Basic Theory of Fractional Differential Equations

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BASIC THEORY OF

FRACTIONAL DIFFERENTIAL EQUATIONS

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BASIC THEORY OF

FRACTIONAL DIFFERENTIAL EQUATIONS Yo n g Z h o u Xiangtan University, China

World Scientific NEW JERSEY

9069hc_9789814579896_tp.indd 2

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LONDON

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TA I P E I

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CHENNAI

17/4/14 3:55 pm

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 Zhou, Yong, 1964– Basic theory of fractional differential equations / by Yong Zhou (Xiangtan University, China). pages cm Includes bibliographical references and index. ISBN 978-9814579896 (hardcover : alk. paper) 1. Fractional differential equations. I. Title. QA372.Z47 2014 515'.352--dc23 2014009125

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

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BC: 9069 – Basic Theory of Fractional Diff.Eqn.

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Preface

The concept of fractional derivative appeared for the first time in a famous correspondence between G.A. de L’Hospital and G.W. Leibniz, in 1695. Many mathematicians have further developed this area and we can mention the studies of L. Euler (1730), J.L. Lagrange (1772), P.S. Laplace (1812), J.B.J. Fourier (1822), N.H. Abel (1823), J. Liouville (1832), B. Riemann (1847), H.L. Greer (1859), H. Holmgren (1865), A.K. Gr¨ unwald (1867), A.V. Letnikov (1868), N.Ya. Sonin (1869), H. Laurent (1884), P.A. Nekrassov (1888), A. Krug (1890), J. Hadamard (1892), O. Heaviside (1892), S. Pincherle (1902), G.H. Hardy and J.E. Littlewood (1917), H. Weyl (1919), P. Lévy (1923), A. Marchaud (1927), H.T. Davis (1924), A. Zygmund (1935), E.R. Love (1938), A. Erdélyi (1939), H. Kober (1940), D.V. Widder (1941), M. Riesz (1949) and W. Feller (1952). In the past sixty years, fractional calculus had played a very important role in various fields such as physics, chemistry, mechanics, electricity, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, fitting of experimental data, etc. In the last decade, fractional calculus has been recognized as one of the best tools to describe long-memory processes. Such models are interesting for engineers and physicists but also for pure mathematicians. The most important among such models are those described by differential equations containing fractional-order derivatives. Their evolutions behave in a much more complex way than in the classical integer-order case and the study of the corresponding theory is a hugely demanding task. Although some results of qualitative analysis for fractional differential equations can be similarly obtained, many classical methods are hardly applicable directly to fractional differential equations. New theories and methods are thus required to be specifically developed, whose investigation becomes more challenging. Comparing with classical theory of differential equations, the researches on the theory of fractional differential equations are only on their initial stage of development. This monograph is devoted to a rapidly developing area of the research for the qualitative theory of fractional differential equations. In particular, we are interested in the basic theory of fractional differential equations. Such basic theory should be the starting point for further research concerning the dynamics, control, numerical analysis and applications of fractional differential equations. The book v

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Basic Theory of Fractional Differential Equations

is divided into six chapters. Chapter 1 introduces preliminary facts from fractional calculus, nonlinear analysis and semigroup theory. In Chapter 2, we present a unified framework to investigate the basic existence theory for discontinuous fractional functional differential equations with bounded delay, unbounded delay and infinite delay. Chapter 3 is devoted to the study of fractional differential equations in Banach spaces via measure of noncompactness method, topological degree method and Picard operator technique. In Chapter 4, we first present some techniques for the investigation of fractional evolution equations governed by C0 -semigroup, then we discuss fractional evolution equations with almost sectorial operators. In Chapter 5, by using critical point theory, we give a new approach to study boundary value problems of fractional differential equations. And in the last chapter, we present recent advances on theory for fractional partial differential equations including fractional Euler-Lagrange equations, time-fractional diffusion equations, fractional Hamiltonian systems and fractional Schrödinger equations. The material in this monograph are based on the research work carried out by the author and other experts during the past four years. The book is self-contained and unified in presentation, and it provides the necessary background material required to go further into the subject and explore the rich research literature. Each chapter concludes with a section devoted to notes and bibliographical remarks and all abstract results are illustrated by examples. The tools used include many classical and modern nonlinear analysis methods. This book is useful for researchers and graduate students for research, seminars, and advanced graduate courses, in pure and applied mathematics, physics, mechanics, engineering, biology, and related disciplines. I would like to thank Professors D. Baleanu, K. Balachandran, M. Benchohra, L. Bourdin, Y.Q. Chen, I. Vasundhara Devi, M. Feˇckan, N.J. Ford, W. Jiang, V. Kiryakova, F. Liu, J.A.T. Machado, M.M. Meerschaert, S. Momani, G.M. N’Guérékata, J.J. Nieto, V.E. Tarasov, J.J. Trujillo, A.S. Vatsala and M. Yamamoto for their support. I also wish to express my appreciation to my colleagues, Professors Z.B. Bai, Y.K. Chang, H.R. Sun, J.R. Wang, R.N. Wang, S.Q. Zhang and my graduate students H.B. Gu, F. Jiao, Y.H. Lan and L. Zhang for their help. Finally, I thank the editorial assistance of World Scientific Publishing Co., especially Ms. L.F. Kwong. I acknowledge with gratitude the support of National Natural Science Foundation of China (11271309, 10971173), the Specialized Research Fund for the Doctoral Program of Higher Education (20114301110001) and Key Projects of Hunan Provincial Natural Science Foundation of China (12JJ2001). Yong Zhou October 2013, Xiangtan, China

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Contents

Preface

v

1. Preliminaries

1

1.1 1.2 1.3

1.4

1.5

Introduction . . . . . . . . . . . . . . . . Some Notations, Concepts and Lemmas Fractional Calculus . . . . . . . . . . . . 1.3.1 Definitions . . . . . . . . . . . . 1.3.2 Properties . . . . . . . . . . . . Some Results from Nonlinear Analysis . 1.4.1 Sobolev Spaces . . . . . . . . . . 1.4.2 Measure of Noncompactness . . 1.4.3 Topological Degree . . . . . . . 1.4.4 Picard Operator . . . . . . . . . 1.4.5 Fixed Point Theorems . . . . . . 1.4.6 Critical Point Theorems . . . . Semigroups . . . . . . . . . . . . . . . . 1.5.1 C0 -Semigroup . . . . . . . . . . 1.5.2 Almost Sectorial Operators . . .

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2. Fractional Functional Differential Equations 2.1 2.2

2.3

2.4

Introduction . . . . . . . . . . . . . . . . Neutral Equations with Bounded Delay 2.2.1 Introduction . . . . . . . . . . . 2.2.2 Existence and Uniqueness . . . . 2.2.3 Extremal Solutions . . . . . . . p-Type Neutral Equations . . . . . . . . 2.3.1 Introduction . . . . . . . . . . . 2.3.2 Existence and Uniqueness . . . . 2.3.3 Continuous Dependence . . . . . Neutral Equations with Infinite Delay . vii

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viii

2.5

2.6

2.4.1 Introduction . . . . . . . . . . . . 2.4.2 Existence and Uniqueness . . . . . 2.4.3 Continuation of Solutions . . . . . Iterative Functional Differential Equations 2.5.1 Introduction . . . . . . . . . . . . 2.5.2 Existence . . . . . . . . . . . . . 2.5.3 Data Dependence . . . . . . . . . 2.5.4 Examples and General Cases . . . Notes and Remarks . . . . . . . . . . . . .

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3. Fractional Ordinary Differential Equations in Banach Spaces 3.1 3.2

3.3

3.4

3.5

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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . Cauchy Problems via Measure of Noncompactness Method 3.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . Cauchy Problems via Topological Degree Method . . . . . . 3.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Qualitative Analysis . . . . . . . . . . . . . . . . . . Cauchy Problems via Picard Operators Technique . . . . . 3.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Results via Picard Operators . . . . . . . . . . . . . 3.4.3 Results via Weakly Picard Operators . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . .

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4. Fractional Abstract Evolution Equations 4.1 4.2

4.3

4.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . Evolution Equations with Riemann-Liouville Derivative 4.2.1 Introduction . . . . . . . . . . . . . . . . . . . . 4.2.2 Definition of Mild Solutions . . . . . . . . . . . 4.2.3 Preliminary Lemmas . . . . . . . . . . . . . . . 4.2.4 Compact Semigroup Case . . . . . . . . . . . . . 4.2.5 Noncompact Semigroup Case . . . . . . . . . . . Evolution Equations with Caputo Derivative . . . . . . 4.3.1 Introduction . . . . . . . . . . . . . . . . . . . . 4.3.2 Definition of Mild Solutions . . . . . . . . . . . 4.3.3 Preliminary Lemmas . . . . . . . . . . . . . . . 4.3.4 Compact Semigroup Case . . . . . . . . . . . . . 4.3.5 Noncompact Semigroup Case . . . . . . . . . . . Nonlocal Cauchy Problems for Evolution Equations . . 4.4.1 Introduction . . . . . . . . . . . . . . . . . . . . 4.4.2 Definition of Mild Solutions . . . . . . . . . . . 4.4.3 Existence . . . . . . . . . . . . . . . . . . . . . .

53 55 62 66 66 66 72 74 80

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Contents

4.5

4.6

ix

Abstract Cauchy Problems with Almost Sectorial 4.5.1 Introduction . . . . . . . . . . . . . . . . 4.5.2 Preliminaries . . . . . . . . . . . . . . . . 4.5.3 Properties of Operators . . . . . . . . . . 4.5.4 Linear Problems . . . . . . . . . . . . . . 4.5.5 Nonlinear Problems . . . . . . . . . . . . 4.5.6 Applications . . . . . . . . . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . . . .

Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5. Fractional Boundary Value Problems via Critical Point Theory 5.1 5.2

5.3

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5.6

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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Existence of Solution for BVP with Left and Right Fractional Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Fractional Derivative Space . . . . . . . . . . . . . . . . . . 5.2.3 Variational Structure . . . . . . . . . . . . . . . . . . . . . 5.2.4 Existence under Ambrosetti-Rabinowitz Condition . . . . . 5.2.5 Superquadratic Case . . . . . . . . . . . . . . . . . . . . . 5.2.6 Asymptotically Quadratic Case . . . . . . . . . . . . . . . Multiple Solutions for BVP with Parameters . . . . . . . . . . . . 5.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . Infinite Solutions for BVP with Left and Right Fractional Integrals 5.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . Existence of Solutions for BVP with Left and Right Fractional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Variational Structure . . . . . . . . . . . . . . . . . . . . . 5.5.3 Existence of Weak Solutions . . . . . . . . . . . . . . . . . 5.5.4 Existence of Solutions . . . . . . . . . . . . . . . . . . . . . Notes and Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .

6. Fractional Partial Differential Equations 6.1 6.2

6.3

Introduction . . . . . . . . . . . . . . Fractional Euler-Lagrange Equations 6.2.1 Introduction . . . . . . . . . 6.2.2 Functional Spaces . . . . . . 6.2.3 Variational Structure . . . . 6.2.4 Existence of Weak Solution . Time-Fractional Diffusion Equations 6.3.1 Introduction . . . . . . . . .

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6.4

6.5

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6.3.2 Regularity and Unique Existence . Fractional Hamiltonian Systems . . . . . . 6.4.1 Introduction . . . . . . . . . . . . 6.4.2 Fractional Derivative Space . . . . 6.4.3 Existence and Multiplicity . . . . Fractional Schrödinger Equations . . . . . 6.5.1 Introduction . . . . . . . . . . . . 6.5.2 Existence and Uniqueness . . . . . Notes and Remarks . . . . . . . . . . . . .

Bibliography

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Chapter 1

Preliminaries

1.1

Introduction

In this chapter, we introduce some notations and basic facts on fractional calculus, nonlinear analysis and semigroup which are needed throughout this book.

1.2

Some Notations, Concepts and Lemmas

As usual N denotes the set of positive integer numbers and N0 the set of nonnegative integer numbers. R denotes the real numbers, R+ denotes the set of nonnegative reals and R+ the set of positive reals. Let C be the set of complex numbers. We recall that a vector space X equipped with a norm | · | is called a normed vector space. A subset E of a normed vector space X is said to be bounded if there exists a number K such that |x| ≤ K for all x ∈ E. A subset E of a normed vector space X is called convex if for any x, y ∈ E, ax + (1 − a)y ∈ E for all a ∈ [0, 1]. A sequence {xn } in a normed vector space X is said to converge to the vector x in X if and only if the sequence {|xn − x|} converges to zero as n → ∞. A sequence {xn } in a normed vector space X is called a Cauchy sequence if for every ε > 0 there exists an N = N (ε) such that for all n, m ≥ N (ε), |xn − xm | < ε. Clearly a convergent sequence is also a Cauchy sequence, but the converse may not be true. A space X where every Cauchy sequence of elements of X converges to an element of X is called a complete space. A complete normed vector space is said to be a Banach space. Let E be a subset of a Banach space X. A point x ∈ X is said to be a limit point of E if there exists a sequence of vectors in E which converges to x. We say a subset E is closed if E contains all of its limit points. The union of E and its limit ¯ Let X, F be normed points is called the closure of E and will be denoted by E. vector spaces, and E be a subset of X. An operator T : E → F is continuous at a point x ∈ E if and only if for any ε > 0 there is a δ > 0 such that |T x − T y| < ε for all y ∈ E with |x − y| < δ. Further, T is continuous on E, or simply continuous, if it is continuous at all points of E. 1

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We say that a subset E of a Banach space X is compact if every sequence of vectors in E contains a subsequence which converges to a vector in E. We say that E is relatively compact in X if every sequence of vectors in E contains a subsequence ¯ is compact. which converges to a vector in X, i.e., E is relatively compact in X if E Let J = [a, b] (−∞ < a < b < ∞) be a finite interval of R. We assume that X is a Banach space with the norm | · |. Denote C(J, X) be the Banach space of all continuous functions from J into X with the norm kxk = sup |x(t)|, t∈J

n

where x ∈ C(J, X). C (J, X) (n ∈ N0 ) denotes the set of mappings having n times continuously differentiable on J, AC(J, X) is the space of functions which are absolutely continuous on J and AC n (J, X) (n ∈ N0 ) is the space of functions f such that f ∈ C n−1 (J, X) and f (n−1) ∈ AC(J, X). In particular, AC 1 (J, X) = AC(J, X). Let 1 ≤ p ≤ ∞. Lp (J, X) denotes the Banach space of all measurable functions f : J → X. Lp (J, X) is normed by Z p1    1 ≤ p < ∞, |f (t)|p dt , J kf kLpJ =    inf { sup |f (t)|}, p = ∞. µ(J¯)=0 t∈J\J¯

In particular, L1 (J, X) is the Banach space of measurable functions f : J → X with the norm Z kf kLJ = |f (t)|dt, J

∞

and L (J, X) is the Banach space of measurable functions f : J → X which are bounded, equipped with the norm kf kL∞J = inf{c > 0 : |f (t)| ≤ c, a.e. t ∈ J}. Lemma 1.1 (H¨ older inequality). Assume that p, q ≥ 1, and f ∈ Lp (J, X), g ∈ Lq (J, X), then for 1 ≤ p ≤ ∞, f g ∈ L1 (J, X) and

1 p

+

1 q

= 1. If

kf gkLJ ≤ kf kLpJ kgkLq J . A family F in C(J, X) is called uniformly bounded if there exists a positive constant K such that |f (t)| ≤ K for all t ∈ J and all f ∈ F . Further, F is called equicontinuous, if for every ε > 0 there exists a δ = δ(ε) > 0 such that |f (t1 ) − f (t2 )| < ε for all t1 , t2 ∈ J with |t1 − t2 | < δ and all f ∈ F . Lemma 1.2 (Arzela-Ascoli’s theorem). If a family F = {f (t)} in C(J, R) is uniformly bounded and equicontinuous on J, then F has a uniformly convergent subsequence {fn (t)}∞ n=1 . If a family F = {f (t)} in C(J, X) is uniformly bounded and equicontinuous on J, and for any t∗ ∈ J, {f (t∗ )} is relatively compact, then F has a uniformly convergent subsequence {fn (t)}∞ n=1 .

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The Arzela-Ascoli’s Theorem is the key to the following result: A subset F in C(J, R) is relatively compact if and only if it is uniformly bounded and equicontinuous on J. Lemma 1.3 (Lebesgue’s dominated convergence theorem). Let E be a measurable set and let {fn } be a sequence of measurable functions such that limn→∞ fn (x) = f (x) a.e. in E, and for every n ∈ N, |fn (x)| ≤ g(x) a.e. in E, where g is integrable on E. Then Z Z lim fn (x)dx = f (x)dx. n→∞

E

E

Finally, we state the Bochner’s theorem. Lemma 1.4 (Bochner’s theorem). A measurable function f : (a, b) → X is Bochner integrable if |f | is Lebesgue integrable. 1.3

Fractional Calculus

The gamma function Γ(z) is defined by Z ∞ Γ(z) = tz−1 e−t dt (Re(z) > 0), 0

where tz−1 = e(z−1) log(t) . This integral is convergent for all complex z ∈ C (Re(z) > 0). For this function the reduction formula Γ(z + 1) = zΓ(z) holds. In particular, if z = n ∈ N0 , then

Γ(n + 1) = n!

(Re(z) > 0) (n ∈ N0 )

with (as usual) 0! = 1. Let us consider some of the starting points for a discussion of fractional calculus. One development begins with a generalization of repeated integration. Thus if f is locally integrable on (c, ∞), then the n-fold iterated integral is given by Z s1 Z sn−1 Z t −n f (sn )dsn ds1 ds2 · · · c Dt f (t) = c c c Z t 1 (t − s)n−1 f (s)ds = (n − 1)! c for almost all t with −∞ ≤ c < t < ∞ and n ∈ N. Writing (n − 1)! = Γ(n), an immediate generalization is the integral of f of fractional order α > 0, Z t 1 −α D f (t) = (t − s)α−1 f (s)ds (left hand) c t Γ(α) c and similarly for −∞ < t < d ≤ ∞ Z d 1 −α (s − t)α−1 f (s)ds (right hand) D f (t) = t d Γ(α) t

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4

both being defined for suitable f . A number of definitions for the fractional derivative has emerged over the years, we refer the reader to Diethelm, 2010; Hilfer, 2006; Kilbas, Srivastava and Trujillo, 2006; Miller and Ross, 1993; Podlubny, 1999. In this book, we restrict our attention to the use of the Riemann-Liouville and Caputo fractional derivatives. In this section, we introduce some basic definitions and properties of the fractional integrals and fractional derivatives which are used further in this book. The materials in this section are taken from Kilbas, Srivastava and Trujillo, 2006. 1.3.1

Definitions

Definition 1.5 (Left and right Riemann-Liouville fractional integrals). Let J = [a, b] (−∞ < a a, α > 0 (1.1) D f (t) = a t Γ(α) a and

−α t Db f (t)

1 = Γ(α)

Z

b

t

(s − t)α−1 f (s)ds,

t < b, α > 0,

(1.2)

respectively, provided the right-hand sides are pointwise defined on [a, b]. When α = n ∈ N, the definitions (1.1) and (1.2) coincide with the n-th integrals of the form Z t 1 −n (t − s)n−1 f (s)ds a Dt f (t) = (n − 1)! a

and

−n t Db f (t) =

1 (n − 1)!

Z

t

b

(s − t)n−1 f (s)ds.

Definition 1.6 (Left and right Riemann-Liouville fractional derivatives). The left and right Riemann-Liouville fractional derivatives a Dtα f (t) and t Dbα f (t) of order α ∈ R+ , are defined by α a Dt f (t)

and α t Db f (t)

dn −(n−α) f (t) aD dtn t Z t n d 1 n−α−1 (t − s) f (s)ds , = Γ(n − α) dtn a

=

dn −(n−α) f (t) tD dtn b Z b 1 dn = (−1)n n (s − t)n−α−1 f (s)ds , Γ(n − α) dt t

(1.3) t>a

= (−1)n

(1.4) t < b,

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5

respectively, where n = [α] + 1, [α] means the integer part of α. In particular, when α = n ∈ N0 , then 0 a Dt f (t) n a Dt f (t)

= t Db0 f (t) = f (t),

= f (n) (t) and

n t Db f (t)

= (−1)n f (n) (t),

where f (n) (t) is the usual derivative of f (t) of order n. If 0 < α < 1, then Z t 1 d α −α (t − s) f (s)ds , t > a a Dt f (t) = Γ(1 − α) dt a and

α t Db f (t)

d 1 =− Γ(1 − α) dt

Z

b

t

−α

(s − t)

f (s)ds ,

t < b.

Remark 1.7. If f ∈ C([a, b], RN ), it is obvious that Riemann-Liouville fractional integral of order α > 0 exists on [a, b]. On the other hand, following Lemma 2.2 in Kilbas, Srivastava and Trujillo, 2006, we know that the Riemann-Liouville fractional derivative of order α ∈ [n − 1, n) exists almost everywhere on [a, b] if f ∈ AC n ([a, b], RN ). The left and right Caputo fractional derivatives are defined via above RiemannLiouville fractional derivatives. Definition 1.8 (Left and right Caputo fractional derivatives). The left and α C α right Caputo fractional derivatives C a Dt f (t) and t Db f (t) of order α ∈ R+ are defined by n−1 X f (k) (a) k α C α (1.5) (t − a) a Dt f (t) = a Dt f (t) − k! k=0

and

C α t Db f (t)

=

α t Db

f (t) −

respectively, where

n−1 X k=0

f (k) (b) k (b − t) , k!

n = [α] + 1 for α 6∈ N0 ; n = α for α ∈ N0 .

(1.6)

(1.7)

In particular, when 0 < α < 1, then C α a Dt f (t)

= a Dtα (f (t) − f (a))

C α t Db f (t)

= t Dbα (f (t) − f (b)).

and

The Riemann-Liouville fractional derivative and the Caputo fractional derivative are connected with each other by the following relations.

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Property 1.9. (i) If α 6∈ N0 and f (t) is a function for which the Caputo fractional derivatives C α C α + exist together with the Riemanna Dt f (t) and t Db f (t) of order α ∈ R α Liouville fractional derivatives a Dt f (t) and t Dbα f (t), then C α a Dt f (t)

= a Dtα f (t) −

n−1 X

f (k) (a) (t − a)k−α Γ(k − α + 1)

= t Dbα f (t) −

n−1 X

f (k) (b) (b − t)k−α , Γ(k − α + 1)

and C α t Db f (t)

k=0

k=0

where n = [α] + 1. In particular, when 0 < α < 1, we have C α a Dt f (t)

= a Dtα f (t) −

f (a) (t − a)−α Γ(1 − α)

C α t Db f (t)

= t Dbα f (t) −

f (b) (b − t)−α . Γ(1 − α)

and

(ii) If α = n ∈ N0 and the usual derivative f (n) (t) of order n exists, then n and C t Db f (t) are represented by C n a Dt f (t)

= f (n) (t) and

C n t Db f (t)

= (−1)n f (n) (t).

C n a Dt f (t)

(1.8)

Property 1.10. Let α ∈ R+ and let n be given by (1.7). If f ∈ AC n ([a, b], RN ), α C α then the Caputo fractional derivatives C a Dt f (t) and t Db f (t) exist almost everywhere on [a, b]. α (i) If α 6∈ N0 , C a Dt f (t) and

and

C α t Db f (t)

are represented by Z t 1 C α n−α−1 (n) D f (t) = (t − s) f (s)ds a t Γ(n − α) a

C α t Db f (t)

=

(−1)n Γ(n − α)

Z

t

b

(s − t)n−α−1 f (n) (s)ds

(1.9)

(1.10)

respectively, where n = [α] + 1. In particular, when 0 < α < 1 and f ∈ AC([a, b], RN ), Z t 1 C α −α ′ D f (t) = (t − s) f (s)ds (1.11) a t Γ(1 − α) a and

C α t Db f (t)

(ii) If α = n ∈ N0 then ular,

=−

C α a Dt f (t)

1 Γ(1 − α)

Z

b t

(s − t)−α f ′ (s)ds .

(1.12)

α and C t Db f (t) are represented by (1.8). In partic-

C 0 a Dt f (t)

0 =C t Db f (t) = f (t).

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7

Remark 1.11. If f is an abstract function with values in Banach space X, then integrals which appear in above definitions are taken in Bochner’s sense. The fractional integrals and derivatives, defined on a finite interval [a, b] of R, are naturally extended to whole axis R. Definition 1.12 (Left and right Liouville-Weyl fractional integrals on the real axis). The left and right Liouville-Weyl fractional integrals −∞ Dt−α f (t) and −α t D+∞ f (t) of order α > 0 on the whole axis R are defined by Z t 1 −α (t − s)α−1 f (s)ds (1.13) D f (t) = −∞ t Γ(α) −∞

and

−α t D+∞ f (t)

1 = Γ(α)

Z

t

∞

(s − t)α−1 f (s)ds,

(1.14)

respectively, where t ∈ R and α > 0. Definition 1.13 (Left and right Liouville-Weyl fractional derivatives on the real axis). The left and right Liouville-Weyl fractional derivatives −∞ Dtα f (t) α and t D+∞ f (t) of order α on the whole axis R are defined by dn −(n−α) (−∞ Dt f (t)) dtn Z t n 1 d n−α−1 = (t − s) f (s)ds Γ(n − α) dtn −∞

(1.15)

dn −(n−α) (t D+∞ f (t)) dtn Z ∞ 1 dn = (−1)n n (s − t)n−α−1 f (s)ds , Γ(n − α) dt t

(1.16)

α −∞ Dt f (t)

and α t D+∞ f (t)

=

= (−1)n

respectively, where n = [α] + 1, α ≥ 0 and t ∈ R. In particular, when α = n ∈ N0 , then 0 −∞ Dt f (t)

n −∞ Dt f (t)

= f (n) (t)

0 = t D+∞ f (t) = f (t),

and

n t D+∞ f (t)

= (−1)n f (n) (t),

where f (n) (t) is the usual derivative of f (t) of order n. If 0 < α < 1 and t ∈ R, then Z t d 1 −α α (t − s) f (s)ds −∞ Dt f (t) = Γ(1 − α) dt −∞ Z ∞ α f (t) − f (t − s) = ds Γ(1 − α) 0 sα+1

and

Z ∞ d 1 (s − t)−α f (s)ds Γ(1 − α) dt t Z ∞ f (t) − f (t + s) α ds. = Γ(1 − α) 0 sα+1

α t D+∞ f (t) = −

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Formulas (1.11) and (1.12) can be used for the definition of the Caputo fractional derivatives on the whole axis R. Definition 1.14 (Left and right Caputo fractional derivatives on the real α C α axis). The left and right Caputo fractional derivatives C −∞ Dt f (t) and t D+∞ f (t) of order α (with α > 0 and α 6∈ N) on the whole axis R are defined by Z t 1 n−α−1 (n) C α (t − s) f (s)ds (1.17) D f (t) = −∞ t Γ(n − α) −∞ and C α t D+∞ f (t)

(−1)n = Γ(n − α)

Z

∞

t

n−α−1 (n)

(s − t)

f

(s)ds ,

(1.18)

respectively. When 0 < α < 1, the relations (1.17) and (1.18) take the following forms Z t 1 C α −α ′ (t − s) f (s)ds −∞ Dt f (t) = Γ(1 − α) −∞ and C α t D+∞ f (t)

=−

1 Γ(1 − α)

Z

t

∞

(s − t)−α f ′ (s)ds .

Now we present the Fourier transform properties of the fractional integral and fractional differential operators. Definition 1.15. The Fourier transform of a function f (t) of real variable t ∈ R is defined by Z ∞ fˆ(w) = e−it·w f (t)dt (w ∈ R). −∞

Let f (t) be defined on (−∞, ∞) and 0 < α < 1. Then the Fourier transform of the Liouville-Weyl integral and differential operator satisfies −α \ −∞ Dt f (t)(w) −α \ t D∞ f (t)(w)

= (iw)−α fˆ(w), = (−iw)−α fˆ(w),

α \ −∞ Dt f (t)(w)

= (iw)α fˆ(w), αˆ α \ t D∞ f (t)(w) = (−iw) f (w).

1.3.2

Properties

We present here some properties of the fractional integral and derivative operators that will be useful throughout this book. Property 1.16. If α ≥ 0 and β > 0, then −α a Dt (t

− a)β−1 =

Γ(β) (t − a)β+α−1 (α > 0), Γ(β + α)

(1.19)

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α a Dt (t

− a)β−1 =

and −α t Db (b

− t)β−1 =

9

Γ(β) (t − a)β−α−1 (α ≥ 0) Γ(β − α)

(1.20)

Γ(β) (b − t)β+α−1 (α > 0), Γ(β + α)

(1.21)

Γ(β) (b − t)β−α−1 (α ≥ 0). (1.22) Γ(β − α) In particular, if β = 1 and α ≥ 0, then the Riemann-Liouville fractional derivatives of a constant are, in general, not equal to zero: (t − a)−α (b − t)−α α , t Dbα 1 = . (1.23) a Dt 1 = Γ(1 − α) Γ(1 − α) On the other hand, for j = 1, 2, ..., [α] + 1, α t Db (b

− t)β−1 =

α a Dt (t

− a)α−j = 0,

α t Db (b

− t)α−j = 0.

(1.24) −α a Dt

−α t Db

The semigroup property of the fractional integral operators and are given by the following result. Property 1.17. If α > 0 and β > 0, then the equations −β −α−β −β −α −α = D f (t) and D = t Db−α−β f (t) (1.25) D D f (t) D f (t) a t a t a t t b t b

are satisfied at almost every point t ∈ [a, b] for f ∈ Lp ([a, b], RN ) (1 ≤ p < ∞). If α + β > 1, then the relations in (1.25) hold at any point of [a, b]. Property 1.18.

(i) If α > 0 and f ∈ Lp ([a, b], RN ) (1 ≤ p ≤ ∞), then the following equalities −α −α α α (1.26) a Dt a Dt f (t) = f (t) and t Db t Db f (t) = f (t) (α > 0)

hold almost everywhere on [a, b]. (ii) If α > β > 0, then, for f ∈ Lp ([a, b], RN ) (1 ≤ p ≤ ∞), the relations −α+β β −α f (t) and t Dbβ t Db−α f (t) = t Db−α+β f (t) (1.27) a Dt a Dt f (t) = a Dt hold almost everywhere on [a, b].

In particular, when β = k ∈ N and α > k, then −α −α+k k f (t) and a Dtk t Db−α f (t) = (−1)k t Db−α+k f (t). a Dt a Dt f (t) = a Dt

(1.28) To present the next property, we use the spaces of functions a Dt−α (Lp ) and −α p t Db (L ) defined for α > 0 and 1 ≤ p ≤ ∞ by −α p a Dt (L )

= {f : f = a Dt−α ϕ, ϕ ∈ Lp ([a, b], RN )}

−α p t Db (L )

= {f : f = t Db−α φ, φ ∈ Lp ([a, b], RN )},

and

respectively. The composition of the fractional integration operator a Dt−α with the fractional differentiation operator a Dtα is given by the following result. −(n−α) Property 1.19. Let α > 0, n = [α] + 1 and let fn−α (t) = a Dt f (t) be the fractional integral (1.1) of order n − α.

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(i) If 1 ≤ p ≤ ∞ and f ∈ a Dt−α (Lp ), then −α α a Dt a Dt f (t) = f (t).

(1.29)

(ii) If f ∈ L1 ([a, b], RN ) and fn−α ∈ AC n ([a, b], RN ), then the equality n (n−j) X fn−α (a) −α α (t − a)α−j , a Dt a Dt f (t) = f (t) − Γ(α − j + 1)

(1.30)

j=1

holds almost everywhere on [a, b].

−(n−α)

Property 1.20. Let α > 0 and n = [α] + 1. Also let gn−α (t) = t Db the fractional integral (1.2) of order n − α. (i) If 1 ≤ p ≤ ∞ and g ∈ t Db−α (Lp ), then −α α t Db t Db g(t) = g(t). 1

N

n

g(t) be

(1.31)

N

(ii) If g ∈ L ([a, b], R ) and gn−α ∈ AC ([a, b], R ), then the equality n (n−j) X (−1)n−j gn−α (a) −α α D D g(t) = g(t) − (b − t)α−j , t b t b Γ(α − j + 1) j=1

(1.32)

holds almost everywhere on [a, b].

In particular, if 0 < α < 1, then g1−α (a) −α α (b − t)α−1 , (1.33) t Db t Db g(t) = g(t) − Γ(α) where g1−α (t) = t Dbα−1 g(t) while for α = n ∈ N, the following equality holds: n−1 (k) X (−1)k gn−α (a) −n n (b − t)k . (1.34) t Db t Db g(t) = g(t) − k! k=0

Property 1.21. Let α > 0 and let y ∈ L∞ ([a, b], RN ) or y ∈ C([a, b], RN ). Then −α −α C α C α (1.35) a Dt a Dt y(t) = y(t) and t Db t Db y(t) = y(t).

Property 1.22. Let α > 0 and let n be given by (1.7). If y ∈ AC n ([a, b], RN ) or y ∈ C n ([a, b], RN ), then n−1 X y (k) (a) −α C α (t − a)k (1.36) D D y(t) = y(t) − a t a t k! k=0

and

−α C α t Db t Db y(t)

= y(t) −

n−1 X k=0

(−1)k y (k) (b) (b − t)k . k!

(1.37)

In particular, if 0 < α ≤ 1 and y ∈ AC([a, b], RN ) or y ∈ C([a, b], RN ), then −α C α −α C α (1.38) a Dt a Dt y(t) = y(t) − y(a) and t Db t Db y(t) = y(t) − y(b).

On the other hand, we have the following property of fractional integration. Property 1.23. Let α > 0, p ≥ 1, q ≥ 1, and 1p + q1 ≤ 1 + α (p 6= 1 and q 6= 1 in the case when p1 + 1q = 1 + α).

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11

(i) If ϕ ∈ Lp ([a, b], RN ) and ψ ∈ Lq ([a, b], RN ), then Z b Z b ϕ(t) a Dt−α ψ(t)dt = ψ(t) t Db−α ϕ(t)dt. a

(1.39)

a

(ii) If f ∈ t Db−α (Lp ) and g ∈ a Dt−α (Lq ), then Z b Z f (t) a Dtα g(t)dt = a

a

b

g(t) t Dbα f (t)dt.

(1.40)

Then applying Property 1.9, we can derive the integration by parts formula for the left and right Riemann-Liouville fractional derivatives looks as follows. Property 1.24. Z b Z b α α 0 < α ≤ 1, a Dt f (t) · g(t)dt = t Db g(t) · f (t)dt, a

a

provided the boundary conditions

f (a) = f (b) = 0, f ′ ∈ L∞ ([a, b], RN ), g ∈ L1 ([a, b], RN ), or g(a) = g(b) = 0, g ′ ∈ L∞ ([a, b], RN ), f ∈ L1 ([a, b], RN ) are fulfilled. Remark 1.25. If f , g are abstract functions with values in Banach space X, then integrals which appear in above properties are taken in Bochner’s sense. 1.4

Some Results from Nonlinear Analysis

1.4.1

Sobolev Spaces

We refer to Cazenave and Haraux, 1998, for the definitions and results given below. Consider an open subset Ω of RN . D(Ω) is the space of C ∞ (real-valued or complex valued) functions with compact support in Ω and D′ (Ω) is the space of distributions on Ω. A distribution T ∈ D′ (Ω) is said to belong to Lp (Ω) (1 ≤ p ≤ ∞) if there exists a function f ∈ Lp (Ω) such that Z hT, ϕi = f (x)ϕ(x)dx, Ω

for all ϕ ∈ D(Ω). In that case, it is well known that f is unique. Let m ∈ N and let p ∈ [1, ∞]. Define W m,p (Ω) = {f ∈ Lp (Ω); Dα f ∈ Lp (Ω) for all α ∈ NN such that |α| ≤ m}. W m,p (Ω) is a Banach space when equipped with the norm defined by X kf kW m,p (Ω) = kDα f kLp , |α|≤m

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for all f ∈ W m,p (Ω). For all m, p as above, we denote by W0m,p (Ω) the closure of D(Ω) in W m,p (Ω). If p = 2, one sets W m,2 (Ω) = H m (Ω), W0m,2 (Ω) = H0m (Ω) and one equips H m (Ω) with the following equivalent norm:   12 X kf kH m =  kDα uk2L2  . |α|≤m

m

Then H (Ω) is a Hilbert space with the scalar product X Z hu, viH m = Dα u · Dα vdx. |α|≤m

Ω

If Ω is bounded, there exists a constant C(Ω) such that kukL2 ≤ C(Ω)k▽ukL2 , H01 (Ω)

for all u ∈ (this is Poincaré’s inequality). It may be more convenient to 1 equip H0 (Ω) with the following scalar product Z hu, vi = ▽u · ▽vdx, Ω

which defines an equivalent norm to k · kH 1 on the closed space H01 (Ω). 1.4.2

Measure of Noncompactness

We recall here some definitions and properties of measure of noncompactness. Assume that X is a Banach space with the norm | · |. The measure of noncompactness α is said to be: (i) Monotone if for all bounded subsets B1 , B2 of X, B1 ⊆ B2 implies α(B1 ) ≤ α(B2 ); (ii) Nonsingular if α({x} ∪ B) = α(B) for every x ∈ X and every nonempty subset B ⊆ X; (iii) Regular α(B) = 0 if and only if B is relatively compact in X. One of the most important examples of measure of noncompactness is the Hausdorff measure of noncompactness α defined on each bounded subset B of X by α(B) = inf{ε > 0 : B ⊂

m [

j=1

Bε (xj ) where xj ∈ X},

(1.41)

where Bε (xj ) is a ball of radius ≤ ε centered at xj , j = 1, 2, ..., m. Without confusion, Kuratowski measure of noncompactness α1 defined on each bounded subset B of X by α1 (B) = inf{ε > 0 : B ⊂

m [

j=1

Mj and diam(Mj ) ≤ ε},

(1.42)

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13

where the diameter of Mj is defined by diam(Mj ) = sup{|x − y| : x, y ∈ Mj }, j = 1, 2, ..., m. It is well known that the Hausdorff measure of noncompactness α and Kuratowski measure of noncompactness α1 enjoy the above properties (i)-(iii) and other properties. We refer the reader to Bana`s and Goebel, 1980; Deimling, 1985; Heinz, 1983; Lakshmikantham and Leela, 1969. (iv) α(B1 + B2 ) ≤ α(B1 ) + α(B2 ), where B1 + B2 = {x + y : x ∈ B1 , y ∈ B2 }; (v) α(B1 ∪ B2 ) ≤ max{α(B1 ), α(B2 )}; (vi) α(λB) ≤ |λ|α(B) for any λ ∈ R. In particular, the relationship of Hausdorff measure of noncompactness α and Kuratowski measure of noncompactness α1 is given by (vii) α(B) ≤ α1 (B) ≤ 2α(B). Let J = [0, a], a ∈ R+ , For any W ⊂ C(J, X), we define Z t Z t W (s)ds = u(s)ds : u ∈ W , for t ∈ [0, a], 0

0

where W (s) = {u(s) ∈ X : u ∈ W }. We present here some useful properties. Property 1.26. If W ⊂ C(J, X) is bounded and equicontinuous, then coW ⊂ C(J, X) is also bounded and equicontinuous. Property 1.27. (Guo, Lakshmikantham and Liu, 1996) If W ⊂ C(J, X) is bounded and equicontinuous, then t → α(W (t)) is continuous on J, and Z t Z t α(W ) = max α(W (t)), α W (s)ds ≤ α(W (s))ds, for t ∈ [0, a]. t∈J

0

0

{un }∞ n=1

Property 1.28. (Mönch, 1980) Let be a sequence of Bochner integrable functions from J into X with |un (t)| ≤ m(t) ˜ for almost all t ∈ J and every n ≥ 1, + where m ˜ ∈ L(J, R+ ), then the function ψ(t) = α({un (t)}∞ n=1 ) belongs to L(J, R ) and satisfies Z t Z t α un (s)ds : n ≥ 1 ≤2 ψ(s)ds. 0

0

Property 1.29. (Bothe, 1998) If W is bounded, then for each ε > 0, there is a sequence {un }∞ n=1 ⊂ W , such that α(W ) ≤ 2α({un }∞ n=1 ) + ε.

1.4.3

Topological Degree

For a minute description of the following notions we refer the reader to Bana`s and Goebel, 1980; Deimling, 1985; Heinz, 1983; Lakshmikantham and Leela, 1969. Definition 1.30. Consider Ω ⊂ X and F : Ω → X a continuous bounded mapping. We say that F is α-Lipschitz if there exists k ≥ 0 such that α(F (B)) ≤ kα(B) (∀) B ⊂ Ω bounded.

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If, in addition, k < 1, then we say that F is a strict α-contraction. We say that F is α-condensing if α(F (B)) < α(B) (∀) B ⊂ Ω bounded with α(B) > 0. In other words, α(F (B)) ≥ α(B) implies α(B) = 0. The class of all strict αcontractions F : Ω → X is denoted by SCα (Ω) and the class of all α-condensing mappings F : Ω → X is denoted by Cα (Ω). We remark that SCα (Ω) ⊂ Cα (Ω) and every F ∈ Cα (Ω) is α-Lipschitz with constant k = 1. We also recall that F : Ω → X is Lipschitz if there exists k > 0 such that |F x − F y| ≤ k|x − y| (∀) x, y ∈ Ω and that F is a strict contraction if k < 1. Next, we collect some properties of the applications defined above. Property 1.31. If F , G : Ω → X are α-Lipschitz mappings with constants k, k ′ , respectively, then F + G : Ω → X is α-Lipschitz with constant k + k ′ . Property 1.32. If F : Ω → X is compact, then F is α-Lipschitz with constant k = 0. Property 1.33. If F : Ω → X is Lipschitz with constant k, then F is α-Lipschitz with the same constant k. The theorem below asserts the existence and the basic properties of the topological degree for α-condensing perturbations of the identity. For more details, see Isaia, 2006. Let T = (I − F , Ω, y) : Ω ⊂ X open and bounded, F ∈ Cα (Ω), y ∈ X \ (I − F )(∂Ω) be the family of the admissible triplets. Theorem 1.34. There exists one degree function D : T → N0 which satisfies the properties:

(i) Normalization D(I, Ω, y) = 1 for every y ∈ Ω; (ii) Additivity on domain For every disjoint, open sets Ω1 , Ω2 ⊂ Ω and every y does not belong to (I − F )(Ω\(Ω1 ∪ Ω2 )) we have D(I − F , Ω, y) = D(I − F , Ω1 , y) + D(I − F , Ω2 , y); (iii) Invariance under homotopy D(I − H(t, ·), Ω, y(t)) is independent of t ∈ [0, 1] for every continuous, bounded mapping H : [0, 1] × Ω → X which satisfies α(H([0, 1] × B)) < α(B) (∀) B ⊂ Ω with α(B) > 0 and every continuous function y : [0, 1] → x which satisfies y(t) 6= x − H(t, x) (∀) t ∈ [0, 1], (∀) x ∈ ∂Ω; (iv) Existence D(I − F , Ω, y) 6= 0 implies y ∈ (I − F )(Ω);

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15

(v) Excision D(I − F , Ω, y) = D(I − F , Ω1 , y) for every open set Ω1 ⊂ Ω and every y does not belong to (I − F )(Ω\Ω1 ). Having in hand a degree function defined on T , we collect the usability of the a priori estimate method by means of this degree. Theorem 1.35. Let F : X → X be α-condensing and S = {x ∈ X : (∃) λ ∈ [0, 1] such that x = λF x} .

If S is a bounded set in X, so there exists r > 0 such that S ⊂ Br (0), then D(I − λF , Br (0), 0) = 1

(∀) λ ∈ [0, 1].

Consequently, F has at least one fixed point and the set of the fixed points of F lies in Br (0). 1.4.4

Picard Operator

Let (X, d) be a metric space and A : X → X an operator. We shall use the following notations: P (X) = {Y ⊆ X | Y 6= ∅}; FA = {x ∈ X | A(x) = x}−the fixed point set of A; I(A) = {Y ∈ P (X) | A(Y ) ⊆ Y }; OA (x) = {x, A(x), A2 (x), ..., An (x), ...}−the A-orbit of x ∈ X; H : P (X) × P (X) → R+ ∪ {+∞}; H(Y, Z) = max{sup inf d(a, b), sup inf d(a, b)}−the Pompeiu-Hausdorff funca∈Y b∈Z

b∈Z a∈Y

tional on P (X). Definition 1.36. (Rus, 1987) Let (X, d) be a metric space. An operator A : X → X is a Picard operator if there exists x∗ ∈ X such that FA = {x∗ } and the sequence (An (x0 ))n∈N converges to x∗ for all x0 ∈ X. Definition 1.37. (Rus, 1993) Let (X, d) be a metric space. An operator A : X → X is a weakly Picard operator if the sequence (An (x0 ))n∈N converges for all x0 ∈ X and its limit (which may depend on x0 ) is a fixed point of A. If A is a weakly Picard operator, then we consider the operator A∞ : X → X, A∞ (x) = lim An (x). n→∞

The following results are useful in what follows. Theorem 1.38. (Rus, 1979) Let (Y, d) be a complete metric space and A, B : Y → Y two operators. Suppose that: (i) A is a contraction with contraction constant ρ and FA = {x∗A }; (ii) B has fixed points and x∗B ∈ FB ; (iii) There exists η > 0 such that d(A(x), B(x)) ≤ η, for all x ∈ Y . η Then d(x∗A , x∗B ) ≤ 1−ρ . Theorem 1.39. (Rus and Mure¸san, 2000) Let (X, d) be a complete metric space and A, B : X → X two orbitally continuous operators. Assume that:

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(i) there exists ρ ∈ [0, 1) such that

d(A2 (x), A(x)) ≤ ρd(x, A(x)),

d(B 2 (x), B(x)) ≤ ρd(x, B(x))

for all x ∈ X; (ii) there exists η > 0 such that d(A(x), B(x)) ≤ η for all x ∈ X. η , where H denotes the Pompeiu-Hausdorff functional. Then H(FA , FB ) ≤ 1−ρ Theorem 1.40. (Rus, 1993) Let (X, d) be a metric space. Then A : X → X is a S weakly Picard operator if and only if there exists a partition X = λ∈Λ Xλ of X such that

(i) Xλ ∈ I(A); (ii) A |Xλ : Xλ → Xλ is a Picard operator, for all λ ∈ Λ. 1.4.5

Fixed Point Theorems

In this subsection, we present some fixed point theorems which will be used in the following chapters. Theorem 1.41 (Banach contraction mapping principle). Let (X, d) be a complete metric space, and T : Ω → Ω a contraction mapping: d(T x, T y) ≤ kd(x, y), where 0 < k < 1, for each x, y ∈ Ω. Then, there exists a unique fixed point x of T in Ω: T x = x. Theorem 1.42 (Schauder’s fixed point theorem). Let X be a Banach space and Ω ⊂ X a convex, closed and bounded set. If T : Ω → Ω is a continuous operator such that T Ω ⊂ X, T Ω is relatively compact, then T has at least one fixed point in Ω. Theorem 1.43 (Schaefer’s fixed point theorem). Let X be a Banach space and let F : X → X be a completely continuous mapping. Then either (i) the equation x = λF x has a solution for λ = 1, or (ii) the set {x ∈ X : x = λF x for some λ ∈ (0, 1)} is unbounded. Theorem 1.44 (Darbo-Sadovskii’s fixed point theorem). If Ω is bounded closed and convex subset of a Banach space X, the continuous mapping T : Ω → Ω is an α-contraction, then the mapping T has at least one fixed point in Ω. Theorem 1.45 (Krasnoselskii’s fixed point theorem). Let X be a Banach space, let Ω be a bounded closed convex subset of X and let S , T be mappings of Ω into X such that S z + T w ∈ Ω for every pair z, w ∈ Ω. If S is a contraction and T is completely continuous, then the equation S z + T z = z has a solution on Ω. Theorem 1.46 (O’Regan’s fixed point theorem). Let U be an open set in a closed, convex set C of X. Assume 0 ∈ U , T (U ) is bounded and T : U → C is given

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17

by T = T1 + T2 where T1 : U → X is completely continuous, and T2 : U → X is a nonlinear contraction. Then either (i) T has a fixed point in U , or (ii) there is a point x ∈ ∂U and λ ∈ (0, 1) with x = λT (x). A non-empty closed set K in a Banach space X is called a cone if: (i) K + K ⊆ K; (ii) λK ⊆ K for λ ∈ R, λ ≥ 0; (iii) {−K} ∩ K = {0}, where 0 is the zero element of X. We introduce an order relation “ ≤ ” in X as follows. Let z, y ∈ X. Then z ≤ y if and only if y − z ∈ K. A cone K is called normal if the norm k · kX is semi-monotone increasing on K, that is, there is a constant N > 0 such that kzkX ≤ N kykX for all z, y ∈ K with z ≤ y. It is known that if the cone K is normal in X, then every order-bounded set in X is norm-bounded. Similarly, the cone K in X is called regular if every monotone increasing (resp. decreasing) order bounded sequence in X converges in norm. For any a, b ∈ X, a ≤ b, the order interval [a, b] is a set in X given by [a, b] = {z ∈ X : a ≤ z ≤ b}.

Let X and Y be two ordered Banach spaces. A mapping T : X → Y is said to be nondecreasing or monotone increasing if z ≤ y implies T z ≤ T y for all z, y ∈ [a, b]. Theorem 1.47 (Hybrid fixed point theorem). (Dhage, 2006) Let X be a Banach space and A, B, C : X → X be three monotone increasing operators such that (i) (ii) (iii) (iv)

A is a contraction with contraction constant k < 1; B is completely continuous; C is totally bounded; there exist elements a and b in X such that a ≤ Aa + Ba + Ca and b ≥ Ab + Bb + Cb with a ≤ b.

Further if the cone K in X is normal, then the operator equation Az +Bz +Cz = z has a least and a greatest solution in [a, b]. 1.4.6

Critical Point Theorems

Let H be a real Banach space and C 1 (H, RN ) denotes the set of functionals that are Fréchet differentiable and their Fréchet derivatives are continuous on H. We need to use the critical point theorems to consider the fractional boundary value problems and fractional Hamiltonian systems. For the reader’s convenience, we state some necessary definitions and theorems and skip the proofs. Definition 1.48. (Rabinowitz, 1986) Let ψ ∈ C 1 (H, RN ). If any sequence

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{uk } ⊂ H for which {ψ(uk )} is bounded and ψ ′ (uk ) → 0 as k → ∞ possesses a convergent subsequence, then we say ψ satisfies Palais-Smale condition (denoted by (PS) condition for short). Definition 1.49. (Mawhin and Willem, 1989) Let H be a real Banach space, ψ : H → R is differentiable and c ∈ R. We say that ψ satisfies the (PS)c condition if the existence of a sequence {uk } in H such that ψ(uk ) → c,

ψ ′ (uk ) → 0

as k → ∞, implies that c is a critical value of ψ. Theorem 1.50. (Mawhin and Willem, 1989) Let H be a real reflexive Banach space. If the functional ψ : H → RN is weakly lower semi-continuous and coercive, i.e. lim|z|→∞ ψ(z) = +∞, then there exists z0 ∈ H such that ψ(z0 ) = inf z∈H ψ(z). Moreover, if ψ is also Fréchet differentiable on H, then ψ ′ (z0 ) = 0. Let Br be the open ball in H with the radius r and centered at 0 and ∂Br denote its boundary. Let us recall two critical point results due to Rabinowitz, 1986. Theorem 1.51 (Mountain pass theorem). (Rabinowitz, 1986) Let H be a real Banach space and I ∈ C 1 (H, R) satisfy Palais-Smale condition. Suppose that I satisfies the following conditions: (i) I(0) = 0, (ii) there exist constants ρ, β > 0 such that I|∂Bρ (0) ≥ β, (iii) there exist e ∈ H \ B ρ (0) such that I(e) ≤ 0. Then I possesses a critical value c ≥ β given by c = inf max I(g(s)), g∈Γ s∈[0,1]

where Bρ (0) is an open ball in H of radius ρ centered at 0, and Γ = {g ∈ C([0, 1], H) : g(0) = 0, g(1) = e}. Theorem 1.52. (Rabinowitz, 1986) Let H be a real Banach space and I ∈ C 1 (H, R) with I even. Suppose that I satisfies (PS) condition, (i), (ii) of Theorem 1.51 and the following condition: (iii′ ) For each finite dimensional subspace H ′ ⊂ H, there is r = r(H ′ ) > 0 such that I(u) ≤ 0 for u ∈ H ′ \ Br (0). Then I possesses an unbounded sequence of critical values. Remark 1.53. A deformation lemma can be proved with condition (C) replacing the usual (PS) condition, and it turns out that Theorem 1.52 holds true under condition (C). We say I satisfies condition (C), i.e., for every sequence {un } ⊂ H, {un } has a convergent subsequence if I(un ) is bounded and (1 + |un |)|I ′ (un )| → 0 as n → +∞. Let X be a reflexive and separable Banach space, then there are ej ∈ X and e∗j ∈ X ∗ such that X = span{ej : j = 1, 2, ...} and X ∗ = span{e∗j : j = 1, 2, ...},

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Preliminaries

and he∗i , ej i

=

1, 0,

19

i = j, i 6= j.

(1.43)

For convenience, we write Xj := span{ej },

Yk :=

k M

Xj ,

Zk :=

j=1

∞ M

Xj .

(1.44)

j=k

And let Bk := {u ∈ Yk : |u| ≤ ρk },

Nk := {u ∈ Zk : |u| = γk }.

(1.45)

Theorem 1.54 (Fountain Theorem). (Bartsch, 1993) Suppose: (H1) X is a Banach space, ϕ ∈ C 1 (X, R) is an even functional, the subspace Xk , Yk and Zk are defined by (1.44). If for every k ∈ N, there exist ρk > rk > 0 such that (H2) ak := max ϕ(u) ≤ 0; u∈Yk

|u|=ρk

(H3) bk := inf ϕ(u) → ∞, as k → ∞; u∈Zk

|u|=rk

(H4) ϕ satisfies the (PS)c condition for every c > 0. Then ϕ has an unbounded sequence of critical values. Theorem 1.55 (Dual Fountain Theorem). (Bartsch, 1993) Assume (H1) is satisfied, and there is a k0 > 0 so as to for each k ≥ k0 , there exist ρk > rk > 0 such that (H5) dk := inf ϕ(u) → 0, as k → ∞; u∈Zk

|u|≤ρk

(H6) ik := max ϕ(u) < 0; u∈Yk

|u|=rk

(H7) inf ϕ(u) ≥ 0; u∈Zk

|u|=ρk

(H8) ϕ satisfies the (PS)∗c condition for every c ∈ [dk0 , 0). Then ϕ has a sequence of negative critical values converging to 0. Remark 1.56. ϕ satisfies the (PS)∗c condition means that: if any sequence {unj } ⊂ ′ X such that nj → ∞, unj ∈ Ynj , ϕ(unj ) → c and (ϕ|Ynj ) (unj ) → 0, then {unj } contains a subsequence converging to a critical point of ϕ. It is obvious that if ϕ satisfies the (PS)∗c condition, then ϕ satisfies the (PS)c condition. ˜ : X → R be two functionals. For r, r1 , r2 , r3 ∈ Let X be a nonempty set and Φ, Ψ R with r1 < supX Φ, r2 > inf X Φ, r2 > r1 , and r3 > 0, we define ϕ(r) :=

inf −1

u∈Φ

(−∞,r)

˜ ˜ supu∈Φ−1 (−∞,r) Ψ(u) − Ψ(u) r − Φ(u)

,

(1.46)

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β(r1 , r2 ) :=

γ(r2 , r3 ) :=

˜ ˜ Ψ(v) − Ψ(u) , 1 ) v∈Φ−1 [r1 ,r2 ) Φ(v) − Φ(u) sup

(1.47)

˜ supu∈Φ−1 (−∞,r2 +r3 ) Ψ(u) , r3

(1.48)

inf

u∈Φ−1 (−∞,r

α(r1 , r2 , r3 ) := max {ϕ(r1 ), ϕ(r2 ), γ(r2 , r3 )} .

(1.49)

Lemma 1.57. (Averna and Bonanno, 2009; Bonanno and Candito, 2008) Let X be a reflexive real Banach space, Φ : X → R be a convex, coercive, and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous ˜ : X → R be a continuously Gâteaux differentiable functional inverse on X ∗ , Ψ whose Gâteaux derivative is compact, such that ˜ (i) inf X Φ = Φ(0) = Ψ(0) = 0; ˜ 1 ) ≥ 0 and Ψ(u ˜ 2 ) ≥ 0, one has (ii) for every u1 , u2 satisfying Ψ(u ˜ (tu1 + (1 − t)u2 ) ≥ 0. inf Ψ

t∈[0,1]

Assume further that there exist three positive constants r1 , r2 and r3 , with r1 < r2 , such that (iii) α(r1 , r2 , r3 )) < β(r1 , r2 ). ˜ has three Then, for each λ ∈ 1/β(r1 , r2 ), 1/α(r1 , r2 , r3 ) , the functional Φ − λΨ −1 distinct critical points u1 , u2 and u3 such that u1 ∈ Φ (−∞, r1 ), u2 ∈ Φ−1 [r1 , r2 ) and u3 ∈ Φ−1 (−∞, r2 + r3 ). 1.5 1.5.1

Semigroups C0 -Semigroup

Let X be a Banach space and B(X) be the Banach space of linear bounded operators. Definition 1.58. A semigroup is a one parameter family {T (t) : t ≥ 0} ⊂ B(X) satisfying the conditions: (i) T (t)T (s) = T (t + s), for t, s ≥ 0; (ii) T (0) = I. Here I denotes the identity operator in X. Definition 1.59. A semigroup {T (t)}t≥0 is uniformly continuous if lim kT (t) − T (0)kB(X) = 0,

t→0+

that is if lim kT (t) − T (s)kB(X) = 0.

|t−s|→0

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Definition 1.60. We say that the semigroup {T (t)}t≥0 is strongly continuous (or a C0 semigroup) if the map t → T (t)x is strongly continuous, for each x ∈ X, i.e. lim T (t)x = x, ∀x ∈ X.

t→0+

Definition 1.61. Let T (t) be a C0 -semigroup defined on X. The infinitesimal generator A of T (t) is the linear operator defined by A(x) = lim

t→0+

where D(A) = {x ∈ X : limt→0+ 1.5.2

T (t)x − x , for x ∈ D(A), t

T (t)x−x t

exists in X}.

Almost Sectorial Operators

We firstly introduce some special functions and classes of functions which will be used in the following, for more details, we refer to Markus, 2006; Periago and Straub, 2002. Let Sµ0 with 0 < µ < π be the open sector {z ∈ C\{0} : | arg z| < µ} and Sµ be its closure, that is Sµ = {z ∈ C\{0} : | arg z| ≤ µ} ∪ {0}. We state the concept of almost sectorial operators as follows. Definition 1.62. (Periago and Straub, 2002) Let −1 < p < 0 and 0 < ω < π/2. By Θpω (X) we denote the family of all linear closed operators A : D(A) ⊂ X → X which satisfy: (i) σ(A) ⊂ Sω ; (ii) for every ω < µ < π there exists a constant Cµ such that kR(z; A)kB(X) ≤ Cµ |z|p , for all z ∈ C \ Sµ ,

(1.50)

where R(z; A) = (zI − A)−1 , z ∈ ρ(A), which are bounded linear operators the resolvent of A. A linear operator A will be called an almost sectorial operator on X if A ∈ Θpω (X). Remark 1.63. Let A ∈ Θpω (X). Then the definition implies that 0 ∈ ρ(A). We denote the semigroup associated with A by {Q(t)}t≥0 . For t ∈ S 0π −ω 2 Z 1 Q(t) = e−tz (A) = e−tz R(z; A)dz, 2πi Γθ

where the integral contour Γθ = {R+ eiθ } ∪ {R+ e−iθ } is oriented counter-clockwise and ω < θ < µ < π/2 − | arg t|, forms an analytic semigroup of growth order 1 + p. Remark 1.64. From Periago and Straub, 2002, note that if A ∈ Θpω (X), then A generates a semigroup Q(t) with a singular behavior at t = 0 in a sense, called

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semigroup of growth 1 + p. Moreover, the semigroup Q(t) is analytic in an open sector of the complex plane C, but the strong continuity fails at t = 0 for data which are not sufficiently smooth. Property 1.65. (Periago and Straub, 2002) Let A ∈ Θpω (X) with −1 < p < 0 and 0 < ω < π2 . Then the following properties remain true. dn Q(t) = (−A)n Q(t) (t ∈ S 0π2 −ω ); 2 dtn the functional equation Q(s + t) = Q(s)Q(t) for all s, t ∈ S 0π −ω holds; 2 there is a constant C0 = C0 (p) > 0 such that kQ(t)kB(X) ≤ C0 t−p−1 (t > 0); if β > 1 + p, D(Aβ ) ⊂ ΣQ = {x ∈ X : limt→0+ Q(t)x = x}; R ∞then −λt R(λ, A) = 0 e Q(t)dt for every λ ∈ C with Re(λ) > 0.

(i) Q(t) is analytic in S 0π −ω and

(ii) (iii) (iv) (v)

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Chapter 2

Fractional Functional Differential Equations

2.1

Introduction

The main objective of this chapter is to present a unified framework to investigate the basic existence theory for a variety of fractional functional differential equations with applications. As far as we know, many complex processes in nature and technology are described by functional differential equations which are dominant nowadays because the functional components in equations allow one to consider prehistory or after-effect influence. Various classes of functional differential equations are of fundamental importance in many problems arising in bionomics, epidemiology, electronics, theory of neural networks, automatic control, etc. Quite long ago delay differential equations had shown their efficiency in the study of the behavior of real populations. One can show that even though the delay terms occurring in the equations are unbounded, the domain of the initial data (past history or memory) may be finite or infinite. Consequently, those two cases need to be discussed independently. Moreover, one can consider functional differential equations so that the delay terms also occur in the derivative of the unknown solution. Since the general formulation of such a problem is difficult to state, a special kind of equations called neutral functional differential equations has been introduced. On the other hand, fractional calculus is one of the best tools to characterize long-memory processes and materials, anomalous diffusion, long-range interactions, long-term behaviors, power laws, allometric scaling laws, and so on. So the corresponding mathematical models are fractional differential equations. Their evolutions behave in a much more complicated way so to study the corresponding dynamics is much more difficult. Although the existence theorems for the fractional differential equations can be similarly obtained, not all the classical theory of differential equation can be directly applied to the fractional differential equations. Hence, a somewhat theoretical frame needs to be established. In Section 2.2, we discuss existence and uniqueness of solutions and existence of extremal solutions of initial value problem for the fractional neutral differential equations with bounded delay. Section 2.3 is devoted to study of basic existence theory for fractional p-type neutral differential equations with unbounded delay 23

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but finite memory. In Section 2.4, we present a unified treatment of fundamental existence theory of fractional neutral differential equations with infinite memory. In Section 2.5, we consider a fractional order iterative functional differential equation with parameter. Some theorems to prove existence of the iterative series solutions are presented under some nature conditions. 2.2 2.2.1

Neutral Equations with Bounded Delay Introduction

Let I0 = [−τ, 0], τ > 0, t0 ≥ 0 and I = [t0 , t0 + σ], σ > 0 be two closed and bounded intervals in R. Denote J = [t0 − τ, t0 + σ]. Let C = C(I0 , Rn ) be the space of continuous functions on I0 . For any element ϕ ∈ C, define the norm kϕk∗ = sup |ϕ(θ)|. θ∈I0

n

If z ∈ C(J, R ), then for any t ∈ I define zt ∈ C by zt (θ) = z(t + θ), θ ∈ [−τ, 0]. Consider the initial value problems (IVP for short) of fractional neutral functional differential equations with bounded delay of the form C α t0 Dt (x(t) − k(t, xt )) = F (t, xt ), a.e. t ∈ (t0 , t0 + a], (2.1) xt0 = ϕ, α n where C t0 Dt is Caputo fractional derivative of order 0 < α < 1, F : I × C → R is a given function satisfying some assumptions that will be specified later, and ϕ ∈ C. In Subsection 2.2.2, we establish the existence and uniqueness theorems of IVP (2.1). In Subsection 2.2.3, we discuss the existence of extremal solutions for IVP 1 1 (2.1). We firstly give the definitions of L β -Carathéodory, L γ -Chandrabhan and 1 L δ -Lipschitz, where β, γ, δ are some given numbers. Next, we apply the hybrid fixed point theorem to prove the existence results of extremal solutions for IVP 1 1 1 (2.1) under L β -Carathéodory, L γ -Chandrabhan and L δ -Lipschitz conditions. We do not require the continuity of the nonlinearities involved in the equation (2.1). In the end, we will present an example to illustrate our main results.

2.2.2

Existence and Uniqueness

Let A(σ, γ) = {x ∈ C([t0 −τ, t0 +σ], Rn ) : xt0 = ϕ, supt0 ≤t≤t0 +σ |x(t)−ϕ(0)| ≤ γ}, where σ, γ are positive constants. Before stating and proving the main results, we introduce the following hypotheses: (H1) F (t, ϕ) is measurable with respect to t on I; (H2) F (t, ϕ) is continuous with respect to ϕ on C(I0 , Rn );

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1

(H3) there exist α1 ∈ (0, α) and a real-valued function m(t) ∈ L α1 I such that for any x ∈ A(σ, γ), |F (t, xt )| ≤ m(t), for t ∈ I0 ; (H4) for any x ∈ A(σ, γ), k(t, xt ) = k1 (t, xt ) + k2 (t, xt ); (H5) k1 is continuous and for any x′ , x′′ ∈ A(σ, γ), t ∈ I |k1 (t, x′t ) − k1 (t, x′′t )| ≤ lkx′ − x′′ k, where l ∈ (0, 1); (H6) k2 is completely continuous and for any bounded set Λ in A(σ, γ), the set {t → k2 (t, xt ) : x ∈ Λ} is equicontinuous in C(I, Rn ). Lemma 2.1. If there exist σ ∈ (0, a) and γ ∈ (0, ∞) such that (H1)-(H3) are satisfied, then for t ∈ (t0 , t0 + σ], IVP (2.1) is equivalent to the following equation  Z t 1  (t − s)α−1 F (s, xs )ds, t ∈ I0 , x(t) = ϕ(0) − k(t0 , ϕ) + k(t, xt ) + (2.2) Γ(α) t0  xt0 = ϕ. Proof. First, it is easy to obtain that F (t, xt ) is Lebesgue measurable on I according to conditions (H1) and (H2). A direct calculation gives that (t − s)α−1 ∈ 1 L 1−α1 ([t0 , t], R), for t ∈ I. In the light of Hölder inequality and (H3), we obtain that (t − s)α−1 F (s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ I0 and x ∈ A(σ, γ), and Z t 1 |(t − s)α−1 F (s, xs )|ds ≤ k(t − s)α−1 k 1−α kmk α1 . (2.3) t0

L

1

[t0 ,t]

L

1

I

According to Definitions 1.5 and 1.8, it is easy to see that if x is a solution of the IVP (2.1), then x is a solution of the equation (2.2). On the other hand, if (2.2) is satisfied, then for every t ∈ (t0 , t0 + σ], we have Z t 1 C α C α α−1 (x(t) − k(t, x )) = D ϕ(0) − k(t , ϕ) + D (t − s) F (s, x )ds t 0 s t0 t t0 t Γ(α) t0 1 Z t α α−1 =C D (t − s) F (s, x )ds s t0 t Γ(α) t0 −α α =C D D F (t, x ) t t0 t t0 t (t − t0 )−α = t0 Dtα t0 Dt−α F (t, xt ) − [t0 Dt−α F (t, xt )]t=t0 Γ(1 − α) −α (t − t0 ) . = F (t, xt ) − [t0 Dt−α F (t, xt )]t=t0 Γ(1 − α)

According to (2.3), we know that [t0 Dt−α F (t, xt )]t=t0 = 0, which means that − k(t, xt )) = F (t, xt ), t ∈ (t0 , t0 + σ], and this completes the proof.

C α t0 Dt (x(t)

Theorem 2.2. Assume that there exist σ ∈ (0, a) and γ ∈ (0, ∞) such that (H1)(H6) are satisfied. Then the IVP (2.1) has at least one solution on [t0 , t0 + η] for some positive number η.

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Proof. tion

According to (H4), the equation (2.2) is equivalent to the following equa x(t) = ϕ(0) − k1 (t0 , ϕ) − k2 (t0 , ϕ) + k1 (t, xt ) + k2 (t, xt )   Z t  1 (t − s)α−1 F (s, xs )ds, t ∈ I, +  Γ(α) t0   xt0 = ϕ.

Let ϕ e ∈ A(σ, γ) be defined as ϕ et0 = ϕ, ϕ(t e 0 + t) = ϕ(0) for all t ∈ [0, σ]. If x is a solution of the IVP (2.1), let x(t0 + t) = ϕ(t e 0 + t) + y(t), t ∈ [−τ, σ], then we have xt0 +t = ϕ et0 +t + yt , t ∈ [0, σ]. Thus y satisfies the equation

y(t) = −k1 (t0 , ϕ) − k2 (t0 , ϕ) + k1 (t0 + t, yt + ϕ et0 +t ) + k2 (t0 + t, yt + ϕ et0 +t ) Z t (2.4) 1 + (t − s)α−1 F (t0 + s, ys + ϕ et0 +s )ds, t ∈ [0, σ]. Γ(α) 0

Since k1 , k2 are continuous and xt is continuous in t, there exists σ ′ > 0, when 0 < t < σ′ , γ |k1 (t0 + t, yt + ϕ et0 +t ) − k1 (t0 , ϕ)| < , (2.5) 3 and γ |k2 (t0 + t, yt + ϕ et0 +t ) − k2 (t0 , ϕ)| < . (2.6) 3 Choose 1 γΓ(α)(1 + β)1−α1 (1+β)(1−α1 ) ′ (2.7) η = min σ, σ , 3M where β =

α−1 1−α1

∈ (−1, 0) and M = kmk

Define E(η, γ) as follows

1

L α1 I

.

E(η, γ) = {y ∈ C([−τ, η], Rn ) : y(s) = 0 for s ∈ [−τ, 0] and kyk ≤ γ}. Then E(η, γ) is a closed bounded and convex subset of C([−τ, σ], Rn ). On E(η, γ) we define the operators S and U as follows 0, t ∈ [−τ, 0], (Sy)(t) = −k1 (t0 , ϕ) + k1 (t0 + t, yt + ϕ et0 +t ), t ∈ [0, η],   t ∈ [−τ, 0],  0, −k (t , ϕ) + k (t + t, y + ϕ e ) (U y)(t) = 2 0 2 0 t t0 +t   + 1 R t (t − s)α−1 F (t0 + s, ys + ϕ et0 +s )ds, t ∈ [0, η]. Γ(α) 0

It is easy to see that the operator equation

y = Sy + U y

(2.8)

has a solution y ∈ E(η, γ) if and only if y is a solution of the equation (2.4). Thus x(t0 + t) = y(t) + ϕ(t e 0 + t) is a solution of the equation (2.1) on [0, η]. Therefore,

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the existence of a solution of the IVP (2.1) is equivalent that (2.8) has a fixed point in E(η, γ). Now we show that S + U has a fixed point in E(η, γ). The proof is divided into three steps. Step I. Sz + U y ∈ E(η, γ) for every pair z, y ∈ E(η, γ). In fact, for every pair z, y ∈ E(η, γ), Sz + U y ∈ C([−τ, η], Rn ). Also, it is obvious that (Sz + U y)(t) = 0, t ∈ [−τ, 0]. Moreover, for t ∈ [0, η], by (2.5)-(2.7) and the condition (H3), we have |(Sz)(t) + (U y)(t)|

≤ | − k1 (t0 , ϕ) + k1 (t0 + t, zt + ϕ et0 +t )| + | − k2 (t0 , ϕ) + k2 (t0 + t, yt + ϕ et0 +t )| Z t 1 |(t − s)α−1 F (t0 + s, ys + ϕ et0 +s )|ds + Γ(α) 0 1−α1 Z t0 +t α1 Z t α−1 1 2γ 1 1−α1 α1 ≤ ds (m(s)) ds + (t − s) 3 Γ(α) t0 0 Z t 1−α1 Z t0 +σ α1 α−1 1 1 2γ + (t − s) 1−α1 ds ≤ (m(s)) α1 ds 3 Γ(α) 0 t0 2γ M η (1+β)(1−α1 ) + 3 Γ(α)(1 + β)1−α1 ≤ γ. ≤

Therefore, kSz + U yk = sup |(Sz)(t) + (U y)(t)| ≤ γ, t∈[0,η]

which means that Sz + U y ∈ E(η, γ) for any z, y ∈ E(η, γ). Step II. S is a contraction on E(η, γ). For any y ′ , y ′′ ∈ E(η, γ), yt′ + ϕ et0 +t , yt′′ + ϕ et0 +t ∈ A(δ, γ). So by (H5), we get that |(Sy ′ )(t) − (Sy ′′ )(t)| = |k1 (t0 + t, yt′ + ϕ et0 +t ) − k1 (t0 + t, yt′′ + ϕ et0 +t )| ≤ lky ′ − y ′′ k,

which implies that

kSy ′ − Sy ′′ k ≤ lky ′ − y ′′ k. In view of 0 < l < 1, S is a contraction on E(η, γ). Step III. Now we show that U is a completely continuous operator. Let 0, t ∈ [−τ, 0], (U1 y)(t) = −k2 (t0 , ϕ) + k2 (t0 + t, yt + ϕ et0 +t ), t ∈ [0, η] and

(U2 y)(t) =

(

0,

1 Γ(α)

t ∈ [−τ, 0], α−1 (t − s) F (t + s, y + ϕ e )ds, t ∈ [0, η]. 0 s t +s 0 0

Rt

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Clearly, U = U1 + U2 . Since k2 is completely continuous, U1 is continuous and {U1 y : y ∈ E(η, γ)} is uniformly bounded. From the condition that the set {t → k2 (t, xt ) : x ∈ Λ} is equicontinuous for any bounded set Λ in A(σ, γ), we can conclude that U1 is a completely continuous operator. On the other hand, for any t ∈ [0, η], we have Z t 1 (t − s)α−1 |F (t0 + s, ys + ϕ et0 +s )|ds |(U2 y)(t)| ≤ Γ(α) 0 1−α1 Z t0 +t α1 Z t α−1 1 1 1−α1 α1 ≤ ds (m(s)) ds (t − s) Γ(α) t0 0

M η (1+β)(1−α1 ) . Γ(α)(1 + β)1−α1 Hence, {U2 y : y ∈ E(η, γ)} is uniformly bounded. Now, we will prove that {U2 y : y ∈ E(η, γ)} is equicontinuous. For any 0 ≤ t1 < t2 ≤ η and y ∈ E(η, γ), we get that ≤

|(U2 y)(t2 ) − (U2 y)(t1 )| Z t1 1 [(t2 − s)α−1 − (t1 − s)α−1 ]F (t0 + s, ys + ϕ et0 +s )ds = Γ(α) 0 Z t2 1 α−1 (t2 − s) F (t0 + s, ys + ϕ et0 +s )ds + Γ(α) t1 Z t1 1 α−1 α−1 ≤ [(t1 − s) − (t2 − s) ]|F (t0 + s, ys + ϕ et0 +s )|ds Γ(α) 0 Z t2 1 + (t2 − s)α−1 |F (t0 + s, ys + ϕ et0 +s )|ds Γ(α) t1 1−α1 Z t1 1 M α−1 α−1 1−α 1 ds ≤ [(t1 − s) − (t2 − s) ] Γ(α) 0 1−α1 Z t2 1 M + [(t2 − s)α−1 ] 1−α1 ds Γ(α) t1 Z t1 Z t2 1−α1 1−α1 M M β β β ≤ (t1 − s) − (t2 − s) ds + (t2 − s) ds Γ(α) Γ(α) 0 t1 1−α1 M 1+β 1+β 1+β ≤ t − t + (t − t ) 2 1 2 Γ(α)(1 + β)1−α1 1 M + (t2 − t1 )(1+β)(1−α1 ) Γ(α)(1 + β)1−α1 2M (t2 − t1 )(1+β)(1−α1 ) , ≤ Γ(α)(1 + β)1−α1 which means that {U2 y : y ∈ E(η, γ)} is equicontinuous. Moreover, it is clear that U2 is continuous. So U2 is a completely continuous operator. Then U = U1 + U2 is a completely continuous operator.

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Therefore, Krasnoselskii’s fixed point theorem shows that S +U has a fixed point on E(η, γ), and hence the IVP (2.1) has a solution x(t) = ϕ(0) + y(t − t0 ) for all t ∈ [t0 , t0 + η]. In the case where k1 ≡ 0, we get the following result. Corollary 2.3. Assume that there exist σ ∈ (0, a) and γ ∈ (0, ∞) such that (H1)(H3) hold and (H5)′ k is continuous and for any x′ , x′′ ∈ A(σ, γ), t ∈ I |k(t, x′t ) − k(t, x′′t )| ≤ lkx′ − x′′ k, where l ∈ (0, 1). Then IVP (2.1) has at least one solution on [t0 , t0 + η] for some positive number η. In the case where k2 ≡ 0, we have the following result. Corollary 2.4. Assume that there exist σ ∈ (0, a) and γ ∈ (0, ∞) such that (H1)(H3) hold and (H6)′ k is completely continuous and for any bounded set Λ in A(σ, γ), the set {t → k(t, xt ) : x ∈ Λ} is equicontinuous on C(I, Rn ). Then IVP (2.1) has at least one solution on [t0 , t0 + η] for some positive number η. 2.2.3

Extremal Solutions

Define the order relation “ ≤ ” by the cone K in C(J, Rn ), given by K = {z ∈ C(J, Rn ) | z(t) ≥ 0 for all t ∈ J}. Clearly, the cone K is normal in C(J, Rn ). Note that the order relation “ ≤ ” in C(J, Rn ) also induces the order relation in the space C which we also denote by “ ≤ ” itself when there is no confusion. We give the following definitions in the sequel. 1 Definition 2.5. A mapping f : I × C → Rn is called L δ -Lipschitz if (i) t 7→ f (t, z) is Lebesgue measurable for each z ∈ C; 1 (ii) there exist a constant δ ∈ [0, α) and a function l ∈ L δ (I, R+ ) such that |f (t, z) − f (t, y)| ≤ l(t)kz − yk∗ , a.e. t ∈ I for all z, y ∈ C. Definition 2.6. A mapping g : I × C → Rn is said to be Carathéodory if (i) t 7→ g(t, z) is Lebesgue measurable for each z ∈ C; (ii) z → 7 g(t, z) is continuous almost everywhere for t ∈ I. 1 Furthermore, a Carathéodory function g(t, z) is called L β -Carathéodory if (iii) for each real number r > 0, there exist a constant β ∈ [0, α) and a function 1 mr ∈ L β (I, R+ ) such that |g(t, z)| ≤ mr (t), a.e. t ∈ I for all z ∈ C with kzk∗ ≤ r.

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Definition 2.7. A mapping h : I × C → Rn is said to be Chandrabhan if (i) t 7→ h(t, z) is Lebesgue measurable for each z ∈ C; (ii) z → 7 h(t, z) is nondecreasing almost everywhere for t ∈ I. 1 Furthermore, a Chandrabhan function h(t, z) is called L γ -Chandrabhan if (iii) for each real number r > 0, there exist a constant γ ∈ [0, α) and a function 1 wr ∈ L γ (I, R+ ) such that |h(t, z)| ≤ wr (t), a.e. t ∈ I for all z ∈ C with kzk∗ ≤ r. Definition 2.8. A function x ∈ C(J, Rn ) is called a solution of IVP (2.1) on J if (i) the function [x(t) − k(t, xt )] is absolutely continuous on I; (ii) xt0 = ϕ, and (iii) x satisfies the equation in (2.1). Definition 2.9. A function a ∈ C(J, Rn ) is called a lower solution of IVP (2.1) on J if the function [a(t) − k(t, at )] is absolutely continuous on I, and C α t0 Dt (a(t) − k(t, at )) ≤ F (t, at ), a.e. t ∈ (t0 , t0 + σ] at0 ≤ ϕ. Again, a function b ∈ C(J, Rn ) is called an upper solution of IVP (2.1) on J if the function [b(t) − k(t, bt )] is absolutely continuous on I, and [b(t) − k(t, bt )] is absolutely continuous on I, and C α t0 Dt (b(t) − k(t, bt )) ≥ F (t, bt ), a.e. t ∈ (t0 , t0 + σ] bt0 ≥ ϕ.

Finally, a function x ∈ C(J, Rn ) is a solution of IVP (2.1) on J if it is a lower as well as an upper solution of IVP (2.1) on J. Definition 2.10. A solution xM of IVP (2.1) is said to be maximal if for any other solution x to IVP (2.1), one has x(t) ≤ xM (t) for all t ∈ J. Again, a solution xm of IVP (2.1) is said to be minimal if xm (t) ≤ x(t) for all t ∈ J, where x is any solution for IVP (2.1) on J. We need the following hypotheses in the sequel. (F1) F (t, zt ) = f (t, zt ) + g(t, zt ) + h(t, zt ), where f, g, h : I × C → Rn ; (F2) IVP (2.1) has a lower solution a and an upper solution b with a ≤ b; (k0) k(t, z) is continuous with respect to t on I for any z ∈ C; (k1) |k(t, z) − k(t, y)| ≤ k0 kz − yk∗ , for z, y ∈ C, t ∈ I, where k0 > 0; (k2) k(t, z) is nondecreasing with respect to z for any z ∈ C and almost all t ∈ I; 1 1 (f1) f is L δ -Lipschitz, and there exists η ∈ [0, α) such that |f (t, 0)| ∈ L η (I, R+ ); (f2) f (t, z) is nondecreasing with respect to z for any z ∈ C and almost all t ∈ I; 1 (g1) g is L β -Carathéodory; (g2) g(t, z) is nondecreasing with respect to z for any z ∈ C and almost all t ∈ I; 1 (h1) h is L γ -Chandrabhan.

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For any positive constant r, let Br = {z ∈ C(J, Rn ) : kzk ≤ r}. Set q1 =

α−1 ∈ (−1, 0), 1−δ

L = klk

1

Lδ I

and q2 =

α−1 ∈ (−1, 0), 1−β

Mr = kmr k

1

Lβ I

.

In order to prove our main results, we need the following lemma. Lemma 2.11. Assume that the hypotheses (F1), (f1), (g1) and (h1) hold. x ∈ C(J, Rn ) is a solution for IVP (2.1) on J if and only if x satisfies the following relation  Z t 1  x(t) = ϕ(0) + k(t, xt ) − k(t0 , ϕ) + (t − s)α−1 F (s, xs )ds, for t ∈ I, Γ(α) t0  x(t0 + θ) = ϕ(θ), for θ ∈ I0 . (2.9) Proof. For any positive constant r and x ∈ Br , since xt is continuous in t, according to (g1 ) and Definition 2.6 (i)-(ii), g(t, xt ) is a measurable function on I. 1 Direct calculation gives that (t − s)α−1 ∈ L 1−β [t0 , t], for t ∈ I and β ∈ [0, α). By using Lemma 1.1 (Hölder inequality) and Definition 2.6 (iii), for t ∈ I, we obtain that Z t 1−β Z t α−1 |(t − s)α−1 g(s, xs )|ds ≤ (t − s) 1−β ds kmr k β1 t0

=

Z

t0 t

t0

≤

(t − s)q2 ds

1−β

L

kmr k

[t0 ,t]

1

(2.10)

L β [t0 ,t]

Mr σ (1+q2 )(1−β) , (1 + q2 )1−β

which means that (t − s)α−1 g(s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ I and x ∈ Br . According to (f1), for t ∈ I and x ∈ Br , we get that |f (t, xt )| ≤ l(t)kxt k∗ + |f (t, 0)| ≤ l(t)r + |f (t, 0)|.

