Functional Dynamic Equations on Time Scales [1st ed.] 978-3-030-15419-6;978-3-030-15420-2

Table of contents :
Front Matter ....Pages i-x
Calculus on Time Scales (Svetlin G. Georgiev)....Pages 1-36
Dynamic Systems (Svetlin G. Georgiev)....Pages 37-98
Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness Theorems (Svetlin G. Georgiev)....Pages 99-139
Linear Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 141-159
Stability for First-Order Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 161-254
Oscillations of First-Order Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 255-345
Oscillations of Second-Order Linear Functional Dynamic Equations with a Single Delay (Svetlin G. Georgiev)....Pages 347-374
Nonoscillations of Second-Order Functional Dynamic Equations with Several Delays (Svetlin G. Georgiev)....Pages 375-406
Oscillations of Second-Order Nonlinear Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 407-468
Oscillations of Third-Order Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 469-547
Oscillations of Fourth-Order Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 549-664
Oscillations of Higher-Order Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 665-748
Shift Operators (Svetlin G. Georgiev)....Pages 749-780
Impulsive Functional Dynamic Equations (Svetlin G. Georgiev)....Pages 781-817
Linear Impulsive Dynamic Systems (Svetlin G. Georgiev)....Pages 819-835
Back Matter ....Pages 837-885

Citation preview

Svetlin G. Georgiev

Functional Dynamic Equations on Time Scales

Functional Dynamic Equations on Time Scales

Svetlin G. Georgiev

Functional Dynamic Equations on Time Scales

123

Svetlin G. Georgiev Faculty of Mathematics & Informatics Sofia University St. Kliment Ohridski Sofia, Bulgaria Sorbonne University, Paris, France

ISBN 978-3-030-15419-6 ISBN 978-3-030-15420-2 (eBook) https://doi.org/10.1007/978-3-030-15420-2 Library of Congress Control Number: 2019935571 Mathematics Subject Classification: 39A10, 34A60, 34B37, 34C10, 34K20, 34K30 © Springer Nature Switzerland AG 2019 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, express 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

Preface

This book is devoted to qualitative theory of functional dynamic equations on time scales and summarizes the most recent contributions in this area. The book is intended for senior undergraduate students and beginning graduate students of engineering and science courses. There are fifteen chapters in this book, which in the book are pedagogically organized. Each chapter is concluded by a section with practical problems. In Chapter 1, the basic definitions of forward and backward jump operators are provided, as well as the graininess function. They are introduced the definitions for delta derivative and delta integral and they are given some of their properties. They are deducted some of the basic properties of the power series. Chapter 2 introduces dynamic systems on time scales. They are considered dynamic systems with constant coefficients. They are given some applications of the nonlinear dynamic equations on time scales to certain real life problems. Chapter 3 deals with the classification of functional dynamic equations, the method of steps, there are formulated and proved existence and uniqueness theorems. Continuous dependence on initial data is established. Chapter 4 discusses the properties of linear functional dynamic equations. Chapter 5 deals with the stability of the first-order functional dynamic equations. Oscillations of the first-order functional dynamic equations are investigated in Chapter 6. They are proved iterated oscillation criteria and criteria for nonoscillations of first-order functional dynamic equations with several delays. Chapter 7 discusses the oscillations of the second-order linear functional dynamic equations with a single delay. They are given criteria for oscillations using the Riccati transformation technique and Kamenev-type oscillation criteria. Nonoscillations of second-order functional dynamic equations with several delays are dealt with in Chapter 8. Chapter 9 is devoted to oscillations of the second-order nonlinear functional dynamic equations. Oscillation criteria for the third-order and fourthorder functional dynamic equations are discussed in Chapter 10 and Chapter 11, respectively. Chapter 12 is devoted to oscillation criteria for higher-order functional dynamic equations. They are deducted some Kamenev-type oscillation criteria. Chapter 13 deals with shift operators and introduces the delay function which proved some criteria for stability and instability. Chapter 14 is devoted to impulsive v

vi

Preface

dynamic equations on time scales. They are given criteria for oscillations of first-order, second-order, and higher-order impulsive dynamic equations. Linear impulsive dynamic systems are introduced in Chapter 15. Some criteria for stability of the solutions of homogeneous and nonhomogeneous linear impulsive dynamic systems are deducted. This book is addressed to a wide audience of specialists, such as mathematicians, physicists, engineers, and biologists. It can be used as a textbook at the graduate level and as a reference book for several disciplines. Finally, the author would like to thank Dr. Ba¸sak Karpuz, Afyon Kocatepe University, Turkey, for a careful reading of the manuscript. Paris, France August 2018

Svetlin G. Georgiev

Contents

1

Calculus on Time Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Forward and Backward Jump Operators, Graininess Function . . . . 1.2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Hyperbolic and Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 5 14 19 28 29 35

2

Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Structure of Dynamic Systems on Time Scales . . . . . . . . . . . . . . . . . . . . . 2.2 Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 The Logistic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Tumor Growth Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 The Predator-Prey Model of Lotka-Volterra . . . . . . . . . . . . . . . . 2.3.4 The Ricker Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.5 The Beverton-Holt Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.6 Competition Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.7 Single Species Population Growth Models . . . . . . . . . . . . . . . . . 2.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 70 83 83 85 89 90 91 92 95 97

3

Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Classification of Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . 3.2 The Method of Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 The Picard-Lindelöf Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Existence and Uniqueness Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Continuous Dependence on Initial Data. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99 99 103 123 131 135 137

vii

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Contents

4

Linear Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Some Properties of Linear Functional Dynamic Equations . . . . . . . . 4.2 First-Order Linear Functional Dynamic Equations . . . . . . . . . . . . . . . . . 4.3 Second-Order Linear Functional Dynamic Equations . . . . . . . . . . . . . . 4.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

141 141 146 152 158

5

Stability for First-Order Functional Dynamic Equations . . . . . . . . . . . . . . 5.1 Uniform Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Uniformly Asymptotical Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Global Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Asymptotic Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Exponential Stability I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Exponential Stability II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161 161 177 183 201 223 232 253

6

Oscillations of First-Order Functional Dynamic Equations . . . . . . . . . . . 6.1 Positive Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations with Several Delays. . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Nonoscillations of First-Order Functional Dynamic Equations with Several Delays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

255 255

7

8

286 318 328 343

Oscillations of Second-Order Linear Functional Dynamic Equations with a Single Delay. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Reduction to First-Order Linear Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 An Oscillation Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Riccati Transformation Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Kamenev-Type Oscillation Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

347 353 355 364 373

Nonoscillations of Second-Order Functional Dynamic Equations with Several Delays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Representation of the Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 A Nonoscillation Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Comparison Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Explicit Nonoscillation and Oscillation Results . . . . . . . . . . . . . . . . . . . . 8.5 Positive Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

375 375 378 388 397 403 405

347

Contents

9

10

11

12

13

Oscillations of Second-Order Nonlinear Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay Dynamic Equations and a Class of Second-Order Nonlinear Dynamic Equations . . . . . . . . . . . . . . . . . . . 9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic Equations I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic Equations II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Oscillation of Second-Order Nonlinear Neutral Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

407

407 421 438 450 461 466

Oscillations of Third-Order Functional Dynamic Equations . . . . . . . . . . 10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with Distributed Deviating Arguments . . . . . . . . . . . . . . . . . . 10.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

469

Oscillations of Fourth-Order Functional Dynamic Equations . . . . . . . . 11.1 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a Nonlinear Middle Term . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

549

Oscillations of Higher-Order Functional Dynamic Equations . . . . . . . . 12.1 Oscillations of Higher-Order Delay Dynamic Equations . . . . . . . . . . 12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Advanced Practical Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

665 665

Shift Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Definition for Shift Operators. Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 The Delay Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Stability Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

749 749 757 761

469 508 528 546

549 584 609 662

698 717 737 747

x

Contents

13.4 A Criterion for Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775 13.5 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779 14

15

Impulsive Functional Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Existence of Solutions for First-Order Nonlinear Impulsive Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Stability of the Solutions of First-Order Impulsive Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Oscillations of First-Order Impulsive Dynamic Equations . . . . . . . . . 14.4 Oscillations of Second-Order Impulsive Dynamic Equations. . . . . . 14.5 Oscillations of Higher-Order Linear Impulsive Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

781 781 791 795 800 808 816

Linear Impulsive Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Homogeneous Linear Impulsive Dynamic Systems . . . . . . . . . . . . . . . . 15.2 Nonhomogeneous Linear Impulsive Dynamic Systems . . . . . . . . . . . . 15.3 Boundedness and Stability of Linear Impulsive Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

819 819 827

A

Fréchet Derivatives and Gâteaux Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . A.1 Remainders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Definition and Uniqueness of the Fréchet Derivative. . . . . . . . . . . . . . . A.3 The Gâteaux Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

837 837 840 847

B

Pötzsche’s Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 B.1 Measure Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 B.2 Pötzsche’s Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851

C

The Knaster-Tarski Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855

D

Some Elementary Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 857

E

Knesser’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865

829 834

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 883

Chapter 1

Calculus on Time Scales

This chapter is devoted to a brief exposition of the time scale calculus. A detailed discussion of the time scale calculus is beyond the scope of this book; for this reason the author confines to outlining a minimal set of properties needed in the further proceeding. The presentation in this chapter follows the books and papers [49, 50, 52, 53, 56, 133, 134, 144–146, 150, 158].

1.1 Forward and Backward Jump Operators, Graininess Function Definition 1.1. A time scale is an arbitrary nonempty closed subset of the real numbers. We will denote a time scale by the symbol T. We suppose that a time scale T has the topology that inherits from the real numbers with the standard topology. Example 1.1. [1, 2],

R,

N are time scales.

Definition 1.2. For t ∈ T we define the forward jump operator σ : T → T as follows σ (t) = inf{s ∈ T : s > t}. We note that σ (t) ≥ t for any t ∈ T. Definition 1.3. For t ∈ T we define the backward jump operator ρ : T → T by ρ(t) = sup{s ∈ T : s < t}. We note that ρ(t) ≤ t for any t ∈ T. © Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_1

1

2

1 Calculus on Time Scales

Definition 1.4. We set inf Ø = sup T,

sup Ø = inf T.

Definition 1.5. For t ∈ T we have the following cases. 1. 2. 3. 4. 5.

If σ (t) > t, then we say that t is right-scattered. If t < sup T and σ (t) = t, then we say that t is right-dense. If ρ(t) < t, then we say that t is left-scattered. If t > inf T and ρ(t) = t, then we say that t is left-dense. If t is left-scattered and right-scattered at the same time, then we say that t is isolated. 6. If t is left-dense and right-dense at the same time, then we say that t is dense. √ √ Example 1.2. Let T = { 2n + 1 : n ∈ N}. If t = 2n + 1 for some n ∈ N, then n=

and

For n = 1 we have

t2 − 1 2

√ √ σ (t) = inf{l ∈ N : 2l + 1 > 2n + 1} √ = √2n + 3 = t 2 + 2 for n ∈ N, √ √ ρ(t) = sup{l ∈ N : 2l + 1 < 2n + 1} √ = √2n − 1 = t 2 − 2 for n ∈ N, n ≥ 2. √ ρ( 3) = sup Ø = inf T √ = 3.

Since

  t2 − 2 < t < t2 + 2 for n ≥ 2, √ 2n + 1, n ∈ N, n ≥ 2, is right-scattered and leftwe conclude that every point √ scattered, i.e., every point 2n + 1, n ∈ N, n ≥ 2, is isolated. Because √ √ 3 = ρ( 3) √ < σ ( 3) √ = 5, √ we have that the point 3 is right-scattered.

1.1 Forward and Backward Jump Operators, Graininess Function

3

√ 3 Exercise 1.1. Classify each point t ∈ T = { 2n − 1 : n ∈ N0 } as left-dense, leftscattered, right-dense, or right-scattered. Definition 1.6. The numbers H0 = 0,

Hn =

n  1 k=1

k

,

n∈N

will be called harmonic numbers. Exercise 1.2. Let H = {Hn : n ∈ N0 }. Prove that H is a time scale. Find σ (t) and ρ(t). Definition 1.7. The graininess function μ : T → [0, ∞) is defined by μ(t) = σ (t) − t.   Example 1.3. Let T = 2n+1 : n ∈ N . Let also, t = 2n+1 ∈ T for some n ∈ N. Then   σ (t) = inf 2l+1 : 2l+1 > 2n+1 , l ∈ N = 2n+2 = 2t. Hence, μ(t) = σ (t) − t = 2t − t =t or  μ 2n+1 = 2n+1 , Exercise 1.3. Let T =

√ 3

n ∈ N.

 n + 2 : n ∈ N0 . Find μ(t), t ∈ T.

4

1 Calculus on Time Scales

Definition 1.8. If f : T → R is a function, then we define the function f σ : T → R by f σ (t) = f (σ (t)) for any

t ∈ T,

i.e.,

f σ = f ◦ σ.

  Example 1.4. Let T = t = 2n+2 : n ∈ N , f (t) = t 2 + t − 1. Then   σ (t) = inf 2l+2 : 2l+2 > 2n+2 , l ∈ N = 2n+3 = 2t,

t ∈ T.

Hence, f σ (t) = f (σ (t)) = (σ (t))2 + σ (t) − 1 = (2t)2 + 2t − 1

Exercise 1.4. Let f (σ (t)), t ∈ T.

= 4t 2 + 2t − 1, t ∈ T.   √ 3 T = t = n + 2 : n ∈ N , f (t) = 1 − t 3 ,

Definition 1.9. We define the set

T\(ρ(sup T), sup T] κ T = T otherwise.

Example 1.5. Let T =

if

sup T < ∞

 1 : n ∈ N ∪ {0}. Then sup T = 1 and n

ρ(1) = sup =

 1 1 , 0 : , 0 < 1, l ∈ N l l

1 . 2

Therefore 1 Tκ = T\ , 1 2 

1 : n ∈ N, n ≥ 2 ∪ {0}. = n

t ∈ T.

Find

1.2 Differentiation

5

Definition 1.10. We assume that a ≤ b. We define the interval [a, b] in T by [a, b] = {t ∈ T : a ≤ t ≤ b}. Open intervals, half-open intervals, and so on are defined accordingly. Example 1.6. Let [a, b] be an interval in T and b be a left-dense point. Then sup[a, b] = b and since b is a left-dense point, we have that ρ(b) = b. Hence, [a, b]κ = [a, b]\(b, b] = [a, b]\Ø = [a, b].

Exercise 1.5. Let T =

 1 : n ∈ N ∪ {0}. Find Tκ . 2n + 1

1.2 Differentiation Definition 1.11. Assume that f : T → R is a function and let t ∈ Tκ . We define f Δ (t)to be the number, provided it exists, as follows: for any  > 0 there is a neighborhood U of t, U = (t − δ, t + δ) ∩ T for some δ > 0, such that |f (σ (t)) − f (s) − f Δ (t)(σ (t) − s)| ≤ |σ (t) − s|

for

all

s ∈ U,

s = σ (t).

We say f Δ (t) is the delta or Hilger derivative of f at t. We say that f is delta or Hilger differentiable, shortly differentiable, in T κ if f Δ (t) exists for all t ∈ Tκ . The function f Δ : T → R is said to be delta derivative or Hilger derivative, shortly derivative, of f in T κ . Remark 1.1. If T = R, then the delta derivative coincides with the classical derivative. Theorem 1.1. The delta derivative is well defined. Remark 1.2. Let us assume that sup T < ∞ and f Δ (t) is defined at a point t ∈ T\Tκ with the same definition as given in Definition 1.11. Then the unique point t ∈ T\Tκ is sup T. Hence, for any  > 0 there is a neighborhood U = (t − δ, t + δ) ∩ (T\Tκ ), for some δ > 0, such that f (σ (t)) = f (s) = f (σ (sup T)) = f (sup T), s ∈ U.

6

1 Calculus on Time Scales

Therefore for any α ∈ R and s ∈ U we have |f (σ (t)) − f (s) − α(σ (t) − s)| = |f (sup T) − f (sup T) − α(sup T − sup T)| ≤ |σ (t) − s|, i.e., any α ∈ R is the delta derivative of f at the point t ∈ T\Tκ . Example 1.7. Let f (t) = α ∈ R. We will prove that f Δ (t) = 0 for any t ∈ Tκ . Really, for t ∈ Tκ and for any  > 0 there is a δ > 0 such that s ∈ (t − δ, t + δ) ∩ T, s = σ (t), implies |f (σ (t)) − f (s) − 0(σ (t) − s)| = |α − α| ≤ |σ (t) − s|. Exercise 1.6. Let f (t) = t ∈ Tκ , t > 0.

√ 1 t, t ∈ T, t > 0. Prove that f Δ (t) = √ for √ t + σ (t)

Theorem 1.2. Assume f : T → R is a function and let t ∈ Tκ . Then we have the following. 1. If f is differentiable at t, then f is continuous at t. 2. If f is continuous at t and t is right-scattered, then f is differentiable at t with f Δ (t) =

f (σ (t)) − f (t) . μ(t)

3. If t is right-dense, then f is differentiable iff the limit lim

s−→t

f (t) − f (s) t −s

exists as a finite number. In this case f Δ (t) = lim

s−→t

f (t) − f (s) . t −s

4. If f is differentiable at t, then f (σ (t)) = f (t) + μ(t)f Δ (t).

1.2 Differentiation

7

 1 : n ∈ N0 ∪ {0}, f (t) = σ (t), t ∈ T. We will find Example 1.8. Let T = 2n + 1 1−t 1 ,n= , n ≥ 1, we have f Δ (t), t ∈ T. For t ∈ T, t = 2n + 1 2t

 1 1 1 σ (t) = inf ,0 : ,0 > , l ∈ N0 2l + 1 2l + 1 2n + 1

= =

1 2n − 1 1 2 1−t 2t − 1

t 1 − 2t > t,

=

i.e., any point t =

1 , n ≥ 1, is right-scattered. At these points 2n + 1 f Δ (t) =

f (σ (t)) − f (t) σ (t) − t

=

σ (σ (t)) − σ (t) σ (t) − t

=2

(σ (t))2 (1 − 2σ (t))(σ (t) − t)  2

= 2

=2 =2 =

t 1−2t

1−

2t 1−2t



t 1−2t

−t



t2 (1−2t)2 1−4t 2t 2 1−2t 1−2t

t2 2t 2 (1 − 4t)

1 . 1 − 4t

Let n = 0, i.e., t = 1. Then

1 1 σ (1) = inf ,0 : , 0 > 1, l ∈ N0 2l + 1 2l + 1 = inf Ø



8

1 Calculus on Time Scales

= sup T = 1, i.e., t = 1 is a right-dense point. Also, lim

s→1

f (1) − f (s) σ (1) − σ (s) = lim s→1 1−s 1−s = lim

1−

s 1−2s

1−s 1 − 3s = lim s→1 (1 − s)(1 − 2s) = +∞. s→1

Therefore σ (1) doesn’t exist. Let now t = 0. Then

σ (0) = inf

1 1 ,0 : , 0 > 0, l ∈ N0 2l + 1 2l + 1



= 0. Consequently t = 0 is right-dense. Also, σ (h) − σ (0) = lim h−→0 h−→0 h lim

Therefore σ (0) = 1. Exercise 1.7. Let T = { 1. 2. 3. 4.

h 1−2h

−0

h 1 = lim h−→0 1 − 2h = 1.

√ 2n + 3 : n ∈ N0 }. Find f Δ (t), t ∈ Tκ , if

11

f (t) = t 3 + t, f (t) = tt+1 4 +1 , √ 3 f (t) = t + 4, t+1 f (t) = √ , 7 t+t+10

5. f (t) = t 4 − 3t 3 − t +

√

t.

Theorem 1.3. Assume f, g : T → R are differentiable at t ∈ Tκ . Then 1. the sum f + g : T → R is differentiable at t with (f + g)Δ (t) = f Δ (t) + g Δ (t).

1.2 Differentiation

9

2. for any constant α, αf : T → R is differentiable at t with (αf )Δ (t) = αf Δ (t). 3. if f (t)f (σ (t)) = 0, we have that

1 : T → R is differentiable at t and f

Δ 1 f Δ (t) . (t) = − f f (t)f (σ (t)) 4. if g(t)g(σ (t)) = 0, we have that

f : T → R is differentiable at t with g

Δ f f Δ (t)g(t) − f (t)g Δ (t) (t) = . g g(t)g(σ (t)) 5. the product f g : T → R is differentiable at t with (f g)Δ (t) = f Δ (t)g(t) + f (σ (t))g Δ (t) = f (t)g Δ (t) + f Δ (t)g(σ (t)). Example 1.9. Let f, g, h : T → R be differentiable at t ∈ Tκ . Then (f gh)Δ (t) = ((f g)h)Δ (t) = (f g)Δ (t)h(t) + (f g)(σ (t))hΔ (t) = (f Δ (t)g(t) + f (σ (t))g Δ (t))h(t) + f σ (t)g σ (t)hΔ (t) = f Δ (t)g(t)h(t) + f σ (t)g Δ (t)h(t) + f σ (t)g σ (t)hΔ (t).  κ 2 Definition 1.12. Let f : T → R and t ∈ T κ = Tκ . We define the second derivative of f at t, provided it exists, by  Δ 2 2 f Δ = f Δ : Tκ → R. n

n

Similarly we define higher-order derivatives f Δ : Tκ → T. (n)

Theorem 1.4 (Leibniz Formula). Let Sk be the set consisting of all possible strings of length n, containing exactly k times σ and n − k times Δ. If fΛ

exists

for

all Λ ∈ Sk(n) ,

10

1 Calculus on Time Scales

then

⎛ n

(f g)Δ =

n  k=0

⎞

⎜  Λ ⎟ Δk f ⎠g . ⎝ (n)

Λ∈Sk

Example 1.10. Let μ is differentiable at t ∈ Tκ and t is right-scattered. Then  σ f Δσ (t) = f Δ (t) 

f (σ (t)) − f (t) σ = σ (t) − t =

f (σ (σ (t))) − f (σ (t)) σ (σ (t)) − σ (t)

=

f (σ (σ (t))) − f (σ (t)) σ (t) − t

 Δ = f σ (t) = f σ Δ (t)

1 σ (σ (t))−σ (t) σ (t)−t

1 σ Δ (t)

1 , 1 + μΔ (t)

i.e., f σ Δ (t) = (1 + μΔ (t))f Δσ (t). Also,  σ Δ (t) f σ σ Δ (t) = f σ

σ  = (1 + μΔ (t)) f σ Δ (t)

= (1 + μΔ (t))f σ Δσ (t),  Δσ f σ Δσ (t) = f σ (t)   σ Δ (t) = fσ  σ = (1 + μΔ (t))(f Δ )σ (t)   = 1 + μΔσ (t) f Δσ σ (t). Theorem 1.5. (Chain Rule) Assume g : R → R is continuous, g : T → R is delta differentiable on Tκ , and f : R → R is continuously differentiable. Then there exists c ∈ [t, σ (t)] with (f ◦ g)Δ (t) = f (g(c))g Δ (t).

1.2 Differentiation

11

Example 1.11. Let T = Z, f (t) = 2t 2 + 1, g(t) = t 2 . We have that g : R → R is continuous, g : T → R is delta differentiable on Tκ , f : R → R is continuously differentiable, and σ (t) = t + 1. Then g Δ (t) = σ (t) + t, (f ◦ g)Δ (1) = f (g(c))g Δ (1) = 4g(c)(σ (1) + 1) = 4g(c)(2 + 1) = 12g(c) = 12c2 . Here c ∈ [1, σ [1]] = [1, 2]. Also, f ◦ g(t) = f (g(t)) = 2(g(t))2 + 1 = 2(t 2 )2 + 1 = 2t 4 + 1,  (f ◦ g)Δ (t) = 2 (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3  = 2 (t + 1)3 + t (t + 1)2 + t 2 (t + 1) + t 3  = 2 t 3 + 3t 2 + 3t + 1 + t (t 2 + 2t + 1) + t 3 + t 2 + t 3  = 2 3t 3 + 4t 2 + 3t + 1 + t 3 + 2t 2 + t  = 2 4t 3 + 6t 2 + 4t + 1 , (f ◦ g)Δ (1) = 2(4 + 6 + 4 + 1) = 30. Hence and from (1.1), we get 30 = 12c2

or

c2 =

and  c=

5 ∈ [1, 2]. 2

5 , 2

(1.1)

12

1 Calculus on Time Scales

Exercise 1.8. Let T = Z, f (t) = 3t 2 − t + 1, and g(t) = 2t 3 + t 2 . Find a constant c ∈ [1, σ (1)] such that (f ◦ g)Δ (1) = f (g(c))g Δ (1). Theorem 1.6 (Chain Rule). Assume v : T → R is strictly increasing and T˜ = ˜ v(T ) is a time scale. Let w : T˜ → R. If v Δ (t) and w Δ (v(t)) exist for t ∈ Tκ , then ˜

(w ◦ v)Δ = (w Δ ◦ v)v Δ .   Example 1.12. Let T = 22n : n ∈ N0 , v(t) = t 2 , and w(t) = t 2 + 1. Then   v : T → R is strictly increasing, T˜ = v(T ) = 24n : n ∈ N0 is a time scale. For t ∈ T, t = 22n , n ∈ N0 , we have   σ (t) = inf 22l : 22l > 22n , l ∈ N0 = 22n+2 = 4t, v (t) = σ (t) + t Δ

= 5t. ˜ t = 24n , and n ∈ N0 , we have For t ∈ T,   σ˜ (t) = inf 24l : 24l > 24n , l ∈ N0 = 24n+4 = 16t. Also, for t ∈ T, we have (w ◦ v)(t) = w(v(t)) = (v(t))2 + 1 = t 4 + 1, (w ◦ v)Δ (t) = (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 = 64t 3 + 16t 3 + 4t 3 + t 3 = 85t 3 , ˜

w Δ ◦ v(t) = σ˜ (v(t)) + v(t)

1.2 Differentiation

13

= 16v(t) + v(t) = 17v(t) = 17t 2 ,  ˜ w Δ ◦ v(t) v Δ (t) = 17t 2 (5t) = 85t 3 . Consequently ˜

(w ◦ v)Δ (t) = (w Δ ◦ v(t))v Δ (t),

t ∈ Tκ .

  Exercise 1.9. Let T = 28n : n ∈ N0 , v(t) = t 4 + 7, and w(t) = t 2 . Prove ˜

(w ◦ v)Δ (t) = (w Δ ◦ v(t))v Δ (t),

t ∈ Tκ .

Theorem 1.7 (Derivative of the Inverse). Assume v : T → R is strictly increasing and T˜ := v(T) is a time scale. Then 1

˜

(v −1 )Δ ◦ v(t) =

v Δ (t)

for any t ∈ Tκ such that v Δ (t) = 0. Example 1.13. Let T = 4N0 , v(t) = 3t 2 + 2t + 3. Then σ (t) = t + 4,

t ∈ T,

v : T → R is strictly increasing and v Δ (t) = 3(σ (t) + t) + 2 = 3(t + 4 + t) + 2 = 3(2t + 4) + 2 = 6t + 12 + 2 = 6t + 14,

t ∈ T.

Hence, ˜

(v −1 )Δ ◦ v(t) = =

1 v Δ (t) 1 , 6t + 14

t ∈ T.

14

1 Calculus on Time Scales

Exercise 1.10. Let T = {2n + 19 : n ∈ N0 } and v(t) = 3t 2 + 4t − 15. Find Δ˜  v −1 ◦ v(t).

1.3 Integration Definition 1.13. A function f : T → R is called regulated provided its right-sided limits exist (finite) at all right-dense points in T and its left-sided limits exist (finite) at all left-dense points in T. Example 1.14. Let T = N and f (t) =

3t 2 + 7t + 5 , t −1

g(t) =

2t 4 + t 3 + t + 7 , 3t 8 + 19

t ∈ T.

We note that all points of T are right-scattered. The points t ∈ T, t = 1, are leftscattered. The point t = 1 is left-dense. Also, lim f (t) is not finite and lim g(t) t−→1−

t−→1−

exists and it is finite. Therefore the function f is not regulated and the function g is regulated. Exercise 1.11. Let T = R and  t 2 + t + 1 for t = 4 f (t) = t 7 +t 5 +t 3 +t+1 for t ∈ R\{4}. 2t−8 Determine if f is regulated. Definition 1.14. A continuous function f : T → R is called pre-differentiable with region of differentiation D, provided 1. D ⊂ Tκ , 2. Tκ \D is countable and contains no right-scattered elements of T, 3. f is differentiable at each t ∈ D. Example 1.15. Let

T = Pa,b =

∞ 

[k(a + b), k(a + b) + a]

k=0

for

a > b > 0,

f : T → R be defined by ⎧ ∞ ⎨ 0 if t ∈ k=0 [k(a + b), k(a + b) + b] f (t) = ⎩ t − (a + b)k − b if t ∈ [(a + b)k + b, (a + b)k + a]. Then f is pre-differentiable with D\

∞  k=0

{(a + b)k + b}.

1.3 Integration

15

Exercise 1.12. Let T = R and ⎧ ⎪ ⎨ 4 if t = −4 f (t) = t + 7 if t = 4 ⎪ ⎩ t 3 +t 2 +t+1 if R\{−4, 4}. t 2 −16 Check if f : T → R is pre-differentiable and if it is, find the region of differentiation. Definition 1.15. A function f : T → R is called rd-continuous provided it is continuous at right-dense points in T and its left-sided limits exist (finite) at leftdense points in T. The set of rd-continuous functions f : T → R will be denoted by Crd (T). The set of functions f : T → R that are differentiable and whose derivative is rd1 continuous is denoted by Crd (T). If f is continuous on T and it is differentiable on κ T except at t1 , . . . , tm ∈ Tκ such that f Δ is regulated on Tκ , we abbreviate this piecewise rd-continuous differentiability by writing f ∈ Cp (T). Some results concerning rd-continuous and regulated functions are contained in the following theorem. Since its statements follow directly from the definitions, we leave its proof to the reader. Theorem 1.8. Assume f : T → R. 1. 2. 3. 4. 5.

If f is continuous, then f is rd-continuous. If f is rd-continuous, then f is regulated. The jump operator σ is rd-continuous. If f is regulated or rd-continuous, then so is f σ . Assume f is continuous. If g : T → R is regulated or rd-continuous, then f ◦ g has that property.

Theorem 1.9. Every regulated function on a compact interval is bounded. Theorem 1.10 (Induction Principle). Let t0 ∈ T and assume that {S(t) : t ∈ [t0 , ∞)} is a family of statements satisfying (i) S(t0 ) is true, (ii) If t ∈ [t0 , ∞) is right-scattered and S(t) is true, then S(σ (t)) is true, (iii) If t ∈ [t0 , ∞) is right-dense and S(t) is true, then there is a neighborhood U of t such that S(s) is true for all s ∈ U ∩ (t, ∞), (iv) If t ∈ (t0 , ∞) is left-dense and S(s) is true for s ∈ [t0 , t), then S(t) is true. Then S(t) is true for all t ∈ [t0 , ∞).

16

1 Calculus on Time Scales

Theorem 1.11 (Dual Version of Induction Principle). Let t0 ∈ T and assume that {S(t) : t ∈ (−∞, t0 ]} is a family of statements satisfying (i) S(t0 ) is true, (ii) If t ∈ (−∞, t0 ] is left-scattered and S(t) is true, then S(ρ(t)) is true, (iii) If t ∈ (−∞, t0 ] is left-dense and S(t) is true, then there is neighborhood U of t such that S(s) is true for all s ∈ U ∩ (−∞, t), (iv) If t ∈ (−∞, t0 ) is right-dense and S(s) is true for s ∈ (t, t0 ), then S(t) is true. Then S(t) is true for all t ∈ (−∞, t0 ]. Theorem 1.12. Let f and g be real-valued functions defined on T, both predifferentiable with D. Then |f Δ (t)| ≤ |g Δ (t)| for

all

t ∈D

|f (s) − f (r)| ≤ g(s) − g(r) for

all

r, s, ∈ T,

implies r ≤ s.

Theorem 1.13. Suppose f : T → R is pre-differentiable with D. If U is a compact interval with endpoints r, s ∈ T, then 

 |f (s) − f (r)| ≤

sup |f (t)| |s − r|. Δ

t∈U k ∩D

Theorem 1.14. Let f is pre-differentiable with D. If f Δ (t) = 0 for all t ∈ D, then f (t) is a constant function. Theorem 1.15. Let f and g are pre-differentiable with D and f Δ (t) = g Δ (t) for all t ∈ D. Then g(t) = f (t) + C

for all

t ∈ T,

where C is a constant. Theorem 1.16. Suppose fn : T → R is pre-differentiable with D for each n ∈ N. Assume that for each t ∈ Tκ there exists a compact interval U (t) such that the sequence {fnΔ }n∈N converges uniformly on U (t) ∩ D. (i) If {fn }n∈N converges at some t0 ∈ U (t) for some t ∈ Tκ , then it converges uniformly on U (t).

1.3 Integration

17

(ii) If {fn }n∈N converges at some t0 ∈ T, then it converges uniformly on U (t) for all t ∈ Tκ . (iii) The limit mapping f = lim fn is pre-differentiable with D and we have n−→∞

f Δ (t) = lim fnΔ (t) for

all

n−→∞

t ∈ D.

Theorem 1.17. Let t0 ∈ T, x0 ∈ R, f : Tκ → R be given regulated map. Then there exists exactly one pre-differentiable function F satisfying F Δ (t) = f (t) for

all

t ∈ D,

F (t0 ) = x0 .

Definition 1.16. Assume f : T → R is a regulated function. Any function F by Theorem 1.17 is called a pre-antiderivative of f . We define the indefinite integral of a regulated function f by f (t)Δt = F (t) + c, where c is an arbitrary constant and F is a pre-antiderivative of f . We define the Cauchy integral by s

f (t)Δt = F (s) − F (τ ) for

τ, s ∈ T.

all

τ

A function F : T → R is called an antiderivative of f : T → R provided F Δ (t) = f (t) holds

for

all

t ∈ Tκ .

Example 1.16. Let T = 2N0 and f (t) = 7t 2 − 9t − 1, g(t) = t 3 − 3t 2 − t + 4,

t ∈ T.

Then σ (t) = 2t, g Δ (t) = (σ (t))2 + tσ (t) + t 2 − 3(σ (t) + t) − 1 = (2t)2 + 2t 2 + t 2 − 3(2t + t) − 1 = 4t 2 + 3t 2 − 9t − 1 = 7t 2 − 9t − 1,

t ∈ T.

Hence, we conclude that (7t 2 − 9t − 1)Δt = t 3 − 3t 2 − t + 4,

t ∈ T.

18

1 Calculus on Time Scales

Exercise 1.13. Let T = 3N0 . Prove that (4t + 1)Δt = t 2 + t,

t ∈ T.

Theorem 1.18. Every rd-continuous function f : T → R has an antiderivative. In particular, if t0 ∈ T, then F defined by t

F (t) =

t ∈ T,

for

f (τ )Δτ t0

is an antiderivative of f . Theorem 1.19. If f ∈ Crd (T) and t ∈ Tκ , then σ (t)

f (τ )Δτ = μ(t)f (t).

t

Theorem 1.20. If f Δ ≥ 0, then f is nondecreasing. Theorem 1.21. If a, b, c ∈ T, α ∈ R and f, g ∈ Crd (T), then b

(i) a

b

(ii) a

b

(iii) a

b

(iv) a

b

(v) a

b

(vi) a

(vii)

a

b

(f (t) + g(t))Δt = b

(αf )(t)Δt = α

g(t)Δt, a

a

f (t)Δt, c

f (t)Δt =

a

f (t)Δt, a

f (t)Δt = −

b

f (t)Δt +

b

b

f (t)Δt +

f (t)Δt,

a

c

b

f (σ (t))g Δ (t)Δt = (f g)(b) − (f g)(a) − b

f (t)g (t)Δt = (f g)(b) − (f g)(a) − Δ

f Δ (t)g(t)Δt,

a

f Δ (t)g(σ (t))Δt,

a

f (t)Δt = 0,

a

(viii) if |f (t)| ≤ g(t) on [a, b], then ! ! ! !

b a

! ! f (t)Δt !! ≤ b

(ix) if f (t) ≥ 0 for all a ≤ t < b, then a

b

g(t)Δt, a

f (t)Δt ≥ 0.

1.4 The Exponential Function

19

Exercise 1.14. Let a, b ∈ T and f ∈ Crd (T). (i) If T = R, prove b

b

f (t)Δt =

a

f (t)dt, a

where the integral on the right is the usual Riemann integral from calculus. (ii) If [a, b] consists only isolated points, then b a

⎧" ⎪ ⎨ t∈[a,b) μ(t)f (t) if a < b f (t)Δt = 0 if a = b ⎪ ⎩−" t∈[b,a) μ(t)f (t) if a > b.

Definition 1.17 (Improper Integral). If a ∈ T, sup T = ∞, and f is rdcontinuous on [a, ∞), then we define improper integral by ∞

b

f (t)Δt := lim

b−→∞ a

a

f (t)Δt

provided this limit exists, and we say that the improper integral converges in this case. If this limit does not exist, then we say that the improper integral diverges. Example 1.17. Let T = q N0 , q > 1. Then σ (t) = qt. Since all points are isolated, then ∞ 1

 1 1 Δt = μ(t) t2 t2 N t∈q

=

0

 q −1 t N

t∈q

0

= ∞.

1.4 The Exponential Function Definition 1.18. Let h > 0. 1. The Hilger complex numbers we define as follows 1 Ch = {z ∈ C : z = − }. h

20

1 Calculus on Time Scales

2. The Hilger real axis we define as follows 1 Rh = {z ∈ C : z > − }. h 3. The Hilger alternative axis we define as follows 1 Ah = {z ∈ C : z < − }. h 4. The Hilger imaginary circle we define as follows ! ! 

! 1 !! 1 ! . I h = z ∈ C : !z + ! = h h For h = 0 we set C0 = C,

R0 = R,

A0 = Ø,

I0 = iR.

Definition 1.19. Let h > 0 and z ∈ Ch . We define the Hilger real part of z by Reh (z) =

|zh + 1| − 1 h

and the Hilger imaginary part by Imh (z) =

Arg(zh + 1) , h

where Arg(z) denotes the principal argument of z, i.e., −π < Arg(z) ≤ π. We note that −

1 < Reh (z) < ∞ and h

−

π π < Imh (z) < . h h

In particular, Reh (z) ∈ Rh . π π Definition 1.20. Let − < w ≤ . We define the Hilger purely imaginary numh h ber by ◦

iw=

eiwh − 1 . h

1.4 The Exponential Function

21 ◦

Theorem 1.22. Let z ∈ Ch . Then i Imh (z) ∈ Ih . Theorem 1.23. We have ◦

lim [Reh (z)+ i Imh (z)] = Re(z) + iIm(z).

h→0

Theorem 1.24. Let −

π π < w ≤ . Then h h ◦

| i w|2 =

wh 4 . sin2 2 h2

Definition 1.21. The “circle plus” addition ⊕ on Ch is defined by z ⊕ w = z + w + zwh.

Theorem 1.25. (Ch , ⊕) is an Abelian group. Example 1.18. Let z ∈ Ch and w ∈ C. We will simplify the expression A=z⊕

w . 1 + hz

We have w zw + h 1 + hz a + hz (1 + hz)w =z+ 1 + hz = z + w.

A =z+

Theorem 1.26. For z ∈ Ch we have ◦

z = Reh (z)⊕ i Imh (z). Definition 1.22. Let n ∈ N and z ∈ Ch . We define “circle dot” multiplication  by n  z = z ⊕ z ⊕ z ⊕ · · · ⊕ z. Theorem 1.27. Let n ∈ N and z ∈ Ch . Then nz=

(zh + 1)n − 1 . h

22

1 Calculus on Time Scales

Definition 1.23. Let z ∈ Ch . The additive inverse of z under the operation ⊕ is defined as follows z =

−z . 1 + zh

Theorem 1.28. Let z ∈ Ch . Then (z) = z. Definition 1.24. Let z, w ∈ Ch . We define “circle minus” subtraction as follows z  w = z ⊕ (w). For z, w ∈ Ch we have z  w = z ⊕ (w) = z + (w) + z(w)h zwh w − 1 + wh 1 + wh z + zwh − w − zwh = 1 + wh z−w , = 1 + wh

=z−

i.e., zw =

z−w . 1 + wh

Theorem 1.29. Let z ∈ Ch . Then z = z iff z ∈ Ih . π π Theorem 1.30. Let − < w ≤ . Then h h ◦

◦

(i w) = i w. Definition 1.25. Let z ∈ Ch . The generalized square of z is defined as follows z = −z(z).

1.4 The Exponential Function

23

We have z = −z =

−z 1 + zh

z2 . 1 + zh

Theorem 1.31. For z ∈ Ch we have (z) = z . Theorem 1.32. For z ∈ Ch we have 1 + zh =

z2 . z

Theorem 1.33. For z ∈ Ch we have z + (z) = z h. Theorem 1.34. For z ∈ Ch we have z ⊕ z = z + z2 . Theorem 1.35. Let −

π π < w ≤ . Then h h ◦

−(i ) =

4 sin2 h2

 wh . 2

Exercise 1.15. Let z ∈ Ch . Prove that z ∈ R

]iff

z ∈ Rh ∪ Ah ∪ Ih .

For h > 0, we define the strip Zh := {z ∈ C : −

π π < Im(z) ≤ }. h h

For h = 0, we set Z0 := C. Definition 1.26. For h > 0, we define the cylindrical transformation ξh : Ch → Zh by ξh (z) :=

1 Log(1 + zh), h

24

1 Calculus on Time Scales

where Log is the principal logarithm function. For h = 0, we define ξ0 (z) = z for all z ∈ C. We note that ξh−1 (z) =

ezh − 1 h

for z ∈ Zh . Definition 1.27. We say that a function f : T → R is regressive provided 1 + μ(t)p(t) = 0 for

all

t ∈ Tκ

holds. The set of all regressive and rd-continuous functions f : T → R will be defined by R(T) or R. In R we define “circle plus” addition as follows (f ⊕ g)(t) = f (t) + g(t) + μ(t)f (t)g(t). Exercise 1.16. Prove that (R, ⊕) is an Abelian group. Definition 1.28. The group (R, ⊕) will be called the regressive group. Definition 1.29. For f ∈ R we define (f )(t) = −

f (t) 1 + μ(t)f (t)

for

all

t ∈ Tκ .

Exercise 1.17. Let f ∈ R. Prove that (f )(t) ∈ R for all t ∈ Tκ . Definition 1.30. We define “circle minus” subtraction  on R as follows (f  g)(t) = (f ⊕ (g))(t) for

all

For f, g ∈ R, we have f  g = f ⊕ (g) 

g =f ⊕ − 1 + μg =f − =

g μf g − 1 + μg 1 + μg

f −g . 1 + μg

t ∈ Tκ .

1.4 The Exponential Function

25

Theorem 1.36. Let f, g ∈ R. Then 1. 2. 3. 4. 5. 6.

f  f = 0, (f ) = f, f  g ∈ R, (f  g) = g  f, (f ⊕ g) = (f ) ⊕ (g), g f⊕ = f + g. 1 + μf

Definition 1.31. If f ∈ R, then we define the generalized exponential function by ef (t, s) = e

#t s

ξμ(τ ) (f (τ ))Δτ

s, t ∈ T.

for

In fact, using the definition for the cylindrical transformation we have ef (t, s) = e

#t

1 s μ(τ ) Log(1+μ(τ )f (τ ))Δτ

for

s, t ∈ T.

Theorem 1.37 (Semigroup Property). If f ∈ R, then ef (t, r)ef (r, s) = ef (t, s)

for

all t, r, s ∈ R.

Exercise 1.18. Let f ∈ R. Prove that e0 (t, s) = 1 and

ef (t, t) = 1.

Theorem 1.38. Let f ∈ R and fix t0 ∈ T. Then efΔ (t, t0 ) = f (t)ef (t, t0 ). Corollary 1.1. Let f ∈ R and fix t0 ∈ T. Then ef (t, t0 ) is a solution to the Cauchy problem y Δ (t) = f (t)y(t),

y(t0 ) = 1.

(1.2)

Corollary 1.2. Let f ∈ R and fix t0 ∈ T. Then ef (t, t0 ) is the unique solution of the problem (1.2). Theorem 1.39. Let f ∈ R. Then ef (σ (t), s) = (1 + μ(t)f (t))ef (t, s). Theorem 1.40. Let f ∈ R. Then ef (t, s) =

1 = ef (s, t). ef (s, t)

26

1 Calculus on Time Scales

Theorem 1.41. Let f, g ∈ R. Then ef (t, s)eg (t, s) = ef ⊕g (t, s). Theorem 1.42. Let f, g ∈ R. Then ef (t, s) = ef g (t, s). eg (t, s) Theorem 1.43. Let f ∈ R. Then 1 ef (t, r). 1 + μ(s)f (s)

ef (t, σ (s))ef (s, r) = Theorem 1.44. Let f, g ∈ R. Then efΔg (t, t0 ) =

(f (t) − g(t))ef (t, t0 ) . eg (σ (t), t0 )

Theorem 1.45. Let f ∈ R and a, b, c ∈ T. Then (ef (c, t))Δ = −f (t)(ef (c, t))σ = −f (t)ef (c, σ (t)) and b

f (t)ef (c, σ (t))Δt = ef (c, a) − ef (c, b).

a

5 2 = 0, 1 + μ(t) = 0 for all t ∈ T ∩ (0, ∞). Let t t also, t0 ∈ T ∩ (0, ∞). Evaluate the integral

Exercise 1.19. Assume 1 + μ(t)

I=

t

e 5 (s, t0 )

t0

seσ2 (s, t0 )

s

Δs.

s

Solution. We have

5 2 − s s

 e 5 (s, t ) 0 s

eσ (s, t 2 s

0)

=

3 e 5s (s, t0 ) s eσ2 (s, t0 )

=

eΔ 5 2 (s, t0 ) ss

s

= eΔ

(s, t0 )

= eΔ

(s, t0 ).

3 s 1+μ(s) 2s 3 s+2μ(s)

1.4 The Exponential Function

27

Hence, I =

1 3

t

eΔ

3 s+2μ(s)

t0

(s, t0 )Δs

!s=t 1 ! e 3 (s, t0 )! s=t0 3 s+2μ(s) 1 1 = e 3 (t, t0 ) − . 3 t+2μ(t) 3 =

Exercise 1.20. Let α ∈ R. Let also, the exponentials e α2 t

− (α−1) σ (t)

2

(t, t0 )

and

e αt (t, t0 )

exist for all t ∈ T ∩ (0, ∞). Prove that 1. e α2 t

− (α−1) σ (t)

2

(t, t0 ) = e α−1 (t, t0 ),

e αt (t, t0 )

σ (t)

2. e α−1 (t, t0 )

= e−

σ (t)

e αt (t, t0 )

1 σ (t)

(t, t0 ) =

t0 t

for all t, t0 ∈ T ∩ (0, ∞). Let α : T → R be regulated and 1 + α(t)μ(t) = 0 for all t ∈ T. Let also, t0 , t ∈ T, t0 < t. 1. T = hZ, Then

h > 0. Every point in T is isolated and μ(t) = h for every t ∈ T. eα (t, t0 ) = e =e =e =

#t

1 t0 μ(τ ) Log(1+α(τ )μ(τ ))Δτ

" "

1 s∈[t0 ,t) μ(s) Log(1+α(s)μ(s))μ(s) s∈[t0 ,t) Log(1+hα(s))

$

(1 + hα(s)).

s∈[t0 ,t)

If α is a constant, then eα (t, t0 ) =

$

(1 + hα)

s∈[t0 ,t)

= (1 + hα)t−t0 .

28

1 Calculus on Time Scales

2. T = q N0 , q > 1. Every point of T is isolated and μ(t) = (q − 1)t for all t ∈ T. Then eα (t, t0 ) = e =e =e =e =

#t

1 t0 μ(τ ) Log(1+α(τ )μ(τ ))Δτ

" " "

1 s∈[t0 ,t) μ(s) Log(1+α(s)μ(s))μ(s) s∈[t0 ,t) Log(1+α(s)μ(s)) s∈[t0 ,t) Log(1+(q−1)sα(s))

$

(1 + (q − 1)sα(s)).

s∈[t0 ,t)

Exercise 1.21. Let T = q N0 ,

0 < q < 1. Prove that  $ 1−q 1+ α(s)s . q

eα (t, t0 ) =

s∈[t0 ,t)

3. T = Nk0 ,

k ∈ N. Every point of T is isolated and μ(t) =

√ k

k t + 1 − t.

Then eα (t, t0 ) = e =e =e =

#t

1 t0 μ(τ ) Log(1+α(τ )μ(τ ))Δτ

" "

1 s∈[t0 ,t) μ(s) Log(1+α(s)μ(s))μ(s) s∈[t0 ,t) Log(1+α(s)μ(s))

$

(1 + α(s)μ(s))

s∈[t0 ,t)

=

$ 

1+

√ k

k s + 1 − s α(s) .

s∈[t0 ,t)

1.5 Hyperbolic and Trigonometric Functions Definition 1.32 (Hyperbolic Functions). If f ∈ Crd and −μf 2 ∈ R, then we define the hyperbolic functions coshf and sinhf by coshf =

ef + e−f 2

and

sinhf =

ef − e−f . 2

1.6 Power Series

29

Theorem 1.46. Let f ∈ Crd . If −μf 2 ∈ R, then we have 1. coshΔ f = f sinhf , 2. sinhΔ f = f coshf , 3. cosh2f − sinh2f = e−μf 2 . Definition 1.33 (Trigonometric Functions). If f ∈ Crd and μf 2 ∈ R, then we define the trigonometric functions cosf and sinf by cosf =

eif + e−if 2

sinf =

and

eif − e−if . 2i

Theorem 1.47. Let f ∈ Crd and −μf 2 ∈ R. Prove that 1. cosΔ f = −f sinf . Δ 2. sinf = f cosf . 3. cos2f + sin2f = eμf 2 . Exercise 1.22. Let f ∈ Crd and μf 2 ∈ R. Show Euler’s formula eif = cosf +i sinf .

1.6 Power Series We introduce the generalized monomials hk : T × T → R, k ∈ N0 , defined recursively by h0 (t, s) = 1, t

hk (t, s) =

hk−1 (τ, s)Δτ,

k ∈ N,

t, s ∈ T.

s

Then t

h1 (t, s) =

Δτ s

= t − s, t hΔ k (t, s) = hk−1 (t, s),

k ∈ N,

t, s ∈ T.

Example 1.19. Let T = R. Then hk (t, s) =

(t − s)k , k!

k ∈ N,

t, s ∈ T.

30

1 Calculus on Time Scales

Theorem 1.48. We have hk+m+1 (t, t0 ) =

t

t, t0 ∈ T,

hk (t, σ (s))hm (s, t0 )Δs, t0

k, m ∈ N0 . We define g0 (t, s) = 1,

t

gk+1 (t, s) =

gk (σ (τ ), s)Δτ,

k ∈ N0 ,

t, s ∈ T.

s

Lemma 1.1. Let n ∈ N. If f is n−times differentiable and pk , 0 ≤ k ≤ n − 1, are differentiable at some t ∈ T with Δ (t) = pkσ (t) for pk+1

all 0 ≤ k ≤ n − 2,

n ≥ 2,

then we have %n−1 

&Δ k

(−1) f

Δk

(t)pk (t)

n

σ = (−1)n−1 f Δ (t)pn−1 (t) + f (t)p0Δ (t).

k=0

Lemma 1.2. The functions gn (t, s) satisfy for all t ∈ T the relationship gn (ρ k (t), t) = 0

for all

n ∈ N and

all

0 ≤ k ≤ n − 1.

Lemma 1.3. Let n ∈ N and suppose that f is (n − 1)−times differentiable at ρ n−1 (t). Then f (t) =

n−1  k (−1)k f Δ (ρ n−1 (t))gk (ρ n−1 (t), t). k=0

Theorem 1.49 (Taylor’s Formula). Let n ∈ N. Suppose f is n−times differenn n−1 tiable on Tκ . Let α ∈ Tκ , t ∈ T. Then f (t) =

n−1  k (−1)k gk (α, t)f Δ (α) k=0 ρ n−1 (t)

+ α

n

(−1)n−1 gn−1 (σ (τ ), t)f Δ (τ )Δτ.

1.6 Power Series

31

Theorem 1.50. The functions gn and hn satisfy the relationship hn (t, s) = (−1)n gn (s, t) n

for all t ∈ T and all s ∈ Tκ . From Theorem 1.49 and Theorem 1.50, it follows the following theorem. Theorem 1.51 (Taylor’s Formula). Let n ∈ N. Suppose f is n−times differenn n−1 tiable on Tκ . Let also, α ∈ Tκ , t ∈ T. Then f (t) =

n−1 

hk (t, α)f

Δk

ρ n−1 (t)

(α) +

n

hn−1 (t, σ (τ ))f Δ (τ )Δτ.

α

k=0

Now we will formulate an other variant of Taylor’s formula. Theorem 1.52 (Taylor’s Formula). Let n ∈ N. Suppose that the function f is n+1 n+1 n + 1-times differentiable on Tκ . Let α ∈ Tκ , t ∈ T, and t > α. Then f (t) =

n 

t

k

hk (t, α)f Δ (α) +

n+1

hn (t, σ (τ ))f Δ

(τ )Δτ.

α

k=0

Let ρ n−1 (t)

Rn (t, α) =

n

hn−1 (t, σ (τ ))f Δ (τ )Δτ.

α

Theorem 1.53. Let t ∈ T, t ≥ α and  ! n ! ! ! Mn (t) = sup !f Δ (τ )! : τ ∈ [α, t] . Then |Rn (t, α)| ≤ Mn (t)

(t − α)n . (n − 1)!

If a function f : T → R is infinitely many Δ-differentiable at a point α ∈ T∞ = '∞ κn n=1 T , then we can formally write ∞ 

k

2

hk (t, α)f Δ (α) = f (α) + h1 (t, α)f Δ (α) + h2 (t, α)f Δ (α) + · · · .

(1.3)

k=0

Definition 1.34. The series (1.3) is called Taylor’s series for the function f at the point α.

32

1 Calculus on Time Scales

For given values of α and t, Taylor’s series can be convergent or divergent. Taylor’s series (1.3) is convergent if and only if the remainder of Taylor’s formula f (t) =

n−1 

k

hk (t, α)f Δ (α) + Rn (t, α)

k=0

tend to zero as n → ∞, that is, limn→∞ Rn (t, α) = 0. It may turn out that the series (1.3) is convergent for some values of t but its sum is not equal to f (t). Theorem 1.54. For all z ∈ C and α, R ∈ T, and R > α, the initial value problem y Δ = zy,

y(α) = 1,

t ∈ [α, R]

(1.4)

has a unique solution y that is represented in the form y(t) =

∞ 

t ∈ [α, R],

zk hk (t, α),

k=0

and satisfies the inequality |y(t)| ≤ e|z|(t−α) ,

t ∈ [α, R].

Let y be the solution of the problem (1.4). Then 2

y Δ = zy Δ = z(zy) = z2 y. Assume that k

y Δ = zk y for some k ∈ N. We will prove that k+1

yΔ

= zk+1 y.

Really, we have k+1

yΔ

 k Δ = yΔ  Δ = zk y = zk+1 y.

(1.5)

1.6 Power Series

33

Therefore (1.5) holds for any k ∈ N. Note that ez (t, α), t ∈ [α, R], coincides with the unique solution of the problem (1.4). Therefore ez (t, α) =

∞ 

t ∈ [α, R].

zk hk (t, α),

k=0

Hence, ez (t, α) + e−z (t, α) 2 & %∞ ∞  1  k k z hk (t, α) + (−z) hk (t, α) = 2

coshz (t, α) =

k=0

= =

1 2

∞  

k=0

zk + (−z)k hk (t, α)

k=0

∞ 

z2k h2k (t, α),

k=0

ez (t, α) − e−z (t, α) 2 & %∞ ∞  1  k k z hk (t, α) − (−z) hk (t, α) = 2

sinhz (t, α) =

k=0

= =

1 2

∞  

k=0



zk − (−z)k hk (t, α)

k=0

∞ 

z2k+1 h2k+1 (t, α),

k=0

eiz (t, α) + e−iz (t, α) 2 ∞  (iz)2k h2k (t, α) =

cosz (t, α) =

k=0

=

∞  (−1)k z2k h2k (t, α), k=0

sinz (t, α) =

eiz (t, α) − e−iz (t, α) 2i

34

1 Calculus on Time Scales ∞

=

1 (iz)2k+1 h2k+1 (t, α) i k=0

=

∞  (−1)k z2k+1 h2k+1 (t, α),

t ∈ [α, R].

k=0

Definition 1.35. Assume that sup T = ∞ and let t0 ∈ T is fixed. A series of the form ∞ 

ak hk (t, t0 ) = a0 + a1 h1 (t, t0 ) + a2 h2 (t, t0 ) + · · · ,

k=0

where ak ∈ C, k ∈ N0 and t ∈ T, is called a power series on the time scale T. The numbers ak , k ∈ N0 , being referred to as its coefficients. We denote by P the set of all functions f : [t0 , ∞) → C of the form f (t) =

∞ 

ak hk (t, t0 ),

t ∈ [t0 , L],

(1.6)

k=0

where the coefficients ak and k ∈ N0 satisfy |ak | ≤ MR k ,

k ∈ N0 ,

(1.7)

for some constants M > 0 and R > 0, depending only on the series (1.6). Here L ∈ T, L > t0 . Note that, under the condition (1.7), it converges uniformly on [t0 , L]. Next, that P is a linear space. Also, any given function f ∈ P can be represented in the form of a power series (1.6) uniquely. Indeed, Δ-differentiating the series (1.6) n times term by term, we get n

f Δ (t) = an + an+1 h1 (t, t0 ) + · · · ,

t ∈ [t0 , L],

whereupon n

f Δ (t0 ) = an ,

n ∈ N0 .

(1.8)

Thus, the coefficients of the power series (1.6) are defined uniquely by the formula (1.8).

1.7 Advanced Practical Problems

35

1.7 Advanced Practical Problems √ 4 Problem 1.1. Classify each points t ∈ T = { 7n : n ∈ N0 } as left-dense, leftscattered, right-dense, or right-scattered.  √ 4 Problem 1.2. Let T = n + 7 : n ∈ N0 . Find μ(t), t ∈ T.   √ 4 Problem 1.3. Let T = t = 2n + 1 : n ∈ N , f (t) = 1 + 2t 4 , t ∈ T. Find f (σ (t)), t ∈ T.

 2 Problem 1.4. Let T = : n ∈ N ∪ {0}. Find Tκ . 4n + 3 Problem 1.5. Let f (t) = t + t 2 + t 3 , t ∈ T. Prove that f Δ (t) = 1 + t + t 2 + (1 + t)σ (t) + (σ (t))2 ,

t ∈ Tκ .

Problem 1.6. Let T = {n3 : n ∈ N0 }, f (t) = t 2 + 2t, t ∈ T. Find f Δ (t), t ∈ Tκ . Problem 1.7. Let T = {n + 2 : n ∈ N0 }, f (t) = t 2 + 2, g(t) = t 2 . Find a constant c ∈ [2, σ (2)] such that (f ◦ g)Δ (2) = f (g(c))g Δ (2).   Problem 1.8. Let T = 24n+2 : n ∈ N0 , v(t) = t 3 , w(t) = t 2 + t. Prove ˜

(w ◦ v)Δ (t) = (w Δ ◦ v(t))v Δ (t),

t ∈ Tκ .

 Δ˜ Problem 1.9. Let T = {n + 9 : n ∈ N0 } and v(t) = t 2 + 7t + 8. Find v −1 ◦ v(t). Problem 1.10. Let T = R and  f (t) =

1 10 t−2

for t = 2 for t ∈ R\{2}.

Determine if f is regulated. Problem 1.11. Let T = R and  f (t) =

0 if t = 5 1 t−5 if R\{5}.

Check if f : T → R is pre-differentiable and if it is, find the region of differentiation.

36

1 Calculus on Time Scales

Problem 1.12. Let T = 3N . Prove that −

1 sin t sin 2tΔt = cos t + c. t

Problem 1.13. Let p : T → R be rd-continuous and 1 + μ(t)p(t) = 0 for all t ∈ T. Let also, φ(t) be a nontrivial solution to the equation φ Δ − p(t)φ = 0. Prove that

1 is a solution to the equation φ ψ Δ + p(t)ψ σ = 0.

Chapter 2

Dynamic Systems

The results of this chapter are partly published in the monographs and papers [4, 8, 21, 22, 48, 57, 58, 60, 84, 86, 91, 98–100, 110, 113, 150, 166, 173, 178, 180, 183, 184, 186, 204, 219, 222, 224]. Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

2.1 Structure of Dynamic Systems on Time Scales Definition 2.1. The first-order linear dynamic equation y Δ = p(t)y,

t ∈ T,

(2.1)

is called regressive if p ∈ R. Theorem 2.1. Suppose that p ∈ R, t0 ∈ T and y0 ∈ R. Then the unique solution of the IVP y Δ = p(t)y,

y(t0 ) = y0 ,

(2.2)

is given by y(t) = ep (t, t0 )y0 ,

t ∈ T.

Proof. Let y be a solution of the IVP (2.2). We have y Δ (t)ep (σ (t), t0 ) = p(t)ep (σ (t), t0 )y(t)  Δ = p(t) ep (t, t0 ) + μ(t)ep (t, t0 ) y(t)

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_2

37

38

2 Dynamic Systems

  = p(t) ep (t, t0 ) + μ(t)(p)(t)ep (t, t0 ) y(t)

 μ(t)p(t) = p(t) 1 − ep (t, t0 )y(t) 1 + μ(t)p(t) =

p(t) ep (t, t0 )y(t) 1 + μ(t)p(t)

= −(p)(t)ep (t, t0 )y(t) Δ (t, t0 )y(t), = −ep

t ∈ T,

whereupon Δ (t, t0 )y(t) = 0, y Δ (t)ep (σ (t), t0 ) + ep

t ∈ T,

or Δ  yep (·, t0 ) (t) = 0,

t ∈ T.

Then y(t)ep (t, t0 ) = c,

t ∈ T,

where c is a constant. Since y(t0 ) = y0 , we get y0 = c, i.e., y(t)ep (t, t0 ) = y0 ,

t ∈ T.

Consequently y(t) = y0 ep (t, t0 ),

t ∈ T.

This completes the proof. Definition 2.2. If p ∈ R and f : T → R is rd-continuous, then the dynamic equation y Δ = p(t)y + f (t), is called regressive.

t ∈ T,

(2.3)

2.1 Structure of Dynamic Systems on Time Scales

39

Theorem 2.2. Suppose that (2.3) is regressive. Let t0 ∈ T and x0 ∈ R. The unique solutions of the IVPs x Δ = −p(t)x σ + f (t),

x(t0 ) = x0 ,

t ∈ T,

(2.4)

and x Δ = p(t)x + f (t),

x(t0 ) = x0 ,

t ∈ T,

(2.5)

are given by t

x(t) = ep (t, t0 )x0 +

ep (t, τ )f (τ )Δτ,

t ∈ T,

(2.6)

ep (t, σ (τ ))f (τ )Δτ,

t ∈ T,

(2.7)

t0

and t

x(t) = ep (t, t0 )x0 +

t0

respectively. Proof. 1. Consider x, defined by (2.6). We will prove that it satisfies (2.4). We have x(t0 ) = ep (t0 , t0 )x0 = x0 , and x Δ (t) = (p) (t)ep (t, t0 )x0 +

t t0

(p) (t)ep (t, τ )f (τ )Δτ

+ep (σ (t), t)f (t) =− +

p(t) p(t) ep (t, t0 )x0 − 1 + μ(t)p(t) 1 + μ(t)p(t) f (t) , 1 + μ(t)p(t)

x σ (t) = ep (σ (t), t0 )x0 +

t ∈ T, σ (t) t0

ep (σ (t), τ )f (τ )Δτ

= ep (t, t0 )x0 (1 + (p) (t)μ(t)) t

+

t0

ep (σ (t), τ )f (τ )Δτ

σ (t)

+ t

ep (σ (t), τ )f (τ )Δτ

t t0

ep (t, τ )f (τ )Δτ

40

2 Dynamic Systems

=

1 1 ep (t, t0 )x0 + 1 + μ(t)p(t) 1 + μ(t)p(t)

t t0

ep (t, τ )f (τ )Δτ

+μ(t)ep (σ (t), t)f (t) =

1 1 ep (t, t0 )x0 + 1 + μ(t)p(t) 1 + μ(t)p(t) +

μ(t) f (t), 1 + μ(t)p(t)

t t0

ep (t, τ )f (τ )Δτ

t ∈ T.

Therefore x Δ (t) + p(t)x σ (t) = − −

p(t) ep (t, t0 )x0 1 + μ(t)p(t) p(t) 1 + μ(t)p(t)

t t0

ep (t, τ )f (τ )Δτ +

+

p(t) ep (t, t0 )x0 1 + μ(t)p(t)

+

p(t) 1 + μ(t)p(t)

= f (t),

t t0

ep (t, τ )f (τ )Δτ +

f (t) 1 + μ(t)p(t)

μ(t)p(t) f (t) 1+μ(t)p(t)

t ∈ T.

Now we multiply the equation (2.4) by ep (t, t0 ) and we get ep (t, t0 )x Δ (t) + p(t)ep (t, t0 )x σ (t) = ep (t, t0 )f (t),

t ∈ T,

or Δ  ep (·, t0 )x (t) = ep (t, t0 )f (t),

t ∈ T,

which we integrate from t0 to t and we obtain ep (t, t0 )x(t) − ep (t0 , t0 )x(t0 ) =

t

ep (τ, t0 )f (τ )Δτ,

t ∈ T,

t0

or ep (t, t0 )x(t) = x0 +

t

ep (τ, t0 )f (τ )Δτ,

t ∈ T,

t0

or x(t) = x0 ep (t, t0 ) + = x0 ep (t, t0 ) +

t t0

ep (t0 , τ )ep (t, t0 )f (τ )Δτ

t t0

ep (t, τ )f (τ )Δτ,

t ∈ T.

2.1 Structure of Dynamic Systems on Time Scales

41

2. Let now x be defined by (2.7). Then x Δ (t) = p(t)ep (t, t0 )x0 + p(t)

t

ep (t, σ (τ ))f (τ )Δτ t0

+ep (σ (t), σ (t))f (t) = p(t)x(t) + f (t),

t ∈ T.

Now we multiply the equation (2.5) by ep (t, t0 ) and we get ep (t, t0 )x Δ (t) − p(t)ep (t, t0 )x(t) = ep (t, t0 )f (t),

t ∈ T,

or p(t) 1 ep (t, t0 )x Δ (t) − ep (t, t0 )x(t) 1 + μ(t)p(t) 1 + μ(t)p(t) =

1 ep (t, t0 )f (t), 1 + μ(t)p(t)

t ∈ T,

or Δ (t, t0 )x(t) = ep (σ (t), t0 )f (t), ep (σ (t), t0 )x Δ (t) + ep

t ∈ T,

or Δ  ep (·, t0 )x (t) = ep (σ (t), t0 )f (t),

t ∈ T,

which we integrate from t0 to t and we obtain ep (t, t0 )x(t) − ep (t0 , t0 )x(t0 ) =

t t0

ep (σ (τ ), t0 )f (τ )Δτ,

or ep (t, t0 )x(t) = x0 +

t

ep (t0 , σ (τ ))f (τ )Δτ,

t ∈ T,

t0

or x(t) = x0 ep (t, t0 ) +

t

ep (t, t0 )ep (t0 , σ (τ ))f (τ )Δτ t0

= x0 ep (t, t0 ) +

t

ep (t, σ (τ ))f (τ )Δτ, t0

This completes the proof.

t ∈ T.

t ∈ T,

42

2 Dynamic Systems

Suppose that A is a m × n-matrix on T, A = (aij )1≤i≤m,1≤j ≤n , shortly A = (aij ), aij : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Definition 2.3. We say that A is differentiable on T if each entry of A is differentiable on T and  AΔ = aijΔ . Example 2.1. Let T = Z,

A(t) =

 t +1 t2 + t , 2t − 3 2t 2 − 3t + 2

t ∈ T.

We will find AΔ (t), t ∈ T. We have σ (t) = t + 1, a11 (t) = t + 1, a12 (t) = t 2 + t, a21 (t) = 2t − 3, a22 (t) = 2t 2 − 3t + 2,

t ∈ T.

Then Δ (t) = 1, a11 Δ a12 (t) = σ (t) + t + 1

= t +1+t +1 = 2t + 2, Δ a21 (t) Δ a22 (t)

= 2, = 2(σ (t) + t) − 3 = 2(t + 1 + t) − 3 = 4t + 2 − 3 = 4t − 1,

t ∈ T.

Therefore

A (t) = Δ

 1 2t + 2 , 2 4t − 1

t ∈ T.

2.1 Structure of Dynamic Systems on Time Scales

43

Example 2.2. Let T = 2N0 , ⎞ t 3 + t t+1 t t+2 A(t) = ⎝ 1 −t 2 t 2 + t ⎠ , 1 2 t3 ⎛

t ∈ T.

We will find AΔ (t), t ∈ T. We have σ (t) = 2t, a11 (t) = t 3 + t, t +1 , t +2 a13 (t) = t,

a12 (t) =

a21 (t) = 1, a22 (t) = −t 2 , a23 (t) = t 2 + t, a31 (t) = 1, a32 (t) = 2, a33 (t) = t 3 ,

t ∈ T.

Then Δ (t) = (σ (t))2 + tσ (t) + t 2 + 1 a11

= (2t)2 + 2t 2 + t 2 + 1 = 4t 2 + 3t 2 + 1 = 7t 2 + 1, Δ a12 (t) =

=

t + 2 − (t + 1) (t + 2)(σ (t) + 2) 1 , 2(t + 1)(t + 2)

Δ a13 (t) = 1, Δ a21 (t) = 0, Δ a22 (t) = −(σ (t) + t)

= −(2t + t) = −3t,

44

2 Dynamic Systems Δ a23 (t) = σ (t) + t + 1

= 2t + t + 1 = 3t + 1, Δ a31 (t) = 0, Δ a32 (t) = 0, Δ (t) = (σ (t))2 + tσ (t) + t 2 a33

= 4t 2 + 2t 2 + t 2 = 7t 2 ,

t ∈ T.

Therefore ⎛

7t 2 + 1 ⎜ Δ A (t) = ⎝ 0 0

1 2(t+1)(t+2)

−3t 0

⎞ 1 ⎟ 3t + 1 ⎠ , 7t 2

Example 2.3. Let T = N20 , % A(t) =

1 t 2 + 1 t+1 2 3t

& ,

t ∈ T.

We will find AΔ (t), t ∈ T. We have σ (t) =

√ 2 t +1 ,

a11 (t) = t 2 + 1, 1 , t +1 a21 (t) = 2, a12 (t) =

a22 (t) = 3t,

t ∈ T.

Then Δ (t) = σ (t) + t a11 √ 2 = t +1 +t √ = t +2 t +1+t √ = 2t + 2 t + 1,

t ∈ T.

2.1 Structure of Dynamic Systems on Time Scales

Δ a12 (t) = −

45

1 (t + 1)(σ (t) + 1)

=−

1 √ (t + 1)(( t + 1)2 + 1)

=−

1 , √ (t + 1)(t + 2 t + 2)

Δ a21 (t) = 0, Δ a22 (t) = 3,

Therefore

% A (t) = Δ

t ∈ T.

& √ 1 √ 2t + 2 t + 1 − (t+1)(t+2 t+2) 0

Exercise 2.1. Let T = 2Z,

A(t) =

3

 t2 t3 , 2t + 4 t − 1

,

t ∈ T.

t ∈ T.

Find AΔ (t), t ∈ T. Answer.

A (t) = Δ

Definition 2.4. Define

 3t 2 + 6t + 4 2t + 2 , 2 1

t ∈ T.

 Aσ = aijσ .

Theorem 2.3. If A is differentiable on Tκ , then Aσ (t) = A(t) + μ(t)AΔ (t),

t ∈ Tκ .

Proof. We have  Aσ (t) = aijσ (t)  = aij (t) + μ(t)aijΔ (t)  = (aij (t)) + μ(t) aijΔ (t) = A(t) + μ(t)AΔ (t), This completes the proof.

t ∈ Tκ .

46

2 Dynamic Systems

Below we suppose that B = (bij )1≤i≤m,1≤j ≤n , bij : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Theorem 2.4. Let A and B be differentiable on Tκ . Then (A + B)Δ = AΔ + B Δ

on

Tκ .

Proof. We have (A + B)(t) = (aij (t) + bij (t)),  Δ (A + B)Δ (t) = aijΔ (t) + bij (t)   Δ (t) = aijΔ (t) + bij = AΔ (t) + B Δ (t),

t ∈ Tκ .

This completes the proof. Theorem 2.5. Let α ∈ R and A be differentiable on Tκ . Then (αA)Δ = αAΔ

on

Tκ .

Proof. We have  Δ αaij (t)  = αaijΔ (t)  = α aijΔ (t)

(αA)Δ (t) =

= αAΔ (t),

t ∈ Tκ .

This completes the proof. Theorem 2.6. Let m = n and A, B be differentiable on Tκ . Then (AB)Δ (t) = AΔ (t)B(t) + Aσ (t)B Δ (t) = AΔ (t)B σ (t) + A(t)B Δ (t),

t ∈ Tκ .

Proof. We have (AB)(t) =

% n  k=1

& aik (t)bkj (t) ,

t ∈ T.

2.1 Structure of Dynamic Systems on Time Scales

47

Then ⎛% (AB)Δ (t) = ⎝

n 

⎞

&Δ

(t)⎠

aik bkj

k=1

% n &  Δ = (aik bkj ) (t) k=1

% n &  Δ σ Δ = aik (t)bkj (t) + aik (t)bkj (t) k=1

=

% n 

&

% n 

+

Δ aik (t)bkj (t)

k=1

& σ Δ aik (t)bkj (t)

k=1

= AΔ (t)B(t) + Aσ (t)B Δ (t) % n &  Δ σ Δ aik (t)bkj (t) + aik (t)bkj (t) = k=1

=

% n 

&

% n 

Δ σ aik (t)bkj (t) +

k=1

& Δ aik (t)bkj (t)

k=1

= A (t)B (t) + A(t)B (t), Δ

σ

Δ

t ∈ Tκ .

This completes the proof. Example 2.4. Let T = 2N0 ,

A(t) =

 t t −1 , 2 3t + 1

B(t) =

 1 t , t +1 t −1

Then

(AB)(t) =

=

t t −1 2 3t + 1



1 t t +1 t −1



t 2 + t − 1 2t 2 − 2t + 1 3t 2 + 4t + 3 3t 2 − 1

= C(t) = (cij (t)), We have σ (t) = 2t,

t ∈ T.



t ∈ T.

48

2 Dynamic Systems

a11 (t) = t, a12 (t) = t − 1, a21 (t) = 2, a22 (t) = 3t + 1, b11 (t) = 1, b12 (t) = t, b21 (t) = t + 1, b22 (t) = t − 1, c11 (t) = t 2 + t − 1, c12 (t) = 2t 2 − 2t + 1, c21 (t) = 3t 2 + 4t + 3, c22 (t) = 3t 2 − 1,

t ∈ T.

Then Δ (t) = 1, a11 Δ a12 (t) = 1, Δ a21 (t) = 0, Δ a22 (t) = 3, Δ b11 (t) = 0, Δ b12 (t) = 1, Δ b21 (t) = 1, Δ b22 (t) = 1, Δ c11 (t) = σ (t) + t + 1

= 2t + t + 1 = 3t + 1, Δ c12 (t) = 2(σ (t) + t) − 2

= 2(2t + t) − 2 = 6t − 2, Δ c21 (t)

= 3(σ (t) + t) + 4 = 3(2t + t) + 4

2.1 Structure of Dynamic Systems on Time Scales

49

= 9t + 4, Δ c22 (t)

= 3(σ (t) + t) = 3(2t + t) = 9t,

t ∈ T.

Therefore

(AB)Δ (t) =

 3t + 1 6t − 2 , 9t + 4 9t

t ∈ T.

Also,

AΔ (t)B(t) = = Aσ (t) = = Aσ (t)B Δ (t) = = AΔ (t)B(t) + Aσ (t)B Δ (t) = =



 1 t t +1 t −1

 t + 2 2t − 1 , 3t + 3 3t − 3

 σ (t) σ (t) − 1 2 3σ (t) + 1

 2t 2t − 1 , 2 6t + 1

  2t 2t − 1 01 2 6t + 1 11

 2t − 1 4t − 1 , 6t + 1 6t + 3

  t + 2 2t − 1 2t − 1 4t − 1 + 3t + 3 3t − 3 6t + 1 6t + 3

 3t + 1 6t − 2 , t ∈ T. 9t + 4 9t 11 03

Consequently (AB)Δ (t) = AΔ (t)B(t) + Aσ (t)B Δ (t),

t ∈ T.

Next,

B σ (t) =

1 σ (t) σ (t) + 1 σ (t) − 1



50

2 Dynamic Systems

 1 2t = , 2t + 1 2t − 1

  11 1 2t AΔ (t)B σ (t) = 03 2t + 1 2t − 1

 2t + 2 4t − 1 = , 6t + 3 6t − 3

  t t −1 01 Δ A(t)B (t) = 2 3t + 1 11

 t − 1 2t − 1 = , 3t + 1 3t + 3

  2t + 2 4t − 1 t − 1 2t − 1 + AΔ (t)B σ (t) + A(t)B Δ (t) = 6t + 3 6t − 3 3t + 1 3t + 3

 3t + 1 6t − 2 = , t ∈ T. 9t + 4 9t Therefore (AB)Δ (t) = AΔ (t)B σ (t) + A(t)B Δ (t),

t ∈ T.

Exercise 2.2. Let T = 3Z,

A(t) =

 t2 + 1 t − 2 , 2t − 1 t + 1

B(t) =

 t 2t + 1 , t t −1

t ∈ T.

Prove (AB)Δ (t) = AΔ (t)B σ (t) + A(t)B Δ (t), Theorem 2.7. Let m = n and A−1 exists on T. Then  σ −1  −1 σ = A on T. A Proof. For any t ∈ T we have A(t)A−1 (t) = I,

t ∈ T.

Then σ  Aσ (t) A−1 (t) = I,

t ∈ Tκ .

2.1 Structure of Dynamic Systems on Time Scales

51

whereupon σ   σ −1 A (t) = A−1 (t),

t ∈ T.

This completes the proof. Example 2.5. Let T = lN0 , l > 0,

A(t) =

 t +1 t +2 , 1 t +3

t ∈ T.

Then

A (t) = σ

σ (t) + 1 σ (t) + 2 1 σ (t) + 3



 t +l+1 t +l+2 , 1 t +l+3

 1 t + l + 3 −t − l − 2 , (Aσ )−1 (t) = −1 t +l+1 (t + l)(t + l + 3) + 1 =

t ∈ T.

Next, A−1 (t) =

1 t (t + 3) + 1

 t + 3 −t − 2 , −1 t + 1

t ∈ T,

whereupon 

−1

A

σ

 σ (t) + 3 −σ (t) − 2 −1 σ (t) + 1

 1 t + l + 3 −t − l − 2 , = −1 t +l+1 (t + l)(t + l + 3) + 1

1 (t) = σ (t)(σ (t) + 3) + 1

Consequently  σ −1 (t) = (A−1 )σ (t), A

t ∈ T.

Exercise 2.3. Let T = 2N0 and % A(t) =

t +2 t2 + 1

1 t+1 1 t+2

& ,

t ∈ T.

t ∈ T.

52

2 Dynamic Systems

Prove that (Aσ )−1 (t) = (A−1 )σ (t),

t ∈ T.

Theorem 2.8. Let m = n, A be differentiable on Tκ and A−1 , (Aσ )−1 exist on T. Then  Δ  −1 A−1 = −A−1 AΔ Aσ  −1 Δ −1 = − Aσ A A

on Tκ .

Proof. We have I = AA−1

on

T,

whereupon, using Theorem 2.6, we get O = IΔ = (AA−1 )Δ = AΔ (A−1 )σ + A(A−1 )Δ = AΔ (Aσ )−1 + A(A−1 )Δ = AΔ A−1 + Aσ (A−1 )Δ

on Tκ .

Hence, A(A−1 )Δ = −AΔ (Aσ )−1 ,

Aσ (A−1 )Δ = −AΔ A−1

on

Tκ ,

and (A−1 )Δ = −A−1 AΔ (Aσ )−1 ,

(A−1 )Δ = −(Aσ )−1 AΔ A−1

on Tκ .

This completes the proof. Exercise 2.4. Let m = n, A and B be differentiable on Tκ , B −1 and (B σ )−1 exist on T. Prove  (AB −1 )Δ = AΔ − AB −1 B Δ (B σ )−1 σ   = AΔ − AB −1 B Δ B −1 on Tκ . Definition 2.5. We say that the matrix A is rd-continuous on T if each entry of A is rd-continuous. The class of such rd-continuous m × n matrix-valued functions on T is denoted by

2.1 Structure of Dynamic Systems on Time Scales

53

Crd = Crd (T) = Crd (T, R m×n ). Below we suppose that A and B are n × n matrix-valued functions. Definition 2.6. We say that the matrix A is regressive with respect to T provided I + μ(t)A(t) is

invertible

f or

t ∈ Tκ .

all

The class of such regressive and rd-continuous functions is denoted, similar to the scalar case, by R = R(T) = R(T, Rn×n ).

Theorem 2.9. The matrix-valued function A is regressive if and only if the eigenvalues λi (t) of A(t) are regressive for all 1 ≤ i ≤ n. Proof. Let j ∈ {1, . . . , n} be arbitrarily chosen and λj (t) is an eigenvalue corresponding to the eigenvector y(t). Then   1 + μ(t)λj (t) y(t) = Iy(t) + μ(t)λj (t)y(t) = Iy(t) + μ(t)A(t)y(t) = (I + μ(t)A(t)) y(t), whereupon it follows the assertion. This completes the proof. Example 2.6. Let T = 3N0 and

A(t) =

 t +2 1 , t2 + 1 t

t ∈ T.

Consider the equation

det

t + 2 − λ(t) 1 2 t − λ(t) t +1

 = 0,

t ∈ T.

We have (λ(t) − t)(λ(t) − t − 2) − t 2 − 1 = 0,

t ∈ T,

or (λ(t))2 − 2(t + 1)λ(t) + t 2 + 2t − t 2 − 1 = 0,

t ∈ T,

54

2 Dynamic Systems

or (λ(t))2 − 2(t + 1)λ(t) + 2t − 1 = 0,

t ∈ T.

Therefore λ1,2 (t) = t + 1 ± = t +1± = t +1±

  

(t + 1)2 − 2t + 1 t 2 + 2t + 1 − 2t + 1 t 2 + 2,

t ∈ T,

and 1 + 3λ1,2 (t) = 0  1 + 3 t + 1 ± t2 + 2 = 0 

⇐⇒

⇐⇒  3t + 4 = ∓3 t 2 + 2

9t + 24t + 16 = 9t + 18 2

2

⇐⇒ ⇐⇒

12t = 1. Consequently 1 + μ(t)λ1,2 (t) = 0,

t ∈ T,

i.e., the matrix A is regressive. Theorem 2.10. Let A be 2 × 2 matrix-valued function. Then A is regressive if and only if trA + μdetA is regressive. Here trA denotes the trace of the matrix A. Proof. Let

A(t) =

 a11 (t) a12 (t) , a21 (t) a22 (t)

t ∈ T.

Then

 μ(t)a11 (t) μ(t)a12 (t) μ(t)a21 (t) μ(t)a22 (t) 

1 + μ(t)a11 (t) μ(t)a12 (t) , t ∈ T. = μ(t)a21 (t) 1 + μ(t)a22 (t)

I + μ(t)A(t) =

10 01



+

2.1 Structure of Dynamic Systems on Time Scales

55

We get det (I + μ(t)A(t)) = (1 + μ(t)a11 (t))(1 + μ(t)a22 (t)) − (μ(t))2 a12 (t)a21 (t) = 1 + μ(t)a22 (t) + μ(t)a11 (t) + (μ(t))2 a11 (t)a22 (t) − (μ(t))2 a12 (t)a21 (t) = 1 + μ(t)(tr(A))(t) + (μ(t))2 (detA)(t) = 1 + μ(t) ((trA)(t) + μ(t)(detA)(t)) , t ∈ T (2.8) 1. Let A is regressive. Then det (I + μ(t)A(t)) = 0,

t ∈ T.

Hence from (2.8), we obtain 1 + μ(t) ((trA)(t) + μ(t)(detA)(t)) = 0,

t ∈ Tκ ,

(2.9)

i.e., trA + μdetA is regressive. 2. Let trA + μdetA is regressive. Then (2.9) holds. Hence from (2.8), we conclude that A is regressive. This completes the proof. Definition 2.7. Assume that A and B are regressive on T. Then we define A ⊕ B, A, and A  B by (A ⊕ B)(t) = A(t) + B(t) + μ(t)A(t)B(t), (A)(t) = −(I + μ(t)A(t))−1 A(t), (A  B)(t) = (A ⊕ (B))(t),

t ∈ T,

respectively. Example 2.7. Let T = 2N and

A(t) =

1 t 2 3t



,

B(t) =

 t 1 , 2t 3

t ∈ T.

56

2 Dynamic Systems

Here σ (t) = t + 2, μ(t) = 2,

t ∈ T.

Then (A ⊕ B)(t) = A(t) + B(t) + μ(t)A(t)B(t)

  1 t t 1 = + 2 3t 2t 3

  1 t t 1 +2 2 3t 2t 3 



1+t 1+t t + 2t 2 1 + 3t = +2 2t + 6t 2 2 + 9t 2(1 + t) 3(1 + t) 

1 + 3t + 4t 2 3 + 7t , = 2 + 6t + 12t 2 7 + 21t (B ⊕ A)(t) = A(t) + B(t) + μ(t)B(t)A(t)

  1 t t 1 = + 2 3t 2t 3

  t 1 1 t +2 2t 3 2 3t 



1+t 1+t t + 2 t 2 + 3t = +2 2t + 6 2t 2 + 9t 2(1 + t) 3(1 + t) 

 1+t 1+t 2t + 4 2t 2 + 6t = + 4t + 12 4t 2 + 18t 2 + 2t 3 + 3t 

5 + 3t 1 + 7t + 2t 2 , = 14 + 6t 3 + 21t + 4t 2



 10 t 1 I + μ(t)B(t) = +2 01 2t 3

 1 + 2t 2 = , 4t 7 det (I + μ(t)B(t)) = 7 + 14t − 8t = 7 + 6t,

2.1 Structure of Dynamic Systems on Time Scales

(I + μ(t)B(t))

−1

1 = 7 + 6t %

7 7+6t 4t − 7+6t

=

57

7 −2 −4t 1 + 2t & 2 − 7+6t , 1+2t



7+6t

(B)(t) = −(I + μ(t)B(t))−1 B(t)

  1 7 −2 t 1 =− 2t 3 7 + 6t −4t 1 + 2t

 1 3t 1 =− 7 + 6t 2t 2t + 3 & % 3t 1 − 7+6t − 7+6t , = 2t − 7+6t − 2t+3 7+6t (A  B)(t) = (A ⊕ (B))(t) = A(t) + (B)(t) + μ(t)A(t)(B)(t) &

 % 3t 1 − 7+6t − 7+6t 1 t = + 2t − 7+6t − 2t+3 2 3t 7+6t &

% 3t 1 − 7+6t − 7+6t 1 t +2 2t − 7+6t − 2t+3 2 3t 7+6t % & 2 6t +7t−1 7+3t 7+6t 7+6t 14+10t 18t 2 +19t−3 7+6t 7+6t

=

%

+2 % =

6t 2 +7t−1 7+3t 7+6t 7+6t 14+10t 18t 2 +19t−3 7+6t 7+6t

% + % =

−2t 2 −3t −2t 2 −3t−1 7+6t 7+6t −6t 2 −6t −6t 2 −9t−2 7+6t 7+6t

&

&

−4t 2 −6t−2 −4t 2 −6t 7+6t 7+6t −12t 2 −12t −12t 2 −18t−4 7+6t 7+6t

2t 2 +t−3 7−3t−4t 2 7+6t 7+6t 2 −12t −2t+14 6t 2 +t−7 7+6t 7+6t

&

&

,

t ∈ T.

 34 , 10

t ∈ T.

Exercise 2.5. Let T = 2N0 ,

A(t) =

 1 1 , 2 −1

B(t) =

58

2 Dynamic Systems

Find (A ⊕ B)(t),

(A)(t),

t ∈ T.

Answer. % (A)(t) =

−1+3t 1−3t 2 2 − 1−3t 2

1 − 1−3t 2

&

(A ⊕ B)(t) =

,

1+3t 1−3t 2

4 + 4t 5 + 4t 3 + 5t −1 + 8t

Theorem 2.11. (R, ⊕) is a group. Proof. Let A, B, C ∈ (R, ⊕). Then (I + μA)−1 ,

(I + μB)−1 ,

(I + μC)−1

exist and I + μ(A ⊕ B) = I + μ(A + B + μAB) = I + μA + μB + μ2 AB = I + μA + (I + μA)μB = (I + μA)(I + μB). Therefore (I + μ(A ⊕ B))−1 exists. Also, O ⊕A = A⊕O = A. Next, A ⊕ (−(I + μA)−1 A) = A − (I + μA)−1 A −μ(I + μA)−1 A2 = A − (I + μA)−1 (I + μA)A = A−A = O,

 ,

t ∈ T.

2.1 Structure of Dynamic Systems on Time Scales

59

i.e., the additive inverse of A under the addition ⊕ is −(I + μA)−1 A. Note that I + μ(−(I + μA)−1 A) = (I + μA)−1 (I + μA) −(I + μA)−1 μA = (I + μA)−1 and then −(I + μA)−1 A ∈ R. Also, (A ⊕ B) ⊕ C = (A ⊕ B) + C + μ(A ⊕ B)C = A + B + μAB + C + μ(A + B + μAB)C = A + B + μAB + C + μAC + μBC + μ2 ABC, A ⊕ (B ⊕ C) = A + (B ⊕ C) + μA(B ⊕ C) = A + B + C + μBC + μA(B + C + μBC) = A + B + C + μBC + μAB + μAC + μ2 ABC. Consequently (A ⊕ B) ⊕ C = A ⊕ (B ⊕ C), i.e., in (R, ⊕) the associative law holds. This completes the proof. With A we will denote the conjugate matrix of A, with AT we will denote the  T transpose matrix of A and A∗ = A is the conjugate transpose of A. Theorem 2.12. Let A and B be regressive. Then 1. A∗ is regressive, 2. A∗ = (A)∗ , 3. (A ⊕ B)∗ = B ∗ ⊕ A∗ . Proof. Since A is regressive, there exists (I + μA)−1 . 1. We have (I + μA)(I + μA)−1 = I (I + μA)(I + μA)−1 = I  ∗ (I + μA)−1 (I + μA∗ ) = I.

⇒ ⇒

Therefore (I + μA∗ )−1

and

(I + μAT )−1

60

2 Dynamic Systems

exist and ∗  (I + μA∗ )−1 = (I + μA)−1 , T  (I + μAT )−1 = (I + μA)−1 . Consequently A∗ is regressive. 2. We have  ∗ (A)∗ = − (I + μA)−1 A  ∗ = −A∗ (I + μA)−1 = −A∗ (I + μA∗ )−1 = A∗ . 3. We have (A ⊕ B)∗ = (A + B + μAB)∗ = A∗ + B ∗ + μB ∗ A∗ = B ∗ + A∗ + μB ∗ A∗ = B ∗ ⊕ A∗ . This completes the proof. Definition 2.8 (Matrix Exponential Functional). Let A ∈ R and t0 ∈ T. The unique solution of the IVP Y Δ = A(t)Y,

Y (t0 ) = I,

is called the matrix exponential function. It is denoted by eA (·, t0 ). Theorem 2.13. Let A, B ∈ R and t, s, r ∈ T. Then 1. 2. 3. 4. 5.

e0 (t, s) = I , eA (t, t) = I , eA (σ (t), s) = (I + μ(t)A(t))eA (t, s), −1 ∗ eA (s, t) = eA (t, s) = eA ∗ (t, s), eA (t, s)eA (s, r) = eA (t, r), eA (t, s)eB (t, s) = eA⊕B (t, s) if eA (t, s) and B commute.

Proof. 1. Consider the IVP Y Δ = O,

Y (s) = I.

2.1 Structure of Dynamic Systems on Time Scales

61

Then its unique solution is eO (t, s) = I. Now we consider the IVP Y Δ = A(t)Y,

Y (s) = I.

By the definition of eA (·, s) we obtain eA (s, s) = I. 2. By Theorem 2.3 and the definition of eA (·, s), we have Δ (t, s) eA (σ (t), s) = eA (t, s) + μ(t)eA

= eA (t, s) + μ(t)A(t)eA (t, s) = (I + μ(t)A(t))eA (t, s). 3. Let  ∗ −1 Z(t) = eA (t, s) . Then Z (t) = Δ



−1 eA (t, s)

Δ ∗

∗  −1 Δ = − eA (t, s)eA (t, s) (eA (σ (t), s))−1 ∗  −1 = − eA (t, s)A(t)eA (t, s) (eA (σ (t), s))−1 ∗  = − A(t) (eA (σ (t), s))−1 ∗  = − A(t) ((I + μ(t)A(t))eA (t, s))−1 ∗  −1 = − A(t)eA (t, s) (I + μ(t)A(t))−1   ∗ −1 = eA (t, s) − (I + μ(t)A(t))−1 A(t)  ∗ −1 = eA (t, s)(A)(t)  ∗ −1 = (A)∗ (t) eA (t, s)   = A∗ (t)Z(t).

62

2 Dynamic Systems

Therefore Z(t) = eA∗ (t, s), i.e.,  ∗ −1 eA∗ (t, s) = eA (t, s) , or −1 ∗ (t, s) = eA eA ∗ (t, s).

4. Let Z(t) = eA (t, s)eA (s, r). Then Δ (t, s)eA (s, r) Z Δ (t) = eA

= A(t)eA (t, s)eA (s, r) = A(t)Z(t), Z(r) = eA (r, s)eA (s, r) −1 = eA (s, r)eA (s, r)

=I = eA (r, r). Therefore eA (t, s)eA (s, r) = eA (t, r). 5. Let Z(t) = eA (t, s)eB (t, s). Then Δ (t, s)eBσ (t, s) + eA (t, s)eBΔ (t, s) Z Δ (t) = eA

= A(t)eA (t, s)(I + μ(t)B(t))eB (t, s) +B(t)eA (t, s)eB (t, s) = A(t)(I + μ(t)B(t))eA (t, s)eB (t, s) + B(t)eA (t, s)eB (t, s)

2.1 Structure of Dynamic Systems on Time Scales

63

= (A(t) + B(t) + μ(t)A(t)B(t))eA (t, s)eB (t, s) = (A ⊕ B)(t)Z(t), Z(s) = eA (s, s)eB (s, s) = I. Consequently eA⊕B (t, s) = eA (t, s)eB (t, s). This completes the proof. Theorem 2.14. Let A ∈ R and t, s ∈ T. Then (eA (s, t))Δ t = −eA (s, σ (t))A(t). Proof. We have  Δ −1 = e (t, s) (eA (s, t))Δ t A t

= = = = = =

 ∗ Δ eA∗ (t, s) t ∗  (eA∗ (t, s))Δ t  ∗ (A∗ )(t)eA∗ (t, s)   ∗ −1 ∗ − I + μ(t)A∗ (t) A (t)eA∗ (t, s)  ∗  −1 −A∗ (t) I + μ(t)A∗ (t) eA∗ (t, s)  ∗ −A∗ (t)eA∗ (σ (t), s)

∗ = −eA ∗ (σ (t), s)A(t)

= −eA (s, σ (t))A(t). This completes the proof. Theorem 2.15 (Variation of Constants). Let A ∈ R and f : T → Rn be rdcontinuous. Let also, t0 ∈ T and y0 ∈ Rn . Then the IVP y Δ = A(t)y + f (t),

y(t0 ) = y0 ,

(2.10)

has unique solution y : T → Rn and this solution is given by y(t) = eA (t, t0 )y0 +

t

eA (t, σ (τ ))f (τ )Δτ. t0

(2.11)

64

2 Dynamic Systems

Proof. 1. Let y be given by (2.11). Then

y(t) = eA (t, t0 ) y0 +

t

 eA (t0 , σ (τ ))f (τ )Δτ ,

t0

y Δ (t) = A(t)eA (t, t0 ) y0 +



t

eA (t0 , σ (τ ))f (τ )Δτ t0

+eA (σ (t), t0 )eA (t0 , σ (t))f (t) = A(t)y(t) + f (t), y(t0 ) = y0 , i.e., y satisfies the IVP (2.10). 2. Let y˜ be another solution of the IVP (2.10). Let also, ˜ v(t) = eA (t0 , t)y(t). Then y(t) ˜ = eA (t, t0 )v(t) and ˜ + f (t) A(t)eA (t, t0 )v(t) + f (t) = A(t)y(t) = y˜ Δ (t) = A(t)eA (t, t0 )v(t) + eA (σ (t), t0 )v Δ (t), whereupon eA (σ (t), t0 )v Δ (t) = f (t) or v Δ (t) = eA (t0 , σ (t))f (t). Hence, t

v(t) = v(t0 ) +

eA (t0 , σ (τ ))f (τ )Δτ t0

= y0 +

t

eA (t0 , σ (τ ))f (τ )Δτ t0

2.1 Structure of Dynamic Systems on Time Scales

65

and y(t) ˜ = eA (t, t0 )v(t)

= eA (t, t0 ) y0 +



t

eA (t0 , σ (τ ))f (τ )Δτ , t0

i.e., y˜ = y, where y is given by (2.11). This completes the proof. Theorem 2.16. Let A ∈ R and C be n × n differentiable matrix. If C is a solution of the dynamic equation C Δ = A(t)C − C σ A(t), then C(t)eA (t, s) = eA (t, s)C(s).

(2.12)

Proof. We fix s ∈ T. Let F (t) = C(t)eA (t, s) − eA (t, s)C(s). Then Δ Δ (t, s) − eA (t, s)C(s) F Δ (t) = C Δ (t)eA (t, s) + C(σ (t))eA

= (A(t)C(t) − C(σ (t))A(t)) eA (t, s) +C(σ (t))A(t)eA (t, s) − A(t)eA (t, s)C(s) = A(t)C(t)eA (t, s) − A(t)eA (t, s)C(s) = A(t) (C(t)eA (t, s) − eA (t, s)C(s)) = A(t)F (t), F (s) = C(s)eA (s, s) − eA (s, s)C(s) = C(s) − C(s) = 0, i.e., F Δ (t) = A(t)F (t),

F (s) = 0.

Hence from Theorem 2.15, we get F (t) = 0 and (2.12) holds. This completes the proof.

66

2 Dynamic Systems

Corollary 2.1. Let A ∈ R and C is an n × n constant matrix that commutes with A. Then C commutes with eA . Proof. We have Cσ = C

and

C Δ = A(t)C − CA(t).

Hence from Theorem 2.16, it follows that C commutes with eA . This completes the proof. Corollary 2.2. Let A ∈ R be a constant matrix. Then A commutes with eA . Proof. We apply Theorem 2.16 for C = A. This completes the proof. Theorem 2.17 (Variation of Constants). Let A ∈ R and f : T → Rn be rdcontinuous. Let also, t0 ∈ T and x0 ∈ Rn . Then the IVP x Δ = −A∗ (t)x σ + f (t),

x(t0 ) = x0 ,

has a unique solution x : T → Rn and this solution is given by x(t) = eA∗ (t, t0 )x0 +

t t0

eA∗ (t, τ )f (τ )Δτ.

Proof. The equation (2.13) can be rewritten in the form   x Δ (t) = −A∗ (t) x(t) + μ(t)x Δ (t) + f (t) = −A∗ (t)x(t) − μ(t)A∗ (t)x Δ (t) + f (t), whereupon (I + μ(t)A∗ (t))x Δ (t) = −A∗ (t)x(t) + f (t), and x Δ (t) = −A∗ (t)(I + μ(t)A∗ (t))−1 x(t) + (I + μ(t)A∗ (t))−1 f (t) = (A∗ )(t)x(t) + (I + μ(t)A∗ (t))−1 f (t). Hence from Theorem 2.15, we obtain x(t) = eA∗ (t, t0 )x0 + = eA∗ (t, t0 )x0 +

t t0 t t0

eA∗ (t, σ (τ ))(I + μ(τ )A∗ (τ ))−1 f (τ )Δτ ∗  ∗ eA (σ (τ ), t) (I + μ(τ )A(τ ))−1 f (τ )Δτ

(2.13)

2.1 Structure of Dynamic Systems on Time Scales t

= eA∗ (t, t0 )x0 + = eA∗ (t, t0 ) + = eA∗ (t, t0 ) +

67

 ∗ (I + μ(τ )A(τ ))−1 eA (σ (τ ), t) f (τ )Δτ

t0 t

(eA (τ, t))∗ f (τ )Δτ

t0 t t0

eA∗ (t, τ )f (τ )Δτ.

This completes the proof. Theorem 2.18 (Liouville’s Formula). Let A ∈ R be a 2 × 2 matrix-valued function and assume that X is a solution of the equation XΔ = A(t)X. Then X satisfies Liouville’s formula detX(t) = etrA+μdetA (t, t0 )detX(t0 ),

t ∈ T.

Proof. By Theorem 2.10, it follows that trA + μdetA is regressive. Let

A(t) =

 a11 (t) a12 (t) , a21 (t) a22 (t)

X(t) =

 x11 (t) x12 (t) . x21 (t) x22 (t)

Then

Δ    Δ (t) x11 (t) x12 a11 (t) a12 (t) x11 (t) x12 (t) Δ (t) x Δ (t) = a (t) a (t) x21 x21 (t) x22 (t) 21 22 22 

a11 (t)x11 (t) + a12 (t)x21 (t) a11 (t)x12 (t) + a12 (t)x22 (t) , = a21 (t)x11 (t) + a22 (t)x21 (t) a21 (t)x12 (t) + a22 (t)x22 (t) whereupon ⎧ Δ (t) = a11 (t)x11 (t) + a12 (t)x21 (t) ⎪ ⎪ x11 ⎨ Δ x12 (t) = a11 (t)x12 (t) + a12 (t)x22 (t) Δ ⎪ ⎪ x21 (t) = a21 (t)x11 (t) + a22 (t)x21 (t) ⎩ Δ (t) = a (t)x (t) + a (t)x (t). x22 21 12 22 22

68

2 Dynamic Systems

Then detX(t) = x11 (t)x22 (t) − x12 (t)x21 (t), Δ σ Δ (detX)Δ (t) = x11 (t)x22 (t) + x11 (t)x22 (t)

= =

=

=

Δ σ Δ −x12 (t)x21 (t) − x12 (t)x21 (t) & % & % Δ (t) x Δ (t) σ (t) x σ (t) x11 x11 12 12 + det det Δ (t) x Δ (t) x21 (t) x22 (t) x21 22 & % a11 (t)x11 (t) + a12 (t)x21 (t) a11 (t)x12 (t) + a12 (t)x22 (t) det x21 (t) x22 (t) & % Δ (t) x (t) + μ(t)x Δ (t) x11 (t) + μ(t)x11 12 12 +det Δ (t) Δ (t) x21 x22  a11 (t)x11 (t)x22 (t) + a12 (t)x21 (t)x22 (t) −a11 (t)x12 (t)x21 (t) − a12 (t)x21 (t)x22 (t) & % & % Δ (t) x Δ (t) x11 x11 (t) x12 (t) 12 + μ(t)det +det Δ (t) x (t) Δ (t) x Δ (t) x21 x21 22 22 & % x11 (t) x12 (t) a11 (t)detX(t) + det Δ (t) x Δ (t) x21 22 & % Δ Δ x11 (t) x12 (t) +μ(t)det Δ (t) x Δ (t) x21 22

= a11 (t)detX(t) %

& x11 (t) x12 (t) +det a21 (t)x11 (t) + a22 (t)x21 (t) a21 (t)x12 (t) + a22 (t)x22 (t) & % a11 (t)x11 (t) + a12 (t)x21 (t) a11 (t)x12 (t) + a12 (t)x22 (t) +μ(t)det Δ (t) Δ (t) x21 x22  = a11 (t)detX(t) + a21 (t)x11 (t)x12 (t) + a22 (t)x11 (t)x22 (t) −a21 (t)x11 (t)x12 (t) − a22 (t)x21 (t)x12 (t) & %  x11 (t) x12 (t) +μ(t) a11 (t)det Δ (t) x Δ (t) x21 22 & % x21 (t) x22 (t) +a12 (t)det Δ (t) x Δ (t) x21 22 = a11 (t)detX(t) + a22 (t)detX(t)

2.1 Structure of Dynamic Systems on Time Scales

 +μ(t) a11 (t)det

69

&

%

x12 (t) x11 (t) a21 (t)x11 (t) + a22 (t)x21 (t) a21 (t)x12 (t) + a22 (t)x22 (t) & % x22 (t) x21 (t) +a12 (t)det a21 (t)x11 (t) + a22 (t)x21 (t) a21 (t)x12 (t) + a22 (t)x22 (t)

= trA(t)detX(t)   +μ(t) a11 (t) a21 (t)x12 (t)x11 (t) + a22 (t)x22 (t)x11 (t) −a21 (t)x11 (t)x12 (t) − a22 (t)x12 (t)x21 (t)  +a12 (t) a21 (t)x21 (t)x12 (t) + a22 (t)x22 (t)x21 (t) −a21 (t)x11 (t)x22 (t) − a22 (t)x21 (t)x22 (t) = trA(t)detX(t) + μ(t) (a11 (t)a22 (t)detX(t) − a12 (t)a21 (t)detX(t)) = trA(t)detX(t) + μ(t)detA(t)detX(t) = (trA(t) + μ(t)detA(t)) detX(t),

i.e., (detX)Δ (t) = (trA(t) + μ(t)detA(t)) detX(t). Therefore detX(t) = etrA+μdetA (t, t0 )detX(t0 ). This completes the proof. Example 2.8. Let T = lZ, l > 0. Consider the IVP

X (t) = Δ

 3t 4t + 1 X(t), 3 2+t

X(0) =

 10 , 11

Here

A(t) = μ(t) = l,

 3t 4t + 1 , 3 2+t t ∈ T.

Then detA(t) = 3t (2 + t) − 3(4t + 1) = 6t + 3t 2 − 12t − 3

t ∈ T,

t > 0.

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2 Dynamic Systems

= 3t 2 − 6t − 3, trA(t) = 3t + 2 + t = 2 + 4t, trA(t) + μ(t)detA(t) = 2 + 4t + l(3t 2 − 6t − 3) = 3lt 2 + (4 − 6l)t + 2 − 3l,

t ∈ T.

Let f (t) = 3lt 2 + (4 − 6l)t + 2 − 3l,

t ∈ T.

Then, using Liouville’s formula, we get detX(t) = ef (t, 0)detX(0) = ef (t, 0) =

t−l $

(1 + μ(s)f (s))

s=0

=

t−l   $ 1 + l 3ls 2 + (4 − 6l)s + 2 − 3l s=0

t−l  $ 3l 2 s 2 + (4l − 6l 2 )s − 3l 2 + 2l + 1 , =

t ∈ T,

t > 0.

s=0

Exercise 2.6. Let A ∈ R be a n × n matrix-valued function and assume that X is a solution of the equation XΔ = A(t)X. Prove that X satisfies Liouville’s formula detX(t) = etrA+μdetA (t, t0 )detX(t0 ),

t ∈ T.

Hint. Use Theorem 2.18 and induction.

2.2 Constant Coefficients In this section we suppose that A is a n × n constant matrix and A ∈ R. Let t0 ∈ T. Consider the equation x Δ = Ax.

(2.14)

2.2 Constant Coefficients

71

Theorem 2.19. Let λ, ξ be an eigenpair of A. Then x(t) = eλ (t, t0 )ξ,

t ∈ Tκ ,

is a solution of the equation (2.14). Proof. We have Aξ = λξ. Then x Δ (t) = eλΔ (t, t0 )ξ = λeλ (t, t0 )ξ = eλ (t, t0 )(λξ ) = eλ (t, t0 )Aξ = A (eλ (t, t0 )ξ ) = Ax(t),

t ∈ Tκ .

This completes the proof. Example 2.9. Consider the system

x1Δ = −3x1 − 2x2 x2Δ = 3x1 + 4x2 .

Here

A=

 −3 −2 . 3 4

Then

0 = det

−3 − λ −2 3 4−λ

= (λ − 4)(λ + 3) + 6 = λ2 − λ − 6 and λ1 = 3,

λ2 = −2.



72

2 Dynamic Systems

The considered system is regressive for any time scale for which −2 ∈ R. Note that

ξ1 =

 1 , −3

ξ2 =

−2 1



are eigenvalues corresponding to λ1 and λ2 , respectively. Therefore x(t) = c1 e3 (t, t0 )ξ1 + c2 e−2 (t, t0 )ξ2

  1 −2 = c1 e3 (t, t0 ) + c2 e−2 (t, t0 ) , −3 1 where c1 and c2 are real constants, is a solution of the considered system for any time scale for which −2 ∈ R. Example 2.10. Consider the system ⎧ Δ ⎨ x1 (t) = x1 (t) − x2 (t) x Δ (t) = −x1 (t) + 2x2 (t) − x3 (t) ⎩ 2Δ x3 (t) = −x2 (t) + x3 (t), t ∈ Tκ . Here ⎞ 1 −1 0 A = ⎝ −1 2 −1 ⎠ . 0 −1 1 ⎛

Then 0 = det (A − λI ) ⎞ ⎛ 1 − λ −1 0 = det ⎝ −1 2 − λ −1 ⎠ 0 −1 1 − λ = −(λ − 1)2 (λ − 2) + (λ − 1) + (λ − 1) = (λ − 1) (−(λ − 1)(λ − 2) + 2)  = (λ − 1) −λ2 + 3λ = −λ(λ − 1)(λ − 3). Therefore λ1 = 0,

λ2 = 1,

λ3 = 3.

2.2 Constant Coefficients

73

Note that the matrix A is regressive for any time scale and ⎛ ⎞ 1 ξ1 = ⎝ 1 ⎠ , 1

⎞ 1 ξ2 = ⎝ 0 ⎠ , −1 ⎛

⎞ 1 ξ3 = ⎝ −2 ⎠ 1 ⎛

are eigenvalues corresponding to λ1 , λ2 and λ3 , respectively. Consequently x(t) = c1 ξ1 + c2 e1 (t, t0 )ξ2 + c3 e3 (t, t0 )ξ3 ⎞ ⎞ ⎛ ⎛ ⎛ ⎞ 1 1 1 = c1 ⎝ 1 ⎠ + c2 e1 (t, t0 ) ⎝ 0 ⎠ + c3 e3 (t, t0 ) ⎝ −2 ⎠ , −1 1 1 where c1 , c2 , and c3 are constants, is a general solution of the considered system. Example 2.11. Consider the system ⎧ Δ x (t) = −x1 (t) + x2 (t) + x3 (t) ⎪ ⎪ ⎨ 1Δ x2 (t) = x2 (t) − x3 (t) + x4 (t) ⎪ x Δ (t) = 2x3 (t) − 2x4 (t) ⎪ ⎩ 3Δ x4 (t) = 3x4 (t), t ∈ Tκ . Here ⎛

−1 ⎜ 0 A=⎜ ⎝ 0 0

1 1 0 0

1 −1 2 0

⎞ 0 1 ⎟ ⎟. −2 ⎠ 3

Then 0 = det (A − λI ) ⎛ ⎞ −1 − λ 1 1 0 ⎜ 0 1 − λ −1 1 ⎟ ⎟ = det ⎜ ⎝ 0 0 2 − λ −2 ⎠ 0 0 0 3−λ = (λ + 1)(λ − 1)(λ − 2)(λ − 3) and λ1 = −1,

λ2 = 1,

λ3 = 2,

λ4 = 3.

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2 Dynamic Systems

The matrix A is regressive for any time scale for which −1 ∈ R. Note that ⎛ ⎞ 1 ⎜0⎟ ⎟ ξ1 = ⎜ ⎝0⎠,

⎛ ⎞ 0 ⎜1⎟ ⎟ ξ2 = ⎜ ⎝0⎠,

0

⎛ ⎞ 0 ⎜0⎟ ⎟ ξ3 = ⎜ ⎝1⎠

0

and

⎛ ⎞ 0 ⎜0⎟ ⎟ ξ4 = ⎜ ⎝0⎠

0

1

are eigenvectors corresponding to λ1 , λ2 , λ3 , and λ4 , respectively. Consequently x(t) = c1 e−1 (t, t0 )ξ1 + c2 e1 (t, t0 )ξ2 + c3 e2 (t, t0 )ξ3 + c4 e3 (t, t0 )ξ4 ⎛ ⎞ ⎛ ⎞ 1 0 ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎟ = c1 e−1 (t, t0 ) ⎜ ⎝ 0 ⎠ + c2 e1 (t, t0 ) ⎝ 0 ⎠ 0 0 ⎛ ⎞ ⎛ ⎞ 0 0 ⎜0⎟ ⎜0⎟ ⎜ ⎟ ⎟ +c3 e2 (t, t0 ) ⎜ ⎝ 1 ⎠ + c4 e3 (t, t0 ) ⎝ 0 ⎠ , 0

1

where c1 , c2 , c3 , and c4 are real constants, is a general solution of the considered system. Exercise 2.7. Find a general solution of the system

x1Δ (t) = x2 (t) x2Δ (t) = x1 (t),

t ∈ Tκ .

Answer. x(t) = c1 e1 (t, t0 )



 1 1 + c2 e−1 (t, t0 ) , 1 −1

where c1 , c2 ∈ R, for any time scale for which −1 ∈ R. Theorem 2.20. Assume that A ∈ R. If x(t) = u(t) + iv(t),

t ∈ Tκ ,

is a complex vector-valued solution of the system (2.14), where u(t) and v(t) are real vector-valued functions on T, then u(t) and v(t) are real vector-valued solutions of the system (2.14) on T.

2.2 Constant Coefficients

75

Proof. We have x Δ (t) = A(t)x(t) = A(t) (u(t) + iv(t)) = A(t)u(t) + iA(t)v(t) = uΔ (t) + iv Δ (t),

t ∈ Tκ .

Equating real and imaginary parts, we get uΔ (t) = A(t)u(t), v Δ (t) = A(t)v(t),

t ∈ Tκ .

This completes the proof. Example 2.12. Consider the system

x1Δ (t) = x1 (t) + x2 (t) x2Δ (t) = −x1 (t) + x2 (t),

t ∈ Tκ .

Here

A=

 1 1 . −1 1

Then 0 = det (A − λI )

 1−λ 1 = det −1 1 − λ = (λ − 1)2 + 1 = λ2 − 2λ + 1 + 1 = λ2 − 2λ + 2, whereupon λ1,2 = 1 ± i. Note that

 1 ξ= i

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2 Dynamic Systems

is an eigenvector corresponding to the eigenvalue λ = 1 + i. We have

 1 i

x(t) = e1+i (t, t0 ) 

= e1 (t, t0 ) cos ⎛⎛ = e1 (t, t0 ) ⎝⎝ ⎛ = e1 (t, t0 ) ⎝

(t, t0 ) + i sin

1 1+μ

cos

i cos

cos

1 1+μ

1 1+μ

− sin

(t, t0 )

1 1+μ

(t, t0 )

(t, t0 )

⎛

⎠+⎝

i sin

(t, t0 )

i 1 1+μ

− sin

⎞

(t, t0 )

1 1+μ

⎞

1 1+μ

1

1 1+μ

⎛

⎠ + ie1 (t, t0 ) ⎝

⎞⎞

(t, t0 ) (t, t0 )

sin

⎞

(t, t0 )

1 1+μ

cos

⎠⎠

1 1+μ

(t, t0 )

⎠.

Consequently ⎛ e1 (t, t0 ) ⎝

cos

1 1+μ

− sin

⎞

(t, t0 )

1 1+μ

(t, t0 )

⎠

⎛ and

e1 (t, t0 ) ⎝

sin cos

1 1+μ 1 1+μ

(t, t0 ) (t, t0 )

⎞ ⎠

are solutions of the considered system. Therefore ⎛ x(t) = c1 e1 (t, t0 ) ⎝

cos

1 1+μ

− sin

(t, t0 )

1 1+μ

(t, t0 )

⎞

⎛

⎠ + c2 e1 (t, t0 ) ⎝

sin cos

1 1+μ 1 1+μ

where c1 , c2 ∈ R, is a general solution of the considered system. Example 2.13. Consider the system ⎧ Δ ⎨ x1 (t) = x2 (t) x Δ (t) = x3 (t) ⎩ 2Δ x3 (t) = 2x1 (t) − 4x2 (t) + 3x3 (t), Here ⎞ 0 1 0 A = ⎝0 0 1⎠. 2 −4 3 ⎛

Then 0 = det (A − λI )

t ∈ Tκ .

(t, t0 ) (t, t0 )

⎞ ⎠,

2.2 Constant Coefficients

77

⎞ −λ 1 0 = det ⎝ 0 −λ 1 ⎠ 2 −4 3 − λ ⎛

= −λ2 (λ − 3) + 2 − 4λ  = − λ3 − 3λ2 + 4λ − 2 = −(λ − 1)(λ2 − 2λ + 2), whereupon λ1 = 1,

λ2,3 = 1 ± i.

Note that ⎛ ⎞ 1 ξ1 = ⎝ 1 ⎠ 1

⎞ 1 ξ2 = ⎝ 1 + i ⎠ 2i ⎛

and

are eigenvectors corresponding to the eigenvalues λ1 = 1 and λ2 = 1 + i, respectively. Note that ⎞ ⎞ ⎛

 1 1 ⎟ ⎟ ⎜ ⎜ e1+i (t, t0 ) ⎝ 1 + i ⎠ = e1 (t, t0 ) cos 1 (t, t0 ) + i sin 1 (t, t0 ) ⎝ 1 + i ⎠ 1+μ 1+μ 2i 2i ⎛ ⎛⎛ ⎞ ⎞⎞ sin 1 (t, t0 ) cos 1 (t, t0 ) 1+μ 1+μ ⎜ ⎜⎜ ⎟ ⎟⎟ ⎜ (1 + i) sin 1 (t, t ) ⎟⎟ ⎜⎜ (1 + i) cos 1 (t, t ) ⎟ = e1 (t, t0 ) ⎜⎜ 0 ⎟+i⎜ 0 ⎟⎟ 1+μ 1+μ ⎝ ⎝⎝ ⎠ ⎠⎠ 2i cos 1 (t, t0 ) 2i sin 1 (t, t0 ) ⎛

1+μ

⎛⎛

cos

1+μ

1 (t, t0 ) 1+μ

⎞

⎛

i sin

1 (t, t0 ) 1+μ

⎞⎞

⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ = e1 (t, t0 ) ⎜⎜ (1 + i) cos 1 (t, t0 ) ⎟ + ⎜ (−1 + i) sin 1 (t, t0 ) ⎟⎟ 1+μ 1+μ ⎠ ⎝ ⎠⎠ ⎝⎝ 2i cos 1 (t, t0 ) −2 sin 1 (t, t0 ) 1+μ

⎛⎛

cos

1+μ

1 (t, t0 ) 1+μ

⎞

⎛

sin

1 (t, t0 ) 1+μ

⎞⎞

⎜ ⎟ ⎟⎟ ⎜⎜ ⎜ ⎟ ⎟⎟ ⎜⎜ = e1 (t, t0 ) ⎜⎜ cos 1 (t, t0 ) − sin 1 (t, t0 ) ⎟ + i ⎜ cos 1 (t, t0 ) + sin 1 (t, t0 ) ⎟⎟ . 1+μ 1+μ 1+μ 1+μ ⎝ ⎠ ⎠⎠ ⎝⎝ −2 sin 1 (t, t0 ) 2 cos 1 (t, t0 ) 1+μ

1+μ

78

2 Dynamic Systems

Consequently ⎛ ⎞ ⎛ ⎞ cos 1 (t, t0 ) 1 1+μ ⎜ ⎜ ⎟ ⎜ cos 1 (t, t0 ) − sin 1 (t, t0 ) ⎟ ⎝ ⎠ x(t) = e1 (t, t0 ) ⎜ ⎝c1 1 + c2 ⎝ ⎠ 1+μ 1+μ 1 −2 sin 1 (t, t0 ) ⎛

⎞⎞

⎛

1+μ

sin 1 (t, t0 ) 1+μ ⎜ ⎟⎟ cos 1 (t, t0 ) + sin 1 (t, t0 ) ⎟⎟ , +c3 ⎜ ⎝ ⎠⎠ 1+μ 1+μ 2 cos 1 (t, t0 ) 1+μ

where c1 , c2 , c3 ∈ R, is a general solution of the considered system. Exercise 2.8. Find a general solution of the system ⎧ Δ ⎨ x1 (t) = x1 (t) − 2x2 (t) + x3 (t) x Δ (t) = −x1 (t) + x3 (t) ⎩ 2Δ x3 (t) = x1 (t) − 2x2 (t) + x3 (t),

t ∈ Tκ .

