , ⎨ × [(1 − s) − (1 − 3s) 3 J2 (s) = 1 ⎪ ⎪ ⎪ (1 − s)5/4 (1 − (1 − s)3/2 ) ⎪ ⎪ Γ(15/4) ⎪ ⎪ ⎪ ⎪ 2 1 ⎪ ⎪ ⎩ + 31/20 (1 − s)5/4 , ≤ s ≤ 1. 3 3 Δ2 Γ(51/20) Now, we choose c1 = 1/4 and c2 = 3/4, and then we deduce γ1 = i = ∞, (1/4)4/3 , γ2 = (1/4)11/4 , γ = γ2 . In addition, we have f0s = 2a , f∞ s i g0 = 0, g∞ = ∞, A ≈ 1.35668478, B ≈ 2.51926854. By Theorem 4.1.1(7), for any λ ∈ (0, L2 ) and μ ∈ (0, ∞) with L2 = 1/(f0s B 3 ), the problem (4.28), (4.29) has a positive solution (u(t), v(t)), t ∈ [0, 1]. For example, if a = 2, we obtain L2 ≈ 0.0156357. We can also use Theorem 4.1.4, because f (t, u, v) ≥ (5/4)a (u + v)3 for all t ∈ [1/4, 3/4] and u, v ≥ 0, that is m1 = (5/4)a . If a = 2, we deduce

0 = 1/(m1 (γγ1 A)3 ) ≈ 6.0810421 × 106 , and then we conclude that for λ

0 and μ > 0, the boundary value problem (4.28), (4.29) has no every λ > λ positive solution.

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4.1.5

189

A relation between two supremum limits

In this section, we will prove that if f (t, u, v) s = f∞ , t∈[0,1] (u + v)r1 −1

lim sup max u+v→∞

u,v≥0

then f ∗ (t, x) s = f∞ , t∈[0,1] xr1 −1

lim sup max x→∞

where f : [0, 1]×[0, ∞)×[0, ∞) → [0, ∞) is a continuous function, f ∗ (t, x) = max{f (t, u, v), u + v ≤ x, u, v ≥ 0} for t ∈ [0, 1] and x ≥ 0, and r1 > 1. s ∈ (0, ∞), from the characterization theorem of supre(I) In the case f∞ mum limit, we have (a) ∀ε > 0, ∃M (ε) > 0 such that ∀u, v ≥ 0, u + v > M (ε) we have max

t∈[0,1]

f (t, u, v) s < f∞ + ε, (u + v)r1 −1

(b) ∀ε > 0 ∀M  > 0 then ∃(u0 , v0 ), u0 ≥ 0, v0 ≥ 0, u0 + v0 > M  such that max

t∈[0,1]

f (t, u0 , v0 ) s > f∞ − ε. (u0 + v0 )r1 −1

The second relation (b) is verified for an arbitrary (u, v) with u+v > M  s if ε > f∞ , because f has nonnegative values. From (a), for ε > 0 arbitrary, but fixed for the moment, there exists M1 = M (ε/2) > 0 such that for all u, v ≥ 0, u + v > M1 we have ε f (t, u, v) s < f∞ + , (u + v)r1 −1 2

s  and then f (t, u, v) < f∞ + 2ε (u + v)r1 −1 for all t ∈ [0, 1]. Then for ε > 0, there exists M1 > 0 such that for all x > M1 and t ∈ [0, 1], we obtain max

t∈[0,1]

f ∗ (t, x) =

max

0≤u+v≤x

f (t, u, v) ≤

= f ∗ (t, M1 ) +

sup

max

0≤u+v≤M1

M1 0 such that f ∗ (t, x) KM1 ε s ≤ r1 −1 + f∞ + , r −1 1 x x 2

∀x > M1 , ∀t ∈ [0, 1],

and so KM1 ε f ∗ (t, x) s ≤ r1 −1 + f∞ + , x 2 t∈[0,1] xr1 −1 max

∀x > M1 .

Because limx→∞ xr11−1 = 0, then for ε > 0 there exists M2 ≥ M1 such that xr11−1 < 2KεM for all x > M2 . 1 So, we conclude f ∗ (t, x) ε ε s < + f∞ + 2 2 t∈[0,1] xr1 −1

∀ε > 0, ∃M2 > 0 such that max s + ε, = f∞

∀x > M2 .

(4.30)

From relation (b) we deduce that for any ε > 0 and any M  > 0 there exists x0 = u0 + v0 > 0 such that f ∗ (t, x0 ) f (t, u0 , v0 ) f (t, u0 , v0 ) s ≥ max = max > f∞ − ε. r r 1 −1 1 −1 t∈[0,1] x t∈[0,1] t∈[0,1] (u0 + v0 )r1 −1 x 0 0 max

Then we obtain ∀ε > 0, ∀M  > 0,

f ∗ (t, x0 ) s > f∞ − ε. r −1 t∈[0,1] x 1 0 (4.31)

∃x0 > M  such that max

By relations (4.30), (4.31) and the characterization theorem for supre∗ s mum limit, we conclude that lim supx→∞ maxt∈[0,1] fxr(t,x) = f∞ . 1 −1 f (t,u,v) s (II) If f∞ = 0, then lim supu+v→∞ maxt∈[0,1] (u+v)r1 −1 = 0 is equivalent to limu+v→∞ maxt∈[0,1] u,v≥0

u,v≥0 f (t,u,v) (u+v)r1 −1

= 0, because f has nonnega-

tive values. Also, lim supx→∞ maxt∈[0,1] ∗ limx→∞ maxt∈[0,1] fxr(t,x) = 0. 1 −1

f ∗ (t,x) xr1 −1

= 0 is equivalent to

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191

In the same manner as used in the case (I) (for the implication (a) ⇒ (4.30)), we can show that relation ∀ε > 0 ∃M > 0 such that ∀u, v ≥ 0, u + v > M we have 0 ≤ max

t∈[0,1]

f (t, u, v) < ε, (u + v)r1 −1

implies the relation f ∗ (t, x) < ε, t∈[0,1] xr1 −1

 > 0 such that ∀x > M  we have 0 ≤ max ∀ε > 0 ∃M ∗

that is, limx→∞ maxt∈[0,1] fxr(t,x) = 0. 1 −1 s = ∞, then by the characterization theorem, we have (III) If f∞ ∀M1 > 0 ∀M2 > 0 ∃(u, v), u ≥ 0, v ≥ 0, u + v > M1 such that max

t∈[0,1]

f (t, u, v) > M2 . (u + v)r1 −1

Then we deduce that for any M1 > 0 and any M2 > 0 there exists x = u + v > M1 such that f ∗ (t, x) f (t, u, v) f (t, u, v) ≥ max = max > M2 . t∈[0,1] xr1 −1 t∈[0,1] xr1 −1 t∈[0,1] (u + v)r1 −1 max

So we obtain that lim supx→∞ maxt∈[0,1]

f ∗ (t,x) xr1 −1

= ∞.

Remark 4.1.4. The results presented in this section under the assumptions p1 ∈ [1, n − 2] and p2 ∈ [1, m − 2] instead of p1 ∈ [1, β1 − 1) and p2 ∈ [1, β2 − 1) were published in [79].

4.2

Systems of Fractional Differential Equations with Coupled Multi-Point Boundary Conditions

We consider the system of nonlinear ordinary fractional differential equations with r1 -Laplacian and r2 -Laplacian operators ⎧ β1 α1 ⎨D0+ (ϕr1 (D0+ u(t))) + λf (t, u(t), v(t)) = 0, t ∈ (0, 1), (4.32) ⎩Dα2 (ϕ (Dβ2 v(t))) + μg(t, u(t), v(t)) = 0, t ∈ (0, 1), r2 0+ 0+

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with the coupled multi-point boundary conditions ⎧ β1 ⎪ u(j) (0) = 0, j = 0, . . . , n − 2; D0+ u(0) = 0, ⎪ ⎪ ⎪ ⎪ N ⎪  ⎪ ⎪ p1 q1 ⎪ ai D0+ v(ξi ), ⎪ ⎨ D0+ u(1) = i=1

β2 D0+ v(0) = 0,

⎪ v (j) (0) = 0, j = 0, . . . , m − 2; ⎪ ⎪ ⎪ ⎪ M ⎪  ⎪ ⎪ p2 q2 ⎪ D v(1) = bi D0+ u(ηi ), ⎪ 0+ ⎩

(4.33)

i=1

where α1 , α2 ∈ (0, 1], β1 ∈ (n − 1, n], β2 ∈ (m − 1, m], n, m ∈ N, n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, β1 − 1), p2 ∈ [1, β2 − 1), q1 ∈ [0, p2 ], q2 ∈ [0, p1 ], ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi , bi ∈ R for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, r1 , r2 > 1, 1 1 ϕri (s) = |s|ri −2 s, ϕ−1 ri = ϕi , ri + i = 1, i = 1, 2, λ, μ > 0, f, g ∈ k C([0, 1] × [0, ∞) × [0, ∞), [0, ∞)), and D0+ denotes the Riemann-Liouville derivative of order k (for k = α1 , β1 , α2 , β2 , p1 , q1 , p2 , q2 ). Under sufficient conditions on the functions f and g, we present intervals for the parameters λ and μ such that problem (4.32), (4.33) has positive solutions. By a positive solution of problem (4.32), (4.33) we mean a pair of functions (u, v) ∈ (C([0, 1], R+ ))2 , satisfying (S) and (BC) with u(t) > 0 for all t ∈ (0, 1], or v(t) > 0 for all t ∈ (0, 1]. We also investigate the nonexistence of positive solutions for the above problem. 4.2.1

Preliminary results

We consider the system of fractional differential equations  α1 β1 u(t))) + h(t) = 0, t ∈ (0, 1), D0+ (ϕr1 (D0+ β2 α2 D0+ (ϕr2 (D0+ v(t))) + k(t) = 0,

t ∈ (0, 1),

(4.34)

with the coupled multi-point boundary conditions (4.33), where h, k ∈ C[0, 1]. β1 β2 u(t)) = x(t) and ϕr2 (D0+ v(t)) = y(t), then the If we denote by ϕr1 (D0+ problem (4.34), (4.33) is equivalent to the following three problems:  α D0+1 x(t) + h(t) = 0, 0 < t < 1, (I) x(0) = 0,  (II)

α2 y(t) + k(t) = 0, D0+

y(0) = 0,

0 < t < 1,

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and  (III)

β1 u(t) = ϕ1 (x(t)), D0+

t ∈ (0, 1),

β2 v(t) D0+

t ∈ (0, 1),

= ϕ2 (y(t)),

with the boundary conditions ⎧ ⎪ ⎪ ⎪ ⎪ u(j) (0) = 0, j = 0, . . . , n − 2; ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ (j) ⎪ ⎪ ⎩v (0) = 0,

p1 D0+ u(1) =

j = 0, . . . , m − 2;

p2 D0+ v(1) =

N  i=1

q1 ai D0+ v(ξi ),

M 

q2 bi D0+ u(ηi ).

i=1

For the first two problems (I) and (II), the functions α1 x(t) = −I0+ h(t) = −

1 Γ(α1 )

α2 k(t) = − y(t) = −I0+

1 Γ(α2 )



t

0

(t − s)α1 −1 h(s) ds,

t ∈ [0, 1],

(4.35)

(t − s)α2 −1 k(s) ds,

t ∈ [0, 1],

(4.36)

and  0

t

are the unique solutions x, y ∈ C[0, 1] for (I) and (II), respectively. For the third problem (III), if Γ(β1 )Γ(β2 ) Γ(β1 )Γ(β2 ) − Δ := Γ(β1 − p1 )Γ(β2 − p2 ) Γ(β1 − q2 )Γ(β2 − q1 ) M   β −q −1 × bi ηi 1 2 = 0,



N  i=1

 ai ξiβ2 −q1 −1

i=1

and x, y ∈ C[0, 1], then by Lemma 3.2.2 from Section 3.2, we deduce that the pair of functions  ⎧ ⎪ ⎪u(t) = − ⎪ ⎨  ⎪ ⎪ ⎪ ⎩v(t) = −

1

0

0

1

1 (t, s)ϕ1 (x(s)) ds − G

3 (t, s)ϕ2 (y(s)) ds − G

 

1

0

0

1

2 (t, s)ϕ2 (y(s)) ds, G

t ∈ [0, 1],

4 (t, s)ϕ1 (x(s)) ds, G

t ∈ [0, 1] (4.37)

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is the unique solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (III). Here, the

i , i = 1, . . . , 4 (see Lemma 3.2.2) are defined by Green functions G tβ1 −1 Γ(β2 )

1 (t, s) =

G g1 (t, s) + ΔΓ(β2 − q1 ) β1 −1 Γ(β2 )

2 (t, s) = t G ΔΓ(β2 − p2 )



N 

N  i=1

ai ξiβ2 −q1 −1



 M 

bi g 2 (ηi , s) ,

i=1

 ai g 3 (ξi , s) ,

i=1

tβ2 −1 Γ(β1 )

3 (t, s) =

G g4 (t, s) + ΔΓ(β1 − q2 ) β2 −1 Γ(β1 )

4 (t, s) = t G ΔΓ(β1 − p1 )





M 



M  i=1

 bi ηiβ1 −q2 −1

N 

 ai g 3 (ξi , s) ,

i=1

 bi

g2 (ηi , s) ,

∀t, s ∈ [0, 1],

i=1

(4.38) where 1 g 1 (t, s) = Γ(β1 )



tβ1 −1 (1 − s)β1 −p1 −1 − (t − s)β1 −1 , tβ1 −1 (1 − s)β1 −p1 −1 ,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1,

⎧ β1 −q2 −1 t (1 − s)β1 −p1 −1 − (t − s)β1 −q2 −1 , ⎪ ⎪ ⎨ 1 g 2 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(β1 − q2 ) ⎪ ⎪ ⎩ β1 −q2 −1 (1 − s)β1 −p1 −1 , 0 ≤ t ≤ s ≤ 1. t ⎧ β2 −q1 −1 (1 − s)β2 −p2 −1 − (t − s)β2 −q1 −1 , ⎪ ⎪t ⎨ 1 g 3 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(β2 − q1 ) ⎪ ⎪ ⎩ β2 −q1 −1 t (1 − s)β2 −p2 −1 , 0 ≤ t ≤ s ≤ 1. 1 g 4 (t, s) = Γ(β2 )



tβ2 −1 (1 − s)β2 −p2 −1 − (t − s)β2 −1 , t

β2 −1

β2 −p2 −1

(1 − s)

,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1. (4.39)

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Therefore, by (4.35)–(4.37), we obtain the following lemma. Lemma 4.2.1. If Δ = 0, then the unique solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (4.34), (4.33) is given by ⎧  1 ⎪ ⎪

1 (t, s)ϕ1 (I α1 h(s)) ds ⎪u(t) = G ⎪ 0+ ⎪ ⎪ 0 ⎪ ⎪ ⎪  ⎪ 1 ⎪ ⎪ ⎪

2 (t, s)ϕ2 (I α2 k(s)) ds, t ∈ [0, 1], G + ⎪ 0+ ⎨ 0 (4.40)  1 ⎪ ⎪ ⎪

3 (t, s)ϕ2 (I α2 k(s)) ds ⎪ G v(t) = ⎪ 0+ ⎪ ⎪ 0 ⎪ ⎪ ⎪  1 ⎪ ⎪ ⎪ ⎪

4 (t, s)ϕ1 (I α1 h(s)) ds, t ∈ [0, 1]. + G ⎩ 0+ 0

Based on the properties of the functions g i , i = 1, . . . , 4 given by (4.39) (see Lemma 3.2.3) we obtain the following properties of the Green functions

i , i = 1, . . . , 4 that will be used in the next sections. G Lemma 4.2.2. Assume that Δ > 0, ai ≥ 0 for all i = 1, . . . , N, and bi ≥ 0

i , i = 1, . . . , 4, given by (4.38) for all i = 1, . . . , M . Then the functions G have the properties (a) (b)

i : [0, 1] × [0, 1] → [0, ∞), i = 1, . . . , 4 are continuous functions; G

G1 (t, s) ≤ J 1 (s), ∀(t, s) ∈ [0, 1] × [0, 1], where N   Γ(β2 ) β2 −q1 −1



J1 (s) = h1 (s) + ai ξi ΔΓ(β2 − q1 ) i=1  M  bi

g2 (ηi , s) , ∀s ∈ [0, 1], × i=1

and

h1 (s) = Γ(β1 1 ) (1 −

1 (t, s) ≥ tβ1 −1 J 1 (s), G

s)β1 −p1 −1 (1 − (1 − s)p1 ), s ∈ [0, 1];

∀(t, s) ∈ [0, 1] × [0, 1]; (c)

2 (t, s) ≤ J 2 (s), ∀(t, s) ∈ [0, 1] × [0, 1], where (d) G J 2 (s) =

N

 Γ(β2 ) ai

g3 (ξi , s), ∀s ∈ [0, 1]; ΔΓ(β2 − p2 ) i=1

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2 (t, s) = tβ1 −1 J 2 (s), ∀(t, s) ∈ [0, 1] × [0, 1]; (e) G

3 (t, s) ≤ J 3 (s), ∀(t, s) ∈ [0, 1] × [0, 1], where (f) G M   N  β −q −1  ) Γ(β 1 J 3 (s) =

h4 (s) + bi ηi 1 2 ai g 3 (ξi , s) , ΔΓ(β1 − q2 ) i=1 i=1 and

h4 (s) = Γ(β1 2 ) (1 − s)β2 −p2 −1 (1 − (1 − s)p2 ), s ∈ [0, 1];

3 (t, s) ≥ tβ2 −1 J 3 (s), ∀(t, s) ∈ [0, 1] × [0, 1]; (g) G

4 (t, s) ≤ J 4 (s), ∀(t, s) ∈ [0, 1] × [0, 1], where (h) G J 4 (s) =

M

 Γ(β1 ) bi g 2 (ηi , s), ∀s ∈ [0, 1]; ΔΓ(β1 − p1 ) i=1

4 (t, s) = tβ2 −1 J 4 (s), ∀(t, s) ∈ [0, 1] × [0, 1]. (i) G 4.2.2

Existence of positive solutions

In this section, we investigate the existence of positive solutions of problem (4.32), (4.33) under some assumptions on the functions f and g, by establishing in the same time various intervals for the positive parameters λ and μ. We present the assumptions that we will use in the sequel. (I1) α1 , α2 ∈ (0, 1], β1 ∈ (n − 1, n], β2 ∈ (m − 1, m], n, m ≥ 3, p1 , p2 , q1 , q2 ∈ R, p1 ∈ [1, β1 − 1), p2 ∈ [1, β2 − 1), q1 ∈ [0, p2 ], N q2 ∈ [0, p1 ], ξi ∈ R, ai ≥ 0 for all i = 1, . . . , N (N ∈ N), i=1 ai > 0, 0 < ξ < · · · < ξN ≤ 1, ηi ∈ R, bi ≥ 0 for all i = 1, . . . , M (M ∈ N), M 1 Γ(β1 )Γ(β2 ) i=1 bi > 0, 0 < η1 < · · · < ηM ≤ 1, λ, μ > 0, Δ = Γ(β1 −p1 )Γ(β2 −p2 ) −     N M Γ(β1 )Γ(β2 ) β2 −q1 −1 β1 −q2 −1 > 0, ri > 1, i=1 ai ξi i=1 bi ηi Γ(β1 −q2 )Γ(β2 −q1 ) ri ϕri (s) = |s|ri −2 s, ϕ−1 ri = ϕi , i = ri −1 , i = 1, 2. (I2) The functions f , g : [0, 1] × [0, ∞) × [0, ∞) → [0, ∞) are continuous.

For [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, we introduce the following extreme limits: f (t, u, v) , (u + v)r1 −1 u+v→0+ t∈[0,1]

g(t, u, v) , (u + v)r2 −1 u+v→0+ t∈[0,1]

f0s = lim sup max

g0s = lim sup max

u,v≥0

u,v≥0

f0i = lim inf min u+v→0+

u,v≥0

t∈[c1 ,c2 ]

f (t, u, v) , (u + v)r1 −1

g0i = lim inf min u+v→0+

u,v≥0

t∈[c1 ,c2 ]

g(t, u, v) , (u + v)r2 −1

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s f∞ = lim sup max u+v→∞

u,v≥0

t∈[0,1]

f (t, u, v) , (u + v)r1 −1

i f∞ = lim inf min u+v→∞

u,v≥0

t∈[c1 ,c2 ]

f (t, u, v) , (u + v)r1 −1

s g∞ = lim sup max u+v→∞

u,v≥0

t∈[0,1]

g(t, u, v) , (u + v)r2 −1

i g∞ = lim inf min u+v→∞

u,v≥0

197

t∈[c1 ,c2 ]

g(t, u, v) . (u + v)r2 −1

By using Lemma 4.2.1 (relations (4.40)), (u, v) is solution of problem (4.32), (4.33) if and only if (u, v) is solution of the nonlinear system of integral equations ⎧  1 ⎪ ⎪  −1 1

1 (t, s)ϕ1 (I α1 f (s, u(s), v(s))) ds ⎪u(t) = λ G ⎪ 0+ ⎪ ⎪ 0 ⎪ ⎪ ⎪  ⎪ 1 ⎪ ⎪ 2 −1 ⎪

2 (t, s)ϕ2 (I α2 g(s, u(s), v(s))) ds, t ∈ [0, 1], G +μ ⎪ 0+ ⎨ 0  1 ⎪ ⎪ ⎪ 2 −1

3 (t, s)ϕ2 (I α2 g(s, u(s), v(s))) ds ⎪ G v(t) = μ ⎪ 0+ ⎪ ⎪ 0 ⎪ ⎪ ⎪  1 ⎪ ⎪ ⎪ 1 −1 ⎪

4 (t, s)ϕ (I α1 f (s, u(s), v(s))) ds, t ∈ [0, 1]. G +λ ⎩ 1

0

0+

We consider the Banach space X = C[0, 1] with the supremum norm · , and the Banach space Y = X ×X with the norm (u, v) Y = u + v . We define the cones P1 = {u ∈ X, u(t) ≥ tβ1 −1 u , ∀t ∈ [0, 1]} ⊂ X, P2 = {v ∈ X, v(t) ≥ tβ2 −1 v , ∀t ∈ [0, 1]} ⊂ X, and P = P1 × P2 ⊂ Y . We define now the operators Q1 , Q2 : Y → X and Q : Y → Y by 1 −1

Q1 (u, v)(t) = λ

1

0

+μ Q2 (u, v)(t) = μ



2 −1

2 −1



1

0



1

0 1 −1

1 (t, s)ϕ1 (I α1 f (s, u(s), v(s))) ds G 0+

2 (t, s)ϕ2 (I α2 g(s, u(s), v(s))) ds, G 0+

t ∈ [0, 1],

3 (t, s)ϕ2 (I α2 g(s, u(s), v(s))) ds G 0+



+λ

0

1

4 (t, s)ϕ1 (I α1 f (s, u(s), v(s))) ds, G 0+

t ∈ [0, 1],

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and Q(u, v) = (Q1 (u, v), Q2 (u, v)), (u, v) ∈ Y . Then (u, v) is a solution of problem (4.32), (4.33) if and only if (u, v) is a fixed point of operator Q. Lemma 4.2.3. If (I1)–(I2) hold, then Q : P → P is a completely continuous operator. Proof. Let (u, v) ∈ P be an arbitrary element. Because Q1 (u, v) and Q2 (u, v) satisfy the problem (4.34), (4.33) for h(t) = λf (t, u(t), v(t)) and k(t) = μg(t, u(t), v(t)), t ∈ [0, 1], then by Lemma 4.2.2, we obtain 1 −1

Q1 (u, v)(t) ≤ λ

1

0

+μ Q2 (u, v)(t) ≤ μ



2 −1

2 −1

α1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds



1

α2 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds,

0



1

α2 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

0

+ λ1 −1

∀t ∈ [0, 1],



1

α1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds,

0

∀t ∈ [0, 1],

and so 1 −1



Q1 (u, v) ≤ λ

1

0

+ μ2 −1 Q2 (u, v) ≤ μ2 −1

α1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

 0



1 0

+ λ1 −1

1

α2 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds,

α2 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds



1

0

α1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds.

Therefore, we conclude that 1 −1

Q1 (u, v)(t) ≥ λ

+μ

 0

2 −1

1

α1 tβ1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

 0

1

α2 tβ1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

page 198

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199

≥ tβ1 −1 Q1 (u, v) , ∀t ∈ [0, 1],  1 α2 tβ2 −1 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds Q2 (u, v)(t) ≥ μ2 −1 0

+ λ1 −1

 0

1

α1 tβ2 −1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

≥ tβ2 −1 Q2 (u, v) ,

∀t ∈ [0, 1].

Hence, Q(u, v) = (Q1 (u, v), Q2 (u, v)) ∈ P , and then Q(P ) ⊂ P . By

i , i = 1, . . . , 4, and the Arzela–Ascoli the continuity of the functions f, g, G theorem, we can show that Q1 and Q2 are completely continuous operators, and then Q is a completely continuous operator.  For [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, we denote by A= B=

1 (Γ(α1 + 1))1 −1 1 (Γ(α2 + 1))2 −1

1 C= (Γ(α2 + 1))2 −1 1 D= (Γ(α1 + 1))1 −1 1

= A (Γ(α1 + 1))1 −1

= B

1 (Γ(α2 + 1))2 −1

= C

1 (Γ(α2 + 1))2 −1

= D

1 (Γ(α1 + 1))1 −1



1

0



1

0



1

0



1

0



sα1 (1 −1) J 1 (s) ds, sα2 (2 −1) J 2 (s) ds, sα2 (2 −1) J 3 (s) ds, sα1 (1 −1) J 4 (s) ds,

c2

c1



c2

c1



c2

c1



c2

c1

(4.41) (s − c1 )α1 (1 −1) J 1 (s) ds, (s − c1 )α2 (2 −1) J 2 (s) ds, (s − c1 )α2 (2 −1) J 3 (s) ds, (s − c1 )α1 (1 −1) J 4 (s) ds,

where J i , i = 1, . . . , 4 are defined in Lemma 4.2.2.

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i i First, for f0s , g0s , f∞ , g∞ ∈ (0, ∞) and numbers γ1 , γ2 ∈ [0, 1], γ3 , γ4 ∈ (0, 1), a ∈ [0, 1] and b ∈ (0, 1), we define the numbers

 L1 = max  L2 = min

1 f0s

 L3 = max  L4 = min  L2

= min 

L4

= min

1 i f∞



1 i g∞

1 g0s 1 f0s 1 g0s



bγ3 A





aγ1

θθ1 A

r1 −1

r1 −1

1 , s f0

a(1 − γ1 )

θθ1 B

b(1 − γ3 ) B

1 , i f∞ 

(1 − a)γ2

θθ2 D

(1 − b)γ4 D

r2 −1

r2 −1



1 , i g∞

1 , s g0





r1 −1 

r1 −1 

,

,

(1 − a)(1 − γ2 )

θθ2 C

(1 − b)(1 − γ4 ) C

r2 −1 

r2 −1 

,

,

 r1 −1  r −1  b 1 1−b 1 , s , A f0 D 

b B

r2 −1

1 , s g0



1−b C

r2 −1 

,

where θ1 = cβ1 1 −1 , θ2 = cβ1 2 −1 , θ = min{θ1 , θ2 }. Theorem 4.2.1. Assume that (I1) and (I2) hold, [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, γ1 , γ2 ∈ [0, 1], γ3 , γ4 ∈ (0, 1), a ∈ [0, 1] and b ∈ (0, 1). i i (1) If f0s , g0s , f∞ , g∞ ∈ (0, ∞), L1 < L2 and L3 < L4 , then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ ∈ (0, ∞) and L3 < L4 , then for each λ ∈ (L1 , ∞) (2) If f0s = 0, g0s , f∞  and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ ∈ (0, ∞) and L1 < L2 , then for each λ ∈ (L1 , L2 ) (3) If g0s = 0, f0s , f∞ and μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ ∈ (0, ∞), then for each λ ∈ (L1 , ∞) and (4) If f0s = g0s = 0, f∞ μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33).

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201

i i (5) If f0s , g0s ∈ (0, ∞) and at least one of f∞ , g∞ is ∞, then for each λ ∈ (0, L2 ) and μ ∈ (0, L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ is ∞, then for each (6) If f0s = 0, g0s ∈ (0, ∞) and at least one of f∞  λ ∈ (0, ∞) and μ ∈ (0, L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ is ∞, then for each (7) If f0s ∈ (0, ∞), g0s = 0 and at least one of f∞ λ ∈ (0, L2 ) and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). i i , g∞ is ∞, then for each λ ∈ (0, ∞) (8) If f0s = g0s = 0 and at least one of f∞ and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33).

Proof. We consider the above cone P ⊂ Y and the operators Q1 , Q2 and Q. Because the proofs of the above cases are similar, in what follows, we will prove two of them, namely Cases (1) and (7). i i , g∞ ∈ (0, ∞), L1 < L2 and L3 < L4 . Let Case (1). We have f0s , g0s , f∞ i λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ). We consider ε > 0 such that ε < f∞ , i ε < g∞ and  r1 −1 r −1    aγ1 (1 − a)γ2 1 1 1 , i ≤λ max i −ε



f∞ f∞ − ε θθ1 A θθ2 D   r −1  r −1  bγ3 1 (1 − b)γ4 1 1 1 , ≤ min , s f0s + ε A f0 + ε D  r −1 r −1    1 1 a(1 − γ1 ) 2 (1 − a)(1 − γ2 ) 2 max , i ≤μ i −ε



g∞ g∞ − ε θθ1 B θθ2 C   r −1  r −1  b(1 − γ3 ) 2 (1 − b)(1 − γ4 ) 2 1 1 ≤ min , s . g0s + ε B g0 + ε C

By using (I2) and the definition of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ (f0s + ε)(u + v)r1 −1 ,

g(t, u, v) ≤ (g0s + ε)(u + v)r2 −1 ,

for all t ∈ [0, 1] and u, v ≥ 0, u + v ≤ R1 . We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. Now, let (u, v) ∈ P ∩ ∂Ω1 , that is, (u, v) ∈ P with (u, v) Y = R1 , or equivalently u + v = R1 . Then u(t) + v(t) ≤ R1 for all t ∈ [0, 1], and by

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Lemma 4.2.2, we obtain   1 1 −1

J1 (s)ϕ1 Q1 (u, v)(t) ≤ λ

1 Γ(α1 ) 0 × f (τ, u(τ ), v(τ )) dτ ds

+ μ2 −1



1 0



J 2 (s)ϕ2



1

J 1 (s)ϕ1





1

 0



1 +μ Γ(α 2) 0 × (u(τ ) + v(τ ))r2 −1 dτ ds 2 −1

1 −1

≤λ

(f0s

J 2 (s)ϕ2

1 −1

+ ε)



1

0

× ( u + v )r1 −1 dτ +μ

2 −1

(g0s

2 −1

(s − τ )α1 −1



s

0

(s − τ )α2 −1

ds

1 Γ(α 1) 0 × (u(τ ) + v(τ ))r1 −1 dτ ds

≤ λ1 −1

s

0

1 Γ(α2 )



× g(τ, u(τ ), v(τ )) dτ



× ( u + v )r2 −1 dτ

(s − τ )α1 −1 (f0s + ε)



J 1 (s)ϕ1

s

0

(s − τ )α2 −1 (g0s + ε)





1 Γ(α1 )



s

0

(s − τ )α1 −1

ds



+ ε)

s

0

1

J 2 (s)ϕ2





1 Γ(α2 )

ds

= λ1 −1 (f0s + ε)1 −1 (u, v) Y  1 1 J 1 (s) × sα1 (1 −1) ds 1 −1 (Γ(α + 1)) 1 0 + μ2 −1 (g0s + ε)2 −1 (u, v) Y

 0

s

(s − τ )α2 −1

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 ×

1

0

J 2 (s)

page 203

203

1 sα2 (2 −1) ds (Γ(α2 + 1))2 −1

= [λ1 −1 (f0s + ε)1 −1 A + μ2 −1 (g0s + ε)2 −1 B] (u, v) Y ≤ [bγ3 + b(1 − γ3 )] (u, v) Y = b (u, v) Y ,

∀t ∈ [0, 1].

Therefore, Q1 (u, v) ≤ b (u, v) Y . In a similar manner, we conclude 

1



1 Γ(α2 ) 0 × g(τ, u(τ ), v(τ )) dτ ds

Q2 (u, v)(t) ≤ μ2 −1

1 −1

J 3 (s)ϕ2



+λ

1

0

J 4 (s)ϕ1





1





J 3 (s)ϕ2

1

J 4 (s)ϕ1

≤ μ2 −1 (g0s + ε)2 −1



r2 −1

× ( u + v ) 1 −1

+λ

(f0s



1

0

× ( u + v )

(s − τ )α1 −1

s

0

(s − τ )α2 −1 (g0s + ε)



s

0

(s − τ )α1 −1 (f0s + ε)





0

1

J 4 (s)ϕ1

dτ

1 Γ(α2 )

 0

s

(s − τ )α2 −1

ds



+ ε)

r1 −1

s

0

J 3 (s)ϕ2

dτ

1 −1





1 Γ(α 1) 0 × (u(τ ) + v(τ ))r1 −1 dτ ds

+ λ1 −1

(s − τ )α2 −1

ds

1 ≤μ Γ(α 2) 0 × (u(τ ) + v(τ ))r2 −1 dτ ds 2 −1

s

0

1 Γ(α1 )



× f (τ, u(τ ), v(τ )) dτ



ds

= μ2 −1 (g0s + ε)2 −1 (u, v) Y



1 Γ(α1 )

 0

s

(s − τ )α1 −1

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 ×

1

0

J 3 (s)

1 sα2 (2 −1) ds (Γ(α2 + 1))2 −1

+ λ1 −1 (f0s + ε)1 −1 (u, v) Y  1 1 J 4 (s) × sα1 (1 −1) ds (Γ(α1 + 1))1 −1 0 = [μ2 −1 (g0s + ε)2 −1 C + λ1 −1 (f0s + ε)1 −1 D] (u, v) Y ≤ [(1 − b)(1 − γ4 ) + (1 − b)γ4 ] (u, v) Y = (1 − b) (u, v) Y ,

∀t ∈ [0, 1].

Hence, Q2 (u, v) ≤ (1 − b) (u, v) Y . Then for (u, v) ∈ P ∩ ∂Ω1 , we deduce Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤ b (u, v) Y + (1 − b) (u, v) Y = (u, v) Y .

(4.42)

i i By the definition of f∞ and g∞ , there exists R2 > 0 such that i − ε)(u + v)r1 −1 , f (t, u, v) ≥ (f∞

i g(t, u, v) ≥ (g∞ − ε)(u + v)r2 −1 ,

for all t ∈ [c1 , c2 ] and u, v ≥ 0, u + v ≥ R2 . We consider R2 = max{2R1 , R2 /θ} and we define the set Ω2 = {(u, v) ∈ Y , (u, v) Y < R2 }. Then for (u, v) ∈ P ∩ ∂Ω2 , we obtain u(t) + v(t) ≥ min tβ1 −1 u + min tβ2 −1 v = cβ1 1 −1 u + cβ1 2 −1 v t∈[c1 ,c2 ]

t∈[c1 ,c2 ]

= θ1 u + θ2 v ≥ θ (u, v) Y = θR2 ≥ R2 ,

∀ t ∈ [c1 , c2 ].

Therefore, by Lemma 4.2.2, we conclude  1 α1 cβ1 1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds Q1 (u, v)(c1 ) ≥ λ1 −1 0

+ μ2 −1 ≥



1 0

λ1 −1 cβ1 1 −1

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds



c2

c1

J 1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ

ds



1 Γ(α1 )



s

c1

(s − τ )α1 −1

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+ μ2 −1 cβ1 1 −1



c2

c1

J 2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ ≥





1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds 

 s 1 (s − τ )α1 −1 Γ(α1 ) c1 c1 i r1 −1 × (f∞ − ε)(u(τ ) + v(τ )) dτ ds

λ1 −1 cβ1 1 −1

c2



J 1 (s)ϕ1



 s 1 (s − τ )α2 −1 Γ(α2 ) c1 c1 i × (g∞ − ε)(u(τ ) + v(τ ))r2 −1 dτ ds + μ2 −1 cβ1 1 −1



c2

J 2 (s)ϕ2



 s 1 (s − τ )α1 −1 Γ(α1 ) c1 c1 i r1 −1 × (f∞ − ε)(θ (u, v) Y ) dτ ds

≥ λ1 −1 cβ1 1 −1

c2



J 1 (s)ϕ1

c2



1 Γ(α 2) c1 i × (g∞ − ε)(θ (u, v) Y )r2 −1 dτ ds + μ2 −1 cβ1 1 −1

J 2 (s)ϕ2



s

c1

(s − τ )α2 −1

i − ε)1 −1 (u, v) Y = θθ1 λ1 −1 (f∞  c2 1 J 1 (s) × (s − c1 )α1 (1 −1) ds (Γ(α + 1))1 −1 1 c1 i − ε)2 −1 (u, v) Y + θθ1 μ2 −1 (g∞  c2 1 J 2 (s) × (s − c1 )α2 (2 −1) ds (Γ(α2 + 1))2 −1 c1 i

+ θθ1 μ2 −1 = [θθ1 λ1 −1 (f∞ − ε)1 −1 A i

v) Y ×(g∞ − ε)2 −1 B] (u,

≥ [aγ1 + a(1 − γ1 )] (u, v) Y = a (u, v) Y .

So, Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ a (u, v) Y .

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In a similar manner, we deduce  1 α2 Q2 (u, v)(c1 ) ≥ μ2 −1 cβ1 2 −1 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds 0

+ λ1 −1 ≥



1

0

μ2 −1 cβ1 2 −1

α1 cβ1 2 −1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds



c2 c1

J 3 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ + λ1 −1 cβ1 2 −1



c2

c1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds

J 4 (s)ϕ1





× f (τ, u(τ ), v(τ )) dτ 



1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

 s 1 (s − τ )α2 −1 Γ(α ) 2 c1 c1 i × (g∞ − ε)(u(τ ) + v(τ ))r2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1

c2



J 3 (s)ϕ2



 s 1 (s − τ )α1 −1 Γ(α1 ) c1 c1 i × (f∞ − ε)(u(τ ) + v(τ ))r1 −1 dτ ds + λ1 −1 cβ1 2 −1

≥



c2

J 4 (s)ϕ1



 s 1 (s − τ )α2 −1 Γ(α2 ) c1 c1 i r2 −1 × (g∞ − ε)(θ (u, v) Y ) dτ ds μ2 −1 cβ1 2 −1

c2



J 3 (s)ϕ2

c2



1 Γ(α 1) c1 i × (f∞ − ε)(θ (u, v) Y )r1 −1 dτ ds + λ1 −1 cβ1 2 −1

J 4 (s)ϕ1



s

c1

(s − τ )α1 −1

i = θθ2 μ2 −1 (g∞ − ε)2 −1 (u, v) Y  c2 1 J 3 (s) × (s − c1 )α2 (2 −1) ds (Γ(α2 + 1))2 −1 c1 i + θθ2 λ1 −1 (f∞ − ε)1 −1 (u, v) Y

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 ×

c2

c1

J 4 (s)

page 207

207

1 (s − c1 )α1 (1 −1) ds (Γ(α1 + 1))1 −1

i

+ θθ2 λ1 −1 = [θθ2 μ2 −1 (g∞ − ε)2 −1 C i

v) Y ×(f∞ − ε)1 −1 D] (u,

≥ [(1 − a)(1 − γ2 ) + (1 − a)γ2 ] (u, v) Y = (1 − a) (u, v) Y . So, Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ (1 − a) (u, v) Y . Hence, for (u, v) ∈ P ∩ ∂Ω2 we obtain Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≥ a (u, v) Y + (1 − a) (u, v) Y = (u, v) Y .

(4.43)

By using (4.42), (4.43), Lemma 4.2.3 and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2), we deduce that Q has a fixed point (u, v) ∈ P ∩ (Ω2 \ Ω1 ) such that R1 ≤ u + v ≤ R2 , u(t) ≥ tβ1 −1 u , v(t) ≥ tβ2 −1 v for all t ∈ [0, 1]. If u > 0 then u(t) > 0 for all t ∈ (0, 1] and if v > 0 then v(t) > 0 for all t ∈ (0, 1]. So (u, v) is a positive solution for problem (4.32), (4.33). i = ∞. Let λ ∈ Case (7). We consider here g0s = 0, f0s ∈ (0, ∞) and g∞  (0, L2 ) and μ ∈ (0, ∞). Instead of the numbers γ3 , γ4 ∈ (0, 1) used in the 

 D first case, we choose γ

3 ∈ (λf0s )1 −1 Ab , 1 and γ

4 ∈ (λf0s )1 −1 1−b ,1 .

r1 −1 The choice of

γ3 and

γ4 is possible because λ < f1s Ab and λ < 0

1−b r1 −1 1 . Let ε > 0 be such that f0s D   r −1  r −1  1 b

γ3 1 (1 − b)

γ4 1 1 λ ≤ min , s , f0s + ε A f0 + ε D    r2 −1 r −1 1

3 ) 2 1 b(1 − γ ε ≤ μ ≤ min ,

ε B θθ1 B  r −1 

4 ) 2 1 (1 − b)(1 − γ . ε C

By using (I2) and the definition of f0s and g0s , we deduce that there exists R1 > 0 such that f (t, u, v) ≤ (f0s + ε)(u + v)r1 −1 , for all t ∈ [0, 1] and u, v ≥ 0, u + v ≤ R1 .

g(t, u, v) ≤ ε(u + v)r2 −1

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We define the set Ω1 = {(u, v) ∈ Y, (u, v) Y < R1 }. In a similar manner as in the proof of Case (1), for any (u, v) ∈ P ∩ ∂Ω1 , we obtain Q1 (u, v)(t) ≤ [λ1 −1 (f0s + ε)1 −1 A + μ2 −1 ε2 −1 B] (u, v) Y ≤ [b

γ3 + b(1 − γ

3 )] (u, v) Y = b (u, v) Y , Q2 (u, v)(t) ≤ [μ

2 −1 2 −1

ε

1 −1

C+λ

(f0s

1 −1

+ ε)

∀t ∈ [0, 1],

D] (u, v) Y

≤ [(1 − b)(1 − γ

4 ) + (1 − b)

γ4 ] (u, v) Y = (1 − b) (u, v) Y ,

∀t ∈ [0, 1],

and so Q(u, v) Y ≤ (u, v) Y . i , there exists For the second part of the proof, by the definition of g∞ R2 > 0 such that g(t, u, v) ≥

1 (u + v)r2 −1 , ε

∀t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≥ R2 .

We consider R2 = max{2R1 , R2 /θ} and we define Ω2 = {(u, v) ∈ Y , (u, v) Y < R2 }. Then for (u, v) ∈ P ∩ ∂Ω2 , we deduce as in Case (1) that u(t) + v(t) ≥ θR2 ≥ R2 for all t ∈ [c1 , c2 ]. Then by Lemma 4.2.2, we have  1 α1 1 −1 cβ1 1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds Q1 (u, v)(c1 ) ≥ λ 0

+μ

2 −1

≥ μ2 −1



 0

1

0 1

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

≥ μ2 −1 cβ1 1 −1



c2

c1

J 2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ ≥

μ2 −1 cβ1 1 −1



c2

c1



1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds

J 2 (s)ϕ2



1 Γ(α2 )



s

c1

(s − τ )α2 −1

1 r2 −1 × (u(τ ) + v(τ )) dτ ds ε   c2  s 1 J 2 (s)ϕ2 ≥ μ2 −1 cβ1 1 −1 (s − τ )α2 −1 Γ(α ) 2 c1 c1

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209

1 (θ (u, v) Y )r2 −1 dτ ds ε  2 −1 1 2 −1 (u, v) Y = θθ1 μ ε  c2 1 J 2 (s) × (s − c1 )α2 (2 −1) ds (Γ(α2 + 1))2 −1 c1  2 −1 1 2 −1

≥ (u, v) Y . = θθ1 μ (u, v) Y B ε ×

So, we conclude that Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q1 (u, v) ≥ (u, v) Y . Therefore, we deduce the conclusion of the theorem.  s s , g∞ ∈ (0, ∞) and numbers γ1 , γ2 ∈ [0, 1], In what follows, for f0i , g0i , f∞ γ3 , γ4 ∈ (0, 1), a ∈ [0, 1] and b ∈ (0, 1), we define the numbers   r1 −1 r1 −1   aγ (1 − a)γ 1 1 1 2

1 = max L , i ,



f0i θθ1 A f0 θθ2 D    r −1 r −1  bγ3 1 (1 − b)γ4 1 1 1

L2 = min , s , s f∞ A f∞ D   r −1 r −1   1 a(1 − γ1 ) 2 1 (1 − a)(1 − γ2 ) 2

, L3 = max , i



g0i g0 θθ1 B θθ2 C    r2 −1 r2 −1  b(1 − γ (1 − b)(1 − γ ) ) 1 1 3 4

4 = min L , s , s g∞ B g∞ C   r1 −1  r1 −1  b 1 − b 1 1 

2 = min L , s , s f∞ A f∞ D   r2 −1  r −1  b 1−b 2 1 1 

L4 = min , s . s g∞ B g∞ C

Theorem 4.2.2. Assume that (I1) and (I2) hold, [c1 , c2 ] ⊂ [0, 1] with 0 < c1 < c2 ≤ 1, γ1 , γ2 ∈ [0, 1], γ3 , γ4 ∈ (0, 1), a ∈ [0, 1] and b ∈ (0, 1). s s

1 < L

2 and L

3 < L

4 , then for each , g∞ ∈ (0, ∞), L (1) If f0i , g0i , f∞





λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33).

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s s

1 < L

 , then for each λ ∈ (L

1, L

 ) (2) If f0i , g0i , f∞ ∈ (0, ∞), g∞ = 0 and L 2 2

3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] and μ ∈ (L for (4.32), (4.33). s s

3 < L

4 , then for each λ ∈ (L

1 , ∞) ∈ (0, ∞), f∞ = 0 and L (3) If f0i , g0i , g∞ 



and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). s s

1 , ∞) and = g∞ = 0, then for each λ ∈ (L (4) If f0i , g0i ∈ (0, ∞), f∞

μ ∈ (L3 , ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). s s , g∞ ∈ (0, ∞) and at least one of f0i , g0i is ∞, then for (5) If f∞

4 ) there exists a positive solution

2 ) and μ ∈ (0, L each λ ∈ (0, L (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). s s ∈ (0, ∞), g∞ = 0 and at least one of f0i , g0i is ∞, then (6) If f∞ 

) and μ ∈ (0, ∞) there exists a positive solution for each λ ∈ (0, L 2 (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). s s = 0, g∞ ∈ (0, ∞) and at least one of f0i , g0i is ∞, then (7) If f∞

4 ) there exists a positive solution for each λ ∈ (0, ∞) and μ ∈ (0, L (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33). s s = g∞ = 0 and at least one of f0i , g0i is ∞, then for each λ ∈ (0, ∞) (8) If f∞ and μ ∈ (0, ∞) there exists a positive solution (u(t), v(t)), t ∈ [0, 1] for (4.32), (4.33).

Proof. We consider the cone P ⊂ Y and the operators Q1 , Q2 and Q defined at the beginning of this section. Because the proofs of the above cases are similar, in what follows, we will prove two of them, namely Cases (1) and (7). s s

1 < L

2 and L

3 < L

4 . Let , g∞ ∈ (0, ∞), L Case (1). We have f0i , g0i , f∞ i





λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ). We consider ε > 0 such that ε < f0 , ε < g0i and

 max

1 f0i − ε 

≤ min



aγ1

θθ1 A

1 s f∞ + ε

r1 −1 

bγ3 A

1 , i f0 − ε

r1 −1



(1 − a)γ2

θθ2 D

1 , s f∞ + ε



r1 −1 

(1 − b)γ4 D

≤λ

r1 −1 

,

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 max

1 i g0 − ε 

≤ min



a(1 − γ1 )

θθ1 B

1 s g∞ + ε



r2 −1

1 , i g0 − ε

b(1 − γ3 ) B

r2 −1



(1 − a)(1 − γ2 )

θθ2 C

1 , s g∞ + ε



r2 −1 

(1 − b)(1 − γ4 ) C

211

≤μ r2 −1 

.

By using (I2) and the definition of f0i and g0i , we deduce that there exists R3 > 0 such that f (t, u, v) ≥ (f0i − ε)(u + v)r1 −1 ,

g(t, u, v) ≥ (g0i − ε)(u + v)r2 −1 ,

for all t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≤ R3 . We denote by Ω3 = {(u, v) ∈ Y, (u, v) Y < R3 }. Let (u, v) ∈ P with (u, v) Y = R3 , that is u + v = R3 . Because u(t)+v(t) ≤ u + v = R3 for all t ∈ [0, 1], then by Lemma 4.2.2, we obtain Q1 (u, v)(c1 ) ≥

λ1 −1 cβ1 1 −1



c2

c1

J 1 (s)ϕ1



c2

c1



s c1

(s − τ )α1 −1

ds

J 2 (s)ϕ2



1 Γ(α2 )



s c1



× g(τ, u(τ ), v(τ )) dτ ≥

1 Γ(α1 )



× f (τ, u(τ ), v(τ )) dτ + μ2 −1 cβ1 1 −1



(s − τ )α2 −1

ds 

 s 1 (s − τ )α1 −1 Γ(α1 ) c1 c1 i r1 −1 × (f0 − ε)(u(τ ) + v(τ )) dτ ds

λ1 −1 cβ1 1 −1



c2



J 1 (s)ϕ1

c2



1 Γ(α2 ) c1 i r2 −1 × (g0 − ε)(u(τ ) + v(τ )) dτ ds + μ2 −1 cβ1 1 −1

J 2 (s)ϕ2



s c1

(s − τ )α2 −1

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 s 1 (s − τ )α1 −1 Γ(α ) 1 c1 c1 × (f0i − ε)(θ (u, v) Y )r1 −1 dτ ds

≥ λ1 −1 cβ1 1 −1



c2



J 1 (s)ϕ1

c2



1 Γ(α2 ) c1 i r2 −1 × (g0 − ε)(θ (u, v) Y ) dτ ds + μ2 −1 cβ1 1 −1

J 2 (s)ϕ2



s c1

(s − τ )α2 −1

= θθ1 λ1 −1 (f0i − ε)1 −1 (u, v) Y  c2 1 J 1 (s) × (s − c1 )α1 (1 −1) ds (Γ(α1 + 1))1 −1 c1 + θθ1 μ2 −1 (g0i − ε)2 −1 (u, v) Y  c2 1 J 2 (s) × (s − c1 )α2 (2 −1) ds 2 −1 (Γ(α + 1)) 2 c1

+ θθ1 μ2 −1 (g0i − ε)2 −1 B] (u,

v) Y = [θθ1 λ1 −1 (f0i − ε)1 −1 A ≥ [aγ1 + a(1 − γ1 )] (u, v) Y = a (u, v) Y . Then, Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ a (u, v) Y . In a similar manner, we deduce Q2 (u, v)(c1 ) ≥ μ2 −1 cβ1 2 −1



c2

c1

J 3 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ + λ1 −1 cβ1 2 −1



c2

c1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds

J 4 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ 







1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds 

 s 1 (s − τ )α2 −1 ≥ Γ(α2 ) c1 c1 × (g0i − ε)(u(τ ) + v(τ ))r2 −1 dτ ds μ2 −1 cβ1 2 −1

c2

J 3 (s)ϕ2

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c2



1 Γ(α 1) c1 × (f0i − ε)(u(τ ) + v(τ ))r1 −1 dτ ds + λ1 −1 cβ1 2 −1

≥



J 4 (s)ϕ1



s

c1

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213

(s − τ )α1 −1



 s 1 (s − τ )α2 −1 Γ(α2 ) c1 c1 i r2 −1 × (g0 − ε)(θ (u, v) Y ) dτ ds μ2 −1 cβ1 2 −1

c2



J 3 (s)ϕ2

c2



1 Γ(α 1) c1 × (f0i − ε)(θ (u, v) Y )r1 −1 dτ ds + λ1 −1 cβ1 2 −1

J 4 (s)ϕ1



s

c1

(s − τ )α1 −1

= θθ2 μ2 −1 (g0i − ε)2 −1 (u, v) Y  c2 1 J 3 (s) × (s − c1 )α2 (2 −1) ds (Γ(α + 1))2 −1 2 c1 + θθ2 λ1 −1 (f0i − ε)1 −1 (u, v) Y  c2 1 J 4 (s) × (s − c1 )α1 (1 −1) ds (Γ(α1 + 1))1 −1 c1

+ θθ2 λ1 −1 (f0i − ε)1 −1 D] (u,

v) Y = [θθ2 μ2 −1 (g0i − ε)2 −1 C ≥ [(1 − a)(1 − γ2 ) + (1 − a)γ2 ] (u, v) Y = (1 − a) (u, v) Y . So, Q2 (u, v) ≥ Q2 (u, v)(c1 ) ≥ (1 − a) (u, v) Y . Hence, for (u, v) ∈ P ∩ ∂Ω3 , we obtain Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≥ a (u, v) Y + (1 − a) (u, v) Y = (u, v) Y .

(4.44)

Now, we define the functions f ∗ , g ∗ : [0, 1] × [0, ∞) → [0, ∞) by f ∗ (t, x) = max0≤u+v≤x f (t, u, v), g ∗ (t, x) = max0≤u+v≤x g(t, u, v), for all t ∈ [0, 1] and x ∈ [0, ∞). Then f (t, u, v) ≤ f ∗ (t, x),

g(t, u, v) ≤ g ∗ (t, x),

∀t ∈ [0, 1], u, v ≥ 0, u + v ≤ x.

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The functions f ∗ (t, ·), g ∗ (t, ·) are nondecreasing for every t ∈ [0, 1] and they satisfy the conditions f ∗ (t, x) s = f∞ , t∈[0,1] xr1 −1

lim sup max x→∞

g ∗ (t, x) s = g∞ . t∈[0,1] xr2 −1

lim sup max x→∞

Therefore, for ε > 0 there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1] we have f ∗ (t, x) f ∗ (t, x) s ≤ lim sup max + ε = f∞ + ε, r −1 r −1 x1 x→∞ t∈[0,1] x 1 g ∗ (t, x) g ∗ (t, x) s ≤ lim sup max + ε = g∞ + ε, r −1 xr2 −1 x→∞ t∈[0,1] x 2 s s + ε)xr1 −1 and g ∗ (t, x) ≤ (g∞ + ε)xr2 −1 . and so f ∗ (t, x) ≤ (f∞ We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y , (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definition of f ∗ and g ∗ we conclude

f (t, u(t), v(t)) ≤ f ∗ (t, (u, v) Y ), g(t, u(t), v(t)) ≤ g ∗ (t, (u, v) Y ), Then for all t ∈ [0, 1] we obtain   1 J 1 (s)ϕ1 Q1 (u, v)(t) ≤ λ1 −1

1 Γ(α 1) 0 × f (τ, u(τ ), v(τ )) dτ ds

+ μ2 −1



1 0

J 2 (s)ϕ2



× g(τ, u(τ ), v(τ )) dτ 

1



+μ

2 −1

J 1 (s)ϕ1



1 0

× g ∗ (τ, (u, v) Y ) dτ

(s − τ )α1 −1



s

0

(s − τ )α2 −1

ds



J 2 (s)ϕ2

s

0

1 Γ(α2 )

1 Γ(α1 ) 0 ∗ × f (τ, (u, v) Y ) dτ ds

≤ λ1 −1



∀t ∈ [0, 1].



 0

1 Γ(α2 ) ds

s

(s − τ )α1 −1

 0

s

(s − τ )α2 −1

(4.45)

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 s 1 ≤λ (s − τ )α1 −1 Γ(α1 ) 0 0 s × (f∞ + ε) (u, v) rY1 −1 dτ ds 1 −1



1

J 1 (s)ϕ1



 s 1 (s − τ )α2 −1 Γ(α ) 2 0 0 s × (g∞ + ε) (u, v) rY2 −1 dτ ds + μ2 −1



1

J 2 (s)ϕ2

s + ε)1 −1 (u, v) Y = λ1 −1 (f∞  1 1 × sα1 (1 −1) J 1 (s) ds (Γ(α + 1))1 −1 1 0 s + ε)2 −1 (u, v) Y + μ2 −1 (g∞  1 1 × sα2 (2 −1) J 2 (s) ds 2 −1 0 (Γ(α2 + 1)) s s + ε)1 −1 A + μ2 −1 (g∞ + ε)2 −1 B] (u, v) Y = [λ1 −1 (f∞

≤ [bγ3 + b(1 − γ3 )] (u, v) Y = b (u, v) Y , and so Q1 (u, v) ≤ b (u, v) Y . In a similar manner, we conclude 

1



1 Q2 (u, v)(t) ≤ μ Γ(α 2) 0 × g(τ, u(τ ), v(τ )) dτ ds 2 −1

+ λ1 −1

J 3 (s)ϕ2



1

0

J 4 (s)ϕ1



× f (τ, u(τ ), v(τ )) dτ 

1



J 3 (s)ϕ2



s

0

1 Γ(α1 )

(s − τ )α2 −1

 0

s

(s − τ )α1 −1

ds

1 Γ(α2 ) 0 × g ∗ (τ, (u, v) Y ) dτ ds

≤μ

2 −1



 0

s

(s − τ )α2 −1

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215

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1 −1



+λ

1

0

J 4 (s)ϕ1

× f ∗ (τ, (u, v) Y ) dτ



1 Γ(α1 )

 0

s

(s − τ )α1 −1

ds



 s 1 (s − τ )α2 −1 Γ(α ) 2 0 0 s × (g∞ + ε) (u, v) rY2 −1 dτ ds

≤ μ2 −1



1



J 3 (s)ϕ2



 s 1 (s − τ )α1 −1 Γ(α1 ) 0 0 r1 −1 s × (f∞ + ε) (u, v) Y dτ ds + λ1 −1

1

J 4 (s)ϕ1

s = μ2 −1 (g∞ + ε)2 −1 (u, v) Y  1 1 × sα2 (2 −1) J 3 (s) ds 2 −1 (Γ(α + 1)) 2 0 s + ε)1 −1 (u, v) Y + λ1 −1 (f∞  1 1 × sα1 (1 −1) J 4 (s) ds 1 −1 0 (Γ(α1 + 1)) s s = [μ2 −1 (g∞ + ε)2 −1 C + λ1 −1 (f∞ + ε)1 −1 D] (u, v) Y

≤ [(1 − b)(1 − γ4 ) + (1 − b)γ4 ] (u, v) Y = (1 − b) (u, v) Y ,

∀t ∈ [0, 1].

Hence, Q2 (u, v) ≤ (1 − b) (u, v) Y . Then for (u, v) ∈ P ∩ ∂Ω4 , we deduce Q(u, v) Y = Q1 (u, v) + Q2 (u, v) ≤ b (u, v) Y + (1 − b) (u, v) Y = (u, v) Y .

(4.46)

By using (4.44), (4.46), Lemma 4.2.3 and the Guo–Krasnosel’skii fixed point theorem (Theorem 1.2.2), we conclude that Q has a fixed point (u, v) ∈ P ∩ (Ω4 \ Ω3 ) which is a positive solution of problem (4.32), (4.33). s s = 0, g∞ ∈ (0, ∞) and f0i = ∞. Let λ ∈ (0, ∞) Case (7). We consider f∞

 ). Instead of the numbers γ3 , γ4 ∈ (0, 1) used in the first and μ ∈ (0, L 4  

 s 2 −1 B s 2 −1 C case, we choose γ

3 ∈ 0, 1 − (μg∞ ) ∈ 0, 1 − (μg ) and γ

4 ∞ b 1−b .

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The choice of γ

3 and

γ4 is possible because μ
0 such that gs C ∞

 ε

1

θθ1 A

μ ≤ min

r1 −1 

1 s g∞

b r2 −1 B

217

and μ
0 such that f (t, u, v) ≥

1 (u + v)r1 −1 , ε

∀t ∈ [c1 , c2 ], u, v ≥ 0, u + v ≤ R3 .

We denote by Ω3 = {(u, v) ∈ Y, (u, v) Y < R3 }. Let (u, v) ∈ P ∩ ∂Ω3 , that is u + v = R3 . Because u(t) + v(t) ≤ u + v = R3 for all t ∈ [0, 1], then by Lemma 4.2.2, we obtain  1 α1 1 −1 cβ1 1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds Q1 (u, v)(c1 ) ≥ λ 0

+μ

2 −1



1 0

≥ λ1 −1 cβ1 1 −1

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds



c2

c1

J 1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ ≥

λ1 −1 cβ1 1 −1



c2

c1



1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds

J 1 (s)ϕ1



1 Γ(α1 )



s

c1

(s − τ )α1 −1

1 r1 −1 × (u(τ ) + v(τ )) dτ ds ε   c2  s 1 J 1 (s)ϕ1 ≥ λ1 −1 cβ1 1 −1 (s − τ )α1 −1 Γ(α ) 1 c1 c1 1 × (θ (u, v) Y )r1 −1 dτ ds ε  1 −1 1

≥ (u, v) Y . = θθ1 λ1 −1 (u, v) Y A ε

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Hence, we get Q1 (u, v) ≥ Q1 (u, v)(c1 ) ≥ (u, v) Y and Q(u, v) Y ≥ Q1 (u, v) ≥ (u, v) Y . For the second part of the proof, we consider the functions f ∗ and g ∗ from Case (1), which satisfy in this case the conditions f ∗ (t, x) = 0, x→∞ t∈[0,1] xr1 −1 lim max

g ∗ (t, x) s = g∞ . t∈[0,1] xr2 −1

lim sup max x→∞

Then for ε > 0 there exists R4 > 0 such that for all x ≥ R4 and t ∈ [0, 1] we have f ∗ (t, x) f ∗ (t, x) ≤ lim max + ε = ε, r −1 x→∞ t∈[0,1] xr1 −1 x1 g ∗ (t, x) g ∗ (t, x) s ≤ lim sup max + ε = g∞ + ε, r −1 r −1 x2 x→∞ t∈[0,1] x 2 s and so f ∗ (t, x) ≤ εxr1 −1 and g ∗ (t, x) ≤ (g∞ + ε)xr2 −1 . We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v) ∈ Y, (u, v) Y < R4 }. Let (u, v) ∈ P ∩ ∂Ω4 . By the definition of f ∗ and g ∗ , we deduce the relation (4.45). In addition, in a similar manner as in the proof of Case (1), we conclude s + ε)2 −1 B (u, v) Y Q1 (u, v)(t) ≤ λ1 −1 ε1 −1 A (u, v) Y + μ2 −1 (g∞

≤ [b

γ3 + b(1 − γ

3 )] (u, v) Y = b (u, v) Y ,

∀t ∈ [0, 1],

s + ε)2 −1 C (u, v) Y + λ1 −1 ε1 −1 D (u, v) Y Q2 (u, v)(t) ≤ μ2 −1 (g∞

≤ [(1 − b)(1 − γ

4 ) + (1 − b)

γ4 ] (u, v) Y = (1 − b) (u, v) Y ,

∀t ∈ [0, 1],

and so Q(u, v) Y ≤ b (u, v) Y + (1 − b) (u, v) Y = (u, v) Y . Therefore, we obtain the conclusion of the theorem. 4.2.3



Nonexistence of positive solutions

In this section, we present intervals for λ and μ for which our problem (4.32), (4.33) has no positive solutions.

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219

Theorem 4.2.3. Assume that (I1) and (I2) hold. If there exist positive numbers M1 , M2 such that f (t, u, v) ≤ M1 (u + v)r1 −1 , g(t, u, v) ≤ M2 (u + v)r2 −1 ,

∀t ∈ [0, 1], u, v ≥ 0,

(4.47)

then there exist positive constants λ0 and μ0 such that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ) the boundary value problem (4.32), (4.33) has no positive solution.   1 1 and μ0 = Proof. We define λ0 = min M1 (4A) r1 −1 , M (4D)r1 −1 1   1 1 min M2 (4B) , where A, B, C, D are given in (4.41). r2 −1 , M (4C)r2 −1 2 We will prove that for every λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ), the problem (4.32), (4.33) has no positive solution. Let λ ∈ (0, λ0 ) and μ ∈ (0, μ0 ). We suppose that the problem (4.32), (4.33) has a positive solution (u(t), v(t)), t ∈ [0, 1]. Then we obtain   1  s 1 J 1 (s)ϕ1 (s − τ )α1 −1 u(t) = Q1 (u, v)(t) ≤ λ1 −1 Γ(α1 ) 0 0 × f (τ, u(τ ), v(τ )) dτ ds   s 1

J2 (s)ϕ2 +μ (s − τ )α2 −1 Γ(α2 ) 0 0 × g(τ, u(τ ), v(τ )) dτ ds 2 −1





1



 s 1 ≤λ (s − τ )α1 −1 Γ(α1 ) 0 0 × M1 (u(τ ) + v(τ ))r1 −1 dτ ds 1 −1

1

J 1 (s)ϕ1



 s 1 (s − τ )α2 −1 Γ(α ) 2 0 0 × M2 (u(τ ) + v(τ ))r2 −1 dτ ds + μ2 −1



1

J 2 (s)ϕ2

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1 −1

≤λ



M11 −1

1

J 1 (s)ϕ1

0

× ( u + v )r1 −1 dτ +μ

2 −1

M22 −1



1





× ( u + v )r2 −1 dτ

s

0

(s − τ )α1 −1

ds

J 2 (s)ϕ2

0



1 Γ(α1 )





1 Γ(α2 )



s

0

(s − τ )α2 −1

ds

= λ1 −1 M11 −1 A (u, v) Y + μ2 −1 M22 −1 B (u, v) Y ,

∀t ∈ [0, 1],

and v(t) = Q2 (u, v)(t) ≤ μ

2 −1



1

0

J 3 (s)ϕ2



× g(τ, u(τ ), v(τ )) dτ

1 Γ(α2 )



s

0

(s − τ )α2 −1

ds 

 s 1 (s − τ )α1 −1 f (τ, u(τ ), v(τ )) dτ ds Γ(α1 ) 0 0   1  s 1 2 −1

J3 (s)ϕ2 ≤μ (s − τ )α2 −1 Γ(α2 ) 0 0 × M2 (u(τ ) + v(τ ))r2 −1 dτ ds + λ1 −1



1



1

J 4 (s)ϕ1



 s 1 (s − τ )α1 −1 Γ(α1 ) 0 0 × M1 (u(τ ) + v(τ ))r1 −1 dτ ds 1 −1

+λ

J 4 (s)ϕ1

≤ μ2 −1 M22 −1

 0

1

J 3 (s)ϕ2

r2 −1

× ( u + v )

dτ

ds



1 Γ(α2 )

 0

s

(s − τ )α2 −1

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1 −1

+λ

M11 −1



1 0

J 4 (s)ϕ1

× ( u + v )r1 −1 dτ





1 Γ(α1 )



s

0

221

(s − τ )α1 −1

ds

= μ2 −1 M22 −1 C (u, v) Y + λ1 −1 M11 −1 D (u, v) Y ,

∀t ∈ [0, 1].

