Lyapunov Inequalities and Applications 3030690288, 9783030690281

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Ravi P. Agarwal Martin Bohner Abdullah Özbekler

Lyapunov Inequalities and Applications

Lyapunov Inequalities and Applications

Ravi P. Agarwal • Martin Bohner ¨ Abdullah Ozbekler

Lyapunov Inequalities and Applications

Ravi P. Agarwal Department of Mathematics Texas A&M University–Kingsville Kingsville, TX, USA ¨ Abdullah Ozbekler Department of Mathematics Atilim University Ankara, Turkey

Martin Bohner Department of Mathematics and Statistics Missouri University of Science and Technology Rolla, MO, USA

ISBN 978-3-030-69028-1 ISBN 978-3-030-69029-8 (eBook) https://doi.org/10.1007/978-3-030-69029-8 Mathematics Subject Classification: 26Dxx, 26Exx, 30-XX, 34Nxx, 35-XX, 39-XX © Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To our mothers:

Godawari Agarwal Monika Bohner Vesile Özbekler

Preface

Inequalities play a fundamental rôle in all areas of mathematics and their applications. Among several well-known inequalities, Lyapunov-type inequalities have been studied extensively in recent years, and research continues. This is due to the fact that derivations of these inequalities require classical analysis and find applications in a wider spectrum of mathematics. In this book, we provide the reader with an up-to-date account of Lyapunov-type inequalities. We study these inequalities for various scenarios, namely for ordinary differential equations and systems of different types, partial differential equations, fractional differential equations, difference equations, and dynamic equations on time scales. The study of Lyapunov inequalities goes back to 1892, when Aleksandr Mikha˘ılovich Lyapunov (also romanized as Liapunov, Liapounoff, Ljapunov, or Ljapunow) born in Yaroslavl, Russia, 1857, died in Odesa, Ukraine, 1918 proved the following result (see also [202]): Let q be a real-valued and continuous function defined on the interval [a, b]. If the so-called Hill differential equation x  (t) + q(t)x(t) = 0,

t ∈ (a, b)

has a nontrivial solution that vanishes at two points of [a, b], then q satisfies the inequality  a

b

|q(t)|dt >

4 . b−a

This remarkable inequality is sharp, in the sense that the constant “4” cannot be replaced by a larger number. Moreover, one may show that if q is a realvalued function such that the second-order Hill differential equation has a nontrivial solution having two distinct zeros on [a, b], then the nonnegative part q + = max{0, q} must satisfy the so-called “Lyapunov inequality”

vii

viii

Preface



b a

q + (t)dt >

4 . b−a

Since then and until today, research on Lyapunov’s inequality and many of its generalizations has far exceeded 5000 publications, and numerous applications to many branches of mathematics have been given. Lyapunov-type inequalities have proved to be useful tools in the studies of oscillation theory, disconjugacy, eigenvalue problems, and numerous other applications in the theories of ordinary differential equations, difference equations, impulsive differential equations, Hamiltonian systems, and dynamic equations on time scales. As far as applications of Lyapunovtype inequalities to eigenvalue problems are concerned, we refer to the recent monograph [239] by Juan P. Pinasco. Moreover, we address two surveys, one of which is presented by Sui-Sun Cheng [91] on Lyapunov inequalities for differential and difference equations and disconjugacy, and the other one is presented by Aydın Tiryaki [271] on recent developments of Lyapunov-type inequalities. Many authors, including Chen [81], Cheng [83], Eliason [123, 124], Hartman [155], Hochstadt [172], Kwong [193], Nehari [220, 221], Reid [248, 249], and Singh [258], made some contributions to the theory of the celebrated Lyapunov inequality. One of the improvements of Lyapunov’s inequality is the so-called “Hartman inequality” 

b

(b − t)(t − a)q + (t)dt > b − a.

a

In this book, first we give a survey of the most basic results related to Lyapunovtype inequalities, and then we sketch some recent developments related to this type of inequalities. This book on the subject of Lyapunov-type inequalities summarizes, organizes, and presents a variety of related integral and sum inequalities. The book is primarily intended for senior undergraduate students and graduate students of mathematics, engineering, and science courses. Students in mathematical and physical sciences will find many sections of direct relevance. Some basic background in calculus, ordinary and partial differential equations, and difference equations will be helpful for the reader. This book consists of nine chapters. In Chap. 1, we present preliminaries and basic inequalities. Moreover, we also give some basic applications of Lyapunov’s inequality related to some important concepts such as the number of zeros, the distance between zeros, disconjugacy, disfocality, eigenvalue problems, and the oscillation of solutions. In Chap. 2, we discuss even-order, odd-order, and, more generally, higher-order boundary value problems with different kinds of boundary conditions such as Dirichlet, Lidstone, clamped-free, multiple point, and conjugate boundary conditions. As far as half-linear equations are considered, Lyapunov and Hartman-type inequalities for second-order, third-order, and higher-order equations are discussed in Chap. 3. Chapter 4 studies systems of linear, nonlinear, and quasilinear differential equations involving the (p1 , . . . , pn )-Laplacian, antiperiodic, and clamped-free periodic boundary conditions. In Chap. 5, some recent

Preface

ix

developments in Lyapunov-type inequalities for fractional differential equations under Dirichlet, fractional, and Robin boundary conditions are presented. Chapter 6 is devoted to partial differential equations and two-dimensional nonlinear systems of partial differential equations. Now changing the subject to discrete equations, Chap. 7 discusses second-order and even-order linear difference equations, linear Hamiltonian difference systems, quasilinear and nonlinear difference systems, and partial difference systems. In this chapter, some applications will also be given such as disconjugacy criteria, stability, and some properties of Green’s function. Finally, in Chap. 8, we discuss Lyapunov-type inequalities for linear, half-linear, and nonlinear dynamic equations on time scales, as well as linear Hamiltonian dynamic systems. The aim of this book is to present a clear and well-organized treatment of the concept of Lyapunov-type inequalities behind the development of mathematics as well as its applications techniques. The text material of this book is presented in a readable and mathematically solid format. Finally, this project was initiated when the third author was on academic leave between September 2014 and September 2015, visiting the first author at Texas A&M University–Kingsville. The authors also acknowledge partial support by ˙ TÜBITAK (the Scientific and Technological Research Council of Turkey). The authors would like to thank Professor Svetlin Georgiev (France), Professor Veysel Fuat Hatipo˘glu (Turkey), and Professor A˘gacık Zafer (Kuwait) for carefully reading the manuscript. Moreover, the authors are grateful to the Springer Mathematics Editor, Robinson Nelson dos Santos (Springer São Paulo). Kingsville, TX, USA Rolla, MO, USA Ankara, Turkey October 2020

Ravi P. Agarwal Martin Bohner Abdullah Özbekler

Contents

1 Lyapunov-Type Inequalities for Second-Order Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Basic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Number of Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Distance between Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Oscillation of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Disconjugacy and Disfocality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Eigenvalues of Sturm–Liouville Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 An Inequality of Nehari . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3

1 1 1 6 11 20 22 28 44 49 53

Lyapunov-Type Inequalities for Higher-Order Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Even-Order Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Dirichlet Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Lidstone Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Clamped-Free Boundary Value Problems . . . . . . . . . . . . . . . . . . . . 2.3 Odd-Order Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Third-Order Differential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 General Odd-Order Differential Equations. . . . . . . . . . . . . . . . . . . 2.4 General Higher-Order Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Multiple-Point Boundary Value Problems . . . . . . . . . . . . . . . . . . . 2.4.2 Conjugate Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55 55 55 56 71 87 95 95 114 117 118 123 130

Lyapunov-Type Inequalities for Half-Linear Differential Equations . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Second-Order Half-Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Lower Bounds for Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

133 133 133 152 xi

xii

Contents

3.3

Third-Order Half-Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Lyapunov-Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 The Linear Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Applications to Boundary Value Problems . . . . . . . . . . . . . . . . . . . Higher-Order Half-Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

157 158 169 172 172 181 186

Lyapunov-Type Inequalities for Nonlinear Differential Systems . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Nonlinear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Quasilinear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Lower Bounds for Generalized Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Nonlinear Systems with Anti-periodic Boundary Conditions . . . . . . . 4.7 Quasilinear Systems with Clamped-Free Boundary Conditions . . . . 4.8 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

189 189 189 209 225 254 274 284 290

5

Lyapunov-Type Inequalities for Fractional Differential Equations . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Linear FDEs with Dirichlet Boundary Conditions . . . . . . . . . . . . . . . . . . . 5.3 Linear FDEs with Fractional Boundary Conditions. . . . . . . . . . . . . . . . . . 5.4 Linear FDEs with Robin Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . 5.5 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

293 293 293 306 314 320

6

Lyapunov-Type Inequalities for Partial Differential Equations . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Linear PDEs with Neumann Boundary Conditions . . . . . . . . . . . . . . . . . . 6.2.1 Lyapunov-Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 The Subcritical Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 The Supercritical Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 The Critical Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.5 Qualitative Properties of βp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.6 Nonlinear Resonant Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Two-Dimensional Nonlinear Systems of PDEs . . . . . . . . . . . . . . . . . . . . . . 6.3.1 An Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Multivariate Lyapunov Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Linear and Quasilinear Elliptic Differential Operators . . . . . . . . . . . . . . 6.5.1 Lyapunov-Type Inequality for p > N . . . . . . . . . . . . . . . . . . . . . . . . 6.5.2 Lyapunov-Type Inequality for p < N . . . . . . . . . . . . . . . . . . . . . . . . 6.5.3 Applications to Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . 6.5.4 Eigenvalues of the p-Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

321 321 322 325 326 329 333 334 338 340 345 347 360 367 371 373 378 380

7

Lyapunov-Type Inequalities for Difference Equations . . . . . . . . . . . . . . . . . . 381 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 7.2 Second-Order Linear Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 381

3.4 3.5 4

Contents

7.3 7.4 7.5

7.6 7.7 7.8

7.9 7.10 8

xiii

Lyapunov-Type Finite Difference Inequalities . . . . . . . . . . . . . . . . . . . . . . . Even-Order Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Discrete Linear Hamiltonian Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 A Disconjugacy Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.3 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Quasilinear Difference Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Discrete Nonlinear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.1 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial Difference Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Nets and Discrete Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . 7.8.2 Green’s Functions and Lyapunov-Type Inequalities . . . . . . . . . 7.8.3 Maxima of Green’s Functions on Straight Nets. . . . . . . . . . . . . . 7.8.4 Maxima of Green’s Functions on Circular Nets . . . . . . . . . . . . . 7.8.5 Maxima of Green’s Functions on Rectangular Nets . . . . . . . . . 7.8.6 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two-Dimensional Nonlinear Systems of Partial Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Lyapunov-Type Inequalities for Dynamic Equations on Time Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Sturm–Liouville Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Linear Hamiltonian Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Higher-Order Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Stability Theory for Hill’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.1 Auxiliary Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6.2 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Linear Hamiltonian Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.1 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.2 A Disconjugacy Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Planar Linear Hamiltonian Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.1 Time Scales Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.2 Lyapunov-Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.3 Disconjugacy Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Nonlinear Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9.1 Preliminaries and Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9.2 Lyapunov-Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10 Notes and References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

406 408 414 439 440 450 453 466 469 490 494 495 497 500 502 503 505 506 510 513 513 516 520 523 530 535 538 541 548 552 560 561 563 566 575 576 578 580 589

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605

Chapter 1

Lyapunov-Type Inequalities for Second-Order Linear Differential Equations

1.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for second-order linear differential equations of the form x  + q(t)x = 0 and sketch some important developments related to this type of inequalities. In Sect. 1.2, we give some preliminaries about the historical developments of Lyapunov-type inequalities for second-order linear differential equations, and we also introduce some basic definitions. In Sect. 1.3, we present the classical Hartman and Lyapunov inequalities with different kinds of proofs. In the next sections, we offer some important applications of Hartman and Lyapunov inequalities such as studies of the number of zeros (in Sect. 1.4), the distance between zeros (in Sect. 1.5), oscillation of solutions (in Sect. 1.6), disconjugacy and disfocality (in Sect. 1.7), and eigenvalues of Sturm–Liouville problems (in Sect. 1.8). In Sect. 1.9, we give a well-known inequality of Nehari, and we present how it improves and extends the classical results of Lyapunov to self-adjoint differential equations and differential equations with damping term.

1.2 Preliminaries Let q be a real and continuous function defined on the interval [a, b]. If the so-called Hill differential equation x  + q(t)x = 0

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_1

(1.1)

1

2

1 Second-Order Linear Differential Equations

has a nontrivial solution x that vanishes at two points of [a, b], then q satisfies the inequality 

b

|q(t)|dt >

a

4 . b−a

(1.2)

This result is originally due to A. M. Lyapunov [202], and it is sharp in the sense that the constant 4 cannot be replaced by a larger number. Let η be a real-valued absolutely continuous function on [a, b] such that η is square integrable and η(a) = η(b) = 0. Then, for s ∈ (a, b), we have 

b

  2 η (t) dt ≥

a

4 [η(s)]2 . b−a

(1.3)

Inequality (1.3) is a very useful tool for the study of the qualitative nature of solutions of second-order ordinary differential equations. If η is nontrivial on [a, b], then equality holds in (1.3) only if s = (a + b)/2 and     2t − a − b   .  η(t) = η(s) 1 −  b−a  In particular, with the aid of (1.3), one may show that if q is a real-valued function such that (1.1) has a nontrivial solution having two distinct zeros on [a, b], then q + must satisfy Lyapunov’s inequality 

b

a

q + (t)dt >

4 , b−a

(1.4)

where q + (t) = max{0, q(t)} is the nonnegative part of q(t). This result is originally due to Lyapunov [202]. The Lyapunov inequality and many of its generalizations have proved to be useful tools in the studies of oscillation theory, disconjugacy, eigenvalue problems, and numerous other applications for the theories of ordinary differential equations, difference equations, impulsive differential equations, Hamiltonian systems, and dynamic equations on time scales. In this section, first we give a survey of the most basic results on Lyapunovtype inequalities, and next we sketch some developments related to this type of inequalities. Several authors, including Chen [81], Cheng [83], Eliason [123, 124], Hartman [155], Hochstadt [172], Kwong [193], Nehari [220, 221], Reid [248, 249], and Singh [258], made some attributions to the celebrated paper of the Russian mathematician Lyapunov [202], who proved the following remarkable result. Now, we consider the boundary value problem consisting of (1.1) and the Dirichlet boundary conditions x(a) = x(b) = 0,

(1.5)

1.2 Preliminaries

3

i.e., 

x  + q(t)x = 0 on

[a, b],

(1.6)

x(a) = x(b) = 0.

Theorem 1.1 (Lyapunov’s Inequality) Let q be a nontrivial, continuous, and nonnegative function defined on [a, b], where a, b ∈ R with a < b. If x is a nontrivial solution of (1.6) then the so-called “Lyapunov inequality” 

b

q(t)dt > a

4 b−a

(1.7)

holds. We are obligated to mention here that although (1.7) is known as the classical Lyapunov inequality, it was pointed out by Cheng [91] that Lyapunov neither stated nor proved Theorem 1.1, but rather in [202], he only claims the following. Theorem 1.2 Let q be a nontrivial, continuous, and nonnegative function with period ω and let 

ω

q(t)dt ≤

0

4 . ω

(1.8)

Then, the roots of the characteristic equation corresponding to Hill’s equation x  + q(t)x = 0 on

R,

(1.9)

are purely imaginary with unity modulus. Here, the characteristic equation is s 2 − 2As + 1 = 0,

where

A=

f (ω) + g  (ω) , 2

(1.10)

where f and g are solutions of (1.9) satisfying the initial conditions f (0) = 1,

f  (0) = 0

and

g(0) = 0,

g  (0) = 1.

Theorem 1.2 implies (we refer to Floquet theory) the following stability theorem. Theorem 1.3 Under the conditions of Theorem 1.2, all solutions of (1.9) remain bounded as t → ±∞, i.e., (1.9) is stable. For original proofs of Theorems 1.2 and 1.3, the reader is referred to Lyapunov [202], Forsyth [135], and Ince [174]. Another proof of Theorem 1.3 has been given by Zhukovski˘ı [310], who also proved that (1.8) is sharp in the sense that if (1.8) is changed to

4

1 Second-Order Linear Differential Equations



ω

q(t)dt ≤

0

4 +ε ω

with ε > 0,

then the conclusion of Theorem 1.3 is no longer true. The sharpness of (1.8) was also demonstrated by Kampen and Wintner [182] and Borg [59]. Improvements and extensions of Theorem 1.3 were later given by a number of authors. For a comprehensive exposition of these results and additional information, the reader is referred to the remarkable report by Cesari [78]. Note that Lyapunov’s [202] proof requires the calculation of series expansions of the functions f and g and term-by-term calculations of the coefficients of the series with quite involved derivations. Venturing to obtain alternative proofs, Borg [58, 59] quoted and proved in [59] the inequality 

b

a

    x (t)  4    x(t)  dt > b − a

(1.11)

due to Beurling [43], and then, in view of (1.1), observed (1.2), see [59] and Theorem 1.7. After that, Borg [59] continued a simplified proof of Theorem 1.3 and investigated also the acuity of (1.11) relating to the stability of (1.9). In 1949, Hartman and Wintner [156] noticed the result of Beurling [43] quoted by Borg in his earlier paper [58]. Theorem 1.4 Let q be a real-valued and continuous function defined in [a, b]. Then, no nontrivial solution of (1.1) can have more than one zero in [a, b] unless (1.2) holds. In fact, the result given in Theorem 1.4 is known as a disconjugacy criterion for (1.1), see also Definition 1.6. As it was first noticed by Wintner [290] and subsequently by several other authors, an application of Sturm’s comparison theorem allows the replacement of |q(t)| in (1.2) by q + (t). In this remarkable result, Wintner introduced not only an alternative proof of Theorem 1.4 but also a sharper result than it. His proof was based on the Riccati equation y  + y 2 + q(t) = 0 associated with (1.1), and by means of Sturm’s comparison theorem, he observed that (1.2) can be replaced by a more general inequality, i.e., (1.4). Theorem 1.5 (Wintner) Let q be a real-valued and continuous function defined on [a, b]. A necessary condition for (1.1) to have a nontrivial solution possessing two zeros is that (1.4) holds. The constant 4 in (1.4) cannot be replaced by a larger number. In view of the above developments, it seems more appropriate to attribute Theorem 1.1 or its variant Theorem 1.5 to Lyapunov–Beurling–Borg–Wintner. Commonly, we shall still use the term Lyapunov inequality (or Lyapunov-type

1.2 Preliminaries

5

inequality) for any inequality which plays an analogous rôle for related differential or difference boundary value problems as (1.4) plays for (1.1). Inequality (1.2) may be applied in two ways. If λ is the smallest eigenvalue of the 2-point boundary value problem   x + λq(t)x = 0 on [a, b], (1.12) x(a) = x(b) = 0, then one, from (1.2), can have



b

λ > 4 (b − a)

−1 |q(t)|dt

.

(1.13)

a

This allows a lower bound. If more is known about the potential function q, then (1.13) can be improved, see [215, Chapter 9]. For the second interpretation, we consider a generalization to a scalar higher-order linear differential operator of the form L x(t) := x (n) + q1 (t)x (n−1) + · · · + qn (t)x with summable coefficients. Definition 1.6 The differential operator L is said to be disconjugate on an interval (a, b) if no nontrivial solution of L x = 0 has n zeros on the interval (a, b). Disconjugacy is connected with solvability of boundary value problems since L is a linear differential operator. Consider x (i) (tj ) = cij

for

j = 0, 1, . . . , Ji

and

i = 1, 2, . . . , I,

(1.14)

where tj ∈ (a, b) and J1 + J2 + · · · + JI = n. The forced equation L x(t) = f (t)

(1.15)

has a unique solution satisfying (1.14) for each choice of cij if and only if the homogenous equation corresponding to (1.15), i.e., L x(t) = 0 has a unique solution satisfying (1.14) with all cij . If the operator L is disconjugate on (a, b), then this is satisfied for all choices of tj ∈ (a, b). In the special case of Lyapunov’s inequality, we may conclude that if  a

b

|q(t)|dt ≤

4 , b−a

then the operator L x := x  + qx

(1.16)

6

1 Second-Order Linear Differential Equations

is disconjugate on (a, b), and we may solve all 2-point boundary value problems on (a, b). In particular, sufficiently short intervals are intervals of disconjugacy.

1.3 Basic Results Inequality (1.2) may be strengthened by replacing |q(t)| by q + (t) or by writing 

b

(b − t)(t − a)q + (t)dt > b − a,

(1.17)

a

see Hartman [155, p. 345] (see also Theorem 1.10 below). This discussion is followed up in later sections. Inequality (1.2) may also be obtained by using the integral inequality 

     u(t)  dt > 4,

1  u (t) 

0

(1.18)

where u ∈ C([0, 1]) and u(0) = u(1) = 0. As also observed by Cheng [91], as an extension of (1.18), Borg [59] proved the following result of Beurling [43]. Theorem 1.7 Suppose that a, b ∈ R with a < b are consecutive zeros of a solution of (1.12). If x(t) > 0 for all t ∈ (a, b), then (1.11) holds. Here, we simply sketch Borg’s [59] proof. He starts with the inequalities  a

b

    b  x (t)      dt > 1  x (t) dt  x(t)  x∞ a  t2    1 x (t) dt > x∞ t1    x (t2 ) − x  (t1 ) > x∞

(1.19)

for arbitrary a ≤ t1 < t2 ≤ b. Now, let x∞ = x(t∗ ). By Rolle’s theorem, we can choose t1 ∈ (a, t∗ ) and t2 ∈ (t∗ , b) such that x  (t1 ) =

1 x∞ t∗ − a

and

− x  (t2 ) =

1 x∞ . b − t∗

(1.20)

Using (1.19) and (1.20), we can obtain  a

b

    x (t)  1 1 4    x(t)  dt > b − t + t − a > b − a , ∗ ∗

(1.21)

1.3 Basic Results

7

where the last inequality, i.e., the estimate 4/(b − a), is simply obtained by minimization of the middle part of (1.21). This seems to be the first proof of Lyapunov’s inequality to appear in the literature. Borg [59] went on to use (1.11) to give a new and much shorter proof of the Lyapunov stability theorem. Hartman and Wintner [157] have given a proof of (1.4). Their proof was based on the fact that the graph of the positive solution x of (1.12) on (a, b) is concave down when q ≥ 0. Thus, t −a 1 ≥  >0 x(t) x (a)

b−t 1 ≥−  > 0 for x(t) x (b)

and

t ∈ (a, b).

Let c ∈ (a, b) such that x  (c) = 0. Then, using this and the last inequality, one has 

b

−

(b − t)(t − a)

a



c

≥−

x  (t) dt x(t)

(b − t)(t − a)

a

≥−

x  (t) dt − x(t)



b

(b − t)(t − a)

c

x  (t) dt x(t)

(b − c)  (c − a)  {x (c) − x  (a)} +  {x (b) − x  (c)} x  (a) x (b)

= b − a, which implies the inequality 

b

(b − t)(t − a)q(t)dt > b − a.

(1.22)

1 (b − a)2 4

(1.23)

a

Since max {(b − t)(t − a)} =

a≤t≤b

and q + (t) ≥ q(t) for all t ∈ R, (1.22) implies (1.4). The Green function method plays an important rôle for boundary value problems. Nehari [219, 221] observed that the solution of the second-order boundary value problem 

z (t) = −f (t), z(a) = z(b) = 0

can be represented by

(1.24)

8

1 Second-Order Linear Differential Equations



b

z(t) =

G (t, s)f (s)ds,

(1.25)

a

where 1 G (t, s) = b−a



(b − t)(s − a)

if a ≤ s ≤ t,

(t − a)(b − s)

if t ≤ s ≤ b

(1.26)

is Green’s function for (1.24). It is clear that 0 ≤ G (t, s) ≤

(b − s)(s − a) b−a

for s ∈ (a, b).

(1.27)

Without loss of generality, it may be assumed that x > 0 on (a, b). (In fact, if x < 0 on (a, b), one can consider −x, which is also a solution.) Then, by using (1.24) and (1.25), x can be expressed as 

b

x(t) =

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

(1.28)

a

Choosing c so that x(c) = maxt∈(a,b) x(t), we have, by (1.27) and (1.28), 1 1≤ b−a



b

(b − t)(t − a)q(t)dt,

a

which implies (1.22). It is well known that some of the qualitative properties of ordinary linear differential equations are related to variational principles. Using such a variational lemma [250, pp. 224–230], Reid proved that if x ∈ C2 [a, b] satisfies (1.5), then 

b

  2 x (t) dt >

a

4 [x(t)]2 b−a

for

t ∈ (a, b).

In particular, if x is a nontrivial solution of (1.12), then, by using the same variational lemma, the so-called Wirtinger-type inequality 

b

 2 x  (t) − q(t)[x(t)]2 dt ≤ 0,

a

(1.22) follows by taking |x(s)| = max{|x(t)| : a ≤ t ≤ b}. It can be shown easily that if (1.1) has a nontrivial solution satisfying the boundary condition x(a) = x  (c) = 0, then 

c a

(t − a)q(t)dt > 1,

1.3 Basic Results

9

and similarly if (1.1) has a nontrivial solution satisfying the boundary condition x  (c) = x(b) = 0, then 

b

(b − t)q(t)dt > 1.

c

Thus, 

b



c

(b − t)(t − a)q(t)dt ≥ (b − c)

a



b

(t − a)q(t)dt + (c − a)

a

(b − t)q(t)dt

c

> b − a, which is (1.22), see also [39, 96, 97, 193]. Remark 1.8 In fact, (1.22) and its variant (1.17) are known as Hartman-type inequalities, see Theorem 1.10 below. Note that a nontrivial solution of (1.12) with x > 0 is an eigenfunction corresponding to the eigenvalue 1. Moreover, if q ≥ 0, then 1 is indeed the lowest eigenvalue of the eigenvalue problem   x + λq(t)x = 0

on

x(a) = x(b) = 0,

[a, b],

x>0

on (a, b).

It can be found in Courant and Hilbert [99] that among the functions q such that 

b

q(t)dt = M,

a

the lowest eigenvalue λ(q) satisfies the inequality λ(q) ≥

4 . M(b − a)

(1.29)

In [192], Kre˘ın used a functional analytic method together with the well-known minimax principle for eigenvalues of Sturm–Liouville problems to extend (1.29) and to prove (1.7). Later, some of the elementary proofs of Lyapunov’s inequality were obtained. One of the most interesting proofs was given by Leighton [198]. Theorem 1.9 Let q be a continuous function on [a, b]. If there exists a nontrivial solution x of (1.1) that has two zeros on [a, b], then (1.4) holds. Proof Consider the differential equation x  + q + (t)x = 0.

10

1 Second-Order Linear Differential Equations x(t)

0

6

.. .. ... .. ...... ... ..... .. . ...... .. ...... .. .. . . . ............ .. . . . . . . .. .... .... ........... .. . . . . .. . . ...... . . .... . . .. . . . . . . . . . ...... . .... .. .. . . . . . . . . . . . ...... .. . .... . .. . . . . . . . .. . . . ...... . . .... . . . .. . . . . . . . . ...... . .. .. ..... . . . . . . . . . . ...... . . ... . . . . . . . . . . . . . ...... ... . .. ......... . . . . . . ...... ... . .. .......... . ...... . . . . . ....... ...... .. . . . . ........ . .. . ....... ........... .... ...... ....... ....... ....... . . ......... . . ... .. .. . 0 .. ..

........... ............. .................... . . . . . . ...... .. ..... ..... . . . .... . . . .... . .. . .... . ... . a t c.

.. .. . .. ....b

-

t

Fig. 1.1 Illustration of the proof of Theorem 1.9

Since q + ≥ q on [a, b], there exists a solution x of (1.1) such that x(a) = x(c) = 0, with x > 0 on (a, c) ⊂ [a, b]. Let t = t0 be a point on (a, c) at which x attains its absolute maximum. We note that since x  ≤ 0 on (a, c), x possesses no relative minima on that open interval (it may possess a line of relative minima, however). Draw the two chords that connect the point [t0 , x(t0 )] with the points (a, 0) and (c, 0). Then, according to Fig. 1.1, it is clear that x  (a) >

x(t0 ) t0 − a

and −x  (c) >

x(t0 ) x(t0 ) ≥ . c − t0 b − t0

It follows that −

b−a x  (c) − x  (a) > , x(t0 ) (b − t0 )(t0 − a)

that is, 1 − x(t0 )



c a

1 x (t)dt = x(t0 ) 



c a

q + (t)x(t)dt >

b−a . (b − t0 )(t0 − a)

1.4 Number of Zeros

11

Since 

1 x(t0 )

c



+

c

q (t)x(t)dt < a

q + (t)dt,

a

we have 

b

q + (t)dt ≥

a



c

q + (t)dt >

a

b−a . (b − t0 )(t0 − a)

(1.30)

As the last term of (1.30) is readily seen to be greater than or equal to 4/(b − a), with equality only if t0 is the midpoint of (a, b), the proof is complete. 

1.4 Number of Zeros A remarkable approach was made in this area by Hartman [155] who was concerned with zeros of real-valued solutions of (1.12) and gave estimates of the number of zeros of its solutions. Theorem 1.10 (Hartman’s Inequality) Let q be a real-valued and continuous function on [a, b]. Let m be a nonnegative and continuous function on [a, b] and put m(t) . t∈(a,b) (b − t)(t − a)

γm := inf

(1.31)

If the nontrivial and real-valued solution x of (1.12) has two zeros in [a, b], then 

b

m(t)q + (t)dt > γ (b − a).

(1.32)

a

In particular, the so-called Hartman inequality 

b

(b − t)(t − a)q + (t)dt > b − a

(1.33)

a

holds. Proof Suppose that (1.12) has a nontrivial solution with two zeros in [a, b]. Since q + ≥ q, the equation z + q + (t)z = 0

12

1 Second-Order Linear Differential Equations

is a Sturm majorant for (1.12) and hence has a nontrivial solution z with two zeros t1 , t2 ∈ [a, b]. Since z = −q + z, it follows that  (t2 − t1 )z(t) = (t2 − t)

t

(s − t1 )q + (s)z(s)ds + (t − t1 )



t1

t2

(t2 − s)q + (s)z(s)ds.

t

(1.34)

Suppose that t1 , t2 are consecutive zeros of z and that z > 0 on (a, b). Choose t = t0 so that z0 (t) = max z(t). t∈(t1 ,t2 )

The right-hand side of (1.34) is increased if z(s) is replaced by z0 (t). Thus, dividing both sides of (1.34) by z0 (t) > 0 gives  t2 − t1 < (t2 − t)

t

(s − t1 )q + (s)ds + (t − t1 )



t2

(t2 − s)q + (s)ds,

t

t1

where t = t0 . Since t2 − t ≤ t2 − s for t ≥ s and t − t1 ≤ s − t1 for s ≥ t,  t2 − t 1
0. Since (b − t)(t − a) ≤

1 (b − a)2 , 4

(1.36)

the choice of m(t) ≡ 1 in Theorem 1.10 gives the well-known Lyapunov inequality. Theorem 1.12 (Lyapunov’s Inequality) Let q be a real-valued and continuous function in [a, b]. If x is a nontrivial solution of (1.6), where a, b ∈ R with a < b are consecutive zeros, then the inequality

1.4 Number of Zeros

13



b

4 b−a

q + (t)dt >

a

(1.37)

holds. Corollary 1.13 Let q ≥ 0 be continuous in [a, b] and suppose that (1.6) has a solution x > 0 on (a, b). Then, the inequality 

b

2M A

q(t)dt > a

holds, where  M := max x(t) t∈(a,b)

A :=

and

b

x(t)dt. a

Moreover, the factor 2 of M/A cannot be replaced by a larger constant. Corollary 1.14 Let p, q ≥ 0 be real-valued continuous functions on [0, b] and let z be a solution of the equation z + p(t)z + q(t)z = 0 satisfying the boundary conditions z(0) = z(b) = 0. Then, the inequality  b
1. 2 6 Moreover, this inequality can be improved to M1 b M 2 b 2 + >1 π π2 by using Wirtinger’s inequality 

b 0

[z(t)]2 dt ≤

b2 π2

 0

b

  2 z (t) dt,

14

1 Second-Order Linear Differential Equations

which can be proved by assuming b = π , expanding z into a Fourier sine series, and applying Parseval’s relation for z and z . It can further be improved to M1 b M2 b 2 + ≥ 1, π π2 see Opial [222]. An analogous result for higher-order differential equations is as follows. Suppose the differential equation z(n) + q1 (t)z(n−1) + · · · + qn (t)z = 0 has continuous coefficients on [0, b] and a nontrivial solution z with n zeros in [0, b]. Assume |qj (t)| ≤ Mj for all t ∈ [0, b], j = 1, . . . , n. Then, 1


u

4 v−u

(1.39)

if u = tk and v = tk+1 , k = 1, . . . , N − 1. Since the harmonic mean of N − 1 positive numbers is majorized by their arithmetic mean, we obtain 

N −1 1 1 N −1 tk+1 − tk k=1

−1 ≤

N −1 1 t N − t1 . (tk+1 − tk ) = N −1 N −1 k=1

1.4 Number of Zeros

15

Thus, summing (1.39) for k = 1, . . . , N − 1 gives 

tN

q + (t)dt >

t1

4(N − 1)2 4(N − 1)2 ≥ , tN − t1 T 

completing the proof.

Remark 1.16 Using (1.32) with m(t) = t − a in place of (1.37), it can be seen that N also satisfies the inequality 

T

N
0 be a continuous function and of bounded variation in [0, T ]. Suppose x is a nontrivial real-valued solution of (1.1) and N is the number of its zeros on (0, T ]. Then,    Nπ − 

T

0



   1 T |dq(t)|  . q(t)dt  ≤ π + 4 0 q(t)

(1.40)

Proof In terms of x, define a continuous function ϕ by √ ϕ(t) := arctan

q(t)x(t) x  (t)

and

0 ≤ ϕ(0) < π.

Then, by Hartman [155, p. 333], we have  ϕ(T ) = ϕ(0) + 0

T



q(t)dt +

1 4



T

sin(2ϕ(t))d(log q(t)). 0

By Hartman [155, Lemma 3.1], N is the greatest integer not exceeding ϕ(T )/π , and hence Nπ ≤ ϕ(T ) ≤ (N + 1)π. This implies (1.40).



Corollary 1.18 Let q be a continuous function in [0, T ]. Suppose x is a real-valued nontrivial solution of (1.1) and N is the number of its zeros on (0, T ]. Then,

16

1 Second-Order Linear Differential Equations



T

|Nπ − T | ≤ π +

|1 − q(t)|dt.

0

If, in addition, q > 0 has a continuous second derivative, then    N π − 

0

T



   q(t)dt  ≤ π +

0

T

    5 q  (t)2 q  (t)   −   dt.  16[q(t)]5/2 4[q(t)]3/2 

Corollary 1.19 Let q > 0 be a continuous function and of bounded variation in [0, T ] for every T > 0. Suppose also that 

T 0

1 |dq(t)| = o q(t)



T



as

q(t)dt

T → ∞,

(1.41)

0

i.e., suppose that q has a continuous derivative q  satisfying   q  (t) = o [q(t)]3/2

as

t → ∞.

(1.42)

Suppose x is a real-valued nontrivial solution of (1.1) and N(T ) is the number of its zeros on (0, T ]. Then,  π N (T ) ≈

T



q(t)dt

as

T → ∞.

(1.43)

0

Corollary 1.19 is clear from (1.41) and (1.40) in Theorem 1.17. It should be mentioned that if, e.g., q is monotone and q(t) → ∞ as t → ∞, then (1.42) imposes no restriction on the rapidity of growth of q but is a condition on regularity of growth. This can be seen from the fact that the integral 

T

q  (t) 2 dt = √ + const [q(t)]3/2 q(T )

tends to a limit as T → ∞; thus, in general, q  (t)[q(t)]−3/2 is small for large t. The conditions of Corollary 1.19 for validity of (1.43) can be lightened as follows. Corollary 1.20 Let q > 0 be a continuous function defined on [0, ∞) such that  

−1   t  log q(t)    1+ sup  q(τ )dτ → 0 as q(s)  s≤t 0.

(1.45)

18

1 Second-Order Linear Differential Equations

Let N be the number of zeros of x1 on (0, T ]. Then,    T   dt   ≤ π.   Nπ − c 2 2   0 r(t) [x1 (t)] + [x2 (t)] 

(1.46)

Proof Let α be an arbitrary real number. Consider the solutions  xˆ1 (t) = x1 (t) cos α + x2 (t) sin α, xˆ2 (t) = −x1 (t) sin α + x2 (t) cos α of (1.45). They satisfy (x1 (t))2 + (x2 (t))2 = (xˆ1 (t))2 + (xˆ2 (t))2

(1.47)

  r(t) xˆ1 (t)xˆ2 (t) − xˆ1 (t)xˆ2 (t) ≡ c > 0.

(1.48)

and

Choose α so that xˆ1 (0) = 0 and let N ∗ be the number of zeros of xˆ1 (t) on (0, T ]. Since (1.47) and (1.48) imply that xˆ1 (t) and xˆ2 (t) are linearly independent, they have no common zeros. Hence, it is possible to define a continuous function by ϕ(t) := arctan

xˆ1 (t) xˆ2 (t)

and

ϕ(0) = 0.

This function is continuously differentiable and, by (1.47) and (1.48), satisfies ϕ  (t) =

c   > 0. r(t) [x1 (t)]2 + [x2 (t)]2

(1.49)

Hence, ϕ is increasing; also ϕ(t) = 0 mod π if and only if x1 (t) = 0. Thus, N ∗ is the greatest integer not exceeding ϕ(T )/π , and integration of (1.49) gives N ∗π ≤ c



T 0

dt  < (N ∗ + 1)π.  r(t) [x1 (t)]2 + [x2 (t)]2

Sturm’s separation theorem implies N ∗ ≤ N ≤ N ∗ +1, and thus (1.46) follows.



Corollary 1.23 Let r, q, x1 , x2 , and N be as in Theorem 1.22 and, in addition, let q ≥ 0. Then,      T   dt Nπ − c     2  2  ≤ 2π.  0 r(t) xˆ 1 (t) + xˆ 2 (t)  

(1.50)

1.4 Number of Zeros

19

If q > 0, then (1.46) and (1.50) are particular cases of “duality”, in which (x1 (t), x1 (t), q(t), dt) are replaced by r(t)x1 (t), −x1 (t),

1 , q(t)dt , q(t)

see [155, Lemma 3.1, p. 512]. Corollary 1.24 Let q be continuous on [0, ∞). If all solutions of (1.1) are bounded, then 1 t



t

q + (s)ds ≥ c0

with c0 ∈ R+

0

for large t by (1.38) and (1.46). Remark 1.25 Replacing x1 and x2 in (1.46) by x1 /ε and εx2 , respectively, and if, in addition, a nontrivial solution x satisfies x(t) → 0 as t → ∞, then 1 t



t

q + (s)ds → ∞ as

t → ∞.

0

Corollary 1.26 Let q ≥ 0 on [0, ∞). If the first derivatives of all solutions of (1.1) are bounded, then 1 t



t

q + (s)ds ≤ c1

with

c1 ∈ R

0

for large t by (1.38) and (1.50). Remark 1.27 If, in addition, x  (t) → 0 as t → ∞ for some nontrivial solution x, then  1 t + q (s)ds → 0 as t → ∞. t 0 Remark 1.28 Corollaries 1.24 and 1.26 can be generalized for the case when (1.1) is replaced by (1.45) and the assumption that solutions (or derivatives of solutions) are bounded is replaced by the assumption that all solutions satisfy x(t) = O (1/Φ(t))

or

where Φ is a positive continuous function.

x  (t) = O (1/Φ(t)) ,

20

1 Second-Order Linear Differential Equations

1.5 Distance between Zeros Let x be a solution of (1.1) with two consecutive zeros at a and b. A simple inequality is proven that relates not only a and b to the integral of q + (t) but also any point c ∈ (a, b) where |x(t)| is maximized. As a consequence, it is shown by Patula [231] that if (1.1) is oscillatory and q + ∈ Lp [0, ∞), p ∈ [1, ∞), then the distance between zeros becomes unbounded. Lemma 1.29 Suppose that a, b with a < b are consecutive zeros of a nontrivial solution x of (1.1). Let c ∈ (a, b) be such that |x(c)| = max{|x(t)| : a < t < b}. Then,  c 1 (i) q + (t)dt > , c−a a b 1 , q + (t)dt > (ii) b − c c b b−a . (iii) q + (t)dt > (b − c)(c − a) a Proof Integrating both sides of (1.1) from c to t yields x  (t) − x  (c) =



t

q − (s)x(s)ds −



c

t

q + (s)x(s)ds.

c

Note that x  (c) = 0. Another integration from c to t gives 

t

x(t) − x(c) =

(t − s)q − (s)x(s)ds −



c

t

(t − s)q + (s)x(s)ds.

(1.51)

c

Let t = b so that x(t) = 0. Equation (1.51) implies 

b

x(c) +

t (b − s)q − (s)x(s)ds =

c



b

(b − s)q + (s)x(s)ds.

c

Without loss of generality assuming that x ≥ 0 on [a, b], it can be obtained that  x(c) ≤

b

(b − s)q + (s)x(s)ds < (b − c)x(c)

c



b

q + (s)ds,

c

which implies 

b

1 < (b − c)

q + (s)ds.

c

This proves (ii). Part (i) follows in a similar manner, except that in (1.51), t is replaced by a. The sum of (i) and (ii) yields (iii), which completes the proof. 

1.5 Distance between Zeros

21

We remark that Lemma 1.29 can also be found in Cohn [97]. One way to view Lemma 1.29 is that it imposes some restrictions on the location of the point c and b thus the maximum of |x(t)| in [a, b]. That is, a q + (t)dt is a finite number. But lim

c→a +

b−a b−a = lim = ∞. (b − c)(c − a) c→b− (b − c)(c − a)

Thus, c cannot be “too close” to a or b. Also, it is interesting to note that b−a 4 ≥ (b − c)(c − a) b−a

for c ∈ (a, b).

Thus, Theorem 1.12 follows from the hypothesis of Lemma 1.29. As a consequence of Lemma 1.29 (also Theorems 1.10 and 1.12), the following result comes forward. Theorem 1.30 (Patula [231]) Suppose that q + ∈ Lp [0, ∞), p ∈ [1, ∞). If (1.1) is oscillatory and if x is a solution of it, then the distance between consecutive zeros of x must tend to infinity. Proof Suppose the statement is not true. Then, there exists a solution x of (1.1) with its sequence of zeros {tn }, which has a subsequence {tnk } such that  t n

k+1

 − tn k  ≤ M < ∞

nk , tnk+1 ) where |x(t)| is maximized. Then, for all k ∈ N. Let {snk } be a point in (t sn − tn  < M for all k ∈ N. Since q + ∈ Lp [0, ∞), p ∈ [1, ∞), choose k so large k k that



∞

+

q (s)

p

1/p ds

≤ M −(1+1/r) ,

(1.52)

tnk

where r is the Hölder conjugate of p, i.e., 1/p + 1/r = 1. From (i) of Lemma 1.29, we have  sn k 1 q + (t)dt > . s n k − tn k tnk Using this and (1.52) and applying Hölder’s inequality, one has  1 < (snk − tnk )

snk tnk

 < (snk − tnk )

q + (τ )dτ

snk

tnk

 + p q (τ ) dτ

1/p (snk − tnk )1/r

22

1 Second-Order Linear Differential Equations

 < (snk − tnk )

∞

1+1/r

+

q (τ )

p

1/p dτ

tnk

< M 1+1/r M −(1+1/r) = 1, which is a contradiction. This completes the proof.



1.6 Oscillation of Solutions Concerning (1.1), there is the following well-known oscillation theorem of Wintner [289]. Theorem 1.31 (Wintner [289]) If  lim

t→∞ 0

t

q(s)ds = ∞,

(1.53)

then (1.1) is oscillatory. Condition (1.53) of Theorem 1.31 can be used to give the following simple example. Example 1.32 Consider the equation x  +

1 x = 0 on t +1

[0, ∞).

(1.54)

Condition (1.53) guarantees that (1.54) is oscillatory. Since q defined by q(t) = 1/(t + 1) satisfies q ∈ L2 [0, ∞), Theorem 1.30 affirms that the distance between zeros of any solution of (1.54) must become unbounded. Definition 1.33 For t ∈ [0, ∞), (1.1) is called limit point, L.P., if there exists at least one solution x ∈ L2 [0, ∞). Theorem 1.30 can also be used to derive a known L.P. result given in Patula and Wong [232, p. 10]. Definition 1.34 If any two linearly independent solutions (i.e., all solutions) are square integrable, then (1.1) is called limit circle, L.C., see Coddington and Levinson [95, p. 225]. The following lemma [232, p. 11] is well known. Lemma 1.35 (Patula and Wong) If (1.1) is L.C., then it is oscillatory, and the distance between any consecutive zeros of any solution of (1.1) tends to zero as t → ∞.

1.6 Oscillation of Solutions

23

Corollary 1.36 If q + ∈ Lp [0, ∞), p ∈ [1, ∞), then (1.1) is in the limit point classification. Proof Suppose the contrary of the statement. Then, (1.1) is L.C. Let x be any solution of (1.1). By Lemma 1.35, x is oscillatory, and the distance between consecutive zeros of any solution of (1.1) tends to zero as t → ∞. However, Theorem 1.30 asserts that if x is oscillatory, then the distance between consecutive zeros must become unbounded, a contradiction. Thus, (1.1) must be L.P.  We note that when p = ∞, Theorem 1.30 does not hold; the simple differential equation x  + x = 0 is a nice example for this. However, it would be interesting to know whether Theorem 1.30 is true for p ∈ (0, 1). If it were, then Corollary 1.36 could also be extended to the case p ∈ (0, 1). This would answer a question posed by Everitt et al. [131, p. 346] as to whether or not (1.1) is L.P. for q + ∈ Lp [0, ∞), p ∈ (0, 1). In 1951, Hartman and Wintner [157] considered (1.1) assuming that q is a realvalued and continuous function on a closed interval [0, T ]. They considered only real-valued and nontrivial solutions x of (1.1). If no such solution has more than one zero on [0, T ], we call (1.1) disconjugate on [0, T ]. A general approach to disconjugate equations of type (1.1) was considered in [290]; it easily leads to various criteria of explicit nature, in particular to the following criterion, which goes back to Lyapunov (see [59]): If  T T |q(t)|dt ≤ 4, (1.55) 0

then (1.1) is disconjugate on [0, T ]. It is also known (see [59, 182]) that the number 4 in (1.55) cannot be improved to any 4 + ε > 4 (if q is unspecified). But it will turn out below that (1.55) can be greatly improved in another direction. First, if the factor T in (1.55) is written inside the integral and is then diminished to t ∈ [0, T ], then it is natural to ask whether there exists a positive absolute constant having the property that (1.1) must be disconjugate on [0, T ] whenever the value of  T t|q(t)|dt (1.56) 0

does not exceed that absolute constant. It turns out that such an absolute constant exists, and that its best value is 1. In other words, (1.1) must or need not be disconjugate on [0, T ] according as the value of (1.56) does not or does exceed 1. It will also follow that if (1.56) is replaced by  0

T

t α |q(t)|dt,

(1.57)

24

1 Second-Order Linear Differential Equations

then there does not exist for any T > 0 and for any α > 1 a positive constant C(α, T ) having the property that (1.1) must be disconjugate on [0, T ] whenever the value of (1.57) is less than C(α, T ). Clearly, neither is the sufficiency of 

T

t|q(t)|dt ≤ 1

(1.58)

0

contained in that of (1.55) nor is the converse true. But the sufficiency of both (1.55) and (1.58) is contained in (a) and the final nature of the respective absolute constants 4 and 1 in (b) of the following theorem. Theorem 1.37 Let r be a positive and continuous function on the open interval (0, T ). (a) Whenever q is a continuous function on the closed interval [0, T ] satisfying 

T−

0+



 T −1 , r(t)|q(t)|dt ≤ T inf r(t) 0 1, α = 1, or α < 1. In the more symmetric case where r(t) = t α is replaced by r(t) = t α (T − t)α , t ∈ [0, T ], the bound on the right-hand side of (1.59) has the value 0 or 41−α T 2α−1 according to α > 1 or α ≤ 1. Thus, the assertion made after (1.57), according to which there cannot exist C(α, T ) > 0 if α > 1, is obvious. It is clear that (1.59) implies the inequality 

T−

0+

r(t)q + (t)dt ≤ T

inf

0 1

(1.64)

0

must hold whenever (1.1) possesses some nontrivial solution x which has at least two zeros on [0, 1]. Suppose that there exists such a solution x, and let t = a and t = b, where 0≤a0 x(t) x (a)

and

1−t 1 ≥−  >0 x(t) x (b)

(1.67)

if t ∈ (a, b). Finally, if c is any point of this interval, then it is seen from (1.65) that 1 − t > 1 − c or t > c according to whether t ∈ (a, c) or t ∈ (c, b). It follows therefore from (1.67) that the integral on the right-hand side of (1.66) is greater than (not equal to) −

(1 − c) x  (a)



c a

x  (t)dt +

c x  (b)



b

x  (t)dt.

(1.68)

c

Hence, this lower estimate holds for the integral on the left-hand side of (1.66). This lower estimate is valid for every choice of c ∈ (a, b). Let c now be chosen so that x  (c) = 0 (such a choice is possible as x(a) = x(b) = 0). Then, (1.68) reduces to −

(1 − c)(0 − x  (a)) c(x  (b) − 0) + = 1 − c + c = 1. x  (a) x  (b)

Hence, the lower estimate, mentioned after (1.68), becomes 

b

t (1 − t)q(t)dt > 1.

(1.69)

a

Since q ≥ 0, it is seen from (1.65) that (1.69) implies (1.64). This completes the proof of (a).

1.6 Oscillation of Solutions

27

Finally, we prove (b). Assume (1.61), (1.62), and (1.63). Let s and ε be arbitrary numbers satisfying 0 < ε < s < 1 − ε, and let x be any function defined on [0, 1] having the following properties: x has a continuous and nonpositive second derivative on [0, 1], x is positive on (0, 1), and 1 x(t) = 1−s−ε



(1 − s − ε)t

if

0 ≤ t ≤ s − ε,

(1 − t)(s − ε)

if

s + ε ≤ t ≤ 1.

(1.70)

In terms of this x = x(·; s, ε), define a continuous q = q(·; s, ε) by (1.1), by putting q = x  /x. Then, q is nonnegative on [0, 1] and vanishes identically on both subintervals [0, s − ε] and [s + ε, 1] of [0, 1]. Since, by (1.70), the solution x of (1.1) has two zeros on [0, 1] (but does not vanish identically), the assertion of (b) will be proved if it is shown that the value of the integral on the left of (1.61) can be made less than any number which exceeds (1.62), provided s and ε are suitably chosen. Now, (1.70) and (1.1) show that the integral on the left-hand side of (1.61) is identical with  s+ε r(t)x  (t) − dt. (1.71) x(t) s−ε But the value of (1.71) is less than 1/(s − ε) times the value of the integral of −r(t)x  (t) over the interval [s − ε, s + ε], since x(t) ≥ s − ε on this interval. Hence, if m = m(s, ε) denotes the maximum of r on [s − ε, s + ε], then the value of (1.71) is less than m/(s − ε) times  −

s+ε

x  (t)dt = x  (s − ε) − x  (s + ε) = 1 +

s−ε

s−ε , 1−s−ε

where we have used (1.70) for the last equality. Consequently, the value of (1.71) is less than that of the ratio m(1 − 2ε) , (s − ε)(1 − s − ε)

(1.72)

where m = m(s, ε). Since r is continuous at s, it is clear from the definition of m that the ratio (1.72) can be brought arbitrarily close to r(s)/[s(1 − s)] by choosing ε small enough. Since s ∈ (0, 1) is arbitrary, it follows that, if ε and s are suitably chosen, then the value of (1.72) will come arbitrarily close to (1.62). Consequently, the value of (1.72) can be made less than any number that is greater than (1.62). This proves (b). 

28

1 Second-Order Linear Differential Equations

1.7 Disconjugacy and Disfocality The classical result of Lyapunov is usually formulated in connection with disconjugacy. Hence, the converse of (1.2), i.e.,  b 4 , |q(t)|dt ≤ b−a a implies that (1.1) is disconjugate in [a, b]. Kwong [193] considered, however, a setting involving disfocality. In this setting, the result can be stated in a stronger form from which the classical one follows immediately. This idea was definitely not original at that time, but it had not been emphasized in the literature. In [193], stronger results are established by exploiting two different ideas. The first one is elementary. The second one makes use of a comparison theorem, which itself is an improvement over a theorem of Taam [266]. The required comparison theorem is proved using the theory of integral inequalities. Definition 1.38 Equation (1.1) is said to be right (left) disfocal in [a, b] if any solution of it satisfying x  (a) = 0 (x  (b) = 0) has no zeros in [a, b]. Moreover, (1.1) is disconjugate in [a, b] if and only if there exists a point c ∈ [a, b] such that (1.1) is right disfocal in [c, b] and left disfocal in [a, c]. Thus, Lyapunov’s result follows from the following stronger result. If (1.1) is not disfocal in an interval [a, c], then  c 1 . (1.73) q + (t)dt > c−a a Inequality (1.73) is given in Wong [292]. It is also implied by the following lemma. Lemma 1.39 If (1.1) has a solution such that x  (0) = x(c) = 0 with c > 0, then  c  c Q+ (t)dt = (c − t)q + (t)dt > 1, (1.74) 0

0

where 

+

Q (t) =

t

q + (s)ds.

0

Proof Changing the order of the double integral on the left, it can easily be seen that the two integrals in (1.74) are equal. There are two cases. First, assume that x has no zeros in [0, c). Suppose that (1.74) has been shown for this case. In the case that x has zeros in [0, c), let c¯ be the smallest zero. Then, we have  0

c¯

Q+ (t)dt > 1,

(1.75)

1.7 Disconjugacy and Disfocality

29

from which (1.74) follows. Next, we may assume that q ≥ 0 so that q + = q. In the contrary case, we consider the equation z + q + (t)z = 0

(1.76)

and one of its solutions z such that z (0) = 0. It follows from the Sturm comparison theorem (notice that the potential q + of (1.76) dominates that of (1.1)) that z has a zero c¯ ∈ (0, c). The result for positive potentials then gives, for (1.76), (1.75), from which (1.74) follows. We henceforth assume q ≥ 0 and x is positive in [0, c). Integrating both sides of (1.1) from 0 to t gives 



− x (t) =

t

q(s)x(s)ds ≤ x(0)Q+ (t).

(1.77)

0

Integrating (1.77) over [0, c] immediately yields (1.74).



Remark 1.40 The converse of (1.74) implies right disfocality in [0, c). Remark 1.41 Lemma 1.39 has been formulated over (0, c] just for the convenience of stating (1.74). A simple translation enables to apply the result to any interval. A reflection also enables to treat left disfocality. Remark 1.42 Inequality (1.74) obviously implies the weaker inequality Q+ (1) =



c 0

q + (t)dt >

1 . c

Remark 1.43 Since a quite rough upper bound for x(s), i.e., x(s) ≤ x(0), was used in deriving (1.77), there is still room for further improvements. It amounts to a search for better upper bounds for x(s), and an attempt has been made. Since the final result is somewhat too complicated to state, we just outline the idea here. From the assumption q ≥ 0, it can be seen that x is concave up in [0, c], and hence x(t) ≥ x(c)(1−t/c), from which we obtain x  (t) ≤ −x(c)(1−t/c)q(t). Integrating this over [0, t] gives an upper bound for x  (t). Integrating over [0, t] once more gives an upper bound for x(t). Remark 1.44 Lemma 1.39 can be interpreted in the following manner. Let x be the solution as described in the hypotheses. Suppose y  (c) is negative, meaning that the tangent line at 0 points upward in the direction of decreasing t. As we move along the curve in the direction of decreasing t, the tangent line will bend downward if q(t) is positive or upward if q(t) is negative. In order that the tangent line at t = 0 be horizontal, so that the overall motion of the tangent line, from t = c to t = 0, is downward, q(t) must be sufficiently positive. The appearance of the weight (c − t) in the second integral in (1.74) indicates that the part of q + (t) near 0 counts more than that further away from 0.

30

1 Second-Order Linear Differential Equations

Remark 1.45 If a disconjugacy criterion is desired, then the following corollaries of Lemma 1.39 can be used. An analysis will set out that these are further extensions of Hartman’s improvement of Lyapunov’s original result [155]. Corollary 1.46 If the inequality 1 t −a



t

(s − a)q + (s)ds +

a

1 b−t



b

(b − s)q + (s)ds ≤

t

1 1 + t −a b−t

holds for all t ∈ [a, b], then (1.1) is disconjugate in (a, b). Corollary 1.47 If the inequalities 

c



(t − a)q + (t)dt ≤ 1 and

a

b

(b − t)q + (t)dt ≤ 1

c

hold for some point c ∈ [a, b], then (1.1) is disconjugate in [a, b]. The following lemma is a simple consequence of the general theory of integral inequalities. It can also be established directly without much complexity. Lemma 1.48 Suppose that Q and P are continuous functions defined on [0, c] such that Q ≤ P on [0, c]. Suppose furthermore that    t  t   + 2   + 2 Q(t) +  ≤ P (t) + Q P (s) ds (s) ds   0

(1.78)

0

for all t ∈ [0, c]. Let x1 and x2 be solutions defined on subintervals [0, c1 ) and [0, c2 ) of [0, c], respectively, of the two integral equations 

t

x1 (t) = Q(t) +

[x1 (s)]2 ds

(1.79)

[x2 (s)]2 ds.

(1.80)

0

and 

t

x2 (t) = P (t) + 0

Then, |x1 (t)| ≤ x2 (t) for all t in the common domain [0, c1 ) ∩ [0, c2 ). Proof We first show that −x2 (t) ≤ x1 (t) < x2 (t) for all t ∈ [0, c1 ) ∩ [0, c2 ) provided we have strict inequality Q(t) < P (t) in the hypotheses. A continuity argument then completes the proof. From (1.79), we have x1 (t) ≥ Q+ (t). When this inequality is substituted into (1.79), we have 

t

x1 (t) ≥ Q(t) + 0

 + 2 Q (s) ds.

1.7 Disconjugacy and Disfocality

31

Similarly, 

t

x2 (t) ≥ P (t) +

 + 2 P (s) ds.

0

Thus, we have, due to (1.79), −x2 (t) ≤ x1 (t), proving one half of the required inequality. Suppose that the other half is false. Then, there must be a smallest t, say t0 > 0, such that x1 (t0 ) = x2 (t0 ) but x1 (t) < x2 (t) if t < t0 . We have (x1 (t))2 ≤ (x2 (t))2 if t < t0 . It follows that 

t0



t0

[x1 (s)] ds ≤ 2

0

[x2 (s)]2 ds.

0

Considering (1.79) and (1.80) when t = t0 yields x1 (t0 ) < x2 (t0 ), a contradiction.  The same consideration proves the following lemma. Lemma 1.49 In addition to the assumptions of Lemma 1.48, assume x1 ≥ 0. Then, the conclusion of Lemma 1.48 still holds even in the absence of (1.78). Theorem 1.50 (Kwong [193]) Let x be a solution of (1.1) such that x  (0) = 0, x(c) = 0, c > 0. Let z be a solution of z + p(t)z = 0

(1.81)

such that z (0) = 0. If furthermore the functions 



t

Q(t) =

q(s)ds

P (t) =

and

0

t

p(s)ds 0

satisfy the hypotheses of Lemma 1.48, then z has a zero in [0, c]. Proof Without loss of generality, we may assume that x has no other zeros in [0, c). Let x1 = −x  /x and x2 = −z /z. It is well known that x1 and x2 are solutions of (1.79) and (1.80). The fact x(c) = 0 implies that x1 (t) → ∞ as t → c. It follows from Lemma 1.48 that x2 (t) → ∞ (since x2 (t) ≥ |x1 (t)|) as t approaches some point in [0, c]. Then, z has a zero at that point.  This result is a refinement of a comparison theorem due to Taam [266], see also Levin [199]. Corollary 1.51 If instead of (1.78), the inequality  − P (t) ≤ Q(t) +

t

 + 2 Q (s) ds

0

is satisfied, then the conclusion of Theorem 1.50 still holds.

(1.82)

32

1 Second-Order Linear Differential Equations

Proof Inequality (1.78) is formed from the two inequalities  Q(t) +

 + 2 Q (s) ds ≤ P (t) +

t



0

t

 + 2 P (s) ds

(1.83)

0

and 

t

− P (t) −

 + 2 P (s) ds ≤ Q(t) +



0

t

 + 2 Q (s) ds.

(1.84)

0

Inequality (1.83) is a direct consequence of Q ≤ P , while (1.84) follows from (1.82).  Define  Q(t) :=

t

q(s)ds

  S(t) := min Q(t∗ ) +

 := Q(t)

⎧ 2Q(t) ⎪ ⎪ ⎪ ⎪ ⎪ 1⎨ 2⎪ ⎪ ⎪ ⎪ ⎪ ⎩

t ∈ [0, c],

for

0

t∗ ∈[0,t]

t∗

  + 2 Q (s) ds ,

0

if S(t) ≥ 0, or if S(t) ≥ Q(t), or if S(t) < 0

|Q(t) − S(t)|

if S(t) < 0

and

and

Q(t) ≥ −S(t),

S(t) < Q(t) < −S(t),

and    : t ∈ [0, c] . A := max Q(t) Lemma 1.52 If (1.1) has a solution such that x  (0) = x(c) = 0 with c > 0, then A>

1 . c

Proof We prove for the time being a weaker form of the statement with the conclusion A > l/c. The stronger form will be included in Theorem 1.54. We define the function  Q(t) if S(t) ≥ −A, P (t) := (1.85) Q(t) − S(t) if S(t) < −A. In other words, P (t) = Q(t) + [−S(t) − A]+ . Since −S is a continuous monotonic increasing function, so is [−S − A]+ . Hence, P is the integral of p = q + s, where s ≥ 0 is the derivative of [−S − A]+ . Thus, p ≥ q. By the classical Sturm

1.7 Disconjugacy and Disfocality

33

comparison theorem, the solution z of (1.81) such that z (0) = 0 must have a zero in [0, c]. Let c¯ be the smallest one. Define 

t

T (t) := P (t) +

 + 2 P (s) ds.

0

Since P = Q + [−S − A]+ ≥ Q, we obtain  T (t) ≥ P (t) +

t

 + 2 Q (s) ds

t

 + 2 Q (s) ds + [−S(t) − A]+

0

 = Q(t) +

0

≥ S(t) + [−S(t) − A]+ ≥ −A. We claim that P ≤ A. Let P attain its maximum at a point t1 . Suppose that at t1 , [−S(t) − A]+ = 0 so that P (t1 ) = Q(t1 ). It is not difficult to see from the definition of Q1 (t) that Q ≤ Q1 ≤ A. Thus, P (t1 ) ≤ A. Suppose [−S(t1 ) − A]+ > 0. Then, −S(t1 ) > A and

P (t1 ) = Q(t1 ) − S(t1 ) − A ≤ 2Q1 (t1 ) − A ≤ A.

Let us now show that Ac¯ ≥ 1, from which follows the weaker form of the statement. To this end, we compare, using Lemma 1.48, the Riccati integral equation associated with (1.81), namely (1.80), with the equation  x¯2 (t) = A +

t

[x¯2 (s)]2 ds.

(1.86)

0

With P and A instead of Q and P , the hypotheses of Lemma 1.48 are satisfied. Thus, x2 (t) ≤ x¯2 (t) in the common domain of definition. Suppose Ac¯ < 1. The solution of (1.86) is x¯2 (t) =

A , 1 − At

which is defined in [0, c]. ¯ In particular, x¯2 (c) ¯ < ∞. Since |x2 (t)| ≤ x¯2 (t), x2 (t) can be continued to be defined throughout [0, c], ¯ thus contradicting the fact that z has a zero at c. ¯  Corollary 1.53 Assume the same hypotheses as in Lemma 1.49. If min Q(s) ≥ −Q(t),

s∈[0,t]

then let Q2 (t) = Q(t). If

34

1 Second-Order Linear Differential Equations

min Q(s) < −Q(t),

s∈[0,t]

then let   1 Q(t) − min Q(s) . Q2 (t) = s∈[0,t] 2 Then, max Q2 (t) >

t∈[0,c]

1 . c

(1.87)

As a consequence, if min Q(t) ≥ − max Q2 (t)

t∈[0,c]

t∈[0,c]

or if 1 min Q(t) ≥ − , c

t∈[0,c]

then max Q(t) >

t∈[0,c]

1 , c

and if min Q(t) < − max Q(t),

t∈[0,c]

t∈[0,c]

then max Q(t) − min Q(t) >

t∈[0,c]

t∈[0,c]

2 . c

(1.88)

Inequality (1.87) is still stronger than (1.73) but (1.88) may not imply (1.73) in some cases. However, (1.88) has the advantage of being much easier to apply to many interesting examples. Lemmas 1.39 and 1.52 are different extensions of the classical result. The two different approaches can be combined to give better results, but unfortunately, we have not succeeded in obtaining one that extends both lemmas. Define Q∗ (t) := max |P (s)|, s∈[0,t]

1.7 Disconjugacy and Disfocality

35

where P is as in (1.85). By definition, Q∗ is nondecreasing and hence is the derivative of a positive function q ∗ . Theorem 1.54 If (1.1) has a solution x such that x  (0) = 0, x(c) = 0, c > 0, then 

c

Q∗ (t)dt > 1.

(1.89)

z + q ∗ (t)z = 0,

(1.90)

0

Proof By comparing (1.1) with

using Theorem 1.50, we see that a solution z of (1.90) satisfying z (0) = 0 has a zero in [0, c], say at c. ¯ Lemma 1.39 when applied to (1.90) gives 

c¯

Q∗ (t)dt > 1,

0



from which (1.89) follows. Remark 1.55 The proof of Lemma 1.52 can be completed by observing that   A = max Q∗ (t) : t ∈ [0, c] . Hence,  Ac ≥

c

Q∗ (t)dt > 1.

0

Remark 1.56 Theorem 1.54 does not include Lemma 1.39, since although P ≤ Q+ , we may not have |P | ≤ Q+ . The following similar result does include Lemma 1.39 but not Lemma 1.52. Instead of using the function P , we use the function Q2 as defined in Corollary 1.53. Define Q∗∗ (t) := max |Q2 (s)| . s∈[0,t]

It is not hard to see that Q∗∗ ≤ Q+ . The proof of the following theorem is similar to that of Theorem 1.54. Theorem 1.57 Assume the same hypotheses as in Theorem 1.54. Then, 

c 0

Q∗∗ (t)dt > 1.

36

1 Second-Order Linear Differential Equations

In 1997, Brown and Hinton [64] considered (1.1) on [a, b], where q is a real measurable function on [a, b] satisfying 

b

q(t)dt < ∞.

a

Two problems of interest are: (i) Obtain lower bounds for the spacing of zeros of a solution of (1.1). (ii) Obtain lower bounds for the spacing β − α, where x is a solution of (1.1) satisfying x(α) = x  (β) = 0 or x  (α) = x(β) = 0. Of particular interest in this work is when q is oscillatory, and this behavior affects the bounds. Brown and Hinton are motivated by a paper of Harris and Kong [153]. Two of their results [153, Theorem 2.1 and Theorem 2.2] state that if x is a solution of (1.1) with no zeros in (α, β) and such that x  (α) = x(β) = 0, then  t    (β − α) max  q(s)ds  > 1. α≤t≤β

(1.91)

α

If instead x(α) = x  (β) = 0, then   (β − α) max  α≤t≤β

t

β

  q(s)ds  > 1.

(1.92)

Finally, if there are no extreme values of x in (α, β), then in either (1.91) or (1.92), the absolute value may be dropped. Their method of proof is to use Riccati equation techniques. Here, by using Opial’s inequality, they proved several results which relate to problems (i) and (ii) above. In particular, they obtained (1.91) or (1.92) as a consequence of Opial’s inequality. A special case of an inequality obtained by Beesack and Das [41] is the following (see also [214, p. 119]. Theorem 1.58 Suppose f is absolutely continuous on [a, b] with f (a) = 0. If ψ ∈ L2 (a, b), then 

b

  |f (t)| f  (t) ψ(t)dt ≤ κ

a



b

  2 f (t) dt,

(1.93)

a

where 1 κ := √ 2



b

1/2 [ψ(t)] (t − a)dt 2

a

with equality if and only if f = 0 (or f is linear and ψ is constant).

(1.94)

1.7 Disconjugacy and Disfocality

37

If we replace f (a) = 0 in Theorem 1.58 by f (b) = 0, then (1.93) holds, where κ in (1.94) is given by 1 κ := √ 2



b

1/2 [ψ(t)]2 (b − t)dt

(1.95)

.

a

Beesack and Das also use another Opial inequality which is a special case of a more general result due to Boyd [60]. Theorem 1.59 Suppose f is absolutely continuous on [a, b] with f (a) = 0 or f (b) = 0. If 1 ≤ α ≤ 2, then 

b

α  |f (t)| f  (t) dt ≤ Kα (b − a)



α

a

b

  2 f (t) dt

α (1.96)

,

a

where ⎧ 1 ⎪ ⎪ ⎪ ⎪ ⎨ 1 (2 − α)α 2−2α I −α /α Kα := 2⎪ ⎪ 8 ⎪ ⎪ ⎩ π2

if α = 1, if 1 < α < 2,

(1.97)

if α = 2

with 

1

1+

I := 0

2(α − 1) t 2−α

−2

{1 + (α − 1)t}1/α−1 dt.

For α = 1, equality holds in (1.96). Theorem 1.59 has immediate applications to the case when f (a) = f (b) = 0. Choose c = (a + b)/2 and apply (1.96) to [a, c] and [c, b] and then add to obtain 

b

α  |f (t)| f  (t) dt ≤ Kα



α

a

≤ Kα

b−a 2 b−a 2

 

c

  2 f (t) dt

α +

a



b

  2 f (t) dt



b

  2 f (t) dt

α 

c

α .

a

(1.98) For α = 1, (1.98) is strict unless f is linear in each of the subintervals [a, c] and [c, b]. Theorem 1.60 Let q ∈ L(a, b). Suppose x is a nontrivial solution of (1.1) satisfying x(a) = x  (b) = 0. Then,

38

1 Second-Order Linear Differential Equations



b

2

[Q1 (t)]2 (t − a)dt > 1,

(1.99)

a

where 

b

Q1 (t) :=

(1.100)

q(s)ds. t

If x  (a) = x(b) = 0, then 

b

2

[Q2 (t)]2 (b − t)dt > 1,

(1.101)

a

where 

t

Q2 (t) :=

(1.102)

q(s)ds. a

Proof We first establish (1.99). Multiplying (1.1) by x(t) and integrating by parts gives 

b

  2 x (t) dt =



a

b

a

q(t)[x(t)]2 dt



b

=− a



b

=2 

Q1 (t)[x(t)]2 dt

Q1 (t)x(t)x  (t)dt

(1.103)

a b

≤2 a

1 ≤√ 2

  |Q1 (t)| x(t)x  (t) dt



b

1/2 

b

[Q1 (t)] (t − a)dt 2

a

  2 x (t) dt

a

by (1.93) and (1.94) of Theorem 1.58. The inequality is strict since linearity of x implies x = 0 as x(a) = x  (b) = 0. Dividing both sides of (1.103) by 

b

  2 x (t) dt

a

and squaring, we obtain (1.99). The proof of (1.101) is similar using integration by parts and (1.93) of Theorem 1.58 and (1.95) instead of (1.94).  Remark 1.61 By using the maximum of |Q1 | and |Q2 | on [a, b] in (1.99) and (1.101), integrating, and then taking the square root, we see that

1.7 Disconjugacy and Disfocality

39

  (b − a) max  a≤t≤b

b

t

  q(s)ds  > 1

(1.104)

when x(a) = x  (b) = 0, and

 t    (b − a) max  q(s)ds  > 1 a≤t≤b

(1.105)

a

when x  (a) = x(b) = 0, which are the inequalities obtained by Harris and Kong [153]. Note also that if x has no extreme values in (a, b), then x(t)x  (t) > 0 + in (1.103). It follows that Q1 ≤ Q+ 1 , and we can replace Q1 by Q1 in the derivation of (1.104). This means that the absolute value signs may be omitted in (1.104). With minor changes in the argument (xx  is now negative on (a, b)), the same conclusion applies to (1.105) if x has no extreme value on (a, b). Results similar to Theorem 1.60 may be obtained by an application of Boyd’s [60] theorem. Theorem 1.62 Suppose x is a nontrivial solution of (1.1) with x(a) = x  (b) = 0. Assume 1 ≤ α ≤ 2. Let α ∗ be the Hölder conjugate of α, i.e., 1/α + 1/α ∗ = 1. Then,  2Kα1/α (b

− a)

b

1/α

α∗

|Q1 (t)|

1/α ∗ ≥ 1.

dt

(1.106)

a

If x  (a) = x(b) = 0, then (1.106) is true with Q2 (t), where Qk , k = 1, 2, are defined as in (1.100) and (1.102). In either case, Kα is given by (1.97). For α = 1, (1.106) is strict. For α = 1, the α ∗ -norm of Q1 in (1.106) becomes maxa≤t≤b |Q1 (t)|. Proof In the case x(a) = x  (b) = 0, from the proof of Theorem 1.60, we have that  b  b     2 |Q1 (t)| |x(t)| x  (t) dt. (1.107) x (t) dt ≤ 2 a

a

By an application of Hölder’s inequality and Theorem 1.59 to (1.107), we get 

b

  2 x (t) dt ≤ 2

a



b

α∗

|Q1 (t)|

1/α ∗  dt

a

≤

  x(t)x  (t)α dt

1/α

a

 2Kα1/α (b

b

− a)

b

1/α

α∗

|Q1 (t)|

1/α ∗  dt

a

with strict inequality for α = 1. Dividing both sides of (1.108) by 

b

a

b

  2 x (t) dt (1.108)

  2 x (t) dt

a

yields (1.106). A similar argument yields (1.106) with Q2 when x  (a) = x(b) = 0. 

40

1 Second-Order Linear Differential Equations

Note that, for α = 1, (1.106) in the x(a) = x  (b) = 0 case is the same as (1.104) and in the x  (a) = x(b) = 0 case the same as (1.105). Theorems 1.60 and 1.62 yield sufficient conditions for disfocality of (1.1), i.e., sufficient conditions so that there does not exist a nontrivial solution x of (1.1) satisfying either x(a) = x  (b) = 0 or x  (a) = x(b) = 0. Application of (1.98) allows the use of an arbitrary antiderivative Q in the above arguments. From this, new Lyapunov-type inequalities can be derived. Theorem 1.63 Suppose x is a nontrivial solution of (1.1) satisfying x(a) = x(b) = 0, 1 ≤ α ≤ 2. If Q = q on [a, b], then 2Kα1/α

b−a 2

1/α 

b

1/α ∗

α∗

|Q(t)| dt

≥ 1,

(1.109)

a

where Kα is defined as in (1.97). For α = 1, (1.109) is strict. Proof As in the proof of Theorem 1.60, multiplying (1.1) by x(t) and integrating by parts yields 

b



  2 x (t) dt =

a

b



b

q(t)[x(t)]2 dt = −2

a

Q(t)x(t)x  (t)dt.

(1.110)

a

By an application of Hölder’s inequality and (1.98) to (1.110), we get 

b

  2 x (t) dt ≤ 2



a

b

1/α ∗ 

α∗

|Q(t)| dt

a

  x(t)x  (t)α dt

1/α

a

≤

b

2Kα1/α

b−a 2

1/α 

b

α∗

1/α ∗ 

|Q(t)| dt

a

b

  2 x (t) dt,

a

from which (1.109) follows. For α = 1, (1.109) is strict since a solution of (1.1) cannot be linear on each of the intervals [a, (a + b)/2], [(a + b)/2, b] as this implies discontinuity of x  .  We examine (1.109) further in the cases α = 1, 2. For α = 1, we take 

t

Q(t) =

q(s)ds + μ.

a

Then, since K1 = 1/2, we have

b−a 2



 t    max  q(s)ds + μ > 1. a≤t≤b

(1.111)

a

We now minimize (1.111) over μ. This is seen to occur with μ = −(M + m)/2, where

1.7 Disconjugacy and Disfocality

41

 M = max



t

a≤t≤b a

m = min

q(s)ds,

t

a≤t≤b a

q(s)ds

so that  t 

  1  inf max  q(s)ds + μ = (M − m). μ a≤t≤b a 2

We have therefore the following corollary of Theorem 1.63. Corollary 1.64 If x is a nontrivial solution of (1.1) satisfying x(a) = x(b) = 0, then there exist t1 , t2 ∈ [a, b] such that  t   2  4  q(t)dt  > . (1.112)  b−a t1 For α = 1, we square (1.109) to obtain 8 (b − a) π2



b



a

t

2 q(s)ds + μ dt ≥ 1.

a

Again we minimize with respect to μ and find that 1 μ=− b−a



b





t

q(s)ds dt. a

a

Using this value of μ yields another corollary. Corollary 1.65 If x is a nontrivial solution of (1.1) satisfying x(a) = x(b) = 0, then π2 ≤ 8(b − a)



b



2

t

dt −

q(s)ds a

 =

a b 

2

t

q(s)ds a

a

1 b−a

1 dt − b−a



b

a





t

2 q(s)ds dt

a b

2 (b − t)q(t)dt

.

a

Example 1.66 Consider the equation x  + λ cos(kt)x = 0

on

[0, B],

where

k > 0.

(1.113)

Let x be a nontrivial solution of (1.113) satisfying x(a) = x(b) = 0. By (1.112),  t   2  1 4 2|λ| <  λ cos(kt)dt  = |λ(sin(kt2 ) − sin(kt1 ))| ≤ b−a k k t1

42

1 Second-Order Linear Differential Equations

Table 1.1 The first positive zero

k λ=1 λ=5

1 7.37 1.77

5 22.25 4.33

10 44.4 8.88

20 88.85 17.77

so that 2k < |λ|(b − a) ≤ |λ|B.

(1.114)

If we view (1.113) as an eigenvalue problem with indefinite weight function and boundary conditions x(0) = x(π ) = 0, then (1.114) says that a real eigenvalue λ has |λ| large if k is large. More generally, it is an inequality which gives a lower bound for the spacing of zeros in the form b − a > 2k/|λ|. With λ ∈ {1, 5} and k ∈ {1, 5, 10, 20}, we can compute the first positive zero b of the solution of (1.113) satisfying y(0) = 0, y  (0) = 1. The results (rounded to 2 decimal places) are given in Table 1.1. Thus, (1.114) gives a lower bound for b which for large k is consistently slightly less than 50% of the true value. However, this is still much better than the estimates for (1.113) obtainable by the classical Lyapunov inequality which for λ = 1, a = 0 is 

b

[cos(kt)]+ dt >

0

4 . b

This inequality gives a lower bound for the spacing of zeros only of order than k as in (1.114).

√ k rather

Theorem 1.63 also allows for counting the number of zeros of a solution of (1.1). If a solution x of (1.1) has consecutive zeros a = a0 < a1 < · · · < an = b, then (1.109) yields that for Q = q, we have  1/α 21−1/α Kα1/α aj − aj −1



aj

1/α ∗ α∗

|Q(t)| dt

≥ 1.

(1.115)

aj −1

If we sum (1.115) and apply Hölder’s inequality, then we obtain n + 1 = number of zeros of x in [a, b]  b

1/α ∗ 1−1/α 1/α 1/α α∗ ≤2 Kα (b − a) |Q(t)| dt + 1. a

Example 1.67 Here, we show how Theorem 1.63 may be applied to yield a lower bound for the first eigenvalue λ0 of the equation

1.7 Disconjugacy and Disfocality

43



−x  + q(t)x = λx,

(1.116)

x(a) = x(b) = 0.

Let x be the eigenfunction of (1.116) corresponding to λ0 . Now, choose μ < λ0 and let Q = q − μ. Proceeding as in the proof of Theorem 1.63, we get 

b

(λ0 − μ)

 [x(t)]2 dt =

a

b

  2 x (t) dt +



a

 =

b

b

  2 x (t) dt − 2



a

 ≥

Q (t)[x(t)]2 dt

a

b

Q(t)x(t)x  (t)dt

a b

  2 x (t) dt

a



b

−2

1/α ∗ 

α∗

b

|Q(t)| dt

a

  x(t)x  (t)α dt

1/α .

a

Since the first eigenvalue of 

−x  = λx

[a, b],

on

(1.117)

x(a) = x(b) = 0 is π 2 /(b − a)2 , we have the Wirtinger-type inequality 

b

[x(t)]2 dt ≤ (b − a)2 π −2

a



b

  2 x (t) dt.

(1.118)

a

Using now (1.98) and (1.118) in (1.117), we obtain (λ0 − μ)(b − a)2 ≥ 1 − 2Kα1/α π2



b−a 2

1/α 

b

α∗

|Q(t)| dt

1/α ∗ ,

a

where 1 ≤ α ≤ 2 and Kα is given by (1.97). Thus, a lower bound for λ0 is obtained. The results given here may be applied to the equation   r(t)x  + q(t)x = 0 on

[a, b]

(1.119)

by a change of variable. If x is a solution of (1.119) satisfying x(a) = x(b) = 0, then let y(s) = x(t), where 

t

s= a

dτ . r(τ )

44

1 Second-Order Linear Differential Equations

Then, y(s) satisfies d2 y(s) + q(s)y(s) ˜ = 0, ds 2

y(0) = y(s0 ) = 0,

b where q(s) ˜ = r(t)q(t) and s0 = a dτ/r(τ ). Then, Corollary 1.64 says that if x is a nontrivial solution of (1.119) satisfying x(a) = x(b) = 0, then there exist t1 , t2 such that    

t2 t1

    q(t)dt  > 4

b a

dt r(t)

−1 .

1.8 Eigenvalues of Sturm–Liouville Problems In 1998, Ha [146] considered the eigenvalue problem 

x  + λx + q(t)x = 0 x(0) = x(π ) = 0,

on

(0, π ), (1.120)

where q ∈ L1 (0, π ). It follows from the classical Lyapunov inequality that if λ = 0 is an eigenvalue of (1.120), then necessarily qL1 >

π . 4

(1.121)

Over the years, there have appeared a number of improvements and extensions of this interesting result (see, e.g., [78, 134]). The purpose of this section is to extend the original Lyapunov inequality in yet another direction. The problem is considered when λ is any nonnegative eigenvalue of (1.120). It is assumed throughout that q ∈ L1 (0, π ), qL1 > 0, and either q ≥ 0 or q ≤ 0 a.e. on (0, π ). In this section, some lower bounds for qL1 are obtained in terms of the nonnegative eigenvalues of (1.120). We also derive some upper bounds for qL1 which ensure an eigenvalue λ ∈ [0, 1] of (1.120) to be the smallest eigenvalue. For the proofs, some basic properties of some boundary value problems and their Green functions are used (see, e.g., [263]). It was observed in [219] that (1.121) could be proved using the Green’s function method. We refer to [243, Chapter 2] for other interesting results relating to the eigenvalue problem for (1.120). We are motivated by a study on the solvability of some nonlinear boundary value problems (see [147, 148]). Let n ∈ N and consider the cases λ = n2 and λ = n2 separately. Suppose first that λ = n2 . We recall that for any h ∈ L1 (0, π ), the boundary value problem

1.8 Eigenvalues of Sturm–Liouville Problems



45

z + λz + h(t) = 0 on

(0, π ), (1.122)

z(0) = z(π ) = 0 has a unique solution z. More precisely, if λ > 0, then 

π

z(t) =

G (t, ξ, λ)h(ξ )dξ,

(1.123)

0

where √  ⎧ √  ⎪ λξ sin λ(π − t) sin ⎨ 1 G (t, ξ, λ) = √ √ √  √  λ sin( λπ ) ⎪ ⎩sin λt sin λ(π − ξ )

if 0 ≤ ξ ≤ t, if t ≤ ξ ≤ π

is Green’s function for (1.122). It follows by a simple calculation that ⎧ √  ⎪ λπ ⎪ ⎪tan 1 ⎨ 2 |G (t, ξ, λ)| < √   2 λ⎪ ⎪ √ ⎪ ⎩2 csc( λπ )

if 0 < λ < 1, (1.124) if λ > 1

holds a.e. on the square 0 ≤ ξ , t ≤ π . Theorem 1.68 If λ > 0 is an eigenvalue of (1.120), λ = n2 , then

qL1

√  ⎧ ⎪ λπ ⎪ ⎪ √ ⎨2 cot 2 > λ ⎪   ⎪ √ ⎪ ⎩sin( λπ )

if

0 < λ < 1, (1.125)

if

λ > 1.

Proof We assume that q ≥ 0 a.e. on (0, π ). The other case can be proved similarly. Let z be an eigenfunction of (1.120) corresponding to the eigenvalue λ. Since z is a nontrivial solution of (1.122) with h = qz, it follows from (1.123) that 

π

z(t) =

G (t, ξ, λ)q(ξ )z(ξ )dξ

0

holds. Taking the inner product in L2 (0, π ) of (1.126) with qz, we have 

π



π

q(t)|z(t)|2 dt =

0



0

>

1 m(λ)

π

G (t, ξ, λ)q(ξ )z(ξ )q(t)z(t)dξ dt

0



2

π

q(t)z(t)dt 0

,

(1.126)

46

1 Second-Order Linear Differential Equations

where m(λ) is the function defined by the right-hand side of (1.125), which is reciprocal to that of the right-hand side of (1.124). Hence, 

π



1/2 

π

q(t)z(t)dt ≤

q(t)|z(t)| dt

0

0

1 2n. Proof We assume that q ≥ 0 a.e. on (0, π ). The other case can be proved similarly. Let z be an eigenfunction of (1.120) corresponding to the eigenvalue λ = n2 . Since z is a nontrivial solution of (1.128) with h = qz, it follows from the remark above that  π  π 2 z(t) = sin(nt) sin(nξ )dξ + G (t, ξ, n2 )q(ξ )z(ξ )dξ (1.129) π 0 0 holds. Moreover, 

π

q(t)z(t) sin(nt)dt = 0,

(1.130)

0

and so taking the inner product in L2 (0, π ) of (1.129) with qz, we have 

π



0



π

q(t)|z(t)|2 dt = 0

1 = n

π

G (t, ξ, n2 )q(ξ )z(ξ )q(t)z(t)dξ dt

0



π



t

sin(nξ ) cos(nt)q(ξ )z(ξ )q(t)z(t)dξ dt 0

+

1 n

0 π





π

cos(nξ ) sin(nt)q(ξ )z(ξ )q(t)z(t)dξ dt. 0

t

It follows again by (1.130) that  0

π



t 0



π

+

π



cos(nξ ) sin(nt)q(ξ )z(ξ )q(t)z(t)dξ dt 0

t



π

= 0



π 0

cos(nξ ) sin(nt)q(ξ )z(ξ )q(t)z(t)dξ dt = 0

48

1 Second-Order Linear Differential Equations

and hence 

π

q(t)|z(t)|2 dt =

0


0 is essential for Theorem 1.69 to hold. Clearly, z(t) = sin(nt) is an eigenfunction of (1.120) corresponding to n2 when q = 0 a.e. on (0, π ). It is also possible to obtain some upper bounds of qL1 which are sufficient for an eigenvalue λ ∈ [0, 1] of (1.120) to be the smallest eigenvalue. Recall that if λ is the smallest eigenvalue of (1.120), then the eigenfunction z corresponding to λ has no zero in (0, π ). This is the property the proof is based on. Theorem 1.70 Let λ ∈ [0, 1]. If λ is an eigenvalue of (1.120) and

qL1

⎧ 16 ⎪ ⎪ ⎪ ⎨π

√  ≤ √ ⎪ λπ ⎪ ⎪ ⎩4 λ cot 4

if

λ = 0,

if

0 < λ ≤ 1,

(1.131)

then λ is the smallest eigenvalue. Proof Let z be an eigenfunction of (1.120) corresponding to λ. We suppose on the contrary that z has a zero  ∈ (0, π ) and define



t (π − )t , x2 (t) = z + . x1 (t) = z π π Then, x1 and x2 are nontrivial solutions of (1.120) with λ replaced by 2  λ π

and

1−

 π

2 λ

1.9 An Inequality of Nehari

49

and q(t) replaced by 2

 t q π π

and

 1− π

2

(π − )t + , q π

respectively. If 0 < λ ≤ 1, then we apply Theorem 1.68 to obtain  0



 √  √  λ |q(t)|dt > 2 λ cot , 2



π



 √  √ (π − ) λ |q(t)|dt > 2 λ cot . 2

It follows by a simple calculation that 

π 0

  √   √  √  λ (π − ) λ |q(t)|dt > 2 λ cot + cot 2 2 √  √ λπ ≥ 4 λ cot , 4

contradicting (1.131). If λ = 0, then we apply (1.121) to obtain 

π





|q(t)|dt =

0

 |q(t)|dt +

0

|q(t)|dt





1 1 + >4  π − ≥

π



16 , π

again contradicting (1.131). This completes the proof.



1.9 An Inequality of Nehari Nehari [221, Theorem I] claims that if [a, b] contains n zeros of a nontrivial solution of x (n) + qn (t)x (n−1) + · · · + q1 (t)x = 0, then n

k=1

 2 (b − a) n

n−k a

b

|qk (t)| dt > 2n+1 .

(1.132)

50

1 Second-Order Linear Differential Equations

In a private communication with one of the authors of [134], Nehari has indicated that (1.132) is undecided since the argument given in [221] that [221, Theorem II] implies [221, Theorem I] is incomplete. It is the purpose of this section to show that (1.132) is correct for n = 2. In fact, Fink and Mary [134] proved a stronger result for the equation x  + p(t)x  + q(t)x = 0.

(1.133)

If a and b are consecutive zeros, then there exists c ∈ (a, b) such that x  (c) = 0. Nehari showed  c  c |q(t)|dt + |p(t)|dt > 1, (1.134) (c − a) a

a

and a similar inequality for the interval (c, b). The trick is to combine the two to get (1.139). Fink and Mary [134] started with an inequality which is stronger than (1.134). They considered the equation   r(t)x  + q(t)x ˆ =0 (1.135) with integrable functions r and qˆ such that r > 0. Lemma 1.71 (See [262]) Let x be a nontrivial solution of (1.135). If x(a) = 0 and x  (c) = 0, a < c, then 

c

a

dt r(t)



c

qˆ + (t)dt

> 1.

a

Proof Let |x(c)| = max |x(t)|. Then, 

c

[x(c)]2 ≤

x  (t)dt

2

a

 =

c

  1/r(t) r(t)x  (t)dt

c

dt r(t)

c

dt r(t)

a


1.

(1.136)

a

Writing (1.136) as a double integral, we see 

c





c

τ

exp a

a

 p(s)ds q + (τ )dτ dt
1

a

and thus  (c − a)

c

+





q (t)dt exp a

c

|p(t)|dt

> 1.

(1.137)

a

A similar result holds for the interval (c, b) when x(b) = 0. In order to motivate a later inequality, we observe that (1.137) implies (1.134) with |q(t)| replaced by q + (t). Indeed, letting 

c

A0 =

 |p(t)|dt

and

a

A21 = (c − a)

c

q + (t)dt,

a

this claim is the statement that A21 > exp(−A0 ) implies A21 + A0 > 1. This follows from et + t ≥ 1 for all t ≥ 0. Now, define 

b

B0 = c

 |p(t)|dt

and

B12 = (b − c)

b

q + (t)dt.

c

Then, we have B12 > exp(−B0 ). Now, the inequality that is related to (1.139) as (1.137) is to (1.134), is gotten from 4e−t/2 ≥ −2t + 4, t ≥ 0.

52

1 Second-Order Linear Differential Equations

Theorem 1.72 Let a and b be consecutive zeros of a nontrivial solution of (1.133), where p and q are integrable. Then, 



 1 b q + (t)dt − 4 exp − |p(t)|dt > 0 2 a

b

(b − a) a

(1.138)

and 

b

(b − a)

q + (t)dt + 2



a

b

|p(t)|dt > 4.

(1.139)

a

Proof By elementary calculus, we have 

b

q + (t)dt =

a

A21 B12 (A1 + B1 )2 + ≥ . c−a b−c b−a

In fact, the right-hand side is the minimum of the middle as function of c. Thus, the left-hand side of (1.138) is greater than

A0 + B0 (A1 + B1 ) − 4 exp − 2 2

A0 + B0 = A21 + B12 + 2A1 B1 − 4 exp − 2

A0 + B0 > exp(−A0 ) + exp(−B0 ) − 2 exp − 2 



2 A0 B0 = exp − + exp − 2 2 ≥ 0.

This proves (1.138). Inequality (1.139) follows from 4e−t/2 + 2t − 4 ≥ 0 for all b t ≥ 0, where t is replaced by a |p(t)|dt.  Remark 1.73 We remark that both (1.137) and (1.138) are more enlightening than (1.134) and (1.139). In particular, they show that 

b

(b − a)

q + (t)dt

a

cannot be small unless  a

b

|p(t)|dt

1.10 Notes and References

53

is very large. In fact, as 

b

(b − a)

+

q (t)dt → 0,

a



b

we have

|p(t)|dt → ∞.

a

This does not follow from (1.139). Finally, (1.138) is sharp since it reduces to Lyapunov’s inequality when p(t) = 0, and this is known to be sharp, see [206].

1.10 Notes and References In Sect. 1.2, we gave a survey of the most basic results on Lyapunov-type inequalities, adopted from the works of several authors, including Chen [81], Cheng [83], Eliason [123, 124], Hartman [155], Hochstadt [172], Kwong [193], Nehari [220, 221], Reid [248, 249], and Singh [258], who gave extensions of the celebrated paper of the Russian mathematician Lyapunov [202]. It was pointed out by Cheng [91] that Lyapunov neither stated nor proved Theorem 1.1, but rather in [202], he only claims the inequality (1.8) in Theorem 1.2 and its direct conclusion, the stability theorem, i.e., Theorem 1.3. For original proofs of Theorems 1.2 and 1.3, the reader is referred to Lyapunov [202], Forsyth [135], and Ince [174]. Another proof of Theorem 1.3 has been given by Zhukovski˘ı [310], who also proved that (1.8) is sharp. The sharpness of (1.8) was also demonstrated by Kampen and Wintner [182] and Borg [59]. Improvements and extensions of Theorem 1.3 were later given by many authors. For additional information, the reader is referred to the report by Cesari [78]. Note that Lyapunov’s [202] proof requires more calculations. Venturing to obtain alternative proofs, Borg [58, 59] quoted and proved a handy inequality in [59] due to Beurling [43]. In 1949, Hartman and Wintner [156] noticed the result of Beurling [43] quoted by Borg in his earlier paper [58]. As it was first noticed by Wintner [290] and subsequently by several other authors, an application of Sturm’s comparison theorem allows the replacement of |q(t)| in (1.2) by q + (t). In this remarkable result, Wintner introduced not only an alternative proof of Theorem 1.4, but also a sharper result than it. By Theorem 1.5, Wintner [290] also proved that the constant 4 in (1.4) cannot be replaced by a larger number. Theorem 1.7 in Sect. 1.3 seems to be the first proof of Lyapunov’s inequality to appear in the literature, which was given by Borg [59] following the result of Beurling [43]. Under the assumption that q ≥ 0, Hartman and Wintner [157] gave a proof of (1.4), which is based on the fact that the graph of the positive solution x of (1.12) on (a, b) is concave down. Under the same assumption on q, Nehari [219, 221] and Reid proved Hartman’s inequality by employing the Green function method and a variational lemma [250, pp. 224–230], a so-called Wirtinger-type inequality, respectively, see also [39, 96, 97, 193]. In the 1950s, Courant and Hilbert [99] and Kre˘ın [192] proved (1.7) by using the minimax principle for eigenvalues

54

1 Second-Order Linear Differential Equations

of Sturm–Liouville problems. Theorem 1.9 can be counted as one of the most interesting proofs of Lyapunov’s inequality (see Leighton [198]), which probably is the only geometric proof of it. All results of Sect. 1.4 are taken from [155, pp. 344–350]. In this section, the proofs of Corollaries 1.13, 1.14, 1.18, 1.20, and 1.21 are presented as [155, Exercise 5.2, p. 346], [155, Exercise 5.3, p. 346], [155, Exercise 5.5, p. 348], [155, Exercise 5.6, p. 348], and [155, Exercise 5.7, p. 349] in Hartman’s book, respectively. In addition to these, Corollary 1.23 is presented as [155, Exercise 5.8, p. 350], Corollary 1.24 and Remark 1.25 are presented as [155, Exercise 5.9 (a), p. 350], Corollary 1.26 and Remark 1.27 are presented as [155, Exercise 5.9 (b), p. 350], and Remark 1.28 is presented as [155, Exercise 5.9 (c), p. 350]. The results of Sects. 1.5 and 1.6 (until the end of Corollary 1.36) are taken from [231]. The remaining part of Sect. 1.6 is adopted from [157]. Next, Sect. 1.7 is adopted from [193] and [64]. All results of Sect. 1.8 are taken from [146]. Finally, the last section of Chap. 1 is taken from the paper of Fink and Mary [134], and its contents is an elementary proof of (1.138) and the observation that (1.139) follows from (1.138).

Chapter 2

Lyapunov-Type Inequalities for Higher-Order Linear Differential Equations

2.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for higher-order linear differential equations and sketch some recent developments related to this type of inequalities. In Sect. 2.2, we deal with Hartman-type and Lyapunov-type inequalities for even-order linear differential equations satisfying Dirichlet (2-point) boundary conditions, and we also present some developments for Lidstone boundary value problems. In particular, we show how the classical Hartman and Lyapunov inequalities for second-order linear differential equations can be extended to even-order ones by using Green’s function as a tool. In Sect. 2.3, we present related results for oddorder linear differential equations satisfying some different boundary conditions. Section 2.4 deals with Hartman-type and Lyapunov-type inequalities for higherorder linear differential equations, some of which generalize the classical results of Hartman and Lyapunov. Particularly, we consider higher-order linear differential equations satisfying n-point boundary value problems and (k, n − k) conjugate boundary value problems in Sect. 2.4.1 and in Sect. 2.4.2, respectively.

2.2 Even-Order Differential Equations In this section, we consider the even-order differential equation x (2n) + (−1)n−1 q(t)x = 0,

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_2

(2.1)

55

56

2 Higher-Order Linear Differential Equations

where n ∈ N and q is a locally Lebesgue integrable real-valued function defined on R. First, we study (2.1) satisfying Dirichlet (2-point) boundary conditions, i.e., 

x (2n) + (−1)n−1 q(t)x = 0

on

[a, b],

x (k) (a) = x (k) (b) = 0 for

k = 0, 1, . . . , n − 1.

(2.2)

Next, we consider (2.1) satisfying Lidstone boundary conditions, i.e.,  (2n) x + (−1)n−1 q(t)x = 0 on x (2k) (a) = x (2k) (b) = 0 for

[a, b], k = 0, 1, . . . , n − 1.

(2.3)

2.2.1 Dirichlet Boundary Value Problems It appears that the first generalization of Hartman’s result was obtained by Das and Vatsala [105, Theorem 3.1] for (2.2). They obtained a necessary condition for a solution of (2.1) to satisfy the boundary conditions in (2.2) with a and b replaced by α and β, respectively, where a ≤ α < β ≤ b. For special classes of q, namely linear, convex, and concave functions, the results are similar to those of Bradley [61] for the fourth-order case. There is a natural restriction to be imposed on q in view of the following. If for a nontrivial solution x of (2.1), ϕx (t) =

n−1

(−1)n x (k) (t)x (2n−k−1) (t), k=0

then it is easy to check that ϕx (t) vanishes at t0 whenever x is a solution of (2.1) which together with its first n − 1 derivatives vanishes at t0 . Noting that ϕx (t)

 = (−1)

n−1

x

(n)

(t)

!2

 − q(t)[x(t)]

2

,

we conclude that ϕx (t) is a monotone function of x if q < 0, and therefore a nontrivial solution x of (2.1) can have at most one nth-order zero. Hence, we assume that q(t) is positive somewhere in [a, b]. The following identity is needed. Lemma 2.1 Let k ∈ N. Then,

2.2 Even-Order Differential Equations



(b − s)(t − a) b−a

57



k

j k−1 k−1+j k−1−j (b − t)(s − a) (s − t) j b−a j =0

− (−1)k−1 (t − s)2k−1 =

(b − t)(s − a) b−a

k k−1 j =0



j k−1+j k−1−j (b − s)(t − a) . (t − s) j b−a (2.4)

Proof Equation (2.4) is immediate for k = 1. Assume now that it is true for k = m ∈ N. Then, in view of the identity (b − s)(t − a) (b − t)(s − a) =t −s+ , b−a b−a the left-hand side of (2.4) for k = m + 1 can be written as ⎧



j +1 m (b − s)(t − a) m ⎨ m + j m−j (b − t)(s − a) (s − t) ⎩ j b−a b−a j =0

⎫

j ⎬

m

m+j (b − t)(s − a) (s − t)m+1−j − − (−1)m (t − s)2m+1 ⎭ b−a j j =0

=

(b − s)(t − a) b−a

m 

2m (b − t)(s − a) m+1 m b−a

⎫



j ⎬ m

m−1+j (b − t)(s − a) (s − t)m+1−j − − (−1)m (t − s)2m+1 ⎭ j b−a j =0



2m (b − t)(s − a) m+1 (b − s)(t − a) m = m b−a b−a



2m − 1 (b − t)(s − a) m − (s − t) b−a m

m m−1

j

m − 1 + j

(b − t)(s − a) m+1−j (b − s)(t − a) (t − s) − j b−a b−a

j =0

(2.5) using the induction hypothesis. Again writing

m−1+j j

=





m+j m−1+j − for j j −1

j = 1, . . . , m − 1

58

2 Higher-Order Linear Differential Equations

and rearranging in (2.5), we get

(b − t)(s − a) b−a

m+1 

2m (b − s)(t − a) m m b−a +

⎫



j ⎬ m+j (b − s)(t − a) (t − s)m−j , ⎭ j b−a

m−1

j =0

which is equal to

(b − t)(s − a) b−a

m+1 +



m

m+j (b − s)(t − a) j (t − s)m−j . j b−a j =0



This completes the proof.

Now, we consider the equation −x (2n) = 0 subject to the 2-point boundary conditions in (2.2), i.e., 

−x (2n) = 0 on

[a, b],

x (k) (a) = x (k) (b) = 0 for

k = 0, 1, . . . , n − 1.

(2.6)

Theorem 2.2 The Green function G2n (t, s) for (2.6) is given by (−1)n−1 G2n (t, s) = (2n − 1)! ×



(t − a)(b − s) b−a

(b − t)(s − a) b−a



n n−1 n−1+j (s − t)n−1−j j j =0

j for

(2.7)

t ≤s≤b

and (−1)n−1 G2n (t, s) = (2n − 1)! ×



(s − a)(b − t) (b − a)

(t − a)(b − s) (b − a)



n n−1 n−1+j (t − s)n−1−j j j =0

j for

(2.8)

a ≤ s ≤ t.

Proof It is easy to show that G2n (t, s) satisfies the following three conditions given by Beesack [40, p. 801].

2.2 Even-Order Differential Equations

59

(i) ∂j G2n (t, s) ∂t j

for j = 0, 1, . . . , n − 2

are continuous functions of (t, s) on the square a ≤ t, s ≤ b, while ∂ n−1 G2n (t, s) ∂t n−1 is a continuous function of (t, s) in each of the two triangles a ≤ t ≤ s ≤ b and a ≤ s ≤ t ≤ b with ∂ n−1 ∂ n−1 + G (s , s) − G2n (s − , s) ≡ −1, 2n ∂t n−1 ∂t n−1

a < s < b.

(ii) ∂n G2n (t, s) ≡ 0 ∂t n in each of the two triangles given in (i). (iii) For each s, with a < s < b, G2n (t, s) satisfies (2.6).



Definition 2.3 Two distinct values a and b on an interval I are said to be conjugate with respect to (2.1) if there exists a nontrivial solution x of (2.2). Theorem 2.4 (Hartman-Type Inequality) If there exists a pair of conjugate points on [a, b] with respect to (2.1), then the inequality 

b

(b − t)2n−1 (t − a)2n−1 q + (t)dt ≥ (2n − 1)[(n − 1)!]2 (b − a)2n−1

(2.9)

a

holds. Proof Consider the auxiliary equation x (2n) + (−1)n−1 q + (t)x = 0.

(2.10)

By Hinton [171, Corollary 2.1], (2.10) has a pair of conjugate points on [a, b], say a1 , b1 . Now, consider the eigenvalue problem 

z(2n) + (−1)n−1 λq + (t)z = 0, z(k) (a1 ) = z(k) (b1 ) = 0

for k = 0, 1, . . . , n − 1.

(2.11)

60

2 Higher-Order Linear Differential Equations

Problem (2.11) is equivalent to the integral equation 

b

u(t) = λ

K (t, s)u(s)ds,

a

where  K (t, s) = (−1)n−1 G2n (t, s) q + (t)q + (s) and  u(t) = z(t) q + (t), and where G2n (t, s) is Green’s function for (2.6) with a = a1 , b = b1 . Since the kth eigenvalue λk of (2.11) is the minimum of 

b1

w

(n)

(t)

!2



b1

dt

+

−1 2

q (t)w (t)dt

a1

,

a1

where w is n times differentiable on [a1 , b1 ] and satisfies the boundary conditions in (2.11) as well as the orthogonality conditions related to Courant’s [99] wellknown max/min principle, K (t, s) is nonnegative definite. Thus, we may appeal to Mercer’s result [99]. This leads to (note that λ1 = 1)  b1 ∞

1 |G2n (t, t)| q + (t)dt ≥ 1. = λk a1

(2.12)

k=1

In view of Theorem 2.2 and [a1 , b1 ] ⊆ [a, b], (2.12) yields (2.9).



Remark 2.5 If in (2.1), q is positive, then the eigenvalues of (2.11) are all positive, and thus the kernel K (t, s) is such that 0 is not an eigenvalue of  μu(t) =

b1

K (t, s)u(s)ds.

a1

Moreover, each eigenvalue λ (different from 0) is of finite multiplicity [251], since K (t, s) defines a completely continuous symmetric operator. Thus, in place of (2.12), we have  b1 |G2n (t, t)| q(t)dt > 1, a1

and hence (2.9) holds with strict inequality. Note that (2.9) reduces to the classical Hartman inequality when n = 1.

2.2 Even-Order Differential Equations

61

Remark 2.6 (Lyapunov-Type Inequality) The inequality  b 42n−1 (2n − 1)[(n − 1)!]2 q + (t)dt > , (b − a)2n−1 a

(2.13)

due to Levin [200], follows from (2.9) in view of the arithmetic-geometric mean inequality, i.e., (1.23) for t ∈ [a, b]. Now, we consider (2.1) when q is either a linear, convex, or concave function. Proposition 2.7 Let q(t) = αt + β in (2.1) be positive on [a, b]. If there exists a pair of conjugate points with respect to (2.1) on [a, b], then α

a+b 2

+β >

2n−1 2n (n − 1)! % (2j + 1). (b − a)2n

(2.14)

j =n−1

Proof In this special case, in view of the strict inequality in (2.9), we have  b (b − t)2n−1 (t − a)2n−1 (αt + β)dt > (2n − 1)[(n − 1)!]2 (b − a)2n−1 .

(2.15)

a

The value of the integral on the left-hand side of (2.15) is

[(2n − 1)!]2 α(a + b) +β (b − a)4n−1 . 2 (4n − 1)! 

Inequality (2.14) now follows by simplification. Remark 2.8 If q(a) = αa + β > 0 and q(b) = αb + β < 0, then α

β +a α

⎧

2n+1 ⎨2n−1

⎩

j =0

j! (2n + j + 1)!



β +a α

j (b − a)2n−1−j

⎫ n ⎬% ⎭

(2n − j )

j =2

≥ (n − 1)!(b − a)2n−1 .

(2.16)

Proposition 2.9 Let [a, b] have a pair of conjugate points with respect to (2.1), where q is continuous, positive, and convex. Then, q(a) + q(b) >

2n−1 2n+1 (n − 1)! % (2j + 1). (b − a)2n

(2.17)

j =n−1

If, however, q(a) > 0 and q(b) < 0, then  j % n 2n−1

[q(a)]2n q(a) j! (n − 1)! (2n − j ) ≥ . (2n + j + 1)! q(b) − q(a) [q(b) − q(a)]2n (b − a)2n+1 j =0

j =2

(2.18)

62

2 Higher-Order Linear Differential Equations

Proof Consider the auxiliary equation y (2n) + (−1)n−1 q(t)y = 0,

(2.19)

where q(t) = αt + β with α=

q(b) − q(a) b−a

and

β=

bq(a) − aq(b) . b−a

Then, by Hinton [171, Corollary 2.1], [a, b] has a pair of conjugate points with respect to (2.19). Hence, (2.17) and (2.18) follow from (2.14) and (2.16) for appropriate values of α and β.  Proposition 2.10 Let q > 0 in (2.1) be concave on [a, b]. If [a, b] has a pair of conjugate points with respect to (2.1), then q

a+b 2

>

2n−1 2n (n − 1)! % (2j + 1). (b − a)2n

(2.20)

j =n−1

Proof Consider (2.19) with α, β such that the graph of p is a supporting line for the graph of q. As in Proposition 2.9, we get p(a) + p(b) >

2n−1 2n+1 (n − 1)! % (2j + 1). (b − a)2n

(2.21)

j =n−1

Since (2.21) is true for any supporting line, it is also true for that supporting line which is the graph of p for which p(a) + p(b) is a minimum. Then, for the set L of all linear functions p, we have p(t) = (t − t1 )q  (t1 ) + q(t1 ) for some t1 ∈ [a, b]. Thus, 

min {p(a) + p(b)} = min {(a + b − 2t1 )q (t1 ) + 2q(t1 )} = 2q t1 ∈[a,b]

q∈L



a+b . 2 

This implies (2.20).

Remark 2.11 On an account of the alternating term (−1)n−1 in Green’s function G2n (t, s) for (2.6) defined in (2.7) and (2.8), Hartman-type and Lyapunov-type inequalities for the 2-point boundary value problem 

x (2n) + q(t)x = 0

on

[a, b],

x (k) (a) = x (k) (b) = 0 for

k = 0, 1, . . . , n − 1

(2.22)

2.2 Even-Order Differential Equations

63

can be obtained by replacing the function q + by |q| in (2.9) and (2.13), respectively. Corollary 2.12 If x is a nontrivial solution of (2.22), where a, b ∈ R with a < b are consecutive zeros, then the Hartman-type inequality 

b

(b − t)2n−1 (t − a)2n−1 |q(t)| dt > (2n − 1)[(n − 1)!]2 (b − a)2n−1

(2.23)

a

and the Lyapunov-type inequality 

b

|q(t)| dt >

a

42n−1 (2n − 1)[(n − 1)!]2 (b − a)2n−1

(2.24)

hold. Remark 2.13 When n = 1, (2.9), (2.23) and (2.13), (2.24) reduce to the classical Hartman and Lyapunov inequalities for (1.9) satisfying (1.5), respectively. In 2010, Yang and Lo [299] considered the even-order differential equation of the form (r2n−1 (t)(r2n−2 (t)(· · · (r2 (t)(r1 (t)x  ) ) · · · ) ) ) + q(t)x = 0,

(2.25)

where rk ∈ C2n−k ([a, b], (0, ∞)) for all k = 1, 2, . . . , 2n − 1 and q ∈ C([a, b], R). If we define functions xk for all k = 1, 2, . . . , 2n − 1 by x1 = r1 x  ,

x2 = r2 x1 ,

...,

 x2n−1 = r2n−1 x2n−2 ,

then (2.25) is equivalent to the first-order differential system ⎧ ⎪ x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪x1 ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ x2n−2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩  x2n−1

x1 , r1 (t) x2 , = r2 (t) =

.. . =

(2.26) x2n−1 , r2n−1 (t)

= −q(t)x.

We need the following auxiliary result [299, Lemma 1]. Lemma 2.14 Assume n ≥ 3 and let x be a nontrivial solution of (2.25) satisfying the 2-point boundary conditions in (2.2), where a, b ∈ R with a < b are consecutive zeros. Let c ∈ (a, b) be such that

64

2 Higher-Order Linear Differential Equations

x  (c) = 0 and

|x(c)| = max |x(t)| > 0. t∈(a,b)

Let {xk (t)} be defined as above. Then, for each k ∈ {1, 2, . . . , n − 1}, xk (a) = j j xk (b) = 0, and there exist {tk } ⊂ (a, b), j = 1, 2, . . . , k + 1, with {tki } < {tk } if j > i and {tk1 } ∈ (a, c), {tkk+1 } ∈ (c, b). For k = n, n + 1, . . . , 2n − 1, xk has at j least 2n − k zeros {tk } ⊂ (a, b), j = 1, 2, . . . , 2n − k. Moreover, {tn1 } ∈ (a, c) and n {tn } ∈ (c, b). Theorem 2.15 (Lyapunov-Type Inequality) If (2.25) has a nontrivial solution satisfying the 2-point boundary conditions in (2.2), where a, b ∈ R with a < b are consecutive zeros, then we have for n = 1, 

b



b

|q(t)| dt > 4

a

a

dt r1 (t)

−1 (2.27)

,

for n = 2, 

b

|q(t)| dt >

a

1 min H2 (c), 2 c∈[a,b]

and for n ≥ 3,  a

b

1 |q(t)| dt > min Hn (c) 2 c∈[a,b]

 2n−1  %

b

k=n+2 a

dt rk (t)

−1 ,

where Hn (c) =

n+1  % k=1 a

c

dt rk (t)

−1 +

n+1  %

b

k=1 c

dt rk (t)

−1 .