Theorem 2.21 (Putzer’s Algorithm). Let A ∈ R be a constant n × n matrix and t0 ∈ T. If λ1 , λ2 , . . ., λn are the eigenvalues of A, then eA (t, t0 ) =

n−1 

rk+1 (t)Pk ,

k=0

where ⎛

⎞ r1 (t) ⎜ ⎟ r(t) = ⎝ ... ⎠ rn (t) is the solution of the IVP ⎛

r

Δ

λ1 ⎜1 ⎜ ⎜ =⎜ 0 ⎜ . ⎝ ..

0 λ2 1 .. .

0 0 λ3 .. .

... ... ... .. .

0 0 0 .. .

⎞ ⎟ ⎟ ⎟ ⎟ r, ⎟ ⎠

0 0 0 . . . λn

⎛ ⎞ 1 ⎜0⎟ ⎜ ⎟ ⎜ ⎟ r(t0 ) = ⎜ 0 ⎟ , ⎜.⎟ ⎝ .. ⎠ 0

P0 = I, Pk+1 = (A − λk+1 I )Pk ,

0 ≤ k ≤ n − 1.

(2.15)

2.2 Constant Coefficients

79

Proof. Since A is regressive, we have that all eigenvalues of A are regressive. Therefore the IVP (2.15) has unique solution. We set X(t) =

n−1 

rk+1 (t)Pk .

k=0

We have P1 = (A − λ1 I )P0 = (A − λ1 I ), P2 = (A − λ2 I )P1 = (A − λ2 I )(A − λ1 I ), .. . Pn = (A − λn I )Pn−1 = (A − λn I ) . . . (A − λ1 I ) = 0. Therefore XΔ (t) =

n−1 

Δ rk+1 (t)Pk ,

k=0

XΔ (t) − AX(t) =

n−1 

Δ rk+1 (t)Pk − A

k=0

n−1 

rk+1 (t)Pk

k=0

= r1Δ (t)P0 +

n−1 

Δ rk+1 (t)Pk

k=1

−A

n−1 

rk+1 (t)Pk

k=0

= λ1 r1 (t)P0 +

n−1 

(rk (t) + λk+1 rk+1 (t)) Pk

k=1

−

n−1  k=0

rk+1 (t)APk

80

2 Dynamic Systems

=

n−1 

rk (t)Pk + λ1 r1 (t)P0

k=1

+

n−1 

λk+1 rk+1 (t)Pk −

k=1

=

n−1 

rk+1 (t)APk

k=0

rk (t)Pk +

k=1

−

n−1 

n−1 

λk+1 rk+1 (t)Pk

k=0

n−1 

rk+1 (t)APk

k=0

=

n−1 

rk (t)Pk −

k=1

=

n−1 

n−1  (A − λk+1 I )rk+1 (t)Pk k=0

rk (t)Pk −

k=1

n−1 

rk+1 (t)Pk+1

k=0

= −rn (t)Pn = 0,

t ∈ Tκ .

Also, X(t0 ) =

n−1 

rk+1 (t0 )Pk

k=0

= r1 (t0 )P0 = I. This completes the proof. Example 2.14. Consider the system ⎧ Δ ⎨ x1 (t) = 2x1 (t) + x2 (t) + 2x3 (t) x Δ (t) = 4x1 (t) + 2x2 (t) + 4x3 (t) ⎩ 2Δ x3 (t) = 2x1 (t) + x2 (t) + 2x3 (t), t ∈ Tκ . Here ⎞ 212 A = ⎝4 2 4⎠. 212 ⎛

2.2 Constant Coefficients

81

Then 0 = det (A − λI ) ⎞ ⎛ 2−λ 1 2 = det ⎝ 4 2 − λ 4 ⎠ 2 1 2−λ = −(λ − 2)3 + 8 + 8 + 4(λ − 2) + 4(λ − 2) + 4(λ − 2) = −(λ − 2)3 + 12(λ − 2) + 16  = − λ3 − 6λ2 + 12λ − 8 − 12λ + 24 − 16  = − λ3 − 6λ2 = −λ2 (λ − 6), whereupon λ1 = 0,

λ2 = 0,

λ3 = 6.

Consider the IVPs r1Δ (t) = 0,

r1 (t0 ) = 1,

r2Δ (t) = r1 (t), r3Δ (t)

r2 (t0 ) = 0,

= r2 (t) + 6r3 (t),

r3 (t0 ) = 0,

t ∈ Tκ .

We have r1 (t) = 1, r2Δ (t) = 1,

t ∈ Tκ , r2 (t0 ) = 0,

t ∈ Tκ .

Then r2 (t) = t − t0 ,

t ∈ Tκ ,

and r3Δ (t) = t − t0 + 6r3 (t),

r3 (t0 ) = 0.

Therefore r3 (t) =

t t0

e6 (t, σ (τ ))(τ − t0 )Δτ,

t ∈ Tκ .

82

2 Dynamic Systems

Next, ⎞ 100 P0 = ⎝ 0 1 0 ⎠ , 001 ⎛

P1 = (A − λ1 I )P0 = AP0 =A ⎛

⎞ 212 = ⎝4 2 4⎠, 212 P2 = (A − λ1 I )(A − λ2 I ) = A2 I = A2 ⎞⎛ ⎛ 21 212 = ⎝4 2 4⎠⎝4 2 21 212 ⎞ ⎛ 12 6 12 = ⎝ 24 12 24 ⎠ , 12 6 12

⎞ 2 4⎠ 2

P3 = 0. Therefore eA (t, t0 ) = r1 (t)P0 + r2 (t)P1 + r3 (t)P2 ⎞ ⎞ ⎛ ⎛ 100 212 = ⎝ 0 1 0 ⎠ + (t − t0 ) ⎝ 4 2 4 ⎠ 001 212

+

t t0

⎞ 12 6 12 e6 (t, σ (τ ))(τ − t0 )Δτ ⎝ 24 12 24 ⎠ , 12 6 12 

⎛

and ⎛ ⎞ ⎞ c1 x1 (t) ⎝ x2 (t) ⎠ = eA (t, t0 ) ⎝ c2 ⎠ , c3 x3 (t) ⎛

where c1 , c2 , c3 ∈ R, is a general solution of the considered system.

2.3 Examples

83

Exercise 2.9. Using Putzer’s algorithm, find eA (t, t0 ), where 1.

A=

 1 2 , −1 3

2. ⎛

⎞ 1 −1 1 A = ⎝ 1 0 2⎠. −1 1 1

2.3 Examples In this section we introduce some applications of the nonlinear dynamic equations on time scales to certain real life problems. The main focus here is on populations which are basic unit in ecology. A population is defined as a group of individuals of the same species within a limited area. Mathematical models are used to predict the size or density (population size per unit area) of a population at any time in the future. Most plants, insects, mammals, and organisms reproduce seasonally or they reproduce only once (semelparous species) (multiple reproductions: Iteroparity species). In these situations, we measure the size of a population at periodic intervals of time, or from one generation to the next. Let N (t) be the size(density) of a population at time t and N (t + 1) be the size(density) of a population at time t + 1. The function NN(t+1) (t) is called the fitness function for the population or the rate of population growth, or the net reproduction rate. In the following subsections we will discuss various population growth models for cases of single species or two different species living in the same environment.

2.3.1 The Logistic Model One of the simplest population models is the logistic model which is given by the equation

 N(t) N (t + 1) = N(t) 1 + r 1 − , t ∈ Z, (2.16) K where r and K are given functions. We can rewrite the equation (2.16) in the form 

N (t) . N(t + 1) − N(t) = rN (t) 1 − K

(2.17)

84

2 Dynamic Systems

If T is an arbitrary time scale with forward jump operator and delta differentiation operator σ and Δ, respectively, we can generalize the equation (2.17) as follows N Δ (t) =

 rN (t) N(t) 1− , μ(t) K

t ∈ [0, T ].

(2.18)

When T = Z, from the equation (2.18) we obtain the equation (2.17). If N (t0 ) = N0 , where N0 is a given constant, for some t0 ∈ T, integrating both sides of the equation (2.18), we obtain N (t) = N0 +

t t0

 N(y) rN (y) 1− Δy, μ(y) K

t0 ∈ T,

(2.19)

which is a nonlinear Volterra integral equation of the second kind. Example 2.15. As a particular example we consider a model where T = Z, t0 = 0 and t ∈ [0, 10], the functions r and K are positive constants and the initial population is some positive constant N0 . In fact, the constant r represents the proportionality constant which is large for quickly growing species like bacteria and small for slowly growing populations like elephants. The constant K stands for the carrying capacity of the system, that is, the size of the population that the environment can long term sustain. On the time scale Z the equation (2.19) takes form 

t N(y) Δy, t ∈ [0, 10]. (2.20) rN (y) 1 − N (t) = N0 + K 0 In Figures 2.1, 2.2, 2.3, and 2.4 we give the graph of the solution N for different values of the positive constants r and K. It can be observed that for large values of r the solution N grows faster, as expected.

Fig. 2.1 Population N on [0, 10] with parameters r = 0.8, K = 50, and N0 = 5

200 Population growth for r=0.8, K=50, N(0)=5

180 160 140

N(t)

120 100 80 60 40 20 0 0

1

2

3

4

5 t

6

7

8

9

10

2.3 Examples

85

Fig. 2.2 Population N on [0, 10] with parameters r = 0.4, K = 50, and N0 = 5

60 Population growth for r=0.4, K=50, N(0)=5 50

N(t)

40

30

20

10

0 0

1

2

3

4

5

6

7

8

9

10

t

Fig. 2.3 Population N on [0, 10] with parameters r = 0.4, K = 100, and N0 = 5

80 Population growth for r=0.4, K=100, N(0)=5 70 60

N(t)

50 40 30 20 10 0 0

1

2

3

4

5

6

7

8

9

10

t

Fig. 2.4 Population N on [0, 10] with parameters r = 0.2, K = 100, and N0 = 5

30 Population growth for r=0.2, K=100, N(0)=5 25

N(t)

20

15

10

5

0 0

1

2

3

4

5

6

7

8

9

10

t

2.3.2 Tumor Growth Models The growth of certain tumors can also be modeled by nonlinear integral equations. These models are usually described by systems of differential equations. However, it is possible to reconsider the models in the form of integral equations on time scales. In this section we introduce two types of tumor growth models. Let N be the

86

2 Dynamic Systems

total number of tumor cells which is divided into a population P which proliferates by splitting and a population Q which remains quiescent. However, the proliferating cells can make a transition to quiescent state with a rate of r(N ) which increases with the overall size of the tumor. The model is then given by the system of dynamic equations P Δ (x) = cP (x) − r(N )P (x) QΔ (x) = r(N )P (x).

(2.21)

Since for the total number of tumor cells which is N = P + Q we have N Δ (x) = P Δ (x) + QΔ (x), then the total number of tumor cells increases only when the proliferating cells split and we have P Δ (x) = cP (x) − r(N )P (x) N Δ (x) = cP (x). Integrating both equations on [0, t], we obtain a system of Volterra integral equations of the second kind, #t P (t) = P (0) + #0 (cP (x) − r(N (x))P (x))Δx (2.22) t N (t) = N(0) + 0 cP (x)Δx. Example 2.16. We consider the tumor growth model on the time scale T = Z. The system (2.22) becomes t

P (t) = P (0) +

(cP (x) − r(N (x))P (x))Δx

0 t

N(t) = N(0) +

cP (x)Δx, 0

where we take t ∈ [0, 10]. Figures 2.5–2.8 show the graphs of P and N for certain values of the initial data P0 and N0 and the constant α. We see that both the P and the total number of cells N increase slowly. However, in Figure 2.7, where the initial number of P0 is big, after some time the number proliferating cells P exceeds the total number of cells N which shows again that the series approximation is not reliable as the distance from the center of the series increases. In the following, we give another tumor growth model. Suppose that  a, rand k are positive constants, and F : T → R is defined as F (P (t)) = rP (t) 1 − P k(t) a . Let b and c represent the transfer rate from proliferating state to quiescent state and from quiescent to proliferating state respectively. Let d > 0 represent the death rate of the cells. The model is then given by the system P Δ (x) = F (P ) − bP (x) + cQ(x) QΔ (x) = bP (x) − cQ(x) − dQ(x).

(2.23)

2.3 Examples

87

Fig. 2.5 Functions P and N on [0, 10] with α = 0.05, c = 0.3, P0 , and N0 = 8

25 denotes P(t) values denotes N(t) values

P(t) and N(t) values

20

15

10

5

0

Fig. 2.6 Functions P and N on [0, 10] with α = 0.05, c = 0.3, P0 = 5, and N0 = 8

0

1

2

3

4

5 6 t values

7

8

9

10

25 denotes P(t) values denotes N(t) values

P(t) and N(t) values

20

15

10

5

0

Fig. 2.7 Functions P and N on [0, 10] with α = 0.08, c = 0.3, P0 = 5, and N0 = 8

0

1

2

3

4

6 5 t values

7

8

9

10

50 denotes P(t) values denotes N(t) values

45

P(t) and N(t) values

40 35 30 25 20 15 10 5 0

0

1

2

3

4

5 6 t values

7

8

9

10

Integration from 0 to t yields t

P (t) = P (0) + 0

Q(t) = Q(0) +

t

(F (P ) − bP (x) + cQ(x))Δx (2.24) (bP (x) − cQ(x) − dQ(x))Δx,

0

which is a system of Volterra integral equations of the second kind.

88

2 Dynamic Systems

Fig. 2.8 Functions P and N on [0, 10] with α = 0.08, c = 0.3, P0 = 2, and N0 = 8

20 denotes P(t) values denotes N(t) values

P(t) and N(t) values

15

10

5

0

−5

0

1

2

3

4

5

6

7

8

9

10

t values

20 denotes P(t) values denotes N(t) values

15 P(t) and N(t) values

Fig. 2.9 Functions P and Q on [0, 10] with r = 0.2, a = 0.3, b = 0.15, c = 0.2, d = 0.01, K = 20 P0 = 4, and Q0 = 4

10

5

0

−5

0

1

2

3

4

5 6 t values

7

8

9

10

Example 2.17. We consider this model on the time scale T = Z. The system (2.24) becomes #t P (t) = P (0) + #0 (F (P ) − bP (x) + cQ(x))Δx t Q(t) = Q(0) + 0 (bP (x) − cQ(x) − dQ(x))Δx,   a into where t ∈ [0, 10]. We substitute the function F (P ) = rP (x) 1 − P (x) k the first equation which yields #t   2 P (t) = P (0) + 0 (r − b)P (x) − ar k (P (x)) + cQ(x) Δx #t Q(t) = Q(0) + 0 (bP (x) − cQ(x) − dQ(x))Δx, Figures 2.9 and 2.10 show the graphs of P and Q for certain values of the initial data P (0) and Q(0) and the constants a, b, c, d, and r.

2.3 Examples

89

Fig. 2.10 Functions P (t) and Q(t) on [0, 10] with r = 0.2, a = 0.3, b = 0.15, c = 0.2, d = 0.01, K = 20 P0 = 2, and Q0 = 6

25 denotes P(t) values denotes N(t) values

P(t) and Q(t) values

20 15 10 5 0 −5

0

1

2

3

4

5 6 t values

7

8

9

10

2.3.3 The Predator-Prey Model of Lotka-Volterra In this subsection, we introduce the Predator-Prey model of Lotka-Volterra on an arbitrary time scales T. We assume that x1 and x2 denote the prey population and the predator population respectively. Let A, B, C, and D be positive constants. Then, the model is given by the system of dynamic equations x1Δ (t) = Ax1 (t) − Bx1 (t)x2 (t) x2Δ (t) = −Cx2 (t) + Dx1 (t)x2 (t).

(2.25)

Integrating both equations from 0 to t we obtain the system of nonlinear integral equations of the form t

x1 (t) = x1 (0) + 0

x2 (t) = x2 (0) +

t

(Ax1 (x) − Bx1 (x)x2 (x))Δx (2.26) (−Cx2 (x) + Dx1 (x)x2 (x))Δx.

0

In the following example we study this problem on the time scales Z. Example 2.18. Let T = Z, so that μ(t) = 1. The system (2.26) becomes t

x1 (t) = x1 (0) + 0

x2 (t) = x2 (0) +

t

(Ax1 (x) − Bx1 (x)x2 (x))Δx (−Cx2 (x) + Dx1 (x)x2 (x))Δx.

0

By choosing some particular values for the constants A, B, C, and D we sketch the graphs of the prey population x1 and the predator population x2 obtained approximately in Figures 2.11–2.14. The last figure, that is, Figure 2.14, shows the case of an equilibrium solution. The number of predator and prey populations remains constant in time.

90

2 Dynamic Systems 20 denotes y(t)(prey) values denotes x(t)(predator) values

18 16 x(t) and y(t) values

Fig. 2.11 Predator x2 and prey x1 populations on [0, 10] with A = 0.3, B = 0.1, C = 0.5, D = 0.15 and x1 (0) = 4 and x2 (0) = 4

14 12 10 8 6 4 2 0

1

2

3

4

6 5 t values

7

8

9

10

20 denotes y(t)(prey) values denotes x(t)(predator) values

18 16 x(t) and y(t) values

Fig. 2.12 Predator x2 and prey x1 populations on [0, 10] with A = 0.3, B = 0.1, C = 0.5, D = 0.15 and x1 (0) = 3 and x2 (0) = 5

0

14 12 10 8 6 4 2 0

1

2

3

4

5 6 t values

7

8

9

10

20 denotes y(t)(prey) values denotes x(t)(predator) values

15 x(t) and y(t) values

Fig. 2.13 Predator x2 and prey x1 populations on [0, 10] with A = 0.3, B = 0.1, C = 0.5, D = 0.15 and x1 (0) = 5 and x2 (0) = 3

0

10

5

0

−5

0

1

2

3

4

5 6 t values

7

8

9

10

2.3.4 The Ricker Model The Ricker model is given by the equation N(t + 1) = N(t)e

 r 1− N(t) K

,

t ∈ Z,

(2.27)

2.3 Examples

91

Fig. 2.14 Predator x2 and prey x1 populations on [0, 10] with A = 0.4, B = 0.1, C = 0.6, D = 0.15 and x1 (0) = 4 and x2 (0) = 4

20 denotes y(t)(prey) values denotes x(t)(predator) values

x(t) and y(t) values

15

10

5

0

−5

0

1

2

3

4

5 6 t values

7

8

9

10

where r and K are given functions. The equation (2.27) can be rewritten in the form

 N(t)  r 1− K N(t + 1) − N(t) = e −1 ,

t ∈ Z.

On an arbitrary time scale T with a forward jump operator and delta differentiation operator σ and Δ, respectively, we can generalize the Ricker model as follows

 N(t)  r 1− K N Δ (t) = e −1 ,

t ∈ T.

(2.28)

When T = Z, from the dynamic equation (2.28), we obtain the equation (2.27). If N(t0 ) = N0 , where N0 is a given constant, for some t0 ∈ T, integrating both sides of (2.28), we get the equation t

N (t) = N0 +

 N(y)  r 1− K e − 1 Δy,

t ∈ T,

t0

which is a nonlinear Volterra integral equation of the second kind.

2.3.5 The Beverton-Holt Model The Beverton-Holt model is given by the equation N(t + 1) =

rKN(t) , K + (r − 1)N (t)

t ∈ T,

(2.29)

92

2 Dynamic Systems

where r and K are given functions. The equation (2.29) can be rewritten in the form rKN(t) − N (t) K + (r − 1)N (t) 

rK −1 = N(t) K + (r − 1)N (t)

N (t + 1) − N(t) =

= N(t)

rK − K − (r − 1)N (t) K + (r − 1)N (t)

= N(t)

(r − 1)K − (r − 1)N (t) K + (r − 1)N (t)

= N (t)(r − 1)

K − N (t) , K + (r − 1)N (t)

t ∈ Z.

If T is an arbitrary time scale with forward jump operator and delta differentiation operator σ and Δ, respectively, then the Beverton-Holt model can be generalized as follows N Δ (t) = N (t)(r − 1)

K − N(t) , K + (r − 1)N (t)

t ∈ T.

(2.30)

When T = Z, from the equation (2.30), we get the Beverton-Holt model. If N (t0 ) = N0 , where N0 is a given constant, for some t0 ∈ T, integrating both sides of (2.30), we obtain N (t) = N0 +

t

N (y)(r − 1)

t0

K − N (y) Δy, K + (r − 1)N (y)

t ∈ T,

which is a nonlinear Volterra integral equation of the second kind.

2.3.6 Competition Models There are two types of competition, one occurs among individuals of the same species (interspecific), and the other occurs among two (or more) species (interspecific competition). *Interspecific competition was accounted for in one-dimensional models as we have already seen. *Interspecific competition will be our focus here. G.F. Gause (1935) conducted an experiment on three different species of paramecium, P. aurelia, P. caudatum, and P. bursarua; T. Park (1954) conducted a similar experiment on two species of the flour beetles, Tribolium castaneum and T. confusum. Based on these experiments, the competition exclusion principle was established: If two species are very similar (such as sharing the same food ecological niche, etc.), they cannot coexist. Let x(t) and y(t) be the population densities of

2.3 Examples

93

species x and y. The Leslie-Gower Competition model is given by the system 

x(t + 1) = y(t + 1) =

r1 K1 x(t) K1 +(r1 −1)x(t)+x1 y(t) r2 K2 y(t) K2 +(r2 −1)y(t)+c2 x(t) ,

(2.31)

t ∈ Z,

where r1 i, ki , ci , i = 1, 2 are given functions. Note that r1 K1 x(t) − x(t) K1 + (r1 − 1)x(t) + c1 y(t) 

r 1 K1 −1 = x(t) K1 + (r1 − 1)x(t) + c1 y(t)

x(t + 1) − x(t) =

= x(t)

r1 K1 − K1 − (r1 − 1)x(t) − c1 y(t) K1 + (r1 − 1)x(t) + c1 y(t)

= x(t)

(r1 − 1)K1 − (r1 − 1)x(t) − c1 y(t) K1 + (r1 − 1)x(t) + c1 y(t)

= x(t)

(r1 − 1)(K1 − x(t)) − c1 y(t) , K1 + (r1 − 1)x(t) + c1 y(t)

r2 K2 y(t) − y(t) K2 + (r2 − 1)y(t) + c2 x(t)

 r 2 K2 = y(t) −1 K2 + (r2 − 1)y(t) + c2 x(t)

y(t + 1) − y(t) =

= y(t)

r2 K2 − K2 − (r2 − 1)y(t) − c2 x(t) K2 + (r2 − 1)y(t) + c2 x(t)

= y(t)

(r2 − 1)K2 − (r2 − 1)y(t) − c2 x(t) K2 + (r2 − 1)y(t) + c2 x(t)

= y(t)

(r2 − 1)(K2 − y(t)) − c2 x(t) , K2 + (r2 − 1)y(t) + c2 x(t)

t ∈ Z.

Therefore the system (2.31) can be rewritten in the form 

1 −x(t))−c1 y(t) x(t + 1) − x(t) = x(t) (rK11−1)(K +(r1 −1)x(t)+c1 y(t)

2 −y(t))−c2 x(t) y(t + 1) − y(t) = y(t) (rK22−1)(K +(r2 −1)y(t)+c2 x(t) ,

t ∈ Z.

If T is an arbitrary time scale with forward jump operator and delta differentiation operator σ and Δ, respectively, the system (2.31) can be generalized as follows 

1 −x(t))−c1 y(t) x Δ (t) = x(t) (rK11−1)(K +(r1 −1)x(t)+c1 y(t)

2 −y(t))−c2 x(t) y Δ (t) = y(t) (rK22−1)(K +(r2 −1)y(t)+c2 x(t) ,

t ∈ T.

(2.32)

94

2 Dynamic Systems

When T = Z, from the system (2.32), we obtain the Leslie-Gower model (2.31). If x(t0 ) = x0 , y(t0 ) = y0 , where x0 and y0 are given constants, for some t0 ∈ T, integrating both sides of (2.32), we get 

x(t) = x0 + y(t) = y0 +

#t #tt0 t0

1 −x(z))−c1 y(z) x(t) (rK11−1)(K +(r1 −1)x(z)+c1 y(z) Δz

2 −y(z))−c2 x(z) y(z) (rK22−1)(K +(r2 −1)y(z)+c2 x(z) Δz,

t ∈ T,

which is a system of nonlinear Volterra integral equations of the second kind. The Ricker Competition model is given by the system ⎧ ⎨ ⎩

x(t + 1) = x(t)e y(t + 1) = y(t)e

 r1 1− x(t) K1 −c1 y(t)  r2 1− y(t) −c2 x(t) K 2

(2.33) ,

t ∈ Z,

where ri , Ki , ci , i = 1, 2 are given functions. The system (2.33) can be rewritten in the form ⎧

 x(t)  r1 1− K −c1 y(t) ⎪ ⎪ 1 −1 ⎨ x(t + 1) − x(t) = x(t) e

 y(t)  r2 1− K −c2 x(t) ⎪ ⎪ 2 − 1 , t ∈ Z, ⎩ y(t + 1) − y(t) = y(t) e If T is an arbitrary time scale with forward jump operator and delta differentiation operator σ and Δ, respectively, the system (2.31) can be generalized as follows ⎧

 x(t)  r1 1− K −c1 y(t) ⎪ Δ ⎪ 1 −1 ⎨ x (t) = x(t) e

 y(t)  r2 1− K −c2 x(t) ⎪ Δ ⎪ 2 −1 , ⎩ y (t) = y(t) e

(2.34) t ∈ T.

When T = Z, from the system (2.34), we obtain the Leslie-Gower model (2.33). If x(t0 ) = x0 , y(t0 ) = y0 , where x0 and y0 are given constants, for some t0 ∈ T, integrating both sides of (2.34), we get ⎧

 x(z)  #t r1 1− K −c1 y(z) ⎪ ⎪ 1 − 1 Δz ⎨ x(t) = x0 + t0 x(z) e

 y(z)  #t r2 1− K −c2 x(z) ⎪ ⎪ 2 − 1 Δz, ⎩ y(t) = y0 + t0 y(z) e

t ∈ T,

which is a system of nonlinear Volterra integral equations of the second kind.

2.3 Examples

95

2.3.7 Single Species Population Growth Models In the following, we introduce population growth models for a single species. These models have been given for both continuous and discrete cases. Let T be a time scale with forward jump operator σ and delta differentiation operator Δ. Let N (t) denote the size of the population of a certain species at time t, where t ∈ T. In the following + three models we assume that N(t) = exp(φ(t)). Let a : T # → R, b, c : T → R 1 w and g : T → T be w-periodic functions. Let also a˜ = w 0 a(t)Δt > 0 and p, q be positive constants. The model is given as N (σ (t)) = N(t) exp(a(t) + b(t)N p (g(t)) − cN q (g(t))).

(2.35)

Using N (t) = exp(φ(t)) we rewrite the equation as exp(φ(σ (t))) = exp(φ(t)) exp(a(t) + b(t) exp(pφ(g(t))) − c exp(qφ(g(t)))), from which we obtain exp(φ(σ (t))) = exp(φ(t) + a(t) + b(t) exp(pφ(g(t))) − c exp(qφ(g(t)))), or φ(σ (t)) = φ(t) + a(t) + b(t) exp(pφ(g(t))) − c exp(qφ(g(t))), Then we deduce φ Δ (t) =

1 (a(t) + b(t) exp(pφ(g(t))) − c exp(qφ(g(t)))) , μ(t)

which upon integration from 0 to x yields x

φ(x) = 0

1 (a(t) + b(t) exp(pφ(g(t))) − c exp(qφ(g(t)))) Δt, μ(t)

which is a Volterra integral equation of the second # wkind. Let r, K : T → R, g : T → T, be w-periodic functions. Let also r˜ = w1 0 r(t)Δt > 0, K(t) > 0 and θ be a positive constant. The model is given as %

N(σ (t)) = N(t) exp r(t) −

N (g(t)) K(t)

θ & .

Using N (t) = eφ(t) we rewrite the equation as ⎛ eφ(σ (t)) = eφ(t) exp ⎝r(t) −

%

eφ(g(t)) K(t)

&θ ⎞ ⎠,

(2.36)

96

2 Dynamic Systems

from which we obtain %

eθφ(g(t)) exp(φ(σ (t))) = exp φ(t) + r(t) − (K(t))θ

& ,

or φ(σ (t)) = φ(t) + r(t) −

eθφ(g(t)) . (K(t))θ

Now, we rewrite this equation as & % 1 eθφ(g(t)) φ (t) = . r(t) − μ(t) (K(t))θ Δ

Integrating from 0 to x we obtain the following Volterra integral equation of the second kind & % x 1 eθφ(g(t)) φ(x) = Δt. r(t) − K(t)θ 0 μ(t) Let r, c, K : T → R+ , g, h : T → T, be w-periodic functions. The population model is given by the dynamic equation 

K(t) − N (g(t)) . N (σ (t)) = N(t) exp r(t) K(t) + c(t)N(h(t))

(2.37)

Employing again N(t) = eφ(t) we rewrite the equation as %

eφ(σ (t))

K(t) − eφ(g(t)) = eφ(t) exp r(t) K(t) + c(t)eφ((h(t)))

&

which can be written as φ(σ (t)) = φ(t) + r(t)

K(t) − eφ(g(t)) . K(t) + c(t)eφ((h(t)))

Now, we have φ Δ (t) =

K(t) − eφ(g(t)) 1 r(t) . μ(t) K(t) + c(t)eφ((h(t)))

,

2.4 Advanced Practical Problems

97

Integrating from 0 to x we obtain the following Volterra integral equation of the second kind x

φ(x) = 0

K(t) − eφ(g(t)) 1 r(t) Δt. μ(t) K(t) + c(t)eφ((h(t)))

2.4 Advanced Practical Problems Problem 2.1. Let T = 3N0 , % A(t) =

t 2 +2 t+1

t 2 + 3t 4t − 1 3t

& t ∈ T.

,

Find AΔ (t), t ∈ T. Answer. % AΔ (t) =

3t 2 +4t−2 (t+1)(3t+1)

4

4t + 3 3

& ,

t ∈ T.

Problem 2.2. Let T = 3N0 ,

A(t) =

 t + 10 t 2 − 2t + 2 , t t2 + t + 1

B(t) =

 t −2 t +1 , t2 t3 − 1

t ∈ T.

Prove (AB)Δ (t) = AΔ (t)B σ (t) + A(t)B Δ (t), Problem 2.3. Let T = 3N0 and % & t 2 + 2t + 2 t + 1 A(t) = , 1 t2 + 2 t+1

t ∈ Tκ .

t ∈ T.

Prove that (Aσ )−1 (t) = (A−1 )σ (t),

t ∈ T.

√ Problem 2.4. Let T be a time scale with μ(t) = t + 2 t + 1, t ∈ T, and

A(t) =

 1 −1 , 0 1

B(t) =

 2 0 , −1 1

t ∈ T.

98

2 Dynamic Systems

Find (A ⊕ B)(t),

t ∈ T.

Answer.

 √ √ 3(t + 2 t + 2) −(t + 2 t + 2) √ √ , −(t + 2 t + 2) t + 2 t + 3

t ∈ T.

Problem 2.5. Find a general solution of the system

x1Δ (t) = 2x1 (t) + 3x2 (t) x2Δ (t) = x1 (t) + 4x2 (t), t ∈ Tκ .

Answer.

3 x(t) = c1 e1 (t, t0 ) −1



 1 + c2 e5 (t, t0 ) , 1

t ∈ Tκ ,

where c1 , c2 ∈ R. Problem 2.6. Find a general solution of the system ⎧ Δ ⎨ x1 (t) = −x1 (t) − x2 (t) − x3 (t) x Δ (t) = x1 (t) − x2 (t) + 3x3 (t) ⎩ 2Δ x3 (t) = x1 (t) − x2 (t) + 4x3 (t),

t ∈ Tκ .

Problem 2.7. Using Putzer’s algorithm, find eA (t, t0 ), where 1.

A=

 2 3 , 1 −4

2. ⎞ −1 2 3 A = ⎝ 1 1 −4 ⎠ . 1 −1 2 ⎛

Chapter 3

Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness Theorems

The material of this chapter follows the papers and monographs [1, 3, 26, 28–31, 67– 69, 71, 77, 78, 90, 111, 129, 136, 142, 151, 159, 172, 198, 200, 223, 244]. Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively, such that (0, ∞) ⊂ T. Let t0 ∈ T.

3.1 Classification of Functional Dynamic Equations Definition 3.1. Functional dynamic equations (FDEs) or dynamic equations with a deviating argument are dynamic equations in which the unknown function and its derivatives enter under different values of the argument. Consider the functional dynamic equation x Δ (t) = f (t, x(t), x(τ (t))),

t ≥ t0 ,

(3.1)

where f : T × R × R → C is a given function and τ : T → T is an increasing and unbounded function on T such that τ (t) ≤ t, t ∈ T. Obviously τ is called delay. The basic initial value problem (IVP) for the equation (3.1) consists in the determination of a function x : [τ (t0 ), ∞) → C that satisfies (3.1) for t > t0 and x(t) = φ(t),

t ∈ [τ (t0 ), t0 ],

(3.2)

where φ is a given function on [τ (t0 ), t0 ]. Definition 3.2. The function φ will be called the initial function for the IVP for the equation (3.1). Definition 3.3. The closed interval [τ (t0 ), t0 ] will be called the initial set for the IVP for the equation (3.1). © Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_3

99

100

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Assume that φ ∈ Crd ([τ (t0 ), t0 ]). Definition 3.4. A function x : [τ (t0 ), ∞) → R is said to be a solution of the 1 ([t , ∞)) and x satisfies (3.1) equation (3.1) provided x ∈ Crd ([τ (t0 ), ∞)), x ∈ Crd 0 for all t ≥ t0 . Moreover, if x satisfies (3.2), we say that it is a solution of the IVP (3.1) and (3.2). Example 3.1. Let T = Z. Consider the equation x Δ (t) = x(t − 1) + 5t − t 2 ,

t ≥ 1.

We will prove that x(t) = t 2 − t − 2,

t ∈ T,

σ (t) = t + 1,

t ∈ T,

is its solution. Here

x (t) = σ (t) + t − 1 Δ

= t +1+t −1 = 2t, x(t − 1) = (t − 1)2 − (t − 1) − 2 = t 2 − 2t + 1 − t + 1 − 2 = t 2 − 3t, x(t − 1) + 5t − t 2 = t 2 − 3t + 5t − t 2 = 2t = x Δ (t),

t > 1.

Example 3.2. Let T = 2N0 . Consider the equation 12 x x (t) = t Δ

x(t) = t 2 ,

 t , 2

t ≥ 2.

t ∈ [1, 2].

We will prove that x(t) = t 2 ,

t ∈ [1, ∞),

is its solution. Here σ (t) = 2t,

t ∈ T.

3.1 Classification of Functional Dynamic Equations

101

For t ∈ [2, ∞), we get x Δ (t) = σ (t) + t = 2t + t = 3t,

 t t2 x = , 2 4



 t 12 t 2 12 x = t 2 t 4 = 3t = x Δ (t). Exercise 3.1. Let T = 3N0 . Prove that x(t) = t 3 ,

t ∈ T,

is a solution of the equation x Δ (t) =

2 t 9477 x , 4 3 t

t ≥ 3.

Remark 3.1. If in the equation (3.1) and in the initial conditions x, f , and φ are vector functions, then we obtain the basic IVP for a system of equations. Definition 3.5. Dynamic equations  m x Δ 0 (t) = f t, x(t), . . . , x Δm0 −1 (t), m

x(τ1 (t)), . . . , x Δ 1 (τ1 (t)), m x(τ2 (t)), . . . , x Δ 2 (τ2 (t)), ··· m

x(τl (t)), . . . , x Δ l (τl (t)) ,

(3.3) t ≥ t0 ,

where τi : T → T are increasing and unbounded functions on T such that τi (t) ≤ t, t ∈ T, i ∈ {1, . . . , l}, n = max mi , i∈{0,...,l}

are called functional dynamic equations of nth order.

102

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Let Iti0 = [τi (t0 ), t0 ], i ∈ {1, . . . , l}. Denote It0 =

l 

Iti0

i=1

and m = max mi . i∈{1,...,l}

Assume that on It0 are given the functions φk ∈ C m (It0 ) and k ∈ {0, . . . , m}. Let x0k = φk (t0 ),

k ∈ {0, . . . , m}.

If m < n − 1, suppose that x0m+1 , . . ., x0n−1 are given. The basic IVP for theequation '(3.3) consists of the determination of a solution x ∈ C m (It0 [t0 , ∞)) C n ([t0 , ∞)) such that k

x Δ (t0 + 0) = x0k , k ∈ {0, . . . , n − 1}, k x Δ (τi (t)) = φk (τi (t)) if τi (t) ≤ t0 ,

k ∈ {0, . . . , m},

i ∈ {1, . . . , l}. (3.4)

Let λ = m0 − m. Definition 3.6. 1. If λ > 0, then the equation (3.3) is called a functional dynamic equation with a retarded argument or a delay dynamic equation. 2. If λ = 0, then the equation (3.3) is called a functional dynamic equation of neutral type. 3. If λ < 0, then the equation (3.3) is called a functional dynamic equation of advanced type. Example 3.3. Let T = Z. The functional dynamic equation   2 x Δ (t) = f t, x(t), x Δ (t − 1) ,

t ≥ 1,

is a functional dynamic equation with a retarded argument. Example 3.4. Let T = 2N0 . The functional dynamic equation x

Δ2

(t) = f



 t Δ2 t ,x , t, x(t), x 2 2 Δ

is a functional dynamic equation of neutral type.

t ≥ 2,

3.2 The Method of Steps

103

Example 3.5. Let T = Z. The functional dynamic equation  2 2 3 x Δ (t) = f t, x(t), x Δ (t − 2), x Δ (t − 3), x Δ (t − 4) ,

t ≥ 4,

is a functional dynamic equation of advanced type. Exercise 3.2. Classify each of the following equations as a delay dynamic equation, a functional dynamic equation of neutral type or a functional dynamic equation of advanced type.   1. T = Z, x Δ (t) = f t, x(t −1), x Δ (t − 2) , t > 2,   2 t2 , x Δ 4t , t > 4, 2. T = 4N0 , x Δ (t) = f t, x 16  3 5 3. T = 2Z, x Δ (t) = f t, x(t), x Δ (t − 2) , t > 2. Definition 3.7. A solution x of a FDE is called eventually positive if there exists a t1 ∈ [t0 , ∞) such that x > 0 on [t1 , ∞), and if (−x) is eventually positive, then x is called eventually negative. If an FDE has a solution which is either eventually positive or eventually negative, then it is called nonoscillatory. A solution, which is neither eventually positive nor eventually negative, is called oscillatory. An FDE is called oscillatory provided that each of its solution is oscillatory.

3.2 The Method of Steps Consider the basic IVP (3.1), (3.2) when the initial function φ ∈ Crd (T). The method of steps consists of determination of a solution x from the dynamic equation without retardation x Δ (t) = f (t, x(t), φ(τ (t))) , x(t0 ) = φ(t0 ),

t ∈ [t0 , γ1 (t0 )],

(3.5)

where γ1 (t0 ) ∈ T is the largest number so that τ (t) is less than t0 when t ≥ t0 . Assume that the problem (3.5) has a solution x = φ1 , defined on the whole interval [t0 , γ1 (t0 )]. Let

x1 (t) =

φ(t) for t ∈ [τ (t0 ), t0 ] φ1 (t) for t ∈ [t0 , γ1 (t0 )].

Then we obtain the following IVP x Δ (t) = f (t, x(t), x1 (τ (t))), x(γ1 (t0 )) = x1 (γ1 (t0 )),

t ∈ [γ1 (t0 ), γ2 (t0 )],

(3.6)

104

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

where γ2 (t0 ) ∈ T is the largest number so that τ (t) is less than γ1 (t0 ) when t ≥ γ1 (t0 ). Let φ2 is a solution of the IVP (3.6), and so on. Then ⎧ φ(t) for t ∈ [τ (t0 ), t0 ] ⎪ ⎪ ⎪ ⎨ φ1 (t) for t ∈ [t0 , γ1 (t0 )] x(t) = φ (t) for t ∈ [γ (t ), γ (t )] 2 1 0 2 0 ⎪ ⎪ ⎪ ⎩ .. . is a solution of the IVP (3.1), (3.2). Example 3.6. Let T = Z. Consider the IVP x Δ (t) = x(t − 1), x(t) = 1,

t ≥ 1,

0 ≤ t ≤ 1.

Here σ (t) = t + 1,

μ(t) = 1,

t ∈ T.

Firstly, we will solve the IVP x Δ (t) = 1,

t ∈ [1, 2],

x(1) = 1. We have φ1 (t) = t + c1 ,

t ∈ [1, 2],

where c1 is a constant. Hence, 1 = x(1) = 1 + c1 , whereupon c1 = 0. Therefore φ1 (t) = t,

t ∈ [1, 2].

Next, consider the IVP x Δ (t) = φ1 (t − 1), x(2) = φ1 (2),

t ∈ [2, 3],

3.2 The Method of Steps

105

or x Δ (t) = t − 1,

t ∈ [2, 3],

x(2) = 2. Let g1 (t) =

1 2 (t − 3t), 2

t ∈ [2, 3].

Then 1 (σ (t) + t − 3) 2 1 = (t + 1 + t − 3) 2 = t − 1, t ∈ [2, 3].

g1Δ (t) =

Therefore φ2 (t) =

(t − 1)Δt + c2

=

g1Δ (t)Δt + c2

= g1 (t) + c2 =

1 2 (t − 3t) + c2 , 2

t ∈ [2, 3],

where c2 is a constant. We have 2 = φ2 (2) 1 (4 − 6) + c2 2 = −1 + c2 , =

from where c2 = 3. Consequently φ2 (t) =

1 2 (t − 3t) + 3, 2

t ∈ [2, 3].

106

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

We have 1 (t − 1)2 − 3(t − 1) + 3 2 12 t − 2t + 1 − 3t + 3 + 3 = 2 12 t − 5t + 5, t ∈ [3, 4]. = 2

φ2 (t − 1) =

Now we consider the IVP x Δ (t) = φ2 (t − 1),

t ∈ [3, 4],

x(3) = φ2 (3), or 1 2 (t − 5t) + 5, 2 x(3) = 3.

x Δ (t) =

t ∈ [3, 4],

Let g2 (t) =

1 3 3 2 19 t − t + t, 6 2 3

t ∈ [3, 4].

Then g2Δ (t) = = = = = =

3 19 1 (σ (t))2 + tσ (t) + t 2 − (σ (t) + t) + 6 2 3  19 3 1 (t + 1)2 + t (t + 1) + t 2 − (t + 1 + t) + 6 2 3  19 3 1 2 t + 2t + 1 + t 2 + t + t 2 − (2t + 1) + 6 2 3  1 3 19 3t 2 + 3t + 1 − 3t − + 6 2 3 1 29 1 2 1 t + t + − 3t + 2 2 6 6 1 2 (t − 5t) + 5, t ∈ [3, 4]. 2

3.2 The Method of Steps

107

Consequently φ3 (t) = =

  1 2 t − 5t + 5 Δt + c3 2 g2Δ (t)Δt + c3

= g2 (t) + c3 =

1 3 3 2 19 t − t + t + c3 , 6 2 3

t ∈ [3, 4],

where c3 is a constant. We have 3 = φ3 (3) 27 1 (27) − + 19 + c3 6 2 9 27 = − + 19 + c3 2 2 = −9 + 19 + c3 =

= 10 + c3 , whereupon c3 = −7. Therefore φ3 (t) =

1 3 3 2 19 t − t + t − 7, 6 2 3

t ∈ [3, 4],

and so on. We get ⎧ ⎪ ⎪ 1 if t ∈ [0, 1] ⎪ ⎪ ⎪ ⎨ t if t ∈ [1, 2] 1 2 x(t) = 2 (t − 3t) + 3 if t ∈ [2, 3] ⎪ 1 3 3 2 19 ⎪ ⎪ 6 t − 2 t + 3 t − 7 if t ∈ [3, 4] ⎪ ⎪ ⎩ .. . is a solution of the considered IVP.

108

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Example 3.7. Let T = 2N0 . Consider the IVP

 t x (t) = x , 2

t ≥ 2,

Δ

x(t) = 2t,

1 ≤ t ≤ 2.

Here σ (t) = 2t,

μ(t) = t,

t ∈ T.

Firstly, consider the IVP x Δ (t) = t,

2 ≤ t ≤ 4,

x(2) = 4. Let g1 (t) =

1 2 t , 3

2 ≤ t ≤ 4.

Then 1 (σ (t) + t) 3 1 = (2t + t) 3 = t, 2 ≤ t ≤ 4.

g1Δ (t) =

Therefore φ1 (t) =

tΔt + c1

= g1 (t) + c1 =

1 2 t + c1 , 3

2 ≤ t ≤ 4,

where c1 is a constant. We have 4 = φ1 (2) =

4 + c1 , 3

3.2 The Method of Steps

109

whereupon c1 =

8 3

and φ1 (t) =

1 2 (t + 8), 3

2 ≤ t ≤ 4.

 t , 2

4 ≤ t ≤ 8,

Now we consider the IVP x Δ (t) = φ1

x(4) = φ1 (4), or x Δ (t) =

1 3

 t2 +8 , 4

4 ≤ t ≤ 8,

x(4) = 8. Let g2 (t) =

1 3 8 t + t, 84 3

4 ≤ t ≤ 8.

Then 8 1  (σ (t))2 + tσ (t) + t 2 + 84 3 8 1  2 4t + 2t 2 + t 2 + = 84 3 1 2 8 t + , 4 ≤ t ≤ 8. = 12 3

g2Δ (t) =

Hence, φ2 (t) = =

1 3

 t2 + 8 Δt + c2 4

g2Δ (t)Δt + c2

= g2 (t) + c2 =

1 3 8 t + t + c2 , 84 3

4 ≤ t ≤ 8,

110

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

where c2 is a constant. We have 8 = φ2 (4) = = = = =

8 1 (64) + (4) + c2 84 3 16 32 + + c2 21 3 16 + 224 + c2 21 240 + c2 21 80 + c2 , 7

whereupon 80 7 24 =− 7

c2 = 8 −

and φ2 (t) =

24 1 3 8 t + t− , 84 3 7

4 ≤ t ≤ 8.

Consequently ⎧ ⎪ 2t if t ∈ [1, 2] ⎪ ⎪ ⎪ t2 8 ⎨ 3 + 3 if t ∈ [2, 4] x(t) = t 3 8t 24 if t ∈ [4, 8] ⎪ 84 + 3 − 7 ⎪ ⎪ ⎪ . ⎩. . is a solution of the considered IVP. Example 3.8. Let

U= T=U

 1 {0}, :n∈N 2n

     (1 − U ) (1 + U ) (2 − U ) (2 + U ) ....

3.2 The Method of Steps

111

Consider 1 t≥ , 2 ) ( 1 , x(t) = f (t), t ∈ 0, 2 * where f is a given continuous function on 0, 12 . We have x Δ (t) = x(ρ(t)),

  

 x σ 12 − x 12 1  xΔ = 2 σ 12 − 12   x 34 − f 12 = 3 1 4 − 2   x 34 − f 12 = 1 4

 1 =x ρ 2

 1 =x 4

 1 , =f 4 i.e.,  x

3 4

−f 1 4

 1 2

 1 , =f 4

whereupon x





 1 1 1 3 =f + f . 4 2 4 4

Next,   

 x σ 34 − x 34 3  = xΔ 4 σ 34 − 34   x 78 − x 34 = 7 3 8 − 4

112

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

 =

x

7 8

−x

 3 4

1 8

 3 =x ρ 4

 1 =x 2

 1 , =f 2 or x





 7 1 3 1 −x = f , 8 4 8 2

or x





 1 3 1 7 =x + f 8 4 8 2





 1 1 1 1 1 + f + f =f 2 4 4 8 2  

 1 1 1 1 + f , = 1+ 3 f 2 4 4 2

or   



1 1 1 1 1 + f . x 1− 3 = 1+ 3 f 2 4 4 2 2 Next,   

 x σ 78 − x 78 7  = xΔ 8 σ 78 − 78   7 x 15 − x 16 8 = 15 7 16 − 8   7 x 15 − x 16 8 = 1 16

3.2 The Method of Steps

113

 7 =x ρ 8

 3 , =x 4 or

x

15 16





 3 7 1 + x 8 16 4  







1 1 1 1 1 1 1 1 + f + f + f = 1+ 3 f 2 4 4 16 2 64 4 2    

1 1 1 1 1 1 f + + , = 1+ 3 + 4 f 2 4 43 4 2 2

=x

or

     1 1 1 1 1 1 1 x 1− 4 = 1+ 3 + 4 f + 3 f + , 2 4 4 4 2 2 2 and so on. Note that there exist constants M1 and p1 such that p1 =

*inf x(ρ(t)), t∈ 12 ,1

M1 = sup x(ρ(t)). * t∈

1 2 ,1

Now we apply the first mean value theorem (for integrals) and we get that there exists a Λ1 ∈ [p1 , M1 ] such that 1 1 2

x(ρ(s))Δs =

1 Λ1 . 2

Hence, 1 1 2

x Δ (s)Δs =

1 Λ1 2

or

 1 1 + Λ1 2 2

 1 1 + Λ1 . =f 2 2

x(1) = x

114

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Let  > 0 be arbitrarily chosen. There exists an N ∈ N, N > 1, such that − + x(1) ≤ x(ρ(t)) ≤  + x(1),

t =1+

1 , 2n

n ≥ N.

Hence, for t = 1 + 21n , n ≥ N , there exists a Λn ∈ [p2 , M2 ], where p2 = min{p1 , − + x(1)}, M2 = max{M1 ,  + x(1)}, so that 1+ 21n 1 2

x(ρ(s))Δs = Λn

 1 1 , + 2n 2

n ≥ N.

Therefore 1+ 21n 1 2

x Δ (s)Δs = Λn

 1 1 + , 2n 2

n ≥ N,

or 





1 1 1 1 + Λn x 1+ n = x + 2 2 2n 2



 1 1 1 + Λn , =f + 2 2n 2

n ≥ N.

In particular,





 1 1 1 1 x 1 + N = ΛN +f . + 2 2N 2 2 From here,

xΔ

1 1+ N 2

 =

 x 1+



1 2N−1 1 2N−1

 −x 1+

− 

1 = x 1 + N +1 . 2

1 2N



1 2N

Then

x 1+

1 2N −1





 1 = +x 1+ N 2 2N −1 2N +1 



 1 1 1 1 1 Λ = + f − + N +1 2N 2 2 2N −1 2N +1



 1 1 1 + f , +ΛN + N 2 2 2 1

1 − N 2

 x 1+

1



3.2 The Method of Steps

115

* * and so on, we determine x on 1, 32 . Then we determine x on 32 , 2 , and then we determine x at t = 2, as in the case t = 1, applying the first mean value theorem for integrals. Exercise 3.3. Let T = 2Z. Find a solution of the IVP x Δ (t) = 2x(t − 2) + t, x(t) = 1 − t,

t > 2,

0 ≤ t ≤ 2.

Remark 3.2. If f and φ in the IVP (3.5) have continuous derivatives of all orders, the solution of the IVP (3.5), constructed by the method of steps solution, generally speaking, has a discontinuity of the first kind of the kth order derivative at the point γk−1 (t0 ), but the lower derivatives will be continuous at this point. At the point t0 the first derivative x Δ has, generally speaking, a discontinuity of the first kind at the point t0 since we have to have f (t0 , x(t0 ), φ(τ (t0 ))) = φ Δ (t0 ). Example 3.9. Consider the IVP in Example 3.6. We have x Δ (1+) = 1 and x(1+) = 1 and

x Δ (1−) = 0, x(1−) = 1.

Example 3.10. Consider the IVP in Example 3.7. We have 2

2

x Δ (2+) = 1 and

x Δ (2−) = 0,

x Δ (2+) = 2 and

x Δ (2−) = 2,

x(2+) = 4 and

x(2−) = 4.

We may apply the method of steps to the IVP for the equations of neutral type   x Δ (t) = f t, x(t), x(τ (t)), x Δ (τ (t)) , x(t) = φ(t) on [τ (t0 ), t0 ],

t ≥ t0 ,

(3.7)

where the delay τ : T → T is an increasing and unbounded function on T such that τ (t) ≤ t, t ∈ T, and the initial function φ ∈ C 1 ([τ (t0 ), t0 ]). We determine a solution φ1 of the IVP   x Δ (t) = f t, x(t), φ(τ (t)), φ Δ (τ (t)) , x(t0 ) = φ(t0 ),

t ∈ [t0 , γ0 (t0 )],

116

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

where γ0 (t0 ) ∈ T is the largest number such that τ (t) ≤ t0 when t ≥ t0 . Then we determine a solution φ2 of the IVP   x Δ (t) = f t, x(t), φ1 (τ (t)), φ1Δ (τ (t)) ,

t ∈ [γ0 (t0 ), γ1 (t0 )],

x(γ0 (t0 )) = φ1 (γ0 (t0 )), where γ1 (t0 ) ∈ T is the largest number such that t0 ≤ τ (t) ≤ γ0 (t0 ) when t ≥ γ0 (t0 ), and so on. Then ⎧ φ(t) if t ∈ [τ (t0 ), t0 ], ⎪ ⎪ ⎪ ⎨ φ1 (t) if t ∈ [t0 , γ0 (t0 )], x(t) = φ (t) if t ∈ [γ (t ), γ (t )], 2 0 0 1 0 ⎪ ⎪ ⎪ ⎩ .. . is a solution of the considered IVP. Example 3.11. Let T = 2N0 . Consider the IVP x Δ (t) = x x(t) = t,



 t t + xΔ + 1, 2 2

t ≥ 2,

t ∈ [1, 2].

Here σ (t) = 2t, φ(t) = t,

t ∈ T.

Then φ Δ (t) = 1, t ∈ [1, 2],

 t t = , φ 2 2

 t = 1, t ∈ [2, 4]. φΔ 2 Consider the IVP x Δ (t) = φ



 t t + φΔ + 1, 2 2

x(2) = 2,

t ∈ [2, 4],

3.2 The Method of Steps

117

or t + 2, 2 x(2) = 2.

x Δ (t) =

t ∈ [2, 4],

Let g1 (t) =

t2 + 2t, 6

t ∈ [2, 4].

Then 1 (σ (t) + t) + 2 6 1 = (2t + t) + 2 6 1 = t + 2, t ∈ [2, 4]. 2

g1Δ (t) =

Hence,

φ1 (t) = =

 t + 2 Δt + c1 2

g1Δ (t)Δt + c1

= g1 (t) + c1 =

t2 + 2t + c1 , 6

t ∈ [2, 4],

where c1 is a constant. We have 2 = φ1 (2) 4 + 4 + c1 6 2 = + 4 + c1 3 14 + c1 , = 3 =

whereupon c1 = −

8 3

118

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

and φ1 (t) =

8 t2 + 2t − , 6 3

t ∈ [2, 4].

Note that 16 8 +8− 6 3 8 8 = +8− 3 3 =8

φ1 (4) =

and φ1Δ (t) =

 t = φ1 2

 Δ t = φ1 2

t + 2, 2

t ∈ [2, 4],

8 t2 +t − , 24 3 t + 2, 4

t ∈ [4, 8].

Now we consider the IVP



 t Δ t + φ1 + 1, x (t) = φ1 2 2 Δ

t ∈ [4, 8],

x(4) = 8, or 5 1 t2 + t+ , 24 4 3 x(4) = 8.

x Δ (t) =

t ∈ [4, 8],

Let g2 (t) =

5 t3 1 + t 2 + t, 168 12 3

t ∈ T.

Then g2Δ (t) =

1  (σ (t))2 + tσ (t) + t 2 168 1 5 + (σ (t) + t) + 12 3

3.2 The Method of Steps

119

=

5 1  2 1 4t + 2t 2 + t 2 + (3t) + 168 12 3

=

t2 5 1 + t+ , 24 4 3

t ∈ [4, 8].

Hence,

φ2 (t) = =

 t2 5 1 Δt + c3 + t+ 24 4 3

g2Δ (t)Δt + c3

= g2 (t) + c3 =

5 t3 1 + t 2 + t + c3 , 168 12 3

t ∈ [4, 8],

where c3 is a constant. We have 8 = φ2 (4) 5 4 64 + (16) + + c3 168 12 3 20 4 8 + + + c3 = 21 3 3 8 + 8 + c3 = 21 176 + c3 , = 21 =

whereupon c3 = 8 − =−

176 21

8 , 21

and φ3 (t) =

5 8 t3 1 + t2 + t − , 168 12 3 21

t ∈ [4, 8].

120

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Consequently ⎧ ⎪ t if t ∈ [1, 2] ⎪ ⎪ ⎪ ⎨ t 2 + 2t − 8 if t ∈ [2, 4] 6 3 x(t) = t3 5 2 1 8 + t ⎪ ⎪ 168 12 + 3 t − 21 if t ∈ [4, 8] ⎪ ⎪ . ⎩. . is a solution of the considered IVP. Exercise 3.4. Let T = 3N0 . Find a solution of the following IVP



 t t x Δ (t) = 2x(t) + x + xΔ , t ≥ 27, 3 3 x(t) = t 2 ,

t ∈ [1, 27].

We may apply the method of steps to the IVP for the functional dynamic equations of advanced type  k x Δ (t) = f t, x(t), x(τ (t)), x Δ (τ (t)), . . . , x Δ (τ (t)) , t ≥ t0 , (3.8) x(t) = φ(t) on [τ (t0 ), t0 ], where k ∈ N, k > 1, the delay τ : T → T is an increasing and unbounded function on T such that τ (t) ≤ t, t ∈ T, and the initial function φ ∈ C k ([τ (t0 ), t0 ]). We determine a solution φ1 of the IVP  x Δ (t) = f t, x(t), φ(τ (t)), φ Δ (τ (t)),

k . . . , φ Δ (τ (t)) ,

t ∈ [t0 , γ0 (t0 )],

x(t0 ) = φ(t0 ), where γ0 (t0 ) ∈ T is the largest number such that τ (t) ≤ t0 when t ≥ t0 . If φ1 ∈ C k ([t0 , γ0 (t0 )]), then we determine a solution φ2 of the IVP  k x Δ (t) = f t, x(t), φ1 (τ (t)), φ1Δ (τ (t)), . . . , φ1Δ (τ (t)) , t ∈ [γ0 (t0 ), γ1 (t0 )], x(γ0 (t0 )) = φ1 (γ0 (t0 )), where γ1 (t0 ) ∈ T is the largest number such that t0 ≤ τ (t) ≤ γ0 (t0 ) when t ≥ γ0 (t0 ), and so on. Then ⎧ ⎪ ⎪ φ(t) if t ∈ [τ (t0 ), t0 ], ⎪ ⎨ φ1 (t) if t ∈ [t0 , γ0 (t0 )], x(t) = φ (t) if t ∈ [γ (t ), γ (t )], 2 0 0 1 0 ⎪ ⎪ ⎪ ⎩ .. . is a solution of the IVP (3.8).

3.2 The Method of Steps

121

Example 3.12. Let T = 2N0 . Consider the IVP x Δ (t) = x





 t t t 2 + xΔ + xΔ , 2 2 2

x(t) = t 2 + t,

t ≥ 2,

t ∈ [1, 2].

Here σ (t) = 2t,

t ∈ T,

φ(t) = t + t, 2

t ∈ [1, 2],

k = 2. We have φ(2) = 4 + 2 = 6, φ (t) = σ (t) + t + 1 Δ

= 2t + t + 1 = 3t + 1, 2

φ Δ (t) =

 t = φ 2

 t = φΔ 2

 Δ2 t = φ 2

3,

t ∈ [1, 2],

t t2 + , 4 2 t 3 + 1, 2 3,

t ∈ [2, 4].

Consider the IVP x Δ (t) = φ





 t t t 2 + φΔ + φΔ , 2 2 2

x(2) = 6,

t ∈ [2, 4],

122

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

or x Δ (t) =

t t t2 + +3 +1+3 4 2 2

t2 + 2t + 4, 4 x(2) = 6. =

t ∈ [2, 4],

Let g1 (t) =

2t 2 t3 + + 4t, 28 3

t ∈ [2, 4].

Then 1  (σ (t))2 + tσ (t) + t 2 28 2 + (σ (t) + t) + 4 3 2 1  2 = 4t + 2t 2 + t 2 + (3t) + 4 28 3 1 2 = t + 2t + 4, t ∈ [2, 4]. 4

g1Δ (t) =

Therefore

φ1 (t) = =

 t2 + 2t + 4 Δt + c1 4

g1Δ (t)Δt + c1

= g1 (t) + c1 =

2 t3 + t 2 + 4t + c1 , 28 3

where c1 is a constant. Also, 6 = φ1 (2) 8 8 + + 8 + c1 28 3 2 8 = + + 8 + c1 7 3 =

t ∈ [2, 4],

3.3 The Picard-Lindelöf Theorem

123

= = c1 = = =

6 + 56 + 168 + c1 21 230 + c1 , 21 230 6− 21 126 − 230 21 104 . − 21

Hence, φ1 (t) =

2 t3 104 + t 2 + 4t − , 28 3 21

t ∈ [2, 4],

and so on. Thus ⎧ 2 ⎪ ⎨ t 3 + t if t ∈ [1, 2], t 2 2 104 x(t) = 28 + 3 t + 4t − 21 if t ∈ [2, 4], ⎪ ⎩ .. . is a solution of the considered IVP. Exercise 3.5. Let T = 3N0 . Find a solution of the IVP



 t t 2 + 3x Δ , x Δ (t) = −x(t) + 2x Δ 3 3 x(t) = t,

t ≥ 3,

t ∈ [1, 3].

3.3 The Picard-Lindelöf Theorem Consider the IVP x Δ (t) = f (t, x(t), x(τ (t))), x(t) = φ(t),

t ∈ [t0 , γ (t0 )],

t ∈ [τ (t0 ), t0 ],

(3.9) (3.10)

where τ : T → T is an increasing and unbounded function on T such that τ (t) ≤ t, t ∈ T, γ (t0 ) ∈ T is the largest number such that τ (t) is less than t0 when t ≥ t0 , f ∈ Crd (T × R × R), φ ∈ Crd ([τ (t0 ), t0 ]).

124

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Theorem 3.1. The IVP (3.9) and (3.10) is equivalent to the following integral equation  x(t) =

#t φ(t0 ) + t0 f (s, x(s), φ(τ (s))) Δs φ(t) for t ∈ [τ (t0 ), t0 ].

for t ∈ [t0 , γ (t0 )]

(3.11)

Proof. 1. Let x be a solution of the IVP (3.9) and (3.10). Then, for t ∈ [τ (t0 ), t0 ], using the initial condition (3.10), we get x(t) = φ(t). Let t ∈ [t0 , γ (t0 )]. We integrate both sides of (3.9) from t0 to t and using that x(t0 ) = φ(t0 ), we get t

x(t) = φ(t0 ) +

f (s, x(s), x(τ (s)))Δs t0 t

= φ(t0 ) +

f (s, x(s), φ(τ (s))) Δs. t0

2. Now we suppose that x satisfies (3.11). For t ∈ [τ (t0 ), t0 ], we have x(t) = φ(t), i.e., x satisfies (3.10). We differentiate (3.11) with respect to t, t ∈ [t0 , γ (t0 )], and we obtain x Δ (t) = f (t, x(t), φ(τ (t))) = f (t, x(t), x(τ (t))). This completes the proof. Let X be a Banach space, endowed with a norm  · . For  > 0 and x0 ∈ X, we introduce the notations B(x0 , ) = {x ∈ X : x − x0  ≤ } ,  I (φ, ) = B(φ(t), ), t∈It0

where It0 = [τ (t0 ), t0 ]. When we write max[a, b] 'T we have in mind that max[a, b]T ∈ T and max[a, b]T ≥ s for any s ∈ [a, b] T. Theorem 3.2 (The Picard-Lindelöf Theorem). Assume that for some  > 0, f ∈ Crd ([t0 , γ (t0 )] × I (φ, ) × I (φ, )), f : [t0 , γ (t0 )] × I (φ, ) × I (φ, ) → X, and that for some M > 0, f (t, u, v) ≤ M for any t ∈ [t0 , γ (t0 )] and for any u, v ∈ I (φ, ), and for some L > 0 the function f satisfies the Lipschitz condition f (t, u1 , v1 ) − f (t, u2 , v2 ) ≤ L (u1 − u2  + v1 − v2 )

3.3 The Picard-Lindelöf Theorem

125

for all t ∈ [t0 , γ (t0 )]κ and for all u1 , u 2 , v1 , v2 ∈ I (φ, ). Then the IVP (3.9), (3.10) has a unique solution x on [τ (t ), t ] [t0 , σ (ζ )], where ζ = max[t0 , t0 + δ]T and 0 0    δ = min γ (t0 ) − t0 , M . Proof. If t0 = γ (t0 ), then the unique solution of the IVP (3.9) and (3.10) is φ. Suppose that t0 < γ (t0 ). Define the space Ω of functions x satisfying the following two properties.  1 ([t , ζ ]), 1. x : [τ (t0 ), t0 ] [t0 , ζ ] → X, x ∈ Crd ([τ (t0 ), t0 ]), x ∈ Crd 0 2. x(t) = φ(t) for all t ∈ [τ (t0 ), t0 ] and x(t) ∈ B(φ(t0 ), ) for all t ∈ [t0 , ζ ]. For x ∈ Ω we define the operator  (T x)(t) =

#t φ(t0 ) + t0 f (s, x(s), x(τ (s)))Δs φ(t) for t ∈ [τ (t0 ), t0 ].

for

t ∈ [t0 , ζ ]

We have that T x : [τ (t0 ), t0 ]

 [t0 , ζ ] → X,

T x ∈ Crd ([τ (t0 ), t0 ]),

1 T x ∈ Crd ([t0 , ζ ]).

Also, for t ∈ [t0 , ζ ], we have + + (T x)(t) − φ(t0 ) = + +

t t0 t

≤

+ + f (s, x(s), x(τ (s)))Δs + +

f (s, x(s), x(τ (s)))Δs

t0

≤M

t

Δs t0

= M(t − t0 ) ≤ M(ζ − t0 ) ≤

Mδ

≤ . Therefore T : Ω ' → Ω. Note that Ω is a closed subset of the Banach 1 ([t , ζ ]) endowed with the supremum norm. Define the space Crd ([τ (t0 ), t0 ]) Crd 0 sequence of functions {xk }k∈N as follows. xk+1 (t) = (T xk )(t),

t ∈ [τ (t0 ), t0 ]

 [t0 , ζ ],

k ∈ N,

where

x0 (t) =

φ(t0 ) for t ∈ [t0 , ζ ] φ(t) for t ∈ [τ (t0 ), t0 ].

(3.12)

126

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

We have x1 (t) − x0 (t) = φ(t) − x0 (t) = 0, + + x1 (t) − x0 (t) = + +

t ∈ [τ (t0 ), t0 ], t t0 t

≤

+ + f (s, x0 (s), x0 (τ (s)))Δs + +

f (s, x0 (s), x0 (τ (s)))Δs

t0

≤M

t

Δs t0

= Mh1 (t, t0 ),

t ∈ [t0 , ζ ].

Assume that xk (t) − xk−1 (t) = 0 for

t ∈ [τ (t0 ), t0 ]

and xk (t) − xk−1 (t) ≤ MLk−1 hk (t, t0 ) for

t ∈ [t0 , ζ ],

for some k ∈ N. We will prove that xk+1 (t) − xk (t) = 0 for

t ∈ [τ (t0 ), t0 ]

and xk+1 (t) − xk (t) ≤ MLk hk+1 (t, t0 ) for

t ∈ [t0 , ζ ].

Really, for t ∈ [τ (t0 ), t0 ], we have xk+1 (t) − xk (t) = φ(t) − φ(t) =0 and xk+1 (t) − xk (t) = φ(t0 ) +

t

f (s, xk (s), xk (τ (s)))Δs t0

−φ(t0 ) − + + =+ +

t

f (s, xk−1 (s), xk−1 (τ (s)))Δs t0

t t0

+ + (f (s, xk (s), xk (τ (s))) − f (s, xk−1 (s), xk−1 (τ (s)))) Δs + +

3.3 The Picard-Lindelöf Theorem t

≤

127

f (s, xk (s), xk (τ (s))) − f (s, xk−1 (s), xk−1 (τ (s))) Δs

t0

≤L

t

xk (s) − xk−1 (s)Δs

t0

≤ MLk

t

hk (s, t0 )Δs t0

= MLk hk+1 (t, t0 ),

t ∈ [t0 , ζ ].

Next, for t ∈ [t0 , ζ ], we have + + + +  + + + x xk (t) = + (t) + (t) − x (t)) (x l+1 l + 0 + + + l∈[0,k) + + + + +  + + ≤ x0 (t) + + (xl+1 (t) − xl (t))+ + +l∈[0,k) +  ≤ x0 (t) + xl+1 (t) − xl (t) l∈[0,k)

≤ x0 (t) + M



Ll hl+1 (t, t0 )

l∈[0,k)

M (eL (t, t0 ) − 1) L M ≥ φ(t0 ) + (eL (ζ, t0 ) − 1) . L ≤ x0 (t) +

Hence, from the Weierstrass M-test, we conclude that  x0 (t) + (xl+1 (t) − xl (t)) l∈N0

is uniformly convergent on t ∈ [τ (t0 ), t0 ] x(t) = lim xk (t), k→∞



[t0 , ζ ]. Let

t ∈ [τ (t0 ), t0 ]

 [t0 , ζ ].

Note that for t ∈ [t0 , ζ ], we have f (t, xk (t), xk (τ (t))) − f (t, x(t), x(τ (t))) ≤ Lxk (t) − x(t) →0

as

k → ∞.

128

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Hence, letting k → ∞ in (3.12), we get that x is a fixed point of the operator T , i.e., x = Tx

on

[τ (t0 ), t0 ]

 [t0 , ζ ].

From here and from Theorem 3.1, we obtain that x is a solution of the IVP (3.9) and (3.10), defined on [τ (t0 ), t0 ] [t0 , ζ ]. Now  we suppose that the IVP (3.9) and (3.10) has two solutions x and y on [τ (t0 ), t0 ] [t0 , ζ ]. Then x(t) − y(t) = 0 for

t ∈ [τ (t0 ), t0 ]

and + + x(t) − y(t) = + +

t0 t

≤

+ + (f (s, x(s), x(τ (s))) − f (s, y(s), y(τ (s)))) Δs + +

t

f (s, x(s), x(τ (s))) − f (s, y(s), y(τ (s)))Δs

t0 t

≤L

x(s) − y(s)Δs,

t ∈ [t0 , ζ ].

t0

By the last inequality and by the Gronwall’s inequality, we get x(t) = y(t),

t ∈ [t0 , ζ ].

If ζ = σ (ζ ), then the proof is completed. Now we suppose that ζ is right-scattered  and x be the solution of the IVP (3.9) and (3.10), defined on [τ (t0 ), t0 ] [t0 , ζ ]. We define y : [τ (t0 ), t0 ] [t0 , σ (ζ )] → X as follows.

y(t) =

 x(t) for t ∈ [τ (t0 ), t0 ] [t0 , ζ ] φ(t0 ) + μ(ζ )f (ζ, x(ζ ), x(τ (ζ ))) .

Thus, y is the unique solution of the IVP (3.9) and (3.10) on [τ (t0 ), t0 ] This completes the proof. Example 3.13. Let T = 2N0 . Consider the IVP   x 2t x(t) + x (t) =   2 , 1 + (x(t))2 1 + x 2t Δ

x(t) = t,

t ∈ [1, 4].

Here σ (t) = 2t,

t ∈ T,

φ(t) = t,

t ∈ [1, 4],

t ∈ [4, 8],

 [t0 , σ (ζ )].

3.3 The Picard-Lindelöf Theorem

129

t0 = 4, γ (t0 ) = 8, f (t, u, v) =

u v + , 1 + u2 1 + v2

t ∈ [1, ∞),

u, v ∈ R.

Take x0 = φ(4) =4 and  = 1. Let X = Crd ([1, 4])

,

1 Crd ([4, ∞))

endowed with the supremum norm. Note that ! ! ! u v !! |f (t, u, v)| = !! + 1 + u2 1 + v2 ! |u| |v| + 2 1+u 1 + v2 ≤ 1+1 ≤

= 2,

t ∈ [1, ∞),

u, v ∈ R.

Then M = 2. Let g(u) =

u , 1 + u2

u ∈ R.

We have g (u) =

1 + u2 − 2u2 (1 + u2 )2

1 − u2 , (1 + u2 )2 ! ! ! ! ! 1 − u2 ! !g (u)! = ! ! ! (1 + u2 )2 ! =

≤

1 + u2 (1 + u2 )2

1 1 + u2 ≤ 1, u ∈ R.

=

130

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Then ! ! |g(u1 ) − g(u2 )| = !g (ξ )! |u1 − u2 | ≤ |u1 − u2 | for any u1 , u2 ∈ R and for some ξ between u1 and u2 . Hence, ! ! ! u v1 u2 v2 !! 1 ! |f (t, u1 , v1 ) − f (t, u2 , v2 )| = ! + − − ! ! 1 + u21 1 + v12 1 + u22 1 + v22 ! ! ! ! ! ! u u2 !! !! v1 v2 !! 1 ! ≤! − − !+! ! ! 1 + u21 1 + u22 ! ! 1 + v12 1 + v22 ! ≤ |u1 − u2 | + |v1 − v2 | for any t ∈ [4, ∞), u1 , u2 , v1 , v2 ∈ R. Therefore L = 1. Next,    δ = min γ (t0 ) − t0 , M 

1 = min 8 − 4, 2 

1 = min 4, 2 1 , 2 ζ = max[t0 , t0 + δ] ( ) 1 = max 4, 4 + 2 ) ( 9 = max 4, 2 =

= 4, σ (ζ ) = σ (4) = 8. Hence, by the Picard-Lindelöf theorem, it follows that the considered IVP has a unique solution x on [1, 8].

3.4 Existence and Uniqueness Theorems

131

Exercise 3.6. Let T = hZ, h > 0. Prove that the IVP x Δ (t) = x(t) =

x(t) x(t − h) +3 , 4 1 + (x(t)) 7 + (x(t − h))8 1 , 1 + t4

t ∈ [h, 2h],

t ∈ [0, h],

has a unique solution x on [0, 2h].

3.4 Existence and Uniqueness Theorems Theorem 3.3. Let γ (t0 ) < ∞, [t0 , γ (t0 )] contains infinite many points of T, f ∈ C (T × R × R), |f (t, u, v)| ≤ M for any t ∈ T, u, v ∈ R and for some positive constant M. Let also, φ ∈ C ([τ (t0 ), t0 ]). Then the IVP (3.9) and (3.10) has at least one solution x defined on [τ (t0 ), t0 ] [t0 , γ (t0 )]. Proof. Define the sequence of functions {xm }m∈N on [t0 , γ (t0 )] as follows. xm (t) = φ(t0 ) +

t

f (s, xm−1 (s), φ(τ (s)))Δs,

t ∈ [t0 , γ (t0 )],

t0

where x0 (t) = φ(t0 ),

t ∈ [t0 , γ (t0 )].

Let  > 0 be arbitrarily chosen and t1 , t2 ∈ [t0 , γ (t0 )] be such that |t1 − t2 |
0 for any z ∈ (0, a] and a

lim

→0+ 

g Δ (t) Δt = ∞ w(g(t))

for any g ∈ C 1 ([0, a]) such that g(x) > 0 for any x ∈ (0, a], g(0) = 0, g Δ (0) = 0, g Δ (x) > 0 for any x ∈ (0, a]. Let u be a nonnegative continuous function on [0, a]. Then the inequality x

u(x) ≤

w(u(t))Δt, 0

implies that u(x) = 0 for any x ∈ [0, a].

0 < x ≤ a,

134

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Proof. Let v(x) = max0≤t≤x u(t). Assume that v(x) > 0 for some 0 < x ≤ a. Then u(x) ≤ v(x) and there exists x1 ∈ [0, x] such that u(x1 ) = v(x). Hence, v(x) = u(x1 ) x1

≤

w(u(t))Δt 0 x

≤

w(u(t))Δt. 0

We set x

v(x) =

w(v(t))Δt. 0

Then v(0) = 0, v(x) ≤ v(x), and v Δ (x) = w(v(x)) ≤ w(v(x)), v (0) = 0. Δ

From the last inequality, for 0 <  < a, we get a 

v Δ (x) Δx ≤ a −  w(v(x)) < a.

From here, we obtain a

lim

→0+ 

v Δ (x) Δx ≤ a, w(v(x))

which is a contradiction. Therefore v(x) = 0 for any x ∈ [0, a] and from here, u(x) = 0 for x ∈ [0, a]. This completes the proof. Theorem 3.5. Let t0 be a right-dense point, γ (t0 ) < ∞, φ ∈ C ([τ (t0 ), t0 ]), f ∈ C (T × R × R), |f (t, u1 , v) − f (t, u2 , v)| ≤ w (|u1 − u2 |) for any t ∈ [t0 , γ (t0 )], u1 , u2 , v ∈ R, where w is as inTheorem 3.4. Then the IVP (3.9) and (3.10) has at most one solution in [τ (t0 ), t0 ] [t0 , γ (t0 )].

3.5 Continuous Dependence on Initial Data

135

Proof. Suppose that the IVP (3.9) and (3.10) has two solutions y1 and y2 defined on  [τ (t0 ), t0 ] [t0 , γ (t0 )]. If t ∈ [τ (t0 ), t0 ], then y1 (t) = y2 (t) = φ(t). Let t ∈ [t0 , γ (t0 )]. Then t

y1 (t) = φ(t0 ) +

f (s, y1 (s), φ(τ (s)))Δs, t0 t

y2 (t) = φ(t0 ) +

f (s, y2 (s), φ(τ (s)))Δs. t0

Hence, ! ! |y1 (t) − y2 (t)| = !φ(t0 ) + −φ(t0 ) − ! ! = !! ≤

t

f (s, y1 (s), φ(τ (s)))Δs t0 t t0

t t0 t

! ! f (s, y2 (s), φ(τ (s)))Δs !

! ! (f (s, y1 (s), φ(τ (s))) − f (s, y2 (s), φ(τ (s)))) Δs !!

|f (s, y1 (s), φ(τ (s))) − f (s, y2 (s), φ(τ (s)))| Δs

t0

≤

t

w (|y1 (s) − y2 (s)|) Δs.

t0

From the last inequality and from Theorem 3.4, we conclude that y1 (t) = y2 (t) for any t ∈ [t0 , γ (t0 )]. This completes the proof.

3.5 Continuous Dependence on Initial Data Consider the IVPs x1Δ (t) = f (t, x1 (t), x1 (τ (t))) , t ∈ [t0 , γ (t0 )], x1 (t) = φ1 (t), t ∈ [τ (t0 ), t0 ],

(3.14)

x2Δ (t) = f (t, x2 (t), x2 (τ (t))) , t ∈ [t0 , γ (t0 )], x2 (t) = φ2 (t), t ∈ [τ (t0 ), t0 ],

(3.15)

and

136

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

where φ1 , φ2 ∈ C ([τ (t0 ), t0 ]), f ∈ C (T × R × R), τ : T → T is an increasing and unbounded function on T such that τ (t) ≤ t, t ∈ T, γ (t0 ) ∈ T is the largest number such that τ (t) is less than t0 when t ≥ t0 . Theorem 3.6. Let γ (t0 ) < ∞ and τ , f , φ1 , φ2 be as in above, and |f (t, u1 , v1 ) − f (t, u2 , v2 )| ≤ L (|u1 − u2 | + |v1 − v2 |) for any t ∈ T, u1 , u2 , v1 , v2 ∈ R, for some positive constant L. If x1 and x2 be solutions of the IVPs (3.14) and (3.15), respectively, then %

γ (t0 )

|x1 (t) − x2 (t)| ≤ |φ1 (t0 ) − φ2 (t0 )| + L

t0

× 1+L

t

& |φ1 (τ (s)) − φ2 (τ (s))|Δs



eL (σ (s), t)Δs ,

t0

t ∈ [t0 , γ (t0 )].

Proof. Let t ∈ [t0 , γ (t0 )]. We have t

x1 (t) = φ1 (t0 ) +

f (s, x1 (s), φ1 (τ (s)))Δs, t0 t

x2 (t) = φ2 (t0 ) +

f (s, x2 (s), φ2 (τ (s)))Δs. t0

Hence, ! ! |x1 (t) − x2 (t)| = !φ1 (t0 ) + −φ2 (t0 ) −

t

f (s, x1 (s), φ1 (τ (s)))Δs t0 t t0

! ! f (s, x2 (s), φ2 (τ (s)))Δs !

≤ |φ1 (t0 ) − φ2 (t0 )| ! t ! ! ! ! + ! (f (s, x1 (s), φ1 (τ (s))) − f (s, x2 (s), φ2 (τ (s)))) Δs !! t0

≤ |φ1 (t0 ) − φ2 (t0 )| +

t

|f (s, x1 (s), φ1 (τ (s))) − f (s, x2 (s), φ2 (τ (s)))| Δs

t0

≤ |φ1 (t0 ) − φ2 (t0 )| + L

t

|φ1 (τ (s)) − φ2 (τ (s))|Δs

t0

+L

t t0

|x1 (s) − x2 (s)|Δs

3.6 Advanced Practical Problems

137 γ (t0 )

≤ |φ1 (t0 ) − φ2 (t0 )| + L

|φ1 (τ (s)) − φ2 (τ (s))|Δs

t0 t

+L

|x1 (s) − x2 (s)|Δs.

t0

From the last inequality and from Gronwall’s inequality, we get the desired result. This completes the proof.

3.6 Advanced Practical Problems Problem 3.1. Classify each of the following equations as a functional dynamic equation of a retarded argument, a functional dynamic equation of neutral type, and a functional dynamic equation of advanced type.  2 2 1. T = 2Z, x Δ (t) = f t, x(t), x Δ (t − 2) , t ≥ 2,   3 2. T = 3Z, x Δ (t) = f t, x(t), x Δ (t − 3) , t ≥ 3, 2 3. T = 6Z, x Δ (t) = f t, x(t − 6), x Δ (t − 12) , t ≥ 12. Problem 3.2. Let T = 3Z. Find a solution of the IVP x Δ (t) = x (t − 3) + 1, x(t) = 2,

t ≥ 3,

0 ≤ t ≤ 3.

Problem 3.3. Let T = 3N0 . Find a solution of the IVP

 t Δ + 1, t ≥ 27, x (t) = 2x 3 x(t) = 1,

t ∈ [1, 27].

Problem 3.4. Let T = 2N0 . Find a solution of the following IVP

 t , x (t) = tx(t) + x 2 Δ

Δ

x(t) = t 2 + t,

t ∈ [1, 4].

t ≥ 4,

Problem 3.5. Let T = 3Z. Find a solution of the following IVP 3

x Δ (t) = (t + 1)x(t) + x Δ (t − 3), x(t) = t + 2,

t ∈ [0, 3].

t ≥ 3,

138

3 Functional Dynamic Equations. Basic Concepts, Existence, and Uniqueness. . .

Problem 3.6. Let T = 3N0 . Prove that the IVP   x 3t (x(t))4 Δ −4 x (t) = 7   4 , 1 + (x(t))4 1+ x t

t ∈ [3, 9],

3

x(t) =

1 , 1 + t4

t ∈ [1, 3],

has a unique solution x on [1, 9]. Problem 3.7. Let T = 3Z. Investigate the following IVP for existence and uniqueness. x Δ (t) = t (x(t − 3))2 , x(t) = 1,

t ∈ [3, 30],

t ∈ [0, 3].

Problem 3.8. Let 

2+ T = [1, 2]

1 1 + 2n2

 . n∈N0

Prove that the IVP   x 2t 7x(t) x (t) = −3   6 , 1 + 64(x(t))2 1 + 7 x 2t

t ∈ 2+

Δ

x(t) =

1 , 7 + t4

1 1 + 2n2

 , n∈N0

t ∈ [1, 2],

has at least one solution x on [1, 2]

 2+



1 . 1+2n2 n∈N0

Problem 3.9. Let T = hZ, h > 0. Consider the IVPs x Δ (t) =

(x(t))2 1+(x(t))4

x(t) = t + 1,

2

(x(t−h)) + 2 1+(x(t−h)) 6,

t ∈ [h, 2h],

(3.16)

t ∈ [h, 2h],

(3.17)

t ∈ [0, h],

and x Δ (t) = x(t) =

(x(t))2 1+(x(t))4 t 2 + t,

2

(x(t−h)) + 2 1+(x(t−h)) 6,

t ∈ [0, h].

Find an estimate of |x1 (t) − x2 (t)|, t ∈ [0, 2h], where x1 and x2 are the solutions of the IVPs (3.16) and (3.17), respectively.

3.6 Advanced Practical Problems

139

Problem 3.10. Let T = 3N0 . Consider the IVPs   x1 3t Δ x1 (t) =   2 + t, 1 + x1 3t x1 (t) = t,

t ∈ [3, 27],

t ∈ [1, 3],

and x2Δ (t)

  x2 3t =   2 + t, 1 + x2 3t

x2 (t) = t +

1 , 100

t ∈ [3, 27],

t ∈ [1, 3].

Find an estimate of |x1 (t) − x2 (t)| for t ∈ [3, 27].

Chapter 4

Linear Functional Dynamic Equations

Our presentation in this chapter follows the papers and monographs [111, 136, 198]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

4.1 Some Properties of Linear Functional Dynamic Equations Definition 4.1. Functional dynamic equations which have the form m n  

p

apj (t)x Δ (τj (t)) = f (t),

t ∈ [t0 , γ (t0 )),

(4.1)

p=0 j =0

where apj : T → R, p ∈ {0, . . . , n}, j ∈ {0, . . . , m}, ank ≡ 0 on [t0 , γ (t0 )) for some k ∈ {0, . . . , m}, f : T → R are given functions, τj : T → T, j ∈ {0, . . . , m}, are increasing unbounded above functions such that τj (t) ≤ t, t ∈ T, j ∈ {0, . . . , m}, γ (t0 ) > t0 , will be called linear functional dynamic equations of nth order. When f ≡ 0 on [t0 , γ (t0 )), the equations (4.1) will be called homogeneous linear functional dynamic equations of nth order. Example 4.1. Let T = Z. The equation 3

2

(t + 1)x Δ (t − 3) + tx Δ (t) + x Δ (t − 4) + (t − 1)x(t − 5) = t,

t ≥5

is a linear functional dynamic equation of third order.

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_4

141

142

4 Linear Functional Dynamic Equations

Example 4.2. Let T = 2N0 . The equation 2 Δ4

t x

(t) + x

Δ3





 t Δ2 t Δ t + (t + 1)x + e1 (t, 1)x = 0, 2 4 8

t ≥ 8,

is a homogeneous linear functional dynamic equation of fourth order. n

For x ∈ C n (Tκ ), we define the operator L(x(t)) =

m n  

p

n

t ∈ Tκ .

apj (t)x Δ (τj (t)),

p=0 j =0

The equation (4.1) can be rewritten in the form L(x(t)) = f (t),

t ∈ [t0 , γ (t0 )).

(4.2)

The corresponding homogeneous equation is L(x(t)) = 0,

t ∈ [t0 , γ (t0 )).

(4.3)

Theorem 4.1. The operator L is a linear operator. n

Proof. Let x1 , x2 ∈ C n (Tκ ) and c1 , c2 ∈ C. Then L(c1 x1 (t) + c2 x2 (t)) =

m n  

p

apj (t) (c1 x1 + c2 x2 )Δ (τj (t))

p=0 j =0

=

m n  

p

apj (t)(c1 x1 )Δ (τj (t))

p=0 j =0

+

m n  

p

apj (t)(c2 x2 )Δ (τj (t))

p=0 j =0

= c1

n  m 

p

apj (t)x1Δ (τj (t))

p=0 j =0

+c2

n  m 

p

apj (t)x2Δ (τj (t))

p=0 j =0

= c1 L(x1 (t)) + c2 L(x2 (t)), This completes the proof.

n

t ∈ Tκ .

4.1 Some Properties of Linear Functional Dynamic Equations

143

j

Let It0 = [τj (t0 ), t0 ], j ∈ {0, . . . , m}. Let also, It0 =

m 

j

It0 .

j =0

Assume that on It0 are given functions φk , k ∈ {0, . . . , m}. Let x0k = φk (t0 ),

k ∈ {0, . . . , m}.

If m < n − 1, suppose that x0m+1 , . . ., x0n−1 are given. The basic IVP for the equation (4.2) is to find a solution x of the equation (4.2) such that k

x Δ (t0 + 0) = x0k , k ∈ {0, . . . , n − 1}, k x Δ (τj (t)) = φk (τj (t)), τj (t) ≤ t0 , k ∈ {0, . . . , n},

j ∈ {0, . . . , m}. (4.4)

Example 4.3. Let T = Z. The problem 2

x Δ (t − 3) + tx Δ (t) + (t + 1)x(t − 6) = 0,

t ≥ 6,

x(6+) = 4, x (6+) = 0, Δ

x(t) = t,

t ∈ [0, 3],

x (t) = t + 2, Δ

x

Δ2

t ∈ [0, 3],

(t) = t − t, 2

x(t) = t,

t ∈ [3, 6],

x (t) = 3t + 2t 2 , Δ 2

t ∈ [0, 3], t ∈ [3, 6],

x Δ (t) = 1 − t 2 + 4t, is a basic IVP.

t ∈ [3, 6],

 Definition 4.2. A function x : It0 [t0 , γ (t0 )) → R is said to be a solution of the n (I [t , γ (t ))) and it satisfies (4.2) IVP (4.2) and (4.4) provided that x ∈ Crd t0 0 0 and (4.4).

144

4 Linear Functional Dynamic Equations

Example 4.4. Let T = 2N0 . Consider the IVP −5x

Δ2



 t t + 6x = 0, (t) − (t − 1)x 2 2 Δ

t ∈ [2, 4],

x(2+) = 6, x Δ (2+) = 7, x(t) =

4 2 8 t −t + , 3 3

x Δ (t) = 4t − 1,

t ∈ [1, 2].

We will prove that

x(t) =

t 2 + t for t ∈ [2, 4] 4 2 8 3 t − t + 3 for t ∈ [1, 2],

is its solution. We have σ (t) = 2t,

t ∈ T,

x (t) = σ (t) + t + 1 Δ

= 2t + t + 1 = 3t + 1, Δ2

(t) = 3,

 t 8 t2 t =4 − + x 2 12 2 3 x

t 8 t2 − + , 3 2 3



 t t xΔ =4 −1 2 2 =

= 2t − 1,

t ∈ [2, 4].

Hence,



 t t + 6x 2 2 

2 t 8 t − + = −5(3) − (t − 1)(2t − 1) + 6 3 2 3 2

−5x Δ (t) − (t − 1)x Δ

= −15 − 2t 2 + t + 2t − 1 + 2t 2 − 3t + 16 = 0,

t ∈ [2, 4].

t ∈ [1, 2],

4.1 Some Properties of Linear Functional Dynamic Equations

145

Corollary 4.1. If x1 , . . ., xr , r ∈ N, are solutions of the equation (4.3), then

r "

ci xi

i=1

is a solution of the equation (4.3) for any constants ci ∈ C, i ∈ {1, . . . , r}.  Theorem 4.2. Let u + iv, where u, v : It0 [t0 , γ (t0 )) → R, be a solution of the IVP (4.2) and (4.4) with initial data x0k + iy0k , x0k , y0k ∈ R, φk + iψk , φk , ψk : It0 → R, k ∈ {0, . . . , n}. Then u and v are solutions of the IVPs L(x(t)) = Ref (t), x

Δk

x

(t0 + 0) = x0k ,

Δk

t ∈ [t0 , γ (t0 )),

k ∈ {0, . . . , n − 1},

(τj (t)) = φk (τj (t)),

τj (t) ≤ t0 ,

k ∈ {0, . . . , n},

j ∈ {0, . . . , m},

k ∈ {0, . . . , n},

j ∈ {0, . . . , m},

and L(x(t)) = Imf (t), x

Δk

x

(t0 + 0) = y0k ,

Δk

t ∈ [t0 , γ (t0 )),

k ∈ {0, . . . , n − 1},

(τj (t)) = ψk (τj (t)),

τj (t) ≤ t0 ,

respectively. Proof. We have L(u(t) + iv(t)) =

m n  

p

apj (t)(u + iv)Δ (τj (t))

p=0 j =0

=

m n  

 p p apj (t) uΔ (τj (t)) + iv Δ (τj (t))

p=0 j =0

=

m n  

p

apj (t)uΔ (τj (t)) + i

p=0 j =0

= Ref (t) + iImf (t),

m n   p=0 j =0

t ∈ [t0 , γ (t0 )).

Hence, m n  

p

t ∈ [t0 , γ (t0 )),

p

t ∈ [t0 , γ (t0 )).

apj (t)uΔ (τj (t)) = Ref (t),

p=0 j =0 m n   p=0 j =0

apj (t)v Δ (τj (t)) = Imf (t),

p

apj (t)v Δ (τj (t))

146

4 Linear Functional Dynamic Equations

Next, k

k

k

(u + iv)Δ (t0 + 0) = uΔ (t0 + 0) + iv Δ (t0 + 0) = x0k + iy0k ,

k ∈ {0, . . . , n − 1},

whereupon k

uΔ (t0 + 0) = x0k , k

v Δ (t0 + 0) = y0k ,

k ∈ {0, . . . , n − 1}.

Also, k

k

k

(u + iv)Δ (τj (t)) = uΔ (τj (t)) + iv Δ (τj (t)) = φk (τj (t)) + iψk (τj (t)),

τj (t) ≤ t0 ,

k ∈ {0, . . . , n}, j ∈ {0, . . . , m}, from where k

uΔ (τj (t)) = φk (τj (t)), k

v Δ (τj (t)) = ψk (τj (t)),

τj (t) ≤ t0 ,

k ∈ {0, . . . , n}, j ∈ {0, . . . , m}. This completes the proof.

4.2 First-Order Linear Functional Dynamic Equations Consider the IVP x Δ (t) = f (t)x(t) + g(t)x(τ (t)) + h(t), x(t) = φ(t),

t ≥ t0 ,

t ∈ [τ (t0 ), t0 ],

(4.5) (4.6)

where f ∈ R, g, h ∈ Crd (T), τ : T → T is an increasing unbounded above function such that τ (t) ≤ t, t ∈ T, φ ∈ C ([τ (t0 ), t0 ]). We will find a solution of the IVP (4.5) and (4.6). Let [t0 , γ1 (t0 )] be the largest interval beginning with t0 and containing those numbers of T such that τ (t) is less than t0 when t ≥ t0 and γ1 (t0 ) ∈ T. Consider the IVP x Δ (t) = f (t)x(t) + g(t)x(τ (t)) + h(t), x(t0 ) = φ(t0 ).

t ∈ [t0 , γ1 (t0 )],

4.2 First-Order Linear Functional Dynamic Equations

147

Since τ (t) ≤ t0 for any t ∈ [t0 , γ1 (t0 )], we can rewrite the last IVP in the form x Δ (t) = f (t)x(t) + g(t)φ(τ (t)) + h(t),

t ∈ [t0 , γ1 (t0 )],

x(t0 ) = φ(t0 ). Its solution is given by the expression x1 (t) = ef (t, t0 )φ(t0 )+

t

ef (t, σ (s)) (g(s)φ(τ (s)) + h(s)) Δs,

t ∈ [t0 , γ1 (t0 )].

t0

Let [γ1 (t0 ), γ2 (t0 )] be the largest interval beginning with γ1 (t0 ) and containing those numbers of T for which τ (t) is less than γ1 (t0 ) when t ≥ γ1 (t0 ) and γ2 (t0 ) ∈ T. We set

x1 (t) for t ∈ [t0 , γ1 (t0 )] φ1 (t) = φ(t) for t ∈ [τ (t0 ), t0 ]. Consider the IVP x Δ (t) = f (t)x(t) + g(t)x(τ (t)) + h(t),

t ∈ [γ1 (t0 ), γ2 (t0 )],

x(γ1 (t0 )) = φ1 (γ1 (t0 )). Since for t ∈ [γ1 (t0 ), γ2 (t0 )] we have that t0 ≤ τ (t) ≤ γ1 (t0 ), we can rewrite the last IVP in the following form x Δ (t) = f (t)x(t) + g(t)x1 (τ (t)) + h(t),

t ∈ [γ1 (t0 ), γ2 (t0 )],

x(γ1 (t0 )) = x1 (γ1 (t0 )). Its solution is given by the expression x2 (t) = ef (t, γ1 (t0 ))x1 (γ1 (t0 ))+

t γ1 (t0 )

ef (t, σ (s)) (g(s)φ1 (τ (s)) + h(s)) Δs,

and so on. Then ⎧ φ(t) for t ∈ [τ (t0 ), t0 ] ⎪ ⎪ ⎪ ⎨ x1 (t) for t ∈ [t0 , γ1 (t0 )] x(t) = x (t) for t ∈ [γ (t ), γ (t )] 2 1 0 2 0 ⎪ ⎪ ⎪ ⎩ .. . is a solution of the considered IVP.

t ∈ [γ1 (t0 ), γ2 (t0 )],

148

4 Linear Functional Dynamic Equations

Example 4.5. Let T = Z. Consider the IVP x Δ (t) = tx(t) + x(t − 1) + 1, x(t) = t,

t ≥ 1,

t ≤ 1.

We have x(t + 1) − x(t) = tx(t) + x(t − 1) + 1,

t ≥ 1,

x(t + 1) = (1 + t)x(t) + x(t − 1) + 1,

t ≥ 1.

or

Then x(2) = 2x(1) + x(0) + 1 = 2+1 = 3, x(3) = 3x(2) + x(1) + 1 = 9+1+1 = 11, x(4) = 4x(3) + x(2) + 1 = 4 · 11 + 3 + 1 = 44 + 4 = 48 and so on. Therefore ⎧ ⎪ ⎪ t for ⎪ ⎪ ⎪ ⎨ 3 for x(t) = 11 for ⎪ ⎪ 48 for ⎪ ⎪ ⎪ ⎩ .. . is a solution of the considered IVP.

t ≤1 t =2 t =3 t =4

4.2 First-Order Linear Functional Dynamic Equations

149

Exercise 4.1. Let T = 2N0 . Find a solution of the IVP

 t + t 2, x Δ (t) = (3t + 1)x(t) + tx 2 x(t) = 3t + 2,

t ≥ 2,

t ∈ [1, 2].

Consider the IVP x Δ (t) = −f (t)x σ (t) + g(t)x(τ (t)) + h(t), x(t) = φ(t),

t ≥ t0 ,

t ∈ [τ (t0 ), t0 ],

(4.7) (4.8)

where f ∈ R, g, h ∈ Crd (T), τ : T → T is an increasing unbounded above function such that τ (t) ≤ t, t ∈ T, φ ∈ C ([τ (t0 ), t0 ]). We will find a solution of the IVP (4.7) and (4.8). Let [t0 , γ1 (t0 )] be the largest interval beginning with t0 and containing those numbers of T such that τ (t) is less than t0 when t ≥ t0 and γ1 (t0 ) ∈ T. Consider the IVP x Δ (t) = −f (t)x σ (t) + g(t)x(τ (t)) + h(t),

t ∈ [t0 , γ1 (t0 )],

x(t0 ) = φ(t0 ). Since τ (t) ≤ t0 for any t ∈ [t0 , γ1 (t0 )], we can rewrite the last IVP in the form x Δ (t) = −f (t)x σ (t) + g(t)φ(τ (t)) + h(t),

t ∈ [t0 , γ1 (t0 )],

x(t0 ) = φ(t0 ). Its solution is given by the expression x1 (t) = ef (t, t0 )φ(t0 )+

t t0

ef (t, s) (g(s)φ(τ (s)) + h(s)) Δs,

t ∈ [t0 , γ1 (t0 )].

Let [γ1 (t0 ), γ2 (t0 )] be the largest interval beginning with γ1 (t0 ) and containing those numbers of T such that τ (t) is less than γ1 (t0 ) when t ≥ γ1 (t0 ) and γ2 (t0 ) ∈ T. We set

x1 (t) for t ∈ [t0 , γ1 (t0 )] φ1 (t) = φ(t) for t ∈ [τ (t0 ), t0 ]. Consider the IVP x Δ (t) = −f (t)x σ (t) + g(t)x(τ (t)) + h(t), x(γ1 (t0 )) = φ1 (γ1 (t0 )).

t ∈ [γ1 (t0 ), γ2 (t0 )],

150

4 Linear Functional Dynamic Equations

Since for t ∈ [γ1 (t0 ), γ2 (t0 )] we have that t0 ≤ τ (t) ≤ γ1 (t0 ), we can rewrite the last IVP in the following form x Δ (t) = −f (t)x σ (t) + g(t)x1 (τ (t)) + h(t),

t ∈ [γ1 (t0 ), γ2 (t0 )],

x(γ1 (t0 )) = x1 (γ1 (t0 )). Its solution is given by the expression x2 (t) = ef (t, γ1 (t0 ))x1 (γ1 (t0 ))+

t γ1 (t0 )

ef (t, s) (g(s)x1 (τ (s)) + h(s)) Δs,

and so on. Then ⎧ φ(t) for t ∈ [τ (t0 ), t0 ] ⎪ ⎪ ⎪ ⎨ x1 (t) for t ∈ [t0 , γ1 (t0 )] x(t) = x (t) for t ∈ [γ (t ), γ (t )] 2 1 0 2 0 ⎪ ⎪ ⎪ ⎩ .. . is a solution of the considered IVP. Example 4.6. Let T = 2N0 . We will find a solution of the IVP

 t x (t) = −tx (t) + x + 1, 8 Δ

σ

x(t) = t,

t ∈ [1, 8].

t ≥ 8,

Here σ (t) = 2t, μ(t) = t,

t ∈ T.

Consider the IVP

 t x (t) = −tx (t) + x + 1, 8 Δ

σ

t ∈ [8, 64],

x(8) = 8. Since x(t) = t for t ∈ [1, 8], and x

t 8

∈ [1, 8] for t ∈ [8, 64], we get

 t t = , 8 8

t ∈ [8, 64].

t ∈ [γ1 (t0 ), γ2 (t0 )],

4.2 First-Order Linear Functional Dynamic Equations

151

Therefore we can rewrite the last IVP in the following form x Δ (t) = −tx σ (t) +

t + 1, 8

t ∈ [8, 64],

x(8) = 8. Let f (t) = t,

t ∈ [8, 64].

Then f (t) 1 + μ(t)f (t) t , =− 1 + t2

(f )(t) = −

ef (t, s) = e =e =e =e

#t

1 s μ(r)

#t

1 s r

#t

1 s r

" 2t

log(1+μ(r)(f )(r))Δr

 log 1− log

r=s

r2 1+r 2

Δr

1 Δr 1+r 2

log

1 1+r 2

t

=

2 $

r=s

1 , 1 + r2

s≤

t , 2

t ∈ [8, 64].

Hence, x1 (t) = 8ef (t, 8) + t 2

=8

$ r=8 t 2

=8

$ r=8

t 8

ef (t, s)

s 8

+ 1 Δs

t 2

r  1 + 1 + re (t, r) f 8 1 + r2 r=8

⎞ t ⎛ t

2  2 2  $ r 1 1 ⎝ ⎠ +r , + 8 1 + r2 1 + s2 s=r r=8

and so on for t ∈ [64, 512], t ∈ [512, 4096], . . ..

t ∈ [8, 64],

152

4 Linear Functional Dynamic Equations

Exercise 4.2. Let T = 2Z. Find a solution of the IVP x Δ (t) = −x σ (t) + tx(t − 16) + t 2 , x(t) = t , 2

t ≥ 16,

t ∈ [0, 16].

4.3 Second-Order Linear Functional Dynamic Equations Consider the equation 2

x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) + a3 (t)x(τ (t)) = f (t),

t > t0 ,

(4.9)

where τ : T → T is an increasing unbounded above function such that τ (t) ≤ t, t ∈ T, a1 , a2 , a3 , f ∈ Crd ([t0 , ∞)). Define the operator 2 L2 : Crd ([t0 , ∞))

,

Crd ([τ (t0 ), ∞)) → Crd ([τ (t0 ), ∞))

as follows 2

L2 x(t) = x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) + a3 (t)x(τ (t)),

t ∈ [t0 , ∞).

Then we can rewrite the equation (4.9) in the following way. L2 x(t) = f (t),

t > t0 .

2 ([τ (t ), t ]) and Let φ0 ∈ Crd 0 0

x0 = φ0Δ (t0 ). The initial conditions for the equation (4.9) are x(t0 +) = φ0 (t0 ), x Δ (t0 +) = x0 , x(t) = φ0 (t), t ∈ [τ (t0 ), t0 ].

(4.10)

Suppose that x1 and x2 form a fundamental system of solutions of the homogeneous equation 2

x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) = 0,

t ∈ [t0 , ∞).

Let γ (t0 ) is the largest number of T such that τ (t) is less than t0 when t ≥ t0 . We will find a solution of the IVP (4.9) and (4.10) for t ∈ [t0 , γ (t0 )]. Since for t ∈ [t0 , γ (t0 )] we have that τ (t) ∈ [τ (t0 ), t0 ]. Then, for t ∈ [t0 , γ (t0 )], we get

4.3 Second-Order Linear Functional Dynamic Equations

153

x(τ (t)) = φ0 (τ (t)). Hence, for t ∈ [t0 , γ (t0 )], we obtain 2

x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) + a3 (t)φ0 (τ (t)) = f (t). Let f1 (t) = f (t) − a3 (t)φ0 (τ (t)),

t ∈ [t0 , γ (t0 )].

Therefore we get the IVP 2

x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) = f1 (t), t ∈ [t0 , γ (t0 )], = φ0 (t0 ), x Δ (t0 ) = x0 . x(t0 )

(4.11)

We will search a solution for the problem (4.11) in the form x(t) = b1 (t)x1 (t) + b2 (t)x2 (t),

t ∈ [t0 , γ (t0 )],

where b1 , b2 ∈ C 1 ([t0 , γ (t0 )]) will be determined below. We have x Δ (t) = b1Δ (t)x1σ (t) + b1 (t)x1Δ (t) + b2Δ (t)x2σ (t) + b2 (t)x2Δ (t) = b1 (t)x1Δ (t) + b2 (t)x2Δ (t),

t ∈ [t0 , γ (t0 )],

provided that b1 and b2 satisfy b1Δ (t)x1σ (t) + b2Δ (t)x2σ (t) = 0,

t ∈ [t0 , γ (t0 )].

Hence, 2

2

x Δ (t) = b1Δ (t)x1Δσ (t) + b1 (t)x1Δ (t) 2

+b2Δ (t)x2Δσ (t) + b2 (t)x2Δ (t),

t ∈ [t0 , γ (t0 )].

From here, 2

f1 (t) = x Δ (t) + a1 (t)x Δ (t) + a2 (t)x(t) 2

2

= b1Δ (t)x1Δσ (t) + b1 (t)x1Δ (t) + b2Δ (t)x2Δσ (t) + b2 (t)x2Δ (t) +a1 (t)b1 (t)x1Δ (t) + a1 (t)b2 (t)x2Δ (t) +a2 (t)b1 (t)x1 (t) + a2 (t)b1 (t)x2 (t)

154

4 Linear Functional Dynamic Equations

= b1Δ (t)x1Δσ (t) + b2Δ (t)x2Δσ (t)  2 +b1 (t) x1Δ (t) + a1 (t)x1Δ (t) + a2 (t)x1 (t)  2 +b2 (t) x2Δ (t) + a1 (t)x2Δ (t) + a2 (t)x2 (t) = b1Δ (t)x1Δσ (t) + b2Δ (t)x2Δσ (t),

t ∈ [t0 , γ (t0 )].

Thus, we get the system

b1Δ (t)x1σ (t) + b2Δ (t)x2σ (t) = 0 b1Δ (t)x1Δσ (t) + b2Δ (t)x2Δσ (t) = f1 (t),

t ∈ [t0 , γ (t0 )].

We can rewrite the last system in the following form.

x1σ (t) x2σ (t) x1Δσ (t) x2Δσ (t)



b1Δ (t) b2Δ (t)



=

 0 , f1 (t)

t ∈ [t0 , γ (t0 )],

whereupon

b1Δ (t) b2Δ (t)

%

 =

x σ (t)f (t)

− W σ2(x1 ,x12 )(t) x1σ (t)f1 (t) W σ (x1 ,x2 )(t)

& ,

t ∈ [t0 , γ (t0 )].

Then t

b1 (t) = −

t0

x1σ (s)f1 (s) Δs + c2 , W σ (x1 , x2 )(s)

t

b2 (t) =

x2σ (s)f1 (s) Δs + c1 , W σ (x1 , x2 )(s)

t0

t ∈ [t0 , γ (t0 )],

where c1 and c2 are constants. From here,

x(t) = c1 −

+

 x2σ (s)f1 (s) Δs x1 (t) σ t0 W (x1 , x2 )(s)  x1σ (s)f1 (s) Δs + c 2 x2 (t), W σ (x1 , x2 )(s) t

t t0

t ∈ [t0 , γ (t0 )].

4.3 Second-Order Linear Functional Dynamic Equations

155

The general solution of the equation (4.11) is y(t) = c3 x1 (t) + c4 x2 (t) + x(t)

 t x σ (s)f (s) 1 2 = c5 − Δs x1 (t) σ t0 W (x1 , x2 )(s) 

t x1σ (s)f1 (s) + Δs + c6 x2 (t), σ t0 W (x1 , x2 )(s)

t ∈ [t0 , γ (t0 )].

Here c3 and c4 are constants, c5 = c1 + c3 , c6 = c2 + c4 . We will find c5 and c6 using the initial conditions (4.11). We have y(t0 ) = c5 x1 (t0 ) + c6 x2 (t0 ) = φ0 (t0 ),



t x σ (s)f (s) x2σ (t)f1 (t) σ 1 2 x (t) + c5 − Δs x1Δ (t) y (t) = − σ σ W (x1 , x2 )(t) 1 t0 W (x1 , x2 )(s)

 t x σ (s)f (s) x1σ (t)f1 (t) σ 1 1 + σ x (t) + c6 + Δs x2Δ (t) σ W (x1 , x2 )(t) 2 t0 W (x1 , x2 )(s)

 t x σ (s)f (s) 1 2 = c5 − Δs x1Δ (t) σ t0 W (x1 , x2 )(s)

 t x σ (s)f (s) 1 1 + c6 + Δs x2Δ (t), t ∈ [t0 , γ (t0 )]. σ t0 W (x1 , x2 )(s) Δ

Hence, y Δ (t0 ) = c5 x1Δ (t0 ) + c6 x2Δ (t0 ) = x0 . Thus we get the system

c5 x1 (t0 ) + c6 x2 (t0 ) = φ0 (t0 ) c5 x1Δ (t0 ) + c6 x2Δ (t0 ) = x0 .

By the last system, we find c5 =

φ0 (t0 )x2Δ (t0 ) − x0 x2 (t0 ) , W (x1 , x2 )(t0 )

c6 =

x0 x1 (t0 ) − x1Δ (t0 )φ0 (t0 ) . W (x1 , x2 )(t0 )

156

4 Linear Functional Dynamic Equations

Consequently % y(t) =

φ0 (t0 )x2Δ (t0 ) − x0 x2 (t0 ) W (x1 , x2 )(t0 ) %

+

t

x1σ (s)f1 (s)

σ t0 W (x1 , x2 )(s)

−

Δs +

x2σ (s)f1 (s)

t

σ t0 W (x1 , x2 )(s)

& Δs x1 (t)

x0 x1 (t0 ) − x1Δ (t0 )φ0 (t0 ) W (x1 , x2 )(t0 )

& x2 (t),

t ∈ [t0 , γ (t0 )]

is the solution of the IVP (4.11). Example 4.7. Consider the IVP 2

x Δ (t) − 2ax Δ (t) + a 2 x(t) + x(τ (t)) = f (t),

t > t0 ,

x(t0 +) = φ0 (t0 ), x Δ (t0 +) = x0 , x(t) = φ0 (t),

t ∈ [τ (t0 ), t0 ],

where φ0 ∈ Crd ([τ (t0 ), t0 ]), a ∈ R. Let γ (t0 ) is the largest number of T such that τ (t) is less than t0 when t ≥ t0 . For t ∈ [t0 , γ (t0 )] we have τ (t) ∈ [τ (t0 ), t0 ] and x(τ (t)) = φ0 (τ (t)). Thus we get the IVP 2

x Δ (t) − 2ax Δ (t) + a 2 x(t) = f (t) − φ0 (τ (t)), x(t0 ) = φ0 (t0 ),

t ∈ [τ (t0 ), t0 ],

x (t0 ) = x0 . Δ

We can rewrite the last IVP in the form

Δ −a Δt

2 x(t) = f (t) − φ0 (τ (t)), x(t0 ) = φ0 (t0 ),

t ∈ [t0 , γ (t0 )],

x Δ (t0 ) = x0 .

Let

y(t) =

 Δ − a x(t), Δt

t ∈ [t0 , γ (t0 )].

We have y(t0 ) = x Δ (t0 ) − x(t0 ) = x0 − φ0 (t0 ).

4.3 Second-Order Linear Functional Dynamic Equations

157

Therefore we obtain the following IVP y Δ (t) = ay(t) + f (t) − φ0 (τ (t)),

t ∈ [t0 , γ (t0 )],

y(t0 ) = x0 − φ0 (t0 ). For its solution we have the representation t

y(t) = ea (t, t0 ) (x0 − φ0 (t0 )) +

ea (t, σ (s)) (f (s) − φ0 (τ (s))) Δs,

t0

t ∈ [t0 , γ (t0 )]. Hence, we get the following IVP x Δ (t) = ax(t) + ea (t, t0 ) (x0 − φ0 (t0 )) +

t

ea (t, σ (s)) (f (s) − φ0 (τ (s))) Δs,

t ∈ [t0 , γ (t0 )],

t0

x(t0 ) = φ0 (t0 ).

Consequently x(t) = ea (t, t0 )φ0 (t0 ) + (x0 − φ0 (t0 ))

t

ea (t, σ (s))ea (s, t0 )Δs t0

+

t

s

ea (t, σ (s)) t0

 ea (s, σ (z)) (f (z) − φ0 (τ (z))) Δz Δs,

t0

t ∈ [t0 , γ (t0 )]. Exercise 4.3. Find the solution of the following IVP 2

x Δ (t) − 6x Δ (t) + 8x(t) + x(τ (t)) = e1 (t, t0 ),

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x (t0 +) = x0 , Δ

x(t) = φ0 (t), 1 ([τ (t ), t ]). where φ0 ∈ Crd 0 0

t ∈ [τ (t0 ), t0 ],

158

4 Linear Functional Dynamic Equations

4.4 Advanced Practical Problems Problem 4.1. Let T = 3N0 . Find a solution of the IVP

 t + t 2 + t + 1, x Δ (t) = −x(t) − x 3 x(t) = 1,

t ≥ 3,

t ∈ [1, 3].

Problem 4.2. Let T = 3N0 . Find a solution of the IVP

  t Δ 2 + t 2, x (t) = − t + 1 x(t) + x 81 x(t) = 1 − t,

t ≥ 81,

t ∈ [1, 81].

Problem 4.3. Find the solution of the following IVPs 1. 2

x Δ (t) + 5x Δ (t) + 6x(t) + x(τ (t)) = e1 (t, t0 ),

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x Δ (t0 +) = x0 , x(t) = φ0 (t),

t ∈ [τ (t0 ), t0 ],

2. 2

x Δ (t) + 8x Δ (t) + 7x(t) + 4x(τ (t)) = e1 (t, t0 ),

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x (t0 +) = x0 , Δ

x(t) = φ0 (t),

t ∈ [τ (t0 ), t0 ],

3. 2

x Δ (t) + 10x Δ (t) + 9x(t) − 4x(τ (t)) = 2e2 (t, t0 ) + t 2 ,

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x Δ (t0 +) = x0 , x(t) = φ0 (t),

t ∈ [τ (t0 ), t0 ],

4.4 Advanced Practical Problems

159

4. 2

x Δ (t) − 10x Δ (t) + 21x(t) − 5x(τ (t)) = t 3 + t 2 − 2t + 4,

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x Δ (t0 +) = x0 , x(t) = φ0 (t),

t ∈ [τ (t0 ), t0 ],

5. 2

x Δ (t) − x Δ (t) − 12x(t) + 5x(τ (t)) = t − 2e3 (t, t0 ),

t ≥ t0 ,

x(t0 +) = φ0 (t0 ), x (t0 +) = x0 , Δ

x(t) = φ0 (t), 1 ([τ (t ), t ]), x = φ Δ (t ). where φ0 ∈ Crd 0 0 0 0 0

t ∈ [τ (t0 ), t0 ],

Chapter 5

Stability for First-Order Functional Dynamic Equations

The results contained in this chapter can be found in the papers and monographs [18, 19, 24, 25, 27, 45, 46, 63, 72, 93, 148, 149, 164, 182, 208, 209, 211, 234, 239, 240].

5.1 Uniform Stability Let T be an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also, β = inf{t : t ∈ T} and r > 0. Consider the equation x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ T,

(5.1)

where p ∈ Crd (T) is a positive function, and τ : T → T is an increasing unbounded above function such that t − r ≤ τ (t) < t, t ∈ T. Assume that t0 ∈ T and t0 ≥ β. Define Tt0 = [t0 − r, t0 ]. Suppose that φ : Tt0 → R is an rd-continuous function. Definition 5.1. We say that x is a solution of the equation [t0 , ∞) through  (5.1) on ' 1 ([t , ∞)) (t0 , φ) and we will denote it by x(t, t0 , φ) if x ∈ Crd (Tt0 [t0 , ∞)) Crd 0 such that x(t) = φ(t) for t ∈ T and x satisfies (5.1) on [t , ∞). Note that x(t) =0 t 0 0  for t ∈ Tt0 [t0 , ∞) is a solution of the equation (5.1) and it is called the zero solution of the equation (5.1). Theorem 5.1. For t0 , t ∈ T, we have the following equality %

t

σ (s)

p(s) t0

t0

& p(z)Δz Δs =

1 2

t

2 p(s)Δs

t0

+

1 t (p(s))2 μ(s)Δs, 2 t0

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_5

t ∈ T.

161

162

5 Stability for First-Order Functional Dynamic Equations

Proof. Define the function F (t) =

%

t

&

σ (s)

1 p(z)Δz Δs − 2

p(s) t0

t0

2

t

p(s)Δs

t ∈ T.

,

t0

We have % 1 p(s)Δs − p(t) 2

σ (t)

F Δ (t) = p(t)

t0

1 = p(t) 2 = = =

1 p(t) 2 1 p(t) 2

%

σ (t)

p(s)Δs −

t0

σ (t) t0

p(s)Δs + p(t)

p(s)Δs t0

&

t

&

t

p(s)Δs t0

σ (t)

p(s)Δs t σ (t)

p(s)Δs t

1 (p(t))2 μ(t), 2

t ∈ T.

Hence, using that F (t0 ) = 0, we get F (t) =

1 2

t

(p(s))2 μ(s)Δs,

t ∈ T.

t0

This completes the proof. Definition 5.2. The zero solution of the equation (5.1) is said to be uniformly stable if for each  > 0 and t0 ∈ T, there exists a δ > 0, independent of t0 , such that if φ ∈ Crd (Tt0 ) and supt∈Tt |φ(t)| < δ, then |x(t, t0 , φ)| <  for all t ∈ [t0 , ∞). 0 ' Note that if t ∈ T [β + r, ∞), then τ (σ (t)) ∈ T. If we assume that t < τ (σ (t)), then t < τ (σ (t)) < σ (t), which is a contradiction with the definition of the operator σ . Therefore t ≥ τ (σ (t)),

t ∈T

,

[β + r, ∞).

5.1 Uniform Stability

163

From here, t ≥ σ (t) − r or μ(t) = σ (t) − t ≤ r, i.e., μ(t) ∈ [0, r] when t ∈ T

'

[β + r, ∞). Define

m = inf μ(t),

M = sup (μ(t) + r) .

t∈T

t∈T

Theorem 5.2. Suppose that σ (t)

p(s)Δs ≤

t−r

m 3 + 2 4M

for

t ∈ [β + r, ∞)

holds. Then the zero solution of the equation (5.1) is uniformly stable. Proof. Let t0 ∈ T and  > 0. Take δ=e

 m − 32 + 4M

.

If supt∈Tt |φ(t)| = 0, then the assertion is evident. Suppose that 0

sup |φ(t)| = 0 and

sup |φ(t)| < δ.

t∈Tt0

t∈Tt0

For convenience, we denote x(t, t0 , φ) by x(t) and g(t) =

sup |x(s)|. s∈[t−r,t]

By the equation (5.1), we get ! Δ ! !x (t)! = |−p(t)x(τ (t))| = p(t) |x(τ (t))| ≤ p(t)g(t),

t ∈ [t0 , t0 + r].

Hence, x Δ (t) ≤ p(t)g(t),

t ∈ [t0 , t0 + r],

164

5 Stability for First-Order Functional Dynamic Equations

and x Δ (t) ≥ −p(t)g(t),

t ∈ [t0 , t0 + r].