(2.11)

Using the similar argument and noting that (f1) and (h1), we can get that (t − s)α−1 f (s, xs ) and (t−s)α−1 h(s, xs ) are Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ I and x ∈ Br . Thus, according to (F1), we get that (t − s)α−1 F (s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ I and x ∈ Br . Let G(θ, s) = (t − θ)−α |θ − s|α−1 mr (s). Since G(θ, s) is a nonnegative, measurable function on D = [t0 , t] × [t0 , t] for t ∈ I, we have Z t Z t Z Z tZ t G(θ, s)ds dθ = G(θ, s)dsdθ = G(θ, s)dθ ds t0

t0

D

t0

t0

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and

Z

G(θ, s)dsdθ =

D

Z tZ t0 t

= =

Z

t0 Z t t0

+

t

t0

G(θ, s)ds dθ −α

(t − θ)

(t − θ)−α

Z

t

t0

Z

t

t0 Z θ t0

(t − θ)−α

|θ − s|

α−1

mr (s)ds dθ

(θ − s)α−1 mr (s)ds dθ

Z

t

θ

(s − θ)α−1 mr (s)ds dθ

Z t 2Mr (1+q2 )(1−β) σ (t − θ)−α dθ ≤ (1 + q2 )1−β t0 2Mr ≤ σ (1+q2 )(1−β)+1−α . (1 − α)(1 + q2 )1−β

Therefore, G1 (θ, s) = (t − θ)−α (θ − s)α−1 g(s, xs ) is a Lebesgue integrable function on D = [t0 , t] × [t0 , t], then we have Z t Z θ Z t Z t dθ G1 (θ, s)ds = ds G1 (θ, s)dθ. t0

t0

t0

s

We now prove that −α α t0 Dt t0 Dt F (t, xt ) = F (t, xt ), for t ∈ (t0 , t0 + σ].

Indeed, we have −α α D g(t, x ) = D t0 t t0 t t

1 d Γ(1 − α)Γ(α) dt

Z

d 1 Γ(1 − α)Γ(α) dt

t0 Z t

1 d = Γ(1 − α)Γ(α) dt =

t

t0 t

Z

t0 Z t

(t − θ)−α dθ ds

Z

θ

t0 Z t

Z

Similarly, we can get −α α t0 Dt t0 Dt f (t, xt ) = f (t, xt ),

which implies

G1 (θ, s)dθ

s

−α α t0 Dt t0 Dt h(t, xt )

−α α t0 Dt t0 Dt F (t, xt )

t0

(θ − s)α−1 g(s, xs )ds dθ

G1 (θ, s)ds

1 d g(s, xs )ds Γ(1 − α)Γ(α) dt t0 Z d t = g(s, xs )ds dt t0 = g(t, xt ) for t ∈ (t0 , t0 + σ]. =

θ

Z

s

t

(t − θ)−α (θ − s)α−1 dθ

= h(t, xt ), for t ∈ (t0 , t0 +σ],

= F (t, xt ), for t ∈ (t0 , t0 + σ].

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If x satisfies the relation (2.9), then we get that x(t) − k(t, xt ) is absolutely continuous on I. In fact, for any disjoint family of open intervals {(ai , bi )}1≤i≤n on Pn I with i=1 (bi − ai ) → 0, we have Z Z ai n X 1 bi α−1 α−1 (bi − s) g(s, xs )ds − (ai − s) g(s, xs )ds Γ(α) ≤

i=1 n X i=1

Z 1 Γ(α)

t0

t0

bi (bi − s)α−1 g(s, xs )ds

ai

Z Z ai 1 ai α−1 α−1 (bi − s) g(s, xs )ds − (ai − s) g(s, xs )ds + Γ(α) t0 t0 i=1 Z n b i X 1 (bi − s)α−1 mr (s)ds ≤ Γ(α) a i i=1 Z ai n X 1 + ((ai − s)α−1 − (bi − s)α−1 )mr (s)ds Γ(α) t 0 i=1 1−β Z bi n X α−1 1 1−β ds kmr k β1 ≤ (bi − s) L I Γ(α) ai i=1 Z ai n 1−β X α−1 α−1 1 1−β 1−β + − (bi − s) ds kmr k β1 (ai − s) L I Γ(α) t0 i=1 n X

≤

n X (bi − ai )(1+q2 )(1−β) i=1

Γ(α)(1 + q2 )1−β

kmr k

1

Lβ I

n 2 2 X (a1+q − b1+q + (bi − ai )1+q2 )1−β i i kmr k β1 + L I Γ(α)(1 + q2 )1−β i=1

≤2

n X (bi − ai )(1+q2 )(1−β) i=1

Γ(α)(1 + q2 )1−β

Mr → 0.

P Using the similar method, as ni=1 (bi − ai ) → 0, we can get that Z Z ai n X 1 bi α−1 α−1 (bi − s) f (s, xs )ds − (ai − s) f (s, xs )ds → 0 Γ(α) i=1

and

t0

t0

Z Z ai 1 bi α−1 α−1 → 0. (b − s) h(s, x )ds − (a − s) h(s, x )ds i s i s Γ(α) t0 t0 i=1 Pn Pn Hence, i=1 kx(bi ) − k(bi , xbi ) − x(ai ) + k(ai , xai )k → 0, as i=1 (bi − ai ) → 0. Therefore, x(t) − k(t, xt ) is absolutely continuous on I which implies that x(t) − k(t, xt ) is differentiable for almost all t ∈ I. According to the argument above, for almost all t ∈ (t0 , t0 + σ], we have Z t i h 1 C α C α (t − s)α−1 F (s, xs )ds t0 Dt (x(t) − k(t, xt )) = t0 Dt ϕ(0) − k(t0 , ϕ) + Γ(α) t0 n X

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i h 1 Z t (t − s)α−1 F (s, xs )ds = Γ(α) t0 −α α =C D D F (t, x ) t t0 t t0 t (t − t0 )−α = t0 Dtα t0 Dt−α F (t, xt ) − [t0 Dt−α F (t, xt )]t=t0 Γ(1 − α) −α (t − t0 ) . = F (t, xt ) − [t0 Dt−α F (t, xt )]t=t0 Γ(1 − α) Since (t−s)α−1 F (s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ I, q we know that t0 Dt−α F (t, xt )]t=t0 = 0, which means that C t0 Dt x(t) = F (t, xt ), a.e. t ∈ n (t0 , t0 + σ]. Hence, x ∈ C(J, R ) is a solution of IVP (2.1). On the other hand, it is obvious that if x ∈ C(J, Rn ) is a solution of IVP (2.1), then x satisfies the relation (2.9), and this completes the proof. C α t0 Dt

Theorem 2.12. Assume that the hypotheses (F1), (F2), (k0)-(k2), (f1), (f2), (g1), (g2) and (h1) hold. Then IVP (2.1) has a minimal and a maximal solution in [a, b] defined on J provided that Lσ (1+q1 )(1−δ) < 1. (2.12) Γ(α)(1 + q1 )1−δ Proof. Define three operators A, B and C on C(J, Rn ) as follows  Z t 1  (Ax)(t) = k(t, xt ) − k(t0 , ϕ) + (t − s)α−1 f (s, xs )ds, for t ∈ I, Γ(α) t0  (Ax)(t0 + θ) = 0, for θ ∈ I0 ,  Z t 1  (Bx)(t) = ϕ(0) + (t − s)α−1 g(s, xs )ds, for t ∈ I, Γ(α) t0  (Bx)(t0 + θ) = ϕ(θ), for θ ∈ I0 , k0 +

and

 

Z t 1 (t − s)α−1 h(s, xs )ds, for t ∈ I, Γ(α) t0  (Cx)(t0 + θ) = 0, for θ ∈ I0 , (Cx)(t) =

where x ∈ C(J, Rn ). Obviously, Ax+ Bx+ Cx ∈ C(J, Rn ) for every x ∈ C(J, Rn ). From Lemma 2.11, we get that IVP (2.1) is equivalent to the operator equation (Ax)(t) + (Bx)(t) + (Cx)(t) = x(t) for t ∈ J. Now we show that the operator equation Ax+Bx+Cx = x has a least and a greatest solution in [a, b]. The proof is divided into three steps. Step I. A is a contraction in C(J, Rn ). For any x, y ∈ C(J, Rn ) and t ∈ I, according to (k1) and (f1), we have Z t 1 (t − s)α−1 |f (s, xs ) − f (s, ys )|ds |(Ax)(t) − (Ay)(t)| ≤ |k(t, xt ) − k(t, yt )| + Γ(α) t0 Z t 1 (t − s)α−1 l(s)kxs − ys k∗ ds ≤ k0 kxt − yt k∗ + Γ(α) t0

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≤ k0 kx − yk +

1 Γ(α)

Z

t

α−1

(t − s) 1−δ ds

t0 (1+q1 )(1−δ)

1−δ

klk

1

L δ [t0 ,t]

35

kx − yk

Lσ kx − yk ≤ k0 kx − yk + Γ(α)(1 + q1 )1−δ Lσ (1+q1 )(1−δ) kx − yk, = k0 + Γ(α)(1 + q1 )1−δ Lσ(1+q1 )(1−δ) which implies that kAx − Ayk ≤ k0 + Γ(α)(1+q kx − yk. Therefore, A is a 1−δ 1) contraction in C(J, Rn ) according to (2.12). Step II. B is a completely continuous operator and C is a totally bounded operator. For any x ∈ C(J, Rn ), we can choose a positive constant r such that kxk ≤ r. Firstly, we will prove that B is continuous on Br . For xn , x ∈ Br , n = 1, 2, ... with limn→∞ kxn − xkC = 0, we get lim xns = xs , for s ∈ I.

n→∞

Thus, by (g1) and Definition 2.6 (ii), and noting that xs is continuous with respect to s on I, we have lim g(s, xns ) = g(s, xs ), a.e s ∈ I.

n→∞

On the other hand, noting that |g(s, xns ) − g(s, xs )| ≤ 2mr (s), by Lebesgue’s dominated convergence theorem, we have Z t 1 |(Bxn )(t) − (Bx)(t)| ≤ (t − s)α−1 |g(s, xns ) − g(s, xs )|ds → 0, as n → ∞, Γ(α) t0 which implies kBxn − Bxk → 0 as n → ∞. This means that B is continuous. Next, we will show that for any positive constant r, {Bx : x ∈ Br } is relatively compact. It suffices to show that the family of functions {Bx : x ∈ Br } is uniformly bounded and equicontinuous. For any x ∈ Br and t ∈ I, we have Z t 1 (t − s)α−1 |g(s, xs )|ds |(Bx)(t)| ≤ |ϕ(0)| + Γ(α) t0 Z t 1−β α−1 1 (t − s) 1−β ds ≤ |ϕ(0)| + kmr k β1 L [t0 ,t] Γ(α) 0 Z t 1−β Mr ≤ |ϕ(0)| + (t − s)q2 ds Γ(α) 0 ≤ |ϕ(0)| +

Mr σ (1+q2 )(1−β) . Γ(α)(1 + q2 )1−β

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For θ ∈ I0 , we have |(Bx)(t0 + θ)| = |ϕ(θ)|. Thus {Bx : x ∈ Br } is uniformly bounded. In the following, we will show that {Bx : x ∈ Br } is a family of equicontinuous functions. For any x ∈ Br and t0 ≤ t1 < t2 ≤ t0 + σ, we get

|(Bx)(t2 ) − (Bx)(t1 )| Z Z t2 1 t1 α−1 α−1 α−1 = ((t2 − s) − (t1 − s) )g(s, xs )ds + (t2 − s) g(s, xs )ds Γ(α) t0 t1 Z t1 1 α−1 α−1 |((t2 − s) − (t1 − s) )g(s, xs )|ds ≤ Γ(α) t0 Z t2 1 α−1 |(t2 − s) g(s, xs )|ds + Γ(α) t1 Z t1 Z t2 1 1 α−1 α−1 α−1 ((t1 − s) (t2 − s) ≤ − (t2 − s) )mr (s)ds + mr (s)ds Γ(α) t0 Γ(α) t1 Z t1 1−β 1 1 α−1 α−1 1−β ((t1 − s) − (t2 − s) ) ≤ ds kmr k β1 L [t0 ,t1 ] Γ(α) t0 1−β Z t2 1 1 + kmr k β1 ((t2 − s)α−1 ) 1−β ds L [t1 ,t2 ] Γ(α) t1 Z t1 1−β Z t2 1−β Mr Mr ((t1 − s)q2 − (t2 − s)q2 )ds + (t2 − s)q2 ds ≤ Γ(α) Γ(α) t0 t1 1−β Mr 1+q2 1+q2 1+q2 ≤ (t1 − t0 ) − (t2 − t0 ) + (t2 − t1 ) Γ(α)(1 + q2 )1−β Mr + (t2 − t1 )(1+q2 )(1−β) Γ(α)(1 + q2 )1−β 2Mr ≤ (t2 − t1 )(1+q2 )(1−β) . Γ(α)(1 + q2 )1−β As t2 − t1 → 0, the right-hand side of the above inequality tends to zero independently of x ∈ Br . In view of the continuity of ϕ, we can get that {Bx : x ∈ Br } is a family of equicontinuous functions. Therefore, {Bx : x ∈ Br } is relatively compact by Ascoli-Arzela Theorem. Using the similar argument, we can get that {Cx : x ∈ Br } is also relatively compact, which means that C is totally bounded. Step III. A, B and C are three monotone increasing operators. Since x, y ∈ C(J, Rn ) with x ≤ y implies that xt ≤ yt for t ∈ I, according to (k2) and (f2), we have Z t 1 (t − s)α−1 f (s, xs )ds (Ax)(t) = k(t, xt ) − k(t0 , ϕ) + Γ(α) t0 Z t 1 (t − s)α−1 f (s, ys )ds ≤ k(t, yt ) − k(t0 , ϕ) + Γ(α) t0 = (Ay)(t).

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Hence A is a monotone increasing operator. Similarly, we can conclude that B and C are also monotone increasing operators according to (g2), (h1) and Definition 2.7 (ii). Clearly, K is a normal cone. From (F2) and Definition 2.9, we have that a ≤ Aa + Ba + Ca and b ≥ Ab + Bb + Cb with a ≤ b. Thus the operators A, B and C satisfy all the conditions of Theorem 1.47 and hence the operator equation Ax + Bx + Cx = x has a least and a greatest solution in [a, b]. Therefore, IVP (2.1) has a minimal and a maximal solution on J. Example 2.13. Consider the following IVP of scalar discontinuous fractional functional differential equation  1 C 2   0 Dt x(t) = F (t, xt ) = f (t) + ζ(t)x(t)   1 (2.13) + 1/3 x(t − 1) + ζ(t)h(x(t)), a.e. t ∈ (0, σ],   t   x(θ) = 0, θ ∈ [−1, 0],

1 )2 = π1 and we take functions f (t), ζ(t) and h(x(t)) as follows where 0 < σ ≤ ( 2Γ(3/2) t, 0 ≤ t ≤ σ2 , 0, 0 ≤ t ≤ σ2 , f (t) = ζ(t) = 0, σ2 < t ≤ σ, 1, σ2 < t ≤ σ,

and h(x(t)) =

x(t), x(t) ≥ 0, x(t) − 1, x(t) < 0.

Evidently, the function F (t, ϕ) = f (t, ϕ) + g(t, ϕ) + h(t, ϕ), ϕ ∈ C([−1, 0], R), where f (t, ϕ) = f (t) + ζ(t)ϕ(0),

g(t, ϕ) =

1 t1/3

ϕ(−1) and h(t, ϕ) = ζ(t)h(ϕ(0)).

One can easily check that a(t) = 0 is a lower solution of IVP (2.13). On the other hand, let t, t ∈ [0, σ], b(t) = 0, t ∈ [−1, 0].

Then, b ∈ C([−1, σ], R) is a upper solution of IVP (2.13). In fact, direct calculation gives that 1 1 t2 t, 0 < t ≤ σ2 C 2 b(t) = ≥ 2t ≥ F (t, b ) = for t ∈ (0, σ]. D t t 0 2t, σ2 < t ≤ σ Γ( 32 ) √

Moreover, noting that Γ( 32 ) = 2π , it is easy to verify that conditions (k0)-(k2), (f1)-(f2), (g1)-(g2), (h1) and (2.12) are satisfied. Therefore, Theorem 2.12 allows us to conclude that IVP (2.13) has a minimal and a maximal solution in [0, b] defined on [−1, σ].

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2.3 2.3.1


ws-book975x65


p-Type Neutral Equations Introduction

Let C = C([−1, 0], Rn ) denote the space of continuous functions on [−1, 0]. For any element ϕ ∈ C, define the norm kϕk∗ = supθ∈[−1,0] |ϕ(θ)|. Consider the IVP of p-type fractional neutral functional differential equations of the form C q t0 Dt g(t, xt )

xt0 = ϕ,

= f (t, xt ),

(2.14)

(t0 , ϕ) ∈ Ω,

(2.15)

q where C t0 Dt is Caputo fractional derivative of order 0 < q < 1, Ω is an open subset of [0, ∞) × C and g, f : Ω → Rn are given functionals satisfying some assumptions that will be specified later. xt ∈ C is defined by xt (θ) = x(p(t, θ)), where −1 ≤ θ ≤ 0, p(t, θ) is a p-function. Definition 2.14. (Lakshmikantham, Wen and Zhang, 1994) A function p ∈ C(J × [−1, 0], R) is called a p-function if it has the following properties:

(i) p(t, 0) = t; (ii) p(t, −1) is a nondecreasing function of t; (iii) there exists a σ ≥ −∞ such that p(t, θ) is an increasing function for θ for each t ∈ (σ, ∞); (iv) p(t, 0) − p(t, −1) > 0 for t ∈ (σ, ∞). In the following, we suppose t ∈ (σ, ∞). Definition 2.15. (Lakshmikantham, Wen and Zhang, 1994) Let t0 ≥ 0, A > 0 and x ∈ C([p(t0 , −1), t0 + A], Rn ). For any t ∈ [t0 , t0 + A], we define xt by xt (θ) = x(p(t, θ)),

−1 ≤ θ ≤ 0,

so that xt ∈ C = C([−1, 0], Rn ). Note that the frequently used symbol “xt ” (in Hale, 1977; Lakshmikantham, 2008; Lakshmikantham, Wen and Zhang, 1994, xt (θ) = x(t + θ), where −τ ≤ θ ≤ 0, r > 0, r = const) in the theory of functional differential equations with bounded delay is a partial case of the above definition. Indeed, in this case we can put p(t, θ) = t + rθ, θ ∈ [−1, 0]. Definition 2.16. A function x is said to be a solution of IVP (2.14)-(2.15) on [p(t0 , −1), t0 + α], if there are t0 ≥ 0, α > 0, such that (i) x ∈ C([p(t0 , −1), t0 + α], Rn ) and (t, xt ) ∈ Ω, for t ∈ [t0 , t0 + α]; (ii) xt0 = ϕ; (iii) g(t, xt ) is differentiable and (2.14) holds almost everywhere on [t0 , t0 + α]. We need the following lemma relative to p-function before we proceed further, which is taken from Lakshmikantham, Wen and Zhang, 1994.

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Lemma 2.17. (Lakshmikantham, Wen and Zhang, 1994) Suppose that p(t, θ) is a p-function. For A > 0, τ ∈ (σ, ∞) (τ may be σ if σ > −∞), let x ∈ C([p(τ, −1), τ + A], Rn ) and ϕ ∈ C([−1, 0], Rn ). Then we have (i) xt is continuous in t on [τ, τ +A] and p˜(t, θ) = p(τ +t, θ)−τ is also a p-function; (ii) if p(τ + t, −1) < τ for t > 0, then there exists −1 < s(τ, t) < 0 such that p(τ + t, s(τ, t)) = τ and p(τ + t, −1) ≤ p(τ + t, θ) ≤ τ, for − 1 ≤ θ ≤ s(τ, t), τ ≤ p(τ + t, θ) ≤ τ + t, for s(τ, t) ≤ θ ≤ 0. Moreover, s → 0 uniformly in τ as t → 0; (iii) there exists a function η ∈ C([p(τ, −1), τ ], Rn ) such that η(p(τ, θ)) = ϕ(θ) for − 1 ≤ θ ≤ 0. It is well known that a neutral functional differential equation (NFDE for short) is one in which the derivatives of the past history or derivatives of functionals of the past history are involved as well as the present state of the system. In other words, in order to guarantee that the equation (2.14) is NFDE, the coefficient of x(t) that is contained in g(t, xt ) can not be equal to zero. Then we need introduce the concept of atomic. Let g ∈ C(R+ × C, Rn ) and g(t, ϕ) be linear in ϕ. Then Riesz representation theorem shows that there exists a n × n matrix function η(t, θ) of bounded variation such that Z 0 g(t, ϕ) = [dθ η(t, θ)]ϕ(θ). −γ

For t0 ≥ 0 and θ0 ∈ (−γ, 0), if

det[η(t0 , θ0+ ) − η(t0 , θ0− )] 6= 0,

then we say that g(t, ϕ) is atomic at θ0 for t0 . Similarly, one can define g(t, ϕ) to be atomic at the endpoints −r and 0 for t0 . If for every t ≥ 0, g(t, ϕ) is atomic at θ0 for t, then we say that g(t, ϕ) is atomic at θ0 for R+ . If g(t, ϕ) is not linear in ϕ, suppose that g(t, ϕ) has a Frechet derivative with respect to ϕ, then gϕ′ (t, ϕ)ψ ∈ Rn for (t, ϕ) ∈ R+ × C and ψ ∈ C, where gϕ′ denote the Fréchet derivative of g with respect to ϕ. Then gϕ′ (t, ϕ) is a linear mapping from C into Rn and therefore Z 0 gϕ′ (t, ϕ)ψ = [dθ µ(t, ϕ, θ)]ψ(θ), −γ

where µ(t, ϕ, θ) is a matrix function of bounded variation. As before, if + − det[µ(t0 , ϕ0 , θ0 ) − µ(t0 , ϕ0 , θ0 )] 6= 0, for t0 ≥ 0, then we say that, the nonlinear g(t, ϕ) is atomic at θ0 for (t0 , ϕ0 ). If g(t, ϕ) is atomic at θ0 , for every (t, ϕ), then we say that g(t, ϕ) is atomic at θ0 for R+ × C. For a detailed discussion on atomic concept we refer the reader to the books Hale, 1977; Lakshmikantham, Wen and Zhang, 1994.

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Lemma 2.18. (Hale, 1977; Lakshmikantham, Wen and Zhang, 1994) Suppose that g(t, ϕ) is atomic at zero on Ω. Then there are a continuous n × n matrix function A(t, ϕ) with detA(t, ϕ) 6= 0 on Ω and a functional L(t, ϕ, ψ) which is linear in ψ such that gϕ′ (t, ϕ)ψ = A(t, ϕ)ψ(0) + L(t, ϕ, ψ). Moreover, there exists a continuous function γ : Ω × [0, 1] → R+ with γ(t, ϕ, 0) = 0 such that for every s ∈ [0, 1] and ψ with (t, ψ) ∈ Ω, ψ(θ) = 0 for −1 ≤ θ ≤ −s, |L(t, ϕ, ψ)| ≤ γ(t, ϕ, s)kψk∗ . 2.3.2


Assume that the functional f : Ω → Rn satisfies the following conditions. (H1) f (t, ϕ) is Lebesgue measurable with respect to t for any (t, ϕ) ∈ Ω; (H2) f (t, ϕ) is continuous with respect to ϕ for any (t, ϕ) ∈ Ω; 1 (H3) there exist a constant q1 ∈ (0, q) and a L q1 -integrable function m such that |f (t, ϕ)| ≤ m(t) for any (t, ϕ) ∈ Ω. For each (t0 , ϕ) ∈ Ω, let p˜(t, θ) = p(t0 + t, θ) − t0 . Define the function η ∈ C([˜ p(0, −1), ∞), Rn ) by η(˜ p(0, θ)) = ϕ(θ), for θ ∈ [−1, 0], η(t) = ϕ(0), for t ∈ [0, ∞). Let x ∈ C([p(t0 , −1), t0 + α], Rn ), α < A and let x(t0 + t) = η(t) + z(t) for p˜(0, −1) ≤ t ≤ α. (2.16) Lemma 2.19. x(t) is a solution of IVP (2.14)-(2.15) on [p(t0 , −1), t0 + α] if and only if z(t) satisfies the relation  Z t 1  (t − s)q−1 f (t0 + s, η˜s + z˜s )ds, for t ∈ [0, α], g(t0 + t, η˜t + z˜t ) − g(t0 , ϕ) = Γ(q) 0  z˜0 = 0, (2.17) where η˜t (θ) = η(˜ p(t, θ)), z˜t (θ) = z(˜ p(t, θ)), for −1 ≤ θ ≤ 0. Proof. Since xt is continuous in t, xt is a measurable function, therefore according to conditions (H1) and (H2), f (t, xt ) is Lebesgue measurable on [t0 , t0 + α]. Direct 1 calculation gives that (t − s)q−1 ∈ L 1−q1 [t0 , t], for t ∈ [t0 , t0 + α] and q1 ∈ (0, q). In light of Hölder inequality, we obtain that (t − s)q−1 f (s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ [t0 , t0 + α], and Z t 1 kmk q1 . |(t − s)q−1 f (s, xs )|ds ≤ k(t − s)q−1 k 1−q t0

L

1

[t0 ,t]

L

1

[t0 ,t0 +α]

Hence x(t) is the solution of IVP (2.14)-(2.15) if and only if it satisfies the relation  Z t 1  (t − u)q−1 f (u, xu )du, for t ∈ [t0 , t0 + α], g(t, xt ) − g(t0 , xt0 ) = Γ(q) t0  xt0 = ϕ,

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or setting u = t0 + s,  Z t 1  (t − s)q−1 f (t0 + s, xt0 +s )ds, for t ∈ [0, α], g(t0 + t, xt0 +t ) − g(t0 , xt0 ) = Γ(q) 0  xt0 = ϕ. (2.18) In view of (2.16), we have xt0 +t (θ) = x(p(t0 + t, θ)) = x(˜ p(t, θ) + t0 ) = η(˜ p(t, θ)) + z(˜ p(t, θ)) = η˜t (θ) + z˜t (θ), for t ∈ [0, α]. In particular xt0 (θ) = η˜0 (θ) + z˜0 (θ). Hence xt0 = ϕ if and only if z˜0 = 0 according to η˜ = ϕ. It is clear that x(t) satisfies (2.18) if and only if z(t) satisfies (2.17). For any σ, ξ > 0, let E(σ, ξ) = {z ∈ C([˜ p(0, −1), σ], Rn ) : z˜0 = 0, k˜ zt k∗ ≤ ξ for t ∈ [0, σ]}, which is a bounded closed convex subset of the Banach space C([˜ p(0, −1), σ], Rn ) endowed with supremum norm k · k. Lemma 2.20. Suppose Ω ⊆ R × C is open, W ⊂ Ω is compact. For any a neighborhood V ′ ⊂ Ω of W , there is a neighborhood V ′′ ⊂ V ′ of W and there exist positive numbers δ and ξ such that (t0 + t, η˜t + λ˜ zt ) ∈ V ′ with 0 ≤ λ ≤ 1 for any ′′ (t0 , ϕ) ∈ V , t ∈ [0, σ] and z ∈ E(σ, ξ). The proof of Lemma 2.20 is similar to that of (iii) of Lemma 2.1.8 in Lakshmikantham, Wen and Zhang, 1994, thus it is omitted. Suppose g is atomic at 0 on Ω. Define two operators S and T on E(α, β) as follows  for t ∈ [˜ p(0, −1), 0],  (Sz)(t) = 0, (2.19) A(t0 + t, η˜t )(Sz)(t) = g(t0 , ϕ) − g(t0 + t, η˜t + z˜t )  + gϕ′ (t0 + t, η˜t )˜ zt − L(t0 + t, η˜t , z˜t ), for t ∈ [0, α] and  (T z)(t) = 0, for t ∈ [˜ p(0, −1), 0],    A(t0 + t, η˜t )(T z)(t) Z t  1   (t − s)q−1 f (t0 + s, η˜s + z˜s )ds, for t ∈ [0, α], = Γ(q) 0

(2.20)

where A(t0 + t, η˜t ), L(t0 + t, η˜t , z˜t ) are functions described in Lemma 2.18. It is clear that the operator equation z = Sz + T z

(2.21)

has a solution z ∈ E(α, β) if and only if z is a solution of (2.17). Therefore the existence of a solution of IVP (2.14)-(2.15) is equivalent to determining α, β > 0 such that S + T has a fixed point on E(α, β). We are now in a position to prove the following existence results, and the proof is based on Krasnoselskii’s fixed point theorem.

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Theorem 2.21. Suppose g : Ω → Rn is continuous together with its first Fréchet derivative with respect to the second argument, and g is atomic at 0 on Ω. f : Ω → Rn satisfies conditions (H1)-(H3). W ⊂ Ω is a compact set. Then there exist a neighborhood V ⊂ Ω of W and a constant α > 0 such that for any (t0 , ϕ) ∈ V , IVP (2.14)-(2.15) has a solution which exists on [p(t0 , −1), t0 + α]. Proof. As we have mentioned above, we only need to discuss operator equation (2.21). For any (t, ϕ) ∈ Ω, the property of the matrix function A(t, ϕ) which is nonsingular and continuous on Ω implies that its inverse matrix A−1 (t, ϕ) exists and is continuous on Ω. Let V0 ⊂ Ω be the neighborhood of W , suppose that there is an M > 0 such that |A−1 (t0 , ϕ)| ≤ M

for every (t0 , ϕ) ∈ V0 .

(2.22)

1 q1

Note the complete continuity of the function (m(t)) , hence, for a given positive number N , there must exist a number α0 > 0 satisfying q1 Z t0 +α0 1 ≤ N. (2.23) (m(s)) q1 ds t0

Due to the continuity of functions γ and gϕ′ described in Lemma 2.18, there exist a neighborhood V1 ⊂ Ω of W and constants h1 > 0, h2 ∈ (0, 1] such that |γ(t0 + t, η˜t , −s)| = |γ(t0 + t, η˜t , −s) − γ(t0 + t, η˜t , 0)|
0, there exist q2 ∈ (0, q) and a real-valued function ℓ1 ∈ L q2 [t0 , t0 + A] such that for any (t, xt ), (t, yt ) ∈ Ω, t ∈ [t0 , t0 + A], we have |f (t, xt ) − f (t, yt )| ≤ ℓ1 (t) sup kxs − ys k∗ . t0 ≤s≤t

Then the result of Theorem 2.24 holds. Proof. It suffices to prove that the condition (H5) of Theorem 2.24 holds. Note 1 that ℓ1 ∈ L q2 [t0 , t0 + A], let K = kℓ1 k q1 . Then for any (t, xt ), (t, yt ) ∈ Ω L

2

[t0 , t0 +A]

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we have

Z t q−1 (t − s) [f (s, x ) − f (s, y )]ds s s t0 Z t ≤ (t − s)q−1 |f (s, xs ) − f (s, ys )|ds t Z 0t ≤ (t − s)q−1 ℓ1 (s) ds sup kxs − ys k∗ t0 ≤s≤t

t0

K (t − t0 )(1+b1 )(1−q2 ) sup kxs − ys k∗ , ≤ (1 + b1 )1−q2 t0 ≤s≤t

where b1 =

q−1 1−q2

∈ (−1, 0). Let

K (t − t0 )(1+b1 )(1−q2 ) . (1 + b1 )1−q2 Obviously, ℓ : [0, A] → [0, ∞) continuous at t = 0 and ℓ(0) = 0. Then the condition (H5) of Theorem 2.24 holds. ℓ(t − t0 ) =

The next result is concerned with the uniqueness of solutions. Theorem 2.26. Suppose that g is defined as in Theorem 2.21 and the condition (H5)′ of Corollary 2.25 holds. If x is a solution of IVP (2.14)-(2.15), then x is unique. Proof. Suppose (for contradiction) x and y are the solutions of IVP (2.14)-(2.15) on [p(t0 , −1), t0 + A] with x 6= y, let t1 = inf{t ∈ [t0 , t0 + A] : x(t) 6= y(t)}.

Then t0 ≤ t1 < t0 + A and

x(t) = y(t) for p(t0 , −1) ≤ t < t1 ,

which implies that

xt (θ) = x(p(t, θ)) = y(p(t, θ)) = yt (θ), t0 ≤ t < t1 , − 1 ≤ θ ≤ 0.

(2.35)

Choose α > 0 such that t1 + α < t0 + A. According to (i) of Definition 2.16, we have {(t, xt ), t1 ≤ t ≤ t1 + α} ∪ {(t, yt ), t1 ≤ t ≤ t1 + α} ⊂ Ω.

On the one hand, x and y satisfy (2.14)-(2.15) on [t0 , t0 + A], thus from (2.35) and the condition (H5)′ , for t ∈ [t0 , t1 + α], we have Z 1 t q−1 |g(t, xt ) − g(t, yt )| ≤ (t − s) [f (s, x ) − f (s, y )]ds s s Γ(q) t0 Z 1 t q−1 (t − s) [f (s, xs ) − f (s, ys )]ds = Γ(q) t1 (2.36) Z t 1 q−1 ≤ (t − s) ℓ1 (s) ds sup kxs − ys k∗ Γ(q) t1 t0 ≤s≤t ≤

K α(1+b1 )(1−q2 ) sup kxs − ys k∗ , Γ(q)(1 + b1 )1−q2 t1 ≤s≤t1 +α

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Fractional Functional Differential Equations q−1 1−q2

where b1 =

∈ (−1, 0), K = kℓ1 k

1

L q2 [t0 , t0 +A]

49

.

On the other hand, since g(t, ϕ) is continuously differentiable in ϕ, we have g(t, xt ) − g(t, yt ) = gϕ′ (t, yt )(xt − yt ) + kkxt − yt k∗

(2.37)

gϕ′ (t, yt )ψ = A(t, yt )ψ(0) + L(t, yt , ψ).

(2.38)

with k → 0 as kxt − yt k∗ → 0. By the hypothesis that g(t, ϕ) is atomic at 0 on Ω, there exist a nonsingular continuous matrix function A(t, yt ) and a function L(t, yt , ψ) which is linear in ψ such that Moreover, there is a positive real-valued continuous function γ(t, yt , −s) such that for every s ∈ [−1, 0], |L(t, yt , ψ)| ≤ γ(t, yt , −s)kψk∗

(2.39)

if ψ(θ) = 0 for −1 ≤ θ ≤ s. Hence for every t ∈ [t1 , t1 +α], by (ii) of Lemma 2.17, there is s(t1 , t−t1 ) ∈ [−1, 0] with s(t1 , t − t1 ) → 0 as t → t1 such that |L(t, yt , xt − yt )| ≤ γ(t, yt , −s(t1 , t − t1 ))kxt − yt k∗ .

From (2.37)-(2.39), it follows that

g(t, xt ) − g(t, yt ) = A(t, yt )(x(t) − y(t)) + L(t, yt , xt − yt ) + kkxt − yt k∗ ,

therefore

|x(t) − y(t)| ≤ |A−1 (t, yt )|{|g(t, xt ) − g(t, yt )| −1

+ γ(t, yt , −s(t1 , t − t1 ))kxt − yt k∗ + kkxt − yt k∗ }.

Let M1 = max{|A (t, yt )| : t1 ≤ t ≤ t1 + α}. Then by relation (2.36), for t ∈ [t1 , t1 + α], we have

where c1 = M1 Noting that

|x(t) − y(t)| ≤ c1

sup

t1 ≤s≤t1 +α

kxs − ys k∗ ,

K (1+b1 )(1−q2 ) α + γ(t, y , −s(t , t − t )) + k . t 1 1 Γ(q)(1 + b1 )1−q2

sup t1 ≤s≤t1 +α

kxs − ys k∗ = =

sup

sup |x(p(s, θ)) − y(p(s, θ))|

t1 ≤s≤t1 +α −1≤θ≤0

sup

sup

t1 ≤s≤t1 +α p(s,−1)≤ρ≤s

=

sup p(t1 ,−1)≤s≤t1 +α

|x(ρ) − y(ρ)|

|x(s) − y(s)|,

we have sup p(t1 ,−1)≤s≤t1 +α

|x(s) − y(s)| ≤ c1

sup p(t1 ,−1)≤s≤t1 +α

|x(s) − y(s)|.

Choose α so small that c1 < 1. Thus sup p(t1 ,−1)≤s≤t1 +α

|x(s) − y(s)| = 0,

contradicting the definition of t1 .

i.e. x(t) ≡ y(t), for t1 ≤ t ≤ t1 + α,

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50

2.3.3

ws-book975x65

Continuous Dependence

The following lemma is introduced in Lakshmikantham, Wen and Zhang, 1994. However, for the sake of completeness, we outline its proof here. Lemma 2.27. Assume x ∈ C([p(0, −1), A], Rn ). Then for every t ∈ [0, A] kxt k ≤ sup |x(s)| + kx0 k. 0≤s≤t

Proof.

By definition, kx0 k = sup−1≤θ≤0 |x(p(0, θ))|. If p(t, −1) ≥ 0, then 0 ≤ p(t, θ) ≤ t

for − 1 ≤ θ ≤ 0.

Thus, sup |x(p(t, θ))| ≤ sup |x(s)| ≤ sup |x(s)| + kx0 k. 0≤s≤t

−1≤θ≤0

0≤s≤t

If p(t, −1) < 0, then by Lemma 2.17, there exists an s ∈ [−1, 0] such that p(t, −1) ≤ p(t, θ) ≤ p(0, θ) for − 1 ≤ θ ≤ s, while 0 ≤ p(t, θ) ≤ t

for s ≤ θ ≤ 0.

Hence sup |x(p(t, θ))| ≤

−1≤θ≤0

≤

sup |x(p(t, θ))| + sup |x(p(t, θ))|

−1≤θ≤s

s≤θ≤0

−1≤θ≤0

s≤θ≤0

sup |x(p(0, θ))| + sup |x(p(t, θ))|

= kx0 k + sup |x(s)|, 0≤s≤t

completing the proof.

We can now prove the following result on continuous dependence. Theorem 2.28. Let (t0 , ϕ) ∈ Ω be given. Suppose that the solution x = x(t0 , ϕ) of (2.14) through (t0 , ϕ) defined on [t0 , A] is unique. Then for every ǫ > 0, there exists a δ(ǫ) > 0 such that (s, ψ) ∈ Ω, |s − t0 | < δ and kψ − ϕk < ϕ imply kxt (s, ψ) − xt (t0 , ϕ)k < ǫ

for all t ∈ [σ, A],

where x(s, ψ) is the solution of (2.14) through (s, ψ) and σ = max{s, t0 }. Proof. In order to prove the theorem, it is enough to show that if {(tk , ϕk )} ⊂ Ω, with tk → t0 and ϕk → ϕ as k → ∞, then there is a natural number N such that each solution xk = x(tk , ϕk ) with k ≥ N of (2.14) through (tk , ϕk ) exists on [p(tk , −1), A] and xk (t) → x(t) uniformly on [p(σ, −1), A], where σ = sup{t0 , tk : k ≥ N }. Since xt (t0 , ϕ) is continuous in t ∈ [t0 , A], the set W = {(t, xt (t0 , ϕ)) : t ∈ [t0 , A]} is compact in Ω. By Theorem 2.21, there exist a neighborhood V of W and number α > 0 such that for any (s, ψ) ∈ V , there is a solution x(s, ψ) of (2.14) through (s, ψ) which exists at least on [s, s + α]. Without loss of generality, we let V = V (W, r),

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51

choose N so large that |tk − t0 | < 2r and kϕk − ϕk < r2 , so that (tk , ϕk ) ∈ V for k ≥ N . Thus xk = x(tk , ϕk ) exists at least on [tk , tk + α]. For convenience, we shall denote ϕ = ϕ0 , x = x0 and xk = x(tk , ϕk ), k = 0, 1, ... . Let pk (t, θ) = p(tk + t, θ) − tk . Define η k , y k the same way as in Lemma 2.19. Recalling the proof of Lemma 2.19, we see that y k satisfy: Z t 1 k k k (t−s)q−1 f (tk +s, ηsk +ysk )ds, t ∈ [0, α] (2.40) g(tk +t, ηt +yt )−g(tk , ϕ ) = Γ(q) 0 if and only if xk is the solution of (2.14) on [p(tk , −1), tk +α], where ηtk = η k (pk (t, θ)), ytk = y k (pk (t, θ)). Set y¯k = y k |[0,α] , the restriction of y k to [0, α]. Let Λ = {¯ y k : k = 0, 1, 2, ...}. k For every zk = (tk , ϕ ), define operators S(zk ) : Λ → C([0, α], Rn ) and T (zk ) : Λ → C([0, α], Rn ) as follows: S(zk )z(t) = A−1 (tk + t, ηtk )[g(tk , ϕk ) − g(tk + t, ηtk + zt ) +gϕ′ (tk + t, ηtk )zt − L(tk + t, ηtk + zt )],

and T (zk )z(t) = A−1 (tk + t, ηtk )

1 Γ(q)

Z

0

t

0 ≤ t ≤ α,

(t − s)q−1 f (tk + s, ηsk + zs ),

0 ≤ t ≤ α,

where zt (θ) = z(¯ pk (t, θ)) with z0 = 0. It is easy to see that {T (zk )¯ y k } is compact in C([0, α], Rn ). Recalling the Theorem 2.21, we see that there exists a constant γ ∈ [0, 1) which is independent to zk such that kSz − Syk ≤ γkz − yk for any z, y ∈ Λ. (2.41) ¯ Denote the Kuratowskii measure of A ⊂ C([0, α], Rn ) Let {zk : k = 0, 1, 2, ...} = Λ. by α(A). Then (2.41) implies that [ α( S(zk )(Λ)) ≤ γα(Λ). ¯ zk ∈Λ

Let R = S + T . Thus y¯k = R(zk )¯ yk . By the well-known properties of Kuratowskii measure α, we immediately obtain that α(Λ) = α({R(zk )¯ yk }) ≤ α({S(zk )¯ y k }) + α({T (zk )¯ y k }) [ = α({S(zk )¯ y k }) ≤ α( S(zk )(Λ)) ≤ γα(Λ). ¯ zk ∈Λ

This means that α(Λ) = 0 which implies Λ is relatively compact in C([0, α], Rn ). Hence there exists a subsequence of Λ, say {¯ y ki }, which converges uniformly on [0, α]. Assume that y¯ki (t) → y¯∗ (t)

∗

uniformly on [0, α].

Define a function y : [p0 (0, −1), α] → Rn by ∗ y (t) = y¯∗ (t), for 0 ≤ t ≤ α, y0∗ = 0,

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52

where p0 is such a p-function that p0 (t, θ) = p(t + t0 , θ) − t0 . Let δ = inf{pk (0, −1) : k = 0, 1, 2, ...} and yˆk denote the extension of y k to [δ, α] which is defined by k yˆ i (t) = y ki (t) for 0 ≤ t ≤ α, yˆki (t) = 0 for δ ≤ t ≤ 0.

Obviously, {ˆ y ki (t)} converges uniformly on [δ, α] as ki → ∞. Consequently, {ˆ y ki } is a relatively compact set. We claim that ytki → yt∗ uniformly in t ∈ [0, α]. In fact, |pki (t, θ) − p0 (t, θ)| = |[p(tki + t, θ) − tki ] − [p(t0 + t, θ) − t0 ]|

≤ |[p(tki + t, θ) − p(t0 + t, θ)| + |t0 − tki |.

Hence, for every µ > 0 there exists a number L such that |[pki (t, θ) − p0 (t, θ)| < µ whenever ki ≥ L. We have the inequality kytki − yt∗ k = =

sup |ˆ y ki (pki (t, θ)) − y ∗ (p0 (t, θ))|

−1≤θ≤0

sup |ˆ y ki (pki (t, θ)) − yˆki (p0 (t, θ)) + yˆki (p0 (t, θ)) − y ∗ (p0 (t, θ))|

−1≤θ≤0

≤

sup |ˆ y ki (pki (t, θ)) − yˆki (p0 (t, θ))|

−1≤θ≤0

+ sup |ˆ y ki (p0 (t, θ)) − y ∗ (p0 (t, θ))|. −1≤θ≤0

By Lemma 2.27, we get sup |ˆ y ki (p0 (t, θ)) − y ∗ (p0 (t, θ))| ≤ sup |y ki (t) − y ∗ (t)| + kˆ y0ki − y0∗ k

−1≤θ≤0

0≤θ≤α

= sup |y ki (t) − y ∗ (t)|. 0≤θ≤α

For every ǫ > 0 there exists a number L1 such that ǫ for ki ≥ L1 , sup |y ki (t) − y ∗ (t)| < 2 0≤θ≤α by the definition of y ∗ . On the other hand, since {ˆ y ki } is an equi-continuous set, for the given ǫ, there exists a µ > 0 such that ǫ |ˆ y ki (t) − yˆki (τ )| < for |t − τ | < µ. (2.42) 2 We can choose L ≥ L1 so that |pki (t, θ) − p0 (t, θ)| < µ. Thus (2.42) holds as long as ki ≥ L. Furthermore, kytki − yt k < ǫ whenever ki ≥ L,

which is just our claim. A similar argument shows that ηtki → ηt uniformly in t ∈ [0, α]. The limiting process upon (2.40) yields  Z t 1  g(t0 + t, ηt + yt∗ ) − g(t0 , ϕ) = (t − s)q−1 f (t0 + s, ηs + ys∗ )ds, 0 ≤ t ≤ α, Γ(q) 0  ∗ y0 = 0,

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53

which demonstrates that y ∗ as well as y 0 is a solution of the initial problem  Z t 1  (t − s)q−1 f (t0 + s, ηs + ys )ds, 0 ≤ t ≤ α, g(t0 + t, ηt + yt∗ ) − g(t0 , ϕ) = Γ(q) 0  y0 = 0.

The hypothesis that x(t0 , ϕ) is unique, that is, y 0 is unique implies that y ∗ = y 0 . Thus y ki (t) → y 0 (t) uniformly on [0, α]. The verified fact that every subsequence of sequence {y k } has a convergent subsequence with a same limit y 0 implies that the entire sequence {y k } converges to y 0 . Translating these remarks back into xk , we have indeed obtained the result stated in this theorem for the interval [p(σ, −1), σ + α]. Let b = σ + α. If b < A, (b, xb ) ∈ W , we can choose N1 ≥ N such that (b, xkb ) ∈ V as long as k ≥ N1 . By Theorem 2.21, for every point (b, xkb ) the solution xk (b, xkb ) exists at least on [p(b, −1), b + α]. The above argument can be adapted to this interval which yields the assertion that xk (b, xkb )(t) → x0 (t0 , ϕ)(t) uniformly on the same interval. The conclusion stated in theorem can be verified by successive steps of finite intervals of length α. Hence the proof is completed. 2.4 2.4.1

Neutral Equations with Infinite Delay Introduction

In Section 2.4, we consider the initial value problem of fractional neutral functional differential equations with infinite delay of the form C q t0 Dt g(t, xt )

xt0 = ϕ,

= f (t, xt ),

t ∈ [t0 , ∞),

(2.43)

(t0 , ϕ) ∈ [0, ∞) × Ω,

(2.44)

C q t0 Dt

where is Caputo fractional derivative of order 0 < q < 1, Ω is an open subset of B and g, f : [t0 , ∞) × Ω → Rn are given functionals satisfying some assumptions that will be specified later. B is called a phase space that will be defined later. If x : (−∞, A) → Rn , A ∈ (0, ∞), then for any t ∈ [0, A) define xt by xt (θ) = x(t + θ), for θ ∈ (−∞, 0]. Denote by BC(J, Rn ) the Banach space of all continuous and bounded functions from J into Rn with the norm k · k. To describe fractional neutral functional differential equations with infinite delay, we need to discuss a phase space B in a convenient way. We shall provide a general description of phase spaces of neutral differential equations with infinite delay which is taken from Lakshmikantham, Wen and Zhang, 1994. Let B be a real vector space either (i) of continuous functions that map (−∞, 0] to Rn with ϕ = ψ if ϕ(s) = ψ(s) on (−∞, 0] or

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(ii) of measurable functions that map (−∞, 0] to Rn with ϕ = ψ (or ϕ is equivalent to ψ) in B if ϕ(s) = ψ(s) almost everywhere on (−∞, 0], and ϕ(0) = ψ(0). Let B be endowed with a norm k · kB such that B is complete with respect to k · kB . Thus B equipped with norm k · kB is a Banach space. We denote this space by (B, k · kB ) or simply by B, whenever no confusion arises. Let 0 ≤ a < A. If x : (−∞, A) → Rn is given such that xa ∈ B and x ∈ [a, A) → n R is continuous, then xt ∈ B for all t ∈ [a, A). This is a very weak condition that the common admissible phase spaces and BC satisfy. For more details of the phase spaces, we refer the reader to Hino, Murakami and Naito, 1991; Lakshmikantham, Wen and Zhang, 1994. Definition 2.29. A function x : (−∞, t0 + σ) → Rn (t0 ∈ [0, ∞), σ > 0) is said to be a solution of IVP (2.43)-(2.44) through (t0 , ϕ) on [t0 , t0 + σ), if (i) (ii) (iii) (iv)

xt0 = ϕ; x is continuous on [t0 , t0 + σ); g(t, xt ) is absolutely continuous on [t0 , t0 + σ); (2.43) holds almost everywhere on [t0 , t0 + σ).

Let Ω ⊆ B be an open set such that for any (t0 , ϕ) ∈ [0, ∞) × Ω, there exist constants σ1 , γ1 > 0 so that xt ∈ Ω provided that x ∈ A(t0 , ϕ, σ1 , γ1 ) and t ∈ [t0 , t0 + σ1 ], where A(t0 , ϕ, σ1 , γ1 ) is defined as |x(t) − ϕ(0)| ≤ γ1 . A(t0 , ϕ, σ1 , γ1 ) = x : (−∞, t0 + σ1 ] → Rn , xt0 = ϕ, sup t0 ≤t≤t0 +σ1

In order to guarantee that equation (2.43) is NFDE, the coefficient of x(t) that is contained in g(t, xt ) cannot be equal to zero. Then we need to introduce the generalized atomic concept. Definition 2.30. (Lakshmikantham, Wen and Zhang, 1994) The functional g : [0, ∞) × Ω → Rn is said to be generalized atomic on Ω, if g(t, ϕ) − g(t, ψ) = K(t, ϕ, ψ)[ϕ(0) − ψ(0)] + L(t, ϕ, ψ) where (t, ϕ, ψ) ∈ [0, ∞)×Ω×Ω, K : [0, ∞)×Ω×Ω → Rn×n and L : [0, ∞)×Ω×Ω → Rn satisfy (i) detK(t, ϕ, ϕ) 6= 0 for all (t, ϕ) ∈ [0, ∞) × Ω; (ii) for any (t0 , ϕ) ∈ [0, ∞) × Ω, there exist constants δ1 , γ1 > 0, and k1 , k2 > 0, with 2k2 + k1 < 1 such that for all x, y ∈ A(t0 , ϕ, σ1 , γ1 ), g(t, xt ), K(t, xt , yt ) and L(t, xt , yt ) are continuous in t ∈ [t0 , t0 + σ1 ], and |K −1 (t0 , ϕ, ϕ)L(t, xt , yt )| ≤ k1 sup |x(s) − y(s)|, t0 ≤s≤t

|K −1 (t0 , ϕ, ϕ)K(t, xt , yt ) − I| ≤ k2 , where I is the n × n unit matrix.

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For a detailed discussion on the atomic concept we refer the reader to the books Hale, 1977; Lakshmikantham, Wen and Zhang, 1994. In Subsection 2.4.2, we shall discuss existence and uniqueness of solutions for IVP (2.43)-(2.44) on a class of comparatively comprehensive phase spaces. We establish various criteria on existence and uniqueness of solutions for IVP (2.43)(2.44). 2.4.2


The following existence result for IVP (2.43)-(2.44) is based on Krasnoselskii’s fixed point theorem. Theorem 2.31. Assume that g is generalized atomic on Ω, and that for any (t0 , ϕ) ∈ [0, ∞) × Ω, there exist constants σ1 , γ1 ∈ (0, ∞), q1 ∈ (0, q) and a real1 valued function m(t) ∈ L q1 [t0 , t0 + σ1 ] such that (H1) for any x ∈ A(t0 , ϕ, σ1 , γ1 ), f (t, xt ) is measurable; (H2) for any x ∈ A(t0 , ϕ, σ1 , γ1 ), |f (t, xt )| ≤ m(t), for t ∈ [t0 , t0 + σ1 ]; (H3) f (t, φ) is continuous with respect to φ on Ω. Then IVP (2.43)-(2.44) has a solution. Proof. We know that f (t, xt ) is Lebesgue measurable in [t0 , t0 + σ1 ] according 1 to conditions (H1). Direct calculation gives that (t − s)q−1 ∈ L 1−q1 [t0 , t], for t ∈ [t0 , t0 + σ1 ]. In light of the Hölder inequality and the condition (H2), we obtain that (t − s)q−1 f (s, xs ) is Lebesgue integrable with respect to s ∈ [t0 , t] for all t ∈ [t0 , t0 + σ1 ], and Z t 1 |(t − s)q−1 f (s, xs )|ds ≤ k(t − s)q−1 k 1−q kmk q1 . (2.45) t0

L

1

[t0 ,t]

L

1

[t0 ,t0 +σ1 ]

According to Definition 2.29, IVP (2.43)-(2.44) is equivalent to the following equation Z t 1 g(t, xt ) = g(t0 , ϕ) + (t − s)q−1 f (s, xs )ds for t ∈ [t0 , t0 + σ1 ]. (2.46) Γ(q) t0 Let ϕˆ ∈ A(t0 , ϕ, σ1 , γ1 ) be defined as ϕˆt0 = ϕ, ϕ(t ˆ 0 + t) = ϕ(0) for all t ∈ [0, σ1 ]. If x is a solution of IVP (2.43)-(2.44), let x(t0 + t) = ϕ(t ˆ 0 + t) + z(t), t ∈ (−∞, σ1 ], then we have xt0 +t = ϕˆt0 +t + zt , t ∈ [0, σ1 ]. Thus (2.46) implies that z satisfies the equation Z t 1 (t − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds (2.47) g(t0 + t, ϕˆt0 +t + zt ) = g(t0 , ϕ) + Γ(q) 0 for 0 ≤ t ≤ σ1 . Since g is generalized atomic on Ω, there exist positive constant α > 1 and a positive function σ2 (γ) defined in (0, γ1 ], such that for any γ ∈ (0, γ1 ], when 0 ≤ t ≤ σ2 (γ), we have α(2k2 + k1 ) < 1,

(2.48)

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|K −1 (t0 , ϕ, ϕ)K(t0 + t, xt0 +t , yt0 +t ) − I| ≤ k2 ,

(2.49)

|I − K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)| ≤ min{αk2 , α − 1}, |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )||g(t0 + t, ϕˆt0 +t ) − g(t0 , ϕ)| ≤

(2.50)

1 − α(2k2 + k1 ) γ. (2.51) 2

1

Note the completely continuity of the function (m(t)) q1 . Hence, for a given positive number M , there must exist a number h > 0, satisfying Z t0 +h 1 (m(s)) q1 ds ≤ M. t0

For a given γ ∈ (0, γ1 ], choose (

σ = min σ1 , σ2 (γ), h, (1 + β)

1 1+β

[1 − α(2k2 + k1 )]Γ(q)γ 2α|K −1 (t0 , ϕ, ϕ)|M q1

(1−q

1 1 )(1+β)

)

,

(2.52)

q−1 where β = 1−q ∈ (−1, 0). 1 For any (t0 , ϕ) ∈ [0, ∞) × Ω, define E(σ, γ) as follows:

E(σ, γ) = {z : (−∞, σ) → Rn is continuous; z(s) = 0 for s ∈ (−∞, 0] and kzk ≤ γ} where kzk = sup0≤s≤σ |z(t)|. Then E(σ, γ) is a closed bounded and convex subset of Banach space BC((−∞, σ1 ], Rn ). Now, on E(σ, γ) define two operators S and U as follows:  t ∈ (−∞, 0],   0, (Sz)(t) = K −1 (t0 + t, ϕˆt +t , ϕˆt +t )[−g(t0 + t, ϕˆt +t + zt ) + g(t0 , ϕ) 0 0 0   +K(t0 + t, ϕˆt0 +t , ϕˆt0 +t )z(t)], t ∈ [0, σ],

and

 0, t ∈ (−∞, 0],     −1 (U z)(t) = K (t0Z+ t, ϕˆt0 +t , ϕˆt0 +t ) t   1  × (t − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds, t ∈ [0, σ], Γ(q) 0

where z ∈ E(σ, γ). It is easy to see that the operator equation

z = Sz + U z

(2.53)

has a solution z ∈ E(σ, γ) if and only if z is a solution of the equation (2.47). Thus, xt+t0 = ϕˆt0 +t + zt is a solution of the equation (2.43) on [0, σ]. Therefore, the existence of a solution of IVP (2.43)-(2.44) is equivalent to determining σ, γ > 0 such that (2.53) has a fixed point in E(σ, γ). Now we show that S + U has a fixed point in E(σ, γ). The proof is divided into three steps. Step I. Sz + U w ∈ E(σ, γ) for every pair z, w ∈ E(σ, γ).

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Obviously, for every pair z, w ∈ E(σ, γ), (Sz)(t) and (U w)(t) are continuous in t ∈ [0, σ], and for t ∈ [0, σ], by using the Hölder inequality and (2.50), we have

|(U w)(t)| ≤ |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) K(t0 , ϕ, ϕ) K −1 (t0 , ϕ, ϕ)| Z 1 t q−1 (t − s) f (t + s, ϕ ˆ + w )ds × 0 t0 +s s Γ(q) 0 1−q1 Z t 1 1 [(t − s)q−1 ] 1−q1 ds ≤ α|K −1 (t0 , ϕ, ϕ)| Γ(q) q1 0 Z t0 +σ (2.54) 1 × (m(s)) q1 ds t0 1−q1 M q1 1 1+β −1 ≤ α|K (t0 , ϕ, ϕ)| σ Γ(q) 1 + β 1 − α(2k2 + k1 ) γ, ≤ 2 q−1 where β = ∈ (−1, 0), and 1 − q1 |(Sz)(t)| = K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )[−g(t0 + t, ϕˆt0 +t + zt ) + g(t0 + t, ϕˆt0 +t ) −g(t0 + t, ϕˆt0 +t ) + g(t0 , ϕ) + K(t0 + t, ϕˆt0 +t , ϕˆt0 +t )z(t)] = K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )[−K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t )z(t) −L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t ) − g(t0 + t, ϕˆt0 +t ) + g(t0 , ϕ) +K(t0 + t, ϕˆt0 +t , ϕˆt0 +t )z(t)] = K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )[K(t0 + t, ϕˆt0 +t , ϕˆt0 +t )

−K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t )]z(t) + K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) ×[−L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t ) − g(t0 + t, ϕˆt0 +t ) + g(t0 , ϕ)] = K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ) ×[K −1 (t0 , ϕ, ϕ)K(t0 + t, ϕˆt0 +t , ϕˆt0 +t ) − I]z(t) −K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)

×[K −1 (t0 , ϕ, ϕ)K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t ) − I]z(t)

+K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )[−L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t ) −g(t0 + t, ϕˆt0 +t ) + g(t0 , ϕ)]

≤ |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)|

×[(|K −1 (t0 , ϕ, ϕ)K(t0 + t, ϕˆt0 +t , ϕˆt0 +t ) − I|

+|K −1 (t0 , ϕ, ϕ)K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t ) − I|)|z(t)|

+|K −1 (t0 , ϕ, ϕ)L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t )| ]

+|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )||g(t0 + t, ϕˆt0 +t ) − g(t0 , ϕ)|.

According to (2.48)-(2.51), we have |(Sz)(t)| ≤ α(2k2 + k1 )γ +

1 + α(2k2 + k1 ) 1 − α(2k2 + k1 ) γ= γ. 2 2

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Therefore, |(Sz)(t)+(U w)(t)| ≤ γ for t ∈ [0, σ]. This means that Sz +U w ∈ E(σ, γ) whenever z, w ∈ E(σ, γ). Step II. S is a contraction mapping on E(σ, γ). For any z, w ∈ E(σ, γ), we obtain |(Sz)(t) − (Sw)(t)|

≤ |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )||K(t0 + t, ϕˆt0 +t , ϕˆt0 +t ) −K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t + wt )||z(t) − w(t)|

+|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t + wt )| ≤ [I − K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)] −[K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)]

×[K −1 (t0 , ϕ, ϕ)K(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t + wt ) − I] |z(t) − w(t)|

+|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )K(t0 , ϕ, ϕ)K −1 (t0 , ϕ, ϕ) ×L(t0 + t, ϕˆt0 +t + zt , ϕˆt0 +t + wt )|

≤ (αk2 + αk2 )|z(t) − w(t)| + αk1 sup |z(s) − w(s)| 0≤s≤t

≤ α(2k2 + k1 ) sup |z(s) − w(s)|, 0≤s≤t

where α(2k2 + k1 ) < 1, and therefore S is a contraction mapping on E(σ, γ). Step III. Now we show that U is a completely continuous operator. For any z ∈ E(σ, γ), 0 ≤ τ < t ≤ σ, we get |(U z)(t) − (U z)(τ )| Z t 1 (t − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds = K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) Γ(q) 0 Z τ 1 (τ − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds −K −1 (t0 + τ, ϕˆt0 +τ , ϕˆt0 +τ ) Γ(q) 0 Z t −1 1 = K (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) (t − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) τ Z τ 1 +K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) (t − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) 0 Z τ 1 −K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) (τ − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) 0 Z τ 1 +K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) (τ − s)q−1 f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) 0 Z τ 1 q−1 −1 (τ − s) f (t0 + s, ϕˆt0 +s + zs )ds −K (t0 + τ, ϕˆt0 +τ , ϕˆt0 +τ ) Γ(q) 0 Z |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| t q−1 ≤ (t − s) f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) τ |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| + Γ(q)

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Z τ q−1 q−1 × [(t − s) − (τ − s) ]f (t0 + s, ϕˆt0 +s + zs )ds 0

|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) − K −1 (t0 + τ, ϕˆt0 +τ , ϕˆt0 +τ )| + Γ(q) Z τ q−1 × (τ − s) f (t0 + s, ϕˆt0 +s + zs )ds 0

|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| (I1 + I2 ) = Γ(q) |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) − K −1 (t0 + τ, ϕˆt0 +τ , ϕˆt0 +τ )| + I3 , Γ(q) where

Z t q−1 I1 = (t − s) f (t0 + s, ϕˆt0 +s + zs )ds , Zτ τ q−1 q−1 I2 = [(t − s) − (τ − s) ]f (t0 + s, ϕˆt0 +s + zs )ds , Z0 τ q−1 (τ − s) f (t0 + s, ϕˆt0 +s + zs )ds . I3 = 0

By using an analogous argument presented in (2.54), we can conclude that h i1−q1 M q1 1+β I1 ≤ (t − τ ) , 1−q (1 + β) 1 1−q1 M q1 1+β τ , I3 ≤ (1 + β)1−q1

and

I2 ≤

Z

τ

0

≤ M q1

q−1

|(t − s)

Z

0

q−1

− (τ − s)

|

1 1−q1

τ

((τ − s)β − (t − s)β )ds

ds

1−q1 Z

1−q1

t0 +τ

t0

|f (s, xs )|

1 q1

ds

q1

h i1−q1 M q1 1+β 1+β 1+β ≤ τ − t + (t − τ ) (1 + β)1−q1 h i1−q1 M q1 1+β ≤ (t − τ ) , (1 + β)1−q1

where β =

q−1 1−q1

∈ (−1, 0). Therefore

h i1−q1 |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| 2M q1 1+β (t − τ ) 1−q Γ(q) (1 + β) 1 1−q1 −1 −1 |K (t0 + t, ϕˆt0 +t , ϕˆt0 +t ) − K (t0 + τ, ϕˆt0 +τ , ϕˆt0 +τ )| M q1 1+β + τ . 1−q Γ(q) (1 + β) 1

|(U z)(t) − (U z)(τ )| ≤

Since the property of the matrix function K(t0 +t, ϕˆt0 +t , ϕˆt0 +t ) which is nonsingular and continuous in t ∈ [0, σ] implies that its inverse matrix K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )

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exists and is continuous in t ∈ [0, σ], then {U z : z ∈ E(σ, γ)} is equicontinuous. On the other hand, U is continuous from condition (H3) and {U z : z ∈ E(σ, γ)} is uniformly bounded from (2.54), thus U is a completely continuous operator by the Ascoli-Arzela Theorem. Therefore, Krasnoselskii’s fixed point theorem shows that S +U has a fixed point on E(σ, γ), and hence IVP (2.43)-(2.44) has a solution x(t) = ϕ(0) + z(t − t0 ) for all t ∈ [t0 , t0 + σ]. Remark 2.32. If we replace condition (H1) by (H1)′ f (t, φ) is measurable with respect to t on [t0 , t0 + σ1 ]. Then we can also conclude that the result of Theorem 2.31 holds. In fact, for any x ∈ A(t0 , ϕ, σ1 , γ1 ), suppose xt0 +t = ϕˆt0 +t + zt , t ∈ [0, σ1 ], then, according to the definition of ϕˆt0 +t and zt , we know that xt0 +t is a measurable function. It follows that from (H1)′ and (H3), f (t, xt ) is measurable in t, where x ∈ A(t0 , ϕ, σ1 , γ1 ) and satisfies xt0 +t = ϕˆt0 +t + zt , t ∈ [0, σ1 ]. Remark 2.33. If we replace condition (H3) by a weaker condition (H3)′ for any x, y ∈ A(t0 , ϕ, σ, γ) with sup |x(s) − y(s)| → 0, t0 ≤s≤t0 +σ

Z t q−1 → 0, (t − s) [f (s, x ) − f (s, y )]ds s s t0

t ∈ [t0 , t0 + σ]

where σ satisfy (2.51), then we can also conclude that the result of Theorem 2.31 holds. The following existence and uniqueness result for IVP (2.43)-(2.44) is based on the Banach’s contraction principle. Theorem 2.34. Assume that g is generalized atomic on Ω, and that for any (t0 , ϕ) ∈ [0, ∞) × Ω, there exist constants σ1 , γ1 ∈ (0, ∞), q1 ∈ (0, q) and a real1 valued function m(t) ∈ L q1 [t0 , t0 + σ1 ] such that conditions (H1)-(H2) of Theorem 2.31 hold. Further assume that: (H4) there exists a nonnegative function ℓ : [0, σ1 ] → [0, ∞) continuous at t = 0 and ℓ(0) = 0 such that for any x, y ∈ A(t0 , ϕ, σ1 , γ1 ) we have Z t q−1 (t − s) [f (s, xs ) − f (s, ys )]ds ≤ ℓ(t − t0 ) sup |x(s) − y(s)|, t ∈ [t0 , t0 + σ1 ], t ≤s≤t t0

0

then IVP (2.43)-(2.44) has a unique solution.

Proof. According to the argument of Theorem 2.31, it suffices to prove that S + U has a unique fixed point on E(σ, γ), where σ, γ > 0 are sufficiently small. Now, choose σ ∈ (0, σ1 ), γ ∈ (0, γ1 ], such that (2.52) holds and that c = α(2k2 + k1 ) + sup

0≤s≤σ

|K −1 (t0 + s, ϕˆt0 +s , ϕˆt0 +s )||ℓ(s)| < 1. Γ(q)

(2.55)

Obviously, S + U is a mapping from E(σ, γ) into itself. Using the same argument as that of Theorem 2.31, for any z, w ∈ E(σ, γ), we get |(Sz)(t) − (Sw)(t)| ≤ α(2k2 + k1 ) sup |z(s) − w(s)|, 0≤s≤σ

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and |(U z)(t) − (U w)(t)|

Zt |K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| q−1 ≤ (t − s) f (t0 + s, ϕˆt0 +s + zs )ds Γ(q) 0 Z t q−1 − (t − s) f (t0 + s, ϕˆt0 +s + ws )ds 0

|K −1 (t0 + t, ϕˆt0 +t , ϕˆt0 +t )| ≤ |ℓ(t)| sup |z(s) − w(s)| Γ(q) 0≤s≤t

≤

sup |K −1 (t0 + s, ϕˆt0 +s , ϕˆt0 +s )||ℓ(s)|

0≤s≤σ

Γ(q)

sup |z(s) − w(s)|.

0≤s≤σ

Therefore |(S + U )z(t) − (S + U )w(t)| |K −1 (t0 + s, ϕˆt0 +s , ϕˆt0 +s )| |ℓ(s)| sup |z(s) − w(s)| ≤ α(2k2 + k1 ) + sup Γ(q) 0≤s≤σ 0≤s≤σ = c sup |z(s) − w(s)|. 0≤s≤σ

Hence, we have k(S + U )z − (S + U )wk ≤ ckz − wk, where c < 1. By applying the Banach contraction principle, we know that S + U has a unique fixed point on E(σ, γ). Corollary 2.35. If the condition (H4) of Theorem 2.34 is replaced by the following condition: 1 (H4)′ there exist q2 ∈ (0, q) and a function ℓ1 ∈ L q2 [t0 , t0 + σ1 ], such that for any x, y ∈ A(t0 , ϕ, σ1 , γ1 ) we have |f (t, xt ) − f (t, yt )| ≤ ℓ1 (t) sup |x(s) − y(s)|, t ∈ [t0 , t0 + σ1 ]. t0 ≤s≤t

Then the result of Theorem 2.34 holds.

Proof. It suffices to prove that the condition (H4) of Theorem 2.34 holds. Note 1 that ℓ1 ∈ L q2 [t0 , t0 + σ1 ], hence, there must exist a positive number N , such that N = kℓ1 k q1 . Then for any x, y ∈ A(t0 , ϕ, σ1 , γ1 ) we have L 2 [t0 , t0 +σ1 ] Z t q−1 (t − s) [f (s, xs ) − f (s, ys )]ds ≤ ≤

Z

t0 t

t0 t

Z

t0

(t − s)q−1 |f (s, xs ) − f (s, ys )|ds

(t − s)q−1 ℓ1 (s) ds sup |x(s) − y(s)| t0 ≤s≤t

′ N (t − t0 )(1+β )(1−q2 ) sup |x(s) − y(s)|, ≤ ′ 1−q 2 (1 + β ) t0 ≤s≤t

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where β ′ =

q−1 1−q2

∈ (−1, 0). Let

ℓ(t − t0 ) =

′ N (t − t0 )(1+β )(1−q2 ) , t ∈ [t0 , t0 + σ1 ]. ′ 1−q 2 (1 + β )

Obviously, ℓ : [0, σ1 ] → [0, ∞) continuous at t = 0 and ℓ(0) = 0. Then the condition (H4) of Theorem 2.34 holds. 2.4.3

Continuation of Solutions

For any t0 , ϕ ∈ [0, ∞) × Ω, ω ⊂ Ω and positive constants σ, γ > 0, define Bω (t0 , ϕ, σ, γ) as the set of all maps x : (−∞, t0 + σ) → Rn such that xt0 = ϕ, x : [t0 , t0 + σ) → Rn is continuous with |x(t) − ϕ(0)| ≤ γ and xt ∈ ω for all t ∈ [t0 , t0 + σ). In the following theorem, W is a set of all subsets of Ω such that for any (t0 , ϕ) ∈ [0, ∞) × Ω, constants σ, γ > 0 and a set ω ∈ W , if x ∈ Bω (t0 , ϕ, σ, γ) and if x(t0 + σ) = lim x(t) exists , then xt0 +σ ∈ Ω. t→(t0 +σ)−

Theorem 2.36. Let all conditions of Theorem 2.31 hold. Besides, suppose that σ ∈ (0, σ1 ], γ ∈ (0, γ1 ] and for any x ∈ Bω (t0 , ϕ, σ, γ), 1 (H5) there exist constants qω ∈ (0, q) and a real-valued function mω (t) ∈ L qω [t0 , t0 + σ] such that f (t, xt ) is measurable and |f (t, xt )| ≤ mω (t) for t ∈ [t0 , t0 + σ); (H6) lim+ [g(t, xt−τ ) − g(t − τ, xt−τ )] = 0 uniformly for t ∈ [t0 + τ, t0 + σ); τ →0

(H7) K(t, xt , xt ) − K(t, xt , xt−τ ) → 0 uniformly for t ∈ [t0 + τ, t0 + σ) as τ → 0+ and as sup |x(s) − x(s − τ )| → 0; t0 +τ ≤s≤t

(H8) there exists a constant H such that |K −1 (t, xt , xt )| ≤ H for all t ∈ [t0 , t0 + σ); (H9) there exists a continuous function ℓω : [0, ∞) → [0, ∞) with ℓω (0) = 0 such that |L(t, xt , xt−τ ) − L∗b (t, xt , xt−τ )| ≤ ℓω (b) sup |x(t + θ) − x(t − τ + θ)| −b≤θ≤0

where for a given b > 0, lim+ L∗b (t, xt , xt−τ ) = 0 uniformly for t ∈ [t0 + τ, t0 + σ). τ →0

Then for any ω ∈ W and any γ > 0, if x(t) is a noncontinuable solution of IVP (2.43)-(2.44) defined on [t0 , t0 + σ), there exists a t∗ ∈ [t0 , t0 + σ) such that |x(t∗ ) − ϕ(0)| > γ or xt∗ 6∈ ω. Proof. By way of contradiction, if there exists a noncontinuable solution x(t) of IVP (2.43)-(2.44) on [t0 , t0 + σ) such that |x(t) − ϕ(0)| ≤ γ and xt ∈ ω for all t ∈ [t0 , t0 +σ), that is, x ∈ Bω (t0 , ϕ, σ, γ), then first, x(t) is not uniformly continuous on [t0 , t0 + σ). Otherwise, x(t0 + σ) = lim − x(t) exists and thus xt0 +σ ∈ Ω. By t→(t0 +σ)

Theorem 2.31, x(t) can be continued beyond t0 + σ. Therefore, there exist a sufficiently small constant ε > 0, and sequences {tk } ⊆ [t0 , t0 + σ), {∆k } with ∆k → 0+ as k → ∞, such that |x(tk ) − x(tk − ∆k )| ≥ ε, for all k = 1, 2, ... .

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Now choose a constant H > 0 so that |K −1 (t, xt , xt )| ≤ H, for all t ∈ [t0 , t0 + σ). For given H and ε > 0, by (H5)-(H7) and (H9), we can find positive constants b and σ0 so that 1−qω ε 2HMω 1+βω σ < , 0 Γ(q)(1 + βω )1−qω 5 qω Z t0 +σ 1 q−1 q ω , where βω = 1−qω ∈ (−1, 0), Mω = (mω (s)) ds t0

H|K(t, xt , xt ) − K(t, xt , xt−τ )|
0 so that if kϕ − ψkB < σ, then |x(t) − y(t)| < ε for t ∈ [t0 , t0 + σ], where y(t) is a solution of (2.43) through (t0 , ψ). Proof. By way of contradiction, if the conclusion above is not true, then there exist ε > 0, sequences {tk } ⊆ [t0 , t0 + σ] and {ϕk } ⊆ Ω such that 1 kϕk − ϕkB < , k |y k (tk ) − x(tk )| = ε and

|y k (t) − x(t)| < ε for t ∈ [t0 , tk ),

where y k (t) is a solution of following IVP C α t0 Dt g(t, yt )

= f (t, yt ), yt0 = ϕk . (2.56) Without loss of generality, we may assume tk → t¯ ∈ [t0 , t0 + σ]. Now define a sequence of functions {z k } as follows: k y (t) for t ∈ [t0 , tk ], z k (t) = k y (tk ) for t ∈ [tk , t¯ ], if tk < t¯.

Using the same argument as that of Theorem 2.31, we can assume that {z k } is equicontinuous in t ∈ [t0 , t¯ ]. By Ascoli-Arzela theorem, without loss of generality, we can find a function y : (−∞, t¯ ] → Rn such that lim sup |z k (s) − y(s)| = 0 k→∞ t0 ≤s≤t¯

and y(s) = ϕ(s) for s ≤ t0 . Now considering the equation (2.56), we get Z t 1 k k (t − s)q−1 f (s, ysk )ds, for t ∈ [t0 , t¯ ]. g(t, yt ) − g(t0 , ϕ ) = Γ(q) t0 By (H10) and Lebesgue’s dominated convergence theorem and let k → ∞, we obtain Z t 1 g(t, yt ) − g(t0 , ϕ) = (t − s)q−1 f (s, ys )ds, for t ∈ [t0 , t¯ ]. Γ(q) t0 This means that y(t) = x(t) for t ∈ [t0 , t¯ ] by the uniqueness assumption of the solutions of IVP(2.43)-(2.44). This is contrary to |y k (tk ) − x(tk )| = ε

and lim

sup |z k (s) − y(s)| = 0.

k→∞ t0 ≤s≤t¯

The proof is therefore complete.

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66

2.5 2.5.1

ws-book975x65

Iterative Functional Differential Equations Introduction

In Section 2.5, we consider the following fractional iterative functional differential equations with parameter C q v  a Dt x(t) = f (t, x(t), x(x (t))) + λ, t ∈ [a, b], v ∈ R \ {0}, q ∈ (0, 1), λ ∈ R, x(t) = ϕ(t), t ∈ [a1 , a],  x(t) = ψ(t), t ∈ [b, b1 ], (2.57) q where C D is Caputo fractional derivative of order q and a t (C1) a1 ≤ a < b ≤ b1 , a1 ≤ av1 and bv1 ≤ b1 ; (C2) f ∈ C([a, b] × [a1 , b1 ]2 , R); (C3) ϕ ∈ C([a1 , a], [a1 , b1 ]) and ψ ∈ C([b, b1 ], [a1 , b1 ]). Definition 2.40. A function x ∈ C([a1 , b1 ], [a1 , b1 ]) is said to be a solution of the q v problem (2.57) if x satisfies the equation C a Dt x(t) = f (t, x(t), x(x (t))) + λ on [a, b], and the conditions x(t) = ϕ(t), t ∈ [a1 , a], x(t) = ψ(t), t ∈ [b, b1 ]. The purpose of this section is to determine the pair (x, λ), x ∈ C([a1 , b1 ], [a1 , b1 ]) (or CqL ([a1 ,b1 ], [a1 , b1 ])), λ ∈ R, which satisfies the problem (2.57). In Subsection 2.5.2, by using the Schauder’s fixed point theorem, we establish existence theorems in C([a1 , b1 ], [a1 , b1 ]) and CLq ([a1 , b1 ], [a1 , b1 ]) respectively. Unfortunately, uniqueness results can not be obtained since the solution operator is not Lipschitz continuous but only Hölder continuous. Meanwhile, data dependence results of solutions and parameters provide possible way to describe the error estimates between explicit and approximative solutions for such problems. In Subsection 2.5.4, We make some examples to illustrate our results and conclude some possible extensions to general parametrized iterative fractional functional differential equations. 2.5.2

Existence

We first give existence result in C([a1 , b1 ], [a1 , b1 ]). Let (x, λ) be a solution of the problem (2.57). Then this problem is equivalent to the following fixed point equation  ϕ(t), for t ∈ [a1 , a],   Z t   1   (t − s)q−1 f (s, x(s), x(xv (s)))ds  ϕ(a) + Γ(q) a x(t) = (2.58)  λ q   + (t − a) , for t ∈ [a, b],   Γ(q + 1)   ψ(t), for t ∈ [b, b1 ]. From the condition of continuity of x in t = b, we have that Z b q Γ(q + 1)(ψ(b) − ϕ(a)) − (b − s)q−1 f (s, x(s), x(xv (s)))ds. (2.59) λ= (b − a)q (b − a)q a

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Now we consider the operator A : C([a1 , b1 ], [a1 , b1 ]) → C([a1 , b1 ], R), where

(Ax)(t) :=

 ϕ(t), for t ∈ [a1 , a],     q q   (t − a) (t − a)   ϕ(a) + (ψ(b) − ϕ(a)) −  q  (b − a) Γ(q)(b − a)q   Z b

           

(b − s)q−1 f (s, x(s), x(xv (s)))ds a Z t 1 + (t − s)q−1 f (s, x(s), x(xv (s)))ds, for t ∈ [a, b], Γ(q) a ψ(t), for t ∈ [b, b1 ]. ×

(2.60)

(2.61)

It is clear that (x, λ) is a solution of the problem (2.57) if and only if x is a fixed point of the operator A and λ is given by (2.58). So, the problem is to study the fixed point equation x = A(x). Now, we are ready to state our first result in this section. Theorem 2.41. We suppose that (i) conditions (C1)-(C3) are satisfied; (ii) there are mf , Mf ∈ R such that mf ≤ f (t, u, w) ≤ Mf , ∀ t ∈ [a, b], u, w ∈ [a1 , b1 ], along with

and

mf (b − a)q Mf (b − a)q + min 0, , a1 ≤ min(ϕ(a), ψ(b)) − max 0, Γ(q + 1) Γ(q + 1)

mf (b − a)q max(ϕ(a), ψ(b)) − min 0, Γ(q + 1)

Mf (b − a)q + max 0, ≤ b1 . Γ(q + 1)

Then problem (2.57) has a solution in C([a1 , b1 ], [a1 , b1 ]). Proof. In what follow we consider on C([a1 , b1 ], R) with the Chebyshev norm k · kC . Condition (ii) assures that the set C([a1 , b1 ], [a1 , b1 ]) is an invariant subset for the operator A, that is, we have A(C([a1 , b1 ], [a1 , b1 ])) ⊂ C([a1 , b1 ], [a1 , b1 ]). Indeed, for t ∈ [a1 , a] ∪ [b, b1 ], we have A(x)(t) ∈ [a1 , b1 ]. Furthermore, we obtain a1 ≤ A(x)(t) ≤ b1 , ∀ t ∈ [a, b], if and only if a1 ≤ min A(x)(t) t∈[a,b]

(2.62)

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and max A(x)(t) ≤ b1

t∈[a,b]

(2.63)

hold. Since

mf (b − a)q Mf (b − a)q + min 0, , min A(x)(t) ≥ min(ϕ(a), ψ(b)) − max 0, Γ(q + 1) Γ(q + 1) t∈[a,b]

and

mf (b − a)q Mf (b − a)q max A(x)(t) ≤ max(ϕ(a), ψ(b)) − min 0, + max 0, , t∈[a,b] Γ(q + 1) Γ(q + 1)

respectively, the requirements (2.62) and (2.63) are equivalent with the conditions appearing in (ii). So, in the above conditions we have a selfmapping operator A : C([a1 , b1 ], [a1 , b1 ]) → C([a1 , b1 ], [a1 , b1 ]). Further, we check A is a completely continuous operator. Let {xn } be a sequence such that xn → x in C([a1 , b1 ], [a1 , b1 ]). Then for each t ∈ [a1 , b1 ], we have that  0, for t ∈ [a1 , a],     q |(Axn )(t) − (Ax)(t)| ≤ 2(b − a) kf (·, xn (xv (·))) − f (·, x(xv (·)))kC , for t ∈ [a, b], n   Γ(q + 1)   0, for t ∈ [b, b1 ].

Since f ∈ C([a, b] × [a1 , b1 ]2 , R), we have that

kAxn − AxkC → 0 as n → ∞. Now, consider a1 ≤ t1 < t2 ≤ a. Then, |(Ax)(t2 ) − (Ax)(t1 )| = |ϕ(t2 ) − ϕ(t1 )|. Similarly, for b ≤ t1 < t2 ≤ b1 , |(Ax)(t2 ) − (Ax)(t1 )| = |ψ(t2 ) − ψ(t1 )|. On the other hand, for a ≤ t1 < t2 ≤ b, |(Ax)(t2 ) − (Ax)(t1 )| ≤

(t2 − t1 )q |ψ(b) − ϕ(a)| (b − a)q 4(t2 − t1 )q max{|mf |, |Mf |} . + Γ(q + 1)

(2.64)

Together with the Arzela-Ascoli theorem and A is a continuous operator, we can conclude that A is a completely continuous operator. It is obvious that the set C([a1 , b1 ], [a1 , b1 ]) ⊆ C([a1 , b1 ], R) is a bounded convex closed subset of the Banach space C([a1 , b1 ], R). Thus, the operator A has a fixed point due to Schauder’s fixed point theorem. This completes the proof.

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In the following, we present the existence and estimate results in CLq ([a1 , b1 ], [a1 , b1 ]). Let L > 0 and I ⊂ R be a compact interval, and introduce the following notation: CLq (I, R) = {x ∈ C(I, R) : |x(t1 ) − x(t2 )| ≤ L|t1 − t2 |q }

for all t1 , t2 ∈ I. Remark that CLq (I, R) ⊆ C(I, R) is a complete metric space. Then (2.64) implies that under assumptions of Theorem 2.41 any solution of problem (2.57) belongs to CLq ∗ ([a, b], R) for L∗ =

|ψ(b) − ϕ(a)| 4 max{|mf |, |Mf |} + . (b − a)q Γ(q + 1)

(2.65)

Now we present our second result in this section. Theorem 2.42. We suppose that (i) conditions of Theorem 2.41 hold and ϕ ∈ CLq ϕ ([a1 , a], [a1 , b1 ]), ψ ∈ CLq ψ ([b, b1 ], [a1 , b1 ]) for some Lϕ , Lψ ≥ 0. Then problem (2.57) has a solution in X = CLq ([a1 , b1 ], [a1 , b1 ]) and all its solution belongs to X for p p 1−q p , L = 1−q Lϕ + 1−q Lψ + 1−q L∗ where L∗ is defined by (2.65). Assume in addition (ii) there exist Lu > 0 and Lw > 0 such that

|f (t, u1 , w1 ) − f (t, u2 , w2 )| ≤ Lu |u1 − u2 | + Lw |w1 − w2 |, for ∀ t ∈ [a, b], ui , wi ∈ [a1 , b1 ], i = 1, 2. Then two solutions x1 and x2 of problem (2.57) satisfy 1

kx1 − x2 kC ≤ LA1−q min{1,v} for LA :=

(2.66)

2(b − a)q 1−q min{1,v} q max{v−1,0} (Lu + Lw )b1 + max{1, v q }b1 Lw L . (2.67) Γ(q + 1)

If in addition

Γ(q + 1) > 2(b − a)q Lu ,

(2.68)

then kx1 − x2 kC 1  1−q min{1,v}  1−q min{1,v} q max{v−1,0} 2(b − a)q Lw b1 + max{1, v q }b1 L (2.69)  ≤ . q Γ(q + 1) − 2(b − a) Lu

Proof.