(2.28)

Proof Let n = 1 in (2.26). Then, (2.25) is equivalent to the system ⎧ x ⎨x  = 1 , r1 (t) ⎩  x1 = −q(t)x. Since x(a) = x(b) = 0, we see that there exists c ∈ (a, b) such that x  (c) = 0 and

|x(c)| = max |x(t)| > 0. t∈(a,b)

(2.29)

2.2 Even-Order Differential Equations

65

This implies x1 (c) = 0. Integrating the first equation in (2.29) from a to c, we get  x(c) =

c

x1 (t) dt, r1 (t)

c

|x1 (t)| dt. r1 (t)

a

which implies  |x(c)| ≤ a

(2.30)

From x1 (c) = 0 and the second equation in (2.29), we get 

t

x1 (t) = c

x1 (s)ds = −



t

q(s)x(s)ds c

for t ∈ [a, c], which implies 

c

|x1 (t)| ≤ a

   x (t) dt = 1



c

 |q(t)x(t)| dt ≤ |x(c)|

a

c

|q(t)| dt.

(2.31)

a

From (2.30) and (2.31), and by using |x(c)| > 0, we obtain 

c

 |q(t)| dt ≥

a

c

a

dt r1 (t)

−1 (2.32)

,

and equality holds only if q = 0 on [a, c]. Similarly, we can obtain 

b

 |q(t)| dt ≥

c

c

b

dt r1 (t)

−1 (2.33)

,

and equality holds only if q = 0 on [c, b]. Since x is a nontrivial solution of (2.25), it is easy to see that at least one strict inequality holds either in (2.32) or in (2.33). Summation of (2.32) and (2.33) yields 

b a

 |q(t)| dt ≥ a

c

dt r1 (t)

−1



b

+ c

dt r1 (t)

−1



b

≥4 a

dt r1 (t)

−1 ,

which is equivalent to (2.27). The case n = 2 can be treated similarly. For n ≥ 3, by Lemma 2.14, for k = 1, 2, . . . , n − 1 and from xk (a) = 0, we get for t ∈ [a, c] and 

t

xk (t) = a

xk (s)ds

66

2 Higher-Order Linear Differential Equations

the estimates ⎧ ⎪ ⎪ |x(c)| ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ |x1 (t)| ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ |x2 (t)| ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩|xn−1 (t)|



|x1 (t)| dt, a r1 (t)   c    x (t) dt = ≤ 1 c

≤



a

a

.. .



c

≤

|x2 (t)| dt, r2 (t)

c

|x3 (t)| dt, r3 (t)

a

c

≤

c

a

   x (t) dt = 2



a

   x (t) dt = n−1



c a

(2.34)

|xn (t)| dt. rn (t)

j

By Lemma 2.14, xn has n zeros tn , j = 1, 2, . . . , n, in (a, b), and tn1 ∈ (a, c), tnn ∈ (c, b). For t ∈ [a, c], we have    t x (s)   c |x (t)| n+1 n+1   |xn (t)| ≤  ds  ≤ dt.  tn1 rn+1 (s)  a rn+1 (t)

(2.35)

Using (2.34) and (2.35), we obtain  |x(c)| ≤

n  % k=1 a

c

dt rk (t)

 a

c

|xn+1 (t)| dt. rn+1 (t)

(2.36)

1 Since by Lemma 2.14, xn+k+1 has at least one zero tn+k+1 in (a, b), k = 1, 2, . . . , n − 2, we obtain for t ∈ [a, c],   ⎧  t x (s)   b |x (t)| ⎪ n+2 n+2   ⎪ ⎪ |xn+1 (t)| =  ds  ≤ dt, ⎪ ⎪  t 1 rn+2 (s)  r (t) n+2 ⎪ a n+1 ⎪ ⎪ ⎪   ⎪ ⎪  t x (s)   b |x (t)| ⎪ ⎪ n+3 n+3   ⎪ ⎪ |xn+2 (t)| =  ds  ≤ dt, ⎪ ⎪ 1   r (s) r (t) ⎪ n+3 n+3 t a ⎪ n+2 ⎪ ⎪ ⎨ .. . ⎪ ⎪ ⎪    ⎪ ⎪   t x b |x ⎪ (s) ⎪ 2n−1 2n−1 (t)|   ⎪ ⎪|x2n−2 (t)| =  ds  ≤ dt, ⎪ ⎪ 1   r (s) r ⎪ 2n−1 (t) t2n−2 2n−1 a ⎪ ⎪ ⎪     ⎪ ⎪  b   ⎪ t b ⎪   ⎪ ⎪ |q(t)x(t)| dt < |x(c)| |q(t)| dt. q(s)x(s)ds  ≤ ⎩|x2n−1 (t)| = −  t1  a a 2n−1 (2.37)

2.2 Even-Order Differential Equations

67

Using (2.36) and (2.37), we obtain |x(c)| ≤ |x(c)|

n+1  %

c

k=1 a

dt rk (t)

  2n−1  %

b

k=n+2 a

dt rk (t)



b

|q(t)| dt.

(2.38)

a

Dividing both sides of (2.38) by |x(c)| > 0, we get n+1  %

1≤

c

k=1 a

dt rk (t)

  2n−1  %

b

k=n+2 a

dt rk (t)



b

|q(t)| dt.

(2.39)

|q(t)| dt.

(2.40)

a

Similarly, we can show n+1  %

1≤

k=1 c

b

dt rk (t)

  2n−1  %

b

k=n+2 a

dt rk (t)



b a

Since x is a nontrivial solution of (2.25), we see that at least one strict inequality either in (2.39) or in (2.40) holds, and from (2.39) and (2.40), we obtain 2

 2n−1  %

b

k=n+2 a

dt rk (t)



b

|q(t)| dt > Hn (c).

a

It is easy to see that Hn is positive and continuous on (a, b) and lim Hn (c) = lim Hn (c) = ∞,

c→a +

c→b−

and hence Hn (c) achieves its minimum in [a, b]. The proof is complete. C2n−1 ([a, b], (0, ∞))



Corollary 2.16 Assume rj = r ∈ for all j = 1, 2, . . . , n + 1 and q ∈ C([a, b], R). If (2.25) has a nontrivial solution satisfying (2.2), where a, b ∈ R with a < b are consecutive zeros, then we have for n = 2, 

b



b

|q(t)| dt > 8

a

a

dt r(t)

−3 ,

and for n ≥ 3,  a

b



b

|q(t)| dt > 2n+1 a

dt r(t)

−(n+1)  2n−1 %  k=n+2 a

b

dt rk (t)

−1 .

68

2 Higher-Order Linear Differential Equations

Proof From Theorem 2.15 and some simple calculations using  b  c  1 b dt dt dt = = , 2 a r(t) a r(t) c r(t) we have 

b

min Hn (c) = 2n+1

c∈[a,b]

a

dt r(t)

−(n+1) . 

This completes the proof.

Corollary 2.17 Let n ∈ N \ {1}. Assume rj (t) ≡ 1 for all j = 1, 2, . . . , n + 1, and q ∈ C([a, b], R). Then, (2.25) reduces to the equation (r2n−1 (t)(r2n−2 (t)(· · · (rn+2 (t)x  ) · · · ) ) ) + q(t)x = 0.

(2.41)

If (2.41) has a nontrivial solution satisfying (2.2), where a, b ∈ R with a < b are consecutive zeros, then we have for n = 2,  b n!2n+1 |q(t)| dt > , (b − a)n+1 a and for n ≥ 3,  a

b

 2n−1  %

n!2n+1 |q(t)| dt > (b − a)n+1

k=n+2 a

b

dt rk (t)

−1 .

Proof Since in this case, we have xk = x (k) , k = 1, 2, . . . , n + 1, it follows from x (k) (a) = 0, k = 0, 1, . . . , n − 1 that x(c) =

1 (n − 1)!



c

(c − t)n−1 x (n) (t)dt,

a

and for t ∈ [a, c],    t    |xn (t)| =  xn+1 (s)ds   tn1   c |xn+1 (s)| ds ≤ a

 ≤

a

c

 a

b



|xn+2 (t)| dt ds rn+2 (t) b

= (c − a) a

|xn+2 (t)| dt. rn+2 (t)

2.2 Even-Order Differential Equations

69

Moreover, by using     b  t  b  |xn+k+1 (t)|    x  (t) dt = |xn+k (t)| =  dt xn+k (s)ds  ≤ n+k 1  t  rn+k+1 (t) a a n+k and    t  |x2n−1 (t)| = −  t1

2n−1

   b  b  |q(t)x(t)| dt < |x(c)| |q(t)| dt q(s)x(s)ds  ≤  a a

for k = 2, 3, . . . , n − 2, we obtain |x(c)| ≤ = ≤ ≤

1 (n − 1)! (c − a)n n!



c

a



a

− a)n+1 n!

b

|xn+1 (t)| dt

a b

(c − a)n+1 n! (c

 (c − t)n−1 dt |xn+1 (t)| dt



b

a



b

a

|xn+2 (t)| dt rn+2 (t)  b |xn+3 (t)| dt dt rn+2 (t) a rn+3 (t)

(2.42)

.. . (c − a)n+1 < n!

 2n−1  % k=n+2 a

b

  b dt |q(t)|dt. |x(c)| rk (t) a

Dividing both sides of (2.42) by |x(c)| > 0, we obtain (c − a)n+1 1< n!

 2n−1  %

b

k=n+2 a

dt rk (t)



b

|q(t)|dt.

(2.43)

|q(t)|dt.

(2.44)

a

Similarly, we can show (b − c)n+1 1< n!

 2n−1  % k=n+2 a

b

dt rk (t)



b a

Using (2.43), (2.44), and the fact  min

c∈[a,b]

1 1 + n+1 (b − c) (c − a)n+1

 =

2n+2 , (b − a)n+1

70

2 Higher-Order Linear Differential Equations



we complete the proof. As a direct consequence of Corollary 2.17, we have the following result.

Corollary 2.18 Assume rj (t) ≡ 1 for all j = 1, 2, . . . , 2n − 1 and q ∈ C([a, b], R). Then, (2.25) reduces to the differential equation in (2.22). If (2.22) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

|q(t)|dt >

a

n!2n+1 (b − a)2n−1

holds. Example 2.19 Consider the eigenvalue problem (r2n−1 (t)(r2n−2 (t)(· · · (r2 (t)(r1 (t)x  ) ) · · · ) ) ) + λq(t)x = 0, where rk , k = 1, 2, . . . , 2n − 1, and q satisfy the assumptions of Theorem 2.15 and λ is the eigenvalue. Then, we have

−1 1 dt for n = 1, a r1 (t)

−1  b 1 |q(t)|dt min H2 (c) for n = 2, |λ| > c∈[a,b] 2 a 

|λ| > 4

b

and 1 |λ| > 2

 2n−1  % k=n+2 a

b

dt rk (t)

−1 

b

−1 |q(t)|dt

min Hn (c)

c∈[a,b]

a

for n ≥ 3,

where Hn (c) is given in (2.28). Example 2.20 Consider the equation x (2n) + λq(t)x = 0,

(2.45)

where q ∈ C([a, b], R). If (2.45) has a nontrivial solution satisfying the 2-point boundary conditions in (2.2), where a, b ∈ R with a < b are consecutive zeros, then by Corollary 2.18, we have n!2n+1 |λ| > (b − a)2n−1

 a

b

−1 |q(t)|dt

.

2.2 Even-Order Differential Equations

71

2.2.2 Lidstone Boundary Value Problems In this section, dissipating the alternating term (−1)n−1 in (2.1), we consider the Lidstone boundary value problem  (2n) + q(t)x = 0 on [a, b], x (2.46) x (2k) (a) = x (2k) (b) = 0 for k = 0, 1, . . . , n − 1. It appears that the first generalization of Lyapunov’s result for (2.46) is the following result obtained by Yang [295, Theorem 4]. Theorem 2.21 (Lyapunov-Type Inequality) If (2.46) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality  b 2n |q(t)|dt ≥ (2.47) (b − a)n a holds. Proof First, we prove   b − a  b    (2i)   (2i+2)  (t) dt x (t) ≤ x 2 a

(2.48)

for all i = 0, 1, . . . , n − 1. In fact,   t    t    (2i)    (2i+1)  (2i+1) (s)ds  ≤ (s) ds x (t) ≤  x x a

(2.49)

a

and     (2i)   x (t) ≤ 

t

b

x

(2i+1)

   (s)ds  ≤

b t

   (2i+1)  (s) ds. x

(2.50)

Therefore,  b     (2i)  1  (2i+1)  (t) dt. x (t) ≤ x 2 a

(2.51)

Since x (2i) (a) = x (2i) (b) = 0 for

i = 0, 1, . . . , n − 1,

there exists ti ∈ (a, b) such that x (2i+1) (ti ) = 0 for all i = 0, 1, . . . , n − 1, and hence   t  b    t     (2i+1)    (2i+2)   (2i+2)  (2i+2)  (t) =  x (s)ds  ≤ (s) ds ≤ (t) dt. x x x ti

ti

a

72

2 Higher-Order Linear Differential Equations

This and (2.51) yield

 b   b  b   

  b−a  (2i+2)   (2i+2)   (2i)  1 (s) ds dt = (t) dt. x x x (t) ≤ 2 a 2 a a Inequality (2.48) is therefore proved. Let |x(c)| = max |x(t)|. Then, by (2.48), we t∈(a,b)

have |x(c)| ≤

b−a 2



b

   x (t) dt

a

≤ ···

 b − a n b  (2n)  ≤ x (t) dt 2 a

 b−a n b = |q(t)x(t)|dt 2 a

 b b−a n |x(c)| |q(t)|dt. ≤ 2 a

(2.52)

Dividing both sides of (2.52) by |x(c)| > 0, we obtain (2.47), and this completes the proof.  Note that (2.47) does not generalize Lyapunov’s classical inequality since when n = 1, (2.47) reduces to 

b

|q(t)|dt >

a

2 . b−a

(2.53)

Thereafter, Çakmak [68, Theorem 2] improved Yang’s result as follows. Theorem 2.22 (Lyapunov-Type Inequality) If (2.46) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality  a

b

|q(t)|dt ≥

4n (b − a)2n−1

(2.54)

holds. Proof Before starting the proof, we remark that the proof of Theorem 2.22 is similar to that of Theorem 2.21 with small changes in some steps. First, we prove  b − a  b     (2i+2)   (2i)  (t) dt x x (t) ≤ 4 a

(2.55)

2.2 Even-Order Differential Equations

73

for all i = 0, 1, . . . , n − 1. Using (2.49) and (2.50), we have (2.51). Since x (2i) (a) = x (2i) (b) = 0 for

i = 0, 1, . . . , n − 1,

there exists ti ∈ (a, b) such that x (2i+1) (ti ) = 0 for all i = 0, 1, . . . , n − 1, and hence   t  b    t     (2i+2)   (2i+2)   (2i+1)   (2i+2) (t) =  x (s)ds  ≤ (s) ds ≤ (s) ds x x x ti

ti

ti

and     (2i+1)   x (t)  =

t

ti

   x (2i+2) (s)ds  ≤

    (2i+2)  x ds ≤ (s)  

ti t

ti a

   (2i+2)  (s) ds. x

Summing up the last two inequalities gives  b     (2i+1)  1  (2i+2)  x ≤ (t) (t) dt.  x  2 a

(2.56)

Using (2.51) and (2.56), we have  b  b    

b − a b  (2i+2)  1  (2i)  1  (2i+2)  (s) ds dt = (t) dt. x (t) ≤ x x 2 a 2 a 4 a Inequality (2.55) is therefore is proved. Let |x(c)| = max |x(t)|. Then, by (2.55), t∈(a,b)

we have |x(c)| ≤ ≤

b−a 4 (b



b a

− a)3 24

   x (t) dt



b

a

≤ ··· (b − a)2n−1 ≤ 22n =

(b

− a)2n−1 22n

   (4)  x (t) dt 

b a



b

   (2n)  x (t) dt

(2.57)

|q(t)x(t)|dt

a

(b − a)2n−1 ≤ |x(c)| 22n



b

|q(t)|dt.

a

Dividing both sides of (2.57) by |x(c)| > 0, we obtain (2.54), which completes the proof. 

74

2 Higher-Order Linear Differential Equations

Remark 2.23 When n = 1, (2.54) in Theorem 2.22 reduces to the classical Lyapunov inequality for (1.7) satisfying (1.5). Example 2.24 We give an application of (2.54) for the eigenvalue problem 

x (2n) ± λq(t)x = 0 on

[a, b],

x (2k) (a) = x (2k) (b) = 0 for

k = 0, 1, . . . , n − 1.

(2.58)

If (2.58) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then 4n |λ| > (b − a)2n−1



b

−1 |q(t)|dt

.

a

In 2012, He and Tang [158] improved and generalized Theorems 2.21 and 2.22 to (2.3). They obtained the following inequality. Theorem 2.25 (Hartman-Type Inequality) Let n ∈ N \ {1}. If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b



2 +

(b − t) (t − a) q (t)dt > 2

a

32 45

[1−(−1)n ]/4

3n−1 10n/2−1 (b − a)2n−5

(2.59)

holds. The proof of Theorem 2.25 is based on the following lemma. Lemma 2.26 Assume that h is a continuous and real-valued function on [a, b] satisfying h(a) = h(b) = 0, where a, b ∈ R with a < b are consecutive zeros, and h ∈ L2 ([a, b], R). Then, |h(t)| ≤

(b − t)(t − a) √ 3(b − a)



b

  2 h (s) ds

1/2 (2.60)

a

for all t ∈ (a, b). Shortly after He and Tang’s [158] results, motivated by the results given in [68, 158, 295], Zhang and He [303] improved and generalized Theorems 2.21, 2.22, and 2.25 to (2.3). Lemma 2.27 (See [30]) Assume that g and g  are continuous on [α, β] such that g(α) = g(β) = 0 and  α

β

g(t)dt = 0.

2.2 Even-Order Differential Equations

75

Then, 



β

|g(t)| dt ≤ 2

α

b−a 2π

2 

  2 g (t) dt.

β

α

Lemma 2.28 Assume that f is a continuous and real-valued function on [a, b] satisfying f (a) = f (b) = 0, where a, b ∈ R with a < b are consecutive zeros, and f  , f  ∈ L2 ([a, b], R). Then, 

b

|f (t)| dt ≤ 2

a

b−a π

2 

  2 f (t) dt

(2.61)

  2 f (t) dt.

(2.62)

b a

and 

b

|f (t)|2 dt ≤

a

b−a π

2 

b

a

Proof First, define the function F by F (t) :=

 f (t)

if t ∈ [a, b],

−f (2a − t)

if t ∈ [2a − b, a].

Let α = 2a − b and β = b. Since f (a) = f (b) = 0 and a, b are consecutive zeros, taking into account the definition of F , we can easily obtain 

β

F (t)dt = 0 and

F (α) = F (β) = 0.

α

Moreover, since f is a continuous and real-valued function on [a, b], it is obvious that F  and F  are continuous on [α, β]. Hence, it follows from Lemma 2.27 that 

β

|F (t)| dt ≤ 2

α

2(b − a) 2π

2 

β

  2 F (t) dt.

(2.63)

α

Since 

β





a

|F (t)|2 dt =

2a−b

α

b

| − f (2a − t)|2 dt +



b

|f (t)|2 dt = 2

a

a

|f (t)|2 dt (2.64)

and  α

β

  2 F (t) dt =



a 2a−b

   −f (2a − t)2 dt +



b a

  2 f (t) dt = 2



b a

  2 f (t) dt, (2.65)

76

2 Higher-Order Linear Differential Equations

we have from (2.63), (2.64), and (2.65) that

  b 2(b − a) 2 β   2 2 f (t) dt 2 |f (t)| dt ≤ 2 2π a α holds, and this implies (2.61). Next, we prove that (2.62) holds. For convenience, we only consider the special case a = 0. In this case, the interval [α, β] reduces to [−b, b]. It follows from the definition of F that F is an odd function on [−b, b], so we have F (−t) = −F (t). Then, according to the definition of the derivative, we have F (t − b) − F (−b) F (b + s) − F (b) = lim = F− (b). t s s→0− (2.66) It follows from (1.81) that F  (α) = F  (β). Furthermore, we can easily obtain  β F  (t)dt = 0 F+ (−b) = lim

t→0+

α

from the property that F is an odd function on [α, β]. The condition that F  is continuous on [α, β] implies that F  is continuous on [α, β], too. Then, by a method similar to the proof of (2.61), together with Lemma 2.27, we can obtain (2.62) immediately. For the other cases, i.e., a = 0, we only need to move the interval [α, β] evenly such that this interval is symmetric about the origin. Then, similar to the proof of (2.66), we can verify the condition F  (α) = F  (β), and the other conditions in Lemma 2.27 are all satisfied. Hence, it also follows from Lemma 2.27 that (2.62) holds.  Theorem 2.29 (Lyapunov-Type Inequality) Let n ∈ N \ {1}. If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

|q(t)|2 dt ≥

a

48π 4(n−1) (b − a)4n−1

(2.67)

holds. Proof Let |x(c)| = max |x(t)|. Then, by Lemmas 2.26 and 2.28, we have t∈(a,b)

(b − a)3 |x(c)| ≤ 48



b

2

≤

(b

− a)3

a



48

(b − a)3 = 48

  2 x (t) dt



(b − a)4 π4 (b − a)4 π4

n−1 

b a

n−1  a

b

   (2n) 2 x (t) dt |q(t)|2 |x(t)|2 dt.

(2.68)

2.2 Even-Order Differential Equations

77

Dividing both sides of (2.68) by |x(c)| > 0, we obtain (2.67), which completes the proof.  Theorem 2.30 (Hartman-Type Inequality) Let n ∈ N \ {1}. If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

3π 2n−4 (b − a)2n−5

(b − t)2 (t − a)2 q + (t)dt ≥

a

(2.69)

holds. Proof Multiplying both sides of (2.1) by x(t), integrating from a to b, using the boundary conditions in (2.3), and integrating by parts gives 

b



b

x (2n) (t)x(t)dt = (−1)n

a

a

   (n) 2 x (t) dt.

(2.70)

Using (2.70) and (2.1), we have 

b



b

q(t)|x(t)|2 dt =

a

a

   (n) 2 x (t) dt.

(2.71)

Now, we consider two cases. In the first case, let n be even, i.e., n = 2m for some m ∈ N. By (2.71), we have 

b

 q(t)|x(t)|2 dt =

a

a

b

   (2m) 2 (t) dt. x

(2.72)

From (2.3), (2.60), and (2.62), we get |x(t)|2 ≤

(b − t)2 (t − a)2 3(b − a)



b

  2 x (t) dt

(2.73)

a

for all t ∈ [a, b] and  a

b

   (b − a)4 b  (2i+2) 2  (2i) 2 (t) dt x (t) dt ≤ x π4 a

for i = 1, 2, . . . , m − 1. Using (2.72), (2.73), and (2.74), we have

(2.74)

78



2 Higher-Order Linear Differential Equations b

q + (t)|x(t)|2 dt ≤

a

≤ ≤ = ≤



 b   2 (b − t)2 (t − a)2 + x (t) dt q (t)dt 3(b − a) a a  b   (b − a)4m−5 b  (2m) 2 2 2 + x (b − t) (t − a) q (t)dt (t)   dt 3π 4m−4 a a   b  (b − a)4m−3 b  (2m+1) 2 2 2 + (b−t) (t−a) q (t)dt (t)  dt x 3π 4m−2 a a  b  (b − a)4m−3 b 2 2 + (b − t) (t − a) q (t)dt q(t)|x(t)|2 dt 3π 4m−2 a a  b  (b − a)4m−3 b 2 2 + (b − t) (t − a) q (t)dt q + (t)|x(t)|2 dt. 3π 4m−2 a a (2.75) b

Now, we claim that 

b

q + (t)|x(t)|2 dt > 0.

(2.76)

q + (t)|x(t)|2 dt = 0.

(2.77)

a

If (2.76) is not true, then 

b

a

From (2.72) and (2.77), we obtain 

b

0≤ a

    (2m) 2 (t) dt = x

b

 q(t)|x(t)|2 dt ≤

a

b

q + (t)|x(t)|2 dt = 0,

a

which implies 

b

a

   (2m) 2 (t) dt = 0. x

(2.78)

Then, from (2.78), we infer x (2m) (t) ≡ 0 for t ∈ [a, b], contradicting the assumptions. So, (2.76) holds, and dividing both sides of (2.75) by the factor 

b

q + (t)|x(t)|2 dt,

a

we obtain 

b a

that is, (2.69).

(b − t)2 (t − a)2 q + (t)dt ≥

3π 4(m−1) 3π 2n−4 = , (b − a)4m−5 (b − a)2n−5

(2.79)

2.2 Even-Order Differential Equations

79

In the second case, let n be odd, i.e., n = 2m + 1 for some m ∈ N. By (2.71), we have 

b



b

q(t)|x(t)|2 dt =

a

a

   (2m+1) 2 (t) dt. x

(2.80)

It follows from (2.3), (2.60), and (2.62) that (2.73) and (2.74) hold. On the other hand, since x (2m) (a) = x (2m) (b) = 0, it follows from (2.61) that 

b a

   (b − a)2 b  (2m+1) 2  (2m) 2 (t) dt ≤ (t) dt x x π2 a

(2.81)

holds. Using (2.72), (2.73), (2.80), and (2.81), we have  b   2 (b − t)2 (t − a)2 + x (t) dt q (t)dt q (t)|x(t)| dt ≤ 3(b − a) a a a  b   (b − a)4m−5 b  (2m) 2 2 2 + x ≤ (b − t) (t − a) q (t)dt (t)   dt 3π 4(m−1) a a  b  (b − a)4m−5 b 2 2 + = (b − t) (t − a) q (t)dt q(t)|x(t)|2 dt 3π 4(m−1) a a  b  (b − a)4m−5 b 2 2 + ≤ (b − t) (t − a) q (t)dt q + (t)|x(t)|2 dt. 3π 4(m−1) a a (2.82) Now, dividing both sides of (2.82) by the term (2.79), we obtain  b 3π 4m−2 3π 2n−4 (b − t)2 (t − a)2 q + (t)dt ≥ = , (b − a)4m−3 (b − a)2n−5 a 

b

+



2

b



that is, (2.69). This completes the proof.

Remark 2.31 In view of the forms of (2.59) and (2.69), we can easily find that (2.69) is simpler than (2.59). Moreover, by induction, we can verify that (2.69) is sharper than (2.59). Corollary 2.32 (Lyapunov-Type Inequality) Let n ∈ N \ {1}. If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality  a

holds.

b

q + (t)dt ≥

48π 2n−4 (b − a)2n−1

(2.83)

80

2 Higher-Order Linear Differential Equations

Note that it is possible to improve Theorems 2.21 and 2.22 obtaining Lyapunovtype inequalities for (2.46) better than (2.47) and (2.54). The proofs are based on Green’s function for the equation x (2n) = 0

(2.84)

satisfying the Lidstone boundary conditions x (2k) (0) = x (2k) (1) = 0

for

k = 0, 1, . . . , n − 1

(2.85)

x (2k) (0) = x (2k) (T ) = 0 for

k = 0, 1, . . . , n − 1,

(2.86)

or

see [4, 18, 19] and the references therein. By making use of the transformation t→

t −a b−a

on (2.84) and (2.85), the Green function Gn (t, s) for the Lidstone boundary value problem 

(−1)n x (2n) = 0 on

[a, b],

x (2k) (a) = x (2k) (b) = 0

for k = 0, 1, . . . , n − 1

(2.87)

can be expressed as  Gn (t, s) =

b

G1 (t, τ )Gn−1 (τ, s)dτ,

n ∈ N \ {1},

(2.88)

a

where 1 G1 (t, s) = b−a



(b − t)(s − a)

if a ≤ s ≤ t ≤ b,

(t − a)(b − s)

if a ≤ t ≤ s ≤ b.

(2.89)

It is also known from [18, 19] that Gn (t, s) ≥ 0

(2.90)

Gn (t, s) = Gn (s, t)

(2.91)

and

for (t, s) ∈ [a, b] × [a, b].

2.2 Even-Order Differential Equations

81

We need the following auxiliary result. Lemma 2.33 For (t, s) ∈ [a, b] × [a, b] and n ∈ N, we have Gn (t, s) ≤

(b − a)2n−3 (b − s)(s − a) , 6n−1

(2.92)

where Gn (t, s) is Green’s function for (2.87). Proof This can be extracted easily from that of [16, Lemma 2.1] by making the transformation t→

t −a T b−a 

on (2.84) and (2.86). In fact, in view of (2.90), (2.91), and (2.92), we have that 0 ≤ Gn (t, s) ≤

(b − a)2n−3 (b − t)(t − a) 6n−1

(2.93)

for (t, s) ∈ [a, b] × [a, b]. Moreover, (2.92) and (2.93) are both strict for (t, s) ∈ (a, b) × (a, b). The Green function Gn (t, s) for (2.87) helps to get Hartman-type and Lyapunovtype inequalities for (2.3). Moreover, dissipating the alternating term (−1)n−1 in (2.1), it is possible to obtain an inequality for (2.46) which is sharper than both (2.47) and (2.54). Theorem 2.34 (Hartman-Type Inequality) If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

(b − t)(t − a)q + (t)dt >

a

6n−1 (b − a)2n−3

(2.94)

holds. Proof Let x be a nontrivial solution of (2.3), where a, b ∈ R with a < b are consecutive zeros. Without loss of generality, we may assume that x(t) > 0 for t ∈ (a, b). In fact, if x(t) < 0 for t ∈ (a, b), then we can consider −x, which is also a solution. Then, by using Green’s function for (2.87), x(t) can be expressed as 

b

x(t) =

Gn (t, s)q(s)x(s)ds.

a

Let x(c) = max x(t). Using (2.95), we obtain t∈(a,b)

 x(c) = a

b



b

Gn (c, s)q(s)x(s)ds < a

Gn (c, s)q + (s)x(c)ds,

(2.95)

82

2 Higher-Order Linear Differential Equations

which implies the inequality 

b

Gn (c, s)q + (s)ds > 1.

a

Finally, using (2.92) with t = c, we complete the proof.



Next, we prove the following result. Theorem 2.35 (Lyapunov-Type Inequality) If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

q + (t)dt >

a

2n+1 3n−1 (b − a)2n−1

(2.96)

holds. Proof In view of (1.23), (2.94) immediately implies (2.96).



We have the following result for (2.46). Corollary 2.36 If (2.46) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the Hartman-type inequality 

b

(b − t)(t − a)|q(t)|dt >

a

6n−1 (b − a)2n−3

(2.97)

and the Lyapunov-type inequality 

b a

|q(t)|dt >

2n+1 3n−1 (b − a)2n−1

(2.98)

hold. Remark 2.37 When n = 1, (2.94), (2.96), (2.97), and (2.98) give the classical results. Remark 2.38 Corollary 2.36 is remarkable since 2n+1 3n−1 > 4n for n ∈ N \ {1} implies that (2.98) is sharper than both inequalities obtained in Yang [295, Theorem 4] and Çakmak [68, Theorem 2], i.e., (2.47) and (2.54). On the other hand, (2.97) is the first Hartman-type inequality obtained for even-order equations of the form (2.46). We recall the well-known inequality [20, p. 3] 2 1 sin π t ≤ t (1 − t) ≤ sin π t π π2

for t ∈ [0, 1].

(2.99)

2.2 Even-Order Differential Equations

83

By making the transformations t→

t −a b−a

t→

b−t b−a

and

in (2.99), we obtain the handy inequalities



t −a t −a 2(b − a)2 (b − a)2 π ≤ (b − t)(t − a) ≤ sin π sin b−a π b−a π2

(2.100)

and



b−t b−t 2(b − a)2 (b − a)2 π ≤ (b − t)(t − a) ≤ sin π , sin b−a π b−a π2 respectively, for t ∈ [a, b], which play an important rôle in later development. The following lemmas give upper and lower bounds for Green’s function Gn (t, s). They are extracted from results in Agarwal and Wong [20, Lemma 2.1, Lemma 2.2], Lemma 2.39 For (t, s) ∈ [a, b] × [a, b], we have

(b − a)2n s−a Gn (t, s) ≤ sin π , b−a π 2n−1

(2.101)

where Gn (t, s) is Green’s function for (2.87). Proof For (t, s) ∈ [a, b] × [a, b] and from (2.89), we obtain G1 (t, s) ≤

(b − s)(s − a) . b−a

Using this, (2.90), (2.91), and (2.100) on (2.88) yields for (t, s) ∈ [a, b] × [a, b],  Gn (t, s) =

b

G1 (t, τ )Gn−1 (τ, s)dτ

a

≤ ≤

1 b−a



(b − a) π

b

Gn−1 (s, τ )(b − τ )(τ − a)dτ

a



b a

Gn−1 (s, τ ) sin

τ −a π dτ. b−a

(2.102)

84

2 Higher-Order Linear Differential Equations

Now, consider the integral 

b

In (t) := a



s−a Gn (t, s) sin π ds, b−a

n ∈ N.

For n = 1, we have

s−a π ds b−a a



 t  b s−a s−a = (b − t) (s−a) sin (b − s) sin π ds + (t − a) π ds b−a b−a a t   (t−a)π (b−t)π   (b−a) (b−a) (b − a)2 u sin udu + (t − a) u sin udu (b − t) = π2 0 0

(b − a)3 t −a = π . sin b−a π2 

b

I1 (t) :=



G1 (t, s) sin

By induction, we can show  In (t) := a

b



(b − a)2n+1 s−a t −a π ds = π Gn (t, s) sin sin b−a b−a π 2n

(2.103)

for t ∈ [a, b] and n ∈ N. Now, applying (2.103) to (2.102), we get Gn (t, s) ≤

1 (b − a)In−1 (s), π 

and hence (2.101) is immediate.

Lemma 2.40 Let ρ ∈ (0, 1) be given. For (t, s) ∈ [a + (b − a)ρ, b − (b − a)ρ] × [a, b], we have Gn (t, s) ≥

2ρ(b − a)2n s−a π , sin b−a π 2n

(2.104)

where Gn (t, s) is Green’s function for (2.87). Proof For (t, s) ∈ [a + (b − a)ρ, b − (b − a)ρ] × [a, b] and from (2.89), we obtain 1 G1 (t, s) ≥ b−a



ρ(b − a)(s − a)

if a ≤ s ≤ t ≤ b,

ρ(b − a)(b − s)

if a ≤ t ≤ s ≤ b

ρ ≥ (b − s)(s − a). b−a

(2.105)

2.2 Even-Order Differential Equations

85

Then, using (2.105), (2.90), (2.91), and (2.100) on (2.88) yields  Gn (t, s) =

b

G1 (t, τ )Gn−1 (τ, s)dτ

a

≥ ≥

ρ b−a



b a

2ρ(b − a) π2

Gn−1 (s, τ )(b − τ )(τ − a)dτ 

b

Gn−1 (s, τ ) sin

a

τ −a π dτ b−a

2ρ(b − a) = In−1 (s) π2 for (t, s) ∈ [a + (b − a)ρ, b − (b − a)ρ] × [a, b]. This completes the proof.



Remark 2.41 Since (s − a)π (b − s)π + = π, b−a b−a the term (s − a)π/(b − a) in (2.101) and (2.104) in Lemmas 2.39 and 2.40 can be replaced by the term (b − s)π/(b − a). Remark 2.42 The bounds for Green’s function obtained in Lemmas 2.39 and 2.40 are sharper than those obtained previously. As an application of Lemma 2.39, we have the following interesting result. Theorem 2.43 If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequalities 



b

sin a

t −a π 2n−1 π q + (t)dt > b−a (b − a)2n

(2.106)

b−t π 2n−1 π q + (t)dt > b−a (b − a)2n

(2.107)

and 



b

sin a

hold. Proof Following steps similar to the proof of Theorem 2.34 and using (2.101) in the last step, we get (2.106). Inequality (2.107) follows from (2.106) by the reason given in Remark 2.41.  Corollary 2.44 If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequalities

86

2 Higher-Order Linear Differential Equations



b

(t − a)q + (t)dt >

π 2n−2 (b − a)2n−1

(2.108)

(b − t)q + (t)dt >

π 2n−2 (b − a)2n−1

(2.109)

a

and 

b a

hold. Proof From the mean value theorem, we have 0 ≤ sin

t −a π b−a

≤

t −a π b−a

and

0 ≤ sin

b−t π b−a

≤

b−t π b−a

for all t ∈ [a, b]. Using (2.106) and (2.107), it is easy to obtain (2.108) and (2.109).  Following Theorem 2.43, we have a remarkable result as an alternative form of Theorem 2.35. Theorem 2.45 (Lyapunov-Type Inequality) If (2.3) has a nontrivial solution, where a, b ∈ R with a < b are consecutive zeros, then the inequality 

b

q + (t)dt >

a

π 2n−1 (b − a)2n

(2.110)

holds. Proof Note that since (t − a)π/(b − a) ∈ [0, π ] (and (b − t)π/(b − a) ∈ [0, π ]) for all t ∈ [a, b], sin ((t − a)π/(b − a)) ∈ [0, π ] (and sin ((b − t)π/(b − a)) ∈ [0, 1]), and hence (2.106) (or (2.107)) implies (2.110).  Remark 2.46 Clearly, (2.110) is sharper than (2.96) for large values of n, more precisely, for all n ∈ N satisfying n>

log(b − a) + log(2π/3) . log(π 2 /6)

Moreover, (2.110) is sharper than (2.83) given by Zhang and He [303] if the distance between the zeros of the solution of (2.3) is less than π 3 /48, i.e., b−a


22n−1 (2n − 1)[(n − 1)!]2 . s 2n−1

In 2012, Watanabe et al. [287] considered a necessary condition for the existence of a nontrivial solution of (2.111) under yet another boundary condition x (i) (±s) = x (i+n) (±s) = 0 for

i = 0, 1, . . . , n − 1.

(2.114)

Now, let us introduce the Lp -type Sobolev inequality

   p  (m)  sup x (t) ≤ C

−s≤t≤s

s −s

   (n) p x (t) dt,

(2.115)

where x belongs to  W(n, p) := x : x (n) ∈ Lp (−s, s), x (i) (±s) = 0, i = 0, 1, . . . , n − 1 , and p > 1, m runs over the range 0 ≤ m ≤ n − 1, and x (i) is the ith derivative of x in a distributional sense. We denote by CCF (n, m, p) the best constant of (2.115). Here, we note that in [267], Takemura obtained the best constant for p = 2 and m = 2 by constructing Green’s function for the clamped-free boundary value problem. Although for the proof of Theorem 2.52 we simply need the value CCF (n, 0, 2), we would like to compute CCF (n, m, p) for general p and m since the proof presented does not depend on special values of p and m and substantially simplifies the proof of [267, Theorem 1]. Now, we offer the following lemmas. Lemma 2.48 Suppose there exists a function u∗ ∈ W(n, p) which attains the best constant C(n, m, p) of (2.115). Then,       (m)   = max u(m) (t) (s) u  . ∗ ∗

−s≤t≤s

Proof Suppose we have       (m)   = max u(m) (t) (a) u  , ∗ ∗

−s≤t≤s

where a = s. Further, let us define  u(t) ˜ :=

0

if

− s ≤ t ≤ −a,

u∗ (t + a − s)

if

− a ≤ t ≤ s.

2.2 Even-Order Differential Equations

89

Then, u˜ ∈ W(n, p),            (m)  max u˜ (m) (t) = max u(m) ∗ (t) = u∗ (a) ,

−s≤t≤s

−s≤t≤s

and & & & (n) & &u˜ &

Lp (−s,s)

& & & & < &u(n) ∗ &

Lp (−s,s)

.

Hence,     (m)  p p   max−s≤t≤s u∗ (t) max−s≤t≤s u˜ (m) (t) & & & & C(n, m, p) = < . & (n) & &u˜ (n) & p &u∗ & p L (−s,s) L (−s,s)

This contradicts the assumption that C(n, m, p) is the best constant of (2.115).



Lemma 2.49 Define Hm (t) :=

(−1)n−m−1 (t − s)n−m−1 . (n − m − 1)!

Then, for x ∈ W(n, p), we have  x (m) (s) =

s −s

x (m) (t)Hm (t)dt.

(2.116) 

Proof An integration by parts yields the result. Proposition 2.50 The best constant of (2.115) is ⎛ ⎞p−1 p(n−m)−1 p−1 (p − 1)(2s) 1 ⎝ ⎠ CCF (n, m, p) = , [(n − m − 1)!]p p(n − m) − 1 and it is attained by  x∗ (t) =

t −s

(t − τ )n−1 (n − 1)!



(s − τ )n−m−1 (n − m − 1)!

p−1 dτ.

Proof From Lemma 2.49, we see that if the function attains the best constant C(n, m, p), it belongs to W∗ (n, m, p) ⊂ W(n, p), where  W∗ (n, m, p) = x ∈ W(n, p) :

     (m)   (m)  max x (t) = x (s) .

−s≤t≤s

90

2 Higher-Order Linear Differential Equations

Let x ∈ W∗ (n, m, p). Then, applying Hölder’s inequality to (2.116), we obtain  & &     & &    max x (m) (t) = x (m) (s) = Hm Lq (−s,s) &x (n) &

−s≤t≤s

Lp (−s,s)

(2.117)

,

where q satisfies 1/p + 1/q = 1. Hence, if there exists a function x∗ ∈ W∗ (n, m, p) which attains equality of (2.117), then p

C(n, m, p) = Hm Lq (−s,s) . On the contrary, we see that equality holds for (2.117) if and only if x satisfies x (n) (t) = sgn{Hm (t)} |Hm (t)|q−1 .

(2.118)

It is easy to see that  x∗ (t) =

−s

 =

t

t

−s

(t − τ )m−1 sgn{Hm (τ )} |Hm (τ )|q−1 dτ (n − 1)!  1/(p−1) (t − τ )m−1 (s − τ )n−m−1 dτ (n − 1)! (n − m − 1)!

satisfies (2.118) and belongs to W∗ (n, m, p). Thus, we have shown p

C(n, m, p) = Hm Lq (−s,s) . p

Now, we compute Hm Lq (−s,s) , i.e., p

Hm Lq (−s,s) =

1 [(n − m − 1)!]p

 ⎛

s

−s

p/q (s − t)q(n−m−1) dt p(n−m)−1

p−1 1 ⎝ (p − 1)(2s) = p [(n − m − 1)!] p(n − m) − 1

⎠

. 

This completes the proof. Proposition 2.51 then

⎞p−1

If there exists a C2n [−s, s]-solution of 

s −s

q + (t)dt >

Moreover, the estimate is sharp.

1 . CCF (n, 0, 2)

(2.111) satisfying (2.114),

2.2 Even-Order Differential Equations

91

Proof Let x be a solution of (2.111). Since x satisfies (2.114), multiplying (2.111) by x(t) and integrating it over [−s, s], we have 

s −s

x (n) (t)

!2

 dt = =

s

−s  s −s

(−1)n x(t)x (2n) (t)dt q(t)[x(t)]2 dt

≤

2 

sup |x(t)|

−s≤t≤s

≤ C(n, 0, 2)



s −s

(2.119)

s −s

+

q (t)dt

q + (t)dt



s −s

x (n) (t)

!2

dt.

Here, if x (n) (t) ≡ 0, then there exist ai ∈ R, i = 0, 1, . . . , n − 1, such that x(t) =

n−1

ai t i .

i=0

Since x satisfies the boundary condition at t = −s, we have x(t) ≡ 0. This contradicts the assumption that x is a nontrivial solution of (2.111). So, cancelling & (n) &2 &x & , we obtain 2 

s

−s

q + (t)dt ≥

1 . CCF (n, 0, 2)

(2.120)

Next, we show that (2.120) is strict. To see this, we note that in (2.119), if the equality holds for the first inequality, then x is a constant. But, again from the boundary condition at t = −s, we have x(t) ≡ 0. Thus, the inequality is strict.  Watanabe et al. [287] proved the following result. Theorem 2.52 (Lyapunov-Type Inequality) Suppose there exists a nontrivial solution x of (2.111) satisfying (2.114). Then, the inequality 

s

−s

q + (t)dt >

(2n − 1)[(n − 1)!]2 (2s)2n−1

(2.121)

holds. Moreover, the estimate is sharp in the sense that there exists a function q, and for this q, the solution x of (2.111) exists such that the right-hand side is arbitrarily close to the left-hand side. Proof Clearly, Theorem 2.52 is obtained from Propositions 2.50 and 2.51. It is enough only to prove (2.121) is sharp. For this purpose, let us define the functional

92

2 Higher-Order Linear Differential Equations

 s  (n) 2 φ (t) dt , J (φ) :=  s−s 2 dt ˜ −s q(t)|φ(t)|

φ ∈ W(n, 2),

φ = 0,

where q˜ ∈ C([−s, s], R+ ) is defined later. By the standard argument of the variational method, J has the minimizer u ∈ W(n, 2) (see, for example, [285, Lemma 16]), i.e., λ1 :=

min

φ∈W(m,2) φ=0

J (φ) = J (u).

Hence, x satisfies the Euler–Lagrange equation (as a classical solution by the regularity argument) (−1)n x (2n) = λ1 q(t)x ˜

[−s, s].

on

(2.122)

Furthermore, we have  λ1 =

min

φ∈W(m,2) φ=0

s −s

s

−s



1

>

    (n) 2 φ (t) dt s

q(t)|φ(t)| ˜ dt 2

    (n) 2 φ (t) dt

(sup−s≤t≤s |φ(t)|)2 −s  s

−1 1 ≥ q(t)dt ˜ . CCF (n, 0, 2) −s

s

−s

−1 

q(t)dt ˜

−1

(2.123)

Here, let us fix q˜ as  1 t −s q(t) ˜ := δ 0

if

s − δ < t ≤ s,

if

− s ≤ t ≤ s − δ.

For such q, ˜ let us substitute φ = x∗ (of Proposition 2.50) into (2.123). It is easy to see that x∗ takes its maximum at t = s. Hence, by taking δ > 0 sufficiently small, we see that the right-hand side of (2.123) can be arbitrarily close to the left-hand side, i.e., for a small positive ε1 , λ1 can be written as λ1 =

1 CCF (n, 0, 2)



s −s

q(t)dt ˜

−1

+ ε1 .

˜ we see from (2.122) that a solution x of Putting q = λ1 q, (−1)n x (2n) = q(t)x

on

[−s, s]

2.2 Even-Order Differential Equations

93

exists, and from (2.122), q satisfies 

s

−s

q(t)dt =



1 + ε1 CCF (n, 0, 2)

s −s

q(t)dt ˜ =

1 + ε2 . CCF (n, 0, 2) 

Hence, (2.121) is sharp. Corollary 2.53 (Lyapunov-Type Inequality) solution x of the nonlinear equation

Suppose there exists a nontrivial

(−1)n xx (2n) − q(t) x (m)

!2

=0

(2.124)

satisfying (2.114). Then, the inequality 

s

−s

q + (t)dt >

(2n − 2m − 1)[(n − m − 1)!]2 (2s)2n−2m−1

(2.125)

holds, where m ∈ {1, . . . , n − 1}. Proof Integrating (2.124), we obtain 

s

−s

x (n) (t)

!2

 dt =

s

−s

≤

q(t) x (m) (t)

!2

dt

  2   (m) 2 sup x (t)

−s≤t≤s

≤ C(n, m, 2)



s

−s

s −s

x

(n)

(t)

!2

q + (t)dt  dt

s −s

(2.126)

q + (t)dt.

& &2 As in the proof of Proposition 2.51, dividing both sides of (2.126) by &x (n) &2 , we have  s 1 . q + (t)dt ≤ C (n, m, 2) CF −s Next, we show that (2.125) is strict. To see this, we note that in (2.126), the equality holds for the first inequality if and only if x (m) is a constant. Hence, from the boundary condition at t = −s, we have x (m) (t) ≡ 0. So, there exist ai ∈ R, i = 0, 1, . . . , m − 1, such that x(t) =

m−1

i=0

ai t i .

94

2 Higher-Order Linear Differential Equations

But, again from the boundary condition at t = −s, we have x(t) ≡ 0. Thus, (2.125) is strict.  In the following, we provide some examples of Theorem 2.52 and Corollary 2.53. Example 2.54 This example corresponds to the case n = 1 and

q(t) = −

−11s 2

6 + 2st + t 2

of (2.111) with clamped-free boundary conditions, i.e., ⎧ ⎪ ⎨−x  (t) +

6 x(t) = 0 −11s 2 + 2st + t 2

on [−s, s],

⎪ ⎩x(−s) = x  (s) = 0.

(2.127)

It is easy to see that x(t) = −(s + t)(11s 2 − 2st − t 2 ) is the solution of (2.127). Moreover, we have 

s

−s



+

q (t)dt = 6

s

dt = 2 11s − 2st − t 2

−s

√ √ 1 1 3 log(2 + 3) > = . 2s CCF (1, 0, 2) 2s

Example 2.55 This example corresponds to the case n = 2,

m = 1,

and

q(t) =

3(17s 2 − 6st + t 2 ) 2(7s 2 − 4st + t 2 )2

of (2.124) with clamped-free boundary conditions, i.e., ⎧ 2 2   ⎪ ⎨x(t)x (4) (t) − 3(17s − 6st + t ) x  (t) 2 = 0 2 2 2 2(7s − 4st + t ) ⎪ ⎩  x(−s) = x (−s) = x  (s) = x  (s) = 0.

on

[−s, s],

(2.128)

It is easy to see that x(t) = (s + t)2 (17s 2 − 6st + t 2 ) is the solution of (2.128). Moreover, we have 

s

−s

+

q (t)dt =



s −s

√ 1 1 3(17s 2 − 6st + t 2 ) 3 + 2 3π > = . dt = 2 2 2 12s CCF (2, 1, 2) 2s 2(7s − 4st + t )

2.3 Odd-Order Differential Equations

95

2.3 Odd-Order Differential Equations In this section, we first consider third-order linear differential equations of the form x  + q(t)x = 0,

(2.129)

where q ∈ C([0, ∞), R). In Sect. 2.3.1, we give some Lyapunov-type inequalities for (2.129) satisfying (1.5) and the 3-point boundary conditions x(a) = x(b) = x(c) = 0.

(2.130)

In Sect. 2.3.2, we establish a Lyapunov-type inequality for odd-order differential equations of the form x (2n+1) + q(t)x = 0,

n ∈ N,

(2.131)

satisfying (1.5).

2.3.1 Third-Order Differential Equations In 1999, Parhi and Panigrahi [230] established a Lyapunov-type inequality for thirdorder differential equations of the form (2.129), where q ∈ C([0, ∞), R). Their result reads as follows. Theorem 2.56 (Lyapunov-Type Inequality) Let (2.129) have a nontrivial solution satisfying (1.5), where a, b ∈ R with a < b are consecutive zeros. If there exists t∗ ∈ (a, b) such that x  (t∗ ) = 0, then the inequality 

b

|q(t)|dt >

a

4 (b − a)2

holds. Proof Let m0 = max |x(t)| = |x(c)|, where c ∈ (a, b). Thus, t∈[a,b]

  m0 = 

c a

   x (t)dt  ≤ 

c

   x (t) dt

b

   x (t) dt

a

and   m0 = 

c

b

   x  (t)dt  ≤

c

(2.132)

96

2 Higher-Order Linear Differential Equations

imply that 

b

2m0 ≤

   x (t) dt.

(2.133)

a

Using the Cauchy–Schwarz inequality and then integrating by parts, (2.133) implies  4m20 ≤ (b − a)

b

  2 x (t) dt

a



b

= −(b − a)

x(t)x  (t)dt

(2.134)

a



b

≤ (b − a)

  x(t)x  (t) dt.

a

Integrating (2.129) from t∗ to t (a ≤ t∗ < t or t < t∗ ≤ b), we get x  (t) = −



t

q(s)x(s)ds, t∗

that is,         t x (t) =  q(s)x(s)ds  ≤   t∗

b

|q(t)x(t)|dt.

a

Thus, (2.134) yields  4m20 ≤ (b − a)

b a

< (b − a)2 m20

 |x(t)|dt

b

|q(t)x(t)|dt

a



b

|q(t)|dt,

a

from which (2.132) follows. Hence, the proof is complete.



Remark 2.57 If either x  (a) = 0 or x  (b) = 0, then there exists t∗ ∈ (a, b) such that x  (t∗ ) = 0. Example 2.58 If z(t) = sin t, t ≥ 0, then we notice that z(0) = z(π ) = 0, z(t) = 0 on t ∈ (0, π ), z (0) = 0, z (π ) = 0, and z (0) = z (π ) = 0. Example 2.59 If z(t) = t (t − 1)(t 2 − t + 1/3), t ≥ 0, then z(0) = z(1) = 0, z(t) = 0 on t ∈ (0, 1), z (0) = 0, z (1) = 0, and z (1/3) = 0. Theorem 2.60 (Lyapunov-Type Inequality) Let (2.129) have a nontrivial solution satisfying (2.130), where a, b, c ∈ R with a < b < c are consecutive zeros. If x  (t) = 0 for t ∈ (a, b), then the inequality

2.3 Odd-Order Differential Equations



97

c

4 (c − a)2

|q(t)|dt >

a

(2.135)

holds. Proof From Rolle’s theorem, it follows that there exist t1 ∈ (a, b) and t2 ∈ (b, c) such that x  (t1 ) = 0 and x  (t2 ) = 0. Further application of Rolle’s theorem yields that there exists t3 ∈ (t1 , t2 ) such that x  (t3 ) = 0. Setting m1 = max |x(t)| = t∈[a,c]

|x(t4 )|, where t4 ∈ (a, b) ∪ (b, c), we obtain   m1 = 

t4 a

   x (t)dt  ≤

t4



   x (t) dt

a

and   m1 = 

c t4

   x (t)dt  ≤ 

c

   x (t) dt.

t4

Then, proceeding as in the proof of Theorem 2.56, one may get  < (c − a)

4m21

2

m21

c

|q(t)|dt.

a



This completes the proof.

Example 2.61 If z(t) = (t 2 − 1)(4 − t 2 ), t ≥ 1, then z(1) = z(2) = 0, z(t) = 0 on t ∈ (1, 2), and z (t) < 0 for t ∈ [1, 2]. This possibility forces us to consider Theorem 2.60. Remark 2.62 If all the conditions of Theorem 2.60 hold for c < a < b, then the result is still valid, and in this case we have the inequality 

b

|q(t)|dt >

c

4 . (c − b)2

Remark 2.63 When x  (t) = 0 for t ∈ (a, b), then t3 ∈ (t1 , t2 ) implies that t3 ∈ (b, c] in Theorem 2.60. Hence, from Theorem 2.56, it follows that 

c

|q(t)|dt >

b

4 . (c − b)2

Thus, (2.135) yields  a

b

|q(t)|dt >

4 − (c − a)2



c b

|q(t)|dt.

98

2 Higher-Order Linear Differential Equations

We may note that 4

1 1 − 2 (c − a) (c − b)2

>

4 − (c − a)2



c

|q(t)|dt.

b

If 

b

a



1 1 |q(t)|dt > 4 − (c − a)2 (c − b)2

,

then the lower bound of the distance between a and b is obtained. If

 b 1 1 |q(t)|dt ≤ 4 − , (c − a)2 (c − b)2 a then the upper bound of the distance between a and b is derived. In the following, we consider a simple example in order to demonstrate the usefulness of (2.135). Example 2.64 Consider the equation x  + x = 0.

(2.136)

The basis of the solution space of (2.136) is given by  t/2

e

√  √   3t 3t t/2 −t cos . , e sin ,e 2 2

√ The zeros of x1 (t) = et/2 cos( 3t/2) are given by tn =

2n − 1 √ π, 3

n ∈ N.

It is easy to see that x1 (tn ) and x1 (tn+1 ) are of opposite signs, and hence there exists tn∗ ∈ (tn , tn+1 ) such that x1 (tn∗ ) = 0. Thus, from Theorem 2.56, it follows √ that (tn+1 − tn )3√> 4, which is true because tn+1 − tn = 2π/ 3. The zeros of x2 (t) = et/2 sin( 3t/2) are given by tn =

2(n − 1) π, √ 3

n ∈ N,

and the preceding statements hold for x2 (t). Further, the zeros of an oscillatory solution of the form z(t) = α1 x1 (t) + α2 x2 (t) with α1 α2 = 0

2.3 Odd-Order Differential Equations

99

of (2.136) are given by

  α1 2 − nπ , tn = √ arctan − α2 3

n ∈ Z.

It is not easy to verify that z (tn∗ ) = 0 for some tn∗ ∈ [tn , tn+1 ] as in the case of x1 (t) = 0 and x2 (t) = 0. In this case, Theorem 2.60 √ is employed to obtain |tn+2 − tn |3 > 4. This is true because |tn+2 − tn | = 4π/ 3. Next, we show that a solution of the type x(t) = α1 x1 (t) + α2 x2 (t) + α3 e−t

(2.137)

of (2.136) is oscillatory, where αj ∈ R, j ∈ {1, 2, 3}, such that α12 + α22 = 0. On the contrary, suppose that x is nonoscillatory. Hence, α3 = 0, and we may assume that α3 > 0. If x(t) > 0 for t ≥ t0 for some t0 > 0, then √   √  3t 3t α1 cos + α2 sin > −α3 2 2

 e3t/2

(2.138)

for t ≥ t0 . This is impossible because the left-hand side of (2.138) oscillates between −∞ and ∞. If x(t) < 0 for t ≥ t0 , then √   √  3t 3t α1 cos + α2 sin < −α3 < 0, 2 2

 3t/2

e

√ √ and hence α1 cos( 3t/2) + α2 sin( 3t/2) is nonoscillatory, a contradiction. Thus, x given in (2.137) is oscillatory. The zeros of x are given by the transcendental equation √ sin

3(θ + tn ) 2

 + α3 e−3tn /2 = 0,

n ∈ N,

where

2 α1 tn = √ arctan . α2 3 As the previous equation cannot be solved for tn , no lower bound for |tn+2 − tn | can be found. However, from Theorem 2.60, it follows that |tn+2 − tn | > 4. Now, suppose there exists t∗ ∈ (a, b) such that x  (t∗ ) = 0, and let max |x(t)| = |x(c)|,

t∈[a,b]

100

2 Higher-Order Linear Differential Equations

where c ∈ (a, b) as in Theorem 2.56. Hence, x  (c) = 0. In the following, we establish a result related to this case. Theorem 2.65 The point c cannot be very close to a as well as to b. Proof In the first case, suppose a ≤ t∗ < c. Clearly, 

c

x(c) =

x  (t)dt

a

implies  [x(c)]2 =

c

2

x  (t)dt

a



c

≤ (c − a)

  2 x (t) dt

a



c

= −(c − a)

(2.139) x(t)x  (t)dt

a



c

≤ (c − a)

  |x(t)| x  (t) dt,

a

where we have used the Cauchy–Schwarz inequality and the integration by parts formula. For a < t < c, we integrate (2.129) from t∗ to t to obtain 



x (t) = −

t

q(s)x(s)ds. t∗

Hence,    x (t) ≤



t

|q(s)||x(s)|ds.

t∗

Consequently, (2.139) yields 

c

[x(c)]2 ≤ (c − a)

 |x(t)|

a

≤ (c − a) [x(c)] 2

 2

t

|q(s)||x(s)|dsdt

t∗ b

|q(t)|dt.

a

Thus, 

b a

|q(t)|dt ≥

1 . (c − a)2

(2.140)

2.3 Odd-Order Differential Equations

101

Since 

b

|q(t)|dt < ∞ and

lim

c→a +

a

1 = ∞, (c − a)2

it follows that c cannot be very close to a. Next, we show that c cannot be very close to b either. Integrating (2.129) from t∗ to t (t∗ < t ≤ b), we obtain 



t

x (t) +

q(s)x(s)ds = 0.

t∗

Successive integration of the foregoing identity first from c to t (c < t ≤ b) and then from c to b yields 1 x(c) = (b − c)2 2





c

b

q(s)x(s)ds +

t∗



c

θ

(θ − s)q(s)x(s)dsdθ.

c

Hence, 

1 (b − c)2 + (b − c)2 |x(c)| ≤ |x(c)| 2



b

|q(t)|dt,

a

that is, 

b

|q(t)|dt ≥

a

2 . 3(b − c)2

Since 

b

|q(t)|dt < ∞ and

lim

c→b−

a

1 = ∞, (b − c)2

it follows that c cannot be very close to b. In the second case, suppose c ≤ t∗ ≤ b. Integration of (2.129) from t to t∗ (a ≤ t ≤ t∗ ) yields 



t∗

x (t) =

q(s)x(s)ds. t

Further, integrating the preceding identity successively first from t to c (a ≤ t < c) and then from a to c, we obtain  x(c) =



c

θ a

θ

t∗



t∗

q(s)x(s)dsdθ −c(c−a) c



c

q(s)x(s)ds −



c

sq(s)x(s)dsdθ. a

θ

102

2 Higher-Order Linear Differential Equations

Hence, 

b

|x(c)| ≤ 3c(c − a)|x(c)|

|q(t)|dt,

a

that is, 

b

|q(t)|dt ≥

a

1 . 3c(c − a)

From this inequality, it follows that c cannot be very close to a. Next, we show that c cannot be very close to b. Clearly,   |x(c)| = 

b

c

   x (t)dt  ≤

b



   x (t) dt

c

implies  [x(c)]2 ≤

b

   x (t) dt

c



b

≤ (b − c)



2

2 x  (t) dt

c



b

= −(b − c) c



b

≤ (b − c) 

x(t)x  (t)dt

  |x(t)| x  (t) dt

c b

≤ (b − c)



c

≤ (b − c) [x(c)] 2

t

|x(t)|

|q(s)||x(s)|dsdt

t∗



b

2

|q(t)|dt.

a

Hence, 

b a

|q(t)|dt ≥

1 . (b − c)2

Thus, c cannot be very close to b. The proof is complete.

(2.141) 

In case x  (t) = 0 for t ∈ [a, b], let a, b, c ∈ R with a < b < c be consecutive zeros of x. Hence, there exist t1 ∈ (a, b) and t2 ∈ (b, c) such that x  (t1 ) = 0 and x  (t2 ) = 0. Thus, there exists t3 ∈ (t1 , t2 ) such that x  (t3 ) = 0. Since x  (t) = 0 for t ∈ [a, b], t3 ∈ (b, c). Let

2.3 Odd-Order Differential Equations

103

|x(t4 )| = max |x(t)|,

t4 ∈ (a, b) ∪ (b, c).

where

t∈[a,c]

If t4 ∈ (b, c), then |x(t4 )| = max{|x(t)| : t ∈ [b, c]}, and hence, from Theorem 2.65, it follows that t4 cannot be very close to b as well as to c. Suppose t4 ∈ (a, b). Then, we have the following theorem. Theorem 2.66 The point t4 cannot be very close to a as well as to b. Proof Clearly, x  (t4 ) = 0, and hence 

t4

x(t4 ) =

x  (t)dt

a

implies 

t4

[x(t4 )] ≤ 2

2



x (t)dt a



t4





t4

≤ (t4 − a)

2 x  (t) dt

a

= −(t4 − a) a t4

 ≤ (t4 − a)

x(t)x  (t)dt

  |x(t)| x  (t) dt.

a

Integration of (2.129) from t to t3 yields 



t3

x (t) =

q(s)x(s)ds. t

Hence,    x (t) ≤



t3

|q(s)||x(s)|ds,

t

and therefore 

t4

[x(t4 )]2 ≤ (t4 − a)

 |x(t)|

a



≤ (t4 − a) [x(t4 )] 2

2 a

t c

t3

|q(s)||x(s)|dsdt

|q(t)|dt.

104

2 Higher-Order Linear Differential Equations

Thus, 

c

|q(t)|dt ≥

a

1 . (t4 − a)2

(2.142)

From (2.142), t4 cannot be very close to a. Similarly, starting with   |x(t4 )| = 

b t4

   x (t)dt  ≤ 

b

   x (t) dt,

t4

one may show that 

c

|q(t)|dt ≥

a

1 . (b − t4 )2

(2.143)

 Hence, t4 cannot be very close to b. This completes the proof.  ∗  ∗ Remark 2.67 If max{|x(t)| : t ∈ [a, b]} = x(t4 ), t4 ∈ (a, b), then it is not known if t4∗ could be very close to a or b. It seems that this is because x  (t) = 0 for t ∈ [a, b]. Theorem 2.68 Let q ∈ Lσ ([0, ∞), R), where 1 ≤ σ < ∞. If {tn } is an increasing sequence of zeros of an oscillatory solution x of (2.129), then (tn+1 − tn ) → ∞ or

(tn+2 − tn ) → ∞

as

n → ∞.

Proof Suppose that x  (t) = 0 for some t ∈ [tn , tn+1 ] for every large n. We claim that in this case, (tn+1 −tn ) → ∞ as n → ∞. If not, then there exists a subsequence {tnk } of the sequence {tn } such that (tnk+1 − tnk ) ≤ ε for every k, where ε > 0 is a constant. Let     max |x(t)| : t ∈ [tnk , tnk+1 ] = x(snk ) ,

where

snk ∈ (tnk , tnk+1 ).

Since q ∈ Lσ ([0, ∞), R), we have 

∞

|q(t)|σ dt < ε1/μ+2

!−σ

(2.144)

tnk

for large k, where the constant μ is the Hölder conjugate of σ , i.e., 1/μ + 1/σ = 1. Let x  (dnk ) = 0 for dnk ∈ [tnk , tnk+1 ]. If tnk ≤ dnk < snk , then from (2.140), it follows that  tn k+1 1 |q(t)|dt ≥ . (2.145) (snk − tnk )2 tnk

2.3 Odd-Order Differential Equations

105

From (2.145), by Hölder’s inequality, we get  tn k+1 2 1 ≤ (snk − tnk ) |q(t)|dt tnk



≤ (tnk+1 − tnk ) (tnk+1 − tnk ) 2

 ≤ (tnk+1 − tnk )

1/μ+2

< ε1/μ+2 ε1/μ+2

!−1

∞

1/μ

tnk+1

1/σ |q(t)| dt σ

tnk

1/σ

|q(t)| dt σ

tnk

= 1, a contradiction. If snk ≤ dnk ≤ tnk+1 , then from (2.141), we obtain  tn k+1 1 |q(t)|dt ≥ . (tnk+1 − snk )2 tnk Then, proceeding as before, we arrive at a contradiction. Next, suppose that x  (t) = 0 for t ∈ [tn , tn+1 ] for some large n. In this case, we consider three consecutive zeros tn < tn+1 < tn+2 , and we show (tn+2 − tn ) → ∞ as n → ∞. On the contrary, assume that there exists a subsequence {tnk } of {tn } such that (tnk+2 − tnk ) ≤ ε for every k, where ε > 0 is a constant for which (2.144) holds for large k and x  (t) = 0 for t ∈ [tnk , tnk+2 ]. (If x  (t) = 0 for some t ∈ [tnk , tnk+2 ], then, proceeding as earlier, one may arrive at a contradiction.) Since x(tnk+2 ) = 0, there exists dnk ∈ (tnk , tnk+2 ) such that x  (dnk ) = 0. Let     max |x(t)| : t ∈ [tnk , tnk+1 ] = x(snk ) ,

snk ∈ (tnk , tnk+1 )∪(tnk+1 , tnk+2 ).

where

If snk ∈ (tnk+1 , tnk+2 ), then, proceeding as in the first part of this proof, one may obtain a contradiction. If snk ∈ (tnk , tnk+1 ), then, from (2.142), it follows that  tn k+2 1 |q(t)|dt ≥ . (2.146) (snk − tnk )2 tnk From (2.146), by Hölder’s inequality, we have  1 ≤ (snk − tnk )

2

tnk+2

|q(t)|dt

tnk

≤ (tnk+2 − tnk ) (tnk+2 − tnk ) 2

 1/μ

tnk+2

tnk

1/σ |q(t)| dt σ

106

2 Higher-Order Linear Differential Equations

 ≤ε

1/μ+2

∞

1/σ |q(t)| dt σ

tnk

< 1, 

a contradiction. Thus, the proof is complete. Remark 2.69 One could have used (2.143) in place of (2.142). Example 2.70 Consider the equation x  +

6 x = 0 on t3

[1, ∞).

(2.147)

The basis of the solution space of (2.147) is given by   √ √ 1 2 2 t cos( 2 log t), t sin( 2 log t), . t 1 Here, q(t) = 6/t 3 , and √ hence q ∈ L ([1, ∞), R). The zeros of the oscillatory 2 solution x1 (t) = t cos( 2 log t) of (2.147) are given by

2n − 1 tn = exp √ π , n ∈ N. 2 2 √ √ Since x1 (t) = −3 2 sin( 2 log t), we have x1 (tn ) = 0 and x1 (tn+1 ) = 0 for any n ∈ N. Moreover, the zeros of x1 are given by √ 2

τn = e(n−1)π/

,

n ∈ N.

Since τn < tn < τn+1 ≤ tn+1 , we have x1 (t) = 0 for some t ∈ (tn , tn+1 ). Clearly, tn+1 − tn = e(2n−1)π/(2

√ 2)

√

(eπ/

2

− 1) → ∞ as

n → ∞.

√ A similar conclusion holds for the oscillatory solution x2 (t) = t 2 sin( 2 log t) of (2.147). In the following, we obtain an estimate for the number of zeros of an oscillatory solution of (2.129) on an interval [0, T ]. Theorem 2.71 (i) If x is an oscillatory solution of (2.129) with x(tk ) = 0, k = 1, 2, . . . , N , and x  (tk∗ ) = 0 for some tk∗ ∈ [tk , tk+1 ] for k = 1, 2, . . . , N − 1, where 0 < t1 < t2 < · · · < tN ≤ T , then the inequality 4 −N

(N − 1) 2 holds.

T2 < 16

 0

T

|q(t)|dt

(2.148)

2.3 Odd-Order Differential Equations

107

(ii) If x is an oscillatory solution of (2.129) with x(tk ) = 0, k = 1, 2, . . . , 2N + 1, and x  (tk∗ ) = 0 for t ∈ [t2k−1 , t2k ] for k = 1, 2, . . . , N, where 0 < t1 < t2 < · · · < t2N +1 ≤ T , then the inequality 4 −N

N 2

T2 < 8



T

|q(t)|dt

0

holds. Proof We first prove (i). From Theorem 2.56, it follows that 

tk+1

|q(t)|dt >

tk

4 (tk+1 − tk )2

for k = 1, 2, . . . , N − 1. Hence, 

T



tN

|q(t)|dt ≥

0

|q(t)|dt > 4

t1

N −1

k=1

1 . (tk+1 − tk )2

From the two well-known inequalities (a1 + a2 + · · · + an )2 ≤ 2n−1 (a12 + a22 + · · · + an2 ) and ⎞−1 n n

1 1 ⎠ ≤ (a1 a2 . . . an )1/n ≤ 1 ⎝ aj , n aj n ⎛

j =1

j =1

where aj > 0, j = 1, 2, . . . , n, it follows that N −1

k=1

1 ≥ 2−(N −2) (tk+1 − tk )2 −(N −2)

≥2

N −1

k=1

1 tk+1 − tk

2

−2 N −1

(N − 1) (tk+1 − tk ) 4

k=1 −(N −2)

=2

(N − 1) (tN − t1 )−2 4

> 2−(N −2) (N − 1)4 T −2 .

108

2 Higher-Order Linear Differential Equations

Hence, 

T

T2

|q(t)|dt > 16 × 2−N (N − 1)4 .

0

Next, we prove (ii). Using Theorem 2.60, we obtain 

t2k+1

|q(t)|dt >

t2k−1

4 (t2k+1 − t2k−1 )2

for k = 1, 2, . . . , N . Thus, 

T

 |q(t)|dt ≥

0

t2N+1

|q(t)|dt

t1

>4

N

1 (t2k+1 − t2k−1 )2

k=1

−(N −1)

≥4×2

N

k=1

 −(N −3)

≥2

N

1 t2k+1 − t2k−1

4

N

(t2k+1 − t2k−1 )

4

−2

2

−2

k=1 −(N −3)

>2

N T

,

that is,  T

T

2

|q(t)|dt > 8 × 2−N N 4 .

0

Hence, the proof is complete.



Remark 2.72 Theorem 2.71 (ii) remains valid if x  (t) = 0 for t ∈ [t2k , t2k+1 ], k = 1, 2, . . . , N. Remark 2.73 In all other cases, we separate all consecutive pairs of subintervals on one of which x  (t) = 0 for some t and all those subintervals on each of which x  (t) = 0 for some t, and we apply the earlier two estimates separately. √ Example 2.74 Consider the oscillatory solution x1 (t) = et/2 cos( 3t/2) of (2.136) √ in Example 2.64. The interval [0, 11π/ 3] contains six simple zeros of x1 . Since x1 (tj∗ ) = 0 for some tj∗ ∈ (tj , tj +1 ), j ∈ {1, 2, 3, 4, 5}, from Theorem 2.71 (i), it follows that

2.3 Odd-Order Differential Equations

16 × 54 × 2−6
0.

(2.163)

  (t2 − s)(s − t1 ) x  (s) ds = 0.

(2.164)

t1

If (2.163) is not true, then 

t2 t1

It follows from (2.153) that x(t) ≡ 0 for t ∈ (t1 , t2 ), which contradicts (2.150) since x(t) = 0 for t ∈ (a1 , a2 ) ∪ (a2 , a3 ). Thus, by using (2.163) in (2.162), we get (2.160). This completes the proof.  By using an inequality similar to (1.23) in (2.160), we obtain the following result. Corollary 2.77 If x is a nontrivial solution of (2.151), then the inequality 

t2

|q(s)| ds ≥

t1

16 (t2 − t1 )2

holds, where t1 and t2 are given in (2.152). Now, we give a lower bound for the distance between the end points in the three consecutive zeros of the solution of (2.149). It is easy to see that from Lemma 2.75, we have the inequality 

1 |x(t)| ≤ 4

a3

  (a3 − s)(s − a1 ) x  (s) ds.

(2.165)

a1

Thus, by using (2.165), we have the following main result, whose proof is the same as that of Theorem 2.76 and hence is omitted. Theorem 2.78 If x is a nontrivial solution of (2.151), then the inequality 

a3

a1

(a3 − s)(s − a1 ) |q(s)| ds ≥ 4

(2.166)

2.3 Odd-Order Differential Equations

113

holds. If we use (1.23) in (2.166), then we have the following result. Corollary 2.79 If x is a nontrivial solution of (2.151), then the inequality 

a3

|q(s)| ds ≥

a1

16 (a3 − a1 )2

(2.167)

holds. Remark 2.80 It is easy to see that (2.167) is sharper than (2.132) and (2.200) with n = 3 in the sense that (2.132) and (2.200) with n = 3 follow from (2.167), but not conversely. Therefore, the results obtained improve Theorems 2.56 and 2.88 with n = 3. Remark 2.81 To the best of our knowledge, the inequalities obtained in [29] are the best results in the literature for (2.151). Now, we give some applications of the obtained Lyapunov-type inequalities. We take a different point of view and use (2.167) to obtain an extension of the following oscillation criterion originally due to Lyapunov (see [59]): If x  and x  /x are continuous on [a1 , a3 ] such that x(a1 ) = x(a3 ) = 0, then  a3     x (s)  4    x(s)  ds ≥ a − a 3 1 a1 from (1.2). Thus, (2.167) leads to the following result: If x  and x  /x are continuous on [a1 , ar ] such that x has three zeros (counting multiplicity) including a1 and ar on [a1 , ar ], then  ar     x (s)  16    x(s)  ds ≥ (a − a )2 . r 1 a1 Here, we give another application of (2.167) for the eigenvalue problem 

x  + λh(t)x = 0,

(2.168)

x(a1 ) = x(a2 ) = x(a3 ) = 0. Namely, if there exists a nontrivial solution x of (2.168), then 16 |λ| ≥ (a3 − a1 )2



a3

a1

|h(s)| ds

−1 .

114

2 Higher-Order Linear Differential Equations

2.3.2 General Odd-Order Differential Equations In 2003, Yang [295] improved the results of Parhi and Panigrahi [230] and established a Lyapunov-type inequality for general odd-order differential equations of the form (2.131), where q is a real-valued and continuous function on R. Theorem 2.82 (Lyapunov-Type Inequality) Let (2.131) have a nontrivial solution x satisfying the 2-point boundary conditions in (2.2), where a, b ∈ R with a < b are consecutive zeros. If there exists t∗ ∈ (a, b) such that x (2n) (t∗ ) = 0, then the inequality 

b

|q(t)|dt >

a

n!2n+1 , (b − a)2n

n∈N

(2.169)

holds. Proof Since if x is a solution of (2.131), then so is −x by homogeneity, and thus, without loss of generality, we may assume that x(t) > 0. Since x(a) = x(b) = 0, there exists c ∈ (a, b) such that x  (c) = 0 and x(c) = max x(t). Thus t∈[a,b]

 0 < x(c) =

c

x  (t)dt ≤

a



c

   x (t) dt.

(2.170)

a

Since x  (a) = x  (c) = 0, there exists t21 ∈ (a, c) such that x  (t21 ) = 0, and from x  (c) = x  (b) = 0, there exists t22 ∈ (c, b) such that x  (t22 ) = 0. In this way, we see j that there exist {tjk }k=1 ∈ (a, b) such that x (j ) (tjk ) = 0, k = 1, 2, . . . , j and j +1

j

0 < tj1+1 < tj1 < c < tj < tj +1 < b,

j = 1, 2, . . . , n.

n−j

k ) = 0, k = 1, 2, . . . , n − Hence, there exist {tjk+n }k=1 ∈ (a, b) such that x (n+j ) (tn+j  j , j = 1, 2, . . . , n − 1. From x (a) = 0, we have

x  (t) =



t

x  (s)ds.

a

Hence,    x (t) ≤



t

   x (s) ds

for

t ∈ [a, c].

a

By (2.170), we have 

c

x(c) ≤ a

   x (t) dt ≤



c a

 a

t

   x (s) ds dt.

(2.171)

2.3 Odd-Order Differential Equations

115

By using the equality 

c



a

 f (t, s)ds dt =

t

c



c

f (t, s)dt ds

a

a

s

c

 c      x (s) dt ds = (c − s) x  (s) ds.

in (2.171), we get  x(c) ≤

c



a

s

(2.172)

a

Since for 1 ≤ k ≤ n − 1,   t    t    (k)    (k+1)  (k+1) (s)ds  ≤ (s) ds, x (t) =  x x a

a

(2.172) yields 

c

x(c) ≤ a

(c − t)n  (n)  x (t) dt. (n − 1)!

(2.173)

n−j

From the above analysis, there exist {tjk+n }k=1 , j = 1, 2, . . . , n − 1, and tn1 ∈ (a, c), k ) = 0, k = 1, 2, . . . , n − j . Therefore, tnn ∈ (c, b), such that x(tn+j     t   (n)   (n+1)  (s) ds ≤ x (t) ≤ x tn1

a

c

   (n+1)  (s) ds x

for t ∈ [a, c]

and     (n+1)  (t) ≤ x 

t 1 tn+1

b

≤ 

a b

≤ a

   (n+2)  (s) ds x

   (n+2)  (s) ds x 

b a

   (n+3)  (s) dsdτ1 x

(2.174)

≤ ···  b  b b   (2n)  ··· ≤ x (s) dsdτ1 . . . dτn−2 . a + a a,. n−1

On the other hand, by assumption, there exists t∗ ∈ (a, b) such that x (2n) (t∗ ) = 0. This gives

116

2 Higher-Order Linear Differential Equations

  t    (2n)   (2n+1)  (s) ds x (t) ≤ x 

t∗

b

≤ a



b

=

   (2n+1)  (s) ds x (2.175) |q(s)x(s)| ds

a



b

≤ x(c)

|q(s)| ds.

a

Using (2.173), (2.174), and (2.175), we obtain  x(c) ≤

(b − a)n (c − a) (n − 1)!



c

 (c − t)

a

(b − a)n−1 (c − a)n+1 ≤ x(c) n!

n−1



b

dt x(c)

|q(t)|dt

a



b

|q(t)|dt,

a

and dividing both sides of this inequality by x(c) > 0, we get 

b

|q(t)|dt ≥

a

n! . (b − a)n−1 (c − a)n+1

(2.176)

n! . (b − a)n−1 (b − c)n+1

(2.177)

Similarly, we may prove that 

b

|q(t)|dt ≥

a

Since the function f (u) = u−(n+1) is convex (note f  (u) > 0) for u > 0, we have 1 1 2 2n+2 + > = . (b − c)n+1 (c − a)n+1 [(b − a)/2]n+1 (b − a)n+1 Finally, (2.176), (2.177), and (2.178) imply (2.169), completing the proof.

(2.178) 

Remark 2.83 When n = 1, Theorem 2.82 reduces to the result of Parhi and Panigrahi [230], i.e., Theorem 2.56. Moreover, from (2.176) and (2.177), it can be seen that c cannot be very close to a as well as to b, which is also a generalization of Theorem 2.56. Corollary 2.84 Let (2.131) have a nontrivial solution x satisfying the N -point boundary conditions x(tk ) = 0

for k = 1, 2, . . . , N,

2.4 General Higher-Order Differential Equations

117

where tk ∈ (a, b) with a ≤ t1 < t2 < · · · < tN ≤ b are consecutive zeros. If there exist tˆk ∈ (tk , tk+1 ) for each k = 1, 2, . . . , N − 1 such that x (2n) (tˆk ) = 0, then the inequality (N − 1)2n+1


tk

n!2n+1 (tk+1 − tk )2n

for

k = 1, 2, . . . , N − 1.

Hence, 

b

|q(t)|dt ≥

a

N −1  tk+1

k=1

|q(t)|dt > n!2

n+1

tk

N −1

k=1

1 . (tk+1 − tk )2n

(2.180)

Since f (u) = u−2n is convex for u > 0, we have for xk = tk+1 − tk > 0, k = 1, 2, . . . , N − 1, N −1

 f (xk ) > (N − 1)f

k=1

 N −1 1 xk , N −1 k=1

that is, N −1

k=1

1 (N − 1)2n+1 (N − 1)2n+1 > ≥ . 2n 2n (tk+1 − tk ) (tN − t1 ) (b − a)2n

Hence, (2.180) and (2.181) give (1.60). This completes the proof.

(2.181) 

Remark 2.85 When n = 1, (2.179) is sharper than (2.148) given by Parhi and Panigrahi [230] in Theorem 2.71 for N ≥ 3.

2.4 General Higher-Order Differential Equations In this section, we first consider nth order differential equations of the form x (n) + q(t)x = 0,

n ∈ N \ {1}

(2.182)

118

2 Higher-Order Linear Differential Equations

satisfying the n-point boundary conditions x(ak ) = 0 for

k = 1, 2, . . . , n,

(2.183)

where n ∈ N \ {1} and q is a locally Lebesgue integrable and real-valued function defined on R, i.e.,  (n) x + q(t)x = 0, x(ak ) = 0

for

n ∈ N \ {1},

(2.184)

k = 1, 2, . . . , n.

Secondly, we consider (2.182) satisfying the n-point boundary conditions x(ai ) = x  (ai ) = x  (ai ) = · · · = x (ki ) (ai ) = 0 for

i = 1, 2, . . . , r,

(2.185)

where a1 < a2 < · · · < ar and r+

r

kj = n,

ki ≥ 0 for all

i = 1, 2, . . . , r,

j =1

i.e., 

x (n) + q(t)x = 0,

n ∈ N \ {1},

x(ai ) = x  (ai ) = x  (ai ) = · · · = x (ki ) (ai ) = 0 for

i = 1, 2, . . . , r. (2.186)

Finally, we consider the equation x (n) + (−1)n−k−1 q(t)x = 0 with n ∈ N \ {1}

(2.187)

satisfying the (k, n − k)-conjugate boundary conditions x (i) (a) = 0,

i = 0, 1, . . . , k − 1 and

x (j ) (b) = 0,

j = 0, 1, . . . , n − k − 1. (2.188) The proofs are based on Green’s functions for the related problems.

2.4.1 Multiple-Point Boundary Value Problems Lyapunov-type and Hartman-type inequalities and their several generalizations to higher-order nonlinear equations having different kinds of nonlinearities have been studied by many authors. In 1975, Singh [258] considered nth-order forced linear differential equations of the form

2.4 General Higher-Order Differential Equations

119

 (n−1) r(t)x  + q(t)x = f (t),

n ∈ N \ {1},

(2.189)

where q, f, r are continuous and real-valued functions on R and r is positive. He proved that under the conditions 

∞



∞

t n−2 |q(t)|dt < ∞,

 t n−2 |f (t)|dt < ∞,

∞

and

1 dt < ∞, r(t) (2.190)

every oscillatory solution of (2.189) tends to zero as t → ∞. He also proved that, under the first and last condition in (2.190), every solution of (2.189) is nonoscillatory provided f (t) ≡ 0. To obtain these oscillation and nonoscillation results, he made use of the Lyapunov-type inequality 4 1 ≤ L (n − 2)!

 a

b

dt r(t)



b

(t − a)n−2 |q(t)|dt ≤

a

1 , L

where L = sup{|f (x(t))| : x(t) ∈ [0, ∞)}, and a and b are two consecutive zeros of x with a < b. However, the Lyapunov-type or Hartman-type inequalities presented in these works depend on the number L, which obviously we do not know in advance. It appears that Green’s function for − x (n) = 0,

n ∈ N \ {1}

(2.191)

satisfying (2.183), i.e.,  (n) −x = 0, x(ak ) = 0

n ∈ N \ {1}, for

k = 1, 2, . . . , n,

(2.192)

was first given by Beesack [40] (see also [104]), and he proved that |Gn (t, s)| ≤

n % 1 |t − ak | , (n − 1)!(an − a1 )

(2.193)

k=1

and for t ∈ [a, b],

n % 1 n−1 (an − a1 )n−1 1 |Gn (t, s)| ≤ |t − ak | ≤ 1 − . (n − 1)!(an − a1 ) n n! k=1 (2.194) In 2003, by using (2.194), Yang [295] established the following Lyapunov-type inequality for (2.184).

120

2 Higher-Order Linear Differential Equations

Theorem 2.86 (Lyapunov-Type Inequality) Let (2.184) have a nontrivial solution, where ak ∈ R, k = 1, 2, . . . n, and a1 < a2 < · · · < an−1 < an are consecutive zeros. Then, the inequality 

b

|q(t)|dt >

a

(n − 2)!nn−1 , (n − 1)n−2 (b − a)n−1

n ∈ N \ {1}

(2.195)

holds, where a = a1 and b = an . Proof Let x be a nontrivial solution of (2.184), where ak ∈ R, k = 1, 2, . . . n, and a = a1 < a2 < · · · < an−1 < an = b are consecutive zeros. Then, by using Green’s function for (2.184), x can be expressed as 

b

x(t) =

Gn (t, s)q(s)x(s)ds.

(2.196)

a

Now, let |x(c)| = max |x(t)|. Then, by (2.194) and (2.196), we obtain t∈(a,b)

 |x(c)| ≤

b

|Gn (c, s)||q(s)||x(s)|ds

a

(n − 1)n−2 (b − a)n−1 |x(c)| < nn−1 (n − 2)!



b

(2.197) |q(s)|ds.

a

Dividing both sides of (2.197) by |x(c)| > 0 completes the proof.



Remark 2.87 When n = 3, (2.195) turns into 

b

|q(t)|dt >

a

9 , 2(b − a)2

(2.198)

which is sharper than (2.135) given by Parhi and Panigrahi [230] in Theorem 2.60. However, when n = 2, (2.195) turns into  a

b

|q(t)|dt >

2 , b−a

(2.199)

which is weaker than the classical Lyapunov inequality. In 2010, Çakmak obtained the following theorem (see [68, Theorem 1]), the proof of which, after careful examination, is almost the same as that of Theorem 2.86 of Yang [295], and hence it is omitted. Theorem 2.88 (Lyapunov-Type Inequality) Let (2.184) have a nontrivial solution, where ak ∈ R, k = 1, 2, . . . n, and a1 < a2 < · · · < an−1 < an are consecutive zeros. Then, the inequality

2.4 General Higher-Order Differential Equations



b

121

(n − 2)!nn , (n − 1)n−1 (b − a)n−1

|q(t)|dt >

a

n ∈ N \ {1}

(2.200)

holds, where a = a1 and b = an . Remark 2.89 When n = 3, (2.200) turns into 

b

|q(t)|dt >

a

27 , 4(b − a)2

(2.201)

which is sharper than the results of Parhi and Panigrahi [230] and Yang [295], i.e., Theorems 2.60 and 2.86, respectively. Moreover, when n = 2, (2.200) turns into the classical Lyapunov inequality. Secondly, we consider (2.186). In 1962, Beesack [40] proved that Green’s function Gn (t, s) for (2.191) satisfying (2.185), i.e., 

−x (n) = 0,

n ∈ N \ {1},

x(ai ) = x  (ai ) = x  (ai ) = · · · = x (ki ) (ai ) = 0

for i = 1, 2, . . . , r, (2.202)

satisfies the inequality % 1 |t − ai |ki +1 (n − 1)!(ar − a1 ) r

|Gn (t, s)| ≤

(2.203)

i=1

for a1 < s < ar and −∞ < t < ∞. He proved also that

1 n−1 (ar − a1 )n−1 |Gn (t, s)| ≤ 1 − n n!

(2.204)

for a1 < t, s < ar by using the inequality r % i=1

1 n−1 (ar − a1 )n |t − ai |ki +1 ≤ 1 − n n

(2.205)

for a1 < t < ar . In case r = n, (2.204) reduces to (2.194), and it also leads to an extension of oscillation criterion of Lyapunov for the case n = 2. In 1983, Agarwal [3] improved (2.205). He obtained the inequality r % i=1

|t − ai |ki +1 ≤

(n − σ − 1)n−σ −1 (σ + 1)σ +1 (ar − a1 )n , nn

(2.206)

122

2 Higher-Order Linear Differential Equations

where σ = min{k1 , kr } for r ∈ N\{1}. Now, by using (2.203) and (2.206), the Green function Gn (t, s) for (2.202) satisfies a better inequality than (2.204), namely, |Gn (t, s)| ≤

(n − σ − 1)n−σ −1 (σ + 1)σ +1 (ar − a1 )n−1 . nn−1 n!

(2.207)

We note that when r = n, σ becomes zero, and hence (2.207) reduces to (2.194). In 2015, by using (2.207), Agarwal and Özbekler [17] established the following Lyapunov-type inequality for (2.186). Theorem 2.90 (Lyapunov-Type Inequality) Suppose that x solves (2.186). If x(t) = 0 in (aj , aj +1 ), j = 1, 2, . . . , r − 1, then the inequality 

ar

|q(t)|dt >

a1

nn−1 (σ + 1)−(σ +1) n! × (n − σ − 1)n−σ −1 (ar − a1 )n−1

(2.208)

holds, where σ = min{k1 , kr }, r ∈ N \ {1}. Proof Let (2.186) have a nontrivial solution. If x(t) = 0 for t ∈ (aj , aj +1 ), j = 1, 2, . . . , r −1, then, by using Green’s function for (2.202), x(t) can be expressed as  x(t) =

ar

Gn (t, s)q(s)x(s)ds.

(2.209)

a1

Now, let |x(c)| = max |x(t)|. Then, by (2.207) and (2.209), we obtain t∈(a1 ,ar )

 |x(c)| ≤

ar

|Gn (c, s)||q(s)||x(s)|ds

a1

(n − σ − 1)n−σ −1 < (σ + 1)σ +1 (ar − a1 )n−1 |x(c)| nn−1 n!



Dividing both sides of (2.210) by |x(c)| > 0 completes the proof.

ar a1

|q(s)|ds. (2.210) 

When r = n, (2.186) reduces to (2.184) with a1 = a and ar = an = b. Then, Theorem 2.90 yields the following result. Corollary 2.91 Let x be a solution of (2.184), where ak ∈ R, k = 1, 2, . . . n, and a1 < a2 < · · · < an−1 < an are consecutive zeros. If x(t) = 0 for t ∈ (aj , aj +1 ), j = 1, 2, . . . , n − 1, then the inequality  a

b

|q(t)|dt >

n n−1

n−1

holds, where a = a1 and b = an .

n! , (b − a)n−1

n ∈ N \ {1}

(2.211)

2.4 General Higher-Order Differential Equations

123

Proof Since σ = 0, when r = n, (2.208) immediately implies (2.211) with a1 = a and ar = b.  Remark 2.92 Corollary 2.91 is remarkable since (2.211) is sharper than both inequalities obtained in Yang [295, Theorem 2] and Çakmak [68, Theorem 1], i.e., (2.195) and (2.200), for all n ∈ N \ {1}. Remark 2.93 When n = 3, (2.211) turns into 

b

|q(t)|dt >

a

27 , 2(b − a)2

which is sharper than the results of Parhi and Panigrahi [230], Yang [295], and Çakmak [68], i.e., (2.132), (2.198), and (2.201). Moreover, when n = 2, (2.211) turns into the classical Lyapunov inequality.

2.4.2 Conjugate Boundary Value Problems When r = 2, k1 = k − 1, and k2 = n − k − 1, (2.185) turns into (2.188), where a = a1 and b = a2 . The following result is an analogue of Theorem 2.90. Theorem 2.94 (Lyapunov-Type Inequality) Suppose x is a solution of (2.182) satisfying (2.188), where a < b. If x(t) = 0 in (a, b), then the inequality  a

b

|q(t)|dt >

nn−1 n! × k k (n − k)n−k (b − a)n−1

(2.212)

holds. Proof Since σ = min{k − 1, n − k − 1}, there are two cases, one is σ = k − 1 and the other is σ = n − k − 1. In both cases, we have (σ + 1)−(σ +1) 1 = k . n−σ −1 k (n − k)n−k (n − σ − 1) Using this in (2.208) in Theorem 2.90, we obtain (2.212) immediately.



In 1976, Gustafson [145] obtained Green’s function Gn (t, s) for the (k, n − k)conjugate boundary value problem ⎧ −x (n) = 0, ⎪ ⎪ ⎨ x (i) (a) = 0, ⎪ ⎪ ⎩ (j ) x (b) = 0,

n ∈ N \ {1}, i = 0, 1, . . . , k − 1, j = 0, 1, . . . , n − k − 1

(2.213)

124

2 Higher-Order Linear Differential Equations

as ⎧ ⎫



−1 k−1 ⎨k−j

n−k+i−1 t − a i ⎬ (t − a)j (a − s)n−j −1 Gn (t, s) = ⎩ i b−a ⎭ j !(n − j − 1)! j =0

×

i=0

b−t b−a

k a≤s≤t ≤b

for

(2.214)

and ⎧

Gn (t, s) = −

n−k−1

⎨n−k−j

−1 j =0

×

⎩

t −a b−a

k+i−1 i

i=0



b−t b−a

⎫

i ⎬

(t − b)j (b − s)n−j −1 ⎭ j !(n − j − 1)!

k a ≤ t ≤ s ≤ b.

for

(2.215)

It was shown by Yang [295] that for a ≤ s, t ≤ b, simply  |Gn (t, s)| ≤

ψ1 (s)

if a ≤ s ≤ t ≤ b,

ψ2 (s)

if a ≤ t ≤ s ≤ b,

(2.216)

where ψ1 (s) =

(b − s)n−k (s − a)n−k {(b − a) + (s − a)}k−1 , (k − 1)!(n − k)!(b − a)n−k

(b − s)k (s − a)k {(b − a) + (b − s)}n−k−1 . ψ2 (s) = k!(n − k − 1)!(b − a)k

(2.217)

In 2003, Yang [295] established the following Hartman-type and Lyapunov-type inequalities for (2.187) satisfying (2.188) by employing Green’s function for (2.213) given by Gustafson [145]. Theorem 2.95 (Hartman-Type Inequality) Let x be a solution of (2.187) satisfying (2.188), where a < b. If x(t) = 0 for t ∈ (a, b), then the inequality 

b

Ψ (t)|q(t)|dt > 1

(2.218)

a

holds, where Ψ (t) := max{ψ1 (t), ψ2 (t)} and ψ1 and ψ2 are given by (2.217). Proof Let x be a nontrivial solution of (2.187) satisfying (2.188), where a < b. If x(t) = 0 for t ∈ (a, b), then, by using Green’s function for (2.213), x(t) can be expressed as

2.4 General Higher-Order Differential Equations



b

x(t) =

125

Gn (t, s)q(s)x(s)ds.

(2.219)

a

Now, let |x(c)| = max |x(t)|. Then, by (2.216) and (2.219), we obtain t∈(a,b)



b

|x(c)| ≤

|Gn (c, s)||q(s)||x(s)|ds

a



(2.220)

b

< |x(c)|

Ψ (s)|q(s)|ds. a

Dividing both sides of (2.220) by |x(c)| > 0, we complete the proof.



Theorem 2.96 (Lyapunov-Type Inequality) Let (2.187) have a solution x satisfying (2.188), where a < b. If x(t) = 0 for t ∈ (a, b), then the inequality  a

b

|q(t)|dt > max{ank , bnk }

(2.221)

holds, where ank =

(k − 1)!(n − k)!22n−3k+1 (b − a)2n−1

bnk =

and

k!(n − k − 1)!23k−n+1 . (b − a)n−1

Proof We have (b − a)2 ≥ (b − t)(t − a), 4 (b − a) + (t − a) ≤ 2(b − a), and (b − a) + (b − t) ≤ 2(b − a), 

and hence (2.218) immediately implies (2.221).

However, the function ψ1 given in (2.217) is not complete. Agarwal and Özbekler [17] obtained the modified form of ψ1 (s). For this, they need the following lemma. Lemma 2.97 Let the function Gn (t, s) be defined as in (2.214) and (2.215). Then, |Gn (t, s)| ≤

 θ1 (t, s)

if

a ≤ s ≤ t ≤ b,

θ2 (t, s)

if

a ≤ t ≤ s ≤ b,

126

2 Higher-Order Linear Differential Equations

where θ1 (t, s) =

(b − t)k (s − a)n−k {t + s − 2a}k−1 (k − 1)!(n − k)!(b − a)k

(2.222)

(b − s)k (t − a)k {2b − t − s}n−k−1 . k!(n − k − 1)!(b − a)k

(2.223)

and θ2 (t, s) =

Proof For a ≤ s ≤ t ≤ b, we get ⎧

|Gn (t, s)| ≤

(b

− t)k (s

k−1 k−j −1 − a)n−k ⎨ n − k

(b − a)k ×

j =0

⎩

i=0

⎫



i ⎬ +i−1 t −a i b−a ⎭

(t − a)j (s − a)k−j −1 . j !(n − j − 1)!

Since (t − a)/(b − a) ≤ 1, we have k−j −1

i=0



k−j

−1 n − k + i − 1 n − j − 1

n−k+i−1 t −a i = . ≤ i i k−j −1 b−a i=0

The last two inequalities imply that |Gn (t, s)| ≤

k−1 (b − t)k (s − a)n−k n − j − 1 (t − a)j (s − a)k−j −1 k−j −1 (b − a)k j !(n − j − 1)! j =0

k−1

k−1 (b − t)k (s − a)n−k (t − a)j (s − a)k−1−j = j (b − a)k (n − k)!(k − 1)! j =0

=

− t)k (s

− a)n−k

(b (t + s − 2a)k−1 (b − a)k (n − k)!(k − 1)!

= θ1 (t, s) for a ≤ s ≤ t ≤ b. In a similar way, it can be shown that |Gn (t, s)| ≤ θ2 (t, s) holds for a ≤ t ≤ s ≤ b.



2.4 General Higher-Order Differential Equations

127

We note that

k n (b − s)k (s − a)n−k (s + b − 2a)k−1 n! k (b − a)k

(2.224)



n − k n (b − s)k (s − a)k (2b − a − s)n−k−1 n! k (b − a)k

(2.225)

θ1 (s) := θ1 (t, s) ≤ / and θ2 (t, s) ≤ / θ2 (s) :=

for a ≤ t ≤ b. Employing the handy inequalities in (2.224) and (2.225), Agarwal and Özbekler [17] improved and extended the results of Yang [295] for (k, n − k)-conjugate boundary value problems consisting of (2.187) and (2.188) as follows. Theorem 2.98 (Hartman-Type Inequality) Let x be a solution of (2.187) satisfying (2.188), where a < b. If x(t) = 0 for t ∈ (a, b), then the inequality 

b

(b − t)k Φ(t)|q(t)|dt > k!(n − k)!(b − a)k

(2.226)

a

holds, where Φ(t) := max

t∈[a,b]



k(t − a)n−k (t + b − 2a)k−1 , (n − k)(t − a)k (2b − a − t)n−k−1 . (2.227)

Proof Let x be a positive solution of (2.187) satisfying (2.188), where a < b. Since if x is a solution of (2.187), then so is −x by homogeneity, we may, without loss of generality, assume that x(t) > 0. Then, by using Green’s function Gn (t, s) for (2.213), x(t) can be expressed as  x(t) =

b

(−1)n−k Gn (t, s)q(s)x(s)ds.

(2.228)

a

Let |x(c)| = max |x(t)|. Using this, (2.224), and (2.225) in (2.228), we obtain t∈(a,b)

 b    n−k  |x(c)| =  (−1) Gn (c, s)q(s)x(s)ds  

a

≤

b

|Gn (c, s)||q(s)||x(s)|ds

(2.229)

a



 b n 1 |x(c)| ≤ (b − s)k Φ(s)|q(s)|ds. n!(b − a)k k a Dividing both sides of (2.229) by |x(c)| > 0 completes the proof.



128

2 Higher-Order Linear Differential Equations

In the sequel, the following lemma is used. Lemma 2.99 (See [17, Lemma 3.1]) Let k and n be positive constants. If n > k, then (b − t)k (t − a)n−k ≤

k k (n − k)n−k (b − a)n nn

(2.230)

for all t ∈ [a, b] with equality holding if and only if t = (k/n)a + (1 − k/n)b. Proof Clearly, when t = a or t = b, (2.230) is obvious. On the other hand, for t ∈ (a, b), using (2.206) with r = 2, a1 = a, ar = a2 = b, k1 = k − 1 and k2 = n − k − 1, we obtain (b − t)k (t − a)n−k ≤

(n − σ − 1)n−σ −1 (σ + 1)σ +1 (b − a)n , nn

(2.231)

where σ = min{k − 1, n − k − 1}. In both cases σ = k − 1 or σ = n − k − 1, (2.231) yields (2.230). Moreover, if we define the function f (t) := (b − t)k (t − a)n−k ,

t ∈ [a, b],

n > k > 0,

then it is easy to see that f attains its maximum at t0 = (k/n)a + (1 − k/n)b, and  hence equality in (2.230) holds if and only if f (t) = f (t0 ) for all t ∈ [a, b]. Note that if k = 0 or k = n, then (2.230) is still true for all t ∈ [a, b] with the convention that 00 = 1. Moreover, when n = 2k = 2, (2.230) turns into (1.23). Note also that the inequality (b − t)n−k (t − a)k ≤

k k (n − k)n−k (b − a)n , nn

t ∈ [a, b]

(2.232)

is true for n > k. Inequality (2.230) allows to show the following Lyapunov-type inequality. Theorem 2.100 (Lyapunov-Type Inequality) Suppose x is a solution of (2.187) satisfying (2.188), where a < b. If x(t) = 0 for t ∈ (a, b), then the inequality  a

b

|q(t)|dt >

2nn k!(n − k)! 2 k k (n − k)n−k (b − a)n−1 Φnk

(2.233)

holds, where  Φnk := max k2k , (n − k)2n−k ,

k = 1, . . . , n − 1.

(2.234)

2.4 General Higher-Order Differential Equations

129

Proof In view of (2.222) and (2.223), we have (b − t)k (t − a)n−k (2t − 2a)k−1 (k − 1)!(n − k)!(b − a)k

2k−1 k n (b − t)k (t − a)n−k = (t − a)k−1 n! k (b − a)k

2k k n (b − t)k (t − a)n−k ≤ 2n! k b−a

(2.235)

(b − t)k (t − a)k (2b − 2t)n−k−1 k!(n − k − 1)!(b − a)k

2n−k−1 (n − k) n (b − t)n−k (t − a)k = (b − t)k−1 n! k (b − a)k

2n−k (n − k) n (b − t)n−k (t − a)k ≤ 2n! k b−a

(2.236)

θ1 (t, s) ≤

and θ2 (t, s) ≤

for a ≤ t ≤ b. Using (2.230) and (2.232), (2.235) and (2.236) turn into

2k k n k k (n − k)n−k θ1 (t, s) ≤ (b − a)n−1 2n! k nn

(2.237)

and θ2 (t, s) ≤



2n−k (n − k) n k k (n − k)n−k (b − a)n−1 , 2n! k nn

(2.238)

respectively. Then, using (2.237) and (2.238) as in the proof of Theorem 2.98 in (2.229), we obtain the inequality  |x(c)| ≤

b

|Gn (c, s)||q(s)||x(s)|ds

a

k  b n k (n − k)n−k n−1 ≤ Φnk (b − a) |x(c)| |q(s)|ds. k 2n!nn a Dividing both sides of (2.239) by x(c) > 0 completes the proof.

(2.239)



Remark 2.101 Theorems 2.98 and 2.100 are remarkable since when n = 2 and k = 1, (2.187) reduces to (1.1), and (2.226) and (2.233) reduce to the classical inequalities

130

2 Higher-Order Linear Differential Equations



b

(b − t)(t − a)|q(t)|dt > b − a

a

and 

b

|q(t)|dt >

a

4 , b−a

respectively.

2.5 Notes and References The results of Sect. 2.2 are taken from [105] and [299]. Theorem 2.21 in Sect. 2.2.2 is adopted from Yang’s paper [295, Theorem 4]. Thereafter, as an of improvement of Yang’s result, Theorem 2.22 is taken from Çakmak [68, Theorem 2]. In 2012, He and Tang [158] improved and generalized Theorems 2.21 and 2.22 to (2.1)–(2.3), and they obtained the inequality given in Theorem 2.25, using Lemma 2.26. Shortly afterwards, motivating by the results given in [68, 158, 295], Zhang and He [303] improved and generalized Theorems 2.21, 2.22, and 2.25 by applying Lemma 2.27 by Almansi [30]. We advise the reader that Lemma 2.28, Theorems 2.29 and 2.30, and Corollary 2.32 can be found in [303]. On the other hand, in 2015, Theorems 2.34 and 2.35 may be improved by extracting [16, Lemma 2.1] by Agarwal to Lemma 2.33. Moreover, the remainder of Sect. 2.2.2 is taken from [20, page 3, Lemma 2.1, Lemma 2.2]. The results of Sect. 2.2.3 are adopted from the paper by Watanabe et al. [287]. In 1999, Parhi and Panigrahi [230] established Lyapunov-type inequalities for third-order differential equations. These results are presented in Sect. 2.3.1 as Theorems 2.56–2.71. The remainder of Sect. 2.3.1 is adopted from the paper by Akta¸s et al. [29]. In 2003, Yang [295] improved the results of Parhi and Panigrahi [230] and established Lyapunov-type inequalities for general odd-order differential equations, see Theorem 2.82 and Corollary 2.84 in Sect. 2.3.2. Section 2.4 is devoted to general higher-order differential equations. In 1975, Singh [258], making use of Lyapunov-type inequalities for nth-order forced linear differential equations of the form (2.189), obtained some oscillation and nonoscillation results. However, the Lyapunov-type or Hartman-type inequalities presented in these works depend on the number L = sup{|f (x(t))| : x(t) ∈ [0, ∞)}, which obviously we do not know in advance.

2.5 Notes and References

131

In 2003, by using properties of Green’s function for the nth-order, n-point boundary conditions, first given by Beesack [40], Yang [295] established a Lyapunov-type inequality for (2.182)–(2.183), see Theorem 2.86 in Sect. 2.4.1. In Theorem 2.88 in Sect. 2.4.1, Çakmak [68, Theorem 1] proved a small extension of Theorem 2.86 of Yang [295]. Using the property (2.207) of Green’s function, which is improved by Agarwal [3] from the result of Beesack [40], Agarwal and Özbekler [17] established a Lyapunov-type inequality for (2.182)–(2.185), see Theorem 2.90 and Corollary 2.91, in which the inequalities are sharper than the results of Parhi and Panigrahi [230], Yang [295], and Çakmak [68]. Section 2.4.2 collects some recent Lyapunov-type inequalities for conjugate boundary value problems. In 2003, Yang [295] established Hartman and Lyapunovtype inequalities for (k, n − k)-conjugate boundary value problems by employing Green’s function given by Gustafson [145], see Theorems 2.95 and 2.96, respectively. However, the function ψ1 given in (2.217) is not complete. Agarwal and Özbekler [17] obtained a modified form of ψ1 by proving Lemma 2.97, and using it and Lemma 2.99, obtained Theorems 2.98 and 2.100, respectively. At the end of Sect. 2.4.2, it is also noted that Theorems 2.98 and 2.100 reduce to the classical Hartman and Lyapunov inequalities, respectively.

Chapter 3

Lyapunov-Type Inequalities for Half-Linear Differential Equations

3.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for second-order, third-order, and higher-order half-linear differential equations and sketch some recent developments related to this type of inequalities. In Sect. 3.2, we deal with Hartman-type and Lyapunov-type inequalities for second-order half-linear differential equations and their applications to disconjugacy. Moreover, we present some developments for second-order half-linear differential equations with damping term. In particular, we show how the classical Hartman and Lyapunov inequalities for second-order linear differential equations can be extended to half-linear ones without using Green’s function as a tool. In Sect. 3.3, we present related results for third-order half-linear differential equations satisfying some different boundary conditions. In that section, we also present some extensions of the obtained results to more general forms of half-linear equations, and we summarize the results for the linear case. In Sect. 3.3.4, we apply the results on Lyapunov-type inequalities obtained in Sect. 3.3.1 to study nonexistence, uniqueness, and existence-uniqueness for solutions of certain third-order boundary value problems. Finally, Sect. 3.4 contains some results on higher-order half-linear differential equations.

3.2 Second-Order Half-Linear Equations In this section, we first consider the half-linear differential equation   r(t)Φα (x  ) + q(t)Φα (x) = 0

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_3

on

[a, b]

(3.1)

133

134

3 Half-Linear Differential Equations

with (1.5), where Φα (s) = |s|α−1 s

with α > 0

and r, q are real-valued functions with r(t) > 0. We shall assume that the potential q ∈ L1 [a, b] and the function r ∈ L−1/α [a, b], α > 0. Moreover, it is not assumed that the potential q is of definite sign. In the proof of the results, the following lemma plays a crucial rôle. Lemma 3.1 (Jensen’s Inequality [215]) Let f be a convex function defined on an interval J ⊂ R. If uk ∈ J and ck ∈ R+ , k = 1, . . . , n, with n

ck = 1,

k=1

then the inequality f

 n

 ck uk

n

≤

k=1

(3.2)

ck f (uk )

k=1

holds. It appears that the first generalization of Lyapunov’s result for half-linear equations was obtained in 2003 by Yang [294] for   r(t)Φα (x  ) + q(t)Φα (x) = 0 on

[a, b],

x(a) = x(b) = 0.

(3.3)

He showed that there is a striking similarity between linear and half-linear equations and obtained the following result. Theorem 3.2 (Lyapunov-Type Inequality) Let x be a solution of (3.3), where a < b. If x(t) = 0 for all t ∈ (a, b), then the inequality 

b

+

 α+1

q (t)dt > 2 a

b

[r(t)]

−1/α

−α dt

(3.4)

a

holds. Proof Let x be a nontrivial solution of (3.3), where a < b. If x(t) = 0 for all t ∈ (a, b), then we may assume that x(t) > 0 for all t ∈ (a, b). In fact, if x(t) < 0 for all t ∈ (a, b), then we can consider −x, which is also solution. Let c ∈ (a, b) be the least point of the local maxima of x in (a, b), i.e., x  (c) = 0 and x  (t) > 0 on [a, c). Then, we have

3.2 Second-Order Half-Linear Equations



c

[x(c)]α+1 =

x  (t)dt

135

α+1

a



c

= 

[r(t)]−1/(α+1) [r(t)]1/(α+1) x  (t)dt

a c

≤

[r(t)]−1/α dt

α 

a

α+1 (3.5)

 α+1 r(t) x  (t) dt

c a

by Hölder’s inequality. On the other hand, multiplying (3.1) by x(t) and integrating the resulting equation from a to c by parts, we obtain 

c



 α+1 r(t) x  (t) dt =

a

c



c

q(t)[x(t)]α+1 dt ≤ [x(c)]α+1

a

q + (t)dt.

(3.6)

a

Then, by using (3.5) and (3.6), we obtain 

c

[r(t)]

−1/α

α 

c

dt

a

q + (t)dt ≥ 1.

(3.7)

a

Similarly, if d is the greatest point of the local maxima of x in (a, b), i.e., x  (d) = 0 and x  (t) < 0 on (d, b), then 

b

[r(t)]

−1/α

α 

b

dt

d

q + (t)dt ≥ 1.

(3.8)

d

On the other hand, since the function f (u) = u−α is convex for u > 0, (3.2) in Lemma 3.1 with 

c

u1 =

[r(t)]

−1/α

 dt,

u2 =

a

b

[r(t)]−1/α dt

d

and c1 = c2 = 1/2 implies 

c

[r(t)]

−1/α

−α dt



b

+

a

[r(t)]

−1/α

−α dt

d



c

≥2α+1 

[r(t)]−1/α dt +

a b

≥2

α+1 a

[r(t)]

−1/α



−α dt

b

d

.

[r(t)]−1/α dt

−α (3.9)

136

3 Half-Linear Differential Equations

Finally, using (3.7), (3.8), and (3.9), we obtain 

b



+

c

q (t)dt ≥

a



+

q (t)dt +

a

b



+

q (t)dt ≥ 2

α+1

d

b

[r(t)]

−1/α

−α dt

,

a



which completes the proof.

Indeed, the same Lyapunov-type inequality with (3.5) was obtained by Došlý ˇ and Rehák [115, p. 190], Lee et al. [195], and Pinasco [235, 237]. As a result of Theorem 3.2, we have the following corollary. Corollary 3.3 (Disconjugacy Criterion) If 

b



q + (t)dt ≤ 2α+1

a

b

[r(t)]−1/α dt

−α (3.10)

,

a

then (3.1) is disconjugate on [a, b]. Before stating the next theorem, we introduce a generalized version of the Opial inequality due to Beesack and Das [41]. Lemma 3.4 Let γβ > 0, γ + β > 1. Suppose s is a nonnegative, measurable function on (a, b). Let c be the unique solution of K(γ , β) = K1 (c, γ , β) = K2 (c, γ , β) < ∞, where K1 (c, γ , β) =

β γ +β

β/(γ +β) 

c

[s(t)]

1+β/γ

(t − a)

γ +β−1

γ /(γ +β) dt

a

and K2 (c, γ , β) =

β γ +β

β/(γ +β) 

b

[s(t)]

1+β/γ

γ +β−1

(b − t)

γ /(γ +β) dt

.

c

If x is an absolutely continuous function on [a, b] with x(a) = x(b) = 0, then 

b

β  s(t)|x(t)|γ x  (t) dt ≤ K(γ , β)

a



b

  γ +β x (t) dt.