Then t

x(t) ≤ x(t0 ) +

p(s)g(s)Δs t0

t

−δ −

p(s)g(s)Δs,

t ∈ [t0 , t0 + r].

p(s)g(s)Δs,

t ∈ [t0 , t0 + r].

t0

Consequently t

|x(t)| < δ +

t0

Hence, if t ∈ [t0 , t0 + r] and s ∈ [t0 , t], we get |x(s)| < δ +

s

p(z)g(z)Δz t0

≤δ+

t

p(z)g(z)Δz. t0

From here, g(t) =

sup |x(s)| s∈[t−r,t] t

≤δ+

p(z)g(z)Δz,

t ∈ [t0 , t0 + r].

t0

By the last inequality and Gronwall’s inequality, we have

g(t) ≤ δ 1 +

t

 ep (t, σ (s))p(s)Δs ,

t0

t ∈ [t0 , t0 + r].

5.1 Uniform Stability

165

Because

1+

Δ

t t0

t

=

ep (t, σ (s))p(s)Δs

p(t)ep (t, σ (s))p(s)Δs t0

t

+ep (σ (t), σ (t))p(t)

 t = p(t) 1 + ep (t, σ (s))p(s)Δs , t0

we conclude that g(t) ≤ δep (t, t0 ),

t ∈ [t0 , t0 + r].

Hence, g(t) ≤ δep (t, t0 ) = δe ≤ δe = δe ≤ δe

#t t0

#t t0

#t t0

log(1+μ(s)p(s)) Δs μ(s) μ(s)p(s) μ(s) Δs

p(s)Δs

# σ (t) t−r 3

p(s)Δs

m

≤ δe 2 + 4M =e

 m − 32 + 4M

= ,

3

m

e 2 + 4M

t ∈ [t0 , t0 + r].

Since |x(t)| ≤ g(t),

t ∈ [t0 , t0 + r],

we get |x(t)| < ,

t ∈ [t0 , t0 + r].

|x(t)| < ,

t ∈ (t0 + r, ∞),

Suppose that

is false. Then there exists a ξ1 ∈ (t0 + r, ∞) such that if ξ ∗ = σ (ξ1 ), then |x(t)| < 

for t ∈ [t0 , ξ1 )

166

5 Stability for First-Order Functional Dynamic Equations

and |x(ξ1 )| ≤ ,

|x(ξ ∗ )| ≥ .

Without loss of generality, we assume that x(ξ ∗ ) > 0. Hence, x(σ (ξ1 )) ≥  and x(σ (ξ1 )) − x(ξ1 ) ≥ 0. Therefore x Δ (ξ1 ) ≥ 0. Also, x Δ (ξ1 ) = −p(ξ1 )x(τ (ξ1 )) ≥ 0, whereupon x(τ (ξ1 )) ≤ 0. Therefore there exists a t1 ∈ [ξ1 − r, ξ ∗ ] such that x(t1 ) ≤ 0 and x(σ (t1 )) ≥ 0 and x(t) > 0 for all t ∈ (σ (t1 ), ξ ∗ ]. Consequently x Δ (t1 ) ≥ 0. Next, x(σ (t1 )) + x Δ (t1 )(t1 − σ (t1 )) = x(t1 ) + x Δ (t1 )(σ (t1 ) − t1 ) +x Δ (t1 )(t1 − σ (t1 )) = x(t1 ) ≤ 0, and x(σ (t1 )) + x Δ (t1 )(t1 − s) ≥ 0,

s ≤ t1 .

5.1 Uniform Stability

167

So, there exists a ξ ∈ (t1 , σ (t1 )] such that x(σ (t1 )) + x Δ (t1 )(t1 − ξ ) = 0.

(5.2)

Also, x Δ (t) = −p(t)x(τ (t)) ≤ p(t)|x(τ (t))| ≤ p(t),

t ∈ (t0 , ξ ∗ ].

Now, we will prove that % − x(t) < 

σ (t1 )

& p(s)Δs − p(t1 )(ξ − t1 )

(5.3)

t−r

for all t ∈ (t0 , ξ1 ]. 1. Let t ∈ [t0 , σ (t1 )). Then # σ (t ) −x(σ (t )) + t 1 x Δ (s)Δs # σ (t1 ) 1Δ x (s)Δs − x Δ (t1 )(ξ − t1 ) #tt1 Δ # σ (t ) x (s)Δs + t1 1 x Δ (s)Δs − x Δ (t1 )(ξ − t1 ) #tt1 Δ Δ Δ #tt1 x Δ (s)Δs + x Δ (t1 )μ(t1 ) − x (t1 )(ξΔ− t1 ) #tt1 x Δ (s)Δs + x Δ (t1 )(σ (t1 ) − t1 ) − x (t1 )(ξ − t1 ) #tt1 x Δ (s)Δs + x (t1 )(σ (t1 ) − ξ ) t# x (s)Δs + p(t1 )(σ (t1 ) − ξ ) t  t 1 p(s)Δs − p(t1 )(ξ − σ (t1 )) # t1 # σ (t )  t p(s)Δs +  t1 1 p(s)Δs # σ (t1 ) − t1 p(s)Δs − p(t1 )(ξ − σ (t1 )) # σ (t1 ) = t p(s)Δs − p(t1 )(σ (t1 ) − t1 ) −p(t1 )(ξ − σ (t1 )) # σ (t ) =  t 1 p(s)Δs − p(t1 )(ξ − t1 ) # σ (t ) =  t 1 p(s)Δs − p(t1 )(ξ − t1 ) .

−x(t) = = = = = = ≤ ≤ =

2. Let t ∈ (σ (t1 ), ξ1 ]. Then x(t) > 0. Since t1 ∈ [ξ1 − r, ξ ∗ ], we get t − r ≤ ξ1 − r ≤ t1 ≤ σ (t1 ). Then, as we have deducted (5.4), we get (5.3).

(5.4)

168

5 Stability for First-Order Functional Dynamic Equations

3. Let t = σ (t1 ). a. If x(σ (t1 )) = 0, then, using (5.2), we have x Δ (t1 )(t1 − ξ ) = 0 and t1 is right-dense. Hence, as we have deducted (5.4), we obtain (5.3). b. Let x(σ (t1 )) > 0. Hence, using (5.2), we have that t1 is right-scattered and as we have deducted (5.4), we get (5.3). Consequently (5.3) holds for all t ∈ [t0 , ξ1 ]. For t ∈ [t1 , ξ1 ), we have x Δ (t) = −p(t)x(τ (t))   ≤ p(t) max # 0, sups∈[t−r,t] (−x(s)) σ (t ) < p(t) t−r1 p(s)Δs − p(t1 )(ξ − t1 ) .

(5.5)

Now we consider the following two cases. 1. Let ξ∗

d=

p(t)Δt + p(t1 )(ξ − t1 ) ≤ 1.

σ (t1 )

Then, using (5.2) and (5.5), we get ξ∗

x(ξ ∗ ) = x(σ (t1 )) +

x Δ (t)Δt

σ (t1 )

= x Δ (t1 )(ξ − t1 ) + % < p(t1 )(ξ − t1 ) +

%

ξ∗

x Δ (t)Δt

σ (t1 ) σ (t1 ) t1 −r σ (t1 )

p(t) σ (t1 )

%

= p(t1 )(ξ − t1 ) +

ξ∗

%

ξ∗

p(s)Δs − p(t1 )(ξ − t1 ) & p(s)Δs − p(t1 )(ξ − t1 ) Δt

t−r σ (t1 ) t1 −r σ (t)

p(t) σ (t1 )

&

t−r

& p(s)Δs − p(t1 )(ξ − t1 )

p(s)Δs +

σ (t1 ) σ (t)

& p(s)Δs − p(t1 )(ξ − t1 ) Δt

5.1 Uniform Stability

169

 m 3 + − p(t1 )(ξ − t1 ) ≤ p(t1 )(ξ − t1 ) 2 4M % &  ξ∗ σ (t) 3 m − + p(t) p(s)Δs − p(t1 )(ξ − t1 ) Δt + 2 4M σ (t1 ) σ (t1 )

 m 3 + − p(t1 )(ξ − t1 ) = p(t1 )(ξ − t1 ) 2 4M  ξ∗

ξ∗ 3 m p(t)Δt − p(t1 )(ξ − t1 ) p(t)Δt + + 2 4M σ (t1 ) σ (t1 ) % & ∗ ξ

−

σ (t)

p(t) σ (t1 )

=

m 3 + 2 4M

p(s)Δs Δt

 % 

σ (t1 )

&

ξ∗

p(t)Δt + p(t1 )(ξ − t1 )

σ (t1 )

−(p(t1 )) (ξ − t1 ) − p(t1 )(ξ − t1 ) 2

2

%

ξ∗

−

p(t) σ (t1 )

=

m 3 + 2 4M %

p(t)Δt σ (t1 )

&

σ (t)

ξ∗

p(s)Δs Δt σ (t1 )



%

ξ∗

d −

p(t) σ (t1 )

− p(t1 )(ξ − t1 )

σ (t)

ξ∗

& p(s)Δs Δt

σ (t1 )

&

p(t)Δt + (p(t1 ))2 (ξ − t1 )2 .

σ (t1 )

Now, using Theorem 5.1, we get  # ∗ 2 # ξ∗ ξ d − 12 σ (t ) p(t)Δt − 12 σ (t ) μ(t)(p(t))2 Δt 1 1  # ξ∗ − p(t1 )(ξ − t1 ) σ (t ) p(t)Δt + (p(t1 ))2 (ξ − t1 )2

 1 ∗ 2  3 + m d − 1 #ξ p(t)Δt + p(t )(ξ − t ) = 1 1 2 4M 2 σ (t1 )  # ∗ ξ 1 1 2 2 2 − 2 σ (t ) (p(t)) μ(t)Δt + 2 (p(t1 )) (ξ − t1 ) 1  ∗ m d − 1 d2 − 1 # ξ 2 μ(t)Δt − 1 (p(t ))2 (ξ − t )2 . =  32 + 4M (p(t)) 1 1 2 2 σ (t ) 2

x(ξ ∗ ) < 



3 + m 2 4M



1

(5.6)

170

5 Stability for First-Order Functional Dynamic Equations

By Cauchy’s inequality, we have %

ξ∗

&2



p(s) μ(s)Δs

% ≤

σ (t1 )

&%

ξ∗

2

(p(s)) μ(s)Δs σ (t1 )

  = ξ ∗ − σ (t1 )

&

ξ∗

Δs σ (t1 )

%

ξ∗

&

(p(s))2 μ(s)Δs .

σ (t1 )

So, ξ∗

1 (p(s)) μ(s)Δs ≥ ∗ ξ − σ (t1 ) σ (t1 )

%

ξ∗

2



p(s) μ(s)Δs

&2 .

σ (t1 )

As in above, if σ (t1 ) > t1 we get σ (t1 ) t1

(p(t1 ))2 (ξ −t1 )2 1 μ(s)Δs ≥ 2 σ (t1 )−t1 (σ (t1 )−t1 )

σ (t1 ) t1

p(t1 )(ξ − t1 )  μ(s)Δs σ (t1 )−t1

2 .

Hence, if σ (t1 ) > t1 , we have ξ∗

(p(s))2 μ(s)Δs + (p(t1 ))2 (ξ − t1 )2 σ (t1 ) # ξ∗ # σ (t ) 2 2 1 )) (ξ −t1 ) = σ (t1 ) (p(s))2 μ(s)Δs + t1 1 (p(t μ(s)Δs (σ (t1 )−t1 )2 # ∗ 2 # √ 2 2 ξ σ (t ) 1 1 )) (ξ −t1 ) ≥ ξ ∗ −t + t+ 1 (p(t μ(s)Δs σ (t1 ) p(s) μ(s)Δs (σ (t1 )−t1 )2 1 # ∗  2 2 # σ (t1 ) p(t1 )(ξ −t1 ) √ √ ξ 1 1 ≥ ξ ∗ −t p(s) μ(s)Δs + μ(s)Δs ∗ t1 σ (t1 ) ξ −t1 (σ (t1 )−t1 ) 1 # ∗ 2 # √ √ ξ σ (t ) p(t )(ξ −t ≥ 2(ξ ∗1−t1 ) σ (t1 ) p(s) μ(s)Δs + t1 1 σ (t1 1 )−t11 ) μ(s)Δs (5.7) m ≥ 2(ξ ∗ − t1 ) ≥

%

ξ∗

&2 p(s)Δs + p(t1 )(ξ − t1 )

σ (t1 )

m 2 d . 2M

From here and from (5.6), we obtain

5.1 Uniform Stability

171

m m 2 d − 12 d 2 − 4M + 4M d    m m d − 12 + 4M d2 =  d + 12 + 4M   m (d − d 2 ) =  d + 12 + 4M 

x(ξ ∗ ) < 

3 2

(5.8)

− d 2)

≤ (d + d = (2d − d 2 ) ≤ ,

which is a contradiction. Let t1 = σ (t1 ). Then, using Cauchy’s inequality, we obtain %

ξ∗

&2



% ≤

p(s) μ(s)Δs t1

&%

ξ∗

&

ξ∗

2

(p(s)) μ(s)Δs

Δs

t1

t1 ξ∗

  = ξ ∗ − t1

(p(s))2 μ(s)Δs,

t1

whereupon ξ∗ t1

1 (p(s)) μ(s)Δs ≥ ∗ ξ − t1 2

%

ξ∗

 p(s) μ(s)Δs

&2 .

t1

Hence, ≥

m ξ ∗ −t1

≥

m M

# ∗ ξ t1

 # ξ∗ t1

p(s)Δs

p(s)Δs

2

2

+ (p(t1 ))2 (ξ − t1 )2 

+ (p(t1

))2 (ξ

− t1

)2 (5.9)

#

ξ∗ t1

≥

m 2M

=

m 2 2M d .

p(s)Δs + p(t1 )(ξ − t1 )

2

From here, from (5.6), and from (5.8), we obtain x(ξ ∗ ) < 



≤ . This is a contradiction.

m 3 + 2 4M



1 m 2 d − d2 − d 2 4M



172

5 Stability for First-Order Functional Dynamic Equations

2. Let ξ∗

d=

p(t)Δt + p(t1 )(ξ − t1 ) > 1.

σ (t1 )

There exists a t2 ∈ [t1 , ξ ∗ ] such that ξ∗

ξ∗

p(s)Δs ≤ 1

and

σ (t2 )

p(t)Δt ≥ 1.

t2

Therefore there exists a number η ∈ [t2 , σ (t2 )] such that ξ∗

p(s)Δs + p(t2 )(σ (t2 ) − η) = 1.

σ (t2 )

Let I1 = x(σ (t1 )) +

σ (t2 )

x Δ (t)Δt + x Δ (t2 )(η − σ (t2 )),

σ (t1 ) ξ∗

I2 =

x Δ (t)Δt + x Δ (t2 )(σ (t2 ) − η).

σ (t2 )

Then x(ξ ∗ ) = I1 + I2 . If t2 > t1 , then σ (t2 )

I1 =

x Δ (t)Δt + x Δ (t1 )(ξ − t1 ) + x Δ (t2 )(η − σ (t2 ))

σ (t1 )

%

≤ %

σ (t2 )

& p(t)Δt + p(t1 )(ξ − t1 ) + p(t2 )(η − σ (t2 ))

σ (t1 ) ξ∗

×

& p(t)Δt + p(t2 )(σ (t2 ) − η) ,

σ (t2 )

I2 ≤ 

ξ∗

%

σ (t1 )

p(t) σ (t2 )

t−r

+p(t2 )(σ (t2 ) − η)

& p(s)Δs − p(t1 )(ξ − t1 ) Δt

%

σ (t1 ) t2 −r

& p(t)Δt − p(t1 )(ξ − t1 ) .

5.1 Uniform Stability

173

Therefore %

σ (t2 )

x(ξ ∗ ) ≤  %

σ (t1 )

p(t)Δt + p(t2 )(σ (t2 ) − η)

σ (t2 )

+

p(t)Δt + p(t1 )(ξ − t1 ) + p(t2 )(η − σ (t2 )) &

ξ∗

×

&

%

ξ∗

σ (t1 )

p(t) σ (t2 )

t−r

%

+p(t2 )(σ (t2 ) − η) % =

+

σ (t2 ) ξ∗ σ (t2 )

%

ξ∗

ξ∗ σ (t2 )

t−r

%

ξ∗

σ (t1 ) t2 −r

σ (t2 )

p(t) σ (t2 )

t−r

%

t2 −r

σ (t2 )

p(t) t−r

ξ∗

p(t)Δt σ (t2 )

p(t)Δt − p(t1 )p(t2 )(σ (t2 ) − η)(ξ − t1 ) &

ξ∗

p(t)Δt σ (t2 )

p(t)Δt − (p(t2 ))2 (η − σ (t2 ))2 &

σ (t2 ) t2 −r

& p(t)Δt − p(t2 )(σ (t2 ) − η)

&  σ (t) 3 m − + ≤ p(t) p(s)Δs − p(t2 )(σ (t2 ) − η) Δt 2 4M σ (t2 ) σ (t2 )

 3 m +p(t2 )(σ (t2 ) − η) + − p(t2 )(σ (t2 ) − η) 2 4M ξ∗

%

− p(t1 )(ξ − t1 )

p(s)Δs − p(t2 )(σ (t2 ) − η) Δt

%

+p(t2 )(σ (t2 ) − η)

&

p(s)Δs Δt + p(t2 )(η − σ (t2 )) σ (t2 )

+p(t2 )(σ (t2 ) − η) ξ∗

&

σ (t1 )

+p(t2 )(σ (t2 ) − η)

p(t)Δt σ (t1 )

p(t)Δt − (p(t2 ))2 (η − σ (t2 ))2

p(s)Δs Δt

σ (t2 )

σ (t2 )

+ p(t2 )(σ (t2 ) − η)

p(t)Δt + p(t1 )p(t2 )(ξ − t1 )(σ (t2 ) − η)

p(t)

σ (t2 )

&

ξ∗

σ (t1 )

%

& p(t)Δt − p(t1 )(ξ − t1 )

p(t)Δt

+p(t2 )(η − σ (t2 ))

=

σ (t1 )

p(t)Δt

+p(t1 )(ξ − t1 )

=

p(s)Δs − p(t1 )(ξ − t1 ) Δt

t2 −r

&%

σ (t2 )

&

174

5 Stability for First-Order Functional Dynamic Equations

% =

3 m + 2 4M

 −

%

3 m + = ⎝ 2 4M



1 − 2

%

3 m + = ⎝ 2 4M %



1 − 2

p(s)Δs Δt

ξ∗ σ (t2 )

&2

ξ∗

%

ξ∗ σ (t2 )

&

p(t)Δt + (p(t2 ))2 (σ (t2 ) − η)2

p(t)Δt

− p(t2 )(σ (t2 ) − η)

&

σ (t2 )

σ (t2 )

% ⎛

&

σ (t)

p(t) σ (t2 )

− p(t2 )(σ (t2 ) − η) ⎛

%

ξ∗

⎞ ∗ 1 ξ − μ(t)(p(t))2 Δt ⎠ 2 σ (t2 ) &

p(t)Δt + (p(t2 ))2 (σ (t2 ) − η)2

ξ∗ σ (t2 )

&2 ⎞ p(t)Δt + p(t2 )(σ (t2 ) − η) ⎠

∗

1 ξ 1 − (p(t))2 μ(t)Δt + (p(t2 ))2 (σ (t2 ) − η)2 2 σ (t2 ) 2  

3 m 1 m + − − σ (t1 ) + t1 . Hence, x(ξ ∗ ) = x(σ (t1 )) +

ξ∗

x Δ (t)Δt

σ (t1 )

= x(σ (t1 )) + x Δ (t1 )(η − σ (t1 )) +

ξ∗

x Δ (t)Δt + x Δ (t1 )(σ (t1 ) − η)

σ (t1 )

= x(σ (t1 )) + x Δ (t1 )(η − σ (t1 )) +

σ (t2 ) σ (t1 )

+

ξ∗ σ (t2 )

x Δ (t)Δt + x Δ (t1 )(σ (t1 ) − η).

x Δ (t)Δt

5.1 Uniform Stability

175

Let σ (t2 )

J1 = x(σ (t1 )) + x Δ (t1 )(η − σ (t1 )) +

x Δ (t)Δt,

σ (t1 )

J2 =

ξ∗

x Δ (t)Δt + x Δ (t1 )(σ (t1 ) − η).

σ (t2 )

Then x(ξ ∗ ) = J1 + J2 . As we have deducted the estimates for I1 and I2 , we get x(ξ ∗ ) < , which is a contradiction. This completes the proof. Example 5.1. Let T = 2N0 . Consider the equation x Δ (t) +

1 x(t − 2) = 0, (t + 3)(t + 5)

t > 2.

Here σ (t) = t + 2, μ(t) = 2,

t ∈ T,

τ (t) = t − 2, β = 0, r = 2, m = 2, M = 4, 3 m 3 + = + 2 4M 2 3 = + 2 13 , = 8 p(t) =

2 16 1 8

1 , (t + 3)(t + 5)

t ∈ T.

176

5 Stability for First-Order Functional Dynamic Equations

Let g(t) =

t +1 , t +3

t ∈ T.

Then g Δ (t) =

(t + 3) − (t + 1) (t + 3)(σ (t) + 3)

=

t +3−t −1 (t + 3)(t + 2 + 3)

=

2 (t + 3)(t + 5)

= 2p(t),

t ∈ T.

Hence, σ (t) t−r

p(s)Δs =

1 2

t+2

g Δ (s)Δs

t−2

!s=t+2 1 ! g(s)! s=t−2 2 1 s + 1 !!s=t+2 = ! 2 s + 3 s=t−2

 1 t +3 t −1 − = 2 t +5 t +1 =

=

(t + 3)(t + 1) − (t − 1)(t + 5) 2(t + 1)(t + 5)

=

t 2 + 4t + 3 − t 2 − 4t + 5 2(t + 1)(t + 5)

=

4 (t + 1)(t + 5)

4 5 13 < 8 m 3 . = + 2 4M ≤

Consequently the zero solution of the considered equation is uniformly stable.

5.2 Uniformly Asymptotical Stability

177

Exercise 5.1. Let T = 3N0 . Prove that the zero solution of the equation x Δ (t) +

t2 + 1 x(t − 3) = 0, 70(t 2 + 2)(t 2 + 6t + 11)

t > 3,

is uniformly stable.

5.2 Uniformly Asymptotical Stability Suppose that the time scale T is as in Section 5.1. Definition 5.3. The zero solution of the equation (5.1) is said to be uniformly attractive if there is a δ > 0 independent of t0 such that supt∈Tt |φ(t)| < δ implies 0 x(t, t0 , φ) → 0 as t → ∞. Definition 5.4. The zero solution of the equation (5.1) is said to be uniformly asymptotically stable if it is uniformly attractive and uniformly stable. Theorem 5.3. Suppose that there exists a positive constant α such that σ (t)

p(s)Δs ≤ α
0 and t0 ∈ T, there exist ξ ∈ (t0 , ∞) and λ ∈ 1 + such that σ (t)

p(s)Δs ≤ λ

for t ∈ T,

t−r

and x −  < x(τ (t)) < x + 

for t ∈ [ξ, ∞).

m 3 2M , 2

+

m 4M



5.2 Uniformly Asymptotical Stability

179

Thus, for t ∈ [ξ, ∞), we have x Δ (t) = −p(t)x(τ (t))   < −x +  p(t) and x Δ (t) = −p(t)x(τ (t)) > − (x + ) p(t). Since x is oscillatory, there exists a sequence {ti }i∈N ⊂ T such that ti , τ (ti ) ∈ [ξ, ∞), x(σ (ti )) > 0, x Δ (ti ) ≥ 0 and limi→∞ x(σ (ti )) = x, where ti → ∞ as i → ∞. For each i ∈ N, using that x Δ (ti ) ≥ 0 and (5.1), we conclude that x(τ (ti )) ≤ 0. Therefore there is a ti∗ ∈ [ti − r, σ (ti )] such that x(ti∗ ) ≤ 0, x(σ (ti∗ )) ≥ 0 and x(t) > 0 for all t ∈ (σ (ti∗ ), σ (ti )]. Note that x(σ (ti∗ )) = x(ti∗ ) + μ(ti∗ )x Δ (ti∗ ). Consequently x Δ (ti∗ ) ≥ 0 and x(σ (ti∗ )) + (ti∗ − σ (ti∗ ))x Δ (ti∗ ) = x(ti∗ ) ≤ 0. Hence, there exists a ξi ∈ [ti∗ , σ (ti∗ )] such that x(σ (ti∗ )) + x Δ (ti∗ )(ti∗ − ξi ) = 0.

(5.10)

Now we will prove that   − x(t) ≤ −x + 

%

σ (ti∗ )

& p(s)Δs

t−r

− p(ti∗ )(ξi

for all t ∈ [ξ, ti ]. If t ∈ [ξ, σ (ti∗ )], then −x(t) = −x(σ (ti∗ )) + σ (ti∗ )

= t

σ (ti∗ )

x Δ (s)Δs

t

x Δ (s)Δs − x Δ (ti∗ )(ξi − ti∗ )

− ti∗ )

(5.11)

180

5 Stability for First-Order Functional Dynamic Equations ti∗

= t

ti∗

= t

x Δ (s)Δs + x Δ (ti∗ )μ(ti∗ ) − x Δ (ti∗ )(ξi − ti∗ ) x Δ (s)Δs + x Δ (ti∗ )(σ (ti∗ ) − ξi )



≤ −x +  

< −x + 

 

% %

σ (ti∗ ) t σ (ti∗ )

& p(s)Δs − p(ti∗ )(ξi − ti∗ ) & p(s)Δs

t−r

− p(ti∗ )(ξi

− ti∗ )

.

 If t ∈ σ (ti∗ ), ti , then −x(t) < 0. When ti∗ ∈ [ti − r, σ (ti )] we have that ti − r ≤ ti∗ and as in above (5.11) holds. If t = σ (ti∗ ), then a. for x(σ (ti∗ )) = 0, using (5.10), we have that t is right-dense and (5.11) holds. b. for x(σ (ti∗ )) > 0, then x Δ (ti∗ ) > 0 and using that x(σ (ti∗ )) + x Δ (ti∗ )(ti∗ − σ (ti∗ )) = x(ti∗ ), we conclude that ξi = σ (ti∗ ) and (5.11) holds. Thus, for t ∈ [ξ, ti ], we have that (5.11) holds. For t ∈ [σ (ti∗ ), σ (ti )], we get x Δ (t) = −p(t)x(τ (t)) 



≤ p(t) max 0, sup (−x(s)) s∈[t−r,t]

  < −x +  p(t)

%

σ (ti∗ ) t−r

& p(s)Δs − p(ti∗ )(ξi − ti∗ ) .

We will consider the following two cases. a. Let di =

σ (ti ) σ (ti∗ )

p(s)Δs + p(ti∗ )(ξi − ti∗ ) ≤ 1.

Now, using Theorem 5.1 and using that σ (ti ) σ (ti∗ )

 2 m 2 d , (p(s))2 μ(s)Δs + p(ti∗ ) (ξi − ti∗ )2 ≥ 2M i

5.2 Uniformly Asymptotical Stability

181

(for the proof of the last inequality see (5.7) and (5.9)), we get σ (ti )

x(σ (ti )) = x(σ (ti∗ )) + =

σ (ti ) σ (ti∗ )

σ (ti∗ )

x Δ (t)Δt

x Δ (t)Δt + x Δ (ti∗ )(ξi − ti∗ )

  < −x +  

+ −x + 

%

σ (ti ) σ (ti∗ )



p(t)

σ (ti∗ )

& p(s)Δs

t−r

%

p(ti∗ )(ξi

− ti∗ )

σ (ti∗ ) ti∗ −r

− p(ti∗ )(ξi

− ti∗ )

Δt &

p(s)Δs

− p(ti∗ )(ξi

− ti∗ )

 1 1 σ (ti ) ≤ (−x + ) λdi − di2 − (p(t))2 μ(t)Δt 2 2 σ (ti∗ ) 2 1 − p(ti∗ ) (ξi − ti∗ )2 2 

  1 m 2 di ≤ −x +  λdi − di2 − 2 4M 

  m 1 ≤ −x +  λ − − . 2 4M b. Let di =

σ (ti ) σ (ti∗ )

p(s)Δs + p(ti∗ )(ξi − ti∗ ) > 1.

We may choose si ∈ [ti∗ , σ (ti )] such that σ (ti )

σ (ti )

p(s)Δs ≤ 1 and

σ (si )

p(s)Δs ≥ 1.

si

Hence, there exists ηi ∈ [si , σ (si )] such that σ (ti )

p(s)Δs + p(si )(σ (si ) − ηi ) = 1.

σ (si )

As we have proved the part of the proof of Theorem 5.2 corresponding to the case d > 1, we get 

  m 1 x(σ (ti )) ≤ −x +  λ − − . 2 4M

182

5 Stability for First-Order Functional Dynamic Equations

Because  > 0 was arbitrarily chosen, by the last inequality, we get 

  m 1 . x = lim x(σ (ti )) ≤ −x λ − − i→∞ 2 4M As in above,

m 1 −x ≤ x λ − − 2 4M Because λ −

1 2

−

m 4M

 .

< 1, we get x = x = 0.

This completes the proof. Example 5.2. Let T = 3N0 . Consider the equation 1 x Δ (t) + x(t − 6) = 0, 9

t > 6.

Here σ (t) = t + 3, μ(t) = 3, τ (t) = t − 6, 1 , 9 β = 0,

p(t) =

t ∈ T,

r = 6, m = 3, M = 9, 3 m 3 + = + 2 4M 2 3 = + 2 19 . = 12

3 36 1 12

5.3 Global Stability

183

Then σ (t)

p(s)Δs =

t−r

1 9

t+3

Δs t−6

1 (t + 3 − (t − 6)) 9 =1 < 19 12 m 3 , t ∈ T, = + 2 4M =

and t

lim

t→∞ β

p(s)Δs =

1 lim 9 t→∞

t

Δs 0

= ∞. Therefore the zero solution of the considered equation is uniformly asymptotically stable. Exercise 5.2. Let T = 12 N0 . Prove that the zero solution of the equation x Δ (t) +

1 x(t − 1) = 0, 100

t > 1,

is uniformly asymptotically stable.

5.3 Global Stability Let T be an unbounded time scale with forward jump operator, backward jump operator, and delta differentiation operator σ , ρ, and Δ, respectively. Let t0 ∈ T. Consider the following nonlinear functional dynamic equation " x Δ (t) = − nj=0 aj (t)fj (x(τj (t))), t ∈ [t0 , ∞), x(t) = ψ(t), t ∈ [τn (t0 ), t0 ], ψ ∈ Crd ([τn (t0 ), t0 ]),

(5.12)

where τj : T → T, j ∈ {0, 1, . . . , n}, are right-dense continuous strictly increasing functions unbounded above and such that τn (T) = T,

184

5 Stability for First-Order Functional Dynamic Equations

and there exists a positive constant M such that ρ n (t) − M ≤ τn (t) < τn−1 (t) < . . . < τ0 (t) = t, fj : R → R are increasing continuous functions, fj (0) = 0, j ∈ {0, 1, . . . , n}, aj : T → R are right-dense continuous functions for which n 

aj (t) ≥ 0,

aj (t) > 0,

n 

∞

aj (t)Δt = ∞. j ∈ {0, 1, . . . , n}.

j =0 t0

j =0

(5.13) Suppose that the delta derivative in (5.12) at t = t0 is from the right. Assume that there exists a strictly increasing continuous function f on (−∞, ∞) such that f (0) = 0, fj (x) ≥ f (x), x = 0, limx→−∞ f (x) is finite if f (x) = x.

j ∈ {0, 1, . . . , n},

(5.14)

Theorem 5.4. Let x be an eventually positive (negative) solution of the equation (5.12). Then there exists an s ∈ [t0 , ∞) such that for some s1 ∈ [s, ∞) the solution x is eventually decreasing (increasing) on [s1 , ∞) and limt→∞ x(t) = 0. Proof. Assume that x is an eventually positive solution of the equation (5.12). Then there exists an s ∈ [t0 , ∞) such that x(t) > 0 for all t ∈ [s, ∞). Hence, there exists an s1 ∈ [s, ∞) such that x(τj (t)) ≥ 0 for

any

t ∈ [s1 , ∞)

and

for

any

j ∈ {0, 1, . . . , n}.

From here, f (x(τj (t))) ≥ 0

for

any

t ∈ [s1 , ∞)

and

for any

j ∈ {0, 1, . . . , n}.

From the last inequality and from (5.14), we conclude that fj (x(τj (t))) ≥ 0

t ∈ [s1 , ∞),

for any

j ∈ {0, 1, . . . , n}.

Therefore −

n 

aj (t)fj (x(τj (t))) ≤ 0

for

any

t ∈ [s1 , ∞),

j =0

and x Δ (t) = −

n 

aj (t)fj (x(τj (t)))

j =0

≤ 0 for

any

t ∈ [s1 , ∞).

5.3 Global Stability

185

Consequently x is decreasing on [s1 , ∞). Define α = lim x(t). t→∞

Assume that α > 0. Then there is a t1 ∈ [t0 , ∞) such that x(τj (t)) ≥ α

for

t ∈ [t1 , ∞),

any

j ∈ {0, 1, . . . , n}.

Hence, for t ∈ [t1 , ∞), we have f (x(τj (t))) ≥ f (α) > f (0) = 0, fj (x(τj (t))) ≥ f (x(τj (t))) ≥ f (α), −aj (t)fj (x(τj (t))) ≤ −aj (t)f (α), −

n 

aj (t)fj (x(τj (t))) ≤ −f (α)

j =0

n 

aj (t).

j =0

Then x Δ (t) = −

n 

aj (t)fj (x(τj (t)))

j =0

≤ −f (α)

n 

aj (t),

t ∈ [t1 , ∞).

j =0

From here, x(t) ≤ x(t1 ) − f (α)

n 

t

aj (s)Δs,

t ∈ [t1 , ∞).

j =0 t1

From the last inequality, using (5.13), we get lim x(t) = −∞,

t→∞

which is a contradiction. Therefore α = 0. We leave to the reader as an exercise the case when x is an eventually negative solution of the equation (5.12). This completes the proof.

186

5 Stability for First-Order Functional Dynamic Equations

Example 5.3. Let T = Z. Consider the equation x Δ (t) +

x(t − 2) x(t − 3) x(t − 4) x(t − 1) + + + = 0, 1 + x(t − 1) 1 + x(t − 2) 1 + x(t − 3) 1 + x(t − 4)

Here t0 = 5, a0 (t) = 0, a1 (t) = 1, a2 (t) = 1, a3 (t) = 1, a4 (t) = 1,

t ∈ T, x , f1 (x) = 1+x x f2 (x) = . 1+x x f3 (x) = , 1+x x f4 (x) = , x ∈ R, 1+x τ1 (t) = t − 1, τ2 (t) = t − 2, τ3 (t) = t − 3, τ4 (t) = t − 4,

t ∈ [5, ∞).

We have f1 (0) = 0, f2 (0) = 0, f3 (0) = 0, f4 (0) = 0, τ4 (t) < τ3 (t) < τ2 (t) < τ1 (t), We have ρ 4 (t) = t − 4,

t ∈ T.

t ∈ T.

t ∈ [5, ∞).

5.3 Global Stability

187

Then we take M = 1 and we get ρ 4 (t) − M = t − 4 − 1 = t −5 t ∈ T.

< τ4 (t), Let f (x) =

x , 1+x

x ∈ R.

Then lim f (x) = lim

x→−∞

x→−∞

x 1+x

=1 and 4 

∞

aj (t)Δt = ∞.

j =0 t0

Therefore, if x is an eventually positive (negative) solution of the considered equation, then there exists an s ∈ [t0 , ∞) such that for some s1 ∈ [s, ∞) the solution x is eventually decreasing (increasing) on [s1 , ∞) and limt→∞ x(t) = 0. Exercise 5.3. Let T = Z and x be an eventually positive solution of the equation x Δ (t) + t 2



 (x(t − 4))2 + 1 e4x(t−4) − 1 + t 4 (x(t − 2) + 1)2 e4x(t−2) − 1 = 0, t ∈ [4, ∞), x(t) = t, t ∈ [0, 4].

Prove that lim x(t) = 0.

t→∞

Theorem 5.5. Let x be a solution of the equation (5.12). If f (x) = x and λ=

sup t≥τn−1 (t0 )

⎛ n  ⎝

⎞ σ (t)

j =0 τn (t)

then x is bounded above and below.

aj (s)Δs ⎠ < ∞,

188

5 Stability for First-Order Functional Dynamic Equations

Proof. Assume that f (x) = x and lim f (x) = −β > −∞.

x→−∞

Since f is strictly increasing on (−∞, ∞) and f (0) = 0, we have that β > 0 and f (x) ≥ −β,

x ∈ (−∞, ∞).

Hence, fj (x(τj (t))) ≥ f (x(τj (t))) ≥ −β

for any

t ∈ [t0 , ∞),

j ∈ {0, 1, . . . , n}.

Therefore " x Δ (t) = − nj=0 aj (t)fj (x(τj (t))) "n ≤ β j =0 aj (t), t ∈ [t0 , ∞).

(5.15)

Suppose that lim sup x(t) = ∞. t→∞

Then there exists a strictly increasing sequence {t k }k∈N of elements of [t0 , ∞) such that lim t k = ∞,

k→∞

lim

max

k→∞ t0 ≤t≤σ (t k )

x(t) = ∞,

max

t0 ≤t≤σ (t k )

x(t) > 0,

x Δ (t k ) ≥ 0,

k ∈ N.

Then, using (5.12), we obtain 0 ≤ x Δ (t k ) =−

n 

aj (t k )fj (x(τj (t k ))).

j =0

Hence, n  j =0

aj (t k )fj (x(τj (t k ))) ≤ 0.

k ∈ N,

5.3 Global Stability

189

Because f is strictly increasing on (−∞, ∞) and f (0) = 0, by the last inequality, it follows that there exists a ξk ∈ [τn (t k ), σ (t k )] such that x(ξk ) ≤ 0. Now we integrate the inequality (5.15) from ξk to σ (t k ) and we get x σ (t k ) ≤ x(ξk ) + β

n 

σ (t k )

aj (t)Δt

j =0 ξk

≤ x(ξk ) + β

n 

σ (t k )

aj (t)Δt

j =0 τn (t k )

⎛ ≤β

sup

⎝

t≥τn−1 (t0 )

n 

⎞

σ (t)

aj (s)Δs ⎠

j =0 τn (t)

= βλ. Hence, lim x σ (t k ) ≤ βλ,

k→∞

which is a contradiction. Thus x is bounded above and x(t) ≤ βλ for all t ∈ [t0 , ∞). Now we will show that x is bounded below. Assume that lim inf x(t) = −∞. t→∞

Then there exists a strictly increasing sequence {t k }k∈N of elements of [t0 , ∞) such that lim t k = ∞,

k→∞

min

t0 ≤t≤σ (tk )

x(t) < 0,

x σ (t k ) → −∞ as x Δ (t k ) ≤ 0,

k → ∞,

k ∈ N.

Hence from (5.12), we obtain 0 ≥ x Δ (t k ) =−

n 

aj (t k )fj (x(τj (t k ))),

k ∈ N.

j =0

Therefore there is an ηk ∈ [τn (t k ), σ (t k )] such that x(ηk ) ≥ 0. Let M=

max

j ∈{0,1,...,n}

{fj (βλ)}.

190

5 Stability for First-Order Functional Dynamic Equations

Then, for t large enough, we obtain fj (x(τj (t))) ≤ fj (βλ) ≤M and using the equation (5.12), we have x Δ (t) = −

n 

aj (t)fj (x(τj (t)))

j =0

≥ −M

n 

aj (t),

t ∈ [t0 , ∞).

j =0

We integrate the last inequality from ηk to σ (t k ) and we get x σ (t k ) ≥ x(ηk ) − M

n σ (t k )  ηk

aj (t)Δt

j =0

≥ x(ηk ) − Mλ ≥ −Mλ. Therefore lim inf x σ (t k ) ≥ −Mλ, k→∞

which is a contradiction. Consequently x is bounded below and x(t) ≥ −Mλ,

t ∈ [t0 , ∞).

Let f (x) = x. Assume that x is unbounded above solution. Take A > 0 arbitrarily. Then there is a t1 ∈ [t0 , ∞) such that x(t) ≥ A,

x(τj (t)) ≥ A

for any t ∈ [t1 , ∞) and for any j ∈ {0, 1, . . . , n}. Hence, fj (x(τj (t))) ≥ fj (A) ≥A

5.3 Global Stability

191

for any t ∈ [t1 , ∞) and for any j ∈ {0, 1, . . . , n}. Hence from (5.12), we obtain x Δ (t) = −

n 

aj (t)fj (x(τj (t)))

j =0

≤ −A

n 

aj (t),

t ∈ [t1 , ∞).

j =0

Then x(t) ≤ x(t1 ) − A

n 

t

aj (s)Δs

j =0 t1

→ −∞

t → ∞.

as

This is a contradiction. Therefore x is bounded above. Now we assume that x is unbounded below. Take B < 0 arbitrarily. Then there exists a t2 ∈ [t0 , ∞) so that x(τj (t)) ≤ B

x(t) ≤ B,

for any t ∈ [t2 , ∞) and for any j ∈ {0, 1, . . . , n}. Hence, fj (x(τj (t))) ≤ fj (B) t ∈ [t2 , ∞),

< 0,

j ∈ {0, 1, . . . , n}.

Let B˜ =



max

j ∈{0,1,...,n}

 fj (B) .

Then, using (5.12), we obtain x Δ (t) = −

n 

aj (t)fj (x(τj (t)))

j =0

≥ −B˜

n 

aj (t),

t ∈ [t2 , ∞).

j =0

From here, x(t) ≥ x(t2 ) − B˜

n 

t

aj (s)Δs

j =0 t2

→ ∞ as

t → ∞.

This is a contradiction. Therefore x is bounded below. This completes the proof.

192

5 Stability for First-Order Functional Dynamic Equations

Corollary 5.1. Suppose that all the conditions of Theorem 5.5 hold. Let f (x) = x and −β = lim f (x), x→−∞

M=

max {fj (βλ)}.

j ∈{0,...,n}

If x is a solution of the equation (5.12), then t ∈ [t0 , ∞).

−Mλ ≤ x(t) ≤ βλ,

Example 5.4. Let T = Z. Consider the equation x Δ (t) +

sign(x(t − 2)) (x(t − 2))2 sign(x(t − 1)) (x(t − 1))2 + = 0, 2 (t + 1)(t + 2) 1 + (x(t − 2)) (t + 3)(t + 4) 1 + (x(t − 1))2 x(t) = t,

t ∈ [2, ∞), t ∈ (−∞, 2].

Here t0 = 2, a0 (t) = 0, a1 (t) =

1 , (t + 3)(t + 4)

a2 (t) =

1 , (t + 1)(t + 2)

τ1 (t) = t − 1, τ2 (t) = t − 2,

t ∈ T,

f1 (x) = f2 (x) =

x 2 sign(x) , 1 + x2

x ∈ R.

f (x) =

x 2 sign(x) , 1 + x2

x ∈ R.

We take

We have

λ = sup t≥4

= sup t≥4

t+1 t−2 t+1 t−2

a1 (s)Δs +



t+1

a2 (s)Δs t−2

1 Δs + (s + 3)(s + 4)

t+1 t−2

1 Δs (s + 1)(s + 2)



5.3 Global Stability

193

 1 !!s=t+1 1 !!s=t+1 = sup − − ! ! s + 3 s=t−2 s + 1 s=t−2 t≥4

 1 1 1 1 + − + = sup − t +4 t +1 t +2 t −1 t≥4

 3 3 = sup + (t − 1)(t + 2) t≥4 (t + 1)(t + 4) 3 3 + 40 18 1 3 + = 40 6 29 , = 120 β = − lim f (x) =

t→−∞

x2 x→−∞ 1 + x 2 = 1,

= lim

f1 (βλ) = f2 (βλ) = f1 (λ)

 29 = f1 120  2 =

=

29 120

1+



29 120

2

841 . 15241

Therefore M=

841 . 15241

If x is a solution of the considered equation, then −

29 24389 ≤ x(t) ≤ , 1828920 120

t ∈ [2, ∞).

194

5 Stability for First-Order Functional Dynamic Equations

Exercise 5.4. Let T = 2N0 and x be a solution of the equation x Δ (t) +

        t 2   t 4 2sign x 2t sign x 4t x 2 x 4   t 2 +    = 0, (1 + t)(1 + 2t) 1 + x (1 + 3t)(1 + 6t) 3 + x t 4 2

t ∈ [8, ∞),

4

x(t) = t 2 ,

t ∈ [1, 8].

Find an estimate of x. Theorem 5.6. Let x be a solution of the equation (5.12). If there exists a point s ∈ [τn−1 (t0 ), ∞) such that x Δ (s) > 0 and x σ (s) > 0, then there exists g(s) ∈ [τn (s), σ (s)] such that x(g(s)) =

min

t∈[τn (s),σ (s)]

x(t) < 0.

If there exists a point s ∈ [τn−1 (t0 ), ∞) such that x Δ (s) < 0 and x σ (s) < 0, then there exists g(s) ∈ [τn (s), σ (s)] such that x(g(s)) =

max

t∈[τn (s),σ (s)]

x(t) > 0.

Proof. 1. Let there exists a point s ∈ [τn−1 (t0 ), ∞) such that x Δ (s) > 0 and x σ (s) > 0. Assume that x(t) ≥ 0 for all t ∈ [τn (s), σ (s)]. Then, using (5.12), we get x Δ (s) = −

n 

aj (s)fj (x(τj (s)))

j =0

≤−

n 

aj (s)f (x(τj (s)))

j =0

≤ 0, which is a contradiction. 2. Let there exists a point s ∈ [τn−1 (t0 ), ∞) such that x Δ (s) < 0 and x σ (s) < 0. Assume that x(t) ≤ 0 for all t ∈ [τn (s), σ (s)]. Then, using (5.12) and fj (x(τj (s))) ≤ 0,

j ∈ {0, 1, . . . , n},

we get x Δ (s) = −

n 

aj (s)fj (x(τj (s)))

j =0

≥ 0, which is a contradiction. This completes the proof.

5.3 Global Stability

195

Let r1 =

r2 =

n 

sup

σ (s)

−1 s≥τn−1 (t0 ) j =0 τj (τn (s))

n 

sup

τj−1 (τn (s))

s≥τn−1 (t0 ) j =0 τn (s)

φ(x) = x − r1 f (x),

aj (t)Δt,

aj (t)Δt,

x ∈ (−∞, ∞).

If f (x) = x and r1 + r2 < ∞, by Theorem 5.5, we get that the solution x of the equation (5.12) is bounded above and below and hence, any solution of the equation (5.12) that oscillates about 0 is bounded above and below. For L < 0, we define RL = max φ(x) − r2 f (L), L≤x≤0

SL =

min

0≤x≤|RL |

φ(x) − r2 f (RL ).

Theorem 5.7. Suppose that x is a solution of the equation (5.12) . that oscillates  about 0. If for some real number L < 0 there exists a point tL ∈ τn−1 (t0 ), ∞ such that x(t) ≥ L for t ∈ [tL , ∞), then RL > 0 and SL < 0, and the solution x satisfies the inequalities x σ (t) ≤ RL

for

t ∈ [T2 , ∞)

x σ (t) ≥ SL

for

t ∈ [T3 , ∞),

and

where T2 ≥ τn−2 (tL ) and T3 ≥ τn−4 (tL ) are such that there exists a point s ∈ [tL , Ti ] such that (−1)i x σ (s) > 0 and (−1)i x Δ (s) > 0, i ∈ {2, 3}.Furthermore, if SL > L for any L < 0, then limt→∞ x(t) = 0. Proof. Assume that x(t) ≥ L for any t ∈ [tL , ∞). Since the solution x oscillates about 0, there exists a point s ∈ [tL , ∞) such that x σ (s) > 0 and x Δ (s) > 0. Hence from Theorem 5.6, there exists a point g(s) ∈ [τn (s), σ (s)] such that x(g(s)) =

min

t∈[τn (s),σ (s)]

x(t) < 0.

Now we integrate the equation (5.12) from g(s) to σ (s) and we get 0 < x σ (s) # σ (s) "n = x(g(s)) − g(s) j =0 aj (t)fj (x(τj (t)))Δt.

(5.16)

196

5 Stability for First-Order Functional Dynamic Equations

Note that τn (g(s)) ≥ τn2 (s). There exists a collection of points {t j (s)}nj=0 from the interval [g(s), σ (s)] with t 0 (s) = g(s) and  t j (s) =

τj−1 (τn (s)) if τj−1 (τn (s)) > g(s) g(s) otherwise,

such that τj (t) ∈ [τn (s), σ (s)] and fj (x(τj (t))) ≥ f (x(g(s))),

j ∈ {0, 1, . . . , n},

for all t ∈ [t j (s), σ (s)] and for all t ∈ [g(s), σ (s)] we have that fj (x(τj (t))) ≥ f (L),

j ∈ {0, 1, . . . , n}.

Hence from (5.16), we obtain 0 < x σ (s) = x(g(s)) −

n 

σ (s)

aj (t)fj (x(τj (t)))Δt

j =0 g(s)

= x(g(s)) −

n 

t j (s)

aj (t)fj (x(τj (t)))Δt

j =0 g(s)

−

n 

σ (s)

aj (t)fj (x(τj (t)))Δt

j =0 t j (s)

≤ x(g(s)) −

−

n  j =0

%

n  j =0

%

&

t j (s)

aj (t)Δt f (x(g(s))) g(s)

&

σ (s)

aj (t)Δt f (L) t j (s)

≤ x(g(s)) − r1 f (x(g(s))) − r2 f (L) = φ(x(g(s))) − r2 f (L) ≤ RL .

5.3 Global Stability

197

If there exists a point t ≥ T2 such that x σ (t) > RL , then there exists an s ≥ T2 so that x σ (s) > 0, x Δ (s) > 0 and x σ (s) > RL . This is a contradiction. Therefore x σ (t) ≤ RL for any t ∈ [T2 , ∞). Also, there exists an s ∈ [tL , ∞) such that x σ (s) < 0, x Δ (s) < 0. Hence from Theorem 5.6, it follows that there exists a point g(s) ∈ [τn (s), σ (s)] such that x(g(s)) =

max

t∈[τn (s),σ (s)]

x(t)

> 0. There exists a collection of points {t j (s)}nj=0 from [g(s), σ (s)] with t 0 (s) = g(s) and  τj−2 (τn (s)) if τj−2 (τn (s)) > g(s) t j (s) = g(s) otherwise, j ∈ {0, 1, . . . , n}, such that τj (t) ∈ [τn (s), σ (s)], j ∈ {0, 1, . . . , n}, and fj (x(τj (t))) ≤ f (x(g(s))),

j ∈ {0, 1, . . . , n},

for all t ∈ [t j (s), σ (s)], j ∈ {0, 1, . . . , n}, and for all t ∈ [g(s), σ (s)] we have that fj (x(τj (t))) ≤ f (RL ), j ∈ {0, 1, . . . , n}. Hence, 0 > x σ (s) n σ (s) 

= x(g(s)) −

g(s)

= x(g(s)) −

n 

aj (t)fj (x(τj (t)))Δt

j =0 t j (s)

aj (t)fj (x(τj (t)))Δt

j =0 g(s)

−

n 

σ (s)

aj (t)fj (x(τj (t)))Δt

j =0 t j (s)

≥ x(g(s)) −

−

n  j =0

%

n  j =0 σ (s)

%

&

τj (s)

aj (t)Δt f (x(g(s))) g(s)

& aj (t)Δt f (RL )

τj (s)

≥ x(g(s)) − r1 f (x(g(s))) − r2 f (RL ) = φ(x(g(s))) − r2 f (RL ) ≥ SL .

198

5 Stability for First-Order Functional Dynamic Equations

If there exists a point t ≥ T3 such that x σ (t) < SL , then there exists an s ≥ τn−4 (tL ) such that x σ (s) < 0, x Δ (s) < 0 and x σ (s) ≥ SL for any t ∈ [T3 , ∞). Suppose that SL > L for any L < 0 and we set L = lim inf x(t). t→∞

If L < 0, then SL > L and there exists t L ≥ τn−4 (tL ) such that x(t) ≥ SL > L for any t ∈ [t L , ∞). This is a contradiction. Therefore L = 0 and limt→∞ x(t) = 0. This completes the proof. Definition 5.5. The zero solution of the equation (5.12) is said to be uniformly stable if for any  > 0 and t ∗ ∈ [t0 , ∞), there exists a δ = δ() > 0 such that sup

s∈[τn (t ∗ ),σ (t ∗ )]

|x(s)| < δ

implies |x(t)| < ,

t ∈ [t ∗ , ∞).

Definition 5.6. The zero solution of the equation (5.12) is said to be globally attractive if every solution of the equation (5.12) tends to zero as t → ∞. Definition 5.7. The zero solution of the equation (5.12) is said to be globally asymptotically stable if it is uniformly stable and globally attractive. Theorem 5.8. Assume that f (x) = x, n σ (t) 

sup

t≥τn−1 (t0 ) τn (t) j =0

aj (s)Δs < ∞

and the hypotheses of Theorem 5.4 and Theorem 5.7 hold. Then the zero solution of the equation (5.12) is uniformly stable. Moreover, limt→∞ x(t) = 0 implies that the zero solution of the equation (5.12) is globally asymptotically stable. Proof. Assume that there exists an s ∈ [t0 , ∞) such that x(t) > 0 for any t ∈ [s, ∞). Hence from Theorem 5.4, we have that there exists an s1 ∈ [s, ∞) such that x is decreasing on [s1 , ∞) and limt→∞ x(t) = 0. Therefore, for any  > 0 and T ≥ t0 , there is a δ = δ() > 0 such that sup

j ∈{0,1,...,n}

|x(τj (t))| < δ.

5.3 Global Stability

199

Then |x(t)| < 

t ∈ [T , ∞).

for

We leave to the reader as an exercise the case when there exists an s ∈ [t0 , ∞) such that x(t) < 0 for any t ∈ [s, ∞). Let x oscillates about 0. Then by Theorem 5.4 and Theorem 5.7, we have that for any  > 0 there is a δ > 0 such that  = max{δ, Rδ } > 0 and sup

j ∈{0,1,...,n}

|x(τj (t))| < δ,

imply that for L = −δ < 0 the solution x of the equation (5.12) satisfies − ≤ −L ≤ x(t) ≤ RL ≤  for t ∈ [T , ∞). Consequently the zero solution of the equation (5.12) is uniformly stable. Hence, limt→∞ x(t) = 0 implies that the zero solution of the equation (5.12) is globally asymptotically stable. This completes the proof. Theorem 5.9. Assume that f (x) = x and sup

n σ (t) 

t≥τn−1 (t0 ) τn (t) j =0

aj (s)Δs < ∞

and φ is monotone on (−∞, ∞), and for any L < 0

−r2 f (−r2 f (L)) > L if φ is increasing φ (φ(L) − r2 f (L)) − r2 f (φ(L) − r2 f (L)) > L

if φ

is

decreasing.

Then the zero solution of the equation (5.12) is globally asymptotically stable. Proof. By Theorem 5.4, it follows that it is enough to consider the case when x is −2 oscillatory about 0. If for some L < 0 there . exists tL ∈ [τn (t0 ), ∞) such that x(t) ≥ L for t ∈ [tL , ∞), then for any t ∈ τn−2 (tL ), ∞ we have x σ (t) ≤ RL

for

* t ∈ τn−2 (tL ), ∞

and x σ (t) ≥ SL .  for t ∈ τn−4 (tL ), ∞ . Assume that φ is increasing on (−∞, ∞). Then max φ(x) = 0 and

L≤x≤0

min φ(x) = 0.

0≤x≤RL

200

5 Stability for First-Order Functional Dynamic Equations

Consequently RL = −r2 f (L) and

SL = −r2 f (RL ).

Therefore SL > L. Hence from Theorem 5.7, we get lim x(t) = 0.

t→∞

From here and from Theorem 5.8, it follows that the zero solution of the equation (5.12) is globally asymptotically stable. The case when φ is decreasing on (−∞, ∞) we leave to the reader as an exercise. This completes the proof. Example 5.5. Let T be an isolated time scale unbounded above and t0 ∈ T. Suppose that sup μ(s) = sup (s − ρ(s)) = 1. s≥σ (t0 )

s≥σ (t0 )

Consider the IVP x Δ (t) = −a1 arctan(x(ρ(t))), x(t) = ψ(t),

t ∈ [t0 , ∞),

t ∈ [ρ(t0 ), t0 ],

ψ ∈ Crd ([ρ(t0 ), t0 ]),

where a1 ∈ (0, 1] is a constant. Here a0 = 0, τ1 (t) = ρ(t),

t ∈ [ρ(t0 ), ∞).

Then r1 = sup

σ (s)

a1 Δt

s≥σ (t0 ) s

= a1 sup μ(s) s≥σ (t0 )

= a1 ≤ 1, r2 = sup

s

a1 Δt

s≥σ (t0 ) ρ(s)

= a1 sup (s − ρ(s)) s≥σ (t0 )

= a1 ≤ 1.

5.4 Asymptotic Stability

201

In our case, φ(x) = x − a1 arctan(x),

x ∈ R.

We have that φ is strictly increasing on R and −r2 f (−r2 f (L)) = −a1 arctan (−a1 arctan(L)) >L for any L < 0. Hence from Theorem 5.9, we conclude that the zero solution of the considered IVP is globally asymptotically stable. Exercise 5.5. Let T be an isolated time scale unbounded above and t0 ∈ T. Suppose that sup μ(s) = sup (s − ρ(s)) = 1. s≥σ (t0 )

s≥σ (t0 )

Prove that the zero solution of the IVP x Δ (t) = −a1 sin(x(ρ(t))), x(t) = ψ(t),

t ∈ [t0 , ∞),

t ∈ [ρ(t0 ), t0 ],

ψ ∈ Crd ([ρ(t0 ), t0 ]),

 where a1 ∈ 0, √1 , is globally asymptotically stable. 2

5.4 Asymptotic Stability Suppose that T is an unbounded above time scale with forward jump operator, backward jump operator, and delta differentiation operator σ , ρ, and Δ, respectively. Consider the equation x Δ (t) = a(t)x(t) + b(t)x(τ (t)),

t ∈ T,

(5.17)

where a and b are nonzero functions on T and τ is an increasing and unbounded function on T such that τ (t) < t and τ (t) ∈ T for any t ∈ T. Firstly, we will introduce some auxiliary results which will be useful for the investigation of the asymptotic behavior of the solutions of the equation (5.17). Consider the equation ψ(τ (t)) = ψ(t) − 1,

t ∈ T,

(5.18)

202

5 Stability for First-Order Functional Dynamic Equations

and the inequality |b(t)|φ(τ (t)) ≤ |a(t)|φ(t),

t ∈ T.

(5.19)

Definition 5.8. The equation (5.18) is said to be the Abel equation. 1 (T) be such that τ (t) < t for all t ∈ T, τ Δ is positive Theorem 5.10. Let τ ∈ Crd and nonincreasing on [t0 , ∞) for some t0 ∈ T and τ (T) = T. Then there exists an 1 ([t , ∞)) of the equation (5.18) with a positive and unbounded solution ψ ∈ Crd 0 nonincreasing delta derivative on [τ (t0 ), ∞). 1 ([τ (t ), t ]) be a given function such that ψ Δ (t) > 0 for Proof. Let ψ0 ∈ Crd 0 0 0 t ∈ [τ (t0 ), t0 ] and ψ0Δ is nonincreasing on [τ (t0 ), t0 ]. Now we differentiate the equation (5.18) and we get

ψ Δ (τ (t))τ Δ (t) = ψ Δ (t),

t ∈ T.

Therefore ψ0 can be extended by the step method to a solution on [τ (t0 ), ∞) of the equation (5.18) that has the required properties. This completes the proof. Remark 5.1. If a and b are constants, then the inequality (5.19) takes the form ! ! !b! ! ! φ(τ (t)) ≤ φ(t), !a !

t ∈ T.

Let ψ be a solution of the equation (5.18) and ! !ψ(t) !b! φ(t) = !! !! , a

t ∈ T.

Hence, ! !ψ(τ (t)) !b! φ(τ (t)) = !! !! a ! !ψ(t)−1 !b! = !! !! , a ! ! ! !ψ(t)−1 ! ! ! !! ! !b! ! ! φ(τ (t)) = ! b ! ! b ! !a ! !a ! !a ! ! !ψ(t) !b! = !! !! a = φ(t),

t ∈ T.

5.4 Asymptotic Stability

203

Theorem 5.11. Let a, b, τ ∈ Crd (T), a(t) > 0, b(t) = 0 for any t ∈ T, τ : T → T be increasing on T such that τ (t) < t for any t ∈ T and limt→∞ τ (t) = ∞. Let also, ∞

|b(t)|ea (τ (t), σ (t))Δt < ∞,

t0 ∈ T.

(5.20)

t0

Then for any solution x of the equation (5.17) there exists a constant L ∈ R such that lim x(t)ea (t0 , t) = L.

(5.21)

t→∞

Proof. Let x be a solution of the equation (5.17), t0 ∈ T. We set z(t) =

x(t) , ea (t, t0 )

t ∈ [t0 , ∞).

Then zΔ (t) = =

x Δ (t)ea (t, t0 ) − a(t)x(t)ea (t, t0 ) ea (t, t0 )ea (σ (t), t0 ) x Δ (t) − a(t)x(t) , ea (σ (t), t0 )

t ∈ [t0 , ∞),

whereupon x Δ (t) = ea (σ (t), t0 )zΔ (t) + a(t)x(t),

t ∈ [t0 , ∞).

Hence from (5.17), we get a(t)x(t) + b(t)x(τ (t)) = ea (σ (t), t0 )zΔ (t) + a(t)x(t), t ∈ [t0 , ∞), or ea (σ (t), t0 )zΔ (t) = b(t)x(τ (t)) = b(t)ea (τ (t), t0 )z(τ (t)),

t ∈ [t0 , ∞).

Note that ea (τ (t), t0 ) = ea (τ (t), t0 )ea (t0 , σ (t)) ea (σ (t), t0 ) = ea (τ (t), σ (t)),

t ∈ T.

(5.22)

204

5 Stability for First-Order Functional Dynamic Equations

Hence from (5.22), we obtain zΔ (t) = b(t)ea (τ (t), σ (t))z(τ (t)),

t ∈ [t0 , ∞).

Let tm+1 = sup{t ∈ T : τ (t) ≤ tm },

m ∈ N0 ,

and Im+1 = [tm , tm+1 ],

m ∈ N0 .

Then [t0 , ∞) =

∞ 

Im .

m=1

Since sup T = ∞, we have that any Im , m ∈ N, is nontrivial. Denote Bm = sup{|z(t)| : t ∈

m 

Ij }.

j =1

Let t ∗ ∈ Im+1 , m ∈ N. Now we integrate (5.23) from tm to t ∗ and we get t∗

z(t ∗ ) = z(tm ) +

b(t)ea (τ (t), σ (t))z(τ (t))Δt, tm

whereupon ! ! ! |z(t )| = !z(tm ) + !

t∗

∗

tm

! ! ! ≤ |z(tm )| + ! ! ≤ |z(tm )| +

! ! ! b(t)ea (τ (t), σ (t))z(τ (t))Δt ! !

t∗ tm t∗

! ! ! b(t)ea (τ (t), σ (t))z(τ (t))Δt ! !

|b(t)|ea (τ (t), σ (t))|z(τ (t))|Δt

tm

≤ Bm + B m

t∗ tm

|b(t)|ea (τ (t), σ (t))Δt.

(5.23)

5.4 Asymptotic Stability

205

Since t ∗ ∈ Im+1 was arbitrarily chosen, by the last inequality we conclude that %

&

t∗

Bm+1 ≤ Bm 1 +

|b(t)|ea (τ (t), σ (t))Δt

tm

%

&

t∗

≤ Bm−1 1 +

|b(t)|ea (τ (t), σ (t))Δt

tm−1

%

t∗

× 1+

&

|b(t)|ea (τ (t), σ (t))Δt

tm

≤ ··· ≤ B1

m $

%

&

t∗

1+

|b(t)|ea (τ (t), σ (t))Δt ,

m ∈ N.

tl

l=1

By the condition (5.20), we conclude that ∞ $

%

t∗

1+

& |b(t)|ea (τ (t), σ (t))Δt

< ∞.

tl

l=1

Therefore z is bounded and there exists a constant K > 0 such that |z(t)| ≤ K,

t ∈ [t0 , ∞).

Let now t 1 , t 2 ∈ [t0 , ∞), t 1 < t 2 , be arbitrarily chosen. We integrate (5.23) from t 1 to t 2 and we get z(t 2 ) − z(t 1 ) =

t2

b(t)ea (τ (t), σ (t))z(τ (t))Δt,

t1

whereupon ! ! ! |z(t ) − z(t )| = ! ! 2

≤

! ! ! b(t)ea (τ (t), σ (t))z(τ (t))Δt ! !

t2

1

t1 t2 t1

≤K

|b(t)|ea (τ (t), σ (t))|z(τ (t))|Δt t2

t1

|b(t)|ea (τ (t), σ (t))Δt.

206

5 Stability for First-Order Functional Dynamic Equations

Let  > 0 be arbitrarily chosen. Then, by the condition (5.20), it follows that there exist t 3 , t 4 ∈ [t0 , ∞), t 3 < t 4 , large enough, so that t4 t3

|b(t)|ea (τ (t), σ (t))Δt
0,

4t (t + 1)(4t + 1)

≥ 0, b(t) =

4 (t + 1)(2t + 1)

= 0, t τ (t) = 2

t ∈ T,

 t , 2

t ≥ 2,

5.4 Asymptotic Stability

207

t > 0,

< t, lim τ (t) = ∞,

t→∞

t0 = 2. We will show that x(t) = 4t + 1,

t ∈ T,

is a solution of the considered equation. For t ≥ 2, we have x Δ (t) = 4,

 t = 2t + 1, x 2 4t 4t x(t) = , (t + 1)(4t + 1) t +1

 t 4 4 x Δ (t) − x = 4− (t + 1)(2t + 1) 2 t +1 4t + 4 − 4 t +1 4t = t +1 4t x(t), = (t + 1)(4t + 1)

=

Next, ea (τ (t), σ (t)) = e =e

# τ (t)

1 σ (t) μ(s)

−

≤ 1,

# 2t

1 t s 2

log(1+μ(s)a(s))Δs

 4s 2 log 1+ (s+1)(4s+1) Δs

t ∈ T.

Let g(t) = −

1 , t +1

t ∈ T.

t ∈ [2, ∞).

208

5 Stability for First-Order Functional Dynamic Equations

Then g Δ (t) =

1 (t + 1)(σ (t) + 1)

=

1 (t + 1)(2t + 1)

=

1 b(t), 4

t > 0.

Hence, ∞

∞

|b(t)|ea (τ (t), σ (t))Δt ≤ 2

t0

4 Δt (t + 1)(2t + 1)

∞

=4

g Δ (t)Δt

2

!t=∞ ! = 4g(t)! t=2

4 !!t=∞ =− ! t + 1 t=2 4 = 3 < ∞. Then, using Theorem 5.11, it follows that there exists a constant L ∈ R such that lim (x(t)ea (2, t)) = L.

t→∞

Exercise 5.6. Let T = 3N0 equation x Δ (t) =

  1 N  3

{0}. Prove that for any solution x of the

8 7t x(t) + x (t + 2)(5t + 4) (t + 7)(4t + 9)

x(t) = 5t 2 + t,

t ∈ [0, 3],

there exists a constant L ∈ R such that lim x(t)ea (3, t) = L.

t→∞

 t , 3

t ∈ [3, ∞),

5.4 Asymptotic Stability

209

Theorem 5.12. Let a > 0 be a constant, τ be a delta differentiable on T such that τ (t) < t for all t ∈ T and 0 < τ Δ ≤ λ < 1 on T. Then lim ea (τ (t), t) = 0.

t→∞

Proof. Let t ∈ T, t > t0 , be arbitrarily chosen. By the Mean Value Theorem, it follows that there exists ξ ∈ [τ (t), t] such that eaΔ (ξ, t0 ) ≤

ea (t, t0 ) − ea (τ (t), t0 ) , t − τ (t)

whereupon (t − τ (t))eaΔ (ξ, t0 ) ≤ ea (t, t0 ) − ea (τ (t), t0 ) or ea (t, t0 ) ≥ (t − τ (t))eaΔ (ξ, t0 ) + ea (τ (t), t0 ). Note that ea (·, t0 ) is an increasing function on T. Then ea (ξ, t0 ) ≥ ea (τ (t), t0 ) and ea (t, τ (t)) = ea (t, t0 )ea (t0 , τ (t)) 0) = eae(τa (t,t (t),t0 ) eΔ (ξ,t )

≥ 1 + ea a(τ (t),t0 0 ) (t − τ (t)) a (ξ,t0 ) = 1 + a eae(τ (t),t0 ) (t − τ (t)) ≥ 1 + a(t − τ (t)).

(5.24)

Now we apply the Mean Value Theorem for τ on [t0 , t]. Therefore there exists a θ ∈ [t0 , t] such that τ Δ (θ ) ≥

τ (t) − τ (t0 ) , t − t0

from where λ≥

τ (t) − τ (t0 ) t − t0

or τ (t) − τ (t0 ) ≤ λ(t − t0 ),

210

5 Stability for First-Order Functional Dynamic Equations

or τ (t) ≤ τ (t0 ) + λ(t − t0 ) and −τ (t) ≥ −τ (t0 ) − λ(t − t0 ). Hence from (5.24), we conclude that ea (t, τ (t)) ≥ 1 + a (t − τ (t0 ) − λ(t − t0 )) = 1 + a ((1 − λ)t − τ (t0 ) + λt0 ) = 1 + a(1 − λ)t + aλt0 − aτ (t0 ). Note that for t > t0 large enough, we have that 1 + a(1 − λ)t + aλt0 − aτ (t0 ) > 0. Therefore, for t > t0 large enough, we get ea (τ (t), t) ≤

1 1 + a(1 − λ)t + aλt0 − aτ (t0 )

and hence, 0 ≤ lim ea (τ (t), t) t→∞

≤ lim

t→∞

1 1 + a(1 − λ)t + aλt0 − aτ (t0 )

= 0. This completes the proof. Theorem 5.13. Consider the equation x Δ (t) = ax(t) + bx(τ (t)),

t ∈ T,

(5.25)

where a > 0, b = 0 are real constants and τ is a delta differentiable on T such that τ (t) < t for any t ∈ T, τ (T) = T, and 0 < τ Δ ≤ λ < 1 on T. Let x be a solution of the equation (5.25) defined on [t0 , ∞) and x(t) = o(ea (t, t0 ))

as

t → ∞.

(5.26)

5.4 Asymptotic Stability

211

If φ is a positive solution of (5.19), then x(t) = O(φ(t)) as

t → ∞.

Proof. We have

x(·) ea (·, t0 )

Δ (t) =

x Δ (t)ea (t, t0 ) − x(t)eaΔ (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

=

x Δ (t)ea (t, t0 ) − ax(t)ea (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

=

x Δ (t) − ax(t) ea (σ (t), t0 )

=

bx(τ (t)) , ea (σ (t), t0 )

t ∈ [t0 , ∞).

Hence, using (5.26), we get ∞ t

x(·) ea (·, t0 )

Δ

∞

(s)Δs = t

bx(τ (s)) Δs, ea (σ (s), t0 )

t ≥ t0 ,

or ∞

x(t) x(s) − =b s→∞ ea (s, t0 ) ea (t, t0 ) lim

t

x(τ (s)) Δs, ea (σ (s), t0 )

t ≥ t0 ,

or ∞

x(t) = −bea (t, t0 ) t

x(τ (s)) Δs, ea (σ (s), t0 )

t ≥ t0 .

Let now tm+1 = sup{t ∈ T : τ (t) ≤ tm }, Im+1 = [tm , tm+1 ],

m ∈ N0 ,

m ∈ N0 .

By (5.26), it follows that there exists a positive constant M such that |x(t)| ≤ Mea (t, t0 ),

t ≥ t0 ,

t ∈ T.

(5.27)

212

5 Stability for First-Order Functional Dynamic Equations

Hence from (5.27), we obtain ! ! #∞ ! ! (s)) |x(t)| = !−bea (t, t0 ) t eax(τ Δs ! (s),t0 ) # ∞ |x(τ(σ(s))| ≤ |b|ea (t, t0 ) t ea (σ (s),t0 ) Δs # ∞ (τ (s),t0 ) ≤ M|b|ea (t, t0 ) t eeaa (σ (s),t0 ) Δs #∞ = M|b|ea (t, t0 ) t ea (τ (s), σ (s))Δs for any t ≥ t1 . Note that (ea (τ (·), ·))Δ (t) = (ea (τ (·), t0 )ea (t0 , ·))Δ (t) = (ea (τ (·), t0 ))Δ (t)ea (t0 , σ (t)) +ea (τ (t), t0 )eaΔ (t0 , t) = aea (τ (t), t0 )ea (t0 , σ (t))τ Δ (t) Δ

1 +ea (τ (t), t0 ) ea (t, t0 ) = aτ Δ (t)ea (τ (t), σ (t)) +ea (τ (t), t0 )

−eaΔ (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

= aτ Δ (t)ea (τ (t), σ (t)) −aea (τ (t), t0 )

ea (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

= aτ Δ (t)ea (τ (t), σ (t)) − aea (τ (t), t0 )ea (t0 , σ (t)) = aτ Δ (t)ea (τ (t), σ (t)) − aea (τ (t), σ (t))   = −a 1 − τ Δ (t) ea (τ (t), σ (t)), t ∈ T, i.e., ea (τ (t), σ (t)) = − ≤−

(ea (τ (·), ·))Δ (t)   a 1 − τ Δ (t) (ea (τ (·), ·))Δ (t) , a(1 − λ)

t ∈ T.

Hence, from (5.28) and Theorem 5.12, we get ∞

|x(t)| ≤ M|b|ea (t, t0 )

ea (τ (s), σ (s))Δs t

≤−

M|b|ea (t, t0 ) a(1 − λ)

∞ t

(ea (τ (·), ·))Δ (s)Δs

(5.28)

5.4 Asymptotic Stability

213

=

M|b|ea (t, t0 ) ea (τ (t), t) a(1 − λ)

=

M|b|ea (τ (t), t0 ) , a(1 − λ)

t ≥ t1 .

Let t ≥ t2 . Then τ (t) ≥ t1 and ∞

x(t) = −bea (t, t0 ) t

x(τ (s)) Δs, ea (σ (s), t0 )

t ≥ t1 ,

and ! ! #∞ ! ! (s)) |x(t)| = !−bea (t, t0 ) t eax(τ (σ (s),t0 ) Δs ! #∞ (s))| ≤ |b|ea (t, t0 ) t ea|x(τ (σ (s),t0 ) Δs # 2 2 (s),t ) ∞ |b| 0 ≤ M a(1−λ) ea (t, t0 ) t eeaa(τ(σ (s),t Δs 0) # 2 ∞ M|b| 2 = a(1−λ) ea (t, t0 ) t ea (τ (s), σ (s))Δs,

(5.29) t ≥ t2 .

Note that  Δ  Δ ea (τ 2 (·), ·) (t) = ea (τ 2 (·), t0 )ea (t0 , ·) (t)  Δ = ea (τ 2 (·), t0 ) (t)ea (t0 , σ (t))

1 +ea (τ (t), t0 ) ea (·, t0 )

Δ

2

(t)

= aea (τ 2 (t), t0 )ea (t0 , σ (t))τ Δ (τ (t))τ Δ (t) −ea (τ 2 (t), t0 )

eaΔ (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

= aea (τ 2 (t), t0 )ea (t0 , σ (t))τ Δ (τ (t))τ Δ (t) −aea (τ 2 (t), t0 )

ea (t, t0 ) ea (t, t0 )ea (σ (t), t0 )

= aea (τ 2 (t), σ (t))τ Δ (τ (t))τ Δ (t) −aea (τ 2 (t), t0 )ea (t0 , σ (t)) = aea (τ 2 (t), σ (t))τ Δ (τ (t))τ Δ (t) −aea (τ 2 (t), σ (t))   = −a 1 − τ Δ (τ (t))τ Δ (t) ea (τ 2 (t), σ (t)),

t ∈ T,

214

5 Stability for First-Order Functional Dynamic Equations

i.e., Δ  ea (τ 2 (·), ·) (t)  ea (τ (t), σ (t)) = −  a 1 − τ Δ (τ (t))τ Δ (t) Δ  ea (τ 2 (·), ·) (t) , t ∈ T. ≤− a(1 − λ2 ) 2

Hence, from (5.29) and Theorem 5.12, we get |x(t)| ≤

M|b|2 ea (t, t0 ) a(1 − λ)

≤−

∞

ea (τ 2 (s), σ (s))Δs

t

M|b|2 ea (t, t0 ) a 2 (1 − λ)(1 − λ2 )

∞

ea (τ 2 (·), ·)

Δ

t

=

M|b|2 ea (t, t0 )ea (τ 2 (t), t) a 2 (1 − λ)(1 − λ2 )

=

M|b|2 ea (τ 2 (t), t0 ), a 2 (1 − λ)(1 − λ2 )

t ≥ t2 .

The repetition leads to the estimate |x(t)| ≤

M|b|m / e (τ m (t), t0 ), l) a am m (1 − λ l=1

Note that, if t ≤ tm+1 , then τ (t) ≤ tm , τ 2 (t) ≤ τ (tm ) ≤ tm−1 , τ (t) ≤ τ (tm−1 ) 3

≤ tm−2 , .. . τ m (t) ≤ t1 ≤ τ −1 (t0 ),

t ≤ tm+1 .

Therefore for all t ≤ tm+1 , we have ea (τ m (t), t0 ) ≤ ea (τ −1 (t0 ), t0 ).

t ≥ tm .

(s)Δs

5.4 Asymptotic Stability

215

Consequently |x(t)| ≤

M|b|m / e (τ −1 (t0 ), t0 ), l) a am m (1 − λ l=1

t ∈ Im+1 .

Since ∞  $

1 − λl < ∞,

l=1

there exists a positive constant K such that m  $ l=1

M 1 − λl ≥ ea (τ −1 (t0 ), t0 ) K

for any m ∈ N. Then |x(t)| ≤ K

|b|m , am

t ∈ Im+1 .

Let N = inf{φ(t) : t ∈ T}. Then N > 0 and |b| φ(τ (t)) a |b| ≥ N , t ∈ I2 , a |b| φ(t) ≥ φ(τ (t)) a

φ(t) ≥

≥N

|b|2 , a2

t ∈ I3 ,

|b|m , am

t ∈ Im+1 .

.. . φ(t) ≥ N

From the last inequality and from (5.30), we conclude that x(t) = O(φ(t)) as This completes the proof.

t → ∞.

(5.30)

216

5 Stability for First-Order Functional Dynamic Equations

Theorem 5.14. Consider the equation (5.25), where a > 0 and b = 0 are real constants and τ is delta differentiable function on T such that τ (t) < t for all t ∈ T, τ (T) = T, and 0 < τ Δ ≤ λ < 1 on T. Then any solution x of the equation (5.25) satisfies (5.21). Moreover, if x1 and x2 are solutions of (5.25) satisfying (5.21) with the same constant L, then x1 (t) − x2 (t) = O(φ(t)) as

t → ∞,

where φ is a positive solution of (5.19). Proof. By the proof of Theorem 5.13, we have ea (τ (t), σ (t)) ≤ −

(ea (τ (·), ·))Δ (t) , a(1 − λ)

t ∈ T.

Then, using Theorem 5.12, we get ∞

∞

|b|ea (τ (t), σ (t))Δt = |b|

t0

ea (τ (t), σ (t))Δt t0

≤−

|b| a(1 − λ)

∞

(ea (τ (·), ·))Δ (t)Δt

t0

|b| ea (τ (t0 ), t0 ) = a(1 − λ) < ∞. Hence from Theorem 5.11, we conclude that for any solution x of the equation (5.25) there exists a constant L such that (5.21) holds. Let now x1 and x2 be two solutions of the equation (5.25) that satisfy (5.21) with the same constant L. Then x1 − x2 is a solution of the equation (5.25) and lim (x1 (t) − x2 (t)) ea (t0 , t) = 0,

t→∞

i.e., x1 (t) − x2 (t) = o(ea (t0 , t))

as

t → ∞.

From here and from Theorem 5.13, we get x1 (t) − x2 (t) = O(φ(t)) as

t → ∞.

This completes the proof. Theorem 5.15. Consider the equation (5.17) with a ∈ R + (T), b ∈ Crd (T), τ ∈ 1 (T), a(t) < 0, b(t) = 0 for any t ∈ T and τ be an increasing function on Crd

5.4 Asymptotic Stability

217

1 (T) be T such that τ (t) < t for all t ∈ T and τ (T) = T. Let also, ψ ∈ Crd an increasing solution of (5.18) such that ψ(T) is a time scale with forward jump ˜ respectively. Assume that there operator and delta differentiation operator σ˜ and Δ, 1 (T) of (5.19) such that exists a positive nonincreasing solution φ ∈ Crd

φ , ea (·, t0 )

Δ φ− −a

Δ φ− −a(φ ◦ τ −1 )

and

are nonincreasing on T and ∞ t0

Δ (t)ψ Δ (t) φ− Δt < ∞, −a(t)φ(τ −1 (t))

t0 ∈ T,

 1 !! Δ !! φ (t) − φ Δ (t) , 2

t ∈ T.

where Δ φ− (t) =

Then x(t) = O(φ(t)) as

t →∞

for any solution x of (5.17). Proof. Let x be a solution of the equation (5.17) on [t0 , ∞), t0 ∈ T. We set z(t) =

x(t) , φ(t)

t ≥ t0 .

Then x(t) = z(t)φ(t),

t ≥ t0 ,

and x Δ (t) = zΔ (t)φ(t) + z(σ (t))φ Δ (t),

t ≥ t0 .

Hence, the equation (5.17) takes the form zΔ (t)φ(t) + z(σ (t))φ Δ (t) = a(t)z(t)φ(t) +b(t)z(τ (t))φ(τ (t)),

t ≥ t0 .

From here, (zφ)Δ (t) − a(t)z(t)φ(t) b(t)z(τ (t))φ(τ (t)) = , ea (σ (t), t0 ) ea (σ (t), t0 )

t ≥ t0 ,

218

5 Stability for First-Order Functional Dynamic Equations

or

zφ ea (·, t0 )

Δ (t) =

b(t)z(τ (t))φ(τ (t)) , ea (σ (t), t0 )

t ≥ t0 .

(5.31)

Let tm = τ −m (t0 ),

m ∈ N,

Im+1 = [tm , tm+1 ],

m ∈ N0 ,

Bm = sup{|z(t)| : t ∈ Im },

m ∈ N.

Let t ∗ ∈ Im+1 , m ∈ N, be arbitrarily chosen. Then we integrate (5.31) from t0 to t ∗ and we get z(t ∗ )φ(t ∗ ) z(tm )φ(tm ) − = ea (t ∗ , t0 ) ea (tm , t0 )

t∗ tm

b(t)z(τ (t))φ(τ (t)) Δt. ea (σ (t), t0 )

From here, z(t ∗ )φ(t ∗ ) z(tm )φ(tm ) = + ea (t ∗ , t0 ) ea (tm , t0 )

t∗ tm

b(t)z(τ (t))φ(τ (t)) Δt ea (σ (t), t0 )

and z(t ∗ ) = z(tm )

φ(tm )ea (t ∗ , t0 ) φ(t ∗ )ea (tm , t0 )

ea (t ∗ , t0 ) + φ(t ∗ )

t∗ tm

b(t)z(τ (t))φ(τ (t)) Δt. ea (σ (t), t0 )

Therefore ! ∗ ! m )ea (t ,t0 ) |z(t ∗ )| = !z(tm ) φ(t φ(t ∗ )ea (tm ,t0 ) ! ! (t ∗ ,t0 ) # t ∗ b(t)z(τ (t))φ(τ (t)) + eaφ(t Δt ! ∗) tm ea!(σ (t),t0 ) ! ! φ(tm )ea (t ∗ ,t0 ) ! ≤ !z(tm ) φ(t ∗ )ea (tm ,t0 ) ! ! ∗ # ∗ ! ! (t ,t0 ) t b(t)z(τ (t))φ(τ (t)) ! + ! eaφ(t Δt ! ∗) tm ea (σ (t),t0 ) (t ∗ ,t

m )ea 0) ≤ |z(tm )| φ(t φ(t ∗ )ea (tm ,t0 ) # ∗ ∗ (t ,t0 ) t |b(t)||z(τ (t))||φ(τ (t))| + eaφ(t Δt ∗) tm ea (σ (t),t0 ) ∗

m )ea (t ,t0 ) ≤ Bm φ(t φ(t ∗ )ea (tm ,t0 )

(5.32)

5.4 Asymptotic Stability

+Bm ≤ Bm

ea (t ∗ , t0 ) φ(t ∗ )

t∗ tm

|b(t)|φ(τ (t)) Δt ea (σ (t), t0 )

(t ∗ , t

φ(tm )ea 0) ∗ φ(t )ea (tm , t0 )

+Bm = Bm

219

ea (t ∗ , t0 ) φ(t ∗ )

t∗ tm

−a(t)φ(t) Δt ea (σ (t), t0 )

(t ∗ , t

φ(tm )ea 0) ∗ φ(t )ea (tm , t0 )

ea (t ∗ , t0 ) +Bm φ(t ∗ )

t∗

tm

1 ea (·, t0 )

Δ

φ(tm )ea (t ∗ , t0 ) φ(t ∗ )ea (tm , t0 ) % ea (t ∗ , t0 ) φ(t) !!t=t ∗ − +Bm ! φ(t ∗ ) ea (t, t0 ) t=tm

(t)φ(t)Δt

= Bm

t∗ tm

& 1 Δ φ (t)Δt . ea (σ (t), t0 )

Since φ is nonincreasing on T and a < 0 on T, then φ(tm ) ≥ φ(t ∗ ) and

ea (t ∗ , t0 ) ≤ ea (tm , t0 ).

Therefore, using (5.32), we obtain !t=t ∗  φ(t) ! (t ∗ ,t0 ) |z(t ∗ )| ≤ Bm + Bm eaφ(t ∗) ea (t,t0 ) !t=t m # t∗ 1 Δ (t)Δt . − tm ea (σ (t),t φ 0) Δ , we get By the definition of φ−

 1 Δ |φ (t)| − φ Δ (t) 2  1 Δ −φ (t) − φ Δ (t) ≥ 2

Δ φ− (t) =

= −φ Δ (t),

t ≥ t0 .

Hence from (5.33), we obtain ea (t ∗ , t0 )  φ(t) !!t=t ∗ ! φ(t ∗ ) ea (t, t0 ) t=tm Δ (t) φ− Δt ea (σ (t), t0 )

|z(t ∗ )| ≤ Bm + Bm +

t∗ tm

(5.33)

220

5 Stability for First-Order Functional Dynamic Equations

ea (t ∗ , t0 )  φ(t) !!t=t ∗ ! φ(t ∗ ) ea (t, t0 ) t=tm t ∗ φ Δ (t) −a(t) − Δt + tm −a(t) ea (σ (t), t0 ) ea (t ∗ , t0 )  φ(t) !!t=t ∗ ≤ B m + Bm ! φ(t ∗ ) ea (t, t0 ) t=tm ∗ φ Δ (tm ) t −a(t) Δt + − −a(tm ) tm ea (σ (t), t0 ) ea (t ∗ , t0 )  φ(t) !!t=t ∗ = B m + Bm ! φ(t ∗ ) ea (t, t0 ) t=tm

! ∗ φ Δ (tm ) 1 !t=t + − ! −a(tm ) ea (t, t0 ) t=tm ea (t ∗ , t0 )  φ(t ∗ )ea (tm , t0 ) − φ(tm )ea (t ∗ , t0 ) = Bm + Bm φ(t ∗ ) ea (t ∗ , t0 )ea (tm , t0 )

 !t=t ∗ φ Δ (tm ) 1 ! + − ! −a(tm ) ea (t, t0 ) t=tm ! ∗ Δ (t ) 1 ea (t ∗ , t0 ) φ− m !t=t ≤ Bm + Bm ! ∗ φ(t ) −a(tm ) ea (t, t0 ) t=tm  Δ (t ) 1 1 ea (t ∗ , t0 ) φ− m − = B m + Bm φ(t ∗ ) −a(tm ) ea (t ∗ , t0 ) ea (tm , t0 ) = B m + Bm

≤ Bm + Bm

Δ (t ) 1 ea (t ∗ , t0 ) φ− m ∗ φ(t ) −a(tm ) ea (t ∗ , t0 )

= Bm + Bm

Δ (t ) φ− m −a(tm )φ(t ∗ )

%

& Δ (t ) φ− m = Bm 1 + −a(tm )φ(t ∗ ) % & Δ (t ) φ− m ≤ Bm 1 + . −a(tm )φ(tm+1 ) Since t ∗ ∈ Im+1 was arbitrarily chosen, from the last inequality we get % Bm+1 ≤ Bm

& Δ (t ) φ− m 1+ . −a(tm )φ(tm+1 )

5.4 Asymptotic Stability

221

Hence, Bm+1 ≤ B1

m $ k=1

%

& Δ (t ) φ− k 1+ . −a(tk )φ(τ −1 (tk ))

Let s = ψ(t),

sk = ψ(tk ),

k ∈ N0 .

Then s1 = ψ(t1 ) = ψ(τ −1 (t0 )) = ψ(t0 ) + 1 = s0 + 1, s2 = ψ(t2 ) = ψ(τ −1 (t1 )) = ψ(t1 ) + 1 = s1 + 1 = s0 + 2, and so on. sk = s0 + k,

k ∈ N.

Note that m  k=1

=

Δ (t ) φ− k −a(tk )φ(τ −1 (tk ))

m  k=1

= ≤

m 

Δ (ψ −1 (s + k)) φ− 0 −1 −a(ψ (s0 + k))φ(τ −1 (ψ −1 (s0 + k))) s0 +k

k=1 s0 +k−1 m 

s0 +k

k=1 s0 +k−1

−a(ψ −1 (s

Δ (ψ −1 (s + k)) φ− 0 ˜ Δs −1 (ψ −1 (s + k))) 0 + k))φ(τ 0

Δ (ψ −1 (s)) φ− ˜ Δs −a(ψ −1 (s))φ(τ −1 (ψ −1 (s)))

222

5 Stability for First-Order Functional Dynamic Equations

=

m 

tk

k=1 tk−1

=

tm t0

Δ (t)ψ Δ (t) φ− Δt −a(t)φ(τ −1 (t))

Δ (t)ψ Δ (t) φ− Δt. −a(t)φ(τ −1 (t))

From here, we conclude that % ∞ $ 1+ k=1

Δ (t ) φ− k −a(tk )φ(τ −1 (tk ))

& < ∞.

Therefore the sequence {Bm }m∈N is bounded and then This completes the proof.

x(t) φ(t)

is bounded as t → ∞.

Definition 5.9. A solution x of the equation (5.17) is said to be asymptotically stable if lim x(t) = 0.

t→∞

Theorem 5.16. Consider the equation (5.17), where a ∈ R + (T), b ∈ Crd (T), 1 (T), a is nonincreasing on T, a(t) ≤ α < 0, 0 = |b(t)| ≤ β, β < −α, τ ∈ Crd τ (t) < t for any t ∈ T, τ (T) = T, τ Δ is positive and nonincreasing on T and 0 < τ Δ ≤ α < 1 on T. Then any solution x defined on [t0 , ∞) for which x(t) = o(ea (t, t0 ))

as

t → ∞,

is asymptotically stable. Proof. Let x be a solution of the equation (5.17) defined on [t0 , ∞) for which x(t) = o(ea (t, t0 ))

as

t → ∞.

By Theorem 5.10, it follows that the equation (5.18) has a solution ψ for which lim ψ(t) = ∞.

t→∞

Note that

φ(t) =

β −α

ψ(t) ,

t ∈ T,

5.5 Exponential Stability I

223

is a solution of the inequality (5.19). Indeed, we have

 β ψ(t)−1 |b(t)|φ(τ (t)) ≤ β −α  

β ψ(t)−1 β = (−α) −α −α ψ(t)

β = −α −α ≤ −a(t)φ(t) = |a(t)|φ(t),

t ∈ T.

Since β < −α and limt→∞ ψ(t) = ∞, we have lim φ(t) = 0.

t→∞

By Theorem 5.15, we have x(t) = O(φ(t)) as

t → ∞.

Thus lim x(t) = 0.

t→∞

This completes the proof.

5.5 Exponential Stability I In this section we suppose that T is a time scale that contains 0 and positive real numbers. We consider the equation x Δ (t) = G(t, x(τ (t)))

on

T,

(5.34)

n n is a given where x : T → Rn is delta differentiable, G : [0, ∞) '× R → R ' continuous function, G(t, 0) = 0, t ∈ T, τ : [0, ∞) T → [0, ∞) T is an increasing function, τ (t) < t for any t ∈ (0, ∞), τ (0) = 0. Let t0 ∈ (0, ∞) and γ0 be the largest value τ (t) such that τ (t) ≤ t0 when t ≥ t0 . Let also,

x(t) = φ(t),

t ∈ [0, t0 ].

(5.35)

224

5 Stability for First-Order Functional Dynamic Equations

Definition 5.10. We say that V : [0, ∞) × Rn → [0, ∞) is a type I Lyapunov functional on [0, ∞) × Rn when V (t, x) =

n 

(Vi (xi ) + Ui (t)) ,

i=1

where each Vi : R → R and Ui : [0, ∞) → R are continuously differentiable. We extend the definition of the derivative of a type I Lyapunov function to type I Lyapunov functional. Let V be a type I Lyapunov functional and x is a solution of the equation (5.34). Then, using Pötzsche’s chain rule (see the appendix of this book), we get (V (t, x))Δ =

n 

(Vi (xi (t)) + Ui (t))Δ

i=1

=

n 

UiΔ (t) +

i=1

=

n 

=

UiΔ (t) +

=

i=1

n  i=1

UiΔ (t) +

i=1 n 

(Vi (xi (t)))Δ

i=1

i=1 n 

n 

UiΔ (t) +

1

  ∂ Vi xi (t) + hμ(t)xiΔ (t) dhxiΔ (t) ∂xi

1

∂ Vi (xi (t) + hμ(t)Gi (t, x(τ (t)))) dhGi (t, x(τ (t))) ∂xi

0

n  i=1

1

0

∇V (x(t) + hμ(t)G(t, x(τ (t)))) · G(t, x(τ (t)))dh

0

" =

"n Vi (xi (t)+μ(t)Gi (t,x(τ (t))))−Vi (xi (t)) n Δ i=1 Ui (t) + i=1 μ(t) " ∇V (x) · G(t, x(τ (t))) + ni=1 UiΔ (t) when μ(t)

when

μ(t) = 0

= 0.

Example 5.7. Let T = 2N0 . Consider the system   x1Δ (t) = tx2 2t   x2Δ (t) = t 2 + x1 2t , t ≥ 4, x1 (t) = t 2 , x2 (t) = 2t + 1,

(5.36) t ∈ [1, 4].

Let V (t, x) = x1 + x2 + t + t 2 ,

t ∈ T,

Here σ (t) = 2t, μ(t) = t,

t ∈ T,

x ∈ R2 ,

x = (x1 , x2 ).

5.5 Exponential Stability I

225

G1 (t, x) = tx2 , G2 (t, x) = t 2 + x1 , V1 (x1 ) = x1 , V2 (x2 ) = x2 , U1 (t) = t, U2 (t) = t 2 , U1Δ (t) = 1, U2Δ (t) = σ (t) + t = 2t + t = 3t.

t ∈ T,

x ∈ R2 .

Then, if x is a solution of the equation (5.36), we have (V (t, x))Δ = U1Δ (t) + U2Δ (t)     V1 x1 (t) + 2tG1 t, x 2t − V1 (x1 (t)) + 2t     V2 x2 (t) + 2tG2 t, x 2t − V2 (x2 (t)) + 2t   2 x1 (t) + 2t x2 2t − x1 (t) = 1 + 3t + 2t   2 x2 (t) + 2t t + x1 2t − x2 (t) + 2t



 t t 2 + tx2 , t ∈ T. = 1 + 3t + t + x1 2 2 For t ∈ [4, 8], we get (V (t, x))Δ = 1 + 3t + t 2 +

t2 + t (t + 1) 4

= 1 + 3t +

5t 2 + t2 + t 4

= 1 + 4t +

9t 2 . 4

226

5 Stability for First-Order Functional Dynamic Equations

Exercise 5.7. Let T = 3N0 . Consider the system   2 x1Δ (t) = t 2 + t + (t − 1) x2 3t   x2Δ (t) = 2 + t + t 2 x1 3t , t ≥ 3, x1 (t) = t + t 2 , x2 (t) = t 2 + 1, t ∈ [1, 3].

(5.37)

Let V (t, x) = x12 + x22 + 4t + 3t 2 + t 3 ,

t ∈ T,

x ∈ R2 ,

x = (x1 , x2 ).

Find (V (t, x(t)))Δ provided x(t) = (x1 (t), x2 (t)), t ∈ T, is a solution of the system (5.37). Theorem 5.17. Assume that φ(·, ·) is rd-continuous with respect to the second variable and it is delta differentiable with respect to the first variable and let t

V (t, x) = (x(t))2 +

φ(t, s)W (|x(s)|)Δs. 0

If x is a solution of (5.34), then (V (t, x))Δ = 2x(t) · G(t, x(τ (t))) + μ(t) (G(t, x(τ (t))))2 t

+

φ Δ (t, s)W (|x(s)|)Δs + φ(σ (t), t)W (|x(t)|),

0

Proof. We have (V (t, x))Δ = x Δ (t)x(t) + x Δ (t)x(σ (t)) t

+

φ Δ (t, s)W (|x(s)|)Δs

0

+φ(σ (t), t)W (|x(t)|)   = x Δ (t)x(t) + x Δ (t) x(t) + μ(t)x Δ (t) t

+

φ Δ (t, s)W (|x(s)|)Δs

0

+φ(σ (t), t)W (|x(t)|)  2 = 2x(t)x Δ (t) + μ(t) x Δ (t) t

+

φ Δ (t, s)W (|x(s)|)Δs

0

+φ(σ (t), t)W (|x(t)|)

t ∈ T.

5.5 Exponential Stability I

227

= 2x(t) · G(t, x(τ (t))) + μ(t) (G(t, x(τ (t))))2 t

+

φ Δ (t, s)W (|x(s)|)Δs + φ(σ (t), t)W (|x(t)|),

t ∈ T.

0

This completes the proof. Definition 5.11. We say that a type I Lyapunov functional V : [0, ∞) × Rn → [0, ∞) is negative definite if V (t, x) = 0 for x = 0, x ∈ Rn , V (t, x) = 0 for x = 0 and along the solutions of (5.34) we have (V (t, x))Δ ≤ 0.

(5.38)

If the condition (5.38) does not hold for all (t, x) ∈ T × Rn , then the Lyapunov functional is said to be nonnegative definite. Definition 5.12. We say that the zero solution of the equation (5.34) is exponentially asymptotically stable on [0, ∞) if there exist a positive constant δ and a constant C ≥ 0 and an M > 0 such that for any solution x(t, t0 , φ) of the equation (5.34) we have |x(t, t0 , φ)| ≤ C(|φ|, t0 ) (eM (t, t0 ))d ,

t ∈ [t0 , ∞),

where C(|φ|, t0 ) is a constant depending on |φ| and t0 . We say that the zero solution of the equation (5.34) is uniformly asymptotically stable on [0, ∞) if C is independent of t0 . Theorem 5.18. Let D ⊂ Rn contains the origin. Suppose that there exists a type I Lyapunov functional V : [0, ∞)×D → [0, ∞) such that for all (t, x) ∈ [0, ∞)×D λ1 (t)W1 (|x|) ≤ V (t, x) ≤ λ2 (t)W2 (|x|) + λ2 (t)

#t 0

φ1 (t, s)W3 (|x(s)|)Δs (5.39)

and (V (t, x))Δ ≤

−λ3 (t)W4 (|x(t)|) − λ3 (t)

#t 0

φ2 (t, s)W5 (|x(s)|)Δs − L(M  δ)eδ (t, 0) 1 + μ(t) λλ23 (t) (t)

,

(5.40) where λ1 , λ2 , and λ3 are positive continuous functions, M and δ are positive constants, L is a nonnegative constant, λ1 is nondecreasing, and φi (t, s) ≥ 0 is rd-continuous with respect to the second variable for 0 ≤ s ≤ t < ∞, i = 1, 2, such that t

W2 (|x(t)|) − W4 (|x(t)|) +

(φ1 (t, s)W3 (|x(s)|) − φ2 (t, s)W5 (|x(s)|)) Δs ≤ 0.

0

(5.41)

228

5 Stability for First-Order Functional Dynamic Equations

#t If 0 φ1 (t, s)Δs ≤ B for some positive constant B, then the zero solution of the equation (5.34) is exponentially asymptotically stable. Proof. Let λ3 (t) >0 t≥0 λ2 (t)

M = inf

and set δ > M. Let x be a solution of the equation (5.34). Since λ3 (t) , λ2 (t)

M≤

t ≥ 0,

then σ (t, t ) + MV (t, x(t))e (t, t ) (V (t, x(t))eM (t, t0 ))Δ = (V (t, x(t)))Δ eM 0 M 0

  = (V (t, x(t)))Δ (1 + Mμ(t)) + MV (t, x(t)) eM (t, t0 ),

t ≥ 0.

(5.42) By (5.39) and (5.40), we get t

(1 + Mμ(t)) (V (t, x(t)))Δ ≤ −λ3 (t)W4 (|x(t)|) − λ3 (t)

φ2 (t, s)W5 (|x(s)|)Δs, 0

−L(M  δ)eδ (t, 0),

t ≥ 0.

Hence from (5.42), we obtain (V (t, x(t))eM (t, t0 ))Δ ≤



− λ3 (t)W4 (|x(t)|) t

−λ3 (t) 0

φ2 (t, s)W5 (|x(s)|)Δs − L(M  δ)eδ (t, 0)

+Mλ2 (t)W2 (|x(t)|) t

+Mλ2 (t) ≤



φ1 (t, s)W3 (|x(s)|)Δs eM (t, t0 )

0

− λ3 (t)W4 (|x(t)|) t

−λ3 (t) 0

φ2 (t, s)W5 (|x(s)|)Δs − L(M  δ)eδ (t, 0)

+λ3 (t)W2 (|x(t)|) t

+λ3 (t)

φ1 (t, s)W3 (|x(s)|)Δs eM (t, t0 )

0

≤ −L(M  δ)eM (t, t0 )eδ (t, 0) = −L(M  δ)eMδ (t, 0)eM (0, t0 ),

t ≥ 0.

5.5 Exponential Stability I

229

Integrating both sides of the last inequality from t0 to t, we get t

(V (s, x(s))eM (s, t0 ))Δ Δs ≤ −L(M  δ)

t0

t

eMδ (s, 0)eM (0, t0 )Δs t0

or V (t, x(t))eM (t, t0 ) − V (t0 , x(t0 )) ≤ −LeMδ (t, 0)eM (0, t0 ) +LeMδ (t0 , 0)eM (0, t0 ) = −LeM (0, t0 )eM (t, 0)eδ (t0 , 0) +LeM (t0 , 0)eM (0, t0 )eδ (t0 , 0) = −LeM (t, t0 )eδ (t0 , 0) + Leδ (t0 , 0),

t ≥ 0,

or V (t, x(t))eM (t, t0 ) ≤ V (t0 , φ) − LeM (t, t0 )eδ (t0 , 0) +Leδ (t0 , 0) ≤ V (t0 , φ) + L,

t ≥ t0 .

Hence, V (t, x(t)) ≤ (V (t0 , φ) + L) eM (t, t0 ),

t ≥ t0 .

From the inequality (5.39), we obtain W1 (|x(t)|) ≤ ≤ ≤

1 λ1 (t) V (t, x(t)) 1 (t , φ) + L) eM (t, t0 ) λ1 (t) (V  0 1 λ1 (t0 ) λ2 (t0 )W2 (|φ(t0 )|) #t +λ2 (t0 )W3 (|φ(t0 )|) 00 φ1 (t0 , s)Δs

+ L eM (t, t0 ),

t ≥ t0 . (5.43)

Hence, |x(t)| ≤ W1−1



1  λ2 (t0 )W2 (|φ(t0 )|) λ1 (t0 ) t0

+λ2 (t0 )W3 (|φ(t0 )|) 0

This completes the proof.

φ1 (t0 , s)Δs + L eM (t, t0 ) ,

t ≥ t0 .

230

5 Stability for First-Order Functional Dynamic Equations

Theorem 5.19. Suppose that the hypotheses of Theorem 5.18 holds except the condition λ1 is nondecreasing is replaced by: there exists a positive constant γ < M such that λ1 (t) ≥ eγ (t, t0 ),

t ≥ t0 ≥ 0.

Then the zero solution of the equation (5.34) is exponentially asymptotically stable. Proof. By (5.43), we have W1 (|x(t)|) ≤ ≤

1 (V (t0 , φ) + L) eM (t, t0 ) λ1 (t) 1 (V (t0 , φ) + L) eM (t, t0 ) eγ (t, t0 )

= (V (t0 , φ) + L) eγ (t, t0 )eM (t, t0 ) = (V (t0 , φ) + L) eγ M (t, t0 )  ≤ λ2 (t0 )W2 (|φ(t0 )|) t0

+λ2 (t0 )W3 (|φ(t0 )|) 0

φ1 (t0 , s)Δs + L eγ M (t, t0 )

for all t ≥ t0 ≥ 0. From here, |x(t)| ≤ W1−1

 λ2 (t0 )W2 (|φ(t0 )|) t0

+λ2 (t0 )W3 (|φ(t0 )|) 0

 φ1 (t0 , s)Δs + L eγ M (t, t0 )

for all t ≥ t0 . This completes the proof. Theorem 5.20. Assume that D ⊂ Rn contains the origin and there exists a type I Lyapunov functional V : [0, ∞)×D → [0, ∞) such that for all (t, x) ∈ [0, ∞)×D λ1 xp ≤ V (t, x), (V (t, x))Δ ≤

−λ2 V (t, x) − L(  δ)eδ (t, 0) , 1 + μ(t)

where λ1 , λ2 , p > 0, δ > 0, and L ≥ 0 are constants and 0 <  < min{λ2 , δ}. Then the zero solution of the equation (5.34) is exponentially asymptotically stable.

5.5 Exponential Stability I

231

Proof. Note that e (t, 0) > 0 for any t ≥ t0 ≥ 0. Then (V (t, x(t))e (t, t0 ))Δ = (V (t, x(t)))Δ eσ (t, t0 ) + V (t, x(t))e (t, t0 ) ≤

−λ2 V (t, x(t)) − L(  δ)eδ (t, 0) σ e (t, t0 ) 1 + μ(t) +V (t, x(t))e (t, t0 )

= (−λ2 V (t, x(t)) − L(  δ)eδ (t, 0)) e (t, t0 ) +V (t, x(t))e (t, t0 ) = e (t, t0 ) (V (t, x(t)) − λ2 V (t, x(t)) − L(  δ)eδ (t, 0)) ≤ −L(  δ)e (t, t0 )eδ (t, 0) = −L(  δ)e (t, 0)e (0, t0 )eδ (t, 0) = −L(  δ)eδ (t, 0)e (0, t0 ). Hence, integrating both sides of the last inequality from t0 to t, we get V (t, x(t))e (t, t0 ) − V (t0 , x(t0 ))e (t0 , t0 ) ≤ −L

t t0

 (  δ)(s)eδ (s, 0)Δs e (0, t0 )

  = −L eδ (t, 0) − eδ (t0 , 0) e (0, t0 ) ≤ Le (0, t0 )eδ (t0 , 0) = Le (0, t0 )e (t0 , 0)eδ (t0 , 0) = Leδ (t0 , 0),

t ≥ t0 .

Hence, V (t, x(t))e (t, t0 ) ≤ V (t0 , φ) + Leδ (t0 , 0) ≤ V (t0 , φ) + L,

t ≥ t0 ,

and V (t, x(t)) ≤ (V (t0 , φ) + L) e (t, t0 ),

t ≥ t0 .

Therefore xp ≤

(V (t0 , φ) + L) e (t, t0 ) , λ1

t ≥ t0 ,

232

5 Stability for First-Order Functional Dynamic Equations

or

x ≤

(V (t0 , φ) + L) e (t, t0 ) λ1

1

p

,

t ≥ t0 .

This completes the proof.

5.6 Exponential Stability II In this section we suppose that T is an unbounded above time scale and t0 ∈ T. Consider the equation x Δ (t) +

n 

Al (t)x(αl (t)) = 0,

t ∈ [t0 , ∞),

(5.44)

l=1

where n ∈ N, αl , Al ∈ Crd ([t0 , ∞)), αl : [t0 , ∞) → T and limt→∞ αl (t) = ∞, αl (t) ≤ t, l ∈ {1, . . . , n}, and t ∈ [t0 , ∞). Let αmin (t) = t−1 =

min αl (t),

l∈{1,...,n}

inf

t∈[t0 ,∞)

t ∈ [t0 , ∞),

αmin (t).

1 ([t , ∞)) → C ([t , ∞)) as follows Define the operator L : Crd 0 rd 0

L x(t) = x Δ (t) +

n 

Al (t)x(αl (t)),

t ∈ [t0 , ∞).

l=1

Now we consider the delay dynamic IVP L x(t) = f (t), t ∈ [t0 , ∞), x(t0 ) = x0 , x(t) = φ(t), t ∈ [t−1 , t0 ),

(5.45)

where x0 ∈ R, φ ∈ Crd ([t−1 , t0 )) and if t0 is left-dense we suppose that φ has a finite left-sided limit at t0 . Definition 5.13. A function x : [t−1 , ∞) → R which is rd-continuous on [t−1 , t0 ) and delta differentiable on [t0 , ∞) is called a solution of (5.45) provided that it satisfies the equality L x = f identically on [t0 , ∞) and x(t0 ) = x0 , x = φ on [t−1 , t0 ). Note that the problem (5.45) admits a unique solution.

5.6 Exponential Stability II

233

Definition 5.14. The zero solution of the equation (5.44) is said to be uniformly exponentially stable if there exist constants M > 0, λ ≥ 0, such that for any s ∈ [t0 , ∞) the solution of the problem

L x(t) = 0 for t ∈ [s, ∞), x(s) = x0 and x(t) = φ(t),

t ∈ [s−1 , s),

where s−1 = inft∈[s,∞) αmin (t), satisfies %

&

|x(t)| ≤ Meλ (t, s) |x0 | +

|φ(η)| ,

sup

t ∈ [s, ∞).

η∈[s−1 ,s]

Definition 5.15. Let s ∈ [t0 , ∞). The solution χ = χ (·, s) : [s−1 , ∞) → R of the IVP

L x(t) = 0 for t ∈ [s, ∞), x(s) = κ{s} (t) and x(t) = φ(t), t ∈ [s−1 , s], is called the fundamental solution of (5.44). Here κ{s} (t) is the characteristic function of {s}. 1 ([t , ∞)) we will denote the space With Crd,0 0

  1 1 Crd,0 ([t0 , ∞)) = x ∈ Crd ([t0 , ∞)) : x(t0 ) = 0 . 1 ([t , ∞)) defined by Definition 5.16. The operator C : Crd ([t0 , ∞)) → Crd,0 0

Cf (t) =

t

χ (t, σ (η))f (η)Δη,

f ∈ Crd ([t0 , ∞)),

t ∈ [t0 , ∞),

t0

where χ is the fundamental solution of the equation (5.44), is called the Cauchy operator of the equation (5.44). Theorem 5.21. Let x be the solution of the IVP (5.45). Then x(t) = χ (t, t0 )x0 + Cf (t) − C

% n 

& Al φ(αl (·)) (t)

l=1

= χ (t, t0 )x0 +

t

χ (t, σ (η))f (η)Δη t0

−

t

χ (t, σ (η)) t0

% n  l=1

& Al (η)φ(αl (η)) Δη.

234

5 Stability for First-Order Functional Dynamic Equations

Proof. Let x be the solution of the equation (5.45). Set I (t) = {j ∈ {1, . . . , n} : κ[t0 ,∞) (αj (t)) = 1}, J (t) = {j ∈ {1, . . . , n} : κ[t−1 ,t0 ) (αj (t)) = 1},

t ∈ [t0 , ∞).

Then t

x Δ (t) = x0 χ Δ (t, t0 ) +

χ Δ (t, σ (η))f (η)Δη

t0

+χ (σ (t), σ (t))f (t) % n & t  Δ − χ (t, σ (η)) Al (η)φ(αl (η)) Δη t0

l=1

−χ (σ (t), σ (t))

n 

Al (t)φ(αl (t))

l=1

= −x0

n 

n 

+f (t) −

Al (t)φ(αl (t))

l=1

−

n 

%

Al (t)χ (αl (t), σ (η))

t0 l=1

=−

n 

n 

& Am (η)φ(αm (η)) Δη

m=1

Al (t) x0 χ (αl (t), t0 ) −

t

χ (αl (t), σ (η))f (η)Δη t0

l=1

−

Al (t)χ (αl (t), σ (η))f (η)Δη

t0 l=1

l=1

t

n 

t

Al (t)χ (αl (t), t0 ) −

%

t

χ (αl (t), σ (η)) t0

+f (t) −



n 

&



Am (η)φ(αm (η)) Δη

m=1

Aj (t)φ(αj (t))

j ∈J (t)

=−



Al (t) x0 χ (αl (t), t0 ) − %

αl (t)

χ (αl (t), σ (η)) t0

+f (t) −

χ (αl (t), σ (η))f (η)Δη t0

l∈I (t)

−

αl (t)

 j ∈J (t)

n 

m=1

Aj (t)φ(αj (t))

& Am (η)φ(αm (η)) Δη



5.6 Exponential Stability II

=−



235

Al (t)x(αl (t)) −

n 

Aj (t)x(αj (t)) + f (t)

j ∈J (t)

l∈I (t)

=−



Al (t)x(αl (t)) + f (t),

t ∈ [t0 , ∞).

l=1

This completes the proof. Theorem 5.22. Let 1 + μ(t)f (t) > 0 for any t ∈ T. Then 0 < ef (t, s) ≤ e

#t s

f (τ )Δτ

for

any

t, s ∈ T,

t ≥ s.

Proof. Since 1 + μ(τ )f (τ ) > 0, τ ∈ T, we have that Log(1 + μ(τ )f (τ )) ∈ R, τ ∈ T, and ef (t, s) > 0, t, s ∈ T, t ≥ s. If μ(τ ) = 0, then ξμ(τ ) (f (τ )) = f (τ ) and if μ(τ ) > 0, then ξμ(τ ) (f (τ )) = ≤

1 log(1 + μ(τ )f (τ )) μ(τ ) μ(τ )f (τ ) μ(τ )

= f (τ ). Hence, for t ≥ s, we get ef (t, s) = e ≤e

#t s

#t s

ξμ(τ ) (f (τ ))Δτ f (τ )Δτ

.

This completes the proof. Theorem 5.23. Suppose that lim sup μ(t) < ∞, t→∞

λ ≥ 0, ν ≥ 0, 1 + (λ) (t)μ(t) ≥ 0, t ∈ T, ν =

λ 1+λμ∗ ,

e−λ(t−s) ≤ eλ (t, s) ≤ e−ν (t, s) ≤ e−ν(t−s) ,

μ∗ = supη∈T μ(η). Then s, t ∈ T,

Proof. By Theorem 5.22, we have eλ (t, s) ≤ eλ(t−s) ,

s, t ∈ T,

t ≥ s,

t ≥ s.

236

5 Stability for First-Order Functional Dynamic Equations

whereupon e−λ(t−s) ≤ eλ (t, s),

s, t ∈ T,

t ≥ s.

Next, (λ) (t) = − ≤−

λ 1 + λμ(t) λ 1 + λμ∗

= −ν,

t ∈ T.

Hence, 0 ≤ 1 + (λ) (t)μ(t) ≤ 1 − νμ(t),

t ∈ T.

Therefore eλ (t, s) ≤ e−ν (t, s),

t, s ∈ T,

t ≥ s.

t, s ∈ T,

t ≥ s.

By Theorem 5.22, we get e−ν (t, s) ≤ e−ν(t−s) , This completes the proof. Theorem 5.24. Suppose that lim sup t→∞

n 

|Al (t)| < ∞

(5.46)

l=1

and lim sup(t − αmin (t)) < ∞. t→∞

(5.47)

5.6 Exponential Stability II

237

Then the following statements are equivalent. (i) The trivial solution of the equation (5.44) is uniformly exponentially stable. (ii) There exist λ ≥ 0, M ≥ 0 such that |χ (t, s)| ≤ Meλ (t, s)

(5.48)

for all t ∈ [s, ∞) and for all s ∈ [t0 , ∞). Proof. 1. Suppose that the trivial solution of the equation (5.44) is uniformly exponentially stable. Then, using the definition for uniform exponential stability, we get (5.48). 2. Suppose that there are λ ≥ 0, M ≥ 0 such that (5.48) holds. Define γ (t) = inf{η ∈ [t0 , ∞) : αmin (η) ≥ t},

t ∈ [t0 , ∞).

By (5.46) and (5.47), it follows that there are positive constants K and L such that n 

|Al (t)| ≤ K

|γ (t) − t| ≤ L,

and

t ∈ [t0 , ∞).

l=1

Then ! ! ! ! t n n γ (s)   ! ! !≤ ! e (σ (η), s) A (η)κ (α (η))Δη eλ (σ (η), s) |Al (η)|Δη λ l [s−1 ,s) l ! ! s ! ! s l=1 l=1 ≤K

γ (s)

eλ(σ (η)−s) Δη

s

≤ KLeλL ,

t, s ∈ [t0 , ∞).

Hence from Theorem 5.21, we obtain ! ! ! |x(t)| = !χ(t, t0 )x0 − !

%

t

χ(t, σ (η)) t0

≤ |x0 ||χ(t, t0 )| +

n  l=1

t

|χ(t, σ (η))|

t0

! ! ! Al (η)φ(αl (η)) Δη! !

n 

&

|Al (η)||φ(αl (η))|Δη

l=1

≤ Meλ (t, t0 )|x0 | + M

t t0

eλ (t, σ (η))

sup η∈[t−1 ,t0 )

|Al (η)||φ(αl (η))|Δη

l=1

% ≤ Meλ (t, t0 ) |x0 | +

n 

|φ(η)|

t

eλ (σ (η), t0 ) t0

n  l=1

& |Al (η)|κ[t−1 ,t0 ) (αl (η))Δη

238

5 Stability for First-Order Functional Dynamic Equations %

&

≤ Meλ (t, t0 ) |x0 | + KLe 

≤ M 1 + KLe

λL



λL

|φ(η)|

sup η∈[t−1 ,t0 )

%

eλ (t, t0 ) |x0 | +

& sup

|φ(η)| ,

t ∈ [t0 , ∞).

η∈[t−1 ,t0 )

This completes the proof. 1 ([t , ∞)) → C ([t , ∞)) as follows. Define the operator M : Crd 0 rd 0

M x(t) = x Δ (t) +

n 

Bl (t)x(βl (t)),

1 x ∈ Crd ([t0 , ∞)),

l=1

t ∈ [t0 , ∞), βl , Bl ∈ Crd ([t0 , ∞)), limt→∞ βl (t) = ∞, βl (t) ≤ σ (t), t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Consider the equation x Δ (t) +

n 

Bl (t)x(βl (t)) = 0,

t ∈ [t0 , ∞).

(5.49)

l=1 1 ([t , ∞)) the set of all bounded rd-continuous We define BCrd ([t0 , ∞)) and BCrd 0 functions and the set of all bounded differentiable functions with a bounded rdcontinuous derivative, respectively. Define

x =

sup |x(η)|,

x ∈ BCrd ([t0 , ∞)),

η∈[t0 ,∞)

x1 = x Δ  + x,

1 x ∈ BCrd ([t0 , ∞)).

  1 ([t , ∞)),  · 1 are Banach spaces. Then (BCrd ([t0 , ∞)),  · ) and BCrd 0 1 ([t , ∞)) → C ([t , ∞)) be defined as Theorem 5.25. Let L , M : Crd 0 rd 0 1 ([t , ∞)) be the Cauchy operators for the above, C, D : Crd ([t0 , ∞)) → Crd 0 equation (5.44) and (5.49), respectively. If CBCrd ([t0 , ∞)) ⊂ BCrd,0 ([t0 , ∞)) and (M C)−1 BCrd ([t0 , ∞)) ⊂ BCrd ([t0 , ∞)), then

DBCrd ([t0 , ∞)) ⊂ BCrd ([t0 , ∞)). Proof. For any g ∈ BCrd ([t0 , ∞)), the function x = C(M C)−1 g is bounded and satisfies M x(t) = M C (M C)−1 g(t) = g(t), x(t) = 0,

t ∈ [t0 , ∞),

t ∈ [t−1 , t0 ].

5.6 Exponential Stability II

239

1 ([t , ∞)) and Consequently D = C(M C)−1 : Crd ([t0 , ∞)) → Crd,0 0 DBCrd ([t0 , ∞)) ⊂ BCrd ([t0 , ∞)). This completes the proof.