Then we deduce u ≤ λ1 −1 M11 −1 A (u, v) Y + μ2 −1 M22 −1 B (u, v) Y < λ01 −1 M11 −1 A (u, v) Y + μ02 −1 M22 −1 B (u, v) Y ≤

1 1 1 (u, v) Y + (u, v) Y = (u, v) Y , 4 4 2

v ≤ μ2 −1 M22 −1 C (u, v) Y + λ1 −1 M11 −1 D (u, v) Y < μ02 −1 M22 −1 C (u, v) Y + λ01 −1 M11 −1 D (u, v) Y ≤

1 1 1 (u, v) Y + (u, v) Y = (u, v) Y , 4 4 2

and so (u, v) Y = u + v < (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.32), (4.33) has no positive solution.  s s , g∞ < ∞, then there exist positive constants Remark 4.2.1. If f0s , g0s , f∞ M1 , M2 such that relation (4.47) holds, and then we obtain the conclusion of Theorem 4.2.3.

Theorem 4.2.4. Assume that (I1) and (I2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m1 > 0 such that f (t, u, v) ≥ m1 (u + v)r1 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.48)

0 such that for every λ > λ

0 and then there exists a positive constant λ μ > 0, the boundary value problem (4.32), (4.33) has no positive solution.   1 1

0 = min

and D

, where A , Proof. We define λ r −1 r −1   1 1 m (θθ A) m (θθ D) 1

1

1

2

are given by (4.41).

0 and μ > 0 the problem (4.32), (4.33) We will show that for every λ > λ

0 and μ > 0. We suppose that the has no positive solution. Let λ > λ problem (4.32), (4.33) has a positive solution (u(t), v(t)), t ∈ [0, 1].

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0 =

≥ θ2 D,

then λ If θ1 A

1  r1 −1 , m1 (θθ1 A)

1 −1



u(c1 ) = Q1 (u, v)(c1 ) ≥ λ +μ

2 −1



1

0

≥ λ1 −1 cβ1 1 −1

1

0

and therefore, we obtain

α1 cβ1 1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds



c2

c1

J 1 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ 

c2



1 Γ(α1 )

≥

J 1 (s)ϕ1

λ1 −1 cβ1 1 −1 m11 −1



s

c1

(s − τ )α1 −1

ds 

1 Γ(α 1) c1 × m1 (u(τ ) + v(τ ))r1 −1 dτ ds

≥ λ1 −1 cβ1 1 −1



c2

c1

× (θ (u, v) Y )r1 −1 dτ

J 1 (s)ϕ1







s

c1

(s − τ )α1 −1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds

v) Y . = (λm1 )1 −1 θθ1 A (u, Then we conclude

v) Y u ≥ u(c1 ) ≥ (λm1 )1 −1 θθ1 A (u,

0 m1 )1 −1 θθ1 A (u,

v) Y = (u, v) Y , > (λ and so (u, v) Y = u + v ≥ u > (u, v) Y , which is a contradiction. 1

0 =

< θ2 D,

then λ If θ1 A  r1 −1 , and therefore, we deduce m (θθ D) 1

v(c1 ) = Q2 (u, v)(c1 ) ≥ μ2 −1 + λ1 −1 ≥



1

0

λ1 −1 cβ1 2 −1



2

1

0

α2 cβ1 2 −1 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

α1 cβ1 2 −1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds



c2

c1

J 4 (s)ϕ1

× f (τ, u(τ ), v(τ )) dτ

ds



1 Γ(α1 )



s

c1

(s − τ )α1 −1

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c2



1 Γ(α 1) c1 × m1 (u(τ ) + v(τ ))r1 −1 dτ ds

≥ λ1 −1 cβ1 2 −1

≥

J 4 (s)ϕ1

λ1 −1 cβ1 2 −1 m11 −1



c2 c1

× (θ (u, v) Y )r1 −1 dτ

J 4 (s)ϕ1







s

c1

page 223

223

(s − τ )α1 −1

1 Γ(α1 )



s

c1

(s − τ )α1 −1

ds

v) Y . = (λm1 )1 −1 θθ2 D (u, Then we conclude

v ≥ v(c1 ) ≥ (λm1 )1 −1 θθ2 D (u, v) Y

0 m1 )1 −1 θθ2 D (u,

v) Y = (u, v) Y , > (λ and so (u, v) Y = u + v ≥ v > (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.32), (4.33) has no positive solution.  i > 0 Remark 4.2.2. If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have f0i , f∞ and f (t, u, v) > 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.48) holds, and we obtain the conclusion of Theorem 4.2.4.

Theorem 4.2.5. Assume that (I1) and (I2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m2 > 0 such that g(t, u, v) ≥ m2 (u + v)r2 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.49)

then there exists a positive constant μ

0 such that for every μ > μ

0 and λ > 0, the boundary value problem (4.32), (4.33) has no positive solution. Proof. We define μ

0 = min



1 1  r2 −1 , m2 (θθ2 C)  r2 −1 m2 (θθ1 B)



and C

, where B

are given by (4.41). We will show that for every μ > μ

0 and λ > 0, the problem (4.32), (4.33) has no positive solution. Let μ > μ

0 and λ > 0. We suppose that the problem (4.32), (4.33) has a positive solution (u(t), v(t)), t ∈ [0, 1].

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≥ θ2 C,

then μ

0 = If θ1 B

1  r2 −1 , m2 (θθ1 B)

1 −1



u(c1 ) = Q1 (u, v)(c1 ) ≥ λ + μ2 −1 ≥



1

0

μ2 −1 cβ1 1 −1

1

0

and therefore, we obtain

α1 cβ1 1 −1 J 1 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds

α2 cβ1 1 −1 J 2 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds



c2

c1

J 2 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ 

c2



1 Γ(α2 )

≥

J 2 (s)ϕ2

μ2 −1 cβ1 1 −1 m22 −1



s

c1

(s − τ )α2 −1

ds 

1 Γ(α 2) c1 × m2 (u(τ ) + v(τ ))r2 −1 dτ ds

≥ μ2 −1 cβ1 1 −1



c2 c1

× (θ (u, v) Y )r2 −1 dτ

J 2 (s)ϕ2







s

c1

(s − τ )α2 −1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds

v) Y . = (μm2 )2 −1 θθ1 B (u, Then we conclude

v) Y u ≥ u(c1 ) ≥ (μm2 )2 −1 θθ1 B (u,

v) Y = (u, v) Y , > (

μ0 m2 )2 −1 θθ1 B (u, and so (u, v) Y = u + v ≥ u > (u, v) Y , which is a contradiction.

< θ2 C,

then μ If θ1 B

0 = m (θθ 1C)  r2 −1 , and therefore, we deduce 2

v(c1 ) = Q2 (u, v)(c1 ) ≥ μ2 −1 + λ1 −1 ≥



1

0

μ2 −1 cβ1 2 −1



2

1

0

α2 cβ1 2 −1 J 3 (s)ϕ2 (I0+ g(s, u(s), v(s))) ds

α1 cβ1 2 −1 J 4 (s)ϕ1 (I0+ f (s, u(s), v(s))) ds



c2

c1

J 3 (s)ϕ2

× g(τ, u(τ ), v(τ )) dτ

ds



1 Γ(α2 )



s

c1

(s − τ )α2 −1

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c2



1 Γ(α 2) c1 × m2 (u(τ ) + v(τ ))r2 −1 dτ ds

≥ μ2 −1 cβ1 2 −1

≥

J 3 (s)ϕ2

μ2 −1 cβ1 2 −1 m22 −1



r2 −1

× (θ (u, v) Y )

c2

c1

dτ

J 3 (s)ϕ2







s

c1

page 225

225

(s − τ )α2 −1

1 Γ(α2 )



s

c1

(s − τ )α2 −1

ds

v) Y . = (μm2 )2 −1 θθ2 C (u, Then we conclude

v ≥ v(c1 ) ≥ (μm2 )2 −1 θθ2 C (u, v) Y

v) Y = (u, v) Y , > (

μ0 m2 )2 −1 θθ2 C (u, and so (u, v) Y = u + v ≥ v > (u, v) Y , which is a contradiction. Therefore, the boundary value problem (4.32), (4.33) has no positive solution.  i > 0 Remark 4.2.3. If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have g0i , g∞ and g(t, u, v) > 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.49) holds, and we obtain the conclusion of Theorem 4.2.5.

Theorem 4.2.6. Assume that (I1) and (I2) hold. If there exist positive numbers c1 , c2 with 0 < c1 < c2 ≤ 1 and m1 , m2 > 0 such that f (t, u, v) ≥ m1 (u + v)r1 −1 , g(t, u, v) ≥ m2 (u + v)r2 −1 ,

∀t ∈ [c1 , c2 ], u, v ≥ 0,

(4.50)

ˆ 0 and ˆ0 and μ ˆ0 such that for every λ > λ then there exist positive constants λ μ>μ ˆ0 , the boundary value problem (4.32), (4.33) has no positive solution. 1 ˆ0 =

Proof. We define λ

0 = m (2θθ1C)  r1 −1 and μ  r2 −1 , where A m1 (2θθ1 A) 2 2

are given by (4.41). Then for every λ > λ ˆ 0 and μ > μ ˆ0 , the and C ˆ0 and problem (4.32), (4.33) has no positive solution. Indeed, let λ > λ μ>μ ˆ0 . We suppose that the problem (4.32), (4.33) has a positive solution (u(t), v(t)), t ∈ [0, 1]. In a similar manner as that used in the proofs of

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Theorems 4.2.4 and 4.2.5, we obtain

u ≥ u(c1 ) ≥ (λm1 )1 −1 θθ1 A (u, v) Y ,

v) Y , v ≥ v(c1 ) ≥ (μm2 )2 −1 θθ2 C (u, and so

v) Y (u, v) Y = u + v ≥ (λm1 )1 −1 θθ1 A (u,

v) Y + (μm2 )2 −1 θθ2 C (u, ˆ 0 m1 )1 −1 θθ1 A (u,



v) Y + (ˆ v) Y > (λ μ0 m2 )2 −1 θθ2 C (u, =

1 1 (u, v) Y + (u, v) Y = (u, v) Y , 2 2

which is a contradiction. Therefore, the boundary value problem (4.32), (4.33) has no positive solution. 1 ˆ =

0 = m (2θθ1B) We can also define λ 0  r1 −1 and μ  r2 −1 , where m1 (2θθ2 D) 2 1  ˆ and μ > μ

and D

are given by (4.41). Then for every λ > λ B ˆ0 , the 0 ˆ and problem (4.32), (4.33) has no positive solution. Indeed, let λ > λ 0

μ>μ ˆ0 . We suppose that the problem (4.32), (4.33) has a positive solution (u(t), v(t)), t ∈ [0, 1]. In a similar manner as that used in the proofs of Theorems 4.2.4 and 4.2.5, we obtain

v) Y , v ≥ v(c1 ) ≥ (λm1 )1 −1 θθ2 D (u,

v) Y , u ≥ u(c1 ) ≥ (μm2 )2 −1 θθ1 B (u, and so

v) Y (u, v) Y = u + v ≥ (λm1 )1 −1 θθ2 D (u,

v) Y + (μm2 )2 −1 θθ1 B (u, ˆ  m1 )1 −1 θθ2 D (u,



v) Y + (ˆ v) Y > (λ μ0 m2 )2 −1 θθ1 B (u, 0 =

1 1 (u, v) Y + (u, v) Y = (u, v) Y , 2 2

which is a contradiction. Therefore, the boundary value problem (4.32), (4.33) has no positive solution. 

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227

i Remark 4.2.4. If for c1 , c2 with 0 < c1 < c2 ≤ 1, we have f0i , f∞ , g0i , i g∞ > 0 and f (t, u, v) > 0, g(t, u, v) > 0 for all t ∈ [c1 , c2 ] and u, v ≥ 0 with u + v > 0, then the relation (4.50) holds, and we obtain the conclusion of Theorem 4.2.6.

4.2.4

An example

Let α1 = 1/3, α2 = 1/4, β1 = 7/2, n = 4, β2 = 14/3, m = 5, p1 = 4/3, p2 = 5/2, q1 = 5/4, q2 = 2/3, N = 2, ξ1 = 1/4, ξ2 = 3/5, a1 = 2, a2 = 1/3, M = 1, η1 = 1/2, b1 = 4, r1 = 5, 1 = 5/4, ϕr1 (s) = s|s|3 , ϕ1 (s) = s|s|−3/4 , r2 = 3, 2 = 3/2, ϕr2 (s) = s|s|, ϕ2 (s) = s|s|−1/2 . We consider the system of fractional differential equations ⎧ 1/3 7/2 ⎨D0+ (ϕ5 (D0+ u(t))) + λ(t + 1)a (u5 (t) + v 5 (t)) = 0, t ∈ (0, 1), ⎩

1/4

14/3



2

D0+ (ϕ3 (D0+ v(t))) + μ(2 − t)b (e(u(t)+v(t)) − 1) = 0,

t ∈ (0, 1), (4.51)

with the coupled multi-point boundary conditions ⎧ 7/2 ⎪ u(0) = u (0) = u (0) = 0, D0+ u(0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ 4/3 1 3 1 5/4 ⎪ 5/4 ⎪ ⎪ ⎨D0+ u(1) = 2D0+ v 4 + 3 D0+ v 5 , ⎪ ⎪ v(0) = v  (0) = v  (0) = v  (0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ 1 ⎪ 5/2 2/3 ⎪ ⎩D0+ u(1) = 4D0+ v , 2

14/3

D0+ v(0) = 0,

(4.52)

where

a, b > 0. 2  Here, we have f (t, u, v) = (t+1)a (u5 +v 5 ), g(t, u, v) = (2−t)b (e(u+v) −1) for all t ∈ [0, 1] and u, v ≥ 0. Then we obtain Δ ≈ 39.98272963 > 0, and so the assumptions (I1) and (I2) are satisfied. In addition, we deduce  5/2 t (1 − s)7/6 − (t − s)5/2 , 0 ≤ s ≤ t ≤ 1, 1 g1 (t, s) =

Γ(7/2) t5/2 (1 − s)7/6 , 0 ≤ t ≤ s ≤ 1,  11/6 (1 − s)7/6 − (t − s)11/6 , 0 ≤ s ≤ t ≤ 1, t 1 g2 (t, s) =

Γ(17/6) t11/6 (1 − s)7/6 , 0 ≤ t ≤ s ≤ 1,

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 t29/12 (1 − s)7/6 − (t − s)29/12 , 0 ≤ s ≤ t ≤ 1, 1 g3 (t, s) =

Γ(41/12) t29/12 (1 − s)7/6 , 0 ≤ t ≤ s ≤ 1,  t11/3 (1 − s)7/6 − (t − s)11/3 , 0 ≤ s ≤ t ≤ 1, 1 g4 (t, s) =

Γ(14/3) t11/3 (1 − s)7/6 , 0 ≤ t ≤ s ≤ 1,    29/12   29/12 1 1 4t5/2 Γ(14/3) 1 3

G1 (t, s) = g 1 (t, s) + 2 ,s , + g 2 ΔΓ(41/12) 4 3 5 2     1 3 1 t5/2 Γ(14/3)

G2 (t, s) = , s + g 3 ,s , 2

g3 ΔΓ(13/6) 4 3 5     11/3 1 3 1 Γ(7/2)(1/2)11/6

3 (t, s) = g 4 (t, s) + 4t G 2

g3 ,s +

g3 ,s , ΔΓ(17/6) 4 3 5  11/3 1 Γ(7/2)

4 (t, s) = 4t G g 2 , s , ∀t, s ∈ [0, 1]. ΔΓ(13/6) 2 For the functions

h1 ,

h4 and J i , i = 1, . . . , 4, we obtain

h1 (s) =

1 (1 − s)7/6 (1 − (1 − s)4/3 ), Γ(7/2)

1 (1 − s)7/6 (1 − (1 − s)5/2 ), Γ(14/3) ⎧ 1 ⎪ ⎪ (1 − s)7/6 (1 − (1 − s)4/3 ) ⎪ ⎪ Γ(7/2) ⎪ ⎪ ⎪ ⎪   ⎪  29/12  ⎪ 29/12 ⎪ Γ(14/3) 1 1 3 ⎪ ⎪ ⎪ + 2 + ⎪ ⎪ ΔΓ(41/12) 4 3 5 ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ 11/6   ⎪ 11/6 ⎪ 1 1 4 ⎪ 7/6 ⎪ −s (1 − s) − , ⎪ ⎨ × Γ(17/6) 2 2

J1 (s) = ⎪ ⎪ 1 ⎪ ⎪ (1 − s)7/6 (1 − (1 − s)4/3 ) ⎪ ⎪ Γ(7/2) ⎪ ⎪ ⎪ ⎪   ⎪  29/12  ⎪ 29/12 ⎪ Γ(14/3) 1 3 1 ⎪ ⎪ ⎪ + 2 + ⎪ ⎪ ΔΓ(41/12) 4 3 5 ⎪ ⎪ ⎪ ⎪ ⎪  11/6 ⎪ ⎪ 4 1 1 ⎪ ⎪ × ≤ s ≤ 1, (1 − s)7/6 , ⎩ Γ(17/6) 2 2

h4 (s) =

1 0 ≤ s< , 2

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229

 ⎧  Γ(14/3) 2  ⎪ ⎪ (1 − s)7/6 − (1 − 4s)29/12 ⎪ ⎪ 29/12 ΔΓ(13/6)Γ(41/12) 4 ⎪ ⎪ ⎪  ⎪ ⎪  29/12  ⎪ 1 1 ⎪ + 7/6 29/12 ⎪ 3 , 0≤s< , (1 − s) − (3 − 5s) ⎪ 29/12 ⎪ 4 3 · 5 ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ 1 Γ(14/3) 2 ⎪ 7/6 ⎪ ⎨ ΔΓ(13/6)Γ(41/12) 429/12 (1 − s) + 3 · 529/12 J 2 (s) = ⎪ 3 1 ⎪ ⎪ ≤s< , × [329/12 (1 − s)7/6 − (3 − 5s)29/12 ] , ⎪ ⎪ 4 5 ⎪ ⎪ ⎪   ⎪  29/12  ⎪ 29/12 ⎪ ⎪ Γ(14/3) 1 1 3 ⎪ ⎪ 2 + ⎪ ⎪ ΔΓ(13/6)Γ(41/12) 4 3 5 ⎪ ⎪ ⎪ ⎪ ⎪ 3 ⎪ ⎩ × (1 − s)7/6 , ≤ s ≤ 1, 5 ⎧ 1 Γ(7/2)21/6 ⎪ ⎪ (1 − s)7/6 (1 − (1 − s)5/2 ) + ⎪ ⎪ ⎪ Γ(14/3) ΔΓ(17/6)Γ(41/12) ⎪ ⎪  ⎪ ⎪ ⎪ 2 1 ⎪ ⎪ × [(1 − s)7/6 − (1 − 4s)29/12 ] + ⎪ 29/12 29/12 ⎪ 4 3 · 5 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ × [329/12 (1 − s)7/6 − (3 − 5s)29/12 ] , 0 ≤ s < , ⎪ ⎪ ⎪ 4 ⎪ ⎪ ⎪ ⎪ ⎪ 1 Γ(7/2)21/6 ⎪ ⎪ ⎪ (1 − s)7/6 (1 − (1 − s)5/2 ) + ⎪ ⎨ Γ(14/3) ΔΓ(17/6)Γ(41/12)  J 3 (s) = 2 1 ⎪ ⎪ (1 − s)7/6 + [329/12 (1 − s)7/6 ⎪ × ⎪ 29/12 29/12 ⎪ 4 3 · 5 ⎪ ⎪ ⎪ ⎪ ⎪ 3 1 ⎪ ⎪ ≤s< , −(3 − 5s)29/12 ] , ⎪ ⎪ 4 5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 Γ(7/2)21/6 ⎪ ⎪ (1 − s)7/6 (1 − (1 − s)5/2 ) + ⎪ ⎪ Γ(14/3) ΔΓ(17/6)Γ(41/12) ⎪ ⎪ ⎪   ⎪  ⎪ 29/12 ⎪ ⎪ 2 1 3 3 ⎪ ⎪ ≤ s ≤ 1, (1 − s)7/6 , ⎩ × 429/12 + 3 5 5

J 4 (s) =

⎧   ⎪ Γ(7/2)21/6 ⎪ ⎪ (1 − s)7/6 − (1 − 2s)11/6 , ⎪ ⎨ ΔΓ(13/6)Γ(17/6) ⎪ ⎪ ⎪ ⎪ ⎩

Γ(7/2)21/6 (1 − s)7/6 , ΔΓ(13/6)Γ(17/6)

1 ≤ s ≤ 1. 2

0≤s
0 and μ > μ

0 , the boundary value problem (4.51), (4.52) has no positive solution.

Remark 4.2.5. The results presented in this section under the assumptions p1 ∈ [1, n−2] and p2 ∈ [1, m−2] instead of p1 ∈ [1, β1 −1) and p2 ∈ [1, β2 −1) were published in [80]. Remark 4.2.6. The existence of positive solutions for the system of Riemann-Liouville fractional differential equations ⎧ α ⎨D0+ u(t) + λf (t, u(t), v(t)) = 0,

t ∈ (0, 1),

⎩Dβ v(t) + μg(t, u(t), v(t)) = 0, 0+

t ∈ (0, 1),

(4.53)

subject to the coupled integral boundary conditions ⎧ ⎪  (n−2) ⎪ (0) = 0, ⎪ ⎨u(0) = u (0) = · · · = u

u (1) =

⎪ ⎪ ⎪ ⎩v(0) = v  (0) = · · · = v (m−2) (0) = 0,

v  (1) =





1

0



0

1

v(s) dH(s), u(s)dK(s),

where α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n, m ≥ 3, λ, μ are positive parameters, and H, K are nondecreasing functions, was investigated in [51] (in the case where f, g are nonnegative functions) and in [44] (in the case where f, g are sign-changing functions). The system (4.53) with f, g sign-changing functions, supplemented with the coupled integral boundary

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conditions ⎧ ⎪ ⎪u(0) = u (0) = · · · = u(n−2) (0) = 0, ⎪ ⎨

p D0+ u(1) =

⎪ ⎪ ⎪ ⎩v(0) = v  (0) = · · · = v (m−2) (0) = 0,

q D0+ v(1) =



1

0



0

1

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231

v(s) dH(s), u(s) dK(s),

where p, q ∈ R, p ∈ [1, n − 2], q ∈ [1, m − 2], and H, K are nondecreasing functions, was studied in [45].

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Chapter 5

Systems of Three Riemann–Liouville Fractional Differential Equations with Parameters and Multi-Point Boundary Conditions

In this chapter, we study the existence and nonexistence of positive solutions for systems of three Riemann-Liouville fractional differential equations with positive parameters subject to uncoupled multi-point boundary conditions, which contain fractional derivatives, and the nonlinearities of systems are nonnegative functions. 5.1

Systems of Fractional Differential Equations with Uncoupled Multi-Point Boundary Conditions

We consider the system of nonlinear ordinary fractional differential equations ⎧ α ⎪ u(t) + λf (t, u(t), v(t), w(t)) = 0, t ∈ (0, 1), D0+ ⎪ ⎪ ⎨ β (5.1) D0+ v(t) + μg(t, u(t), v(t), w(t)) = 0, t ∈ (0, 1), ⎪ ⎪ ⎪ ⎩ Dγ w(t) + νh(t, u(t), v(t), w(t)) = 0, t ∈ (0, 1), 0+

233

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with the multi-point boundary conditions ⎧ N  ⎪ p1 q1 ⎪ (j) ⎪ u (0) = 0, j = 0, . . . , n − 2; D0+ u(1) = ai D0+ u(ξi ), ⎪ ⎪ ⎪ ⎪ i=1 ⎪ ⎪ ⎪ ⎪ ⎪ M ⎨  p2 q2 (j) v (0) = 0, j = 0, . . . , m − 2; D0+ v(1) = bi D0+ v(ηi ), ⎪ ⎪ i=1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ L ⎪  ⎪ p3 q3 ⎪ (j) ⎪ ci D0+ w(ζi ), ⎩ w (0) = 0, j = 0, . . . , l − 2; D0+ w(1) =

(5.2)

i=1

where λ, μ, ν > 0, α, β, γ ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], γ ∈ (l − 1, l], n, m, l ∈ N, n, m, l ≥ 3, p1 , p2 , p3 , q1 , q2 , q3 ∈ R, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), p3 ∈ [1, γ − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], q3 ∈ [0, p3 ], ξi , ai ∈ R for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi , bi ∈ R for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, ζi , ci ∈ R for all i = 1, . . . , L k denotes the Riemann-Liouville (L ∈ N), 0 < ζ1 < · · · < ζL ≤ 1, and D0+ derivative of order k (for k = α, β, γ, p1 , q1 , p2 , q2 , p3 , q3 ). Under some assumptions on f, g and h, we give intervals for the parameters λ, μ and ν such that positive solutions of (5.1), (5.2) exist. By a positive solution of problem (5.1), (5.2) we mean a triplet of functions (u, v, w) ∈ (C([0, 1], R+ ))3 , satisfying (5.1) and (5.2) with u(t) > 0 for all t ∈ (0, 1], or v(t) > 0 for all t ∈ (0, 1], or w(t) > 0 for all t ∈ (0, 1]. The nonexistence of positive solutions for the above problem is also studied. The results obtained in this section improve and extend the results from [98], where only a few cases are presented for the existence of positive solutions for a system of integral equations, and, as an application, for a system with three fractional equations subject to some boundary conditions in points t = 0 and t = 1 (Application 4.3 from [98]). 5.1.1

Auxiliary results

We present firstly some auxiliary results from Section 2.1.1 that will be used to prove our main results. We consider the fractional differential equation α u(t) + x(t) = 0, D0+

0 < t < 1,

(5.3)

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with the multi-point boundary conditions u(j) (0) = 0,

j = 0, . . . , n − 2;

p1 D0+ u(1) =

N 

q1 ai D0+ u(ξi ),

(5.4)

i=1

where α ∈ (n − 1, n], n ∈ N, n ≥ 3, ai , ξi ∈ R, i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, p1 , q1 ∈ R, p1 ∈ [1, α − 1), q1 ∈ [0, p1 ], and Γ(α) Γ(α) N α−q1 −1 − Γ(α−q . x ∈ C[0, 1]. We denote by Δ1 = Γ(α−p i=1 ai ξi 1) 1) Lemma 5.1.1. If Δ1 = 0, then the unique solution u ∈ C[0, 1] of problem (5.3), (5.4) is  1 G1 (t, s)x(s) ds, t ∈ [0, 1], (5.5) u(t) = 0

where the Green function G1 is given by G1 (t, s) = g1 (t, s) + and 1 g1 (t, s) = Γ(α)



N tα−1  ai g2 (ξi , s), Δ1 i=1

∀ (t, s) ∈ [0, 1] × [0, 1],

tα−1 (1 − s)α−p1 −1 − (t − s)α−1 , t

α−1

(5.6)

0 ≤ s ≤ t ≤ 1,

α−p1 −1

(1 − s) , 0 ≤ t ≤ s ≤ 1, ⎧ α−q1 −1 ⎪ (1 − s)α−p1 −1 − (t − s)α−q1 −1 , ⎪t ⎨ 1 g2 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(α − q1 ) ⎪ ⎪ ⎩ tα−q1 −1 (1 − s)α−p1 −1 , 0 ≤ t ≤ s ≤ 1.

(5.7)

Lemma 5.1.2. The functions g1 and g2 given by (5.7) have the properties: (a) g1 (t, s) ≤ h1 (s) for all t, s ∈ [0, 1], where h1 (s) =

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 ), s ∈ [0, 1]; Γ(α)

(b) g1 (t, s) ≥ tα−1 h1 (s) for all t, s ∈ [0, 1]; α−1 (c) g1 (t, s) ≤ tΓ(α) , for all t, s ∈ [0, 1]; (d) g2 (t, s) ≥ tα−q1 −1 h2 (s) for all t, s ∈ [0, 1], where h2 (s) = (e) g2 (t, s) ≤

1 (1 − s)α−p1 −1 (1 − (1 − s)p1 −q1 ), s ∈ [0, 1]; Γ(α − q1 )

1 α−q1 −1 Γ(α−q1 ) t

for all t, s ∈ [0, 1];

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(f) The functions g1 and g2 are continuous on [0, 1] × [0, 1]; g1 (t, s) ≥ 0, g2 (t, s) ≥ 0 for all t, s ∈ [0, 1]; g1 (t, s) > 0, g2 (t, s) > 0 for all t, s ∈ (0, 1). Lemma 5.1.3. Assume that ai ≥ 0 for all i = 1, . . . , N and Δ1 > 0. Then the function G1 given by (5.6) is a nonnegative continuous function on [0, 1] × [0, 1] and satisfies the inequalities: (a) G1 (t, s) ≤ J1 (s) for all t, s ∈ [0, 1], where J1 (s) = h1 (s) + 1 N i=1 ai g2 (ξi , s), s ∈ [0, 1]; Δ1 (b) G1 (t, s) ≥ tα−1 J1 (s) for all t, s ∈ [0, 1]; 1 1 + Δ1 Γ(α−q (c) G1 (t, s) ≤ σ1 tα−1 , for all t, s ∈ [0, 1], where σ1 = Γ(α) 1) N α−q1 −1 × i=1 ai ξi . Lemma 5.1.4. Assume that ai ≥ 0 for all i = 1, . . . , N, Δ1 > 0, x ∈ C[0, 1] and x(t) ≥ 0 for all t ∈ [0, 1]. Then the solution u of problem (5.3), (5.4) given by (5.5) satisfies the inequality u(t) ≥ tα−1 u(t ) for all t, t ∈ [0, 1]. We can also formulate similar results as Lemmas 5.1.1–5.1.4 for the fractional boundary value problems β D0+ v(t) + y(t) = 0,

v (j) (0) = 0,

j = 0, . . . , m − 2;

0 < t < 1,

p2 D0+ v(1) =

M 

(5.8) q2 bi D0+ v(ηi ),

(5.9)

i=1

and γ D0+ w(t) + z(t) = 0,

w(j) (0) = 0,

j = 0, . . . , l − 2;

0 < t < 1,

p3 D0+ w(1) =

L 

q3 ci D0+ w(ζi ),

(5.10)

(5.11)

i=1

where β ∈ (m − 1, m], γ ∈ (l − 1, l], m, l ∈ N, m, l ≥ 3, bi , ηi ∈ R, i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, ci , ζi ∈ R, i = 1, . . . , L (L ∈ N), 0 < ζ1 < · · · < ζL ≤ 1, p2 , q2 , p3 , q3 ∈ R, p2 ∈ [1, β − 1), q2 ∈ [0, p2 ], p3 ∈ [1, γ − 1), q3 ∈ [0, p3 ], and y, z ∈ C[0, 1]. We denote by Δ2 , g3 , g4 , G2 , h3 , h4 , J2 and σ2 , and Δ3 , g5 , g6 , G3 , h5 , h6 , J3 and σ3 the corresponding constants and functions for problem (5.8),

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(5.9) and problem (5.10), (5.11), respectively, defined in a similar manner as Δ1 , g1 , g2 , G1 , h1 , h2 , J1 and σ1 , respectively. More precisely, we have Δ2 = g3 (t, s) =

M Γ(β)  β−q2 −1 Γ(β) − bi ηi , Γ(β − p2 ) Γ(β − q2 ) i=1 tβ−1 (1 − s)β−p2 −1 − (t − s)β−1 , 1

t

β−1

0 ≤ s ≤ t ≤ 1,

β−p2 −1

(1 − s) , 0 ≤ t ≤ s ≤ 1, ⎧ ⎪ tβ−q2 −1 (1 − s)β−p2 −1 − (t − s)β−q2 −1 , ⎪ ⎨ 1 g4 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(β − q2 ) ⎪ ⎪ ⎩ tβ−q2 −1 (1 − s)β−p2 −1 , 0 ≤ t ≤ s ≤ 1, Γ(β)

G2 (t, s) = g3 (t, s) +

M tβ−1  bi g4 (ηi , s), Δ2 i=1

∀ (t, s) ∈ [0, 1] × [0, 1],

h3 (s) =

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 ), s ∈ [0, 1], Γ(β)

h4 (s) =

1 (1 − s)β−p2 −1 (1 − (1 − s)p2 −q2 ), s ∈ [0, 1], Γ(β − q2 )

J2 (s) = h3 (s) +

M 1  bi g4 (ηi , s), s ∈ [0, 1], Δ2 i=1 M

 β−q −1 1 1 σ2 = + bi ηi 2 , Γ(β) Δ2 Γ(β − q2 ) i=1 and L Γ(γ)  γ−q3 −1 Γ(γ) − ci ζ , Γ(γ − p3 ) Γ(γ − q3 ) i=1 i tγ−1 (1 − s)γ−p3 −1 − (t − s)γ−1 , 0 ≤ s ≤ t ≤ 1, 1 g5 (t, s) = Γ(γ) tγ−1 (1 − s)γ−p3 −1 , 0 ≤ t ≤ s ≤ 1, ⎧ ⎪ tγ−q3 −1 (1 − s)γ−p3 −1 − (t − s)γ−q3 −1 , ⎪ ⎨ 1 g6 (t, s) = 0 ≤ s ≤ t ≤ 1, Γ(γ − q3 ) ⎪ ⎪ ⎩ tγ−q3 −1 (1 − s)γ−p3 −1 , 0 ≤ t ≤ s ≤ 1,

Δ3 =

L tγ−1  ci g6 (ζi , s), G3 (t, s) = g5 (t, s) + Δ3 i=1

∀ (t, s) ∈ [0, 1] × [0, 1],

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1 (1 − s)γ−p3 −1 (1 − (1 − s)p3 ), s ∈ [0, 1], Γ(γ) 1 (1 − s)γ−p3 −1 (1 − (1 − s)p3 −q3 ), s ∈ [0, 1], h6 (s) = Γ(γ − q3 ) h5 (s) =

J3 (s) = h5 (s) +

L 1  ci g6 (ζi , s), s ∈ [0, 1], Δ3 i=1 L

σ3 =

 γ−q −1 1 1 3 + ci ζ . Γ(γ) Δ3 Γ(γ − q3 ) i=1 i

The inequalities from Lemmas 5.1.3 and 5.1.4 for the functions G2 , G3 , v and w are the following G2 (t, s) ≤ J2 (s), G2 (t, s) ≥ tβ−1 J2 (s), G2 (t, s) ≤ σ2 tβ−1 , G3 (t, s) ≤ J3 (s), G3 (t, s) ≥ tγ−1 J3 (s), G3 (t, s) ≤ σ3 tγ−1 for all t, s ∈ [0, 1], and v(t) ≥ tβ−1 v(t ), w(t) ≥ tγ−1 w(t ) for all t, t ∈ [0, 1]. 5.1.2

Existence of positive solutions

In this section, we give sufficient conditions on λ, μ, ν, f, g and h such that positive solutions with respect to a cone for our problem (5.1), (5.2) exist. We present the assumptions that we shall use in the sequel. (H1) α, β, γ ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], γ ∈ (l − 1, l], n, m, l ∈ N, n, m, l ≥ 3, p1 , p2 , p3 , q1 , q2 , q3 ∈ R, p1 ∈ [1, α − 1), p2 ∈ [1, β − 1), p3 ∈ [1, γ − 1), q1 ∈ [0, p1 ], q2 ∈ [0, p2 ], q3 ∈ [0, p3 ], ξi ∈ R, ai ≥ 0 for all i = 1, . . . , N (N ∈ N), 0 < ξ1 < · · · < ξN ≤ 1, ηi ∈ R, bi ≥ 0 for all i = 1, . . . , M (M ∈ N), 0 < η1 < · · · < ηM ≤ 1, and ζi ∈ R, ci ≥ 0 for all i = 1, . . . , L (L ∈ N), 0 < ζ1 < Γ(α) Γ(α) N α−q1 −1 − Γ(α−q > · · · < ζL ≤ 1; λ, μ, ν > 0, Δ1 = Γ(α−p i=1 ai ξi 1) 1) Γ(β) Γ(β) M β−q2 −1 Γ(γ) > 0, Δ3 = Γ(γ−p3 ) − 0, Δ2 = Γ(β−p2 ) − Γ(β−q2 ) i=1 bi ηi Γ(γ) L γ−q3 −1 > 0. i=1 ci ζi Γ(γ−q3 ) (H2) The functions f, g, h : [0, 1] × R+ × R+ × R+ → R+ are continuous. For σ ∈ (0, 1), we introduce the following extreme limits: f0s = hs0 =

lim sup

max

f (t, u, v, w) s g(t, u, v, w) , g0 = lim sup max , u+v+w t∈[0,1] u+v+w u+v+w→0+

lim sup

max

h(t, u, v, w) f (t, u, v, w) , f0i = lim inf min , u+v+w→0+ t∈[σ,1] u + v + w u+v+w

u+v+w→0+ t∈[0,1]

u+v+w→0+ t∈[0,1]

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lim inf

min

g(t, u, v, w) i h(t, u, v, w) , h0 = lim inf min , u+v+w→0+ t∈[σ,1] u + v + w u+v+w

s f∞ = lim sup

max

f (t, u, v, w) s g(t, u, v, w) , g∞ = lim sup max , u+v+w u+v+w t∈[0,1] u+v+w→∞

hs∞ = lim sup

max

h(t, u, v, w) f (t, u, v, w) i , f∞ , = lim inf min u+v+w→∞ t∈[σ,1] u + v + w u+v+w

i g∞ =

min

g(t, u, v, w) i h(t, u, v, w) , h∞ = lim inf min , u+v+w→∞ t∈[σ,1] u + v + w u+v+w

g0i =

u+v+w→0+ t∈[σ,1]

u+v+w→∞ t∈[0,1]

u+v+w→∞ t∈[0,1]

lim inf

u+v+w→∞ t∈[σ,1]

In the definition of the extreme limits above, the variables u, v and w are nonnegative. By using the Green functions Gi , i = 1, 2, 3 from Section 5.1.1, we consider the following nonlinear system of integral equations: ⎧  1 ⎪ ⎪ u(t) = λ G1 (t, s)f (s, u(s), v(s), w(s)) ds, t ∈ [0, 1], ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪  1 ⎨ v(t) = μ G2 (t, s)g(s, u(s), v(s), w(s)) ds, t ∈ [0, 1], ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪  1 ⎪ ⎪ ⎪ ⎩ w(t) = ν G3 (t, s)h(s, u(s), v(s), w(s)) ds, t ∈ [0, 1]. 0

By Lemma 5.1.1 and the corresponding lemmas for problems (5.8), (5.9) and (5.10), (5.11), we deduce that (u, v, w) is a solution of the above system if and only if (u, v, w) is a solution of problem (5.1), (5.2). We consider the Banach space X = C[0, 1] with supremum norm  ·  and the Banach space Y = X × X × X with the norm (u, v, w)Y = u + v + w. We define the cones P1 = {u ∈ X, u(t) ≥ tα−1 u, ∀ t ∈ [0, 1]} ⊂ X, P2 = {v ∈ X, v(t) ≥ tβ−1 v, ∀ t ∈ [0, 1]} ⊂ X, P3 = {w ∈ X, w(t) ≥ tγ−1 w, ∀ t ∈ [0, 1]} ⊂ X, and P = P1 × P2 × P3 ⊂ Y .

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For λ, μ, ν > 0, we define now the operator Q : P → Y by Q(u, v, w) = (Q1 (u, v, w), Q2 (u, v, w), Q3 (u, v, w)) with  1 Q1 (u, v, w)(t) = λ G1 (t, s)f (s, u(s), v(s), w(s)) ds, 0

t ∈ [0, 1], (u, v, w) ∈ P,  1 Q2 (u, v, w)(t) = μ G2 (t, s)g(s, u(s), v(s), w(s)) ds, 0

t ∈ [0, 1], (u, v, w) ∈ P,  1 Q3 (u, v, w)(t) = ν G3 (t, s)h(s, u(s), v(s), w(s)) ds, 0

t ∈ [0, 1], (u, v, w) ∈ P. Lemma 5.1.5. If (H1) and (H2) hold, then Q : P → P is a completely continuous operator. Proof. Let (u, v, w) ∈ P be an arbitrary element. Because Q1 (u, v, w), Q2 (u, v, w) and Q3 (u, v, w) satisfy the problem (5.3), (5.4) for x(t) = λf (t, u(t), v(t), w(t)), t ∈ [0, 1], the problem (5.8), (5.9) for y(t) = μg(t, u(t), v(t), w(t)), t ∈ [0, 1], and the problem (5.10), (5.11) for z(t) = νh(t, u(t), v(t), w(t)), t ∈ [0, 1], respectively, then by Lemma 5.1.4 and the corresponding ones for problems (5.8), (5.9) and (5.10), (5.11), we obtain Q1 (u, v, w)(t ) ≥ tα−1 Q1 (u, v, w)(t ), Q2 (u, v, w)(t ) ≥ tβ−1 Q2 (u, v, w)(t ), Q3 (u, v, w)(t ) ≥ tγ−1 Q3 (u, v, w)(t ), ∀ t, t ∈ [0, 1], (u, v, w) ∈ P, and so Q1 (u, v, w)(t) ≥ tα−1 max Q1 (u, v, w)(t ) = tα−1 Q1 (u, v, w), t ∈[0,1]

∀ t ∈ [0, 1], (u, v, w) ∈ P, Q2 (u, v, w)(t) ≥ tβ−1 max Q2 (u, v, w)(t ) = tβ−1 Q2 (u, v, w), t ∈[0,1]

∀ t ∈ [0, 1], (u, v, w) ∈ P, Q3 (u, v, w)(t) ≥ tγ−1 max Q3 (u, v, w)(t ) = tγ−1 Q3 (u, v, w), t ∈[0,1]

∀ t ∈ [0, 1], (u, v, w) ∈ P.

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Therefore, Q(u, v, w) = (Q1 (u, v, w), Q2 (u, v, w), Q3 (u, v, w)) ∈ P , and then Q(P ) ⊂ P . By using standard arguments, we can easily show that Q1 , Q2 and Q3 are completely continuous, and then Q is a completely continuous operator.  The triplet (u, v, w) ∈ P is a fixed point of operator Q if and only if (u, v, w) is solution of problem (5.1), (5.2). So, we will investigate the existence of fixed points of operator Q.

1 1 For σ ∈ (0, 1), we denote by A = σ J1 (s) ds, B = 0 J1 (s) ds, C =

1

1

1

1 σ J2 (s) ds, D = 0 J2 (s) ds, E = σ J3 (s) ds, F = 0 J3 (s) ds, where J1 , J2 and J3 are defined in Section 5.1.1. i i , g∞ , hi∞ ∈ (0, ∞) and numbers α1 , α2 , α3 ≥ 0 First, for f0s , g0s , hs0 , f∞ with α1 +α2 +α3 = 1, α 1 , α 2 , α 3 > 0 with α 1 + α 2 + α 3 = 1, α 2 , α 3 > 0 with 3 = 1, α 1 , α 3 > 0 with α 1 + α 3 = 1, α     α 2 + α 1 , α 2 > 0 with α 1 +α 2 = 1, we define the numbers L1 =

α1 , i A θσ α−1 f∞

α2 , i C θσ β−1 g∞

L5 =

α3 , θσ γ−1 hi∞ E

L2 =

α 1 , f0s B

L4 =

α 2 , g0s D

L6 =

α 3 , hs0 F

L4 =

α 2 , g0s D

L6 =

α 3 , hs0 F

L2 =

α 1 , f0s B

L6 =

α 3 , hs0 F

L 2 =

L 4 =

α  2 , g0s D

2 = L

1 , f0s B

4 = L

1 , g0s D

6 = L

L3 =

α  1 , f0s B

1 , hs0 F

where θ = min{σ α−1 , σ β−1 , σ γ−1 }. Theorem 5.1.1. Assume that (H1) and (H2) hold, σ ∈ (0, 1), α1 , α2 , α3 ≥ 0 with α1 + α2 + α3 = 1, α 1 , α 2 , α 3 > 0 with α 1 + α 2 + α 3 = 1, α 2 , α 3 > 0         3 = 1, α 1 , α 3 > 0 with α 1 + α 3 = 1, α 1 , α 2 > 0 with α  with α 2 + α 1 +  α 2 = 1. i i , g∞ , hi∞ ∈ (0, ∞), L1 < L2 , L3 < L4 and L5 < L6 , (1) If f0s , g0s , hs0 , f∞ then for each λ ∈ (L1 , L2 ), μ ∈ (L3 , L4 ), ν ∈ (L5 , L6 ) there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ ∈ (0, ∞), L3 < L4 and L5 < L6 , then (2) If f0s = 0, g0s , hs0 , f∞ for each λ ∈ (L1 , ∞), μ ∈ (L3 , L4 ), ν ∈ (L5 , L6 ) there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2).

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i i (3) If g0s = 0, f0s , hs0 , f∞ , g∞ , hi∞ ∈ (0, ∞), L1 < L2 and L5 < L6 , then  for each λ ∈ (L1 , L2 ), μ ∈ (L3 , ∞), ν ∈ (L5 , L6 ) there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i  , g∞ , hi∞ ∈ (0, ∞), L1 < L (4) If hs0 = 0, f0s , g0s , f∞ 2 and L3 < L4 ,   then for each λ ∈ (L1 , L2 ), μ ∈ (L3 , L4 ), ν ∈ (L5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i  6 , then for each , g∞ , hi∞ ∈ (0, ∞), L5 < L (5) If f0s = g0s = 0, hs0 , f∞  6 ), there exists a positive solution λ ∈ (L1 , ∞), μ ∈ (L3 , ∞), ν ∈ (L5 , L (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i  4 , then for each , g∞ , hi∞ ∈ (0, ∞), L3 < L (6) If f0s = hs0 = 0, g0s , f∞  4 ), ν ∈ (L5 , ∞), there exists a positive solution λ ∈ (L1 , ∞), μ ∈ (L3 , L (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i  2 , then for each , g∞ , hi∞ ∈ (0, ∞), L1 < L (7) If g0s = hs0 = 0, f0s , f∞  λ ∈ (L1 , L2 ), μ ∈ (L3 , ∞), ν ∈ (L5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ ∈ (0, ∞), then for each λ ∈ (8) If f0s = g0s = hs0 = 0, f∞ (L1 , ∞), μ ∈ (L3 , ∞), ν ∈ (L5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (9) If f0s , g0s , hs0 ∈ (0, ∞) and at least one of f∞ for each λ ∈ (0, L2 ), μ ∈ (0, L4 ), ν ∈ (0, L6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (10) If f0s = 0, g0s , hs0 ∈ (0, ∞) and at least one of f∞   for each λ ∈ (0, ∞), μ ∈ (0, L4 ), ν ∈ (0, L6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (11) If g0s = 0, f0s , hs0 ∈ (0, ∞) and at least one of f∞   for each λ ∈ (0, L2 ), μ ∈ (0, ∞), ν ∈ (0, L6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (12) If hs0 = 0, f0s , g0s ∈ (0, ∞) and at least one of f∞  for each λ ∈ (0, L 2 ), μ ∈ (0, L4 ), ν ∈ (0, ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (13) If f0s = g0s = 0, hs0 ∈ (0, ∞) and at least one of f∞  6 ), there exists a positive for each λ ∈ (0, ∞), μ ∈ (0, ∞), ν ∈ (0, L solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (14) If f0s = hs0 = 0, g0s ∈ (0, ∞) and at least one of f∞  for each λ ∈ (0, ∞), μ ∈ (0, L4 ), ν ∈ (0, ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). i i , g∞ , hi∞ is ∞, then (15) If g0s = hs0 = 0, f0s ∈ (0, ∞) and at least one of f∞  2 ), μ ∈ (0, ∞), ν ∈ (0, ∞), there exists a positive for each λ ∈ (0, L solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2).

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i i (16) If f0s = g0s = hs0 = 0 and at least one of f∞ , g∞ , hi∞ is ∞, then for each λ ∈ (0, ∞), μ ∈ (0, ∞), ν ∈ (0, ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2).

Proof. We consider the above cone P ⊂ Y and the operators Q1 , Q2 , Q3 and Q. We will prove some illustrative cases of this theorem. i i , g∞ , hi∞ ∈ (0, ∞). Let λ ∈ (L1 , L2 ), Case (1). We consider f0s , g0s , hs0 , f∞ μ ∈ (L3 , L4 ) and ν ∈ (L5 , L6 ). We choose ε > 0 a positive number such i i , ε < g∞ , ε < hi∞ and that ε < f∞

α 1 ≥ λ, (f0s + ε)B α1 ≤ λ, α−1 i − ε)A θσ (f∞

α 2 α 3 ≥ μ, ≥ ν, (g0s + ε)D (hs0 + ε)F α2 α3 ≤ μ, ≤ ν. β−1 i γ−1 θσ (g∞ − ε)C θσ (hi∞ − ε)E

By using (H2) and the definition of f0s , g0s and hs0 , we deduce that there exists R1 > 0 such that f (t, u, v, w) ≤ (f0s + ε)(u + v + w), g(t, u, v, w) ≤ (g0s + ε)(u + v + w), h(t, u, v, w) ≤ (hs0 + ε)(u + v + w), for all t ∈ [0, 1] and u, v, w ≥ 0 with u + v + w ≤ R1 . We define the set Ω1 = {(u, v, w) ∈ Y, (u, v, w)Y < R1 }. Now, let (u, v, w) ∈ P ∩ ∂Ω1 , that is, (u, v, w)Y = R1 or equivalently u + v + w = R1 . Then u(t) + v(t) + w(t) ≤ R1 for all t ∈ [0, 1], and by Lemma 5.1.3, we obtain  1 J1 (s)f (s, u(s), v(s), w(s)) ds Q1 (u, v, w)(t) ≤ λ 0

 ≤λ ≤

1

0

λ(f0s

J1 (s)(f0s + ε)(u(s) + v(s) + w(s)) ds  + ε)

1

0

J1 (s)(u + v + w) ds

= λ(f0s + ε)B(u, v, w)Y ≤ α 1 (u, v, w)Y ,  1 Q2 (u, v, w)(t) ≤ μ J2 (s)g(s, u(s), v(s), w(s)) ds

∀ t ∈ [0, 1],

0

 ≤μ

0

1

J2 (s)(g0s + ε)(u(s) + v(s) + w(s)) ds

≤ μ(g0s + ε)

 0

1

J2 (s)(u + v + w) ds

= μ(g0s + ε)D(u, v, w)Y ≤ α 2 (u, v, w)Y ,

∀ t ∈ [0, 1],

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 Q3 (u, v, w)(t) ≤ ν

0

 ≤ν

1

0

1

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)(hs0 + ε)(u(s) + v(s) + w(s)) ds

≤ ν(hs0 + ε)



1

0

J3 (s)(u + v + w) ds

= ν(hs0 + ε)F (u, v, w)Y ≤ α 3 (u, v, w)Y ,

∀ t ∈ [0, 1].

Therefore, Q1 (u, v, w) ≤ α 1(u, v, w)Y , Q2 (u, v, w) ≤ α 2(u, v, w)Y , 3 (u, v, w)Y . Q3 (u, v, w) ≤ α Then for (u, v, w) ∈ P ∩ ∂Ω1 we deduce Q(u, v, w)Y = Q1 (u, v, w) + Q2 (u, v, w) + Q3 (u, v, w) ≤ ( α1 + α 2 + α 3 )(u, v, w)Y = (u, v, w)Y .

(5.12)

i i By the definition of f∞ , g∞ and hi∞ , there exists R2 > 0 such that i i f (t, u, v, w) ≥ (f∞ − ε)(u + v + w), g(t, u, v, w) ≥ (g∞ − ε)(u + v + w), i h(t, u, v, w) ≥ (h∞ − ε)(u + v + w) for all u, v, w ≥ 0 with u + v + w ≥ R2 and t ∈ [σ, 1]. We consider R2 = max{2R1 , R2 /θ} and we define Ω2 = {(u, v, w) ∈ Y, (u, v, w)Y < R2 }. Then for (u, v, w) ∈ P with (u, v, w)Y = R2 , we obtain

u(t) + v(t) + w(t) ≥ σ α−1 u + σ β−1 v + σ γ−1 w ≥ θ(u + v + w) = θ(u, v, w)Y = θR2 ≥ R2 ,

∀ t ∈ [σ, 1].

Then by Lemma 5.1.3, we conclude  1 Q1 (u, v, w)(t) ≥ λ tα−1 J1 (s)f (s, u(s), v(s), w(s)) ds 0

≥ λσ α−1 ≥ λσ α−1 ≥ λσ

α−1

 

1 σ 1 σ

J1 (s)f (s, u(s), v(s), w(s)) ds i J1 (s)(f∞ − ε)(u(s) + v(s) + w(s)) ds

i θ(f∞

 − ε)

1

σ

J1 (s)(u, v, w)Y ds

i = λσ α−1 θ(f∞ − ε)A(u, v, w)Y

≥ α1 (u, v, w)Y , ∀ t ∈ [σ, 1],

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 Q2 (u, v, w)(t) ≥ μ

0

≥ μσ

1

tβ−1 J2 (s)g(s, u(s), v(s), w(s)) ds

β−1



1

σ

≥ μσ β−1



1

σ

J2 (s)g(s, u(s), v(s), w(s)) ds i J2 (s)(g∞ − ε)(u(s) + v(s) + w(s)) ds

i − ε) ≥ μσ β−1 θ(g∞



1

σ

J2 (s)(u, v, w)Y ds

i − ε)C(u, v, w)Y = μσ β−1 θ(g∞

≥ α2 (u, v, w)Y , ∀ t ∈ [σ, 1],  1 Q3 (u, v, w)(t) ≥ ν tγ−1 J3 (s)h(s, u(s), v(s), w(s)) ds 0

≥ νσ

γ−1

≥ νσ γ−1



1

σ



1

σ

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)(hi∞ − ε)(u(s) + v(s) + w(s)) ds

≥ νσ γ−1 θ(hi∞ − ε)



1

σ

J3 (s)(u, v, w)Y ds

= νσ γ−1 θ(hi∞ − ε)F (u, v, w)Y ≥ α3 (u, v, w)Y , ∀ t ∈ [σ, 1]. So, Q1 (u, v, w) ≥ Q1 (u, v, w)(σ) ≥ α1 (u, v, w)Y , Q2 (u, v, w) ≥ Q2 (u, v, w)(σ) ≥ α2 (u, v, w)Y , Q3 (u, v, w) ≥ Q3 (u, v, w)(σ) ≥ α3 (u, v, w)Y . Hence, for (u, v, w) ∈ P ∩ ∂Ω2 , we obtain Q(u, v, w)Y = Q1 (u, v, w) + Q2 (u, v, w) + Q3 (u, v, w) ≥ (α1 + α2 + α3 )(u, v, w)Y = (u, v, w)Y .

(5.13)

By using Lemma 5.1.5, Theorem 1.2.2 (i), and the relations (5.12), (5.13), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω2 \ Ω1 ), u(t) ≥ tα−1 u, v(t) ≥ tβ−1 v, w(t) ≥ tγ−1 w for all t ∈ [0, 1], and R1 ≤ u + v + w ≤ R2 . If u > 0 then u(t) > 0 for all t ∈ (0, 1], if v > 0 then v(t) > 0 for all t ∈ (0, 1], and if w > 0 then w(t) > 0 for all t ∈ (0, 1]. So, (u, v, w) is a positive solution for our problem (5.1), (5.2). 

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i i Case (10). We consider f0s = 0, f∞ = ∞, g0s , hs0 , g∞ , hi∞ ∈ (0, ∞). Let   λ ∈ (0, ∞), μ ∈ (0, L4 ) and ν ∈ (0, L6 ). We choose ε > 0 a positive number such that ε ≤ λθσ α−1 A and

ε≤

1 − μg0s D − νhs0 F , 2λB

ε≤

α 2 − μg0s D , 2μD

ε≤

α 3 − νhs0 F . 2νF

The numerators of the above fractions are positive, because μ < α 

α 2 g0s D ,

that is, α 2 > μg0s D, ν < hs3F , that is, α 3 > νhs0 F , and 1 − μg0s D − νhs0 F = 0 3 − μg0s D − νhs0 F = ( α2 − μg0s D) + ( α3 − νhs0 F ) > 0. α 2 + α s By using (H2) and the definition of f0 , g0s , hs0 , we deduce that there exists R1 > 0 such that f (t, u, v, w) ≤ ε(u+v+w), g(t, u, v, w) ≤ (g0s +ε)(u+ v + w), h(t, u, v, w) ≤ (hs0 + ε)(u + v + w) for all t ∈ [0, 1], u, v, w ≥ 0 with u + v + w ≤ R1 . We define the set Ω1 = {(u, v, w) ∈ Y, (u, v, w)Y < R1 }. Now, let (u, v, w) ∈ P ∩ ∂Ω1 , that is, (u, v, w)Y = R1 . Then u(t) + v(t) + w(t) ≤ R1 for all t ∈ [0, 1], and by Lemma 5.1.3, we obtain  Q1 (u, v, w)(t) ≤ λ  ≤λ

1 0 1 0

 ≤ λε

J1 (s)f (s, u(s), v(s), w(s)) ds J1 (s)ε(u(s) + v(s) + w(s)) ds 1

0

J1 (s)(u + v + w) ds

= λεB(u, v, w)Y ≤  Q2 (u, v, w)(t) ≤ μ

0

 ≤μ ≤

1

1

0

μ(g0s

1 (1 − μg0s D − νhs0 F )(u, v, w)Y , 2

J2 (s)g(s, u(s), v(s), w(s)) ds J2 (s)(g0s + ε)(u(s) + v(s) + w(s)) ds 

+ ε)

0

1

J2 (s)(u + v + w) ds

= μ(g0s + ε)D(u, v, w)Y

α  − μg0s D ≤ μ g0s + 2 D(u, v, w)Y 2μD =

1 (μg0s D + α 2 )(u, v, w)Y , 2

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 Q3 (u, v, w)(t) ≤ ν

1

0

 ≤ν

1

0

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)(hs0 + ε)(u(s) + v(s) + w(s)) ds

≤ ν(hs0 + ε)



1

0

J3 (s)(u + v + w) ds

= ν(hs0 + ε)F (u, v, w)Y

α  − νhs0 F ≤ ν hs0 + 3 F (u, v, w)Y 2νF =

1 (νhs0 F + α 3 )(u, v, w)Y , 2

∀ t ∈ [0, 1].

Therefore, Q1 (u, v, w) ≤

1 (1 − μg0s D − νhs0 F )(u, v, w)Y , 2

Q2 (u, v, w) ≤

1 (μg0s D + α 2 )(u, v, w)Y , 2

Q3 (u, v, w) ≤

1 (νhs0 F + α 3 )(u, v, w)Y . 2

Then for (u, v, w) ∈ P ∩ ∂Ω1 , we conclude Q(u, v, w)Y = Q1 (u, v, w) + Q2 (u, v, w) + Q3 (u, v, w) ≤

1 (1 − μg0s D − νhs0 F + μg0s D + α 2 + νhs0 F + α 3 ) 2 × (u, v, w)Y = (u, v, w)Y .

(5.14)

i , there exists R2 > 0 such that f (t, u, v, w) ≥ By the definition of f∞ for all u, v, w ≥ 0 with u + v + w ≥ R2 and t ∈ [σ, 1]. We consider R2 = max{2R1 , R2 /θ} and we define Ω2 = {(u, v, w) ∈ Y, (u, v, w)Y < R2 }. Then for (u, v, w) ∈ P with (u, v, w)Y = R2 , we obtain u(t) + v(t) + w(t) ≥ θ(u, v, w)Y = θR2 ≥ R2 for all t ∈ [σ, 1]. Then by Lemma 5.1.3, we deduce  1 tα−1 J1 (s)f (s, u(s), v(s), w(s)) ds Q1 (u, v, w)(t) ≥ λ 1 ε (u + v + w)

0

≥ λσ

α−1



1

σ

J1 (s)f (s, u(s), v(s), w(s)) ds

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1

1 J1 (s) (u(s) + v(s) + w(s)) ds ε σ  1 1 ≥ λσ α−1 θ J1 (s)(u, v, w)Y ds ε σ ≥ λσ

α−1

1 = λσ α−1 θ A(u, v, w)Y ≥ (u, v, w)Y , ∀ t ∈ [σ, 1]. ε Then for (u, v, w) ∈ P ∩ ∂Ω2 we find Q1 (u, v, w) ≥ Q1 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q1 (u, v, w) ≥ (u, v, w)Y .

(5.15)

By using Lemma 5.1.5, Theorem 1.2.2 (i) and the inequalities (5.14), (5.15), we conclude that Q has a fixed point (u, v, w) ∈ P ∩ (Ω2 \ Ω1 ), which is a positive solution of problem (5.1), (5.2). i i Case (15). We consider g0s = hs0 = 0, g∞ = ∞, f0s , f∞ , hi∞ ∈ (0, ∞). Let  λ ∈ (0, L2 ), μ ∈ (0, ∞), ν ∈ (0, ∞). We choose ε > 0 a positive number such that ε ≤ μθσ β−1 C and 1 − λf0s B 1 − λf0s B 1 − λf0s B , ε≤ , ε≤ . ε≤ 2λB 4μD 4νF

The numerator of the above fractions is positive because λ < f s1B that 0 is 1 − λf0s B > 0. By using (H2) and the definition of f0s , g0s , hs0 , we deduce that there exists R1 > 0 such that f (t, u, v, w) ≤ (f0s + ε)(u + v + w), g(t, u, v, w) ≤ ε(u + v + w), h(t, u, v, w) ≤ ε(u + v + w) for all t ∈ [0, 1], u, v, w ≥ 0 with u + v + w ≤ R1 . We define the set Ω1 = {(u, v, w) ∈ Y, (u, v, w)Y < R1 }. Now, let (u, v, w) ∈ P ∩ ∂Ω1 , that is (u, v, w)Y = R1 . Then u(t) + v(t) + w(t) ≤ R1 for all t ∈ [0, 1], and by Lemma 5.1.3, we obtain  1 Q1 (u, v, w)(t) ≤ λ J1 (s)f (s, u(s), v(s), w(s)) ds  ≤λ ≤

0

1 0

λ(f0s

J1 (s)(f0s + ε)(u(s) + v(s) + w(s)) ds  + ε)

0

1

J1 (s)(u + v + w) ds



1 − λf0s B = λ(f0s + ε)B(u, v, w)Y ≤ λ f0s + 2λB ×B(u, v, w)Y =

1 (λf0s B + 1)(u, v, w)Y , 2

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 Q2 (u, v, w)(t) ≤ μ

1

J2 (s)g(s, u(s), v(s), w(s)) ds

0

 ≤μ

1

J2 (s)ε(u(s) + v(s) + w(s)) ds

0

 ≤ με

1

0

J2 (s)(u + v + w) ds = μεD(u, v, w)Y

1 − λf0s B 1 D(u, v, w)Y = (1 − λf0s B)(u, v, w)Y , 4μD 4  1 Q3 (u, v, w)(t) ≤ ν J3 (s)h(s, u(s), v(s), w(s)) ds ≤μ

0

 ≤ν

1

0

 ≤ νε ≤ν =

J3 (s)ε(u(s) + v(s) + w(s)) ds 1

0

J3 (s)(u + v + w) ds = νεF (u, v, w)Y

1 − λf0s B F (u, v, w)Y 4νF

1 (1 − λf0s B)(u, v, w)Y , 4

∀ t ∈ [0, 1].

Therefore, Q1 (u, v, w) ≤

1 (λf0s B + 1)(u, v, w)Y , 2

Q2 (u, v, w) ≤

1 (1 − λf0s B)(u, v, w)Y , 4

Q3 (u, v, w) ≤

1 (1 − λf0s B)(u, v, w)Y . 4

Then for (u, v, w) ∈ P ∩ ∂Ω1 , we deduce Q(u, v, w)Y = Q1 (u, v, w) + Q2 (u, v, w) + Q3 (u, v, w) ≤

1 (2 + 2λf0s B + 1 − λf0s B + 1 − λf0s B) 4 × (u, v, w)Y = (u, v, w)Y .

(5.16)

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i By the definition of g∞ , there exists R2 > 0 such that g(t, u, v, w) ≥ for all u, v, w ≥ 0 with u+v+w ≥ R2 and t ∈ [σ, 1]. We consider R2 = max{2R1 , R2 /θ} and we define Ω2 = {(u, v, w) ∈ Y, (u, v, w)Y < R2 }. Then for (u, v, w) ∈ P with (u, v, w)Y = R2 , we obtain u(t) + v(t) + w(t) ≥ θ(u, v, w)Y = θR2 ≥ R2 , for all t ∈ [σ, 1]. Then by Lemma 5.1.3, we conclude 1 ε (u+v+w)

 Q2 (u, v, w)(t) ≥ μ ≥ μσ

0

1

tβ−1 J2 (s)g(s, u(s), v(s), w(s)) ds

β−1



1

σ



J2 (s)g(s, u(s), v(s), w(s)) ds

1

1 J2 (s) (u(s) + v(s) + w(s)) ds ε σ  1 1 J2 (s)(u, v, w)Y ds ≥ μσ β−1 θ ε σ

≥ μσ

β−1

1 = μσ β−1 θ C(u, v, w)Y ≥ (u, v, w)Y , ε

∀ t ∈ [σ, 1].

Then for (u, v, w) ∈ P ∩ ∂Ω2 , we find Q2 (u, v, w) ≥ Q2 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q2 (u, v, w) ≥ (u, v, w)Y .

(5.17)

By using Lemma 5.1.5, Theorem 1.2.2 (i) and the inequalities (5.16), (5.17), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω2 \ Ω1 ) which is a positive solution of problem (5.1), (5.2). i i Case (16). We consider f0s = g0s = hs0 = 0, hi∞ = ∞, f∞ , g∞ ∈ (0, ∞). Let λ ∈ (0, ∞), μ ∈ (0, ∞) and ν ∈ (0, ∞). We choose ε > 0 such that

ε ≤ νθσ γ−1 E,

ε≤

1 , 3λB

ε≤

1 , 3μD

ε≤

1 . 3νF

By using (H2) and the definition of f0s , g0s , hs0 , we deduce that there exists R1 > 0 such that f (t, u, v, w) ≤ ε(u + v + w), g(t, u, v, w) ≤ ε(u + v + w), h(t, u, v, w) ≤ ε(u + v + w) for all t ∈ [0, 1], u, v, w ≥ 0 with u + v + w ≤ R1 . We define the set Ω1 = {(u, v, w) ∈ Y, (u, v, w)Y < R1 }.