(3.11)

a

Theorem 3.5 Let x be a solution of (Φα (x  )) + q(t)Φα (x) = 0

(3.12)

3.2 Second-Order Half-Linear Equations

137

satisfying (1.5), where a < b. If x(t) = 0 for all t ∈ (a, b), then there exists a (unique) c ∈ (a, b) such that 

c



c

|Q(t)|1+1/α (t − a)α dt =

a

|Q(t)|1+1/α (b − t)α dt ≥

b

1 , α+1

(3.13)

where 

or

t

Q(t) =

q(s)ds

 Q(t) =

a

b

q(s)ds .

t

Proof Let x be a nontrivial solution of (3.12) satisfying (1.5), where a < b. If x(t) = 0 in t ∈ (a, b), then multiplying (3.12) by x(t) and integrating by parts over [a, b] gives 

b

  α+1 x (t) dt =

a



b

q(t)|x(t)|α+1 dt

a



b

= −(α + 1)

Q(t)|x(t)|α−1 x(t)x  (t)dt

a



b

≤ (α + 1)

  |Q(t)||x(t)| x  (t) dt

a

(3.14)

α



b

≤ (α + 1)K(α, 1)

  α+1 x (t) dt,

a

where we have used (3.11) of Lemma 3.4. Since x is a nontrivial solution of (3.12), dividing both sides of (3.14) by 

b

  α+1 x (t) dt > 0

a



yields (3.13).

Remark 3.6 When α = 1, Theorem 3.5 reduces to the results of Brown and Hinton [64] (see also Theorem 1.60 in Sect. 1.7). In 2004, Lee et al. [195] considered half-linear differential equations with damping term of the form (φα (x  )) + p(t)φα (x  ) + q(t)φα (x) = 0

on

[a, b]

(3.15)

with (1.5), where (i) φα (s) = |s|α−2 s, α > 1, (ii) p and q are integrable on [a, b]. Moreover, it is not assumed that the potentials p and q are of definite sign.

138

3 Half-Linear Differential Equations

In order to prove their main result, they use the following lemma for the halflinear differential equation   r(t)φα (x  ) + u(t)φα (x) = 0 on

[a, b],

(3.16)

where r and u are integrable on [a, b] with r(t) > 0 on [a, b]. Lemma 3.7 Let x be a solution of (3.16) satisfying (1.5), where a < b. If x(t) = 0 for all t ∈ (a, b), then there exists c ∈ (a, b) such that 

c

a

ds ∗ [r(s)]α /α

1/α ∗ 

c

u+ (s)ds

1/α >1

(3.17)

> 1,

(3.18)

a

and 

b c

ds ∗ [r(s)]α /α

1/α ∗ 

b

u+ (s)ds

1/α

c

where α ∗ is the Hölder conjugate of α, i.e., 1/α + 1/α ∗ = 1. Proof It follows from (1.5) and Rolle’s theorem that there exists c ∈ (a, b) such that x  (c) = 0. Clearly, |x(c)| = max |x(t)| = 0. Using Hölder’s inequality, we t∈[a,b]

obtain   |x(c)| =   ≤

c a c

  x  (s)ds 

   x (s) ds

a



  1 1/α    ds x [r(s)] (s) 1/α a [r(s)]  c 1/α ∗  c 1/α   α ds   < r(s) x (s) ds α ∗ /α a [r(s)] a 1/α ∗  c  c 1/α ds α = u(s)|x(s)| ds α ∗ /α a [r(s)] a 1/α ∗  c  c 1/α ds + ≤ |x(c)| u (s)ds . α ∗ /α a [r(s)] a =

c

(3.19)

Hence, dividing both sides of (3.19) by |x(c)| = 0, we obtain (3.17). Similarly,

3.2 Second-Order Half-Linear Equations

139

|x(c)| = | − x(c)|   b     x (s)ds  = 

c

≤

b

   x (s) ds

c



  1 [r(s)]1/α x  (s) ds 1/α [r(s)] c  b 1/α ∗  b 1/α   α ds   < x r(s) (s) ds α ∗ /α c [r(s)] c 1/α ∗  b  b 1/α ds α = u(s)|x(s)| ds ∗ α /α c [r(s)] c 1/α ∗  b  b 1/α ds + ≤ |x(c)| u (s)ds , α ∗ /α c [r(s)] c

=

b

(3.20)

and dividing both sides of (3.20) by |x(c)| = 0, we obtain (3.18). Thus, the proof is complete.  Theorem 3.8 (Lyapunov-Type Inequality) Let x be a solution of (3.15) satisfying (1.5), where a < b. If x(t) = 0 for all t ∈ (a, b), then the inequalities (b − a)α/α

∗



b

 q + (t)dt − 4 exp −

a

b

|p(t)|dt

>0

(3.21)

|p(t)|dt > 4

(3.22)

a

and 

α/α ∗

b

(b − a)



+

b

q (t)dt + 4

a

a

hold, where α ≥ 2 and α ∗ is the Hölder conjugate of α, i.e., 1/α + 1/α ∗ = 1. Proof Using 



t

r(t) = exp

and

p(s)ds

u(t) = r(t)q(t)

a

or  r(t) = exp −



b

p(s)ds

and

u(t) = r(t)q(t),

t

(3.15) reduces to (3.16). By Lemma 3.7, there exists c ∈ (a, b) such that

140

3 Half-Linear Differential Equations



c

 t

α ∗ /α 1/α ∗  exp − p(s)ds dt

a

a

c

q + (t) exp



a

t

1/α p(s)ds dt

a

>1 and  c

b

 exp

b

α ∗ /α 1/α ∗  b  + p(s)ds dt q (t) exp −

t

c

b

1/α p(s)ds dt

t

> 1. Thus, α/α ∗



c

(c − a)



 q (t)dt exp 2 +

a

c

|p(t)|dt

>1

(3.23)

> 1.

(3.24)

a

and α/α ∗



(b − c)

b



 q (t)dt exp 2 +

c

b

|p(t)|dt

c

Let A1 := (c − a)α/(2α

∗)



c

q + (t)dt

1/2 ,

a

B1 := (b − c)

α/(2α ∗ )

b

+

q (t)dt

1/2 ,

c



c

A0 := 2 



|p(t)|dt,

a b

B0 := 2

|p(t)|dt,

c

and K := (A1 /B1 )2/α . These and (3.23), (3.24) imply A21 > e−A0 > 1 − A0

(3.25)

B12 > e−B0 > 1 − B0 .

(3.26)

and

3.2 Second-Order Half-Linear Equations

141

Clearly, it follows from α/α ∗ = α − 1 ≥ 1 that 

b



+

c

q (t)dt =

a



+

b

q (t)dt +

a

q + (t)dt

c ∗

= A21 (c − a)−α/α + B12 (b − c)−α/α ≥ A21 {(Kb + a)/(K + 1) − a}−α/α

∗

∗ ∗

+ B12 {b−(Kb + a)/(K+1)}−α/α     ∗ ∗ ∗ ≥ 1 + K −α/α (b − a)−α/α A21 + K α/α B12   ∗ 2/α ∗ 2/α 2/α 2/α ∗ + B12 . = (b − a)−α/α A21 + A1 B1 + A1 B1

(3.27)

It follows from (3.25), (3.26), (3.27), and the inequality e−x + e−y ≥ 2e−(x+y)/2

for

x, y ∈ R

that (b − a)

α/α ∗



  q (t)dt − 4 exp −

b

+

a

c

 |p(t)|dt +

a

2/α ∗

2/α

≥ A21 + A1

2/α

> e−A0 + e−(A0 /α

2/α ∗

+ A1 B1

B1

∗ +B /α) 0

b

 |p(t)|dt

c

+ B12 − 4e−(A0 +B0 )/2 ∗

+ e−(A0 /α+B0 /α ) + e−B0 − 4e−(A0 +B0 )/2

≥ e−A0 + e−B0 − 2e−(A0 +B0 )/2 2  = e−A0 /2 − e−B0 /2 ≥ 0. Hence, (3.21) holds. It follows from the inequality e−x > 1 − x

for x > 0, 

that (3.22) holds. Thus, the proof is complete.

Remark 3.9 If α = 2 and p(t) = 0, then (3.22) reduces to the classical Lyapunov inequality. Theorem 3.10 (Lyapunov-Type Inequality) Let x be a solution of (3.15) satisfying (1.5), where a < b. If x(t) = 0 for all t ∈ (a, b), then the inequalities (b − a)

α/α ∗

 a

b

 q (t)dt − 2 exp − +

b

α

a

|p(t)|dt

>0

(3.28)

142

3 Half-Linear Differential Equations

and (b − a)α/α

∗



b

q + (t)dt + 2α



a

b

|p(t)|dt > 2α

(3.29)

a

hold, where α ∈ (1, 2] and α ∗ is the Hölder conjugate of α, i.e., 1/α + 1/α ∗ = 1. Proof It follows from 1 < α < 2 that 0 < α/α ∗ = α − 1 < 1. Therefore, the inequality ∗

(1 + x)α/α ≥ 2α/α

∗ −1

∗

(1 + x α/α ),

x≥0

holds. Then, 

b

q + (t)dt ≥ 2α/α

a

∗ −1



1 + K −α/α

∗



(b − a)−α/α

∗



∗

A21 + K α/α B12



  ∗ ∗ A21 + K α/α B12 + K −α/α A21 + B12   ∗ ∗ 2/α ∗ 2/α 2/α 2/α ∗ + B12 , = 2α/α −1 (b − a)−α/α A21 + A1 B1 + A1 B1

= 2α/α

∗ −1

(b − a)−α/α

∗

where K, A1 , B1 are defined as in the proof of Theorem 3.8. Thus, 1−α/α ∗

2

(b − a)

α/α ∗



b

  q (t)dt − 4 exp −

c

+

a

 |p(t)|dt +

a

b

 |p(t)|dt

> 0.

c



Hence, (3.28) and (3.29) hold. This completes the proof.

Remark 3.11 Since α ∗ is the Hölder conjugate of α, (3.28), (3.21) and (3.29), (3.22) can be replaced by the inequalities  (b − a)

b

α−1

 + q (t)dt > exp −

a

b



|p(t)|dt ×

a

2α

if 1 < α ≤ 2,

4

if α ≥ 2 (3.30)

and 

b

(b − a)

α−1

  q (t)dt > 1 − exp −

b

+

a

 |p(t)|dt

a

×

 α 2 4

if 1 < α ≤ 2, if α ≥ 2,

respectively. In [274], Tiryaki et al. further improved (3.30) to 

b

(b − a)α−1 a

 q + (t)dt > 2α exp −

b a

|p(t)|dt ,

(3.31)

3.2 Second-Order Half-Linear Equations

143

see Corollary 4.8 in Sect. 4.2 below. However, when α = 2, both (3.30) and (3.31) reduce to  b

 b (b − a) q + (t)dt > 4 exp − |p(t)|dt , a

a

which is not a stronger condition than that of Fink and Mary [134], i.e., (1.138). Thus, neither (3.30) nor (3.31) is a generalization of (1.138) to (3.15). Motivated by Yang [294], Lee et al. [195], Tiryaki et al. [274], and Pachpatte [229], Wang [282] generalized (1.33) for (1.1) to the half-linear differential equation   r(t)φα (x  ) + q(t)φα (x) = 0.

(3.32)

As a simple corollary of his main results, Wang obtained two new Lyapunov inequalities different from Yang’s result [294], i.e., (3.4): 

b



+

b

q (t)dt



+

q (t)

a

a

t

[r(s)]

−1/(α−1)



b

ds

a

[r(s)]

−1/(α−1)

α−1 ds

dt > 4

t

(3.33)

and 

b

q + (t)



a

t

[r(s)]−1/(α−1) ds



a

b

[r(s)]−1/(α−1) ds

t



b

>

[r(t)]

−1/(α−1)



α−1 dt

×

a

α−1 dt

1

if 1 < α < 2,

22−α

if α ≥ 2.

(3.34)

Since 

t

[r(s)]

−1/(α−1)



b

ds

a

[r(s)]

t

−1/(α−1)

1 ds ≤ 4



b

[r(t)]

−1/(α−1)

2 dt

,

a

it is easy to see that (3.33) yields (3.4). Moreover, (3.34) also reproduces (1.33) if α = 2 and r(t) ≡ 1. On the other hand, applying the results to (3.15), the inequality 

b

+



b

q (t)dt a

+

q (t) [(b − t)(t − a)]

a

α−1

 dt > exp −

b

|p(t)|dt

a

can be obtained. It follows that 

b

(b − a)α−1 a



 1 b q + (t)dt > 2α exp − |p(t)|dt , 2 a

144

3 Half-Linear Differential Equations

which is clearly stronger than (3.31) and reproduces (1.138) if α = 2. Thus, Wang’s results obtained in [282] generalized and extended the related results in [134, 155, 195, 229, 274, 294]. In fact, they almost generalize and extend all related results in [83, 91, 124, 134, 143, 155, 172, 193, 195, 202, 220, 229, 248, 249, 258, 274, 294]. Let u(t) = r(t)φα (x  (t)). Then, (3.32) is equivalent to the system 

x  = r˜ (t)φα ∗ (u),

(3.35)

u = −q(t)φα (x), where r˜ (t) = [r(t)]−1/(α−1) and α ∗ is the Hölder conjugate of α.

Theorem 3.12 (Lyapunov-Type Inequality) Let x be a solution of (3.32) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then there exists c ∈ (a, b) such that the inequality  t

α−1  b

α−1  c  b q + (t) [r(s)]−1/(α−1) ds dt = q + (t) [r(s)]−1/(α−1) ds dt a

a

c

t

>1

(3.36)

holds. Proof Multiplying the first equation in (3.35) by u(t) and the second one by x(t) and then adding the results, we obtain ∗

(xu) (t) = r˜ (t)|u(t)|α − q(t)|x(t)|α .

(3.37)

Integrating (3.37) from a to b and taking into account x(a) = x(b) = 0, we get 

b



α∗

b

r˜ (t)|u(t)| dt =

a

q(t)|x(t)|α dt.

(3.38)

a

Integrating the first equation in (3.35) from a to t and then from t to b and taking into account x(a) = x(b) = 0, we have  x(t) =

t



b

r˜ (s)φ (u(s))ds = − α∗

a

r˜ (s)φα ∗ (u(s))ds,

a < t < b.

t

It follows from the Cauchy–Schwarz inequality that  |x(t)| ≤ a

t

1/α ∗  r˜ (s)ds a

t

α∗

r˜ (s)|u(s)| ds

1/α ,

a 0

a

holds, which, together with (3.41), implies 

c

q + (t)dt > 0 and



a

b

q + (t)dt > 0.

(3.42)

c

It follows from (3.35), (3.39), (3.42), and Hölder’s inequality that 

c

q + (t)|x(t)|α dt
1,

(3.46)

t



that is, (3.36). The proof is complete.

Theorem 3.13 (Lyapunov-Type Inequality) Let x be a solution of (3.32) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then the inequality 

b

+



q (t) a

a

t

[r(s)]

−1/(α−1)

 ds

b

[r(s)]

−1/(α−1)

ds

dt

t

 > C(α) a

holds, where

α−1

b

r −1/(α−1) (t)dt

α−1 (3.47)

3.2 Second-Order Half-Linear Equations

 C(α) =

147

1

if 1 < α < 2,

22−α

if α ≥ 2.

(3.48)

Proof Set r˜ (t) = [r(t)]−1/(α−1) . Then, by the proof of Theorem 3.12, (3.46) holds. By (3.46), we have 

c



+

t

q (t) a

 r˜ (s)ds

a

α−1

b

r˜ (s)ds



α−1

b

r˜ (t)dt

dt >

t

(3.49)

c

and 

b



+

t

q (t) c

 r˜ (s)ds

a

b

α−1 r˜ (s)ds



c

dt >

t

α−1 r˜ (t)dt

(3.50)

.

a

For γ , β ≥ 0, we have  γ

+ β ≥ (γ + β) ×

m

m

m

1

if 0 < m ≤ 1,

21−m

if m > 1.

(3.51)

In (3.51), the first inequality can be deduced by the monotonicity of the function f (t) = (1 + t)m − t m − 1, and the second inequality is a simple corollary of Jensen’s inequality, see Lemma 3.1. Adding (3.49) and (3.50) and using (3.51), we have 

b

+



q (t) a

t



b

r˜ (s)ds

a

α−1 r˜ (s)ds



b

dt >

t

α−1 r˜ (t)dt



c

+

c

α−1 r˜ (t)dt

a



b

≥ C(α)

α−1 r˜ (t)dt

,

a



which implies that (3.47) holds. The proof is complete.

Theorem 3.14 (Lyapunov-Type Inequality) Let x be a solution of (3.32) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then the inequality 

b a

q + (t)dt



b

q + (t)



a

t

[r(s)]−1/(α−1) ds



a

b

[r(s)]−1/(α−1) ds

t

α−1 dt > 4 (3.52)

holds. Proof Set r˜ (t) = [r(t)]−1/(α−1) . Then, by the proof of Theorem 3.12, (3.46) holds. It follows from (3.46) that (3.49) and (3.50) hold and that  a

c

q + (t)dt

 a

c

α−1 r˜ (t)dt

>1

(3.53)

148

3 Half-Linear Differential Equations

and 

b



+

b

q (t)dt c

α−1 r˜ (t)dt

> 1.

(3.54)

c

Adding (3.49) and (3.50) and using (3.53) and (3.54), we obtain 

b



q + (t)

a

t



a



b

>

b

r˜ (s)ds

α−1 r˜ (s)ds

t

q + (t)dt



b

α−1

b

r˜ (t)dt





c

+

q + (t)dt

c

+

c

−1

c

=

 dt >

α−1 r˜ (t)dt

a

−1

a

q + (t)dt



b

q + (t)dt

a

c



−1

b

≥4

q + (t)dt



c

q + (t)dt

−1

a

,

a



which implies that (3.52) holds. The proof is complete. Notice that 

t

[r(s)]−1/(α−1) ds



a

b

[r(s)]−1/(α−1) ds ≤

t

1 4



b

[r(t)]−1/(α−1) dt

2 .

(3.55)

a

By (3.47) and (3.55), we conclude 

b

C(α)

[r(t)]

−1/(α−1)

α−1 dt

a



b


4α−1 C(α).

(3.56)

3.2 Second-Order Half-Linear Equations

149

On the other hand, by (3.52) and (3.55), we have  4
2α .

(3.57)

a

Corollary 3.15 Let x be a solution of (3.32) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then (3.57) holds. Remark 3.16 Corollary 3.15 is the main result in Yang [294] (see [294, Theorem 2.1]). When α ≥ 2, (3.56) reproduces (3.57). Moreover, (3.47) reduces to (1.33) if α = 2 and r(t) ≡ 1. Thus, Theorems 3.13 and 3.14 generalize and extend the main results in Hartman [155] and in Yang [294], respectively, and so Theorem 3.12 generalizes and extends the main results both in Hartman [155] and in Yang [294]. In 2011, Wang [282] considered the half-linear differential equation with damping term, i.e., (3.15) with (1.5). It is clear that if we multiply both sides of (3.15) by the function 

t

exp

p(s)ds ,

a

then (3.15) reduces to (3.32) with 



t

r(t) = exp

p(s)ds

 and



t

q(t) → q(t) exp

p(s)ds .

a

a

Thus, we can deduce Lyapunov inequalities for (3.15) by the results obtained previously by Wang [282]. Theorem 3.17 (Lyapunov-Type Inequality) Let x be a solution of (3.15) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then the inequality 

b

a

holds.

q + (t)dt



b a

 q + (t)[(b − t)(t − a)]α−1 dt > 4 exp −

b a

|p(t)|dt

(3.58)

150

3 Half-Linear Differential Equations

Proof By Theorem 3.12, we have  t

 t   c + ∗ q (t) exp p(s)ds exp (1 − α ) 1< a



a b

=

q + (t) exp



a



t

a

p(s)ds dτ

 exp (1 − α ∗ )

b t

τ



c

=

q + (t)

a



a

t

a

c

≤

α−1 p(s)ds dτ dt. (3.59)

τ

α−1

p(s)ds dτ

dt

a



α−1  t exp (1 − α ∗ ) p(s)ds dτ dt

a



dt

a

From (3.59), we have  t

 t  c  + ∗ 1< q (t) exp p(s)ds exp (1 − α ) a

α−1

a

p(s)ds

c



τ

τ



+

q (t)(t − a)

α−1



c

dt exp

a

|p(t)|dt .

a

It follows that 

c

+

q (t)(t − a)

α−1

 dt > exp −

a



c

|p(t)|dt .

(3.60)

a

Similarly, it follows from (3.59) that 

b

1
exp −

c

|p(t)|dt .

b

(3.61)

c

From (3.60) and (3.61), we have  b q + (t)[(b − t)(t − a)]α−1 dt a



c

=

q + (t)[(b − t)(t − a)]α−1 dt +

a



b

q + (t)[(b − t)(t − a)]α−1 dt

c



c

≥ (b − c)

α−1 a

+

q (t)(t − a)

α−1



dt + (c − a)

α−1 c

b

q + (t)(b − t)α−1 dt

3.2 Second-Order Half-Linear Equations



c

≥ (b − c)α−1

151

q + (t)(t − a)α−1 dt + (c − a)α−1

a



b

≥

q + (t)dt

−1 

c

b

c



c

+

c

 

q (t)(t − a)

a

b

−1

+

 +

q (t)dt

>

−1 

+

c

q (t)dt

−1 

+

+

a



b

q + (t)dt

−1



b

dt



+

q (t)(b − t)

α−1

dt

c

 × exp −

b

|p(t)|dt

a

−1 

b

q (t)dt

c

≥4

q + (t)(t − a)α−1 dt

a

b

=

c

α−1

q (t)dt

c



c



+

q (t)dt a

q + (t)(b − t)α−1 dt

a

−1 

+

b

c



q + (t)(b − t)α−1 dt





 q (t)dt exp − +

a

 exp −

a

b

b

|p(t)|dt

a

|p(t)|dt .

a



This shows that (3.58) holds. The proof is complete.

Theorem 3.18 (Lyapunov-Type Inequality) Let x be a solution of (3.15) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then the inequality  b  b + q (t)dt q + (t)[(b − t)(t − a)]α−1 dt a

a

> C(α)(b − a)

α−1

 exp −

b

|p(t)|dt

(3.62)

a

holds, where C(α) is given by (3.48). Proof In view of the proof of Theorem 3.17, (3.60) and (3.61) are satisfied. It follows from (3.60) and (3.61) that  b q + (t)[(b − t)(t − a)]α−1 dt a



c

=

q + (t)[(b − t)(t − a)]α−1 dt +

a



c

 exp −

c

|p(t)|dt + (c − a)

≥ (b − c)

+ (c − a)

α−1

≥ C(α)(b − a)

α−1

!

 exp −

b a

 exp −

b a



b

q + (t)(b − t)α−1 dt

c



a α−1

q + (t)[(b − t)(t − a)]α−1 dt

q + (t)(t − a)α−1 dt + (c − a)α−1

a

≥ (b − c)

b c

≥ (b − c)α−1 α−1



|p(t)|dt .

α−1

 exp −

|p(t)|dt

b c

|p(t)|dt

152

3 Half-Linear Differential Equations



This shows that (3.62) holds. The proof is complete. It follows from (1.23) and (3.58) that  (b − a)

b

α−1 a



 1 b q (t)dt > 2 exp − |p(t)|dt 2 a +

α

(3.63)

holds. Corollary 3.19 Let x be a solution of (3.32) satisfying (1.5), where a < b. If x(t) > 0 for all t ∈ (a, b), then (3.63) holds. Remark 3.20 Corollary 3.19 improves the main results in Lee et al. [195] and Tiryaki et al. [274]. Moreover, (3.63) reduces to (1.33) if α = 2. Thus, Theorem 3.17 generalizes and extends the main results in Fink and Mary [134], Lee et al. [195], and Tiryaki et al. [274].

3.2.1 Lower Bounds for Eigenvalues In [192], Kre˘ın obtained sharp lower bounds for eigenvalues of weighted secondorder Sturm–Liouville differential operators with zero Dirichlet boundary conditions. In 2004, Pinasco [235] presented a new proof of this result, and he extended it to the one-dimensional p-Laplacian ⎧    ⎨− x  p−2 x  = λr(t)|x|p−2 x

on

(a, b),

⎩ x(a) = x(b) = 0,

(3.64)

where λ is a real parameter, p > 1, and r is a bounded positive function. The method of proof is based on a suitable generalization of the Lyapunov inequality to the nonlinear case and on some elementary inequalities. Eigenvalue problems for quasilinear operators of p-Laplacian type like (3.64) have received considerable attention in the last years (see, e.g., [31, 32, 109, 116, 140, 278]). The asymptotic behavior of eigenvalues was obtained in [55, 138]. First, we consider the quasilinear 2-point boundary value problem ⎧    ⎨− x  p−2 x  = r(t)|x|p−2 x, ⎩ x(a) = x(b) = 0,

(3.65)

where r is a bounded positive function and p > 1. By a solution of (3.65), we 1,p understand a real-valued function x ∈ W0 (a, b) such that

3.2 Second-Order Half-Linear Equations



b

153



  p−2  x (t) x (t)v  (t)dt =

a

b

|x(t)|p−2 x(t)v(t)dt

a

1,p

for each v ∈ W0 (a, b). The regularity results of [113] imply that the solutions x are at least of class C1,α loc and satisfy the differential equation almost everywhere in (a, b). The first result provides an estimate of the location of the maxima of a solution in (a, b). We make use of the following lemma. Lemma 3.21 Let r : [a, b] → R be a bounded positive function. Suppose x is a solution of (3.65) and let c ∈ (a, b) be such that |x(t)| is maximized at c. Then, the inequalities 

c

r(t)dt ≥ (c − a)



−p/q

b

r(t)dt ≥ (b − c)−p/q

and

a

(3.66)

c

hold, where q is the Hölder conjugate of p, that is, 1/p + 1/q = 1. Proof Clearly, by using Hölder’s inequality, we obtain  x(c) =

c



x  (t)dt ≤ (c − a)1/q

a

c

  p x (t) dt

1/p .

(3.67)

a

We note that x(c) = 0. So, integrating by parts in (3.65) after multiplying by x gives 

c

  p x (t) dt =



a

c

r(t)|x(t)|p dt.

(3.68)

a

Thus, (3.67) and (3.68) yield  x(c) ≤ (c − a)1/q

c

a

≤ (c − a)

1/q

1/p r(t)|x(t)|p dt 

|x(c)|

(3.69)

1/p

c

r(t)dt

.

a

The first inequality in (3.66) follows after cancelling x(c) on both sides of (3.69), while the second one is proved in a similar fashion.  Remark 3.22 The sum of both inequalities in (3.66) shows that c cannot be too close to a or b. We have  a

b

r(t)dt < ∞,

154

3 Half-Linear Differential Equations

but lim

c→a +



   (c − a)−p/q + (b − c)−p/q = lim (c − a)−p/q + (b − c)−p/q = ∞. c→b−

The next result restates the Lyapunov inequality. Theorem 3.23 (Lyapunov-Type Inequality) Let r : [a, b] → R be a bounded positive function, let x be a nontrivial solution of (3.65), and let q be the Hölder conjugate of p ∈ (1, ∞). Then, the inequality 

b

r(t)dt ≥

a

2p (b − a)p/q

(3.70)

holds. Proof For every c ∈ (a, b), we have   2|x(c)| = 

c

a

    x  (t)dt  + 

b

c

   x  (t)dt  ≤

b

   x (t) dt.

(3.71)

a

By using Hölder’s inequality in (3.71), we obtain 

b

2|x(c)| ≤ (b − a)

1/q

  p x (t) dt

1/p

a



b

= (b − a)1/q

(3.72)

1/p r(t)|x(t)|p dt

.

a

We now choose c ∈ (a, b) such that |x(t)| is maximized. Then, (3.72) yields 

1/p

b

2|x(c)| ≤ (b − a)1/q |x(c)|

r(t)dt

(3.73)

.

a

After cancelling |x(c)| > 0 on both sides of (3.73), we obtain (3.70), and the proof is complete.  Remark 3.24 We note that, for p = q = 2, (3.70) coincides with Lyapunov’s classical inequality, i.e., (1.4). Now, consider (3.64), where r ∈ L∞ (a, b) is a positive function, λ is a real parameter, and p > 1. Remark 3.25 Eigenvalues may be characterized in a variational way as  

  p x (t) dt

λk (Ω) = inf sup F ∈CΩ k x∈F

Ω

 p

r(t)|x(t)| dt Ω

−1  ,

(3.74)

3.2 Second-Order Half-Linear Equations

155

where   Ω CΩ k = C ⊂ M : C compact, C = −C, γ (C) ≥ k ,      p 1,p x (t) dt = 1 , MΩ = x ∈ W0 (Ω) : Ω

and γ : Σ → N ∪ {∞} defined by  γ (A) = min k ∈ N : there exists f ∈ C(A, Rk \ {0}), f (t) = −f (−t) is the Krasnosel ski˘ı genus. The spectrum of (3.64) consists of a countable sequence of nonnegative eigenvalues λ1 < λ2 < · · · < λk < · · · , and it coincides with the eigenvalues obtained by the Lyusternik–Shnirel man theory. Now, we prove the lower bound for the eigenvalues of (3.64) for every p ∈ (1, ∞). We first present Pinasco’s [235] result. Theorem 3.26 Let λn be the nth eigenvalue of (3.64). Then, λn ≥

2p np (b − a)p−1



−1

b

r(t)dt a

holds, where C(α) is given by (3.48). Proof Let λn be the nth eigenvalue of (3.64), and let xn be a corresponding eigenfunction. As in the linear case, un has n nodal domains in [a, b] (see [32, 278]). Applying (3.70) in each nodal domain, we obtain 

b

r(t)dt ≥ λn

a

n 

k=1



tk

r(t)dt

≥

tk−1

n

k=1

2p , (tk − tk−1 )p/q

where a = t0 < t1 < · · · < tn = b are the zeros of xn in [a, b]. Now, the sum on the left-hand side is minimized when all summands are the same, which gives the lower bound 

b

λn a



n r(t)dt ≥ 2 n b−a p

The proof is complete.

p/q . 

Finally, we prove that the lower bound is sharp. Theorem 3.27 For each n ∈ N and ε > 0, there exist weight functions rn,ε such that

156

3 Half-Linear Differential Equations

 lim

ε→0+

2p np λn,ε − (b − a)p−1



−1 

b

= 0,

rn,ε (t)dt a

where λn,ε is the nth eigenvalue of ⎧    ⎨− x  p−2 x  = λrn,ε (t)|x|p−2 x ⎩

on

(a, b),

x(a) = x(b) = 0.

Proof We begin with the first eigenvalue λ1 . We put 

b

r(t)dt = M

c=

and

a

a+b . 2

Let r1 = Mδc , where δc is the delta function. We obtain   λ1 = min

1,p

x∈W0

b

  p x (t) dt

a

−1 

b

p

δc (t)[x(t)] dt a

2 = min 1,p Mx p (c) x∈W 0





c a

  p x (t) dt = 2μ1 , M

where μ1 is the first Steklov eigenvalue of ⎧    ⎪ ⎨− x  p−2 x  = 0, ⎪ ⎩x  (c)p−2 x  (c) = μ|x(c)|p−2 x(c),

x(a) = 0

in [a, c]. A direct computation gives μ1 =

2p−1 . (b − a)p−1

Now, we define the functions r1,ε by

r1,ε

⎧ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎨ M = 2ε ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩0

if if if

  a+b t ∈ a, −ε , 2   a+b a+b − ε, +ε , t∈ 2 2   a+b + ε, b , t∈ 2

3.3 Third-Order Half-Linear Equations

157

and the result follows by testing, in (3.74), the first Steklov eigenfunction

x(t) =

⎧ ⎪ ⎪t − a ⎪ ⎨ ⎪ ⎪ ⎪ ⎩b − t

if if

  a+b t ∈ a, , 2   a+b ,b . t∈ 2

Thus, the inequality is sharp for n = 1. We now consider the case n ≥ 2. We divide the interval [a, b] in n subintervals Ii of equal length and let ci be the midpoint of the ith subinterval. By using a symmetry argument, the nth eigenvalue corresponding to the weight rn =

n M δc i , n i=1

restricted to Ii , is the first eigenvalue in this interval, that is, λn =

2p np 2nμ1 = . M M(b − a)p−1

The proof is now completed.



3.3 Third-Order Half-Linear Equations Second-order half-linear differential equations have been widely studied in recent years, and there is a nice overview in the monograph [115]. Less literature is dealing with such equations of higher-order (in particular, odd-order differential equations), but one can see, for example [217]. Lyapunov-type inequalities have not been well developed for higher-order half-linear differential equations. This is due to the complexity caused by the nonlinear nature of the Laplacian operators in the equation. In this section, we establish Lyapunov-type inequalities for third-order half-linear equations. We point out that the inequalities obtained utilize integrals of both q + and q − , the positive and negative parts of q. This is different from most existing results for the third-order linear case, where integrals of |q| are involved. As a special case, the results obtained improve the results in Theorems 2.56 and 2.60 for the linear case given by Parhi and Panigrahi [230] in 1999. While the Lyapunov inequality and its generalizations have been used as a successful tool in oscillation, disconjugacy, the study of eigenvalue problems, and many other areas of differential equations, applications of Lyapunov-type inequalities to boundary value problems are not known. In this section, we apply the obtained Lyapunov-type inequalities to show the nonexistence of nontrivial

158

3 Half-Linear Differential Equations

solutions of third-order half-linear boundary value problems and the uniqueness of solutions of third-order linear boundary value problems. Furthermore, by employing “uniqueness implies existence” theorems given by Jackson [175, 176] and by Jackson and Schrader [177], we establish an existence-uniqueness result for several classes of third-order linear boundary value problems. It is expected that the Lyapunov-type inequality approach can be further applied to study general higherorder boundary value problems. For detailed discussion on the “uniqueness implies existence” criteria for general boundary value problems, the reader is referred to [92, 126–128, 139, 164–166]. In 2013, Kisel ák [187] gave a proof of a Lyapunov-type inequality for thirdorder half-linear differential equations. He generalized the Lyapunov inequality given by Parhi and Panigrahi [230] for (2.129) to a class of third-order half-linear differential equations. Some applications, e.g., studies of the distance between consecutive zeros of a solution, were done with the help of the inequality.

3.3.1 Lyapunov-Type Inequalities Kisel ák [187] was concerned with the third-order half-linear differential equation

1 Φα r2 (t) 2



1 Φα (x  ) r1 (t) 1





+ q(t)Φβ (x) = 0,

(3.75)

where α1 , α2 > 0, q ∈ C([a, b], R), and Φα (s) := |s|α−1 s is known as the signedpower function. Moreover, we assume that 1/rk ∈ C3−k ([a, b], (0, ∞)), k = 1, 2, and in order to preserve the mentioned homogeneity property, we also demand that β = α1 α2 . Equation (3.75) can be written by means of quasi-derivatives with respect to the coefficients ri and functions Φαi , i = 1, 2. We denote them by D(0) x(t) = x(t),

d (i−1) 1 Φαi D D(i) x(t) = x(t) ri (t) dt D(3) x(t) =

for

i ∈ {1, 2},

d (2) D x(t). dt

A solution of (3.75) is said to be oscillatory (nonoscillatory) if it has (has not) a sequence of zeros converging to infinity. Equation (3.75) is oscillatory if all its solutions are oscillatory. Otherwise, it is nonoscillatory. If a solution of (3.75) has two consecutive zeros a < b, then two cases can occur. Either there exists d ∈ [a, b] such that d (1) D x(d) = 0 dt

or

d (1) D x(t) = 0 dt

for

t ∈ [a, b].

3.3 Third-Order Half-Linear Equations

159

The first case is considered in the following assertion. Theorem 3.28 (Lyapunov-Type Inequality) Let x be a solution of (3.75) satisfying (1.5), where a < b. If x(t) = 0 for all t ∈ (a, b) and there exists d ∈ [a, b] such that d (1) D x(d) = 0, dt then the inequality 

b

1/β |q(t)|dt

>

a

1 min h(c) 2 c∈[a,b]

(3.76)

holds, where 

c

h(c) =

1/α1

[r1 (t)]

−1  dt

c

1/α2

[r2 (t)]

a

−1/α1 dt

a



b

+

−1  [r1 (t)]1/α1 dt

c

b

−1/α1 [r2 (t)]1/α2 dt

.

c

Proof We first define functions xi , i ∈ {0, 1, 2}, by x0 (t) := D(0) x(t),

d 1 Φαi xi−1 (t) xi (t) := D(i) x(t) = ri (t) dt

for

i = 1, 2.

(3.77)

Equation (3.75) is then equivalent to the differential system 

(xi+1 ) for xi = [ri+1 (t)]1/(αi +1) Φα−1 i

x2

i = 0, 1,

= −q(t)Φβ (x0 ).

(3.78)

The condition x0 (a) = x0 (b) = 0 gives us the existence of c ∈ (a, b) such that x0 (c) = 0 and |x0 (c)| = max |x0 (t)|. It follows from the latter that x1 (c) = 0. By t∈[a,b]

integrating (3.78) from a to c, we obtain 

c

x0 (t) = a

(x1 (t))dt, [r1 (t)]1/α1 Φα−1 1

which implies 

c

|x0 (t)| ≤ a

[r1 (t)]1/α1 |x1 (t)|1/α1 dt.

(3.79)

160

3 Half-Linear Differential Equations

Now, let t ∈ [a, c]. From the fact that x1 (c) = 0, we get 

t

x1 (t) = c

x1 (s)ds,

and hence 

c

|x1 (t)| ≤

[r2 (t)]1/α2 |x2 (t)|1/α2 dt.

a

Further, from the second condition of the assumptions and (3.77) or (3.78), we know that x2 (d) = 0, which implies 

t

x2 (t) =

q(s)Φβ (x0 (s))ds d

for t ∈ [a, c]. Moreover, we have  |x2 (t)| < |x0 (c)|

b

β

|q(t)|dt.

(3.80)

a

Using (3.79) and (3.80), we get 

c

|x0 (c)| < |x0 (c)|



c

[r1 (t)]1/α1 dt

a

[r2 (t)]1/α2 dt

1/α1 

a

b

β |q(t)|dt

,

a

which implies 

c

 1/α1

[r1 (t)]

c

dt

1/α2

[r2 (t)]

a

1/α1 

b

dt

a

β |q(t)|dt

> 1.

(3.81)

> 1.

(3.82)

a

Similarly, we can get 

b



b

[r1 (t)]1/α1 dt

c

1/α1  [r2 (t)]1/α2 dt

c

b

β |q(t)|dt

a

Hence, (3.81) and (3.82) both imply (3.76). Moreover, it is obvious that h takes its minimum in (a, b), since it is continuous there and lim h(c) = lim h(c) = ∞.

c→a +

This completes the proof.

c→b−



3.3 Third-Order Half-Linear Equations

161

In the case that d (1) D x(t) = 0 for dt

t ∈ [a, b],

we consider three consecutive zeros of x(t). We give only a sketch of the proof as it is almost similar to the previous one. Theorem 3.29 (Lyapunov-Type Inequality) Let x be a solution of (3.75) satisfying the 3-point boundary conditions x(a) = x(d) = x(c) = 0, where a < d < b. If x(t) = 0 for all t ∈ (a, d) ∪ (d, b) and d (1) D x(t) = 0 for dt

t ∈ [a, d],

then (3.76) holds. Proof The conditions x0 (a) = x0 (d) = x0 (c) = 0 give us the existence of c1 ∈ (a, d) and c2 ∈ (d, b) such that x1 (c1 ) = x1 (c2 ) = 0. Further application of Rolle’s theorem provides the existence of c3 ∈ (c1 , c2 ) such that x2 (c3 ) = 0. Denoting by c ∈ (a, d) ∪ (d, b) a point where max |x0 (t)| = |x0 (c)| and using the same t∈[a,b]

procedure as in the proof of Theorem 3.28, it can be proved that (3.76) holds. Notice that there cannot be a problem with continuity of h on (a, b).  Remark 3.30 Put r1 (t) = r2 (t) = 1. Since h attains its minimum at c0 = (a + b)/2, we find that (3.76) reduces to

α2 (α1 +1)  b 2 |q(t)|dt > . b−a a Notice that in case α1 = α2 = 1, this inequality reduces to (2.135) given by Parhi and Panigrahi [230] in Theorem 2.60. Remark 3.31 Put r1 (t) = r2 (t) = r(t) and α1 = α2 = α. Then, (3.76) reduces to the inequality 

b

 b

−α(α+1) |q(t)|dt > 2 [r(t)]1/α dt ,

a

a

since 

c

 [r(t)]1/α dt =

a

c

b

[r(t)]1/α dt =

1 2



b a

must hold for c in order to minimize h on (a, b), where

1 [R(a) + R(b)] c = R −1 2 and R is an antiderivative of r 1/α .

[r(t)]1/α dt

162

3 Half-Linear Differential Equations

In 2014, Dhar and Kong [112] considered the reduced equation (i.e., r1 (t) = r2 (t) = 1 in (3.75)), namely the third-order half-linear differential equation  Φα2

   Φα1 (x  ) + q(t)Φα1 α2 (x) = 0,

(3.83)

where q ∈ C(R, R), Φα (s) := |s|α−1 s, and α1 , α2 > 0. For simplicity, we denote α := (α1 + 1)α2 . Theorem 3.32 (Lyapunov-Type Inequality) Assume x is a solution of (3.83) satisfying (1.5) and x(t) = 0 for all t ∈ (a, b). Suppose that there exists ξ ∈ [a, b] such that (Φα1 (x  )) (ξ ) = 0. Then, the inequality 

ξ



−

b

q (t)dt +

a



+

q (t)dt > ξ

2 b−a

α (3.84)

holds. Proof Without loss of generality, we may assume x > 0 on (a, b). Then, there exists d ∈ [a, b] such that m := x(d) = max x(t). It follows that t∈[a,b]



d

m=

x  (t)dt ≤



a

d

   x (t) dt

a

and 

b

m=−

x  (t)dt ≤



d

b

   x (t) dt.

d

Therefore,  2m ≤

b

   x (t) dt.

(3.85)

a

Applying Hölder’s inequality 

b

 |f (t)g(t)|dt ≤

a

b

1/p  |f (t)| dt p

a

b

p∗

1/p∗

|g(t)| dt

(3.86)

a

with f = x  , g(t) ≡ 1, p = α1 + 1 and p∗ = 1 + 1/α1 , we obtain 

b a

   x (t) dt ≤ (b − a)α1 /(α1 +1)

 a

b

  α1 +1 x (t) dt

1/(α1 +1) .

3.3 Third-Order Half-Linear Equations

163

Hence, by (3.85), we have (2m)

α1 +1

 ≤ (b − a)

b

α1

  α1 +1 x (t) dt = (b − a)α1

a



b a

x  (t)Φα1 (x  (t))dt.

Note that x(a) = x(b) = 0. Integration by parts leads to (2m)α1 +1 ≤ (b − a)α1



b

a

x(t)(−Φα1 (x  )) (t)dt.

(3.87)

For t ∈ [a, b], by integrating (3.83) from ξ to t and using the fact that (Φα1 (x  )) (ξ ) = 0, we see that   Φα2 (−Φα1 (x  )) (t) =



t ξ

q(s)Φα1 α2 (x(s))ds.

Since −q − (t) ≤ q(t) ≤ q + (t), we have ⎧ t + ⎪ ⎪ ⎪  ⎨ q (s)Φα1 α2 (x(s))ds t

ξ

q(s)Φα1 α2 (x(s))ds ≤

if t ≥ ξ,

ξ

 t ⎪ ⎪ ⎪ ⎩− q − (s)Φα1 α2 (x(s))ds

(3.88)

(3.89) if t < ξ.

ξ

Define  ∗

q (s) =

q + (s)

if ξ ≤ s ≤ b,

−q − (s)

if a ≤ s ≤ ξ.

(3.90)

From (3.89), we see that for t ∈ [a, b], we have  t  t q(s)Φα1 α2 (x(s))ds ≤ q ∗ (s)Φα1 α2 (x(s))ds. ξ

ξ

Then, from (3.88),   Φα2 (−Φα1 (x  )) (t) ≤



t ξ

q ∗ (s)Φα1 α2 (x(s))ds.

It follows that   −Φα1 (x  ) (t) ≤ Φα−1 2

 ξ

t

∗



q (s)Φα1 α2 (x(s))ds .

(3.91)

164

3 Half-Linear Differential Equations

Using (3.87), (3.91), and the facts that 0 ≤ x(t) ≤ m, x(t) ≡ m on [a, b], and 

t

q ∗ (s)ds ≥ 0 for all

t ∈ [a, b],

ξ

we have 

(2m)α1 +1 ≤ (b − a)α1

b

a

< mα1 +1



b

α1 +1



ξ



t

q ∗ (s)ds dt

ξ



Φα−1 2

a

q ∗ (s)Φα1 α2 (x(s))ds dt

t

ξ

Φα−1 2

a

=m



x(t)Φα−1 2

t



∗



q (s)ds dt +

ξ

ξ

b

Φα−1 2



t

∗





q (s)ds dt . ξ

This together with (3.90) yields 2α1 +1 < (b − a)α1



ξ

a



ξ

= 

a

≤

ξ

a

 t

 − − q (s)ds dt +

Φα−1 2



Φα−1 2 Φα−1 2

ξ



ξ

 q − (s)ds dt +

t ξ



q − (s)ds dt +

a



ξ b

ξ b ξ

b

Φα−1 2

Φα−1 2 Φα−1 2



t

+



q (s)ds dt

 

ξ t

q + (s)ds dt

b

q + (s)ds dt.

ξ

ξ

Therefore, 2α1 +1 < (ξ −a)Φα−1 2 (b − a)α1



ξ a



q − (s)ds +(b−ξ )Φα−1 2

b

q + (s)ds .

(3.92)

ξ

We discuss the two cases α2 ≥ 1 and 0 < α2 < 1 separately. In the first case, consider α2 ≥ 1. In this case, Φα2 is a concave-up function on [0, ∞). Hence, for any x1 , x2 ∈ [0, ∞) and t ∈ [0, 1], we have Φα2 (tx1 + (1 − t)x2 ) ≤ tΦα2 (x1 ) + (1 − t)Φα2 (x2 ).

(3.93)

Now, dividing both sides of (3.92) by b − a and applying Φα2 , we get

2 b−a



α = Φα2

< Φα2

ξ −a b−a



2 b−a 

Φα−1 2

α1 +1 

ξ a

 b



b−ξ + Φα−1 q − (s)ds + q (s)ds . 2 b−a ξ

3.3 Third-Order Half-Linear Equations

165

Note that b−ξ ξ −a + = 1. b−a b−a Then, using (3.93) with t = (ξ − a)/(b − a), we obtain

2 b−a

α






ξ

2 b−a

α ,

i.e., (3.84) holds. In the second case, consider 0 < α2 < 1. In this case, Φα2 is a concave-down function on [0, ∞). Hence, for any x1 , x2 ∈ [0, ∞), we have Φα2 (x1 + x2 ) ≤ Φα2 (x1 ) + Φα2 (x2 ).

(3.94)

From (3.92), we get  ξ  b 

 2α1 +1 −1 − −1 + < (b − a) Φ q (s)ds + Φ q (s)ds . α2 α2 (b − a)α1 a ξ Hence,

2 b−a

α1

< Φα−1 2



ξ a



q − (s)ds + Φα−1 2

b

q + (s)ds .

(3.95)

ξ

Applying Φα2 to both sides of (3.95) and using (3.94), we obtain 

ξ

−



b

q (s)ds +

+

q (s)ds >

a

ξ



2 b−a

α , 

i.e., (3.84) holds.

Theorem 3.33 (Lyapunov-Type Inequality) Let a < b < c. Assume x is a solution of (3.83) with x(a) = x(b) = x(c) = 0 and x(t) = 0 for all t ∈ (a, b) ∪ (b, c). Then, either  max

ξ ∈[a,b]

a

ξ

q − (s)ds +



b ξ

 q + (s)ds >

2 b−a

α (3.96)

166

3 Half-Linear Differential Equations

or 

ξ

c

q (s)ds +

max

ξ ∈[b,c]



−



+



q (s)ds >

b

ξ

2 b−a

α .

(3.97)

.

(3.98)

Moreover, 

ξ

c

q (s)ds +

max

ξ ∈[a,c]



−



+



q (s)ds >

a

ξ

2 b−a

α

Proof By Rolle’s theorem, there exist ξ1 ∈ (a, b) and ξ2 ∈ (b, c) such that Φα1 (x  )(ξ1 ) = 0 and

Φα1 (x  )(ξ2 ) = 0.

A further application of Rolle’s theorem yields that there exists ξ ∈ (ξ1 , ξ2 ) such that (Φα1 (x  )) (ξ ) = 0. Clearly, either ξ ∈ (a, b] or ξ ∈ (b, c]. If ξ ∈ (a, b], then, by applying Theorem 3.32 to the interval [a, b], we obtain 

ξ

q − (s)ds +



a

b

q + (s)ds >



ξ

2 b−a

α ,

and hence (3.96) holds. If ξ ∈ [b, c), a similar argument leads to (3.97). It is easy to see that (3.96) leads to (3.98) and that (3.97) leads to (3.98) as well.  Corollary 3.34 (Lyapunov-Type Inequality) (a) Let a < b. Assume x is a solution of (3.83) satisfying (1.5) and x(t) = 0 for all t ∈ (a, b). If there exists ξ ∈ [a, b] such that (Φα1 (x  )) (ξ ) = 0, then the inequality 

b

|q(t)|dt >

a

2 b−a

α (3.99)

holds. (b) Let a < b < c. If x is a solution of (3.83) with x(a) = x(b) = x(c) = 0 and x(t) = 0 for all t ∈ (a, b) ∪ (b, c), then 

b

either

|q(t)|dt >

a

2 b−a

α



c

or

|q(t)|dt >

b

Moreover,  a

c

|q(t)|dt >

2 c−a

α .

2 c−b

α .

3.3 Third-Order Half-Linear Equations

167

Remark 3.35 Although the inequalities in Corollary 3.34 are simpler, they are not as sharp as those in Theorems 3.32 and 3.33. This is due to the fact that |q(t)| = q + (t) + q − (t). For instance, if a = ξ , then (3.83) becomes 

b

q + (t)dt >



a

2 b−a

α ,

and if b = ξ , then (3.83) becomes 

b

q − (t)dt >



a

2 b−a

α .

Both are sharper than (3.99). Furthermore, under the assumptions of Theorem 3.32, we never expect that  q(t)

≥0

if

t ∈ [a, ξ ],

≤0

if

t ∈ [ξ, b]

could happen. However, this cannot be observed from Corollary 3.34. In the following, as direct applications of Theorem 3.33, we study the number of zeros and the distances between consecutive zeros of a nontrivial solution of (3.83) on a given interval. The first result gives us an estimate for the number of zeros. Theorem 3.36 (Lyapunov-Type Inequality) Let x = 0 be a solution of (3.83). +1 Let {tk }2N k=1 , N ≥ 1, be an increasing sequence of zeros of x in a compact interval I with length . Then,  ξk α/(α+1) 0 11/(α+1)  t2k+1 N  − + N< max q (t)dt + q (t)dt . ξk ∈[t2k−1 ,t2k+1 ] 2 t2k−1 ξk k=1 (3.100) Proof For k = 1, 2, . . . , N , we apply Theorem 3.33 to the interval [t2k−1 , t2k+1 ] ⊆ I . It follows from (3.98) that 

ξk

q − (t)dt +

max

ξk ∈[t2k−1 ,t2k+1 ]



t2k−1

t2k+1

 q + (t)dt >

ξk

2 t2k+1 − t2k−1

α .

Taking the sum on both sides for k from 1 to N, we get N

k=1



ξk

max

ξk ∈[t2k−1 ,t2k+1 ]

t2k−1

q − (t)dt +



t2k+1 ξk

 N

q + (t)dt > 2α (t2k+1 − t2k−1 )−α . k=1

(3.101)

168

3 Half-Linear Differential Equations

Note that ak = t2k+1 − t2k−1 > 0 for 1 ≤ k ≤ N . Then, ψ(x) := x −α is a concave-up function for α > 0 on (0, ∞). Thus, applying N 1 ψ(ak ) ≥ ψ N



k=1

N 1 ak N



k=1

to the right-hand side of (3.101), we obtain 

N

k=1

ξk

max

ξk ∈[t2k−1 ,t2k+1 ]

 α

>2 N

q − (t)dt +



t2k−1

t2k+1

q + (t)dt



ξk

N 1 (t2k+1 − t2k−1 ) N

−α

k=1

=2 N (t2N +1 − t1 )−α α 2 ≥ N α+1 .  α

α+1



Then, (3.100) follows.

Next, we show how the distances between consecutive zeros of an oscillatory solution of (3.83) may change. Theorem 3.37 (Lyapunov-Type Inequality) Let x be an oscillatory solution of (3.83) with {tn }∞ n=1 its increasing sequence of zeros in [0, ∞). Assume there exists σ ≥ 1 such that for any M > 0, we have 

t+M

|q(s)|ds → 0 as

t → ∞.

(3.102)

t

Then, tn+2 − tn → ∞ as n → ∞. Proof It suffices to prove the statement for the case when σ = 1. In fact, if (3.102) holds for some σ > 1, then, by (3.86) with p = σ and p∗ = σ/(σ − 1), we obtain 

t+M



t+M

|q(s)|ds ≤

t

1/α |q(s)|σ ds

M 1−1/σ → 0

as

t → ∞,

t

and hence (3.102) holds for σ = 1. Assume the contrary. Then, there exist M > 0 and a subsequence {tnk }∞ k=1 of {tn }∞ such that t − t ≤ M for all large k. By the assumption, nk +2 nk n=1 

tnk +2

tnk

 |q(t)|dt ≤

tnk +M tnk

|q(t)|dt → 0

as

k → ∞.

3.3 Third-Order Half-Linear Equations

169

Applying Corollary 3.34 to the interval [tnk , tnk +2 ], we have 



tnk +2

|q(t)|dt >

tnk

2 tnk +2 − tnk

α ,

i.e., −α

1 0, Φα (s) := |s|α−1 s, and α1 , α2 > 0. For simplicity, as before, we denote α = (α1 + 1)α2 . Theorem 3.40 (Lyapunov-Type Inequality) Assume x is a solution of (3.103) with x(a) = x(b) = 0 and x(t) = 0 for all t ∈ (a, b). Suppose that there exists ξ ∈ [a, b] such that (r1 Φα1 (x  )) (ξ ) = 0. Then, the inequality 

ξ

−



b

 α

q (t)dt > 2

q (t)dt+ a

+

ξ

b

−1/α1

[r1 (t)] a

−α1 α2  b

−α2 −1/α2 dt dt [r2 (t)] a

holds. Proof This proof is essentially the same as that of Theorem 3.32. We only give an outline of the proof here. Without loss of generality, we may assume x > 0 on

170

3 Half-Linear Differential Equations

(a, b). Then, there exists d ∈ [a, b] such that m := x(d) = max x(t). As in the t∈[a,b]

proof of Theorem 3.32, we have 

b

2m ≤

   x (t) dt =



a

b

   [r1 (t)]−1/(α1 +1) [r1 (t)]1/(α1 +1) x  (t) dt.

a

Applying (3.86) to the right-hand side of the inequality with p = α1 + 1 and

p∗ = 1 +

1 , α1

we get (2m)



α1 +1

b

≤

−1/α1

[r1 (t)]

α1 

a



 α +1 r1 (t) x  (t) 1 dt

b

    x (t) r1 (t)Φα (x  (t)) dt. 1

a

b

=

b

dt

[r1 (t)]−1/α1 dt

α1 

a

a

Using integration by parts for the second integral on the right-hand side, we have (2m)α1 +1



b

[r1 (t)]−1/α1 dt

−α1



b

≤

a

a

x(t)(−r1 (t)Φα1 (x  (t))) dt.

The rest of the proof is similar to that of Theorem 3.32 and hence is omitted.



Theorem 3.41 (Lyapunov-Type Inequality) Let a < b < c. Assume x is a solution of (3.103) with x(a) = x(b) = x(c) = 0 and x(t) = 0 for all t ∈ (a, b) ∪ (b, c). Then, either 

ξ

max

ξ ∈[a,b]

q − (s)ds +



a

b

q + (s)ds



ξ



b

α

>2

−1/α1

[r1 (t)]

−α1 α2 

b

dt

−1/α2

[r2 (t)]

a

−α2 dt

(3.104)

a

or  max

ξ ∈[b,c]

ξ

q − (s)ds +

b



> 2α b



c

q + (s)ds

ξ c

[r1 (t)]−1/α1 dt



−α1 α2  b

c

[r2 (t)]−1/α2 dt

−α2 .

(3.105)

3.3 Third-Order Half-Linear Equations

171

Moreover, 

ξ

max

ξ ∈[a,c]



−

c

q (s)ds +

a

q (s)ds ξ



c

> 2α



+

[r1 (t)]−1/α1 dt

−α1 α2 

a

c

[r2 (t)]−1/α2 dt

−α2 (3.106)

.

a

Proof The proof is similar to that of Theorem 3.33. By Rolle’s theorem, there exist ξ1 ∈ (a, b) and ξ2 ∈ (b, c) such that (r1 Φα1 (x  ))(ξ1 ) = 0 and (r1 Φα1 (x  ))(ξ2 ) = 0. Again by Rolle’s theorem, there exists ξ ∈ (ξ1 , ξ2 ) such that (r1 Φα1 (x  )) (ξ ) = 0. Clearly, either ξ ∈ (a, b] or ξ ∈ (b, c]. If ξ ∈ (a, b], then, by applying Theorem 3.40 to the interval (a, b], we obtain 

ξ



−

b

q (s)ds +

a

q + (s)ds

ξ



b

> 2α

[r1 (t)]−1/α1 dt

−α1 α2 

a

b

[r2 (t)]−1/α2 dt

−α2 .

a

It follows that 

ξ

max

ξ ∈[a,b]

q − (s)ds +

a



b

q + (s)ds



ξ



b

α

>2

−1/α1

[r1 (t)] a

−α1 α2  dt

b

−1/α2

[r2 (t)]

−α2 dt

,

a

i.e., (3.104) holds. Similarly, if ξ ∈ [b, c), then (3.105) holds. It is easy to see that (3.104) leads to (3.106) and that (3.105) leads to (3.106) as well.  In 2013, Kisel ák [187] considered (3.103) under the stronger assumption that rk ∈ C3−k ([a, b], (0, ∞)), k = 1, 2. By changing the equation to a system of two equations, he obtained a Lyapunov-type inequality, i.e., (3.76) in Theorem 3.28. We note that (3.76) is much more complicated than (3.84) and (3.98) given in Theorems 3.32 and 3.33, respectively, and it is not comparable in form with (3.4) given in Theorem 3.2 for the second-order case. Moreover, since (3.76) uses |q(t)| rather than q ± (t), it is not as sharp as those in (3.84) and (3.98). In Theorem 3.45 below, Kisel ák also gave an estimate for the number of zeros of an oscillatory solution of (3.83) for the case α ≥ 1. However, the case 0 < α < 1 was left there as an open problem. In Theorem 3.36, we solve the problem for the general case including the case 0 < α < 1, which improves the estimate for the number of zeros even for the case α ≥ 1.

172

3 Half-Linear Differential Equations

3.3.3 The Linear Case Now, for convenience, we summarize the results for the linear case. Consider the third-order linear differential equation x  + q(t)x = 0.

(3.107)

Note that α1 = α2 = 1 and α = 2 in this case. By Theorem 3.32, we obtain the following result. Corollary 3.42 (Lyapunov-Type Inequality) Assume x is a solution of (3.107) with x(a) = x(b) = 0 and x(t) = 0 for all t ∈ (a, b). Suppose that there exists ξ ∈ [a, b] such that x  (ξ ) = 0. Then,  ξ  b 4 q − (t)dt + q + (t)dt > . (b − a)2 a ξ The next result follows from Theorem 3.33. Corollary 3.43 (Lyapunov-Type Inequality) Let a < b < c. Assume x is a solution of (3.107) with x(a) = x(b) = x(c) = 0 and x(t) = 0 for all t ∈ (a, b) ∪ (b, c). Then, either  ξ   b 4 − + max q (s)ds + q (s)ds > ξ ∈[a,b] (b − a)2 a ξ or 

ξ

max

ξ ∈[b,c]

q − (s)ds +



b

c

 q + (s)ds >

ξ

4 . (c − b)2

Moreover,  max

ξ ∈[a,c]

a

ξ

−



c

q (s)ds +

+



q (s)ds > ξ

4 . (c − a)2

It is easy to see that these corollaries provide sharper results for the linear case than those by Parhi and Panigrahi [230] as summarized in Theorem 3.2.

3.3.4 Applications to Boundary Value Problems Now, we introduce some applications of the previous results for (3.83). Theorem 3.44 Assume q ∈ Lμ ([0, ∞], R) with μ ∈ [1, ∞). Let x be an oscillatory solution of (3.83) with an increasing sequence of zeros {tk }∞ k=1 . Then, the distances between consecutive zeros {tk+1 − tk } or {tk+2 − tk } tend to infinity as k → ∞.

3.3 Third-Order Half-Linear Equations

173

Proof In a proof by contradiction, we suppose that, in the case of d (1) D x(t) = 0 dt for some t ∈ [tk , tk+1 ] for every large k, it is not true that {tk+1 − tk } → ∞. Hence, there exists a subsequence {tkn }∞ n=1 such that (tkn+1 − tkn ) < K for every n ∈ N, with some K > 0. Let d (1) D x(ckn ) = 0 dt   for ckn ∈ tkn , tkn+1 and max

! |x(t)|

  = x(dkn ) ,

t∈ tkn ,tkn+1

  where dkn ∈ tkn , tkn+1 . Without loss of generality, we can assume ckn < dkn . Then, we get 

tkn+1

|q(t)|dt >

2 − tkn

dkn

tkn

β+α2 .

From integrability of q, we have 

∞

|q(t)|μ dt >

μ

2β+α2 ∗

K β+α2 +1/μ

tkn

for sufficiently large n and 1/μ + 1/μ∗ = 1. Therefore, using Hölder’s inequality, we obtain 1
0, and   t ∈ tkn , tkn+1 .

d (1) D x(t) = 0 for dt

Since x(tkn+2 ) = 0, there exists ckn ∈ [tkn+1 , tkn+2 ] such that d (1) D x(ckn ) = 0. dt Now, set max

! |x(t)|

  = x(dkn ) ,

t∈ tkn ,tkn+2

      where dkn ∈ tkn , tkn+1 ∪ tkn+1 , tkn+2 . If dkn ∈ tkn+1 , tkn+2 , then we can proceed as in the previous part of the proof. If dkn ∈ (tkn , tkn+1 ), then it follows that 

tkn+2

|q(t)|dt >

tkn

dkn

2 − tkn

β+α2 .

Therefore, using Hölder’s inequality, we obtain a contradiction as in the first part of the proof.  The following theorems give an estimate (upper bound) for the number of zeros of an oscillatory solution of (3.83) on a bounded interval [0, T ]. Theorem 3.45 Let x be an oscillatory solution of (3.83), with zeros 0 < t1 < t2 < · · · < tN ≤ T and d (1) D x(ek ) = 0 dt

for some

ek ∈ [tk , tk+1 ],

k = 1, 2, . . . , N − 1. Moreover, assume β + α2 ≥ 1. Then,  T

β+α2 0

T

|q(t)|dt > 2β+α2 (N − 1)β+α2 +1 .

3.3 Third-Order Half-Linear Equations

175

Proof We know that 

tk+1

|q(t)|dt >

tk

2 tk+1 − tk

β+α2

for k = 1, 2, . . . , N − 1. Thus, 

T



tN

|q(t)|dt ≥

0

|q(t)|dt >

t1

N −1

k=1

2 tk+1 − tk

β+α2 .

Now, using a well-known inequality for the power mean with exponent β + α2 and the arithmetic-geometric mean inequality  n 1/n  n −1 n % 1 1 1 Ai ≥ Ai ≥ , n n Ai i=1

i=1

Ai > 0,

i=1

we obtain N −1

k=1

1 tk+1 − tk



β+α2

N −1 1 1 ≥ (N − 1) N −1 tk+1 − tk

β+α2

k=1

≥ (N − 1)β+α2 +1

N −1

−β−α2 (tk+1 − tk )

k=1

= (N − 1)β+α2 +1 (tN − t1 )−β−α2 >

(N − 1)β+α2 +1 . T β+α2 

This completes the proof.

Theorem 3.46 Let x be an oscillatory solution of (3.83) with zeros 0 < t1 < t2 < · · · < t2N +1 ≤ T and d (1) D x(t) = 0 dt

for

t ∈ [t2k−1 , t2k ],

k = 1, 2, . . . , N. Moreover, assume β + α2 ≥ 1. Then,  T β+α2

T

|q(t)|dt > 2β+α2 N β+α2 +1 .

0

The proof of Theorem 3.46 can be omitted as one can proceed similarly as in the proof of Theorem 3.45. We leave the case β + α2 ∈ (0, 1) as an open problem.

176

3 Half-Linear Differential Equations

Example 3.47 For simplicity, we consider the exponents α1 , α2 to be the quotients of two odd integers. Moreover, let the coefficients of the quasi-derivatives be identically constant and consider the generalized Euler differential equation ((((x  )α1 ) )α2 ) +

γ x β = 0. (t + 1)β+α2 +1

(3.108)

We can proceed using an analogue of the linear Euler differential equation. If we denote D = (α1 + α2 )2 + 4β(β + α2 ), then the roots of an algebraic (indical) equation corresponding to a solution tλ are λ± =

α1 + 2β(1 + α1 ) + α2 ± 2α1 (β + α2 + 1)

√ D

.

Although it has not been proven yet, Kisel ák (see [186] and [187, Example 3.1]) conjectures that the constants β

γ± = λ± (λ± − 1)α2 α1α2 (β(λ± − 1) − α2 ) decide whether (3.108) is oscillatory or not (notice that in the linear case γ± = √ 2 3/9). To be more precise, a conjecture states that if γ ∈ [γ− , γ+ ], then (3.108) is nonoscillatory; otherwise it is oscillatory. Thus, we can state at least an estimate for γ . So, from the previous considerations, we have |γ |(1 − (T + 1)−β−α2 )T β+α2 > 2β+α2 (N − 1)β+α2 +1 . β + α2 Example 3.48 We give an application of Theorem 3.28 for the eigenvalue problem 

D(3) x ± λq(t)ϕβ (x) = 0, x(a) = x(c) = x(b) = 0,

a < c < b.

Let the assumptions of Theorem 3.28 be fulfilled. Then, it follows that Hβ |λ| > β 2



b

−1 |q(t)|dt

,

a

where H = min h(c) and h is the function defined in Theorem 3.28. In particular, c∈(a,b)

for the reduced problem (r1 (t) = r2 (t) = 1), we obtain 2β+α2 |λ| > (b − a)β+α2



b a

−1 |q(t)|dt

.

3.3 Third-Order Half-Linear Equations

177

Hereafter, we apply the results on the Lyapunov-type inequalities obtained in Sect. 3.3.1 to study the nonexistence, uniqueness, and existence-uniqueness for solutions of certain third-order boundary value problems. We now consider two boundary value problems consisting of (3.83), namely    Φα2 Φα1 (x  ) + q(t)Φα1 α2 (x) = 0, (Φα1 (x  )) (ξ ) = 0,

x(a) = x(b) = 0,

where

a≤ξ ≤b

(3.109)

and    Φα2 Φα1 (x  ) + q(t)Φα1 α2 (x) = 0, x(a) = x(b) = x(c) = 0,

where

(3.110)

a < b < c.

The first result is on the nonexistence of solutions of (3.109) and (3.110). Theorem 3.49 (Lyapunov-Type Inequality) If the inequality 

ξ



q − (s)ds +

a

b

q + (s)ds ≤



ξ

2 b−a

α (3.111)

holds, then (3.109) has no nontrivial solution. If the inequalities 

ξ

b

q (s)ds +

max

ξ ∈[a,b]



−

a

+





q (s)ds ≤

ξ

2 b−a

α (3.112)

and 

ξ

q − (s)ds +

max

ξ ∈[b,c]



b

c

ξ

α  2 q + (s)ds ≤ . c−b

(3.113)

hold, then (3.110) has no nontrivial solution. Proof We first prove the first statement. Assume the contrary, i.e., (3.109) has a nontrivial solution x. Then, there exist consecutive zeros t1 , t2 of x such that a ≤ t1 < t2 ≤ b and ξ ∈ [t1 , t2 ]. By Theorem 3.32, we have 

ξ

q − (s)ds +



t2

q + (s)ds >



ξ

t1

2 t2 − t1

α .

It follows that 

ξ a

−



q (s)ds + ξ

b

+

q (s)ds ≤



2 b−a

α .

178

3 Half-Linear Differential Equations

This contradicts (3.111). Now, we prove the second statement. Assume the contrary, i.e., (3.110) has a nontrivial solution x. Then, there exist three consecutive zeros t1 , t2 , t3 of x such that a ≤ t1 < t2 < t3 ≤ b and ξ ∈ [t1 , t2 ]. By Theorem 3.29, we have either  ξ

α   t2 2 − + q (s)ds + q (s)ds > max ξ ∈[t1 ,t2 ] t2 − t1 ξ t1 or 

ξ

max

ξ ∈[t2 ,t3 ]

q − (s)ds +



 q + (s)ds >

t3 ξ

t2

2 t3 − t2

α .

It follows that either 

ξ

max

ξ ∈[a,b]



−

b

q (s)ds +



+



q (s)ds >

a

ξ

2 b−a

α

or  max

ξ ∈[b,c]

ξ



−

c

q (s)ds +

b

+





q (s)ds > ξ

2 c−b

α . 

This contradicts (3.112) and (3.113).

As a direct consequence of the first statement of Theorem 3.49, we have the following corollary. Corollary 3.50 (Lyapunov-Type Inequality) Assume  q(t)

≥0

if t ∈ [a, ξ ],

≤0

if t ∈ [ξ, b].

Then, (3.109) has no nontrivial solution. In particular, if either a=ξ

and

q(t) ≤ 0 for all t ∈ [a, b]

ξ =b

and

q(t) ≥ 0 for all t ∈ [a, b],

or

then (3.109) has no nontrivial solution. Next, we consider the third-order nonhomogeneous linear equation x  + q(t)x = h(t)

(3.114)

3.3 Third-Order Half-Linear Equations

179

on (A, B), where q, h ∈ C((A, B), R), and let and

A < t1 < t 2 < t 3 < B

k1 , k2 , k3 ∈ R.

(3.115)

Now, we present an existence and uniqueness criterion for a nontrivial solution of the three problems 

x  + q(t)x = h(t) on x(t1 ) = k1 ,



x(t2 ) = k2 ,

x  + q(t)x = h(t) on x  (t1 ) = k2 ,

x(t1 ) = k1 ,

(A, B), (3.116)

x  (t2 ) = k3 , (A, B),

(3.117)

x(t2 ) = k3 ,

and 

x  + q(t)x = h(t) on x(t1 ) = k1 ,

x(t2 ) = k2 ,

(A, B), (3.118)

x  (t3 ) = k3 .

Theorem 3.51 (Lyapunov-Type Inequality) Assume 

ξ

max

ξ ∈[A,B]

q − (s)ds +



A

B

 q + (s)ds ≤

ξ

4 b−a

2 .

(3.119)

Then, each of (3.116), (3.117), and (3.118) has a unique solution on (A, B) for any t1 , t2 , t3 and k1 , k2 , k3 satisfying (3.115). To prove Theorem 3.51, we first introduce the following results by Jackson and Schrader [177] and Jackson [176] on the general third-order equation x  = f (t, x, x  , x  ),

(3.120)

where (a) f ∈ C((A, B) × R3 , R), and (b) any initial value problem associated with (3.120) has a unique solution which exists on the whole interval (A, B). We consider the three problems 

x  = f (t, x, x  , x  ) x(t1 ) = k1 ,



x(t2 ) = k2 ,

x  = f (t, x, x  , x  ) x(t1 ) = k1 ,

on



on

x (t1 ) = k2 ,

(A, B), x  (t2 ) = k3 ,

(3.121)

(A, B), x(t2 ) = k3 ,

(3.122)

180

3 Half-Linear Differential Equations

and 

x  = f (t, x, x  , x  ) x(t1 ) = k1 ,

on

x(t2 ) = k2 ,

(A, B), (3.123)

x  (t3 ) = k3 .

Lemma 3.52 Assume (a) and (b) are satisfied and (3.123) has at most one solution on (A, B). Then, each of (3.121), (3.122), and (3.123) has a unique solution on (A, B). Under the assumptions (a) and (b), Jackson and Schrader [177, Theorem 2 and Theorem 3] showed that if the solution of (3.123), when it exists, is unique on (A, B), then each of (3.121), (3.122), and (3.123) has at least one solution on (A, B). Then, Jackson [176, Theorem 1 and Theorem 2] further proved that the uniqueness of the solution of (3.123) guarantees the uniqueness of the solutions of (3.121) and (3.122). Based on Lemma 3.52, we now give the proof for Theorem 3.51. Proof of Theorem 3.51 Note that assumptions (a) and (b) are satisfied by (3.114) since it is linear. First we show that (3.118) has at most one solution for any t1 , t2 , t3 and k1 , k2 , k3 satisfying (3.115). Assume the contrary, i.e., (3.118) has two solutions x1 and x2 in (A, B). Define x = x1 − x2 . Then, x is a solution of the third-order homogeneous linear boundary value problem 

x  + q(t)x = 0,

(3.124)

x(t1 ) = x(t2 ) = x(t3 ) = 0. On the other hand, from (3.119), we have 

ξ

max

ξ ∈[t1 ,t2 ]

−



t2

q (s)ds +

t1

+





q (s)ds ≤

ξ

4 t2 − t1

2

and 

ξ

max

ξ ∈[t2 ,t3 ]

t2

q − (s)ds +



t3 ξ

 q + (s)ds ≤

4 t3 − t2

2 .

Then, by applying the second statement of Theorem 3.49 with α = 2, we see that (3.124) has no nontrivial solution in (A, B). Hence, x(t) ≡ 0 from which it follows that x1 (t) ≡ x2 (t). Thus, the uniqueness of the solution of (3.118) is established. By Lemma 3.52, we conclude that each of (3.116), (3.117), and (3.118) has a unique solution on (A, B). 

3.4 Higher-Order Half-Linear Equations

181

3.4 Higher-Order Half-Linear Equations In this section, we establish several Lyapunov-type inequalities for certain halflinear higher-order differential equations under anti-periodic boundary conditions. Anti-periodic problems have received considerable attention as they appear in numerous situations. Examples include anti-periodic trigonometric polynomials in the study of interpolation problems [111], anti-periodic wavelets [79], difference equations [67], ordinary, partial, and abstract differential equations [136, 218, 257, 260, 280, 283, 300], and impulsive differential equations [24, 205]. For some more applications of anti-periodic boundary conditions in physics, see [82] and the references therein. Throughout this section, we assume that p > 1 and p∗ is the Hölder conjugate of p, i.e., 1/p + 1/p∗ = 1. In 2012, Wang [284] considered the boundary value problem consisting of an (m + 1)-order half-linear differential equation and antiperiodic boundary conditions of the form ⎧ 

 p−2 ⎪ (m) ⎨ u(m)  u + q(t)|u|p−2 u = 0 on ⎪ ⎩

u (a) + u (b) = 0 for (i)

(i)

(a, b), (3.125)

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

To prove the main results, we use the following lemmas. Lemma 3.53 If u is a nonzero solution of (3.125), then, for i = 0, 1, . . . , m, we have

  b  b − a p b  (i+1) p  (i) p (t) dt. (3.126) u (t) dt ≤ u 2 a a Proof Let us define  1 1 H (x, y) = 2 −1

if a ≤ x ≤ y, if y ≤ x ≤ b.

(3.127)

Then, by the anti-periodic boundary conditions in (3.125), we have  u(i) (y) = a

b

u(i+1) (x)H (x, y)dx,

where

a ≤ y ≤ b.

(3.128)

182

3 Half-Linear Differential Equations

So, using Hölder’s inequality in (3.128), we have     (i)  u (y) ≤

b

a

 ≤ a

   (i+1)  (x) |H (x, y)|dx u b

1/p     (i+1) p (x) dx u

b

p∗

1/p∗

|H (x, y)| dx

(3.129)

a

1 ∗ = (b − a)1/p 2

 a

b

1/p    (i+1) p (x) dx , u

and thus  b    1  (i+1) p  (i) p (x) dx, u u (y) ≤ p (b − a)p−1 2 a which, after integrating over [a, b], yields 

b

a

   b − a p b  (i+1) p  (i) p (x) dx. u (y) dy ≤ u 2 a

Therefore, (3.126) holds.

(3.130) 

Lemma 3.54 If u is a nonzero solution of (3.125), then |u(t)|

 u

p−1  (m−1)

 1 b − a m(p−1)  b    (m) p  (t) ≤ u (x) dx. 2 2 a

(3.131)

Proof By (3.129) and (3.130), we have  b     (m) p  (m−1)  1 1/p∗ (y) ≤ (b − a) u (x) dx, u 2 a  b

1/p   p 1 ∗ u (x) dx |u(y)| ≤ (b − a)1/p , 2 a

(3.132) (3.133)

and  a

b

 b 

1/p

1/p    b−a  (i) p  (i+1) p ≤ (x) dx . u (y) dy u 2 a

(3.134)

3.4 Higher-Order Half-Linear Equations

183

Then, by (3.133) and (3.134), we get |u(y)| ≤

1 ∗ (b − a)1/p 2

1 ∗ ≤ (b − a)1/p 2 ≤

1 ∗ (b − a)1/p 2



b

  p u (x) dx

1/p

a



b−a 2 b−a 2



  p u (x) dx

b

1/p

a

2 

b

  p u (x) dx

1/p

a

≤ ··· 1 ∗ ≤ (b − a)1/p 2



b−a 2

m−1  a

b

1/p    (m) p . u (x) dx

Therefore, |u(y)|

p−1

≤

1 2p−1

(b − a)

(p−1)/p∗



b−a 2

(m−1)(p−1)  a

b

1−1/p    (m) p . u (x) dx (3.135) 

Multiplying (3.132) and (3.135), we obtain (3.131). Now, we give Wang’s [284] first result.

Theorem 3.55 (Lyapunov-Type Inequality) If u is a nonzero solution of (3.125), then 

b

|q(t)|dt > 2

a

2 b−a

m(p−1) .

(3.136)

Proof Multiplying the differential equation in (3.125) by u(m−1) (t) and integrating over [a, b], we obtain  a

b



    (m) p−2 (m) (m−1) u (t)u (t)dt + u 

b

q(t)|u(t)|p−2 u(t)u(m−1) (t)dt = 0.

a

Using integration by parts in the first integral of the former equation and employing the anti-periodic boundary conditions in (3.125), we have  a

b

    (m) p u (t) dt =

a

b

q(t)|u(t)|p−2 u(t)u(m−1) (t)dt.

(3.137)

184

3 Half-Linear Differential Equations

By Lemma 3.54, (3.137) yields  a

b

    (m) p u (t) dt = 

b

q(t)|u(t)|p−2 u(t)u(m−1) (t)dt

a b

≤ a

    |q(t)||u(t)|p−1 u(m−1) (t) dt

≤ 1 ≤ 2

max |u(t)|

a≤t≤b



2 b−a

 u

p−1  (m−1)

m(p−1) 

b a

   (t)

b

(3.138) |q(t)|dt

a

    (m) p u (t) dt

b

|q(t)|dt.

a

Now, we claim 

b

a

   (m) p u (t) dt > 0.

(3.139)

In fact, if (3.139) is not true, then 

b a

   (m) p u (t) dt = 0.

Then, u(m) (t) = 0 for all t ∈ [a, b]. By the anti-periodic boundary conditions in (3.125), we obtain u(t) = 0 for all t ∈ [a, b], which contradicts u(t) ≡ 0, t ∈ [a, b]. Thus, (3.139) holds, and dividing both sides of (3.138) by 

b a

   (m) p u (t) dt > 0, 

we obtain (3.136), completing the proof. Now, we give an application of Theorem 3.55 to an eigenvalue problem. Theorem 3.56 If λ is an eigenvalue of the equation 

   (m) p−2 (m) u + λq(t)|u|p−2 u = 0 u 

on

(a, b)

satisfying the anti-periodic boundary conditions in (3.125), then |λ| > 2

2 b−a

m(p−1)  a

b

−1 |q(t)|dt

.

3.4 Higher-Order Half-Linear Equations

185

For the linear case p = 2 with m = 2n − 1, 2n, n − 1, we have the following results. Corollary 3.57 If u is a nonzero solution of the boundary value problem  (2n) u + q(t)u = 0, u(i) (a) + u(i) (b) = 0

for i = 0, 1, . . . , 2n − 1,

then the inequality 

b

|q(t)|dt >

a

22n (b − a)2n−1

holds. Corollary 3.58 If u is a nonzero solution of the boundary value problem 

u(2n+1) + q(t)u = 0, u(i) (a) + u(i) (b) = 0 for

i = 0, 1, . . . , 2n,

then the inequality 

b

22n+1 (b − a)2n

|q(t)|dt >

a

holds. Corollary 3.59 If u is a nonzero solution of the boundary value problem 

u(n) + q(t)u = 0, u(i) (a) + u(i) (b) = 0 for

i = 0, 1, . . . , n − 1,

then the inequality  a

holds.

b

|q(t)|dt >

2n (b − a)n−1

186

3 Half-Linear Differential Equations

For the nonlinear case with m = 1, 2, we give the following results. Corollary 3.60 If u is a nonzero solution of the boundary value problem ⎧   ⎨ u p−2 u + q(t)|u|p−2 u = 0, ⎩

u(a) + u(b) = 0,

u (a) + u (b) = 0,

then the inequality 

b

|q(t)|dt >

a

2p (b − a)p−1

(3.140)

holds. Corollary 3.61 If u is a nonzero solution of the boundary value problem ⎧   ⎨ u p−2 u + q(t)|u|p−2 u = 0, ⎩ u(a) + u(b) = u (a) + u (b) = u (a) + u (b) = 0, then the inequality 

b

|q(t)|dt >

a

22p−1 (b − a)2p−2

holds.

3.5 Notes and References It appears that, for half-linear equations, the first generalization of Lyapunov’s result was obtained in 2003 by Yang [294]. Showing that there is a striking similarity between linear and half-linear equations, Yang proved Theorem 3.2 in Sect. 3.2 by using Jensen’s inequality [215] given in Lemma 3.1. Indeed, the same Lyapunovˇ type inequality was also obtained by Došlý and Rehák [115, p. 190], Lee et al. [195], and Pinasco [235, 237]. As a direct conclusion of Theorem 3.2, we have a disconjugacy criterion, namely Corollary 3.3, for which we can refer to the ˇ monograph by Došlý and Rehák [115, p. 191]. As an extension of the results of Brown and Hinton [64] (see also Theorem 1.60 in Sect. 1.7), a generalized version of the Opial inequality is given as Theorem 3.5 by Beesack and Das [41]. In 2004, Lee et al. [195] considered the half-linear differential equation with damping term. Lemma 3.7 and Theorems 3.8 and 3.10 are due to Lee et al. [195]. The remainder of Sect. 3.2 concerning Lyapunov-type inequalities for half-linear equations is taken from Wang [282]. Section 3.2.1 is based on Pinasco [235].

3.5 Notes and References

187

Section 3.3.1 is due to Kisel ák [187] and Dhar and Kong [112], who were concerned with third-order half-linear differential equations of the forms (3.75) and (3.83), respectively. Moreover, for Sects. 3.3.2 and 3.3.4, we refer to the paper by Dhar and Kong [112]. Some applications to boundary value problems obtained in the previous subsections are presented in Sect. 3.3.4, see Kisel ák [187] and Dhar and Kong [112]. Section 3.4 is due to Wang [284], who considered (m + 1)-order half-linear differential equations with anti-periodic boundary conditions, i.e., (3.125).

Chapter 4

Lyapunov-Type Inequalities for Nonlinear Differential Systems

4.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for second-order nonlinear systems of differential equations under some boundary conditions. We also sketch some recent developments related to this type of inequalities. In Sect. 4.2, we deal with Lyapunov-type inequalities for second-order nonlinear systems subject to Dirichlet boundary conditions. In Sect. 4.3, we give some recent results about a special kind of nonlinear systems, called quasilinear systems, subject to Dirichlet boundary conditions. In Sect. 4.4, extending the results given in Sect. 4.3, we state and prove a generalized Lyapunov-type inequality for Dirichlet problems associated to quasilinear systems involving (p1 , p2 , . . . , pn )-Laplacian. Section 4.5 deals with eigenvalue problems for quasilinear operators of p-Laplace type under Dirichlet boundary conditions. In this section, some lower bounds are obtained for generalized eigenvalues. In Sects. 4.6 and 4.7, we are concerned with Lyapunov-type inequalities for cycled systems with anti-periodic boundary conditions and quasilinear systems with clamped-free boundary conditions.

4.2 Nonlinear Systems In this section, we are concerned with the problem of finding Lyapunov-type inequalities for the nonlinear system of differential equations in the form   x = α1 (t)x + β1 (t)|u|γ −2 u, u = −β2 (t)|x|β−2 x − α1 (t)u,

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_4

(4.1)

189

190

4 Nonlinear Differential Systems

whose special cases contain the well-known equations of Emden–Fowler type and half-linear equations. We recall that a nontrivial solution (x, u) of (4.1) defined on some infinite interval [t0 , ∞) is said to be proper if sup {|x(t)| + |u(t)| : t ≤ s < ∞} > 0 for any t ≥ t0 . A proper solution (x, u) of (4.1) is called weakly oscillatory if at least one component has a sequence of zeros tending to ∞. This solution is said to be oscillatory if both components have sequences of zeros tending to ∞. If both components (at least one component) are different from zero for large t, then the solution (x, u) is called nonoscillatory (weakly nonoscillatory). System (4.1) is said to be oscillatory if all solutions are oscillatory. Recently, much attention has been paid to the existence of proper solutions in both general and special cases of (4.1). For a comprehensive treatment of the ˇ subject, we refer the reader to the books by Došlý and Rehák [115], Kiguradze and Chanturia [183], and Mirzov [213] and the paper by Kitano and Kusano [188]. Since our attention focuses on Lyapunov-type inequalities for the nonlinear system of differential equations, existence of nontrivial solutions (x, u) of (4.1) shall be assumed. We state the basic hypotheses (i) γ > 1, β > 1 are real constants, (ii) β1 , β2 ∈ C([t0 , ∞), R), β1 (t) > 0 for t ≥ t0 , (iii) α1 ∈ C([t0 , ∞), R). A linear Hamiltonian system, in the case of two scalar linear differential equations, has the form (see, for example [191, 293]) y  = J H (t)y

on

R,

(4.2)

where

y y= 1 , y2



0 1 J = , −1 0



h11 h12 H = h21 h22



with real-valued and continuous functions hj k , j, k = 1, 2, and h12 = h21 . Setting x = y1 ,

u = y2 ,

β2 = h11 ,

α1 = h12 = h21 ,

and

β1 = h22

in (4.2), one can easily obtain the linear system 

x  = α1 (t)x + β1 (t)u, u = −β2 (t)x − α1 (t)u,

(4.3)

which is a special case of the linear counterpart of (4.1) with γ = 2 and β = 2.

4.2 Nonlinear Systems

191

We remark that the second-order linear differential equation   r(t)x  + q(t)x = 0 and the Emden–Fowler-type equations   r(t)Φα (x  ) + q(t)Φβ (x) = 0

(4.4)

and (Φα (x  )) + g(t)Φα (x  ) + f (t)Φβ (x) = 0,

(4.5)

where Φγ (s) = |s|γ −2 s for γ > 1, and r, q, g, f are real functions with r(t) > 0 for all t ∈ R, can be clearly written as an equivalent Hamiltonian system in the form (4.3) and also as (4.1). Since Lyapunov inequalities have found applications in the study of various properties of solutions such as oscillation theory, disconjugacy, and eigenvalue problems, we present numerous generalizations. Several authors, including Cheng [83], Eliason [124], Hartman [155], Hochstadt [172], Kwong [193], Reid [248, 249], and Singh [258], have contributed to the presented results. Although there is an extensive literature on Lyapunov-type inequalities, there is not much done for (4.3). We refer the reader to the introductory paper by Guseinov and Kaymakçalan [143] in this direction. The principal aim of this section is to state and prove Lyapunov-type inequalities for (4.1) and some special cases. Motivated by the papers of Guseinov and Kaymakçalan [143], Lee et al. [195], and Pachpatte [229], Tiryaki et al. [274] derived a Lyapunov-type inequality for (4.1), where the first component of the solution (x, u) has zeros at the points a and b in I = [t0 , ∞) ∈ R. For some special cases of (4.1), they also derived some Lyapunov-type inequalities which not only relate the points a and b in I at which the first component of the solution (x, u) has zeros but also any point in (a, b), where the first component of the solution (x, u) is maximized. Their inequalities for (4.4) and (4.5) contain better lower bounds than those bounds given in [195] and [229], respectively. The inequalities that we present in this section can be used as a handy tool in the study of the qualitative nature of solutions. Finally, we also give some applications to show the importance of the results of Tiryaki et al. [274]. Now, we present the main results of this section. Theorem 4.1 (Lyapunov-Type Inequality) Let the hypotheses (i)–(iii) hold. If (4.1) has a real solution (x, u) such that x(a) = x(b) = 0 and x = 0 on [a, b], where a,b ∈ R with a < b, then the inequality  a

b

 |α1 (t)| dt + M β/α−1

1/γ 

b

b

β1 (t)dt a

a

holds, where 1/α + 1/γ = 1 and M = max |x(t)|. a 0 on (t1 , t2 ). Choose τ ∈ (t1 , t2 ) such that M = |x(τ )| = max{|x(t)| : t1 < t < t2 } > M1 . Clearly, the inequalities in Theorem 4.4 are satisfied on (t1 , t2 ). Because of (4.43), one can choose T ≥ t0 large enough so that for every t1 ≥ T , we have 

∞

β1 (t)dt < M

−(β−α)/(α−1)



∞

and

t1

|β2 (t)| dt < 1.

t1

Substituting (4.44) into (4.16) in Theorem 4.4, we obtain  2α < M β−α

t2

β1 (t)dt 

≤ M β−α

α−1 

t2 t1

t1

α−1 

∞

∞

β1 (t)dt t1

β2+ (t)dt

t1

|β2 (t)| dt < 1,

(4.44)

204

4 Nonlinear Differential Systems

where α > 1. This contradiction shows that |x(t)| is bounded on I . Hence, there exists K > 0 such that |x(t)| ≤ K for all t ∈ I . To show that |u(t)| is bounded, we integrate the second equation in (4.13) from τ ≤ t ≤ t2 to t to obtain 

t

u(t) = −

β2 (s)|x(s)|β−2 x(s)ds,

τ

where we used the fact that u(τ ) = 0 in view of the explanation made in the proof of Theorem 4.4. Hence, it follows from the last equality that 

t

|u(t)| ≤

 |β2 (s)| |x(s)|β−1 ds ≤

τ

∞

 |β2 (s)| |x(s)|β−1 ds ≤ K β−1

τ

∞

|β2 (s)| ds,

τ

and hence |u(t)| is bounded on I since 

∞

|β2 (s)| ds < ∞.

τ

It follows from lim sup{|x(t)| + |u(t)|} ≤ lim sup |x(t)| + lim sup |u(t)| that lim sup{|x(t)| + |u(t)|} is bounded on I , which completes the proof. β2+



Lμ [t

Theorem 4.13 (Lyapunov-Type Inequality) Suppose that ∈ 0 , ∞) with 1 ≤ μ < ∞. If (x, u) is any weakly oscillatory proper solution of (4.13) with β1 (t) = 1, then the distance between consecutive zeros of the first component of (x, u) tends to infinity as t → ∞. Proof We first assume that (x, u) is a nontrivial weakly oscillatory proper solution of (4.13) with β1 (t) = 1 on I such that x has a sequence of zeros tending to ∞ and the conclusion is not true. Then, the sequence {tn } of zeros of x has a subsequence  {tnm } such that tnm+1 − tnm  ≤ N < ∞ for  ) be such  all m. Let snm ∈ (tnm , tnm+1 that |x(t)| is maximized. Then, snm − tnm  < N for all m. Let M = x(snm ) and μ∗ be the Hölder conjugate of μ, i.e., 1/μ∗ + 1/μ = 1. Suppose β2+ ∈ Lμ [t0 , ∞), 1 ≤ μ < ∞, for m large enough so that 

∞

1/μ [β2 (s)]μ ds

∗

≤ M α−β N −(α−1+1/μ ) .

(4.45)

tnm

By using (4.14) with β1 (t) ≡ 1, we obtain  M β−α (snm − tnm )α−1

snm

tnm

β2+ (t)dt ≥ 1.

(4.46)

4.2 Nonlinear Systems

205

Using Hölder’s inequality with Hölder conjugates μ and μ∗ on the right-hand side of (4.46) yields  1 ≤ M β−α (snm − tnm )α−1

snm

 ≤ M β−α (snm − tnm )α−1

≤M

≤M

β−α

β−α

(snm − tnm )

(snm − tnm )

β2+ (t)dt

tnm

snm

tnm

α−1+1/μ∗

 + μ β2 (t) dt



tnm α−1+1/μ∗

dt

∞

tnm

1/μ∗

snm

tnm

 + μ β2 (t) dt

snm



1/μ 

μ β2+ (t) dt

1/μ

(4.47)

1/μ .

Substituting (4.45) into (4.47), we get 1≤M

β−α

(snm − tnm )

< M β−α N

α−1+1/μ∗

α−1+1/μ∗



∞ tnm

μ β2+ (t) dt

M α−β N −(α−1+1/μ

1/μ

∗)

= 1. 

This contradiction completes the proof.

In 2012, Tiryaki et al. [272] stated and proved some new Lyapunov-type inequalities for a class of nonlinear system, special cases of which contain the wellknown Hamiltonian systems, Emden–Fowler, half-linear, and linear differential equations of second-order. Their results improve and generalize these types of inequalities related to all existing ones. They are concerned with the problem of finding Lyapunov-type inequalities of the nonlinear system given by the differential equations of the form (4.1). It is not difficult to see that if we let α1 (t) ≡ 0 in (4.1), then we get the wellknown Emden–Fowler equation   α−2   x + q(t)|x|β−2 x = 0, r(t) x  

(4.48)

where 1/α + 1/γ = 1, β1 = r 1−γ , and β2 = q. Moreover, if we make the transformation x = h(t)y

and

u=

1 v, h(t)

206

4 Nonlinear Differential Systems

where h = α1 (t)h, i.e.,  h(t) = exp



t

α1 (s)ds t0

in (4.1) with β = α, then we have the system 

y  = β1 (t)|h(t)|−γ |v|γ −2 v,

(4.49)

v  = −β2 (t)|h(t)|α |y|α−2 y,

which does not have the diagonal terms, i.e., α1 and −α1 . Furthermore, (4.49) is equivalent to the half-linear equation of the form (3.32). Equation (4.48) is called super-half-linear if β > α and sub-half-linear if β < α (see [15, 115]). Although there is a large body of literature concerning Lyapunov-type inequalities for (3.32), to the best of our knowledge, there are only a few studies which are interested in (3.32) with α > 2 or 1 < α < 2, except for Lee et al. [195] and Wang [282]. The aim is to obtain Lyapunov-type inequalities for (4.1) that are better than the previously existing results in the literature for (3.32) with α > 2 or 1 < α < 2. In 2010, Wang [281] obtained the following theorem. Theorem 4.14 If (4.3) has a real nontrivial solution (x, u) such that x(a) = x(b) = 0 where a, b ∈ R with a < b are consecutive zeros, and x is not identically zero on [a, b], then there exists some point τ ∗ ∈ (a, b) such that the inequality 

b

 β1 (s) exp −2

a

s τ∗

 α1 (u)du ds a

b

β2+ (s)ds ≥ 4

holds. The principal aim of Tiryaki et al. [272] is to prove Lyapunov-type inequalities for the nonlinear systems of the form (4.1) by adopting the methods used in Tang and He [269] and Wang [281]. For some special cases, their results are improvements and generalizations of the above mentioned results. In what follows, Theorem 4.15 for (3.32) coincides with Theorem 4.87 below. Theorem 4.15 below generalizes Theorem 4.14 to the nonlinear systems of the form (4.1). Denote now

 t  z ⎧ ⎪ ⎪ β1 (z) exp −γ α1 (v)dv dz, ⎪ ⎨ha (t) :=  ⎪ ⎪ ⎪ ⎩hb (t) :=

a

t

b t

 β1 (z) exp −γ

z

α1 (v)dv dz

t

for all t ∈ R. By using a technique similar to that of Tang and He [269], Tiryaki et al. [272] obtained the following result, which relates only points a and b in I at which the first component of the solution (x, u) of (4.1) with β = α has consecutive

4.2 Nonlinear Systems

207

zeros. We note that [270, Theorem 2.1] in Tang et al. and Theorem 4.15 have the same results. Since the proof of the theorem given in [272] differs from theirs, we provide the proof of the theorem given in [272]. Theorem 4.15 Let the hypotheses (i) γ > 1, β > 1 are real constants, (ii) α1 , β1 , β2 are real-valued continuous functions hold. If (4.1) with β1 (t) > 0 and β = α has a real nontrivial solution (x, u) such that x(a) = x(b) = 0, where a, b ∈ R with a < b are consecutive zeros, and x is not identically zero on [a, b], then the inequality 

β2+ (s)

b

[ha (s)]1−α + [hb (s)]1−α

a

ds ≥ 1

(4.50)

holds, where 1/α + 1/γ = 1. Proof Let x(a) = x(b) = 0, where a, b ∈ R with a < b are consecutive zeros, and x is not identically zero on [a, b]. From (4.1), d dt

 t

 t

x(t) exp − α1 (s)ds = β1 (t) exp − α1 (s)ds |u(t)|γ −2 u(t), a

a

(4.51)

and by integrating (4.51) from a to t, we have 

t

|x(t)| ≤

 β1 (s) exp −

a

s

α1 (v)dv |u(s)|γ −1 ds

(4.52)

t

for all t ∈ R. By using Hölder’s inequality with Hölder conjugates α and γ on the integral of the right-hand side of (4.52), we obtain  |x(t)| ≤

t





β1 (s) exp −γ

a

1/γ 



s

α1 (v)dv ds

1/α

t

γ

β1 (s)|u(s)| ds

t

a

and  |x(t)| [ha (t)]

1−α

α

t

≤

β1 (s)|u(s)|γ ds,

(4.53)

a

where 1/α + 1/γ = 1. Similarly, from (4.1), d dt

 x(t) exp



b

α1 (s)ds t

 = β1 (t) exp t

b



α1 (s)ds |u(t)|γ −2 u(t), (4.54)

208

4 Nonlinear Differential Systems

and by integrating (4.54) from t to b, we get  |x(t)| [hb (t)]

1−α

α

≤

b

β1 (s)|u(s)|γ ds,

(4.55)

t

where 1/α + 1/γ = 1. Adding (4.53) and (4.55), we have |x(t)|α ≤



1 1−α

[ha (t)]

(t) + [hb (t)]

1−α

b

β1 (s)|u(s)|γ ds

(4.56)

a

for all t ∈ R. On the other hand, multiplying the first equation in (4.1) with β = α by u(t) and the second equation in (4.1) by x(t) and then adding the results, we obtain (xu) (t) = β1 (t)|u(t)|γ − β2 (t)|x(t)|α .

(4.57)

Integrating (4.57) from a to b by taking into account x(a) = x(b) = 0 yields 



b

b

β1 (s)|u(s)| ds = γ

a

β2 (s)|x(s)|α ds.

(4.58)

a

Thus, substituting (4.58) in (4.56), we get |x(t)|α ≤



1 1−α

[ha (t)]

+ [hb (t)]

1−α

b a

β2+ (s)|x(s)|α ds

(4.59)

for all t ∈ R. Multiplying both sides of (4.59) by β2+ (t) and then integrating from a to b, we obtain (4.50), which completes the proof.  Remark 4.16 Let α1 (t) ≡ 0 in (4.1). In this case, Theorem 4.15 reduces to Theorem 4.87 below. Moreover, (4.50) is better than (3.47) by using [282, (2.16)] in the sense that (3.47) follows from (4.50), but not conversely. The following is [272, Theorem 2.1]. Theorem 4.17 Let the hypotheses (i) and (ii) hold. If (4.1) with β1 (t) > 0 and β = α has a real nontrivial solution (x, u) such that x(a) = 0 = x(b), where a, b ∈ R with a < b are consecutive zeros, and x is not identically zero on [a, b], then the inequality  1≤ a

b

β2+ (s)2α−2



1 1 + ha (s) hb (s)

holds, where α and β2+ are as in Theorem 4.15.

1−α ds

(4.60)

4.3 Quasilinear Systems

209

Remark 4.18 Theorem 4.17 is not correct. Inequalities [272, (2.17) and (2.18) on p. 1808] are on the intervals [a, c] and [c, b], respectively, and they cannot just be added to get [272, (2.19)].

4.3 Quasilinear Systems In 2006, de Nápoli and Pinasco [107], and in 2008, Bonder and Pinasco [56] studied the problem of finding lower and upper bounds for the eigenvalues of a quasilinear elliptic system. In [107], the authors have also obtained a Lyapunov-type inequality for the quasilinear system  ⎧    p−2  ⎪ u = f1 (x)|u|α−2 u|v|β , ⎨− u  (4.61)

  ⎪ ⎩− v  q−2 v   = f (x)|u|α |v|β−2 v, 2

where 1 < p, q < ∞, the positive parameters α and β satisfy α/p + β/q = 1, and f1 , f2 are real-valued positive continuous functions, as follows. Theorem 4.19 (Lyapunov-Type Inequality) If (4.61) has a real nontrivial solution (u, v) such that u(a) = u(b) = v(a) = v(b) = 0, where a, b ∈ R with a < b are consecutive zeros of both u and v, and u and v are not identically zero on [a, b], then the inequality  2α+β ≤ (b − a)α+β−1

α/p 

b

a

β/q

b

f1 (x)dx

f2 (x)dx a

holds. As an application of Theorem 4.19, the authors have improved the lower bounds on the eigenvalues problem of the one-dimensional system  ⎧    p−2  ⎪ u = λαr(x)|u|α−2 u|v|β , ⎨− u  (4.62)

  ⎪ ⎩− v  q−2 v   = μβr(x)|u|α |v|β−2 v as follows.

Theorem 4.20 There exists a function h such that h(λ) ≤ μ for every generalized eigenvalue (λ, μ) of (4.62), namely 1 h(λ) = β



C λα/p



−1 q/β

b

r(x)dx a

,

210

4 Nonlinear Differential Systems

and the constant C is given by C = 2α+β α −α/p (b − a)1−α−β . The principal aim of this section is to state and prove a Lyapunov-type inequality for quasilinear systems in the form ⎧    p−2   ⎪ u = f1 (x)|u|α−2 u|v|β , ⎨− r1 (x) u    ⎪ ⎩− r (x) v  q−2 v   = f (x)|u|θ |v|γ −2 v, 2 2

(4.63)

whose special cases contain well-known equations such as half-linear and linear equations. The result of Çakmak and Tiryaki [73] is an extension of a result by de Nápoli and Pinasco [107]. By using this inequality, we prove a better lower bound than de Nápoli and Pinasco for the first eigenvalue of (4.62). Their motivation comes from the papers of Lee et al. [195], de Nápoli and Pinasco [107], Pachpatte [229], and Tiryaki et al. [274]. Now, we derive a Lyapunov-type inequality for quasilinear systems of the form (4.63), where both components of the solution (u, v) have consecutive zeros at the points a, b ∈ R with a < b in I = [t0 , ∞) ⊂ R. For the special cases of (4.63), we also derive some Lyapunov-type inequalities which not only relate points a and b in I at which both components of the solution (u, v) have consecutive zeros but also any point in (a, b), where both components of the solution (u, v) are maximized. Since our attention is restricted to the Lyapunov-type inequality for the quasilinear system of differential equations, we shall assume the existence of a nontrivial solution (u, v) of (4.63). Our basic hypothesis with respect to (4.63) are (i) r1 , r2 , f1 , f2 are real-valued continuous functions such that r1 (x) > 0 and r2 (x) > 0 for all x ∈ R, (ii) p > 1, q < ∞, and α, β, θ, γ > 0 satisfy α β + = 1 and p q

θ γ + = 1. p q

For the sake of convenience, we define the integral operator 

t

M(t, s, μ) = a

[s(x)]

−μ∗ /μ

1−μ dx

 +

b

[s(x)]

−μ∗ /μ

1−μ dx

,

(4.64)

t

where t ∈ (a, b), s is a real-valued continuous function such that s(x) > 0 for all x ∈ R, and μ∗ is the Hölder conjugate of μ ∈ (1, ∞), that is, 1/μ + 1/μ∗ = 1. For a given s and μ, set F (t) = M(t, s, μ) for t ∈ (a, b). F attains its minimum at the point t ∈ (a, b) such that

4.3 Quasilinear Systems

211



t

∗ /μ

[s(x)]−μ



b

dx =

a

∗ /μ

[s(x)]−μ

dx

(4.65)

= M(t, s, μ).

(4.66)

t

holds. Thus, we have  F (t) ≥ Fmin (t) = 2

t

∗ /μ

[s(x)]−μ

1−μ dx

a

The main result of Çakmak and Tiryaki [73] is the following theorem, which is an extension of Theorem 4.19. Theorem 4.21 (Lyapunov-Type Inequality) Let the hypothesis (i) and (ii) hold. If (4.63) has a real nontrivial solution (u, v) such that u(a) = u(b) = v(a) = v(b) = 0, where a, b ∈ R with a < b are consecutive zeros, and u and v are not identically zero on [a, b], then the inequality 

b

[M(c, r1 , p)]θ/p [M(d, r2 , q)]β/q ≤ a

f1+ (x)dx

θ/p  a

b

f2+ (x)dx

β/q (4.67)

holds, where |u(c)| = max |u(x)| a 0 for all x ∈ R reduces to (4.62). By using techniques similar to de Nápoli and Pinasco [107], we obtain the following result, which gives better lower bounds for the first eigencurve in the generalized spectra. The proof of this result is rather elementary, based on the above generalization of the Lyapunov-type inequality, exactly as in that of de Nápoli and Pinasco [107, Theorem 1.4]. Theorem 4.29 There exists a function h1 such that μ ≥ h1 (λ) for every generalized eigenvalue (λ, μ) of (4.62), namely 1 h1 (λ) = β



D λα/p



−1 q/β

b

r(x)dx

,

a

and the constant D is given by  α/p  β/q D = α −α/p (c − a)1−p + (b − c)1−p (d − a)1−q + (b − d)1−q , where c and d are defined as in Theorem 4.21. Proof Let (λ, μ) be a generalized eigenpair and (u, v) be the corresponding nontrivial solution of (4.62). For all x ∈ R, by substituting f1 (x) = λαr(x) > 0 and f2 (x) = μβr(x) > 0 in (4.67), we obtain  α/p  β/q (c − a)1−p + (b − c)1−p (d − a)1−q + (b − d)1−q 

α/p 

b

≤

β/q

b

λαr(x)dx

μβr(x)dx

a

.

(4.85)

a

By rearranging terms in (4.85) and using the condition α/p + β/q = 1, we obtain  α/p  β/q (c − a)1−p + (b − c)1−p (d − a)1−q + (b − d)1−q  ≤ (λα)

α/p

(μβ)

β/q

b

r(x)dx ,

a

which gives 

D λα/p



−1 q/β

b

r(x)dx a

≤ μβ,

(4.86)

4.3 Quasilinear Systems

217



and the proof is completed. 0+ ,

Remark 4.30 Since h1 is a continuous function and h1 (λ) → ∞ as λ → there exists a ball centered in the origin such that the generalized spectrum is contained in its exterior. Also, by rearranging terms in (4.86), we obtain β/q α/p

μ

λ

D ≥ β/q β



−1

b

r(x)dx

.

(4.87)

a

It is clear that when the interval collapses, the right-hand side of (4.87) goes to infinity. Thus, we obtain the desired generalizations of de Nápoli and Pinasco’s [107] result for one-dimensional nonlinear systems. Remark 4.31 If we compare Theorem 4.29 with Theorem 4.20, then we obtain D ≥ C since (4.79) and (4.80) hold. Thus, the inequality h1 (λ) ≥ h(λ) holds. Therefore, Theorem 4.29 presents a better lower bound than Theorem 4.20. In 2014, Tiryaki et al. [273] proved generalized Lyapunov-type inequalities for (4.63) with p = p1 and q = p2 , i.e., for system ⎧   ⎨− (r1 (x)φp1 (u1 )  = f1 (x)φα1 (u1 ) |u2 |α2 , (4.88) ⎩− (r (x)φ (u ) = f (x)φ (u ) |u |β1 , 2 p2 2 2 β2 2 1 where φγ (u) = |u|γ −2 u, γ > 1, rk , fk ∈ C([a, b], R), rk (x) > 0 for k = 1, 2 and x ∈ R, (u1 , u2 ) is a real nontrivial solution of (4.88) such that uk (a) = uk (b) = 0 for k = 1, 2, a, b ∈ R with a < b are consecutive zeros, uk for k = 1, 2 are not identically zero on [a, b], 1 < pk < ∞, and α2 ≥ 0, β1 ≥ 0 satisfy α1 α2 + = 1 and p1 p2

β1 β2 + = 1. p1 p2

(4.89)

For the sake of brevity, we denote

ξk (x)ηk (x) pk −1 Dk (x) = , Ek (x) = 2 , ξk (x) + ηk (x) [ξk (x)]pk −1 + [ηk (x)]pk −1  b

pk −1 −pk pk −1 −pk 1/(1−pk ) Fk = 2 =2 ds , [ξk (x) + ηk (x)] [rk (s)] [ξk (x)ηk (x)]pk −1

pk −2

a

where  ξk (x) = a

x

 [rk (s)]1/(1−pk ) ds,

b

ηk (x) = x

[rk (s)]1/(1−pk ) ds

218

4 Nonlinear Differential Systems

for k = 1, 2, . . . , n. Now, we present some inequalities for Dk (x), Ek (x), and Fk for k = 1, 2, . . . , n, which are useful in the comparison of the results. We know that since the function h(x) = x pk −1 is concave for x > 0 and 1 < pk < 2, Jensen’s inequality (see (4.78)) with ω=

1 ξk (x)

v=

and

1 ηk (x)

implies Dk (x) ≥ Ek (x)

(4.90)

for 1 < pk < 2, k = 1, 2, . . . , n. If pk > 2 for k = 1, 2, . . . , n, then the function h(x) = x pk −1 is convex for x > 0. Thus, (4.90) is reversed, i.e., Dk (x) ≤ Ek (x)

(4.91)

for pk > 2, k = 1, 2, . . . , n. In addition, since the function l(x) = x 1−pk is convex for x > 0 and pk > 1, Jensen’s inequality (see (4.78)) with ω = ξk (x) and v = ηk (x) implies Dk (x) ≤ Fk

(4.92)

for k = 1, 2, . . . , n. By using (4.40) with u = ξk (x) > 0 and v = ηk (x) > 0 for k = 1, 2, . . . , n in Ek (x), we obtain the inequality Ek (x) ≤ Fk

(4.93)

for k = 1, 2, . . . , n. For (4.88) and (4.89), the first result of Tiryaki et al. [273] is the following theorem. Theorem 4.32 (Hartman-Type Inequality) If fk ∈ C([a, b], R) for k = 1, 2 and (u1 , u2 ) is a solution of (4.88) and (4.89) on [a, b], where a, b ∈ R with a < b are consecutive zeros, and u1 and u2 are not identically zero on [a, b], then the inequality  a

b

f1+ (s) [D1 (s)]α1 /p1 [D2 (s)]α2 /p2 ds 

b

× a

holds.

β1 /p1

f2+ (s) [D1 (s)]β1 /p1

α2 /p2 [D2 (s)]

β2 /p2

ds

>1

(4.94)

4.3 Quasilinear Systems

219

Proof Let uk (a) = 0 = uk (b) for k = 1, 2, where a, b ∈ R with a < b are consecutive zeros, and uk for k = 1, 2 are not identically zero on [a, b]. Multiplying the first equation of (4.88) by u1 and the second equation in (4.88) by u2 , integrating from a to b, and taking into account uk (a) = 0 = uk (b) for k = 1, 2, we get 

b a

 p r1 (s) u1 (s) 1 ds =



b

f1 (s) |u1 (s)|α1 |u2 (s)|α2 ds

(4.95)

f2 (s) |u1 (s)|β1 |u2 (s)|β2 ds.