Theorem 5.26. Suppose that lim sup t→∞

n 

|Al (t)| < ∞.

l=1

1 ([t , ∞)) → BC ([t , ∞)) is a bijective conThen the operator L : BCrd,0 0 rd 0 tinuous linear operator. Moreover, the Cauchy operator C : BCrd,0 ([t0 , ∞)) → 1 ([t , ∞)) of the equation (5.44) is the inverse of L and it is bounded. BCrd 0 1 ([t , ∞)) be arbitrarily chosen and a, b ∈ R. There Proof. Let x1 , x2 ∈ BCrd,0 0 exists a positive constant L such that n 

|Al (t)| ≤ L,

t ∈ [t0 , ∞).

l=1

We have L x1 ∈ BCrd ((t0 , ∞)) and L x1  ≤ x1 1 + Lx1 1 = (1 + L)x1 1 . Therefore L  ≤ 1 + L 1 ([t , ∞)) → BC ([t , ∞)) is a bounded operator. Also, and L : BCrd,0 0 rd 0

L (ax1 + bx2 )(t) = (ax1 + bx2 )Δ (t) +

n 

Al (t) (ax1 + bx2 ) (αl (t))

l=1

= ax1Δ (t) + bx2Δ (t) + a

n 

Al (t)x1 (αl (t))

l=1

+b % =a

n 

Al (t)x2 (αl (t))

l=1

x1Δ (t) +

n  l=1

& Al (t)x1 (αl (t))

240

5 Stability for First-Order Functional Dynamic Equations

% +b

x2Δ (t) +

n 

& Al (t)x2 (αl (t))

l=1

= aL x1 (t) + bL x2 (t),

t ∈ [t0 , ∞).

1 ([t , ∞)) → BC ([t , ∞)) is a linear bounded operator. Consequently L : BCrd,0 0 rd 0 Note that

0[t0 ,∞) (t) = 0,

t ∈ [t0 , ∞),

satisfies L x(t) = 0,

t ∈ [t0 , ∞),

x(t) = 0,

t ∈ [t−1 , t0 ]

and it is the unique solution of the last IVP. Let f ∈ BCrd ([t0 , ∞)). Then L Cf (t) = (Cf )Δ (t) +

n 

Al (t)(Cf )(αl (t))

l=1

Δ

t

=

χ (t, σ (η))f (η)Δη t0

+

n 

αl (t)

Al (t)

χ (αl (t), σ (η))f (η)Δη t0

l=1

= χ (σ (t), σ (t))f (t) +

t

χ Δ (t, σ (η))f (η)Δη

t0

+

n 

αl (t)

Al (t)

l=1

=

t

χ (αl (t), σ (η))f (η)Δη t0

%

χ Δ (t, σ (η)) +

t0

n 

& Al (t)χ (αl (t), σ (η)) f (η)Δη

l=1

+f (t) = f (t),

t ∈ [t0 , ∞).

Therefore L C = I d, where I d is the identity operator in the space BCrd ([t0 , ∞)). By the Inverse Mapping Theorem, the operator C : BCrd ([t0 , ∞)) → 1 ([t , ∞)) is invertible and C −1 = L . This completes the proof. BCrd 0

5.6 Exponential Stability II

241

Definition 5.17. The equation (5.44) is said to have the Perron property if for every f ∈ BCrd ([t0 , ∞)) the solution x = Cf of the nonhomogeneous equation L x(t) = f (t), t ∈ [t0 , ∞), x(t) = 0, t ∈ [t−1 , t0 ],

(5.50)

is bounded. Note that the Perron property is equivalent to CBCrd ([t0 , ∞)) ⊂ BCrd ([t0 , ∞)). Theorem 5.27. Assume that the trivial solution of the equation (5.44) is uniformly exponentially stable. Then the following statements are equivalent. (i) The equation (5.44) has the Perron property. (ii) The graininess function satisfies lim sup μ(t) < ∞.

(5.51)

t→∞

Proof. Since the trivial solution of the equation (5.44) is uniformly exponentially stable, using Theorem 5.24, there exist constants M > 1 and λ > 1 such that |χ (t, s)| ≤ Meλ (t, s) for any t ∈ [s, ∞) and for any s ∈ [t0 , ∞). (i)⇒ (ii) Assume that (5.51) does not hold. Note that for any f ∈ BCrd ([t0 , ∞)) we have Cf ∈ BCrd,0 ([t0 , ∞)). We take an increasing sequence {tk }k∈N ⊂ [t0 , ∞) such that tk → ∞ as

k → ∞,

{μ(tk )}k∈N is an increasing sequence and μ(tk ) → ∞ as

k → ∞.

For any t ∈ [t0 , tk ] and for any k ∈ N we have eλ (σ (tk ), σ (t)) =

1 + λμ(t) eλ (tk , t) 1 + λμ(tk )

≤ eλ (tk , t),

242

5 Stability for First-Order Functional Dynamic Equations

and hence, ! # !# ! ! tk t ! t0 χ (σ (tk ), σ (η))Δη! ≤ t0k |χ (σ (tk ), σ (η))|Δη #t ≤ M t0k eλ (σ (tk ), σ (η))Δη #t ≤ M t0k eλ (tk , η)Δη #t ≤ λM t0k eλ (η, tk )Δη = M (eλ (tk , tk ) − eλ (t0 , tk )) = M (1 − eλ (t0 , tk )) = M (1 − eλ (tk , t0 )) ≤ M, k ∈ N.

(5.52)

Now we consider the bounded function 1[t0 ,∞) defined by 1[t0 ,∞) (t) = 1,

t ∈ [t0 , ∞).

Then 

 C1[t0 ,∞) (σ (tk )) =

σ (tk )

χ (σ (tk ), σ (η))Δη t0 tk

=

χ (σ (tk ), σ (η))Δη +

t0 tk

=

σ (tk )

χ (σ (tk ), σ (η))Δη tk

χ (σ (tk ), σ (η))Δη + μ(tk )χ (σ (tk ), σ (tk ))

t0 tk

=

χ (σ (tk ), σ (η))Δη + μ(tk ),

k ∈ N.

t0

Hence,   C1[t0 ,∞) (σ (tk )) → ∞ as

k → ∞,

which contradicts with (5.52). Therefore (5.51) holds. (ii)⇒(i) Assume that (5.51) holds. Then ! ! |Cf (t)| = !! ≤

t t0 t

! ! χ (t, σ (η))f (η)Δη!!

|χ (t, σ (η))||f (η)|Δη

t0

≤ Mf 

t t0

eλ (t, σ (η))Δη

5.6 Exponential Stability II

243

= Mf  %

t t0

(1 + λμ(η)) eλ (t, η)Δη &

≤ M 1 + λ sup μ(η) f  η∈[t0 ,∞)

%

&

t t0

eλ (t, η)Δη

≤ M 1 + λ sup μ(η) f  (1 − eλ (t, t0 )) η∈[t0 ,∞)

%

&

≤ M 1 + λ sup μ(η) f , η∈[t0 ,∞)

i.e., x = Cf is bounded and (5.44) has the Perron property. This completes the proof. Theorem 5.28. Suppose that lim sup t→∞

n 

|Al (t)| < ∞

l=1

and the equation (5.44) has the Perron property. Then the Cauchy operator C : 1 ([t , ∞)) of the equation (5.44) satisfies Crd ([t0 , ∞)) → Crd,0 0 1 ([t0 , ∞)). CBCrd ([t0 , ∞)) ⊂ BCrd,0

Proof. We apply Theorem 5.21 for x0 = 0 and φ = 0 and we get that the solution of (5.50) has the representation x = Cf

on

[t0 , ∞).

By the Perron property, it follows that for any f ∈ BCrd ([t0 , ∞)) we have x ∈ BCrd,0 ([t0 , ∞)). Hence from lim sup t→∞

n 

|Al (t)| < ∞,

l=1

we obtain that x Δ ∈ BCrd ([t0 , ∞)) for any f ∈ BCrd ([t0 , ∞)). Therefore x ∈ 1 ([t , ∞)). This completes the proof. BCrd 0 Theorem 5.29. Assume that lim sup t→∞

n  l=1

|Al (t)| < ∞

244

5 Stability for First-Order Functional Dynamic Equations

and the equation (5.44) has the Perron property. Then for any s ∈ [t0 , ∞), the shifted equation L x(t) = f (t),

t ∈ [s, ∞),

(5.53)

has the Perron property. Proof. We take s ∈ (t0 , ∞) and f ∈ BCrd ([s, ∞)). Let 1 ([s, ∞)) D : BCrd ([s, ∞)) → BCrd,0

be the Cauchy operator of the equation (5.53). If s is left-scattered, then the solution x = Df of the equation (5.53) coincides with the bounded solution of L x = g on [t0 , ∞), where g ∈ BCrd ([t0 , ∞)) is defined by

g(t) =

f (t), t ∈ [s, ∞), 0, t ∈ [t0 , s).

If s ∈ (t0 , ∞) is left-dense, then we take a sequence {tk }k∈N ⊂ [t0 , s) such that tk → s as k → ∞. Now we define the sequence {gk }k∈N ⊂ Crd ([t0 , ∞)) such that ⎧ ⎪ ⎨ f (t), t ∈ [s, ∞), t−tk , t ∈ [tk , s), gk (t) = f (s) s−t ⎪ ⎩ 0, t ∈k [t , t ). 0 k Note that for any f ∈ BCrd ([s, ∞)) we have gk ∈ BCrd ([t0 , ∞)) and gk  = f  and there exists a constant M > 0 such that Cgk  ≤ Mgk  = Mf ,

k ∈ N.

Also, we have Cgk (t) = Df (t) +

s

χ (t, σ (η))gk (η)Δη,

t ∈ [s, ∞).

tk

Hence, + + lim Cgk  = lim + +Df +

k→∞

k→∞

= Df  ≤ Mf .

s tk

+ + χ (·, σ (η))gk (η)Δη+ +

5.6 Exponential Stability II

245

Then the solution of (5.53) is bounded on [s, ∞) for any f ∈ BCrd ([s, ∞)). This completes the proof. Theorem 5.30. Assume that lim sup t→∞

n 

|Al (t)| < ∞

l=1

and lim sup (σ (t) − αmin (t)) < ∞. t→∞

If the equation (5.44) has the Perron property, then there exist M > 0 and λ ∈ (0, 1) such that the fundamental solution χ of the equation (5.44) satisfies (5.48). Proof. Let s ∈ [t0 , ∞). Consider the equation (5.53). Let y(t) = eλ (t, s)x(t),

t ∈ [s, ∞),

1 ([s, ∞)). We have where λ > 0, x ∈ Crd

x(t) = eλ (t, s)y(t),

t ∈ [s, ∞),

and L x(t) = x Δ (t) +

n 

Al (t)x(αl (t))

l=1

= (eλ (·, s)y(·))Δ (t) +

n 

Al (t)eλ (αl (t), s)y(αl (t))

l=1

= eλ (σ (t), s)y Δ (t) + (λ) (t)eλ (t, s)y(t) +

n 

Al (t)eλ (αl (t), s)y(αl (t))

l=1

= eλ (σ (t), s)y Δ (t) − λeλ (σ (t), s)y(t) +

n 

Al (t)eλ (σ (t), s)eλ (αl (t), σ (t))y(αl (t))

l=1

= eλ (σ (t), s)y Δ (t) − λeλ (σ (t), s)y(t) +

n  l=1

Al (t)eλ (σ (t), s)eλ (σ (t), αl (t))y(αl (t))

246

5 Stability for First-Order Functional Dynamic Equations

% = eλ (σ (t), s) y (t) − λy(t) + Δ

n 

& Al (t)eλ (σ (t), αl (t))y(αl (t))

l=1 n   = eλ (σ (t), s) y Δ (t) + Al (t)y(αl (t)) l=1

−λy(t) +

n 

Al (t) (eλ (σ (t), αl (t)) − 1) y(αl (t))



l=1

= eλ (σ (t), s) (L y(t) + K y(t)) , where K y(t) = −λy(t) +

n 

Al (t) (eλ (σ (t), αl (t)) − 1) y(αl (t)),

(5.54)

l=1

t ∈ [s, ∞). Let M =L +K . Then we get L x(t) = eλ (σ (t), s)M y(t),

t ∈ [s, ∞).

Note that the Cauchy operator C of the shifted equation (5.53) acts from 1 ([s, ∞)). Hence, it follows that L : BC 1 ([s, ∞)) → BCrd ([s, ∞)) to BCrd,0 rd,0 1 ([s, ∞)) are bounded. BCrd ([s, ∞)) and its inverse C : BCrd ([s, ∞)) → BCrd,0 Since lim sup (σ (t) − αmin (t)) < ∞, t→∞

using (5.54), we conclude that K : BCrd ([s, ∞)) → BCrd ([s, ∞)) is bounded, and for x ∈ BCrd ([s, ∞)) we have n 

K x ≤ λx +

+ + + + Al  +eλ(σ (·)−αl (·)) − 1[s,∞) + x.

l=1

Then K  ≤ λ +

n 

+ + + + Al  +eλ(σ (·)−αl (·)) − 1[s,∞) +

l=1

→0

as

λ → 0.

5.6 Exponential Stability II

247

Therefore M C = (L + K ) C = I d + K C. Note that K C ≤ K C & % n + +  + + λ(σ (·)−αl (·)) ≤ λ+ Al  +e − 1[s,∞) + C l=1

0,

l=1

and lim sup t→∞

n 

n σ (t) 

Al (t)

Aj (η)Δη < lim inf t→∞

αl (t) j =1

l=1

n 

Al (t).

l=1

Then the trivial solution of the equation (5.44) is uniformly exponentially stable. Proof. Let A(t) =

n 

Al (t),

t ∈ [t0 , ∞).

l=1 1 ([t , ∞)) define the operators For x ∈ BCrd,0 0

M x(t) = x Δ (t) + A(t)x σ (t),

t ∈ [t0 , ∞),

and J x(t) = −

n  l=1

Al (t)

n σ (t)  αl (t) j =1

Aj (η)x(αj (η))Δη,

t ∈ [t0 , ∞).

5.6 Exponential Stability II

251

1 ([t , ∞)) be a solution of the equation Let x ∈ BCrd,0 0

L x = 0 on

[t0 , ∞).

Then 0 = L x(t) = x Δ (t) +

n 

Al (t)x(αl (t))

l=1

= x (t) + A(t)x (t) + Δ

σ

n 

  Al (t) x(αl (t)) − x σ (t)

l=1

= x Δ (t) + A(t)x σ (t) −

n 

σ (t)

Al (t)

l=1

= x Δ (t) + A(t)x σ (t) +

n 

n σ (t) 

Al (t)

Aj (η)x(αj (η))Δη

αl (t) j =1

l=1

= M x(t) − J x(t),

x Δ (η)Δη

αl (t)

t ∈ [t0 , ∞).

Therefore kerL ⊂ ker (M − J ) . Without loss of generality, we suppose that there exist α0 > 0 and β0 ∈ (0, α0 ) such that A(t) ≥ α0 ,

t ∈ [t0 , ∞),

and n  l=1

σ (t)

Al (t)

A(η)Δη ≤ β0 ,

t ∈ [t0 , ∞).

αl (t)

We have β0 ≥

n 

σ (t)

Al (t)

A(η)Δη αl (t)

l=1 σ (t)

≥ A(t)

A(η)Δη t

= (A(t))2 μ(t) ≥ α02 μ(t),

t ∈ [t0 , ∞),

252

5 Stability for First-Order Functional Dynamic Equations

i.e., μ(t) ≤

β0 , α02

t ∈ [t0 , ∞).

Also, we have that A ∈ BCrd ([t0 , ∞)), A(t) > 0, and t ∈ [t0 , ∞). For the Cauchy operator D of the equation [t0 , ∞)

M x = 0 on we have the representation Df (t) =

t t0

eA (t, η)f (η)Δη,

t ∈ [t0 , ∞),

for any f ∈ BCrd ([t0 , ∞)). Note that t t0

t

eA (t, η)Δη ≤

t0

eα0 (t, η)Δη

=

 1  1 − eα0 (t, t0 ) α0

≤

1 , α0

t ∈ [t0 , ∞).

Let K : BCrd ([t0 , ∞)) → BCrd ([t0 , ∞)) be defined as follows K = (M − L )D. We have K = JD and for all x ∈ BCrd ([t0 , ∞)), we get K x(t) = J Dx(t) =−

=−

n 

Al (t)

n σ (t) 

l=1

αl (t) j =1

n 

n σ (t) 

l=1

Al (t)

αl (t) j =1

Aj (η)Dx(αj (η))Δη %

αj (η)

Aj (η) t0

& eA (αj (η), ζ )x(ζ )Δζ Δη,

5.7 Advanced Practical Problems

253

t ∈ [t0 , ∞). Hence, K x ≤

β0 x α0

< x,

x ∈ BCrd ([t0 , ∞)).

Therefore K  < 1. Hence from Theorem 5.34, we get that the trivial solution of the equation (5.44) is uniformly exponentially stable. This completes the proof. Exercise 5.8. Let T = Z. Check if the trivial solution of the equation x Δ (t) +

t +1 x(t − 1) = 0, 4t + 7

t ∈ [1, ∞),

is uniformly asymptotically stable.

5.7 Advanced Practical Problems Problem 5.1. Let T = N0 . Prove that the zero solution of the equation 2t + 1   x(t − 1) = 0, x Δ (t) +  2 t + 2t + 2 t 2 + 1

t ∈ T,

t ≥ 1,

is uniformly stable. Problem 5.2. Let T = 2N0 . Check if the zero solution of the equation 2t + 2   x(t − 2) = 0, x Δ (t) +  2 t + 4t + 5 t 2 + 1

t ∈ T,

t ≥ 2,

is uniformly asymptotically stable.   1 N  {0}. Prove that for any solution x of the Problem 5.3. Let T = 2N0 2 equation 1 x (t) = 2x(t) − 2 x 2t Δ

 t , 2

t ∈ T,

t ≥ 2,

254

5 Stability for First-Order Functional Dynamic Equations

there exists a constant L such that lim x(t)e2 (2, t) = L.

t→∞

Problem 5.4. Let T = 2N0 equation



2N



x Δ (t) = x(t) + x

{0}. Prove that for any solution x of the

 t , 2

t ∈ T,

t ≥ 2,

there exists a constant L such that lim x(t)e1 (2, t) = L.

t→∞

Problem 5.5. Let T = 4N0 . Consider the system   3 x1Δ (t) = t + (t − 1) x2 4t   2 x2Δ (t) = t + x1 4t , t ≥ 4, x1 (t) = t, x2 (t) = 2t + 1, t ∈ [1, 4].

(5.56)

Let V (t, x) = 2x14 + 3x2 + t + t 2 − 4t 3 ,

t ∈ T,

x ∈ R2 ,

x = (x1 , x2 ).

Find (V (t, x(t)))Δ provided x(t) = (x1 (t), x2 (t)), t ∈ T, is a solution of the system (5.56). Problem 5.6. Let T = 2Z. Check if the trivial solution of the equation x Δ (t) +

t + 14 t + 10 x(t − 2) + x(t − 4) = 0, 7t + 21 9t + 131

is uniformly asymptotically stable.

t ∈ [6, ∞),

Chapter 6

Oscillations of First-Order Functional Dynamic Equations

The results on this chapter are taken from [14, 59, 61, 64, 65, 85, 87, 97, 104, 168– 170, 175–177, 185, 215, 221, 230–233, 243, 245, 247, 248, 253].

6.1 Positive Solutions Let T be a time scale that is unbounded above with forward jump operator and delta differentiation operator σ and Δ, respectively. Consider the equation x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ T,

(6.1)

where p ∈ Crd (T), p(t) > 0, t ∈ T, the delay τ : T → T satisfies τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. Theorem 6.1. Let p(t) ≥ 0, t ∈ T, 1 − μ(t)p(t) > 0, t ∈ T. Then t

1− s

p(u)Δu ≤ e−p (t, s) ≤ e−

#t s

p(u)Δu

,

t, s ∈ T,

t ≥ s.

Proof. Fix s ∈ T. Let t

y(t) = −

p(u)Δu,

t ∈ T,

t ≥ s.

s

We have y(s) = 0

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_6

255

256

6 Oscillations of First-Order Functional Dynamic Equations

and y(t) ≤ 0, y (t) = −p(t) Δ

≤ −p(t) − p(t)y(t),

t ∈ T,

t ≥ s.

We put f = y Δ + py + p

on

T.

Then y Δ (t) = −p(t) = −p(t)y(t) + f (t) − p(t),

t ∈ T,

t ≥ s.

Hence, y(t) = e−p (t, s)y(s) +

t s

e−p (t, σ (u)) (f (u) − p(u)) Δu

t

= s

e−p (t, σ (u))f (u)Δu t

−

e−p (t, σ (u))p(u)Δu

s t

≤−

e−p (t, σ (u))p(u)Δu

s t

=−

e(−p) (σ (u), t)p(u)Δu

s t

=− s

t

=− s

t

=− s

(1 + μ(u)((−p))(u)) e(−p) (u, t)p(u)Δu

1+

 p(u)μ(u) e(−p) (u, t)p(u)Δu 1 − p(u)μ(u)

p(u) e(−p) (u, t)Δu 1 − μ(u)p(u)

t

=−

((−p))(u)e(−p) (u, t)Δu

s t

=− s

Δ e(−p) (u, t)Δu

6.1 Positive Solutions

257

!u=t ! = −e(−p) (u, t)!

u=s

= −1 + e(−p) (s, t) = −1 + e−p (t, s),

t ∈ T,

t ≥ s.

Therefore t

1− s

p(u)Δu ≤ e−p (t, s),

t ∈ T,

t ≥ s.

Next, if μ(u) = 0, then ξμ(u) (−p(u)) = −p(u), and if μ(u) > 0, we get ξμ(u) (−p(u)) = =

Log(1 − μ(u)p(u)) μ(u) log(1 − μ(u)p(u)) μ(u)

= −p(u) +

p(u)μ(u) + log(1 − μ(u)p(u)) μ(u)

≤ −p(u). Hence, e−p (t, s) = e

#t s

≤ e−

ξμ(u) (−p(u))Δu #t s

p(u)Δu

,

t ∈ T,

t ≥ s.

This completes the proof. Theorem 6.2. Let p(t) ≥ 0, t ∈ T, p is rd-continuous on T. Then t

1+

p(u)Δu ≤ ep (t, s) ≤ e

#t s

p(u)Δu

,

t, s ∈ T,

s

Proof. Fix s ∈ T. Let t

y(t) =

p(u)Δu, s

t ∈ T,

t ≥ s.

t ≥ s.

258

6 Oscillations of First-Order Functional Dynamic Equations

We have y(s) = 0 and y(t) ≥ 0, y (t) = p(t) Δ

≤ p(t) + p(t)y(t),

t ∈ T,

t ≥ s.

We put f = y Δ − py − p

on

T.

Then f (t) ≤ 0,

t ∈ T,

t ≥ s,

and y Δ (t) = p(t) = p(t)y(t) + f (t) + p(t),

t ∈ T,

t ≥ s.

Hence, t

y(t) = ep (t, s)y(s) +

ep (t, σ (u)) (f (u) + p(u)) Δu

s t

=

ep (t, σ (u))f (u)Δu s t

+

ep (t, σ (u))p(u)Δu s t

≤

ep (t, σ (u))p(u)Δu s t

=

ep (σ (u), t)p(u)Δu

s t

= s

t

= s

(1 + μ(u)(p)(u)) ep (u, t)p(u)Δu

1−

 p(u)μ(u) ep (u, t)p(u)Δu 1 + p(u)μ(u)

6.1 Positive Solutions

259 t

= s

p(u) ep (u, t)Δu 1 + μ(u)p(u) t

=−

(p)(u)ep (u, t)Δu

s t

=− s

Δ ep (u, t)Δu

!u=t ! = −ep (u, t)!

u=s

= −1 + ep (s, t) = −1 + ep (t, s),

t ∈ T,

t ≥ s.

Therefore t

1+

p(u)Δu ≤ ep (t, s),

t ∈ T,

t ≥ s.

s

The second part of the assertion follows from Theorem 5.22. This completes the proof. Theorem 6.3. Let p : T → [0, ∞), 1 − μ(t)p(t) > 0, and t ∈ T. Then for all λ ∈ (0, 1] we have 1 ≤ lim inf t→∞ 4α

t

p(s)Δs ≤

τ (t)

1 , eα

where α=

sup λ>0,1−λμp>0

  λe−λp (t, s) ,

t, s ∈ T,

t ≥ s.

Proof. Let t, s ∈ T, t ≥ s. Define t

P =

p(u)Δu. s

By Theorem 6.1, for any λ ∈ (0, 1], we have 1 − λP ≤ e−λp (t, s) ≤ e−λP , whereupon λ − λ2 P ≤ λe−λp (t, s) ≤ λe−λP .

(6.2)

260

6 Oscillations of First-Order Functional Dynamic Equations

Let g(λ) = λe−λP ,

f (λ) = λ − λ2 P ,

λ ∈ (0, ∞).

We have f (λ) = 1 − 2λP ,

λ ∈ (0, ∞),

and f (λ) = 0

⇐⇒

λ=

1 . 2P

Hence,

sup f (λ) = f λ∈(0,∞)

1 2P

1 − 2P 1 − = 2P 1 . = 4P =



1 P 4P 2 1 4P

Next, g (λ) = e−λP − λP e−λP = (1 − λP )e−λP ,

λ ∈ (0, ∞).

Then g (λ) = 0

⇐⇒

λ=

and sup g(λ) = λ>0

=

1 −1 e P 1 . eP

Hence from (6.2), we get 1 1 ≤α≤ 4P eP

1 P

6.1 Positive Solutions

261

or 1 1 ≤P ≤ , 4α eα or 1 ≤ 4α

t

p(u)Δu ≤

s

1 . eα

Since s, t ∈ T, t ≥ s, was arbitrarily chosen, we get 1 ≤ 4α

t

p(u)Δu ≤

τ (t)

1 eα

and 1 ≤ lim inf t→∞ 4α

t

p(u)Δu ≤

τ (t)

1 . eα

This completes the proof. Theorem 6.4. Let p is rd-continuous and positive, τ : T → T satisfies τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. If the equation (6.1) has an eventually positive solution, then α = lim sup

sup

t→∞ λ>0,1−μλp>0



 λe−λp (t, τ (t)) ≥ 1.

Proof. Let y solves (6.1) and be eventually positive. Assume that α < 1. Let β ∈  1, α1 . Then there exists a T0 ∈ T such that β≤

1  , supλ>0,1−μλp>0 λe−λp (t, τ (t))

for all t ≥ T0 . Since y is eventually positive, then it is eventually decreasing and we take t0 large enough so that y is decreasing on [t0 , ∞). Hence, y(τ (t)) ≥ y(t),

t ∈ [t0 , ∞).

From here, 0 = y Δ (t) + p(t)y(τ (t)) ≥ y Δ (t) + p(t)y(t),

t ∈ [t0 , ∞).

262

6 Oscillations of First-Order Functional Dynamic Equations

Hence from Gronwall’s inequality, we get y(t) ≤ e−p (t, τ (t))y(τ (t)),

t ∈ [t0 , ∞),

or 1 y(τ (t)) ≥ , y(t) e−p (t, τ (t))

t ∈ [t0 , ∞).

Then there exists a t1 ∈ [t0 , ∞) such that y(τ (t)) ≥ β, y(t)

t ∈ [t1 , ∞).

Thus, 0 = y Δ (t) + p(t)y(τ (t)) ≥ y Δ (t) + βp(t)y(t),

t ∈ [t1 , ∞).

Hence from Gronwall’s inequality, there exists t2 ∈ [t1 , ∞) such that y(τ (t)) 1 ≥ y(t) e−βp (t, τ (t)) =

β βe−βp (t, τ (t))

≥ β 2,

t ∈ [t2 , ∞).

Proceeding this procedure in this manner, we obtain a sequence {tn }n∈N ⊂ T such that y(τ (t)) ≥ β n, y(t)

t ∈ [tn , ∞).

Because β > 1, we conclude that lim inf t→∞

Let M ∈



1 1 4 , 4α

y(τ (t)) = ∞. y(t)

. By Theorem 6.3, we get that there exists T ∈ T such that t τ (t)

p(s)Δs ≥

1 4α

≥ M,

t ∈ [T , ∞).

(6.3)

6.1 Positive Solutions

263

Hence, σ (t)

t

p(s)Δs ≥

p(s)Δs

τ (t)

τ (t)

≥ M,

t ∈ [T , ∞).

Consider the function f : T → R defined by u

f (u) =

p(s)Δs −

τ (t)

M , 2

t ∈ [T , ∞).

We find f (τ (t)) = −

M 2

< 0, t

f (t) =

p(s)Δs −

τ (t)

≥M−

M 2

M 2

M 2 > 0, t ≥ T . =

Then there exists a t ∗ ∈ [τ (t), t), t ∈ [T , ∞), such that f (t ∗ ) = 0

f (t ∗ ) < 0

or

and

f (σ (t ∗ )) > 0.

Hence, σ (t ∗ )

p(s)Δs =

τ (t)

≥

M + f (σ (t ∗ )) 2 M 2

and σ (t) t∗

p(s)Δs =

τ (t) t∗

=−

p(s)Δs +

σ (t)

p(s)Δs τ (t)

t∗ τ (t)

p(s)Δs +

σ (t)

p(s)Δs τ (t)

264

6 Oscillations of First-Order Functional Dynamic Equations

 M ∗ + f (t ) + M ≥− 2 M − f (t ∗ ) 2 M , t ∈ [T , ∞). ≥ 2

=

Then y(t ∗ ) ≥ y(t ∗ ) − y(σ (t)) σ (t)

=− =

t∗

y Δ (s)Δs

σ (t)

p(s)y(τ (s))Δs

t∗

σ (t)

≥ y(τ (t))

p(s)Δs

t∗

M y(τ (t)) 2  M ≥ y(τ (t)) − y(σ (t ∗ )) 2 ≥

=− =

σ (t ∗ )

M 2

M 2

σ (t ∗ )

p(s)y(τ (s))Δs τ (t)

M ≥ y(τ (t ∗ )) 2 =

y Δ (s)Δs

τ (t)

σ (t ∗ )

p(s)Δs τ (t)

M2 y(τ (t ∗ )). 4

Hence, y(τ (t ∗ )) 4 ≤ 2. y(t ∗ ) M Then lim inf t→∞

y(τ (t)) < ∞, y(t)

which contradicts with (6.3). Therefore α ≥ 1. This completes the proof.

6.1 Positive Solutions

265

Example 6.1. Let T = {tn : n ∈ Z}, where {tn }n∈Z is a strictly increasing sequence of real numbers such that T is closed, p satisfies all the conditions of Theorem 6.4. Suppose that the equation y Δ (t) + p(t)y(ρ(t)) = 0,

t ∈ T,

has an eventually positive solution. We have e−λp (tn , ρ(tn )) = e =e

# tn

log(1−λμ(s)p(s))Δs

# tn

log(1−λμ(s)p(s))Δs

1 ρ(tn ) μ(s) 1 tn−1 μ(s)

= elog(1−λμ(tn−1 )p(tn−1 )) = 1 − λμ(tn−1 )p(tn−1 ), λe−λp (tn , ρ(tn )) = λ − λ2 μ(tn−1 )p(tn−1 ). Let g(λ) = λ − λ2 μ(tn−1 )p(tn−1 ),

λ > 0.

Then g (λ) = 1 − 2λμ(tn−1 )p(tn−1 ), g (λ) = 0

⇐⇒

λ=

1 . 2μ(tn−1 )p(tn−1 )

Therefore sup g(λ) = λ>0

=

1 1 − μ(tn−1 )p(tn−1 ) 2μ(tn−1 )p(tn−1 ) 4 (μ(tn−1 )p(tn−1 ))2 1 . 4μ(tn−1 )p(tn−1 )

By Theorem 6.4, it follows that lim sup t→∞

1 ≥1 4μ(t)p(t)

or lim inf (μ(t)p(t)) ≤ t→∞

1 . 4

266

6 Oscillations of First-Order Functional Dynamic Equations

Exercise 6.1. Let T = {4n : n ∈ N0 } and p satisfies Theorem 6.4. Suppose that the equation y(4t) = y(t) − μ(t)p(t)y

 t 4

has an eventually positive solution. Prove that   1 . lim inf 4n p(4n ) ≤ n→∞ 12 Theorem 6.5. Let T = {tn : n ∈ Z}, where {tn }n∈Z is a strictly increasing sequence of real numbers such that T is closed, p satisfies all the conditions of Theorem 6.4. Let k ∈ N and τ (tn ) = tn−k . If (6.1) has an eventually positive solution, then

t

p(s)Δs ≤

lim inf t→∞

τ (t)

k k+1

k+1 (6.4)

.

Proof. Assume that (6.4) does not hold. We have λe−λp (tn , τ (tn )) = λe−λp (tn , tn−k ) = λe = λe =λ

# tn

1 tn−k μ(s)

"n−1 l=n−k

n−1 $

log(1−λp(s)μ(s))Δs

log(1−λp(tl )μ(tl ))

(1 − λp(tl )μ(tl ))

l=n−k

%

# tn

≤ λ 1−λ

τ (tn ) p(s)Δs

k

= λ(1 − λS)k , where S=

1 k

tn

p(s)Δs. τ (tn )

We have "n−1 S=

l=n−k (tl+1

− tl )p(tl )

k

.

Let f (λ) = λ(1 − λS)k ,

λ > 0.

&k

6.1 Positive Solutions

267

Then f (λ) = (1 − λS)k − kλS(1 − λS)k−1 = (1 − λS)k−1 (1 − λS − kλS) = (1 − λS)k−1 (1 − λ(k + 1)S) = 0 ⇐⇒ 1 1 . λ= , λ= S (k + 1)S Hence,

 1 f (λ) ≤ f (1 + k)S

k 1 S = 1− (1 + k)S (1 + k)S =

kk S k (1 + k)S(1 + k)k S k

kk S(1 + k)k+1 k+1

1 k . ≤ k+1 S

=

From here, lim sup

sup

t→∞ λ>0,1−λμp>0

  λe−λp (t, τ (t)) ≤

k k+1

k+1

1 lim inft→∞ S

< 1, which contradicts with Theorem 6.4. This completes the proof. Corollary 6.1. Let p is rd-continuous and positive, and 1 − λμp > 0 on T, τ : T → T satisfies τ (t) < t, t ∈ T, limt→∞ τ (t) = ∞. If lim sup

sup

t→∞ λ>0,1−λμp>0

  λe−λp (t, τ (t)) < 1,

then all the solutions of (6.1) are oscillatory.

268

6 Oscillations of First-Order Functional Dynamic Equations

Example 6.2. Let T = N. Consider the equation x Δ (t) +

t +1 x(t − 1) = 0, (t + 2)(t + 3) x(0) =

t ≥ 1,

1 . 2

Here p(t) =

t +1 , (t + 2)(t + 3)

σ (t) = t + 1,

t ∈ T,

k = 1. We will check that x(t) =

1 , t +2

t ∈ T,

is its solution. We have x Δ (t) = − =−

1 (t + 2)(σ (t) + 2) 1 , (t + 2)(t + 3)

t ∈ T.

Then 1 t +1 t +1 x (t) + x(t − 1) = − + (t + 2)(t + 3) (t + 2)(t + 3) (t + 2)(t + 3) Δ

=−

1 t +1

1 1 + (t + 2)(t + 3) (t + 2)(t + 3)

= 0,

t ≥ 1.

Therefore the considered equation has an eventually positive solution. Next, t t−1

s + 1 !!s=t s+1 Δs = ! (s + 2)(s + 3) s + 2 s=t−1 =

t +1 t − t +2 t +1

=

(t + 1)2 − t (t + 2) (t + 1)(t + 2)



6.1 Positive Solutions

269

=

t 2 + 2t + 1 − t 2 − 2t (t + 1)(t + 2)

=

1 . t ∈ T, (t + 1)(t + 2)

and t

lim inf t→∞

t−1

s+1 1 Δs = lim inf t→∞ (t + 1)(t + 2) (s + 2)(s + 3) =0 1 < 4 k+1

k . = k+1

Example 6.3. Let T = {q k : k ∈ Z} equation



{0}, q > 1, and k ∈ N. Suppose that the

      x q n+1 = x q n − (q − 1)q n p q n x q n−k , has an eventually positive solution. Then qn q n−k

p(s)Δs =

n−1 

  p ql μ ql

l=n−k

 qlp ql .

n−1 

= (q − 1)

l=n−k

Hence from Theorem 6.5, we get n  l=n−k

 qlp ql ≤



k k+1

k+1

q −1

.

Example 6.4. Let T = hZ, where h > 0. Consider the equation x Δ (t) + x(t − kh) = 0, where k ∈ N is such that kk < h. (k + 1)k+1

t ∈ T,

n ∈ Z,

270

6 Oscillations of First-Order Functional Dynamic Equations

Here σ (t) = t + h, p(t) = 1, τ (t) = t − kh,

t ∈ T.

Then 1 − λμ(t)p(t) = 1 − λh >0



if λ < h1 . For λ ∈ 0, h1 , we have λe−λp (t, τ (t)) = λe = λe = λe

#t

1 t−kh μ(s)

#t

1 t−kh h

log(1−λμ(s)p(s))Δs

log(1−λh)Δs

"t−h

s=t−kh log(1−λh)

= λek log(1−λh) = λ(1 − λh)k . Let  1 . λ ∈ 0, h

g(λ) = λ(1 − λh) , k

Then g (λ) = (1 − λh)k − kλh(1 − λh)k−1 = (1 − λh)k−1 (1 − λh − kλh) = (1 − λh)k−1 (1 − λ(k + 1)h) =0 if λ =

1 (k+1)h .

Hence, sup g(λ) ≤ 

λ∈ 0, h1

k h 1 1− (k + 1)h (k + 1)h

=

k k+1

k

1 (k + 1)h

6.1 Positive Solutions

271

k k+1


0 and A > 0 such that 1 − λμ(t)p(t) > 0,

t ∈ T,

λe−λp (t, τ (t)) ≥ 1,

t > A,

then the equation (6.1) has a positive solution. Proof. Define  x(t) =

1

if t < A

1 λe−λpx (t,τ (t))

if t ∈ [A, ∞).

We have 0 < x(t) ≤ 1,

t ∈ T.

Let z(t) = 1 − p(t)e−λpx (τ (t), t),

t ∈ T,

t ∈ [A, ∞).

Define

y(t) =

1 if t < A ez−1 (t, A) if t ∈ [A, ∞).

272

6 Oscillations of First-Order Functional Dynamic Equations

We have y(t) > 0, t ∈ T, and z(t) − 1 = −p(t)e−λpx (τ (t), t) =−

p(t) e−λpx (t, τ (t))

=−

λp(t) λe−λpx (t, τ (t))

= −λp(t)x(t),

t ∈ [A, ∞),

and y(t) = e−λpx (t, A),

t ∈ [A, ∞).

Also, y Δ (t) = (z(t) − 1)ez−1 (t, A) = (z(t) − 1)y(t),

t ∈ [A, ∞).

Hence, 1+

y Δ (t) = 1 + z(t) − 1 y(t) = z(t) = 1 − p(t)e−λpx (τ (t), t) = 1 − p(t)e−λpx (τ (t), A)e−λpx (A, t) = 1 − p(t)

e−λpx (τ (t), A) e−λpx (t, A)

= 1 − p(t)

y(τ (t)) , y(t)

t ∈ [A, ∞),

whereupon y Δ (t) y(τ (t)) = −p(t) , y(t) y(t)

t ∈ [A, ∞),

y Δ (t) = −p(t)y(τ (t)),

t ∈ [A, ∞).

or

This completes the proof.

6.1 Positive Solutions

273

Example 6.5. Let T = Z. Consider the equation x Δ (t) +

1 x(t − 1) = 0, t +1

t ∈ Z.

Here σ (t) = t + 1, μ(t) = 1, 1 , t +1 τ (t) = t − 1, t ∈ T.

p(t) =

Take λ ∈ [3, 4] and A = 20. Then 1 t +1 4 ≥ 1− 21 17 = 21 > 0, t ∈ [20, ∞).

1 − λμ(t)p(t) = 1 − λ

Next, e−λp (t, τ (t)) = e =e =e

#t

1 t−1 μ(s)

#t



t−1 log

log(1−λp(s)μ(s))Δs λ 1− s+1 Δs

 log 1− λt

λ = 1− , t 

λ λe−λp (t, τ (t)) = λ 1 − t λ2 t 16 ≥ 3− 20 4 = 3− 5 = λ−

274

6 Oscillations of First-Order Functional Dynamic Equations

11 5 > 1, t ∈ [20, ∞),

=

λ ∈ [3, 4].

Therefore the considered equation has a positive solution. Exercise 6.3. Let T = 3Z. Prove that the equation 1 x(t − 3) = 0, t2 + 2

t ∈ T,

x Δ (t) + p(t)x(τ (t)) = q(t),

t ∈ T,

x Δ (t) + has a positive solution. Consider the equation

(6.5)

where p, q ∈ Crd (T), p(t) > 0, t ∈ T, τ : T → T satisfies τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. Define Q : T → T as follows QΔ (t) = q(t),

t ∈ T,

and Q+ (t) = max{0, Q(t)},

Q− (t) = max{−Q(t), 0}.

Theorem 6.7. Suppose that there exists T ∈ T such that ∞ T

p(t)Q+ (τ (t))Δt =

∞ T

p(t)Q− (τ (t))Δt = ∞.

Then all the solutions of (6.5) are oscillatory. Proof. Assume that x is an eventually positive solution of the equation (6.5). Then there exists a number t0 ∈ T such that x(t) > 0 for any t ∈ [t0 , ∞), t ∈ T. By (6.5), we get x Δ (t) + p(t)x(τ (t)) = QΔ (t),

t ∈ T,

(x − Q)Δ (t) = −p(t)x(τ (t)),

t ∈ T.

or

There exists a t1 ∈ (t0 , ∞) such that (x − Q)Δ (t) < 0,

t ∈ [t1 , ∞).

(6.6)

6.1 Positive Solutions

275

Hence, I d −Q is nonincreasing on [t1 , ∞). Assume that there is a t2 ∈ [t1 , ∞) such that x(t) − Q(t) ≤ 0,

t ∈ [t2 , ∞).

Then Q(t) > 0,

t ∈ [t2 , ∞),

and Q+ (t) = Q(t),

Q− (t) = 0,

t ∈ [t2 , ∞).

Hence, ∞ T

p(t)Q− (τ (t))Δt < ∞,

which is a contradiction. Therefore x(t) − Q(t) > 0 for any t ∈ [t1 , ∞). From here, using that I d − Q is nonincreasing, we conclude that lim (x(t) − Q(t))

t→∞

exists and it is finite. Let T1 ∈ [T , ∞) be such that x(τ (t)) > Q(τ (t)),

t ∈ [T1 , ∞).

By (6.6), we get ∞

(x − Q)Δ (t)Δt = −

T1

∞

p(t)x(τ (t))Δt, T1

whereupon ∞

p(t)x(τ (t))Δt < ∞

T1

and then ∞ T1

p(t)Q+ (τ (t))Δt ≤

∞

p(t)x(τ (t))Δt T1

< ∞. This is a contradiction. Therefore (6.5) has no eventually positive solutions. Suppose that the equation (6.5) has an eventually negative solution x. Then there is an s ∈ T

276

6 Oscillations of First-Order Functional Dynamic Equations

such that x(t) < 0 for any t ∈ [s, ∞). Hence from (6.6), it follows that there exists an s1 ∈ [s, ∞) such that t ∈ [s1 , ∞),

(x − Q)Δ (t) > 0 for

i.e., I d − Q is increasing on [s1 , ∞). If there exists an s2 ∈ [s1 , ∞) such that t ∈ [s2 , ∞),

x(t) − Q(t) ≥ 0, then Q(t) ≤ x(t) ≤ 0,

t ∈ [s2 , ∞),

and Q+ (t) = 0,

t ∈ [s2 , ∞).

Hence, there exists an s3 ∈ [s2 , ∞) so that Q+ (τ (t)) = 0,

t ∈ [s3 , ∞).

Then ∞ s3

p(t)Q+ (τ (t))Δt = 0.

This is a contradiction. Therefore x(t) − Q(t) < 0,

t ∈ [s1 , ∞),

and lim (x(t) − Q(t))

t→∞

exists and it is finite. By (6.6), it follows that −

∞

p(t)x(τ (t))Δt < ∞.

s1

Take s4 ∈ [s1 , ∞) so that −Q(τ (t)) < −x(τ (t)),

t ∈ [s4 , ∞).

6.1 Positive Solutions

277

Then ∞ s1

p(t)Q− (τ (t))Δt < −

∞

p(t)x(τ (t))Δt s4

< ∞. This is a contradiction. Therefore the equation (6.5) has no eventually negative solutions. This completes the proof. Consider the equation x Δ (t) + p(t)x(τ (σ (t))) = 0,

t ∈ T,

(6.7)

where p ∈ Crd (T), p(t) > 0, t ∈ T, the delay τ : T → T satisfies τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. For λ > 0 define

∗

α = lim inf inf

t→∞ λ>0

 eλp (t, τ (σ (t))) . λ

Theorem 6.8. Let (6.7) has an eventually positive solution. Then α ∗ ≤ 1. Proof. Let x be an eventually positive solution. Assume that α ∗ > 1. We take β ∗ ∈ (1, α ∗ ). Then there exists a T0 ∈ T such that

inf

λ>0

eλp (t, τ (t)) λ



≥ β ∗,

t ≥ T0 .

Because x is an eventually positive solution of the equation (6.7), using that p(t) > 0, t ∈ T, we conclude that x is an eventually decreasing function. Then there exists a T ∈ [T0 , ∞) such that x(τ (σ (t))) ≥ x(σ (t)),

t ∈ [T , ∞),

and 0 = x Δ (t) + p(t)x(τ (σ (t))) ≥ x Δ (t) + p(t)x(σ (t)),

t ∈ [T , ∞).

Then x(τ (σ (t))) ≥ ep (σ (t), τ (σ (t)))x(σ (t)),

t ∈ [T , ∞),

278

6 Oscillations of First-Order Functional Dynamic Equations

or x(τ (σ (t))) ≥ ep (σ (t), τ (σ (t))), x(σ (t))

t ∈ [T , ∞).

Hence, there exists a T1 ≥ T0 such that ep (σ (t), τ (σ (t))) ≥ β ∗ ,

t ≥ T1 .

From here, x(τ (σ (t))) ≥ β ∗, x(σ (t))

t ≥ T1 .

Therefore 0 = x Δ (t) + p(t)x(τ (σ (t))) ≥ x Δ (t) + β ∗ p(t)x(σ (t)),

t ≥ T1 .

Hence, x(τ (σ (t))) ≥ eβ ∗ p (σ (t), τ (σ (t)))x(σ (t)),

t ≥ T1 ,

or x(τ (σ (t))) ≥ eβ ∗ p (σ (t), τ (σ (t))), x(σ (t))

t ≥ T1 .

There exists a T2 ≥ T1 such that eβ ∗ p (σ (t), τ (σ (t))) ≥ (β ∗ )2 ,

t ≥ T2 .

Then x(τ (σ (t))) ≥ (β ∗ )2 , x(σ (t))

t ≥ T2 .

Proceeding in this manner, we obtain a sequence {Tn }n∈N ⊂ T such that x(τ (σ (t))) ≥ (β ∗ )n , x(σ (t))

t ≥ Tn .

Because β ∗ > 1, we conclude that lim inf t→∞

x(τ (t)) = ∞. x(t)

(6.8)

6.1 Positive Solutions

279

Next, there exists a constant M > 0 such that t

p(s)Δs ≥ M,

t ≥ T.

τ (σ (t))

Hence, σ (t)

t

p(s)Δs ≥

p(s)Δs

τ (σ (t))

τ (σ (t))

≥ M,

t ≥ T.

Let f : T → R be defined by u

f (u) =

p(s)Δs −

τ (σ (t))

M , 2

t ≥ T.

We have f (τ (σ (t))) = −

M 2

< 0, f (σ (t)) =

σ (t)

p(s)Δs −

τ (σ (t))

M 2

≥M− =

M 2

M , 2

t ≥ T.

Then there exists a t ∗ ∈ [τ (σ (t)), σ (t)), t ≥ T , such that f (t ∗ ) = 0

or

f (t ∗ ) < 0

and

f (σ (t ∗ )) > 0.

Hence, σ (t ∗ )

p(s)Δs =

τ (σ (t))

≥

M + f (σ (t ∗ )) 2 M , 2

t ≥ T,

280

6 Oscillations of First-Order Functional Dynamic Equations

and σ (t) t∗

τ (σ (t))

p(s)Δs =

t∗

σ (t)

p(s)Δs +

p(s)Δs τ (σ (t))

t∗

=−

τ (σ (t))

≥−

σ (t)

p(s)Δs +

p(s)Δs τ (σ (t))

 M + f (t ∗ ) + M 2

M − f (t ∗ ) 2 M ≥ , t ≥ T. 2

=

Then x(t ∗ ) ≥ x(t ∗ ) − x(σ (t)) =

σ (t) t∗

p(s)x(τ (σ (s)))Δs

≥ x(τ (σ (t)))

σ (t)

p(s)Δs

t∗

M x(τ (σ (t))) 2  M x(τ (σ (t))) − x(σ (t ∗ )) ≥ 2 ≥

=

M 2

σ (t ∗ )

p(s)x(τ (σ (s)))Δs τ (σ (t))

M ≥ x(τ (σ (t ∗ ))) 2

σ (t ∗ )

p(s)Δs τ (σ (t))

M2 x(τ (σ (t ∗ ))), 4

t ≥ T.

x(τ (σ (t ∗ ))) 4 ≤ 2, x(t ∗ ) M

t ≥ T,

≥ Therefore

6.1 Positive Solutions

281

and lim inf t→∞

x(τ (t)) < ∞, x(t)

which contradicts with (6.8). Consequently α ∗ ≤ 1. This completes the proof. For λ > 0 define

α˜ = lim inf inf

t→∞ λ>0

 eλp (t, τ (t)) . λ

Theorem 6.9. Assume that there exists a constant M > 0 such that t

p(s)Δs ≥ M

(6.9)

τ (σ (t))

for all large t ∈ T. If (6.7) has an eventually positive solution, then α˜ ≤ 1. Proof. Let x be an eventually positive solution. Assume that α˜ > 1. We take β ∗ ∈ (1, α). ˜ Then there exists T0 ∈ T such that

inf

λ>0

eλp (t, τ (t)) λ



≥ β ∗,

t ∈ [T0 , ∞).

Because x is an eventually positive solution of the equation (6.7), using that p(t) > 0, t ∈ T, we conclude that x is an eventually decreasing function. Then there exists a T ∈ [T0 , ∞) such that x(τ (σ (t))) ≥ x(σ (t)),

t ∈ [T , ∞),

and 0 = x Δ (t) + p(t)x(τ (σ (t))) ≥ x Δ (t) + p(t)x(σ (t)),

t ∈ [T , ∞).

x(τ (t)) ≥ ep (t, τ (t))x(t),

t ∈ [T , ∞),

Then

or x(τ (t)) ≥ ep (t, τ (t)), x(t)

t ∈ [T , ∞).

282

6 Oscillations of First-Order Functional Dynamic Equations

Hence, there exists a T1 ∈ [T , ∞) such that ep (t, τ (t)) ≥ β ∗ ,

t ∈ [T1 , ∞).

From here, x(τ (t)) ≥ β ∗, x(t)

t ∈ [T1 , ∞).

Therefore 0 = x Δ (t) + p(t)x(τ (σ (t))) ≥ x Δ (t) + β ∗ p(t)x(σ (t)),

t ∈ [T1 , ∞).

x(τ (t)) ≥ eβ ∗ p (t, τ (t))x(t),

t ∈ [T1 , ∞),

Hence,

or x(τ (t)) ≥ eβ ∗ p (t, τ (t)), x(t)

t ∈ [T1 , ∞).

There exists T2 ∈ [T1 , ∞) such that eβ ∗ p (t, τ (t)) ≥ (β ∗ )2 ,

t ∈ [T2 , ∞).

Then x(τ (t)) ≥ (β ∗ )2 , x(t)

t ∈ [T2 , ∞).

Proceeding in this manner, we obtain a sequence {Tn }n∈N ⊂ T such that x(τ (t)) ≥ (β ∗ )n , x(t)

t ∈ [Tn , ∞).

Because β ∗ > 1, we conclude that lim inf t→∞

x(τ (t)) = ∞. x(t)

Since, there exists a constant M > 0 such that t τ (σ (t))

p(s)Δs ≥ M,

t ∈ [T , ∞),

(6.10)

6.1 Positive Solutions

283

we get σ (t)

t

p(s)Δs ≥

p(s)Δs

τ (σ (t))

τ (σ (t))

≥ M,

t ∈ [T , ∞).

Let f : T → R be defined by u

f (u) =

p(s)Δs −

τ (σ (t))

M , 2

t ∈ [T , ∞).

We have f (τ (σ (t))) = −

M 2

< 0, σ (t)

f (σ (t)) =

p(s)Δs −

τ (σ (t))

≥M− =

M , 2

M 2

M 2 t ∈ [T , ∞).

Then there exists a t ∗ ∈ [τ (σ (t)), σ (t)), t ∈ [T , ∞), such that f (t ∗ ) = 0

or

f (t ∗ ) < 0

and

f (σ (t ∗ )) > 0.

Hence, σ (t ∗ )

M + f (σ (t ∗ )) 2

p(s)Δs =

τ (σ (t))

M , 2

≥

t ∈ [T , ∞),

and σ (t) t∗

p(s)Δs =

τ (σ (t)) t∗

=−

p(s)Δs +

σ (t)

p(s)Δs τ (σ (t))

t∗ τ (σ (t))

p(s)Δs +

σ (t)

p(s)Δs τ (σ (t))

284

6 Oscillations of First-Order Functional Dynamic Equations

 M ∗ + f (t ) + M =− 2 M − f (t ∗ ) 2 M , t ∈ [T , ∞). ≥ 2

=

Then x(τ (σ (t))) ≥ x(τ (σ (t))) − x(t) t

=

p(s)x(τ (σ (s)))Δs τ (σ (t)) t

≥ x(τ (t))

p(s)Δs τ (σ (t))

≥ Mx(τ (t)) for all large t ∈ T. Hence, x(t ∗ ) ≥ x(t ∗ ) − x(σ (t)) =

σ (t) t∗

p(s)x(τ (σ (s)))Δs σ (t)

≥ x(τ (σ (t)))

t∗

p(s)Δs

M x(τ (σ (t))) 2  M ≥ x(τ (σ (t))) − x(σ (t ∗ )) 2 ≥

M = 2 ≥

σ (t ∗ )

p(s)x(τ (σ (s)))Δs τ (σ (t))

M x(τ (σ (t ∗ ))) 2

≥

M2 x(τ (σ (t ∗ ))) 4

≥

M3 x(τ (t ∗ )) 4

σ (t ∗ )

p(s)Δs τ (σ (t))

6.1 Positive Solutions

285

for all large t ∈ T, and 4 x(τ (t ∗ )) ≤ 3 ∗ x(t ) M for all large t ∈ T. Therefore lim inf t→∞

x(τ (t)) < ∞, x(t)

which contradicts with (6.10). Then α˜ ≤ 1. This completes the proof. Example 6.6. Let T = Z, τ (t) = t − 2, and t ∈ Z. Then #t

eλp (t, τ (t)) e = λ = =

e

1 τ (t) μ(s)

log(1+λμ(s)p(s))Δs

λ

#t

t−2 log(1+λp(s))Δs

λ elog(1+λp(t−2))+log(1+λ p(t−1))

λ (1 + λp(t − 2))(1 + λp(t − 1)) , = λ

λ > 0.

Let g(λ) =

λ2 p(t − 1)p(t − 2) + λ(p(t − 1) + p(t − 2)) + 1 , λ

λ > 0.

Then g(λ) = λp(t − 1)p(t − 2) + (p(t − 1) + p(t − 2)) + and g (λ) = p(t − 1)p(t − 2) − =0 λ = ±√

1 λ2

⇐⇒ 1 , p(t − 1)p(t − 2)

t ∈ T.

1 , λ

λ>0

286

6 Oscillations of First-Order Functional Dynamic Equations

Then  1 inf g(λ) = g √ λ>0 p(t − 1)p(t − 2)   p(t−2) 1 + √p(t−1)p(t−2) 1+ = 1

√

=



p(t − 1) +

√

p(t−1) p(t−1)p(t−2)



p(t−1)p(t−2)



p(t − 2)

2

for all large t ∈ T. Note that t

p(s)Δs =

τ (σ (t))

t

p(s)Δs t−1

= p(t − 1),

t ∈ T.

Then if there exists a constant K > 0 such that p(n) ≥ K for large n ∈ N and (6.7) has an eventually positive solution, we have  2  p(n) + p(n + 1) > 1 for all large n ∈ N.

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations Theorem 6.10. Let x be a nonoscillatory solution of (6.1). If t

lim sup t→∞

p(η)Δη > 0, τ (t)

then lim inf t→∞

x(τ (t)) < ∞. x(t)

(6.11)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

287

Proof. Suppose that x is an eventually positive solution of the equation (6.1). Then x is eventually decreasing. Let t1 ∈ T be such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By (6.11), it follows that there exists an  > 0 and an increasing divergent sequence {ξn }n∈N ⊂ [t1 , ∞) such that σ (ξn )

ξn

p(η)Δη ≥

τ (ξn )

p(η)Δη ≥ ,

n ∈ N.

τ (ξn )

Define the functions fn : [τ (ξn ), σ (ξn )) → R as follows fn (t) =

t

 p(η)Δη − , 2 τ (ξn )

n ∈ N.

We have fn (τ (ξn )) = −

 2

< 0, fn (ξn ) =

ξn

p(η)Δη −

τ (ξn )

≥− =

 2

 2

 . 2

Therefore there exists ζn ∈ [τ (ξn ), ξn ) such that fn (ζn ) ≤ 0

and

fn (σ (ζn )) ≥ 0,

n ∈ N.

Hence, σ (ζn ) τ (ξn )

σ (ζn )   + p(η)Δη − 2 2 τ (ξn )  = + fn (σ (ζn )) 2  ≥ 2

p(η)Δη =

288

6 Oscillations of First-Order Functional Dynamic Equations

and σ (ξn )

σ (ξn )

p(η)Δη =

p(η)Δη +

τ (ξn )

ζn

p(η)Δη ζn

σ (ξn )

=

τ (ξn )

p(η)Δη −

τ (ξn )

ζn

p(η)Δη τ (ξn )

  p(η)Δη − fn (ζn ) + 2 τ (ξn )   ≥  − fn (ζn ) + 2  = − fn (ζn ) 2  ≥ . 2 σ (ξn )

=

Therefore x(ζn ) ≥ x(ζn ) − x(σ (ξn )) σ (ξn )

=

p(η)x(τ (n))Δη ζn σ (ξn )

≥ x(τ (ξn ))

p(η)Δη ζn

 x(τ (ξn )) 2  ≥ (x(τ (ξn )) − x(σ (ζn ))) 2

≥

=

 2

σ (ζn )

p(η)x(τ (η))Δη τ (ξn )

σ (ζn )  x(τ (ζn )) p(η)Δη 2 τ (ξn )   2 ≥ x(τ (ζn )). 2

≥

Therefore x(τ (ζn )) ≤ x(ζn )

2 2 

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

and lim inf t→∞

x(τ (t)) < ∞. x(t)

This completes the proof. Example 6.7. Let T = 2N0 . Consider the equation 7 x x (t) + 64t Δ

 t = 0, 2

t ∈ [2, ∞),

x(1) = 1. Here σ (t) = 2t, 7 , 8t t τ (t) = , t ∈ T. 2

p(t) =

We will prove that x(t) =

1 , t3

t ∈ T,

is its solution. We have x Δ (t) = −

(σ (t))2 + tσ (t) + t 2 t 3 (σ (t))3

=−

(2t)2 + 2t 2 + t 2 t 3 (2t)3

=−

4t 2 + 2t 2 + t 2 8t 6

7t 2 8t 6 7 = − 4, 8t

 1 t =  3 x t 2 =−

2

=

8 , t3

289

290

6 Oscillations of First-Order Functional Dynamic Equations



 t 7 8 7 =− 4 + 2 64t t 3 8t

7 x x (t) + 64t Δ

7 7 + 4 4 8t 8t = 0, t ∈ [2, ∞). =−

Next, t

p(s)Δs =

τ (t)

t

7 Δs 64s

t 2

t ∈ [4, ∞),

> 0, and lim inf t→∞

  x 2t x(τ (t)) = lim inf t→∞ x(t) x(t) = lim inf t→∞

8 t3 1 t3

= 8. Let t0 ∈ T. Define  αn (t) =

1 if

n=0 inf

λ>0,1−λpμαn=1 ∈R + ([τ (t),t))

1 λe−λpαn−1 (t,τ (t))

if n ∈ N

for t ∈ [s, ∞), where τ n (s) ∈ [t0 , ∞). Theorem 6.11. Let x be a nonoscillatory solution of the equation (6.1). If there exists an n0 ∈ N such that lim inf αn0 (t) > 1, t→∞

then x(τ (t)) = ∞. t→∞ x(t) lim

Proof. Without loss of generality, we suppose that x is an eventually positive solution. Then x is an eventually nonincreasing solution of the equation (6.1). Therefore there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

291

Hence, using that x is eventually nonincreasing, we can increase t1 so that t ∈ [t1 , ∞).

x(τ (t)) ≥ x(t), Then x(τ (t)) ≥ 1, x(t)

t ∈ [t1 , ∞).

We rewrite the equation (6.1) in the form x Δ (t) +

x(τ (t)) p(t)x(t) = 0, x(t)

t ∈ [t1 , ∞).

(6.12)

Integrating both sides of the equation (6.12) from t to σ (t), t ∈ [t1 , ∞), we get σ (t)

0=

σ (t)

x Δ (s)Δs +

t

t

= x(σ (t)) − x(t) + μ(t)

x(τ (s)) p(s)x(s)Δs x(s)

x(τ (t)) p(t)x(t) x(t)

x(τ (t)) p(t)x(t) x(t)

 x(τ (t)) = −x(t) 1 − μ(t) p(t) , x(t) ≥ −x(t) + μ(t)

t ∈ [t1 , ∞),

whereupon 1 − μ(t)

x(τ (t)) p(t) ≥ 0, x(t)

t ∈ [t1 , ∞).

Let yx (t) =

x(τ (t)) , x(t)

t ∈ [t1 , ∞).

Then 1 − μyx p ∈ R + ([t1 , ∞)). By (6.12), we get x Δ (t) = −yx (t)p(t)x(t),

t ∈ [t1 , ∞),

292

6 Oscillations of First-Order Functional Dynamic Equations

and hence, t ∈ [t1 , ∞).

x(t) = x(t1 )e−yx p (t, t1 ),

We take t2 ∈ [t1 , ∞) such that τ (t2 ) ∈ [t1 , ∞). Then x(t) = x(τ (t))e−yx p (t, τ (t)),

t ∈ [t2 , ∞),

and 1 , e−yx p (t, τ (t))

yx (t) =

t ≥ t2 .

Define

zn (t) =

yx (t) if n = 0 inf{zn−1 (η) : η ∈ [τ (t), t)} if n ∈ N.

We have yx (η) ≥ z1 (t),

η ∈ [τ (t), t),

t ∈ [t2 , ∞).

Hence, −z1 p ∈ R + ([τ (t), t)),

t ∈ [t2 , ∞).

Therefore yx (t) =

1 e−yx p (t, τ (t))

≥

1 e−z1 p (t, τ (t))

=

z1 (t) z1 (t)e−z1 p (t, τ (t))

≥ α1 (t)z1 (t),

t ∈ [t2 , ∞).

Therefore, using (6.12), we get 0 = x Δ (t) + yx (t)p(t)x(t) ≥ x Δ (t) + z1 (t)p(t)α1 (t)x(t),

t ∈ [t2 , ∞).

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

293

Integrating both sides from t to σ (t), t ∈ [t2 , ∞), we get σ (t)

0≥

σ (t)

x Δ (s)Δs +

t

z1 (s)p(s)α1 (s)x(s)Δs t

= x(σ (t)) − x(t) + μ(t)z1 (t)p(t)α1 (t)x(t) ≥ −x(t) + μ(t)z1 (t)p(t)α1 (t)x(t) = −x(t) (1 − μ(t)z1 (t)p(t)α1 (t)) ,

t ∈ [t2 , ∞),

whereupon −z1 pα1 ∈ R + ([t2 , ∞)). Because z2 (t) ≤ y2 (η),

η ∈ [τ (t), t),

t ∈ [t2 , ∞),

we conclude that there exists t3 ∈ [t2 , ∞) such that τ (t3 ) ≥ t2 and −z2 pα1 ∈ R + ([τ (t), t)), t ∈ [t3 , ∞). Thus yx (t) =

1 e−yx p (t, τ (t))

≥

1 e−z1 pα1 (t, τ (t))

≥

1 e−z2 pα2 (t, τ (t))

=

z2 (t) z2 (t)e−z2 pα2 (t, τ (t))

≥ α2 (t)z2 (t),

t ∈ [t3 , ∞).

Proceeding in this manner, we get an increasing sequence {tn }n∈N of points of T such that yx (t) ≥ zn (t)αn (t),

t ∈ [tn+1 , ∞).

Now we assume that lim inf yx (t) < ∞. t→∞

Then lim inf yx (t) ≥ lim inf (zn (t)αn (t)) t→∞

t→∞

≥ lim inf zn (t) lim inf αn (t) t→∞

t→∞

= lim inf yx (t) lim inf αn (t), t→∞

t→∞

n ∈ N.

294

6 Oscillations of First-Order Functional Dynamic Equations

Therefore lim inf αn (t) ≤ 1, t→∞

n ∈ N.

This is a contradiction. This completes the proof. Theorem 6.12. Assume that t

p(η)Δη > 0

lim sup t→∞

τ (t)

and there exists an n0 ∈ N such that lim inf αn0 (t) > 1. t→∞

Then every solution of (6.1) oscillates on [t0 , ∞). Proof. Assume that the equation (6.1) has a nonoscillatory solution x. Then, using Theorem 6.10, we have lim inf t→∞

x(τ (t)) < ∞. x(t)

On the other hand, using Theorem 6.11, we get lim inf t→∞

x(τ (t)) = ∞, x(t)

which is a contradiction. This completes the proof. Define βn (t) = sup{αn−1 (η) : η ∈ [τ (t), t)} for n ∈ N, t ∈ [s, ∞), where τ n (s) ∈ [t0 , ∞). Theorem 6.13. If there exists an n0 ∈ N such that

 1 1 1− > 0, lim sup αn0 (t) t→∞ βn0 (t) then t

p(η)Δη > 0.

lim sup t→∞

τ (t)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

295

Proof. We have that there exists a t1 ∈ [t0 , ∞) such that −pαn0 −1 ∈ R + ([t1 , ∞)). Then αn0 (t) ≤ ≤ ≤

1 e−pαn0 −1 (t, τ (t)) 1−

1

#t

τ (t) p(η)αn0 −1 (η)Δη

1 − βn0 (t)

1 #t

,

t ∈ [t1 , ∞),

1 , αn0 (t)

t ∈ [t1 , ∞),

p(η)Δη,

t ∈ [t1 , ∞),

τ (t) p(η)Δη

whereupon t

1 − βn0 (t)

p(η)Δη ≤

τ (t)

or 1−

1 ≤ βn0 (t) αn0 (t)

t τ (t)

or

 1 1 1− , p(η)Δη ≥ βn0 (t) αn0 (t) τ (t) t

t ∈ [t1 , ∞).

Therefore t

p(η)Δη > 0,

t ∈ [t1 , ∞).

τ (t)

This completes the proof. Theorem 6.14. Assume that there exists an n0 ∈ N such that

 1 1 1− >0 lim sup αn0 (t) t→∞ βn0 (t) and let lim inf αn0 (t) > 1. t→∞

Then every solution of (6.1) is oscillatory on [t0 , ∞).

296

6 Oscillations of First-Order Functional Dynamic Equations

Proof. Assume that (6.1) has a nonoscillatory solution x on [t0 , ∞). Then, using Theorem 6.13, we have t

p(η)Δη > 0.

lim sup t→∞

τ (t)

Hence from Theorem 6.10, we conclude that lim inf t→∞

x(τ (t)) < ∞. x(t)

On the other hand, using lim inf αn0 (t) > 1 t→∞

and Theorem 6.11, we obtain that lim

t→∞

x(τ (t)) = ∞, x(t)

which is a contradiction. This completes the proof. Example 6.8. Let T = Z, p ∈ Crd (T), p > 0 on T, τ (t) = t − τ , τ ∈ N, t ∈ T,

pn (t) =

⎧ ⎪ ⎨1 ⎪ ⎩

if n = 0

t−1 "

η=t−τ

p(η)pn−1 (η) if n ∈ N.

Then α1 (t) =

inf λ>0 1 − λp(η) > 0 η ∈ [t − τ, t − 1]

=

1 λe−λp (t, t − τ )

inf λe λ>0 1 − λp(η) > 0 η ∈ [t − τ, t − 1]

=

inf λe λ>0 1 − λp(η) > 0 η ∈ [t − τ, t − 1]

1

#t t−τ

log(1−λp(s))Δs

"t−1 η=t−τ

1 log(1−λp(η))

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

=

/t−1

inf

 " t−1 1

λ λ>0 1 − λp(η) > 0 η ∈ [t − τ, t − 1]

≥

λ λ>0 1 − λp(η) > 0 η ∈ [t − τ, t − 1] = inf

λ>0

1

inf

η=t−τ (1 − λp(η))

τ

1

η=t−τ (1 − λp(η))

τ

1  τ . λ λ 1 − τ p1 (t)

Let −τ

1 λ g(λ) = , 1 − p1 (t) λ τ

λ > 0.

Then

 −τ −τ −1 λ p1 (t) λ τ 1 − p1 (t) − 1 − p1 (t) − τ λ τ τ

−τ −τ −1 λ λ 1 p1 (t) 1 − p1 (t) = − 2 1 − p1 (t) + τ λ τ λ

−τ −1   1 λ λ 1 − 1 − p1 (t) 1 − p1 (t) + p1 (t) = λ τ λ τ

−τ −1  λ λ 1 1 − p1 (t) − λp1 (t) = − 2 1 − p1 (t) τ τ λ

 −τ −1

λ 1 1 1 − λp1 (t) 1 + , λ > 0. = − 2 1 − p1 (t) τ τ λ

1 g (λ) = − 2 λ

Hence, g (λ) = 0

⇐⇒

λ=

τ , (τ + 1)p1 (t)

λ=

τ . p1 (t)

297

298

6 Oscillations of First-Order Functional Dynamic Equations

Then

−τ 1 (τ + 1)p1 (t) 1− τ τ +1

 (τ + 1)p1 (t) τ + 1 τ = τ τ τ +1

τ +1 = p1 (t) τ

g(λ) ≥

and

α1 (t) ≥

τ +1 τ

τ +1 p1 (t).

Now we will estimate α2 (t). We have α2 (t) =

inf λ>0 1 − λp(η)α1 (η) > 0 η ∈ [t − τ, t − 1]

=

1 λe−λpα1 (t, τ (t))

inf λ>0 1 − λp(η)α1 (η) > 0 η ∈ [t − τ, t − 1]

=

λe

inf λe λ>0 1 − λp(η)α1 (η) > 0 η ∈ [t − τ, t − 1]

=

inf λ>0 1 − λp(η)α1 (η) > 0 η ∈ [t − τ, t − 1]

≥

λ

1

#t t−τ

log(1−λp(η)α1 (η))Δη

"t−1 η=t−τ

1 log(1−λp(η)α1 (η))

1

/t−1

η=t−τ (1 − λp(η)α1 (η))

inf λ>0 1−λ



τ +1 τ

τ +1

λ p(η)p1 (η) > 0

η ∈ [t − τ, t − 1]

/t−1 η=t−τ

1−λ



1 τ +1 τ

τ +1

 p1 (η)p(η)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

≥

inf λ>0 1−λ



τ +1 τ

τ +1

λ

1 τ

"t−1 η=t−τ

1−λ

1 

τ +1 τ

τ +1

299

τ p1 (η)p(η)

p(η)p1 (η) > 0

η ∈ [t − τ, t − 1] ≥ inf

λ>0

λ 1−λ



1 τ +1 τ

τ +1

τ . p2 (t)

Let &−τ %

 1 λ τ + 1 τ +1 g1 (λ) = p2 (t) , 1− λ τ τ

λ > 0.

Then 1 g1 (λ) = − 2 λ

%

λ 1− τ

τ +1 τ

&−τ

τ +1 p2 (t)

% &−τ −1 %

 &

 τ λ τ + 1 τ +1 1 τ + 1 τ +1 − p2 (t) p2 (t) 1 − − λ τ τ τ τ &−τ −1 %

 1 λ τ + 1 τ +1 =− p2 (t) 1− λ τ τ % % & &

  1 τ + 1 τ +1 λ τ + 1 τ +1 × p2 (t) − p2 (t) , 1− λ τ τ τ λ > 0. Hence, g1 (λ) = 0

λ=τ

τ τ +1

⇐⇒ τ +1

1 p2 (t)

300

6 Oscillations of First-Order Functional Dynamic Equations

and 1 1 − λ τ

τ +1 τ

τ +1

1 − λ



p2 (t) −

τ +1 τ

λ=

τ +2

τ τ +1

τ +1 τ

τ +1

p2 (t) = 0

τ +2

p2 (t) = 0

⇒

⇒

1 . p2 (t)

From here,

τ +1 g1 (λ) ≥ p2 (t) τ

= p2 (t)

τ +1 τ

τ +1 = p2 (t) τ

τ +2 %

τ +2  &−τ τ + 1 τ +1 τ 1 1− τ τ +1 τ τ +2 1− 2(τ +1)

1 τ +1

−τ

.

Then

α2 (t) ≥

τ +1 τ

2(τ +1) p2 (t).

Assume that

αn−1 (t) ≥

τ +1 τ

(n−1)(τ +1) pn−1 (t)

for some n ∈ N. We will prove that

αn (t) ≥

τ +1 τ

We have αn (t) =

inf λ>0 1 − λp(η)αn−1 (η) > 0 η ∈ [t − τ, t − 1]

1 λe−λpαn−1 (t, τ (t))

n(τ +1) pn (t).

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations =

inf λe

λ>0

301

1

#t t−τ

log(1−λp(η)αn−1 (η))Δη

1 − λp(η)αn−1 (η) > 0 η ∈ [t − τ, t − 1] =

inf λe

λ>0

1

"t η=t−τ

log(1−λp(η)αn−1 (η))

1 − λp(η)αn−1 (η) > 0 η ∈ [t − τ, t − 1] =

inf

λ

λ>0

/t−1

1

η=t−τ (1 − λp(η)αn−1 (η))

1 − λp(η)αn−1 (η) > 0 η ∈ [t − τ, t − 1] ≥

inf λ>0 1−λ



τ +1 τ

(n−1)(τ +1)

λ

/t−1 η=t−τ

1

  (n−1)(τ +1) τ +1 1−λ τ pn−1 (η)p(η)

pn−1 (η)p(η) > 0

η ∈ [t − τ, t − 1] ≥

inf λ>0 1−λ



τ +1 τ

(n−1)(τ +1)

λ

1 τ

"t−1 η=t−τ

1−λ



τ +1 τ

1 (n−1)(τ +1)

τ pn−1 (η)p(η)

pn−1 (η)p(η) > 0

η ∈ [t − τ, t − 1] ≥ inf

λ>0

λ 1−

λ τ



τ +1 τ

1 (n−1)(τ +1)

τ . pn (t)

Let &−τ %

 1 λ τ + 1 (n−1)(τ +1) pn (t) , g2 (λ) = 1− λ τ τ

λ > 0.

Then g2 (λ)

1 =− 2 λ

%

λ 1− τ

τ +1 τ

(n−1)(τ +1)

&−τ pn (t)

&% &−τ −1 %

  τ + 1 (n−1)(τ +1) τ 1 τ + 1 (n−1)(τ +1) − pn (t) 1−λ pn (t) − λ τ τ τ

302

6 Oscillations of First-Order Functional Dynamic Equations

&−τ −1 %

 1 λ τ + 1 (n−1)(τ +1) =− pn (t) 1− λ τ τ % % & &

  1 τ + 1 (n−1)(τ +1) λ τ + 1 (n−1)(τ +1) × pn (t) − pn (t) , 1− λ τ τ τ

λ > 0. Hence, g2 (λ) = 0 1 1 − λ τ

τ +1 τ

(n−1)(τ +1)

1 1+τ − λ τ



pn (t) −

τ +1 τ

λ=

⇐⇒ τ +1 τ

(n−1)(τ +1)

τ τ +1

(n−1)(τ +1)

pn (t) = 0

(n−1)τ +n

pn (t) = 0

⇒

⇒

1 pn (t)

and

λ=

τ τ +1

(n−1)(τ +1)

τ . pn (t)

Then

g2 (λ) ≥

τ +1 τ %

(n−1)τ +n pn (t)

1 × 1− τ

=

=

=

τ

τ τ +1

(n−1)τ +n

(n−1)τ +n

1 pn (t)

τ +1 τ

1−

and

αn (t) ≥

τ +1 τ

&−τ

(n−1)(τ +1)

−τ 1 pn (t) τ +1

  + 1 (n−1)τ +n τ + 1 τ pn (t) τ τ  + 1 n(τ +1) pn (t), τ

τ +1 τ τ

n(τ +1) pn (t).

pn (t)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

303

Therefore every solution of the equation Δx(t) + p(t)x(t − τ ) = 0 oscillates on [t0 , ∞) if there exists an n0 ∈ N such that

τ +1 lim inf t→∞ τ

n0 (τ +1) pn0 (t) > 1

or

lim inf pn0 (t) > t→∞

τ τ +1

n0 (τ +1) .

Example 6.9. Let T = q N0 , τ (t) = q −τ t, q > 1, τ ∈ N, p ∈ Crd (T), p(t) > 0, t ∈ T, ⎧ ⎨ 1 if n = 0   τ " pn (t) = t t t p p η η ⎩ (q − 1) n−1 q q qη

if n ∈ N.

η=1

Then #t   e−p t, q −τ t = e q −τ t "τ

1 (q−1)s

log(1−(q−1)sp(s))Δs

  1−(q−1)p qtη qtη

=e

  τ $ t t 1 − (q − 1)p . = qη qη η=1

η=1

Hence,

  τ $   t t 1 − λ(q − 1)p λe−λp t, q −τ t = λ η q qη η=1

⎛

⎞

  τ τ  1 t t ⎠ 1 − λ(q − 1)p ≤ λ⎝ τ qη qη ⎛

η=1

⎞τ

 τ  t t λ ⎠ = λ ⎝1 − (q − 1) p τ qη qη η=1

τ λ = λ 1 − p1 (t) , τ

λ > 0.

304

6 Oscillations of First-Order Functional Dynamic Equations

Let

τ λ g1 (λ) = λ 1 − p1 (t) , τ

λ > 0.

Then 

τ

τ −1 λ λ p1 (t) g1 (λ) = 1 − p1 (t) + λ − τ 1 − p1 (t) τ τ τ

τ −1  λ λ 1 − p1 (t) − λp1 (t) = 1 − p1 (t) τ τ

τ −1  τ +1 λ 1− λp1 (t) , λ > 0. = 1 − p1 (t) τ τ Hence,

τ τ 1 τ p1 (t) 1− g1 (λ) ≤ (τ + 1)p1 (t) τ (τ + 1)p1 (t)

τ 1 τ 1− = (τ + 1)p1 (t) τ +1 τ +1

1 τ . = τ +1 p1 (t) Therefore

α1 (t) ≥

τ +1 τ

τ +1 p1 (t).

Now we will make an estimation of α2 . We have

   τ $ t t t 1 − λ(q − 1)p α1 λe−λpα1 (t, τ (t)) = λ qη qη qη η=1

⎛

⎞

   τ τ  1 t t t ⎠ 1 − λ(q − 1)p α1 ≤ λ⎝ τ qη qη qη ⎛

η=1

⎞τ

  τ  t t t λ ⎠ α1 = λ ⎝1 − (q − 1)p τ qη qη qη η=1

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

305

⎛

⎞τ τ +1 

 τ  τ +1 t t ⎠ t λ ≤ λ ⎝1 − (q − 1)p p1 τ qη τ qη qη η=1

⎛

λ = λ ⎝1 − τ %

λ = λ 1− τ

τ +1 τ τ +1 τ

⎞τ

  τ +1  τ t t t ⎠ (q − 1)p p1 qη qη qη η=1

&τ

τ +1 p2 (t)

λ > 0.

,

Let %

λ g2 (λ) = λ 1 − τ

τ +1 τ

&τ

τ +1 p2 (t)

,

λ > 0.

Then %

λ g2 (λ) = 1 − τ

τ +1 τ

τ +1 −λ τ %

λ = 1− τ %

λ × 1− τ %

λ = 1− τ

p2 (t)

τ +1

τ +1 τ

%

λ p2 (t) 1 − τ

τ +1 τ

&τ −1

τ +1 p2 (t)

&τ −1

τ +1 p2 (t)

τ +1 τ

τ +1 τ

&τ

τ +1

τ +1

τ +1 p2 (t) − λ τ &τ −1 %

τ +1 p2 (t)

&

τ +1

p2 (t)

τ +1 1−λ τ

τ +2

& p2 (t) ,

λ > 0. Hence,

g2 (λ) ≤

=

=

τ τ +1 τ τ +1 τ τ +1

τ +2

&τ %

τ +2

 τ τ + 1 τ +1 1 1 1 p2 (t) 1− p2 (t) τ τ +1 p2 (t) τ

τ +2

τ 1 1 1− p2 (t) τ +1

2(τ +1)

1 p2 (t)

306

6 Oscillations of First-Order Functional Dynamic Equations

and

α2 (t) ≥

τ +1 τ

2(τ +1) p2 (t).

Assume that

αn−1 (t) ≥

τ +1 τ

(n−1)(τ +1) pn−1 (t)

for some n ∈ N. We will prove that

αn (t) ≥

τ +1 τ

n(τ +1) pn (t).

We have λe−λpαn−1 (t, τ (t)) = λ

τ $

1 − λ(q − 1)p

η=1

t qη



αn−1

t qη



t qη



⎛

⎞



  τ τ  1 t t t ⎠ ≤ λ⎝ 1 − λ(q − 1)p αn−1 τ qη qη qη η=1

⎛

 λ = λ ⎝1 − (q − 1) p τ τ

η=1

t qη



αn−1

t qη



⎞τ t ⎠ qη

⎛

⎞τ





(n−1)(τ +1)  τ λ t t t τ + 1 ⎠ ≤ λ ⎝1 − (q − 1) p pn−1 τ τ qη qη qη η=1

% = λ 1−

λ τ

τ +1 τ

&τ

(n−1)(τ +1)

pn (t)

λ > 0.

,

Let %

λ g3 (λ) = λ 1 − τ

τ +1 τ

&τ

(n−1)(τ +1) pn (t)

,

λ > 0.

Then %

g3 (λ)

λ = 1− τ



τ +1 τ

τ +1 −λ τ

&τ

(n−1)(τ +1)

(n−1)(τ +1)

pn (t) %

λ pn (t) 1 − τ

τ +1 τ

&τ −1

(n−1)(τ +1) pn (t)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

%

λ = 1− τ %

λ × 1− τ

τ +1 τ

307

&τ −1

(n−1)(τ +1) pn (t)

τ +1 τ

(n−1)(τ +1)

τ +1 pn (t) − λ τ

&

(n−1)(τ +1) pn (t)

%

&τ −1

 λ τ + 1 (n−1)(τ +1) = 1− pn (t) τ τ % & 

τ + 1 (n−1)(τ +1)+1 × 1−λ pn (t) , τ

λ > 0.

Then

g3 (λ) ≤

τ τ +1 %

(n−1)(τ +1)+1

1 × 1− τ

=

=

τ τ +1 τ τ +1

τ τ +1

(n−1)(τ +1)+1

(n−1)(τ +1)+1 n(τ +1)

1 pn (t) 1 pn (t)

τ +1 τ

τ 1 1 1− pn (t) τ +1

1 pn (t)

and

αn (t) ≥

τ +1 τ

n(τ +1) pn (t).

Hence, every solution of the equation

x (t) + p(t)x Δ

t qτ

 =0

is oscillatory on [t0 , ∞) if there exists an n0 ∈ N such that

lim inf pn0 (t) > t→∞

τ τ +1

n0 (τ +1) .

&τ

(n−1)(τ +1) pn (t)

308

6 Oscillations of First-Order Functional Dynamic Equations

Exercise 6.4. Let T = {ξm : m ∈ N}, τ (ξm ) = ξm−τ , where {ξm }m∈N is an increasing divergent sequence and τ ∈ N. Let also, p ∈ Crd (T), p(t) > 0, t ∈ T, ⎧ ⎨ 1 if n = 0 m−1  "  pn (ξm ) = ⎩ ξη+1 − ξη p(ξη )pn−1 (ξη )

if

n=m−τ

n ∈ N.

1. Prove that m−1 $

λe−λp (ξm , ξm−1 ) = λ

  1 − λ(ξη+1 − ξη )p(ξη ) ,

λ > 0.

η=m−τ

2. Prove that

αn (ξm ) ≥

τ τ +1

n(τ +1)

n ∈ N.

pn (ξm ),

3. Prove that every solution of the equation x Δ (ξm ) + p(ξm )x(ξm−τ ) = 0 is oscillatory on [ξτ , ∞) if there exists an n0 ∈ N such that

lim sup pn0 (ξm ) ≥ t→∞

τ τ +1

n0 (τ +1) .

Below we suppose that τ : T → T is nondecreasing, τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. Theorem 6.15. Assume that f : T → R is rd-continuous, and g : T → R is nonincreasing. If v < u, then σ (u)

σ (u)

f (s)g(τ (s))Δs ≥ g(τ (u))

v

f (s)Δs. v

Proof. Since v < u, we have σ (u) v

u

f (s)g(τ (s))Δs =

σ (u)

f (s)g(τ (s))Δs +

v

f (s)g(τ (s))Δs u

u

≥ g(τ (u)) v

σ (u)

f (s)Δs +

f (s)g(τ (s))Δs u

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations u

= g(τ (u))

309

f (s)Δs + μ(u)f (u)g(τ (u))

v u

= g(τ (u))

σ (u)

f (s)Δs + g(τ (u))

v

f (s)Δs u

%

u

= g(τ (u))

σ (u)

f (s)Δs +

v

&

f (s)Δs u

σ (u)

= g(τ (u))

f (s)Δs. v

This completes the proof. Theorem 6.16. Assume that f, g : T → R are rd-continuous and τ : T → T is rd-continuous. Then σ (u)

σ (u)

f (s)g(τ (s))Δs = g(τ (u))

u

f (s)Δs,

u ∈ T.

u

Proof. Let u ∈ T. Then σ (u)

f (s)g(τ (s))Δs = μ(u)f (u)g(τ (u))

u σ (u)

= g(τ (u))

f (s)Δs. u

This completes the proof. Theorem 6.17. Let p ∈ Crd (T), p(t) > 0, and t ∈ T. Assume that x is an eventually positive solution of the equation (6.1) and let there exists a positive constant M such that t

lim inf t→∞

p(s)Δs > M. τ (t)

Then 4 x(τ (t)) ≤ 2 x(t) M for all large t.

310

6 Oscillations of First-Order Functional Dynamic Equations

Proof. Let t0 ∈ T be sufficiently large such that x(t) > 0 for t ∈ [t0 , ∞). Let t1 = τ −1 (t0 ). Then x Δ (t) ≤ 0 for t ∈ [t1 , ∞). Let t2 ∈ [τ −1 (t1 ), ∞) be chosen so that t

p(s)Δs ≥ M,

t ∈ [t2 , ∞).

τ (t)

Using that τ −1 (t) > t,

t ∈ [t2 , ∞),

we get τ −1 (t)

p(s)Δs ≥ M,

t ∈ [t2 , ∞).

t

Define r

G(r) =

p(s)Δs −

t

M , 2

r ∈ [t, τ −1 (t)],

t ∈ [t2 , ∞).

We have G : [t, τ −1 (t)] → R, t ∈ [t2 , ∞), is continuous and nondecreasing. Also, G(t) = −

M 2

< 0,  G τ −1 (t) =

τ −1 (t)

p(s)Δs −

t

≥M−

M 2

M 2

M 2 > 0, t ∈ [t2 , ∞). =

Then there exists a t∗ ∈ [t, τ −1 (t)], t ∈ [t2 , ∞), such that G(t∗ ) ≤ 0,

G(σ (t∗ )) > 0.

Then 0 ≥ G(t∗ ) t∗

= t

p(s)Δs −

M , 2

t ∈ [t2 , ∞),

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

or t∗

p(s)Δs ≤

t

M , 2

t ∈ [t2 , ∞),

and 0 < G(σ (t∗ )) σ (t∗ )

=

p(s)Δs −

t

M , 2

t ∈ [t2 , ∞),

or σ (t∗ )

p(s)Δs > t

M , 2

t ∈ [t2 , ∞).

From here, σ (t)

p(s)Δs ≥

τ (t∗ )

t

p(s)Δs τ (t∗ )

=

t∗

t

p(s)Δs +

p(s)Δs

τ (t∗ )

=

t∗

t∗ t∗

p(s)Δs −

p(s)Δs

τ (t∗ )

≥M− =

M , 2

t

M 2 t ∈ [t2 , ∞).

Hence, σ (t)

p(s)x(τ (s))Δs ≥ x(τ (t))

τ (t∗ )

σ (t)

p(s)Δs τ (t∗ )

≥

M x(τ (t)), 2

t ∈ [t2 , ∞),

and σ (t∗ ) t

p(s)x(τ (s))Δs ≥ x(τ (t∗ )) >

σ (t∗ )

p(s)Δs t

M x(τ (t∗ )), 2

t ∈ [t2 , ∞).

311

312

6 Oscillations of First-Order Functional Dynamic Equations

Then x(t) ≥ x(t) − x(σ (t∗ )) σ (t∗ )

=−

x Δ (s)Δs

t σ (t∗ )

=

p(s)x(τ (s))Δs t

M x(τ (t∗ )) 2 M ≥ (x(τ (t∗ )) − x(σ (t))) 2

>

=− = ≥

σ (t)

M 2

M 2

x Δ (s)Δs

τ (t∗ ) σ (t)

p(s)x(τ (s))Δs τ (t∗ )

M2 x(τ (t)), 4

t ∈ [t2 , ∞).

Hence, 4 x(τ (t)) ≤ 2, x(t) M

t ∈ [t2 , ∞).

This completes the proof. Example 6.10. Let T = Z. Consider the equation x Δ (t) +

t2 + 1 x(t − 2) = 0, t2 + 5

Here σ (t) = t + 1, t2 + 1 t2 + 5 > 0,

p(t) =

τ (t) = t − 2,

t ∈ T.

t ∈ Z.

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

313

Then t

p(s)Δs =

τ (t)

t t−2

s2 + 1 Δs s2 + 5

=

(t − 2)2 + 1 (t − 1)2 + 1 + (t − 2)2 + 5 (t − 1)2 + 5

=

t 2 − 4t + 5 t 2 − 2t + 2 + , t 2 − 4t + 9 t 2 − 2t + 6

t ∈ T,

and

t

p(s)Δs = lim inf

lim inf t→∞

t→∞

t−2

t 2 − 4t + 5 t 2 − 2t + 2 + t 2 − 4t + 9 t 2 − 2t + 6



= 1+1 = 2. Let M ∈ (0, 2) be arbitrarily chosen and x is an eventually positive solution of the considered equation. Then t

lim inf t→∞

p(s)Δs > M t−2

and 4 x(t − 2) ≤ 2 x(t) M for all large t. Example 6.11. Let T = 2N0 . Consider the equation

x Δ (t) + t 2 x

t 22

 = 0,

t ∈ [4, ∞).

Here p(t) = t 2 , σ (t) = 2t, t τ (t) = 2 , 2

t ∈ [4, ∞).

314

6 Oscillations of First-Order Functional Dynamic Equations

Then t

t

p(s)Δs =

t 22

τ (t)

s 2 Δs

t 22

2

=

t 22

=

t3 t3 + 26 23

=

9t 3 , 64

+

t 2

2 t 2

t ∈ T.

Hence, t

lim inf t→∞

t 22

p(s)Δs = ∞.

Therefore, if x is an eventually positive solution of the considered equation, then for any positive constant M there exists a t1 ∈ T such that  x 2t2 4 ≤ 2 x(t) M for any t ∈ [t1 , ∞). Exercise 6.5. Let T = 2N0 . Consider the equation

 t t = 0, t ∈ [2, ∞), x x Δ (t) + 2 2 t +1 x(t) = 1,

t ∈ [1, 2).

 Suppose that x is an eventually positive solution. Prove that for any M ∈ 0, 12 there exists t1 ∈ T such that   x 2t 4 ≤ 2 x(t) M for any t ∈ [t1 , ∞). Theorem 6.18. If σ (t)

lim sup t→∞

then the equation (6.1) is oscillatory.

p(s)Δs > 1, τ (t)

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

315

Proof. Assume that the equation (6.1) is not oscillatory. Let x be a nonoscillatory solution of the equation (6.1). Without loss of generality, we assume that x is an eventually positive solution. Then there exists t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Hence, x Δ (t) < 0,

t ∈ [t1 , ∞).

Integrating both sides of the equation (6.1) from τ (t) to σ (t), we get σ (t)

x Δ (s)Δs +

τ (t)

σ (t)

p(s)x(τ (s))Δs = 0,

t ∈ [t1 , ∞),

τ (t)

whereupon 0 = x(σ (t)) − x(τ (t)) +

σ (t)

p(s)x(τ (s))Δs τ (t)

≥ x(σ (t)) − x(τ (t)) + x(τ (t)) % = x(σ (t)) + x(τ (t))

σ (t)

p(s)Δs τ (t)

σ (t)

&

p(s)Δs − 1 ,

t ∈ [t1 , ∞)

τ (t)

This is a contradiction. This completes the proof. Example 6.12. Let T = 2Z. Consider the equation x Δ (t) + (t 2 + t)x(t − 4) = 0,

t ∈ T.

Here σ (t) = t + 2, τ (t) = t − 4, p(t) = t 2 + t,

t ∈ T.

Then σ (t) τ (t)

p(s)Δs =

t+2 t−4

(s 2 + s)Δs

   = 2 (t − 4)2 + (t − 4) + 2 (t − 2)2 + t − 2 + 2 t 2 + t

316

6 Oscillations of First-Order Functional Dynamic Equations

  = 2 t 2 − 8t + 16 + t − 4 + 2 t 2 − 4t + 4 + t − 2  +2 t 2 + t    = 2 t 2 − 7t + 12 + 2 t 2 − 3t + 2 + 2 t 2 + t = 2t 2 − 14t + 24 + 2t 2 − 6t + 4 + 2t 2 + 2t = 6t 2 − 18t + 28,

t ∈ T.

Hence, σ (t)

lim sup t→∞

p(s)Δs > 1. τ (t)

Therefore the considered equation is oscillatory. Exercise 6.6. Let T = 4Z. Prove that the equation  x Δ (t) + t 2 − t + 3 x(t − 8) = 0,

t ∈ T,

is oscillatory. Theorem 6.19. Assume that there exists a positive constant M such that t

lim inf t→∞

p(s)Δs > M τ (t)

and t

lim sup t→∞

p(s)Δs > 1 −

τ (t)

M2 . 4

Then the equation (6.1) is oscillatory. Proof. Assume that x is an eventually positive solution of the equation (6.1). Then there exists a t1 ∈ T such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Theorem 6.17, we have that there exists t2 ∈ [t1 , ∞) such that x(τ (t)) 4 ≤ 2, x(t) M

t ∈ [t2 , ∞).

6.2 Iterated Oscillation Criteria for First-Order Functional Dynamic Equations

Integrating both sides of the equation (6.1) from τ (t) to t, we get 0=

t

x Δ (s)Δs +

τ (t)

t

p(s)x(τ (s))Δs τ (t) t

= x(t) − x(τ (t)) +

p(s)x(τ (s))Δs τ (t) t

≥ x(t) − x(τ (t)) + x(τ (t))

p(s)Δs τ (t)

t

x(t) + x(τ (t))

= x(τ (t))

M2

≥ x(τ (t))

4

+

 p(s)Δs − 1

τ (t)

 p(s)Δs − 1 ,

t

t ∈ [t2 , ∞).

τ (t)

This is a contradiction. This completes the proof. Example 6.13. Let T = N. Consider the equation x Δ (t) +

2(t + 2) x(t − 1) = 0, t +3

t ∈ [3, ∞).

Here σ (t) = t + 1, τ (t) = t − 1, p(t) =

2(t + 2) , t +3

t ∈ T.

We have t

p(s)Δs = 2

τ (t)

t t−1

s+2 Δs s+3

t +1 t +2 3 > , t ∈ [3, ∞). 2

=2

Take M=

3 . 2

317

318

6 Oscillations of First-Order Functional Dynamic Equations

We have t

p(s)Δs = lim sup

lim sup t→∞

t→∞

τ (t)

2t + 2 t +3

=2 7 > 16 = 1−

9 16

= 1−

M2 . 4

Therefore the considered equation is oscillatory. Exercise 6.7. Let T = hN, h > 0, a > 0, k ∈ N, k > 1, akh > 1. Prove that the equation 

k−1 x (t) + ax t − = 0, h Δ

 k ,∞ , t∈ h (

is oscillatory.

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations with Several Delays Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Consider the equation y Δ (t) +

n 

pl (t)y(τl (t)) = 0 on

T,

l=1

where (H1) (H2)

pl , l ∈ {1, . . . , n}, are positive rd-continuous functions on T, τl : T → T, τl (t) < t, t ∈ T, l ∈ {1, . . . , n}.

Let t0 ∈ T, τmax (t) =

max τl (t),

l∈{1,...,n}

t ∈ [t0 , ∞).

(6.13)

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations. . .

319

Then [τmax (t1 ), t] =

n ,

[τl (t1 ), t]

l=1

for any t > t1 > τmax (t1 ). Let also, p(t) =

n 

pl (t),

t ∈ T.

l=1

Theorem 6.20. Suppose (H 1), (H 2), and μ(t)p(t) < 1,

t ∈ [t0 , ∞).

Then 1−

"n

n 

t

l=1

τl (t)

l=1 pl (t) , p (t)e −pl (τl (t), t) l=1 l

pl (s)Δs ≤ "n

t ∈ [t0 , ∞).

Proof. By Theorem 6.1, it follows that % n 

t

1−

τmax (t)

& pl (s) Δs ≤ e−p (t, τmax (t)),

t ∈ [t0 , ∞).

l=1

Since 1−

n 

t

l=1

τl (t)

t

pl (s)Δs ≤ 1 −

n 

pl (s)Δs,

t ∈ [t0 , ∞),

τmax (t) l=1

and e−p (t, τmax (t)) =

1

e−p (τmax (t), t) "n l=1 pl (t) = "n l=1 pl (t)e−p (τmax (t), t) "n l=1 pl (t) , t ∈ [t0 , ∞). ≤ "n l=1 pl (t)e−p (τl (t), t)

(6.14)

320

6 Oscillations of First-Order Functional Dynamic Equations

Hence from (6.14), we get 1−

n 

t

l=1

τl (t)

pl (s)Δs ≤ e−p (t, τmax (t)) "n

l=1 pl (t)

≤ "n

l=1 pl (t)e−p (τl (t), t)

,

t ∈ [t0 , ∞).

This completes the proof. Theorem 6.21. Assume (H 1), (H 2) and assume that there exists a λ > 0 such that 1 − λp(t)μ(t) > 0,

t ∈ T,

and there is an A > 0 so that λp(t) ≥ 1, p (t)e −λp (τl (t), t) l=1 l

"n

t ∈ (A, ∞).

Then the equation (6.13) has a positive solution. Proof. Define x(t) =

⎧ ⎨ 1 if ⎩

"n

t < #A

l=1 pl (t)e

− τt (t) ξμ(s) (−λp(s)x(s))Δs l

λp(t)

if

t ∈ [A, ∞).

Then 0 < x(t) ≤ 1,

t ∈ T.

Let z(t) = 1 −

n 

pl (t)e

−

#t

τl (t) ξμ(s) (−λp(s)x(s))Δs

,

l=1

Then z(t) = 1 − λp(t)x(t),

t ∈ [A, ∞).

Since 0 < x(t) ≤ 1,

t ∈ T,

t ∈ T.

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations. . .

we have 1 + μ(t) (z(t) − 1) = 1 − λμ(t)p(t)x(t) ≥ 1 − λμ(t)p(t) t ∈ T.

> 0, Define

y(t) =

1 if t < A ez−1 (t, A) if t ∈ [A, ∞).

Then y(t) = e

#t

A ξμ(s) (−λp(s)x(s))Δs

,

t ∈ [A, ∞).

Hence, y Δ (t) = (z(t) − 1)y(t), y Δ (t) = z(t) − 1, y(t) 1+

y Δ (t) = z(t) y(t) = 1−

n 

pl (t)e

−

#t

τl (t) ξμ(s) (−λp(s)x(s))Δs

l=1

= 1− ×e

n 

 # τl (t) pl (t) e A ξμ(s) (−λp(s)x(s))Δs

l=1 #t − A ξμ(s) (−λp(s)x(s))Δs

= 1−

n  l=1

pl (t)

y(τl (t)) , y(t)

t ∈ [A, ∞),

whereupon  y Δ (t) y(τl (t)) =− , pl (t) y(t) y(t) n

l=1

t ∈ [A, ∞),

321

322

6 Oscillations of First-Order Functional Dynamic Equations

or y Δ (t) = −

n 

pl (t)y(τl (t)),

t ∈ [A, ∞).

l=1

This completes the proof. Theorem 6.22. Assume that (H 1) and (H 2) hold. Let also, the equation (6.13) has an eventually positive solution. Then α = lim sup t0 →∞ t>t0

λp(t) "n ≥ 1. sup p (t)e l −λp (τl (t), t) l=1 λ>0 1 − λμp > 0

Proof.  Let y solves (6.13) and be eventually positive. Assume that α < 1. Let β ∈ 1, α1 . Define τmin (t) =

min τl (t),

l∈{1,...,n}

t ∈ T.

Then e−λp (τl (t), t) ≤ e−λp (τmin (t), t),

t ∈ T,

l ∈ {1, . . . , n}.

Hence, n 

pl (t)e−λp (τl (t), t) ≤ e−λp (τmin (t), t)p(t),

t ∈ T.

l=1

Therefore lim sup

t0 →∞ t>t0

λp(t) ≥ lim sup t0 →∞ t>t0 p (t)e l −λp (τl (t), t) l=1

"n

sup λ>0 1 − λμp > 0

sup λ>0

λp(t) p(t)e−λp (τmin (t), t)

1 − λμp > 0 = lim sup t0 →∞ t>t0

sup

  λe−λp (t, τmin (t))

λ>0 1 − λμp > 0

and then lim sup

t0 →∞ t>t0

  sup λe−λp (t, τmin (t)) < 1. λ>0 1 − λμp > 0

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations. . .

There exists a T0 ∈ T such that β≤

1  , supλ>0,1−μλp>0 λe−λp (t, τmax (t))

and y(t) > 0 for all t ∈ [T0 , ∞). Hence, y(τl (t)) ≥ y(t),

t ∈ [T0 , ∞).

From here, 0 = y (t) + Δ

n 

pl (t)y(τl (t))

l=1

≥ y Δ (t) + p(t)y(t),

t ∈ [T0 , ∞).

Hence from Gronwall’s inequality, we get y(t) ≤ e−p (t, τmax (t))y(τmax (t)),

t ∈ [T0 , ∞),

or y(τmax (t)) 1 ≥ , y(t) e−p (t, τmax (t))

t ∈ [T0 , ∞).

Then there exists T1 ∈ [T0 , ∞) such that y(τmax (t)) ≥ β, y(t)

t ∈ [T1 , ∞).

Thus 0 = y Δ (t) +

n 

p(t)y(τl (t))

l=1

≥ y Δ (t) + βp(t)y(t),

t ∈ [T1 , ∞).

Hence from Gronwall’s inequality, there exists T2 ∈ [T1 , ∞) such that y(τmax (t)) 1 ≥ y(t) e−βp (t, τmax (t)) =

β βe−βp (t, τmax (t))

≥ β 2,

t ∈ [T2 , ∞).

323

324

6 Oscillations of First-Order Functional Dynamic Equations

Proceeding this procedure in this manner, we obtain a sequence {Tn }n∈N ⊂ T such that y(τmax (t)) ≥ β n, y(t)

t ∈ [Tn , ∞).

Because β > 1, we conclude that lim inf t→∞

Let M ∈



1 1 4 , 4α

y(τ (t)) = ∞. y(t)

(6.15)

. By Theorem 6.3, we get that there exists a T ∈ T such that t

p(s)Δs ≥

τmax (t)

1 4α

≥ M,

t ∈ [T , ∞).

Hence, σ (t)

p(s)Δs ≥

τmax (t)

t

p(s)Δs τmax (t)

≥ M,

t ∈ [T , ∞).

Consider the function f : T → R defined by f (u) =

u

p(s)Δs −

τmax (t)

M , 2

t ∈ [T , ∞).

We find f (τmax (t)) = −

M 2

< 0, f (t) =

t

p(s)Δs −

τmax (t)

≥M−

M 2

M 2 > 0, t ∈ [T , ∞). =

M 2

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations. . .

Then there exists t ∗ ∈ [τmax (t), t), t ∈ [T , ∞), such that f (t ∗ ) = 0

f (t ∗ ) < 0

or

and

f (σ (t ∗ )) > 0.

Hence, σ (t ∗ )

M + f (σ (t ∗ )) 2

p(s)Δs =

τmax (t)

M 2

≥ and σ (t) t∗

τmax (t)

p(s)Δs =

t∗

p(s)Δs +

p(s)Δs τmax (t)

t∗

=−

σ (t)

p(s)Δs +

τmax (t)

≥−

σ (t)

p(s)Δs τmax (t)

 M + f (t ∗ ) + M 2

M − f (t ∗ ) 2 M , t ∈ [T , ∞). ≥ 2

=

Then y(t ∗ ) ≥ y(t ∗ ) − y(σ (t)) σ (t)

=− =

t∗

y Δ (s)Δs

n σ (t)  t∗

pl (s)y(τl (s))Δs

l=1

≥ y(τmax (t))

σ (t) t∗

p(s)Δs

M y(τmax (t)) 2  M y(τmax (t)) − y(σ (t ∗ )) ≥ 2 ≥

=−

M 2

σ (t ∗ ) τmax (t)

y Δ (s)Δs

325

326

6 Oscillations of First-Order Functional Dynamic Equations

=

σ (t ∗ )

M 2

n 

M y(τmax (t ∗ )) ≥ 2 =

pl (s)y(τl (s))Δs

τmax (t) l=1 σ (t ∗ )

p(s)Δs τmax (t)

M2 y(τmax (t ∗ )). 4

Hence, y(τmax (t ∗ )) 4 ≤ 2. y(t ∗ ) M Then lim inf t→∞

y(τmax (t)) < ∞, y(t)

which contradicts with (6.15). Therefore α ≥ 1. This completes the proof. Example 6.14. Let T = N. Consider the equation y Δ (t) +

(t 2

t 2 − 2t + 2 2t (t 2 − 4t + 5) y(t − 1) + 2 y(t − 2) = 0, 2 + 1)(t + 2t + 2) (t + 1)(t 2 + 2t + 2) y(t) =

Here σ (t) = t + 1, τ1 (t) = t − 1, τ2 (t) = t − 2, p1 (t) = p2 (t) =

(t 2

t 2 − 2t + 2 , + 1)(t 2 + 2t + 2)

(t 2

2t (t 2 − 4t + 5) , + 1)(t 2 + 2t + 2)

We will prove that y(t) =

1 , t2 + 1

t ∈ T,

t ∈ T.

t ∈ [4, ∞),

1 , t2 + 1

t ∈ [1, 4].

6.3 Oscillations of the Solutions of First-Order Functional Dynamic Equations. . .

327

is a solution of the considered equation. We have y Δ (t) = − =− =− y(t − 1) =

(t 2

σ (t) + t + 1)((σ (t))2 + 1)

t +1+t (t 2 + 1)((t + 1)2 + 1) (t 2

2t + 1 , + 1)(t 2 + 2t + 2)

1 (t − 1)2 + 1

1 , t 2 − 2t + 2 1 y(t − 2) = (t − 2)2 + 1 =

=

t2

1 , − 4t + 5

t ∈ [4, ∞).

Hence, 2t (t 2 − 4t + 5) t 2 − 2t + 2 y(t − 1) + y(t − 2) (t 2 + 1)(t 2 + 2t + 2) (t 2 + 1)(t 2 + 2t + 2)

 2t + 1 1 t 2 − 2t + 2 =− 2 + (t + 1)(t 2 + 2t + 2) (t 2 + 1)(t 2 + 2t + 2) t 2 − 2t + 2

 1 2t (t 2 − 4t + 5) + 2 (t + 1)(t 2 + 2t + 2) t 2 − 4t + 5

y Δ (t) +

=− = 0,

(t 2

1 2t 2t + 1 + 2 + 2 2 2 + 1)(t + 2t + 2) (t + 1)(t + 2t + 2) (t + 1)(t 2 + 2t + 2)

t ∈ [4, ∞).

Consequently lim sup

t0 →∞ t>t0

λ(p1 (t) + p2 (t)) ≥ 1. sup p1 (t)e−λp (t − 1, t) + p2 (t)e−λp (t − 2, t) λ>0 1 − λμp > 0

328

6 Oscillations of First-Order Functional Dynamic Equations

Corollary 6.2. Let (H 1), (H 2) hold. If λp(t) "n < 1, sup l=1 pl (t)e−λp (τl (t), t) λ>0 1 − λμp > 0

lim sup

t0 →∞ t>t0

then all the solutions of the equation (6.13) are oscillatory. Exercise 6.8. Let T = 2N0 . Prove that all the solutions of the equation



 t t + (t + 1)x =0 x (t) + tx 2 4 Δ

are oscillatory.

6.4 Nonoscillations of First-Order Functional Dynamic Equations with Several Delays Consider the equation " x Δ (t) + nl=1 pl (t)x(τl (t)) = f (t), = φ(t), x(t0 ) = x0 , x(t)

t ∈ [t0 , ∞), t ∈ [t−1 , t0 ),

(6.16)

where n ∈ N, t0 ∈ T is the initial point, x0 > 0 is the initial value, φ ∈ Crd ([t−1 , t0 )) is the initial function such that φ has a finite limit at t0 provided that t0 is left-dense, f ∈ Crd ([t0 , ∞)), pl ∈ Crd ([t0 , ∞)), l ∈ {1, . . . , n}, τl is the delay function which satisfies τl ∈ Crd ([t0 , ∞)), limt→∞ τl (t) = ∞, τl (t) ≤ t, t ∈ [t0 , ∞), and t−1 = minl∈{1,...,n} {inft∈[t0 ,∞) τl (t)}. Note that t−1 is finite because limt→∞ τl (t) = ∞, l ∈ {1, . . . , n}. Let D ⊂ R and f : D → R. Then it is understood that f (t) = κD (t)f (t), t ∈ R, where κD is the characteristic function of D. Let χ be the fundamental solution of the equation (6.16). By Theorem 5.21, we have x(t) = x0 χ (t, t0 ) +

t

χ (t, σ (η))f (η)Δη t0

−

n 

t

l=1

t0

χ (t, σ (η))pl (η)φ(τl (η))Δη,

t ∈ [t0 , ∞).

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

329

Theorem 6.23. Let n ∈ N, t0 ∈ T and τl , αl ∈ Crd ([t0 , ∞)), τl (t), αl (t) ≤ t, t ∈ [t0 , ∞), Kl ∈ Crd (T × T) and Kl ≥ 0 on T × T, l ∈ {1, . . . , n}. Let also, f, g ∈ Crd ([t0 , ∞)) satisfy f (t) =

n 

t

l=1

τl (t)

Kl (t, η)f (αl (η))Δη + g(t),

t ∈ [t0 , ∞).

Then the nonnegativity of g on [t0 , ∞) implies the same for f . Proof. Assume the contrary. Set t1 = sup{t ∈ [t0 , ∞) : f (η) ≥ 0,

η ∈ [t0 , t]}.

Suppose that t1 is right-scattered. Then σ (t1 ) > t1 . We have f (t) ≥ 0, t ∈ [t0 , t1 ], and f σ (t1 ) < 0. Hence, f σ (t1 ) = =

n 

σ (t1 )

l=1

τl (σ (t1 ))

n  l=1

+

t1

Kl (σ (t1 ), η)f (αl (η))Δη τl (σ (t1 ))

n  l=1

=

n  l=1

+

Kl (σ (t1 ), η)f (αl (η))Δη + g(σ (t1 ))

σ (t1 )

Kl (σ (t1 ), η)f (αl (η))Δη + g(σ (t1 ))

t1 t1

Kl (σ (t1 ), η)f (αl (η))Δη τl (σ (t1 ))

n 

μ(t1 )Kl (σ (t1 ), t1 )f (αl (t1 )) + g(σ (t1 ))

l=1

≥ 0, which is a contradiction. Therefore t1 is right-dense. Hence, every right neighborhood of t1 contains some points for which f is negative. Thus inf f (η) < 0

η∈[t1 ,t)

for all t ∈ (t1 , ∞). Pick t3 ∈ (t1 , ∞). Then, for any l ∈ {1, . . . , n}, there exists a constant Ml > 0 such that Kl (t, s) ≤ Ml , t ∈ [t1 , t3 ], s ∈ [τl (t), t]. Set M=

n  l=1

Ml .

330

6 Oscillations of First-Order Functional Dynamic Equations

Since f is rd-continuous, we have lim f (t) = f (t1 ).

t→t1+

Then, we may find a t2 ∈ (t1 , t3 ] with τl (t2 ) ≤ t1 , t2 − t1 ≤

l ∈ {1, . . . , n},

1 3M

and inf

η∈[t1 ,t2 )

f (η) ≥ 2f (t2 ),

f (t2 ) < 0.

Since f (t) ≥ 0, t ∈ [t0 , t1 ), we have f (η) =

inf

η∈[t0 ,t2 )

inf f (η).

t∈[t1 ,t2 )

Then f (t2 ) = ≥

n 

t2

l=1

τl (t2 )

n 

t2

l=1

≥

Kl (t2 , η)f (αl (η))Δη t1

% n  l=1

Kl (t2 , η)f (αl (η))Δη + g(t2 )

&

t2

Ml Δη t1

= M(t2 − t1 ) ≥

inf

η∈[t0 ,t2 )

inf

η∈[t0 ,t2 )

f (η)

f (η)

2 f (t2 ), 3

which is a contradiction. This completes the proof. Now we consider the equation x Δ (t) +

n  l=1

pl (t)x(τl (t)) = 0,

t ∈ [t0 , ∞),

(6.17)

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

331

and the following inequalities n 

x Δ (t) +

pl (t)x(τl (t)) ≤ 0,

t ∈ [t0 , ∞),

(6.18)

pl (t)x(τl (t)) ≥ 0,

t ∈ [t0 , ∞),

(6.19)

l=1 n 

x Δ (t) +

l=1

under the assumptions which were formulated for the equation (6.16). Theorem 6.24. Suppose that τl , pl ∈ Crd ([t0 , ∞)), pl : [t0 , ∞) → (0, ∞), l ∈ {1, . . . , n}. Then the following statements are equivalent. (i) The equation (6.17) has an eventually positive solution. (ii) The inequality (6.18) has an eventually positive solution and/or the inequality (6.19) has an eventually negative solution. (iii) There exists a t1 ∈ [t0 , ∞) and a Λ ∈ Crd ([t0 , ∞)), Λ > 0 on [t1 , ∞) and 1 − μ(t)Λ(t) > 0,

t ∈ [t1 , ∞),

and for all t ∈ [t1 , ∞) Λ(t) ≥

n 

pl (t)e−Λ (t, τl (t)).

(6.20)

l=1

(iv) The fundamental solution χ is eventually positive, i.e., there exists a t1 ∈ [t0 , ∞) such that χ (·, s) > 0 on [s, ∞) for any s ∈ [t1 , ∞). Moreover, if (6.20) holds for all t ∈ [t1 , ∞), for some t1 ∈ [t0 , ∞), then χ (·, s) > 0 holds on [s, ∞) for any s ∈ [t1 , ∞). Proof. (i) ⇒ (ii) Let x be an eventually positive solution of the equation (6.17). Then it is an eventually positive solution of (6.18) and −x is an eventually negative solution of (6.19). (ii) ⇒ (iii) Let x be an eventually positive solution of (6.18). Then there exists a t1 ∈ [t0 , ∞) such that x(τl (t)) > 0,

x(t) > 0,

t ∈ [t1 , ∞),

l ∈ {1, . . . , n}.

From (6.18), it follows that x Δ (t) ≤ 0,

t ∈ [t1 , ∞).

Set Λ(t) = −

x Δ (t) , x(t)

t ∈ [t1 , ∞).

332

6 Oscillations of First-Order Functional Dynamic Equations

We have Λ ∈ Crd ([t1 , ∞)) and Λ(t) > 0, t ∈ [t1 , ∞). Also, x Δ (t) + Λ(t)x(t) = 0,

t ∈ [t1 , ∞).

x(t) = x(t1 )e−Λ (t, t1 ),

t ∈ [t1 , ∞).

Then

We put x in (6.18) and we get −Λ(t)x(t1 )e−Λ (t, t1 ) +

n 

pl (t)x(t1 )e−Λ (τl (t), t1 ) ≤ 0,

t ∈ [t1 , ∞),

l=1

whereupon, using that x(t1 ) > 0, we obtain −Λ(t)e−Λ (t, t1 ) +

n 

pl (t)e−Λ (τl (t), t1 ) ≤ 0,

t ∈ [t1 , ∞),

l=1

or Λ(t) ≥

n  l=1

=

n 

pl (t)

e−Λ (τl (t), t1 ) e−Λ (t, t1 )

pl (t)e−Λ (t1 , τl (t))e−Λ (t, t1 )

l=1

=

n 

pl (t)e−Λ (t, τl (t)),

t ∈ [t1 , ∞).

l=1

(iii)–(iv) Let t1 ≥ t0 be such that τmin (t) ≥ t0 , t ∈ [t1 , ∞). Let also, Λ ∈ Crd ([t0 , ∞)), Λ(t) > 0,

1 − μ(t)Λ(t) > 0,

t ∈ [t1 , ∞),

and (6.20) holds. Consider the IVP x Δ (t) + x(t)

"n

l=1 pl (t)x(τl (t))

= f (t), t ∈ [t1 , ∞), = 0, t ∈ [t0 , t1 ].

Let x be its solution. Set g(t) = x Δ (t) + Λ(t)x(t),

t ∈ [t1 , ∞).

(6.21)

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

333

Then we get the following IVP x Δ (t) + Λ(t)x(t) = g(t),

t ∈ [t1 , ∞),

x(t1 ) = 0, which has the unique solution t

x(t) =

e−Λ (t, σ (η))g(η)Δη,

t1

t ∈ [t1 , ∞).

(6.22)

Substituting (6.22) into (6.21), we get f (t) = −Λ(t)

t t1

e−Λ (t, σ (η))g(η)Δη

+e−Λ (σ (t), σ (t))g(t) +

n 

τl (t)

pl (t)e−Λ (t, τl (t))

l=1

= −Λ(t)

t t1

+

n 

n 

e−Λ (t, σ (η))g(η)Δη

e−Λ (t, σ (η))g(η)Δη + g(t) τl (t)

pl (t)e−Λ (t, τl (t))

l=1

=−

t1

t

Yl (t)

l=0

τl (t)

t1

e−Λ (t, σ (η))g(η)Δη

e−Λ (t, σ (η))g(η)Δη + g(t),

t ∈ [t1 , ∞),

where τ0 (t) = t1 , Yl (t) = pl (t)e−Λ (t, τl (t)) ≥ 0, Y0 (t) = Λ(t) −

n 

t ≥ t1 ,

pl (t)e−Λ (t, τl (t)) ≥ 0,

l ∈ {1, . . . , n}, t ∈ [t1 , ∞).

l=1

By Theorem 6.23, it follows that the nonnegativity of f on [t1 , ∞) implies the nonnegativity of g on [t1 , ∞) and the nonnegativity of g on [t1 , ∞) implies the nonnegativity of x on [t1 , ∞). On the other hand, x has the following representation x(t) =

t

χ (t, σ (η))f (η)Δη, t1

t ∈ [t1 , ∞).

334

6 Oscillations of First-Order Functional Dynamic Equations

Assume that x ≥ 0 on [t1 , ∞) and χ is not nonnegative on [t1 , ∞). Then there exist a t2 ∈ [t1 , ∞) and an s ∈ [t1 , t2 ) such that χ (t2 , σ (s)) < 0. Let f (t) = − min{χ (t2 , σ (s)), 0} ≥ 0,

t ∈ [t1 , ∞).