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Now, let (u, v, w) ∈ P ∩ ∂Ω1 , that is (u, v, w)Y = R1 . Then u(t) + v(t) + w(t) ≤ R1 for all t ∈ [0, 1], and by Lemma 5.1.3, we obtain  Q1 (u, v, w)(t) ≤ λ

1

J1 (s)f (s, u(s), v(s), w(s)) ds

0

 ≤λ

1

J1 (s)ε(u(s) + v(s) + w(s)) ds

0

 ≤ λε

1

0

J1 (s)(u + v + w) ds

= λεB(u, v, w)Y ≤  Q2 (u, v, w)(t) ≤ μ

1

J2 (s)g(s, u(s), v(s), w(s)) ds

0

 ≤μ

1

J2 (s)ε(u(s) + v(s) + w(s)) ds

0

 ≤ με

1

0

J2 (s)(u + v + w) ds

= μεD(u, v, w)Y ≤  Q3 (u, v, w)(t) ≤ ν

0

 ≤ν

1

1

0

 ≤ νε

0

1 (u, v, w)Y , 3

1 (u, v, w)Y , 3

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)ε(u(s) + v(s) + w(s)) ds 1

J3 (s)(u + v + w) ds

= νεF (u, v, w)Y ≤

1 (u, v, w)Y , 3

∀ t ∈ [0, 1].

Therefore, Q1 (u, v, w) ≤ 13 (u, v, w)Y , Q2 (u, v, w) ≤ 13 (u, v, w)Y , Q3 (u, v, w) ≤ 13 (u, v, w)Y . Then for (u, v, w) ∈ P ∩ ∂Ω1 , we conclude Q(u, v, w)Y ≤ (u, v, w)Y .

(5.18)

By the definition of hi∞ , there exists R2 > 0 such that h(t, u, v, w) ≥ for all u, v, w ≥ 0 with u+v+w ≥ R2 and t ∈ [σ, 1]. We consider

1 ε (u+v+w)

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R2 = max{2R1 , R2 /θ} and we define Ω2 = {(u, v, w) ∈ Y, (u, v, w)Y < R2 }. Then for (u, v, w) ∈ P with (u, v, w)Y = R2 , we obtain u(t) + v(t) + w(t) ≥ θ(u, v, w)Y = θR2 ≥ R2 , for all t ∈ [σ, 1]. Then by Lemma 5.1.3, we deduce  1 tγ−1 J3 (s)h(s, u(s), v(s), w(s)) ds Q3 (u, v, w)(t) ≥ ν 0

≥ νσ γ−1



1

σ



J3 (s)h(s, u(s), v(s), w(s)) ds

1

1 J3 (s) (u(s) + v(s) + w(s)) ds ε σ  1 1 ≥ νσ γ−1 θ J3 (s)(u, v, w)Y ds ε σ

≥ νσ γ−1

1 = νσ γ−1 θ E(u, v, w)Y ≥ (u, v, w)Y , ε

∀ t ∈ [σ, 1].

Then for (u, v, w) ∈ P ∩ ∂Ω2 we find Q3 (u, v, w) ≥ Q3 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q3 (u, v, w) ≥ (u, v, w)Y .

(5.19)

By using Lemma 5.1.5, Theorem 1.2.2 (i) and the inequalities (5.18), (5.19), we conclude that Q has a fixed point (u, v, w) ∈ P ∩ (Ω2 \ Ω1 ) which is a positive solution of problem (5.1), (5.2). Remark 5.1.1. Each of the cases (9)–(16) of Theorem 5.1.1 contains 7 i i i i = ∞, g∞ , hi∞ ∈ (0, ∞)}, or {g∞ = ∞, f∞ , hi∞ ∈ cases as follows: {f∞ i i i i i i (0, ∞)}, or {h∞ = ∞, f∞ , g∞ ∈ (0, ∞)}, or {f∞ = g∞ = ∞, h∞ ∈ (0, ∞)}, i i i i or {f∞ = hi∞ = ∞, g∞ ∈ (0, ∞)}, or {g∞ = hi∞ = ∞, f∞ ∈ (0, ∞)}, or i i i {f∞ = g∞ = h∞ = ∞}. So, the total number of cases from Theorem 5.1.1 is 64, which we grouped in 16 cases. Each of the cases (1)–(8) contains 4 subcases because α1 , α2 , α3 ∈ (0, 1), or α1 = 1 and α2 = α3 = 0, or α2 = 1 and α1 = α3 = 0, or α3 = 1 and α1 = α2 = 0. Remark 5.1.2. In [98], the authors present only 15 cases (Theorems 2.1– 2.15 from [98]) from 64 cases, namely the first 9 cases of our Theorem 5.1.1. They did not study the cases when some extreme limits are 0 and other are ∞. Besides, our intervals for parameters λ, μ, ν presented in Theorem 5.1.1 (our cases (2)–(7) and (9)) are better than the corresponding ones from [98, Theorems 2.2–2.7 and 2.9–2.15]. In addition, the cone used in [98]

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implies the existence of nonnegative solutions, which satisfy the condition inf t∈[ξ,η] (u(t) + v(t) + w(t)) > 0, which is different than our definition of positive solutions. Remark 5.1.3. One can formulate existence results for the general case of the system of n fractional differential equations α

D0+j uj (t) + λj fj (t, u1 (t), . . . , un (t)) = 0,

j = 1, . . . , n,

(5.20)

with the boundary conditions ⎧ (k) ⎪ u (0) = 0, k = 0, . . . , mj − 2, j = 1, . . . , n, ⎪ ⎨ j Nj  pj qj ⎪ ⎪ ajk D0+ uj (ξjk ), j = 1, . . . , n, ⎩ D0+ uj (1) =

(5.21)

k=1

where αj ∈ (mj −1, mj ], mj ∈ N, mj ≥ 3; ξjk , ajk ∈ R for all k = 1, . . . , Nj , (Nj ∈ N); 0 < ξj1 < ξj2 < · · · ≤ ξjNj , pj ∈ [1, αj − 1), qj ∈ [0, pj ], j = 1, . . . , N . s = lim supu1 +···+un →0+ supt∈[0,1] According to the values of fj0 fj (t,u1 ,...,un ) u1 +···+un

f (t,u ,...,u )

1 n i ∈ [0, ∞), and fj∞ = lim inf u1 +···+un →∞ inf t∈[σ,1] ju1 +···+u ∈ n 2n n+1 cases. (0, ∞], j = 1, . . . , n we have 2 cases, which can be grouped in 2

s s , g∞ , hs∞ ∈ (0, ∞) and numbers In what follows, for f0i , g0i , hi0 , f∞ 1 , α 2 , α 3 > 0 with α 1 + α 2 + α 3 = 1, α1 , α2 , α3 ≥ 0 with α1 + α2 + α3 = 1, α          α 2 , α 3 > 0 with α 2 + α 3 = 1, α 1 , α 3 > 0 with α 1 + α 3 = 1, α 1 , α  2 > 0   2 = 1, we define the numbers with α 1 + α

M1 =

α1 , θσ α−1 f0i A

M2 =

α 1 , s B f∞

M4 =

α 2 , s g∞ D

M6 =

M6 =

α 3 , hs∞ F

M2 =

α 1 , s B f∞

M6 =

M4 =

α  2 , s D g∞

2 = M

M3 =

1 s B f∞

α2 , θσ β−1 g0i C

,

4 = M

M5 =

α 3 , s h∞ F α 3 , hs∞ F 1 s D g∞

,

α3 , θσ γ−1 hi0 E

M4 =

α 2 , s D g∞

M2 = 6 = M

α  1 , s B f∞ 1

hs∞ F

,

where θ = min{σ α−1 , σ β−1 , σ γ−1 }. Theorem 5.1.2. Assume that (H1) and (H2) hold, σ ∈ (0, 1), α1 , α2 , 1 , α 2 , α 3 > 0 with α 1 + α 2 + α 3 = 1, α3 ≥ 0 with α1 + α2 + α3 = 1, α          3 > 0 with α 2 + α 3 = 1, α 1 , α 3 > 0 with α 1 + α 3 = 1, α 1 , α  α 2 , α 2 > 0   2 = 1. with α 1 + α

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s s (1) If f0i , g0i , hi0 , f∞ , g∞ , hs∞ ∈ (0, ∞), M1 < M2 , M3 < M4 and M5 < M6 , then for each λ ∈ (M1 , M2 ), μ ∈ (M3 , M4 ), ν ∈ (M5 , M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = 0, g∞ , hs∞ , f0i , g0i , hi0 ∈ (0, ∞), M3 < M4 and M5 < M6 , (2) If f∞ then for each λ ∈ (M1 , ∞), μ ∈ (M3 , M4 ), ν ∈ (M5 , M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = 0, f∞ , hs∞ , f0i , g0i , hi0 ∈ (0, ∞), M1 < M2 and M5 < M6 , (3) If g∞ then for each λ ∈ (M1 , M2 ), μ ∈ (M3 , ∞), ν ∈ (M5 , M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s , g∞ , f0i , g0i , hi0 ∈ (0, ∞), M1 < M2 and M3 < M4 , (4) If hs∞ = 0, f∞ then for each λ ∈ (M1 , M2 ), μ ∈ (M3 , M4 ), ν ∈ (M5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s 6 , then for each = g∞ = 0, hs∞ , f0i , g0i , hi0 ∈ (0, ∞), M5 < M (5) If f∞  λ ∈ (M1 , ∞), μ ∈ (M3 , ∞), ν ∈ (M5 , M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s 4 , then for each = hs∞ = 0, g∞ , f0i , g0i , hi0 ∈ (0, ∞), M3 < M (6) If f∞  λ ∈ (M1 , ∞), μ ∈ (M3 , M4 ), ν ∈ (M5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s 2 , then for each = hs∞ = 0, f∞ , f0i , g0i , hi0 ∈ (0, ∞), M1 < M (7) If g∞  λ ∈ (M1 , M2 ), μ ∈ (M3 , ∞), ν ∈ (M5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = g∞ = hs∞ = 0, f0i , g0i , hi0 ∈ (0, ∞), then for each λ ∈ (8) If f∞ (M1 , ∞), μ ∈ (M3 , ∞), ν ∈ (M5 , ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s , g∞ , hs∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then for (9) If f∞ each λ ∈ (0, M2 ), μ ∈ (0, M4 ), ν ∈ (0, M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = 0, g∞ , hs∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then (10) If f∞ for each λ ∈ (0, ∞), μ ∈ (0, M4 ), ν ∈ (0, M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = 0, f∞ , hs∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then (11) If g∞ for each λ ∈ (0, M2 ), μ ∈ (0, ∞), ν ∈ (0, M6 ), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s , g∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then (12) If hs∞ = 0, f∞  for each λ ∈ (0, M2 ), μ ∈ (0, M4 ), ν ∈ (0, ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2).

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255

s s (13) If f∞ = g∞ = 0, hs∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then 6 ), there exists a positive for each λ ∈ (0, ∞), μ ∈ (0, ∞), ν ∈ (0, M solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = hs∞ = 0, g∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then (14) If f∞ 4 ), ν ∈ (0, ∞), there exists a positive for each λ ∈ (0, ∞), μ ∈ (0, M solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = hs∞ = 0, f∞ ∈ (0, ∞) and at least one of f0i , g0i , hi0 is ∞, then (15) If g∞ 2 ), μ ∈ (0, ∞), ν ∈ (0, ∞), there exists a positive for each λ ∈ (0, M solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2). s s = g∞ = hs∞ = 0 and at least one of f0i , g0i , hi0 is ∞, then for (16) If f∞ each λ ∈ (0, ∞), μ ∈ (0, ∞), ν ∈ (0, ∞), there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] for problem (5.1), (5.2).

Proof. We consider again the above cone P ⊂ Y and the operators Q1 , Q2 , Q3 and Q. We will also prove for this theorem some illustrative cases. s s , g∞ , hs∞ ∈ (0, ∞). Let λ ∈ (M1 , M2 ), Case (1). We consider f0i , g0i , hi0 , f∞ μ ∈ (M3 , M4 ), ν ∈ (M5 , M6 ). We choose ε > 0 a positive number such that ε < f0i , ε < g0i , ε < hi0 and

α1 α−1 θσ (f0i

− ε)A

α2 β−1 θσ (g0i

≤ λ,

− ε)C

α 2 ≥ μ, s + ε)D (g∞

α 1 ≥ λ, s + ε)B (f∞

≤ μ,

α3 γ−1 θσ (hi0

− ε)E

≤ ν,

α 3 ≥ ν. (hs∞ + ε)F

By using (H2) and the definition of f0i , g0i , hi0 , we deduce that there exists R3 > 0 such that f (t, u, v, w) ≥ (f0i − ε)(u + v + w), g(t, u, v, w) ≥ (g0i − ε)(u + v + w), h(t, u, v, w) ≥ (hi0 − ε)(u + v + w) for all u, v, w ≥ 0 with u + v + w ≤ R3 and t ∈ [σ, 1]. We denote by Ω3 = {(u, v, w) ∈ Y, (u, v, w)Y < R3 }. Let (u, v, w) ∈ P ∩ ∂Ω3 , that is (u, v, w)Y = R3 or equivalently u + v + w = R3 . Because u(t) + v(t) + w(t) ≤ R3 for all t ∈ [0, 1], then by Lemma 5.1.3 we obtain for all t ∈ [σ, 1]  1 tα−1 J1 (s)f (s, u(s), v(s), w(s)) ds Q1 (u, v, w)(t) ≥ λ 0

≥ λσ α−1



1

σ

J1 (s)f (s, u(s), v(s), w(s)) ds

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≥ λσ

α−1

≥ λσ

α−1



1

J1 (s)(f0i − ε)(u(s) + v(s) + w(s)) ds

σ

θ(f0i

 − ε)

1

σ

J1 (s)(u, v, w)Y ds

= λσ α−1 θ(f0i − ε)A(u, v, w)Y ≥ α1 (u, v, w)Y ,  1 Q2 (u, v, w)(t) ≥ μ tβ−1 J2 (s)g(s, u(s), v(s), w(s)) ds 0

≥ μσ β−1 ≥ μσ

β−1

≥ μσ

β−1



1

J2 (s)g(s, u(s), v(s), w(s)) ds

σ



1

J2 (s)(g0i − ε)(u(s) + v(s) + w(s)) ds

σ

θ(g0i

 − ε)

1

σ

J2 (s)(u, v, w)Y ds

= μσ β−1 θ(g0i − ε)C(u, v, w)Y ≥ α2 (u, v, w)Y ,  1 Q3 (u, v, w)(t) ≥ ν tγ−1 J3 (s)h(s, u(s), v(s), w(s)) ds 0

≥ νσ γ−1 ≥ νσ γ−1



1

σ



1

σ

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)(hi0 − ε)(u(s) + v(s) + w(s)) ds

≥ νσ γ−1 θ(hi0 − ε)



1

σ

J3 (s)(u, v, w)Y ds

= νσ γ−1 θ(hi0 − ε)E(u, v, w)Y ≥ α3 (u, v, w)Y . So Q1 (u, v, w) ≥ Q1 (u, v, w)(σ) ≥ α1 (u, v, w)Y , Q2 (u, v, w) ≥ Q2 (u, v, w)(σ) ≥ α2 (u, v, w)Y , Q3 (u, v, w) ≥ Q3 (u, v, w)(σ) ≥ α3 (u, v, w)Y . Then for an arbitrary element (u, v, w) ∈ P ∩ ∂Ω3 , we deduce Q(u, v, w)Y ≥ (α1 + α2 + α3 )(u, v, w)Y = (u, v, w)Y .

(5.22)

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257

Now, we define the functions f ∗ , g ∗ , h∗ : [0, 1] × R+ → R+ , f (t, x) = max0≤u+v+w≤x f(t, u, v, w), g ∗(t, x) = max0≤u+v+w≤x g(t, u, v, w), h∗ (t, x) = max0≤u+v+w≤x h(t, u, v, w), t ∈ [0, 1], x ∈ R+ . Then f (t, u, v, w) ≤ f ∗ (t, x), g(t, u, v, w) ≤ g ∗ (t, x), h(t, u, v, w) ≤ h∗ (t, x) for all t ∈ [0, 1], u, v, w ≥ 0 and u + v + w ≤ x. The functions f ∗ (t, ·), g ∗ (t, ·), h∗ (t, ·) are nondecreasing for every t ∈ [0, 1], and they satisfy the conditions ∗

f ∗ (t, x) s = f∞ , t∈[0,1] x

lim sup max x→∞

g ∗ (t, x) s = g∞ , t∈[0,1] x

lim sup max x→∞

∗

lim sup max

x→∞ t∈[0,1]

h (t, x) = hs∞ . x

Therefore, for ε > 0, there exists R4 > 0 such that for all x ≥ R4 and s s + ε)x, g ∗ (t, x) ≤ (g∞ + ε)x, h∗ (t, x) ≤ t ∈ [0, 1], we have f ∗ (t, x) ≤ (f∞ s (h∞ + ε)x. We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v, w) ∈ Y, (u, v, w)Y < R4 }. Let (u, v, w) ∈ P ∩ ∂Ω4 . By the definition of f ∗ , g ∗ , h∗ we conclude f (t, u(t), v(t), w(t)) ≤ f ∗ (t, (u, v, w)Y ), g(t, u(t), v(t), w(t)) ≤ g ∗ (t, (u, v, w)Y ), h(t, u(t), v(t), w(t)) ≤ h∗ (t, (u, v, w)Y ), ∀ t ∈ [0, 1]. Then for all t ∈ [0, 1], we obtain  1 J1 (s)f (s, u(s), v(s), w(s)) ds Q1 (u, v, w)(t) ≤ λ 0

 ≤λ ≤

1

0

s λ(f∞

 Q2 (u, v, w)(t) ≤ μ

0

 ≤μ ≤

1

1

0

s μ(g∞

J1 (s)f ∗ (s, (u, v, w)Y ) ds  + ε)

1

0

J1 (s)(u, v, w)Y ds ≤ α 1 (u, v, w)Y ,

J2 (s)g(s, u(s), v(s), w(s)) ds J2 (s)g ∗ (s, (u, v, w)Y ) ds  + ε)

0

1

J2 (s)(u, v, w)Y ds ≤ α 2 (u, v, w)Y ,

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 Q3 (u, v, w)(t) ≤ ν

1

0

 ≤ν

1

0

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)h∗ (s, (u, v, w)Y ) ds

≤ ν(hs∞ + ε)

 0

1

J3 (s)(u, v, w)Y ds ≤ α 3 (u, v, w)Y .

Therefore, we deduce Q1 (u, v, w) ≤ α 1 (u, v, w)Y , Q2 (u, v, w) ≤ 3 (u, v, w)Y . α 2 (u, v, w)Y , Q3 (u, v, w) ≤ α Hence, for (u, v, w) ∈ P ∩ ∂Ω4 , we conclude that α1 + α 2 + α 3 )(u, v, w)Y = (u, v, w)Y . Q(u, v, w)Y ≤ (

(5.23)

By using Lemma 5.1.5, Theorem 1.2.2 (ii), the relations (5.22), (5.23), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω4 \ Ω3 ), which is a positive solution for our problem (5.1), (5.2). s s Case (11). We consider g∞ = 0, hi0 = ∞, f∞ , hs∞ , f0i , g0i ∈ (0, ∞). Let   λ ∈ (0, M2 ), μ ∈ (0, ∞), ν ∈ (0, M6 ). We choose ε > 0 such that ε ≤ νθσ γ−1 E and

ε≤

s α 1 − λf∞ B , 2λB

s 1 − λf∞ B − νhs∞ F , 2μD

ε≤

ε≤

α 3 − νhs∞ F . 2νF

The numerators of the above fractions are positive, because λ < α 

α  1 s B f∞

s s that is α 1 > λf∞ B, ν < hs 3F that is α 3 > νhs∞ F , and 1−λf∞ B −νhs∞ F = ∞   s s  s  s 3 − λf∞ B − νh∞ F = ( α1 − λf∞ B) + ( α3 − νh∞ F ) > 0. α 1 + α By using (H2) and the definition of hi0 , we deduce that there exists R3 > 0 such that h(t, u, v, w) ≥ 1ε (u+v+w) for all u, v, w ≥ 0 with u+v+w ≤ R3 and t ∈ [σ, 1]. We denote by Ω3 = {(u, v, w) ∈ Y, (u, v, w)Y < R3 }. Let (u, v, w) ∈ P ∩ ∂Ω3 , that is (u, v, w)Y = R3 . Because u(t) + v(t) + w(t) ≤ R3 for all t ∈ [0, 1], then by using Lemma 5.1.3, we obtain  1 tγ−1 J3 (s)h(s, u(s), v(s), w(s)) ds Q3 (u, v, w)(t) ≥ ν 0

≥ νσ γ−1 ≥ νσ γ−1



1

σ



1

σ

J3 (s)h(s, u(s), v(s), w(s)) ds 1 J3 (s) (u(s) + v(s) + w(s)) ds ε

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≥ νσ

γ−1

1 θ ε



1

σ

J3 (s)(u, v, w)Y ds

1 = νσ γ−1 θ E(u, v, w)Y ≥ (u, v, w)Y , ε

∀ t ∈ [σ, 1].

Then for (u, v, w) ∈ P ∩ ∂Ω3 we find Q3 (u, v, w) ≥ Q3 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q3 (u, v, w) ≥ (u, v, w)Y .

(5.24)

Now, using the functions f ∗ , g ∗ , h∗ defined in the proof of Case (1), we have f ∗ (t, x) s = f∞ , t∈[0,1] x

g ∗ (t, x) = 0, x→∞ t∈[0,1] x

lim sup max x→∞

lim max

h∗ (t, x) = hs∞ . t∈[0,1] x

lim sup max x→∞

Therefore for ε > 0, there exists R4 > 0 such that for all x ≥ R4 and s +ε)x, g ∗ (t, x) ≤ εx, h∗ (t, x) ≤ (hs∞ +ε)x. t ∈ [0, 1], we deduce f ∗ (t, x) ≤ (f∞ We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v, w) ∈ Y, (u, v, w)Y < R4 }. Let (u, v, w) ∈ P ∩ ∂Ω4 . Then for all t ∈ [0, 1], we obtain  1 Q1 (u, v, w)(t) ≤ λ J1 (s)f (s, u(s), v(s), w(s)) ds  ≤λ

0

1 0

J1 (s)f ∗ (s, (u, v, w)Y ) ds

s + ε) ≤ λ(f∞

≤λ

s f∞

 0

1

J1 (s)(u, v, w)Y ds

s B α  − λf∞ + 1 2λB

B(u, v, w)Y

1 s (λf∞ B+α 1 )(u, v, w)Y , 2  1 Q2 (u, v, w)(t) ≤ μ J2 (s)g(s, u(s), v(s), w(s)) ds =

0

 ≤μ

0

1

J2 (s)g ∗ (s, (u, v, w)Y ) ds

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 ≤ με ≤μ

0

1

J2 (s)(u, v, w)Y ds

s 1 − λf∞ B − νhs∞ F D(u, v, w)Y 2μD

1 s (1 − λf∞ B − νhs∞ F )(u, v, w)Y , 2  1 Q3 (u, v, w)(t) ≤ ν J3 (s)h(s, u(s), v(s), w(s)) ds =

 ≤ν

0

1 0

J3 (s)h∗ (s, (u, v, w)Y ) ds

≤ ν(hs∞ + ε)

 0

1

J3 (s)(u, v, w)Y ds



α  − νhs∞ F ≤ ν hs∞ + 3 F (u, v, w)Y 2νF =

1 (νhs∞ F + α 3 )(u, v, w)Y . 2

Therefore, 1 s (λf∞ B+α 1 )(u, v, w)Y , 2 1 s Q2 (u, v, w) ≤ (1 − λf∞ B − νhs∞ F )(u, v, w)Y , 2 1 Q3 (u, v, w) ≤ (νhs∞ F + α 3 )(u, v, w)Y . 2 Q1 (u, v, w) ≤

Then for (u, v, w) ∈ P ∩ ∂Ω4 , we conclude that Q(u, v, w)Y ≤

1 s s (λf∞ B+α 1 + 1 − λf∞ B − νhs∞ F + νhs∞ F + α 3 ) 2 (5.25) × (u, v, w)Y = (u, v, w)Y .

By using Lemma 5.1.5, Theorem 1.2.2 (ii) and relations (5.24), (5.25), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω4 \ Ω3 ), which is a positive solution for our problem (5.1), (5.2). s s = hs∞ = 0, g0i = ∞, g∞ , f0i , hi0 ∈ (0, ∞). Case (14). We consider f∞  Let λ ∈ (0, ∞), μ ∈ (0, M4 ), ν ∈ (0, ∞). We choose ε > 0 such that

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ε ≤ μθσ β−1 C and ε≤

s 1 − μg∞ D , 4λB

ε≤

s 1 − μg∞ D , 2μD

ε≤

s 1 − μg∞ D . 4νF

The numerator of the above fractions is positive because μ < gs1D , that 0 s D > 0. is, 1 − μg∞ By using (H2) and the definition of g0i , we deduce that there exists R3 > 0 such that g(t, u, v, w) ≥ 1ε (u+v+w) for all u, v, w ≥ 0 with u+v+w ≤ R3 and t ∈ [σ, 1]. We denote by Ω3 = {(u, v, w) ∈ Y, (u, v, w)Y < R3 }. Let (u, v, w) ∈ P ∩ ∂Ω3 , that is (u, v, w)Y = R3 . Because u(t) + v(t) + w(t) ≤ R3 for all t ∈ [0, 1], then by using Lemma 5.1.3, we obtain  Q2 (u, v, w)(t) ≥ μ ≥ μσ

0

1

tβ−1 J2 (s)g(s, u(s), v(s), w(s)) ds

β−1



1

σ



J2 (s)g(s, u(s), v(s), w(s)) ds

1

1 J2 (s) (u(s) + v(s) + w(s)) ds ε σ  1 1 ≥ μσ β−1 θ J2 (s)(u, v, w)Y ds ε σ

≥ μσ β−1

1 = μσ β−1 θ C(u, v, w)Y ≥ (u, v, w)Y , ε

∀ t ∈ [σ, 1].

Then for (u, v, w) ∈ P ∩ ∂Ω3 we find Q2 (u, v, w) ≥ Q2 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q2 (u, v, w) ≥ (u, v, w)Y .

(5.26)

Now, using the functions f ∗ , g ∗ , h∗ defined in the proof of Case (1), we have f ∗ (t, x) = 0, x→∞ t∈[0,1] x lim max

g ∗ (t, x) s = g∞ , x t∈[0,1]

lim sup max x→∞

h∗ (t, x) = 0. x→∞ t∈[0,1] x lim max

Therefore, for ε > 0, there exists R4 > 0 such that for all x ≥ R4 and s + ε)x, h∗ (t, x) ≤ εx. t ∈ [0, 1] we deduce f ∗ (t, x) ≤ εx, g ∗ (t, x) ≤ (g∞

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We consider R4 = max{2R3 , R4 }, and we denote by Ω4 = {(u, v, w) ∈ Y, (u, v, w)Y < R4 }. Let (u, v, w) ∈ P ∩∂Ω4 . Then for all t ∈ [0, 1] we obtain  Q1 (u, v, w)(t) ≤ λ

1

J1 (s)f (s, u(s), v(s), w(s)) ds

0

 ≤λ

1

J1 (s)f ∗ (s, (u, v, w)Y ) ds

0

 ≤ λε

1

0

J1 (s)(u, v, w)Y ds

= λεB(u, v, w)Y ≤ λ

s D 1 − μg∞ B(u, v, w)Y 4λB

1 s (1 − μg∞ D)(u, v, w)Y , 4  1 Q2 (u, v, w)(t) ≤ μ J2 (s)g(s, u(s), v(s), w(s)) ds =

0

 ≤μ ≤

1

J2 (s)g ∗ (s, (u, v, w)Y ) ds

0

s μ(g∞

 + ε)

0

1

J2 (s)(u, v, w)Y ds

s = μ(g∞ + ε)D(u, v, w)Y

s D 1 − μg∞ s ≤ μ g∞ + D(u, v, w)Y 2μD

1 s (μg∞ D + 1)(u, v, w)Y , 2  1 Q3 (u, v, w)(t) ≤ ν J3 (s)h(s, u(s), v(s), w(s)) ds =

0

 ≤ν

1

0

 ≤ νε

0

J3 (s)h∗ (s, (u, v, w)Y ) ds 1

J3 (s)(u, v, w)Y ds

= νεF (u, v, w)Y ≤ ν =

s 1 − νg∞ D F (u, v, w)Y 4νF

1 s D)(u, v, w)Y . (1 − μg∞ 4

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Therefore, 1 s (1 − μg∞ D)(u, v, w)Y , 4 1 s Q2 (u, v, w) ≤ (1 + μg∞ D)(u, v, w)Y , 2 1 s Q3 (u, v, w) ≤ (1 − μg∞ D)(u, v, w)Y . 4

Q1 (u, v, w) ≤

Then for (u, v, w) ∈ P ∩ ∂Ω4 , we conclude that Q(u, v, w)Y ≤

1 s s s (1 − μg∞ D + 2 + 2μg∞ D + 1 − μg∞ D) 4 × (u, v, w)Y = (u, v, w)Y .

(5.27)

By using Lemma 5.1.5, Theorem 1.2.2 (ii), the relations (5.26) and (5.27), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω4 \ Ω3 ), which is a positive solution for our problem (5.1), (5.2). s s = g∞ = hs∞ = 0, f0i = g0i = ∞ and hi0 ∈ Case (16). We consider f∞ (0, ∞). Let λ ∈ (0, ∞), μ ∈ (0, ∞), ν ∈ (0, ∞). We choose ε > 0 such that

ε ≤ λθσ α−1 A,

ε≤

1 , 3λB

ε≤

1 , 3μD

ε≤

1 . 3νF

By using (H2) and the definition of f0i , we deduce that there exists R3 > 0 such that f (t, u, v, w) ≥ 1ε (u+v+w) for all u, v, w ≥ 0 with u+v+w ≤ R3 and t ∈ [σ, 1]. We denote by Ω3 = {(u, v, w) ∈ Y, (u, v, w)Y < R3 }. Let (u, v, w) ∈ P ∩ ∂Ω3 , that is (u, v, w)Y = R3 . Because u(t) + v(t) + w(t) ≤ R3 for all t ∈ [0, 1], then by using Lemma 5.1.3, we obtain  1 Q1 (u, v, w)(t) ≥ λ tα−1 J1 (s)f (s, u(s), v(s), w(s)) ds 0

≥ λσ α−1



1

σ



J1 (s)f (s, u(s), v(s), w(s)) ds

1

1 J1 (s) (u(s) + v(s) + w(s)) ds ε σ  1 1 ≥ λσ α−1 θ J1 (s)(u, v, w)Y ds ε σ

≥ λσ

α−1

1 = λσ α−1 θ A(u, v, w)Y ≥ (u, v, w)Y , ε

∀ t ∈ [σ, 1].

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Then for (u, v, w) ∈ P ∩ ∂Ω3 we find Q1 (u, v, w) ≥ Q1 (u, v, w)(σ) ≥ (u, v, w)Y , and Q(u, v, w)Y ≥ Q1 (u, v, w) ≥ (u, v, w)Y .

(5.28)

Now, using the functions f ∗ , g ∗ , h∗ defined in the proof of Case (1), we have f ∗ (t, x) g ∗ (t, x) = 0, lim max = 0, lim max x→∞ t∈[0,1] x→∞ t∈[0,1] x x h∗ (t, x) = 0. x→∞ t∈[0,1] x lim max

Therefore, for ε > 0, there exists R4 > 0 such that f ∗ (t, x) ≤ εx, g ∗ (t, x) ≤ εx, h∗ (t, x) ≤ εx for all x ≥ R4 and t ∈ [0, 1]. We consider R4 = max{2R3 , R4 } and we denote by Ω4 = {(u, v, w) ∈ Y, (u, v, w)Y < R4 }. Let (u, v, w) ∈ P ∩ ∂Ω4 . Then for all t ∈ [0, 1], we obtain  1 Q1 (u, v, w)(t) ≤ λ J1 (s)f (s, u(s), v(s), w(s)) ds 0

 ≤λ

1

0

 ≤ λε

J1 (s)f ∗ (s, (u, v, w)Y ) ds 1

0

J1 (s)(u, v, w)Y ds

= λεB(u, v, w)Y ≤  Q2 (u, v, w)(t) ≤ μ

1

0

 ≤μ

1

0

 ≤ με

0

J2 (s)g(s, u(s), v(s), w(s)) ds J2 (s)g ∗ (s, (u, v, w)Y ) ds 1

J2 (s)(u, v, w)Y ds

= μεD(u, v, w)Y ≤  Q3 (u, v, w)(t) ≤ ν

0

 ≤ν

1

0

1

1 (u, v, w)Y , 3

1 (u, v, w)Y , 3

J3 (s)h(s, u(s), v(s), w(s)) ds J3 (s)h∗ (s, (u, v, w)Y ) ds

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 ≤ νε

0

1

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265

J3 (s)(u, v, w)Y ds

= νεF (u, v, w)Y ≤

1 (u, v, w)Y . 3

Therefore Q1 (u, v, w) ≤ 13 (u, v, w)Y , Q2 (u, v, w) ≤ 13 (u, v, w)Y , Q3 (u, v, w) ≤ 13 (u, v, w)Y . Then for (u, v, w) ∈ P ∩ ∂Ω4 , we conclude that Q(u, v, w)Y ≤ (u, v, w)Y .

(5.29)

By using Lemma 5.1.5, Theorem 1.2.2 (ii), the relations (5.28) and (5.29), we deduce that Q has a fixed point (u, v, w) ∈ P ∩ (Ω4 \ Ω3 ), which is a positive solution for our problem (5.1), (5.2).  Remark 5.1.4. Each of the cases (9)–(16) of Theorem 5.1.2 contains 7 cases as follows: {f0i = ∞, g0i , hi0 ∈ (0, ∞)}, or {g0i = ∞, f0i , hi0 ∈ (0, ∞)}, or {hi0 = ∞, f0i , g0i ∈ (0, ∞)}, or {f0i = g0i = ∞, hi0 ∈ (0, ∞)}, or {f0i = hi0 = ∞, g0i ∈ (0, ∞)}, or {g0i = hi0 = ∞, f0i ∈ (0, ∞)}, or {f0i = g0i = hi0 = ∞}. So, the total number of cases from Theorem 5.1.2 is 64, which we grouped in 16 cases. Each of the cases (1)–(8) contains 4 subcases because α1 , α2 , α3 ∈ (0, 1), or α1 = 1 and α2 = α3 = 0, or α2 = 1 and α1 = α3 = 0, or α3 = 1 and α1 = α2 = 0. Remark 5.1.5. In [98], the authors present only 15 cases (Theorems 2.16– 2.30 from [98]) from 64 cases, namely the first 9 cases of our Theorem 5.1.2. They did not study the cases when some extreme limits are 0 and other are ∞. Besides, our intervals for parameters λ, μ, ν presented in Theorem 5.1.2 (our cases (2)–(7) and (9)) are better than the corresponding ones from [98, Theorems 2.17–2.22 and 2.24–2.30]. Remark 5.1.6. One can formulate existence results for the general case of the system of n fractional differential equations (5.20) with the boundary conditions (5.21) from Remark 5.1.3. According to the values of f (t,u1 ,...,un ) s i = lim supu1 +···+un →∞ supt∈[0,1] ju1 +···+u ∈ [0, ∞), and fj0 = fj∞ n f (t,u ,...,u )

1 n lim inf u1 +···+un →0 inf t∈[σ,1] ju1 +···+u ∈ (0, ∞], j = 1, . . . , n we have 22n n n+1 cases, which can be grouped in 2 cases.

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Nonexistence of positive solutions

We present in this section intervals for λ, μ and ν for which there exist no positive solutions of problem (5.1), (5.2). Theorem 5.1.3. Assume that (H1) and (H2) hold. If there exist positive numbers A1 , A2 , A3 such that f (t, u, v, w) ≤ A1 (u + v + w),

g(t, u, v, w) ≤ A2 (u + v + w),

h(t, u, v, w) ≤ A3 (u + v + w),

∀ t ∈ [0, 1], u, v, w ≥ 0,

(5.30)

then there exist positive constants λ0 , μ0 , ν0 such that for every λ ∈ (0, λ0 ), μ ∈ (0, μ0 ), ν ∈ (0, ν0 ) the boundary value problem (5.1), (5.2) has no positive solution. Proof. We define λ0 = 3A11 B , μ0 = 3A12 D , ν0 = 3A13 F , where B =

1

1

1 J (s) ds, D = 0 J2 (s) ds, F = 0 J3 (s) ds. We will show that for any 0 1 λ ∈ (0, λ0 ), μ ∈ (0, μ0 ), ν ∈ (0, ν0 ), the problem (5.1), (5.2) has no positive solution. Let λ ∈ (0, λ0 ), μ ∈ (0, μ0 ), ν ∈ (0, ν0 ). We suppose that the problem (5.1), (5.2) has a positive solution (u(t), v(t), w(t)), t ∈ [0, 1]. Then, we have  1 G1 (t, s)f (s, u(s), v(s), w(s)) ds u(t) = Q1 (u, v, w)(t) = λ 0

 ≤λ

1 0

J1 (s)f (s, u(s), v(s), w(s)) ds



≤ λA1

1

0

J1 (s)(u(s) + v(s) + w(s)) ds 

≤ λA1 (u + v + w)

0

1

J1 (s) ds

= λA1 B(u, v, w)Y , ∀ t ∈ [0, 1],  1 G2 (t, s)g(s, u(s), v(s), w(s)) ds v(t) = Q2 (u, v, w)(t) = μ  ≤μ

0

≤ μA2

0

1

J2 (s)g(s, u(s), v(s), w(s)) ds

 0

1

J2 (s)(u(s) + v(s) + w(s)) ds

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 ≤ μA2 (u + v + w)

1

0

J2 (s) ds

= μA2 D(u, v, w)Y , ∀ t ∈ [0, 1],  1 w(t) = Q3 (u, v, w)(t) = ν G3 (t, s)h(s, u(s), v(s), w(s)) ds  ≤ν

0

≤ νA3

0

1

J3 (s)h(s, u(s), v(s), w(s)) ds

 0

1

J3 (s)(u(s) + v(s) + w(s)) ds 

≤ νA3 (u + v + w) = νA3 F (u, v, w)Y ,

0

1

J3 (s) ds

∀ t ∈ [0, 1].

Therefore, we conclude 1 (u, v, w)Y , 3 1 v ≤ μA2 D(u, v, w)Y < μ0 A2 D(u, v, w)Y = (u, v, w)Y , 3 1 w ≤ νA3 F (u, v, w)Y < ν0 A3 F (u, v, w)Y = (u, v, w)Y . 3

u ≤ λA1 B(u, v, w)Y < λ0 A1 B(u, v, w)Y =

Hence, we deduce (u, v, w)Y = u + v + w < (u, v, w)Y , which is a contradiction. So, the boundary value problem (5.1), (5.2) has no positive solution.  Remark 5.1.7. In the proof of Theorem 5.1.3, we can also define λ0 = α1 α2 α3 A1 B , μ0 = A2 D , ν0 = A3 F with α1 , α2 , α3 > 0 and α1 + α2 + α3 = 1. s s , g∞ , hs∞ < ∞, then there exist positive Remark 5.1.8. If f0s , g0s , hs0 , f∞ constants A1 , A2 , A3 such that (5.30) holds, and then we obtain the conclusion of Theorem 5.1.3.

Theorem 5.1.4. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m1 > 0 such that f (t, u, v, w) ≥ m1 (u + v + w),

∀ t ∈ [σ, 1], u, v, w ≥ 0,

(5.31)

0 such that for every λ > λ 0 , μ > 0 then there exists a positive constant λ and ν > 0, the boundary value problem (5.1), (5.2) has no positive solution.

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1 0 = α−11 Proof. We define λ θσ m1 A where A = σ J1 (s) ds. We will show 0 , μ > 0 and ν > 0, the problem (5.1), (5.2) has no that for every λ > λ positive solution. 0 , μ > 0 and ν > 0. We suppose that the problem (5.1), (5.2) Let λ > λ has a positive solution (u(t), v(t), w(t)), t ∈ [0, 1]. Then we obtain  1 u(t) = Q1 (u, v, w)(t) = λ G1 (t, s)f (s, u(s), v(s), w(s)) ds ≥ λtα−1

0



≥ λσ α−1

1

σ



J1 (s)f (s, u(s), v(s), w(s)) ds

1

σ

J1 (s)m1 (u(s) + v(s) + w(s)) ds

≥ λθσ α−1 m1



1

σ

J1 (s)(u + v + w) ds

= λθσ α−1 m1 A(u, v, w)Y . Therefore, we deduce 0 θσ α−1 m1 A(u, v, w)Y u ≥ u(σ) ≥ λθσ α−1 m1 A(u, v, w)Y > λ = (u, v, w)Y , and so, (u, v, w)Y = u + v + w > (u, v, w)Y , which is a contradiction. Therefore, the boundary value problem (5.1), (5.2) has no positive solution.  In a similar manner, we obtain the following theorems. Theorem 5.1.5. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m2 > 0 such that g(t, u, v, w) ≥ m2 (u + v + w),

∀ t ∈ [σ, 1], u, v, w ≥ 0,

(5.32)

0 then there exists a positive constant μ 0 such that for every λ > 0, μ > μ and ν > 0, the boundary value problem (5.1), (5.2) has no positive solution.

1 In Theorem 5.1.5, we define μ 0 = θσβ−11 m2 C , where C = σ J2 (s) ds. Theorem 5.1.6. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m3 > 0 such that h(t, u, v, w) ≥ m3 (u + v + w),

∀ t ∈ [σ, 1], u, v, w ≥ 0,

(5.33)

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then there exists a positive constant ν0 such that for every λ > 0, μ > 0 and ν > ν0 , the boundary value problem (5.1), (5.2) has no positive solution.

1 In Theorem 5.1.6, we define ν0 = θσγ−11 m3 E , where E = σ J3 (s) ds. Remark 5.1.9. i > 0 and f (t, u, v, w) > 0 for all t ∈ [σ, 1] and (a) If for σ ∈ (0, 1), f0i , f∞ u, v, w ≥ 0 with u + v + w > 0, then relation (5.31) holds and we obtain the conclusion of Theorem 5.1.4. i > 0 and g(t, u, v, w) > 0 for all t ∈ [σ, 1] and (b) If for σ ∈ (0, 1), g0i , g∞ u, v, w ≥ 0 with u + v + w > 0, then relation (5.32) holds and we obtain the conclusion of Theorem 5.1.5. (c) If for σ ∈ (0, 1), hi0 , hi∞ > 0 and h(t, u, v, w) > 0 for all t ∈ [σ, 1] and u, v, w ≥ 0 with u + v + w > 0, then relation (5.33) holds and we obtain the conclusion of Theorem 5.1.6.

Theorem 5.1.7. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m1 , m2 > 0 such that f (t, u, v, w) ≥ m1 (u + v + w),

g(t, u, v, w) ≥ m2 (u + v + w),

∀ t ∈ [σ, 1], u, v, w ≥ 0,

(5.34)

   0 and μ 0 , then there exist positive constants λ 0 such that for every λ > λ  μ>μ 0 and ν > 0, the boundary value problem (5.1), (5.2) has no positive solution.       μ 0 λ0 1 1  0 = = and μ  = Proof. We define λ = α−1 0 2θσ m1 A 2 2 . 2θσβ−1 m2 C   0 , μ > μ Then for every λ > λ 0 and ν > 0, the problem (5.1), (5.2) has no   0 , μ > μ positive solution. Indeed, let λ > λ  and ν > 0. We suppose that 0

the problem (5.1), (5.2) has a positive solution (u(t), v(t), w(t)), t ∈ [0, 1]. Then in a similar manner as in the proof of Theorem 5.1.4, we deduce u ≥ λθσ α−1 m1 A(u, v, w)Y ,

v ≥ μθσ β−1 m2 C(u, v, w)Y ,

and so (u, v, w)Y = u + v + w ≥ u + v ≥ (λθσ α−1 m1 A + μθσ β−1 m2 C)(u, v, w)Y   0 θσ α−1 m1 A + μ > (λ 0 θσ β−1 m2 C)(u, v, w)Y

1 1 + = (u, v, w)Y = (u, v, w)Y , 2 2

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which is a contradiction. Therefore, the boundary value problem (5.1), (5.2) has no positive solution.   0 = Remark 5.1.10. In the proof of Theorem 5.1.7, we can also define λ α 1 α   0 = θσβ−12m2 C with α 1 , α 2 > 0 with α 1 + α 2 = 1. θσα−1 m1 A , μ In a similar manner, we obtain the following theorems. Theorem 5.1.8. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m1 , m3 > 0 such that f (t, u, v, w) ≥ m1 (u + v + w), h(t, u, v, w) ≥ m3 (u + v + w), ∀ ∈ t ∈ [σ, 1], u, v, w ≥ 0,

(5.35)

      and ν  , then there exist positive constants λ 0 such that for every λ > λ 0 0  μ > 0 and ν >  ν , the boundary value problem (5.1), (5.2) has no positive 0

solution.       λ0 1  = and ν 0 = In Theorem 5.1.8, we define λ 0 2θσα−1 m1 A = 2   ν0    α 1 α 2 1   = α−1 = , or in general λ 0 = θσγ−1 γ−1 0 2θσ m3 E 2 θσ m1 A and ν m3 E with 2 > 0, α 1 + α 2 = 1. α 1 , α Theorem 5.1.9. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m2 , m3 > 0 such that g(t, u, v, w) ≥ m2 (u + v + w),

h(t, u, v, w) ≥ m3 (u + v + w),

∀ ∈ t ∈ [σ, 1], u, v, w ≥ 0,

(5.36)

   then there exist positive constants μ 0 and  ν0 such that for every λ > 0,    μ>μ  and ν >  ν , the boundary value problem (5.1), (5.2) has no positive 0

0

solution.     1  = μ20 and ν 0 = In Theorem 5.1.9, we define μ 0 = 2θσβ−1 m2 C  ν    α 1 α 2 1 0   0 = θσγ−1 0 = θσβ−1 2θσγ−1 m3 E = 2 , or in general μ m3 E with m2 C and ν α 1 , α 2 > 0, α 1 + α 2 = 1.

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271

Remark 5.1.11. i i , g0i , g∞ > 0 and f (t, u, v, w) > 0, g(t, u, v, w) > 0 (a) If for σ ∈ (0, 1), f0i , f∞ for all t ∈ [σ, 1] and u, v, w ≥ 0 with u + v + w > 0, then the relation (5.34) holds, and we obtain the conclusion of Theorem 5.1.7. i , hi0 , hi∞ > 0 and f (t, u, v, w) > 0, h(t, u, v, w) > 0 (b) If for σ ∈ (0, 1), f0i , f∞ for all t ∈ [σ, 1] and u, v, w ≥ 0 with u + v + w > 0, then the relation (5.35) holds, and we obtain the conclusion of Theorem 5.1.8. i , hi0 , hi∞ > 0 and g(t, u, v, w) > 0, h(t, u, v, w) > 0 (c) If for σ ∈ (0, 1), g0i , g∞ for all t ∈ [σ, 1] and u, v, w ≥ 0 with u + v + w > 0, then the relation (5.36) holds, and we obtain the conclusion of Theorem 5.1.9.

Theorem 5.1.10. Assume that (H1) and (H2) hold. If there exist positive numbers σ ∈ (0, 1) and m1 , m2 , m3 > 0 such that f (t, u, v, w) ≥ m1 (u + v + w), g(t, u, v, w) ≥ m2 (u + v + w), h(t, u, v, w) ≥ m3 (u + v + w), ∀ t ∈ [σ, 1], u, v, w ≥ 0,

(5.37)

ˆ0, ˆ0 , μ ˆ0 and νˆ0 such that for every λ > λ then there exist positive constants λ μ>μ ˆ0 and ν > νˆ0 , the boundary value problem (5.1), (5.2) has no positive solution. 1 1 1 ˆ0 = Proof. We define λ ˆ0 = 3θσβ−1 , νˆ0 = 3θσγ−1 3θσα−1 m1 A , μ m3 E . m2 C ˆ0 , μ > μ ˆ0 , ν > νˆ0 , the problem (5.1), (5.2) has no Then for every λ > λ ˆ0, μ > μ ˆ0 and ν > νˆ0 . We suppose that positive solution. Indeed, let λ > λ

the problem (5.1), (5.2) has a positive solution (u(t), v(t), w(t)), t ∈ [0, 1]. Then in a similar manner as in the proof of Theorem 5.1.7, we deduce u ≥ λθσ α−1 m1 A(u, v, w)Y , v ≥ μθσ β−1 m2 C(u, v, w)Y , w ≥ νθσ γ−1 m3 E(u, v, w)Y , and so (u, v, w)Y = u + v + w ≥ (λθσ α−1 m1 A + μθσ β−1 m2 C + νθσ γ−1 m3 E)(u, v, w)Y   ˆ 0 θσ α−1 m1 A + μ > λ ˆ0 θσ β−1 m2 C + νˆ0 θσ γ−1 m3 E × (u, v, w)Y = (u, v, w)Y , which is a contradiction. Therefore, the boundary value problem (5.1), (5.2) has no positive solution. 

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ˆ0 = Remark 5.1.12. In the proof of Theorem 5.1.10, we can also define λ α1 α2 α3    ˆ0 = θσβ−1 m2 C , νˆ0 = θσγ−1 m3 F , where α1 , α2 , α3 > 0 with θσα−1 m1 A , μ   α1 + α2 + α3 = 1. i i , g0i , g∞ , hi0 , hi∞ > 0 and Remark 5.1.13. If for σ ∈ (0, 1), f0i , f∞ f (t, u, v, w) > 0, g(t, u, v, w) > 0, h(t, u, v, w) > 0 for all t ∈ [σ, 1], u, v, w ≥ 0, u + v + w > 0, then the relation (5.37) holds, and we have the conclusion of Theorem 5.1.10.

Remark 5.1.14. The conclusions of Theorems 5.1.1–5.1.10 remain valid for general systems of Hammerstein integral equations of the form ⎧  1 ⎪ ⎪ u(t) = λ G1 (t, s)f (s, u(s), v(s), w(s)) ds, t ∈ [0, 1], ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪  1 ⎨ v(t) = μ G2 (t, s)g(s, u(s), v(s), w(s)) ds, t ∈ [0, 1], ⎪ ⎪ 0 ⎪ ⎪ ⎪  1 ⎪ ⎪ ⎪ ⎩ w(t) = ν G3 (t, s)h(s, u(s), v(s), w(s)) ds, t ∈ [0, 1],

(5.38)

0

with positive parameters λ, μ, ν, and instead of assumptions (H1) − (H2), the following assumptions are satisfied  The functions G1 , G2 , G3 : [0, 1]×[0, 1] → R are continuous and there (H1) exist the continuous functions J1 , J2 , J3 : [0, 1] → R and σ ∈ (0, 1), α, β, γ > 2 such that (a) 0 ≤ Gi (t, s) ≤ Ji (s), ∀ t, s ∈ [0, 1], i = 1, 2, 3; (b) G1 (t, s) ≥ tα−1 J1 (s), G2 (t, s) ≥ tβ−1 J2 (s), G3 (t, s) tγ−1 J (s), ∀ t, s ∈ [0, 1];  1 3 Ji (s) ds > 0, i = 1, 2, 3. (c)

≥

σ

 The functions f, g, h : [0, 1] × R+ × R+ × R+ → R+ are continuous. (H2) 5.1.4

Examples

10 1 7 Let n = 3, m = 5, l = 4, α = 52 , β = 17 4 , γ = 3 , p1 = 1, q1 = 2 , p2 = 3 , q2 = 32 , p3 = 74 , q3 = 23 , N = 2, M = 1, L = 3, ξ1 = 13 , ξ2 = 23 , a1 = 2, a2 = 12 , η1 = 12 , b1 = 4, ζ1 = 14 , ζ2 = 12 , ζ3 = 34 , c1 = 3, c2 = 2, c3 = 1.

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We consider the system of fractional differential equations ⎧ 5/2 D0+ u(t) + λf (t, u(t), v(t), w(t)) = 0, t ∈ (0, 1), ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

17/4

D0+ v(t) + μg(t, u(t), v(t), w(t)) = 0, 10/3

D0+ w(t) + νh(t, u(t), v(t), w(t)) = 0,

t ∈ (0, 1),

(5.39)

t ∈ (0, 1),

with the multi-point boundary conditions ⎧ 1 2 1 1/2 ⎪ 1/2   ⎪ u(0) = u D (0) = 0, u (1) = 2D u u + , ⎪ 0+ 0+ ⎪ ⎪ 3 2 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ v(0) = v  (0) = v  (0) = v  (0) = 0, D7/3 v(1) = 4D3/2 v 1 , 0+ 0+ 2 ⎪   ⎪ ⎪ (0) = w (0) = 0, w(0) = w ⎪ ⎪ ⎪ ⎪ ⎪ 1 1 3 ⎪ 7/4 2/3 2/3 2/3 ⎪ + 2D0+ w + D0+ w . ⎩ D0+ w(1) = 3D0+ w 4 2 4

(5.40)

√ 1/4 Γ(17/4) Γ(17/4) 6−3 π ≈ 0.17065961 > 0, Δ2 = Γ(23/12) − 2 Γ(11/4) 4 Γ(10/3) Γ(10/3) −(3+28/3 +35/3 ) 45/3 ≈ 0.25945301 > Δ3 = Γ(19/12) Γ(8/3)

We have Δ1 =

2.43672831 > 0, So assumption (H1) is satisfied. Besides, we deduce t3/2 (1 − s)1/2 − (t − s)3/2 , 0 ≤ s ≤ t ≤ 1, 1 g1 (t, s) = Γ(5/2) t3/2 (1 − s)1/2 , 0 ≤ t ≤ s ≤ 1, g2 (t, s) =

t(1 − s)1/2 − (t − s),

0 ≤ s ≤ t ≤ 1,

1/2

t(1 − s) , 0 ≤ t ≤ s ≤ 1, t13/4 (1 − s)11/12 − (t − s)13/4 , 0 ≤ s ≤ t ≤ 1, 1 g3 (t, s) = Γ(17/4) t13/4 (1 − s)11/12 , 0 ≤ t ≤ s ≤ 1, t7/4 (1 − s)11/12 − (t − s)7/4 , 0 ≤ s ≤ t ≤ 1, 1 g4 (t, s) = Γ(11/4) t7/4 (1 − s)11/12 , 0 ≤ t ≤ s ≤ 1,

1 g5 (t, s) = Γ(10/3) 1 g6 (t, s) = Γ(8/3)





t7/3 (1 − s)7/12 − (t − s)7/3 , t

7/3

7/12

(1 − s)

,

0 ≤ t ≤ s ≤ 1,

t5/3 (1 − s)7/12 − (t − s)5/3 , t

5/3

7/12

(1 − s)

,

0 ≤ s ≤ t ≤ 1,

0 ≤ s ≤ t ≤ 1,

0 ≤ t ≤ s ≤ 1.

≈ 0.

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Then we obtain

1 2 1 , s + g2 ,s , 2g2 3 2 3

1 4t13/4 G2 (t, s) = g3 (t, s) + ,s , g4 Δ2 2



1 1 3 t7/3 G3 (t, s) = g5 (t, s) + , s + 2g6 , s + g6 ,s , 3g6 Δ3 4 2 4 G1 (t, s) = g1 (t, s) +

t3/2 Δ1

4 1 (1 − s)11/12 (1 − (1 − s)7/3 ), h1 (s) = √ s(1 − s)1/2 , h3 (s) = Γ(17/4) 3 π h5 (s) =

1 (1 − s)7/12 (1 − (1 − s)7/4 ), Γ(10/3)



1 2 4 1 1 2g2 , s + g2 ,s J1 (s) = √ s(1 − s)1/2 + Δ1 3 2 3 3 π ⎧   4 1 1 ⎪ ⎪ √ s(1 − s)1/2 + 2(1 − s)1/2 + 5s − 2 , 0 ≤ s < , ⎪ ⎪ 3 π 2Δ1 3 ⎪ ⎪ ⎪ ⎪   ⎨ 4 2 1 1 √ s(1 − s)1/2 + 6(1 − s)1/2 + 3s − 2 , ≤s< , = 6Δ 3 3 3 π ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ 4 1 2 ⎪ ⎪ ⎩ √ s(1 − s)1/2 + ≤ s ≤ 1. (1 − s)1/2 , 3 π Δ1 3

1 1 4 11/12 7/3 (1 − s) ,s J2 (s) = (1 − (1 − s) ) + g4 Γ(17/4) Δ2 2 ⎧ 1 ⎪ ⎪ (1 − s)11/12 (1 − (1 − s)7/3 ) ⎪ ⎪ Γ(17/4) ⎪ ⎪ ⎪ ⎪ ⎪ 21/4 1 ⎪ ⎪ [(1 − s)11/12 − (1 − 2s)7/4 ], 0 ≤ s < , + ⎨ Δ2 Γ(11/4) 2 = 1 ⎪ 11/12 7/3 ⎪ (1 − s) (1 − (1 − s) ) ⎪ ⎪ ⎪ Γ(17/4) ⎪ ⎪ ⎪ ⎪ 21/4 1 ⎪ ⎪ ⎩ (1 − s)11/12 , ≤ s ≤ 1. + Δ2 Γ(11/4) 2 J3 (s) =

1 (1 − s)7/12 (1 − (1 − s)7/4 ) Γ(10/3)



1 1 1 3 , s + 2g6 , s + g6 ,s + 3g6 Δ3 4 2 4

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⎧ 1 ⎪ ⎪ (1 − s)7/12 (1 − (1 − s)7/4 ) ⎪ ⎪ Γ(10/3) ⎪ ⎪ ⎪  ⎪ 1 ⎪ ⎪ ⎪ + 10/3 (3 + 28/3 + 35/3 )(1 − s)7/12 ⎪ ⎪ 2 Δ Γ(8/3) 3 ⎪ ⎪ ⎪  ⎪ 1 ⎪ ⎪ ⎪ − 3(1 − 4s)5/3 − 28/3 (1 − 2s)5/3 − (3 − 4s)5/3 , 0 ≤ s < , ⎪ ⎪ 4 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ (1 − s)7/12 (1 − (1 − s)7/4 ) ⎪ ⎪ Γ(10/3) ⎪ ⎪ ⎪  ⎪ ⎪ 1 ⎪ ⎪ (3 + 28/3 + 35/3 )(1 − s)7/12 + ⎪ 10/3 Δ Γ(8/3) ⎪ 2 ⎪ 3 ⎪ ⎪  1 ⎨ 1 ≤s< , −28/3 (1 − 2s)5/3 − (3 − 4s)5/3 , = 4 2 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ (1 − s)7/12 (1 − (1 − s)7/4 ) ⎪ ⎪ Γ(10/3) ⎪ ⎪ ⎪  ⎪ ⎪ 1 ⎪ ⎪ (3 + 28/3 + 35/3 )(1 − s)7/12 + ⎪ 10/3 ⎪ 2 Δ Γ(8/3) ⎪ 3 ⎪ ⎪  1 ⎪ ⎪ 3 ⎪ 5/3 ⎪ , ≤s< , −(3 − 4s) ⎪ ⎪ 2 4 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ (1 − s)7/12 (1 − (1 − s)7/4 ) ⎪ ⎪ Γ(10/3) ⎪ ⎪ ⎪ ⎪ ⎪ 1 3 ⎪ ⎪ ⎩ + 10/3 ≤ s ≤ 1. (3 + 28/3 + 35/3 )(1 − s)7/12 , 4 2 Δ3 Γ(8/3) Now, we choose σ = 14 ∈ (0, 1) and then θ = 2−13/2 ≈ 0.01104854. We

1

1 also obtain A = 1/4 J1 (s) ds ≈ 2.42142749, B = 0 J1 (s) ds ≈ 2.80487506,

1

1 C = 1/4 J2 (s) ds ≈ 0.11093116, D = 0 J2 (s) ds ≈ 0.13771787, E =

1

1 J (s) ds ≈ 1.49070723, F = 0 J3 (s) ds ≈ 1.80167568. 1/4 3 Example 1. We consider the functions q1 + sin v) (2t + 1)[ p1 (u + v + w) + 1](u + v + w)( , u+v+w+1 √ t + 1[ p2 (u + v + w) + 1](u + v + w)( q2 + cos w) , g(t, u, v, w) = u+v+w+1

f (t, u, v, w) =

h(t, u, v, w) =

t2 [ p3 (u + v + w) + 1](u + v + w)( q3 + sin u) , u+v+w+1

for t ∈ [0, 1], u, v, w ≥ 0, where p1 , p2 , p3 > 0, q1 , q2 , q3 > 1.

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√ i i We have f0s = 3 q1 , g0s = 2( q2 + 1), hs0 = q3 , f∞ = 32 p1 ( q1 − 1), g∞ = 5 1 1 i 2 ( q2 −1), h∞ = 16 p3 ( q3 −1). For α1 = α2 = α3 = α 1 = α 2 = α 3 = 3 , we 2 p

√

obtain L1 =

221/2 9 p1 ( q1 −1)A ,

L2 =

1 9 q1 B ,

L3 =

91/6

14 √ 2 , 3 5 p2 ( q2 −1)C

L4 =

√ 1 , 3 2( q2 +1)D

2 1 L5 = 3 p3 ( q3 −1)E , and L6 = 3 q3 F . The conditions L1 < L2 , L3 < L4 and L5 < L6 become

q1 − 1) p1 ( 221/2 B , > q1 A

p2 ( 229/2 D q2 − 1) > 1/2 , q2 + 1 5 C

p3 ( q3 − 1) 291/6 F . > q3 E

q2 −1) For example, if p1 (qq11−1) ≥ 1678, p2q( ≥ 12865 and p3 (qq33−1) ≥ 44454, 2 +1 then the above conditions are satisfied. As an example, we consider q1 = 2, q2 = 3, q3 = 4, p1 = 3356, p2 = 25730, p3 = 59272, and then the inequalities L1 < L2 , L3 < L4 and L5 < L6 are satisfied. In this case, L1 ≈ 0.01980063, L2 ≈ 0.01980678, L3 ≈ 0.42784885, L4 ≈ 0.4278716, L5 ≈ 0.04625271, L6 ≈ 0.04625324. By Theorem 5.1.1 (1) we deduce that for every λ ∈ (L1 , L2 ), μ ∈ (L3 , L4 ) and ν ∈ (L5 , L6 ) there exists a positive solution (u(t), v(t), w(t)), t ∈ [0, 1] of problem (5.39), (5.40). √ √ s s q1 , f∞ = 3 p1 ( q1 +1), g0s = 2( q2 +1), g∞ = 2 p2 ( q2 +1), Because f0s = 3 q3 + 1), then by Theorem 5.1.3 and Remark 5.1.8, we hs0 = q3 , hs∞ = p3 ( conclude that for any λ ∈ (0, λ0 ), μ ∈ (0, μ0 ) and ν ∈ (0, ν0 ), the problem (5.39), (5.40) has no positive solution, where λ0 = 3A11 B , μ0 = 3A12 D , ν0 = 3A13 F . If we consider as above p1 = 3356, q1 = 2, p2 = 36386, q2 = 3, √ p3 = 59272, q3 = 4, then A1 = 30204, A2 = 102920 2 ≈ 145551, A3 = 296360. Therefore we obtain λ0 ≈ 3.9346 × 10−6 , μ0 ≈ 1.6629 × 10−5 , ν0 ≈ 6.24284 × 10−7 . i i , g0i , g∞ , hi0 , hi∞ > 0 and f (t, u, v, w) > 0, g(t, u, v, w) > Because f0i , f∞ 0, h(t, u, v, w) > 0 for all t ∈ [1/4, 1] and u, v, w ≥ 0 with u + v + w > 0, we 1 ˆ0 = can also apply Theorem 5.1.10 and Remark 5.1.13. Here, λ 3θσα−1 m1 A , 1 1 ˆ0 = 3θσγ−1 . For the functions f, g, h presented μ ˆ0 = 3θσβ−1 m m2 C and ν √3 E ˆ 0 ≈ 33.22545838, μ above, we have m1 = 3, m2 = 2 5, m3 = 1 , λ ˆ0 ≈ 4

5504.275396, νˆ0 ≈ 2056.117822. So, if λ > 33.23, μ > 5504.28 and ν > 2056.12, the problem (5.39), (5.40) has no positive solution. Example 2. We consider the functions f (t, u, v, w) = ta (u2 + v 2 + w2 ), h(t, u, v, w) = (u + v + w)c ,

g(t, u, v, w) = (2 − t)b (eu+v+w − 1),

t ∈ [0, 1], u, v, w ≥ 0,

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i i where a, b > 0, c > 1. We have f0s = 0, f∞ = ∞, g0s = 2b , g∞ = ∞, hs0 = 0, i h∞ = ∞.  4 ) and ν ∈ (0, ∞), By Theorem 5.1.1 (14), for any λ ∈ (0, ∞), μ ∈ (0, L 1  with L4 = 2b D , the problem (5.39),(5.40) has a positive solution. Here,

1 4 = 1 ≈ D = 0 J2 (s) ds ≈ 0.13771787. For example, if b = 2, we obtain L 4D 1.8153054. We can also use Theorem 5.1.5, because g(t, u, v, w) ≥ u + v + w for all 0 = θσβ−11 m2 C ≈ t ∈ [1/4, 1] and u, v, w ≥ 0, that is m2 = 1. Because μ 73847.6037, we deduce that for every λ > 0, μ > 73847.61 and ν > 0, the boundary value problem (5.39), (5.40) has no positive solution.

Remark 5.1.15. The results presented in this section under the assumptions p1 ∈ [1, n−2], p2 ∈ [1, m−2] and p3 ∈ [1, l−2] instead of p1 ∈ [1, α−1), p2 ∈ [1, β − 1) and p3 ∈ [1, γ − 1) were published in [77].

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Chapter 6

Existence of Solutions for Riemann–Liouville Fractional Boundary Value Problems

In this chapter, we investigate the existence of solutions for Riemann– Liouville fractional differential equations and systems of Riemann–Liouville fractional differential equations with integral terms, subject to nonlocal boundary conditions which contain fractional derivatives and Riemann– Stieltjes integrals.