(4.96)

a

and 

b a

 p r2 (s) u2 (s) 2 ds =



b

a

By using uk (a) = 0 and Hölder’s inequality, we get 

   u (s) ds k

x

|uk (x)| ≤ a



x

≤

[rk (s)]

1/(1−pk )

(pk −1)/pk  ds

a

x

 p rk (s) u (s) k ds

a

1/pk

k

for k = 1, 2 and x ∈ [a, b]. Thus, we get 

x

|uk (x)|pk [ξk (x)]1−pk ≤ a

 p rk (s) uk (s) k ds

(4.97)

for k = 1, 2. Similarly, by using uk (b) = 0 and Hölder’s inequality, we get  |uk (x)|pk [ηk (x)]1−pk ≤ x

b

 p rk (s) uk (s) k ds

(4.98)

for k = 1, 2 and x ∈ [a, b]. Adding (4.97) and (4.98), we have 

b

|uk (x)|pk ≤ Dk (x) a

 p rk (s) uk (s) k ds

(4.99)

for k = 1, 2 and x ∈ [a, b]. After that, by using a technique similar to the one in the proof of Theorem 4.86 given by Tang and He [269], it can be shown that the equality case in (4.99) does not hold. Thus, we get  |uk (x)|

pk

b

< Dk (x) a

 p rk (s) uk (s) k ds

(4.100)

220

4 Nonlinear Differential Systems

for k = 1, 2 and x ∈ (a, b). For k = 1 in (4.100), we get 

b

|u1 (x)|p1 < D1 (x)

f1 (s) |u1 (s)|α1 |u2 (s)|α2 ds

(4.101)

a

from (4.95). If we take the (α1 /p1 )th and (β1 /p1 )th powers of both sides of (4.101), then 

b

|u1 (x)|α1 < [D1 (x)]α1 /p1 a

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds

α1 /p1 (4.102)

and 

b

|u1 (x)|β1 < [D1 (x)]β1 /p1 a

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds

β1 /p1 ,

(4.103)

respectively. Multiplying both sides of (4.102) by f1+ (x) |u2 (x)|α2 and integrating from a to b, we have 

b a

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds

1−α1 /p1



b

< a

f1+ (s) |u2 (s)|α2 [D1 (s)]α1 /p1 ds. (4.104)

Similarly, for k = 2 in (4.100), we get 

b

|u2 (x)|p2 < D2 (x)

f2 (s) |u1 (s)|β1 |u2 (s)|β2 ds

(4.105)

a

from (4.96). If we take the (α2 /p2 )th and (β2 /p2 )th powers of both sides of (4.105), then 

b

|u2 (x)|α2 < [D2 (x)]α2 /p2 a

f2+ (s) |u1 (s)|β1 |u2 (s)|β2 ds

α2 /p2 (4.106)

and  |u2 (x)|β2 < [D2 (x)]β2 /p2

b

a

f2+ (s) |u1 (s)|β1 |u2 (s)|β2 ds

β2 /p2 ,

(4.107)

respectively. Multiplying both sides of (4.107) by f2+ (x) |u1 (x)|β1 and integrating from a to b, we have 

b a

f2+ (s) |u1 (s)|β1 |u2 (s)|β2 ds

1−β2 /p2



b

< a

f2+ (s) |u1 (s)|β1 [D2 (s)]β2 /p2 ds. (4.108)

4.3 Quasilinear Systems

221

By using (4.106) in (4.104) and (4.103) in (4.108), we have 

b

a

1−α1 /p1

f1+ (s) |u1 (s)|α1

|u2 (s)|

α2

ds



b

< M1 a

f2+ (s) |u1 (s)|β1 |u2 (s)|β2 ds

α2 /p2 (4.109)

and  a

b

1−β2 /p2

f2+ (s) |u1 (s)|β1

|u2 (s)| ds β2



b

< M2 a

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds

β1 /p1 ,

(4.110)

where 

b

M1 = a

f1+ (s) [D1 (s)]α1 /p1 [D2 (s)]α2 /p2 ds

and 

b

M2 = a

f2+ (s) [D1 (s)]β1 /p1 [D2 (s)]β2 /p2 ds,

respectively. If we take the e1 th and e2 th powers of both sides of (4.109) and (4.110) and multiply the resulting equations, then we obtain  

b

a

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds

 

b

× a





f2+ (s) |u1 (s)|β1 b

< M1 a



1−β2 /p2 e2 |u2 (s)| ds

f2+ (s) |u1 (s)|β1



b

× M2 a

1−α1 /p1 e1

β2

α2 /p2 e1

(4.111)

|u2 (s)| ds

f1+ (s) |u1 (s)|α1

β2

β1 /p1 e2 |u2 (s)|

α2

ds

.

It is easy to see that, by using a technique similar to the one in the proof of Theorem 4.86 given by Tang and He [269], we obtain the inequalities

222

 a

b

4 Nonlinear Differential Systems

f1+ (s) |u1 (s)|α1 |u2 (s)|α2 ds > 0

 and a

b

f2+ (s) |u1 (s)|β1 |u2 (s)|β2 ds > 0.

(4.112) Now, we choose e1 and e2 such that (4.112) cancels out in (4.111), i.e., as solutions of the homogeneous linear system

⎧ α1 β1 ⎪ ⎪ ⎪ ⎨ 1 − p1 e1 − p1 e2 = 0,

⎪ β2 α2 ⎪ ⎪ ⎩ e1 − 1 − e2 = 0. p2 p2 We observe that by the hypotheses given in (4.89), this system admits a nontrivial solution; indeed, both equations are equivalent to

α1 β1 1− e1 = e2 p1 p1

and

α2 β2 e2 . e1 = 1 − p2 p2

Hence, we may take e1 = β1 /p1 and e2 = α2 /p2 , and we get (4.94), which completes the proof.  For (4.88), another result is the following theorem. Theorem 4.33 (Hartman-Type Inequality) If fk ∈ C([a, b], R) for k = 1, 2 and (u1 , u2 ) is a solution of (4.88) and (4.89) on [a, b], where a, b ∈ R with a < b are consecutive zeros, and u1 and u2 are not identically zero on [a, b], then the inequality  a

b

f1+ (s) [E1 (s)]α1 /p1 

b

× a

β1 /p1 [E2 (s)]

α2 /p2

ds

f2+ (s) [E1 (s)]β1 /p1 [E2 (s)]β2 /p2 ds

α2 /p2 >1

(4.113)

holds. Proof Let uk (a) = 0 = uk (b) for k = 1, 2, where a, b ∈ R with a < b are consecutive zeros, and uk for k = 1, 2 are not identically zero on [a, b]. As in the proof of Theorem 4.32, we have (4.95), (4.96), (4.97), and (4.98). If we take k = 1 in (4.97) and (4.98), then |u1 (x)|p1 ≤ [ξ1 (x)]p1 −1



x a

 p r1 (s) u1 (s) 1 ds

(4.114)

4.3 Quasilinear Systems

223

and |u1 (x)|p1 ≤ [η1 (x)]p1 −1



b

x

 p r1 (s) u1 (s) 1 ds

(4.115)

for x ∈ [a, b]. Multiplying (4.114) and (4.115) by [η1 (x)]p1 −1 and [ξ1 (x)]p1 −1 , respectively, we obtain [η1 (x)]p1 −1 |u1 (x)|p1 ≤ [ξ1 (x)η1 (x)]p1 −1



 p r1 (s) u1 (s) 1 ds

x a

(4.116)

and [ξ1 (x)]p1 −1 (x) |u1 (x)|p1 ≤ [ξ1 (x)η1 (x)]p1 −1



b x

 p r1 (s) u1 (s) 1 ds

(4.117)

for x ∈ [a, b]. Thus, adding (4.116) and (4.117), we have    |u1 (x)|p1 [ξ1 (x)]p1 −1 + [η1 (x)]p1 −1 ≤ [ξ1 (x)η1 (x)]p1 −1

b

a

 p r1 (s) u1 (s) 1 ds (4.118)

for x ∈ [a, b]. It is easy to see that [ξ1 (x)]p1 −1 + [η1 (x)]p1 −1 takes the minimum value at c1 ∈ (a, b) such that ξ1 (c1 ) = η1 (c1 ). Thus, by (4.118), we get    |u1 (x)|p1 [ξ1 (c1 )]p1 −1 + [η1 (c1 )]p1 −1 ≤ [ξ1 (x)η1 (x)]p1 −1 a

b

 p r1 (s) u1 (s) 1 ds. (4.119)

Since ξ1 (c1 ) + η1 (c1 ) = ξ1 (x) + η1 (x)

for all

x, c1 ∈ (a, b)

and ξ1 (c1 ) =

1 ξ1 (x) + η1 (x) = 2 2



b a

1/(1−p1 )

r1

(s)ds,

(4.119) yields   ! p −1 |u1 (x)|p1 22−p1 [ξ1 (x) + η1 (x)]p1 −1 = |u1 (x)|p1 2ξ1 1 (c1 ) ≤ [ξ1 (x)η1 (x)]p1 −1

 a

b

 p r1 (s) u1 (s) 1 ds,

224

4 Nonlinear Differential Systems

and hence from (4.95),  |u1 (x)|

p1

b

≤ E1 (x)

f1 (s) |u1 (s)|α1 |u2 (s)|α2 ds

(4.120)

a

for x ∈ [a, b]. After that, by using a technique similar to the one in the proof of Theorem 4.86 given by Tang and He [269], it can be shown that the equality case in (4.120) does not hold. Thus, we get  b |u1 (x)|p1 < E1 (x) f1 (s) |u1 (s)|α1 |u2 (s)|α2 ds a

for x ∈ (a, b). The rest of the proof is the same as in the proof of Theorem 4.32, and hence it is omitted.  Remark 4.34 It is easy to see from (4.90) that if we take 1 < pk < 2 for k = 1, 2, then (4.113) is weaker than (4.94). Hence, Theorem 4.33 is better than Theorem 4.32. Similarly, from (4.91), if pk > 2 for k = 1, 2, then Theorem 4.32 is better than Theorem 4.33. In addition, if pk = 2 for k = 1, 2, then Theorem 4.32 coincides with Theorem 4.33. In the general case, the results of Tang and He [269] cannot be compared with our results, but they can be compared with each other in the special cases as follows. Remark 4.35 Let α2 = 0 and

α1 = p1

β1 = 0 and

β2 = p2

or

hold. Thus, Theorem 4.32 coincides with Theorem 4.86 (or Theorem 4.87 given in Tang and He [269]) for (3.3). Moreover, from (4.90), if we take 1 < p1 < 2, then Theorem 4.33 gives a better result than Theorem 4.86 for (3.3). By using (4.92) in (4.94) or (4.93) in (4.113), we obtain the following result from Theorem 4.32 or Theorem 4.33. Corollary 4.36 (Lyapunov-Type Inequality) If fk ∈ C([a, b], R) for k = 1, 2 and (u1 , u2 ) is a solution of (4.88) and (4.89) on [a, b], where a, b ∈ R with a < b are consecutive zeros, and u1 and u2 are not identically zero on [a, b], then the inequality 

b

β1 (p1 −1)/p1 

a

α2 (p2 −1)/p2 [r2 (s)]1/(1−p2 ) ds

a



b

× a

holds.

b

[r1 (s)]1/(1−p1 ) ds f1+ (s)ds

β1 /p1  a

b

f2+ (s)ds

α2 /p2

> 2α2 +β1

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian

225

Remark 4.37 It is clear that Corollary 4.36 coincides with Corollary 4.24 given by Çakmak and Tiryaki [73]. In fact, Theorems 4.32 and 4.33 are generalizations of Corollary 4.24.

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian In 2006, de Nápoli and Pinasco [107] studied the problem of finding Lyapunovtype inequalities for quasilinear systems involving (p, q)-Laplacian operators, i.e., (4.61), see Theorem 4.19. This result was used by proving Theorem 4.20, which improves the lower bounds on the eigenvalues of (4.62). In this section, by using a technique similar to that of de Nápoli and Pinasco [107], we state and prove a generalized Lyapunov-type inequality for a Dirichlet problem associated to the quasilinear system involving (p1 , p2 , . . . , pn )-Laplacian operators ⎧    ⎪ u p1 −2 u ⎪ − 1 1 ⎪ ⎪ ⎪ ⎪ ⎪  ⎪   p2 −2   ⎪ ⎪ u2 ⎨− u2  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ ⎪ ⎩− u pn −2 u n n

= f1 (x) |u1 |α1 −2 u1 |u2 |α2 · · · |un |αn , = f2 (x) |u1 |α1 |u2 |α2 −2 u2 |u3 |α3 · · · |un |αn , (4.121)

.. . = fn (x) |u1 |α1 · · · |un−1 |αn−1 |un |αn −2 un ,

where n ∈ N, fi are real valued continuous functions defined on R, the exponents satisfy 1 < pi < ∞, and the positive parameters αi , i = 1, 2, . . . , n, satisfy n

αi = 1. pi

(4.122)

i=1

If n = 1, then (4.121) reduces to the equation    u p1 −2 u + f1 (x) |u1 |p1 −2 u1 = 0, 1 1 which is known as the half-linear equation. Similarly, if n = 2, then (4.121) reduces to the system  ⎧    p1 −2  ⎪ u1 = f1 (x) |u1 |α1 −2 u1 |u2 |α2 , ⎨− u1  ⎪   p2 −2   ⎩ u2 = f2 (x) |u1 |α1 |u2 |α2 −2 u2 , − u2

226

4 Nonlinear Differential Systems

which is the same system as (4.62). The aim of this section is to extend and generalize Theorems 4.19 and 4.20 of de Nápoli and Pinasco [107] given in Sect. 4.4 to the general case. Motivated from the paper by Afrouzi and Heidarkhani [2], Çakmak and Tiryaki [72] derived a Lyapunov-type inequality for quasilinear systems of the form (4.121), where all components of the solution (u1 , u2 , . . . , un ) have consecutive zeros at the points a, b ∈ R with a < b in I = [t0 , ∞) ⊂ R. For some special cases of (4.121), we also derive certain Lyapunov-type inequalities, which relate not only points a and b in I at which all components of the solution (u1 , u2 , . . . , un ) have consecutive zeros but also any point in (a, b), where all components of the solution (u1 , u2 , . . . , un ) are maximized. Since our attention is restricted to Lyapunov-type inequalities for the quasilinear system of differential equations, we shall assume the existence of a nontrivial solution (u1 , u2 , . . . , un ) of (4.121). The main result of Çakmak and Tiryaki [72] is the following theorem. Theorem 4.38 (Lyapunov-Type Inequality) If (4.121) has a real nontrivial solution u such that ui (a) = ui (b) = 0 for i = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and ui for i = 1, 2, . . . , n are not identically zero on [a, b], then the inequality n  n  α /p % % 1−pi 1−pi i i (ci − a) + (b − ci ) ≤ i=1

b a

i=1

fi+ (x)dx

αi /pi (4.123)

holds, where u = (u1 , u2 , . . . , un )

(4.124)

and |ui (ci )| = max |ui (x)| a 1 for i = 1, 2, . . . , n, Jensen’s inequality (see (4.78)) with ω = ci − a and v = b − ci for i = 1, 2, . . . , n implies (ci − a)1−pi + (b − ci )1−pi ≥ 2pi (b − a)1−pi

(4.137)

for i = 1, 2, . . . , n. Thus, by using (4.137), Theorem 4.38 reduces to the following result. Corollary 4.40 (Lyapunov-Type Inequality) If (4.121) has a real nontrivial solution u such that ui (a) = ui (b) = 0 for i = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and ui for i = 1, 2, . . . , n are not identically zero on [a, b], then the inequality 2An (b − a)1−An ≤

n  % i=1

b

a

fi+ (x)dx

αi /pi (4.138)

holds, where u is defined in (4.124) and An =

n

αi .

(4.139)

i=1

Remark 4.41 Let n = 1. If we compare (4.130) and (4.131) with inequalities in Pinasco [235, (2.3) in Lemma 2.1], respectively, then it is easy to see that the restricted condition, i.e., a bounded positive function, on the function r in that result can be dropped. Thus, Theorem 4.38 and Corollary 4.40 generalize and extend Pinasco [235, Theorem 2.3]. Remark 4.42 Let n = 2. If we compare Theorem 4.38 with Theorem 4.19 of de Nápoli and Pinasco [107], since (4.137) holds, we conclude that Theorem 4.38 is more general than Theorem 4.19.

230

4 Nonlinear Differential Systems

Remark 4.43 Corollary 4.40 with n = 2 and fi (x) > 0 for i = 1, 2 reduces to Theorem 4.19. Remark 4.44 When αi = pi for i = 1, 2, . . . , n, and for j = i, αj ≡ 0 for j = 1, 2, . . . , n, then we obtain the result for the case of a single equation from Theorem 4.38 or Corollary 4.40. Remark 4.45 Since f + (x) ≤ |f (x)|, the integrals 

b

a

fi+ (x)dx

for i = 1, 2, . . . , n in the above results can also be replaced by 

b

|fi (x)| dx

a

for i = 1, 2, . . . , n, respectively. Now, we present an application of the obtained Lyapunov-type inequality for (4.121). Let λi for i = 1, 2, . . . , n be generalized eigenvalues of (4.121), and r be a positive function for all defined on R. Therefore, (4.121) with fi (x) = λi αi r(x) > 0 for i = 1, 2, . . . , n and all x ∈ R reduces to the system ⎧    ⎪ u p1 −2 u ⎪ − 1 1 ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ p2 −2   ⎪ ⎪ u2 ⎨− u2  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ ⎪ ⎩− u pn −2 u n n

= λ1 α1 r(x) |u1 |α1 −2 u1 |u2 |α2 · · · |un |αn , = λ2 α2 r(x) |u1 |α1 |u2 |α2 −2 u2 |u3 |α3 · · · |un |αn , (4.140)

.. . = λn αn r(x) |u1 |α1 · · · |un−1 |αn−1 |un |αn −2 un .

By using techniques similar to those in de Nápoli and Pinasco [107], we obtain the following result, which gives lower bounds for the nth eigenvalue of λn . The proof of this theorem is based on the above generalization of the Lyapunov-type inequality, as in that of de Nápoli and Pinasco [107, Theorem 1.4]. Theorem 4.46 There exists a function h1 such that λn ≥ h1 (λ1 , λ2 , . . . , λn−1 ) for every generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.140), where |ui (ci )| = max |ui (x)| a 0 and 1 < pk < 2, Jensen’s inequality (see (4.78)) with ω=

1 ck − a

v=

and

1 b − ck

for k = 1, 2, . . . , n implies  22−pk

1 1 + ck − a b − ck

pk −1

≥

1 1 + = m1 (ck ) p −1 k (ck − a) (b − ck )pk −1 (4.144)

for 1 < pk < 2, k = 1, 2, . . . , n. If pk > 2 for k = 1, 2, . . . , n, then the function h(x) = x pk −1 is convex for x > 0. Thus, (4.144) is reversed, i.e., 1 1 + ≥ 22−pk p −1 k (ck − a) (b − ck )pk −1



1 1 + ck − a b − ck

pk −1

= m2 (ck ) (4.145)

for pk > 2, k = 1, 2, . . . , n. Moreover, if we get the minimum of the right-hand sides of (4.144) and (4.145) for ck ∈ (a, b), k = 1, 2, . . . , n, then it is easy to see that min mi (ck ) = mi

a

from Theorem 4.56.

2α1 +α2 −α1 /p1 −1 (b − a)α1 +α2 −1

from Corollary 4.58. Note that the lower bounds found by using (4.167) in Theorem 4.57 coincide with those of [72, Theorem 9]. Now, we present lower bounds by using (4.166) in Theorem 4.56, which gives a better lower bound for the eigenvalues of the following system than that of [72, Theorem 9] when 1 < pk < 2 for k = 1, 2, . . . , n. Let λk for k = 1, 2, . . . , n be generalized eigenvalues of (4.121) and r be a positive function defined on R. Therefore, (4.121) with fk (t) = λk αk r(t) > 0 for k = 1, 2, . . . , n and t ∈ R reduces to the system ⎧    p −2 ⎪ ⎪ − x1  1 x1 ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ p2 −2   ⎪ ⎪ x2 ⎨− x2  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪    ⎪ ⎪ ⎩− x  pn −2 x  n

n

= λ1 α1 r(t) |x1 |α1 −2 x1 |x2 |α2 . . . |xn |αn , = λ2 α2 r(t) |x1 |α1 |x2 |α2 −2 x2 . . . |xn |αn , .. .

(4.170)

= λn αn r(t) |x1 |α1 |x2 |α2 . . . |xn |αn −2 xn .

By using techniques similar to those in [72, Theorem 9], we obtain the following result, which gives lower bounds for the nth eigenvalue λn . The proof of this theorem is based on the above generalization of the Lyapunov-type inequality, as in that of [72, Theorem 9], and hence it is omitted.

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian

241

Theorem 4.64 There exists a function k1 such that λn ≥ k1 (λ1 , λ2 , . . . , λn−1 )

(4.171)

for every generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.170), where |xk (ck )| = max |xk (t)| a 0 for k = 1, 2, . . . , n and x ∈ R. Let u = (u1 , u2 , . . . , un ) be a real nontrivial solution of (4.173) such that uk (a) = uk (b) = 0 for

k = 1, 2, . . . , n

(4.174)

for a, b ∈ R with a < b are consecutive zeros of uk , and uk are not identically zero on [a, b], 1 < pk < ∞, and αki are nonnegative constants for k, i = 1, 2, . . . , n. In 2013, Yang et al. [298] obtained the following inequality for (4.173). Theorem 4.67 (Hartman-Type Inequality) Assume that there exist nontrivial solutions (e1 , e2 , . . . , en ) of the linear homogeneous system

n

αik αkk ek 1 − ei = 0 for − pk pk

k = 1, 2, . . . n,

i=1,i=k

where ek ≥ 0 for k = 1, 2, . . . , n and n

ek2 > 0.

k=1

If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.173), then the inequality  n % Fk k=1

a

b

fk+ (s)ds

ek >1

holds. Akta¸s [25] stated and proved new generalized Lyapunov-type inequalities for (4.173) under the condition αki = αik for k, i = 1, 2, . . . , n. Since our attention is restricted to Lyapunov-type inequalities for the quasilinear systems of differential equations, we shall assume the existence of a nontrivial solution of (4.173). For readers interested in the existence of the solutions of these types of systems, we refer to the paper by Afrouzi and Heidarkhani [2]. Theorem 4.68 (Hartman-Type Inequality) Assume that there exist nontrivial solutions (e1 , e2 , . . . , en ) of the linear homogeneous system

n

αkk αki ek 1 − − ei = 0 for pk pk i=1,=k

where ek ≥ 0 for k = 1, 2, . . . , n and

k = 1, 2, . . . n,

(4.175)

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian n

243

ek2 > 0.

k=1

If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.173) with αki = αik for k, i = 1, 2, . . . , n, then the inequality  n %

b

a

k=1

fk+ (s)

n %

ek [Di (s)]

αki /pi

ds

>1

(4.176)

i=1

holds. Proof Let uk (a) = uk (b) = 0 for k = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and uk for k = 1, 2, . . . , n are not identically zero on [a, b]. By using uk (a) = 0 and Hölder’s inequality, we get  |uk (x)| ≤

x

   u (s) ds k

a

 ≤

x

[rk (s)]1/(1−pk ) ds

a

= [ξk (x)]

1−1/pk

1−1/pk 

x

a



x a

 p rk (s) uk (s) k ds

 p rk (s) uk (s) k ds

1/pk

1/pk

for k = 1, 2, . . . , n and x ∈ [a, b]. Thus, we have  |uk (x)|

pk

[ξk (x)]

1−pk

x

≤ a

 p rk (s) uk (s) k ds

(4.177)

for k = 1, 2, . . . , n and x ∈ [a, b]. Similarly, by using uk (b) = 0 and Hölder’s inequality, we obtain  |uk (x)|

pk

[ηk (x)]

1−pk

b

≤ x

 p rk (s) uk (s) k ds

(4.178)

for k = 1, 2, . . . , n and x ∈ [a, b]. Adding (4.177) and (4.178), we have  |uk (x)|

pk

b

≤ Dk (x) a

 p rk (s) uk (s) k ds

(4.179)

for k = 1, 2, . . . , n and x ∈ [a, b]. After that, by using a technique similar to the one in [269, Theorem 3.1], it can be shown that the equality case in (4.179) does not hold. Thus, we get |uk (x)|pk < Ak Dk (x),

x ∈ (a, b),

(4.180)

244

4 Nonlinear Differential Systems

where 

b

Ak = a

 p rk (s) uk (s) k ds

for k = 1, 2, . . . , n. If we take the (αki /pk )th power of both sides of (4.180), then we obtain α /pk

|uk (x)|αki < Ak ki

[Dk (x)]αki /pk

(4.181)

for k, i = 1, 2, . . . , n. Multiplying both sides of (4.181) with i = k by n %

fk+ (x)

|ui (x)|αki

i=1,i=k

for k = 1, 2, . . . , n and integrating from a to b, we have 

b

fk (s) a

n %

α /pk



|ui (s)|αki ds < Ak kk

i=1

b a

fk+ (s) [Dk (s)]αkk /pk

n %

|ui (s)|αki ds

i=1,i=k

(4.182) for k = 1, 2, . . . , n. On the other hand, multiplying the kth equation in (4.173) by uk and integrating from a to b, we get 

b

Ak = a

 p rk (s) uk (s) k ds =



b

fk (s) a

n %

|ui (s)|αki ds

(4.183)

i=1

for k = 1, 2, . . . , n. By using (4.183) in (4.182), we have α /p Ak kk k

 a

b

n %

fk+ (s) [Dk (s)]αkk /pk

|ui (s)|αki ds > Ak ,

i=1,i=k

and hence, from αki = αik for k, i = 1, 2, . . . , n, we get α /pk

Ak kk



b a

fk+ (s) [Dk (s)]αkk /pk

n %

|ui (s)|αik ds > Ak

(4.184)

i=1,i=k

for k = 1, 2, . . . , n. Therefore, by using (4.181) in (4.184), we obtain  a

b

fk+ (s) [Dk (s)]αkk /pk

n % i=1,i=k

α /pi

Ai ik

1−αkk /pk

[Di (s)]αik /pi ds > Ak

,

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian

245

and hence n %

α /pi



Ai ik

b a

i=1,i=k

fk+ (s)

n %

1−αkk /pk

[Di (s)]αik /pi ds > Ak

(4.185)

i=1

for k = 1, 2, . . . , n. Raising both sides of (4.185) to the power ek for each k = 1, 2, . . . , n, respectively, and multiplying the resulting inequalities side by side, we get n % k=1

⎡

⎤e k

n %

⎣

α /p Ai ik i ⎦

i=1,i=k

and hence 0 n 7 n % i=1,i=k Ak

 n % a

k=1

αki pk ei

k=1

1

b

 n %

b a

k=1

fk+ (s)

n %

 ek αik /pi

[Di (s)]

ds

>

i=1

fk+ (s)

n %

n %

e (1−αkk /pk )

Akk

k=1

 ek αik /pi

[Di (s)]

ds

i=1

>

n %

e (1−αkk /pk )

Akk

.

k=1

It is easy to see that, by using a technique similar to the one in [269, Theorem 3.1], we obtain the inequalities Ak > 0 for k = 1, 2, . . . , n. Thus, we have ek  n n n b % % % + αki /pi fk (s) ds > Aθkk , (4.186) [Di (s)] k=1

a

i=1

k=1

where θk = ek

αkk 1− pk



n

αki − ei pk i=1,i=k

for k = 1, 2, . . . , n. By assumption, (4.175) has nonzero solutions (e1 , . . . , en ) such that θk = 0 for k = 1, 2, . . . , n, where ek ≥ 0 for k = 1, 2, . . . , n and at least one ej > 0 for j ∈ {1, 2, . . . , n}. Choosing one of the solutions (e1 , . . . , en ), we obtain from (4.186) that (4.176) holds. This completes the proof.  Another result is the following theorem. Theorem 4.69 (Hartman-Type Inequality) Assume that there exist nontrivial solutions (e1 , . . . , en ) of (4.175). If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.173) with αki = αik for k, i = 1, 2, . . . , n, then the inequality  ek  n n b % % fk+ (s) >1 (4.187) [Ei (s)]αki /pi ds k=1

holds.

a

i=1

,

246

4 Nonlinear Differential Systems

Proof Let uk (a) = uk (b) = 0 for k = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and uk for k = 1, 2, . . . , n are not identically zero on [a, b]. As in the proof of Theorem 4.68, we have (4.177) and (4.178). Multiplying (4.177) and (4.178) by [ηk (x)]pk −1 and [ξk (x)]pk −1 , k = 1, 2, . . . , n, respectively, we obtain [ηk (x)]pk −1 |uk (x)|pk ≤ [ηk (x)]pk −1 [ξk (x)]pk −1



x

a

 p rk (s) uk (s) k ds (4.188)

and [ξk (x)]pk −1 |uk (x)|pk ≤ [ξk (x)]pk −1 [ηk (x)]pk −1

 x

b

 p rk (s) uk (s) k ds (4.189)

for k = 1, 2, . . . , n and x ∈ [a, b]. Thus, adding (4.188) and (4.189), we have    |uk (x)|pk [ξk (x)]pk −1 + [ηk (x)]pk −1 ≤ [ξk (x)ηk (x)]pk −1

b

a

 p rk (s) uk (s) k ds (4.190)

for k = 1, 2, . . . , n and x ∈ [a, b]. It is easy to see that [ξk (x)]pk −1 + [ηk (x)]pk −1 take the minimum values at ck ∈ (a, b) such that ξk (ck ) = ηk (ck ) for k = 1, 2, . . . , n. Thus, by (4.190), we obtain    |uk (x)|pk [ξk (ck )]pk −1 + [ηk (ck )]pk −1 ≤ [ξk (x)ηk (x)]pk −1

b a

 p rk (s) uk (s) k ds (4.191)

for k = 1, 2, . . . , n. Since ξk (ck ) + ηk (ck ) = ξk (x) + ηk (x) and 1 ξk (x) + ηk (x) = ξk (ck ) = 2 2

for all



b

x, ck ∈ (a, b)

[rk (s)]1/(1−pk ) ds,

a

(4.191) yields  p −1 |uk (x)|pk 22−pk [ξk (x) + ηk (x)]pk −1 = 2 |uk (x)|pk ξk k (ck ) ≤ [ξk (x)ηk (x)]

pk −1

 a

and hence



b

|uk (x)|pk ≤ Ek (x) a

 p rk (s) uk (s) k ds

b

 p rk (s) uk (s) k ds,

(4.192)

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian

247

for k = 1, 2, . . . , n and x ∈ [a, b]. After that, by using a technique similar to the one in [269, Theorem 3.1], it can be shown that the equality case in (4.192) does not hold. Thus, we have |uk (x)|pk < Ak Ek (x)

for

x ∈ (a, b),

where 

b

Ak = a

 p rk (s) uk (s) k ds,

k = 1, 2, . . . , n.

The rest of the proof is the same as in the proof of Theorem 4.68, and hence it is omitted.  Remark 4.70 It is easy to see from (4.90) that if we take 1 < pk < 2 for k = 1, 2, . . . , n, then (4.187) is better than (4.176) in the sense that (4.176) follows from (4.187), but not conversely. Similarly, from (4.91), if pk > 2 for k = 1, 2, . . . , n, then (4.176) is better than (4.187) in the sense that (4.187) follows from (4.176), but not conversely. By using (4.92) in Theorem 4.68 or (4.93) in Theorem 4.69, we obtain the following result. Corollary 4.71 (Hartman-Type Inequality) Assume that there exist nontrivial solutions (e1 , . . . , en ) of (4.175). If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.173) with αki = αik for k, i = 1, 2, . . . , n, then the inequality  n % Fk k=1

a

b

fk+ (s)ds

ek >1

holds. Remark 4.72 Note that Theorem 4.68 or 4.69 yields a Lyapunov-type inequality, which is not covered by Yang et al. [298, Theorem 1.4]. It is easy to see that Corollary 4.71 coincides with that theorem under the condition αki = αik for k, i = 1, 2, . . . , n. Remark 4.73 Since |f (x)| ≥ f + (x), the functions fk+ (x) for k = 1, 2, . . . , n in the above results can also be replaced by |fk (x)| for k = 1, 2, . . . , n. Remark 4.74 Now, we give an application of the obtained Lyapunov-type inequalities for the eigenvalue problem n %   |ui |αki − rk (x)φpk (uk ) = λk h(x)φαkk (uk ) i=1,i=k

uk (a) = uk (b) = 0,

248

4 Nonlinear Differential Systems

where h(x) > 0. Thus, if there exist nontrivial solutions (e1 , . . . , en ) of (4.175), then n−1 %



 n %

λekk

k=1

n %

b

h(s)

a

k=1

ek −1/en αki /pi

[Di (s)]

ds

< λn

i=1

or n−1 %



 n %

λekk

k=1

b

h(s)

a

k=1

n %

ek −1/en [Ei (s)]αki /pi ds

< λn .

i=1

In 2014, Tiryaki et al. [273] also considered the system ⎧    p1 −2   ⎪ u  ⎪ − r (t) u1 1 1 ⎪ ⎪ ⎪ ⎪ ⎪   p2 −2   ⎪ ⎪ ⎪ u2 ⎨− r2 (t) u2  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪   ⎪ ⎩− rn (t) u pn −2 u n n

= f1 (t) |u1 |α1 −2 u1 |u2 |α2 · · · |un |αn , = f2 (t) |u1 |α1 |u2 |α2 −2 u2 |u3 |α3 · · · |un |αn , .. . = fn (t) |u1 |α1 · · · |un−1 |αn−1 |un |αn −2 un . (4.193)

Their result is the following theorem. Theorem 4.75 (Hartman-Type Inequality) Assume (4.122). If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution of (4.193) on [a, b] satisfying (4.174), then the inequality  n %

b

a

k=1

fk+ (s)

n %

αk /pk [Di (s)]

αi /pi

ds

>1

(4.194)

i=1

holds. Proof Let uk (a) = uk (b) = 0 for k = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and uk for k = 1, 2, . . . , n are not identically zero on [a, b]. By using uk (a) = 0 and Hölder’s inequality, we get 

x

|uk (x)| ≤ a



   u (s) ds k x

≤

[rk (s)] a

1/(1−pk )

(pk −1)/pk  ds a

x

 p rk (s) u (s) k ds k

1/pk

4.4 Dirichlet Quasilinear Systems Involving the (p1 , . . . , pn )-Laplacian

249

for k = 1, 2, . . . , n and x ∈ [a, b]. Thus, we get 

x

|uk (x)|pk [ξk (x)]1−pk ≤ a

 p rk (s) uk (s) k ds

(4.195)

for k = 1, 2, . . . , n. Similarly, by using uk (b) = 0 and Hölder’s inequality, we get 

b

|uk (x)|pk [ηk (x)]1−pk ≤ x

 p rk (s) uk (s) k ds

(4.196)

for k = 1, 2, . . . , n and x ∈ [a, b]. Adding (4.195) and (4.196), we have 

b

|uk (x)|pk ≤ Dk (x) a

 p rk (s) uk (s) k ds

(4.197)

for k = 1, 2, . . . , n and x ∈ [a, b]. After that, by using a technique similar to the proof of Theorem 4.93 given by Tang and He [269], it can be shown that the equality case in (4.197) does not hold. Thus, we get p /αk

|uk (x)|pk < Ak k

(4.198)

Dk (x),

where x ∈ (a, b) and 

b

Ak =

 p rk (s) u (s) k ds

αk /pk

k

a

for k = 1, 2, . . . , n. If we take the (αk /pk )th power of both sides of (4.198), then we obtain |uk (x)|αk < Ak [Dk (x)]αk /pk .

(4.199)

Multiplying both sides of (4.199) by fk+ (x)

n %

|ui (x)|αi ,

k = 1, 2, . . . , n

i=1,i=k

and then integrating from a to b, we have 

b a

fk+ (s)

n % i=1



b

|ui (s)| ds < αi

αk /pk

Ak [Dk (s)] a

fk+ (s)

n %

|ui (s)|αi ds

i=1,i=k

(4.200)

250

4 Nonlinear Differential Systems

for k = 1, 2, . . . , n. On the other hand, multiplying the kth equation in (4.193) by uk and then integrating from a to b, we get 

 p rk (s) uk (s) k ds =

b

a



b

fk (s) a

n %



b

|ui (s)| ds ≤ αi

a

i=1

fk+ (s)

n %

|ui (s)|αi ds

i=1

(4.201) for k = 1, 2, . . . , n. By using (4.201) in (4.200), we have 

b

a



 p rk (s) uk (s) k ds
2 for k = 1, 2, . . . , n, and Theorem 4.76 is better than Theorem 4.75 when 1 < pk < 2 for k = 1, 2, . . . , n. In addition, if pk = 2 for k = 1, 2, . . . , n, then Theorem 4.75 coincides with Theorem 4.76. Remark 4.78 Let n = 1. Thus, Theorem 4.75 coincides with Theorem 4.93 given by Tang and He [269] for (3.3). Moreover, from (4.90), if we take 1 < p1 < 2, then Theorem 4.76 gives a better result than Theorem 4.93 for (3.3). Remark 4.79 Note that (4.88) is a special case of (4.193), where n = 2, β1 = α1 , and β2 = α2 . Under these conditions, we can see that (4.194) (or (4.206)) of Theorem 4.75 (or 4.76) reduces to (4.94) (or (4.113)) of Theorem 4.32 (or Theorem 4.33). If we use the second mean value theorem for integrals in (4.194) and (4.206), then we obtain the following corollaries from Theorems 4.75 and 4.76, respectively. Corollary 4.80 (Hartman-Type Inequality) If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.193) satisfying (4.174), then there exist some points dk ∈ (a, b) for k = 1, 2, . . . , n such that the inequality  n n % % k=1

αk /pk [Di (dk )]

−αi /pi

i=1


0

t ∈ (a, b)

for all

i = 1, 2, (4.217)

and

where 1 < p1 , p2 < ∞ and α1 , α2 > 0 satisfy α1 α2 + =1 p1 p2 and q is a real-valued positive continuous function defined on R. Moreover, they found a hyperbolic-type function h defining a region which contains all the generalized eigenvalues (λ1 , λ2 ) of (4.216) and (4.217): 1 λ2 ≥ h(λ1 ) := α2



2α1 +α2 (λ1 α2 )−p1 /α1 (b − a)α1 +α2 −1



−1 p2 /α2

b

q(t)dt a

.

(4.218)

256

4 Nonlinear Differential Systems

The proof of (4.218) is based on the extension of the Lyapunov inequality (b − a)α1 +α2 −1



α1 /p1 

b

f1 (t)dt a

α2 /p2

b

f2 (t)dt

≥ 2α1 +α2 ,

a

which holds provided the quasilinear system  ⎧    p1 −2  ⎪ u1 = f1 (t) |u1 |α1 −2 |u2 |α2 u1 , ⎨− u1    ⎪ ⎩− u p2 −2 u  = f (t) |u |α1 |u |α2 −2 u 2 1 2 2 2 2 has a solution (u1 , u2 ) satisfying (4.217), where p1 , p2 , α1 , α2 are the same as those of (4.216) and f1 , f2 are real-valued positive continuous functions defined on R. More related works can be found in [56, 57] and the references therein. Motivated by de Nápoli and Pinasco [107] and Çakmak and Tiryaki [72, 73], Tang and He [269] improved (4.81) and (4.138) for (4.61) and (4.193), more general than (4.140), by using a method different from de Nápoli and Pinasco [107]. As a byproduct, we derive a better Lyapunov-type inequality than (3.4), namely  a

b

 t  t

a [r(τ )]

1/(1−p) dτ

1/(1−p) dτ a [r(τ )]

p−1 

p−1

+

p−1 b 1/(1−p) dτ t [r(τ )]



p−1 b 1/(1−p) dτ t [r(τ )]

q + (t)dt > 1

(4.219)

for second-order half-linear differential equations of the form (3.1). In particular, (4.219) reproduces (1.33) for (1.1) when p = 2 and r(t) ≡ 1. As an application of their Lyapunov-type inequalities, Tang and He [269] obtained a lower bound for the eigenvalues of the generalized eigenvalue problem consisting of (4.214) and (4.215). When n = 2, they found another hyperbolic-type function g defining a region which contains all the generalized eigenvalues (λ1 , λ2 ) of (4.216) and (4.217): 1 λ2 > g(λ1 ) := α2



 λ1 α1 a

b

1−p2 /α2 [(t − a)(b − t)]p1 −1 q(t)dt (t − a)p1 −1 + (b − t)p1 −1  b

−1 [(t − a)(b − t)]p2 −1 × q(t)dt . p −1 + (b − t)p2 −1 a (t − a) 2

We can show that g(λ1 ) > h(λ1 ). Tang and He [269] established some new Lyapunov-type inequalities for (4.61). Denote

4.5 Lower Bounds for Generalized Eigenvalues



t

ζ1 (t) :=

[r1 (τ )]

1/(1−p1 )

p1 −1 dτ

257

 η1 (t) :=

,

p1 −1

b

1/(1−p1 )

[r1 (τ )]

a

dτ

t

(4.220)

and 

t

ζ2 (t) :=

1/(1−p2 )

[r2 (τ )]

p2 −1 dτ

 η2 (t) :=

,

a

p2 −1

b

[r2 (τ )]

1/(1−p2 )

dτ

t

. (4.221)

Theorem 4.86 (Hartman-Type Inequality) Assume the hypotheses (H 1 ) r1 , r2 , f1 , f2 are real-valued continuous functions defined on R and r1 (t) > 0 and r2 (t) > 0 for all t ∈ R; (H 2 ) 1 < p1 , p2 < ∞ and α1 , α2 , β1 , β2 > 0 satisfy α2 α1 + =1 p1 p2

(4.222)

β2 β1 + = 1. p1 p2

(4.223)

and

If (4.61) has a solution (u1 , u2 ) satisfying (4.217), then the inequality 

b

ζ1 (t)η1 (t) + f1 (t)dt ζ a 1 (t) + η1 (t)

α1 β1 /p2  1

b

ζ1 (t)η1 (t) + f2 (t)dt ζ a 1 (t) + η1 (t)

α2 β1 /(p1 p2 )

 α2 β1 /(p1 p2 )  α2 β2 /p2 2 b ζ (t)η (t) b ζ (t)η (t) 2 2 2 2 + + f1 (t)dt f2 (t)dt × >1 a ζ2 (t) + η2 (t) a ζ2 (t) + η2 (t) (4.224)

holds. Proof By (4.61) and (4.217), we obtain 

b

a

 p r1 (t) u1 (t) 1 dt =



b

f1 (t) |u1 (t)|α1 |u2 (t)|α2 dt

(4.225)

f2 (t) |u1 (t)|β1 |u2 (t)|β2 dt.

(4.226)

a

and  a

b

 p r2 (t) u2 (t) 2 dt =



b a

258

4 Nonlinear Differential Systems

It follows from (4.220), (4.217), and Hölder’s inequality that  t p1   p1   |u1 (t)| =  u1 (τ )dτ  a



t

≤ a

[r1 (τ )]1/(1−p1 ) dτ 

= ζ1 (t)

p1 −1 

t

a t

a

 p r1 (τ ) u1 (τ ) 1 dτ,

 p r1 (τ ) u1 (τ ) 1 dτ

(4.227)

a≤t ≤b

and |u1 (t)|

p1

  = 

b

t



p1 

u1 (τ )dτ 

b

≤ t

p1 −1  [r1 (τ )]1/(1−p1 ) dτ t



b

= η1 (t) t

 p r1 (τ ) u1 (τ ) 1 dτ,

b

 p r1 (τ ) u1 (τ ) 1 dτ

(4.228)

a≤t ≤b

hold. From (4.227) and (4.228), we have |u1 (t)|p1 ≤

ζ1 (t)η1 (t) ζ1 (t) + η1 (t)



b

a

 p r1 (τ ) u1 (τ ) 1 dτ,

a ≤ t ≤ b.

(4.229)

 p r1 (τ ) u1 (τ ) 1 dτ,

a ≤ t ≤ b.

(4.230)

We claim that |u1 (t)|p1
0 for τ ∈ [a, b], and then it follows from (4.227) that u1 (b) = 0, 1 which contradicts the fact that u1 (b) = 0. Therefore, (4.230) holds. Now, it follows from (4.217), (4.225), (4.230), (H2 ), and Hölder’s inequality that 

b

a

f1+ (t) |u1 (t)|p1 dt



 b  p ζ1 (t)η1 (t) + f1 (t)dt r1 (t) u1 (t) 1 dt a ζ1 (t) + η1 (t) a  b = M11 f + (t) |u1 (t)|α1 |u2 (t)|α2 dt b


1.

(4.247) 

It follows from (4.238), (4.242), (4.247) that (4.224) holds.

When α1 = β2 = p1 = p2 = p, α2 = β1 = 0, r1 = r2 = r, and f1 = f2 = q, (4.61) reduces to the second-order half-linear differential equation of the form (3.1). Hence, we can derive the following Lyapunov-type inequality for (3.1) from (4.236) and (4.243). Theorem 4.87 (Hartman-Type Inequality) Suppose that p > 1 and r > 0. If (3.1) has a solution u satisfying the boundary conditions u(a) = u(b) = 0

u(t) = 0 for all

and

t ∈ (a, b),

(4.248)

then the inequality 

 t

p−1  p−1 b 1/(1−p) dτ 1/(1−p) dτ [r(τ )] [r(τ )] a t +  p−1  p−1 q (t)dt > 1 t b 1/(1−p) dτ + t [r(τ )]1/(1−p) dτ a [r(τ )]

b

a

(4.249)

holds. Since 

t

p−1 [r(τ )]

1/(1−p)

dτ



b

+

a

p−1 [r(τ )]

1/(1−p)

dτ

t

 ≥2 a

t

 [r(τ )]1/(1−p) dτ

b

(p−1)/2 [r(τ )]1/(1−p) dτ

t

it follows from Theorem 4.87 that the following corollary holds.

,

4.5 Lower Bounds for Generalized Eigenvalues

263

Corollary 4.88 (Hartman-Type Inequality) Suppose that p > 1 and r > 0. If (3.1) has a solution u satisfying (4.248), then the inequality 

b

q + (t)



a

t



b

[r(τ )]1/(1−p) dτ

a

(p−1)/2 [r(τ )]1/(1−p) dτ

dt > 2

(4.250)

t

holds. Remark 4.89 It is easy to see that (4.249) and (4.250) are better than (3.4). Moreover, the Lyapunov-type inequality (4.249) reproduces (1.33) for (1.1) when p = 2 and r(t) ≡ 1. Corollary 4.90 (Hartman-Type Inequality) Suppose that hypotheses (H1 ) and (H2 ) are satisfied. If (4.61) has a solution (u1 , u2 ) satisfying (4.217), then the inequality 

b

a

×

 f1+ (t) ξ1 (t)η1 (t) dt

 b a

α1 β1 /p2 



b

1

a

f1+ (t) ξ2 (t)η2 (t) dt

 f2+ (t) ξ1 (t)η1 (t) dt

α2 β1 /(p1 p2 ) 

b

a

α2 β1 /(p1 p2 )



f2+ (t) ξ2 (t)η2 (t) dt

α2 β2 /p2 2

> 2α2 /p2 +β1 /p1 (4.251)

holds. Proof Since ζi (t) + ηi (t) ≥ 2 [ζi (t)ηi (t)]1/2 ,

i = 1, 2, 

it follows from (4.224) and (H2 ) that (4.251) holds.

Corollary 4.91 (Hartman-Type Inequality) Suppose that hypotheses (H1 ) and (H2 ) are satisfied. If (4.61) has a solution (u1 , u2 ) satisfying (4.217), then the inequality 

b

β1 (p1 −1)/p1  1/(1−p1 )

[r1 (t)]

dt

a

1/(1−p2 )

[r2 (t)]

dt

a



b

× a

holds.

α2 (p2 −1)/p2

b

f1+ (t)dt

β1 /p1 

b a

f2+ (t)dt

α2 /p2

> 2α2 +β1

(4.252)

264

4 Nonlinear Differential Systems

Proof Since 

t

[ζi (t)ηi (t)]1/2 =

 [ri (τ )]1/(1−pi ) dτ

a

≤

b

(pi −1)/2 [ri (τ )]1/(1−pi ) dτ

t



1 2pi −1

pi −1

b

1/(1−pi )

[ri (t)]

dt

,

i = 1, 2,

a



it follows from (4.251) and (H2 ) that (4.252) holds.

Remark 4.92 Obviously, (4.252) in Corollary 4.91 is still better than (4.81), which is the main result in [42]. Now, let us consider (4.193). Denote 

t

ζi (t) :=

1/(1−pi )

[ri (τ )]

pi −1 dτ

,

i = 1, 2, . . . , n

(4.253)

,

i = 1, 2, . . . , n.

(4.254)

a

and 

pi −1

b

ηi (t) :=

[ri (τ )]

1/(1−pi )

dτ

t

Theorem 4.93 (Hartman-Type Inequality) Assume the hypothesis (H 3 ) ri , fi are real-valued continuous functions defined on R and ri (t) > 0 for i = 1, 2, . . . , n, and 1 < pi < ∞ and αi > 0 satisfy (4.122). If (4.193) has a solution u satisfying the boundary conditions ui (a) = ui (b) = 0,

ui (t) = 0

for all

t ∈ (a, b), i = 1, 2, . . . , n,

(4.255)

then the inequality n  n % % i=1 j =1

a

b

ζi (t)ηi (t) + f (t)dt ζi (t) + ηi (t) j

αi αj /(pi pj ) >1

(4.256)

holds. Proof By (4.193), (H3 ), and (4.255), we obtain 

b a

 p ri (t) ui (t) i dt =



b

fi (t) a

n % k=1

|uk (t)|αk dt,

i = 1, 2, . . . , n.

(4.257)

4.5 Lower Bounds for Generalized Eigenvalues

265

From (4.253), (4.255), and Hölder’s inequality, we have  t pi   pi   |ui (t)| =  ui (τ )dτ  a



t

≤ a

[ri (τ )]1/(1−pi ) dτ

pi −1 

t a



t

= ζi (t) a

 p ri (τ ) ui (τ ) i dτ,

 p ri (τ ) ui (τ ) i dτ

(4.258)

a≤t ≤b

for i = 1, 2, . . . , n. Similarly, from (4.254), (4.255), and Hölder’s inequality, we get  b pi   |ui (t)|pi =  ui (τ )dτ  t

 ≤

pi −1 

b

1/(1−pi )

[ri (τ )] t



= ηi (t) t

b

dτ t

b

 p ri (τ ) ui (τ ) i dτ,

 p ri (τ ) ui (τ ) i dτ

(4.259)

a≤t ≤b

for i = 1, 2, . . . , n. From (4.258) and (4.259), we have  b  p ζi (t)ηi (t) |ui (t)|pi ≤ ri (τ ) ui (τ ) i dτ, a ≤ t ≤ b ζi (t) + ηi (t) a

(4.260)

for i = 1, 2, . . . , n. We claim that |ui (t)|pi
0 for τ ∈ [a, b], and then it follows from (4.258) that ui0 (b) = 0, which contradicts the fact that ui0 (b) = 0. Therefore, (4.261) holds. Now, by (4.255), (4.257), (4.258), (4.261), (H3 ), and the generalized Hölder inequality, we get  a

b

 b  p ζi (t)ηi (t) + fj (t)dt ri (t) ui (t) i dt a ζi (t) + ηi (t) a  b n % |uk (t)|αk dt = Mij fi (t)

fj+ (t) |ui (t)|pi dt
0,

i, j = 1, 2, . . . , n.

(4.268)

4.5 Lower Bounds for Generalized Eigenvalues

267

If (4.268) is not true, then there exist i0 , j0 ∈ {1, 2, . . . , n} such that 

b a

  uj (t)pj0 dt = 0. fi+ (t) 0 0

(4.269)

From (4.257) and (4.269), we have 

b

0≤ a

 =

b

fi0 (t)

a

 ≤ a

 p ri0 (t) ui0 (t) i0 dt n %

|uk (t)|αk dt

k=1 b

fi0 (t)

n %

|uk (t)|αk dt = 0.

k=1

From the fact that ri0 (t) > 0, we get ui0 (t) ≡ 0,

a ≤ t ≤ b.

(4.270)

Combining (4.258) with (4.270), we obtain ui0 (t) ≡ 0 for a ≤ t ≤ b, which contradicts (4.255). Therefore, (4.268) holds. From (4.267), (4.268), and (H3 ), we have n n % %

α αj /(pi pj )

Miji

> 1,

i=1 j =1

from which it follows by (4.238), (4.242), and (4.247) that (4.224) holds.



Corollary 4.94 (Hartman-Type Inequality) Suppose that the hypothesis (H3 ) is satisfied. If (4.193) has a solution u satisfying (4.255), then the inequality n  n % % i=1 j =1

b a

fj+ (t) [ζi (t)ηi (t)]1/2 dt

αi αj /(pi pj ) >2

(4.271)

holds. Proof Since ζi (t) + ηi (t) ≥ 2 [ζi (t)ηi (t)]1/2 , it follows from (4.256) and (H3 ) that (4.271) holds.

i = 1, 2, . . . , n, 

Corollary 4.95 (Hartman-Type Inequality) Suppose that the hypothesis (H3 ) is satisfied. If (4.193) has a solution u satisfying (4.255), then the inequality

268

4 Nonlinear Differential Systems

n  % i=1

b

1/(1−pi )

[ri (t)]

dt

αi (pi −1)/pi % n 

a

b

fj+ (t)dt

a

j =1

αj /pj > 2An

(4.272)

holds, where An is defined in (4.139). Proof Since  1/2

[ζi (t)ηi (t)]



t

=

[ri (τ )]

1/(1−pi )

dτ

a

≤

(pi −1)/2

b

[ri (τ )]

1/(1−pi )

dt

t



1 2pi −1

pi −1

b

1/(1−pi )

[ri (t)]

dτ

,

i = 1, 2, . . . , n,

a

it follows from (4.271) and (H3 ) that (4.272) holds.



From Corollary 4.95, we obtain the following corollary immediately. Corollary 4.96 (Lyapunov-Type Inequality) Suppose that fi , i = 1, 2, . . . , n, are real-valued continuous functions defined on R, and 1 < pi < ∞ and αi > 0 satisfy (4.122). If (4.121) has a solution u satisfying (4.255), then the inequality n  % i=1

b a

fi+ (t)dt

αi /pi > 2An (b − a)1−An

holds, where An is defined in (4.139). Remark 4.97 Obviously, Corollary 4.96 coincides with [73, Corollary 3]. Remark 4.98 When n = 1, Theorem 4.93 and Corollary 4.94 reduce to Theorem 4.87 and Corollary 4.88, respectively. Therefore, Theorem 4.93 generalizes and improves all related existing Lyapunov-type inequalities. Now, we apply the Lyapunov-type inequalities to obtain a lower bound for the generalized eigenvalues of (4.214) and (4.215). Let (λ1 , λ2 , . . . , λn ) be a generalized eigenvalue of (4.214) and (4.215) and u be the eigenfunction associated with (λ1 , λ2 , . . . , λn ). Then, u is a solution of (4.193) satisfying (4.255), where ri (t) ≡ 1 and fi (t) = λi αi q(t) > 0 for i = 1, 2, . . . , n. By using techniques similar to those of de Nápoli and Pinasco [107], we can obtain the following theorem. Theorem 4.99 Assume that 1 < pi < ∞ and αi > 0 satisfy (4.122) and that q is a real-valued positive continuous function defined on R. Then, there exists a function g such that λn > g(λ1 , λ2 , . . . , λn−1 )

4.5 Lower Bounds for Generalized Eigenvalues

269

for every generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.214) and (4.215), namely ⎧ n−1 α /p 1 ⎨%  g(λ1 , λ2 , . . . , λn−1 ) = λj αj j j αn ⎩ j =1

×

n−1 %



i=1

b

a

[(t − a)(b − t)]pi −1 q(t)dt (t − a)pi −1 + (b − t)pi −1

αi /pi −pn /αn .

(4.273)

Proof For (4.193), ri (t) ≡ 1, and fi (t) = λi αi q(t) > 0 for all t ∈ R, i = 1, 2, . . . , n. Thus, it follows from (4.253), (4.254), and (4.256) that 1
λj αj q(t)dt . p −1 p −1 ⎭ αn ⎩ a (t − a) i +(b−t) i j =1

i=1



This completes the proof.

When n = 2, (4.214) and (4.215) reduce to the one-dimensional quasilinear elliptic system of resonant type (4.216) with Dirichlet boundary conditions of the form (4.217). Applying Theorem 4.99 to (4.216) and (4.217), we immediately have the following corollary. Corollary 4.100 Assume that 1 < p1 , p2 < ∞ and α1 , α2 > 0 satisfy (4.222) and that q is a real-valued positive continuous function defined on R. Then, there exists a function g such that λ2 > g(λ1 ) for every generalized eigenvalue (λ1 , λ2 ) of (4.216) and (4.217), namely

270

4 Nonlinear Differential Systems

1 g(λ1 ) = α2

 λ1 α1

b a

1−p2 /α2 [(t − a)(b − t)]p1 −1 q(t)dt (t − a)p1 −1 + (b − t)p1 −1  b [(t − a)(b − t)]p2 −1 × q(t)dt. p −1 + (b − t)p2 −1 a (t − a) 2

Since 

(b − a)p1 −1 [(t − a)(b − t)]p1 −1 < 2p1 (t − a)p1 −1 + (b − t)p1 −1

for all

a+b t ∈ [a, b] \ 2

(b − a)p2 −1 [(t − a)(b − t)]p2 −1 < p −1 p −1 2p2 (t − a) 2 + (b − t) 2

for all

t ∈ [a, b] \



and 

 a+b , 2

we have

1−p2 /α2  b [(t − a)(b − t)]p1 −1 1 λ1 α1 q(t)dt p −1 + (b − t)p1 −1 α2 a (t − a) 1  b [(t − a)(b − t)]p2 −1 × q(t)dt p −1 + (b − t)p2 −1 a (t − a) 2  −p2 /α2  1 (λ1 α1 )α1 /p1 (b − a)α1 +α2 −1 b q(t)dt > α2 2α1 +α2 a

g(λ1 ) =

(4.274)

= h(λ1 ), where h is defined by (4.218). Remark 4.101 Inequality (4.274) shows that λ2 > g(λ1 ) gives the better lower bound g(λ1 ) than (4.218) (h(λ1 )). Remark 4.102 It is not difficult to see that (4.273) also gives a better lower bound than [72, (30) in Theorem 9]. In 2012, Akta¸s et al. [27] established new Lyapunov-type inequalities for two classes of one-dimensional quasilinear elliptic systems of resonant type, which improve the results of Tang and He [269] when 1 < pi < 2 for i = 1, 2, . . . , n. They were concerned with the problem of finding new Lyapunov-type inequalities for the systems ⎧   ⎨− (r1 (t)φp1 (u1 )  = f1 (t)φα1 (u1 ) |u2 |α2 , ⎩− (r (t)φ (u ) = f (t)φ (u ) |u |β1 2 p2 2 2 β2 2 1

(4.275)

4.5 Lower Bounds for Generalized Eigenvalues

271

and n %   |uk |αk − ri (t)φpi (ui ) = fi (t)φαi (ui )

for i ∈ {1, 2, . . . , n},

(4.276)

k=1 k=i

where n ∈ N and φγ (u) = |u|γ −2 u with γ > 1, under the following hypotheses: (H1 ) ri , fi are real-valued continuous functions defined on R, ri (t) > 0 and 1 < pi < ∞ for i = 1, 2, α2 , β1 ≥ 0 satisfy (4.222) and (4.223). (H2 ) ri , fi are real-valued continuous functions defined on R, ri (t) > 0, 1 < pi < ∞ and αi ≥ 0 for i = 1, 2, . . . , n satisfy (4.122). Akta¸s et al. [27] obtained new Lyapunov-type inequalities for (4.275) and (4.276) such that their results are better than the previously existing results in the literature when 1 < pi < 2 for i = 1, 2, . . . , n as described below. For some related works on Lyapunov-type inequalities, the reader is referred to [68, 72, 73, 107, 143, 193, 195, 202, 237, 269, 271, 274, 276, 277, 281, 282, 296, 297, 299]. Denote 



t

ξi (t) =

[ri (τ )]

1/(1−pi )

a

dτ

and

b

ηi (t) =

[ri (τ )]1/(1−pi ) dτ

(4.277)

t

for i = 1, 2, . . . , n. It is easy to see that when αi = pi for fixed i = 1, 2, . . . , n, and for k = i, αk = 0 for k = 1, 2, . . . , n, we obtain the half-linear differential equation 

 ri (t)φpi (ui ) + fi (t)φpi (ui ) = 0

from (4.276) with the hypothesis (H2 ). For example if n = 1, then we get   r1 (t)φp1 (u1 ) + f1 (t)φp1 (u1 ) = 0.

(4.278)

Equation (4.278) is called sub-linear, linear and super-linear if 1 < p1 < 2, p1 = 2, and p1 > 2, respectively. Although there is a large body of literature concerning Lyapunov-type inequalities for (4.278), to the best of our knowledge, before [27], there was almost no study for the super-linear or the sub-linear equation of the form (4.278) except for Lee et al. [195] and Wang [282]. Motivated from the paper by Tang and He [269], Akta¸s et al. [27] proved new Lyapunov-type inequalities for (4.275) and (4.276) such that their results are better than Theorems 4.86, 4.93, and 4.99 when 1 < pi < 2 for i = 1, 2, . . . , n, respectively. Since our attention is restricted to Lyapunov-type inequalities for quasilinear systems of differential equations, we shall assume the existence of a nontrivial solution of (4.275) or (4.276). For authors who contributed to the existence of the solution of these types of systems, we refer to the paper by Afrouzi and Heidarkhani [2].

272

4 Nonlinear Differential Systems

Theorem 4.103 (Hartman-Type Inequality) Let (H1 ) hold. If (4.275) has a real nontrivial solution (u1 , u2 ) such that ui (a) = ui (b) = 0 for i = 1, 2, where a, b ∈ R with a < b are consecutive zeros, and ui for i = 1, 2 are not identically zero on [a, b], then the inequality 

b



p1 −2

2 a



b

×

ζ1 (t)η1 (t) ζ1 (t) + η1 (t)

2p1 −2



a



b

×

2p2 −2



a



b

×

2p2 −2



a

p1 −1

ζ1 (t)η1 (t) ζ1 (t) + η1 (t) ζ2 (t)η2 (t) ζ2 (t) + η2 (t) ζ2 (t)η2 (t) ζ2 (t) + η2 (t)

α1 β1 /p2 1

f1+ (t)dt

p1 −1

p2 −1

p2 −1

β1 α2 /(p1 p2 ) f2+ (t)dt β1 α2 /(p1 p2 )

(4.279)

f1+ (t)dt α2 β2 /p2 f2+ (t)dt

2

>1

holds. The proof of Theorem 4.103 can be easily obtained by combining some ingredients of the methods used in Tang and He [269] with a slight modification. Therefore, it is left to reader. The following is [27, Theorem 2.2]. Theorem 4.104 (Hartman-Type Inequality) Let (H2 ) hold. If (4.276) has a real nontrivial solution u such that ui (a) = ui (b) = 0 for i = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b are consecutive zeros, and ui for i = 1, 2, . . . , n are not identically zero on [a, b], then the inequality  n n % % i=1 j =1

b

pi −2

2 a



ζi (t)ηi (t) ζi (t) + ηi (t)

pi −1

αi αj /(pi pj ) fj+ (t)dt

>1

(4.280)

holds. Remark 4.105 For the same reason as Theorem 4.17 (see Remark 4.18), Theorem 4.104 is also not correct. Inequalities [27, (2.9) and (2.10) on p. 4870] are on the intervals [a, ci ] and [ci , b], respectively, and they cannot just be added to get [27, (2.11)]. In 2014, Tiryaki et al. [273] presented an application of the Lyapunov-type inequality obtained for (4.193). They obtained the following result which gives lower bounds for the nth component of any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of the system

4.5 Lower Bounds for Generalized Eigenvalues

273

⎧ n %   ⎪ ⎪ ⎨− rk (x)φpk (uk ) = λk αk r(x)φαk (uk ) |ui |αi ⎪ ⎪ ⎩

n

αk = 1, pk

with

i=1,i=k

uk (a) = uk (b) = 0 for

k=1

k = 1, 2, . . . , n. (4.281)

The proof of the following theorem is based on the above generalization of the Lyapunov-type inequality, as in that of Theorem 4.46 of Çakmak and Tiryaki [72], and hence it is omitted. Theorem 4.106 There exists a function h1 such that (4.282)

h1 (λ1 , λ2 , . . . , λn−1 ) < λn for any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.281), namely 1 h1 (λ1 , λ2 , . . . , λn−1 ) = αn

n−1 %



b

αk /pk

[λk αk ]

r(s) a

k=1

n %

−pn /αn αi /pi

[Di (s)]

ds

i=1

(4.283) from (4.194), or 1 h1 (λ1 , λ2 , . . . , λn−1 ) = αn

n−1 %

 [λk αk ]

αk /pk

b

r(s) a

k=1

n %

−pn /αn αi /pi

[Ei (s)]

ds

i=1

(4.284) from (4.206). Remark 4.107 By using (4.90) and (4.91), it is easy to see that Theorem 4.106 with (4.283) gives a better lower bound than Theorem 4.106 with (4.284) when pk > 2 for k = 1, 2, . . . , n, and Theorem 4.106 with (4.284) gives a better lower bound than Theorem 4.106 with (4.283) when 1 < pk < 2 for k = 1, 2, . . . , n. In addition, if pk = 2 for k = 1, 2, . . . , n, then Theorem 4.106 with (4.283) is exactly the same as Theorem 4.106 with (4.284). Remark 4.108 Note that, after the second mean value theorem for integrals in (4.283) and (4.284) is used, if we take rk (x) ≡ 1 and dk = ci , where |ui (ci )| = max |ui (x)| a1

(4.292)

a

i=1

holds, where pn+1 := p1 . Proof Let (4.289) be satisfied for m = 0, 1 and i = 1, 2, . . . , n, where a, b ∈ R with a < b and ui for i = 1, 2, . . . , n are not identically zero on [a, b]. By using (4.290) in Lemma 4.110, we have 1/pi

|ui (x)| < Ai Ri

(4.293)

,

where  Ri =

b

a

  pi u (s) ds i

and

Ai = 2−1 (b − a)1−1/pi

for i = 1, 2, . . . , n, and hence 1/p

|ui+1 (x)| < Ai+1 Ri+1i+1

(4.294)

for i = 0, 1, . . . , n − 1. Next, we prove that Ri > 0 for i = 1, 2, . . . , n. If the inequality Ri > 0 is not true, then Ri = 0 for i = 1, 2, . . . , n. If  b   pi u (s) ds = 0, Ri = a

i

then it follows that ui (x) ≡ 0

on

[a, b]

(4.295)

for i = 1, 2, . . . , n. Combining (4.293) with (4.295), we obtain ui (x) ≡ 0 on [a, b], which contradicts ui (x) ≡ 0 on [a, b] and i = 1, 2, . . . , n. Therefore, Ri > 0 for i = 1, 2, . . . , n. On the other hand, multiplying the ith equation in (4.287) by ui , integrating from a to b, and taking into account (4.289), (4.293), and (4.294), we have  a

b

  pi u (s) ds = i



b

fi (s)ui (s) |ui+1 (s)|αi −2 ui+1 (s)ds

a

and  Ri ≤

b

|fi (s)| |ui (s)| |ui+1 (s)|αi −1 ds

a


0 cancel out for i = 1, 2, . . . , n in (4.299). Therefore, we obtain (4.292). This completes the proof.  Corollary 4.112 (Hartman-Type Inequality) Let the hypothesis n % αi − 1 =1 pi − 1

(4.300)

i=1

be satisfied. If (4.287) has a real nontrivial solution u satisfying (4.289) with m = 0, 1 and i = 1, 2, . . . , n, then the inequality  n  % −αi 1−1/pi +(pi+1 −1)(αi −1)/pi+1 2 (b − a) i=1

holds, where pn+1 := p1 .

b a

1/(αi −1) |fi (s)| ds

>1

(4.301)

278

4 Nonlinear Differential Systems

Proof We proceed as in the proof of Theorem 4.111 to obtain (4.299). Now, we find a relation between αi and pi for i = 1, 2, . . . , n such that Ri > 0 for i = 1, 2, . . . , n cancel out in (4.299), i.e., solve (4.291). We observe that by the hypothesis given in (4.300), this system admits a nontrivial solution. Hence, we may take ei = e1 (p1 − 1)/(αi − 1), where e1 > 0 and i = 1, 2, . . . , n. Therefore, we obtain (4.301). This completes the proof.  Remark 4.113 If we take αi = pi = 2 for i = 1, 2, . . . , n, fi (x) ≡ −1 for i = 1, . . . , n − 1, and fn = g1 in (4.287) and (4.289) with m = 0, 1 and i = 1, 2, . . . , n, then Theorem 4.111 reduces to Corollary 3.57 given by Wang [284]. Another result is the following theorem for solutions of (4.288) satisfying antiperiodic boundary conditions. Theorem 4.114 (Hartman-Type Inequality) If (4.288) has a real nontrivial solution u satisfying (4.289) with m = 0, 1 and i = 1, 2, . . . , n, then the inequality  n 

2−pi (b − a)pi −1

 |fi (s)| ds > 1

b

(4.302)

a

i=1

holds. Proof Let (4.289) hold for m = 0, 1, i = 1, 2, . . . , n, where a, b ∈ R with a < b and ui for i = 1, 2, . . . , n are not identically zero on [a, b]. By using (4.290) in Lemma 4.110, we have 1/pi

|ui (x)| < Ai Ri

(4.303)

,

and hence p −1

|ui (x)|pi −1 < Ai i

1−1/pi

Ri

,

(4.304)

where 

b

Ri = a

  pi u (s) ds i

and

Ai =

1 (b − a)1−1/pi 2

for i = 1, 2, . . . , n. It is easy to see that, by using a technique similar to the one in the proof of Theorem 4.111, we obtain Ri > 0 for i = 1, 2, . . . , n. On the other hand, multiplying the ith equation in (4.288) by ui , integrating from a to b, and taking into account (4.289), (4.303), and (4.304), we have 

b a

  pi u (s) ds = i



b

fi (s)ui (s) a

n

  uj (s)pj −2 uj (s)ds j =1

4.6 Nonlinear Systems with Anti-periodic Boundary Conditions

279

and  Ri ≤

b

n

  uj (s)pj −1 ds |fi (s)| |ui (s)|

a

j =1 1/pi

< Ai Ri

n

p −1

Aj j



1−1/pj

Rj

b

|fi (s)| ds,

a

j =1

and hence 1−1/pi

Ri

< Ai

n

p −1

Aj j

1−1/pj



Rj

b

|fi (s)| ds

p −1

for i = 1, 2, . . . , n. Multiplying (4.305) by Ai i inequalities side by side, we have n

i=1

p −1 1−1/pi Ai i Ri


0 and summing the resulting

 n

p Ai i

i=1



b

 |fi (s)| ds .

a

Therefore, we obtain (4.302). This completes the proof.



Remark 4.115 If we take n = 1 in (4.287) with α1 = p1 or (4.288), then Theorems 4.111 and 4.114 reduce to Corollary 3.60 given by Wang [284]. Remark 4.116 The results do not compare with some results in the literature because of the anti-periodic boundary conditions of the form (4.289) except for the results of Wang [284] in some special cases. In 2014, Akta¸s et al. [28] proved a new Lyapunov-type inequality for the system n %   |ui |αki − φpk (uk ) = fk (x)φαkk (uk )

for

k = 1, 2, . . . , n,

(4.306)

i=1 i=k

where n ∈ N, φγ (u) = |u|γ −2 u, γ > 1, fk ∈ C([a, b], R) for k = 1, 2, . . . , n, u = (u1 , u2 , . . . , un ) is a real nontrivial solution of (4.306) satisfying (4.289), and 1 < pk < ∞ and αki for k, i = 1, 2, . . . , n are nonnegative constants. As an application, they have also investigated the lower bounds on the eigenvalues of the following problem. Let λk for k = 1, 2, . . . , n be the generalized eigenvalues of ⎧ n % ⎪   ⎪   ⎪ |ui |αki for k = 1, 2, . . . , n, − φ (u ) = f (x)φ (u ) ⎪ pk k k αkk k ⎨ i=1 (4.307) i=k ⎪ ⎪ ⎪ ⎪ ⎩ (m) ui (a) + u(m) i (b) = 0

280

4 Nonlinear Differential Systems

and r be a function defined on R. Then, (4.307) with fk (x) = λk r(x) for k = 1, 2, . . . , n and x ∈ R reduces to the problem ⎧ n % ⎪   ⎪   ⎪ |ui |αki − φ (u ) = λ r(x)φ (u ) ⎪ p k α k k kk k ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

for k = 1, 2, . . . , n,

i=1 i=k (m) u(m) k (a) + uk (b) = 0

(4.308)

m = 0, 1

for

and

k = 1, 2, . . . , n.

Now, we give the result of Akta¸s et al. [28]. Theorem 4.117 (Hartman-Type Inequality) Assume there exists a nontrivial solution (e1 , e2 , . . . , en ) of the linear homogeneous system

n αkk αik ek 1 − − ei = 0 for pk pk

k = 1, 2, . . . , n,

(4.309)

i=1 i=k

where ek ≥ 0 for k = 1, 2, . . . , n. If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.307), then the inequality  n  % Bn Cn 2 (b − a)

b a

k=1

fk+ (s)ds

 ek >1

(4.310)

holds, where Bn = −

n

αki

Cn =

and

i=1

n

(pi − 1) αki . pi

(4.311)

i=1

Proof Let (4.289) hold for m = 0, 1, k = 1, 2, . . . , n, where n ∈ N, a, b ∈ R with a < b, and uk for k = 1, 2, . . . , n are not identically zero on [a, b]. Multiplying the kth equation in (4.306) by uk , integrating from a to b, and then using (4.289), we get 

b a



  pk u (s) ds = k

b

fk (s) a

n %

|ui (s)|αki ds

i=1

for k = 1, 2, . . . , n. By using (4.290) in Lemma 4.110, (4.312) yields 

b a

  pk u (s) ds = k



b

fk (s) a

 a

|ui (s)|αki ds

i=1 b

≤

n %

fk+ (s)

n % i=1

|ui (s)|αki ds

(4.312)

4.6 Nonlinear Systems with Anti-periodic Boundary Conditions

< max

a≤x≤b

n %



b

|ui (x)|αki a

i=1

≤ 2 (b − a) Bn

Cn

n  %

b a

i=1

281

fk+ (s)ds

  pi u (s) ds i

αki /pi 

b a

fk+ (s)ds,

and hence 

b

  pk u (s) ds

1−αkk /pk

k

a

< 2Bn (b − a)Cn

n  %

b a

i=1 i=k

  pi u (s) ds i

αki /pi  a

b

fk+ (s)ds

(4.313)

for k = 1, 2, . . . , n. Raising both sides of (4.313) to the power ek for each k = 1, 2, . . . , n, respectively, and multiplying the resulting inequalities side by side, we obtain n  %

b

  pk u (s) ds k

a

k=1

(1−αkk /pk )ek

⎡
0 (4.314) k a

for k = 1, 2, . . . , n. By (4.314), we have n  % k=1

a

b

  pk u (s) ds k

θk
0 for j ∈ {1, 2, . . . , n}. Choosing one of the solutions (e1 , . . . , en ), we obtain (4.310) from (4.315). This completes the proof.  The proof of the following result proceeds along the lines of that of Yang et al. [298, Corollary 1]. Hence, it is omitted. Corollary 4.118 (Hartman-Type Inequality) Assume that n

αik = pk

i=1

for k = 1, 2, . . . , n. If fk ∈ C([a, b], R) for k = 1, 2, . . . , n and u is a nontrivial solution on [a, b] for (4.307), then the inequality  n  % 2Bn (b − a)Cn a

k=1

b

fk+ (s)ds

 >1

(4.316)

holds, where Bn and Cn are defined in (4.311). Remark 4.119 We do not know if the right-hand side of (4.316) can be replaced by a smaller one arbitrarily close to 1. This is an open problem for the readers. Remark 4.120 If we compare Theorems 1.1, 4.38, and 4.67 with Theorem 4.117 (or Corollary 4.118), then it is easy to see that the main difference between these results are the boundary conditions on the solution u. Therefore, they are different from each other. Remark 4.121 Since f + ≤ |f |, the integrals  a

b

fk+ (s)ds

for k = 1, 2, . . . , n in the above results can also be replaced by

(4.317)

4.6 Nonlinear Systems with Anti-periodic Boundary Conditions



b

283

|fk (s)| ds

a

for k = 1, 2, . . . , n, respectively. Remark 4.122 If we take n = 1 in (4.307), then we have the inequality  b 2p1 < f1+ (s)ds (b − a)p1 −1 a

(4.318)

from (4.316) in Corollary 4.118. It is easy to see from (4.317) that (4.318) is better than (3.140) in the sense that (3.140) follows from (4.318), but not conversely. Therefore, Corollary 4.118 with n = 1 improves Corollary 3.60 given by Wang [284]. Now, we present an application of the Lyapunov-type inequality obtained for (4.306). We get the following result which gives lower bounds for the nth component of any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.308). The proof of the following theorem is based on the above generalization of the Lyapunov-type inequality, as in that of Theorem 4.46 of Çakmak et al. [72], and hence it is omitted. Theorem 4.123 Assume that there exist nontrivial solutions (e1 , e2 , . . . , en ) of (4.309) and a function h1 such that h1 (λ1 , λ2 , . . . , λn−1 ) < |λn |

(4.319)

for any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.308), namely h1 (λ1 , λ2 , . . . , λn−1 ) =

0n−1 %

10 |λk |

k=1

ek

 n % 2Bn (b−a)Cn

b

ek 1−1/en |r(s)|ds

,

a

k=1

where Bn and Cn are defined in (4.311). Remark 4.124 Since h1 is a continuous function, h1 (λ1 , λ2 , . . . , λn−1 ) → ∞ as any eigenvalue λk → 0 for k = 1, 2, . . . , n − 1. Therefore, there exists a ball centered in the origin such that the generalized spectrum is contained in its exterior. Also, by rearranging terms in (4.319), we obtain  n  % 2Bn (b − a)Cn k=1

b a

−ek |r(s)|ds


1, fk , rk ∈ C([−s, s], R), rk (x) > 0 for k = 1, 2, . . . , n and x ∈ R, u = (u1 , u2 , . . . , un ) is a real nontrivial solution of (4.321) such that the boundary conditions uk (−s) = uk (s) = 0

(4.322)

for k = 1, 2, . . . , n are satisfied, uk for k = 1, 2, . . . , n are not identically zero on [−s, s], 1 < pk < ∞, and αki ≥ 0 for k, i = 1, 2, . . . , n. As an application, we also investigate lower bounds on the generalized eigenvalue (λ1 , λ2 , . . . , λn ) of the problem n %   |ui |αki = 0 rk (x)φpk (uk ) + λk r(x)φαkk (uk )

(4.323)

i=1 i=k

with the boundary conditions given in (4.322) for k = 1, 2, . . . , n and r ∈ C([−s, s], R). As usual, it is easier to find upper bounds for eigenvalues than lower bounds. In fact, they can be obtained by using elementary inequalities. Finding the estimated lower bounds is based on giving a suitable Lyapunov inequality for the corresponding systems. For authors who contributed to the existence of the generalized eigenvalues for the special case of (4.323), we refer to the paper by de Nápoli and Pinasco [107]. Note that, if we take αkk = pk , k = 1, 2, . . . , n, and for i = k, αki = 0 for i = 1, 2, . . . , n, then we obtain uncoupled equations, i.e., the half-linear secondorder differential equations   rk (x)φpk (uk ) + fk (x)φpk (uk ) = 0

for k = 1, 2, . . . , n

(4.324)

from (4.321). However, (2.111), which was considered by Watanabe et al. [287], does not reduce to (4.324). Moreover, when n = 1 in

4.7 Quasilinear Systems with Clamped-Free Boundary Conditions

285

⎧ n %   ⎪   ⎪ ⎪ |ui |αki = 0 for r (x)φ (u ) + f (x)φ (u ) pk k k αkk k ⎨ k ⎪ ⎪ ⎪ ⎩

k = 1, 2, . . . , n,

i=1 i=k

uk (−s) = uk (s) = 0

(4.325) with r1 (x) ≡ 1 and p1 = 2 or (2.111) and (2.114), we have the linear problem   u1 + f1 (x)u1 = 0

[−s, s],

on

u1 (−s) = u1 (s) = 0. Thus, we obtain the inequality 

s −s

f1+ (z)dz >

1 2s

(4.326)

from Theorem 2.52 with n = 1 given by Watanabe et al. [287]. In 2014, motivated by the recent studies of Çakmak and Tiryaki [72], Yang et al. [298] and Watanabe et al. [287], Çakmak [70] proved a new Lyapunov-type inequality for (4.325). Çakmak [70] proved a lemma which will be used in the proof of the main result. Lemma 4.125 If u is a nontrivial solution of (4.325), then the inequality  |uk (s)|
1

(4.331)

holds. Proof Let uk (−s) = 0 = uk (s) for k = 1, 2, . . . , n, where n ∈ N and uk for k = 1, 2, . . . , n are not identically zero on [−s, s]. Multiplying the kth equation in (4.321) by uk , integrating from −s to s, and then using (4.322), we get 

s −s

 p rk (z) uk (z) k dz =



s

−s

fk (z)

n %

|ui (z)|αki dz

i=1

for k = 1, 2, . . . , n. By using (4.327) in (4.332), we obtain

(4.332)

4.7 Quasilinear Systems with Clamped-Free Boundary Conditions



s −s

 p rk (z) uk (z) k dz ≤


0

(4.334)

for k = 1, 2, . . . , n. If (4.334) is not true, then 

s

−s

 p rk (z) uk (z) k dz = 0

for k = 1, 2, . . . , n. But then it follows that uk (x) ≡ 0

[−s, s]

on

(4.335)

for k = 1, 2, . . . , n. Combining (4.328) with (4.335), we obtain uk (s) ≡ 0 for k = 1, 2, . . . , n, which contradicts uk (s) ≡ 0 for k = 1, 2, . . . , n. Therefore, (4.334) holds for k = 1, 2, . . . , n. Thus, from (4.333) and (4.334), we have 

s

−s

 p rk (z) u (z) k dz

1−αkk /pk




z

−z

[r1 (v)]1/(1−p1 ) dv

1−p1 (4.341)

from (4.339) in Corollary 4.127. In addition to this, if we take p1 = 2 and r1 (x) ≡ 1 in (4.340), then (4.341) reduces to (4.326) given by Watanabe et al. [287]. Now, we present an application of the Lyapunov-type inequality obtained for (4.321). We obtain the following result, which gives lower bounds for the nth component of any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.323). The proof of this theorem is based on the above generalization of the Lyapunov-type inequality, as in that of Theorem 4.46 of Çakmak and Tiryaki [72], and hence it is omitted. Theorem 4.129 Assume that there exist nontrivial solutions (e1 , . . . , en ) of (4.330). Then, there exists a function h1 such that h1 (λ1 , λ2 , . . . , λn−1 ) < |λn |

(4.342)

for any generalized eigenvalue (λ1 , λ2 , . . . , λn ) of (4.323) with the boundary conditions given in (4.322), namely

h1 (λ1 , λ2 , . . . , λn−1 ) =

0n−1 %

1 |λk |

ek

k=1

0 ×

 n % k=1

s

−s

|r(z)|

n  % i=1

z

−z

[ri (v)]1/(1−pi ) dv

ek 1−1/en

αki (pi −1)/pi dz

.

Remark 4.130 Since h1 is a continuous function, h1 (λ1 , λ2 , . . . , λn−1 ) → ∞ as any eigenvalue λk → 0 for k = 1, 2, . . . , n − 1. Therefore, there exists a ball

290

4 Nonlinear Differential Systems

centered in the origin such that the generalized spectrum is contained in its exterior. Also, by rearranging terms in (4.342), we obtain 0 n % k=1

s

−s

|r(z)|

n  % i=1

z

−z

[ri (v)]

1/(1−pi )

1−ek

αki (pi −1)/pi dv

dz


0 for s < (a + b)/2 and f  (s) < 0 for s > (a + b)/2. The proof is complete.  We now give the main result, i.e., the Lyapunov inequality for the fractional case. Theorem 5.5 (Lyapunov-Type Inequality) Let x be a solution of (5.2), where a < b. If x(t) = 0 for all t ∈ (a, b), then the inequality 

b a

holds.



4 |q(t)|dt > Γ (α) b−a

α−1 (5.6)

5.2 Linear FDEs with Dirichlet Boundary Conditions

297

Proof Let B = C([a, b]) be the Banach space endowed with norm u = sup |u(t)|.

(5.7)

t∈[a,b]

It follows from Lemma 5.3 that a solution of (5.2) satisfies the integral equation 

b

x(t) =

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

(5.8)

a

Thus  x ≤ x max

b

t∈[a,b] a

G (t, s)|q(s)|ds,

i.e.,  1 ≤ max

b

t∈[a,b] a

G (t, s)|q(s)|ds.

Appealing now to the properties of Green’s function G (t, s) provided in Lemma 5.4, we get 1 Γ (α)



b−a 4

α−1 

b

|q(t)|dt > 1,

a



from which (5.6) follows. Remark 5.6 Note that if we let α = 2 in (5.6), then we obtain (1.2) for (1.1).

Now, we present an application of Theorem 5.5. More specifically, we show how (5.6) can be used to determine intervals for the real zeros of the Mittag–Leffler function Eα (z) =

∞

k=0

zk , Γ (kα + α)

where z ∈ C and Re(α) > 0. Let now a = 0 and b = 1 for simplicity and consider the fractional Sturm–Liouville eigenvalue problem (0 D α x)(t) + λx(t) = 0

(5.9)

satisfying Dirichlet boundary conditions x(0) = x(1) = 0.

(5.10)

298

5 Fractional Differential Equations

In contrast with the classical case (when α = 2), there is not a solid (spectral) theory for this problem when α < 2. We refer the reader to the works [216, 242], where some results about the existence of eigenvalues to fractional Sturm–Liouville problems are presented, although not necessarily for (5.9). Now, by Kilbas et al. [184, Corollary 5.1], we know that the eigenvalues λ ∈ R of  (0 D α x)(t) + λx(t) = 0 on [0, 1], (5.11) x(0) = x(1) = 0 are the solutions of Eα (−λ) = 0,

(5.12)

and the corresponding eigenfunctions are given by x(t) = t α−1 Eα (−λt α ) = 0,

(5.13)

where t ∈ [0, 1]. To be more precise, we should mention that [184, Corollary 5.1] provides only the explicit formula for the solution of (5.9). It is nevertheless true that the eigenfunctions given in (5.13) are nontrivial since they are not differentiable at t = 0. By Theorem 5.5, if λ ∈ R is an eigenvalue of (5.11), i.e., λ is a zero of (5.12), then |λ| > Γ (λ)4α−1 . Theorem 5.7 Let 1 < α ≤ 2. Then, the Mittag–Leffler function Eα (z) has no real zeros for |z| ≤ Γ (λ)4α−1 . In 2014, Ferreira [133] obtained a Lyapunov-type inequality for the case when one is dealing with a fractional differential boundary value problem. He then used that result to obtain an interval in which a certain Mittag–Leffler function has no real zeros. Let us start by introducing the concepts of the Caputo fractional derivative of order α. Definition 5.8 The Caputo fractional derivative of order α ≥ 0 is defined explicitly by 0 (C a D f )(t) = f (t)

and α (C a D f )(t)

= (a I

m−α

1 D f )(t) = Γ (m − α)



t

m

a

(t − s)m−α−1 f (m) (s)ds

5.2 Linear FDEs with Dirichlet Boundary Conditions

299

for t ∈ [a, b], where α > 0, and m is the smallest integer greater or equal than α. We generalize and derive some consequences of the well-known Lyapunov theorem (cf. [202]). More formally, the following result is proved below. To state the result, let us consider the fractional equation α (C a D x)(t) + q(t)x(t) = 0

(5.14)

together with Dirichlet boundary conditions, i.e., 

α (C a D x)(t) + q(t)x(t) = 0 on

[a, b],

x(a) = x(b) = 0.

(5.15)

Theorem 5.9 (Lyapunov-Type Inequality) Suppose q ∈ C([a, b], R). Let x be a solution of (5.15), where a < b and 1 < α ≤ 2. If x(t) = 0 for all t ∈ (a, b), then the inequality 

b

|q(t)|dt >

a

α α Γ (α) [(α − 1)(b − a)]α−1

(5.16)

holds. The search for Lyapunov-type inequalities in which the differential equation depends on a fractional differential operator was done first in one research paper on the subject [132]. There, a Lyapunov-type inequality was obtained for the case when the differential equation depends on the Riemann–Liouville fractional derivative, i.e., for (5.2). Let us point out that (5.2) is not equivalent to (5.15): By Kilbas et al. [184, page 91], we know that α α (C a D x)(t) = (a D x)(t) −

x(a) x  (a) (t − a)−α − (t − a)1−α . Γ (1 − α) Γ (2 − α)

α α Since x(a) = 0, in order that (C a D x)(t) = (a D x)(t), it is necessary that  y (a) = 0. But, in such a case, the unique solution of (5.14) (or (5.1)) is the trivial one (cf. [184, Section 3.5.1]), and we recall that we are interested only in nontrivial solutions. The interest of this section does not lie only in the fact that the problems are not equivalent. The best strategy known so far to deduce the Lyapunov inequality within this fractional differential setting seems to be to convert (5.15) into an equivalent integral form and then to find the maximum value of its Green function. It turns out that deducing this maximum value for the case where the differential equation depends on the Riemann–Liouville derivative is very similar to the classical (ordinary derivative) case [132], while the picture is quite different for the Caputo derivative case. By elementary analysis, it takes much more effort to obtain the Lyapunov inequality in the latter case. Although it is harder to obtain the aforementioned inequality for the Caputo case, the result also leads to different

300

5 Fractional Differential Equations

applications (specifically, a different Mittag–Leffler function from the one in [132] is considered (cf. Theorem 5.13 below)), as was expected in view of the discussion (that the problems are not equivalent) made above. We now write (5.15) in its equivalent integral form. Lemma 5.10 x ∈ C([a, b]) is a solution of (5.15) if and only if x satisfies the integral equation 

b

x(t) =

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

a

where

G (t, s) =

⎧ t −a α−1 ⎪ ⎪ − (t − s)α−1 ⎨ b − a (b − s)

if a ≤ s ≤ t ≤ b,

1 t −a Γ (α) ⎪ ⎪ ⎩ (b − s)α−1 b−a

if a ≤ t ≤ s ≤ b. (5.17)

Proof It is a standard result within the fractional calculus theory involving the Caputo differential operator that (see [309, Section 2]) x is a solution of (5.14) if and only if x(t) = c0 + c1 (t − a) −

1 Γ (α)



t

(t − s)α−1 q(s)x(s)ds

(5.18)

a

for some real constants c0 and c1 . Since x(a) = 0, we immediately get that c0 = 0. Now, x(b) = 0 is equivalent to c1 (b − a) −

1 Γ (α)



b

(b − s)α−1 q(s)x(s)ds = 0,

a

i.e., c1 =

1 (b − a)Γ (α)



b

(b − s)α−1 q(s)x(s)ds.

a

Hence, (5.18) becomes t −a x(t) = (b − a)Γ (α)



b

(b − s)

a

which concludes the proof.

α−1

1 q(s)x(s)ds − Γ (α)



t

(t − s)α−1 q(s)x(s)ds,

a



The following result contains the most important feature (for our purposes) of Green’s function G (t, s) defined in (5.17).

5.2 Linear FDEs with Dirichlet Boundary Conditions

301

Lemma 5.11 The Green function G (t, s) defined in (5.17) satisfies the inequality |G (t, s)| ≤

(α − 1)α−1 (b − a)α−1 , α α Γ (α)

(5.19)

where (t, s) ∈ [a, b] × [a, b]. Equality holds if and only if t =s=

b + (α − 1)a . α

Proof Let us start defining two functions g1 (t, s) :=

t −a (b − s)α−1 − (t − s)α−1 b−a

for a ≤ s ≤ t ≤ b and g2 (t, s) :=

t −a (b − s)α−1 b−a

for a ≤ t ≤ s ≤ b. We begin with the function g2 , which is easier to treat. Obviously, g2 satisfies the inequalities 0 ≤ g2 (t, s) ≤ g2 (s, s). Now, we differentiate g2 (s, s) on (a, b) and we obtain, after some simplifications, g2 (s, s) =

1 (b − s)α−2 {b − s − (α − 1)(s − a)}. b−a

Note that g2 (s, s) has a unique zero, attained in the point s∗ =

b + (α − 1)a , α

and it is easily seen that g2 (s, s) > 0 on (a, s ∗ ) and g2 (s, s) < 0 on (s ∗ , b). By the continuity of g2 (t, s), we conclude that max g2 (s, s) = g2 (s ∗ , s ∗ ).

s∈[a,b]

Moreover, evaluating g2 (s ∗ , s ∗ ), we finally get g2 (t, s) ≤ g2 (s, s) ≤ g2 (s ∗ , s ∗ )



b + (α − 1)a b + (α − 1)a α−1 1 −a b− = b−a α α

302

5 Fractional Differential Equations



b + (α − 1)a − aα bα − b − (α − 1)a α−1 α(b − a) α

α−1 1 (α − 1)(b − a) = α α =

=

(α − 1)α−1 (b − a)α−1 . αα

Now, we turn our attention to the function g1 . We start by fixing an arbitrary s ∈ [a, b). Differentiating g1 (t, s) with respect to t, we get g1 (t, s) =

(b − s)α−1 − (α − 1)(t − s)α−2 , b−a

s < t.

(5.20)

It follows from (5.20) that g1 (t ∗ , s) = 0 if and only if 

(b − s)α−1 t =s+ (α − 1)(b − a) ∗

1/(α−2)

provided t ∗ ≤ b, i.e., as long as s ≤ b − (α − 1)(b − a). So, if s > b − (α − 1)(b − a) (i.e., g1 (t, s) has no zeros), then g1 (t, s) < 0, i.e., g1 (t, s) is strictly decreasing, and, since g1 (b, s) = 0, we conclude that max g1 (t, s) = g1 (s, s) = g2 (s, s),

t∈[s,b]

where s ∈ (b − (α − 1)(b − a), b]. It is easy to check that s∗ =

b + (α − 1)a ∈ (b − (α − 1)(b − a), b), α

and hence |g1 (t, s)| ≤

(α − 1)α−1 (b − a)α−1 αα

for b − (α − 1)(b − a) < s ≤ t ≤ b. It remains to verify the result when s ≤ b − (α − 1)(b − a), i.e., when t ∗ ≤ b. It is easily seen that g1 (t, s) ≥ 0 for t ≥ t ∗ . This, together with the fact that g1 (b, s) = 0, implies g1 (t ∗ , s) ≤ 0, and, therefore, in order to complete the proof, we only have to show that   ∗  (α − 1)α−1 g (t , s) < (b − a)α−1 1 αα

5.2 Linear FDEs with Dirichlet Boundary Conditions

303

for s ∈ [a, b − (α − 1)(b − a)]. We have  ⎞ ⎛ 1   α−1  α−2    1 (b − s) g1 (t ∗ , s) =  ⎝ − a ⎠ (b − s)α−1  b − a s + (α − 1)(b − a)  ⎞α−1   (b  − ⎝s + − s⎠   (α − 1)(b − a)  ⎞ ⎛ 1   α−1  α−2  α−2 (b − s)α−1 (b − s)α−1 ⎝ (b − s)α−1 = − −a ⎠ s+ (α − 1)(b − a) b−a (α − 1)(b − a) ⎛

= =



(b − s) (α − 1)

α−1 α−2

(α−1)2 α−2

(b − a)

α−1 α−2

2−α (α − 1)

α−1 α−2

1  α−2

− s)α−1

(b − a)

α−1 α−2

−

(b − s) (α − 1)

(b − s)

1 α−2

(α−1)2 α−2

(α−1)2 α−2

(b − a)

−

α−1 α−2

−

s−a (b − s)α−1 b−a

s−a (b − s)α−1 . b−a

Now, we define a function h by h(s) :=

2−α (α − 1)

α−1 α−2

(b − s) α−1

(b − a) α−2

(α−1)2 α−2

−

s−a (b − s)α−1 b−a

for s ∈ [a, b − (α − 1)(b − a)]. In order to analyze the monotonicity of h, we differentiate and check the sign of h in the interior of [a, b − (α − 1)(b − a)]. We get h (s) = − − =


1.

On the other hand, the function f defined on [1, 2) by f (α) = (2 − α)α α satisfies f  (α) = −α α + α α (2 − α)(log α + 1) = α α {(2 − α)(log α + 1) − 1} < α α {(2 − α)α − 1} = −α α (α − 1)2 < 0, where we have used the elementary inequality log x < x − 1 for x > 1. Since f (1) = 1, we conclude that f (α) < 1 for all α ∈ (1, 2). Therefore, (5.21) is shown to be true, and that completes the proof.  Now, we have the necessary tools to prove the Lyapunov-type inequality. Proof of Theorem 5.9 Let B = C([a, b]) be the Banach space endowed with the norm defined in (5.7). It follows from Lemma 5.3 that a solution of (5.15) satisfies (5.8) for t ∈ [a, b]. Clearly, q cannot be the zero function on [a, b] as otherwise x would be the trivial solution. Proceeding, we obtain

5.2 Linear FDEs with Dirichlet Boundary Conditions

 |x(t)| ≤

b

305

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

a

for t ∈ [a, b]. Now, an application of Lemma 5.4 yields x (α − 1)α−1 x < (b − a)α−1 Γ (α) αα



b

|q(s)|ds,

a



from which (5.16) follows. Remark 5.12 Note that one obtains (1.2) when α = 2 in (5.16). Consider now the two parameter Mittag–Leffler function Eα,β (z) =

∞

k=0

zk , Γ (kα + β)

(5.22)

where z, β ∈ C and Re(α) > 0. We can use Theorem 5.9 to obtain an interval in which the Mittag–Leffler function given in (5.22) with β = 2 has no real zeros. Formally, we prove the following theorem. Theorem 5.13 Let 1 < α ≤ 2. Then, the Mittag–Leffler function Eα,2 has no real zeros for  x ∈ −Γ (α)

αα , 0 . (α − 1)α−1

Proof Let a = 0 and b = 1 and consider (5.11). By Kilbas et al. [184, Corollary 5.7], we know that the eigenvalues λ ∈ R of (5.11) are the solutions of Eα,2 (−λ) = 0,

(5.23)

and the corresponding eigenfunctions (normalized by y  (0) = 1) are given by x(t) = tEα,2 (−λt α ),

t ∈ [0, 1].

It is clear that the zeros of (5.23) are positive (if they exist). By Theorem 5.9, if a real eigenvalue λ of (5.11) exists, i.e., λ is a zero of (5.23), then λ > Γ (α) which concludes the proof.

αα , (α − 1)α−1 

306

5 Fractional Differential Equations

5.3 Linear FDEs with Fractional Boundary Conditions In [181], Jleli and Samet considered (5.14) with mixed boundary conditions x(a) = x  (b) = 0

(5.24)

x  (a) = x(b) = 0.

(5.25)

or

For boundary conditions of the form (5.24) and (5.25), two Lyapunov-type inequalities were established, respectively, as  b Γ (α) (5.26) (b − t)α−2 |q(t)|dt ≥ max{α − 1, 2 − α}(b − a) a and 

b

(b − t)α−1 |q(t)|dt ≥ Γ (α).

(5.27)

a

Motivated by the above works, Rong and Bai [254] considered a Caputo fractional differential equation under boundary conditions involving the Caputo fractional derivative. More precisely, they considered the Caputo fractional differential equation of the form (5.14) satisfying the boundary conditions β x(a) = C a D x(b) = 0,

(5.28)

where 1 < α ≤ 2, 0 < β ≤ 1, and q : [a, b] → R is a continuous function. We write C α (a D x)(t) + q(t)x(t) = 0 on [a, b], (5.29) β x(a) = C a D x(b) = 0 as an equivalent integral equation, and then, by using some properties of its Green function, we are able to get a corresponding Lyapunov-type inequality. After that, we show that this inequality can be used to obtain a real interval in which a certain Mittag–Leffler function has no real zeros. The results obtained generalize the main results of Jleli and Samet [181]. The following results are standard within the fractional calculus theory involving the Caputo differential operator. Lemma 5.14 (See [241, Chapter 2]) Let γ > α > 0 and f ∈ C([a, b]). Then, C α γ a D (a I f )(t)

for t ∈ [a, b].

=a I γ −α f (t)

5.3 Linear FDEs with Fractional Boundary Conditions

307

Lemma 5.15 (See [309, Section 2]) Let x ∈ C([a, b]) and 1 < α ≤ 2. Then, aI

α C α (a D x)(t)

= x(t) + c0 + c1 (t − a)

for some real constants c0 and c1 . We begin by writing (5.29) in its equivalent integral form. Lemma 5.16 x ∈ C([a, b]) is a solution of (5.29) if and only if x satisfies the integral equation 

b

x(t) =

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

a

where G (t, s) = H (t, s)(b − s)α−β−1

(5.30)

and ⎧ Γ (2 − β)(t − a) ⎪ ⎪ ⎪ ⎪ Γ (α − β)(b − a)1−β ⎪ ⎪ ⎪ ⎪ ⎨ (t − s)α−1 (b − s)1+β−α H (t, s) = − ⎪ Γ (α) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Γ (2 − β)(t − a) ⎪ ⎩ Γ (α − β)(b − a)1−β

if

a ≤ s ≤ t ≤ b,

if

a ≤ t ≤ s ≤ b.

(5.31)

Proof From (5.29) and Lemma 5.15, we obtain x(t) = c0 + c1 (t − a) −

1 Γ (α)



t

(t − s)α−1 q(s)x(s)ds,

a

where c0 and c1 are some real constants. By the boundary condition x(a) = 0, we can obtain c0 = 0. Thus, we have x(t) = c1 (t − a) − (a I α qx)(t).

(5.32)

By (5.32), we get C β a D x(t)

c1 1 (t − a)1−β − = Γ (2 − β) Γ (α − β)



b

(t − s)α−β−1 q(s)x(s)ds.

a

β Since C a D x(b) = 0, we have from (5.33) that

Γ (2 − β) c1 = Γ (α − β)(b − a)1−β

 a

b

(b − s)α−β−1 q(s)x(s)ds

(5.33)

(5.34)

308

5 Fractional Differential Equations

holds. Substituting (5.34) into (5.32), we obtain  b Γ (2 − β) (b − s)α−β−1 (t − a)q(s)x(s)ds x(t) = Γ (α − β)(b − a)1−β a  t 1 (t − s)α−1 q(s)x(s)ds, − Γ (α) a 

which concludes the proof. Lemma 5.17 If 1 < α < 2 and 0 < β < 1, then Γ (α)
0. dt (t + k)2

(5.38)

k=0

Since α < 2, we get by (5.36) and (5.38) that ψ(α − t) < ψ(2 − t), i.e., Γ  (α − t) Γ  (2 − t) < . Γ (α − t) Γ (2 − t) Define a function f by f (t) =

Γ (α − t) , Γ (2 − t)

0 < t < α.

By (5.39), we get f  (t) =

−Γ  (α − t)Γ (2 − t) + Γ (α − t)Γ  (2 − t) > 0. Γ 2 (2 − t)

(5.39)

5.3 Linear FDEs with Fractional Boundary Conditions

309

Thus, f (0) < f (β) < f (1) for 0 < β < 1, which implies that (5.35) holds. Lemma 5.18 Assume 0 < β ≤ 1 and 1 < α ≤ 1 + β. Then, ⎧ Γ (2 − β) ⎪ ⎨ if a ≤ s ≤ t ≤ b, β Γ (α − β) |H (t, s)| ≤ (b − a) ⎪ ⎩ λ if a ≤ t ≤ s ≤ b,



(5.40)

where  λ = max

 Γ (2 − β) Γ (2 − β) 2 − α Γ (2 − β) 1 − , , × . Γ (α) Γ (α − β) Γ (α − β) α − 1 Γ (α − β)

(5.41)

Proof Throughout the proof, we consider β < 1 since when β = 1, it is reduced to the case given in [181]. For a ≤ t ≤ s ≤ b, we easily find |H (t, s)| ≤

Γ (2 − β)(s − a) Γ (2 − β)(b − a) Γ (2 − β)(b − a)β . ≤ ≤ 1−β 1−β Γ (α − β) Γ (α − β)(b − a) Γ (α − β)(b − a)

For convenience, define ψ(t, s) =

Γ (2 − β)(t − a) 1 − (t − s)α−1 (b − s)β−α+1 Γ (α) Γ (α − β)(b − a)1−β

for a ≤ s ≤ t ≤ b. Fixing an arbitrary s ∈ [a, b) and differentiating ψ(t, s) with respect to t, we obtain ψt (t, s) =

Γ (2 − β) (t − s)α−2 (b − s)β−α+1 , − 1−β Γ (α − 1) Γ (α − β)(b − a)

s < t.

(5.42)

From (5.42), we easily get that ψt (t ∗ , s) = 0 if and only if t∗ = s +



Γ (2 − β)Γ (α − 1) (b − s)α−β−1 × Γ (α − β) (b − a)1−β

1  α−2

(5.43)

provided t ∗ ≤ b, i.e., as long as s ≤ b − , where  =

Γ (α − β) Γ (2 − β)Γ (α − 1)



1 1−β

(b − a) < b − a

by (5.35). Hence, if s > b − , then ψt (t, s) < 0,

t ∈ (s, b).

(5.44)

310

5 Fractional Differential Equations

On the other hand, we have lim ψ(t, s) =

t→s +

Γ (2 − β)(s − a) Γ (α − β)(b − a)1−β

and ψ(b, s) =

Γ (2 − β) (b − s)β (b − a)β − . Γ (α − β) Γ (α)

Thus, we obtain |ψ(s, s)|
0. Obviously, Eα,γ (z) > 0 for all z ≥ 0. Hence, the real zeros of Eα,γ (z), if they exist, must be negative real numbers. Theorem 5.23 Let 0 < β ≤ 1 and 1 < α ≤ 1 + β. Then, the Mittag–Leffler function Eα,2−β has no real zeros for  α−β t∈ − ,0 , λ where the constant λ is defined in (5.41). Proof Let a = 0 and b = 1. Consider the fractional Sturm–Liouville eigenvalue problem 

C Dα x 0

+ λ0 x = 0 on

x(0) =

C D β x(1) 0

(0, 1),

= 0.

(5.53)

By the Laplace transform method as in [117, 212, 241], the general solution of the fractional differential equation in (5.53) can be written as x(t) = c0 Eα,1 (−λ0 t α ) + c1 tEα,2 (−λ0 t α ).

(5.54)

In the following, we use the general solution as in (5.54) and its fractional Caputo derivative C α 0 D x(t)

= c0 t −β Eα,1−β (−λ0 t α ) + c1 t 1−β Eα,2−β (−λ0 t α ).

(5.55)

By (5.54), (5.55), and the boundary conditions in (5.53), we obtain c0 = 0 and

c1 Eα,2−β (−λ0 ) = 0.

Thus, the eigenvalues λ0 ∈ R of (5.53) are the solutions of Eα,2−β (−λ0 ) = 0,

(5.56)

and the corresponding eigenfunctions are given by x(t) = tEα,2−β (−λ0 t α ),

t ∈ [0, 1].

By Theorem 5.19, if a real eigenvalue λ0 of (5.53) exists, i.e., −λ0 is a zero of (5.56), then

314

5 Fractional Differential Equations



1

λ0

(1 − t)α−β−1 dt ≥

0

1 , λ

that is, λ0 ≥

α−β , λ

where the constant λ is defined in (5.41), and this concludes the proof.



Remark 5.24 If β = 1, then Theorem 5.23 reduces to [181, Theorem 3.5].

5.4 Linear FDEs with Robin Boundary Conditions In [180], Jleli, Ragoub, and Samet considered Caputo fractional differential equations of the form (5.14) under Robin boundary conditions. More precisely, they considered (5.14) satisfying the boundary conditions x(a) − x  (a) = x(b) + x  (b) = 0,

(5.57)

and they obtained a corresponding Lyapunov-type inequality. This result is then used to obtain a real interval in which a linear combination of certain Mittag–Leffler functions has no real zeros. Throughout this section, we shall use the notations h1 (t, s) = 1 + t − a −

γ (t − s)α−1 , (b − s)α−1 + (α − 1)(b − s)α−2

h2 (t, s) = 1 + t − a, γ = b − a + 2. Now, let us write 

α (C a D x)(t) + q(t)x(t) = 0 on

[a, b],

x(a) − x  (a) = x(b) + x  (b) = 0

(5.58)

in its equivalent integral form. Lemma 5.25 x ∈ C([a, b]) is a solution of (5.58) if and only if x satisfies the integral equation 

b

x(t) = a

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

5.4 Linear FDEs with Robin Boundary Conditions

315

where (b − s)α−2 (b − s + α − 1) H (t, s) γ Γ (α)

G (t, s) =

(5.59)

with H (t, s) =

 h1 (t, s)

if a ≤ s ≤ t ≤ b,

h2 (t, s)

if a ≤ t ≤ s ≤ b.

(5.60)

Proof From Lemma 5.15, we have x(t) = c0 + c1 (t − a) −

1 Γ (α)



t

(t − s)α−1 q(s)x(s)ds,

(5.61)

a

where c0 and c1 are some real constants. Then, 1 x (t) = c1 − Γ (α) 



t

(α − 1)(t − s)α−2 q(s)x(s)ds.

a

Since x(a) = x  (a), we obtain c0 = c1 .

(5.62)

From the boundary condition x(b) = −x  (b), we get c1 =

1 γ Γ (α)



b

  (α − 1)(b − s)α−2 + (b − s)α−1 q(s)x(s)ds.

(5.63)

a

Using (5.61), (5.62), and (5.63), we obtain the desired result.



Lemma 5.26 For all (t, s) ∈ [a, b] × (a, b), we have   2−α |H (t, s)| ≤ max 1 + b − a, (b − a) − 1 . α−1

(5.64)

Proof It is easy to see that, for a ≤ t ≤ s ≤ b, we get 0 ≤ h2 (t, s) ≤ 1 + b − a. On the other hand, for a ≤ s ≤ t ≤ b, we have (α − 1)γ (t − s)α−2 ∂h1 (t, s) = 1 − . ∂t (b − s)α−1 + (α − 1)(b − s)α−2

(5.65)

316

5 Fractional Differential Equations

Hence, lim

t→s +

∂h1 (t, s) ∈ [−∞, 0). ∂t

(5.66)

Now, for fixed s ∈ (a, b), we study the variation of the function t → h1 (t, s) for t ∈ [s, b]. First, we have   ∂h1 (α − 1)γ =1− (t, s) . ∂t α−1+b−s t=b Let a ∗ = b + (1 − γ )(α − 1). In the remainder of the proof, we distinguish two possible cases according to the value of a ∗ . Case 1. If a ∗ ≤ a, then   ∂h1 (t, s) ≤ 0, s ∈ (a, b). (5.67) ∂t t=b From (5.65), (5.66), and (5.67), we deduce ∂h1 (t, s) ≤ 0, ∂t

s < t,

which yields h1 (b, s) = 1+b −a −

γ (b − s) ≤ h1 (t, s) ≤ h1 (s, s) ≤ 1+b −a. α−1+b−s

(5.68)

Observe that, in this case, we have 1 2−α ≤ . α−1 b−a

(5.69)

Using (5.69), we obtain (1 + b − a)(b − s) + (α − 1)(1 + b − a) − γ (b − s) = s − b + (α − 1)(1 + b − a) ≥ (b − a)(α − 2) + α − 1 ≥ 0, which implies h1 (b, s) ≥ 0.

(5.70)

5.4 Linear FDEs with Robin Boundary Conditions

317

From (5.68) and (5.70), we obtain 0 ≤ h1 (t, s) ≤ 1 + b − a. Case 2. If a < a ∗ ≤ b, then we consider two further possibilities. Subcase (i). If a ∗ ≤ s < b, then   ∂h1 (t, s) ≤ 0. ∂t t=b Therefore, we conclude h1 (b, s) ≤ h1 (t, s) ≤ h1 (s, s) ≤ 1 + b − a.

(5.71)

On the other hand, we have (1+b−a)(b−s+α−1)−γ (b−s) = (b−s)(1+b−a−γ ) + (α − 1)(1 + b − a) = s − b + (1 + b − a)(α − 1) ≥ (2 + b − a − γ )(α − 1) ≥ 0, which implies h1 (b, s) ≥ 0.

(5.72)

From (5.71) and (5.72), we deduce 0 ≤ h1 (t, s) ≤ 1 + b − a. Subcase (ii) If a < s < a ∗ , then   ∂h1 (t, s) > 0. ∂t t=b Hence, there exists t ∗ ∈ (s, b) such that

  ∂h1 (t, s) = 0. ∂t t=t ∗

As mentioned above, it is easy to verify that h1 (s, s) ≥ 0 and h1 (b, s) ≤ 0. This yields h1 (t ∗ , s) ≤ h1 (b, s) ≤ h1 (s, s) ≤ 1 + b − a.

318

5 Fractional Differential Equations

Then,   |h1 (t, s)| ≤ max −h1 (t ∗ , s), 1 + b − a .

(5.73)

Observe that   ∂h1 (t, s) =0 ∂t t=t ∗ is equivalent to γ (α − 1)(t ∗ − s)α−1 = (b − s)α−1 + (α − 1)(b − s)α−2 . Therefore, we get α−2 ∗ s t + +1−a α−1 α−1 α−2 a ≥ b+ +1−a α−1 α−1 α−2 + 1. = (b − a) α−1

h1 (t ∗ , s) =

(5.74)

Finally, using (5.73) and (5.74), we obtain   2−α |h1 (t, s)| ≤ max 1 + b − a, (b − a) − 1 , α−1 

which completes the proof. The following is the next main result.

Theorem 5.27 Let x be a solution of (5.58), where a < b. If x(t) = 0 for all t ∈ (a, b), then the inequality 

b

(b − t)α−2 (b − t + α − 1)|q(t)|dt ≥

a

γ Γ (α) M0

holds, where   2−α (b − a) − 1 . M0 = max 1 + b − a, α−1 Proof Let B = C([a, b]) be the Banach space endowed with the norm x∞ = max |x(t)|, t∈[a,b]

x ∈ B.

(5.75)

5.4 Linear FDEs with Robin Boundary Conditions

319

From Lemma 5.25, the solution of (5.58) can be written as x(t) =

1 γ Γ (α)



b

(b − s)α−2 (b − s + α − 1)H (t, s)q(s)x(s)ds.

a

Now, an application of Lemma 5.15 yields x∞ ≤ x∞

M0 γ Γ (α)



b

(b − s)α−2 (b − s + α − 1)|q(s)|ds,

a



from which the proof is complete.