Then x(t2 ) < 0, which is a contradiction. Now we set

x(t) =

t ∈ [t1 , ∞),

χ (t, s) − e−Λ (t, s), 0, t ∈ [t0 , t1 ),

(6.23)

where s ∈ [t1 , ∞) is arbitrarily chosen. Then, x, defined by (6.23), satisfies (6.21) with a nonnegative forcing term f . Since x is nonnegative on [s, ∞), we conclude that χ (·, s) ≥ e−Λ (·, s) >0 on [s, ∞) for any s ∈ [t1 , ∞). (iv) ⇒ (i) Let χ be eventually positive. Then χ is an eventually positive solution of the equation (6.17). This completes the proof. Example 6.15. Let T = hZ, h > 0, λ ∈ (0, 1] be such that 1 − λh > 0,

1−λ≥h

n 

pl (t)λ−

t−τl (t) h

,

t ∈ [t1 , ∞),

l=1

where t1 ∈ [t0 , ∞) is sufficiently large. Let also, Λ = −Λ = =

Λ 1 − hΛ 1−λ h 1 − h 1−λ h 1−λ h

=

1−1+λ 1−λ , = hλ

e−Λ (t, τl (t)) = e

#t

1 τl (t) h

 log 1+ 1−λ Δs λ

1−λ h .

Then

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

=e

#t

1 τl (t) h

=e

335

log λ1 Δs

 1 − log λ h (t−τl (t))

= λ−

t−τl (t) h

,

t ∈ [t1 , ∞),

l ∈ {1, . . . , n}.

Hence, n 

pl (t)e−Λ (t, τl (t)) =

l=1

n 

pl (t)e−

t−τl (t) h

l=1

1−λ h = Λ. ≤

Therefore the equation (6.17) has an eventually positive solution.   1 N  Exercise 6.9. Consider the equation (6.17). Let T = 2N0 {0}, τl (t) = 2 t , ql

t ∈ T, l ∈ {1, . . . , n}, λ ∈ (0, t], and t1 ∈ [t0 , ∞) be sufficiently large so that 1−λ≥t

n 

pl (t)λ−l ,

t ∈ [t1 , ∞).

l=1

Prove that the equation (6.17) has an eventually positive solution. Consider the equation x Δ (t) +

n 

ql (t)x(αl (t)) = 0,

t ∈ [t0 , ∞),

(6.24)

l=1

where n ∈ N, ql ∈ Crd ([t0 , ∞)), αl ∈ Crd ([t0 , ∞)), limt→∞ αl (t) = ∞, αl (t) ≤ t, t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Let Y be the fundamental solution of the equation (6.24). Theorem 6.25. Suppose that ql ∈ Crd ([t0 , ∞)), ql : [t0 , ∞) → [0, .∞), pl ≥ ql , τl ≤ αl on [t1 , ∞), l ∈ {1, . . . , n}, for some t1 ∈ [t0 , ∞). If the fundamental solution χ of the equation (6.17) is eventually positive, then the fundamental solution Y of (6.24) is eventually positive. Proof. Since χ is eventually positive, by Theorem 6.24, it follows that there exists a t2 ∈ [t1 , ∞) and Λ ∈ Crd ([t2 , ∞)), Λ > 0 on [t2 , ∞), 1 − Λμ > 0 on [t2 , ∞), and for all t ∈ [t2 , ∞)

336

6 Oscillations of First-Order Functional Dynamic Equations

Λ(t) ≥

n 

pl (t)e−Λ (t, τl (t)).

l=1

Note that e−Λ (t, s) is nondecreasing in s. Hence, e−Λ (t, s) =

1 e−Λ (t, s)

is nonincreasing in s. Then Λ(t) ≥

n 

pl (t)e−Λ (t, τl (t))

l=1

≥

n 

t ∈ [t2 , ∞).

ql (t)e−Λ (t, αl (t)),

l=1

Hence from Theorem 6.24, it follows that Y (·, s) > 0 on [s, ∞) for any s ∈ [t2 , ∞), i.e., the equation (6.24) has an eventually positive solution. This completes the proof. Corollary 6.3. Assume that all the conditions of Theorem 6.25 hold. If (6.24) is oscillatory, then so is (6.17). Proof. Assume that (6.17) is nonoscillatory. Without loss of generality, suppose that the equation (6.17) has an eventually positive solution. Hence from Theorem 6.25, it follows that (6.24) has an eventually positive solution. This is a contradiction. This completes the proof. Now we consider the equation x Δ (t) +

n 

ql (t)x(τl (t)) = 0,

t ∈ [t0 , ∞),

(6.25)

l=1

where ql ∈ Crd ([t0 , ∞)) and τl are the same as in (6.16). With Y we will denote the fundamental solution of the equation (6.25). Theorem 6.26. Suppose that pl , ql ∈ Crd ([t0 , ∞)), pl : [t0 , ∞) → [0, ∞), pl ≥ ql on [t1 , ∞), l ∈ {1, . . . , n}, for some t1 ∈ [t0 , ∞). Let also, χ (·, s) > 0 on [s, ∞) for any s ∈ [t1 , ∞). Then Y (·, s) ≥ χ (·, s) on [s, ∞) for any s ∈ [t1 , ∞). Proof. Fix s ∈ [t1 , ∞). Then for all t ∈ [s, ∞), we obtain Y Δ (t, s) +

n  l=1

pl (t)Y (τl (t), s) = Y Δ (t) +

n  l=1

ql (t)Y (τl (t), s)

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

+

n 

337

(pl (t) − ql (t)) Y (τl (t), s)

l=1

=

n 

(pl (t) − ql (t)) Y (τl (t), s).

l=1

Hence, Y (t, s) = χ (t, s) +

n  l=1

t

χ (t, σ (η)) (pl (η) − ql (η)) Y (τl (η), s)Δs,

(6.26)

s

t ∈ [s, ∞). Since χ (·, s) > 0 on [s, ∞), using Theorem 6.23, we conclude that Y (·, s) > 0 on [s, ∞). Hence from (6.26), we get Y (t, s) ≥ χ (t, s) for all t ∈ [s, ∞). This completes the proof. Corollary 6.4. Suppose that the inequality x Δ (t) +

n 

pl+ (t)x(τl (t)) ≤ 0,

t ∈ [t0 , ∞),

l=1

where pl+ (t) = max{pl (t), 0}, t ∈ [t0 , ∞), l ∈ {1, . . . , n}, and pl , τl are the same as in (6.17), l ∈ {1, . . . , n}, has an eventually positive solution. Then so does (6.17). Proof. By Theorem 6.24, it follows that the fundamental solution of the equation x Δ (t) +

n 

pl+ (t)x(τl (t)) = 0,

t ∈ [t0 , ∞),

l=1

is eventually positive. Hence, applying Theorem 6.26, the fundamental solution of the equation (6.17) is eventually positive. This completes the proof. Now we consider the following IVP " x Δ (t) + nl=1 ql (t)x(τl (t)) = g(t), = φ(t), x(t0 ) = x0 , x(t)

t ∈ [t0 , ∞), t ∈ [t−1 , t0 ),

(6.27)

where n ∈ N, x0 , φ, τl , l ∈ {1, . . . , n}, are the same as in (6.16), ql , g ∈ Crd ([t0 , ∞)), l ∈ {1, . . . , n}. Theorem 6.27. Suppose that pl ≥ ql , l ∈ {1, . . . , n}, g ≥ f on [t0 , ∞), χ (·, s) > 0 on [s, ∞), s ∈ [t0 , ∞). Let x be a solution of the equation (6.16). Then y ≥ x holds on [t0 , ∞), where y is a solution of (6.27). Proof. By Theorem 6.24 and Theorem 6.26, we obtain Y (·, s) ≥ χ (·, s)

on

[s, ∞)

338

6 Oscillations of First-Order Functional Dynamic Equations

for any s ∈ [t0 , ∞). Also, by (6.16), we get x Δ (t) +

n 

ql (t)x(τl (t)) = x Δ (t) +

l=1

n 

pl (t)x(τl (t))

l=1

+

n 

(ql (t) − pl (t)) x(τl (t))

l=1

= f (t) +

n 

(ql (t) − pl (t)) x(τl (t)),

t ∈ [t0 , ∞).

l=1

Therefore ⎛ x(t) = x0 Y (t, s)+

−

n 

t

l=1 t0

n 

t

l=1 t0

= y(t),

t0

Y (t, σ (η)) ⎝f (η)−

n 

⎞ (pl (η)−ql (η)) κ[t0 ,∞) (τl (η))x(τl (η))⎠Δη

l=1

Y (t, σ (η))pl (η)φ(τl (η))Δη

≤ x0 Y (t, t0 ) + −

t

t t0

Y (t, σ (η))g(η)Δη

Y (t, σ (η))ql (η)φ(τl (η))Δη

t ∈ [t0 , ∞).

This completes the proof. √ 3 Example 6.16. Let T = N. Consider the following IVPs x Δ (t) +

2  3 3 3 x t − 2 = 3, t3 2t x(t) = 1,

√ 3 t ∈ [ 3, ∞), t ∈ [1,

√ 3 3],

and y Δ (t) +

1  3 3 3 y t − 2 = 3, 3 t t y(t) = 1,

√ 3 t ∈ [ 3, ∞), t ∈ [1,

Then, by Theorem 6.27, we obtain that x(t) ≤ y(t),

√ 3 t ∈ [ 3, ∞).

√ 3

3].

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

Exercise 6.10. Let T = 2N0 . Consider the IVPs

  4 4 t = , x Δ (t) + 2 + 1 x 2 t t x(t) = 2,

y Δ (t) +

3 y t4

 7 t = , 2 t

339

t ∈ [4, ∞), t ∈ [1, 4],

t ∈ [4, ∞),

y(t) = 2,

t ∈ [1, 4].

Prove that x(t) ≤ y(t), t ∈ [4, ∞). Theorem 6.28. Let pl satisfies all the conditions of Theorem 6.27, l ∈ {1, . . . , n}. Let also, x be a solution of (6.17), and Y (·, s) > 0 on [s, ∞) for any s ∈ [t0 , ∞) be the fundamental solution of the equation x (t) + Δ

n 

pl+ (t)x(τl (t)) = 0,

t ∈ [t0 , ∞),

l=1

and y > 0 on [t0 , ∞) be a solution of this equation, where pl+ (t) = max{pl (t), 0}, t ∈ [t0 , ∞). If x ≡ y on [t−1 , t0 ], then x ≥ y on [t0 , ∞). Proof. Since pl (t) ≤ pl+ (t), t ∈ [t0 , ∞), using Theorem 6.27, we get x ≥ y on [t0 , ∞). This completes the proof. Theorem 6.29. Let pl , l ∈ {1, . . . , n}, be as in Theorem 6.27. Suppose that there exist a t1 ∈ [t0 , ∞) and a Λ ∈ Crd ([t0 , ∞)), Λ : [t0 , ∞) → [0, ∞) such that 1 − μΛ > 0 on

[t1 , ∞)

and for all t ∈ [t1 , ∞) n 

pl+ (t) ≤ Λ(t) 1 −

l=1

t

 Λ(η)Δη .

τmin (t)

Then the equation (6.17) has an eventually positive solution. Proof. Since Λ ∈ Crd ([t0 , ∞)), 1 − μΛ > 0 on [t1 , ∞), we have that e−Λ (t, s) is nondecreasing in s and e−Λ (t, s) =

1 e−Λ (t, s)

340

6 Oscillations of First-Order Functional Dynamic Equations

is nonincreasing in s. Then Λ(t) ≥

n  l=1

≥

n  l=1

=

n 

1−

#t

pl+ (t)

τmin (t) Λ(η)Δη

pl+ (t) e−Λ (t, τmin (t)) pl+ (t)e−Λ (t, τmin (t))

l=1

≥

n 

pl+ (t)e−Λ (t, τl (t)),

t ∈ [t1 , ∞).

l=1

Hence from Theorem 6.24, it follows that the inequality x Δ (t) +

n 

pl+ (t)x(τl (t)) ≤ 0,

t ∈ [t1 , ∞),

l=1

has an eventually positive solution. From here and from Corollary 6.4, we get that the equation (6.17) has an eventually positive solution. This completes the proof. Theorem 6.30. Let pl , l ∈ {1, . . . , n}, be as in Theorem 6.27. Suppose that M, λ > 0 with λ(1 − Mλ) ≥ 1 and t1 ∈ [t0 , ∞) be such that 1 − λμ

n 

pl+ > 0

on

[t1 , ∞)

l=1

and t

n 

τmin (t) l=1

pl+ (η)Δη ≤ M,

t ∈ [t2 , ∞),

where t2 ∈ [t1 , ∞) satisfies τmin (t) ≥ t1 for all t ∈ [t2 , ∞). Then (6.17) has an eventually positive solution.

6.4 Nonoscillations of First-Order Functional Dynamic Equations with. . .

341

Proof. Let Λ(t) = λ

n 

pl+ (t),

t ∈ [t1 , ∞).

l=1

Then [t1 , ∞),

1 − μΛ > 0 on and n 

t

λM ≥ λ

τmin (t) l=1

pl+ (η)Δη

t

=

Λ(η)Δη, τmin (t)

1 − λM ≤ 1 −

t

Λ(η)Δη, τmin (t)

1 ≤ λ(1 − λM)

t ≤ λ 1−

 Λ(η)Δη ,

τmin (t)

n 

pl+ (t) ≤ λ

l=1

n 

pl+ (t) 1 −

l=1

= Λ(t) 1 −



t

Λ(η)Δη τmin (t)



t

Λ(η)Δη ,

t ∈ [t1 , ∞).

τmin (t)

Hence from Theorem 6.29, it follows that the equation (6.17) has an eventually positive solution. This completes the proof. Example 6.17. Let T = 2Z. Consider the equation x Δ (t) +

1 50(t 4

+ 1)

x(t − 2) +

1 50(t 6

We have σ (t) = t + 2, μ(t) = 2, τ1 (t) = t − 2, τ2 (t) = t − 4,

+ 1)

x(t − 4) = 0,

t ∈ [4, ∞).

342

6 Oscillations of First-Order Functional Dynamic Equations

τmin (t) = t − 4, p1 (t) = p2 (t) = p1+ (t) = p2+ (t) =

1 , 50(t 4 + 1) 1 50(t 6

+ 1)

,

+ 1)

,

1 50(t 4

1 , 50(t 6 + 1)

t ∈ [4, ∞).

Then  +  1 p1 (η) + p2+ (η) Δη = 50 τmin (t) t

t

t−4

 1 1 + Δη η4 + 1 η6 + 1

 1 1 + (t − 4)2 + 1 (t − 4)6 + 1

 1 1 2 + + 50 (t − 2)4 + 1 (t − 2)6 + 1

2 = 50

4 4 + 50 50 4 . = 25 ≤

Let M =

4 25 .

Then 

4 λ 1− λ ≥1 25

(

⇐⇒

) 5 λ∈ ,5 . 4

Also,   1 − 2λ p1+ (t) + p2+ (t) ≥ 1 − 2λ = 1−

1 1 + 50 50



2λ 25

>0 if λ
0, T = hZ, p satisfies all the conditions of Theorem 6.4. Suppose that the equation y(t + h) = y(t) − hp(t)y(t − 2h),

t ∈ T,

has an eventually positive solution. Prove that lim inf t→∞

(p(t) + M(t))(p(t + h) + M(t)) ≥ h, (p(t) + p(t + h) + M(t))3

where M(t) =

0

(p(t))2 + (p(t + h))2 − p(t)p(t + h),

t ∈ T.

344

6 Oscillations of First-Order Functional Dynamic Equations

Problem 6.3. Let T = 2Z. Prove that the equation x Δ (t) +

t4

1 x(t − 2) = 0, +5

t ∈ T,

has a positive solution. Problem 6.4. Let T = 2N0 . Prove that the equation 1 x x (t) + 2 t +7 Δ

 t = 0, 2

t ≥ 8,

has a positive solution. Problem 6.5. Let T = 3N0 . Consider the equation t x x (t) + 2 t +1 Δ

 t = 0, 9

t ∈ T.

 Suppose that x is an eventually positive solution. Prove that for any M ∈ 0, 89 there exists t1 ∈ T such that   x 9t 4 ≤ 2 x(t) M for any t ∈ [t1 , ∞). Problem 6.6. Let T = 3N0 . Prove that the equation x Δ (t) +

t3 + t x t +3

 t = 0, 9

t ∈ [27, ∞),

is oscillatory. Problem 6.7. Let T = N, p2n =

8 , 100

p2n+1 =

 nπ 746 8 , + sin2 100 1000 2

n ∈ N.

Prove that the equation x(n + 1) − x(n) + pn x(n − 3) = 0, is oscillatory.

n ∈ N,

6.5 Advanced Practical Problems

345

Problem 6.8. Let T = 3N0 . Prove that all the solutions of the equation

  t  t 3 2 + (2t + 1)x =0 x (t) + t + t + t x 3 9 Δ

are oscillatory. Problem 6.9. Consider the equation (6.17). Let T = 3N0 , τl (t) = {1, . . . , n}, λ ∈ (0, 1] and t1 ≥ t0 be sufficiently large so that 1 − λ ≥ 2t

n 

pl (t)λ−2l ,

t , q 2l

∈ T, l ∈

t ∈ [t1 , ∞).

l=1

Prove that the equation (6.17) has an eventually positive solution. Problem 6.10. Let T = 2N0 . Consider the IVPs

  4 1 t Δ = , x (t) + 2 + 10 x 9 t t x(t) = 2, 1 y (t) + 4 y t + t2 + 5 Δ

 t 7 = , 9 t y(t) = 2,

t ∈ [9, ∞), t ∈ [1, 9], t ∈ [9, ∞), t ∈ [1, 9].

Prove that x(t) ≤ y(t), t ∈ [9, ∞). Problem 6.11. Let T = 3N0 . Prove that the equation x Δ (t) +

1 x 210t 2 + 1



 1 t t + = 0, x 3 9 700t 3 + t 2

has an eventually positive solution.

t ∈ [9, ∞),

Chapter 7

Oscillations of Second-Order Linear Functional Dynamic Equations with a Single Delay

The material of this chapter are based on results of the papers and monographs [36, 82, 88, 89, 92, 95, 152, 188, 193, 203, 213, 214, 249]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

7.1 Reduction to First-Order Linear Functional Dynamic Equations We will investigate the oscillatory behavior of the following second-order linear functional dynamic equation. 2

x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ T,

(7.1)

where p ∈ Crd (T), p(t) > 0, t ∈ T, τ : T → T, τ (t) < t, t ∈ T, and limt→∞ τ (t) = ∞. We will start with the following useful lemma. Lemma 7.1. If x is an eventually positive solution of the equation (7.1), then there exists a t0 ∈ T such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0 and

2

x Δ (t) < 0

for all t ≥ t0 . Proof. Since x is an eventually positive solution of the equation (7.1), there exists a t1 ∈ T such that x(t) > 0,

x(τ (t)) > 0

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_7

347

348

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

for any t ≥ t1 . Hence from (7.1), we get 2

x Δ (t) = −p(t)x(τ (t)) 0 for any t ≥ t1 . Then there exists a t0 ≥ t1 so that x(t) ≥ x(τ (t)) for any t ≥ t0 . This completes the proof. Theorem 7.1. Suppose that the equation (7.1) has an eventually positive solution. Then α(c) ≥ 1

for all c ∈ (0, 1),

(7.2)

where α(c) = lim sup t→∞

sup

  λe−λcpτ (t, τ (t)) .

λ>0 1 − μλcpτ > 0

Proof. Let x be an eventually positive solution of the equation (7.1). Then, there exists a t0 ∈ T such that x(t) ≥ x(τ (t)),

x Δ (t) > 0 and

2

x Δ (t) < 0

7.1 Reduction to First-Order Linear Functional DynamicEquations

349

for all t ≥ t0 . Define y = x Δ on T. Then y is decreasing on [t0 , ∞) and x(t) = x(t0 ) +

t

y(s)Δs t0

≥ x(t0 ) + (t − t0 )y(t) ≥ (t − t0 )y(t), Let c ∈ (0, 1) and t ∗ =

t0 1−c .

t ≥ t0 .

Then, for t ≥ t ∗ , we have (1 − c)t ≥ t0 ,

or t ≥ t0 + ct, or t − t0 ≥ ct. Also, for t ≥ t ∗ , we get #t x(t) = x(t0 ) + t0 y(s)Δs ≥ y(t)(t − t0 ) ≥ cty(t).

(7.3)

Pick t1 ≥ t ∗ so that τ (t) ≥ t ∗ for any t ≥ t1 . Hence from (7.3), we obtain x(τ (t)) ≥ cτ (t)y(τ (t)) for any t ≥ t1 . From here and from (7.1), we get 2

y Δ (t) = x Δ (t) = −p(t)x(τ (t)) ≤ −p(t)cτ (t)y(τ (t)),

t ≥ t1 .

Using the last inequality and Theorem 6.24, it follows that the equation zΔ (t) + p(t)cτ (t)z(τ (t)) = 0 has an eventually positive solution. Then, applying Theorem 6.4, we get (7.2). This completes the proof.

350

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Example 7.1. Let T = 2N0 . Consider the equation

 t 3(t + 4) Δ2 = 0, x x (t) + 2 2 4(t + 1)(t + 2) (2t + 1) x(t) =

t ∈ [4, ∞),

t +1 , t +2

t ∈ [1, 4).

Here σ (t) = 2t, t τ (t) = , 2 p(t) =

3(t + 4) , 4(t + 1)(t + 2)2 (2t + 1)

t ∈ T.

We will show that x(t) =

t +1 , t +2

t ∈ T,

is its solution. We have x Δ (t) =

t + 2 − (t + 1) (t + 2)(σ (t) + 2)

=

t +2−t −1 (t + 2)(2t + 2)

=

1 2(t + 1)(t + 2)

=

1 , 2(t 2 + 3t + 2)

2

x Δ (t) = − =− =− =− x

 t = 2 =

t 2 t 2

σ (t) + t + 3 2(t + 1)(t + 2)(σ (t) + 1)(σ (t) + 2) 2t + t + 3 2(t + 1)(t + 2)(2t + 1)(2t + 2) 4(t

3t + 3 + 2)(2t + 1)

+ 1)2 (t

3 , 4(t + 1)(t + 2)(2t + 1) +1 +2

t +2 , t +4

t ∈ T.

7.1 Reduction to First-Order Linear Functional DynamicEquations

351

Hence, 2

x Δ (t) +

 t 2

3(t + 4) x 4(t + 1)(t + 2)2 (2t + 1)

3(t + 4) 3 + =− 4(t + 1)(t + 2)(2t + 1) 4(t + 1)(t + 2)2 (2t + 1) =− = 0,

t +2 t +4



3 3 + 4(t + 1)(t + 2)(2t + 1) 4(t + 1)(t + 2)(2t + 1) t ∈ T.

Therefore the considered equation has an eventually positive solution and α(c) ≥ 1 for any c ∈ (0, 1). Corollary 7.1. If α(c) < 1 for some c ∈ (0, 1), then all the solutions of (7.1) are oscillatory. Proof. Assume the contrary. Without loss of generality, we suppose that the equation (7.1) has an eventually positive solution x. Then α(c) ≥ 1 for any c ∈ (0, 1). This is a contradiction. This completes the proof. Example 7.2. Let T = {tn : n ∈ Z} be a time scale such that σ (tn ) = tn+1 and let τ (tn ) = tn−k for any n ∈ Z and for some k ∈ N. Then, using the arithmeticgeometric inequality, we get e−λcpτ (t, τ (t)) = e =e =e =

#t

1 τ (t) μ(s)

log(1−μ(s)cλp(s)τ (s))Δs

# tn

1 tn−k μ(s)

log(1−μ(s)cλp(s)τ (s))Δs

"k

l=1 log(1−μ(tn−l )cλp(tn−l )τ (tn−l ))

k $

(1 − μ(tn−l )cλp(tn−l )τ (tn−l ))

l=1

% ≤

1 (1 − μ(tn−l )cλp(tn−l )τ (tn−l )) k k

&k

l=1

%

cλ  = 1− μ(tn−l )p(tn−l )τ (tn−l ) k k

l=1

cλ = 1− k

k

t

p(s)τ (s)Δs τ (t)

= (1 − λS)k ,

t ∈ T,

&k

352

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

where S=

t

c k

t ∈ T.

p(s)τ (s)Δs, τ (t)

Hence, λe−λcpτ (t, τ (t)) = λ(1 − λS)k ,

t ∈ T.

Let g(λ) = λ(1 − λS)k ,

λ > 0.

Then g (λ) = −λkS(1 − λS)k−1 + (1 − λS)k = (1 − λS)k−1 (1 − λS − kλS) = (1 − λS)k−1 (1 − (k + 1)λS) , Therefore

λ > 0.

 1 g(λ) ≤ g (k + 1)S

k 1 S = 1− (k + 1)S (k + 1)S

=

kk S k 1 (k + 1)S (k + 1)k S k

=

kk , (k + 1)k+1 S

λ > 0.

Consequently if t

lim inf t→∞

p(s)τ (s)Δs ≥

τ (t)

k k+1

k+1

1 c

for some c ∈ (0, 1), all the solutions of (7.1) are oscillatory. Exercise 7.1. Let T = 2N0 . Prove that all the solutions of the equation x are oscillatory.

Δ2

1 x (t) + 2 t +1

 t = 0, 2

t ≥ 2,

7.2 An Oscillation Criteria

353

7.2 An Oscillation Criteria Theorem 7.2. If t

lim sup

p(s)τ (s)Δs >

t→∞

τ (t)

1 c

for some c ∈ (0, 1), then every solution of the equation (7.1) is oscillatory. Proof. Suppose that the equation (7.1) has a nonoscillatory solution x. Without loss of generality, we suppose that x is an eventually positive solution of the equation (7.1). Then there exists a t0 ∈ T such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0 and

2

x Δ (t) < 0

for any t ≥ t0 . Proceeding as in Theorem 7.1, there exists a t1 ≥ t0 such that y Δ (t) ≤ −cp(t)τ (t)y(τ (t)) for any t ≥ t1 , where y = x Δ . We pick t2 ≥ t1 so that t

p(s)τ (s)Δs > 1

c τ (t)

for any t ≥ t2 . Then 0≥

t

 Δ  y (s) + cp(s)τ (s)y(τ (s)) Δs

τ (t) t

= y(t) − y(τ (t)) + c

p(s)τ (s)y(τ (s))Δs τ (t)

≥ y(t) − y(τ (t)) + cy(τ (t))

= y(t) + y(τ (t)) c

t

p(s)τ (s)Δs τ (t)

t

 p(s)τ (s)Δs − 1

τ (t)

>0 for any t ≥ t2 , which is a contradiction. This completes the proof. Example 7.3. Let T = 2Z. Consider the equation  2 x Δ (t) + t 2 + 1 x(t − 4) = 0,

t ∈ T.

354

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Here σ (t) = t + 2, μ(t) = 2, p(t) = t 2 + 1, τ (t) = t − 4,

t ∈ T.

Then t

p(s)τ (s)Δs =

τ (t)

t



s 2 + 1 (s − 4)Δs

t−4

 = 2 (t − 4)2 + 1 (t − 4 − 4)  +2 (t − 2)2 + 1 (t − 2 − 4)  = 2 t 2 − 8t + 17 (t − 8)  +2 t 2 − 4t + 5 (t − 6)  = 2 t 3 − 8t 2 − 8t 2 + 64t + 17t − 136 +t 3 − 6t 2 − 4t 2 + 24t + 5t − 30  = 2 2t 3 − 26t 2 + 110t − 166 → ∞ as

t → ∞.

Therefore there exists a c ∈ (0, 1) such that t

lim sup t→∞

p(s)τ (s)Δs > τ (t)

1 . c

Hence from Theorem 7.2, it follows that every solution of the considered equation is oscillatory. Exercise 7.2. Let T = 2N0 . Prove that every solution of the equation  t  2 = 0, x Δ (t) + t 2 + t x 4 is oscillatory.

t ∈ T,

t ≥ 4,

7.3 The Riccati Transformation Technique

355

7.3 The Riccati Transformation Technique Throughout this section we suppose that p ∈ Crd (T), p(t) > 0, t ∈ T, τ : T → T, τ (t) ≤ t, t ∈ T, limt→∞ τ (t) = ∞ and ∞

σ (s)p(s)Δs = ∞

t0

for some t0 ∈ T. Theorem 7.3. Every bounded solution of (7.1) is oscillatory. Proof. Assume the contrary. Without loss of generality, assume that the equation (7.1) has an eventually positive solution x which is bounded. Then there exist a t1 ≥ t0 such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0,

and

2

x Δ (t) < 0

for any t ≥ t1 . There exists α, β > 0, and α < β, such that α < x(τ (t)) ≤ x(t) < β for any t ≥ t1 . Set X(t) = tx Δ (t),

t ≥ t1 .

Then X(t) = X(t1 ) +

t

XΔ (s)Δs

t1

= X(t1 ) +

t

 2 x Δ (s) + σ (s)x Δ (s) Δs

t1

= X(t1 ) +

t

 Δ  x (s) − σ (s)p(s)x(τ (s)) Δs

t1 t

= X(t1 ) + x(t) − x(t1 ) −

σ (s)p(s)x(τ (s))Δs t1

≤ X(t1 ) + β − x(t1 ) − α

t

σ (s)p(s)Δs t1

→ −∞ as

t → ∞.

356

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Hence, there exists a positive constant M such that X(t) ≤ −M for any t ≥ t2 , for some t2 ≥ t1 . From here, x Δ (t) ≤ −

M t

for t ≥ t2 . From the last inequality, we get ∞

∞

x Δ (s)Δs ≤ −M

t2

t2

Δs s

or ∞

lim x(t) ≤ x(t2 ) − M

t→∞

t2

Δs . s

Then lim x(t) = −∞,

t→∞

which is a contradiction. This completes the proof. Example 7.4. Consider the Euler delay dynamic equation 2

x Δ (t) +

1 x(τ (t)) = 0, tσ (t)

t ≥ 1,

t ∈ T.

Here p(t) =

1 , tσ (t)

t > 1,

t ∈ T.

We have ∞

σ (s)p(s)Δs =

t0

∞

σ (s) t0

=

∞ t0

1 Δs sσ (s)

Δs s

= ∞. Hence from Theorem 7.3, we conclude that every bounded solution of the Euler delay dynamic equation is oscillatory.

7.3 The Riccati Transformation Technique

357

Exercise 7.3. Let T = 2N0 . Prove that every bounded solution of the equation x

Δ2

 t = 0, (t) + t x 2 2

t ∈ T,

t ≥ 2,

is oscillatory. Theorem 7.4. Every eventually positive solution x of the equation (7.1) satisfies eventually x(t) t

x(t) ≥ tx Δ (t) and

is

nonincreasing.

Proof. Since x is an eventually positive solution of the equation (7.1), then there exists a t0 ∈ T such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0 and

2

x Δ (t) < 0

for any t ≥ t0 . We set X(t) = x(t) − tx Δ (t),

t ≥ t0 .

Then X(t) = X(t0 ) +

t

XΔ (s)Δs

t0

= X(t0 ) +

t

 2 x Δ (s) − x Δ (s) − σ (s)x Δ (s) Δs

t0

= X(t0 ) −

t

2

σ (s)x Δ (s)Δs

t0

= X(t0 ) +

t

p(s)σ (s)x(τ (s))Δs t0

≥ X(t0 ) + x(τ (t0 ))

t

p(s)σ (s)Δs t0

→ ∞ as

t → ∞.

Hence, there exists a t1 ≥ t0 such that x(t) ≥ tx Δ (t)

358

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

for any t ≥ t1 . Let x(t) , t

Y (t) =

t ≥ t1 .

Then Y Δ (t) =

tx Δ (t) − x(t) tσ (t)

≤0 for any t ≥ t1 . Therefore

x(t) t

is eventually nonincreasing. This completes the proof.

Theorem 7.5. If

lim

∞

t

t→∞

p(s) t

τ (s) Δs s

 = ∞,

then every solution of the equation (7.1) is oscillatory. Proof. Suppose the contrary. Without loss of generality, we assume that there exists a t0 ∈ T such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0 and

2

x Δ (t) < 0

for any t ≥ t0 . For any T ≥ t ≥ t0 , we have T

T

p(s)x(τ (s))Δs = −

t

2

x Δ (s)Δs

t

= x Δ (t) − x Δ (T ) ≤ x Δ (t). Hence, ∞

p(s)x(τ (s))Δs ≤ x Δ (t)

t

for any t ≥ t0 . By Theorem 7.4, it follows that there exists a t1 ≥ t0 such that x(t) ≥ tx Δ (t) and

x(t) t

for any t ≥ t1 . We take t2 ≥ t1 so that τ (t) ≥ t1

is nonincreasing

7.3 The Riccati Transformation Technique

359

for any t ≥ t2 . Therefore x(t) ≥ tx Δ (t) ∞

≥t

p(s)x(τ (s))Δs t ∞

=t

p(s)τ (s)

x(τ (s)) Δs τ (s)

p(s)τ (s)

x(s) Δs s

t ∞

≥t t

≥ x(t) t

∞ t

τ (s) Δs p(s) s



for any t ≥ t2 . Consequently ∞

p(s)

t t

τ (s) Δs ≤ 1 s

for any t ≥ t2 . This is a contradiction. This completes the proof. Example 7.5. Let T = hZ, where h > 0. Consider the equation (7.1). Suppose that t0 ∈ T. Then ∞

nh

σ (t)p(t)Δt = lim

σ (t)p(t)Δt

n→∞ t 0

t0

= h lim

(n−1)h 

n→∞

= h lim

(n−1)h 

n→∞

= h2 lim

n→∞

(m + h)p(m)

m=t0 n−1 

n→∞

∞ nh

τ (s) Δs p(s) s

(ν + 1)p(νh)

t ν= h0

and

lim nh

σ (m)p(m)

m=t0

%

 = lim

n→∞

nh %

2

∞  m=nh

τ (m) p(m) m

&

& τ (νh) p(νh) = lim nh2 n→∞ νh ν=n % ∞ &  τ (νh) p(νh) = h lim n . n→∞ ν ν=n ∞ 

360

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Therefore, if n 

lim

n→∞

and

% lim

n→∞

(ν + 1)p(νh) = ∞

t ν= h0

∞ 

τ (νh) p(νh) n ν ν=n

& = ∞,

every solution of the equation (7.1) is oscillatory. Exercise 7.4. Let T = 2N0 . Prove that every solution of the equation

 t 2 x Δ (t) + t 2 x = 0, t ≥ 2 2 is oscillatory. Theorem 7.6. If z and x are differentiable on T with x(t) = 0, t ∈ T, then

xΔ

z2 x

Δ

 2  2 z Δ = zΔ − xx σ . x

Proof. We have  Δ z2 = zzΔ + zΔ zσ   = zΔ z + zσ   = zΔ z + z + μzΔ   = zΔ 2z + μzΔ on

T.

Hence,

xΔ

z2 x

Δ + xx σ

 2 z Δ x

 Δ x z2 − x Δ z2 xx σ

Δ Δ 2 σ z x − zx +xx xx σ   Δ 2z + μzΔ − z2 x Δ Δ xz =x xx σ 2  Δ z x − zx Δ + xx σ = xΔ

7.3 The Riccati Transformation Technique

361

 2 1 2xx Δ zzΔ + μxx Δ zΔ = xx σ   2  2 − zx Δ + zΔ x − zx Δ =

 2 1 2xzx Δ zΔ + μxx Δ zΔ σ xx 

− zx

 Δ 2

2  2  + zΔ x − 2xzzΔ x Δ + zx Δ



 1  Δ 2  Δ 2  σ z x + z x x −x σ xx 1  Δ 2  Δ 2 σ  Δ 2 2 z x + z xx − z x = xx σ  2 = zΔ . =

This completes the proof. Theorem 7.7. If there exist a t0 ∈ T and a differentiable function z on T such that t

lim

t→∞ t 0

  Δ 2 (z(σ (s)))2 Δs = ∞, p(s)τ (s) − z (s) σ (s)

then every solution of the equation (7.1) is oscillatory on [t0 , ∞). Proof. Assume the contrary. Without loss of generality, suppose that the equation (7.1) has an eventually positive solution x. Then there exists a t1 ≥ t0 such that x(t) ≥ x(τ (t)) > 0,

x Δ (t) > 0 and

2

x Δ (t) < 0

for any t ≥ t1 . Let w=

z2 x Δ x

on

[t1 , ∞).

Then  2 x − z2 x Δ w = xx σ   2  Δ 2 (zσ )2 x Δ + z2 x Δ x − z2 x Δ = xx σ 

Δ

z2 x Δ

Δ

362

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

on [t1 , ∞). Hence, −w Δ = = = = ≥ = =

 2   Δ 2 z2 x Δ − (zσ )2 x Δ + z2 x Δ x xx σ   2  Δ z2 x Δ − − (zσ )2 p(x ◦ τ ) + z2 x Δ x xx σ

 2 Δ 2 Δ x (zσ )2 p(x ◦ τ ) Δz x − z + x xσ xx σ

2 Δ (zσ )2 p(x ◦ τ ) Δ z − x xσ x

 2 Δ (zσ )2 px σ τ Δ z − x σ xσ x

 2 Δ (zσ )2 pτ Δ z −x σ x %

 2 &  Δ 2 z Δ (zσ )2 pτ σ − z − xx σ x

=

 2  2 z Δ (zσ )2 pτ + xx σ − zΔ σ x

≥

(zσ )2 pτ  Δ 2 − z σ

on [t1 , ∞). From here, for sufficiently large t ≥ t1 , we get w(t1 ) ≥ w(t1 ) − w(t) t

= −

t1

≥

t

t1

→∞

w Δ (s)Δs   2 z(σ (s))2 Δs p(s)τ (s) − zΔ (s) σ (s) t → ∞,

as

which is a contradiction. This completes the proof. Corollary 7.2. If there exists a t0 ∈ T such that t

lim

t→∞ t 0

p(s)

τ (s) Δs = ∞, σ (s)

then every solution of the equation (7.1) is oscillatory.

7.3 The Riccati Transformation Technique

363

Proof. We apply Theorem 7.7 for the function z(t) = 1, t ∈ T. This completes the proof. Corollary 7.3. If there exists a t0 ∈ T such that t

%

1 p(s)τ (s) − √ 2 √ s + σ (s)

lim

t→∞ t 0

& Δs = ∞,

then every solution of the equation (7.1) is oscillatory on [t0 , ∞). √ Proof. We apply Theorem 7.7 for z(t) = t, t ∈ T. This completes the proof. Corollary 7.4. If there exists a t0 ∈ T such that t

lim

t→∞ t 0

 1 p(s)τ (s) − Δs = ∞, 4s

then every solution of the equation (7.1) is oscillatory on [t0 , ∞). Proof. Since σ (s) ≥ s, the assertion follows from Corollary 7.3. This completes the proof. Example 7.6. Consider the equation 2

x Δ (t) +

γ x(τ (t)) = 0, tτ (t)

t ≥ 1,

where 4γ > 1. Then for any t ∈ T, t ≥ 1, we have t

t

σ (s)p(s)Δs =

1

1

γ σ (s) Δs sτ (s)

t

γ Δs 1 s → ∞ as t → ∞. ≥

Next, t

lim

t→∞ 1

 1 τ (s)p(s) − Δs = lim t→∞ 4s

t 1

γ 1 − s 4s

4γ − 1 lim 4 t→∞ = ∞.

t

=

1

 Δs

Δs s

Therefore any solution of the considered equation is oscillatory on [1, ∞).

364

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Exercise 7.5. Let T = 2N0 . Prove that every solution of the equation x

Δ2

10 (t) + 2 x t

 t = 0, 2

t ≥ 2,

is oscillatory on [2, ∞).

7.4 The Kamenev-Type Oscillation Criteria In this section we consider the equation Δ  r(t)x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ T,

(7.4)

1 (T), p > 0, r > 0 on T, τ : T → T satisfies τ (t) ≤ t, where p ∈ Crd (T), r ∈ Crd t ∈ T, limt→∞ τ (t) = ∞. Let t0 ∈ T. Define R to be the set of all functions H : [t0 , ∞) × [t0 , ∞) → R such that H (t, t) ≥ 0, t ≥ t0 , H (t, s) > 0, t > s ≥ t0 , H Δs (t, s) ≤ 0, t ≥ s ≥ t0 , and for each fixed t the function H (t, ·) is rd-continuous.

Theorem 7.8. Let ∞ t0

Δt = ∞, r(t)

∞

τ (s)p(s)Δs = ∞ and

r Δ (t) ≥ 0,

t ∈ [t0 , ∞).

t0

(7.5) Let also, the equation (7.4) has a positive solution x on [t0 , ∞). Then there exists a T ∈ [t0 , ∞) sufficiently large so that 2

(i) x Δ (t) > 0, x Δ (t) < 0, x(t) > tx Δ (t) for all t ∈ [T , ∞), (ii) x is strictly increasing and x(t) t is strictly decreasing on [T , ∞). Proof. Pick t1 ∈ [t0 , ∞) so that t1 > t0 and x(τ (t)) > 0,

t ∈ [t1 , ∞).

Then Δ  r(t)x Δ (t) = −p(t)x(τ (t)) < 0,

t ∈ [t1 , ∞).

From here, we conclude that r(t)x Δ (t) is strictly decreasing on [t1 , ∞). Assume that there is a t2 ∈ [t1 , ∞) such that r(t2 )x Δ (t2 ) = c < 0.

7.4 The Kamenev-Type Oscillation Criteria

365

Then r(t)x Δ (t) ≤ r(t2 )x Δ (t2 ) = c,

t ∈ [t2 , ∞).

c , r(t)

t ∈ [t2 , ∞).

Therefore x Δ (t) ≤

Integrating both sides of the last inequality from t2 to t, we get t

x(t) ≤ x(t2 ) + c

Δs r(s)

t2

→ −∞ as

t → ∞,

which is a contradiction. Hence, r(t)x Δ (t) > 0,

t ∈ [t1 , ∞).

Because r > 0 on [t1 , ∞), we get x Δ (t) > 0,

t ∈ [t1 , ∞).

Therefore x is strictly increasing on [t1 , ∞). Next, Δ  0 > r(t)x Δ (t) 2

= r Δ (t)x Δ (t) + r σ (t)x Δ (t),

t ∈ [t1 , ∞),

whereupon 2

x Δ (t)r σ (t) < −r Δ (t)x Δ (t),

t ∈ [t1 , ∞),

or 2

x Δ (t) < − ≤ 0,

r Δ (t)x Δ (t) r σ (t) t ∈ [t1 , ∞).

Let X(t) = x(t) − tx Δ (t),

t ∈ [t1 , ∞).

366

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Then 2

XΔ (t) = x Δ (t) − x Δ (t) − σ (t)x Δ (t) 2

= −σ (t)x Δ (t) t ∈ [t1 , ∞).

> 0,

Therefore X is strictly increasing on [t1 , ∞). Assume that X(t) < 0 on [t1 , ∞). Then

 x(t) Δ tx Δ (t) − x(t) = t tσ (t) =−

X(t) tσ (t) t ∈ [t1 , ∞).

> 0,

Hence, we conclude that x(t) t is strictly increasing on [t1 , ∞). Pick t3 ∈ [t1 , ∞) so that τ (t) ≥ τ (t1 ) for any t ≥ t3 . Then x(τ (t1 )) x(τ (t)) ≥ τ (t) τ (t1 ) =d > 0. So, x(τ (t)) ≥ dτ (t),

t ≥ t3 .

Now, integrating both sides of the equation (7.4) from t3 to t, t ≥ t3 , we get r(t)x Δ (t) − r(t3 )x Δ (t3 ) +

t

p(s)x(τ (s))Δs = 0,

t3

whereupon r(t3 )x Δ (t3 ) = r(t)x Δ (t) +

t

p(s)x(τ (s))Δs t3

≥

t

p(s)x(τ (s))Δs t3

≥ d

t

p(s)τ (s)Δs t3

→ ∞ as

t → ∞,

7.4 The Kamenev-Type Oscillation Criteria

367

which is a contradiction. Therefore there exists a t4 ∈ [t1 , ∞) so that X(t4 ) > 0. Because X is strictly increasing on [t1 , ∞), we obtain X(t) > 0 on [t4 , ∞). Consequently

x(t) t

Δ =−

X(t) tσ (t)

< 0, Thus

x(t) t

t ∈ [t4 , ∞).

is strictly decreasing on [t4 , ∞). This completes the proof.

Theorem 7.9. Assume (7.5) holds, H ∈ R and for t > s let H Δs (t, s) . h(t, s) = − √ H (t, s) If there exists a positive delta differentiable function δ such that for every t1 ∈ [t0 , ∞) 1 lim sup t→∞ H (t, t1 )

∞ t1

& τ (s) r(s) (δ σ (s))2 2 − H (t, s) δ(s)p(s) (p(t, s)) Δs = ∞, s 4δ(s) %

(7.6) where b(s) h(t, s) − σ , p(t, s) = √ H (t, s) δ (s)

b(t) = max{0, δ Δ (t)},

then every solution of the equation (7.4) is oscillatory on [t0 , ∞). Proof. Suppose the contrary. Then there exists a solution x of the equation (7.4) and a t2 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t2 , ∞).

Let w(t) = δ(t)

r(t)x Δ (t) , x(t)

t ∈ [t2 , ∞).

(7.7)

368

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

By the proof of Theorem 7.8, it follows that there exists a T ≥ t2 so that w(t) > 0, t ∈ [T , ∞). Also, we have w Δ (t) = = = =

Δ  σ Δ δ(t) δ(t)  r(t)x Δ (t) + x(t) r(t)x Δ (t) x(t) σ δ Δ (t)x(t)−x Δ (t)δ(t)  δ(t) r(t)x Δ (t) − x(t) p(t)x(τ (t)) x(t)x σ (t) σ  Δ x(τ (t)) δ (t) Δ −δ(t)p(t) x(t) + x σ (t) r(t)x (t) σ Δ (t)δ(t)  Δ − xx(t)x σ (t) r(t)x (t) (t)) δ Δ (t) σ δ(t)x Δ (t) σ −δ(t)p(t) x(τ x(t) + δ σ (t) w (t) − x(t)δ σ (t) w (t), 

(7.8)

t ∈ [T , ∞). By Theorem 7.8, for T ≥ t2 large enough, it follows that x Δ (t) > 0, and

x(t) t

 Δ r(t)x Δ (t) < 0,

t ∈ [T , ∞),

is strictly decreasing for t ∈ [T , ∞). Hence,  Δ  σ rx (t) ≥ rx Δ (t),

x(σ (t)) ≥ x(t),

x(t) x(τ (t)) ≥ , τ (t) t

t ∈ [T , ∞). Hence from (7.8), we obtain b(t) τ (t) δ(t)r(t)x Δ (t) σ + σ w σ (t) − w (t) t δ (t) r(t)x(t)δ σ (t) σ  δ(t) rx Δ (t) σ τ (t) b(t) σ ≤ −δ(t)p(t) + σ w (t) − w (t) t δ (t) r(t)x σ (t)δ σ (t)  σ 2 b(t) τ (t) δ(t) + σ w σ (t) − = −δ(t)p(t) w (t) , 2 t δ (t) r(t) (δ σ (t))

w Δ (t) ≤ −δ(t)p(t)

t ∈ [T , ∞), whereupon δ(t)p(t)

 σ 2 b(t) δ(t) τ (t) ≤ −w Δ (t) + σ w σ (t) − w (t) , 2 σ t δ (t) r(t) (δ (t))

t ∈ [T , ∞),

and H (t, s)δ(s)p(s)

τ (s) b(s) ≤ −H (t, s)w Δ (s) + H (t, s) σ w σ (s) s δ (s)  σ 2 δ(s) −H (t, s) w (s) , 2 σ r(s) (δ (s))

7.4 The Kamenev-Type Oscillation Criteria

369

t ≥ s ≥ T . Therefore #t T

#t Δ H (t, s)δ(s)p(s) τ (s) s Δs ≤ − #T H (t, s)w (s)Δs t b(s) σ + T H (t, s) δ σ (s) w (s)Δs #t 2 σ − T H (t, s) δ(s) 2 (w (s)) Δs, σ

(7.9)

r(s)(δ (s))

t ≥ T . Note that t T

!s=t ! H (t, s)w Δ (s)Δs = H (t, s)w(s)! − s=T

t

H Δs (t, s)w σ (s)Δs

T

= H (t, t)w(t) − H (t, T )w(T ) t

−

H Δs (t, s)w σ (s)Δs

T t

≥ −H (t, T )w(T ) −

H Δs (t, s)w σ (s)Δs,

T

t ≥ T . From the last inequality and from (7.9), we get t

H (t, s)δ(s)p(s) T

τ (s) Δs ≤ H (t, T )w(T ) s t

−

 h(t, s) H (t, s)w σ (s)Δs

T t

+

H (t, s) T t

−

H (t, s) T

b(s) σ w (s)Δs δ σ (s) 

δ(s) r(s) (δ σ (s))2

2 w σ (s) Δs

= H (t, T )w(T )  t  b(s) w σ (s)Δs h(t, s) H (t, s) − H (t, s) σ − δ (s) T t

−

H (t, s) T



δ(s) r(s) (δ σ (s))2

2 w σ (s) Δs

= H (t, T )w(T )  t  b(s) w σ (s)Δs h(t, s) H (t, s) − H (t, s) σ − δ (s) T t

−

H (t, s) T



δ(s) r(s) (δ σ (s))

2

2 w σ (s) Δs

370

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

t

− T

r(δ )

t

+

⎛ √ 2 ⎞ h H − H δbσ ⎜ ⎟ ⎝ ⎠ (t, s)Δs 4 Hσδ 2

T

%

(δ σ )2 Hr 4δ

h b √ − σ δ H

2 & (t, s)Δs

= H (t, T )w(T ) ⎛1 ⎞2 √ b t H − H h H δ σ δ ⎠ ⎝ 0 wσ + (t, s)Δs − r (δ σ )2 T 2 Hσδ 2 r(δ )

t

+

H (t, s) T

r(s) (δ σ (s))2 4δ(s)

h(t, s) b(s) − σ √ δ (s) H (t, s)

2 Δs

≤ H (t, T )w(T ) t

+

H (t, s) T

r(s) (δ σ (s))2 (p(t, s))2 Δs, 4δ(s)

t ≥ T,

whereupon t T

%

& τ (s) r(s) (δ σ (s))2 2 H (t, s) δ(s)p(s) − (p(t, s)) Δs ≤ H (t, T )w(T ), s 4δ(s)

t ≥ T,

or 1 H (t, T )

t T

& τ (s) r(s) (δ σ (s))2 2 H (t, s) δ(s)p(s) − (p(t, s)) Δs ≤ w(T ), s 4δ(s) %

t ≥ T,

which is a contradiction. This completes the proof. Definition 7.1. The substitution (7.7) is called the Riccati substitution. Corollary 7.5. Suppose that (7.5) holds, H ∈ R and for t > s let H Δs (t, s) h(t, s) = − √ . H (t, s) If there exists a positive delta differentiable function δ such that for any t1 ∈ [t0 , ∞) lim sup t→∞

1 H (t, t1 )

t

H (t, s)δ(s)p(s) t1

τ (s) Δs = ∞ s

(7.10)

7.4 The Kamenev-Type Oscillation Criteria

371

and lim sup t→∞

1 H (t, t1 )

t

H (t, s) t1

r(s) (δ σ (s))2 (p(t, s))2 Δs < ∞, 4δ(s)

(7.11)

then every solution of the equation (7.4) is oscillatory on [t1 , ∞). Proof. By the conditions (7.10) and (7.11), it follows that the condition (7.6) holds. Hence from Theorem 7.9, we conclude that every solution of the equation (7.4) is oscillatory on [t1 , ∞). This completes the proof. Example 7.7. Consider the equation 2

x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ T.

We take H (t, s) = 1,

δ(t) = t,

t, s ∈ T.

Then (7.6) takes the from t

lim sup t→∞

p(s)τ (s) −

t1

1 4s

 Δs = ∞.

Theorem 7.10. Let ∞ t0

1 Δt < ∞ r(t)

and ∞ t0

1 r(t)

t

p(s)ΔsΔt = ∞.

t0

Suppose that x is a nonoscillatory solution of the equation (7.4) such that there exists a t1 ∈ T with x(t)x Δ (t) < 0,

t ≥ t1 .

Then lim x(t) = 0.

t→∞

372

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Proof. Without loss of generality, we assume that x is an eventually positive solution of the equation (7.4). Then there exists a t2 ≥ t1 so that x(t) > 0,

t ≥ t2 .

x(τ (t)) > 0,

Hence, x Δ (t) < 0 for any t ≥ t2 . Therefore x is decreasing on [t2 , ∞). Assume that lim x(t) = b > 0.

t→∞

Pick t3 ≥ t2 so that x(τ (t)) ≥ b for any t ≥ t3 . Integrating both sides of the equation (7.4) from t3 to t, t ≥ t3 , we get t

r(t)x Δ (t) = r(t3 )x Δ (t3 ) −

p(s)x(τ (s))Δs t3 t

≤ r(t3 )x Δ (t3 ) − b

p(s)Δs,

t ≥ t3 ,

p(s)Δs,

t ≥ t3 .

t3

whereupon x Δ (t) ≤

b r(t3 )x Δ (t3 ) − r(t) r(t)

t t3

Now we integrate the last inequality from t3 to t, t ≥ t3 , and we get t

x(t) ≤ x(t3 ) + r(t3 )x Δ (t3 )

t3 t

−b

t3

1 r(s)

→ −∞ as

Δs r(s)

s

p(η)Δη t3

t → ∞.

This is a contradiction. Then lim x(t) = 0.

t→∞

This completes the proof. Theorem 7.11. Let ∞ t0

1 Δt < ∞ and r(t)

∞ t0

1 r(t)

t t0

p(s)ΔsΔt = ∞.

7.5 Advanced Practical Problems

373

Let also, H ∈ R and H Δs (t, s) , h(t, s) = − √ H (t, s)

t > s.

If there exists a positive delta differentiable function δ such that for any t1 ≥ t0 the condition (7.6) holds, then every solution of the equation (7.4) is oscillatory on [t1 , ∞) or converges to zero as t → ∞. Proof. Assume that x is a nonoscillatory solution of the equation (7.4). Without loss of generality, we suppose that x is an eventually positive solution on [t0 , ∞). Then there exists a t2 ≥ t0 so that x(t) > 0,

x(τ (t)) > 0,

t ≥ t2 .

Then Δ  r(t)x Δ (t) = −p(t)x(τ (t)) < 0,

t ≥ t2 .

Hence, rx Δ is decreasing on [t2 , ∞). If x Δ is eventually positive on [t2 , ∞), then x is eventually increasing on [t2 , ∞). This is a contradiction with Theorem 7.9. Let x Δ is eventually negative on [t2 , ∞). Then, by Theorem 7.10, it follows that lim x(t) = 0.

t→∞

This completes the proof.

7.5 Advanced Practical Problems Problem 7.1. Let T = 3N0 . Check if all the solutions of the equation x

Δ2

are oscillatory. Problem 7.2. Let T = 2

x Δ (t) + are oscillatory.

1 x (t) + 8 t + t6 + 1

 t = 0, 9

t ≥ 9,

√ k N0 , k ∈ N. Check if all the solutions of the equation

 1 k k x t − 2 = 0, t 22 + t 4 + t 2 + 3

t≥

√ k 2,

374

7 Oscillations of Second-Order Linear Functional Dynamic Equations. . .

Problem 7.3. Let T = 3N0 . Prove that every solution of the equation x

Δ2

t  = 0, (t) + t + t x 9 

4

2

t ∈ T, t ≥ 9,

is oscillatory. Problem 7.4. Let T = 3N0 . Prove that every bounded solution of the equation x

Δ2

t  = 0, (t) + t + 3t + 7t + 10 x 9 

4

2

t ∈ T, t ≥ 9,

is oscillatory. Problem 7.5. Let T = hZ, h > 0. Prove that every bounded solution of the equation 2

x Δ (t) +

t 8 + 3t 6 + 7t + 10 x (t − 9) = 0, t2 + 1

t ∈ T,

is oscillatory. Problem 7.6. Let T = 4N0 . Prove that every solution of the equation  t  2 x Δ (t) + t 4 + t 2 + 2 x = 0, 8

t ≥ 8,

is oscillatory. Problem 7.7. Let T = 3N0 . Prove that every solution of the equation 2

x Δ (t) + is oscillatory.

100 x t2

 t = 0, 9

t ≥ 9,

Chapter 8

Nonoscillations of Second-Order Functional Dynamic Equations with Several Delays

The results contained in this chapter can be found in the papers and monographs [47, 55, 62, 96]. Suppose that T is a time scale that is unbounded above with forward jump operator and delta differentiation operator σ and Δ, respectively.

8.1 Representation of the Solutions Consider the following delay dynamic equation Δ  " p0 x Δ (t) + nl=1 pl (t)x(τl (t)) = f (t), x(t0 ) = x1 , x Δ (t0 ) = x2 , x(t) = φ(t),

t ∈ [t0 , ∞), t ∈ [t−1 , t0 ),

(8.1)

where n ∈ N, t0 ∈ T, x1 , x2 ∈ R, φ ∈ Crd ([t−1 , t0 )), f ∈ Crd ([t0 , ∞)), pl ∈ Crd ([t0 , ∞)), l ∈ {0, 1, . . . , n}, p0 > 0 on [t0 , ∞), τl ∈ Crd ([t0 , ∞)), τl (t) ≤ σ (t), t ∈ [t0 , ∞), limt→∞ τl (t) = ∞, t−1 =

inf

t∈[t0 ,∞)

τmin (t),

τmin (t) =

min τl (t),

l∈{1,...,n}

t ∈ [t0 , ∞).

For convenience and simplicity, we suppose that the functions vanish out of their specific domains, i.e., if h : D → R be defined for some D ⊂ R, then it is always understood that h(t) = κD (t)h(t), t ∈ R. 1 ([t , ∞)) and a Definition 8.1. A function x : [t−1 , ∞) → R with x ∈ Crd 0 1 Δ derivative satisfying p0 x ∈ Crd ([t0 , ∞)) is called a solution of (8.1) if it satisfies the equation in the first line of (8.1) identically on [t0 , ∞) and the initial conditions in the second line of (8.1).

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_8

375

376

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Note that for a given function φ ∈ Crd ([t−1 , t0 )) with a finite left-sided limit at the initial point t0 provided that it is left-dense and x1 , x2 ∈ R the equation (8.1) admits a unique solution. Definition 8.2. Let s ∈ T and s−1 = inft∈[s,∞) τmin (t). The solutions X1 = X1 (·, s) and X2 = X2 (·, s) of the problems Δ  " p0 x Δ (t) + nl=1 pl (t)x(τl (t)) = 0, = 0, x Δ (s) = p01(s) , x(t)

t ∈ [s, ∞), t ∈ [s−1 , s],

(8.2)

 Δ " p0 x Δ (t) + nl=1 pl (t)x(τl (t)) = 0, t ∈ [s, ∞), = κ{s} (t), t ∈ [s−1 , s], x Δ (s) = 0, x(t)

(8.3)

1 ([s, ∞)), are called the first and the second which satisfy X1 (·, s), X2 (·, s) ∈ Crd fundamental solutions of the equation (8.1), respectively.

Theorem 8.1. Let x be a solution of (8.1). Then x can be written in the form x(t) = x2 X1 (t, t0 ) + x1 X2 (t, t0 ) % t

+

X1 (t, σ (η)) f (η) −

t0

n 

& pl (η)φ(τl (η)) Δη

l=1

for t ∈ [t0 , ∞). Proof. For t ∈ [t−1 , ∞), we define # y(t) =

  " X1 (t, σ (η)) f (η) − nl=1 pl (η)φ(τl (η)) Δη, φ(t), t ∈ [t−1 , t0 ). t t0

t ∈ [t0 , ∞),

For t ∈ [t0 , ∞), we have y (t) = Δ

t t0

% X1Δ (t, σ (η))

f (η) −

+X1 (σ (t), σ (t)) f (t) −

=

t0

pl (η)φ(τl (η)) Δη

n 

& pl (t)φ(τl (t))

l=1

% X1Δ (t, σ (η))

&

l=1

%

t

n 

f (η) −

n 

& pl (η)φ(τl (η)) Δη.

l=1

Hence, y(t0 ) = 0 and

y Δ (t0 ) = 0.

8.1 Representation of the Solutions

377

Also, y=φ

on

[t−1 , t0 ).

Next, 

⎛



t

p0 y Δ (t) =

t0

 Δ p0 y Δ (t) =

p0 (t)X1Δ (t, σ (η)) ⎝f (η) −

t t0

p0 X1Δ (·, σ (η))

Δ

⎛

n 

⎞ pl (η)φ(τl (η))⎠ Δη,

l=1

(t) ⎝f (η) −

+p0 (σ (t))X1Δ (σ (t), σ (t)) ⎝f (t) −

=−

⎛ τl (t)

pl (t)

t0

l=1

+f (t) −

n 

⎞ pl (η)φ(τl (η))⎠ Δη

l=1

⎛

n 

n 

n 

⎞ pl (t)φ(τl (t))⎠

l=1

X1 (τl (t), σ (η)) ⎝f (η) −

n 

⎞ pl (η)φ(τl (η))⎠ Δη

l=1

pl (t)φ(τl (t)),

t ∈ [t0 , ∞).

l=1

For t ∈ [t0 , ∞), we define   I (t) = l ∈ {1, . . . , n} : κ[t0 ,∞) (τl (t)) = 1 ,   J (t) = l ∈ {1, . . . , n} : κ[t−1 ,t0 ) (τl (t)) = 1 . Then 

⎛

⎞

n Δ τl (t)   p0 y Δ (t) = − pl (t) X1 (τl (t), σ (η)) ⎝f (η) − pl (η)φ(τl (η))⎠ Δη t0 l=1 l∈I (t)

−



⎛

τl (t)

pl (t)

t0

l∈J (t)

+f (t) −

n 

X1 (τl (t), σ (η)) ⎝f (η) −

=−

pl (η)φ(τl (η))⎠ Δη

pl (t)φ(τl (t)) ⎛ τl (t)

pl (t)

l∈I (t)

+f (t) −

⎞

l=1

l=1



n 

t0

 l∈J (t)

X1 (τl (t), σ (η)) ⎝f (η) −

pl (t)φ(τl (t))

n  l=1

⎞ pl (η)φ(τl (η))⎠ Δη

378

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

=−



pl (t)y(τl (t)) −

l∈I (t)

=−

n 



pl (t)y(τl (t)) + f (t)

l∈J (t)

pl (t)y(τl (t)) + f (t),

t ∈ [t0 , ∞).

l=1

Consequently y solves the equation n  Δ  pl (t)y(τl (t)) = f (t), p0 y Δ (t) +

t ∈ [t0 , ∞),

l=1

y(t0 ) = 0,

y Δ (t0 ) = 0,

y(t) = φ(t),

t ∈ [t−1 , t0 ).

z(t) = x2 X1 (t, t0 ) + x1 X2 (t, t0 ) + y(t),

t ∈ [t0 , ∞),

Then

is a solution of the equation (8.1). This completes the proof.

8.2 A Nonoscillation Criteria Consider the delay dynamic equation n  Δ  pl (t)x(τl (t)) = 0, p0 x Δ (t) +

t ∈ [t0 , ∞),

(8.4)

t ∈ [t0 , ∞),

(8.5)

t ∈ [t0 , ∞).

(8.6)

l=1

and the delay dynamic inequalities n    Δ Δ (t) + pl (t)x(τl (t)) ≤ 0, p0 x l=1 n  Δ  pl (t)x(τl (t)) ≥ 0, p0 x Δ (t) + l=1

Theorem 8.2. Suppose (A1) (A2) (A3)

p0 ∈ Crd ([t0 , ∞)), p0 > 0 on [t0 , ∞), pl ∈ Crd ([t0 , ∞)), pl ≥ 0 on [t0 , ∞), l ∈ {1, . . . , n}, τl ∈ Crd ([t0 , ∞)), τl : T → T, τl (t) ≤ σ (t), t ∈ [t0 , ∞), limt→∞ τl (t) = ∞, l ∈ {1, . . . , n}.

Then the following assertions are equivalent.

8.2 A Nonoscillation Criteria

379

(i) The equation (8.4) has a nonoscillatory solution. (ii) The inequality (8.5) has an eventually positive solution and/or the inequality (8.6) has an eventually negative solution. 1 ([t , ∞)) with 1 + μ Λ > 0 (iii) There exists a t1 ∈ [t0 , ∞) and a function Λ ∈ Crd 1 p0 on [t1 , ∞) and  1 Λσ (t)Λ(t) + pl (t)e Λ (t, τl (t)) ≤ 0, p0 p0 (t) n

ΛΔ (t) +

(8.7)

l=1

t ∈ [t1 , ∞). (iv) The first fundamental solution X1 of the equation (8.4) is eventually positive, i.e., there exists a t1 ∈ [t0 , ∞) such that X1 (t, s) > 0 for any t ∈ (s, ∞) and any s ∈ [t1 , ∞). Proof. (i) ⇒ (ii) Suppose that the equation (8.4) has an eventually positive solution x. Then x is an eventually positive solution of the inequality (8.5) and −x is an eventually negative solution of the inequality (8.6). (ii) ⇒ (iii) Let x be an eventually positive solution of the inequality (8.5). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0 for any t ∈ [t1 , ∞). Without x(t) loss of generality, we suppose that x(t1 ) = 1. Otherwise, we will consider x(t , 1) t ∈ [t1 , ∞). Let Λ(t) = p0 (t)

x Δ (t) , x(t)

t ∈ [t1 , ∞).

1 ([t , ∞)) and We have Λ ∈ Crd 1

1 + μ(t)

Λ(t) x Δ (t) = 1 + μ(t)p0 (t) p0 (t) x(t)p0 (t) = 1+

μ(t)x Δ (t) x(t)

= 1+

x σ (t) − x(t) x(t)

=

x σ (t) x(t)

> 0,

t ∈ [t1 , ∞).

Therefore e Λ (·, t1 ) is well defined and it is positive on [t1 , ∞). We have p0

x Δ (t) =

Λ(t)x(t) , p0 (t)

x(t1 ) = 1.

t ∈ [t1 , ∞),

380

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Therefore x(t) = e Λ (t, t1 ), p0

t ∈ [t1 , ∞).

Also,

Λ 1+μ  p0

 = 1+μ = =

− pΛ0 1 + μ pΛ0

1 1 + μ pΛ0 p0 p0 + μΛ

> 0 on

[t1 , ∞).

Therefore e Λ (t, t1 ) and e Λ (t, τl (t)) are well defined for any t ∈ [t1 , ∞) and p0

p0

they are positive. Then

Δ   Δ Δ Δ (t) = p0 e Λ (·, t1 ) (t) p0 x p0

Δ

= Λe Λ (·, t1 ) (t) p0

= Λ (t)e Λ (t, t1 ) + Λσ (t)eΔΛ (t, t1 ) Δ

p0

p0

= ΛΔ (t)e Λ (t, t1 ) + Λσ (t) p0

Λ(t) e Λ (t, t1 ), p0 (t) p0

t ∈ [t1 , ∞). Hence, (8.5) takes the form 0 ≥ ΛΔ (t)e Λ (t, t1 ) + Λσ (t) p0

+

n  l=1

Λ(t) e Λ (t, t1 ) p0 (t) p0

pl (t)e Λ (τl (t), t1 ) p0

⎛

⎞ n e Λ (τl (t), t1 )  Λ(t) p ⎠ + pl (t) 0 = e Λ (t, t1 ) ⎝ΛΔ (t) + Λσ (t) p0 p0 (t) e Λ (t, t1 ) l=1

Λ(t) = e Λ (t, t1 ) ΛΔ (t) + Λσ (t) p0 p0 (t)

p0

8.2 A Nonoscillation Criteria

+

n 

381

 pl (t)e Λ (t1 , τl (t))e Λ (t, t1 ) p0

l=1

p0

%

& n Λ(t)  + pl (t)e Λ (t, τl (t)) , = e Λ (t, t1 ) Λ (t) + Λ (t) p0 p0 p0 (t) Δ

σ

l=1

t ∈ [t1 , ∞). Since e Λ (t, t1 ) > 0, p0

t ∈ [t1 , ∞),

we get Λ(t)  + pl (t)e Λ (t, τl (t)) ≤ 0, p0 p0 (t) n

ΛΔ (t) + Λσ (t)

l=1

t ∈ [t1 , ∞). (iii) ⇒ (iv) Consider the IVP  Δ " p0 x Δ (t) + nl=1 pl (t)x(τl (t)) = f (t), t ∈ [t1 , ∞), = 0, t ∈ [t−1 , t1 ]. x Δ (t1 ) = 0, x(t)

(8.8)

We set g(t) = p0 (t)x Δ (t) − Λ(t)x(t),

t ∈ [t1 , ∞),

where x is a solution of (8.8) and Λ is a solution of the inequality (8.7). Then x Δ (t) =

g(t) Λ(t) x(t) + , p0 (t) p0 (t)

t ∈ [t1 , ∞),

x(t1 ) = 0. The last IVP has a unique solution, represented in the form x(t) =

t

e Λ (t, σ (η)) t1

p0

g(η) Δη, p0 (η)

t ∈ [t1 , ∞).

Then x(t) =

t

e Λ (t, σ (η)) p0

t1

= e Λ (σ (t), t) p0

g(η) Δη p0 (η) t

e Λ (σ (t), σ (η)) t1

p0

g(η) Δη p0 (η)

382

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

= e Λ (σ (t), t) p0

σ (t)

e Λ (σ (t), σ (η))

−e Λ (σ (t), t) p0

p0

t1

g(η) Δη p0 (η)

σ (t)

e Λ (σ (t), σ (η)) p0

t

g(η) Δη p0 (η)

= e Λ (σ (t), t)x σ (t) p0

−μ(t)e Λ (σ (t), t)e Λ (σ (t), σ (t)) p0

p0

g(t) p0 (t)

g(t) p0 (t)μ(t) p0 (t) x σ (t) − p0 (t) + μ(t)Λ(t) p0 (t) + μ(t)Λ(t) p0 (t)

 g(t) p0 (t) x σ (t) − μ(t) = p0 (t) + μ(t)Λ(t) p0 (t)

=

=

  1 p0 (t)x σ (t) − μ(t)g(t) , p0 (t) + μ(t)Λ(t)

t ∈ [t1 , ∞).

Next, τl (t)

x(τl (t)) =

e Λ (τl (t), σ (η)) p0

t1

g(η) Δη p0 (η)

τl (t)

=

e Λ (τl (t), σ (t))e Λ (σ (t), σ (η)) p0

t1

p0

= e Λ (σ (t), τl (t)) p0

σ (t)

−

τl (t)

g(η) Δη p0 (η)

σ (t)

e Λ (σ (t), σ (η)) p0

t1

g(η) Δη e Λ (σ (t), σ (η)) p0 p0 (η) %

= e Λ (σ (t), τl (t)) x (t) − σ

p0

= e Λ (σ (t), τl (t))x σ (t) − p0

σ (t)

τl (t)



g(η) Δη p0 (η)

g(η) Δη e Λ (σ (t), σ (η)) p0 p0 (η)

σ (t)

e Λ (τl (t), σ (η)) τl (t)

p0

g(η) Δη, p0 (η)

t ∈ [t1 , ∞), l ∈ {1, . . . , n}. Also, Δ  p0 x Δ (t) = (Λx + g)Δ (t) = ΛΔ (t)x σ (t) + Λ(t)x Δ (t) + g Δ (t) = ΛΔ (t)x σ (t) +

g(t) (Λ(t))2 x(t) + Λ(t) + g Δ (t), p0 (t) p0 (t)

&

8.2 A Nonoscillation Criteria

383

t ∈ [t1 , ∞). Therefore n  Δ  f (t) = p0 x Δ (t) + pl (t)x(τl (t)) l=1

= ΛΔ (t)x σ (t) + +g Δ (t) +

pl (t)x(τl (t))

l=1

% =

n 

Λ(t) (Λ(t))2 x(t) + g(t) p0 (t) p0 (t)

ΛΔ (t)x σ (t) +

& n  (Λ(t))2 pl (t)x(τl (t)) x(t) + p0 (t) l=1

+ =

Λ(t) g(t) + g Δ (t) p0 (t)

ΛΔ (t)x σ (t) + − +

μ(t) (Λ(t))2 g(t) p0 (t) p0 (t) + μ(t)Λ(t) n 

pl (t)e Λ (σ (t), τl (t))x σ (t) p0

l=1

−

n 

σ (t)

pl (t)

p0

g(η) Δη p0 (η)

Λ(t) g(t) + g Δ (t) p0 (t) Δ

% −

l=1

 μ(t) (Λ(t))2 pl (t) g(t) + p0 (t) (p0 (t) + μ(t)Λ(t)) n

l=1

+ % =



& n  (Λ(t))2 pl (t)e Λ (σ (t), τl (t)) x σ (t) + Λ (t) + p0 p0 (t) + μ(t)Λ(t)

% =

e Λ (τl (t), σ (η)) τl (t)

l=1

+

(Λ(t))2 x σ (t) p0 (t) + μ(t)Λ(t)

σ (t)

e Λ (τl (t), σ (η)) τl (t)

Λ(t) g(t) + g Δ (t) p0 (t)

ΛΔ (t) +

& n  (Λ(t))2 + pl (t)e Λ (σ (t), τl (t)) p0 p0 (t) + μ(t)Λ(t) l=1

 Λ(t) × 1 + μ(t) p0 (t) −

n  l=1

+

p0

σ (t)

e Λ (t, σ (η)) t1

p0

σ (t)

pl (t)

e Λ (τl (t), σ (η)) τl (t)

p0

Λ(t) g(t) + g Δ (t), p0 (t) + μ(t)Λ(t)

g(η) Δη p0 (η)

g(η) Δη p0 (η) t ∈ [t1 , ∞).

g(η) Δη p0 (η)

&

384

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Let %

 (Λ(t))2 Y (t) = Λ (t) + pl (t)e Λ (σ (t), τl (t)) + p0 p0 (t) + μ(t)Λ(t) l=1 

Λ(t) , × 1 + μ(t) p0 (t) n

Δ

t ∈ [t1 , ∞). We have 

1 Λ(t) ΛΔ (t) + Y (t) = − 1 + μ(t) (Λ(t))2 p0 (t) p0 (t)  n  + pl (t)e Λ (t, τl (t)) p0

l=1

%

Λσ (t)Λ(t)  + pl (t)e Λ (t, τl (t)) = − Λ (t) + p0 p0 (t) n

Δ

l=1

≥ 0,

t ∈ [t1 , ∞).

Also, g Δ (t) = −

Λ(t) g(t) p0 (t) + μ(t)Λ(t)

+Y (t)

σ (t)

e Λ (t, σ (η)) t1

+

n 

p0

g(η) Δη p0 (η)

σ (t)

pl (t)

l=1

+f (t),

e Λ (τl (t), σ (η)) τl (t)

p0

t ∈ [t1 , ∞).

Also, g(t1 ) = 0.

g(η) Δη p0 (η)

&

&

8.2 A Nonoscillation Criteria

385

We set t

H g(t) =

e−

t1

n 

+

Λ p0 +μΛ

(t, σ (η)) Y (η)

σ (η)

e Λ (σ (η), σ (ζ )) p0

t1

 g(ζ ) Δζ Δη, e Λ (τl (η), σ (ζ )) p0 p0 (ζ )

σ (η)

pl (η) τl (η)

l=1 t

h(t) =

e−

t1

Λ p0 +μΛ

g(ζ ) Δζ p0 (ζ )

t ∈ [t1 , ∞).

(t, σ (ζ ))f (ζ )Δζ,

Then g =H g+h

on

[t1 , ∞).

Note that f (t) ≥ 0, t ∈ [t1 , ∞), implies h(t) ≥ 0, t ∈ [t1 , ∞). We set K1 (t, s) =

t 1 e (t, σ (η))Y (η)e Λ (σ (η), σ (s))Δη, Λ p0 (s) s − p0 +μΛ p0

K2 (t, s) =

n t  1 e− Λ (t, σ (η)) pl (η)κ[τl (η),∞) (s)e Λ (τl (η), σ (s))Δη, p0 (s) s p0 +μΛ p0 l=1

s, t ∈ [t1 , ∞). Note that t

σ (η)

t1

e−

t1 t

=

Λ p0 +μΛ

(t, σ (η))Y (η)e Λ (σ (η), σ (ζ )) p0

t

t1

ζ

e−

Λ p0 +μΛ

(t, σ (η))Y (η)e Λ (σ (η), σ (ζ )) p0

t

=

g(ζ ) Δζ Δη p0 (ζ ) g(ζ ) ΔηΔζ p0 (ζ )

t ∈ [t1 , ∞),

K1 (t, ζ )g(ζ )Δζ, t1

and t t1

=

σ (η) t1 t t1

=

e−

Λ p0 +μΛ

(t, σ (η))

ζ

e−

Λ p0 +μΛ

pl (η)κ[τl (η),∞) (ζ )e Λ (τl (η), σ (ζ )) p0

l=1

t

(t, σ (η))

t

K2 (t, ζ )g(ζ )Δζ, t1

n 

n 

g(ζ ) Δζ Δη p0 (ζ )

pl (η)κ[τl (η),∞) (ζ )e Λ (τl (η), σ (ζ ))

l=1

t ∈ [t1 , ∞).

p0

g(ζ ) ΔηΔζ p0 (ζ )

386

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Therefore H g(t) =

t

(K1 (t, η) + K2 (t, η)) g(η)Δη,

t ∈ [t1 , ∞).

t1

By Theorem 6.23, we get g ≥ 0 on [t1 , ∞) when h ≥ 0 on [t1 , ∞). Therefore x ≥ 0 on [t1 , ∞) if f ≥ 0 on [t1 , ∞). On the other hand, we have for x the representation x(t) =

t

X1 (t, σ (η)f (η)Δη,

t ∈ [t1 , ∞).

(8.9)

t1

Assume that there is a t2 ∈ [t1 , ∞) and an s ∈ [t1 , t2 ) such that X1 (t2 , σ (s)) < 0. Let f (t) = − min{X1 (t2 , σ (t)), 0},

t ∈ [t1 , ∞).

Then f (t) ≥ 0,

t ∈ [t1 , ∞).

Hence from (8.9), we get x(t2 ) < 0, which is a contradiction. We set t ∈ [t0 , ∞),

x(t) = X1 (t, s),

s ∈ [t1 , ∞).

 Δ Then x ≥ 0 and p0 x Δ ≤ 0 on [s, ∞). Therefore p0 x Δ is nonincreasing on [s, ∞). We take t1 ∈ [t0 , ∞) large enough such that x Δ is of fixed sign on [s, ∞) ⊂ [t1 , ∞). Since x Δ (s) =

1 , p0 (s)

we get x Δ > 0 on [s, ∞). Therefore x(t) = X1 (t, s) > X1 (s, s) = 0,

t ∈ (s, ∞) ⊂ [t1 , ∞).

(iv) ⇒ (i) We have that X1 (·, t0 ) is an eventually positive solution of the equation (8.4). This completes the proof.

8.2 A Nonoscillation Criteria

387

Definition 8.3. The inequality (8.7) is called the Riccati inequality. Theorem 8.3. Let (A1) and (A2) hold. Let also, ∞ t0

1 Δη = ∞. p0 (η)

Then for every nonoscillatory solution x of the equation (8.4) there exists a t1 ∈ [t0 , ∞) such that x(t)x Δ (t) ≥ 0 for any t ∈ [t1 , ∞). Proof. Let x be a nonoscillatory solution of the equation (8.4). Without loss of generality, assume that x is an eventually positive solution of the equation (8.4). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Assume that there is a t2 ∈ [t1 , ∞) such that x Δ (t2 ) < 0. We integrate the equation (8.4) from t2 to t, t ≥ t2 , and we get p0 (t)x Δ (t) = p0 (t2 )x Δ (t2 ) −

n 

t

l=1

t2

pl (ζ )x(τl (ζ ))Δζ,

whereupon % n  1 1 − x (t) = p0 (t2 )x (t2 ) p0 (t) p0 (t) Δ

l=1

&

t

Δ

pl (ζ )x(τl (ζ ))Δζ . t2

Now we integrate the last equation from t2 to t, t ≥ t2 , and we get t

x(t) = x(t2 ) + p0 (t2 )x Δ (t2 ) −

t t2

% n  1 p0 (η) l=1

t2

Δη p0 (η)

η

pl (ζ )x(τl (ζ ))Δζ Δη. t2

Hence, x(t) → −∞

&

as

t → ∞.

388

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

This is a contradiction. Therefore x Δ (t) ≥ 0,

t ∈ [t1 , ∞).

This completes the proof. Example 8.1. Let T = hZ, h > 0. Then %

#t

1 τl (t) μ(s)

e Λ (t, τl (t)) = e

log 1−μ(s)

p0

=e =e =e =

#t

1 τl (t) μ(s)

#t

1 τl (t) h

"t−h

 log 

log

Λ(s) p0 (s) 1+μ(s) Λ(s) p0 (s)

p0 (s) p0 (s)+μ(s)Λ(s)

p0 (s) p0 (s)+hΛ(s)



m=τl (t)

Δs

Δs

Δs

p0 (m) m=τl (t) log p0 (m)+hΛ(m)

t−h $

&



p0 (m) , p0 (m) + hΛ(m)

t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Then the Riccati inequality (8.7) takes the form ΛΔ (t) +

n t−h  $ p0 (m) 1 Λσ (t)Λ(t) + ≤ 0, pl (t) p0 (t) p0 (m) + hΛ(m) l=1

m=τl (t)

t ∈ [t1 , ∞). Exercise 8.1. Let T = 2Z . Prove that the Riccati inequality (8.7) takes the form t

n 2  $ p0 (m) 1 ΛΔ (t) + Λσ (t)Λ(t) + ≤ 0, pl (t) p0 (t) p0 (m) + mΛ(m) l=1

m=τl (t)

t ∈ [t1 , ∞).

8.3 Comparison Theorems Consider the equation n    Δ Δ (t) + ql (t)x(τl (t)) = 0, p0 x l=1

where ql ∈ Crd ([t0 , ∞)), l ∈ {1, . . . , n}.

t ∈ [t0 , ∞),

(8.10)

8.3 Comparison Theorems

389

Theorem 8.4. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞ p0 (η)

and ql ∈ Crd ([t0 , ∞)), pl (t) ≥ ql (t), t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Assume that 1 ([t , ∞)) with 1 + μ Λ ≥ 0 the Riccati inequality (8.7) admits a solution Λ ∈ Crd 1 p0 on [t1 , ∞) for some t1 ∈ [t0 , ∞). Then the first fundamental solution Y1 of the equation (8.10) satisfies Y1 (t, s) ≥ X1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). Proof. Consider the IVP n   Δ p0 x Δ (t) + ql (t)x(τl (t)) = f (t),

t ∈ [t0 , ∞),

l=1

x Δ (t0 ) = 0,

x(t) = 0,

t ∈ [t−1 , t0 ],

where f ∈ Crd ([t0 , ∞)). Let g ∈ Crd ([t1 , ∞)). Define the function x as follows. x(t) =

t

X1 (t, σ (η))g(η)Δη,

t ∈ [t1 , ∞).

t1

Then x Δ (t) =

t t1

=

t t1

  p0 x Δ (t) =

t t1

 Δ p0 x Δ (t) =

t t1

X1Δ (t, σ (η))g(η)Δη + X1 (σ (t), σ (t))g(t) X1Δ (t, σ (η))g(η)Δη,   p0 X1Δ (·, σ (η)) (t)g(η)Δη, Δ  p0 X1Δ (·, σ (η)) (t)g(η)Δη

+p0 (σ (t))X1Δ (σ (t), σ (t))g(t) =

t t1

 Δ p0 X1Δ (·, σ (η)) (t)g(η)Δη + g(t),

Therefore n   Δ f (t) = p0 x Δ (t) + ql (t)x(τl (t)) l=1

=

t t1

 Δ p0 X1Δ (·, σ (η)) (t)g(η)Δη

t ∈ [t1 , ∞).

390

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

+

n 

τl (t)

ql (t)

X1 (τl (t), σ (η))g(η)Δη

t1

l=1

+g(t) =−

n 

τl (t) t1

l=1

+

n 

τl (t)

ql (t)

=

X1 (τl (t), σ (η))g(η)Δη + g(t)

t1

l=1 n 

X1 (τl (t), σ (η))g(η)Δη

pl (t)

X1 (τl (t), σ (η))g(η)Δη + g(t)

t1

l=1

=

τl (t)

(ql (t) − pl (t))

n  (ql (t) − pl (t))

t

X1 (τl (t), σ (η))g(η)Δη + g(t),

t1

l=1

t ∈ [t1 , ∞). Let t

H g(t) =

n 

X1 (τl (t), σ (η)) (pl (t) − ql (t)) g(η)Δη,

t1 l=1

t ∈ [t1 , ∞). Then g =H g+f

on

[t1 , ∞).

When f is nonnegative on [t1 , ∞), by Theorem 6.23, it follows that g is nonnegative on [t1 , ∞) and hence, x is nonnegative on [t1 , ∞). On the other hand, x can be represented in the form x(t) =

t

Y1 (t, σ (η))f (η)Δη,

t ∈ [t0 , ∞).

t0

Assume that there is a t2 ∈ [t1 , ∞) and an s ∈ [t1 , t2 ) so that Y1 (t2 , σ (s)) < 0. Take f (t) = − min {Y1 (t2 , σ (t)), 0} ,

t ∈ [t1 , ∞).

Then f (t) ≥ 0,

t ∈ [t1 , ∞).

(8.11)

8.3 Comparison Theorems

391

Hence from (8.11), we get that x(t2 ) < 0, which is a contradiction. Then Y1 (t, s) ≥ 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). For any fixed s ∈ [t1 , ∞) and for any t ∈ [s, ∞), we have n n    Δ p0 Y1Δ (·, s) (t) + pl (t)Y1 (τl (t), s) = (pl (t) − ql (t)) Y1 (τl (t), s). l=1

l=1

Hence, t

Y1 (t, s) = X1 (t, s) +

X1 (t, σ (η))

s

n 

(pl (η) − ql (η)) Y1 (τl (η), s)Δη

l=1

≥ X1 (t, s) for all t ∈ [s, ∞). By Theorem 8.2, (iv), it follows that X1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). This completes the proof. Corollary 8.1. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞ p0 (η)

and ql ∈ Crd ([t0 , ∞)), pl (t) ≥ ql (t) for any t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Let also, the equation (8.4) has a nonoscillatory solution on [t1 , ∞) ⊂ [t0 , ∞). Then (8.10) admits a nonoscillatory solution on [t2 , ∞) ⊂ [t1 , ∞). Proof. Let x be an eventually positive solution of the equation (8.4) on [t1 , ∞) ⊂ [t0 , ∞). Then, by Theorem 8.2, it follows that the first fundamental solution X1 of the equation (8.4) is eventually positive and there exists a t2 ∈ [t1 , ∞) such that 1 ([t , ∞)) with 1 + μ Λ > 0 on [t , ∞) and (8.7) holds. Hence there is a Λ ∈ Crd 1 2 p0 from Theorem 8.4, it follows that the fundamental solution Y1 of the equation (8.10) satisfies Y1 (t, s) ≥ X1 (t, s) > 0 for any t ∈ (s, ∞) and for any s ∈ [t2 , ∞). Therefore the equation (8.10) admits an eventually positive solution on [t2 , ∞). This completes the proof. Corollary 8.2. Assume (A1) and (A3) hold, and ∞ t0

1 Δη = ∞. p0 (η)

392

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

(i) If the dynamic inequality n   Δ p0 x Δ (t) + pl+ (t)x(τl (t)) ≤ 0,

t ∈ [t0 , ∞),

l=1

where pl+ (t) = max{pl (t), 0},

t ∈ [t0 , ∞),

l ∈ {1, . . . , n},

has a positive solution on [t0 , ∞), then the equation (8.4) has a positive solution on [t1 , ∞) ⊂ [t0 , ∞). 1 ([t , ∞)), (ii) If there exists a sufficiently large t1 ∈ [t0 , ∞) and a function Λ ∈ Crd 1 Λ 1 + μ p0 ≥ 0 on [t1 , ∞), satisfying the inequality  1 Λσ (t)Λ(t) + pl+ (t)e Λ (t, τl (t)) ≤ 0, p0 p0 (t) n

ΛΔ (t) +

(8.12)

l=1

t ∈ [t1 , ∞), then the first fundamental solution X1 of the equation (8.4) satisfies X1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). Proof. (i) Consider the dynamic equation n   Δ p0 x Δ (t) + pl+ (t)x(τl (t)) = 0,

t ∈ [t0 , ∞).

(8.13)

l=1

By Theorem 8.2, it follows that this equation has an eventually positive solution. Since pl+ (t) ≥ pl (t),

t ∈ [t0 , ∞),

l ∈ {1, . . . , n},

by Corollary 8.1, it follows that the equation (8.4) admits an eventually positive solution. (ii) By Theorem 8.4, it follows that the first fundamental solution Z1 of the equation (8.13) satisfies X1 (t, s) ≥ Z1 (t, s) > 0 for any t ∈ (s, ∞) and for any s ∈ [t1 , ∞). This completes the proof. Now we consider the following IVP  Δ " p0 x Δ (t) + nl=1 ql (t)x(τl (t)) = g(t), t ∈ [t0 , ∞), x(t0 ) = y1 , x Δ (t0 ) = y2 , x(t) = ψ(t), t ∈ [t−1 , t0 ),

(8.14)

8.3 Comparison Theorems

393

where ql , g ∈ Crd ([t0 , ∞)), l ∈ {1, . . . , n}, ψ ∈ Crd ([t−1 , t0 )), ψ has a finite left-sided limit at the initial point t0 provided that it is left-dense, y1 , y2 ∈ R. Theorem 8.5. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

and ql ∈ Crd ([t0 , ∞)), pl (t) ≥ ql (t), t ∈ [t0 , ∞), l ∈ {1, . . . , n}. Let also, f, g ∈ Crd ([t0 , ∞)), φ, ψ ∈ Crd ([t−1 , t0 )) satisfy f (t) −

n 

ql (t)φ(τl (t)) ≤ g(t) −

l=1

n 

ql (t)ψ(τl (t)),

l=1

t ∈ [t0 , ∞). Moreover, let (8.1) has a positive solution x on [t0 , ∞), y1 = x1 and y2 ≥ x2 . Then the solution y of (8.14) satisfies y(t) ≥ x(t),

t ∈ [t0 , ∞).

1 ([t , ∞)), 1 + μ Λ on Proof. By Theorem 8.2, it follows that there exists a Λ ∈ Crd 0 p0 [t0 , ∞) that satisfies the Riccati inequality (8.7). Then, using that

ql (t) ≤ pl (t),

t ∈ [t0 , ∞),

l ∈ {1, . . . , n},

we get  1 Λσ (t)Λ(t) + ql (t)e Λ (t, τl (t)) ≤ 0, Λ (t) + p0 p0 (t) n

Δ

l=1

t ∈ [t0 , ∞). Hence from Theorem 8.2, it follows that the first fundamental solution Y1 of (8.14) satisfies Y1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t0 , ∞). We have that n n    Δ Δ  p0 x Δ (t) + ql (t)x(τl (t)) = p0 x Δ (t) + pl (t)x(τl (t)) l=1

l=1

−

n 

(pl (t) − ql (t)) x(τl (t))

l=1

= f (t) −

n  l=1

(pl (t) − ql (t)) x(τl (t)),

394

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

t ∈ [t0 , ∞). From here and from Theorem 8.1, we get x(t) = x2 Y1 (t, s) + x1 Y2 (t, s) t

+

Y1 (t, σ (η)) ×

t0

n  × f (η) − (pl (η) − ql (η)) κ[t0 ,∞) (τl (η))x(τl (η)) l=1

−

n 

 ql (η)φ(τl (η)) Δη

l=1

≤ x2 Y1 (t, t0 ) + x1 Y2 (t, t0 ) % +

t

Y1 (t, σ (η)) f (η) −

t0

+

& ql (η)φ(τl (η)) Δη

l=1

≤ y2 Y1 (t, t0 ) + y1 Y2 (t, t0 ) % t

n 

Y1 (t, σ (η)) g(η) −

t0

n 

& ql (η)ψ(τl (η)) Δη

l=1

t ∈ [t0 , ∞).

= y(t), This completes the proof.

Corollary 8.3. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

and that (8.4) is nonoscillatory. Then for f ∈ Crd ([t0 , ∞)), f ≥ 0 on [t0 , ∞), the dynamic equation 

p0 x Δ

Δ

(t) +

n 

pl (t)x(τl (t)) = f (t),

t ∈ [t0 , ∞),

(8.15)

l=1

is also nonoscillatory. Proof. Since, −

n  l=1

pl (t)x(τl (t)) ≤ f (t) −

n 

pl (t)x(τl (t)),

t ∈ [t0 , ∞),

l=1

by Theorem 8.5, it follows that the equation (8.15) is nonoscillatory. This completes the proof.

8.3 Comparison Theorems

395

Now we consider the dynamic equation n Δ  " pl (t)x(τl (t)) p0 x Δ (t) + l=1 x Δ (t

x(t0 ) = y1 ,

0)

= y2 ,

= g(t),

t ∈ [t0 , ∞),

x(t) = ψ(t),

(8.16)

t ∈ [t−1 , t0 ),

where the parameters are the same as in (8.14). Corollary 8.4. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

f, g ∈ Crd ([t0 , ∞)), φ, ψ ∈ Crd ([t−1 , t0 )) satisfy f (t) −

n 

pl (t)φ(τl (t)) ≤ g(t) −

l=1

n 

pl (t)ψ(τl (t)),

t ∈ [t0 , ∞).

(8.17)

l=1

If x is a positive solution of the equation (8.1) on [t0 , ∞) with x1 = y1 and y2 ≥ x2 , then the solution y of (8.16) satisfies y(t) ≥ x(t),

t ∈ [t0 , ∞).

Proof. By Corollary 8.2 and Theorem 8.2, it follows that the first fundamental solution X1 associated with the equation (8.1) satisfies X1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t0 , ∞). Hence from the representation formula, we obtain x(t) = x2 X1 (t, t0 ) + x1 X2 (t, t0 ) % +

t

X1 (t, σ (η)) f (η) −

t0

+

& pl (η)φ(τl (η)) Δη

l=1

≤ y2 X1 (t, t0 ) + y1 X2 (t, t0 ) % t

n 

X1 (t, σ (η)) g(η) −

t0

= y(t), This completes the proof.

n  l=1

t ∈ [t0 , ∞).

& pl (η)ψ(τl (η)) Δη

396

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Now we consider the equation n   Δ p0 x Δ (t) + ql (t)x(γl (t)) = 0,

t ∈ [t0 , ∞),

(8.18)

l=1

where ql , γl ∈ Crd ([t0 , ∞)), l ∈ {1, . . . , n}. Theorem 8.6. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

ql ∈ Crd ([t0 , ∞)), pl (t) ≥ ql (t), t ∈ [t0 , ∞), l ∈ {1, . . . , n}, and γl ∈ Crd ([t0 , ∞)), γl (t) ≤ τl (t), t ∈ [t0 , ∞), limt→∞ γl (t) = ∞, l ∈ {1, . . . , n}. Let 1 ([t , ∞)), 1 + μ Λ the Riccati dynamic inequality (8.7) has a solution Λ ∈ Crd 1 p0 on [t1 , ∞) for some t1 ∈ [t0 , ∞). Then the first fundamental solution Y1 of (8.18) satisfies Y1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). Proof. Since pl (t) ≥ ql+ (t),

t ∈ [t0 , ∞),

l ∈ {1, . . . , n},

we have  (Λ(t))2 + pl (t)e Λ (σ (t), τl (t)) p0 p0 (t) + μ(t)Λ(t) n

0 ≥ ΛΔ (t) +

l=1

≥ ΛΔ (t) +

 (Λ(t)) + ql+ (t)e Λ (σ (t), τl (t)), p0 p0 (t) + μ(t)Λ(t) 2

n

l=1

t ∈ [t1 , ∞). Hence from Corollary 8.2, it follows that the first fundamental solution Y1 of (8.18) satisfies Y1 (t, s) > 0 for all t ∈ (s, ∞) and for all s ∈ [t1 , ∞). This completes the proof. We introduce the notation τmax (t) =

max {τl (t)} ,

l∈{1,...,n}

t ∈ [t0 , ∞).

8.4 Explicit Nonoscillation and Oscillation Results

397

Corollary 8.5. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

ql ∈ Crd ([t0 , ∞)), pl (t) ≥ ql (t), t ∈ [t0 , ∞), l ∈ {1, . . . , n}. If the equation % n &    Δ Δ (t) + pl (t) x(τmax (t)) = 0, p0 x

t ∈ [t0 , ∞),

(8.19)

l=1

has a nonoscillatory solution, then the equation (8.10) has a nonoscillatory solution. Proof. Without loss of generality, we suppose that the equation (8.19) has an eventually positive solution. Then, by Theorem 8.2, it follows that there exists a 1 ([t , ∞)) such that 1 + μ Λ on [t , ∞) and t1 ∈ [t0 , ∞) and Λ ∈ Crd 1 1 p0  1 Λσ (t)Λ(t) + pl (t)e Λ (t, τmax (t)) ≤ 0, p0 p0 (t) n

ΛΔ (t) +

l=1

t ∈ [t1 , ∞). From here,  1 Λ (t) + pl (t)e Λ (t, τl (t)) ≤ 0, Λσ (t)Λ(t) + p0 p0 (t) n

Δ

l=1

t ∈ [t1 , ∞). Then, using Theorem 8.6, it follows that the first fundamental solution associated with the equation (8.10) is eventually positive. Again applying Theorem 8.2, we conclude that the equation (8.10) has a nonoscillatory solution. This completes the proof.

8.4 Explicit Nonoscillation and Oscillation Results Theorem 8.7. Suppose that (A1), (A2), and (A3) hold, and  σ (t) + 2tσ (t) pl (t)e 1 (σ (t), τl (t)) ≤ 1 2I dp0 2tp0 (t) + μ(t) n

l=1

for any t ∈ [t1 , ∞) and for some t1 > 0. Then the equation (8.4) has a nonoscillatory solution. Here I d is the identity function on T.

398

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Proof. Take t1 > 1 arbitrarily. Let Λ(t) =

1 , 2t

t ∈ [t1 , ∞).

Then ΛΔ (t) = −

1 , 2tσ (t)

e Λ (t, τl (t)) = e Λ (t, σ (t))e Λ (σ (t), τl (t)) p0

p0

p0

= e Λ (σ (t), t)e Λ (σ (t), τl (t)) p0

p0

=

p0 (t) + μ(t)Λ(t) e Λ (σ (t), τl (t)) p0 p0 (t)

=

2tp0 (t) + μ(t) e Λ (σ (t), τl (t)), p0 2tp0 (t)

t ∈ [t1 , ∞). Therefore  1 Λσ (t)Λ(t) + pl (t)e Λ (t, τl (t)) p0 p0 (t) n

ΛΔ (t) +

l=1

= ΛΔ (t) +

 1  μ(t)ΛΔ (t) + Λ(t) Λ(t) p0 (t)

2tp0 (t) + μ(t)  pl (t)e 1 (σ (t), τl (t)) 2I dp0 2tp0 (t) n

+

l=1

=

p0 (t) + μ(t)Λ(t) Δ (Λ(t))2 Λ (t) + p0 (t) p0 (t) 2tp0 (t) + μ(t)  pl (t)e 1 (σ (t), τl (t)) 2I dp0 2tp0 (t) n

+

l=1

1 2tp0 (t) + μ(t) + 2 =− 2 4t σ (t)p0 (t) 4t p0 (t) 2tp0 (t) + μ(t)  + pl (t)e 1 (σ (t), τl (t)) 2I dp0 2tp0 (t) n

l=1

8.4 Explicit Nonoscillation and Oscillation Results

2tp0 (t) + μ(t) = 4t 2 σ (t)p0 (t) +2tσ (t)

n 

2tp0 (t) + μ(t) ≤ 4tσ (t)p0 (t) n 

−1+

pl (t)e

l=1

+2tσ (t)

−1+

pl (t)e

≤ 0,

σ (t) 2tp0 (t) + μ(t)  (σ (t), τl (t))

1 2I dp0

l=1

399

σ (t) 2tp0 (t) + μ(t)  (σ (t), τl (t))

1 2I dp0

t ∈ [t1 , ∞),

i.e., Λ solves the Riccati inequality (8.7). Hence from Theorem 8.2, it follows that the equation (8.4) has a nonoscillatory solution. This completes the proof. Example 8.2. Let T = 2N0 . Consider the equation Δ  p0 x Δ (t) +

1 x 200t 4



 1 t t + = 0, x 7 2 4 300t

where p0 (t) = t. Here σ (t) = 2t, μ(t) = t, 1 , 200t 4 1 , p2 (t) = 300t 7 t τ1 (t) = , 2 t τ2 (t) = , 4

p1 (t) =

e

1 2I dp0

(σ (t), τ1 (t)) =

t $ m= 2t

=

t $ m= 2t

2mp0 (m) μ(m) + 2mp0 (m) 2m2 m + 2m2

t ∈ [8, ∞),

400

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

=

t $ m= 2t

= =

2m 1 + 2m

t 1+t

  2t · 1 + 2t

2t 2 (1 + t)(1 + 2t)

≤ 1, e

1 2I dp0

(σ (t), τ2 (t)) =

t $ m= 4t

=

t $ m= 4t

%

= =

2mp0 (m) μ(m) + 2mp0 (m) 2m 1 + 2m &

t 2

1+

t 2

t 1+t



2t 1 + 2t



2t 3 (2 + t)(1 + t)(1 + 2t)

≤ 1,

t ∈ [8, ∞).

Hence,

 σ (t) + 2tσ (t) p1 (t)e 1 (σ (t), τ1 (t)) + p2 (t)e 1 (σ (t), τ2 (t)) 2I dp0 2I dp0 2tp0 (t) + μ(t) σ (t) + 2tσ (t) (p1 (t) + p2 (t)) 2tp0 (t) + μ(t)

 1 1 2t + 4t 2 + ≤ 2 2t + t 200t 4 300t 7 ≤

2 + 3 2 = + 3 53 = 75 < 1, ≤

8 200 1 25

t ∈ [8, ∞).

Then the considered equation has a nonoscillatory solution.

8.4 Explicit Nonoscillation and Oscillation Results

401

Exercise 8.2. Let T = 3N0 . Prove that the equation 

p0 x Δ

Δ

(t) +

1 x 110t 7 + 25t 4 + 100



 t t 1 x + = 0, 3 9 100t 8 + 200t 5 + 300t 2 + 500

t ∈ [27, ∞), where p0 (t) = t 2 +2t +27, t ∈ [27, ∞), has a nonoscillatory solution. Define the function t

p(t, s) = s

1 Δη, p0 (η)

t, s ∈ [t0 , ∞).

Theorem 8.8. Suppose that (A1), (A2), and (A3) hold, and for every t1 ∈ [t0 , ∞) the dynamic equation Δ  p0 x Δ (t) +

& % n  1 pl (t)p(τl (t), t1 ) x(τmax (t)) = 0, p(τmax (t), t1 )

(8.20)

l=1

t ∈ [t2 , ∞), is oscillatory, where t2 ∈ [t1 , ∞) satisfies τmin (t) > t1 for all t ∈ [t2 , ∞). Then the equation (8.4) is also oscillatory. Proof. Assume that the equation (8.4) has a nonoscillatory solution. Without loss of generality, we suppose that the equation (8.4) has an eventually positive solution x. Then there exists a t1 ∈ [t0 , ∞) such that l ∈ {1, . . . , n},

x(τl (t)) > 0,

x(t) > 0,

for any t ∈ [t1 , ∞). Hence from (8.4), we get Δ  p0 x Δ (t) ≤ 0,

t ∈ [t1 , ∞).

Therefore p0 x Δ is nonincreasing on [t1 , ∞). Then x(t) ≥ x(t) − x(t1 ) =

t t1

1 p0 (η)x Δ (η)Δη p0 (η)

≥ p0 (t)x Δ (t)

t t1

1 Δη p0 (η)

= p(t, t1 )p0 (t)x Δ (t),

t ∈ [t1 , ∞),

or x(t) − p(t, t1 )p0 (t)x Δ (t) ≥ 0,

t ∈ [t1 , ∞).

(8.21)

402

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

Let x(t) , p(t, t1 )

ψ(t) =

t ∈ [t1 , ∞).

Then ψ Δ (t) = = =

x Δ (t)p(t, t1 ) − x(t)ptΔ (t, t1 ) p(t, t1 )p(σ (t), t1 ) x Δ (t)p(t, t1 ) − x(t) p01(t) p(t, t1 )p(σ (t), t1 ) x Δ (t)p0 (t)p(t, t1 ) − x(t) p0 (t)p(t, t1 )p(σ (t), t1 )

≤ 0,

t ∈ [t1 , ∞).

Therefore ψ is nonincreasing on (t1 , ∞). Hence, x(τmax (t)) = ψ(τmax (t)) p(τmax (t), t1 ) ≤ ψ(τl (t)) =

x(τl (t)) , p(τl (t), .t1 )

t ∈ [t2 , ∞),

l ∈ {1, . . . , n},

where t2 ∈ [t1 , ∞) is chosen so that τmin (t) > t1 for any t ∈ [t2 , ∞). Hence, we get  Δ p0 x Δ (t) +

& % n  1 pl (t)p(τl (t), t1 ) x(τmax (t)) p(τmax (t), t1 ) l=1

 Δ ≤ p0 x Δ (t) +

n  l=1

pl (t)

x(τl (t)) p(τl (t), t1 ) p(τl (t), t1 )

n   Δ = p0 x Δ (t) + pl (t)x(τl (t)) l=1

= 0,

t ∈ [t2 , ∞).

Hence from Theorem 8.2, it follows that (8.20) has a nonoscillatory solution. This is a contradiction. This completes the proof. Theorem 8.9. Assume (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

8.5 Positive Solutions

403

and n ∞ t2

pl (η)e

l=1

1 p0 p(·,t1 )

(σ (η), τl (η))Δη = ∞,

where t2 ∈ (t1 , ∞). Then every solution of the equation (8.4) is oscillatory. Proof. Assume that the equation (8.4) has a nonoscillatory solution. Without loss of generality, we suppose that x is an eventually positive solution of the equation (8.4). Then, by Theorem 8.2, it follows that there exists a sufficiently large t1 ∈ [t0 , ∞) 1 ([t , ∞)) so that 1 + μ Λ on [t , ∞) and it satisfies (8.7). and a function Λ ∈ Crd 1 1 p0 By the proof of Theorem 8.2, we have Λ(t) = p0 (t)

x Δ (t) , x(t)

t ∈ [t1 , ∞).

Hence from (8.21), we obtain 1 , p(t, t1 )

Λ(t) ≤

t ∈ [t2 , ∞).

Therefore ΛΔ (t) +

n 

pl (t)e

l=1

1 p0 p(·,t1 )

(σ (t), τl (t)) ≤ 0,

t ∈ [t2 , ∞). From here, t

Λ(t) − Λ(t2 ) +

n 

t2 l=1

pl (η)e

1 p0 p(·,t1 )

(σ (η), τl (η))Δη ≤ 0,

t ∈ [t2 , ∞), which is a contradiction. This completes the proof.

8.5 Positive Solutions Theorem 8.10. Suppose that (A1), (A2), and (A3) hold, ∞ t0

1 Δη = ∞, p0 (η)

f ∈ Crd ([t0 , ∞)), f ≥ 0 on [t0 , ∞), and the Riccati inequality (8.7) has a solution 1 ([t , ∞)), Λ > 0 on [t , ∞). Moreover, suppose that there are x , x ≥ 0 Λ ∈ Crd 0 0 1 2

404

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . . Λ(t0 )x1 p0 (t0 ) .

such that φ(t) ≤ x1 e Λ (t, t0 ), t ∈ [t−1 , t0 ), x2 ≥ p0

Then the equation (8.1)

admits a positive solution x such that x(t) ≥ x1 , t ∈ [t0 , ∞). Proof. Let y be the solution of the following IVP n   Δ p0 y Δ (t) + pl (t)y(τl (t)) = 0,

t ∈ [t0 , ∞),

l=1

y Δ (t0 ) =

Λ(t0 ) x1 , p0 (t0 )

y(t) = x1 e Λ (t, t0 ), p0

t ∈ [t−1 , t0 ].

Set t ∈ [t−1 , ∞).

z(t) = x1 e Λ (t, t0 ), p0

Let n   Δ g(t) = p0 zΔ (t) + pl (t)z(τl (t)),

t ∈ [t0 , ∞).

l=1

Then zΔ (t) =

Λ(t) x1 e Λ (t, t0 ), p0 p0 (t)

p0 (t)zΔ (t) = x1 Λ(t)e Λ (t, t0 ), p0

Δ  Λ(t) e Λ (t, t0 ), p0 zΔ (t) = x1 ΛΔ (t)e Λ (t, t0 ) + x1 Λσ (t) p0 p0 (t) p0 t ∈ [t0 , ∞). Hence, 

p0 z Δ

Δ

(t) +

n 

pl (t)z(τl (t)) = x1 ΛΔ (t)e Λ (t, t0 ) p0

l=1

+x1 Λσ (t) +

n 

Λ(t) e Λ (t, t0 ) p0 (t) p0

pl (t)x1 e Λ (τl (t), t0 ) p0

l=1

= x1 ΛΔ (t)e Λ (t, t0 ) p0

+x1 Λσ (t)

Λ(t) e Λ (t, t0 ) p0 (t) p0

8.6 Advanced Practical Problems

405

+e Λ (t, t0 )x1 p0

n 

pl (t)e Λ (t0 , τl (t))e Λ (t, t0 ) p0

l=1

p0

Λ(t) = x1 e Λ (t, t0 ) ΛΔ (t) + Λσ (t) p0 (t) p0  n  + pl (t)e Λ (t, τl (t)) p0

l=1

t ∈ [t0 , ∞).

≤ 0,

Therefore g(t) ≤ 0,

t ∈ [t0 , ∞).

Note that z is the solution of the following IVP 

p0 z Δ

Δ

(t) +

n 

pl (t)z(τl (t)) = g(t),

t ∈ [t0 , ∞),

l=1

zΔ (t0 ) =

Λ(t0 ) x1 , p0 (t0 )

z(t) = x1 e Λ (t, t0 ), p0

t ∈ [t−1 , t0 ].

Hence from Theorem 8.5, we get y(t) ≥ z(t) ≥ x1 ,

t ∈ [t0 , ∞).

By Corollary 8.4, we obtain x(t) ≥ y(t) ≥ x1 ,

t ∈ [t0 , ∞).

This completes the proof.

8.6 Advanced Practical Problems Problem 8.1. Let T = q Z . Prove that the Riccati inequality (8.7) takes the form ΛΔ (t) +

n t−h  $ p0 (m) 1 Λσ (t)Λ(t) + ≤ 0, pl (t) p0 (t) p0 (m) + (q − 1)mΛ(m) l=1

t ∈ [t1 , ∞).

m=τl (t)

406

8 Nonoscillations of Second-Order Functional Dynamic Equations with Several. . .

1 ([t , ∞)), Problem 8.2. Suppose that (A1), (A2), and (A3) hold, and Λ ∈ Crd 1 1 + μ pΛ0 > 0 on [t1 , ∞). Prove that the Riccati inequality is equivalent to the inequality

 1 pl (t)e Λ (σ (t), τl (t)) ≤ 0 (Λ(t))2 + p0 p0 (t) + μ(t)Λ(t) n

ΛΔ (t) +

l=1

for any t ∈ [t1 , ∞). Problem 8.3. Let T = 5N0 . Prove that the equation 

p0 x Δ

Δ

(t) +

1 x 10t 10 + 125t 4 + 120t 2 + 100



 1 t t + = 0, x 5 25 150t 6 + 2000t 4 + 3t 2 + 9

t ∈ [25, ∞), where p0 (t) = 7t 7 + t 2 + 2t + 27, t ∈ [25, ∞), has a nonoscillatory solution. Problem 8.4. Let T = 7N0 . Prove that the equation Δ  p0 x Δ (t) +

1 x t 12 + 1000



 1 t t + 6 = 0, x 7 49 t + 2000

t ∈ [49, ∞), where p0 (t) = 70t 17 , t ∈ [49, ∞), has a nonoscillatory solution. Problem 8.5. Let T = hZ, h > 0. Prove that the equation 

p0 x Δ

Δ

(t) +

1 1 x (t − h) + x (t − 2h) = 0, 101t 4 + 25t 2 + 2220 500t 6 + 200t 4 + 300t 2 + 9

t ∈ T, where p0 (t) = 65t 12 +t 8 +200t 2 +270, t ∈ T, has a nonoscillatory solution. Problem 8.6. Let T = Z. Suppose that (A1), (A2), and (A3) hold, and there exists t0 ∈ [t0 , ∞) such that n t  $ t +1 2mp0 (m) + 2t (t + 1) ≤ 1, pl (t) 2tp0 (t) + 1 2mp0 (m) + 1 l=1

m=τl (t)

Prove that the equation (8.4) has a nonoscillatory solution.

t ∈ [t1 , ∞).

Chapter 9

Oscillations of Second-Order Nonlinear Functional Dynamic Equations

The results contained in this chapter can be found in the papers and monographs [9, 20, 106–109, 139, 190, 216, 235, 246, 252]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay Dynamic Equations and a Class of Second-Order Nonlinear Dynamic Equations In this section we consider the second-order nonlinear delay dynamic equation 2

x Δ (t) + p(t)f (x(t − τ )) = 0,

(9.1)

and the second-order nonlinear dynamic equation 2

x Δ (t) + p(t)f (x(σ (t))) = 0,

(9.2)

where (A)

τ ∈ R, t − τ ∈ T, t ∈ T, p : T → [0, ∞) is a real-valued rd-continuous function, f : R → R is continuous, nondecreasing real-valued function, f (−u) = −f (u) for any u ∈ R, and uf (u) > 0 for any u ∈ R, u = 0.

Theorem 9.1. Assume that (A) holds. A necessary and sufficient condition for the equation (9.2) to oscillate is that the inequality 2

x Δ (t) + p(t)f (x(σ (t))) ≤ 0

(9.3)

has no eventually positive solutions. © Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_9

407

408

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Proof. 1. Sufficiency. Let the inequality (9.3) has no eventually positive solutions. Then (9.2) has no eventually positive solutions. Suppose that (9.2) has an eventually negative solution x. Then 2

0 = x Δ (t) + p(t)f (x(σ (t))) 2

= − (−x)Δ (t) − p(t)f (−x(σ (t)))  2 = − (−x)Δ (t) + p(t)f (−x(σ (t))) or 2

(−x)Δ (t) + p(t)f (−x(σ (t))) = 0, whereupon we conclude that the inequality (9.3) has an eventually positive solution. This is a contradiction. Therefore the equation (9.2) oscillates. 2. Necessity. Let the equation (9.2) oscillates. Assume that the inequality (9.3) has an eventually positive solution x. Then there exists a t1 ≥ t0 so that x(t) > 0,

t ∈ [t1 , ∞).

Since x(σ (t))f (x(σ (t))) > 0,

t ∈ [t1 , ∞),

we get t ∈ [t1 , ∞).

f (x(σ (t))) > 0, Consequently 2

x Δ (t) ≤ −p(t)f (x(σ (t))) ≤ 0, t ∈ [t1 , ∞). Assume that there is a t2 ∈ [t1 , ∞) so that x Δ (t2 ) < 0. Integrating (9.4) from t2 to t, t ≥ t2 , we get x Δ (t) ≤ x Δ (t2 ) < 0,

t ∈ [t2 , ∞).

(9.4)

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

409

Hence, x(t) = x(t2 ) +

t

x Δ (s)Δs,

t ∈ [t2 , ∞),

t2

and then x(t) → −∞

t → ∞.

as

Therefore t ∈ [t1 , ∞).

x Δ (t) > 0,

Now, integrating (9.3) from t to s, s ≥ t ≥ t1 , we get s

x Δ (s) − x Δ (t) +

p(u)f (x(σ (u)))Δu t

≤ 0,

s ≥ t ≥ t1 ,

i.e., s

x Δ (t) ≥ x Δ (s) +

s ≥ t ≥ t1 .

p(u)f (x(σ (u)))Δu,

(9.5)

t

Since x Δ (t) > 0, t ≥ t1 , and x Δ is nonincreasing on [t1 , ∞), the limit lim x Δ (t) = k

t→∞

≥0 exists. Hence, letting s → ∞, by (9.5), we get ∞

x Δ (t) ≥ k +

p(u)f (x(σ (u)))Δu t

∞

≥

p(u)f (x(σ (u)))Δu,

t ∈ [t1 , ∞).

t

Now we integrate the last inequality from t1 to t, t ≥ t1 , and we get x(t) ≥ x(t1 ) +

∞

t

p(u)f (x(σ (u)))ΔuΔs, t1

t ∈ [t1 , ∞).

s

Let Ω = {w ∈ C ([t0 , ∞)) : 0 ≤ w(t) ≤ 1,

t ∈ [t0 , ∞)}.

(9.6)

410

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Let also, Ω be endowed with the usual pointwise ordering ≤, namely, if w1 , w2 ∈ Ω, then w1 ≤ w2 if and only if w1 (t) ≤ w2 (t) for any t ∈ [t0 , ∞). Note that any A ⊆ Ω has a supremum that belongs to Ω. Define the mapping  Sw(t) =

1, t0 ≤ t ≤ t1 , #t #∞ 1 x(t) x(t1 ) + t1 s p(u)f (x(σ (u))w(σ (u)))ΔuΔs ,

t ∈ [t1 , ∞).

Using (9.6), we conclude that SΩ ⊂ Ω. Also, S is nondecreasing. Therefore, using the Knaster-Tarski fixed point theorem (see the appendix), S has a fixed point w ∈ Ω. Hence, w(t) =

1 x(t1 ) + x(t)

∞

t t1

 p(u)f (x(σ (u))w(σ (u)))ΔuΔs ,

s

t ∈ [t1 , ∞). We have w(t) ≥

x(t1 ) x(t)

> 0,

t ∈ [t1 , ∞).

We set z(t) = w(t)x(t),

t ∈ [t1 , ∞).

Then t ∈ [t1 , ∞),

z(t) > 0, z is continuous on [t1 , ∞) and z(t) = x(t1 ) +

∞

t

p(u)f (z(σ (u)))ΔuΔs, t1

s

t ∈ [t1 , ∞), whereupon 2

zΔ (t) + p(t)f (z(σ (t))) = 0,

t ∈ [t1 , ∞).

Therefore the equation (9.2) has an eventually positive solution. This is a contradiction. This completes the proof.

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

411

Theorem 9.2. Assume that (A) holds. A necessary and sufficient condition for the equation (9.1) to oscillate is that the inequality 2

x Δ (t) + p(t)f (x(t − τ )) ≤ 0

(9.7)

has no eventually positive solutions. Proof. 1. Sufficiency. Let the inequality (9.7) has no eventually positive solutions. Then (9.1) has no eventually positive solutions. Suppose that (9.1) has an eventually negative solution x. Then 2

0 = x Δ (t) + p(t)f (x(t − τ )) 2

= − (−x)Δ (t) − p(t)f (−x(t − τ ))  2 = − (−x)Δ (t) + p(t)f (−x(t − τ )) or 2

(−x)Δ (t) + p(t)f (−x(t − τ )) = 0, whereupon we conclude that the inequality (9.7) has an eventually positive solution. This is a contradiction. Therefore the equation (9.1) oscillates. 2. Necessity. Let the equation (9.1) oscillates. Assume that the inequality (9.7) has an eventually positive solution x. Then there exists a t ≥ t0 so that x(t) > 0,

t ∈ [t , ∞).

Let t1 ≥ t be chosen so that t − τ ≥ t ,

t ∈ [t1 , ∞).

Since x(t − τ )f (x(t − τ )) > 0,

t ∈ [t1 , ∞),

we get f (x(t − τ )) > 0,

t ∈ [t1 , ∞).

Consequently 2

x Δ (t) ≤ −p(t)f (x(t − τ )) ≤ 0, t ∈ [t1 , ∞).

(9.8)

412

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Assume that there is a t2 ∈ [t1 , ∞) so that x Δ (t2 ) < 0. Integrating (9.8) from t2 to t, t ≥ t2 , we get x Δ (t) ≤ x Δ (t2 ) < 0,

t ∈ [t2 , ∞).

Hence, t

x(t) = x(t2 ) +

x Δ (s)Δs,

t ∈ [t2 , ∞),

t2

and then x(t) → −∞

as

t → ∞.

Therefore x Δ (t) > 0,

t ∈ [t1 , ∞).

Now, integrating (9.7) from t to s, s ≥ t ≥ t1 , we get s

x Δ (s) = x Δ (t) +

p(u)f (x(u − τ ))Δu ≤ 0,

s ≥ t ≥ t1 ,

t

i.e., s

x Δ (t) ≥ x Δ (s) +

p(u)f (x(u − τ ))Δu,

s ≥ t ≥ t1 .

t

Since x Δ (t) > 0, t ≥ t1 , and x Δ is nonincreasing on [t1 , ∞), the limit lim x Δ (t) = k ≥ 0

t→∞

exists. Hence, letting s → ∞, by (9.9), we get ∞

x Δ (t) ≥ k +

p(u)f (x(u − τ ))Δu

t ∞

≥ t

p(u)f (x(u − τ ))Δu,

t ∈ [t1 , ∞).