6.1

Riemann–Liouville Fractional Differential Equations with Nonlocal Boundary Conditions

We consider the nonlinear fractional differential equation p α u(t) + f (t, u(t), I0+ u(t))) = 0, D0+

t ∈ (0, 1),

with the nonlocal boundary conditions ⎧  (n−2) (0) = 0, ⎪ ⎨ u(0) = u (0) = · · · = u  m 1  β0 βi ⎪ D0+ u(t) dHi (t), ⎩ D0+ u(1) = i=1

(6.1)

(6.2)

0

where α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 2, βi ∈ R for all i = 0, . . . , m, k denotes the Riemann– 0 ≤ β1 < β2 < · · · < βm < α−1, β0 ∈ [0, α−1), D0+ p is the Liouville derivative of order k (for k = α, β0 , β1 , . . . , βm ), p > 0, I0+ Riemann–Liouville integral of order p, f is a nonlinear function, and the 279

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integrals from the boundary condition (6.2) are Riemann–Stieltjes integrals with Hi , i = 1, . . . , m functions of bounded variation. We present conditions for the nonlinearity f such that problem (6.1), (6.2) has at least one solution u ∈ C[0, 1]. In the proofs of the main existence results, we use some theorems from the fixed point theory. 6.1.1

Preliminary results

We consider the fractional differential equation α u(t) + y(t) = 0, D0+

t ∈ (0, 1),

(6.3)

with the boundary conditions (6.2), where y ∈ C(0, 1)∩L1 (0, 1). We denote by  1 m  Γ(α) Γ(α) − sα−βi −1 dHi (s). Δ0 = Γ(α − β0 ) i=1 Γ(α − βi ) 0 In a similar manner as we proved Lemma 2.4.1 from Section 2.4, we obtain the following lemma. Lemma 6.1.1. If Δ0 = 0, then the unique solution u ∈ C[0, 1] of problem (6.2), (6.3) is given by  t 1 (t − s)α−1 y(s) ds u(t) = − Γ(α) 0  1 tα−1 (1 − s)α−β0 −1 y(s) ds + Δ0 Γ(α − β0 ) 0 (6.4)  1  s m tα−1  1 (s − τ )α−βi −1 y(τ ) dτ dHi (s), − Δ0 i=1 Γ(α − βi ) 0 0 t ∈ [0, 1]. By using standard computations, we obtain the following result. Lemma 6.1.2. If x ∈ C[0, 1] then for θ > 0, we have θ |I0+ x(t)| ≤

where x = supt∈[0,1] |x(t)|.

x , Γ(θ + 1)

∀ t ∈ [0, 1],

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6.1.2

Existence of solutions

We introduce firstly the assumptions that we will use in our main existence theorems for problem (6.1), (6.2). (H1) α ∈ R, α ∈ (n − 1, n], n, m ∈ N, n ≥ 2, βi ∈ R for all i = 0, . . . , m, 0 ≤ β1 < β2 < · · · < βm < α − 1, β0 ∈ [0, α − 1), p > 0, Hi : [0, 1] → R, i = 1, . . . , m are functions of bounded variation, and

m 1 α−βi −1 Γ(α) Γ(α) − i=1 Γ(α−β dHi (s) = 0. Δ0 = Γ(α−β 0 s 0) i) 2 (H2) The function f : [0, 1] × R → R is continuous and there exists L1 > 0 such that |f (t, x, y) − f (t, x1 , y1 )| ≤ L1 (|x − x1 | + |y − y1 |), for all t ∈ [0, 1], x, y, x1 , y1 ∈ R. (H3) There exists a function g ∈ C([0, 1], [0, ∞)) such that |f (t, x, y)| ≤ g(t),

∀ (t, x, y) ∈ [0, 1] × R2 .

(H4) The function f : [0, 1] × R2 → R is continuous and there exist real constants a0 > 0, a1 ≥ 0, a2 ≥ 0 such that |f (t, x, y)| ≤ a0 + a1 |x| + a2 |y|,

∀ t ∈ [0, 1], x, y ∈ R.

(H5) The function f : [0, 1] × R2 → R is continuous and there exist the constants b0 , b1 , b2 ≥ 0 with at least one nonzero, and l1 , l2 ∈ (0, 1) such that |f (t, x, y)| ≤ b0 + b1 |x|l1 + b2 |y|l2 ,

∀ t ∈ [0, 1], x, y ∈ R.

(H6) The function f : [0, 1] × R2 → R is continuous and there exist c0 , c1 , c2 ≥ 0 with at least one nonzero, and nondecreasing functions h1 , h2 ∈ C([0, ∞), [0, ∞)) such that |f (t, x, y)| ≤ c0 + c1 h1 (|x|) + c2 h2 (|y|),

∀ t ∈ [0, 1], x, y ∈ R.

We denote by M1 =

1 1 + Γ(α + 1) |Δ0 |Γ(α − β0 + 1)  1 m 1  1 α−βi , + s dH (s) i |Δ0 | i=1 Γ(α − βi + 1) 0

M2 = M1 −

1 , Γ(α + 1)

L0 = 1 +

1 . Γ(p + 1)

(6.5)

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We consider the Banach space X = C([0, 1]) with the supremum norm u = supt∈[0,1] |u(t)|, and define the operator A : X → X by (Au)(t) = −

1 Γ(α)



t

0

p (t − s)α−1 f (s, u(s), I0+ u(s)) ds

α−1

+ −

t Δ0 Γ(α − β0 )

1

0

p (1 − s)α−β0 −1 f (s, u(s), I0+ u(s)) ds

m tα−1  1 Δ0 i=1 Γ(α − βi )

 ×



0

1



0

s

α−βi −1

(s − τ )

(6.6) p f (τ, u(τ ), I0+ u(τ )) dτ

dHi (s),

t ∈ [0, 1]. By using Lemma 6.1.1, we see that u is a fixed point of operator A if and only if u is a solution of problem (6.1), (6.2). Therefore, next, we will investigate the existence of fixed points of operator A. Theorem 6.1.1. Assume that (H1) and (H2) hold. If Ξ := L1 L0 M1 < 1, then problem (6.1), (6.2) has a unique solution on [0, 1], where L0 and M1 are given by (6.5). Proof. Let us fix r > 0 such that r ≥ M0 M1 (1 − L1 L0 M1 )−1 , where M0 = supt∈[0,1] |f (t, 0, 0)|. We consider the set B r = {u ∈ X, uX ≤ r} and we show firstly that A(B r ) ⊂ B r . Let u ∈ B r . By p u(t)), we obtain the following using (H2) and Lemma 6.1.2, for f (t, u(t), I0+ inequalities p p u(t))| ≤ |f (t, u(t), I0+ u(t)) − f (t, 0, 0)| + |f (t, 0, 0)| |f (t, u(t), I0+ p ≤ L1 (|u(t)| + |I0+ u(t)|)  1 u + M0 + M0 ≤ L1 u + Γ(p + 1)  1 = L1 1 + u Γ(p + 1)

+ M0 ≤ L1 L0 r + M0 ,

∀ t ∈ [0, 1].

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Then by the definition of operator A from (6.6), we deduce  t 1 (t − s)α−1 (L1 L0 r + M0 ) ds |(Au)(t)| ≤ Γ(α) 0  1 tα−1 (1 − s)α−β0 −1 (L1 L0 r + M0 ) ds + |Δ0 |Γ(α − β0 ) 0 m

tα−1  1 |Δ0 | i=1 Γ(α − βi )  1  s α−βi −1 (s − τ ) (L1 L0 r + M0 ) dτ dHi (s) × +

0

0

tα−1 tα−1 tα + + Γ(α + 1) |Δ0 |Γ(α − β0 + 1) |Δ0 |   1 m  1 α−βi , ∀ t ∈ [0, 1]. s dH (s) × i Γ(α − βi + 1) 0 i=1

= (L1 L0 r + M0 )

Therefore, we conclude



1 1 + Γ(α + 1) |Δ0 |Γ(α − β0 + 1)  1  m 1 1  sα−βi dHi (s) + |Δ0 | i=1 Γ(α − βi + 1) 0

Au ≤ (L1 L0 r + M0 )

= (L1 L0 r + M0 )M1 ≤ r. So, we deduce that A maps B r into itself. Now, for u, v ∈ B r we have  t 1 p |(Au)(t) − (Av)(t)| ≤ − (t − s)α−1 f (s, u(s), I0+ u(s)) ds Γ(α) 0  t tα−1 1 p + (t − s)α−1 f (s, v(s), I0+ v(s)) ds + Γ(α) 0 |Δ0 |Γ(α − β0 )  1 p p (1 − s)α−β0 −1 |f (s, u(s), I0+ u(s)) − f (s, v(s), I0+ v(s))| ds × 0

 1  s m  1 t p (s − τ )α−βi −1 |f (τ, u(τ ), I0+ u(τ )) |Δ0 | i=1 Γ(α − βi ) 0 0 p − f (τ, v(τ ), I0+ v(τ ))| dτ dHi (s) α−1

+

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L1 ≤ Γ(α)



t

0

(t − s)α−1 [|u(s) − v(s)| tα−1 L1 |Δ0 |Γ(α − β0 )

p p u(s) − I0+ v(s)|] ds + + |I0+

 ×

+

1

0

p p (1 − s)α−β0 −1 [|u(s) − v(s)| + |I0+ u(s) − I0+ v(s)|] ds

 1  s m 1 tα−1 L1  (s − τ )α−βi −1 [|u(τ ) − v(τ )| |Δ0 | i=1 Γ(α − βi ) 0 0

p p u(τ ) − I0+ v(τ )|] dτ + |I0+

≤

L1 Γ(α) +

 0

t



dHi (s)

 (t − s)α−1 u − v +

tα−1 L1 |Δ0 |Γ(α − β0 )



1 0

 1 u − v ds Γ(p + 1)

 (1 − s)α−β0 −1 u − v +

 1 u − v ds Γ(p + 1)

 1  s L1 1 + (s − τ )α−βi −1 |Δ0 | i=1 Γ(α − βi ) 0 0   1 u − v dτ dHi (s) × u − v + Γ(p + 1)  tα−1 tα + = L1 L0 u − v Γ(α + 1) |Δ0 |Γ(α − β0 + 1) t

α−1

m 

 1  m 1 tα−1  α−β i + s dHi (s) , |Δ0 | i=1 Γ(α − βi + 1) 0 Hence, we obtain



∀ t ∈ [0, 1].

1 1 + Γ(α + 1) |Δ0 |Γ(α − β0 + 1)  1  m 1  1 α−β i + s dHi (s) |Δ0 | i=1 Γ(α − βi + 1) 0

Au − Av ≤ L1 L0 u − v

= Ξu − v.

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By using the condition Ξ < 1, we deduce that operator A is a contraction. Then by Theorem 1.2.1, we conclude that operator A has a unique fixed point u ∈ B r , which is the unique solution of problem (6.1), (6.2) on [0, 1].  1 < 1, Theorem 6.1.2. Assume that (H1)–(H3) hold. If Ξ1 := L1 L0 Γ(α+1) then problem (6.1), (6.2) has at least one solution on [0, 1].

Proof. Let us fix r1 > 0 such that r1 ≥ M1 g. We consider the set B r1 = {u ∈ X, u ≤ r1 }, and we introduce the operators A1 , A2 : B r1 → X defined by  t 1 p (A1 u)(t) = − (t − s)α−1 f (s, u(s), I0+ u(s)) ds, t ∈ [0, 1], Γ(α) 0  1 tα−1 p (A2 u)(t) = (1 − s)α−β0 −1 f (s, u(s), I0+ u(s)) ds Δ0 Γ(α − β0 ) 0 m

1 tα−1  Δ0 i=1 Γ(α − βi )  1  s p α−βi −1 × (s − τ ) f (τ, u(τ ), I0+ u(τ )) dτ dHi (s), −

0

0

(6.7) for all t ∈ [0, 1] and u ∈ B r1 . By using (H3), we obtain for all u, v ∈ B r1 A1 u + A2 v ≤ A1 u + A2 v ≤ 

1 g Γ(α + 1)

m 1 1  1 + |Δ0 |Γ(α − β0 + 1) |Δ0 | i=1 Γ(α − βi + 1)  1  α−β i × s dHi (s) g = M1 g ≤ r1 .

+

0

Hence, A1 u + A2 v ∈ B r1 for all u, v ∈ B r1 . The operator A1 is a contraction, because A1 u − A1 v ≤ L1 L0 and Ξ1 < 1.

1 u − v = Ξ1 u − v, Γ(α + 1)

∀ u, v ∈ B r1 ,

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The continuity of f implies that the operator A2 is continuous on B r1 . We prove next that A2 is compact. The operator A2 is uniformly bounded on B r1 , because  1 A2 u ≤ |Δ0 |Γ(α − β0 + 1)  1  m 1 1  α−β i s dHi (s) g = M2 g, + |Δ0 | i=1 Γ(α − βi + 1) 0 for all u ∈ B r1 . Now, we prove that A2 is equicontinuous on B r1 . We denote by   1 r1 . Λr1 = sup |f (t, x, y)|, t ∈ [0, 1], |x| ≤ r1 , |y| ≤ Γ(p + 1) (6.8) Then for u ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t2 , we obtain  ) 1 (tα−1 − tα−1 1 (1 − s)α−β0 −1 Λr1 ds |(A2 u)(t2 ) − (A2 u)(t1 )| ≤ 2 |Δ0 |Γ(α − β0 ) 0 m

− tα−1 ) 1 (tα−1 2 1 |Δ0 | Γ(α − βi ) i=1  1  s (s − τ )α−βi −1 Λr1 dτ dHi (s) × +

0

≤

0



1 |Δ0 |Γ(α − β0 + 1)  1  m 1  1 α−β i s + dHi (s) |Δ0 | i=1 Γ(α − βi + 1) 0

Λr1 (tα−1 2

−

tα−1 ) 1

− tα−1 ). = Λr1 M2 (tα−1 2 1 Therefore, we conclude |(A2 u)(t2 ) − (A2 u)(t1 )| → 0, as t2 → t1 , uniformly with respect to u ∈ B r1 . We deduce that A2 is equicontinuous on B r1 , and so, by using the Arzela-Ascoli theorem, the set A2 (B r1 ) is relatively compact. We conclude that operator A2 is compact on B r1 . Thus, all assumptions of Theorem 1.2.3 are satisfied, and then by Theorem 1.2.3 we deduce that there exists a fixed point of operator A1 +A2 , which is a solution of the boundary value problem (6.1), (6.2) on [0, 1]. 

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Theorem 6.1.3. Assume that (H1)–(H3) hold. If Ξ2 := L1 L0 M2 < 1, then problem (6.1), (6.2) has at least one solution on [0, 1]. Proof. We consider again a positive number r1 ≥ M1 g and the operators A1 , A2 defined on B r1 given by (6.7). In a similar manner as in the proof of Theorem 6.1.2, we obtain that A1 u + A2 v ∈ B r1 for all u, v ∈ B r1 . The operator A2 is a contraction because  1 A2 u − A2 v ≤ L1 L0 |Δ0 |Γ(α − β0 + 1)  1  m 1  1 sα−βi dHi (s) + |Δ0 | i=1 Γ(α − βi + 1) 0 × u − v = L1 L0 M2 u − v = Ξ2 u − v, ∀ u, v ∈ B r1 , with Ξ2 < 1. Then the continuity of f implies that the operator A1 is continuous on B r1 . We prove now that A1 is compact. The operator A1 is uniformly bounded on B r1 because A1 u ≤

1 g, Γ(α + 1)

∀ u ∈ B r1 .

Now, we show that A1 is equicontinuous on B r1 . By using Λr1 (defined in the proof of Theorem 6.1.2), we obtain for u ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t 2 |(A1 u)(t2 ) − (A1 u)(t1 )|  t2 1 p = − (t2 − s)α−1 f (s, u(s), I0+ u(s)) ds Γ(α) 0  t1 1 p α−1 + (t1 − s) f (s, u(s), I0+ u(s)) ds Γ(α) 0  t1 1 p [(t2 − s)α−1 − (t1 − s)α−1 ]f (s, u(s), I0+ u(s)) ds = − Γ(α) 0  t2 1 p α−1 (t2 − s) f (s, u(s), I0+ u(s)) ds − Γ(α) t1

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Λ r1 ≤ Γ(α) +



t1

0

Λ r1 Γ(α)



[(t2 − s)α−1 − (t1 − s)α−1 ] ds t2

t1

(t2 − s)α−1 ds

Λ r1 Λ r1 α [−(t2 − t1 )α + tα (t2 − t1 )α = 2 − t1 ] + Γ(α + 1) Γ(α + 1) ≤

Λ r1 (tα − tα 1 ). Γ(α + 1) 2

Then we deduce |(A1 u)(t2 ) − (A1 u)(t1 )| → 0, as t2 → t1 , uniformly with respect to u ∈ B r1 . We conclude that A1 is equicontinuous on B r1 , and by using the ArzelaAscoli theorem, the set A1 (B r1 ) is relatively compact. We deduce that operator A1 is compact on B r1 . By Theorem 1.2.3, we obtain that there exists a fixed point of operator A1 + A2 , which is a solution of the boundary value problem (6.1), (6.2) on [0, 1].  Theorem 6.1.4. Assume that (H1) and (H4) hold. If Ξ3 := M1 (a1 + a2 Γ(p+1) ) < 1, then the boundary value problem (6.1), (6.2) has at least one solution on [0, 1]. Proof. We consider the operator A : X → X defined in (6.6). We firstly prove that A is completely continuous. By the continuity of f we deduce that A is a continuous operator. We show next that A is a compact operator. Let Ω ⊂ X be a bounded set. Then there exist a positive constant L2 such that p u(t))| ≤ L2 , |f (t, u(t), I0+

∀ t ∈ [0, 1], u ∈ Ω.

Therefore, we obtain as in the proof of Theorem 6.1.1 that |(Au)(t)| ≤ L2 M1 , for all t ∈ [0, 1] and u ∈ Ω. So, A(Ω) is uniformly bounded. We will show next that A(Ω) is equicontinuous. Let u ∈ Ω and t1 , t2 ∈ [0, 1] with t1 < t2 . Then by using the operators A1 and A2 defined on Ω (given by (6.7)), and based on a similar approach as that used in the

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proof of Theorems 6.1.2 and 6.1.3, we obtain |(Au)(t2 ) − (Au)(t1 )| = |(A1 u)(t2 ) − (A1 u)(t1 ) + (A2 u)(t2 ) − (A2 u)(t1 )| ≤ |(A1 u)(t2 ) − (A1 u)(t1 )| + |(A2 u)(t2 ) − (A1 u)(t1 )| L2 α−1 (tα − tα − tα−1 ). 1 ) + L2 M2 (t2 1 Γ(α + 1) 2 Then |(Au)(t2 ) − (Au)(t1 )| → 0 as t2 → t1 uniformly with respect to u ∈ Ω. Thus, A(Ω) is equicontinuous. By the Arzela-Ascoli theorem, we deduce that A(Ω) is relatively compact, and so A is compact. Therefore, A is completely continuous. Now, we will prove that the set F = {u ∈ X, u = νA(u), 0 < ν < 1} is bounded. Let u ∈ F , that is u = νA(u) for some ν ∈ (0, 1). Then we have ≤

|u(t)| = |ν(Au)(t)| ≤ |(Au)(t)|,

∀ t ∈ [0, 1].

By (H4), we obtain |u(t)| ≤ |(Au)(t)|  t 1 p (t − s)α−1 [a0 + a1 |u(s)| + a2 |I0+ u(s)|] ds ≤ Γ(α) 0  1 tα−1 p (t − s)α−β0 −1 [a0 + a1 |u(s)| + a2 |I0+ u(s)|] ds + |Δ0 |Γ(α − β0 ) 0  1  s m 1 tα−1  (s − τ )α−βi −1 [a0 + a1 |u(τ )| + |Δ0 | i=1 Γ(α − βi ) 0 0 p + a2 |I0+ u(τ )|] dτ dHi (s)   tα tα−1 a2 ≤ a0 + a1 u + u + Γ(p + 1) Γ(α + 1) |Δ0 |Γ(α − β0 + 1)   m 1 α−β tα−1  1 i + s dHi (s) . |Δ0 | i=1 Γ(α − βi + 1) 0 Therefore, we deduce u ≤ M1



a2 u . a0 + a1 u + Γ(p + 1)

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Because Ξ3 < 1, we obtain

 −1 M 1 a2 . u ≤ M1 a0 1 − M1 a1 − Γ(p + 1)

Hence, we deduce that the set F is bounded. By using Theorem 1.2.5, we conclude that the operator A has at least one fixed point, which is a solution for our problem (6.1), (6.2).  Theorem 6.1.5. Assume that (H1) and (H5) hold. Then the problem (6.1), (6.2) has at least one solution. Proof. Let B R = {u ∈ X, u ≤ R}, where 

1  1−l 2 1 3b M 2 1 . R ≥ max 3b0 M1 , (3b1 M1 ) 1−l1 , (Γ(p + 1))l2 We prove now that A : B R → B R . For u ∈ B R , we deduce  b2 R R R l1 l2 + + = R, R |(Au)(t)| ≤ b0 + b1 R + M1 ≤ (Γ(p + 1))l2 3 3 3 ∀ t ∈ [0, 1], and then Au ≤ R, which implies that A(B R ) ⊂ B R . From the continuity of the function f , we can easily show that the operator A is continuous. The functions from A(B R ) are uniformly bounded and equicontinuous. Indeed, by using the notation (6.8), with r1 replaced by R, we obtain for any u ∈ B R and t1 , t2 ∈ [0, 1], t1 < t2 that |(Au)(t2 ) − (Au)(t1 )| ≤

ΛR α−1 (tα − tα − tα−1 ). 1 ) + ΛR M2 (t2 1 Γ(α + 1) 2

Therefore, |(Au)(t2 ) − (Au)(t1 )| → 0 as t2 → t1 uniformly with respect to u ∈ B R . By the Arzela-Ascoli theorem, we conclude that A(B R ) is relatively compact, and then A is a completely continuous operator. By Theorem 1.2.4, we deduce that operator A has at least one fixed point u in B R which is a solution of our problem (6.1), (6.2).  Theorem 6.1.6. Assume that (H1) and (H6) hold. If there exists Ξ0 > 0 such that   Ξ0 (6.9) c0 + c1 h1 (Ξ0 ) + c2 h2 M1 < Ξ0 , Γ(p + 1) where c0 , c1 , c2 , h1 , h2 are given in (H6), then problem (6.1), (6.2) has at least one solution on [0, 1].

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Proof. We consider the set B Ξ0 = {u ∈ X, u ≤ Ξ0 }, where Ξ0 is given in the assumptions of the theorem. We will show that A : B Ξ0 → B Ξ0 . For u ∈ B Ξ0 and t ∈ [0, 1], we obtain   Ξ0 |(Au)(t)| ≤ c0 + c1 h1 (Ξ0 ) + c2 h2 M1 < Ξ0 . Γ(p + 1) Then A(B Ξ0 ) ⊂ B Ξ0 . In a similar manner as in the proof of Theorem 6.1.5 we can show that operator A is completely continuous. We suppose now that there exists u ∈ ∂BΞ0 such that u = νA(u) for some ν ∈ (0, 1). We obtain as above that u ≤ Au < Ξ0 , which is a contradiction, because u ∈ ∂BΞ0 . Then by Theorem 1.2.6, we conclude that operator A has a fixed point u ∈ B Ξ0 , and so problem (6.1), (6.2) has at least one solution.  6.1.3

Examples

6 1 3 2 Let α = 52 (n = 3), p = 10 3 , m = 2, β0 = 5 , β1 = 3 , β2 = 4 , H1 (t) = t for all t ∈ [0, 1], H2 (t) = {0, if t ∈ [0, 1/2); 3, if t ∈ [1/2, 1]}. We consider the fractional differential equation 5/2

10/3

D0+ u(t) + f (t, u(t), I0+ u(t)) = 0, with the boundary conditions 

u(0) = u (0) = 0,

6/5 D0+ u(1)

 =2 0

1

0 < t < 1,

1/3 tD0+ u(t) dt

+

3/4 3D0+ u

(6.10)  1 . 2 (6.11)

We obtain here Δ0 ≈ −1.87462428 = 0, L0 ≈ 1.1079852, M1 ≈ 1.16312084 and M2 ≈ 0.86221973. So, assumption (H1) is satisfied. Example 1. We consider the function t 3t |x| − arctan y − 2 , f (t, x, y) = 2 2(t + 1) (1 + |x|) 4 t +4 t ∈ [0, 1], x, y ∈ R. Here, we have |f (t, x, y) − f (t, x1 , y1 )| ≤

1 (|x − x1 | + |y − y1 |), 2 ∀ t ∈ [0, 1], x, y, x1 , y1 ∈ R,

so L1 = 12 , and then Ξ ≈ 0.64436034 < 1. Therefore, assumption (H2) is satisfied, and by Theorem 6.1.1, we deduce that problem (6.10), (6.11) has a unique solution u(t), t ∈ [0, 1].

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Example 2. We consider the function 1 1 |x| − sin2 y, f (t, x, y) = √ sin t + 3(2 + |x|) 2(1 + t) 4 + t2 t ∈ [0, 1], x, y ∈ R. In this case, we have |f (t, x, y) − f (t, x1 , y1 )| ≤ |x − x1 | + |y − y1 |, ∀ t ∈ [0, 1], x, y, x1 , y1 ∈ R, so L1 = 1, and |f (t, x, y)| ≤ g(t) for all t ∈ [0, 1] and x, y ∈ R, where | sin t| 1 1 g(t) = √ , + + 2 3 2(1 + t) 4+t

∀ t ∈ [0, 1].

Then assumptions (H2) and (H3) are satisfied, and, in addition, we obtain Ξ1 ≈ 0.33339398 < 1. Therefore, by Theorem 6.1.2, we conclude that problem (6.10), (6.11) has at least one solution on [0, 1]. Example 3. We consider the function  t 1 1 f (t, x, y) = 2 y, 4 cos t + sin x − t +1 2 (t + 1)3 ∀ t ∈ [0, 1], x, y ∈ R. Because we have 1 |f (t, x, y)| ≤ 2 + |x| + |y|, 4

∀ t ∈ [0, 1], x, y ∈ R,

the assumption (H4) is satisfied with a0 = 2, a1 = 14 and a2 = 1. In addition, we obtain Ξ3 ≈ 0.41638005 < 1, and then by Theorem 6.1.4 we deduce that problem (6.10), (6.11) has at least one solution on [0, 1]. Example 4. We consider the function f (t, x, y) =

1 1 e−t arctan y 1/5 , − x2/3 + 1 + t3 3 4(3 + t)

t ∈ [0, 1], x, y ∈ R.

Because we obtain 1 1 |f (t, x, y)| ≤ 1 + |x|2/3 + |y|1/5 , 3 12

∀ t ∈ [0, 1], x, y ∈ R,

1 then assumption (H5) is satisfied with b0 = 1, b1 = 13 , b2 = 12 , l1 = 23 , l2 = 15 . Then by Theorem 6.1.5, we deduce that problem (6.10), (6.11) has at least one solution on [0, 1].

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Example 5. We consider the function (1 − t)x2 t4 y 3 (1 − t)2 + − , 2 10 15(1 + x ) 5

f (t, x, y) =

∀ t ∈ [0, 1], x, y ∈ R.

Because we have 1 1 1 + |x|2 + |y|3 , t ∈ [0, 1], x, y ∈ R, 10 15 5 then assumption (H6) is satisfied with h1 (x) = x2 and h2 (x) = x3 for 1 1 , c1 = 15 and c2 = 15 . For Ξ0  = 2, the condition (6.9) x ∈ [0, ∞), c0 = 10 |f (t, x, y)| ≤

is satisfied, because

c0 + c1 h1 (2) + c2 h2

2 Γ(p+1)

M1 ≈ 0.428821 < 2.

Therefore, by Theorem 6.1.6, we conclude that problem (6.10), (6.11) has at least one solution on [0, 1]. Remark 6.1.1. The results presented in this section were published in [82]. 6.2

Systems of Riemann–Liouville Fractional Differential Equations with Uncoupled Boundary Conditions

We consider the nonlinear system of fractional differential equations

θ1 σ1 α u(t) + f (t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, t ∈ (0, 1), D0+ β θ2 σ2 v(t) + g(t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, D0+

t ∈ (0, 1),

(6.12)

with the uncoupled nonlocal boundary conditions ⎧ u(0) = u (0) = · · · = u(n−2) (0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p  1 ⎪  ⎪ ⎪ γ0 γi ⎪ D0+ u(1) = D0+ u(t) dHi (t), ⎪ ⎨ i=1

0

i=1

0

⎪ v(0) = v  (0) = · · · = v (m−2) (0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ q  1 ⎪  ⎪ ⎪ δ0 δi ⎪ ⎪ D0+ v(t) dKi (t), D0+ v(1) = ⎩

(6.13)

where α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n ≥ 2, m ≥ 2, θ1 , θ2 , σ1 , σ2 > 0, p, q ∈ N, γi ∈ R for all i = 0, . . . , p, 0 ≤ γ1 < γ2 < · · · < γp < α − 1, γ0 ∈ [0, α − 1), δi ∈ R for all i = 0, . . . , q, 0 ≤ δ1 < δ2 < · · · < k δq < β − 1, δ0 ∈ [0, β − 1), D0+ denotes the Riemann–Liouville derivative ζ is of order k (for k = α, β, γ0 , γi , i = 1, . . . , p, δ0 , δi , i = 1, . . . , q), I0+ the Riemann–Liouville integral of order ζ (for ζ = θ1 , σ1 , θ2 , σ2 ), f and g are nonlinear functions, and the integrals from the boundary conditions

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(6.13) are Riemann–Stieltjes integrals with Hi for i = 1, . . . , p and Ki for i = 1, . . . , q functions of bounded variation. Based on some theorems from the fixed point theory, we give conditions for the nonlinearities f and g such that problem (6.12), (6.13) has at least one solution (u, v) ∈ (C[0, 1])2 . 6.2.1

Auxiliary results

We consider the fractional differential equation α D0+ u(t) + h(t) = 0,

t ∈ (0, 1),

with the boundary conditions ⎧  (n−2) (0) = 0, ⎪ ⎨ u(0) = u (0) = · · · = u  p 1  γ0 γi ⎪ D0+ u(t) dHi (t), ⎩ D0+ u(1) = i=1

(6.14)

(6.15)

0

where h ∈ C(0, 1) ∩ L1 (0, 1). We denote by p

 Γ(α) Γ(α) − Δ1 = Γ(α − γ0 ) i=1 Γ(α − γi )

 0

1

sα−γi −1 dHi (s).

In a similar manner as we proved Lemma 2.4.1 from Section 2.4, we obtain the following lemma. Lemma 6.2.1. If Δ1 = 0, then the unique solution u ∈ C[0, 1] of problem (6.14), (6.15) is given by  t 1 (t − s)α−1 h(s) ds u(t) = − Γ(α) 0  1 tα−1 (1 − s)α−γ0 −1 h(s) ds + Δ1 Γ(α − γ0 ) 0  1  s p tα−1  1 − (s − τ )α−γi −1 h(τ ) dτ dHi (s), Δ1 i=1 Γ(α − γi ) 0 0 t ∈ [0, 1].

(6.16)

We also consider the fractional differential equation β D0+ v(t) + k(t) = 0,

t ∈ (0, 1),

(6.17)

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with the boundary conditions ⎧  (m−2) (0) = 0, ⎪ ⎨ v(0) = v (0) = · · · = v  q  1 δ0 δi ⎪ D0+ v(t) dKi (t), ⎩ D0+ v(1) = i=1

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(6.18)

0

where k ∈ C(0, 1) ∩ L1 (0, 1). We denote by q

Δ2 =

 Γ(β) Γ(β) − Γ(β − δ0 ) i=1 Γ(β − δi )



1 0

sβ−δi −1 dKi (s).

Lemma 6.2.2. If Δ2 = 0, then the unique solution v ∈ C[0, 1] of problem (6.17), (6.18) is given by  t 1 (t − s)β−1 k(s) ds v(t) = − Γ(β) 0  1 tβ−1 (1 − s)β−δ0 −1 k(s) ds + Δ2 Γ(β − δ0 ) 0  1  s q 1 tβ−1  (s − τ )β−δi −1 k(τ ) dτ dKi (s), − Δ2 i=1 Γ(β − δi ) 0 0 t ∈ [0, 1].

(6.19)

We denote by (I1) the following basic assumptions for problem (6.12), (6.13) that will be used in the main theorems. (I1) α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n ≥ 2, m ≥ 2, θ1 , θ2 , σ1 , σ2 > 0, p, q ∈ N, γi ∈ R for all i = 0, . . . , p, 0 ≤ γ1 < γ2 < · · · < γp < α − 1, γ0 ∈ [0, α − 1), δi ∈ R for all i = 0, . . . , q, 0 ≤ δ1 < δ2 < · · · < δq < β − 1, δ0 ∈ [0, β − 1), Hi : [0, 1] → R, i = 1, . . . , p and Kj : [0, 1] → R, j = 1, . . . , q are functions of bounded variation, Δ1 = 0, Δ2 = 0. We introduce the following constants: M1 = 1 +

1 , Γ(θ1 + 1)

M2 = 1 +

1 , Γ(σ1 + 1)

M3 = 1 +

1 , Γ(θ2 + 1)

M4 = 1 +

1 , Γ(σ2 + 1)

M5 = max{M1 , M2 },

M6 = max{M3 , M4 },

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M7 =

1 1 + Γ(α + 1) |Δ1 |Γ(α − γ0 + 1)  1 p 1  1 α−γi + s dHi (s) , |Δ1 | Γ(α − γi + 1)

M9 =

(6.20)

0

i=1

1 1 + Γ(β + 1) |Δ2 |Γ(β − δ0 + 1)  1 q 1  1 β−δi , s dK (s) + i |Δ2 | i=1 Γ(β − δi + 1) 0

M8 = M7 −

1 , Γ(α + 1)

M10 = M9 −

1 . Γ(β + 1)

We consider the Banach space X = C[0, 1] with supremum norm u = supt∈[0,1] |u(t)|, and the Banach space Y = X × X with the norm (u, v)Y = u + v. We introduce the operator A : Y → Y defined by A(u, v) = (A1 (u, v), A2 (u, v)) for (u, v) ∈ Y , where the operators A1 , A2 : Y → X are given by  t 1 θ1 σ1 A1 (u, v)(t) = − (t − s)α−1 f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds Γ(α) 0  1 tα−1 + (1 − s)α−γ0 −1 Δ1 Γ(α − γ0 ) 0 θ1 σ1 × f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds  1 s p α−1  1 t − (s − τ )α−γi −1 Δ1 i=1 Γ(α − γi ) 0 0 θ1 σ1 × f (τ, u(τ ), v(τ ), I0+ u(τ ), I0+ v(τ )) dτ dHi (s),

1 A2 (u, v)(t) = − Γ(β)

 0

t

θ2 σ2 (t − s)β−1 g(s, u(s), v(s), I0+ u(s), I0+ v(s)) ds

tβ−1 + Δ2 Γ(β − δ0 )



1

(1 − s)β−δ0 −1

0 θ2 σ2 ×g(s, u(s), v(s), I0+ u(s), I0+ v(s)) ds   q 1 s 1 tβ−1  β−δi −1

−

Δ2

i=1

Γ(β − δi )

0

0

(s − τ )

θ2 σ2 × g(τ, u(τ ), v(τ ), I0+ u(τ ), I0+ v(τ )) dτ dKi (s), ∀ t ∈ [0, 1], (u, v) ∈ Y.

(6.21)

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By using Lemmas 6.2.1 and 6.2.2, we note that (u, v) is a solution of problem (6.12), (6.13) if and only if (u, v) is a fixed point of operator A. 6.2.2

Existence of solutions

In this section, we present some conditions for the nonlinearities f and g such that operator A has at least one fixed point, which is a solution of problem (6.12), (6.13). Theorem 6.2.1. Assume that (I1) and (I2) The functions f, g : [0, 1] × R4 → R are continuous and there exist L1 , L2 > 0 such that |f (t, x1 , x2 , x3 , x4 ) − f (t, x 1 , x 2 , x 3 , x 4 )| ≤ L1

4 

|xi − x i |,

i=1

|g(t, y1 , y2 , y3 , y4 ) − g(t, y1 , y2 , y3 , y4 )| ≤ L2

4 

|yi − yi |,

i=1

i , yi ∈ R, i = 1, . . . , 4, for all t ∈ [0, 1], xi , yi , x hold. If Ξ := L1 M5 M7 + L2 M6 M9 < 1, then problem (6.12), (6.13) has a unique solution (u(t), v(t)), t ∈ [0, 1], where M5 , M6 , M7 , M9 are given by (6.20). Proof. We consider the positive number r given by 0 M9 )(1 − L1 M5 M7 − L2 M6 M9 )−1 , r = (M0 M7 + M 0 = sup where M0 = supt∈[0,1] |f (t, 0, 0, 0, 0)|, M t∈[0,1] |g(t, 0, 0, 0, 0)|. We define the set B r = {(u, v) ∈ Y, (u, v)Y ≤ r} and first we show that A(B r ) ⊂ B r . Let (u, v) ∈ B r . By using (I2) and Lemma 6.1.2, for θ1 σ1 u(t), I0+ v(t)), we deduce the following inequalities: f (t, u(t), v(t), I0+ θ1 σ1 u(t), I0+ v(t))| |f (t, u(t), v(t), I0+ θ1 σ1 u(t), I0+ v(t)) − f (t, 0, 0, 0, 0)| ≤ |f (t, u(t), v(t), I0+ θ1 + |f (t, 0, 0, 0, 0)| ≤ L1 (|u(t)| + |v(t)| + |I0+ u(t)| σ1 + |I0+ v(t)|) + M0  ≤ L1 u + v +

v u + Γ(θ1 + 1) Γ(σ1 + 1)

+ M0

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= L1

 1+

1 Γ(θ1 + 1)



 u + 1 +

1 Γ(σ1 + 1)



v + M0

= L1 (M1 u + M2 v) + M0 ≤ L1 M5 (u, v)Y + M0 ≤ L1 M5 r + M0 ,

∀ t ∈ [0, 1].

In a similar manner, we have θ2 σ2 |g(t, u(t), v(t), I0+ u(t), I0+ v(t))| θ2 σ2 u(t), I0+ v(t)) − g(t, 0, 0, 0, 0)| ≤ |g(t, u(t), v(t), I0+ θ2 + |g(t, 0, 0, 0, 0)| ≤ L2 (|u(t)| + |v(t)| + |I0+ u(t)| σ2 0 + |I0+ v(t)|) + M  ≤ L2 u + v +

v u 0 + +M Γ(θ2 + 1) Γ(σ2 + 1)   1 1 0 = L2 1+ u + 1 + v + M Γ(θ2 + 1) Γ(σ2 + 1) 0 = L2 (M3 u + M4 v) + M 0 ≤ L2 M6 r + M 0 , ≤ L2 M6 (u, v)Y + M

∀ t ∈ [0, 1].

Then by (6.21) (the definition of operators A1 and A2 ), we obtain  t 1 |A1 (u, v)(t)| ≤ (t − s)α−1 (L1 M5 r + M0 ) ds Γ(α) 0  1 tα−1 (1 − s)α−γ0 −1 (L1 M5 r + M0 ) ds + |Δ1 |Γ(α − γ0 ) 0 p

tα−1  1 |Δ1 | i=1 Γ(α − γi )  1  s × (s − τ )α−γi −1 (L1 M5 r + M0 ) dτ dHi (s) +

0

0



tα−1 tα + Γ(α + 1) |Δ1 |Γ(α − γ0 + 1)  1  p tα−1  1 α−γi s dHi (s) , + |Δ1 | i=1 Γ(α − γi + 1) 0

= (L1 M5 r + M0 )

∀ t ∈ [0, 1].

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Therefore, we conclude



1 1 + Γ(α + 1) |Δ1 |Γ(α − γ0 + 1)  1  p 1 1  sα−γi dHi (s) + |Δ1 | i=1 Γ(α − γi + 1) 0

A1 (u, v) ≤ (L1 M5 r + M0 )

= (L1 M5 r + M0 )M7 .

(6.22)

Arguing as before, we find  t 1 0 ) ds (t − s)β−1 (L2 M6 r + M |A2 (u, v)(t)| ≤ Γ(β) 0  1 tβ−1 0 ) ds + (1 − s)β−δ0 −1 (L2 M6 r + M |Δ2 |Γ(β − δ0 ) 0 q

1 tβ−1  |Δ2 | i=1 Γ(β − δi )  1  s β−δi −1  (s − τ ) (L2 M6 r + M0 ) dτ dKi (s) × +

0

0



tβ−1 tβ + Γ(β + 1) |Δ2 |Γ(β − δ0 + 1)  1  q tβ−1  1 β−δ + s i dKi (s) , |Δ2 | i=1 Γ(β − δi + 1) 0

0 ) = (L2 M6 r + M

∀ t ∈ [0, 1]. Then we have



1 1 + Γ(β + 1) |Δ2 |Γ(β − δ0 + 1)  1  q 1  1 β−δi + s dKi (s) |Δ2 | i=1 Γ(β − δi + 1) 0

0 ) A2 (u, v) ≤ (L2 M6 r + M

0 )M9 . = (L2 M6 r + M By relations (6.22) and (6.23), we deduce A(u, v)Y = A1 (u, v) + A2 (u, v) 0 )M9 = r, ≤ (L1 M5 r + M0 )M7 + (L2 M6 r + M for all (u, v) ∈ B r , which implies that A(B r ) ⊂ B r .

(6.23)

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Next, we prove that operator A is a contraction. For (ui , vi ) ∈ B r , i = 1, 2, and for each t ∈ [0, 1], we obtain |A1 (u1 , v1 )(t) − A1 (u2 , v2 )(t)|  t  1 θ1 σ1 ≤ − (t − s)α−1 f (s, u1 (s), v1 (s), I0+ u1 (s), I0+ v1 (s)) Γ(α) 0  θ1 σ1 u2 (s), I0+ v2 (s)) ds − f (s, u2 (s), v2 (s), I0+ +

tα−1 |Δ1 |Γ(α − γ0 )



1

0

(1 − s)α−γ0 −1

θ1 σ1 ×|f (s, u1 (s), v1 (s), I0+ u1 (s), I0+ v1 (s)) θ1 σ1 u2 (s), I0+ v2 (s))| ds − f (s, u2 (s), v2 (s), I0+  1  s p tα−1  1 + (s − τ )α−γi −1 |Δ1 | i=1 Γ(α − γi ) 0 0 θ1 σ1 ×|f (τ, u1 (τ ), v1 (τ ), I0+ u1 (τ ), I0+ v1 (τ ))

− ≤

θ1 σ1 u2 (τ ), I0+ v2 (τ ))| dτ f (τ, u2 (τ ), v2 (τ ), I0+

L1 Γ(α)

 0

t



dHi (s)

 (t − s)α−1 |u1 (s) − u2 (s)| + |v1 (s) − v2 (s)|

 θ1 θ1 σ1 σ1 + |I0+ u1 (s) − I0+ u2 (s)| + |I0+ v1 (s) − I0+ v2 (s)| ds +

tα−1 L1 |Δ1 |Γ(α − γ0 )

 0

1

(1 − s)α−γ0 −1 [|u1 (s) − u2 (s)|

θ1 θ1 u1 (s) − I0+ u2 (s)| + |v1 (s) − v2 (s)| + |I0+  σ1 σ1 + |I0+ v1 (s) − I0+ v2 (s)| ds p

1 tα−1 L1  |Δ1 | i=1 Γ(α − γi )  1  s  (s − τ )α−γi −1 |u1 (τ ) − u2 (τ )| + |v1 (τ ) − v2 (τ )| ×

+

0

+

0

θ1 |I0+ u1 (τ )

−

θ1 I0+ u2 (τ )|

+

σ1 |I0+ v1 (τ )



−

σ1 I0+ v2 (τ )|

dτ

dHi (s)

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≤

L1 Γ(α)

 0

t

 (t − s)α−1 u1 − u2  + v1 − v2 

 1 1 u1 − u2  + v1 − v2  ds + Γ(θ1 + 1) Γ(σ1 + 1)   1 α−1 t L1 + (1 − s)α−γ0 −1 u1 − u2  + v1 − v2  |Δ1 |Γ(α − γ0 ) 0  1 1 + u1 − u2  + v1 − v2  ds Γ(θ1 + 1) Γ(σ1 + 1) p

tα−1 L1  1 |Δ1 | i=1 Γ(α − γi )  1  s (s − τ )α−γi −1 [u1 − u2  + v1 − v2  ×

+

0

0

 1 1 u1 − u2  + v1 − v2  dτ dHi (s) + Γ(θ1 + 1) Γ(σ1 + 1)  L1 M5 t ≤ (t − s)α−1 (u1 − u2  + v1 − v2 ) ds Γ(α) 0  1 tα−1 L1 M5 + (1 − s)α−γ0 −1 (u1 − u2  + v1 − v2 ) ds |Δ1 |Γ(α − γ0 ) 0  1  s p 1 tα−1 L1 M5  α−γi −1 (s − τ ) dτ dHi (s) + |Δ1 | Γ(α − γi ) 0 0 i=1 ×(u1 − u2  + v1 − v2 )  tα tα−1 + = L1 M5 Γ(α + 1) |Δ1 |Γ(α − γ0 + 1)  1  p tα−1  1 α−γ i s dHi (s) + |Δ1 | i=1 Γ(α − γi + 1) 0 × (u1 − u2  + v1 − v2 ),

∀ t ∈ [0, 1].

Then we conclude A1 (u1 , v1 ) − A1 (u2 , v2 ) ≤ L1 M5 M7 (u1 − u2  + v1 − v2 ). (6.24)

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By similar computation, we also find A2 (u1 , v1 ) − A2 (u2 , v2 ) ≤ L2 M6 M9 (u1 − u2  + v1 − v2 ). (6.25) Therefore, by (6.24) and (6.25), we obtain A(u1 , v1 ) − A(u2 , v2 )Y = A1 (u1 , v1 ) − A1 (u2 , v2 ) + A2 (u1 , v1 ) − A2 (u2 , v2 ) ≤ (L1 M5 M7 + L2 M6 M9 )(u1 − u2  + v1 − v2 ) = Ξ(u1 , v1 ) − (u2 , v2 )Y . By using the condition Ξ < 1, we deduce that operator A is a contraction. By Theorem 1.2.1, we conclude that operator A has a unique fixed point (u, v) ∈ B r , which is the unique solution of problem (6.12), (6.13) on [0, 1].  Theorem 6.2.2. Assume that (I1) and (I3) The functions f, g : [0, 1] × R4 → R are continuous and there exist real constants ci , di ≥ 0, i = 0, . . . , 4, and at least one of c0 and d0 is positive, such that |f (t, x1 , x2 , x3 , x4 )| ≤ c0 +

4 

ci |xi |,

i=1

|g(t, y1 , y2 , y3 , y4 )| ≤ d0 +

4 

di |yi |,

i=1

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, c3 Γ(θ1 +1) )M7 + (d1 + d4 Γ(σ2 +1) )M9 , then the

hold. If Ξ1 := max{M11 , M12 } < 1, where M11 = (c1 + d3 Γ(θ2 +1) )M9

and M12 = (c2 + Γ(σc14+1) )M7 + (d2 + boundary value problem (6.12), (6.13) has at least one solution (u(t), v(t)), t ∈ [0, 1].

Proof. We prove that operator A is completely continuous. By the continuity of functions f and g, we obtain that operators A1 and A2 are continuous, and then A is a continuous operator. Next, we prove that A is a compact operator. Let Ω ⊂ Y be a bounded set. Then there exist positive constants

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L3 and L4 such that θ1 σ1 u(t), I0+ v(t))| ≤ L3 , |f (t, u(t), v(t), I0+ θ2 σ2 u(t), I0+ v(t))| ≤ L4 , |g(t, u(t), v(t), I0+

for all (u, v) ∈ Ω and t ∈ [0, 1]. Therefore, we deduce as in the proof of Theorem 6.2.1 that |A1 (u, v)(t)| ≤ L3 M7 ,

|A2 (u, v)(t)| ≤ L4 M9 ,

∀ t ∈ [0, 1], (u, v) ∈ Ω. So, we obtain A1 (u, v) ≤ L3 M7 ,

A2 (u, v) ≤ L4 M9 ,

A(u, v)Y ≤ L3 M7 + L4 M9 ,

∀(u, v) ∈ Ω,

and then A(Ω) is uniformly bounded. Next, we will prove that the functions from A(Ω) are equicontinuous. Let (u, v) ∈ Ω and t1 , t2 ∈ [0, 1] with t1 < t2 . Then we have |A1 (u, v)(t2 ) − A1 (u, v)(t1 )|  t2 1 θ1 σ1 ≤ − (t2 − s)α−1 f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds Γ(α) 0  t1 1 θ1 σ1 + (t1 − s)α−1 f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds Γ(α) 0  1 tα−1 − tα−1 1 (1 − s)α−γ0 −1 + 2 |Δ1 |Γ(α − γ0 ) 0 tα−1 − tα−1 θ1 σ1 1 × f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds + 2 |Δ1 |  1  s p  1 × (s − τ )α−γi −1 Γ(α − γ ) i 0 0 i=1 θ1 σ1 × f (τ, u(τ ), v(τ ), I0+ u(τ ), I0+ v(τ )) dτ dHi (s) L3 ≤ Γ(α)

 0

t1

α−1

[(t2 − s)

L3 (tα−1 − tα−1 ) 2 1 + |Δ1 |Γ(α − γ0 )

 0

α−1

− (t1 − s) 1

L3 ]ds + Γ(α)

(1 − s)α−γ0 −1 ds +



t2 t1

(t2 − s)α−1 ds

L3 (tα−1 − tα−1 ) 2 1 |Δ1 |

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×

p  i=1

=

 1  s 1 α−γi −1 (s − τ ) dτ dHi (s) Γ(α − γi ) 0 0

L3 α−1 (tα − tα − tα−1 ). 1 ) + L3 M8 (t2 1 Γ(α + 1) 2

Then |A1 (u, v)(t2 ) − A1 (u, v)(t1 )| → 0, as t2 → t1 , uniformly with respect to (u, v) ∈ Ω. In a similar manner, we find |A2 (u, v)(t2 ) − A2 (u, v)(t1 )| ≤

L4 − tβ−1 ), (tβ − tβ1 ) + L4 M10 (tβ−1 2 1 Γ(β + 1) 2

and so |A2 (u, v)(t2 ) − A2 (u, v)(t1 )| → 0, as t2 → t1 , uniformly with respect to (u, v) ∈ Ω. Thus, A1 (Ω) and A2 (Ω) are equicontinuous, and then A(Ω) is also equicontinuous. Hence, by the Arzela-Ascoli theorem, we conclude that A(Ω) is relatively compact, and then A is compact. Therefore, we deduce that A is completely continuous. We will show next that the set V = {(u, v) ∈ Y, (u, v) = νA(u, v), 0 < ν < 1} is bounded. Let (u, v) ∈ V , that is, (u, v) = νA(u, v) for some ν ∈ (0, 1). Then for any t ∈ [0, 1], we have u(t) = νA1 (u, v)(t), v(t) = νA2 (u, v)(t). Hence, we find |u(t)| ≤ |A1 (u, v)(t)| and |v(t)| ≤ |A2 (u, v)(t)| for all t ∈ [0, 1]. By (I3), we obtain |u(t)| ≤ |A1 (u, v)(t)|  t  1 ≤ (t − s)α−1 c0 + c1 |u(s)| + c2 |v(s)| Γ(α) 0  θ1 σ1 + c3 |I0+ u(s)| + c4 |I0+ v(s)| ds +

tα−1 |Δ1 |Γ(α − γ0 )

θ1 + c3 |I0+ u(s)|

 0

1

 (1 − s)α−γ0 −1 c0 + c1 |u(s)| + c2 |v(s)|

 σ1 + c4 |I0+ v(s)| ds

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 1  s p 1 tα−1  + (s − τ )α−γi −1 [c0 + c1 |u(τ )| |Δ1 | i=1 Γ(α − γi ) 0 0  θ1 σ1 +c2 |v(τ )| + c3 |I0+ u(τ )| + c4 |I0+ v(τ )| dτ dHi (s)  ≤ c0 + c1 u + c2 v + 

c4 c3 u + v Γ(θ1 + 1) Γ(σ1 + 1)



tα tα−1 + Γ(α + 1) |Δ1 |Γ(α − γ0 + 1)  1  p 1 tα−1  α−γi s dHi (s) . + |Δ1 | i=1 Γ(α − γi + 1) 0

×

Then we deduce   u ≤ c0 + c1 +

c3 Γ(θ1 + 1)



 u + c2 +

c4 Γ(σ1 + 1)



 v M7 .

In a similar manner, we have     d3 d4 v ≤ d0 + d1 + u + d2 + v M9 , Γ(θ2 + 1) Γ(σ2 + 1) and therefore (u, v)Y ≤ c0 M7 + d0 M9 + M11 u + M12 v ≤ c0 M7 + d0 M9 + Ξ1 (u, v)Y . Because Ξ1 < 1, we obtain (u, v)Y ≤ (c0 M7 + d0 M8 )(1 − Ξ1 )−1 ,

∀ (u, v) ∈ V.

So, we conclude that the set V is bounded. By Theorem 1.2.5, we deduce that operator A has at least one fixed point, which is a solution for our problem (6.12), (6.13).  Theorem 6.2.3. Assume that (I1), (I2) and (I4) There exist the functions φ1 , φ2 ∈ C([0, 1], [0, ∞)) such that |f (t, x1 , x2 , x3 , x4 )| ≤ φ1 (t),

|g(t, x1 , x2 , x3 , x4 )| ≤ φ2 (t),

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, 1 1 hold. If Ξ2 := L1 M5 Γ(α+1) + L2 M6 Γ(β+1) < 1, then problem (6.12), (6.13) has at least one solution on [0, 1].

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Proof. We fix r1 > 0 such that r1 ≥ M7 φ1 +M9 φ2 . We consider the set B r1 = {(u, v) ∈ Y, (u, v)Y ≤ r1 }, and we introduce the operators D = (D1 , D2 ) : B r1 → Y and E = (E1 , E2 ) : B r1 → Y , where D1 , D2 , E1 , E2 : B r1 → X are defined by D1 (u, v)(t) = − E1 (u, v)(t) =

1 Γ(α)

 0

t

θ1 σ1 (t − s)α−1 f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds,

α−1

t Δ1 Γ(α − γ0 )



1 0

(1 − s)α−γ0 −1

θ1 σ1 × f (s, u(s), v(s), I0+ u(s), I0+ v(s)) ds   p 1 s tα−1  1 − (s − τ )α−γi −1 Δ1 i=1 Γ(α − γi ) 0 0 θ1 σ1 × f (τ, u(τ ), v(τ ), I0+ u(τ ), I0+ v(τ )) dτ dHi (s),

1 D2 (u, v)(t) = − Γ(β) E2 (u, v)(t) =

 0

t

(6.26) β−1

(t − s)

θ2 σ2 g(s, u(s), v(s), I0+ u(s), I0+ v(s)) ds,

tβ−1 Δ2 Γ(β − δ0 )  1 θ2 σ2 × (1 − s)β−δ0 −1 g(s, u(s), v(s), I0+ u(s), I0+ v(s)) ds 0

 1  s q  1 t (s − τ )β−δi −1 Δ2 i=1 Γ(β − δi ) 0 0 θ2 σ2 × g(τ, u(τ ), v(τ ), I0+ u(τ ), I0+ v(τ )) dτ dKi (s),

−

β−1

for t ∈ [0, 1] and (u, v) ∈ B r1 . So A1 = D1 + E1 , A2 = D2 + E2 and A = D + E. By using (I4), we obtain for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 that D(u1 , v1 ) + E(u2 , v2 )Y ≤ D(u1 , v1 )Y + E(u2 , v2 )Y = D1 (u1 , v1 ) + D2 (u1 , v1 ) + E1 (u2 , v2 ) + E2 (u2 , v2 ) ≤

1 1 φ1  + φ2  Γ(α + 1) Γ(β + 1)

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307

p

1  1 1 + |Δ1 |Γ(α − γ0 + 1) |Δ1 | i=1 Γ(α − γi + 1)   1 α−γ i s dHi (s) φ1  ×

+



0

q

1 1  1 + + |Δ2 |Γ(β − δ0 + 1) |Δ2 | i=1 Γ(β − δi + 1)  1 sβ−δi dKi (s) φ2  × 0

= M7 φ1  + M9 φ2  ≤ r1 . Hence, D(u1 , v1 ) + E(u2 , v2 ) ∈ B r1 for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 . The operator D is a contraction, because D(u1 , v1 ) − D(u2 , v2 )Y = D1 (u1 , v1 ) − D1 (u2 , v2 ) + D2 (u1 , v1 ) − D2 (u2 , v2 )  1 1 ≤ L1 M5 + L2 M6 (u1 − u2  + v1 − v2 ) Γ(α + 1) Γ(β + 1) = Ξ2 (u1 , v1 ) − (u2 , v2 )Y , for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 , and Ξ2 < 1. The continuity of f and g implies that operator E is continuous on B r1 . We prove in what follows that E is compact. The functions from E(B r1 ) are uniformly bounded, because E(u, v)Y = E1 (u, v) + E2 (u, v)  p 1  1 1 + ≤ |Δ1 |Γ(α − γ0 + 1) |Δ1 | i=1 Γ(α − γi + 1)  1 sα−γi dHi (s) φ1  × 

0

q

1 1  1 + + |Δ2 |Γ(β − δ0 + 1) |Δ2 | i=1 Γ(β − δi + 1)  1 sβ−δi dKi (s) φ2  × 0

= M8 φ1  + M10 φ2 ,

∀ (u, v) ∈ B r1 .

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We prove now that the functions from E(B r1 ) are equicontinuous. We denote by  Ψr1 = sup |f (t, u, v, x, y)|, t ∈ [0, 1], |u| ≤ r1 , |v| ≤ r1 ,  r1 r1 , |y| ≤ |x| ≤ , Γ(θ1 + 1) Γ(σ1 + 1)

Θ r1

 = sup |g(t, u, v, x, y)|, t ∈ [0, 1], |u| ≤ r1 , |v| ≤ r1 , |x| ≤

(6.27)

 r1 r1 , |y| ≤ . Γ(θ2 + 1) Γ(σ2 + 1)

Then for (u, v) ∈ B r1 , and t1 , t2 ∈ [0, 1] with t1 < t2 , we obtain |E1 (u, v)(t2 ) − E1 (u, v)(t1 )|  1 tα−1 − tα−1 1 ≤ 2 (1 − s)α−γ0 −1 Ψr1 ds |Δ1 |Γ(α − γ0 ) 0 p

 1 tα−1 − tα−1 1 + 2 |Δ1 | Γ(α − γi ) i=1  1  s (s − τ )α−γi −1 Ψr1 dτ dHi (s) × 0

≤

0



p

1  1 1 + |Δ1 |Γ(α − γ0 + 1) |Δ1 | i=1 Γ(α − γi + 1)   1 α−γ i s dHi (s) × Ψr1 (tα−1 2

−

tα−1 ) 1

0

= Ψr1 M8 (tα−1 − tα−1 ), 2 1 |E2 (u, v)(t2 ) − E2 (u, v)(t1 )|  1 − tβ−1 tβ−1 2 1 (1 − s)β−δ0 −1 Θr1 ds ≤ |Δ2 |Γ(β − δ0 ) 0 q  − tβ−1 1 tβ−1 2 1 |Δ2 | Γ(β − δi ) i=1  1  s β−δi −1 (s − τ ) Θr1 dτ dKi (s) ×

+

0

0

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≤

page 309

309

q

1  1 1 + |Δ2 |Γ(β − δ0 + 1) |Δ2 | i=1 Γ(β − δi + 1)  1  β−δi × s dKi (s) Θr1 (tβ−1 2

−

tβ−1 ) 1

0

= Θr1 M10 (tβ−1 − tβ−1 ). 2 1 Hence, we find |E1 (u, v)(t2 ) − E1 (u, v)(t1 )| → 0,

|E2 (u, v)(t2 ) − E2 (u, v)(t1 )| → 0,

as t2 → t1 uniformly with respect to (u, v) ∈ B r1 . So, E1 (B r1 ) and E2 (B r1 ) are equicontinuous, and then E(B r1 ) is also equicontinuous. By using the Arzela-Ascoli theorem, we deduce that the set E(B r1 ) is relatively compact. Therefore E is a compact operator on B r1 . By Theorem 1.2.3, we conclude that there exists a fixed point of operator D + E(= A), which is a solution of problem (6.12), (6.13).  Theorem 6.2.4. Assume that (I1), (I2) and (I4) hold. If Ξ3 := L1 M5 M8 + L2 M6 M10 < 1, then problem (6.12), (6.13) has at least one solution (u, v) on [0, 1]. Proof. We consider again a positive number r1 ≥ M7 φ1  + M9 φ2 , and the operators D and E defined on B r1 given by (6.26). As in the proof of Theorem 6.2.3, we obtain that D(u1 , v1 ) + E(u2 , v2 ) ∈ B r1 for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 . The operator E is a contraction, because E(u1 , v1 ) − E(u2 , v2 )Y = E1 (u1 , v1 ) − E1 (u2 , v2 ) + E2 (u1 , v1 ) − E2 (u2 , v2 ) ≤ L1 M5 M8 (u1 − u2  + v1 − v2 ) + L2 M6 M10 (u1 − u2  + v1 − v2 ) = (L1 M5 M8 + L2 M6 M10 )(u1 , v1 ) − (u2 , v2 )Y = Ξ3 (u1 , v1 ) − (u2 , v2 )Y , for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 , with Ξ3 < 1. Next, the continuity of f and g implies that operator D is continuous on B r1 . We show now that D is compact. The functions from D(B r1 ) are

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uniformly bounded, because D(u, v)Y = D1 (u, v) + D2 (u, v) ≤ +

1 φ2 , Γ(β + 1)

1 φ1  Γ(α + 1)

∀ (u, v) ∈ B r1 .

Now, we prove that D(B r1 ) is equicontinuous. By using Ψr1 and Θr1 defined in (6.27), we find for (u, v) ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t2 that |D1 (u, v)(t2 ) − D1 (u, v)(t1 )| ≤

Ψ r1 (tα − tα 1 ), Γ(α + 1) 2

|D2 (u, v)(t2 ) − D2 (u, v)(t1 )| ≤

Θ r1 (tβ − tβ1 ). Γ(β + 1) 2

Then we obtain |D1 (u, v)(t2 ) − D1 (u, v)(t1 )| → 0,

|D2 (u, v)(t2 ) − D2 (u, v)(t1 )| → 0,

as t2 → t1 uniformly with respect to (u, v) ∈ B r1 . We deduce that D1 (B r1 ) and D2 (B r1 ) are equicontinuous, and so D(B r1 ) is equicontinuous. By using the Arzela-Ascoli theorem, we conclude that the set D(B r1 ) is relatively compact. Then D is a compact operator on B r1 . By Theorem 1.2.3, we deduce that there exists a fixed point of operator D + E(= A), which is a solution of problem (6.12), (6.13).  Theorem 6.2.5. Assume that (I1) and (I5) The functions f, g : [0, 1] × R4 → R are continuous and there exist the constants ai ≥ 0, i = 0, . . . , 4 with at least one nonzero, the constants bi ≥ 0, i = 0, . . . , 4 with at least one nonzero, and li , mi ∈ (0, 1), i = 1, . . . , 4 such that |f (t, x1 , x2 , x3 , x4 )| ≤ a0 +

4 

ai |xi |li ,

i=1

|g(t, y1 , y2 , y3 , y4 )| ≤ b0 +

4 

bi |yi |mi ,

i=1

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, hold. Then problem (6.12), (6.13) has at least one solution.

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311

Proof. Let B R = {(u, v) ∈ Y, (u, v)Y ≤ R}, where

1

1

R ≥ max 10a0 M7 , (10a1 M7 ) 1−l1 , (10a2 M7 ) 1−l2 , 

10a3 M7 (Γ(θ1 + 1))l3

1 1−l

3

 ,

10a4 M7 (Γ(σ1 + 1))l4

1

1 1−l

4

,

1

10b0 M9 , (10b1 M9 ) 1−m1 , (10b2 M9 ) 1−m2 ,  1 1   1−m 1−m 3 4 10b3 M9 10b4 M9 . , (Γ(θ2 + 1))m3 (Γ(σ2 + 1))m4 We prove that A : B R → B R . For (u, v) ∈ B R , we deduce  R l3 |A1 (u, v)(t)| ≤ a0 + a1 Rl1 + a2 Rl2 + a3 (Γ(θ1 + 1))l3 R l4 R + a4 M7 ≤ , (Γ(σ1 + 1))l4 2  R m3 |A2 (u, v)(t)| ≤ b0 + b1 Rm1 + b2 Rm2 + b3 (Γ(θ2 + 1))m3 R R m4 + b4 M9 ≤ , m 4 (Γ(σ2 + 1)) 2 for all t ∈ [0, 1]. Then we obtain A(u, v)Y = A1 (u, v) + A2 (u, v) ≤ R, ∀ (u, v) ∈ B R , which implies that A(B R ) ⊂ B R . Because the functions f and g are continuous, we conclude that operator A is continuous on B R . In addition, the functions from A(B R ) are uniformly bounded and equicontinuous. Indeed, by using the notations (6.27) with r1 replaced by R, we find for any (u, v) ∈ B R and t1 , t2 ∈ [0, 1], t1 < t2 that |A1 (u, v)(t2 ) − A1 (u, v)(t1 )| ≤

ΨR (tα − tα 1) Γ(α + 1) 2 + ΨR M8 (tα−1 − tα−1 ), 2 1

|A2 (u, v)(t2 ) − A2 (u, v)(t1 )| ≤

ΘR (tβ − tβ1 ) Γ(β + 1) 2 + ΘR M10 (tβ−1 − tβ−1 ). 2 1

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Therefore, |A1 (u, v)(t2 ) − A1 (u, v)(t1 )| → 0, |A2 (u, v)(t2 ) − A2 (u, v)(t1 )| → 0, as t2 → t1 , uniformly with respect to (u, v) ∈ B R . By the Arzela-Ascoli theorem, we deduce that A(B R ) is relatively compact, and then A is a compact operator. By Theorem 1.2.4, we conclude that operator A has at least one fixed point  (u, v) in B R , which is a solution of our problem (6.12), (6.13). Theorem 6.2.6. Assume that (I1) and (I6) The functions f, g : [0, 1] × R4 → R are continuous and there exist pi ≥ 0, i = 0, . . . , 4 with at least one nonzero, qi ≥ 0, i = 0, . . . , 4 with at least one nonzero, and nondecreasing functions hi , ki ∈ C([0, ∞), [0, ∞)) i = 1, . . . , 4 such that |f (t, x1 , x2 , x3 , x4 )| ≤ p0 +

4 

pi hi (|xi |),

i=1

|g(t, y1 , y2 , y3 , y4 )| ≤ q0 +

4 

qi ki (|yi |),

i=1

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, hold. If there exists Ξ0 > 0 such that   p0 + p1 h1 (Ξ0 ) + p2 h2 (Ξ0 ) + p3 h3  + p4 h4

Ξ0 Γ(σ1 + 1)



Ξ0 Γ(θ1 + 1)

M7

  + q0 + q1 k1 (Ξ0 ) + q2 k2 (Ξ0 ) + q3 k3  + q4 k4

Ξ0 Γ(σ2 + 1)





Ξ0 Γ(θ2 + 1)

M9 < Ξ0 ,



(6.28)

then problem (6.12), (6.13) has at least one solution on [0, 1]. Proof. We consider the set B Ξ0 = {(u, v) ∈ Y, (u, v)Y ≤ Ξ0 }, where Ξ0 is given in the theorem. We will prove that A : B Ξ0 → B Ξ0 . For (u, v) ∈ B Ξ0

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and t ∈ [0, 1] we obtain  |A1 (u, v)(t)| ≤

 p0 + p1 h1 (Ξ0 ) + p2 h2 (Ξ0 ) + p3 h3 

+ p4 h4  |A2 (u, v)(t)| ≤

Ξ0 Γ(σ1 + 1)





Ξ0 Γ(σ2 + 1)



M7 , 

q0 + q1 k1 (Ξ0 ) + q2 k2 (Ξ0 ) + q3 k3 + q4 k4

Ξ0 Γ(θ1 + 1)



Ξ0 Γ(θ2 + 1)



M9 ,

and then for all (u, v) ∈ B Ξ0 we have   A(u, v)Y ≤ p0 + p1 h1 (Ξ0 ) + p2 h2 (Ξ0 ) + p3 h3  + p4 h4 +

Ξ0 Γ(σ1 + 1)



Ξ0 Γ(θ1 + 1)

M7

  q0 + q1 k1 (Ξ0 ) + q2 k2 (Ξ0 ) + q3 k3 

+ q4 k4

Ξ0 Γ(σ2 + 1)





Ξ0 Γ(θ2 + 1)



M9 < Ξ0 .