Consider the operator A given by (2.61). From Theorem 2.41, we have A : C([a1 , b1 ], [a1 , b1 ]) → C([a1 , b1 ], [a1 , b1 ])

and A has a fixed point in C([a1 , b1 ], [a1 , b1 ]).

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Now, consider a1 ≤ t1 < t2 ≤ a. Then,

|(Ax)(t2 ) − (Ax)(t1 )| = |ϕ(t2 ) − ϕ(t1 )| ≤ Lϕ |t1 − t2 |q ≤ L∗ |t1 − t2 |q

as ϕ ∈ CLq ϕ ([a1 , a], [a1 , b1 ]), due to (i). Similarly, for b ≤ t1 < t2 ≤ b1 ,

|(Ax)(t2 ) − (Ax)(t1 )| = |ψ(t2 ) − ψ(t1 )| ≤ Lψ |t1 − t2 |q ≤ L∗ |t1 − t2 |q

that follows from (i), too. On the other hand, for a ≤ t1 < t2 ≤ b, we already know (see (2.64)) |(Ax)(t2 ) − (Ax)(t1 )| ≤ L∗ |t1 − t2 |q .

Next, if a1 ≤ t1 ≤ a ≤ t2 ≤ b, then by the Hölder inequality with q ′ = 1 p′ = 1−q (note q ′ , p′ > 1)

1 q

and

|(Ax)(t2 ) − (Ax)(t1 )| ≤ |(Ax)(a) − (Ax)(t1 )| + |(Ax)(t2 ) − (Ax)(a)| ≤ Lϕ (a − t1 )q + L∗ (t2 − a)q q q ′ ′ ′ p′ ≤ Lpϕ + Lp∗ q (a − t1 )qq′ + (t2 − a)qq′ ≤ L|t1 − t2 |q .

Furthermore, if a1 ≤ t1 ≤ a < b ≤ t2 ≤ b1 , then again by the Hölder inequality 1 with q ′ = 1q and p′ = 1−q |(Ax)(t2 ) − (Ax)(t1 )|

≤ |(Ax)(a) − (Ax)(t1 )| + |(Ax)(b) − (Ax)(a)| + |(Ax)(t2 ) − (Ax)(b)| ≤ Lϕ (a − t1 )q + L∗ (b − a)q + Lψ (t2 − b)q q q ′ ′ ′ ′ ′ ≤ p Lpϕ + Lp∗ + Lpψ q (a − t1 )qq′ + (b − a)qq′ + (t2 − b)qq′ = L|t1 − t2 |q .

Therefore, the function A(x)(t) belongs to X. This proves the first statement. Take x1 , x2 ∈ X. Then for all t ∈ [a1 , a] ∪ [b, b1 ], we have |A(x1 )(t) − A(x2 )(t)| = 0. Moreover, for t ∈ [a, b], from our conditions, we get |A(x1 )(t) − A(x2 )(t)| Z b (t − a)q ≤ (b − s)q−1 |f (s, x1 (s), x1 (xv1 (s))) − f (s, x2 (s), x2 (xv2 (s)))| ds Γ(q)(b − a)q a Z t 1 + (t − s)q−1 |f (s, x1 (s), x1 (xv1 (s))) − f (s, x2 (s), x2 (xv2 (s)))| ds Γ(q) a Z b 1 (b − s)q−1 [Lu |x1 (s) − x2 (s)| + Lw |x1 (xv1 (s)) − x2 (xv2 (s))|] ds ≤ Γ(q) a Z t 1 + (t − s)q−1 [Lu |x1 (s) − x2 (s)| + Lw |x1 (xv1 (s)) − x2 (xv2 (s))|] ds Γ(q) a

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71

h (b − s)q−1 Lu |x1 (s) − x2 (s)| + Lw |x1 (xv1 (s)) − x1 (xv2 (s))| a i +Lw |x1 (xv2 (s)) − x2 (xv2 (s))| ds Z t h 1 + (t − s)q−1 Lu |x1 (s) − x2 (s)| + Lw |x1 (xv1 (s)) − x1 (xv2 (s))| Γ(q) a i +Lw |x1 (xv2 (s)) − x2 (xv2 (s))| ds Z b h i 1 q (b − s)q−1 (Lu + Lw )kx1 − x2 kC + Lw L |xv1 (s) − xv2 (s)| ds ≤ Γ(q) a Z t h i 1 q + (t − s)q−1 (Lu + Lw )kx1 − x2 kC + Lw L |xv1 (s) − xv2 (s)| ds Γ(q) a Z b h 1 (b − s)q−1 (Lu + Lw )kx1 − x2 kC ≤ Γ(q) a i q max{v−1,0} q min{1,v} + max{1, v q }b1 Lw Lkx1 − x2 kC ds Z t h 1 + (t − s)q−1 (Lu + Lw )kx1 − x2 kC Γ(q) a i q max{v−1,0} q min{1,v} + max{1, v q }b1 Lw Lkx1 − x2 kC ds b

≤

1 Γ(q)

≤

2(b − a)q ((Lu + Lw )kx1 − x2 kC Γ(q + 1) q max{v−1,0}

q min{1,v}

+ max{1, v q }b1 Lw Lkx1 − x2 kC ), where we use the inequality rv − sv ≤ max{1, v}rmax{v−1,0} (r − s)min{1,v} for any r ≥ s ≥ 0 and v > 0. From kx1 − x2 kC ≤ b1 we get 2(b − a)q 1−q min{1,v} kA(x1 ) − A(x2 )kC ≤ (Lu + Lw )b1 Γ(q + 1) q max{v−1,0} q min{1,v} + max{1, v q }b1 Lw L kx1 − x2 kC

(2.70)

q min{1,v}

= LA kx1 − x2 kC . So A is Hölder continuous but never Lipschitz continuous, since q min{1, v} ≤ q < 1. If x1 and x2 are fixed points of A then q min{1,v} kx1 − x2 kC = kA(x1 ) − A(x2 )kC ≤ LA kx1 − x2 kC which implies (2.66). In general, we have 2(b − a)q kx1 − x2 kC ≤ Lu kx1 − x2 kC Γ(q + 1) 2(b − a)q Lw 1−q min{1,v} + b1 Γ(q + 1) q max{v−1,0} q min{1,v} + max{1, v q }b1 L kx1 − x2 kC , which implies (2.69) under (2.68). The proof is completed.

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We do not know about uniqueness. But this is not so surprising, since A is not Lipschitzian in general. So we cannot apply metric fixed point theorems, only topological one. This can be simply illustrated on a simpler problem p 1 C 2 x(0) = 0, t ∈ [0, 1]. (2.71) 0 Dt x(t) = x( x(t)),

Rewriting (2.71) as

1 x(t) = B(x)(t) = Γ( 12 )

Z

t

0

p x( x(s)) √ ds, t−s

it follows that B : C 12 ([0, 1], [0, 1]) → C 12 ([0, 1], R) satisfies 1p 2 kx1 − x2 kC + kB(x1 ) − B(x2 )kC ≤ √ kx1 − x2 kC , π 2

so it is not Lipschitzian. Hence (2.71) should have a nonzero solution, and it does have x(t) = π4 t. 2.5.3

Data Dependence

Consider the following two problems C q v  a Dt x(t) = fi (t, x(t), x(x (t))) + λi , t ∈ [a, b], v ∈ (0, 1], q ∈ (0, 1), x(t) = ϕi (t), t ∈ [a1 , a],  x(t) = ψi (t), t ∈ [b, b1 ],

(2.72)

where fi , λi , ϕi and ψi , i = 1, 2 be as in the Theorem 2.42. Consider the operators

Ai : C([a1 , b1 ], [a1 , b1 ]) → C([a1 , b1 ], [a1 , b1 ]) given by (2.61) when ϕ, ψ, f and λ are replaced by ϕi , ψi , fi and λi , respectively. We are ready to state the third result in this section. Theorem 2.43. Suppose conditions of the Theorem 2.42, and, moreover (i) there exists η1 > 0 such that |ϕ1 (t) − ϕ2 (t)| ≤ η1 , ∀ t ∈ [a1 , a], and |ψ1 (t) − ψ2 (t)| ≤ η1 , ∀ t ∈ [b, b1 ]; (ii) there exists η2 > 0 such that |f1 (t, u, w) − f2 (t, u, w)| ≤ η2 , ∀ t ∈ [a, b], u, w ∈ [a1 , b1 ]. Let r∗ be a positive root of equation q min{1,v}

r∗ = L∗ r∗

+ 3η1 +

2(b − a)q η2 , Γ(q + 1)

(2.73)

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where L∗ = min{LA1 , LA2 } (see (2.67)). Then kx∗1 − x∗2 kC ≤ r∗ , (2.74) and Γ(q + 1) L∗ q min{1,v} |λ∗1 − λ∗2 | ≤ 2η + r + η2 , (2.75) 1 ∗ (b − a)q 2 where (x∗i , λ∗i ), i = 1, 2 are solutions of the corresponding problems (2.72). Note r∗ is uniquely defined. Proof. Using the condition (i), it is easy to see that for x ∈ C([a1 , b1 ], [a1 , b1 ]) and t ∈ [a1 , a] ∪ [b, b1 ], we have kA1 (x) − A2 (x)kC ≤ η1 . On the other hand, for t ∈ [a, b], using the condition (ii), we obtain |A1 (x)(t) − A2 (x)(t)| (t − a)q (|ψ1 (b) − ψ2 (b)| + |ϕ1 (a) − ϕ2 (a)|) ≤ |ϕ1 (a) − ϕ2 (a)| + (b − a)q Z b (t − a)q + (b − s)q−1 |f1 (s, x(s), x(xv (s))) − f2 (s, x(s), x(xv (s)))| ds Γ(q)(b − a)q a Z t 1 + (t − s)q−1 |f1 (s, x(s), x(xv (s))) − f2 (s, x(s), x(xv (s)))| ds Γ(q) a 2(b − a)q ≤ 3η1 + η2 . Γ(q + 1) So, we have 2(b − a)q η2 . kA1 (x) − A2 (x)kC ≤ 3η1 + Γ(q + 1) Next, (2.70) holds for both Ai with Lfi . Without loss of generality, we may suppose that L∗ = LA1 = min{LA1 , LA2 }. Consequently, we obtain kx∗1 − x∗2 kC = kA1 (x∗1 ) − A2 (x∗2 )kC ≤ kA1 (x∗1 ) − A1 (x∗2 )kC + kA1 (x∗2 ) − A2 (x∗2 )kC 2(b − a)q q min{1,v} ≤ L∗ kx∗1 − x∗2 kC + 3η1 + η2 , Γ(q + 1) which implies (2.74). Moreover, we get |λ∗1 − λ∗2 | Γ(q + 1)(|ψ1 (b) − ψ2 (b)| + |ϕ1 (a) − ϕ2 (a)|) ≤ (b − a)q Z b q ∗ ∗ ∗v + (b − s)q−1 |f1 (s, x∗1 (s), x∗1 (x∗v 1 (s))) − f1 (s, x2 (s), x2 (x2 (s)))| ds (b − a)q a Z b q ∗ ∗ ∗v (b − s)q−1 |f1 (s, x∗2 (s), x∗2 (x∗v + 2 (s))) − f2 (s, x2 (s), x2 (x2 (s)))| ds (b − a)q a Γ(q + 1) L∗ q min{1,v} ≤ 2η1 + r∗ + η2 . (b − a)q 2 The proof is completed.

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2.5.4

Examples and General Cases

Example 2.44. Consider the following problem:  1 2  C 0 Dt x(t) = µx(x(t)) + λ, t ∈ [0, 1], µ > 0, λ ∈ R, x(t) = 0, t ∈ [−h, 0], h > 0,   x(t) = 1, t ∈ [1, 1 + h],

(2.76)

where x ∈ C([−h, 1 + h], [−h, 1 + h]). Proposition 2.45. Suppose that

Γ( 32 )h . 1 + 2h Then the problem (2.76) has a solution in C([−h, 1 + h], [−h, 1 + h]). µ≤

Proof. First of all notice that accordingly to the Theorem 2.41 we have v = 1, q = 21 , a = 0, b = 1, ψ(b) = 1, ϕ(a) = 0 and f (t, u1 , u2 ) = µu2 . Moreover, a1 = −h and b1 = 1 + h can be taken. Therefore, from the relation mf ≤ f (t, u1 , u2 ) ≤ Mf , ∀ t ∈ [0, 1], u1 , u2 ∈ [−h, 1 + h], we can choose mf = −hµ and Mf = (1+h)µ. For these data it can be easily verified that the conditions (ii) from the Theorem 2.41 are equivalent to the relation µ≤

Γ( 32 )h , 1 + 2h

consequently we complete the proof.

Example 2.46. Consider the following problem:  1 2 2  C 2h Dt x(t) = µx (x(t)) + λ, t ∈ [2h, 3h], µ > 0, λ ∈ R, 1 x(t) = , t ∈ [h, 2h],   x(t) = 21 , t ∈ [3h, 4h], h ∈ [ 18 , 12 ], 2

(2.77)

where x ∈ C([h, 4h], [h, 4h]). Note 21 ∈ [h, 4h] for h ∈ [ 18 , 12 ]. Proposition 2.47. We suppose that √ 1 1 (−1 + 8h) π , for h ∈ ( , ], 0 0 such that

|ϕ1 (t) − ϕ2 (t)| ≤ η1 , ∀ t ∈ [h, 2h],

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and |ψ1 (t) − ψ2 (t)| ≤ η1 , ∀ t ∈ [3h, 4h]. Then kx∗1 − x∗2 kC ≤ r∗ , and 2 |λ∗1 − λ∗2 | ≤ √ πh

L∗ √ r∗ , 2η1 + 2

where L∗ and r∗ are given by (2.80) and (2.81), respectively. Proof. Results follow √ from Theorem 2.43 as follows. By Proposition 2.47, we √ 2 3 √ have L = 3L∗ = 128µh and then (see (2.67)) π √ √ 64h2 µ(64 3h3/2 µ + π)) ∗ . (2.80) L = L A1 = L A2 = π Realizing that now η2 = 0, equation (2.73) has the form √ r∗ = L∗ r∗ + 3η1 , which has the positive solution 3 1 25165824h7µ4 + 64h2 µπ 2 + η1 π 2 π2 ! √ q 3 14 8 9 5 7 4 2 +8 2 4947802324992h µ + 25165824h µ π 2 + 393216h µ η1 π .

r∗ =

(2.81)

The estimate for |λ∗1 − λ∗2 | follows directly from (2.75). The proof is finished.

We conclude this section by considering a general fractional order iterative functional differential equations with parameter given by C q v  a Dt x(t) = f (t, x(t), x(x (t)), λ), t ∈ [a, b], v ∈ (0, 1], q ∈ (0, 1), λ ∈ J, (2.82) x(t) = ϕ(t), t ∈ [a1 , a],  x(t) = ψ(t), t ∈ [b, b1 ],

when J ⊂ R is an open interval, conditions (C1), (C3) are supposed and (C2) is extended to (C4) f ∈ C([a, b] × [a1 , b1 ]2 × J, R). Then by (2.61) we have an operator A(λ, x). It is easy to see that A(λ, x) = A(x) for the problem (2.57). Supposing the assumptions of Theorem 2.41 for the problem (2.82) uniformly with respect to λ ∈ J, we can find its fixed point x∗ (λ, ·) ∈ C([a1 , b1 ], [a1 , b1 ]). In order to get a solution of the problem (2.82), we need to solve Z b Υ(λ) = Γ(q)(ψ(b) − ϕ(a)) − (b − s)q−1 f (s, x∗ (λ, s), x∗ (λ, x∗ (λ, s)v ))ds = 0. a

(2.83)

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If there is an λ0 ∈ J solving (2.83) then x∗ (λ0 , t) is a solution of the problem (2.82). Since x∗ (λ, ·) is not unique in general, function Υ(λ) is multivalued. Consequently this way is not very useful. We propose another approach. The problem (2.82) is equivalent to the following fixed point equation  ϕ(t), for t ∈ [a1 , a],   Z t  1 q−1 v x(t) = ϕ(a) + (t − s) f (s, x(s), x(x (s)), λ)ds, for t ∈ [a, b], (2.84)  Γ(q) a   ψ(t), for t ∈ [b, b1 ]. From the condition of continuity of x in t = b, we have that Z b 1 ψ(b) = ϕ(a) + (b − s)q−1 f (s, x(s), x(xv (s)), λ)ds. Γ(q) a

(2.85)

Now we consider the operator A : C b ([a1 , b1 ], [a1 , b1 ]) × J → C b ([a1 , b1 ], R)

(2.86)

where C b ([a1 , b1 ], [a1 , b1 ]) = {x ∈ C([a1 , b], [a1 , b1 ]) ∩ C b ((b, b1 ], [a1 , b1 ]) : ∃ lim x(s)}, s→b+

b

b

C ([a1 , b1 ], R) = {x ∈ C([a1 , b], R) ∩ C ((b, b1 ], R) : ∃ lim x(s)} s→b+

and  ϕ(t), for t ∈ [a1 , a],   Z t  1 A(x, λ)(t) := ϕ(a) + (t − s)q−1 f (s, x(s), x(xv (s)), λ)ds, for t ∈ [a, b],  Γ(q) a   ψ(t), for t ∈ (b, b1 ]. (2.87) Now, we are ready to state the following result. Theorem 2.50. Suppose that (i) conditions (C1), (C3) and (C4) are satisfied; (ii) there are mf , Mf ∈ R such that mf ≤ f (t, u, w, λ) ≤ Mf , ∀ t ∈ [a, b], u, w ∈ [a1 , b1 ], λ ∈ J along with mf (b − a)q , a1 ≤ ϕ(a) + min 0, Γ(q + 1) and

Mf (b − a)q ϕ(a) + max 0, ≤ b1 . Γ(q + 1)

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Then operator A(x, λ) has a fixed point in C b ([a1 , b1 ], [a1 , b1 ]) for any λ ∈ J. Proof. Like in the proof of Theorem 2.41, condition (ii) assures that the set C b ([a1 , b1 ], [a1 , b1 ]) is an invariant subset for the operator A, that is, we have A(C b ([a1 , b1 ], [a1 , b1 ]) × J) ⊂ C b ([a1 , b1 ], [a1 , b1 ]).

(2.88)

Similarly, A is a completely continuous operator. It is obvious that the set C b ([a1 , b1 ], [a1 , b1 ]) ⊆ C b ([a1 , b1 ], R) is a bounded convex closed subset of the Banach space C b ([a1 , b1 ], R). Thus, the operator A(x, λ) has a fixed point due to Schauder’s fixed point theorem. This completes the proof. We still do not have uniqueness result. For this purpose, we suppose (C5 ) f is nonnegative and nondecreasing, i.e. mf ≥ 0 and 0 ≤ f (s1 , u1 , v1 , λ) ≤ f (s2 , u2 , v2 , λ) for any s1 ≤ s2 ∈ [a, b], u1 ≤ u2 , v1 ≤ v2 ∈ [a1 , b1 ] and λ ∈ J. We introduce the Banach space b Cm ([a1 , b1 ], [a1 , b1 ]) = x ∈ C b ([a1 , b1 ], [a1 , b1 ]) | x is nondecreasing on [a1 , b1 ] .

Now, we have the next result. Theorem 2.51. We suppose conditions (i), (ii) of Theorem 2.50, (C5 ) as well and ϕ(t), ψ(t) are nondecreasing with ϕ(a) ≤ ψ(b). Then operator A(x, λ) is b monotone nondecreasing in x on Cm ([a1 , b1 ], [a1 , b1 ]) for any λ ∈ J. Consequently it b has a unique smallest and largest fixed points xm (λ), xM (λ) in Cm ([a1 , b1 ], [a1 , b1 ]). k Moreover, a nondecreasing sequence {A (a1 , λ)(t)}k≥1 and a nonincreasing sequence {Ak (b1 , λ)(t)}k≥1 satisfy a1 ≤ Ak (a1 , λ)(t) ≤ xm (λ)(t) ≤ xM (λ)(t) ≤ Ak (b1 , λ)(t) ≤ b1 k

t∈J

k

for any k ≥ 1 and limk→∞ A (a1 , λ)(t) = xm (λ)(t) and limk→∞ A (b1 , λ)(t) = xM (λ)(t) uniformly on [a1 , b1 ]. Proof. We already know (2.88). Let x ∈ Cm ([a1 , b1 ], [a1 , b1 ]) then clearly A(x, λ)(t1 ) ≤ A(x, λ)(t2 ) for t1 ≤ t2 ∈ [a1 , a] and t1 ≤ t2 ∈ (b, b1 ]. Next for s1 ≤ s2 ∈ [a, b] we have x(s1 ) ≤ x(s2 ), xv (s1 ) ≤ xv (s2 ) and x(xv (s1 )) ≤ x(xv (s2 )), which imply f (s1 , x(s1 ), x(xv (s1 )), λ) ≤ f (s2 , x(s2 ), x(xv (s2 )), λ).

(2.89)

Furthermore, for t1 ≤ t2 ∈ [a, b], following El-Sayed, 1995 and Darwish, 2008, and using (2.89), we derive A(x, λ)(t2 ) − A(x, λ)(t1 ) Z t2 1 (t2 − s)q−1 f (s, x(s), x(xv (s)), λ)ds = Γ(q) a Z t1 1 − (t1 − s)q−1 f (s, x(s), x(xv (s)), λ)ds Γ(q) a Z t1 1 = (t2 − s)q−1 − (t1 − s)q−1 f (s, x(s), x(xv (s)), λ)ds Γ(q) a Z t2 1 + (t2 − s)q−1 f (s, x(s), x(xv (s)), λ)ds Γ(q) t1

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≥

1 f (t1 , x(t1 ), x(xv (t1 )), λ) Γ(q) Z t1 Z × (t2 − s)q−1 − (t1 − s)q−1 ds + a

t2

t1

(t2 − s)q−1 ds

1 f (t1 , x(t1 ), x(xv (t1 )), λ) ((t2 − a)q − (t1 − a)q ) = Γ(q + 1) ≥ 0.

79

Consequently, we obtain b b A(Cm ([a1 , b1 ], [a1 , b1 ]), λ) ⊂ Cm ([a1 , b1 ], [a1 , b1 ])

for any λ ∈ J. b Next, if x1 , x2 ∈ Cm ([a1 , b1 ], [a1 , b1 ]) with x1 (t) ≤ x2 (t), t ∈ [a1 , b1 ] then clearly we have A(x1 , λ)(t) ≤ A(x2 , λ)(t) for t ∈ [a1 , a] ∪ (b, b1 ]. For s ∈ [a, b], we have x1 (s) ≤ x2 (s), xv1 (s) ≤ xv2 (s) and x1 (xv1 (s)) ≤ x2 (xv2 (s)), which imply f (s, x1 (s), x1 (xv1 (s)), λ) ≤ f (s, x2 (s), x2 (xv2 (s)), λ).

Then for t ∈ [a, b] we have A(x2 , λ)(t) − A(x1 , λ)(t) =

Z

(2.90)

(t − s)q−1 f (s, x2 (s), x2 (xv2 (s)), λ) a −f (s, x1 (s), x1 (xv1 (s)), λ) ds ≥ 0. 1 Γ(q)

t

This means that operator A(x, λ) is monotone nondecreasing in x on b Cm ([a1 , b1 ], [a1 , b1 ]) for any λ ∈ J. We also know that A is a completely continuous operator. Then results follow from the general theory of nondecreasing compact operators in Banach spaces (see, e.g., Deimling, 1985). The proof is completed. To get continuous solution, we need to solve either Z b 1 (b − s)q−1 f (s, xm (λ)(s), xm (λ)(xvm (λ)(s)), λ)ds = 0 Υm (λ) = ψ(b) − ϕ(a) − Γ(q) a or Z b 1 (b−s)q−1 f (s, xM (λ)(s), xM (λ)(xvM (λ)(s)), λ)ds = 0. ΥM (λ) = ψ(b)−ϕ(a)− Γ(q) a We can use to handle these equations also an analytical-numerical method like in Ronto, 2009. This means that first successive approximation is used xn+1 (λ, t) = A(xn (λ, t), λ) for up to some order j with either x0 (λ, t) = a1 or x0 (λ, t) = b1 . Then approximations Z b 1 Υj (λ) = ψ(b) − ϕ(a) − (b − s)q−1 f (s, xj (λ, s), xj (λ, xvj (λ, s)), λ)ds (2.91) Γ(q) a of Υm and ΥM are numerically drawn to check if they change the sign over J.

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2.6


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Notes and Remarks

The results in Section 2.2 are taken from Agarwal, Zhou and He, 2010. The main results in Section 2.3 are adopted from Zhou, Jiao and Li, 2009a. The material in Section 2.4 due to Zhou, Jiao and Li, 2009b. The results in Section 2.5 are taken from Wang, Fe˘ckan and Zhou, 2013c.

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Chapter 3

Fractional Ordinary Differential Equations in Banach Spaces

3.1

Introduction

In Chapter 3, we discuss the Cauchy problem of fractional ordinary differential equations in Banach spaces under hypotheses based on Carathéodory condition. The tools used include some classical and modern nonlinear analysis methods such as fixed point theory, measure of noncompactness method, topological degree method and Picard operators technique, etc. Firstly, we give an example which show that the criteria on existence of solutions for the initial value problem of fractional differential equations in finitedimensional spaces may not be true in infinite-dimensional cases. It is well known that Peano’s theorem of integer order ordinary differential equations is not true in infinite-dimensional Banach spaces. The first result in this direction was obtained by Dieudonne, 1950. In Dieudonne, 1950, he produced an example which showed that Peano’s theorem is not true in the space c0 of sequences which converge to zero. In fact, Peano’s theorem of fractional differential equations is also not true in infinite-dimensional Banach spaces. In the following, we shall show that the existence result of nonlocal Cauchy problem for fractional abstract differential equations which has been obtained in N’Guerekata, 2009 is not true in the space c0 . Example 3.1. Let E = c0 = {z = (z1 , z2 , z3 , ...) n → 0 p : zp pas n → ∞} with the norm kzk = supn≥1 |zn | and f (z) = 2( |z1 |, |z2 |, |z3 |, ...) with z = (z1 , z2 , z3 , ...) ∈ c0 . Consider the nonlocal Cauchy problem for fractional differential equations given by C q 0 Dt x(t)

= f (x(t)), x(0) = ξ, t ∈ (0, t0 ]

(3.1)

q where C 0 Dt is Caputo fractional 1 derivative of order 0 < q < 1, ξ = 2 (1, 1/2 , 1/32, ...) ∈ c0 , t0 < ( Γ(1+q) )q . 2 It is obvious that f : c0 → c0 is continuous. According to N’Guerekata, 2009, Γ(1+q) there exists a constant k ∗ = Γ(1+q)−2t q , such that IVP (3.1) possesses at least one 0 continuous solution x ∈ C([0, t0 ], c0 ) and x(t) = (x1 (t), x2 (t), x3 (t), ...) ∈ c0 on [0, t0 ] with supt∈[0,t0 ] kx(t)k ≤ k ∗ . According to the definition of the norm of c0 , we can

81

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conclude that

p 1 (3.2) |xn (t)|, xn (0) = 2 , t ∈ (0, t0 ], n = 1, 2, 3, ..., n where xn satisfies that xn ∈ C([0, t0 ], R) with supt∈[0,t0 ] |xn (t)| ≤ k ∗ . Let us consider Eq. (3.2) which can be written as the following equivalent form Z t p p 1 2 1 −q xn (t) = 2 + 20 Dt (t − s)q−1 |xn (s)|ds, t ∈ [0, t0 ]. |xn (t)| = 2 + n n Γ(q) 0 (3.3) q−1 Since (t − s) > 1 with s ∈ [0, t) for t ∈ (0, t0 ], we have by (3.3) Z tp 2 1 |xn (s)|ds, t ∈ [0, t0 ], n = 1, 2, 3, ... . (3.4) xn (t) ≥ 2 + n Γ(q) 0 Assume that yn ∈ C([0, t0 ], R) is a solution of the following integral equation Z tp 1 2 yn (t) = 2 + |yn (s)|ds, t ∈ [0, t0 ], n = 1, 2, 3, ... . (3.5) Γ(q) 0 4n Then, we get C q 0 Dt xn (t)

=2

xn (t) ≥ yn (t), t ∈ [0, t0 ], n = 1, 2, 3, ... .

(3.6)

xn (t1 ) = yn (t1 ), xn (t) > yn (t) t ∈ [0, t1 ), n = 1, 2, 3, ... .

(3.7)

In fact, suppose (for contraction) that the conclusion (3.6) is not true. Then, because of the continuity of x and y, and that xn (0) > yn (0), it follows that there exists a t1 ∈ (0, t0 ] such that Then using (3.4) and (3.7), we get

Z t1 p 2 1 + |yn (s)|ds Γ(q) 0 4n2 Z t1 p 2 1 < 2+ |xn (s)|ds n Γ(q) 0 ≤ xn (t1 ), n = 1, 2, 3, ...,

yn (t1 ) =

which is a contraction in view of (3.7). Hence the conclusion (3.6) is valid. Since the integral (3.5) is equivalent to the following IVP 1 2 p |yn (t)|, yn (0) = 2 , t ∈ [0, t0 ], n = 1, 2, 3, ..., yn′ (t) = (3.8) Γ(q) 4n and noting yn (t) > 0, t ∈ [0, t0 ], we can conclude that IVP (3.8) has a continuous solution 2 t 1 yn (t) = + , t ∈ [0, t0 ], n = 1, 2, 3, ..., Γ(q) 2n which means that 2 1 t + , t ∈ [0, t0 ], n = 1, 2, 3, ... . (3.9) xn (t) ≥ yn (t) = Γ(q) 2n Therefore, for t ∈ (0, t0 ], limn→∞ xn (t) 6= 0 by (3.9), contracting x(t) ∈ c0 . Hence IVP (3.1) has no nonlocal solution in c0 .

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3.2 3.2.1

83

Cauchy Problems via Measure of Noncompactness Method Introduction

In Section 3.2, we assume that X is a Banach space with the norm | · |. Let J ⊂ R. Denote C(J, X) be the Banach space of continuous functions from J into X. Let r > 0 and C = C([−r, 0], X) be the space of continuous functions from [−r, 0] into X. For any element z ∈ C, define the norm kzk∗ = supθ∈[−r,0] |z(θ)|. Consider the initial value problem (IVP) for fractional functional differential equation given by C q t ∈ (0, a), 0 Dt x(t) = f (t, xt ), (3.10) x0 = ϕ ∈ C,

q where C 0 Dt is Caputo fractional derivative of order 0 < q < 1, f : [0, a) × C → X is a given function satisfying some assumptions and define xt by xt (θ) = x(t + θ), for θ ∈ [−r, 0]. In this section, we shall discuss the existence of the solutions for IVP (3.10) under assumptions that f satisfies Carathéodory condition and the condition on measure of noncompactness. Then, we give an example to illustrate the application of our abstract results. Definition 3.2. A function x ∈ C([−r, T ], X) is a solution for IVP (3.10) on [−r, T ] for T ∈ (0, a) if

(i) the function x(t) is absolutely continuous on [0, T ]; (ii) x0 = ϕ; (iii) x satisfies the equation in (3.10). 3.2.2

Existence

We are now ready to prove the existence of the solutions for IVP (3.10) under the following hypotheses: (H1) For almost all t ∈ [0, a), the function f (t, ·) : C → X is continuous and for each z ∈ C, the function f (·, z) : [0, a) → X is strongly measurable; 1 (H2) for each τ > 0, there exist a constant q1 ∈ [0, q) and m1 ∈ L q1 ([0, a), R+ ) such that |f (t, z)| ≤ m1 (t) for all z ∈ C with kzk∗ ≤ τ and almost all t ∈ [0, a); 1 (H3) there exist a constant q2 ∈ (0, q) and m2 ∈ L q2 ([0, a), R+ ) such that α(f (t, B)) ≤ m2 (t)α(B) for almost all t ∈ [0, a) and B a bounded set in C. In order to prove the existence theorem, we need the following lemma. Lemma 3.3. Assume that the hypotheses (H1) and (H2) hold. x ∈ C([−r, T ], X) is a solution for IVP (3.10) on [−r, T ] for T ∈ (0, a) if and only if x satisfies the following relation  for θ ∈ [−r, 0],  x(θ) = ϕ(θ), Z t (3.11) 1  x(t) = ϕ(0) + (t − s)q−1 f (s, xs )ds, for t ∈ [0, T ]. Γ(q) 0

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Proof. Since xt is continuous in t ∈ [0, a), according to (H1), f (t, xt ) is a mea1 surable function in [0, a). Direct calculation gives that (t − s)q−1 ∈ L 1−q1 [0, t] for t ∈ (0, a) and q1 ∈ [0, q). Let b1 =

q−1 ∈ (−1, 0), 1 − q1

M = km1 k

1

L q1 [0,a)

.

By using Hölder inequality and (H2), for t ∈ (0, a), we obtain that 1−q1 Z t Z t q−1 q−1 1−q1 ds km1 k q1 |(t − s) f (s, xs )|ds ≤ (t − s) 0

L

0

≤

1

[0,t]

M a(1+b1 )(1−q1 ) . (1 + b1 )1−q1

(3.12)

Thus, |(t − s)q−1 f (s, xs )| is Lebesgue integrable with respect to s ∈ [0, t) for all t ∈ (0, a). From Lemma 1.4 (Bochner’s theorem), it follows that (t − s)q−1 f (s, xs ) is Bochner integrable with respect to s ∈ [0, t) for all t ∈ (0, a). Let L(τ, s) = (t−τ )−q |τ −s|q−1 m1 (s). Since L(τ, s) is a nonnegative, measurable function on D = [0, t] × [0, t], then we have Z t Z t Z Z t Z t L(τ, s)dsdτ = L(τ, s)dτ ds L(τ, s)ds dτ = 0

and

D

0

Z

0

L(τ, s)ds dτ 0 0 Z t Z t = (t − τ )−q |τ − s|q−1 m1 (s)ds dτ 0 Z0 τ Z t = (t − τ )−q (τ − s)q−1 m1 (s)ds dτ 0 0Z t Z t (s − τ )q−1 m1 (s)ds dτ + (t − τ )−q 0 τ Z t 2M (1+b1 )(1−q1 ) a (t − τ )−q dτ ≤ (1 + b1 )1−q1 0 2M ≤ a(1+b1 )(1−q1 )+1−q . (1 − q)(1 + b1 )1−q1

L(τ, s)dsdτ =

D

Z tZ

0

t

Therefore, L1 (τ, s) = (t − τ )−q (τ − s)q−1 f (s, xs ) is a Bochner integrable function on D = [0, t] × [0, t], then we have Z t Z τ Z t Z t dτ L1 (τ, s)ds = ds L1 (τ, s)dτ. 0

0

0

We now prove that q −q 0 Dt 0 Dt f (t, xt ) = f (t, xt ),

s

for t ∈ (0, T ],

where 0 Dtq is Riemann-Liouville fractional derivative.

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Indeed, we have q −q 0 Dt 0 Dt f (t, xt ) = = =

d 1 Γ(1 − q)Γ(q) dt 1 d Γ(1 − q)Γ(q) dt 1 d Γ(1 − q)Γ(q) dt

Z

t −q

(t − τ )

0

Z

t

dτ

0

Z

t

ds

0

Z

τ

0 Z t

Z

0

85

τ q−1

(τ − s)

f (s, xs )ds dτ

L1 (τ, s)ds L1 (τ, s)dτ

s

Z Z t 1 d t f (s, xs )ds (t − τ )−q (τ − s)q−1 dτ Γ(1 − q)Γ(q) dt 0 s Z d t = f (s, xs )ds dt 0 = f (t, xt ) for t ∈ (0, T ].

=

If x satisfies the relation (3.11), then we can get that x(t) is absolutely continuous on [0, T ]. In fact, for any disjoint family of open intervals {(ci , di )}1≤i≤n in [0, T ] P with ni=1 (di − ci ) → 0, we have n X i=1 n X

|x(di ) − x(ci )|

Z Z ci 1 di q−1 q−1 (di − s) f (s, xs )ds − (ci − s) f (s, xs )ds = Γ(q) 0 0 i=1 Z di n X 1 (di − s)q−1 f (s, xs )ds ≤ Γ(q) ci i=1 Z ci n X 1 Z ci q−1 q−1 (d − s) f (s, x )ds − (c − s) f (s, x )ds + i s i s Γ(q) 0 0 i=1 Z di n X 1 ≤ (di − s)q−1 m1 (s)ds Γ(q) c i i=1 Z ci n X 1 ((ci − s)q−1 − (di − s)q−1 )m1 (s)ds + Γ(q) 0 i=1 Z di 1−q1 n X q−1 1 (di − s) 1−q1 ds ≤ km1 k q1 L 1 [0,T ] Γ(q) ci i=1 1−q1 Z n ci X q−1 q−1 1 1−q1 1−q1 + − (di − s) ds km1 k q1 (ci − s) L 1 [0,T ] Γ(q) 0 i=1 =

n X (di − ci )(1+b1 )(1−q1 ) i=1

+

Γ(q)(1 + b1 )1−q1

km1 k

1

L q1 [0,T ]

n 1 1 X (c1+b − d1+b + (di − ci )1+b1 )1−q1 i i km1 k q1 L 1 [0,T ] Γ(q)(1 + b1 )1−q1 i=1

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86

≤2

n X (di − ci )(1+b1 )(1−q1 ) i=1

Γ(q)(1 + b1 )1−q1

km1 k

1

L q1 [0,T ]

→ 0.

Therefore, x(t) is absolutely continuous on [0, T ], which implies that x(t) is differentiable a.e. on [0, T ]. According to the argument above and Definition 1.8, for t ∈ (0, T ], we have Z t 1 C q C q q−1 (t − s) f (s, xs )ds 0 Dt x(t) = 0 Dt ϕ(0) + Γ(q) 0 Z t 1 q q−1 (t − s) f (s, x )ds =C D s 0 t Γ(q) 0 q −q =C 0 Dt 0 Dt f (t, xt ) t−q = 0 Dtq 0 Dt−q f (t, xt ) − [0 Dt−q f (t, xt )]t=0 Γ(1 − q) −q t = f (t, xt ) − [0 Dt−q f (t, xt )]t=0 . Γ(1 − q)

Since (t − s)q−1 f (s, xs ) is Lebesgue integrable with respect to s ∈ [0, t) for all q t ∈ (0, T ], we know that [0 Dt−q f (t, xt )]t=0 = 0, which means that C 0 Dt x(t) = f (t, xt ), for t ∈ (0, T ]. Hence, x ∈ C([−r, T ], X) is a solution of IVP (3.10). On the other hand, it is obvious that if x ∈ C([−r, T ], X) is a solution of IVP (3.10), then x satisfies the relation (3.11), and this completes the proof. Theorem 3.4. Assume that hypotheses (H1)-(H3) hold. Then, for every ϕ ∈ C, there exists a solution x ∈ C([−r, T ], X) for IVP (3.10) with some T ∈ (0, a). Proof.

Let k > 0 be any number and we can choose T ∈ (0, a) such that T (1+b1 )(1−q1 ) ≤k km1 k q1 L 1 [0,T ] Γ(q)(1 + b1 )1−q1

(3.13)

T (1+b2 )(1−q2 ) km2 k q1 < 1, L 2 [0,T ] Γ(q)(1 + b2 )1−q2

(3.14)

and

q−1 ∈ (−1, 0), i = 1, 2. 1 − qi Consider the set Bk defined as follows Bk = x ∈ C([−r, T ], X) : x0 = ϕ, sup |x(s) − ϕ(0)| ≤ k .

where bi =

s∈[0,T ]

Define the operator F on Bk as follows  for θ ∈ [−r, 0],  F x(θ) = ϕ(θ), Z t 1  F x(t) = ϕ(0) + (t − s)q−1 f (t, xs )ds, for t ∈ [0, T ], Γ(q) 0

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87

where x ∈ Bk . We prove that the operator equation x = F x has a solution x ∈ Bk , which means that x is a solution of IVP (3.10). Firstly, we observe that for every y ∈ Bk , (F y)(t) is continuous on t ∈ [−r, T ] and for t ∈ [0, T ], by (3.13) and Hölder inequality, we have Z

1 Γ(q)

t

|(t − s)q−1 f (s, ys )|ds 1−q1 Z t q−1 1 km1 k q1 (t − s) 1−q1 ds ≤ L 1 [0,T ] Γ(q) 0 (1+b1 )(1−q1 ) T km1 k q1 ≤ L 1 [0,T ] Γ(q)(1 + b1 )1−q1 ≤ k,

|(F y)(t) − ϕ(0)| ≤

0

(3.15)

q−1 where b1 = 1−q ∈ (−1, 0). Thus, supt∈[0,T ] |(F y)(t) − ϕ(0)| ≤ k, which implies that 1 F : Bk → Bk . Further, we prove that F is a continuous operator on Bk . Let {y n } ⊆ Bk with n y → y on Bk . Then by (H1) and the fact that ytn → yt , t ∈ [0, T ], we have

f (s, ysn ) → f (s, ys ), a.e. s ∈ [0, T ], as n → ∞. Noting that (t − s)q−1 |f (s, ysn ) − f (s, ys )| ≤ (t − s)q−1 2m1 (s), by Lebesgue’s dominated convergence theorem, as n → ∞, we have |(F y n )(t) − (F y)(t)| ≤

1 Γ(q)

Z

0

t

(t − s)q−1 |f (s, ysn ) − f (s, ys )|ds → 0.

Therefore F y n → F y as n → ∞ which implies that F is continuous. For each n ≥ 1, we define a sequence {xn : n ≥ 1} in the following way n

x (t) =

 0  ϕ (t),

 ϕ(0) +

1 Γ(q)

Z

t− T n

0

for t ∈ [−r, Tn ], (t − s)q−1 f (t, xns )ds, for t ∈ [ Tn , T ],

where ϕ0 ∈ C([−r, a), X) denotes the function defined by 0

ϕ (t) =

ϕ(t), for t ∈ [−r, 0], ϕ(0), for t ∈ [0, a).

Using the similar method as we did in (3.15), we get that xn ∈ Bk for all n ≥ 1. Let A = {xn : n ≥ 1}. It follows that the set A is uniformly bounded. Further, we show that the set A is equicontinuous on [−r, T ]. If −r ≤ t1 < t2 ≤ Tn , then for each xn ∈ A, we have limt1 →t2 |xn (t2 ) − xn (t1 )| = limt1 →t2 |ϕ0 (t2 ) − ϕ0 (t1 )| = 0 independently of xn ∈ A. Next, if −r ≤ t1 ≤ Tn
0, we can find δ sufficiently small such that δ (1+b1 )(1−q1 ) ǫ km1 k q1 < . L 1 [0,T ] Γ(q)(1 + b1 )1−q1 2 Therefore, for each t ∈ (0, T ], we can choose Nδ ≥ 1 such that Tn ≤ δ for n ≥ Nδ . Then we obtain that Z t 1 q−1 n (t − s) f (s, xs )ds : n ≥ Nδ α Γ(q) t− Tn Z t 2 ≤ sup (t − s)q−1 m1 (s)ds Γ(q) n≥Nδ t− Tn < ǫ, for each t ∈ (0, T ]. Hence, by the properties (iii) and (v) of the measure of noncompactness, it follows that Z t 1 (t − s)q−1 f (s, xns )ds : n ≥ 1 < ǫ. α Γ(q) t− Tn Then, we obtain that Z t 1 α(A(t)) ≤ α (t − s)q−1 f (s, xns )ds : n ≥ 1 + ǫ, Γ(q) 0 for t ∈ (0, T ]. By Property 1.28 and (H3), we have that Z t 2 (t − s)q−1 α(f (s, As ))ds + ǫ α(A(t)) ≤ Γ(q) 0 Z t 2 ≤ (t − s)q−1 m2 (s)α(As )ds + ǫ, Γ(q) 0

where t ∈ (0, T ]. Since xn (θ) = ϕ(θ), θ ∈ [−r, 0], we have α({xn (θ) : n ≥ 1}) = 0 for θ ∈ [−r, 0]. Moreover, by Property 1.27, for s ∈ [0, t] with t ∈ (0, T ], we deduce that α(As ) = max α({xns (θ) : n ≥ 1}) ≤ sup α({xn (s) : n ≥ 1}) = sup α(A(s)). θ∈[−r,0]

s∈[0,t]

s∈[0,t]

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Since ǫ is arbitrary, we have that α(A(t)) ≤

2T (1+b2 )(1−q2 ) km2 k q1 sup α(A(s)), L 2 [0,T ] s∈[0,t] Γ(q)(1 + b2 )1−q2

q−1 where t ∈ (0, T ] and b2 = 1−q ∈ (−1, 0). 2 n Since (3.14) and x0 = ϕ, we must have that α(A(t)) = 0 for every t ∈ [−r, T ]. Then, by Property 1.27, we have that α(A) = supt∈[−r,T ] α(A(t)) = 0. Therefore, A is a relatively compact subset of Bk . Then, there exists a subsequence if necessary, we may assume that the sequence {xn }n≥1 converges uniformly on [−r, T ] to a continuous function x ∈ Bk with x(θ) = ϕ(θ), θ ∈ [−r, 0]. Moreover, for t ∈ [0, Tn ], we have Z Tn Z Tn 1 1 (t − s)q−1 |f (t, xns )|ds ≤ (t − s)q−1 m1 (s)ds |(F xn )(t) − xn (t)| ≤ Γ(q) 0 Γ(q) 0

and for t ∈ [ Tn , T ], we have

Z Z t− T n 1 t q−1 n q−1 n (t − s) f (t, x )ds − (t − s) f (t, x )ds s s Γ(q) 0 0 Z t 1 (t − s)q−1 f (t, xns )ds = Γ(q) t− Tn Z t 1 ≤ (t − s)q−1 m1 (s)ds. Γ(q) t− Tn

|(F xn )(t) − xn (t)| =

Therefore, it follows that sup |(F xn )(t) − xn (t)| → 0 as n → ∞.

(3.16)

t∈[0,T ]

Since sup |(F x)(t) − x(t)| ≤ sup |(F x)(t) − (F xn )(t)|

t∈[0,T ]

t∈[0,T ]

+ sup |(F xn )(t) − xn (t)| + sup |xn (t) − x(t)|, t∈[0,T ]

t∈[0,T ]

then, by (3.16) and the fact that F is a continuous operator, we obtain that supt∈[0,T ] |(F x)(t) − x(t)| = 0. It follows that x(t) = (F x)(t) for every t ∈ [0, T ]. Hence  for t ∈ [−r, 0],  ϕ(t), Z t x(t) = 1  ϕ(0) + (t − s)q−1 f (t, xs )ds, for t ∈ [0, T ] Γ(q) 0 solve IVP (3.10), and this completes the proof.

Corollary 3.5. Assume that hypotheses (H1)-(H3) hold. Then, for every ϕ ∈ C, there exist T ∈ (0, a) and a sequence of continuous function xn : [−r, T ] → X, such that

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(i) xn (t) are absolutely continuous on [0, T ]; (ii) xn0 = ϕ, for every n ≥ 1, and (iii) extracting a subsequence which is labeled in the same way such that xn (t) → x(t) uniformly on [−r, T ] and x : [−r, T ] → X is a solution for IVP (3.10). We now give an example to illustrate the application of our abstract results. Example 3.6. Consider the infinite system of fractional functional differential equations  1 1 2  x2 (t − r), for t ∈ (0, a), C 0 Dt xn (t) = nt1/3 n (3.17)   x (θ) = ϕ(θ) = θ , for θ ∈ [−r, 0], n = 1, 2, 3, ... . n n

Let E = c0 = {x = (x1 , x2 , x3 , ...) : xn → 0} with norm |x| = supn≥1 |xn |. Then the infinite system (3.17) can be regarded as a IVP of form (3.10) in E. In this situation, q = 21 , x = (x1 , ..., xn , ...), xt = x(t − r) = (x1 (t − r), ..., xn (t − r), ...), ϕ(θ) = (θ, 2θ , ..., nθ , ...) for θ ∈ [−r, 0] and f = (f1 , ..., fn , ...), in which fn (t, xt ) =

1 x2 (t − r). nt1/3 n

(3.18)

It is obvious that conditions (H1) and (H2) are satisfied. Now, we check the condition (H3) and the argument is similar to Section 2.4. Let t ∈ (0, a), R > 0 be (m) (m) given and {w(m) } be any sequence in f (t, B), where w(m) = (w1 , ..., wn , ...) and B = {z ∈ C: kzk∗ ≤ R} is a bounded set in C. By (3.18), we have 0 ≤ wn(m) ≤

R2 , nt1/3

n, m = 1, 2, 3, ... .

(3.19)

(m)

So, {wn } is bounded and, by the diagonal method, we can choose a subsequence {mi } ⊂ {m} such that wn(mi ) → wn as i → ∞,

n = 1, 2, 3, ...,

(3.20)

which implies by virtue of (3.19) that 0 ≤ wn ≤

R2 , nt1/3

n = 1, 2, 3, ... .

(3.21)

Hence w = (w1 , ..., wn , ...) ∈ c0 . It is easy to see from (3.19)-(3.21) that |w(mi ) − w| = sup |wn(mi ) − wn | → 0 n

as i → ∞.

(3.22)

Thus, we have proved that f (t, B) is relatively compact in c0 for t ∈ (0, a), which means that f (t, B) = 0 for almost all t ∈ [0, a) and B a bounded set in C. Hence, the condition (H3) is satisfied. Finally, from Theorem 3.4, we can conclude that the infinite system (3.17) has a continuous solution.

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3.3 3.3.1


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Cauchy Problems via Topological Degree Method Introduction

It is well known that the topological methods proved to be a powerful tool in the study of various problems which appear in nonlinear analysis. Particularly, a priori estimate method has been often used together with the Brouwer degree, the Leray-Schauder degree or the coincidence degree in order to prove the existence of solutions for some boundary value problems and bifurcation problems for nonlinear differential equations or nonlinear partial differential equations. See, for example, Fe˘ckan, 2008; Mawhin, 1979. In Section 3.3, we will consider the following nonlocal problem via a coincidence degree for condensing mapping in a Banach space X C q 0 Dt u(t) = f (t, u(t)), t ∈ J := [0, T ], (3.23) u(0) + g(u) = u0 ,

q where C 0 Dt is Caputo fractional derivative of order q ∈ (0, 1), u0 is an element of X, f : J × X → X is continuous. The nonlocal term g : C(J, X) → X is a given function, here C(J, X) is the Banach space of all continuous functions from J into X with the norm kuk := supt∈J |u(t)| for u ∈ C(J, X).

3.3.2

Qualitative Analysis

This subsection deals with existence of solutions for the nonlocal problem (3.23). Definition 3.7. A function u ∈ C 1 (J, X) is said to be a solution of the nonlocal q problem (3.23.) if u satisfies the equation C 0 Dt u(t) = f (t, u(t)) a.e. on J, and the condition u(0) + g(u) = u0 . Lemma 3.8. A function u ∈ C(J, X) is a solution of the fractional integral equation Z t 1 (t − s)q−1 f (s, u(s))ds, (3.24) u(t) = u0 − g(u) + Γ(q) 0 if and only if u is a solution of the nonlocal problem (3.23). We make some following assumptions: (H1) for arbitrary u, v ∈ C(J, X), there exists a constant Kg ∈ [0, 1) such that |g(u) − g(v)| ≤ Kg ku − vk; (H2) for arbitrary u ∈ C(J, X), there exist Cg , Mg > 0, q1 ∈ [0, 1) such that |g(u)| ≤ Cg kukq1 + Mg ;

(H3) for arbitrary (t, u) ∈ J × X, there exist Cf , Mf > 0, q2 ∈ [0, 1) such that |f (t, u)| ≤ Cf |u|q2 + Mf ;

(H4) for any r > 0, there exists a constant βr > 0 such that α(f (s, M)) ≤ βr α(M),

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for all t ∈ J, M ⊂ Br := {kuk ≤ r : u ∈ C(J, X)} and 2T q βr < 1. Γ(q + 1) Under the assumptions (H1)-(H4), we will show that fractional integral equation (3.24) has at least one solution u ∈ C(J, X). Define operators F : C(J, X) → C(J, X), (F u)(t) = u0 − g(u), t ∈ J, Z t 1 G : C(J, X) → C(J, X), (Gu)(t) = (t − s)q−1 f (s, u(s))ds, t ∈ J, Γ(q) 0 T : C(J, X) → C(J, X), Tu = F u + Gu.

It is obvious that T is well defined. Then, fractional integral equation (3.24) can be written as the following operator equation u = Tu = F u + Gu.

(3.25)

Thus, the existence of a solution for the nonlocal problem (3.23) is equivalent to the existence of a fixed point for operator T. Lemma 3.9. The operator F : C(J, X) → C(J, X) is Lipschitz with constant Kg . Consequently F is α-Lipschitz with the same constant Kg . Moreover, F satisfies the following growth condition: for every u ∈ C(J, X).

kF uk ≤ |u0 | + Cg kukq1 + Mg ,

(3.26)

Proof. Using (H1), we have kF u − F vk ≤ |g(u) − g(v)| ≤ Kg ku − vk, for every u, v ∈ C(J, X). By Property 1.33, F is α-Lipschitz with constant Kg . Relation (3.26) is a simple consequence of (H2). Lemma 3.10. The operator G : C(J, X) → C(J, X) is continuous. Moreover, G satisfies the following growth condition: T q (Cf kukq2 + Mf ) , (3.27) kGuk ≤ Γ(q + 1) for every u ∈ C(J, X). Proof. For that, let {un } be a sequence of a bounded set BK ⊆ C(J, X) such that un → u in BK (K > 0). We have to show that kGun − Guk → 0. It is easy to see that f (s, un (s)) → f (s, u(s)) as n → ∞ due to the continuity of f . On the one hand, using (H3), we get for each t ∈ J, (t − s)q−1 |f (s, un (s)) − f (s, u(s))| ≤ (t − s)q−1 2(Cf K q2 + Mf ). On the other hand, using the fact that the function s → (t − s)q−1 2(Cf K q2 + Mf ) is integrable ∈ [0, t], t ∈ J, R t for sq−1 the Lebesgue’s dominated convergence theorem yields 0 (t − s) |f (s, un (s)) − f (s, u(s))|ds → 0 as n → ∞. Then, for all t ∈ J, Z t 1 |(Gun )(t) − (Gu)(t)| ≤ (t − s)q−1 |f (s, un (s)) − f (s, u(s))|ds → 0. Γ(q) 0 Therefore, Gun → Gu as n → ∞ which implies that G is continuous. Relation (3.27) is a simple consequence of (H3).

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Lemma 3.11. The operator G : C(J, X) → C(J, X) is compact. Consequently G is α-Lipschitz with zero constant. Proof. In order to prove the compactness of G, we consider a bounded set M ⊆ C(J, X) and the key step is to show that G(M) is relatively compact in C(J, X). Let {un } be a sequence on M ⊂ BK , for every un ∈ M. By relation (3.27), we have T q (Cf K q2 + Mf ) =: r, kGun k ≤ Γ(q + 1) for every un ∈ M, so G(M) is bounded in Br . Now we prove that {Gun } is equicontinuous. For 0 ≤ t1 < t2 ≤ T , we get |(Gun )(t1 ) − (Gun )(t2 )| Z t1 1 ≤ ((t1 − s)q−1 − (t2 − s)q−1 )|f (s, un (s))|ds Γ(q) 0 Z t2 1 + (t2 − s)q−1 |f (s, un (s))|ds Γ(q) t1 Z t1 1 ≤ ((t1 − s)q−1 − (t2 − s)q−1 )(Cf |un (s)|q2 + Mf )ds Γ(q) 0 Z t2 1 + (t2 − s)q−1 (Cf |un (s)|q2 + Mf )ds Γ(q) t1 tq (t2 − t1 )q (t2 − t1 )q (Cf K q2 + Mf ) tq1 − 2+ + ≤ Γ(q) q q q q q2 q 2(Cf K + Mf )(t2 − t1 ) . ≤ Γ(q + 1) As t2 → t1 , the right-hand side of the above inequality tends to zero. Therefore {Gun } is equicontinuous. Consider a bounded set Z t 1 q−1 M(t) := vn (t) : vn (t) = (t − s) f (s, vn (s))ds ⊂ Br . Γ(q) 0

Applying Property 1.27, we know that the function t → α(M(t)) is continuous on J. Moreover, (t − s)q−1 |f (s, vn (s))| ≤ (t − s)q−1 (Cf rq2 + Mf ) ∈ L1 (J, R+ ), f or s ∈ [0, t], t ∈ J. Using (H4) and Property 1.28, we have Z t 1 q−1 (t − s) f (s, M(s))ds α(M(t)) ≤ α Γ(q) 0 Z t 2 (t − s)q−1 α (f (s, M(s))) ds ≤ Γ(q) 0 Z t 2βr ≤ (t − s)q−1 α(M(s))ds, Γ(q) 0

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which implies that

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Z t 2βr (t − s)q−1 ds α(M) Γ(q) 0 2T q βr α(M) ≤ Γ(q + 1) < α(M),

α(M) ≤

due to the condition

2T q βr < 1. Γ(q + 1) Then we can deduce that α(M) = 0. Therefore, G(M) is a relatively compact subset of C(J, X), and so, there exists a subsequence vn which converge uniformly on J to some v∗ ∈ C(J, X) together with the Arzela-Ascoli theorem. By Property 1.32, G is α-Lipschitz with zero constant. Theorem 3.12. Assume that (H1)-(H4) hold, then the nonlocal problem (3.23) has at least one solution u ∈ C(J, X) and the set of the solutions of system (3.23) is bounded in C(J, X). Proof. Let F, G, T : C(J, X) → C(J, X) be the operators defined in the beginning of this section. They are continuous and bounded. Moreover, F is α-Lipschitz with constant Kg ∈ [0, 1) and G is α-Lipschitz with zero constant (see Lemmas 3.9-3.11). Property 1.31 shows that T is a strict α-contraction with constant Kg . Set S0 = {u ∈ C(J, X) : (∃) λ ∈ [0, 1] such that u = λTu}. Next, we prove that S0 is bounded in C(J, X). Consider u ∈ S0 and λ ∈ [0, 1] such that u = λTu. It follows from (3.26) and (3.27) that kuk = λkTuk ≤ λ(kF uk + kGuk) (3.28) T q (Cf kukq2 +Mf ) ≤ |u0 | + Cg kukq1 + Mg + . Γ(q+1) This inequality (3.28), together with q1 < 1 and q2 < 1, shows that S0 is bounded in C(J, X). If not, we suppose by contradiction, ρ := kuk → ∞. Dividing both sides of (3.28) by ρ, and taking ρ → ∞, we have T q (Cf ρq2 +Mf ) |u0 | + Cg ρq1 + Mg + Γ(q+1) = 0. (3.29) 1 ≤ lim ρ→∞ ρ This is a contradiction. Consequently, by Theorem 1.35 we deduce that T has at least one fixed point and the set of the fixed points of T is bounded in C(J, X). Remark 3.13.

(i) If the growth condition (H2) is formulated for q1 = 1, then the conclusions of Theorem 3.12 remain valid provided that Cg < 1; (ii) If the growth condition (H3) is formulated for q2 = 1, then the conclusions of T q Cf Theorem 3.12 remain valid provided that Γ(q+1) < 1; (iii) If the growth conditions (H2) and (H3) are formulated for q1 = 1 and qq2 = 1, T Cf < then the conclusions of Theorem 3.12 remain valid provided that Cg + Γ(q+1) 1.

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3.4 3.4.1


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Cauchy Problems via Picard Operators Technique Introduction

Assume that (X, | · |) is a Banach space, and J := [0, T ], T > 0. Let C(J, X) be the Banach space of all continuous functions from J into X with the norm kxk := sup{|x(t)| : t ∈ J} for x ∈ C(J, X). Consider the following Cauchy problem of fractional differential equation C q 0 Dt x(t) = f (t, x(t)), a.e. t ∈ J, (3.30) x(0) = x0 ∈ X,

q where C 0 Dt is Caputo fractional derivative of order q ∈ (0, 1), the function f : J × X → X satisfies some assumptions that will be specified later. To our knowledge, Picard and weakly Picard operators technique due to Rus 1979, 1987, 1993, 2003; Rus and Muresan, 2000 have been used to study the existence for the solutions of some integer differential equations (see, Mure¸san, 2004; S¸erban, Rus and Petru¸sel, 2010). In the present section we consider suitable Bielecki norms in some convenient spaces and obtain existence, uniqueness and data dependence results for the solutions of the fractional Cauchy problem (3.30) via Picard and weakly Picard operators technique. Definition 3.14. A function x ∈ C 1 (J, X) is said to be a solution of the fractional q Cauchy problem (3.30) if x satisfies the equation C 0 Dt x(t) = f (t, x(t)) a.e. on J, and the condition x(0) = x0 .

3.4.2

Results via Picard Operators

Consider a Banach space (X, | · |), let k · kB and k · kC be the Bielecki and Chebyshev norms on C(J, X) defined by kxkB = max |x(t)|e−τ t (τ > 0) and kxkC = max |x(t)| t∈J

t∈J

and denote by dB and dC their corresponding metrics. We consider the set n o ∗ ∗ CLq−q (J, X) = x ∈ C(J, X) : |x(t1 ) − x(t2 | ≤ L|t1 − t2 |q−q for all t1 , t2 ∈ J where L > 0, q ∗ ∈ (0, q), and ¯ 1 − t2 |q for all t1 , t2 ∈ J CLq¯ (J, X) = x ∈ C(J, X) : |x(t1 ) − x(t2 )| ≤ L|t

¯ > 0, and where L ¯ 1 − t2 |q for all t1 , t2 ∈ J CLq¯ (J, BR ) = x ∈ C(J, BR ) : |x(t1 ) − x(t2 )| ≤ L|t

where BR = {x ∈ X : |x| ≤ R} with R > 0. ∗ If d ∈ {dC , dB }, then (C(J, X), d), (CLq−q (J, X), d), (CLq¯ (J, X), d) and (CLq¯ (J, BR ), d) are complete metric spaces. 1

Let qi ∈ (0, q), i = 1, 2, 3 and the functions m(t) ∈ L q1 (J, R+ ), η(t) ∈ 1 1 L q2 (J, R+ ), µ(t) ∈ L q3 (J, R+ ) and l(t) ∈ C(J, R+ ).

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For brevity, let M = kmk

1

L q1 J

, N = kηk

1

L q2 J

, V = kµk

1

L q2 J

, L0 = max{l(t)}, t∈J

q−1 q−1 q−1 β= ∈ (−1, 0), γ = ∈ (−1, 0), ν = ∈ (−1, 0). 1 − q1 1 − q2 1 − q3 Lemma 3.15. A function x ∈ C(J, X) is a solution of the fractional integral equation Z t 1 x(t) = x0 + (t − s)q−1 f (s, x(s))ds, (3.31) Γ(q) 0 if and only if x is a solution of the fractional Cauchy problem (3.30). Theorem 3.16. Suppose the following conditions hold: (C1) f ∈ C(J × X, X); 1 (C2) there exist a constant q1 ∈ (0, q) and function m(·) ∈ L q1 (J, R+ ) such that |f (t, x)| ≤ m(t) for all x ∈ X and all t ∈ J; (C3) There exists a constant L > 0 such that L≥

2M ; Γ(q)(1 + β)1−q1

(C4) there exists a function l(·) ∈ C(J, R+ ) such that |f (t, u1 ) − f (t, u2 )| ≤ l(t)|u1 − u2 | for all ui ∈ X (i = 1, 2) and all t ∈ J; (C5) there exist constants q1 and τ such that L0 T (1+β)(1−q1 ) q1 q1 < 1. Γ(q) (1 + β)1−q1 τ

Then the fractional Cauchy problem (3.30) has a unique solution x∗ in CLq−q1 (J, X), and this solution can be obtained by the successive approximation method, starting from any element of CLq−q1 (J, X). Proof.

Consider the operator A : (CLq−q1 (J, X), k · kB ) → (CLq−q1 (J, X), k · kB )

defined by Ax(t) = x0 +

1 Γ(q)

Z

0

t

(t − s)q−1 f (s, x(s)) ds.

It is easy to see the operator A is well defined due to (C1). Firstly, we check that Ax ∈ C(J, X) for every x ∈ CLq−q1 (J, X).

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For any δ > 0, every x ∈ CLq−q1 (J, X), by (C2) and Hölder inequality, |(Ax)(t + δ) − (Ax)(t)| Z t 1 ≤ ((t − s)q−1 − (t + δ − s)q−1 )|f (s, x(s))|ds Γ(q) 0 Z t+δ 1 (t + δ − s)q−1 |f (s, x(s))|ds + Γ(q) t Z t Z t+δ 1 1 ≤ ((t − s)q−1 − (t + δ − s)q−1 )m(s)ds + (t + δ − s)q−1 m(s)ds Γ(q) 0 Γ(q) t Z t 1−q1 Z t q1 1 1 1 q−1 q−1 1−q1 q [(t − s) − (t + δ − s) ] ≤ ds (m(s)) 1 ds Γ(q) 0 0 !1−q1 Z !q1 Z t+δ t+δ 1 1 1 q−1 1−q [(t + δ − s) ] 1 ds + (m(s)) q1 ds Γ(q) t t !1−q1 Z t Z t+δ 1−q1 M M β β β (t − s) − (t + δ − s) ds + (t + δ − s) ds ≤ Γ(q) Γ(q) 0 t

1−q1 M M |t1+β − (t + δ)1+β | + δ 1+β + δ (1+β)(1−q1 ) Γ(q)(1 + β)1−q1 Γ(q)(1 + β)1−q1 2M M ≤ δ (1+β)(1−q1 ) + δ (1+β)(1−q1 ) Γ(q)(1 + β)1−q1 Γ(q)(1 + β)1−q1 3M ≤ δ (1+β)(1−q1 ) . Γ(q)(1 + β)1−q1 ≤

It is easy to see that the right-hand side of the above inequality tends to zero as δ → 0. Therefore Ax ∈ C(J, X). Secondly, we show that Ax ∈ CLq−q1 (J, X). Without loss of generality, for any t1 < t2 , t1 , t2 ∈ J, applying (C2) and Hölder inequality, we have |(Ax)(t2 ) − (Ax)(t1 )| Z Z t2 1 t1 q−1 q−1 q−1 [(t − s) − (t − s) ]f (s, x(s))ds + (t − s) f (s, x(s))ds ≤ 2 1 2 Γ(q) 0 t1 Z t1 1 ≤ [(t1 − s)q−1 − (t2 − s)q−1 ]|f (s, x(s))|ds Γ(q) 0 Z t2 1 (t2 − s)q−1 |f (s, x(s))|ds + Γ(q) t1 Z t1 Z t2 1 1 ≤ [(t1 − s)q−1 − (t2 − s)q−1 ]m(s)ds + (t2 − s)q−1 m(s)ds Γ(q) 0 Γ(q) t1 1−q1 Z t1 q1 Z t1 1 1 1 ≤ (m(s)) q1 ds [(t1 − s)q−1 − (t2 − s)q−1 ] 1−q1 ds Γ(q) 0 0

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1−q1 Z t2 q1 Z t2 1 1 1 (m(s)) q1 ds [(t2 − s)q−1 ] 1−q1 ds Γ(q) t1 t1 Z t1 Z t2 1−q1 1−q1 M M (t1 − s)β − (t2 − s)β ds + (t2 − s)β ds ≤ Γ(q) Γ(q) 0 t1 1−q1 M ≤ t1+β − t1+β + (t2 − t1 )1+β 2 Γ(q)(1 + β)1−q1 1 M + (t2 − t1 )(1+β)(1−q1 ) Γ(q)(1 + β)1−q1 2M ≤ |t1 − t2 |(1+β)(1−q1 ) Γ(q)(1 + β)1−q1 2M |t1 − t2 |q−q1 . ≤ Γ(q)(1 + β)1−q1 Similarly, for any t1 > t2 , t1 , t2 ∈ J, we also have the above inequality. This implies that Ax is belong to CLq−q1 (J, X) due to (C3). Thirdly, A is continuous. For that, let {xn } be a sequence of BR such that xn → x in BR . Then, f (s, xn (s)) → f (s, x(s)) as n → ∞ due to (C1). On the one hand, by using (C2), we get for each s ∈ [0, t], |f (s, xn (s)) − f (s, x(s))| ≤ 2m(s). On the other hand, using the fact that the function s → 2(t − s)q−1 m(s) is integrable on [0, t], the Lebesgue’s dominated convergence theorem yields Z t (t − s)q−1 |f (s, xn (s)) − f (s, x(s))|ds → 0. +

0

For all t ∈ J, we have

Z t 1 (t − s)q−1 |f (s, xn (s)) − f (s, x(s))|ds. Γ(q) 0 Thus, Axn → Ax as n → ∞ which implies that A is continuous. Moreover, for all x, z ∈ CLq−q1 (J, X), using (C4) and Hölder inequality we have Z t 1 |(Ax)(t) − (Az)(t)| ≤ (t − s)q−1 |f (s, x(s)) − f (s, z(s))|ds Γ(q) 0 Z t 1 (t − s)q−1 l(s)|x(s) − z(s)|ds ≤ Γ(q) 0 Z t 1 ≤ (t − s)q−1 max {l(s)} |x(s) − z(s)|e−τ s eτ s ds Γ(q) 0 s∈[0,t] Z t L0 ≤ kx − zkB (t − s)q−1 eτ s ds Γ(q) 0 q1 Z t 1−q1 Z t τs L0 β q1 kx − zkB (t − s) ds e ds ≤ Γ(q) 0 0 L0 T (1+β)(1−q1 ) q1 q1 τ t e kx − zkB . ≤ Γ(q) (1 + β)1−q1 τ |(Axn )(t) − (Ax)(t)| ≤

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It follows that |(Ax)(t) − (Az)(t)|e−τ t ≤ for all t ∈ J. So we have

L0 T (1+β)(1−q1 ) q1 q1 kx − zkB Γ(q) (1 + β)1−q1 τ

L0 T (1+β)(1−q1 ) q1 q1 kx − zkB Γ(q) (1 + β)1−q1 τ for all x, z ∈ CLq−q1 (J, X). The operator A is of Lipschitz type with constant L0 T (1+β)(1−q1 ) q1 q1 (3.32) LA = Γ(q) (1 + β)1−q1 τ and 0 < LA < 1 due to (C5). By applying the Contraction Principle to this operator we obtain that A is a Picard operator. This completes the proof. kAx − AzkB ≤

Example 3.17. Consider the fractional Cauchy problem C q 1 0 Dt x(t) = x(t), q = 2 , (3.33) x(0) = 0 ∈ X, on [0, 1]. Set L0 = 1, T = 1, q1 = 13 , then β = − 34 . Indeed qL0 T (1+β)(1−q1 ) q1 q1 L0 T (1+β)(1−q1 ) q1 q1 < 1 ⇐⇒ < 1, Γ(q) (1 + β)1−q1 τ Γ(q + 1) (1 + β)1−q1 τ 1 13 1 which implies that we must choose a suitable τ0 > 0 such that Γ(23 ) 11 2 τ30 < 1. Γ( 23 )

2

√

π 2 ,

16 9

(4)3

16 √ 3 π3

we have the condition (C5) in Theorem Noting that = for τ0 = > 3.16. Theorem 3.18. Suppose the following conditions hold: (C1) f ∈ C(J × X, X); ¯ > 0 such that |f (t, x)| ≤ M ¯ for all x ∈ X and all (C2)′ there exists a constant M t ∈ J; ¯ > 0 such that L ¯ ≥ 2M¯ ; (C3)′ there exists a constant L Γ(q+1) ¯ 0 > 0 such that |f (t, u1 ) − f (t, u2 )| ≤ L ¯ 0 |u1 − u2 | for (C4)′ there exists a constant L all ui ∈ X (i = 1, 2) and all t ∈ J; 1) q1 q1 ¯ A¯ = L¯ 0 T (1+β)(1−q < 1. (C5)′ there exist constants q1 and τ such that L Γ(q) (1+β)1−q1 τ q ∗ Then the fractional Cauchy problem (3.30) has a unique solution x in CL¯ (J, X), and this solution can be obtained by the successive approximation method, starting from any element of CLq¯ (J, X). Proof.

Consider the following continuous operator A¯ : (C q¯ (J, X), k · kB ) → (C q¯ (J, X), k · kB ) L

defined by

L

Z t 1 (t − s)q−1 f (s, x(s)) ds. Γ(q) 0 As the proof in Theorem 3.16, applying the given conditions one can verify that ¯ 0 T (1+β)(1−q1 ) q1 q1 L ¯ ¯ kx − zkB kA(x) − A(z)k B ≤ Γ(q) (1 + β)1−q1 τ q for all x, z ∈ CL¯ (J, X). So, the operator A¯ is a Picard operator. ¯ Ax(t) = x0 +

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Similarly, we can prove Theorem 3.19. Suppose the following conditions hold: (C1)′ f ∈ C(J × BR , X); ¯ (R) > 0 such that |f (t, x)| ≤ M ¯ (R) for all x ∈ BR (C2)′′ there exists a constant M q ¯ M(R)T and all t ∈ J with R ≥ |x0 | + Γ(q+1) ; ¯ > 0 such that L ¯ ≥ 2M¯ (R) ; (C3)′′ there exists a constant L Γ(q+1) ¯ 0 > 0 such that |f (t, u1 ) − f (t, u2)| ≤ L ¯ 0 |u1 − u2 | for (C4)′′ there exists a constant L

all ui ∈ BR (i = 1, 2) and all t ∈ J; 1) q1 q1 ¯ A¯ = L¯ 0 T (1+β)(1−q (C5)′ there exist constants q1 and τ such that L < 1. Γ(q) (1+β)1−q1 τ q ∗ Then the fractional Cauchy problem (3.30) has a unique solution x in CL¯ (J, BR ), and this solution can be obtained by the successive approximation method, starting from any element of CLq¯ (J, BR ). Consider the following fractional Cauchy problem C q 0 Dt x(t) = g(t, x(t)), t ∈ J, (3.34) x(0) = y0 ∈ X, where g ∈ C(J × X, X). By Lemma 3.15, a function x ∈ C(J, X) is a solution of the fractional integral equation Z t 1 x(t) = y0 + (t − s)q−1 g (s, x(s)) ds, (3.35) Γ(q) 0 if and only if x is a solution of the fractional Cauchy problem (3.34). Now, we consider both fractional integral equation (3.31) and (3.35). Theorem 3.20. Suppose the following: (D1) all conditions in Theorem 3.16 are satisfied and x∗ ∈ CLq−q1 (J, X) is the unique solution of the fractional integral equation (3.31); (D2) with the same function m(·) as in Theorem 3.16, |g(t, x)| ≤ m(t) for all x ∈ X and all t ∈ J; (D3) with the same function l(·) as in Theorem 3.16, |g(t, u1 ) − g(t, u2 )| ≤ l(t)|u1 − u2 | for all ui ∈ X (i = 1, 2) and all t ∈ J; 2M (D4) L ≥ Γ(q)(1+β) 1−q1 ; 1

(D5) there exists a constant q2 ∈ (0, q) and function η(·) ∈ L q2 (J, R+ ) such that |f (t, u) − g(t, u)| ≤ η(t) for all u ∈ X and all t ∈ J. If y ∗ is the solution of the fractional integral equation (3.35), then ∗

∗

kx − y kB ≤

|x0 − y0 | +

N T (1+γ)(1−q2 ) Γ(q)(1+γ)1−q2

1 − LA

,

where LA is given by (3.32) with τ = τ0 > 0 such that 0 < LA < 1. Proof.

Consider the following two operators A, B : (CLq−q1 (J, X), k · kB ) → (CLq−q1 (J, X), k · kB )

(3.36)

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defined by 1 Ax(t) = x0 + Γ(q) 1 Bx(t) = y0 + Γ(q) on J. We have

Z

t

0

Z

1 |Ax(t) − Bx(t)| ≤ |x0 − y0 | + Γ(q)

(t − s)q−1 f (s, x(s)) ds,

t

(t − s)q−1 g (s, x(s)) ds,

0

Z

t

0

Z

(t − s)q−1 |f (s, x(s)) − g (s, x(s)) |ds

t

≤ |x0 − y0 | +

1 Γ(q)

≤ |x0 − y0 | +

NT , Γ(q)(1 + γ)1−q2

(t − s)q−1 η(s)ds

0 (1+γ)(1−q2 )

for t ∈ J. It follows that kAx − BxkB ≤ |x0 − y0 | +

N T (1+γ)(1−q2 ) . Γ(q)(1 + γ)1−q2

So we can apply Theorem 1.38 to obtain (3.36) which completes the proof.

Remark 3.21. All the results obtained in Theorem 3.16 hold even if the condition (C2) is replaced by the following: 1 (C2-E) there exist a constant q1 ∈ [0, q) and function m(·) ∈ L q1 (J, R+ ) such that |f (t, x)| ≤ m(t) for all x ∈ X and all t ∈ J. In fact, we only need extend the space Lp (J, R+ ) (1 < p < ∞) to Lp (J, R+ ) (1 ≤ p ≤ ∞) where Lp (J, R+ ) (1 ≤ p ≤ ∞) be the Banach space of all Lebesgue measurable functions φ : J → R+ with kφkLp J < ∞. 3.4.3

Results via Weakly Picard Operators

Now, we consider another fractional integral equation Z t 1 (t − s)q−1 f (s, x(s)) ds x(t) = x(0) + Γ(q) 0

(3.37)

on J, where f ∈ C(J × X, X) is as in the fractional Cauchy problem (3.30). Theorem 3.22. Suppose that for the fractional integral equation (3.37) the same conditions as in Theorem 3.16 are satisfied. Then this equation has solutions in CLq−q1 (J, X). If S ⊂ CLq−q1 (J, X) is its solutions set, then card S = card X. Proof.

Consider the operator A∗ : (CLq−q1 (J, X), k · kB ) → (CLq−q1 (J, X), k · kB )

defined by A∗ x(t) = x(0) +

1 Γ(q)

Z

0

t

(t − s)q−1 f (s, x(s)) ds.

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This is a continuous operator, but not a Lipschitz one. We can write [ CLq−q1 (J, X) = Xα , Xα = x ∈ CLq−q1 (J, X) : x(0) = α . α∈X

We have that Xα is an invariant set of A∗ and we apply Theorem 3.16 to A∗ |Xα . By using Theorem 1.40 we obtain that A∗ is a weakly Picard operator. Consider the operator q−q1 n A∞ (J, X) → CLq−q1 (J, X), A∞ ∗ : CL ∗ x = lim A∗ x.

From

An+1 (x) ∗

=

A∗ (An∗ (x)) and the continuity of q−q1 A∞ (J, X)) = FA∗ = S ∗ (CL

So, card S = card X.

n→∞ ∞ A∗ , A∗ (x) ∈

FA∗ . Then

and S = 6 ∅.

Theorem 3.23. Suppose that for the fractional integral equation (3.37) the same conditions as in Theorem 3.18 are satisfied. Then this equation has solutions in CLq¯ (J, X). If S ⊂ CLq¯ (J, X) is its solutions set, then card S = card X. Proof. As the proof in Theorem 3.22, we need to consider the continuous operator (but not a Lipschitz one) A¯∗ : (C q¯ (J, X), k · kB ) → (C q¯ (J, X), k · kB ) L

L

defined by

Z t 1 (t − s)q−1 f (s, x(s)) ds. Γ(q) 0 S ¯α, X ¯ α = x ∈ C q¯ (J, X) : x(0) = α . We have We can write CLq¯ (J, X) = α∈X X L ¯ α is an invariant set of A¯∗ and we apply Theorem 3.18 to A¯∗ |Xα . By using that X Theorem 1.40 we obtain that A¯∗ is a weakly Picard operator. Consider the operator q q ¯∞ ¯n ¯n+1 (x) = A¯∗ (A¯n∗ (x)) A¯∞ ¯ (J, X) → CL ¯ (J, X), A∗ (x) = limn→∞ A∗ (x). From A∗ ∗ : CL q ∞ ∞ ¯ ¯ ¯ and the continuity of A∗ , A∗ (x) ∈ FA¯ . Then A∗ (C ¯ (J, X)) = FA¯ = S and S 6= A¯∗ x(t) = x(0) +

∅. So, card S = card X.

∗

L

∗

Similarly as above, we can prove Theorem 3.24. Suppose that for the fractional integral equation (3.37) the same conditions as in Theorem 3.19 are satisfied. Then this equation has solutions in CLq¯ (J, BR ). If S ⊂ CLq¯ (J, BR ) is its solutions set, then card S = card BR . In order to study data dependence for the solutions set of the fractional integral equation (3.37), we consider both (3.37) and the following fractional integral equation Z t 1 (t − s)q−1 g (s, x(s)) ds x(t) = x(0) + Γ(q) 0 on J where g ∈ C(J × X, X). Let S1 be the solutions set of this equation. Theorem 3.25. Suppose the following conditions: (E1) there exists a function l(t) ∈ C(J, R+ ) such that |f (t, u1 ) − f (t, u2 )| ≤ l(t)|u1 − u2 | and |g(t, u1 ) − g(t, u2 )| ≤ l(t)|u1 − u2 |

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for all ui ∈ X (i = 1, 2) and all t ∈ J; 1 1 (E2) there exist q1 , q3 ∈ (0, q) and functions m(t) ∈ L q1 (J, R+ ), µ(t) ∈ L q3 (J, R+ ) such that |f (t, x)| ≤ m(t) and |g(t, x)| ≤ µ(t) for all x ∈ X and all t ∈ J; (E3) there exists a constant L > 0 such that L≥

2 max{M, V } ; Γ(q) min {(1 + β)1−q1 , (1 + ν)1−q3 } 1

(E4) there exist a constant q2 ∈ (0, q) and function η ∈ L q2 (J, R+ ) |f (t, u) − g(t, u)| ≤ η(t) for all u ∈ X and all t ∈ J; L0 T q (E5) Γ(q+1) < 1. Then Hk·kC (S, S1 ) ≤

qN T (1+γ)(1−q2 ) (Γ(q + 1) − L0 T q )(1 + γ)1−q2

where by Hk·kC we denote the Pompeiu-Hausdorff functional with respect to k · kC on CLq−q1 (J, X). Proof.

Consider the operator B∗ : (CLq−q1 (J, X), k · kB ) → (CLq−q1 (J, X), k · kB )

defined by 1 B∗ x(t) = x(0) + Γ(q)

Z

0

t

(t − s)q−1 g (s, x(s)) ds, for t ∈ J.

Because of (E1)-(E3), A∗ , B∗ : (CLq−q1 (J, X), k · kB ) → (CLq−q1 (J, X), k · kB ) are two orbitally continuous operators. Moreover, we have Z t L0 |A2∗ x(t) − A∗ x(t)| ≤ (t − s)q−1 |A∗ x(s) − x(s)|ds Γ(q) 0 L0 T q ≤ kA∗ x − xkC , Γ(q + 1) for all x ∈ CLq−q1 (J, X). Similarly, |B∗2 x(t) − B∗ x(t)| ≤ for all x ∈ CLq−q1 (J, X). It follows that

L0 T q kB∗ x − xkC Γ(q + 1)

L0 T q kA∗ x − xkC , Γ(q + 1) L0 T q kB∗2 x − B∗ xkC ≤ kB∗ x − xkC . Γ(q + 1) kA2∗ x − A∗ xkC ≤

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Because of (E4), 1 Γ(q)

kA∗ x − B∗ xkC ≤

Z

t

(t − s)q−1 η(s)ds

0 (1+γ)(1−q2 )

NT , Γ(q)(1 + γ)1−q2

≤

for all x ∈ CLq−q1 (J, X). By (E5) and applying Theorem 1.39, we obtain qN T (1+γ)(1−q2 ) (Γ(q + 1) − L0 T q )(1 + γ)1−q2

Hk·kC (FA∗ , FB∗ ) ≤ and the theorem is proved.