Now, we give a nonexistence result of real zeros for a linear combination of certain Mittag–Leffler functions. Let α and β be fixed positive numbers. The complex function Eα,β (z), see (5.22), is analytic in the whole complex plane; according to [207, 208, 241], we refer to it as the Mittag–Leffler function with parameters (α, β). At this stage, using (5.75), we give an interval in which a linear combination of Mittag–Leffler functions has no real zeros. More precisely, we prove the following result. Theorem 5.28 Let 1 < α ≤ 2. Then, Eα,2 (z) + Eα,1 (z) + zEα,α (z) has no real zeros for z∈

 −3αΓ (α) ,0 , (1 + α)N0

where   2−α −1 . N0 = max 2, α−1 Proof Let (a, b) = (0, 1) and consider the fractional Sturm–Liouville eigenvalue problem C 0

D α x + λ1 x = 0

on

(0, 1),

x(0) − x  (0) = x(1) + x  (1) = 0.

(5.76)

The real values of λ1 for which a nontrivial solution of (5.76) exists are called eigenvalues of (5.76), and the corresponding solutions are called eigenfunctions. As established in [117], the eigenvalues of (5.76) must be positive; moreover, the positive number λ1 is an eigenvalue of (5.76) if and only if Eα,2 (−λ1 ) + Eα,1 (−λ1 ) − λ1 Eα,α (−λ1 ) = 0.

320

5 Fractional Differential Equations

Thanks to Theorem 5.27, if a positive real eigenvalue λ1 of (5.76) exists, then 

1

λ1

(1 − t)α−2 (α − t)dt ≥

0

3Γ (α) . N0

Hence,

1 3Γ (α) ≥ λ1 1 + , α N0 which concludes the proof.



5.5 Notes and References As far as fractional differential equations are concerned, the first Lyapunov-type inequality was given in 2013 in a paper by Rui Ferreira [132]. This is one of the results given in Sect. 5.2. The reader is invited to consult the monograph by Kilbas et al. [184] for some basic definitions related to fractional differential equations. To obtain a Lyapunov-type inequality for (5.2), we present an equivalent integral equation, and then, by making use of some properties of its Green function, such an inequality is obtained. After that, it is shown how this inequality can be used to prove that a Mittag–Leffler type function has no real zeros on a determined interval. The Green function for (5.2) was deduced by Bai and Lü [38] for a = 0 and b = 1. The results given in this section, up to Theorem 5.7, are taken from the paper by Ferreira [132]. In 2014, Ferreira [133] obtained a Lyapunov-type inequality for the case when one is dealing with a Caputo fractional differential boundary value problem. He then used that result to obtain an interval in which a certain Mittag–Leffler function has no real zeros. All the given results from Theorem 5.9 to Theorem 5.13 are taken from [133]. Motivated by the works of Jleli and Samet [181], who handled (5.14) with mixed boundary conditions of the form (5.24) or (5.25), Rong and Bai [254] considered a Caputo fractional differential equation under boundary conditions involving the Caputo fractional derivative, i.e., they obtained a Lyapunov-type inequalities for (5.14) satisfying boundary conditions of the form (5.28) (see Sect. 5.3). Some basic results for the theory of fractional calculus involving the Caputo differential operator are given in Lemmas 5.14 and 5.15, and they are taken from [241, Chapter 2] and [309, Section 2], respectively. The rest of the proofs are all from the paper [254] by Rong and Bai. Jleli et al. [180] considered Caputo fractional differential equations of the form (5.14) under Robin boundary conditions of the form (5.57), and they obtained a corresponding Lyapunov-type inequality. Their result is then used to obtain a real interval in which a linear combination of certain Mittag–Leffler functions has no real zeros. Section 5.4 is devoted to the paper [180].

Chapter 6

Lyapunov-Type Inequalities for Partial Differential Equations

6.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for partial differential equations. We also sketch some recent developments related to this type of inequalities. Section 6.2 is devoted to the study of Lp -Lyapunov-type inequalities (p ≥ 1) for linear partial differential equations under Neumann boundary conditions on bounded and regular domains in RN . It is proved how the relation between the quantities p and N/2 plays a crucial rôle by considering the subcritical (p ∈ [1, N/2)), supercritical (p ∈ (N/2, ∞)), and critical (p = N/2) cases. In the last part of Sect. 6.2, we present some results on the existence and uniqueness of solutions of nonlinear resonant problems in a domain Ω ⊂ RN . In Sect. 6.3, we consider two-dimensional nonlinear systems of partial differential equations. In Sect. 6.4, some basic univariate Lyapunov inequalities are transferred to the multivariate setting of a shell via the polar method by considering partial differential equations involving radial derivatives of functions on the closure of a spherical shell A ⊆ RN , N > 1. It is also emphasized that Lyapunov-type inequalities obtained in this section generalize and complement some classical results in the literature. Finally, in Sect. 6.5, a Lyapunov inequality for linear and quasilinear elliptic differential operators in N-dimensional domains is presented. Singular and degenerate elliptic problems involving the p-Laplace operator with zero Dirichlet boundary conditions are considered as well. As an application of the obtained inequalities, lower bounds for the first eigenvalue of the p-Laplacian are derived and compared with the usual ones in the literature.

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_6

321

322

6 Partial Differential Equations

6.2 Linear PDEs with Neumann Boundary Conditions This section is devoted to the study of Lp -Lyapunov-type inequalities (1 ≤ p ≤ ∞) for linear partial differential equations. More precisely, we treat the case of Neumann boundary conditions on bounded and regular domains in RN . It is proved that the relation between the quantities p and N/2 plays a crucial rôle. This fact shows a deep difference with respect to the case of ordinary differential equations. The linear study is combined with Schauder’s fixed point theorem to provide conditions about the existence and uniqueness of solutions for resonant nonlinear problems. The well-known Lyapunov inequality states that if q ∈ L1 (a, b), then a necessary condition for (1.1) with (1.5) to have nontrivial solutions is that (1.4) holds (see Sect. 1.2). An analogous result is true for Neumann boundary conditions. In fact, if we consider the linear problem   x + q(t)x = 0 on 

(a, b),



x (a) = x (b) = 0,

(6.1)

where q ∈ Λ0 and Λ0 is defined by the set of functions q ∈ L1 (a, b) \ {0} such that 

b

q(t)dt ≥ 0,

(6.2)

a

and (6.1) has nontrivial solutions, then 

b

q + (t)dt ≥

a

4 b−a

for any function a ∈ Λ0 (see [65, 74, 75, 279]). In the case of Neumann boundary conditions, (6.2) is necessary in order to obtain this result (see [74, Remark 4]). In [74], Cañada et al. generalized this result by considering, for each p with 1 ≤ p ≤ ∞, the quantity βp :=

inf

q∈Λ0 ∩Lp (Ω)

Ip (q),

where Ω = (a, b) and & & Ip (q) = &q + &p =



b

 + p q (t) dt

1/p ,

q ∈ Λ0 ∩ Lp (Ω), 1 ≤ p < ∞,

a

I∞ (q) = ess sup(q),

q ∈ Λ0 ∩ L∞ (Ω),

6.2 Linear PDEs with Neumann Boundary Conditions

323

and obtaining an explicit expression for βp as a function of p, a, and b. One of the main applications of Lyapunov inequalities is to give optimal nonresonance conditions for the existence (and uniqueness) of solutions of nonlinear boundary value problems at resonance [74, 210, 279]. To the best of our knowledge, similar results for partial differential equations have only been proved by Cañada et al. [76] in 2006. In this section, we carry out a complete qualitative study of this question, pointing out the important rôle played by the dimension of the problem. More precisely, we consider the linear problem ⎧ ⎪ ⎨−Δu(x) = q(x)u(x) ∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

(6.3)

where Ω ⊂ RN with N ≥ 2 is a bounded and regular domain, and the function q : Ω → R belongs to the set Λ defined as   N/2 q(x)dx ≥ 0 and (6.3) has nontrivial solutions Λ = q ∈ L (Ω) \ {0} : Ω

if N ≥ 3, and  ∗ Λ = q ∈ Lp (Ω) \ {0} for some p∗ ∈ [1, ∞) :  q(x)dx ≥ 0 and (6.3) has nontrivial solutions Ω

if N = 2. Since the positive eigenvalues of the eigenvalue problem ⎧ ⎪ ⎨−Δu(x) = λu(x) ∂u ⎪ ⎩ (x) = 0 ∂n

if

x ∈ Ω,

if

x ∈ ∂Ω

(6.4)

belong to Λ, the quantity βp :=

infp

q∈Λ∩L (Ω)

& +& &q & , p

1 ≤ p ≤ ∞,

(6.5)

is well defined and it is a nonnegative real number. (Here, ·p denotes the usual Lp -norm.) The first novelty of this section is that β1 = 0 for each N ≥ 2. Moreover, we prove that if N = 2, then βp > 0 for all p ∈ (1, ∞] and that if N ≥ 3, then βp > 0 if and only if p ≥ N/2. Also, for each N ≥ 2, βp is attained if p > N/2. These results show a great difference with respect to the ordinary case, where βp > 0 for each p ∈ [1, ∞]. Moreover, we prove some qualitative properties of βp > 0 such as the continuity and monotonicity with respect to p. It seems

324

6 Partial Differential Equations

difficult to obtain explicit expressions for βp , as a function of p, Ω and N, at least for general domains  (see [74, 100, 102] for the case N = 1). As in the ordinary case, we have imposed Ω q ≥ 0 in the definition of the set Λ. This is not a technical but a natural assumption for Neumann boundary conditions. In fact, under this positivity  condition on Ω q, there is no positive solution of (6.3) (see Remark 6.12). The section finishes with an application of the main linear result to nonlinear boundary value problems of the form ⎧ ⎪ ⎨−Δu(x) = f (x, u(x)) ∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

(6.6)

where Ω ∈ RN , N ≥ 2, is a bounded and regular domain and the function f : Ω¯ × R → R, (x, u) → f (x, u), satisfies the condition (H) f , fu are Carathéodory functions and fu (x, u) ≥ 0 in Ω¯ × R. The existence of a solution of (6.6) implies  f (x, s0 )dx = 0

(6.7)

Ω

for some s0 ∈ R. Trivially, conditions (H) and (6.7) are not sufficient for the existence of solutions of (6.6). Indeed, consider the problem ⎧ ⎪ ⎨−Δu(x) = λ1 u(x) + ϕ1 (x) ∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

(6.8)

where ϕ1 is a nontrivial eigenfunction associated to λ1 . Here, λ1 is the first positive eigenvalue of (6.4). The function f (x, u) = λ1 u(x) + ϕ1 (x) satisfies (H) and (6.7), but the Fredholm alternative theorem shows that there is no solution of (6.8). If, in addition to (H) and (6.7), f satisfies a nonuniform nonresonance condition of the type (H1 ) fu (x, u) ≤ β(x) in Ω¯ × R with β(x) ≤ λ1 in Ω and β(x) < λ1 in a subset of Ω of positive measure, then it has been proved in [210] that (6.6) has a solution. Let us observe that the supplementary condition (H1 ) is given in terms of β∞ . In this section, we provide supplementary conditions in terms of β∞ , where N/2 < p ≤ ∞, obtaining a generalization of [210, Theorem 2]. Moreover, we consider other situations, where the condition fu (x, u) ≤ 0 in Ω¯ × R is not necessary (see Theorem 6.14 below).

6.2 Linear PDEs with Neumann Boundary Conditions

325

6.2.1 Lyapunov-Type Inequalities This subsection is concerned with the existence of nontrivial solutions of (6.3), where Ω ⊂ RN , N ≥ 2, is a bounded and regular domain and the function q : Ω → R belongs to the set Λ. Obviously, the positive eigenvalues of (6.4) belong to Λ. Therefore, Λ is not empty, and (6.5) is a well-defined real number. Now, we state the main result of this section. Theorem 6.1 The following statements hold: (1) If N = 2, then βp > 0 if and only if 1 < p ≤ ∞. If N = 3, then βp > 0 if and only if N/2 ≤ p ≤ ∞. (2) If N/2 < p ≤ ∞, then βp is attained. In this case, any function q ∈ Λ∩Lp (Ω) in which βp is attained is of the form (i) q(x) ≡ λ1 if p = ∞, where λ1 is the first strictly positive eigenvalue of (6.4). (ii) q(x) ≡ |u(x)|2/(p−1) if N/2 < p < ∞, where u is a solution of the problem ⎧ 2/(p−1) ⎪ u(x) ⎨−Δu(x) = |u(x)|

if x ∈ Ω,

∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ ∂Ω.

(6.9)

(3) The mapping (N/2, ∞) → R, p → βp , is continuous and the mapping (N/2, ∞) → R,

p → |Ω|−1/p βp ,

is strictly increasing. (4) There always exist the limits lim βp

p→∞

and

lim

p→(N/2)+

βp ,

and they take the values (i) limp→∞ βp = β∞ if N ≥ 2, (ii) limp→(N/2)+ βp ≥ βN/2 > 0 if N ≥ 3, and limp→1+ βp = 0 if N = 3. Remark 6.2 Since any nontrivial solution of (6.9) is a C1 (Ω)-function with sign changes, we deduce that, for N/2 < p < ∞, any function q ∈ Λ ∩ Lp (Ω) in which βp is attained is a continuous nonnegative function which vanishes at some point of Ω.

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6 Partial Differential Equations

For the proof of Theorem 6.1, we will distinguish three cases: The subcritical case (1 ≤ p < N/2 if N ≥ 3, and p = 1 if N = 2), the supercritical case (p > N/2 if N ≥ 2), and the critical case (p = N/2 if N ≥ 3).

6.2.2 The Subcritical Case In this section, we study the subcritical case, i.e., 1 ≤ p < N/2 if N ≥ 3, and p = 1 if N = 2. In all those cases, we prove that βp = 0. Roughly speaking if, for instance, N ≥ 3, then the main idea of the proof is to take first a function u and to calculate the corresponding function q for which u is a solution of (6.3). Obviously, if u is smooth enough, then we must impose two conditions: ∂u (x) = 0 on x ∈ ∂Ω, ∂n (ii) the zeros of u are also zeros of Δu. For instance, if Ω = B(0, 1) (the open ball in RN of center zero and radius one), then we can take radial functions u(x) = f (|x|) of the form (i)

f (r) = αr −a − βr −b ,

r ∈ (ε, 1]

for certain a, b, α, β such that the two mentioned conditions are satisfied. Lemma 6.3 Let N ≥ 3 and 1 ≤ p < N/2. Then, βp = 0. Proof First of all, note that if we define Ω + x0 = {x + x0 : x ∈ Ω} for arbitrary x0 ∈ RN , then βp (Ω + x0 ) = βp (Ω). On the other hand, if we define rΩ = {rx : x ∈ Ω} for arbitrary r ∈ R+ , then βp (rΩ) = r N/p−2 βp (Ω). Hence, βp (rΩ) = 0 if and only if

βp (rΩ + x0 ) = 0.

¯ Then, we can suppose without loss of generality that B(0, 1) ∈ Ω. Take now arbitrary real numbers a > b > 0 satisfying a + b = N − 2 and choose 0 0 2

β = ε−a b

and

 b + 1 − ε−b a +1 . 2 2

a

Then, it is easy to check that u is a solution of (6.3), where q : Ω → R is the radial function defined by

q(x) =

⎧ 2Nα{−α|x|2 + β}−1 ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

if |x| ≤ ε,

ab|x|−2

if ε < |x| < 1,

0

if |x| ≥ 1,

x ∈ Ω.

It is easily seen that q(x) ≥ 0 and q ∈ L∞ (Ω). Hence, q ∈ Λ. Let us estimate the ¯ ε) Lp -norm of q. To this aim, taking into account that the maximum of q(x) in B(0, is attained at |x| = ε, we have

 qp ≤

B(0,ε)



wN = εN N

2Nα −αε2 + β



p



 + B(0,1)\B(0,ε)

Nab(ε−a−2 − ε−b−2 ) bε−a − aε−b

p

ab |x|2

p 1/p

(ab)N wN (1 − εN −2p ) + N − 2p

1/p . (6.10)

Then, βp is smaller than this expression. But (for fixed real numbers a > b > 0 with a + b = N − 2), we can take the limit when ε tends to zero in (6.10). Taking into account p < N/2, this gives βp ≤

ab(wN )1/p . (N − 2p)1/p

Finally, taking the limit when b tends to zero in the last formula, we get βp = 0. Lemma 6.4 Let N = 2 and p = 1. Then, β1 = 0. Proof As we have argued in Lemma 6.3, it is easy to check that β1 (rΩ + x0 ) = β1 (Ω)



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6 Partial Differential Equations

for every x0 ∈ R2 , r ∈ R+ . Then, we can suppose again without loss of generality ¯ that B(0, 1) ⊂ Ω. Take now an arbitrary real number K > log(4) and ε > 0 satisfying log(ε2 ) + K < 0. Define u : Ω → R as the radial function ⎧ 2 2 |x| /ε + log(ε2 ) + K − 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎨log(|x| ) + K u(x) = ⎪ ⎪ ⎪ −4(1 − |x|2 ) + 1 + K − log 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 + K − log 4

if

|x| ≤ ε,

if

ε < |x| ≤

if

1 < |x| < 1, 2

if

|x| ≥ 1,

1 , 2

x ∈ Ω.

Then, it is easy to check that u is a solution of (6.3), where q : Ω → R is the radial function defined by ⎧

−1 ⎪ |x|2 ⎪ 2 ⎪ −4 + log(ε ) + K − 1 ⎪ ⎪ ε2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨0 q(x)= ⎪

⎪ −1 ⎪ 8  ⎪ 2 ⎪ ⎪ 16 − −4(1 − |x| ) + 1 + K − log 4 ⎪ ⎪ |x| ⎪ ⎪ ⎪ ⎩ 0

if |x| ≤ ε, if ε < |x| ≤ if

1 , 2

1 < |x| < 1, 2

if |x| ≥ 1,

x ∈ Ω.

It is easily seen that q(x) ≥ 0 and q ∈ L∞ (Ω). Hence, q ∈ Λ. Let us estimate the L1 -norm of q as  q1 =

 q(x)dx +

B(0,ε)



ε

= 2π 0

q(x)dx B(0,1)\B(0,1/2)

−4rdr + 2π 2 2 r + ε log(ε2 ) + K − 1



1

1/2

(16r − 8)dr . −4(1 − r)2 + 1 + K − log 4

It is possible to evaluate the first integral and to estimate the second one to arrive at q1 ≤ 4π log

 1 (16 × 1−8)dr ε2 log(ε2 )+K − 1 +2π . 2 2 2 2 ε +ε log(ε )+K−1 1/2 −4(1−1/2) +1 + K − log 4

Then, β1 is smaller than this expression. But, for a fixed real number K > log 4, we can take the limit when ε tends to zero in this formula. This gives

6.2 Linear PDEs with Neumann Boundary Conditions

β1 ≤

329

8π . K − log 4

Finally, taking the limit when K tends to ∞, we conclude β1 = 0.



6.2.3 The Supercritical Case In this section, we study the supercritical case, i.e., p > N/2 if N ≥ 2. In all those cases, we prove that the positive quantity βp is attained. We begin by studying the case p = ∞. Lemma 6.5 β∞ is attained in a unique element q∞ ∈ Λ. Moreover, q∞ (x) ≡ λ1 , where λ1 is the first strictly positive eigenvalue of the Neumann eigenvalue problem. Proof If q ∈ Λ ∩ L∞ (Ω) and u ∈ H1 (Ω) is a nontrivial solution of (6.3), then 

 ∇u · ∇v = Ω

for all v ∈ H1 (Ω).

quv Ω

In particular, we have 





|∇u|2 = Ω

qu = 0.

qu2 , Ω

Ω

Therefore, for each k ∈ R, we have      |∇(u + k)|2 = |∇u|2 = qu2 ≤ qu2 + k 2 q Ω



Ω



Ω

Ω

qu2 + k 2

=

Ω



 q + 2k

Ω

 qu =

Ω

Ω

q(u + k)2 Ω

q + (u + k)2 .

≤ Ω

This implies  Ω

& & |∇(u + k)|2 ≤ &q + &∞

 (u + k)2 . Ω

Also, since u is a nonconstant solution of (6.3), u + k is a nontrivial function. Consequently

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6 Partial Differential Equations

 & +& &q &

≥

∞

Ω 

|∇(u + k)|2 . (u + k)2 Ω

Now, choose k0 ∈ R satisfying  (u + k0 ) = 0. Ω

Then,  & +& &q & ≥ ∞

Ω 

 |∇(u + k0 )|2 (u + k0 )

≥

2

inf

v∈X∞ \{0}

Ω 

Ω

|∇v|2 v2

= λ1 ,

q ∈ Λ,

(6.11)

Ω

where X∞

   1 = v ∈ H (Ω) : v=0 . Ω

Hence, β∞ ≥ λ1 . Since the constant function & λ1& is an element of Λ, we deduce β∞ = λ1 . Furthermore, if q ∈ Λ is such that &q + &∞ = λ1 , then all the inequalities of the previous proof become equalities. In particular, it follows from (6.11) that  |∇(u + k0 )|2

Ω 

(u + k0 )2

= λ1 .

Ω

The variational characterization of λ1 (this constant is the second eigenvalue of (6.4)) implies that u + k0 is an eigenfunction associated to λ1 . Therefore, −Δ(u + k0 ) = λ1 (u + k0 ) = q(x)u. Multiplying by u + k0 , we obtain  (λ1 − q)(u + k0 )2 ≤ 0. Ω

& & Since &q + &∞ = λ1 , we deduce  (λ1 − q)(u + k0 )2 = 0. Ω

6.2 Linear PDEs with Neumann Boundary Conditions

331

The unique continuation property of the eigenfunctions implies that u(x) + k0 vanishes in a set of measure zero, and therefore q(x) ≡ λ1 . This completes the proof.  Next, we concentrate on the case N/2 < p < ∞. We need some auxiliary lemmas. Lemma 6.6 Assume N/2 < p < ∞ and let    Xp = u ∈ H1 (Ω) : |u|2/(p−1) u = 0 . Ω

If Jp : Xp \ {0} → R is defined by 



Jp (u) =

|∇u|

1/p−1 |u|

2

Ω

2p/(p−1)

Ω

and mp := infXp \{0} Jp , then mp is attained. Moreover, if up ∈ Xp \ {0} is a minimizer, then up satisfies the problem ⎧  2/(p−1) ⎪ up (x) ⎨−Δup (x) = Ap (up ) up  ∂u ⎪ ⎩ p (x) = 0 ∂n

if

x ∈ Ω,

if

x ∈ ∂Ω,

(6.12)

where 

 2p/(p−1) up 

Ap (up ) = mp

−1/p .

Ω

Proof It is clear that for any u ∈ H1 (Ω), there exists some constant k ∈ R such that u + k ∈ Xp . Hence, mp is well defined. Now, let {un } ⊂ Xp \ {0} be a minimizing sequence. Since the sequence {kn un }, kn = 0, is also a minimizing sequence, we can assume without loss of generality that 

 2p/(p−1) up  =1 Ω

holds. Then, have

 2 is also bounded. On the other hand, since p > N/2, we Ω |∇un |



2
 0 such that q(x) ≡ M |u(x) + k0 |2/(p−1) . Hence, q(x) ≥ 0, and, consequently, Ω q > 0. Therefore, since   qu2 = q(u + k0 )2 , Ω

Ω

we deduce k0 = 0. Finally, if we define w(x) = M (p−1)/2 u(x), then |w(x)|2/(p−1) = q(x). Moreover, since u is a solution of (6.3) and w is a multiple of u, also w is a solution of (6.3), and, consequently, a solution of (6.9). The proof is complete. 

6.2.4 The Critical Case In this section, we study the critical case, i.e., p = N/2 if N ≥ 3. We prove that βp > 0 is attained.

334

6 Partial Differential Equations

Lemma 6.9 If N ≥ 3 and p = N/2, then βp > 0. Proof As in Lemma 6.5, if q ∈ Λ and u ∈ H1 (Ω) is a nontrivial solution of (6.3), then for each k ∈ R, we have    |∇(u + k)|2 ≤ q(u + k)2 ≤ q + (u + k)2 . Ω

Ω

Ω

It follows from the Hölder inequality that 

& & & & & & |∇(u + k)|2 ≤ &q + &N/2 &(u + k)2 &

N/(N −2)

Ω

holds. Also, since u is a nonconstant solution of (6.3), u + k is a nontrivial function. Consequently,

N/2

≥

N/2

 ≥ Ω

≥

2N/(N −2)

Ω

Now, choose k0 ∈ R satisfying & +& &q &

& &−1 & & |∇(u + k)|2 &(u + k)2 &



& +& &q &



Ω (u + k0 )

.

= 0. Then,

& &−1 & & |∇(u + k0 )|2 &(u + k0 )2 &

2N/(N −2)



inf

v∈X∞ \{0}

Ω

& &−2 & & |∇v|2 &(u + k0 )2 &

2N/(N −2)

=C

for all q ∈ Λ, where X∞

   1 = v ∈ H (Ω) : v=0 . Ω

Finally, the continuous inclusions X∞ ⊂ H1 (Ω) ⊂ L2N/(N −2) (Ω) give us C > 0, which completes the proof.



6.2.5 Qualitative Properties of βp In this section, we study some qualitative aspects of the function p → βp . Specifically, we prove some results of continuity, monotonicity and behavior of βp when p is near N/2 and ∞.

6.2 Linear PDEs with Neumann Boundary Conditions

335

Proof of (3) and (4) of Theorem 6.1 We first prove the continuity of βp in (N/2, ∞). To this aim, consider a sequence {pn } → & p& ∈ (N/2, ∞). Take & & a nonnegative function qp ∈ Λ ∩ Lp (Ω) such that &qp+ & = βp . By (2) of p

p Theorem & 6.1, & and &using & standard regularity arguments, we have qp ∈ L (Ω). & & & & Hence, qp p → qp p , and it follows that n

& & & & lim sup βpn ≤ lim sup &qp &p = &qp &p = βp n

holds. In order to obtain the inverse inequality, and using that βp = mp , consider a nonzero sequence {upn } ⊂ Xpn

   1 2/(pn −1) = u ∈ H (Ω) : |u| u=0 Ω

and Jpn (upn ) = βpn . We can suppose without loss of generality that & & &up & n

2pn /(pn −1)

=1

& & holds (and consequently, &upn &p∗ is bounded for some p∗ < 2N/(N − 2)). Hence,  Ω

  ∇up 2 = βp , n n

and we have that {upn } is bounded in H1 (Ω). Therefore, there exists u0 ∈ H1 (Ω) ∗ such that, up to a subsequence, {upn }  u0 in H1 (Ω) and {upn } → u0 in Lp (Ω) for every p∗ < 2N/(N − 2). So, u0 ∈ Xp . Using these facts, we have  lim inf βpn = lim inf 

Ω

  ∇up 2 n



|∇u0 |2

≥ Ω

 Ω

 2pn /(pn −1) up 

1/pn −1

n

1/p−1

|u0 |2p/(p−1) Ω

≥ βp , and the continuity of βp is proved. We now prove that the mapping [N/2, ∞) → R, p → |Ω|−1/p βp is strictly increasing in [N/2, ∞). To do this, take N/2 ≤ p∗ < p < ∞. Taking into account ∗

|Ω|−1/p f p∗ ≤ |Ω|−1/p f p for every f ∈ Lp (Ω) (with strict inequality if |f | is not constant), we have

336

6 Partial Differential Equations

& & & ∗ ∗ & |Ω|−1/p βp∗ ≤ |Ω|−1/p &qp &p∗ ≤ |Ω|−1/p &qp &p = |Ω|−1/p βp .   Since qp  is not constant, we have that the above inequality is strict. On the other hand, similar arguments of the continuity of βp in [N/2, ∞) give us lim βp = β∞ .

p→∞

To study the behavior of βp for p near N/2, let us observe that, since |Ω|−1/p βp is strictly increasing in [N/2, ∞), there exists the limit lim

p→(N/2)+

|Ω|−1/p βp ≥ |Ω|−2/N βN/2 .

Hence, there exists the limit lim

p→(N/2)+

βp ≥ βN/2 > 0

if N ≥ 3. Finally, let us consider the case N = 2. If we fix q ∈ Λ ∩ L∞ (Ω), then & & & & lim sup βp ≤ lim sup &q + &p = &q + &1 . p→1+

p→1+

But, when we prove β1 = 0, for N = 2, we have used nonnegative minimizing functions q ∈ Λ ∩ L∞ (Ω). Then, we can conclude lim βp = 0.

p→1+



This completes the proof. As an application of Theorem 6.1 to the linear problem ⎧ ⎪ ⎨−Δu(x) = q(x)u(x) + f (x) ∂u ⎪ ⎩ (x) = 0 ∂n

if

x ∈ Ω,

if

x ∈ ∂Ω,

(6.14)

we have the following corollary (see [74, Corollary 2.11] and [279, Theorem 3] for the ordinary case). & &  Corollary 6.10 Let q ∈ Lp (Ω)\{0} for some p > N/2, Ω q(x) ≥ 0, &q + &p < βp & & (or &q + &p = βp is not a minimizer of the Lp -norm in Λ). Then, for each f ∈ Lp (Ω), (6.14) has a unique solution.

6.2 Linear PDEs with Neumann Boundary Conditions

337

Remark 6.11 Let us observe that to prove Theorem 6.1, we have chosen nonnegative minimizing sequences in all the cases N = 2 and N ≥ 3, p = N/2. Therefore, if we define β˜p :=

inf

q∈Λ∩Lp (Ω)

qp ,

N =2

1 ≤ p ≤ ∞,

and

N ≥ 3 and

p = N/2,

then it is easily seen that β˜p = βp .

 Remark 6.12 In the definition of the set Λ, we have imposed Ω q ≥ 0. This is not a technical but a natural assumption for Neumann boundary conditions. Otherwise, the corresponding infimum is zero. To see this, note that if u ∈ H1 (Ω) is a positive nonconstant solution of (6.3) and we consider v = 1/u as test function in the weak formulation, then we obtain   1 1 = ∇u · ∇ qu , u u Ω Ω which implies 

 q=− Ω

Ω

|∇u|2 < 0. u2

With this in mind, if we take a nonconstant u0 ∈ C2 (Ω) such that ∂u0 /∂n(x) = 0 for all x ∈ ∂Ω, then, for large n ∈ N, we have that un = u0 + n is a positive qn p → 0 as nonconstant solution of (6.3) with qn = −Δu0 /(u0 + n). Clearly,  n → ∞ for every 1 ≤ p ≤ ∞, and, as we have seen before, Ω qn < 0. Remark 6.13 In this section, we have considered Neumann boundary conditions. In the case of Dirichlet conditions, it is possible to obtain analogous results in an easier way. To be more precise, consider the linear problem 

−Δu(x) = q(x)u(x)

if

x ∈ Ω,

u(x) = 0

if

x ∈ ∂Ω,

(6.15)

where Ω ⊂ RN , N ≥ 2, is a bounded and regular domain and the function q : Ω → R belongs to the set ΛD defined as the set of functions q ∈ LN/2 (Ω) \ {0} such that (6.15) has nontrivial solutions if N ≥ 3 and the set of functions q : Ω → R ∗ such that there exists p∗ ∈ [1, ∞) with q ∈ Lp (Ω) and (6.15) has nontrivial solutions if N = 2. Then, we can define the value βpD ≡

inf p

q∈ΛD ∩L (Ω)

& +& &q & , p

1 ≤ p ≤ ∞,

and it is possible to prove that all the assertions of Theorem 6.1 remain true if we replace βp by βpD and Neumann boundary conditions in (6.9) by Dirichlet

338

6 Partial Differential Equations

conditions. In fact, as in the Neumann case, it is possible to obtain a variational characterization of βpD for N/2 < p < ∞ as  βpD =

 |∇u|2

inf

u∈H10 (Ω)\{0}

Ω

1/p−1 |u|2p/(p−1)

.

Ω

If Ω is, moreover, a radial domain, then the previous minimization problem is related to a more general one which involves the Rayleigh quotient 

 |∇u|2

1/p−1 ρ(x)|u|2p/(p−1)

Ω

,

Ω

∗

where ρ ∈ Lp (Ω), q = N(p − 1)/(2p − N), is a positive function. This has been used in the study of the existence of nonsymmetric ground states of symmetric problems for nonlinear partial differential equations (see [62, 63, 259]).

6.2.6 Nonlinear Resonant Problems In this subsection, we give some results on the existence and uniqueness of solutions of (6.6) in a domain Ω ⊂ RN . As is seen below, Theorem 6.14 is a generalization of [210, Theorem 2] in the sense that the main hypothesis of f (x, u) in [210] is given in terms of an L∞ -restriction, while we give here a more general Lp -restriction for N/2 < p ≤ ∞. In the proof, the basic idea is to combine the results obtained in the previous subsections with Schauder’s fixed point theorem. In fact, once we have the results on the linear problem, the procedure is standard and may be seen, for example, in [74, 279]. Theorem 6.14 Let Ω ⊂ RN , N ≥ 2, be a bounded and regular domain. Suppose f : Ω¯ × R → R, (x, u) → f (x, u), satisfies (1) f, fu are Carathéodory functions and f (·, 0) ∈ Lp (Ω) for some N/2 < p ≤ ∞, (2) there exist functions α, β ∈ Lp (Ω) such that α(x) ≤ fu (x, u) ≤ β(x)

in Ω¯ × R

& & & & with &β + &p < βp (or &β + &p = βp and β(x) is not a minimizer of the Lp norm in Λ, where βp is given by Theorem 6.1). Moreover, we assume one of the conditions  α ≥ 0 with α ≡ 0 (6.16) Ω

6.2 Linear PDEs with Neumann Boundary Conditions

339

and 

¯ f (x, s0 )dx = 0 and fu (x, u(x)) ≡ 0, u ∈ C(Ω).

α ≡ 0, ∃s0 ∈ R : Ω

(6.17)

Then, (6.6) has a unique solution. Proof We first prove uniqueness. Let u1 and u2 be two solutions of (6.6). Then, the function v = u1 − u2 is a solution of the problem ⎧ ⎪ ⎨−Δv(x) = q(x)v(x) ∂v ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

where 

1

q(x) =

fv (x, u2 (x) + θ v(x))dθ.

0

& & Hence, α(x) ≤ q(x) ≤ β(x), and we deduce q ∈ Λ and &q + &p ≤ βp . Applying Theorem 6.1, we obtain v ≡ 0. Next, we prove existence. First, we write (6.6) in the equivalent form ⎧ ⎪ ⎨−Δu(x) = b(x, u(x))u(x) + f (x, 0) ∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

where the function b : Ω × R → R is defined by  b(x, z) =

1

fu (x, θ z)dθ. 0

Hence, α(x) ≤ b(x, z) ≤ β(x) for all (x, z) ∈ Ω × R, and our hypothesis permits us to apply Corollary 6.10 in order to have a well-defined operator T : X → X defined by T y = uy with uy being the unique solution of the linear problem ⎧ ⎪ ⎨−Δu(x) = b(x, y(x))u(x) + f (x, 0) ∂u ⎪ ⎩ (x) = 0 ∂n

if x ∈ Ω, if x ∈ ∂Ω,

¯ with the uniform norm. We now show that T is completely where X = C(Ω) continuous and that T (X) is bounded. The Schauder fixed point theorem then provides a fixed point for T , which is a solution of (6.6). The fact that T is completely continuous is a consequence of the compact embedding of the Sobolev

340

6 Partial Differential Equations

¯ It remains to prove that T (X) is bounded. Suppose, space W2,p (Ω) ⊂ C(Ω). contrary to our claim, that T (X) & is & not bounded. In this case, there would exist a sequence {yn } ⊂ X such that &uyn &X → ∞. Passing to a subsequence if necessary, we may assume that the sequence of functions {b(·, yn (·))} is weakly convergent in Lp (Ω) to a function q0 satisfying α(x) ≤ q0 (x) ≤ β(x) a.e. in Ω. If uy zn := & n& , &uy & n

X

then, passing to a subsequence if necessary, we may assume that zn → z0 strongly ¯ with z0 in X, where we have used again the compact embedding W2,p (Ω) ⊂ C(Ω) being a nonzero function satisfying ⎧ ⎪ ⎨−Δz0 (x) = q0 (x)z0 (x) ∂z ⎪ ⎩ 0 (x) = 0 ∂n

if

x ∈ Ω,

if

x ∈ ∂Ω.

Now, if we are assuming (6.16), then q0 ≡ 0, q0 ∈ Λ, and we obtain a contradiction with Theorem 6.1. If we assume (6.17), then there is no loss of generality if we suppose s0 = 0. (Otherwise, we can do the change of variables u(x) = v(x) + s0 and obtain a similar problem with the same original hypothesis.) Then, for every n ∈ N, we get 

 Ω

b(x, yn (x))uyn (x)dx = −

f (x, 0)dx = 0. Ω

Therefore, for each n ∈ N, the function uyn has a zero in Ω¯ and hence so does z0 .  Thus, q0 ≡ 0, q0 ∈ Λ, and we obtain again a contradiction with Theorem 6.1.

6.3 Two-Dimensional Nonlinear Systems of PDEs In this section, the two-dimensional nonlinear system of partial differential equations ⎧ ⎪ ∂ 2 x(s, t) ⎪ ⎪ = α1 (s, t)x(s, t) + β1 (s, t)|u(s, t)|γ −2 u(s, t), ⎨ ∂s∂t (6.18) ⎪ ⎪ ∂ 2 u(s, t) ⎪ β−2 ⎩ = −β2 (s, t)|x(s, t)| x(s, t) − α1 (s, t)u(s, t) ∂s∂t is considered. We shall assume the existence of a nontrivial solution (x, u) of (6.18) and furthermore, the conditions

6.3 Two-Dimensional Nonlinear Systems of PDEs

341

(i) γ > 1 and β > 1 are real constants, (ii) β1 , β2 : [s0 , ∞) × [t0 , ∞) ⊂ R2 → R are continuous functions such that β1 (s, t) > 0 for all (s, t) ∈ [s0 , ∞) × [t0 , ∞), (iii) α1 : [s0 , ∞) × [t0 , ∞) → R is a continuous function. In 2010, Chen et al. [80] established the following Lyapunov-type inequality for (6.18). Theorem 6.15 (Lyapunov-Type Inequality) If (6.18) has a real solution (x, u) such that

Let the hypotheses (i)–(iii) hold.

x(a, t) = x(b, t) = x(s, c) = x(s, d) = 0 for all

(s, t) ∈ [a, b] × [c, d] (6.19)

and ∂u(s, t) ∂x(s, t) ∂u(s, t) ∂x(s, t) × + × = 0, ∂s ∂t ∂t ∂s

x(s, t) ≡ 0

(6.20)

on ∈ [a, b] × [c, d], where a, b, c, d ∈ R with a < b and c < d, then 

b

a

+M



d

|α1 (s, t)| dtds

c



b

β/α−1



1/γ 

d

b



β1 (s, t)dtds a

c

a

c

d

β2+ (s, t)dtds

1/α ≥ 2,

(6.21)

where γ and α are Hölder conjugates of each other, i.e., 1/γ + 1/α = 1, M = max |x(s, t)|, a 1 and β2+ (s, t) ≤ |β2 (s, t)|. This contradiction shows that |x(s, t)| is bounded on I = [s0 , ∞) × [t0 , ∞). Therefore, there exists a positive constant K such that |x(s, t)| ≤ K for all (s, t) ∈ I . On the other hand, integrating the second equation in (6.33) over the first variable from τ to s and over the second variable from σ to t, respectively, we obtain u(s, t) − u(s, σ ) − u(τ, t) + u(τ, σ ) = −

 t σ

s

β2 (ξ, η)|x(ξ, η)|β−2 x(ξ, η)dξ dη.

τ

(6.37)

Notice that u(τ, t) is bounded on [t0 , ∞), u(s, σ ) is bounded on [s0 , ∞), and in view of the triangle inequality, (6.37) yields |u(s, t)| ≤ |u(τ, t) + u(s, σ ) − C| +

 t σ

s τ

|β2 (ξ, η)| |x(ξ, η)|β−1 dξ dη



≤ |u(τ, t) + u(s, σ ) − C| + K β−1 σ

∞ ∞ τ

|β2 (ξ, η)| dξ dη, (6.38)

6.4 Multivariate Lyapunov Inequalities

347

where C = u(τ, σ ) is a constant. Equation (6.38) implies that |u(s, t)| is bounded on I = [s0 , ∞) × [t0 , ∞) since 

∞ ∞

τ

|β2 (s, t)| dtds < ∞.

σ

It follows from lim sup{|x(s, t)| + |u(s, t)|} ≤ lim sup |x(s, t)| + lim sup |u(s, t)| that lim sup{|x(s, t)| + |u(s, t)|} is bounded on I = [s0 , ∞) × [t0 , ∞). This completes the proof. 

6.4 Multivariate Lyapunov Inequalities In this section, some basic univariate Lyapunov inequalities are transferred to the multivariate setting of a shell by using the polar method. Let A ⊆ RN , N ∈ N, be a spherical shell, i.e., A := B(0, R2 ) − B(0, R1 ),

0 < R1 < R2 .

Here, we have the ball B(0, R) := {x ∈ RN : |x| < R},

R > 0,

where | · | is the Euclidean norm. Also, S N −1 := {x ∈ RN : |x| = 1}, is the unit sphere in RN with surface area ωN :=

2π N/2 , Γ (N/2)

i.e.,  S N−1

dω =

2π N/2 , Γ (N/2)

where Γ (·) is the Gamma function. For x ∈ RN \ {0}, one can write uniquely x = rω, where r > 0, ω ∈ S N −1 , see [255, pp. 149–150]. Here, r = |x| > 0 and ¯ we have ω = x/r ∈ S N −1 . For F ∈ C(A),

348

6 Partial Differential Equations





 F (x)dx =

R2

F (rω)r S N−1

A

N −1

dr dω.

(6.39)

R1

¯ n ∈ N, then, for a fixed In particular, A¯ = [R1 , R2 ] × S N −1 . If f ∈ Cn (A), N −1 n ω∈S , f (·ω) ∈ C ([R1 , R2 ]). In 2011, Anastassiou [33] dealt with partial differential equations involving ¯ using the polar coordinates r, ω. The following radial derivatives of functions on A, multivariate Lyapunov inequality is the first in the literature. Theorem 6.18 (Lyapunov-Type Inequality) solution of the PDE

If f

∂ 2 f (x) + q(x)f (x) = 0, ∂r 2

¯ is a nontrivial ∈ C2 (A)

¯ x ∈ A,

(6.40)

¯ with where q ∈ C(A), f (∂B(0, R1 )) = f (∂B(0, R2 )) = 0, and f (x) = 0 for any x ∈ A, then  |q(x)|dx > A

8π N/2 R1N −1 . Γ (N/2)(R2 − R1 )

(6.41)

Proof One can rewrite (6.40) as ∂ 2 f (rω) + q(rω)f (rω) = 0, ∂r 2

(r, ω) ∈ [R1 , R2 ] × S N −1 ,

where q(·ω) ∈ C([R1 , R2 ]), ω ∈ S N −1 , such that f (R1 ω) = f (R2 ω) = 0,

ω ∈ S N −1 .

Also f (rω) = 0 for any r ∈ (R1 , R2 ) and ω ∈ S N −1 . So, for fixed ω ∈ S N −1 , by (1.2), we get 4 < R2 − R1 =

 

R2

|q(rω)|dr

R1 R2

r 1−N r N −1 |q(rω)|dr

R1

 ≤

R2

r R1

N −1

|q(rω)|dr R11−N ,

6.4 Multivariate Lyapunov Inequalities

349

that is, 

R2

r N −1 |q(rω)|dr >

R1

for all ω ∈ S N −1 and  R2  r S N−1

N −1



|q(rω)|dr dω >

R1

4R1N −1 R2 − R1

4R1N −1 R2 − R1



2π N/2 , Γ (N/2) 

which, by (6.39), proves (6.41).

Remark 6.19 It was noticed by Wintner [290] that in (1.2), we can replace |q| by q + , the nonnegative part of q. The same can happen in (6.41), that is, we have 

q + (x)dx > A

8π N/2 R1N −1 . Γ (N/2)(R2 − R1 )

The following result is analogous to that of Parhi and Panigrahi [230], i.e., Theorem 2.56. Theorem 6.20 (Lyapunov-Type Inequality) solution of the PDE

If f

∂ 3 f (x) + q(x)f (x) = 0, ∂r 3

¯ is a nontrivial ∈ C3 (A)

¯ x ∈ A,

(6.42)

¯ with f (∂B(0, R1 )) = f (∂B(0, R2 )) = 0, and f (x) = 0 for any where q ∈ C(A), x ∈ A, and if there exists d ∈ (R1 , R2 ) such that ∂ 2 f (∂B(0, d)) = 0, ∂r 2 then  |q(x)|dx > A

8π N/2 R1N −1 . Γ (N/2)(R2 − R1 )2

Proof One can rewrite (6.42) as ∂ 3 f (rω) + q(rω)f (rω) = 0, ∂r 3

(r, ω) ∈ [R1 , R2 ] × S N −1 ,

where q(·ω) ∈ C([R1 , R2 ]), ω ∈ S N −1 , such that f (R1 ω) = f (R2 ω) = 0,

ω ∈ S N −1 .

(6.43)

350

6 Partial Differential Equations

Also f (rω) = 0 for any r ∈ (R1 , R2 ) and ω ∈ S N −1 . Furthermore there exists d ∈ (R1 , R2 ) such that ∂ 2 f (dω) = 0, ∂r 2

ω ∈ S N −1 .

So, for a fixed ω ∈ S N −1 , by (2.132) given in Theorem 2.56, we get 4 < (R2 − R1 )2 =

 

R2

|q(rω)|dr

R1 R2

r 1−N r N −1 |q(rω)|dr

R1

 ≤

R2

R1

r N −1 |q(rω)|dr R11−N ,

that is, 

R2

r N −1 |q(rω)|dr >

R1

4R1N −1 (R2 − R1 )2

for all ω ∈ S N −1 and 



R2

r S N−1

N −1





|q(rω)|dr dω >

R1

4R1N −1 (R2 − R1 )2



2π N/2 , Γ (N/2) 

which, by (6.39), proves (6.43). The following result is analogous to that of Yang [295], i.e., Theorem 2.82.

¯ Suppose Theorem 6.21 (Lyapunov-Type Inequality) Let n ∈ N and q ∈ C(A). that there exists d ∈ (R1 , R2 ) such that ∂ 2n f (∂B(0, d)) = 0, ∂r 2n where f is a solution of the PDE ∂ 2n+1 f (x) + q(x)f (x) = 0, ∂r 2n+1

x ∈ A¯

(6.44)

satisfying ∂if ∂if (∂B(0, R )) = (∂B(0, R2 )) = 0, 1 ∂r i ∂r i

i = 0, 1, . . . , n − 1,

(6.45)

6.4 Multivariate Lyapunov Inequalities

351

and f (x) = 0 for any x ∈ A. Then,  |q(x)|dx > A

2n+2 π N/2 R1N −1 n! . Γ (N/2)(R2 − R1 )2n

(6.46)

Proof One can rewrite (6.44) as ∂ 2n+1 f (rω) + q(rω)f (rω) = 0, ∂r 2n+1

(r, ω) ∈ [R1 , R2 ] × S N −1 ,

where q(·ω) ∈ C([R1 , R2 ]), ω ∈ S N −1 , such that for d ∈ (R1 , R2 ), we have ∂ 2n f (dω) = 0, ∂r 2n

ω ∈ S N −1 .

Furthermore, we get ∂ i f (R1 ω) ∂ i f (R2 ω) = = 0, ∂r i ∂r i

ω ∈ S N −1 ,

i = 0, 1, . . . , n − 1, with f (rω) = 0, for any r ∈ (R1 , R2 ) and ω ∈ S N −1 . So, for a fixed ω ∈ S N −1 , by (2.169) given in Theorem 2.82, we get n!2n+1 < (R2 − R1 )2n =

 

R2

|q(rω)|dr

R1 R2

r 1−N r N −1 |q(rω)|dr

R1

 ≤

R2

R1

r N −1 |q(rω)|dr R11−N ,

that is, 

R2

r N −1 |q(rω)|dr >

R1

n!2n+1 R1N −1 (R2 − R1 )2n

for all ω ∈ S N −1 and 



R2

r S N−1

N −1

|q(rω)|dr dω >

R1



n!2n+1 R1N −1 (R2 − R1 )2n



2π N/2 , Γ (N/2)

which, by (6.39), proves (6.46). The following result is analogous to that of Çakmak [68], i.e., Theorem 2.88.



352

6 Partial Differential Equations

¯ Theorem 6.22 (Lyapunov-Type Inequality) Let n ∈ N, n ≥ 2 and q ∈ C(A). Suppose that the PDE ∂ n f (x) + q(x)f (x) = 0, ∂r n

x ∈ A¯

has a solution satisfying f (∂B(0, R1 )) = f (∂B(0, t2 )) = · · · = f (∂B(0, tn−1 )) = f (∂B(0, R2 )) = 0, where R1 = t1 < t2 < · · · < tn−1 < tn = R2 and f (tω) = 0, for all ω ∈ S N −1 and t ∈ (tk , tk+1 ), k = 1, 2, . . . , n − 1. Then, 

 |q(x)|dx > A

(n − 2)!nn R1N −1 (n − 1)n−1 (R2 − R1 )n−1



2π N/2 . Γ (N/2)

Proof The proof is similar to those of earlier theorems in this section and based on Theorem 2.88.  The following result is analogous to another result of Çakmak [68], i.e., Theorem 2.22. Theorem 6.23 (Lyapunov-Type Inequality) Let us consider on A¯ the PDE ∂ 2n f (x) + q(x)f (x) = 0, ∂r 2n

x ∈ A¯

(6.47)

satisfying the boundary conditions ∂ 2i f ∂ 2i f (∂B(0, R1 )) = (∂B(0, R2 )) = 0 2i ∂r ∂r 2i

(6.48)

¯ is a solution of (6.47) satisfying f (x) = 0 for i = 1, 2, . . . , n − 1. If f ∈ C2n (A) for any x ∈ A, then 

 |q(x)|dx > A

22n+1 R1N −1 (R2 − R1 )2n−1

Proof The proof is based on Theorem 2.22.



π N/2 . Γ (N/2) 

In 2012, He and Tang [158] obtained the following inequality for (2.46), which we will use below. Theorem 6.24 (Lyapunov-Type Inequality) Let x be a solution of (2.46) with a < b. If x(t) = 0 for all t ∈ (a, b), then the inequality

6.4 Multivariate Lyapunov Inequalities



b

353

|q(t)|dt >

a

23n−1 . (b − a)2n−1

(6.49)

holds. In 2014, motivated by the results of Anastassiou [33] and He and Tang [158], Akta¸s [26] transferred some known univariate Lyapunov-type inequalities to the multivariate setting of a shell via the polar method. His results are better than some results of Anastassiou for even-order partial differential equations. ¯ be a solution Theorem 6.25 (Lyapunov-Type Inequality) Let f ∈ C2n (A) of (6.47) satisfying (6.48). If f (x) = 0 for any x ∈ A, then the inequality 

 |q(x)|dx > A

23n−1 R1N −1 (R2 − R1 )2n−1



2π N/2 Γ (N/2)

(6.50)

holds. Proof Equation (6.47) can be rewritten as ∂ 2n f (rω) + q(rω)f (rω) = 0, ∂r 2n

(r, ω) ∈ [R1 , R2 ] × S N −1 ,

where q(·ω) ∈ C([R1 , R2 ]), ω ∈ S N −1 , such that the boundary conditions ∂ 2i f (R1 ω) ∂ 2i f (R2 ω) = =0 2i ∂r ∂r 2i are satisfied for i = 0, 1, . . . , n − 1 and ω ∈ S N −1 . In addition, f (rω) = 0 holds for any r ∈ (R1 , R2 ) and ω ∈ S N −1 . Thus, from (6.49) given in Theorem 6.24, we get 23n−1 < (R2 − R1 )2n−1 =

 

R2

|q(rω)|dr

R1 R2

r 1−N r N −1 |q(rω)|dr

R1

 ≤

R2

r

N −1

R1

|q(rω)|dr R11−N

for fixed ω ∈ S N −1 . Therefore, we have the inequality 

R2

R1

r N −1 |q(rω)|dr >

23n−1 R1N −1 (R2 − R1 )2n−1

354

6 Partial Differential Equations

for all ω ∈ S N −1 and 



R2

r S N−1

N −1



|q(rω)|dr dω >

R1

23n−1 R1N −1 (R2 − R1 )2n−1



2π N/2 , Γ (N/2) 

which, by (6.39), proves (6.50).

¯ be a solution Theorem 6.26 (Lyapunov-Type Inequality) Let f ∈ C2n (A) of (6.47) satisfying (6.45). If f (x) = 0 for any x ∈ A, then the inequality 

 |q(x)|dx > A

42n−1 (2n − 1)[(n − 1)!]2 R1N −1 (R2 − R1 )2n−1



2π N/2 Γ (N/2)



holds. Proof The proof can be obtained easily by using the same approach as Anastassiou [33]. Therefore, it is left to reader.  In 2014, Ji and Fan [178] generalized Theorems 6.25 and 6.26 to a more general class of even-order partial differential equations. Moreover, as we shall see by the end of this section, their results improve Theorem 6.25 significantly. Ji and Fan [178] considered the even-order partial differential equation ∂ 2n y(x) ∂ k y(x) + q (x) = 0, k ∂r k ∂r 2n n

(6.51)

k=0

¯ qk ∈ C(A), ¯ k = 0, 1, 2, . . . , n, and x ∈ RN . For the proofs of where y ∈ C2n (A), their results, they first considered the even-order linear ordinary differential equation x (2n) (t) +

n

qk (t)x (k) (t) = 0,

(6.52)

k=0

where qk ∈ C([a, b]), k = 0, 1, 2, . . . , n. In order to prove the main results, we need the following lemmas. Lemma 6.27 (See [288, Proposition 2.1]) Let M ∈ N and  HC = u : u(M) ∈ L2 (a, b), u(2k) (a) = u(2k) (b) = 0, 0 ≤ k ≤ (M − K)/2 . Then, there exists a positive constant C such that, for any u ∈ HC , the Sobolev inequality 2

 sup |u(t)| a≤t≤b



b

≤C a

   (M) 2 u (t) dt

6.4 Multivariate Lyapunov Inequalities

355

holds. Moreover, the best constant C = C(M) is C(M) =

(22M − 1)(2M)(b − a)2M−1 . 22M−1 π 2M

Lemma 6.28 (See [286, Theorem 1.2, Corollary 1.3]) Let M ∈ N and  HD = u : u(M) ∈ L2 (a, b), u(k) (a) = u(k) (b) = 0, 0 ≤ k ≤ M − 1 . Then, there exists a positive constant D such that, for any u ∈ HD , the Sobolev inequality 2

 sup |u(t)|

 ≤D a

a≤t≤b

b

   (M) 2 u (t) dt

holds. Moreover, the best constant D = D(M) is D(M) =

(b − a)2M−1 . (2M − 1)[(M − 1)!]2 42M−1

In the following, we need the Riemann zeta function defined by ζ (s) =

∞

1 , ns

s > 1.

(6.53)

n=1

Table 6.1 gives the first seven values of ζ (2n), C(n), and D(n). Table 6.1 The first seven values of ζ (2n), C(n), and D(n)

n 1 2 3 4 5 6 7

ζ (2n) π2 6 π4 90 π6 945 π8 9450 π 10 93555 691π 12 638512875 2π 14 18243225

C(n) b−a 4 (b − a)3 48 (b − a)5 480 17(b − a)7 80640 31(b − a)9 1451520 691(b − a)11 9123840 (214 − 1)(b − a)13 213 × 18243225

D(n) b−a 4 (b − a)3 192 (b − a)5 20480 (b − a)7 4128768 (b − a)9 1358954496 (b − a)11 664377753600 (b − a)13 13(6!)2 413

356

6 Partial Differential Equations

Proposition 6.29 (Lyapunov-Type Inequality) If (6.52) has a nonzero solution x satisfying the Lidstone boundary conditions in (2.3), then the inequality :

+

(22n − 1)ζ (2n)(b − a)2n−1 22n−1 π 2n

; n−1

(b − a)2n−k−1 k=0

22n−k−1 π 2n−k



b

1/2 [qn (t)]2 dt

a



b

(22n −1)(22(n−k) −1)ζ (2n)ζ (2(n − k))

|qk (t)| dt > 1

a

(6.54) holds, where ζ is the Riemann zeta function defined by (6.53). Proof Multiplying both sides of (6.52) by x(t) and integrating by parts from a to b and using the boundary value conditions in (2.3), we obtain 



b

x

(2n)

b

(t)x(t)dt = (−1)

n

a

=−

n 

x (n) (t)

!2

dt

a b

qk (t)x (k) (t)x(t)dt.

k=0 a

This yields 

b

x (n) (t)

!2

dt ≤

a

n 

k=0 a

 = a

b

b

    |qk (t)| x (k) (t)x(t) dt

n−1   

  |qn (t)| x (n) (t)x(t) dt +

b

k=0 a

    |qk (t)| x (k) (t)x(t) dt. (6.55)

Now, by using Lemma 6.27, for any t ∈ [a, b], k = 1, 2, . . . , n − 1, we get |x(t)| ≤





b

x

C(n)

(n)

(t)

!2

1/2 dt

(6.56)

a

and     (k)   x (t) ≤ C(n − k)

b

x

(n)

a

Substituting (6.56) and (6.57) into (6.55), we obtain

(t)

!2

1/2 dt

.

(6.57)

6.4 Multivariate Lyapunov Inequalities



b

x

(n)

(t)

!2

357





b

dt ≤ C(n)

x

a

(n)

(t)

!2

1/2 

a

+

n−1 

b

dt a



b

C(n)C(n − k)

    |qn (t)| x (n) (t) dt 

a

k=0

b

|qk (t)| dt

x (n) (t)

!2

dt.

a

(6.58) Now, by applying the Hölder inequality, we get 

    (n)  |qn (t)| x (t) dt ≤

b

a

1/2 

b

b

2

[qn (t)] dt

x

a

(n)

(t)

!2

1/2 dt

(6.59)

.

a

Substituting (6.59) into (6.58) and by using the fact that x is not a constant function, we obtain the strict inequality 

b

x (n) (t)

!2

  dt < C(n)

a

b

1/2  [qn (t)]2 dt

a

+

b

x (n) (t)

!2

dt

a

 n−1 

C(n)C(n − k)

b



a

k=0

b

|qk (t)| dt

x (n) (t)

!2

dt.

a

(6.60) b 2 Dividing both sides of (6.60) by a x (n) (t) dt, which can be proved to be positive by using the boundary conditions in (2.3) and the assumption that x(t) ≡ 0, we obtain   C(n)

b

1/2 2

[qn (t)] dt

+

a

 n−1 

C(n)C(n − k)

b

|qk (t)| dt > 1.

a

k=0



This is equivalent to (6.54). Thus, the proof is complete.

Proposition 6.30 (Lyapunov-Type Inequality) If (6.52) has a nonzero solution x satisfying the 2-point boundary conditions in (2.2), then the inequality : 1 (n − 1)!22n−1 +

n−1

k=0

holds.

(b − a)2n−1 2n − 1



b

1/2 [qn (t)]2 dt

a

(b − a)2n−k−1 √ (n − 1)!(n − k − 1)!42n−k−1 (2n − 1)(2n − 2k − 1)

 a

b

|qk (t)| dt > 1

358

6 Partial Differential Equations

Proof Since the proof is similar to that of Proposition 6.29, it is left to the reader.  Lemma 6.31 For any f ∈ C(A), we have  |f (x)|dx ≥ A

R1N −1 2π N/2 Γ (N/2)



R2

|f (rω)|dr.

R1

Proof Similar to the proofs given in [26] and [33], we have 

R2

 |f (rω)|dr =

R1

R2 R1

r 1−N r N −1 |f (rω)|dr ≤ R11−N



R2

r N −1 |f (rω)|dr,

R1

which implies 

 |f (x)|dx = A

 ≥ =

S N−1

S N−1



R2

|f (rω)|dr dω

R1



R11−N

R1N −1 2π N/2 Γ (N/2)

R2

|f (rω)|dr dω

R1



R2

|f (rω)|dr.

R1



This completes the proof. We have the following theorem.

Theorem 6.32 (Lyapunov-Type Inequality) If y is a nonzero solution of (6.51) satisfying (6.48), then the inequality :

(22n − 1)ζ (2n)(R2 − R1 )2n−1 Γ (N/2)



22n π 2n+N/2 R1N −1

1/2 [qn (x)]2 dx

A

; n−1

(R2 − R1 )2n−k−1 Γ (N/2) + (22n − 1)(22(n−k) − 1)ζ (2n)ζ (2(n − k)) 2n−k R N −1 π N/2 (2π ) 1 k=0  |qk (x)| dx > 1 × A

(6.61)

holds, where ζ is the Riemann zeta function defined in (6.53). Proof It follows from (6.54) and Lemma 6.31 that for any fixed ω ∈ S N −1 , we have :

(22n − 1)ζ (2n)(R2 − R1 )2n−1 1< 22n−1 π 2n



R2

1/2 2

[qn (rω)] dr R1

6.4 Multivariate Lyapunov Inequalities

+

; n−1

(R2 − R1 )2n−k−1 k=0



R2

×

22n−k−1 π 2n−k

359

(22n − 1)(22(n−k) − 1)ζ (2n)ζ (2(n − k))

|qk (rω)| dr

R1

:

(22n − 1)ζ (2n)(R2 − R1 )2n−1 Γ (N/2)

≤

 [qn (x)] dx

22n π 2n+N/2 R1N −1 +

A

; n−1

(R2 − R1 )2n−k−1 Γ (N/2) k=0



(2π )2n−k R1N −1 π N/2

1/2 2

(22n − 1)(22(n−k) − 1)ζ (2n)ζ (2(n − k))

|qk (x)| dx,

× A



which is (6.61). This completes the proof.

Theorem 6.33 (Lyapunov-Type Inequality) If y is a nonzero solution of (6.51) satisfying (6.48), then the inequality 1 (n − 1)!22n−1 +

n−1

k=0



:

(R2 − R1 )2n−1 Γ (N/2) (2n − 1)R1N −1 2π N/2



1/2 2

[qn (x)] dx A

(R2 − R1 )2n−k−1 Γ (N/2) √ (n − 1)!(n − k − 1)!42n−k−1 R1N −1 2π N/2 (2n − 1)(2n − 2k − 1)

|qk (x)| dx > 1

× A

holds. Proof The proof is similar to that of Theorem 6.32, so we omit it for simplicity.  Let us compare Theorems 6.32 and 6.33 with Theorems 6.25 and 6.26. It is evident that Theorem 6.33 is a natural generalization of Theorem 6.26. If we let qn (x) = qn−1 (x) = · · · q1 (x) = 0 and q0 (x) = q(x) for all x ∈ A, then (6.61) reduces to the inequality  |q(x)|dx > A

22n−1 π 2n 2π N/2 N −1 R × . Γ (N/2) 1 (22n − 1)ζ (2n)(R2 − R1 )2n−1

Let us compare the right-hand sides of (6.50) and (6.62): If we denote δn =

(22n

22n−1 π 2n , − 1)ζ (2n)23n−1

(6.62)

360

6 Partial Differential Equations

Table 6.2 The first eight values of δn 1 1.00

n δn

2 1.50

3 1.42

4 2.32

5 2.86

6 3.53

7 4.35

8 5.37

then π 2n π 2n > 3n = δn = n 2n 2 (2 − 1)ζ (2n) 2 ζ (2n)



π2 8

n

1 → ∞ as ζ (2n)

n → ∞,

since ζ (2n) → 1 as n → ∞. Table 6.2 gives the first eight values of δn . From Table 6.2, we see that δn increases very quickly, so Theorem 6.32 improves Theorem 6.25 significantly even in the special case of (6.47).

6.5 Linear and Quasilinear Elliptic Differential Operators In this section, a Lyapunov inequality for linear and quasilinear elliptic differential operators in N-dimensional domains Ω is presented. Singular and degenerate elliptic problems with Ap coefficients involving the p-Laplace operator with zero Dirichlet boundary condition are also considered. As an application of the inequalities obtained, lower bounds for the first eigenvalue of the p-Laplacian are derived and compared with the usual ones in the literature. In his classical work [202], Lyapunov proved that, given a continuous periodic and positive function w with period L, the solution u of the ordinary differential equation u + w(t)u = 0, in (−∞, ∞), was stable if 

L

w(t)dt < 4.

L 0

Then, Borg in [59], introduced the Lyapunov inequality in his proof of stability criteria for sign-changing weights w. He showed that the inequality  0

L

w(t)dt ≥

4 L

(6.63)

must be satisfied in order to have a nontrivial solution in [0, L] ⊂ R of the problem 

u + w(t)u = 0, u(0) = u(L) = 0.

(6.64)

Since then, this was rediscovered and generalized many times. Inequality (6.63) was applied in stability problems, oscillation theory, a priori estimates, other inequalities,

6.5 Linear and Quasilinear Elliptic Differential Operators

361

and eigenvalue bounds for ordinary differential equations. Different proofs of this inequality have appeared in the literature: The proof of Patula [231] by direct integration, or the one of Nehari [219] showing the relationship with Green’s functions, among several others. See the survey [65] for other proofs. In the nonlinear setting, the inequality 

L

w(t)dt ≥

0

2p Lp−1

(6.65)

generalized Lyapunov inequality (6.63) to p-Laplacian problems,   p−2   (|u | u ) + w(t)|u|p−2 u = 0, u(0) = u(L) = 0. Here, w ∈ L1 and 1 < p < ∞. For p = 2, we recover the linear problem (6.64). Several proofs were given in the last years, see [195, 228, 235, 294], although it seems to be derived first by Elbert [122]. Later, this was extended in [106] to nonlinear operators in Orlicz spaces generalizing the p-Laplacian, − (ϕ(u )) = λr(t)ϕ(u),

(6.66)

where ϕ is a convex nondecreasing function, such that sϕ(s) satisfies the Δ2 condition. Moreover, in [107], this was also extended to systems of resonant type (see [46]) involving p-Laplacians and q-Laplacians. The interested reader is referred to the book [240] for a review of developments in these problems. Besides the one-dimensional case, there are few works devoted to similar inequalities for partial differential equations. An exception is the work of Cañada, Montero, and Villegas [76], where they considered the problem ⎧ ⎪ ⎨Δu + w(x)u = 0 ∂u ⎪ ⎩ ∂η

if x ∈ Ω, if x ∈ ∂Ω,

(6.67)

and a nonexistence result was obtained for general domains. The authors gave some bounds involving the second Neumann eigenvalue μ2 . However, it is well known that μ2 fails to reflect geometric properties of Ω and can be made arbitrarily close to zero by adding a slight perturbation of the domain as in [99]. Also, some results of Egorov and Kondriatev, included in their book [120], contain Lyapunov-type inequalities for higher-order linear differential operators. In 2016, de Nápoli and Pinasco [108] proved a Lyapunov inequality for N-dimensional (linear and quasilinear) elliptic operators with zero Dirichlet boundary conditions, reflecting more geometric information than the measure of the domain. The toy model is the p-Laplace operator, and we consider here the problem

362

6 Partial Differential Equations

 Δp u(x) + w(x)|u(x)|p−2 u(x) = 0

if x ∈ Ω,

u(x) = 0

if x ∈ ∂Ω.

(6.68)

As usual, we denote Δp u = div(|∇u|p−2 ∇u) for any 1 < p < ∞, and the weight w ∈ Ls for some s depending on p and N . At the end of this section, in Sect. 6.5.4, we include some useful facts about the eigenvalues of the p-Laplace operator. Denote rΩ = max dΩ (x), x∈Ω

where dΩ (x) = d(x, Ω c ) = inf |x − y| x∈∂Ω

is the distance from x ∈ Ω to the boundary. Let us note that the length L of the interval in inequality (6.65) can be thought of as the measure of the interval, but it can be understood also as twice the inner radius of the interval, by rewriting the inequality as 

L

q(t)dt ≥ 2

0

p−1 2 . L

The main objective here is to derive some Lyapunov-type inequalities involving the inner radius of the domain and norms of the weight w. This section is divided in two main parts. In the first part, we cover the case p > N and prove the existence of a Lyapunov inequality involving the L1 -norm of the weight and the inner radius of the domain. We also consider singular problems and prove a Morrey theorem for Ap weights. In the second part, the case p < N is analyzed. It is shown that there are Lyapunov-type inequalities involving the Ls norm for s > N/p. The case p = N is not considered. For p = N = 2, we mention the following two interesting results by Osserman [225]. Theorem 6.34 (Osserman [225]) Given a domain Ω ⊂ R2 of connectivity k ≥ 2, the first Dirichlet eigenvalue of the problem 

−Δu(x) = λu(x)

if

x ∈ Ω,

u(x) = 0

if

x ∈ ∂Ω

satisfies λ1 ≥

1 2 k 2 rΩ

.

6.5 Linear and Quasilinear Elliptic Differential Operators

363

Theorem 6.35 (Osserman [225]) Let Ω ⊂ R2 and Ωε be the domain obtained by removing from Ω a finite number of disjoint disks of radius ε centered at a fixed set E of points in Ω. Then lim λ1 (Ωε ) = λ1 (Ω).

ε→0

Clearly, Theorems 6.34 and 6.35 are enough to conclude that we cannot expect to obtain a general inequality involving the inner radius of the domain when p = N, although it would be very interesting to find a related inequality. Finally, the optimality of the bounds is shown, and we apply them to eigenvalue problems. We compare them with Sturmian and isoperimetric bounds. Also, some related inequalities of Anane [31] and Cuesta [101] are considered, which involves the measure of the set Ω. In Sect. 6.5.1, the case p > N is considered. There, we will prove the following result. Theorem 6.36 (Lyapunov-Type Inequality) Let Ω ⊂ R2 be an open set, let w ∈ 1,p L1 (Ω) be a nonnegative weight, and let u ∈ W0 (Ω) with p > N be a nontrivial solution of (6.68). Then N −p

wL1 (Ω) ≥ CrΩ

(6.69)

,

where C is an universal constant depending only on p and N . Let us note that the constant C is the same for any Ω ⊂ RN , since it is related to the constant given by Morrey’s theorem. Maybe it can be improved for particular domains. However, the power of the inner radius is optimal. Then, we consider the problem − div(v(x)|∇u|p−2 ∇u) = w(x)|u|p−2 u, where now v is a singular or degenerate weight, typically a power of the distance to the boundary or powers of |x| (as in Henon equations, and Caffarelli–Kohn– Nirenberg inequalities). Here, the problem is more subtle since we need the density of continuous functions in the weighted Sobolev space  1,p W0 (RN , v, w):= u ∈ L1loc (RN ) : w 1/p u ∈ Lp (RN ) and u1/p ∇u ∈ [Lp (RN )]N , where ∇u is a distributional gradient in the sense of Schwartz. Following [185], this is true when v = w belong to the Muckenhoupt class Ap , that is, v is a nonnegative function in L1loc (RN ) and there exists a constant cp,v such that 

 v(x)dx B

B

[v(x)]−1/(p−1) dx

p−1 ≤ cp,v |B|

(6.70)

364

6 Partial Differential Equations

for every ball B ∈ RN . The same argument applies for different weights v, w ∈ Ap , as it is shown in Lemma 6.45 below. So, the weights v, w ∈ At with t < p/N will be restricted, and in this case, the following Lyapunov-type inequality is proved. Theorem 6.37 (Lyapunov-Type Inequality) Let Ω ⊂ RN , and let v ∈ At (RN ) with t < p/N and v ≥ 0. Let us define 

[v(x)]−1/(t−1) dx.

g(rΩ ) = sup x∈Ω B(x,rΩ ) 1,p

Let u ∈ W0 (Ω) be a nontrivial solution of 

− div(v(x)|∇u(x)|p−2 ∇u(x)) = w(x)|u(x)|p−2 u(x)

if

x ∈ Ω,

u(x) = 0

if

x ∈ ∂Ω.

Then, we have the Lyapunov-type inequality p−tN

1 ≤ C(p, t, N)rΩ

 [g(rΩ )]t−1

w(z)dz,

(6.71)

Ω

where the constant C(p, t, N ) depends only on p, t, and N. Theorem 6.37 is based on the fact that At ⊂ Ap whenever t < p. Briefly, we will bound u by the fractional integral (or Riesz potential) of its gradient, and after adding the corresponding power of the coefficient, we wish to use Hölder’s inequality with exponents p in the gradient and an exponent close to p in | · |1−N . Remark 6.38 Theorem 6.37 can be thought as a Morrey embedding with Ap weights. No such result was proved before for the case p > N. For p < N, we refer the interested reader to the book of Turesson [275]. Although the terms in the Lyapunov inequality (6.71) seem difficult to compute, in certain interesting cases they are rather simple to compute. We choose as an example a coefficient which is a power of the distance to the boundary, v(x) = γ dΩ (x), and in this case, we obtain the very clean bound p−N −γ

1 ≤ CrΩ

 w(z)dz,

(6.72)

Ω

where C depends only on N, p, and γ . Of course, γ is restricted by the At condition. γ Let us recall that dΩ (x) ∈ At for −1 < γ < t − 1, see [223]. For 1 < p < N , a similar inequality cannot hold for arbitrary domains. Perhaps the easiest way to understand this is to remove a discrete set of points with zero capacity from a ball, and the first eigenvalue remains the same. So, in Sect. 6.5.2, the following weaker inequality will be proved.

6.5 Linear and Quasilinear Elliptic Differential Operators

365

Theorem 6.39 (Lyapunov-Type Inequality) Let Ω ⊂ RN be a smooth domain, 1,p N/p < s, and w ∈ Ls (Ω). Let u ∈ W0 (Ω) be a nontrivial solution of (6.68). Then, we have the Lyapunov inequality N/s−p

wLs (Ω) ≥ CrΩ

.

(6.73)

The constant C depends on p, N, and the capacity of RN \ Ω. The proof of Theorem 6.39 is based on the Sobolev immersion with critical exponent and Hardy’s inequality, and for this reason the p-capacity of RN \ Ω appears in the constant. Although the constant is domain dependent, for certain classes of sets, we can give a uniform constant, i.e., for Lipschitz or convex domains, we have an explicit constant depending only on p and N (see the details below at the end of Sect. 6.5.2). Remark 6.40 Singular problems when p < N are not considered. Similar results as in Sect. 6.5.1 can be obtained by combining the results in [275] with Hardy-type inequalities involving Ap weights, see the book of Opic and Kufner [223], following the proof of Theorem 6.39. Let us note that we have the following lower bounds for the first eigenvalue of the p-Laplacian with zero Dirichlet boundary conditions. Corollary 6.41 Let λ1 be the first eigenvalue of − Δp u = λw(x)|u|p−2 u

(6.74)

in Ω with zero Dirichlet boundary conditions in ∂Ω. Then, • for p > N and w as in Theorem 6.36, we have C p−N rΩ

w(x)1

≤ λ1 ,

• for p < N and w as in Theorem 6.39, we have C p−N/s rΩ

w(x)s

≤ λ1 .

(6.75)

Corollary 6.41 follows directly from Theorems 6.36 and 6.39, by replacing w with λ1 w. In Sect. 6.5.3, the bounds of Corollary 6.41 are applied to eigenvalue problems. First, it is shown that the powers of the inner radius appearing in Theorems 6.36 and 6.39 are optimal.

366

6 Partial Differential Equations

Proposition 6.42 Let B(0, R) be the ball of radius R centered at the origin and let  γ =

p−N

if

p > N,

p − N/s

if

p < N.

• Let R > 1. For any β < γ and C fixed, there exists a nonnegative weight w and 1,p a solution uβ ∈ W0 (B(0, R)) of  −Δp u(x) = w(x)|u(x)|p−2 u(x) u(x) = 0

if x ∈ B(0, R), if x ∈ ∂B(0, R)

(6.76)

such that the inequality C ≤ w(x)L1 (B(0,R)) Rβ

(6.77)

does not hold. • Let R < 1. For any β > γ and C fixed, there exists a nonnegative weight w and 1,p a solution uβ ∈ W0 (B(0, R)) of (6.76) such that inequality (6.77) does not hold. The result follows by computing a bound for the first eigenvalue of the pLaplacian on a ball with a radial weight restricted to a small ball of radius ε for a suitable ε. Finally, the lower bounds for the first eigenvalue of the p-Laplacian in Corollary 6.41 are compared with the ones obtained with other techniques. A classical tool for problems without weights is the Faber–Krahn inequality λ1 (B) ≤ λ1 (Ω), where B is the ball with Lebesgue measure |B| = |Ω|. Several proofs of this inequality for the p-Laplacian appeared in the literature, and they are based on the ideas of Talenti. Some improvements involving measures of the asymmetry of the domain Ω are known, see [45, 137]. For bounded weights, a Sturmian comparison argument combined with the variational characterization of the first eigenvalue (see (6.84)) enables us to replace w with the norm wL∞ , obtaining new lower bounds for λ1 . For arbitrary weights, there are few inequalities involving their norms and the measure of the domain, namely the works of Anane [31] and Cuesta [101]. It is shown that for certain domains and weights, the bounds given by the Lyapunov inequality are better.

6.5 Linear and Quasilinear Elliptic Differential Operators

367

6.5.1 Lyapunov-Type Inequality for p > N Let us recall first Morrey’s inequality. Theorem 6.43 (Morrey) If p > n, then there exists a constant C(N, p) such that 1,p for all u ∈ W0 (Ω), |u(x) − u(y)| ≤ C(N, p) ∇uLp |x − y|p for all x, y ∈ Ω¯ and α = 1 − N/p. We now prove Theorem 6.36. 1,p

Proof of Theorem 6.36 Let u ∈ W0 (Ω) be a nontrivial solution of −Δp u = w(x)|u|p−2 u with Dirichlet boundary conditions. Multiplying by u and integrating by parts, we obtain   |∇u|p = w(x)|u|p . Ω

Ω

¯ where |u(x)| Since p > N, u is continuous, and let us choose c ∈ Ω, a point of Ω, ¯ achieves its maximum. Then, for y = c and x ∈ Ω, we have that

1/p

 |∇u| dx

|u(c)| ≤ C(N, p)

|x − c|α .

p

Ω

By using that |x − c| ≤ rΩ , the inner radius of Ω, we get  |u(c)| ≤ C(N, p)

1/p w(x)|u|p dx

α rΩ .

Ω

Hence, 

1/p

|u(c)| ≤ C(N, p)|u(c)|

w(x)dx Ω

α rΩ ,

and canceling out |u(c)|, we have the Lyapunov inequality 1 α ≤ C(N, p) rΩ



with α = 1 − N/p. The proof is complete.

1/p w(x)dx

Ω



368

6 Partial Differential Equations

Remark 6.44 In particular, let λ1 be the first eigenvalue of (6.74) in Ω with zero Dirichlet boundary conditions in ∂Ω. We have C −p (N, p) p−N rΩ

w(x)1

≤ λ1 ,

(6.78)

which gives the lower bound for λ1 in Corollary 6.41. The following lemma extends the results in [185] for different weights in the function and its distributional gradient. 1,p

Lemma 6.45 For v, w ∈ Ap , the space W0 (RN , v, w) is the completion of C∞ (RN ) with the norm 1/p  p p ·p,v,w := ∇·[Lp (RN ,v)]N + ·Lp (RN ,w) . 1,p

Proof The proof follows by taking u ∈ W0 (RN , v, w) and regularizing it by convolution with a mollifier ηj . Now, we have from [185, Lemma 1.5], ηj ∗ u → u in Lp (RN , w), ∇(ηj ∗ u) = ηj ∗ ∇u → ∇u

in [Lp (RN , v)]N ,

1,p

that is, ηj ∗ u → u in W0 (RN , v, w).



We now prove Theorem 6.37. Proof of Theorem 6.37 Thanks to Lemma 6.45, we can choose a smooth function u. Now, given x, y ∈ Ω¯ such that r = |x−y| ≤ rΩ , let us put A = B(x, r)∩B(y, r). Hence,   1 1 |u(x) − u(y)| ≤ |u(x) − u(z)|dz + |u(y) − u(z)|dz |A| A |A| A   |∇u(z)| |∇u(z)| dz + C dz ≤C N −1 |x − z| |y − z|N −1 B(x,r) B(y,r) = I1 + I2 , where the constant C depends only on N, see for instance, Evans [130]. Let us bound now I1 . We need to include the coefficient v appearing in the equation, and let us put B = B(x, r). By using Hölder’s inequality, we get 

|∇u(z)| 1/p −1/p v v dz |x − z|N −1 B 

1/p 

1/q 

1/s 1 p −s/p ≤ v|∇u(z)| dz dz v dz , q(N −1) B B |x − z| B

I1 = C

6.5 Linear and Quasilinear Elliptic Differential Operators

369

where 1 1 1 + + = 1, p q s

s=

p . t −1

Now, we have following bounds: 

 v(z)|∇u(z)| dz ≤ p

B

 B

w(z)|u(z)|p dz,

(6.79)

Ω

1 q−qN +N dz ≤ crΩ , |x − z|q(N −1)  v −s/p (z)dz ≤ g(rΩ ).

(6.80) (6.81)

B

We have used that v is positive, and by integrating the equation multiplied by u in Ω by parts, we get the first inequality. The second one follows by integrating in polar coordinates in a bigger ball of radius rΩ , the constant c can be computed explicitly and depends only on N, p and q. The last one was defined in this way in the hypotheses. The bound for I2 is almost identical, although we need first to 1,p impose some extra condition on u. Since we are working in W0 , we can extend any function by zero outside of Ω, and we can take a smooth function u supported in Ω. So, we can integrate only over B(y, r) ∩ Ω in the first inequality (6.79), and we get 1−N +N/q

|u(x) − u(y)| ≤ CrΩ



1/p

[g(rΩ )]1/s

w(z)dz

,

Ω

where C is a universal constant depending only on N, p and q. We are able to choose yet the points x and y, and this is the last step of the proof. Let x be the point where |u| is maximized, and let y be one of the points in ∂Ω which minimizes |x − y|. So, u(y) = 0 and |x − y| < rΩ . After bounding |u(z)| ≤ |u(x)| at the right-hand side and canceling out with the one in the left-hand side, we get 1≤

p−pN +pN/q [g(rΩ )]p/s C(p, t, N )rΩ

 w(z)dz. Ω

Finally, let us observe that the relationship between the Hölder exponents implies that p = p − t, q This completes the proof.

p = t − 1. s 

370

6 Partial Differential Equations

Remark 6.46 Let us note that (6.80) holds when q − qN + N > 0 and q ≥ p in Hölder’s inequality. That is, N p N > 1. On the other hand, the bigger the q, the bigger is s. When q → N/(N −1), we have that s → pN/(p −N), and the integral in (6.81) is well defined when v ∈ At with t < p/N. As an application of Theorem 6.37, we have the following result for quasilinear problems involving the distance to the boundary. Proposition 6.47 Suppose Ω ⊂ RN is a bounded and open set, let p > N, and let 1,p u ∈ W0 (Ω, d γ , w) be a nontrivial solution of γ

− div(dΩ (x)|∇u|p−2 ∇u) = w(x)|u|p−2 u in Ω with zero Dirichlet boundary conditions in ∂Ω, where dΩ (x) is the distance to the boundary. Then, (6.72) holds, where C depends only on N, p, and γ . In order to prove Proposition 6.47, we can repeat the proof of Theorem 6.37, γ although only (6.81) depends on dΩ . So, we will improve this bound by integrating in B(x, dΩ (x)) instead of B(x, rΩ ). Proof of Proposition 6.47 We divide the proof in two cases, depending on the sign of γ . First, we consider γ < 0. Given z ∈ Ω, we choose y ∈ ∂Ω with r = |x −y| = dΩ (x). Clearly, we have r ≤ rΩ . After a translation, if necessary, we can suppose that y = 0, and we have dΩ (z) ≤ |z|. Then, −sγ /p

dΩ

(z) ≤ |z|−sγ /p .

Hence, we can estimate g(rΩ ) by computing 

−sγ /p

B(x,r)

dΩ



|z|−sγ /p dz=r N −sγ /p

(z) ≤



B(x,r)

N −sγ /p

B(x/r,1)

|η|−sγ /p dη ≤ CrΩ

where in the last step we changed variables, η = z/r. So, we can bound  B(x,r)

−sγ /p

dΩ

N −sγ /p

(z) ≤ CrΩ

.

Let us consider now γ > 0. Given z ∈ Ω and y ∈ ∂Ω with r = |x − y| = dΩ (x) ≤ rΩ as before, clearly, we have r ≤ rΩ . After a translation, if necessary, we can

,

6.5 Linear and Quasilinear Elliptic Differential Operators

371

suppose that x = 0, and we have dΩ (z) ≥ d∂B(0,r) (z), the distance to the boundary of the ball. Then, since γ > 0, −sγ /p

dΩ

−sγ /p

(z) ≤ d∂B(0,r) (z)

and  B(0,r)

−sγ /p dΩ (z)



(r − |z|)−sγ /p (z)dz

≤ B(0,r)



= cN

r

(r − ρ)−sγ /p ρ N −1 dρ

0

= cN r

N −sγ /p



1

(1 − ρ) ˜ −sγ /p ρ˜ N −1 dρ˜

0

= Cr

N −sγ /p

.

Again, we have the bound 

−sγ /p

B(0,r)

dΩ

N −sγ /p

(z) ≤ CrΩ

.

The last step is to replace this bound instead of the power of g(rΩ ) in Lyapunov’s inequality given by Theorem 6.36. By using that p/s = t − 1, we have (6.72), and the proof is complete. 

6.5.2 Lyapunov-Type Inequality for p < N Let us now prove Theorem 6.39. Proof of Theorem 6.39 Let us define q = αp + (1 − α)p∗ , where p∗ is the Sobolev conjugate exponent and α ∈ (0, 1) will be chosen later. Then, we have   1 |u|q q |u| dx ≤ dx, pα pα rΩ Ω Ω d(x) where d(x) is the distance from x to the boundary. Now, Hölder’s inequality with exponents 1/α and 1/(1 − α) gives

372

6 Partial Differential Equations



∗

Ω

|u|pα |u|(1−α)p dx ≤ d(x)pα



|u|p dx d(x)p

Ω

α 

1−α

∗

|u|p dx

.

(6.82)

Ω

Let us recall the Hardy and Sobolev inequalities 

 |u|p dx ≤ Ch |∇u|p dx, p Ω d(x) Ω 

p∗ /p  p∗ p |u| dx ≤ C∗ |∇u| dx , Ω

Ω

and by using them in (6.82), we get  Ω

|u|p dx d(x)p

α 



1−α

∗

|u|p dx

α+(1−α)p∗ /p

≤ Chs

|∇u|p dx

Ω

,

Ω

where Chs is a constant depending only on Ch and Cs , the constants involved in the Hardy and Sobolev inequalities. Hence, by using the weak formulation for the equation −Δp u = w(x)|u|p−2 u and by applying again Hölder’s inequality with exponents s and s  , we obtain 

α+(1−α)p∗ /p |∇u|p dx

Ω

 = Ω

α/s+(1−α)p∗ /(ps) 

 ≤

α+(1−α)p∗ /p w(x)|u|p dx |u|

s

w (x)dx Ω

ps 

α/s  +(1−α)p∗ /(ps  ) dx

Ω

We choose now α such that ps  = q. Let us observe that α (1 − α)p∗ + = 1, s ps  (1 − α)p∗ s α + = , s ps s and α=

p∗ − ps  = 1. p∗ − p

Finally, we get 1 pα rΩ

 |u| dx ≤ q

Ω

 wsLs

 Ω

|u|q dx, d(x)pα

.

6.5 Linear and Quasilinear Elliptic Differential Operators

373



and the proof is complete. Remark 6.48 A tedious computation shows that pα ps − N α p∗ − ps  = . = · ∗   s s p −p s Since s > N/p, the exponent is positive.

Remark 6.49 The constant C depends on the constant H appearing in the Hardy inequality. When Ω is convex, we have H =

p N −p

p .

For other domains, the constant depends on the capacity of RN \ Ω. For Lipschitz domains, the constant is close to 1/2, see [149, 197] for details. Remark 6.50 In particular, let λ1 be the first eigenvalue of (6.74) in Ω with zero Dirichlet boundary conditions in ∂Ω. We have (6.75), which gives the lower bound for λ1 in Corollary 6.41.

6.5.3 Applications to Eigenvalue Problems We close this section with a discussion about the optimality of the lower bounds and its application to eigenvalue problems. We show that in certain cases, the new bounds are better than the known ones.

6.5.3.1

Optimality of the Bounds

Let us show the optimality of the power of the inner radius appearing in the inequality. Proof of Proposition 6.42 For brevity, we will consider only the case p > N and R > 1 since the remaining ones follow exactly in the same way. Fix R > 1, and let us show that the bound (6.78) from Remark 6.44 cannot hold for some power β < p − N and w(r) = χ[0,ε] (r)r 1−N , where χ[0,ε] (r) is the characteristic function of [0, ε]. Clearly, w1 = ωN −1 ε, where ωN −1 ε is the surface measure of the unit ball, since

374

6 Partial Differential Equations



 B(0,R)

χ[0,ε] (|x|)|x|

1−N



ε

dx = ωN−1

r 1−N r N −1 drdθ.

0

and λ(ε) Let λ(R) 1 1 be the first eigenvalues of the p-Laplacian problem (6.74) with Dirichlet boundary conditions in B(0, R) and B(0, ε), respectively. We have 1,p (R) (ε) λ1 < λ1 , since extending the functions by zero, we have W0 (B(0, ε)) ⊂ 1,p W0 (B(0, R)), and the inequality follows by using the variational characterization  λ(R) 1

=



inf

B(0,R) |∇u|

{u∈W0 (B(0,R)):u≡0} B(0,R) χ[0,ε] (|x|)|x| 1,p



(ε) λ1

=

p dx

 B(0,ε)

inf

|∇u|p dx

{u∈W0 (B(0,ε)):u≡0} B(0,ε) |x| 1,p

1−N dx

1−N dx

,

.

Since the first eigenfunction in a ball is radial, ε (R) λ1

≤

(ε) λ1

=

inf

1,p

{u∈W0 (0,ε):u(ε)=0, u≡0}

≤ εN −1

0

r N −1 |u |p dr ε p 0 |u| dr

p

πp . εp

Then, C ≤ λ1 ωN −1 ε. Rβ Let ε = R α . If we can choose α < 1 such that β − α(p − N) < 0, then we arrive at the contradiction R α(p−N ) ≤ R β . However, this is equivalent to finding α satisfying 0
γ is of no interest when the inner radius is greater than 1, since we get a worse bound instead of an improvement. Similar observations hold for the remaining cases.

6.5.3.2

Comparison with Other Estimates

Let us consider the eigenvalue problem  −Δp u(x) = λw(x)|u(x)|p−2 u(x) u=0

if x ∈ Ω, if x ∈ ∂Ω.

(6.83)

There are few ways to obtain lower bounds for the eigenvalues of the p-Laplacian. In the constant coefficient case, we can use symmetrization and then compare with the first eigenvalue of a ball with the same measure as Ω, since the Faber–Krahn inequality implies λ1 (B) ≤ λ1 (Ω). For weighted problems, a Sturmian-type comparison theorem is available, that is, if w1 (x) ≤ w2 (x), then λk (w2 ) ≤ λk (w1 ), since the eigenvalues are computed with the Rayleigh quotient. Also, Anane and Cuesta obtained some inequalities that will be reviewed below. In the sequel, we compare those bounds with the one obtained from Corollary 6.41 when p > N and N = 2. Similar results hold for p < N and higher dimensions. Let us discuss Faber–Krahn bounds. In order to compare the Faber–Krahn inequality and the Lyapunov inequality (6.69), we can expect that the former will be worse in thin domains. So, let us take the family of domains in R2   1 2 ΩR = (x, y) ∈ R : 0 ≤ x ≤ R, 0 ≤ y ≤ R with 0 < R ≤ 1. Since |ΩR | = 1, Faber–Krahn gives a fixed lower bound for any ΩR . However, the Lyapunov inequality (with w ≡ 1) implies C −p (2, p) p−2 rΩR w(x)1

≤

C −p (2, p) C = p−2 ≤ λ1 . (R/2)p−2 R

376

6 Partial Differential Equations

Now, from (6.85), when R → 0,  p p πp πp p p , λ1 = p + πp R = O R Rp and by using (6.86), we get 

p

πp λ1 = O Rp

 .

Lyapunov’s inequality is better for small R, although it is not optimal in this family of sets. The Faber–Krahn inequality can be improved as in [45, 137]. Following Fusco, Maggi and Pratelli, we have   A2+p (Ω) , λ1 (Ω) ≥ λ1 (B) 1 + C(N, p) where C(N, p) is a fixed constant and A(E) is the Fraenkel asymmetry of a set E with finite measure, i.e.,   |EΔ(x0 + rB(0, 1))| A(E) := inf : x0 ∈ RN , r N |B(0, 1)| = |E| . |E| Since A is bounded above by 2, the maximum constant that can be involved in the lower bound is independent of R for the previous family of sets. Now, we discuss Sturm-type bounds. Intuitively, this kind of bounds can be improved because by adding a highly concentrated spike with very low mass in a given weight, we can change slightly the eigenvalue, and the supremum norm of the weight can be made arbitrarily big. The proof follows easily by using the eigenfunction of the unperturbed weight as a test function. However, the improvement can be better, even for domains with an inner radius of the same order than the diameter of the domain. Suppose that 0 ≤ ω ≤ M, Ω = [0, R] × [0, R], and R  1, with  w(x) = 1. Ω

The variational characterization of the first eigenvalue, together with (6.85) and (6.86), implies p

2πp ≤ λ1 . MR p

6.5 Linear and Quasilinear Elliptic Differential Operators

377

Now, the Lyapunov inequality gives the bound C R p−2

≤ λ1 .

Let us observe that the difference between them does not depend only on M, but also on a factor MR 2 . Indeed, we always have  R

w(x)dx ≤ MR p .

p−2 Ω

Finally, we discuss bounds involving norms of the weights. For arbitrary weights, there are few estimates involving their norms and the measure of the domain. First, Anane obtained in [31] the estimate |Ω|σ

C ≤ λ, w(x)∞

where  σ =

p/N

if

1 < p ≤ N,

1/2

if

N < p.

Also, Cuesta proved in [101] the inequality C

≤ λ,

|Ω|p/N −1/s σ

w(x)s

s > N/p

if 1 < p ≤ N,

s=1

if N < p.

where 

Clearly, these are Lyapunov-type inequalities, involving the measure of the domain instead of the inner radius. Those inequalities were widely used to show that the first eigenvalue is isolated, since any other eigenfunction has at least two nodal domains and one of them must shrink, but the inequality implies that the first eigenvalue of the shrinking domain cannot converge to the first eigenvalue of the full domain. Let us observe that |Ω|1/N ≥ CrΩ with equality only when Ω is a ball, so Corollary 6.41 gives in general better bounds.

378

6 Partial Differential Equations

6.5.4 Eigenvalues of the p-Laplacian We say that a function u is an eigenfunction of (6.83) corresponding to the eigenvalue λ if   p−2 |∇u| ∇u∇ϕdx = λ w(x)|u|p−2 uϕdx Ω

Ω 1,p

for any test-function ϕ ∈ W0 (Ω). The existence of infinitely many eigenvalues was proved by García Azorero and Peral Alonso in [138] by using the critical point theory of Lyusternik–Shnirel man and the variational characterization given by the Rayleigh quotient  |∇u|p dx , (6.84) λk = inf sup  Ω C∈Ck u∈C Ω w(x)|u|p dx 1,p

where Ck is the class of compact symmetric (C = −C) subsets of W0 (Ω) of Krasnosel ski˘ı genus greater than or equal to k, see [245] for details. It is well known that the first eigenfunction is positive and simple, see for instance [31]. Indeed, this result holds for more general operators, including socalled pseudo-p-Laplacian operators −Δˆ p := −

N

∂ ∂xi i=1

    ∂v p−2 ∂v   ,  ∂x  ∂xi i

and the proof is exactly the same, the simplicity follows by a Picone-type identity, and the positivity by considering |u1 | as a test function, where u1 is the first eigenfunction. We use the pseudo-p-Laplacian in order to control the eigenvalues of the pLaplacian. The equivalence of norms in RN , |x|q ≤ Cp,q |x|p , enables us to compare the first eigenvalue of each problem, since both can be defined by p λˆ 1 = inf |∇u|p p ; u∈B

p

λ1 = inf |∇u|2 p , u∈B

where    1,p B = u ∈ W0 (Ω) : w(x)|u|p dx . Ω

Clearly, 

λˆ 1 ≤ λ1 ≤ N p/2−1 λˆ 1

if 2 < p,

N p/2−1 λˆ 1 ≤ λ1 ≤ λˆ 1

if p < 2.

(6.85)

6.5 Linear and Quasilinear Elliptic Differential Operators

379

The first eigenvalue of the one-dimensional problem with w ≡ 1 

−(|u |p−2 u ) = λ|u|p−2 u = 0

in (0, L),

u(0) = u(L) = 0 can be computed explicitly with the help of the function sinp (x), defined implicitly as  x=

sinp (x)

0

p−1 1 − tp

1/p dt

and its first zero πp 1

 πp = 2 0

p−1 1 − tp

1/p dt.

We have p

λ1 =

πp . Lp

Also, for the mixed boundary condition u (0) = u(L) = 0, the first eigenvalue is given by p

λ1 =

2p πp . Lp

We refer the interested reader to the work of Del Pino et al. [110], for more details about the one-dimensional case. Finally, for w ≡ 1, the first eigenvalue λˆ 1 and the corresponding eigenfunction uˆ 1 of the pseudo-p-Laplacian in a cube Q = [0, L]N ⊂ RN can be computed explicitly. Following [57], we have λˆ 1 =

p

πp N , Lp

uˆ 1 (x) =

N %

sinp

j =1

π x  p j , L

which combined with inequalities (6.85) gives upper and lower bounds for the first eigenvalue 2of the p-Laplacian in Q with w(x) ≡ 1. A similar computation gives, for Ω = N j =1 [0, Li ], λˆ 1 =

N p

πp

p,

j =1

Lj

uˆ 1 (x) =

N % j =1

sinp

πp xj Lj

.

(6.86)

380

6 Partial Differential Equations

6.6 Notes and References The well-known Lyapunov inequality states that if q ∈ L1 (a, b), then a necessary condition for (1.1) with (1.5) to have nontrivial solutions is that (1.4) holds (see Sect. 1.2). It is shown that an analogous result is true for (1.1) with Neumann boundary conditions, i.e., (6.1) (see [65, 74, 75, 279]). One of the main applications of Lyapunov inequalities is to give optimal nonresonance conditions for the existence (and uniqueness) of solutions of nonlinear boundary value problems at resonance [74, 210, 279]. To the best of our knowledge, similar results for partial differential equations have only been proved by Cañada et al. [76] in 2006, see also the monograph [77] by Cañada and Villegas. Section 6.2 is devoted to the study of Lp -Lyapunov-type inequalities (p ≥ 1) for linear partial differential equations under Neumann boundary conditions on bounded and regular domains in RN . It is proved that how the relation between the quantities p and N/2 plays a crucial rôle by considering the subcritical (p ∈ [1, N/2)), supercritical (p ∈ (N/2, ∞)), and critical (p = N/2) cases. In the last part of Sect. 6.2, we present some results on the existence and uniqueness of solutions of nonlinear resonant problems in a domain Ω ⊂ RN . The results given in Sect. 6.2 are taken from the paper by Cañada et al. [76], see also the monograph [77]. In Sect. 6.3, we consider two-dimensional nonlinear systems of partial differential equations. The results given in this section are from a recent paper by Chen et al. [80]. In Sect. 6.4, some basic univariate Lyapunov inequalities are transferred to the multivariate setting of a shell via the polar method by considering partial differential equations involving radial derivatives of functions on the closure of a spherical shell A ⊆ RN , N > 1. In Sect. 6.4, Theorems 6.18–6.23 can be found in the paper [33] by Anastassiou. Then, motivated by the results of Anastassiou [33] and He and Tang [158], Akta¸s [26] transferred some known univariate Lyapunovtype inequalities to the multivariate setting of a shell via the polar method. His results are better than some results of Anastassiou for even-order partial differential equations. Theorems 6.25 and 6.26 are given in the paper [26] by Akta¸s. The rest of this section contains the very recent results of Ji and Fan [178]. They generalized Theorems 6.25 and 6.26 to a more general class of even-order partial differential equations. Moreover, as we see by the end of this section, their results improve Theorem 6.25 significantly. The last section of this chapter, Sect. 6.5, contains the recent results of de Nápoli and Pinasco [108]. They proved a Lyapunov inequality for N-dimensional (linear and quasilinear) elliptic operators with zero Dirichlet boundary conditions, reflecting more geometric information than the measure of the domain. It appears that the first generalization of Lyapunov’s result for half-linear equations was obtained in 2003 by Yang [294] (see Theorem 3.2 in Sect. 3.1). In 2005, Napoli and Pinasco extended it to nonlinear operators in Orlicz spaces [106] and, in 2006, to systems of resonant type (see [46]) involving the p-Laplacian and the q-Laplacian [107]. Cañada et al. [66, 76] considered the Neumann problem. We refer the interested reader to the book by Pinasco [240] for further recent developments.

Chapter 7

Lyapunov-Type Inequalities for Difference Equations

7.1 Introduction In this chapter, we give a survey of the most basic results on Lyapunov-type inequalities for difference equations, discrete systems, and partial difference systems. We sketch some recent developments related to this type of inequalities. Sections 7.2 and 7.3 are devoted to the study of Lyapunov-type inequalities for second-order linear difference equations under the usual boundary conditions motivated by a classical result of Lyapunov [202]. In Sect. 7.4, Lyapunov-type inequalities are presented for even-order difference equations. Section 7.5 discusses recent results related to Lyapunov-type inequalities for discrete linear Hamiltonian systems. In this section, a disconjugacy criterion and stability criteria are also given as some applications of the Lyapunov-type inequalities. In Sects. 7.6 and 7.7, some recent developments related to Lyapunov-type inequalities for quasilinear and nonlinear difference systems under Dirichlet boundary conditions are presented and some applications of the results obtained are also given. In Sect. 7.8, some basic results about Lyapunov-type inequalities for partial difference systems are offered. In this section, some concepts such as discrete harmonic functions, Green’s functions and maxima of Green’s functions on straight nets, circular nets, and rectangular nets are also presented. In Sect. 7.9, Lyapunov-type inequalities are given for two-dimensional nonlinear systems of partial difference equations.

7.2 Second-Order Linear Difference Equations This section is concerned with the second-order linear difference equation Δ2 x(k − 1) + q(k)x(k) = 0,

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8_7

(7.1)

381

382

7 Difference Equations

where q is a real-valued function defined on a set of consecutive integers to be specified later. Here and throughout, the forward difference operator Δ is defined by Δx(k) = x(k + 1) − x(k)

for

k ∈ Z.

The results obtained in this section are motivated by a classical result of Lyapunov [202] which states that if q ∈ L1 (a, b), then a necessary condition for (1.1) satisfying (1.5) to have nontrivial solutions is (1.4) (see Sect. 1.2). In 1983, in view of the obvious similarity between (7.1) and (1.1), Cheng [83] found discrete analogues of (1.4) which are necessary for the existence of a nontrivial solution of (7.1) satisfying certain boundary conditions. Throughout this section, we shall assume that q is a nonnegative function defined on the set {1, 2, . . . , N }. We derive a condition which is necessary for (7.1) to have a nontrivial solution x satisfying x(0) = 0 and x(N + 1) = 0. Under the same assumption on q, we then derive a more general condition which is necessary for the same equation to have a nontrivial solution x satisfying 

x(0) + σ x(1) = 0, x(N + 1) + λx(N) = 0,

where σ and λ are nonnegative real numbers. We could have omitted one part of this section entirely but include it here for contrasting the principles and computations involved. In the final part, we use a comparison theorem to deal with the case when q can take on nonpositive values. In the sequel, the smallest integer which is larger than or equal to the real number t will be denoted by t + . Let An = (a(i, j )) be the n×n tridiagonal matrix defined by ⎧ 2 ⎪ ⎪ ⎨ a(i, j ) = −1 ⎪ ⎪ ⎩ 0

if

i = j,

if

|i − j | = 1,

otherwise.

Let Gn = (g(i, j )) be the n × n matrix defined by  g(i, j ) =

(n − i + 1)j

if 1 ≤ j ≤ i,

(n − j + 1)i

if i ≤ j ≤ n.

The elements of Gn are clearly positive. Furthermore, we may easily verify that     max {g(i, j )} = g (n/2)+ , (n/2)+ = n − (n/2)+ + 1 (n/2)+

1≤i,j ≤n

and that (n + 1)−1 Gn is the inverse of An . If q is a nonnegative function defined on {1, 2, . . . , N } and if x is a nontrivial solution of (7.1) defined on {0, 1, . . . , N + 1}

7.2 Second-Order Linear Difference Equations

383

satisfying x(0) = 0 and x(N + 1) = 0, then the vector x = col(x(1), x(2), . . . , x(N)) satisfies the matrix equation AN x − diag(q(1), q(2), . . . , q(N))x = 0.

(7.2)

Multiplying (7.2) by (N + 1)−1 GN , we obtain x = (N + 1)−1 GN diag(q(1), q(2), . . . , q(N))x. Let i ∈ {1, 2, . . . , N} be such that |x(i)| = max |x(j )|. 1≤j ≤N

Then, |x(i)| ≤ (N + 1)−1

N

g(i, j )q(j )|x(j )|

j =1 −1

≤ |x(i)|(N + 1)

max GN

N

q(j ),

j =1

i.e., q(1) + q(2) + · · · + q(N) ≥ μ(N),

(7.3)

where

μ(N) =

⎧ (2m + 1) ⎪ ⎪ ⎨ m(m + 1)

N +1 = ⎪ (N − (N/2)+ + 1)(N/2)+ ⎪ ⎩

1 m+1

if N = 2m, if N = 2m + 1.

Inequality (7.3) is best possible in the sense that for any N , we can find nonnegative q and nontrivial x such that q(1) + · · · + q(N) = μ(N) and x is a solution of (7.1) for 0 ≤ k ≤ N + 1. To see this, we first suppose N = 2m + 1 for some m ∈ Z. Define  k if 0 ≤ k ≤ m + 1, x(k) = −k + 2 if m + 1 ≤ k ≤ 2m + 2

384

7 Difference Equations

and q(k) = −

Δ2 x(k − 1) , x(k)

1 ≤ k ≤ 2m + 1.

Then, x satisfies (7.1) on {1, 2, . . . , N}, x(0) = 0, x(N + 1) = 0, and ⎧ if 1 ≤ k ≤ m, ⎪ ⎪0 2 ⎨ 1 if k = m + 1, q(k) = m+1⎪ ⎪ ⎩ 0 if m + 2 ≤ k ≤ 2m + 1 as required. Next, suppose N = 2m for some m ∈ Z. Define ⎧ ⎪ ⎨k x(k) = (m + 1)(2m − k + 1) ⎪ ⎩ m

if

0 ≤ k ≤ m + 1,

if

m + 1 ≤ k ≤ 2m + 1

and q(k) = −

Δ2 x(k − 1) , x(k)

1 ≤ k ≤ 2m.

It can similarly be verified that x and q are the desired functions. After a change of the independent variable k in (7.1), the above conclusions can be summarized as follows. Proposition 7.1 (Lyapunov-Type Inequality) If q is a nonnegative function defined on the set of consecutive integers {a, a + 1, . . . , b} and if Δ2 x(k − 1) + q(k)x(k) = 0,

a≤k≤b

(7.4)

has a nontrivial solution x which satisfies x(a − 1) = 0 and x(b + 1) = 0, then the inequality q(a) + · · · + q(b) ≥ μ(b − a + 1) holds, and the inequality is sharp. Note that μ(N) is a strictly decreasing function of N. It follows from Proposition 7.1 that if q(a) + q(a + 1) + · · · + q(b) < μ(b − a + 1),

7.2 Second-Order Linear Difference Equations

385

then (7.4) cannot have a nontrivial solution x which satisfies x(c − 1) = 0

and

x(d + 1) = 0,

where a − 1 ≤ c − 1 ≤ d + 1 ≤ b + 1. For otherwise b

j =a

q(j ) ≥

d

q(j ) ≥ μ(d − c + 1) ≥ μ(b − a + 1) >

j =c

b

q(j ),

j =a

which is a contradiction. This principle can be applied to the problem  2 Δ x(k − 1) + q(k)x(k) = 0 for k = 1, 2, . . . , N, x(0) + σ x(1) = 0,

x(N + 1) + λx(N) = 0 with λ, σ ≥ 0,

(7.5)

where q is a nonnegative function defined on {1, 2, . . . , N}. If x is a nontrivial solution of (7.5) defined on {0, 1, . . . , N + 1}, then x = col(x(1), x(2), . . . , x(N)) satisfies BN x − diag(q(1), . . . , q(N))x = 0, where BN = (b(i, j )) is the matrix defined by ⎧ 2+σ ⎪ ⎪ ⎨ b(i, j ) = 2 + λ ⎪ ⎪ ⎩ a(i, j )

if

i = j = 1,

if

i = j = N,

otherwise.

It can be verified that the inverse of BN is the matrix {N + 1 + Nσ + Nλ + (N − 1)σ λ}−1 HN where HN = (h(i, j )) is defined by ⎧ h(1, j ) = (N − j + 1) + (N − j )λ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ h(N, j ) = j + (j − 1)σ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨h(i, 1) = (N − i + 1) + (N − i)λ ⎪ h(i, j ) = i + (i − 1)σ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ h(i, j ) = h(i, 1)h(N, j ) ⎪ ⎪ ⎪ ⎪ ⎩ h(i, j ) = h(i, N)h(1, j )

if

1 ≤ j ≤ N,

if

1 ≤ j ≤ N,

if

1 ≤ i ≤ N,

if

1 ≤ i ≤ N,

if

1 ≤ j ≤ i ≤ N,

if

1 ≤ i ≤ j ≤ N.

386

7 Difference Equations

Note that the matrix HN has positive components and that its interior components are products of boundary components. With these observations, we may arrive at the fact that  h(m + 1, m + 1) if N = 2m + 1, max HN = max {h(m, m), h(m + 1, m + 1)} if N = 2m, where h(m + 1, m + 1) = (m + 1 + mλ)(m + 1 + mσ ) and max {h(m, m), h(m + 1, m + 1)} = max {(m + 1 + mλ)(m + (m − 1)σ ), (m + 1 + mσ )(m + (m − 1)λ)} . We may now proceed to obtain the inequality q(1) + q(2) + · · · + q(N) ≥ μ(N, σ, λ),

(7.6)

where μ(N, σ, λ) =

N + 1 + Nσ + Nλ + (N − 1)σ λ . max HN

Inequality (7.6) is sharp. To see this, we first assume that N = 2m + 1 for some m ∈ Z. Let x1∗ be the linear function whose graph passes through the points (0, −σ ) and (1, 1). Let x(k) = x1∗ (k) for k ∈ {0, 1, . . . , m + 1}. The linear function whose graph passes through the points (N, 1) and (N + 1, −λ) has a zero N + 1/(1 + λ) in (N, N + 1]. Let x2∗ be the linear function whose graph passes through the points (m + 1, x(m + 1)) and (N + 1/(1 + λ), 0), and let x(k) = x2∗ (k) for k ∈ {m + 1, m + 2, . . . , 2m + 2. If we now set q(k) = −

Δ2 x(k − 1) , x(k)

1 ≤ k ≤ N,

then clearly x satisfies (7.5) and q(k) = 0 for 1 ≤ k ≤ m and m + 2 ≤ k ≤ 2m + 1. Moreover, q(m + 1) = =

slope of x1∗ slope of x2∗ − x(m + 1) x(m + 1) 1+σ σ − (1 + σ )(m + 1) − (1 + σ )(m + 1) − σ (m + 1/(1 + λ))((1 + σ )(m + 1) − σ )

7.2 Second-Order Linear Difference Equations

=

387

2m + 2 + (2m + 1)λ + (2m + 1)σ + 2mσ λ (m + 1 + mσ )(m + 1 + mλ)

= μ(2m + 1, σ, λ). Next, we suppose N = 2m for some m ∈ Z and that λ ≥ σ . Let x1∗ be the linear function whose graph passes through the points (0, −σ ) and (1, 1). Let x1∗ (k) = x(k) for k ∈ {0, 1, . . . , m}. Let x2∗ be the linear function whose graph passes through the points (m, x(m)) and (2m + 1/(1 + λ), 0). Let x(k) = x2∗ (k) for k ∈ {m, m + 1, . . . , 2m + 1}. If we now set q(k) = −

Δ2 x(k − 1) , x(k)

1 ≤ k ≤ 2m,

then clearly x satisfies (7.5) and q(k) = 0 for 1 ≤ k ≤ m − 1 and m + 1 ≤ k ≤ 2m. Furthermore, q(m) = =

1+σ σ − m(1 + σ ) − m(1 + σ ) − σ (m(1 + σ ) − σ )(2m + 1/(1 + λ)) 2m + 1 + 2mσ + 2mλ + (2m − 1)σ λ (m + 1 + mλ)(m + (m − 1)σ )

= μ(2m, σ, λ). The case N = 2m and σ ≥ λ can be dealt with similarly. We summarize our results as follows. Proposition 7.2 (Lyapunov-Type Inequality) If q is a nonnegative function defined on the set of consecutive integers {a, a + 1, . . . , b} and if the system 

Δ2 x(k − 1) + q(k)x(k) = 0 for

x(a − 1) + σ x(a) = 0,

a ≤ k ≤ b,

x(b − 1) + λx(b) = 0

with σ, λ ≥ 0

has a nontrivial solution, then the inequality q(a) + q(a + 1) + · · · + q(b) ≥ μ(b − a + 1, σ, λ) holds, and the inequality is sharp. Proposition 7.3 The function μ has the following properties. (i) (ii) (iii) (iv) (v)

μ(N, 0, 0) = μ(N), if σ  ≥ σ and λ ≥ λ, then μ(N, σ  , λ ) ≥ μ(N, σ, λ), if σ → ∞, then μ(N, σ, λ) → μ(N − 1, 0, λ), if λ → ∞, then μ(N, σ, λ) → μ(N − 1, σ, 0), and if σ, λ → ∞, then μ(N, σ, λ) → μ(N − 2, 0, 0).