Then A(B Ξ0 ) ⊂ B Ξ0 . In a similar manner used in the proof of Theorem 6.2.5, we can prove that operator A is completely continuous. We suppose now that there exists (u, v) ∈ ∂BΞ0 such that (u, v) = νA(u, v) for some ν ∈ (0, 1). We obtain as above that (u, v)Y ≤ A(u, v)Y < Ξ0 , which is a contradiction, because (u, v) ∈ ∂BΞ0 . Then by Theorem 1.2.6, we conclude that operator A has a fixed point (u, v) ∈ B Ξ0 , and so problem (6.12), (6.13) has at least one solution. 

6.2.3

Examples

1 9 16 25 Let α = 52 (n = 3), β = 10 3 (m = 4), θ1 = 3 , σ1 = 4 , θ2 = 5 , σ2 = 6 , 4 1 3 11 1 15 3 γ0 = 3 , γ1 = 2 , γ2 = 4 , δ0 = 5 , δ1 = 6 , δ2 = 7 , H1 (t) = t , t ∈ [0, 1], H2 (t) = {0, t ∈ [0, 13 ); 2, t ∈ [ 13 , 1]}, K1 (t) = {0, t ∈ [0, 12 ); 4, t ∈ [ 12 , 1], }, K2 (t) = t2 , t ∈ [0, 1].

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We consider the system of fractional differential equations

5/2 1/3 9/4 t ∈ (0, 1), D0+ u(t) + f (t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, 10/3

16/5

25/6

D0+ v(t) + g(t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, t ∈ (0, 1), (6.29) with the boundary conditions ⎧   1 1 ⎪ 4/3 3/4  2 1/2 ⎪ u(0) = u (0) = 0, D0+ u(1) = 3 t D0+ u(t) dt + 2D0+ u , ⎪ ⎪ ⎪ 3 0 ⎪ ⎪  ⎨ 1 11/5 1/6 v(0) = v  (0) = v  (0) = 0, D0+ v(1) = 4D0+ v ⎪ 2 ⎪ ⎪  1 ⎪ ⎪ ⎪ 15/7 ⎪ ⎩ tD0+ v(t) dt. +2 0

(6.30) We obtain Δ1 ≈ −0.83314732 = 0 and Δ2 ≈ −0.85088584 = 0. So, assumption (I1) is satisfied. In addition, we have M1 ≈ 2.11984652, M2 ≈ 1.39227116, M3 ≈ 1.12892098, M4 ≈ 1.03231866, M5 = M1 , M6 = M3 , M7 ≈ 1.98819306, M8 ≈ 1.68729195, M9 ≈ 1.95523852, M10 ≈ 1.84725332. Example 1. We consider the functions 1 |x1 | f (t, x1 , x2 , x3 , x4 ) = √ + 2 7(t + 1)3 (1 + |x1 |) 4+t − g(t, y1 , y2 , y3 , y4 ) =

t t2 1 arctan x2 + cos x3 − sin2 x4 , 8 t+9 2(t + 10)

1 2t |y2 | − sin y1 + t2 + 9 10 4(2 + |y2 |)

1 t arctan y3 − cos2 y4 , 12 t + 20 for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. We find the inequalities +

4

|f (t, x1 , x2 , x3 , x4 ) − f (t, x 1 , x 2 , x 3 , x 4 )| ≤

1 |xi − x i |, 7 i=1 4

1 |g(t, y1 , y2 , y3 , y4 ) − g(t, y1 , y2 , y3 , y4 )| ≤ |yi − yi |, 8 i=1 for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. So we have L1 = 17 , L2 = 1 8 , and Ξ ≈ 0.878 < 1. Therefore, assumption (I2) is satisfied, and by Theorem 6.2.1 we deduce that problem (6.29), (6.30) has a unique solution (u(t), v(t)), t ∈ [0, 1].

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315

Example 2. We consider the functions t+1 f (t, x1 , x2 , x3 , x4 ) = 2 t +3 − g(t, y1 , y2 , y3 , y4 ) =

 1 3 sin t + sin x1 4

1 t x2 + arctan x3 − cos x4 , 2 (t + 3) 4

e−t 1 1 − sin y2 + cos2 y3 + arctan y4 , 2 1+t 3 5

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. Because we have 5 1 1 1 + |x1 | + |x2 | + |x3 |, 2 8 9 4 1 1 |g(t, y1 , y2 , y3 , y4 )| ≤ 2 + |y2 | + |y4 |, 3 5

|f (t, x1 , x2 , x3 , x4 )| ≤

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, then assumption (I3) is satisfied with c0 = 52 , c1 = 18 , c2 = 19 , c3 = 14 , c4 = 0, d0 = 2, d1 = 0, d2 = 13 , d3 = 0, d4 = 15 . Besides, we obtain M11 ≈ 0.805142, M12 ≈ 0.885295 and Ξ1 = M12 < 1. Then by Theorem 6.2.2 we conclude that problem (6.29), (6.30) has at least one solution (u(t), v(t)), t ∈ [0, 1]. Example 3. We consider the functions 1 3/5 1 2/3 arctan x4 , f (t, x1 , x2 , x3 , x4 ) = − x1 + 4 2(1 + t) g(t, y1 , y2 , y3 , y4 ) =

e−t 1 − |y2 |1/2 + sin |y3 |3/4 , 1 + t4 3

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. Because we obtain 1 1 |x1 |3/5 + |x4 |2/3 , 4 2 1 |g(t, y1 , y2 , y3 , y4 )| ≤ 1 + |y2 |1/2 + |y3 |3/4 , 3

|f (t, x1 , x2 , x3 , x4 )| ≤

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, then assumption (I5) is satisfied with a0 = 0, a1 = 14 , a2 = 0, a3 = 0, a4 = 12 , b0 = 1, b1 = 0, b2 = 13 , b3 = 1, b4 = 0, l1 = 35 , l4 = 23 , m2 = 12 , m3 = 34 . Therefore, by Theorem 6.2.5, we deduce that problem (6.29), (6.30) has at least one solution (u(t), v(t)), t ∈ [0, 1].

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Example 4. We consider the functions f (t, x1 , x2 , x3 , x4 ) =

1/3

e−t x32 t2 x3 (1 − t)3 + − , 10 20(1 + x21 ) 5

1 − t2 2 t2 1 1/5 + y − y , 20 25 1 30 4 for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. Because we have g(t, y1 , y2 , y3 , y4 ) =

1 1 1 + |x2 |3 + |x3 |1/3 , 10 20 5 1 1 1 |g(t, y1 , y2 , y3 , y4 )| ≤ + |y1 |2 + |y4 |1/5 , 20 25 30 for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, then assumption (I6) is sat1 1 , p1 = 0, p2 = 20 , p3 = 15 , p4 = 0, h1 (x) = 0, isfied with p0 = 10 1 1 3 1/3 , q1 = 25 , q2 = 0, h2 (x) = x , h3 (x) = x , h4 (x) = 0, q0 = 20 1 2 q3 = 0, q4 = 30 , k1 (x) = x , k2 (x) = 0, k3 (x) = 0, k4 (x) = x1/5 . For Ξ0 = 2, the condition (6.28) is satisfied, because (p0 + p1 h1 (2) + p2 h2 (2) + p3 h3 ( Γ(θ12+1) )+p4 h4 ( Γ(σ12+1) ))M7 + (q0 +q1 k1 (2)+q2 k2 (2)+q3 k3 ( Γ(θ22+1) )+ q4 k4 ( Γ(σ22+1) ))M9 ≈ 1.96264 < 2. Therefore, by Theorem 6.2.6 we conclude that problem (6.29), (6.30) has at least one solution (u(t), v(t)), t ∈ [0, 1]. |f (t, x1 , x2 , x3 , x4 )| ≤

Remark 6.2.1. The results presented in this section were published in [81]. 6.3

Systems of Riemann–Liouville Fractional Differential Equations with Coupled Boundary Conditions

We consider the nonlinear system of fractional differential equations

θ1 σ1 α D0+ u(t) + f (t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, t ∈ (0, 1), β θ2 σ2 v(t) + g(t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, D0+

t ∈ (0, 1),

with the coupled nonlocal boundary conditions ⎧ u(0) = u (0) = · · · = u(n−2) (0) = 0, ⎪ ⎪ ⎪ p  1 ⎪ ⎪  ⎪ γ0 γi ⎪ ⎪ u(1) = D0+ v(t) dHi (t), D ⎪ ⎨ 0+ 0 i=1

⎪ v(0) = v  (0) = · · · = v (m−2) (0) = 0, ⎪ ⎪ ⎪ ⎪ q  1 ⎪  ⎪ ⎪ δ0 δi ⎪ v(1) = D0+ u(t) dKi (t), D ⎩ 0+ i=1

0

(6.31)

(6.32)

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where α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n ≥ 2, m ≥ 2, θ1 , θ2 , σ1 , σ2 > 0, p, q ∈ N, γi ∈ R for all i = 0, . . . , p, 0 ≤ γ1 < γ2 < · · · < γp < β − 1, γ0 ∈ [0, α − 1), δi ∈ R for all i = 0, . . . , q, 0 ≤ δ1 < δ2 < · · · < k denotes the Riemann–Liouville derivative δq < α − 1, δ0 ∈ [0, β − 1), D0+ ζ is of order k (for k = α, β, γ0 , γi , i = 1, . . . , p, δ0 , δi , i = 1, . . . , q), I0+ the Riemann–Liouville integral of order ζ (for ζ = θ1 , σ1 , θ2 , σ2 ), f and g are nonlinear functions, and the integrals from the boundary conditions (6.32) are Riemann–Stieltjes integrals with Hi for i = 1, . . . , p and Ki for i = 1, . . . , q functions of bounded variation. By using some theorems from the fixed point theory, we will give conditions for the nonlinearities f and g such that problem (6.31), (6.32) has at least one solution (u, v) ∈ (C[0, 1])2 . 6.3.1

Preliminary results

We consider the system of fractional differential equations

α D0+ u(t) + h(t) = 0, t ∈ (0, 1), β v(t) + k(t) = 0, D0+

t ∈ (0, 1),

(6.33)

with the boundary conditions (6.32), where h, k ∈ C(0, 1) ∩ L1 (0, 1). We denote by Δ=

Γ(α)Γ(β) Γ(α − γ0 )Γ(β − δ0 )   p  Γ(β)  1 β−γi −1 s dHi (s) − Γ(β − γi ) 0 i=1  q   Γ(α)  1 × sα−δi −1 dKi (s) . Γ(α − δ ) i 0 i=1

Lemma 6.3.1. If Δ = 0, then the unique solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (6.33), (6.32) is given by u(t) = −

1 Γ(α) 

 0

t

(t − s)α−1 h(s) ds +

Γ(β) × Γ(α − γ0 )Γ(β − δ0 )

 0

1

tα−1 Δ

(1 − s)α−γ0 −1 h(s) ds

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−

1 Γ(β)  Γ(β − δ0 ) i=1 Γ(β − γi ) 

×



0

 +  ×

 −

1

0

p 

(s − τ )β−γi −1 k(τ ) dτ

Γ(β) Γ(β − γi )

i=1

1 Γ(β − δ0 ) p 





1

0

Γ(β) Γ(β − γi )

i=1

×

s

q  i=1

1 Γ(α − δi )





1

0

s

(1 − s) 1

0

 0

1

dHi (s) 

β−γi −1

dHi (s)

β−δ0 −1





k(s) ds 

sβ−γi −1 dHi (s)  0

s

α−δi −1

(s − τ )



h(τ ) dτ

 dKi (s)

, (6.34)

t Γ(α) tβ−1 1 (t − s)β−1 k(s) ds + Γ(β) 0 Δ Γ(α − γ0 )Γ(β − δ0 )  1 Γ(α) (1 − s)β−δ0 −1 k(s) ds − × Γ(α − γ0 ) 0  1  s q  1 × (s − τ )α−δi −1 h(τ ) dτ dKi (s) Γ(α − δi ) 0 0 i=1   q  1  Γ(α) α−δi −1 s dKi (s) + Γ(α − δi ) 0 i=1   1 1 α−γ0 −1 × (1 − s) h(s) ds Γ(α − γ0 ) 0  q   Γ(α)  1 α−δi −1 − s dKi (s) Γ(α − δi ) 0 i=1  p   1  s  1 β−γi −1 × (s − τ ) k(τ ) dτ dHi (s) . Γ(β − γi ) 0 0 i=1

v(t) = −

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Proof. The solutions (u, v) ∈ (C(0, 1) ∩ L1 (0, 1))2 of system (6.33) are α h(t) + a1 tα−1 + a2 tα−2 + · · · + an tα−n , u(t) = −I0+ β v(t) = −I0+ k(t) + b1 tβ−1 + b2 tβ−2 + · · · + bm tβ−m ,

with a1 , a2 , . . . , an , b1 , b2 , . . . , bm ∈ R. By using the boundary conditions u(0) = u (0) = · · · = u(n−2) (0) = 0 and v(0) = v  (0) = · · · = v (m−2) (0) = 0, we obtain a2 = · · · = an = 0 and b2 = · · · = bm = 0. So the above solutions become  t 1 (t − s)α−1 h(s) ds + a1 tα−1 , u(t) = − Γ(α) 0 (6.35)  t 1 β−1 β−1 (t − s) k(s) ds + b1 t . v(t) = − Γ(β) 0 For the obtained functions u and v, we have γi D0+ v(t) = b1

Γ(β) β−γi tβ−γi −1 − I0+ k(t), Γ(β − γi )

γ0 u(t) = a1 D0+

Γ(α) α−γ0 h(t), tα−γ0 −1 − I0+ Γ(α − γ0 )

δi D0+ u(t) = a1

Γ(α) α−δi −1 α−δi t − I0+ h(t), Γ(α − δi )

δ0 D0+ v(t) = b1

Γ(β) β−δ0 tβ−δ0 −1 − I0+ k(t), Γ(β − δ0 )

i = 1, . . . , p,

i = 1, . . . , q,

Γ(α) Γ(α) α−γ0 − I0+ h(1) = a1 Γ(α − γ0 ) Γ(α − γ0 )  1 1 − (1 − s)α−γ0 −1 h(s) ds, Γ(α − γ0 ) 0

γ0 D0+ u(1) = a1

Γ(β) Γ(β) β−δ0 − I0+ k(1) = b1 Γ(β − δ0 ) Γ(β − δ0 )  1 1 − (1 − s)β−δ0 −1 k(s) ds. Γ(β − δ0 ) 0

p  1 γi γ0 By imposing the boundary conditions D0+ u(1) = i=1 0 D0+ v(t) dHi (t)

q  1 δi δ0 and D0+ v(1) = i=1 0 D0+ u(t) dKi (t), we deduce the following system δ0 D0+ v(1) = b1

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in a1 and b1 ⎧ p  1  ⎪ Γ(α) Γ(β) ⎪ ⎪ a − b tβ−γi −1 dHi (t) ⎪ 1 1 ⎪ ⎪ Γ(α − γ ) Γ(β − γ ) 0 i ⎪ i=1 0 ⎪ ⎪  1 ⎪ ⎪ ⎪ 1 ⎪ ⎪ = (1 − s)α−γ0 −1 h(s) ds ⎪ ⎪ Γ(α − γ ) ⎪ 0 0 ⎪ ⎪ ⎪  t p  1 ⎪ ⎪  1 ⎪ β−γi −1 ⎪ ⎪ (t − s) k(s) ds dHi (t), − ⎪ ⎨ Γ(β − γi ) 0 i=1 0 q  1 ⎪  Γ(β) Γ(α) α−δi −1 ⎪ ⎪ ⎪ − a1 t dKi (t) b1 ⎪ ⎪ Γ(β − δ ) Γ(α − δi ) 0 ⎪ ⎪ i=1 0 ⎪ ⎪  1 ⎪ ⎪ 1 ⎪ ⎪ = (1 − s)β−δ0 −1 k(s) ds ⎪ ⎪ ⎪ Γ(β − δ ) 0 0 ⎪ ⎪ ⎪  t ⎪ q  1 ⎪  ⎪ 1 ⎪ α−δi −1 ⎪ (t − s) h(s) ds dKi (t). ⎪ ⎩ − Γ(α − δi ) 0 i=1 0

(6.36)

The above system in the unknowns a1 and b1 has the determinant Δ, which, by the assumption of lemma, is nonzero. So, the system (6.36) has the unique solution  1 Γ(β) (1 − s)α−γ0 −1 h(s) ds a1 = ΔΓ(α − γ0 )Γ(β − δ0 ) 0 p

 1 Γ(β) ΔΓ(β − δ0 ) i=1 Γ(β − γi )  1  t β−γi −1 × (t − s) k(s) ds dHi (t) −

0

0

 p   1 1  Γ(β) β−γi −1 + t dHi (t) Δ i=1 Γ(β − γi ) 0   1 1 × (1 − s)β−δ0 −1 k(s) ds Γ(β − δ0 ) 0  p   1 1  Γ(β) − tβ−γi −1 dHi (t) Δ i=1 Γ(β − γi ) 0  q   1  t  1 α−δi −1 × (t − s) h(s) ds dKi (t) , Γ(α − δi ) 0 0 i=1

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Γ(α) b1 = ΔΓ(α − γ0 )Γ(β − δ0 )

 0

1

(1 − s)β−δ0 −1 k(s) ds

page 321

321

(6.37)

q

 1 Γ(α) ΔΓ(α − γ0 ) i=1 Γ(α − δi )  1  t α−δi −1 (t − s) h(s) ds dKi (t) ×

−

0

0

 q   1 1  Γ(α) α−δi −1 + t dKi (t) Δ i=1 Γ(α − δi ) 0   1 1 (1 − s)α−γ0 −1 h(s) ds × Γ(α − γ0 ) 0  q   1 1  Γ(α) − tα−δi −1 dKi (t) Δ i=1 Γ(α − δi ) 0  p   1  t  1 β−γi −1 × (t − s) k(s) ds dHi (t) . Γ(β − γi ) 0 0 i=1 Now, replacing the constants a1 and b1 given by (6.37) in (6.35) we find the solution (u, v) ∈ C[0, 1] × C[0, 1] of problem (6.33), (6.32) presented in (6.34). Conversely, we can easily prove that the functions u, v given by (6.34) satisfy the problem (6.33), (6.32).  We introduce now the assumption (J1) for problem (6.31), (6.32) that will be used in our main results. (J1) α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n ≥ 2, m ≥ 2, θ1 , θ2 , σ1 , σ2 > 0, p, q ∈ N, γi ∈ R for all i = 0, . . . , p, 0 ≤ γ1 < γ2 < · · · < γp < β − 1, γ0 ∈ [0, β − 1), δi ∈ R for all i = 0, . . . , q, 0 ≤ δ1 < δ2 < · · · < δq < α − 1, δ0 ∈ [0, α − 1), Hi : [0, 1] → R, i = 1, . . . , p and Kj : [0, 1] → R, j = 1, . . . , q are functions of bounded variation, and Δ = 0. We introduce the following constants: M1 = 1 +

1 , Γ(θ1 + 1)

M2 = 1 +

1 , Γ(σ1 + 1)

M3 = 1 +

1 , Γ(θ2 + 1)

M4 = 1 +

1 , Γ(σ2 + 1)

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M5 = max{M1 , M2 }, M7 =

1 Γ(β) + Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   Γ(β)  1 1 + sβ−γi −1 dHi (s) |Δ| i=1 Γ(β − γi ) 0 ×

 q  i=1

M10 =

M6 = max{M3 , M4 },

 1  1 α−δi s dKi (s) , Γ(α − δi + 1) 0

 1 p  1 Γ(β) β−γi s dH (s) i |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1) 0 1 + |Δ|Γ(β − δ0 + 1)



p  i=1

  Γ(β) 1 β−γi −1 s dHi (s) , Γ(β − γi ) 0

 1 q  1 Γ(α) α−δi s dKi (s) M9 = |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0  q   Γ(α)  1 1 α−δi −1 + s dKi (s) , |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0 M8 =

Γ(α) 1 + Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q   Γ(α)  1 1 + sα−δi −1 dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1   1 β−γi × s dHi (s) , Γ(β − γi + 1) 0 i=1

M11 = M7 −

1 , Γ(α + 1)

M12 = M8 −

1 . Γ(β + 1) (6.38)

We consider the Banach space X = C[0, 1] with supremum norm u = supt∈[0,1] |u(t)|, and the Banach space Y = X × X with the norm (u, v)Y = u + v. We introduce the operator Q : Y → Y defined by Q(u, v) = (Q1 (u, v), Q2 (u, v)) for (u, v) ∈ Y , where the operators Q1 ,

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Q2 : Y → X are given by  t 1 Q1 (u, v)(t) = − (t − s)α−1 fˆuv (s) ds Γ(α) 0  1 tα−1 Γ(β) + (1 − s)α−γ0 −1 fˆuv (s) ds ΔΓ(α − γ0 )Γ(β − δ0 ) 0 p

1 tα−1 Γ(β)  − ΔΓ(β − δ0 ) i=1 Γ(β − γi )  1  s (s − τ )β−γi −1 gˆuv (τ ) dτ dHi (s) × 0

0



  1 Γ(β) β−γi −1 + s dHi (s) Δ Γ(β − γi ) 0 i=1   1 1 β−δ0 −1 × (1 − s) gˆuv (s) ds Γ(β − δ0 ) 0  p   1 tα−1  Γ(β) β−γi −1 − s dHi (s) Δ Γ(β − γi ) 0 i=1  q   1  s  1 α−δi −1 ˆ × (s − τ ) fuv (τ ) dτ dKi (s) , Γ(α − δi ) 0 0 i=1 t

α−1

1 Q2 (u, v)(t) = − Γ(β)

p 



t

0

(6.39) (t − s)β−1 gˆuv (s) ds

tβ−1 Γ(α) + ΔΓ(α − γ0 )Γ(β − δ0 )

 0

1

(1 − s)β−δ0 −1 gˆuv (s) ds

q Γ(α)  1 t − ΔΓ(α − γ0 ) i=1 Γ(α − δi )  1  s α−δi −1 ˆ (s − τ ) fuv (τ ) dτ dKi (s) × β−1

0

0



  1 q tβ−1  Γ(α) α−δi −1 + s dKi (s) Δ Γ(α − δi ) 0 i=1   1 1 α−γ0 −1 ˆ (1 − s) fuv (s) ds × Γ(α − γ0 ) 0

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 q   1 tβ−1  Γ(α) α−δi −1 − s dKi (s) Δ Γ(α − δi ) 0 i=1  p   1  s  1 β−γi −1 × (s − τ ) gˆuv (τ ) dτ dHi (s) , Γ(β − γi ) 0 0 i=1 for t ∈ [0, 1] and (u, v) ∈ Y , where fˆuv (s) = f (s, u(s), v(s), θ1 σ1 θ2 σ2 u(s), I0+ v(s)), gˆuv (s) = g(s, u(s), v(s), I0+ u(s), I0+ v(s)) for s ∈ [0, 1]. I0+ By using Lemma 6.3.1, we see that (u, v) is a solution of problem (6.31), (6.32) if and only if (u, v) is a fixed point of operator Q. 6.3.2

Existence of solutions

In this section, we will give some existence results for the solutions of our problem (6.31), (6.32). Theorem 6.3.1. Assume that (J1) and (J2) The functions f, g : [0, 1] × R4 → R are continuous and there exist L1 , L2 > 0 such that |f (t, x1 , x2 , x3 , x4 ) − f (t, y1 , y2 , y3 , y4 )| ≤ L1

4 

|xi − yi |,

i=1

|g(t, x1 , x2 , x3 , x4 ) − g(t, y1 , y2 , y3 , y4 )| ≤ L2

4 

|xi − yi |,

i=1

for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4, hold. If Ξ := L1 M5 (M7 + M9 ) + L2 M6 (M8 + M10 ) < 1, then problem (6.31), (6.32) has a unique solution (u(t), v(t)), t ∈ [0, 1], where M5 , . . . , M10 are given by (6.38). Proof. We consider the positive number r given by 0 (M8 + M10 )] r = [M0 (M7 + M9 ) + M ×[1 − L1 M5 (M7 + M9 ) − L2 M6 (M8 + M10 )]−1 , 0 = supt∈[0,1] |g(t, 0, 0, 0, 0)|. We where M0 = supt∈[0,1] |f (t, 0, 0, 0, 0)|, M define the set B r = {(u, v) ∈ Y, (u, v)Y ≤ r} and we show firstly that

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325

Q(B r ) ⊂ B r . Let (u, v) ∈ B r . By using (J2) and Lemma 6.1.2, for fˆuv (t), we deduce the following inequalities θ1 σ1 u(t), I0+ v(t)) − f (t, 0, 0, 0, 0)| |fˆuv (t)| ≤ |f (t, u(t), v(t), I0+

+ |f (t, 0, 0, 0, 0)|   θ1 σ1 ≤ L1 |u(t)| + |v(t)| + |I0+ u(t)| + |I0+ v(t)| + M0  v u + ≤ L1 u + v + + M0 Γ(θ1 + 1) Γ(σ1 + 1) = L1 (M1 u + M2 v) + M0 ≤ L1 M5 (u, v)Y + M0 ≤ L1 M5 r + M0 ,

∀ t ∈ [0, 1].

Arguing as before, we find θ2 σ2 |ˆ guv (t)| ≤ |g(t, u(t), v(t), I0+ u(t), I0+ v(t)) − g(t, 0, 0, 0, 0)|

+ |g(t, 0, 0, 0, 0)|   θ2 σ2 0 ≤ L2 |u(t)| + |v(t)| + |I0+ u(t)| + |I0+ v(t)| + M  u v 0 ≤ L2 u + v + + +M Γ(θ2 + 1) Γ(σ2 + 1) 0 = L2 (M3 u + M4 v) + M 0 ≤ L2 M6 r + M 0 , ≤ L2 M6 (u, v)Y + M

∀ t ∈ [0, 1].

Then by the definition of operators Q1 and Q2 , we conclude  t 1 (t − s)α−1 (L1 M5 r + M0 ) ds |Q1 (u, v)(t)| ≤ Γ(α) 0 tα−1 Γ(β) |Δ|Γ(α − γ0 )Γ(β − δ0 )  1 (1 − s)α−γ0 −1 (L1 M5 r + M0 ) ds × +

0

p

1 tα−1 Γ(β)  |Δ|Γ(β − δ0 ) i=1 Γ(β − γi )  1  s β−γi −1  × (s − τ ) (L2 M6 r + M0 ) dτ dHi (s) +

0

0

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 p   tα−1  Γ(β) 1 β−γi −1 + s dHi (s) |Δ| i=1 Γ(β − γi ) 0   1 1 β−δ0 −1  × (1 − s) (L2 M6 r + M0 ) ds Γ(β − δ0 ) 0  p   tα−1  Γ(β) 1 β−γi −1 + s dHi (s) |Δ| i=1 Γ(β − γi ) 0  q  1 × Γ(α − δi ) i=1  1  s α−δi −1 (s − τ ) (L1 M5 r + M0 ) dτ dKi (s) × 0

0



1 = (L1 M5 r + M0 ) Γ(α)



t

0

(t − s)α−1 ds

 1 tα−1 Γ(β) (1 − s)α−γ0 −1 ds |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0  p   tα−1  Γ(β) 1 β−γi −1 + s dHi (s) |Δ| i=1 Γ(β − γi ) 0

+



×

q  i=1

 1  s  1 α−δi −1 (s − τ ) dτ dKi (s) Γ(α − δi ) 0 0 

0 ) + (L2 M6 r + M  ×

1

0

tα−1 + |Δ|  ×

 

0

s

p

1 tα−1 Γ(β)  |Δ|Γ(β − δ0 ) i=1 Γ(β − γi )

β−γi −1

(s − τ )

p  i=1



dτ

dHi (s)

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0

1 Γ(β − δ0 )



= (L1 M5 r + M0 )

1

0



β−δ0 −1

(1 − s)

 ds

tα−1 Γ(β) tα + Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )

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tα−1 + |Δ| ×

p  i=1

 q  i=1

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0

 1  1 sα−δi dKi (s) Γ(α − δi + 1) 0 

0 ) + (L2 M6 r + M  ×

1

0

p

tα−1 Γ(β)  1 |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1)

sβ−γi dHi (s)

tα−1 + |Δ|Γ(β − δ0 + 1)



p  i=1

  Γ(β) 1 β−γi −1 , s dHi (s) Γ(β − γi ) 0

∀ t ∈ [0, 1]. Therefore, we obtain



Q1 (u, v) ≤ (L1 M5 r + M0 ) 1 + |Δ| ×



p  i=1

 q  i=1

Γ(β) 1 + Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0

 1  1 α−δi s dKi (s) Γ(α − δi + 1) 0 

0 ) + (L2 M6 r + M  ×

1

0

 ×

s

p  i=1

β−γi

p

 Γ(β) 1 |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1)

dHi (s) +

1 |Δ|Γ(β − δ0 + 1)

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0

0 )M10 . = (L1 M5 r + M0 )M7 + (L2 M6 r + M

(6.40)

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In a similar manner, we deduce |Q2 (u, v)(t)| ≤

1 Γ(β)

 0

t

0 ) ds (t − s)β−1 (L2 M6 r + M

β−1

Γ(α) t |Δ|Γ(α − γ0 )Γ(β − δ0 )  1 0 ) ds × (1 − s)β−δ0 −1 (L2 M6 r + M +

0

q

1 tβ−1 Γ(α)  |Δ|Γ(α − γ0 ) i=1 Γ(α − δi )  1  s (s − τ )α−δi −1 (L1 M5 r + M0 ) dτ dKi (s) × +

0

0

  t Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0   1 1 α−γ0 −1 × (1 − s) (L1 M5 r + M0 ) ds Γ(α − γ0 ) 0  q   tβ−1  Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1  s   1 β−γ −1 i 0 ) dτ dHi (s) × (s − τ ) (L2 M6 r + M Γ(β − γi ) 0 0 i=1  q 1 tβ−1 Γ(α)  = (L1 M5 r + M0 ) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi )  1  s × (s − τ )α−δi −1 dτ dKi (s) 0 0  q  β−1  Γ(α)  1 t α−δi −1 s dKi (s) + |Δ| i=1 Γ(α − δi ) 0    1 1 α−γ0 −1 × (1 − s) ds Γ(α − γ0 ) 0   t 1 tβ−1 Γ(α)  + (L2 M6 r + M0 ) (t − s)β−1 ds + Γ(β) 0 |Δ|Γ(α − γ0 )Γ(β − δ0 ) β−1



q 

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 ×

0

1

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329

(1 − s)β−δ0 −1 ds

 q   tβ−1  Γ(α) 1 α−δi −1 s dKi (s) + |Δ| i=1 Γ(α − δi ) 0  p  1  s   1 β−γ −1 i × (s − τ ) dτ dHi (s) Γ(β − γi ) 0 0 i=1   1 q 1 tβ−1 Γ(α)  α−δi s dKi (s) = (L1 M5 r + M0 ) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0  q   Γ(α)  1 tβ−1 α−δi −1 s dKi (s) + |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0  tβ tβ−1 Γ(α)  + + (L2 M6 r + M0 ) Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q   tβ−1  Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1   1 β−γ i s dHi (s) , × Γ(β − γi + 1) 0 i=1 for all t ∈ [0, 1]. Then we find 

q

 1 Γ(α) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 1 α−δi s dKi (s)

Q2 (u, v) ≤ (L1 M5 r + M0 )  ×

0

 q  Γ(α) 1 + |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi )   1 sα−δi −1 dKi (s) × 0

(6.41) Γ(α) 1 0 ) + (L2 M6 r + M + Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1) 

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 q   Γ(α)  1 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1   1 β−γ i × s dHi (s) Γ(β − γi + 1) 0 i=1 0 )M8 . = (L1 M5 r + M0 )M9 + (L2 M6 r + M By relations (6.40) and (6.41), we conclude Q(u, v)Y = Q1 (u, v) + Q2 (u, v) ≤ (L1 M5 r + M0 )(M7 + M9 ) 0 )(M8 + M10 ) = r, + (L2 M6 r + M for all (u, v) ∈ B r , which implies that Q(B r ) ⊂ B r . Next, we prove that operator Q is a contraction. For (ui , vi ) ∈ B r , i = 1, 2, and for each t ∈ [0, 1], we have |Q1 (u1 , v1 )(t) − Q1 (u2 , v2 )(t)|  t 1 ≤ (t − s)α−1 |fˆu1 v1 (s) − fˆu2 v2 (s)| ds Γ(α) 0  1 tα−1 Γ(β) (1 − s)α−γ0 −1 + |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0 × |fˆu1 v1 (s) − fˆu2 v2 (s)| ds p

1 tα−1 Γ(β)  + |Δ|Γ(β − δ0 ) i=1 Γ(β − γi )  1  s (s − τ )β−γi −1 |ˆ gu1 v1 (τ ) − gˆu2 v2 (τ )| dτ dHi (s) × 0

0

(6.42)   Γ(β) 1 β−γi −1 tα−1 s dHi (s) + |Δ| i=1 Γ(β − γi ) 0   1 1 β−δ0 −1 × (1 − s) |ˆ gu1 v1 (s) − gˆu2 v2 (s)| ds Γ(β − δ0 ) 0  p   tα−1  Γ(β) 1 β−γi −1 + s dHi (s) |Δ| Γ(β − γi )  p 

i=1

0

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×

Because

 q 

 1  s 1 (s − τ )α−δi −1 |fˆu1 v1 (τ ) Γ(α − δ ) i 0 0 i=1  − fˆu2 v2 (τ )| dτ dKi (s) .

 |fˆu1 v1 (s) − fˆu2 v2 (s)| ≤ L1 |u1 (s) − u2 (s)| + |v1 (s) − v2 (s)|  θ1 θ1 σ1 σ1 u1 (s) − I0+ u2 (s)| + |I0+ v1 (s) − I0+ v2 (s)| + |I0+  1 u1 − u2  ≤ L1 u1 − u2  + v1 − v2  + Γ(θ1 + 1) 1 v1 − v2  + Γ(σ1 + 1) = L1 (M1 u1 − u2  + M2 v1 − v2 ) ≤ L1 M5 (u1 , v1 ) − (u2 , v2 )Y , ∀ s ∈ [0, 1],  |ˆ gu1 v1 (s) − gˆu2 v2 (s)| ≤ L2 |u1 (s) − u2 (s)| + |v1 (s) − v2 (s)|  θ2 θ2 σ2 σ2 u1 (s) − I0+ u2 (s)| + |I0+ v1 (s) − I0+ v2 (s)| +|I0+  1 ≤ L2 u1 − u2  + v1 − v2  + u1 − u2  Γ(θ2 + 1) 1 v1 − v2  + Γ(σ2 + 1) = L2 (M3 u1 − u2  + M4 v1 − v2 ) ≤ L2 M6 (u1 , v1 ) − (u2 , v2 )Y ,

∀ s ∈ [0, 1],

the inequality (6.42) gives us |Q1 (u1 , v1 )(t) − Q1 (u2 , v2 )(t)| ≤ L1 M5 (u1 , v1 ) − (u2 , v2 )Y  tα tα−1 Γ(β) × + Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   tα−1  Γ(β) 1 β−γi −1 s dHi (s) + |Δ| i=1 Γ(β − γi ) 0

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 1  1 α−δi × s dKi (s) Γ(α − δi + 1) 0 i=1  p 1 tα−1 Γ(β)  + L2 M6 (u1 , v1 ) − (u2 , v2 )Y |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1)  1 tα−1 sβ−γi dHi (s) + × |Δ|Γ(β − δ0 + 1) 0  1  p  Γ(β) sβ−γi −1 dHi (s) , × Γ(β − γ ) i 0 i=1 

q 

∀ t ∈ [0, 1]. Therefore, we obtain Q1 (u1 , v1 ) − Q1 (u2 , v2 ) 

Γ(β) 1 + ≤ L1 M5 Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   Γ(β)  1 1 β−γ −1 i + s dHi (s) |Δ| i=1 Γ(β − γi ) 0  q  1   1 α−δ i s dKi (s) × Γ(α − δi + 1) 0 i=1   1 p  1 Γ(β) β−γi s dHi (s) + L2 M6 |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1) 0   p  Γ(β) 1 β−γi −1 1 s dHi (s) + |Δ|Γ(β − δ0 + 1) i=1 Γ(β − γi ) 0 ×(u1 , v1 ) − (u2 , v2 )Y = (L1 M5 M7 + L2 M6 M10 )(u1 , v1 ) − (u2 , v2 )Y . (6.43) In a similar manner, we deduce |Q2 (u1 , v1 )(t) − Q2 (u2 , v2 )(t)|  t 1 (t − s)β−1 |ˆ gu1 v1 (s) − gˆu2 v2 (s)| ds ≤ Γ(β) 0  1 tβ−1 Γ(α) (1 − s)β−δ0 −1 |ˆ gu1 v1 (s) − gˆu2 v2 (s)| ds + |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0

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333

q

tβ−1 Γ(α)  1 |Δ|Γ(α − γ0 ) i=1 Γ(α − δi )  1  s     ×  (s − τ )α−δi −1 |fˆu1 v1 (τ ) − fˆu2 v2 (τ )| dτ dKi (s) +

0

0

 q 

   t Γ(α)  1 α−δi −1 + s dKi (s)  |Δ| i=1 Γ(α − δi ) 0    1 1 (1 − s)α−γ0 −1 |fˆu1 v1 (s) − fˆu2 v2 (s)| ds × Γ(α − γ0 ) 0  q    tβ−1  Γ(α)  1 α−δi −1 s dKi (s) +  |Δ| i=1 Γ(α − δi ) 0  p   1  s     1 β−γi −1  × (s − τ ) |ˆ gu1 v1 (τ ) − gˆu2 v2 (τ )| dτ dHi (s)  Γ(β − γ i) 0 0 i=1 tβ tβ−1 Γ(α) ≤ L2 M6 (u1 , v1 ) − (u2 , v2 )Y + Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q     tβ−1  Γ(α)  1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi )  0  p 

 1    1 β−γ i  × s dHi (s)  Γ(β − γi + 1) 0 i=1 q 1 tβ−1 Γ(α)  + L1 M5 (u1 , v1 ) − (u2 , v2 )Y |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) β−1

  × 

  sα−δi dKi (s) +

q  tβ−1 Γ(α) |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0  1 

  α−δi −1  × s dKi (s) , ∀ t ∈ [0, 1]. 1

0

Hence, we find Q2 (u1 , v1 ) − Q2 (u2 , v2 )

 1 Γ(α) ≤ L2 M6 + Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q   Γ(α)  1 1 α−δi −1 s dKi (s) + |Δ| i=1 Γ(α − δi ) 0

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 1  1 β−γi × s dHi (s) Γ(β − γi + 1) 0 i=1   1 q  1 Γ(α) α−δi + L1 M5 s dK (s) i |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0   q  Γ(α) 1 α−δi −1 1 s dKi (s) + |Δ|Γ(α − γ0 + 1) Γ(α − δi )  p 

i=1

0

× (u1 , v1 ) − (u2 , v2 )Y = (L1 M5 M9 + L2 M6 M8 )(u1 , v1 ) − (u2 , v2 )Y . Then by using relations (6.43) and (6.44), we obtain Q(u1 , v1 ) − Q(u2 , v2 )Y = Q1 (u1 , v1 ) − Q1 (u2 , v2 )

(6.44)

+ Q2(u1 , v1 ) − Q2 (u2 , v2 ) ≤ [L1 M5 (M7 + M9 ) + L2 M6 (M8 + M10 )] × (u1 , v1 ) − (u2 , v2 )Y = Ξ(u1 , v1 ) − (u2 , v2 )Y . By using the condition Ξ < 1, we deduce that operator Q is a contraction. By Theorem 1.2.1, we conclude that operator Q has a unique fixed point (u, v) ∈ B r , which is the unique solution of problem (6.31), (6.32) on [0, 1].  Theorem 6.3.2. Suppose that (J1) and (J3) The functions f, g : [0, 1] × R4 → R are continuous and there exist real constants ai , bi ≥ 0, i = 0, . . . , 4, and at least one of a0 and b0 is positive, such that 4  ai |xi |, |f (t, x1 , x2 , x3 , x4 )| ≤ a0 + i=1

|g(t, x1 , x2 , x3 , x4 )| ≤ b0 + for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4,

4 

bi |xi |,

i=1

hold. If Ξ1 := max{M13 , M14 } < 1, where M13 = (a1 + M9 ) + (b1 +

b3 Γ(θ2 +1) )(M8

b4 Γ(σ2 +1) )(M8

+ M10 ) and M14 = (a2 +

a3 Γ(θ1 +1) )(M7

a4 Γ(σ1 +1) )(M7

+

+ M9 ) +

+ M10 ), then the boundary value problem (6.31), (6.32) (b2 + has at least one solution (u(t), v(t)), t ∈ [0, 1].

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Proof. We show that operator Q is completely continuous. Because the functions f and g are continuous, we deduce that the operators Q1 and Q2 are continuous, and then Q is a continuous operator. We will prove next that Q is a compact operator, that is, it maps bounded sets into relatively compact sets. Let Ω ⊂ Y be a bounded set. Then there exist positive guv (t)| ≤ L4 for all constants L3 and L4 such that |fˆuv (t)| ≤ L3 and |ˆ t ∈ [0, 1] and (u, v) ∈ Ω. Hence, we obtain as in the proof of Theorem 6.3.1 that |Q1 (u, v)(t)| ≤ L3 M7 + L4 M10 ,

|Q2 (u, v)(t)| ≤ L3 M9 + L4 M8 ,

for all t ∈ [0, 1] and (u, v) ∈ Ω. So, we find Q1 (u, v) ≤ L3 M7 + L4 M10 ,

Q2 (u, v) ≤ L3 M9 + L4 M8 ,

Q(u, v)Y ≤ L3 (M7 + M9 ) + L4 (M8 + M10 ),

∀ (u, v) ∈ Ω,

and then Q(Ω) is uniformly bounded. We show now that Q(Ω) are equicontinuous. Let (u, v) ∈ Ω and t1 , t2 ∈ [0, 1] with t1 < t2 . Then we have |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )|  t2  t1 1 1 ≤ − (t2 − s)α−1 fˆuv (s) ds + (t1 − s)α−1 fˆuv (s) ds Γ(α) 0 Γ(α) 0  1 α−1 α−1 (t2 − t1 )Γ(β) + (1 − s)α−γ0 −1 |fˆuv (s)| ds |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0 p

)Γ(β)  1 (tα−1 − tα−1 1 + 2 |Δ|Γ(β − δ0 ) Γ(β − γi ) i=1  1  s (s − τ )β−γi −1 |ˆ guv (τ )| dτ dHi (s) × 0

+

− tα−1 1 |Δ|

 × +

0

tα−1 2

1 Γ(β − δ0 )

tα−1 2

tα−1 1

− |Δ|

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0 i=1  1 β−δ0 −1 (1 − s) |ˆ guv (s)| ds 

p 

0



p  i=1

  Γ(β) 1 β−γi −1 s dHi (s) Γ(β − γi ) 0

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 1  s  1 α−δi −1 ˆ × (s − τ ) |fuv (τ )| dτ dKi (s) Γ(α − δi ) 0 0 i=1  t1  t2 L3 L3 α−1 α−1 ≤ [(t2 − s) − (t1 − s) ] ds + (t2 − s)α−1 ds Γ(α) 0 Γ(α) t1  1 L3 (tα−1 − tα−1 )Γ(β) 2 1 + (1 − s)α−γ0 −1 ds |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0  q 

p

− tα−1 )Γ(β)  1 L4 (tα−1 2 1 |Δ|Γ(β − δ0 ) Γ(β − γi ) i=1  1  s (s − τ )β−γi −1 dτ dHi (s) × 0 0  p  α−1  Γ(β)  1 L4 (tα−1 − t ) 2 1 + sβ−γi −1 dHi (s) |Δ| Γ(β − γ ) i 0 i=1   1 1 × (1 − s)β−δ0 −1 ds Γ(β − δ0 ) 0  p   − tα−1 )  Γ(β) 1 β−γi −1 L3 (tα−1 2 1 + s dHi (s) |Δ| Γ(β − γi ) 0 i=1  q  1  s   1 α−δ −1 i × (s − τ ) dτ dKi (s) Γ(α − δi ) 0 0 i=1  L3 Γ(β) α−1 α−1 α α = (t − t1 ) + L3 (t2 − t1 ) Γ(α + 1) 2 |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   Γ(β)  1 1 β−γi −1 s dHi (s) + |Δ| i=1 Γ(β − γi ) 0  q  1   1 α−δi × s dKi (s) Γ(α − δi + 1) 0 i=1   1 p  1 Γ(β) α−1 α−1 sβ−γi dHi (s) + L4 (t2 − t1 ) |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1) 0  1  p  Γ(β) 1 + sβ−γi −1 dHi (s) |Δ|Γ(β − δ0 + 1) Γ(β − γi ) +

i=1

=

0

L3 α−1 (tα − tα − tα−1 ). 1 ) + (L3 M11 + L4 M10 )(t2 1 Γ(α + 1) 2

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Hence, we infer |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )| → 0,

as t2 → t1 , uniformly with respect to

(u, v) ∈ Ω. In a similar manner, for (u, v) ∈ Ω and t1 , t2 ∈ [0, 1] with t1 < t2 , we obtain |Q2 (u, v)(t2 ) − Q2 (u, v)(t1 )|

 Γ(α) L4 β β β−1 β−1 (t − t1 ) + L4 (t2 − t1 ) ≤ Γ(β + 1) 2 |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q   Γ(α)  1 1 α−δi −1 s dKi (s) + |Δ| Γ(α − δi )  p 

i=1

0

 1  1 β−γ i × s dHi (s) Γ(β − γi + 1) 0 i=1   1 q  1 Γ(α) β−1 β−1 sα−δi dKi (s) + L3 (t2 − t1 ) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0  1  q  Γ(α) 1 α−δi −1 + s dKi (s) |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0

=

L4 (tβ − tβ1 ) + (L4 M12 + L3 M9 )(tβ−1 − tβ−1 ). 2 1 Γ(β + 1) 2

So, we deduce |Q2 (u, v)(t2 ) − Q2 (u, v)(t1 )| → 0,

as t2 → t1 , uniformly with respect to

(u, v) ∈ Ω. Then Q1 (Ω) and Q2 (Ω) is equicontinuous, and so Q(Ω) is also equicontinuous. Therefore, by the Arzela-Ascoli theorem, we conclude that Q(Ω) is relatively compact, and then Q is compact. We infer that operator Q is completely continuous. Next, we will show that the set U = {(u, v) ∈ Y, (u, v) = νQ(u, v), 0 < ν < 1} is bounded. Let (u, v) ∈ U , that is (u, v) = νQ(u, v). Then for any t ∈ [0, 1] we get u(t) = νQ1 (u, v)(t), v(t) = νQ2 (u, v)(t). We denote the

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following functions θ1 σ1 u(s)| + a4 |I0+ v(s)|, Fuv (s) = a0 + a1 |u(s)| + a2 |v(s)| + a3 |I0+ θ2 σ2 u(s)| + b4 |I0+ v(s)|, Guv (s) = b0 + b1 |u(s)| + b2 |v(s)| + b3 |I0+

s ∈ [0, 1], s ∈ [0, 1].

By (J3), we find |u(t)| ≤ |Q1 (u, v)(t)| ≤

1 Γ(α)

α−1

+

 0

t

(t − s)α−1 Fuv (s) ds

Γ(β) t |Δ|Γ(α − γ0 )Γ(β − δ0 )

 0

1

(1 − s)α−γ0 −1 Fuv (s) ds

p 

α−1

Γ(β) 1 t |Δ|Γ(β − δ0 ) i=1 Γ(β − γi )  1  s (s − τ )β−γi −1 Guv (τ ) dτ dHi (s) ×

+

0

0

  p  t Γ(β) 1 β−γi −1 + s dHi (s) |Δ| i=1 Γ(β − γi ) 0   1 1 β−δ0 −1 × (1 − s) Guv (s) ds Γ(β − δ0 ) 0  p   tα−1  Γ(β) 1 β−γi −1 + s dHi (s) |Δ| i=1 Γ(β − γi ) 0  q  1  s   1 (s − τ )α−δi −1 Fuv (τ ) dτ dKi (s) × Γ(α − δi ) 0 0 i=1  a4 a3 ≤ a0 + a1 u + a2 v + u + v Γ(θ1 + 1) Γ(σ1 + 1)  tα tα−1 Γ(β) × + Γ(α + 1) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   tα−1  Γ(β) 1 β−γi −1 s dHi (s) + |Δ| i=1 Γ(β − γi ) 0  q  1   1 α−δi s dKi (s) × Γ(α − δi + 1) 0 i=1 α−1



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b4 b3 u + v Γ(θ2 + 1) Γ(σ2 + 1)   p 1 β−γ tα−1 Γ(β)  1 i × s dHi (s) |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1) 0   p  tα−1 Γ(β) 1 β−γi −1 + s dHi (s) , |Δ|Γ(β − δ0 + 1) i=1 Γ(β − γi ) 0

 + b0 + b1 u + b2 v +

∀ t ∈ [0, 1]. Therefore, we deduce  u ≤ a0 + a1 u + a2 v +

a4 a3 u + v M7 Γ(θ1 + 1) Γ(σ1 + 1)  b3 b4 + b0 + b1 u + b2 v + u + v M10 . Γ(θ2 + 1) Γ(σ2 + 1) (6.45)

In a similar manner, we obtain  t 1 |v(t)| ≤ |Q2 (u, v)(t)| ≤ (t − s)β−1 Guv (s) ds Γ(β) 0  1 tβ−1 Γ(α) (1 − s)β−δ0 −1 Guv (s) ds + |Δ|Γ(α − γ0 )Γ(β − δ0 ) 0 q

tβ−1 Γ(α)  1 |Δ|Γ(α − γ0 ) i=1 Γ(α − δi )  1  s α−δi −1 × (s − τ ) Fuv (τ ) dτ dKi (s) +

0



0

  q tβ−1  Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0   1 1 (1 − s)α−γ0 −1 Fuv (s) ds × Γ(α − γ0 ) 0  q   tβ−1  Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1  s   1 β−γi −1 (s − τ ) Guv (τ ) dτ dHi (s) × Γ(β − γi ) 0 0 i=1

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≤

 a0 + a1 u + a2 v +

a4 a3 u + v Γ(θ1 + 1) Γ(σ1 + 1)   1 q tβ−1 Γ(α)  1 α−δi × s dKi (s) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0



  q  tβ−1 Γ(α) 1 α−δi −1 + s dKi (s) |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0  b4 b3 u + v + b0 + b1 u + b2 v + Γ(θ2 + 1) Γ(σ2 + 1)  tβ tβ−1 Γ(α) × + Γ(β + 1) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)  q   tβ−1  Γ(α) 1 α−δi −1 + s dKi (s) |Δ| i=1 Γ(α − δi ) 0  p  1   1 β−γ i × , ∀ t ∈ [0, 1]. s dHi (s) Γ(β − γi + 1) 0 i=1 Then we have  v ≤ a0 + a1 u + a2 v +

a4 a3 u + v M9 Γ(θ1 + 1) Γ(σ1 + 1)  b4 b3 u + v M8 . + b0 + b1 u + b2 v + Γ(θ2 + 1) Γ(σ2 + 1) (6.46)

By (6.45) and (6.46), we infer (u, v)Y = u + v ≤ a0 (M7 + M9 ) + b0 (M8 + M10 )  a3 (M7 + M9 ) + b1 (M8 + M10 ) + a1 (M7 + M9 ) + Γ(θ1 + 1)  b3 (M8 + M10 ) u + Γ(θ2 + 1)  a4 + a2 (M7 + M9 ) + (M7 + M9 ) + b2 (M8 + M10 ) Γ(σ1 + 1)  b4 (M8 + M10 ) v + Γ(θ2 + 1) = a0 (M7 + M9 ) + b0 (M8 + M10 ) + M13 u + M14 v ≤ a0 (M7 + M9 ) + b0 (M8 + M10 ) + Ξ1 (u, v)Y .

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Because Ξ1 < 1, we find (u, v)Y ≤ [a0 (M7 + M9 ) + b0 (M8 + M10 )](1 − Ξ1 )−1 ,

∀ (u, v) ∈ U.

So, we deduce that the set U is bounded. By using Theorem 1.2.5, we conclude that operator Q has at least one fixed point, which is a solution for problem (6.31), (6.32).  Theorem 6.3.3. Assume that (J1), (J2) and (J4) There exist the functions ψ1 , ψ2 ∈ C([0, 1], [0, ∞)) such that |f (t, x1 , x2 , x3 , x4 )| ≤ ψ1 (t),

|g(t, x1 , x2 , x3 , x4 )| ≤ ψ2 (t),

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, 1 1 + L2 M6 Γ(β+1) < 1, then problem (6.31), (6.32) hold. If Ξ2 := L1 M5 Γ(α+1) has at least one solution on [0, 1].

Proof. We fix r1 > 0 such that r1 ≥ (M7 + M9 )ψ1  + (M8 + M10 )ψ2 . We consider the set B r1 = {(x, y) ∈ Y, (x, y)Y ≤ r1 }, and we introduce the operators D = (D1 , D2 ) : B r1 → Y and E = (E1 , E2 ) : B r1 → Y , where D1 , D2 , E1 , E2 : B r1 → X are defined by  t 1 D1 (u, v)(t) = (t − s)α−1 fˆuv (s) ds, Γ(α) 0  1 tα−1 Γ(β) (1 − s)α−γ0 −1 fˆuv (s) ds E1 (u, v)(t) = ΔΓ(α − γ0 )Γ(β − δ0 ) 0 p

1 tα−1 Γ(β)  ΔΓ(β − δ0 ) i=1 Γ(β − γi )  1  s (s − τ )β−γi −1 gˆuv (τ ) dτ dHi (s) × −

0

0



  1 p tα−1  Γ(β) + sβ−γi −1 dHi (s) Δ Γ(β − γi ) 0 i=1   1 1 (1 − s)β−δ0 −1 gˆuv (s) ds × Γ(β − δ0 ) 0  p   1 tα−1  Γ(β) β−γi −1 − s dHi (s) Δ Γ(β − γi ) 0 i=1

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  1  s 1 α−δi −1 ˆ × (s − τ ) fuv (τ ) dτ dKi (s) , Γ(α − δi ) 0 0 i=1  t 1 D2 (u, v)(t) = − (t − s)β−1 gˆuv (s) ds Γ(β) 0  1 tβ−1 Γ(α) (1 − s)β−δ0 −1 gˆuv (s) ds E2 (u, v)(t) = ΔΓ(α − γ0 )Γ(β − δ0 ) 0 q 

q

tβ−1 Γ(α)  1 ΔΓ(α − γ0 ) i=1 Γ(α − δi )  1  s α−δi −1 ˆ × (s − τ ) fuv (τ ) dτ dKi (s) −

0

0



  1 q tβ−1  Γ(α) α−δi −1 + s dKi (s) Δ Γ(α − δi ) 0 i=1   1 1 × (1 − s)α−γ0 −1 fˆuv (s) ds Γ(α − γ0 ) 0  q   1 tβ−1  Γ(α) − sα−δi −1 dKi (s) Δ Γ(α − δ ) i 0 i=1  p   1  s  1 β−γi −1 × (s − τ ) gˆuv (τ ) dτ dHi (s) , Γ(β − γi ) 0 0 i=1 (6.47) for all t ∈ [0, 1] and (u, v) ∈ B r1 . So Q1 = D1 + E1 , Q2 = D2 + E2 and Q = D + E. By using (J4), we find for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 as in the proof of Theorem 6.3.1 that D(u1 , v1 ) + E(u2 , v2 )Y ≤ D(u1 , v1 )Y + E(u2 , v2 )Y = D1 (u1 , v1 ) + D2 (u1 , v1 ) + E1 (u2 , v2 ) + E2 (u2 , v2 ) ≤

1 1 ψ1  + ψ2  + (M11 ψ1  + M10 ψ2 ) Γ(α + 1) Γ(β + 1) + (M9 ψ1  + M12 ψ2 )

= (M7 + M9 )ψ1  + (M8 + M10 )ψ2  ≤ r1 . So, D(u1 , v1 ) + E(u2 , v2 ) ∈ B r1 for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 .

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The operator D is a contraction, because D(u1 , v1 ) − D(u2 , v2 )Y = D1 (u1 , v1 ) − D1 (u2 , v2 ) + D2 (u1 , v1 ) − D2 (u2 , v2 )  1 1 + L2 M6 ≤ L1 M5 Γ(α + 1) Γ(β + 1) × (u1 − u2  + v1 − v2 ) = Ξ2 (u1 , v1 ) − (u2 , v2 )Y , for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 , and Ξ2 < 1. Because the functions f and g are continuous, we obtain that operator E is continuous on B r1 . We show next that E is compact. The functions from E are uniformly bounded on B r1 , because E(u, v)Y = E1 (u, v) + E2 (u, v) ≤ (M11 + M9 )ψ1  +(M10 + M12 )ψ2 ,

∀ (u, v) ∈ B r1 .

We prove next that the functions from E(B r1 ) are equicontinuous. We denote by  Ψr1 = sup |f (t, u, v, x, y)|, t ∈ [0, 1], |u| ≤ r1 , |v| ≤ r1 ,  r1 r1 , |y| ≤ |x| ≤ , Γ(θ1 + 1) Γ(σ1 + 1)

Θ r1

 = sup |g(t, u, v, x, y)|, t ∈ [0, 1], |u| ≤ r1 , |v| ≤ r1 , |x| ≤

 r1 r1 , |y| ≤ . Γ(θ2 + 1) Γ(σ2 + 1)

Then for (u, v) ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t2 , we deduce |E1 (u, v)(t2 ) − E1 (u, v)(t1 )| − tα−1 )Γ(β) (tα−1 2 1 ≤ |Δ|Γ(α − γ0 )Γ(β − δ0 ) p

+

 0

1

(1 − s)α−γ0 −1 Ψr1 ds

− tα−1 )Γ(β)  1 (tα−1 2 1 |Δ|Γ(β − δ0 ) Γ(β − γi ) i=1

(6.48)

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 ×

1



0

0

s

β−γi −1

(s − τ )

Θr1 dτ

dHi (s)

 p   Γ(β)  1 tα−1 − tα−1 2 1 β−γi −1 + s dHi (s) |Δ| Γ(β − γi ) 0 i=1   1 1 β−δ0 −1 × (1 − s) Θr1 ds Γ(β − δ0 ) 0  p   Γ(β)  1 − tα−1 tα−1 2 1 β−γ −1 i + s dHi (s) |Δ| Γ(β − γi ) 0 i=1  q  1  s   1 × (s − τ )α−δi −1 Ψr1 dτ dKi (s) Γ(α − δi ) 0

i=1

  α−1 = Ψr1 tα−1 − t 2 1



0

Γ(β) |Δ|Γ(α − γ0 + 1)Γ(β − δ0 )  p   Γ(β)  1 1 β−γ −1 i s dHi (s) + |Δ| i=1 Γ(β − γi ) 0  p  1   1 × sα−δi dKi (s) Γ(α − δi + 1) 0 i=1  p    1 Γ(β) α−1 α−1 + Θ r 1 t2 − t1 |Δ|Γ(β − δ0 ) i=1 Γ(β − γi + 1)  ×

dHi (s) +

p

 Γ(β) 1 |Δ|Γ(β − δ0 + 1) i=1 Γ(β − γi ) 0   1 sβ−γi −1 dHi (s) × 1

0

s

β−γi

    α−1 α−1 α−1 + M t , = M11 Ψr1 tα−1 − t Θ − t 10 r 1 2 1 2 1   tβ−1 − tβ−1 Γ(α) 2 1 |E2 (u, v)(t2 ) − E2 (u, v)(t1 )| ≤ |Δ|Γ(α − γ0 )Γ(β − δ0 )    1 tβ−1 Γ(α) − tβ−1 2 1 × (1 − s)β−δ0 −1 Θr1 ds + |Δ|Γ(α − γ0 ) 0

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×

q  i=1

×

1 Γ(α − γ0 )

tβ−1 − tβ−1 1 + 2 |Δ| ×

345

 1  s 1 α−δi −1 (s − τ ) Ψr1 dτ dKi (s) Γ(α − δi ) 0 0

tβ−1 − tβ−1 1 + 2 |Δ| 

page 345

 p  i=1



q  i=1



1

0



  Γ(α) 1 α−δi −1 s dKi (s) Γ(α − δi ) 0 α−γ0 −1

(1 − s)

q  i=1

Ψr1 ds

  Γ(α) 1 α−δi −1 s dKi (s) Γ(α − δi ) 0

 1  s  1 β−γi −1 (s − τ ) Θr1 dτ dHi (s) Γ(β − γi ) 0 0

  β−1 = Ψr1 tβ−1 − t 2 1



 1 q  Γ(α) 1 sα−δi dKi (s) |Δ|Γ(α − γ0 ) i=1 Γ(α − δi + 1) 0

  q  1 Γ(α) 1 α−δi −1 + s dKi (s) |Δ|Γ(α − γ0 + 1) i=1 Γ(α − δi ) 0  + Θ r1

tβ−1 2

−

tβ−1 1





Γ(α) |Δ|Γ(α − γ0 )Γ(β − δ0 + 1)

 q   Γ(α)  1 1 α−δ −1 i s dKi (s) + |Δ| i=1 Γ(α − δi ) 0  p  1   1 β−γ i s dHi (s) × Γ(β − γi + 1) 0 i=1     = M9 Ψr1 tβ−1 + M12 Θr1 tβ−1 . − tβ−1 − tβ−1 2 1 2 1 Therefore, we infer |E1 (u, v)(t2 ) − E1 (u, v)(t1 )| → 0,

|E2 (u, v)(t2 ) − E2 (u, v)(t1 )| → 0,

as t2 → t1 uniformly with respect to (u, v) ∈ B r1 . Then E1 (B r1 ) and E2 (B r1 ) are equicontinuous, and so E(B r1 ) is also equicontinuous. By applying the Arzela-Ascoli theorem, we conclude that the set E(B r1 ) is relatively compact. Hence, E is a compact operator on B r1 . By using

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Theorem 1.2.3, we deduce that there exists a fixed point of operator D + E(= Q), which is a solution of problem (6.31), (6.32).  Theorem 6.3.4. Suppose that (J1), (J2) and (J4) hold. If Ξ3 := L1 M5 (M9 + M11 ) + L2 M6 (M10 + M12 ) < 1, then problem (6.31), (6.32) has at least one solution (u, v) on [0, 1]. Proof. We consider again a positive number r1 ≥ (M7 + M9 )ψ1  + (M8 + M10 )ψ2  and the operators D and E defined on B r1 given by (6.47). As in the proof of Theorem 6.3.3, we have D(u1 , v1 ) + E(u2 , v2 ) ∈ B r1 for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 . The operator E is a contraction, because E(u1 , v1 ) − E(u2 , v2 )Y = E1 (u1 , v1 ) − E1 (u2 , v2 ) + E2 (u1 , v1 ) − E2 (u2 , v2 ) ≤ (L1 M5 M11 + L2 M6 M10 )(u1 , v1 ) − (u2 , v2 )Y + (L1 M5 M9 + L2 M6 M12 )(u1 , v1 ) − (u2 , v2 )Y = (L1 M5 (M9 + M11 ) + L2 M6 (M10 + M12 ))(u1 , v1 ) − (u2 , v2 )Y = Ξ3 (u1 , v1 ) − (u2 , v2 )Y , for all (u1 , v1 ), (u2 , v2 ) ∈ B r1 , with Ξ3 < 1. In what follows, the continuity of functions f and g implies that operator D is continuous on B r1 . We prove now that D is a compact operator. The functions from D(B r1 ) are uniformly bounded, because 1 D(u, v)Y = D1 (u, v) + D2 (u, v) ≤ ψ1  Γ(α + 1) +

1 ψ2 , Γ(β + 1)

∀ (u, v) ∈ B r1 .