Theorem 3.26. Suppose the following conditions: (E1)′ there exists a constant L∗ > 0 such that |f (t, u1 ) − f (t, u2 )| ≤ L∗ |u1 − u2 | and |g(t, u1 ) − g(t, u2 )| ≤ L∗ |u1 − u2 | for all ui ∈ X (i = 1, 2) and all t ∈ J; (E2)′ there exists a constant M∗ > 0 such that |f (t, x)| ≤ M∗ and |g(t, x)| ≤ M∗ for all x ∈ X and all t ∈ J; ¯ > 0 such that (E3)′ there exists a constant L ¯≥ L

2M∗ ; Γ(q + 1)

(E4)′ there exists a constant η∗ > 0 such that |f (t, u) − g(t, u)| ≤ η∗ for all u ∈ X and all t ∈ J; L∗ T q < 1. (E5)′ Γ(q+1) Then we have ¯ k·k (S, S1 ) ≤ H C

η∗ T q Γ(q + 1) − L∗ T q

¯ k·k we denote the Pompeiu-Hausdorff functional with respect to k · kC where by H C on CLq¯ (J, X). Proof.

Consider the operator ¯∗ : (C q¯ (J, X), k · kB ) → (C q¯ (J, X), k · kB ) B L L

defined by ¯∗ x(t) = x(0) + B

1 Γ(q)

Z

0

t

(t − s)q−1 g (s, x(s)) ds, for t ∈ J.

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1 1 ¯∗ : (C q−q Applying (E1)′ -(E3)′ , A¯∗ , B (J, X), k · kB ) → (CLq−q (J, X), k · kB ) are two ¯ ¯ L orbitally continuous operators. Moreover, we have L∗ T q kA¯∗ (x) − xkC , |A¯2∗ x(t) − A¯∗ x(t)| ≤ Γ(q + 1) q ¯∗2 x(t) − B ¯∗ x(t)| ≤ L∗ T kB ¯∗ (x) − xkC , |B Γ(q + 1) for all x ∈ CLq¯ (J, X). It follows that

L∗ T q kA¯∗ (x) − xkC Γ(q + 1) q ¯ 2 (x) − B ¯∗ (x)kC ≤ L∗ T kB ¯∗ (x) − xkC . kB ∗ Γ(q + 1) Because of (E4)′ , we obtain Z t η∗ T q ¯∗ (x)kC ≤ 1 (t − s)q−1 η∗ ds ≤ , kA¯∗ (x) − B Γ(q) 0 Γ(q + 1) for all x ∈ CLq¯ (J, X). By (E5)′ and applying Theorem 1.39, we obtain the result and the theorem is proved. kA¯2∗ (x) − A¯∗ (x)kC ≤

Similarly, we can prove Theorem 3.27. Suppose the following: (E1)′′ there exists a constant L∗ > 0 such that |f (t, u1 ) − f (t, u2 )| ≤ L∗ |u1 − u2 | and |g(t, u1 ) − g(t, u2 )| ≤ L∗ |u1 − u2 |

for all ui ∈ BR (i = 1, 2) and all t ∈ J; (E2)′′ there exists a constant M∗ (R) > 0 such that

|f (t, x)| ≤ M∗ (R) and |g(t, x)| ≤ M∗ (R)

for all x ∈ BR and all t ∈ J with

R ≥ |x(0)| +

M∗ (R)T q ; Γ(q + 1)

¯ > 0 such that (E3)′′ there exists a constant L ¯ ≥ 2M∗ (R) ; L Γ(q + 1) ′′ (E4) there exists a constant η∗ > 0 such that for all u ∈ BR and all t ∈ J; L∗ T q < 1. (E5)′′ Γ(q+1) Then

|f (t, u) − g(t, u)| ≤ η∗

η∗ T q Γ(q + 1) − L∗ T q we denote the Pompeiu-Hausdorff functional with respect to k · kC ¯ k·k (S, S1 ) ≤ H C

¯ k·k where by H C on CLq¯ (J, BR ).

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3.5

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Notes and Remarks

The results in Sections 3.1 and 3.2 are taken from Zhou, Jiao and Pecaric, 2013. The material in Section 3.3 due to Wang, Zhou and Medved, 2012. The main results in Section 3.4 are adopted from Wang, Zhou and Wei, 2013.

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icmp12-master

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Chapter 4

Fractional Abstract Evolution Equations

4.1

Introduction

The existence of mild solutions for the Cauchy problem of fractional evolution equations has been considered in several recent papers (see, e.g., Agarwal and Shmad, 2011; Belmekki and Benchohra, 2010; Chang, Kavitha and Mallika, 2009; Darwish, Henderson and Ntouyas, 2009; Hernandez, O’Regan and Balachandran, 2010; Hu, Ren and Sakthivel, 2009; Kumar and Sukavanam, 2012; Li, Peng and Jia, 2012; Shu, Lai and Chen, 2011; Wang, Chen and Xiao, 2012; Wang and Zhou, 2011; Wang, Feˇckan and Zhou, 2011; Zhou and Jiao, 2010), much less is known about the fractional evolution equations with Riemann-Liouville derivative. In most of the existing articles, Schauder’s fixed point theorem, Krasnoselskii’s fixed point theorem or Darbo’s fixed point theorem, Kuratowski measure of noncompactness are employed to obtain the fixed points of the solution operator of the Cauchy problems under some restrictive conditions. In order to show that the solution operator is compact, a very common approach is to use Arzela-Ascoli’s theorem. However, it is difficult to check the relatively compactness of the solution operator and the equicontinuity of certain family of functions which is given by the solution operator. In this chapter, we discuss the existence of mild solutions of fractional abstract evolution equations. The suitable mild solutions of fractional evolution equations with Riemann-Liouville derivative and Caputo derivative are introduced respectively. In Sections 4.2 and 4.3, by using the theory of Hausdorff measure of noncompactness, we investigate the existence of mild solutions for the Cauchy problems in the cases C0 -semigroup is compact or noncompact. Section 4.4 devoted to study the evolution equations with almost sectorial operators. In Section 4.5, the existence results of mild solutions of nonlocal problem of fractional evolution equations are presented.

109

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110

4.2 4.2.1


ws-book975x65


Evolution Equations with Riemann-Liouville Derivative Introduction

Assume that X is a Banach space with the norm | · |. Let a ∈ R+ , J = [0, a] and J ′ = (0, a]. Denote C(J, X) be the Banach space of continuous functions from J into X with the norm kxk = sup |x(t)|, t∈[0,a]

where x ∈ C(J, X), and B(X) be the space of all bounded linear operators from X to X with the norm kQkB(X) = sup{|Q(x)| : |x| = 1}, where Q ∈ B(X) and x ∈ X. Consider the following nonlocal Cauchy problem of fractional evolution equation with Riemann-Liouville derivative q 0 Dt x(t) = Ax(t) + f (t, x(t)), a.e. t ∈ [0, a], (4.1) q−1 x(0) + g(x) = x0 , 0 Dt

where 0 Dtq is Riemann-Liouville derivative of order q, 0 Dtq−1 is Riemann-Liouville integral of order 1 − q, 0 < q < 1, A is the infinitesimal generator of a strongly continuous semigroup of bounded linear operators (i.e. C0 -semigroup) {Q(t)}t≥0 in Banach space X, f : J × X → X is a given function, g : C(J, X) → L(J, X) is a given operator satisfying some assumptions and x0 is an element of the Banach space X. A strong motivation for investigating the nonlocal Cauchy problem (4.1) comes from physics. For example, fractional diffusion equations are abstract partial differential equations that involve fractional derivatives in space and time. They are useful to model anomalous diffusion, where a plume of particles spreads in a different manner than the classical diffusion equation predicts. The time fractional diffusion equation is obtained from the standard diffusion equation by replacing the first-order time derivative with a fractional derivative of order q ∈ (0, 1), namely ∂tq u(z, t) = Au(z, t),

∂zβ1 ,

t ≥ 0,

z ∈ R.

We can take A = for β1 ∈ (0, 1], or A = ∂z + ∂zβ2 for β2 ∈ (1, 2], where q ∂t , ∂zβ1 , ∂zβ2 are the fractional derivatives of order q, β1 , β2 respectively. We refer the interested reader to Eidelman and Kochubei, 2004; Hanyga, 2002; Hayashi, Kaikina and Naumkin, 2005; Meerschaert, Benson, Scheffler et al., 2002; Schneider and Wayes, 1989; Zaslavsky, 1994 and the references therein for more details. The nonlocal conditions 0 Dtq−1 x(0) + g(x) = x0 and x(0) + g(x) = x0 can be applied in physics with better effect than the classical initial conditions 0 Dtq−1 x(0) = x0 and x(0) = x0 respectively. For example, g(x) may be given by m X g(x) = ci x(ti ), i=1

where ci (i = 1, 2, . . . , n) are given constants and 0 < t1 < t2 < · · · < tn ≤ a. Nonlocal conditions were initiated by Byszewski, 1991, which he proved the existence

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and uniqueness of mild and classical solutions for nonlocal Cauchy problems. As remarked by Byszewski and Lakshmikantham, 1991, the nonlocal condition can be more useful than the standard initial condition to describe some physical phenomena. In this section, we study the nonlocal Cauchy problems of fractional evolution equations with Riemann-Liouville derivative by considering an integral equation which is given in terms of probability density. By using the theory of Hausdorff measure of noncompactness, we establish various existence theorems of mild solutions for the Cauchy problem (4.1) in the cases C0 -semigroup is compact or noncompact. Subsection 4.2.2 is devoted to obtain the appropriate definition on the mild solutions of the problem (4.1) by considering an integral equation which is given in terms of probability density. In Subsection 4.2.3, we give some preliminary lemmas. Subsection 4.2.4 provides various existence theorems of mild solutions for the Cauchy problem (4.1) in the case C0 -semigroup is compact. In Subsection 4.2.5, we establish various existence theorems of mild solutions for the Cauchy problem (4.1) in the case C0 -semigroup is noncompact. 4.2.2

Definition of Mild Solutions

Definition 4.1. (Mainardi, Paraddisi and Forenflo, 2000) The Wright function Mq (̺) is defined by ∞ X (−̺)n−1 , 0 < q < 1, ̺ ∈ C. Mq (̺) = (n − 1)!Γ(1 − qn) n=1 It is known that Mq (̺) satisfies the following equality (see Mönch, 1980) Z ∞ Γ(1 + δ) θδ Mq (θ)dθ = , for δ ≥ 0. Γ(1 + qδ) 0 Lemma 4.2. (Tazali, 1982) (i) Let ξ, η ∈ R such that η > −1. If t > 0, then ( tξ+η tη , if ξ + η 6= −n −ξ = Γ(ξ+η+1) (n ∈ N). 0 Dt Γ(η + 1) 0, if ξ + η = −n

(ii) Let ξ > 0 and ϕ ∈ L((0, a), X). Define

Gξ (t) = 0 Dt−ξ ϕ, for t ∈ (0, a),

then −η 0 Dt Gξ (t)

−(ξ+η)

= 0 Dt

ϕ(t), η > 0, almost all t ∈ [0, a].

Lemma 4.3. The nonlocal Cauchy problem (4.1) is equivalent to the integral equation Z t 1 tq−1 (x0 − g(x)) + (t − s)q−1 [Ax(s) + f (s, x(s))]ds, for t ∈ (0, a], x(t) = Γ(q) Γ(q) 0 (4.2)

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provided that the integral in (4.2) exists. Suppose (4.2) is true, then q−1 Z t 1 q−1 q−1 t q−1 D x(t) = D (x − g(x)) + (t − τ ) [Ax(τ ) + f (τ, x(τ ))]dτ , 0 t 0 t 0 Γ(q) Γ(q) 0 Proof.

applying Lemma 4.2 we obtain that Z t q−1 x(t) = x0 − g(x) + [Ax(s) + f (s, x(s))]ds, almost all t ∈ [0, a], 0 Dt 0

and this proves that q 0 Dt x(t)

q−1 x(t) 0 Dt

is absolutely continuous on [0, a]. Then we have

d q−1 x(t) = Ax(t) + f (t, x(t)), almost all t ∈ [0, a] 0 Dt dt

=

and q−1 x(0) 0 Dt

+ g(x) = x0 .

The proof of the converse is given as follows. Suppose (4.1) is true, then −q q 0 Dt (0 Dt x(t))

= 0 Dt−q (Ax(t) + f (t, x(t))).

Since −q q 0 Dt (0 Dt x(t))

tq−1 q−1 x(0) 0 Dt Γ(q) tq−1 = x(t) − (x0 − g(x)), for t ∈ (0, a], Γ(q) = x(t) −

then we have tq−1 x(t) = (x0 − g(x)) + 0 Dt−q (Ax(t) + f (t, x(t))) Γ(q) Z t tq−1 1 = (x0 − g(x)) + (t − s)q−1 [Ax(s) + f (s, x(s))]ds, for t ∈ (0, a]. Γ(q) Γ(q) 0 The proof is completed.

Before giving the definition of mild solution of (4.1), we firstly prove the following lemma. Lemma 4.4. If Z t 1 tq−1 (x0 − g(x)) + (t − s)q−1 [Ax(s) + f (s, x(s))]ds, for t > 0 (4.3) x(t) = Γ(q) Γ(q) 0

holds, then we have x(t) = t where

q−1

Pq (t)(x0 − g(x)) +

Z

0

Pq (t) =

t

(t − s)q−1 Pq (t − s)f (s, x(s))ds, for t > 0, (4.4) Z

0

∞

qθMq (θ)Q(tq θ)dθ.

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Let λ > 0. Applying the Laplace transform Z ∞ Z ∞ −λs ν(λ) = e x(s)ds and ω(λ) = e−λs f (s, x(s))ds, for λ > 0

Proof.

0

0

to (4.3), we have

1 1 1 (x0 − g(x)) + q Aν(λ) + q ω(λ) λq λ λ = (λq I − A)−1 (x0 − g(x)) + (λq I − A)−1 ω(λ) Z ∞ Z ∞ q q = e−λ s Q(s)(x0 − g(x))ds + e−λ s Q(s)ω(λ)ds,

ν(λ) =

0

(4.5)

0

provided that the integrals in (4.5) exist, where I is the identity operator defined on X. Set q ψq (θ) = q+1 Mq (θ−q ), θ whose Laplace transform is given by Z ∞ q e−λθ ψq (θ)dθ = e−λ , where q ∈ (0, 1). (4.6) 0

Using (4.6), we get Z ∞ Z ∞ q −λq s e Q(s)(x0 − g(x))ds = qtq−1 e−(λt) Q(tq )(x0 − g(x))dt 0 0 Z ∞Z ∞ −(λtθ) q q−1 = qψq (θ)e Q(t )t (x0 − g(x))dθdt 0 0 q q−1 Z ∞Z ∞ (4.7) t t = qψq (θ)e−λt Q q (x − g(x))dθdt 0 θ θq q q−1 Z0 ∞ 0 Z ∞ t t −λt (x0 − g(x))dθ dt, = e q ψq (θ)Q q θ θq 0 0 Z ∞ q e−λ s Q(s)ω(λ)ds 0 Z ∞Z ∞ q = qtq−1 e−(λt) Q(tq )e−λs f (s, x(s))dsdt Z0 ∞ Z0 ∞ Z ∞ = qψq (θ)e−(λtθ) Q(tq )e−λs tq−1 f (s, x(s))dθdsdt (4.8) 0 0 0 q q−1 Z ∞Z ∞Z ∞ t t = qψq (θ)e−λ(t+s) Q q f (s, x(s))dθdsdt θ θq 0 0 0Z t Z ∞ Z ∞ (t − s)q (t − s)q−1 −λt f (s, x(s))dθds dt. = e q ψq (θ)Q θq θq 0 0 0 According to (4.7) and (4.8), we have Z ∞ q q−1 Z ∞ t t ν(λ) = e−λt q ψq (θ)Q q (x0 − g(x))dθ θ θq 0 0 Z tZ ∞ (t − s)q (t − s)q−1 f (s, x(s))dθds dt. +q ψq (θ)Q θq θq 0 0

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Now we can invert the last Laplace transform to get Z ∞ x(t) = q θtq−1 Mq (θ)Q(tq θ)(x0 − g(x))dθ 0 Z tZ ∞ +q θ(t − s)q−1 Mq (θ)Q((t − s)q θ)f (s, x(s))dθds 0 0 Z t = tq−1 Pq (t)(x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds. 0

The proof is completed.

Due to Lemma 4.4, we give the following definition of the mild solution of (4.1). Definition 4.5. By the mild solution of the nonlocal Cauchy problem (4.1), we mean that the function x ∈ C(J ′ , X) which satisfies Z t x(t) = tq−1 Pq (t)(x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds, for t ∈ (0, a]. 0

Suppose that A is the infinitesimal generator of a C0 semigroup {Q(t)}t≥0 of uniformly bounded linear operators on Banach space X. This means that there exists M > 1 such that M = supt∈[0,∞) kQ(t)kB(X) < ∞. Proposition 4.6. (Zhou and Jiao, 2010a) For any fixed t > 0, Pq (t) is linear and bounded operator, i.e., for any x ∈ X M |x|. |Pq (t)x| ≤ Γ(q)

Proposition 4.7. (Zhou and Jiao, 2010a) Operator {Pq (t)}t>0 is strongly continuous, which means that, for ∀x ∈ X and 0 < t′ < t′′ ≤ a, we have |Pq (t′′ )x − Pq (t′ )x| → 0 as t′′ → t′ .

Proposition 4.8. (Zhou and Jiao, 2010a) Assume that {Q(t)}t>0 is compact operator. Then {Pq (t)}t>0 is also compact operator. Proposition 4.9. (Pazy, 1983) Assume that {Q(t)}t>0 is compact operator. Then {Q(t)}t>0 is equicontinuous. 4.2.3 Define

Preliminary Lemmas n o X (q) (J ′ ) = x ∈ C(J ′ , X) : lim t1−q x(t) exists and is finite . t→0+

For any x ∈ X

(q)

′

(J ), let the norm k · kq defined by

kxkq = sup {t1−q |x(t)|}. t∈(0,a]

Then (X (q) (J ′ ), k · kq ) is a Banach space. (q) For r > 0, define a closed subset Br (J ′ ) ⊂ X (q) (J ′ ) as follows Br(q) (J ′ ) = {x ∈ X (q) (J ′ ) : kxkq ≤ r}.

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(q)

Thus, Br (J ′ ) is a bounded closed and convex subset of X (q) (J ′ ). Let B(J) be the closed ball of the space C(J, X) with radius r and center at 0, that is B(J) = {y ∈ C(J, X) : kyk ≤ r}. Thus B(J) is a bounded closed and convex subset of C(J, X). We introduce the following hypotheses: (H0) Q(t)(t > 0) is equicontinuous, i.e., Q(t) is continuous in the uniform operator topology for t > 0; (H1) for each t ∈ J ′ , the function f (t, ·) : X → X is continuous and for each x ∈ X, the function f (·, x) : J ′ → X is strongly measurable; (H2) there exists a function m ∈ L(J ′ , R+ ) such that −q 0 Dt m

∈ C(J ′ , R+ ),

lim t1−q 0 Dt−q m(t) = 0,

t→0+

and |f (t, x)| ≤ m(t) for all x ∈ Br(q) (J ′ ) and almost all t ∈ [0, a];

′ (H3) there exists a constant L ∈ (0, Γ(q) M ) such that the operator g : C(J , X) → ′ L(J , X) satisfies

|g(x1 ) − g(x2 )| ≤ Lkx1 − x2 kq , for x1 , x2 ∈ Br(q) (J ′ ); (H4) there exists a constant r > 0 such that Z t M |x0 | + |g(0)| + sup t1−q (t − s)q−1 m(s)ds ≤ r; Γ(q) − M L t∈(0,a] 0 (H3)′ the operator g : C(J ′ , X) → L(J ′ , X) is a continuous and compact map, and there exist positive constants L1 , L2 such that L1 ∈ (0, Γ(q) M ) and |g(x)| ≤ (q) ′ L1 kxkq + L2 for all x ∈ Br (J ); (H4)′ there exists a constant r > 0 such that Z t M |x0 | + L2 + sup t1−q (t − s)q−1 m(s)ds ≤ r. Γ(q) − M L1 t∈(0,a] 0 (q)

For any x ∈ Br (J ′ ), define an operator T as follows (T x)(t) = (T1 x)(t) + (T2 x)(t), where (T1 x)(t) = tq−1 Pq (t)(x0 − g(x)), for t ∈ (0, a], Z t (T2 x)(t) = (t − s)q−1 Pq (t − s)f (s, x(s))ds, for t ∈ (0, a]. 0

It is easy to see that limt→0+ t1−q (T x)(t) =

x0 −g(x) Γ(q) .

For any y ∈ B(J), set

x(t) = tq−1 y(t), for t ∈ (0, a].

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(q)

Then, x ∈ Br (J ′ ). Define T as follows (T y)(t) = (T1 y)(t) + (T2 y)(t), where

  t1−q (T1 x)(t), (T1 y)(t) = x0 − g(x) ,  Γ(q) 1−q t (T2 x)(t), (T2 y)(t) = 0,

for t ∈ (0, a], for t = 0, for t ∈ (0, a], for t = 0. (q)

Obviously, x is a mild solution of (4.1) in Br (J ′ ) if and only if the operator (q) equation x = T x has a solution x ∈ Br (J ′ ). Before giving the main results, we firstly prove the following lemmas. Lemma 4.10. Assume that (H0)-(H4) hold, then {T y : y ∈ B(J)} is equicontinuous. Proof. Step I. {T1 y : y ∈ B(J)} is equicontinuous. For any y ∈ B(J), let (q) x(t) = tq−1 y(t), t ∈ (0, a]. Then x ∈ Br (J ′ ). For t1 = 0, 0 < t2 ≤ a, we get x0 − g(x) |(T1 y)(t2 ) − (T1 y)(0)| ≤ Pq (t2 )(x0 − g(x)) − Γ(q) 1 )(x0 − g(x)) ≤ (Pq (t2 ) − Γ(q) 1 ≤ (Pq (t2 ) − ) (|x0 | + Lkxkq + |g(0)|) Γ(q) 1 ) (|x0 | + Lr + |g(0)|) ≤ (Pq (t2 ) − Γ(q) → 0, as t2 → 0.

For 0 < t1 < t2 ≤ a, we get

|(T1 y)(t2 ) − (T1 y)(t1 )| ≤ |Pq (t2 )(x0 − g(x)) − Pq (t1 )(x0 − g(x))| ≤ |(Pq (t2 ) − Pq (t1 ))(x0 − g(x))|

≤ |(Pq (t2 ) − Pq (t1 ))|(|x0 | + Lkxkq + |g(0)|)

≤ |(Pq (t2 ) − Pq (t1 ))|(|x0 | + Lr + |g(0)|)

→ 0, as t2 → t1 .

Hence, {T1 y : y ∈ B(J)} is equicontinuous. Step II. {T2 y : y ∈ B(J)} is equicontinuous. For any y ∈ B(J), let x(t) = (q) tq−1 y(t), t ∈ (0, a]. Then x ∈ Br (J ′ ). For t1 = 0, 0 < t2 ≤ a, we get Z 1−q t2 q−1 |(T2 y)(t2 ) − (T2 y)(0)| = t2 (t2 − s) Pq (t2 − s)f (s, x(s))ds 0 Z M 1−q t2 t2 (t2 − s)q−1 m(s)ds → 0, as t2 → 0. ≤ Γ(q) 0

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For 0 < t1 < t2 ≤ a, we have Z t2 1−q q−1 |(T2 y)(t2 ) − (T2 y)(t1 )| ≤ t2 (t2 − s) Pq (t2 − s)f (s, x(s))ds t1 Z t1 + t1−q (t2 − s)q−1 Pq (t2 − s)f (s, x(s))ds 2 0 Z t1 1−q q−1 − t1 (t1 − s) Pq (t2 − s)f (s, x(s))ds Z0 t1 + t1−q (t1 − s)q−1 Pq (t2 − s)f (s, x(s))ds 1 0 Z t1 1−q q−1 − t1 (t1 − s) Pq (t1 − s)f (s, x(s))ds 0 Z M t2 1−q q−1 t (t2 − s) m(s)ds ≤ Γ(q) t1 2 Z t1 h i M t1−q (t1 − s)q−1 − t1−q (t2 − s)q−1 m(s)ds + 1 2 Γ(q) Z t1 0 1−q q−1 + t1 (t1 − s) [Pq (t2 − s)f (s, x(s)) − Pq (t1 − s)f (s, x(s))]ds 0

≤ I1 + I2 + I3 ,

where Z Z t1 M t2 1−q 1−q q−1 q−1 , t (t − s) m(s)ds − t (t − s) m(s)ds 2 1 2 1 Γ(q) 0 0 Z t1 h i 2M t1−q (t1 − s)q−1 − t1−q (t2 − s)q−1 m(s)ds, I2 = 1 2 Γ(q) 0 Z t1 1−q q−1 I3 = t1 (t1 − s) [Pq (t2 − s) − Pq (t1 − s)]f (s, x(s))ds .

I1 =

0

One can reduce that limt2 →t1 I1 = 0, since 0 Dt−q m ∈ C(J ′ , R+ ). Noting that h i 1−q q−1 q−1 t1−q (t − s) − t (t − s) m(s) ≤ t1−q (t1 − s)q−1 m(s), 1 2 1 2 1

Rt and 0 1 t1−q (t1 − s)q−1 m(s)ds exists (s ∈ [0, t1 ]), then by Lebesgue’s dominated 1 convergence theorem, we have Z

0

t1

i h t1−q (t1 − s)q−1 − t1−q (t2 − s)q−1 m(s)ds → 0, as t2 → t1 , 1 2

then one can deduce that limt2 →t1 I2 = 0.

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For ε > 0 be enough small, we have Z t1 −ε I3 ≤ t11−q (t1 − s)q−1 kPq (t2 − s) − Pq (t1 − s)kB(X) |f (s, x(s))|ds 0 Z t1 + t1−q (t1 − s)q−1 kPq (t2 − s) − Pq (t1 − s)kB(X) |f (s, x(s))|ds 1 ≤

t1 −ε Z t1 1−q t1 (t1 0

− s)q−1 m(s)ds

sup s∈[0,t1 −ε]

kPq (t2 − s) − Pq (t1 − s)kB(X)

Z t1 2M + t1−q (t1 − s)q−1 m(s)ds Γ(q) t1 −ε 1 ≤ I31 + I32 + I33 , where

rΓ(q) sup kPq (t2 − s) − Pq (t1 − s)kB(X) , M s∈[0,t1 −ε] Z Z t1 −ε 2M t1 1−q q−1 1−q q−1 t (t1 − s) m(s)ds − (t1 − ε) (t1 − ε − s) m(s)ds , = Γ(q) 0 1 0 Z t1 −ε 2M = [(t1 − ε)1−q (t1 − ε − s)q−1 − t1−q (t1 − s)q−1 ]m(s)ds. 1 Γ(q) 0

I31 = I32 I33

By (H0), it is easy to see that I31 → 0 as t2 → t1 . Similar to the proof that I1 , I2 tend to zero, we get I32 → 0 and I33 → 0 as ε → 0. Thus, I3 tends to zero independently of y ∈ B(J) as t2 → t1 , ε → 0. Therefore, |(T2 y)(t2 ) − (T2 y)(t1 )| tends to zero independently of y ∈ B(J) as t2 → t1 , which means that {T2 y : y ∈ B(J)} is equicontinuous. Therefore, {T y : y ∈ B(J)} is equicontinuous. Lemma 4.11. Assume that (H1)-(H4) hold. Then T maps B(J) into B(J), and T is continuous in B(J). Proof. Step I. T maps B(J) into B(J). For any y ∈ B(J), let x(t) = tq−1 y(t). (q) Then x ∈ Br (J ′ ). For t ∈ [0, a], by (H1)-(H4), we have Z t |(T y)(t)| ≤ |Pq (t)(x0 − g(x))| + t1−q (t − s)q−1 Pq (t − s)f (s, x(s))ds 0 Z M M t1−q t ≤ (|x0 | + Lkxkq + |g(0)|) + (t − s)q−1 |f (s, x(s))|ds Γ(q) Γ(q) 0 Z t M 1−q q−1 ≤ |x0 | + Lr + |g(0)| + sup t (t − s) m(s)ds Γ(q) t∈[0,a] 0 ≤ r.

Hence, kT yk ≤ r, for any y ∈ B(J).

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Step II. T is continuous in B(J). For any ym , y ∈ B(J), m = 1, 2, ..., with limm→∞ ym = y, we have lim ym (t) = y(t) and lim tq−1 ym (t) = tq−1 y(t), for t ∈ (0, a].

m→∞

m→∞

Then by (H1), we have f (t, xm (t)) = f (t, tq−1 ym (t)) → f (t, tq−1 y(t)) = f (t, x(t)), as m → ∞,

where xm (t) = tq−1 ym (t) and x(t) = tq−1 y(t). On the one hand, using (H2), we get for each t ∈ J ′ ,

(t − s)q−1 |f (s, xm (s)) − f (s, x(s))| ≤ (t − s)q−1 2m(s), a.e. in [0, t].

On the other hand, the function s → (t − s)q−1 2m(s) is integrable for s ∈ [0, t] and t ∈ J. By Lebesgue’s dominated convergence theorem, we get Z t (t − s)q−1 |f (s, xm (s)) − f (s, x(s))|ds → 0, as m → 0. 0

For t ∈ [0, a]

|(T ym )(t) − (T y)(t)| = |t1−q (T xm (t) − T x(t))| Z t 1−q q−1 ≤ |Pq (t)(g(xm ) − g(x))| + t (t − s) Pq (t − s)(f (s, xm (s)) − f (s, x(s)))ds 0 Z M t1−q t ML kxm − xkq + (t − s)q−1 |f (s, xm (s)) − f (s, x(s))|ds ≤ Γ(q) Γ(q) 0 Z ML M t1−q t ≤ kym − yk + (t − s)q−1 |f (s, xm (s)) − f (s, x(s))|ds. Γ(q) Γ(q) 0 Therefore, T ym → T y pointwise on J as m → ∞, by which Lemma 4.10 implies that T ym → T y uniformly on J as m → ∞ and so T is continuous.

Lemma 4.12. Assume that (H0)-(H2), (H3)′ and (H4)′ hold. Then {T y : y ∈ B(J)} is equicontinuous. Proof.

For any y ∈ B(J), for t1 = 0, 0 < t2 ≤ a, then, we get

|(T y)(t2 ) − (T y)(0)| Z x0 − g(x) 1−q t2 q−1 + t2 (t2 − s) Pq (t2 − s)f (s, x(s))ds ≤ Pq (t2 )(x0 − g(x)) − Γ(q) 0 Z t2 M x − g(x) 0 1−q + t (t2 − s)q−1 m(s)ds ≤ Pq (t2 )(x0 − g(x)) − Γ(q) Γ(q) 2 0 → 0, as t2 → 0.

For any y ∈ B(J) and 0 < t1 < t2 ≤ a, we get

|(T y)(t2 ) − (T y)(t1 )| ≤ |(T1 y)(t2 ) − (T1 y)(t1 )| + |(T2 y)(t2 ) − (T2 y)(t1 )| ≤ |(Pq (t2 ) − Pq (t1 ))(x0 − g(x))| + I1 + I2 + I3 ,

where I1 , I2 and I3 are defined as in the proof of Lemma 4.10. According to Proposition 4.7, we know that |(T y)(t2 )−(T y)(t1 )| tends to zero independently of y ∈ B(J) as t2 → t1 , which means that {T y : y ∈ B(J)} is equicontinuous.

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Lemma 4.13. Assume that (H1), (H2), (H3)′ and (H4)′ hold. Then T maps B(J) into B(J), and T is continuous in B(J). Proof.

For any y ∈ B(J), we have |(T y)(t)| ≤

|x0 | + L1 r + L2 ≤ r, for t = 0 Γ(q)

and |(T y)(t)| = t1−q |(T x)(t)| ≤ r, for t ∈ (0, a]. Hence, kT ykB ≤ r, for any y ∈ B(J). Using the similar argument as that we did in the proof of Lemma 4.11, we know that T is continuous in B(J). 4.2.4

Compact Semigroup Case

In the following, we suppose that the operator A generates a compact C0 -semigroup {Q(t)}t≥0 on X, that is, for any t > 0, the operator Q(t) is compact. Theorem 4.14. Assume that Q(t)(t > 0) is compact. Furthermore assume that (H1)-(H4) hold. Then the nonlocal Cauchy problem (4.1) has at least one mild (q) solution in Br (J ′ ). (q)

Proof. Obviously, x is a mild solution of (4.1) in Br (J ′ ) if and only if y is a fixed point of y = T y in B(J), where x(t) = tq−1 y(t). So, it is enough to prove that y = T y has a fixed point in B(J). For any y1 , y2 ∈ B(J), according to (H3), we have |T1 y1 (t) − T1 y2 (t)| = t1−q |(T1 x1 )(t) − (T1 x2 )(t)| M ≤ |g(x1 ) − g(x2 )| Γ(q) ML ≤ kx1 − x2 kq Γ(q) ML ky1 − y2 k = Γ(q)

which implies that kT1 y1 − T1 y2 k ≤

ML Γ(q) ky1

α(T1 (B(J))) ≤

− y2 k. Thus, we obtain that

ML α(B(J)). Γ(q)

(4.9)

Next, we will show that for any t ∈ [0, a], V (t) = {(T2 y)(t), y ∈ B(J)} is relatively compact in X. Obviously, V (0) is relatively compact in X. Let t ∈ (0, a] be fixed. For ∀ ε ∈ (0, t) and ∀ δ > 0, define an operator Tε,δ on B(J) by the formula Z t−ε Z ∞ (Tε,δ y)(t) = qt1−q θ(t − s)q−1 Mq (θ)Q((t − s)q θ)f (s, x(s))dθds 0 δ Z t−ε Z ∞ = qt1−q Q(εq δ) θ(t − s)q−1 Mq (θ)Q((t − s)q θ − εq δ)f (s, x(s))dθds, 0

δ

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121

(q)

where x ∈ Br (J ′ ). Then from the compactness of Q(εq δ)(εq δ > 0), we obtain that the set Vε,δ (t) = {(Tε,δ y)(t), y ∈ B(J)} is relatively compact in X for ∀ ε ∈ (0, t) and ∀ δ > 0. Moreover, for every y ∈ B(J), we have |(T2 y)(t) − (Tε,δ y)(t)| Z Z 1−q t δ q−1 q ≤ qt θ(t − s) Mq (θ)Q((t − s) θ)f (s, x(s))dθds 0 0 Z Z ∞ 1−q t q−1 q + qt θ(t − s) Mq (θ)Q((t − s) θ)f (s, x(s))dθds ≤ qM t

1−q

Z

0

+qM t1−q ≤ qM t1−q

t−ε t

Z

Z

0

q−1

(t − s) t

t−ε

t

δ

m(s)ds

Z

δ

θMq (θ)dθ Z ∞ (t − s)q−1 m(s)ds θMq (θ)dθ 0

(t − s)q−1 m(s)ds

→ 0, as ε → 0, δ → 0.

Z

0

0

δ

θMq (θ)dθ +

M 1−q t Γ(q)

Z

t

t−ε

(t − s)q−1 m(s)ds

Therefore, there are relatively compact sets arbitrarily close to the set V (t), t > 0. Hence the set V (t), t > 0 is also relatively compact in X. Therefore, {(T2 y)(t), y ∈ B(J)} is relatively compact by Ascoli-Arzela Theorem. Thus, we (q) have α(T2 (Br (J ′ ))) = 0. By (4.9), we have α(T (B(J))) ≤ α(T1 (B(J))) + α(T2 (B(J))) ML α(B(J)). ≤ Γ(q) Thus, the operator T is an α-contraction in B(J). By Lemma 4.11, we know that T is continuous. Hence, Theorem 1.44 shows that T has a fixed point y ∗ ∈ B(J). Let x∗ (t) = tq−1 y ∗ (t). Then x∗ is a mild solution of (4.1). Theorem 4.15. Assume that Q(t)(t > 0) is compact. Furthermore assume that (H1), (H2), (H3)′ and (H4)′ hold. Then the nonlocal Cauchy problem (4.1) has at (q) least one mild solution in Br (J ′ ). Proof. Since Proposition 4.9, Q(t)(t > 0) is equicontinuous, which implies (H0) is satisfied. Then, by Lemmas 4.10-4.11, we know that T : B(J) → B(J) is bounded, continuous and {T y : y ∈ B(J)} is equicontinuous. According to the argument of Theorem 4.14, we only need prove that for any t ∈ J, the set V1 (t) = {(T1 y)(t), y ∈ B(J)} is relatively compact in X. Obviously, V1 (0) is relatively compact in X. Let 0 < t ≤ a be fixed. For ∀ δ > 0, define an operator T1δ on B(J) by the formula Z ∞ (T1δ y)(t) = q θMq (θ)Q(tq θ)(x0 − g(x))dθ δ Z ∞ = qQ(tq δ) θMq (θ)Q(tq θ − tq δ)(x0 − g(x))dθ, δ

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where x(t) = tq−1 y(t), t ∈ (0, a]. From the compactness of Q(tq δ) (tq δ > 0), we obtain that the set V1δ (t) = {(T1δ y)(t), y ∈ B(J)} is relatively compact in X for ∀ δ > 0. Moreover, for any y ∈ B(J), we have Z δ δ q |(T1 y)(t) − (T1 y)(t)| = q θMq (θ)Q(t θ)(x0 − g(x))dθ 0 Z δ ≤ qM (|x0 | + L1 r + L2 ) θMq (θ)dθ. 0

Therefore, there are relatively compact sets arbitrarily close to the set V1 (t), t > 0. Hence the set V1 (t), t > 0 is also relatively compact in X. Moreover, {T y : y ∈ B(J)} is uniformly bounded by Lemmas 4.13. Therefore, {(T y)(t), y ∈ B(J)} is relatively compact by Ascoli-Arzela Theorem. Hence, Theorem 1.44 shows that T has a fixed point y ∗ ∈ B(J). Let x∗ (t) = tq−1 y ∗ (t). Then x∗ is a mild solution of (4.1).

Remark 4.16. If g is not a compact map, we use another method given in Zhu and Li, 2008 to consider the following integral equations Z t 1 (x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds, t ∈ (0, a]. x(t) = tq−1 Pq t + n 0 (4.10) For any n ∈ N, noticing that the operator Q( n1 ) is compact, one can easily derive the relative compactness of V (0) and V (t)(t > 0). Then, (4.10) has a solution in (q) Br (J ′ ). By passing the limit, as n → ∞, one obtains a mild solution of the nonlocal Cauchy problem (4.1). However, because Q(t) is replaced by Q( n1 ), one needs a more restrictive condition than (H4)′ , such as (H4)′′ there exists a constant r > 0 such that Z t Mε |x0 | + L1 r + L2 + sup t1−q (t − s)q−1 m(s)ds ≤ r, Γ(q) t∈(0,a] 0 where Mε = supt∈[0,a+ε] kQ(t)kB(X) , ε is a small constant. Remark 4.17. The condition (H2) of Theorems 4.14-4.15 can be replaced by the following condition. 1 (H2)′ There exist a constant q1 ∈ (0, q) and m ∈ L q1 (J, R+ ) such that |f (t, x)| ≤ m(t) for all x ∈ Br(q) (J ′ ) and almost all t ∈ [0, a].

In fact, if (H2)′ holds, by using the Hölder inequality, for any t1 , t2 ∈ J ′ and t1 < t2 , we obtain |0 Dt−q m(t2 ) − 0 Dt−q m(t1 )| Z Z t2 1 t1 q−1 q−1 q−1 = ((t2 − s) − (t1 − s) )m(s)ds + (t2 − s) m(s)ds Γ(q) 0 t 1−q11 Z t1 q1 Z t1 1 1 1 q−1 q−1 1−q1 q1 ds (m(s)) ds ((t1 − s) − (t2 − s) ) ≤ Γ(q) 0 0 1−q1 Z t2 q1 Z t2 1 1 1 q−1 1−q q + (m(s)) 1 ds ((t2 − s) ) 1 ds Γ(q) t1 t1

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1−q1 q−1 q−1 kmk q1 ((t1 − s) 1−q1 − (t2 − s) 1−q1 )ds L 1 0 Z t2 1−q1 q−1 1 (t2 − s) 1−q1 ds + kmk q1 L 1 Γ(q) t1 1−q1 q−q1 kmk q1 1 − q 1−q1 q−q1 q−q1 1 1−q1 1−q1 L 1 1−q1 + (t2 − t1 ) ≤ − t2 t1 Γ(q) q − q1 1−q1 kmk q1 1 − q 1−q1 q−q1 1 L 1 (t2 − t1 ) 1−q1 + Γ(q) q − q1 2kmk q1 1 − q 1−q1 1 L 1 (t2 − t1 )q−q1 → 0, as t2 → t1 , ≤ Γ(q) q − q1 ≤

1 Γ(q)

Z

123

t1

(4.11)

where kmk

1 L q1

=

Z

0

a

1

(m(t)) q1 dt

q1

.

Furthermore, Z

t

(t − s)q−1 m(s)ds 1−q1 Z t q1 0Z t q−1 1 1−q 1−q1 q1 ds (m(s)) ds ≤t (t − s) 0 0 1−q1 1 − q1 ≤ t1−q1 kmk q1 → 0, as t → 0. L 1 q − q1 t1−q

(4.12)

Thus, (4.11) and (4.12) mean that 0 Dt−q m ∈ C(J ′ , R+ ), and limt→0+ t1−q 0 Dt−q m(t) = 0. Hence, (H2) holds. Example 4.18. Let X = L2 ([0, π], R). Consider the following fractional partial differential equations.  q ∂ u(t, z) = ∂z2 u(t, z) + ∂z G(t, u(t, z)), z ∈ [0, π], t ∈ (0, a],    t  u(t, 0) = u(t, π) = 0, t ∈ (0, a], n Z π (4.13) X   u(0, z) + k(z, y)u(t , y)dy = u (z), z ∈ [0, π],  i 0  i=0

0

where ∂tq is Riemann-Liouville fractional partial derivative of order 0 < q < 1, a > 0, G is a given function, n is a positive integer, 0 < t0 < t1 < · · · < tn ≤ a, u0 (z) ∈ X = L2 ([0, π], R), k(z, y) ∈ L2 ([0, π] × [0, π], R+ ). We define an operator A by Av = v ′′ with the domain D(A) = {v(·) ∈ X : v, v ′ absolutly continuous, v ′′ ∈ X, v(0) = v(π) = 0}.

Then A generates a strongly continuous semigroup {Q(t)}t≥0 which is compact, analytic and self-adjoint. Clearly the nonlocal Cauchy problem (4.2) and (H1) are satisfied.

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The system (4.13) can be reformulated as the following nonlocal Cauchy problem in X q 0 Dt x(t) = Ax(t) + f (t, x(t)), almost all t ∈ [0, a], q−1 x(0) + g(x) = x0 , 0 Dt where x(t) = u(t, ·), that is x(t)(z) = u(t, z), t ∈ (0, a], z ∈ [0, π]. The function f : J ′ × X → X is given by f (t, x(t))(z) = ∂z G(t, u(t, z)), and the operator g : C(J ′ , X) → L(J ′ , X) is given by g(x)(z) =

n X

Kg x(ti )(z),

i=0

where Kg v(z) =

Z

π

0

k(z, y)v(y)dy, for v ∈ X = L2 ([0, π], R), z ∈ [0, π].

We can take q = 1/3 and f (t, x(t)) = t−1/4 sin x(t), and choose Z m(t) = t−1/4 , L = (n + 1)

0

π

Z

π

k 2 (z, y)dydz

0

12

and Γ( 13 )Γ( 34 ) 3 M r= |x0 | + g(0) + a4 . Γ( 13 ) − M L Γ( 12 13 ) Then, (H1)-(H4) are satisfied (noting that Kg : X → X is completely continuous). (1/3) According to Theorem 4.14, system (4.13) has a mild solution in Br ((0, a]) proML vided that Γ(1/3) < 1. 4.2.5

Noncompact Semigroup Case

If Q(t) is noncompact, we give an assumption as follows. (H5) There exists a constant ℓ > 0 such that for any bounded D ⊂ X, α(f (t, D)) ≤ ℓα(D). Theorem 4.19. Assume that (H0)-(H5) hold. Then the nonlocal Cauchy problem (q) (4.1) has at least one mild solution in Br (J ′ ). Proof. By Lemmas 4.11-4.12, we know that T2 : B(J) → B(J) is bounded, continuous and {T2 y : y ∈ B(J)} is equicontinuous. Next, we will show that T2 is compact in a subset of B(J). For each bounded subset B0 ⊂ B(J), set T 1 (B0 ) = T2 (B0 ), T n (B0 ) = T2 co(T n−1 (B0 )) , n = 2, 3, ... .

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(1)

Then, from Propositions 1.28-1.29, for any ε > 0, there is a sequence {yn }∞ n=1 ⊂ B0 such that α(T 1 (B0 (t))) = α(T2 (B0 (t))) Z t ≤ 2α t1−q (t − s)q−1 Pq (t − s)f (s, {sq−1 yn(1) (s)}∞ )ds +ε n=1 0

≤

4M 1−q t Γ(q)

4M ℓ 1−q ≤ t Γ(q)

Z

t

0

Z

(t − s)q−1 α f (s, {sq−1 yn(1) (s)}∞ ) ds + ε n=1

t

0

(t − s)q−1 α({sq−1 yn(1) (s)}∞ n=1 )ds + ε

Z 4M ℓα(B0 ) 1−q t ≤ t (t − s)q−1 sq−1 ds + ε Γ(q) 0 4M ℓΓ(q)tq α(B0 ) ≤ + ε. Γ(2q)

Since ε > 0 is arbitrary, we have α(T 1 (B0 (t))) ≤

4M ℓΓ(q)tq α(B0 ). Γ(2q) (2)

From Propositions 1.28-1.29, for any ε > 0, there is a sequence {yn }∞ n=1 ⊂ co(T 1 (B0 )) such that α(T 2 (B0 (t))) = α(T2 (co(T 1 (B0 (t))))) Z t ≤ 2α t1−q (t − s)q−1 Pq (t − s)f (s, {sq−1 yn(2) (s)}∞ )ds +ε n=1 4M t1−q ≤ Γ(q)

Z

4M ℓt1−q ≤ Γ(q) ≤

4M ℓt1−q Γ(q)

0 t

0

Z

(t − s)q−1 α (f (s, {sq−1 yn(2) (s)}∞ n=1 ) ds + ε t

(t − s)q−1 α({sq−1 yn(2) (s)}∞ n=1 )ds + ε

0

Z

0 2 1−q

t

(t − s)q−1 sq−1 α({yn(2) (s)}∞ n=1 )ds + ε

Z t (4M ℓ) t (t − s)q−1 s2q−1 ds + ε Γ(2q) 0 (4M ℓ)2 Γ(q) 2q = t α(B0 ) + ε. Γ(3q)

≤

It can be shown, by mathematical induction, that for every n ¯ ∈ N, α(T n¯ (B0 (t))) ≤

(4M ℓ)n¯ Γ(q) n¯ q t α(B0 ). Γ((¯ n + 1)q)

Since (4M ℓaq )n¯ Γ(q) = 0, n ¯ →∞ Γ((¯ n + 1)q) lim

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there exists a positive integer n ˆ such that (4M ℓaq )nˆ Γ(q) (4M ℓ)nˆ Γ(q) nˆ q t ≤ = k < 1. Γ((ˆ n + 1)q) Γ((ˆ n + 1)q) Then α(T nˆ (B0 (t))) ≤ kα(B0 ).

We know from Proposition 1.26, T nˆ (B0 (t)) is bounded and equicontinuous. Then, from Proposition 1.27, we have α(T nˆ (B0 )) = max α(T nˆ (B0 (t))). t∈[0,a]

Hence α(T nˆ (B0 )) ≤ kα(B0 ). Let D0 = B(J), D1 = co(T nˆ (D)), ..., Dn = co(T nˆ (Dn−1 )), n = 2, 3, ... . Then, we can get (i) D0 ⊃ D1 ⊃ D2 ⊃ · · · ⊃ Dn−1 ⊃ Dn ⊃ · · · ; (ii) limn→∞ α(Dn ) = 0. ˆ = T∞ Dn is a nonempty, compact and convex subset in B(J). Then D n=0 ˆ ⊂ D. ˆ Firstly, we show We will prove T2 (D) 1

T2 (Dn ) ⊂ Dn , n = 0, 1, 2, ... .

(4.14)

1

From T (D0 ) = T2 (D0 ) ⊂ D0 , we know co(T (D0 )) ⊂ D0 . Therefore T 2 (D0 ) = T2 (co(T 1 (D0 ))) ⊂ T2 (D0 ) = T 1 (D0 ),

T 3 (D0 ) = T2 (co(T 2 (D0 ))) ⊂ T2 (co(T 1 (D0 ))) = T 2 (D0 ), ... T nˆ (D0 ) = T2 (co(T

n ˆ −1

(D0 ))) ⊂ T2 (co(T

n ˆ −2

(D0 ))) = T nˆ −1 (D0 ).

Hence, D1 = co(T nˆ (D0 )) ⊂ co(T nˆ −1 (D0 )), so T (D1 ) ⊂ T (co(T nˆ −1 (D0 ))) = T nˆ (D0 ) ⊂ co(T nˆ (D0 )) = D1 . Employing the same method, we can prove T∞ ˆ ⊂ T2 (Dn ) ⊂ Dn (n = 0, 1, 2, ...). By (4.14), we get T2 (D) n=0 T2 (Dn ) ⊂ T∞ ˆ ˆ ˆ D = D. Then T ( D) is compact. Hence, α(T ( D)) = 0. 2 2 n=0 n ˆ and t ∈ (0, a], according to (H3), we have On the other hand, for any y1 , y2 ∈ D |T1 y1 (t) − T1 y2 (t)| = t1−q |(T1 x1 )(t) − (T1 x2 )(t)| M ≤ |g(x1 ) − g(x2 )| Γ(q) ML kx1 − x2 kq ≤ Γ(q) ML ky1 − y2 k, = Γ(q)

(4.15)

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which implies that kT1 y1 − T1 y2 k ≤

ML Γ(q) ky1

127

− y2 k. Thus, we obtain that

ˆ ≤ M L α(D). ˆ α(T1 (D)) Γ(q)

(4.16)

By (4.16), we have ˆ ≤ α(T1 (D)) ˆ + α(T2 (D)) ˆ α(T (D)) ML ˆ ≤ α(D). Γ(q) ˆ By Lemma 4.11, we know that T Thus, the operator T is an α-contraction in D. is continuous. Hence, Theorem 1.44 shows that T has a fixed point y ∗ ∈ B(J). Let x∗ (t) = tq−1 y ∗ (t). Then x∗ is a mild solution of (4.1). Theorem 4.20. Assume that (H0)-(H2), (H3)′ , (H4)′ and (H5) hold, then the (q) nonlocal Cauchy problem (4.1) has at least one mild solution in Br (J ′ ). Proof. Since g(x) is compact and Pq (t) is bounded, for every t > 0, {(T1 y)(t), y ∈ B(J)} is relatively compact. Thus, we have α(T1 (B(J))) = 0. ˆ ⊂ B(J) such that By the proof of Theorem 4.19, we know that there exists a D ˆ ˆ T2 (D) is relatively compact, i.e., α(T2 (D)) = 0. Hence, we have ˆ ≤ α(T1 (D)) ˆ + α(T2 (D)) ˆ = 0. α(T (D)) Hence, Theorem 1.44 shows that T has a fixed point y ∗ ∈ B(J). Let x∗ (t) = tq−1 y ∗ (t). Then x∗ is a mild solution of (4.1). 4.3 4.3.1

Evolution Equations with Caputo Derivative Introduction

Consider the following nonlocal Cauchy problems of fractional evolution equation with Caputo derivative C q 0 Dt x(t) = Ax(t) + f (t, x(t)), a.e. t ∈ [0, a], (4.17) x(0) + g(x) = x0 , q where C 0 Dt is Caputo derivative of order q, 0 < q < 1, A is the infinitesimal generator of a strongly continuous semigroup of bounded linear operators (i.e. C0 semigroup) {Q(t)}t≥0 in Banach space X, f : J × X → X , g : C(J, X) → L(J, X) are given operators satisfying some assumptions and x0 is an element of the Banach space X. In this section, by using the theory of Hausdorff measure of noncompactness and fixed point theorems, we study the nonlocal Cauchy problem (4.17) in the cases Q(t) is compact or noncompact. Subsection 4.3.2 is devoted to obtain the appropriate definition on the mild solutions of the problem (4.17) by considering a integral equation which is given in terms of probability density. In Subsection 4.3.3, we give

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some preliminary lemmas. Subsection 4.3.4 provides various existence theorems of mild solutions for the Cauchy problem (4.17) in the case Q(t) is compact. In Subsection 4.3.5, we establish various existence theorems of mild solutions for the Cauchy problem (4.17) in the case Q(t) is noncompact. 4.3.2


Lemmas 4.21. If x(t) = x0 − g(x) +

1 Γ(q)

Z

t

0

(t − s)q−1 [Ax(s) + f (s, x(s))]ds, for t ≥ 0

(4.18)

holds, then we have x(t) = Sq (t)(x0 − g(x)) + where Sq (t) =

Z

∞

Z

0

t

(t − s)q−1 Pq (t − s)f (s, x(s))ds, for t ≥ 0,

Mq (θ)Q(tq θ)dθ, Pq (t) =

0

Proof.

Z

∞

(4.19)

qθMq (θ)Q(tq θ)dθ.

0

Let λ > 0. Applying the Laplace transform Z ∞ Z ∞ −λs ν(λ) = e x(s)ds and ω(λ) = e−λs f (s, x(s))ds, λ > 0 0

0

to (4.18), we have 1 1 1 (x0 − g(x)) + q Aν(λ) + q ω(λ) λ λ λ q −1 −1 = λq−1 (λ I − A) (x − g(x)) + (λq I − 0 Z Z A) ω(λ)

ν(λ) =

∞

q−1

=λ

0

e

−λq s

Q(s)(x0 − g(x))ds +

∞

e

−λq s

(4.20)

Q(s)ω(λ)ds,

0

provided that the integrals in (4.20) exist, where I is the identity operator defined on X. Using (4.6) and (4.20), we get Z ∞ q q−1 λ e−λ s Q(s)(x0 − g(x))ds Z ∞ 0 q = q(λt)q−1 e−(λt) Q(tq )(x0 − g(x))dt Z0 ∞ 1 d −(λt)q = − [e ]Q(tq )(x0 − g(x))dt (4.21) λ dt 0 Z ∞Z ∞ = θψq (θ)e−λtθ Q(tq )(x0 − g(x))dθdt q Z0 ∞ 0 Z ∞ t = e−λt ψq (θ)Q q (x0 − g(x))dθ dt. θ 0 0

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According to (4.8), (4.20) and (4.21), we have Z ∞ q Z ∞ t ν(λ) = e−λt ψq (θ)Q q (x0 − g(x))dθ θ 0 Z t Z ∞0 (t − s)q (t − s)q−1 +q ψq (θ)Q f (s, x(s)) dθds dt. θq θq 0 0 Now we can invert the last Laplace transform to get q Z ∞ t x(t) = ψq (θ)Q q (x0 − g(x))dθ θ 0 Z tZ ∞ (t − s)q (t − s)q−1 f (s, x(s)) dθds +q ψq (θ)Q q θ θq 0 0 Z ∞ = Mq (θ)Q(tq θ)(x0 − g(x))dθ 0 Z tZ ∞ +q θ(t − s)q−1 Mq (θ)Q((t − s)q θ)f (s, x(s))dθds 0 0 Z t = Sq (t)(x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds. 0

The proof is completed.

Due to Lemma 4.21, we give the following definition of the mild solution of (4.17). Definition 4.22. By the mild solution of the nonlocal Cauchy problem (4.17), we mean that the function x ∈ C(J, X) which satisfies Z t x(t) = Sq (t)(x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds, for t ∈ [0, a]. 0

Suppose that A is the infinitesimal generator of a C0 -semigroup {Q(t)}t≥0 of uniformly bounded linear operators on Banach space X. This means that there exists M > 1 such that M = supt∈[0,∞) kQ(t)kB(X) < ∞. Proposition 4.23. (Zhou and Jiao, 2010a) For any fixed t > 0, {Sq (t)}t>0 and {Pq (t)}t>0 are linear and bounded operators, i.e., for any x ∈ X |Sq (t)x| ≤ M |x|, |Pq (t)x| ≤

M |x|. Γ(q)

Proposition 4.24. (Zhou and Jiao, 2010a) Operators {Sq (t)}t>0 and {Pq (t)}t>0 are strongly continuous, which means that, for ∀x ∈ X and 0 < t′ < t′′ ≤ a, we have |Sq (t′′ )x − Sq (t′ )x| → 0, |Pq (t′′ )x − Pq (t′ )x| → 0, as t′′ → t′ . Proposition 4.25. (Zhou and Jiao, 2010a) Assume that {Q(t)}t>0 is compact operator. Then {Sq (t)}t>0 and {Pq (t)}t>0 are also compact operators.

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4.3.3

Preliminary Lemmas

For r > 0, let Br (J) be the closed ball of the space C(J, X) with radius r and center at 0, that is, Br (J) = {x ∈ C(J, X) : kxk ≤ r}, where kxk = sup |x(t)|. t∈[0,a]

We introduce the following hypotheses: (H0) Q(t)(t > 0) is equicontinuous, i.e., Q(t) is continuous in the uniform operator topology for t > 0; (H1) for each t ∈ J, the function f (t, ·) : X → X is continuous and for each x ∈ X, the function f (·, x) : J → X is strongly measurable; (H2) there exists a function m ∈ L(J, R+ ) such that −q 0 Dt m

∈ C(J, R+ ),

lim 0 Dt−q m(t) = 0

t→0+

and |f (t, x)| ≤ m(t) for all x ∈ Br (J) and almost all t ∈ [0, a]; 1 (H3) there exists a constant L ∈ (0, M ), the operator g : C(J, X) → L(J, X) satisfies

|g(x1 ) − g(x2 )| ≤ Lkx1 − x2 k, for x1 , x2 ∈ Br (J); (H4) there exists a constant r > 0 such that Z t M 1 q−1 ≤ r; |x0 | + |g(0)| + sup (t − s) m(s)ds 1 − ML t∈[0,a] Γ(q) 0 (H3)′ the operator g : C(J, X) → L(J, X) is a continuous and compact map, and there exist positive constants L1 , L2 such that |g(x)| ≤ L1 kxk+L2 for all x ∈ Br (J); (H4)′ there exists a constant r > 0 such that Z t 1 M q−1 |x0 | + L2 + sup (t − s) m(s)ds ≤ r. 1 − M L1 t∈[0,a] Γ(q) 0 For any x ∈ Br (J), we define an operator T as follows (T x)(t) = (T1 x)(t) + (T2 x)(t), where (T1 x)(t) = Sq (t)(x0 − g(x)), for t ∈ [0, a], Z t (T2 x)(t) = (t − s)q−1 Pq (t − s)f (s, x(s))ds, for t ∈ [0, a]. 0

Obviously, x is a mild solution of (4.17) in Br (J) if and only if the operator equation x = T x has a solution x ∈ Br (J). Lemma 4.26. Assume that (H0)-(H3) hold. Then {T2 x : x ∈ Br (J)} is equicontinuous.

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131

For any x ∈ Br (J), for t1 = 0, 0 < t2 ≤ a, by (H2), we get Z t2 q−1 |(T2 x)(t2 ) − (T2 x)(0)| = (t2 − s) Pq (t2 − s)f (s, x(s))ds 0

M ≤ Γ(q)

Z

t2

0

(t2 − s)q−1 m(s)ds → 0 as t2 → 0.

For 0 < t1 < t2 ≤ a, we have |(T2 x)(t2 ) − (T2 x)(t1 )| Z t2 Z t1 = (t2 − s)q−1 Pq (t2 − s)f (s, x(s))ds − (t1 − s)q−1 Pq (t1 − s)f (s, x(s))ds 0 0 Z t2 (t2 − s)q−1 Pq (t2 − s)f (s, x(s))ds = t Z1 t1 Z t1 q−1 q−1 (t2 − s) Pq (t2 − s)f (s, x(s))ds − (t1 − s) Pq (t2 − s)f (s, x(s))ds + 0 0 Z t1 Z t1 q−1 q−1 + (t1 − s) Pq (t2 − s)f (s, x(s))ds − (t1 − s) Pq (t1 − s)f (s, x(s))ds 0

Z

Z

t2

0

t1

M M (t2 − s)q−1 m(s)ds + [(t1 − s)q−1 − (t2 − s)q−1 ]m(s)ds Γ(q) t1 Γ(q) 0 Z t1 + (t1 − s)q−1 |Pq (t2 − s)f (s, x(s)) − Pq (t1 − s)f (s, x(s))|ds 0 Z Z t1 M t2 q−1 q−1 (t2 − s) m(s)ds − (t1 − s) m(s)ds ≤ Γ(q) 0 0 Z t1 2M [(t1 − s)q−1 − (t2 − s)q−1 ]m(s)ds + Γ(q) 0 Z t1 + (t1 − s)q−1 |Pq (t2 − s) − Pq (t1 − s)|m(s)ds

≤

0

=: I1 + I2 + I3 .

Since 0 Dt−q m ∈ C(J, R+ ), thus I1 → 0 as t2 → t1 . For t1 < t2 , Z t1 2M I2 ≤ (t1 − s)q−1 m(s)ds, Γ(q) 0 then by Lebesgue’s dominated convergence theorem, we have that I2 → 0 as t2 → t1 . For ε > 0 be small enough, we have Z t1 −ε I3 ≤ (t1 − s)q−1 kPq (t2 − s) − Pq (t1 − s)kB(X) m(s)ds 0

+

Z

t1

t1 −ε

(t1 − s)q−1 kPq (t2 − s) − Pq (t1 − s)kB(X) m(s)ds

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≤

Z

t1 0

(t1 − s)q−1 m(s)ds

sup s∈[0,t1 −ε]

kPq (t2 − s) − Pq (t1 − s)kB(X)

Z t1 2M (t1 − s)q−1 m(s)ds + Γ(q) t1 −ε Z t1 ≤ (t1 − s)q−1 m(s)ds sup kPq (t2 − s) − Pq (t1 − s)kB(X) s∈[0,t1 −ε]

0

Z Z t1 −ε 2M t1 q−1 q−1 (t1 − s) m(s)ds − (t1 − ε − s) m(s)ds + Γ(q) 0 0 Z t1 −ε 2M + [(t1 − ε − s)q−1 − (t1 − s)q−1 ]m(s)ds. Γ(q) 0 =: I31 + I32 + I33 . By (H0), it is easy to see that I31 → 0 as t2 → t1 . Similar to the proof that I1 , I2 tend to zero, we get I32 → 0 and I33 → 0 as ε → 0. Thus, I3 tends to zero independently of x ∈ Br (J) as t2 → t1 , ε → 0. Therefore, |(T2 x)(t1 ) − (T2 x)(t2 )| tends to zero independently of x ∈ Br (J) as t2 → t1 , which means that {T2 x : x ∈ Br (J)} is equicontinuous. Lemma 4.27. Assume that (H1)-(H4) hold. Then T maps Br (J) into Br (J), and T is continuous in Br (J). Proof. Step I. T maps Br (J) into Br (J). For any x ∈ Br (J) and t ∈ J, by using (H1)-(H4), we have |(T x)(t)| = |(T1 x)(t) + (T2 x)(t)| Z t q−1 ≤ |Sq (t)(x0 − g(x))| + (t − s) Pq (t − s)f (s, x(s))ds 0

Z t M (t − s)q−1 |f (s, x(s))|ds ≤ M (|x0 | + Lkx − 0k + |g(0)|) + Γ(q) 0 Z t 1 q−1 (t − s) m(s)ds ≤ M |x0 | + Lr + |g(0)| + sup t∈[0,a] Γ(q) 0

≤ r. Hence, kT xk ≤ r for any x ∈ Br (J). Step II. T is continuous in Br (J). For any {xm }∞ m=1 ⊆ Br (J), x ∈ Br (J) with limn→∞ kxm − xk = 0, by the condition (H1), we have lim f (s, xm (s)) = f (s, x(s)). m→∞

On the one hand, using (H2), we get for each t ∈ J, (t − s)q−1 |f (s, xm (s)) − f (s, x(s))| ≤ (t − s)q−1 2m(s). On the other hand, the function s → (t − s)q−1 2m(s) is integrable for s ∈ [0, t] and t ∈ J. By Lebesgue’s dominated convergence theorem, we get Z t (t − s)q−1 |f (s, xm (s)) − f (s, x(s))|ds → 0, as m → ∞. 0

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Then, for t ∈ J, |(T xm )(t) − (T x)(t)|

Z t q−1 ≤ |Sq (t)(g(xm ) − g(x))| + (t − s) Pq (t − s)[f (s, xm (s)) − f (s, x(s))]ds 0 Z t M (t − s)q−1 |f (s, xm (s)) − f (s, x(s))|ds, as m → ∞. ≤ M Lkxm − xk + Γ(q) 0

Therefore, T xm → T x pointwise on J as m → ∞, by which Lemma 4.26 implies that T xm → T x uniformly on J as m → ∞ and so T is continuous. Lemma 4.28. Assume that there exists a constant r > 0 such that the conditions (H0)-(H2) and (H3)′ are satisfied. Then {T x : x ∈ Br (J)} is equicontinuous. Proof.

For any x ∈ Br (J) and 0 ≤ t1 < t2 ≤ a, we get |(T x)(t2 ) − (T x)(t1 )| ≤ |(Sq (t2 ) − Sq (t1 ))(x0 − g(x))| + I1 + I2 + I3 ,

where I1 , I2 and I3 are defined as in the proof of Lemma 4.26. According to Proposition 4.24, we know that |(T x)(t2 ) − (T x)(t1 )| tends to zero independently of x ∈ Br (J) as t2 → t1 , which means that {T x, x ∈ Br (J)} is equicontinuous. Lemma 4.29. Assume that there exists a constant r > 0 such that the conditions (H1), (H2), (H3)′ and (H4)′ are satisfied. Then T maps Br (J) into Br (J), and T is continuous in Br (J). Proof. have

For any x ∈ Br (J) and t ∈ J, by using (H1), (H2), (H3)′ and (H4)′ , we

|(T x)(t)| = |(T1 x)(t) + (T2 x)(t)|

Z t q−1 ≤ |Sq (t)(x0 − g(x))| + (t − s) Pq (t − s)f (s, x(s))ds 0 Z t M ≤ M (|x0 | + L1 r + L2 ) + (t − s)q−1 |f (s, x(s))|ds Γ(q) 0 Z t 1 (t − s)q−1 m(s)ds ≤ M |x0 | + L1 r + L2 + sup t∈[0,a] Γ(q) 0 ≤ r.

(4.22)

Hence, kT xk ≤ r for any x ∈ Br (J). Using the similar argument as that we did in the proof of Lemma 4.27, we know that T is continuous in Br (J). 4.3.4

Compact Semigroup Case

In the following, we suppose that the operator A generates a compact C0 -semigroup {Q(t)}t≥0 on X, that is, for any t > 0, the operator Q(t) is compact. Theorem 4.30. Assume that Q(t)(t > 0) is compact. Furthermore assume that there exists a constant r > 0 such that the conditions (H1)-(H4) are satisfied. Then the nonlocal Cauchy problem (4.17) has at least one mild solution in Br (J).

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Proof. Since Proposition 4.9, Q(t)(t > 0) is equicontinuous, which implies (H0) is satisfied. For any x1 , x2 ∈ Br (J) and t ∈ J, according to (H3), we have |(T1 x1 )(t) − (T1 x2 )(t)| ≤ M |g(x1 ) − g(x2 )| ≤ M Lkx1 − x2 k,

which implies that kT1 x1 − T1 x2 k ≤ M Lkx1 − x2 k. Thus, we obtain that α(T1 Br (J)) ≤ M Lα(Br (J)).

(4.23)

Next, we will show that {T2 x, x ∈ Br (J)} is relatively compact, i.e. α(T2 (Br (J))) = 0. It suffices to show that the family of functions {T2 x : x ∈ Br (J)} is uniformly bounded and equicontinuous, and for any t ∈ J, {(T2 x)(t), x ∈ Br (J)} is relatively compact in X. By Lemma 4.27, we have kT2 xk ≤ r, for any x ∈ Br (J), which means that {T2 x : x ∈ Br (J)} is uniformly bounded. By Lemma 4.26, {T2 x : x ∈ Br (J)} is equicontinuous. It remains to prove that for any t ∈ J, V (t) = {(T2 x)(t), x ∈ Br (J)} is relatively compact in X. Obviously, V (0) is relatively compact in X. Let 0 < t ≤ a be fixed. For ∀ ε ∈ (0, t) and ∀ δ > 0, define an operator Tεδ on Br (J) by the formula Z t−ε Z ∞ (Tεδ x)(t) = q θ(t − s)q−1 Mq (θ)Q((t − s)q θ)f (s, x(s))dθds 0 δ Z t−ε Z ∞ = q Q(εq δ) θ(t − s)q−1 Mq (θ)Q((t − s)q θ − εq δ)f (s, x(s))dθds, 0

δ

where x ∈ Br (J). Then from the compactness of Q(εq δ)(εq δ > 0), we obtain that the set Vεδ (t) = {(Tεδ x)(t), x ∈ Br (J)} is relatively compact in X for ∀ ε ∈ (0, t) and ∀ δ > 0. Moreover, for any x ∈ Br (J), we have Z t Z δ δ q−1 q |(T2 x)(t) − (Tε x)(t)| ≤ q θ(t − s) Mq (θ)Q((t − s) θ)f (s, x(s))dθds 0 0 Z t Z ∞ q−1 q + q θ(t − s) Mq (θ)Q((t − s) θ)f (s, x(s))dθds t−ε t

δ

Z δ (t − s)q−1 m(s)ds θMq (θ)dθ 0 0 Z t M (t − s)q−1 m(s)ds → 0, as ε → 0, δ → 0. + Γ(q) t−ε

≤ qM

Z

Therefore, there are relatively compact sets arbitrarily close to the set V (t), t > 0. Hence the set V (t), t > 0 is also relatively compact in X. Therefore, {(T2 x)(t), x ∈ Br (J)} is relatively compact by Ascoli-Arzela Theorem. Thus, we have α(T2 (Br (J))) = 0. By (4.23), we have α(T (Br (J))) ≤ α(T1 (Br (J))) + α(T2 (Br (J))) ≤ M Lα(Br (J)).

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Thus, the operator T is an α-contraction in Br (J). By Lemma 4.27, we know that T is continuous. Hence, Theorem 1.44 shows that T has a fixed point in Br (J). Therefore, the nonlocal Cauchy problem (4.17) has a mild solution in Br (J). Theorem 4.31. Assume that Q(t)(t > 0) is compact. Furthermore assume that (H1), (H2), (H3)′ and (H4)′ hold. Then the nonlocal Cauchy problem (4.17) has at least a mild solution in Br (J). Proof. Since Proposition 4.9, Q(t)(t > 0) is equicontinuous, which implies (H0) is satisfied. By Lemma 4.29, we have kT xk ≤ r, for any x ∈ Br (J), which means that {T x : x ∈ Br (J)} is uniformly bounded. By Lemmas 4.28-4.29, we know that T is continuous, {T x, x ∈ Br (J)} is equicontinuous. It remains to prove that for t ∈ J, the set {(T x)(t), x ∈ Br (J)} is relatively compact in X. According to the argument of Theorem 4.30, we only need to prove that for any t ∈ J, the set V1 (t) = {(T1 x)(t), x ∈ Br (J)} is relatively compact in X. Obviously, V1 (0) is relatively compact in X. Let 0 < t ≤ a be fixed. For ∀ δ > 0, define an operator T1δ on Br (J) by the formula Z ∞ (T1δ x)(t) = Mq (θ)Q(tq θ)(x0 − g(x))dθ δ Z ∞ q = Q(t δ) Mq (θ)Q(tq θ − tq δ)(x0 − g(x))dθ δ

where x ∈ Br (J). From the compactness of Q(tq δ)(tq δ > 0), we obtain that the set V1δ (t) = {(T1δ x)(t), x ∈ Br (J)} is relatively compact in X for ∀ δ > 0. Moreover, for every x ∈ Br (J), we have |(T1 x)(t) − (T1δ x)(t)| Z ∞ Z ∞ q q Mq (θ)Q(t θ)(x0 − g(x))dθ Mq (θ)Q(t θ)(x0 − g(x))dθ − = δ 0 Z δ = Mq (θ)Q(tq θ)(x0 − g(x))dθ 0

≤ M (|x0 | + L1 r + L2 )

Z

δ

Mq (θ)dθ.

0

Therefore, there are relatively compact sets arbitrarily close to the set V1 (t), t > 0. Hence the set V1 (t), t > 0 is also relatively compact in X. Moreover, {T x : x ∈ Br (J)} is uniformly bounded by Lemma 4.27. Therefore, {T x, x ∈ Br (J)} is relatively compact by Ascoli-Arzela Theorem. Therefore, α(T (Br (J))) = 0. Hence, Theorem 1.44 shows that T has a fixed point in Br (J), which means that the nonlocal Cauchy problem (4.17) has a mild solution. Remark 4.32. If g is not a compact mapping, we consider the following integral equations Z t 1 (x0 − g(x)) + (t − s)q−1 Pq (t − s)f (s, x(s))ds, t ∈ (0, a]. x(t) = tq−1 Pq t + n 0 (4.24)

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For any n ∈ N, noticing that the operator Q( n1 ) is compact, one can easily derive the relative compactness of V (0) and V (t)(t > 0). Then, (4.24) has a solution in (q) Br (J ′ ). By passing the limit, as n → ∞, one obtains a mild solution of the nonlocal Cauchy problem (4.17). However, because Q(t) is replaced by Q( n1 ), one needs a more restrictive condition than (H4)′ , such as (H4)′′ there exists a constant r > 0 such that Z t 1 q−1 (t − s) m(s)ds ≤ r, Mε |x0 | + L1 r + L2 + sup t∈[0,a] Γ(q) 0 where Mε = supt∈[0,a+ε] kQ(t)kB(X) , ε is a small constant. Remark 4.33. The condition (H2) of Theorems 4.30-4.31 can be replaced by the following condition. 1 (H2)′ There exist a constant q1 ∈ (0, q) and m ∈ L q1 (J, R+ ) such that |f (t, x)| ≤ m(t) for all x ∈ Br (J) and almost all t ∈ [0, a]. We emphasize that (H2) is more weak than the condition (H2)′ . 4.3.5

Noncompact Semigroup Case

If Q(t) is noncompact, we give an assumption as follows. (H5) There exists ℓ > 0 such that for any bounded D ⊂ X, α(f (t, D)) ≤ ℓα(D). Theorem 4.34. Assume that (H0)-(H5) hold. Then the nonlocal Cauchy problem (4.17) has at least one mild solution in Br (J). Proof. By Lemmas 4.26-4.27, we know that T : Br (J) → Br (J) is bounded, continuous and {T2 x : x ∈ Br (J)} is equicontinuous. For each bounded subset B0 ⊂ Br (J), set T 1 (B0 ) = T2 (B0 ), T n (B0 ) = T2 co(T n−1 (B0 )) , n = 2, 3, ... . (1)

Then for any ε > 0, there is a sequence {xn }∞ n=1 such that

α(T 1 (B0 (t))) = α(T2 (B0 (t))) Z t q−1 (1) ∞ ≤ 2α (t − s) Pq (t − s)f (s, {xn (s)}n=1 )ds + ε 0

4M ≤ Γ(q)

Z

t

0

∞ (t − s)q−1 α f (s, {x(1) (s)} ) ds + ε n n=1

Z 4M ℓα(B0 ) t (t − s)q−1 ds + ε Γ(q) 0 4M ℓtq α(B0 ) = + ε. Γ(q + 1)

≤

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Since ε > 0 is arbitrary, we have α(T 1 (B0 (t))) ≤

4M ℓtq α(B0 ). Γ(q + 1) (2)

From Propositions 1.28-1.29, for any ε > 0, there is a sequence {xn }∞ n=1 ⊂ co(T 1 (B0 )) such that α(T 2 (B0 (t))) = α(T2 (co(T 1 (B0 (t)))) Z t q−1 (2) ∞ ≤ 2α (t − s) Pq (t − s)f (s, {xn (s)}n=1 )ds + ε 0

4M ≤ Γ(q)

4M ℓ ≤ Γ(q)

Z

t

0

Z

∞ (t − s)q−1 α f (s, {x(2) (s)} n n=1 ds + ε

t

0 2

∞ (t − s)q−1 α({x(2) n (s)}n=1 )ds + ε

Z (4M ℓ) α(B0 ) t (t − s)q−1 sq ds + ε Γ(q)Γ(q + 1) 0 (4M ℓ)2 t2q α(B0 ) = + ε. Γ(2q + 1)

≤

It can be shown, by mathematical induction, that for every n ¯ ∈ N, α(T n¯ (B0 (t))) ≤

(4M ℓ)n¯ tn¯ q α(B0 ) . Γ(¯ nq + 1)

Since (4M ℓaq )n¯ = 0, n ¯ →∞ Γ(¯ nq + 1) lim

there exists a positive integer n ˆ such that (4M ℓaq )nˆ (4M ℓ)nˆ tnˆ q ≤ = k < 1. Γ(ˆ nq + 1) Γ(ˆ nq + 1) Then α(T nˆ (B0 (t))) ≤ kα(B0 ).

We know from Proposition 1.26, T nˆ (B0 (t)) is bounded and equicontinuous, Then, from Proposition 1.27, we have α(T nˆ (B0 )) = max α(T nˆ (B0 (t))). t∈[0,a]

Hence α(T nˆ (B0 )) ≤ kα(B0 ). By using the similar method as in the proof of Theorem 4.19, we can prove that there exists a D ⊂ Br (J) such that α(T2 (D)) = 0.

(4.25)

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On the other hand, for any x1 , x2 ∈ D and t ∈ J, according to (H3), we have |(T1 x1 )(t) − (T1 x2 )(t)| ≤ M |g(x1 ) − g(x2 )| ≤ M Lkx1 − x2 k,

which implies that kT1 x1 − T1 x2 k ≤ M Lkx1 − x2 k. Thus, we obtain that α(T1 D) ≤ M Lα(D).

(4.26)

By (4.25) and (4.26), we have α(T (D)) ≤ α(T1 (D)) + α(T2 (D)) ≤ M Lα(T (D)). Thus, the operator T is an α-contraction in D. By Lemma 4.27, we know that T is continuous. Hence, Theorem 1.44 shows that T has a fixed point in D ⊂ Br (J). Therefore, the nonlocal Cauchy problem (4.17) has a mild solution in Br (J). Theorem 4.35. Assume that (H0)-(H2), (H3)′ , (H4)′ and (H5) hold, then the nonlocal Cauchy problem (4.17) has at least a mild solution in Br (J). Proof. By the proof of Theorem 4.34, we know that there exists a D ⊂ Br (J) such that T2 (D) is relatively compact, i.e., α(T2 (D)) = 0. Clearly, α(T1 (D)) = 0, since g(x) is compact and Sq (t) is bounded. Hence, we have α(T (D)) ≤ α(T1 (D)) + α(T2 (D)) = 0. Therefore, Theorem 1.44 shows that T has a fixed point in D ⊂ Br (J). Therefore, the nonlocal Cauchy problem (4.17) has a mild solution in Br (J). 4.4 4.4.1

Nonlocal Cauchy Problems for Evolution Equations Introduction

The nonlocal condition has a better effect on the solution and is more precise for physical measurements than the classical condition alone. In this section, we discuss the existence of mild solutions of Cauchy problem for fractional order evolution equations with nonlocal conditions C α = Ax(t) + f (t, x(t)) , α ∈ (0, 1), t ∈ J = [0, 1], 0 Dt x(t) Pm (4.27) x(0) = k=1 ak x(tk ), k = 1, 2, ..., m,

where A : D(A) ⊂ X → X is the generator of a C0 -semigroup {Q(t), t ≥ 0} on a Banach space X, f : J × X → X is a given function and ak (k = 1, 2, ..., m) are real numbers with Σm k=1 ak 6= 1 and tk , k = 1, 2, ...m are given points satisfying 0 ≤ t1 ≤ t2 ≤ · · · ≤ tm < 1. A suitable definition of mild solution of the equation (4.27) will be introduced by defining a bounded operator B. Meanwhile, two sufficient conditions are given to guarantee such B exists. We mention that our first existence result relies on a growth condition on J and second existence result relies on a growth condition involving two parts, one for [0, tm ], and the other for [tm , 1].

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Assume that Pα , Sα are defined as in Subsection 4.3.2. Suppose that there exists a bounded operator B : X → X given by " #−1 m X B= I− ak Sα (tk ) . (4.28) k=1

We can give two sufficient conditions to guarantee such B exists and is bounded. Lemma 4.36. The operator B defined in (4.28) exists and is bounded, if one of the following two conditions holds: (C1) there exist some real numbers ak such that M

m X

k=1

|ak | < 1,

(4.29)

where M = supt∈(0,∞) ||Q(t)||B(X) < ∞. (C2) Q(t) is compact for each t > 0 and homogeneous linear nonlocal problems C α = Ax(t), α ∈ (0, 1), t ∈ J, 0 Dt x(t) Pm (4.30) x(0) = k=1 ak x(tk ), has no non-trivial mild solutions. Proof.

For (C1), it is easy to see

m

X

ak Sα (tk )

k=1

B(X)

≤M

m X

k=1

|ak | < 1,

where Proposition 4.23 and (4.29) are used. Thus by Neumann theorem, B defined by (4.28) exists and it is bounded. For (C2), it is obvious that the mild solutions of the problem (4.30) is given by x(t) = Sα (t)x(0), which implies that x(0) =

m X

ak x(tk ) =

k=1

m X

ak Sα (tk )x(0).

k=1

By Proposition 4.25, Sα (tk ) is compact for each tk > 0, k = 1, 2, ..., m. Then Pm k=1 ak Sα (tk ) is also compact. Since the problem (4.30) has no non-trivial mild solutions, one can obtain the desired result via Fredholm alternative theorem. Now we introduce the following definition of mild solutions of the equation (4.27). Definition 4.37. By a mild solution of the equation (4.27), we mean a function x ∈ C(J, X) satisfying x(t) = Sα (t)

m X

k=1

ak B(g(tk )) + g(t), t ∈ J,

(4.31)

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where g(tk ) =

Z

tk

0

and g(t) =

Z

t

0

(tk − s)α−1 Pα (tk − s)f (s, x(s))ds,

(t − s)α−1 Pα (t − s)f (s, x(s))ds, t ∈ J.

(4.32)

(4.33)

Remark 4.38. To explain the formula (4.31), we note that a mild solution of the fractional evolution equation (4.27) with the initial condition is just x(t) = Sα (t)x(0) + g(t), so taking into account also the nonlocal condition, we get x(0) =

m X

ak Sα (tk )x(0) +

k=1

m X

ak g(tk ).

k=1

Pm Pm So x(0) = k=1 ak B(g(tk )) and hence x(t) = Sα (t) k=1 ak B(g(tk )) + g(t) which is just (4.31). 4.4.3

Existence

Our first existence result is based on the well-known Schaefer fixed point theorem. In this subsection, we will study our problem under the following assumptions: (H1) f : J × X → X satisfies the Carathéodory type conditions; (H2) there exists a function h such that 0 Dt−α h(t) exists for all t ∈ J and 0 D·−α h(·) ∈ C((0, 1], R+ ) with limt→0+ 0 Dt−α h(t) = 0 and a nondecreasing continuous function Ω : R+ → R+ such that |f (t, x)| ≤ h(t)Ω(|x|) for all x ∈ X and all t ∈ J; (H3) the inequality lim sup ρ→∞

M 2 BΩ(ρ)

ρ >1 −α −α |a | D h(t k ) + M Ω(ρ) supt∈J 0 Dt h(t) k=1 k 0 t

Pm

hold; (H4) Q(t) is compact for each t > 0. We begin to consider the following problem C α = Ax(t) + λf (t, x(t)) , α ∈ (0, 1], λ, t ∈ J, 0 Dt x(t) Pm x(0) = k=1 ak x(tk ). Define an operator F : C(J, X) → C(J, X) as follows

(F x)(t) = (F1 x)(t) + (F2 x)(t), t ∈ J, where Fi : C(J, X) → C(J, X), i = 1, 2 are given by the formulas (F1 x)(t) = Sα (t)

m X

k=1

ak B(g(tk )), (F2 x)(t) = g(t),

(4.34)

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where B is the operator defined in (4.28), g(tk ) is defined in (4.32) and g(t) is defined in (4.33). Obviously, a mild solution of the equation (4.34) is a solution of the operator equation x = λF x

(4.35)

and conversely. Thus, we can apply Schaefer’s fixed point theorem to derive the existence results of solutions of the equation (4.27). Lemma 4.39. Let x be any solution of the equation (4.35). Then, there exists R∗ > 0 such that kxkC ≤ R∗ which is independent of the parameter λ ∈ J. Proof. Denote R0 := kxk. Taking into accounts our conditions and Lemma Proposition 4.23 and Proposition 4.24, it follows from (4.31) that

Note that

|x(t)| ≤ |(F1 x)(t)| + |(F2 x)(t)| Pm ≤ M k=1 |ak |kBkB(X) |g(tk )| + |g(t)|, t ∈ J. Z

(4.36)

tk

(tk − s)α−1 kPα (tk − s)kB(X) |f (s, x(s))|ds Z tk M ≤ (tk − s)α−1 h(s)Ω(kxk)ds Γ(α) 0 Z M Ω(R0 ) tk (tk − s)α−1 h(s)ds ≤ Γ(α) 0 = M Ω(R0 )0 Dt−α h(tk ), k = 1, 2, ..., m, k

(4.37)

Z M Ω(R0 ) t (t − s)α−1 h(s)ds Γ(α) 0 = M Ω(R0 ) sup 0 Dt−α h(t), t ∈ J.