(9.9)

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

413

Now we integrate the last inequality from t1 to t, t ≥ t1 , and we get x(t) ≥ x(t1 ) +

∞

t t1

p(u)f (x(u − τ ))ΔuΔs,

t ∈ [t1 , ∞).

s

Using the proof of Theorem 9.1, we get a function z ∈ C ([t0 , ∞)) such that z(t) = x(t1 ) +

∞

t t1

p(u)f (z(u − τ ))ΔuΔs,

s

t ∈ [t1 , ∞), whereupon 2

zΔ (t) + p(t)f (z(t − τ )) = 0,

t ∈ [t1 , ∞).

Therefore the equation (9.1) has an eventually positive solution. This is a contradiction. This completes the proof. Theorem 9.3. Assume (A) and μ are bounded. Then the oscillation of (9.1) and (9.2) is equivalent. Proof. Since μ is bounded, there exists a positive constant m such that μ(t) ≤ m, t ∈ T. 1. Sufficiency. Let the equation (9.2) oscillates. Assume that the equation (9.1) does not oscillate. Without loss of generality, we suppose that the equation (9.1) has an eventually positive solution x. Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x Δ (t) > 0,

2

x Δ (t) ≤ 0,

x(t − τ ) > 0,

t ∈ [t1 , ∞).

Then x is increasing on [t1 , ∞), x Δ is nonincreasing on [t1 , ∞) and lim x Δ (t) = k ≥ 0

t→∞

exists. (a) Let τ < 0. Then σ (t) ≤ t − τ and x(σ (t)) ≤ x(t − τ ),

t ∈ [t1 , ∞).

Hence, 2

2

x Δ (t) + p(t)f (x(σ (t))) ≤ x Δ (t) + p(t)f (x(t − τ )) = 0, This contradicts with Theorem 9.1.

t ∈ [t1 , ∞).

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9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

(b) Let τ > 0 and k > 0. Then, using that x Δ is nonincreasing on [t1 , ∞) and x is increasing on [t1 , ∞), there exist  ≥ k and t ≥ t so that x Δ (t) ≤ ,

x(t) > 2τ,

t ∈ [t , ∞).

Then t

x(t) − x(t − τ ) =

x Δ (s)Δs

t−τ

≤ τ,

t ∈ [t + τ, ∞),

x(t − τ ) ≥ x(t) − τ,

t ∈ [t + τ, ∞),

(9.10)

t ∈ [t + τ, ∞).

(9.11)

i.e.,

and x(σ (t) − τ ) ≥ x(σ (t)) − τ, Since t ∈ [t + τ, ∞),

t ≥ σ (t) − τ, we have

x(t) ≥ x(σ (t) − τ ),

t ∈ [t + τ, ∞).

Therefore, using (9.10) and (9.11), we get x(t − τ ) ≥ x(t) − τ ≥ x(σ (t) − τ ) − τ ≥ x(σ (t)) − 2τ, t ∈ [t + τ, ∞). Let z(t) = x(t) − 2τ,

t ∈ [t + τ, ∞).

Then z(t) > 0,

t ∈ [t + τ, ∞).

(9.12)

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

415

Also, z(σ (t)) = x(σ (t)) − 2τ t ∈ [t + τ, ∞),

≤ x(t − τ ), and 2

2

zΔ (t) + p(t)f (z(σ (t))) = x Δ (t) + p(t)f (z(σ (t))) 2

≤ x Δ (t) + p(t)f (x(t − τ )) t ∈ [t + τ, ∞).

= 0,

This contradicts with Theorem 9.1. (c) Let τ > 0 and k = 0. Since x is increasing on [t1 , ∞) and x Δ is nonincreasing on [t1 , ∞), then there exist an  > 0 and a t2 ∈ [t1 , ∞) so that x(t) > 2τ ,

x Δ (t) ≤ ,

t ∈ [t2 , ∞).

Then t

x(t) − x(t − τ ) =

x Δ (s)Δs

t−τ

≤ τ,

t ∈ [t2 + τ, ∞),

x(t − τ ) ≥ x(t) − τ,

t ∈ [t2 + τ, ∞),

i.e.,

and x(σ (t) − τ ) ≥ x(σ (t)) − τ,

t ∈ [t2 + τ, ∞).

Then x(t − τ ) ≥ x(t) − τ ≥ x(σ (t) − τ ) − τ ≥ x(σ (t)) − 2τ,

t ∈ [t2 + τ, ∞).

Let z(t) = x(t) − 2τ,

t ∈ [t2 + τ, ∞).

416

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then z(t) > 0,

z(σ (t)) ≤ x(t − τ ),

t ∈ [t2 + τ, ∞),

and 2

2

zΔ (t) + p(t)f (z(σ (t))) = x Δ (t) + p(t)f (z(σ (t))) 2

≤ x Δ (t) + p(t)f (x(t − τ )) = 0,

t ∈ [t2 + τ, ∞).

This contradicts with Theorem 9.1. (d) Let τ = 0 and k > 0. Then there exist  > 0 and t2 ∈ [t1 , ∞) so that x Δ (t) ≤ ,

t ∈ [t2 , ∞).

x(t) > m,

Hence, σ (t)

x(σ (t)) − x(t) =

x Δ (s)Δs

t

≤ μ(t) ≤ m,

t ∈ [t2 , ∞).

Let z(t) = x(t) − m,

t ∈ [t2 , ∞).

Then z(t) > 0,

z(σ (t)) ≤ x(t),

t ∈ [t2 , ∞),

and 2

2

zΔ (t) + p(t)f (z(σ (t))) = x Δ (t) + p(t)f (z(σ (t))) 2

≤ x Δ (t) + p(t)f (x(t)) = 0,

t ∈ [t2 , ∞).

This contradicts with Theorem 9.1. (e) Let τ = 0 and k = 0. Then there exist an  > 0 and a t2 ∈ [t1 , ∞) so that x Δ (t) ≤ ,

x(t) > m,

t ∈ [t2 , ∞).

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

417

Then, σ (t)

x(σ (t)) − x(t) =

x Δ (s)Δs

t

≤ μ(t) ≤ m,

t ∈ [t2 , ∞),

x(t) ≥ x(σ (t)) − m,

t ∈ [t2 , ∞).

i.e.,

Let t ∈ [t2 , ∞).

z(t) = x(t) − m, Then z(t) > 0,

z(σ (t)) ≤ x(t),

t ∈ [t2 , ∞).

Hence, 2

2

zΔ (t) + p(t)f (z(σ (t))) = x Δ (t) + p(t)f (z(σ (t))) 2

≤ x Δ (t) + p(t)f (x(t)) = 0,

t ∈ [t2 , ∞).

This contradicts with Theorem 9.1. 2. Necessity. Let (9.1) oscillates. Assume that the equation (9.2) does not oscillate. Without loss of generality, we suppose that the equation (9.2) has an eventually positive solution x. Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x Δ (t) > 0,

2

x Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let τ ≥ 0. We choose t2 ∈ [t1 , ∞) so that t − τ ≥ t1 for any t ∈ [t2 , ∞). Then σ (t) ≥ t − τ,

t ∈ [t2 , ∞),

and x(σ (t)) ≥ x(t − τ ),

t ∈ [t2 , ∞).

418

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Hence, 2

2

x Δ (t) + p(t)f (x(t − τ )) ≤ x Δ (t) + p(t)f (x(σ (t))) = 0,

t ∈ [t2 , ∞).

This contradicts with Theorem 9.2. (b) Let τ < 0 and k > 0. Then there exist an  > 0 and a t2 ∈ [t1 , ∞) so that x Δ (t) ≤ ,

x(t) ≥ −τ,

t ∈ [t2 , ∞).

Hence, t−τ

x(t − τ ) − x(t) =

x Δ (s)Δs

t

≤ −τ,

t ∈ [t2 , ∞),

or x(σ (t)) ≥ x(t) ≥ x(t − τ ) + τ,

t ∈ [t2 , ∞).

Let z(t) = x(t) + τ,

t ∈ [t2 , ∞).

Then z(t) > 0,

z(t − τ ) ≤ x(σ (t)),

t ∈ [t2 , ∞),

and 2

2

zΔ (t) + p(t)f (z(t − τ )) = x Δ (t) + p(t)f (z(t − τ )) 2

≤ x Δ (t) + p(t)f (x(σ (t))) = 0,

t ∈ [t2 , ∞).

This contradicts with Theorem 9.2. (c) Let τ < 0 and k = 0. Then there exist an  > 0 and a t2 ∈ [t1 , ∞) so that x Δ (t) ≤ ,

x(t) ≥ −τ,

t ∈ [t2 , ∞).

9.1 Oscillation Equivalence of a Class of Second-Order Nonlinear Delay. . .

419

Then t−τ

x(t − τ ) − x(t) =

x Δ (s)Δs

t

≤ −τ,

t ∈ [t2 , ∞).

Hence, x(σ (t)) ≥ x(t) ≥ x(t − τ ) + τ,

t ∈ [t2 , ∞).

Let z(t) = x(t) + τ,

t ∈ [t2 , ∞).

Then z(t) > 0,

z(t − τ ) ≤ x(σ (t)),

t ∈ [t2 , ∞),

and 2

2

zΔ (t) + p(t)f (z(t − τ )) = x Δ (t) + p(t)f (z(t − τ )) 2

≤ x Δ (t) + p(t)f (x(σ (t))) = 0,

t ∈ [t2 , ∞).

This contradicts with Theorem 9.2. This completes the proof. Example 9.1. Let T = hZ, h > 0. Then μ(t) = h, t ∈ T, and the oscillation of the equation  2 x Δ (t) + t 2 + 1 (x(t − h))3 = 0,

t ∈ T,

 2 x Δ (t) + t 2 + 1 (x(t + h))3 = 0,

t ∈ T,

and the equation

is equivalent. Exercise 9.1. Let T = hZ, h > 0. Then μ(t) = h, t ∈ T, and the oscillation of the equation  2 x Δ (t) + 5t 4 + 10t 2 + 1 (x(t − h))5 = 0,

t ∈ T,

420

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

and the equation  2 x Δ (t) + 5t 4 + 10t 2 + 1 (x(t + h))5 = 0,

t ∈ T,

is equivalent. Now we consider the equation 2

x Δ (t) + q(t)g(x(σ (t))) = 0,

(9.13)

where (B)

q : T → [0, ∞) is a real-valued rd-continuous function, g : R → R is continuous, nondecreasing real-valued function, g(−u) = −g(u) for any u ∈ R, and ug(u) > 0 for any u ∈ R, u = 0.

Theorem 9.4. Assume (A) and (B). Let also, μ is bounded and q(t) ≤ p(t), t ∈ T, g(u) ≤ f (u), u ∈ R. Then the oscillation of the equation (9.13) implies the oscillation of the equation (9.1). Proof. Let the equation (9.13) oscillates. Assume that the equation (9.1) does not oscillate. Without loss of generality, we suppose that the equation (9.1) has an eventually positive solution. By Theorem 9.3, it follows that the equation (9.2) has an eventually positive solution x. Then there exists a t1 ∈ [t0 , ∞) so that x(t) > 0 for any t ∈ [t1 , ∞). Hence, g(x(σ (t))) > 0,

t ∈ [t1 , ∞),

and 2

2

x Δ (t) + q(t)g(x(σ (t)) ≤ x Δ (t) + p(t)g(x(σ (t))) 2

≤ x Δ (t) + p(t)f (x(σ (t))) = 0,

t ∈ [t1 , ∞).

This contradicts with Theorem 9.1. This completes the proof. Example 9.2. Let T = hZ, h > 0. Then the oscillation of the equation 2

x Δ (t) +

12 t + 10 (x(t + h))3 = 0, 2

t ∈ T,

implies the oscillation of the equation  2 x Δ (t) + t 2 + 100 (x(t − h))3 = 0,

t ∈ T.

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

421

Exercise 9.2. Let T = hZ, h > 0. Prove that the oscillation of the equation 14 2 t + t 2 + 1 (x(t + h))5 = 0, t ∈ T, x Δ (t) + 7 implies the oscillation of the equation  2 x Δ (t) + t 4 + 2t 2 + 10 (x(t − h))5 = 0,

t ∈ T.

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic Equations I In this section we consider the following second-order nonlinear delay dynamic equation 

rx Δ

Δ

(t) + p(t)f (x(τ (t))) = 0,

t ∈ [t0 , ∞),

(9.14)

where (C)

r, p ∈ Crd (T), r, p > 0 on T, τ ∈ Crd ([t0 , ∞)), τ (t) ≤ t for any t ∈ [t0 , ∞), limt→∞ τ (t) = ∞, f : R → R is a continuous function satisfying uf (u) > 0 for any u ∈ R, u = 0, |f (u)| ≥ K|u| for any u ∈ R.

Let R be the set of all functions H defined for t0 ≤ s ≤ σ (t), t, s ∈ [t0 , ∞), H (t, s) ≥ 0, H (σ (t), t) = 0, H Δs (t, s) ≤ 0 for t ≥ s ≥ t0 and for each fixed t, H Δs (t, s) is delta integrable with respect to s. Theorem 9.5. Assume (C) and ∞

Δt = ∞, r(t)

(9.15)

τ (t)p(t)Δt = ∞.

(9.16)

t0 ∞ t0

Let the equation (9.14) has a positive solution x on [t0 , ∞). Then there exists a T ∈ [t0 , ∞) such that (i) x Δ (t) > 0, x(t) > tx Δ (t), t ∈ [T , ∞), (ii) x is strictly increasing on [T , ∞) and x(t) t is strictly decreasing on [T , ∞). Proof. Let t1 ∈ [t0 , ∞) be chosen so that x(τ (t)) > 0 for any t ∈ [t1 , ∞). Then, by (9.14), we get  Δ Δ (t) = −p(t)f (x(τ (t))) rx < 0,

t ∈ [t1 , ∞).

422

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Therefore rx Δ is strictly decreasing on [t1 , ∞). Assume that there is a t2 ∈ [t1 , ∞) so that r(t2 )x Δ (t2 ) = c < 0. Then r(t)x Δ (t) ≤ r(t2 )x Δ (t2 ) =c < 0,

t ∈ [t2 , ∞),

c , r(t)

t ∈ [t2 , ∞).

and x Δ (t) ≤ Hence, t

x(t) = x(t2 ) +

x Δ (s)Δs

t2 t

≤ x(t2 ) + c

t2

→ −∞

Δs r(s)

t → ∞.

as

This is a contradiction. Therefore r(t)x Δ (t) > 0,

t ∈ [t1 , ∞).

Hence, x Δ (t) > 0,

t ∈ [t1 , ∞),

and x is strictly increasing on [t1 , ∞). Let X(t) = x(t) − tx Δ (t),

t ∈ [t1 , ∞).

Let also, t2 ∈ [t1 , ∞) be such that X(t) < 0,

t ∈ [t2 , ∞).

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

Then

x(t) t

Δ =

tx Δ (t) − x(t) tσ (t)

=−

X(t) tσ (t) t ∈ [t2 , ∞).

> 0,

Pick t3 ∈ [t2 , ∞) so that τ (t) ≥ τ (t3 ) ≥ t2 for t ∈ [t3 , ∞). Then x(τ (t)) x(τ (t3 )) ≥ τ (t) τ (t3 ) =d t ∈ [t3 , ∞),

> 0, i.e., x(τ (t)) ≥ dτ (t),

t ∈ [t3 , ∞).

Now we integrate both sides of the equation (9.14) from t3 to t and we get t

0 = r(t)x Δ (t) − r(t3 )x Δ (t3 ) +

p(s)f (x(τ (s)))Δs t3 t

≥ r(t)x Δ (t) − r(t3 )x Δ (t3 ) + K

p(s)x(τ (s))Δs, t3

t ∈ [t3 , ∞). Therefore r(t3 )x Δ (t3 ) ≥ r(t)x Δ (t) + K

t

p(s)x(τ (s))Δs t3

≥ K

t

p(s)x(τ (s))Δs t3

≥ dK

t

p(s)τ (s)Δs t3

→ ∞ as

t → ∞,

423

424

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

which is a contradiction. Hence, there is a t2 ∈ [t1 , ∞) such that X(t) > 0 for any t ∈ [t2 , ∞). Consequently

x(t) t

Δ =−

X(t) tσ (t)

< 0, Therefore

x(t) t

t ∈ [t2 , ∞).

is strictly decreasing on [t2 , ∞). This completes the proof.

Theorem 9.6. Assume (C), (9.15), and (9.16). Let H ∈ R and there are a function a and a positive differentiable function δ such that lim sup t→∞

1 H (σ (t), t1 )

t

H (t, s)δ σ (s) (ψ(s) − φ(t, s)) Δs = ∞

(9.17)

t1

for sufficiently large t1 , where H (t, s) = H (σ (t), σ (s)),

 1 δ(s) 2 r(s) (A(t, s))2 , φ(t, s) = 4 δ σ (s) C(s) t C(t) = , σ (t) ψ(s) = A(t, s) = C1 (s) =

sr(s)(a(s))2 Kp(s)τ (s) − (ar)Δ (s) + , σ (s) σ (s) δ σ (s)C1 (s) H Δs (σ (t), s) + , δ(s) H (t, s) sa(s) δ Δ (s) +2 . δ σ (s) σ (s)

Then every solution of the equation (9.14) is oscillatory on [t0 , ∞). Proof. Suppose that the equation (9.14) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we assume that the solution x is eventually positive on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Theorem 9.5, it follows that there is a t2 ∈ [t1 , ∞) so that x Δ (t) > 0,

x(t) > tx Δ (t),

t ∈ [t2 , ∞),

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

and

x Id

is strictly decreasing on [t2 , ∞). Note that  Δ Δ rx (t) < 0,

t ∈ [t2 , ∞).

Define the function w by the generalized Riccati substitution

 r(t)x Δ (t) w(t) = δ(t) + r(t)a(t) , x(t)

t ∈ [t2 , ∞).

We have

 r(t)x Δ (t) w (t) = δ (t) + r(t)a(t) x(t) Δ

Δ rx σ +δ (t) + ra (t) x Δ

Δ

=

=

=

=

≤

δ Δ (t) w(t) + δ σ (t)(ra)Δ (t) δ(t)

Δ Δ rx σ +δ (t) (t) x δ Δ (t) w(t) + δ σ (t)(ra)Δ (t) δ(t) %   2 & Δ x(t) rx Δ (t) − r(t) x Δ (t) σ +δ (t) x(t)x σ (t) δ Δ (t) w(t) + δ σ (t)(ra)Δ (t) δ(t) 2  Δ Δ  (t) rx r(t) x Δ (t) σ σ − δ (t) +δ (t) x σ (t) x(t)x σ (t) δ Δ (t) w(t) + δ σ (t)(ra)Δ (t) δ(t) 2  r(t) x Δ (t) p(t)f (x(τ (t))) σ − δ σ (t) −δ (t) σ x(t)x (t) x σ (t) δ Δ (t) w(t) + δ σ (t)(ra)Δ (t) δ(t) 2  r(t) x Δ (t) p(t)x(τ (t)) σ − δ σ (t)K , −δ (t) x(t)x σ (t) x σ (t)

425

426

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

t ∈ [t2 , ∞). By the definition of the function w, we get x Δ (t) w(t) = + a(t), δ(t)r(t) x(t)

t ∈ [t2 , ∞),

and hence,

Since

x Id

x Δ (t) x(t)

2

=

w(t) − a(t) δ(t)r(t)

2 ,

t ∈ [t2 , ∞).

is strictly decreasing on [t2 , ∞), there is a t3 ∈ [t2 , ∞) so that τ (t) ≥ t2 ,

t ∈ [t3 , ∞),

and x σ (t) x(τ (t)) ≥ , τ (t) σ (t)

t ∈ [t3 , ∞),

x(τ (t)) τ (t) ≥ , σ x (t) σ (t)

t ∈ [t3 , ∞),

or

and t x(t) ≥ σ x (t) σ (t) = C(t),

t ∈ [t3 , ∞).

Therefore w Δ (t) ≤ −δ σ (t)K

p(t)τ (t) δ Δ (t) + w(t) σ (t) δ(t)

+δ σ (t)(ra)Δ (t) −δ σ (t)r(t)C(t) = −δ σ (t)K

w(t) − a(t) δ(t)r(t)

2

p(t)τ (t) δ Δ (t) + w(t) σ (t) δ(t)

+δ σ (t)(ra)Δ (t) −δ (t)r(t)C(t) σ

%

w(t) δ(t)r(t)

2

a(t)w(t) + (a(t))2 −2 δ(t)r(t)

&

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

= −δ σ (t)K

p(t)τ (t) δ Δ (t) + w(t) σ (t) δ(t)

+δ σ (t)(ra)Δ (t) − δ σ (t)r(t)C(t)

w(t) δ(t)r(t)

427

2

C(t) − δ σ (t)r(t)C(t)(a(t))2 δ(t) 

δ σ (t) δ Δ (t) + 2a(t)C(t) w(t) = −δ σ (t)ψ(t) + δ(t) δ σ (t) +2δ σ (t)a(t)w(t)

−

C(t)δ σ (t) (w(t))2 r(t)(δ(t))2

= −δ σ (t)ψ(t) + −

δ σ (t) C1 (t)w(t) δ(t)

C(t)δ σ (t) (w(t))2 , r(t)(δ(t))2

t ∈ [t3 , ∞).

Hence, w Δ (s) ≤ −δ σ (s)ψ(s) + −

δ σ (s) C1 (s)w(s) δ(s)

C(s)δ σ (s) (w(s))2 , r(s)(δ(s))2

s ∈ [t3 , ∞),

and H (σ (t), σ (s))w Δ (s) ≤ −δ σ (s)ψ(s)H (σ (t), σ (s)) + −H (σ (t), σ (s))

δ σ (s) C1 (s)H (σ (t), σ (s))w(s) δ(s)

C(s)δ σ (s) (w(s))2 , r(s)(δ(s))2

s ∈ [t3 , ∞),

or δ σ (s)ψ(s)H (σ (t), σ (s)) ≤ −H (σ (t), σ (s))w Δ (s) + −H (σ (t), σ (s))

δ σ (s) C1 (s)H (σ (t), σ (s))w(s) δ(s)

C(s)δ σ (s) (w(s))2 , r(s)(δ(s))2

s ∈ [t3 , ∞).

428

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then t

δ σ (s)ψ(s)H (σ (t), σ (s))Δs ≤ −

t3

t

H (σ (t), σ (s))w Δ (s)Δs

t3

+

t t3

−

δ σ (s) C1 (s)H (σ (t), σ (s))w(s)Δs δ(s)

t

H (σ (t), σ (s)) t3

= H (σ (t), t3 )w(t3 ) +

C(s)δ σ (s) (w(s))2 Δs r(s)(δ(s))2 t

H Δs (σ (t), s)w(s)Δs

t3

+

t t3

−

δ σ (s) C1 (s)H (σ (t), σ (s))w(s)Δs δ(s)

t

H (σ (t), σ (s)) t3

= H (σ (t), t3 )w(t3 ) +

C(s)δ σ (s) (w(s))2 Δs r(s)(δ(s))2 t t3

+

t t3

−

H Δs (σ (t), s) H (t, s)

H (σ (t), σ (s))w(s)Δs

δ σ (s) C1 (s)H (σ (t), σ (s))w(s)Δs δ(s)

t

H (σ (t), σ (s)) t3

= H (σ (t), t3 )w(t3 ) +

C(s)δ σ (s) (w(s))2 Δs r(s)(δ(s))2 t

A(t, s)H (σ (t), σ (s))w(s)Δs t3

−

t

H (σ (t), σ (s)) t3

= H (σ (t), t3 )w(t3 ) +

C(s)δ σ (s) (w(s))2 Δs r(s)(δ(s))2 t

A(t, s)H (σ (t), σ (s))w(s)Δs t3

−

t

H (σ (t), σ (s)) t3

−

t t3

+

t t3

C(s)δ σ (s) (w(s))2 Δs r(s)(δ(s))2

(A(t, s))2 r(s)(δ(s))2 H (t, s)Δs 4C(s)δ σ (s) (A(t, s))2 r(s)(δ(s))2 H (t, s)Δs 4C(s)δ σ (s)

≤ H (σ (t), t3 )w(t3 ) +

t t3

(A(t, s))2 r(s)(δ(s))2 H (t, s)Δs, 4C(s)δ σ (s)

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

429

t ∈ [t3 , ∞). Therefore t

1 H (σ (t), t3 )

%

t3

(A(t, s))2 r(s)(δ(s))2 δ (s)ψ(s) − 4C(s)δ σ (s)

&

σ

H (t, s)Δs ≤ w(t3 ),

t ∈ [t3 , ∞). Hence, 1 lim sup H (σ (t), t3 ) t→∞

t t3

%

(A(t, s))2 r(s)(δ(s))2 δ (s)ψ(s) − 4C(s)δ σ (s) σ

& H (t, s)Δs ≤ w(t3 ),

(9.18) which is a contradiction. This completes the proof. Now we define the function h(t, s) as follows.  H Δs (σ (t), s) = −h(t, s) H (σ (t), σ (s)).

(9.19)

Then δ σ (s)C1 (s) H Δs (σ (t), s) + δ(s) H (t, s) √ δ σ (s)C1 (s) h(t, s) H (σ (t), σ (s)) − = δ(s) H (σ (t), σ (s))

A(t, s) =

=

h(t, s) δ σ (s)C1 (s) −√ δ(s) H (σ (t), σ (s))

=

h(t, s) δ σ (s)C1 (s) − . δ(s) H (t, s)

Corollary 9.1. Assume (C), (9.15), and (9.16). Let t

lim sup t→∞

t1

  r Δ 5r(s) Kp(s)τ (s) − σ (s) Δs = ∞. (s) − Id 4s

(9.20)

Then every solution of the equation (9.14) is oscillatory. Proof. Define H (t, s) for t0 ≤ s ≤ σ (t) by H (σ (t), t) = 0, H (t, s) = 1 otherwise. Let δ(t) = t,

a(t) =

1 . t

430

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then δ Δ (s) = 1, δ σ (s) = σ (s), C1 (s) =

sa(s) δ Δ (s) +2 δ σ (s) σ (s)

=

s1 1 +2 s σ (s) σ (s)

=

2 1 + σ (s) σ (s)

=

3 , σ (s)

H Δs (σ (t), s) = 0, H (t, s) = H (σ (t), σ (s)) = 1, A(t, s) = =

δ σ (s)C1 (s) H Δs (σ (t), s) + δ(s) H (t, s) 3 σ (s) σ (s)

s

3 = , s ψ(s) = = C(s) = φ(t, s) =

= =

Kp(s)τ (s)  r Δ − (s) + σ (s) Id Kp(s)τ (s)  r Δ − (s) + σ (s) Id s , σ (s)

 1 δ(s) 2 r(s)(A(t, s))2 4 δ σ (s) C(s)  2

2 r(s) 3 s s 1 s 4 σ (s) σ (s) 9 1 s r(s) 2 4 σ (s) s

sr(s) s12 σ (s) r(s) , sσ (s)

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

= ψ(s) − φ(t, s) =

9r(s) , 4sσ (s)

Kp(s)τ (s)  r Δ r(s) (s) + − σ (s) Id sσ (s)

− =

431

9r(s) 4sσ (s)

5r(s) Kp(s)τ (s)  r Δ − . (s) − σ (s) Id 4sσ (s)

Hence,  Kp(s)τ (s)  r Δ 5r(s) − (s) − σ (s) Id 4sσ (s)  r Δ 5r(s) . = Kp(s)τ (s) − σ (s) (s) − Id 4s

H (t, s)δ σ (s) (ψ(s) − φ(t, s)) = σ (s)

From here and from (9.17), we get (9.20). This completes the proof. Example 9.3. Let T = 2N0 . Consider the equation  Δ Δ rx (t) + t 4 x

%

2 & t t = 0, 1+ x 2 2

where r(t) =

1 , t5

t ∈ T.

Here t0 = 2, σ (t) = 2t, p(t) = t 4 , t τ (t) = , t ∈ T, 2  f (u) = u 1 + u2 ,

u ∈ R.

t ≥ 2,

432

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then ∞ 2

∞

1 Δt = r(t)

t 5 Δt

2 t

= lim

t→∞ 2

s 5 Δs

= = = ∞

τ (t)p(t)Δt =

2

=

 1 6 !!s=t lim s ! t→∞ 63 s=2

6  t 64 − lim t→∞ 63 63 ∞, ∞ t  t 4 Δt 2 2 1 lim 2 t→∞

t

s 5 Δs

2

!s=t 1 ! lim s 6 ! 126 t→∞ s=2  1 lim t 6 − 64 = 126 t→∞ = ∞. =

Next,  |f (u)| = |u| 1 + u2 ≥ |u|,

u ∈ R,

uf (u) = u2 (1 + u2 ) > 0,

u ∈ R.

g(t) = t 6 ,

t ∈ T.

Then K = 1. Let

Then

Δ  r Δ 1 (t) = (t) Id g =−

g Δ (t) g(t)g σ (t)

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

=−

433

63t 5 t 6 (2t)6

63t 5 64t 12 63 = − 7, 64t =−

and

t ∈ T,

  r Δ 5r(s) lim sup Kp(s)τ (s) − σ (s) Δs (s) − Id 4s t→∞ 2  

t s5 5 63 − 6 Δs − 2s − = lim sup 2 64s 7 4s t→∞ 2  t s5 5 63 = lim sup − 6 Δs + 6 2 32s 4s t→∞ 2  t s5 23 Δs + = lim sup 2 32s 6 t→∞ 2

 

1 6 !!s=t 23 32 !!s=t = lim sup s ! − ! 126 s=2 32 31s 5 s=2 t→∞

6  t 32 23 23 − − = lim sup + 126 63 31t 5 31.32 t→∞ = ∞. t

Therefore every solution of the considered equation is oscillatory. Exercise 9.3. Let T = hZ, h > 0. Prove that every solution of the equation    Δ Δ (t) + t 8 + 1 x(t − h) 1 + (x(t − h))4 = 0, rx

t ∈ T,

where r(t) =

t 10

1 , +1

t ∈ T,

is oscillatory on T. Corollary 9.2. Assume (C), (9.15), and (9.16). Let H ∈ R and there is a positive differentiable function δ such that 1 lim sup t→∞ H (σ (t), t1 )

t t1

%

σ (s) (δ(s))2 r(s)(h(t, s))2 H (t, s)δ (s)ψ(s) − 4sδ σ (s) σ

& Δs = ∞

(9.21)

434

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

for sufficiently large t1 , where h is defined by (9.19), H , φ, C, ψ, A, and C1 are as in Theorem 9.6. Then every solution of the equation (9.14) is oscillatory on [t0 , ∞). Proof. Let a(t) = − Then

σ (t)δ Δ (t) , 2tδ σ (t)

t ≥ t0 .

(9.22)

 (s)δ Δ (s) s − σ2sδ σ (s) δ Δ (s) +2 C1 (s) = σ δ (s) σ (s) =

δ Δ (s) δ Δ (s) − σ δ σ (s) δ (s)

= 0, A(t, s) =

H Δs (σ (t), s)

=−

H (t, s)  h(t, s) H (t, s) H (t, s)

h(t, s) , = − H (t, s)

 1 δ(s) 2 r(s)(A(t, s))2 φ(t, s) = 4 δ σ (s) C(s)

2 1 δ(s) r(s) (h(t, s))2 = s 4 δ σ (s) H (t, s) σ (s)

 1 δ(s) 2 r(s)σ (s) (h(t, s))2 = , 4 δ σ (s) sH (t, s)

t, s ∈ [t0 , ∞).

Hence, H (t, s)δ σ (s) (ψ(s) − φ(t, s)) = H (t, s)δ σ (s)ψ(s) −H (t, s)δ σ (s)φ(t, s) = H (t, s)δ σ (s)ψ(s) −

H (t, s) (δ(s))2 σ r(s)σ (s) (h(t, s))2 δ (s) 4 (δ σ (s))2 sH (t, s)

= H (t, s)δ σ (s)ψ(s) −

(δ(s))2 r(s)σ (s) (h(t, s))2 , 4sδ σ (s)

t, s ∈ [t0 , ∞).

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

435

From here and from (9.21), we get (9.17). Then, using Theorem 9.6, we conclude that every solution of the equation (9.14) is oscillatory on [t0 , ∞). This completes the proof. Corollary 9.3. Assume (C), (9.15) and (9.16). Let also, the function a be defined by (9.22). If t

lim sup t→∞

δ σ (s)ψ(s)Δs = ∞

(9.23)

t1

for sufficiently large t1 , where ψ(s) =

Kp(s)τ (s) sr(s)(a(s))2 − (ra)Δ (s) + , σ (s) σ (s)

s ∈ [t0 , ∞),

then every solution of the equation (9.14) is oscillatory on [t0 , ∞). Proof. We define H (t, s) for t0 ≤ s ≤ σ (t) by H (σ (t), t) = 0 and H (t, s) = 1 otherwise. Then h(t, s) = 0, C1 (s) = 0,

t, s ∈ [t0 , ∞).

Hence, H (t, s)δ σ (s)ψ(s) −

σ (s) (δ(s))2 r(s) (h(t, s))2 = δ σ (s)ψ(s), 4sδ σ (s)

s ∈ [t0 , ∞). From here and from (9.23), we get (9.21). Then, using Corollary 9.2, it follows that every solution of the equation (9.14) is oscillatory on [t0 , ∞). This completes the proof. Corollary 9.4. Assume (C), (9.15), and (9.16). Let t

lim sup t→∞

t1

 Kp(s)τ (s) 1  r Δ r(s) + Δs = ∞ σ (s) (s) + σ (s) 2 Id 4sσ (s)

(9.24)

for sufficiently large t1 . Then every solution of the equation (9.14) is oscillatory on [t0 , ∞). Proof. Let δ(t) = t,

a(t) = −

1 , 2t

t ∈ [t0 , ∞).

436

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then  2 1 sr(s) − 2s Kp(s)τ (s)  r Δ + (s) + ψ(s) = σ (s) 2I d σ (s)  Δ Kp(s)τ (s) r r(s) = + , s ∈ [t0 , ∞). (s) + σ (s) 2I d 4sσ (s) Hence from (9.24), we get (9.23). From here and from Corollary 9.3, it follows that every solution of the equation (9.14) is oscillatory on [t0 , ∞). This completes the proof. Corollary 9.5. Assume (C), (9.15), and (9.16). Let %

t

lim sup t→∞

2

(σ (s)) t1

Kp(s)τ (s) + σ (s)

(I d + σ )r 2I dσ

Δ

r(s)(s + σ (s))2 (s) + 4s(σ (s))3

& Δs = ∞

(9.25) for sufficiently large t1 . Then every solution of the equation (9.14) is oscillatory on [t0 , ∞). Proof. Let δ(t) = t 2 ,

a(t) = −

t + σ (t) , 2tσ (t)

t ∈ [t0 , ∞).

Then ψ(s) =

Kp(s)τ (s) + σ (s)

(I d + σ )r 2I dσ

Δ (s) +

r(s)(s + σ (s))2 , 4s(σ (s))3

s ∈ [t0 , ∞).

Hence from (9.25), we get (9.23). From here and from Corollary 9.3, it follows that every solution of the equation (9.14) is oscillatory on [t0 , ∞). This completes the proof. Theorem 9.7. Assume (C) and ∞ t0

1 Δt < ∞, r(t)

∞ t0

1 r(t)

t

p(s)ΔsΔt = ∞.

(9.26)

t0

Let x be a nonoscillatory solution of the equation (9.14) such that there exists a t1 ∈ [t0 , ∞) with x(t)x Δ (t) < 0,

t ∈ [t1 , ∞).

9.2 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

437

Then lim x(t) = 0.

t→∞

Proof. Without loss of generality, we suppose that x is an eventually positive solution of the equation (9.14). Then there exists a t1 ∈ [t0 , ∞) so that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Hence, x Δ (t) < 0,

t ∈ [t1 , ∞).

Let b = lim x(t). t→∞

Assume that b > 0. Then there exists a t2 ∈ [t1 , ∞) so that x(t) ≥ b,

t ∈ [t2 , ∞).

x(τ (t)) ≥ b,

Now we integrate (9.14) from t2 to t, t ≥ t2 , and we get t

r(t)x Δ (t) = r(t2 )x Δ (t2 ) −

p(s)f (x(τ (s)))Δs t2 t

≤ r(t2 )x Δ (t2 ) − K

p(s)x(τ (s))Δs t2 t

≤ r(t2 )x Δ (t2 ) − Kb

p(s)Δs. t2

Hence, x Δ (t) ≤ r(t2 )x Δ (t2 )

t

1 1 − Kb r(t) r(t)

p(s)Δs,

t ∈ [t2 , ∞).

t2

We integrate the last inequality from t2 to t, t ≥ t2 , and we get x(t) ≤ x(t2 ) + r(t2 )x Δ (t2 )

t t2

−Kb

t t2

→ −∞ as

1 r(s)

1 Δs r(s)

s

p(η)ΔηΔs t2

t → ∞.

This is a contradiction. Therefore b = 0. This completes the proof.

438

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Theorem 9.8. Assume (C), (9.26). Let H ∈ R and there are a function a and a positive differentiable function δ such that (9.17) holds. Then every solution of the equation (9.14) is oscillatory or converges to zero as t → ∞. Proof. Let x be a nonoscillatory solution of the equation (9.14). Without loss of generality, we suppose that x is an eventually positive solution of the equation (9.14). Then there exists a t1 ∈ [t0 , ∞) so that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Then  Δ Δ (t) = −p(t)f (x(τ (t))) rx ≤ −Kp(t)x(τ (t)) < 0,

t ∈ [t1 , ∞).

Hence, rx Δ is strictly decreasing on [t1 , ∞). If x Δ is eventually positive on [t1 , ∞), then x is eventually increasing on [t1 , ∞). Hence from the proof of Theorem 9.6, we get the contradiction (9.18). Therefore x Δ is eventually negative on [t1 , ∞). From here and from Theorem 9.7, it follows that lim x(t) = 0.

t→∞

This completes the proof.

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic Equations II In this section we consider the equation 

 γ Δ r(t) x Δ (t) + p(t)f (x(τ (t))) = 0,

t ∈ [t0 , ∞),

where (H1) (H2)

γ ≥ 1 is the ratio of two positive odd integers. r and p are positive real-valued rd-continuous functions and ∞ t0

1 r(t)

1 γ

Δt = ∞.

(9.27)

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

(H3)

τ : T → T is a strictly increasing and differentiable function such that τ (t) ≤ t,

(H4)

439

τ (σ (t)) = σ (τ (t)),

t ∈ T,

lim τ (t) = ∞.

t→∞

f : R → R is a continuous function such that f (x) ≥ Lx γ ,

x ∈ R,

x = 0,

for some positive constant L. Theorem 9.9. Suppose (H 1)–(H 4). Let x be an eventually positive solution of the equation (9.27). Then there exists a t1 ∈ [t0 , ∞) such that x Δ (t) > 0,

  γ Δ r(t) x Δ (t) < 0,

t ∈ [t1 , ∞).

Proof. Since x is an eventually positive solution of the equation (9.27), there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

Hence from (9.27), we get   γ Δ r(t) x Δ (t) = −p(t)f (x(τ (t))) ≤ −Lp(t) (x(τ (t)))γ < 0,

t ∈ [t1 , ∞).

γ  Therefore r(t) x Δ (t) is strictly decreasing on [t1 , ∞). Suppose that there exists a t2 ∈ [t1 , ∞) so that  γ r(t2 ) x Δ (t2 ) = c < 0. Hence,  γ  γ r(t) x Δ (t) ≤ r(t2 ) x Δ (t2 ) = c,

t ∈ [t2 , ∞).

c , r(t)

t ∈ [t2 , ∞).

Then 

x Δ (t)

γ

≤

440

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

From here, using that γ is a ratio of two positive odd integers, we obtain 1

x Δ (t) ≤ c γ

1 r(t)

1 γ

t ∈ [t2 , ∞).

,

Thus 1

x(t) ≤ x(t2 ) + c γ

t

t2

→ −∞ as

1 r(s)

1 γ

Δs

t → ∞.

This is a contradiction. Consequently t ∈ [t2 , ∞).

x Δ (t) > 0, This completes the proof. Theorem 9.10. Suppose (H 1)–(H 4). Let H : {(t, s) : t ≥ s ≥ t0 ,

t, s ∈ [t0 , ∞)} → R

be an rd-continuous function such that H (t, t) = 0,

t ≥ t0 ,

H (t, s) > 0,

t > s ≥ t0 ,

H has a non-positive continuous delta partial derivative H Δs (t, s) with respect to the second variable. Let also, there exists a positive delta differentiable function 1 ([t , ∞)) such that δ ∈ Crd 0 ⎛

1 lim sup t→∞ H (t, t1 )

t t1

γ +1 ⎞  r(τ (s)) δ Δ (s) +  γ ⎠ Δs = ∞ H (t, s) ⎝Lδ(s)p(s) − (γ + 1)γ +1 δ(s)τ Δ (s) (9.28)

for sufficiently large t1 . Then every solution of the equation (9.27) is oscillatory on [t0 , ∞). Here  Δ  δ (s) + = max{0, δ Δ (s)},

s ∈ [t0 , ∞).

Proof. Let x be a nonoscillatory solution of the equation (9.27). Without loss of generality, we assume that x is an eventually positive solution of the equation (9.27) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

x Δ (t) > 0,

t ∈ [t1 , ∞).

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

441

Let γ  r(t) x Δ (t) w(t) = δ(t) , (x(τ (t)))γ

t ∈ [t1 , ∞).

We have w(t) > 0,

t ∈ [t1 , ∞).

Also, w Δ (t) =

  γ Δ r(t) x Δ (t)  γ δ Δ (t)(x(τ (t)))γ −δ(t)((x(τ (t)))γ )Δ +r(σ (t)) x Δ (σ (t)) (x(τ (t)))γ (x(τ (σ (t))))γ δ(t) (x(τ (t)))γ

Δ

δ (t) = −p(t)f (x(τ (t))) (x(τδ(t) (t)))γ + w(σ (t)) δ(σ (t))  Δ

δ(t)r(σ (t))((x(τ (t)))γ ) x Δ (σ (t)) (x(τ (t)))γ (x(τ (σ (t))))γ δ Δ (t) −Lp(t)δ(t) + w(σ (t)) δ(σ (t))   γ δ(t)r(σ (t))((x(τ (t)))γ )Δ x Δ (σ (t)) − (x(τ (t)))γ (x(τ (σ (t))))γ

−

≤

γ

(9.29)

,

t ∈ [t0 , ∞). Using the chain rule, we have 

(x(τ (t)))γ

Δ

γ −1 (x(τ (t)))Δ dh h(x(τ (t)))σ + (1 − h)x(τ (t))

1

=γ 0

1

≥γ

(hx(τ (t)) + (1 − h)x(τ (t)))γ −1 (x(τ (t)))Δ dh

0

= γ (x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t),

t ∈ [t0 , ∞).

Hence from (9.29), we get δ Δ (t) w(σ (t))  Δδ(σ (t))γ Δ γ δ(t)r(σ (t)) x (σ (t)) x (τ (t))τ Δ (t) − , (x(τ (σ (t))))γ +1

w Δ (t) ≤ −Lp(t)δ(t) +

t ∈ [t1 , ∞). Since  γ Δ  < 0, r(t) x Δ (t)

t ∈ [t1 , ∞),

we have γ  γ  r(τ (t)) x Δ (τ (t)) ≥ r(σ (t)) x Δ (σ (t)) ,

t ∈ [t1 , ∞),

(9.30)

442

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

whereupon γ  γ  Δ r(σ (t)) x Δ (σ (t)) , x (τ (t)) ≥ r(τ (t))

t ∈ [t1 , ∞),

and

x (τ (t)) ≥ Δ

r(σ (t)) r(τ (t))

1 γ

x Δ (σ (t)),

t ∈ [t1 , ∞).

Hence from (9.30), we obtain δ Δ (t) w(σ (t)) δ(σ (t)) γ +1  γ +1 γ δ(t)τ Δ (t)(r(σ (t))) γ x Δ (σ (t))

w Δ (t) ≤ −Lp(t)δ(t) + −

1

(r(τ (t))) γ (x(τ (σ (t))))γ +1 δ Δ (t) w(σ (t)) δ(σ (t)) γ +1 γ +1  γ +1 γ (δ(σ (t))) γ (r(σ (t))) γ x Δ (σ (t)) δ(t)τ Δ (t)

= −Lp(t)δ(t) + −

(δ(σ (t))) = −Lp(t)δ(t) + −

γ +1 γ

1

(x(τ (σ (t))))γ +1 (r(τ (t))) γ

δ Δ (t) w(σ (t)) δ(σ (t))

γ δ(t)τ Δ (t)(w(σ (t))) (δ(σ (t)))

γ +1 γ

γ +1 γ

1

(r(τ (t))) γ  Δ  δ (t) + w(σ (t)) ≤ −Lp(t)δ(t) + δ(σ (t)) −

γ δ(t)τ Δ (t) (δ(σ (t)))

γ +1 γ

(r(τ (t)))

1 γ

(w(σ (t)))

γ +1 γ

,

t ∈ [t0 , ∞). Let λ=

γ +1 . γ

Then  Δ  δ (t) + δ(σ (t)) w(σ (t)) γ δ(t)τ Δ (t) λ − (r(τ (t))) λ−1 (δ(σ (t)))λ (w(σ (t))) ,

w Δ (t) ≤ −Lp(t)δ(t) +

(9.31)

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

443

t ∈ [t0 , ∞). Let

 λ1 γ δ(t)τ Δ (t) X= w(σ (t)), (r(τ (t)))λ−1 (δ(σ (t)))λ ⎛ ⎞ 1  − λ1 λ−1 Δ (t) δ Δ (t) + γ δ(t)τ ⎠ Y =⎝ , λδ(σ (t)) (r(τ (t)))λ−1 (δ(σ (t)))λ t ∈ [t1 , ∞). Now, using the inequality (see the appendix) λXY λ−1 − Xλ ≤ (λ − 1)Y λ ,

X ≥ 0,

λ > 1,

Y ≥ 0,

we get % λ

 δ Δ (t)

&1

γ δ(t)τ Δ (t)

λ

(r(τ (t)))λ−1 (δ(σ (t)))λ

w(σ (t))

%

+

λδ(σ (t))

γ δ(t)τ Δ (t) (r(τ (t)))λ−1 (δ(σ (t)))λ

γ δ(t)τ Δ (t) (w(σ (t)))λ (r(τ (t)))λ−1 (δ(σ (t)))λ ⎛ ⎞ λ &− 1 λ−1 % δ Δ (t) λ Δ γ δ(t)τ (t) ⎜ ⎟ + ≤ (λ − 1) ⎝ , ⎠ λδ(σ (t)) (r(τ (t)))λ−1 (δ(σ (t)))λ −

t ∈ [t1 , ∞), or  Δ  δ (t) + δ(σ (t))

γ δ(t)τ Δ (t) (w(σ (t)))λ (r(τ (t)))λ−1 (δ(σ (t)))λ % Δ  & λ 1 − λ−1 δ (t) + λ−1 γ δ(t)τ Δ (t) δ(σ (t)) (r(τ (t)))λ−1 (δ(σ (t)))λ

w(σ (t)) − λ

≤ (λ − 1)λ− λ−1

λ   λ−1 Δ (t) δ r(τ (t)) λ + = (λ − 1)λ− λ−1  1 , 1  γ λ−1 δ(t)τ Δ (t) λ−1

t ∈ [t1 , ∞). Let λ

1

C = (λ − 1)λ− λ−1 γ − λ−1 .

&− 1

λ

444

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Then γ +1

γ   1 γ + 1 − γ γ+1 −1 − γ γ+1 −1 γ +1 −1 γ C= γ γ

−(γ +1) 1 γ +1 = γ −γ γ γ

=

γγ (γ + 1)γ +1 γ γ

=

1 . (γ + 1)γ +1

Then γ +1  r(τ (t)) δ Δ (t) + w (t) ≤ −Lδ(t)p(t) +  γ −1 , (γ + 1)γ +1 δ(t)τ Δ (t) Δ

t ∈ [t1 , ∞). Hence, γ +1  r(τ (s)) δ Δ (s) + Δ Lp(s)δ(s) −  γ −1 ≤ −w (s), (γ + 1)γ +1 δ(s)τ Δ (s) s ∈ [t1 , ∞). From here, for t > s ≥ t1 , we get  γ +1 r(τ (s)) δ Δ (s) + Δ Lp(s)δ(s)H (t, s) − H (t, s)  γ −1 ≤ −w (s)H (t, s). γ +1 Δ δ(s)τ (s) (γ + 1) Therefore #t t1

⎛

⎞  γ +1 r(τ (s)) δ Δ (s) + H (t, s) ⎝Lp(s)δ(s) − γ −1 ⎠ Δs ≤ −  (γ + 1)γ +1 δ(s)τ Δ (s)

!s=t ! = −w(s)H (t, s)! + s=t1

= w(t1 )H (t, t1 ) +

t

t

H Δs (t, s)w(σ (s))Δs

t1

H Δs (t, s)w(σ (s))Δs

t1

≤ w(t1 )H (t, t1 ),

t ∈ [t1 , ∞).

t t1

wΔ (s)H (t, s)Δs

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

445

Hence, ⎛

1 H (t, t1 )

t t1

⎞ γ +1  r(τ (s)) δ Δ (s) + H (t, s) ⎝Lp(s)δ(s) −  γ −1 ⎠ Δs ≤ w(t1 ), (γ + 1)γ +1 δ(s)τ Δ (s)

t ∈ [t1 , ∞). Consequently ⎞ γ +1  r(τ (s)) δ Δ (s) + H (t, s) ⎝Lp(s)δ(s) −  γ −1 ⎠ Δs ≤ w(t1 ), (γ + 1)γ +1 δ(s)τ Δ (s) ⎛

1 lim sup H (t, t1 ) t→∞

t t1

(9.32) which is a contradiction. This completes the proof. Corollary 9.6. Suppose (H 1)–(H 4) and t

lim sup t→∞

t1

%

r(τ (s))  γ Lsp(s) − (γ + 1)γ +1 sτ Δ (s)

& Δs = ∞

(9.33)

for sufficiently large t1 . Then every solution of the equation (9.27) is oscillatory. Proof. Define H (t, s) for t0 ≤ s ≤ t by H (t, t) = 0 and H (t, s) = 1 otherwise. Let δ(t) = t, Then

t ∈ T.

γ +1 ⎞  r(τ (s)) δ Δ (s) +  γ ⎠ H (t, s) ⎝Lδ(s)p(s) − (γ + 1)γ +1 δ(s)τ Δ (s) ⎛

= Lsp(s) −

r(τ (s))  γ . (γ + 1)γ +1 sτ Δ (s)

t ≥ s ≥ t0 .

Hence from (9.33), we get (9.28). From here and from Theorem 9.10, we conclude that every solution of the equation (9.27) is oscillatory on [t0 , ∞). This completes the proof. Example 9.4. Let T = 2N0 . Consider the equation

1  Δ 3 x (t) t

Δ

3 %

3 & t t +t x 1+ x = 0, 2 2 2

446

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

t ∈ [2, ∞). Here σ (t) = 2t, 1 , t t τ (t) = , 2 r(t) =

p(t) = t 2 ,

t ∈ T,

γ = 3,

 f (u) = u3 1 + u3 ,

u ∈ R.

We have f (u) ≥ u3 ,

u ∈ R.

Then L = 1. Also, τ Δ (s) = ∞ t0

1 r(t)

1 γ

1 , 2 ∞

Δt =

1

t 3 Δt 2

= ∞, 1

Lsp(s) −

r(τ (s)) τ (s)  γ = s 3 −  3 4 (γ + 1)γ +1 sτ Δ (s) 4 2s = s3 −

2 s 256 3 8 s

16 256s 4 1 = s3 − , 16s 4 = s3 −

Let g(s) =

1 4 1 , s − 15 14s 3

s ∈ T.

s ∈ T.

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

447

Then g Δ (s) =

1  (σ (s))3 + s(σ (s))2 + s 2 σ (s) + s 3 15

1 (σ (s))2 + sσ (s) + s 2 14 s 3 (σ (s))3 1  3 8s + 4s 3 + 2s 3 + s 3 = 15 −

1 4s 2 + 2s 2 + s 2 14 8s 6 1 = s3 − , s ∈ T. 16s 4 −

Hence, for sufficiently large t1 , we get t t1

%

r(τ (s))  γ Lsp(s) − (γ + 1)γ +1 sτ Δ (s)

&

t

Δs =

s3 −

t1 t

=

1 16s 4

 Δs

g Δ (s)Δs

t1

!s=t ! = g(s)!

s=t1

=

1 4 1 s − 15 14s 3

=

1 4 1 t − 15 14t 3 1 1 − t14 + 15 14t13

→ ∞ as

! !s=t !

s=t1

t → ∞.

Therefore every solution of the considered equation is oscillatory on [2, ∞). Exercise 9.4. Let T = 3N0 . Prove that the equation

1  Δ 5 x (t) t3

Δ

%

5 & t 5 t + t + t + 10 x 1+ x = 0, 3 3 

4

2

t ∈ [3, ∞), is oscillatory on [3, ∞).

448

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Theorem 9.11. Suppose (H 1), (H 3), (H 4), p, and r are positive real-valued rdcontinuous functions on [t0 , ∞) such that ∞ t0

1 r(s)

1

1

s

γ

γ

Δs = ∞.

p(η)Δη

(9.34)

t0

Let x be a nonoscillatory solution of the equation (9.27) with t ∈ [t0 , ∞).

x(t)x Δ (t) < 0, Then

lim x(t) = 0.

t→∞

Proof. Without loss of generality, suppose that x is an eventually positive solution of the equation (9.27) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) so that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Then x Δ (t) < 0,

t ∈ [t1 , ∞),

i.e., x is a strictly decreasing function on [t1 , ∞). Let b = lim x(t). t→∞

Assume that b > 0. Then there exists a t2 ∈ [t1 , ∞) so that x(t) ≥ b,

t ∈ [t2 , ∞).

x(τ (t)) ≥ b,

Now we integrate the equation (9.27) from t2 to t, t ≥ t2 , and we get  γ  γ r(t) x Δ (t) = r(t2 ) x Δ (t2 ) −

t

p(s)f (x(τ (s)))Δs t2

 γ ≤ r(t2 ) x Δ (t2 ) − L

t

p(s) (x(τ (s)))γ Δs

t2

 γ ≤ r(t2 ) x Δ (t2 ) − Lbγ

t

p(s)Δs t2

≤ −Lbγ

t

p(s)Δs, t2

t ∈ [t2 , ∞).

9.3 Oscillation Criteria for a Class Second-Order Nonlinear Delay Dynamic. . .

449

Hence, 

x Δ (t)

γ

t

1 r(t)

≤ −Lbγ

t ∈ [t2 , ∞),

p(s)Δs, t2

and

1 x (t) ≤ −L b r(t) Δ

1 γ

1

 γ1

t

γ

p(s)Δs

,

t ∈ [t2 , ∞).

t2

From here, 1

x(t) ≤ x(t2 ) − L γ b

t t2

→ −∞ as

1 r(s)

1 γ

1

s

γ

p(η)Δη

Δs

t2

t → ∞.

This is a contradiction. Therefore b = 0. This completes the proof. Theorem 9.12. Assume (H 1), (H 3), (H 4), r, and p are positive real-valued rdcontinuous functions such that (9.34) holds, Let H be as in Theorem 9.10 and (9.28) holds. Then every solution of the equation (9.27) is oscillatory or converges to zero as t → ∞. Proof. Let x be a nonoscillatory solution of the equation (9.27) on [t0 , ∞). Without loss of generality, we assume that x is an eventually positive solution of the equation (9.27) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t),

x(τ (t)) > 0,

t ∈ [t1 , ∞).

Hence,   γ Δ r(t) x Δ (t) = −p(t)f (x(τ (t))) ≤ −Lp(t) (x(τ (t)))γ < 0,

t ∈ [t1 , ∞).

Then γ  r(t) x Δ (t) is strictly decreasing on [t1 , ∞). If x Δ is eventually positive on [t1 , ∞), then x is eventually increasing on [t1 , ∞). Hence from the proof of Theorem 9.10, we get the contradiction (9.32). Therefore x Δ is eventually negative on [t1 , ∞). From here and

450

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

from Theorem 9.11, it follows that lim x(t) = 0.

t→∞

This completes the proof.

9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic Equations In this section we consider the equation Δ     + p(t)φ x Δ (t) + q(t)f (φ(x(τ (t)))) = 0, a(t)φ x Δ (t)

(9.35)

t ∈ [t0 , ∞), where (T1) (T2) (T3)

φ(s) = |s|γ −2 s, s ∈ R, γ > 2. a, p, q : T → R are positive rd-continuous functions such that 1 − μ pa > 0 on [t0 , ∞). τ : T → T is a strictly increasing and differentiable function such that τ (t) ≤ t,

(T4)

lim τ (t) = ∞.

τ (T) = T,

t→∞

f : R → R is a continuous function such that f (x) ≥ Lx,

x ∈ R,

x = 0,

for some positive constant L. Theorem 9.13. Assume (T 1) − −(T 4) and ∞ t0

1 e p (t, t0 ) a(t) − a



1 γ −1

Δt = ∞.

(9.36)

Let x be an eventually positive solution of the equation (9.35) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x Δ (t) > 0,

  Δ a(t)φ x Δ (t) < 0,

t ∈ [t1 , ∞).

Proof. Since x is an eventually positive solution of the equation (9.35) on [t0 , ∞), then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic. . .

451

Hence from (9.35), we get 

Δ    + p(t)φ x Δ (t) = −q(t)f (φ(x(τ (t)))) a(t)φ x Δ (t) ≤ −Lq(t)φ(x(τ (t))) = −Lq(t)|x(τ (t))|γ −2 x(τ (t)) < 0,

t ∈ [t1 , ∞).

Hence, %

 &Δ 

  Δ a(t)φ x Δ (t) 1 a(t)φ x Δ (t) = e− p (t, t0 ) a e− p (t, t0 ) e− p (t, t0 )e− p (σ (t), t0 ) a a a  p(t) a(t)φ(x Δ (t))e− p (t, t0 ) + a a(t)

  Δ 1 a(t)φ x Δ (t) e− p (t, t0 ) = a e− p (t, t0 )e− p (σ (t), t0 ) a a    +p(t)φ x Δ (t) e− p (t, t0 ) a


0,

x Δ (t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞),

and 

 γ −1 Δ a(t) x Δ (t) < 0,

t ∈ [t1 , ∞).

Then x(τ (t)) ≤ x(τ (σ (t))),  Δ γ −1  γ −1 a(t) x (t) ≥ a(σ (t)) x Δ (σ (t)) , γ −1  Δ γ −1  Δ ≥ a(σ (t)) x (σ (t)) , a(τ (t)) x (τ (t))

t ∈ [t1 , ∞).

Hence, 1 1 ≥ , x(τ (t)) x(τ (σ (t)))

t ∈ [t1 , ∞),

and 1

x (τ (t)) ≥ Δ

(a(σ (t))) γ −1 x Δ (σ (t)) 1

,

(a(τ (t))) γ −1

t ∈ [t1 , ∞).

By the chain rule, we have 

(x(τ (t)))γ −1

Δ

1

= (γ − 1) (x(τ (t)))Δ

(hx(τ (σ (t))) + (1 − h)x(τ (t)))γ −2 dh

0 1

≥ (γ − 1)(x(τ (t)))Δ

(hx(τ (t)) + (1 − h)x(τ (t)))γ −2 dh

0

= (γ − 1)(x(τ (t))) (x(τ (t)))γ −2 Δ

= (γ − 1)x Δ (τ (t))τ Δ (t)(x(τ (t)))γ −2 ,

t ∈ [t1 , ∞).

454

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Therefore  Δ (x(τ (t)))γ −1 (γ − 1)x Δ (τ (t))τ Δ (t)(x(τ (t)))γ −2 ≥ (x(τ (t)))γ −1 (x(τ (t)))γ −1 =

(γ − 1)x Δ (τ (t))τ Δ (t) x(τ (t))

≥

(γ − 1)x Δ (τ (t))τ Δ (t) , x(τ (σ (t)))

t ∈ [t1 , ∞).

Let γ −1  a(t) x Δ (t) , w(t) = z(t) (x(τ (t)))γ −1

t ∈ [t1 , ∞).

We have w(t) > 0,

t ∈ [t1 , ∞),

and   Δ γ −1 Δ z(t) a(t) x (t) (x(τ (t)))γ −1  γ −1 +a(σ (t)) x Δ (σ (t)) Δ  (x(τ (t)))γ −1 zΔ (t) − z(t) (x(τ (t)))γ −1 × (x(τ (t)))γ −1 (x(τ (σ (t))))γ −1   q(t)z(t)f (x(τ (t)))γ −1 =− (x(τ (t)))γ −1 γ −1  p(t)z(t) x Δ (t) − (x(τ (t)))γ −1  γ −1 +a(σ (t)) x Δ (σ (t)) Δ  (x(τ (t)))γ −1 zΔ (t) − z(t) (x(τ (t)))γ −1 × (x(τ (t)))γ −1 (x(τ (σ (t))))γ −1   q(t)z(t)f (x(τ (t)))γ −1 =− (x(τ (t)))γ −1

w Δ (t) =

−

zΔ (t) p(t) w(t) + w(σ (t)) a(t) z(σ (t))

9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic. . .

Δ γ −1   (x(τ (t)))γ −1 a(σ (t))z(t) x Δ (σ (t)) − (x(τ (t)))γ −1 (x(τ (σ (t))))γ −1 zΔ (t) p(t) w(t) + w(σ (t)) a(t) z(σ (t))  Δ γ −1  a(σ (t))z(t) x Δ (σ (t)) (x(τ (t)))γ −1 − (x(τ (t)))γ −1 (x(τ (σ (t))))γ −1

≤ −Lq(t)z(t) −

zΔ (t) p(t) w(t) + w(σ (t)) a(t) z(σ (t))  γ −1  Δ  (γ − 1)z(t)a(σ (t)) x Δ (σ (t)) x (τ (t)) τ Δ (t) − (x(τ (σ (t))))γ

≤ −Lq(t)z(t) −

zΔ (t) p(t) w(t) + w(σ (t)) a(t) z(σ (t)) γ  γ (γ − 1)z(t)τ Δ (t)(a(σ (t))) γ −1 x Δ (σ (t))

≤ −Lq(t)z(t) − −

1

(a(τ (t))) γ −1 (x(τ (σ (t))))γ zΔ (t) p(t)z(t) w(σ (t)) + w(σ (t)) a(t)z(σ (t)) z(σ (t)) γ  γ (γ − 1)z(t)τ Δ (t)(a(σ (t))) γ −1 x Δ (σ (t))

≤ −Lq(t)z(t) − −

1

(a(τ (t))) γ −1 (x(τ (σ (t))))γ

≤ −Lq(t)z(t) +

|a(t)zΔ (t) − p(t)z(t)| w(σ (t)) a(t)z(σ (t)) γ

−

(γ − 1)z(t)τ Δ (t)(w(σ (t))) γ −1 1

γ

(a(τ (t))) γ −1 (z(σ (t))) γ −1

,

t ∈ [t1 , ∞).

Let γ , γ −1  λ1

(γ − 1)z(t)τ Δ (t) X(t) = w(σ (t)), (a(τ (t)))λ−1 (z(σ (t)))λ ⎞ 1 ⎛! ! 1 !a(t)zΔ (t) − p(t)z(t)! (γ − 1)z(t)τ Δ (t) − λ λ−1 ⎠ Y (t) = ⎝ , λa(t)z(σ (t)) (a(τ (t)))λ−1 (z(σ (t)))λ λ=

455

456

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

t ∈ [t1 , ∞). Then w Δ (t) ≤ −Lq(t)z(t) + λX(t) (Y (t))λ−1 − (X(t))λ , t ∈ [t1 , ∞). Let λ

1

C = (λ − 1)λ− λ−1 (γ − 1)− λ−1 . Then C=

1 . γγ

We have (see the appendix) λX(t)(Y (t))λ−1 − (X(t))λ ≤ (λ − 1)(Y (t))λ ! λ ! ! a(t)zΔ (t) − p(t)z(t) ! λ−1 ! = C !! ! a(t)z(σ (t)) 1 − λ−1 z(t)τ Δ (t) × (a(τ (t)))λ−1 (z(σ (t)))λ ! λ ! ! λ−1 ! Δ a(τ (t)) p(t) ! z(t)!! = C !z (t) −   1 a(t) z(t)τ Δ (t) λ−1 !γ ! ! ! Δ a(τ (t)) p(t) ! , ! z z(t) = (t) −  γ −1 ! ! a(t) γ γ z(t)τ Δ (t)

t ∈ [t1 , ∞). Then w (t) ≤ −Lq(t)z(t) + Δ

a(τ (t))

 γ −1 γ γ z(t)τ Δ (t)

!γ ! ! ! Δ !z (t) − p(t) z(t)! , ! ! a(t)

t ∈ [t1 , ∞). Consequently %

a(τ (t)) Lq(t) −  γ −1 γ γ τ Δ (t)

! & ! Δ ! z (t) p(t) !γ Δ ! ! ! z(t) − a(t) ! z(t) ≤ −w (t),

9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic. . .

457

t ∈ [t1 , ∞). Hence, %

! & ! Δ ! z (s) p(s) !γ a(τ (s)) ! z(s)Δs ≤ − ! Lq(s) − −  γ −1 ! z(s) a(s) ! γ γ τ Δ (s)

t t1

t

w Δ (s)Δs

t1

= w(t1 ) − w(t) ≤ w(t1 ) − w(t) ≤ w(t1 ),

t ∈ [t1 , ∞),

which contradicts with (9.37). This completes the proof. Corollary 9.7. Suppose (T 1)–(T 4) and (9.36). If ⎛ t

lim sup t→∞

t0

! !γ ⎞ ! p(s) ! (s) a(τ (s)) ! s+σ − a(s) ! ⎟ 2 s 2 +1 ⎜ ⎝Lq(s) − ⎠ (s + 1)Δs = ∞,  γ −1 γ γ τ Δ (s)

(9.38)

then every solution of the equation (9.35) is oscillatory on [t0 , ∞). Proof. Let z(s) = s 2 + 1,

s ∈ T.

Then zΔ (s) = s + σ (s),

s ∈ T,

and ⎛

⎛ ! ⎞ !γ ⎞ ! Δ ! ! (s) p(s) !γ ! p(s) ! (s) − a(s) ! − a(τ (s)) ! zz(s) a(τ (s)) ! s+σ 2 a(s) ! ⎟ 2 s +1 ⎜ ⎟ ⎜ ⎝Lq(s) − ⎠ z(s) = ⎝Lq(s) − ⎠ (s + 1),   γ −1 γ −1 γ γ τ Δ (s) γ γ τ Δ (s)

s ∈ [t0 , ∞). Hence from (9.38), we get (9.37). From here and from Theorem 9.14, we conclude that every solution of the equation (9.35) is oscillatory on [t0 , ∞). This completes the proof. Example 9.5. Let T = hZ, h > 0. Consider the equation ! ! ! ! 2 Δ t + 1 !x!Δ (t)! x Δ (t)! + (2t + h) !x!Δ (t)! x Δ (t)!   +(t + 1) !x Δ (t − h)! x Δ (t − h) 1 + !x Δ (t − h)! x Δ (t − h) = 0,

458

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

t ∈ [h, ∞). Here γ = 3, σ (t) = t + h, a(t) = t 2 + 1, p(t) = 2t + h, q(t) = t + 1,

t ∈ T,

f (u) = u(1 + u),

u ∈ R.

Then f (u) ≥ u,

u ∈ R,

u = 0.

Hence, L = 1. Next, t + σ (t) p(t) 2t + h 2t + h = 2 − − 2 a(t) t2 + 1 t +1 t +1 = 0,

t ∈ T.

From here, ⎛ t t0

! !γ ⎞ ! p(s) ! (s) a(τ (s)) ! s+σ − 2 a(s) ! ⎟ 2 s +1 ⎜ ⎝Lq(s) − ⎠ (s + 1)Δs  γ −1 γ γ τ Δ (s) t

≥

q(s)Δs h t

=

(s + 1)Δs,

t ∈ [h, ∞).

h

Let g(t) =

12 t + (2 − h)t , 2

t ∈ T.

9.4 Oscillation Criteria for a Class Second-Order Half-Linear Delay Dynamic. . .

459

Then 1 (σ (t) + t + 2 − h) 2 1 = (t + h + t + 2 − h) 2 = t + 1, t ∈ T.

g Δ (t) =

Therefore ⎛ t t0

! !γ ⎞ ! p(s) ! (s) a(τ (s)) ! s+σ − a(s) ! ⎟ 2 s 2 +1 ⎜ ⎝Lq(s) − ⎠ (s + 1)Δs  γ −1 γ γ τ Δ (s) t

≥

g Δ (s)Δs

h

!s=t ! = g(s)!

s=h

= = = →

!s=t 1 2 ! s + (2 − h)s ! s=h 2 1 12 t + (2 − h)t − h2 + (2 − h)h 2 2 12 t + (2 − h)t − h 2 ∞ as t → ∞.

Hence from Corollary 9.7, we conclude that every solution of the considered equation is oscillatory on [h, ∞). Exercise 9.5. Let T = 2N0 . Prove that every solution of the equation 

! ! ! Δ !3 Δ Δ 1 !x (t)! x (t) + (t + 1) !x Δ (t)!3 x Δ (t) t 4 +3t 2 +1 !  !3       !  !3 + t 20 + t 10 + t 8 + 1 !x Δ 2t ! x Δ 2t 1 + !x Δ 2t ! x Δ 2t

t ∈ [2, ∞), is oscillatory on [2, ∞). Theorem 9.15. Assume (T 1)–(T 4) and (9.36). Let H : {(t, s) : t ≥ s ≥ t0 } → R be a rd-continuous function such that H (t, t) = 0,

t ≥ t0 ,

H (t, s) > 0,

t > s ≥ t0 ,

= 0,

460

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

and H has a non-positive continuous Δ-partial derivative H Δs (t, s) with respect to the second variable. If there exists a positive differentiable function z : T → R such that ⎛ ! Δ ! ⎞ ! z (s) p(s) !γ t a(τ (s)) − ! z(s) a(s) ! ⎟ 1 ⎜ lim sup H (t, s) ⎝Lq(s) − ⎠ z(s)Δs = ∞,  γ −1 t→∞ H (t, t1 ) t0 γ γ τ Δ (s) for sufficiently large t1 , then every solution of the equation (9.35) is oscillatory on [t0 , ∞). Proof. Suppose that the equation (9.35) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

x Δ (t) > 0,

t ∈ [t1 , ∞),

and 

 γ −1 Δ a(t) x Δ (t) < 0,

t ∈ [t1 , ∞).

As in the proof of Theorem 9.14, we get ⎛ ! Δ ! ⎞ ! (s) p(s) !γ a(τ (s)) ! zz(s) − a(s) ! ⎜ ⎟ Δ ⎝Lq(s) − ⎠ z(s) ≤ −w (s),  γ −1 γ Δ γ τ (s) s ∈ [t1 , ∞). Hence, ! Δ ! ⎞ ! (s) p(s) !γ a(τ (s)) ! zz(s) − a(s) ! ⎟ ⎜ Δ H (t, s) ⎝Lq(s) − ⎠ z(s) ≤ −H (t, s)w (s),  γ −1 γ Δ γ τ (s) ⎛

s, t ∈ [t1 , ∞), t > s. Then ! ⎞ ! Δ ! (s) p(s) !γ − a(s) ! a(τ (s)) ! zz(s) ⎟ ⎜ H (t, s) ⎝Lq(s) − ⎠ z(s)Δs ≤ −  γ −1 γ Δ γ τ (s) ⎛

t t1

t

H (t, s)w Δ (s)Δs

t1

!s=t ! = −H (t, s)w(s)!

s=t1

+

t

H Δs (t, s)w(σ (s))Δs

t1

≤ H (t, t1 )w(t1 ),

t ∈ [t1 , ∞).

9.5 Oscillation of Second-Order Nonlinear Neutral Delay Dynamic Equations

461

Therefore ! Δ ! ⎞ ! (s) p(s) !γ a(τ (s)) ! zz(s) − a(s) ! ⎟ ⎜ H (t, s) ⎝Lq(s) − ⎠ z(s)Δs ≤ w(t1 ), γ −1  γ γ τ Δ (s) ⎛

lim sup t→∞

1 H (t, t1 )

t t1

which is a contradiction. This completes the proof.

9.5 Oscillation of Second-Order Nonlinear Neutral Delay Dynamic Equations In this section we consider the equation   γ Δ r(t) y Δ (t) + f (t, x(δ(t))) = 0,

t ∈ [t0 , ∞),

(9.39)

where y(t) = x(t) + p(t)x(τ (t)), (U1) (U2)

γ is a quotient of two odd integers, r, p ∈ Crd (T) are positive real-valued functions such that 0 ≤ p(t) < 1,

∞ t0

(U3)

1 γ

Δt = ∞,

f : T × R → R is a continuous function such that uf (t, u) > 0,

u ∈ R,

|f (t, u)| ≥ q(t)|u| , γ

(U4)

1 r(t)

u = 0, u ∈ R,

u = 0,

τ, δ : T → T, τ (t) ≤ t, δ(t) < t, t ∈ T, lim τ (t) = lim δ(t) = ∞.

t→∞

t→∞

Theorem 9.16. Suppose (U 1)–(U 4) and δ(t) < t, t ∈ T. Let E = {λ : λ > 0,

1 − λg(t)μ(t) > 0,

  lim sup sup λe−λg (t, δ(t)) < 1, t→∞ λ>0

t ∈ T} , (9.40)

462

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

where g(t) = (1 − p(δ(t)))γ q(t),

t ∈ T.

Then every solution of the equation (9.39) is oscillatory on [t0 , ∞). Proof. Assume that the equation (9.39) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (9.39) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

x(δ(t)) > 0,

t ∈ [t1 , ∞).

Hence, y(t) > 0,

t ∈ [t1 , ∞).

Suppose that y Δ (t2 ) ≤ 0 for some t2 ∈ [t1 , ∞). Then   γ Δ r(t) y Δ (t) = −f (t, x(δ(t))) ≤ −q(t) (x(δ(t)))γ < 0,

t ∈ [t2 , ∞).

Hence, γ  γ  r(t) y Δ (t) < r(t2 ) y Δ (t2 ) ≤ 0,

t ∈ (t2 , ∞).

Therefore y Δ (t) < 0,

t ∈ (t2 , ∞).

Take t3 ∈ (t2 , ∞) and set  γ r(t3 ) y Δ (t3 ) = d. We have d < 0 and γ  γ  r(t) y Δ (t) < r(t3 ) y Δ (t3 ) = d,

t ∈ [t3 , ∞),

9.5 Oscillation of Second-Order Nonlinear Neutral Delay Dynamic Equations

463

and  Δ γ d , y (t) < r(t)

t ∈ [t3 , ∞).

Consequently Δ

y (t) < d

1 γ

1 r(t)

1 γ

t ∈ [t3 , ∞).

,

From the last inequality, we get y(t) < y(t3 ) + d

t

1 γ

t3

→ −∞

1 r(s)

1 γ

Δs

t → ∞.

as

This is a contradiction. Therefore t ∈ [t1 , ∞).

y Δ (t) > 0, Hence,

y(t) = x(t) + p(t)x(τ (t)) ≤ x(t) + p(t)y(τ (t)) ≤ x(t) + p(t)y(t),

t ∈ [t1 , ∞),

x(t) ≥ (1 − p(t)) y(t),

t ∈ [t1 , ∞).

or

Then t

y(t) = y(t1 ) +

y Δ (s)Δs

t1

≥

t

y Δ (s)Δs

t1

=

t t1

1 (r(s)) 1

1

1 γ

(r(s)) γ y Δ (s)Δs

≥ (r(t)) γ y Δ (t)

t t1

1 1

(r(s)) γ

Δs,

t ∈ [t1 , ∞).

(9.41)

464

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Since ∞ t0

1 r(t)

1 γ

Δt = ∞,

we have for sufficiently large t ∈ [t1 , ∞) the following inequality t

t1

1 r(s)

1 γ

Δs > 1.

Then for sufficiently large t ∈ [t1 , ∞) we have 1

y(t) ≥ (r(t)) γ y Δ (t). Take t2 ∈ [t1 , ∞) so that δ(t) ≥ t1

for t ∈ [t2 , ∞).

Then 1

y(δ(t)) ≥ (r(δ(t))) γ y Δ (δ(t)),

t ∈ [t2 , ∞).

From here and from (9.41), we obtain x(δ(t)) ≥ (1 − p(δ(t))) y(δ(t)) 1

≥ (1 − p(δ(t))) (r(δ(t))) γ y Δ (δ(t)) for sufficiently large t ∈ [t2 , ∞). Let γ  z(t) = r(t) y Δ (t) ,

t ∈ [t2 , ∞).

Then, using (9.39), we obtain zΔ (t) + q(t) (x(δ(t)))γ ≤ 0,

t ∈ [t2 , ∞).

From the last inequality and from (9.42), we get 0 ≥ zΔ (t) + q(t) (x(δ(t)))γ ≥ zΔ (t) + q(t) (1 − p(δ(t)))γ z(δ(t)) = zΔ (t) + g(t)z(δ(t))

(9.42)

9.5 Oscillation of Second-Order Nonlinear Neutral Delay Dynamic Equations

465

for sufficiently large t ∈ [t2 , ∞), Consequently, the inequality zΔ (t) + g(t)z(δ(t)) ≤ 0,

t ∈ [t2 , ∞),

has an eventually positive solution on [t2 , ∞). This is a contradiction. This completes the proof. Example 9.6. Let T = 2N0 and t0 = 8, p(t) =

1 , t

q(t) = t 5 ,

r(t) =

1 , t8

τ (t) =

t t , δ(t) = , 4 2

Then

3 t g(t) = 1 − p q(t) 2 

2 3 5 = 1− t t = (t − 2)3 t 2 , #t

e−λg (t, δ(t)) = e

1 t μ(s) Log(1−λμ(s)g(s))Δs 2

t t = eLog(1−λμ( 2 )g ( 2 ))

  t t g , = 1 − λμ 2 2

t ∈ [8, ∞).

Let h(λ) = λe−λg (t, δ(t)),

t ∈ [8, ∞).

Then h(λ) = = h (λ) = h(λ) ≤ =

  t t g λ 1 − λμ 2 2

  t t g , λ − λ2 μ 2 2

  t t g , 1 − 2λμ 2 2

 1 1     1− 2 2μ 2t g 2t 1 t  t  4μ 2 g 2

γ = 3.

466

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

= =

2t

1

3  t 2 −2 2

t 2

t 3 (t

< 1,

16 − 4)3 λ > 0,

t ∈ [8, ∞).

Also, 0 ≤ p(t) < 1,

t ∈ [8, ∞),

and ∞ t0

1 r(t)

1 γ

∞

Δt =

8

t 3 Δt 8

≥ lim

t

t→∞ 8

Δs

= ∞. Therefore every solution of the equation (9.39) is oscillatory on [8, ∞). Exercise 9.6. Let T = 3N0 and t0 = 27, p(t) =

1 , t2

q(t) = t 9 ,

r(t) =

1 , t 13 + 1

τ (t) =

t , 9

δ(t) =

t , 3

γ = 5.

Prove that every solution of the equation (9.39) is oscillatory on [27, ∞).

9.6 Advanced Practical Problems Problem 9.1. Let T = hZ, h > 0. Prove that the oscillation of the equation  2 x Δ (t) + t 6 + t 4 + t 2 + 1 (x(t − h))5 = 0,

t ∈ T,

and the equation  2 x Δ (t) + t 6 + t 4 + t 2 + 1 (x(t + h))5 = 0, is equivalent.

t ∈ T,

9.6 Advanced Practical Problems

467

Problem 9.2. Let T = hZ, h > 0. Prove that the oscillation of the equation 2

x Δ (t) +

2 2 t + 1 (x(t + h))15 = 0, 13

t ∈ T,

implies the oscillation of the equation  2 x Δ (t) + t 4 + 2t 2 + 10 (x(t − h))15 = 0,

t ∈ T.

Problem 9.3. Let T = 3N0 . Prove that every solution of the equation   Δ Δ rx (t) + t 10 + t 8 + t 2 x

%

6 & t t = 0, 1+ x 3 3

t ≥ 3,

where r(t) =

1 , t 18

t ∈ T,

is oscillatory on [3, ∞). Problem 9.4. Let T = q N0 , q > 1. Prove that every solution of the equation 

rx

 Δ Δ



(t) + t

5q

+t

6q

%

4 6 & t  t t 7+ x +t + x = 0, x q q q 4q

t ≥ q,

where r(t) =

1 t 8q

,

t ∈ T,

is oscillatory on [q, ∞). Problem 9.5. Let T = 4N0 . Prove that the equation

 Δ 9 1 x (t) 4 2 t + t + 12

Δ

%

9 &  t 9 t 1+ x = 0, + t 8 + 10t 4 + 12t 2 + 10 x 4 4

t ∈ [4, ∞), is oscillatory on [4, ∞). Problem 9.6. Let T = 3N0 . Prove that every solution of the equation 

! Δ !7 Δ Δ ! Δ !7 Δ !x (t)! x (t) + !x (t)! x (t) !  !7   !  !7     + 120t 11 + 7t 9 + t 4 + 10 !x Δ 3t ! x Δ 2t 1 + !x Δ 3t ! x Δ 3t = 0, 1

3t 2 +1

t ∈ [3, ∞), is oscillatory on [3, ∞).

468

9 Oscillations of Second-Order Nonlinear Functional Dynamic Equations

Problem 9.7. Let T = 4N0 and t0 = 64, 1 p(t) = 4 , t + t2

q(t) = t 9 +t 7 +t 5 ,

1 r(t) = 10 , t + t 5 + 10

τ (t) =

t , 16

δ(t) =

t , 4

γ = 7.

Prove that every solution of the equation (9.39) is oscillatory on [64, ∞). Problem 9.8. Let T = 5N0 and t0 = 125, p(t) =

1 , t 50

q(t) = t 20 +10t 9 +10t 7 +t 5 ,

r(t) =

1 , t 20

τ (t) =

t , 25

δ(t) =

t , 5

Prove that every solution of the equation (9.39) is oscillatory on [125, ∞).

γ = 11.

Chapter 10

Oscillations of Third-Order Functional Dynamic Equations

The material of this chapter are based on the results of the papers and monographs [11–13, 34, 35, 79, 101–103, 105, 115, 119, 120, 126–128, 137, 140, 141, 157, 187, 191, 192, 236–238]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations Consider the equation   Δ γ Δ a(t) r(t)x Δ (t) + f (t, x(τ (t))) = 0 on

T,

(10.1)

where (A1) (A2) (A3) (A4)

γ ≥ 1 is a quotient of odd positive integers, a, r ∈ Crd (T) are positive functions, τ : T → T, τ (t) ≤ t, τ (σ (t)) = σ (τ (t)), τ Δ (t) ≥ 0, t ∈ T, limt→∞ τ (t) = ∞, f ∈ C (T × R, R) satisfies uf (t, u) > 0,

u = 0,

t ∈ T,

and there exists a positive function p ∈ Crd (T) such that f (t, u) ≥ p(t), uγ

u = 0,

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_10

t ∈ T.

469

470

10 Oscillations of Third-Order Functional Dynamic Equations

Definition 10.1. By a solution of (10.1) we mean a nontrivial real-valued function 1 ([T , ∞)), T ≥ t , such that x ∈ Crd x x 0 rx Δ ,

a

 Δ γ rx Δ ∈ Crd ([Tx , ∞)),

and x satisfies (10.1) on [Tx , ∞). Definition 10.2. A solution x of (10.1) is said to be oscillatory if it is neither eventually positive nor eventually negative. Otherwise, it is called nonoscillatory. Theorem 10.1. Assume (A1)–(A4), ∞ t0

Δt (a(t))

∞

= ∞,

1 γ

t0

Δt = ∞, r(t)

(10.2)

and ∞ t0

1 r(t)

∞ t

1 a(s)

∞

1 γ p(u)Δu ΔsΔt = ∞.

(10.3)

s

Suppose that the equation (10.1) has a positive solution x on [t0 , ∞). Then there exists a T ∈ [t0 , ∞), sufficiently large, so that   Δ γ Δ a(t) r(t)x Δ (t) < 0, Δ  r(t)x Δ (t) > 0,

t ∈ [T , ∞),

t ∈ [T , ∞),

and either x Δ (t) > 0 on [T , ∞) or limt→∞ x(t) = 0. Proof. Let t1 ∈ [t0 , ∞) be such that x(τ (t)) > 0,

t ∈ [t1 , ∞).

By (10.1) and (A4), we get   Δ γ Δ = −f (t, x(τ (t))) a(t) r(t)x Δ (t) ≤ −p(t) (x(τ (t)))γ < 0, t ∈ [t1 , ∞). Therefore a(t)

 Δ γ r(t)x Δ (t)

(10.4)

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

471

 Δ is strictly decreasing on [t1 , ∞). Then r(t)x Δ (t) is eventually of one sign. Assume that there exists a t2 ∈ [t1 , ∞), sufficiently large, such that 

r(t)x Δ (t)

Δ

t ∈ [t2 , ∞).

< 0,

Now we integrate (10.4) from t2 to t, t ∈ [t2 , ∞), and we get a(t)

  γ Δ Δ γ r(t)x Δ (t) ≤ a(t2 ) rx Δ (t2 ) ,  γ Δ γ a(t2 )  Δ Δ rx r(t)x Δ (t) ≤ (t2 ) , a(t) 1

 Δ (a(t2 )) γ  Δ Δ r(t)x Δ (t) rx ≤ (t2 ), 1 (a(t)) γ r(t)x Δ (t) ≤ r(t2 )x Δ (t2 ) 1  Δ +(a(t2 )) γ rx Δ (t2 )

t t2

→ −∞

t → ∞.

as

Therefore there is a t3 ∈ [t2 , ∞) such that t ∈ [t3 , ∞),

r(t)x Δ (t) < 0,

and since r(t)x Δ (t) is strictly decreasing on [t3 , ∞), we get r(t)x Δ (t) ≤ r(t3 )x Δ (t3 ),

t ∈ [t3 , ∞),

1 , r(t)

t ∈ [t3 , ∞).

and x Δ (t) ≤ r(t3 )x Δ (t3 ) Hence, x(t) ≤ x(t3 ) + r(t3 )x Δ (t3 )

t t3

→ −∞ as

Δs r(s)

t → ∞.

This is a contradiction. Therefore 

r(t)x Δ (t)

Δ

> 0,

t ∈ [t1 , ∞).

Δs 1

(a(s)) γ

472

10 Oscillations of Third-Order Functional Dynamic Equations

Thus x Δ (t) is eventually of one sign. Then either there exists a t3 ∈ [t1 , ∞) such that x Δ (t) > 0 on

[t1 , ∞)

x Δ (t) < 0

or

on

[t1 , ∞).

If x Δ (t) < 0 on [t1 , ∞), we obtain lim x(t) = l1

t→∞

≥0 and   lim r(t)x Δ (t) = l2

t→∞

≤ 0. Assume that l1 > 0. Then there is a t4 ∈ [t1 , ∞) such that t ∈ [t4 , ∞).

x(τ (t)) ≥ l1 , Hence from (10.1), we get −a(t)

 Δ γ ≤− r(t)x Δ (t)

∞

f (s, x(τ (s)))Δs t ∞

≤−

p(s)(x(τ (s)))γ Δs

t ∞

γ

≤ −l1

p(s)Δs,

t ∈ [t4 , ∞).

t

From here, −

 ∞ Δ γ γ 1 r(t)x Δ (t) ≤ −l1 p(s)Δs, a(t) t

1 ∞ γ  Δ 1 Δ − r(t)x (t) ≤ −l1 p(s)Δs , a(t) t

t ∈ [t4 , ∞),

and ∞

r(t)x Δ (t) ≤ −l1 t

∞

≤ −l1 t

1 a(s) 1 a(s)

∞

1 γ p(u)Δu Δs + l2

∞

1 γ p(u)Δu Δs,

s

s

t ∈ [t4 , ∞),

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

473

and ∞

l1 x (t) ≤ − r(t) Δ

t

∞

1 a(s)

1 γ p(u)Δu Δs,

t ∈ [t4 , ∞).

s

Consequently x(t) ≤ x(t4 ) − l1

t t4

→ −∞ as

%

1 r(s)

∞ s

1 a(u)

1

∞

&

γ

p(t)Δt

Δu Δs

u

t → ∞.

This is a contradiction and we conclude that l1 = 0. This completes the proof. Example 10.1. Let T = 2N0 . Consider the equation

 t 21 Δ3 = 0, t ∈ [2, ∞), x x (t) + 3 2 128t x(t) =

1 , t

t ∈ [1, 2].

Here σ (t) = 2t, a(t) = 1, r(t) = 1,

t ∈ T,

γ = 1, t0 = 2, 21 u, t ∈ T, u ∈ R, 128t 3 uf (t, u) > 0, u = 0, t ∈ T, f (t, u) =

p(t) =

21 , 128t 3

t ∈ T.

We will prove that x(t) =

1 , t

t ∈ T,

is a solution of the considered equation. We have x Δ (t) = − =−

1 tσ (t) 1 , 2t 2

474

10 Oscillations of Third-Order Functional Dynamic Equations 2

x Δ (t) = =

σ (t) + t 2t 2 (σ (t))2 2t + t 2t 2 (2t)2

3t 8t 4 3 = 3 8t > 0, =

x

Δ3

  3 (σ (t))2 + tσ (t) + t 2 (t) = − 8t 3 (σ (t))3   2 3 4t + 2t 2 + t 2 =− 8t 3 (8t 3 ) 21t 2 64t 6 21 =− 4 64t < 0, t ∈ T, =−

and x

Δ3

21 (t) + x 128t 3



 2 21 t 21 =− 4 + 2 64t 128t 3 t = 0,

t ∈ [2, ∞).

Next, ∞ 2 ∞

∞ 2

1 r(t)

∞ t

2

1 a(s)

∞ s

Δt = ∞, a(t) Δt = ∞, r(t)

 p(u)Δu ΔsΔt =

∞

∞

2

=−

t ∞

7 32

7 = 32

∞

2 ∞ 2

21 ΔuΔsΔt 128u3 s ∞ 1 !u=∞ ! ΔsΔt 2 !u=s u t ∞

t

1 ΔsΔt s2

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

=−

∞

7 16

7 = 16 = ∞.

2 ∞ 2

475

1 !!s=∞ Δt ! s s=t

Δt t

Also, lim x(t) = lim

t→∞

t→∞

1 t

= 0. For all sufficiently large T1 ∈ [t0 , ∞) for any T > T1 such that τ (T ) > T1 , we define, for t ∈ [T , ∞), δ1 (t, T1 ) =

t

Δs

T1

(a(s)) γ

1

,

t

δ1 (s, T1 ) Δs, r(s) T1  γ −1 δ(t, T1 ) = δ1 (t, T1 ) δ2σ (t, T1 ) .

δ2 (t, T1 ) =

Theorem 10.2. Assume that x is a positive solution of the equation (10.1) such that  Δ Δ rx (t) > 0, x Δ (t) > 0 on [t1 , ∞), for some t1 ∈ [t0 , ∞). Then there exists a t∗ ∈ [t1 , ∞) such that x Δ (t) ≥

1  Δ δ1 (t, t∗ ) (a(t)) γ rx Δ (t), r(t)

t ∈ [t∗ , ∞),

(10.5)

and 1  Δ x(t) ≥ δ2 (t, t∗ ) (a(t)) γ rx Δ (t),

t ∈ [t∗ , ∞).

Proof. Note that there exists a t∗ ∈ [t1 , ∞) so that  Δ γ a(t) rx Δ (t) is strictly decreasing on [t∗ , ∞). Hence, r(t)x Δ (t) ≥ r(t)x Δ (t) − r(t∗ )x Δ (t∗ ) =

t t∗

 Δ Δ (s)Δs rx

(10.6)

476

10 Oscillations of Third-Order Functional Dynamic Equations

t

=

  Δ γ γ1 a(s) rx Δ (s) Δs

1

t∗

(a(s)) γ   Δ γ γ1 ≥ a(t) rx Δ (t)

t

Δs 1

t∗

(a(s)) γ   Δ γ γ1 = δ1 (t, t∗ ) a(t) rx Δ (t) , t ∈ [t∗ , ∞), whereupon  Δ γ γ δ1 (t, t∗ )  a(t) rx Δ (t) , r(t) 1

x Δ (t) ≥ t ∈ [t∗ , ∞). From here, x(t) ≥ x(t∗ ) +

t t∗

 Δ γ γ δ1 (s, t∗ )  a(s) rx Δ (s) Δs r(s) 1

  Δ γ γ1 ≥ a(t) rx Δ (t)

t t∗

δ1 (s, t∗ ) Δs r(s)

  Δ γ γ1 = δ2 (t, t∗ ) a(t) rx Δ (t) , t ∈ [t∗ , ∞). This completes the proof. 1 ([t , ∞)) be such that Lemma 10.1. Let τ satisfies (A3) and x ∈ Crd 0

x(t) > 0,

x(τ (t)) > 0,

x Δ (t) > 0,

x Δ (τ (t)) > 0,

t ∈ [t0 , ∞).

Then for any n ∈ N we have Δ  (x(τ (t)))n ≥ n(x(τ (t)))n−1 x Δ (τ (t))τ Δ (t), t ∈ [t0 , ∞). Proof. We will use the principle of the mathematical induction. 1. Let n = 2. Then  Δ (x(τ (t)))2 = (x(τ (t)))Δ x(τ (σ (t))) +x(τ (t))(x(τ (t)))Δ = x Δ (τ (t))τ Δ (t)x(τ (σ (t)))

(10.7)

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

477

+x(τ (t))x Δ (τ (t))τ Δ (t) ≥ x Δ (τ (t))τ Δ (t)x(τ (t)) +x Δ (τ (t))τ Δ (t)x(τ (t)) = 2x Δ (τ (t))τ Δ (t)x(τ (t)),

t ∈ [t0 , ∞).

2. Assume that (10.7) holds for some n ∈ N. 3. We will prove that 

(x(τ (t)))n+1

Δ

≥ (n + 1)(x(τ (t)))n x Δ (τ (t))τ Δ (t),

t ∈ [t0 , ∞).

Really, we have 

(x(τ (t)))n+1

Δ

Δ  = x(τ (t))(x(τ (t)))n Δ  = x(τ (σ (t))) (x(τ (t)))n +x Δ (τ (t))τ Δ (t)(x(τ (t)))n ≥ nx(τ (t))(x(τ (t)))n−1 τ Δ (t)x Δ (τ (t)) +x Δ (τ (t))τ Δ (t)(x(τ (t)))n = (n + 1)(x(τ (t)))n τ Δ (t)x Δ (τ (t)),

t ∈ [t0 , ∞).

This completes the proof. Lemma 10.2. Let γ be a quotient of two positive odd integers, τ satisfies (A3), and 1 ([t , ∞)) be such that x ∈ Crd 0 x(t) > 0,

x(τ (t)) > 0,

x Δ (t) > 0,

x Δ (τ (t)) > 0,

t ∈ [t0 , ∞).

Then Δ  ≥ γ (x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t), (x(τ (t)))γ t ∈ [t0 , ∞). Proof. By the chain rule we have  Δ (x(τ (t)))γ =γ

1

x(τ (t)) + hμ(t)(x(τ (t)))Δ

γ −1

 dh (x(τ (t)))Δ

0

≥ γ (x(τ (t)))γ −1 (x(τ (t)))Δ = γ (x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t), This completes the proof.

t ∈ [t0 , ∞).

478

10 Oscillations of Third-Order Functional Dynamic Equations

For a given function φ such that aφ is a differentiable function and for a given positive differentiable function Φ, and for all sufficiently large T1 ∈ [t0 , ∞), and for any T ≥ T1 such that τ (T ) > T1 , we will denote 2γ τ Δ (t)δ(τ (t), T1 )Φ(t)(a(t)φ(t))σ , r(τ (t)) 1  γ +1 Δ η2 (t, T1 ) = Φ Δ (t) + τ (t)δ1 (τ (t), T1 )Φ(t) (a(t)φ(t))σ γ , r(τ (t)) 

2 γ τ Δ (t)δ(τ (t), T1 )  , (a(t)φ(t))σ ψ1 (t, T1 ) = Φ(t) p(t) − (a(t)φ(t))Δ + r(τ (t))

 1+ 1 τ Δ (t)δ1 (τ (t), T1 )  γ ψ2 (t, T1 ) = Φ(t) p(t) − (a(t)φ(t))Δ + (a(t)φ(t))σ . r(τ (t)) η1 (t, T1 ) = Φ Δ (t) +

Theorem 10.3. Assume (A1)–(A4) and (10.2) and (10.3) hold. Suppose that for sufficiently large T1 ∈ [t0 , ∞) there is a T > T1 such that τ (t) > T1 and t

lim sup t→∞

ψ1 (s, T1 ) −

T

 r(τ (s))(η1 (s, T1 ))2 Δs = ∞, 4γ Φ(s)τ Δ (s)δ(τ (s), T1 )

where φ is a nonnegative function. Then any solution of the equation (10.1) is either oscillatory or tends to zero. Proof. Assume that the equation (10.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that the solution x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

Hence from Theorem 10.1, we get that there exists a t2 ∈ [t1 , ∞) such that   Δ γ Δ a(t) rx Δ (t) < 0,

 Δ Δ rx (t) > 0,

t ∈ [t2 , ∞),

and either x Δ (t) > 0 for t ∈ [t2 , ∞) or limt→∞ x(t) = 0. Suppose that x Δ (t) > 0 on [t2 , ∞). We consider the generalized Riccati substitution & %%  Δ &γ rx Δ (t) + φ(t) w(t) = Φ(t)a(t) x(τ (t)) =

 γ  Φ(t)  Δ Δ a(t) rx (t) (x(τ (t)))γ +Φ(t)a(t)φ(t),

t ∈ [t2 , ∞).

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

Then   γ Δ  Φ(t) Δ Δ a(t) rx w (t) = (t) (x(τ (t)))γ

Δ   γ σ  Φ(t) Δ Δ a(t) rx + (t) (x(τ (t)))γ

Δ

+Φ(t)(a(t)φ(t))Δ + Φ Δ (t)(a(t)φ(t))σ   Δ γ Δ a(t) rx Δ (t) = Φ(t) (x(τ (t)))γ +Φ(t)(a(t)φ(t))Δ + Φ Δ (t)(a(t)φ(t))σ % & Φ Δ (t) Φ(t) ((x(τ (t)))γ )Δ + − (x(τ (t)))γ σ (x(τ (t)))γ (x(τ (t)))γ σ   Δ γ σ × a(t) rx Δ (t) = −Φ(t) +

f (t, x(τ (t))) + Φ(t)(a(t)φ(t))Δ (x(τ (t)))γ

Φ Δ (t) σ w (t) Φ σ (t)

((x(τ (t)))γ )Δ −Φ(t) (x(τ (t)))γ

%

% Δ &γ &σ rx Δ (t) a(t) x(τ (t))

Φ Δ (t) σ w (t) Φ σ (t) % % Δ &γ &σ rx Δ (t) ((x(τ (t)))γ )Δ , −Φ(t) a(t) (x(τ (t)))γ x(τ (t))

≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ +

t ∈ [t2 , ∞). Note that (x(τ (t)))Δ = x Δ (τ (t))τ Δ (t),

t ∈ [t2 , ∞),

and by Lemma 10.2, we have  Δ (x(τ (t)))γ ≥ γ (x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t),

479

480

10 Oscillations of Third-Order Functional Dynamic Equations

t ∈ [t2 , ∞). Therefore w Δ (t) ≤ −Φ(t)p(t) + Φ(t) (a(t)φ(t))Δ Δ (t) σ +Φ Φ σ (t) w (t)

 Δ γ σ Δ (t)) rx Δ (t) −γ Φ(t)τ Δ (t) xx(τ(τ(t)) a(t) x(τ (t)) ,

(10.8)

t ∈ [t2 , ∞). Take t3 ∈ [t2 , ∞) such that τ (t) ≥ t2 , Then, using (10.5) and using that a(t) [t3 , ∞), we obtain

t ∈ [t3 , ∞).  Δ γ rx Δ (t) is strictly decreasing on

1  Δ x Δ (τ (t)) δ1 (τ (t), t2 ) (a(τ (t))) γ r(τ (t))x Δ (τ (t)) ≥ x σ (τ (t)) r(τ (t)) x σ (τ (t))  1  Δ σ Δ γ δ1 (τ (t), t2 ) (a(t)) r(t)x (t) ≥ r(τ (t)) x σ (τ (t)) ⎛ ⎞γ −1 ⎞σ % Δ &γ ⎛ Δ rx (t) x(τ (t)) δ1 (τ (t), t2 ) ⎜ ⎟ ⎝ ⎠ = ⎝a(t) ⎠ , 1   Δ r(τ (t)) x(τ (t)) (a(t)) γ rx Δ (t)

t ∈ [t3 , ∞). Hence from (10.6), we obtain  γ −1 δ1 (τ (t), t2 ) δ2σ (τ (t), t2 ) x Δ (τ (t)) ≥ x σ (τ (t)) r(τ (t)) % % Δ &γ &σ rx Δ (t) × a(t) , x(τ (t))

t ∈ [t3 , ∞).

From the last inequality and from (10.8), we get Φ Δ (t) σ w (t) Φ σ (t) ⎛% % Δ &γ &σ ⎞2 σ rx Δ (t) γ Φ(t)τ Δ (t)δ(τ (t), t2 ) ⎝ ⎠ x (τ (t)) − a(t) r(τ (t)) x(τ (t)) x(τ (t))

w Δ (t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ +

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

Φ Δ (t) σ w (t) Φ σ (t) ⎛% % Δ &γ &σ ⎞2 rx Δ (t) γ Φ(t)τ Δ (t)δ(τ (t), t2 ) ⎝ ⎠ , − a(t) r(τ (t)) x(τ (t))

≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ +

t ∈ [t3 , ∞). Observe that ⎛%

% Δ &γ &σ ⎞2 2 σ rx Δ (t) ⎝ a(t) ⎠ = w (t) − (a(t)φ(t))σ x(τ (t)) Φ σ (t) (w σ (t))2

=

(Φ σ (t))2

−2

2 (a(t)φ(t))σ w σ (t)  + (a(t)φ(t))σ , σ Φ (t)

t ∈ [t3 , ∞). Then w Δ (t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ +

Φ Δ (t) σ w (t) Φ σ (t)

−

(w σ (t))2 γ Φ(t)τ Δ (t)δ(τ (t), t2 ) r(τ (t)) (Φ σ (t))2

+

2(a(t)φ(t))σ w σ (t) γ Φ(t)τ Δ (t)δ(τ (t), t2 ) Φ σ (t) r(τ (t))

 2 γ Φ(t)τ Δ (t)δ(τ (t), t2 ) − (a(t)φ(t))σ r(τ (t)) = −ψ1 (t, t2 ) + −

η1 (t, t2 ) σ w (t) Φ σ (t)

(w σ (t))2 γ Φ(t)τ Δ (t)δ(τ (t), t2 ) r(τ (t)) (Φ σ (t))2

γ Φ(t)τ Δ (t)δ(τ (t), t2 )  σ 2 w (t) r(τ (t)) (Φ σ (t))2  r(τ (t))Φ σ (t)η1 (t, t2 )w σ (t) − γ Φ(t)τ Δ (t)δ(τ (t), t2 )

= −ψ1 (t, t2 ) −

= −ψ1 (t, t2 ) − +

γ Φ(t)τ Δ (t)δ(τ (t), t2 ) r(τ (t)) (Φ σ (t))2



r(τ (t))Φ σ (t)η1 (t, t2 ) 2 σ w (t) − 2γ Φ(t)τ Δ (t)δ(τ (t), t2 )

γ Φ(t)τ Δ (t)δ(τ (t), t2 ) (r(τ (t)))2 (Φ σ (t))2 (η1 (t, t2 ))2   r(τ (t)) (Φ σ (t))2 4γ 2 (Φ(t))2 τ Δ (t) 2 (δ(τ (t), t2 ))2

481

482

10 Oscillations of Third-Order Functional Dynamic Equations

= −ψ1 (t, t2 ) − +

γ Φ(t)τ Δ (t)δ(τ (t), t2 ) r(τ (t)) (Φ σ (t))2

 r(τ (t))Φ σ (t)η1 (t, t2 ) 2 w σ (t) − 2γ Φ(t)τ Δ (t)δ(τ (t), t2 )

r(τ (t))(η1 (t, t2 ))2 4γ Φ(t)τ Δ (t)δ(τ (t), t2 )

≤ −ψ1 (t, t2 ) +

r(τ (t))(η1 (t, t2 ))2 , 4γ Φ(t)τ Δ (t)δ(τ (t), t2 )

t ∈ [t3 , ∞). Hence, ψ1 (t, t2 ) − t ∈ [t3 , ∞), and t ψ1 (t, t2 ) − t3

r(τ (t))(η1 (t, t2 ))2 ≤ −w Δ (t), 4γ Φ(t)τ Δ (t)δ(τ (t), t2 )

 r(τ (t))(η1 (t, t2 ))2 Δt ≤ w(t3 ) − w(t) 4γ Φ(t)τ Δ (t)δ(τ (t), t2 ) ≤ w(t3 ), t ∈ [t3 , ∞),

which is a contradiction. Therefore lim x(t) = 0.

t→∞

Otherwise, the solution x is oscillatory on [t0 , ∞). This completes the proof. Denote η+ = max{0, η},

η− = max{0, −η}.

Theorem 10.4. Assume (A1)–(A4), (10.2), and (10.3) hold. If for all sufficiently large T1 ∈ [t0 , ∞) there is a T > T1 such that τ (T ) > T1 and t

lim sup t→∞

T

%

γ +1

(r(τ (s)))γ (η2 (s, T1 ))+  γ ψ2 (s, T1 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 )

& Δs = ∞,

where φ is a nonnegative function, then every solution of the equation (10.1) is either oscillatory or tends to zero. Proof. Assume that the equation (10.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (10.1) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

483

By Theorem 10.1, it follows that there exists a t2 ∈ [t1 , ∞) such that   Δ γ Δ a(t) r(t)x Δ (t) < 0,

 Δ r(t)x Δ (t) > 0,

t ∈ [t1 , ∞),

and either x Δ (t) > 0 for t ∈ [t2 , ∞) or limt→∞ x(t) = 0. Suppose that x Δ (t) > 0 for t∈ [t2 , ∞). Let t3 ∈ [t2 , ∞) be such that τ (t) > t2 for t ∈ [t3 , ∞). Since Δ γ  Δ is strictly decreasing on [t3 , ∞), using (10.5), we have a(t) r(t)x (t)  Δ γ γ δ1 (τ (t), t2 )  a(τ (t)) r(τ (t))x Δ (τ (t)) r(τ (t)) 1  Δ σ δ1 (τ (t), t2 )  (a(t)) γ r(t)x Δ (t) ≥ , r(τ (t))

1

x Δ (τ (t)) ≥

t ∈ [t3 , ∞). Let w be the generalized Riccati substitution which is used in the proof of Theorem 10.3. Then, using (10.8), we get w Δ (t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ Δ (t) σ +Φ Φ σ (t) w (t)

 Δ γ σ Δ (t)) rx Δ (t) −γ Φ(t)τ Δ (t) xx(τ(τ(t)) a(t) x(τ (t)) Δ

(10.9)

(t) σ ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ + Φ Φ σ (t) w (t) %

 Δ 1+γ &σ rx Δ (t) 1+ 1 (τ (t),t2 ) −γ Φ(t)τ Δ (t) δ1r(τ (a(t)) γ (t)) x(τ (t))

t ∈ [t3 , ∞). Now, using the inequality (u − v)

1+ γ1

1+ γ1

≥u

+



1 1 1+ γ1 1 v v γ u, − 1+ γ γ

where u and v are constants, we get ⎛ ⎝(a(t))

1+ γ1

%

Δ &γ +1 ⎞σ 1+ 1

σ γ rx Δ (t) ⎠ = w (t) − (a(t)φ(t))σ σ x(τ (t)) Φ (t) ≥

1+ γ1

(w σ (t))

1+ γ1

+

1+ 1 1 γ (a(t)φ(t))σ γ

(Φ σ (t)) 1 

1 ((a(t)φ(t))σ ) γ w σ (t) , − 1+ γ Φ σ (t)

t ∈ [t3 , ∞).

484

10 Oscillations of Third-Order Functional Dynamic Equations

Hence from (10.9), we obtain w Δ (t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))Δ + (τ (t),t2 ) (w −γ Φ(t)τ Δ (t) δ1r(τ (t))

1 σ (t))1+ γ

Φ Δ (t) σ w (t) Φ σ (t) (10.10)

1+ γ1

(Φ σ (t))

−Φ(t)τ Δ (t)

1+ 1 δ1 (τ (t), t2 )  γ (a(t)φ(t))σ r(τ (t)) 1

+(γ + 1)Φ(t)τ Δ (t) ≤ −ψ2 (t, t2 ) +

(η2 (t, t2 ))+ σ w (t) Φ σ (t)

−γ Φ(t)τ Δ (t) where λ =

γ +1 γ .

δ1 (τ (t), t2 ) ((a(t)φ(t))σ ) γ w σ (t) r(τ (t)) Φ σ (t)

δ1 (τ (t), t2 ) (w σ (t))λ , r(τ (t)) (Φ σ (t))λ

t ∈ [t3 , ∞),

Define Aλ =

γ Φ(t)τ Δ (t)δ1 (τ (t), t2 )  σ λ w (t) , r(τ (t)) (Φ σ (t))λ 1

B λ−1 =

(r(τ (t))) λ (η2 (t, t2 ))+

 1 , λ γ Φ(t)τ Δ (t)δ1 (τ (t), t2 ) λ

t ∈ [t3 , ∞).

Note that A and B are nonnegative. Now, using the inequality (see the appendix) λAB λ−1 − Aλ ≤ (λ − 1)B λ , we get

γ Φ(t)τ Δ (t)δ1 (τ (t), t2 ) λ r(τ (t))

 λ1

γ Φ(t)τ Δ (t)δ1 (τ (t), t2 ) − λ σ

1

(r(τ (t))) λ (η2 (t, t2 ))+ w σ (t)  σ Φ (t) λ γ Φ(t)τ Δ (t)δ (τ (t), t ) λ1 1

2

 σ λ w (t)

r(τ (t)) (Φ (t)) ⎛ ⎞ λ λ−1 1 λ (η2 (t, t2 ))+ (r(τ (t))) ⎠ ≤ (λ − 1) ⎝  , 1 λ γ Φ(t)τ Δ (t)δ1 (τ (t), t2 ) λ

t ∈ [t3 , ∞),

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

485

or (η2 (t, t2 ))+ σ γ Φ(t)τ Δ (t)δ1 (τ (t), t2 )  σ λ w (t) (t) − w Φ σ (t) r(τ (t)) (Φ σ (t))λ γ +1

≤

(r(τ (t)))γ (η2 (t, t2 ))+  γ , (γ + 1)γ +1 Φ(t)τ Δ (t)δ1 (t, t2 )

t ∈ [t3 , ∞).

Hence from (10.10), we get γ +1

w Δ (t) ≤

(r(τ (t)))γ (η2 (t, t2 ))+  γ − ψ2 (t, t2 ), (γ + 1)γ +1 Φ(t)τ Δ (t)δ1 (t, t2 )

t ∈ [t3 , ∞), or γ +1

ψ2 (t, t2 ) −

(r(τ (t)))γ (η2 (t, t2 ))+  γ ≤ −w Δ (t), (γ + 1)γ +1 Φ(t)τ Δ (t)δ1 (t, t2 )

t ∈ [t3 , ∞). By the last inequality, we obtain % & γ +1 t (r(τ (s)))γ (η2 (s, t2 ))+ γ Δs ≤ w(t3 ) − w(t)  ψ2 (s, t2 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (s, t2 ) t3 ≤ w(t3 ),

t ∈ [t3 , ∞),

which is a contradiction. Therefore limt→∞ x(t) = 0. Otherwise, the solution x is oscillatory on [t0 , ∞). This completes the proof. Corollary 10.1. Suppose (A1)–(A4), (10.2), and (10.3) hold. If ∞

p(t)Δt = ∞,

t0

then every solution of the equation (10.1) is oscillatory or tends to zero. Proof. Let φ(t) = 0,

Φ(t) = 1,

t ∈ [t0 , ∞).

Then, for any T1 ∈ [t0 , ∞), we have η1 (t, T1 ) = 0, η2 (t, T1 ) = 0, ψ1 (t, T1 ) = p(t), ψ2 (t, T1 ) = p(t),

t ∈ [T1 , ∞).

486

10 Oscillations of Third-Order Functional Dynamic Equations

Hence,

ψ1 (s, T1 ) −

t

lim sup t→∞

T t

= lim sup t→∞

= lim

%

T

 r(τ (s))(η1 (s, T1 ))2 Δs 4γ Φ(s)τ Δ (s)δ(τ (s), T1 ) γ +1

(r(τ (s)))γ (η2 (s, T1 ))+  γ ψ2 (s, T1 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 )

& Δs

t

t→∞ t 1

p(s)Δs

= ∞. Then, using Theorem 10.3 or Theorem 10.4, we conclude that every solution of the equation (10.1) is oscillatory or tends to zero. This completes the proof. Corollary 10.2. Assume (A1)–(A4) and (10.2) and (10.3) hold. Suppose that for sufficiently large T1 ∈ [t0 , ∞) there is a T > T1 such that τ (t) > T1 and t

%

& 2  r(τ (s)) Φ Δ (s) p(s)Φ(s) − Δs = ∞, 4γ Φ(s)τ Δ (s)δ(τ (s), T1 )

lim sup t→∞

T

then any solution of the equation (10.1) is either oscillatory or tends to zero. Proof. Let φ(t) = 0,

t ∈ [t0 , ∞).

Then, for any T1 ∈ [t0 , ∞), we have η1 (t, T1 ) = Φ Δ (t), η2 (t, T1 ) = Φ Δ (t), ψ1 (t, T1 ) = p(t)Φ(t), ψ2 (t, T1 ) = p(t)Φ(t),

t ∈ [T1 , ∞).

Hence,

ψ1 (s, T1 ) −

 r(τ (s))(η1 (s, T1 ))2 Δs 4γ Φ(s)τ Δ (s)δ(τ (s), T1 ) t→∞ T % & 2  t r(τ (s)) Φ Δ (s) = lim sup p(s)Φ(s) − Δs 4γ Φ(s)τ Δ (s)δ(τ (s), T1 ) t→∞ T t

lim sup

= ∞.

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

487

Then, using Theorem 10.3, we conclude that every solution of the equation (10.1) is oscillatory or tends to zero. This completes the proof. Corollary 10.3. Assume (A1)–(A4), (10.2) and (10.3) hold. If for all sufficiently large T1 ∈ [t0 , ∞) there is a T > T1 such that τ (T ) > T1 and ⎛ ⎞  γ +1 t (r(τ (s)))γ Φ Δ (s) + ⎝p(s)Φ(s) −  γ ⎠ Δs = ∞, lim sup (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 ) t→∞ T then every solution of the equation (10.1) is either oscillatory or tends to zero. Proof. Let φ(t) = 0,

t ∈ [t0 , ∞).

Then, for any T1 ∈ [t0 , ∞), we have η1 (t, T1 ) = Φ Δ (t), η2 (t, T1 ) = Φ Δ (t), ψ1 (t, T1 ) = p(t)Φ(t), ψ2 (t, T1 ) = p(t)Φ(t),

t ∈ [T1 , ∞).

Hence, & γ +1 (r(τ (s)))γ (η2 (s, T1 ))+  γ Δs lim sup ψ2 (s, T1 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 ) t→∞ T ⎛ ⎞  γ +1 t (r(τ (s)))γ Φ Δ (s) + ⎝p(s)Φ(s) −  γ ⎠ Δs = lim sup (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 ) t→∞ T t

%

= ∞. Then, using Theorem 10.4, we conclude that every solution of the equation (10.1) is oscillatory or tends to zero. This completes the proof. Example 10.2. Let T be a time scale with forward jump operator σ such that σ (t) ≤ kt, t ∈ T, for some positive constant k. Let t0 ∈ T. Consider the equation   2 γ Δ β t γ x Δ (t) + γ +1 (x(τ (t)))γ = 0, t

t ∈ [t0 , ∞),

where γ ≥ 1 is a quotient of two odd positive integers and β is a positive constant

β>

γ γ +1

γ +1

kγ

2 −1

.

488

10 Oscillations of Third-Order Functional Dynamic Equations

Here a(t) = t γ , r(t) = 1, β

p(t) =

t γ +1

t ∈ T.

,

Then ∞ t0 ∞ t0

∞ Δt = Δt r(t) t0 = ∞,

Δt (a(t))

1 γ

∞

=

t0

Δt t

= ∞, ∞

∞

Δu γ +1 u s

 1 Δ β ∞ − γ ≥ Δu γ s u

p(u)Δu = β

s

β , s ∈ [t0 , ∞), γ sγ 1

1 ∞ ∞ 1 ∞ γ 1 β γ p(u)Δu Δs ≥ Δs 2 a(s) γ s t s t

1 ∞  β γ 1 Δ ≥ − Δs γ s t

1 1 β γ = , t ∈ [t0 , ∞), t γ =

and ∞ t0

1 r(t)

∞ t

1 a(s)

∞ s

1

1 γ β γ p(u)Δu ΔsΔt ≥ γ = ∞.

Take Φ(t) = t γ ,

t ∈ T.

∞ t0

1 Δt t

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

489

Then  Δ Φ Δ (t) = t γ 1

=γ

(t + hμ(t))γ −1 dh

0 1

=γ

((1 − h)t + hσ (t))γ −1 dh

0

≤ γ (σ (t))γ −1 ,

t ∈ T.

We take T ∈ [t0 , ∞) and T1 ∈ [T , ∞) sufficiently large so that T1 ≥ 1, τ Δ (t)δ1 (τ (t), T1 ) ≥ 1, t ∈ [T1 , ∞), τ (t) ≥ T1 , and t ∈ [t0 , ∞). Hence, ⎛

⎞  γ +1 (r(τ (s)))γ Φ Δ (s) ⎜ ⎟ + lim sup  γ ⎠ Δs ⎝p(s)Φ(s) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 ) t→∞ T t

⎛ ∞

≥ lim sup t→∞

T1 t

= lim sup t→∞

T1



γ k γ −1 s

γ +1 ⎞

⎜β ⎟ ⎝ − ⎠ Δs s (γ + 1)γ +1 s γ

%

β − s

& γ +1 2 −1 1 γ γ k Δs γ +1 s

= ∞.

From here and from Theorem 10.4, we conclude that every solution of the considered equation is oscillatory or tends to zero. Exercise 10.1. Let T = Z. Prove that every solution of the equation

 5 Δ 240 2 t 5 x Δ (t) + 6 (x(t − 1))5 = 0 on [1, ∞) t is oscillatory or tends to zero. Example 10.3. Consider the equation

t

γ −1

 Δ γ Δ 1 Δ γ t x (t) +

1 t 3−2γ

(x(τ (t)))γ = 0,

t ∈ [t0 , ∞), where γ and τ satisfy (A1) and (A3), respectively. Here a(t) = t γ −1 , 1

r(t) = t γ , p(t) =

1 , t 3−2γ

t ∈ T.

490

10 Oscillations of Third-Order Functional Dynamic Equations

Then ∞ t0

Δt = r(t)

∞ t0

Δt 1

tγ

= ∞, ∞ t0

Δt (a(t))

1 γ

∞

=

t0

Δt t

1− γ1

= ∞, ∞

∞

p(t)Δt =

t0

t0

Δt t 3−2γ

= ∞. Hence from Corollary 10.1, it follows that every solution of the considered equation is oscillatory or tends to zero. Exercise 10.2. Let T = 2N0 . Prove that every solution of the equation 1 3 x Δ (t) + x t

 t = 0, 2

t ∈ [2, ∞),

is oscillatory or tends to zero. Theorem 10.5. Assume (A1)–(A4), (10.2), and (10.3) hold. Suppose that there exist real-valued functions G, g ∈ Crd (D), D = {(t, s) : t ≥ s ≥ t0 } , such that G(t, t) = 0,

t ≥ t0 ,

G(t, s) > 0,

t > s ≥ t0 ,

and G has a nonpositive continuous Δ-partial derivative GΔs (t, s) with respect to the second variable, and for all sufficiently large T1 ∈ [t0 , ∞) there is a T > T1 such that τ (T ) > T1 , − GΔs (t, s) −

g(t, s)  η1 (s, T1 ) G(t, s) = σ G(t, s), σ Φ (s) Φ (s)

(10.11)

and lim sup t→∞

1 G(t, T )

∞

ψ1 (s, T1 )G(t, s) −

T

 (g− (t, s))2 r(τ (s)) Δs = ∞. 4γ Φ(s)τ Δ (s)δ(τ (s), T1 )

Then every solution of the equation (10.1) is oscillatory or tends to zero.