Now, we show that the functions from D(B r1 ) are equicontinuous. By using Ψr1 and Θr1 defined by (6.48), we deduce that for (u, v) ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t2 that Ψ r1 (tα − tα |D1 (u, v)(t2 ) − D1 (u, v)(t1 )| ≤ 1 ), Γ(α + 1) 2 |D2 (u, v)(t2 ) − D2 (u, v)(t1 )| ≤

Θ r1 (tβ − tβ1 ). Γ(β + 1) 2

Therefore, we conclude |D1 (u, v)(t2 ) − D1 (u, v)(t1 )| → 0,

|D2 (u, v)(t2 ) − D2 (u, v)(t1 )| → 0,

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347

as t2 → t1 uniformly with respect to (u, v) ∈ B r1 . We infer that D1 (B r1 ) and D2 (B r1 ) are equicontinuous, and so D(B r1 ) is equicontinuous. By using the Arzela-Ascoli theorem, we deduce that the set D(B r1 ) is relatively compact. Then D is a compact operator on B r1 . By using Theorem 1.2.3, we conclude that there exists a fixed point of operator D + E(= Q), which is a solution of problem (6.31), (6.32).  Theorem 6.3.5. Assume that (J1) and (J5) The functions f, g : [0, 1] × R4 → R are continuous and there exist the constants ci ≥ 0, i = 0, . . . , 4 with at least one nonzero, the constants di ≥ 0, i = 0, . . . , 4 with at least one nonzero, and li , mi ∈ (0, 1), i = 1, . . . , 4 such that |f (t, x1 , x2 , x3 , x4 )| ≤ c0 +

4 

ci |xi |li ,

i=1

|g(t, x1 , x2 , x3 , x4 )| ≤ d0 +

4 

di |xi |mi ,

i=1

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, hold. Then problem (6.31), (6.32) has at least one solution. Proof. Let B R = {(x, y) ∈ Y, (x, y)Y ≤ R}, where

1

1

R ≥ max 20c0 M7 , (20c1 M7 ) 1−l1 , (20c2 M7 ) 1−l2 , 

20c3 M7 (Γ(θ1 + 1))l3

1 1−l

3

 ,

20c4 M7 (Γ(σ1 + 1))l4

1

1 1−l

4

,

1

20d0 M10 , (20d1 M10 ) 1−m1 , (20d2 M10 ) 1−m2 , 1 1   1−m 1−m 3 4 20d4 M10 20d3 M10 , , m m 3 4 (Γ(θ2 + 1)) (Γ(σ2 + 1)) 1

1

20c0 M9 , (20c1 M9 ) 1−l1 , (20c2 M9 ) 1−l2 , 1 1  1−l 1−l  3 4 20c4 M9 20c3 M9 , , (Γ(θ1 + 1))l3 (Γ(σ1 + 1))l4

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1

20d0 M8 , (20d1 M8 ) 1−m1 , (20d2 M8 ) 1−m2 ,  1 1   1−m 1−m 3 4 20d4 M8 20d3 M8 , . (Γ(θ2 + 1))m3 (Γ(σ2 + 1))m4 We prove that Q : B R → B R . For (u, v) ∈ B R , we have

 R l3 R l4 M7 + c 4 (Γ(θ1 + 1))l3 (Γ(σ1 + 1))l4   R Rm3 Rm4 M10 ≤ , + d0 + d1 Rm1 + d2 Rm2 + d3 + d 4 (Γ(θ2 + 1))m3 (Γ(σ2 + 1))m4 2   l3 l4 R R |Q2 (u, v)(t)| ≤ c0 + c1 Rl1 + c2 Rl2 + c3 + c4 M9 (Γ(θ1 + 1))l3 (Γ(σ1 + 1))l4   R Rm3 Rm4 m1 m2 + d0 + d1 R + d2 R + d3 + d4 M8 ≤ , (Γ(θ2 + 1))m3 (Γ(σ2 + 1))m4 2 

|Q1 (u, v)(t)| ≤

c0 + c1 Rl1 + c2 Rl2 + c3

for all t ∈ [0, 1]. Then we obtain Q(u, v))Y = Q1 (u, v) + Q2 (u, v) ≤ R,

∀ (u, v) ∈ B R ,

which implies that Q(B R ) ⊂ B R . By using the fact that the functions f and g are continuous, we deduce that operator Q is continuous on B R . Besides, the functions from Q(B R ) are uniformly bounded and equicontinuous. Indeed, by using the notations (6.48) with r1 replaced by R, we find for any (u, v) ∈ B R and t1 , t2 ∈ [0, 1], t1 < t2 that |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )| ≤

ΨR (tα − tα 1) Γ(α + 1) 2 − tα−1 ), + (ΨR M11 + ΘR M10 )(tα−1 2 1

|Q2 (u, v)(t2 ) − Q2 (u, v)(t1 )| ≤

ΘR (tβ − tβ1 ) Γ(β + 1) 2 + (ΨR M9 + ΘR M12 )(tβ−1 − tβ−1 ). 2 1

Therefore, we obtain |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )| → 0, |Q2 (u, v)(t2 ) − Q2 (u, v)(t1 )| → 0,

as t2 → t1 ,

uniformly with respect to (u, v) ∈ B R . By the Arzela-Ascoli theorem, we conclude that Q(B R ) is relatively compact, and then Q is a compact operator. By using Theorem 1.2.4, we infer that operator

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Existence of Solutions for Riemann–Liouville Fractional BVPs

Q has at least one fixed point (u, v) in B R , which is a solution of our problem (6.31), (6.32).  Theorem 6.3.6. Suppose that (J1) and (J6) The functions f, g : [0, 1] × R4 → R are continuous and there exist pi ≥ 0, i = 0, . . . , 4 with at least one nonzero, qi ≥ 0, i = 0, . . . , 4 with at least one nonzero, and nondecreasing functions ξi , ηi ∈ C([0, ∞), [0, ∞)) i = 1, . . . , 4 such that |f (t, x1 , x2 , x3 , x4 )| ≤ p0 +

4 

pi ξi (|xi |),

i=1

|g(t, x1 , x2 , x3 , x4 )| ≤ q0 +

4 

qi ηi (|xi |),

i=1

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, hold. If there exists Ξ0 > 0 such that  p0 + p1 ξ1 (Ξ0 ) + p2 ξ2 (Ξ0 )  + p3 ξ3

Ξ0 Γ(θ1 + 1)



+ p4 ξ4

 + q0 + q1 η1 (Ξ0 ) + q2 η2 (Ξ0 )  + q3 η3

Ξ0 Γ(θ2 + 1)



+ q4 η4

Ξ0 Γ(σ1 + 1)

Ξ0 Γ(σ2 + 1)

(M7 + M9 ) (6.49) (M8 + M10 ) < Ξ0 ,

then problem (6.31), (6.32) has at least one solution on [0, 1]. Proof. We consider the set B Ξ0 = {(x, y) ∈ Y, (x, y)Y ≤ Ξ0 }, where Ξ0 is given in the theorem. We will show that Q : B Ξ0 → B Ξ0 . For (u, v) ∈ B Ξ0 and t ∈ [0, 1] we obtain  |Q1 (u, v)(t)| ≤

p0 + p1 ξ1 (Ξ0 ) + p2 ξ2 (Ξ0 )

 Ξ0 M7 Γ(σ1 + 1)      Ξ0 Ξ0 + q0 + q1 η1 (Ξ0 ) + q2 η2 (Ξ0 ) + q3 η3 + q4 η4 M10 , Γ(θ2 + 1) Γ(σ2 + 1) 

+ p 3 ξ3

Ξ0 Γ(θ1 + 1)





+ p 4 ξ4

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 |Q2 (u, v)(t)| ≤ 

p0 + p1 ξ1 (Ξ0 ) + p2 ξ2 (Ξ0 ) 

 Ξ0 M9 Γ(σ1 + 1)      Ξ0 Ξ0 + q0 + q1 η1 (Ξ0 ) + q2 η2 (Ξ0 ) + q3 η3 + q4 η4 M8 , Γ(θ2 + 1) Γ(σ2 + 1) + p 3 ξ3

Ξ0 Γ(θ1 + 1)



+ p 4 ξ4

and then for all (u, v) ∈ B Ξ0 we find   Q(u, v)Y ≤ p0 + p1 ξ1 (Ξ0 ) + p2 ξ2 (Ξ0 ) + p3 ξ3  + p4 ξ4

Ξ0 Γ(σ1 + 1)



Ξ0 Γ(θ1 + 1)

(M7 + M9 )

  + q0 + q1 η1 (Ξ0 ) + q2 η2 (Ξ0 ) + q3 η3  + q4 η4

Ξ0 Γ(σ2 + 1)





Ξ0 Γ(θ2 + 1)



(M8 + M10 ) < Ξ0 .

Hence, Q(B Ξ0 ) ⊂ B Ξ0 . Using a similar approach as in the proof of Theorem 6.3.5, we can show that operator Q is completely continuous. We suppose now that there exists (u, v) ∈ ∂BΞ0 such that (u, v) = νQ(u, v) for some ν ∈ (0, 1). Arguing as above we deduce (u, v)Y ≤ Q(u, v)Y < Ξ0 , which is a contradiction, because (u, v) ∈ ∂BΞ0 . Then by using Theorem 1.2.6, we conclude that operator Q has a fixed point (u, v) ∈ B Ξ0 , and so problem (6.31), (6.32) has at least one solution.  6.3.3

Examples

Let α = 32 (n = 2), β = 73 (m = 3), θ1 = 14 , σ1 = 65 , θ2 = 17 4 , σ2 = 1 1 3 8 1 1 , p = 1, q = 2, γ = , γ = , δ = , δ = , δ = , 0 1 0 1 2 3 6 4 7 5 3 H1 (t) = {0, t ∈ [0, 12 ); 3, t ∈ [ 12 , 1]}, K1 (t) = −t2 , t ∈ [0, 1], K2 (t) = {0, t ∈ [0, 13 ); 4, t ∈ [ 13 , 1]}. We consider the system of fractional differential equations

3/2 1/4 6/5 D0+ u(t) + f (t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, t ∈ (0, 1), (6.50) 7/3 17/4 1/3 D0+ v(t) + g(t, u(t), v(t), I0+ u(t), I0+ v(t)) = 0, t ∈ (0, 1),

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with the boundary conditions  ⎧ 1 1/6 3/4 ⎪ ⎪ u(0) = 0, D u(1) = 3D v , 0+ 0+ ⎨ 2   1 ⎪ 1 8/7 1/5 1/3 ⎪ ⎩v(0) = v  (0) = 0, D0+ v(1) = −2 tD0+ u(t) dt + 4D0+ u . 3 0 (6.51) We obtain Δ ≈ −4.92715202 = 0. So assumption (J1) is satisfied. In addition, we have M1 ≈ 2.10326265, M2 ≈ 1.90760368, M3 ≈ 1.02839972, M4 ≈ 2.11984652, M5 = M1 , M6 = M4 , M7 ≈ 1.81109405, M10 ≈ 0.68108088, M9 ≈ 0.9999811, M8 ≈ 1.12515265, M11 ≈ 1.05884127, M12 ≈ 0.76520198. Example 1. We consider the functions 1 t |x2 | arctan x1 + f (t, x1 , x2 , x3 , x4 ) = √ − 4 3 10 (t + 2) (1 + |x2 |) 9+t + g(t, x1 , x2 , x3 , x4 ) =

1 t2 sin2 x3 − cos x4 , 3(t + 8) t + 12

|x1 | 1 3t − + sin x2 t2 + 4 6(2 + |x1 |) 15 +

t 1 cos2 x3 − arctan x4 , t + 24 12

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4. We find the inequalities |f (t, x1 , x2 , x3 , x4 ) − f (t, y1 , y2 , y3 , y4 )| ≤

1 1 1 |x1 − y1 | + |x2 − y2 | + |x3 − y3 | 10 16 12 4

+

1 1  |x4 − y4 | ≤ |xi − yi |, 13 10 i=1

|g(t, x1 , x2 , x3 , x4 ) − g(t, y1 , y2 , y3 , y4 )| ≤

1 1 2 |x1 − y1 | + |x2 − y2 | + |x3 − y3 | 12 15 25 4

+

1  1 |x4 − y4 | ≤ |xi − yi |, 12 12 i=1

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1 1 for all t ∈ [0, 1], xi , yi ∈ R, i = 1, . . . , 4. So we have L1 = 10 , L2 = 12 and Ξ = L1 M5 (M7 + M9 ) + L2 M6 (M8 + M10 ) ≈ 0.91032 < 1. Therefore, assumption (J2) is satisfied, and by Theorem 6.3.1, we deduce that problem (6.50), (6.51) has a unique solution (u(t), v(t)), t ∈ [0, 1].

Example 2. We consider the functions  t+2 1 2 sin t + cos x1 f (t, x1 , x2 , x3 , x4 ) = 2 t +5 5 − g(t, x1 , x2 , x3 , x4 ) =

t 1 1 x2 − arctan x3 + sin x4 , (t + 5)2 6 7

e−t 1 1 1 + cos2 x2 − sin x3 + arctan x4 , 2 + t3 4 5 9

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4. Because we have 1 11 1 1 + |x2 | + |x3 | + |x4 |, 10 25 6 7 3 1 1 |g(t, x1 , x2 , x3 , x4 )| ≤ + |x3 | + |x4 |, 4 5 9 for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, the assumption (J3) is satisfied with 1 1 1 3 1 a0 = 11 10 , a1 = 0, a2 = 25 , a3 = 6 , a4 = 7 , b0 = 4 , b1 = b2 = 0, b3 = 5 , 1 b4 = 9 . In addition, we obtain M13 ≈ 0.52715168, M14 ≈ 0.70166538 and Ξ1 = max{M13 , M14 } = M14 < 1. Then by Theorem 6.3.2, we conclude that problem (6.50), (6.51) has at least one solution (u(t), v(t)), t ∈ [0, 1]. |f (t, x1 , x2 , x3 , x4 )| ≤

Example 3. We consider the functions 1 1 arctan |x3 |1/2 , f (t, x1 , x2 , x3 , x4 ) = − |x2 |3/4 + 5 3(1 + t2 ) g(t, x1 , x2 , x3 , x4 ) =

e−t 1 4/5 2/3 − x1 + sin x4 , 1 + t3 3

for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4. Because we obtain 1 1 |x2 |3/4 + |x3 |1/2 , 5 3 1 |g(t, x1 , x2 , x3 , x4 )| ≤ 1 + |x1 |4/5 + |x4 |2/3 , 3 for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, then assumption (J5) is satisfied with c0 = c1 = 0, c2 = 15 , c2 = 13 , c4 = 0, d0 = 1, d1 = 13 , d2 = d3 = 0, d4 = 1, l2 = 34 , l3 = 12 , m1 = 45 , m4 = 23 . Therefore, by Theorem 6.3.5, we deduce that problem (6.50), (6.51) has at least one solution (u(t), v(t)), t ∈ [0, 1]. |f (t, x1 , x2 , x3 , x4 )| ≤

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353

Example 4. We consider the functions f (t, x1 , x2 , x3 , x4 ) =

1/3

e−t x41 t2 x4 t3 + − , 25 20(1 + x22 ) 10

1 − t2 2 (1 − t)4 1 2/5 − x2 − x3 , 20 15 25 for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4. Because we have g(t, x1 , x2 , x3 , x4 ) =

1 1 1 + |x1 |4 + |x4 |1/3 , 25 20 10 1 1 1 |g(t, x1 , x2 , x3 , x4 )| ≤ + |x2 |2 + |x3 |2/5 , 20 15 25 for all t ∈ [0, 1], xi ∈ R, i = 1, . . . , 4, then assumption (J6) is satisfied 1 1 1 1 1 , p1 = 20 , p2 = p3 = 0, p4 = 10 , q0 = 20 , q1 = 0, q2 = 15 , with p0 = 25 1 4 1/3 2 2/5 q3 = 25 , q4 = 0, ξ1 (x) = x , ξ4 (x) = x , η2 (x) = x , η3 (x) = x 1 for x ≥ 0. For Ξ0 = 1, the condition (6.49) is satisfied, because ( 25 + 1 1 1 1 1 1 1 1/3 2/5 + ( ) )(M + M ) + ( + + ( ) )(M + M ) ≈ 7 9 8 10 20 10 Γ(11/5) 20 15 25 Γ(21/4) 0.75328 < 1. Then by Theorem 6.3.6, we conclude that problem (6.50), (6.51) has at least one solution (u(t), v(t)), t ∈ [0, 1]. |f (t, x1 , x2 , x3 , x4 )| ≤

Remark 6.3.1. The results presented in this section will be published in [84].

b2530   International Strategic Relations and China’s National Security: World at the Crossroads

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Chapter 7

Existence of Solutions for Caputo Fractional Boundary Value Problems

In this chapter, we study the existence of solutions for some Caputo fractional differential equations and inclusions, and systems of Caputo fractional differential equations subject to nonlocal boundary conditions which contain Riemann-Stieltjes integrals. 7.1

Sequential Caputo Fractional Differential Equations and Inclusions with Nonlocal Boundary Conditions

In this section, we consider the nonlinear Caputo type sequential fractional integro-differential equation and inclusion: p q α−1 α + λcD0+ )u(t) = f (t, u(t),c D0+ u(t), I0+ u(t)), (cD0+ q α−1 α (cD0+ + λcD0+ )u(t) ∈ F (t, u(t), I0+ u(t)),

t ∈ (0, 1),

subject to the nonlocal boundary conditions u(0) = h(u),

u (0) = u (0) = 0,

t ∈ (0, 1),

β aI0+ u(ξ) =



1 0

u(s) dH(s),

(7.1) (7.2)

(7.3)

α where α ∈ (3, 4], p ∈ (0, 1), q > 0, λ > 0, ξ ∈ (0, 1], a ∈ R, β > 0, cD0+ , α−1 c p D0+ and D0+ are Caputo fractional derivatives of orders α, α − 1 and p, q β and I0+ are Riemann-Liouville fractional integrals of order respectively, I0+

c

355

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Boundary Value Problems for Fractional Differential Equations and Systems

q and β, respectively, H is a bounded variation function, h is a nonlinear function defined on C[0, 1], f is a nonlinear function and F is a nonlinear multivalued function which satisfy some assumptions. We give conditions for the nonlinearities f and F , and for the functions h and H, such that the problems (7.1), (7.3) and (7.2), (7.3) have at least one solution. For the proofs of our main theorems, we use the contraction mapping principle (Theorem 1.2.1) and the Krasnosel’skii fixed point theorem for the sum of two operators (Theorem 1.2.3) in the case of fractional equations, and the nonlinear alternative of Leray-Schauder type for Kakutani maps (Theorem 1.2.14) and the Covitz–Nadler fixed point theorem (Theorem 1.2.15) in the case of fractional inclusions. In the last few years, nonlocal boundary value problems for sequential fractional differential equations, integro-differential equations and inclusions, and systems of such equations have been studied by many researchers. In [10], by using the fixed point theory, the authors investigated the existence of solutions for the sequential fractional integro-differential equation β α−1 α + k cD0+ )u(t) = pf (t, u(t)) + qI0+ g(t, u(t)), (cD0+

0 < t < 1,

with the boundary conditions u(0) = 0, u(1) = 0, or u (0) + ku(0) = a, α denotes the u(1) = b, a, b ∈ R, or u(0) = a, u (0) = u (1), a ∈ R, where cD0+ β Caputo fractional derivative of order α ∈ (1, 2], I0+ denotes the RiemannLiouville fractional integral of order β ∈ (0, 1), f, g are given continuous functions, k = 0 and p, q are real constants. The word “sequential” is used in α−1 α +k cD0+ can be written as the composition the sense that the operator cD0+ α−1 of operators cD0+ and D + k. In [16], the authors studied the existence of solutions for the sequential fractional differential equation α−1 α + k cD0+ )x(t) = f (t, x(t)), (cD0+

t ∈ [0, 1],

with the boundary conditions x(0) = 0,

x (0) = 0,

β x(ζ) = aI0+ x(η),

(7.4)

where α ∈ (2, 3], β > 0, 0 < η < ζ < 1, f is a given continuous function, and k, a are appropriate positive real constants. They use the Banach contraction mapping principle, the Krasnosel’skii fixed point theorem and the nonlinear alternative of Leray-Schauder type. In [12], by using some fixed point theorems, the authors investigated the existence of solutions for the Caputo type sequential fractional differential inclusion α−1 α + k cD0+ )x(t) ∈ F (t, x(t)), (cD0+

t ∈ [0, 1],

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with the nonlocal Riemann-Liouville fractional integral boundary conditions (7.4), where α ∈ (2, 3], F : [0, 1] × R → P(R) is a multivalued map, and P(R) is the family of all nonempty subsets of R. In [14], the authors studied the existence of solutions for the sequential fractional differential equations and inclusions q q−1 γ δ + k cD0+ )x(t) = f (t, x(t),c D0+ x(t), I0+ x(t)), (cD0+ q D0+

c

(

+k

q−1 D0+ )x(t)

c

∈ F (t, x(t),

c

γ δ D0+ x(t), I0+ x(t)),

t ∈ [0, 1], t ∈ [0, 1],

supplemented with semi-periodic and nonlocal integro-multi-point boundary conditions involving Riemann-Liouville integral given by x(0) = x(1),

x (0) = 0,

m 

β ai x(ζi ) = λI0+ x(η),

i=1

where q ∈ (2, 3], δ, γ ∈ (0, 1), k > 0, β > 0, 0 < η < ζ1 < · · · < ζm < 1, f : [0, 1]×R3 → R is a given continuous function, F : [0, 1]×R3 → P(R) is a multivalued map, and λ, ai , i = 1, . . . , m are real constants. Some standard fixed point theorems for single-valued and multivalued maps are applied in [14]. 7.1.1

Auxiliary results

We consider the linear sequential fractional differential equation α−1 α + λcD0+ )u(t) = x(t), (cD0+

t ∈ (0, 1),

supplemented with the nonlocal integral boundary conditions  1 β u(0) = u0 , u (0) = u (0) = 0, aI0+ u(ξ) = u(s) dH(s),

(7.5)

(7.6)

0

where x ∈ C[0, 1], α ∈ (3, 4], ξ ∈ (0, 1], λ > 0, a ∈ R, u0 ∈ R, and H is a bounded variation function. In the sequel, we denote by  ξ a Δ= (ξ − s)β−1 (λ2 s2 − 2λs + 2 − 2e−λs ) ds Γ(β) 0  1 (7.7) (λ2 s2 − 2λs + 2 − 2e−λs ) dH(s), − 0

λ2 t2 − 2λt + 2 − 2e−λt γ(t) = , Δ

t ∈ [0, 1],

for

Δ = 0.

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Lemma 7.1.1. If x ∈ C[0, 1] and Δ = 0, then the unique solution u ∈ C 4 [0, 1] of the boundary value problem (7.5), (7.6) is given by 

 e

u(t) = u0

−λt

+ γ(t)

 + γ(t)

1



0

0 s

0

1

e

−λs

a dH(s) − Γ(β)

α−1 e−λ(s−τ ) I0+ x(τ ) dτ

 0

ξ

 β−1 −λs

(ξ − s)

e

ds

dH(s)

  s  ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ x(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ x(s) ds,

(7.8)

0

where Δ and γ are given by (7.7). Proof. Equation (7.5) can equivalently be written as c

α D0+ (u(t) + λD−1 u(t)) = x(t),

t 1 where D−1 u(t) = I0+ u(t) = 0 u(s) ds. Then, by Lemma 1.1.4, the general solution of problem (7.5), (7.6) is  u(t) = −λ

t

0

u(s) ds +

1 Γ(α)



t

0

+ a 0 + a 1 t + a 2 t2 + a 3 t3 ,

(t − s)α−1 x(s) ds t ∈ [0, 1],

(7.9)

where ai ∈ R (i = 0, 1, 2, 3) are unknown arbitrary constants. Differentiating (7.9), we obtain 1 u (t) = −λu(t) + Γ(α − 1) 

 0

t

(t − s)α−2 x(s) ds + a1 + 2a2 t + 3a3 t2 ,

which can alternatively be written as (eλt u(t)) =

eλt Γ(α − 1)

 0

t

(t − s)α−2 x(s) ds + a1 eλt + 2a2 teλt + 3a3 t2 eλt . (7.10)

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Integrating (7.10) from 0 to t, we get c1 c2 u(t) = e−λt c0 + (1 − e−λt ) + 2 (λt − 1 + e−λt ) λ λ c3 2 2 −λt + 3 (λ t − 2λt + 2 − 2e ) λ  t α−1 + e−λ(t−s) I0+ x(s) ds, t ∈ [0, 1],

(7.11)

0

where c0 = u(0), c1 = a1 , c2 = 2a2 , c3 = 3a3 . Using the boundary conditions u(0) = u0 , u (0) = u (0) = 0 in (7.11), we find that c0 = u0 , c1 = 0 and c2 = 0 and consequently (7.11) becomes c3 u(t) = u0 e−λt + 3 (λ2 t2 − 2λt + 2 − 2e−λt ) λ  t α−1 e−λ(t−s) I0+ x(s) ds, t ∈ [0, 1]. (7.12) + 0

1 β Now using the boundary condition aI0+ u(ξ) = 0 u(s) dH(s), we obtain   ξ c3 a (ξ − s)β−1 (λ2 s2 − 2λs + 2 − 2e−λs ) ds λ3 Γ(β) 0   1

−

0

 = u0 

(λ2 s2 − 2λs + 2 − 2e−λs ) dH(s) 1

0 1

e

−λs



s

+ 0

−

a Γ(β)

0



0

ξ

a dH(s) − Γ(β)

 0

ξ

 (ξ − s)

α−1 e−λ(s−τ ) I0+ x(τ ) dτ

(ξ − s)β−1



s 0

β−1 −λs

e

ds

dH(s)

α−1 e−λ(s−τ ) I0+ x(τ ) dτ

ds,

which, in view of the notation Δ = 0 given by (7.7), takes the form    ξ 1 c3 u0 a = e−λs dH(s) − (ξ − s)β−1 e−λs ds λ3 Δ Γ(β) 0 0   s 1 1 α−1 + e−λ(s−τ ) I0+ x(τ ) dτ dH(s) Δ 0 0  s  ξ a α−1 (ξ − s)β−1 e−λ(s−τ ) I0+ x(τ ) dτ ds. − ΔΓ(β) 0 0

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Substituting the above value of c3 /λ3 in (7.12) and using the expression for γ(t) given by (7.7), we obtain     ξ 1 a −λt −λs β−1 −λs + γ(t) u0 e dH(s) − (ξ − s) e ds u(t) = u0 e Γ(β) 0 0  1  s −λ(s−τ ) α−1 e I0+ x(τ ) dτ dH(s) + 0

0

  s ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ x(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ x(s) ds. 

0

The converse of the lemma follows by direct computation. This completes the proof.  7.1.2

Existence of solutions for problem (7.1), (7.3)

In this section, we investigate the existence and uniqueness of solutions for the fractional differential equation (7.1) with the boundary conditions (7.3). p u ∈ C[0, 1]} equipped We consider the space X = {u ∈ C[0, 1], cD0+ p u, where w = supt∈[0,1] |w(t)| for w ∈ C[0, 1]. with uX = u + cD0+ Obviously (X,  · X ) is a Banach space. By Lemma 7.1.1, we introduce the operator A : X → X defined by   1 −λt (Au)(t) = h(u) e + γ(t) e−λs dH(s) 0



ξ a (ξ − s)β−1 e−λs ds − Γ(β) 0   

+ γ(t)

1

0

s

0



α−1 ˆ e−λ(s−τ ) I0+ fu (τ ) dτ

dH(s)

  s  ξ a β−1 −λ(s−τ ) α−1 ˆ − (ξ − s) e I0+ fu (τ ) dτ ds Γ(β) 0 0  t α−1 ˆ + e−λ(t−s) I0+ fu (s) ds, 0

p q for t ∈ [0, 1] and u ∈ X, where fˆu (s) = f (s, u(s),c D0+ u(s), I0+ u(s)), s ∈ [0, 1].

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Note that the function u is a solution of problem (7.1), (7.3) if and only if u is a fixed point of operator A. Now, we enlist the assumptions that we need in the sequel. (H1) α ∈ (3, 4], p ∈ (0, 1), q > 0, λ > 0, ξ ∈ (0, 1], a ∈ R, β > 0, Δ = 0 (given by (7.7)), and H : [0, 1] → R is a function of bounded variation. (H2) The function f : [0, 1] × R3 → R is continuous and there exists L1 > 0 such that |f (t, x, y, z) − f (t, x1 , y1 , z1 )| ≤ L1 (|x − x1 | + |y − y1 | + |z − z1 |), for all t ∈ [0, 1], x, y, z, x1 , y1 , z1 ∈ R. (H3) The function f : [0, 1] × R3 → R is continuous and there exists a function g ∈ C([0, 1], R+ ) such that |f (t, x, y, z)| ≤ g(t),

∀(t, x, y, z) ∈ [0, 1] × R3 .

(H4) For the function h : C[0, 1] → R, there exists L3 > 0 such that |h(x) − h(y)| ≤ L3 x − y,

∀x, y ∈ C[0, 1].

(H5) The function h : C[0, 1] → R is continuous and there exists L2 > 0 such that |h(x)| ≤ L2 x,

∀x ∈ C[0, 1].

Furthermore, we set the notations: l = sup |γ(t)| = t∈[0,1]

λ2 − 2λ + 2 − 2e−λ > 0, |Δ|

2λ(λ − 1 + e−λ ) > 0, |Δ| t∈[0,1]     ξ  1 −λs  |a| β−1 −λs   M1 = 1 + l  e dH(s) + (ξ − s) e ds , Γ(β) 0 0    ξ  |a| 1  1 −λs  ) dH(s) + (ξ − s)β−1 (1 − e−λs ) ds, M2 =  (1 − e λ 0 λΓ(β) 0     ξ  1 −λs  |a| ∗ β−1 −λs M1 = λ + m  e dH(s) + (ξ − s) e ds , Γ(β) 0 0 m = sup |γ  (t)| =

L0 = 1 +

1 . Γ(q + 1)

(7.13)

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Theorem 7.1.1. Assume that (H1), (H2), (H4) and (H5) hold. If   1 1 L1 L0 ∗ −λ M Λ := max{L2 , L3 } M1 + + lM2 + (1 − e ) Γ(2 − p) 1 Γ(α) λ +

 L1 L0 mM2 + 2 − e−λ < 1, Γ(2 − p)Γ(α)

(7.14)

then problem (7.1), (7.3) has a unique solution on [0, 1], where l, m, L0 , M1 , M2 , M1∗ are given by (7.13). Proof. Let us fix r > 0 such that  

  M0 1 1 −λ −λ 1−e mM2 + 2 − e r≥ + lM2 + Γ(α) λ Γ(2 − p)      1 1 L1 L0 ∗ −λ M + × 1 − L2 M1 + lM2 + 1−e Γ(2 − p) 1 Γ(α) λ 

 −1 L1 L0 −λ mM2 + 2 − e + , (7.15) Γ(2 − p)Γ(α) where M0 = supt∈[0,1] |f (t, 0, 0, 0)|. We first consider the set B r = {u ∈ X, uX ≤ r} and show that A(B r ) ⊂ B r . Let u ∈ B r . For fˆu , we obtain p q u(t), I0+ u(t)| |fˆu (t)| = |f (t, u(t),c D0+ p q ≤ |f (t, u(t),c D0+ u(t), I0+ u(t)) − f (t, 0, 0, 0)| + |f (t, 0, 0, 0)| p q u(t)| + |I0+ u(t)|) + M0 ≤ L1 (|u(t)| + |cD0+  1 p u + M0 ≤ L1 u + cD0+ u + Γ(q + 1)  1 ≤ L1 1 + uX + M0 ≤ L1 L0 r + M0 , Γ(q + 1)

Then we have



|(Au)(t)| ≤ |h(u)| e |a| + Γ(β)

−λt

   + |γ(t)| 

0



ξ

0

  + |γ(t)| 

1

∀t ∈ [0, 1].

  e−λs dH(s) 

(ξ − s)β−1 e−λs ds

0

1

 0

s

     α−1 ˆ  e−λ(s−τ ) I0+ fu (τ ) dτ dH(s)

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363

  s  ξ |a| β−1 −λ(s−τ ) α−1 ˆ + (ξ − s) e |I0+ fu (τ )| dτ ds Γ(β) 0 0  t α−1 ˆ + fu (s)| ds e−λ(t−s) |I0+ 0



   ≤ L2 u 1 + l 

1

0

|a| + Γ(β)



ξ

0

 0

ξ

 β−1 −λs

(ξ − s)

l(L1 L0 r + M0 ) + Γ(α) |a| + Γ(β)

  e−λs dH(s) e

    

1

ds



0

β−1

(ξ − s)

s 0

 0

s

e

e

−λ(s−τ )

−λ(s−τ )

dτ

  dH(s)







dτ

ds

 L1 L0 r + M0 t −λ(t−s) + e ds Γ(α) 0      ξ  1 −λs  |a| β−1 −λs   ≤ L2 r 1 + l  e dH(s) + (ξ − s) e ds Γ(β) 0 0      l(L1 L0 r + M0 )  1 1 −λs  1 − e + dH(s)   Γ(α) 0 λ   ξ |a| β−1 1 −λs (1 − e (ξ − s) ) ds + Γ(β) 0 λ L1 L0 r + M0 (1 − e−λt ), ∀t ∈ [0, 1], λΓ(α) which, on taking the norm for t ∈ [0, 1], yields      ξ  1 −λs  |a| β−1 −λs Au ≤ L2 r 1 + l  e dH(s) + (ξ − s) e ds Γ(β) +

0

0

    l(L1 L0 r + M0 ) 1  1 −λs  + (1 − e ) dH(s)   Γ(α) λ 0   ξ |a| + (ξ − s)β−1 (1 − e−λs ) ds λΓ(β) 0

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L1 L0 r + M0 (1 − e−λ ) = rL2 M1 λΓ(α)  L1 L0 r + M0 1 − e−λ + lM2 + , Γ(α) λ +

where l, L0 , M1 and M2 are given by (7.13). Furthermore, one can find that   1

(Au) (t) = h(u) −λe−λt + γ  (t) 

a − Γ(β)

ξ

0





+ γ (t)

1





0

s

0

ξ



(ξ − s)β−1 e−λs ds

0

a − Γ(β)

0

e−λs dH(s)

α−1 ˆ e−λ(s−τ ) I0+ fu (τ ) dτ

β−1



(ξ − s)

α−1 ˆ + I0+ fu (t)

0

 −λ

s

t

0

dH(s)

α−1 ˆ e−λ(s−τ ) I0+ fu (τ ) dτ

α−1 ˆ e−λ(t−s) I0+ fu (s) ds,

−λt

−λt

) where γ  (t) = 2λ(λt−1+e , with |γ  (t)| = 2λ(λt−1+e Δ |Δ|  and supt∈[0,1] |γ (t)| = m (m is given by (7.13)). Then, we obtain      1 −λs   −λt  + |γ (t)|  e dH(s) |(Au) (t)| ≤ |h(u)| λe 0

|a| + Γ(β)



ξ

0

   + |γ (t)| 

1

0

 0

ξ

ds

∀t ∈ (0, 1),

)

for all t ∈ [0, 1],



(ξ − s)β−1 e−λs ds



|a| + Γ(β)







s

0

e

−λ(s−τ )

β−1

(ξ − s)

α−1 ˆ + |I0+ fu (t)|

0

 +λ



0

t

s

α−1 ˆ |I0+ fu (τ )| dτ

e

−λ(s−τ )



  dH(s)

α−1 ˆ |I0+ fu (τ )| dτ

α−1 ˆ e−λ(t−s) |I0+ fu (s)| ds



ds

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  ≤ L2 u λ + m 

1

0

|a| + Γ(β)



ξ

e

−λs

 e

m(L1 L0 r + M0 ) + Γ(α)

    

|a| + Γ(β)

 0

ξ

  dH(s)

β−1 −λs

(ξ − s)

0

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ds



1

0

(ξ − s)β−1

s

0

 0

s

e

−λ(s−τ )

e−λ(s−τ ) dτ

dτ



  dH(s)

 ds +

L1 L0 r + M0 Γ(α)



t L1 L0 r + M0 λ e−λ(t−s) ds Γ(α) 0      ξ 1   |a| −λs β−1 −λs   ≤ L2 r λ + m  e dH(s) + (ξ − s) e ds Γ(β) 0 0     m(L1 L0 r + M0 ) 1  1 −λs + (1 − e ) dH(s)  Γ(α) λ 0   ξ |a| β−1 −λs + (ξ − s) (1 − e ) ds λΓ(β) 0

+

+

L1 L0 r + M0 (2 − e−λt ), Γ(α)

∀t ∈ (0, 1).

Hence, we conclude (Au)  ≤ rL2 M1∗ +

L1 L0 r + M0 (mM2 + 2 − e−λ ), Γ(α)

where M1∗ is given by (7.13). By the definition of Caputo fractional derivative with p ∈ (0, 1), we deduce  t (t − s)−p c p |(Au) (s)| ds | D0+ (Au)(t)| ≤ 0 Γ(1 − p)  t  (t − s)−p L1 L0 r + M0 ∗ −λ (mM2 + 2 − e ) ds ≤ rL2 M1 + Γ(α) 0 Γ(1 − p)   1 L1 L0 r + M0 = (mM2 + 2 − e−λ ) , ∀t ∈ [0, 1]. rL2 M1∗ + Γ(2 − p) Γ(α)

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Therefore, by using the definition of r from (7.15), we obtain  L1 L0 r + M0 1 − e−λ c p AuX = Au +  D0+ (Au) ≤ rL2 M1 + lM2 + Γ(α) λ   1 L1 L0 r + M0 (mM2 + 2 − e−λ ) ≤ r. + rL2 M1∗ + Γ(2 − p) Γ(α) So, we conclude that A maps B r into itself. Now, for u, v ∈ B r we have |(Au)(t) − (Av)(t)| ≤ |h(u) − h(v)|      ξ 1   |a| × e−λt + |γ(t)|  e−λs dH(s) + (ξ − s)β−1 e−λs ds Γ(β) 0 0     s  1  α−1 ˆ −λ(s−τ ) α−1 ˆ  + |γ(t)|  e |I0+ fu (τ ) − I0+ fv (τ )| dτ dH(s) 0

0

  s  ξ |a| α−1 ˆ β−1 −λ(s−τ ) α−1 ˆ + (ξ − s) e |I0+ fu (τ ) − I0+ fv (τ )| dτ ds Γ(β) 0 0  t α−1 ˆ α−1 ˆ + e−λ(t−s) |I0+ fu (s) − I0+ fv (s)| ds 0

  ≤ L3 u − v 1 + l  +

l Γ(α)

|a| + Γ(β) +



   ξ  |a| β−1 −λs e dH(s) + (ξ − s) e ds Γ(β) 0 0   1  s   −λ(s−τ ) ˆ ˆ  fu − fv   e dτ dH(s)  0

0

ξ

β−1

(ξ − s)

1 fˆu − fˆv  Γ(α)

1

−λs

 0

 0

t

0

s

e

−λ(s−τ )

e−λ(t−s) ds,



dτ

ds

∀t ∈ [0, 1].

Observe that p |fˆu (s) − fˆv (s)| ≤ L1 (|u(s) − v(s)| + |cD0+ u(s) p q q − cD0+ v(s)| + |I0+ u(s) − I0+ v(s)|)

 p p q q u −c D0+ v + I0+ u − I0+ v ≤ L1 u − v + cD0+

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 p p ≤ L1 u − v + cD0+ u −c D0+ v +  ≤ L1 1 +

1 Γ(q + 1)



1 u − v Γ(q + 1)



u − vX = L1 L0 u − vX ,

∀s ∈ [0, 1]. Thus, we obtain |(Au)(t) − (Av)(t)|

    lL1 L0 1  1 −λs  ≤ L3 M1 u − v + u − vX (1 − e ) dH(s)   Γ(α) λ 0   ξ |a| L1 L0 u − vX + (1 − e−λt ) (ξ − s)β−1 (1 − e−λs ) ds + λΓ(β) 0 λΓ(α)

≤ L3 M1 u − v +

lL1 L0 M2 L1 L0 u − vX u − vX + (1 − e−λ ), Γ(α) λΓ(α)

∀t ∈ [0, 1], which implies that



 L1 L0 lL1 L0 M2 −λ Au − Av ≤ L3 M1 + + (1 − e ) u − vX . Γ(α) λΓ(α)

Also, for all t ∈ (0, 1), we have |(Au) (t) − (Av) (t)| 

  ≤ |h(u) − h(v)| λ + m 

0

|a| + Γ(β)

 0

ξ

1

  e−λs dH(s) 

(ξ − s)β−1 e−λs ds

   s  1  m ˆ −λ(s−τ ) ˆ  fu − fv   + e dτ dH(s) Γ(α) 0 0    ξ s |a| β−1 −λ(s−τ ) (ξ − s) e dτ ds + Γ(β) 0 0   t 1 −λ(t−s) ˆ ˆ + fu − fv  1 + λ e ds Γ(α) 0

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≤

L3 M1∗ u

+

+

|a| λΓ(β)

mL1 L0 u − vX − vX + Γ(α)  0

ξ

    1  1 −λs (1 − e ) dH(s)  λ 0 

(ξ − s)β−1 (1 − e−λs ) ds

L1 L0 u − vX L1 L0 (1 − e−λt ) + u − vX . Γ(α) Γ(α)

Thus, we have

 L1 L0 (2 − e−λ ) mL1 L0 M2 ∗ (Au) − (Av)  ≤ L3 M1 + + u − vX , Γ(α) Γ(α) 



which implies that p p |cD0+ (Au)(t) −c D0+ (Av)(t)|  t (t − s)−p |(Au) (s) − (Av) (s)| ds ≤ 0 Γ(1 − p)  L1 L0 (2 − e−λ ) 1 mL1 L0 M2 ∗ + u − vX , ≤ L3 M1 + Γ(α) Γ(α) Γ(2 − p)

∀t ∈ [0, 1], and p p (Au) −c D0+ (Av) cD0+  L1 L0 (2 − e−λ ) mL1 L0 M2 1 ∗ ≤ L3 M1 + + u − vX . Γ(α) Γ(α) Γ(2 − p)

From the above inequalities, we obtain p p (Au) −c D0+ (Av) Au − AvX = Au − Av + c D0+    M1∗ 1 L1 L0 −λ ≤ L3 M1 + + lM2 + (1 − e ) Γ(2 − p) Γ(α) λ  L1 L0 − vX . (mM2 + 2 − e−λ ) u − vX = Λu + Γ(2 − p)Γ(α)

< 1, we deduce that A is a In view of the condition (7.14), that is, Λ contraction. Then we conclude by the contraction mapping principle (Theorem 1.2.1) that the operator A has a unique fixed point u ∈ B r , which corresponds to the unique solution u of problem (7.1), (7.3) on [0, 1]. This completes the proof. 

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369

Theorem 7.1.2. Assume that (H1), (H2), (H3) and (H5) hold. If 1 L2 M1∗ < 1, Γ(2 − p)

1 := L2 M1 + Λ and 2 := L1 L0 Λ Γ(α)

 1 L1 L0 −λ (mM2 + 2 − e−λ ) < 1, lM2 + (1 − e ) + λ Γ(2 − p)Γ(α)

then problem (7.1), (7.3) has at least one solution on [0, 1]. Proof. Let us fix r1 > 0 such that   1 1 − e−λ mM2 + 2 − e−λ g r1 ≥ lM2 + g + Γ(α) λ Γ(2 − p)Γ(α)   −1 1 × 1 − L2 M1 + L2 M1∗ , (7.16) Γ(2 − p) 

where g is the function given in the assumption (H3). We consider the set B r1 = {u ∈ X, uX ≤ r1 } and define the operators A1 , A2 : B r1 → X as  (A1 u)(t) = h(u) e a − Γ(β)  (A2 u)(t) = γ(t)

−λt



0

0 1

ξ

 + γ(t)

1

0

e−λs dH(s)

β−1 −λs

(ξ − s)

 0

s

e

 ds

,

α−1 ˆ e−λ(s−τ ) I0+ fu (τ ) dτ

dH(s)

(7.17)

  s  ξ a α−1 ˆ fu (τ ) dτ ds − (ξ − s)β−1 e−λ(s−τ ) I0+ Γ(β) 0 0  t α−1 ˆ + e−λ(t−s) I0+ fu (s) ds, 0

p q u(s), I0+ u(s)), for all t ∈ [0, 1] and u ∈ B r1 , where fˆu (s) = f (s, u(s),c D0+ s ∈ [0, 1].

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Using the assumptions (H3), (H5) and applying the arguments employed in the proof of Theorem 7.1.1, for all u, v ∈ B r1 , we obtain A1 u + A2 v ≤ A1 u + A2 v ≤ r1 L2 M1 + +

(1 − e−λ )g , λΓ(α)

(A1 u) + (A2 v)  ≤ r1 L2 M1∗ + 

p D0+ (A1 u

c

lM2 g Γ(α)

g (mM2 + 2 − e−λ ), Γ(α)

  1 g ∗ −λ (mM2 + 2 − e ) . + A2 v) ≤ r1 L2 M1 + Γ(2 − p) Γ(α)

From the above inequalities and the definition of r1 , we find that p A1 u + A2 vX = A1 u + A2 v + cD0+ (A1 u + A2 v)  1 1 − e−λ ≤ r1 L2 M1 + lM2 + g Γ(α) λ   mM2 + 2 − e−λ 1 g ≤ r1 , + r1 L2 M1∗ + Γ(2 − p) Γ(α)

∀u, v ∈ B r1 . Hence, A1 u + A2 v ∈ B r1 for all u, v ∈ B r1 . Next, we show that the operator A2 is a contraction. As in the previous theorem, we can obtain 

 lL1 L0 M2 L1 L0 + (1 − e−λ ) u − vX , Γ(α) λΓ(α)  mL1 L0 M2 L1 L0 (2 − e−λ )   + (A2 u) − (A2 v)  ≤ u − vX , Γ(α) Γ(α)  mL1 L0 M2 L1 L0 (2 − e−λ ) 1 p p + cD0+ (A2 u) −c D0+ (A2 v) ≤ Γ(2 − p) Γ(α) Γ(α) A2 u − A2 v ≤

× u − vX ,

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which imply that



A2 u − A2 vX ≤

lL1 L0 M2 L1 L0 + (1 − e−λ ) Γ(α) λΓ(α)

L1 L0 (2 − e−λ ) mL1 L0 M2 + + Γ(2 − p)Γ(α) Γ(2 − p)Γ(α)

page 371



2 u − vX , × u − vX = Λ 2 < 1. This shows that A2 is a contraction. for all u, v ∈ B r1 , with Λ Continuity of h implies that the operator A1 is continuous on B r1 . Next, we prove that A1 is compact. The operator A1 is uniformly bounded on B r1 as A1 u ≤ r1 L2 M1 , p cD0+ (A1 u) ≤

(A1 u)  ≤ r1 L2 M1∗ ,

1 r1 L2 M1∗ , Γ(2 − p)

In consequence, we get A1 uX ≤ r1 L2 M1 +

1 r1 L2 M1∗ , Γ(2 − p)

∀u ∈ B r1 .

Next, for u ∈ B r1 , t1 , t2 ∈ [0, 1], t1 < t2 , we obtain |(A1 u)(t2 ) − (A1 u)(t1 )|  ≤ L2 r1

+

|e

|a| Γ(β)

−λt2



ξ 0

−e

−λt1

   | + |γ(t2 ) − γ(t1 )| 

0

 (ξ − s)β−1 e−λs ds

,

p p |cD0+ (A1 u)(t2 ) −c D0+ (A1 u)(t1 )|     L2 r1 M1∗  t1 −p −p ≤ [(t2 − s) − (t1 − s) ] ds  Γ(1 − p) 0    L2 r1 M1∗  t2 −p + (t2 − s) ds  Γ(1 − p) t1

≤

L2 r1 M1∗ [2(t2 − t1 )1−p + t1−p − t1−p ]. 2 1 Γ(2 − p)

1

  e−λs dH(s)

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p p Clearly, |(A1 u)(t2 )−(A1 u)(t1 )| → 0 and |cD0+ (A1 u)(t2 )−cD0+ (A1 u)(t1 )| → 0, as t2 → t1 uniformly with respect to u ∈ B r1 . This shows that A1 is equicontinuous on B r1 , and so, by using Arzela-Ascoli theorem, the set A1 (B r1 ) is relatively compact. Thus, the operator A1 is compact on B r1 . As all the assumptions of Theorem 1.2.3 are satisfied, the conclusion of Theorem 1.2.3 implies that there exists a fixed point of operator A1 + A2 , which is a solution of problem (7.1), (7.3) on [0, 1]. This completes the proof. 

Theorem 7.1.3. Assume that (H1), (H3), (H4) and (H5) hold. If  1 3 := max{L2 , L3 } M1 + Λ M1∗ < 1 Γ(2 − p) then problem (7.1), (7.3) has at least one solution on [0, 1]. Proof. We consider the number r1 given by (7.16), and the operators A1 , A2 defined on B r1 and given by (7.17). As in the proof of Theorem 7.1.2, we can obtain that A1 u + A2 v ∈ B r1 for all u, v ∈ B r1 . In order to establish that the operator A1 is a contraction, one can find that A1 u − A1 v ≤ L3 M1 u − vX , p p (A1 u) −c D0+ (A1 v) ≤ cD0+

In consequence, we have  A1 u − A1 vX ≤ L3 M1 +

(A1 u) − (A1 v)  ≤ L3 M1∗ u − vX ,

1 L3 M1∗ u − vX . Γ(2 − p)

1 ∗ 3 u − vX , L3 M1 u − vX ≤ Λ Γ(2 − p)

3 < 1. This implies that A1 is a contraction. for all u, v ∈ B r1 , with Λ It follows from the continuity of f that the operator A2 is continuous on B r1 . Next, we prove that A2 is compact. Note that the operator A2 is uniformly bounded on B r1 as A2 u ≤ (A2 u)  ≤ p cD0+ (A2 u) ≤

lM2 g (1 − e−λ )g + , Γ(α) λΓ(α) g (mM2 + 2 − e−λ ), Γ(α) g (mM2 + 2 − e−λ ). Γ(2 − p)Γ(α)

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Consequently, g lM2 g (1 − e−λ )g + + (mM2 + 2 − e−λ ), A2 uX ≤ Γ(α) λΓ(α) Γ(2 − p)Γ(α) ∀u ∈ B r1 . Now, we prove that A2 is equicontinuous on B r1 . We denote by Λr1 = sup{|f (t, x, y, z)|, t ∈ [0, 1], |x|, |y|, |z| ≤ r1 }. Then, for u ∈ B r1 and t1 , t2 ∈ [0, 1] with t1 < t2 , we obtain |(A2 u)(t2 ) − (A2 u)(t1 )| ≤ Λr1 |γ(t2 ) − γ(t1 )|     s  1 −λ(s−τ )  ×  e 0

0

|a| + Γ(β)   + 

0

Λ r1 ≤ Γ(α)



|a| + Γ(β) 

t1

≤

Λ r1 Γ(α)





ξ

0

s

0

e−λ(s−τ )



τ

0

 (τ − ζ)α−2 dζ dτ ds Γ(α − 1)

β−1

0



(ξ − s)



1

0

s

0

e

s

e−λ(s−τ ) τ α−1 dτ

−λ(s−τ ) α−1

τ

dτ



 1  |γ(t2 ) − γ(t1 )| α

0



ξ 0

t2 t1



  dH(s)





  e−λ(t1 −s) − e−λ(t2 −s) sα−1 ds +

|a| + αΓ(β) +

(ξ − s)β−1

   |γ(t2 ) − γ(t1 )| 



+ 0

0



  α−1 ˆ e−λ(t2 −s) I0+ fu (s) ds

t2

t1

ξ

0

    α−1 ˆ −λ(t2 −s) −λ(t1 −s) e I0+ fu (s) ds −e

t1

  + 



  (τ − ζ)α−2 dζ dτ dH(s) Γ(α − 1)

τ

ds e−λ(t2 −s) sα−1 ds



 1  sα dH(s)

 β−1 α

(ξ − s)

s ds

 tα−1    tα−1 1 1 − e−λt1 − e−λ(t2 −t1 ) + e−λt2 + 2 1 − e−λ(t2 −t1 ) . λ λ

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In addition, we have mΛr1 |(A2 u) (t)| ≤ Γ(α)

    1  1 −λs (1 − e ) dH(s) λ

|a| + λΓ(β) +

0

 0

ξ

 β−1

(ξ − s)

(1 − e

−λs

) ds

Λ r1 Λ r1 mΛr1 M2 (2 − e−λ ) = + (2 − e−λ ) =: Λ0 , Γ(α) Γ(α) Γ(α)

and p p |cD0+ (A2 u)(t2 ) −c D0+ (A2 u)(t1 )|  t2   t1   (t2 − s)−p (t1 − s)−p    = (A2 u) (s) ds − (A2 u) (s) ds Γ(1 − p) Γ(1 − p) 0 0   t2   t1     Λ0 −p −p −p     ≤ [(t − s) − (t − s) ] ds (t − s) ds + 2 1 2     Γ(1 − p) 0 t1

≤

Λ0 [2(t2 − t1 )1−p + t1−p − t1−p ]. 2 1 Γ(2 − p)

Clearly, |(A2 u)(t2 ) − (A2 u)(t1 )| → 0,

p p and |cD0+ (A2 u)(t2 ) −cD0+ (A2 u)(t1 )| → 0,

as t2 → t1 uniformly with respect to u ∈ B r1 . From the foregoing arguments, we deduce that A2 is equicontinuous on B r1 , and so, by using Arzela-Ascoli theorem, the set A2 (B r1 ) is relatively compact. In consequence, operator A2 is compact on B r1 . Thus, all the assumptions of Theorem 1.2.3 are satisfied (with the roles of A1 and A2 interchanged). Hence, by Theorem 1.2.3, we deduce that there exists a fixed point of operator A1 + A2 , which is a solution of the boundary value problem (7.1), (7.3) on [0, 1]. This completes the proof.  7.1.3

Existence of solutions for problem (7.2), (7.3)

In this section, we investigate the existence of solutions for the fractional inclusion (7.2) with the boundary conditions (7.3).

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375

Definition 7.1.1. A function u ∈ C 4 [0, 1] is a solution of the boundary β u(ξ) = value problem (7.2), (7.3) if u(0) = h(u), u (0) = u (0) = 0, aI0+

1 0 u(s) dH(s), and there exists a function v ∈ SF,u such that 



  ξ a u(t) = h(u) e−λt + γ(t) e−λs dH(s) − (ξ − s)β−1 e−λs ds Γ(β) 0 0  1  s α−1 +γ(t) e−λ(s−τ ) I0+ v(τ ) dτ dH(s) 0

1

0

  s a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ v(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ v(s) ds, ∀t ∈ [0, 1]. 

ξ

0

In the above definition, the set of selections SF,u of F is defined by q u(t)) for a.a. t ∈ [0, 1]}. SF,u = {v ∈ L1 (0, 1), v(t) ∈ F (t, u(t), I0+ 7.1.3.1

The upper semicontinuous case

Here, we prove the existence of solutions for problem (7.2), (7.3) when the multivalued map F has convex values by means of the nonlinear alternative of Leray-Schauder type for Kakutani maps (Theorem 1.2.14). In the next result, we need the following assumptions. (A1) F : [0, 1]×R2 → P(R) is L1 -Carath´eodory and has nonempty compact and convex values. (A2) There exist a function φ ∈ C([0, 1], R+ ) and a nondecreasing, subhomogeneous function Ω : R+ → R+ (that is, Ω(μx) ≤ μΩ(x) for all μ ≥ 1 and x ∈ R+ ) such that def

F (t, x, y)P = sup{|w|, w ∈ F (t, x, y)} ≤ φ(t)Ω(|x| + |y|) for each (t, x, y) ∈ [0, 1] × R2 . (A3) There exists a constant M > 0 such that −1   L0 φΩ(M ) 1 > 1, M M L2 M1 + lM2 + (1 − e−λ ) Γ(α) λ where l, L0 , M1 , M2 are given by (7.13).

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Theorem 7.1.4. Suppose that the conditions (H1), (H5) and (A1)–(A3) hold. Then the boundary value problem (7.2), (7.3) has at least one solution on [0, 1]. Proof. We define the operator ΩF : C[0, 1] → P(C[0, 1]) by ΩF (u) = {w ∈ C[0, 1], w(t) ∈ N (u)(t), ∀t ∈ [0, 1]}, where



N (u)(t) =

 h(u) e

−λt

 + γ(t)



0

1

e−λs dH(s)

ξ a (ξ − s)β−1 e−λs ds − Γ(β) 0    1

+ γ(t)

s

0

0

u ∈ C[0, 1], (7.18)



α−1 e−λ(s−τ ) I0+ v(τ ) dτ

dH(s)

  s  ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ v(τ ) dτ ds Γ(β) 0 0   t −λ(t−s) α−1 + e I0+ v(s) ds, v ∈ SF,u , t ∈ [0, 1], u ∈ C[0, 1]. 0

We will show that ΩF satisfies the hypothesis of Theorem 1.2.14 in several steps. First of all, we observe that ΩF (u) is convex for each u ∈ C[0, 1] as SF,u is convex (F has convex values). In the next step, we show that ΩF maps bounded sets (balls) into bounded sets in C[0, 1]. For a positive number r, let B r = {u ∈ C[0, 1], u ≤ r} be a bounded ball in C[0, 1]. Then, for u ∈ B r and for each w ∈ ΩF (u), there exits v ∈ SF,u such that w(t) ∈ N (u)(t) for all t ∈ [0, 1]. Thus     ξ 1 a −λt −λs β−1 −λs + γ(t) e dH(s) − (ξ − s) e ds w(t) = h(u) e Γ(β) 0 0    + γ(t)

1

0

s

0

α−1 e−λ(s−τ ) I0+ v(τ ) dτ

dH(s)

  s  ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ v(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ v(s) ds, ∀t ∈ [0, 1]. 0

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Then, for t ∈ [0, 1], we have     |w(t)| ≤ L2 u 1 + |γ(t)| 

1

0

|a| + Γ(β)

 0

ξ

e

−λs

β−1 −λs

(ξ − s)

   + |γ(t)| 

0

1

 0

s

e

e

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377

  dH(s) 

ds

−λ(s−τ ) vτ

α

Γ(α)

dτ

  dH(s)

  s  ξ α |a| β−1 −λ(s−τ ) vτ + dτ ds (ξ − s) e Γ(β) 0 Γ(α) 0  t vsα + e−λ(t−s) ds Γ(α) 0      ξ  1 −λs  |a| β−1 −λs ≤ L2 r 1 + l  e dH(s) + (ξ − s) e ds Γ(β) 0 0     s  1  −λ(s−τ ) 1  +l  dτ dH(s) v e Γ(α) 0 0   s  ξ |a| β−1 −λ(s−τ ) 1 dτ dsv (ξ − s) e + Γ(β) 0 Γ(α) 0  t v + e−λ(t−s) ds Γ(α) 0     1  1 l −λs ≤ L2 rM1 + (1 − e ) dH(s)  Γ(α) λ 0   ξ |a| + (ξ − s)β−1 (1 − e−λs ) ds λΓ(β) 0 1 (1 − e−λ )L0 φΩ(uX ) λΓ(α) l 1 M2 L0 φΩ(r) + (1 − e−λ )L0 φΩ(r), ≤ L2 rM1 + Γ(α) λΓ(α) which yields 1 lM2 L0 φΩ(r) + (1 − e−λ )L0 φΩ(r), w ≤ L2 rM1 + Γ(α) λΓ(α) × L0 φΩ(uX ) +

for all w ∈ ΩF (u) and all u ∈ B r . So ΩF (B r ) is bounded.

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Now, we show that ΩF maps bounded sets into equicontinuous sets of C[0, 1]. Let B r = {u ∈ C[0, 1], u ≤ r} be a bounded ball in C[0, 1], and let u ∈ B r and t1 , t2 ∈ [0, 1] with t1 < t2 . For each w ∈ ΩF (u), and q u(t)), t ∈ [0, 1], we obtain v(t) ∈ F (t, u(t), I0+  |w(t2 ) − w(t1 )| ≤ L2 ue−λt2 − e−λt1 + (γ(t2 ) − γ(t1 ))  ×

1

0

e

−λs

a dH(s) − Γ(β)

   + |γ(t2 ) − γ(t1 )| 

1



0

|a| + Γ(β)   + 

t2

0



ξ

0

β−1

(ξ − s)

 0 s

0

 0

s

e

e

α−1 e−λ(t2 −s) I0+ v(s) ds

β−1 −λs

(ξ − s)

−λ(s−τ )

−λ(s−τ )

 −

0

t1

e

   ds  

α−1 |I0+ v(τ )| dτ

α−1 |I0+ v(τ )| dτ



  dH(s) 

ds

 

α−1 e−λ(t1 −s) I0+ v(s) ds

 M1 − 1 ≤ L2 r |e |γ(t2 ) − γ(t1 )| −e |+ l    1  s   τ α−1 + v |γ(t2 ) − γ(t1 )|  dτ dH(s) e−λ(s−τ ) Γ(α) 0 0   s  ξ |a| β−1 −λ(s−τ ) α−1 + (ξ − s) e τ dτ ds Γ(β) 0 0 +



ξ

1 Γ(α)

−λt2

 0

t1

−λt1

  e−λ(t1 −s) − e−λ(t2 −s) sα−1 ds

 t2 1 −λ(t2 −s) α−1 + e s ds Γ(α) t1   M1 − 1 L0 φΩ(uX ) ≤ L2 r |e−λt2 − e−λt1 | + |γ(t2 ) − γ(t1 )| + l Γ(α)       ξ  |a| 1  1 α β−1 α × |γ(t2 ) − γ(t1 )| s dH(s) + (ξ − s) s ds α 0 αΓ(β) 0 

+

  tα−1   tα−1 1 1 − e−λt1 − e−λ(t2 −t1 ) + e−λt2 + 2 1 − e−λ(t2 −t1 ) . λ λ

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As the right-hand side of the last inequality tends to zero as t2 → t1 , independently of u ∈ B r , so ΩF (u) is a equicontinuous set of C[0, 1]. In view of the foregoing arguments, it follows by the Arzela-Ascoli theorem that ΩF : C[0, 1] → P(C[0, 1]) is compact. In order to establish that ΩF is upper semicontinuous, it is sufficient to show that ΩF has a closed graph (by [29, Proposition 1.2]). Let un → u∗ , wn ∈ ΩF (un ) and wn → w∗ . Then we will show that w∗ ∈ ΩF (u∗ ). For wn ∈ ΩF (un ), there exists vn ∈ SF,un such that for each t ∈ [0, 1], we have   1 −λt + γ(t) e−λs dH(s) wn (t) = h(un ) e 0

a − Γ(β)



ξ

0

 + γ(t)



t

0

(ξ − s)

1



0



a − Γ(β) +

 β−1 −λs

ξ

0

s

0

e

ds

α−1 e−λ(s−τ ) I0+ vn (τ ) dτ

β−1



(ξ − s)

s

0

dH(s)

α−1 e−λ(s−τ ) I0+ vn (τ ) dτ



ds

α−1 e−λ(t−s) I0+ vn (s) ds.

We will show that there exists v∗ ∈ SF,u∗ such that for each t ∈ [0, 1],   1 −λt w∗ (t) = h(u∗ ) e + γ(t) e−λs dH(s) 0

a − Γ(β)

 0

 + γ(t)

ξ

0

1

 β−1 −λs

(ξ − s)  0

s

e

ds

α−1 e−λ(s−τ ) I0+ v∗ (τ ) dτ

dH(s)

 ξ a (ξ − s)β−1 − Γ(β) 0  s  α−1 × e−λ(s−τ ) I0+ v∗ (τ ) dτ ds  + 0

0

t

α−1 e−λ(t−s) I0+ v∗ (s) ds.

(7.19)

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We consider the linear operator Ψ : L1 (0, 1) → C[0, 1] given by    1 s α−1 Ψ(w)(t) = γ(t) e−λ(s−τ ) I0+ w(τ ) dτ dH(s) 0

0

  s ξ a α−1 − (ξ − s)β−1 e−λ(s−τ ) I0+ w(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ w(s) ds, 

0

for t ∈ (0, 1), w ∈ L1 (0, 1). ∗ ∈ L1 (0, 1) Notice that the operator Ψ is continuous. Indeed, for w n , w 1 ∗ in L (0, 1), we obtain with w n → w |Ψ(w n )(t) − Ψ(w ∗ )(t)|     s  1  −λ(s−τ ) α−1  e |I0+ (w n (τ ) − w ∗ (τ ))| dτ dH(s) ≤l  0

0

  s  ξ |a| β−1 −λ(s−τ ) α−1 (ξ − s) e |I0+ (w n (τ ) − w ∗ (τ ))| dτ ds + Γ(β) 0 0  t α−1 + e−λ(t−s) |I0+ (w n (s) − w ∗ (s))| ds, ∀t ∈ [0, 1], 0

which implies that Ψ(w n ) → Ψ(w ∗ ) in C[0, 1]. Then, as argued in [71] (see also Lemma 4.3 from [14]), we deduce that Ψ ◦ SF,u has a closed graph. For the above functions un and wn , we have   wn − h(un ) e−λt + γ(t) a − Γ(β)



ξ

0

1

0

β−1 −λs

(ξ − s)

e

e−λs dH(s) 

ds

Since un → u, there exists v∗ ∈ SF,u∗ such that   1 −λt w∗ (t) − h(u∗ ) e + γ(t) e−λs dH(s) 0

a − Γ(β)

 0

ξ

β−1 −λs

(ξ − s)

e

 ds

∈ Ψ(SF,un ).

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 = γ(t)

1



0

0

s

α−1 e−λ(s−τ ) I0+ v∗ (τ ) dτ

page 381

381

dH(s)

 s   ξ a β−1 −λ(s−τ ) α−1 (ξ − s) e I0+ v∗ (τ ) dτ ds − Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ v∗ (s) ds, 0

or equivalently, we deduce the relation (7.19). Finally, we show that there exists an open set U ⊂ C[0, 1] with u ∈ θΩF (u) for any θ ∈ (0, 1) and all u ∈ ∂U . Let θ ∈ (0, 1) and u ∈ θΩF (u). Then there exists v ∈ L1 (0, 1) with v ∈ SF,u such that for all t ∈ [0, 1], we obtain 1 lM2 L0 φΩ(u) + (1 − e−λ )L0 φΩ(u), u ≤ uL2M1 + Γ(α) λΓ(α) which can be written as −1   L0 φΩ(u) 1 ≤ 1. u uL2M1 + lM2 + (1 − e−λ ) Γ(α) λ In view of assumption (A3), there exists M such that u = M . Let us define the set U = {u ∈ C[0, 1], u < M }. The operator ΩF : U → P(C[0, 1]) is upper semicontinuous and compact. From the definition of U , there exists no u ∈ ∂U such that u ∈ θΩF (u) for some θ ∈ (0, 1). Therefore, by the nonlinear alternative of Leray-Schauder type (Theorem 1.2.14), we conclude that ΩF has a fixed point u ∈ U which is a solution of problem (7.2), (7.3).  7.1.3.2

The Lipschitz case

We discuss here the existence of solutions for problem (7.2), (7.3) with a nonconvex valued right-hand side by applying the fixed point theorem due to Covitz and Nadler (Theorem 1.2.15) for multivalued functions. Before proceeding for the main result, let us state the assumptions needed for it. (I1) F : [0, 1] × R2 → Pcp (R) satisfies the condition F (·, x, y) : [0, 1] → Pcp (R) is measurable for each (x, y) ∈ R2 . ¯| + |y − y¯|], for a.a. t ∈ [0, 1] and (I2) Hd (F (t, x, y), F (t, x¯, y¯)) ≤ ϕ(t)[|x − x for all x, y, x ¯, y¯ ∈ R with ϕ ∈ C([0, 1], R+ ), and d(0, F (t, 0, 0)) ≤ ϕ(t) for a.a. t ∈ [0, 1].