(4.38)

|g(tk )| ≤

0

and

|g(t)| ≤

t∈J

In view of (4.36)-(4.38), one can obtain m X R0 := kxk ≤ M 2 kBkB(X) Ω(R0 ) |ak |0 Dt−α h(tk ) + M Ω(R0 ) sup 0 Dt−α h(t), t ∈ J, k k=1

t∈J

which implies that

M 2 kBkB(X) Ω(R0 )

R0 ≤ 1. (4.39) −α −α |a | D k=1 k 0 tk h(tk ) + M Ω(R0 ) supt∈J 0 Dt h(t)

Pm

However, it follows (H3) that there exists a R∗ > 0 such that for all R > R∗ we can derive that R Pm > 1. (4.40) −α 2 M kBkB(X) Ω(R) k=1 |ak |0 Dtk h(tk ) + M Ω(R) supt∈J 0 Dt−α h(t)

Now, comparing the equalities (4.39) and (4.40), we claim that R0 ≤ R∗ . As a result, we find that kxk ≤ R∗ which independents the parameter λ. This completes the proof.

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Let BR∗ = {x ∈ C(J, X) : kxk ≤ R∗ }.

Then BR∗ is a bounded closed and convex subset in C(J, X). By Lemma 4.39, we can derive the following result. Lemma 4.40. The operator F maps BR∗ into itself. Lemma 4.41. The operator F : BR∗ → BR∗ is completely continuous. Proof. For our purpose, we only need to check that Fi : BR∗ → BR∗ , i = 1, 2 is completely continuous. Firstly, by repeating the same producers of Lemma 4.27 and Theorem 4.30, one can obtain F2 : BR∗ → BR∗ is completely continuous. Secondly, one can check that F1 : BR∗ → BR∗ is continuous (by (H1), (H2) and Proposition 4.23) and F1 : BR∗ → BR∗ is compact in viewing of Sα (t) is compact for each t > 0 (by (H4) and Proposition 4.25). The proof is completed. Now, we can state the main result of this section. Theorem 4.42. Assume that (H1)-(H4) hold and the condition (C1) (or (C2)) is satisfied. Then the equation (4.27) has at least one solution u ∈ C(J, X) and the set of the solutions of the equation (4.27) is bounded in C(J, X). Proof. Obviously, the set {x ∈ C(J, X) : x = λF x, 0 < λ < 1} is bounded due to Lemma 4.40. Now we can apply Theorem 1.43 to derive that F has a fixed point in BR∗ which is just the mild solution of the equation (4.27). This completes the proof. Our second existence result is based on O’Regan fixed point theorem. In addition to (H1), (H4) and (C1) (or (C2)), motivated by Boucherif and Precup, 2003, 2007, we introduce the following two assumptions: (H5) There exists a function h such that 0 Dt−α h(t) exists for all t ∈ [0, tm ] and −α −α + 0 D· h(·) ∈ C((0, tm ], R ) with limt→0+ 0 Dt h(t) = 0 and a nondecreasing contin+ + uous function Ω : R → R such that |f (t, x)| ≤ h(t)Ω(|x|) for all x ∈ X, and for all t ∈ [tm , 1] there exists a function l such that exists and tm D·−α l(·) ∈ C([tm , 1], R+ ) such that |f (t, x)| ≤ l(t),

for all x ∈ X. Moreover, Ω has the property ! m X r > M Ω(r) |ak |kBkB(X) + 1 sup k=1

t∈[0,tm ]

−α 0 Dt h(t)

−α tm Dt l(t)

(4.41)

(4.42)

for all r > R1∗ > 0; (H6) there exists a function q such that tm Dt−α q(t) exists for all t ∈ [tm , 1] and −α −α + tm D· q(·) ∈ C([tm , 1], R ) with M supt∈[tm ,1] 0 Dt q(t) ≤ 1 and a nondecreasing + + continuous function Ψ : R → R with Ψ(r) < r for r > 0 such that |f (t, x) − f (t, y)| ≤ q(t)Ψ(|x − y|)

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for all t ∈ [tm , 1] and x, y ∈ X. Consider the equation (4.34) again and the equivalent equation x = λT x

(4.43)

where T : C(J, X) → C(J, X) is defined by (T x)(t) = (T1 x)(t) + (T2 x)(t), t ∈ J, where Ti : C(J, X) → C(J, X), i = 1, 2 given by  m X    S (t) ak B(g(tk )) + g(t),  α    k=1   m X (T1 x)(t) = Sα (t) ak B(g(tk ))    k=1  Z  tm    + (t − s)α−1 Pα (t − s)f (s, x(s))ds,  0

and

   0, (T2 x)(t) = Z t  (t − s)α−1 Pα (t − s)f (s, x(s))ds,  tm

t ∈ [0, tm ),

t ∈ [tm , 1],

t ∈ [0, tm ), t ∈ [tm , 1].

We first prove that any solutions of the equation (4.43) are a priori bounded. Lemma 4.43. Let x be any solution of the equation (4.43). Then, there exist Ri∗ > 0 i = 1, 2 such that kxkC([0,tm ],X) ≤ R1∗ and kxkC([tm ,1],X) ≤ R2∗ . In other words, kxk ≤ R∗ = max{R1∗ , R2∗ } which is independent of the parameter λ. Proof. Case I. We prove that there exists R1∗ > 0 such that kxkC([0,tm ],X) ≤ R1∗ . For t ∈ [0, tm ] and λ ∈ J, denote R[0,tm ] := kxkC([0,tm ],X) , we have |x(t)| ≤ λ|(T1 x)(t)| + |(T2 x)(t)| m X ≤M |ak |kBkB(X) |g(tk )| + |g(t)| ≤M

k=1 m X k=1

|ak |kBkB(X)

M Γ(α)

Z

M Γ(α)

Z

0

tk

(tk − s)α−1 h(s)Ω(R[0,tm ] )ds

t

(t − s)α−1 h(s)Ω(R[0,tm ] )ds ! m X ≤ M Ω(R[0,tm ] ) |ak |kBkB(X) + 1 sup +

0

t∈[0,tm ]

k=1

which implies that

R[0,tm ] ≤ M Ω(R[0,tm ] )

m X

k=1

|ak |kBkB(X) + 1

!

−α 0 Dt h(t),

sup t∈[0,tm ]

−α 0 Dt h(t).

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From (4.42) we find that there exists R1∗ ≥ R[0,tm ] > 0 such that kxkC([0,tm ],X) ≤ R1∗ . Case II. We prove that there exists R2∗ > 0 such that kxkC([tm ,1],X) ≤ R2∗ . For t ∈ [tm , 1] and λ ∈ J, keeping in mind our assumptions, we find that Z tk m X M (tk − s)α−1 h(s)Ω(R1∗ )ds |x(t)| ≤ M |ak |kBkB(X) Γ(α) 0 k=1 Z tm Z t M M + (t − s)α−1 h(s)Ω(R1∗ )ds + (t − s)α−1 h(s)ds Γ(α) 0 Γ(α) tm ! m X ∗ ≤ M Ω(R1 ) |ak |kBkB(X) + 1 sup 0 Dt−α h(t) t∈[0,tm ]

k=1

+M sup

t∈[tm ,1]

−α tm Dt l(t),

which implies that where R2∗

=M

"

kxkC([tm ,1],X) ≤ R2∗ ,

Ω(R1∗ )

m X

|ak |kBkB(X) + 1

k=1 max{R1∗ , R2∗ }.

!

sup t∈[0,tm ]

−α 0 Dt h(t)

+ sup t∈[tm ,1]

#

−α tm Dt l(t)

.

Let R∗ = Then we will find that any possible solutions of the equation (4.43) satisfy kxk ≤ R∗ which are independent of the parameter λ. This completes the proof. Denote D = {x ∈ C(J, X) : kxk < R∗ + 1}. We can proceed as in the proof of Lemma 4.43 to derive the following result. Lemma 4.44. T (D) is bounded. One can proceed as in the proof of Lemma 4.41 to obtain the following result. Lemma 4.45. The operator T1 : D → C(J, X) is completely continuous. Next, we show the following result. Lemma 4.46. The operator T2 : D → C(J, X) is nonlinear contraction. Proof. It follows from the definition of T2 we only need to show T2 : D → C([tm , 1], X) is a nonlinear contraction. In fact, for any x, y ∈ D and t ∈ [tm , 1], we have Z t |(T2 x)(t) − (T2 y)(t)| ≤ (t − s)α−1 kPα (t − s)[f (s, x(s)) − f (s, y(s))]k ds tm

Z

t

≤

M Γ(α)

≤

M Ψ(kx − yk) Γ(α)

≤

tm

(t − s)α−1 q(s)Ψ(|x(s) − y(s)|)ds

M sup t∈[tm ,1]

Z

t

(t − s)α−1 q(s)ds tm !

−α tm Dt q(t)

Ψ(kx − yk),

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which implies that This completes the proof.

kT2 x − T2 yk ≤ Ψ(kx − yk).

Now, we are ready to present the main result of this section. Theorem 4.47. Assume that (H1), (H4), (H5) and (H6) hold and the condition (C1) (or (C2)) is satisfied. Then the equation (4.27) has at least a solution u ∈ C(J, X). Proof. By Lemma 4.43 we see that (ii) of Theorem 1.46 does not hold. Thus, there is no solution of the equation (4.43) with x ∈ ∂D. Therefore, one can apply Theorem 1.46 to derive that T has a fixed point in D which is just the mild solution of the equation (4.27). This completes the proof. Remark 4.48. Theorem 4.47 also holds even if (H5) and (H6) are replaced by the following conditions: (H5)′ Condition (H5) is assumed without (4.41); ≤ 1, condition (H6) is assumed in addition (H6)′ denoting δ := limr→+∞ inf Ψ(r) r with M δ sup tm Dt−α q(t) < 1. t∈[tm ,1]

Indeed, we can modify Case II in the proof of Lemma 4.43 as follows Z tk m X M (tk − s)α−1 h(s)Ω(R1∗ )ds |x(t)| ≤ M |ak |kBkB(X) Γ(α) 0 k=1 Z tm M + (t − s)α−1 h(s)Ω(R1∗ )ds Γ(α) 0 Z t Z t M M + (t − s)α−1 |f (s, 0)|ds + (t − s)α−1 q(s)Ψ(|x(s)|)ds Γ(α) tm Γ(α) tm ! m X ∗ ≤ M Ω(R1 ) |ak |kBkB(X) + 1 sup tm Dt−α h(t) t∈[0,tm ]

k=1

α

M supt∈[tm ,t] |f (t, 0)|(1 − tm ) Γ(α + 1) Z t M + (t − s)α−1 q(s)(δ|x(s)| + δ1 )ds, Γ(α) tm for some δ1 ≥ 0. Then we have 1 R2∗ = 1 − M δ supt∈[tm ,1] tm Dt−α q(t) ! m X ∗ × M Ω(R1 ) |ak |kBkB(X) + 1 sup +

k=1

t∈[0,tm ]

−α tm Dt h(t)

! M supt∈[tm ,t] |f (t, 0)|(1 − tm )α −α + M δ1 sup tm Dt q(t) . + Γ(α + 1) t∈[tm ,1]

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4.5 4.5.1


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Abstract Cauchy Problems with Almost Sectorial Operators Introduction

Let X be a complex Banach space with norm | · |. As usual, for a linear operator A, we denote by D(A) the domain of A, by σ(A) its spectrum, while ρ(A) := C − σ(A) is the resolvent set of A, and denote by the family R(z; A) = (zI − A)−1 , z ∈ ρ(A) of bounded linear operators the resolvent of A. Moreover, we denote by L (X, Y ) the space of all bounded linear operators from Banach space X to Banach space Y with the usual operator norm k · kL (X,Y ) , and we abbreviate this notation to L (X) when Y = X, and write kT kL (X) as kT k for every T ∈ L (X). When dealing with parabolic evolution equations, it is usually assumed that the partial differential operator in the linear part is a sectorial operator, stimulated by the fact that this class of operators appears very often in the applications. For example, one can find from Henry, 1981; Lunardi, 1995 and Pazy, 1983 that many elliptic differential operators equipped homogeneous boundary conditions are sectorial when they are considered in the Lebesgue spaces (e.g. Lp -spaces) or in the space of continuous functions. We here mention that the operator Aε in Example 4.49, which acts on a domain of “dumb-bell with a thin handle”, is sectorial on Vǫp . However, as presented in Example 4.49 and Example 4.51, though the resolvent set of some partial differential operators considered in some special domains such as the limit “domain” of dumb-bell with a thin handle or in some spaces of more regular functions such as the space of Hölder continuous functions, contains a sector, but for which the resolvent operators do not satisfy the required estimate to be a sectorial operator. Example 4.49. In this notation the “dumb-bell with a thin handle” has the form Ωε = D 1 ∪ Q ε ∪ D 2

(ε ∈ (0, 1]; small),

where D1 and D2 are mutually disjoint bounded domains in RN (N ≥ 2) with smooth boundaries, joined by a thin channel, Qε (which is not required to be cylindrical), which degenerates to a 1-dim line segment Q0 as ε approaches zero. This implies that passing to the limit as ε → 0, the limit “domain” of Ωε consists of the fixed part D1 , D2 and the line segment Q0 . Without loss of generality, we may assume that Q0 = {(x, 0, . . . , 0); 0 < x < 1}. Let P0 = (0, 0, . . . , 0), P1 = (1, 0, . . . , 0) be the points where the line segment touches the boundary of D1 and D2 . Put Ω = D1 ∪ D2 . Firstly, consider the evolution equation of parabolic type equipped with Neumann boundary condition in the form   ut − ∆u + u = f (u), x ∈ Ωǫ , t > 0, (4.44) ∂u  = 0, x ∈ ∂Ωǫ , ∂n where ∆ stands for the Laplacian operator with respect to the spatial variable ∂ denotes the outward normal derivative on x ∈ Ωǫ , ∂Ωǫ is the boundary of Ωǫ , ∂n

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∂Ωǫ and f : R → R is a nonlinearity. Let Vǫp (1 ≤ p < ∞) denote the family of spaces based on Lp (Ωǫ ), equipped with the norm Z Z 1/p 1 p p |u|p . kukVǫ = |u| + N −1 ε Qǫ Ω Define the linear operator Aε : D(Aε ) ⊂ Vǫp 7→ Vǫp by ∂u D(Aε ) = u ∈ W 2,p (Ωǫ ); ∆u ∈ Vǫp , = 0 , ∂n ∂Ωǫ Aε u = −∆u + u,

u ∈ D(Aε ).

It follows from a standard argument that the operator Aε generates an analytic semigroup on Vǫp . Moreover, the following estimate holds kR(λ; −Aε )kL (Lp (Ωǫ )) ≤

C , for λ ∈ Σ′θ , |λ|

where Σ′θ = {λ ∈ C; | arg(λ − 1)| ≤ θ} with θ > π2 , and C is a constant that does not depend on ε (see, e.g., Henry, 1981 and Pazy, 1983). The limit problem of (4.44) as ε → 0 is the following problem studied in Carvalho, Dlotko and Neseimento, 2008  wt − ∆w + w = f (w), x ∈ Ω, t > 0,    ∂w    = 0, x ∈ ∂Ω, ∂n 1   vt − (gvx )x + v = f (v), x ∈ Q0 = (0, 1),   g   v(0) = w(P0 ) , v(1) = w(P1 ),

where w is a function that lives in Ω and v lives in the line segment Q0 , the function g : [0, 1] → (0, ∞) is a smooth function related to the geometry of the channel Qε , more exactly, on the way the channel Qε collapses to the segment line Q0 . Observe that the vector (w, v) is continuous in the junction between Ω and Q0 and the variable w does not depend on the variable v, but v depends on w. We identify V0p with Lp (Ω) ⊕ Lpg (0, 1) (1 ≤ p < ∞) endowed with the norm Z 1 Z 1/p k(w, v)kV0p = |w|p + g|v|p . Ω

0

Consider the operator A0 : D(A0 ) ⊂ V0p 7→ V0p defined by

D(A0 ) = {(w, v) ∈ V0P ; w ∈ D(∆Ω ), v ∈ Lpg (0, 1), w(P0 ) = v(0), w(P1 ) = v(1)},

1 (4.45) − ∆w + w, − (gv ′ )′ + v , (w, v) ∈ V0p , g where ∆Ω is the Laplace operator with homogeneous Neumann boundary conditions in Lp (Ω) and ∂u 2,p D(∆Ω ) = u ∈ W (Ω); =0 . ∂n A0 (w, v) =

∂Ω

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As pointed out by Arrieta, Carvalho and Lozada-Cruz, 2009a, the operator A0 defined by (4.45) is not a sectorial operator. Its spectrum is all real and, therefore, it is contained in a sector but the resolvent estimate is different from the case of sectorial operator. More precisely, the operator A0 has the following properties (see also Arrieta, Carvalho and Lozada-Cruz, 2006, 2009a): (a) the domain D(A0 ) is dense in V0P , (b) if p > N2 , then A0 is a closed operator, (c) A0 has compact resolvent, (d) for some µ ∈ (0, π2 ), Σµ := {λ ∈ C \ {0}; | arg λ| ≤ π − µ} ∪ {0} ⊂ ρ(−A0 ), and for N2 < q ≤ p, the following estimate holds: kR(λ; −A0 )kL (V0q ,V0p ) ≤

C , 1 + |λ|γ ′

λ ∈ Σµ ,

(4.46)

N − 12 ( 1q − 1p ) < 1, where C is a positive constant. for each 0 < γ ′ < 1 − 2q Remark 4.50. In fact, it is easy to prove that the estimate (4.46) with p = q > is equivalent to

kR(λ; −A0 )kL (V0p ) ≤

e C , |λ|γ ′

N 2

λ ∈ Σµ \ {0},

N e is a positive constant. for 0 < γ ′ < 1 − 2p , where C We refer to Section 2 of Arrieta, Carvalho and Lozada-Cruz, 2006 for a complete and rigorous definition of the dumb-bell domain, and to Arrieta, 1995; Arrieta, Carvalho and Lozada-Cruz, 2006, 2009a,b; Dancer and Daners, 1997; Gadyl’shin, 2005; Jimbo, 1989 for related studies of partial differential equations involving dumb-bell domain. Example 4.51. Assume that Ω is a bounded domain in RN (N ≥ 1) with boundary ∂Ω of class C 4m (m ∈ N). Let C l (Ω), l ∈ (0, 1), denote the usual Banach space with norm | · |l . Consider the elliptic differential operator A′ : D(A′ ) ⊂ C l (Ω) 7→ C l (Ω) in the form

D(A′ ) = {u ∈ C 2m+l (Ω); Dβ u|∂Ω = 0, |β| ≤ m − 1}, A′ u =

X

aβ (x)Dβ u(x),

|β|≤2m

where β is a multiindex in (N |β| =

S

u ∈ D(A′ ),

{0})n ,

n X

βj ,

Dβ =

j=1

n Y

j=1

(−i

∂ βj ) . ∂xj

The coefficients aβ : Ω 7→ C of A′ are assumed to satisfy (i) aβ ∈ C l (Ω) for all |β| ≤ 2m, (ii) aβ (x) ∈ R for all x ∈ Ω and |β| = 2m,

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(iii) there exists a constant M > 0 such that X M −1 |ξ|2 ≤ aβ (x)ξ β ≤ M |β|2 , |β|=2m

for all ξ ∈ RN

149

and x ∈ Ω.

Then, the following statements hold. (a) A′ is not densely defined in C l (Ω), (b) there exist ν, ε > 0 such that σ(A′ + ν) ⊂ S π2 −ε = {λ ∈ C \ {0}; | arg λ| ≤ kR(λ; A′ + ν)kL (C l (Ω)) ≤

C l |λ|1− 2m

,

π − ε} ∪ {0}, 2

λ ∈ C \ S π2 −ε ,

l − 1 ∈ (−1, 0) is sharp. In particular, the operator A′ + ν is not (c) the exponent 2m sectorial. Notice in particular that the Laplace operator satisfies the conditions (a)-(c) in Example 4.51. For more details we refer to Wahl, 1972. Observe that from Example 4.49 and Remark 4.50, if p > N2 , then A0 ∈ N −γ ′ ) and µ ∈ (0, π2 ), that is, A0 is an almost secΘµ (V0p ) for some γ ′ ∈ (0, 1 − 2p p torial operator on V0 . Also, from Example 4.51 one can find that (A′ + ν) ∈ l

−1

Θ π2m−ε (C l (Ω)), which implies that A′ + ν is an almost sectorial operator on C l (Ω). 2 In this section, motivated by the above consideration, we are interested in studying the Cauchy problem for the linear evolution equation C α t > 0, 0 Dt u(t) + Au(t) = f (t), (4.47) u(0) = u0 ,

as well as the Cauchy problem for the corresponding semilinear fractional evolution equation C α t > 0, 0 Dt u(t) + Au(t) = f (t, u(t)), (4.48) u(0) = u0 α in X, where C 0 Dt , 0 < α < 1, is Caputo fractional derivative of order α and A is an almost sectorial operator, that is, A ∈ Θγω (X) (−1 < γ < 0, 0 < ω < π/2). The main purpose is to study the existence and uniqueness of mild solutions and classical solutions of Cauchy problems (4.47) and (4.48). To do this, we construct two operator families based on the generalized Mittag-Leffler-type functions and the resolvent operators associated with A, present deep anatomy on basic properties for these families consisting on the study of the compactness, and prove that, under natural assumptions, reasonable concept of solutions can be given to problems (4.47) and (4.48), which in turn is used to find solutions to the Cauchy problems. Remark 4.52. We make no assumption on the density of the domain of A.

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Remark 4.53. (i) M. M. Dzhrbashyan and A. B. Nersessyan in Dzhrbashyan and Nersessyan, 1968 (see also Miller and Ross, 1993) showed that the solution of the Cauchy problem C α t > 0, 0 Dt u(t) + λu(t) = 0, u(0) = 1, 0 < α < 1, has the form u(t) = Eα (−λtα ), where Eα is the known Mittag-Leffler function. This result issues a warning to us that no matter how smooth the data u0 is, it is inappropriate to define the mild solution of problem (4.47) as follows Z t 1 (t − s)α−1 T (t − s)f (s)ds, u(t) = T (t)u0 + Γ(α) 0 where T (t) is the semigroup generated by A (see Remark 1.64 (i)), though this fashion was used in some situations of previous research (see, e.g., Jaradat, Ao-Omari and Momani, 2008). (ii) Let us point out that in the treatment of problems (4.47) and (4.48), one of the difficult points is to give reasonable concept of solutions (see also the works of Zhou and Jiao, 2010a; Hernandez, O’Regan and Balachandran, 2010). Another is that even though the operator A generates a semigroup T (t) in X, it will not be continuous at t = 0 for nonsmooth initial data u0 . (iii) It is worth mentioning that if it is the case when A is a matrix (or even bounded linear operators) then Kilbas, Srivastava and Trujill, 2006, obtained an explicit representation of mild solution to problem (4.47). Let us now give a short summary of this section, which is organized in a way close to that given by Carvalho, Dlotko and Nescimento, 2008. In Subsection 4.5.2 we give brief overview of the construction of functional calculus about almost sectorial operators, state some results about the analytic semigroups of growth order 1 + γ, and summarize some properties on two special functions. In Subsection 4.5.3, we construct a pair of families of operators and present deep anatomy on the properties for these families. Based on the families of operators defined in Subsection 4.5.3, a reasonable concept of solution will be given in Subsection 4.5.4 to problems (4.47), which in turn is used to analyze the existence of mild solutions and classical solutions to the Cauchy problem. The corresponding semilinear problem (4.48) is studied in Subsection 4.5.5. We investigate the existence of mild solutions, and then the existence of classical solutions. Finally, based mainly in Carvalho, Dlotko and Nescimento, 2008; Periago and Straub, 2002, we present three examples in Subsection 4.5.6 to illustrate our results. Remark 4.54. Let us note that results in this section can be easily extended to the case of (general) sectorial operators. 4.5.2

Preliminaries

We first introduce some special functions and classes of functions which will be used in the following, for more details, we refer to Markus, 2006; Periago and

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Straub, 2002. Let −1 < γ < 0, and let Sµ0 with 0 < µ < π be the open sector {z ∈ C \ {0}; | arg z| < µ} and Sµ be its closure, that is Sµ = {z ∈ C \ {0}; | arg z| ≤ µ} ∪ {0}. Set [ [ F0γ (Sµ0 ) = Ψγs (Sµ0 ) Ψ0 (Sµ0 ), s k, one easily sees that ψnk ∈ F0γ (Sµ0 ). Assume that A ∈ Θγω (X) with −1 < γ < 0 and 0 < ω < π/2. Following Periago and Straub, 2002 (see also McIntosh, 1986; Cowling, Doust, McIntosh et al., 1996), a closed linear operator f → f (A) can be constructed for every f ∈ F (Sµ0 ) via a extended functional calculus. In the following we give a short overview to this construction. For f ∈ F0γ (Sµ0 ), via the Dunford-Riesz integral, the operator f (A) is defined by Z 1 f (z)R(z; A)dz, (4.49) f (A) = 2πi Γθ

where the integral contour Γθ := {R+ eiθ } ∪ {R+ e−iθ }, is oriented counter-clockwise and ω < θ < µ < π. It follows that the integral is absolutely convergent and defines a bounded linear operator on X, and its value does not depend on the choice of θ. Notice in particular that for k, n ∈ N ∪ {0} with n > k, ψnk (A) = Ak (A + 1)−n

and the operator ψnk (A) is injective. Notice also that if f ∈ F (Sµ0 ), then there exist k, n ∈ N such that f ψnk ∈ F0γ (Sµ0 ). Hence, for f ∈ F(Sµ0 ), one can define a closed linear operator, still denoted by f (A), D(f (A)) = {x ∈ X; (f ψnk )(A)x ∈ D(A(n−1)k )}, f (A) = (ψnk (A))−1 (f ψnk )(A),

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and the definition of f (A) does not depend on the choice of k and n. We emphasize that f (A) is indeed an extension of the original and the triple (F0γ (Sµ0 ), F (Sµ0 ), f (A)) is called an abstract functional calculus on X (see Markus, 2006). With respect to this construction we collect some basic properties. For more details, we refer to Periago and Straub, 2002. Proposition 4.55. The following assertions hold. (i) αf (A) + βg(A) = (αf + βg)(A), (f g)(A) = f (A)g(A) for ∀ f, g ∈ F0γ (Sµ0 ), α, β ∈ C; (ii) f (A)g(A) ⊂ (f g)(A) for ∀ f, g ∈ F (Sµ0 ), and (iii) f (A)g(A) = (f g)(A), provided that g(A) is bounded or D((f g)(A)) ⊂ D(g(A)). Since for each β ∈ C, z β ∈ F(Sµ0 ) (z ∈ C \ (−∞, 0], 0 < µ < π), one can define, via the triple (F0γ (Sµ0 ), F (Sµ0 ), f (A)), the complex powers of A which are closed by Aβ = z β (A),

β ∈ C,

However, in difference to the case of sectorial operators, having 0 ∈ ρ(A) does not imply that the complex powers A−β with Re(β) > 0, are bounded. The operator A−β belongs to L (X) whenever Re(β) > 1+γ. So, in this situation, the linear space X β := D(Aβ ), β > 1 + γ, endowed with the graph norm |x|β = |Aβ x|, x ∈ X β , is a Banach space. Next, we turn our attention to the semigroup associated with A. Since given t ∈ S 0π −ω , e−tz ∈ H∞ (Sµ0 ) satisfies the conditions (a) and (b) of Lemma 2.13 of 2 Periago and Straub, 2002, the family Z 1 e−tz R(z; A)dz, t ∈ S 0π2 −ω (4.50) T (t) = e−tz (A) = 2πi Γθ here ω < θ < µ < π2 − | arg t|, forms a analytic semigroup of growth order 1 + γ. For more properties on T (t), please see the following proposition. Proposition 4.56. (Periago and Straub, 2002) Let A ∈ Θγω (X) with −1 < γ < 0 and 0 < ω < π/2. Then the following properties remain true. (i) T (t) is analytic in S 0π −ω and 2

dn T (t) = (−A)n T (t), for all t ∈ S 0π2 −ω ; dtn (ii) The functional equation T (s + t) = T (s)T (t) for all s, t ∈ S 0π −ω holds; 2 (iii) There exists a constant C0 = C0 (γ) > 0 such that kT (t)kL (X) ≤ C0 t−γ−1 , for all t > 0;

(iv) The range R(T (t)) of T (t), t ∈ S 0π −ω is contained in D(A∞ ). Particularly, 2 R(T (t)) ⊂ D(Aβ ) for all β ∈ C with Re(β) > 0, Z 1 Aβ T (t)x = z β e−tz R(z; A)xdz, for all x ∈ X, 2πi Γθ and hence there exists a constant C ′ = C ′ (γ, β) > 0 such that kAβ T (t)kL (X) ≤ C ′ t−γ−Re(β)−1 , for all t > 0;

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(v) If β > 1 + γ, then D(Aβ ) ⊂ ΣT , where ΣT is the continuity set of the semigroup {T (t)}t≥0 , that is, ΣT = x ∈ X; lim T (t)x = x . t→0; t>0

Remark 4.57. We note that the condition (ii) of the proposition does not satisfy for t = 0 or s = 0. Recall that semigroups of growth 1 + γ were investigated earlier in deLaubenfels, 1994 and Toropova, 2003. The relation between the resolvent operators of A and the semigroup T (t) is characterized by Proposition 4.58. (Periago and Straub, 2002) Let A ∈ Θγω (X) with −1 < γ < 0 0 < ω < π/2. Then for every λ ∈ C with Re(λ) > 0, one has R(λ, −A) = Rand ∞ −λt e T (t)dt. 0 At the end of this section, we present some properties of two special functions. Definition 4.59. (Miller and Ross, 1993; Podlubny, 1999) The generalized MittagLeffler special function Eα,β is defined by Z ∞ X 1 λα−β eλ zk = dλ, α, β > 0, z ∈ C, Eα,β (z) := Γ(αk + β) 2πi Υ λα − z k=0

where Υ is a contour which starts and ends as −∞ and encircles the disc |λ| ≤ |z|1/α counter-clockwise. If 0 < α < 1, β > 0, then the asymptotic expansion of Eα,β as z → ∞ is given by   1 z (1−β)/α exp(z 1/α ) + εα,β (z), | arg z| ≤ 1 απ, 2 Eα,β (z) = α (4.51)  εα,β (z), | arg(−z)| < (1 − 12 α)π,

where

εα,β (z) = − For short, set

N −1 X n=1

z −n + O(|z|−N ), as z → ∞. Γ(β − αn)

Eα (z) := Eα,1 (z),

eα (z) := Eα,α (z).

Then one has C α α 0 Dt E(ωt )

= ωE(ωtα ),

α−1 α−1 (t eα (ωtα )) 0 Dt

= Eα (ωtα ).

Consider the function of Wright-type ∞ X (−z)n Ψα (z) : = n!Γ(−αn + 1 − α) n=0 =

∞ 1 X (−z)n Γ(nα) sin(nπα), π n=1 (n − 1)!

z∈C

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154

with 0 < α < 1. For −1 < r < ∞, λ > 0, the following results hold. Property 4.60. (W1) Ψα (t) ≥ 0, t > 0; Z ∞ α 1 α Ψα ( α )e−λt dt = e−λ ; (W2) α+1 t t 0 Z ∞ Γ(1 + r) (W3) Ψα (t)tr dt = ; Γ(1 + αr) 0 Z ∞ (W4) Ψα (t)e−zt dt = Eα (−z), z ∈ C; (W5)

Z

4.5.3

0

∞

0

αtΨα (t)e−zt dt = eα (−z), z ∈ C.

Properties of Operators

Throughout this subsection we let A be an operator in the class Θγω (X) and −1 < γ < 0, 0 < ω < π/2. In the sequel, we will succeed in defining two families of operators based on the generalized Mittag-Leffler-type functions and the resolvent operators associated with A. They will be two families of linear and bounded operators. In order to check the properties of the families, we will need a third object, namely the semigroup associated with A. We stress that these families will be used very frequently throughout the rest of this section. Below the letter C will denote various positive constants. Define operator families {Sα (t)}|t∈S 0π , {Pα (t)}|t∈S 0π by −ω −ω 2 2 Z 1 Sα (t) := Eα (−ztα )(A) = Eα (−ztα )R(z; A)dz, 2πi Γθ Z 1 eα (−ztα )R(z; A)dz, Pα (t) := eα (−ztα )(A) = 2πi Γθ

where the integral contour Γθ := {R+ eiθ } ∪ {R+ e−iθ }, is oriented counter-clockwise and ω < θ < µ < π2 − | arg t|. We need some basic properties of these families which are used further in this section. Theorem 4.61. For each fixed t ∈ S 0π −ω , Sα (t) and Pα (t) are linear and bounded 2 operators on X. Moreover, there exists constants Cs = C(α, γ) > 0, Cp = C(α, γ) > 0 such that for all t > 0, Proof.

kSα (t)k ≤ Cs t−α(1+γ) ,

kPα (t)k ≤ Cp t−α(1+γ) .

(4.52)

Note, from the asymptotic expansion of Eα,β that for each fixed t ∈ S 0π −ω , 2

Eα (−ztα ), eα (−ztα ) ∈ F0γ (Sµ0 ).

Therefore, by (4.49), the operators families {Sα (t)}|t∈S 0π 0

2

−ω

, {Pα (t)}|t∈S 0π 2

−ω

are

well-defined, and for each t ∈ S π −ω , Sα (t) and Pα (t) are linear bounded operators 2

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155

on X. So, to prove the theorem, it is sufficient to prove that the estimates in (4.52) hold. Let T (t), t ∈ S 0π −ω , be the semigroup defined by (4.50). Then by (W4) and the 2 Fubini Theorem, we get Z 1 Sα (t)x = Eα (−ztα )R(z; A)xdz 2πi Γθ Z ∞ Z α 1 (4.53) Ψα (λ) e−λzt R(z; A)xdzdλ = 0 Γθ Z2πi ∞ = Ψα (s)T (stα )xds, t ∈ S 0π2 −ω , x ∈ X. 0

A similar argument shows that Z ∞ Pα (t)x = αsΨα (s)T (stα )xds,

t ∈ S 0π2 −ω , x ∈ X.

0

(4.54)

Hence, by (4.53), (4.54), Proposition 4.56 (iii), (W1) and (W3), we have Z ∞ |Sα (t)x| ≤ C0 Ψα (s)s−(1+γ) t−α(1+γ) |x|ds 0

≤ C0

Γ(−γ) t−α(1+γ) |x|, t > 0, x ∈ X, Γ(1 − α(1 + γ))

|Pα (t)x| ≤ αC0

Z

∞

0

Ψα (s)s−γ t−α(1+γ) |x|ds

Γ(1 − γ) −α(1+γ) ≤ αC0 t |x| t > 0, x ∈ X. Γ(1 − αγ) Therefore, the estimates in (4.52) hold. This completes the proof.

From now on, we will frequently use the representations (4.53) and (4.54) for operators Sα (t) and Pα (t), respectively. Theorem 4.62. For t > 0, Sα (t) and Pα (t) are continuous in the uniform operator topology. Moreover, for every r > 0, the continuity is uniform on [r, ∞). Proof. Let ǫ > 0 be given. For choose δ1 , δ2 > 0 such that Z δ1 2C0 Ψα (s)s−(1+γ) ds ≤ rα(1+γ) 0

every r > 0, it follows from (W3) that we may ǫ , 3

2C0 rα(1+γ)

Z

∞

δ2

Ψα (s)s−(1+γ) ds ≤

ǫ . 3

(4.55)

Then we deduce, by Proposition 4.56 (i), that there exists a positive constant δ such that Z δ2 ǫ α Ψα (s)kT (tα (4.56) 1 s) − T (t2 s)kds ≤ , 3 δ1 for t1 , t2 ≥ r and |t1 − t2 | < δ.

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On the other hand, using (4.55), (4.56) and Theorem 4.61, we get Z δ1 α |Sα (t1 )x − Sα (t2 )x| ≤ Ψα (s)(kT (tα 1 s)k + kT (t2 s)k)|x|ds 0

+ +

Z

δ2

δ Z 1∞ δ2

≤

2C0

α Ψα (s)kT (tα 1 s) − T (t2 s)k|x|ds α Ψα (s)(kT (tα 1 s)k + kT (t2 s)k)|x|ds

Z

δ1

Ψα (s)s−(1+γ) |x|ds rα(1+γ) 0 Z δ2 α + Ψα (s)kT (tα 1 s) − T (t2 s)k|x|ds δ1 Z ∞ 2C0 Ψα (s)s−(1+γ) |x|ds + α(1+γ) r δ2

≤ ǫ|x|, for any x ∈ X, that is, kSα (t1 ) − Sα (t2 )k ≤ ǫ,

which implies that Sα (t) is uniformly continuous on [r, ∞) in the uniform operator topology and hence, by the arbitrariness of r > 0, Sα (t) is continuous in the uniform operator topology for t > 0. A similar argument enables us to give the characterization of continuity on Pα (t). This completes the proof. Theorem 4.63. Let 0 < β< 1 − γ. Then (i) The range R(Pα (t)) of Pα (t) for t > 0, is contained in D(Aβ ); (ii) S ′ α (t)x = −tα−1 APα (t)x (x ∈ X), and S ′ α (t)x for x ∈ D(A) is locally integrable on (0, ∞); (iii) for all x ∈ D(A) and t > 0, kASα (t)xk ≤ Ct−α(1+γ) kAxk, here C is a constant depending on γ, α. Proof. It follows from Proposition 4.56 (iv) that for all x ∈ X, t > 0, T (t)x ∈ D(Aβ ) with β > 0. Therefore, in view of (4.54), Proposition 4.56 (iv) and (W3) we have Z ∞ β |A Pα (t)x| ≤ αsΨα (s)kAβ T (tα s)k|x|ds 0 Z ∞ ≤ αC ′ t−α(γ+β+1) Ψα (s)s−(β+γ) ds|x| 0

≤ αC ′

Γ(1 − β − γ) t−α(1+β+γ) |x|, Γ(1 − α(β + γ + 1))

which implies that the assertion (i) holds. From (i), it is easy to see that for all x ∈ X,

Sα′ (t)x = −tα−1 APα (t)x.

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Moreover, for every x ∈ D(A), one has by Proposition 4.56 (iv), Z ∞ |tα−1 APα (t)x| ≤ tα−1 αsΨα (s)kT (tα s)k|Ax|ds 0

Γ(1 − γ) −αγ−1 t |Ax|. ≤ αC0 Γ(1 − αγ)

Since −αγ −1 > −1, this shows that S ′ α (t)x for each x ∈ D(A), is locally integrable on (0, ∞), that is , (ii) is true. Moreover, Proposition 4.56 (iv) and (4.53) imply that Z ∞ |ASα (t)x| ≤ C0 t−α(1+γ) Ψα (s)s−1−γ ds|Ax| 0

Γ(−γ) ≤ C0 t−α(1+γ) |Ax|, x ∈ D(A). Γ(1 − α(1 + γ))

This means that (iii) holds, and completes the proof.

Remark 4.64. Particularly, from the proof of Theorem 4.63 (i), we can conclude that kAPα (t)k ≤ Ct−α(2+γ) , where C is a constant depending on γ, α. Moreover, using a similar argument with that in Theorem 4.62, we have that APα (t) for t > 0 is continuous in the uniform operator topology. Theorem 4.65. The following properties hold. (i) (ii) (iii) (iv)

Let β > 1 + γ. For all x ∈ D(AβR), limt→0;t>0 Sα (t)x = x, t for all x ∈ D(A),(Sα (t) − I)x = 0 −sα−1 APα (s)xds, α for all x ∈ D(A), t > 0, 0 Dt Sα (t)x = −ASα (t)x, for all t > 0, Sα (t) = 0 Dα−1 (tα−1 Pα (t)). t

Proof.

For any x ∈ X, note by (4.53) and (W3 ) that Z ∞ Sα (t)x − x = Ψα (s)(T (tα s)x − x)ds. 0

On the other hand, by Proposition 4.56 (v) it follows that D(Aβ ) ⊂ ΣT in view of β > 1 + γ. Therefore, we deduce, using Proposition 4.56 (iii), that for any x ∈ D(Aβ ), there exists a function η(s) ∈ L1 (0, +∞) depending on Ψα (s) such that k Ψα (s)(T (tα s)x − x) k≤ η(s).

Hence, by means of the Lebesgue’s dominated convergence theorem we obtain that Sα (t)x − x → 0, as t → 0, that is, the assertion (i) remains true. From (i) and Theorem 4.63 (ii) we get for all x ∈ D(A), Z t −λα−1 APα (λ)xdλ, (Sα (t) − I)x = lim (Sα (t)x − Sα (s)x) = s→0

0

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which implies that the assertion (ii) holds. To prove (iii), first it is easy to see that ϕ10 ∈ F(Sµ0 ) and the operator ϕ0 (A) is injective. Taking x ∈ D(A), by Proposition 4.55 (iii) one has 1 α α (A)x. Sα (t)x = Eα (−zt )(A)x = (Eα (−zt )ϕ0 )(A) ϕ0 Moreover, by (4.51) we have supz→∞ |ztα Eα (−ztα )| < ∞, which implies that |zEα (−ztα )(1+ z)−1 | ≤ C|z|−1 t−α , as z → ∞,

where C is a constant which is independent of t. Consequently, −zEα (−ztα )(1 + z)−1 ∈ F0γ (Sµ0 ).

(4.57)

Notice also that C α α −1 0 Dt Eα (−zt )(1+z) R(z; A)

= (−z)Eα (−ztα )(1+z)−1 R(z; A).

Combining Proposition 4.55 (ii) and (4.57), we get Z 1 C α α β −1 D ((E (−zt )(1 + z ) )(A)) = (−z)Eα (−ztα )(1+ z)−1 R(z; A)dz α 0 t 2πi Γθ = (−z)(A) Eα (−ztα )(1 + z)−1 (A) = −A Eα (−ztα )(1+ z)−1 (A). Hence, we obtain

α 0 Dt Sα (t)x

= −A Eα (−ztα )(1 + z)−1 (A)(1 + z)(A)x

= −A(Eα (−ztα ))(A)x = −ASα (t)x.

This proves (iii). For (iv), by a similar argument with (iii), one can prove that tα−1 eα (−ztα ) belongs to F0γ (Sµ0 ) for t > 0 and hence α−1 α−1 (t Pα (t)) 0 Dt

=

α−1 ((tα−1 eα (−ztα )(A)) 0 Dt

in view of α−1 0 Dt

This completes the proof.

= (Eα (−ztα ))(A) = Sα (t),

tα−1 eα (−ztα ) = Eα (−ztα ).

Before proceeding with our theory further, we present the following result. Lemma 4.66. If R(λ, −A) is compact for every λ > 0, then T (t) is compact for every t > 0. Proof. Note first that as a consequence of Theorem 3.13 Periago and Straub, R ∞in −λs 2002., for every λ ∈ C with Re(λ) > 0, R(λ; −A) = 0 e T (s)ds defines a bounded linear operator on X. Therefore, we obtain Z ∞ λR(λ; −A)T (t) − T (t) = λ e−λs (T (t + s) − T (t))ds. (4.58) 0

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Let ǫ > 0 be given. For every λ > 0 and t > 0, it follows from Theorem 4.62 that there exists a ν > 0 such that ǫ sup kT (s + t) − T (t)k ≤ . 2 s∈[0,ν] So λ

Z

ν

0

e−sλ kT (t + s) − T (t)kds ≤

ǫ . 2

(4.59)

On the other hand, by Proposition 4.56 (iii) we get Z ∞

Z ∞

λ e−sλ (T (s + t) − T (t))ds ≤ λC e−sλ ((t + s)−1−γ + t−γ−1 )ds ν

ν

≤ 2Ct−γ−1 e−λν ,

which implies that there exists a λ0 > 0 such that

Z ∞

ǫ −sλ

e (T (s + t) − T (t))ds λ

≤ 2 , λ ≥ λ0 . ν Thus, for all λ ≥ λ0 , using (4.58), (4.59) and (4.60) we deduce that Z ν kλR(λ; −A)T (t) − T (t)k ≤ λ e−sλ kT (t + s) − T (t)kds 0 Z ∞ +λ e−sλ kT (s + t) − T (t)kds

(4.60)

ν

≤ ǫ.

It follows from the arbitrariness of ν > 0 that lim kλR(λ; −A)T (t) − T (t)k = 0.

λ→∞

Since λR(λ; −A)T (t) is compact for every λ > 0 and t > 0, T (t) is compact for every t > 0. With the help of this lemma we now show the following result. Theorem 4.67. If R(λ, −A) is compact for every λ > 0, then Sα (t), Pα (t) are compact for every t > 0. Proof.

Let ǫ > 0 be arbitrary. Put Z ∞ Z ςǫ (t) = Ψα (s)T (stα − ǫtα )ds, ζǫ (t) = ǫ

∞

Ψα (s)T (stα )ds.

ǫ

Then, one has ζǫ (t) = T (ǫtα )ςǫ (t), and it is easy to prove that for every t > 0, ςǫ (t) is bounded linear operators on X. Therefore, from the compactness of T (t), t > 0, we see that ζǫ (t) is compact for every t > 0. On the other hand, note that Z ǫ

Z ǫ

α −α(1+γ) kζǫ (t) − Sα (t)k ≤ Ψα (s)T (st )ds ≤ C0 t Ψα (s)s−1−γ ds. 0

0

Hence, it follows from the compactness of ζǫ (t), t > 0 that Sα (t) is compact for every t > 0. By a similar technique we can conclude that Pα (t) is compact for every t > 0. The proof is completed.

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4.5.4

Linear Problems

Let A ∈ Θγω (X) with −1 < γ < 0 and 0 < ω < π/2. We discuss the existence and uniqueness of mild solution and classical solutions for inhomogeneous linear abstract Cauchy problem (4.47). We assume the following condition. (H∗ ) u ∈ C([0, T ]; X), g1−α ∗ u ∈ C 1 ((0, T ]; X), u(t) ∈ D(A) for t ∈ (0, T ], Au ∈ L1 ((0, T ); X), and u satisfies (4.47). Then, by Definitions 1.5 and 1.8, one can rewrite (4.47) as Z t Z t 1 1 α−1 (t − s) Au(s)ds + (t − s)α−1 f (s)ds, (4.61) u(t) = u0 − Γ(α) 0 Γ(α) 0

for t ∈ [0, T ]. Before presenting the definition of mild solution of problem (4.47), we first prove the following lemma. Lemma 4.68. If u : [0, T ] → X is a function satisfying the assumption (H∗ ), then u(t) satisfies the following integral equation Z t u(t) = Sα (t)u0 + (t − s)α−1 Pα (t−s)f (s)ds, t ∈ (0, T ]. 0

Proof. Note that the Laplace transform of a abstract function f ∈ L1 (R+ , X) is R∞ defined by fb(λ) = 0 e−λt f (t)dt, λ > 0. Applying Laplace transfer to (4.61) we get u b(λ) =

u0 λ

−

1 u(λ) λα Ab

+

fb(λ) λα ,

that is,

u b(λ) = λα−1 (λα + A)−1 u0 + (λα + A)−1 fb(λ).

On the other hand, using Proposition 4.58 and (W2) we deduce that λα−1 (λα + A)−1 u0 + (λα + A)−1 fb(λ) Z ∞ Z ∞ α −λα t α−1 e T (t)u0 dt + e−λ t T (t)fb(λ)dt =λ Z ∞ 0 Z0 ∞ Z ∞ α d −(λt)α α = e T (t )u0 dt + αtα−1 e−(λt) t T (tα )f (s)e−sλ dsdt dλ 0 0 Z0 ∞ Z ∞ αt 1 −λtτ α = Ψα α e T (t )dτ dt τα τ 0 0 Z ∞Z ∞Z ∞ 1 tα α α−1 t Ψ e−λt T ( α )f (s)e−sλ dτ dsdt + 2α α τ τ τ 0 0 0 Z ∞Z ∞ α α 1 t = Ψ e−λt T ( α )dτ dt α+1 α τ α τ τ 0 Z ∞0Z ∞ Z ∞ + ατ tα−1 Ψ(τ )T (tα τ )f (s)e−(s+t)λ dτ dsdt 0 Z ∞0 Z ∞0 −λt = e Ψα (τ )T (tα τ )dτ 0 0 Z ∞ Z ∞ Z t + e−tλ (t − s)α−1 f (s) ατ Ψ(τ )T ((t − s)α τ )dτ dsdt 0

0

0

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= =

Z

Z

∞

0 ∞

0

This implies that u b(λ) =

e

−λt

Sα (t)dt +

Z

∞

e

−λt

0

Z

0

t

161

(t − s)α Pα (t − s)f (s)dsdt

Z t e−λt Sα (t)u0 + (t − s)α−1 Pα (t − s)f (s)ds dt. 0

Z

0

∞

Z t e−λt Sα (t)u0 + (t − s)α−1 Pα (t − s)f (s)ds dt. 0

Now using the uniqueness of the Laplace transform (cf. Theorem 1.1.6 of Xiao and Liang, 1998), we deduce that the assertion of lemma holds. This completes this proof. Motivated by Lemma 4.68, we adopt the following concept of mild solution to problem (4.47). Definition 4.69. By a mild solution of problem (4.47), we mean a function u ∈ C((0, T ]; X) satisfying Z t u(t) = Sα (t)u0 + (t − s)α−1 Pα (t−s)f (s)ds, t ∈ (0, T ]. 0

Remark 4.70. It is to be noted that (a) unlike the case of strongly continuous operator semigroups, we do not require the mild solution of problem (4.47) to be continuous at t = 0. Moreover, in general, since the operator Sα (t) is singular at t = 0, solutions to problem (4.47) are assumed to have the same kind of singularity at t = 0 as the operator Sα (t). This is the case, for instance, if f ≡ 0 so that we have that u(t) = Sα (t)u0 , which presents a discontinuity at the initial time; (b) When u0 ∈ D(Aβ ), β > 1 + γ, it follows from Theorem 4.65 (i) that the mild solution is continuous at t = 0. For f ∈ L1 ((0, T ); X), the initial problem (4.47) has an unique mild solution for every u0 ∈ X. We will now be interested in imposing further condition on f and u0 so that the mild solution will become a classical solution. To this end we first introduce the following definition. Definition 4.71. By a classical solution to problem (4.47), we mean a function α u(t) ∈ C([0, T ]; X) with C 0 Dt u(t) ∈ C((0, T ]; X), which, for all t ∈ (0, T ], takes values in D(A) and satisfies (4.47). We are now ready to state our main result in this subsection. Theorem 4.72. Let A ∈ Θγω (X) with 0 < ω < π2 . Suppose that f (t) ∈ D(A) for all 0 < t ≤ T , Af (t) ∈ L∞ ((0, T ); X), and f (t) is Hölder continuous with an exponent θ′ > α(1 + γ), that is, ′

|f (t) − f (s)| ≤ K|t − s|θ , for all 0 < t, s ≤ T. Then, for every u0 ∈ D(A), there exists a classical solution to problem (4.47) and this solution is unique.

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Proof. For u0 ∈ D(A), let u(t) = Sα (t)u0 (t > 0). Then it follows from Theorem 4.65 (i, iii) that u(t) is a classical solution of the following problem C α 0 < t ≤ T, 0 Dt u(t) + Au(t) = 0, (4.62) u(0) = u0 . Moreover, from Lemma 4.68, it is easy to see that u(t) is the only solution to problem (4.62). Put Z t w(t) = (t − s)α−1 Pα (t − s)f (s)ds, 0 < t ≤ T. 0

Then from the assumptions on f and Theorem 4.61 we obtain Z t kAw(t)k ≤ (t − s)α−1 kPα (t − s)kkAf (t)kL∞ ((0,T );X) ds 0

1 −γα t , −αγ which implies that w(t) ∈ D(A) for all 0 < t ≤ T . α Next, we show C w(0) = 0 and hence 0 Dt w(t) ∈ C((0, T ; X). Since Z t d α−1 C α 1 w(t) = Sα (t − s)f (s)ds, (4.63) 0 Dt w(t) = 0 Dt 0 Dt dt 0 Rt in view of Properties 1.17, 1.21, 1.22 and Theorem 4.65 (iv). Let v(t) = 0 Sα (t − s)f (s)ds, it remains to prove v(t) ∈ C 1 ((0, T ]; X). Let h > 0 and h ≤ T − t. Then it is easy to verify the identity Z t v(t + h) − v(t) Sα (t + h − s) − Sα (t − s) = f (s)ds h h 0 Z 1 t+h Sα (t + h − s)f (s)ds. + h t Again by the assumptions on f and Theorem 4.61, we have, for t > 0 fixed, k(t − s)α−1 APα (t − s)f (s)k ≤ Cp (t − s)−αγ−1 kAf (s)k ∈ L1 ((0, T ); X), for all s ∈ [0, t). Therefore, using Theorem 4.63 (ii) and the Dominated Convergence Theorem we get Z t Sα (t + h−s) − Sα (t − s) lim f (s)ds h→0 0 h Z t (4.64) = (t − s)α−1 (−A)Pα (t − s)f (s)ds ≤ Cp kAf (t)kL∞ ((0,T );X)

0

= −Aw(t). Furthermore, note that Z 1 t+h Sα (t + h−s)f (s)ds h t Z 1 h Sα (s)f (t + h − s)ds = h 0 Z 1 h = Sα (s)(f (t + h−s) − f (t − s))ds h 0 Z Z 1 h 1 h Sα (s)(f (t − s) − f (t))ds + Sα (s)f (t)ds. + h 0 h 0

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From Theorem 4.61 and the H¨ older continuity on f we have Z h Cs Khθ′ −α(1+γ) 1 ≤ S (s)(f (t + h−s) − f (t−s))ds , α h 0 1 − α(1 + γ)

and

Z ′ Cs Khθ −α(1+γ) 1 h S (s)(f (t − s) − f (t))ds ≤ α 1 + θ − α(1 + γ) . h 0 Rh And since f (t) ∈ D(A) (0 < t ≤ T ), limh→0 h1 0 Sα (s)f (t)ds = f (t) in view of Theorem 4.65 (i). Hence, Z 1 t+h Sα (t + h−s)f (s)ds → f (t), as h → 0+ . (4.65) h t Combining (4.64) and (4.65) we deduce that v is differentiable from the right at t ′ and v+ (t) = f (t) − Aw(t), t ∈ (0, T ]. By a similar argument with the above one ′ has that v is differentiable from the left at t and v− (t) = f (t) − Aw(t), t ∈ (0, T ]. Next, we prove Aw(t) ∈ C((0, T ]; X). To the end, let Aw(t) = I1 (t) + I2 (t), where Z t (t − s)α−1 APα (t − s) f (s) − f (t) ds, I1 (t) = 0 Z t I2 (t) = A(t−s)α−1Pα (t−s)f (t)ds. 0

By Theorem 4.65 (ii), we obtain I2 (t) = −(Sα (t) − I)f (t). So, by the assumption of f and Theorem 4.62 note that I2 (t) is continuous for 0 < t ≤ T . To prove the same conclusion for I1 (t), we let 0 < h ≤ T − t and write I1 (t + h) − I1 (t) Z t = (t + h−s)α−1APα (t + h−s) − (t−s)α−1 APα (t−s) f (s)−f (t) ds 0

+

Z

t

0

+

Z

t

(t + h−s)α−1APα (t + h−s)(f (t) − f (t + h))ds

t+h

(t + h−s)α−1APα (t + h − s) f (s) − f (t + h) ds

=: h1 (t) + h2 (t) + h3 (t).

For h1 (t), on the one hand, it follows from Theorem 4.62 that lim (t + h − s)α−1APα (t + h−s)(f (s) − f (t))

h→0

= (t − s)α−1 APα (t−s)(f (s) − f (t)).

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On the other hand, for t ∈ (0, T ] fixed, by Remark 4.64 and the assumption on f , we get |(t + h − s)α−1 APα (t + h − s) f (s) − f (t) | ≤ Cp′ K(t + h − s)−α(1+γ)−1 (t − s)θ

′

′

≤ Cp′ K(t − s)(θ −α−αγ)−1 ∈ L1 ((0, t); X).

Thus, by mean of the Dominated Convergence Theorem one has Z t lim (t+ h− s)α−1APα (t + h−s) f (s)−f (t) ds h→0

=

Z

0

t

0

(t − s)α−1APα (t − s) f (s)−f (t) ds,

which implies that h1 (t) → 0 as h → 0+ . For h2 (t), using Theorem 4.63 (i), Remark 4.64, Z t (t+ h−s)α−1kAPα (t+h−s)k|f (t)−f (t + h)|ds 0 Z t ′ ≤ Cp′ K(t + h − s)−α(1+γ)−1 hθ ds 0

′

Cp′ Khθ (h−α(1+γ) − (h + t)−α(1+γ) ). = α(1 + γ)

This yields h2 (t) → 0 as h → 0+ . Moreover,h3(t) → 0 as h → 0+ by the following estimate Z t+h α−1 (t + h−s) Pα (t + h − s) Af (s) − Af (t + h) ds t

2Cp ≤ kAf (s)kL∞ (0,T ;X) h−αγ −αγ in view of Af (s) ∈ L∞ ((0, T ); X) and Theorem 4.62. The same reasoning establishes I1 (t − h) − I1 (h) → 0 as h → 0+ . Consequently, Aw ∈ C((0, T ]; X), which implies that v ′ ∈ C((0, T ]; X), provided that f is continα uous on (0, T ]. Then, by (4.63) we have C 0 Dt w ∈ C((0, T ]; X). Hence, we prove that u + w is a classical solution to problem (4.47), and Lemma 4.68 implies that it is unique. This completes the proof. 4.5.5

Nonlinear Problems

In this subsection we apply the theory developed in the previous sections to nonlinear abstract Cauchy problem (4.48). Definition 4.73. By a mild solution to problem (4.48), we mean a function u ∈ C((0, T ]; X) satisfying Z t u(t) = Sα (t)u0 + (t − s)α−1 Pα (t−s)f (s, u(s))ds, t ∈ (0, T ]. 0

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Theorem 4.74. Let A ∈ Θγω (X) with −1 < γ < − 12 and 0 < ω < π2 . Suppose that the nonlinear mapping f : (0, T ] × X → X is continuous with respect to t and there exist constants M, N > 0 such that |f (t, x) − f (t, y)| ≤ M (1 + |x|ν−1 + |y|ν−1 )|x − y|, |f (t, x)| ≤ N (1 + |x|ν ), γ for all t ∈ (0, T ] and for each x, y ∈ X, where ν is a constant in [1, − 1+γ ). Then, for every u0 ∈ X, there exists a T0 > 0 such that the problem (4.48) has a unique mild solution defined on (0, T0 ].

For fixed r > 0, we introduce the metric space

Proof.

Fr (T, u0 ) = {u ∈ C((0, T ]; X); ρT (u, Sα (t)u0 ) ≤ r}, ρT (u1 , u2 ) = sup |u1 (t) − u2 (t)|. t∈(0,T ]

It is not difficult to see that, with this metric, Fr (T, u0 ) is a complete metric space. Take L = T α(1+γ) r + Cs ku0 k, then for any u ∈ Fr (T, u0 ), we have |sα(1+γ) u(s)| ≤ sα(1+γ) |u − Sα (t)u0 | + sα(1+γ) |Sα (t)u0 | ≤ L. Choose 0 < T0 ≤ T such that

T0−αγ −α(ν(1+γ)+γ) + Cp N Lν T0 β(−γα, 1 − να(1 + γ)) ≤ r, −αγ

(4.66)

T0−αγ 1 −α(γ+(1+γ)(ν−1)) + 2Lρ−1 T0 β(−αγ, 1 − α(1 + γ)(ν − 1)) ≤ , −αγ 2

(4.67)

Cp N

M Cp

where β(η1 , η2 ) with ηi > 0, i = 1, 2, denotes the Beta function. Assume that u0 ∈ X. Consider the mapping Γα given by Z t α (Γ u)(t) = Sα (t)u0 + (t − s)α−1 Pα (t − s)f (s, u(s))ds, u ∈ Fr (T0 , u0 ). 0

By the assumptions on f , Theorem 4.61 and Theorem 4.62, we see that (Γα u)(t) ∈ C((0, T ]; X) and Z t α (t − s)−αγ−1 (1 + ku(s)kν )ds |(Γ u)(t) − Sα (t)u0 | ≤ Cp N 0

T −αγ + ≤ Cp N 0 −αγ

≤ Cp N ≤ r,

Z

0

t

Cp N Lν (t − s)−αγ−1 s−να(1+γ) ds

T0−αγ −α(ν(1+γ)+γ) + Cp N Lν T0 β(−γα, 1 − να(1 + γ)) −αγ

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in view of (4.66). So, Γα maps Fr (T0 , u0 ) into itself. Next, for any u, v ∈ Fr (T0 , u0 ), by the assumptions on f and Theorem 4.61, we have |(Γα u)(t) − (Γα v)(t)| Z t = (t − s)α−1 Pα (t − s) f (s, u(s)) − f (s, v(s)) ds 0

Z

t

(t − s)−αγ−1 (1 + |u(s)|ρ−1 + |v(s)|ρ−1 )|u(s) − v(s)|ds Z t ≤ Cp M ρt (u, v) (t − s)−αγ−1 (1 + 2Lν−1 s−α(ν−1)(1+γ) )ds ≤ Cp M

0

0 −α(γ+(1+γ)(ν−1)) ≤ 2L T0 β(−αγ, 1 −αγ T ρT0 (u, v). + M Cp 0 ρ−1

− α(1 + γ)(ν − 1))ρT0 (u, v)

−αγ This yields that Γα is a contraction on Fr (T0 , u0 ) due to (4.67). So, Γα has a unique fixed point u ∈ Fr (T0 , u0 ) in view of the Banach Fixed Point Theorem, this means that u is a mild solution to problem (4.48) defined on (0, T0 ]. The proof is completed. By a similar argument with the proof of Theorem 4.74 we have: Corollary 4.75. Assume that A ∈ Θγω (X) with −1 < γ < − 32 and 0 < ω < π2 . Suppose in addition that the nonlinear mapping f : (0, T ] × X β → X, β ∈ (1 + γ, −1 − 2γ), is continuous with respect to t and there exist constants M, N > 0 such that |f (t, x) − f (t, y)| ≤ M (1 + |x|ν−1 + |y|ν−1 )|x − y|β , β β |f (t, x)| ≤ N (1 + |x|νβ ),

for all t ∈ (0, T ] and for each x, y ∈ X β , where ν is a constant in [1, − γ+β 1+γ ). Then, β for every u0 ∈ X , there exists a T0 > 0 such that the problem (4.48) has a unique mild solution u ∈ C((0, T0 ]; X β ). Remark 4.76. If A ∈ Θγω (X) with −1 < γ < 0 and 0 < ω < π2 , then we can derive the local existence and uniqueness of mild solutions to problem (4.48), under the conditions: (i) u0 ∈ X β with β > 1 + γ; (ii) the nonlinear mapping f : [0, T ] × X → X is continuous with respect to t and there exists a continuous function Lf (·) : R+ → R+ such that |f (t, x) − f (t, y)| ≤ Lf (r)|x − y|,

for all 0 ≤ t ≤ T and for each x, y ∈ X satisfying |x|, |y| ≤ r. Indeed, for r > such that

Cp T0−αγ −αγ

supt∈[0,T ] |f (t, u0 )| fixed, we may choose 0 < T0 ≤ T

sup k(Sα (t) − I)u0 k +

t∈[0,T0 ]

Cp T0−αγ Lf (r)r + sup |f (t, u0 )| < r −αγ t∈[0,T0 ]

(4.68)

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in view of Theorem 4.65 (i). Assume that the map Γα is defined the same as in Theorem 4.74 and the space Fr (T0 , u0 ) is replaced by the following Banach space: Fr′ (T0 , u0 ) = {u ∈ C([0, T0 ]; X); u(0) = u0 and

sup |u − u0 | ≤ r}.

t∈[0,T0 ]

Then, it is easy to verify, thanks to the assumptions on f and (4.68), that Γα maps Fr′ (T0 , u0 ) into itself and is a contraction on Fr′ (T0 , u0 ), which implies that the problem (4.48) has a unique mild solution defined on [0, T0 ]. Since 1 > 1 + γ (−1 < γ < − 12 ), X 1 = D(A) is a Banach space endowed with the graph norm |x|X 1 = |Ax|, for x ∈ X 1 . The following is the existence of X 1 -smooth solutions. Theorem 4.77. Let A ∈ Θγω (X) with −1 < γ < − 12 and 0 < ω < π2 and u0 ∈ X 1 . Assume that there exists a continuous function Mf (·) : R+ → R+ and a constant Nf > 0 such that the nonlinear mapping f : (0, T ] × X 1 → X 1 satisfies |f (t, x) − f (t, y)|X 1 ≤ Mf (r)|x − y|X 1 , |f (t, Sα (t)u0 )|X 1 ≤ Nf (1 + t−α(1+γ) |u0 |X 1 ), for all 0 < t ≤ T and for each x, y ∈ X 1 satisfying supt∈(0,T ] |x(t) − Sα (t)u0 |X 1 ≤ r, supt∈(0,T ] |y(t) − Sα (t)u0 |X 1 ≤ r. Then there exists a T0 > 0 such that the problem (4.48) has a unique mild solution defined on (0, T0 ]. Proof.

For u0 ∈ X 1 and r > 0, set Fr′′ (T, u0 ) = {u ∈ C((0, T ]; X 1 ); sup |u − Sα (t)u0 |X 1 ≤ r}. t∈(0,T ]

For any u ∈

Fr′′ (T, u0 ),

by the assumptions on f and Theorem 4.61 we have

|(Γα u)(t) − Sα (t)u0 |X 1 Z t ≤ (t − s)α−1 kPα (t − s)k|f (s, u(s)) − f (s, Sα (t)u0 )|X 1 ds 0 Z t + (t − s)α−1 kPα (t − s)k|f (s, Sα (t)u0 ))|X 1 ds 0 Z t ≤ Cp (t − s)−αγ−1 (Mf (r)r + Nf + Nf s−α(1+γ) |u0 |)ds 0

≤ Cp (Mf (r)r + Nf )

T −αγ + Cp Nf T −α(1+2γ) β(−γα, 1 − α(1 + γ))|u0 |. −αγ

Using this result, it follows from an analogous idea with Theorem 4.74 that the claim of theorem follows. Here we omit the details. Next, we will derive mild solutions under the condition of compactness on the resolvent of A. Theorem 4.78. Let A ∈ Θγω (X) with −1 < γ < 0 and 0 < ω < π2 . Let (H1) R(λ, −A) is compact for every λ > 0;

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(H2) f : [0, T ] × X → X is a Carath´ eodory function and for any r > 0, there exists 1 such that a function mr (t) ∈ Lp ((0, T ); R+ ) with p > − αγ |f (t, x)| ≤ mr (t), and lim inf r→+∞

|mr (t)|Lp (0,T ) =σ 1 + γ, the problem (4.48) has at least a mild solution, provided that T 1−(1+αγ) q 1/q Cp σ < 1, (4.69) 1 − (1 + αγ)q where q = p/(p − 1). Proof.

Assume that u0 ∈ D(Aβ ). On C([0, T ]; X) define the map Z t (Γα u)(t) = Sα (t)u0 + (t − s)α−1 Pα (t − s)f (s, u(s))ds. 0

From our assumptions it is easy to see that Γµ is well defined and maps C([0, T ]; X) into itself. Put Ωr = {u ∈ C([0, T ]; X); kuk ≤ r, for all 0 ≤ t ≤ T }, for r > 0 as selected below. We seek for solutions in Ωr . We claim that there exists an integer r > 0 such that Γα maps Ωr into Ωr . In fact, if this is not the case, then for each r > 0, there would exist ur ∈ Ωr and tr ∈ [0, T ] such that k(Γα ur )(tr )k > r. On the other hand, by (H2) and Theorem 4.61 we get r < |(Γα ur )(tr )| ≤ |Sα (tr )u0 | +

Z

tr

|(tr − s)α−1 Pα (tr − s)f (s, u(s))|ds 0 Z tr ≤ sup |Sα (t)u0 | + Cp (tr − s)−1−αγ mr (s)ds t∈[0,T ] 0 p1 Z tr 1q Z tr −(1+αγ)q mpr (s)ds ≤ sup |Sα (t)u0 | + Cp s ds t∈[0,T ]

0

0

T 1−(1+αγ) q q1 ≤ sup |Sα (t)u0 | + Cp kmr kLp (0,T ) , 1 − (1 + αγ)q t∈[0,T ]

where q = p/(p − 1). Dividing on both sides by r and taking the lower limit as r → ∞, one has T 1−(1+αγ) q 1/q , 1 ≤ Cp σ 1 − (1 + αγ)q which contradicts (4.69). Hence for some positive integer r, Γα (Ωr ) ⊂ Ωr . The rest of the proof is divided into three steps. Step I. Γα is continuous on Ωr .

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Take {un }∞ n=1 ⊂ Ωr with un → u in C([0, T ]; X). Then by the continuity of f with respect to the second argument we deduce that f (s, un (s)) → f (s, u(s)) a.e. s ∈ [0, T ]. Moreover, observe from (H2) and Theorem 4.61, that for fixed 0 < t ≤ T , (t − s)α−1 |Pα (t − s)f (s, un (s))| ≤ Cp (t − s)−1−αγ mr (s). Thus, by means of the Lebesgue’s dominated convergence theorem we obtain that Z

0

t

(t − s)α−1 kPα (t − s)k|f (s, un (s)) − f (s, u(s))|ds → 0,

which means that limn→∞ kΓα un − Γα uk∞ = 0, that is, Γα is continuous on Ωr . Step II. P = {(Γα u)(·); · ∈ [0, T ], u ∈ Ωr } is equicontinuous. For 0 < t1 < t2 ≤ T and δ > 0 small enough, we have |(Γα u)(t1 ) − (Γα u)(t2 )| ≤ I1 + I2 + I3 + I4 + I5 , where I1 = |Sα (t1 )u0 − Sα (t2 )u0 |, Z t2 (t2 − s)α−1 kPα (t2 − s)f (s, u(s))kds, I2 = I3 = I4 =

t1 t1 −δ

Z

Z

0 t1

t1 −δ t1

I5 =

Z

0

(t1 − s)α−1 kPα (t2 − s) − Pα (t1 − s)k|f (s, u(s))|ds,

(t1 − s)α−1 kPα (t2 − s) − Pα (t1 − s)k|f (s, u(s))|ds,

|(t2 − s)α−1 − (t1 − s)α−1 |kPα (t2 − s)k|f (s, u(s))|ds.

From Theorem 4.62 and Theorem 4.65 (i) it is easy to see that I1 → 0 when t1 → t2 . Moreover, using (H2) and Theorem 4.61 we get I2 ≤ Cp I3 ≤ ≤

(t2 − t1 )1−(1+αγ)q 1 − (1 + αγ)q

sup s∈[0,t1 −δ]

sup s∈[0,t1 −δ]

1/q

kmr kLp (0,T ) ,

kPα (t2 − s) − Pα (t1 − s)k kPα (t2 − s) − Pα (t1 − s)k

Z

0

t1 −δ

(t1 − s)qα−q qds

1/q

t1+q(α−1) − δ 1+q(α−1) 1

1 + q(α − 1)

kmr kLp (0,T )

kmr kLp (0,T ) ,

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I4 ≤ Cp

Z

≤ 2Cp I5 ≤

Z

t1

0

Z

t1

(t1 − s)α−1 t1 −δ 1−(1+αγ)q

· 2(t1 − s)−α(γ+1) mr (s)ds

δ kmr kLp (0,T ) , 1 − (1 + αγ)q Cp ((t1 − s)α−1 − (t2 − s)α−1 )(t2 − s)−α(1+γ) mr (s)ds

t1

Cp ((t1 − s)−γα−1 − (t2 − s)−αγ−1 )mr (s)ds 0 Z t1 1/q ≤ Cp ((t1 − s)−q(γα+1) − (t2 − s)−q(αγ+1) )ds kmr kLp (0,T )

≤

0

1−(1+αγ)q 1−(1+αγ)q 1/q (t − t )1−(1+αγ)q t − t2 2 1 + 1 kmr kLp (0,T ) . = Cp 1 − (1 + αγ)q 1 − (1 + αγ)q

It follows from Theorem 4.62 that Ii (i = 2, 3, 4, 5) tends to zero independent of u ∈ Ωr as t2 − t1 → 0, δ → 0. Hence, we can conclude that k(Γα u)(t1 ) − (Γα u)(t2 )k → 0, as t2 − t1 → 0, and the limit is independent of u ∈ Ωr . For the case when 0 = t1 < t2 ≤ T , since Z

0

t2

(t2 − s)α−1 kP (t2 − s)f (s, u(s))kds ≤ Cp

1−q(αγ+1)

t2 1 − q(αγ + 1)

1/q

kmr kLp (0,T ) ,

in view of (H2) and Theorem 4.61, k(Γα u)(t2 )k can be made small when t2 is small independently of u ∈ Ωr . Thus, we prove that the assertion in Step II holds. Step III. For each t ∈ [0, T ], {(Γα u)(t); u ∈ Ωr } is precompact in X. For the case when t = 0, it is not difficult to see that {(Γα u)(0); u ∈ Ωr } = {u0 : u ∈ Ωr } is compact. Let t ∈ (0, T ] be fixed and ǫ, δ > 0. For u ∈ Ωr , define the map Γα ǫ,δ by (Γα ǫ,δ u)(t) = Sα (t)u0 +

Z

0

t−ǫ Z ∞ δ

ατ (t − s)α−1 Ψα (τ )T ((t − s)α τ )f (s, u(s))dτ ds.

Since A has compact resolvent, {T (t)}t>0 is compact in view of Theorem 4.67. Thus, for each t ∈ (0, T ], {(Fǫ,δ u)(t); u ∈ Ωr , δ > 0, 0 < ǫ < t} is precompact in X. On the other hand, using (H2) and Theorem 4.61, a direct calculation yields |(Γα u)(t) − (Γα ǫ,δ u)(t)| Z t Z δ α−1 α ≤ ατ (t − s) Ψα (τ )T ((t − s) τ )f (s, u(s))dτ ds 0 0 Z t Z ∞ α−1 α + ατ (t − s) Ψα (τ )T ((t − s) τ )f (s, u(s))dτ ds t−ǫ

δ

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≤

Z

t

0

+

Z

Cp (t − s)−1−αγ mr (s)ds t

t−ǫ

Z

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δ

τ −γ Ψα (τ )dτ Z ∞ Cp (t − s)−1−αγ mr (s)ds τ −γ Ψα (τ )dτ 0

δ

Z δ T 1−(1+αγ)q 1/q p kmr kL (0,T ) τ −γ Ψα (τ )dτ ≤ Cp 1 − (1 + αγ)q 0 ǫ1−(1+αγ)q 1/q Γ(1 − γ) kmr kLp (0,T ) . + Cp 1 − (1 + αγ)q Γ(1 − γα)

Using the total boundedness we have that for each t ∈ (0, T ] {(Γα u)(t); u ∈ Ωr } is precompact in X. Therefore, for each t ∈ [0, T ], {(Γα u)(t); u ∈ Ωr } is precompact in X. Finally, by Steps I-III and the Arzela-Ascoli theorem, we conclude that Γα is a compact operator. Hence, from Schauder’s second fixed point theorem it follows that Γα has a fixed point, which gives rise to a mild solution. This completes the proof. Theorem 4.79. Let A ∈ Θγω (X) with 0 < ω < π2 and −1 < γ < − 12 . Suppose that there exists a continuous function Mf′ (·) : R+ → R+ and a constant κ > α(1 + γ) such that the nonlinear mapping f : [0, T ] × X → X satisfies |f (t, x) − f (s, y)| ≤ Mf′ (r)(|t − s|κ + |x − y|), for all 0 ≤ t ≤ T and x, y ∈ X satisfying |x|, |y| ≤ r. In addition, let the assumptions of Theorem 4.77 be satisfies and u be a mild solution corresponding to u0 , defined on [0, T0 ]. Then u is in fact the unique classical solution to problem (4.48), existing on [0, T0 ], provided that u0 ∈ D(A) with Au0 ∈ D(Aβ ), β > (1 + γ). Proof. In order to prove that u is a classical solution, by Theorem 4.72 and the condition on f , we only have to verify that u is Hölder continuous with an exponent ς > α(1 + γ) on (0, T0 ]. For fixed t ∈ (0, T0 ], take 0 < h < 1 such that h + t ≤ T0 . We estimate the difference |u(t + h) − u(t)| ≤ |Sα (t + h)u0 − Sα (t)u0 | Z h α−1 + (t + h − s) P(t + h − s)f (s, u(s))ds 0 Z t + (t − s)α−1 P(t − s)[f (s + h, u(s + h)) − f (s, u(s))]ds 0

= I1 + I2 + I3 .

According to Theorem 4.61, Theorem 4.63 (ii) and the assumptions on f we obtain Z t Cp (t + h)−αγ − t−αγ , I1 = −sα−1 APα (s)u0 ds ≤ −αγ 0

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′

Z

t

(t − s)−αγ−1 (|h|κ + |u(s + h) − u(s)|)ds Z t M ′ Cp −αγ κ ≤ T0 h + M ′ Cp (t − s)−αγ−1 |u(s + h) − u(s)|ds. −αγ 0

I3 ≤ M Cp

Put N2 =

0

sup |f (t, u(t))|. Then, it follows from Theorem 4.61 that

t∈(0,T0 )

I2 ≤ Cp ≤

Z

0

h

(t + h − s)−αγ−1 |f (s, u(s))|ds

Cp N2 (t + h)−αγ − t−αγ . −αγ

Collecting these estimates and using the inequality (t+h)−αγ −t−αγ ≤ h−αγ (0 < −αγ < 1) we have |u(t + h) − u(t)|

Mp′ −αγ κ Cp N2 + Cp (t + h)−αγ − t−αγ + T h −αγ −αγ 0 Z t + M ′ Cp (t − s)−αγ−1 |u(s + h) − u(s)|ds 0 Z t Cp N2 + Cp + M ′ Cp ς h + M ′ Cp (t − s)−αγ−1 |u(s + h) − u(s)|ds, ≤ −αγ 0 ≤

where ς = min{κ, −αγ} > α(γ + 1). Now, it follows from the usual Gronwall’s inequality that u has Hölder continuity on (0, T0 ]. This completes the proof of theorem. 4.5.6

Applications

In this subsection, we present three examples (Examples 4.80-4.82) motivated from physics, which do not aim at generality but indicate how our theorems can be applied to concrete problems. Examples 4.80 and 4.81 are inspired directly from the work of Carvalho, Dlotko and Nescimento, 2008, and they describe anomalous diffusion on fractals (physical objects of fractional dimension, like some amorphous semiconductors or strongly porous materials; see Anh and Leonenko, 2001; Metzler and Klafter, 2000 and references therein). Example 4.80 is the limit problem of certain fractional diffusion equations in complex systems on domains of “dumb-bell with a thin handle”(see, e.g., Anh and Leonenko, 2001; Metzler and Klafter, 2000). Example 4.81 displays anomalous dynamical behavior of anomalous transport processes (see, e.g., Anh and Leonenko, 2001; Metzler and Klafter, 2000). Example 4.82 is a modified fractional Schrödinger equation with fractional Laplacians whose physical background is statistical physics and fractional quantum mechanics (see, e.g., Hu and Kallianpur, 2000; Podlubny, 1999). We refer the reader to Kirane, Laskri and Tatar, 2005 and references therein for more research results related to

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fractional Laplacians. Example 4.80. Consider the system of fractional partial differential equations in the form C α x ∈ Ω, t > 0,  0 Dt w − ∆w + w = f (w),    ∂w   = 0, x ∈ ∂Ω,   ∂n 1 C α (4.70) x ∈ (0, 1), 0 Dt v − (gvx )x + v = f (v),   g    v(0) = w(P0 ) , v(1) = w(P1 ),    w(x, 0) = w0 (x) x ∈ Ω, v(x, 0) = v0 (x), x ∈ (0, 1),

where Ω = D1 ∪ D2 and D1 and D2 are mutually disjoint bounded domains in α RN (N ≥ 2) with smooth boundaries, joined by the line segment Q0 , and C 0 Dt , 0 < α < 1, is the regularized Caputo fractional derivative of order α, that is, ∂ Z t 1 −α −α α (t − s) u(s, x)ds − t u(0, x) . (4.71) (C D u)(t, x) = 0 t Γ(1 − α) ∂t 0

When α = 1, we regard (4.70) as the limit problem of (4.44) as ε → 0, which is described in more detail in Example 4.49. Here, our objective is to show that system (4.70) is well posed in V0p = Lp (Ω) ⊕ Lpg (0, 1) (1 ≤ p < ∞). Let the operators A0 : D(A0 ) ⊂ V0p 7→ V0p be defined by D(A0 ) = {(w, v) ∈ V0P ; w ∈ D(∆Ω ), v ∈ Lpg (0, 1), w(P0 ) = v(0), w(P1 ) = v(1)}, 1 A0 (w, v) = (−∆w + w, − (gv ′ )′ + v), g

(w, v) ∈ V0p ,

where ∆Ω is the Laplace operator with homogeneous Neumann boundary conditions in Lp (Ω) and ∂u 2,p =0 . D(∆Ω ) = u ∈ W (Ω); ∂n ∂Ω ′

N From Example 4.49, if p > N2 , then A0 ∈ Θµ−γ (V0p ) for some γ ′ ∈ (0, 1 − 2p ) and π µ ∈ (0, 2 ). Therefore, system (4.70) can be seen as an abstract evolution equation in the form C α t > 0, 0 Dt u + A0 u = f (u), (4.72) u(0) = u0 = (w0 , v0 ) ∈ V0p .

We assume that the nonlinearity f : R → R is globally Lipschitz continuous. It can define a Nemitskiˇı operator from V0p into itself by f (w, v) = (fΩ (w), fI (v)) with fΩ (w)(x) = f (w(x)), x ∈ Ω and fI (v)(x) = f (v(x)), x ∈ (0, 1) such that |f (u) − f (u′ )|V0p ≤ L′′ (r)|u − u′ |V0p

for all u, u′ ∈ V0p satisfying |u|V0p , |u′ |V0p ≤ r. Hence, from Remark 4.76, (4.72) (that is, (4.70)) has a unique mild solution provided that u0 ∈ D(Aβ0 ) with β > 1 − γ ′ (in particular, u0 ∈ D(A0 )).

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Example 4.81. Let Ω be a bounded domain in RN (N ≥ 1) with boundary ∂Ω of class C 4 . Consider the fractional initial-boundary value problem of form  C α  (0 Dt u)(t, x) − ∆u(t, x) = f (u(t, x)), t > 0, x ∈ Ω, (4.73) u| = 0,  ∂Ω u(0, x) = u0 (x), x ∈ Ω,

in the space C l (Ω) (0 < l < 1), where ∆ stands for the Laplacian operator with α respect to the spatial variable and C 0 Dt , representing the regularized Caputo fractional derivative of order α (0 < α < 1), is given by (4.71). Set e = −∆, A

e = {u ∈ C 2+l (Ω); u = 0 on ∂Ω}. D(A)

e+ν ∈ It follows from Example 4.51 that there exist ν, ε > 0 such that A l l 2 −1 Θ π −ε (C (Ω)). Then, problem (4.73) can be written abstractly as 2

C α 0 Dt u(t)

e + Au(t) = f (u),

t > 0.