(7.7)

388

7 Difference Equations

Proof The verifications of these properties are straightforward, and thus they are omitted.  Now, let f be a real function defined on a set of consecutive integers {a, a + 1, · · · , b}. If the points (k, f (k)) for a ≤ k ≤ b are joined by straight line segments to form a broken line, then this broken line gives a representation of a continuous function, henceforth denoted by f ∗ , such that f (k) = f ∗ (k) for k ∈ {a, a + 1, . . . , b}. The zeros of f ∗ are called the nodes of f . Note that x is a nontrivial solution of (7.7) if and only if x is a nontrivial solution of the difference equation in (7.7) with nodes a − 1/(1 + σ ) and b + 1/(1 + λ). Note also that if α and β are consecutive nodes of a nontrivial solution of the difference equation in (7.7), then β + > α + 1. Proposition 7.4 (Lyapunov-Type Inequality) Let σ and λ be two nonnegative real numbers. Let q be a nonnegative function defined on the set of consecutive integers {a, a + 1, · · · , b}. If q(a) + q(a + 1) + · · · + q(b) < μ(b − a + 1, σ, λ), then the difference equation in (7.7) cannot have a nontrivial solution which has nodes ξ and δ satisfying a−

1 1 ≤ξ

j =a

b

q(j ),

j =a

a contradiction. If ξ = a, δ = b + 1/(1 + λ ), then q(a + 1) + · · · + q(b) ≥ μ(b − a, 0, λ ) =

lim

σ  ≥σ,λ ≥λ;σ  λ →∞

μ(b − a + 1, σ  , λ )

≥ μ(b − a + 1, σ, λ) > q(a) + · · · + q(b), again a contradiction. Similarly, we can show that the other cases are also impossible. 

7.2 Second-Order Linear Difference Equations

389

The following comparison theorem [87, Lemma 2] shall be needed in proving our discrete analogue of the inequality of Lyapunov. Theorem 7.5 (Comparison Theorem) Suppose x and y, defined on {a − 1, a, . . . , b + 1}, are respectively nontrivial solutions of the equations Δ2 x(k − 1) + f (k)x(k) = 0,

a≤x≤b

Δ2 y(k − 1) + g(k)y(k) = 0,

a ≤ x ≤ b.

and

If x has two consecutive nodes α and β in [a − 1, b + 1] and if g(k) ≥ f (k) for a ≤ k ≤ b, then y has a node in (α, β]. Theorem 7.6 (Lyapunov-Type Inequality) Let σ and λ be two nonnegative real numbers. Let q be a real function defined on the set of consecutive integers {a, a + 1, · · · , b}. If b

max{q(k), 0} < μ(b − a + 1, σ, λ),

(7.8)

k=a

then the difference equation in (7.7) cannot have a nontrivial solution with two distinct nodes in [a − 1/(1 + σ ), b + 1/(1 + λ)]. Inequality (7.8) is sharp. Proof Assume to the contrary that x is a nontrivial solution of the difference equation in (7.7) that has two consecutive nodes ξ and β in [a − 1/(1 + σ ), b + 1/(1 + λ)]. Then, the comparison result Theorem 7.5 asserts that the system 

Δ2 y(k − 1) + max{q(k), 0}y(k) = 0

for a ≤ k ≤ b,

y ∗ (ξ ) = 0 has a nontrivial solution y which has a node δ in (ξ, β]. Since a−

1 1 ≤ξ ≤δ ≤β ≤b+ , 1+σ 1+λ

by Propositions 7.4 and 7.3, b

k=a

max{q(k), 0} ≥ μ(b − a + 1, σ, λ) >

b

max{q(k), 0},

k=a

which is the desired contradiction. The sharpness of (7.8) has been shown previously. 

390

7 Difference Equations

In 1989, Lin and Yang [203] generalized (7.3) and (7.6). They considered the system 

Δ2 x(k − 1) + q2 (k)Δx(k − 1) + q1 (k)x(k) = 0 for

1 ≤ k ≤ N,

(7.9)

x(0) = x(N + 1) = 0. Let AN = (a(i, j )) and BN = (b(i, j )) be the N × N matrices defined by ⎧ 2 ⎪ ⎪ ⎨ a(i, j ) = −1 ⎪ ⎪ ⎩ 0

if

i = j,

if

|i − j | = 1,

otherwise.

and ⎧ 2 + σ1 ⎪ ⎪ ⎨ b(i, j ) = 2 + σ2 ⎪ ⎪ ⎩ a(i, j )

if i = j = 1, if i = j = N, otherwise.

−1 Then, A−1 N = (α(i, j )) and BN = (b(i, j )) by [83], where

α(i, j ) =

β(i, j ) =

 (N − i + 1)j/(N + 1)

if

1 ≤ j ≤ i ≤ N,

(N − j + 1)i/(N + 1)

if

1 ≤ i ≤ j ≤ N,

⎧ [(N − j + 1)j + (N − j )σ2 ] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (j + (j − 1)σ1 ) ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎨((N − i + 1)j + (N − i)σ2 ) b⎪ (i + (i − 1)σ1 ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ β(i, 1)β(N, j ) ⎪ ⎪ ⎪ ⎪ ⎩ β(i, N )β(1, j )

if i = 1, 1 ≤ j ≤ N, if i = N, 1 ≤ j ≤ N, if j = 1, 1 ≤ i ≤ N, if j = N, 1 ≤ i ≤ N, if 1 ≤ j ≤ i ≤ N, if 1 ≤ i ≤ j ≤ N,

and b = N + 1 + Nσ1 + Nσ2 + (N − 1)σ1 σ1 . Theorem 7.7 (Lyapunov-Type Inequality) If q1 and q2 are nonnegative functions defined on {1, 2, . . . , N } and if (7.9) has a nontrivial solution, then the inequality N

j =1

(2q2 (j ) + q1 (j )) − q2 (1) ≥

1 max A−1 N

(7.10)

7.2 Second-Order Linear Difference Equations

391

holds, where max A−1 N = max α(i, j ). 1≤i,j ≤N

Proof Let x = col(x(1), . . . , x(N )) and y = col(0, x(1), . . . , x(N − 1)). Then, AN x − diag(q1 (1) + q2 (1), . . . , q1 (N ) + q2 (N ))x + diag(q2 (1), . . . , q2 (N ))y = 0. Multiplying this equation by A−1 N , we obtain   x = A−1 N diag(q1 (1) + q2 (1), . . . , q1 (N ) + q2 (N ))x − diag(q2 (1), . . . , q2 (N ))y . Let |x(i)| = max |x(j )|. 1≤j ≤N

Then,     N

 N  α(i, j ) (q1 (j ) + q2 (j )) x(j ) − α(i, j )q2 (j )x(j − 1) |x(i)| =   j =1 j =1   N  N

  =  α(i, j ) (q1 (j ) + q2 (j )) x(j ) − α(i, j )q2 (j )x(j − 1) j =1  j =2 ⎫ ⎧ N ⎬ ⎨ ≤ |x(i)| max A−1 (j ) + 2q (j )) − q (1) . (q 1 2 2 N ⎩ ⎭ j =1

Hence, N

(q1 (j ) + 2q2 (j )) − q2 (1) ≤

j =1

This completes the proof.

1 max A−1 N

. 

392

7 Difference Equations

Remark 7.8 We note that (7.3) is a special case of (7.10) when q2 (j ) = 0, 1 ≤ j ≤ N. Theorem 7.9 (Lyapunov-Type Inequality) If q1 and q2 are nonnegative functions defined on {1, . . . , N } and if the problem 

Δ2 x(k − 1) + q2 (k)Δx(k − 1) + q1 (k)x(k) = 0 for

k = 1, 2, . . . , N,

x(0) + σ1 x(1) = 0,

σ1 , σ2 ≥ 0

x(N + 1) + σ2 x(N) = 0 with

has a nontrivial solution, then the inequality N

(2q2 (j ) + q1 (j )) − (1 − σ1 )q2 (1) ≥

j =1

1 −1 max BN

.

(7.11)

holds. Proof Let x = col(x(1), . . . , x(N )) and y = col(−σ1 x(1), x(1), . . . , x(N − 1)). Then, −BN x+diag(q1 (1)+q2 (1), . . . , q1 (N )+q2 (N ))x−diag(q2 (1), . . . , q2 (N ))y = 0. −1 , we obtain Multiplying this equation by BN

 −1  x = BN diag(q1 (1) + q2 (1), . . . , q1 (N ) + q2 (N ))x − diag(q2 (1), . . . , q2 (N ))y . Let |x(i)| = max |x(j )|. 1≤j ≤N

Then,   N β(i, j ) (q1 (j ) + q2 (j )) x(j ) + β(i, 1)q2 (1)σ1 x(1) |x(i)| =  j =1   N

 β(i, j )q2 (j )x(j − 1) −  j =1

7.2 Second-Order Linear Difference Equations

−1 ≤|x(i)| max BN

⎧ N ⎨ ⎩

393

(q1 (j ) + 2q2 (j )) + q2 (1)σ1 +

j =1

N

j =2

⎫ ⎬ q2 (j ) . ⎭

Hence, N

(q1 (j ) + 2q2 (j )) − (1 − σ1 )q2 (1) ≤

j =1

1 −1 max BN

. 

This is the desired inequality.

Remark 7.10 We note that (7.6) is a special case of (7.11) when q2 (j ) = 0, 1 ≤ j ≤ N. If n, k ∈ Z, then we define, in the usual way, the binomial coefficient 

1 n n! = k k!(n − k)! 0

if 0 ≤ k ≤ n, otherwise.

Then,



n−1 n−1 n + = k k−1 k if 0 ≤ k ≤ n and n ≥ 1. Also, as usual, we define j

ar = 0

for j < i.

r=i

In order to establish more general results, we use the following lemmas. Lemma 7.11 If m ∈ {2, 3, . . . , N } and q ∈ N, then

q

m−1+q −r r+1 m (−1) = 0. r m−1 r=0

Proof We have

q

m m−1+q −r (−1)r+1 r m−1 r=0

=





 q

m m+q −r m−1+q −r − (−1)r+1 r m m r=0

(7.12)

394

=

7 Difference Equations







q q

q m+q −r q −1 m+q −1−r − (−1)r+1 (−1)r+1 r m r q −1 r=0

=

r=0







q −1 q −1 m+q −r (−1)r+1 + r r −1 m

q

r=0





q

m+q −1−r r+1 q − 1 − (−1) r q −1 r=0

=



q−1

(−1)

r+1

r=0

r=0



q−1

+

(−1)r+2

r=0

=







q−1 q −1 m+q −r m+q −1−r r+2 q − 1 + (−1) r q r q



q−1

(−1)

r+1

r=0



q −1 m+q −1−r r q −1







q−1 q −1 m+q −r m+q −r r+2 q − 1 + (−1) r q r q r=0

= 0, and this shows (7.12).



Lemma 7.12 If m ∈ {2, 3, . . . , N }, k ∈ {1, 2, . . . , m}, and q ∈ N, then q

(−1)r+1

r=0







k−1+q −r m+q −r m = (−1)q+1 . k−1 m−k k+q

(7.13)

Proof Let k = 1. Then, by using (7.12), we obtain

q

m+q −r r+1 m (−1) r m−1 r=0





q+1

m+q −r m r+1 m q+2 − (−1) = (−1) r m−1 q +1 r=0

m . = (−1)q+1 q +1 Suppose (7.13) is true for 1 ≤ k ≤ m − 1, i.e.,





q

m k−1+q −r m+q −r m = (−1)q+1 . (−1)r+1 r k−1 m−k k+q r=0

(7.14)

7.2 Second-Order Linear Difference Equations

395

Then,





m k+q −r m+q +1−r m = (−1)q . r k−1 m−k k+q +1 r=0 (7.15) Now, by using (7.14) and (7.15), we obtain q+1

q

(−1)r+1

(−1)

r+1

r=0

=





m k+q −r m+q −r r k m−k−1







q

k+q +1−r k+q −r m+q −r − (−1)r+1 k k−1 m−k−1 r=0





q

m−1 m+q −r r+1 m = (−1) r k m−1 r=0

−







 q

m k+q −r m+q +1−r m+q −r − (−1)r+1 r k−1 m−k m−k r=0





q m+q −r m−1 r+1 m (−1) = r m−1 k r=0

+

q

(−1)

r+1

r=0

−





m m−1 m+q −r r k−1 m−1





q

m k+q −r m+q +1−r (−1)r+1 r k−1 m−k r=0









q m+q −r m m m r+1 m q+1 + (−1) = (−1) r m−1 q +1 k−1 k r=0





m k+q −r m+q +1−r − (−1) r k−1 m−k r=0







m m m m + (−1)q+2 = (−1)q+1 − (−1)q k+q +1 q +1 k k

m . = (−1)q+1 k+q +1 q+1

r+1

This completes the proof.



396

7 Difference Equations

Lemma 7.13 For 2 ≤ m ≤ N, let FNm = (f (i, j )) and Gm N = (g(i, j )) be the N × N matrices defined by

1 m × m−1−j +i 0

i+j

f (i, j )=(−1)

if 1 ≤ i, j ≤ N, −1 ≤ j −i ≤ m−1, otherwise

and m−1+N −i m−2+j  g(i, j ) =

m−1 m−1 m−1+N  m−1

⎧ ⎪ ⎨0 − j − i + m − 2

⎪ ⎩ m−1

if 1 ≤ j ≤ i ≤ N, otherwise.

Then, (FNm )−1 = Gm N. Proof Suppose FNm Gm N = (e(i, j )). Then, e(i, j ) =

N

f (i, r)g(r, j ).

r=1

First, we let i = 1. Then, using Lemma 7.11, we have e(1, j ) =

m

f (1, r)g(r, j )

r=1

=



m

m g(r, j ) (−1)r+1 m−r r=1



j −1

m = (−1)r−1 m−r

 m−1+N −r m−2+j 

r=1



m

m r−1 + (−1) m−r r=j

m−1 m−1 m−1+N  m−1



 j −r +m−2 − m−1

m−1+N −r m−2+j  m−1 m−1 m−1+N  m−1

m−2+j 

=

 m  m

m−1 m−r r−1 (−1) m−1+N  m−1+N −r  m−1 m−1 r=1



j −1

m j −r +m−2 r−1 (−1) − m−r m−1 r=1

m−2+j 

=

N

m−1+N m−1 r+1 m (−1) m−1+N  r m−1 m−1 r=0

−r



m−1+N m−1

7.2 Second-Order Linear Difference Equations

397



N

m−1+j −r −1 r+1 m − (−1) r m−1 r=1





j −1 m−2+j −r m−2+j r−1 m (−1) = − r m−1 m−1 r=1  1 if j = 1, = 0 if j > i. Similarly, for 2 ≤ i ≤ N + 1 − m, we have e(i, j ) =

m+i−1

(−1)

r−i

r=i−1

=

m

(−1)

r−1

(−1)

r+1

m g(r + i − 1, j ) m−r

r=0

=

j −i



m g(r, j ) m−1−r +i



m m−r

r=0



m

+

(−1)r+1

r=j −i+1

m−2+j 

m−1 = m−1+N  m−1

−

j −i

=



m

m m−i+N −r (−1)r+1 r m−1 r=0

(−1)r+1



m m−1+j −i−r r m−1

r+1



m m−1+j −i−r r m−1

if

j ≤ i − 1,

if

j = i,

if

i + 1 ≤ j ≤ N.

(−1)

r=0

⎧ 0 ⎪ ⎪ ⎨ = 1 ⎪ ⎪ ⎩ 0

m−1

m−i+N −r m−2+j  m m−1 m−1 m−1+N  m−r m−1

r=0 j −i

 m−i+N −r m−2+j 

 j −i−r +m−1 m−1 m−1 − m−1+N  m−1

Finally, for i = N + 1 − m + t, 1 ≤ t ≤ m − 1, we have e(N + 1 − m + t, j ) =

N

r=N −m+t

f (N − m + t + 1, r)g(r, j )

398

=

7 Difference Equations m−t

f (N − m + t + 1, N − m + t + r)g(N − m + t + r, j )

r=0

=

m−t

(−1)r−1

r=0

=

m g(N − m + t + r, j ) m−r

m−t+j −N −1

(−1)

r+1

m−1

r=0

2m−1−t−r m−2+j  m m−1 m−1 m−1+N  m−r



m−t

+

  2m−1−t−r m−2+j  2m−2+j −N−t−r m m−1 m−1 − m−1+N  m−1 r

(−1)r+1

r=m+j −N −t

m−1

m−2+j 



m−t

2m − 1 − t − r m−1 r+1 m = m−1+N  (−1) r m−1 m−1

−

r=0

m+j −N

−t−1

(−1)

⎧ 0 ⎪ ⎪ ⎨ = 1 ⎪ ⎪ ⎩ 0

r+1

r=0



m 2m − 2 + j − N − t − r r m−1

if j ≤ i − 1 = N − m + t, if j = i, if j ≤ i + 1. 

This completes the proof. Remark 7.14 Note that g(i, j ) > 0 for all 1 ≤ i, j ≤ N . Note also that

FNm

= AN .

For m, N ∈ N with 2 ≤ m ≤ N, we define



k

m

N +k−3 N +m−2 % N +m−1 bN = σr + k−1 m−k m−1 k=1

r=1

+

m

k=2



k N +k−2 N +m−1 % σr . k−1 m−k r=2

Then, we have the following result. m = (d(i, j )), H m = (h(i, j )) be the N × N matrices Lemma 7.15 Let DN N defined by

7.2 Second-Order Linear Difference Equations

399

⎧

m ⎪ ⎪ + σ1 if i = j = 1, ⎪ ⎪ ⎪ m−1 ⎪ ⎪ ⎪ ⎪   ⎪

k−N m ⎪

%+i ⎨ m m N −i σr + d(i, j ) = (−1) m−N +i−1 k ⎪ k=N −i+2 r=2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ if j = N, N − m + 2 ≤ i ≤ N, ⎪ ⎪ ⎪ ⎪ ⎩f (i, j ) otherwise, ⎧





k m ⎪ N+m − 1−i N +k−2−i N +m−1−i % ⎪ ⎪ ⎪ + σr ⎪ ⎪ m−1 k−1 m−k ⎪ ⎪ k=2 r=2 ⎪ ⎪ ⎪ ⎪ ⎪ if j = 1, 1 ≤ i ≤ N, ⎪ ⎪ ⎪ 1 ⎨ j + m − 2 j + m − 3

h(i, j ) = m + σ1 if i = 1, 1 ≤ j ≤ N, bN ⎪ ⎪ m−1 m−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪h(i, 1)h(N, j ) if 1 ≤ j ≤ i ≤ N, ⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎪ ⎪h(i, 1)h(N, j ) − j − i + m − 2 bm if 1 ≤ j ≤ i ≤ N. ⎩ N m−1 m )−1 = H m . Then, (DN N

Proof By using Lemmas 7.11 and 7.12, the proof follows by an argument similar to the one used in the proof of Lemma 7.13.  m =B . Remark 7.16 Note that h(i, j ) > 0 for all 1 ≤ i, j ≤ N. Note also that DN N

Theorem 7.17 (Lyapunov-Type Inequality) For 2 ≤ m ≤ N , 1 ≤ i ≤ m, if qi is a nonnegative function defined on {1, 2, . . . , N } and if the problem 

Δm x(k − 1) + qm (k)Δm−1 x(k − 1) + · · · + q2 (k)Δx(k − 1) + q1 (k)x(k) = 0, x(0) = x(N + 1) = x(N + 2) = · · · = x(N + m − 1) = 0

has a nontrivial solution, then the inequality m N

2r−1 qr (j ) − Um ≥

j =1 r=1

1 max(FNm )−1

holds, where Um =

m

r=2

qr (1) +

m

r=3



 s

r −1 qr (N − r + 3 + s) . t t=0

400

7 Difference Equations

Proof The case m = 2 is Theorem 7.7. Hence, we may assume 3 ≤ m ≤ N. The vector x = col(x(1), x(2), . . . , x(N)) satisfies (−1)m+1 FNm x ⎞ ⎛   7m−1 qm (1) t=0 (−1)t m−1 t x(m − 1 − t) + · · · + q1 (1)x(1) ⎟ ⎜ .. ⎟ ⎜ . ⎟ ⎜ m−1 7m−1 ⎟ ⎜ t + ⎜ qm (j ) t=0 (−1) t x(j + m − 2 − t) + · · · + q1 (j )x(j ) ⎟ = 0. ⎟ ⎜ .. ⎟ ⎜ ⎠ ⎝ .   7m−1 (N )x(N ) x(N + m − 2 − t) + · · · + q qm (N ) t=0 (−1)t m−1 1 t Multiplying this equation by (FNm )−1 = Gm N = (g(i, j )), we obtain x = (−1)m (FNm )−1 ⎞ ⎛   7m−1 qm (1) t=0 (−1)t m−1 (1)x(1) x(m − 1 − t) + · · · + q 1 t ⎟ ⎜ .. ⎟ ⎜ . ⎟ ⎜   7m−1 ⎟ ⎜ × ⎜ qm (j ) t=0 (−1)t m−1 ⎟. x(j + m − 2 − t) + · · · + q (j )x(j ) 1 t ⎟ ⎜ .. ⎟ ⎜ ⎠ ⎝ .   7m−1 m−1 t qm (N ) t=0 (−1) t x(N + m − 2 − t) + · · · + q1 (N )x(N ) Let |x(i)| = max |x(j )|. 1≤j ≤N

Then,

 0 

m−1 N

 m t m−1  |x(i)| = (−1) x(j + m − 2 − t) g(i, j ) qm (j ) (−1) t  j =1 t=0 1

1 

 t 1 + · · · + q2 (j ) x(j − t) + q1 (j )x(j )  (−1)  t ≤

N

0

t=0

g(i, j ) qm (j )

j =1

m−1

t=0

m−1 |x(j + m − 2 − t)| t

1 1

1 |x(j − t)| + q1 (j )|x(j )| + · · · + q2 (j ) t t=0

7.2 Second-Order Linear Difference Equations

401

so that ⎧⎡ ⎤ N m−1 ⎨

m − 1

|x(j + m − 2 − t)|⎦ |x(i)| ≤ max(FNm )−1 ⎣ qm (j ) ⎩ t j =1

t=0

⎡ ⎤ ⎡ ⎤⎫ N 1

N ⎬

1 + ··· + ⎣ |x(j − t)|⎦ + ⎣ q2 (j ) q1 (j )|x(j )|⎦ . ⎭ t j =1

(7.16)

j =1

t=0

For 3 ≤ r ≤ m, consider N

qr (j )

j =1

r−1

r −1 |x(j + r − 2 − t)| t t=0



r−2 N −r+2 r−1

r −1 r −1 = qr (1) |x(r − 1 − t)| + |x(j + r − 2 − t)| qr (j ) t t j =2

t=0

t=0

r−3 r−1

r −1 |x(N + s + 1 − t)| + qr (N − r + 3 + s) t s=0 t=s+1 ⎧ 0 r−1 1 N −r+2

r−1 ⎨

r − 1 r − 1

r −1 − ≤ |x(i)| qr (1) qr (j ) + ⎩ t r −1 t j =2

t=0

t=0

0 r−1

1 r−3 s

r − 1 r −1 − + qr (N − r + 3 + s) t t s=0 t=0 t=0 ⎧ ⎫

0

1 N r−1 r−3 s ⎨

r −1 r −1 ⎬ − qr (1)+ ≤ |x(i)| qr (j ) qr (N−r+3+s) . ⎩ ⎭ t t j =1

t=0

s=0

t=0

It follows from (7.16) that ⎧ ⎡ m N ⎨

⎣ 2r−1 qr (j ) − qr (1) |x(i)| ≤ |x(i)| max(FNm )−1 ⎩ r=3

−

r−3

s=0

Hence,

j =1

⎫ ⎤ ⎡

1 s N N ⎬

r −1 qr (N − r + 3 + s) 2q2 (j ) − q2 (1)⎦ + q1 (j ) . +⎣ ⎭ t t=0

j =1

j =1

402

7 Difference Equations

0 m N

j =1

1 r−1

2

qr (j ) −

0 m

r=1

qr (1) +

r=2

r−3 m

r=3 s=0

1 s

r −1 qr (N − r + 3 + s) t t=0

≥

1 , max(FNm )−1

i.e., 0 m N

j =1

1 r−1

2

qr (j ) − Um ≥

r=1

1 . max(FNm )−1 

The proof is complete.

Remark 7.18 We note that Theorem 7.7 is a special case of Theorem 7.17 when m = 2. Theorem 7.19 (Lyapunov-Type Inequality) For 2 ≤ m ≤ N, 1 ≤ i ≤ m, if qi is a nonnegative function defined on {1, 2, . . . , N } and if the problem ⎧ m Δ x(k − 1) + qm (k)Δm−1 x(k − 1) + · · · + q2 (k)Δx(k − 1) + q1 (k)x(k) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x(0) + σ1 x(1) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨x(N + 1) + σ2 x(N) = 0, x(N + 2) + σ3 x(N + 1) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ .. ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x(N + m − 1) + σm x(N + m − 2) = 0, where σi ≥ 0, has a nontrivial solution, then the inequality m N

2k−1 qk (j ) − Vm ≥

j =1 k=1

1 m )−1 max(DN

holds, where

Vm = (1 − σ1 )

+

 r−3 m

r=3

s=0

m

r=2

qr (1) 0



s+2−t s s

r −1 r −1 % − qr (N − r + 3 + s) σk t t t=0

t=0

k=2

1 .

7.2 Second-Order Linear Difference Equations

403

Proof The case m = 2 is Theorem 7.9. Hence, we may assume 3 ≤ m ≤ N. The vector x = col(x(1), x(2), . . . , x(N)) satisfies m (−1)m+1 DN x ⎞ ⎛   7m−1 qm (1) t=0 (−1)t m−1 t x(m − 1 − t) + · · · + q1 (1)x(1) ⎟ ⎜ .. ⎟ ⎜ . ⎟ ⎜ m−1 7m−1 ⎟ ⎜ t + ⎜ qm (j ) t=0 (−1) t x(j + m − 2 − t) + · · · + q1 (j )x(j ) ⎟ = 0. ⎟ ⎜ .. ⎟ ⎜ ⎠ ⎝ .   7m−1 x(N + m − 2 − t) + · · · + q (N )x(N ) qm (N ) t=0 (−1)t m−1 1 t m )−1 = H m = (h(i, j )), we obtain Multiplying this equation by (DN N m −1 x = (−1)m (DN ) ⎞ ⎛   7m−1 qm (1) t=0 x(m − 1 − t) + · · · + q (−1)t m−1 (1)x(1) 1 t ⎟ ⎜ .. ⎟ ⎜ . ⎟ ⎜   7m−1 ⎟ ⎜ × ⎜ qm (j ) t=0 (−1)t m−1 ⎟. x(j + m − 2 − t) + · · · + q (j )x(j ) 1 t ⎟ ⎜ .. ⎟ ⎜ ⎠ ⎝ .   7m−1 m−1 t qm (N ) t=0 (−1) t x(N + m − 2 − t) + · · · + q1 (N )x(N )

Let |x(i)| = max |x(j )|. 1≤j ≤N

Then,   

m−1 N

 m−1 m x(j + m − 2 − t) |x(i)| = (−1) h(i, j ) qm (j ) (−1)t t  j =1 t=0

r−1

r −1 + · · · + qr (j ) x(j + r − 1 − t) + · · · + q1 (j )x(j ) (−1)t t t=0

   , 

404

7 Difference Equations

i.e., ⎧ N ⎨

m −1 ) ≤ max(DN ⎩

|x(i)|

qm (j )

j =1

m−1

t=0

m−1 |x(j + m − 2 − t)| t

N m−2

m − 2

|x(j + m − 3 − t)| + · · · + qm−1 (j ) t j =1 t=0 ⎫ N 1

N ⎬

1 + |x(j − t)| + q2 (j ) q1 (j )|x(j )| . ⎭ t j =1

(7.17)

j =1

t=0

For 3 ≤ r ≤ m ≤ N, N

qr (j )

j =1

r−1

r −1 |x(j + r − 2 − t)| t t=0

 r−2 

r − 1

= qr (1) |x(r − 1 − t)| + σ1 |x(1)| t t=0

+

N −r+2 j =2

+

r−3



r−1

r −1 qr (j ) |x(j + r − 2 − t)| t t=0



qr (N − r + 3 + s)

s=0

r−1

r −1 |x(N + s + 1 − t)| t

t=s+1

0



s+2−t 1 s

r −1 % + σk |x(N)| t t=0 k=2 ⎧ 0 r−1



1 N −r+2 r−1 ⎨

r − 1

r −1 r −1 ≤ |x(i)| qr (1) + σ1 − qr (j ) + ⎩ t t r −1 j =2

t=0

t=0

0 r−1

s+2−t

1 r−3 s s

r − 1

r−1 % r−1 + + qr (N − r + 3 + s) σk − t t t s=0 t=0 t=0 k=2 t=0 ⎧

 

N r−1 ⎨

r −1 r −1 − qr (1) = |x(i)| qr (j ) − σ1 ⎩ t r −1 j =1

−

r−3

s=0

t=0

0



s+2−t 1 s s

r −1 r −1 % − qr (N − r + 3 + s) σk t t t=0

t=0

k=2

7.2 Second-Order Linear Difference Equations

405

so that, for 3 ≤ r ≤ m ≤ N,

r−1

r −1 |x(j + r − 2 − t)| t j =1 t=0 ⎧ N ⎨ ≤ |x(i)| 2r−1 qr (j ) − qr (1)(1 − σ1 ) ⎩

N

qr (j )

j =1

−

r−3

s=0

0



s+2−t 1 s s

r −1 r −1 % − qr (N − r + 3 + s) σk . t t t=0

t=0

k=2

It follows from (7.17) that ⎧⎡ N ⎨ m −1 ⎣ ) 2m−1 qm (j ) − qm (1)(1 − σ1 ) |x(i)| ≤|x(i)| max(DN ⎩ j =1

−

m−3

qm (N − m + 3 + s)

 s

m − 1

s=0

t

t=0

−

⎡ N

+⎣ 2m−2 qm−1 (j ) − qm−1 (1)(1 − σ1 ) j =1

−

m−4

qm−1 (N − m + 4 + s)

 s

m − 2

s=0

t

t=0

s+2−t s

m−1 % t=0

−

t=0



t

k=2



s+2−t s s

r −1 r −1 % − qr (N − r + 3 + s) σk − t t s=0 t=0 t=0 k=2 ⎡ N

+ ··· + ⎣ 22 q3 (j ) − q3 (1)(1 − σ1 ) r−3

j =1

−

0

q3 (N + s)

s=0

⎡ +⎣

N

j =1

 s

2 t=0

t

−

s s+2−t

2 % t=0

⎤

2q2 (j ) − q2 (1)(1 − σ1 )⎦ +

t N

j =1

k=2

σk

k=2

s+2−t s

m−2 %

⎡ N

+ ··· + ⎣ 2r−1 qr (j ) − qr (1)(1 − σ1 ) j =1

t

1

1 σk ⎫ ⎬

q1 (j ) . ⎭

1

1 σk

406

7 Difference Equations

Hence, N

j =1

2m−1 qm (j ) + 2m−2 qm−1 (j ) + · · · + 22 q3 (j ) + 2q2 (j ) + q1 (j ) 

− (1 − σ1 )

m

qr (1)

r=2

+

r−3 m

r=3 s=0

≥

!

0



s+2−t s s

r −1 r −1 % − qr (N − r + 3 + s) σk t t t=0

t=0

1

k=2

1 m )−1 , max(DN

i.e., 0 m N

j =1

k=1

1 k−1

2

qk (j ) − Vm ≥

1 m )−1 . max(DN 

The proof is complete.

Remark 7.20 If σi = 0 for all 1 ≤ i ≤ m, then = Vm = Um . Hence, Theorem 7.17 is a special case of Theorem 7.19. We also note that Theorem 7.9 is a special case of Theorem 7.19 when m = 2. m DN

FNm ,

7.3 Lyapunov-Type Finite Difference Inequalities The main purpose of this section is to establish a Lyapunov-type inequality for the second-order linear finite difference equation Δ(r(n)Δx(n)) + q(n)x(n) = 0,

(7.18)

where n ∈ Z[a, b] = {a, a + 1, a + 2, . . . , b}, a and b = a + m, m ≥ 2, are integers. It is assumed that p and r are real-valued functions defined on Z[a, b] and r(n) > 0 for n ∈ Z[a, b]. In 1990, Pachpatte [226] made a remarkable approach which was more direct and elementary, and the result provided a new estimate in the Lyapunov-type inequality for (7.18). Theorem 7.21 (Lyapunov-Type Inequality) If x is a solution of (7.18) such that x(a) = x(b) = 0 and x ≡ 0 on Z(a, b) = {a + 1, a + 2, . . . , b − 1}, then the inequality

7.3 Lyapunov-Type Finite Difference Inequalities

407

 b−1  b−1

1

|q(n)| ≥ 4 r(n) n=a n=a

(7.19)

holds. Proof Let k ∈ Z(a, b) be such that |x(n)| is maximized. Let M = |x(k)|. It is obvious that x(k) =

k−1

Δx(n),

k ∈ Z[a, b]

(7.20)

n=a

and x(k) = −

b−1

Δx(n),

k ∈ Z[a, b].

(7.21)

n=k

From (7.20) and (7.21), we observe that 2M ≤

b−1

|Δx(n)|.

(7.22)

n=a

Now, squaring both sides of (7.22) and using the Schwarz inequality, the summation by parts formula n−1

u(s)Δv(s) = u(n)v(n) − u(0)v(0) −

s=0

n−1

v(s + 1)Δu(s),

s=0

the facts that x(a) = x(b) = 0, and (7.18), we observe that 4M 2 ≤

b−1

2 r −1/2 (n)r 1/2 (n)|Δx(n)|

n=a

b−1   b−1

1

≤ (r(n)Δx(n))Δx(n) r(n) n=a n=a b−1    b−1

1

= x(n + 1)Δ(r(n)Δx(n)) − r(n) n=a n=a b−1   b−1

1

= x(n + 1)q(n)x(n) r(n) n=a n=a  b−1  b−1

1

2 ≤M |q(n)| . r(n) n=a n=a

(7.23)

408

7 Difference Equations

Dividing both sides of (7.23) by M 2 , we get the desired inequality in (7.19). This completes the proof.  It is interesting to note that in the special case when r(n) ≡ 1, (7.19) reduces to the inequality b−1

|q(n)| ≥

n=a

4 . b−a

(7.24)

Inequality (7.24) yields the implicit lower bound on the distance between consecutive zeros of a nontrivial solution of (7.18).

7.4 Even-Order Difference Equations In this section, we consider the even-order difference equation Δ2k x(n) + (−1)k−1 q(n)x(n + 1) = 0,

(7.25)

where k ∈ N, n ∈ Z, and q is a real-valued function defined on Z. When k = 1, (7.25) reduces to Δ2 x(n) + q(n)x(n + 1) = 0.

(7.26)

In 2012, Zhang and Tang [307] established two discrete Lyapunov-type inequalities for (7.25) under the boundary conditions Δ2i x(a) = Δ2i x(b) = 0

for i = 0, 1, . . . , k − 1

(7.27)

with x ≡ 0 on Z[a, b], where Z[a, b] = {a, a + 1, a + 2, . . . , b − 1, b}. Furthermore, applying their Lyapunov-type inequalities to the eigenvalue problem 

Δ2k x(n) + (−1)k−1 λq(n)x(n + 1) = 0,

Δ2i x(a) = Δ2i x(b) = 0 for

i = 0, 1, . . . , k − 1,

(7.28)

they obtained two different lower bounds for the eigenvalue λ. In 1983, Cheng [83] first obtained the Lyapunov inequality F (b − a)

b−2

n=a

q(n) ≥ 4,

(7.29)

7.4 Even-Order Difference Equations

409

which is a discrete analogue of (1.2), where a, b ∈ Z and  2 1 m −1 F (m) = m m2

if m − 1

is even,

if m − 1

is odd

(7.30)

provided (7.26) has a real solution x such that x(a) = x(b) = 0

(7.31)

and x ≡ 0 on Z[a, b], where and in the sequel, a, b ∈ N and Z(a, b) = {a + 1, a + 2, . . . , b − 2, b − 1}, see Sect. 7.2. The constant 4 in (7.29) cannot be replaced by a larger number. For more discrete Lyapunov-type inequalities, we refer the reader to Cheng [84, 88, 89], Clark and Hinton [93, 94], Guseinov and Kaymakçalan [143], Lin and Yang [203], and Zhang and Tang [305]. Zhang and Tang [307] established some Lyapunov-type inequalities for (7.25). Lemma 7.22 Assume that x is a real-valued function defined on Z[a, b], x(a) = x(b) = 0, and x ≡ 0 on Z[a, b]. Then, the inequalities b−1  (b − n)(n − a)  2  x(n) ≤ Δ x(s) b−a s=a

(7.32)

for all n ∈ Z(a, b − 1), b−1

x(s) ≤

s=a

b−1   1   (s − a + 1)(b − s − 1) Δ2 x(s) 2 s=a

b−1 (b − a)2 ≤ |Δx(s)|2 , 8 s=a

x(n) ≤

(b − n)(n − a)[2(b − n)(n − a) + 1] 6(b − a)

(7.33)



1/2  b−1  2 1/2  2  Δ x(s) s=a

(7.34) for all n ∈ Z[a, b − 1], and b−1

n=a

hold.

|x(n)|2 ≤

b−1 2 [(b − a)2 − 1][2(b − a)2 + 7]  2  Δ x(s) 180 s=a

(7.35)

410

7 Difference Equations

Proof Since x(a) = x(b) = 0, it is easy to verify that x(n) = −

b−1 1 G(n, s)Δ2 x(s), b − a s=a

(7.36)

where  G(n, s) =

(s + 1 − a)(b − n)

if

s ≤ n,

n, s ∈ Z[a, b],

(n − a)(b − s − 1)

if

n ≤ s,

n, s ∈ Z[a, b − 1].

(7.37)

Since G(n, s) ≤ (n − a)(b − n),

n, s ∈ Z[a, b − 1],

it follows from (7.36) that |x(n)| ≤

≤

b−1   1   G(n, s) Δ2 x(s) b − a s=a b−1  (n − a)(b − n)  2  Δ x(s) b−a s=a

holds for all n ∈ Z(a, b − 1), which implies (7.32). Next, by (7.36) and (7.37), we have b−1

|x(n)| ≤

b−1 b−1   1

  G(n, s) Δ2 x(s) b − a n=a s=a

=

b−1 b−1  1  2  G(n, s) Δ x(s) b − a s=a n=a

=

b−1   1   (s + 1 − a)(b − s − 1) Δ2 x(s) 2 s=a

≤

b−1  

1   2 (b − a)2 Δ x(s) , 8 s=a

n=a

which implies (7.33). By a relatively simple sum computation, we have b−1

s=a

[G(n, s)]2 =

[(b − a)2 − 1][2(b − a)2 + 7] . 6(b − a)

(7.38)

7.4 Even-Order Difference Equations

411

Hence, by (7.36), (7.37), (7.38), and the Cauchy inequality, we have 1 |x(n)|2 ≤ (b − a)2 1 ≤ (b − a)2 ≤

=

1 (b − a)2

b−1 2     G(n, s)Δ2 x(s)   s=a  b−1

2 2

G(n, s)Δ x(s)

s=a b−1 b−1  2

 2  2 [G(n, s)] Δ x(s) s=a

(7.39)

s=a

b−1 2 (b − n)(n − a)[2(b − n)(n − a) + 1]  2  Δ x(s) 6(b − a) s=a

for all n ∈ Z[a, b − 1], which implies (7.34). Summing (7.39) from a to b − 1, we obtain b−1

|x(n)|2 ≤

n=a

=

b−1 b−1 b−1  2

1  2  2 [G(n, s)] x(s) Δ  (b − a)2 n=a s=a s=a b−1 2 [(b − a)2 − 1][2(b − a)2 + 7]  2  Δ x(s) . 180 s=a



Hence, (7.35) holds.

Theorem 7.23 (Hartman-Type Inequality) Suppose k ∈ N and q is a realvalued function defined on Z. If (7.25) has a nontrivial solution x satisfying (7.27), then the inequality b−1

|q(n)|(b − n − 1)(n − a + 1) ≥

n=a

23(k−1) (b − a)2k−3

(7.40)

holds. Proof Choose c ∈ Z[a, b] such that |x(c)| = maxn∈[a,b] |x(n)|. Because of (7.27), it follows from Lemma 7.22 that we have |x(c)| ≤

b−1 b−1   b−a  (b − c)(c − a)  2    2 ≤ x(n)  Δ Δ x(n) , b−a 4 n=a n=a

b−1  b−1   (b − a)2 

 2i  2i+2   x(n) , Δ x(n) ≤ Δ 8 n=a n=a

i = 1, 2, . . . , k − 2,

(7.41)

(7.42)

412

7 Difference Equations

and b−1  b−1  1  

 2k−2    x(n) ≤ (b − n − 1)(n − a + 1) Δ2k x(n) . Δ 2 n=a n=a

(7.43)

From (7.25), (7.41), (7.42), and (7.43), we obtain |x(c)| ≤

≤

≤

b−1  b − a  2  Δ x(n) 4 n=a

b−a 4 b−a 8





(b − a)2 8 (b − a) 8

k−2 b−1    2k−2  x(n) Δ n=a

b−1 2 k−2

    (b − n − 1)(n − a + 1) Δ2k x(n)

(7.44)

n=a

b−1 (b − a)2k−3 = (b − n − 1)(n − a + 1)|q(n)x(n + 1)| 23(k−1) n=a

≤

b−1

(b − a)2k−3 |x(c)| (b − n − 1)(n − a + 1)|q(n)|. 23(k−1) n=a

Since |x(c)| > 0, it follows from (7.44) that (7.40) holds.



Corollary 7.24 (Lyapunov-Type Inequality) Suppose k ∈ N and q is a realvalued function defined on Z. If (7.25) has a nontrivial solution x satisfying (7.27), then the inequality b−1

n=a

|q(n)| ≥

23k−1 (b − a)2k−1

holds. Theorem 7.25 (Lyapunov-Type Inequality) Suppose k ∈ N and q is a realvalued function defined on Z. If (7.25) has a nontrivial solution x satisfying (7.27), then the inequality b−1

n=a

holds.

|q(n)|2 ≥

2k+3 32k−1 5k−1  k−1  k−1 (b − a)2 + 7/2 (b − a) (b − a)2 + 2 (b − a)2 − 1 (7.45) 

7.4 Even-Order Difference Equations

413

Proof Choose c ∈ Z[a, b] such that |x(c)| = maxn∈[a,b] |x(n)|. Because of (7.27), from Lemma 7.22 and (7.25), we have |x(c)|2 ≤

b−1 2 (b − c)(c − a)[2(b − c)(c − a) + 1]  2  Δ x(s) 6(b − a) s=a

  k−1  k−1 b−1   2

 2(b−a)2 +7 (b − a)2 −1 (b−a) (b−a)2 +2  2 · ≤ x(s)  Δ k−1 48 180 s=a  k−1  k−1  2(b − a)2 + 7 (b − a) (b − a)2 + 2 (b − a)2 − 1 = 48 × 180k−1 ×

b−1

|q(n)|2 |x(n + 1)|2

n=a

≤|x(c)| ×

2 (b

b−1

  k−1  k−1 2(b − a)2 + 7 − a) (b − a)2 + 2 (b − a)2 − 1 48 × 180k−1

|q(n)|2

n=a

=|x(c)| ×

2 (b

b−1

  k−1  k−1 (b − a)2 + 7/2 − a) (b − a)2 + 2 (b − a)2 − 1 2k+3 32k−1 5k−1

|q(n)|2 .

n=a

Since |x(c)| > 0, it follows from (7.46) that (7.45) holds.

(7.46) 

Example 7.26 Now, we give an application of (7.40) and (7.45) for (7.28). We have |λ| ≥

(b − a)

7b−1 2k−3

23(k−1)

n=a (b

− n − 1)(n − a + 1)|q(n)|

and |λ| ≥ 2(k+3)/2 3k−1/2 5(k−1)/2  −1/2 k−1 b−1 ! !k−1  2 2 2 7 2 (b−a) + × (b−a) (b−a) +2 (b − a) −1 |q(n)| . 2 n=a

414

7 Difference Equations

7.5 Discrete Linear Hamiltonian Systems Recall from Chap. 4 that the continuous Hamiltonian system, in the case of two scalar linear differential equations, has the form (see (4.2) and, for example, [191, 293]) y  (t) = J H (t)y,

t ∈ R,

(7.47)

in which y=



y1 , y2

J =

0 1 , −1 0

H =

h11 h12 h21 h22



with real-valued and piecewise continuous functions hj k , j, k = 1, 2, and h12 = h21 . Equation (7.47) can be written as y1 = h21 (t)y1 + h22 (t)y2 ,

y2 = −h11 (t)y1 − h12 (t)y2

or setting y1 = x, y2 = u, a = h12 = h21 , b = h22 , c = h11 (t), it can be written as x  = a(t)x + b(t)u,

u = −c(t)x − a(t)u,

t ∈ R.

(7.48)

We remark that the second-order differential equation (p(t)x  ) + q(t)x = 0,

t ∈ R,

(7.49)

in which p, q are real-valued functions and p(t) = 0 for all t ∈ R, can be written as an equivalent Hamiltonian system of type (7.48). Indeed, let x be a solution of (7.49) and set u(t) = p(t)x  (t). Then, we have x =

1 u, p(t)

u = −q(t)x.

So, (7.49) is equivalent to (7.48) with a(t) ≡ 0,

b(t) =

1 , p(t)

c(t) = q(t).

The main concern of this section is to obtain a discrete analogue of the following continuous result, which we prove here for completeness and comparison. Theorem 7.27 (Lyapunov-Type Inequality) Assume that b(t) ≥ 0 for all t ∈ R and suppose (7.48) has a real solution (x, u) such that x(α) = x(β) = 0 and x

7.5 Discrete Linear Hamiltonian Systems

415

is not identically zero on [α, β], where α, β ∈ R with α < β. Then, the Lyapunov inequality 

β



β

|a(t)|dt +

α



β

b(t)dt ·

α

c+ (t)dt

1/2 ≥2

(7.50)

α

holds, where c+ = max{c, 0} is the nonnegative part of c. Proof Multiplying the first equation in (7.48) by u and the second one by x and then adding the results, we obtain (xu) (t) = b(t)[u(t)]2 − c(t)[x(t)]2 . Integrating this equation from α to β and taking into account x(α) = x(β) = 0, we get 

β



β

b(t)[u(t)] dt = 2

α

c(t)[x(t)]2 dt.

(7.51)

α

Choose τ ∈ (α, β) such that |x(τ )| = max |x(t)|. α≤t≤β

Since x is not identically zero on [α, β], we have |x(τ )| > 0. Integrating the first equation in (7.48) initially from α to τ and then from τ to β and taking into account x(α) = x(β) = 0, we get, respectively,  x(τ )=



τ

α



τ

a(t)x(t)dt+

b(t)u(t)dt,



β

−x(τ )=

β

a(t)x(t)dt+

α

τ

b(t)u(t)dt. τ

Hence, employing the triangle inequality gives 

τ

|x(τ )| ≤



τ

|a(t)||x(t)|dt +

b(t)|u(t)|dt

α

α

and 

β

|x(τ )| ≤



β

|a(t)||x(t)|dt +

τ

b(t)|u(t)|dt. τ

Adding these last two inequalities gives rise to 

β

2|x(τ )| ≤ α

 |a(t)||x(t)|dt +

β

b(t)|u(t)|dt. α

(7.52)

416

7 Difference Equations

On the other hand, applying the Cauchy–Schwarz inequality and using (7.51), we have 

β



1/2 

β

b(t)|u(t)|dt ≤ α

α



1/2 

β

=

β

b(t)dt α



1/2 c(t)[x(t)]2 dt

α

1/2 

β

≤

2

b(t)[u(t)] dt

b(t)dt

α

1/2

β

β

1/2

+

2

c (t)[x(t)] dt

b(t)dt α

.

α

Therefore, an application of (7.52) yields 

β

2|x(τ )| ≤



α



≤ |x(τ )|

1/2 

β

|a(t)||x(t)|dt + α

β

 |a(t)|dt +

α

β

c+ (t)[x(t)]2 dt

b(t)dt



β

α

b(t)dt α

β

c+ (t)dt

1/2

 .

α

Dividing the latter estimate by |x(τ )|, we get (7.50).



Now, we move to discrete linear Hamiltonian systems. As it is well known (see [7, 22, 129]), an adequate form of the discrete Hamiltonian system corresponding to (7.48) is 

Δx(t) = a(t)x(t + 1) + b(t)u(t), Δu(t) = −c(t)x(t + 1) − a(t)u(t),

(7.53)

where t ∈ Z, and the coefficient function a satisfies the condition 1 − a(t) = 0 for all

t ∈ Z.

(7.54)

Notice that the second-order difference equation Δ(p(t)Δx(t)) + q(t)x(t + 1) = 0,

t ∈ Z,

(7.55)

where p(t) = 0 for all t ∈ Z, can be written as an equivalent discrete Hamiltonian system of type (7.53). Indeed, let x be a solution of (7.55) and set u = pΔx. Then, we have Δx(t) =

1 u(t), p(t)

Δu(t) = −q(t)x(t + 1).

7.5 Discrete Linear Hamiltonian Systems

417

So, (7.55) is equivalent to (7.53) with a(t) ≡ 0,

b(t) =

1 , p(t)

c(t) = q(t).

Concerning (7.53), besides taking into account (7.54), we also assume that a, b, c are real-valued functions defined on Z and b(t) ≥ 0,

t ∈ Z.

In the discrete case, instead of the usual zero, the concept of generalized zero, which is due to Hartman [154], is used. A function f : Z → R is said to have a generalized zero at t0 ∈ Z provided either f (t0 ) = 0 or f (t0 − 1)f (t0 ) < 0. In 2003, Guseinov and Kaymakçalan [143] established some Lyapunov inequalities for (7.53). The following theorems are their results. Theorem 7.28 (Lyapunov-Type Inequality) Let α, β ∈ Z with α ≤ β − 2. Assume (7.53) has a real solution (x, u) such that x(α) = x(β) = 0 and x is not identically zero on [α, β]. Then, the inequality β−2

⎛ |a(t)| + ⎝

t=α

β−1

b(t)

t=α

β−2

⎞1/2 c+ (t)⎠

≥2

(7.56)

t=α

holds. Proof Multiplying the first equation in (7.53) by u(t) and the second equation in (7.53) by x(t + 1) and then adding, we get Δ(x(t)u(t)) = b(t)[u(t)]2 − c(t)[x(t + 1)]2 .

(7.57)

Summing the last equation from α to β − 1 and taking into account x(α) = x(β) = 0, we obtain β−1

b(t)[u(t)] − 2

t=α

β−1

c(t)[x(t + 1)]2 = 0.

t=α

Since x(β) = 0, we have β−1

t=α

b(t)[u(t)]2 =

β−2

c(t)[x(t + 1)]2 ≤

t=α

β−2

c+ (t)[x(t + 1)]2 .

t=α

Choose τ ∈ [α + 1, β − 1] such that |x(τ )| =

max

α+1≤t≤β−1

|x(t)|.

(7.58)

418

7 Difference Equations

Then, |x(τ )| > 0. Summing the first equation in (7.53) at first from α to τ − 1 and then from τ to β − 1, we get, respectively, x(τ ) =

τ −1

a(t)x(t +1)+

t=α

τ −1

b(t)u(t),

−x(τ ) =

t=α

β−2

a(t)x(t +1)+

t=τ

β−1

b(t)u(t).

t=τ

Passing here to the modulus, we have |x(τ )| ≤

τ −1

τ −1

|a(t)||x(t + 1)| +

t=α

b(t)|u(t)|

t=α

and |x(τ )| ≤

β−2

|a(t)||x(t + 1)| +

β−1

t=τ

b(t)|u(t)|.

t=τ

Adding these inequalities gives 2|x(τ )| ≤

β−2

|a(t)||x(t + 1)| +

t=α

β−1

(7.59)

b(t)|u(t)|.

t=α

On the other hand, applying the Cauchy–Schwarz inequality and using (7.58), we have β−1

⎛

β−1

b(t)|u(t)| ≤ ⎝

t=α

⎞1/2 ⎛ b(t)⎠

β−1

⎝

t=α

≤⎝

b(t)[u(t)]2 ⎠

t=α

⎛

β−1

⎞1/2

⎞1/2 ⎛ b(t)⎠

t=α

β−1

⎝

⎞1/2 c+ (t)[x(t + 1)]2 ⎠

.

t=α

Therefore, an application of (7.59) gives

2|x(τ )| ≤

β−2

t=α

⎛

β−1

|a(t)||x(t + 1)| + ⎝ ⎧ ⎪ ⎨β−2

⎞1/2 ⎛ b(t)⎠

t=α

β−1

⎝

⎞1/2 c+ (t)[x(t + 1)]2 ⎠

t=α

⎞1/2 ⎫ ⎪ ⎬ + ⎝ ⎠ ≤ |x(τ )| |a(t)| + b(t) c (t) . ⎪ ⎪ ⎩ t=α ⎭ t=α t=α ⎛

β−1

β−2

Dividing the latter inequality by |x(τ )|, we obtain (7.56).



7.5 Discrete Linear Hamiltonian Systems

419

Theorem 7.29 (Lyapunov-Type Inequality) Suppose 1 − a(t) > 0

and

for all t ∈ Z

b(t) > 0

(7.60)

and let α, β ∈ Z with α ≤ β − 2. Assume (7.53) has a real solution (x, u) such that x(α) = 0 and x(β − 1)x(β) < 0. Then, the inequality β−2

⎛

β−2

|a(t)| + ⎝

t=α

b(t)

t=α

β−2

⎞1/2 c+ (t)⎠

>1

(7.61)

t=α

holds. Proof Choose τ ∈ [α + 1, β − 1] such that |x(τ )| =

max

α+1≤t≤β−1

|x(t)|.

Then, |x(τ )| > 0. Summing the first equation in (7.53) from α to τ − 1 and taking into account x(α) = 0, we get x(τ ) =

τ −1

a(t)x(t + 1) +

t=α

τ −1

b(t)u(t).

t=α

Hence, |x(τ )| ≤

τ −1

|a(t)||x(t + 1)| +

t=α

≤

β−2

≤

b(t)|u(t)|

t=α

|a(t)||x(t + 1)| +

t=α β−2

τ −1

β−2

b(t)|u(t)|

t=α

⎛ |a(t)||x(t + 1)| + ⎝

t=α

β−2

(7.62)

⎞1/2 ⎛ b(t)⎠

⎝

t=α

β−2

⎞1/2 b(t)[u(t)]2 ⎠

.

t=α

Now, summing (7.57) from α to β − 2 and taking into account x(α) = 0, we obtain x(β − 1)u(β − 1) =

β−2

t=α

b(t)[u(t)]2 −

β−2

c(t)[x(t + 1)]2 .

t=α

Further, from the first equation in (7.53), we have, for t = β − 1 [1 − a(β − 1)]x(β) = x(β − 1) + b(β − 1)u(β − 1).

(7.63)

420

7 Difference Equations

Multiplying this by x(β − 1) yields [1 − a(β − 1)]x(β − 1)x(β) = [x(β − 1)]2 + b(β − 1)x(β − 1)u(β − 1). Since x(β − 1)x(β) < 0, in view of (7.60), the above latter equality gives rise to x(β − 1)u(β − 1) < 0. Therefore, from (7.63), the inequality β−2

b(t)[u(t)]2
0. Summing the first equation in (7.53) from τ to β − 1 and taking into account x(β) = 0, we obtain x(τ ) = −

β−2

t=τ

a(t)x(t + 1) −

β−1

t=τ

b(t)u(t).

7.5 Discrete Linear Hamiltonian Systems

421

Hence, |x(τ )| ≤

β−2

|a(t)||x(t + 1)| +

t=τ

≤

β−2

≤

b(t)|u(t)|

t=τ

|a(t)||x(t + 1)| +

t=α β−2

β−1

β−1

b(t)|u(t)|

t=α

⎛ |a(t)||x(t + 1)| + ⎝

t=α

β−1

(7.65)

⎞1/2 ⎛ b(t)⎠

⎝

β−1

t=α

⎞1/2 b(t)[u(t)]2 ⎠

.

t=α

Now, summing (7.57) from α − 1 to β − 1 and taking into account x(β) = 0, we obtain −x(α − 1)u(α − 1) =

β−1

β−2

b(t)[u(t)]2 −

t=α−1

c(t)[x(t + 1)]2 ,

t=α−1

i.e., − u(α − 1)[x(α − 1) + b(α − 1)u(α − 1)] =

β−1

b(t)[u(t)]2 −

t=α

β−2

c(t)[x(t + 1)]2 .

t=α−1

(7.66) Further, from the first equation in (7.53), we have, for t = α − 1, [1 − a(α − 1)]x(α) = x(α − 1) + b(α − 1)u(α − 1).

(7.67)

Multiplying this by x(α − 1) gives [1 − a(α − 1)]x(α − 1)x(α) = [x(α − 1)]2 + b(α − 1)x(α − 1)u(α − 1). Since x(α − 1)x(α) < 0, by (7.60), it follows from the above latter equality that x(α − 1)u(α − 1) < 0.

(7.68)

u(α − 1)[x(α − 1) + b(α − 1)u(α − 1)] > 0

(7.69)

Now, our aim is to show that

holds. Indeed, multiplying (7.67) by u(α − 1) gives [1 − a(α − 1)]x(α)u(α − 1) = u(α − 1)[x(α − 1) + b(α − 1)u(α − 1)].

(7.70)

422

7 Difference Equations

On the other hand, it follows from x(α −1)x(α) < 0 and (7.68) that x(α)u(α −1) > 0. Therefore, the left-hand side of (7.70) is positive, and hence, (7.69) is true. By virtue of (7.69), the string of inequalities β−1

b(t)[u(t)]2
0,

c(t) > 0

for all t ∈ Z,

(7.71)

and let α, β ∈ Z with α ≤ β − 1. Assume (7.53) has a real solution (x, u) such that x(α − 1)x(α) < 0 and x(β − 1)x(β) < 0. Then, the inequality ⎛

β−2

|a(t)| + ⎝

t=α−1

β−1

b(t)

t=α−1

β−2

⎞1/2 c(t)⎠

>1

(7.72)

t=α−1

holds. Proof In the first part of this proof, we assume that x(t) = 0 for all t ∈ [α, β − 1]. Denote by β0 the smallest integer in [α, β] such that β0 = α and x(β0 − 1)x(β0 ) < 0. Then, x does not have any generalized zero in [α + 1, β0 − 1], and without loss of generality, we may assume that x(t) > 0 for all

t ∈ [α, β0 − 1].

Then, we have x(α − 1) < 0 and

x(β0 ) < 0.

7.5 Discrete Linear Hamiltonian Systems

423

Let s ∈ [α − 1, β0 − 1]. Summing the second equation in (7.53) first from α − 1 to s − 1 and then from s to β0 − 2, we get s−1

u(s) − u(α − 1) = −

c(t)x(t + 1) −

t=α−1

s−1

a(t)u(t)

(7.73)

a(t)u(t),

(7.74)

t=α−1

and u(β0 − 1) − u(s) = −

β 0 −2

c(t)x(t + 1) −

t=s

β 0 −2 t=s

respectively. Notice that for s = α − 1, we write solely (7.74), and for s = β0 − 1, only (7.73) is written. Let us now show that u(α − 1) > 0 and

u(β0 − 1) < 0.

(7.75)

Indeed, from the first equation in (7.53), we have [1 − a(t)]x(t + 1) = x(t) + b(t)u(t). Multiplying this last equation by x(t) gives [1 − a(t)]x(t)x(t + 1) = [x(t)]2 + b(t)x(t)u(t), where setting t = α − 1 and t = β0 − 1, respectively, yields 

[1 − a(α − 1)] x(α − 1)x(α) = [x(α − 1)]2 + b(α − 1)x(α − 1)u(α − 1), [1 − a(β0 − 1)] x(β0 − 1)x(β0 ) = [x(β0 − 1)]2 + b(β0 − 1)x(β0 − 1)u(β0 − 1).

Using the inequalities x(α − 1)x(α) < 0, x(β0 − 1)x(β0 ) < 0, and (7.71), we get from the above latter equalities the estimates x(α − 1)u(α − 1) < 0 and

x(β0 − 1)u(β0 − 1) < 0.

(7.76)

Hence, taking into account x(α − 1) < 0 and x(β0 − 1) > 0, we obtain (7.75). Employing (7.73) if u(s) < 0, using (7.74) whenever u(s) > 0, and also taking into account (7.75), we get

424

7 Difference Equations

|u(s)| ≤

β 0 −2

c(t)|x(t + 1)| +

t=α−1

⎛ ≤⎝

β 0 −2

|a(t)||u(t)|

t=α−1

⎞1/2 ⎛

β 0 −2

c(t)⎠

⎝

t=α−1

β 0 −2

⎞1/2 c(t)[x(t + 1)]2 ⎠

+

t=α−1

β 0 −2

|a(t)||u(t)|.

t=α−1

(7.77) Next, summing (7.57) from α − 1 to β0 − 1 gives x(β0 )u(β0 ) − x(α − 1)u(α − 1) =

β 0 −1

b(t)[u(t)]2 −

t=α−1

β 0 −1

c(t)[x(t + 1)]2 ,

t=α−1

i.e., x(β0 ) [u(β0 ) + c(β0 − 1)x(β0 )] − x(α − 1)u(α − 1) =

β 0 −1

β 0 −2

b(t)[u(t)] − 2

t=α−1

c(t)[x(t + 1)]2 .

(7.78)

t=α−1

We proceed to show that x(β0 ) [u(β0 ) + c(β0 − 1)x(β0 )] > 0

(7.79)

holds. Indeed, from the second equation in (7.53), we have, for β0 − 1, [1 − a(β0 − 1)] u(β0 − 1) = u(β0 ) + c(β0 − 1)x(β0 ), which upon multiplication by x(β0 ) yields [1 − a(β0 − 1)] x(β0 )u(β0 − 1) = x(β0 ) [u(β0 ) + c(β0 − 1)x(β0 )] .

(7.80)

On the other hand, from the inequalities x(β0 − 1)x(β0 ) < 0 and

x(β0 − 1)u(β0 − 1) < 0,

it follows that u(β0 − 1)x(β0 ) > 0. Therefore, (7.79) follows from (7.80). By virtue of (7.76) and (7.79), from (7.78), the inequality β 0 −2 t=α−1

c(t)[x(t + 1)] < 2

β 0 −1 t=α−1

b(t)[u(t)]2

7.5 Discrete Linear Hamiltonian Systems

425

follows. In view of (7.77), the last estimate above yields ⎛ |u(s)| < ⎝

β 0 −2

⎞1/2 ⎛ c(t)⎠

⎝

t=α−1

⎞1/2

β 0 −1

b(t)[u(t)]2 ⎠

β 0 −2

+

t=α−1

|a(t)||u(t)|

(7.81)

t=α−1

for all s ∈ [α − 1, β0 − 1]. Choose s0 ∈ [α − 1, β0 − 1] such that |u(s0 )| =

max

α−1≤s≤β0 −1

|u(s)|.

Then, |u(s0 )| > 0, and from (7.81), we have ⎫ ⎧⎛ ⎞1/2 ⎪ ⎪ β β 0 −2 0 −1 0 −2 ⎬ ⎨ β ⎠ ⎝ |u(s0 )| < |u(s0 )| c(t) b(t) + |a(t)| . ⎪ ⎪ ⎭ ⎩ t=α−1 t=α−1 t=α−1 Hence, dividing by |u(s0 )|, we get ⎛ ⎝

β 0 −2

c(t)

t=α−1

⎞1/2

β 0 −1

b(t)⎠

+

t=α−1

β 0 −2

|a(t)| > 1.

t=α−1

Since β0 ≤ β, from the latter inequality, (7.72) follows. In the second part of this proof, we consider the case when x(t0 ) = 0 for some t0 ∈ [α + 1, β − 2]. In this case, applying Theorem 7.29 to the points t0 and β, we get the inequality β−2

⎛

β−2

|a(t)| + ⎝

t=t0

c(t)

t=t0

β−2

⎞1/2 b(t)⎠

> 1.

t=t0



Therefore, (7.72) holds in this case as well. Combining Theorems 7.28, 7.29, 7.30, and 7.31 yields the following corollary. Corollary 7.32 (Lyapunov-Type Inequality) Suppose 1 − a(t) > 0,

b(t) > 0,

c(t) > 0 for all

t ∈Z

and let α, β ∈ Z with α ≤ β − 2. Assume (7.53) has a real solution (x, u) such that x has generalized zeros at α and β and x is not identically zero on [α, β]. Then, the inequality

426

7 Difference Equations

⎛

β−2

|a(t)| + ⎝

t=α−1

β−1

β−2

b(t)

t=α−1

⎞1/2 c(t)⎠

>1

t=α−1

holds. In 2012, Zhang and Tang [306] considered the linear Hamiltonian system 

Δx(n) = α(n)x(n + 1) + β(n)y(n), Δy(n) = −γ (n)x(n + 1) − α(n)y(n),

(7.82)

where α, β, γ are real-valued functions defined on Z. They obtained a Lyapunovtype inequality better than (7.72) given in Theorem 7.31 by using some simpler methods different from those used by Guseinov and Kaymakçalan [143]. Under the assumption β(n) ≥ 0

and

1 − α(n) > 0 for all

n ∈ Z,

(7.83)

Zhang and Tang [306] obtained the inequality b−1

 +

γ (n) +

n

β(τ )

τ =a

n=a

b

1/2 ≥2

β(τ )

τ =n+1

b−1 %

Θ[α(n)],

(7.84)

n=a

where   −1 Θ[α(n)] = min 1 − α + (n), 1 + α − (n)

(7.85)

and α ± (n) = max{±α(n), 0}. When the right end-point b is a usual zero, we obtain a better Lyapunov-type inequality than (7.84), namely b−2

n=a

 +

γ (n) +

n

β(τ )

τ =a

b−1

τ =n+1

1/2 β(τ )

≥2

b−2 % n=a

Applying (7.86) to the equation Δ2 x(n) + q(n)x(n + 1) = 0,

Θ[α(n)].

(7.86)

7.5 Discrete Linear Hamiltonian Systems

427

we can obtain the Lyapunov-type inequality b−2

 q(n) (n + 1 − a)(b − n − 1) ≥ 2.

(7.87)

n=a

It is easy to see that (7.87) is better than (7.29) given by Cheng [83]. To illustrate that (7.84) is better than (7.72), note that 

n

β(τ )

τ =a

1/2

b

1 β(τ ) 2 τ =a n

≤

β(τ )

τ =n+1

and b−1 %

b−1 %

Θ[α(n)] ≥

n=a

 [1 − |α(n)|] ≥ 1 −

n=a

b−1





|α(n)| ≥ 1 −

n=a

b−1

2 |α(n)|

n=a

when b−1

|α(n)| ≤ 1.

n=a

Let us define ζ (n) :=

n

n %

β(τ )

τ =a

[1 − α(s)]−2

(7.88)

s=τ

and η(n) :=

b

τ% −1

β(τ )

τ =n+1

[1 − α(s)]2 ,

(7.89)

s=n+1

and for λ ∈ [0, 1), we define ζλ (n) := (1 − λ)β(a)

n %

[1 − α(s)]−2 +

s=a

n

β(τ )

τ =a+1

n %

[1 − α(s)]−2

(7.90)

s=τ

and ηλ (n) := λβ(a)

b−1 % s=n+1

[1 − α(s)]2 +

b−1

τ =n+1

β(τ )

τ% −1 s=n+1

[1 − α(s)]2 .

(7.91)

428

7 Difference Equations

Theorem 7.33 (Hartman-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that x(a) = 0

or

x(a)x(a + 1) < 0,

x(b) = 0 or

x(b)x(b + 1) < 0

(7.92)

hold. If max |x(n)| > 0,

(7.93)

b−1

ζ (n)η(n) + γ (n) ≥ 1 ζ (n) + η(n) n=a

(7.94)

a≤n≤b

then the inequality

holds. Proof It follows from (7.92) and (7.93) that there exist ξ1 , ξ2 ∈ [0, 1) such that (1 − ξ1 )x(a) + ξ1 x(a + 1) = 0

(7.95)

(1 − ξ2 )x(b) + ξ2 x(b + 1) = 0.

(7.96)

and

Multiplying the first equation in (7.82) by y(n) and the second equation in (7.82) by x(n + 1) and then adding, we get Δ(x(n)y(n)) = β(n)[y(n)]2 − γ (n)[x(n + 1)]2 .

(7.97)

Summing (7.97) from a to b − 1, we obtain x(b)y(b) − x(a)y(a) =

b−1

β(n)[y(n)]2 −

n=a

b−1

γ (n)[x(n + 1)]2 .

(7.98)

n=a

From the first equation in (7.82), we have [1 − α(n)]x(n + 1) = x(n) + β(n)y(n).

(7.99)

Combining (7.99) with (7.95), we have x(a) = −

ξ1 β(a) y(a). 1 − (1 − ξ1 )α(a)

(7.100)

7.5 Discrete Linear Hamiltonian Systems

429

Similarly, it follows from (7.99) and (7.96) that x(b) = −

ξ2 β(b) y(b) 1 − (1 − ξ2 )α(b)

(7.101)

holds. Substituting (7.100) and (7.101) into (7.98), we have b−1

β(n)[y(n)]2 −

n=a

b−1

γ (n)[x(n + 1)]2

n=a

=−

ξ2 β(b) ξ1 β(a) [y(b)]2 + [y(a)]2 , 1 − (1 − ξ2 )α(b) 1 − (1 − ξ1 )α(a)

which implies b−1

(1 − ξ1 )[1 − α(a)] ξ2 β(b) β(a)[y(a)]2 + [y(b)]2 β(n)[y(n)]2 + 1 − (1 − ξ1 )α(a) 1 − (1 − ξ2 )α(b) n=a+1

=

b−1

γ (n)[x(n + 1)]2 .

(7.102)

ξ2 β(b) 1 − (1 − ξ2 )α(b)

(7.103)

n=a

Denote ˜ β(a) =

(1 − ξ1 )[1 − α(a)] β(a), 1 − (1 − ξ1 )α(a)

˜ β(b) =

and ˜ β(n) = β(n),

a + 1 ≤ n ≤ b − 1.

(7.104)

Then, we can rewrite (7.102) as b

n=a

2 ˜ β(n)[y(n)] =

b−1

γ (n)[x(n + 1)]2 .

n=a

From (7.99), (7.100), (7.103), and (7.104), we obtain

(7.105)

430

7 Difference Equations

x(n + 1) =x(a)

n %

[1 − α(s)]−1 +

n %

[1 − α(s)]−1

s=τ

n %

ξ1 β(a) y(a) [1 − α(s)]−1 1 − (1 − ξ1 )α(a) s=a +

n

β(τ )y(τ )

τ =a

=

β(τ )y(τ )

τ =a

s=a

=

n

n

n %

(7.106)

[1 − α(s)]−1

s=τ

˜ )y(τ ) β(τ

τ =a

n %

[1 − α(s)]−1

s=τ

for a ≤ n ≤ b − 1. Similarly, from (7.99), (7.101), (7.103), and (7.104), we have x(n + 1) =x(b)

b−1 %

β(τ )y(τ )

τ =n+1

s=n+1

=−

b−1

[1 − α(s)] −

τ% −1

[1 − α(s)]

s=n+1

b−1 % ξ2 β(b) [1 − α(s)] y(b) 1 − (1 − ξ2 )α(b) s=n+1

−

b−1

β(τ )y(τ )

τ =n+1

=−

b

τ% −1

(7.107)

[1 − α(s)]

s=n+1

˜ )y(τ ) β(τ

τ =n+1

τ% −1

[1 − α(s)]

s=n+1

for a ≤ n ≤ b − 1. Since ˜ 0 ≤ β(n) ≤ β(n),

a ≤ n ≤ b,

(7.108)

it follows from (7.88), (7.106), and the Cauchy inequality that  [x(n + 1)]2 =  ≤

˜ )y(τ ) β(τ

τ =a n

˜ ) β(τ

s=τ

n

n %

β(τ )

τ =a

= ζ (n)

n %

2 [1 − α(s)]−1

s=τ n %

τ =a

 ≤

n

n

τ =a

[1 − α(s)]

 −2

 [1 − α(s)]−2

s=τ

˜ )[y(τ )]2 β(τ

n

˜ )[y(τ )]2 β(τ

τ =a n

τ =a

(7.109) ˜ )[y(τ )]2 β(τ

7.5 Discrete Linear Hamiltonian Systems

431

holds for a ≤ n ≤ b − 1. Similarly, it follows from (7.89), (7.107), (7.108), and the Cauchy inequality that  [x(n + 1)] = 2

 ≤

τ =n+1 b

τ% −1

˜ )y(τ ) β(τ

˜ ) β(τ

τ% −1 s=n+1

b

τ% −1

β(τ )

τ =n+1

= η(n)

2 [1 − α(s)]

s=n+1

τ =n+1

 ≤

b

[1 − α(s)]

 [1 − α(s)]2

s=n+1

b

 2

b

˜ )[y(τ )]2 β(τ

τ =n+1 b

(7.110) ˜ )[y(τ )]2 β(τ

τ =n+1

˜ )[y(τ )]2 β(τ

τ =n+1

holds for a ≤ n ≤ b − 1. From (7.109) and (7.110), we obtain b ζ (n)η(n) ˜ )[y(τ )]2 [x(n + 1)] ≤ β(τ ζ (n) + η(n) τ =a 2

(7.111)

for a ≤ n ≤ b − 1. Combining (7.111) with (7.105), we have b−1

γ + (n)[x(n + 1)]2 ≤

n=a

=

≤

b−1 b

ζ (n)η(n) + 2 ˜ β(n)[y(n)] γ (n) ζ (n) + ζ (n) n=a n=a b−1 b−1

ζ (n)η(n) + γ (n)[x(n + 1)]2 γ (n) ζ (n) + ζ (n) n=a n=a b−1 b−1

ζ (n)η(n) + γ (n) γ + (n)[x(n + 1)]2 . ζ (n) + ζ (n) n=a n=a (7.112)

We claim that b−1

γ + (n)[x(n + 1)]2 > 0.

(7.113)

γ + (n)[x(n + 1)]2 = 0.

(7.114)

n=a

If (7.113) is not true, then b−1

n=a

432

7 Difference Equations

From (7.105) and (7.114), we have 0≤

b

2 ˜ β(n)[y(n)] =

n=a

b−1

γ (n)[x(n + 1)]2 =

n=a

b−1

γ + (n)[x(n + 1)]2 = 0.

n=a

It follows that ˜ β(n)y(n) = 0,

a ≤ n ≤ b.

(7.115)

Combining (7.106) with (7.115), we obtain x(a + 1) = x(a + 2) = · · · = x(b) = 0, which, together with (7.95), implies x(a) = 0. This contradicts (7.92). Therefore, (7.113) holds. Hence, it follows from (7.112) and (7.113) that (7.94) holds.  ˜ In the case x(b) = 0, i.e., ξ2 = 0, and so β(b) = 0, we have the equation b−1

2 ˜ β(n)[y(n)] =

n=a

b−2

γ (n)[x(n + 1)]2

n=a

and inequality  [x(n + 1)] = 2

 ≤

b−1

˜ )y(τ ) β(τ

τ =n+1 b−1

≤

˜ ) β(τ

b−1

[1 − α(s)]

τ% −1



[1 − α(s)]

2

s=n+1

β(τ )

τ =n+1

= η0 (n)

2

s=n+1

τ =n+1



τ% −1

b−1

τ% −1

 [1 − α(s)]2

s=n+1

˜ )[y(τ )]2 , β(τ

b−1

˜ )[y(τ )]2 β(τ

τ =n+1 b−1

˜ )[y(τ )]2 β(τ

τ =n+1

a ≤n≤b−1

τ =n+1

instead of (7.105) and (7.110), respectively. Similar to the proof of Theorem 7.33, we can show the following result. Theorem 7.34 (Hartman-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 2. Suppose (7.82) has a real solution (x, y) such that x(a) = 0

or

x(a)x(a + 1) < 0,

hold. If x(n) ≡ 0 in [a, b], then the inequality

x(b) = 0

(7.116)

7.5 Discrete Linear Hamiltonian Systems

433

b−2

ζ (n)η0 (n) + γ (n) ≥ 1 ζ (n) + η0 (n) n=a

holds, where ζ and η0 are defined by (7.88) and (7.91), respectively. Theorem 7.35 (Hartman-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that x(a) = 0

or

x(a)x(a+1) < 0,

(x(b), y(b)) = (μ1 x(a), μ2 y(a))

(7.117)

with 0 < μ21 ≤ μ1 μ2 ≤ 1. If x(n) ≡ 0 in [a, b], then there exists θ ∈ [0, 1) such that the inequality b−1

ζθ (n)ηθ (n) + γ (n) ≥ 1 ζ (n) + ηθ (n) n=a θ

(7.118)

holds, where ζθ and ηθ are defined by (7.90) and (7.91), respectively. Proof It follows from (7.92) and (7.93) that there exists ξ1 ∈ [0, 1) such that (7.95) holds. Further, by the proof of Theorem 7.33, (7.97), (7.98), (7.99), and (7.100) hold. Since (x(b), y(b)) = (μ1 x(a), μ2 y(a)), by (7.98), we have (μ1 μ2 − 1)x(a)y(a) =

b−1

β(n)[y(n)]2 −

n=a

b−1

γ (n)[x(n + 1)]2 .

(7.119)

n=a

Substituting (7.100) into (7.119), we have b−1

β(n)[y(n)]2 −

n=a

b−1

γ (n)[x(n + 1)]2 =

n=a

(1 − μ1 μ2 )ξ1 β(a) [y(a)]2 , 1 − (1 − ξ1 )α(a)

which implies κβ(a)[y(a)]2 +

b−1

n=a+1

β(n)[y(n)]2 =

b−1

γ (n)[x(n + 1)]2 ,

(7.120)

n=a

where κ=

1 − (1 − μ1 μ2 )ξ1 − (1 − ξ1 )α(a) . 1 − (1 − ξ1 )α(a)

(7.121)

434

7 Difference Equations

From (7.99) and (7.100), we obtain n %

x(n + 1) =x(a)

[1 − α(s)]−1 +

+

n %

β(τ )y(τ )

τ =a

s=a

=−

n

[1 − α(s)]−1

s=τ

n %

ξ1 β(a) [1 − α(s)]−1 y(a) 1 − (1 − ξ1 )α(a) s=a n

β(τ )y(τ )

τ =a

n %

[1 − α(s)]−1

s=τ n %

=κ1 β(a)y(a)

n

[1 − α(s)]−1 +

β(τ )y(τ )

τ =a+1

s=a

n %

[1 − α(s)]−1

s=τ

(7.122)

for a ≤ n ≤ b − 1, where (1 − ξ1 )[1 − α(a)] . 1 − (1 − ξ1 )α(a)

κ1 =

(7.123)

Similarly, from (7.99), (7.100), and the fact that x(b) = μ1 x(a), we have x(n + 1) = x(b)

b−1 %

[1 − α(s)] −

b−1 %

[1 − α(s)] − b−1 %

b−1

[1 − α(s)] −

s=n+1

τ% −1

[1 − α(s)]

s=n+1

β(τ )y(τ )

τ =n+1

s=n+1

= κ2 β(a)y(a)

β(τ )y(τ )

τ =n+1

s=n+1

= μ1 x(a)

b−1

b−1

τ% −1

[1 − α(s)]

s=n+1

β(τ )y(τ )

τ =n+1

τ% −1

[1 − α(s)]

s=n+1

(7.124)

for a ≤ n ≤ b − 1, where κ2 =

μ1 ξ1 . 1 − (1 − ξ1 )α(a)

It follows from (7.122) and the Cauchy inequality that

(7.125)

7.5 Discrete Linear Hamiltonian Systems

 [x(n+1)] = κ1 β(a)y(a) 2

n %

435

[1 − α(s)]

−1

+

n %

≤ κ1 β(a)

n

[1 − α(s)]−2 +

× κ1 β(a)[y(a)] + 2

n

n %

β(τ )



[1 − α(s)] 

[1 − α(s)]−2



β(τ )[y(τ )]

n

2 −1

s=τ

2

τ =a+1

= ζ˜ (n) κ1 β(a)[y(a)]2 +

n % s=τ

τ =a+1

s=a



β(τ )y(τ )

τ =a+1

s=a



n

 β(τ )[y(τ )]

2

τ =a+1

(7.126)

holds for a ≤ n ≤ b − 1, where ζ˜ (n) = κ1 β(a)

n %

n

[1 − α(s)]−2 +

β(τ )

τ =a+1

s=a

n %

[1 − α(s)]−2 .

(7.127)

s=τ

Similarly, it follows from (7.124) and the Cauchy inequality that  [x(n+1)]2 = κ2 β(a)y(a)

b−1 %

[1 − α(s)] +

β(τ )y(τ )

τ =n+1

s=n+1



b−1

τ% −1

2 [1 − α(s)]

s=n+1

b−1 b−1 τ% −1 %

κ2 2 ≤ β(a) [1 − α(s)] + β(τ ) [1 − α(s)]2 μ1 τ =n+1

s=n+1



× μ1 κ2 β(a)[y(a)]2 + 

b−1

s=n+1



β(τ )[y(τ )]2

τ =n+1

=η(n) ˜ μ1 κ2 β(a)[y(a)]2 +



b−1

 β(τ )[y(τ )]2

τ =n+1

(7.128)

holds for a ≤ n ≤ b − 1, where η(n) ˜ =

b−1 b−1 τ% −1 %

κ2 β(a) [1 − α(s)]2 + β(τ ) [1 − α(s)]2 . μ1 s=n+1

τ =n+1

(7.129)

s=n+1

Since 0 < μ21 ≤ μ1 μ2 ≤ 1, it follows from (7.121), (7.123), and (7.125) that κ1 + μ1 κ2 ≤ κ. Hence, from (7.126) and (7.128), we obtain

436

7 Difference Equations

[x(n + 1)] ≤ 2



ζ˜ (n)η(n) ˜ ζ˜ (n) + η(n) ˜

b−1

κβ(a)[y(a)] + 2

 β(τ )[y(τ )]

2

(7.130)

τ =a+1

for a ≤ n ≤ b − 1. Combining (7.130) with (7.120), we have b−1

n=a

  b−1 ˜ b−1

ζ (n) η(n) ˜ γ + (n) κβ(a)[y(a)]2 + γ + (n)[x(n+1)]2 ≤ β(τ )[y(τ )]2 ζ˜ (n)+η(n) ˜ τ =a+1

n=a

=

b−1 ˜ b−1

ζ (n)η(n) ˜ γ (n)[x(n + 1)]2 γ + (n) ζ˜ (n) + η(n) ˜

n=a

≤

n=a

b−1

ζ˜ (n)η(n) ˜

n=a

ζ˜ (n) + η(n) ˜

γ + (n)

b−1

γ + (n)[x(n + 1)]2 ,

n=a

which together with (7.113) implies b−1 ˜

ζ (n)η(n) ˜ γ + (n) ≥ 1. ˜ζ (n) + η(n) ˜

(7.131)

n=a

Let κ2 /μ1 = θ with θ ∈ [0, 1). Since κ1 + κ2 /μ1 = 1, it follows from (7.127) and (7.129) that ζ˜ (n) = ζθ (n) and η(n) ˜ = ηθ (n) hold. Substituting these into (7.131), we obtain (7.118).  Corollary 7.36 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that (7.92) holds. If max |x(n)| > 0,

a≤n≤b

then the inequality b−1

 +

γ (n) +

n=a

n

τ =a

β(τ )

b

1/2 β(τ )

τ =n+1

≥2

b−1 %

Θ[α(n)]

n=a

holds, where Θ[α(n)] is defined in (7.85). Proof Since ζ (n) + η(n) ≥ 2[ζ (n)η(n)]1/2 , we have

(7.132)

7.5 Discrete Linear Hamiltonian Systems

1≤

437

b−1

ζ (n)η(n) + γ (n) ζ (n) + η(n) n=a

1 [ζ (n)η(n)]1/2 γ + (n) 2 n=a b−1

≤

 n 1/2 b−1 n τ% −1 b

%

1 + −2 2 = γ (n) β(τ ) [1 − α(s)] β(τ ) [1−α(s)] 2 n=a τ =a s=τ τ =n+1

s=n+1



b−1 n n τ% −1 b

%   −2 2 1 + 1 − α + (s) 1−α − (s) ≤ γ (n) β(τ ) β(τ ) 2 n=a τ =a s=τ τ =n+1



1/2

b−1 n b

1 + ≤ γ (n) β(τ ) β(τ ) 2 n=a τ =a τ =n+1

1/2

s=n+1

n % %   −1 b−1  + 1 − α (s) 1 − α − (s) s=a

s=n+1



1/2 b−1 b−1 n b

% 1 + γ (n) β(τ ) β(τ ) {Θ[α(n)]}−1 , ≤ 2 n=a τ =a s=a τ =n+1



which implies (7.132). Since 

n

b

β(τ )

τ =a

1/2 β(τ )

1 β(n), 2 n=a b

≤

τ =n+1

it follows from (7.132) that the following corollary is true. Corollary 7.37 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that (7.92) holds. If max |x(n)| > 0,

a≤n≤b

then the inequality b

n=a

β(n)

b−1

γ + (n) ≥ 4

n=a

b−1 %

Θ[α(n)]

n=a

holds, where Θ[α(n)] is defined in (7.85). By a fashion similar as in the proofs of Corollaries 7.36 and 7.37, we can prove the following corollaries using Theorems 7.34 and 7.35 instead of Theorem 7.33, respectively.

438

7 Difference Equations

Corollary 7.38 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 2. Suppose (7.82) has a real solution (x, y) such that (7.116) holds. If x(n) ≡ 0 in [a, b], then the inequality b−2

 +

γ (n)

n

β(τ )

τ =a

n=a

b−1

1/2 ≥2

β(τ )

τ =n+1

b−2 %

Θ[α(n)]

n=a

holds, where Θ[α(n)] is defined in (7.85). Corollary 7.39 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 2. Suppose (7.82) has a real solution (x, y) such that (7.116) holds. If x(n) ≡ 0 in [a, b], then the inequality b−1

β(n)

n=a

b−2

γ + (n) ≥ 4

n=a

b−2 %

Θ[α(n)]

n=a

holds, where Θ[α(n)] is defined in (7.85). Corollary 7.40 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that (7.117) holds. If x(n) ≡ 0 in [a, b], then there exists θ ∈ [0, 1) such that the inequality b−1

0 +

γ (n)

(1 − θ )β(n) +

n

10 β(τ )

θβ(n) +

τ =a+1

n=a

b−1n

11/2 β(τ )

τ =n+1

≥2

b−1 %

Θ[α(n)]

(7.133)

n=a

holds, where Θ[α(n)] is defined in (7.85). Corollary 7.41 (Lyapunov-Type Inequality) Assume (7.83) and let a, b ∈ Z with a ≤ b − 1. Suppose (7.82) has a real solution (x, y) such that (7.117) holds. If x(n) ≡ 0 in [a, b], then the inequality b−1

n=a

β(n)

b−1

γ + (n) ≥ 4

n=a

b−1 %

Θ[α(n)]

n=a

holds, where Θ[α(n)] is defined in (7.85). As applications of Lyapunov inequalities, we obtain in Sect. 7.5.1 a disconjugacy criterion, and in Sect. 7.5.2, instability and stability criteria for discrete Hamiltonian systems are provided. Related references include [34, 35, 47, 48, 91, 93, 94, 129, 142, 143, 150, 306].

7.5 Discrete Linear Hamiltonian Systems

439

7.5.1 A Disconjugacy Criterion Let α, β ∈ Z with α ≤ β − 2. Consider the discrete linear Hamiltonian system 

Δx(t) = a(t)x(t + 1) + b(t)u(t), Δu(t) = −c(t)x(t + 1) − a(t)u(t),

(7.134)

where t ∈ [α, β] ∩ Z. We assume that the coefficients a, b, c are real-valued functions defined on [α, β] and 1 − a(t) > 0 and

b(t) > 0 for all

t ∈ [α, β].

(7.135)

Note that each solution (x, u) of (7.134) is a vector-valued function defined on [α, β + 1]. Now, we define the concept of a relatively generalized zero for the component x of a real solution (x, u) of (7.134) and also the concept of disconjugacy of this system on [α, β + 1]. The definition is relative to the interval [α, β + 1], and the left end-point α is treated separately. We say x (or (x, u)) has a relatively generalized zero at α if and only if x(α) = 0, while we say x has a relatively generalized zero at t0 > α provided either x(t0 ) = 0 or x(t0 − 1)x(t0 ) < 0. Finally, we say that (7.134) is disconjugate on [α, β + 1] provided there is no real solution (x, u) of this system with x nontrivial and having two (or more) relatively generalized zeros in [α, β +1]. Notice that under (7.135), the given definitions of a relatively generalized zero and of disconjugacy are equivalent to those given in [47] and [22, page 354]. Theorem 7.42 Assume (7.135). If β−1

⎛ ⎞1/2 β β−1

|a(t)| + ⎝ b(t) c+ (t)⎠ ≤ 1,

t=α

t=α

(7.136)

t=α

then (7.134) is disconjugate on [α, β + 1]. Proof Suppose, on the contrary, that (7.134) is not disconjugate on [α, β +1]. Then, there exists (see [47] and [22, Theorem 9.29]) a real solution (x, u) of (7.134) with x nontrivial and such that x(α) = 0 and that x has a generalized zero β0 in [α + 1, β + 1]. We have β0 > α + 1 and either x(β0 ) = 0 or x(β0 − 1)x(β0 ) < 0. Therefore, applying Theorems 7.28 and 7.29, we get β 0 −2 t=α

contradicting (7.136).

⎛

β 0 −1

|a(t)| + ⎝

t=α

b(t)

β 0 −2

⎞1/2 c+ (t)⎠

> 1,

t=α



440

7 Difference Equations

7.5.2 Stability Criteria Consider the discrete linear Hamiltonian system 

Δx(t) = a(t)x(t + 1) + b(t)u(t), Δu(t) = −c(t)x(t + 1) − a(t)u(t),

(7.137)

where t ∈ Z. Let the coefficients a, b, c be real-valued functions defined on Z and suppose 1 − a(t) = 0 for all

t ∈ Z.

In addition, we assume that the coefficients of (7.137) are periodic, i.e., a(t + N ) = a(t),

b(t + N) = b(t),

c(t + N) = c(t) for all

t ∈ Z, (7.138)

where N ≥ 2 is the fixed integer period. We first present some facts about (7.137) and (7.138) that will be necessary for the subsequent discussions. Setting

0 1 J = , −1 0



ca H = ab



and

x ϕ= , u



x(t + 1) , ϕ (t) = u(t) σ

we can write (7.137) in the vector form t ∈ Z.

Δϕ(t) = J H (t)ϕ σ (t),

(7.139)

Let us seek a nonzero complex number ρ and a nontrivial solution ϕ of (7.139) such that ϕ(t + N) = ρϕ(t),

t ∈ Z.

Denote ϕ1 =



x1 u1

and

ϕ2 =



x2 , u2

(7.140)

7.5 Discrete Linear Hamiltonian Systems

441

where x1 and x2 are the solutions of (7.139) satisfying the initial conditions x1 (0) = 1,

u1 (0) = 0

and

x2 (0) = 0,

u2 (0) = 1,

(7.141)

and set

  x1 x2 . Φ = ϕ1 ϕ2 = u1 u2

(7.142)

Then, 

ΔΦ(t) = J H (t)Φ σ (t) for

t ∈ Z,

Φ(0) = I,

(7.143)

where

x1 (t + 1) x2 (t + 1) Φ (t) = u2 (t) u1 (t) σ

and



10 I= . 01

The general solution ϕ of (7.139) has the form ϕ(t) = c1 ϕ1 (t) + c2 ϕ2 (t) = Φ(t)c,

(7.144)

where c1 , c2 are arbitrary complex constants and c is a column vector with the components c1 and c2 . Substituting (7.144) in (7.140), we obtain Φ(t + N)c = ρΦ(t)c,

t ∈ Z.

(7.145)

On the other hand, Φ(t + N) = Φ(t)Φ(N),

t ∈ Z.

(7.146)

Indeed, because H (t + N) = H (t), both the left-hand side and the right-hand side of (7.146) are solutions of the matrix system in (7.143), and for t = 0, they coincide. Then, by the uniqueness of solutions, (7.146) holds. Since det Φ(t) = 1, the matrix Φ(t) is invertible for all t ∈ Z. Therefore, from (7.145), by (7.146), we get Φ(N)c = ρc. Thus, in order that the vector function ϕ defined by (7.144) is a nontrivial solution of (7.139) satisfying (7.140), it is necessary and sufficient that ρ is an eigenvalue and c is a corresponding eigenvector of the matrix Φ(N).

442

7 Difference Equations

The matrix Φ(N) is called the monodromy matrix of (7.137). The eigenvalues of the matrix Φ(N), i.e., the roots of the algebraic equation det[Φ(N) − ρI ] = 0,

(7.147)

are called the multipliers of (7.137). Equation (7.147) can be written as the quadratic equation ρ 2 − Dρ + 1 = 0,

(7.148)

D = x1 (N ) + u2 (N ).

(7.149)

where

The roots of (7.148) are given by ρ1,2 =

  1 D ± D2 − 4 . 2

(7.150)

Definition 7.43 System (7.137) is said to be 1. unstable if all nontrivial solutions are unbounded on Z, 2. conditionally stable if there exists a nontrivial solution which is bounded on Z, 3. stable if all solutions are bounded on Z. Since the coefficient of (7.139) and the initial conditions given in (7.141) are real, the solutions ϕ1 , ϕ2 are real, and hence, the number D defined by (7.149) is real. The following statement can be proved in a standard way (see [121, 191, 293]). Lemma 7.44 System (7.137) is unstable if |D| > 2. The system is stable if |D| < 2. If |D| = 2, then (7.137) is stable in the case u1 (N ) = x2 (N ) = 0, but conditionally stable or not stable otherwise. Theorem 7.45 Assume 1 − a(t) > 0,

b(t) ≥ 0,

c(t) ≤ 0 for all

t ∈Z

(7.151)

 N  % b(t)c(t) 1 − a(t) − > 1. 1 − a(t)

(7.152)

and N % t=1

1 ≥ 1, 1 − a(t)

t=1

Then, (7.137) is unstable. Proof Our aim is to show that, under the hypotheses of the theorem, the inequalities x1 (N ) ≥ 1 and

u2 (N ) > 1

(7.153)

7.5 Discrete Linear Hamiltonian Systems

443

hold. Then, D = x1 (N ) + u2 (N ) > 2, and therefore, by Lemma 7.44, (7.137) is unstable. The matrix system in (7.143) can be written as Φ(t + 1) = M(t)Φ(t), where

t ∈ Z,

 1 b m11 m12 1−a = 1−ac M= m21 m22 − 1−a 1 − a −

(7.154)

 (7.155)

.

bc 1−a

Conditions (7.138), (7.151), and (7.152) are, respectively, equivalent to mij (t + N) = mij (t), m12 (t) ≥ 0,

m11 (t) > 0,

i, j = 1, 2,

m21 (t) ≥ 0,

m22 (t) > 0,

(7.156)

and N %

m11 (t) ≥ 1,

t=1

N %

m22 (t) > 1.

t=1

From (7.154) and taking into account the initial condition in (7.143), we have Φ(N) = M(N − 1)M(N − 2) · · · M(0). Hence, by (7.155) and (7.142), we get x1 (N ) ≥

N −1 %

m11 (t) =

t=0

N %

m11 (t),

u2 (N ) ≥

t=1

N −1 %

m22 (t) =

t=0

N %

m22 (t).

t=1

Consequently, by (7.156), (7.153) hold. The proof is therefore complete.



Theorem 7.46 Assume b(t) > 0,

c(t) > 0,

b(t)c(t) − [a(t)]2 ≥ 0,

b(t)c(t) − [a(t)]2 ≡ 0,



1/2

t ∈Z (7.157)

and N

t=1

|a(t)| +

b0 +

N

t=1

 b(t)

N

c(t)

≤ 1,

(7.158)

t=1

where b0 = max{b(1), b(2), . . . , b(N )}.

(7.159)

444

7 Difference Equations

Then, (7.137) is stable. Proof It is sufficient by Lemma 7.44 to show that D 2 < 4. Assuming on the contrary that D 2 ≥ 4 leads to a contradiction.  We prove the following lemma. Lemma 7.47 If D 2 ≥ 4, then (7.137) has a real nontrivial solution (x, u) such that x has a generalized zero in [1, N]. Proof If D 2 ≥ 4, then it follows from (7.150) that (7.137) has a real nontrivial solution (x, u) satisfying x(t + N) = ρx(t),

u(t + N) = ρu(t),

t ∈ Z,

(7.160)

where ρ is a nonzero real number. Now, we show that x must have at least one generalized zero in the segment [1, N]. If not, then by (7.160), x does not have any generalized zero in Z, so x(t) = 0 and x(t − 1)x(t) > 0 for all t ∈ Z. Multiplying the first equation in (7.137) by u(t) and the second equation in (7.137) by x(t) and then subtracting, we get u(t)Δx(t) − x(t)Δu(t) = a(t)x(t + 1)u(t) + b(t)[u(t)]2 + c(t)x(t)x(t + 1) + a(t)x(t)u(t). Hence, substituting x(t + 1) =

b(t) 1 x(t) + u(t) 1 − a(t) 1 − a(t)

on the right-hand side of the previous equation and then dividing both sides by x(t)x(t + 1), we obtain

u(t) −Δ x(t)



  c(t)[x(t)]2 + 2a(t) − [a(t)]2 + b(t)c(t) x(t)u(t) + b(t)[u(t)]2 = . [1 − a(t)]x(t)x(t + 1)

Summing the latter equation from 1 to N and taking into account that, by (7.160), u(N + 1) u(1) ρu(1) u(1) − = − =0 x(N + 1) x(1) ρx(1) x(1) holds, we get   N

c(t)[x(t)]2 + 2a(t) − [a(t)]2 + b(t)c(t) x(t)u(t) + b(t)[u(t)]2 = 0. [1 − a(t)]x(t)x(t + 1) t=1 (7.161)

7.5 Discrete Linear Hamiltonian Systems

445

From (7.158) and b(t) > 0, c(t) > 0, we obtain |a(t)| < 1,

0 < b(t)c(t) < 1,

t ∈ Z.

Therefore, 1 − a(t) > 0, and the denominator of the fraction under the sum sign in (7.161) is positive. The numerator is equal to cx 2 + dxu + bu2 =

1  (dx + 2bu)2 + (4bc − d 2 )x 2 , 4b

where d = 2a − a 2 + bc. On the other hand, ! 4bc − d 2 = (bc − a 2 ) (2 − a)2 − bc . Hence, taking into account bc − a 2 ≥ 0,

2 − a > 1,

0 < bc < 1,

we get 4bc − d 2 ≥ 0 and 4bc − d 2 = 0. Therefore, the numerator of the fraction under the summation sign in (7.161) is nonnegative and not identically zero, since (x, u) is nontrivial. Consequently, (7.161) leads to a contradiction, and hence, the proof is complete.  Continuation of the proof of Theorem 7.46 Let (x, u) be the solution of (7.137) indicated in Lemma 7.47. So, x has at least one generalized zero α in [1, N]. From (7.160), we get that x also has a generalized zero at α + N. Applying Corollary 7.32 to the solution (x, u) and the points α and β = α + N, we get α+N

−2 t=α−1

|a(t)| +

α+N −1

b(t)

t=α−1

α+N

−2

1/2 c(t)

> 1.

(7.162)

t=α−1

Next, noticing that, for any N-periodic function f defined on Z, the equality t0 +N

−1 t=t0

f (t) =

N

t=1

f (t)

446

7 Difference Equations

holds for all t0 ∈ Z, we have α+N

−2

|a(t)| =

N

t=α−1

α+N

−2

|a(t)|,

t=1

c(t) =

t=α−1

N

c(t),

t=1

and α+N

−1

b(t) = b(α − 1) +

α+N

−1

b(t) = b(α − 1) +

t=α

t=α−1

N

b(t) ≤ b0 +

N

t=1

b(t),

t=1

where b0 is defined by (7.159). Consequently, by (7.162), we have N

 |a(t)| +

b0 +

t=1

N

 b(t)

t=1

N

1/2 c(t)

> 1,

t=1

contradicting (7.158). Therefore, the inequality D 2 ≥ 4 cannot be true. Thus, D 2 < 4, and (7.137) is stable. The proof is therefore complete.  Furthermore, Zhang and Tang [306], applying the theorems that they obtained in Sect. 7.5.1, greatly improved (7.157) and (7.158) by the following conditions: There exists a nonnegative function θ such that |α(n)| ≤ θ (n)β(n),

n ∈ [1, N],

(7.163)

N  

  γ (n) − [θ (n)]2 β(n) > 0,

(7.164)

n=1

and N

β(n)

n=1

N

γ + (n) < 4

n=1

N %

(7.165)

Θ[α(n)].

n=1

To this end, it is assumed that (7.82) is N-periodic, i.e., the coefficients α, β, γ satisfy the periodicity conditions α(n + N ) = α(n),

β(n + N) = β(n),

γ (n + N) = γ (n),

Let u = (x, y)T and  A=

1 1−α γ − 1−α

β 1−α

1−α−

 βγ 1−α

.

n ∈ Z.

(7.166)

7.5 Discrete Linear Hamiltonian Systems

447

Then, det A(n) = 1 for n ∈ Z, and (7.82) can be written as u(n + 1) = A(n)u(n).

(7.167)

Let Φ with Φ(0) = I be a fundamental matrix solution of (7.167). Then, Φ(n + 1) = A(n)Φ(n),

Φ(n + N) = Φ(n)Φ(N),

n ∈ Z.

It follows that det Φ(n) = det Φ(0) = 1 for n ∈ Z. The Floquet multipliers (real or complex) of (7.167) are the roots of det(λI − Φ(N)) = 0, which is equivalent to λ2 − ρλ + 1 = 0, where ρ is the trace of the matrix Φ(N). Let λ1 and λ2 be the Floquet multipliers. Then, λ1 + λ2 = ρ,

λ1 λ2 = 1.

It follows from Floquet theory [7, 22, 121] that corresponding to each (complex) root λk , k = 1, 2, there exists a solution uk = (xk , yk )T of (7.167) (or (7.82)) with xk (n) ≡ 0 such that uk (n + N) = λk uk (n),

k = 1, 2

for all

n ∈ Z.

(7.168)

These are the so-called Floquet solutions of (7.167) (or (7.82)). Lemma 7.48 Assume (7.166). Suppose that there exists a nonnegative function θ such that (7.163) and (7.164) hold. If ρ 2 ≥ 4, then (7.82) has a nonzero solution (x, y) such that x has a generalized zero in [1, N]. Proof Suppose that |ρ| ≥ 2. Then, one has real Floquet multipliers λk and real Floquet solutions uk = (xk , yk )T , k = 1, 2. Let us consider any Floquet solution, say u1 = (x1 , y1 )T . We assert that x1 must have at least one generalized zero in the segment [1, N]. Otherwise, one may assume that x1 (n) > 0 for all n ∈ [1, N], and so x1 (n) > 0 for all n ∈ Z. Define z := y1 /x1 . Due to (7.168), one sees that z is N-periodic. From (7.82), we have

448

7 Difference Equations

Δz(n) =

x1 (n)Δy1 (n) − y1 (n)Δx1 (n) x1 (n)x1 (n + 1)

−γ (n)x1 (n)x1 (n + 1) − α(n) [x1 (n) + x1 (n + 1)] y1 (n) − β(n)y12 (n) x1 (n)x1 (n + 1)     y1 (n) y1 (n) y1 (n) y1 (n) + − β(n) · = −γ (n) − α(n) x1 (n) x1 (n + 1) x1 (n) x1 (n + 1)     y1 (n) y1 (n) − β(n)z(n) . = −γ (n) − α(n) z(n) + x1 (n + 1) x1 (n + 1) (7.169) Since x1 (n) > 0 for all n ∈ Z, from (7.99), we have =

1 + β(n)z(n) = 1 + β(n)

x1 (n + 1) y1 (n) = [1 − α(n)] > 0, x1 (n) x1 (n)

which yields y1 (n) [1 − α(n)]z(n) = . x1 (n + 1) 1 + β(n)z(n)

(7.170)

Substituting (7.170) into (7.169), we obtain   −2α(n) + [α(n)]2 z(n) − β(n)[z(n)]2 . Δz(n) = −γ (n) + 1 + β(n)z(n)

(7.171)

If β(n) > 0, then it is easy to verify that   −2α(n) + [α(n)]2 z(n) − β(n)[z(n)]2 [α(n)]2 ≤ ≤ [θ (n)]2 β(n) 1 + β(n)z(n) β(n)

(7.172)

holds. If β(n) = 0, then it follows from (7.163) that α(n) = 0, and hence   −2α(n) + [α(n)]2 z(n) − β(n)[z(n)]2 = 0 = [θ (n)]2 β(n). 1 + β(n)z(n)

(7.173)

Combining (7.172) with (7.173), we have 

 −2α(n) + [α(n)]2 z(n) − β(n)[z(n)]2 ≤ [θ (n)]2 β(n). 1 + β(n)z(n)

(7.174)

Substituting (7.174) into (7.171), we obtain Δz(n) ≤ −γ (n) + [θ (n)]2 β(n).

(7.175)

7.5 Discrete Linear Hamiltonian Systems

449

Summing (7.175) from 1 to N and noticing that z(n) is N-periodic, we get −

N  

γ (n) − [θ (n)]2 β(n) ≥ 0, n=1



contradicting (7.164).

Theorem 7.49 Assume (7.166), (7.163), (7.164), and (7.165). Then, (7.82) is stable. Proof Because of (7.163) and (7.164), if |ρ| ≥ 2, then one has real Floquet multipliers λk and real Floquet solutions uk = (xk , yk )T , k = 1, 2, such that (7.168) holds for k = 1, 2. Due to λ1 λ2 = 1, we have 0 ≤ min{λ21 , λ22 } ≤ 1. Suppose λ21 ≤ 1. Then, by Lemma 7.48, (7.82) has a nonzero solution (x1 , y1 ) such that x1 has a generalized zero in [1, N], say n1 . It follows from (7.168) that n1 + N is also a generalized zero of x1 and (x1 (n1 + N), y1 (n1 + N)) = λ1 (x1 (n1 ), y1 (n1 )). Applying Corollary 7.41 to the solution (x1 , y1 ) with a = n1 , b = n1 + N, and μ1 = μ2 = λ1 , we get n1 +N

−1

β(n)

n1 +N

−1

n=n1

γ + (n) ≥ 4

n=n1

n1 +N %−1

Θ[α(n)].

(7.176)

n=n1

Next, note that for any N-periodic function f defined on Z, the equations n0 +N

−1

f (n) =

n=n0

N

f (n)

n=1

and n0 +N %−1

Θ[f (n)] =

n=n0

N

Θ[f (n)]

n=1

hold for all n0 ∈ Z. It follows from (7.176) that N

n=1

β(n)

N

n=1

+

γ (n) ≥ 4

N %

Θ[α(n)]

n=1

holds, contradicting (7.175). Thus, |ρ| < 2, and hence, (7.82) is stable.



450

7 Difference Equations

7.5.3 Concluding Remarks Remark 7.50 Consider (7.48), in which a, b, c are real-valued piecewise continuous functions defined on R and periodic with period ω > 0, i.e., a(t + ω) = a(t),

b(t + ω) = b(t),

c(t + ω) = c(t),

t ∈ R.

Applying Theorem 7.27, we can prove the following statement: Assume b(t) > 0,

b(t)c(t) − [a(t)]2 ≥ 0,

c(t) ≥ 0,

t ∈ R,

b(t)c(t) − [a(t)]2 ≡ 0,

(7.177) (7.178)

and 

ω



0



ω

|a(t)|dt +

b(t)dt 0

1/2

ω

c(t)dt

< 2.

(7.179)

0

Then, (7.48) is stable. Indeed, if (7.48) is not stable, then this system has a real nontrivial solution (x, u) such that x(t + ω) = ρx(t),

u(t + ω) = ρu(t),

t ∈R

(7.180)

where ρ is a real nonzero number. Now, we show that x must have at least one zero in the segment [0, ω]. If not, then by (7.180), x(t) = 0 for all t ∈ R. Multiplying the first equation in (7.48) by u(t) and the second one by x(t) and then subtracting, we get u(t)x  (t) − x(t)u (t) = c(t)[x(t)]2 + 2a(t)x(t)u(t) + b(t)[u(t)]2 . Hence, dividing both sides by [x(t)]2 , we obtain −

 u  x

(t) =

c(t)[x(t)]2 + 2a(t)x(t)u(t) + b(t)[u(t)]2 . [x(t)]2

Integrating the latter equation from 0 to ω and taking into account that, by (7.180), u(ω) u(0) ρu(0) u(0) − = − =0 x(ω) x(0) ρx(0) x(0) holds, we get  0

ω

c(t)[x(t)]2 + 2a(t)x(t)u(t) + b(t)[u(t)]2 dt = 0. [x(t)]2

(7.181)

7.5 Discrete Linear Hamiltonian Systems

451

On the other hand, cx 2 + 2axu + bu2 =

1 (ax + bu)2 + (bc − a 2 )x 2 . b

Therefore, (7.181) cannot be true, since (x, u) is nontrivial. Thus, x has at least one zero α in [0, ω]. From (7.180), we get that x also has a zero at α + ω. Applying Theorem 7.27 to the points α and β = α + ω and taking into account the fact that for any ω-periodic function f defined on R, the equality 

t0 +ω



ω

f (t)dt =

f (t)dt 0

t0

holds for all t0 ∈ R, we obtain 

ω



0



ω

|a(t)|dt + 0

1/2

ω

b(t)dt

≥ 2,

c(t)dt 0

contradicting (7.179). Therefore, (7.48) must be stable under (7.177), (7.178), and (7.179). Remark 7.51 Consider the discrete linear Hamiltonian system with constant coefficients Δx(t) = ax(t + 1) + bu(t),

Δu(t) = −cx(t + 1) − au(t),

t ∈ Z,

(7.182)

where a, b, c ∈ R are constants and a = 1. This case corresponds to the value N = 1 of the period. System (7.182) can be written as the vector equation ϕ(t + 1) = Mϕ(t),

t ∈ Z,

(7.183)

where

x ϕ= , u

 M=

1 1−a c − 1−a

b 1−a

1−a−

 bc 1−a

.

By (7.183), we have ϕ(t) = M t ϕ(0),

t ∈ Z.

Therefore, if the matrix M has two distinct eigenvalues with modulus 1, then all solutions of (7.182) are bounded on Z. The eigenvalues of the matrix M are given as the roots of the quadratic equation λ2 + Aλ + 1 = 0,

(7.184)

452

7 Difference Equations

where A=

bc − a 2 − 2. 1−a

(7.185)

Since the roots of (7.184) are given by λ1,2 =

  1 −A ∓ A2 − 4 , 2

the matrix M has two distinct eigenvalues with modulus 1 provided − 2 < A < 2.

(7.186)

From (7.185), it follows that (7.186) is satisfied if bc − a 2 > 0 and

|a| +

√

bc < 2.

(7.187)

So, if (7.187) is satisfied, then (7.182) is stable. This result is better than the result given by Theorem 7.46. Remark 7.52 Systems (7.48) and (7.53), in general, may have, for any interval, a nontrivial solution (x, u) such that x is identically zero on that interval. Indeed, as an example for the continuous case, consider a(t) ≡ −1,

c(t) ≡ 0,

⎧ 2(t − β)e−2t ⎪ ⎪ ⎨ b(t) = 0 ⎪ ⎪ ⎩ −2t 2te

u(t) = et ,

t ∈ R,

if t > β, if 0 ≤ t ≤ β, if t < 0,

and ⎧ (t − β)2 e−t ⎪ ⎪ ⎨ x(t) = 0 ⎪ ⎪ ⎩ 2 −t t e

if t > β, if 0 ≤ t ≤ β, if t < 0,

where β > 0 is arbitrary. Then, (7.48) is satisfied. The statement of Theorem 7.27 is not true for the solution (x, u) and the points 0 and β if β < 2. Also, as an example for the discrete case, consider a(t) ≡ −1,

c(t) ≡ 0,

u(t) = 2t ,

t ∈ Z,

7.6 Quasilinear Difference Systems

453

⎧ t 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2−β+1 ⎪ ⎪ ⎨ b(t) = 0 ⎪ ⎪ ⎪ ⎪ −2 ⎪ ⎪ ⎪ ⎪ ⎩ −t 2

if

t ≥ β + 1,

if

t = β,

if

0 ≤ t ≤ β − 1,

if

t = −1,

if

t ≤ −2,

and  x(t) =

0

if 0 ≤ t ≤ β,

1

if t ∈ Z \ [0, β],

where β ∈ N \ {1} is arbitrary. Then, (7.53) is satisfied. The statement of Theorem 7.28 is not true for the solution (x, u) and the points 0 and β if β = 2. However, if b(t) = 0 for all t, then for any nontrivial solution (x, u), the component x cannot be identically zero on any interval containing two or more points.