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In assumption (I1), the set Pcp (R) = {Y ∈ P(R), Y is compact}, and in assumption (I2), Hd is the Hausdorff metric, where d is the Euclidean metric in R defined by d(x, y) = |x − y| for x, y ∈ R. Theorem 7.1.5. Assume that the hypotheses (H1), (H4), (I1) and (I2) hold. If  L0 ϕ 1 −λ θ0 := L3 M1 + (7.20) lM2 + (1 − e ) < 1, Γ(α) λ where l, L0 , M1 , M2 are given by (7.13), then the boundary value problem (7.2), (7.3) has at least one solution on [0, 1]. Proof. Observe that the set SF,u is nonempty for each u ∈ C[0, 1] by assumption (I1), and thus F has a measurable selection (see [25, Theorem III.6]). We now show that the operator ΩF : C[0, 1] → P(C[0, 1]) defined by (7.18) satisfies the assumptions of Theorem 1.2.15. To establish that ΩF (u) ∈ Pcl (C[0, 1]) for each u ∈ C[0, 1], let {wn }n≥0 ⊂ ΩF (u) be such that wn → w as n → ∞ in C[0, 1]. Then w ∈ C[0, 1] and there exists vn ∈ SF,u such that for each t ∈ [0, 1], we have     ξ 1 a e−λs dH(s) − (ξ − s)β−1 e−λs ds wn (t) = h(u) e−λt + γ(t) Γ(β) 0 0    1 s α−1 + γ(t) e−λ(s−τ ) I0+ vn (τ ) dτ dH(s) 0

0

  s a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ vn (τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ vn (s) ds, 

ξ

0

q with vn (t) ∈ F (t, u(t), I0+ u(t)), t ∈ [0, 1]. Since F has compact values, therefore, we can pass onto a subsequence (denoted in a same way) to obtain that vn converges to v in L1 (0, 1). Thus, v ∈ SF,u and for each t ∈ [0, 1], we have wn (t) → w(t), where     ξ 1 a −λt −λs β−1 −λs + γ(t) e dH(s) − (ξ − s) e ds w(t) = h(u) e Γ(β) 0 0    1

s

+γ(t) 0

0

α−1 e−λ(s−τ ) I0+ v(τ ) dτ

dH(s)

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  s  ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ v(τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ v(s) ds. 0

Hence, w ∈ ΩF (u). Next, we show that ΩF is a contraction, that is, u)) ≤ θ0 u − u ¯, Hd1 (ΩF (u), ΩF (¯

∀u, u ¯ ∈ C[0, 1],

where θ0 is defined in (7.20), and d1 is the metric induced by the norm  ·  in C[0, 1]. For this, let u, u ¯ ∈ C[0, 1] and w1 ∈ ΩF (u). Then there exists v1 ∈ SF,u such that for all t ∈ [0, 1], we obtain     ξ 1 a −λt −λs β−1 −λs w1 (t) = h(u) e + γ(t) e dH(s) − (ξ − s) e ds Γ(β) 0 0    1

+ γ(t)

0

s

0

α−1 e−λ(s−τ ) I0+ v1 (τ ) dτ

dH(s)

  s  ξ a α−1 − (ξ − s)β−1 e−λ(s−τ ) I0+ v1 (τ ) dτ ds Γ(β) 0 0  t α−1 + e−λ(t−s) I0+ v1 (s) ds. 0

By (I2), we have q q u(t)), F (t, u¯(t), I0+ u¯(t))) Hd (F (t, u(t), I0+ q q ≤ ϕ(t)[|u(t) − u ¯(t)| + |I0+ u(t) − I0+ u ¯(t)|], q for a.a. t ∈ [0, 1]. Then there exists ψ ∈ F (t, u¯(t), I0+ u ¯(t)) such that q q u(t) − I0+ u ¯(t)|], |v1 (t) − ψ| ≤ ϕ(t)[|u(t) − u¯(t)| + |I0+

a.a. t ∈ [0, 1].

: [0, 1] → P(R) by U (t) = {ψ ∈ R, |v1 (t) − ψ| ≤ We define U q q ¯(t)|]}. As the multivalued operator ϕ(t)[|u(t) − u ¯(t)| + |I0+ u(t) − I0+ u q ¯(t), I0+ u V (t) = U (t) ∩ F (t, u ¯(t)) is measurable (see [25, Proposition III.4]), there exists a function v2 (t) which is a measurable selection for V (t). Hence,

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q v2 (t) ∈ F (t, u ¯(t), I0+ u ¯(t)) for a.a. t ∈ [0, 1], and q q |v1 (t) − v2 (t)| ≤ ϕ(t)[|u(t) − u ¯(t)| + |I0+ u(t) − I0+ u¯(t)|],

a.a. t ∈ [0, 1].

Let us define the function w2 (t), t ∈ [0, 1] by     ξ 1 a u) e−λt + γ(t) e−λs dH(s) − (ξ − s)β−1 e−λs ds w2 (t) = h(¯ Γ(β) 0 0    1

+ γ(t)

s

0

0

α−1 e−λ(s−τ ) I0+ v2 (τ ) dτ

dH(s)

  s  ξ a β−1 −λ(s−τ ) α−1 − (ξ − s) e I0+ v2 (τ ) dτ ds Γ(β) 0 0  t α−1 e−λ(t−s) I0+ v2 (s) ds. + 0

Then we conclude that |w1 (t) − w2 (t)| ≤ |h(u) − h(¯ u)|M1     1 + |γ(t)|  0

s

0

α−1 α−1 e−λ(s−τ ) |I0+ v1 (τ ) − I0+ v2 (τ )| dτ



  dH(s)

  s  ξ |a| α−1 β−1 −λ(s−τ ) α−1 (ξ − s) e |I0+ v1 (τ ) − I0+ v2 (τ )| dτ ds + Γ(β) 0 0  t α−1 α−1 + e−λ(t−s) |I0+ v1 (s) − I0+ v2 (s)| ds 0



¯ + ≤ L3 M1 u − u |a| + Γ(β)

 0

ξ

ϕl Γ(α)

(ξ − s)β−1

    

0

 0

s

1

 0

s

e−λ(s−τ ) τ α−1 dτ

e−λ(s−τ ) τ α−1 dτ



  dH(s)



ds

  ϕ t −λ(t−s) α−1 1 + e s ds 1+ u − u ¯ Γ(α) 0 Γ(q + 1)    L0 ϕ 1 − e−λ lM2 + u − u ¯, ∀t ∈ [0, 1]. ≤ L3 M1 + Γ(α) λ

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Existence of Solutions for Caputo Fractional BVPs

Therefore, w1 − w2  ≤ θ0 u − u ¯. By interchanging the roles of u and u ¯, we obtain a similar relation, and thus we get Hd1 (ΩF (u), ΩF (¯ u)) ≤ θ0 u − u ¯. In view of the condition θ0 < 1 (given by (7.20)), it follows that ΩF is a contraction, and therefore by Theorem 1.2.15, ΩF has a fixed point u, which is a solution of problem (7.2), (7.3). This completes the proof.  7.1.4

Examples

7 3 6 5 1 (I) Let us  1, λ = 4 and  fix α = 2 ,1 p = 4 , q = 51, 2β = 4 , ξ = 2, 2a = H(s) = 1, if s ∈ 0, 3 ; 5, if s ∈ 3 , 3 ; 14, if s ∈ 3 , 1 and consider the following fractional differential equation 7/2

5/2

3/4

6/5

(cD0+ + 4cD0+ )u(t) = f (t, u(t),c D0+ u(t), I0+ u(t)),

t ∈ (0, 1), (7.21)

supplemented with the boundary conditions ⎧   ⎪ ⎨ u(0) = h(u), u (0) = u (0) = 0,  1/4   1/2  1 1 2 1 ⎪

 −s u(s) ds. + 9u = ⎩ 4u 3 3 2 Γ 54 0

(7.22)

Using the given values, it is found that Δ ≈ −34.94855476, (Δ = 0), l ≈ 0.28508672, m ≈ 0.69091627, L0 ≈ 1.90760368, M1 ≈ 1.52905173, M2 ≈ 2.87883365, M1∗ ≈ 5.28217286. Now we set the values of h(u) and f (t, x, y, z) in (7.21) and illustrate the obtained results.

 1 sin u 12 and the function f be defined by (a) Let h(u) = 12  |x| z 5t 1 − arctan y + − 2 , f (t, x, y, z) = √ 9 t +3 t + 81 1 + |x| for t ∈ [0, 1], x, y, z ∈ R. For these functions, we have L2 = L3 = |f (t, x, y, z) − f (t, x1 , y1 , z1 )| ≤

1 12

and

1 (|x − x1 | + |y − y1 | + |z − z1 |), 9

for all t ∈ [0, 1] and x, y, z, x1 , y1 , z1 ∈ R. So L1 = 19 . In addition we obtain ≈ 0.960446 < 1. Thus, all the conditions of Theorem 7.1.1 are satisfied. Λ Therefore, by Theorem 7.1.1, there exists a unique solution for problem (7.21), (7.22) on [0, 1].

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(b) Consider h(u) =

|u(1/2)| 8(1+|u(1/2)|)

and the function f defined by

1 x2/3 y 4/7 1 f (t, x, y, z) = √ sin t + − − cos z 2/5 , 2/3 4(2 + x ) 2(3 + y 4/7 ) 3(1 + t) 4 + t2 for all t ∈ [0, 1] and x, y, z ∈ R. For the above functions, we have L2 = L3 = 18 and |f (t, x, y, z)| ≤ g(t), t| 1 for all t ∈ [0, 1] and x, y, z ∈ R, with g(t) = √| sin + 34 + 3(1+t) , t ∈ 4+t2 3 ≈ 0.919584 < 1. As all the conditions of [0, 1]. In addition we obtain Λ Theorem 7.1.3 are satisfied, we deduce by the conclusion of Theorem 7.1.3 that problem (7.21), (7.22) has at least one solution on [0, 1]. 7 6 5 1 1, (II) Fixing  1 42, ξ = 2 , a  =  λ = 4 and  α = 2 , 1q = 5 , β = H(s) = 1, if s ∈ 0, 3 ; 5, if s ∈ 3 , 3 ; 14, if s ∈ 23 , 1 , we consider the following fractional differential inclusion 7/2

5/2

6/5

(cD0+ + 4cD0+ )u(t) ∈ F (t, u(t), I0+ u(t)),

t ∈ (0, 1),

(7.23)

equipped with the boundary conditions (7.22). With the given data, we find that Δ ≈ −34.94855476, (Δ = 0), l ≈ 0.28508672, L0 ≈ 1.90760368, M1 ≈ 1.52905173, M2 ≈ 2.87883365.

 (a) In order to apply Theorem 7.1.4, we consider h(u) = − 12 cos u 13 , and the multivalued function    1 1 1 F (t, x, y) = √ (sin x − y + 1), √ x + sin y + , 2 400 + t2 900 + t2 for all t ∈ [0, 1], x, y ∈ R. It is easy to find that L2 = 12 , F (t, x, y)P ≤ φ(t)Ω(|x| + |y|) for all 1 and Ω(u) = u + 1. So, Ω t ∈ [0, 1] and x, y ∈ R, where φ(t) = √400+t 2 1 . If we is a nondecreasing and subhomogeneous function, and φ = 20 take M ≥ 0.15, then the condition given by (A3) is satisfied. Thus all the conditions of Theorem 7.1.4 are satisfied, and consequently, there exists at least one solution u for problem (7.23), (7.22) on [0, 1]. (b) For the illustration of Theorem 7.1.5, we introduce the function h(u) = 1 2 arctan u(1), and the multivalued function  F (t, x, y) = 0,

1 10 + t3



 |x| 1 + arctan y + , 4 + |x| 12 + t

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Existence of Solutions for Caputo Fractional BVPs

for t ∈ [0, 1], x, y ∈ R. Clearly L3 = 12 , and Hd (F (t, x, y), F (t, x¯, y¯)) ≤ ϕ(t) (|x − x ¯| + |y − y¯|), 1 for all t ∈ [0, 1], x, y, x ¯, y¯ ∈ R with ϕ(t) = 10+t t ∈ [0, 1]. Furthermore, 3, 1 ϕ = 10 , d(0, F (t, 0, 0)) ≤ ϕ(t) for t ∈ [0, 1], and θ0 ≈ 0.825722 < 1. Hence all the assumptions of Theorem 7.1.5 are satisfied, and consequently problem (7.23), (7.22) has at least one solution on [0, 1].

Remark 7.1.1. The results presented in this section were published in [9].

7.2

Sequential Caputo Fractional Integro-Differential Systems with Coupled Integral Boundary Conditions

In this section, we investigate the system of nonlinear Caputo type sequential fractional integro-differential equations c α p1 q1 α−1 )u(t) = f (t, u(t), v(t),c D0+ v(t), I0+ v(t)), t ∈ (0, 1), ( D0+ + λcD0+ β β−1 p2 q2 + μ cD0+ )v(t) = g(t, u(t),c D0+ u(t), I0+ u(t), v(t)), (cD0+

with the coupled boundary conditions ⎧  ⎪   ⎪ ⎪ ⎨u(0) = u (0) = u (0) = 0, u(1) =

1



1

⎪ ⎪ ⎪ ⎩v(0) = v  (0) = v  (0) = 0,

v(1) =

0

0

 u(s) dH1 (s) +  u(s) dK1 (s) +

1

0

0

1

t ∈ (0, 1), (7.24)

v(s) dH2 (s), v(s) dK2 (s), (7.25)

k where α, β ∈ (3, 4], p1 , p2 ∈ (0, 1), q1 , q2 > 0, λ, μ > 0, cD0+ denotes the Caputo derivative of fractional order k (for k = α, α − 1, β, β − 1, p1 , p2 ), j represents the Riemann-Liouville integral of fractional order j (for I0+ j = q1 , q2 ), the integrals from (7.25) are Riemann-Stieltjes integrals, and H1 , H2 , K1 , K2 are functions of bounded variation. Under some assumptions on the functions f and g, we prove the existence, and the existence and uniqueness of solutions for problem (7.24), (7.25), by applying the Leray-Schauder alternative (Theorem 1.2.5) and the Banach contraction principle (Theorem 1.2.1). The system (7.24) with λ = μ, α, β ∈ (2, 3], p1 , p2 ∈ (0, 1), q1 , q2 ∈ (0, 1), with the nonlocal six-point coupled Riemann-Liouville type integral

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boundary conditions  u(0) = u (0) = 0, v(0) = v  (0) = 0,

p a1 u(1) + a2 u(ζ) = aI0+ v(η),

q b1 v(1) + b2 v(z) = bI0+ u(θ),

where p > 0, ζ, η ∈ (0, 1), q > 0, z, θ ∈ (0, 1), a, b, ai , bi , i = 1, 2, are real constants, was investigated in [5]. The system (7.24) with α, β ∈ (2, 3] and the functions f and g dependent only on t, u and v, subject to the boundary conditions u(0) = u (0) = 0, u(1) = av(ξ), v(0) = v  (0) = 0, v(1) = bu(η), ξ, η ∈ (0, 1) has been studied in [61]. The system (7.24) with λ = μ, α, β ∈ (2, 3] and the functions f and g dependent only on t, u and v, subject to the boundary conditions  β1 u(η), u(0) = 0, u (0) = 0, u(ζ) = aI0+ v(0) = 0,

v  (0) = 0,

γ1 v(z) = bI0+ v(θ),

where β1 > 0, 0 < η < ζ < 1, γ1 > 0, 0 < θ < z < 1, and a, b are real constants, was investigated in [11]. 7.2.1

Preliminary results

We consider the fractional differential system c α α−1 )u(t) = x(t), ( D0+ + λ cD0+

t ∈ (0, 1),

β β−1 + μ cD0+ )v(t) = y(t), (cD0+

t ∈ (0, 1),

(7.26)

with the coupled boundary conditions (7.25), where x, y ∈ C[0, 1]. We denote by  1 1 1 (λ2 s2 − 2λs + 2 − 2e−λs ) dH1 (s), a = 3 (λ2 − 2λ + 2 − 2e−λ ) − 3 λ λ 0  1 1 (μ2 s2 − 2μs + 2 − 2e−μs ) dH2 (s), b= 3 μ 0  1 1 (λ2 s2 − 2λs + 2 − 2e−λs ) dK1 (s), c= 3 λ 0 d=

1 2 (μ − 2μ + 2 − 2e−μ ) μ3  1 1 − 3 (μ2 s2 − 2μs + 2 − 2e−μs ) dK2 (s), μ 0

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Δ0 = ad − bc, t ∈ [0, 1], γ2 (t) =

γ1 (t) =

page 389

389

λ2 t2 − 2λt + 2 − 2e−λt , λ3 Δ0

(for Δ0 = 0),

μ2 t2 − 2μt + 2 − 2e−μt , μ3 Δ0

t ∈ [0, 1],

(for Δ0 = 0). (7.27)

Lemma 7.2.1. Let x, y ∈ C[0, 1] and Δ0 = 0. Then the solution (u, v) ∈ (C 4 [0, 1])2 of the boundary value problem (7.26), (7.25) is given by    u(t) = γ1 (t) d − 

0



1

s

+ 0

0



1



s

+ 0

0

  +b −

1



1

s

+ 0

0





1

s

+ 

0

0

t

+ 0

α−1 e−λ(s−τ ) I0+ x(τ ) dτ β−1 e−μ(s−τ ) I0+ y(τ ) dτ

α−1 e−λ(s−τ ) I0+ x(τ ) dτ β−1 e−μ(s−τ ) I0+ y(τ ) dτ

α−1 e−λ(t−s) I0+ x(s) ds,

   v(t) = γ2 (t) a − 

1

0



s

+ 0

0



1



s

+ 0

0

  +c − 

1

0

1



+ 0

α−1 e−λ(1−s) I0+ x(s) ds

0

dH1 (s)

 dH2 (s)

β−1 e−μ(1−s) I0+ y(s) ds

0



1

1

dK1 (s)

 dK2 (s)

t ∈ [0, 1],

β−1 e−μ(1−s) I0+ y(s) ds

α−1 e−λ(s−τ ) I0+ x(τ ) dτ β−1 e−μ(s−τ ) I0+ y(τ ) dτ

dK1 (s)

 dK2 (s)

α−1 e−λ(1−s) I0+ x(s) ds

s

α−1 e−λ(s−τ ) I0+ x(τ ) dτ

dH1 (s)

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1

s

+ 

0 t

+ 0

0

β−1 e−μ(s−τ ) I0+ y(τ ) dτ

β−1 e−μ(t−s) I0+ y(s) ds,



 dH2 (s)

t ∈ [0, 1],

(7.28)

where a, b, c, d, Δ0 and the functions γ1 , γ2 are given by (7.27). Proof. The system (7.26) is equivalent to the following system  c α D0+ (u(t) + λD−1 u(t)) = x(t), t ∈ (0, 1), β D0+ (v(t) + μD−1 v(t)) = y(t),

c

t ∈ (0, 1).

(7.29)

The general solution of problem (7.29), (7.25) can be written as  t 1 u(t) + λD−1 u(t) = (t − s)α−1 x(s) ds + a0 + a1 t + a 2 t2 + a 3 t3 , Γ(α) 0 t ∈ [0, 1],  t 1 v(t) + μD−1 v(t) = (t − s)β−1 y(s) ds + b0 + b1 t + b2 t2 + b3 t3 , Γ(β) 0 t ∈ [0, 1], or equivalently  t u(s) ds + u(t) = −λ 0

 v(t) = −μ

t

0

v(s) ds +

1 Γ(α) 1 Γ(β)

 

t

0

0

t

(t − s)α−1 x(s) ds + a0 + a1 t + a 2 t2 + a 3 t3 ,

(t − s)β−1 y(s) ds + b0 + b1 t + b2 t2 + b3 t3 ,

with ai , bi ∈ R, i = 0, . . . , 3. By differentiating the above relations, we obtain  t 1 (t − s)α−2 x(s) ds + a1 + 2 a2 t + 3 a 3 t2 , u (t) = −λu(t) + Γ(α − 1) 0  t 1 (t − s)β−2 y(s) ds + b1 + 2 b2 t + 3 b3 t2 , v  (t) = −μv(t) + Γ(β − 1) 0 and then eλt (e u(t)) = Γ(α − 1) λt



(eμt v(t)) =

eμt Γ(β − 1)

 

t

0

0

t

(t − s)α−2 x(s) ds + a1 eλt + 2 a2 teλt + 3 a3 t2 eλt , (t − s)β−2 y(s) ds + b1 eμt + 2 b2 teμt + 3 b3 t2 eμt .

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391

Integrating from 0 to t, we have u(t) = c0 e−λt +

c 1 c2 (1 − e−λt ) + 2 (λt − 1 + e−λt ) λ λ

c 3 2 2 (λ t − 2λt + 2 − 2e−λt ) λ3  t α−1 + e−λ(t−s) I0+ x(s) ds,

+

0

d 1 d 2 v(t) = d 0 e−μt + (1 − e−μt ) + 2 (μt − 1 + e−μt ) μ μ d 3 2 2 (μ t − 2μt + 2 − 2e−μt ) μ3  t β−1 + e−μ(t−s) I0+ y(s) ds,

+

0

where c0 = u(0), d 0 = v(0), c1 = a1 , c2 = 2 a2 , c3 = 3 a3 , d 1 = b1 , d 2 = 2 b2 , d3 = 3b3 . Using the boundary conditions u(0) = u (0) = u (0) = 0 and v(0) =  v (0) = v  (0) = 0 we deduce that c0 = c1 = c2 = 0 and d 0 = d 1 = d 2 = 0. Then the above solution becomes  t c3 α−1 e−λ(t−s) I0+ x(s) ds, t ∈ [0, 1], u(t) = 3 (λ2 t2 − 2λt + 2 − 2e−λt ) + λ 0  t d 3 β−1 v(t) = 3 (μ2 t2 − 2μt + 2 − 2e−μt ) + e−μ(t−s) I0+ y(s) ds, t ∈ [0, 1]. μ 0 (7.30)

1 Now by using the boundary conditions u(1) = 0 u(s) dH1 (s) +

1

1

1 0 v(s) dH2 (s), v(1) = 0 u(s) dK1 (s) + 0 v(s) dK2 (s), we obtain the following system in the unknowns c3 and d 3  1 c3 2 α−1 −λ (λ − 2λ + 2 − 2e ) + e−λ(1−s) I0+ x(s) ds λ3 0  c3 1 2 2 = 3 (λ s − 2λs + 2 − 2e−λs ) dH1 (s) λ 0  1  s α−1 e−λ(s−τ ) I0+ x(τ ) dτ dH1 (s) + 0

0

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 d 3 1 2 2 + 3 (μ s − 2μs + 2 − 2e−μs ) dH2 (s) μ 0  1  s β−1 + e−μ(s−τ ) I0+ y(τ ) dτ dH2 (s), 0

0

 1 d 3 2 β−1 −μ (μ − 2μ + 2 − 2e ) + e−μ(1−s) I0+ y(s) ds μ3 0  c3 1 2 2 = 3 (λ s − 2λs + 2 − 2e−λs ) dK1 (s) λ 0  1  s −λ(s−τ ) α−1 e I0+ x(τ ) dτ dK1 (s) + 0

0

 d 3 1 2 2 + 3 (μ s − 2μs + 2 − 2e−μs ) dK2 (s) μ 0  1  s −μ(s−τ ) β−1 + e I0+ y(τ ) dτ dK2 (s), 0

0

or ⎧  1 ⎪ α−1 ⎪ 3 = − ⎪ a c − b d e−λ(1−s) I0+ x(s) ds 3 ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪  1  s ⎪ ⎪ ⎪ −λ(s−τ ) α−1 ⎪ ⎪ + e I x(τ ) dτ dH1 (s) ⎪ 0+ ⎪ ⎪ 0 0 ⎪ ⎪ ⎪  1  s ⎪ ⎪ ⎪ ⎪ −μ(s−τ ) β−1 ⎪ e I0+ y(τ ) dτ dH2 (s), ⎪ ⎨ + 0

0

0

0

 1 ⎪ ⎪ ⎪ β−1 3 = − ⎪ −c c + d d e−μ(1−s) I0+ y(s) ds ⎪ 3 ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪  1  s ⎪ ⎪ ⎪ −λ(s−τ ) α−1 ⎪ e I x(τ ) dτ dK1 (s) + ⎪ 0+ ⎪ ⎪ 0 0 ⎪ ⎪ ⎪ ⎪  1  s ⎪ ⎪ ⎪ −μ(s−τ ) β−1 ⎪ e I0+ y(τ ) dτ dK2 (s), ⎩ +

(7.31)

where a, b, c and d are given by (7.27). The system (7.31) has the determinant Δ0 = ad − bc which, by the assumption of this lemma is different from zero. Then the unique solution

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393

( c3 , d 3 ) of (7.31) is

   1 1 α−1 e−λ(1−s) I0+ x(s) ds c3 = d − Δ0 0  1  s −λ(s−τ ) α−1 + e I0+ x(τ ) dτ dH1 (s) 0

0



1



s

+ 0

0

  +b −

1

1



s

+ 0

0



1



s

+ 0



 dH2 (s)

β−1 e−μ(1−s) I0+ y(s) ds

0



β−1 e−μ(s−τ ) I0+ y(τ ) dτ

0

α−1 e−λ(s−τ ) I0+ x(τ ) dτ β−1 e−μ(s−τ ) I0+ y(τ ) dτ

dK1 (s)

 dK2 (s) ,

   1 1 β−1 d3 = a − e−μ(1−s) I0+ y(s) ds Δ0 0  1  s −λ(s−τ ) α−1 e I0+ x(τ ) dτ dK1 (s) + 0

0



1



s

+ 0

0

  +c −

1

0



1



s



0

1



+ 0

0



 dK2 (s)

α−1 e−λ(1−s) I0+ x(s) ds

+ 0

β−1 e−μ(s−τ ) I0+ y(τ ) dτ

s

α−1 e−λ(s−τ ) I0+ x(τ ) dτ β−1 e−μ(s−τ ) I0+ y(τ ) dτ

dH1 (s)

 dH2 (s) .

Replacing the above constants c3 and d 3 in (7.30), we obtain the solution (u(t), v(t)), t ∈ [0, 1] of problem (7.26), (7.25) given by (7.28).  Lemma 7.2.2. For x, y ∈ C[0, 1] with x = supt∈[0,1] |x(t)| and y = supt∈[0,1] |y(t)|, we have the following inequalities:  1    1 α−1 (1 − e−λ )x; (a)  e−λ(1−s) I0+ x(s) ds ≤ λΓ(α) 0

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(b) (c) (d) (e)

(f)

(g)

(h)

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 1    1 −μ(1−s) β−1  (1 − e−μ )y; e I0+ y(s) ds ≤  μΓ(β) 0  t    1 −λ(t−s) α−1  (1 − e−λ )x, ∀t ∈ [0, 1]; e I0+ x(s) ds ≤  λΓ(α) 0  t    1 −μ(t−s) β−1  (1 − e−μ )y, ∀t ∈ [0, 1]; e I0+ y(s) ds ≤  μΓ(β) 0   1  s   −λ(s−τ ) α−1   e I x(τ ) dτ dH (s) 1 0+   0 0  1   x  ≤ sα−1 (1 − e−λs )dH1 (s) ;  λΓ(α) 0  1  s    −λ(s−τ ) α−1   e I x(τ ) dτ dK (s) 1 0+   0 0  1   x  sα−1 (1 − e−λs )dK1 (s) ; ≤  λΓ(α) 0  1  s    −μ(s−τ ) β−1   e I y(τ ) dτ dH (s) 2 0+   0 0  1   y  ≤ sβ−1 (1 − e−μs )dH2 (s) ;  μΓ(β) 0  1  s    −μ(s−τ ) β−1   e I y(τ ) dτ dK (s) 2 0+   0 0  1   y  sβ−1 (1 − e−μs )dK2 (s). ≤  μΓ(β) 0

Proof. For the inequality from (a), we have  1    −λ(1−s) α−1  e I0+ x(s) ds  0

  1 (s − τ )α−2 x(τ ) dτ ds 0 0 Γ(α − 1)  s  1 1 −λ(1−s) α−2 ≤ x (s − τ ) e dτ ds 0 0 Γ(α − 1)  1 1 x (1 − e−λ )x. e−λ(1−s) sα−1 ds ≤ = Γ(α) 0 λΓ(α)   = 

1

e−λ(1−s)



s

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For the inequality from (c), we get   t   −λ(t−s) α−1  e I0+ x(s) ds  0  t   s   1 −λ(t−s) α−2  (s − τ ) e x(τ ) dτ ds = 0 0 Γ(α − 1)  s  t 1 −λ(t−s) α−2 ≤ x (s − τ ) e dτ ds 0 0 Γ(α − 1)  t x e−λ(t−s) sα−1 ds = Γ(α) 0 ≤

1 1 (1 − e−λt )x ≤ (1 − e−λ )x, λΓ(α) λΓ(α)

∀t ∈ [0, 1].

For the inequality from (e), we have    

1

0

 0

s

α−1 e−λ(s−τ ) I0+ x(τ ) dτ



  dH1 (s)

  s   x  1 −λ(s−τ ) α−1 ≤ e τ dτ dH1 (s) Γ(α)  0 0  1  s   x  −λ(s−τ ) α−1 e s dτ dH1 (s) ≤ Γ(α)  0 0  1   x  α−1 −λs ≤ s (1 − e )dH1 (s) . λΓ(α)  0

In a similar manner, we also deduce the inequalities (b), (d), (f), (g) and (h). The inequalities (a)–(d) were also used in [5].  7.2.2

Existence of solutions

p2 We consider the spaces X = {u ∈ C[0, 1], cD0+ u ∈ C[0, 1]} and Y = {v ∈ c p1 C[0, 1], D0+ v ∈ C[0, 1]} equipped respectively with the norms uX = p2 p1 u and vY = v + cD0+ v, where  ·  is the supremum u + cD0+ norm, that is w = supt∈[0,1] |w(t)| for w ∈ C[0, 1]. The spaces (X ,  · X ) and (Y,  · Y ) are Banach spaces, and the product space X × Y endowed with the norm (u, v)X ×Y = uX + vY is also a Banach space. By using Lemma 7.2.1, we introduce the operator T : X × Y → X × Y defined by T (u, v) = (T1 (u, v), T2 (u, v)) for (u, v) ∈ X × Y, where the

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operators T1 : X × Y → X and T2 : X × Y → Y are given by    1 T1 (u, v)(t) = γ1 (t) d − e−λ(1−s) 0

p1 q1 α−1 × I0+ f (s, u(s), v(s),c D0+ v(s), I0+ v(s)) ds



1



s

+ 0

0

α−1 e−λ(s−τ ) I0+ f (τ, u(τ ), v(τ ),

p1 q1 D0+ v(τ ), I0+ v(τ )) dτ

c



1



s

+ 0

0

dH1 (s)

β−1 e−μ(s−τ ) I0+ g(τ, u(τ ),

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ )) dτ

c

  +b −

1



 dH2 (s)

β−1 e−μ(1−s) I0+ g(s, u(s),

0 p2 q2 D0+ u(s), I0+ u(s), v(s)) ds

c



1



s

+ 0 c

0

α−1 e−λ(s−τ ) I0+ f (τ, u(τ ), v(τ ),

p1 q1 D0+ v(τ ), I0+ v(τ )) dτ



1



+ 0

0

s

dK1 (s)

β−1 e−μ(s−τ ) I0+ g(τ, u(τ ),

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ )) dτ

c





 dK2 (s)

t

α−1 e−λ(t−s) I0+ f (s, u(s), v(s), 0 q1 c p1 D0+ v(s), I0+ v(s)) ds, t ∈ [0, 1],

+

   T2 (u, v)(t) = γ2 (t) a −

0

1

(u, v) ∈ X × Y,

β−1 e−μ(1−s) I0+ g(s, u(s),

p2 q2 D0+ u(s), I0+ u(s), v(s)) ds  1  s α−1 e−λ(s−τ ) I0+ f (τ, u(τ ), v(τ ), + c

0

0

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p1 q1 D0+ v(τ ), I0+ v(τ )) dτ

c



1



s

+ 0

0

dK1 (s)

β−1 e−μ(s−τ ) I0+ g(τ, u(τ ),

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ )) dτ 1

0

397



c

  +c −

page 397



 dK2 (s)

α−1 e−λ(1−s) I0+ f (s, u(s), v(s),

p1 q1 D0+ v(s), I0+ v(s)) ds  1  s α−1 e−λ(s−τ ) I0+ f (τ, u(τ ), v(τ ), + c

0

0

p1 q1 D0+ v(τ ), I0+ v(τ )) dτ

c



1



+ 0

0

s

dH1 (s)

β−1 e−μ(s−τ ) I0+ g(τ, u(τ ),

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ )) dτ

c





 dH2 (s)

t

β−1 e−μ(t−s) I0+ g(s, u(s), 0 q2 c p2 D0+ u(s), I0+ u(s), v(s)) ds,

+

t ∈ [0, 1], (u, v) ∈ X × Y.

The pair (u, v) is a solution of problem (7.24), (7.25) if and only if (u, v) is a fixed point of operator T . We present now the assumptions that we use in this section. (J1) α, β ∈ R, 3 < α ≤ 4, 3 < β ≤ 4, p1 , p2 ∈ (0, 1), q1 , q2 > 0, Δ0 = 0 (given by (7.27)), H1 , H2 , K1 , K2 : [0, 1] → R are functions of bounded variation. (J2) The functions f, g : [0, 1] × R4 → R are continuous and there exist real constants ai , bi ≥ 0, i = 1, . . . , 4 and a0 > 0, b0 > 0 such that |f (t, x1 , x2 , x3 , x4 )| ≤ a0 + a1 |x1 | + a2 |x2 | + a3 |x3 | + a4 |x4 |, |g(t, x1 , x2 , x3 , x4 )| ≤ b0 + b1 |x1 | + b2 |x2 | + b3 |x3 | + b4 |x4 |, for all t ∈ [0, 1] and xi ∈ R,

i = 1, . . . , 4.

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(J3) The functions f, g : [0, 1] × R4 → R are continuous and there exist positive constants c0 , d0 such that |f (t, x1 , x2 , x3 , x4 ) − f (t, y1 , y2 , y3 , y4 )| ≤ c0 (|x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 |), |g(t, x1 , x2 , x3 , x4 ) − g(t, y1 , y2 , y3 , y4 )| ≤ d0 (|x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 |), for all t ∈ [0, 1] and xi , yi ∈ R, i = 1, . . . , 4. We denote by l1 = sup |γ1 (t)| =

λ2 − 2λ + 2 − 2e−λ > 0, λ3 |Δ0 |

l2 = sup |γ2 (t)| =

μ2 − 2μ + 2 − 2e−μ > 0, μ3 |Δ0 |

t∈[0,1]

t∈[0,1]

m1 = sup |γ1 (t)| = t∈[0,1]

2λ2 − 2λ + 2λe−λ > 0, λ3 |Δ0 |

2μ2 − 2μ + 2μe−μ > 0, μ3 |Δ0 |  1      −λ α−1 −λs  s (1 − e )dH1 (s) |d| (1 − e ) + 

m2 = sup |γ2 (t)| = t∈[0,1]

M1 =

l1 λΓ(α)   + |b| 

0

1

0

  sα−1 (1 − e−λs )dK1 (s) +

1 (1 − e−λ ), λΓ(α)  1  β−1 −μs s (1 − e )dH2 (s)

   l1 M2 = |d|  μΓ(β) 0    + |b| (1 − e−μ ) +  M1∗

m1 = λΓ(α)   + |b| 

1 0

  sβ−1 (1 − e−μs )dK2 (s) ,

    −λ |d| (1 − e ) + 

0

0

1

1

s

α−1

(1 − e

  sα−1 (1 − e−λs )dK1 (s) +

−λs

  )dH1 (s)

1 (2 − e−λ ), Γ(α)

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M2∗

page 399

   1   m1 β−1 −μs  = s (1 − e )dH2 (s) |d|  μΓ(β) 0  1     −μ β−1 −μs  + |b| (1 − e ) +  s (1 − e )dK2 (s) ,

1 = M

2 = M

0

  l2 α−1 −λs s (1 − e )dK1 (s) λΓ(α) 0  1     −λ α−1 −λs  + |c| (1 − e ) +  s (1 − e )dH1 (s) ,    |a| 



l2 μΓ(β)   + |c| 

1

0

   −μ |a| (1 − e ) + 

1

0

1

s

β−1

(1 − e

−μs

  )dK2 (s)

  )dH2 (s) +

1 (1 − e−μ ), μΓ(β) 0    1   m2 ∗ α−1 −λs   M1 = s (1 − e )dK1 (s) |a|  λΓ(α) 0  1     −λ α−1 −λs  + |c| (1 − e ) +  s (1 − e )dH1 (s) , ∗ = M 2



m2 μΓ(β)   + |c| 

0

s

β−1

(1 − e

−μs

0

   −μ |a| (1 − e ) + 

0

1

s

β−1

1 + N1 = M 1 + M

(1 − e

−μs

1

s

β−1

(1 − e

  )dH2 (s) +

−μs

  )dK2 (s)

1 (2 − e−μ ), Γ(β)

∗ M M1∗ 1 + , Γ(2 − p2 ) Γ(2 − p1 )

2∗ M M2∗ + , Γ(2 − p2 ) Γ(2 − p1 )  b3 Λ 2 = a1 N 1 + b 1 + b 2 + N2 , Γ(q2 + 1)  a4 Λ 3 = a2 + a3 + N1 + b4 N2 , Γ(q1 + 1)

2 + N2 = M 2 + M

Λ 1 = a0 N 1 + b 0 N 2 ,

1 = 1 +

1 , Γ(1 + q1 )

2 = 1 +

1 . Γ(1 + q2 )

Theorem 7.2.1. Assume that (J1) and (J2) hold. If max{Λ2 , Λ3 } < 1, then the boundary value problem (7.24), (7.25) has at least one solution on [0, 1].

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Proof. We firstly prove that the operator T : X ×Y → X ×Y is completely continuous. By continuity of the functions f and g , we deduce that the operators T1 and T2 are continuous, and then T is a continuous operator. Next we show that T is uniformly bounded. Let Ω ⊂ X ×Y be a bounded set. Then there exist positive constants L1 and L2 such that p1 q1 v(t), I0+ v(t))| ≤ L1 , |f (t, u(t), v(t),c D0+ p2 q2 |g(t, u(t),c D0+ u(t), I0+ u(t), v(t))| ≤ L2 , for all (u, v) ∈ Ω and t ∈ [0, 1]. Then for any (u, v) ∈ Ω and t ∈ [0, 1], we obtain   1 α−1 |T1 (u, v)(t)| ≤ |γ1 (t)| |d| e−λ(1−s) (I0+ |f (s, u(s), v(s), 0

p1 q1 D0+ v(s), I0+ v(s))|)ds  1  s  α−1 +  e−λ(s−τ ) (I0+ |f (τ, u(τ ), v(τ ), c

0

0

p1 q1 D0+ v(τ ), I0+ v(τ ))|) dτ

c

  + 

1



0

s 0



  dH1 (s)

 β−1 e−μ(s−τ ) I0+ |g(τ, u(τ ),

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ ))|

c



dτ

  dH2 (s)

 β−1 e−μ(1−s) I0+ |g(s, u(s), 0  q2 c p2 D0+ u(s), I0+ u(s), v(s))| ds  1  s  α−1 e−λ(s−τ ) (I0+ |f (τ, u(τ ), v(τ ), +  + |b|

1

0

0

p1 q1 D0+ v(τ ), I0+ v(τ ))|) dτ

c

  + 

1

0



s 0



  dK1 (s)

 β−1 e−μ(s−τ ) I0+ |g(τ, u(τ ), 

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ ))|

c



t

dτ

  dK2 (s)

α−1 e−λ(t−s) (I0+ |f (s, u(s), v(s), 0 q1 c p1 D0+ v(s), I0+ v(s))|) ds

+

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  ≤ l1 |d| L1

1 (1 − e−λ ) λΓ(α)    1  1 α−1 −λs  + L1 s (1 − e )dH (s) 1   λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dH (s) 2   μΓ(β) 0  1 + |b| L2 (1 − e−μ ) μΓ(β)    1  1 α−1 −λs + L1 s (1 − e )dK1 (s)  λΓ(α) 0    1  1 β−1 −μs + L2 s (1 − e )dK2 (s)  μΓ(β) 0

+

L1 (1 − e−λ ) = L1 M1 + L2 M2 . λΓ(α)

Then T1 (u, v) ≤ L1 M1 + L2 M2 , for all (u, v) ∈ Ω. From the definition of T1 (u, v), we have |T1 (u, v)(t)|

≤

|γ1 (t)|

  |d|

0

1

α−1 e−λ(1−s) (I0+ |f (s, u(s), v(s),

p1 q1 D0+ v(s), I0+ v(s))|)ds  1  s  α−1 +  e−λ(s−τ ) (I0+ |f (τ, u(τ ), v(τ ), c

0

0

p1 q1 D0+ v(τ ), I0+ v(τ ))|) dτ

c

  + 





  dH1 (s)

 β−1 e−μ(s−τ ) I0+ |g(τ, u(τ ), 0 0    q2 c p2 D0+ u(τ ), I0+ u(τ ), v(τ ))| dτ dH2 (s) 1



+ |b|

0

1

s

 β−1 e−μ(1−s) I0+ |g(s, u(s),

 p2 q2 D0+ u(s), I0+ u(s), v(s))| ds  1  s  α−1 +  e−λ(s−τ ) (I0+ |f (τ, u(τ ), v(τ ), c

0

0

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p1 q1 D0+ v(τ ), I0+ v(τ ))|) dτ

c

  + 

1 0

 0

s



  dK1 (s)

 β−1 e−μ(s−τ ) I0+ |g(τ, u(τ ), 

p2 q2 D0+ u(τ ), I0+ u(τ ), v(τ ))|

c

 +λ 

0

t

dτ

  dK2 (s)

p1 q1 α−1 e−λ(t−s) (I0+ |f (s, u(s), v(s),c D0+ v(s), I0+ v(s))|)ds

t

(t − s)α−2 p1 q1 |f (s, u(s), v(s),c D0+ v(s), I0+ v(s))|ds 0 Γ(α − 1)   1 ≤ m1 |d| L1 (1 − e−λ ) λΓ(α)    1  1 α−1 −λs  + L1 s (1 − e )dH (s) 1   λΓ(α) 0    1  1 β−1 −μs  +L2 s (1 − e )dH (s) 2   μΓ(β) 0  1 + |b| L2 (1 − e−μ ) μΓ(β)    1  1 α−1 −λs  + L1 s (1 − e )dK (s) 1   λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dK (s) 2   μΓ(β) 0 +

+

L1 (2 − e−λ ) = L1 M1∗ + L2 M2∗ . Γ(α)

By the definition of Caputo fractional derivative of order p2 ∈ (0, 1), we deduce 

t

(t − s)−p2  |T1 (u, v)(s)|ds 0 Γ(1 − p2 )  t (t − s)−p2 ds ≤ (L1 M1∗ + L2 M2∗ ) 0 Γ(1 − p2 )

p2 T1 (u, v)(t)| ≤ |cD0+

=

L1 M1∗ + L2 M2∗ , Γ(2 − p2 )

∀t ∈ [0, 1],

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from where we obtain p2 cD0+ T1 (u, v) ≤

page 403

403

L1 M1∗ + L2 M2∗ . Γ(2 − p2 )

Therefore we conclude p2 T1 (u, v) T1 (u, v)X = T1 (u, v) + cD0+

≤ L1 M1 + L2 M2 +

1 (L1 M1∗ + L2 M2∗ ). Γ(2 − p2 )

In a similar manner, we have   |T2 (u, v)(t)| ≤ l2 |a| L2

1 (1 − e−μ ) μΓ(β)    1  1 α−1 −λs  + L1 s (1 − e )dK (s) 1   λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dK (s) 2   μΓ(β) 0  1 + |c| L1 (1 − e−λ ) λΓ(α)    1  1 α−1 −λs + L1 s (1 − e )dH1 (s)  λΓ(α) 0    1  1 β−1 −μs + L2 s (1 − e )dH2 (s)  μΓ(β) 0

1 1 + L2 M 2 , (1 − e−μ ) = L1 M μΓ(β)   1 |T2 (u, v)(t)| ≤ m2 |a| L2 (1 − e−μ ) μΓ(β)    1  1 α−1 −λs  + L1 s (1 − e )dK (s) 1   λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dK (s) 2   μΓ(β) 0  1 + |c| L1 (1 − e−λ ) λΓ(α)    1  1 α−1 −λs + L1 s (1 − e )dH1 (s)  λΓ(α) 0    1  1 β−1 −μs s (1 − e )dH2 (s) + L2  μΓ(β) 0 + L2

(7.32)

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+ p1 T2 (u, v)(t)| ≤ |cD0+

L2 1∗ + L2 M 2∗ , (2 − e−μ ) = L1 M Γ(β)

∗ + L2 M ∗ L1 M 1 2 , Γ(2 − p1 )

∀t ∈ [0, 1].

Then we deduce p1 T2 (u, v) T2 (u, v)Y = T2 (u, v) + cD0+

1 1∗ + L2 M 2∗ ). (L1 M (7.33) Γ(2 − p1 ) From the inequalities (7.32) and (7.33), we conclude that T1 and T2 are uniformly bounded, which implies that the operator T is uniformly bounded. We will show next that T is equicontinuous. Let t1 , t2 ∈ [0, 1], with t1 < t2 . Then we have 1 + L2 M 2 + ≤ L2 M

|T1 (u, v)(t2 ) − T1 (u, v)(t1 )|   ≤ |γ1 (t2 ) − γ1 (t1 )| |d| L1

1 (1 − e−λ ) λΓ(α)    1  1 α−1 −λs  + L1 s (1 − e )dH (s) 1   λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dH (s) 2   μΓ(β) 0     1 1  1 α−1 −λs  + |b| L2 (1 − e−μ ) + L1 s (1 − e )dK (s) 1   μΓ(β) λΓ(α) 0    1  1 β−1 −μs  + L2 s (1 − e )dK (s) 2   μΓ(β) 0  t2  p1 q1 α−1 +  e−λ(t2 −s) I0+ f (s, u(s), v(s),c D0+ v(s), I0+ v(s)) ds  −

t1 0

2

=

0

  p1 q1 α−1 e−λ(t1 −s) I0+ f (s, u(s), v(s),c D0+ v(s), I0+ v(s)) ds

|λ (t22 − t21 ) − 2λ(t2 − t1 ) + 2 − 2e−λt2 − 2 + 2e−λt1 | Λ0 λ3 |Δ0 |  s  t1  1 −λ(t2 −s) −λ(t1 −s)  + (e −e )  (s − τ )α−2 Γ(α − 1) 0 0

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page 405

405

 p1 q1 × |f (τ, u(τ ), v(τ ),c D0+ v(τ ), I0+ v(τ ))| dτ  ds  s  t2  1 + e−λ(t2 −s)  (s − τ )α−2 Γ(α − 1) t1 0   p1 q1 × f (τ, u(τ ), v(τ ),c D0+ v(τ ), I0+ v(τ )) dτ  ds ≤

≤

=

where

λ2 (t22 − t21 ) + 2λ(t2 − t1 ) + 2(e−λt1 − e−λt2 ) Λ0 λ3 |Δ0 |  t1    s L1 −λ(t2 −s) −λ(t1 −s) α−2 e + −e (s − τ ) dτ ds Γ(α − 1) 0 0  s  t2 L1 −λ(t2 −s) α−2 + e (s − τ ) dτ ds Γ(α − 1) t1 0 λ2 (t22 − t21 ) + 2λ(t2 − t1 ) + 2e−λt1 (1 − e−λ(t2 −t1 ) ) Λ0 λ3 |Δ0 |  t1   t2  L1 tα−1 L1 tα−1 1 2 e−λ(t2 −s) − e−λ(t1 −s) ds + + e−λ(t2 −s) ds Γ(α) 0 Γ(α) t1 λ2 (t22 − t21 ) + 2λ(t2 − t1 ) + 2e−λt1 (1 − e−λ(t2 −t1 ) ) Λ0 λ3 |Δ0 |   L1 tα−1 1 e−λ(t2 −t1 ) − e−λt2 − 1 + e−λt1 + λΓ(α)   L1 tα−1 2 1 − e−λ(t2 −t1 ) , + λΓ(α)

   1 1  1 α−1 −λ −λs Λ0 = |d| L1 (1 − e ) + L1 s (1 − e )dH1 (s) λΓ(α) λΓ(α)  0     1  1 β−1 1 −μs  + L2 (1 − e−μ ) s (1 − e )dH2 (s) + |b| L2 μΓ(β)  0 μΓ(β)    1  1 α−1 −λs + L1 s (1 − e )dK1 (s) λΓ(α)  0    1  1 β−1 −μs s (1 − e )dK2 (s) . + L2 μΓ(β)  0 

Then |T1 (u, v)(t2 ) − T1 (u, v)(t1 )| → 0, as t2 → t1 .

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In addition, we obtain p2 p2 T1 (u, v)(t2 ) −c D0+ T1 (u, v)(t1 )| |cD0+  t1 |(t1 − s)p2 − (t2 − s)p2 |  1 |T1 (u, v)(s)| ds ≤ Γ(1 − p2 ) 0 (t1 − s)p2 (t2 − s)p2  t2 1 (t2 − s)−p2 |T1 (u, v)(s)|ds + Γ(1 − p2 ) t1   t1  t2 L1 M1∗ + L2 M2∗ −(t1 − s)p2 + (t2 − s)p2 −p2 ≤ ds + (t − s) ds 2 Γ(1 − p2 ) (t1 − s)p2 (t2 − s)p2 0 t1  L1 M1∗ + L2 M2∗  2 2 = → 0, as t2 → t1 . 2(t2 − t1 )1−p2 − t1−p + t1−p 2 1 Γ(2 − p2 )

In a similar manner, we have |T2 (u, v)(t2 ) − T2 (u, v)(t1 )| → 0

and,

p1 p1 |cD0+ T2 (u, v)(t2 ) −c D0+ T2 (u, v)(t1 )| → 0,

as t2 → t1 .

Thus the operators T1 and T2 are equicontinuous, and then T is also equicontinuous. Hence by Arzela-Ascoli theorem, we deduce that T is compact. Therefore we conclude that T is completely continuous. Now we will prove that the set F = {(u, v) ∈ X × Y, (u, v) = νT (u, v), 0 < ν < 1} is bounded. Let (u, v) ∈ F , that is (u, v) = νT (u, v) for some ν ∈ (0, 1). Then for any t ∈ [0, 1], we have u(t) = νT1 (u, v)(t), v(t) = νT2 (u, v)(t). From these last relations we deduce |u(t)| ≤ |T1 (u, v)(t)| and |v(t)| ≤ |T2 (u, v)(t)| for all t ∈ [0, 1]. Then by (J2), we obtain   1   s 1 −λ(1−s) e (s − τ )α−2 |u(t)| ≤ |γ1 (t)| |d| Γ(α − 1) 0 0 × [a0 + a1 |u(τ )| + a2 |v(τ )| p1 v(τ )| + a3 | D0+ c

+

 q1 a4 |I0+ v(τ )| dτ

ds

 1  s   τ  1 −λ(s−τ )  + e (τ − ζ)α−2 Γ(α − 1) 0 0 0  p1 × a0 + a1 |u(ζ)| + a2 |v(ζ)| + a3 |cD0+ v(ζ)|    q1 + a4 |I0+ v(ζ)| dζ dτ dH1 (s)

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 1  s   τ  1 −μ(s−τ )  + e (τ − ζ)β−2 Γ(β − 1) 0 0 0  p2 q2 × b0 + b1 |u(ζ)| + b2 |cD0+ u(ζ)| + b3 |I0+ u(ζ)|   + b4 |v(ζ)|] dζ dτ dH2 (s)  + |b|

0

1

e

−μ(1−s)



1 Γ(β − 1)

 0

s

(s − τ )β−2

 p2 × b0 + b1 |u(τ )| + b2 |cD0+ u(τ )|  q2 + b3 |I0+ u(τ )| + b4 |v(τ )| dτ ds

 1  s   τ  1 −λ(s−τ )  + e (τ − ζ)α−2 Γ(α − 1) 0 0 0  p1 × a0 + a1 |u(ζ)| + a2 |v(ζ)| + a3 |cD0+ v(ζ)|    q1 + a4 |I0+ v(ζ)| dζ dτ dK1 (s)  1  s   τ  1 +  e−μ(s−τ ) (τ − ζ)β−2 Γ(β − 1) 0 0 0  p2 q2 × b0 + b1 |u(ζ)| + b2 |cD0+ u(ζ)| + b3 |I0+ u(ζ)|   + b4 |v(ζ)|] dζ dτ dK2 (s)   t  s 1 + e−λ(t−s) (s − τ )α−2 Γ(α − 1) 0 0  p1 × a0 + a1 |u(τ )| + a2 |v(τ )| + a3 |cD0+ v(τ )|  q1 v(τ )| dτ ds, ∀t ∈ [0, 1]. + a4 |I0+ Therefore, we deduce   1  −λ(1−s) u ≤ l1 |d| e 0

1 Γ(α − 1)



s 0

(s − τ )

 × a0 + a1 uX + a2 vY + a3 vY +   + 

0

1

 0

s

e

−λ(s−τ )



1 Γ(α − 1)

α−2

dτ

ds

a4 vY Γ(q1 + 1)   τ  α−2 (τ − ζ) dζ dτ dH1 (s) 0

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 × a0 + a1 uX + a2 vY + a3 vY +   + 

0

1

 0

s

e−μ(s−τ )



1 Γ(β − 1)

 × b0 + b1 uX + b2 uX +

a4 vY Γ(q1 + 1)   τ  (τ − ζ)β−2 dζ dτ dH2 (s) 0

 b3 u|X + b4 vY Γ(q2 + 1)  1   s 1 e−μ(1−s) (s − τ )β−2 dτ ds + |b| Γ(β − 1) 0 0  b3 × b0 + b1 uX + b2 uX + uX + b4 vY Γ(q2 + 1)   1  s   τ   1 −λ(s−τ ) α−2  e (τ − ζ) dζ dτ dK1 (s) + Γ(α − 1) 0 0 0  a4 × a0 + a1 uX + a2 vY + a3 vY + vY Γ(q1 + 1)   1  s   τ   1 −μ(s−τ ) β−2  e (τ − ζ) dζ dτ dK2 (s) + Γ(β − 1) 0 0 0   b3 u|X + b4 vY × b0 + b1 uX + b2 uX + Γ(q2 + 1)  t   s 1 + e−λ(t−s) (s − τ )α−2 dτ ds Γ(α − 1) 0 0  a4 × a0 + a1 uX + a2 vY + a3 vY + vY Γ(q1 + 1)    1 −λ ≤ l1 |d| (1 − e ) a0 + a1 uX + a2 vY λΓ(α) a4 vY + a3 vY + Γ(q1 + 1)  1   1  α−1 −λs  a0 + a1 uX s (1 − e )dH (s) + 1   λΓ(α) 0 a4 vY + a2 vY + a3 vY + Γ(q1 + 1)  1   1  β−1 −μs  b0 + b1 uX + b2 uX s (1 − e )dH (s) + 2   μΓ(β) 0

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page 409

409

 b3 uX + b4 vY Γ(q2 + 1)   1 (1 − e−μ ) b0 + b1 uX + b2 uX + |b| μΓ(β) b3 uX + b4 vY + Γ(q2 + 1)    1  1 α−1 −λs + s (1 − e )dK1 (s) a0 + a1 uX λΓ(α)  0 a4 vY + a2 vY + a3 vY + Γ(q1 + 1)  1   1  β−1 −μs s (1 − e )dK2 (s) b0 + b1 uX +  μΓ(β) 0  b3 uX + b4 vY + b2 uX + Γ(q2 + 1)  1 −λ + (1 − e ) a0 + a1 uX + a2 vY λΓ(α) a4 vY + a3 vY + Γ(q1 + 1)    a4 = M1 a0 + a1 uX + a2 + a3 + vY Γ(q1 + 1)    b3 + M2 b0 + b1 + b2 + uX + b4 vY . Γ(q2 + 1) +

In a similar manner, we obtain    ∗ u  ≤ M1 a0 + a1 uX + a2 + a3 +

 a4 vY Γ(q1 + 1)    b3 + M2∗ b0 + b1 + b2 + uX + b4 vY , Γ(q2 + 1)

which gives us 

p2 D0+ u

c

    1 a4 ∗ ≤ M1 a0 + a1 uX + a2 + a3 + vY Γ(2 − p2 ) Γ(q1 + 1)    b3 + M2∗ b0 + b1 + b2 + uX + b4 vY . Γ(q2 + 1)

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Then we conclude uX

 ≤ M1 +

 M1∗ = u +  a0 + a1 uX Γ(2 − p2 )    a4 M2∗ + a2 + a3 + vY + M2 + Γ(q1 + 1) Γ(2 − p2 )    b3 × b0 + b1 + b2 + (7.34) uX + b4 vY . Γ(q2 + 1) p2 D0+ u

c

In a similar manner, we deduce for the function v ! 1∗ M c p1  vY = v +  D0+ v ≤ M1 + a0 + a1 uX Γ(2 − p1 )  + a2 + a3 +





∗ M 2 2 + vY + M Γ(2 − p1 )    b3 × b0 + b1 + b2 + uX + b4 vY . Γ(q2 + 1) a4 Γ(q1 + 1)

!

(7.35)

Therefore, by (7.34) and (7.35), we obtain (u, v)X ×Y = uX + vY 

∗ M M1∗ 1 1 + + ≤ uX a1 M1 + M Γ(2 − p2 ) Γ(2 − p1 )  b3 + b1 + b2 + Γ(q2 + 1) ! ∗ 2∗ M M 2 2 + + × M2 + M Γ(2 − p2 ) Γ(2 − p1 )  a4 + vY a2 + a3 + Γ(q1 + 1) ! ∗ ∗ M M 1 1 1 + + × M1 + M Γ(2 − p2 ) Γ(2 − p1 ) ! ∗ ∗ M M 2 2 2 + + b4 M2 + M + Γ(2 − p2 ) Γ(2 − p1 ) ! ∗ ∗  M M 1 1 1 + + + a0 M 1 + M Γ(2 − p2 ) Γ(2 − p1 )

!

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+ b0

2 + M2 + M

∗ M M2∗ 2 + Γ(2 − p2 ) Γ(2 − p1 )

!

≤ Λ1 + max{Λ2 , Λ3 }(u, v)X ×Y .

(7.36)

So by using the assumption max{Λ2 , Λ3 } < 1, we conclude (u, v)X ×Y ≤

Λ1 . 1 − max{Λ2 , Λ3 }

Hence, we deduce that the set F is bounded. By using Theorem 1.2.5, we conclude that the operator T has at least one fixed point, which is a solution for our problem (7.24), (7.25). This completes the proof.  Next, we will present an existence and uniqueness result for our problem (7.24), (7.25), based on the Banach contraction mapping principle (Theorem 1.2.1). We introduce the constants r1 = sup |f (t, 0, 0, 0, 0)|, t∈[0,1]

r2 = sup |g(t, 0, 0, 0, 0)|, t∈[0,1]

A1 = c0 1 M1 + d0 2 M2 ,

A∗1

= c0 1 M1∗ + d0 2 M2∗ ,

1 + d0 2 M 2 , 1 = c0 1 M A

∗ + d0 2 M ∗ , ∗ = c0 1 M A 1 1 2

D1 = r1 M1 + r2 M2 ,

D1∗ = r1 M1∗ + r2 M2∗ ,

1 + r2 M 2 , 1 = r1 M D

1∗ + r2 M 2∗ . 1∗ = r1 M D

Theorem 7.2.2. Assume that (J1) and (J3) hold. If 1 + A1 + A

∗1 A A∗1 + < 1, Γ(2 − p2 ) Γ(2 − p1 )

(7.37)

then problem (7.24), (7.25) has a unique solution. Proof. We consider the positive number r given by ! ∗ D D1∗ 1 r = D1 + D1 + + Γ(2 − p2 ) Γ(2 − p1 ) × 1−

∗1 A A∗1 1 + + A1 + A Γ(2 − p2 ) Γ(2 − p1 )

!!−1 .

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We will show that T (B r ) ⊂ B r , where B r = {(u, v) ∈ X × Y, (u, v)X ×Y ≤ r}. For (u, v) ∈ B r , we obtain p1 q1 |f (t, u(t), v(t),c D0+ v(t), I0+ v(t))| p1 q1 ≤ |f (t, u(t), v(t),c D0+ v(t), I0+ v(t)) − f (t, 0, 0, 0, 0)| p1 q1 v(t)| + |I0+ v(t)|) + r1 + |f (t, 0, 0, 0, 0)| ≤ c0 (|u(t)| + |v(t)| + |cD0+  1 vY + r1 ≤ c0 uX + vY + Γ(q1 + 1)

= c0 (uX + 1 vY ) + r1 ≤ c0 1 (u, v)X ×Y + r1 ≤ c0 1 r + r1 . In a similar manner, we have p2 q2 u(t), I0+ u(t), v(t))| ≤ d0 2 r + r2 . |g(t, u(t),c D0+

Then |T1 (u, v)(t)| ≤ (c0 1 r + r1 )M1 + (d0 2 r + r2 )M2 = (c0 1 M1 + d0 2 M2 )r + r1 M1 + r2 M2 = A1 r + D1 , ∀t ∈ [0, 1], and |T1 (u, v)(t)| ≤ (c0 1 r + r1 )M1∗ + (d0 2 r + r2 )M2∗ = (c0 1 M1∗ + d0 2 M2∗ )r + r1 M1∗ + r2 M2∗ = A∗1 r + D1∗ , ∀t ∈ [0, 1], which gives us c

|

p2 D0+ T1 (u, v)(t)|

 ≤ ≤

0

t

(t − s)−p2  |T (u, v)(s)|ds Γ(1 − p2 ) 1

1 (A∗ r + D1∗ ), Γ(2 − p2 ) 1

∀t ∈ [0, 1].

Therefore, we deduce p2 T1 (u, v) T1 (u, v)X = T1 (u, v) + cD0+  A∗1 D1∗ . ≤ A1 + r + D1 + Γ(2 − p2 ) Γ(2 − p2 )

(7.38)

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In a similar manner, we obtain 1 r + D 1 , |T2 (u, v)(t)| ≤ A ∗1 r + D 1∗ , |T2 (u, v)(t)| ≤ A  t (t − s)−p1  p1 |T2 (u, v)(s)| ds |cD0+ T2 (u, v)(t)| ≤ 0 Γ(1 − p1 ) ≤

1 ∗ r + D ∗ ), (A 1 Γ(2 − p1 ) 1

∀t ∈ [0, 1].

Then, we conclude p1 T2 (u, v) T2 (u, v)Y = T2 (u, v) + cD0+ ! ∗ ∗ A D 1 1 1 + 1 + ≤ A r+D . Γ(2 − p1 ) Γ(2 − p1 )

(7.39)

By relations (7.38) and (7.39), we deduce T (u, v)X ×Y = T1 (u, v)X + T2 (u, v)Y ∗1 A A∗1 1 + + A1 + A Γ(2 − p2 ) Γ(2 − p1 )

≤

1 + + D1 + D

! r

1∗ D D1∗ + = r, Γ(2 − p2 ) Γ(2 − p1 )

which implies T (B r ) ⊂ B r . Next, we prove that the operator T is a contraction. For (ui , vi ) ∈ B r , i = 1, 2, and for each t ∈ [0, 1], we have |T1 (u1 , v1 )(t) − T1 (u2 , v2 )(t)|   1  −λ(1−s) ≤ |γ1 (t)| |d| e 0

1 Γ(α − 1)

 0

s

(s − τ )α−2

p1 q1 × |f (τ, u1 (τ ), v1 (τ ),c D0+ v1 (τ ), I0+ v1 (τ )) p1 q1 − f (τ, u2 (τ ), v2 (τ ),c D0+ v2 (τ ), I0+ v2 (τ ))| dτ

  + 

0

1

 0

s

e−λ(s−τ )



1 Γ(α − 1)

 0

τ

ds

(τ − ζ)α−2

p1 q1 × |f (ζ, u1 (ζ), v1 (ζ),c D0+ v1 (ζ), I0+ v1 (ζ))

  p1 q1 v2 (ζ), I0+ v2 (ζ))| dζ dτ dH1 (s) − f (ζ, u2 (ζ), v2 (ζ),c D0+

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  + 

1



0

s

0

e



−μ(s−τ )

1 Γ(β − 1)



τ

0

(τ − ζ)β−2

p2 q2 × |g(ζ, u1 (ζ),c D0+ u1 (ζ), I0+ u1 (ζ), v1 (ζ))

− g(ζ, u2 (ζ),  + |b|

1

e

0

c

p2 q2 D0+ u2 (ζ), I0+ u2 (ζ), v2 (ζ))| dζ

−μ(1−s)



1 Γ(β − 1)

 0

s



  dτ dH2 (s)

(s − τ )β−2

p2 q2 × |g(τ, u1 (τ ),c D0+ u1 (τ ), I0+ u1 (τ ), v1 (τ ))

p2 q2 u2 (τ ), I0+ u2 (τ ), v2 (τ ))| dτ ds − g(τ, u2 (τ ),c D0+  1  s   τ  1 +  e−λ(s−τ ) (τ − ζ)α−2 Γ(α − 1) 0 0 0 p1 q1 × |f (ζ, u1 (ζ), v1 (ζ),c D0+ v1 (ζ), I0+ v1 (ζ))

− f (ζ, u2 (ζ), v2 (ζ),   + 

1



0

0

s

e

c

p1 q1 D0+ v2 (ζ), I0+ v2 (ζ))| dζ

−μ(s−τ )



1 Γ(β − 1)

 0

τ

− g(ζ, u2 (ζ),  + 0

t

e

p2 q2 D0+ u2 (ζ), I0+ u2 (ζ), v2 (ζ))|dζ

−λ(t−s)



1 Γ(α − 1)

p1 q1 D0+ v1 (τ ), I0+ v1 (τ ))

c

 0

s

  dτ dK1 (s)

(τ − ζ)β−2

p2 q2 × |g(ζ, u1 (ζ),c D0+ u1 (ζ), I0+ u1 (τ ), v1 (ζ)) c





  dτ dK2 (s)

(s − τ )α−2 |f (τ, u1 (τ ), v1 (τ ),

− f (τ, u2 (τ ), v2 (τ ),

c

p1 q1 D0+ v2 (τ ), I0+ v2 (τ ))| dτ

p1 v1 ≤ M1 c0 (u1 − u2  + v1 − v2  + cD0+ p1 q1 q1 −cD0+ v2  + I0+ v1 − I0+ v2 ) p2 p2 + M2 d0 (u1 − u2  + cD0+ u1 −c D0+ u2  q2 q2 + I0+ u1 − I0+ u2  + v1 − v2 ) p1 p1 ≤ M1 c0 (u1 − u2  + 1 v1 − v2  + cD0+ v1 −c D0+ v2 ) p2 p2 + M2 d0 (2 u1 − u2  + cD0+ u1 −c D0+ u2  + v1 − v2 )

≤ A1 (u1 − u2 X + v1 − v2 Y ).

ds

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415

Then we obtain |T1 (u1 , v1 )(t) − T1 (u2 , v2 )(t)| ≤ A∗1 (u1 − u2 X + v1 − v2 Y ), which gives us p2 p2 |cD0+ T1 (u1 , v1 )(t) −c D0+ T1 (u2 , v2 )(t)|

 ≤ ≤

0

t

(t − s)−p2  |T (u1 , v1 )(s) − T1 (u2 , v2 )(s)| ds Γ(1 − p2 ) 1

A∗1 (u1 − u2 X + v1 − v2 Y ). Γ(2 − p2 )

From the above inequalities, we conclude T1 (u1 , v1 ) − T1 (u2 , v2 )X p2 = T1 (u1 , v1 ) − T1 (u2 , v2 ) + cD0+ T1 (u1 , v1 ) p2 − cD0+ T1 (u2 , v2 )  A∗1 ≤ A1 + (u1 − u2 X + v1 − v2 Y ). Γ(2 − p2 )

(7.40)

In a similar manner, we deduce T2 (u1 , v1 ) − T2 (u2 , v2 )Y ≤

1 + A

∗1 A Γ(2 − p1 )

! (u1 − u2 X + v1 − v2 Y ). (7.41)

Therefore, by (7.40) and (7.41) we obtain T (u1 , v1 ) − T (u2 , v2 )X ×Y = T1 (u1 , v1 ) − T1 (u2 , v2 )X + T2 (u1 , v1 ) − T2 (u2 , v2 )Y ! ∗ ∗ A A 1 1 1 + + (u1 − u2 X + v1 − v2 Y ). ≤ A1 + A Γ(2 − p2 ) Γ(2 − p1 ) By using the condition (7.37), we deduce that T is a contraction. By the Banach fixed point theorem (Theorem 1.2.1), we conclude that the operator T has a unique fixed point, which is the unique solution of problem (7.24), (7.25). This completes the proof. 