With respect to the nonlinearity f , we assume that f : R → R is continuously differentiable and satisfies the condition k(r) |f (x) − f (y)| ≤ |x − y|, |x|, |y| ≤ r (4.74) r for any r > 0. It defines a Nemitskiˇı operator from C l (Ω) into itself by f (u)(x) = f (u(x)) with |f (u) − f (v)|C l (Ω) ≤ k(r)|u − v|C l (Ω) ,

|v|C l (Ω) , |u|C l (Ω) ≤ r.

Noting 2l − 1 ∈ (−1, − 21 ), we then obtain the following conclusion: (i) according eβ ) with β > to Remark 4.76, (4.73) has a unique mild solution for each u0 ∈ D(A l ′ ′′ 2 . Moreover, (ii) if f , f are continuously differentiable functions satisfying the condition (4.74), then one finds that the Nemitskiˇı operator satisfies the assumptions e with of Theorem 4.77 and Theorem 4.79, which implies that for each u0 ∈ D(A) l β e e Au0 ∈ D(A ) (β > 2 ), the corresponding mild solution to (4.73) is also a unique classical solution. Example 4.82. Consider the following fractional Cauchy problem C α (0 Dt y)(t, x) + (−i∆ + σ)1/2 u(t, x) = f (u(t, x)), t > 0, x ∈ R2 , (4.75) u(0, x) = u0 (x), x ∈ R2 , in L3 (R2 ), where σ > 0 is a suitable constant, i∆ is the Schrödinger operator and C α 0 Dt (0 < α < 1) is given by (4.71). Let b = (−i∆ + σ)1/2 , D(A) b = W 1,3 (R2 ) A

(a Sobolev space).

5 Then i∆ generates a β-times integrated semigroup S β (t) with β = 12 on L3 (R2 ) β β c c such that kS (t)kL (L3 (R2 )) ≤ M t for all t ≥ 0 and some constants M > 0 (see Neerven and Straub, 1998). Therefore, by virtue of Theorem 1.3.5 and Definition 1.3.1 for C = I of Xiao and Liang, 1998, we deduce that the operator −i∆ + σ

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(L3 (R2 )), which denotes the family of all linear closed operators belongs to Θβ−1 π 2 A : D(A) ⊂ L3 (R2 ) → L3 (R2 ) satisfying σ(A) ⊂ S π2 = {z ∈ C \ {0}; | arg z| ≤ π π 2 } ∪ {0}, and for every 2 < µ < τ there exists a constant Cµ such that kR(z; A)k ≤ Cµ |z|β−1 . Thus, it follows from Proposition 3.6 of Periago and Straub, 2002 that b ∈ Θ−1+2β A (L3 (R2 )) for some 0 < ω < π2 . Moreover, the system (4.75) can be ω rewritten as follows: ( α b (C t > 0, 0 Dt y)(t, x) + Au = f (u), u(0, x) = u0 ∈ L3 (R2 ).

Assume that f : C → C is globally Lipschitz continuous. Then we have a Nemitskiˇı b operator from L3 (R2 ) to itself given by f (u)(x) = f (u(x)), and for a constant L(r) 3 2 and all u, v ∈ L (R ) such that |u|L3 (R2 ) ≤ r and |v|L3 (R2 ) ≤ r. Consequently, it follows from Remark 4.76 that (4.75) has a unique mild solution provided u0 ∈ b τ with τ > 5 . D(A) 6

4.6

Notes and Remarks

The results in Section 4.2 are taken from Zhou, Zhang and Shen, 2013. The material in Section 4.3 due to Zhou, Shen and Zhang, 2013. The main results in Section 4.4 is taken from Wang, Zhou and Fe˘ckan, 2014. The contents of Section 4.5 are adopted from Wang, Chen and Xiao, 2012.

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Chapter 5

Fractional Boundary Value Problems via Critical Point Theory

5.1

Introduction

The main purpose of this chapter is to present a new approach via critical point theory to study the existence of solutions for the boundary value problem of fractional differential equations. This approach is, to the best of our knowledge, novel and it may open a new approach to deal with some types of nonlinear fractional differential equations with certain boundary conditions. 5.2

5.2.1

Existence of Solution for BVP with Left and Right Fractional Integrals Introduction

In Section 5.2, we consider the fractional boundary value problem (BVP for short) of the following form   d 1 D−β (u′ (t)) + 1 D−β (u′ (t)) + ∇F (t, u(t)) = 0, a.e. t ∈ [0, T ], 0 t t T dt 2 2 (5.1)  u(0) = u(T ) = 0,

where 0 Dt−β and t DT−β are the left and right Riemann-Liouville fractional integrals of order 0 ≤ β < 1 respectively, F : [0, T ] × RN → R is a given function satisfying some assumptions and ∇F (t, x) is the gradient of F at x. In particular, if β = 0, BVP (5.1) reduces to the standard second order BVP. Physical models containing fractional differential operators have recently renewed attention from scientists which is mainly due to applications as models for physical phenomena exhibiting anomalous diffusion. A strong motivation for investigating the fractional BVP (5.1) comes from fractional advection-dispersion equation (ADE for short). A fractional ADE is a generalization of the classical ADE in which the second-order derivative is replaced with a fractional-order derivative. In contrast to the classical ADE, the fractional ADE has solutions that resemble the highly skewed and heavy-tailed breakthrough curves observed in field and labora177

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tory studies (see, Benson, Schumer, Meerschaert et al., 2001; Benson, Wheatcraft and Meerschaert, 2000a), in particular in contaminant transport of ground-water flow (see, Benson, Wheatcraft and Meerschaert, 2000b). Benson et al. stated that solutes moving through a highly heterogeneous aquifer violations violates the basic assumptions of local second-order theories because of large deviations from the stochastic process of Brownian motion. Let φ(t, x) represents the concentration of a solute at a point x at time t in an arbitrary bounded connected set Ω ⊂ RN . According to Benson, Wheatcraft and Meerschaert, 2000a; Fix and Roop, 2004, the N -dimensional form of the fractional ADE can be written as ∂φ = −∇(vφ) − ∇(∇−β (−k∇φ)) + f, in Ω, (5.2) ∂t where v is a constant mean velocity, k is a constant dispersion coefficient, vφ and −k∇φ denote the mass flux from advection and dispersion respectively. The components of ∇−β in (5.2) are linear combination of the left and right Riemann-Liouville fractional integral operators ∂φ −β −β −β (∇ (−k∇φ))i = (q −∞ Dxi + (1 − q) xi D+∞ ) − k , i = 1, ..., N, (5.3) ∂xi where q ∈ [0, 1] describes the skewness of the transport process, and β ∈ [0, 1) is the order of the Liouville-Weyl left and right fractional integral operators on the real line (see Definition 1.12). This equation may be interpreted as stating that the mass flux of a particle is related to the negative gradient via a combination of the left and right fractional integrals. Eq. (5.3) is physically interpreted as a Fick’s law for concentrations of particles with a strong nonlocal interaction. For discussions of Eq. (5.2), see Benson, Wheatcraft and Meerschaert, 2000b; Fix and Roop, 2004. When β = 0, the dispersion operators in (5.2) are identical and the classical ADE is recovered. In a more general version of (5.2), k is replaced by a symmetric positive definite matrix. A special case of the fractional ADE (Eq. (5.2)) describes symmetric transitions. In this case, ∇−β is equivalent to the symmetric operator 1 1 −β + xi D+∞ , i = 1, ..., N. (5.4) (∇−β )i = −∞ Dx−β i 2 2 Combining (5.2) and (5.4) gives the mass balance equation for advection and symmetric fractional dispersion. The fractional ADE has been studied in one dimension (see, e.g., Benson, Wheatcraft and Meerschaert, 2000b), and in three dimension (see Lu, Molz and Fix, 2002), over infinite domains by using the Fourier transform of fractional differential operators to determine a classical solution. Variational methods, especially the Galerkin approximation has been investigated to find the solutions of fractional BVP (see, e.g., Fix and Roop, 2004) and fractional ADE (see, e.g., Ervin and Roop, 2006) on a finite domain by establishing some suitable fractional derivative spaces. A Lagrangian structure for some partial differential equations is obtained by using

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the fractional embedding theory of continuous Lagrangian systems (see, Cresson, 2010). We note that for nonlinear fractional BVP, some fixed point theorems were already applied successfully to investigate the existence of solutions (see, e.g., Agarwal, Benchohra and Hamani, 2010; Ahmad and Nieto, 2009; Benchohra, Hamani and Ntouyas, 2009; Zhang, 2010). However, it seems that fixed point theorem is not appropriate for discussing BVP (5.1) since the equivalent integral equation is not easy to be obtained. On the other hand, there is another effective approach, calculus of variation, which proved to be very useful in determining the existence of solutions for integer order differential equation provided that equation with certain boundary conditions possesses a variational structure on some suitable Sobolev spaces, for example, one can refer to Corvellec, Motreanu and Saccon, 2010; Li, Liang and Zhang, 2005; Mawhin and Willem, 1989; Rabinowitz, 1986; Tang and Wu, 2010 and the references therein for detailed discussions. However, to the best of author’s knowledge, there are few results on the solutions to fractional BVP which were established by the critical point theory, since it is often very difficult to establish a suitable space and variational functional for fractional differential equations with some boundary-conditions. These difficulties are mainly caused by the following properties of fractional integral and fractional derivative operators. These are: (i) the composition rule in general fails to be satisfied by fractional integral and fractional derivative operators (e.g., Lemma 2.21 in Kilbas, Srivastava and Trujillo, 2006); (ii) the fractional integral is a singular integral operator and fractional derivative operator is non-local (see Definitions 1.5, 1.6 and 1.8), and (iii) the adjoint of a fractional differential operator is not the negative of itself (e.g., Lemma 2.7 in Kilbas, Srivastava and Trujillo, 2006). It should be mentioned here that the fractional variational principles were started to be investigated deeply. The fractional calculus of variations was introduced by Riewe, 1996, where he presented a new approach to mechanics that allows one to obtain the equations for a nonconservative system using certain functionals. Klimek, 2002, gave another approach by considering fractional derivatives, and corresponding Euler-Lagrange equations were obtained, using both Lagrangian and Hamiltonian formalisms. Agrawal, 2002, presented Euler-Lagrange equations for unconstrained and constrained fractional variational problems, and as a continuation of Agrawal’s work, the generalized mechanics are considered to obtain the Hamiltonian formulation for the Lagrangian depending on fractional derivative of coordinates (see, Rabei, Nawafleh and Hijjawi et al., 2007). The recent book by Malinowska and Torres, 2012, provides a broad introduction to the important subject of fractional calculus of variations. In Section 5.2, we investigate the existence of solutions for BVP (5.1). The

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technical tool is the critical point theory. In Subsection 5.2.2, we develop a fractional derivative space and some propositions are proven which will aid in our analysis, and in Subsection 5.2.3, we shall exhibit a variational structure for BVP (5.1). The results presented in Subsections 5.2.2 and 5.2.3 are basic, but crucial to limpidly reveal that under some suitable assumptions, the critical points of the variational functional defined on a suitable Hilbert space are the solutions of BVP (5.1). In Subsection 5.2.4, we will introduce some critical point theorems. Also, various criteria on the existence of solutions for BVP (5.1) will be established. As it was already mentioned, if β = 0, then BVP (5.1) reduces to the standard second order BVP of the following form ′′ u (t) + ∇F (t, u(t)) = 0, a.e. t ∈ [0, T ], u(0) = u(T ) = 0,

where F : [0, T ] × RN → R is a given function and ∇F (t, x) is the gradient of F at x. Although many excellent results have been worked out on the existence of solutions for second order BVP (e.g., Li, Liang and Zhang, 2005; Rabinowitz, 1986), it seems that no similar results were obtained in the literature for fractional BVP. The present results in Section 5.2 are to show that the critical point theory is an effective approach to tackle the existence of solutions for fractional BVP. 5.2.2

Fractional Derivative Space

Let us recall that for any fixed t ∈ [0, T ] and 1 ≤ p < ∞, Z T p1 Z t p1 |u(t)|p dt and kuk = max |u(t)|. kukLp[0,t] = |u(ξ)|p dξ , kukLp = 0

0

t∈[0,T ]

The following result yields the boundedness of the Riemann-Liouville fractional integral operators from the space Lp ([0, T ], RN ) to the space Lp ([0, T ], RN ), where 1 ≤ p < ∞. It should be mentioned here that the similar results have been presented in Fix and Roop, 2004; Kilbas, Srivastava and Trujillo, 2006; Samko, Kilbas and Marichev, 1993. Lemma 5.1. Let 0 < α ≤ 1 and 1 ≤ p < ∞. For any f ∈ Lp ([0, T ], RN ), we have tα kf kLp[0,t] , for ξ ∈ [0, t], t ∈ [0, T ]. (5.5) k0 Dξ−α f kLp[0,t] ≤ Γ(α + 1) Proof. Inspired by the proof of the Young’s theorem in Adams, 1975, we can prove (5.5). In fact, if p = 1, we have Z Z 1 t ξ α−1 k0 Dξ−α f kL1 [0,t] = (ξ − τ ) f (τ )dτ dξ Γ(α) 0 0 Z tZ ξ 1 (5.6) (ξ − τ )α−1 |f (τ )|dτ dξ ≤ Γ(α) 0 0 Z t Z t 1 |f (τ )|dτ (ξ − τ )α−1 dξ = Γ(α) 0 τ

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Z t 1 |f (τ )|(t − τ )α dτ = Γ(α + 1) 0 tα kf kL1 [0,t] , for t ∈ [0, T ]. ≤ Γ(α + 1) Now, suppose that 1 < p < ∞ and g ∈ Lq ([0, T ], RN ), where p1 + 1q = 1. We have Z t Z t Z ξ Z ξ α−1 α−1 = g(ξ) (ξ − τ ) f (τ )dτ dξ g(ξ) τ f (ξ − τ )dτ dξ 0 0 0 0 Z t Z ξ ≤ |g(ξ)| τ α−1 |f (ξ − τ )|dτ dξ 0 0 Z t Z t α−1 (5.7) τ dτ |g(ξ)kf (ξ − τ )|dξ = τ

0

≤

Z

t

0 α

τ

α−1

dτ

Z

t

τ

q

|g(ξ)| dξ

1q Z

t

p

|f (ξ − τ )| dξ

τ

p1

t kf kLp[0,t] kgkLq [0,t] , for t ∈ [0, T ]. α For any fixed t ∈ [0, T ], consider the functional Hξ∗f : Lq ([0, T ], RN ) → R Z t Z ξ Hξ∗f (g) = (ξ − τ )α−1 f (τ )dτ g(ξ)dξ. ≤

0

(5.8)

0

According to (5.7), it is obvious that Hξ∗f ∈ (Lq ([0, T ], RN ))∗ , where (Lq ([0, T ], RN ))∗ denotes the dual space of Lq ([0, T ], RN ). Therefore, by (5.7), (5.8) and Riesz representation theorem, there exists h ∈ Lp ([0, T ], RN ) such that Z t Z tZ ξ h(ξ)g(ξ)dξ = (ξ − τ )α−1 f (τ )dτ g(ξ)dξ (5.9) 0

0

0

and

khkLp[0,t] ≤

tα kf kLp[0,t] α

for all g ∈ Lq ([0, T ], RN ). Hence, we have by (5.9) Z ξ 1 1 h(ξ) = (ξ − τ )α−1 f (τ )dτ = 0 Dξ−α f (ξ), Γ(α) Γ(α) 0

(5.10)

for ξ ∈ [0, t],

which means that

k0 Dξ−α f kLp [0,t] =

tα 1 khkLp[0,t] ≤ kf kLp [0,t] Γ(α) Γ(α + 1)

(5.11)

according to (5.10). Combining (5.6) and (5.11), we obtain the inequality (5.5). In order to establish a variational structure for construct appropriate function spaces. Denote by functions h ∈ C ∞ ([0, T ], RN ) with h(0) = h(T ) = for any h ∈ C0∞ ([0, T ], RN ) and 1 < p < ∞, we

BVP (5.1), it is necessary to C0∞ ([0, T ], RN ) the set of all 0. According to Lemma 5.1, have h ∈ Lp ([0, T ], RN ) and

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∈ Lp ([0, T ], RN ). Therefore, one can construct a set of space E0α,p , which depend on Lp -integrability of the Caputo fractional derivative of a function. Definition 5.2. Let 0 < α ≤ 1 and 1 < p < ∞. The fractional derivative space E0α,p is defined by the closure of C0∞ ([0, T ], RN ) with respect to the norm Z T p1 Z T α p , ∀u ∈ E0α,p . (5.12) kukα,p = D u(t)| dt |u(t)|p dt + |C 0 t C α 0 Dt h

0

0

Remark 5.3.

(i) It is obvious that the fractional derivative space E0α,p is the space of functions α u ∈ Lp ([0, T ], RN ) having an α-order Caputo fractional derivative C 0 Dt u ∈ p N L ([0, T ], R ) and u(0) = u(T ) = 0. α (ii) For any u ∈ E0α,p , noting the fact that u(0) = 0, we have C 0 Dt u(t) = α 0 Dt u(t), t ∈ [0, T ] according to Property 1.9. (iii) It is easy to verify that E0α,p is a reflexive and separable Banach space. Proposition 5.4. Let 0 < α ≤ 1 and 1 < p < ∞. The fractional derivative space E0α,p is a reflexive and separable Banach space. Proof. In fact, owing to Lp ([0, T ], RN ) be reflexive and separable, the Cartesian product space Lp2 ([0, T ], RN ) = Lp ([0, T ], RN ) × Lp ([0, T ], RN ) is also a reflexive and separable Banach space with respect to the norm X p1 2 kvkLp2 = kvi kpLp ,

(5.13)

i=1

where v = (v1 , v2 ) ∈ Lp2 ([0, T ], RN ). Consider the space Ω = {(u, 0C Dtα u) : u ∈ E0α,p }, which is a closed subset of p L2 ([0, T ], RN ) as E0α,p is closed. Therefore, Ω is also a reflexive and separable Banach space with respect to the norm (5.13) for v = (v1 , v2 ) ∈ Ω. We form the operator A : E0α,p → Ω as follows A : u → (u,

C α 0 Dt u),

∀u ∈ E0α,p .

It is obvious that kukα,p = kAukLp2 ,

α which means that the operator A : u → (u, C 0 Dt u) is an isometric isomorphic α,p mapping and the space E0 is isometric isomorphic to the space Ω. Thus E0α,p is a reflexive and separable Banach space, and this completes the proof.

Applying Property 1.22 and Lemma 5.1, we now can give the following useful estimates. Proposition 5.5. Let 0 < α ≤ 1 and 1 

1 p

and

1 p

+

1 q

183

= 1, then 1

kuk ≤

T α− p Γ(α)((α − 1)q + 1)

1 q

α kC 0 Dt ukLp .

(5.15)

Proof. For any u ∈ E0α,p , according to (1.38) and noting the fact that u(0) = 0, we have that −α C α 0 Dt (0 Dt u(t))

= u(t),

t ∈ [0, T ].

Therefore, in order to prove inequalities (5.14) and (5.15), we only need to prove that Tα α p ≤ kC Dα ukLp , (5.16) k0 Dt−α (C D u)k L 0 t Γ(α + 1) 0 t where 0 < α ≤ 1 and 1 p1 and 1p + 1q = 1. α p N Firstly, we note that C 0 Dt u ∈ L ([0, T ], R ), the inequality (5.16) follows from (5.5) directly. We are now in a position to prove (5.17). For α > p1 , choose q such that α,p 1 1 p + q = 1. ∀u ∈ E0 , we have Z 1 t α α−1 C α |0 Dt−α (C D u(t))| = (t − s) D u(s)ds 0 t 0 s Γ(α) 0 Z t q1 1 α (t − s)(α−1)q ds kC ≤ 0 Dt ukLp Γ(α) 0 1

≤

T q +α−1

1

α kC 0 Dt ukLp

1

α kC 0 Dt ukLp ,

Γ(α)((α − 1)q + 1) q 1

=

T α− p

Γ(α)((α − 1)q + 1) q

and this completes the proof. According to (5.14), we can consider

E0α,p

α kukα,p = kC 0 Dt ukLp =

Z

0

with respect to the norm p1 T C α p |0 Dt u(t)| dt

(5.18)

in the following analysis. Proposition 5.6. Let 0 < α ≤ 1 and 1 p1 and the sequence {uk } converges weakly to u in E0α,p , i.e., uk ⇀ u. Then uk → u in C([0, T ], RN ), i.e., ku − uk k → 0, as k → ∞.

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Proof. If α > 1p , then by (5.15) and (5.18), the injection of E0α,p into C([0, T ], RN ), with its natural norm k · k, is continuous, i.e., if uk → u in E0α,p , then uk → u in C([0, T ], RN ). Since uk ⇀ u in E0α,p , it follows that uk ⇀ u in C([0, T ], RN ). In fact, For any h ∈ (C([0, T ], RN ))∗ , if uk → u in E0α,p , then uk → u in C([0, T ], RN ), and thus h(uk ) → h(u). Therefore, h ∈ (E0α,p )∗ , which means that (C([0, T ], RN ))∗ ⊆ (E0α,p )∗ . Hence, if uk ⇀ u in E0α,p , then for any h ∈ (C([0, T ], RN ))∗ , we have h ∈ (E0α,p )∗ , and thus h(uk ) → h(u), i.e., uk ⇀ u in C([0, T ], RN ). By the Banach-Steinhaus theorem, {uk } is bounded in E0α,p and, hence, in C([0, T ], RN ). We are now in a position to prove that the sequence {uk } is equiuniformly continuous. Let p1 + 1q = 1 and 0 ≤ t1 < t2 ≤ T . ∀f ∈ Lp ([0, T ], RN ), by using the Hölder inequality and noting that α > p1 , we have |0 Dt−α f (t1 ) − 0 Dt−α f (t2 )| 1 2 Z t1 Z t2 1 α−1 α−1 (t1 − s) f (s)ds − (t2 − s) f (s)ds = Γ(α) Z0 Z0 t1 1 t1 α−1 α−1 ≤ (t1 − s) f (s)ds − (t2 − s) f (s)ds Γ(α) 0Z 0 1 t2 α−1 + (t2 − s) f (s)ds Γ(α) t1 Z t1 1 ((t1 − s)α−1 − (t2 − s)α−1 )|f (s)|ds ≤ Γ(α) 0Z t2 1 + (t2 − s)α−1 |f (s)|ds Γ(α) t1 Z t1 1q 1 α−1 α−1 q ((t1 − s) − (t2 − s) ) ds kf kLp ≤ Γ(α) 0Z t2 1q 1 + (t2 − s)(α−1)q ds kf kLp Γ(α) t1 Z t1 q1 1 ≤ ((t1 − s)(α−1)q − (t2 − s)(α−1)q )ds kf kLp Γ(α) 0Z t2 1q 1 (α−1)q + (t2 − s) ds kf kLp Γ(α) t1 1 kf kLp (α−1)q+1 (α−1)q+1 − t2 + (t2 − t1 )(α−1)q+1 ) q = 1 (t1 Γ(α)(1 + (α − 1)q) q kf kLp (α−1)q+1 1q ) + 1 ((t2 − t1 ) q Γ(α)(1 + (α − 1)q) 2kf kLp α−1+ 1q ≤ 1 (t2 − t1 ) Γ(α)(1 + (α − 1)q) q 1 2kf kLp α− p . = 1 (t2 − t1 ) q Γ(α)(1 + (α − 1)q)

(5.19)

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Therefore, the sequence {uk } is equi-uniformly continuous since, for 0 ≤ t1 < t2 ≤ T , by applying (5.19) and in view of (5.18), we have −α C α α |uk (t1 ) − uk (t2 )| = |0 Dt−α (C 0 Dt1 uk (t1 )) − 0 Dt2 (0 Dt2 uk (t2 ))| 1 1

≤

2(t2 − t1 )α− p

1

α kC 0 Dt u k k L p

1

kuk kα,p

Γ(α)(1 + (α − 1)q) q 1

=

2(t2 − t1 )α− p

Γ(α)(1 + (α − 1)q) q 1

≤ c(t2 − t1 )α− p ,

where 1p + 1q = 1 and c ∈ R+ is a constant. By the Ascoli-Arzela theorem, {uk } is relatively compact in C([0, T ], RN ). By the uniqueness of the weak limit in C([0, T ], RN ), every uniformly convergent subsequence of {uk } converges uniformly on [0, T ] to u. The proof is completed. 5.2.3

Variational Structure

In this section, we will establish a variational structure which enables us to reduce the existence of solutions of BVP (5.1) to the one of critical points of corresponding functional defined on the space E0α,p with p = 2 and 21 < α ≤ 1. First of all, making use of Property 1.17, for any u ∈ AC([0, T ], RN ), BVP (5.1) transforms to  β β β β  d 1 D− 2 ( D− 2 u′ (t)) + 1 D− 2 ( D− 2 u′ (t)) + ∇F (t, u(t)) = 0, 0 t t T 0 t t T dt 2 2 (5.20)  u(0) = u(T ) = 0,

for almost every t ∈ [0, T ], where β ∈ [0, 1). Furthermore, in view of Definition 1.8 and Property 1.10, it is obvious that u ∈ AC([0, T ], RN ) is a solution of BVP (5.20) if and only if u is a solution of the following problem   d 1 Dα−1 (C Dα u(t)) − 1 Dα−1 (C Dα u(t)) + ∇F (t, u(t)) = 0, 0 t t 0 t t T T dt 2 2 (5.21)  u(0) = u(T ) = 0,

for almost every t ∈ [0, T ], where α = 1 − β2 ∈ ( 12 , 1]. Therefore, we seek a solution u of BVP (5.21) which, of course, corresponds to the solutions u of BVP (5.1) provided that u ∈ AC([0, T ], RN ). Let us denote by 1 1 α−1 C α α Dα (u(t)) = 0 Dtα−1 (C (t DT u(t)). (5.22) t DT 0 Dt u(t)) − 2 2 We are now in a position to give a definition of the solution of BVP (5.21). Definition 5.7. A function u ∈ AC([0, T ], RN ) is called a solution of BVP (5.21) if

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(i) Dα (u(t)) is derivable for almost every t ∈ [0, T ], and (ii) u satisfies (5.21). In the sequel, we will treat BVP (5.21) in the Hilbert space E α = E0α,2 with the corresponding norm kukα = kukα,2 which we defined in (5.18). RT α C α α Consider the functional u → − 0 (C 0 Dt u(t), t DT u(t))dt on E . The following estimate is useful for our further discussion. Proposition 5.8. If 21 < α ≤ 1, then for any u ∈ E α , we have Z T 1 α C α |cos(πα)|kuk2α ≤ − (C kuk2α . (5.23) 0 Dt u(t), t DT u(t))dt ≤ |cos(πα)| 0 Proof. Let u ∈ E α and u˜ be the extension of u by zero on R \ [0, T ]. Then supp(˜ u) ⊆ [0, T ]. However, as the left and right fractional derivatives are nonlocal, α supp(−∞ Dtα u˜) ⊆ [0, ∞) and supp(t D+∞ u ˜) ⊆ (−∞, T ].

α Nonetheless, the product (−∞ Dtα u ˜, t D+∞ u ˜) has support in [0, T ]. On the other hand, according to Theorem 2.3 and Lemma 2.4 in Ervin and Roop, 2006, we have Z ∞ Z ∞ α α (−∞ Dt u˜(t), t D+∞ u ˜(t))dt = cos(πα) |−∞ Dtα u ˜(t)|2 dt −∞ −∞ Z ∞ (5.24) α = cos(πα) |t D+∞ u ˜(t)|2 dt, −∞

α where −∞ Dtα and t D+∞ are the Liouville-Weyl fractional derivatives on the real line (see Definition 1.12). Helpful in establishing (5.24) is the Fourier transform of the Liouville-Weyl fractional derivative on the real line (see, Podlubny, 1999). Hence, according to Remark 5.3, (5.24) and noting that cos(πα) ∈ [−1, 0) as α ∈ ( 12 , 1], we have Z T Z T α C α − (C D u(t), D u(t))dt = − (0 Dtα u(t), t DTα u(t))dt 0 t t T 0 0 Z T α =− (−∞ Dtα u˜(t), t D+∞ u˜(t))dt 0 Z ∞ α =− (−∞ Dtα u˜(t), t D+∞ u˜(t))dt −∞ Z ∞ = −cos(πα) |−∞ Dtα u˜(t)|2 dt (5.25) Z−∞ ∞ α 2 = −cos(πα) |0 Dt u˜(t)| dt 0 Z T ≥ −cos(πα) |0 Dtα u(t)|2 dt 0 Z T α 2 = |cos(πα)| |C 0 Dt u(t)| dt 0

= |cos(πα)|kuk2α .

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On the other hand, by using the Young’s inequality, we obtain Z T C α C α (0 Dt u(t), t DT u(t))dt Z0 T (0 Dtα u(t), t DTα u(t))dt = 0 Z T √ 1 √ |0 Dtα u(t)| 2ε |t DTα u(t)|dt ≤ 2ε 0 Z T Z T 1 ≤ |0 Dtα u(t)|2 dt + ε |t DTα u(t)|2 dt 4ε 0 0 Z T Z ∞ 1 C α 2 α = | D u(t)| dt + ε |t D+∞ u ˜(t)|2 dt 4ε 0 0 t 0 Z ∞ 1 2 α |t D+∞ u ˜(t)|2 dt ≤ kukα + ε 4ε −∞ Z ∞ 1 ε 2 α α = kukα + (−∞ Dt u ˜(t), t D+∞ u ˜(t))dt 4ε |cos(πα)| −∞ Z T 1 ε 2 α α = kukα + (0 Dt u(t), t DT u(t))dt 4ε |cos(πα)| 0 Z T ε 1 2 C α C α ( D u(t), t DT u(t))dt . = kukα + 4ε |cos(πα)| 0 0 t

Therefore, by taking ε = |cos(πα)|/2, we have Z T C α C α ≤ ( D u(t), D u(t))dt 0 t t T 0

1 kuk2α . |cos(πα)|

(5.26)

The inequality (5.23) follows then from (5.25) and (5.26), and the proof is complete. Remark 5.9. According to (5.23) and (5.24), for any u ∈ E α , it is obvious that Z T Z ∞ α 2 α |C D u(t)| dt ≤ |t D+∞ u ˜(t)|2 dt t T 0 −∞ Z T C α α (0 Dt u(t), C t DT u(t)) dt =− |cos(πα)| 0 1 ≤ kuk2α , |cos(πα)|2 α 2 N which means that C t DT u ∈ L ([0, T ], R ). In the following, we establish a variational structure on E α with α ∈ ( 12 , 1]. Also, we will show that the critical points of that functional are indeed solutions of BVP(5.21), and therefore, are solutions of BVP (5.1). Theorem 5.10. Let L : [0, T ] × RN × RN × RN → R be defined by

1 L(t, x, y, z) = − (y, z) − F (t, x), 2

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where F : [0, T ] × RN → R satisfies the following assumption: (A) F (t, x) is measurable in t for each x ∈ RN , continuously differentiable in x for almost every t ∈ [0, T ] and there exist m1 ∈ C(R+ , R+ ) and m2 ∈ L1 ([0, T ], R+ ) such that |F (t, x)| ≤ m1 (|x|)m2 (t),

|∇F (t, x)| ≤ m1 (|x|)m2 (t)

for all x ∈ RN and a.e. in t ∈ [0, T ]. If 12 < α ≤ 1, then the functional defined by Z T α C α ϕ(u) = L(t, u(t), C 0 Dt u(t), t DT u(t))dt 0 Z T 1 C α C α = − (0 Dt u(t), t DT u(t)) − F (t, u(t)) dt 2 0

(5.27)

is continuously differentiable on E α , and ∀u, v ∈ E α , we have Z T α hϕ′ (u), vi = (Dx L(t, u(t), 0C Dtα u(t), C t DT u(t)), v(t))dt 0 Z T α C α C α + (Dy L(t, u(t), C 0 Dt u(t), t DT u(t)), 0 Dt v(t))dt 0 Z T α C α + (Dz L(t, u(t), 0C Dtα u(t), C t DT u(t)), t DT v(t))dt 0 Z T 1 C α α C α C α =− [(0 Dt u(t), C t DT v(t)) + (t DT u(t), 0 Dt v(t))]dt 2 0 Z T − (∇F (t, u(t)), v(t))dt.

(5.28)

0

Proof. one has

First, we note that for a.e. t ∈ [0, T ] and every [x, y, z] ∈ RN × RN × RN , 1 |L(t, x, y, z)| ≤ m1 (|x|)m2 (t) + (|y|2 + |z|2 ), 4

(5.29)

|Dx L(t, x, y, z)| ≤ m1 (|x|)m2 (t),

(5.30)

1 1 |z| and |Dz L(t, x, y, z)| ≤ |y|. (5.31) 2 2 Then, inspired by the proof of Theorem 1.4 in Mawhin and Willem, 1989, it suffices to prove that at every point u,ϕ has a directional derivative ϕ′ (u) ∈ (E α )∗ given by (5.28) and that the mapping |Dy L(t, x, y, z)| ≤

ϕ′ : E α → (E α )∗ ,

u → ϕ′ (u)

is continuous. 1) It follows easily from Remark 5.9 and (5.29) that ϕ is everywhere finite on E α . Let us define, for u and v fixed in E α , t ∈ [0, T ], λ ∈ [−1, 1], α G(λ, t) = L(t, u(t) + λv(t), C 0 Dt u(t) + λ

C α C α 0 Dt v(t), t DT u(t)

+λ

C α t DT v(t))

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and ψ(λ) =

Z

189

T

G(λ, t)dt = ϕ(u + λv).

0

We shall apply Leibniz formula of differentiation under integral sign to ψ. By (5.30) and (5.31), we have |Dλ G(λ, t)|

α = |(Dx L(t, u(t) + λv(t), C 0 Dt u(t) + λ α +|(Dy L(t, u(t) + λv(t), C 0 Dt u(t) +

C α C α 0 Dt v(t), t DT u(t) α λC 0 Dt v(t),

+λ

C α t DT v(t)), v(t))|

C α C α t DT v(t)), 0 Dt v(t))| C α α +|(Dz L(t, u(t) + λv(t), C 0 Dt u(t) + λ 0 Dt v(t), C α C α C α t DT u(t) + λ t DT v(t)), t DT v(t))| C α t DT u(t)

+λ

1 α C α Dα u(t) + λ C ≤ m1 (|u(t) + λv(t)|)m2 (t)|v(t)| + |C t DT v(t)k0 Dt v(t)| 2t T 1 α C α + |C Dα u(t) + λ C 0 Dt v(t)kt DT v(t)| 20 t 1 1C α α C α ≤ m0 m2 (t)|v(t)| + |C Dα u(t)kC 0 Dt v(t)| + |0 Dt u(t)kt DT v(t)| 2 t T 2 α C α +|C 0 Dt v(t)kt DT v(t)|, where m0 =

max

(λ,t)∈[−1,1]×[0,T ]

m1 (|u(t) + λv(t)|).

Since m2 ∈ L1 ([0, T ], R+ ), v is continuous on [0, T ], and in view of Remark 5.9, we have |Dλ G(λ, t)| ≤ d(t), where d ∈ L1 ([0, T ], R+ ). Thus Leibniz formula is applicable and Z T d ψ(0) = Dλ G(0, t)dt dλ 0 Z T α C α = (Dx L(t, u(t), C 0 Dt u(t), t DT u(t)), v(t))dt 0

+

Z

T

0

+

Z

0

Moreover,

T

α C α C α (Dy L(t, u(t), C 0 Dt u(t), t DT u(t)), 0 Dt v(t))dt α C α (Dz L(t, u(t), 0C Dtα u(t), C t DT u(t)), t DT v(t))dt.

α |Dx L(t, u(t), 0C Dtα u(t), C t DT u(t))| ≤ m1 (|u(t)|)m2 (t), 1 C α α |Dy L(t, u(t), 0C Dtα u(t), C t DT u(t))| ≤ |t DT u(t)| 2

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and α C α |Dz L(t, u(t), C 0 Dt u(t), t DT u(t))| ≤

1 C α | D u(t)|. 2 0 t

Thus, by Remark 5.9 and (5.15), Z T α C α hϕ′ (u), vi = (Dx L(t, u(t), C 0 Dt u(t), t DT u(t)), v(t))dt 0

+

Z

T

0

+

Z

0

T

α C α (Dy L(t, u(t), 0C Dtα u(t), C t DT u(t)), 0 Dt v(t))dt α C α C α (Dz L(t, u(t), C 0 Dt u(t), t DT u(t)), t DT v(t))dt

α C α ≤ c1 kvk + c2 kC 0 Dt v(t)kL2 + c3 kt DT v(t)kL2 c3 ≤ c1 kvk + c2 kvkα + kvkα |cos(πα)| ≤ c4 kvkα ,

where c1 , c2 , c3 and c4 are some positive constants. Therefore, ϕ has, at u, a directional derivative ϕ′ (u) ∈ (E α )∗ given by (5.28). 2) By a theorem of Krasnoselskii, (5.30) and (5.31) imply that the mapping from E α into L1 ([0, T ], RN ) × L2 ([0, T ], RN ) × L2 ([0, T ], RN ) defined by α C α C α C α C α C α u → (Dx L(·, u, C 0 Dt u, t DT u), Dy L(·, u, 0 Dt u, t DT u), Dz L(·, u, 0 Dt u, t DT u))

is continuous, so that ϕ′ is continuous from E α into (E α )∗ , and the proof is completed. Theorem 5.11. Let 21 < α ≤ 1 and ϕ be defined by (5.27). If assumption (A) is satisfied and u ∈ E α is a solution of corresponding Euler equation ϕ′ (u) = 0, then u is a solution of BVP (5.21) which, of course, corresponding to the solution of BVP (5.1). By Theorem 5.10 and Property 1.23, we have Z T 1 C α α C α C α 0 = hϕ′ (u), vi = − [(0 Dt u(t), C t DT v(t)) + (t DT u(t), 0 Dt v(t))]dt 2 0 Z T − (∇F (t, u(t)), v(t))dt 0 Z T 1 1 α−1 C α α ′ ′ = (0 Dtα−1 (C D u(t)), v (t)) − ( D ( D u(t)), v (t)) dt t T 0 t t T 2 2 0 Z T − (∇F (t, u(t)), v(t))dt

Proof.

0

(5.32)

for all v ∈ E α . Let us define w ∈ C([0, T ], RN ) by Z t w(t) = ∇F (s, u(s))ds, 0

t ∈ [0, T ],

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so that

Z

T

(w(t), v ′ (t))dt =

0

Z

Z

T

0

t 0

ws-book975x65

191

(∇F (s, u(s)), v ′ (t))ds dt.

By the Fubini theorem and noting that v(T ) = 0, we obtain Z T Z T Z T ′ ′ (w(t), v (t))dt = (∇F (s, u(s)), v (t))dt ds 0

=

Z

0

s

T

0

=−

Z

(∇F (s, u(s)), v(T ) − v(s))ds T

(∇F (s, u(s)), v(s))ds.

0

Hence, by (5.32) we have, for every v ∈ E α , Z T 1 1 α−1 C α α−1 C α ′ (5.33) D ( D u(t)) − D ( D u(t)) + w(t), v (t) dt = 0. 0 t t T 0 t t T 2 2 0 If (ej ) denotes the canonical basis of RN , we can choose v ∈ E α such that 2kπt 2kπt v(t) = sin ej or v(t) = ej − cos ej , k = 1, 2, ... and j = 1, ..., N. T T The theory of Fourier series and (5.33) imply that 1 1 α−1 C α α (0 Dt u(t)) − t DTα−1 (C 0 Dt t DT u(t)) + w(t) = C, 2 2 a.e. t ∈ [0, T ], for some C ∈ RN . According to the definition of w ∈ C([0, T ], RN ), we have Z t 1 1 α−1 C α α−1 C α (0 Dt u(t)) − t DT (t DT u(t)) = − ∇F (s, u(s))ds + C, (5.34) 0 Dt 2 2 0 a.e. t ∈ [0, T ], for some C ∈ RN . In view of ∇F (·, u(·)) ∈ L1 ([0, T ], RN ), we shall identify the equivalence class Dα (u(t)) given by (5.22) and its continuous representation 1 1 α−1 C α α (t DT u(t)) Dα (u(t)) = 0 Dtα−1 (C t DT 0 Dt u(t)) − 2Z t 2 (5.35) =−

0

∇F (s, u(s))ds + C

for t ∈ [0, T ]. Therefore, it follows from (5.35) and a classical result of Lebesgue theory that −∇F (·, u(·)) is the classical derivative of Dα (u(t)) a.e. on [0, T ] which means that (i) in Definition 5.7 is verified. Since u ∈ E α implies that u ∈ AC([0, T ], RN ), it remains to show that u satisfies (5.21). In fact, according to (5.35), we can get that d 1 1 d α α−1 C α α−1 C α D (u(t)) = (0 Dt u(t)) − t DT (t DT u(t)) = −∇F (t, u(t)). 0 Dt dt dt 2 2 α Moreover, u ∈ E implies that u(0) = u(T ) = 0, and therefore (5.1) is verified. The proof is completed. 1 2

From now on, ϕ given by (5.27) will be considered as a functional on E α with < α ≤ 1.

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5.2.4


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Existence under Ambrosetti-Rabinowitz Condition

According to Theorem 5.11, we know that in order to find solutions of BVP (5.1), it suffices to obtain the critical points of functional ϕ given by (5.27). We need to use some critical point theorems. First, we use Theorem 1.50 to consider the existence of solutions for BVP (5.1). Assume that condition (A) is satisfied. Recall that, in our setting in (5.27), the corresponding functional ϕ on E α given by Z T 1 α C α ϕ(u) = − (C D u(t), D u(t)) − F (t, u(t)) dt t T 2 0 t 0 is continuously differentiable according to Theorem 5.10 and is also weakly lower semi-continuous functional on E α as the sum of a convex continuous function (see Theorem 1.2 in Mawhin and Willem, 1989) and of a weakly continuous one (see Proposition 1.2 in Mawhin and Willem, 1989). In fact, according to Proposition 5.6, if uk ⇀ u in E α , then uk → u in C([0, T ], RN ). Therefore, F (t, uk (t)) → F (t, Ru(t)) a.e. t ∈ [0, T ]. RBy Lebesgue’s T T dominated convergence theorem, we have 0 F (t, uk (t))dt → 0 F (t, u(t))dt, RT which means that the functional u → 0 F (t, u(t))dt is weakly continuous onR E α . Moreover, the following lemma implies that the functional u → T α C α α − 0 [(C D 0 t u(t), t DT u(t))/2]dt is convex and continuous on E . 1 Lemma 5.12. Let 2 < α ≤ 1 and assumption (A) be satisfied. If u ∈ E α , then the functional H : E α → RN denoted by Z 1 T C α α H(u) = − ( D u(t), C t DT u(t))dt 2 0 0 t is convex and continuous on E α . Proof. The continuity follows from (5.23) and (5.18) directly. We are now in a position to prove the convexity of H. Let λ ∈ (0, 1), u, v ∈ E α and u˜, v˜ be the extension of u and v by zero on R/[0, T ] respectively. Since the Caputo fractional derivative operator is linear operator, we have by Remark 5.9 and (5.24) that H((1 − λ)u + λv) Z 1 T C α α =− ( D ((1 − λ)u(t) + λv(t)), C t DT ((1 − λ)u(t) + λv(t)))dt 2 0 0 t Z 1 T (0 Dtα ((1 − λ)u(t) + λv(t)), t DTα ((1 − λ)u(t) + λv(t)))dt =− 2 0 Z 1 ∞ α (−∞ Dtα ((1 − λ)˜ u(t) + λ˜ v (t)), t D+∞ ((1 − λ)˜ u(t) + λ˜ v (t)))dt =− 2 −∞ Z |cos(πα)| ∞ = |−∞ Dtα ((1 − λ)˜ u(t) + λ˜ v (t))|2 dt 2 −∞ Z |cos(πα)| ∞ ≤ [(1 − λ)|−∞ Dtα u˜(t)|2 + λ|−∞ Dtα v˜(t)|2 ]dt 2 −∞

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193

1−λ λ α α α α = − (−∞ Dt u˜(t), t D+∞ u˜(t)) − (−∞ Dt v˜(t), t D+∞ v˜(t)) dt 2 2 −∞ Z T λ C α 1−λ C α C α C α ( D u(t), t DT u(t)) − (0 Dt v(t), t DT v(t)) dt = − 2 0 t 2 0 = (1 − λ)H(u) + λH(v), Z

∞

which implies that H is a convex functional defined on E α . This completes the proof. According to the arguments above, if ϕ is coercive, by Theorem 1.50, ϕ has a minimum so that BVP (5.1) is solvable. It remains to find conditions under which ϕ is coercive on E α , i.e., limkukα →∞ ϕ(u) = +∞, for u ∈ E α . We shall see that it suffices to require that F (t, x) is bounded by a function for a.e. t ∈ [0, T ] and all x ∈ RN . Theorem 5.13. Let α ∈ ( 21 , 1] and assume that F satisfies condition (A). If |F (t, x)| ≤ a ¯|x|2 + ¯b(t)|x|2−γ + c¯(t), t ∈ [0, T ], x ∈ RN , (5.36)

where a ¯ ∈ [0, |cos(πα)|Γ2 (α + 1)/2T 2α), γ ∈ (0, 2), ¯b ∈ L2/γ ([0, T ], R), and c¯ ∈ L1 ([0, T ], R), then BVP (5.1) has at least one solution which minimizes ϕ on E α .

Proof. According to arguments above, our problem reduces to prove that ϕ is coercive on E α . For u ∈ E α , it follows from (5.23), (5.36) and (5.14) that Z Z T 1 T C α C α ϕ(u) = − ( D u(t), t DT u(t))dt − F (t, u(t))dt 2 0 0 t 0 Z Z T |cos(πα)| T C α |0 Dt u(t)|2 dt − a ¯ |u(t)|2 dt ≥ 2 0 0 Z T Z T ¯b(t)|u(t)|2−γ dt − − c¯(t)dt 0

=

0

|cos(πα)| kuk2α − a ¯kuk2L2 − 2

Z

T

0

¯b(t)|u(t)|2−γ dt − c¯1

Z T γ/2 Z T 1−γ/2 |cos(πα)| ≥ kuk2α − a ¯kuk2L2 − |¯b(t)|2/γ dt |u(t)|2 dt − c¯1 2 0 0 |cos(πα)| = kuk2α − a ¯kuk2L2 − ¯b1 kuk2−γ ¯1 L2 − c 2 2−γ a ¯T 2α Tα |cos(πα)| 2 2 ¯ kukα − 2 kukα − b1 kuk2−γ − c¯1 ≥ α 2 Γ (α + 1) Γ(α + 1) 2−γ |cos(πα)| a ¯T 2α Tα 2−γ = − 2 kuk2α − ¯b1 kukα − c¯1 , 2 Γ (α + 1) Γ(α + 1) RT RT where ¯b1 = ( |¯b(t)|2/γ dt)γ/2 and c¯1 = c¯(t)dt. 0

0

Noting that a ¯ ∈ [0, |cos(πα)|Γ2 (α + 1)/2T 2α) and γ ∈ (0, 2), we have ϕ(u) = +∞ as kukα → ∞,

and hence ϕ is coercive, which completes the proof.

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194

Our task is now to use Theorem 1.51 (Mountain pass theorem) to find a nonzero critical point of functional ϕ on E α . Theorem 5.14. Let α ∈ ( 21 , 1] and suppose that F satisfies condition (A). If (A1) F ∈ C([0, T ] × RN , R) and there exists µ ∈ [0, 12 ) and M > 0 such that 0 < F (t, x) ≤ µ(∇F (t, x), x) for all x ∈ RN with |x| ≥ M and t ∈ [0, T ]; (A2) lim sup|x|→0 F (t, x)/|x|2 < |cos(πα)|Γ2 (α + 1)/2T 2α uniformly for t ∈ [0, T ] and x ∈ RN ; are satisfied, then BVP (5.1) has at least one nonzero solution on E α . Proof. We will verify that ϕ satisfies all conditions of Theorem 1.51. First, we will prove that ϕ satisfies (PS) condition. Since F (t, x)−µ(∇F (t, x), x) is continuous for t ∈ [0, T ] and |x| ≤ M , there exists c ∈ R+ , such that F (t, x) ≤ µ(∇F (t, x), x) + c,

t ∈ [0, T ],

|x| ≤ M.

F (t, x) ≤ µ(∇F (t, x), x) + c,

t ∈ [0, T ],

x ∈ RN .

By assumption (A1), we obtain

Let {uk } ⊂ E α , |ϕ(uk )| ≤ K, k = 1, 2, ..., ϕ′ (uk ) → 0. Notice that Z T α C α hϕ′ (uk ), uk i = − [(C 0 Dt uk (t), t DT uk (t)) + (∇F (t, uk (t)), uk (t))]dt.

(5.37)

(5.38)

0

It follows from (5.37), (5.38) and (5.23) that Z Z T 1 T C α C α K ≥ ϕ(uk ) = − ( D uk (t), t DT uk (t))dt − F (t, uk (t))dt 2 0 0 t 0 Z T Z T 1 α (C Dα uk (t), C (∇F (t, uk (t)), uk (t))dt − cT ≥− t DT uk (t))dt − µ 2 0 0 t 0 Z T 1 α C α ′ (C = µ− 0 Dt uk (t), t DT uk (t))dt + µhϕ (uk ), uk i − cT 2 0 1 − µ |cos(πα)|kuk k2α − µkϕ′ (uk )kα kuk kα − cT, k = 1, 2, ... . ≥ 2 Since ϕ′ (uk ) → 0, there exists N0 ∈ N such that 1 K≥ − µ |cos(πα)|kuk k2α − kuk kα − cT, 2

k > N0 ,

and this implies that {uk } ⊂ E α is bounded. Since E α is a reflexive space, going to a subsequence if necessary, we may assume that uk ⇀ u weakly in E α , thus we have hϕ′ (uk ) − ϕ′ (u), uk − ui = hϕ′ (uk ), uk − ui − hϕ′ (u), uk − ui (5.39) ≤ kϕ′ (uk )kα kuk − ukα − hϕ′ (u), uk − ui → 0,

as k → ∞. Moreover, according to (5.15) and Proposition 5.6, we have uk is bounded in C([0, T ], RN ) and kuk − uk → 0 as k → ∞. Hence, we have Z T Z T ∇F (t, uk (t))dt → ∇F (t, u(t))dt as k → ∞. (5.40) 0

0

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195

Noting that Z

T

α C α hϕ (uk ) − ϕ (u), uk − ui = − (C 0 Dt (uk (t) − u(t)), t DT (uk (t) − u(t)))dt 0 Z T − (∇F (t, uk (t)) − ∇F (t, u(t))), (uk (t) − u(t)) dt 0 Z T 2 ≥ |cos(πα)|kuk − ukα − (∇F (t, uk (t)) − ∇F (t, u(t)))dt kuk − uk. ′

′

0

Combining (5.39) and (5.40), it is easy to verify that kuk − uk2α → 0 as k → ∞, and hence that uk → u in E α . Thus, we obtain the desired convergence property. From lim sup|x|→0 F (t, x)/|x|2 < |cos(πα)|Γ2 (α + 1)/2T 2α uniformly for t ∈ [0, T ], there exists ǫ ∈ (0, |cos(πα)|) and δ > 0 such that F (t, x) ≤ (|cos(πα)| − ǫ)(Γ2 (α + 1)/2T 2α)|x|2 for all t ∈ [0, T ] and x ∈ RN with |x| ≤ δ. 1

Let ρ =

Γ(α)((α−1)/2+1) 2 1 T α− 2

δ and σ = ǫρ2 /2 > 0. Then it follows from (5.15) that 1

kuk ≤

T α− 2 1

Γ(α)((α − 1)/2 + 1) 2

kukα = δ

for all u ∈ E α with kukα = ρ. Therefore, we have Z Z T 1 T C α ϕ(u) = − (0 Dt u(t), tC DTα u(t))dt − F (t, u(t))dt 2 0 0 Z Γ2 (α + 1) T |cos(πα)| kuk2α − (|cos(πα)| − ǫ) |u(t)|2 dt ≥ 2 2T 2α 0 1 |cos(πα)| 2 2 kukα − (|cos(πα)| − ǫ)kukα ≥ 2 2 1 2 = ǫkukα 2 =σ for all u ∈ E α with kukα = ρ. This implies (ii) in Theorem 1.51 is satisfied. It is obvious from the definition of ϕ and (A2) that ϕ(0) = 0, and therefore, it suffices to show that ϕ satisfies (iii) in Theorem 1.51. Since 0 < F (t, x) ≤ µ(∇F (t, x), x) for all x ∈ RN and |x| ≥ M , a simple regularity argument then shows that there exists r1 , r2 > 0 such that F (t, x) ≥ r1 |x|1/µ − r2 ,

x ∈ RN , t ∈ [0, T ].

For any u ∈ E α with u 6= 0, κ > 0 and noting that µ ∈ [0, 21 ) and (5.23), we have Z Z T 1 T C α α ϕ(κu) = − (0 Dt κu(t), C D κu(t))dt − F (t, κu(t))dt t T 2 0 0 Z T κ2 kuk2α − r1 |κu(t)|1/µ dt + r2 T ≤ 2|cos(πα)| 0 κ2 1/µ = kuk2α − r1 κ1/µ kukL1/µ + r2 T → −∞ 2|cos(πα)|

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196

as κ → ∞. Then there exists a sufficiently large κ0 such that ϕ(κ0 u) ≤ 0. Hence (iii) in Theorem 1.51 holds. Lastly noting that ϕ(0) = 0 while for our critical point u, ϕ(u) ≥ σ > 0. Hence u is a nontrivial weak solution of BVP (5.1), and this completes the proof. Corollary 5.15. ∀α ∈ ( 21 , 1], suppose that F satisfies conditions (A) and (A1). If (A2)′ F (t, x) = o(|x|2 ), as |x| → 0 uniformly for t ∈ [0, T ] and x ∈ RN is satisfied, then BVP (5.1) has at least one nonzero solution on E α . 5.2.5

Superquadratic Case

Under the usual Ambrosetti-Rabinowitz condition, it is easy to show that the energy functional associated with the system has the Mountain Pass geometry and satisfies the (PS) condition. However, the A.R. condition is so strong that many potential functions can not satisfy it, then the problem becomes more delicate and complicated. Assume that F : [0, T ] × RN → R satisfies the condition (A) which is assumed as in Subsection 5.2.3. In the following, we introduce the function space E α , where α ∈ ( 21 , 1]. For u ∈ E α , where α 2 N E α := u ∈ L2 (0, T ; RN ) : C 0 Dt u ∈ L (0, T ; R )

is a reflexive Banach space with the norm defined by α kukα = kC 0 Dt ukL2

and kuk = max |u(t)|. t∈[0, T ]

It follows from Theorem 5.10 that the functional ϕ on E α given by Z Th i 1 α C α ϕ(u) = − (C 0 Dt u(t), t DT u(t)) − F (t, u(t)) dt 2 0

is continuously differentiable on E α . Moreover, we have Z T 1 C α α C α C α hϕ′ (u), vi = − (0 Dt u(t), C t DT v(t)) + (t DT u(t), 0 Dt v(t)) dt 0 2 Z T − ∇F t, u(t)), v(t) dt.

(5.41)

(5.42)

0

Recall that a sequence {un } ⊂ E α is said to be a (C) sequence of ϕ if ϕ(un ) is bounded and (1 + kun kα )kϕ(un )kα → 0 as n → ∞. The functional ϕ satisfies condition (C) if every (C) sequence of ϕ has a convergent subsequence. This condition is due to Cerami, 1978.

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197

For the superquadratic case, we make the following assumptions: F (t, x) π2 F (t, x) = 0, lim inf ≥L> uni(A3) lim 2 2 2 |x| |cos(πα)|Γ (2 − α)T 2α (3 − 2α) |x|→∞ |x|→0 |x| formly for some L > 0 and a.e. t ∈ [0, T ]; F (t, x) (A4) lim sup ≤ M < +∞ uniformly for some M > 0 and a.e. t ∈ [0, T ]; r |x|→+∞ |x| (∇F (t, x), x) − 2F (t, x) ≥ Q > 0 uniformly for some Q > 0 and a.e. (A5) lim inf |x|µ |x|→+∞ t ∈ [0, T ], where r > 2 and µ > r − 2. We will first establish the following lemma. Lemma 5.16. Assume (A), (A4), (A5) hold, then the functional ϕ satisfies condition (C). Proof. Let {un } ⊂ E α is a (C) sequence of ϕ, that is ϕ(un ) is bounded and (1 + kun kα )kϕ′ (un )kα → 0 as n → ∞. Then there exists M0 such that and (1 + kun kα )kϕ′ (un )kα ≤ M0

|ϕ(un )| ≤ M0

(5.43)

for all n ∈ N. By (A4), there exist positive constants B1 and M1 such that F (t, x) ≤ B1 |x|r

(5.44)

for all |x| ≥ M1 and a.e. t ∈ [0, T ]. It follows from (A) that |F (t, x)| ≤ max a(s)b(t) s∈[0, M1 ]

for all |x| ≤ M1 and a.e. t ∈ [0, T ]. Therefore, we obtain F (t, x) ≤ B1 |x|r + max a(s)b(t)

(5.45)

s∈[0, M1 ]

for all x ∈ RN and a.e. t ∈ [0, T ]. Combining (5.23) and (5.45), we get Z T |cos(πα)| 2 kun kα ≤ ϕ(un ) + F (t, un (t))dt 2 0 Z T Z ≤ M0 + max a(s) b(t)dt + B1 s∈[0,M1 ]

0

0

T

(5.46) |un (t)|r dt.

On the other hand, by (A5), there exist η > 0 and M2 > 0 such that ∇F (t, x), x − 2F (t, x) ≥ η|x|µ

for a.e. t ∈ [0, T ] and |x| ≥ M2 . By (A), we have

| (∇F (t, x), x) − 2F (t, x)| ≤ (2 + M2 ) max a(s)b(t) s∈[0, M2 ]

for all |x| ≤ M2 and a.e. t ∈ [0, T ].

(5.47)

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198

Therefore, we obtain (∇F (t, x), x) − 2F (t, x) ≥ η|x|µ − (2 + M2 ) max a(s)b(t) s∈[0, M2 ]

(5.48)

for all x ∈ RN and a.e. t ∈ [0, T ]. It follows from (5.43) and (5.48) that 3M0 ≥ 2ϕ(un ) − hϕ′ (un ), un i Z T 1 C α α =2 − (0 Dt un (t), C t DT un (t)) − F (t, un (t)) dt 2 0 Z T α C α − − (C 0 Dt un (t), t DT un (t)) − (∇F (t, un (t)), un (t)) dt =

Z

0

T

0

≥η thus,

RT 0

Z

∇F (t, un (t)), un (t) − 2F (t, un (t)) dt

T

µ

|un (t)| dt − (2 + M2 ) max a(s) s∈[0, M2 ]

0

Z

T

b(t)dt,

0

|un (t)|µ dt is bounded.

If µ > r, then Z

0

T

r

|un (t)| dt ≤ T

µ−r µ

Z

T 0

|un (t)|µ dt

r/µ

,

which combining (5.46) implies that kun kα is bounded. If µ ≤ r, then Z T Z T Z r−µ µ r−µ |un (t)|r dt ≤ kun kr−µ |u (t)| dt ≤ C ku k n n ∞ α 1 0

0

T 0

|un (t)|µ dt,

where

1

C1 :=

T α− 2

1

Γ(α)(2α − 1) 2

by (5.15). Since µ > r − 2, it follows from (5.46) that kun kα is bounded too. Thus kun kα is bounded in E α . By Proposition 5.6, the sequence {un } has a subsequence, also denoted by {un }, such that un ⇀ u in E α

and un → u in C([0, T ], RN ).

(5.49)

Then we obtain un → u in E α by use of the same argument of Theorem 5.14. The proof of Lemma 5.16 is completed. We state our first existence result as follows. Theorem 5.17. Assume that (A3)-(A5) hold and that F (t, x) satisfies the condition (A). Then BVP (5.1) has at least one solution on E α .

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Proof.

ws-book975x65

199

By (A3), there exist ǫ1 ∈ (0, |cos(πα)|) and δ > 0 such that F (t, x) ≤ (|cos(πα)| − ǫ1 )

Γ2 (α + 1) 2 |x| 2T 2α

for a.e. t ∈ [0, T ] and x ∈ RN with |x| ≤ δ. Let 1

ρ=

Γ(α)(2(α − 1) + 1) 2 1 T α− 2

δ

and σ =

ǫ 1 ρ2 > 0. 2

Then it follows from (5.15) that 1

kuk ≤

T α− 2 1

Γ(α)(2(α − 1) + 1) 2

kukα = δ

for all u ∈ E α with kukα = ρ. Therefore, we have Z Th i 1 α C α ϕ(u) = − (C D u(t), D u(t)) − F (t, u(t)) dt t T 2 0 t 0 Z Γ2 (α + 1) T |cos(πα)| kuk2α − (|cos(πα)| − ǫ1 ) |u(t)|2 dt ≥ 2 2T 2α 0 (|cos(πα)| − ǫ1 |cos(πα)| kuk2α − kuk2α ≥ 2 2 ǫ1 = kuk2α 2 =σ

(5.50)

for all u ∈ E α with kukα = ρ. This implies that (ii) in Theorem 1.51 is satisfied. It is obvious from the definition of ϕ and (A3) that ϕ(0) = 0, and therefore, it suffices to show that ϕ satisfies (iii) in Theorem 1.51. By (A3), there exist ǫ2 > 0 and M3 > 0 such that π2 F (t, x) > + ǫ2 |x|2 (5.51) |cos(πα)|Γ2 (2 − α)T 2α (3 − 2α) for all |x| ≥ M3 and a.e. t ∈ [0, T ]. It follows from (A) that |F (t, x)| ≤ max a(s)b(t), s∈[0, M3 ]

for all |x| ≤ M3 and a.e. t ∈ [0, T ]. Therefore, we obtain π2 + ǫ2 (|x|2 − M32 ) F (t, x) ≥ |cos(πα)|Γ2 (2 − α)T 2α (3 − 2α) − maxs∈[0, M3 ] a(s)b(t) for all x ∈ RN and a.e. t ∈ [0, T ].

(5.52)

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Choosing u0 =

T πt sin , 0, ..., 0 π T

ku0 k2L2 =

T3 2π 2

∈ E α , then

and ku0 k2α ≤

T 3−2α . Γ2 (2 − α)(3 − 2α)

(5.53)

For ς > 0 and noting that (5.52) and (5.53), we have Z Th i 1 α C α D ςu (t), D ςu (t)) − F (t, ςu (t)) dt ϕ(ςu0 ) = − (C 0 0 0 t T 2 0 t 0 ς2 ku0 k2α ≤ 2|cos(πα)| Z T ς 2 π2 2 − + ς ǫ2 |u0 (t)|2 dt + C2 (5.54) |cos(πα)|T 2α Γ2 (2 − α)(3 − 2α) 0 2 3−2α ς T ≤ · 2|cos(πα)| Γ2 (2 − α)(3 − 2α) T3 ς 2 ǫ2 T 3 ς 2 π2 · 2− + C2 − 2α 2 |cos(πα)|T Γ (2 − α)(3 − 2α) 2π 2π 2 → −∞ as ς → ∞, where C2 is a positive constant. Then there exists a sufficiently large ς0 such that ϕ(ς0 u0 ) ≤ 0. Hence (iii) in Theorem 1.51 holds. Finally, noting that ϕ(0) = 0 while for critical point u, ϕ(u) ≥ σ > 0. Hence u is a nontrivial solution of BVP (5.1), and this completes the proof. We give an example to illustrate our results. Example 5.18. In BVP (5.1), let F (t, x) = ln(1 + 2|x|2 )|x|2 . These show that all conditions of Theorem 5.17 are satisfied, where r = 2.5,

µ = 2.

By Theorem 5.17, BVP (5.1) has at least one solution u ∈ E α . 5.2.6

Asymptotically Quadratic Case

For the asymptotically quadratic case, we assume: F (t, x) ≤ M < +∞ uniformly for some M > 0 and a.e. t ∈ [0, T ]; (A4)′ lim sup 2 |x|→+∞ |x| (A6) there exists τ (t) ∈ L1 ([0, T ], R+ ) such that (∇F (t, x), x) − 2F (t, x) ≥ τ (t) for all x ∈ RN and a.e. t ∈ [0, T ]; (A7) lim [(∇F (t, x), x) − 2F (t, x)] = +∞ for a.e. t ∈ [0, T ]. |x|→+∞

Theorem 5.19. Assume that F (t, x) satisfies (A), (A3), (A4)′ , (A6) and (A7). Then BVP (5.1) has at least one solution on E α .

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The following lemmas are needed in the proof of Theorem 5.19. Lemma 5.20. Assume that (A7) holds. Then for any ε > 0, there exists a subset Eε ⊂ [0, T ] with α([0, T ]\Eε ) < ε such that lim [(∇F (t, x), x) − 2F (t, x)] = +∞

|x|→∞

uniformly for t ∈ Eε . The proof is similar to that of Lemma 2 in Tang and Wu, 2001, and is omitted. Lemma 5.21. Assume that (A), (A4)′ , (A6) and (A7) hold. Then the functional ϕ satisfies condition (C). Proof. Suppose that {un } ⊂ E α is a (C) sequence of ϕ, that is ϕ(un ) is bounded and (1 + kun kα )kϕ′ (un )kα → 0 as n → ∞. Then we have lim inf [hϕ′ (un ), un i − 2ϕ(un )] > −∞, n→∞

which implies that lim sup n→∞

Z

T

0

[(∇F (t, un ), un ) − 2F (t, un )]dt < +∞.

(5.55)

We only need to show that {un } is bounded in E α . If {un } is unbounded, we may un , assume, without loss of generality, that kun kα → ∞ as n → ∞. Put zn = kun kα we then have kzn kα = 1. Going to a sequence if necessary, we assume that zn ⇀ z in E α , zn → z in C([0, T ], RN ) and L2 ([0, T ], RN ). By (A2)′ , it follows that there exist constants B2 > 0 and M4 > 0 such that F (t, x) ≤ B2 |x|2 for all |x| ≥ M4 and a.e. t ∈ [0, T ]. By assumption (A), it follows that |F (t, x)| ≤ max a(s)b(t) s∈[0, M4 ]

for all |x| ≤ M4 and a.e. t ∈ [0, T ]. Therefore, we obtain F (t, x) ≤ B2 |x|2 + max a(s)b(t) s∈[0, M4 ]

for all x ∈ R

N

and a.e. t ∈ [0, T ]. Therefore, we have Z Th i 1 ϕ(u) = − (c0 Dtα u(t), ct DTα u(t)) − F (t, u(t)) dt 2 0 Z T Z T |cos(πα)| 2 2 kukα − B2 |u| dt − max a(s) b(t)dt, ≥ 2 s∈[0, M4 ] 0 0

from which, it follows that ϕ(un ) |cos(πα)| 1 ≥ − B2 kzn k2L2 − 2 kun kα 2 kun k2α

max a(s)

s∈[0, M4 ]

Z

0

T

b(t)dt.

(5.56)

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Passing to the limit in the last inequality, we get |cos(πα)| − B2 kzk2L2 ≤ 0, 2 which yields z 6= 0. Therefore, there exists a subset E ⊂ [0, T ] with α(E) > 0 such that z(t) 6= 0 on E. 1 By virtue of Lemma 5.20, for ε = α(E) > 0, we can choose a subset Eε ⊂ [0, T ] 2 with α([0, T ]\Eε ) < ε such that lim [(∇F (t, x), x) − 2F (t, x)] = +∞

|x|→∞

(5.57)

uniformly for t ∈ Eε . T T We assert that α(E Eε ) > 0. If not, α(E Eε ) = 0. T S Since E = (E Eε ) (E\Eε ), it follows that T 0 < α(E) = α(E Eε ) + α(E\Eε ) ≤ α([0, T ]\Eε ) < ε = 12 α(E), which leads to a contradiction and establishes the assertion. By (A6), we obtain Z T [(∇F (t, un ), un ) − 2F (t, un )]dt Z0 = [(∇F (t, un ), un ) − 2F (t, un )]dt T E Z Eε + [(∇F (t, un ), un ) − 2F (t, un )]dt T [0,T ]\(E Eε ) Z Z T ≥ [(∇F (t, u ), u ) − 2F (t, u )]dt − |τ (t)|dt. n n n T E

Eε

(5.58)

0

By (5.57), (5.58) and Fatou’s lemma, it follows that Z T lim [(∇F (t, un ), un ) − 2F (t, un )]dt = +∞, n→∞

0

which contradicts (5.55). This contradiction shows that kun kα is bounded in E α and this completes the proof. Theorem 5.22. Assume that F (t, x) satisfies (A), (A3), (A4)′ and the following conditions: (A6)′ there exists τ (t) ∈ L1 (0, T ; R+ ) such that (∇F (t, x), x) − 2F (t, x) ≤ τ (t) for all x ∈ RN and a.e. t ∈ [0, T ]; (A7)′ lim [(∇F (t, x), x) − 2F (t, x)] = −∞ for a.e. t ∈ [0, T ]. |x|→+∞

Then BVP (5.1) has at least one solution on E α . By virtue of Lemma 5.20 and Lemma 5.21, similar to Theorem 5.17, we can complete the proof of Theorem 5.19 by using the similar proof of Theorem 5.17. Theorem 5.22 can be proved similarly.

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We give an example to illustrate our results. Example 5.23. In BVP (5.1), let T = 2π and F (t, x) = κf (x)(2 + sin t) arctan |x|2 , where κ > 0 and f (x) will be specified below. Let f (x) = |x|2 + ln(1 + |x|2 ). Noting that 0 ≤ ln(1 + |x|2 ) ≤ |x|2 , we see that (A) and (A4)′ hold. It is also easy to see that (A3) hold for κ> Furthermore, we have

(2π)1−2α . |cos(πα)|Γ2 (2 − α)(3 − 2α)

(∇f (x), x) − 2f (x) = as |x| → +∞. Therefore, we have

2|x|2 − 2 ln(1 + |x|2 ) → −∞ 1 + |x|2

(∇F (t, x), x) − 2F (t, x) 2|x|2 =κ f (x)(2 + sin t) + κ[(∇f (x), x) − 2f (x)](2 + sin t) arctan |x|2 1 + |x|4 → −∞

uniformly for all t ∈ [0, 2π] as |x| → +∞. Thus (A6)′ and (A7)′ hold. By virtue of Theorem 5.22, we conclude that BVP (5.1) has at least one solution on E α . If f (x) = |x|2 − ln(1 + |x|2 ), then exact the same conclusions as above hold true by Theorem 5.19. 5.3 5.3.1

Multiple Solutions for BVP with Parameters Introduction

In this section, we study the existence of three solutions to fractional BVP of the form  1 −β ′  d 1 −β ′ t ∈ [0, T ], 0 Dt (u (t)) + t DT (u (t)) + λ∇F (t, u(t)) = 0, (5.59) dt 2 2  u(0) = u(T ) = 0,

where T > 0, λ > 0 is a parameter, 0 ≤ β < 1, 0 Dt−β and t DT−β are the left and right Riemann-Liouville fractional integrals of order β, respectively, N ≥ 1 is an integer, F : [0, T ] × RN → R is a given function such that F (t, x) is measurable in t for each x = (x1 , . . . , xN ) ∈ RN and continuously differentiable in x for a.e. t ∈ [0, T ], F (t, 0, . . . , 0) ≡ 0 on [0, T ], and ∇F (t, x) = (∂F/∂x1 , . . . , ∂F/∂xN ) is the gradient of F at x. By a solution of (5.59), we mean an absolutely continuous function u : [0, T ] → RN such that u(t) satisfies both equation for a.e. t ∈ [0, T ] and the boundary conditions in (5.59). We notice that when β = 0, problem (5.59) has the form ′′ u (t) + λ∇F (t, u(t)) = 0, t ∈ [0, T ], (5.60) u(0) = u(T ) = 0,

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which has been extensively studied. The equation in (5.59) is motivated by the steady fractional advection dispersion equation studied in Ervin and Roop, 2006, −D a (p 0 Dt−β + q t DT−β )Du + b(t)Du + c(t)u = f,

(5.61)

where D represents a single spatial derivative, 0 ≤ p, q ≤ 1 satisfying p + q = 1, a > 0 is a constant, and b, c, f are functions satisfying some suitable conditions. The interest in (5.61) arises from its application as a model for physical phenomena exhibiting anomalous diffusion; i.e., diffusion not accurately modeled by the usual advection dispersion equation. Anomalous diffusion has been used in modeling turbulent flow (see, Carreras, Lynch and Zaslavsky, 2001; Shlesinger, West and Klafter, 1987), and chaotic dynamics of classical conservative systems (see, Zaslavsky, Stevens and Weitzner, 1993). The reader may find more background information and applications on (5.61) in Benson, Wheatcraft and Meerschaert, 2000a; Ervin and Roop, 2006. Remark 5.24. When N = 1, problem (5.59) reduces to the scalar BVP   d 1 D−β (u′ (t)) + 1 D−β (u′ (t)) + λf (t, u(t)) = 0, t ∈ [0, T ], 0 t t T dt 2 2 (5.62)  u(0) = u(T ) = 0, where f : [0, T ] × R → R is such that f (t, x) is measurable in t for each x ∈ R and continuous in x for a.e. t ∈ [0, T ]. It is clear that the equation in (5.62) is of the special form of (5.61) with D = d/dt, a = 1, p = q = 21 , b(t) = c(t) = 0, and f = λf (t, u). We also notice that since (5.61) is the steady fractional advection dispersion equation, it has no dependence on the time variable and it just depends on the space variable t (here, the notation t stands for the space variable in (5.61)). Since the space we studied is one dimensional and has the form of an interval, say [0, T ], the boundary conditions in the space reduce to the conditions at the two endpoints t = 0 and t = T of the interval. In our system, we study the Dirichlet type boundary conditions. 5.3.2

Existence

For 0 ≤ β < 1 given in (5.59), let α = 1 − β2 ∈ ( 12 , 1] and define Z 3T /4 Z T 1 T 3−2α 16N + g 2 (t)dt + h2 (t)dt , ρα = 2 2 T Γ (2 − α) 3 − 2α 4 T /4 3T /4

(5.63)

where g(t) = t1−α − (t − T /4)1−α ,

(5.64)

h(t) = t1−α − (t − T /4)1−α − (t − 3T /4)1−α .

(5.65)

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In the remainder of this section, for some c, d, l, m, p ∈ R, let the bold letters c, d, l, m, and p be the constant vectors in RN defined by c = (c, ..., c),

d = (d, ..., d),

l = (l, ..., l),

m = (m, ..., m),

p = (p, ..., p),

and any other bold letter, such as x, is used to denote an arbitrary vector in RN . Let E α be the space of functions u ∈ L2 ([0, T ], RN ) having an α-order Caputo α 2 N fractional derivatives C 0 Dt u ∈ L ([0, T ], R ) and u(0) = u(T ) = 0. Then, by α Remark 5.3 (i) and Proposition 5.4, E is a reflexive and separable Banach space with the norm Z T Z T 21 2 α 2 for any u ∈ E α . kukα = |u(t)| dt + |C 0 Dt u(t)| dt 0

0

We see that the norm kukα is equivalent to the norm defined as the follow Z T 12 α 2 for any u ∈ E α . kukα = |C D u(t)| dt 0 t 0

We recall the norms

kukL2 = α

Z

0

T

|u(t)|2 dt

21

and kuk = max |u(t)|. t∈[0,T ]

For u ∈ E , let the functionals Φ and Ψ be defined as follows Z 1 T C α C α Φ(u) = − 0 Dt u(t), t DT u(t) dt, 2 0 Ψ(u) =

Z

(5.66)

T

F (t, u(t))dt.

(5.67)

0

Then, by Theorem 5.10, we see that Φ and Ψ are continuously differentiable, and for any u, v ∈ E α , we have Z 1 T C α C α C α C α (5.68) hΦ′ (u), vi = − 0 Dt u(t), t DT v(t) + t DT u(t), 0 Dt v(t) dt, 2 0 ′

hΨ (u), vi =

Z

0

T

∇F (t, u(t)), v(t) dt.

(5.69)

Parts (i) and (ii) of Lemma 5.25 below are taken from Lemma 5.12 and Theorem 5.11, respectively. Lemma 5.25. We have that (i) The functional Φ is convex and continuous on E α . (ii) If u ∈ E α is a critical point of the functional Φ − λΨ, then u is a solution of BVP (5.59).

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We now state the results of this section. Theorem 5.26. Assume that there exist four positive constants c, d, l and m, with 1

d < m and c < such that

1

T α− 2 ρα2 d

< |cos(πα)|l < |cos(πα)|m,

1

Γ(α)(2α − 1) 2

F (t, x) ≥ 0, for (t, x) ∈ [0, T ] × [−m, m]N , max F (t, x) ≤ F (t, c),

|x|≤c

RT 0

RT 0

RT 0

max F (t, x) ≤ F (t, l),

F (t, c)dt Γ2 (α)cos2 (πα)(2α − 1) < c2 T 2α−1 ρα d2

Z

F (t, l)dt Γ2 (α)cos2 (πα)(2α − 1) < l2 T 2α−1 ρα d2

Z

F (t, m)dt Γ2 (α)cos2 (πα)(2α − 1) < m2 − l 2 T 2α−1 ρα d2

|x|≤m

F (t, d)dt −

Z

F (t, d)dt −

Z

3T /4 T /4 3T /4

T /4

Z

F (t, d)dt −

F (t, c)dt , (5.73)

T

F (t, c)dt , (5.74)

0

Z

(5.72)

T

0

3T /4

T /4

(5.71)

max F (t, x) ≤ F (t, m),

|x|≤l

(5.70)

0

T

F (t, c)dt . (5.75)

¯ the system (5.59) has at least three solutions u1 , u2 and u3 Then, for each λ ∈ (λ, λ), such that maxt∈[0,T ] |u1 (t)| < c, maxt∈[0,T ] |u2 (t)| < l, and maxt∈[0,T ] |u3 (t)| < m, where ρα d2 λ= (5.76) R 3T /4 RT 2|cos(πα)| T /4 F (t, d)dt − 0 F (t, c)dt and

n 2 2 2 2 ¯ = min Γ (α)(2α −R 1)|cos(πα)|c , Γ (α)(2α −R 1)|cos(πα)|l , λ T T 2T 2α−1 0 F (t, c)dt 2T 2α−1 0 F (t, l)dt Γ2 (α)(2α − 1)|cos(πα)|(m2 − l2 ) o . RT 2T 2α−1 0 F (t, m)dt

(5.77)

Proof. For any x ∈ R, let p(x) = max{−m, min{x, m}}. For any x = ˜ ), where x ˜ = (p(x1 ), . . . , p(xN )). Then, (x1 , . . . , xN ) ∈ E α , let F˜ (t, x) = F (t, x F˜ (t, x) is measurable in t for each x ∈ RN and continuously differentiable in x for a.e. t ∈ [0, T ], and F˜ (t, 0, . . . , 0) = 0 on [0, T ]. Note that −m ≤ p(ui ) ≤ m for any u = (u1 , . . . , uN ) ∈ E α and i = 1, . . . , N . Then, (5.71) implies that F˜ (t, u) ≥ 0, for (t, u) ∈ [0, T ] × E α .

(5.78)

Note that d < m and c < l < m by (5.70). Then, we have F˜ (t, x) = F (t, x), for (t, x) ∈ [0, T ] × RN with |x| < m, F˜ (t, c) = F (t, c), F˜ (t, d) = F (t, d), F˜ (t, l) = F (t, l), F˜ (t, m) = F (t, m). (5.79)

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Let the continuously differentiable functional Φ be given by (5.66) and the func˜ be defined by tional Ψ Z T ˜ Ψ(u) = F˜ (t, u(t))dt, for u ∈ E α . (5.80) 0

Then, by Proposition 5.8 and (5.66), we have

1 1 |cos(πα)| kuk2α ≤ Φ(u) ≤ kuk2α, for u ∈ E α . 2 2|cos(πα)|

(5.81)

˜ is continuously differentiable, and for any u, v ∈ E α , in view of (5.78), Moreover, Ψ we have Z T ′ ˜ ˜ Ψ(u) ≥ 0 and hΨ (u), vi = ∇F˜ (t, u(t)), v(t) dt. (5.82) 0

In the following, we will apply Lemma 1.57 with X = E α to the functionals Φ ˜ and Ψ. We first show that some basic assumptions of Lemma 1.57 are satisfied. The convexity and coercivity of Φ follow from Lemma 5.25 (i) and (5.81), respectively. For any u, v ∈ E α , from Proposition 5.8 and (5.68), hΦ′ (u) − Φ′ (v), u − vi Z i 1 T hC α C α C α C α D u(t), D (u(t) − v(t)) + D u(t), D (u(t) − v(t)) dt =− 0 t t T t T 0 t 2 0 Z i 1 T hC α C α C α C α D v(t), D (u(t) − v(t)) + D v(t), D (u(t) − v(t)) dt + 0 t t T t T 0 t 2Z 0 T C α C α =− 0 Dt (u(t) − v(t)), t DT (u(t) − v(t)) dt 0

≥ |cos(πα)| ku − vk2α .

Thus, Φ′ is uniformly monotone. Hence, by Theorem 26.A (d) in Zeidler, 1990, (Φ′ )−1 : (E α )∗ → E α exists and is continuous. Suppose that un ⇀ u ∈ E α . Then, by Proposition 5.6 un → u in C([0, T ], RN ). Since F˜ (t, x) is continuously differentiable in x for a.e. t ∈ [0, 1], from the derivative formula in (5.82), we have ˜ ′ (un ) → Ψ ˜ ′ (u), i.e., Ψ ˜ ′ is strongly continuous. Therefore, Ψ ˜ ′ is a compact operator Ψ by Proposition 26.2 in Zeidler, 1990. Next, note that the facts that F˜ (t, 0, ..., 0) = 0 on [0, T ] and the inequality in (5.82), from Proposition 5.8, (5.66) and (5.80), we see that conditions (i) and (ii) of Lemma 1.57 are satisfied. Now, we show that condition (iii) of Lemma 1.57 holds. For i = 1, ..., N , let  4d   t, t ∈ [0, T /4),   T  t ∈ [T /4, 3T /4], wi (t) = d,      4d (T − t), t ∈ (3T /4, T ], T

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and w(t) = (w1 (t), ..., wN (t)). Then, w ∈ E α and  1−α , t ∈ [0, T /4), t 4d C α g(t), t ∈ [T /4, 3T /4], 0 Dt wi (t) = T Γ(2 − α)  h(t), t ∈ (3T /4, T ],

(5.83)

where g(t) and h(t) are defined by (5.64) and (5.65). From (5.63) and (5.83), Z T α 2 |C 0 Dt w(t)| dt 0 Z T Z T Z 3T /4 2 C α 2 C α 2 C α =N |0 Dt w1 (t)| dt + |0 Dt w1 (t)| dt + |0 Dt w1 (t)| dt 0 T /4 3T /4 Z Z Z T /4 3T /4 T 16N d2 2−2α 2 2 = 2 2 t dt + g (t)dt + |h(t)| dt T Γ (2 − α) 0 T /4 3T /4 Z 3T /4 Z T 16N d2 1 T 3−2α 2 2 = 2 2 + g (t)dt + h (t)dt T Γ (2 − α) 3 − 2α 4 T /4 3T /4 = ρα d2 . Then, kwk2α = ρα d2 . Thus, from (5.81) with u = w,

1 1 |cos(πα)|ρα d2 ≤ Φ(w) ≤ ρα d2 . 2 2|cos(πα)|

(5.84)

Let r1 =

Γ2 (α)(2α − 1)|cos(πα)| 2 c , 2T 2α−1

r2 =

Γ2 (α)(2α − 1)|cos(πα)| 2 l , 2T 2α−1

(5.85)

Γ2 (α)(2α − 1)|cos(πα)| 2 (m − l2 ). (5.86) 2T 2α−1 Then, from (5.70) and (5.84), we have r1 < Φ(w) < r2 and r3 > 0. For any u ∈ E α , from the first inequality in (5.81), we see that kuk2α ≤ 2Φ(u)/|cos(πα)|. Then, by (5.15) and (5.18), we have r3 =

kuk2 ≤

2T 2α−1 Φ(u) T 2α−1 kuk2α ≤ 2 . − 1) Γ (α)(2α − 1)|cos(πα)|

Γ2 (α)(2α

Thus, by (5.85) and (5.86), we have the following implications Φ(u) < r1 ⇒ kuk < c, Φ(u) < r2 ⇒ kuk < l, Φ(u) < r2 + r3 ⇒ kuk < m.