7.6 Quasilinear Difference Systems In 1964, Atkinson [36] investigated the discrete problem  Δ(r(n)Δu(n)) = λq(n)u(n + 1), u(a) = u(b) = 0 and

u(n) ≡ 0

on

Z[a, b],

(7.188)

and he proved that discrete boundary value problems of the form (7.188) have exactly b−a−1 real and simple eigenvalues, which can be arranged in the increasing order λ1 < λ2 < · · · < λb−a−1 , where a, b ∈ Z with a ≤ b − 2, λ ∈ R, r(n) > 0 and q(n) > 0 for all n ∈ Z. Recall that Z[a, b] = {a, a + 1, a + 2, . . . , b − 1, b}. In 1983, Cheng [83] proved that if (7.26) has a real solution u satisfying the Dirichlet boundary conditions in (7.188), then (7.29) holds, see the proof in Sect. 7.2. In 2008, Ünal et al. [277] established (7.305) below for second-order half-linear difference equations of the form (7.303) below under the same boundary conditions in (7.188), see Corollary 7.80. Applying (7.305) below to (7.30), we can derive the Lyapunov-type inequality

454

7 Difference Equations b−2

n=a

q(n) ≥

4 , b−a

(7.189)

which was also obtained by Guseinov and Kaymakçalan [143], see Sect. 7.5. When b − a − 1 is odd, (7.189) is the same as (7.29). However, (7.189) is worse than (7.29) when b − a − 1 is even. For more discrete cases and continuous cases for Lyapunov-type inequalities, we refer the reader to [84, 85, 88–91, 93, 94, 143, 162, 179, 203, 281, 305]. For a single p-Laplacian equation of the form (7.303), there are many papers in the literature which deal with various dynamics behavior of its solutions. However, there is less research done for p-Laplacian systems. In 2012, Zhang and Tang [308] considered the quasilinear difference system of resonant type 

−Δ(r1 (n)Φp1 (Δu(n))) = f1 (n)Φα1 (u(n + 1))|v(n + 1)|α2 , −Δ(r2 (n)Φp2 (Δv(n))) = f2 (n)Φβ2 (v(n + 1))|u(n + 1)|β1

(7.190)

and the quasilinear difference system involving the (p1 , p2 , . . . , pn )-Laplacian ⎧ ⎪ ⎪ ⎪ −Δ(r1 (n)Φp1 (Δu1 (n))) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −Δ(r2 (n)Φp2 (Δu2 (n))) ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −Δ(rm (n)Φpm (Δum (n))) ⎪ ⎪ ⎪ ⎪ ⎩

= f1 (n)Φα1 (u1 (n + 1))

m %

|ui (n + 1)|αi ,

i=1 i=1

= f2 (n)Φα2 (u2 (n + 1))

m %

|ui (n + 1)|αi ,

i=1 i=2

.. . = fm (n)Φαm (um (n + 1))

m %

|ui (n + 1)|αi ,

i=1 i=m

(7.191) where Φα (w) = |w|α−2 w. For the sake of convenience, we give the following hypotheses (H1 ) and (H2 ) for (7.190) and hypothesis (H3 ) for (7.191): (H1 ) r1 , r2 , f1 , f2 are real-valued functions defined on Z and r1 (n) > 0 and r2 (n) > 0 for all n ∈ Z, (H2 ) 1 < p1 , p2 < ∞, α1 , α2 , β1 , β2 > 0 satisfy α1 α2 + = 1 and p1 p2

β1 β2 + = 1, p1 p2

(7.192)

7.6 Quasilinear Difference Systems

455

(H3 ) ri , fi are real-valued functions defined on Z and ri (n) > 0 for all n ∈ Z, i = 1, 2, . . . , m. Furthermore, 1 < pi < ∞ and αi > 0 satisfy m

αi = 1. pi

(7.193)

i=1

Systems (7.190) and (7.191) are discrete analogues of the two quasilinear differential systems 

−(r1 (t)Φp (u (t))) = f1 (t)Φα1 (u(t))|v(t)|α2 , −(r2 (n)Φq (v  (t))) = f2 (t)Φβ2 (v(t))|u(t)|β1

(7.194)

and ⎧ −[r1 (t)Φp1 (u1 (t))] = f1 (t)Φα1 (u1 (t)) |u2 (t)|α2 . . . |un (t)|αn , ⎪ ⎪ ⎪ ⎪ ⎪  α α ⎪ ⎨−[r2 (t)Φp2 (u2 (t))] = f2 (t) |u1 (t)| 1 Φα2 (u2 (t)) . . . |un (t)| n , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(7.195)

.. . −[rn (t)Φpn (un (t))] = fn (t) |u1 (t)|α1 |u2 (t)|α2 . . . Φαn (un (t)),

respectively. de Nápoli and Pinasco [107], Çakmak and Tiryaki [72, 73], and Tang and He [269] established some Lyapunov-type inequalities for (7.194) and (7.195). Motivated by the above mentioned papers, the purpose of this section is to establish some Lyapunov-type inequalities for (7.190) and (7.191). As a byproduct, we derive a Lyapunov-type inequality that is better than (7.305) given in Corollary 7.80 below, namely  n

b−2

n=a

p−1  [r(s)]

1/(1−p)

s=a

b−1

p−1 [r(s)]

1/(1−p)

+  n p−1  b−1 p−1 q (n) ≥ 1

[r(s)]1/(1−p) + [r(s)]1/(1−p) s=n+1

s=a

(7.196)

s=n+1

for (7.303). In particular, (7.196) produces the Lyapunov-type inequality (Hartmantype inequality) b−2

(n + 1 − a)(b − n − 1)q + (t) ≥ b − a

n=a

for (7.26) when p = 2 and r(t) ≡ 1. It is easy to see that (7.196) is better than (7.29).

456

7 Difference Equations

Denote now  ζ1 (n) :=

n



p1 −1 [r1 (τ )]

1/(1−p1 )

,

η1 (n) :=

τ =a

p1 −1

b−1

[r1 (τ )]

1/(1−p1 )

τ =n+1

(7.197)

and  ζ2 (n) :=

n



p2 −1 1/(1−p2 )

[r2 (τ )]

,

η2 (n) :=

τ =a

b−1

p2 −1 [r2 (τ )]

1/(1−p2 )

τ =n+1

. (7.198)

Theorem 7.53 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b − 2. Suppose that hypotheses (H1 ) and (H2 ) are satisfied. If (7.190) has a solution (u, v) satisfying the boundary conditions u(a) = u(b) = v(a) = v(b) = 0,

u(n) ≡ 0, v(n) ≡ 0 on Z[a, b],

(7.199)

then the inequality α1 β1 /p12 b−2 α2 β1 /(p1 p2 ) b−2

ζ1 (n)η1 (n)

ζ1 (n)η1 (n) + + f (n) f (n) ζ (n) + η1 (n) 1 ζ (n) + η1 (n) 2 n=a 1 n=a 1 b−2 α2 β1 /(p1 p2 ) b−2 α2 β2 /p22

ζ2 (n)η2 (n)

ζ2 (n)η2 (n) f1+ (n) f2+ (n) × ≥1 ζ (n) + η (n) ζ (n) + η (n) 2 2 2 2 n=a n=a (7.200) holds, where fi+ (n) = max{fi (n), 0} for i = 1, 2. Proof By (7.190) and (7.199), we obtain b−1

r1 (n)|Δu(n)|p1 =

n=a

b−2

f1 (n)|u(n + 1)|α1 |v(n + 1)|α2

(7.201)

f2 (n)|u(n + 1)|β1 |v(n + 1)|β2 .

(7.202)

n=a

and b−1

n=a

r2 (n)|Δv(n)|

p2

=

b−2

n=a

7.6 Quasilinear Difference Systems

457

By (7.197), (7.199), and Hölder’s inequality, we have |u(n + 1)|

 n p1     = Δu(τ ) τ =a   n p1

≤ |Δu(τ )|

p1

τ =a

 ≤

n

p1 −1 [r1 (τ )]1/(1−p1 )

τ =a

= ζ1 (n)

n

(7.203) r1 (τ )|Δu(τ )|p1

τ =a n

r1 (τ )|Δu(τ )|p1 ,

a ≤n≤b−1

τ =a

and |u(n + 1)|p1

 b−1 p1     = Δu(τ )   τ =n+1

 ≤

p1

b−1

|Δu(τ )|

τ =n+1

 ≤

p1 −1

b−1

[r1 (τ )]1/(1−p1 )

τ =n+1

= η1 (n)

b−1

(7.204) b−1

r1 (τ )|Δu(τ )|p1

τ =n+1

r1 (τ )|Δu(τ )|p1 ,

a ≤ n ≤ b − 1.

τ =n+1

From (7.203) and (7.204), we get |u(n + 1)|p1 ≤

b−1 ζ1 (n)η1 (n) r1 (τ )|Δu(τ )|p1 , ζ1 (n) + η1 (n) τ =a

a ≤ n ≤ b − 1.

(7.205)

Now, it follows from (7.199), (7.201), (7.205), (H2 ), and Hölder’s inequality that

458

7 Difference Equations b−2

f1+ (n)|u(n + 1)|p1

n=a

≤

 b−2  b−1

ζ1 (n)η1 (n) + f1 (n) r1 (n)|Δu(n)|p1 ζ (n) + η (n) 1 1 n=a n=a

= M11

b−2

f1 (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

≤ M11

b−2

f1+ (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

≤ M11

b−2

f1+ (n)|u(n + 1)|p1

α1 /p1 b−2

n=a

α2 /p2 f1+ (n)|v(n + 1)|p2

n=a

(7.206) and b−2

f2+ (n)|u(n + 1)|p1

n=a

≤

 b−2  b−1

ζ1 (n)η1 (n) + f2 (n) r1 (n)|Δu(n)|p1 ζ (n) + η (n) 1 1 n=a n=a

= M12

b−2

f1 (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

≤ M12

b−2

f1+ (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

≤ M12

b−2

f1+ (n)|u(n + 1)|p1

α1 /p1 b−2

n=a

α2 /p2 f1+ (n)|v(n + 1)|p2

n=a

(7.207) hold, where M11 =

 b−2 

ζ1 (n)η1 (n) + f1 (n) , ζ1 (n) + η1 (n) n=a

M12 =

 b−2 

ζ1 (n)η1 (n) + f2 (n) . ζ1 (n) + η1 (n) n=a (7.208)

7.6 Quasilinear Difference Systems

459

Similarly as in the proof of (7.205), from (7.198) and (7.199), we have |v(n + 1)|p2 ≤

b−1 ζ2 (n)η2 (n) r2 (τ )|Δv(τ )|p2 , ζ2 (n) + η2 (n) τ =a

a ≤ n ≤ b − 1.

(7.209)

It follows from (7.199), (7.202), (7.209), (H2 ), and Hölder’s inequality that b−2

f1+ (n)|v(n + 1)|p2

n=a

≤

 b−2  b−1

ζ2 (n)η2 (n) + f1 (n) r2 (n)|Δv(n)|p2 ζ (n) + η (n) 2 2 n=a n=a

= M21

b−2

f2 (n)|u(n + 1)|β1 |v(n + 1)|β2

n=a

≤ M21

b−2

f2+ (n)|u(n + 1)|β1 |v(n + 1)|β2

n=a

≤ M21

b−2

f2+ (n)|u(n + 1)|p1

n=a

β1 /p1 b−2

α2 /p2 f2+ (n)|v(n + 1)|p2

n=a

(7.210) and b−2

f2+ (n)|v(n + 1)|p2

n=a

≤

 b−2  b−1

ζ2 (n)η2 (n) + f2 (n) r2 (n)|Δv(n)|p2 ζ (n) + η (n) 2 2 n=a n=a

= M22

b−2

f2 (n)|u(n + 1)|β1 |v(n + 1)|β2

n=a

≤ M22

b−2

f2+ (n)|u(n + 1)|β1 |v(n + 1)|β2

n=a

≤ M22

b−2

n=a

f2+ (n)|u(n + 1)|p1

β1 /p1 b−2

n=a

β2 /p2 f2+ (n)|v(n + 1)|p2

460

7 Difference Equations

hold, where M21 =

 b−2 

ζ2 (n)η2 (n) + f1 (n) , ζ2 (n) + η2 (n) n=a

M22 =

 b−2 

ζ2 (n)η2 (n) + f2 (n) . ζ2 (n) + η2 (n) n=a (7.211)

Next, we prove that b−2

f1+ (n)|u(n + 1)|p1 > 0.

(7.212)

f1+ (n)|u(n + 1)|p1 = 0.

(7.213)

n=a

If (7.212) is not true, then b−2

n=a

From (7.201) and (7.213), we have 0≤

b−1

r1 (n)|Δu(n)|p1

n=a

=

b−2

f1 (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

≤

b−2

f1+ (n)|u(n + 1)|α1 |v(n + 1)|α2

n=a

= 0, and thus, it follows from (H1 ) that |Δu(n)| ≡ 0,

a ≤n≤b−1

(7.214)

holds. Combining (7.203) with (7.214), we obtain u(n) ≡ 0 for a ≤ n ≤ b, contradicting (7.199). Therefore, (7.212) holds. Similarly, we have b−2

f2+ (n)|u(n + 1)|p1 > 0,

n=a

b−2

f1+ (n)|v(n + 1)|p2 > 0,

(7.215)

n=a

and b−2

n=a

f2+ (n)|v(n + 1)|p2 > 0.

(7.216)

7.6 Quasilinear Difference Systems

461

From (7.206), (7.207), (7.210), (7.212), (7.215), (7.216), and (H2 ), we have α1 β1 /p12

M11

α β1 /(p1 p2 )

M122

α β1 /(p1 p2 )

M212

α2 β2 /p22

M22

≥ 1.

(7.217) 

It follows from (7.208), (7.211), and (7.217) that (7.200) holds.

Corollary 7.54 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b − 2. Suppose that hypotheses (H1 ) and (H2 ) are satisfied. If (7.190) has a solution (u, v) satisfying (7.199), then the inequality b−2

f1+ (n) [ζ1 (n)η1 (n)]1/2

α1 β1 /p12 b−2

n=a

×

α2 β1 /(p1 p2 ) f2+ (n) [ζ1 (n)η1 (n)]1/2

n=a

b−2

f1+ (n) [ζ2 (n)η2 (n)]1/2

α2 β1 /(p1 p2 ) b−2

n=a

α2 β2 /p22 f2+ (n) [ζ2 (n)η2 (n)]1/2

n=a

≥ 2(p2 β1 +p1 α2 )/(p1 p2 )

(7.218)

holds. Proof Since ζi (n) + ηi (n) ≥ 2 [ζi (n)ηi (n)]1/2 ,

i = 1, 2, 

it follows from (7.200) and (H2 ) that (7.218) holds.

Corollary 7.55 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b − 2. Suppose that hypotheses (H1 ) and (H2 ) are satisfied. If (7.190) has a solution (u, v) satisfying (7.199), then the inequality b−1

1/(1−p1 )

[r1 (n)]

β1 (p1 −1)/p1 b−1

n=a

α2 (p2 −1)/p2 [r2 (n)]

1/(1−p2 )

n=a

×

b−2

f1+ (n)

β1 /p1 b−2

n=a

α2 /p2 f2+ (n)

≥ 2β1 +α2

(7.219)

n=a

holds. Proof Because of  1/2

[ζ1 (n)η1 (n)]

=

n

τ =a

1/(1−p1 )

[r1 (τ )]

b−1

τ =n+1

(p1 −1)/2 1/(1−p1 )

[r1 (τ )]

462

7 Difference Equations

≤

b−1

1 2p1 −1

p1 −1 [r1 (τ )]

1/(1−p1 )

τ =a

and  1/2

[ζ2 (n)η2 (n)]

=

n

1/(1−p2 )

[r2 (τ )]

τ =a

≤

1

(p2 −1)/2

b−1

1/(1−p2 )

[r2 (τ )]

τ =n+1

b−1

2p2 −1

p2 −1 [r2 (τ )]

1/(1−p2 )

,

τ =a



it follows from (7.218) and (H2 ) that (7.219) holds.

When α1 = β2 = p1 = p2 , α2 = β1 = 0, r1 = r2 = r, and f1 = f2 = f , (7.190) reduces to (7.303). Hence, we can directly derive from (7.206) and (7.212) the following Lyapunov-type inequality for (7.303). Theorem 7.56 Let a, b ∈ Z with a ≤ b − 2. Suppose that p > 1 and r(n) > 0 for all n ∈ Z. If (7.303) has a solution u satisfying (7.31), then one has (7.196). Since n

[r(τ )]1/(1−p) +

τ =a

b−1

[r(τ )]1/(1−p)

τ =n+1

 ≥2

n

[r(τ )]

τ =a

1/(1−p)

b−1

(p1 −1)/2 [r(τ )]

1/(1−p)

,

τ =n+1

it follows from Theorem 7.56 that the following corollary holds. Corollary 7.57 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b−2. Suppose that p > 1 and r(n) > 0 for all n ∈ Z. If (7.303) has a solution u satisfying (7.31), then the inequality ⎧  n (p1 −1)/2 ⎫ b−2 ⎨ b−1 ⎬

[r(τ )]1/(1−p) [r(τ )]1/(1−p) q + (n) ≥2 ⎩ ⎭

n=a

τ =a

(7.220)

τ =n+1

holds. Remark 7.58 It is easy to see that the Lyapunov-type inequalities (7.196) and (7.220) are better than (7.305) given in Corollary 7.80.

7.6 Quasilinear Difference Systems

463

Next, we establish some Lyapunov-type inequalities for (7.191). Denote  ζi (n) :=

n

pi −1 [ri (τ )]

1/(1−pi )

i = 1, 2, . . . , m

,

(7.221)

τ =a

and 

pi −1

b−1

ηi (n) :=

[ri (τ )]

1/(1−pi )

i = 1, 2, . . . , m.

,

(7.222)

τ =n+1

Theorem 7.59 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b−2. Suppose that hypothesis (H3 ) is satisfied. If (7.191) has a solution (u1 , u2 , . . . , um ) satisfying the boundary conditions ui (a) = ui (b) = 0,

ui (n) ≡ 0 on Z[a, b],

i = 1, 2, . . . , m,

(7.223)

then the inequality αi αj /(pi pj ) b−2 m m % %

ζi (n)ηi (n) + f (n) ≥1 ζ (n) + ηi (n) j n=a i

(7.224)

i=1 j =1

holds. Proof By (7.191), (H3 ), and (7.223), we obtain b−1

ri (n) |Δui (n)|

pi

=

n=a

b−2

 fi (n)

n=a

m

 |uk (n + 1)|

αk

,

i = 1, 2, . . . , m.

k=1

(7.225) It follows from (7.221), (7.223), and Hölder’s inequality that |ui (n + 1)|pi

 n pi     = Δui (τ )   τ =a

 ≤

n

pi −1 [ri (τ )]

τ =a

= ζi (n)

1/(1−pi )

n

ri (τ ) |Δui (τ )|pi

τ =a n

τ =a

ri (τ ) |Δui (τ )|pi ,

a ≤ n ≤ b − 1,

i = 1, 2, . . . , m (7.226)

464

7 Difference Equations

holds. Similarly, it follows from (7.222), (7.223), and Hölder’s inequality that |ui (n + 1)|

pi

 b−1 pi     = Δui (τ )   τ =n+1

 ≤

pi −1

b−1

[ri (τ )]

τ =n+1

= ηi (n)

b−1

1/(1−pi )

ri (τ ) |Δui (τ )|pi

τ =n+1

b−1

ri (τ ) |Δui (τ )|pi ,

a ≤ n ≤ b − 1,

i=1, 2, . . . , m

τ =n+1

(7.227)

holds. From (7.226) and (7.227), we have |ui (n + 1)|pi ≤

b−1 ζi (n)ηi (n) ri (τ ) |Δui (τ )|pi , ζi (n) + ηi (n) τ =a

a ≤n≤b−1

(7.228)

for i = 1, 2, . . . , m. Now, by (7.223), (7.225), (7.228), (H3 ), and the generalized Hölder inequality, we obtain b−2

 b−2  b−1

ζi (n)ηi (n) + ri (τ ) |Δui (τ )|pi fj (n) ζ (n) + η (n) i i n=a τ =a   b−2 m

% αk |uk (n + 1)| = Mij fj (n)

fj+ (n) |ui (n + 1)|pi ≤

n=a

n=a

k=1

  b−2 m

% |uk (n + 1)|αk ≤ Mij fj+ (n) n=a

≤ Mij

k=1

b−2 m %

k=1

αk /pk fj+ (n) |uk (n + 1)|pk

,

n=a

(7.229) where Mij =

 b−2 

ζi (n)ηi (n) + fj (n) , ζi (n) + ηi (n) n=a

i, j = 1, 2, . . . , m.

(7.230)

Next, we prove that b−2

n=a

fi+ (n) |uk (n + 1)|pk > 0,

i, k = 1, 2, . . . , m.

(7.231)

7.6 Quasilinear Difference Systems

465

If (7.231) is not true, then there exist i0 , k0 ∈ {1, 2, . . . , m} such that b−2

 p fi+ (n) uk0 (n + 1) k0 = 0. 0

(7.232)

n=a

From (7.225) and (7.232), we have 0≤

b−1

 p ri0 (n) Δui0 (n) i0

n=a

=

b−1

fi0 (n)

n=a

≤

|uk (n + 1)|αk

k=1

b−1 m %

k=1

m %

αk /pk fi0 (n) |uk (n + 1)|pk

n=a

= 0, and thus, it follows from the fact that ri0 (n) > 0 that   Δui (n) ≡ 0, 0

a ≤n≤b−1

(7.233)

holds. Combining (7.226) with (7.233), we obtain ui0 (n) ≡ 0 for a ≤ n ≤ b, contradicting (7.223). Therefore, (7.231) holds. From (7.229), (7.231), and (H3 ), we have m m % %

α αj /(pi pj )

≥ 1.

(7.234)

It follows from (7.230) and (7.234) that (7.224) holds.



Miji

i=1 j =1

Corollary 7.60 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b−2. Suppose that hypothesis (H3 ) is satisfied. If (7.191) has a solution (u1 , u2 , . . . , um ) satisfying (7.223), then the inequality b−2 m m % %

i=1 j =1

αi αj /(pi pj ) fj+ (n) [ζi (n)ηi (n)]1/2

≥2

n=a

holds. Proof Since ζi (n) + ηi (n) ≥ 2 [ζi (n)ηi (n)]1/2 ,

i = 1, 2, . . . , m,

(7.235)

466

7 Difference Equations



it follows from (7.224) and (H3 ) that (7.235) holds.

Corollary 7.61 (Hartman-Type Inequality) Let a, b ∈ Z with a ≤ b−2. Suppose that hypothesis (H3 ) is satisfied. If (7.191) has a solution (u1 , u2 , . . . , um ) satisfying (7.223), then the inequality b−2 m %

i=1

αi (pi −1)/pi [ri (n)]

1/(1−pi )

b−2 m %

j =1

n=a

αj /pj fj+ (n)

≥ 2Am

(7.236)

n=a

holds, where Am =

m

αi .

i=1

Proof Since  [ζi (n)ηi (n)]

1/2

=

n

1/(1−pi )

[ri (τ )]

τ =a

≤

1

b−1

(pi −1)/2 [ri (τ )]

1/(1−pi )

τ =n+1

b−1

2pi −1

pi −1 [ri (τ )]1/(1−pi )

,

i = 1, 2, . . . , m,

τ =a

it follows from (7.235) and (H3 ) that (7.236) holds.



7.6.1 Some Applications In this section, some applications of Lyapunov-type inequalities from Sect. 7.6 are given to obtain lower bounds for the first eigencurve in the generalized spectra. Let a, b ∈ Z with a ≤ b − 2. We consider here a quasilinear difference system of the form ⎧ −Δ(Φp1 (Δu1 (n))) = λ1 α1 q(n)Φα1 (u1 (n + 1)) |u2 (n + 1)|α2 . . . |um (n + 1)|αm , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−Δ(Φp2 (Δu2 (n))) = λ2 α2 q(n) |u1 (n + 1)|α1 Φα2 (u2 (n + 1)) . . . |um (n + 1)|αm , ⎨ .. ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎩ −Δ(Φpm (Δum (n))) = λm αm q(n) |u1 (n + 1)|α1 |u2 (n + 1)|α2 . . . Φαm (um (n + 1)), (7.237)

where q(n) > 0, λi ∈ R, pi and αi are the same as those in (7.191), and ui satisfies the Dirichlet boundary conditions

7.6 Quasilinear Difference Systems

ui (a) = ui (b) = 0 and

467

i = 1, 2, . . . , m. (7.238) We define the generalized spectrum S of a nonlinear difference system as the set of vectors (λ1 , λ2 , . . . , λm ) ∈ Rm such that (7.237) with (7.238) admits a nontrivial solution. Problem (7.237) with (7.238) is a generalization of the p-Laplacian difference equation ui (n) > 0

on

Z[a + 1, b − 1],

−Δ(Φp (Δu(n))) = λαq(n)Φp (u(n + 1)) with Dirichlet boundary conditions, i.e.,  −Δ(Φp (Δu(n))) = λαq(n)Φp (u(n + 1)), u(a) = u(b) = 0 and

u(n) > 0 on

Z[a + 1, b − 1],

(7.239)

where p > 1, λ ∈ R, and q(n) > 0 for all n ∈ Z. Atkinson [36, Theorems 4.3.1 and 4.3.5] investigated the existence of eigenvalues for (7.239), see also [283]. Let fi (n) = λi αi q(n) and ri (n) = 1 for i = 1, 2, . . . , m. Then, we can apply Theorem 7.59 to the boundary value problem consisting of (7.237) and (7.238) in order to obtain a lower bound for the first eigencurve in the generalized spectra. Theorem 7.62 Let a, b ∈ Z with a ≤ b − 2. Assume that 1 < pi < ∞, αi > 0 satisfy (7.193) and that q(n) > 0 for all n ∈ Z. Then, there exists a function h such that λm ≥ h(λ1 , . . . , λm−1 ) for every generalized eigenvalue (λ1 , λ2 , . . . , λm ) of the boundary value problem consisting of (7.237) and (7.238), namely h(λ1 , . . . , λm−1 ) ⎧ ⎫ pm b−2  pαi ⎪− αm ⎪ m m α ⎨ ⎬ i  j % [(n − a + 1)(b − n − 1)]pi −1 1 % = λj αj pj . ⎪ αm ⎪ (n − a + 1)pi −1 + (b − n − 1)pi −1 ⎩j =1 ⎭ i=1 n=a Proof For the eigenvalue (λ1 , λ2 , . . . , λm ), (7.237) with (7.238) has a nontrivial solution (u1 , u2 , . . . , um ). That is, (7.191) with ri (n) ≡ 1 and fi (n) = λi αi q(n) > 0 has a solution (u1 , u2 , . . . , um ) satisfying (7.223). It follows from (7.224) that fi (n) = λi αi q(n) > 0 for all n ∈ Z, i = 1, 2, . . . , m, and that

468

7 Difference Equations

 b−2 m m % %

ζi (n)ηi (n) + f (n) 1≤ ζ (n) + ηi (n) j n=a i i=1 j =1

αi /pi b−2 m m %

ζi (n)ηi (n)  αj /pj % = q(n) λj αj ζ (n) + ηi (n) n=a i j =1

i=1

b−2 m m %

 αj /pj % = λj αj j =1

i=1

n=a

[(n − a + 1)(b − n − 1)]pi −1 q(n) (n − a + 1)pi −1 + (b − n − 1)pi −1

αi /pi ,

and hence, we have ⎧ m α /p 1 ⎨%  λm ≥ λj αj j j αm ⎩ j =1

×

b−2 m %

i=1

n=a

− n − 1)]pi −1

[(n − a + 1)(b q(n) (n − a + 1)pi −1 + (b − n − 1)pi −1

αi /pi ⎫−(pm /αm ) ⎬ ⎭

. 

This completes the proof.

When m = 2, the boundary value problem consisting of (7.237) and (7.238) reduces to the simpler form 

−Δ(Φp1 (Δu1 (n))) = λ1 α1 q(n)Φα1 (u1 (n + 1)) |u2 (n + 1)|α2 , −Δ(Φp2 (Δu2 (n))) = λ2 α2 q(n) |u1 (n + 1)|α1 Φα2 (u2 (n + 1))

(7.240)

with Dirichlet boundary conditions ui (a) = ui (b) = 0

and

ui (n) > 0 on

Z[a + 1, b − 1],

i = 1, 2, (7.241)

where 1 < p1 , p2 < ∞, α1 , α2 > 0 satisfy (7.192), and q(n) > 0 for all n ∈ Z. Applying Theorem 7.62 to (7.239) and also to the boundary value problem consisting of (7.240) and (7.241), respectively, we immediately have the following two corollaries. Corollary 7.63 Let a, b ∈ Z with a ≤ b − 2. Assume that 1 < p1 , p2 < ∞, α1 , α2 > 0 satisfy (7.192), and that q(n) > 0 for all n ∈ Z. Then, there exists a function h such that λ2 ≥ h(λ1 )

7.7 Discrete Nonlinear Systems

469

for every generalized eigenvalue (λ1 , λ2 ) of the problem consisting of (7.240) and (7.241), namely 1 h(λ1 ) = α2 ×

 λ1 α1

b−2

n=a

b−2

n=a

(XY )p1 −1 q(n) Xp1 −1 + Y p1 −1

(XY )p2 −1 q(n) Xp2 −1 + Y p2 −1

1−p2 /α2

−1 ,

where X = n − a + 1 and Y = b − n − 1. Corollary 7.64 Let a, b ∈ Z with a ≤ b − 2. Assume that p > 1 and q(n) > 0 for all n ∈ Z. Then, for every eigenvalue λ of (7.239), the inequality 1 λ≥ p

b−2

n=a

[(n − a + 1)(b − n − 1)]p−1 q(n) (n − a + 1)p−1 + (b − n − 1)p−1

−1

holds.

7.7 Discrete Nonlinear Systems In this section, we obtain Lyapunov-type inequalities for discrete nonlinear systems of the form  Δx(t) = α1 (t)x(t + 1) + β1 (t)|u(t)|γ −2 u(t), (7.242) Δu(t) = −β2 (t)|x(t + 1)|β−2 x(t + 1) − α1 (t)u(t), where t ∈ Z, γ , β > 1 are constants, the functions α1 , β1 , β2 are real valued with β1 (t) > 0 and 1 − α1 (t) = 0 for all t ∈ Z, and Δ denotes the usual forward difference operator, that is, Δx(t) = x(t + 1) − x(t). We recall that a nontrivial solution (x, u) of (7.242) defined for t ∈ Z is said to be proper if sup {|x(s)| + |u(s)| : s ≥ t} > 0 for all

t ∈ Z.

In the discrete case, instead of the usual zero, the concept of generalized zero is used. A function f : Z → R is said to have a generalized zero at t0 ∈ Z provided either f (t0 ) = 0 or f (t0 −1)f (t0 ) < 0. A proper solution (x, u) of (7.242) is called weakly oscillatory if at least one component has a sequence of generalized zeros tending to ∞. This solution is said to be oscillatory if both components have sequences of generalized zeros tending to ∞. If both components (at least one component) are eventually positive or negative, then the solution (x, u) is called nonoscillatory

470

7 Difference Equations

(weakly nonoscillatory). System (7.242) is said to be oscillatory if all solutions are oscillatory. Before we give the precise formulation of Lyapunov-type inequalities for (7.242), we mention a few background details, which serve to motivate the presented results. The behavior of solutions for (1.1) and Δ2 x(t) + q(t)x(t + 1) = 0 for

t ∈Z

has been studied extensively in the literature in both the continuous and the discrete settings, respectively. Since an exhaustive list of references is impossible due to the incredible amount of papers devoted to Lyapunov-type inequalities for these equations, we confine ourselves to those papers which contributed to (1.2), namely Cheng [83], Eliason [124], Hartman [155], Hochstadt [172], Kwong [193], Reid [248, 249], and Singh [258]. In fact, the classical Lyapunov inequality has been generalized to secondorder nonlinear differential equations by Eliason [124] and Pachpatte [229], to delay differential equations of second order by Eliason [125] and Dahiya and Singh [103], and to higher-order differential equations by Pachpatte [227]. Lyapunovtype inequalities for Emden–Fowler-type equations can be found in Pachpatte’s paper [229]. Lyapunov-type inequalities for half-linear equations were obtained independently by Lee et al. [195] and by Pinasco [235]. The proof for the halfˇ linear extension can be found in Došlý and Rehák’s book [115, page 190]. A thorough literature review of continuous and discrete Lyapunov inequalities and their applications can be found in the survey paper [91] by Cheng and the references quoted therein. Although there is an extensive literature on Lyapunov-type inequalities for the above mentioned equations, there is not much done for linear Hamiltonian systems 

x  (t) = α1 (t)x(t) + β1 (t)u(t), u (t) = −β2 (t)x(t) − α1 (t)u(t)

(7.243)

and nonlinear systems of differential equations of the form 

x  (t) = α1 (t)x(t) + β1 (t)|u(t)|γ −2 u(t), u (t) = −β2 (t)|x(t)|β−2 x(t) − α1 (t)u(t).

(7.244)

We refer the reader to the introductory papers by Guseinov and Kaymakçalan [143] for (7.243) and by Tiryaki et al. [274] for (7.244), respectively. For a comprehensive ˇ treatment of the subject, we refer the reader to the books by Došlý and Rehák [115], Kiguradze and Chanturia [183], and Mirzov [213], and the paper by Kitano and Kusano [188]. In this section, we obtain a discrete analogue of Lyapunov-type inequalities for nonlinear systems of the form (7.242). We remark that the discrete Hamiltonian system, in case of two scalar linear difference equations, has the form

7.7 Discrete Nonlinear Systems



471

Δx(t) = a(t)x(t + 1) + b(t)u(t), Δu(t) = −c(t)x(t + 1) − a(t)u(t),

(7.245)

where t ∈ Z, and the coefficient a satisfies the condition 1 − a(t) = 0 for all t ∈ Z. Clearly, (7.245) is a special case of (7.242) with γ = β = 2. Also notice that the second-order difference equation Δ (p(t)Δx(t)) + q(t)x(t + 1) = 0,

t ∈ Z,

(7.246)

where p(t) = 0 for all t ∈ Z, can be equivalently written as a discrete Hamiltonian system of the form (7.245). Indeed, let x be a solution of (7.246) and set u = pΔx. Then, we obtain Δx(t) =

1 u(t), p(t)

Δu(t) = −q(t)x(t + 1)

so that (7.246) is equivalent to (7.245) with a(t) ≡ 0,

b(t) =

1 , p(t)

c(t) = q(t).

We also remark that the Emden–Fowler-type difference equation   Δ p(t)|Δx(t)|α−2 Δx(t) + q(t)|x(t + 1)|β−2 x(t + 1) = 0

(7.247)

and the half-linear difference equation   Δ p(t)|Δx(t)|α−2 Δx(t) + q(t)|x(t + 1)|α−2 x(t + 1) = 0,

(7.248)

where α > 1 and β > 1 are constants, p and q are real-valued functions defined on Z, and p(t) > 0 for all t ∈ Z, can clearly be written as a special case of discrete nonlinear systems of the form (7.242). Being motivated from the recent papers by Pachpatte [229], Lee et al. [195], Guseinov and Kaymakçalan [143], Tiryaki et al. [274], and Patula [231], Ünal et al. [277] set up and proved some results for discrete nonlinear systems of the form (7.242). The inequalities proposed in this section can be used as a handy tool in the study of the qualitative nature of solutions. Some applications are given to show the importance of the results (see Sect. 7.7.1). Since our attention is restricted to Lyapunov-type inequalities for discrete nonlinear systems, we shall assume the existence of nontrivial real solutions (x, u) of (7.242). Theorem 7.65 (Hartman-Type Inequality) Suppose β1 (t) > 0 for all t ∈ Z. Let n, m ∈ Z with n ≤ m − 2. Assume (7.242) has a real solution (x, u) such that

472

7 Difference Equations

x(n) = x(m) = 0 and x is not identically zero on [n, m]. Then, the inequality m−2

|α1 (t)| + M

β/α−1

t=n

m−1

β1 (t)

1/γ m−2

t=n

1/α β2+ (t)

≥2

(7.249)

t=n

holds, where α is the Hölder conjugate of the γ , i.e., 1/γ + 1/α = 1, and M=

max

n+1≤t≤m−1

|x(t)|.

Proof Let (x, u) be nontrivial real solution of (7.242) such that x(n) = x(m) = 0 and x is not identically zero on [n, m]. Then, multiplying the first equation in (7.242) by u(t) and the second one by x(t + 1) and then adding them up yields Δ(x(t)u(t)) = β1 (t)|u(t)|γ − β2 (t)|x(t + 1)|β .

(7.250)

Summing (7.250) from n to m − 1 and taking into account x(n) = x(m) = 0, we get 0=

m−1

β1 (t)|u(t)| − γ

t=n

m−1

β2 (t)|x(t + 1)|β .

t=n

Since x(m) = 0, we have m−1

β1 (t)|u(t)|γ =

t=n

m−2

β2 (t)|x(t + 1)|β ≤

m−2

t=n

β2+ (t)|x(t + 1)|β .

(7.251)

t=n

Choose τ ∈ [n + 1, m − 1] such that |x(τ )| =

max

n+1≤t≤m−1

|x(t)| = M.

Hence, |x(τ )| > 0. Summing the first equation in (7.242) at first from n to τ − 1 and then from τ to m − 1, we obtain, respectively, x(τ ) =

τ −1

α1 (t)x(t + 1) +

τ −1

t=n

β1 (t)|u(t)|γ −2 u(t)

t=n

and −x(τ ) =

m−2

t=τ

α1 (t)x(t + 1) +

m−1

t=τ

β1 (t)|u(t)|γ −2 u(t).

7.7 Discrete Nonlinear Systems

473

Taking the absolute value of the above two equalities yields τ −1

|x(τ )| ≤

|α1 (t)| |x(t + 1)| +

t=n

τ −1

β1 (t)|u(t)|γ −1

t=n

and |x(τ )| = | − x(τ )| ≤

m−2

|α1 (t)| |x(t + 1)| +

t=τ

m−1

β1 (t)|u(t)|γ −1 ,

t=τ

respectively. Adding the last two inequalities, we obtain 2|x(τ )| ≤

m−2

|α1 (t)| |x(t + 1)| +

t=n

m−1

β1 (t)|u(t)|γ −1 .

(7.252)

t=n

On the other hand, applying Hölder’s inequality with Hölder conjugates α and γ to the second sum on the right-hand side of (7.252) gives m−1

β1 (t)|u(t)|γ −1 =

t=n

m−1

1/γ +1/α

β1

(t)|u(t)|γ −1

t=n

≤

=

m−1

β1 (t)

1/γ m−1

1/α β1 (t)|u(t)|(γ −1)α

t=n

t=n

m−1

1/γ m−1

β1 (t)

t=n

(7.253)

1/α β1 (t)|u(t)|

γ

,

t=n

since 1/γ + 1/α=1. Hence, using (7.251) and (7.253), we get m−1

β1 (t)|u(t)|

γ −1

≤

t=n

m−1

β1 (t)

1/γ m−2

t=n

1/α β2+ (t)|x(t

+ 1)|

β

.

(7.254)

+ 1)|

β

t=n

Substituting (7.254) into (7.252) yields 2|x(τ )| ≤

m−2

|α1 (t)| |x(t + 1)| +

t=n

≤ |x(τ )|

m−1

β1 (t)

1/γ m−2

t=n m−2

t=n

|α1 (t)| + |x(τ )|

β α

1/α β2+ (t)|x(t

t=n

m−1

t=n

β1 (t)

1/γ m−2

t=n

1/α β2+ (t)

,

474

7 Difference Equations

and dividing this inequality by |x(τ )| > 0, we obtain (7.249).



Theorem 7.66 (Hartman-Type Inequality) Suppose 1−α1 (t) > 0 and β1 (t) > 0 for all t ∈ Z. Let n, m ∈ Z with n ≤ m − 2. Assume (7.242) has a real solution (x, u) such that x(n) = 0 and x(m − 1)x(m) < 0. Then, the inequality m−2

|α1 (t)| + M

β/α−1

m−2

t=n

β1 (t)

1/γ m−2

t=n

1/α β2+ (t)

>1

(7.255)

t=n

holds, where β, γ , α, M, β2+ are defined as before. Proof Choose τ ∈ [n + 1, m − 1] such that M = |x(τ )| =

max

n+1≤t≤m−1

|x(t)|.

Hence, |x(τ )| > 0. Summing the first equation in (7.242) from n to τ − 1 and taking into account x(n) = 0, we obtain x(τ ) =

τ −1

α1 (t)x(t + 1) +

t=n

τ −1

β1 (t)|u(t)|γ −2 u(t),

t=n

and hence, |x(τ )| ≤

τ −1

|α1 (t)| |x(t + 1)| +

t=n

≤

≤

=

m−2

τ −1

β1 (t)|u(t)|γ −1

t=n

|α1 (t)| |x(t + 1)| +

m−2

β1 (t)|u(t)|γ −1

t=n

t=n

m−2

m−2

|α1 (t)| |x(t + 1)| +

β1 (t)

1/γ m−2

t=n

t=n

t=n

m−2

m−2

1/γ m−2

|α1 (t)| |x(t + 1)| +

t=n

β1 (t)

t=n

1/α β1 (t)|u(t)|

(γ −1)α

1/α β1 (t)|u(t)|

γ

.

t=n

(7.256) Now, summing (7.250) from n to m−2 and taking into account x(n) = 0, we obtain x(m − 1)u(m − 1) =

m−2

t=n

β1 (t)|u(t)|γ −

m−2

t=n

β2 (t)|x(t + 1)|β .

(7.257)

7.7 Discrete Nonlinear Systems

475

In addition, from the first equation in (7.242), we have, for t = m − 1, (1 − α1 (m − 1))x(m) = x(m − 1) + β1 (m − 1)|u(m − 1)|γ −2 u(m − 1), and multiplying this equality by x(m − 1) yields (1 − α1 (m − 1))x(m)x(m − 1) = x 2 (m − 1) + β1 (m − 1)|u(m − 1)|γ −2 u(m − 1)x(m − 1).

(7.258)

Since 1 − α1 (t) > 0, β1 (t) > 0 for all t ∈ Z, and x(m)x(m − 1) < 0, (7.258) implies that u(m − 1)x(m − 1) < 0 must hold. Hence, it follows from (7.257) that the inequality m−2

β1 (t)|u(t)|γ
0 and β1 (t) > 0 for all t ∈ Z. Let n, m ∈ Z with n ≤ m − 1. Assume (7.242) has a real solution (x, u) such that x(n − 1)x(n) < 0 and x(m) = 0. Then, the inequality m−2

t=n

β/α−1 |α1 (t)| + M1

m−1

1/γ  β1 (t)

t=n

m−2

t=n−1

holds, where β, γ , α, β2+ are defined as before and M1 =

max

n≤t≤m−1

|x(t)|.

1/α β2+ (t)

>1

(7.259)

476

7 Difference Equations

Theorem 7.68 (Hartman-Type Inequality) Suppose 1 − α1 (t) > 0, β1 (t) > 0, and β2 (t) > 0 for all t ∈ Z. Let n, m ∈ Z with n ≤ m − 1. Assume (7.242) has a real solution (x, u) such that x(n − 1)x(n) < 0

x(m − 1)x(m) < 0,

and

and x(t) = 0 for all t ∈ [n, m − 1]. Then, the inequality m−2

 γ /α−1 |α1 (t)| + M2

t=n−1

m−1

1/α  β1 (t)

t=n−1

m−2

1/β >1

β2 (t)

(7.260)

t=n−1

holds, where α is the Hölder conjugate of β and M2 =

max

n−1≤τ ≤m0 −1

max|u(τ )|.

Proof Suppose that x(t) = 0 for all t ∈ [n, m − 1]. Let m0 denote the smallest integer in [n, m] such that m0 = n and x(m0 − 1)x(m0 ) < 0. Then, x does not have any generalized zero in [n + 1, m0 − 1], and without loss of generality, we may assume that x(t) > 0

for all

t ∈ [n, m0 − 1].

Hence, we must have x(n − 1) < 0

and

x(m0 ) < 0.

Let τ ∈ [n − 1, m0 − 1]. Summing the second equation in (7.242) first from n − 1 to τ − 1 and then from τ to m0 − 2, we obtain τ −1

u(τ )−u(n−1) = −

β2 (t)|x(t +1)|β−2 x(t +1)−

t=n−1

τ −1

α1 (t)u(t)

(7.261)

α1 (t)u(t),

(7.262)

t=n−1

and u(m0 −1)−u(τ ) = −

m 0 −2 t=τ

β2 (t)|x(t +1)|β−2 x(t +1)−

m 0 −2 t=τ

respectively. Here, notice that, for τ = n − 1, we write solely (7.262), and for τ = m0 − 1, only (7.261) is written. Now, we claim that

7.7 Discrete Nonlinear Systems

477

u(n − 1) > 0 and

u(m0 − 1) < 0.

(7.263)

Indeed, from the first equation in (7.242), we have (1 − α1 (t)) x(t + 1) = x(t) + β1 (t)|u(t)|γ −2 u(t). First multiplying this last equation by x(t) and then setting t = n−1 and t = m0 −1 in the obtained equation, respectively, yields (1 − α1 (n − 1)) x(n−1)x(n) = [x(n−1)]2 +β1 (n−1)|u(n−1)|γ −2 u(n−1)x(n−1) and (1 − α1 (m0 − 1)) x(m0 − 1)x(m0 ) = [x(m0 − 1)]2 + β1 (m0 − 1) |u(m0 − 1)|γ −2 u(m0 − 1)x(m0 − 1). Using the inequalities x(n−1)x(n) < 0, x(m0 −1)x(m0 ) < 0, and since 1−α1 (t) > 0 and β1 (t) > 0 for all t ∈ Z, we get from the above latter inequality that u(n − 1)x(n − 1) < 0

u(m0 − 1)x(m0 − 1) < 0

and

(7.264)

hold. Hence, taking into account x(n − 1) < 0 and x(m0 − 1) > 0, we obtain (7.263). Employing (7.261) if u(τ ) < 0 and (7.262) if u(τ ) > 0, and taking into account (7.263), we obtain m 0 −2

|u(τ )| ≤

β2 (t)|x(t + 1)|

β−1

+

t=n−1

m 0 −2

|α1 (t)| |u(t)|.

(7.265)

t=n−1

By using Hölder’s inequality on the first sum of the right-hand side of (7.265) with Hölder conjugates α and β with 1/α + 1/β = 1, we get ⎛ |u(τ )| ≤ ⎝

m 0 −2 t=n−1

⎞1/β ⎛ β2 (t)⎠

⎝

m 0 −2

⎞1/α β2 (t)|x(t + 1)|

β⎠

t=n−1

+

m 0 −2

|α1 (t)| |u(t)|.

t=n−1

(7.266) Next, summing (7.250) from n − 1 to m0 − 1 yields x(m0 )u(m0 ) − x(n − 1)u(n − 1) =

m 0 −1 t=n−1

β1 (t)|u(t)|γ −

m 0 −1 t=n−1

β2 (t)|x(t + 1)|β ,

478

7 Difference Equations

i.e., x(m0 )u(m0 ) + β2 (m0 − 1) |x(m0 )|β − x(n − 1)u(n − 1) =

m 0 −1

β1 (t)|u(t)| − γ

t=n−1

m 0 −2

β2 (t)|x(t + 1)|β .

(7.267)

t=n−1

Now, we claim x(m0 )u(m0 ) + β2 (m0 − 1) |x(m0 )|β > 0.

(7.268)

Indeed, from the second equation in (7.242), we have, for t = m0 − 1, u(m0 ) − u(m0 − 1) = −β2 (m0 − 1) |x(m0 )|β−2 x(m0 ) − α1 (m0 − 1)u(m0 − 1), which upon multiplication by x(m0 ) yields u(m0 )x(m0 ) + β2 (m0 − 1) |x(m0 )|β = (1 − α1 (m0 − 1)) u(m0 − 1)x(m0 ). (7.269) On the other hand, from the inequalities x(m0 − 1)x(m0 ) < 0 and

x(m0 − 1)u(m0 − 1) < 0,

it follows that u(m0 −1)x(m0 ) > 0. Therefore, our claim follows from (7.269) since 1 − α1 (t) > 0 for all t ∈ Z. By virtue of (7.264) and (7.268), from (7.267), the inequality m 0 −2

β2 (t)|x(t + 1)| < β

t=n−1

m 0 −1

β1 (t)|u(t)|γ

t=n−1

follows. Substituting this last inequality into (7.266) yields ⎛ |u(τ )| < ⎝

m 0 −2 t=n−1

⎞1/β ⎛ β2 (t)⎠

⎝

m 0 −1

⎞1/α β1 (t)|u(t)|

γ⎠

+

t=n−1

m 0 −2

|α1 (t)| |u(t)|

t=n−1

(7.270) for all τ ∈ [n − 1, m0 − 1]. Choose τ0 ∈ [n − 1, m0 − 1] such that M2 = |u(τ0 )| =

max

n−1≤τ ≤m0 −1

|u(τ )|.

7.7 Discrete Nonlinear Systems

479

Then, |u(τ0 )| > 0, and from (7.270), we obtain ⎛ |u(τ0 )| < |u(τ0 )|γ /α ⎝

m 0 −2

⎞1/β ⎛ β2 (t)⎠

⎝

t=n−1

m 0 −1

⎞1/α β1 (t)⎠

+ |u(τ0 )|

t=n−1

m 0 −2

|α1 (t)| .

t=n−1

Hence, dividing by |u(τ0 )|, we have ⎛ γ /α−1 ⎝

M2

⎞1/β ⎛

m 0 −2

β2 (t)⎠

⎝

t=n−1

m 0 −1

⎞1/α β1 (t)⎠

m 0 −2

+

t=n−1

|α1 (t)| > 1.

t=n−1

Since m0 ≤ m, from the latter inequality follows (7.260).



Remark 7.69 We should note that (7.260) is valid only if x(t) = 0 for all t ∈ [n, m−1]. If x has a zero on [n+1, m−2], i.e., x(t0 ) = 0 for some t0 ∈ [n+1, m−2], then we have the following two cases: In the case when x(n − 1)x(n) < 0 and x(t0 ) = 0, (7.260) is replaced by (7.259), and in the case when x(t0 ) = 0 and x(m − 1)x(m) < 0, (7.260) is replaced by (7.255). Remark 7.70 If we had imposed γ and β in (7.242) to be Hölder conjugates of each other to begin with, then all of (7.249), (7.255), (7.259), and (7.260) would still be obtained without constants M by using Hölder’s inequalities with Hölder conjugates γ and β in the proofs of Theorems 7.65, 7.66, 7.67, and 7.68. By virtue of Remark 7.70, the following corollary follows from combining Theorems 7.65, 7.66, 7.67, and 7.68. Corollary 7.71 (Hartman-Type Inequality) Suppose 1 − α1 (t) > 0, β1 (t) > 0, and β2 (t) > 0 for all t ∈ Z. Let n, m ∈ Z with n ≤ m − 2. Assume (7.242) with 1/γ + 1/β = 1 has a real solution (x, u) such that x has generalized zeros at n and m, and x is not identically zero on [n, m]. Then, the inequality m−2

t=n−1

 |α1 (t)| +

m−1

t=n−1

1/γ  β1 (t)

m−2

1/β β2 (t)

>1

(7.271)

t=n−1

holds, where γ > 1 and β > 1 are constants. Remark 7.72 Taking β = γ = 2 in (7.242) yields the discrete linear Hamiltonian system  Δx(t) = α1 (t)x(t + 1) + β1 (t)u(t), Δu(t) = −β2 (t)x(t + 1) − α1 (t)u(t),

(7.272)

where t ∈ Z. Hence, all the results presented in this section for (7.242) are also valid for (7.272). Thus, we should remark here that (7.242) may be viewed as a

480

7 Difference Equations

natural generalization of (7.272). When β = γ = 2 in (7.242), it is easy to see that Theorems 7.65, 7.66, 7.67, and 7.68 and Corollary 7.71 reduce to Guseinov and Kaymakçalan [143, Theorems 1.2–1.5 and Corollary 1.6], respectively. Remark 7.73 Now, consider the following two special cases of (7.242), which are equivalent systems for Emden–Fowler-type difference equations of the form (7.247) and for half-linear difference equations of the form (7.248), namely 

Δx(t) = β1 (t)|u(t)|γ −2 u(t), Δu(t) = −β2 (t)|x(t + 1)|β−2 x(t + 1)

(7.273)

and 

Δx(t) = β1 (t)|u(t)|γ −2 u(t), Δu(t) = −β2 (t)|x(t + 1)|α−2 x(t + 1),

(7.274)

respectively, where t ∈ Z, β1 (t) = p1−γ (t), β2 (t) = q(t), and 1/α + 1/γ = 1. Needless to say that all of the above results, with the special case α1 (t) ≡ 0, are also valid for (7.273) and (7.274), and hence for Emden–Fowler-type difference equations of the form (7.247) and for half-linear difference equations of the form (7.248). Remark 7.74 We should also note here that Theorem 7.65 is the discrete analogue of Tiryaki et al. [274, Theorem 1]. In 2011, He and Zhang [159] considered obtaining Lyapunov-type inequalities for discrete nonlinear systems of the form 

Δx(n) = α(n)x(n + 1) + β(n)|y(n)|μ−2 y(n), Δy(n) = −γ (n)|x(n + 1)|ν−2 x(n + 1) − α(n)y(t),

(7.275)

where μ, ν > 1 with 1/μ + 1/ν = 1 and α, β, γ are real-valued functions defined on Z with β(n) ≥ 0 for all n ∈ Z. In [159], by using some simpler methods different from [277], He and Zhang obtained a better Lyapunov-type inequality than (7.271), namely b−1

n=a

 |α(n)| +

b

β(n)

1/μ b−1

n=a

1/ν +

γ (n)

≥ 2,

(7.276)

n=a

only under the assumption 1 − α(n) > 0

for all

n ∈ Z.

When μ = ν = 2, (7.276) reduces to the Lyapunov-type inequality

(7.277)

7.7 Discrete Nonlinear Systems b−1

481

 |α(n)| +

n=a

b

β(n)

n=a

b−1

1/2 +

γ (n)

≥2

n=a

for (7.82), which also greatly improves the main result in [143, Theorem 1.6]. Even if the right endpoint b is not a generalized zero of x, we can still obtain a Lyapunov-type inequality better than (7.271), namely b−1

|α(n)| +

n=a

b−1

β(n)

1/μ b−1

n=a

1/ν +

γ (n)

≥ 2,

(7.278)

n=a

as long as (x(b), y(b)) = (λ1 x(a), λ2 y(a)) with 0 ≤ |λ1 |μ/(μ−1) ≤ λ1 λ2 ≤ 1. Theorem 7.75 Suppose that (7.277) holds and let a ≤ b − 1. Assume (7.275) has a real solution (x, y) such that (7.92) holds. If max |x(n)| > 0,

(7.279)

a+1≤n≤b

then (7.276) holds. Proof It follows from (7.92) and (7.279) that there exist ξ, η ∈ [0, 1) such that (1 − ξ )x(a) + ξ x(a + 1) = 0

(7.280)

(1 − η)x(b) + ηx(b + 1) = 0.

(7.281)

and

Multiplying the first equation in (7.275) by y(n) and the second one by x(n + 1) and then adding, we get Δ(x(n)y(n)) = β(n)|y(n)|μ − γ (n)|x(n + 1)|ν .

(7.282)

Summing (7.282) from a to b − 1, we can obtain x(b)y(b) − x(a)y(a) =

b−1

β(n)|y(n)| −

n=a

μ

b−1

γ (n)|x(n + 1)|ν .

(7.283)

n=a

From the first equation in (7.275), we have [1 − α(n)]x(n + 1) = x(n) + β(n)|y(n)|μ−2 y(n).

(7.284)

482

7 Difference Equations

Combining (7.284) with (7.280), we have x(a) = −

ξβ(a) |y(a)|μ−2 y(a). 1 − (1 − ξ )α(a)

(7.285)

Similarly, it follows from (7.284) and (7.281) that x(b) = −

ηβ(b) |y(b)|μ−2 y(b) 1 − (1 − η)α(b)

(7.286)

holds. Substituting (7.285) and (7.286) into (7.283), we have b−1

β(n)|y(n)| − μ

n=a

b−1

γ (n)|x(n + 1)|ν

n=a

=−

ηβ(b) ξβ(a) |y(b)|μ + |y(a)|μ , 1 − (1 − η)α(b) 1 − (1 − ξ )α(a)

which implies b−1

(1 − ξ )[1 − α(a)] ηβ(b) β(n)|y(n)|μ + β(a)|y(a)|μ + |y(b)|μ−2 y(b) 1−(1−ξ )α(a) 1−(1−η)α(b) n=a

=

b−1

γ (n)|x(n + 1)|ν .

(7.287)

n=a

Denote ˜ β(a) =

(1 − ξ )[1 − α(a)] β(a), 1 − (1 − ξ )α(a)

˜ β(b) =

η β(b), 1 − (1 − η)α(b)

and ˜ β(n) = β(n),

a + 1 ≤ n ≤ b − 1.

Then, we can rewrite (7.287) as b−1

n=a

μ ˜ = β(n)|y(n)|

b−1

n=a

γ (n)|x(n + 1)|ν .

(7.288)

7.7 Discrete Nonlinear Systems

483

On the other hand, summing the first equation in (7.275) from a to τ − 1 and using (7.285), we obtain x(τ ) = x(a) +

τ −1

α(n)x(n + 1) +

n=a

=

=

τ −1

β(n)|y(n)|μ−2 y(n)

n=a

τ −1 τ −1

ξβ(a) |y(a)|μ−2 y(a)+ α(n)x(n + 1)+ β(n)|y(n)|μ−2 y(n) 1−(1−ξ )α(a) n=a n=a τ −1

α(n)x(n + 1) +

n=a

τ −1

μ−2 ˜ β(n)|y(n)| y(n),

a + 1 ≤ τ ≤ b.

n=a

(7.289) Similarly, summing the first equation in (7.275) from τ to b − 1 and using (7.286), we have x(τ ) =x(b) −

b−1

α(n)x(n + 1) −

n=τ

β(n)|y(n)|μ−2 y(n)

n=τ

b−1

ηβ(b) α(n)x(n + 1) |y(b)|μ−2 y(b) − 1 − (1 − η)α(b) n=τ

=−

−

b−1

b−1

β(n)|y(n)|μ−2 y(n)

n=τ

=−

b−1

α(n)x(n + 1) −

n=τ

b

μ−2 ˜ β(n)|y(n)| y(n),

a + 1 ≤ τ ≤ b.

n=τ

(7.290) It follows from (7.289) and (7.290) that |x(τ )| ≤

τ −1

|α(n)||x(n + 1)| +

τ −1

n=a

n=a

b−1

b

μ−1 ˜ β(n)|y(n)| ,

a+1≤τ ≤b

μ−1 ˜ β(n)|y(n)| ,

a+1≤τ ≤b

and |x(τ )| ≤

|α(n)||x(n + 1)| +

n=τ

n=τ

hold. Adding these two inequalities, we have 2|x(τ )| ≤

b−1

n=a

|α(n)||x(n + 1)| +

b

n=a

μ−1 ˜ β(n)|y(n)| ,

a + 1 ≤ τ ≤ b.

(7.291)

484

7 Difference Equations

Let |x(τ ∗ )| = maxa+1≤n≤b |x(n)|. Applying Hölder’s inequality and using (7.288), we have b−1

2|x(τ ∗ )| ≤

|α(n)||x(n + 1)| +

n=a

b

μ−1 ˜ β(n)|y(n)|

n=a



b−1 b

  ˜ ≤ x(τ ∗ ) |α(n)| + β(n) n=a

1/μ 

n=a

1/ν μ ˜ β(n)|y(n)|

n=a



b−1 b

  ˜ β(n) = x(τ ∗ ) |α(n)| + n=a

b

1/μ b−1

n=a

1/ν +

γ (n)|x(n + 1)|

ν

n=a

⎧  b 1/ν ⎫ 1/μ b−1 b−1 ⎬

 ∗  ⎨ ˜ ≤ x(τ ) |α(n)| + γ + (n) . β(n) ⎩ ⎭ n=a

n=a

n=a

(7.292)

Dividing (7.292) by |x(τ ∗ )|, we obtain b−1

 |α(n)| +

n=a

b

˜ β(n)

1/μ b−1

n=a

1/ν +

γ (n)

≥ 2.

(7.293)

n=a

Since ˜ β(n) ≤ β(n),

a ≤ n ≤ b, 

(7.276) follows immediately from (7.293). In the case when x(b) = 0, we have the equation b−1

μ ˜ = β(n)|y(n)|

n=a

b−2

γ (n)|x(n + 1)|ν

n=a

and inequality 2|x(τ )| ≤

b−2

|α(n)||x(n + 1)| +

n=a

b−1

ν−1 ˜ β(n)|y(n)| ,

a+1≤τ ≤b−1

n=a

instead of (7.288) and (7.291), respectively. Similar to the proof of (7.293), we have b−2

n=a

|α(n)| +

b−1

n=a

˜ β(n)

1/μ b−2

n=a

1/ν +

γ (n)

≥ 2.

7.7 Discrete Nonlinear Systems

485

˜ Since β(n) ≤ β(n) for a ≤ n ≤ b, we get b−2

|α(n)| +

b−1

n=a

β(n)

1/μ b−2

n=a

1/ν +

γ (n)

≥ 2.

n=a

Therefore, we have the following theorem. Theorem 7.76 Suppose that (7.277) holds and let a ≤ b − 2. Assume (7.275) has a real solution (x, y) such that x(a) = 0 or x(a)x(a + 1) < 0 and x(b) = 0, and x(n) is not identically zero on [a, b]. Then, (7.110) holds. Remark 7.77 We obtain (7.110), which is the same as (7.249), but under weaker assumptions than those of Theorem 7.65. Theorem 7.78 Suppose that (7.277) holds and let a ≤ b − 1. Assume (7.275) has a real solution (x, y) such that x(a) = 0 or x(a)x(a + 1) < 0 and (x(b), y(b)) = (λ1 x(a), λ2 y(a)) with 0 ≤ |λ1 |μ/(μ−1) ≤ λ1 λ2 ≤ 1, and x(n) is not identically zero on [a, b]. Then, (7.278) holds. Proof It follows from the assumption x(a) = 0 or x(a)x(a + 1) < 0 that there exists ξ ∈ [0, 1) such that (7.280) holds. Further, by the proof of Theorem 7.75, (7.282), (7.283), (7.284), and (7.285) hold. Since (x(b), y(b)) = (λ1 x(a), λ2 y(a)), by (7.283), we have (λ1 λ2 − 1)x(a)y(a) =

b−1

β(n)|y(n)|μ −

n=a

b−1

γ (n)|x(n + 1)|ν .

(7.294)

n=a

Substituting (7.285) into (7.294), we have b−1

β(n)|y(n)|μ −

n=a

b−1

γ (n)|x(n + 1)|ν =

n=a

(1 − λ1 λ2 )ξβ(a) |y(a)|μ , 1 − (1 − ξ )α(a)

which implies κ1 β(a)|y(a)|μ +

b−1

n=a+1

where

β(n)|y(n)|μ =

b−1

n=a

γ (n)|x(n + 1)|ν ,

(7.295)

486

7 Difference Equations

κ1 =

(1 − λ1 λ2 )ξ − (1 − ξ )α(a) . 1 − (1 − ξ )α(a)

(7.296)

On the other hand, summing the first equation in (7.275) from a to τ − 1 and using (7.285), we obtain x(τ ) =x(a) +

τ −1

α(n)x(n + 1) +

n=a

β(n)|y(n)|μ−2 y(n)

n=a

τ −1

ξβ(a) |y(a)|μ−2 y(a) + α(n)x(n + 1) 1 − (1 − ξ )α(a) n=a

=−

+

τ −1

τ −1

β(n)|y(n)|μ−2 y(n)

(7.297)

n=a

=

τ −1

(1 − ξ )[1 − α(a)] β(a)|y(a)|μ−2 y(a) + α(n)x(n + 1) 1 − (1 − ξ )α(a) n=a τ −1

+

a + 1 ≤ τ ≤ b.

β(n)|y(n)|μ−2 y(n),

n=a+1

Similarly, summing the first equation in (7.275) from τ to b − 1 and using (7.285) and the fact that x(b) = λ1 x(a), we have x(τ ) =x(b) −

b−1

α(n)x(n + 1) −

n=τ

=λ1 x(a) −

b−1

b−1

β(n)|y(n)|μ−2 y(n)

n=τ

α(n)x(n + 1) −

n=τ

b−1

β(n)|y(n)|μ−2 y(n)

n=τ

b−1

λ1 ξ β(a)|y(a)|μ−2 y(a) + = α(n)x(n + 1) 1 − (1 − ξ )α(a) n=τ

−

b−1

β(n)|y(n)|μ−2 y(n),

a + 1 ≤ τ ≤ b.

n=τ

It follows from (7.297) and (7.298) that |x(τ )| ≤

τ −1

(1 − ξ )[1 − α(a)] β(a)|y(a)|μ−1 + |α(n)||x(n + 1)| 1 − (1 − ξ )α(a) n=a

+

τ −1

n=a+1

and

β(n)|y(n)|μ−1

(7.298)

7.7 Discrete Nonlinear Systems

|x(τ )| ≤

487

b−1 b−1

|λ1 | ξ β(a)|y(a)|μ−1 + |α(n)||x(n+1)|+ β(n)|y(n)|μ−1 1 − (1 − ξ )α(a) n=τ n=τ

hold for a + 1 ≤ τ ≤ b. Adding these two inequalities, we have 2|x(τ )| ≤ 2κ2 β(a)|y(a)|μ−1 +

b−1

b−1

|α(n)||x(n + 1)| +

n=a

β(n)|y(n)|μ−1

n=a+1

(7.299)

for a + 1 ≤ τ ≤ b, where κ2 =

1 − (1 − |λ1 |)ξ − (1 − ξ )α(a) . 1 − (1 − ξ )α(a)

(7.300)

Let |x(τ ∗ )| = maxa+1≤n≤b |x(n)|. Applying (7.295), (7.299), and Hölder’s inequality, we have b−1 b−1

  2 x(τ ∗ ) ≤ κ2 β(a)|y(a)|μ−1 + |α(n)||x(n + 1)| + β(n)|y(n)|μ−1 n=a

n=a+1

b−1   ≤ x(τ ∗ ) |α(n)| n=a

 +

b−1

μ

κ2

β(a) +

μ−1

κ1

 μ1  κ1 β(a)|y(a)|μ +

β(n)

n=a+1

 ν1

b−1

β(n)|y(n)|μ

n=a+1

b−1   = x(τ ∗ ) |α(n)| n=a

 +

b−1

μ

κ2

β(a) +

μ−1

κ1

β(n)

1/μ b−1

1/ν γ (n)|x(n + 1)|

ν

n=a

n=a+1

b−1   ≤ x(τ ∗ ) |α(n)| n=a

 +

μ

κ2

β(a) +

μ−1

κ1

b−1

β(n)

1/μ b−1

1/ν +

γ (n)|x(n + 1)|

ν

n=a

n=a+1

b−1  ∗    ≤ x(τ ) |α(n)| n=a

 +

μ

κ2

μ−1

κ1

β(a) +

b−1

n=a+1

β(n)

1/μ b−1

n=a

γ + (n)

1/ν ⎫ ⎬ ⎭

. (7.301)

488

7 Difference Equations

Dividing (7.301) by |x(τ ∗ )|, we obtain b−1

 |α(n)| +

μ

κ2

μ−1

κ1

n=a

β(a) +

b−1

β(n)

1/μ b−1

n=a+1

1/ν +

≥ 2.

γ (n)

(7.302)

n=a

Set d = 1 − (1 − ξ )α(a). Since (1 − ξ )[1 − α(a)] > 0, it follows that d > ξ ≥ 0. Let f (t) = [1 − (1 − |λ1 |)t]μ/(μ−1) + (1 − λ1 λ2 )t − 1. Then, f  (t) = −

μ(1 − |λ1 |) [1 − (1 − |λ1 |)t]1/(μ−1) + (1 − λ1 λ2 ) μ−1

and f  (t) =

μ(1 − |λ1 |)2 [1 − (1 − |λ1 |)t](2−μ)/(μ−1) ≥ 0 (μ − 1)2

for all

t ∈ [0, 1].

It follows that  f (t) ≤ max{f (0), f (1)} = max 0, |λ1 |μ/(μ−1) − λ1 λ2 = 0

for all

t ∈ [0, 1],

that is, [1 − (1 − |λ1 |)t]μ/(μ−1) ≤ 1 − (1 − λ1 λ2 )t

for all t ∈ [0, 1],

which implies   ξ μ/(μ−1) ξ 1 − (1 − |λ1 |) ≤ 1 − (1 − λ1 λ2 ) . d d This, together with (7.296) and (7.300), implies

μ

κ2

μ−1

κ1

=

1 − (1 − |λ1 |)ξ − (1 − ξ )α(a) 1 − (1 − ξ )α(a)

×

μ

1 − (1 − λ1 λ2 )ξ − (1 − ξ )α(a) 1 − (1 − ξ )α(a)

1−μ

1 = {d − (1 − |λ1 |)ξ }μ {d − (1 − λ1 λ2 )ξ }1−μ d

7.7 Discrete Nonlinear Systems

489

    ξ μ ξ 1−μ 1 − (1 − λ1 λ2 ) = 1 − (1 − |λ1 |) d d ≤1. 

Substituting this into (7.302), we obtain (7.278).

Applying Theorems 7.75, 7.76, and 7.78 to the second-order half-linear difference equation ! Δ p(n)|Δx(n)|r−2 Δx(n) + q(n)|x(n + 1)|r−2 x(n + 1) = 0,

(7.303)

we have the following corollaries. Corollary 7.79 (Lyapunov-Type Inequality) Let a ≤ b − 1. Suppose that p(n) > 0 for all n ∈ Z.

(7.304)

Assume (7.303) has a real solution x such that (7.92) and (7.279) hold. Then, the inequality 

b

n=a

1 [p(n)]1/(r−1)

1−1/r b−1

1/r q + (n)

≥2

n=a

holds. Corollary 7.80 (Lyapunov-Type Inequality) Suppose that (7.304) holds and let a ≤ b − 2. Assume (7.303) has a real solution x with x(a) = 0 or x(a)x(a + 1) < 0 and x(b) = 0, and x(n) is not identically zero on [a, b]. Then, the inequality b−1

n=a

1 [p(n)]1/(r−1)

1−1/r b−2

1/r +

q (n)

≥2

(7.305)

n=a

holds. Corollary 7.81 (Lyapunov-Type Inequality) Suppose that (7.304) holds and let a ≤ b − 1. Assume (7.303) has a real solution x such that x(a)x(a + 1) < 0, x(b) = λ1 x(a), and p(b)|Δx(b)|r−2 Δx(b) = λ2 p(a)|Δx(a)|r−2 Δx(a), where 0 ≤ |λ1 |r ≤ λ1 λ2 ≤ 1, and x(n) is not identically zero on [a, b]. Then, the inequality

490

7 Difference Equations

b−1

n=a

1 [p(n)]1/(r−1)

1−1/r b−1

1/r +

q (n)

≥2

n=a

holds.

7.7.1 Some Applications Applying the inequalities derived, Ünal et al. [277] established some results related to disconjugacy and boundedness for the solution of (7.242). Let n, m ∈ Z with n ≤ m − 2. Consider the discrete nonlinear system 

Δx(t) = α1 (t)x(t + 1) + β1 (t)|u(t)|γ −2 u(t), Δu(t) = −β2 (t)|x(t + 1)|β−2 x(t + 1) − α1 (t)u(t),

(7.306)

where t ∈ [n, m]. It is assumed that γ , β > 1 are constants and the coefficients α1 , β1 , β2 are real-valued functions defined on [n, m] such that 1 − α1 (t) > 0 and

β1 (t) > 0 for all

t ∈ [n, m].

(7.307)

We also note that each solution (x, u) of (7.306) is a vector-valued function defined on [n, m + 1]. In the sequel, we introduce the concept of a relatively generalized zero for the component x of the real solution (x, u) of (7.306) and also the concept of disconjugacy of the same system on [n, m + 1]. The definition is relative to the interval [n, m + 1], and the left endpoint n is treated separately. Definition 7.82 The component x of the real solution (x, u) of (7.306) has a relatively generalized zero at n if and only if x(n) = 0, while x has a relatively generalized zero at t0 > n provided either x(t0 ) = 0 or x(t0 − 1)x(t0 ) < 0. Definition 7.83 System (7.306) is said to be disconjugate on [n, m + 1] if no real solution (x, u) of this system with x ≡ 0 has two (or more) relatively generalized zero in [n, m + 1]. Remark 7.84 We should remark here, as mentioned in [143], that under (7.307), Definitions 7.82 and 7.83 are equivalent to those given in [22, page 354] and in [47]. We conclude our remark with related references on the subject [6, 10, 121, 129, 154], and the references given therein. Theorem 7.85 Assume (7.307) holds. If the inequality m−1

t=n

|α1 (t)| + M

β/α−1

 m

t=n

β1 (t)

1/γ m−1

t=n

1/α β2+ (t)

≤1

(7.308)

7.7 Discrete Nonlinear Systems

491

holds, where α is the Hölder conjugate of γ and M=

max

n+1≤t≤m−1

|x(t)|,

then (7.306) is disconjugate on [n, m + 1]. Proof Suppose, on the contrary, that (7.306) is not disconjugate on [n, m+1]. Then, by definition, there exists a real solution (x, u) of (7.306) with x is nontrivial and such that x(n) = 0 and that x has a generalized zero m0 ∈ [n + 1, m + 1]. We have m0 > n + 1 and either x(m0 ) = 0 or x(m0 − 1)x(m0 ) < 0. Therefore, applying Theorems 7.65 and 7.66, we obtain m 0 −2

⎞1/γ ⎛

⎛

m 0 −1

|α1 (t)| + M β/α−1 ⎝

t=n

β1 (t)⎠

t=n

m 0 −2

⎝

⎞1/α β2+ (t)⎠

> 1,

t=n



contradicting (7.308).

Remark 7.86 We should mention here that if we impose 1/γ + 1/β = 1 in (7.306), then we could still obtain (7.308) without M. In addition to this, when β = γ = 2 in (7.306), it is easy to see that Theorem 7.85 reduces to Guseinov and Kaymakçalan [143, Theorem 7.1], see Theorem 7.42. Theorem 7.87 If ∞

t=0

β1 (t) < ∞ and

∞

|β2 (t)| < ∞,

(7.309)

t=0

then every weakly oscillatory proper solution (x, u) of (7.273) is bounded on Z. Proof Let n, m ∈ Z with n ≤ m − 2. Let (x, u) be any nontrivial weakly oscillatory proper solution of (7.273) on Z such that x has a sequence of generalized zeros tending to ∞. Suppose, to the contrary, that lim supt→∞ |x(t)| = ∞. Then, given any positive number M3 , we can find N = N(M3 ) ∈ N such that |x(t)| > M3 for all t > N. Since x has a sequence of generalized zeros tending to ∞, there exists an interval [n, m] with n ≥ N such that one of the following cases hold. (i) (ii) (iii) (iv)

x(n) = 0 = x(m), x(n) = 0 and x(m − 1)x(m) < 0, x(n − 1)x(n) < 0 and x(m) = 0, x(n − 1)x(n) < 0 and x(m − 1)x(m) < 0, and x(t) = 0 for all t ∈ [n, m − 1].

Clearly all theorems given for (7.242) with α1 (t) ≡ 0 are valid for (7.273). Notice that each of the above cases corresponds to one theorem given previously. That is, each of the cases (i), (ii), (iii), and (iv) corresponds to Theorems 7.65, 7.66, 7.67, and 7.68, respectively. Here, we only prove the case x(n) = 0 = x(m). The proofs of the other cases can be obtained similarly.

492

7 Difference Equations

Now, choose τ ∈ [n, m] such that M = |x(τ )| = max |x(t)| > M3 . n 0 for w ∈ S, and since G(w|z) = 0 for w ∈ ∂C ∩ ∂S (which is nonempty by Corollary 7.91), G(w|z) cannot be a constant function over C ∪ ∂C. According to the maximum principle, max G(w|z) < max G(w|z) ≤ w∈S

w∈∂C

max G(w|z) = G(z|z),

w∈∂S∪{z}



as desired.

Condition (7.321) is sharp in the sense that we can find a nonnegative function p on S and a nontrivial solution of (7.318) such that

p(w) =

w∈S

1 maxw,z∈S G(w|z)

=

1 . maxz∈S G(z|z)

Indeed, let z∗ ∈ S such that max G(z|z) = G(z∗ |z∗ ). z∈S

Let u(z) = G(z|z∗ ) for z ∈ S ∪ ∂S and let p(z) = −

Du(z) , u(z)

z ∈ S.

Then, (7.318) is clearly satisfied, and

z∈s

p(z) = −

Du(z) z∈S

u(z)

=

δ(z|z∗ ) 1 = , G(z|z∗ ) G(z∗ |z∗ ) z∈S

as required. We summarize the above discussions as follows.

7.8 Partial Difference Systems

499

Theorem 7.95 Suppose p is a nonnegative function defined on a net S. If (7.318) has a nontrivial solution, then the inequality

p(z) ≥ μ(S)

z∈S

holds, where μ(S) =

1 1 = , maxw,z∈S G(z|w) maxz∈S G(z|z)

and the inequality is sharp. Before we turn to the estimations of the constant μ(S) for various nets, note that the following comparison theorem for Green functions holds. Theorem 7.96 Let z0 ∈ ∂S and let S  = S ∪ {z0 }. Let G (z|w) be Green’s function for the system 

DG (z|w) = −δ(z|w)

if z ∈ S  ,

G (z|w) = 0

if z ∈ ∂S  .

Then, G(z|w) < G (z|w) for all z ∈ S. Proof Define h(z) = G(z|w) − G (z|w),

z ∈ S ∪ ∂S.

Then, Dh(z) = 0 for z ∈ S and h(z) = −G (z|w),

z ∈ ∂S.

Since G (z0 |w) > 0 and since G (z|w) = 0 for any z ∈ ∂S ∩ ∂S  , h cannot be a constant function over S ∪ ∂S. By the maximum principle, we have max h < max h ≤ 0, S

as required.

∂S



500

7 Difference Equations

7.8.3 Maxima of Green’s Functions on Straight Nets A net S said to be straight if there are exactly two points in S with degree 1 and the remaining points in S have degree 2. An example of a straight net is the set {(2, 2), (2, 3), (2, 4), (3, 4), (4, 4), (4, 5)} . If a straight net S has N points, then it is not difficult to see that these points can be ordered as a chain z1 , z2 , . . . , zN such that zi and zj are neighbors if and only if |i − j | = 1. As a consequence, (7.318) can be written as 

Du(zi ) + p(zi )u(zi ) = 0

if 1 ≤ i ≤ N,

u(z) = 0

if z ∈ ∂S,

(7.322)

and (7.319) can be written as 

DG(zi |zj ) = −δ(zi |zj )

if

1 ≤ i ≤ N,

G(z|zj ) = 0

if

z ∈ ∂S,

(7.323)

where zj is some point of S. For convenience’s sake, we write G(i, j ) instead of G(zi |zj ). We now find an explicit formula for G(i, j ). In order to do this, let X−1 = 0, X0 = 1, and Xk be defined by the recurrence relation Xk = 4Xk−1 − Xk−2

for

k ∈ N \ {1}.

We can verify that 1 Xk = √ (α k+1 − α −(k+1) ) 2 3

for all

k ∈ N0 ,

with α = 2 +

√ 3.

(7.324)

By means of (7.324), it is easy to check that xk is positive and increasing. Also G(i, j ) =

 −1 Xi−1 XN −j XN

if

1 ≤ i ≤ j,

−1 XN −i Xj −1 XN

if

j ≤i≤N

and max G(i, j ) =

1≤i≤N

 −1 Xm−1 Xm XN 2 −1 Xm XN

if

N = 2m,

if

N = 2m + 1.

7.8 Partial Difference Systems

501

The proof of these assertions amounts to direct verification, which is straightforward. In view of Theorem 7.95 and the above discussions, we have the following result. Theorem 7.97 Let S be a straight net with N points. If (7.322) has a nontrivial solution, then (7.317) holds, where  μ(S) =

−1 −1 XN Xm−1 Xm

if

N = 2m,

−2 XN Xm

if

N = 2m + 1,

and Xk is given by (7.324). The inequality is sharp. The function μ(S) is a decreasing but bounded as can be seen from the following theorem. Theorem 7.98 Let MN = max1≤i≤N G(i, i). Then, {MN }∞ N =1 is an increasing √ sequence and has the limit 3/6. Proof We have −1 M2k+1 = X2k+1 Xk2 √ (2 3)−2 (α k+1 − α −k−1 )2 = √ (2 3)−1 (α 2k+2 − α −2k−2 )

1 − 2α −2k−2 + α −4k−4 √ 2 3(1 − α −4k−4 ) √ 3 . → 6 =

Similarly, M2k →

√

3/6. Next,

−1 −1 Xk+1 Xk − X2k+1 Xk Xk M2k+2 − M2k+1 = X2k+2 −1 −1 = X2k+2 X2k+1 Xk (X2k+1 Xk+1 − X2k+2 Xk ) −1 −1 = X2k+2 X2k+1 Xk {X2k+1 (4Xk − Xk−1 ) − (4X2k+1 − X2k ) Xk } −1 −1 = X2k+2 X2k+1 Xk (X2k Xk − X2k+1 Xk−1 ) −1 −1 X2k+1 Xk (Xk+1 X1 − Xk+2 X0 ) = X2k+2 −1 −1 = X2k+2 X2k+1 Xk2

> 0. Similarly, M2k+1 − M2k > 0. The proof is complete.



502

7 Difference Equations

7.8.4 Maxima of Green’s Functions on Circular Nets A net S is said to be circular if every one of its points has degree 2. An example of a circular net is the set {(1, 1), (1, 2), (1, 3), (1, 4), (2, 4), (3, 4), (3, 3), (4, 3), (4, 2), (4, 1), (3, 1), (2, 1)} . If a circular net S has N points, then it is not difficult to see that these points can be ordered as a closed chain z1 , z2 , . . . , zN such that z1 has only two neighbors zN and z2 , zk has only two neighbors zk−1 and zk+1 , where k = 2, 3, . . . , N − 1, and zN has only two neighbors zN −1 and z1 . Again, (7.318) and (7.319) can be written as (7.322) and (7.323), respectively. We also write G(i, j ) instead of G(zi |zj ) as in Sect. 7.8.3. We need an explicit formula for G(i, j ). In order to do this, let Yk , −1 ≤ k ≤ N, be defined by Yk =

 Xk 4XN −1 − 2XN −2 − 2

if

− 1 ≤ k ≤ N − 1,

if

k = N,

(7.325)

where Xk is defined in Sect. 7.8.3. We have already shown that Xk > 1 for k ∈ N \ {1} and Xk+1 − Xk > 0 for k ∈ N0 . Thus YN = 2(XN −1 − XN −2 ) + 2(XN −1 − 1) > 0. We now assert that G(i, j ) = (Y|i−j |−1 + YN −1−|i−j | )YN−1 . Again, the proof amounts to straightforward verification, and the details are omitted. Note further that G(i, i) = YN −1 YN−1 . In view of Theorem 7.95, we have the following result. Theorem 7.99 Let S be a circular net with N points. If (7.322) has a nontrivial solution, then (7.318) holds, where μ(S) = YN−1−1 YN , and YN −1 , YN are given by (7.325). The inequality is sharp.

√ We show that the sequence {YN −1 YN−1 }∞ N =0 is decreasing and its limit is 3/6. For this purpose, we first show that Xk2 − Xk−1 Xk+1 = 1 for k = 0, 1, . . . , N − 1. Indeed, X02 − X−1 X1 = 1 by definition. Assume that our hypothesis holds for k = m. Then,     2 2 −X Xm+1 −Xm+2 Xm = 4Xm − Xm−1 Xm+1 −Xm 4Xm+1 −Xm =Xm m+1 Xm−1 =1.

7.8 Partial Difference Systems

503

Theorem 7.100 The sequence {YN −1 YN−1 }∞ 0 is decreasing and approaches

√

3/6.

Proof We have YN−1+1 YN − YN−1 YN −1 =YN−1+1 YN−1 {(4XN −1 − 2XN −2 − 2) XN − (4XN − 2XN −1 − 2) XN −1 }  2 =YN−1+1 YN−1 2XN − 2X X + 2X − 2X N N −2 N −1 N −1 = − 2YN−1+1 YN−1 (XN − XN −1 − 1) . Since XN − XN −1 ≥ 3 (see (7.324)), the last term of the above chain of equalities is negative. Also, by means of (7.324) and (7.325), it is easily verified that YN −1 YN−1 √ approaches 3/6. The proof is complete.  √ Remark 7.101 It is interesting to note that μ(A) ≥ 6/ 3 ≥ μ(B) for any circular net A and any straight net B.

7.8.5 Maxima of Green’s Functions on Rectangular Nets A net S is said to be rectangular if S consists of n rows of m lattice points, that is, S = {(i, j ) : 1 ≤ i ≤ m, 1 ≤ j ≤ n with i, j, m, n ∈ N} .

(7.326)

Writing z as (i, j ) and w as (a, b), Green’s function G(z|w) for (7.319) associated with (7.326) is given by (see McCrea and Whipple [211])



m irπ sinh((n + 1 − j )βr ) 2 arπ sin sin m+1 m+1 m+1 sinh((n + 1)βr ) r=1

×

⎧ sinh(jβ ) r ⎪ ⎪ ⎨ sinh(β )

if j ≤ b,

⎪ ⎪ ⎩ sinh(bβr ) sinh(βr )

if j ≥ b,

r

where βr , 1 ≤ r ≤ m, are the roots of the equation

rπ cos m+1

+ cosh(βr ) = 2.

504

7 Difference Equations

As a consequence, 2 G(a, b|a, b) = m+1 m



r=1

arπ sin m+1

2

 sinh(bβr ) sinh((n + 1 − b)βr ) · · . sinh(βr ) sinh((n + 1)βr ) (7.327)

For convenience’s sake, we write g(a, b) instead of G(a, b|a, b). We need to find the maximum of g(a, b) for 1 ≤ a ≤ m, 1 ≤ b ≤ n. Various cases have to be considered. We first consider the case when n is odd. We assert that for any fixed a,

n+1 . max g(a, b) = g a, 1≤b≤n 2 Note first that 1 sinh(bβr ) sinh ((n + 1 − b)βr ) = {cosh ((b + n + 1 − b)βr ) 2 − cosh ((b − n − 1 + b)βr )} 1 = {cosh ((n + 1)βr ) − cosh ((n + 1 − 2b)βr )} . 2 Since cosh x is even and increasing for x ≥ 0, the minimum of sinh(bβr ) sinh ((n + 1 − b)βr ) occurs when 2b = n + 1. In view of (7.327) and the fact that sinh(βr ) sinh ((n + 1)βr ) > 0, our assertion is proved. Similarly, when n is even, then for any fixed a,  n   n = g a, + 1 . max g(a, b) = g a, 1≤b≤n 2 2 By symmetry considerations, we see that when m is odd, then for any b, max g(a, b) = g

1≤a≤m

m+1 ,b , 2

and when m is even, then for any b, max g(a, b) = g

1≤a≤m

The following is now clear.

 m  ,b = g + 1, b . 2 2

m

7.8 Partial Difference Systems

505

Lemma 7.102 For (7.326), the maximum M of the associated Green function G(i, j |a, b) is given by

⎧ m n+1 ⎪ ⎪ g , ⎪ ⎪ 2 2 ⎪ ⎪ ⎪

⎪ ⎪ ⎪ m+1 n ⎪ ⎪ , ⎨g 2 2 M= m n ⎪ ⎪ ⎪ , g ⎪ ⎪ ⎪ 2 2 ⎪ ⎪

⎪ ⎪ ⎪ m+1 n+1 ⎪ ⎩g , 2 2

if

m is even, n is odd,

if

m is odd, n is even, (7.328)

if

m is even, n is even,

if

m is odd, n is odd,

where g(i, j ) = G(i, j |i, j ). Theorem 7.103 Let S be a rectangular net of the form (7.326). If (7.318) has a nontrivial solution, then (7.317) holds, where μ(S) = M −1 , where M is given by (7.328), and the inequality is sharp.

7.8.6 Final Remarks We have discussed three special types of nets. In general, nets can take on various forms. The question then arises whether explicit Green functions can be found for other types of nets, and if not, whether μ(S) can be calculated or estimated. In view of the fact that Green’s function for the rectangular net is implicitly given, it is unlikely that explicit Green’s function for other types of nets can be found. Even if these functions can be found, their values are not vital since (after some computer experimentation) an elementary algorithm for locating the maximal points of Green’s function for an arbitrary net is known. It is unfortunate that a proof for the validity of this algorithm is not yet known. However, given that the maximal points can be found, it is then straightforward to use standard computer packages to calculate the maxima of these functions (by means of (7.319). As for now, we can rely on Theorem 7.96 for estimation purposes. Indeed, if the nets S1 and S2 are related by S1 ⊆ S2 , then by Theorem 7.96, their corresponding Green functions G1 and G2 are related by max G1 ≤ max G2 so that μ(S1 ) ≥ μ(S2 ). In actual applications, we can take S1 to be any of the three types of nets discussed before, or any net whose corresponding μ(S1 ) is known, and get a lower bound μ(S1 ) for

S2

p.

506

7 Difference Equations

7.9 Two-Dimensional Nonlinear Systems of Partial Difference Equations In this section, the two-dimensional nonlinear system of partial difference equations  Δ1 Δ2 x(s, t) = α1 (s, t)x(s + 1, t + 1) + β1 (s, t)|u(s, t)|γ −2 u(s, t), Δ1 Δ2 u(s, t) = −β2 (s, t)|x(s + 1, t + 1)|β−2 x(s + 1, t + 1) − α1 (s, t)u(s, t). (7.329) is considered, where s, t ∈ Z, Δ1 denotes the forward difference operator for s, that is, Δ1 x(s, t) = x(s + 1, t) − x(s, t), and Δ2 denotes the forward difference operator for t, that is, Δ2 x(s, t) = x(s, t + 1) − x(s, t). We assume the existence of a nontrivial solution (x, u) of (7.329). Moreover, we assume that the following hypotheses are satisfied: (H1 ) (H2 ) (H3 )

γ , β > 1 are real constants, β1 , β2 : Z × Z → R are such that β1 (s, t) > 0 for all s, t ∈ Z, α1 : Z × Z → R.

In 2010, Chen et al. [80] established the following Lyapunov-type inequality for (7.329). Theorem 7.104 (Lyapunov-Type Inequality) Let the hypotheses (H1 )–(H3 ) hold. Assume n1 , m1 , n2 , m2 ∈ Z and n1 < m1 − 2, n2 < m2 − 2. If (7.329) has a real solution (x, u) such that x(n1 , t) = x(m1 , t) = x(s, n2 ) = x(s, m2 ) = 0,

(s, t) ∈ [n1 , m1 ] × [n2 , m2 ] (7.330)

and Δ1 x(s, t + 1)Δ2 u(s, t) + Δ2 x(s + 1, t)Δ1 u(s, t) = 0,

x(s, t) ≡ 0

(7.331)

on [n1 , m1 ] × [n2 , m2 ], then the inequality ⎞1/γ ⎛ ⎞1/α ⎛ m m 1 −2 m 2 −2 1 −2 m 2 −2 + β/α−1 ⎝ ⎝ |α1 (s, t)| + M β1 (s, t)⎠ β2 (s, t)⎠ ≥2 s=n1 t=n2 s=n1 t=n2 s=n1 t=n2

m 1 −2 m 2 −2

(7.332)

holds, where γ and α are Hölder conjugates of each other, i.e., 1/γ + 1/α = 1,

7.9 Two-Dimensional Nonlinear Systems of Partial Difference Equations

M=

507

|x(s, t)|,

(7.333)

β2+ (s, t) = max{β2 (s, t), 0}.

(7.334)

max

n1 +1 0, q Z = {q k : k ∈ Z} ∪ {0} for some

q>1

520

8 Dynamic Equations on Time Scales

(which gives rise so-called q-difference equations),   n

1 2 2 : n∈N , N = {k : k ∈ N}, k k=1

>

[2k, 2k + 1],

k∈Z

or the Cantor set.

8.3 Sturm–Liouville Equations We let T ⊂ R be any time scale and q : T → R be an rd-continuous function with q(t) > 0 for all t ∈ T. Consider (8.4) together with the quadratic functional  b F (x) = {(x Δ )2 − q(x σ )2 }Δt. a

Our first auxiliary result reads as follows. Lemma 8.13 If x solves (8.4) and if F (y) is defined, then F (y) − F (x) = F (y − x) + 2(y − x)(b)x Δ (b) − 2(y − x)(a)x Δ (a). Proof Under the above assumptions, we find F (y) − F (x) − F (y − x)  b {(y Δ )2 − q(y σ )2 − (x Δ )2 + q(x σ )2 − (y Δ − x Δ )2 + q(y σ − x σ )2 }Δt = a



b

=

 (y Δ )2 − q(y σ )2 − (x Δ )2 + q(x σ )2 − (y Δ )2 + 2y Δ x Δ − (x Δ )2

a

+ q(y σ )2 − 2qy σ x σ + q(x σ )2 (t)Δt 

b

=2 

a b

=2 

 y Δ x Δ − y σ x ΔΔ − x σ x ΔΔ − (x Δ )2 (t)Δt

a

=2 

 y Δ x Δ − qy σ x σ + q(x σ )2 − (x Δ )2 (t)Δt

b

 Δ Δ yx − xx Δ Δt

b

 Δ (y − x)x Δ Δt

a

=2 a

= 2 {y(b) − x(b)} x Δ (b) − 2 {y(a) − x(a)} x Δ (a),

8.3 Sturm–Liouville Equations

521

where we have used the product rule from Sect. 8.2.



Lemma 8.14 If F (y) is defined, then for any r, s ∈ T with a ≤ r < s ≤ b, we have 

s r

 Δ 2 {y(s) − y(r)}2 . y (t) Δt ≥ s−r

Proof Under the above assumptions, we define x(t) =

sy(r) − ry(s) y(s) − y(r) t+ . s−r s−r

Then, we have x(r) = y(r), x(s) = y(s), x Δ (t) =

y(s) − y(r) , s−r

and x ΔΔ (t) = 0. Hence, x solves (8.4) with q = 0. Therefore, we may apply Lemma 8.13 to F0 defined by  F0 (x) =

s

 Δ 2 x (t) Δt

r

to find F0 (y) = F0 (x) + F0 (y − x) + (y − x)(s)x Δ (s) − (y − x)(r)x Δ (r) = F0 (x) + F0 (y − x) ≥ F0 (x)   s y(s) − y(r) 2 = Δt s−r r =

{y(s) − y(r)}2 , s−r 

and this proves our claim.

Using the above Lemma 8.14, we now can prove Theorem 8.3 as stated in Sect. 8.1. Proof of Theorem 8.3 Suppose x is a solution of (8.4) with x(a) = x(b) = 0. Then, we apply Lemma 8.13 with y = 0 and find  F (x) = a

b

{(x Δ )2 − q(x σ )2 }Δt = 0.

522

8 Dynamic Equations on Time Scales

Since x is nontrivial, we have that M defined by M = max{[x(t)]2 : t ∈ [a, b] ∩ T}

(8.11)

is positive. Let c ∈ [a, b] be such that [x(c)]2 = M. Applying the above as well as Lemma 8.14 twice (once with r = a and s = c and a second time with r = c and s = b), we find  M

b

 q(t)Δt ≥

a

b

 2 q(t) x σ (t) Δt

b

 Δ 2 x (t) Δt

c

 Δ 2 x (t) Δt +

a

 = 

a

= a



b



2 x Δ (t) Δt

c

{x(b) − x(c)}2 {x(c) − x(a)}2 + ≥ c−a b−c   1 1 2 + = [x(c)] c−a b−c =M

b−a f (c)

≥M

b−a , f (d)

where the last inequality holds because of f (d) = max{f (t) : t ∈ [a, b] ∩ T}. Hence, dividing by M > 0 yields the desired inequality.



As an application of Theorem 8.3, we now prove a sufficient criterion for disconjugacy of (8.4). Definition 8.15 Equation (8.4) is called disconjugate on [a, b] if the solution x˜ of (8.4) with x(a) ˜ = 0 and x˜ Δ (a) = 1 satisfies x˜ x˜ σ > 0 on (a, ρ(b)]. Lemma 8.16 Equation (8.4) is disconjugate on [a, b] if and only if  F (x) =

b

{(x Δ )2 − q(x σ )2 }Δt > 0

a

for all nontrivial x with x(a) = x(b) = 0.

8.4 Linear Hamiltonian Systems

523



Proof This is a special case of [8, Theorem 5]. Theorem 8.17 (Sufficient Condition for Disconjugacy of (8.4)) If q satisfies 

b

q(t)Δt < a

b−a , f (d)

(8.12)

then (8.4) is disconjugate on [a, b]. Proof Suppose that (8.12) holds. For the sake of contradiction, we assume that (8.4) is not disconjugate. Then, by Lemma 8.16, there exists a nontrivial x with x(a) = x(b) = 0 such that F (x) ≤ 0. Using this x, we now define M by (8.11) to find  M

b

 q(t)Δt ≥

a

b

 2 q(t) x σ (t) Δt

b

 Δ 2 x (t) Δt

a

 ≥

a

≥

M(b − a) , f (d)

where the last inequality follows precisely as in the proof of Theorem 8.1. Hence, after dividing by M > 0, we arrive at 

b

q(t)Δt ≥

a

b−a , f (d)

which contradicts (8.12) and hence completes the proof.



Remark 8.18 Note that in both (8.6) and (8.12), we could replace (b − a)/f (d) by 4/(b − a), and Theorems 8.3 and 8.17 would remain true. This is because for a ≤ c ≤ b, we have 1 1 (a + b − 2c)2 4 4 + = + ≥ . c−a b−c (b − a)(c − a)(b − c) b − a b−a

8.4 Linear Hamiltonian Systems In this section, we consider (8.10), where A, B, C are rd-continuous n × n-matrixvalued functions on T such that I − μ(t)A(t) is invertible and B(t) and C(t) are positive semidefinite for all t ∈ T. For the continuous case of this theory, we refer to [189] (in particular for Lyapunov inequalities [93]) while [22] is a good reference for the discrete case. A corresponding quadratic functional is given by

524

8 Dynamic Equations on Time Scales



b

F (x, u) =

{u∗ Bu − (x σ )∗ Cx σ }(t)Δt.

(8.13)

a

A pair (x, u) is called admissible if it satisfies the equation of motion x Δ = A(t)x σ + B(t)u. As in Sect. 8.3, we start with the following auxiliary result. Lemma 8.19 If (x, u) solves (8.10) and if (y, v) is admissible, then F (y, v) − F (x, u) = F (y − x, v − u) + 2 Re[(y − x)∗ (b)u(b) − (y − x)∗ (a)u(a)]. Proof Under the above assumptions, we calculate F (y, v) − F (x, u) − F (y − x, v − u)  b {v ∗ Bv − (y σ )∗ Cy σ − u∗ Bu + (x σ )∗ Cx σ − [(v − u)∗ B(v − u) = a

−(y σ − x σ )∗ C(y σ − x σ )]}(t)Δt  b {−2u∗ Bu + v ∗ Bu + u∗ Bv + 2(x σ )∗ Cx σ − (y σ )∗ Cx σ − (x σ )∗ Cy σ }(t)Δt = a

 =

b

{−2u∗ Bu + 2 Re[u∗ Bv] + 2(x σ )∗ Cx σ − 2 Re[(y σ )∗ Cx σ ]}(t)Δt

a



b

= 2 Re 

a

σ ∗



σ ∗

{u (Bv − Bu) + [(x ) − (y ) ]Cx }(t)Δt σ

{u (y − Ay − x + Ax ) + [(x ) − (y ) ][−u − A u]}(t)Δt

b

= 2 Re

∗

∗

Δ

σ

Δ

σ ∗

σ

σ ∗

Δ

∗

a



b

= 2 Re 

b

∗

Δ

∗

Δ

σ ∗ Δ

∗

b

σ

σ ∗ Δ



{u (y −x )+(y −x ) u }(t)Δt Δ

σ

Δ ∗

= 2 Re



{u (y −x )+(u ) (y −x )}(t)Δt Δ

a b

σ ∗

∗

{u (y −x )+(y −x ) u +2i Im[u Ax +(y ) A u]}(t)Δt Δ

a

= 2 Re 

Δ

a

= 2 Re 

∗

∗

σ

σ



{[u (y − x)] }(t)Δt Δ

a

= 2 Re{u∗ (b)[y(b) − x(b)] − u∗ (a)[y(a) − x(a)]} = 2 Re{[y − x]∗ (b)u(b) − [y − x]∗ (a)u(a)},

σ



8.4 Linear Hamiltonian Systems

525



which completes the proof.

For simplicity, denote by W (·, r) the unique (see [37, Section 6]) solution of the initial value problem 

W Δ = −A∗ (t)W, W (r) = I,

(8.14)

where r ∈ [a, b] is given. We also write 

s

F (s, r) =

W ∗ (t, r)B(t)W (t, r)Δt.

(8.15)

r

Observe that W (t, r) ≡ I provided A(t) ≡ 0. Lemma 8.20 Let W and F be as in (8.14) and (8.15). If (y, v) is admissible and if r, s ∈ T with a ≤ r < s ≤ b such that F (s, r) is invertible, then 

s

  ∗  (v ∗ Bv)(t)Δt ≥ W ∗ (s, r)y(s) − y(r) [F (s, r)]−1 W ∗ (s, r)y(s) − y(r) .

r

Proof Under the above assumptions, we define  −1    x(t) = W ∗ (t, r) y(r) + F (t, r)[F (s, r)]−1 W ∗ (s, r)y(s) − y(r) and   u(t) = W (t, r)[F (s, r)]−1 W ∗ (s, r)y(s) − y(r) . Then, we have x(r) = y(r), x(s) = y(s), uΔ (t) = −A∗ (t)u(t), and  −1  −1 ∗ x Δ (t) = W ∗ (σ (t), r) (W Δ )∗ (t, r)x(t) + W ∗ (σ (t), r) W (t, r)B(t)u(t) −1 ∗  ∗ −1 ∗  ∗ W (t, r)A(t)x(t) + W (σ (t), r) W (t, r)B(t)u(t) = W (σ (t), r) ∗  = W (t, r)[W (σ (t), r)]−1 (A(t)x(t) + B(t)u(t)) . Moreover,   W (t, r)[W (σ (t), r)]−1 = W (σ (t), r) − μ(t)W Δ (t, r) [W (σ (t), r)]−1 = I + μ(t)A∗ (t)W (t, r)[W (σ (t), r)]−1 , which yields   I − μ(t)A∗ (t) W (t, r)[W (σ (t), r)]−1 = I.

526

8 Dynamic Equations on Time Scales

Hence, (I − μ(t)A(t)) x Δ (t) = A(t)x(t) + B(t)u(t) and x Δ (t) = A(t)x(t) + μ(t)A(t)x Δ (t) + B(t)u(t) = A(t)x σ (t) + B(t)u(t). Thus, (x, u) solves (8.10) with C = 0, and we may apply Lemma 8.19 to F0 defined by 

s

F0 (x, u) =

(u∗ Bu)(t)Δt

r

to obtain F0 (y, v) =F0 (x, u) − F0 (y − x, v − u) + 2 Re{u∗ (s)[y(s) − x(s)] − u∗ (r)[y(r) − x(r)]} =F0 (x, u) + F0 (y − x, v − u) ≥F0 (x, u)  s = (u∗ Bu)(t)Δt r

  ∗  = W ∗ (s, r)y(s) − y(r) [F (s, r)]−1 W ∗ (s, r)y(s) − y(r) , 

which shows our claim.

Remark 8.21 The assumption in Lemma 8.20 that F (s, r) is invertible if r < s can be dropped in case B is positive definite rather than positive semidefinite. As before in Sect. 8.3, we now may use Lemma 8.20 to derive a Lyapunov inequality for Hamiltonian systems. Theorem 8.22 (Lyapunov’s Inequality) Assume (8.10) has a solution (x, u) such that x is nontrivial and satisfies x(a) = x(b) = 0. With W and F as in (8.14) and (8.15), suppose that F (b, c) and F (c, a) are invertible, where x(c) =

max x(t) .

t∈[a,b]∩T

Let λ be the largest eigenvalue of  F = a

b

W ∗ (t, c)B(t)W (t, c)Δt,

8.4 Linear Hamiltonian Systems

527

and let ν(t) be the largest eigenvalue of C(t). Then, the Lyapunov inequality 

b

ν(t)Δt ≥

a

4 λ

holds. Proof Suppose (x, u) is a solution of (8.10) such that x(a) = x(b) = 0. Then, by Lemma 8.19 applied with y = v = 0, we get 

b

F (x, u) =

{u∗ Bu − (x σ )∗ Cx σ }(t)Δt = 0.

a

Now, we use Lemma 8.20 twice (once with r = a and s = c and a second time with r = c and s = b) to obtain 

b

a



σ ∗

b

[(x ) Cx ](t)Δt =



c

=

σ



∗

b

(u Bu)(t)Δt +

a

(u∗ Bu)(t)Δt

a

(u∗ Bu)(t)Δt

c

≥ x ∗ (c)W (c, a)[F (c, a)]−1 W ∗ (c, a)x(c) + x ∗ (c)[F (b, c)]−1 x(c)   = x ∗ (c) [F (b, c)]−1 − [F (a, c)]−1 x(c) ≥ 4x ∗ (c)F −1 x(c). Here, we have used the relation W (t, r)W (r, s) = W (t, s) (see [37, Theorem 9 (i)]) as well as the inequality M −1 + N −1 > 4(M + N)−1 (see [98, Lemma 11, page 63] or [233]). Now, by applying the Rayleigh–Ritz theorem [173, page 167], we conclude 

b



b

ν(t)Δt ≥

a

ν(t) a

=

1 x(c)2

x σ (t)2

Δt x(c)2  b ν(t)(x σ (t))∗ x σ (t)Δt a

528

8 Dynamic Equations on Time Scales

≥ ≥

x(c)2 1 x(c)2

≥ min x=0

=



1

b

(x σ (t))∗ C(t)x σ (t)Δt

a

4x ∗ (c)F −1 x(c)

x ∗ F −1 x x∗x

4 , λ 

and this finishes the proof. Remark 8.23 If A ≡ 0, then W ≡ I and  F =

b

B(t)Δt. a

If, in addition B ≡ 1, then F = b − a. Note that the Lyapunov inequality 

b

ν(t)Δt ≥

a

4 λ

reduces to 

b

q(t)Δt ≥

a

4 b−a

for the scalar case as discussed in Sect. 8.3. It is possible to provide a slightly better bound than the one given in Theorem 8.22, similarly as in Theorem 8.1, but we shall not do so here. Without introducing the notion of disconjugacy for (8.10), we now state the following corollary of Theorem 8.22, whose proof is similar to the one of Theorem 8.17. For the definition of disconjugacy and the result analogous to Lemma 8.16, we refer to the work of Hilscher [169, 170] (see also [8]). Theorem 8.24 (Sufficient Condition for Disconjugacy of (8.10)) Using the notation introduced in (8.14) and (8.15), if 

b

ν(t)Δt < a

4 , λ

then (8.10) is disconjugate on [a, b]. We conclude this section with a result concerning so-called right-focal boundary conditions, i.e., x(a) = u(b) = 0.

8.4 Linear Hamiltonian Systems

529

Theorem 8.25 (Lyapunov-Type Inequality) Assume (8.10) has a solution (x, u) with x nontrivial and x(a) = u(b) = 0. With the notation introduced in (8.14) and (8.15), the Lyapunov inequality 

b

ν(t)Δt ≥

a

1 λ

holds. Proof Suppose (x, u) is a solution of (8.10) such that x(a) = u(b) = 0 with a < b. Choose the point c in (a, b], where x(t) is maximal. Apply Lemma 8.19 with y = v = 0 to see that F (x, u) = 0. Therefore, 

b



σ ∗

b

[(x ) Cx ](t)Δt = σ

a



(u∗ Bu)(t)Δt

a c

≥

(u∗ Bu)(t)Δt.

a

Using Lemma 8.20 with r = a and s = c, we get 

c

  ∗  (u∗ Bu)(t)Δt ≥ W ∗ (c, a)x(c) − x(a) [F (c, a)]−1 W ∗ (c, a)x(c) − x(a)

a

= x ∗ (c)W (c, a)[F (c, a)]−1 W ∗ (c, a)x(c) = −x ∗ (c)[F (a, c)]−1 x(c)  c

−1 ∗ ∗ = x (c) W (t, c)B(t)W (t, c)Δt x(c) a

≥ x ∗ (c)



b

W ∗ (t, c)B(t)W (t, c)Δt

−1 x(c)

a

= x ∗ (c)F −1 x(c). Hence, 

b

[(x σ )∗ Cx σ ](t)Δt ≥ x ∗ (c)F −1 x(c),

a

and the same arguments as in the proof of Theorem 8.22 lead us to our final conclusion. 

530

8 Dynamic Equations on Time Scales

8.5 Higher-Order Dynamic Equations In this section, we establish a Lyapunov-type inequality for the higher-order dynamic equation SnΔ (t, x(t)) + q(t)[x(t)]α = 0

(8.16)

on an arbitrary time scale T, where n ∈ N, α ≥ 1 is the quotient of two odd positive integers and ⎧ x ⎪ ⎪ ⎨ Δ (t, x) Sk (t, x) = ak (t)Sk−1 ⎪ ⎪  Δ α ⎩ an (t) Sn−1 (t, x)

if k = 0, if 1 ≤ k ≤ n − 1, if k = n

with ak ∈ Crd (T, (0, ∞)), k = 1, 2, . . . , n, and q ∈ Crd (T, R). Lemma 8.26 Assume that u, v ∈ [a, b]T with u < v, q ∈ Crd (T, (0, ∞)) and q(x Δ )2 is integrable on [u, v]. Then, 

v

 2 q(t) x Δ (t) Δt ≥ [x(v) − x(u)]2



u

v

u

Δt q(t)

−1 .

Proof Putting 

v

W = [x(v) − x(u)] u

Δt q(t)

−1 ,

we have  u

v

2   W q(t)x Δ (t) − √ Δt ≥ 0. q(t)

It follows that 

v





2

q(t) x (t) Δt − 2W Δ

u

v

 x (t)Δt + Δ

u

v

u

W2 Δt ≥ 0, q(t)

i.e., 

v

 2 q(t) x Δ (t) Δt ≥ 2W

u



v



u



v

=2W u

v

x Δ (t)Δt − u

W2 Δt q(t) 

x (t)Δt−W [x(v)−x(u)] Δ

u

v

Δt q(t)

−1  u

v

Δt q(t)

8.5 Higher-Order Dynamic Equations

531

= W [x(v) − x(u)]  2 = [x(v) − x(u)] u

v

Δt q(t)

−1 . 

The proof is completed. Now, we state and prove the main results.

Theorem 8.27 (Lyapunov-Type Inequality) Let n = 1. If x is a solution of (8.16) with x(a) = x(b) = 0 and x ≡ 0 on [a, b]T , then 

b

|q(t)|Δt ≥

a

1 f (d)



b a

Δt a1 (t)

(α+1)/2 

(1−α)/2

b

a1 (t)Δt

,

a

where 

c

f (d) = max

Δt a1 (t)

a

 c

b

 Δt : c ∈ (a, b)T . a1 (t)

(8.17)

Proof For n = 1, (8.16) reduces to the second-order boundary value problem (a1 (x Δ )α )Δ (t) + q(t)[x(t)]α = 0 and x(a) = x(b) = 0. Let M be the maximum of |x(t)| on [a, b]T . We see that there exists c ∈ (a, b)T such that |x(t)| = M > 0. Case 1. If α > 1, then by Theorem 8.10 (iv), we obtain  M α+1

b



b

|q(t)|Δt =

a



b

≥ 

MM α |q(t)|Δt

a

|x(σ (t))||x(t)|α |q(t)|Δt

a b

≥ a

x(σ (t))[x(t)]α q(t)Δt



b

=−

 Δ x(σ (t)) a1 (x Δ )α (t)Δt

a

   Δ α  b  + = −x(t) a1 (t) x (t)  

a



b

= a

 α+1 a1 (t) x Δ (t) Δt.

a

b

 α+1 a1 (t) x Δ (t) Δt

532

8 Dynamic Equations on Time Scales

By Theorem 8.11 and Lemma 8.26, we have 

2/(α+1)   2 (α+1)/2 Δt [a1 (t)]2/(α+1) x Δ (t)

b

a



b

×

(α−1)/(α+1)

(α+1)/(α−1)

[a1 (t)]

≥

Δt

b

 2 [a1 (t)]2/(α+1) x Δ (t) [a1 (t)](α−1)/(α+1) Δt

b

 2 a1 (t) x Δ (t) Δt

c

 2 a1 (t) x Δ (t) Δt +

a

= 

(α−1)/(α+1)

a

 



a

= a



c

≥ M2 a

 =M

Δt a1 (t)

b

2 a

b

 2 a1 (t) x Δ (t) Δt

c



−1

Δt a1 (t)



b

+ M2 c

 

c

a

Δt a1 (t)



−1

Δt a1 (t) b

Δt a1 (t)

c

−1 .

Thus, we have  b  α+1 a1 (t) x Δ (t) Δt a

⎡ ≥ ⎣M



b

2 a

 =M

b

α+1 a

Δt a1 (t)

 

Δt a1 (t)

c

a

Δt a1 (t)

  α+1 2

c

a



b

c

Δt a1 (t)

Δt a1 (t) 

b

−1 

a1 (t)Δt

⎤ α+1 2

⎦

a

Δt a1 (t)

c

 1−α 1+α

b

 − α+1 2

 1−α 2

b

a1 (t)Δt a

From above, it is easy to prove that 

b

|q(t)|Δt ≥ M −(α+1)



a

b

 α+1 a1 (t) x Δ (t) Δt

a



b a

Δt a1 (t)



b

≥ ×

(α+1)/2 

c

a

Δt a1 (t)



b c

Δt a1 (t)

−(α+1)/2

(1−α)/2 a1 (t)Δt

a

 ≥

1 f (d)



b a

Δt a1 (t)

(α+1)/2 

(1−α)/2

b

a1 (t)Δt a

.

.

8.5 Higher-Order Dynamic Equations

533

Case 2. If α = 1, then from Case 1, we have 

b

M2

 |q(t)|Δt ≥

a

b

 2 a1 (t) x Δ (t) Δt

c

 2 a1 (t) x Δ (t) Δt +

a

 =

a

≥M



c

2 a

M2

≥

f (d)



b

a

Δt a1 (t)



 2 a1 (t) x Δ (t) Δt

b c



−1

+M

b

2 c

Δt a1 (t)

−1

Δt , a1 (t)

which implies 

b

a

1 |q(t)|Δt ≥ f (d)

 a

b

Δt . a1 (t) 

This completes the proof.