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7.2.3

Examples

Let α = 7/2, β = 10/3, p1 = 1/3, p2 = 1/2, q1 = 3/2, q2 = 8/3, λ = 1, μ = 2, H1 (s) = s2 /2 for all s ∈ [0, 1], ⎧ ⎧ ⎪ ⎪ ⎨0, 0 ≤ s < 1 , ⎨0, 0 ≤ s < 1 , 2 3 H2 (s) = K1 (s) = 1 1 ⎪ ⎪ ⎩3, ⎩2, ≤ s ≤ 1, ≤ s ≤ 1, 2 3 ⎧ 1 ⎪ ⎨0, 0≤s< , 2 K2 (s) = 3 ⎪ ⎩− s , 1 ≤ s ≤ 1. 3 2 We consider the system of fractional differential equations ⎧ 7/2 5/2 1/3 3/2 ⎨(cD0+ +c D0+ )u(t) = f (t, u(t), v(t),c D0+ v(t), I0+ v(t)), t ∈ (0, 1), ⎩(cD10/3 + 2cD7/3 )v(t) = g(t, u(t),c D1/2 u(t), I 8/3 u(t), v(t)), 0+

0+

0+

with the boundary conditions ⎧ ⎪ ⎪ ⎨u(0) = u (0) = u (0) = 0,

0+

t ∈ (0, 1), (7.42)

 1 u(1) = 0 su(s) ds + 3v , 2 

1 1 v(1) = 2u − 1/2 s2 v(s) ds. 3

1

⎪ ⎪ ⎩v(0) = v  (0) = v  (0) = 0,

(7.43)

1 We have a = (1 − 2e−1 ) − 0 (s3 − 2s2 + 2s − 2se−s ) ds ≈ 0.20939002, b = 38 (1 − 2e−1 ) ≈ 0.09909042, c = 29 (13 − 18e−1/3 ) ≈ 0.02276365,

1 d = 14 (1 − e−2 ) + 14 1/2 (2s4 − 2s3 + s2 − s2 e−2s ) ds ≈ 0.25358146, Δ0 ≈ 0.05084177 (Δ0 = 0), γ1 (t) = Δ10 (t2 − 2t + 2 − 2e−t ), γ2 (t) = 1 2 −2t ). Besides, we deduce l1 = Δ10 (1 − 2e−1 ) ≈ 4Δ0 (2t − 2t + 1 − e 1 5.1973236, l2 = 4Δ0 (1 − e−2 ) ≈ 4.25174399, m1 = Δ10 2e−1 ≈ 14.47154415, 1 (1 + e−2 ) ≈ 11.16537977, M1 ≈ 0.49529109, M2 ≈ 0.18317499, m2 = 2Δ 0 1 ≈ 0.03173319, M 2 ≈ 0.32453092, M1∗ ≈ 1.34059325, M2∗ ≈ 0.51003654, M ∗ ∗  ≈ 0.08333359, M  ≈ 1.11476371, N1 ≈ 2.13203307, N2 ≈ 2.34109666. M 1

2

Example 1. We consider the functions  t 1 f (t, x1 , x2 , x3 , x4 ) = 2 3 cos t + sin(x1 + x2 ) t +1 8 −

1 1 arctan x4 , x3 + 8(t + 1)2 10

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g(t, x1 , x2 , x3 , x4 ) =

page 417

  1 x1 −t 5e + 2x + 2 (t + 2)3 2 t − sin(x3 + x4 ), 6

for all t ∈ [0, 1], x1 , x2 , x3 , x4 ∈ R. We obtain the inequalities |f (t, x1 , x2 , x3 , x4 )| ≤

1 3 1 1 1 + |x1 | + |x2 | + |x3 | + |x4 |, 2 16 16 8 10

|g(t, x1 , x2 , x3 , x4 )| ≤

1 5 1 1 1 + |x1 | + |x2 | + |x3 | + |x4 |, 8 16 4 6 6

1 1 , a2 = 16 , for all t ∈ [0, 1], x1 , x2 , x3 , x4 ∈ R. So we have a0 = 32 , a1 = 16 1 1 5 1 1 1 1 a3 = 8 , a4 = 10 , b0 = 8 , b1 = 16 , b2 = 4 , b3 = 6 , b4 = 6 . Because Λ2 ≈ 0.962094 and Λ3 ≈ 0.950322, we deduce that the condition max{Λ2 , Λ3 } < 1 is satisfied, and then by Theorem 7.2.1, we conclude that problem (7.42), (7.43) has at least one solution (u(t), v(t)), t ∈ [0, 1].

Example 2. We consider the functions f (t, x1 , x2 , x3 , x4 ) =

1 t + 2 8(t + 1)2 +

g(t, x1 , x2 , x3 , x4 ) =

t3



|x1 | − x2 1 + |x1 |



t 1 sin2 x3 − arctan x4 , 32 9 1 t2 1 − sin x1 + x2 + 1 16 10

1 |x4 | , + √ cos x3 − 2 6(1 + |x4 |) 8 4+t for all t ∈ [0, 1], x1 , x2 , x3 , x4 ∈ R. We obtain the following inequalities |f (t, x1 , x2 , x3 , x4 ) − f (t, y1 , y2 , y3 , y4 )| ≤

1 1 1 1 |x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 | 8 8 16 9

≤

1 (|x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 |), 8

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|g(t, x1 , x2 , x3 , x4 ) − g(t, y1 , y2 , y3 , y4 )| 1 1 1 1 |x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 | 16 10 16 6 1 ≤ (|x1 − y1 | + |x2 − y2 | + |x3 − y3 | + |x4 − y4 |), 6

≤

for all t ∈ [0, 1] and x1 , x2 , x3 , x4 ∈ R. Here c0 = 18 and d0 = 16 . Besides, we deduce 1 ≈ 1.75225278, 2 ≈ 1 ≈ 0.07452006, A ∗1 ≈ 1.24923974, A1 ≈ 0.14662264, A∗1 ≈ 0.39982527, A ∗ ∗ A A 1 1 1 + + ≈ 0.949622(< 1). Thus all 0.25035388, and A1 + A Γ(2−p2 )

Γ(2−p1 )

the conditions of Theorem 7.2.2 are satisfied. Therefore, by Theorem 7.2.2 we conclude that problem (7.42), (7.43) has a unique solution (u(t), v(t)), t ∈ [0, 1]. Remark 7.2.1. The results presented in this section were published in [7].

7.3

Caputo Fractional Differential Systems with Coupled Nonlocal Boundary Conditions

In this section, we investigate the system of nonlinear fractional differential equations c α p D0+ u(t) = f (t, v(t),c D0+ v(t)), t ∈ (0, 1), (7.44) q c β D0+ v(t) = g(t, u(t),c D0+ u(t)), t ∈ (0, 1), supplemented with the coupled nonlocal boundary conditions  γ u(0) = ϕ(v), u (0) = 0, . . . , u(n−2) (0) = 0, u(1) = λI0+ v(ξ), v(0) = ψ(u),

v  (0) = 0, . . . , v (m−2) (0) = 0,

δ v(1) = μI0+ u(η),

(7.45)

where α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n, m ≥ 2, p, q ∈ (0, 1), λ, μ ∈ R, γ, δ > 0, ξ, η ∈ [0, 1], f, g : [0, 1] × R × R → R, ϕ, ψ : C[0, 1] → R θ denotes the Caputo derivative of fractional are continuous functions, cD0+ γ δ are the Riemann-Liouville fractional order θ (θ = α, β, p, q), and I0+ , I0+ integrals of orders γ and δ, respectively. If n = 2 and m = 2, then the γ v(ξ) boundary conditions (7.45) take the form u(0) = ϕ(v), u(1) = λI0+ δ and v(0) = ψ(u), v(1) = μI0+ u(η), respectively, that is, the derivatives of u and v at the point 0 do not appear in these conditions. Under some assumptions on the functions f and g, we will establish the existence

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419

of solutions for problem (7.44), (7.45) by using the Schauder fixed point theorem (Theorem 1.2.4) and the nonlinear alternative of Leray-Schauder type (Theorem 1.2.6). The system (7.44) with α, β ∈ (1, 2) subject to the uncoupled boundary conditions  θ1 u(η), u(0) = ϕ(u), u(1) = aI0+ v(0) = ψ(v),

θ2 v(ξ), v(1) = bI0+

with ξ, η ∈ [0, 1], θ1 , θ2 > 0, a, b ∈ R, was studied in [94]. In [13], the authors investigated the system of fractional differential equations equipped with nonlocal coupled boundary conditions ⎧c α γ D0+ x(t) = f (t, x(t), y(t),c D0+ y(t)), t ∈ [0, T ], ⎪ ⎪ ⎪ ⎪ ⎪ β ⎪ δ ⎨cD0+ y(t) = g(t, x(t),c D0+ x(t), y(t)), t ∈ [0, T ], (7.46)

T ⎪ ⎪ y(s) ds = μ1 x(η), x(0) = h(y), ⎪ 0 ⎪ ⎪ ⎪

T ⎩ x(s) ds = μ2 y(ξ), y(0) = φ(x), 0 where α, β ∈ (1, 2], γ, δ ∈ (0, 1), η, ξ ∈ (0, T ), f, g : [0, T ] × R × R × R → R, h, φ : C([0, T ], R) → R are continuous functions, μ1 , μ2 are real constants. By using the Banach contraction mapping principle and the Leray-Schauder nonlinear alternative, the authors proved the existence and uniqueness, and the existence of solutions for problem (7.46). 7.3.1

Auxiliary results

We consider the fractional differential system c α D0+ u(t) = x(t),

t ∈ (0, 1),

β D0+ v(t) = y(t),

t ∈ (0, 1),

c

with the coupled boundary conditions  u(0) = u0 , u (0) = 0, . . . , u(n−2) (0) = 0,

γ u(1) = λI0+ v(ξ),

v  (0) = 0, . . . , v (m−2) (0) = 0,

δ v(1) = μI0+ u(η),

v(0) = v0 ,

(7.47)

(7.48)

where α ∈ (n− 1, n], β ∈ (m− 1, m], n, m ∈ N, n, m ≥ 2, λ, μ ∈ R, γ, δ > 0, ξ, η ∈ [0, 1], u0 , v0 ∈ R, and x, y are real functions.

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Boundary Value Problems for Fractional Differential Equations and Systems γ+m−1 δ+n−1 η (m − 1)!(n − 1)! = 1 − λμξ . In the sequel we set Δ Γ(γ + m)Γ(δ + n)

= 0, then the Lemma 7.3.1. If x, y ∈ C[0, 1], u0 , v0 ∈ R and Δ n m unique solution (u, v) ∈ C [0, 1] × C [0, 1] of the boundary value problem (7.47), (7.48) is given by  tn−1 λv0 ξ γ α α −u0 − I0+ x(1) + u(t) = u0 + I0+ x(t) + Γ(γ + 1) Δ β+γ + λI0+ y(ξ) −

λv0 ξ γ+m−1 (m − 1)! Γ(γ + m)

λξ γ+m−1 (m − 1)! β λμu0 ξ γ+m−1 η δ (m − 1)! I0+ y(1) + Γ(γ + m) Γ(δ + 1)Γ(γ + m)  λμξ γ+m−1 (m − 1)! α+δ I0+ x(η) , t ∈ [0, 1], + Γ(γ + m)  tm−1 μu0 η δ β β v(t) = v0 + I0+ −v0 − I0+ y(t) + y(1) + Γ(δ + 1) Δ −

α+δ + μI0+ x(η) −

μu0 η δ+n−1 (n − 1)! Γ(δ + n)

μη δ+n−1 (n − 1)! α λμv0 ξ γ η δ+n−1 (n − 1)! I0+ x(1) + Γ(δ + n) Γ(γ + 1)Γ(δ + n)  λμη δ+n−1 (n − 1)! β+γ I0+ y(ξ) , t ∈ [0, 1]. + Γ(δ + n)

−

(7.49)

α Proof. The solution u ∈ C n [0, 1] of equation cD0+ u(t) = x(t) is α u(t) = c1 + c2 t + · · · + cn tn−1 + I0+ x(t),

t ∈ [0, 1],

where c1 , c2 , . . . , cn ∈ R. By using the conditions u(0) = u0 , u (0) = · · · = (n−2) (0) = 0, we obtain c1 = u 0 , c2 = · · · = cn−1 = 0, and so we deduce u n−1 α cn t + I0+ x(t), t ∈ [0, 1]. u(t) = u0 + β v(t) = In a similar manner the solution v ∈ C m [0, 1] of equation cD0+  (m−2) y(t) satisfying the conditions v(0) = v0 , v (0) = · · · = v (0) = 0 is β y(t), t ∈ [0, 1], with d m ∈ R. v(t) = v0 + d m tm−1 + I0+ γ δ v(ξ) and v(1) = μI0+ u(η) on the Imposing the conditions u(1) = λI0+ above functions u and v, we obtain the following system in the unknown

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constants cn and d m ⎧ λd m ξ γ+m−1 (m − 1)! λv0 ξ γ ⎪ β+γ α ⎪ ⎪ c = −u + λI0+ − − I x(1) + y(ξ), n 0 0+ ⎨ Γ(γ + m) Γ(γ + 1) ⎪ ⎪ μ c η δ+n−1 (n − 1)! μu0 η δ β ⎪ α+δ ⎩− n + dm = −v0 − I0+ + μI0+ y(1) + x(η). Γ(δ + n) Γ(δ + 1) = 0, the unique solution ( In view of the given assumption Δ cn , d m ) of the above system is  1 λv0 ξ γ β+γ α + λI0+ −u0 − I0+ x(1) + y(ξ) cn = Γ(γ + 1) Δ −

λv0 ξ γ+m−1 (m − 1)! λξ γ+m−1 (m − 1)! β − I0+ y(1) Γ(γ + m) Γ(γ + m)

 λμu0 ξ γ+m−1 η δ (m − 1)! λμξ γ+m−1 (m − 1)! α+δ + + I0+ x(η) , Γ(δ + 1)Γ(γ + m) Γ(γ + m)  1 μu0 η δ β α+δ −v0 − I0+ y(1) + x(η) dm = + μI0+ Γ(δ + 1) Δ − +

μu0 η δ+n−1 (n − 1)! μη δ+n−1 (n − 1)! α − I0+ x(1) Γ(δ + n) Γ(δ + n)

 λμv0 ξ γ η δ+n−1 (n − 1)! λμη δ+n−1 (n − 1)! β+γ + I0+ y(ξ) . Γ(γ + 1)Γ(δ + n) Γ(δ + n)

Substituting the values of cn and d m in the expressions for u and v, we obtain the solution (u, v) of problem (7.47), (7.48) given by formula (7.49).  By direct computation, we can obtain the following result. Lemma 7.3.2. If x, y ∈ C[0, 1], then the following inequalities are satisfied α x(t)| ≤ (a) |I0+

x ; Γ(α + 1)

α+δ x(η)| ≤ (c) |I0+ β (e) |I0+ y(1)| ≤

α x(1)| ≤ (b) |I0+

η α+δ x ; Γ(α + δ + 1)

y ; Γ(β + 1)

x ; Γ(α + 1)

β (d) |I0+ y(t)| ≤

β+γ (f) |I0+ y(ξ)| ≤

y ; Γ(β + 1)

ξ β+γ y , Γ(β + γ + 1)

where x = supt∈[0,1] |x(t)| and y = supt∈[0,1] |y(t)|.

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Existence of solutions

q u ∈ C[0, 1]} and Y = {v ∈ We consider the spaces X = {u ∈ C[0, 1], cD0+ p c C[0, 1], D0+ v ∈ C[0, 1]} equipped respectively with the norms uX = q p u and vY = v + cD0+ v, where  ·  is the supremum u + cD0+ norm, that is w = supt∈[0,1] |w(t)| for w ∈ C[0, 1]. The spaces (X,  · X ) and (Y,  · Y ) are Banach spaces, and the product space X × Y endowed with the norm (u, v)X×Y = uX + vY is also a Banach space. By using Lemma 7.3.1, we introduce the operator Q : X × Y → X × Y defined by Q(u, v) = (Q1 (u, v), Q2 (u, v)) for (u, v) ∈ X × Y , where the operators Q1 : X × Y → X and Q2 : X × Y → Y are given by

 tn−1 λμξ γ+m−1 η δ (m − 1)! tn−1 Q1 (u, v)(t) = ϕ(v) 1 − + Δ ΔΓ(δ + 1)Γ(γ + m)   1 tn−1 λξ γ ξ m−1 (m − 1)! + ψ(u) − Γ(γ + 1) Γ(γ + m) Δ  tn−1 p α c p α + I0+ f (t, v(t), D0+ v(t)) + −I0+ f (t, v(t),c D0+ v(t))|t=1 Δ 

+

λμξ γ+m−1 (m − 1)! α+δ p I0+ f (t, v(t),c D0+ v(t))|t=η Γ(γ + m)

β+γ q + λI0+ g(t, u(t),c D0+ u(t))|t=ξ

 λξ γ+m−1 (m − 1)! β q I0+ g(t, u(t),c D0+ u(t))|t=1 , Γ(γ + m)   1 η n−1 (n − 1)! tm−1 μη δ − Q2 (u, v)(t) = ϕ(v) Γ(δ + 1) Γ(δ + n) Δ   tm−1 λμξ γ η δ+n−1 (n − 1)! tm−1 + + ψ(u) 1 − Δ ΔΓ(γ + 1)Γ(δ + n) −

t β q + I0+ g(t, u(t),c D0+ u(t))+ −

m−1

Δ



p α+δ μI0+ f (t, v(t),c D0+ v(t))|t=η

μη δ+n−1 (n − 1)! α p I0+ f (t, v(t),c D0+ v(t))|t=1 Γ(δ + n)

page 422

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423

β q − I0+ g(t, u(t),c D0+ u(t))|t=1

+

 λμη δ+n−1 (n − 1)! β+γ q I0+ g(t, u(t),c D0+ u(t))|t=ξ , Γ(δ + n)

for all (u, v) ∈ X × Y and t ∈ [0, 1]. The pair (u, v) is a solution of problem (7.44), (7.45) if and only if (u, v) is a fixed point of operator Q. Now we enlist the assumptions that we need in this section. (K1) α, β ∈ R, α ∈ (n − 1, n], β ∈ (m − 1, m], n, m ∈ N, n, m ≥ = 1 − 2; p, q ∈ (0, 1); γ, δ > 0; λ, μ ∈ R; ξ, η ∈ [0, 1]; Δ λμξ γ+m−1 η δ+n−1 (m−1)!(n−1)! = 0. Γ(γ+m)Γ(δ+n) (K2) f, g : [0, 1] × R × R → R are continuous functions and there exist the constants ai , bi ≥ 0, i = 0, 1, 2, and lj , mj ∈ (0, 1), j = 1, 2 such that |f (t, u, v)| ≤ a0 + a1 |u|l1 + a2 |v|l2 , |g(t, u, v)| ≤ b0 + b1 |u|

m1

+ b2 |v|

m2

∀t ∈ [0, 1], u, v ∈ R, ,

∀t ∈ [0, 1], u, v ∈ R.

(K3) ϕ, ψ : C[0, 1] → R are continuous functions, ϕ(0) = ψ(0) = 0 and there exist constants L1 , L2 > 0, θ1 , θ2 ∈ (0, 1) such that |ϕ(x)| ≤ L1 xθ1 ,

|ψ(x)| ≤ L2 xθ2 ,

∀x ∈ C[0, 1].

(K4) f, g : [0, 1] × R × R → R are continuous functions, and there exist ci , di ≥ 0, i = 0, 1, 2 and nondecreasing functions hj , kj ∈ C([0, ∞), [0, ∞)), j = 1, 2 such that |f (t, u, v)| ≤ c0 + c1 h1 (|u|) + c2 h2 (|v|),

∀t ∈ [0, 1], u, v ∈ R,

|g(t, u, v)| ≤ d0 + d1 k1 (|u|) + d2 k2 (|v|),

∀t ∈ [0, 1], u, v ∈ R.

(K5) There exists L0 > 0 such that 1 + N1 + N 1 )Lθ1 + (M2 + M 2 + N2 + N 2 )Lθ2 (M1 + M 0 0 3 + N3 + N 3 ) + A2 (M4 + M 4 + N4 + N 4 ) < L0 , +A1 (M3 + M where A1 = c0 + c1 h1 (L0 ) + c2 h2 (L0 ), A2 = d0 + d1 k1 (L0 ) + d2 k2 (L0 ), θ1 , θ2 are given in (K3), ci , di , i = 0, 1, 2 and hj , kj , j = 1, 2 are i , Ni , N i , i = 1, 2, 3, 4 are given given in (K4), and the constants Mi , M below.

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For computational convenience, we introduce the following notations: ! 1 λμξ γ+m−1 η δ (m − 1)! M1 = L1 1 + + , |Δ| |Δ|Γ(δ + 1)Γ(γ + m) L2 λξ γ M2 = |Δ| M3 = M4 =



ξ m−1 (m − 1)! 1 + , Γ(γ + 1) Γ(γ + m)

1 λμξ γ+m−1 η α+δ (m − 1)! 1 + + , Γ(α + 1) |Δ|Γ(α + 1) |Δ|Γ(γ + m)Γ(α + δ + 1) λξ β+γ

+

λξ γ+m−1 (m − 1)! , |Δ|Γ(γ + m)Γ(β + 1)

|Δ|Γ(β + γ + 1)  γ+m−1 δ η (m − 1)! 1 = L1 (n − 1) 1 + λμξ M , Γ(δ + 1)Γ(γ + m) Γ(2 − q)|Δ|  γ 1 ξ m−1 (m − 1)! 2 = L2 (n − 1)λξ M + , Γ(γ + 1) Γ(γ + m) Γ(2 − q)|Δ|

! 1 n−1 (n − 1)λμξ γ+m−1 η α+δ (m − 1)! + + , Γ(α) |Δ|Γ(α + 1) |Δ|Γ(γ + m)Γ(α + δ + 1)  λξ β+γ λξ γ+m−1 (m − 1)! n−1 + = , Γ(β + γ + 1) Γ(γ + m)Γ(β + 1) Γ(2 − q)|Δ|  1 η n−1 (n − 1)! L1 μη δ + = , Γ(δ + 1) Γ(δ + n) |Δ| ! 1 λμξ γ η δ+n−1 (n − 1)! = L2 1 + + , |Δ| |Δ|Γ(γ + 1)Γ(δ + n)

1 3 = M Γ(2 − q) 4 M N1 N2

N3 =

μη α+δ

|Δ|Γ(α + δ + 1)

+

μη δ+n−1 (n − 1)! , |Δ|Γ(δ + n)Γ(α + 1)

1 1 λμξ β+γ η δ+n−1 (n − 1)! + + , Γ(β + 1) |Δ|Γ(β + 1) |Δ|Γ(δ + n)Γ(β + γ + 1)  η n−1 (n − 1)! 1 L1 (m − 1)μη δ + N1 = , Γ(δ + 1) Γ(δ + n) Γ(2 − p)|Δ|  γ δ+n−1 (n − 1)! 2 = L2 (m − 1) 1 + λμξ η N , Γ(γ + 1)Γ(δ + n) Γ(2 − p)|Δ| N4 =

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3 = N



m−1

Γ(2 − p)|Δ|

page 425

μη α+δ μη δ+n−1 (n − 1)! + , Γ(α + δ + 1) Γ(δ + n)Γ(α + 1)

! m−1 1 (m − 1)λμη δ+n−1 ξ β+γ (n − 1)! + + . Γ(β) |Δ|Γ(β + 1) |Δ|Γ(δ + n)Γ(β + γ + 1)

1 4 = N Γ(2 − p)

Theorem 7.3.1. Assume that (K1)–(K3) hold. Then problem (7.44), (7.45) has at least one solution on [0, 1]. Proof. Let B R = {(u, v) ∈ X × Y , (u, v)X×Y ≤ R}, where " 1 1 1 )] 1−θ1 , [8(M2 + M 2 )] 1−θ2 , R ≥ max [8(M1 + M 1

3 ), [24a1 (M3 + M 3 )] 1−l1 , 24a0 (M3 + M 1

3 )] 1−l2 , 24b0(M4 + M 4 ), [24a2 (M3 + M 1

1

4 )] 1−m1 , [24b2 (M4 + M 4 )] 1−m2 , [24b1 (M4 + M 1

1

1 )] 1−θ1 , [8(N2 + N 2 )] 1−θ2 , [8(N1 + N 1

3 ), [24a1 (N3 + N 3 )] 1−l1 , 24a0 (N3 + N 1

3 )] 1−l2 , 24b0 (N4 + N 4 ), [24a2 (N3 + N 1

1

4 )] 1−m1 , [24b2 (N4 + N 4 )] 1−m2 [24b1 (N4 + N

#

.

Let us first show that Q : B R → B R . For (u, v) ∈ B R , it follows by Lemma 7.3.2 that ! 1 λμξ γ+m−1 η δ (m − 1)! |Q1 (u, v)(t)| ≤ |ϕ(v)| 1 + + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 |ψ(u)|λξ γ ξ m−1 (m − 1)! + + Γ(γ + 1) Γ(γ + m) |Δ|  1 p p α α f (t, v(t),c D0+ v(t))|+ f (t, v(t),c D0+ v(t))|t=1 | |I0+ + |I0+ |Δ| +

λμξ γ+m−1 (m − 1)! α+δ p |I0+ f (t, v(t),c D0+ v(t))|t=η | Γ(γ + m)

β+γ q + λ|I0+ g(t, u(t),c D0+ u(t))|t=ξ |

+



λξ γ+m−1 (m − 1)! β q |I0+ g(t, u(t),c D0+ u(t))|t=1 | Γ(γ + m)

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! 1 λμξ γ+m−1 η δ (m − 1)! ≤ L1 v 1+ + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! L2 θ2 γ + + u λξ Γ(γ + 1) Γ(γ + m) |Δ| θ1

 1 p a0 + a1 vl1 + a2 cD0+ vl2 Γ(α + 1) 

 1 1 p a0 + a1 vl1 + a2 cD0+ + vl2 |Δ| Γ(α + 1)

+

+

 λμξ γ+m−1 η α+δ (m − 1)! p a0 + a1 vl1 + a2 c D0+ vl2 Γ(γ + m)Γ(α + δ + 1)

+

 λξ β+γ q b0 + b1 um1 + b2 cD0+ um2 Γ(β + γ + 1)

  λξ γ+m−1 (m − 1)! m1 c q m2 + b0 + b1 u + b2  D0+ u Γ(γ + m)Γ(β + 1) ! 1 λμξ γ+m−1 η δ (m − 1)! θ1 ≤ L1 R 1+ + |Δ| |Δ|Γ(δ + 1)Γ(γ + m) +

L2 Rθ2 λξ γ |Δ|



1 ξ m−1 (m − 1)! + Γ(γ + 1) Γ(γ + m)



1 1 + Γ(α + 1) |Δ|Γ(α + 1) ! λμξ γ+m−1 η α+δ (m − 1)! + |Δ|Γ(γ + m)Γ(α + δ + 1) + (a0 + a1 Rl1 + a2 Rl2 )

+ (b0 + b1 Rm1 + b2 Rm2 ) λξ γ+m−1 (m − 1)! + |Δ|Γ(γ + m)Γ(β + 1)

!

λξ β+γ

|Δ|Γ(β + γ + 1)

= M1 Rθ1 + M2 Rθ2 + M3 (a0 + a1 Rl1 + a2 Rl2 ) + M4 (b0 + b1 Rm1 + b2 Rm2 ),

∀t ∈ [0, 1].

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On the other hand, we have q D0+ Q1 (u, v)(t) =

c

1 Γ(1 − q)



t

0

(Q1 (x, y)) (s) ds, (t − s)q

where (n − 1)tn−2 λμξ γ+m−1 η δ (m − 1)! (n − 1)tn−2 (Q1 (u, v)) (t) = ϕ(v) − + Δ ΔΓ(δ + 1)Γ(γ + m)  1 (n − 1)tn−2 λξ γ ξ m−1 (m − 1)! + ψ(u) − Γ(γ + 1) Γ(γ + m) Δ 

p α−1 f (t, v(t),c D0+ v(t)) + + I0+

(n − 1)tn−2 Δ

 p α × − I0+ f (t, v(t),c D0+ v(t))|t=1 +

λμξ γ+m−1 (m − 1)! α+δ p I0+ f (t, v(t),c D0+ v(t))|t=η Γ(γ + m)

β+γ q + λI0+ g(t, u(t),c D0+ u(t))|t=ξ

 λξ γ+m−1 (m − 1)! β c q I0+ g(t, u(t), D0+ u(t))|t=1 , − Γ(γ + m) ∀t ∈ (0, 1). Then, by Lemma 7.3.2, we obtain

! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! + |(Q1 (u, v)) (t)| ≤ |ϕ(v)| |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! |ψ(u)|(n − 1)λξ γ + + Γ(γ + 1) Γ(γ + m) |Δ| 

p α−1 f (t, v(t),c D0+ v(t))| + + |I0+

n−1 |Δ|

 p α f (t, v(t),c D0+ v(t))|t=1 | × |I0+ +

λμξ γ+m−1 (m − 1)! α+δ p |I0+ f (t, v(t),c D0+ v(t))|t=η | Γ(γ + m)

β+γ q + λ|I0+ g(t, u(t),c D0+ u(t))|t=ξ |



λξ γ+m−1 (m − 1)! β q |I0+ g(t, u(t),c D0+ + u(t))|t=1 | Γ(γ + m)

!

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! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! ≤ L1 v + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! L2 uθ2 (n − 1)λξ γ + + Γ(γ + 1) Γ(γ + m) |Δ| θ1

 1 p a0 + a1 vl1 + a2 cD0+ vl2 Γ(α) 

 1 n−1 p a0 + a1 vl1 + a2 cD0+ + vl2 Γ(α + 1) |Δ| +

+

 λμξ γ+m−1 (m − 1)!η α+δ p a0 + a1 vl1 + a2 cD0+ vl2 Γ(γ + m)Γ(α + δ + 1)

+

 λξ β+γ q b0 + b1 um1 + b2 cD0+ um2 Γ(β + γ + 1)

  λξ γ+m−1 (m − 1)! q b0 + b1 um1 + b2 cD0+ um2 Γ(γ + m)Γ(β + 1) ! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! θ1 ≤ L1 R + |Δ| |Δ|Γ(δ + 1)Γ(γ + m) +

+

L2 Rθ2 (n − 1)λξ γ |Δ|



1 ξ m−1 (m − 1)! + Γ(γ + 1) Γ(γ + m)

n−1 1 + Γ(α) |Δ|Γ(α + 1) ! (n − 1)λμξ γ+m−1 η α+δ (m − 1)! + |Δ|Γ(γ + m)Γ(α + δ + 1) + (a0 + a1 Rl1 + a2 Rl2 )

+ (b0 + b1 Rm1 + b2 Rm2 )

(n − 1)λξ β+γ |Δ|Γ(β + γ + 1)

! (n − 1)λξ γ+m−1 (m − 1)! + , |Δ|Γ(γ + m)Γ(β + 1)

∀t ∈ (0, 1).



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429

In consequence, we obtain q Q1 (u, v)(t)| ≤ |cD0+

≤

1 Γ(1 − q)

 0

t

(t − s)−q (Q1 (u, v)) ds

1 (Q1 (u, v))  Γ(2 − q)

! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! L2 Rθ2 (n − 1)λξ γ + + Γ(γ + 1) Γ(γ + m) |Δ|Γ(2 − q)

L 1 R θ1 ≤ Γ(2 − q)

+

1 (a0 + a1 Rl1 + a2 Rl2 ) Γ(2 − q)

+

(n − 1)λμξ γ+m−1 η α+δ (m − 1)! |Δ|Γ(γ + m)Γ(α + δ + 1)

1 (b0 + b1 Rm1 + b2 Rm2 ) Γ(2 − q) ! (n − 1)λξ γ+m−1 (m − 1)! + |Δ|Γ(γ + m)Γ(β + 1) +

n−1 1 + Γ(α) |Δ|Γ(α + 1) !

(n − 1)λξ β+γ |Δ|Γ(β + γ + 1)

2 Rθ2 + M 3 (a0 + a1 Rl1 + a2 Rl2 ) 1 Rθ1 + M =M 4 (b0 + b1 Rm1 + b2 Rm2 ), +M

∀t ∈ [0, 1].

Thus we have q Q1 (u, v)X = Q1 (u, v) + cD0+ Q1 (u, v)

1 )Rθ1 + (M2 + M 2 )Rθ2 ≤ (M1 + M 3 )(a0 + a1 Rl1 + a2 Rl2 ) + (M3 + M 4 )(b0 + b1 Rm1 + b2 Rm2 ) + (M4 + M ≤

R R R R R + + + = . 8 8 8 8 2

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In a similar manner, we can obtain p Q2 (u, v)Y = Q2 (u, v) + cD0+ Q2 (u, v)

1 )Rθ1 + (N2 + N 2 )Rθ2 ≤ (N1 + N 3 )(a0 + a1 Rl1 + a2 Rl2 ) + (N3 + N 4 )(b0 + b1 Rm1 + b2 Rm2 ) + (N4 + N R R R R R + + + = . 8 8 8 8 2 By the foregoing arguments, we deduce that ≤

Q(u, v)X×Y = Q1 (u, v)X + Q2 (u, v)Y ≤

R R + = R, 2 2

which implies that Q : B R → B R . From the continuity of the functions f, g, ϕ and ψ, we can easily show that the operator Q is continuous. Next we will show that the operator Q : B R → B R is equicontinuous. We denote Λ1 = maxt∈[0,1],|u|≤R,|v|≤R |f (t, u, v)| and Λ2 = maxt∈[0,1],|u|≤R,|v|≤R |g(t, u, v)|. For any (u, v) ∈ B R and t1 , t2 ∈ [0, 1], t1 < t2 , we have |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )|   tn−1 tn−1 λμξ γ+m−1 η δ (m − 1)!  + 2 ≤ ϕ(v) 1 − 2  Δ ΔΓ(δ + 1)Γ(γ + m) ! tn−1 λμξ γ+m−1 η δ (m − 1)!  tn−1 1 1 −1+ −   Δ ΔΓ(δ + 1)Γ(γ + m)     1 ξ m−1 (m − 1)!  tn−1 − tn−1 1 + ψ(u)λξ γ 2 −  Γ(γ + 1) Γ(γ + m) Δ  α  p p α + I0+ f (t, v(t),c D0+ v(t))|t=t2 − I0+ f (t, v(t),c D0+ v(t))|t=t1   − tn−1 tn−1 p 2 1 α f (t, v(t),c D0+ v(t))|t=1 | |I0+ + |Δ| +

λμξ γ+m−1 (m − 1)! α+δ p |I0+ f (t, v(t),c D0+ v(t))|t=η | Γ(γ + m)

β+γ q g(t, u(t),c D0+ u(t))|t=ξ | + λ|I0+



λξ γ+m−1 (m − 1)! β q + |I0+ g(t, u(t),c D0+ u(t))|t=1 | Γ(γ + m)

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 λμξ γ+m−1 η δ (m − 1)! ≤ L1 v 1+ Γ(δ + 1)Γ(γ + m)  n−1 1 ξ m−1 (m − 1)! − tn−1 1 θ 2 t2 γ + λξ + L2 u Γ(γ + 1) Γ(γ + m) |Δ|  t2 1  p + (t2 − s)α−1 f (s, v(s),c D0+ v(s)) ds Γ(α)  0   t1  p − (t1 − s)α−1 f (s, v(s),c D0+ v(s)) ds 0  n−1 t Λ1 λμξ γ+m−1 η α+δ (m − 1)!Λ1 − tn−1 1 + 2 + Γ(α + 1) Γ(γ + m)Γ(α + δ + 1) |Δ|  λξ β+γ Λ2 λξ γ+m−1 (m − 1)!Λ2 + + Γ(β + γ + 1) Γ(γ + m)Γ(β + 1)  n−1 − tn−1 λμξ γ+m−1 η δ (m − 1)! θ 1 t2 1 ≤ L1 R 1+ Γ(δ + 1)Γ(γ + m) |Δ|  n−1 1 ξ m−1 (m − 1)! − tn−1 1 θ 2 t2 γ + + L2 R λξ Γ(γ + 1) Γ(γ + m) |Δ|  t1   1  α−1 α−1 c p  + [(t − s) − (t − s) ]f (s, v(s), D v(s)) ds 2 1 0+   Γ(α) 0    1  t2 α−1 c p  + (t − s) f (s, v(s), D v(s)) ds 2 0+   Γ(α) t1  Λ1 λμξ γ+m−1 η α+δ (m − 1)!Λ1 tn−1 − tn−1 1 + + 2 Γ(α + 1) Γ(γ + m)Γ(α + δ + 1) |Δ|  λξ γ+m−1 (m − 1)!Λ2 λξ β+γ Λ2 + + , Γ(β + γ + 1) Γ(γ + m)Γ(β + 1) n−1 θ 1 t2

− tn−1 1 |Δ|

which further implies that |Q1 (u, v)(t2 ) − Q1 (u, v)(t1 )|  n−1 − tn−1 λμξ γ+m−1 η δ (m − 1)! 1 θ 1 t2 ≤ L1 R 1+ Γ(δ + 1)Γ(γ + m) |Δ|  1 ξ m−1 (m − 1)! tn−1 − tn−1 1 + + L2 Rθ2 λξ γ 2 Γ(γ + 1) Γ(γ + m) |Δ|

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   Λ1  t1 α−1 α−1 + [(t2 − s) − (t1 − s) ] ds  Γ(α) 0    Λ1  t2 α−1 + (t2 − s) ds  Γ(α) t1  tn−1 λμξ γ+m−1 η α+δ (m − 1)!Λ1 Λ1 − tn−1 2 1 + + Γ(α + 1) Γ(γ + m)Γ(α + δ + 1) |Δ|  λξ γ+m−1 (m − 1)!Λ2 λξ β+γ Λ2 + + Γ(β + γ + 1) Γ(γ + m)Γ(β + 1) n−1 n−1  − t1 λμξ γ+m−1 η δ (m − 1)! θ 1 t2 = L1 R 1+ Γ(δ + 1)Γ(γ + m) |Δ|  1 ξ m−1 (m − 1)! tn−1 − tn−1 1 + + L2 Rθ2 λξ γ 2 Γ(γ + 1) Γ(γ + m) |Δ|  Λ1 − tn−1 tn−1 Λ1 α 2 1 (tα + 2 − t1 ) + Γ(α + 1) Γ(α + 1) |Δ| +

λμξ γ+m−1 η α+δ (m − 1)!Λ1 Γ(γ + m)Γ(α + δ + 1)

 λξ β+γ Λ2 λξ γ+m−1 (m − 1)!Λ2 + + . Γ(β + γ + 1) Γ(γ + m)Γ(β + 1) On the other hand we obtain q q Q1 (u, v)(t2 ) −c D0+ Q1 (u, v)(t1 )| |cD0+  t2   t1  (Q1 (u, v)) (s) (Q1 (u, v)) (s)  1  ds − ds =  Γ(1 − q)  0 (t2 − s)q (t1 − s)q 0  t1      1 1 1    ≤ − (u, v)) (s) ds (Q 1   q q Γ(1 − q) 0 (t2 − s) (t1 − s)  t2   1 (Q1 (u, v)) (s)   + ds . Γ(1 − q)  t1 (t2 − s)q

As

! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! |(Q1 (u, v)) (t)| ≤ L1 R + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! n−1 + + L2 Rθ2 λξ γ Γ(γ + 1) Γ(γ + m) |Δ| 

θ1

(7.50)

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 Λ1 n−1 λμξ γ+m−1 η α+δ (m − 1)!Λ1 Λ1 + + + Γ(α) Γ(α + 1) Γ(γ + m)Γ(α + δ + 1) |Δ|  λξ γ+m−1 (m − 1)!Λ2 λξ β+γ Λ2 + + =: D1 , Γ(β + γ + 1) Γ(γ + m)Γ(β + 1) ∀t ∈ (0, 1), we deduce by the above inequalities that q q Q1 (u, v)(t2 ) −c D0+ Q1 (u, v)(t1 )| |cD0+   t1  t2 D1 −q −q −q [(t1 − s) − (t2 − s) ] ds + (t2 − s) ds ≤ Γ(1 − q) 0 t1

=

D1 [2(t2 − t1 )1−q − t1−q + t1−q ]. 2 1 Γ(2 − q)

(7.51)

In a similar manner, we find that |Q2 (u, v)(t2 ) − Q2 (u, v)(t1 )|

 1 η n−1 (n − 1)! tm−1 − tm−1 2 1 + Γ(δ + 1) Γ(δ + n) |Δ|  m−1 − tm−1 λμξ γ η δ+n−1 (n − 1)! θ 2 t2 1 1+ + L2 R Γ(γ + 1)Γ(δ + n) |Δ|  μη α+δ Λ1 tm−1 − tm−1 μη δ+n−1 (n − 1)!Λ1 Λ2 β 1 (t2 − tβ1 ) + 2 + + Γ(β) Γ(α + δ + 1) Γ(δ + n)Γ(α + 1) |Δ|  Λ2 λμη δ+n−1 ξ β+γ (n − 1)!Λ2 + + . (7.52) Γ(β + 1) Γ(δ + n)Γ(β + γ + 1)

≤ L1 Rθ1 μη δ

Using the estimate

 1 η n−1 (n − 1)! m−1 + Γ(δ + 1) Γ(δ + n) |Δ|  λμξ γ η δ+n−1 (n − 1)! θ2 m − 1 + L2 R 1+ Γ(γ + 1)Γ(δ + n) |Δ|  μη α+δ Λ1 m−1 μη δ+n−1 (n − 1)!Λ1 Λ2 + + + Γ(β) Γ(α + δ + 1) Γ(δ + n)Γ(α + 1) |Δ|  Λ2 λμη δ+n−1 ξ β+γ (n − 1)!Λ2 + + =: D2 , Γ(β + 1) Γ(δ + n)Γ(β + γ + 1)

|(Q2 (u, v)) (t)| ≤ L1 Rθ1 μη δ

∀t ∈ (0, 1),

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we get p p |cD0+ Q2 (u, v)(t2 ) −c D0+ Q2 (u, v)(t1 )|

≤

D2 [2(t2 − t1 )1−p − t1−p + t1−p ]. 2 1 Γ(2 − p)

(7.53)

By the relations (7.50)–(7.53), we deduce that Q : B R → B R is equicontinuous. Thus the Arzela-Ascoli theorem applies and that the set Q(B R ) is relatively compact, and thus Q is a completely continuous operator. Therefore, by the Schauder fixed point theorem (Theorem 1.2.4), we deduce that the operator Q has at least one fixed point (u, v) in B R , which is a solution of problem (7.44), (7.45).  Theorem 7.3.2. Assume that (K1), (K3), (K4) and (K5) hold. Then problem (7.44), (7.45) has at least one solution on [0, 1]. Proof. With L0 given by (K5), we consider a set B L0 = {(u, v) ∈ X × Y, (u, v)X×Y ≤ L0 } and show that Q(B L0 ) ⊂ B L0 . For (u, v) ∈ B L0 and t ∈ [0, 1], we obtain ! 1 λμξ γ+m−1 η δ (m − 1)! θ1 |Q1 (u, v)(t)| ≤ L1 L0 1 + + |Δ| |Δ|Γ(δ + 1)Γ(γ + m) λξ γ |Δ|



1 ξ m−1 (m − 1)! + Γ(γ + 1) Γ(γ + m)  1 + (c0 + c1 h1 (L0 ) + c2 h2 (L0 )) Γ(α + 1) + L2 Lθ02



λμξ γ+m−1 η α+δ (m − 1)! + + |Δ|Γ(α + 1) |Δ|Γ(γ + m)Γ(α + δ + 1) 1

+ (d0 + d1 k1 (L0 ) + d2 k2 (L0 )) λξ γ+m−1 (m − 1)! + |Δ|Γ(γ + m)Γ(β + 1)

!

λξ β+γ

|Δ|Γ(β + γ + 1)

!

= M1 Lθ01 + M2 Lθ02 + M3 (c0 + c1 h1 (L0 ) + c2 h2 (L0 )) + M4 (d0 + d1 k1 (L0 ) + d2 k2 (L0 )),

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and q Q1 (u, v)(t)| |cD0+

! n − 1 (n − 1)λμξ γ+m−1 η δ (m − 1)! + |Δ| |Δ|Γ(δ + 1)Γ(γ + m)  1 ξ m−1 (m − 1)! L2 Lθ02 (n − 1)λξ γ + + Γ(γ + 1) Γ(γ + m) |Δ|Γ(2 − q)

L1 Lθ01 ≤ Γ(2 − q)

+

1 (c0 + c1 h1 (L0 ) + c2 h2 (L0 )) Γ(2 − q)

1 Γ(α)

(n − 1)λμξ γ+m−1 η α+γ (m − 1)! + + |Δ|Γ(α + 1) |Δ|Γ(γ + m)Γ(α + δ + 1) n−1

+ ×

1 (d0 + d1 k1 (L0 ) + d2 k2 (L0 )) Γ(2 − q) (n − 1)λξ β+γ (n − 1)λξ γ+m−1 (m − 1)! + |Δ|Γ(β + γ + 1) |Δ|Γ(γ + m)Γ(β + 1)

!

!

2 Lθ2 + M 3 (c0 + c1 h1 (L0 ) + c2 h2 (L0 )) 1 Lθ1 + M =M 0 0 4 (d0 + d1 k1 (L0 ) + d2 k2 (L0 )). +M In view of the above estimates, we find that q Q1 (u, v)X = Q1 (u, v) + cD0+ Q1 (u, v)

1 )Lθ1 + (M2 + M 2 )Lθ2 ≤ (M1 + M 0 0 3 )(c0 + c1 h1 (L0 ) + c2 h2 (L0 )) + (M3 + M

(7.54)

4 )(d0 + d1 k1 (L0 ) + d2 k2 (L0 )). + (M4 + M In a similar manner, we obtain 1 )Lθ1 + (N2 + N 2 )Lθ2 Q2 (u, v)Y ≤ (N1 + N 0 0 3 )(c0 + c1 h1 (L0 ) + c2 h2 (L0 )) + (N3 + N

(7.55)

4 )(d0 + d1 k1 (L0 ) + d2 k2 (L0 )). + (N4 + N Using (7.54) and (7.55), we get 1 + N1 + N 1 )Lθ1 Q(u, v)X×Y ≤ (M1 + M 0 2 + N2 + N 2 )Lθ2 + (M2 + M 0 3 + N3 + N 3 )(c0 + c1 h1 (L0 ) + c2 h2 (L0 )) + (M3 + M

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4 + N4 + N 4 )(d0 + d1 k1 (L0 ) + d2 k2 (L0 )) + (M4 + M 1 + N1 + N 1 )Lθ1 + (M2 + M 2 + N2 + N 2 )Lθ2 = (M1 + M 0 0 3 + N3 + N 3 ) + A1 (M3 + M 4 + N4 + N 4 ) < L0 , + A2 (M4 + M which shows that Q(B L0 ) ⊂ B L0 . As argued in the proof of Theorem 7.3.1, it can be shown that the operator Q is completely continuous. Next we suppose that there exists (u, v) ∈ ∂BL0 such that (u, v) = νQ(u, v) for some ν ∈ (0, 1). Then 1 + N1 + N 1 )Lθ1 (u, v)X×Y ≤ Q(u, v)X×Y ≤ (M1 + M 0 2 + N2 + N 2 )Lθ2 + A1 (M3 + M 3 + N3 + N 3 ) + (M2 + M 0 4 + N4 + N 4 ) < L0 , + A2 (M4 + M which contradicts that (u, v) ∈ ∂BL0 . Thus, by the nonlinear alternative of Leray-Schauder type (Theorem 1.2.6), we deduce that the operator Q has a fixed point (u, v) ∈ B L0 , and so problem (7.44), (7.45) has at least one solution on [0, 1].  7.3.3

Examples

Example 1. Letting α = 10/3 (n = 3), β = 9/2 (m = 4), p = 3/4, q = 1/2, γ = 5/3, δ = 11/5, ξ = 1/2, η = 1/3, λ = 1 and μ = 2, we consider the following system of Caputo fractional differential equations ⎧  2/5 1 1 1 ⎪ 1/3 c 3/4 c 10/3 ⎪ √ (v(t)) D u(t) = sin t + − D v(t) , ⎪ 0+ 0+ ⎪ 4 3(1 + t) ⎪ 4 + t2 ⎪ ⎪ ⎨ t ∈ (0, 1),  1/7 ⎪ e−t 1 1 9/2 1/2 ⎪ ⎪ arctan cD0+ u(t) − (u(t))2/3 + , ⎪cD0+ v(t) = ⎪ 2 1+t 2 5(3 + t) ⎪ ⎪ ⎩ t ∈ (0, 1), (7.56) supplemented with the boundary conditions ⎧ 1/5  1 ⎪ ⎪ 5/3 ⎪ v(t) ds , u (0) = 0, u(1) = I0+ v(1/2), ⎨u(0) = 2 0 1/3  1 ⎪ ⎪ 11/5 ⎪ ⎩v(0) = 4 u(t) dt , v  (0) = v  (0) = 0, v(1) = 2I0+ u(1/3). 0

(7.57)

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Existence of Solutions for Caputo Fractional BVPs

≈ 0.9999958 (Δ = 0) and With the given data, it is found that Δ     1 1 1/3 1 2/5   v  |f (t, u, v)| =  √ sin t + u − 2 4 3(1 + t) 4+t 1 1 1/3 1 2/5 + |u| + |v| , 2 4 3  −t   e  1 2/3 1 1/7  |g(t, u, v)| =  − + ) u arctan(v  2 1+t 2 5(3 + t) ≤

1 1 ≤ 1 + |u|2/3 + |v|1/7 , 2 15 for all t ∈ [0, 1], u, v ∈ R and ϕ(0) = ψ(0) = 0, |ϕ(v)| ≤ 2v1/5 , |ψ(u)| ≤ 4u1/3 , where  ϕ(v) = 2

0

1

1/5 v(t) dt

 ,

ψ(u) = 4

0

1

1/3 u(t) dt

,

∀u, v ∈ C[0, 1].

Here l1 = 1/3, l2 = 2/5, m1 = 2/3, m2 = 1/7, θ1 = 1/5, θ2 = 1/3, L1 = 2 and L2 = 4. Clearly the assumptions (K1)–(K3) are satisfied. Thus, by Theorem 7.3.1, we deduce that the problem (7.56), (7.57) has at least one solution on [0, 1]. Example 2. Let us choose α = 5/2 (n = 2), β = 11/3 (m = 3), p = 1/2, q = 1/3, γ = 9/4, δ = 7/2, λ = 2, μ = 3, ξ = 1/5, η = 1/2, and consider the system of Caputo fractional differential equations ⎧ 3 v 4 (t) t2 c 1/2 (1 + t)3 ⎪ c 5/2 ⎪ − + D u(t) = D v(t) , ⎪ 0+ 0+ ⎨ 200 500(1 + |v(t)|) 400 4 ⎪ ⎪ (1 − t)u2 (t) t3 c 1/3 (1 − t)2 11/3 ⎪ ⎩cD0+ + − v(t) = D u(t) , 0+ 300 400(1 + u2 (t)) 100

(7.58)

with the boundary conditions ⎧ 1/2  ⎪ 1 ⎪ ⎪ u(0) = |v(t)| , max ⎪ ⎨ 300 t∈[0,1] 1/4  ⎪ ⎪ 1 ⎪ ⎪ , max |u(t)| ⎩v(0) = 200 t∈[0,1]

9/4

u(1) = 2I0+ v(1/5), v  (0) = 0,

7/2

v(1) = 3I0+ u(1/2), (7.59)

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where f (t, u, v) =

u4 t2 v 3 (1 + t)3 − + , 200 500(1 + |u|) 400

(1 − t)2 (1 − t)u2 t3 v 4 + − , ∀t ∈ [0, 1], u, v ∈ R, 300 400(1 + u2 ) 100 1/2 1/4   1 1 ϕ(x) = , ψ(x) = , max |x(t)| max |x(t)| 300 t∈[0,1] 200 t∈[0,1]

g(t, u, v) =

∀x ∈ C[0, 1]. Obviously 1 1 1 + |u|4 + |v|3 , 25 500 400 1 1 1 |g(t, u, v)| ≤ + |u|2 + |v|4 , 300 400 100 1 1 |ϕ(x)| ≤ x1/2 , |ψ(x)| ≤ x1/4 , 300 200

|f (t, u, v)| ≤

for all t ∈ [0, 1], u, v ∈ R, x ∈ C[0, 1] and that h1 (x) = x4 , h2 (x) = x3 , 1 1 1 , c1 = 500 , c2 = 400 , k1 (x) = x2 , k2 (x) = x4 . One can notice that c0 = 25 1 1 1 1 1 1 1 d0 = 300 , d1 = 400 , d2 = 100 , θ1 = 2 , θ2 = 4 , L1 = 300 , L2 = 200 . Using ≈ 0.99999969, M1 ≈ 0.00666668, M2 ≈ the given values, we obtain Δ 1 ≈ 0.00369245, 0.00010554, M3 ≈ 0.60180232, M4 ≈ 8.49974 × 10−6 , M    M2 ≈ 0.00011691, M3 ≈ 1.16661255, M4 ≈ 9.41543 × 10−6 , N1 ≈ 1 ≈ 0.00008443, N2 ≈ 0.01000027, N3 ≈ 0.00082728, N4 ≈ 0.13594897, N 3 ≈ 0.00186696, and N 4 ≈ 0.43463894. 2 ≈ 0.01128439, N 0.00019054, N Taking L0 = 3, we find that A1 ≈ 0.26949999 and A2 ≈ 0.83583333, and the assumption (K5) holds true as 1 + N1 + N 1 )Lθ1 + (M2 + M 2 + N2 + N 2 )Lθ2 (M1 + M 0 0 3 + N3 + N 3 ) + A2 (M4 + M 4 + N4 + N 4 ) ≈ 1.001 < 3. +A1 (M3 + M Therefore, the conclusion of Theorem 7.3.2 applies and consequently problem (7.58), (7.59) has at least one solution on [0, 1]. Remark 7.3.1. The results presented in this section were published in [8].

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b3996-index

Index

A absolute continuity of the integral function, 85, 145 approximating fractional differential equation, 44, 59 approximation method, 39, 53 Arzela-Ascoli theorem, 38, 72, 86, 145, 161, 199, 286, 288–290, 304, 309–310, 312, 337, 345, 347–348, 372, 374, 379, 406, 434

302, 334, 358, 374–376, 382, 389, 399, 420 bounded operator, 76 bounded set, 4–5, 32, 77, 85, 87, 109, 112, 130, 132, 140–141, 289, 290, 302, 304–305, 335, 337, 341, 376–378, 400, 406, 411 bounded variation function, 39, 53, 68, 81, 83, 280–281, 294–295, 317, 321, 356–357, 361, 387, 397

B

C

Banach contraction mapping principle, 3, 25, 29, 356, 368, 387, 411, 415, 419 Banach space, 3–5, 17, 45, 59, 69, 103, 129, 139, 160, 197, 239, 282, 296, 322, 360, 395, 422 boundary conditions, 10–11, 23, 26, 38, 44, 50, 53–54, 59, 65, 69, 77, 79, 81–82, 92, 156, 158, 193, 253, 265, 280, 291, 294–295, 314, 317, 319, 351, 356, 359, 374, 385–386, 388, 391, 416, 436–437 boundary of a set, 5 boundary value problem, 18, 21, 23, 52, 68, 78, 86, 88, 90–91, 157, 181, 183–186, 188, 219, 221, 223, 225–226, 230, 266–271, 286, 288,

Caputo fractional boundary value problems, 355 Caputo fractional derivative, 1–2, 26, 355–356, 365, 387, 402, 418 Caputo fractional differential equations, 26, 355 Caputo fractional differential inclusion, 355 characterization theorem of supremum limit, 189–191 closed graph, 380 closed set, 3–4, 7, 32 closure of a set, 5 compact map, 7 compact multivalued map, 379, 381 compact operator, 4, 47, 85–86, 145, 286, 288–289, 302, 304, 307, 309, 447

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448

312, 335, 337, 343, 345–346, 348, 371–372, 374, 406 compact values of a multivalued map, 375, 382 complete metric space, 3, 7 completely continuous operator, 4–6, 18, 38, 45–47, 60, 70–71, 73–74, 84–86, 103–104, 129, 139, 146, 160–161, 198–199, 240–241, 288–289, 291, 304, 313, 335, 337, 350, 400, 406, 434, 436 concave function, 6, 101–102, 105, 110, 113–114 cone, 3–6, 17, 45, 59, 69, 84, 103, 129–130, 139, 146, 149, 160, 163, 172, 197, 201, 210, 239, 243, 255 continuous function, 14–16, 29, 31, 43, 57–58, 69, 83–84, 99, 101, 126–128, 159, 189, 195–196, 199, 236, 238, 272, 281, 297, 302, 310–312, 324, 334, 347, 349, 361, 397–398, 423 continuous operator, 4, 37, 46, 72–73, 85, 146, 286–288, 290, 302, 307, 309, 311, 335, 343, 346, 348, 371–372, 380, 400, 430 contraction mapping, 3–4, 7, 29–30, 37 contraction multivalued map, 383, 385 contraction operator, 285, 287, 300, 302, 307, 309, 330, 334, 343, 346, 368, 370–372, 413, 415 convex set, 3–4, 6–7, 32, 376 convex values of a multivalued map, 375–376 coupled boundary conditions, 387–388 coupled integral boundary conditions, 230, 419 coupled multi-point boundary conditions, 117, 155, 191–192, 227 coupled nonlocal boundary conditions, 316, 418 Covitz–Nadler fixed point theorem, 7, 356, 381

b3996-index

Index

E eigenfunction, 73–74, 86, 89 eigenvalue, 86, 89 equicontinuous family of functions, 37, 46–47, 72, 86, 141, 145, 286–290, 303–304, 308–311, 335, 337, 343, 345–348, 372–374, 378–379, 404, 406, 430, 434 Euclidean metric, 382 existence of nonnegative solutions, 26 existence of positive solutions, 16, 18, 21, 23, 43, 69, 83, 97–98, 101, 117, 128, 138, 154–156, 159, 192, 196, 233–234, 238, 276 existence of solutions, 279, 281, 285, 287–288, 290, 297, 302, 305, 309–310, 312, 315–316, 324, 334, 341, 346–347, 349, 355–357, 360, 369, 372, 374–376, 381, 386–387, 395, 399, 417–419, 422, 425, 434, 437–438 extreme limits, 159, 196, 238 F fixed point, 3–5, 7, 18, 20, 23, 28–30, 45, 47–48, 60, 63, 70, 75, 77, 87–92, 104, 109, 113, 115–116, 129, 133, 136–139, 148, 151, 160, 167, 176, 198, 207, 216, 241, 245, 248, 250, 252, 258, 260, 263, 265, 280, 282, 285–286, 290–291, 294, 297, 302, 305, 309–310, 312–313, 317, 324, 334, 341, 346–347, 349–350, 361, 368, 372, 374, 381, 385, 397, 411, 415, 423, 434, 436 fixed point index, 5–6, 63, 69, 77, 81, 98, 106, 108–109, 112–113, 115, 117, 132–133, 135–136, 138 fractional boundary value problems, 236 fractional differential equation, 10, 17, 23, 38–40, 50, 54, 65, 68, 77, 79, 82, 92, 98, 100, 156, 158, 234, 280, 291, 294, 385 fractional differential inclusion, 386

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Index

fractional differential system, 388, 419 fractional inclusion, 374 fractional integral boundary conditions, 357 fractional integro-differential equation, 25 G Gelfand formula, 77 Green function, 11, 15, 41, 55, 82–83, 98, 100, 103, 121, 126, 157, 159, 194–195, 235, 239 Guo–Krasnosel’skii fixed point theorem, 3–4, 10, 39, 69, 78, 117, 148, 151, 167, 176, 207, 216 H Hausdorff metric, 382 height functions, 53 I integral boundary conditions, 39–40, 53, 81 integral equation, 17, 28, 45, 59 integral terms, 25, 279 integro-multi-point boundary conditions, 357 J Jensen inequality, 105, 110 K Krasnosel’skii fixed point index theorem, 6 Krasnosel’skii fixed point theorem for the sum of two operators, 4, 25, 356 Krein–Rutman theorem, 69, 73, 81 L L1 -Caratheodory map, 375 Lebesgue dominated convergence theorem, 46, 73 Lebesgue integrable function, 28 Lebesgue spaces, 27

b3996-index

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449

Leggett–Williams theorem, 5, 60 Leray–Schauder alternative, 4, 387 Leray-Schauder nonlinear alternative, 419 linear operator, 5, 69–70, 73, 76, 81, 136, 380 linear sequential fractional differential equation, 357 M measurable function, 26 measurable multivalued map, 381 measurable selection, 382–383 monotone function, 105, 110 multi-point boundary conditions, 25, 53, 97–98, 100, 151, 155–156, 187, 233–235, 273 multi-point fractional boundary conditions, 9, 17 multiplicity of positive solutions, 39, 43, 50, 53, 58, 63, 83, 90–91, 97–98, 101, 115, 117, 128, 137–138 multivalued function, 356 multivalued map, 357, 375 N nonconvex values of a multivalued map, 381 nondecreasing functions, 43, 58, 70, 103, 137, 174, 214, 230–231, 257, 281, 312, 349, 375, 423 nondecreasing operator, 88 nonexistence of positive solutions, 154–156, 181, 183–184, 186, 192, 218–219, 221, 223, 225, 233–234, 266–271, 276 nonlinear alternative of Leray-Schauder type, 5, 356, 419, 436 nonlinear alternative of Leray–Schauder type for Kakutani maps, 7, 356, 375, 381 nonlinearities, 128, 138, 317 nonlocal boundary conditions, 9, 68, 279, 355

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Boundary Value Problems for. . . – 9in x 6in

450

nonlocal integral boundary conditions, 357 nonnegative functions, 59, 99, 101, 155, 230, 233, 236 nonnegative solution, 9, 25, 27, 30–31, 38–39, 68 nonsingular function, 25, 97, 117, 128 O open ball, 5 open set, 4–5, 7, 381 operators, 18, 28, 32, 45, 60, 63, 70, 76, 84, 89, 103, 129, 139, 160, 163, 172, 197, 201, 210, 240, 243, 255, 282, 285, 296, 322, 341, 360, 369, 376, 395, 422 P p-Laplacian operator, 155, 191 parameters, 9–10, 68–69, 116, 154–156, 159, 192, 196, 230, 233–234, 272 partial ordering, 4–5 positive solution, 9–10, 17, 20, 23–25, 39, 43, 48–49, 52, 58, 64, 68–69, 74, 77–78, 80–81, 83, 86, 88, 90–92, 95, 98, 101, 104, 109, 113, 116–117, 130, 133, 137, 146, 149, 153, 156, 162, 171–172, 181–186, 188, 192, 200–201, 207, 209–210, 216, 218–219, 221, 223, 225–226, 230, 238, 241–242, 245, 248, 250, 252, 254–255, 258, 260, 263, 265–271, 277 principal characteristic value, 69, 73–74 principal eigenvalue, 73 R relatively compact set, 4, 38, 46–47, 49, 64, 85–86, 140, 145, 286, 288–290, 304, 309–310, 312, 335, 337, 345, 347–348, 372, 374, 434 Riemann–Liouville fractional boundary value problems, 279

b3996-index

Index

Riemann–Liouville fractional derivative, 1, 9, 25, 39, 53, 68, 81, 98, 156, 192, 234, 279, 293, 317 Riemann–Liouville fractional differential equations, 9, 279 Riemann–Liouville fractional integral, 1, 279, 293, 317, 355–356, 387, 418 Riemann–Stieltjes integral boundary conditions, 9 Riemann–Stieltjes integrals, 39, 53, 68, 81, 279–280, 294, 317, 355, 387 S Schauder fixed point theorem, 4, 419, 434 semipositone problem, 10, 39, 77 sequential Caputo fractional differential equation, 355–357 sequential Caputo fractional differential inclusion, 355–357 sequential fractional integro-differential equation, 355–356 sequential fractional integro-differential inclusion, 355 set of selections, 375 sign-changing function, 9–10, 53, 230 singular fractional boundary value problem, 68 singular fractional differential equation, 9, 53, 81 singular function, 9, 16, 39, 53, 69, 77, 81, 97, 117, 138 solution, 10–11, 13, 17, 28, 40–45, 54–55, 57–60, 70, 83, 88, 98–101, 103, 118–119, 121, 125, 127–130, 139, 157–158, 160, 193–195, 197–198, 235–236, 239, 241, 280, 282, 285–286, 288, 290–295, 297, 305, 309–310, 312–313, 315, 317, 321, 324, 341, 346–347, 349–350, 352–353, 356, 358, 361, 372, 374, 375, 381, 385–387, 389, 393, 397, 411, 415, 420–421, 423, 434, 436 spectral radius, 73–74, 81, 86, 89

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b3996-index

Index

strictly increasing function, 101–102, 113–114 subhomogeneous function, 375 supremum limits, 189 supremum norm, 17, 45, 59, 69, 84, 103, 129, 139, 160, 197, 239, 282, 296, 322, 395, 422 system of Caputo fractional differential equations, 355, 418–419, 436–437 system of fractional differential equations, 317, 416 system of integral equations, 103, 128, 160, 197, 239 system of Riemann-Liouville fractional differential equations, 97, 117, 119, 151, 155, 187, 191–192, 227, 230, 233, 253, 265, 273, 293, 314, 316, 350 system of sequential Caputo fractional integro-differential equations, 387

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systems of Hammerstein integral equations, 272 U unbounded set, 4 uncoupled multi-point boundary conditions, 97, 155, 233 uncoupled nonlocal boundary conditions, 293 uniformly bounded family of functions, 37, 46, 71, 86, 286–288, 290, 303, 307, 310–311, 335, 343, 346, 348, 371–372, 400, 404 uniformly continuous function, 72, 85–86 uniqueness of solution, 25, 27, 282, 297, 302, 314, 324, 334, 360, 362, 368, 385, 387, 411, 418 upper semicontinuous map, 7 upper semicontinuous multivalued map, 379, 381