(5.87)

This, together with (5.72) and (5.79), implies Z T Z T Z ˜ sup F (t, u(t))dt ≤ max F (t, x)dt ≤ u∈Φ−1 (−∞,r1 )

0

sup u∈Φ−1 (−∞,r2 )

0

Z

0

T

F˜ (t, u(t))dt ≤

|x|≤c

Z

0

T

T

max F (t, x)dt ≤ |x|≤l

F (t, c)dt,

0

Z

0

T

F (t, l)dt,

(5.88)

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sup u∈Φ−1 (−∞,r2 +r3 )

Z

0

T

F˜ (t, u(t))dt ≤

Z

T 0

max F (t, x)dt ≤

|x|≤m

Z

209

T

F (t, m)dt.

0

Let ϕ, β, γ and α be defined by (1.46)-(1.49). Then, taking into account the fact that 0 ∈ Φ−1 (−∞, ri ), i = 1, 2, from (5.80), (5.85) and (5.86), it follows that RT ˜ supu∈Φ−1 (−∞,r1 ) Ψ(u) 2T 2α−1 0 F (t, c)dt ≤ 2 , (5.89) ϕ(r1 ) ≤ r1 Γ (α)(2α − 1)|cos(πα)|c2 RT ˜ supu∈Φ−1 (−∞,r2 ) Ψ(u) 2T 2α−1 0 F (t, l)dt ϕ(r2 ) ≤ ≤ 2 , (5.90) r2 Γ (α)(2α − 1)|cos(πα)|l2 RT ˜ supu∈Φ−1 (−∞,r2 +r3 ) Ψ(u) 2T 2α−1 0 F (t, m)dt ≤ 2 . (5.91) γ(r2 , r3 ) = r3 Γ (α)(2α − 1)|cos(πα)|(m2 − l2 ) On the other hand, in view of the fact that w(t) = d < m on [T /4, 3T /4] and from (5.78) and (5.79), Z T Z 3T /4 Z 3T /4 F˜ (t, w(t))dt ≥ F˜ (t, w(t))dt = F˜ (t, d)dt. 0

T /4

T /4

Note that w ∈ Φ−1 [r1 , r2 ), from (1.47) and (5.88), we obtain ˜ ˜ ˜ ˜ Ψ(w) − Ψ(u) Ψ(w) − Ψ(u) ≥ inf β(r1 , r2 ) ≥ inf −1 −1 Φ(w) u∈Φ (−∞,r1 ) u∈Φ (−∞,r1 ) Φ(w) − Φ(u) R 3T /4 RT ˜ ˜ T /4 F (t, d)dt − 0 F (t, c)dt ≥ . Φ(w) By (5.84), 1/Φ(w) ≥ 2|cos(πα)|/(ρα d2 ). Then Z Z T 2|cos(πα)| 3T /4 ˜ β(r1 , r2 ) ≥ F˜ (t, c)dt . (5.92) F (t, d)dt − 2 ρα d T /4 0 ¯ defined by (5.76) and (5.77), from (5.73)-(5.75) and (5.89)-(5.92), For λ and λ we have 1 1 ϕ(r1 ) < ¯ < < β(r1 , r2 ), λ λ 1 1 ϕ(r2 ) < ¯ < < β(r1 , r2 ), λ λ 1 1 γ(r2 , r3 ) < ¯ < < β(r1 , r2 ). λ λ ¯ < 1/λ < β(r1 , r2 ); i.e., condition (iii) of In view of (1.49), α(r1 , r2 , r3 ) < 1/λ Lemma 1.57 holds. Hence, all the assumptions of Lemma 1.57 are satisfied. Then, ¯ , the functional Φ − λΨ ˜ has three distinct by Lemma 1.57, for each λ ∈ λ, λ −1 critical points u1 , u2 and u3 such that u1 ∈ Φ (−∞, r1 ), u2 ∈ Φ−1 [r1 , r2 ), and u3 ∈ Φ−1 (−∞, r2 + r3 ). From (5.87), we have ku1 k < c, ku2 k < l, ku3 k < m. ˜ Then, in view of (5.67), (5.79) and (5.80), we have Ψ(u) = Ψ(u). Therefore, u1 , u2 , and u3 are three distinct critical points of the functional Φ − λΨ. Thus, by Proposition 5.25 (ii), u1 , u2 and u3 are three distinct solutions of (5.59). This completes the proof of the theorem.

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The following results are consequences of Theorem 5.26. In particular, Corollaries 5.27 and 5.29 give some conditions for the system (5.60) to have at least three solutions, and Corollary 5.28 provide some relatively simpler existence criteria for the system (5.59). Corollary 5.27. Assume that there exist four positive constants c, d, l and m, with 1

c < (8N ) 2 d < l < m, such that (5.71) and (5.72) hold, and RT Z Z T 1 3T /4 0 F (t, c)dt < F (t, d)dt − F (t, c)dt , c2 8N d2 T /4 0 RT Z Z T F (t, l)dt 1 3T /4 0 < F (t, d)dt − F (t, c)dt , l2 8N d2 T /4 0 RT Z Z T F (t, m)dt 1 3T /4 0 < F (t, d)dt − F (t, c)dt . m2 − l 2 8N d2 T /4 0

¯ 1 ), system (5.60) has at least three solutions u1 , u2 and u3 Then, for each λ ∈ (λ1 , λ such that maxt∈[0,T ] |u1 (t)| < c, maxt∈[0,T ] |u2 (t)| < l, and maxt∈[0,T ] |u3 (t)| < m, where 4N d2 λ1 = R 3T /4 RT , T T /4 F (t, d)dt − 0 F (t, c)dt n o c2 l2 m2 − l 2 ¯ 1 = min λ , , . RT RT RT 2T 0 F (t, c)dt 2T 0 F (t, l)dt 2T 0 F (t, m)dt

Proof. When α = 1, from (5.63), we have ρα = 8N/T . Then, under the assumptions of Corollary 5.27, it is easy to see that all the conditions of Theorem 5.26 hold for α = 1. Note that the system (5.60) is a special case of the system (5.59) with α = 1. The conclusion then follows directly from Theorem 5.26. The proof is completed. Corollary 5.28. Assume that there exist three positive constants c, d and p, with 1

d F (t, d)dt T 2α−1 ρα d2 T /4 Z 3T /4 Γ2 (α)cos4 (πα)(2α − 1) − 2α−1 F (t, c)dt T ρα d2 (1 + cos2 (πα)) T /4 Z 3T /4 Γ2 (α)cos2 (πα)(2α − 1) = 2α−1 F (t, d)dt. T ρα d2 (1 + cos2 (πα)) T /4

(5.102)

By (5.96) and (5.100)-(5.102), we see that (5.73)-(5.74) hold. From (5.76), (5.77), ¯=λ ¯2 . Therefore, the conclusion (5.98), (5.99) and (5.102), we have λ < λ2 and λ now follows from Theorem 5.26. The proof is completed.

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Corollary 5.29. Assume that there exist three positive constants c, d and p, with 1 p c < (8N ) 2 d < √ , (5.103) 2 such that (5.94) and (5.95) hold, and RT Z 3T /4 F (t, c)dt 1 0 < F (t, d)dt, (5.104) c2 16N d2 T /4 and RT Z 3T /4 F (t, p)dt 1 0 < F (t, d)dt. (5.105) p2 32N d2 T /4 ¯ 3 ), system (5.60) has at least three solutions u1 , u2 , and u3 Then, for each λ ∈ (λ3 , λ √ such that maxt∈[0,T ] |u1 (t)| < c, maxt∈[0,T ] |u2 (t)| < p/ 2, and maxt∈[0,T ] |u3 (t)| < p, where 8N d2 λ3 = R 3T /4 , T T /4 F (t, d)dt n o c2 p2 ¯ 3 = min λ , . RT RT 2T 0 F (t, c)dt 4T 0 F (t, p)dt Proof. When α = 1, from (5.63), we have ρα = 8N/T . Under the assumptions of Corollary 5.29, it is easy to see that all the conditions of Corollary 5.28 hold for α = 1. Note that system (5.60) is a special case of system (5.59) with α = 1. The conclusion then follows directly from Corollary 5.28. The proof is completed. Remark 5.30. We want to point out that when F does not depend on t, (5.104) and (5.105) reduce to F (d) F (p) F (d) F (c) < and < , (5.106) c2 32N d2 p2 64N d2 ¯ 3 become and λ3 and λ c2 p2 16N d2 ¯3 = min λ3 = 2 and λ , . (5.107) T F (d) 2T 2 F (c) 4T 2 F (p) Remark 5.31. We observe that, in our results, no asymptotic condition on F is needed and only local conditions on F are imposed to guarantee the existence of solutions. Moreover, in the conclusions of the above results, one of the three solutions may be trivial since ∇F (t, 0, ..., 0) may be zero. In the remainder of this subsection, we give two examples to illustrate the applicability of our results. Example 5.32. Let T > 0. For (t, x, y) ∈ [0, T ] × R2 , let F (t, x, y) = tG(x, y), where G : R2 → R satisfies that G(−x, −y) = G(x, y), and that for x ∈ [0, ∞) and y ∈ R,  3 x + |y|3 , 0 ≤ x ≤ 1, 0 ≤ |y| ≤ 1,    3 x + 2|y|3/2 − 1, 0 ≤ x ≤ 1, |y| > 1, G(x, y) = (5.108) 3/2 3  2x + |y| − 1, x > 1, 0 ≤ |y| ≤ 1,   3/2 2x + 2|y|3/2 − 2, x > 1, |y| > 1.

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It is easy to verify that F : [0, T ] × R2 → R is measurable in t for (x, y) ∈ R2 and continuously differentiable in x and y for t ∈ [0, T ], and F (t, 0, 0) ≡ 0 on [0, T ]. Let 0 ≤ β < 1, α = 1 − β2 ∈ ( 12 , 1], ρα be defined by (5.63), and u(t) = (u1 (t), u2 (t)). We claim that for each ρ (1 + cos2 (πα)) α , ∞ , λ∈ T 2 |cos(πα)| the system   d 1 0 D−β (u′ (t)) + 1 t D−β (u′ (t)) + λ∇F (t, u(t)) = 0, t T dt 2 2  u(0) = u(T ) = 0

t ∈ [0, T ],

(5.109)

has at least three solutions. In fact, system (5.109) is a special case of system (5.59) with N = 2. For 0 < c < 1 and p > 1, in view of (5.108), we have RT RT F (t, c, c)dt 2c3 0 tdt 0 = = T 2 c, (5.110) c2 c2 RT 0

(4p3/2 − 2) F (t, p, p)dt = p2 p2

RT 0

tdt

=

T 2 (2p3/2 − 1) . p2

(5.111)

Choose d = 1. Then, Z

3T /4

F (t, d, d)dt = 2 T /4

Z

3T /4

T /4

tdt =

1 2 T . 2

(5.112)

By (5.110)-(5.112), we see that there exist 0 < c∗ < 1 and p∗ > 1 such that (5.93), (5.96) and (5.97) hold for any 0 < c < c∗ and p > p∗ . Moreover, (5.94) and (5.95) hold for any c, p > 0. Finally, note from (5.98) and (5.99) that ρα (1 + cos2 (πα)) , T 2 |cos(πα)| ¯ 2 → ∞ as c → 0+ and p → ∞. λ λ2 =

Then, the claim follows from Corollary 5.28. Example 5.33. Let F : R2 → R satisfy that F (−x, −y) = F (x, y), and that for x ∈ [0, ∞) and y ∈ R,  3 x , 0 ≤ x ≤ 1, 0 ≤ |y| ≤ 1,    3 3/2 x + 2|y| − 3|y| + 1, 0 ≤ x ≤ 1, |y| > 1, F (x, y) = (5.113) 3/2  2x − 1, x > 1, 0 ≤ |y| ≤ 1,   3/2 2x + 2|y|3/2 − 3|y|, x > 1, |y| > 1. It is easy to verify that F : R2 → R is continuously differentiable in x and y and F (0, 0) = 0.

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Let T > 0 and u(t) = (u1 (t), u2 (t)). We claim that for each λ ∈ (32/T 2 , ∞), the system u′′ (t) + λ∇F (u(t)) = 0, u(0) = u(T ) = 0

t ∈ [0, T ],

(5.114)

has at least three solutions. In fact, the system (5.114) is a special case of the system (5.60) with N = 2. For 0 < c < 1 and p > 1, from (5.113), we have c3 F (c, c) = 2 = c, 2 c c

(5.115) 1

F (p, p) 4p3/2 − 3p 4p 2 − 3 = = . 2 2 p p p

(5.116)

Choose d = 1. Then F (d, d) 1 = 2 32N d 64

and

F (d, d) 1 = . 2 64N d 128

(5.117)

By (5.115)-(5.117), we see that there exist 0 < c∗ < 1 and p∗ > 1 such that (5.103) and (5.106) hold for any 0 < c < c∗ and p > p∗ . Moreover, (5.94) and (5.95) hold for any c, p > 0. Finally, note from (5.107) that λ3 =

32 T2

and

¯3 → ∞ λ

as c → 0+ and p → ∞.

Then, the claim follows from Corollary 5.29 and Remark 5.30. Remark 5.34. As noted in Remark 5.31, one of the three solutions in the conclusions of the above examples may be trivial. 5.4

5.4.1

Infinite Solutions for BVP with Left and Right Fractional Integrals Introduction

In this section, we consider BVP (5.1), i.e.,   d 1 0 D−β (u′ (t)) + 1 t D−β (u′ (t)) + ∇F (t, u(t)) = 0, t T dt 2 2  u(0) = u(T ) = 0,

a.e. t ∈ [0, T ],

where 0 Dt−β and t DT−β are the left and right Riemann-Liouville fractional integrals of order 0 ≤ β < 1 respectively. Assume that F : [0, T ] × RN → R satisfies the condition (A) which is assumed as in Subsection 5.2.3. In particular, if β = 0, BVP (5.1) reduces to the standard second-order BVP. In the Subsection 5.4.2, using variational methods we prove the multiplicity results for the solutions of problem (5.1).

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5.4.2

215

Existence

Making use of the Property 1.17 and Definition 1.8, for any u ∈ AC([0, T ], RN ), BVP (5.1) is equivalent to (5.21). In the following, we will treat BVP (5.21) in the Hilbert space E α = E0α,2 with the corresponding norm kukα = kukα,2. As E α is a reflexive and separable Banach space, then there are ej ∈ E α and ∗ ej ∈ (E α )∗ such that E α = span{ej : j = 1, 2, ...} and (E α )∗ = span{ej ∗ : j = 1, 2, ...}. For k = 1, 2, ..., denote Xj := span{ej },

Yk :=

k M

Xj ,

Zk :=

j=1

∞ M

Xj .

j=k

Theorem 5.35. Assume that F (t, x) satisfies the condition (A), and suppose the following conditions hold: (A1) there exist κ > 2 and r > 0 such that κF (t, x) ≤ (∇F (t, x), x) for a.e. t ∈ [0, T ] and all |x| ≥ r in RN ; (A2) there exist positive constants µ > 2 and Q > 0 such that F (t, x) ≤Q µ |x|→+∞ |x| lim sup

uniformly for a.e. t ∈ [0, T ]; (A3) there exist µ′ > 2 and Q′ > 0 such that F (t, x) ≥ Q′ µ′ |x|→+∞ |x| lim sup

uniformly for a.e. t ∈ [0, T ]; (A4) F (t, x) = F (t, −x) for t ∈ [0, T ] and all x in RN . Then BVP (5.1) has infinite solutions {un } on E α for every positive integer n such that kun k∞ → ∞, as n → ∞. Proof. Let {un } ⊂ E α such that ϕ(un ) is bounded and ϕ′ (un ) → 0 as n → ∞, where ϕ(u) and ϕ′ (u) are defined by (5.27) and (5.28) respectively. First we prove {un } is a bounded sequence, otherwise, {un } would be unbounded sequence, passing to a subsequence, still denoted by {un }, such that kun kα ≥ 1 and kun kα → ∞, as n → ∞. Noting that Z T α C α hϕ′ (un ), un i = − [(C 0 Dt un (t), t DT un (t)) + (∇F (t, un (t)), un (t))]dt. 0

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In view of the condition (A1) and (5.23) that Z T 1 1 1 α C α − (C ϕ(un ) − hϕ′ (un ), un i = 0 Dt un (t), t DT un (t))dt κ κ 2 0 Z T 1 (∇F (t, un (t)), un (t)) − F (t, un (t)) dt + κ 0 1 1 ≥ − |cos(πα)|kun k2α 2 κ Z Z 1 (∇F (t, un (t)), un (t)) − F (t, un (t)) dt + + κ Ω1 Ω2 1 1 ≥ − |cos(πα)|kun k2α − C1 , 2 κ

(5.118)

where Ω1 := {t ∈ [0, T ]; |un (t)| ≤ r}, Ω2 := [0, T ]\Ω1 and C1 is a positive constant. Since ϕ(un ) is bounded, there exists a positive constant C2 , such that |ϕ(un )| ≤ C2 . Hence, we have 1 1 1 C2 ≥ ϕ(un ) ≥ − |cos(πα)|kun k2α + hϕ′ (un ), un i − C1 2 κ κ (5.119) 1 1 1 ≥ − |cos(πα)|kun k2α − kϕ′ (un )kα kun kα − C1 , 2 κ κ

so {un } is a bounded sequence in E α by (5.119). Since E α is a reflexive space, going to a subsequence if necessary, we may assume that un ⇀ u weakly in E α , thus we have hϕ′ (un ) − ϕ′ (u), un − ui = hϕ′ (un ), un − ui − hϕ′ (u), un − ui ≤ kϕ′ (un )kα kun − ukα − hϕ′ (u), un − ui → 0

(5.120)

as n → ∞. Moreover, according to (5.15) and Proposition 5.6, we have un is bounded in C([0, T ], RN ) and kun − uk → 0 as n → ∞. Observing that hϕ′ (un ) − ϕ′ (u), un − ui Z T α C α =− (C 0 Dt (un (t) − u(t)), t DT (un (t) − u(t)))dt 0 Z T − (∇F (t, un (t)) − ∇F (t, u(t)), un (t) − u(t))dt 0

(5.121)

u(t)k2α

≥ |cos(πα)|kun (t) − Z T − (∇F (t, un (t)) − ∇F (t, u(t)))dt kun (t) − u(t)k. 0

Combining this with (5.120), it is easy to verify that kun (t)− u(t)kα → 0 as n → ∞, and hence that un → u in E α . Thus, {un } admits a convergent subsequence. For any u ∈ Yk , let Z T 1/µ′ µ′ kuk∗ := |u(t)| dt , (5.122) 0

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and it is easy to verify that k · k∗ define by (5.122) is a norm of Yk . Since all the norms of a finite dimensional normed space are equivalent, so there exists positive constant C3 such that C3 kukα ≤ kuk∗

for u ∈ Yk .

(5.123)

In view of (A3), there exist two positive constants M1 and C4 such that ′

F (t, x) ≥ M1 |x|µ

(5.124)

for a.e. t ∈ [0, T ] and |x| ≥ C4 . It follows from (5.23), (5.123) and (5.124) that Z T Z T 1 C α α ϕ(u) = − (0 Dt u(t), C D u(t))dt − F (t, u(t))dt t T 0 2 0 Z Z 1 F (t, u(t))dt kuk2α − F (t, u(t))dt − ≤ |2cos(πα)| Ω Ω3 Z 4 Z ′ 1 kuk2α − M1 |u(t)|µ dt − F (t, u(t))dt ≤ |2cos(πα)| Ω3 Ω4 Z T Z Z ′ ′ 1 = kuk2α − M1 |u(t)|µ dt + M1 |u(t)|µ dt − F (t, u(t))dt |2cos(πα)| 0 Ω4 Ω4 ′ ′ 1 ≤ kuk2α − C3µ M1 kukµα + C5 , |2cos(πα)| where Ω3 := {t ∈ [0, T ] : |u(t)| ≥ C4 }, Ω4 := [0, T ]\Ω3 and C5 is a positive constant. Since µ′ > 2, then there exist positive constants dk such that ϕ(u) ≤ 0, for u ∈ Yk , and kukα ≥ dk . For any u ∈ Zk , let kukµ :=

Z

0

T

µ

|u(t)| dt

1/µ

and βk := sup kukµ ,

(5.125)

(5.126)

u∈Zk kukα =1

then we conclude βk → 0 as k → ∞. In fact, it is obvious that βk ≥ βk+1 > 0, so βk → β. For every k ∈ N, there exists uk ∈ Zk such that kuk kα = 1 and kuk kµ > βk /2.

(5.127)

As E α is reflexive, {uk } has a weakly convergent subsequence, still denoted by {uk }, such that uk ⇀ u. We claim u = 0. In fact, for any fm ∈ {fn : n = 1, 2, ...}, we have fm (uk ) = 0, when k > m, so fm (uk ) → 0,

as k → ∞

for any fm ∈ {fn : n = 1, 2, ...}, therefore u = 0. By Proposition 5.6, when uk ⇀ 0 in E α , then uk → 0 strongly in C([0, T ], RN ). So we conclude β = 0 by (5.127).

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In view of (A2), there exist two positive constants M2 and C6 such that F (t, x) ≤ M2 |x|µ

(5.128)

uniformly for a.e. t ∈ [0, T ] and |x| ≥ C6 . We have Z T Z T 1 C α α (0 Dt u(t), C D u(t))dt − F (t, u(t))dt ϕ(u) = − t T 0 0 2 Z Z |cos(πα)| ≥ kuk2α − F (t, u(t))dt − F (t, u(t))dt 2 Ω5 Z ZΩ6 |cos(πα)| kuk2α − M2 |u(t)|µ dt − F (t, u(t))dt ≥ 2 Ω5 Ω6 Z T Z Z |cos(πα)| 2 µ µ kukα − M2 |u(t)| dt + M2 |u(t)| dt − F (t, u(t))dt = 2 0 Ω6 Ω6 |cos(πα)| kuk2α − M2 βk µ kukµα − C7 , ≥ 2 where Ω5 := {t ∈ [0, T ] : |u(t)| ≥ C6 }, Ω6 := [0, T ]\Ω5 and C7 is a positive constant. Choosing rk = 1/βk , it is obvious that rk → ∞ as k → ∞, then bk :=

inf

u∈Zk

kukα =ρk

ϕ(u) → ∞ as k → ∞,

(5.129)

that is, the condition (H3) in Theorem 1.54 is satisfied. In view of (5.125), let ρk := max{dk , rk + 1}, then ak := max ϕ(u) ≤ 0, u∈Yk

kukα =ρk

and this shows the condition of (H2) in Theorem 1.54 is satisfied. We have proved the functional ϕ satisfies all the conditions of Theorem 1.54, then ϕ has an unbounded sequence of critical values cn = ϕ(un ) by Theorem 1.54. We only need to show kun k → ∞ as n → ∞. In fact, since un is a critical point of the functional ϕ, that is Z T ′ α C α hϕ (un ), un i = − [(C 0 Dt un (t), t DT un (t)) + (∇F (t, un (t)), un (t))]dt = 0. 0

Hence, we have

(5.130)

Z

Z T 1 C α α (0 Dt un (t), C D u (t))dt − F (t, un (t))dt, t T n 0 2 0 Z Z T 1 T (∇F (t, un (t)), un (t))dt − F (t, un (t))dt, = 2 0 0 Z Z T 1 T ≤ |∇F (t, un (t))kun (t)|dt + |F (t, un (t))|dt, (5.131) 2 0 0

cn = ϕ(un ) = −

T

since cn → ∞, we conclude kun k → ∞, as n → ∞

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by (5.131). In fact, if not, going to a subsequence if necessary, we may assume that kun k ≤ M3 ,

for all n ∈ N and some positive constant M3 . Combining assumption (A) and (5.131), we have Z Z T 1 T |∇F (t, un (t))kun (t)|dt + |F (t, un (t))|dt cn ≤ 2 0 0 Z T 1 m2 (t)dt, ≤ (M3 + 1) max m1 (s) 0≤s≤M3 2 0 which contradicts the unboundness of cn . This completes the proof of Theorem 5.35. Example 5.36. In BVP (5.1), let F (t, x) = |x|4 , and choose κ = 4,

r = 2,

µ = µ′ = 4 and Q = Q′ = 1,

so it is easy to verify that all the conditions (A1)-(A4) are satisfied. Then by Theorem 5.35, BVP (5.1) has infinite solutions {uk } on E α for every positive integer k such that kuk k → ∞, as k → ∞. Theorem 5.37. Assume that F (t, x) satisfies the following assumption: (A5) F (t, x) := a(t)|x|γ , where a(t) ∈ L∞ ([0, T ], R+ ) and 1 < γ < 2 is a constant. Then BVP (5.1) has infinite solutions {un } on E α for every positive integer n with kun kα bounded. Proof. Let us show that ϕ satisfies conditions in the Theorem 1.55 item by item. First, we show that ϕ satisfies the (PS)∗c condition for every c ∈ R. ′ Suppose nj → ∞, unj ∈ Ynj , ϕ(un ) → c and (ϕ|Ynj ) (unj ) → 0, then {unj } is a bounded sequence, otherwise, {unj } would be unbounded sequence, passing to a subsequence, still denoted by {unj } such that kunj kα ≥ 1 and kunj kα → ∞. Note that Z T γ α (C Dα un (t), C (5.132) hϕ′ (unj ), unj i − γϕ(unj ) = (−1 + ) t DT unj (t))dt. 2 0 0 t j

However, from (5.132), we have γ −γϕ(unj ) ≥ (1 − )|cos(πα)|kunj k2α − k(ϕ|Ynj )′ (unj )kkunj kα , (5.133) 2 thus kunj kα is a bounded sequence in E α . Going, if necessary, to a subsequence, S we can assume that unj ⇀ u in E α . As E α = Ynj , we can choose vnj ∈ Ynj such nj that vnj → u. Hence lim hϕ′ (unj ), unj − ui nj →∞

= lim hϕ′ (unj ), unj − vnj i + lim hϕ′ (unj ), vnj − ui nj →∞

nj →∞

= lim h(ϕ|Ynj )′ (unj ), unj − vnj i nj →∞

= 0.

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So we have lim hϕ′ (unj ) − ϕ′ (u), unj − ui

nj →∞

= lim hϕ′ (unj ), unj − ui − lim hϕ′ (u), unj − ui nj →∞

nj →∞

= 0, and

hϕ′ (unj ) − ϕ′ (u), unj − ui Z T α C α =− (C 0 Dt (unj (t) − u(t)), t DT (unj (t) − u(t)))dt 0 Z T − (∇F (t, unj (t)) − ∇F (t, u(t)), unj (t) − u(t))dt 0

≥ |cos(πα)|kunj (t) − u(t)k2α Z T − (∇F (t, unj (t)) − ∇F (t, u(t)))dt kunj (t) − u(t)k, 0

we can conclude unj → u in E α , furthermore, we have ϕ′ (unj ) → ϕ′ (u). Let us prove ϕ′ (u) = 0 below. Taking arbitrarily wk ∈ Yk , notice when nj ≥ k, we have hϕ′ (u), wk i = hϕ′ (u) − ϕ′ (unj ), wk i + hϕ′ (unj ), wk i = hϕ′ (u) − ϕ′ (unj ), wk i + h(ϕ|Ynj )′ (unj ), wk i. Let nj → ∞ in the right side of above equation. Then hϕ′ (u), wk i = 0,

′

∀wk ∈ Yk ,

so ϕ (u) = 0, this shows that ϕ satisfies the (PS)∗c for every c ∈ R. For any finite dimensional subspace E ⊂ E α , there exists ε > 0 such that α{t ∈ [0, T ] : a(t)|u(t)|γ ≥ εkukγα } ≥ ε,

∀u ∈ E\{0}.

Otherwise, for any positive integer n, there exists un ∈ E\{0} such that 1 1 α{t ∈ [0, T ] : a(t)|un (t)|γ ≥ kun kγα } < . n n Set vn :=

un (t) kun kα

(5.134)

(5.135)

∈ E\{0}, then kvn kα = 1 for all n ∈ N and

1 1 }< . (5.136) n n Since dimE < ∞, it follows from the compactness of the unit sphere of E that there exists a subsequence, denoted also by {vn }, such that {vn } converges to some v0 in E. It is obvious that kv0 kα = 1. By the equivalence of the norms on the finite dimensional space, we have vn → v0 in L2 ([0, T ], RN ), i.e., Z T |vn − v0 |2 dt → 0, as n → ∞. (5.137) α{t ∈ [0, T ] : a(t)|vn (t)|γ ≥

0

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By (5.137) and Hölder inequality, we have 2−γ Z Z T Z T 2 2 γ a(t)|vn − v0 | dt ≤ a(t) 2−γ dt 0

0

2 = kak 2−γ

Z

0

T

0

T

|vn − v0 |2 dt

Thus, there exist ξ1 , ξ2 > 0 such that

2

|vn − v0 | dt

γ2

γ2

221

(5.138)

→ 0, as n → ∞.

α{t ∈ [0, T ] : a(t)|v0 (t)|γ ≥ ξ1 } ≥ ξ2 .

(5.139)

In fact, if not, we have

α{t ∈ [0, T ] : a(t)|v0 (t)|γ ≥

1 }=0 n

for all positive integer n. It implies that Z T C2T T 0≤ a(t)|v0 |γ+2 dt < kv0 k2 ≤ 8 kv0 k2α → 0 n n 0 as n → ∞, where 1

C8 :=

T α− 2

1

Γ(α)(2α − 1) 2 by (5.15). Hence v0 = 0 which contradicts that kv0 kα = 1. Therefore, (5.139) holds. Now let 1 Ω0 = {t ∈ [0, T ] : a(t)|v0 (t)|γ ≥ ξ1 }, Ωn = {t ∈ [0, T ] : a(t)|vn (t)|γ < }, n 1 and Ωcn = [0, T ] \ Ωn = {t ∈ [0, T ] : a(t)|vn (t)|γ ≥ }. n By (5.136) and (5.139), we have α(Ωn ∩ Ω0 ) = α(Ω0 \ (Ωcn ∩ Ω0 ))

≥ α(Ω0 ) − α(Ωcn ∩ Ω0 )

1 n for all positive integer n. Let n be large enough such that 1 1 1 1 1 ξ2 − ≥ ξ2 and γ−1 ξ1 − ≥ γ ξ1 , n 2 2 n 2 then we have Z T Z a(t)|vn − v0 |γ dt ≥ a(t)|vn − v0 |γ dt 0 Ωn ∩ΩZ0 Z 1 γ a(t)|v0 | dt − a(t)|vn |γ dt ≥ γ−1 2 Ωn ∩Ω0 Ωn ∩Ω0 1 1 ≥ ξ − α(Ωn ∩ Ω0 ) 1 2γ−1 n ξ1 ξ2 ξ1 ξ2 = γ+1 > 0 ≥ γ 2 2 2 ≥ ξ2 −

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for all large n, which is a contradiction to (5.138). Therefore, (5.134) holds. For any u ∈ Zk , let Z T 12 2 and γk := sup kuk2 , kuk2 := |u(t)| dt 0

u∈Zk

kukα =1

then we conclude γk → 0 as k → ∞ in the same way as in the proof of Theorem 5.35. Z T Z T 1 C α α (0 Dt u(t), C D u(t))dt − F (t, u(t))dt ϕ(u) = − t T 0 0 2 Z T 1 ≥ |cos(πα)|kuk2α − a(t)|u(t)|γ dt 2 0 (5.140) 2−γ Z T 2 2 1 γ 2 2−γ ≥ |cos(πα)|kukα − dt a(t) kuk2 2 0 Z T 2−γ 2 2 1 2 2−γ a(t) γkγ kukγα . ≥ |cos(πα)|kukα − dt 2 0 1 1 R 2−γ 2−γ 2 T 4cγkγ 2−γ dt , where c := a(t) , it is obvious that Let ρk := |cos(πα)| 0 ρk → 0, as k → ∞. In view of (5.140), we conclude |cos(πα)| 2 inf ϕ(u) ≥ ρk > 0, u∈Zk 4 kukα =ρk

so the condition (H7) in Theorem 1.55 is satisfied. Furthermore, by (5.140), for any u ∈ Zk with kukα ≤ ρk , we have Therefore,

ϕ(u) ≥ −cγkα kukγα .

−cγkγ ργk ≤ So we have

inf

u∈Zk kukα ≤ρk

inf

u∈Zk

kukα ≤ρk

ϕ(u) ≤ 0.

ϕ(u) → 0,

for ρk , γk → 0, as k → ∞. Hence (H5) in Theorem 1.55 is satisfied. For any u ∈ Yk \ {0}, Z T Z T 1 C α α ϕ(u) = − (0 Dt u(t), C D u(t))dt − F (t, u(t))dt t T 0 2 0 Z T 1 kuk2α − a(t)|u(t)|γ dt ≤ 2|cos(πα)| 0 1 2 ≤ kukα − εkukγαα(Ωu ) 2|cos(πα)| 1 kuk2α − ε2 kukγα , ≤ 2|cos(πα)|

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where ε is given in (5.134), and Ωu := {t ∈ [0, T ] : a(t)|u(t)|γ ≥ εkukγα}. 1 Choosing 0 < rk < min{ρk , (|cos(πα)|ε2 ) 2−γ }, we conclude ik := max ϕ(u) < − u∈Yk kukα =rk

1 r2 < 0, 2|cos(πα)| k

∀ k ∈ N,

that is, the condition (H6) in Theorem 1.55 is satisfied. We have proved the functional ϕ satisfies all the conditions of Theorem 1.55, then ϕ has a bounded sequence of negative critical values cn = ϕ(un ) converging to 0 by Theorem 1.55, we only need to show kun kα is bounded as for every positive integer n. Since Z T Z T 1 C α α cn = ϕ(un ) = − (0 Dt un (t), C D u (t))dt − F (t, un (t))dt t T n 0 2 0 Z T Z T 1 C α α =− (0 Dt un (t), C D u (t))dt − a(t)|un (t)|γ dt t T n 2 0 0 (5.141) |cos(πα)| 2 γ kun kα − a0 kun k T ≥ 2 |cos(πα)| kun k2α − a0 T C8γ kun kγα , ≥ 2 where a0 = ess sup{a(t) : t ∈ [0, T ]}, by Theorem 1.55, cn → 0 as n → ∞. If kun kα has an unbounded sequence, then cn is unbounded by (5.141), which is a contradiction. The proof is completed. 3

Example 5.38. In BVP (5.1), let F (t, x) = a(t)|x| 2 , where T, t = 0, a(t) = t, 0 < t ≤ T. By Theorem 5.37, BVP (5.1) has infinite solutions {uk } on E α for every positive integer k with kuk kα bounded. 5.5

5.5.1

Existence of Solutions for BVP with Left and Right Fractional Derivatives Introduction

In Section 5.5, we consider the fractional BVP of the following form α α t DT (0 Dt u(t)) = ∇F (t, u(t)), a.e. t ∈ [0, T ], u(0) = u(T ) = 0,

(5.142)

where t DTα and 0 Dtα are the right and left Riemann-Liouville fractional derivatives of order 0 < α ≤ 1 respectively, F : [0, T ] × RN → R is a given function satisfying some assumptions and ∇F (t, x) is the gradient of F at x.

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In particular, if α = 1, BVP (5.142) reduces to the standard second order boundary value problem of the following form ′′ u (t) + ∇F (t, u(t)) = 0, a.e. t ∈ [0, T ], u(0) = u(T ) = 0, where F : [0, T ]× RN → R is a given function and ∇F (t, x) is the gradient of F at x. Although many excellent results have been worked out on the existence of solutions for second order BVP (e.g., Li, Liang and Zhang, 2005; Nieto and O’Regan, 2009; Rabinowitz, 1986), it seems that no similar results were obtained in the literature for fractional BVP. According to Benson, Wheatcraft and Meerschaert, 2000a, the one-dimensional form of the fractional ADE can be written as ∂C ∂γ C ∂γ C ∂C = −v + Dj γ + D(1 − j) , ∂t ∂x ∂x ∂(−x)γ

(5.143)

where C is the expected concentration, t is time, v is a constant mean velocity, x is distance in the direction of mean velocity, D is a constant dispersion coefficient, 0 ≤ j ≤ 1 describes the skewness of the transport process, and γ is the order of left and right fractional differential operators. For discussions of this equation, see Benson, Wheatcraft and Meerschaert, 2000b; Fix and Roop, 2004, when γ = 2, the dispersion operators are identical and the classical ADE is recovered. Fundamental (Green function) solutions are Lévy’s γ-stable densities. A special case of the fractional ADE (Eq. (5.143)) describes symmetric transitions, where j = 12 . Defining the symmetric operator equivalent to the Riesz potential in Samko, Kilbas and Marichev, 1993, γ γ 2∇γ ≡ D+ + D−

(5.144)

gives the mass balance equation for advection and symmetric fractional dispersion ∂C = −v∇C + D∇γ C. ∂t

(5.145)

In Subsection 5.5.2, we shall establish a variational structure for BVP (5.142). We show that under some suitable assumptions, the critical points of the variational functional defined on a suitable Hilbert space are the solutions of BVP (5.142). In Subsection 5.5.3, the existence of weak solutions for BVP (5.142) with 12 < α ≤ 1 will be established, where α is the order of fractional derivative in BVP (5.142). In Subsection 5.5.4, we will give some existence results of solutions for BVP (5.142). 5.5.2


Proposition 5.39. Let 0 < α ≤ 1 and 1 1p , we have 0 Dt−α (0 Dtα u(t)) = u(t). Moreover, we can get that E0α,p ∈ C0 ([0, T ], RN ).

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Proof. Let 1p + q1 = 1 and 0 ≤ t1 < t2 ≤ T . ∀f ∈ Lp ([0, T ], RN ), by using Hölder inequality and noting that α > p1 , we have

= =

≤

≤

≤

=

≤ =

|0 Dt−α f (t1 ) − 0 Dt−α f (t2 )| 1 Z 2 Z t2 t1 1 α−1 α−1 (t − s) f (s)ds − (t − s) f (s)ds 1 2 Γ(α) 0 0 Z t1 Z t1 1 α−1 α−1 (t − s) f (s)ds − (t − s) f (s)ds 1 2 Γ(α) 0 0 Z t2 1 (t2 − s)α−1 f (s)ds + Γ(α) t1 Z t1 1 ((t1 − s)α−1 − (t2 − s)α−1 )|f (s)|ds Γ(α) 0 Z t2 1 + (t2 − s)α−1 |f (s)|ds Γ(α) t1 Z t1 1q 1 α−1 α−1 q ((t1 − s) − (t2 − s) ) ds kf kLp Γ(α) 0 Z t2 q1 1 (α−1)q + (t2 − s) ds kf kLp Γ(α) t1 Z t1 q1 1 (α−1)q (α−1)q ((t1 − s) − (t2 − s) )ds kf kLp Γ(α) 0 Z t2 q1 1 + (t2 − s)(α−1)q ds kf kLp Γ(α) t1 1 kf kLp (α−1)q+1 (α−1)q+1 (α−1)q+1 q t − t + (t − t ) 2 1 1 1 2 Γ(α)(1 + (α − 1)q) q 1q kf kLp (t2 − t1 )(α−1)q+1 + 1 Γ(α)(1 + (α − 1)q) q 2kf kLp α−1+ 1q 1 (t2 − t1 ) q Γ(α)(1 + (α − 1)q) 2kf kLp α− p1 . 1 (t2 − t1 ) Γ(α)(1 + (α − 1)q) q

(5.146)

For any u ∈ E0α,p , as 0 Dtα u(t) ∈ Lp ([0, T ], RN ), we apply (5.146) to obtain the continuity of the function 0 Dt−α (0 Dtα u(t)) on [0, T ]. We complete the argument by using Properties 1.19-1.20, and we have −α α 0 Dt (0 Dt u(t))

= u(t) + Ctα−1 ,

t ∈ [0, T ],

where C ∈ RN . Since u(0) = 0 and 0 Dt−α (0 Dtα u(t)) is continuous in [0, T ], we can get that C = 0, which means that 0 Dt−α (0 Dtα u(t)) = u(t) and u is continuous in [0, T ]. Remark 5.40. In the case that 1 − α ≥ 1p , for any u ∈ E0α,p , we also have −α α−1 α u(t). According to Properties 0 Dt (0 Dt u(t)) = u(t). In fact, set f (t) = 0 Dt

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1.19-1.20, we only need to prove that f (0) = [0 Dtα−1 u(t)]t=0 = 0. Noting that 1 − α ≥ p1 , by using Hölder inequality, Lemma 5.1 and the similar method in the proof of Lemma 7 in Fix and Roop, 2004, we can obtain the desired result, i.e. f (0) = 0. We skip the proof since it is similar to Lemma 7 in Fix and Roop, 2004. If α > 1p , the following theorem is useful for us to establish the variational structure on the space E0α,p for BVP (5.142). Theorem 5.41. Let 1 < p < ∞, p1 < α ≤ 1 and L : [0, T ] × RN × RN → R, (t, x, y) → L(t, x, y) be measurable in t for each [x, y] ∈ RN × RN and continuously differentiable in [x, y] for almost every t ∈ [0, T ]. If there exist m1 ∈ C(R+ , R+ ), m2 ∈ L1 ([0, T ], R+ ) and m3 ∈ Lq ([0, T ], R+ ), 1 < q < ∞, such that, for a.e. t ∈ [0, T ] and every [x, y] ∈ RN × RN , one has |L(t, x, y)| ≤ m1 (|x|)(m2 (t) + |y|p ), |Dx L(t, x, y)| ≤ m1 (|x|)(m2 (t) + |y|p ), p−1

where

1 p

+

1 q

|Dy L(t, x, y)| ≤ m1 (|x|)(m3 (t) + |y| = 1, then the functional ϕ defined by Z T ϕ(u) = L(t, u(t), 0 Dtα u(t))dt

(5.147) ),

0

is continuously differentiable on E0α,p , and ∀u, v ∈ E0α,p , we have Z T ′ hϕ (u), vi = [(Dx L(t, u(t), 0 Dtα u(t)), v(t))

(5.148) 0 + (Dy L(t, u(t), 0 Dtα u(t)), 0 Dtα v(t))]dt. Proof. It suffices to prove that ϕ has at every point u a directional derivative ϕ′ (u) ∈ (E0α,p )∗ given by (5.148) and that the mapping ϕ′ : E0α,p → (E0α,p )∗ , u → ϕ′ (u) is continuous. We omit the rather technical proof which is similar to the proof of Theorem 1.4 in Mawhin and Willem, 1989. In fact, the only change we need is to replace the weak derivatives for u and v of Theorem 1.4 in Mawhin and Willem, 1989, by 0 Dtα u and 0 Dtα v respectively. The proof is completed. We are now in a position to give the definition for the solution of BVP (5.142). Definition 5.42. A function u : [0, T ] → RN is called a solution of BVP (5.142) if (i) t DTα−1 (0 Dtα u(t)) and 0 Dtα−1 u(t) are differentiable for almost every t ∈ [0, T ], and (ii) u satisfies (5.142). For a solution u ∈ E α of BVP (5.142) such that ∇F (·, u(·)) ∈ L1 ([0, T ], RN ), multiplying (5.142) by v ∈ C0∞ ([0, T ], RN ) yields Z T [(t DTα (0 Dtα u(t)), v(t)) − (∇F (t, u(t)), v(t))]dt 0 Z T (5.149) = [(0 Dtα u(t), 0 Dtα v(t)) − (∇F (t, u(t)), v(t))]dt 0

=0

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after applying (1.40) and Definition 5.42. Therefore, we can give the definition of weak solution for BVP (5.142) as follows. Definition 5.43. By the weak solution of BVP (5.142), we mean that the function u ∈ E α such that ∇F (·, u(·)) ∈ L1 ([0, T ], RN ) and satisfies (5.149) for all v ∈ C0∞ ([0, T ], RN ). Any solution u ∈ E α of BVP (5.142) is a weak solution provided that ∇F (·, u(·)) ∈ L1 ([0, T ], RN ). Our task is now to establish a variational structure on E α with α ∈ ( 12 , 1], which enables us to reduce the existence of weak solutions of BVP (5.142) to the one of finding critical points of corresponding functional. Corollary 5.44. Let L : [0, T ] × RN × RN → R be defined by L(t, x, y) =

1 2 |y| − F (t, x), 2

where F : [0, T ] × RN → R satisfies the following condition (A) which is assumed as in Subsection 5.2.3. If 12 < α ≤ 1 and u ∈ E α is a solution of corresponding Euler equation ϕ′ (u) = 0, where ϕ is defined as Z T 1 α 2 ϕ(u) = |0 Dt u(t)| − F (t, u(t)) dt, for u ∈ E α , (5.150) 2 0 then u is a weak solution of BVP (5.142) with Proof.

1 2

< α ≤ 1.

By Theorem 5.41, we have Z T ′ 0 = hϕ (u), vi = [(0 Dtα u(t), 0 Dtα v(t)) − (∇F (t, u(t)), v(t))]dt 0

α

for all v ∈ E and hence for all v ∈ C0∞ ([0, T ], RN ). Thus, according to Definition 5.43, u is a weak solution of BVP (5.142). The proof is completed. Remark 5.45. Generally speaking, a critical point u of ϕ on E α will be a weak solution of BVP (5.142). However, we shall show that every weak solution is also a solution of BVP (5.142). 5.5.3

Existence of Weak Solutions

According to Corollary 5.44, we know that in order to find weak solutions of BVP (5.142), it suffices to obtain the critical points of functional ϕ given by (5.150). We need to use some critical point theorems. First, we use Theorem 1.50 to consider the existence of weak solutions for BVP (5.142). Assume that the condition (A) is satisfied. Recall that, in our setting in (5.150), the corresponding functional ϕ on E α is continuously differentiable according to Corollary 5.44 and is also weakly lower semi-continuous functional on E α as the sum of a convex continuous function (see Theorem 1.2 in Mawhin and Willem, 1989) and of a weakly continuous one (see Proposition 1.2 in Mawhin and Willem, 1989).

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In fact, according to Proposition 5.6, if uk ⇀ u in E α , then uk → u in C([0, T ], RN ). Therefore, F (t, uk (t)) → F u(t)) a.e. t ∈ [0, R (t, R T T ]. By Lebesgue’s T dominated convergence theorem, we have 0 F (t, uk (t))dt → 0 F (t, u(t))dt, which RT means that the functional u → 0 F (t, u(t))dt is weakly continuous on E α . Moreover, since fractional derivative operator is linear operator, the functional RT u → 0 (|0 Dtα u(t)|2 /2)dt is convex and continuous on E α . If ϕ is coercive, by Theorem 1.50, ϕ has a minimum so that BVP (5.142) is solvable. It remains to find conditions under which ϕ is coercive on E α , i.e. limkukα →∞ ϕ(u) = +∞, for u ∈ E α . We shall see that it suffices to require that F (t, x) is bounded by a function for a.e. t ∈ [0, T ] and all x ∈ RN . Theorem 5.46. Let α ∈ ( 21 , 1] and assume that F satisfies (A). If |F (t, x)| ≤ a ¯|x|2 + ¯b(t)|x|2−γ + c¯(t),

a.e. t ∈ [0, T ], x ∈ RN ,

(5.151)

where a ¯ ∈ [0, Γ2 (α + 1)/2T 2α ), γ ∈ (0, 2), ¯b ∈ L2/γ ([0, T ], R), and c¯ ∈ L1 ([0, T ], R), then BVP (5.142) has at least one weak solution which minimizes ϕ on E α . Proof. According to the arguments above, our problem reduces to prove that ϕ is coercive on E α . For u ∈ E α , it follows from (5.151) and (5.14) that Z Z T 1 T |0 Dtα u(t)|2 dt − F (t, u(t))dt ϕ(u) = 2 0 0 Z Z T Z T Z T 1 T α 2 2 2−γ ¯ ≥ |0 Dt u(t)| dt − a ¯ |u(t)| dt − b(t)|u(t)| dt − c¯(t)dt 2 0 0 0 0 Z T 1 ¯b(t)|u(t)|2−γ dt − c¯1 ¯kuk2L2 − = kuk2α − a 2 0 Z T γ/2 Z T 1−γ/2 1 ¯kuk2L2 − |¯b(t)|2/γ dt |u(t)|2 dt − c¯1 ≥ kuk2α − a 2 0 0 1 = kuk2α − a ¯kuk2L2 − ¯b1 kuk2−γ ¯1 L2 − c 2 2−γ a ¯T 2α Tα 1 2 2 2−γ ¯ kukα − b1 kukα − c¯1 ≥ kukα − 2 2 Γ (α + 1) Γ(α + 1) 2−γ 1 a ¯T 2α Tα = − 2 kuk2α − ¯b1 kuk2−γ − c¯1 , α 2 Γ (α + 1) Γ(α + 1) RT RT where ¯b1 = ( 0 |¯b(t)|2/γ dt)γ/2 and c¯1 = 0 c¯(t)dt. Noting that a ¯ ∈ [0, Γ2 (α + 1)/2T 2α ) and γ ∈ (0, 2), we have ϕ(u) = +∞ as kukα → ∞, and hence ϕ is coercive, which completes the proof. Let a0 = minλ∈[ 21 ,1] {Γ2 (λ + 1)/2T 2λ }. The following result follows immediately from Theorem 5.46. Corollary 5.47. ∀α ∈ ( 21 , 1] and if F satisfies the condition (A) and (5.151) with a ∈ [0, a0 ), then BVP (5.142) has at least one weak solution which minimizes ϕ on Eα.

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Our task is now to use Theorem 1.51(Mountain pass theorem) to find a nonzero critical point of functional ϕ on E α . Theorem 5.48. Let α ∈ ( 12 , 1] and suppose that F satisfies the condition (A). If (A1) F ∈ C([0, T ] × RN , R) and there exists µ ∈ [0, 12 ) and M > 0 such that 0 < F (t, x) ≤ µ(∇F (t, x), x) for all x ∈ RN with |x| ≥ M and t ∈ [0, T ]; (A2) lim sup|x|→0 F (t, x)/|x|2 < Γ2 (α+1)/2T 2α uniformly for t ∈ [0, T ] and x ∈ RN are satisfied, then BVP (5.142) has at least one nonzero weak solution on E α . Proof. We will verify that ϕ satisfies all the conditions of Theorem 1.51. First, we will prove that ϕ satisfies (PS) condition. Since F (t, x)−µ(∇F (t, x), x) is continuous for t ∈ [0, T ] and |x| ≤ M , there exists c ∈ R+ , such that F (t, x) ≤ µ(∇F (t, x), x) + c,

t ∈ [0, T ],

|x| ≤ M.

t ∈ [0, T ],

x ∈ RN .

By assumption (A1), we obtain F (t, x) ≤ µ(∇F (t, x), x) + c,

Let {uk } ⊂ E α , |ϕ(uk )| ≤ K, k = 1, 2, ..., ϕ′ (uk ) → 0. Notice that Z T hϕ′ (uk ), uk i = [(0 Dtα uk (t), 0 Dtα uk (t)) − (∇F (t, uk (t)), uk (t))]dt 0 Z T = kuk k2α − (∇F (t, uk (t)), uk (t))dt.

(5.152)

(5.153)

0

It follows from (5.152) and (5.153) that Z T 1 2 K ≥ ϕ(uk ) = kuk kα − F (t, uk (t))dt 2 0 Z T 1 (∇F (t, uk (t)), uk (t))dt − cT ≥ kuk k2α − µ 2 0 1 = − µ kuk k2α + µhϕ′ (uk ), uk i − cT 2 1 − µ kuk k2α − µkϕ′ (uk )kα kuk kα − cT, k = 1, 2, ... . ≥ 2 Since ϕ′ (uk ) → 0, there exists N0 ∈ N such that 1 K≥ − µ kuk k2α − kuk kα − cT, 2

k > N0 ,

and this implies that {uk } ⊂ E α is bounded. Since E α is a reflexive space, going to a subsequence if necessary, we may assume that uk ⇀ u weakly in E α , thus we have hϕ′ (uk ) − ϕ′ (u), uk − ui = hϕ′ (uk ), uk − ui − hϕ′ (u), uk − ui ≤ kϕ′ (uk )kα kuk − ukα − hϕ′ (u), uk − ui → 0,

(5.154)

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as k → ∞. Moreover, according to (5.15) and Proposition 5.6, we get that uk is bounded in C([0, T ], RN ) and kuk − uk = 0 as k → ∞. Hence, we have Z T Z T ∇F (t, uk (t))dt → ∇F (t, u(t))dt, as k → ∞. (5.155) 0

0

Noting that hϕ′ (uk ) − ϕ′ (u), uk − ui Z T Z T α α 2 = (0 Dt uk (t) − 0 Dt u(t)) dt − (∇F (t, uk (t)) − ∇F (t, u(t)))(uk (t) − u(t))dt 0 0 Z T 2 ≥ kuk − ukα − (∇F (t, uk (t)) − ∇F (t, u(t)))dt kuk − uk. 0

Combining (5.154) and (5.155), it is easy to verify that kuk − uk2α → 0 as k → ∞, and hence uk → u in Eα . Thus, we obtain the desired convergence property. From lim sup|x|→0 F (t, x)/|x|2 < Γ2 (α + 1)/2T 2α uniformly for t ∈ [0, T ], there exists ǫ ∈ (0, 1) and δ > 0 such that F (t, x) ≤ (1 − ǫ)(Γ2 (α + 1)/2T 2α)|x|2 for all t ∈ [0, T ] and x ∈ RN with |x| ≤ δ. 1

Let ρ =

Γ(α)((α−1)/2+1) 2 1 T α− 2

δ and σ = ǫρ2 /2 > 0. Then it follows from (5.15) that 1

kuk ≤

T α− 2 1

Γ(α)((α − 1)/2 + 1) 2 for all u ∈ E α with kukα = ρ. Therefore, we have Z T 1 F (t, u(t))dt ϕ(u) = kuk2α − 2 0

kukα = δ

Z Γ2 (α + 1) T 1 2 |u(t)|2 dt ≥ kukα − (1 − ǫ) 2 2T 2α 0 1 1 ≥ kuk2α − (1 − ǫ)kuk2α 2 2 1 = ǫkuk2α 2 =σ for all u ∈ E α with kukα = ρ. This implies (ii) in Theorem 1.51 is satisfied. It is obvious from the definition of ϕ and (A2) that ϕ(0) = 0, and therefore, it suffices to show that ϕ satisfies (iii) in Theorem 1.51. Since 0 < F (t, x) ≤ µ(∇F (t, x), x) for all x ∈ RN and |x| ≥ M , a simple regularity argument then shows that there exists r1 , r2 > 0 such that F (t, x) ≥ r1 |x|1/µ − r2 , x ∈ RN , t ∈ [0, T ]. For any u ∈ E α with u 6= 0, κ > 0 and noting that µ ∈ [0, 21 ), we have Z T 1 F (t, κu(t))dt ϕ(κu) = kκuk2α − 2 0 Z T 1 ≤ κ2 kuk2α − r1 |κu(t)|1/µ dt + r2 T 2 0 1 2 1/µ 2 1/µ = κ kukα − r1 κ kukL1/µ + r2 T → −∞ 2

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as κ → ∞. Then there exists a sufficiently large κ0 such that ϕ(κ0 u) ≤ 0. Hence (iii) holds. Lastly noting that ϕ(0) = 0 while for our critical point u, ϕ(u) ≥ σ > 0. Hence u is a nontrivial weak solution of BVP (5.142), and this completes the proof. Theorem 5.49. ∀α ∈ ( 12 , 1], suppose that F satisfies conditions (A) and (A1). If (A2)′ F (t, x) = o(|x|2 ), as |x| → 0 uniformly for t ∈ [0, T ] and x ∈ RN is satisfied, then BVP (5.142) has at least one nonzero weak solution on E α . Remark 5.50. The assumptions in Theorem 5.46 and Theorem 5.48 are classical and the examples can be found in many papers which use critical point theory to discuss differential equations, see, e.g., Li, Liang and Zhang, 2005; Mawhin and Willem, 1989; Rabinowitz, 1989 and references therein. 5.5.4

Existence of Solutions

We firstly give the following lemma which is useful for our further discussion. Lemma 5.51. Let 0 < α ≤ 1. If u ∈ E α is a weak solution of BVP (5.142), then there exists a constant C ∈ RN such that α 0 Dt u(t)

= t DT−α ∇F (t, u(t)) + C(T − t)α−1 ,

a.e. t ∈ [0, T ].

(5.156)

Proof. Since u ∈ E α is a weak solution of BVP (5.142), i.e. ∀h ∈ C0∞ ([0, T ], RN ), we have Z T [(0 Dtα u(t), 0 Dtα h(t)) − (∇F (t, u(t)), h(t))]dt = 0. (5.157) 0

Noting that ∇F (·, u(·)) ∈ L1 ([0, T ], RN ), and applying a similar argument as that for (5.5) in the proof of Lemma 5.1, we get that t DT−α ∇F (·, u(·)) ∈ L1 ([0, T ], RN ). Let us define w ∈ L1 ([0, T ], RN ) by w(t) = t DT−α ∇F (t, u(t)),

t ∈ [0, T ],

so that Z

0

T

(w(t), 0 Dtα h(t))dt = = =

Z Z

Z

T

(t DTα w(t), h(t))dt

0 T 0

(t DTα (t DT−α ∇F (t, u(t))), h(t))dt

T

(∇F (t, u(t)), h(t))dt 0

by applying (1.40) and Property 1.18. Hence, by (5.157) we have, for every h ∈ C0∞ ([0, T ], RN ), Z T (0 Dtα u(t) − w(t), 0 Dtα h(t))dt = 0. 0

(5.158)

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According to Properties 1.9 and in view of h ∈ C0∞ ([0, T ], RN ), we have 0 Dtα h(t) = α−1 ′ h (t). Since 0 Dtα u ∈ L2 ([0, T ], RN ) and w ∈ L1 ([0, T ], RN ), using (1.39) and 0 Dt (5.158), we get that Z T (t DTα−1 (0 Dtα u(t) − w(t)), h′ (t))dt = 0. 0

If (ej ) denotes the Canonical basis of RN , we can choose 2kπt 2kπt ej or h(t) = ej − cos ej , k = 1, ... and j = 1, ..., N. h(t) = sin T T In view of t DTα−1 (0 Dtα u − w) ∈ L1 ([0, T ], RN ), and the theory of Fourier series implies that α−1 (0 Dtα u(t) − w(t)) = C˜ (5.159) tD T

a.e. on [0, T ] for some C˜ ∈ RN . Using Property 1.18 and Property 1.16, we can get that α 0 Dt u(t)

= w(t) + C(T − t)α−1 ,

for some C ∈ RN and this completes the proof.

a.e. t ∈ [0, T ],

Remark 5.52. (i) According to (5.159) and Property 1.17, we have = t DTα−1 (t DT−α ∇F (t, u(t))) + C˜ = t DT−1 ∇F (t, u(t)) + C˜ a.e. on [0, T ] for some C˜ ∈ RN . In view of Definition 1.5 and ∇F (·, u(·)) ∈ L1 ([0, T ], RN ), we shall identify the equivalence class t DTα−1 (0 Dtα u) and its continuous representant Z T α−1 (0 Dtα u(t)) = ∇F (s, u(s))ds + C˜ (5.160) t DT α−1 (0 Dtα u(t)) t DT

t

for t ∈ [0, T ]. (ii) It follows from (5.160) and a classical result of Lebesgue theory that −∇F (·, u(·)) is the classical derivative of t DTα−1 (0 Dtα u) a.e. on [0, T ].

We are now in a position to show that every weak solution of BVP (5.142) is also a solution of BVP (5.142). Theorem 5.53. Let 0 < α ≤ 1. If u ∈ E α is a weak solution of BVP (5.142), then u is also a solution of BVP (5.142). Proof. Firstly, We notice that 0 Dtα−1 u(t) is derivative for almost every t ∈ [0, T ] and (0 Dtα−1 u(t))′ = 0 Dtα u(t) ∈ L2 ([0, T ], RN ) as u ∈ E α . On the other hand, Remark 5.52 implies that t DTα−1 (0 Dtα u(t)) is derivative a.e. on [0, T ] and (t DTα−1 (0 Dtα u(t)))′ ∈ L1 ([0, T ], RN ). Therefore, (i) in Definition 5.42 is verified. It remains to show that u satisfies (5.142). In fact, according to Definition 1.6 and (5.160), we can get that α α t DT (0 Dt u(t)) α

= −(t DTα−1 (0 Dtα u(t)))′ = ∇F (t, u(t)),

a.e. t ∈ [0, T ].

Moreover, u ∈ E implies that u(0) = u(T ) = 0, and therefore (5.142) is verified. The proof is completed.

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The conclusions in Subsection 5.5.3 and Theorem 5.53 imply that BVP (5.142) with α ∈ ( 12 , 1] possesses at least one solution if F satisfies some hypotheses. However, we would like to consider the existence of solutions for BVP (5.142) with α = 12 under the same hypotheses. For any given ǫ0 ∈ (0, 12 ), let ǫ ∈ (0, ǫ0 ) and δ = δ(ǫ) = 21 + ǫ. According to Corollary 5.47 and Theorem 5.49, if (A) and (5.151) with a ∈ [0, a0 ), or (A), (A1) and (A2)′ are satisfied, then ∀ǫ ∈ (0, ǫ0 ), the following BVP   t Dδ 0 Dδ u(t) = ∇F (t, u(t)), a.e. t ∈ [0, T ], t T (5.161)  u(0) = u(T ) = 0

has at least a weak solution uǫ ∈ E δ . Moreover, according to Theorem 5.53, uǫ is also the solution of BVP (5.161). Now, our idea is to obtain the solution of BVP (5.142) with δ = 12 by considering the approximation of uǫ as ǫ → 0. Theorem 5.54. Assume that there exists ǫ0 ∈ (0, 21 ) such that ∀ǫ ∈ (0, ǫ0 ) and δ = δ(ǫ) = 21 + ǫ, BVP (5.161) possesses a weak solution uǫ ∈ E δ . Moreover, if the following conditions are satisfied (A3) there exist β > 2 and m ∈ Lβ ([0, T ], R+) such that |∇F (t, uǫ (t))| ≤ m(t); (A4) there exists β1 > 1/( 21 − ǫ0 ) such that 0 Dtδ uǫ ∈ Lβ1 ([0, T ], RN ). Then there exists a sequence {ǫn } such that ǫ1 > ǫ2 > · · · > ǫn → 0 as n → ∞, u(t) = limn→∞ uǫn (t) exists uniformly on [0, T ] and u is a solution of BVP (5.142) with α = 21 . Proof. According to Theorem 5.53, uǫ is also a solution of BVP (5.161). Thus, we have δ δ (5.162) t DT 0 Dt uǫ (t) = ∇F (t, uǫ (t)), a.e. t ∈ [0, T ].

Properties 1.19-1.20 implies that the equation (5.162) is equivalent to the integral equation δ 0 Dt uǫ (t)

= t DT−δ (∇F (t, uǫ (t))) + C(T − t)δ−1 , a.e. t ∈ [0, T ],

(1/Γ(δ))[t DTδ−1 (0 Dtδ uǫ (t))]t=T .

δ 0 Dt u ǫ

(5.163) β1

where C = Noting that ∈ L ([0, T ], RN ) according to (A4), direct calculation gives that Z T 1 δ−1 δ (s − t)−δ |0 Dsδ uǫ (s)|ds DT 0 Dt uǫ (t) ≤ Γ(1 − δ) t t Z T 1−1/β1 −δβ1 1 ≤ (s − t) β1 −1 ds k0 Dtδ uǫ kLβ1 Γ(1 − δ) t ≤ c(T − t)1−δ−1/β1 k0 Dtδ uǫ kLβ1 ,

where c ∈ R+ is a constant. It is obvious that 1−δ−1/β1 > 0 since β1 > 1/( 21 −ǫ0 ) > 1/(1 − δ), then we have C = (1/Γ(δ))[t DTδ−1 (0 Dtδ uǫ (t))]t=T = 0. Therefore, (5.163) can be written as δ 0 Dt uǫ (t)

= t DT−δ (∇F (t, uǫ (t))),

a.e. t ∈ [0, T ].

(5.164)

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According to Proposition 5.39 and in view of the continuity of uǫ ∈ E δ , (5.164) is equivalent to the integral equation uǫ (t) = 0 Dt−δ (t DT−δ ∇F (t, uǫ (t))),

t ∈ [0, T ].

(5.165)

On the other hand, we observe that m ∈ Lβ ([0, T ], R+) and β > 2 in (A3) imply that Z T 1 (s − t)δ−1 |m(s)|ds |t DT−δ m(t)| ≤ Γ(δ) t 1−1/β Z T (δ−1)β 1 β−1 ds kmkLβ ≤ (s − t) Γ(δ) t ≤ c1 T δ−1/β kmkLβ

≤ c1 kmkLβ max {T λ−1/β }, 1 λ∈[ 2 ,1]

t ∈ [0, T ],

where c1 ∈ R+ is a constant. Therefore, there exists a constant M ∈ R+ such that kt DT−δ mk ≤ M , which means that |t DT−δ ∇F (t, uǫ (t))| ≤ M on [0, T ] since |∇F (t, uǫ (t))| ≤ m(t). Set G(t, uǫ (t)) = t DT−δ ∇F (t, uǫ (t)), and we have by (5.165) Z t 1 (t − s)δ−1 |G(s, uǫ (s))|ds |uǫ (t)| ≤ Γ(δ) 0 Z t M (t − s)δ−1 ds ≤ Γ(δ) 0 (5.166) M Tδ ≤ Γ(δ + 1) Tλ ≤ M max , t ∈ [0, T ]. λ∈[ 21 ,1] Γ(λ + 1) The last inequality follows from the continuity of T λ /Γ(λ + 1) with respect to λ > 0 and the fact that Γ(λ) > 0 for λ > 0. Furthermore, letting 0 ≤ t1 < t2 ≤ T , we see that |uǫ (t1 ) − uǫ (t2 )| Z Z t2 1 t1 δ−1 δ−1 = (t1 − s) G(s, uǫ (s))ds − (t2 − s) G(s, uǫ (s))ds Γ(δ) 0 0 Z 1 t1 [(t1 − s)δ−1 − (t2 − s)δ−1 ]G(s, uǫ (s))ds = Γ(δ) 0 Z t2 δ−1 + (t2 − s) G(s, uǫ (s))ds t1 Z Z t2 M t1 δ−1 δ−1 δ−1 ≤ [(t1 − s) − (t2 − s) ]ds + (t2 − s) ds Γ(δ) 0 t1 M δ δ δ [2(t2 − t1 ) + t1 − t2 ] = Γ(δ + 1)

(5.167)

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2M (t2 − t1 )δ Γ(δ + 1) (t2 − t1 )λ ≤ 2M max . Γ(λ + 1) λ∈[ 12 ,1] ≤

It then follows from (5.166) and (5.167) that the family {uǫ } forms an equicontinuous and uniformly bounded functions. Application of the Ascoli-Arzela theorem shows the existence of a sequence {ǫn } such that ǫ1 > ǫ2 > · · · > ǫn → 0 as n → ∞, and u(t) = limn→∞ uǫn (t) exists uniformly on [0, T ]. Since the continuity and boundness of ∇F (t, ·) imply the continuity of t DT−δ ∇F (t, ·), we obtain that −δ t DT ∇F (t, uǫn (t))

→ t DT−δ ∇F (t, u(t)), as n → ∞,

and combining (5.164) yields u(t) = 0 Dt−δ (t DT−δ ∇F (t, u(t))),

t ∈ [0, T ].

This proves that u is a solution of BVP (5.142) by using the Property 1.18 and Lemma 5.1. The proof is completed. Example 5.55. Set F (t, x) = m(t) sin(|x|), where m ∈ Lβ ([0, T ], R+ ) and x ∈ RN . Then (A3) is verified since |F (t, x)| ≤ m(t) for x ∈ RN . If for any ǫ ∈ (0, ǫ0 ), we have uǫ ∈ AC([0, T ], RN ) and u′ǫ ∈ Lβ1 ([0, T ], RN ), then 0 Dtδ uǫ ∈ Lβ1 ([0, T ], RN ) by using Property 1.9 and (5.5). Thus, (A4) is satisfied. 5.6

Notes and Remarks

The results in Subsections 5.2.1-5.2.4 are taken from Jiao and Zhou, 2011. The material in Subsections 5.2.5-5.2.6 due to Chen and Tang, 2012. The results in Section 5.3 are adopted from Kong, 2013. The main results of Section 5.4 are from Chen and Tang, 2013. The material in Section 5.5 due to Jiao and Zhou, 2012.

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Chapter 6

Fractional Partial Differential Equations

6.1

Introduction

The main objective of this chapter is to investigate the existence theory for a variety of fractional partial differential equations with applications. Section 6.2 is devoted to study the existence of a weak solution for Euler-Lagrange equations. In Section 6.3, we investigate the regularity and unique existence of the solution for initial-boundary value problems of diffusion equation with multiple time-fractional derivatives. In Section 6.4, we discuss existence and multiplicity of solutions of fractional Hamiltonian systems. And in the last Subsection, we present some results on existence of mild solution for fractional Schrödinger equations.

6.2 6.2.1

Fractional Euler-Lagrange Equations Introduction

In this section, we consider a < b two reals, d ∈ N and the following Lagrangian functional Z b L(u) = L(u, a Dtα u, t)dt, a

where L is a Lagrangian, i.e., a map of the form:

L : Rd × Rd × [a, b] → R α a Dt

(x, y, t) → L(x, y, t),

where is the left fractional derivative of Riemann-Liouville of order 0 < α < 1 and where the variable u is a function defined almost everywhere on (a, b) with values in Rd . It is well-known that critical points of the functional L are characterized by the solutions of the fractional Euler-Lagrange equation: ∂L ∂L (u, a Dtα u, t) + t Dbα (u, a Dtα u, t) = 0, (6.1) ∂x ∂y where t Dbα is the right fractional derivative of Riemann-Liouville, see detailed proofs in Agrawal, 2002; Baleanu and Muslih, 2005 for example. 237

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For any p ≥ 1, Lp := Lp ((a, b), Rd ) denotes the classical Lebesgue space of pintegrable functions endowed with its usual norm k · kLp . We denote by | · | the Euclidean norm of Rd and C := C([a, b], Rd ) the space of continuous functions endowed with its usual norm k · k. We remind that a function f is an element of AC if and only if f ′ ∈ L1 and the following equality holds Z t ∀t ∈ [a, b], f (t) = f (a) + f ′ (ξ)dξ, a

′

where f denotes the derivative of f . We refer to Kolmogorov, Fomine and Tihomirov, 1974 for more details concerning the absolutely continuous functions. In addition, we denote by Ca (resp. ACa or Ca∞ ) the space of functions f ∈ C (resp. AC or C ∞ ) such that f (a) = 0. In particular, Cc∞ ⊂ Ca∞ ⊂ ACa . Remark 6.1. In the whole section, an equality between functions must be understood as an equality holding for almost all t ∈ (a, b). When it is not the case, the interval on which the equality is valid will be specified. Definition 6.2. A function u is said to be a weak solution of (6.1) if u ∈ C and if u satisfies (6.1) a.e. on [a,b]. In the following, we will provide some properties concerning the left fractional operators of Riemann-Liouville. One can easily derive the analogous versions for the right ones. Property 6.3 is well-known and one can find their proofs in the classical literature on the subject (see Lemma 2.1 in Kilbas, Srivastava and Trujillo, 2006). Property 6.3. For any α > 0 and any p ≥ 1, a Dt−α is linear and continuous from Lp to Lp . Precisely, the following inequality holds ka Dt−α f kLp ≤

(b − a)α kf kLp , for f ∈ Lp . Γ(1 + α)

The following classical property concerns the integration of fractional integrals. It is occasionally called fractional integration by parts: p . Then, for any f ∈ Lp , we have Property 6.4. Let 0 < p1 < α < 1 and q = p−1 (i) a Dt−α is Hölder continuous on [a, b] with exponent α − (ii) lim a Dt−α f (t) = 0.

1 p

> 0;

t→a

Consequently, a Dt−α f (t) can be continuously extended by 0 in t = a. Finally, for any f ∈ Lp , we have a Dt−α f ∈ Ca . Moreover, the following inequality holds 1

ka Dt−α f k ≤

(b − a)α− p

1

Γ(α)((α − 1)q + 1) q

kf kLp , for f ∈ Lp .

Proof. Let us note that this result is mainly proved in Section 5.2. Let f ∈ Lp . We first remind the following inequality (ξ1 − ξ2 )q ≤ ξ1q − ξ2q , for ξ1 ≥ ξ2 ≥ 0.

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Let us prove that a Dt−α f (t) is Hölder continuous on [a, b]. For any a < t1 < t2 ≤ b, using Hölder’s inequality, we have Z Z t1 1 t2 −α −α α−1 α−1 (t2 − ξ) f (ξ)dξ − (t1 − ξ) f (ξ)dξ |a Dt f (t2 ) − a Dt f (t1 )| = Γ(α) a Z a 1 t2 ≤ (t2 − ξ)α−1 f (ξ)dξ Γ(α) t1 Z 1 t1 α−1 α−1 ((t2 − ξ) − (t1 − ξ) )f (ξ)dξ + Γ(α) a Z t2 1q kf kLp (α−1)q (t2 − ξ) dξ ≤ Γ(α) t1 Z t1 q1 kf kLp α−1 α−1 q + ((t1 − ξ) − (t2 − ξ) ) dξ Γ(α) a Z t2 1q kf kLp ≤ (t2 − ξ)(α−1)q dξ Γ(α) t1 Z t1 1q kf kLp (α−1)q (α−1)q + (t1 − ξ) − (t2 − ξ) dξ Γ(α) a 1 2kf kLp α− p ≤ . 1 (t2 − t1 ) q Γ(α)((α − 1)q + 1) The proof of the first point is complete. Let us consider the second point. For any t ∈ [a, b], we can prove in the same manner that 1 kf kLp α− p , as t → 0. |a Dt−α f (t)| ≤ 1 (t − a) q Γ(α)((α − 1)q + 1) The proof is now complete. 6.2.2

Functional Spaces

In order to prove the existence of a weak solution of (6.1) using a variational method, we need the introduction of an appropriate space of functions. This space has to present some properties like reflexivity, see Dacorogna, 2008. For any 0 < α < 1 and any p ≥ 1, we define the following space of functions Eα,p := {u ∈ Lp

satisfying

α a Dt u

We endow Eα,p with the following norm

∈ Lp

and

−α α a Dt (a Dt u)

k · kα,p : Eα,p → R+ , 1

Let us note that

u 7→ (kukpLp + ka Dtα ukpLp ) p . | · |α,p : Eα,p → R+ , u 7→ ka Dtα ukLp

=u

a.e.}.

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is an equivalent norm to k · kα,p for Eα,p . Indeed, Property 6.3 leads to kukLp = ka Dt−α (a Dtα u)kLp ≤

(b − a)α ka Dtα ukLp , for u ∈ Eα,p . Γ(1 + α)

(6.2)

The goal of this section is to prove the following proposition: Proposition 6.5. Assuming 0 < p1 < α < 1, Eα,p is a reflexive separable Banach space and the compact embedding Eα,p # Ca holds. Proof.

Consider that 0
1, (Lp , k · kLp ) is a reflexive separable Banach space and therefore, ((Lp )2 , k · k(Lp )2 ) is also a reflexive separable Banach space. We define Ω := {(u, a Dtα u) : u ∈ Eα,p }. Let us prove that Ω is a closed subspace of ((Lp )2 , k · k(Lp )2 ). Let (un , vn )n∈N ⊂ Ω such that (Lp )2

(un , vn ) −−−→ (u, v). Then, we prove that (u, v) ∈ Ω. For any n ∈ N, (un , vn ) ∈ Ω. Thus, un ∈ Eα,p and vn = a Dtα un . Consequently, we have Lp

un −−→ u

and

Lp α −→ a Dt u n −

v.

For any n ∈ N, since un ∈ Eα,p and a Dt−α is continuous from Lp to Lp , we have Lp

un = a Dt−α (a Dtα un ) −−→ a Dt−α v.

Thus, u = a Dt−α v, a Dtα u = a Dtα (a Dt−α v) = v ∈ Lp and a Dt−α (a Dtα u) = a Dt−α v = u. Hence, u ∈ Eα,p and (u, v) = (u, a Dtα u) ∈ Ω. In conclusion, Ω is a closed subspace of ((Lp )2 , k · k(Lp )2 ) and then Ω is a reflexive separable Banach space. Finally, defining the following operator A : Eα,p → Ω, u 7→ (u, a Dtα u), we prove that Eα,p is isometric isomorphic to Ω. This completes the proof of Step I. Step II. The continuous embedding Eα,p ֒→ Ca . Let u ∈ Eα,p and then a Dtα u ∈ Lp . Since 0 < p1 < α < 1, Property 6.4 leads to a Dt−α (a Dtα u) ∈ Ca . Furthermore, u = a Dt−α (a Dtα u) and consequently, u can be identified to its continuous representative. Finally, Property 6.4 also gives 1

kuk =

ka Dt−α (a Dtα u)k

≤

(b − a)α− p

1

Γ(α)((α − 1)q + 1) q

|u|α,p , for u ∈ Eα,p .

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Since k · kα,p and | · |α,p are equivalent norms, the proof of Step II is complete. Step III. The compact embedding Eα,p # Ca . Since Eα,p is a reflexive Banach space, we only have to prove that ∀(un )n∈N ⊂ Eα,p

Eα,p

such that un ⇀ u,

C

then un − → u.

Let (un )n∈N ⊂ Eα,p such that Eα,p

un ⇀ u. Since Eα,p ֒→ Ca , we have C

un ⇀ u. Since (un )n∈N converges weakly in Eα,p , (un )n∈N is bounded in Eα,p . Consequently, (a Dtα un )n∈N is bounded in Lp by a constant M ≥ 0. Let us prove that (un )n∈N ⊂ Ca is uniformly Lipschitzian on [a, b]. According to the proof of Property 6.4, for ∀n ∈ N, ∀a ≤ t1 < t2 ≤ b, we have, |un (t2 ) − un (t1 )| ≤ |a Dt−α (a Dtα un (t2 )) − a Dt−α (a Dtα un (t1 ))| ≤ ≤

2ka Dtα un kLp

Γ(α)((α − 1)q + 1) 2M Γ(α)((α − 1)q + 1)

1

1 p

(t2 − t1 )α− p

1 p

(t2 − t1 )α− p .

1

Hence, from Ascoli’s theorem, (un )n∈N is relatively compact in C. Consequently, there exists a subsequence of (un )n∈N converging strongly in C and the limit is u by uniqueness of the weak limit. Now, let us prove by contradiction that the whole sequence (un )n∈N converges strongly to u in C. If not, there exist ε > 0 and a subsequence (unk )k∈N such that kunk − uk > ε > 0, for k ∈ N.

(6.3)

Nevertheless, since (unk )k∈N is a subsequence of (un )n∈N , then it satisfies Eα,p

unk ⇀ u. In the same way (using Ascoli’s theorem), we can construct a subsequence of (unk )k∈N converging strongly to u in C which is a contradiction to (6.3). The proof of Step III is now complete. Let us remind the following property −α a Dt ϕ

∈ Ca∞ , for ϕ ∈ Cc∞ .

From this result, we get the following results. Proposition 6.6. Ca∞ is dense in Eα,p . Proof. Indeed, let us first prove that Ca∞ ⊂ Eα,p . Let u ∈ Ca∞ ⊂ Lp . Since u ∈ ACa and u′ ∈ Lp , we have a Dtα u = a Dtα−1 u′ ∈ Lp . Since u ∈ AC, we also have

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−α α a Dt (a Dt u)

= u. Finally, u ∈ Eα,p . Now, let us prove that Ca∞ is dense in Eα,p . Let u ∈ Eα,p , then a Dtα u ∈ Lp . Consequently, there exists (vn )n∈N ⊂ Cc∞ such that Lp

vn −−→ a Dtα u and then

Lp −α −→ a Dt−α (a Dtα u) a Dt vn −

= u,

since a Dt−α is continuous from Lp to Lp . Defining un := a Dt−α vn ∈ Ca∞ for any n ∈ N, we obtain Lp

un −−→ u and

α a Dt u n

Lp

= a Dtα (a Dt−α vn ) = vn −−→ a Dtα u.

Finally, (un )n∈N ⊂ Ca∞ and converges to u in Eα,p . The proof is completed. Proposition 6.7. If Lp }.

1 p

< min(α, 1 − α), then Eα,p = {u ∈ Lp satisfying a Dtα u ∈

Proof. Indeed, let u ∈ Lp satisfying a Dtα u ∈ Lp and let us prove that −α 1 α p α ∞ a Dt (a Dt u) = u. Let ϕ ∈ Cc ⊂ L . Since a Dt u ∈ L , Property 1.23 leads to Z b Z b Z b d −α −α α α (a Dtα−1 u) · t Db−α ϕdt. a Dt (a Dt u) · ϕdt = a Dt u · t Db ϕdt = a a a dt Then, an integration by parts gives Z b Z b −α α−1 α u · t Db1−α udt. a Dt (a Dt u) · ϕdt = a Dt a

−α t Db ϕ(b)

a α−1 u(a) a Dt

Indeed, = 0 since ϕ ∈ and = 0 since u ∈ Lp and p1 < 1−α. Finally, using Property 1.23 again, we obtain Z b Z b Z b −α α−1 1−α α u · t Db (t Db ϕ)dt = u · ϕdt, a Dt (a Dt u) · ϕdt = Cc∞

a

a

a

this completes the proof.

Remark 6.8. In the Proposition 6.7, let us note that such a definition of Eα,p could lead us to name it fractional Sobolev space and to denote it by W α,p . Nevertheless, these notions and notations are already used, see Brezis, 2011. 6.2.3


In this subsection, we assume that Lagrangian L is of class C 1 and we define the Lagrangian functional L on Eα,p (with 0 < p1 < α < 1). Precisely, we define L : Eα,p → R, Z b u 7→ L(u, a Dtα u, t)dt. a

L is said to be Gâteaux-differentiable in u ∈ Eα,p if the map DL(u) : Eα,p → R,

v 7→ DL(u)(v) := lim

h→0

L(u + hv) − L(u) h

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is well-defined for any v ∈ Eα,p and if it is linear and continuous. A critical point u ∈ Eα,p of L is defined by DL(u) = 0. We introduce the following hypotheses: (H1) there exist 0 ≤ d1 ≤ p and r1 , s1 ∈ C(Rd × [a, b], R+ ) such that |L(x, y, t) − L(x, 0, t)| ≤ r1 (x, t)kykd1 + s1 (x, t), for (x, y, t) ∈ Rd × Rd × [a, b];

(H2) there exist 0 ≤ d2 ≤ p and r2 , s2 ∈ C(Rd × [a, b], R+ ) such that

∂L d2 d d

∂x (x, y, t) ≤ r2 (x, t)kyk + s2 (x, t), for (x, y, t) ∈ R × R × [a, b];

(H3) there exist 0 ≤ d3 ≤ p − 1 and r3 , s3 ∈ C(Rd × [a, b], R+ ) such that

∂L

≤ r3 (x, t)kykd3 + s3 (x, t), for (x, y, t) ∈ Rd × Rd × [a, b];

(x, y, t)

∂y

(H4) coercivity condition: there exist γ > 0, 1 ≤ d4 < p, c1 ∈ C(Rd ×[a, b], [γ, ∞)), c2 , c3 ∈ C([a, b], R) such that ∀(x, y, t) ∈ Rd × Rd × [a, b],

L(x, y, t) ≥ c1 (x, t)kykp + c2 (t)kxkd4 + c3 (t);

(H5) convexity condition: ∀t ∈ [a, b], L(·, ·, t) is convex. Hypotheses denoted by (H1)-(H3) are usually called regularity hypotheses (see Cesari, 1983; Dacorogna, 2008). Let us prove the following results. Lemma 6.9. The following implications hold (i) L satisfies (H1) ⇒ for any u ∈ Eα,p , L(u, a Dtα u, t) ∈ L1 and then L(u) exists in R; (ii) L satisfies (H2) ⇒ for any u ∈ Eα,p , ∂L/∂x(u, aDtα u, t) ∈ L1 ; p (iii) L satisfies (H3) ⇒ for any u ∈ Eα,p , ∂L/∂y(u, aDtα u, t) ∈ Lq , where q = p−1 . Proof. Let us assume that L satisfies (H1) and let u ∈ Eα,p ⊂ Ca . Then, ka Dtα ukd1 ∈ Lp/d1 ⊂ L1 and the three maps t → r1 (u(t), t), s1 (u(t), t), |L(u(t), 0, t)| ∈ C([a, b], R+ ) ⊂ L∞ ⊂ L1 . Hypothesis (H1) implies for almost all t ∈ [a, b] |L(u(t), a Dtα u(t), t)| ≤ r1 (u(t), t)ka Dtα u(t)kd1 + s1 (u(t), t) + |L(u(t), 0, t)|.