Theorem 8.28 (Lyapunov-Type Inequality) Let n > 1 and x be a solution of (8.16) with Sn (a, x(a)) = · · · = S2 (a, x(a)) = x(a) = x(b) = 0

(8.18)

and x ≡ 0 on [a, b]T . Then, we have  a

b

Δt a2 (t)



t a

Δτ3 a3 (τ3 )

 a

τ3



Δτ4 ··· a4 (τ4 )

τn−1 a



Δτn an (τn )



τn

1/α |q(τ )|Δτ

a

1 ≥ f (d)



b a

Δt , a1 (t)

where f (d) is defined in (8.17). Proof By integrating (8.16) from a to t and taking into account (8.18), one has  a

t

 SnΔ (τ, x(τ ))Δτ +

t

q(τ )[x(τ )]α Δτ = 0,

a

i.e.,  Δ (t, x(t)) = − Sn−1

1 an (t)

 a

t

1/α q(τ )[x(τ )]α Δτ

.

(8.19)

534

8 Dynamic Equations on Time Scales

Integrating (8.19) from a to t, we similarly have Sn−1 (t, x(t)) = −

 t a

1 an (τ )



1/α

τ

α

q(s)[x(s)] Δs

Δτ.

a

By induction, it is obvious that (a1 x Δ )Δ (t) 1/α  τn−1   t  τ3  τn Δτn 1 Δτ3 Δτ4 ··· + q(τ )[x(τ )]α Δτ = 0. a2 (t) a a3 (τ3 ) a a4 (τ4 ) an (τn ) a a Then, 

1/α  τn Δτn |q(τ )|Δτ an (τn ) a a a a a 1/α  b  τn−1   t  τ3  τn Δτn Δτ3 Δτ4 Δt α = M |q(τ )|M Δτ ··· a2 (t) a a3 (τ3 ) a a4 (τ4 ) an (τn ) a a a  b  t  τ3 Δτ3 Δτ4 Δt ··· ≥ |x(σ (t))| a (t) a (τ ) a 2 3 3 4 (τ4 ) a a a 1/α  τn−1   τn Δτn α ··· |q(τ )||x(τ )| Δτ an (τn ) a a  b  t  τ3 Δτ3 Δτ4 Δt ··· ≥ x(σ (t)) a2 (t) a a3 (τ3 ) a a4 (τ4 ) a 1/α  τn−1   τn Δτn ··· q(τ )[x(τ )]α Δτ an (τn ) a a  b =− x(σ (t))(a1 x Δ )Δ (t)Δt 

M2

b

Δt a2 (t)



t

Δτ3 a3 (τ3 )



τ3

Δτ4 ··· a4 (τ4 )



τn−1

a

b   = −x(t)a1 (t)x Δ (t) + a



c

= a

≥M

b a

 2 a1 (t) x Δ (t) Δt + 

c

2 a



b

= M2 a



b

 2 a1 (t) x Δ (t) Δt

c



−1 Δt +M c a1 (t)

 c

−1  b

−1 Δt Δt Δt , a1 (t) a a1 (t) c a1 (t)

Δt a1 (t)

−1

 2 a1 (t) x Δ (t) Δt

2

b

8.6 Stability Theory for Hill’s Equation

535

where M and c are defined as the proof of Theorem 8.27. From here, we obtain  a

b

Δt a2 (t)



t a

Δτ3 a3 (τ3 )



τ3

a

Δτ4 ··· a4 (τ4 )



τn−1 a



Δτn an (τn )



τn

1/α |q(τ )|Δτ

a

≥

1 f (d)



b a

Δt . a1 (t) 

This completes the proof.

8.6 Stability Theory for Hill’s Equation In this section, we obtain sufficient conditions for instability and stability to hold for second-order linear dynamic equations on periodic time scales with periodic coefficients. Definition 8.29 If there exists a positive number Λ ∈ R such that t + nΛ ∈ T for all t ∈ T and n ∈ Z, then we call T a periodic time scale with period Λ. Suppose T is a Λ-periodic time scale. For the sake of simplicity, we assume 0 ∈ T. Consider the second-order linear dynamic equation (rx Δ )Δ (t) + q(t)x(σ (t)) = 0,

t ∈ T,

(8.20)

where σ is the forward jump operator, and the coefficients r and q are real-valued Λ-periodic functions defined on T, i.e., r(t + Λ) = r(t),

q(t + Λ) = q(t),

t ∈ T.

(8.21)

q ∈ C1rd ([0, Λ]),

(8.22)

Besides, we assume that r(t) > 0,

r ∈ C1rd ([0, Λ]),

where [0, Λ] = {t ∈ T : 0 ≤ t ≤ Λ}. Definition 8.30 Equation (8.20) is said to be unstable if all nontrivial solutions are unbounded on T, conditionally stable if there exists a nontrivial solution which is bounded on T, and stable if all solutions are bounded on T. The main results of this section are the following two theorems [35]. Theorem 8.31 If q(t) ≤ 0 and q ≡ 0, then (8.20) is unstable.

536

8 Dynamic Equations on Time Scales

Theorem 8.32 If 

Λ

q(t)Δt ≥ 0,

q ≡ 0

(8.23)

0

and 



Λ

r0 + 0

Δt r(t)



Λ 0

q+ (t)Δt ≤ 4,

(8.24)

where r0 =

max

t∈[0,ρ(Λ)]

σ (t) − t , r(t)

(8.25)

then (8.20) is stable. We dwell on the three special cases as follows. Example 8.33 If T = R, then we can take as Λ any ω ∈ R, ω > 0. Equation (8.20) takes the form (rx  ) (t) + q(t)x(t) = 0,

t ∈ R,

(8.26)

and (8.21) becomes r(t + ω) = r(t),

q(t + ω) = q(t),

t ∈ R.

Conditions (8.23) and (8.24) of Theorem 8.32 turn into 

ω

q(t)dt ≥ 0,

q ≡ 0

0

and  0

ω

dt r(t)



ω

0

q+ (t)dt ≤ 4.

We note here that the last conditions lead to a well-known result. In [202], Lyapunov has proved that, if a real, continuous and periodic function q of period ω > 0 satisfies the conditions q(t) ≥ 0,

q ≡ 0

and 

ω

ω 0

q(t)dt ≤ 4,

8.6 Stability Theory for Hill’s Equation

537

then the equation x  (t) + q(t)x(t) = 0,

t ∈R

(8.27)

is stable. In [59], Borg extended Lyapunov’s result to functions q of variable sign, showing that if 

ω

q(t)dt ≥ 0,

q ≡ 0

0

and  ω

ω

|q(t)|dt ≤ 4,

(8.28)

0

then (8.27) is stable. In [190], Kre˘ın improved Borg’s result replacing in (8.28) |q(t)| by q+ (t). Notice that (8.26) can be transformed into an equation of the type (8.27) by a change of variable. For direct investigation of (8.26), we refer to [144]. Example 8.34 If T = Z, then we can take any Λ ∈ N. Equation (8.20) takes the form Δ(r(n)Δx(n)) + q(n)x(n + 1) = 0,

n ∈ Z,

where Δ is the forward difference operator defined by Δx(n) = x(n + 1) − x(n). Condition (8.21) can be written as r(n + N) = r(n),

q(n + N) = q(n),

n ∈ Z.

Conditions (8.23) and (8.24) of Theorem 8.32 become N −1

q(n) ≥ 0,

q ≡ 0

n=0

and 0

N −1 1 1 + r r(n) n=0

1 N −1

q+ (n) ≤ 4,

(8.29)

n=0

where r = min{r(0), r(1), . . . , r(N − 1)}, q + (n) = max{q(n), 0}. This result was established in [34]. Example 8.35 Let ω be a positive real number and N ∈ N. Setting Λ = ω + N, consider the time scale T defined by

538

8 Dynamic Equations on Time Scales

T=

>

[{t ∈ R : kΛ ≤ t ≤ kΛ + ω} ∪ {kΛ + ω + n : n = 0, 1, . . . , N − 1}] .

k∈Z

Evidently, the set T defined in such a way is a Λ-periodic time scale. Equation (8.20) takes the form > {t ∈ R : kΛ ≤ t ≤ kΛ + ω} (rx  ) (t) + q(t)x(t) = 0, t ∈ k∈Z

Δ(r(t)Δx(t)) + q(t)x(t + 1) = 0,

t∈

>

{kΛ + ω + n : n = 0, 1, . . . , N − 2},

k∈Z

which incorporates both differential and difference equations. In this case, (8.23) and (8.24) of Theorem 8.32 can be written as 

ω

q(t)dt +

0

N −1

q(ω + n) ≥ 0,

q ≡ 0

n=0

and 0 0

ω

N −1 1 dt 1 + + r(t) r r(ω + n)

1 0

n=0

0

ω

q+ (t)dt +

N −1

1 q+ (ω + n) ≤ 4,

n=0

where r = min{r(ω), r(ω + 1), . . . , r(ω + N − 1)}. Remark 8.36 As a conclusive remark, we note that the assumption 0 ∈ T is not necessary. Instead we can take any fixed point t0 ∈ T in place of zero, and in that case, the integrals in (8.23) and (8.24) of Theorem 8.32 are considered from t0 to t0 + Λ.

8.6.1 Auxiliary Propositions Consider (8.20), where T is a Λ-periodic time scale containing zero, with the coefficients r and q being real valued and satisfying (8.21) and (8.22). Floquet theory applies to (8.20). For details of Floquet theory, we may refer to, for example, [119, 206] for differential equations, and [141, pages 113–115], [34], [121, pages 144–149] for difference equations. For Floquet theory on time scales, we refer to [23]. Let us denote by x [Δ] = rx Δ the quasi-Δ-derivative of x. For arbitrary complex numbers c0 and c1 , (8.20) has a unique solution x satisfying the initial conditions x(0) = c0 ,

x [Δ] (0) = c1 .

8.6 Stability Theory for Hill’s Equation

539

Denote by θ and ϕ the solutions of (8.20) under the initial conditions θ (0) = 1,

θ [Δ] (0) = 0

and

ϕ(0) = 0,

ϕ [Δ] (0) = 1.

(8.30)

There exist a nonzero complex number β and a nontrivial solution ψ of (8.20) such that ψ(t + Λ) = βψ(t),

t ∈ T.

The number β is a root of the quadratic equation β 2 − Dβ + 1 = 0,

(8.31)

D = θ (Λ) + ϕ [Δ] (Λ).

(8.32)

where

The roots of (8.31) are defined by β1,2 =

 1 (D ∓ D 2 − 4). 2

Since the coefficients of (8.20) and the initial conditions in (8.30) are real, the solutions θ , ϕ and hence the number D defined by (8.32) is real. Proposition 8.37 Equation (8.20) is unstable if |D| > 2, and stable if |D| < 2. Proof If the discriminant D 2 − 4 is nonzero, then (8.31) has two distinct roots β1 and β2 , and hence there exist two nontrivial solutions ψ1 and ψ2 of (8.20) such that ψ1 (t + Λ) = β1 ψ1 (t),

ψ2 (t + Λ) = β2 ψ2 (t),

t ∈ T.

(8.33)

It is easy to see that ψ1 and ψ2 are linearly independent. From (8.33), it follows that, for all k ∈ Z, ψ1 (t + kΛ) = β1k ψ1 (t),

ψ2 (t + kΛ) = β2k ψ2 (t),

t ∈ T.

(8.34)

If |D| > 2, then the numbers β1 and β2 are distinct and real. Therefore, from the equality β1 β2 = 1, it follows that |β1 | = 1 and |β2 | = 1, since if this were false, we would get β1 = β2 = ±1. Obviously, if |β1 | > 1, then |β2 | < 1, and if |β1 | < 1, then |β2 | > 1. Consequently, from (8.34) as k → ±∞, it follows that every nontrivial linear combination of ψ1 (t) and ψ2 (t) is unbounded on T, that is, (8.20) is unstable. If |D| < 2, then the numbers β1 and β2 are distinct, nonreal and such that |β1 | = |β2 | = 1. Therefore, from (8.33), we have |ψ1 (t + Λ)| = |ψ1 (t)| ,

|ψ2 (t + Λ)| = |ψ2 (t)| ,

t ∈ T.

540

8 Dynamic Equations on Time Scales

Consequently, ψ1 (t) and ψ2 (t) and hence every solution of (8.20) which is a linear combination of ψ1 (t) and ψ2 (t), are bounded on T, that is, (8.20) is stable. This completes the proof.  Remark 8.38 If |D| = 2, then (8.20) is stable in the case θ [Δ] (Λ) = ϕ(Λ) = 0, but conditionally stable and not stable otherwise. The following mean value result (for the case Z, see [5, page 24] and [34]) plays a significant rôle in the proof of Theorem 8.32. Proposition 8.39 Let a < b be any two points in the time scale T, and let f and g be two real functions continuous on the segment [a, b] = {t ∈ T : a ≤ t ≤ b} and Δ-differentiable on (a, b) = {t ∈ T : a < t < b}. Suppose the function g is increasing on [a, b]. Then, there exist ξ, τ ∈ a, b) such that f Δ (τ ) f (b) − f (a) f Δ (ξ ) ≤ ≤ , g Δ (τ ) g(b) − g(a) g Δ (ξ )

(8.35)

where a, b) = (a, b) if σ (a) = a, and a, b) = [a, b) if σ (a) > a. Proof We prove the right-hand side of (8.35), the proof of the left-hand side being similar. Assume the contrary, i.e., f (b) − f (a) f Δ (t) > Δ g(b) − g(a) g (t)

t ∈ a, b).

for all

Since g(t) is increasing on [a, b], g Δ (t) > 0 on a, b). So, we have f (b) − f (a) Δ g (t) > f Δ (t) for all g(b) − g(a)

t ∈ a, b).

Integrating (in the sense of Δ-integral) both sides from a to b, we arrive at the contradiction f (b) − f (a) > f (b) − f (a), 

which completes the proof. We will also make use of the following formulas, which can easily be verified.

Proposition 8.40 Let a, b ∈ T, a < b and let f be continuous on [a, ρ(b)] ⊂ T. Then, 



b

ρ(b)

f (t)Δt =

a

 a

f (t)Δt + [b − ρ(b)]f (ρ(b)),

a b

f (t)Δt = [σ (a) − a]f (a) +



b

f (t)Δt. σ (a)

8.6 Stability Theory for Hill’s Equation

541

8.6.2 Proofs Proof of Theorem 8.31 Let θ and ϕ be solutions of (8.20) satisfying (8.30). Our aim is to show that, under the hypotheses of the theorem, the inequalities ϕ [Δ] (Λ) > 1

θ (Λ) ≥ 1,

hold. Then, D = θ (Λ)+ϕ [Δ] (Λ) > 2 is obtained and therefore, by Proposition 8.37, (8.20) is unstable. First, we show that θ (t) ≥ 1, θ [Δ] (t) ≥ 0, ϕ(t) ≥ 0, ϕ [Δ] (t) ≥ 1

for all

t ∈ T, t ≥ 0.

(8.36)

For this purpose, we apply the induction principle developed for time scales (see, for example, [37, pages 9–20] and [194]), to the statement, A(t) : θ (t) ≥ 1 and θ [Δ] (t) ≥ 0 for all t ∈ T ∩ [0, ∞).

1. The statement A(0) is true, since θ (0) = 0 and θ [Δ] (0) = 0. 2. Let t be right-scattered and suppose A(t) is true, i.e., θ (t) ≥ 1 and θ [Δ] (t) ≥ 0. We need to show that θ (σ (t)) ≥ 1 and θ [Δ] (σ (t)) ≥ 0. But, in view of the induction assumptions, the required results are immediate from the relations θ (σ (t)) = θ (t) + μ(t)θ Δ (t), θ [Δ] (σ (t)) = θ [Δ] (t) − μ(t)q(t)θ (σ (t)), where the first one follows from the definition of the Δ-derivative and the second one from (8.20) for θ (t). Here, μ(t) = σ (t) − t. 3. Let t0 be right-dense, suppose A(t0 ) is true, and let t ∈ [t0 , t1 ] = {t ∈ T : t0 ≤ t ≤ t1 }, where t1 ∈ T is such that t1 > t0 and is sufficiently close to t0 . We need to prove that A(t) is true for all t ∈ [t0 , t1 ]. From (8.20) with x(t) = θ (t), the equations θ [Δ] (t) = θ [Δ] (t0 ) −



t

q(s)θ (σ (s))Δs

(8.37)

t0

and θ (t) = θ (t0 )+θ [Δ] (t0 )



t t0

Δτ − r(τ )



t t0

1 r(τ )



τ

t0

 q(s)θ (σ (s))Δs Δτ

(8.38)

542

8 Dynamic Equations on Time Scales

follow. To investigate the term θ (t) appearing in (8.38), we consider the equation x(t) = θ (t0 ) + θ [Δ] (t0 )



t t0

Δτ − r(τ )



t

t0

1 r(τ )



τ

 q(s)θ (σ (s))Δs Δτ,

t0

(8.39) where x is a desired solution. Our aim is to show that for t1 sufficiently close to t0 , (8.39) has a unique, continuous (in the topology of T) solution x satisfying the inequality x(t) ≥ θ (t0 ) + θ [Δ] (t0 )



t t0

Δτ , r(τ )

t0 ≤ t ≤ t1 .

(8.40)

We solve (8.39) by the method of successive approximations, setting x0 (t) = θ (t0 ) + θ [Δ] (t0 )



t t0

Δτ r(τ )

(8.41)

and 

t

xj (t) = −

t0

1 r(τ )





τ

q(s)xj −1 (σ (s))Δs Δτ,

t0

j ∈ N.

(8.42)

7 If the series ∞ j =0 xj (t) converges uniformly with respect to t ∈ [t0 , t1 ], then its sum is, obviously, a continuous solution of (8.39). To prove the uniform convergence of this series, we let c0 = θ (t0 ) + θ [Δ] (t0 )



t1

Δτ , r(τ )

t0

 c1 =

t1 t0

1 r(τ )



τ

 |q(s)|Δs Δτ.

t0

Then, the estimate j

0 ≤ xj (t) ≤ c0 c1

(t0 ≤ t ≤ t1 ),

j ∈ N0

(8.43)

can easily be obtained. Indeed, (8.43) evidently holds for j = 0. Suppose it also holds for j = n. Then, from (8.41) and (8.42), applying Proposition 8.40, we get  τ  1 0 ≤xn+1 (t) ≤ |q(s)|xn (σ (s))Δs Δτ t0 r(τ ) t0  τ   ρ(t1 ) 1 = |q(s)|xn (σ (s))Δs Δτ r(τ ) t0 t0  ρ(t1 ) 1 + [t1 − ρ(t1 )] |q(s)|xn (σ (s))Δs r(ρ(t1 )) t0 

t

8.6 Stability Theory for Hill’s Equation

 ≤c0 c1n

ρ(t1 ) t0

543

1 r(τ )



τ

 |q(s)|Δs Δτ

t0

  ρ(t1 ) 1 + [t1 − ρ(t1 )] |q(s)|Δs r(ρ(t1 )) t0  τ   t1 1 =c0 c1n |q(s)|Δs Δτ t0 t0 r(τ ) =c0 c1n+1 . Therefore, by the usual mathematical induction principle, (8.43) holds for all j ∈ N0 . Now, choosing t1 appropriately, we obtain c1 < 1. Then, (8.39) has a continuous solution x(t) =

∞

xj (t) for

t ∈ [t0 , t1 ].

j =0

Since xj (t) ≥ 0, it follows that x(t) ≥ x0 (t), which proves the validity of (8.40). Uniqueness of the solution of (8.39) can be shown in the usual way. From (8.38) and (8.39), in view of the uniqueness of the solution, we get θ ≡ x on [t0 , t1 ]. Therefore, θ (t) ≥ θ (t0 ) + θ [Δ] (t0 )



t t0

Δτ , r(τ )

t0 ≤ t ≤ t1 .

Hence, by making use of the induction hypothesis A(t0 ) being true, we obtain from the above inequality θ (t) ≥ 1 for

t ∈ [t0 , t1 ].

Taking this into account, from (8.37), we also get θ [Δ] (t) ≥ 0 for

t ∈ [t0 , t1 ].

Thus, A(t) is true for all t ∈ [t0 , t1 ]. 4. Let t ∈ T, t > 0 be left-dense and such that A(s) is true for all s < t, i.e., θ (s) ≥ 1, θ [Δ] (s) ≥ 0 for all s < t. Passing here to the limit as s → t, we get by the continuity of θ and θ [Δ] that θ (t) ≥ 1 and θ [Δ] (t) ≥ 0, which verifies the validity of A(t). Consequently, by the induction principle on a time scale, (8.36) holds for θ (t) and θ [Δ] (t), for all t ∈ T, t ≥ 0. The proof for ϕ(t) ≥ 0 and ϕ [Δ] (t) ≥ 1 is similar. From (8.36), in particular, we have θ (Λ) ≥ 1, ϕ [Δ] (Λ) ≥ 1. Actually, ϕ [Δ] (Λ) > 1. Indeed, consider the equations

544

8 Dynamic Equations on Time Scales

ϕ

[Δ]



t

(t) = 1 −

q(s)ϕ(σ (s))Δs

(8.44)

 q(s)ϕ(σ (s))Δs Δτ,

(8.45)

0

and

 ϕ(t) = 0

t

Δτ − r(τ )



t

0

1 r(τ )



τ

0

which follow from (8.20) with x = ϕ using (8.30). If ϕ [Δ] (Λ) = 1, then (8.44) gives rise to  Λ q(s)ϕ(σ (s))Δs = 0. (8.46) 0

On the other hand, since ϕ(t) ≥ 0 by (8.36), from (8.45), it follows that ϕ(t) > 0 for all t > 0. Therefore, (8.46) yields q(s) = 0 for all s ∈ [0, ρ(Λ)]. Hence, employing the Λ-periodicity of q, it follows that q ≡ 0 on T, contradicting the hypothesis. Hence, ϕ [Δ] (Λ) > 1, and the proof is complete.  Now, we aim to prove Theorem 8.32. We start with the following definition and two auxiliary results. Definition 8.41 We say that a function f : T → R has a generalized zero (a node) at t0 ∈ T if either f (t0 ) = 0 or f (ρ(t0 ))f (t0 ) < 0. Lemma 8.42 If D 2 ≥ 4, then (8.20) has a real, nontrivial solution ψ possessing the following properties: There exist two points a, b ∈ T such that 0 ≤ a ≤ ρ(Λ), b > a, b − a ≤ Λ, ψ has generalized zeros at a and b, and ψ(t) > 0 for a < t < b. Proof If D 2 ≥ 4, then it follows from Sect. 8.6.1 that (8.20) has a nontrivial solution x having the property x(t + Λ) = βx(t), t ∈ T, where β is a real nonzero number. Since Re x and Im x are also solutions of (8.20) with the same property, we may assume that (8.20) has a real, nontrivial solution ψ satisfying ψ(t + Λ) = βψ(t),

t ∈ T,

(8.47)

where β is a real nonzero number. First, we show that ψ(t) must have at least one generalized zero a in the segment [0, ρ(Λ)] = {t ∈ T : 0 ≤ t ≤ ρ(Λ)}. If not, then by (8.47), ψ(t) does not have any generalized zero in T, so ψ(t) = 0 and ψ(ρ(t))ψ(t) > 0 for all t in T. Hence, we also have ψ(t)ψ(σ (t)) > 0 for all t ∈ T. From the equation (rψ Δ )Δ (t) + q(t)ψ(σ (t)) = 0,

t ∈ T,

we have  0

Λ

r(t)ψ Δ (t) Δt + ψ(σ (t))

 0

Λ

q(t)Δt = 0.

(8.48)

8.6 Stability Theory for Hill’s Equation

545

Therefore, using the integration by parts formula (Theorem 8.10 (iv)) and noting that, by the periodicity of r(t) and (8.47),  r(t)ψ Δ (t) Λ r(Λ)ψ Δ (Λ) r(0)ψ Δ (0) ψ Δ (0) = − = [r(Λ) − r(0)] = 0, ψ(t) 0 ψ(Λ) ψ(0) ψ(0) we get 

Λ 0

 2  Λ r(t) ψ Δ (t) Δt + q(t)Δt = 0. ψ(t)ψ(σ (t)) 0

Hence, taking into account (8.23) along with the facts that ψ(t)ψ(σ (t)) > 0 and r(t) > 0, we obtain ψ Δ (t) = 0 for all t ∈ [0, ρ(Λ)], i.e., ψ ≡ C on [0, Λ]. Note that C = 0 since ψ is a nontrivial solution of (8.20). Therefore, setting t = 0 in (8.47), we get β = 1. Hence, ψ(t) = C for all t ∈ T. Consequently, from (8.48), we have Cq(t) = 0 on T and hence q ≡ 0 on T, which contradicts (8.23). Thus, ψ(t) has at least one generalized zero a in [0, ρ(Λ)]. From (8.47), we get that ψ(t) also has a generalized zero at a + Λ. It is not difficult to show that on the segment [a, a + Λ], the solution ψ may have only finitely many generalized zeros. Denote by b the smallest generalized zero of ψ(t) lying to the right of a and different from a. Then, b ≤ a + Λ so that b > a, b − a ≤ Λ, and ψ(t) does not have a generalized zero at t for a < t < b. Since ψ(t) must keep a constant sign on (a, b), and together with ψ, the function −ψ is also a solution of (8.20) with the same generalized zeros at a and b, we may assume that ψ(t) > 0 for a < t < b. The proof is therefore complete.  Lemma 8.43 (Lyapunov-Type Inequality) Under the hypothesis of Lemma 8.42, the inequality   r0 +

Λ

Δt r(t)

0



Λ

q + (t)Δt > 4

0

holds, where r0 is defined by (8.25). Proof Let us set 

t

g(t) = 0

Δs , r(s)

G+ = max{G, 0},

G = (rψ Δ )Δ G− = − min{G, 0}.

Evidently G+ + G− = |G|,

G+ − G− = G.

(8.49)

546

8 Dynamic Equations on Time Scales

Let ψ be a solution of (8.20) possessing the properties indicated in Lemma 8.42. There are four possibilities with respect to a and b. Case 1. Case 2. Case 3. Case 4.

ψ(ρ(a))ψ(a) < 0 and ψ(ρ(b))ψ(b) < 0. ψ(ρ(a))ψ(a) < 0 and ψ(b) = 0. ψ(a) = 0 and ψ(b) = 0. ψ(a) = 0 and ψ(ρ(b))ψ(b) < 0.

Suppose one of Cases 1 and 2 holds. Then, ρ(a) < a and ρ(b) ≤ b. Choose c ∈ [a, ρ(b)] such that ψ(c) =

max ψ(t).

a≤t≤ρ(b)

It is easy to see that, in the cases considered, ψ(c) − ψ(ρ(a)) > ψ(c),

ψ(c) − ψ(b) ≥ ψ(c),

with the strict inequality in the latter one occurring in Case 1. Consequently, 1 1 + g(c) − g(ρ(a)) g(b) − g(c) ψ(c) − ψ(ρ(a)) 1 ψ(c) − ψ(b) 1 × + × g(c) − g(ρ(a)) ψ(c) g(b) − g(c) ψ(c)   1 ψ(c) − ψ(ρ(a)) ψ(b) − ψ(c) − . = g(c) − g(ρ(a)) g(b) − g(c) ψ(c)
4. ψ(σ (t))

ξ

(8.51)

Since 

b ρ(a)

Δt , r(t)



b

g(b) − g(ρ(a)) =

G− (t) = q + (t), ψ(σ (t))

t ∈ [ρ(a), ρ(b)),

it follows from (8.51) that

ρ(a)

Δt r(t)



τ

q + (t)Δt > 4

(8.52)

ξ

holds. Next, taking into account ρ(a) ≤ ξ < τ ≤ ρ(b), b ≤ a + Λ, and that for any periodic function f on T with period Λ, the equality 

t0 +Λ



Λ

f (t)Δt =

f (t)Δt 0

t0

holds for all t0 ∈ T, we have 

b ρ(a)

Δt ≤ r(t)



a+Λ

ρ(a)

a − ρ(a) Δt = + r(t) r(ρ(a))



a+Λ

a

Δt ≤ r0 + r(t)

 0

Λ

Δt r(t)

and  ξ

τ

q + (t)Δt ≤



ρ(b)

q + (t)Δt ≤

ρ(a)



ρ(b)+Λ ρ(a)

q + (t)Δt =



Λ

q + (t)Δt,

0

since by the Λ-periodicity of T and r, a − ρ(a) σ (t) − t = max = r0 , a∈[0,ρ(Λ)] r(ρ(a)) t∈[0,ρ(Λ)] r(t) max

and from the inequality b ≤ a + Λ, we get ρ(b) ≤ ρ(a) + Λ. Consequently, (8.49) follows from (8.52). Now, we assume that one of Cases 3 or 4 holds. In these cases, choosing c ∈ (a, b) such that

548

8 Dynamic Equations on Time Scales

ψ(c) =

max

σ (a)≤t≤ρ(b)

ψ(t)

yields ψ(c) − ψ(a) = ψ(c),

ψ(c) − ψ(b) ≥ ψ(c),

with the strict inequality in the second one holding for Case 4. Consequently, 1 1 + g(c) − g(a) g(b) − g(c) ψ(c) − ψ(a) 1 ψ(c) − ψ(b) 1 × + × g(c) − g(a) ψ(c) g(b) − g(c) ψ(c)   1 ψ(c) − ψ(a) ψ(b) − ψ(c) − . = g(c) − g(a) g(b) − g(c) ψ(c) ≤

Finally, reasoning as in the previous case, we obtain the inequality 

Λ 0

Δt r(t)



Λ

q + (t)Δt > 4.

0



Therefore, (8.49) is true in these cases as well. The proof is complete. D2

Proof of Theorem 8.32 It is sufficient, by Proposition 8.37, to show that < 4. Assume on the contrary that D 2 ≥ 4. By Lemma 8.43, we get (8.49). Since (8.49) contradicts (8.24), the inequality D 2 ≥ 4 cannot be true. Thus, D 2 < 4 and (8.20) is stable. The proof is complete. 

8.7 Linear Hamiltonian Systems In this section, we consider Hamiltonian systems (see [12, 52, 53]), which contain two scalar linear dynamic equations 

x Δ (t) = a(t)x(σ (t)) + b(t)u(t), uΔ (t) = −c(t)x(σ (t)) − a(t)u(t)

(8.53)

on an arbitrary time scale T, where a, b, c are real-valued rd-continuous functions on T with the coefficient a satisfying the condition 1 − μ(t)a(t) = 0 for all

t ∈ T.

(8.54)

8.7 Linear Hamiltonian Systems

549

Notice that (8.20) can be written as an equivalent Hamiltonian system of type (8.53). Indeed, let x be a solution of (8.20) and set u = rx Δ . Then, we have x Δ (t) =

1 , r(t)

uΔ (t) = −c(t)x(σ (t)).

So, (8.20) is equivalent to (8.53) with a(t) ≡ 0,

b(t) =

1 , r(t)

c(t) = q(t).

We remark that (8.53) cover the continuous Hamiltonian system (when T = R, see [143, 209]) x  (t) = a(t)x(t) + b(t)u(t),

u (t) = −c(t)x(t) − a(t)u(t),

t ∈ R,

and the discrete Hamiltonian system (when T = Z, see [22, 47, 143]) Δx(t) = a(t)x(t + 1) + b(t)u(t),

Δu(t) = −c(t)x(t + 1) − a(t)u(t),

t ∈ Z. (8.55)

Furthermore, (8.53) extends these classical cases to many cases in between as well, such as so-called q-difference equations, where T = q Z := {q k : k ∈ Z} ∪ {0}

for some

q>1

and difference equations with constant step size, where T = hZ := {hk : k ∈ Z}

for some

h > 0.

Particularly useful for the discretization aspect are time scales of the form T = {tk : k ∈ Z},

where

tk ∈ R, tk < tk+1

for all k ∈ Z.

An introduction to Lyapunov inequalities for continuous and discrete linear Hamiltonian system can be found in the paper [143] by Guseinov. The main purpose of this section is to obtain Lyapunov inequalities for Hamiltonian system on time scales. Concerning (8.53) with (8.54), we also assume that b(t) ≥ 0,

t ∈ T.

For each t ∈ T, let us set c+ (t) = max{c(t), 0}.

(8.56)

550

8 Dynamic Equations on Time Scales

Instead of the usual zero, the concept of generalized zero on time scales is given as follows. Definition 8.44 Let t ∈ T. A vector solution (x, u) of (8.53) has a generalized zero at σ (t) if one of the following two conditions is satisfied: (i) t is dense and x(t) = 0, (ii) t is right-scattered, and x(t)x(σ (t)) < 0 or x(σ (t)) = 0. Note that under (8.56), the definition of generalized zero, a special case of that given in [53], is consistent with what is used for the generalized zero n the discrete case [22, 143]. The main results of this section are the following theorems. The continuous and/or discrete versions of these results may be found in [143], but the following theorems cover all of these results. Theorem 8.45 (Lyapunov-Type Inequality) Let α, β ∈ Tκ with σ (α) < β. Assume that (8.53) has a real solution (x, u) such that x(σ (α)) = x(σ (β)) = 0 and x is not identically zero on [σ (α), β]. Then, the inequality 

β

 |a(t)|Δt +

σ (α)



σ (β)

1/2

β

+

b(t)Δt

≥2

c (t)Δt

σ (α)

(8.57)

σ (α)

holds. Theorem 8.46 (Lyapunov-Type Inequality) Suppose 1 − μ(t)a(t) > 0 and

b(t) > 0 for all t ∈ T

(8.58)

and let α, β ∈ Tκ with σ (α) < β. Assume that (8.53) has a real solution (x, u) such that x(σ (α)) = 0

and

x(β)x(σ (β)) < 0.

Then, the inequality 

β

σ (α)

 |a(t)|Δt +



β

β

b(t)Δt σ (α)

c+ (t)Δt

1/2 >1

(8.59)

σ (α)

holds. Theorem 8.47 (Lyapunov-Type Inequality) Suppose (8.58) holds and let α, β ∈ Tκ with α < β. Assume that (8.53) has a real solution (x, u) such that

8.7 Linear Hamiltonian Systems

551

x(α)x(σ (α)) < 0

x(β) = 0.

and

Then, the inequality 



β

|a(t)|Δt +

σ (α)



σ (β)

β

b(t)Δt σ (α)

1/2 +

>1

c (t)Δt

(8.60)

α

holds. Theorem 8.48 (Lyapunov-Type Inequality) Suppose 1 − μ(t)a(t) > 0,

b(t) > 0,

and

c(t) > 0 for all

t ∈ T,

(8.61)

and let α, β ∈ Tκ with α < β. Assume that (8.53) has a real solution (x, u) such that x(α)x(σ (α)) < 0

and

x(β)x(σ (β)) < 0.

Then, the inequality 

β

 |a(t)|Δt +

α



σ (β)

b(t)Δt α

1/2

β

c(t)Δt

>1

(8.62)

α

holds. Combining Theorems 8.45, 8.46, 8.47, and 8.48 yields the following corollary. Corollary 8.49 (Lyapunov-Type Inequality) Suppose (8.61), and let α, β ∈ Tκ with σ (α) < β. Assume that (8.53) has a real solution (x, u) with generalized zeros in σ (α) and σ (β) such that x is not identically zero on [σ (α), β]. Then, the inequality  α

σ (β)

 |a(t)|Δt +



σ (β)

b(t)Δt α

1/2

σ (β)

c(t)Δt

>1

α

holds. In Sect. 8.7.1, the proofs of Theorems 8.45, 8.46, 8.47, and 8.56 are given successively. As applications of Lyapunov inequalities, we establish a disconjugacy criterion in Sect. 8.7.2.

552

8 Dynamic Equations on Time Scales

8.7.1 Proofs Proof of Theorem 8.45 Multiplying the first equation in (8.53) by u(t) and the second one by x(σ (t)), and then adding the results, we obtain (xu)Δ (t) = b(t)[u(t)]2 − c(t)[x(σ (t))]2 .

(8.63)

Integrating this equation from σ (α) to σ (β) and noticing x(σ (α)) = x(σ (β)) = 0, we have  0=

σ (β)



σ (β)

b(t)[u(t)]2 Δt −

σ (α)

c(t)[x(σ (t))]2 Δt.

σ (α)

Since x(σ (β)) = 0, by Theorem 8.10 (ii), (iii), we have 

σ (β)

 b(t)[u(t)] Δt = 2

σ (α)

β



σ (α)

 =

β

σ (β)

c(t)[x(σ (t))] Δt + 2

c(t)[x(σ (t))]2 Δt

β

c(t)[x(σ (t))]2 Δt + μ(β)c(β)[x(σ (β))]2

σ (α)

 =

β

c(t)[x(σ (t))]2 Δt.

σ (α)

(8.64) Choose τ ∈ (σ (α), σ (β)) such that |x(τ )| =

max

σ (α)≤t≤σ (β)

|x(t)|.

Since x is not identically zero on [σ (α), β], we have |x(τ )| > 0. Integrating the first equation in (8.53) initially from σ (α) to τ and then from τ to σ (β) and observing that x(σ (α)) = x(σ (β)) = 0, we get 



τ

x(τ ) =

τ

a(t)x(σ (t))Δt +

b(t)u(t)Δt

σ (α)

σ (α)

and 

σ (β)

−x(τ ) = 



τ β

= τ

σ (β)

a(t)x(σ (t))Δt + 

a(t)x(σ (t))Δt +

b(t)u(t)Δt τ σ (β)

b(t)u(t)Δt, τ

respectively, where for the second equal sign of the latter equation, we have used Theorem 8.10 (ii), (iii). Hence, employing the triangle inequality and Theorem 8.10

8.7 Linear Hamiltonian Systems

553

(v) gives 



τ

|x(τ )| ≤

τ

|a(t)||x(σ (t))|Δt +

b(t)|u(t)|Δt

σ (α)

σ (α)

and 

β

|x(τ )| ≤



σ (β)

|a(t)||x(σ (t))|Δt +

b(t)|u(t)|Δt.

τ

τ

Adding these two inequalities gives rise to  2|x(τ )| ≤



β

σ (β)

|a(t)||x(σ (t))|Δt +

σ (α)

(8.65)

b(t)|u(t)|Δt. σ (α)

Applying the Cauchy–Schwarz inequality (Theorem 8.12) and (8.64), we have 

σ (β)

 b(t)|u(t)|Δt ≤

1/2 

σ (β)

b(t)Δt

σ (α)

=

σ (α)

1/2 

σ (β)

≤

1/2

β

b(t)Δt

2

c(t)[x(σ (t))] Δt

σ (α)



2

b(t)[u(t)] Δt

σ (α)



1/2

σ (β)

σ (α)

1/2 

σ (β)

β

b(t)Δt σ (α)

c+ (t)[x(σ (t))]2 Δt

1/2 .

σ (α)

Therefore, we get from (8.65)  2|x(τ )| ≤

β

|a(t)||x(σ (t))|Δt

σ (α)



+

1/2 

σ (β)

β

c+ (t)[x(σ (t))]2 Δt

b(t)Δt σ (α)

σ (α)

⎡



≤ |x(τ )| ⎣

β σ (α)

 |a(t)|Δt +



σ (β)

β

b(t)Δt σ (α)

1/2

1/2 ⎤ ⎦. c+ (t)Δt

σ (α)

Dividing the latter estimate by |x(τ )|, we get (8.57). Proof of Theorem 8.46 Integrating (8.63) from σ (α) to β and observing that x(σ (α)) = 0,



554

8 Dynamic Equations on Time Scales

we obtain  u(β)x(β) =



β

β

b(t)[u(t)]2 Δt −

σ (α)

c(t)[x(σ (t))]2 Δt.

(8.66)

σ (α)

Further, by using Theorem 8.9 (iv), we rewrite the first equation in (8.53) and get (1 − μ(t)a(β)) x(σ (t)) = x(t) + μ(t)b(t)u(t).

(8.67)

If t = β, then (1 − μ(β)a(β)) x(σ (β)) = x(β) + μ(β)b(β)u(β). Multiplying this by x(β) yields (1 − μ(β)a(β)) x(β)x(σ (β)) = [x(β)]2 + μ(β)b(β)x(β)u(β). Since x(β)x(σ (β)) < 0, it is easy to see that μ(t) > 0. In view of (8.58), the above latter equality gives rise to x(β)u(β) < 0. Therefore, from (8.66), the inequality 

β



β

b(t)[u(t)]2 Δt
0. Integrating the first equation in (8.53) from σ (α) to τ and noticing that x(σ (α)) = 0, we obtain  x(τ ) =

τ

 a(t)x(σ (t))Δt +

σ (α)

τ

b(t)u(t)Δt. σ (α)

Hence, applying the Cauchy–Schwarz inequality and (8.68), we get  |x(τ )| ≤

τ

 |a(t)||x(σ (t))|Δt +

σ (α)

 ≤

β

σ (α)

 ≤

β

σ (α)

b(t)|u(t)|Δt 

|a(t)||x(σ (t))|Δt +

τ σ (α) β

b(t)|u(t)|Δt σ (α)

 |a(t)||x(σ (t))|Δt +

1/2 

β

β

b(t)Δt σ (α)

1/2 2

b(t)[u(t)] Δt σ (α)

8.7 Linear Hamiltonian Systems



β

555

|a(t)||x(σ (t))|Δt

< σ (α)



1/2 

β

+

β

b(t)Δt σ (α) β

≤|x(τ )|

2

c (t)[x(σ (t))] Δt

σ (α)

0

1/2

+





β

|a(t)|Δt +

β

b(t)Δt

σ (α)

σ (α)

1/2 1

+

c (t)Δt

.

σ (α)

Therefore, dividing the latest estimate by |x(τ )|, we obtain (8.59).



Proof of Theorem 8.47 Choose τ ∈ (σ (α), σ (β)) such that |x(τ )| =

max

σ (α)≤t≤σ (β)

|x(t)|.

Then, |x(τ )| > 0. Integrating the first equation in (8.53) from τ to σ (β) and taking into account x(σ (β)) = 0, we get 

σ (β)

x(τ ) = − 

τ β

=− 

b(t)|u(t)|Δt



σ (β)

 a(t)x(σ (t))Δt −

β β

β

σ (β)

b(t)|u(t)|Δt τ



σ (β)

a(t)x(σ (t))Δt − μ(β)a(β)x(σ (β)) −

τ

=−

σ (β)

τ

a(t)x(σ (t))Δt −

τ

=− 

 a(t)x(σ (t))Δt −

b(t)|u(t)|Δt τ



σ (β)

a(t)x(σ (t))Δt −

b(t)|u(t)|Δt.

τ

τ

Therefore, 

β

|x(τ )| ≤  ≤



≤

b(t)|u(t)|Δt

τ

τ β



β σ (α)

σ (β)

|a(t)||x(σ (t))|Δt +

σ (α)



σ (β)

|a(t)||x(σ (t))|Δt +

b(t)|u(t)|Δt σ (α)

 |a(t)||x(σ (t))|Δt+

1/2 

σ (β)

σ (β)

b(t)Δt σ (α)

1/2 2

b(t)[u(t)] Δt σ (α)

(8.69) Now, integrating (8.63) from α to σ (β) and taking into account x(σ (β)) = 0, we get

.

556

8 Dynamic Equations on Time Scales

−x(α)u(α) = = =

 σ (β) α

 σ (β)

 σ (β)

b(t)[u(t)]2 Δt − b(t)[u(t)]2 Δt−

 β

α

c(t)[x(σ (t))]2 Δt−

 σ (β)

α

 σ (β) α

c(t)[x(σ (t))]2 Δt

α

 β

b(t)[u(t)]2 Δt −

c(t)[x(σ (t))]2 Δt

β

c(t)[x(σ (t))]2 Δt.

α

Applying Theorem 8.10 (iii), we rewrite the above last equality as 

σ (α)

−x(α)u(α)−

 b(t)[u(t)] Δt = 2

α

σ (β)



β

b(t)[u(t)] Δt − 2

σ (α)

c(t)[x(σ (t))]2 Δt.

α

By Theorem 8.10 (ii), it follows that  − u(α)[x(α) + μ(α)b(α)u(α)] =

σ (β)



β

b(t)[u(t)] Δt − 2

σ (α)

c(t)[x(σ (t))]2 Δt.

α

(8.70) Further, from (8.67), we have, for t = α, (1 − μ(α)a(α)) x(σ (α)) = x(α) + μ(α)b(α)u(α).

(8.71)

Multiplying this by x(α) gives that (1 − μ(α)a(α)) x(α)x(σ (α)) = [x(α)]2 + μ(α)b(α)x(α)u(α). Since x(α)x(σ (α)) < 0, it is easy to see that μ(α) > 0 holds. By (8.58) and the above latter equality, we have x(α)u(α) < 0.

(8.72)

u(α)[x(α) + μ(α)b(α)u(α)] > 0

(8.73)

Now, we claim that

holds. Indeed, multiplying (8.71) by u(α) gives (1 − μ(α)a(α)) x(σ (α))u(α) = u(α)[x(α) + μ(α)b(α)u(α)].

(8.74)

On the other hand, it follows from x(α)x(σ (α)) < 0 and (8.72) that x(α)u(α) > 0. Therefore, the left-hand side of (8.74) is positive, and hence, (8.73) is true. By virtue of (8.73), the string of inequalities 

σ (β)

σ (α)

 b(t)[u(t)]2 Δt < α

β

 c(t)[x(σ (t))]2 Δt ≤ α

β

c+ (t)[x(σ (t))]2 Δt

8.7 Linear Hamiltonian Systems

557

follows from (8.70). As a result of these last relations, from (8.69), we obtain 

β

|x(τ )|
0 for all

t ∈ [σ (α), β0 ].

Then, we have x(α) < 0

and

x(σ (β0 )) < 0.

(8.75)

Let s ∈ [α, σ (β0 )]. Integrating the second equality of (8.53) from α to s and then from s to β0 , we get 

s

u(s) − u(α) = −



s

c(t)x(σ (t))Δt −

α

a(t)u(t)Δt

(8.76)

α

and 

β0

u(β0 ) − u(s) = −

 c(t)x(σ (t))Δt −

s

β0

a(t)u(t)Δt,

(8.77)

s

respectively. Noticing that for s = α, we write solely (8.77), and for s = β0 , only (8.76) is written. Now, we aim to show that u(α) > 0,

u(β0 ) < 0.

(8.78)

558

8 Dynamic Equations on Time Scales

Indeed, multiplying (8.67) by x(t) gives (1 − μ(t)a(t)) x(t)x(σ (t)) = [x(t)]2 + μ(t)b(t)x(t)u(t), where setting t = α and t = β0 yields (1 − μ(α)a(α)) x(α)x(σ (α)) = [x(α)]2 + μ(α)b(α)x(α)u(α) and (1 − μ(β0 )a(β0 )) x(β0 )x(σ (β0 )) = [x(β0 )]2 + μ(β0 )b(β0 )x(β0 )u(β0 ), respectively. Using the inequalities x(α)x(σ (α)) < 0 and

x(β0 )x(σ (β0 )) < 0,

μ(α) > 0 and μ(β0 ) > 0 can be obtained easily. Combining (8.61) with the above equalities, we get the estimates x(α)u(α) < 0 and

x(β0 )u(β0 ) < 0.

(8.79)

Observing that x(α) < 0 and x(β0 ) > 0, we obtain (8.78). Employing (8.76) if u(s) < 0 and using (8.77) whenever u(s) > 0, and also taking into account (8.78), we get 

β0

|u(s)| ≤

c(t)|x(σ (t))|Δt α



1/2 

β0

≤

1/2

β0

c(t)Δt

2

c(t)[x(σ (t))] Δt

α



β0

+

α

α

|a(t)||u(t)|Δt. (8.80)

Next, integrating (8.63) from α to σ (β0 ) gives rise to 

σ (β0 )

x(σ (β0 ))u(σ (β0 ))−x(α)u(α) =



σ (β0 )

b(t)[u(t)] Δt − 2

α

c(t)[x(σ (t))]2 Δt.

α

From Theorem 8.10 (iii), we obtain  x(σ (β0 ))u(σ (β0 )) +

σ (β0 )

c(t)[x(σ (t))]2 Δt − x(α)u(α)

β0

 = α

Using Theorem 8.10 (ii), we have

σ (β0 )



β0

b(t)[u(t)] Δt − 2

α

c(t)[x(σ (t))]2 Δt.

8.7 Linear Hamiltonian Systems

559

x(σ (β0 )) (u(σ (β0 )) + μ(β0 )c(β0 )x(σ (β0 ))) − x(α)u(α)  σ (β0 )  β0 2 = b(t)[u(t)] Δt − c(t)[x(σ (t))]2 Δt.

(8.81)

α

α

We claim that x(σ (β0 )) [u(σ (β0 )) + μ(β0 )c(β0 )x(σ (β0 ))] > 0

(8.82)

holds. Indeed, from the second equation in (8.53), we have, for t = β0 , (1 − μ(β0 )a(β0 )) u(β0 ) = u(σ (β0 )) + μ(β0 )c(β0 )x(σ (β0 )). Multiplying this equation by x(σ (β0 )) yields (1 − μ(β0 )a(β0 )) u(β0 )x(σ (β0 )) = x(σ (β0 )) [u(σ (β0 )) + μ(β0 )c(β0 )x(σ (β0 ))] . (8.83) On the other hand, from (8.75) and (8.78), it follows that u(β0 )x(σ (β0 )) > 0. Therefore, from 1 > μ(β0 )a(β0 ) > 0 and (8.83), (8.82) follows. In view of (8.79) and (8.82), the inequality 



β0

2

σ (β0 )

c(t)[x(σ (t))] Δt
1.

⎦.

560

8 Dynamic Equations on Time Scales

Since β0 ≤ β, (8.83) follows. Second, we consider the case when x(t0 ) = 0 for some t0 ∈ (σ (α), β). In this case, applying Theorem 8.46 to the points t0 and β, we get the inequality 

β

 |a(t)|Δt +

t0



β t0

1/2

β

b(t)Δt

> 1.

c(t)Δt t0



Therefore, (8.62) holds in this case as well.

8.7.2 A Disconjugacy Criterion Let α, β ∈ Tκ with σ (α) < β. Consider the linear Hamiltonian dynamic system 

x Δ (t) = a(t)x(σ (t)) + b(t)u(t), uΔ (t) = −c(t)x(σ (t)) − a(t)u(t),

t ∈ [α, β],

(8.85)

where the coefficients a, b, c are real rd-continuous functions defined on [α, β] satisfying 1 − μ(t)a(t) > 0,

b(t) > 0,

and

c(t) > 0 for all

t ∈ [α, β].

(8.86)

Note that each solution (x, u) of (8.85) is a vector-valued function defined on [α, σ (β)]. Now, we give the concept of a relatively generalized zero for the component x of a real solution (x, u) of (8.85) and also the concept of disconjugacy of this system on [α, σ (β)]. The definition is relative to the interval [α, σ (β)], and the left end-point α is treated separately. Definition 8.50 The component x of a solution (x, u) of (8.85) has a relatively generalized zero at α if and only if x(α) = 0, while x has a relatively generalized zero at σ (t0 ) > α provided (x, u) has a generalized zero at σ (t0 ). System (8.85) is called disconjugate on [α, σ (β)] provided there is no real solution (x, u) of this system with x nontrivial and having two (or more) relatively generalized zeros in [α, σ (β)]. Noting that when T = Z, the definitions of relatively generalized zero and that of disconjugacy are equivalent to those given in [22, 143]. Theorem 8.51 Assume (8.86) holds. If 

σ (β) α





σ (β)

|a(t)|Δt +

b(t)Δt α

1/2

σ (β)

c(t)Δt α

≤ 1,

(8.87)

8.8 Planar Linear Hamiltonian Systems

561

then (8.85) is disconjugate on [α, σ (β)]. Proof Suppose, on the contrary, that (8.85) is not disconjugate on [α, σ (β)]. By Definition 8.50, there exists a real solution (x, u) of (8.85) with x nontrivial and such that x has at least two relatively generalized zeros in [α, σ (β)]. Now, we have the following two cases to consider. (I) One of the two relatively generalized zeros is at the end point α, i.e., x(α) = 0, the other is at σ (β0 ) ∈ (α, σ (β)]. Therefore, applying Theorem 8.45 or Theorem 8.46, we get 

σ (β0 )





σ (β0 )

|a(t)|Δt +

b(t)Δt

α

1/2

σ (β0 )

> 1.

c(t)Δt

α

α

This contradicts (8.87). (II) None of the two relatively generalized zeros is at α, that is, x has two generalized zeros σ (α0 ), σ (β0 ) ∈ (α, σ (β)] (σ (α0 ) < σ (β0 )). Therefore, applying Corollary 8.49, we have 

σ (β0 )

 |a(t)|Δt +

α0



σ (β0 )

b(t)Δt α0

1/2

σ (β0 )

c(t)Δt

> 1,

α0

which is contrary to (8.87). The proof of Theorem 8.51 is now completed by combining cases (I) and (II). 

8.8 Planar Linear Hamiltonian Systems In this section, we establish Lyapunov-type inequalities for the planar Hamiltonian system 

x Δ = α(t)x σ + β(t)u, uΔ = −γ (t)x σ − α(t)u,

(8.88)

where α, β, γ are real-valued rd-continuous functions defined on a given arbitrary time scale T. Recently, He et al. [161] obtained several Lyapunov-type inequalities for (8.88), which improved the earlier results given by Jiang and Zhou [179], and hence the related ones in [35, 48, 91, 142, 143]. The following theorem, see [161, Theorem 3.1], seems to be the best result for (8.88) thus far.

562

8 Dynamic Equations on Time Scales

Theorem 8.52 (Lyapunov-Type Inequality) Suppose that 1 − μ(t)α(t) > 0

for all t ∈ T

(8.89)

and β(t) ≥ 0 for all t ∈ T.

(8.90)

Let a, b ∈ Tκ with σ (a) ≤ b. Assume (8.88) has a real solution (x, y) such that x is nontrivial and has generalized zeros at a and b, i.e., either x(a) = 0 or x(a)x σ (a) < 0,

either x(b) = 0 or x(b)x σ (b) < 0.

Then, one has the inequality 

b





σ (b)

|α(t)|Δt +

β(t)Δt

a

1/2

b

+

γ (t)Δt

a

≥ 2.

(8.91)

a

In all Lyapunov-type inequalities given for (8.88) in the literature, (8.89) is a must. Bohner and Zafer [54] showed that this condition can be completely dropped. To do this, they introduced a new definition for a generalized zero, motivated by the one given in [301] for the discrete case. Note that (8.91) is trivial if 

b

|α(t)|Δt ≥ 2.

a

If 

b

|α(t)|Δt < 2,

a

then (8.91) is equivalent to 



σ (b)

β(t)Δt a

b

+





b

γ (t)Δt ≥ 2 −

a

2 |α(t)|Δt

.

(8.92)

a

As an improvement as well as an alternative to (8.92), we also show that if 1 − μ(t)α(t) = 0 for all then a Lyapunov-type inequality of the form

t ∈ T,

(8.93)

8.8 Planar Linear Hamiltonian Systems





σ (b)

b

β(t)Δt a

563

 γ + (t)Δt ≥ exp −

a

b

  ψμ(t) (−α(t)) Δt

(8.94)

a

holds, where ⎧ ⎪ ⎨ log |1 + hz| h ψh (z) = ⎪ ⎩z,

if h = 0, 1 + hz = 0,

(8.95)

if h = 0.

In fact, (8.92) follows from (8.94) under an additional condition implying (8.89), see Remark 8.68 below. Definition 8.53 A real nontrivial solution (x, u) of (8.88) is said to have a relative generalized zero (with respect to x) at t0 ∈ T if either x(t0 ) = 0 or x ∗ (t0 ) < 0, where x ∗ (t) := [1 − μ(t)α(t)]x(t)x(σ (t)). Definition 8.54 System (8.88) is said to be relatively disconjugate (with respect to x) on [a, b]T if there is no real solution (x, u) with x having more than one generalized zero in [a, b]T . The section is organized as follows. Next, we give some properties of the time scale exponential function and introduce some estimates for a time scale exponential bound function (see Definition 8.59). Then, Lyapunov-type inequalities are given in Sect. 8.8.2. The last subsection is devoted to a simple application, namely disconjugacy criteria for linear Hamiltonian systems.

8.8.1 Time Scales Exponential Function In this section, we let p : T → R be rd-continuous and regressive, i.e., 1 + μ(t)p(t) = 0 for all

t ∈ T,

and we let s, t, r ∈ T. Definition 8.55 The time scales exponential function is defined by  ep (t, s) := exp

t

ξμ(t) (p(τ ))Δτ ,

s

where the cylinder transformation ξ is defined by

564

8 Dynamic Equations on Time Scales

⎧ ⎪ ⎨ Log(1 + hz) h ξh (z) = ⎪ ⎩z,

if h = 0, 1 + hz = 0,

(8.96)

if h = 0.

Some of the properties enjoyed by the time scales exponential function are given next. Theorem 8.56 (See [52, Theorem 2.36]) We have ep (t, s) = ep (s, t) =

1 , ep (t, s)

where

ep (t, s)ep (s, r) = ep (t, r), epσ (·, s)

= (1 + μp)ep (·, s),

epΔ (·, s) = pep (·, s),

 p :=

1 , 1 + μp

ep (t, t) = 1, epσ (s, ·)

ep (s, ·) , = 1 + μp

(8.97)

epΔ (s, ·) = −pepσ (s, ·).

The following variation of parameter formula holds. Theorem 8.57 (See [52, Theorem 2.74]) Suppose f : T → R is rd-continuous. Then, x solves x Δ = −p(t)x σ + f (t) if and only if  x(t) = ep (s, t)x(s) +

t

ep (τ, t)f (τ )Δτ. s

Theorem 8.58 (See [50, Theorem 3.4]) We have   ep (t, s) = exp



t

ψμ(t) (p(τ ))Δτ ,

s

where ψ is defined by (8.95). We now introduce a function that serves as a bound for the absolute value of the exponential function on time scales. Definition 8.59 The time scales exponential bound function is defined by  Ep (t, s) := exp

t

  ψμ(t) (p(τ )) Δτ .

s

For later use in this section and also for future reference, some of the properties satisfied by the time scales exponential bound function are gathered next.

8.8 Planar Linear Hamiltonian Systems

565

Theorem 8.60 We have 1 ≤ Ep (t, s) ≤ Ep (t˜, s˜ ) if s˜ ≤ s ≤ t ≤ t˜,   Ep (s, t) ≤ ep (t, s) ≤ Ep (t, s) for t ≥ s,   Ep (t, s) ≤ ep (t, s) ≤ Ep (s, t) for t ≤ s,   Ep (min{s, t}, max{s, t}) ≤ ep (t, s) ≤ Ep (max{s, t}, min{s, t}),

(8.99) (8.100) (8.101)

1 , Ep (s, t)

(8.102)

Ep (t, t) = 1,

(8.103)

Ep (t, s) = Ep (t, s) = Ep (t, s)Ep (s, r) = Ep (t, r),

(8.98)

and  Epσ (·, s) = max |1 + μp|,

 1 Ep (·, s). |1 + μp|

(8.104)

Proof Clearly, (8.98) and (8.103) follow from the definition of E. The second equality in (8.102) follows from (8.103). Now, note that ⎧ 1 ⎪ ⎨ log |1 + μ(t)(p)(t)| if μ(t) = 0, ψμ(t) ((p)(t)) = μ(t) ⎪ ⎩ (p)(t) if μ(t) = 0   ⎧   1 1 ⎪  ⎨ if μ(t) = 0, log  1 + μ(t)p(t)  = μ(t) ⎪ ⎩ −p(t) if μ(t) = 0 = −ψμ(t) (p(t)) implies     ψμ(t) ((p)(t)) = ψμ(t) (p(t)) . This shows the first equality in (8.102). Now, let t ≥ s. Then, we have   ep (t, s) = exp





t

ψμ(t) (p(τ ))Δτ s

 ≤ exp

t

  ψμ(t) (p(τ )) Δτ

= Ep (t, s).

s

This shows the second inequality in (8.99). Moreover, by using (8.97), (8.102), and the second inequality in (8.99), we obtain

566

8 Dynamic Equations on Time Scales

  1 1 1 ep (t, s) =  ≥ = = Ep (s, t). ep (t, s) Ep (t, s) Ep (t, s) This shows the first inequality in (8.99). Next, let t ≤ s. Then, we can use (8.97), the second inequality in (8.99), and (8.102) to obtain   1 ep (t, s) =  1  ≥ = Ep (t, s), ep (s, t) Ep (s, t) which shows the second inequality in (8.100). Moreover, by using (8.97), the second inequality in (8.100), and (8.102), we obtain   1 1 1 ep (t, s) =  ≥ = = Ep (t, s). ep (t, s) Ep (s, t) Ep (s, t) This shows the first inequality in (8.100). Finally, (8.101) follows by combining (8.99) and (8.100). 

8.8.2 Lyapunov-Type Inequalities Theorem 8.61 (Lyapunov-Type Inequality) Assume (8.93) and β(t) ≥ 0,

β(t) ≡ 0,

Let a, b ∈ Tκ with σ (a) ≤ b. t ∈ [a, b]T .

(8.105)

If (8.88) has a real solution (x, u) such that x(a) = 0 and x(b) = 0, and if x(t) = 0 for all t ∈ [a, b]T , then 



b

b

β(t)Δt a

 γ (t)Δt ≥ 4 exp −

b

+

a

  ψμ(t) (−α(t)) Δt .

(8.106)

a

Proof By the variation of parameters formula (Theorem 8.57), we write  x(t) = e−α (s, t)x(s) +

t s

e−α (τ, t)β(τ )u(τ )Δτ.

(8.107)

Put s = a and use x(a) = 0 in (8.107). Then, 

t

|x(t)| ≤ a

|e−α (τ, t)| β(τ )|u(τ )|Δτ.

(8.108)

8.8 Planar Linear Hamiltonian Systems

567

For a ≤ τ < t ≤ b, we use (8.100) and (8.98) to obtain |e−α (τ, t)| ≤ E−α (t, τ ) ≤ E−α (t, a), which together with (8.108) shows  |x(t)| ≤ E−α (t, a)

t

β(τ )|u(τ )|Δτ.

(8.109)

a

Next, putting s = b and using x(b) = 0 in (8.107) leads to 

b

|x(t)| ≤ t

|e−α (τ, t)| β(τ )|u(τ )|Δτ.

(8.110)

For a ≤ t ≤ τ < b, we use (8.99) and (8.98) to obtain |e−α (τ, t)| ≤ E−α (τ, t) ≤ E−α (b, t), which together with (8.110) shows  |x(t)| ≤ E−α (b, t)

b

β(τ )|u(τ )|Δτ.

(8.111)

t

Now, let Q1 =

|x(t)| , E−α (t, a)

Q2 =

|x(t)| . E−α (b, t)

Then, (8.103), the arithmetic-geometric inequality, (8.109), (8.111), and (8.98) yield 

|x(t)| E−α (b, a)

= =



|x(t)| E−α (b, t)E−α (t, a)

Q1 Q2

Q1 + Q2 2 |x(t)| |x(t)| + = 2E−α (t, a) 2E−α (b, t) t b E−α (b, t) t β(τ )|u(τ )|Δτ E−α (t, a) a β(τ )|u(τ )|Δτ + ≤ 2E−α (t, a) 2E−α (b, t)  b 1 = β(s)|u(s)|Δs, 2 a

≤

568

8 Dynamic Equations on Time Scales

and thus, by the Cauchy–Schwarz inequality (see Theorem 8.12), 4[x(t)]2 ≤ E−α (b, a)



2

b





b

≤

β(s)|u(s)|Δs

b

β(s)[u(s)]2 Δs.

β(s)Δs

a

a

(8.112)

a

Next, we use the time scales product rule (see Theorem 8.9 (v)) and (8.88) to calculate (xu)Δ = x Δ u+x σ uΔ = (αx σ +βu)u−(γ x σ +αu)x σ = βu2 −γ (x σ )2 .

(8.113)

Hence, 

b

  2 β(τ )[u(τ )]2 − γ (τ ) x σ (τ ) Δτ = 0,

a

and thus 

b



b

β(τ )[u(τ )]2 Δτ =

a

 2 γ (τ ) x σ (τ ) Δτ ≤

a



b

 2 γ + (τ ) x σ (τ ) Δτ.

a

(8.114)

Using (8.114) in (8.112), we find 4[x(t)]2 ≤ E−α (b, a)





b

β(s)Δs a

b

 2 γ + (s) x σ (s) Δs.

(8.115)

a

Pick now t∗ ∈ [a, σ (b)] such that |x(t∗ )| =

max |x(t)| > 0.

a≤t≤σ (b)

As in [161], by treating b left-scattered and left-dense separately, (8.115) yields 4 [x(t∗ )]2 ≤ [x(t∗ )]2 E−α (b, a)





b

b

β(s)Δs a

 2 γ + (s) x σ (s) Δs,

a



which clearly results in (8.106). Tκ

Theorem 8.62 (Lyapunov-Type Inequality) Let a, b ∈ with σ (a) ≤ b. Assume (8.93) and (8.105). If (8.88) has a real solution (x, u) such that x(a) = 0 and x ∗ (b) < 0, then 



σ (b)

β(t)Δt a

a

b

 γ (t)Δt ≥ 4 exp −

b

+

  ψμ(t) (−α(t)) Δt .

(8.116)

a

Proof We proceed as in the proof of Theorem 8.61 and arrive at (8.109). Replacing s by b in (8.107), we obtain

8.8 Planar Linear Hamiltonian Systems

569

 x(t) = e−α (b, t)x(b) −

b

(8.117)

e−α (τ, t)β(τ )u(τ )Δτ.

t

Multiplying the first equation in (8.88) by μ(t) and using the simple useful formula (see Theorem 8.9 (iv)), we obtain (1 − μ(t)α(t))x(σ (t)) = x(t) + β(t)μ(t)u(t).

(8.118)

Let kb := −

x ∗ (b) > 0. [x(b)]2

Then, (8.118) yields x(b) = −

1 β(b)μ(b)u(b), kb + 1

(8.119)

and hence (8.117) leads to x(t) = −

1 β(b)μ(b)u(b)e−α (b, t) − kb + 1



b

e−α (τ, t)β(τ )u(τ )Δτ,

t

and thus, by (8.99) and (8.98), we get 

1 β(b)μ(b)u(b) + |x(t)| ≤ E−α (b, t) kb + 1  σ (b) βb (τ )|u(τ )|Δτ ≤ E−α (b, t)





b

β(τ )|u(τ )|Δτ t

(8.120)

t

(use Theorem 8.10 (ii)), where ⎧ ⎪ ⎨β(t) βb (t) = 1 ⎪ ⎩ β(b) kb + 1

if

t = b,

if

t = b.

Note that since 1/(kb + 1) < 1, we have βb (t) ≤ β(t)

for all

t ∈ T.

(8.121)

As in the proof of Theorem 8.61, applying the arithmetic-geometric inequality with Q1 =

|x(t)| , E−α (t, a)

Q2 =

|x(t)| E−α (b, t)

570

8 Dynamic Equations on Time Scales

and using (8.103), (8.109), (8.120), (8.98), and the Cauchy–Schwarz inequality, we get 4[x(t)]2 ≤ E−α (b, a)



2

σ (b)

βb (τ )|u(τ )|Δτ a





σ (b)

≤

σ (b)

βb (τ )Δτ 

a



σ (b)

≤

(8.122)

a σ (b)

β(τ )Δτ a

βb (τ )[u(τ )]2 Δτ

βb (τ )[u(τ )]2 Δτ,

a

where we also have used (8.121). On the other hand, integrating (8.113) from a to b and using (8.119) yields 

b

β(τ )[u(τ )]2 Δτ +

a

1 β(b)μ(b)[u(b)]2 = kb + 1



b

 2 γ (τ ) x σ (τ ) (τ )Δτ,

a

and hence 

σ (b)



b

βb (τ )[u(τ )] Δτ = 2

a



 2 γ (τ ) x σ (τ ) (τ )Δτ

a b

≤

+



σ

(8.123)

2

γ (τ ) x (τ ) (τ )Δτ. a



Combining (8.122) and (8.123), we arrive at (8.116). Tκ

Theorem 8.63 (Lyapunov-Type Inequality) Let a, b ∈ with σ (a) ≤ b. Assume (8.93) and (8.105). If (8.88) has a real solution (x, u) such that x ∗ (a) < 0 and x(b) = 0, then (8.106) is satisfied. Proof As in the proof of Theorem 8.61, we see that (8.111) is satisfied. Replacing s by a in (8.107), we obtain  x(t) = e−α (a, t)x(a) +

t a

e−α (τ, t)β(τ )u(τ )Δτ.

(8.124)

Let ka := −

x ∗ (a) > 0. [x(a)]2

From (8.118), we have x(a) = −

1 β(a)μ(a)[u(a)]2 . ka + 1

(8.125)

8.8 Planar Linear Hamiltonian Systems

571

Using (8.125) in (8.124) gives  t 1 β(a)μ(a)u(a)e−α (a, t) + e−α (τ, t)β(τ )u(τ )Δτ ka + 1 a

 t 1 β(a)μ(a)u(a)e−α (a, t) + e−α (τ, t)β(τ )u(τ )Δτ = 1− ka + 1 σ (a)  t = e−α (τ, t)βa (τ )u(τ )Δτ,

x(t) = −

a

(8.126)

where βa (t) =

⎧ ⎪ ⎨β(t)

if

t = a

ka ⎪ ⎩ β(a) ka + 1

if

t = a.

Note that ka /(ka + 1) < 1 implies βa (t) ≤ β(t)

t ∈ T.

for all

(8.127)

From (8.126), using (8.100) and (8.98), we get  |x(t)| ≤ E−α (t, a)

t

βa (τ )|u(τ )|Δτ. a

As before, by employing the arithmetic-geometric inequality with |x(t)| , E−α (t, a)

Q1 =

Q2 =

|x(t)| E−α (b, t)

and then using the Cauchy–Schwarz inequality, we get 4[x(t)]2 ≤ E−α (b, a)



2

b

βb (τ )|u(τ )|Δτ a





b

≤

b

βa (τ )Δτ 

a

b

β(τ )Δτ a

(8.128)

a



b

≤

βa (τ )[u(τ )]2 Δτ

βa (τ )[u(τ )]2 Δτ,

a

where the last inequality follows from (8.127). Now, from (8.113), we see that  a

b

 2 γ (τ ) x σ (τ ) Δτ =

 a

b

β(τ )[u(τ )]2 Δτ −

1 β(a)μ(a)[u(a)]2 ka + 1

572

8 Dynamic Equations on Time Scales

 =

β(τ )[u(τ )] Δτ + 1 −

b

2

σ (a)

 =

b

1 β(a)μ(a)[u(a)]2 ka + 1

βa (τ )[u(τ )]2 Δτ,

a

and hence 

b



b

βa (τ )[u(τ )] Δτ ≤ 2

a

 2 γ + (τ ) x σ (τ ) Δτ.

(8.129)

a



Combining (8.128) and (8.129), we see that (8.106) holds. Tκ

Theorem 8.64 (Lyapunov-Type Inequality) Let a, b ∈ with σ (a) ≤ b. Assume (8.93) and (8.105). If (8.88) has a real solution (x, u) such that x ∗ (a) < 0 and x ∗ (b) < 0, then (8.116) is satisfied. Proof The proof can be easily accomplished by combining the arguments in the proofs of the last two theorems.  From Theorems 8.61, 8.62, 8.63, and 8.64, we easily deduce the following theorem. Theorem 8.65 (Lyapunov-Type Inequality) Let a, b ∈ Tκ with σ (a) ≤ b. Assume (8.93) and (8.105). If (8.88) has a real solution (x, u) with generalized zeros at a and b, and if x(t) = 0 for all t ∈ [a, b]T , then (8.116) is satisfied. By using similar arguments, we next show that (8.91) is valid without (8.89). The result follows from the following counterpart of Theorem 8.65. Since (8.93) is dropped, we deduce that (8.89) in Theorem 8.52 is superfluous. The proof is relatively less complicated because no exponential bound function is involved. The main difference is the use of  t  t x(t) = x(s) + α(τ )x(σ (τ ))Δτ + β(τ )u(τ )Δτ (8.130) s

s

instead of the variation of parameters formula given in (8.107). Equation (8.130) simply follows from integrating the first equation in (8.88). Theorem 8.66 (Lyapunov-Type Inequality) Let a, b ∈ Tκ with σ (a) ≤ b. Assume (8.90). If (8.88) has a real solution (x, u) with generalized zeros at a and b, and if x(t) = 0 for all t ∈ [a, b]T , then 

b a

 α(t)Δt +

1/2 

σ (b)

b

β(t)Δt a

+

γ (t)Δt

1/2 ≥ 2.

(8.131)

a

Proof We only give the proof when x(a) = 0 and x ∗ (b) < 0, i.e., the case contained in Theorem 8.62. From (8.130), we write

8.8 Planar Linear Hamiltonian Systems



t

x(t) =

573

 α(τ )x(σ (τ ))Δτ +

t

(8.132)

β(τ )u(τ )Δτ

a

a

and 

b

x(t) = x(b) −



b

α(τ )x(σ (τ ))Δτ −

(8.133)

β(τ )u(τ )Δτ.

t

t

From (8.132), we have 

t

|x(t)| =



t

|α(τ )||x(σ (τ ))|Δτ +

a

(8.134)

β(τ )|u(τ )|Δτ. a

As in the proof of Theorem 8.62, with kb defined as there, we obtain (8.119). Using (8.119) in (8.133) leads to 1 β(b)μ(b)u(b) − x(t) = − kb + 1





b

b

α(τ )x(σ (τ ))Δτ −

t

β(τ )u(τ )Δτ, t

and hence  b  b 1 β(b)μ(b)|u(b)| + |α(τ )||x(σ (τ ))||Δτ + β(τ )|u(τ )|Δτ kb + 1 t t  σ (b)  b |α(τ )||x(σ (τ ))||Δτ + βb (τ )|u(τ )|Δτ, ≤

|x(t)| ≤

t

t

(8.135) where βb is defined as in the proof of Theorem 8.62. Note that 1/(kb + 1) < 1 implies that (8.121) holds. By using (8.134) and (8.135), (8.121), and the Cauchy– Schwarz inequality, we have 

b

2|x(t)| ≤ 



βb (τ )|u(τ )|Δτ

a

a b

≤ a

σ (b)

|α(τ )||x(σ (τ ))||Δτ + |α(τ )||x(σ (τ ))||Δτ



+

(8.136)

1/2 

σ (b)

σ (b)

β(τ )Δτ a

1/2 βb (τ )[u(τ )]2 Δτ

.

a

On the other hand, (8.123) remains valid. In view of (8.136) and (8.123), we arrive at (8.131).  Remark 8.67 If (8.105) is replaced by β(t) ≥ 0 for all

t ∈ [a, b]T

574

8 Dynamic Equations on Time Scales

with β(t) ≡ 0

J ⊂ [a, b]T ,

on any subinterval

then (8.116) and (8.131) become strict. In case T = R, we thus recover [268, Theorem 2.4] from Theorems 8.65 and 8.66. Remark 8.68 Assume (8.89). If μ(t) = 0, then ψμ(t) (−α(t)) = −α(t), and if μ(t) > 0, then 1 log |1 − μ(t)α(t)| μ(t)

ψμ(t) (−α(t)) =

1 log(1 − μ(t)α(t)) μ(t)  1  log(1 − μ(t)α(t)) + μ(t)α(t) = −α(t) + μ(t) =

≤ −α(t) as log(1 + x) ≤ x

for all

x ≥ −1.

Hence, we conclude ψμ(t) (−α(t)) ≤ −α(t)

t ∈ T.

for all

(8.137)

In case of α(t) ≤ 0 for all t ∈ T, (8.89) is satisfied and (8.137) implies   ψμ(t) (−α(t)) ≤ |α(t)|, and so (8.116) implies 



σ (b)

β(t)Δt a

b

 γ (t)Δt ≥ 4 exp −

b

+

a

a

In view of (2 − η)2 < 4e−η for η ∈ (0, 2), by taking 

b

η= a

|α(t)|Δt,

|α(t)|Δt .

(8.138)

8.8 Planar Linear Hamiltonian Systems

575

we see that (8.131) follows from (8.138). So, we may say in this case that (8.116) is better than (8.131). In the special case T = R, (8.116) implies (8.131) in view of ψμ(t) (−α(t)) = −α(t).

8.8.3 Disconjugacy Criteria In this subsection, we give a simple application. Consider (8.88) on [a, b]T . Theorem 8.69 Let a, b ∈ Tκ with σ (a) ≤ b. Assume (8.93) and (8.105). If 



σ (b)

b

β(t)Δt a

 γ (t)Δt < 4 exp −

b

+

a

  ψμ(t) (−α(t)) Δt ,

(8.139)

a

then (8.88) is relatively disconjugate on [a, b]T . Proof Suppose that (8.88) is not relatively disconjugate on [a, b]T . Then, there exists a real solution (x, u) with x nontrivial and such that x(a) = 0 and that x has a next generalized zero at c ∈ (a, b]T . We have either x(c) = 0 or x ∗ (c) < 0. Applying Theorems 8.61 and 8.62, we see that 



σ (c)

c

β(t)Δt a

 γ + (t)Δt ≥ 4 exp −

a

c

  ψμ(t) (−α(t)) Δt ,

a

and hence 



σ (b)

β(t)Δt a

b

 γ (t)Δt ≥ 4 exp −

b

+

a

  ψμ(t) (−α(t)) Δt .

(8.140)

a



Inequalities (8.139) and (8.140) contradict each other. In a similar manner, we can prove the following theorem. Theorem 8.70 Let a, b ∈ Tκ with σ (a) ≤ b. Assume (8.93) and (8.105). If 

b a

 α(t)Δt +

1/2 

σ (b)

b

β(t)Δt a

+

γ (t)Δt

1/2 < 2,

a

then (8.88) is relatively disconjugate on [a, b]T . Remark 8.71 Note that the second-order equation (px Δ )Δ (t) + q(t)x σ = 0

(8.141)

576

8 Dynamic Equations on Time Scales

can be expressed as an equivalent Hamiltonian system of type (8.88) with α(t) ≡ 0,

β(t) =

1 , p(t)

γ (t) = q(t).

Therefore, one can easily rewrite the corresponding theorems for (8.141). Remark 8.72 In the special case T = R, our results coincide with the corresponding ones in [301], where additionally stability criteria are also given in connection with Lyapunov-type inequalities when the system is periodic. The stability problem for (8.88) on an arbitrary time scale has been studied in [302].

8.9 Nonlinear Dynamic Systems In 2002, Bohner et al. [48] investigated the second-order Sturm–Liouville dynamic equation x ΔΔ + q(t)x σ = 0

(8.142)

on a time scale T under the conditions x(a) = x(b) = 0 (a, b ∈ T with a < b) and q ∈ Crd (T, (0, ∞)) and showed that if x is a solution of (8.142) with maxt∈[a,b]T |x(t)| > 0, then 

b

q(t)Δt ≥

a

b−a , C

where [a, b]T = {t ∈ T : a ≤ t ≤ b} and C = max{(t − a)(b − t) : t ∈ [a, b]T }. When T is the set Z of the integers, (8.142) reduces to the linear difference equation Δ2 x(n) + q(n)x(n + 1) = 0.

(8.143)

In 1983, Cheng [83] showed that if a, b ∈ Z with 0 < a < b and x(n) is a solution of (8.143) satisfying x(a) = x(b) = 0 and maxn∈{a,a+1,...,b} |x(n)| > 0, then ⎧ 4(b − a) ⎪ ⎪ b−2 ⎨ (b − a)2 − 1

|q(n)| ≥ ⎪ ⎪ n=a ⎩ 4 b−a

if

b − a − 1 is even, (8.144)

if

b − a − 1 is odd.

The purpose of this section is to establish several Lyapunov inequalities for the nonlinear dynamic system

8.9 Nonlinear Dynamic Systems

577

⎧  p−2 ⎪ ⎨x Δ (t) = −A(t)x(σ (t)) − B(t)y(t)  B(t)y(t) ,

(8.145)

⎪ ⎩y Δ (t) = C(t)x(σ (t))|x(σ (t))|q−2 + AT (t)y(t)

on a given time scale interval [a, b]T (a, b ∈ T with σ (a) < b), where p, q ∈ (1, ∞) satisfy 1/p + 1/q = 1, A is a real n × n-matrix-valued function on [a, b]T such that I + μ(t)A(t) is invertible for all t ∈ T, B and C are two real symmetric n × n-matrix-valued functions on [a, b]T , B(t) being positive definite, AT (t) is the transpose of A(t), and x, y are two real n-dimensional vector-valued functions on [a, b]T . When n = 1 and p = q = 2, (8.145) reduces to 

x Δ (t) = u(t)x(σ (t)) + v(t)y(t),

(8.146)

y Δ (t) = −w(t)x(σ (t)) − u(t)y(t),

where u, v, w are real-valued rd-continuous functions on T satisfying v(t) ≥ 0 for all t ∈ T. In 2011, He et al. [161] obtained the following result. Theorem 8.73 (Lyapunov-Type Inequality) Suppose 1 − μ(t)u(t) > 0 for all t ∈ T and let a, b ∈ Tκ with σ (a) ≤ b. If (8.146) has a real solution (x, y) such that x(a) = 0 or

x(a)x(σ (a)) < 0,

x(b) = 0 or

x(b)x(σ (b)) < 0,

max |x(t)| > 0,

[a,b]T

then the inequality  a

b





σ (b)

|u(t)|Δt +

b

v(t)Δt a

1/2 w + (t)Δt

≥2

a

holds. In 2016, Liu et al. [204] obtained the following result. Theorem 8.74 (Lyapunov-Type Inequality) Let p = q = 2 and a, b ∈ T with σ (a) ≤ b. If (8.145) has a solution (x, y) such that x(a) = x(b) = 0

and

max x T (t)x(t) > 0,

[a,b]T

then, for any symmetric n × n-matrix-valued function C1 with C1 (t) − C(t) ≥ 0 for all t ∈ T, the inequalities

578

8 Dynamic Equations on Time Scales



b

 |C1 (t)|

a



b

b a



b

|A(t)|Δt +

a

a

 |B(s)| |eA (σ (t), s)| Δs Δt ≥ 4, 2

2 1/2      B(t) Δt

b

1/2 |C1 (t)| Δt

≥ 2,

a

and  a

b



σ (t) a

  |B(s)| |eA (σ (t), s)| Δs 2



b

× a



b

|B(s)| |eA (σ (t), s)| Δs 2

σ (t)

−1 |B(s)| |eA (σ (t), s)|2 Δs

|C1 (t)| Δt ≥ 1

hold. For some other related results on Lyapunov-type inequalities, see, for example, [14, 91, 143, 181, 224, 268].

8.9.1 Preliminaries and Auxiliary Results For any z ∈ Rn and any S ∈ Rn×n (the space of real n × n matrices), we write |z| =



zT z

and

|Sz| , \{0} |z|

|S| = max n z∈R

which are called the Euclidean norm of z and the matrix norm of S, respectively. It is obvious that, for any z ∈ Rn and U, V ∈ Rn×n , |U z| ≤ |U ||z| and

|U V | ≤ |U ||V |.

be the set of all symmetric real n × n matrices. We can show that, for any Let Rn×n s U ∈ Rn×n , s |U | =

max

|λI −U |=0

|λ| and

   2 U  = |U |2 .

A matrix S ∈ Rn×n is said to be positive definite (resp. semipositive definite), s written as S > 0 (resp. S ≥ 0) provided y T Sy > 0 (resp. y T Sy ≥ 0) for any y ∈ Rn with y = 0. If S is positive definite (resp. semipositive definite), then there exists a unique positive definite matrix (resp. semipositive definite matrix), written √ √ !2 as S, satisfying S = S.

8.9 Nonlinear Dynamic Systems

579

In this section, we establish Lyapunov inequalities for (8.145) possessing a solution (x, y) satisfying x(a) = x(b) = 0 and

max x T (t)x(t) > 0.

(8.147)

[a,b]T

We first present the following lemmas. Lemma 8.75 Let a, b ∈ T with a < b. Suppose that α, β, γ , δ ∈ R and p, q ∈ (1, ∞) with α/p+β/q = γ /p+δ/q = 1. Then, for any f, g ∈ Crd ([a, b]T , R\{0}), the inequality 

b



b

|f (t)g(t)|Δt ≤

a

1/p 

b

|f (t)| |g(t)| Δt α

γ

a

1/q |f (t)| |g(t)| Δt β

δ

a

holds. Proof Let M(t) = (|f (t)|α |g(t)|γ )1/p and N(t) = (|f (t)|β |g(t)|δ )1/q . Then, by the Hölder inequality, Theorem 8.11, we have 

b



b

|f (t)g(t)|Δt =

a

M(t)N(t)Δt a



b

≤

1/p  [M(t)] Δt

a



1/q [N(t)] Δt q

a b

=

b

p

1/p  |f (t)|α |g(t)|γ Δt

a

b

1/q |f (t)|β |g(t)|δ Δt

.

a



This completes the proof.

Remark 8.76 Let γ = 0 in Lemma 8.75. Then, for any f, g ∈ Crd ([a, b]T , R \ {0}), we obtain  a

b

1/q 

 |f (t)g(t)|Δt ≤

max |f (t)|

b

β

[a,b]T

1/p  |f (t)| Δt α

a

b

1/q |g(t)| Δt q

.

a

Lemma 8.77 (See [52, Theorem 5.27]) If A ∈ Crd (T, Rn×n ) with invertible I + μ(t)A(t), f ∈ Crd (T, Rn ), t0 ∈ T, and a ∈ Rn , then  x(t) = eA (t, t0 )a +

t

t0

eA (t, τ )f (τ )Δτ

is the unique solution of the initial value problem x Δ (t) = −A(t)x(σ (t)) + f (t),

x(t0 ) = a,

580

8 Dynamic Equations on Time Scales

where (A)(t) = −[I + μ(t)A(t)]−1 A(t) for any t ∈ Tκ , and eA (·, t0 ) is the unique matrix-valued solution of the initial value problem Y Δ (t) = (A)(t)Y (t),

Y (t0 ) = I. ∈

Lemma 8.78 (See [52, Theorem 5.3]) If A, B differentiable, then

Crd (T, Rn×n ) are Δ-

(AB)Δ = Aσ B Δ + AΔ B = AΔ B σ + AB Δ . Lemma 8.79 (See [204]) If f1 , f2 , . . . , fs are Δ-integrable on [a, b]T and x = (f1 , f2 , . . . , fs ), then ⎫1/2   ⎧ n  2 ⎫1/2  ⎧ n  b  b  ⎨ ⎬ ⎬ b b ⎨   2 x(t)Δt  = fi (t)Δt ≤ Δt = |x(t)|Δt. [fi (t)]   a  ⎩ ⎭ ⎭ a a ⎩i=1 a i=1

Lemma 8.80 (See [204]) If A1 , A2 ∈ Rn×n and A1 −A2 ≥ 0, then, for any x ∈ Rn , s  2 (x σ )T A2 x σ ≤ |A1 | x σ  .

8.9.2 Lyapunov-Type Inequalities In this section, we assume that α, β ∈ R and p, q ∈ (1, ∞) satisfy β 1 1 α + = + = 1, p q p q and we put     F (t, τ ) = |eA (σ (t), τ )|  B(τ ) ,   p−2 p     G(t) =  B(t)y(t) y T (t)B(t)y(t) =  B(t)y(t) , 

σ (t)

Φ(σ (t)) =

q/p [F (t, s)]α Δs

,

a

 Ψ (σ (t)) =

b

σ (t)

q/p [F (t, s)] Δs α

,

8.9 Nonlinear Dynamic Systems

581

P (t) = Φ(σ (t))Ψ (σ (t)) max [F (t, τ )]β a≤τ ≤σ (t)

max [F (t, τ )]β ,

σ (t)≤τ ≤b

Q(t) = Φ(σ (t)) max [F (t, τ )]β + Ψ (σ (t)) max [F (t, τ )]β . a≤τ ≤σ (t)

σ (t)≤τ ≤b

Theorem 8.81 (Hartman-Type Inequality) Let a, b ∈ T with σ (a) < b and C1 ∈ Rn×n with C1 (t) − C(t) ≥ 0. If (8.145) has a solution (x, y) with x, y ∈ s Crd (T, Rn ) satisfying (8.147) on the interval [a, b]T , then 

b a

P (t) |C1 (t)| Δt ≥ 1. Q(t)

(8.148)

Proof Since (x, y) is a solution of (8.145), we have  q−2 (y T x)Δ (t) = (x σ (t))T C(t)x σ (t) x σ (t) − G(t).

(8.149)

Integrating (8.149) from a to b and noting that x(a) = x(b) = 0, we obtain 

b

 G(t)Δt =

a

b

 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt.

a

Noting that B(t) > 0, we know that y T (t)B(t)y(t) ≥ 0, t ∈ [a, b]T . We claim that y T (t)B(t)y(t) ≡ 0, t ∈ [a, b]T . Indeed, if y T (t)B(t)y(t) ≡ 0, t ∈ [a, b]T , then  2    B(t)y(t) = y T (t)B(t)y(t) ≡ 0, which implies B(t)y(t) ≡ 0, t ∈ [a, b]T . Thus, the first equation in (8.145) reduces to x Δ (t) = −A(t)x σ (t),

x(a) = 0.

By Lemma 8.77, it follows that x(t) = eA (t, a) · 0 = 0, which is a contradiction to (8.147). Hence, we obtain  b  b  σ q−2 σ T σ x (t) (x (t)) C(t)x (t)Δt = G(t)Δt > 0, a

a

and it follows from Lemma 8.77 that, for t ∈ [a, b]T 

t

x(t) = − 

a t

=− b

 p−2   eA (t, τ )B(τ )y(τ )  B(τ )y(τ ) Δτ  p−2   eA (t, τ )B(τ )y(τ )  B(τ )y(τ ) Δτ,

(8.150)

582

8 Dynamic Equations on Time Scales

which implies 

σ (t)

x (t) = − σ

a



b

=

σ (t)

 p−2   eA (σ (t), τ )B(τ )y(τ )  B(τ )y(τ ) Δτ

 p−2   eA (σ (t), τ )B(τ )y(τ )  B(τ )y(τ ) Δτ

for t ∈ [a, b]T . For a ≤ σ (t) ≤ b, we have   p−2   eA (σ (t), τ )B(τ )y(τ )  B(τ )y(τ )     p−2   ≤ |eA (σ (t), τ )| |B(τ )y(τ )|  B(τ )y(τ )    p−2    ≤ F (t, τ )  B(τ )y(τ )  B(τ )y(τ ) = F (t, τ )[G(τ )]1/q . Then, by Remark 8.76 and Lemma 8.79, we obtain  q  p−2   σ q  σ (t)    x (t) =  eA (σ (t), τ )B(τ )y(τ )  B(τ )y(τ ) Δτ   a  1q 0   p−2  σ (t)      ≤ eA (σ (t), τ )B(τ )y(τ )  B(τ )y(τ )  Δτ a

0

1q

σ (t)

≤

F (t, τ )[G(τ )]

1/q

Δτ

a



σ (t)

≤

q/p  [F (t, τ )] Δτ

a

 [F (t, τ )] G(τ )Δτ β

a

 ≤

σ (t)

α

σ (t)

max [F (t, τ )]β

a≤τ ≤σ (t)

q/p  [F (t, τ )]α Δτ

a

σ (t)

 G(τ )Δτ ,

a

that is,  σ q x (t) ≤

 max [F (t, τ )]β Φ(σ (t))

a≤τ ≤σ (t)

σ (t)

G(τ )Δτ. a

(8.151)

8.9 Nonlinear Dynamic Systems

583

Similarly, for a ≤ σ (t) ≤ b, we have  σ q x (t) ≤

 max [F (t, τ )] Ψ (σ (t)) β

b

(8.152)

G(τ )Δτ.

σ (t)≤τ ≤b

σ (t)

By (8.151) and (8.152), we get  σ q x (t) ≤ P (t) Q(t)



b

G(τ )Δτ. a

Then, by (8.150) and Lemma 8.80, we have 

b

q  |C1 (t)| x σ (t) Δt ≤

a



b a



b

= a



b

=

P (t) |C1 (t)| Δt Q(t) P (t) |C1 (t)| Δt Q(t) |C1 (t)|

a

P (t) Δt Q(t)



b

G(t)Δt a

 

b

 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt

b

q  |C1 (t)| x σ (t) Δt.

a

a

Since 

b

q  |C1 (t)| x σ (t) Δt ≥

a



 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt=

b

a



b

G(t)Δt > 0, a

we get 

b a

|C1 (t)|

P (t) Δt ≥ 1. Q(t) 

This completes the proof.

Corollary 8.82 (Hartman-Type Inequality) Let a, b ∈ T with σ (a) < b and C1 ∈ Rn×n with C1 (t) − C(t) ≥ 0. If (8.145) has a solution (x, y) with x, y ∈ s Crd (T, Rn ) satisfying (8.147) on the interval [a, b]T , then 

b

Q(t) |C1 (t)| Δt ≥ 4.

a

Proof Note that P (t) Q(t) ≤ . Q(t) 4

584

8 Dynamic Equations on Time Scales

By (8.148), we have 

b

Q(t) |C1 (t)| Δt ≥ 1, 4

a

that is, 

b

Q(t) |C1 (t)| Δt ≥ 4.

a



This completes the proof.

Corollary 8.83 (Hartman-Type Inequality) Let a, b ∈ T with σ (a) < b and C1 ∈ Rn×n with C1 (t) − C(t) ≥ 0. If (8.145) has a solution (x, y) with x, y ∈ s Crd (T, Rn ) satisfying (8.147) on the interval [a, b]T , then 

b

 P (t) |C1 (t)| Δt ≥ 2.

a

Proof Note that  Q(t) ≤ 2 P (t). By (8.148), we get 



b



b

P (t) |C1 (t)| Δt ≥ 2

a

a

P (t) |C1 (t)| Δt ≥ 2. Q(t) 

This completes the proof.

Theorem 8.84 (Hartman-Type Inequality) Let a, b ∈ T with σ (a) < b and C1 ∈ Rn×n with C1 (t) − C(t) ≥ 0. If (8.145) has a solution (x, y) with x, y ∈ s Crd (T, Rn ) satisfying (8.147) on the interval [a, b]T , then there exists c ∈ (a, b) such that 

σ (c)

Φ(σ (t)) max [F (t, τ )]β |C1 (t)| Δt ≥ 1 a≤τ ≤σ (t)

a

and 

b

Ψ (σ (t)) max [F (t, τ )]β |C1 (t)| Δt ≥ 1.

c

σ (t)≤τ ≤b

Proof Set U (t) = Φ(σ (t)) max [F (t, τ )]β a≤τ ≤σ (t)

and

V (t) = Ψ (σ (t)) max [F (t, τ )]β . σ (t)≤τ ≤b

8.9 Nonlinear Dynamic Systems

585

Let 

t

f (t) =

 U (s) |C1 (s)| Δs −

a

b

V (s) |C1 (s)| Δs.

t

Then, we have f (a) < 0 and f (b) > 0. Hence, we can choose c ∈ (a, b) such that f (c) ≤ 0 and f (σ (c)) ≥ 0, that is, 

c



b

U (s) |C1 (s)| Δs ≤

a

V (s) |C1 (s)| Δs

c

and 

σ (c)



b

U (s) |C1 (s)| Δs ≥

a

V (s) |C1 (s)| Δs.

(8.153)

σ (c)

By (8.151), we have q  |C1 (t)| x σ (t) ≤ U (t) |C1 (t)|



σ (t)

(8.154)

G(τ )Δτ. a

Integrating (8.154) from a to σ (c), we obtain 

σ (c)

q  |C1 (t)| x σ (t) Δt ≤



a

σ (c)

 U (t) |C1 (t)|

G(τ )Δτ Δt

a



a c

≤



σ (t)

 U (t) |C1 (t)| Δt

a

σ (c)

G(τ )Δτ a



σ (t)

+ U (c) |C1 (c)| (σ (c) − c) 

σ (c)

=



σ (c)

U (t) |C1 (t)| Δt

a

G(τ )Δτ a

G(τ )Δτ. a

Similarly, we obtain from (8.151) and (8.153) that 

b

q  |C1 (t)| x σ (t) Δt ≤

σ (c)

 

b

 V (t) |C1 (t)| Δt

σ (c) σ (c)

≤ a

b

G(τ )Δτ σ (c)

 U (t) |C1 (t)| Δt

b

G(τ )Δτ. σ (c)

586

8 Dynamic Equations on Time Scales

This yields  b a

q  |C1 (t)| x σ (t) Δt ≤ = ≤

 σ (c) a

 σ (c) a

 σ (c) a

U (t) |C1 (t)| Δt

 b G(t)Δt a

 b  σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt

U (t) |C1 (t)| Δt U (t) |C1 (t)| Δt

a

 b a

q  |C1 (t)| x σ (t) Δt.

Since 

b

q  |C1 (t)| x σ (t) Δt ≥

a



b

 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt

a

 =

b

G(t)Δt > 0, a

we have 

σ (c)

U (t) |C1 (t)| Δt ≥ 1.

a

Next, from (8.152), we obtain q  |C1 (t)| x σ (t) ≤ V (t) |C1 (t)|



b

(8.155)

G(τ )Δτ. σ (t)

Integrating (8.155) from c to b, we have 

b

q  |C1 (t)| x σ (t) Δt ≤

a



b

 V (t) |C1 (t)|

G(τ )Δτ Δt

c



σ (t) b

≤



b



V (t) |C1 (t)| Δt

c

b

G(τ )Δτ. σ (c)

Similarly, we obtain 

c

q  |C1 (t)| x σ (t) Δt ≤

a







c

σ (c)

U (t) |C1 (t)| Δt

a

≤ This yields

c

G(τ )Δτ a

b

 V (t) |C1 (t)| Δt

σ (c)  σ (c)

G(τ )Δτ. a

a

8.9 Nonlinear Dynamic Systems



b

q  |C1 (t)| x σ (t) Δt ≤

a



587

b



c

 = 

G(t)Δt a

b

 V (t) |C1 (t)| Δt

c b

≤

b

V (t) |C1 (t)| Δt



 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt

b

q  |C1 (t)| x σ (t) Δt.

a

V (t) |C1 (t)| Δt

c

b

a

Thus, we have 

b

V (t) |C1 (t)| Δt ≥ 1.

c



This completes the proof.

Theorem 8.85 (Lyapunov-Type Inequality) Let a, b ∈ T with σ (a) < b and C1 ∈ Rn×n with C1 (t) − C(t) ≥ 0. If (8.145) has a solution (x, y) with x, y ∈ s Crd (T, Rn ) satisfying (8.147) on the interval [a, b]T , then 

b

 |A(t)|Δt+ max

a≤t≤b

a

 β 1/q     B(t)

b a

 α 1/p     B(t) Δt

b

1/q |C1 (t)| Δt

a

≥ 2. Proof Since x(a) = x(b) = 0, we have 

b

 G(t)Δt =

a

b

 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt.

a

It follows from the first equation in (8.145) that, for all a ≤ t ≤ b,

 t  p−2   σ −A(τ )x (τ ) − B(τ )  B(τ )y(τ ) x(t) = y(τ ) Δτ a

 =

t

b



 p−2   A(τ )x (τ ) + B(τ )  B(τ )y(τ ) y(τ ) Δτ. σ

Thus, we have  t 

p−2      −A(τ )x σ (τ ) − B(τ )  B(τ )y(τ ) |x(t)| =  y(τ ) Δτ  a

  t  p−2     σ  ≤ y(τ ) Δτ A(τ )x (τ ) + B(τ )  B(τ )y(τ ) a

588

8 Dynamic Equations on Time Scales

 t  p−2       Δτ ≤ B(τ )y(τ )  B(τ )y(τ )  a a  t  t       A(τ )x σ (τ ) Δτ + ≤  B(τ ) [G(τ )]1/q Δτ. 

t

  A(τ )x σ (τ ) Δτ +

a

a

Similarly, we have  |x(t)| ≤

b

  A(τ )x σ (τ ) Δτ +

t

 t

b

     B(τ ) [G(τ )]1/q Δτ.

Then, we obtain |x(t)| =

1 2

1 ≤ 2



b

  |A(t)| x σ (t) Δt +

a

0

b

  |A(t)| x σ (t) Δt +

a

×

b

a

0

b

 × 1 ≤ 2



      B(t) [G(t)]1/q Δt

 β 1/q   max  B(t)

a≤t≤b

 α 1/p     B(t) Δt

1/q 1

b

G(t)Δt

a

  |A(t)| x σ (t) Δt +

a

0

b

a



1 = 2



b



β 1/q     max  B(t)

a≤t≤b

 σ q−2 σ x (t) (x (t))T C(t)x σ (t)Δt

a

b

 α 1/p    B(t) Δt

1/q 1

a b

  |A(t)| x σ (t) Δt +

a

 × a

b



 β 1/q   max  B(t)

a≤t≤b

 α 1/p     B(t) Δt

b

q  |C1 (t)| x σ (t) Δt

1/q 1 .

a

Denote M = maxa≤t≤b |x(t)| > 0. Then,   b 1 M M≤ |A(t)|Δt 2 a  β 1/q     + max  B(t) a≤t≤b

a

b

 α 1/p    Mq  B(t) Δt

Dividing by M > 0 completes the proof.

b

1/q 1 |C1 (t)| Δt

.

a



8.10 Notes and References

589

8.10 Notes and References A short introduction to time scales calculus is provided in Sects. 8.1 and 8.2. In Sect. 8.3, we prove Theorem 8.3, and for the proof, several lemmas on quadratic functionals connected to (8.4) are needed. In time scales calculus, the concept of a zero of a function is replaced by a so-called generalized zero, and (as in the classical case), a Lyapunov inequality immediately leads to disconjugacy criteria as presented in Sect. 8.4. Two extensions, which we have not considered in this section, are the cases when q is not necessarily positive valued and when the endpoints are not necessarily zeros but generalized zeros. All results are taken from the paper by Bohner et al. [48]. In Sect. 8.5, a Lyapunov-type inequality for higher-order dynamic equations of the form (8.16) is established for an arbitrary time scale T. The results in this section are adopted from the paper [264] by Sun and Liu. In Sect. 8.6, some sufficient conditions for instability and stability to hold for second-order linear dynamic equations of the form (8.20) with periodic coefficients are obtained for periodic time scales. The results in this section are taken from the paper by Atıcı et al. [35]. In Sect. 8.7, linear Hamiltonian dynamic systems of the form (8.53) are considered, and these systems contain two scalar linear dynamic equations on an arbitrary time scale T. In Sect. 8.7.2, in addition to recalling the Cauchy–Schwarz inequality (see [13, 52]) on a time scale T, a disconjugacy criterion is established in order to illustrate an application of Lyapunov inequalities. All results in Sect. 8.7 are taken from the paper [179] by Jiang and Zhou. In Sect. 8.8, Lyapunov-type inequalities for planar linear Hamiltonian systems of the form (8.88) are established, where α, β, γ are real-valued rd-continuous functions defined on a given arbitrary time scale T. Recently, He et al. [161] obtained several Lyapunov-type inequalities for (8.88), which improved the earlier results given by Jiang and Zhou [179] and hence the related ones in [35, 48, 91, 142, 143]. Theorem 8.52 seems to be the best result for (8.88) thus far. In all Lyapunov-type inequalities given for (8.88) in the literature, (8.89) is a must. Motivated by the one given in [301] for the discrete case, Bohner and Zafer [54] showed that this condition can be completely dropped by introducing a new definition for a generalized zero. In Sect. 8.8.1, some properties of the time scale exponential function are given and some estimates for a time scale exponential bound function (see Definition 8.59) are derived as well. Lyapunov-type inequalities are given in Sect. 8.8.2. The last subsection is devoted to a simple application, namely disconjugacy criteria for linear Hamiltonian systems. All results in Sect. 8.8 are adopted from the paper [54] by Bohner and Zafer. In Sect. 8.9, several Lyapunov inequalities for nonlinear dynamic systems of the form (8.145) on a given time scale interval [a, b]T (a, b ∈ T with σ (a) < b) are established, where p, q ∈ (1, ∞) satisfy 1/p + 1/q = 1, A is a real n × n-matrixvalued function on [a, b]T such that I + μ(t)A(t) is invertible for all t ∈ T, B and C are two real symmetric n × n-matrix-valued functions on [a, b]T , B(t) being

590

8 Dynamic Equations on Time Scales

positive definite. All results in this section are taken from the paper by Sun et al. [265]. We refer to [11, 160, 201, 247, 256, 264, 265, 291, 304] for some further results related to Lyapunov inequalities on time scales. Some of the results presented in this chapter are, if T = Z, already contained in the chapter on difference equations, Chap. 7, as special cases. In order to address experts in difference equations who might not be familiar with time scales calculus, we included those results with self-contained proofs also in Chap. 7. For example, as a special case when T = Z, the Lyapunov inequality given in (8.6) for the Sturm– Liouville dynamic equation (8.4) satisfying the 2-point boundary conditions (8.5) reduces to inequalities (8.8) and (8.9) (see also (7.29), (7.30), (8.144) by Cheng [83]), which can be considered a generalization of the Lyapunov inequality given in (8.3) for the Sturm–Liouville (difference) equation (8.1) satisfying the 2-point boundary condition xa = xb = 0. The proof in the discrete case, i.e., T = N, is given by Cheng [83], we refer the reader to Sect. 7.2. The constant 4 in these Lyapunov inequalities also, as in the continuous case, cannot be replaced by a larger number. For the discrete case of linear Hamiltonian dynamic systems of the form (8.10), the monograph by Ahlbrandt and Peterson [22] is a handy reference for this theory via the discrete version of the functional (8.13). As given in Example 8.34, when T = Z, then (8.20) takes the form (8.29), which is the discrete version of the Sturm– Liouville equation with periodic coefficients, and the Lyapunov inequality for this equation satisfying the 2-point boundary condition x0 = xN = 0 is given in (8.29). This result was established by Atıcı and Guseinov [34], and it can be considered as an extension of the remarkable approach of Pachpattte [226] who obtained the Lyapunov-type inequality (7.19) for the Sturm–Lioville difference equation (7.18) with nonperiodic coefficients, see Theorem 7.21 in Sect. 7.3. We note that (8.53) covers the discrete Hamiltonian system (8.55) when T = Z, i.e., (7.53), see [22, 47, 143]. Furthermore, (8.53) extends these classical cases to many cases in between as well, such as so-called q-difference equations. The discrete versions of the Lyapunov inequalities (8.57), (8.59), (8.60), and (8.62) for Hamiltonian systems containing two linear dynamic equations (8.53) are those in (7.56), (7.61), (7.64), and (7.72) given in Sect. 7.5, respectively, and they also may be found in [143].

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301. A˘gacık Zafer. Discrete linear Hamiltonian systems: Lyapunov type inequalities, stability and disconjugacy criteria. J. Math. Anal. Appl., 396(2):606–617, 2012. 302. A˘gacık Zafer. The stability of linear periodic Hamiltonian systems on time scales. Appl. Math. Lett., 26(3):330–336, 2013. 303. Qi-Ming Zhang and Xiaofei He. Lyapunov-type inequalities for a class of even-order differential equations. J. Inequal. Appl., 2012:5, 7 pages, 2012. 304. Qi-Ming Zhang, Xiaofei He, and Jianchu Jiang. On Lyapunov-type inequalities for nonlinear dynamic systems on time scales. Comput. Math. Appl., 62(11):4028–4038, 2011. 305. Qi-Ming Zhang and Xianhua Tang. Lyapunov inequalities and stability for discrete linear Hamiltonian systems. Appl. Math. Comput., 218(2):574–582, 2011. 306. Qi-Ming Zhang and Xianhua Tang. Lyapunov inequalities and stability for discrete linear Hamiltonian systems. J. Difference Equ. Appl., 18(9):1467–1484, 2012. 307. Qi-Ming Zhang and Xianhua Tang. Lyapunov-type inequalities for even order difference equations. Appl. Math. Lett., 25(11):1830–1834, 2012. 308. Qi-Ming Zhang and Xianhua Tang. Lyapunov-type inequalities for the quasilinear difference systems. Discrete Dyn. Nat. Soc., Art. ID 860598, 16 pages, 2012. 309. Shuqin Zhang. Positive solutions for boundary-value problems of nonlinear fractional differential equations. Electron. J. Differential Equations, No. 36, 12 pages (electronic), 2006. 2 310. Nikola˘ı E. Zhukovski˘ı. Conditions for finiteness of integrals of the equation ddxy2 + py = 0. Mat. Sb., 16(3):582–591, 1892. In Russian.

Index

A Antiderivative, 40, 161, 518 Arithmetic-geometric mean inequality, 61, 175, 567, 569, 571 B Boundary conditions anti-periodic, 181, 183, 274 clamped, 87 clamped-free, 87, 94 conjugate, 118, 123 Dirichlet, 2, 56, 152, 233, 240, 254, 255, 275, 297, 299, 365, 466, 468 Lidstone, 56, 71, 80 Neumann, 322, 337 right-focal, 516, 528 Robin, 314 Bounded variation, 15, 16 C Cauchy–Schwarz inequality, 51, 96, 145, 416, 519, 553, 568 c-interval, 17 Concave, 62, 218, 232 Conjugate, 59, 61 Convex, 26, 61, 116, 117, 134, 196, 214, 218, 232 Convex arch, 26 Cylinder transformation, 563 D Delta differentiable, 517

Dirac delta function, 497 Disconjugate, 5, 23, 28, 136, 439, 490, 522, 528, 560 relatively, 563, 575 Disfocal, 28, 40

E Eigenvalue problem, 9, 44, 59, 70, 74, 113, 176, 247, 297, 313, 319, 323, 329, 378, 408 Emden–Fowler equation, 190, 191, 193, 345 discrete, 471, 480 Equation of motion, 524 Euler constant, 308

F Faber–Krahn inequality, 366, 375, 376 Floquet multipliers, 447, 449 Floquet theory, 3, 447, 538 Fractional calculus Caputo, 298, 306, 313 Riemann–Liouville, 294 Fraenkel asymmetry, 376 Fredhom alternative, 324

G Gamma function, 308, 347 Graininess, 517 Green’s function, 7, 45, 58, 60, 81, 119, 124, 295, 497, 505

© Springer Nature Switzerland AG 2021 R. P. Agarwal et al., Lyapunov Inequalities and Applications, https://doi.org/10.1007/978-3-030-69029-8

605

606 H Hartman inequality difference equation even order, 411 linear Hamiltonian system, 428, 432, 433 nonlinear system, 471, 474–476, 479 quasilinear system, 456, 461–463, 465, 466 dynamic equation on time scales nonlinear system, 581, 583, 584 linear differential equation even order, 59, 74, 77, 81 higher order, 124, 127 second order, 11 system of differential equations nonlinear anti-periodic, 275, 277, 278, 280, 282 quasilinear, 218, 222 quasilinear clamped-free, 288 quasilinear Dirichlet, 242, 245, 247, 248, 251, 253, 254 quasilinear elliptic, 257, 262–264, 267, 272 Hill’s equation, 1, 3, 535 Hölder conjugate, 21, 39, 104, 138, 139, 142, 210, 341, 472, 473, 506 Hölder inequality, 21, 40, 90, 135, 153, 192, 259, 333, 334, 357, 368, 371, 457, 464, 519, 579

J Jensen inequality, 134, 147, 196, 214, 218, 229, 232 Jump operator backward, 517 forward, 517

K Krasnosel ski˘ı genus, 378 Krasnosel ski genus, 155

L Left-dense, 517 Left disfocal, 28, 29 Left-scattered, 517 Limit circle, 22 Limit point, 22, 23 Lyapunov inequality difference equation even order, 412

Index linear Hamiltonian system, 417, 419, 420, 422, 425, 436–438 nonlinear system, 489 partial, 506 second-order linear, 384, 387–390, 392, 399, 402, 406 dynamic equation on time scales higher order, 531, 533 Hill’s equation, 545 linear Hamiltonian system, 526, 529, 550, 551, 562, 566, 568, 570, 572 nonlinear system, 577, 587 fractional differential equation linear, 296, 299, 312 half-linear differential equation higher order, 183 second order, 134, 139, 141, 144, 146, 147, 149, 151, 154 third order, 159, 161, 162, 165–170, 172, 177–179 linear differential equation even order, 61, 64, 71, 72, 76, 79, 82, 86, 87, 91, 93 higher order, 120, 122, 123, 125, 128 odd order, 114 second order, 3, 12 third order, 95, 96 partial differential equation elliptic, 363–365 linear, 327 multivariate, 348–350, 352–354, 356–359 nonlinear system, 341, 344, 345 system of differential equations linear Hamiltonian system, 414 nonlinear, 191, 193, 194, 197, 198, 201–204 quasilinear, 209, 211, 213, 214, 224 quasilinear Dirichlet, 226, 229, 234, 236–238 quasilinear elliptic, 268 Lyusternik–Shnirel man theory, 155, 378

M Minimax principle, 9, 54 Mittag–Leffler function, 297, 298, 305, 313 Monodromy matrix, 442 Morrey’s theorem, 362, 363, 367

N Net circular, 502, 503, 512

Index rectangular, 503, 505, 512 straight, 500, 501, 503, 512

O Opial inequality, 36, 136 Oscillation, 22, 113, 121

P Parseval’s relation, 14 p-Laplacian, 152, 255, 361, 375, 378, 454, 467 Preharmonic function, 496

R Rayleigh quotient, 375, 378 Rayleigh–Ritz theorem, 527 Rd-continuous, 518 Riccati equation, 4, 33, 36 Riemann zeta function, 355, 356, 358 Right-dense, 517 Right disfocal, 28, 29 Right-scattered, 517 Rolle’s theorem, 6, 97, 109, 110, 138, 192, 235

607 S Signed-power function, 158 Simple useful formula, 517 Sobolev exponent, 331 Sobolev inequality, 87, 88, 354, 355 Sobolev space, 340, 363 Steklov eigenvalue, 156 Sturm comparison theorem, 29, 33 Sturm–Liouville problem, 44, 297, 313, 319, 514, 576 Sturm majorant, 11

T Time scale, 516 exponential function, 563 periodic, 535

W Weakly oscillatory, 190, 203, 345, 469, 491 Wintner’s theorem, 22 Wirtinger inequality, 8, 13, 43