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

491

Proof. Let x be a nonoscillatory solution of the equation (10.1) on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (10.1) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Hence from Theorem 10.1, it follows that there is a t2 ∈ [t1 , ∞) such that   Δ γ Δ a(t) r(t)x Δ (t) < 0,

 Δ r(t)x Δ (t) > 0,

t ∈ [t1 , ∞),

and either x Δ (t) > 0 for t ∈ [t2 , ∞) or limt→∞ x(t) = 0. Assume that x Δ (t) > 0 for t ∈ [t2 , ∞). Let w be the generalized Riccati transformation which is used in the proof of Theorem 10.3. Then ψ1 (t, t2 ) ≤ −w Δ (t) + −

η1 (t, t2 ) σ w (t) Φ σ (t)

γ Φ(t)τ Δ (t)δ(τ (t), t2 )  σ 2 w (t) , r(τ (t)) (Φ σ (t))2

t ∈ [t3 , ∞).

Hence, ψ1 (s, t2 ) ≤ −w Δ (s) + −

η1 (s, t2 ) σ w (s) Φ σ (s)

γ Φ(s)τ Δ (s)δ(τ (s), t2 )  σ 2 w (s) , r(τ (s)) (Φ σ (s))2

s ∈ [t3 , ∞),

and G(t, s)ψ1 (s, t2 ) ≤ −G(t, s)w Δ (s) + G(t, s) −

γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 )  σ 2 w (s) , r(τ (s)) (Φ σ (s))2

t ≥ s ≥ t3 . Then, using (10.11), we get t t3

G(t, s)ψ1 (s, t2 )Δs ≤ − +

t t3

η1 (s, t2 ) σ w (s) Φ σ (s)

t

G(t, s)wΔ (s)Δs

t3

η (s, t ) G(t, s) 1 σ 2 wσ (s)Δs Φ (s)

s ∈ [t3 , ∞),

492

10 Oscillations of Third-Order Functional Dynamic Equations

−

t γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t )  2 2 wσ (s) Δs 2 σ r(τ (s)) (Φ (s))

t3

= G(t, t3 )w(t3 ) +

t

GΔs (t, s)wσ (s)Δs

t3

t

+

η (s, t ) G(t, s) 1 σ 2 wσ (s)Δs Φ (s) t3

−

t γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t )  2 2 wσ (s) Δs r(τ (s)) (Φ σ (s))2 t 3

= G(t, t3 )w(t3 ) √ t g(t, s) G(t, s) σ w (s) − + Φ σ (s) t3 −

 γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 )  σ 2 Δs w (s) r(τ (s)) (Φ σ (s))2

≤ G(t, t3 )w(t3 ) √  Δ t g (t, s) G(t, s) − σ (s) − γ G(t, s)Φ(s)τ (s)δ(τ (s), t2 ) w σ (s)2 Δs w + Φ σ (s) r(τ (s)) (Φ σ (s))2 t3 = G(t, t3 )w(t3 )

t γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 )  σ 2 − w (s) + r(τ (s)) (Φ σ (s))2 t3 √  r(τ (s))g− (t, s) G(t, s)Φ σ (s) σ − (s) Δs w γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 ) = G(t, t3 )w(t3 )  2 t γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t ) (r(τ (s)))2 (g− (t, s))2 G(t, s) Φ σ (s) 2 +  2 r(τ (s)) (Φ σ (s))2 t3 4γ 2 (G(t, s))2 (Φ(s))2 τ Δ (s) (δ(τ (s), t2 ))2 −

γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 ) r(τ (s)) (Φ σ (s))2

√ 2  G(t, s)Φ σ (s) Δs 2γ G(t, s)Φ(s)τ Δ (s)δ(τ (s), t2 )

r(τ (s))g− (t, s) wσ (s) −

≤ G(t, t3 )w(t3 ) +

r(τ (s))(g− (t, s))2 Δs, Δ t3 4γ Φ(s)τ (s)δ(τ (s), t2 ) t

t ∈ [t3 , ∞),

whereupon t t3

G(t, s) ψ1 (s, t2 ) −

 r(τ (s))(g− (t, s))2 Δs ≤ G(t, t3 )w(t3 ), 4γ Φ(s)τ Δ (s)δ(τ (s), t2 )

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

493

t ∈ [t3 , ∞), or 1 G(t, t3 )

t

G(t, s) ψ1 (s, t2 ) −

t3

 r(τ (s))(g− (t, s))2 Δs ≤ w(t3 ) 4γ Φ(s)τ Δ (s)δ(τ (s), t2 ) ≤ |w(t3 )|,

t ∈ [t3 , ∞), which is a contradiction. Therefore limt→∞ x(t) = 0. Otherwise, the solution x is oscillatory on [t0 , ∞). This completes the proof. Example 10.4. Suppose that (A1)–(A4), (10.2), and (10.3) hold. Take G(t, s) = 1,

t > s ≥ t0 ,

G(t, t) = 0,

t ∈ [t0 , ∞),

g(t, s) = 1,

t > s ≥ t0 ,

g(t, t) = 0,

t ∈ [t0 , ∞),

φ(t) = 0, Φ(t) = 1,

t ∈ T.

Then η1 (t, s) = 0, η2 (t, s) = 0, ψ1 (t, s) = p(t), ψ2 (t, s) = p(t), g− (t, s) = 0,

t ≥ s ≥ t0 .

Hence, lim sup t→∞

1 G(t, T )

ψ1 (s, T1 )G(t, s)− T

 (g− (t, s))2 r(τ (s)) Δs 4γ Φ(s)τ Δ (s)δ(τ (s), T1 )

t

= lim sup t→∞

∞

p(y)Δy. T

Example 10.5. Let φ(t) = 0, Φ(t) = 1, G(t, s) = t − s, g(t, s) = √

1 , t −s

t ≥ s ≥ t0 , t > s ≥ t0 .

494

10 Oscillations of Third-Order Functional Dynamic Equations

Then η1 (t, s) = 0, η2 (t, s) = 0, ψ1 (t, s) = p(t), ψ2 (t, s) = p(t), GΔs (t, s) = −1, g− (t, s) = 0,

t ≥ s ≥ t0 ,

and −GΔs (t, s) −

η1 (s, T1 ) G(t, s) = 1 Φ σ (s) 1 √ = √ t −s t −s g(t, s)  G(t, s), = σ Φ (s)

t > s ≥ t0 .

Hence, 1 lim sup G(t, T) t→∞ = lim sup t→∞

1 t −T

∞

 (g− (t, s))2 r(τ (s)) ψ1 (s, T1 )G(t, s) − Δs 4γ Φ(s)τ Δ (s)δ(τ (s), T1 )

T t

p(s)(t − s)Δs.

T

Exercise 10.3. Let T = Z. Write an analogue of Theorem 10.5. Theorem 10.6. Assume (A1)–(A4), (10.2), and (10.3) hold. Suppose that there exist functions H, h ∈ Crd (D) such that H (t, t) = 0,

t ≥ t0 ,

H (t, s) > 0,

t > s ≥ t0 ,

and H has a nonpositive continuous Δ-partial derivative H Δs with respect to the second variable and for all sufficiently large T1 ∈ [t0 , ∞), there is a T > T1 such that τ (T ) > T1 , −H Δs (t, s) −

γ h(t, s) η2 (s, T1 ) H (t, s) = σ (H (t, s)) γ +1 , σ Φ (s) Φ (s)

and lim sup t→∞

1 H (t, T )

t T

% ψ2 (s, T1 )H (t, s) −

(h− (t, s))γ +1 (r(τ (s)))γ γ  (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 )

&

Then every solution of the equation (10.1) is oscillatory or tends to zero.

Δs = ∞.

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

495

Proof. Let x be a nonoscillatory solution of the equation (10.1) on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (10.1) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

Hence from Theorem 10.1, it follows that there is a t2 ∈ [t1 , ∞) such that   Δ γ Δ a(t) r(t)x Δ (t) < 0,

 Δ r(t)x Δ (t) > 0,

t ∈ [t1 , ∞),

and either x Δ (t) > 0 for t ∈ [t2 , ∞) or limt→∞ x(t) = 0. Assume that x Δ (t) > 0 for t ∈ [t2 , ∞). Let w be the generalized Riccati transformation which is used in the proof of Theorem 10.3. By (10.10), for λ = γ γ+1 , we get ψ2 (t, t2 ) ≤ −w Δ (t) + −

η2 (t, t2 ) σ w (t) Φ σ (t)

γ Φ(t)τ Δ (t)δ1 (τ (t), t2 )  σ λ w (t) , r(τ (t)) (Φ σ (t))λ

t ∈ [t3 , ∞),

and ψ2 (s, t2 ) ≤ −w Δ (s) + −

η2 (s, t2 ) σ w (s) Φ σ (s)

γ Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ w (s) , r(τ (s)) (Φ σ (s))λ

s ∈ [t3 , ∞),

and H (t, s)ψ2 (s, t2 ) ≤ −H (t, s)w Δ (s) + H (t, s) −

η2 (s, t2 ) σ w (s) Φ σ (s)

γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ w (s) , r(τ (s)) (Φ σ (s))λ

s ∈ [t3 , ∞).

Then t

H (t, s)ψ2 (s, t2 )Δs ≤ −

t3

t

H (t, s)w Δ (s)Δs +

t3

−

t t3

t

H (t, s) t3

η2 (s, t2 ) σ w (s)Δs Φ σ (s)

γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ w (s) Δs r(τ (s)) (Φ σ (s))λ

= H (t, t3 )w(t3 ) +

t t3

H Δs (t, s)w σ (s)Δs

496

10 Oscillations of Third-Order Functional Dynamic Equations t

+

H (t, s) t3 t

−

t3

η2 (s, t2 ) σ w (s)Δs Φ σ (s)

γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ w (s) Δs r(τ (s)) (Φ σ (s))λ

= H (t, t3 )w(t3 ) 1 t h(t, s)(H (t, s)) λ σ − w (s) + Φ σ (s) t3

 γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ Δs, w (s) − r(τ (s)) (Φ σ (s))λ

t ∈ [t3 , ∞). Define γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 ) (w σ (s))λ

Aλ =

r(τ (s)) (Φ σ (s))λ

,

1

B

λ−1

h− (t, s)(r(τ (s))) λ

=

s ∈ [t3 , ∞).

 1 , λ γ Φ(s)τ Δ (s)δ1 (τ (s), t2 ) λ

Note that A and B are positive. Now, using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ , we get λ

 1 γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 ) λ w σ (s) 1 λ

(r(τ (s))) Φ σ (s) −

1

h− (t, s)(r(τ (s))) λ  1 λ γ Φ(s)τ Δ (s)δ1 (τ (s), t2 ) λ

γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 ) (w σ (s))λ r(τ (s)) (Φ σ (s))λ λ

≤ (λ − 1)

1

(h− (t, s)) λ−1 (r(τ (s))) λ−1  1 , λ  λ−1 γ Φ(s)τ Δ (s)δ1 (τ (s), t2 ) λ−1 λ

t ≥ s ≥ t3 , or 1

h− (t, s)(H (t, s)) λ σ γ H (t, s)Φ(s)τ Δ (s)δ1 (τ (s), t2 )  σ λ w (s)− w (s) σ Φ (s) r(τ (s)) (Φ σ (s))λ ≤

(h− (t, s))γ +1 (r(τ (s)))γ γ ,  (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), t2 )

t ≥ s ≥ t3 .

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

497

Then t

H (t, s)ψ2 (s, t2 )Δs ≤ H (t, t3 )w(t3 )

t3 t

+

t3

(h− (t, s))γ +1 (r(τ (s)))γ  γ Δs, (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), t2 )

Hence, % t

t3

(h− (t, s))γ +1 (r(τ (s)))γ  γ H (t, s)ψ2 (s, t2 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), t2 )

t ∈ [t3 , ∞).

& Δs ≤ H (t, t3 )w(t3 ),

t ∈ [t3 , ∞), or 1 H (t, t3 )

t t3

%

(h− (t, s))γ +1 (r(τ (s)))γ  γ H (t, s)ψ2 (s, t2 ) − (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), t2 )

& Δs ≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. Therefore limt→∞ x(t) = 0. Otherwise, the solution x is oscillatory. This completes the proof. Example 10.6. Suppose that (A1)–(A4), (10.2), and (10.3) hold. Take H (t, s) = 1,

t > s ≥ t0 ,

H (t, t) = 0,

t ∈ [t0 , ∞),

h(t, s) = 1,

t > s ≥ t0 ,

h(t, t) = 0,

t ∈ [t0 , ∞),

φ(t) = 0, Φ(t) = 1,

t ∈ T.

Then η1 (t, s) = 0, η2 (t, s) = 0, ψ1 (t, s) = p(t), ψ2 (t, s) = p(t), h− (t, s) = 0,

t ≥ s ≥ t0 .

498

10 Oscillations of Third-Order Functional Dynamic Equations

Hence, lim sup t→∞

1 H (t, T )

% ψ2 (s, T1 )H (t, s) −

T

(h− (t, s))γ +1 (r(τ (s)))γ  γ (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 )

t

= lim sup t→∞

t

p(y)Δy. T

Example 10.7. Let φ(t) = 0, Φ(t) = t, H (t, s) = t − s, h(t, s) =

σ (s) + s − t γ

(t − s) γ +1

h(t, t) = 0,

,

t > s ≥ t0 ,

t ∈ [t0 , ∞).

Then Φ Δ (t) = 1, η1 (t, T1 ) = 1, η2 (t, T1 ) = 1, ψ1 (t, T1 ) = tp(t), ψ2 (t, T1 ) = tp(t),

t ∈ [T1 , ∞).

Next, H Δs (t, s) = −1, −H Δs (t, s) −

t −s η2 (s, T1 ) H (t, s) = 1 − σ Φ (s) σ (s) =

σ (s) + s − t , σ (s)

γ γ σ (s) + s − t h(t, s) γ +1 = γ +1 (H (t, s)) (t − s) γ Φ σ (s) σ (s)(t − s) γ +1

=

σ (s) + s − t , σ (s)

H (t, t) = 0,

t ∈ [t0 , ∞).

t > s ≥ t0 ,

& Δs

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

499

Hence, % & t 1 (h− (t, s))γ +1 (r(τ (s)))γ ψ2 (s, T1 )H (t, s) −  γ Δs (γ + 1)γ +1 Φ(s)τ Δ (s)δ1 (τ (s), T1 ) t→∞ H (t, T ) T ⎛ ⎞ % &γ +1

lim sup

1 = lim sup t→∞ t − T

σ (s)+s−t (r(τ (s)))γ ⎜ γ (t−s) γ +1 − ⎜  γ ⎜s(t − s)p(s) − (γ + 1)γ +1 sτ Δ (s)δ1 (τ (s), T1 ) T ⎜ ⎝ t⎜

⎟ ⎟ ⎟ ⎟ Δs. ⎟ ⎠

Exercise 10.4. Let T = Z. Write an analogue of Theorem 10.6. Now we will consider the equation (10.1) under the assumptions (A2), (A4), and (B1) (B3)

γ > 0 is a quotient of odd positive integers, τ : T → T satisfies τ (t) ≤ t, t ∈ T, limt→∞ τ (t) = ∞.

Let b(t) =

t , σ (t)

t ∈ T,

and

β(t) =

b(t) if 0 < γ ≤ 1 (b(t))γ if γ > 1, t ∈ T.

Theorem 10.7. Assume (A2), (A4), (B1), and (B3). Suppose that x is a positive solution of the equation (10.1) such that  Δ r(t)x Δ (t) > 0,

x Δ (t) > 0,

t ∈ [t1 , ∞),

for some t1 ∈ [t0 , ∞). Furthermore, r Δ (t) ≤ 0, t ∈ [t1 , ∞), ∞

p(t)(τ (t))γ Δt = ∞.

t0

Then there exists a t2 ∈ [t1 , ∞) sufficiently large, so that x(t) > tx Δ (t), x(t) t

is strictly decreasing on [t2 , ∞).

t ∈ [t2 , ∞),

(10.12)

500

10 Oscillations of Third-Order Functional Dynamic Equations

Proof. Note that  Δ 2 r(t)x Δ (t) = r Δ (t)x Δ (t) + r σ (t)x Δ (t) t ∈ [t1 , ∞).

> 0, Therefore 2

t ∈ [t1 , ∞).

x Δ (t) > 0, Let

t ∈ [t1 , ∞).

v(t) = x(t) − tx Δ (t), Hence,

2

v Δ (t) = x Δ (t) − σ (t)x Δ (t) − x Δ (t) 2

= −σ (t)x Δ (t) t ∈ [t1 , ∞).

< 0,

Assume that for any t2 ∈ [t1 , ∞) we have v(t) < 0. Then 

x(t) t

Δ

=

tx Δ (t)−x(t) tσ (t) − tσv(t) (t)

= > 0,

(10.13)

t ∈ [t2 , ∞).

Therefore x(t) t is strictly decreasing on [t1 , ∞). Take t3 ∈ [t2 , ∞) so that τ (t) ≥ τ (t2 ) for t ∈ [t3 , ∞). Let d=

x(τ (t2 )) . τ (t2 )

Then x(τ (t2 )) x(τ (t)) ≥ τ (t) τ (t2 ) = d,

t ∈ [t3 , ∞).

By the equation (10.1), we get   Δ γ Δ a(t) r(t)x Δ (t) = −f (t, x(τ (t))) ≤ −p(t)(x(τ (t)))γ < 0,

t ∈ [t3 , ∞).

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

501

Then a(t)

  Δ γ Δ γ − a(t3 ) r(t3 )x Δ (t3 ) ≤− r(t)x Δ (t)

t

p(s)(x(τ (s)))γ Δs,

t3

or a(t3 )

  Δ γ Δ γ r(t3 )x Δ (t3 ) ≥ a(t) r(t)x Δ (t) t

+

p(s)(x(τ (s)))γ Δs

t3 t

≥

p(s)(x(τ (s)))γ Δs

t3

≥ dγ

t

p(s)(τ (s))γ Δs

t3

→ ∞,

as

t → ∞.

This is a contradiction. Therefore there exists a t2 ∈ [t1 , ∞) so that v(t) > 0, t ∈ [t2 , ∞). Hence from (10.13), we conclude that

i.e.,

x(t) t

x(t) t

Δ < 0,

t ∈ [t2 , ∞),

is strictly decreasing on [t2 , ∞). This completes the proof.

Theorem 10.8. Assume (A2), (A4), (B1), (B3), (10.2), (10.3) and (10.12) hold. 1 ([t , ∞)), for all sufficiently Suppose that there exists a positive function α ∈ Crd 0 large T1 ∈ [t0 , ∞), there is a T > T1 such that ⎛ t

lim sup t→∞

T

⎝α (s)p(s) τ (s) σ (s)

γ

σ

−

 γ +1 (r(s))γ α Δ (s) +

⎞

(γ + 1)γ +1 (β(s)α σ (s)δ1 (s, T1 ))γ

⎠ Δs = ∞.

Then every solution of the equation (10.1) is oscillatory or tends to zero. Proof. Let x be a nonoscillatory solution of the equation (10.1) on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (10.1) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

Hence and from Theorem 10.1, it follows that there is a t2 ∈ [t1 , ∞) such that   Δ γ Δ a(t) r(t)x Δ (t) < 0,

 Δ r(t)x Δ (t) > 0,

t ∈ [t2 , ∞),

502

10 Oscillations of Third-Order Functional Dynamic Equations

and either x Δ (t) > 0 for t ∈ [t2 , ∞) or limt→∞ x(t) = 0. Assume that x Δ (t) > 0 for t ∈ [t2 , ∞). Also, there exists a t3 ∈ [t2 , ∞) so that x(t) t is strictly decreasing on [t3 , ∞). Let w be the generalized Riccati transformation

w(t) = α(t)

a(t)

 Δ γ r(t)x Δ (t) ,

t ∈ [t3 , ∞).

 Δ  γ −1 Δ (x(t))γ ≥ γ x σ (t) x (t),

t ∈ [t3 , ∞),

(x(t))γ

Then, if 0 < γ ≤ 1, using the chain rule, we have

and w Δ (t) = α Δ (t)

 Δ γ a(t) r(t)x Δ (t)

%

+α σ (t)

(x(t))γ  Δ γ a(t) r(t)x Δ (t)

&Δ

(x(t))γ Δ γ a(t) r(t)x Δ (t) 

= α Δ (t)

+α σ (t)

−α σ (t) ≤ −α σ (t)

−α σ (t)

a(t)

γ

 (x(t))  Δ γ Δ a(t) r(t)x Δ (t) (x σ (t))γ

 Δ γ r(t)x Δ (t) ((x(t))γ )Δ (x(t))γ (x σ (t))γ

f (t, x(τ (t))) α Δ (t) w(t) + α(t) (x σ (t))γ  Δ γ a(t) r(t)x Δ (t) ((x(t))γ )Δ (x(t))γ (x σ (t))γ (x(τ (t)))γ α Δ (t) w(t) + γ α(t) (x σ (t))  Δ γ a(t) r(t)x Δ (t) ((x(t))γ )Δ

≤ −α σ (t)p(t)

−α σ (t)

(x(t))γ (x σ (t))γ  τ (t) γ α Δ (t) σ w(t) ≤ −α (t)p(t) + σ (t) α(t)  Δ γ a(t) r(t)x Δ (t) ((x(t))γ )Δ σ −α (t) (x(t))γ (x σ (t))γ

(10.14)

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

 τ (t) γ α Δ (t) ≤ −α (t)p(t) + w(t) σ (t) α(t)  Δ γ a(t) r(t)x Δ (t) x(t)x Δ (t) σ −γ α (t) (x(t))γ +1 x σ (t) 

τ (t) γ α Δ (t) σ w(t) ≤ −α (t)p(t) + σ (t) α(t) 1  Δ w(t)δ1 (t, t2 )(a(t)) γ r(t)x Δ (t) σ −γ α (t) α(t)r(t)x σ (t) γ

τ (t) α Δ (t) w(t) = −α σ (t)p(t) + σ (t) α(t)

σ

1

−γ α σ (t)

w(t)δ1 (t, t2 )(w(t)) γ x(t) 1

(α(t)) γ r(t)x σ (t)α(t) 

τ (t) γ α Δ (t) ≤ −α σ (t)p(t) + w(t) σ (t) α(t) γ +1

t (w(t)) γ δ1 (t, t2 ) , −γ α (t) σ (t) (α(t)) γ γ+1 r(t) σ

t ∈ [t3 , ∞).

If γ > 1, using the chain rule, we have  Δ (x(t))γ ≥ γ (x(t))γ −1 x Δ (t),

t ∈ [t3 , ∞).

Hence, using (10.14), we obtain  τ (t) γ α Δ (t) w (t) ≤ −α (t)p(t) w(t) + σ (t) α(t)  Δ γ a(t) r(t)x Δ (t) ((x(t))γ )Δ σ −α (t) (x(t))γ (x σ (t))γ 

τ (t) γ α Δ (t) w(t) ≤ −α σ (t)p(t) + σ (t) α(t)  Δ γ Δ a(t) r(t)x Δ (t) x (t) −α σ (t)γ γ σ x(t) (x (t)) 

τ (t) γ α Δ (t) w(t) = −α σ (t)p(t) + σ (t) α(t)

Δ

σ

503

504

10 Oscillations of Third-Order Functional Dynamic Equations

w(t)(x(t))γ x Δ (t) x(t)α(t) (x σ (t))γ 

τ (t) γ α Δ (t) σ w(t) ≤ −α (t)p(t) + σ (t) α(t) −γ α σ (t)

1  Δ w(t)(x(t))γ δ1 (t, t2 )(a(t)) γ r(t)x Δ (t) −γ α (t) x(t)r(t)α(t) (x σ (t))γ γ

τ (t) α Δ (t) w(t) = −α σ (t)p(t) + σ (t) α(t)

σ

1

−γ α (t) σ

w(t)(x(t))γ (w(t)) γ δ1 (t, t2 ) 1

r(t)α(t)(α(t)) γ (x σ (t))γ 

τ (t) γ α Δ (t) σ w(t) ≤ −α (t)p(t) + σ (t) α(t)

t −γ α (t) σ (t)

γ

σ

(w(t))

γ +1 γ

δ1 (t, t2 )

r(t)(α(t))

γ +1 γ

,

t ∈ [t3 , ∞).

Therefore

τ (t) w (t) ≤ −α (t)p(t) σ (t) Δ

σ

−γ α σ (t)β(t) where λ =

γ +1 γ .



γ +

α Δ (t)



α(t)

δ1 (t, t2 ) (w(t))λ , r(t) (α(t))λ

+

w(t)

t ∈ [t3 , ∞),

Define Aλ = γ α σ (t)β(t) B λ−1 =

δ1 (t, t2 ) (w(t))λ , r(t) (α(t))λ

 Δ  1 α (t) + (r(t)) λ 1

λ (γβ(t)α σ (t)δ1 (t, t2 )) λ

,

t ∈ [t3 , ∞).

Note that A and B are positive on [t3 , ∞). Now, using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ ,

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

505

we get  Δ  1 1 α (t) + (r(t)) λ  σ 1 1 (δ1 (t, t2 )) λ w(t) λγ α (t) λ (β(t)) λ 1 1 (r(t)) λ α(t) λ (γβ(t)α σ (t)δ1 (t, t2 )) λ 1 λ

δ1 (t, t2 ) (w(t))λ r(t) (α(t))λ  Δ  λ 1 α (t) +λ−1 (r(t)) λ−1

−γ α σ (t)β(t) ≤ (λ − 1)

1

λ

t ∈ [t3 , ∞),

,

λ λ−1 (γβ(t)α σ (t)δ1 (t, t2 )) λ−1

or  Δ  α (t) + α(t) ≤

w(t) − γ α σ (t)β(t)  γ +1 (r(t))γ α Δ (t) +

δ1 (t, t2 ) (w(t))λ r(t) (α(t))λ

(γ + 1)γ +1 (β(t)α σ (t)δ1 (t, t2 ))γ

,

t ∈ [t3 , ∞).

Consequently

τ (t) w (t) ≤ −α (t)p(t) σ (t) Δ

σ

γ +

 γ +1 (r(t))γ α Δ (t) +

(γ + 1)γ +1 (β(t)α σ (t)δ1 (t, t2 ))γ

,

t ∈ [t3 , ∞), or

τ (t) α (t)p(t) σ (t)

γ −

σ

 γ +1 (r(t))γ α Δ (t) +

(γ + 1)γ +1 (β(t)α σ (t)δ1 (t, t2 ))γ

≤ −w Δ (t),

t ∈ [t3 , ∞). Hence, ⎛ t t3

⎝α σ (s)p(s) τ (s) σ (s)

γ −

 γ +1 (r(s))γ α Δ (s) +

(γ + 1)γ +1 (β(s)α σ (s)δ1 (s, t2 ))γ

⎞ ⎠ Δs ≤ −w(t) + w(t3 ) ≤ w(t3 ),

which is a contradiction. Therefore any solution of the equation (10.1) is oscillatory or tends to zero. This completes the proof. Example 10.8. Take α(t) = 1,

t ∈ T.

506

10 Oscillations of Third-Order Functional Dynamic Equations

Then α Δ (t) = 0,

t ∈ T.

Hence, ⎛ t

lim sup t→∞

T

⎜ σ ⎝α (s)p(s)

t

= lim

p(s)

t→∞ T

τ (s) σ (s)

γ

⎞

 γ +1 (r(s))γ α Δ (s) −

+

(γ + 1)γ +1 (β(s)α σ (s)δ1 (s, T1 ))γ

⎟ ⎠ Δs

 τ (s) γ Δs. σ (s)

Example 10.9. Take α(t) = t,

t ∈ T.

Then α Δ (t) = 1,

t ∈ T,

and ⎛ t

lim sup t→∞

T

⎝α σ (s)p(s) τ (s) σ (s) t

= lim sup t→∞

T

γ −

 γ +1 (r(s))γ α Δ (s) +

(γ + 1)γ +1 (β(s)α σ (s)δ1 (s, T1 ))γ

⎞ ⎠ Δs

  τ (s) γ (r(s))γ σ (s)p(s) Δs. − σ (s) (γ + 1)γ +1 (β(s)σ (s)δ1 (s, T1 ))γ

Example 10.10. Consider the equation 3

x Δ (t) +

β x(τ (t)) = 0, tτ (t)

t ∈ [t0 , ∞),

where β is a positive constant. Here a(t) = 1, r(t) = 1, p(t) =

β , tτ (t)

γ = 1.

t ∈ [t0 , ∞),

10.1 Oscillations of Third-Order Nonlinear Delay Dynamic Equations

507

Then ∞ t0

Δt 1

∞ t0 ∞ t0

1 r(t)

∞ t

1 a(s)

∞

= ∞,

(a(t)) γ

Δt = ∞, r(t)

1 γ p(u)Δu ΔsΔt =

s

∞ t0

≥

t ∞ t

∞ t0

∞ s

∞

∞ t0

=

t

∞ t0

≥

∞

∞ t

β ΔuΔsΔt uτ (u)

∞

β ΔuΔsΔt u2 s ∞ β Δ − ΔuΔsΔt u s β ΔsΔt s

= ∞. Take α = t. Then ⎛ t

lim sup t→∞

T

⎝α σ (s)p(s) τ (s) σ (s) t

= lim sup t→∞

T

γ −

 γ +1 (r(s))γ α Δ (s) +

(γ + 1)γ +1 (β(s)α σ (s)δ1 (s, T1 ))γ

⎞ ⎠ Δs

 β 1 − Δs s 4s(s − T1 )

= ∞. Hence from Theorem 10.8, we conclude that any solution of the considered equation is oscillatory on [t0 , ∞) or tends to zero. This completes the proof. Exercise 10.5. Prove that any solution of the equation ⎛% ⎞

Δ &3 Δ 2 1 ⎝ ⎠ + σ (t) (x(τ (t)))2 = 0, x Δ (t) t (τ (t))2 is oscillatory or tends to zero.

t ∈ [t0 , ∞),

508

10 Oscillations of Third-Order Functional Dynamic Equations

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations In this section we will investigate the equation 

 2 γ Δ r(t) (x(t) + p(t)x(τ0 (t)))Δ +q1 (t)(x(τ1 (t)))α + q2 (t)(x(τ2 (t)))β = 0,

(10.15)

t ∈ [t0 , ∞),

where (C1) (C2)

γ , α, β are quotients of positive odd integers, 0 < α < γ < β, r ∈ Crd (T) is a positive function such that ∞ t0

(C3)

1 r(t)

1 γ

Δt = ∞,

q1 , q2 , p ∈ Crd (T) are positive functions and 0 ≤ p(t) ≤ P < 1,

(C4)

t0 ∈ T,

t ∈ T,

where P is a constant, τi : T → T satisfy τi (t) ≤ t, t ∈ T, limt→∞ τi (t) = ∞, i ∈ {0, 1, 2}, and there exists a function τ : T → T so that τ (t) ≤ τ1 (t),

τ (t) ≤ τ2 (t),

t ∈ T,

lim τ (t) = ∞.

t→∞

Let z(t) = x(t) + p(t)x(τ0 (t)),

t ∈ [t0 , ∞).

Then the equation (10.15) can be rewritten in the following form.   2 γ Δ r(t) zΔ (t) + q1 (t) (x(τ1 (t)))α + q2 (t) (x(τ2 (t)))β = 0,

t ∈ [t0 , ∞).

Theorem 10.9. Assume (C1)–(C4). Let x be a solution of the equation (10.15) such that x(t) > 0,

x(τi (t)) > 0,

i ∈ {0, 1, 2},

t ∈ [t0 , ∞).

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

509

Then there exists a t1 ∈ [t0 , ∞) such that z has one of the following two properties. 2

1. z(t) > 0, zΔ (t) > 0, zΔ (t) > 0, t ∈ [t1 , ∞), 2 2. z(t) > 0, zΔ (t) < 0, zΔ (t) > 0, t ∈ [t1 , ∞). Proof. Note that t ∈ [t0 , ∞).

z(t) > x(t) > 0, Then

  2 γ Δ r(t) zΔ (t) = −q1 (t) (x(τ1 (t)))α − q2 (t) (x(τ2 (t)))β < 0,

t ∈ [t0 , ∞).

Therefore  2 γ r(t) zΔ (t) is a strictly decreasing function on [t0 , ∞). Consequently it has eventually one sign 2 on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that zΔ (t) has one sign on [t1 , ∞). 2 Assume that zΔ (t) < 0 for t ∈ [t1 , ∞). Therefore there exists a positive constant M such that  2 γ r(t) zΔ (t) ≤ −M < 0, t ∈ [t1 , ∞). Hence,  2 γ M zΔ (t) ≤ − , r(t)

t ∈ [t1 , ∞),

and 1

z

Δ2

(t) ≤ −

Mγ

1

,

t ∈ [t1 , ∞),

(r(t)) γ and

1

zΔ (t) ≤ zΔ (t1 ) − M γ

t t1

→ −∞ as

1 1

(r(s)) γ t → ∞.

Δs

510

10 Oscillations of Third-Order Functional Dynamic Equations

Therefore there exists a t2 ∈ [t1 , ∞) such that t ∈ [t2 , ∞).

zΔ (t) < 0, Since zΔ (t) < 0,

2

t ∈ [t2 , ∞),

zΔ (t) < 0,

there exists a t3 ∈ [t2 , ∞) such that t ∈ [t3 , ∞).

z(t) < 0, This is a contradiction and consequently 2

t ∈ [t1 , ∞).

zΔ (t) > 0, This completes the proof.

Theorem 10.10. Assume (C1)–(C4). Let x be a positive solution of the equation (10.15) such that z(t) > 0,

zΔ (t) < 0,

∞

∞

2

zΔ (t) > 0,

t ∈ [t0 , ∞).

If ∞ t0

v

1 r(u)

1 (q1 (s) + q2 (s)) Δs

γ

ΔuΔv = ∞,

u

then lim x(t) = lim z(t) = 0.

t→∞

t→∞

Proof. Because zΔ (t) < 0,

t ∈ [t0 , ∞),

there is a nonnegative constant l such that lim z(t) = l.

t→∞

Assume that l > 0. Take  > 0 such that 
l − p(t)z(τ0 (t)) > l − p(t)(l + ) ≥ l − P (l + ),

t ∈ [t1 , ∞).

Let k=

l − P (l + ) . l+

Then x(t) > k(l + ) > kz(t),

t ∈ [t1 , ∞).

Let t2 ∈ [t1 , ∞) be such that τ1 (t) ≥ t1 ,

τ2 (t) ≥ t1 ,

t ∈ [t2 , ∞).

Then   2 γ Δ r(t) zΔ (t) = −q1 (t)(x(τ1 (t)))α − q2 (t)(x(τ2 (t)))β < −k α q1 (t)(z(τ1 (t)))α − q2 (t)k β (z(τ2 (t)))β   k(z(t))α = min k α (z(t))α , k β (z(t))β < −kq1 (t)(z(t))α − kq2 (t)(z(t))β ≤ −kq1 (t)(z(t))α − kq2 (t)(z(t))α = −k(z(t))α (q1 (t) + q2 (t)),

t ∈ [t2 , ∞).

Now we integrate the last inequality from t to ∞, t ∈ [t2 , ∞), and using (C2), we get  2 γ −r(t) zΔ (t) ≤ −k

∞ t

(z(s))α (q1 (s) + q2 (s))Δs,

t ∈ [t2 , ∞),

512

10 Oscillations of Third-Order Functional Dynamic Equations

or  2 γ r(t) zΔ (t) ≥ k

∞

(z(s))α (q1 (s) + q2 (s))Δs,

t ∈ [t2 , ∞),

(z(s))α (q1 (s) + q2 (s))Δs,

t ∈ [t2 , ∞),

t

whereupon  2 γ k zΔ (t) ≥ r(t)

∞ t

and

1

kγ

2

zΔ (t) ≥

(r(t)) 1 γ

k l

≥

1 γ

1 γ

(z(s))α (q1 (s) + q2 (s))Δs

t

α γ

(r(t))

∞

1 γ

∞

1 (q1 (s) + q2 (s))Δs

γ

,

t ∈ [t2 , ∞).

t

We integrate the last inequality from t to ∞ and we obtain 1 γ

−z (t) ≥ k l Δ

∞

α γ

1 1

t

∞

1 γ (q1 (u) + q2 (u))Δu Δs,

s

(r(s)) γ

t ∈ [t2 , ∞). We integrate the last inequality from t2 to ∞ and we obtain 1

α

z(t2 ) ≥ k γ l γ

∞ t2

∞ t

1 1

(r(s)) γ

∞

1 γ (q1 (u) + q2 (u))Δu ΔsΔt

s

= ∞. This is a contradiction. Hence, l = 0. This completes the proof. For t∗ ∈ [t0 , ∞), we denote R(t, t∗ ) =

t t∗

1 r(s)

1 γ

Δs,

t ∈ [t∗ , ∞).

Theorem 10.11. Assume (C1)–(C4). Let x be a positive solution of the equation (10.15) such that 2

zΔ (t) > 0,

zΔ (t) > 0,

t ∈ [t∗ , ∞),

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

513

for some t∗ ∈ [t0 , ∞). Then 1

2

zΔ (t) ≥ R(t, t∗ )(r(t)) γ zΔ (t),

t ∈ [t∗ , ∞).

Proof. Note that there exists a t∗ ∈ [t0 , ∞) such that τ1 (t) ≥ t0 ,

τ2 (t) ≥ t0 ,

t ∈ [t∗ , ∞).

Hence from (10.15), we obtain   2 γ Δ r(t) zΔ (t) = −q1 (t)(x(τ1 (t)))α − q2 (t)(x(τ2 (t)))β t ∈ [t∗ , ∞).

< 0,

 2 γ Then r(t) zΔ (t) is strictly decreasing on [t∗ , ∞). From here, zΔ (t) > zΔ (t) − zΔ (t∗ )   2 γ γ1 t r(s) zΔ (s) = Δs 1 t∗ (r(s)) γ   2 γ γ1 t 1  γ1 Δs ≥ r(t) zΔ (t) r(s) t∗ 1

2

= R(t, t∗ )(r(t)) γ zΔ (t),

t ∈ [t∗ , ∞).

This completes the proof. Theorem 10.12. Assume (C1)–(C4). Let x be a positive solution of the equation (10.15) such that zΔ (t) > 0,

2

zΔ (t) > 0,

t ∈ [t∗ , ∞),

for some t∗ ∈ [t0 , ∞). Let also, ∞

(q1 (s) + q2 (s))(τ (s))α Δs = ∞.

t0

Then there exists a T ∈ [t∗ , ∞), sufficiently large, so that z(t) > tzΔ (t), z(t) t

is strictly decreasing on [T , ∞).

t ∈ [T , ∞),

(10.17)

514

10 Oscillations of Third-Order Functional Dynamic Equations

Proof. Let U (t) = z(t) − tzΔ (t),

t ∈ [t0 , ∞).

Then 2

U Δ (t) = zΔ (t) − σ (t)zΔ (t) − zΔ (t) 2

= −σ (t)zΔ (t) t ∈ [t∗ , ∞).

< 0,

Assume that for any t1 ∈ [t∗ , ∞) we have U (t) < 0 for t ∈ [t1 , ∞). Then

z(t) t

Δ =

tzΔ (t) − z(t) tσ (t)

=−

U (t) tσ (t) t ∈ [t1 , ∞).

> 0, Hence,

z(t) t

is strictly increasing on [t1 , ∞). Take t2 ∈ [t1 , ∞) so that τ (t) ≥ τ (t∗ ),

t ∈ [t2 , ∞).v

Let d=

z(τ (t∗ )) . τ (t∗ )

Then z(τ (t∗ )) z(τ (t)) > τ (t) τ (t∗ ) = d,

t ∈ [t2 , ∞).

Hence, z(τ (t)) > dτ (t),

t ∈ [t2 , ∞).

Note that x(t) = z(t) − p(t)x(τ0 (t)) > z(t) − p(t)z(τ0 (t)) > z(t)(1 − p(t)) ≥ (1 − P )z(t),

t ∈ [t2 , ∞).

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

515

From here,   2 γ Δ r(t) zΔ (t) = −q1 (t)(x(τ1 (t)))α − q2 (t)(x(τ2 (t)))β ≤ −q1 (t)(1 − P )α (z(τ1 (t)))α −q2 (t)(1 − P )β (z(τ2 (t)))β ≤ −q1 (t)(1 − P )α (z(τ (t)))α −q2 (t)(1 − P )β (z(τ (t)))β ≤ −q1 (t)(1 − P )β (z(τ (t)))α −q2 (t)(1 − P )β (z(τ (t)))α = −(1 − P )β (z(τ (t)))α (q1 (t) + q2 (t)),

t ∈ [t2 , ∞).

Now, integrating both sides of the last inequality from t2 to t, t ∈ [t2 , ∞), we have  2 γ  2 γ r(t) zΔ (t) − r(t2 ) zΔ (t2 ) + (1 − P )β

t

(z(τ (s)))α (q1 (s) + q2 (s))Δs ≤ 0,

t2

t ∈ [t2 , ∞). Therefore  2 γ r(t2 ) zΔ (t2 ) ≥ (1 − P )β

∞

(z(τ (s)))α (q1 (s) + q2 (s))Δs

t2 ∞

≥ (1 − P )β d α

(τ (s))α (q1 (s) + q2 (s))Δs

t2

→ ∞ as

t → ∞.

This is a contradiction. Therefore there exists a t1 ∈ [t∗ , ∞) so that U (t) > 0 for t ∈ [t1 , ∞). Consequently

 z(t) Δ tzΔ (t) − z(t) = t tσ (t) =−

U (t) tσ (t)

< 0, Then we have that

z(t) t

t ∈ [t1 , ∞).

is strictly decreasing on [t1 , ∞). This completes the proof.

We introduce the following notations. 

β −α β −α , , φ = min β −γ γ −α   κ = min k α , k β ,

516

10 Oscillations of Third-Order Functional Dynamic Equations

  β−γ   γ −α Φ(t) = φ q1 (t)(1 − P )α β−α q2 (t)(1 − P )β β−α

τ (t) σ (t)

γ ,

for sufficiently large t∗ ∈ [t0 , ∞). Theorem 10.13. Assume (C1)–(C4), (10.16), and (10.17). Let also, there is a T ∈ [t0 , ∞) such that ⎛ t

lim sup t→∞

⎝f σ (s)Φ(s) −

T

⎞

 Δ γ +1 f (s) +

(γ + 1)γ +1 (β(s)f σ (s)R(s, t∗ ))γ

⎠ Δs = ∞,

1 ([t , ∞)) is a nonnegative function. Then every solution of the where f ∈ Crd 0 equation (10.15) is oscillatory on [t0 , ∞) or tends to zero.

Proof. Assume that the equation (10.15) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that x is an eventually positive solution of the equation (10.15) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(τ0 (t)) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞).

By Theorem 10.9 and Theorem 10.10, it follows that there is a t2 ∈ [t1 , ∞) such that   2 γ Δ r(t) zΔ (t) < 0,

2

zΔ (t) > 0,

t ∈ [t1 , ∞),

and either zΔ (t) > 0 for t ∈ [t2 , ∞) or lim z(t) = lim x(t) = 0.

t→∞

t→∞

Suppose that zΔ (t) > 0, t ∈ [t2 , ∞). Consider the generalized Riccati transformation  2 γ r(t) zΔ (t) w(t) = f (t) , t ∈ [t0 , ∞). (z(t))γ Using that 

 2 γ Δ r(t) zΔ (t) ≤ −q1 (t)(1−P )α (z(τ1 (t)))α −q2 (t)(1−P )β (z(τ2 (t)))β ,

t ∈ [t2 , ∞),

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

517

we have w Δ (t)

=

f Δ (t)

= f Δ (t)

 2 γ r(t) zΔ (t) (z(t))γ  2 γ r(t) zΔ (t)

−f σ (t) ≤

(z(t))γ

% + f σ (t) + f σ (t)

 2 γ r(t) zΔ (t)

&Δ

(z(t))γ

(10.18)

  2 γ Δ r(t) zΔ (t) (zσ (t))γ

 2 γ r(t) zΔ (t) ((z(t))γ )Δ (z(t))γ (zσ (t))γ

f Δ (t) (z(τ1 (t)))α w(t) − f σ (t)q1 (t)(1 − P )α f (t) (zσ (t))γ −f σ (t)q2 (t)(1 − P )β

−f σ (t)

(z(τ2 (t)))β (zσ (t))γ

 2 γ r(t) zΔ (t) ((z(t))γ )Δ (z(t))γ (zσ (t))γ

t ∈ [t2 , ∞).

,

By the Young inequality |ab| ≤

1 p 1 q |a| + |b| , p q

a, b ∈ R,

p > 1,

1 1 + = 1, p q

q > 1,

we have β β −γ (z(τ1 (t)))α γ −α β (z(τ2 (t))) q1 (t)(1 − P )α q + (t)(1 − P ) 2 γ γ β −α β −α (zσ (t)) (zσ (t)) γ −α β−γ

α  β−α β  β−α α (z(τ1 (t))) β (z(τ2 (t))) q2 (t)(1 − P ) ≥ q1 (t)(1 − P ) (zσ (t))γ (zσ (t))γ β−γ

γ −α

 β−γ   γ −α (z(τ (t)))α β−α +β β−α  ≥ q1 (t)(1 − P )α β−α q2 (t)(1 − P )β β−α (zσ (t))γ

  β−γ   γ −α z(τ (t)) γ  = q1 (t)(1 − P )α β−α q2 (t)(1 − P )β β−α , zσ (t) t ∈ [t2 , ∞). Hence from (10.18), we get w Δ (t) ≤

(z(τ1 (t)))α f Δ (t) β −γ w(t) − f σ (t) φq1 (t)(1 − P )α f (t) β −α (zσ (t))γ −f σ (t)

(z(τ2 (t)))β γ −α φq2 (t)(1 − P )β β −α (zσ (t))γ

518

10 Oscillations of Third-Order Functional Dynamic Equations

−f σ (t) ≤

 2 γ r(t) zΔ (t) ((z(t))γ )Δ (z(t))γ (zσ (t))γ

f Δ (t) w(t) f (t)  β−γ   γ −α  −φf σ (t) q1 (t)(1 − P )α β−α q2 (t)(1 − P )β β−α

−f σ (t)

 2 γ r(t) zΔ (t) ((z(t))γ )Δ (z(t))γ (zσ (t))γ

z(τ (t)) zσ (t)

γ

,

t ∈ [t2 , ∞). Let 0 < γ ≤ 1. Then, using the chain rule, we have  γ −1 Δ z (t), (z(t))γ ≥ γ zσ (t)

t ∈ [t2 , ∞),

and w Δ (t) ≤

f Δ (t) w(t) f (t)  β−γ   γ −α  −φf σ (t) q1 (t)(1 − P )α β−α q2 (t)(1 − P )β β−α

−f σ t)

 2 γ r(t) zΔ (t) γ (zσ (t))γ −1 zΔ (t) (z(t))γ (zσ (t))γ

≤ −f (t)Φ(t) + σ

−γf σ (t)

 Δ  f (t) +

w(t) f (t)  2 γ r(t) zΔ (t) zΔ (t)

(z(t))γ zσ (t)  Δ  f (t) + w(t) ≤ −f σ (t)Φ(t) + f (t)  2 γ +1 γ +1 (r(t)) γ R(t, t2 ) zΔ (t) −γf σ (t) (z(t))γ zσ (t)  Δ  f (t) + w(t) = −f σ (t)Φ(t) + f (t)  2 γ σ r(t) zΔ (t) (r(t)) γ1 R(t, t )zΔ2 (t) f (t) 2 −γ f (t) f (t) (z(t))γ zσ (t)

z(τ (t)) zσ (t)

γ

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

= −f (t)Φ(t) + σ

 Δ  f (t) + f (t)

w(t) 1

1

2

f σ (t) (f (t)) γ (r(t)) γ zΔ (t)z(t) −γ w(t)R(t, t2 ) 1 f (t) (f (t)) γ z(t)zσ (t)  Δ  f (t) + w(t) = −f σ (t)Φ(t) + f (t) −γf σ (t)R(t, t2 )

(w(t))

γ +1 γ

z(t)

γ +1 γ

(f (t)) zσ (t)  Δ  f (t) + w(t) ≤ −f σ (t)Φ(t) + f (t) γ +1

t (w(t)) γ −γf (t)R(t, t2 ) , σ (t) (f (t)) γ γ+1 σ

t ∈ [t2 , ∞).

Let γ > 1. Then, using the chain rule, we get  Δ (z(t))γ ≥ γ (z(t))γ −1 zΔ (t),

t ∈ [t2 , ∞),

and w (t) ≤ −f (t)Φ(t) + Δ

σ

−f σ (t)γ r(t)

 Δ  f (t) +

w(t) f (t)  2 γ zΔ (t) (z(t))γ −1 zΔ (t)

(z(t))γ (zσ (t))γ  Δ  f (t) + w(t) ≤ −f σ (t)Φ(t) + f (t)  2 γ 1 2 zΔ (t) R(t, t2 )(r(t)) γ zΔ (t)(z(t))γ −1 σ −f (t)γ r(t) (z(t))γ (zσ (t))γ  Δ  f (t) + = −f σ (t)Φ(t) + w(t) f (t) 1

1

2

R(t, t2 )(f (t)) γ (r(t)) γ zΔ (t)(z(t))γ f σ (t) w(t) −γ 1 f (t) (f (t)) γ z(t) (zσ (t))γ  Δ  f (t) + σ w(t) = −f (t)Φ(t) + f (t)

519

520

10 Oscillations of Third-Order Functional Dynamic Equations γ +1

f σ (t)

R(t, t2 )(w(t)) γ (z(t))γ −γ γ +1 (zσ (t))γ (f (t)) γ  Δ  f (t) + σ w(t) ≤ −f (t)Φ(t) + f (t) γ +1 γ

R(t, t2 )(w(t)) γ t −γf σ (t) , γ +1 σ (t) (f (t)) γ t ∈ [t2 , ∞). Let λ=

γ +1 . γ

Then 

 f Δ (t) + f (t) w(t) λ σ −γf (t)β(t)R(t, t2 ) (w(t)) , (f (t))λ

w Δ (t) ≤ −f σ (t)Φ(t) +

t ∈ [t2 , ∞),

(10.19)

γ > 0. Define Aλ = γf σ (t)β(t)R(t, t2 ) B

λ−1

=

 Δ  f (t) +

(w(t))λ , (f (t))λ 1

λ (γf σ (t)β(t)R(t, t2 )) λ

t ∈ [t2 , ∞).

,

Note that A and B are positive on [t2 , ∞). Now, using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ , we get  Δ  f (t) +  σ 1 1 1 w(t) λγ f (t) λ (β(t)) λ (R(t, t2 )) λ f (t) λ (γf σ (t)β(t)R(t, t2 )) λ1 1 λ

(w(t))λ (f (t))λ  Δ  λ f (t) λ−1

−γf σ (t)β(t)R(t, t2 ) ≤ (λ − 1)

λ

1

λ λ−1 (γf σ (t)β(t)R(t, t2 )) λ−1

,

t ∈ [t2 , ∞),

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

521

or  Δ  f (t) + f (t)

w(t) − γf σ (t)β(t)R(t, t2 ) 

≤

f Δ (t)

(w(t))λ (f (t))λ

γ +1 +

(γ + 1)γ +1 (β(t)f σ (t)R(t, t2 ))γ

t ∈ [t2 , ∞).

,

Consequently  w (t) ≤ Δ

f Δ (t)

γ +1 +

(γ + 1)γ +1 (β(t)f σ (t)R(t, t2 ))γ

− f σ (t)Φ(t),

t ∈ [t2 , ∞),

whereupon f (t)Φ(t) − σ

 Δ γ +1 f (t) +

(γ + 1)γ +1 (β(t)f σ (t)R(t, t2 ))γ

Hence, ⎛ t

⎝f σ (s)Φ(s)−

t2

≤ −w Δ (t),

 Δ γ +1 f (s) +

(γ + 1)γ +1 (β(s)f σ (s)R(s, t2 ))γ

t ∈ [t2 , ∞).

⎞ ⎠ Δs ≤ w(t2 ) − w(t) ≤ w(t2 ),

t ∈ [t2 , ∞).

This is a contradiction. Therefore any solution of the equation (10.15) is oscillatory or tends to zero. This completes the proof. Example 10.11. Let T = 2N0 . Consider the equation

1 x(t) + x 3

Δ3

 1

 5 3 3 t t t 1 1 x x + + = 0, 4 t 2 t 2

t ∈ [4, ∞). Here r(t) = 1, 1 , 3 1 q1 (t) = , t 1 q2 (t) = , t γ = 1, p(t) =

522

10 Oscillations of Third-Order Functional Dynamic Equations

1 , 3 5 β= , 3 1 P = , 3 t τ0 (t) = , 4 t τ1 (t) = , 2 t τ2 (t) = , 2 t τ (t) = , 4 σ (t) = 2t, α=

t ∈ T.

Next, 

4 3 2 3

φ = min

,

4 3 2 3



= 2, 2

2

%  1 & 34 %  5 & 32 % t & 1 1 3 3 1 1 3 3 4 Φ(t) = 2 1− 1− t 3 t 3 2t  1  5 =2 =

6

2 3

t 1 1

3

2 3

3 2

3

1 8

,

t 2 26 t β(t) = 2t 1 = , 2 R(t, t∗ ) =

t

Δs t∗

= t − t∗ ,

t ∈ [t∗ , ∞),

t∗ ∈ [4, ∞).

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

523

Then ∞

1 r(t)

t0

1 γ

∞

Δt =

Δt 4

= ∞, and ∞

∞ t0

v

∞

1 r(u)

1 (q1 (s) + q2 (s))Δs

γ

∞

ΔuΔv =

∞

4

u

∞

v

u

2 ΔsΔuΔv s

= ∞, and ∞

4

2  s 13 Δs s 4

1

∞

∞

(q1 (s) + q2 (s))(τ (s))α Δs =

t0

= 23

1 2

Δs

s3

4

= ∞. Take f (t) = t,

t ∈ T,

and t∗ = 4, T = 8. Then ⎛ t

lim sup t→∞

≥

=

∞

T



⎜ σ ⎝f (s)Φ(s) −

⎝

8

%

+

⎞ ⎟ ⎠ Δs

⎞ s 3

1

s 2 26 ∞

γ +1

(γ + 1)γ +1 (β(s)f σ (s)R(s, t∗ ))γ

⎛

8

f Δ (s)

1 1 1 26 s 2

1

 ⎠ Δs 4 12 (2s)(s − 4) & 1 − Δs s(s − 4) −

= ∞.

Therefore any solution of the considered equation is oscillatory or tends to zero. Exercise 10.6. Prove that any solution of the equation

1 t2

Δ3 1 5 1 σ (t) σ (t) x(t) + x(τ0 (t)) + (x(τ2 (t))) 3 = 0, (x(τ1 (t))) 3 + 3 4τ (t) τ (t)

t ∈ [t0 , ∞), is oscillatory or tends to zero.

524

10 Oscillations of Third-Order Functional Dynamic Equations

Theorem 10.14. Assume (C1)–(C4), (10.16), and (10.17) hold. Suppose that there exist functions H, h ∈ Crd (D) such that H (t, t) = 0,

t ≥ t0 ,

H (t, s) > 0,

t > s ≥ t0 ,

and H has a nonpositive continuous Δ-partial derivative H Δs (t, s) with respect to the second variable and satisfies H Δs (σ (t), s) + H (σ (t), σ (s))

γ f Δ (s) h(t, s) =− (H (σ (t), σ (s))) γ +1 f (s) f (s)

(10.20)

and for all sufficiently large t1 ∈ [t0 , ∞), there is a T > t1 such that lim sup t→∞

σ (t)

1 H (σ (t), T )

χ (t, s)Δs = ∞,

T

where f is a positive Δ-differentiable function and χ (t, s) = H (σ (t), σ (s))f σ (s)Φ(s) −

(h− (t, s))γ +1 , (γ + 1)γ +1 (β(s)f σ (s)R(s, t1 ))γ

t ≥ s ≥ t0 .

Then every solution of the equation (10.15) is oscillatory or tends to zero. Proof. Assume that the equation (10.15) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that x is an eventually positive solution of the equation (10.15) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(τ0 (t)) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞).

By Theorem 10.9 and Theorem 10.10, it follows that there is a t2 ∈ [t1 , ∞) such that   2 γ Δ r(t) zΔ (t) < 0,

2

zΔ (t) > 0,

t ∈ [t1 , ∞),

and either zΔ (t) > 0 for t ∈ [t2 , ∞) or lim z(t) = lim x(t) = 0.

t→∞

t→∞

Suppose that zΔ (t) > 0, t ∈ [t2 , ∞). Let t3 ∈ [t2 , ∞) be sufficiently large such that (10.20) holds. Consider the generalized Riccati transformation which is used in the proof of Theorem 10.13. By (10.19), we get

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

f σ (s)Φ(s) ≤ −w Δ (s) +

525

f Δ (s) w(s) f (s)

−γf σ (s)β(s)R(s, t3 )

(w(s))λ , (f (s))λ

s ∈ [t3 , ∞).

Hence, H (σ (t), σ (s))f σ (s)Φ(s) ≤ −H (σ (t), σ (s))w Δ (s) + H (σ (t), σ (s)) −γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

f Δ (s) w(s) f (s)

(w(s))λ , (f (s))λ

t, s ∈ [t3 , ∞),

t ≥ s,

and # σ (t) t3

H (σ (t), σ (s))f σ (s)Φ(s)Δs ≤ −

# σ (t) #tσ3 (t)

H (σ (t), σ (s))w Δ (s)Δs Δ

(s) H (σ (t), σ (s)) ff (s) w(s)Δs # σ (t) λ Δs −γ t3 H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) (w(s)) (f (s))λ # σ (t) Δ s = H (σ (t), t3 )w(t3 ) + t3 H (σ (t), s)w(s)Δs # σ (t) Δ (s) + t3 H (σ (t), σ (s)) ff (s) w(s)Δs # σ (t) λ σ Δs −γ t3 H (σ (t), σ (s))f (s)β(s)R(s, t3 ) (w(s)) (f (s))λ

+

t3

= H (σ (t),t3 )w(t3 ) # σ (t) Δ (s) H Δs (σ (t), s) + H (σ (t), σ (s)) ff (s) w(s)Δs + t3

(10.21)

σ (t)

−γ

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

t3

(w(s))λ Δs (f (s))λ

= H (σ (t), t3 )w(t3 ) −

σ (t) t3

γ h(t, s) (H (σ (t), σ (s))) γ +1 w(s)Δs f (s)

σ (t)

−γ

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

t3

(w(s))λ Δs (f (s))λ

≤ H (σ (t), t3 )w(t3 ) +

σ (t) t3

−γ

γ h− (t, s) (H (σ (t), σ (s))) γ +1 w(s)Δs f (s)

σ (t) t3

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

(w(s))λ Δs, (f (s))λ

526

10 Oscillations of Third-Order Functional Dynamic Equations

t ∈ [t3 , ∞). Let λ=

γ +1 , γ

Aλ = γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) B λ−1 =

h− (t, s) λ (γf σ (s)β(s)R(s, t

t, s ∈ [t3 , ∞),

,

1

(w(s))λ , (f (s))λ

3 )) λ

t ≥ s.

Using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ , we find  1 w(s)  h− (t, s) λ γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) λ f (s) λ (γf σ (s)β(s)R(s, t3 )) λ1 −γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

(w(s))λ (f (s))λ

λ

≤ (λ − 1)

(h− (t, s)) λ−1 λ

λ λ−1 (γf σ (s)β(s)R(s, t3 ))λ−1

,

t, s ∈ [t3 , ∞),

t ≥ s,

or γ (w(s))λ w(s) h− (t, s) (H (σ (t), σ (s))) γ +1 − γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) f (s) (f (s))λ

(h− (t, s))γ +1

≤

1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

,

t, s ∈ [t3 , ∞),

t ≥ s.

From here and from (10.21), we obtain σ (t)

H (σ (t), σ (s))f σ (s)Φ(s)Δs ≤ H (σ (t), t3 )w(t3 )

t3

+

(h− (t, s))γ +1

σ (t) t3

1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

t ∈ [t3 , ∞),

Δs,

whereupon σ (t)

% σ

H (σ (t), σ (s))f (s)Φ(s)− t3

&

(h− (t, s))γ +1 1

(γ +1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

Δs ≤ H (σ (t), t3 )w(t3 ),

10.2 Oscillations of Third-Order Quasilinear Neutral Dynamic Equations

527

t ∈ [t3 , ∞), and 1 H (σ (t), t3 )

σ (t)

% H (σ (t), σ (s))f σ (s)Φ(s)−

t3

&

(h− (t, s))γ +1 1

(γ +1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

Δs ≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. Therefore any solution of the equation (10.15) is oscillatory or tends to zero. This completes the proof. Example 10.12. Let H (t, s) = t − s, f (t) = 1, h(t, s) =

t, s ∈ [t3 , ∞),

t ∈ [t0 , ∞), 1 γ

(σ (t) − σ (s)) γ +1

h(t, t) = 0,

t ≥ s,

,

t, s ∈ [t3 , ∞),

t > s,

t ∈ [t3 , ∞).

Then f Δ (t) = 0, H Δs (t, s) = −1,

t, s ∈ [t3 , ∞),

t ≥ s,

and H Δs (σ (t), s) + H (σ (t), σ (s)) −

f Δ (s) = −1, f (s)

γ h(t, s) (H (σ (t), σ (s))) γ +1 = − f (s)

1

γ

(σ (t) − σ (s))

= −1,

γ γ +1

(σ (t) − σ (s)) γ +1

t, s ∈ [t3 , ∞),

t > s,

and h− (t, s) = 0,

t, s ∈ [t3 , ∞),

t ≥ s.

Hence, χ (t, s) = (σ (t) − σ (s))Φ(s),

t, s ∈ [t3 , ∞),

t ≥ s,

528

10 Oscillations of Third-Order Functional Dynamic Equations

and ⎛ ⎞ γ +1 σ (t) 1 (t, s)) (h − σ ⎝H (σ (t), σ (s))f (s)Φ(s)− ⎠Δs lim sup 1 t→∞ H (σ (t), T ) T (γ +1)γ +1 (f σ (s)β(s)R(s, t3 )) γ 1 t→∞ σ (t) − T

= lim sup

σ (t) T

(σ (t) − σ (s))Φ(s)Δs.

Exercise 10.7. Let T = Z. Write an analogue of Theorem 10.14.

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with Distributed Deviating Arguments Let c, d ∈ [t0 , ∞), c < d. In this section we investigate the equation 

 2 γ Δ r(t) (x(t) + p(t)x(τ (t)))Δ #d + c f (t, x(φ(t, ξ )))Δξ = 0, t ∈ [t0 , ∞),

(10.22)

where (D1) (D2)

γ > 0 is a quotient of two odd positive integers, r ∈ Crd (T) and ∞ t0

(D3) (D4)

1 r(t)

γ

Δt = ∞,

p ∈ Crd (T), 0 ≤ p(t) ≤ P < 1 for some positive constant P , τ : T → T is a strictly increasing and differentiable function such that τ (t) ≤ t,

(D5)

1

lim τ (t) = ∞,

t→∞

φ : [t0 , ∞) × [c, d] → T, φ ∈ Crd ([t0 , ∞) × [c, d]) is not decreasing function for ξ and φ(t, ξ ) ≤ t, and lim min φ(t, ξ ) = ∞,

t→∞ ξ ∈[c,d]

(D6)

f : T × R → R, f ∈ C (T × R), uf (t, u) > 0, u = 0, t ∈ [t0 ∞), and there exists a positive function δ ∈ Crd (T) such that f (t, u) ≥ δ(t)uγ ,

u = 0,

t ∈ [t0 , ∞).

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

529

Let z(t) = x(t) + p(t)x(τ (t)),

t ∈ [t0 , ∞).

Then the equation (10.22) can be rewritten in the form 

 2 γ Δ r(t) zΔ (t) +

d

f (t, x(φ(t, ξ )))Δξ = 0,

t ∈ [t0 , ∞).

c

Theorem 10.15. Assume (D1)–(D6). Let x be a solution of the equation (10.22) such that x(t) > 0,

x(τ (t)) > 0,

x(φ(t, ξ )) > 0,

t ∈ [t0 , ∞),

ξ ∈ [c, d].

Then there exists a t1 ∈ [t0 , ∞) such that z has one of the following two properties. 2

1. z(t) > 0, zΔ (t) > 0, zΔ (t) > 0, t ∈ [t1 , ∞), 2 2. z(t) > 0, zΔ (t) < 0, zΔ (t) > 0, t ∈ [t1 , ∞). Proof. Note that t ∈ [t0 , ∞).

z(t) > x(t) > 0, Then   2 γ Δ =− r(t) zΔ (t)

d

f (t, x(φ(t, ξ )))Δξ c

< 0,

t ∈ [t0 , ∞).

Therefore  2 γ r(t) zΔ (t) is a strictly decreasing function on [t0 , ∞). Consequently it has eventually one sign 2 on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that zΔ (t) has one sign on [t1 , ∞). 2 Assume that zΔ (t) < 0 for t ∈ [t1 , ∞). Therefore there exists a positive constant M such that  2 γ r(t) zΔ (t) ≤ −M < 0, t ∈ [t1 , ∞). Hence,  2 γ M zΔ (t) ≤ − , r(t)

t ∈ [t1 , ∞),

530

10 Oscillations of Third-Order Functional Dynamic Equations

and 1

z

Δ2

(t) ≤ −

Mγ

1

t ∈ [t1 , ∞),

,

(r(t)) γ and

t

1

zΔ (t) ≤ zΔ (t1 ) − M γ

1 1

t1

Δs

(r(s)) γ t → ∞.

→ −∞ as

Therefore there exists a t2 ∈ [t1 , ∞) such that zΔ (t) < 0,

t ∈ [t2 , ∞).

Since zΔ (t) < 0,

2

zΔ (t) < 0,

t ∈ [t2 , ∞),

there exists a t3 ∈ [t2 , ∞) such that z(t) < 0,

t ∈ [t3 , ∞).

This is a contradiction and consequently 2

t ∈ [t1 , ∞).

zΔ (t) > 0, This completes the proof.

Theorem 10.16. Suppose (D1)–(D6). Let x be an eventually positive solution of the equation (10.22) such that there exists a t1 ∈ [t0 , ∞) for which z(t) > 0,

zΔ (t) < 0,

2

zΔ (t) > 0,

t ∈ [t1 , ∞).

Let also, ∞

∞ t1

v

1 r(u)

1

∞

γ

δ(s)Δs

ΔuΔv = ∞.

u

Then lim x(t) = lim z(t) = 0.

t→∞

t→∞

(10.23)

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

Proof. Since zΔ (t) < 0,

z(t) > 0,

t ∈ [t1 , ∞),

there exists an l ≥ 0 such that lim z(t) = l.

t→∞

Suppose that l > 0. Take  > 0 so that 
0. Note that there exists a t2 ∈ [t1 , ∞) such that l < z(t) < l + ,

t ∈ [t2 , ∞),

l < z(τ (t)) < l + ,

t ∈ [t2 , ∞),

l < z(φ(t, ξ )) < l + ,

t ∈ [t2 , ∞),

ξ ∈ [c, d].

Then x(t) = z(t) − p(t)x(τ (t)) > z(t) − p(t)z(τ (t)) > l − P (l + ) = k(l + ) > kz(t),

t ∈ [t2 , ∞),

x(φ(t, ξ )) = z(φ(t, ξ )) − p(t)x(φ(t, ξ )) > z(φ(t, ξ )) − p(t)z(φ(t, ξ )) > l − P (l + ) = k(l + ) > kz(φ(t, ξ )),

t ∈ [t2 , ∞),

ξ ∈ [c, d].

531

532

10 Oscillations of Third-Order Functional Dynamic Equations

Hence,   2 γ Δ r(t) zΔ (t) =−

d

f (t, x(φ(t, ξ )))Δξ c d

≤−

(x(φ(t, ξ )))γ δ(t)Δξ

c d

< −k γ

(z(φ(t, ξ )))γ δ(t)Δξ

c d

≤ −k γ δ(t)

(z(φ(t, d)))γ Δξ

c

= −k γ δ(t)(z(φ(t, d)))γ (d − c),

t ∈ [t2 , ∞).

Now we integrate the last inequality from t to ∞, t ∈ [t2 , ∞), and using (D2), we obtain  2 γ −r(t) zΔ (t) ≤ −k γ (d − c)

∞

δ(s)(z(φ(s, d)))γ Δs

t ∞

≤ −k γ (d − c)l γ

δ(s)Δs,

t ∈ [t2 , ∞).

t

From here,  2 γ r(t) zΔ (t) ≥ k γ (d − c)l γ

∞

δ(s)Δs,

t ∈ [t2 , ∞),

δ(s)Δs,

t ∈ [t2 , ∞),

t

and  2 γ 1 zΔ (t) ≥ k γ l γ (d − c) r(t)

∞ t

and z

Δ2

(t) ≥ kl(d − c)

1 γ

1 r(t)

1

∞

γ

δ(s)Δs

,

t ∈ [t2 , ∞).

t

Then −z (t) ≥ kl(d − c) Δ

∞

1 γ

t

1 r(u)

1

∞

γ

δ(s)Δs u

Δu,

t ∈ [t2 , ∞),

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

533

and z(t2 ) ≥ kl(d − c)

∞

∞

1 γ

t2

t

1

∞

1 r(u)

γ

δ(s)Δs

ΔuΔt

u

= ∞. This is a contradiction. Therefore l = 0. This completes the proof. For t∗ ∈ [t0 , ∞), we set R(t, t∗ ) =

t t∗

1 r(s)

1 γ

t ∈ [t∗ , ∞).

Δs,

Theorem 10.17. Assume (D1)–(D6). Let x be a positive solution of the equation (10.22) such that 2

zΔ (t) > 0,

zΔ (t) > 0,

t ∈ [t∗ , ∞),

for some t∗ ∈ [t0 , ∞). Then 1

2

zΔ (t) ≥ R(t, t∗ )(r(t)) γ zΔ (t),

t ∈ [t∗ , ∞).

Proof. Note that there exists a t∗ ∈ [t0 , ∞) such that φ(t, ξ ) ≥ t0 ,

τ (t) ≥ t0 ,

t ∈ [t∗ , ∞),

ξ ∈ [c, d].

Hence from (10.22), we obtain   2 γ Δ =− r(t) zΔ (t)

d

f (t, x(φ(t, ξ )))Δξ c

< 0, t ∈ [t∗ , ∞).  2 γ Then r(t) zΔ (t) is strictly decreasing on [t∗ , ∞). From here, zΔ (t) > zΔ (t) − zΔ (t∗ )   2 γ γ1 t r(s) zΔ (s) = Δs 1 t∗ (r(s)) γ   2 γ γ1 t 1  γ1 Δs ≥ r(t) zΔ (t) r(s) t∗ 1

2

= R(t, t∗ )(r(t)) γ zΔ (t), This completes the proof.

t ∈ [t∗ , ∞).

534

10 Oscillations of Third-Order Functional Dynamic Equations

Theorem 10.18. Suppose (D1)–(D6) and there exists a t∗ ∈ [t0 , ∞) such that 2

zΔ (t) > 0,

t ∈ [t∗ , ∞).

zΔ (t) > 0,

Let also, ∞

q(s) (φ2 (s))γ Δs = ∞,

(10.24)

t∗

where q(t) = (d − c)δ(t)(1 − p(t))γ , φ2 (t) = φ(t, c),

t ∈ [t∗ , ∞).

Then there exists a T ∈ [t∗ , ∞), sufficiently large, so that t ∈ [T , ∞),

z(t) > tzΔ (t), and

z(t) t

is strictly decreasing on [T , ∞).

Proof. Let U (t) = z(t) − tzΔ (t),

t ∈ [t∗ , ∞).

Then 2

U Δ (t) = zΔ (t) − zΔ (t) − σ (t)zΔ (t) 2

= −σ (t)zΔ (t) < 0,

t ∈ [t∗ , ∞).

Suppose that for any t1 ∈ [t∗ , ∞) we have t ∈ [t1 , ∞).

U (t) < 0, Therefore

z(t) t

Δ =

tzΔ (t) − z(t) tσ (t)

=− > 0,

U (t) tσ (t) t ∈ [t1 , ∞).

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

Then

z(t) t

is strictly increasing on [t1 , ∞). Take a t2 ∈ [t1 , ∞) so that t ∈ [t2 , ∞),

φ(t, ξ ) ≥ φ(t1 , ξ ),

ξ ∈ [c, d].

If m=

z(φ(t1 , ξ )) , φ(t1 , ξ )

then z(φ(t1 , ξ )) z(φ(t, ξ )) ≥ φ(t, ξ ) φ(t1 , ξ ) =m t ∈ [t2 , ∞),

> 0,

ξ ∈ [c, d],

i.e., z(φ(t, ξ )) ≥ mφ(t, ξ ),

t ∈ [t2 , ∞),

ξ ∈ [c, d].

Next, x(t) = z(t) − p(t)x(τ (t)) > z(t) − p(t)z(τ (t)) > z(t)(1 − p(t)),

t ∈ [t2 , ∞),

x(φ(t, ξ )) = z(φ(t, ξ )) − p(t)x(τ (φ(t, ξ ))) > z(φ(t, ξ )) − p(t)z(τ (φ(t, ξ ))) > z(φ(t, ξ ))(1 − p(t)),

t ∈ [t2 , ∞),

ξ ∈ [c, d].

Hence, 

 2 γ Δ r(t) zΔ (t) =−

d

f (t, x(φ(t, ξ )))Δξ c d

≤ −δ(t)

(x(φ(t, ξ )))γ Δξ

c d

≤ −δ(t)(1 − p(t))γ

(z(φ(t, ξ )))γ Δξ

c

≤ −δ(t)(1 − p(t))γ (z(φ(t, c)))γ (d − c) = −q(t)(z(φ2 (t)))γ ,

t ∈ [t2 , ∞).

535

536

10 Oscillations of Third-Order Functional Dynamic Equations

Hence,  2 γ r(t) zΔ (t) +

t

 2 γ q(s)(z(φ2 (s)))γ Δs ≤ r(t2 ) zΔ (t2 ) ,

t2

t ∈ [t2 , ∞), and  2 γ ≥ r(t2 ) zΔ (t2 )

t

q(s)(z(φ2 (s)))γ Δs

t2

≥ m

t

q(s)(φ2 (s))γ Δs

t2

→ ∞ as

t → ∞.

This is a contradiction. Then there is a t1 ∈ [t∗ , ∞) so that U (t) > 0, t ∈ [t1 , ∞). Also,

z(t) t

Δ =

tzΔ (t) − z(t) tσ (t)

=− < 0, Consequently

z(t) t

U (t) tσ (t) t ∈ [t1 , ∞).

is strictly decreasing on [t1 , ∞). This completes the proof.

Let  β(t) =

t if 0 < γ ≤ 1 σ (t) γ t if γ > 1. σ (t)

Theorem 10.19. Assume (D1)–(D6), (10.23), and (10.24). Let also, there is a T ∈ [t0 , ∞) such that ⎛ t

lim sup t→∞

T

⎝f σ (s)q(s) φ2 (s) σ (s)



γ −

f Δ (s)

γ +1 +

(γ + 1)γ +1 (β(s)f σ (s)R(s, t∗ ))γ

⎞ ⎠ Δs=∞,

1 ([t , ∞)) is a nonnegative function. Then every solution of the where f ∈ Crd 0 equation (10.22) is oscillatory on [t0 , ∞) or tends to zero.

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

537

Proof. Assume that the equation (10.22) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that x is an eventually positive solution of the equation (10.22) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

x(φ(t, ξ )) > 0,

By Theorem 10.15 and Theorem 10.16, it follows that there is a t2 ∈ [t1 , ∞) such that   2 γ Δ r(t) zΔ (t) < 0,

2

zΔ (t) > 0,

t ∈ [t1 , ∞),

and either zΔ (t) > 0 for t ∈ [t2 , ∞) or lim z(t) = lim x(t) = 0.

t→∞

t→∞

Suppose that zΔ (t) > 0, t ∈ [t2 , ∞). Consider the generalized Riccati transformation  2 γ r(t) zΔ (t) w(t) = f (t) , t ∈ [t0 , ∞). (z(t))γ Using that   2 γ Δ ≤ −q(t)(z(φ2 (t)))γ , r(t) zΔ (t)

t ∈ [t2 , ∞),

we have w Δ (t) = f Δ (t) = f Δ (t)

 2 γ r(t) zΔ (t) (z(t))γ  2 γ r(t) zΔ (t)

−f σ (t) ≤

(z(t))γ

% + f σ (t) + f σ (t)

 2 γ r(t) zΔ (t)

&Δ

(z(t))γ   2 γ Δ r(t) zΔ (t) (zσ (t))γ

 2 γ r(t) zΔ (t) ((z(t))γ )Δ (z(t))γ (zσ (t))γ

f Δ (t) (z(φ2 (t)))γ w(t) − f σ (t)q(t) f (t) (z(σ (t)))γ  2 γ r(t) zΔ (t) ((z(t))γ )Δ σ −f (t) , t ∈ [t2 , ∞). (z(t))γ (zσ (t))γ

(10.25)

538

10 Oscillations of Third-Order Functional Dynamic Equations

Note that we can take t2 such that from (10.25), we get w Δ (t) ≤

z(t) t

is strictly decreasing on [t2 , ∞). Hence

f Δ (t) (φ2 (t))γ w(t) − f σ (t)q(t) f (t) (σ (t))γ  2 γ r(t) zΔ (t) ((z(t))γ )Δ σ −f (t) , (z(t))γ (zσ (t))γ

t ∈ [t2 , ∞).

Let 0 < γ ≤ 1. Then, using the chain rule, we have  γ −1 Δ z (t), (z(t))γ ≥ γ zσ (t)

t ∈ [t2 , ∞),

and w Δ (t) ≤

f Δ (t) w(t) f (t) (φ2 (t))γ (σ (t))γ  2 γ r(t) zΔ (t) γ (zσ (t))γ −1 zΔ (t)

−f σ (t)q(t)

−f σ t)

(z(t))γ (zσ (t))γ

 Δ  f (t) + (φ2 (t))γ w(t) ≤ −f (t)q(t) + γ (σ (t)) f (t)  2 γ r(t) zΔ (t) zΔ (t) −γf σ (t) (z(t))γ zσ (t)  Δ  f (t) + (φ2 (t))γ σ w(t) ≤ −f (t)q(t) + γ (σ (t)) f (t)  2 γ +1 γ +1 (r(t)) γ R(t, t2 ) zΔ (t) −γf σ (t) (z(t))γ zσ (t)  Δ  f (t) + (φ2 (t))γ σ w(t) = −f (t)q(t) + γ (σ (t)) f (t)  2 γ r(t) zΔ (t) (r(t)) γ1 R(t, t )zΔ2 (t) f σ (t) 2 −γ f (t) f (t) (z(t))γ zσ (t)  Δ  f (t) + (φ2 (t))γ σ w(t) = −f (t)q(t) + γ (σ (t)) f (t) σ

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . . 1

1

2

f σ (t) (f (t)) γ (r(t)) γ zΔ (t)z(t) w(t)R(t, t2 ) −γ 1 f (t) (f (t)) γ z(t)zσ (t)  Δ  f (t) + (φ2 (t))γ σ = −f (t)q(t) + w(t) γ (σ (t)) f (t) −γf (t)R(t, t2 ) σ

(w(t)) (f (t))

≤ −f σ (t)q(t)

(φ2 (t))γ + (σ (t))γ

γ +1 γ

z(t)

γ +1 γ

zσ (t)  Δ  f (t) + f (t)

w(t)

γ +1

t (w(t)) γ −γf (t)R(t, t2 ) , σ (t) (f (t)) γ γ+1

t ∈ [t2 , ∞).

σ

Let γ > 1. Then, using the chain rule, we get  Δ (z(t))γ ≥ γ (z(t))γ −1 zΔ (t),

t ∈ [t2 , ∞),

and  Δ  f (t) + (φ2 (t))γ w (t) ≤ −f (t)q(t) w(t) + γ (σ (t)) f (t)  2 γ zΔ (t) (z(t))γ −1 zΔ (t) σ −f (t)γ r(t) (z(t))γ (zσ (t))γ  Δ  f (t) + (φ2 (t))γ σ w(t) ≤ −f (t)q(t) + γ (σ (t)) f (t)  2 γ 1 2 zΔ (t) R(t, t2 )(r(t)) γ zΔ (t)(z(t))γ −1 −f σ (t)γ r(t) (z(t))γ (zσ (t))γ  Δ  f (t) + (φ2 (t))γ σ w(t) = −f (t)q(t) + γ (σ (t)) f (t) Δ

σ

1

1

2

f σ (t) R(t, t2 )(f (t)) γ (r(t)) γ zΔ (t)(z(t))γ −γ w(t) 1 f (t) (f (t)) γ z(t) (zσ (t))γ  Δ  f (t) + (φ2 (t))γ σ w(t) = −f (t)q(t) + γ (σ (t)) f (t)

539

540

10 Oscillations of Third-Order Functional Dynamic Equations γ +1

f σ (t)

R(t, t2 )(w(t)) γ (z(t))γ −γ γ +1 (zσ (t))γ (f (t)) γ  Δ  f (t) + (φ2 (t))γ σ w(t) ≤ −f (t)q(t) + (σ (t))γ f (t) γ +1 γ

R(t, t2 )(w(t)) γ t −γf σ (t) , γ +1 σ (t) (f (t)) γ t ∈ [t2 , ∞). Let λ=

γ +1 . γ

Then 

 f Δ (t) + f (t) w(t) (w(t))λ σ −γf (t)β(t)R(t, t2 ) (f (t))λ , t ∈ [t2 , ∞), γ

2 (t)) w Δ (t) ≤ −f σ (t)q(t) (φ (σ (t))γ +

(10.26)

γ > 0. Define Aλ = γf σ (t)β(t)R(t, t2 ) B λ−1 =

(w(t))λ , (f (t))λ

f Δ (t) 1

λ (γf σ (t)β(t)R(t, t2 )) λ

t ∈ [t2 , ∞).

,

Note that A and B are positive on [t2 , ∞). Now, using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ , we get 1 1  1 1 w(t) f Δ (t) λγ λ f σ (t) λ (β(t)) λ (R(t, t2 )) λ f (t) λ (γf σ (t)β(t)R(t, t2 )) λ1 (w(t))λ (f (t))λ  Δ  λ f (t) λ−1

−γf σ (t)β(t)R(t, t2 ) ≤ (λ − 1)

λ

1

λ λ−1 (γf σ (t)β(t)R(t, t2 )) λ−1

,

t ∈ [t2 , ∞),

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

541

or  Δ  f (t) + f (t)

w(t) − γf σ (t)β(t)R(t, t2 ) 

≤

f Δ (t)

(w(t))λ (f (t))λ

γ +1 +

(γ + 1)γ +1 (β(t)f σ (t)R(t, t2 ))γ

,

t ∈ [t2 , ∞).

Consequently w (t) ≤ Δ

 Δ γ +1 f (t) +

(γ + 1)γ +1 (β(t)f σ (t)R(t, t2 ))γ

− f σ (t)q(t)

(φ2 (t))γ , (σ (t))γ

t ∈ [t2 , ∞),

whereupon  Δ γ +1 f (t) + (φ2 (t))γ f (t)q(t) − ≤ −w Δ (t), γ γ +1 (σ (t)) (γ + 1) (β(t)f σ (t)R(t, t2 ))γ σ

t ∈ [t2 , ∞).

Hence, ⎛

⎞  Δ γ +1 γ f (s) + (s)) (φ 2 ⎝f (s)q(s) ⎠ Δs ≤ w(t2 ) − w(t) − (σ (s))γ (γ + 1)γ +1 (β(s)f σ (s)R(s, t2 ))γ

t

σ

t2

≤ w(t2 ),

t ∈ [t2 , ∞).

This is a contradiction. Therefore any solution of the equation (10.22) is oscillatory or tends to zero. This completes the proof. Example 10.13. Let T = Z. Consider the equation %

1 2 t 3 x(t) x(t) + x(t − 1)Δ t

3 &Δ 1

Here t0 = 1, c = 1, d = 4, γ = 3, r(t) = t 3 , p(t) =

4

+

1 , t

(x(t + ξ ))3 Δξ = 0,

t ∈ [1, ∞).

542

10 Oscillations of Third-Order Functional Dynamic Equations

δ(t) = 1, 

1 3 q(t) = 3 1 − , t φ(t, ξ ) = t + ξ, φ2 (t) = φ(t, 1) = t + 1,

t ∈ T,

ξ ∈ [1, 4].

Then ∞ t0

1 r(t)

∞

1 γ

Δt = 1

∞

= 1

1 t3

1 3

Δt

1 Δt t

= ∞, ∞

∞ t0

v

1 r(u)

1

∞

γ

δ(s)Δs

ΔuΔv = 1

u

v

= ∞, ∞

∞

∞

∞

1−

q(s)(φ2 (s))γ = 3 1

t0

1 u3

∞

=3 1

1 s

1

∞

3

Δs

ΔuΔv

u

3 (s + 1)3 Δs

(s 2 − 1)3 Δs s3

= ∞. Consequently any solution of the considered equation is oscillatory or tends to zero. Exercise 10.8. Let T = Z. Prove that any solution of the equation %

t x(t) x(t) + 5

1 2 x(t − 1)Δ 2t + 3

5 &Δ

7

+

(x(t + 2ξ ))5 Δξ = 0,

t ∈ [1, ∞),

1

is oscillatory or tends to zero. Theorem 10.20. Assume (D1)–(D6), (10.23), and (10.24) hold. Suppose that there exist functions H, h ∈ Crd (D) such that H (t, t) = 0,

t ≥ t0 ,

H (t, s) > 0,

t > s ≥ t0 ,

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

543

and H has a nonpositive continuous Δ-partial derivative H Δs (t, s) with respect to the second variable and satisfies H Δs (σ (t), s) + H (σ (t), σ (s))

γ f Δ (s) h(t, s) =− (H (σ (t), σ (s))) γ +1 f (s) f (s)

(10.27)

and for all sufficiently large t1 ∈ [t0 , ∞), there is a T > t1 such that lim sup t→∞

σ (t)

1 H (σ (t), T )

χ (t, s)Δs = ∞,

T

where f is a positive Δ-differentiable function and

χ (t, s) = H (σ (t), σ (s))f σ (s)q(s)

φ2 (s) σ (s)

γ −

(γ

(h− (t, s))γ +1 γ +1 + 1) (β(s)f σ (s)R(s, t1 ))γ

,

t ≥ s ≥ t0 .

Then every solution of the equation (10.22) is oscillatory or tends to zero. Proof. Assume that the equation (10.22) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we suppose that x is an eventually positive solution of the equation (10.22) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

x(φ(t, ξ )) > 0,

t ∈ [t1 , ∞),

ξ ∈ [c, d].

By Theorem 10.15 and Theorem 10.16, it follows that there is a t2 ∈ [t1 , ∞) such that   2 γ Δ r(t) zΔ (t) < 0,

2

zΔ (t) > 0,

t ∈ [t1 , ∞),

and either zΔ (t) > 0 for t ∈ [t2 , ∞) or lim z(t) = lim x(t) = 0.

t→∞

t→∞

Suppose that zΔ (t) > 0, t ∈ [t2 , ∞). Let t3 ∈ [t2 , ∞) be sufficiently large such that (10.27) holds. Consider the generalized Riccati transformation which is used in the proof of Theorem 10.19. We have

f σ (s)q(s)

φ2 (s) σ (s)

γ ≤ −w Δ (s) +

f Δ (s) w(s) f (s)

−γf σ (s)β(s)R(s, t3 )

(w(s))λ , (f (s))λ

s ∈ [t3 , ∞).

544

10 Oscillations of Third-Order Functional Dynamic Equations

Hence,

H (σ (t), σ (s))f σ (s)q(s)

φ2 (s) σ (s)

γ ≤ −H (σ (t), σ (s))w Δ (s) + H (σ (t), σ (s)) −γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

f Δ (s) w(s) f (s)

(w(s))λ , (f (s))λ

t, s ∈ [t3 , ∞),

t ≥ s,

and # σ (t) t3

H (σ (t), σ (s))f σ (s)q(s)



φ2 (s) σ (s)

γ

Δs ≤ −

# σ (t) t3

# σ (t)

H (σ (t), σ (s))wΔ (s)Δs Δ

(s) H (σ (t), σ (s)) ff (s) w(s)Δs # σ (t) λ σ Δs −γ t3 H (σ (t), σ (s))f (s)β(s)R(s, t3 ) (w(s)) (f (s))λ # σ (t) = H (σ (t), t3 )w(t3 ) + t3 H Δs (σ (t), s)w(s)Δs # σ (t) Δ (s) w(s)Δs + t3 H (σ (t), σ (s)) ff (s) # σ (t) λ σ Δs −γ t3 H (σ (t), σ (s))f (s)β(s)R(s, t3 ) (w(s)) (f (s))λ

+

t3

(10.28)

= H (σ (t), t3 )w(t3 )  σ (t) f Δ (s) H Δs (σ (t), s) + H (σ (t), σ (s)) w(s)Δs + f (s) t3 σ (t)

−γ

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

t3

(w(s))λ Δs (f (s))λ

= H (σ (t), t3 )w(t3 ) −

σ (t) t3

γ h(t, s) (H (σ (t), σ (s))) γ +1 w(s)Δs f (s)

σ (t)

−γ

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

t3

(w(s))λ Δs (f (s))λ

≤ H (σ (t), t3 )w(t3 ) +

σ (t) t3

−γ

γ h− (t, s) (H (σ (t), σ (s))) γ +1 w(s)Δs f (s)

σ (t) t3

H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

(w(s))λ Δs, (f (s))λ

10.3 Oscillations of Third-Order Nonlinear Neutral Dynamic Equations with. . .

545

t ∈ [t3 , ∞). Let λ=

γ +1 , γ

Aλ = γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) B λ−1 =

h− (t, s) λ (γf σ (s)β(s)R(s, t

1

(w(s))λ , (f (s))λ

t, s ∈ [t3 , ∞),

,

3 )) λ

t ≥ s.

Using the inequality λAB λ−1 − Aλ ≤ (λ − 1)B λ , we find  1 w(s)  h− (t, s) λ γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) λ f (s) λ (γf σ (s)β(s)R(s, t3 )) λ1 −γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 )

(w(s))λ (f (s))λ

λ

≤ (λ − 1)

(h− (t, s)) λ−1 λ

λ λ−1 (γf σ (s)β(s)R(s, t3 ))λ−1

t, s ∈ [t3 , ∞),

,

t ≥ s,

or γ (w(s))λ w(s) h− (t, s) (H (σ (t), σ (s))) γ +1 − γ H (σ (t), σ (s))f σ (s)β(s)R(s, t3 ) f (s) (f (s))λ

(h− (t, s))γ +1

≤

1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

,

t, s ∈ [t3 , ∞),

t ≥ s.

From here and from (10.28), we obtain σ (t)

H (σ (t), σ (s))f σ (s)q(s)

t3

+

σ (t) t3

φ2 (s) σ (s)

γ Δs ≤ H (σ (t), t3 )w(t3 )

(h− (t, s))γ +1 1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

Δs,

t ∈ [t3 , ∞),

546

10 Oscillations of Third-Order Functional Dynamic Equations

whereupon σ (t)

φ2 (s) H (σ (t), σ (s))f (s)q(s) σ (s) & (h− (t, s))γ +1

t3

−

γ

σ

1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

Δs ≤ H (σ (t), t3 )w(t3 ),

t ∈ [t3 , ∞), and 1 H (σ (t), t3 )

σ (t)

%

H (σ (t), σ (s))f σ (s)q(s)

t3

φ2 (s) σ (s)

γ −

&

(h− (t, s))γ +1 1

(γ + 1)γ +1 (f σ (s)β(s)R(s, t3 )) γ

Δs ≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. Therefore any solution of the equation (10.22) is oscillatory or tends to zero. This completes the proof. Exercise 10.9. Let T = 2Z. Write an analogue of Theorem 10.20.

10.4 Advanced Practical Problems Problem 10.1. Let T = 2Z. Prove that every solution of the equation

 3 Δ 10000 2 t 3 x Δ (t) + (x(t − 2))3 = 0 on t4

[2, ∞)

is oscillatory or tends to zero. Problem 10.2. Let T = 3N0 . Prove that every solution of the equation % &

3  1 Δ 3 t 2 Δ t 3 x (t) t = 0, + t3 x 3

t ∈ [3, ∞),

is oscillatory or tends to zero. Problem 10.3. Let T = 2Z. Write an analogue of Theorem 10.5. Problem 10.4. Let T = 3Z. Write an analogue of Theorem 10.6. Problem 10.5. Using Theorem 10.5 and Theorem 10.6, establish sufficient conditions for oscillation of the solutions of the equation 3

x Δ (t) + p(t)x(t) = 0,

t ∈ [t0 , ∞).

10.4 Advanced Practical Problems

547

Problem 10.6. Using Theorem 10.5 and Theorem 10.6, establish sufficient conditions for oscillation of the solutions of the equation 3

x Δ (t) + p(t)x(τ (t)) = 0,

t ∈ [t0 , ∞).

Problem 10.7. Prove that any solution of the equation ⎛% ⎞

Δ &3 Δ 3 1 ⎝ ⎠ + σ (t) (x(τ (t)))3 = 0, x Δ (t) t (τ (t))3

t ∈ [t0 , ∞),

is oscillatory or tends to zero. Problem 10.8. Let T = 2Z. Write an analogue of Theorem 10.8. Problem 10.9. Write an analogue of Theorem 10.8 in the following cases. 1. α(t) = t 2 , t ∈ T, 2. α(t) = t 3 , t ∈ T, 3. α(t) = t 4 , t ∈ T. Problem 10.10. Prove that any solution of the equation

1 3t 3

Δ3 1 7 1 σ (t) σ (t) x(t) + x(τ0 (t)) (x(τ2 (t))) 3 = 0, + (x(τ1 (t))) 3 + 7 4τ (t) 3τ (t)

t ∈ [t0 , ∞), is oscillatory or tends to zero. Problem 10.11. Let T = 2N0 . Write an analogue of Theorem 10.14. Problem 10.12. Let T = Z. Prove that any solution of the equation %

t x(t) x(t) + 7

1 2 x(t−1)Δ 2 2t + t + 9

7 &Δ

9

+

(x(t+2ξ ))7 Δξ =0, t ∈ [1, ∞),

1

is oscillatory or tends to zero. Problem 10.13. Let T = 2N0 . Write an analogue of Theorem 10.20.

Chapter 11

Oscillations of Fourth-Order Functional Dynamic Equations

The material of this chapter are based on results of the papers and monographs [7, 116–118, 122, 153, 205, 206, 212, 217, 220, 242, 251]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

11.1 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations Consider the equation 4

x Δ (t) + p(t)(x(τ (t))γ = 0,

t ∈ [t0 , ∞),

(11.1)

where γ is a quotient of two positive odd integers, p ∈ Crd (T) is a positive function, τ : T → T, τ ∈ Crd (T), τ (t) ≤ t, t ∈ T, and limt→∞ τ (t) = ∞. We will start our investigations with the following useful lemmas. 2 (T) and there exists a T ∈ [t , ∞) such that Lemma 11.1. Let x ∈ Crd 0

x(t) > 0,

x Δ (t) > 0,

2

x Δ (t) < 0,

t ∈ [T , ∞).

Then for each k ∈ (0, 1) there exists a Tk ∈ [T , ∞) such that τ (t) − T τ (t) x(τ (t)) ≥ ≥k , x(σ (t)) σ (t) − T σ (t)

t ∈ [Tk , ∞),

τ (t) − T τ (t) x(τ (t)) ≥ ≥k , x(t) t −T t

t ∈ [Tk , ∞).

and

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_11

549

550

11 Oscillations of Fourth-Order Functional Dynamic Equations

Proof. Let y(t) = x(t) − (t − T )x Δ (t),

t ∈ [T , ∞).

Then 2

y Δ (t) = x Δ (t) − x Δ (t) − (σ (t) − T )x Δ (t) 2

= −(σ (t) − T )x Δ (t) ≥ 0,

t ∈ [T , ∞).

Then y(t) ≥ y(T ) = x(T ) > 0,

t ∈ [T , ∞).

Hence,

x(t) t −T

Δ =

x Δ (t)(t − T ) − x(t) (t − T )(σ (t) − T )

≤ 0, Therefore such that

x(t) t−T

t ∈ [T , ∞).

is decreasing on [T , ∞). Note that there exists a T1 ∈ [T , ∞) τ (t) ≥ T ,

t ∈ [T1 , ∞).

Then x(τ (t)) x(σ (t)) ≥ , τ (t) − T σ (t) − T

t ∈ [T1 , ∞),

and x(τ (t)) x(t) ≥ , τ (t) − T t −T

t ∈ [T1 , ∞).

From here, τ (t) − T x(τ (t)) ≥ x(σ (t)) σ (t) − T ≥

τ (t) − T , σ (t)

t ∈ [T1 , ∞),

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

551

and x(τ (t)) τ (t) − T ≥ x(t) t −T ≥

τ (t) − T , t

t ∈ [T1 , ∞).

Let k ∈ (0, 1). Then there exists a Tk ∈ [T1 , ∞) such that (1 − k)τ (t) ≥ T ,

t ∈ [Tk , ∞),

τ (t) − T ≥ kτ (t),

t ∈ [Tk , ∞).

or

Consequently x(τ (t)) τ (t) − T τ (t) ≥ ≥k , x(σ (t)) σ (t) − T σ (t)

t ∈ [Tk , ∞),

τ (t) − T τ (t) x(τ (t)) ≥ ≥k , x(t) t −T t

t ∈ [Tk , ∞).

and

This completes the proof.  N   {0} 2N , τ (t) = 2t , x(t) = 1t , and t ∈ T. We have Example 11.1. Let T = 12 σ (t) = 2t, x(t) > 0,

t ∈ T, t ≥ 1,

1 x Δ (t) = − tσ (t) 1 2t 2 < 0, t ≥ 1, 

1 σ (t) + t Δ2 x (t) = 2 t 2 (σ (t))2 

1 2t + t = 2 t 2 (2t)2 =−

552

11 Oscillations of Fourth-Order Functional Dynamic Equations

1 = 2

3t 4t 4



3 8t 3 > 0, t ≥ 1. =

Take T = 5. Note that   x 2t x(τ (t)) = x(σ (t)) x(2t) =

2 t 1 2t

= 4,

t ≥ 5,

and τ (t) − T ≤4 σ (t) − T

⇐⇒

−5 ≤ 4 ⇐⇒ 2t − 5 t − 10 ≤ 4 ⇐⇒ 4t − 10 t − 10 ≤ 16t − 40 t 2

30 ≤ 15t

⇐⇒

⇐⇒

2 ≤ t. Hence, x(τ (t)) τ (t) − T ≥ , x(σ (t)) σ (t) − T

t ≥ 5.

Let k ∈ (0, 1) be arbitrarily chosen. Then τ (t) − T τ (t) ≥k σ (t) − T σ (t)

⇐⇒

t −5 ≥k 2 ⇐⇒ 2t − 5 2t t − 10 k ≥ ⇐⇒ 4t − 10 4 t 2

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

4t − 40 ≥ 4kt − 10k

⇐⇒

4(1 − k)t ≥ 10(4 − k)

⇐⇒

t≥

553

5(4 − k) . 2(1 − k)

Take 

5(4 − k) . Tk ≥ max 5, 2(1 − k) Therefore x(τ (t)) τ (t) − T τ (t) ≥ ≥k , x(σ (t)) σ (t) − T σ (t)

t ≥ Tk .

3 (T) and there exists a T ∈ [t , ∞) such that Lemma 11.2. Assume that x ∈ Crd 0

x(t) > 0,

2

x Δ (t) > 0,

3

x Δ (t) > 0,

x Δ (t) < 0,

t ∈ [T , ∞).

Then lim inf t→∞

tx(t) ≥ 1. h2 (t, t0 )x Δ (t)

Proof. Let G(t) = (t − T )x(t) − h2 (t, T )x Δ (t),

t ∈ [T , ∞).

Then GΔ (t) = (σ (t) − T )x Δ (t) + x(t) 2

−h2 (σ (t), T )x Δ (t) − h1 (t, T )x Δ (t) 2

= (σ (t) − T )x Δ (t) + x(t) − h2 (σ (t), T )x Δ (t) − (t − T )x Δ (t) 2

= μ(t)x Δ (t) + x(t) − h2 (σ (t), T )x Δ (t) % & σ (t)

= x σ (t) − T

2

(τ − T )Δτ x Δ (t),

t ∈ [T , ∞).

554

11 Oscillations of Fourth-Order Functional Dynamic Equations

By Taylor’s formula, we get σ (t)

x σ (t) = x(T ) + h1 (σ (t), T )x Δ (T ) +

2

h1 (σ (t), σ (τ ))x Δ (τ )Δτ

T σ (t)

2

≥ x(T ) + h1 (σ (t), T )x Δ (T ) + x Δ (t)

h1 (σ (t), σ (τ ))Δτ T

= x(T ) + h1 (σ (t), T )x Δ (T ) σ (t)

2

+x Δ (t)

(σ (t) − σ (τ ))Δτ

T

= x(T ) + h1 (σ (t), T )x Δ (T )

!τ =σ (t) 2 ! +x Δ (t)(σ (t) − τ )(τ − T )! τ =T

σ (t)

2

+x Δ (t)

(τ − T )Δτ

T

= x(T ) + h1 (σ (t), T )x Δ (T ) σ (t)

2

+x Δ (t)

(τ − T )Δτ,

t ∈ [T , ∞).

T

Hence, σ (t)

2

x σ (t) − x Δ (t)

(τ − T )Δτ = x(T ) + h1 (σ (t), T )x Δ (t)

T

> 0,

t ∈ [T , ∞),

and t ∈ [T , ∞).

GΔ (t) > 0, Therefore G(t) > G(T ) = 0,

t ∈ [T , ∞).

This implies that (t − T )x(t) > h2 (t, T )x Δ (t),

t ∈ [T , ∞),

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

or (t − T )x(t) > 1, h2 (t, T )x Δ (t)

t ∈ [T , ∞).

Since t (t − T )x(t) tx(t) = , h2 (t, T )x Δ (t) h2 (t, T )x Δ (t) t − T

t ∈ (T , ∞),

we get lim inf t→∞

tx(t) t ≥ lim inf t→∞ t − T h2 (t, T ) = 1.

This completes the proof. Example 11.2. Let T = 2N0



{0}

  1 N 2 t

x(t) = 0

3 (T) is defined as follows , x ∈ Crd

s+1 Δs, s+2

t ∈ T.

We have t ∈ T,

x(t) > 0,

t +1 x Δ (t) = t +2 > 0, t ∈ T, 2

x Δ (t) = = =

t + 2 − (t + 1) (t + 2)(σ (t) + 2) 1 2(t + 1)(t + 2) 2(t 2

> 0, x

Δ3

(t) = − =−

1 + 3t + 2) t ∈ T,

σ (t) + t + 3 2(t + 1)(t + 2)(σ (t) + 1)(σ (t) + 2) 2t + t + 3 2(t + 1)(t + 2)(2t + 1)(2t + 2)

555

556

11 Oscillations of Fourth-Order Functional Dynamic Equations

=−

3(t + 1) 4(t + 1)2 (t + 2)(2t + 1)

=−

3 4(t + 1)(t + 2)(2t + 1)

< 0,

t ∈ T.

Take t0 = 0. Note that h1 (t, 0) = t, t

h2 (t, 0) =

h1 (s, 0)Δs 0 t

=

sΔs 0

=

1 2 t , 3

t ∈ T,

and x(t) ≥

t , 2

t ∈ T.

Then lim inf t→∞

tx(t) ≥ lim inf t→∞ h2 (t, 0)x Δ (t) = lim inf t→∞

t2 t2 3

2

t+1 t+2



3(t + 2) 2(t + 1)

≥ 1. Lemma 11.3. Assume that x is an eventually positive solution of the equation (11.1). Then there exists a T ∈ [t0 , ∞) such that there are only the following two cases. 1. x(t) > 0, 2.

2

x Δ (t) > 0,

2

x Δ (t) > 0,

x Δ (t) > 0,

x Δ (t) > 0,

x Δ (t) > 0,

x Δ (t) < 0,

3

x Δ (t) < 0,

4

3

x Δ (t) < 0,

t ∈ [T , ∞), x(t) > 0, t ∈ [T , ∞).

4

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

557

Proof. Since x is an eventually positive solution of the equation (11.1), there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

Hence from (11.1), we obtain 4

x Δ (t) = −p(t)(x(τ (t)))γ < 0,

t ∈ [t1 , ∞).

Suppose that there exist a t2 ∈ [t1 , ∞) and a constant c < 0 such that 3

x Δ (t) ≤ c,

t ∈ [t2 , ∞).

Then 2

2

x Δ (t) ≤ x Δ (t2 ) + c(t − t2 ) → −∞

as

t → ∞.

Hence, there exist a t3 ∈ [t2 , ∞) and a constant c1 < 0 such that 2

x Δ (t) ≤ c1 ,

t ∈ [t3 , ∞).

From here, x Δ (t) ≤ x Δ (t3 ) + c1 (t − t3 ) → −∞ as

t → ∞.

Then there exist a t4 ∈ [t3 , ∞) and a constant c2 < 0 so that x Δ (t) ≤ c2 ,

t ∈ [t4 , ∞).

Hence, x(t) ≤ x(t4 ) + c2 (t − t4 ) → −∞ as

t → ∞.

This is a contradiction. Therefore 3

x Δ (t) > 0,

t ∈ [t1 , ∞).

558

11 Oscillations of Fourth-Order Functional Dynamic Equations

Suppose that there exist a t5 ∈ [t1 , ∞) and a constant c3 < 0 such that 2

x Δ (t) ≤ c3 ,

t ∈ [t5 , ∞).

Assume that there exist a t6 ∈ [t5 , ∞) and a constant c4 < 0 such that x Δ (t) ≤ c4 ,

t ∈ [t6 , ∞).

Then x(t) ≤ x(t6 ) + c4 (t − t6 ) → −∞ as

t → ∞.

This is a contradiction. Therefore t ∈ [t5 , ∞).

x Δ (t) > 0,

Now we suppose that there exists a t7 ∈ [t1 , ∞) such that 2

x Δ (t) > 0,

t ∈ [t7 , ∞).

Assume that there exist a t8 ∈ [t7 , ∞) and a constant c5 < 0 such that x Δ (t) ≤ c5 ,

t ∈ [t8 , ∞).

Then x(t) ≤ x(t8 ) + c5 (t − t8 ) → −∞ as

t → ∞.

This is a contradiction. Consequently x Δ (t) > 0,

t ∈ [t7 , ∞).

This completes the proof. Example 11.3. Let T = 3N0 . Consider the equation x

Δ4

2080 x (t) + 7 4 4 3 t (t − 81)

 t = 0, 3

t ∈ [9, ∞),

1 x(t) = − + t 3 , t

t ∈ [1, 9].

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

We will show that 1 x(t) = − + t 3 , t

t ∈ T,

is its solution. We have x(t) ≥ 0, x Δ (t) =

1 + (σ (t))2 + tσ (t) + t 2 tσ (t)

1 + (3t)2 + 3t 2 + t 2 3t 2 1 = 2 + 9t 2 + 3t 2 + t 2 3t 1 = 2 + 13t 2 3t > 0, =

2

x Δ (t) = − =−

σ (t) + t + 13(σ (t) + t) 3t 2 (σ (t))2 3t + t + 13(3t + t) 3t 2 (3t)2

4t + 52t 33 t 4 4 − 3 3 + 52t 3 t 0,

 4 (σ (t))2 + tσ (t) + t 2 + 52 33 t 3 (σ (t))3

 4 (3t)2 + 3t 2 + t 2 + 52 33 t 3 (3t)3

 4 9t 2 + 3t 2 + t 2 + 52 33 33 t 6

=− = > 3

x Δ (t) = = =

52 + 52 36 t 4 > 0,

 52 (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 Δ4 x (t) = − 6 3 t 4 (σ (t))4 =

559

560

11 Oscillations of Fourth-Order Functional Dynamic Equations

(3t)3 + t (3t)2 + t 2 (3t) + t 3 t 4 (3t)4

 52 27t 3 + 9t 3 + 3t 3 + t 3 =− 6 3 34 t 8

52 =− 6 3



2080 310 t 5 < 0, t ∈ T,

3

 t t 3 x =− + 3 t 3 =−

t3 3 =− + t 27 =

t 4 − 81 , 27t

t ∈ [9, ∞).

Hence, x

Δ4

2080 x (t) + 7 4 4 3 t (t − 81)



4  2080 t t − 81 2080 = − 10 5 + 7 4 4 3 3 t 3 t (t − 81) 33 t = 0,

t ∈ [9, ∞).

Note that x is an eventually positive solution of the considered equation. Lemma 11.4. Assume that x is an eventually positive bounded solution of the equation (11.1). Then x only satisfies the case 2 of Lemma 11.3. Proof. Assume that x satisfies the case 1 of Lemma 11.3. Then there exist a t1 ∈ [t0 , ∞) and a constant c > 0 such that 2

x Δ (t) ≥ c,

t ∈ [t1 , ∞).

Hence, x Δ (t) ≥ x Δ (t1 ) + c(t − t1 ),

t ∈ [t1 , ∞),

and x(t) ≥ x(t1 ) + x Δ (t1 )(t − t1 ) + c

t

(s − t1 )Δs

t1

→∞

as

t → ∞.

This is a contradiction. Therefore x only satisfies the case 2 of Lemma 11.3. This completes the proof.

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

561

Theorem 11.1. Let γ ≥ 1. Assume that there exist positive functions α, β ∈ 1 ([t , ∞)) such that for some k ∈ (0, 1) and for all constants M, P ∈ (0, ∞) Crd 0 and sufficiently large t1 , for t2 > t1 , t3 > t2 , one has τ (t) > t2 for t ≥ t3 ,



t2 − t1 τ (s) γ α σ (s)p(s) kh2 (τ (s), t2 ) τ (s) − t1 σ (s) t→∞ t3 (11.2)  Δ 2  γ  α+ (s) σ (s) Δs = ∞, − 4γ M γ −1 α σ (s) ks t

lim sup

and t

lim sup t→∞

t1

%

ξ k β (ξ ) σ (ξ ) 2γ

γ

σ

 Δ 2 & (ξ ) (σ (ξ ))γ β+ f (ξ ) − Δξ = ∞, 4γ k γ P γ −1 β σ (ξ )ξ γ (11.3)

where ∞

f (ξ ) = ξ

∞ s

τ (v) p(v) v

γ ΔvΔs

is well defined. Then the equation (11.1) is oscillatory. Proof. Suppose that the equation (11.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we assume that the solution x is eventually positive on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

1. Let x satisfies the case 1 of Lemma 11.3. Define 3

x Δ (t) w(t) = α(t)  2 γ , x Δ (t)

t ∈ [t1 , ∞).

Then w(t) > 0, t ∈ [t1 , ∞). Using the chain rule, we have  2 γ Δ 3 x Δ (t) = γ x Δ (t)

1

2

2

hx Δ (σ (t)) + (1 − h)x Δ (t)

0

 2 γ −1 3 ≥ γ x Δ (t) x Δ (t) ,

t ∈ [t1 , ∞),

γ −1

dh

562

11 Oscillations of Fourth-Order Functional Dynamic Equations

and % &Δ 3 3 x Δ (t) x Δ (t) σ w (t) = α (t)  2 γ + α (t)  2 γ x Δ (t) x Δ (t) Δ

Δ

3

4

x Δ (t) x Δ (t) = α Δ (t)  2 γ + α σ (t)  2 γ x Δ (t) x Δ (σ (t))  2 γ Δ 3 x Δ (t) x Δ (t) −α σ (t)  2 γ  2 γ x Δ (t) x Δ (σ (t)) 3

(11.4)

4

x Δ (t) x Δ (t) ≤ α Δ (t)  2 γ + α σ (t)  2 γ x Δ (t) x Δ (σ (t))  3 2 % &γ 2  2 γ −1 x Δ (t) x Δ (t) σ x Δ (t) −γ α (t)  2γ 2 Δ2 (σ (t)) Δ x x (t) 4

α Δ (t) x Δ (t) w(t) + α σ (t)  2 = γ α(t) x Δ (σ (t))  3 2 % &γ 2  2 γ −1 x Δ (t) x Δ (t) σ x Δ (t) −γ α (t)  ,  2 2γ 2 Δ (σ (t)) Δ x x (t) 2

4

t ∈ [t1 , ∞). Since x Δ (t) > 0 and x Δ (t) < 0, t ∈ [t1 , ∞), we obtain 2

2

2

x Δ (t) > x Δ (t) − x Δ (t1 ) t

=

3

x Δ (s)Δs

t1 3

≥ (t − t1 )x Δ (t),

t ∈ [t1 , ∞).

Hence,

2

x Δ (t) h1 (t,t1 )

Δ

3

2

(t−t1 )x Δ (t)−x Δ (t) (t−t1 )(σ (t)−t1 )

=

< 0,

t ∈ [t1 , ∞).

Take t2 ∈ (t1 , ∞). By (11.5), we obtain 2

2

x Δ (t) x Δ (t2 ) ≥ , h1 (t2 , t1 ) h1 (t, t1 )

t ∈ [t2 , ∞),

(11.5)

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

563

or 2

x Δ (t2 ) ≥

t2 − t1 Δ2 x (t), t − t1

t ∈ [t2 , ∞).

Using Taylor’s formula, we obtain x(t) =

2 

hk (t, t2 )x

Δk

ρ 2 (t)

(t2 ) +

3

h2 (t, σ (η))x Δ (η)Δη

t2

k=0 2

≥ h2 (t, t2 )x Δ (t2 ) t2 − t1 Δ2 x (t), ≥ h2 (t, t2 ) t − t1

t ∈ [t2 , ∞),

or x(t) 2

x Δ (t)

≥ h2 (t, t2 )

t − t2 , t − t1

t ∈ [t2 , ∞).

Hence, x(τ (t))

≥ h2 (τ (t), t2 )

2 x Δ (τ (t))

t 2 − t1 , τ (t) − t1

t ∈ [t2 , ∞).

2

We apply Lemma 11.1 for x Δ and we get that there is a t3 ∈ [t2 , ∞) such that 2

x Δ (τ (t)) x

Δ2

(σ (t))

≥k

τ (t) , σ (t)

≥k

t , σ (t)

2

x Δ (t) 2 x Δ (σ (t))

t ∈ [t3 , ∞).

Then γ

 (x(τ (t))) γ 2 x Δ (σ (t))

=

2

x(τ (t)) x Δ (τ (t)) 2 2 Δ x (τ (t)) x Δ (σ (t))

γ

γ  2 −t1 τ (t) ≥ kh2 (τ (t), t2 ) τt(t)−t , 1 σ (t)

(11.6) t ∈ [t3 , ∞).

Also, there is a positive constant M such that  2 γ −1 x Δ (t) ≥ M γ −1 ,

t ∈ [t3 , ∞).

564

11 Oscillations of Fourth-Order Functional Dynamic Equations

Hence from (11.4), we obtain 

α Δ (t) t2 − t1 τ (t) γ w(t) − α σ (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t) 

γ (w(t))2 γ −1 t −γ α σ (t) M k 2 σ (t) (α(t)) 

t2 − t1 τ (t) γ ≤ −α σ (t)p(t) kh2 (τ (t), t2 ) τ (t) − t1 σ (t) 

α Δ (t) kt γ (w(t))2 + + w(t) − γ M γ −1 α σ (t) α(t) σ (t) (α(t))2

 t2 − t1 τ (t) γ ≤ −α σ (t)p(t) kh2 (τ (t), t2 ) τ (t) − t1 σ (t)  Δ 2 γ

α+ (t) σ (t) + , t ∈ [t3 , ∞), γ −1 σ kt 4γ M α (t)

w Δ (t) ≤

or  Δ 2  

α+ (t) σ (t) γ t2 − t1 τ (t) γ α σ (t)p(t) kh2 (τ (t), t2 ) − ≤ −w Δ (t), τ (t) − t1 σ (t) kt 4γ M γ −1 α σ (t)

t ∈ [t3 , ∞). Then t t3

%

 Δ 2   &

α+ (s) σ (s) γ t2 − t1 τ (s) γ − α σ (s)p(s) kh2 (τ (s), t2 ) Δs ≤ −w(t) + w(t3 ) τ (s) − t1 σ (s) ks 4γ M γ −1 α σ (s) ≤ w(t3 ),

This is a contradiction. 2. Assume that x satisfies the case 2 of Lemma 11.3. Define the function u(t) = β(t)

x Δ (t) , (x(t))γ

t ∈ [t1 , ∞).

Then u(t) > 0, t ∈ [t1 , ∞), and by the chain rule, we get  Δ (x(t))γ ≥ γ (x(t))γ −1 x Δ (t),

t ∈ [t1 , ∞).

t ∈ [t3 , ∞).

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

565

Hence, Δ

Δ x Δ (t) x (t) σ u (t) = β (t) + β (t) (x(t))γ (x(t))γ Δ

Δ

2

= β Δ (t)

x Δ (t) x Δ (t) + β σ (t) γ (x(t)) (x(σ (t)))γ

−β σ (t)

x Δ (t) ((x(t))γ )Δ (x(t))γ (x(σ (t)))γ 2

x Δ (t) x Δ (t) σ ≤ β (t) + β (t) (x(t))γ (x(σ (t)))γ

 γ x(t) β σ (t) −γ (x(t))γ −1 (u(t))2 , (β(t))2 x(σ (t)) Δ

t ∈ [t1 , ∞).

By Lemma 11.1, we obtain that there is a t2 ∈ [t1 , ∞) such that x(t) kt ≥ , x(σ (t)) σ (t) x(τ (t)) τ (t) ≥k , x(t) t

t ∈ [t2 , ∞).

Because x Δ (t) > 0, t ∈ [t2 , ∞), there is a constant P > 0 such that (x(t))γ −1 ≥ P γ −1 ,

t ∈ [t2 , ∞).

Thus, 2

Δ (t) β+ x Δ (t) u(t) u (t) ≤ β (t) + (x(σ (t)))γ β(t)

γ t β σ (t) −γ k γ P γ −1 (u(t))2 , (β(t))2 σ (t) Δ

σ

t ∈ [t2 , ∞).

Next, by the equation (11.1), we obtain 3

z

3

0 = x Δ (z) − x Δ (t) +

p(s)(x(τ (s)))γ Δs

t z

τ (s) ≥ x (z) − x (t) + k (x(t)) p(s) s t 

z τ (s) γ 3 ≥ −x Δ (t) + k γ (x(t))γ p(s) Δs, s t Δ3

Δ3

γ

γ

γ Δs

566

11 Oscillations of Fourth-Order Functional Dynamic Equations

z, t ∈ [t2 , ∞), z > t. By letting z → ∞ in the last inequality, we get 

∞ τ (s) γ Δ3 γ γ −x (t) + k (x(t)) p(s) Δs ≤ 0, t ∈ [t2 , ∞). s t Therefore

∞

z

2

x Δ (t) + k γ (x(t))γ

p(v) t

s

  z ∞ 2 2 τ (v) γ τ (v) γ ΔvΔs ≤ −x Δ (z) + x Δ (t) + k γ (x(t))γ p(v) ΔvΔs v v t s ≤ 0,

t, z ∈ [t2 , ∞), z > t. Hence, letting z → ∞ in the last inequality, we obtain

 ∞ ∞ τ (v) γ 2 p(v) ΔvΔs ≤ 0, x Δ (t) + k γ (x(t))γ v t s t ∈ [t2 , ∞). Thus, x

Δ2

(t) ≤ −k (x(t)) γ

∞

γ t

∞ s

τ (v) p(v) v

γ ΔvΔs,

t ∈ [t2 , ∞), and 2

x Δ (t) (x(σ (t)))γ

γ  (x(t))γ # ∞ # ∞ τ (v) ≤ −k γ (x(σ p(v) ΔvΔs γ  (t))) γ #t #s  v γ ∞ ∞ τ (v) t ΔvΔs, ≤ −k 2γ σ (t) t s p(v) v

(11.7)

t ∈ [t2 , ∞), and

γ Δ (t) σ β+ t γ γ −1 β (t) u(t) − γ k P (u(t))2 u (t) ≤ β(t) (β(t))2 σ (t)

γ ∞ ∞ 

t τ (v) γ σ 2γ −β (t)k p(v) ΔvΔs σ (t) v t s

γ ∞ ∞ 

t τ (v) γ σ 2γ ≤ −β (t)k p(v) ΔvΔs σ (t) v t s  Δ 2 (t) (σ (t))γ β+ , t ∈ [t2 , ∞). + γ γ −1 σ 4γ k P β (t)t γ Δ

From here,

σ

β (t)k

2γ

t σ (t)

≤ −uΔ (t),

γ

∞ t

∞ s

t ∈ [t2 , ∞),

τ (v) p(v) v

γ

 Δ 2 (t) (σ (t))γ β+ ΔvΔs − γ γ −1 σ 4γ k P β (t)t γ

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

567

and t

β σ (s)k 2γ

t2

 2 Δ (s)  γ ∞ ∞

 (σ (s))γ β+ s τ (v) γ Δs p(v) ΔvΔξ Δs − σ (s) v 4γ k γ P γ −1 β σ (s)s γ s ξ

≤ −u(t) + u(t2 ) ≤ u(t2 ),

t ∈ [t2 , ∞).

This is a contradiction. This completes the proof. Example 11.4. Let α(t) = β(t) = 1,

t ∈ T.