Hence, L(u, a Dtα u(t), t) ∈ L1 and then L(u) exists in R. We proceed in the same manner in order to prove the second point of Lemma 6.9. Now, assuming that L satisfies (H3), we have ka Dtα ukd3 ∈ Lp/d3 ⊂ Lq for any u ∈ Eα,p . An analogous argument gives the third point of Lemma 6.9. This completes the proof. Lemma 6.10. Assuming that L satisfies Hypotheses (H1)-(H3), L is Gâteauxdifferentiable in any u ∈ Eα,p and Z b ∂L ∂L DL(u)(v) = (u, a Dtα u, t) · v + (u, a Dtα u, t) · a Dtα v dt, for u, v ∈ Eα,p . ∂x ∂y a

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Proof. Let u, v ∈ Eα,p ⊂ Ca . Let ψu,v defined for any h ∈ [−1, 1] and for almost all t ∈ [a, b] by ψu,v (t, h) := L u(t) + hv(t), a Dtα u(t) + ha Dtα v(t), t . Then, we define the following mapping

φu,v : [−1, 1] → R, Z b Z α α h 7→ L u + hv(t), a Dt u + ha Dt v, t dt = a

b

ψu,v (t, h)dt.

a

Our aim is to prove that the following term DL(u)(v) = lim

h→0

L(u + hv) − L(u) φu,v (h) − φu,v (0) = lim = φ′u,v (0) h→0 h h

exists in R. In order to differentiate φu,v , we use the theorem of differentiation under the integral sign. Indeed, we have for almost all t ∈ [a, b], ψu,v (t, ·) is differentiable on [−1, 1] with ∂L ∂ψu,v (t, h) = u(t) + hv(t), a Dtα u(t) + ha Dtα v(t), t · v(t) ∂h ∂x ∂L u(t) + hv(t), a Dtα u(t) + ha Dtα v(t), t · a Dtα v(t). + ∂y

Then, from Hypotheses (H2) and (H3), we have for any h ∈ [−1, 1] and for almost all t ∈ [a, b] ∂ψu,v (t, h) ∂h ≤ r2 (u(t) + hv(t), t)ka Dtα u(t) + ha Dtα v(t)kd2 + s2 (u(t) + hv(t), t) kv(t)k + r3 (u(t) + hv(t), t)ka Dtα u(t) + ha Dtα v(t)kd3 + s3 (u(t) + hv(t), t) ka Dtα v(t)k.

We define

r2,0 :=

max

(t,h)∈[a,b]×[−1,1]

r2 (u(t) + hv(t), t)

and we define similarly s2,0 , r3,0 , s3,0 . Finally, it holds ∂ψu,v α d2 d2 + ka Dtα v(t)kd2 ) kv(t)k +s2,0 kv(t)k ∂h (t, h) ≤ 2 r2,0 |(ka Dt u(t)k {z } | {z } | {z } +2

d3

r3,0 (ka Dtα u(t)kd3 |

+ {z

∈Ca ⊂L∞ ∈Lp/d2 ⊂L1 α α d3 ka Dt v(t)k ) ka Dt v(t)k +s3,0

∈Lp/d3 ⊂Lq

}|

{z

∈Lp

}

∈Ca ⊂L1 α ka Dt v(t)k .

|

{z

∈Lp ⊂L1

}

The right term is then a L1 function independent of h. Consequently, applying the theorem of differentiation under the integral sign, φu,v is differentiable with Z b ∂ψu,v φ′u,v (h) = (t, h)dt, for h ∈ [−1, 1]. h a

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Hence DL(u)(v) = =

φ′u,v (0) Z

b a

=

Z

b

a

∂ψu,v (t, 0)dt h

∂L ∂L α α α (u, a Dt , t) v + (u, a Dt u, t) a Dt v dt. ∂x ∂y

From Lemma 6.9, it holds ∂L (u, a Dtα u, t) ∈ L1 ∂x

and

∂L (u, a Dtα u, t) ∈ Lq . ∂y

Since v ∈ Ca ⊂ L∞ and a Dtα ∈ Lp , DL(u)(v) exists in R. Moreover, we have

∂L

∂L α α

ka Dtα vkLp (u, D u, t) kvk + (u, D u, t) |DL(u)(v)| ≤ a t a t

1

∂y

q

∂x L L

1

∂L (b − a)α− p ∂L α α

≤ (u, a Dt u, t) + (u, a Dt u, t) 1

q |v|α,p . ∂x ∂y q Γ(α)((α − 1)q + 1) L1 L

Consequently, DL(u) is linear and continuous from Eα,p to R. The proof is completed. 6.2.4

Existence of Weak Solution

In this subsection, we will present the existence theorem of weak solution for (6.1). We firstly give two preliminary theorems. Theorem 6.11. Assume that L satisfies Hypotheses (H1)-(H3). If u is a critical point of L, u is a weak solution of (6.1). Let u be a critical point of L. Then, we have in particular Z b ∂L ∂L (u, a Dtα u, t) · v + (u, a Dtα u, t) · a Dtα v dt = 0, for v ∈ Cc∞ . DL(u)(v) = ∂x ∂y a

Proof.

For any v ∈ Cc∞ ⊂ ACa , a Dtα v = a Dtα−1 v ′ ∈ Ca∞ . Since ∂L/∂y(u, aDtα u, t) ∈ Lq , Property 1.23 gives Z b ∂L α−1 ∂L α α ′ (u, a Dt u, t) · v + t Db (u, a Dt u, t) · v dt = 0, for v ∈ Cc∞ . ∂x ∂y a Finally, we define wu (t) =

Z

a

∂L/∂x(u, aDtα , t)

1

t

∂L (u, a Dtα u, t)dt, for t ∈ [a, b]. ∂x

Since ∈ L , wu ∈ ACa and wu′ = ∂L/∂x(u, a Dtα u, t). Then, an integration by parts leads to Z b α−1 ∂L α D (u, D u, t) − w v ′ dt = 0, for v ∈ Cc∞ . t b a t u ∂y a

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Consequently, there exists a constant C ∈ Rd such that α−1 ∂L α (u, a Dt u, t) = C + wu ∈ AC. t Db ∂y

By differentiation, we obtain ∂L α ∂L α − t Db (u, a Dt u, t) = (u, a Dtα u, t), ∂y ∂x and then u ∈ Eα,p ⊂ C satisfies (6.1) a.e. on [a, b]. The proof is completed.

As usual in a variational method, in order to prove the existence of a global minimizer of a functional, coercivity and convexity hypotheses need to be added on the Lagrangian. We have already define Hypotheses (H4) (coercivity) and (H5) (convexity). Next, we introduce two different convexity hypotheses (H5)′ and (H5)′′ : (H5)′ ∀(x, t) ∈ Rd ×[a, b], L(x, ·, t) is convex and (L(·, y, t))(y,t)∈Rd ×[a,b] is uniformly equicontinuous on Rd , i.e., ∀ε > 0, ∃δ > 0, ∀(x1 , x2 ) ∈ (Rd )2 , kx2 − x1 k < δ ⇒ ∀(y, t) ∈ Rd × [a, b], |L(x2 , y, t) − L(x1 , y, t)| < ε. (H5)′′ ∀(x, t) ∈ Rd × [a, b], L(x, ·, t) is convex. Let us note that Hypotheses (H5) and (H5)′ are independent. Hypothesis (H5)′′ is the weakest. Nevertheless, in this case, the detailed proof of Theorem 6.13 is more complicated. Consequently, in the case of Hypothesis (H5)′′ , we do not develop the proof and we use a strong result proved in Dacorogna, 2008. Let us prove the following preliminary result. Lemma 6.12. Assume that L satisfies Hypothesis (H4). Then, L is coercive in the sense that lim

kukα,p →+∞

Let u ∈ Eα,p , we have Z b Z L(u) = L(u, a Dtα u, t)dt ≥

L(u) = +∞.

Proof.

a

a

b

c1 (u, t)ka Dtα ukp + c2 (t)kukd4 + c3 (t)dt.

Eq.(6.2) implies that kukdL4d4

1−

≤ (b − a)

d4 p

kukdL4p

(b − a)α+1− ≤ Γ(α + 1)

d4 p

ka Dtα ukdL4p

(b − a)α+1− = Γ(α + 1)

d4 p

4 |u|dα,p .

Finally, we conclude that L(u) ≥ γka Dtα ukpLp − kc2 kkukdL4d4 − (b − a)kc3 k ≥

γ|u|pα,p

kc2 k(b − a)α+1− − Γ(α + 1)

d4 p

4 |u|dα,p − (b − a)kc3 k, for u ∈ Eα,p .

Since d4 < p and the norms | · |α,p and k · kα,p are equivalent, the proof is completed.

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Theorem 6.13. Assume that L satisfies Hypotheses (H1)-(H4) and one of Hypotheses (H5), (H5)′ or (H5)′′ . Then, L admits a global minimizer. Proof.

Let (un )n∈N be a sequence in Eα,p satisfying L(un ) → inf L(v) =: K. v∈Eα,p

Since L satisfies Hypothesis (H1), L(u) ∈ R for any u ∈ Eα,p . Hence, K < +∞. Let us prove by contradiction that (un )n∈N is bounded in Eα,p . In the negative case, we can construct a subsequence (unk )k∈N satisfying kunk kα,p → ∞. Since L satisfies Hypothesis (H4), Lemma 6.12 gives: K = lim L(unk ) = +∞, k∈N

which is a contradiction. Hence, (un )n∈N is bounded in Eα,p . Since Eα,p is reflexive, there exists a subsequence still denoted by (un )n∈N converging weakly in Eα,p to an element denoted by u ∈ Eα,p . Let us prove that u is a global minimizer of L. Since Eα,p

un ⇀ u and Eα,p # Ca , we have C

un − → u and

Lp α α a Dt un ⇀ a Dt u.

(6.4)

Case L satisfies (H5): by convexity, it holds for any n ∈ N Z b Z b L(un ) = L(un , a Dtα un , t)dt ≥ L(u, a Dtα u, t)dt a a Z b Z b ∂L ∂L (u, a Dtα u, t) (un − u)dt + (u, a Dtα u, t) (a Dtα un − a Dtα u)dt. + a ∂y a ∂x

Since L satisfies Hypotheses (H2) and (H3), ∂L/∂x(u, aDtα u, t) ∈ L1 and ∂L/∂y(u, aDtα u, t) ∈ Lq . Consequently, using (6.4) and making n tend to +∞, we obtain Z b K = inf L(v) ≥ L(u, a Dtα u, t)dt = L(u). v∈Eα,p

a

Consequently, u is a global minimizer of L. Case L satisfies (H5)′ : let ε > 0. Since (un )n∈N converges strongly in C to u, we have ∃N ∈ N, ∀n ≥ N, kun − uk < δ,

where δ is given in the definition of (H5)′ . In consequence, it holds a.e. on [a,b] |L(un (t), a Dtα un (t), t) − L(u(t), a Dtα un (t), t)| < ε, for n ≥ N. Moreover, for any n ≥ N , we have Z b Z b L(un ) = L(u, a Dtα u, t)dt + [L(un , a Dtα un , t) − L(u, a Dtα un , t)] dt a a Z b + [L(u, a Dtα un , t) − L(u, a Dtα u, t)] dt. a

(6.5)

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Then, for any n ≥ N , it holds by convexity Z b Z b α L(un ) ≥ L(u, a Dt u, t)dt − |L(un , a Dtα un , t) − L(u, a Dtα un , t)|dt a a Z b ∂L (u, a Dtα u, t) (a Dtα un − a Dtα u)dt. + a ∂y

And, using Eq. (6.5), we obtain for any n ≥ N Z b Z b ∂L L(un ) ≥ L(u, a Dtα u, t)dt − ε(b − a) + (u, a Dtα u, t) (a Dtα un − a Dtα u)dt. a a ∂y

We remind that ∂L/∂y(u, aDtα u, t) ∈ Lq since L satisfies (H3). Since (a Dtα un )n∈N converges weakly in Lp to a Dtα u we obtain by making n tend to +∞ and then by making ε tend to 0 Z b K = inf L(v) ≥ L(u, a Dtα u, t)dt = L(u). v∈Eα,p

a

Consequently, u is a global minimizer of L. Case L satisfies (H5)′′ : we refer to Theorem 3.23 in Bacorogna, 2008.

Finally, we give the existence theorem of weak solution for (6.1). Theorem 6.14. Let L be a Lagrangian of class C 1 and 0 < p1 < α < 1. If L satisfies the hypotheses denoted by (H1)-(H5). Then (6.1) admits a weak solution. Combining Theorems 6.11 and 6.13, the proof of Theorem 6.14 is obvious. Let us consider some examples of Lagrangian L satisfying Hypotheses of Theorem 6.14. Consequently, the fractional Euler-Lagrange equation (6.1) associated admits a weak solution u ∈ Eα,p . Examples 6.15. The most classical example is the Dirichlet integral, i.e. the Lagrangian functional associated to the Lagrangian L given by 1 kyk2 . 2 In this case, L satisfies Hypotheses (H1)-(H5) for p = 2. Hence, the fractional EulerLagrange equation (6.1) associated admits a weak solution in Eα,p for 21 < α < 1. In a more general case, the following Lagrangian L L(x, y, t) =

L(x, y, t) =

1 kykp + a(x, t), p

where p > 1 and a ∈ C 1 (Rd × [a, b], R+ ), satisfies Hypotheses (H1)-(H4) and (H5)′′ . Consequently, the fractional Euler-Lagrange equation (6.1) associated to L admits a weak solution in Eα,p for any 1p < α < 1. Let us note that if for any t ∈ [a, b], a(·, t) is convex, then L satisfies Hypothesis (H5). In the unidimensional case d = 1, let us take a Lagrangian with a second term linear in its first variable, i.e. L(x, y, t) =

1 p |y| + f (t)x, p

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where p > 1 and f ∈ C 1 ([a, b], R). Then, L satisfies Hypotheses (H1)-(H5). Then, the fractional Euler-Lagrange equation (6.1) associated admits a weak solution in Eα,p for any p1 < α < 1. Theorem 6.14 is a result based on strong conditions on Lagrangian L. Consequently, some Lagrangian do not satisfy all hypotheses of Theorem 6.14. We can cite Bolza’s example in dimension d = 1 given by L(x, y, t) = (y 2 − 1)2 + x4 . L does not satisfy Hypothesis (H4) neither Hypothesis (H5)′′ . Nevertheless, as usual with variational methods, the conditions of regularity, coercivity and/or convexity can often be replaced by weaker assumptions specific to the studied problem. As an example, we can cite Ammi and Torres, 2008 and references therein about higherorder integrals of the calculus of variations. Indeed, in this Subsection, it is proved that calculus of variations is still valid with weaker regularity assumptions. 6.3 6.3.1

Time-Fractional Diffusion Equations Introduction

We assume Ω to be a bounded domain in Rd with sufficiently smooth boundary ∂Ω. We consider an initial-boundary value problem for a diffusion equation with two fractional time derivatives  α1 α  ∂t u(x, t) + q(x)∂t 2 u(x, t) = (−Au)(x, t), x ∈ Ω, t ∈ (0, T ), u(x, t) = 0, x ∈ ∂Ω, t ∈ (0, T ), (6.6)  u(x, 0) = a(x), x ∈ Ω.

Here 0 < α2 < α1 < 1. For α ∈ (0, 1), by ∂tα we denote the Caputo fractional derivative with respect to t Z t d 1 α (t − τ )−α g(τ )dτ ∂t g(t) = Γ(1 − α) 0 dt

and Γ is the Gamma function and q ∈ W 2,∞ (Ω). The space W 2,∞ (Ω) is the usual Sobolev space (see, e.g., Adams, 1999). The operator A denotes a second-order partial differential operator in the following form d X ∂ ∂ aij (x) u(x) + b(x)u(x), x ∈ Ω, (−Au)(x) = ∂xi ∂xj i,j=1 T ¯ 1 ≤ i, j ≤ d, b ∈ for u ∈ H 2 (Ω) H01 (Ω), and we assume that aij = aji ∈ C 1 (Ω), ¯ ¯ C(Ω), b(x) ≤ 0 for x ∈ Ω and that there exists a constant ν > 0 such that ν

d X j=1

ξj2 ≤

d X

j, k=1

ajk (x)ξj ξk ,

¯ x ∈ Ω,

ξ ∈ Rd .

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The classic diffusion models (diffusion equation with integer-order derivative) have played important roles in modelling contaminants diffusion processes. However, in recent two decades, more and more experimental data (e.g., the diffusion process in the highly heterogeneous media) showed that the classical model are inadequate to explain the phenomenon described by the experimental data, e.g., Adams and Gelhar pointed out that the field data in the saturated zone of a highly heterogeneous aquifer indicated the long-tailed profile in the spatial distribution of densities as the time passes, which is very different from the classical one (see, Adams and Gelhar, 1992). The above phenomenon of long-tailed profile has been investigated by many researchers, see Berkowitz, Scher and Silliman, 2000; Giona, Gerbelli and Roman, 1992; Hatano and Hatano, 1980, and the references therein. In the many researches, there is an effective one that being used to explain the longtailed profile phenomenon, that is to replace the first-order time derivative with a fractional derivative of order α ∈ (0, 1) since the fractional derivative possesses the memory effect which leads to the not too fast diffusion. This modified model is presented as a useful approach for the description of transport dynamics in complex system that are governed by anomalous diffusion and non-exponential relaxation patterns, and attracted great attention from different areas. For numerical calculation, see Benson, Schumer, Meerschaert et al., 2001; Meerschaert and Tadjeran et al., 2004; Diethelm and Luchko, 2004 and the references therein. For the theoretics, see Gorenflo, Luchko and Zabrejko, 1999; Hanyaga, 2002; Luchko, 2009a,b, 2010; Luchko and Gorenflo, 1999; Sakamoto and Yamamoto, 2011; Xu, Cheng and Yamamoto, 2011, etc. For the stochastic analysis, one can regard the time-fractional diffusion equation as a macroscopic model derived from the continuous-time random walk. Metzler and Klafter, 2000b demonstrated that a fractional diffusion equation describes a non-Markovian diffusion process with a memory. Roman and Alemany, 1994 investigated continuous-time random walks on fractals and showed that the average probability density of random walks on fractals obeys a diffusion equation with a fractional time derivative asymptotically. As for diffusion equations with multiple fractional time derivatives, see Jiang, Liu, Turner et al., 2012; Daftardar-Gejji and Bhalekar, 2008; Luchko, 2011 and the references therein. In this Section, we consider the case of multiple fractional time derivatives. Such equations can be considered as more feasible model equations than equations with a single fractional time derivative in modeling diffusion in porous media. We apply the perturbation method and the theory of evolution equations to prove regularity as well as unique existence of solution to (6.6).

6.3.2

Regularity and Unique Existence

Let L2 (Ω) be a usual L2 -space with the scalar product (·, ·), and H l (Ω), H0m (Ω) 1 denote the usual Sobolev spaces (e.g., Adams, 1999). We set kakL2 (Ω) = (a, a) 2 .

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We define the operator A in L2 (Ω) by (Au)(x) = (Au)(x),

x ∈ Ω, D(A) = H 2 (Ω) ∩ H01 (Ω).

Then the fractional power Aγ is defined for γ ∈ R (see, e.g., Pazy, 1983), and 1 D(Aγ ) ⊂ H 2γ (Ω), D(A 2 ) = H01 (Ω) for example. We note that kukD(Aγ ) := kAγ ukL2 (Ω) is stronger than kukL2(Ω) for γ > 0. Since −A is a symmetric uniformly elliptic operator, the spectrum of A is entirely composed of eigenvalues and counting according to the multiplicities, we can set 0 < λ1 ≤ λ2 ≤ · · · . By φn ∈ D(A), we denote the orthonormal eigenfunction corresponding to λn : Aφn = λn φn . Then the sequence {φn }n∈N is orthonormal basis in L2 (Ω). Then we see that ∞ n o X 2 D(Aγ ) = ψ ∈ L2 (Ω) : λ2γ n |(ψ, φn )| < ∞ n=1

γ

and that D(A ) is a Hilbert space with the norm X 12 ∞ 2γ 2 kψkD(Aγ ) = λn |(ψ, φn )| . n=1

Henceforth we associate with u(x, t), provided that it is well-defined, a map u(·) : (0, T ) → L2 (Ω) by u(t)(x) = u(x, t), 0 < t < T , x ∈ Ω. Then we can write (6.6) as α1 ∂t u(t) + q∂tα2 u(t) = −Au(t), t > 0 in L2 (Ω), (6.7) u(0) = a ∈ L2 (Ω). Remark 6.16. The interpretation of the initial condition should be made in a suitable function space. In our case, as Theorem 6.17 asserts, we have limt→0 ku(t)− akL2 (Ω) = 0. Moreover we define the Mittag-Leffler function Eα,β (z) by Eα,β (z) :=

∞ X

k=0

zk , z ∈ C, Γ(αk + β)

where α > 0 and β ∈ R are arbitrary constants. By the power series, we can directly verify that Eα,β (z) is an entire function of z ∈ C. Now we define the operator S(t) : L2 (Ω) → L2 (Ω), t ≥ 0, by S(t)a :=

∞ X

(a, φn )Eα1 ,1 (−λn tα1 )φn

in L2 (Ω)

(6.8)

n=1

for a ∈ L2 (Ω). Then we can prove that S(t) : L2 (Ω) → L2 (Ω) is a bounded linear operator for t ≥ 0 (see, e.g., Sakamoto and Yamamoto, 2011). Moreover the term-wise differentiations are possible and give S ′ (t)a := −

∞ X

n=1

λn (a, φn )tα1 −1 Eα1 ,α1 (−λn tα1 )φn

in L2 (Ω)

(6.9)

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and ′′

S (t)a := −

∞ X

n=1

λn (a, φn )tα1 −2 Eα1 ,α1 −1 (−λn tα1 )φn

in L2 (Ω)

(6.10)

for a ∈ L2 (Ω). For F ∈ L2 (Ω×(0, T )) and a ∈ L2 (Ω), there exists a unique solution in a suitable class (see, e.g., Sakamoto and Yamamoto, 2011) to the problem α1 ∂t u(t) = −Au(t) + F, 0 < t < T, (6.11) u(0) = a. This solution is given by Z t u(t) = A−1 S ′ (t − τ )F (τ )dτ + S(t)a,

t > 0.

(6.12)

0 < t < T,

(6.13)

0

In view of (6.12), we mainly discuss the equation Z t u(t) = S(t)a − A−1 S ′ (t − τ )q∂tα2 u(τ )dτ, 0

in order to establish unique existence of solutions to (6.7). Henceforth, C denotes generic positive constants which are independent of a in (6.6), but may depend on T, α1 , α2 and the coefficients of the operator A and q. Now we are ready to state first main result in this section. Theorem 6.17. We assume that u ∈ C((0, T ], L2 (Ω)) satisfy (6.13) and α1 + α2 > 1. Then ku(t)kH 2γ (Ω) ≤ Ct−α1 γ kakL2(Ω) ,

00

(see, e.g., Theorem 1.6 in Podlubny, 1999), we can prove kAγ−1 S ′ (t)k ≤ Ctα1 −1−α1 γ , t > 0

(6.15)

kA−1 S ′′ (t)k ≤ Ctα1 −2 , t > 0.

(6.16)

and

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Now we proceed to the proof of Theorem 6.17. We set Z t v(t) := Aγ−1 S ′ (t − η)q∂tα2 u(η)dη, 0 < t < T. 0

By (6.13), we have Therefore, using

Aγ u(t) = Aγ S(t)a − v(t),

0 < t < T.

ku(t)kH 2γ (Ω) ≤ CkAγ u(t)kL2 (Ω) ,

it is sufficient to estimate kAγ S(t)akL2 (Ω) + kv(t)kL2 (Ω) . First we will estimate kv(t)kL2 (Ω) . Substituting the definition of ∂tα2 u and changing the order of integration, we have Z η Z t 1 (η − τ )−α2 qu′ (τ )dτ dη v(t) = Aγ−1 S ′ (t − η) Γ(1 − α2 ) 0 0 (6.17) Z t 1 ′ H(t, τ )qu (τ )dτ, 0 < t < T. = Γ(1 − α2 ) 0 Here we have set Z t Aγ−1 S ′ (t − η)(η − τ )−α2 dη. H(t, τ ) = τ

Decomposing the integrand and introducing the change of variables η − τ → η we obtain Z t H(t, τ ) = Aγ−1 S ′ (t − η)(η − τ )−α2 dη, τ Z t = Aγ−1 S ′ (t − η)[(η − τ )−α2 − (t − τ )−α2 ]dη τ Z t + Aγ−1 S ′ (t − η)dη(t − τ )−α2 τ Z t−τ = Aγ−1 S ′ (t − η − τ )[η −α2 − (t − τ )−α2 ]dη (6.18) 0 Z t + Aγ−1 S ′ (t − η)dη(t − τ )−α2 τ Z t−τ = Aγ−1 S ′ (t − η − τ )[η −α2 − (t − τ )−α2 ]dη 0

+Aγ−1 S(0)(t − τ )−α2 − Aγ−1 S(t − τ )(t − τ )−α2

=: I1 (t, τ ) + I2 (t, τ ). On the other hand, we have Z t−τ ∂τ I1 (t, τ ) = − Aγ−1 S ′′ (t − η − τ )(η −α2 − (t − τ )−α2 )dη 0 Z t−τ −α2 Aγ−1 S ′ (t − η − τ )(t − τ )−α2 −1 dη 0

− lim Aγ−1 S ′ (t − τ − η)[η −α2 − (t − τ )−α2 ]. η→t−τ

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By the estimate (6.15) we obtain kAγ−1 S ′ (t − τ − η)(η −α2 −(t − τ )−α2 )kL2 (Ω)

≤ C(t − τ − η)α1 −1−α1 γ

|(t − τ )α2 − η α2 | . η α2 (t − τ )α2

According to the mean value theorem, we can choose θ ∈ (η, t − τ ) such that |(t − τ )α2 − η α2 | = |α2 θα2 −1 (t − τ − η)| ≤ α2 η α2 −1 (t − τ − η).

Hence we obtain kAγ−1 S ′ (t − τ − η)(η −α2 − (t − τ )−α2 )kL2 (Ω) ≤ Cα2 η −1 (t − τ )−α2 (t − τ − η)α1 −α1 γ → 0

as η → t − τ

by α1 − α1 γ > 0. This implies Z t−τ ∂τ I1 (t, τ ) = − Aγ−1 S ′′ (t − η − τ )(η −α2 − (t − τ )−α2 )dη 0 Z t−τ −α2 Aγ−1 S ′ (t − η − τ )(t − τ )−α2 −1 dη, 0 < t < T.

(6.19)

0

On the other hand, we have

∂τ I2 (t, τ ) = −α2 Aγ−1 S(t − τ )(t − τ )−α2 −1 + α2 Aγ−1 S(0)(t − τ )−α2 −1 +Aγ−1 S ′ (t − τ )(t − τ )−α2 Z t−τ = α2 Aγ−1 S ′ (t − η − τ )(t − τ )−α2 −1 dη 0

+Aγ−1 S ′ (t − τ )(t − τ )−α2 .

Adding this and (6.19) we obtain Z t−τ ∂τ H(t, τ ) = − Aγ−1 S ′′ (t − η − τ )(η −α2 − (t − τ )−α2 )dη 0

(6.20)

+Aγ−1 S ′ (t − τ )(t − τ )−α2 .

Using (6.20) in (6.17), integrating by parts and using H(t, t) = 0 we obtain Z t (Γ(1 − α2 )) v(t) = H(t, τ )qu′ (τ )dτ 0 Z t Z t−τ = −H(t, 0)qa + Aγ−1 S ′′ (t − η − τ )(η −α2 − (t − τ )−α2 )dη 0 0 −Aγ−1 S ′ (t − τ )(t − τ )−α2 qu(τ )dτ =: I3 (t) + I4 (t).

We set B(α, β) =

Γ(α)Γ(β) , Γ(α + β)

α, β > 0.

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First, by (6.15) and q ∈ W 2,∞ (Ω) we have kI3 (t)kL2 (Ω) = k − H(t, 0)qakL2 (Ω)

Z t

γ−1 ′ −α2

= − A S (t − η)η dηqa

2 0 L (Ω) Z t α1 −α1 γ−1 −α2 ≤ CkakL2 (Ω) (t − η) η dη

(6.21)

0

= CkakL2 (Ω) B(1 − α2 , α1 − α1 γ)tα1 −α1 γ−α2 ,

since 1 − α2 > 0 and α1 − α1 γ > 0. On the other hand, by q ∈ W 2,∞ (Ω) and u|∂Ω = 0, we have

kA(qu(τ ))kL2 (Ω) ≤ Ckqu(τ )kH 2 (Ω) ≤ Cku(τ )kH 2 (Ω) ≤ CkAu(τ )kL2 (Ω) and kqu(τ )kL2 (Ω) ≤ Cku(τ )kL2 (Ω) , that is,

kA0 (qu(τ ))kL2 (Ω) ≤ CkA0 u(τ )kL2 (Ω) .

Hence the interpolation theorem (see, e.g., Theorem 5.1 in Lions and Magenes, 1972) we obtain kAγ (qu(τ ))kL2 (Ω) ≤ CkAγ u(τ )kL2 (Ω) . Therefore by (6.15) and (6.16), the second term of I4 (t) can be estimated as follows: Z t Z t−τ kI4 (t)kL2 (Ω) ≤ C (t − η − τ )α1 −2 (η −α2 − (t − τ )−α2 )dη 0 0 +(t − τ )α1 −1−α2 kAγ (qu(τ ))kL2 (Ω) dτ Z t Z t−τ (t − τ − η)α2 dη ≤C (t − η − τ )α1 −2 α2 η (t − τ )α2 0 0 +(t − τ )α1 −1−α2 kAγ u(τ )kL2 (Ω) dτ Z t Z t−τ ≤C (t − η − τ )α1 +α2 −2 η −α2 dη 0 0 α1 −1−α2 +(t − τ ) kAγ u(τ )kL2 (Ω) dτ Z t =C B(1 − α2 , α1 + α2 − 1)(t − τ )α1 −1 0 +(t − τ )α1 −1−α2 kAγ u(τ )kL2 (Ω) dτ.

For the last equality, we used α1 + α2 > 1. Therefore we have

kΓ(1 − α2 )v(t)kL2 (Ω) ≤ Ckak2L2 (Ω) B(1 − α2 , α1 − α1 γ)tα1 −α1 γ−α2 Z t +C (t − τ )α1 −1−α2 kAγ u(τ )kL2 (Ω) dτ. 0

Thus the estimate of kv(t)kL2 (Ω) is completed.

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Next we estimate kAγ S(t)akL2 (Ω) . By Theorem 1.6 in Podlubny, 1999, we obtain

X

2

∞

γ α1

kAγ S(t)ak2L2 (Ω) = (a, φ )λ E (−λ t )φ n n α1 ,1 n n

L2 (Ω) n=1 !2 ∞ X (λn tα1 )γ ≤C (a, φn )2 t−2α1 γ 1 + λn tα1 n=1 ≤ Ct−2α1 γ kak2L2 (Ω) , and hence kAγ u(t) kL2 (Ω) ≤ CkakL2 (Ω) (t−α1 γ + tα1 −α1 γ−α2 ) Z t +C (t − τ )α1 −1−α2 kAγ u(τ )kL2 (Ω) dτ 0 Z t (t − τ )α1 −1−α2 kAγ u(τ )kL2 (Ω) dτ, 0 < t < T. ≤ CkakL2 (Ω) t−α1 γ + C 0

Therefore by an inequality of Gronwall type (see, Exercise 3 (p.190) in Henry, 1981), we obtain kAγ u(t)kL2 (Ω) ≤ CkakL2 (Ω) t−α1 γ , 0 < t ≤ T. Thus the proof is completed. Remark 6.18. We may be able to remove the condition α1 + α2 > 1. On the other hand, Pr¨ uss established regularity in case γ = 1 for general α1 , α2 ∈ (0, 1) under a strong condition on a ∈ D(A) (see, Pr¨ uss, 1993). On the basis of Theorem 6.17, a standard argument (see, Henry, 1981) yields: Theorem 6.19. For any γ ∈ (0, 1) there exists a mild solution to (6.13) in the T space u ∈ C((0, T ], D(Aγ )) C((0, T ], L2 (Ω)). The above results shall now be extended to the solution of linear diffusion equation with multiple fractional time derivatives l X α ∂tα1 u(t) + qj ∂t j u(t) = −Au(t), t > 0 j=2

and

u(0) = a ∈ L2 (Ω), where 0 < αl < · · · < α2 < α1 < 1 and qj ∈ W 2,∞ (Ω), 2 ≤ j ≤ l. As before the lower-order derivatives are regarded as source terms and we consider Z t l X α u(t) = S(t)a − A−1 S ′ (t − τ ) qj ∂t j u(τ )dτ, 0 < t < T. (6.22) 0

j=2

Similarly to Theorems 6.17 and 6.19, we can prove Theorem 6.20. Assume that u ∈ C((0, T ], L2 (Ω)) satisfies (6.22) and 0 < αl < · · · < α1 , α1 + αl > 1. Then ku(t)kH 2γ (Ω) ≤ Ct−α1 γ kakL2(Ω) , 0 < t ≤ T for any γ ∈ (0, 1). Moreover there exists a mild solution to (6.22) in the space C((0, T ], D(Aγ )) ∩ C([0, T ], L2 (Ω)) with γ ∈ (0, 1).

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6.4 6.4.1

257

Fractional Hamiltonian Systems Introduction

Consider the following fractional Hamiltonian system α α t D∞ (−∞ Dt u(t)) + L(t)u(t) = ∇W (t, u(t)), u ∈ H α (R),

t ∈ R,

(6.23)

α are left and right Liouville-Weyl fractional derivawhere α ∈ ( 21 , 1], −∞ Dtα and t D∞ tives of order 0 < α < 1 on the whole axis R respectively, u ∈ Rn , L(t) is positive definite symmetric matrix for all t ∈ R and W : R × Rn → R is a function that satisfy conditions which will be stated later and ∇W (t, u) is the gradient of W at u. In particular, if α = 1, (6.23) reduces to the standard second order Hamiltonian system of the following form

u′′ (t) − L(t)u(t) + ∇W (t, u(t)) = 0,

t ∈ R.

(6.24)

The existence of homoclinic orbits is one of the most important problems in the history of Hamiltonian systems. It has been intensively studied by many mathematicians (see Ambrosetti and Zelati, 1993; Ding and Jeanjean, 2007; Izydorek and Janczewska, 2005; Makita, 2012; Omana and Willem, 1992; Paturel, 2001; Rabinowitz, 1990; Séré, 1992; Szulkin and Zou, 2001; Zelati, Ekeland and Séré, 1990; Zelati and Rabinowitz, 1991; Zou and Li, 2002). Variational methods to prove the existence of homoclinic orbits for second order Hamiltonian systems were first used by Rabinowitz, 1990, while the first multiplicity result, later improved by Paturel, 2001, is due to Ambrosetti and Zelati, 1993. In the recent years, the existence and multiplicity of homoclinic orbits for the second order Hamiltonian systems have been extensively studied via variational methods in many papers (e.g. Ambrosetti and Zelati, 1993; Ding and Jeanjean, 2007; Izydorek and Janczewska, 2005; Makita, 2012; Paturel, 2001; Rabinowitz, 1990; Zelati and Rabinowitz, 1991; Zou and Li, 2002). It is worth to mention that the fractional Hamiltonian is not uniquely defined and many researchers have explored this area giving new insight into this problem (e.g. Baleanu, Golmankaneh and Golmankaneh, 2009, Tarasov 2010, Toress, 2013). In Subsection 6.4.2, we introduce the fractional space that we use in our work and some proposition are proven which will aid in our analysis. In Subsection 6.4.3, we discuss existence and multiplicity of solutions of fractional Hamiltonian systems. 6.4.2

Fractional Derivative Space

In this section we introduce some fractional spaces, for more detail, see Ervin and Roop, 2006. Let α > 0. Define the semi-norm α |u|I−∞ = k−∞ Dxα ukL2

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and norm 1

α kukI−∞ = (kuk2L2 + |u|2I−∞ α )2,

(6.25)

and let k·kI α

α I−∞ (R) = C0∞ (R, Rn )

−∞

.

α

Now we define the fractional Sobolev space H (R) in terms of the Fourier transform. Let 0 < α < 1, let the semi-norm |u|α = k|w|α u bkL2

and norm

(6.26) 1

kukα = (kuk2L2 + |u|2α ) 2 , and let k·kα

H α (R) = C0∞ (R, Rn ) 2

We note a function u ∈ L (R) belong to Especially

α I−∞ (R)

.

if and only if

|w|α u b ∈ L2 (R).

(6.27)

α |u|I−∞ = k|w|α uˆkL2 .

(6.28)

α Therefore I−∞ (R) and H α (R) are equivalent with equivalent semi-norm and norm. α α Analogous to I−∞ (R) we introduce I∞ (R). Let the semi-norm α α = kx D |u|I∞ ∞ ukL2

and norm 1

2 2 α = (kuk 2 + |u| α ) 2 , kukI∞ I∞ L

(6.29)

and let α I∞ (R) = C0∞ (R, Rn )

α k·kI∞

.

α α Moreover I−∞ (R) and I∞ (R) are equivalent , with equivalent semi-norm and norm (see, Ervin and Roop, 2006). Now we give the proof of the Sobolev lemma. Theorem 6.21. If α > 12 , then H α (R) ⊂ C(R) and there is a constant C = Cα such that

kuk = sup |u(x)| ≤ Ckukα .

(6.30)

x∈R

Proof.

By the Fourier inversion theorem, if u ˆ ∈ L1 (R) then u is continuous and sup |u(x)| ≤ kˆ ukL1 . x∈R

Hence, to prove the theorem it is enough to prove that kˆ ukL1 ≤ kukα,

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so by Schwarz inequality, we have Z Z |ˆ u(w)|dw = (1 + |w|2 )α/2 |ˆ u(w)|

1 dw 2 )α/2 (1 + |w| R Z 21 Z 12 2 α 2 2 −α ≤ (1 + |w| ) |ˆ u(w)| dw (1 + |w| ) dw .

R

R

R

kuk2α ,

so the theorem boils down to the fact The first integral on the right is Z Z ∞ (1 + |w|2 )−α dw = (1 + r2 )−α rn−1 dr < ∞ 0

R

precisely when α >

1 2.

Remark 6.22. If u ∈ H α (R), then u ∈ Lq (R) for all q ∈ [2, ∞], since Z 2 |u(x)|q dx ≤ kukq−2 ∞ kukL2 . R

Now we introduce a new fractional space. Let Z X α = {u ∈ H α (R, Rn ) : |−∞ Dtα u(t)|2 + (L(t)u(t), u(t)) dt < ∞}. R

The space X α is a Hilbert space with the inner product Z hu, viX α = [(−∞ Dtα u(t),−∞ Dtα v(t)) + (L(t)u(t), v(t))] dt R

and the corresponding norm

kuk2X α = hu, uiX α . Before give the main existence theorem, we firstly state the conditions: (L) L(t) is positive definite symmetric matrix for all t ∈ R and there exists an l ∈ C(R, (0, ∞)) such that l(t) → +∞ as t → +∞ and (L(t)x, x) ≥ l(t)|x|2 ,

for all t ∈ R

and x ∈ Rn ;

(W1) W ∈ C 1 (R × Rn , R) and there is a constant µ > 2 such that 0 < µW (t, x) ≤ (x, ∇W (t, x)),

for all t ∈ R

and x ∈ Rn \{0};

(W2) |∇W (t, x)| = o(|x|) as x → 0 uniformly with respect to t ∈ R; (W3) there exists W ∈ C(Rn , R) such that |W (t, x)| + |∇W (t, x)| ≤ W (x),

for every x ∈ Rn

and t ∈ R.

Lemma 6.23. Suppose L satisfies (L). Then X α is continuously embedded in H α (R, Rn ). Proof. Since l ∈ C(R, (0, ∞)) and l is coercive, then lmin = mint∈R l(t) exists, so we have (L(t)u(t), u(t)) ≥ l(t)|u(t)|2 ≥ lmin |u(t)|2 ,

∀t ∈ R.

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Then lmin ku(t)k2α

= lmin ≤ lmin

So

where K =

max{lmin ,1} lmin

Z Z

R

|−∞ Dtα u(t)|2

2

+ |u(t)| dt Z |−∞ Dtα u(t)|2 dt + (L(t)u(t), u(t))dt . R

R

kuk2α ≤ Kkuk2X α , .

Lemma 6.24. Suppose L satisfies (L). Then the imbedding of X compact. Proof.

(6.31)

α

2

in L (R) is

We note first that by Lemma 6.23 and Remark 6.22 we have X α ֒→ L2 (R) is continuous.

Now, let (uk ) ∈ X α be a sequence such that uk ⇀ u in X α . We will show that uk → u in L2 (R). Suppose, without loss of generality, that uk ⇀ 0 in X α . The Banach-Steinhaus theorem implies that A = sup kukX α < +∞ . k

1 Let ǫ > 0, there is T0 < 0 such that l(t) ≤ ǫ for all t such that t ≤ T0 . Similarly, 1 there is T1 > 0, such that l(t) ≤ ǫ for all t ≥ T1 . Sobolev’s theorem (see Stuart, ¯ = [T0 , T1 ], so there is a k0 such that 1995) implies that uk → 0 uniformly on Ω Z |uk (t)|2 dt ≤ ǫ, for all k ≥ k0 . (6.32) Ω

Since

1 l(t)

≤ ǫ on (−∞, T0 ] we have Z T0 Z |uk (t)|2 dt ≤ ǫ

Similarly, since

1 l(t)

−∞

T0

−∞

l(t)|uk (t)|2 dt ≤ ǫA2 .

≤ ǫ on (T1 , +∞], we have Z +∞ |uk (t)|2 dt ≤ ǫA2 .

(6.33)

(6.34)

T1

Combining (6.32), (6.33) and (6.34), we get uk → 0 in L2 (R, Rn ).

Lemma 6.25. There are constants c1 > 0 and c2 > 0 such that W (t, u) ≥ c1 |u|µ , |u| ≥ 1

(6.35)

W (t, u) ≤ c2 |u|µ , |u| ≤ 1.

(6.36)

and

Proof.

By (W1) we note that µW (t, σu) ≤ (σu, ∇W (t, σu)).

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Let f (σ) = W (t, σu), then d (f (σ)σ −µ ) ≥ 0. dσ

(6.37)

Now we consider two cases: Case 1. |u| ≤ 1. In this case we integrate (6.37), from 1 until u )|u|µ . W (t, u) ≤ W (t, |u|

Case 2. |u| ≥ 1. In this case we integrate (6.37), from W (t, u) ≥ |u|µ W (t,

u ). |u|

1 |u|

and we get (6.38)

1 until 1 and we get |u| (6.39)

u ∈ B(0, 1). So, since W is continuous and B(0, 1) is compact, |u| there are c1 > 0 and c2 > 0 such that Now, since u ∈ Rn ,

c1 ≤ W (t, u) ≤ c2 ,

for every u ∈ B(0, 1).

Therefore we get the affirmation of the lemma.

Remark 6.26. By Lemma 6.25, we have W (t, u) = o(|u|2 ),

as u → 0 uniformly in t ∈ R.

(6.40)

n

In addition, by (W2), we have, for any u ∈ R such that |u| ≤ M1 , there exists some constant d > 0 (dependent on (W1)) such that |∇W (t, u(t))| ≤ d|u(t)|.

(6.41)

Similar to Lemma 2 of Zaslavsky, 2002, we can get the following result. Lemma 6.27. Suppose that (L), (W1)-(W2) are satisfied. If uk ⇀ u in X α , then ∇W (t, uk ) → ∇W (t, u) in L2 (R, Rn ). Proof. Assume that uk ⇀ u in X α . Then there exists a constant d1 > 0 such that, by Banach-Steinhaus theorem and (6.30), sup kuk k∞ ≤ d1 , k∈R

kuk∞ ≤ d1 .

By (W2), for any ǫ > 0 there is δ > 0 such that |uk | < δ

implies |∇W (t, uk )| ≤ ǫ|uk |

and by (W3) there is M > 0 such that |∇W (t, uk )| ≤ M,

for all δ < |uk | ≤ d1 .

Therefore, there exists a constant d2 > 0 (dependent on d1 ) such that |∇W (t, uk (t))| ≤ d2 |uk (t)|,

for all k ∈ N and t ∈ R. Hence,

|∇W (t, u(t))| ≤ d2 |u(t)|

|∇W (t, uk (t)) − ∇W (t, u(t))| ≤ d2 (|uk (t)| + |u(t)|) ≤ d2 (|uk (t) − u(t)| + 2|u(t)|).

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Since, by Lemma 6.24, uk → u in L2 (R, Rn ), passing to a subsequence if necessary, it can be assumed that ∞ X kuk − ukL2 < ∞. k=1

But this implies uk → u almost everywhere t ∈ R and ∞ X

k=1

Therefore

|uk (t) − u(t)| = v(t) ∈ L2 (R, Rn ).

|∇W (t, uk (t)) − ∇W (t, u(t))| ≤ d2 (v(t) + 2|u(t)|). Then, using Lebesgue’s dominated convergence theorem, the lemma is proved.

In the following, we consider the case L is uniformly bounded from below. Here we do not need that L satisfies the corecive condition (L). Explicitly, we assume that (L) L ∈ C(R, Rn×n ) is a symmetric and positive definite symmetric matrix for all t ∈ R and there exists a M > 0 such that (L(t)x, x) ≥ M |x|2 ,

for all t ∈ R, x ∈ Rn .

In order to prove our main results, we assume that W (t, x) = b(t)π(x) satisfies the following conditions: (W1) b : R → (0, +∞) is a continuous function such that b(t) → 0 as |t| → ∞; (W2) π ∈ C 1 (Rn , R) and there is a constant µ > 2 such that 0 < µπ(x) ≤ (∇π(x), x),

∀ x ∈ Rn \ {0};

(W3) ∇π(x) = o(|x|) as |x| → 0; (W4) π(−x) = π(x), for all x ∈ Rn ; (W5) For any r > 0, there exists α0 , β0 > 0 and ̺ < 2 such that 1 W (t, x) ≤ (∇W (t, x), x), ∀(t, x) ∈ R×{x ∈ Rn : |x| ≥ r}. 0≤ 2+ α0 + β0 |x|̺

Let Lpb (R, Rn ) denote the weighted space of measurable functions u : R → Rn with the norm Z 1p p . kukp,b = b(t)|u(t)| dt R

Lemma 6.28. Suppose that (L) and (W1) hold. Then the imbedding of X α in L2b (R, Rn ) is compact. Proof. The proof is standard (cf., Ervin and Roop, 2006). For readers convenience, we give the rough proof. It is easy to check that the embedding of X α ֒→ L2b (R, Rn ) is continuous. Next, we prove that the embedding is compact. Let {un }n∈N ⊂ X α be a sequence such that un ⇀ u in X α , we show that un → u in

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L2b (R, Rn ). Suppose, without loss of generality, that un ⇀ 0 in X α . The BanachSteinhaus Theorem implies that A = sup kun k < ∞. n∈N

For any ε > 0, there is T0 < 0 such that b(t) ≤ ε for all t such that t ≤ T0 . Similarly, there is T1 > 0 such that b(t) ≤ ε for all t ≥ T1 . Sobolev’s theorem (see Yuan and ¯ = [T0 , T1 ], so there is a k0 such Zhang, 2012) implies that un → u uniformly on Ω that Z b(t)|un (t)|2 dt < ε, ∀ k ≥ k0 . (6.42) Ω

Since b(t) ≤ ε on (−∞, T0 ] we have Z T0 Z T0 ε 2 ε M |un (t)|2 dt < A . b(t)|un (t)|2 dt ≤ M M −∞ −∞

Similarly, since b(t) ≤ ε on (T1 , +∞), one can get Z +∞ Z +∞ ε ε 2 b(t)|un (t)|2 dt ≤ M |un (t)|2 dt < A . M M T1 T1 Combining (6.42)-(6.44) we get un → 0 in L2b (R, Rn ).

(6.43)

(6.44)

Remark 6.29. From Lemma 6.28, there is a constant Cb such that kuk2,b ≤ Cb kukX α ,

∀ u ∈ X α.

(6.45)

Lemma 6.30. Suppose that (L), (W1) and (W3) are satisfied. If un ⇀ u in X α , then ∇π(un ) → ∇π(u) in L2b (R, Rn ). The proof is similar to Lemma 6.27 and is omitted. Now, from Theorem 6.21, it is well known that for α > 21 , X α ⊂ H α (R, Rn ) ⊂ C(R, Rn ), the space of continuous functions u on R such that u(t) → 0 as |t| → +∞. Lemma 6.31. We have u |u|µ , |u| ≤ 1, (6.46) π(u) ≤ π |u| u π(u) ≥ π |u|µ , |u| ≥ 1. (6.47) |u| The proof is similar to Lemma 6.25 and is omitted.

6.4.3

Existence and Multiplicity

Now we are going to establish the corresponding variational framework to obtain the existence and multiplicity of solutions for (6.23). α Definition 6.32. We say that u ∈ I−∞ (R) is a weak solution of (6.23) if Z ∞ Z ∞ [(−∞ Dtα u(t), −∞ Dtα v(t)) + (L(t)u(t), u(t))]dt = (∇W (t, u(t)), v(t))dt, −∞

−∞

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α for all v ∈ I−∞ (R). α For u ∈ I−∞ (R), we may define the functional I : X α → R by Z h i 1 1 I(u) = |−∞ Dtα u(t)|2 + (L(t)u(t), u(t)) − W (t, u(t)) dt 2 R 2 Z 1 = kuk2X α − W (t, u(t))dt, 2 R

(6.48)

which is of class C 1 . We say that u ∈ X α is a weak solution of (6.23) if u is a critical point of I. Lemma 6.33. Suppose that (L), (W1)-(W2) hold, we have Z I ′ (u)v = [(−∞ Dtα u(t), −∞ Dtα v(t)) + (L(t)u(t), v(t)) − (∇W (t, u(t)), v(t))]dt, R

(6.49)

for all u, v ∈ X α , which yields that I ′ (u)u = kuk2X α −

Z

(∇W (t, u(t)), u(t))dt.

(6.50)

R

Moreover, I is a continuously Fréchet-differentiable functional defined on X α , i.e., I ∈ C 1 (X α , R).

Proof. We firstly show that I : X α → R. By (6.40), there is a δ > 0 such that |u| ≤ δ implies that α

W (t, u) ≤ ǫ|u|2 for t ∈ R.

1 2.

(6.51)

n

Let u ∈ X , α > Then u ∈ C(R, R ), the space of continuous function u ∈ R such that u(t) → 0 as |t| → +∞. Therefore there is a constant R > 0 such that |t| ≥ R implies |u(t)| ≤ δ. Hence, by (6.51), we have Z Z R Z W (t, u(t))dt ≤ W (t, u(t))dt + ǫ |u(t)|2 dt < +∞. (6.52) −R

R

|t|≥R

Combining (6.48) and (6.52), we show that I : X α → R. Now we prove that I ∈ C 1 (X α , R). Rewrite I as follows I = I1 − I2 , where Z Z 1 [|−∞ Dtα u(t)|2 + (L(t)u(t), u(t))]dt, I2 = W (t, u(t))dt. I1 = 2 R R

It is easy to check that I1 ∈ C 1 (X α , R) and Z I1′ (u)v = [(−∞ Dtα u(t), −∞ Dtα v(t)) + (L(t)u(t), v(t))]dt.

(6.53)

R

Thus it is sufficient to show this is the case for I2 . In the process we will see that Z ′ I2 (u)v = (∇W (t, u(t)), v(t))dt, (6.54) R

which is defined for all u, v ∈ X α . For any given u ∈ X α , let us define J(u) : X α → R as follows Z J(u)v = (∇W (t, u(t)), v(t))dt, ∀v ∈ X α . R

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It is obvious that J(u) is linear. Now we show that J(u) is bounded. Indeed, for any given u ∈ X α , by (6.41), there is a constant d3 > 0 such that |∇W (t, u(t))| ≤ d3 |u(t)|,

which yields that, by the Hölder inequality and Lemma 6.23 Z Z |J(u)v| = (∇W (t, u(t)), v(t))dt ≤ d3 |u(t)kv(t)|dt R

R

≤

d3

lmin

(6.55)

kukX α kvkX α .

Moreover, for u and v ∈ X α , by Mean Value theorem, we have Z Z Z W (t, u(t) + v(t))dt − W (t, u(t))dt = (∇W (t, u(t) + h(t)v(t)))dt, R

R

R

where h(t) ∈ (0, 1). Therefore, by Lemma 6.24 and the Hölder inequality, we have Z Z (∇W (t, u(t) + h(t)v(t)), v(t))dt − (∇W (t, u(t)), v(t))dt RZ R (6.56) = (∇W (t, u(t) + h(t)v(t) − ∇W (t, u(t)), v(t))dt → 0 R

as v → 0 in X α . Combining (6.55) and (6.56), we see that (6.54) holds. It remains to prove that I2′ is continuous. Suppose that u → u0 in X α and note that Z ′ ′ sup |I2 (u)v − I2 (u0 )v| = sup (∇W (t, u(t)) − ∇W (t, u0 (t)), v(t))dt kvkX α =1

≤

kvkX α =1

sup

kvkX α =1

R

k∇W (., u(.)) − ∇W (., u0 (.))kL2 kvkL2

1 k∇W (., u(.)) − ∇W (., u0 (.))kL2 . lmin By Lemma 6.24, we obtain that I2′ (u)v − I2′ (u0 )v → 0 as u → u0 uniformly with respect to v, which implies the continuity of I2′ and I ∈ C 1 (X α , R). The proof is completed. ≤ √

Lemma 6.34. Under the conditions of (L) and (W1)-(W2), I satisfies the (PS) condition. Proof. Assume that {uk }k∈N ∈ X α is a sequence such that {I(uk )}k∈N is bounded and I ′ (uk ) → 0 as k → +∞. Then there exists a constant C1 > 0 such that |I(uk )| ≤ C1 ,

kI ′ (uk )k(X α )∗ ≤ C1

(6.57)

for every k ∈ N. We firstly prove that {uk }k∈N is bounded in X α . By (6.48), (6.50) and (W1), we have 1 C1 + kuk kX α ≥ I(uk ) − I ′ (uk )uk µ Z 1 µ (6.58) = ( − 1)kuk k2X α − [(W (t, uk (t)) − (∇W (t, uk (t)), uk (t))]dt 2 µ R µ ≥ ( − 1)kuk k2X α . 2

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Since µ > 2, the inequality (6.58) shows that (uk )k∈N is bounded in X α . So passing to a subsequence if necessary, it can be assumed that uk ⇀ u in X α and hence, by Lemma 6.24, uk → u in L2 (R, Rn ). It follows from the definition of I that Z ′ ′ 2 (I (uk )−I (u))(uk −u) = kuk −ukX α − [∇W (t, uk )−∇W (t, u)](uk −u)dt. (6.59) R

Since uk → u in L2 (R, Rn ), we have (see Lemma 6.27) ∇W (t, uk (t)) → ∇W (t, u(t)) in L2 (R, Rn ). Hence Z (∇W (t, uk ) − ∇W (t, u), uk − u)dt → 0 R

as k → ∞. So (6.59) implies

kuk − ukX α → 0, as k → ∞. The proof is completed.

Theorem 6.35. Suppose that (L), (W1)-(W2) hold, then (6.23) possesses at least one nontrivial solution. Proof. We divide the proof into several steps: Step I. It is clear that I(0) = 0 and I ∈ C 1 (X α , R) satisfies the (PS) condition by Lemma 6.33 and Lemma 6.34. Step II. Now We show that there exist constant ρ > 0 and β > 0 such that I satisfies the condition (i) of Theorem 1.51. By Lemma 6.24, there is a C0 > 0 such that kukL2 ≤ C0 kukX α . On the other hand, by Theorem 6.21 there is Cα > 0 such that kuk ≤ Cα kukX α . By (6.40), for all ǫ > 0, there exists δ > 0 such that

Let ρ =

δ Cα

W (t, u(t)) ≤ ǫ|u(t)|2

wherever |u(t)| < δ.

and kukX α ≤ ρ; we have kuk∞ ≤ |W (t, u(t))| ≤ ǫ|u(t)|2

Integrating on R, we get Z R

So, if kukX α = ρ, then

1 I(u) = kuk2X α − 2

δ Cα

· Cα = δ. Hence

for all t ∈ R.

W (t, u(t))dt ≤ εkuk2L2 ≤ ǫC02 kuk2X α . Z

1 1 W (t, u(t))dt ≥ ( − ǫC02 )kuk2X α = ( − ǫC02 )ρ2 . 2 2 R

And it suffices to choose ǫ =

1 4C02

to get

I(u) ≥

ρ2 = β > 0. 4C02

(6.60)

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Step III. It remains to prove that there exists an e ∈ X α such that kekX α > ρ and I(e) ≤ 0, where ρ is defined in Step II. Consider Z σ2 kuk2X α − W (t, σu(t))dt I(σu) = 2 R for all σ ∈ R. By (6.35), there is c1 > 0 such that W (t, u(t)) ≥ c1 |u(t)|µ

for all |u(t)| ≥ 1.

(6.61)

Take some u ∈ X α such that kukX α = 1. Then there exists a subset Ω of positive measure of R such that u(t) 6= 0 for t ∈ Ω. Take σ > 0 such that σ|u(t)| ≥ 1 for t ∈ Ω. Then by (6.61), we obtain Z σ2 − c1 σ µ |u(t)|µ dt. (6.62) I(σu) ≤ 2 Ω Since c1 > 0 and µ > 2, (6.62) implies that I(σu) < 0 for some σ > 0 with σ|u(t)| ≥ 1 for t ∈ Ω and kσukX α > ρ, where ρ is defined in Step II. By Theorem 1.51, I possesses a critical value c ≥ β > 0 given by c = inf max I(γ(s)), γ∈Γ s∈[0,1]

where Γ = {γ ∈ C([0, 1], X α ) : γ(0) = 0, γ(1) = e}. Hence there is u ∈ X α such that I(u) = c, I ′ (u) = 0.

Similar to Lemma 6.33, we give the following lemma. Lemma 6.36. Suppose that (L), (W1) and (W4) are satisfied. Then the conclusions of Lemma 6.33 hold. In the following, we give the result on multiplicity of solutions of fractional Hamiltonian systems. Theorem 6.37. Assume that L and W satisfy (L) and (W1)-(W3). Then system (6.23) possesses a nontrivial homoclinic orbit. Proof. It is clear that I(0) = 0. We show that I satisfies the hypotheses of the Theorem 1.51. Step I. We show that I satisfies the (PS) condition. Assume that {un }n∈N ⊂ X α is a sequence such that {I(un )}n∈N is bounded and I ′ (un ) → 0 as n → +∞. Then there exists a constant c > 0 such that |I(un )| ≤ c,

kI ′ (un )kX α ≤ c,

for every n ∈ N.

(6.63)

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We firstly prove that {un }n∈N is bounded in X α . By (6.48), (6.49), (6.63) and (W2), one can get c+

c 1 kun kX α ≥ I(un ) − I ′ (un )un µ µ Z 1 1 − |−∞ Dtα un (t)|2 + (L(t)un (t), un (t)) dt = 2 µ R Z Z 1 − W (t, un (t))dt + (∇W (t, un (t)), un (t))dt µ R R Z 1 1 1 = − kun k2X α + [(∇W (t, un (t)), un (t)) − µW (t, un (t))] dt 2 µ µ R 1 1 − kun k2X α , n ∈ N. ≥ 2 µ

Since µ > 2, the above inequality shows that {un }n∈N is bounded in X α , i.e., that there exists a constant θ > 0 such that kun kX α ≤ θ, α

for every n ∈ N.

(6.64)

α

Since X is a reflexive space (X is a Hilbert space), thus passing to a subsequence if necessary, by Lemma 6.28, we may assume that un ⇀ u, weakly in X α , (6.65) un → u, a.e. in L2b (R, Rn ). Thus, (I ′ (un ) − I ′ (u))(un − u) → 0, and by Lemma 6.30 and the H¨ older inequality, one can get Z (∇W (t, un (t)) − ∇W (t, u(t)), un (t) − u(t))dt → 0, R

as n → +∞. On the other hand we have

(I ′ (un ) − I ′ (u))(un − u) = kun − uk2X α Z − (∇W (t, un (t)) − ∇W (t, u(t)), un (t) − u(t))dt. R

Hence, kun − ukX α → 0 as n → +∞. Therefore, I satisfies (PS) condition. Step II. We now show that there exist constants ρ > 0 and α > 0 such that I satisfies assumption (ii) of Theorem 1.51. For any give number ε > 0, from (W3), we can choose δ > 0 such that

δ Cα

π(x) ≤ ε|x|2 ,

for every |x| ≤ δ.

(6.66)

By (6.30), if kukX α = =: ρ, then |u(t)| ≤ Cα · ρ = δ, so π(u(t)) ≤ ε|u(t)|2 for all t ∈ R. Integrating on R and by Remark 3 and Lemma 6.31, we have Z W (t, u(t))dt ≤ εkuk22,b ≤ εCb2 kuk2X α . (6.67) R

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Let 1 β= 4

δ Cα

2

.

For kukX α = ρ ≤ 1, from (6.48) and (6.67), we obtain Z Z 1 α 2 I(u) = |−∞ Dt u(t)| + (b(t)u(t), u(t)) dt − W (t, u(t))dt 2 R R 1 2 2 2 ≥ kukX α − εCb kukX α 2 1 2 = − εCb kuk2X α . 2 Setting ε =

1 , 4Cb2

(6.68)

the inequality (6.68) implies that I|∂Bρ ≥

1 4

δ Cα

2

= β.

(6.69)

Clearly, (6.69) shows that kukX α = ρ implies that I(u) ≥ β, i.e., I satisfies assumption (ii) of Theorem 1.51. Step III. We shall prove (iii) of Theorem 1.51, i.e., there exists an e ∈ X α such that kekX α > ρ and I(e) ≤ 0, where ρ is defined in Step II. By (6.47), there is c1 > 0 such that π(u(t)) ≥ c1 |u(t)|µ , for all |u(t)| ≥ 1.

(6.70)

Take some u ∈ X α such that kukX α = 1. Then there exists a subset Ω of positive measure |Ω| < ∞ of R such that u(t) 6= 0 for t ∈ Ω. Take σ > 0 such that σ|u(t)| ≥ 1 for t ∈ Ω. Then by (6.48) and (6.70), one can get Z σ2 − c1 σ µ b(t)|u(t)|µ dt. (6.71) I(σu) ≤ 2 Ω R Since µ > 2, b(t) > 0 and Ω b(t)|u(t)|µ dt > 0, (6.71) implies that I(σu) < 0 for some σ > 0 with σ|u(t)| ≥ 1 for t ∈ Ω and kσukX α > ρ. Therefore, I possesses a critical value c ≥ β given by c = inf max I(g(s)), g∈Γ s∈[0,1]

where Bρ (0) is an open ball in E of radius ρ centered at 0, and Γ = {g ∈ C([0, 1], X α ) : g(0) = 0, g(1) = e}. Here there is u∗ ∈ X α such that I(u∗ ) = c,

I ′ (u∗ ) = 0.

Since c > 0, u∗ is a nontrivial homoclinic solution. The proof is complete.

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Theorem 6.38. Assume that L and W satisfy (L) and (W1)-(W4). Then there exists an unbounded sequence of homoclinic orbits for system (6.23). Proof. The conditions (W1) and (W4) imply that I is even. In view of the proof of Theorem 6.37, we see that I ∈ C 1 (X α , R), and I satisfies the (PS) condition and assumptions (i) and (ii) of Theorem 1.52. To apply Theorem 1.52, it suffices to prove that I satisfies the condition (iii′ ) of Theorem 1.52. Let H ′ ⊂ X α be a finite dimensional subspace. From Step III of Theorem 6.37, we know that, for any u0 ∈ H ′ ⊂ X α such that ku0 kX α = 1, there is mu0 > 0 such that I(mu0 ) < 0, for all |m| ≥ mu0 > 0.

Since H ′ ⊂ X α is a finite dimensional subspace, we can choose an R = r(H ′ ) > 0 such that I(ω) < 0, for ω ∈ H ′ \ BR (0).

Therefore, by Theorem 1.52, I possesses an unbounded sequence of critical values {cj }j∈N with cj → +∞. Let uj be the critical point of I corresponding to cj , then (6.23) has infinitely many distinct homoclinic solutions. Theorem 6.39. Let α > 12 . Assume that L and W satisfy (L) and (W1), (W3)(W5). Then there exists an unbounded sequence of homoclinic orbits for system (6.23). Proof. In view of the proof of Theorem 6.38, we see that I ∈ C 1 (X α , R), and I satisfies assumptions (i), (ii) and (iii′ ) of Theorem 1.52. To apply Theorem 1.52, it suffices to prove that I satisfies the condition (C). Suppose that {un }n∈N ⊂ X α is a (C) sequence of I, that is, {I(un )} is bounded and (1 + kun kX α )kI ′ (un )k(X α )∗ → 0 as n → ∞. Then, in view of (6.48) and (6.49), For a constant C0 > 0, we have C0 ≥ 2I(un ) − I ′ (un )un Z = [(∇W (t, un (t)), un (t)) − 2W (t, un (t))] dt.

(6.72)

R

Since π(0) = 0, then from (W3) that there exists η ∈ (0, 1) such that 1 |W (t, x)| ≤ b(t)|x|2 , for every t ∈ R, |x| ≤ η. 4 By (W5), we have (∇W (t, x), x) ≥ 2W (t, x) ≥ 0,

∀(t, x) ∈ R × Rn ,

(6.73)

(6.74)

W (t, x) ≤ (α0 +β0 |x|̺ ) [(∇W (t, x), x) − 2W (t, x)] , ∀(t, x) ∈ R×{x ∈ Rn : |x| > η}. (6.75) Now from (6.30), (6.48), (6.72)-(6.75) and Remark 6.29, we get Z 1 kun k2X α = I(un ) + W (t, un (t))dt 2 R Z Z = I(un ) + W (t, un (t))dt + W (t, un (t))dt {t∈R, |un (t)|≤η}

{t∈R, |un (t)|>η}

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1 = I(un ) + 4 Z +

Z

271

b(t)|un (t)|2 dt

{t∈R, |un (t)|≤η}

(α0 + β0 |un (t)|̺ ) [(∇W (t, un (t)), un (t)) − 2W (t, un (t))] dt

{t∈R, |un (t)|>η}Z

1 ≤ C1 + kun k22,b + (α0 + β0 |un (t)|̺ ) [(∇W (t, un (t)), un (t)) − 2W (t, un (t))] dt 4 R 1 2 2 ≤ C1 + Cb kun kX α + C0 (α0 + β0 kun k̺∞ ) 4 1 2 ≤ C1 + Cb kun k2X α + C0 (α0 + β0 Cα̺ kun k̺X α ). (6.76) 4 Since ̺ < 2, it follows that {kun k} is bounded. Next, similar to the proof of Theorem 6.37, we can also prove that {un } has a convergent subsequence in X α . Thus, I satisfies condition (C). Therefore, the proof is complete.

6.5 6.5.1

Fractional Schr¨ odinger Equations Introduction

Schrödinger equations have received a great deal of interest from the mathematicians in the past twenty years or so, due in particular to their applications to optics. Indeed, simplified versions or limits of Zakharov’s system lead to certain Schrödinger equations. Schrödinger equations also arise in quantum field theory, and in particular in the Hartree-Fock theory. From the mathematical point of view, Schrödinger equations appears a delicate problem, since it possesses a mixture of the properties of parabolic and hyperbolic equations. Indeed, it is almost reversible, it has conservation laws and also some dispersive properties like the Klein-Gordon equation, but it has an infinite speed of propagation. On the other hand, Schrödinger equations has a kind of smoothing effect shared by parabolic problems but the timereversibility it from generating an analytic semigroup. For more details, one can see the monographs Cazenave, 2003; Sulem and Sulem, 1999 and the papers Buslaev and Sulem, 2013; Cazenave, 1983; Cazenave and Lions, 1982; Cuccagna, 2001; Eid, Muslih and Baleanu et al., 2009; Fibich, 2011; Floer and Weinstein, 1986; Guo and Wu, 1995; Tsai, 1995; Wang, 2008 and etc. In this section we study the initial value problem of the following fractional Schrödinger equations with potential (1 C α D x(t, y) − ∆x(t, y) + kV (y)x(t, y) = 0, α ∈ (0, 1), y ∈ Ω, t ∈ (0, T ], i 0 t x(0, y) = x0 (y), y ∈ Ω, (6.77) α 2 where C 0 Dt is Caputo fractional derivative of order α in time t, Ω ⊆ R is a bounded domain with a smooth boundary ∂Ω, ∆ denotes the Laplace operator in R2 , x is a complex valued function in [0, T ] × R2 , k := maxt∈[0,T ] |χ(t)| with χ ∈ C(J, R), is a positive constant, and the function V is called potential.

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In Subsection 6.5.2, a suitable concept on a mild solution for our problems is introduced and existence and uniqueness of mild solutions are presented. We recall the following initial value problem for linear Schrödinger equations  1 ∂ x(t, y) − ∆x(t, y) = 0, y ∈ Ω, t ∈ (0, T ], (6.78) i ∂t  x(0, y) = x (y), y ∈ Ω, 0

where Ω ⊆ R2 is a bounded domain with a smooth boundary ∂Ω, ∆ denotes the Laplace operator in R2 , and x is a complex valued function which defined in [0, T ] × R2 . Take X = L2 (Ω), D(A) = H 2 (Ω) ∩ H01 (Ω), x ∈ D(A), define Ax = i∆x. By virtue of the well known Hille-Yosida theorem, it is obvious that A is the infinitesimal generator of a strongly continuous group {S(t), −∞ < t < ∞} in X. Moreover, {S(t), −∞ < t < ∞} can be given by Z i|y−z|2 1 e 4t x(z)dz. (6.79) (S(t)x)(y) = 4πit Ω

In order to derive the expression (6.79), we define x(t, y), y ∈ Ω, x ¯(t, y) = 0, y ∈ R2 \ Ω,

and

x ¯0 (y) =

x0 (y), y ∈ Ω, 0, y ∈ R2 \ Ω.

Then the equation (6.78) can be rewritten as   ∂x ¯(t, y) = i∆¯ x(t, y), y ∈ R2 , t ∈ (0, T ], ∂t  x ¯(0, y) = x ¯0 (y), y ∈ R2 , x¯0 ∈ H 2 (R2 ).

Applying Fourier transformation to the equation (6.82), we obtain   dx ˜(t, z) = −iz 2x˜(t, z), z ∈ R2 , t ∈ (0, T ], dt  x ˜(0, z) = x ˜0 (z), z ∈ R2 .

(6.80)

(6.81)

(6.82)

(6.83)

Thus, the classical solution of the equation (6.83) can be given by 2

x ˜(t, z) = e−i|z| t x˜0 (z). Thus,

Z Z 2 1 e(i(y−τ )z−i|z| t) x ¯0 (τ )dτ dz 4π 2 R2 R2 Z i|y−ξ|2 1 ¯0 (ξ)dξ. = e 4t x 4πit R2 It comes from the inverse of Fourier transformation that Z i|y−z|2 1 x(t, y) = ¯0 (z)dz. e 4t x 4πit Ω x¯(t, y) =

(6.84)

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273

On the other hand, the equation (6.82) has a unique classical solution x ¯(t, y) = S(t)¯ x0 (y), y ∈ R2 . Keeping in mind of (6.80) and (6.81), we have x ¯(t, y) = S(t)¯ x0 (y), y ∈ Ω. Combined (6.84) and (6.85), we obtain (S(t)x0 )(y) =

1 4πit

Z

e

i|y−z|2 4t

(6.85)

x0 (z)dz.

(6.86)

Ω

By Lemma 1.1 in Pazy, 1983, we have the estimation immediately. Lemma 6.40. Let {S(t), t ≥ 0} be the strongly continuous semigroup given by (6.79). Then S(·) can be extended in a unique way to a bounded operator from L2 (Ω) into L2 (Ω) and kS(t)xkL2 (Ω) ≤ kxkL2 (Ω) . 6.5.2


In this section, we study the existence and uniqueness of mild solutions for system (6.77). To achieve our aim, we adopt the idea of Zhou and Jiao, 2010a,b and introduce the following two characteristic solution operators: Z ∞ (T (t)x)(y) := ξα (θ)(S(tα θ)x)(y)dθ 0 Z ∞ Z (6.87) i|y−z|2 1 αθ 4t = ξα (θ) e x(z)dz dθ, 4πitα θ Ω 0 and

Z ∞ θξα (θ)(S(tα θ)x)(y)dθ (S (t)x)(y) := α 0 Z Z ∞ i|y−z|2 1 αθ 4t e x(z)dz dθ, =α θξα (θ) 4πitα θ Ω 0

(6.88)

where ξα (θ) =

1 −1− 1 1 α ̟ (θ − α ) ≥ 0, θ α α

and ̟α (θ) =

∞ Γ(nα + 1) 1X (−1)n−1 θ−αn−1 sin(nπα), π n=1 n!

θ ∈ (0, ∞).

Here, ξα satisfies limθ→∞ ξα (θ) = 0 and it is a probability density function defined on (0, ∞), that is Z ∞ ξα (θ) ≥ 0, θ ∈ (0, ∞) and ξα (θ)dθ = 1. 0

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Then, we can introduce the following definition for our problem. Definition 6.41. By a mild solution of the system (6.77), we mean that a continuous function x : J → L2 (Ω) which satisfies Z ∞ x(t, y) = ξα (θ)(S(tα θ)x0 )(y)dθ 0 Z ∞ Z t α−1 α + (t − s) α θξα (θ)S((t − s) θ)(kV x)(s)(y)dθ ds Z0 Z ∞0 i|y−z|2 1 αθ 4t (6.89) e x0 (z)dz dθ = ξα (θ) 4πitα θ Ω 0 Z t Z ∞ 1 α−1 + (t − s) α θξα (θ) 4πi(t − s)α θ 0 Z0 i|y−z|2 × e 4(t−s)α θ kV (z)x(s, z)dz dθ ds, Ω

for t ∈ J.

Remark 6.42. Keeping in mind of that

Z

∞

ξα (θ)dθ and

0 limθ→∞ ξα (θ)

Z

∞

θξα (θ)dθ are ab-

0

solutely convergence respectively, and = 0 and limθ→∞ θξα (θ) = 0, there exist two constants M1 > 0 and M2 > 0 respectively such that Z ∞ Z ∞ 2 ξα (θ)dθ ≤ M1 , θ2 ξα2 (θ)dθ ≤ M2 . 0

0

The following properties of {T (t), t ≥ 0} and {S (t), t ≥ 0} are widely used in the sequel. Lemma 6.43. Let {T (t), t ≥ 0} and {S (t), t ≥ 0} be two solution operators defined by (6.87) and (6.88) respectively. Then T (t) and S (t) can be extended in a unique way to the bounded operators from L2 (Ω) into L2 (Ω) and p p kT (t)xkL2 (Ω) ≤ M1 kxkL2 (Ω) , kS (t)xkL2 (Ω) ≤ α M2 kxkL2 (Ω) . For any x ∈ L2 (Ω), keeping in mind of Lemma 6.40, we obtain 2 Z Z Z ∞ i|y−z|2 1 4tα θ x(z)dz e dθ dy kT (t)xk2L2 (Ω) = ξ (θ) α α 4πit θ Ω Ω 0 2 Z ∞ Z Z ∞ Z i|y−z|2 1 dθdy 4tα θ x(z)dz ≤ ξα2 (θ)dθ · e 4πitα θ

Proof.

0

Ω

0

Ω

≤ M1 kS(tα )xk2L2 (Ω)

≤ M1 kxk2L2 (Ω) . Thus, one can obtain the first estimation immediately. Using the similar method, one can derive the second estimation. Lemma 6.44. Operators {T (t), t ≥ 0} and {S (t), t ≥ 0} are strongly continuous, which means that for all x ∈ L2 (Ω) and 0 ≤ t′ < t′′ ≤ T , we have kT (t′′ )x − T (t′ )xkL2 (Ω) → 0 and kS (t′′ )x − S (t′ )xkL2 (Ω) → 0, as t′ → t′′ .

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For any x ∈ L2 (Ω) and Z ′′ ′ 2 kT (t )x − T (t )xkL2 (Ω) =

Proof.

275

0 ≤ t′ < t′′ ≤ T , we get that Z ∞ 2 ′′ α ′ α ξα (θ)[S((t ) θ) − S((t ) θ)]x(y)dθ dy

Ω

0

2

≤ M1 k[S((t′′ )q θ − (t′ )q θ) − I]xkL2 (Ω) .

According to the strongly continuity of {S(t), t ≥ 0}, we know that kT (t′′ )x − T (t′ )xkL2 (Ω) tends to zero as t′′ −t′ → 0, which means that {T (t), t ≥ 0} is strongly continuous. Using the similar method, we can also obtain that {S (t), t ≥ 0} is also strongly continuous. Lemma 6.45. Let Ω be a measurable subset of R2 , k = maxt∈[0,T ] |χ(t)| and V ∈ H 2 (Ω). Then, we have kkV xkL2 (Ω) ≤ kkV kL∞ (Ω) kxkL2 (Ω) .

(6.90)

Proof. Since H 2 (Ω) can be continuous embedded in the space C 0 (Ω) = {V ∈ C(Ω) : V ∈ L∞ (Ω)} and V ∈ L∞ (Ω), for arbitrary ε > 0 there exists Ωε ⊂ Ω, such that α(Ωε ) = 0 and sup |V (y)| < kV kL∞ (Ω) + ε.

Ω\Ωε

Thus, kkV xk2L2 (Ω) =

Z

Ωε

Z

|kV (y)x(y)|2 dy +

Z

Ω\Ωε

|kV (y)x(y)|2 dy

|V (y)x(y)|2 dy Ω\Ωε Z 2 ≤ k kV kL∞ (Ω) + ε |x(y)|2 dy

=k

≤ k kV kL∞ (Ω) + ε

2

Ω

kxk2L2 (Ω) .

Let ε → 0 and taking the limit in the above inequality, one can obtain the inequality (6.90) immediately. In what follows, we collect the Henry-Gronwall inequality (see Lemma 7.1.1 in Henry, 1981), which can be used in fractional differential equations and integral equations with singular kernel. Lemma 6.46. Let z, ω : [0, T ) → [0, +∞) be continuous functions where T ≤ ∞. If ω is nondecreasing and there are constants κ ≥ 0 and q > 0 such that Z t z(t) ≤ ω(t) + κ (t − s)q−1 z(s)ds, t ∈ [0, T ), 0

then

# Z t "X ∞ (κΓ(q))n nq−1 z(t) ≤ ω(t) + (t − s) ω(s) ds, t ∈ [0, T ). Γ(nq) 0 n=1

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If ω(t) = a ¯, constant on 0 ≤ t < T , then the above inequality is reduce to z(t) ≤ a ¯Eq (κΓ(q)tq ), 0 ≤ t < T,

where Eq is the Mittag-Leffler function defined by Eβ (y) :=

∞ X

k=0

Remark 6.47.

yk , y ∈ C, Re(β) > 0. Γ(kβ + 1)

(i) There exists a constant Mκ∗ > 0 independent of a ¯ such that z(t) ≤ Mκ∗ a ¯ for all 0 ≤ t < T. (ii) For more generalized Henry-Gronwall inequalities, see Ye, Gao and Ding, 2007. In order to discuss the existence of mild solutions for system (6.77), we need the following important priori estimate. Lemma 6.48. Let V ∈ H 2 (Ω). Suppose system (6.77) has a mild solution on [0, T ], then there exists a constant ρ > 0 such that kx(t)kL2 (Ω) ≤ ρ for all t ∈ [0, T ]. Proof. If x is a mild solution x of system (6.77) on [0, T ], then x satisfies (6.89). Keeping in mind of Lemma 6.43 and Lemma 6.45, we have 2 Z Z Z ∞ i|y−z|2 1 4tα θ x (z)dz e kx(t)k2L2 (Ω) ≤ ξ (θ) dθ dy α 0 α 4πit θ Ω 0 Z ∞ Ω Z Z t 1 + (t − s)α−1 α θξα (θ) 4πi(t − s)α θ Ω 0 0 (6.91) Z 2 i|y−z|2 × e 4(t−s)α θ kV (z)x(s, z)dz dθ ds dy Ω Z t ≤ M1 kx0 k2L2 (Ω) + α2 M2 k 2 kV k2L∞ (Ω) (t − s)α−1 kx(s)k2L2 (Ω) ds. 0

By Lemma 6.46 and Remark 6.47, there exists a constant Mα√M2 kkV kL∞ (Ω) > 0 such that kx(t)k2L2 (Ω) ≤ M1 kx0 k2L2 (Ω) Mα√M2 kkV kL∞ (Ω) := ρ2 , for all t ∈ [0, T ]. The proof is completed.

Theorem 6.49. Let V ∈ H 2 (Ω). System (6.77) has a unique mild solution x ∈ C([0, T ], L2 (Ω)). Proof.

Fixed x0 ∈ L2 (Ω), define n o B(x0 , 1) = x ∈ C([0, T1 ], L2 (Ω)) : kx(t) − x0 kL2Ω ≤ 1, t ∈ [0, T1 ] ,

where T1 will be chosen latter. It is obvious that B(x0 , 1) is a closed and convex subset of C([0, T1 ], L2 (Ω)).

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Define an operator P : B(x0 , 1) → B(x0 , 1) as follows: Z ∞ Z i|y−z|2 1 αθ 4t x0 (z)dz dθ (P x)(t, y) = ξα (θ) e 4πitα θ Ω 0 Z ∞ Z t Z i|y−z|2 1 α−1 αθ 4(t−s) kV (z)x(s, z)dz dθ ds. + (t − s) α θξα (θ) e 4πi(t − s)α θ Ω 0 0

It is easy to see that P is well defined. By Lemma 6.43, for all x ∈ C([0, T1 ], L2 (Ω)),

k(P x)(t) − x0 kL2 (Ω) 2 Z Z ∞ Z i|y−z|2 1 dy 4tα θ x (z)dz ≤ ξ (θ) e dθ − x (y) q 0 0 α 4πit θ Ω Ω 0 Z ∞ Z Z t 1 + (t − s)α−1 α θξα (θ) 4πi(t − s)α θ 0 2 ZΩ 0 2 i|y−z| × e 4(t−s)α θ kV (z)x(s, z)dz dθ ds dy Ω Z t p ≤ kT (t)x0 − x0 kL2 (Ω) + α M2 (t − s)α−1 k(kV x)(s)kL2 (Ω) ds

(6.92)

0

≤ kT (t)x0 − x0 kL2 (Ω) √ +α M2 kkV kL∞ (Ω)

Z

t

(t − s)α−1 (1 + kx0 kL2 (Ω) )ds 0 √ = kT (t)x0 − x0 kL2 (Ω) + tα M2 kkV kL∞ (Ω) (1 + kx0 kL2 (Ω) ).

By Lemma 6.44, {T (t), t ≥ 0} is a strongly continuous operator in L2 (Ω). Thus, there exists a t > 0 such that 1 kT (t)x0 − x0 kL2 (Ω) ≤ , t ≤ t. 2 Let  " # α1    1 T11 = min t, √ ,  2 M2 kkV kL∞ (Ω) (1 + kx0 kL2 (Ω) ) 

then for all t ≤ T11 , it comes from (6.92) that

k(P x)(t) − x0 kL2 (Ω) ≤ 1. Let x1 , x2 ∈ B(x0 , 1), we have k(P x1 )(t) − (P x2 )(t)kL2 (Ω) Z ∞ Z Z t 1 α−1 ≤ (t − s) α θξ (θ) α 4πi(t − s)α θ ΩZ 0 0 2 i|y−z|2 αθ 4(t−s) × e kV (z)[x1 (s, z) − x2 (s, z)]dz dθ ds dy Ω Z t p (t − s)α−1 kx1 (s) − x2 (s)kL2 (Ω) ds ≤ α M2 kkV kL∞ (Ω) 0 p ≤ tα M2 kkV kL∞ (Ω) kx1 − x2 kC([0,T1 ],L2 (Ω)) .

(6.93)

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Let T12

1 = 2

1 √ 2 M2 kkV kL∞ (Ω)

! α1

,

T1 = min{T11 , T12 },

then P is a contraction map on B(x0 , 1). It follows from the contraction mapping principle that P has a unique fixed point x ∈ B(x0 , 1), and x is the unique mild solution of system (6.77) on [0, T1 ]. 6.6

Notes and Remarks

The results in Section 6.2 are adopted from Bourdin, 2013. The results in Section 6.3 are taken from Beckers and Yamamoto, 2013. The material in Subsection 6.4.2 and Theorem 6.35 due to Torres, 2013. Theorems 6.37, 6.38 and 6.39 are adopted Nyamoradi and Zhou, 2014. The results of Section 6.5 are from Wang, Zhou and Wei, 2012d.

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