Then (11.2) and (11.3) take the form t

lim sup t→∞

t3

t2 − t1 τ (s) p(s) kh2 (τ (s), t2 ) τ (s) − t1 σ (s)

γ Δs = ∞

and ∞

lim sup t→∞

t1

k 2γ

ξ σ (ξ )

γ f (ξ )Δξ = ∞,

respectively, provided that f is well defined. Exercise 11.1. Let T = Z. Write an analogue of Theorem 11.1 in the following cases. 1. 2. 3. 4. 5. 6. 7.

α(t) = 1, β(t) = t, t ∈ T, α(t) = t, β(t) = 1, t ∈ T, α(t) = β(t) = t, t ∈ T, α(t) = t 2 , β(t) = t, t ∈ T, α(t) = 1, β(t) = t 2 , t ∈ T, α(t) = t, β(t) = t 2 , t ∈ T, α(t) = β(t) = t 2 , t ∈ T.

Corollary 11.1. Let γ ≥ 1. Assume that there exists a positive function β ∈ 1 ([t , ∞)) such that, for some k ∈ (0, 1), for all constants P ∈ (0, ∞) Crd 0 and sufficiently large t1 , one has (11.3). Then every bounded solution of the equation (11.1) is oscillatory. Theorem 11.2. Let γ > 1. If for all sufficiently large t1 ∈ [t0 , ∞), for t2 > t1 , and for t3 > t2 , one has τ (t) > t2 for t ∈ [t3 , ∞), ∞ t3



t2 − t1 τ (t) γ σ (t)p(t) h2 (τ (t), t2 ) Δt = ∞ τ (t) − t1 σ (t)

568

11 Oscillations of Fourth-Order Functional Dynamic Equations

and ∞ t1

ξ σ (ξ ) σ (ξ )

γ

∞

∞

ξ

s

∞

∞

τ (v) p(v) v

γ

 ΔvΔs Δξ = ∞,

(11.8)

where

p(v) ξ

s

τ (v) v

γ ΔvΔs

is well defined, then the equation (11.1) is oscillatory. Proof. Suppose that the equation (11.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we assume that the solution x is eventually positive on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

1. Suppose that x satisfies the case 1 of Lemma 11.3. Define the function 3

tx Δ (t) w(t) =  2 γ , x Δ (t)

t ∈ [t1 , ∞).

Then w(t) > 0, t ∈ [t1 , ∞), and

  2 −γ Δ Δ3 w (t) = tx (t) x Δ (t) Δ

 3  2 −γ 4 = x Δ (t) + σ (t)x Δ (t) x Δ (σ (t)) 3

+tx Δ (t)

 −γ Δ 2 x Δ (t)

 2  3 −γ = x Δ (t) − σ (t)p(t)(x(τ (t)))γ x Δ (σ (t)) 3

+tx Δ (t)

 −γ Δ 2 x Δ (t) ,

t ∈ [t1 , ∞).

By the chain rule, we get

 −γ Δ 3 Δ2 x (t) = −γ x Δ (t)

1 0

≤ 0,

2

2

hx Δ (σ (t)) + (1 − h)x Δ (t)

t ∈ [t1 , ∞).

−γ −1

dh

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

569

Hence, 

 2 −γ x(τ (t)) γ 3 − σ (t)p(t) , w Δ (t) ≤ x Δ (t) x Δ (σ (t)) 2 x Δ (σ (t)) t ∈ [t1 , ∞). Again we apply the chain rule and we obtain

 1−γ Δ 2 3 x Δ (t) = (1 − γ )x Δ (t)

1

2

2

2

2

hx Δ (σ (t)) + (1 − h)x Δ (t)

−γ

dh

0 1

3

hx Δ (σ (t)) + (1 − h)x Δ (σ (t))

≤ (1 − γ )x Δ (t)

−γ

dh

0

 2 −γ 3 , = (1 − γ )x Δ (t) x Δ (σ (t))

t ∈ [t1 , ∞),

whereupon x

Δ3

 2 −γ (t) x Δ (σ (t)) ≤

1 1−γ

 1−γ Δ Δ2 x (t) ,

t ∈ [t1 , ∞).

Therefore 1 w (t) ≤ 1−γ Δ

 

1−γ Δ x(τ (t)) γ Δ2 x (t) − σ (t)p(t) , 2 x Δ (σ (t))

t ∈ [t1 , ∞). Hence from (11.6), we get that there exist t2 , t3 ∈ [t1 , ∞), t3 > t2 , such that τ (t) > t2 , t ∈ [t3 , ∞), and

x(τ (t)) 2

x Δ (σ (t))

γ



t2 − t1 τ (t) γ ≥ kh2 (τ (t), t2 ) , τ (t) − t1 σ (t)

t ∈ [t3 , ∞).

Consequently 1 w (t) ≤ 1−γ Δ

 1−γ Δ Δ2 x (t)



t2 − t1 τ (t) γ −σ (t)p(t) kh2 (τ (t), t2 ) , τ (t) − t1 σ (t)

t ∈ [t3 , ∞), or 



1−γ Δ t2 − t1 τ (t) γ 1 Δ Δ2 x (t) σ (t)p(t) kh2 (τ (t), t2 ) ≤ −w (t)+ , τ (t) − t1 σ (t) 1−γ

570

11 Oscillations of Fourth-Order Functional Dynamic Equations

t ∈ [t3 , ∞). From here, 

t − t1 τ (s) γ σ (s)p(s) kh2 (τ (s), t2 ) 2 Δs ≤ −w(t) + w(t3 ) τ (s) − t1 σ (s) t3

 1−γ  2 1−γ  2 1 + x Δ (t) − x Δ (t3 ) 1−γ 1−γ  2 x Δ (t3 ) ≤ w(t3 ) + , γ −1 t

t ∈ [t3 , ∞). This is a contradiction. 2. Suppose that x satisfies the case 2 of Lemma 11.3. Define u(t) =

tx Δ (t) , (x(t))γ

t ∈ [t1 , ∞).

Then u(t) > 0, t ∈ [t1 , ∞) and using the chain rule, we obtain Δ  = −γ x Δ (t) (x(t))−γ

1

(hx(σ (t)) + (1 − h)x(t))−γ −1 dh

0

t ∈ [t1 , ∞).

≤ 0, Then

Δ  Δ  tx (t) (x(t))−γ  2 = x Δ (t) + σ (t)x Δ (t) (x(σ (t)))−γ

uΔ (t) =

Δ  +tx Δ (t) (x(t))−γ  2 ≤ x Δ (t) + σ (t)x Δ (t) (x(σ (t)))−γ 2

−γ

= x (t)(x(σ (t))) Δ

x Δ (t) + σ (t) , (x(σ (t)))γ

t ∈ [t1 , ∞). Applying the chain rule, we get  Δ (x(t))1−γ = (1 − γ )x Δ (t)

1

(hx(σ (t)) + (1 − h)x(t))−γ dh

0 1

≤ (1 − γ )x Δ (t)

(hx(σ (t)) + (1 − h)x(σ (t)))−γ dh

0

= (1 − γ )x (t)(x(σ (t)))−γ , Δ

t ∈ [t1 , ∞),

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

571

whereupon −γ

Δ

x (t)(x(σ (t)))

Δ  (x(t))1−γ ≤ , 1−γ

t ∈ [t1 , ∞).

Therefore Δ  2 (x(t))1−γ x Δ (t) + σ (t) u (t) ≤ , 1−γ (x(σ (t)))γ Δ

t ∈ [t1 , ∞). By (11.7), it follows that there exists a t2 ∈ [t1 , ∞) such that 2

x Δ (t) ≤ −k 2γ (x(σ (t)))γ

t σ (t)

γ

∞

∞

t

s

τ (v) p(v) v

γ ΔvΔs,

t ∈ [t2 , ∞).

Consequently Δ 

γ (x(t))1−γ t − σ (t)k 2γ u (t) ≤ 1−γ σ (t)

∞

Δ

∞

p(v) t

s

τ (v) v

γ ΔvΔs,

t ∈ [t2 , ∞), and

σ (t)k

2γ

t σ (t)

γ

∞ t

∞ s

τ (v) p(v) v

γ

Δ  (x(t))1−γ , ΔvΔs ≤ −u (t) + 1−γ Δ

t ∈ [t2 , ∞). Hence,

t

σ (s)k

2γ

t2

s σ (s)

γ

≤ −u(t) + u(t2 ) + ≤ u(t2 ) +

∞ s

∞ v

τ (z) p(z) z

γ ΔzΔvΔs

(x(t))1−γ (x(t2 ))1−γ − γ −1 γ −1

(x(t2 ))1−γ , γ −1

t ∈ [t2 , ∞).

This is a contradiction. This completes the proof. Corollary 11.2. Let γ > 1 and (11.8) holds for all sufficiently large t1 ∈ [t0 , ∞). Then every bounded solution of the equation (11.1) is oscillatory. Example 11.5. Let T = 2N0 . Consider the equation x

Δ4

1 (t) + 3 t

3 t = 0, x 2

t ∈ [2, ∞).

572

11 Oscillations of Fourth-Order Functional Dynamic Equations

Here γ = 3, σ (t) = 2t, 1 , t3 t τ (t) = , t ∈ T. 2

p(t) =

Let 4 , 3t 2 2 f (t) = − , t ∈ T. t g(t) = −

Then g Δ (t) =

4 σ (t) + t 3 t 2 (σ (t))2

4 3t 3 4t 4 1 = 3, t 1 f Δ (t) = 2 tσ (t) =

1 2t 2 1 = 2 , t ∈ T. t =2

Then ∞

∞

p(v) ξ

s

τ (v) v

γ

∞

ΔvΔs = ξ

=

1 8

1 = 8

∞ s ∞

ξ

∞ s

∞ ξ

1 v3

∞ s

v 3 2

v

ΔvΔs

1 ΔvΔs v3 g Δ (v)ΔvΔs

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

=

1 8

∞

!v=∞ ! g(v)! Δs v=s

ξ

 4 !!v=∞ 1 ∞ − 2 ! = Δs v=s 8 ξ 3v = =

1 6 1 6

∞ ξ ∞

1 Δs s2 f Δ (s)Δs

ξ

!s=∞ 1 ! = f (s)! s=ξ 6

 1 2 !!s=∞ = − ! 6 s s=ξ =

1 , 3ξ

ξ ∈ [2, ∞),

and γ ξ σ (ξ ) t1

3 % ∞ ξ = (2ξ ) 2ξ t1 

1 ∞ 1 Δξ = ξ 4 t1 3ξ

∞

∞

∞

σ (ξ )

=

1 12

= ∞,

  τ (v) γ ΔvΔs Δξ v &

v 3 1 2 ΔvΔs Δξ v3 v

p(v) ξ

s ∞

ξ

∞ s

∞

Δξ t1

t1 ∈ [2, ∞).

For t2 ∈ [2, ∞), we define h(t) =

1 2 t − t2 t, 3

t ∈ T.

Then 1 (σ (t) + t) − t2 3 1 = (2t + t) − t2 3 = t − t2 , t ∈ T.

hΔ (t) =

573

574

11 Oscillations of Fourth-Order Functional Dynamic Equations

Hence, h1 (t, t2 ) = t − t2 , t

h2 (t, t2 ) =

(s − t2 )Δs

t2 t

=

hΔ (s)Δs

t2

!s=t ! = h(s)!

=

s=t2

! 1 2 !s=t s − t2 s ! s=t2 3

1 2 t − tt2 − 3 1 = t 2 − tt2 + 3 =

h2

t , t2 2



=

1 2 t + t22 32 2 2 t , 32

t2 1 2 − tt2 + t22 , 12 2 3

t ∈ [2, ∞),

and 

t2 − t1 τ (t) γ σ (t)p(t) h2 (τ (t), t2 ) Δt τ (t) − t1 σ (t) t3 &3 3 %

 2 ∞ t2 − t1 2t t 1 1 2 2 − tt2 + t2 = 2t 3 Δt t 12 2 3 2t t t3 2 − t1 ∞

1 t 1 4 t3 t 2 = ∞. =

t2 1 2 − tt2 + t22 12 2 3

3

t2 − t1 t − 2t1

3 Δt

Therefore the considered equation is oscillatory. Exercise 11.2. Let T = 3N0 . Prove that the equation 4

x Δ (t) +

2 t4

5 t x = 0, 9

t ∈ [9, ∞),

is oscillatory. 1 ([t , ∞)), τ : T → T, τ Δ > 0 Theorem 11.3. Let γ ≥ 1. Suppose that τ ∈ Crd 0 on [t0 , ∞), τ (t) ≤ t, t ∈ T, limt→∞ τ (t) = ∞. Assume that there exist positive

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

575

1 ([t , ∞)) such that for some k ∈ (0, 1), for all constants functions α, β ∈ Crd 0 M, P ∈ (0, ∞) and sufficiently large t1 , for t2 > t1 , one has (11.3) and t

lim sup t→∞

t2

%

&  Δ 2 (s) σ (s) α+ α(s)p(s) − 2 Δs = ∞. 4k γ M γ −1 τ Δ (s)α(s)h2 (τ (s), t0 )τ (s) (11.9)

Then the equation (11.1) is oscillatory. Proof. Suppose that the equation (11.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we assume that the solution x is eventually positive on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x(τ (t)) > 0,

1. Let x satisfies the case 1 of Lemma 11.3. Define the function w(t) =

α(t) 3 x Δ (t), (x(τ (t)))γ

t ∈ [t1 , ∞).

Then w(t) > 0, t ∈ [t1 , ∞), and w Δ (t) = =



α(t) (x(τ (t)))γ

Δ

3

x Δ (σ (t)) +

α(t) (x(τ (t)))γ

4

x Δ (t)

3

α(t) α Δ (t)x Δ (σ (t)) Δ4 (x(τ (t)))γ x (t) + (x(τ (σ (t))))γ 3 x Δ (σ (t))((x(τ (t)))γ )Δ −α(t) (x(τ (t)))γ (x(τ (σ (t))))γ ,

(11.10)

t ∈ [t1 , ∞). By the chain rule, we get ((x(τ (t)))γ )Δ = γ x Δ (τ (t))τ Δ (t) #1 × 0 (hx(τ (σ (t))) + (1 − h)x(τ (t)))γ −1 dh ≥ γ (x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t), t ∈ [t1 , ∞).

(11.11)

By Lemma 11.2, it follows that there exists a t2 ∈ (t1 , ∞) such that x Δ (t) 2 x Δ (t)

≥k

h2 (t, t0 ) , t

t ∈ [t2 , ∞),

for any k ∈ (0, 1). Note that 2

2

2

x Δ (t) ≥ x Δ (t) − x Δ (t1 ) =

t t1

3

x Δ (s)Δs

(11.12)

576

11 Oscillations of Fourth-Order Functional Dynamic Equations 3

≥ (t − t1 )x Δ (t) 3

≥ ktx Δ (t),

t ∈ [t2 , ∞),

for some k ∈ (0, 1). Hence from (11.12), we obtain x Δ (t) 3 x Δ (t)

= ≥ =

2

x Δ (t) x Δ (t) 2 3 x Δ (t) x Δ (t) h2 (t,t0 ) k t (kt) k 2 h2 (t, t0 ),

(11.13) t ∈ [t2 , ∞).

Next, %

3

x Δ (t) t

&Δ

4

3

tx Δ (t) − x Δ (t) = tσ (t) < 0,

t ∈ [t2 , ∞).

Therefore 3

3

x Δ (τ (t)) x Δ (σ (t)) ≥ , τ (t) σ (t)

t ∈ [t2 , ∞),

or 3

x Δ (τ (t)) 3 x Δ (σ (t))

≥

τ (t) , σ (t)

t ∈ [t2 , ∞).

Let M be a positive constant such that (x(τ (σ (t))))γ −1 ≥ M γ −1 ,

t ∈ [t2 , ∞).

Hence from (11.1), (11.10), (11.11), (11.13), and (11.14), we obtain 3

wΔ (t) ≤ −p(t)α(t) +

α Δ (t)x Δ (σ (t)) (x(τ (σ (t))))γ

x Δ (σ (t))(x(τ (t)))γ −1 x Δ (τ (t))τ Δ (t) −γ α(t) (x(τ (t)))γ (x(τ (σ (t))))γ 3

= −p(t)α(t) + 3

α Δ (t) σ w (t) α σ (t)

x Δ (σ (t))x Δ (τ (t))τ Δ (t) −γ α(t) x(τ (t))(x(τ (σ (t))))γ

(11.14)

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

= −p(t)α(t) + −γ

α Δ (t) σ w (t) α σ (t)

 σ 2 (x(τ (σ (t))))γ x Δ (τ (t)) Δ τ (t) w (t) 3 (α σ (t))2 x(τ (t))x Δ (σ (t)) α(t)

≤ −p(t)α(t) + −γ

α Δ (t) σ w (t) α σ (t)

α(t) σ x Δ (τ (t))τ Δ (t) w (t) α σ (t) x(τ (t))

= −p(t)α(t) + −γ

577

α Δ (t) σ w (t) α σ (t)

 σ 2 (x(τ (σ (t))))γ −1 x Δ (τ (t)) Δ τ (t) w (t) 3 (α σ (t))2 x Δ (σ (t)) α(t)

≤ −p(t)α(t) + −γ k 2

α Δ (t) σ w (t) α σ (t)

h2 (τ (t), t0 )τ (t) γ −1 Δ α(t)  σ 2 M w (t) τ (t) σ (t) (α σ (t))2

≤ −p(t)α(t) +

Δ (t) α+ w σ (t) α σ (t)

h2 (τ (t), t0 )τ (t) γ −1 Δ α(t)  σ 2 M w (t) τ (t) σ (t) (α σ (t))2  Δ 2 (t) σ (t) α+ , ≤ −p(t)α(t) + 2 4k γ M γ −1 τ Δ (t)α(t)h2 (τ (t), t0 )τ (t) −γ k 2

t ∈ [t2 , ∞), and  Δ 2 (t) σ (t) α+ ≤ −w Δ (t), p(t)α(t) − 2 γ −1 Δ 4k γ M τ (t)α(t)h2 (τ (t), t0 )τ (t) t ∈ [t2 , ∞). Then t t2

%

&  Δ 2 (s) σ (s) α+ p(s)α(s) − 2 Δs ≤ −w(t) + w(t2 ) 4k γ M γ −1 τ Δ (s)α(s)h2 (τ (s), t0 )τ (s) ≤ w(t2 ),

t ∈ [t2 , ∞). This is a contradiction.

578

11 Oscillations of Fourth-Order Functional Dynamic Equations

2. The proof of the case 2 of Lemma 11.3 is the same as the proof of the case 2 of Theorem 11.1. Therefore we leave it to the reader as an exercise. This completes the proof. 1 ([t , ∞)), τ : T → T, τ Δ > 0 on Corollary 11.3. Let γ ≥ 1. Suppose that τ ∈ Crd 0 [t0 , ∞), τ (t) ≤ t, t ∈ T, limt→∞ τ (t) = ∞. If for some k ∈ (0, 1), for sufficiently large t1 ∈ [t0 , ∞), t2 ∈ (t1 , ∞),

t

lim sup t→∞

k t1

2γ

γ

ξ σ (ξ )

∞

∞

ξ

s

τ (v) p(v) v

γ ΔvΔsΔξ = ∞,

where ∞

∞

p(v) ξ

s

τ (v) v

γ ΔvΔs

is well defined, and t

lim sup t→∞

p(s)Δs = ∞,

t2

then the equation (11.1) is oscillatory. Proof. We take α(t) = 1, t ∈ T, in (11.9), and β(t) = 1, t ∈ T, in (11.3). This completes the proof. Example 11.6. Let T = R+ . Consider the equation  √ 5 t = 0, x (4) (t) + t x

t ∈ R+ .

Here p(t) = t, √ τ (t) = t, σ (t) = t,

t ∈ T,

γ = 5. Then t

lim sup t→∞

t2

p(s)ds = lim sup t→∞

= lim sup t→∞

t

sds t2

s 2 !!s=t ! 2 s=t2

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

%

t2 t2 − 2 = lim sup 2 2 t→∞

579

&

=∞ and t

lim sup t→∞

k 2γ

t1

= lim sup t→∞

t

t→∞

t→∞

t1

∞ ξ

ξ

s

∞ s

∞ ξ

∞

∞ ξ

τ (v) v

γ ΔvΔsΔξ

√ 5 v dvdsdξ v

1 3

dvdsdξ

v2

s

2k 10

∞

p(v)

v

t t1

t

γ

∞

k 10

t1

= lim sup k 10 = lim sup

ξ σ (ξ )

1 1

dsdξ

s2

= ∞. Therefore the considered equation is oscillatory. Exercise 11.3. Let T = R+ . Prove that the equation  √ 3 = 0, x (4) (t) + t 3 x 3 t

t ∈ R+ ,

is oscillatory. Exercise 11.4. Write an analogue of Theorem 11.3 in the following cases. 1. 2. 3. 4. 5.

α(t) = 1, β(t) = t, t ∈ T, α(t) = β(t) = t, t ∈ T, α(t) = t, β(t) = 1, t ∈ T, α(t) = t, β(t) = t 2 , t ∈ T, α(t) = β(t) = t 2 , t ∈ T.

Theorem 11.4. Let γ ≤ 1. Assume that there exist positive functions α, β ∈ 1 ([t , ∞)) such that for some k ∈ (0, 1) and for all constants M, P ∈ (0, ∞) Crd 0 and sufficiently large t1 , for t2 > t1 , t3 > t2 , one has τ (t) > t2 for t ≥ t3 ,



t2 − t1 τ (s) γ σ α (s)p(s) kh2 (τ (s), t2 ) lim sup τ (s) − t1 σ (s) t→∞ t3  Δ 2 γ 

α+ (s) σ (s) Δs = ∞, − γ −1 σ ks 4γ M α (s) t

(11.15)

580

11 Oscillations of Fourth-Order Functional Dynamic Equations

and t

lim sup t→∞

t1

%

ξ k 2γ β σ (ξ ) σ (ξ )

 Δ 2 & (ξ ) σ γ (ξ ) β+ f (ξ ) − Δξ = ∞, 4γ k γ P γ −1 β σ (ξ )ξ γ (11.16)

γ

where ∞

f (ξ ) = ξ

∞ s

τ (v) p(v) v

γ ΔvΔs

is well defined. Then the equation (11.1) is oscillatory. Proof. Suppose that the equation (11.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, we assume that the solution x is eventually positive on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

1. Let x satisfies the case 1 of Lemma 11.3. Define 3

x Δ (t) w(t) = α(t)  2 γ , x Δ (t)

t ∈ [t1 , ∞).

Then w(t) > 0, t ∈ [t1 , ∞). Using the chain rule, we have  2 γ Δ 3 = γ x Δ (t) x Δ (t)

1

2

2

hx Δ (σ (t)) + (1 − h)x Δ (t)

0

 2 γ −1 3 ≥ γ x Δ (t) x Δ (σ (t)) ,

t ∈ [t1 , ∞).

Note that there exists a positive constant M1 such that 3

x Δ (t) ≤ M1 ,

t ∈ [t1 , ∞).

Then, using that 2

2

x Δ (t) = x Δ (t1 ) +

t

3

x Δ (s)Δs,

t ∈ [t1 , ∞),

t1

we conclude that there exists a constant M > 0 such that 2

x Δ (t) ≤ Mt,

t ∈ [t1 , ∞).

γ −1

dh

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

581

Using (11.4) and (11.6), we have w Δ (t) ≤

≤

≤

≤

≤

≤



α Δ (t) t2 − t1 τ (t) γ w(t) − α σ (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t)  γ Δ 3 2 x Δ (t) x Δ (t) −α σ (t)  2 γ  2 γ x Δ (t) x Δ (σ (t)) 

α Δ (t) t2 − t1 τ (t) γ w(t) − α σ (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t)  3 2 x Δ (t) σ −γ α (t)  2 γ  2  x Δ (t) x Δ (σ (t)) 

t2 − t1 τ (t) γ α Δ (t) w(t) − α σ (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t) % &γ 2  2 γ −1 α σ (t) x Δ (t) 2 Δ x −γ (w(t)) (σ (t)) 2 (α(t))2 x Δ (σ (t)) 

α Δ (t) t2 − t1 τ (t) γ σ w(t) − α (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t)

 γ  2 γ −1 kt α σ (t) x Δ (σ (t)) −γ (w(t))2 2 σ (t) (α(t)) 

Δ (t) α+ t2 − t1 τ (t) γ w(t) − α σ (t)p(t) kh2 (τ (t), t2 ) α(t) τ (t) − t1 σ (t)

 γ kt α σ (t) −γ (w(t))2 M γ −1 (σ (t))γ −1 σ (t) (α(t))2 

t2 − t1 τ (t) γ −α σ (t)p(t) kh2 (τ (t), t2 ) τ (t) − t1 σ (t)  Δ 2 

α+ (t) σ (t) γ + , t ∈ [t3 , ∞), kt 4γ (Mσ (t))γ −1 α σ (t)

or  Δ 2

 

α+ (t) σ (t) γ t2 − t1 τ (t) γ α (t)p(t) kh2 (τ (t), t2 ) − ≤ −w Δ (t), τ (t) − t1 σ (t) kt 4γ (Mσ (t))γ −1 α σ (t) σ

582

11 Oscillations of Fourth-Order Functional Dynamic Equations

t ∈ [t3 , ∞). Then ⎛

⎞ 2  Δ (s)  

α+ σ (s) γ ⎟ t2 − t1 τ (s) γ ⎜ σ − ⎝α (s)p(s) kh2 (τ (s), t2 ) ⎠ Δs ≤ −w(t) + w(t3 ) τ (s) − t1 σ (s) 4γ (Mσ (s))γ −1 α σ (s) ks t3 t

≤ w(t3 ),

t ∈ [t3 , ∞).

This is a contradiction. 2. Let x satisfies the case 2 of Lemma 11.3. Define u(t) = β(t)

x Δ (t) , (x(t))γ

t ∈ [t1 , ∞).

By the chain rule, we get  Δ (x(t))γ ≥ γ (x(σ (t)))γ −1 x Δ (t),

t ∈ [t1 , ∞).

Then, using the proof of Theorem 11.1, we obtain uΔ (t) ≤

β Δ (t) u(t) − β σ (t)k 2γ β(t)

t σ (t)

γ f (t)

x Δ (t) ((x(t))γ )Δ (x(t))γ (x(σ (t)))γ

γ t β Δ (t) u(t) − β σ (t)k 2γ ≤ f (t) β(t) σ (t)  Δ 2 x (t) (x(σ (t)))γ −1 σ −γβ (t) (x(t))γ (x(σ (t)))γ γ

t β Δ (t) σ 2γ u(t) − β (t)k = f (t) β(t) σ (t)

 x(t) γ β σ (t) −γ (x(σ (t)))γ −1 (u(t))2 (β(t))2 x(σ (t))

γ t β Δ (t) u(t) − β σ (t)k 2γ ≤ f (t) β(t) σ (t)

 β σ (t) kt γ −γ (x(σ (t)))γ −1 (u(t))2 , t ∈ [t2 , ∞). (β(t))2 σ (t) −β σ (t)

Note that there exists a constant P1 such that x Δ (t) ≤ P1 ,

t ∈ [t1 , ∞).

11.1 Oscillations of Fourth-Order Nonlinear Delay DynamicEquations

583

Hence from x(t) = x(t1 ) +

t

x Δ (s)Δs,

t ∈ [t1 , ∞),

t1

we conclude that there is a positive constant P such that x(t) ≤ P t,

t ∈ [t1 , ∞).

Then

γ t f (t) σ (t)

 kt γ β σ (t) γ −1 −γ (P σ (t)) (u(t))2 σ (t) (β(t))2

γ t ≤ −β σ (t)k 2γ f (t) σ (t)  Δ 2 (t) (σ (t))γ β+ , t ∈ [t2 , ∞). + γ γ −1 4γ k (P σ (t)) β σ (t)t γ

uΔ (t) ≤

Δ (t) β+ u(t) − β σ (t)k 2γ β(t)

Thus

σ

β (t)k

2γ

t σ (t)

γ

 Δ 2 (t) (σ (t))γ β+ f (t) − ≤ −uΔ (t), γ γ −1 4γ k (P σ (t)) β σ (t)t γ

t ∈ [t2 , ∞). Hence, ⎛ t t2

⎜ σ 2γ ⎝β (s)k

⎞  2 Δ (s) γ (σ (s))γ β+ s ⎟ f (s) − ⎠ Δs ≤ −u(t) + u(t2 ) σ (s) 4γ k γ (P σ (s))γ −1 β σ (s)s γ ≤ u(t2 ),

t ∈ [t2 , ∞).

This is a contradiction. This completes the proof. Corollary 11.4. Let γ ≤ 1. Assume that there exists a positive function β ∈ 1 ([t , ∞)) such that, for some k ∈ (0, 1), for all constants P ∈ (0, ∞) Crd 0 and sufficiently large t1 , one has (11.16). Then every bounded solution of the equation (11.1) is oscillatory. Example 11.7. Let α(t) = β(t) = 1,

t ∈ T.

584

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then (11.15) and (11.16) take the form t

lim sup t→∞

t3



t2 − t1 τ (s) γ p(s) kh2 (τ (s), t2 ) Δs = ∞ τ (s) − t1 σ (s)

and

∞

lim sup t→∞

k

2γ

t1

ξ σ (ξ )

γ f (ξ )Δξ = ∞,

respectively, provided that f is well defined. Exercise 11.5. Let T = 2N0 . Investigate if the equation x

Δ4

 1 5 t (t) + t x = 0, 2 5

t ∈ [2, ∞),

is oscillatory. Exercise 11.6. Let T = hZ, h > 0. Write an analogue of Theorem 11.4 in the following cases. 1. 2. 3. 4. 5.

α(t) = 1, β(t) = t, t ∈ T, α(t) = β(t) = t, t ∈ T, α(t) = t, β(t) = 1, t ∈ T, α(t) = t, β(t) = t 2 , t ∈ T, α(t) = β(t) = t 2 , t ∈ T.

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a Nonlinear Middle Term In this section we investigate the equation 

 3 Δ  2 a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t)) + q(t)φβ (x(τ2 (t))) = 0,

t ∈ [t0 , ∞), where (E1) (E2)

φλ (u) = |u|λ sign(u), λ > 0, β ≤ α, a, p, q ∈ Crd ([t0 , ∞)) are positive functions and ∞ t0

1

(a(s))− α Δs = ∞,

(11.17)

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

(E3)

585

1 (T), τ Δ (t) > 0, τ Δ (t) > 0, t ∈ T, τ (t) ≤ t, τ1 , τ2 : T → T, τ1 , τ2 ∈ Crd 1 1 2 τ2 (t) ≤ t, t ∈ T, limt→∞ τ1 (t) = limt→∞ τ2 (t) = ∞.

Consider the following inequality  Δ  a(t)φα x Δ (t) + p(t)φβ (x(τ1 (t))) ≤ 0,

t ∈ [t0 , ∞),

(11.18)

t ∈ [t0 , ∞),

(11.19)

and the following equation  Δ  + p(t)φβ (x(τ1 (t))) = 0, a(t)φα x Δ (t)

where α and β are positive numbers, a, p satisfies (E2), and τ1 satisfies (E3). Theorem 11.5. Suppose (E1)–(E3). If the inequality (11.18) has an eventually positive solution, then the equation (11.19) has an eventually positive solution. Proof. Let x is an eventually positive solution of the inequality (11.18) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞),

and τ1 (t) ≥ t0 ,

t ∈ [t1 , ∞).

Note that  Δ  ≤ −p(t)φβ (x(τ1 (t))) a(t)φα x Δ (t) < 0,

t ∈ [t1 , ∞).

  Then a(t)φα x Δ (t) is strictly decreasing on [t1 , ∞). Assume that x Δ (t) ≤ 0,

t ∈ [t1 , ∞).

Then there is a t2 ∈ [t1 , ∞) and a constant c > 0 such that !α ! − !x Δ (t)! a(t) ≤ −c,

t ∈ [t2 , ∞),

or ! Δ !α !x (t)! ≥ c , a(t)

t ∈ [t2 , ∞),

586

11 Oscillations of Fourth-Order Functional Dynamic Equations

whereupon

1

−x Δ (t) ≥ c α

1 a(t)

1

α

,

t ∈ [t2 , ∞),

,

t ∈ [t2 , ∞).

or x (t) ≤ −c Δ

1 α

1

1 a(t)

α

Hence, t

1

x(t) − x(t2 ) ≤ −c α

t2

→ −∞,

1 a(s)

1

α

Δs

t → ∞.

as

This is a contradiction. Therefore there exists a t3 ∈ [t1 , ∞) such that t ∈ [t3 , ∞).

x Δ (t) > 0, Let

  y(t) = a(t)φα x Δ (t) ,

t ∈ [t0 , ∞).

We take t4 ∈ [t3 , ∞) such that τ (t) ≥ t3 , t ∈ [t4 , ∞). Then t ∈ [t4 , ∞),

y(t) > 0, and x (t) = Δ

φα−1

 y(t) , a(t)

t ∈ [t0 , ∞),

and hence, x(t) = x(t0 ) +

t t0

φα−1

 y(s) Δs, a(s)

t ∈ [t0 , ∞).

Then the inequality (11.18) takes the form % y (t) + p(t)φβ x(t0 ) + Δ

τ1 (t) t0

φα−1

 & y(s) Δs ≤ 0, a(s)

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

587

t ∈ [t4 , ∞). Hence, for v ≥ t ≥ t4 , we get v

0 ≥ y(v) − y(t) + t v

≥ −y(t) +

% p(s)φβ x(t0 ) +

τ1 (s) t0

%

τ1 (s)

p(s)φβ x(t0 ) +

t

t0

φα−1

φα−1

 & y(z) Δz Δs a(z)  &

y(z) Δz Δs. a(z)

From here, ∞

y(t) ≥

%

τ1 (s)

p(s)φβ x(t0 ) +

t

t0

φα−1

 & y(z) Δz Δs, a(z)

t ∈ [t4 , ∞). Let ∞

G(t, y(t)) =

% p(s)φβ x(t0 ) +

t

τ1 (s) t0

φα−1

 & y(z) Δz Δs, a(z)

Then y(t) ≥ G(t, y(t)),

t ∈ [t4 , ∞).

Define the sequence {wj }j ∈N in the following way. w0 (t) = y(t), wj +1 (t) = G(t, wj (t)),

j ∈ N0 ,

t ∈ [t4 , ∞).

We have w1 (t) = G(t, w0 (t)) = G(t, y(t)) ≤ y(t),

t ∈ [t4 , ∞).

Hence, w2 (t) ≤ G(t, w1 (t)) ≤ w1 (t) ≤ y(t),

t ∈ [t4 , ∞).

t ∈ [t4 , ∞).

588

11 Oscillations of Fourth-Order Functional Dynamic Equations

Assume that wj +1 (t) ≤ wj (t) ≤ y(t),

t ∈ [t4 , ∞),

for some j ∈ N, j ≥ 2. We will prove that wj +2 (t) ≤ wj +1 (t) ≤ y(t),

t ∈ [t4 , ∞).

Really, we have wj +2 (t) = G(t, wj +1 (t)) ≤ wj +1 (t) ≤ wj (t) ≤ y(t),

t ∈ [t4 , ∞).

Therefore {wj }j ∈N is nonincreasing and bounded on [t4 , ∞). Then there exists w(t) = lim wj (t), j →∞

t ∈ [t4 , ∞).

We have w(t) ≥ 0,

t ∈ [t4 , ∞),

and t

wj (s)Δs ≤

t4

t

y(s)Δs,

t ∈ [t4 , ∞).

t4

Also, by the Lebesgue dominated theorem, it follows that w(t) = G(t, w(t)),

t ∈ [t4 , ∞).

Hence,   # τ (t) w Δ (t) = −p(t)φβ x(t0 ) + t01 φα−1 w(s) a(s) Δs = −p(t)φβ (m(τ1 (t))),

t ∈ [t4 , ∞),

where m(t) = x(t0 ) +

t t0

φα−1

 w(s) Δs, a(s)

t ∈ [t4 , ∞).

(11.20)

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

589

We have m(t) > 0,

  a(t)φα mΔ (t) = w(t),

t ∈ [t4 , ∞).

From here and from (11.20), we obtain   Δ a(t)φα mΔ (t) + p(t)φβ (m(τ1 (t))) = 0,

t ∈ [t4 , ∞).

This completes the proof. Suppose (E4)

Q ∈ Crd ([t0 , ∞)) is a positive function

and consider the equation   Δ a(t)φα x Δ (t) = Q(t)φβ (x(τ1 (t)),

(11.21)

where (E5)

α and β are ration of positive odd integers,

φα satisfies (E1), and a and τ1 satisfy (E2) and (E3), respectively. Theorem 11.6. Suppose (E1)–(E5). If %

t

lim sup t→∞

τ1 (t)

Q(s) τ1 (t)

τ1 (s)

1 a(z)

& β1

1

α

Δz

Δs >

1 0

if β = α if β < α,

(11.22)

then all bounded solutions of the equation (11.21) are oscillatory. Proof. Suppose that x is a nonoscillatory bounded solution of the equation (11.21). Without loss of generality, we assume that x is an eventually positive solution of the equation (11.21). Then there exists a t1 ∈ [t0 , ∞) such that τ1 (t) > t0 , t ∈ [t1 , ∞), and x(t) > 0,

x(τ1 (t)) > 0,

t ∈ [t1 , ∞).

By (11.21), it follows that a(t)φα (x Δ (t)) is an increasing function on [t1 , ∞). Assume that x Δ (t) ≥ 0, t ∈ [t1 , ∞). Then there exists a positive constant c such that   a(t)φα x Δ (t) ≥ c,

t ∈ [t1 , ∞).

Hence,  Δ α c x (t) ≥ , a(t)

t ∈ [t1 , ∞),

590

11 Oscillations of Fourth-Order Functional Dynamic Equations

and 1

cα

x (t) ≥ Δ

1

t ∈ [t1 , ∞).

,

(a(t)) α

Therefore x(t) ≥ x(t1 ) + c

1 α

t

t1

→∞

as

1 a(s)

1

α

Δs

t → ∞.

This is a contradiction because x is a bounded solution of the equation (11.21). Consequently there is a t2 ∈ [t1 , ∞) so that x(t) > 0,

t ∈ [t2 , ∞).

x Δ (t) < 0,

x(τ1 (t)) > 0,

Hence, for u, v ∈ [t2 , ∞), v ≥ u, we have x(u) ≥ x(u) − x(v) v

=−

x Δ (s)Δs

u v

=− u

   1 (a(s))− α φα−1 a(s)φα x Δ (s) Δs

   ≥ −φα−1 a(v)φα x Δ (v)

v

1

(a(s))− α Δs.

u

From here, for t, s ∈ [t2 , ∞), t ≥ s, we get    x(τ1 (s)) ≥ φα−1 a(τ1 (t))φα −x Δ (τ1 (t))

τ1 (t)

1

(a(z))− α Δz.

τ1 (s)

Then       −a(τ1 (t))φα x Δ (τ1 (t)) ≥ a(t)φα x Δ (t) − a(τ1 (t))φα x Δ (τ1 (t))   Δ a(s)φα x Δ (s) Δs

t

=

τ1 (t) t

=

Q(s)φβ (x(τ1 (s)))Δs τ1 (t)

   ≥ φ β a(τ1 (t))φα −x Δ (τ1 (t)) α

×

%

t

Q(s) τ1 (t)

&β

τ1 (t)

(a(z)) τ1 (s)

− α1

Δz

Δs,

t ∈ [t2 , ∞).

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

591

Therefore  1− βα  ≥ a(τ1 (t))φα −x Δ (τ1 (t))

%

t

&β

τ1 (t)

Q(s)

(a(z))

τ1 (t)

− α1

Δz

Δs,

τ1 (s)

t ∈ [t2 , ∞). Hence,   1− βα > lim sup a(τ1 (t))φα −x Δ (τ1 (t))



t→∞

1 if 0 if

β=α β < α.

Let β = α. Then 1> 1, which is a contradiction. Let β < α. Then, by (11.21), we  obtain that a(t)φα x Δ (t) is an increasing function on [t2 , ∞). Assume that there exist a positive constant c1 and a t3 ∈ [t2 , ∞) such that   a(t)φα x Δ (t) ≤ −c1 , t ∈ [t3 , ∞). Hence,   c1 , φα x Δ (t) ≤ − a(t)

t ∈ [t3 , ∞),

and  Δ α c1 −x (t) ≥ , a(t)

t ∈ [t3 , ∞),

and 1

−x (t) ≥

c1α

Δ

1

,

(a(t)) α

t ∈ [t3 , ∞).

Then 1

x(t) ≤ x(t3 ) − c1α → −∞ as

t

Δs 1

(a(s)) α t → ∞. t3

This is a contradiction and therefore   a(t)φα x Δ (t) → 0 as

t → ∞.

Hence,   a(τ1 (t))φα −x Δ (τ1 (t)) → 0 This is a contradiction. This completes the proof.

as

t → ∞.

592

11 Oscillations of Fourth-Order Functional Dynamic Equations

Theorem 11.7. Suppose (E1)–(E5). If t

lim sup t→∞

τ1 (t)

1 (a(s))

 α1

t

Q(z)Δz

1 α

Δs >

s

if β = α if β < α,

1 0

(11.23)

then all bounded solutions of the equation (11.21) are oscillatory. Proof. Suppose that x is a nonoscillatory bounded solution of the equation (11.21). Without loss of generality, we assume that x is an eventually positive solution of the equation (11.21). Then there exists a t1 ∈ [t0 , ∞) such that τ1 (t) > t0 , t ∈ [t1 , ∞), and x(t) > 0,

t ∈ [t1 , ∞).

x(τ1 (t)) > 0,

By (11.21), it follows that a(t)φαΔ (t) is an increasing function on [t1 , ∞). Assume that x Δ (t) ≥ 0, t ∈ [t1 , ∞). Then there exists a positive constant c such that   a(t)φα x Δ (t) ≥ c, t ∈ [t1 , ∞). Hence,  Δ α c , x (t) ≥ a(t)

t ∈ [t1 , ∞),

and 1

x (t) ≥ Δ

cα 1

t ∈ [t1 , ∞).

,

(a(t)) α

Therefore 1

x(t) ≥ x(t1 ) + c α

t t1

→∞

1 a(s)

1

α

Δs

t → ∞.

as

This is a contradiction because x is a bounded solution of the equation (11.21). Consequently there is a t2 ∈ [t1 , ∞) so that x(t) > 0,

x(τ1 (t)) > 0,

x Δ (t) < 0,

t ∈ [t2 , ∞).

We integrate the equation (11.21) from s to t, s, t ∈ [t2 , ∞), t ≥ s, and we get       −a(s)φα x Δ (s) ≥ a(t)φα x Δ (t) − a(s)φα x Δ (s) t

=

Q(z)φβ (x(τ1 (z)))Δz, s

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

593

whereupon  α − x Δ (s) ≥

t

1 a(s)

s

t

Q(z)φβ (x(τ1 (z)))Δz

or −x (s) ≥

1

Δ

 α1 Q(z)φβ (x(τ1 (z)))Δz

1

(a(s)) α

s

≥ φ β (x(τ1 (t))) α

1

 α1

t

Q(z)Δz

1

(a(s)) α

.

s

Hence, x(τ1 (t)) ≥ −x(t) + x(τ1 (t)) t

≥−

x Δ (s)Δs

τ1 (t) t

≥

φ β (x(τ1 (t))) τ1 (t)

α

= (x(τ1 (t)))

1

Q(z)Δz

1

(a(s)) α 1

τ1 (t)

(a(s)) α

Δs

s

t

β α

 α1

t

 α1

t

Q(z)Δz

1

Δs,

s

t ∈ [t2 , ∞). Then 1− βα

(x(τ1 (t)))

≥

t

1

τ1 (t)

(a(s)) α

 α1

t

Q(z)Δz

1

Δs,

s

t ∈ [t2 , ∞). Hence, β

lim sup (x(τ1 (t)))1− α > t→∞



1 0

if β = α if β < α,

which is a contradiction. This completes the proof. Theorem 11.8. Suppose (E1)–(E5) and τ2 (t) ≤ τ1 (t), t ∈ [t0 , ∞). If the equation (11.21) is oscillatory and (11.22) or (11.23) holds with Q(t) = cβ (τ2 (t))β q(t)(τ1 (t) − τ2 (t))β − p(t) ≥ 0,

t ∈ [t0 , ∞),

for some c ∈ (0, 1), then every bounded solution x of the equation (11.17), either x 2 or x Δ is oscillatory.

594

11 Oscillations of Fourth-Order Functional Dynamic Equations

Proof. Suppose that x is a bounded nonoscillatory solution of the equation (11.17). Then there is a t1 ∈ [t0 , ∞) so that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞).

2

1. Suppose that x Δ (t) > 0, t ∈ [t1 , ∞). By the equation (11.17), we obtain   3 Δ  2 a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t)) ≤ 0,

t ∈ [t1 , ∞).

Hence from Theorem 11.5, it follows that the equation (11.19) has an eventually positive solution. This is a contradiction. 2 2. Suppose that x Δ (t) < 0, t ∈ [t1 , ∞). Assume that x Δ (t) ≤ 0, t ∈ [t1 , ∞). Let c1 > 0 be a constant such that 2

x Δ (t) ≤ −c1 ,

t ∈ [t1 , ∞).

Then x Δ (t) − x Δ (t1 ) =

t

2

x Δ (s)Δs

t1

≤ −c1 (t − t1 ),

t ∈ [t1 , ∞).

Hence, x Δ (t) ≤ −c1 (t − t1 ),

t ∈ [t1 , ∞),

and x(t) ≤ x(t1 ) − c1

t

(s − t1 )Δs

t1

→ −∞

as

t → ∞.

This is a contradiction. Therefore there exists a t2 ∈ [t1 , ∞) such that x Δ (t) > 2 0, t ∈ [t2 , ∞). Since x Δ (t) < 0, t ∈ [t2 , ∞), we have that x Δ is a strictly decreasing function on [t2 , ∞). Then there exists a positive constant d such that x Δ (t) ≤ d,

t ∈ [t2 , ∞).

We choose c ∈ (0, 1) such that x(t2 ) − dct2 > 0.

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

595

Let g(t) = x(t) − ctx Δ (t),

t ∈ [t2 , ∞).

Then 2

g Δ (t) = x Δ (t) − cx Δ (t) − cσ (t)x Δ (t) 2

= (1 − c)x Δ (t) − cσ (t)x Δ (t) t ∈ [t2 , ∞).

> 0, Hence,

g(t) ≥ g(t2 ) = x(t2 ) − ct2 x Δ (t2 ) ≥ x(t2 ) − bct2 > 0,

t ∈ [t2 , ∞).

Therefore x(t) ≥ ctx Δ (t),

t ∈ [t2 , ∞).

From here, there exists a t3 ∈ [t2 , ∞) such that x(τ2 (t)) ≥ cτ2 (t)x Δ (τ2 (t)),

t ∈ [t3 , ∞).

Let y(t) = x Δ (t),

t ∈ [t3 , ∞).

Then, using (11.17), we obtain   2 Δ   0 = a(t)φα y Δ (t) + p(t)φβ y Δ (τ1 (t)) + q(t)φβ (x(τ2 (t)))  2 Δ    ≥ a(t)φα y Δ (t) + p(t)φβ y Δ (τ1 (t)) +q(t)cβ (τ2 (t))β φβ (y(τ2 (t))),

t ∈ [t3 , ∞). (11.24)

Note that y(t) > 0 and

y Δ (t) < 0,

t ∈ [t3 , ∞).

596

11 Oscillations of Fourth-Order Functional Dynamic Equations

Assume that there is a constant c2 > 0 and a t4 ∈ [t3 , ∞) such that 2

y Δ (t) ≤ −c2 ,

t ∈ [t4 , ∞).

Then y Δ (t) ≤ y Δ (t4 ) − c2 (t − t4 ) ≤ −c2 (t − t4 ),

t ∈ [t4 , ∞),

and y(t) ≤ y(t4 ) − c2

t

(s − t4 )Δs

t4

→ −∞

t → ∞.

as

This is a contradiction. Therefore 2

t ∈ [t3 , ∞).

y Δ (t) > 0,

Hence, for u, v ∈ [t3 , ∞), v ≥ u, we have y(u) ≥ y(u) − y(v) v

=−

y Δ (s)Δs

u

  ≥ −y Δ (v) (v − u) and then   y(τ2 (t)) ≥ −y Δ (τ1 (t)) (τ1 (t) − τ2 (t)),

t ∈ [t3 , ∞).

Let w(t) = −y Δ (t),

t ∈ [t3 , ∞).

Then the inequality (11.24) takes the form  Δ  − p(t)φβ (w(τ1 (t))) 0 ≥ − a(t)φα w Δ (t) +q(t)cβ (τ2 (t))β φβ (y(τ2 (t)))  Δ  ≥ − a(t)φα w Δ (t) − p(t)φβ (w(τ1 (t))) +q(t)cβ (τ2 (t))β (τ1 (t) − τ2 (t))β φβ (w(τ1 (t)))

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

597

  Δ = − a(t)φα w Δ (t)   + q(t)cβ (τ2 (t))β (τ1 (t) − τ2 (t))β − p(t) φβ (w(τ1 (t))), t ∈ [t3 , ∞). Therefore   Δ   a(t)φα w Δ (t) ≥ q(t)cβ (τ2 (t))β (τ1 (t) − τ2 (t))β − p(t) ×φβ (w(τ1 (t))) = Q(t)φβ (w(τ1 (t))), t ∈ [t3 , ∞).

(11.25)

(a) Suppose (11.22). Then   −a(τ1 (t))φα w Δ (τ1 (t)) ≥ a(t)φα w Δ (t)  −a(τ1 (t))φα w Δ (τ1 (t)) =

t τ1 (t)

Q(s)φβ (w(τ1 (s)))Δs

  ≥ φ β a(τ1 (t))φα −w Δ (τ1 (t)) α

×

%

t

τ1 (t)

Q(s) τ1 (t)

&β

1

(a(z))− α Δz

t ∈ [t3 , ∞).

Δs,

τ1 (s)

Therefore   1− βα a(τ1 (t))φα −w Δ (τ1 (t)) ≥

%

t

&β

τ1 (t)

Q(s)

(a(z))

τ1 (t)

− α1

Δz

Δs,

τ1 (s)

t ∈ [t3 , ∞). Hence,   1− βα lim sup a(τ1 (t))φα −w Δ (τ1 (t)) > t→∞



1 0

if β = α if β < α.

If β = α, then we get 1 > 1, which is a contradiction. If β < α, then, using that w Δ (t) < 0, t ∈ [t3 , ∞), we get   a(t)φα w Δ (t) → 0

as

t → ∞.

Hence,   a(τ1 (t))φα −w Δ (τ1 (t)) → 0 as This is a contradiction.

t → ∞.

598

11 Oscillations of Fourth-Order Functional Dynamic Equations

(b) Suppose (11.23). By (11.25), we have, for t, s ∈ [t3 , ∞), t ≥ s,       −a(s)φα w Δ (s) ≥ a(t)φα w Δ (t) − a(s)φα w Δ (s)   Δ a(z)φα w Δ (z) Δz

t

= s

t

≥

Q(z)φβ (w(τ1 (z)))Δz, s

α  −w Δ (s) ≥

t

1 a(s)

Q(z)φβ (w(τ1 (z)))Δz, s

1

−w Δ (s) ≥

Q(z)φβ (w(τ1 (z)))Δz

1

(a(s)) α

s

1

≥

 α1

t

1

φ β (w(τ1 (t)))

(a(s)) α

 α1

t

Q(z)Δz

α

.

s

Hence, w(τ1 (t)) ≥ −w(t) + w(τ1 (t)) t

=−

w Δ (s)Δs

τ1 (t)

≥

t

φ β (w(τ1 (t))) τ1 (t)

α

= (w(τ1 (t)))

1

Q(z)Δz

1

(a(s)) α 1

τ1 (t)

(a(s)) α

Δs

s

t

β α

 α1

t

 α1

t

Q(z)Δz

1

Δs,

s

t ∈ [t3 , ∞), and β

(w(τ1 (t)))1− α ≥

t

1 1

(a(s)) α

τ1 (t)

 α1

t

Q(z)Δz s

t ∈ [t3 , ∞). Therefore 1− βα

lim sup (w(τ1 (t))) t→∞

>

1 if β = α 0 if β < α.

Δs,

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

599

If β = α, then we get 1 > 1, which is a contradiction. Let β < α. Then there exist a constant c2 > 0 and a t4 ∈ [t3 , ∞) such that w(t) ≥ c2 ,

w(τ1 (t)) ≥ c2 ,

t ∈ [t4 , ∞).

Hence,       a(t)φα w Δ (t) ≤ a(t)φα w Δ (t) − a(t)φα w Δ (t4 ) t

=

  Δ a(s)φα w Δ (s) Δs

t4 t

≥

Q(s)φβ (w(τ1 (s)))Δs t4 t

≥ cβ

Q(s)Δs,

t ∈ [t4 , ∞),

t4

and ! !α cβ − !w Δ (t)! ≥ a(t)

t

Q(s)Δs,

t ∈ [t4 , ∞).

t4

This is a contradiction. This completes the proof. Example 11.8. Let T = 2N0 . Consider the equation

 3 Δ φ x Δ (t) + 40 5

1 t

 t = 0, +t 4 φ3 x 4

 t 10 Δ2 t φ3 x 16384 2 t ∈ [4, ∞).

Here σ (t) = 2t, a(t) = p(t) =

1 t 40

,

t 10 , 16384

q(t) = t 4 , t τ1 (t) = , 2 t τ2 (t) = , 4

t ∈ T,

600

11 Oscillations of Fourth-Order Functional Dynamic Equations

α = 5, β = 3, t0 = 4. Take c =

1 √ 3 . 2

Then 1 Q(t) = 2

3  t t 3 t 10 4 t − t − 4 2 4 16384

=

t 10 1 t3 4 t3 t − 2 64 64 16384

=

t 10 t 10 − 8192 16384

=

t 10 , 16384

t ∈ [4, ∞).

Next, ∞ t0

= ⎛ lim sup ⎝ t→∞

t

t

1

1

α

Q(τ )Δτ

1

τ1 (t) (a(s)) α

∞ 1 − 5 Δs s 40 1

1

(a(s))− α Δs =

⎞

4

∞

s 8 Δs

4

= ∞,

Δs ⎠ = lim sup ⎝ t→∞

s

⎛ t t 2

% s8

τ 10 Δτ s 16384 t

⎞

&1 4

Δs ⎠

> 0. 2

Therefore any bounded solution x of the considered equation, either x or x Δ , is oscillatory. Exercise 11.7. Let T = 3N0 . Prove that any bounded solution x of the equation, 2 either x or x Δ ,

 3 Δ φ x Δ (t) + 25 5

1 t

 t = 0, +t φ3 x 9 5

is oscillatory.

 t 20 Δ2 t φ3 x 100000 3 t ∈ [9, ∞),

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

601

Now we consider the equation 

 3 Δ  2 a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t)) #c + b q1 (t, τ )φβ (x, g1 (t, τ ))Δτ = 0, t ∈ [t0 , ∞),

(11.26)

where b, c ∈ [t0 , ∞), b < c, (E6) q1 ∈ Crd (T × [b, c]) is a positive function, (E7) g1 : T × [b, c] → T is decreasing with respect to the second variable, g1 (t, τ ) ≤ t, limt→∞ g1 (t, τ ) = ∞, t ∈ [t0 , ∞), τ ∈ [b, c], φ satisfies (E1), α and β satisfy (E5), a, p, q satisfy (E2), and τ1 satisfies (E3). Let c

q(t) =

q1 (t, τ )Δτ, b

τ3 (t) = g1 (t, c), β

Q(t) = c0 q(t)(τ3 (t))β (τ1 (t) − τ3 (t))β − p(t),

t ∈ [t0 , ∞),

for some c0 ∈ (0, 1). Theorem 11.9. Suppose (E1)–(E3), (E5)–(E7), Q(t) ≥ 0, t ∈ [t0 , ∞). Let also, %

t

lim sup

τ1 (t)

Q(s)

t→∞

τ1 (t)

τ1 (s)

1 a(z)

& β1

1

α



Δz

Δs >

 α1



if β = α if β < α,

1 0

(11.27)

or t

lim sup t→∞

1

τ1 (t)

(a(s))

1 α

t

Q(z)Δz

Δs >

s

1 0

if β = α if β < α,

(11.28)

2

Then every bounded solution x of (11.26), either x or x Δ is oscillatory. Proof. Suppose that x is a bounded nonoscillatory solution of the equation (11.26). Then there is a t1 ∈ [t0 , ∞) so that x(t) > 0,

x(τ1 (t)) > 0,

x(g1 (t, τ ))) > 0,

t ∈ [t1 , ∞).

Note that   3 Δ  2 0 ≥ a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t))  # c q (t, τ )Δτ φ (x, g1 (t, c)) +  b 1 3 Δβ  2 = a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t)) +q(t)φβ (x, τ3 (t)),

t ∈ [t1 , ∞).

(11.29)

602

11 Oscillations of Fourth-Order Functional Dynamic Equations 2

1. Suppose that x Δ (t) > 0, t ∈ [t1 , ∞). By (11.29), we obtain   3 Δ  2 a(t)φα x Δ (t) + p(t)φβ x Δ (τ1 (t)) ≤ 0,

t ∈ [t1 , ∞).

Hence from Theorem 11.5, it follows that the equation (11.19) has an eventually positive solution. This is a contradiction. 2 2. Suppose that x Δ (t) < 0, t ∈ [t1 , ∞). Assume that x Δ (t) ≤ 0, t ∈ [t1 , ∞). Let c1 > 0 be a constant such that 2

x Δ (t) ≤ −c1 ,

t ∈ [t1 , ∞).

Then x Δ (t) − x Δ (t1 ) =

t

2

x Δ (s)Δs

t1

≤ −c1 (t − t1 ),

t ∈ [t1 , ∞).

Hence, x Δ (t) ≤ −c1 (t − t1 ),

t ∈ [t1 , ∞),

and t

x(t) ≤ x(t1 ) − c1

(s − t1 )Δs

t1

→ −∞

as

t → ∞.

This is a contradiction. Therefore there exists a t2 ∈ [t1 , ∞) such that x Δ (t) > 0, 2 t ∈ [t2 , ∞). Since x Δ (t) < 0, t ∈ [t2 , ∞), we have that x Δ (t) is a strictly decreasing function on [t2 , ∞). Then there exists a positive constant d such that x Δ (t) ≤ d,

t ∈ [t2 , ∞).

We choose c ∈ (0, 1) such that x(t2 ) − dct2 > 0. Let g(t) = x(t) − ctx Δ (t),

t ∈ [t2 , ∞).

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

603

Then 2

g Δ (t) = x Δ (t) − cx Δ (t) − cσ (t)x Δ (t) 2

= (1 − c)x Δ (t) − cσ (t)x Δ (t) t ∈ [t2 , ∞).

> 0, Hence,

g(t) ≥ g(t2 ) = x(t2 ) − ct2 x Δ (t2 ) ≥ x(t2 ) − dct2 > 0,

t ∈ [t2 , ∞).

Therefore x(t) ≥ ctx Δ (t),

t ∈ [t2 , ∞).

From here, there exists a t3 ∈ [t2 , ∞) such that x(τ3 (t)) ≥ cτ3 (t)x Δ (τ3 (t)),

t ∈ [t3 , ∞).

Let y(t) = x Δ (t),

t ∈ [t3 , ∞).

Then, using (11.29), we obtain   2 Δ   0 ≥ a(t)φα y Δ (t) + p(t)φβ y Δ (τ1 (t)) + q(t)φβ (x(τ3 (t)))  2 Δ    ≥ a(t)φα y Δ (t) + p(t)φβ y Δ (τ1 (t)) +q(t)cβ (τ3 (t))β φβ (y(τ3 (t))),

t ∈ [t3 , ∞). (11.30)

Note that y(t) > 0 and

y Δ (t) < 0,

t ∈ [t3 , ∞).

Assume that there are a constant c2 > 0 and a t4 ∈ [t3 , ∞) such that 2

y Δ (t) ≤ −c2 ,

t ∈ [t4 , ∞).

604

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then y Δ (t) ≤ y Δ (t4 ) − c2 (t − t4 ) ≤ −c2 (t − t4 ),

t ∈ [t4 , ∞),

and y(t) ≤ y(t4 ) − c2

t

(s − t4 )Δs

t4

→ −∞

t → ∞.

as

This is a contradiction. Therefore 2

t ∈ [t3 , ∞).

y Δ (t) > 0,

Hence, for u, v ∈ [t3 , ∞), v ≥ u, we have y(u) ≥ y(u) − y(v) v

=−

y Δ (s)Δs

u

  ≥ −y Δ (v) (v − u) and then   y(τ3 (t)) ≥ −y Δ (τ1 (t)) (τ1 (t) − τ3 (t)),

t ∈ [t3 , ∞).

Let w(t) = −y Δ (t),

t ∈ [t3 , ∞).

Then the inequality (11.30) takes the form  Δ  − p(t)φβ (w(τ1 (t))) 0 ≥ − a(t)φα w Δ (t) +q(t)cβ (τ3 (t))β φβ (y(τ3 (t)))  Δ  − p(t)φβ (w(τ1 (t))) ≥ − a(t)φα w Δ (t) +q(t)cβ (τ3 (t))β (τ1 (t) − τ3 (t))β φβ (w(τ1 (t)))   Δ = − a(t)φα w Δ (t)   + q(t)cβ (τ3 (t))β (τ1 (t) − τ3 (t))β − p(t) φβ (w(τ1 (t))),

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

605

t ∈ [t3 , ∞). Therefore   Δ   a(t)φα w Δ (t) ≥ q(t)cβ (τ3 (t))β (τ1 (t) − τ3 (t))β − p(t) ×φβ (w(τ1 (t))) = Q(t)φβ (w(τ1 (t))), t ∈ [t3 , ∞).

(11.31)

(a) Suppose (11.27). Then   −a(τ1 (t))φα w Δ (τ1 (t)) ≥ a(t)φα w Δ (t)  −a(τ1 (t))φα w Δ (τ1 (t)) =

t τ1 (t)

Q(s)φβ (w(τ1 (s)))Δs

  ≥ φ β a(τ1 (t))φα −w Δ (τ1 (t)) α

×

%

t

τ1 (t)

Q(s) τ1 (t)

&β

1

(a(z))− α Δz

t ∈ [t3 , ∞).

Δs,

τ1 (s)

Therefore   1− βα a(τ1 (t))φα −w Δ (τ1 (t)) ≥

%

t

&β

τ1 (t)

Q(s)

(a(z))

τ1 (t)

− α1

Δz

Δs,

τ1 (s)

t ∈ [t3 , ∞). Hence,   1− βα lim sup a(τ1 (t))φα −w Δ (τ1 (t)) > t→∞



1 0

if β = α if β < α.

If β = α, then we get 1 > 1, which is a contradiction. If β < α, then, using that w Δ (t) < 0, t ∈ [t3 , ∞), we get   a(t)φα w Δ (t) → 0

as

t → ∞.

Hence,   a(τ1 (t))φα −w Δ (τ1 (t)) → 0 as This is a contradiction.

t → ∞.

606

11 Oscillations of Fourth-Order Functional Dynamic Equations

(b) Suppose (11.28). By (11.31), we have, for t, s ∈ [t3 , ∞), t ≥ s,       −a(s)φα w Δ (s) ≥ a(t)φα w Δ (t) − a(s)φα w Δ (s)   Δ a(z)φα w Δ (z) Δz

t

= s

t

≥

Q(z)φβ (w(τ1 (z)))Δz, s

α  −w Δ (s) ≥

t

1 a(s)

Q(z)φβ (w(τ1 (z)))Δz, s

1

−w Δ (s) ≥

Q(z)φβ (w(τ1 (z)))Δz

1

(a(s)) α

s

1

≥

 α1

t

1

φ β (w(τ1 (t)))

(a(s)) α

 α1

t

Q(z)Δz

α

.

s

Hence, w(τ1 (t)) ≥ −w(t) + w(τ1 (t)) t

=−

w Δ (s)Δs

τ1 (t)

≥

t

φ β (w(τ1 (t))) τ1 (t)

α

= (w(τ1 (t)))

1

Q(z)Δz

1

(a(s)) α 1

τ1 (t)

(a(s)) α

Δs

s

t

β α

 α1

t

 α1

t

Q(z)Δz

1

Δs,

s

t ∈ [t3 , ∞), and β

(w(τ1 (t)))1− α ≥

t

1 1

(a(s)) α

τ1 (t)

 α1

t

Q(z)Δz s

t ∈ [t3 , ∞). Therefore 1− βα

lim sup (w(τ1 (t))) t→∞

>

1 if β = α 0 if β < α.

Δs,

11.2 Oscillations of Fourth-Order Nonlinear Delay Dynamic Equations with a. . .

607

If β = α, then we get 1 > 1, which is a contradiction. Let β < α. Then there exist a constant c > 0 and a t4 ∈ [t3 , ∞) such that w(t) ≥ c,

w(τ1 (t)) ≥ c,

t ∈ [t4 , ∞).

Hence,       a(t)φα w Δ (t) ≤ a(t)φα w Δ (t) − a(t)φα w Δ (t4 ) t

=

  Δ a(s)φα w Δ (s) Δs

t4 t

≥

Q(s)φβ (w(τ1 (s)))Δs t4

≥ cβ

t

Q(s)Δs,

t ∈ [t4 , ∞),

t4

and ! !α cβ − !w Δ (t)! ≥ a(t)

t

Q(s)Δs,

t ∈ [t4 , ∞).

t4

This is a contradiction. This completes the proof. Example 11.9. Let T = 2N0 . Consider the equation



  3 Δ 3 · 73 Δ 7 Δ2 t φ5 x (t) + t φ3 x 2 t 15 168

 t t 16 Δτ = 0, t ∈ [16, ∞). + φ3 x 4 4 τ 1

Here a(t) =

1 , t 15

3 · 73 7 t , 168 t q1 (t, τ ) = , 4 t g1 (t, τ ) = , τ σ (t) = 2t, p(t) =

608

11 Oscillations of Fourth-Order Functional Dynamic Equations

t , 2 α = 5,

t ∈ T,

τ1 (t) =

τ ∈ [4, 16],

β = 3, t0 = 16, b = 4, c = 16. Let h= Take c0 =

1 √ 3 . 2

3 · 73 3 · 73 − . 2 · 166 168

Then τ3 (t) = g1 (t, 16) t , = 16 16

q(t) =

q1 (t, τ )Δτ 4

t (12) 4 = 3t, =

t3 1 Q(t) = (3t) 3 2 16 =

t t − 2 16

3 −

3 · 73 7 t 168

3 · 73 7 3 · 73 7 t − t 2 · 166 168

= ht 7 ,

t ∈ [16, ∞).

Hence, %

t

lim sup t→∞

Q(s) τ1 (t)

= lim sup t→∞

> 0.

τ1 (t) τ1 (s)

%

t t 2

hs

t 2

7 s 2

1 a(z)

α

Δz

&1 3

3

z Δz

& β1

1

Δs

Δs

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

609 2

Therefore any bounded solution x of the considered equation, either x or x Δ is oscillatory. Exercise 11.8. Let T = 3N0 . Prove that any bounded solution x of the equation



 t 1 2 t 8 φ3 x Δ 100000 3 t

 27 2 t t Δτ = 0, t ∈ [27, ∞), + τ φ3 x 10 3 τ  3 Δ x Δ (t) φ + 5 75

1

2

either x or x Δ is oscillatory.

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations Consider the equation 

2

r(t)(x(t) + p(t)x(τ1 (t)))Δ

Δ2

+ q(t)G(x(τ2 (t))) = 0,

t ∈ [t0 , ∞), (11.32)

where (F1)

r ∈ Crd ([t0 , ∞)) is a positive function, p ∈ Crd ([t0 , ∞)), q ∈ Crd ([t0 , ∞)) is a nonnegative function, and ∞ t0

(F2)

t Δt = ∞, r(t)

G ∈ Crd (R) is a nondecreasing function such that uG(u) > 0,

(F3)

u = 0,

τ1 , τ2 : T → T are strictly increasing such that τ1 , τ2 ∈ C (T), τ1 (t) ≤ t, τ2 (t) ≤ t, t ∈ T, τ1 (τ2 (t)) = τ2 (τ1 (t)),

t ∈ T,

and lim τ1 (t) = lim τ2 (t) = ∞,

t→∞

τ1−1 exists and τ1−1 ∈ C (T).

t→∞

610

11 Oscillations of Fourth-Order Functional Dynamic Equations

We will start our investigations with the following useful lemmas. Δ2  2 2 ([t , ∞)), ruΔ2 ≤ 0 on [t0 , ∞). Let also, Lemma 11.5. Let u, ruΔ ∈ Crd 0 (F 1) holds. If u > 0 on [t0 , ∞), then there exists a t1 ∈ [t0 , ∞) such that one of the following cases holds. Case 1. 2



2



uΔ (t) > 0,

uΔ (t) > 0,

uΔ (t) > 0,

uΔ (t) < 0,

2

Δ

2

Δ

ruΔ

(t) > 0,

t ∈ [t1 , ∞),

(t) > 0,

t ∈ [t1 , ∞).

Case 2. ruΔ

 2  2 Δ 2 Δ Proof. Since ruΔ (t) ≤ 0, t ∈ [t0 , ∞), then ruΔ is a decreasing function on [t0 , ∞). Assume that  2 Δ ruΔ (t) < 0,

t ∈ [t0 , ∞).

Then there exist a constant c1 > 0 and a t1 ∈ [t0 , ∞) such that 

2

ruΔ

Δ

(t) ≤ −c1 ,

t ∈ [t1 , ∞).

Consider the following cases. 2

1. Let uΔ (t) < 0, uΔ (t) < 0, t ∈ [t1 , ∞). Then uΔ (t) ≤ uΔ (t1 ),

t ∈ [t1 , ∞).

Hence, u(t) ≤ u(t1 ) + uΔ (t1 )(t − t1 ) → −∞

as

t → ∞.

This is a contradiction. 2. Let 2

uΔ (t) < 0,

uΔ (t) > 0,

t ∈ [t1 , ∞).

Then 2

2

r(t)uΔ (t) ≤ r(t1 )uΔ (t1 ) − c1 (t − t1 ) ≤ −c1 (t − t1 ), t ∈ [t1 , ∞).

(11.33)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

Take c2 > 0 such that c2 < c1 . Let 

c1 t1 t2 > max t1 , . c1 − c2 Then, for any t ∈ [t2 , ∞), we have t≥

c1 t1 c1 − c2

and c1 t − c2 t ≥ c1 t1 , −c2 t ≥ −c1 (t − t1 ). Hence from (11.33), we get 2

r(t)uΔ (t) ≤ −c2 t,

t ∈ [t2 , ∞).

t , r(t)

t ∈ [t2 , ∞).

Therefore 2

uΔ (t) ≤ −c2 From here,

t

uΔ (t) ≤ uΔ (t2 ) − c2

t2

→ −∞

as

s Δs r(s)

t → ∞.

This is a contradiction. 2 3. Let uΔ (t) > 0, t ∈ [t1 , ∞). Then   2 Δ 2 Δ (t) ≤ ruΔ (t1 ), ruΔ

t ∈ [t1 , ∞),

and  2 2 2 Δ r(t)uΔ (t) ≤ r(t1 )uΔ (t1 ) + ruΔ (t1 )(t − t1 ) → −∞ This is a contradiction.

as

t → ∞.

611

612

11 Oscillations of Fourth-Order Functional Dynamic Equations

Consequently there exists a t3 ∈ [t1 , ∞) such that  2 Δ ruΔ (t) > 0,

t ∈ [t3 , ∞).

2

1. Let uΔ (t) > 0, t ∈ [t3 , ∞). Assume that uΔ (t) < 0, t ∈ [t3 , ∞). Since  2  2 Δ 2 Δ ruΔ (t) > 0 and ruΔ (t) > 0, t ∈ [t3 , ∞), there exist a constant c3 > 0 and a t4 ∈ [t3 , ∞) such that 

2

ruΔ

Δ

(t) ≥ c3 ,

t ∈ [t4 , ∞).

Hence, 2

2

r(t)uΔ (t) ≥ c3 (t − t4 ) + r(t4 )uΔ (t4 ) ≥ c3 (t − t4 ), t ∈ [t4 , ∞). Take a constant c4 > 0 such that c4 < c3 . Let 

c3 t4 . t5 > max t4 , c3 − c4 Then, for t ∈ [t5 , ∞), we have (c3 − c4 )t ≥ (c3 − c4 )t5 ≥ (c3 − c4 )

c3 t4 c3 − c4

= c3 t4 and c3 (t − t4 ) ≥ c4 t. Hence from (11.34), we get 2

r(t)uΔ (t) ≥ c4 t,

t ∈ [t5 , ∞).

t , r(t)

t ∈ [t5 , ∞),

Then 2

uΔ (t) ≥ c4

(11.34)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

613

and t

uΔ (t) ≥ uΔ (t5 ) + c4

t5

→ ∞ as

s Δs r(s)

t → ∞.

This is a contradiction. Consequently there exists a t6 ∈ [t3 , ∞) such that uΔ (t) > 0, t ∈ [t6 , ∞). 2 2. Let uΔ (t) < 0, t ∈ [t3 , ∞). Assume that uΔ (t) < 0, t ∈ [t3 , ∞). Then uΔ (t) ≤ uΔ (t3 ),

t ∈ [t3 , ∞),

and u(t) ≤ u(t3 ) + uΔ (t3 )(t − t3 ) → −∞

t → ∞.

as

This is a contradiction. Hence, there is a t7 ∈ [t3 , ∞) such that uΔ (t) > 0, t ∈ [t3 , ∞). This completes the proof. Δ2  2 2 ([t , ∞)), ruΔ2 ≤ 0 on [t0 , ∞). Let also, Exercise 11.9. Let u, ruΔ ∈ Crd 0 (F 1) holds. If u < 0 on [t0 , ∞), prove that there exists a t1 ∈ [t0 , ∞) such that one of the following cases holds. Case 1. 2



2



2



2



uΔ (t) > 0,

uΔ (t) < 0,

uΔ (t) < 0,

uΔ (t) < 0,

uΔ (t) < 0,

uΔ (t) < 0,

uΔ (t) < 0,

uΔ (t) > 0,

2

Δ

2

Δ

2

Δ

2

Δ

ruΔ

(t) > 0,

t ∈ [t1 , ∞),

(t) > 0,

t ∈ [t1 , ∞).

(t) < 0,

t ∈ [t1 , ∞).

(t) > 0,

t ∈ [t1 , ∞).

Case 2. ruΔ

Case 3. ruΔ

Case 4. ruΔ

614

11 Oscillations of Fourth-Order Functional Dynamic Equations

Δ2  2 ([t , ∞)), ruΔ2 Lemma 11.6. Let u, ruΔ ∈ Crd ≤ 0 on [t0 , ∞), (F 1) holds 0 and u > 0 on [t0 , ∞). Then there exists a T ∈ [t0 , ∞) such that  Δ 2 u(t) > RT (t) r(t)uΔ (t) ,

t ∈ [T , ∞),

where (t − σ (s))(s − T ) Δs. r(s)

ρ(t)

RT (t) = T

Proof. By Lemma 11.5, it follows that u satisfies Case 1 or Case 2 of Lemma 11.5. 1. Suppose that u satisfies Case 1 of Lemma 11.5. Then there exists a t1 ∈ [t0 , ∞) such that  Δ 2 r(t)uΔ (t) > 0,

2

uΔ (t) > 0,

uΔ (t) > 0,

t ∈ [t1 , ∞).

 2 Δ is a decreasing function on [t1 , ∞). Then Note that ruΔ 2

2

2

r(t)uΔ (t) ≥ r(t)uΔ (t) − r(t1 )uΔ (t1 ) t 2 Δ = ruΔ (s)Δs t1



2

≥ r(t)uΔ (t)

Δ

(t − t1 ),

t ∈ [t1 , ∞).

Hence,  Δ t − t 2 2 1 , uΔ (t) > r(t)uΔ (t) r(t)

t ∈ [t1 , ∞),

and  Δ s − t 2 1 Δs r(s)uΔ (s) r(s) t1 Δ s − t t 2 1 ≥ Δs r(s)uΔ (s) r(s) t1  Δ t s − t 2 1 ≥ r(t)uΔ (t) Δs, t ∈ [t1 , ∞), r(s) t1

uΔ (t) > uΔ (t1 ) +

t

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

and 

t

u(t) > u(t1 ) +

2

r(s)uΔ (s)

Δ

s

t1 t

≥



t1 2

r(s)uΔ (s)

Δ

s

t1

t1



2

Δ

2

Δ

2

Δ

≥ r(t)uΔ (t)

t

s

t1



= r(t)uΔ (t) 

= r(t)uΔ (t)

t1

ρ(t) t1

τ − t1 Δτ Δs r(τ )

τ − t1 Δτ Δs r(τ ) τ − t1 Δτ Δs r(τ )

(t − σ (s))(s − t1 ) Δs r(s) t ∈ [t1 , ∞).

Rt1 (t),

2. Suppose Case 2 of Lemma 11.5. Then there exists a t1 ∈ [t0 , ∞) such that uΔ (t) > 0,

2

uΔ (t) < 0,

 Δ 2 r(t)uΔ (t) > 0,

t ∈ [t1 , ∞).

Then 0≥

t t1

 Δ2 2 Rt1 (s) r(s)uΔ (s) Δs

 Δ !s=t 2 ! = Rt1 (s) r(s)uΔ (s) ! − s=t1

 Δ 2 = Rt1 (t) r(t)uΔ (t) −

t t1

t t1

 2 Δσ RtΔ1 (s) ruΔ (s)Δs

 2 Δσ RtΔ1 (s) ruΔ (s)Δs,

t ∈ [t1 , ∞), whereupon  2 Δ Rt1 (t) ruΔ (t) ≤

t t1 t

≤

t1

 2 Δσ RtΔ1 (s) ruΔ (s)Δs  2 Δ RtΔ1 (s) ruΔ (s)Δs



2

= ruΔ −

!s=t ! (s)RtΔ1 (s)!

s=t1

 2 2 σ RtΔ1 (s) ruΔ (s)Δs

t t1





2

= ruΔ



(t)RtΔ1 (t) −

t t1

 2 2 σ RtΔ1 (s) ruΔ (s)Δs

615

616

11 Oscillations of Fourth-Order Functional Dynamic Equations t

≤−

t1 t

≤−

t1 t

=−

t1 t

=−

 2 2 σ RtΔ1 (s) ruΔ (s)Δs 2

2

RtΔ1 (s)r(s)uΔ (s)Δs s − t1 2 r(s)uΔ (s)Δs r(s) 2

(s − t1 )uΔ (s)Δs

t1

!s=t ! = −(s − t1 )uΔ (s)! + s=t1

t

= −(t − t1 )uΔ (t) +

t

uΔσ (s)Δs

t1

uΔσ (s)Δs

t1 t

≤

uΔσ (s)Δs

t1 t

≤

uΔ (s)Δs

t1

= u(t) − u(t1 ) < u(t),

t ∈ [t1 , ∞).

This completes the proof. Lemma 11.7. Let (F 3) holds and F , H , P : [t0 , ∞) → R be such that F (t) = H (t) + P (t)H (τ1 (t)),

t ∈ [t1 , ∞),

for some t1 ∈ [t0 , ∞) such that τ1 (t) ≥ t0 , t ∈ [t1 , ∞). Assume that there exist numbers p1 , p2 , p3 , p4 ∈ R so that the function P satisfies one of the following possibilities. 1. p1 ≤ P (t) ≤ 0, t ∈ [t0 , ∞), 2. 0 ≤ P (t) ≤ p2 < 1, t ∈ [t0 , ∞), 3. 1 < p3 ≤ P (t) ≤ p4 , t ∈ [t0 , ∞). Suppose that H (t) > 0,

t ∈ [t0 , ∞),

lim inf H (t) = 0, t→∞

lim F (t) = L ∈ R

t→∞

exists. Then L = 0.

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

617

Proof. Note that τ1−1 ∈ C (T), τ1−1 : T → T is an increasing function, lim τ −1 (t) t→∞ 1

= ∞,

and    F τ1−1 (t) − F (t) = H τ1−1 (t) + P τ1−1 (t) H (t) − H (t) −P (t)H (τ1 (t)),

t ∈ [t1 , ∞).

Hence,     lim H τ1−1 (t) + P τ1−1 (t) − 1 H (t) − P (t)H (τ1 (t)) t→∞   = limt→∞ F τ1−1 (t) − F (t) (11.35) = L−L = 0. Since lim inft→∞ H (t) = 0, then there exists a sequence {tn }n∈N ⊂ [t1 , ∞) such that lim H (tn ) = 0.

n→∞

Hence from (11.35), using that the function P is bounded, we obtain     H τ1−1 (tn ) + P τ1−1 (tn ) − 1 H (tn ) − P (tn )H (τ1 (tn )) n→∞   = lim H τ1−1 (tn ) − P (tn )H (τ1 (tn )) n→∞   + lim P τ1−1 (tn ) − 1 H (tn ) n→∞   = lim H τ1−1 (tn ) − P (tn )H (τ1 (tn )) . lim

n→∞

1. Let p1 ≤ P (t) ≤ 0, t ∈ [t0 , ∞). Since  H τ1−1 (tn ) > 0,

n ∈ N,

and P (tn )H (τ1 (tn )) < 0,

n ∈ N,

618

11 Oscillations of Fourth-Order Functional Dynamic Equations

we conclude that lim P (tn )H (τ1 (tn )) = 0

n→∞

and L = limn→∞ (H (tn ) + P (tn )H (τ1 (tn ))) = 0. 2. Let 0 ≤ P (t) ≤ p2 < 1, t ∈ [t0 , ∞). Since  H τ1−1 (tn ) ≤ H (τ1 (tn )),  H τ1−1 (tn ) > 0, 1 − P (tn ) > 1 − p2 > 0,

n ∈ N,

we get   H τ1−1 (tn ) − P (tn )H τ1−1 (tn ) n→∞  = lim H τ1−1 (tn ) (1 − P (tn )) n→∞  ≥ (1 − p2 ) lim H τ1−1 (tn ) .

0 ≥ lim



n→∞

Then  lim H τ1−1 (tn ) = 0

n→∞

and   H τ1−1 (tn ) − P (tn )H (τ1 (tn )) n→∞  = lim H τ1−1 (tn ) − lim P (tn )H (τ1 (tn ))

0 = lim

n→∞

n→∞

= − lim P (tn )H (τ1 (tn )). n→∞

Hence from (11.36), we obtain that L = 0. 3. Let 1 < p3 ≤ P (t) ≤ p4 , t ∈ [t0 , ∞).

(11.36)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

619

Since  H τ1−1 (tn ) ≤ H (τ1 (tn )), 1 − P (tn ) ≤ 1 − p3 < 0, we get 0 ≤ lim (H (τ1 (tn )) − P (tn )H (τ1 (tn ))) n→∞

= lim (1 − P (tn ))H (τ1 (tn )) n→∞

≤ (1 − p3 ) lim H (τ1 (tn )). n→∞

Therefore lim H (τ1 (tn )) = 0

n→∞

and lim P (tn )H (τ1 (tn )) = 0.

n→∞

Hence from (11.36), we get L = 0. This completes the proof. Theorem 11.10. Suppose (F 1)–(F 3), 0 ≤ p(t) ≤ P < ∞, P is a constant, τ2 (t) ≤ τ1 (t), and t ∈ T. Assume that 1. There exists a λ > 0 such that G(u) + G(v) ≥ λG(u + v), 2. G(u)G(v) = G(uv), u, v ∈ R, 3. ! ! ! !

±c 0

u > 0,

v > 0,

! du !! 0, 4. ∞ T +t ∗

G (RT (τ2 (t))) Q(t)Δt = ∞,

T ≥ t0 ,

where Q(t) = min{q(t), q(τ1 (t))}, t ∈ [t1 , ∞), t1 ∈ [t0 , ∞) and t ∗ ∈ [0, ∞) such that T + t ∗ ∈ [T , ∞), τ2 (t) ≥ T for all t ∈ [T + t ∗ , ∞).

620

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then every solution of the equation (11.32) oscillates. Proof. Suppose that the equation (11.32) has a nonoscillatory solution x on [t0 , ∞). 1. Let x be an eventually positive solution of the equation (11.32) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0, x(τ2 (τ1 (t))) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ1 (t))) > 0,

t ∈ [t1 , ∞).

Define the following function z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

We have 0 < z(t) ≤ x(t) + P x(τ1 (t)), 2  Δ 2 r(t)zΔ (t) = −q(t)G(x(τ2 (t))) < 0, z(τ1 (t)) > 0,

t ∈ [t1 , ∞).

Hence, one of the cases of Lemma 11.5 holds. Observe that  Δ2 2 0 = r(t)zΔ (t) + q(t)G(x(τ2 (t))) Δ2  2 +G(P ) r(τ1 (t))zΔ (τ1 (t)) +G(P )q(τ1 (t))G(x(τ2 (τ1 (t))))  Δ2  Δ2 2 2 = r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +q(t)G(x(τ2 (t))) +q(τ1 (t))G(P x(τ2 (τ1 (t))))  Δ2  Δ2 2 2 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +Q(t) (G(x(τ2 (t))) + G(P x(τ2 (τ1 (t)))))

(11.37)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

621

 Δ2  Δ2 2 2 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +λQ(t)G (x(τ2 (t)) + P x(τ2 (τ1 (t))))  Δ2  Δ2 2 2 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +λQ(t)G(z(τ2 (t))),

t ∈ [t1 , ∞).

Take t ∗ ∈ [0, ∞) such that T + t ∗ ∈ [T , ∞)

and

τ2 (t) ≥ T ≥ t1 ,

t ∈ [T + t ∗ , ∞).

By Lemma 11.6, we have  Δ 2 z(τ2 (t)) ≥ RT (τ2 (t)) r(τ2 (t))zΔ (τ2 (t)) , t ∈ [T + t ∗ , ∞). Then, using (F 2), we obtain

 Δ  2 G(z(τ2 (t))) ≥ G RT (τ2 (t)) r(τ2 (t))zΔ (τ2 (t)) = G (RT (τ2 (t))) G

 r(τ2 (t))z

Δ2

Δ  (τ2 (t))

t ∈ [T + t ∗ , ∞). Hence from (11.37), we get  Δ2  Δ2 2 2 0 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t))

 Δ  Δ2 , +λQ(t)G(RT (τ2 (t)))G r(τ2 (t))z (τ2 (t)) t ∈ [T + t ∗ , ∞). Hence,

 Δ  Δ2 λQ(t)G(RT (τ2 (t)))G r(τ2 (t))z (τ2 (t)) Δ2  Δ2  2 2 − G(P ) r(τ1 (t))zΔ (τ1 (t)) , ≤ − r(t)zΔ (t)

,

622

11 Oscillations of Fourth-Order Functional Dynamic Equations

t ∈ [T + t ∗ , ∞), and  λQ(t)G(RT (τ2 (t))) ≤ − G

2

r(t)zΔ (t)

Δ2

 Δ 2 r(τ2 (t))zΔ (τ2 (t))

 Δ2 2 G(P ) r(τ1 (t))zΔ (τ1 (t))  − Δ , 2 G r(τ2 (t))zΔ (τ2 (t)) t ∈ [T + t ∗ , ∞). By Lemma 11.5, it follows that  Δ 2 lim r(τ2 (t))zΔ (τ2 (t))

t→∞

exists. Therefore ∞

λ

T +t ∗

Q(t)G (RT (τ2 (t))) Δt < ∞.

This is a contradiction. 2. Let x is an eventually negative solution of the equation (11.32) on [t0 , ∞). Set y = −x. Then, using that G(−u) = −G(u), u ∈ R, we get that y is an eventually positive solution of the equation (11.32) on [t0 , ∞). Proceeding as above we arrive at a contradiction. This completes the proof. Example 11.10. Let T = 2N0 . Consider the equation % t x(t) +

1 x t +1

1 

Δ2 &Δ2 t t = 0, + t2 3 x 2 4

t ∈ [4, ∞). Here σ (t) = 2t, r(t) = t, p(t) =

1 , t +1

q(t) = t 2 , t τ1 (t) = , 2 t τ2 (t) = , t ∈ T, 4 √ G(u) = 3 u, u ∈ R.

(11.38)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

We have 0 ≤ p(t) ≤ 1, τ2 (t) ≤ τ1 (t), τ2 (τ1 (t)) = = τ1 (τ2 (t)) = = G(u)G(v) = =

τ1 (t) 4 t , 8 τ2 (t) 2 t , t ∈ T, 8 √ √ 3 u3v √ 3 uv

= G(uv), √ uG(u) = u 3 u

u, v ∈ R,

> 0, u ∈ R, u = 0, √ √ √ 3 u + 3 v > 3 u + v, u, v ∈ R, u > 0, λ = 1, ±c 0

du = G(u)

±c 0

du √ 3 u

3 2 !!u=c u3 ! u=0 2 3 2 = c3 2 < ∞, c > 0,   Q(t) = min t 2 , (τ1 (t))2 =



t2 = min t 2 , 4 =

t2 , 4

t ∈ T.

Take T = 4,

t ∗ = 4.

v > 0,

623

624

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then ∞ T +t ∗

8 ∞

=

∞

G (RT (τ2 (t))) Q(t)Δt =

8

t 4

τ 4

s−4 ΔsΔτ s

 13

 t2 3 RT (τ2 (t)) Δt 4

t2 Δt 4

= ∞. Therefore any solution of the considered equation oscillates. Exercise 11.10. Let T = 3N0 . Prove that any solution of the equation % t 2 x(t) +

1 x 2 t +t +1

1 

Δ2 &Δ2 t t = 0, + t7 5 x 3 9

t ∈ [9, ∞), oscillates. Theorem 11.11. Suppose (F 1)–(F 3), 0 ≤ p(t) ≤ P < ∞, and P is a constant. Assume that 1. there exists a constant λ > 0 such that G(u) + G(v) ≥ λG(u + v),

u > 0,

v > 0,

2. G(u)G(v) ≥ G(uv), u, v ∈ R, G(−u) = −G(u), u ∈ R, 3. ∞

Q(t)Δt = ∞,

T

where Q(t) = min{q(t), q(τ1 (t))}, t ∈ [t0 , ∞), for some T ∈ [t0 , ∞). Then any solution of the equation (11.32) oscillates. Proof. Suppose that the equation (11.32) has a nonoscillatory solution x on [t0 , ∞). 1. Let x be an eventually positive solution of the equation (11.32) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0, x(τ2 (τ1 (t))) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ1 (t))) > 0,

t ∈ [t1 , ∞).

Define the following function z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

625

We have 0 < z(t) ≤ x(t) + P x(τ1 (t)),  2 r(t)zΔ (t)

Δ2

= −q(t)G(x(τ2 (t))) < 0,

z(τ1 (t)) > 0,

t ∈ [t1 , ∞).

Hence, one of the cases of Lemma 11.5 holds and then there is a t2 ∈ [t1 , ∞) such that zΔ (t) > 0, t ∈ [t2 , ∞). Consequently z is an increasing function on [t2 , ∞). Hence, there are a t3 ∈ [t2 , ∞) and a k > 0 such that z(τ2 (t)) > k,

z(t) > k,

t ∈ [t3 , ∞).

Now, using (11.37), we obtain  Δ2  Δ2 2 2 0 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +λQ(t)G(z(τ2 (t)))  Δ2  Δ2 2 2 ≥ r(t)zΔ (t) + G(P ) r(τ1 (t))zΔ (τ1 (t)) +λQ(t)G(k),

t ∈ [t3 , ∞).

From here,  Δ2  Δ2 2 2 λQ(t)G(k) ≤ − r(t)zΔ (t) − G(P ) r(τ1 (t))zΔ (τ1 (t)) , t ∈ [t3 , ∞), and  λQ(t) ≤ −

2

r(t)zΔ (t)

Δ2 −

G(k)

 Δ2 2 G(P ) r(τ1 (t))zΔ (τ1 (t)) G(k)

 Δ 2 t ∈ [t3 , ∞). Since limt→∞ r(t)zΔ (t) exists, we get ∞

λ t3

This is a contradiction.

Q(t)Δt < ∞.

,

626

11 Oscillations of Fourth-Order Functional Dynamic Equations

2. Let x is an eventually negative solution of the equation (11.32) on [t0 , ∞). Set y = −x. Then, using that G(−u) = −G(u), u ∈ R, we get that y is an eventually positive solution of the equation (11.32) on [t0 , ∞). Proceeding as above we arrive at a contradiction. This completes the proof. Example 11.11. Consider Example 11.10. Since ∞

∞

Q(s)Δs =

t

t

s2 Δs 4

= ∞, we conclude that any solution of the equation (11.38) oscillates. Theorem 11.12. Suppose (F 1)–(F 3), 0 ≤ p(t) ≤ P < ∞, and P is a constant. Assume that 1. there exists a constant λ > 0 such that G(u) + G(v) ≥ λG(u + v),

u > 0,

v > 0,

2. G(u)G(v) = G(uv), u, v ∈ R, 3. G(x1 ) G(x2 ) ≥ γ γ , x1 x2

x1 ≥ x2 > 0,

γ ≥ 1,

4. ∞ T +t ∗

γ

RT (τ2 (t))Q(t)Δt = ∞,

where Q(t) = min{q(t), q(τ1 (t))}, t ∈ [t0 , ∞),t ∈ [t1 , ∞), t1 ∈ [t0 , ∞), t ∗ ∈ [0, ∞) such that T + t ∗ ∈ [T , ∞), τ2 (t) ≥ t for t ∈ [T + t ∗ , ∞). Then any solution of the equation (11.32) oscillates. Proof. Suppose that the equation (11.32) has a nonoscillatory solution x on [t0 , ∞). 1. Let x be an eventually positive solution of the equation (11.32) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0, x(τ2 (τ1 (t))) > 0,

x(τ1 (t)) > 0, t ∈ [t1 , ∞).

x(τ2 (t)) > 0,

x(τ1 (τ1 (t))) > 0,

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

627

Define the following function z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

We have 0 < z(t) ≤ x(t) + P x(τ1 (t)),  2 r(t)zΔ (t)

Δ2

= −q(t)G(x(τ2 (t))) < 0, t ∈ [t1 , ∞).

z(τ1 (t)) > 0,

Hence, one of the cases of Lemma 11.5 holds and then there is a t2 ∈ [t1 , ∞) such that zΔ (t) > 0, t ∈ [t2 , ∞). Consequently z is an increasing function on [t2 , ∞). Hence, there are a t3 ∈ [t2 , ∞) and a k > 0 such that z(t) > k,

z(τ2 (t)) > k,

t ∈ [t3 , ∞).

By Lemma 11.6, we obtain G(z(τ2 (t))) =

G(z(τ2 (t))) (z(τ2 (t)))γ (z(τ2 (t)))γ

G(k) (z(τ2 (t)))γ kγ

Δ γ γ  G(k)  Δ2 r(τ2 (t))z (τ2 (t)) > γ Rt3 (τ2 (t)) k

Δ γ γ  G(k)  Δ2 r(t)z (t) ≥ γ Rt3 (τ2 (t)) , k ≥

t ∈ [t3 , ∞). Hence from (11.37), we get λ

γ G(k)  Rt3 (τ2 (t)) γ k



2

r(t)zΔ (t)

Δ γ Q(t)

≤ λQ(t)G(z(τ2 (t)))  Δ2  Δ2 2 2 ≤ − r(t)zΔ (t) − G(P ) r(τ1 (t))zΔ (τ1 (t)) ,

628

11 Oscillations of Fourth-Order Functional Dynamic Equations

t ∈ [t3 , ∞). Then  Δ2 Δ2 (t) r(t)z γ G(k)  λ γ Rt3 (τ2 (t)) Q(t) ≤ −  Δ γ 2 k r(t)zΔ (t)  Δ2 2 G(P ) r(τ1 (t))zΔ (τ1 (t))  − Δ γ 2 r(t)zΔ (t)  Δ2 2 r(t)zΔ (t) ≤ −  Δ γ 2 r(t)zΔ (t)  Δ2 2 G(P ) r(τ1 (t))zΔ (τ1 (t)) −  Δ γ , 2 r(τ1 (t))zΔ (τ1 (t))  Δ 2 t ∈ [t3 , ∞). Since limt→∞ r(t)zΔ (t) exists, we conclude that ∞ t3

(Rt3 (τ2 (t)))γ Q(t)Δt < ∞.

This is a contradiction. 2. Let x is an eventually negative solution of the equation (11.32) on [t0 , ∞). Set y = −x. Then, using that G(−u) = −G(u), u ∈ R, we get that y is an eventually positive solution of the equation (11.32) on [t0 , ∞). Proceeding as above we arrive at a contradiction. This completes the proof. Theorem 11.13. Suppose (F 1)–(F 3), −1 < P ≤ p(t) ≤ 0, P is a constant, t ∈ T, and 1. G(u)G(v) = G(uv), u, v ∈ R, 2. ! ±c ! ! du !! ! < ∞, ! G(u) ! 0

c > 0.

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

629

If ∞

q(t)Δt = ∞,

t0

then every solution of the equation (11.32) oscillates or tends to zero. Proof. Suppose that the equation (11.32) has a nonoscillatory solution x on [t0 , ∞). 1. Let x be an eventually positive solution of the equation (11.32) on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0, x(τ2 (τ1 (t))) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ1 (t))) > 0,

t ∈ [t1 , ∞).

Define the following function z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

We have  Δ2 2 r(t)zΔ (t) = −q(t)G(x(τ2 (t))) < 0, t ∈ [t1 , ∞).

Hence, one of the cases of Lemma 11.5 holds and  Δ2 2 r(t)zΔ (t) = −q(t)G(x(τ2 (t))) ≤ 0,

t ∈ [t1 , ∞).

(a) Suppose that there is a t2 ∈ [t1 , ∞) such that z(t) > 0,

t ∈ [t2 , ∞).

By Lemma 11.6, it follows that there is a t3 ∈ [t2 , ∞) such that  Δ 2 z(t) > Rt3 (t) r(t)zΔ (t) ,

t ∈ [t3 , ∞).

630

11 Oscillations of Fourth-Order Functional Dynamic Equations

Take t4 ∈ [t3 , ∞) so that τ2 (t) ≥ t3 , t ∈ [t4 , ∞). Hence, using that z(t) ≤ x(t), we obtain  Δ2 2 r(t)zΔ (t) = −q(t)G(x(τ2 (t))) ≤ −q(t)G(z(τ2 (t)))

 Δ  2 ≤ −q(t)G Rt3 (τ2 (t)) r(τ2 (t))zΔ (τ2 (t))

 Δ  Δ2 = −q(t)G Rt3 (τ2 (t)) G r(τ2 (t))z (τ2 (t)) 



 Δ  Δ2 , ≤ −q(t)G Rt3 (τ2 (t)) G r(t)z (t) 



t ∈ [t4 , ∞), whereupon  Δ2 2 r(t)zΔ (t) q(t)G Rt3 (τ2 (t)) ≤ −  Δ , 2 G r(t)zΔ (t) 



t ∈ [t4 , ∞), and ∞ t4

  q(t)G Rt3 (τ2 (t)) Δt ≤ −

∞ t4

 Δ2 2 r(t)zΔ (t)  Δ Δt. 2 G r(t)zΔ (t)

Because RtΔ4 (t) > 0 and Rt4 (t) > 0, then ∞

q(t)Δt < ∞.

t4

This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that z(t) < 0 for t ∈ [t2 , ∞). Then x(t) + p(t)x(τ1 (t)) < 0,

t ∈ [t2 , ∞),

and x(t) ≤ −p(t)x(τ1 (t)) < x(τ1 (t)),

t ∈ [t2 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

631

From here, we conclude that x is bounded. Then z is bounded. (i) Suppose that there is a t3 ∈ [t2 , ∞) so that 2

zΔ (t) > 0,

zΔ (t) < 0,

 Δ 2 r(t)zΔ (t) > 0,

t ∈ [t4 , ∞).

Then there exists lim z(t) = l.

t→∞

We have −∞ < l ≤ 0. Let l < 0. Then there are a t5 ∈ [t4 , ∞) and an m < 0 such that z(t) < m,

t ∈ [t5 , ∞).

Next, t ∈ [t5 , ∞),

z(t) > P x(τ1 (t)), and hence, 0 < P −1 m < x(τ1 (t)),

t ∈ [t5 , ∞).

Since τ1−1 exists, there is a t6 ∈ [t5 , ∞) such that τ2 (t) ≥ τ1 (t5 ),

t ∈ [t6 , ∞).

Therefore 0 < P −1 m < x(τ2 (t)),

t ∈ [t6 , ∞),

and  Δ2 2 r(t)zΔ (t) ≤ −q(t)G(x(τ2 (t)))  ≤ −q(t)G P −1 m ,

t ∈ [t6 , ∞),

whereupon   Δ2 2 q(t)G P −1 m ≤ − r(t)zΔ (t) ,

t ∈ [t6 , ∞),

632

11 Oscillations of Fourth-Order Functional Dynamic Equations

and  G P −1 m

∞

∞

q(t)Δt ≤ −

t6

2

r(t)zΔ (t)

Δ2 Δt

t6

< ∞. This is a contradiction. Hence, l = 0 and 0 = lim sup z(t) t→∞

= lim sup (x(t) + p(t)x(τ1 (t))) t→∞

≥ lim sup x(t) + lim inf (P x(τ1 (t))) t→∞

t→∞

= lim sup x(t) + P lim sup x(t) t→∞

t→∞

= (1 + P ) lim sup x(t). t→∞

Because 1 + P > 0, we conclude that lim sup x(t) = 0. t→∞

(ii) Let there is a t3 ∈ [t2 , ∞) so that 2

 Δ 2 r(t)zΔ (t) > 0,

t ∈ [t3 , ∞),

2

 Δ 2 r(t)zΔ (t) < 0,

t ∈ [t3 , ∞).

zΔ (t) < 0,

zΔ (t) < 0,

zΔ (t) < 0,

zΔ (t) < 0,

or

Then lim z(t) = −∞.

t→∞

This is a contradiction because z is bounded. (iii) Let there is a t3 ∈ [t2 , ∞) so that zΔ (t) < 0,

2

zΔ (t) > 0,

 Δ 2 r(t)zΔ (t) > 0,

t ∈ [t3 , ∞).

Since z is bounded, there exists limt→∞ z(t). Observe that 2

2

r(t)zΔ (t) ≥ r(t3 )zΔ (t3 ),

t ∈ [t3 , ∞),

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

633

or 2

2

zΔ (t) ≥

r(t3 )zΔ (t3 ) , r(t)

t ∈ [t3 , ∞).

Then 2

r(t3 )zΔ (t3 )

t t3

t

s Δs ≤ r(s)

2

szΔ (s)Δs

t3

!s=t ! = szΔ (s)! − s=t3

t

zΔσ (s)Δs

t3

= tzΔ (t) − t3 zΔ (t3 ) −

t

zΔσ (s)Δs

t3

≤ −t3 zΔ (t3 ) −

t

zΔ (s)Δs

t3

= −t3 zΔ (t3 ) − z(t) + z(t3 ),

t ∈ [t3 , ∞).

Hence from (F 1), we get that z is unbounded. This is a contradiction. 2. Let x is an eventually negative solution of the equation (11.32) on [t0 , ∞). Set y = −x. Then, using that G(−u) = −G(u), u ∈ R, we get that y is an eventually positive solution of the equation (11.32) on [t0 , ∞). Proceeding as above we arrive at a contradiction. This completes the proof. Example 11.12. Let T = 2N0 . Consider the equation % t x(t) −

1 x 2(t 2 + 1)

1 

Δ2 &Δ2 t t 7 3 = 0, +t x 4 8

t ∈ [8, ∞). Here σ (t) = 2t, r(t) = t, p(t) = −

1 , 2(t 2 + 1)

q(t) = t 7 , t τ1 (t) = , 4 t τ2 (t) = , t ∈ T, 8 √ G(u) = 3 u, u ∈ R.

634

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then −

1 1 ≤− 2 2 2(t + 1)

≤ 0, t ∈ T, √ √ G(u)G(v) = 3 u 3 v √ = 3 uv ! ! ! !

±c 0

= G(uv), u, v ∈ R, ! ! ! ! du ! !! 3 2 !!u=±c !! 3 = u ! √ 3 u=0 ! u! !2 3 2 c3 2 < ∞, =

∞

∞

q(t)Δt =

8

t 7 Δt

8

= ∞. Therefore any solution of the considered equation oscillates or tends to zero. Exercise 11.11. Let T = 3N0 . Prove that any solution of the equation % t x(t) −

1 x 7(t 3 + t 2 + t + 1)

1 

Δ2 &Δ2 t t 9 3 = 0, +t x 3 9

t ∈ [9, ∞), oscillates or tends to zero. Theorem 11.14. Let −∞ < p1 ≤ p(t) ≤ p2 < −1 and (F 1)–(F 3) hold, ∞

q(t)Δt = ∞.

t0

Then every bounded solution of the equation (11.32) oscillates or tends to zero as t →∞ Proof. Suppose that the equation (11.32) has a bounded nonoscillatory solution on [t0 , ∞). 1. Let x is an eventually positive solution of the equation (11.32). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

635

Define t ∈ [t1 , ∞).

z(t) = x(t) + p(t)x(τ1 (t)), Then

 Δ2 2 r(t)zΔ (t) = −q(t)G(x(τ2 (t))) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that t ∈ [t2 , ∞).

z(t) > 0,

Then one of the cases of Lemma 11.5 holds. Hence, x(t) > −p(t)x(τ1 (t)) t ∈ [t2 , ∞).

> x(τ1 (t)), Therefore

lim inf x(t) > 0. t→∞

Also, we have

q(t) = −

 Δ2 2 r(t)zΔ (t) G(x(τ2 (t)))

,

t ∈ [t2 , ∞),

and ∞

q(t)Δt = −

t2

∞ t2

 Δ2 2 r(t)zΔ (t) G(x(τ2 (t)))

Δt

< ∞. This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that z(t) < 0 for t ∈ [t2 , ∞). Since x is bounded, then z is bounded. Therefore there exists a t3 ∈ [t2 , ∞) such that zΔ (t) > 0,

2

zΔ (t) < 0,

 Δ2 2 r(t)zΔ (t) > 0,

t ∈ [t3 , ∞).

636

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then there exists lim z(t) = l.

t→∞

We have −∞ < l ≤ 0. Let l < 0. Then there are a t5 ∈ [t4 , ∞) and an m < 0 such that t ∈ [t5 , ∞).

z(t) < m, Next,

t ∈ [t5 , ∞),

z(t) > p1 x(τ1 (t)), and hence, 0 < p1−1 m < x(τ1 (t)),

t ∈ [t5 , ∞).

Since τ1−1 exists, there is a t6 ∈ [t5 , ∞) such that τ2 (t) ≥ τ1 (t5 ),

t ∈ [t6 , ∞).

Therefore 0 < p1−1 m < x(τ2 (t)),

t ∈ [t6 , ∞),

and  Δ2 2 r(t)zΔ (t) ≤ −q(t)G(x(τ2 (t)))  ≤ −q(t)G p1−1 m ,

t ∈ [t6 , ∞),

  Δ2 2 q(t)G p1−1 m ≤ − r(t)zΔ (t) ,

t ∈ [t6 , ∞),

whereupon

and  G p1−1 m

∞

q(t)Δt ≤ −

t6

∞ t6

< ∞.

2

r(t)zΔ (t)

Δ2 Δt

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

637

This is a contradiction. Hence, l = 0 and 0 = lim inf z(t) t→∞

≤ lim inf (x(t) + p2 x(τ1 (t))) t→∞

≤ lim sup x(t) + lim inf (p2 x(τ1 (t))) t→∞

t→∞

≤ lim sup x(t) + p2 lim sup x(τ1 (t)) t→∞

t→∞

= (1 + p2 ) lim sup x(t). t→∞

Hence, using that 1 + p2 < 0, we conclude that lim sup x(t) = 0. t→∞

2. Let x is an eventually negative solution of the equation (11.32) on [t0 , ∞). Set y(t) = −x(t),

t ∈ [t0 , ∞),

and ˜ G(u) = −G(−u),

u ∈ R.

Then y is an eventually positive solution of the equation 2  2 Δ ˜ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ 2 (t))) = 0,

t ∈ [t0 , ∞).

Proceeding as above, we obtain limt→∞ y(t) = 0 and hence, limt→∞ x(t) = 0. This completes the proof. Now we consider the following nonhomogeneous equation  2 2 Δ r(t) (x(t) + p(t)x(τ1 (t)))Δ + q(t)G(x(τ2 (t))) = f (t),

(11.39)

t ∈ [t0 , ∞), where r, p, q, G, τ1 , and τ2 satisfy (F 1)–(F 3) and f ∈ Crd ([t0 , ∞)). Theorem 11.15. Let 0 ≤ p(t) ≤ P < ∞, t ∈ [t0 , ∞), P is a constant, and (F 1)–(F 3) hold, and 1. there exists a λ > 0 such that G(u) + G(v) ≥ λG(u + v),

u > 0,

v > 0,

638

11 Oscillations of Fourth-Order Functional Dynamic Equations

2. G(u)G(v) ≥ G(uv), u > 0, v > 0, 3. G(−u) = −G(u), u ∈ R, 2 ([t , ∞)) such that F changes sign with 4. there exists a function F ∈ Crd 0 −∞ < lim inf F (t) < 0 < lim sup F (t) < ∞, t→∞

t→∞

Δ2  2 2 ([t , ∞)) and rF Δ2 rF Δ ∈ Crd = f. 0 If ∞

∞

  Q(t)G F + (τ2 (t)) Δt = ∞ and

t0

 Q(t)G F −1 (τ2 (t)) Δt = ∞,

t0

then all the solutions of the equation (11.39) oscillate. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞). 1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

Then 0 < z(t) ≤ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then  Δ2   2 Δ2 2 2 r(t)wΔ (t) = r(t) zΔ (t) − F Δ (t)  Δ2  Δ2 2 2 = r(t)zΔ (t) − r(t)F Δ (t)  Δ2 2 = r(t)zΔ (t) − f (t) = −q(t)G(x(τ2 (t))) ≤ 0, t ∈ [t1 , ∞).

(11.40)

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

639

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). Take a t3 ∈ [t2 , ∞) such that t ∈ [t3 , ∞).

w(τ1 (t)) > 0,

Therefore one of the cases of Lemma 11.5 holds. We have  Δ2  Δ2 2 2 0 = r(t)w Δ (t) + G(P ) r(τ1 (t))w Δ (τ1 (t)) +q(t)G(x(τ2 (t))) + G(P )q(τ1 (t))G(x(τ2 (τ1 (t))))  Δ2  Δ2 2 2 ≥ r(t)w Δ (t) + G(P ) r(τ1 (t))w Δ (τ1 (t))

(11.41)

+λQ(t)G (x(τ2 (t)) + P x(τ2 (τ1 (t))))  Δ2  Δ2 2 2 ≥ r(t)w Δ (t) + G(P ) r(τ1 (t))w Δ (τ1 (t)) +λQ(t)G(z(τ2 (t))),

t ∈ [t3 , ∞).

Since w(t) > 0, t ∈ [t2 , ∞), we have that z(t) > F (t), t ∈ [t2 , ∞) and z(t) > F + (t), t ∈ [t2 , ∞). Take a t4 ∈ [t3 , ∞) such that τ2 (t) ≥ t3 , t ∈ [t4 , ∞). Then, using (11.41), we have  Δ2  Δ2 2 2 0 ≥ r(t)w Δ (t) + G(P ) r(τ1 (t))w Δ (τ1 (t))   +λQ(t)G F + (τ2 (t)) , t ∈ [t4 , ∞), whereupon  Δ2   2 λQ(t)G F + (τ2 (t)) ≤ − r(t)w Δ (t)  Δ2 2 −G(P ) r(τ1 (t))w Δ (τ1 (t)) ,

t ∈ [t4 , ∞),

and ∞

λ

  Q(t)G F + (τ2 (t)) Δt ≤ −

t4

∞

2

r(t)w Δ (t)

Δ2 Δt

t4

−G(P )

∞ t4

< ∞. This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0,

2

r(τ1 (t))w Δ (τ1 (t))

t ∈ [t2 , ∞).

Δ2 Δt

640

11 Oscillations of Fourth-Order Functional Dynamic Equations

Take a t3 ∈ [t2 , ∞) so that w(τ1 (t)) < 0, t ∈ [t3 , ∞). Hence, t ∈ [t3 , ∞),

z(t) − F (t) < 0, or

x(t) + p(t)x(τ1 (t)) < F (t),

t ∈ [t3 , ∞),

and t ∈ [t3 , ∞).

x(t) < F (t), Then

lim inf x(t) ≤ lim inf F (t) t→∞

t→∞

< 0. This is a contradiction. 2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain a contradiction. This completes the proof. Theorem 11.16. Let −1 < P ≤ p(t) ≤ 0, t ∈ [t0 , ∞), P is a constant, and 2 ([t , ∞)) such that F changes sign (F 1)–(F 3) hold, there exists a function F ∈ Crd 0 with −∞ < lim inf F (t) < 0 < lim sup F (t) < ∞, t→∞

t→∞

Δ2  2 2 ([t , ∞)), rF Δ2 rF Δ ∈ Crd = f . If 0 ∞ t0

  q(t)G F + (τ2 (t)) Δt = ∞

∞

and t0

  q(t)G F − τ1−1 (τ2 (t)) Δt = ∞,

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

641

and ∞

  q(t)G F − (τ2 (t)) Δt = ∞

∞

and

t0

t0

  q(t)G F + τ1−1 (τ2 (t)) Δt = ∞,

then every solutions of the equation (11.39) oscillate. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞). 1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

Then z(t) ≥ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then, using (11.40), we have that  Δ2 2 r(t)w Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). Take a t3 ∈ [t2 , ∞) such that w(τ1 (t)) > 0,

t ∈ [t3 , ∞).

Therefore one of the cases of Lemma 11.5 holds. Since w(t) > 0, t ∈ [t3 , ∞), we obtain F (t) < z(t) = x(t) + p(t)x(τ1 (t)) < x(t),

t ∈ [t3 , ∞).

642

11 Oscillations of Fourth-Order Functional Dynamic Equations

Hence, x(t) ≥ F + (t),

t ∈ [t3 , ∞).

Take a t4 ∈ [t3 , ∞) so that τ2 (t) ≥ t3 for t ∈ [t4 , ∞). From here and from (11.40), we arrive to  Δ2 2 r(t)w Δ (t) = −q(t)G(x(τ2 (t)))   ≤ −q(t)G F + (τ2 (t)) ,

t ∈ [t4 , ∞),

whereupon  Δ2   2 q(t)G F + (τ2 (t)) ≤ − r(t)w Δ (t) ,

t ∈ [t4 , ∞),

and ∞

  q(t)G F + (τ2 (t)) Δt ≤ −

t4

∞

2

r(t)w Δ (t)

Δ2 Δt

t4

< ∞. This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0, t ∈ [t2 , ∞). (i) Suppose that there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

2

w Δ (t) < 0,

 Δ 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞).

Since w(t) < 0, t ∈ [t3 , ∞), we get F (t) > z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t3 , ∞),

and −F (t) ≤ −x(t) − p(t)x(τ1 (t)) ≤ −p(t)x(τ1 (t)) ≤ x(τ1 (t)),

t ∈ [t3 , ∞).

F − (t) ≤ x(τ1 (t)),

t ∈ [t3 , ∞),

Then

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

643

and  F − τ1−1 (t) ≤ x(t),

t ∈ [t3 , ∞).

Take t4 ∈ [t3 , ∞) such that τ2 (t) ≥ t3 , t ∈ [t4 , ∞). Therefore  F − τ1−1 (τ2 (t)) ≤ x(τ2 (t)),

t ∈ [t4 , ∞).

Hence from (11.40), we obtain  Δ2 2 r(t)w Δ (t) = −q(t)G(x(τ2 (t)))   ≤ −q(t)G F − τ1−1 (τ2 (t)) ,

t ∈ [t4 , ∞),

and    Δ2 2 q(t)G F − τ1−1 (τ2 (t)) ≤ − r(t)w Δ (t) , t ∈ [t4 , ∞), and ∞ t4

  q(t)G F − τ1−1 (τ2 (t)) Δt ≤ −

∞

2

r(t)w Δ (t)

Δ2 Δt

t4

< ∞.

This is a contradiction. (ii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) < 0,

2

 Δ2 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞),

2

 Δ2 2 r(t)w Δ (t) < 0,

t ∈ [t3 , ∞).

w Δ (t) < 0,

or w Δ (t) < 0,

w Δ (t) < 0,

Since w(t) < 0 and t ∈ [t3 , ∞), we have that w is unbounded on [t3 , ∞). Because F is bounded on [t3 , ∞), we conclude that z is unbounded on [t3 , ∞) and then x is unbounded on [t3 , ∞). Hence, there is an increasing sequence {sn }n∈N ⊂ [t3 , ∞) such that sn → ∞, as n → ∞, and x(sn ) = − max{x(t) : t3 ≤ t ≤ sn }.

644

11 Oscillations of Fourth-Order Functional Dynamic Equations

We choose n large enough so that τ1 (sn ) > t3 . Hence, w(sn ) = x(sn ) + p(sn )x(τ1 (sn )) − F (sn ) ≥ x(sn ) + P x(τ1 (sn )) − F (sn ) ≥ x(sn ) + P x(sn ) − F (sn ) = (1 + P )x(sn ) − F (sn ), n ∈ N is large enough. Since 1 + P > 0, x is unbounded and F is bounded on [t3 , ∞), we conclude that w(sn ) > 0 for n ∈ N large enough. This is a contradiction. (iii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) < 0,

 Δ2 2 r(t)w Δ (t) > 0,

2

w Δ (t) > 0,

t ∈ [t3 , ∞).

Since w < 0 on [t3 , ∞), we have that w is bounded on [t3 , ∞). Then there are a t4 ∈ [t3 , ∞) and a constant c > 0 such that 2

r(t)w Δ (t) ≥ c,

t ∈ [t4 , ∞).

Hence, c

t 2 ≤ tw Δ (t), r(t)

t ∈ [t4 , ∞).

Then t

c t4

s Δs ≤ r(s)

t

2

sw Δ (s)Δs

t4

!s=t ! = sw Δ (s)! − s=t4

t

w Δσ (s)Δs

t4

≤ tw Δ (t) − t4 w Δ (t4 ) −

t

w Δ (s)Δs

t4

≤ −t4 w Δ (t4 ) + w(t) − w(t4 ), Because w is bounded on [t4 , ∞), we conclude that ∞ t4

This is a contradiction.

s Δs < ∞. r(s)

t ∈ [t4 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

645

2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain a contradiction. This completes the proof. Theorem 11.17. Let −∞ < P ≤ p(t) ≤ 0, t ∈ [t0 , ∞), P is a constant, and 1. G(u)G(v) = G(uv), u, v ∈ R, 2 ([t , ∞)) such that F changes sign with 2. there exists a function F ∈ Crd 0 −∞ < lim inf F (t) < 0 < lim sup F (t) < ∞, t→∞

t→∞

Δ2  2 2 ([t , ∞)), rF Δ2 rF Δ ∈ Crd = f. 0 If ∞

  q(t)G F + (τ2 (t)) Δt = ∞

∞

and

t0

t0

  q(t)G F − τ1−1 (τ2 (t)) Δt = ∞,

and ∞

  q(t)G F − (τ2 (t)) Δt = ∞

∞

and

t0

t0

  q(t)G F + τ1−1 (τ2 (t)) Δt = ∞,

then every solution of the equation (11.39) oscillates or tends to ±∞ as t → ∞. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞). 1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

646

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then z(t) ≥ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then, using (11.40), we have that  Δ2 2 r(t)w Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). Take a t3 ∈ [t2 , ∞) such that w(τ1 (t)) > 0,

t ∈ [t3 , ∞).

Therefore one of the cases of Lemma 11.5 holds. Since w(t) > 0, t ∈ [t3 , ∞), we obtain F (t) < z(t) = x(t) + p(t)x(τ1 (t)) < x(t),

t ∈ [t3 , ∞).

x(t) ≥ F + (t),

t ∈ [t3 , ∞).

Hence,

Take a t4 ∈ [t3 , ∞) so that τ2 (t) ≥ t3 for t ∈ [t4 , ∞). From here and from (11.40), we arrive to  Δ2 2 r(t)w Δ (t) = −q(t)G(x(τ2 (t)))   ≤ −q(t)G F + (τ2 (t)) ,

t ∈ [t4 , ∞),

whereupon  Δ2   2 q(t)G F + (τ2 (t)) ≤ − r(t)w Δ (t) ,

t ∈ [t4 , ∞),

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

647

and ∞

  q(t)G F + (τ2 (t)) Δt ≤ −

t4

∞

2

r(t)w Δ (t)

Δ2 Δt

t4

< ∞. This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0, t ∈ [t2 , ∞). (i) Suppose that there is a t3 ∈ [t2 , ∞) such that  Δ 2 2 w Δ (t) > 0, w Δ (t) < 0, r(t)w Δ (t) > 0,

t ∈ [t3 , ∞).

Since w(t) < 0, t ∈ [t3 , ∞), we get F (t) > z(t) = x(t) + p(t)x(τ1 (t)) ≥ P x(τ1 (t)),

t ∈ [t3 , ∞).

Hence,  x(t) ≥ P −1 F − τ1−1 (t) ,

t ∈ [t3 , ∞).

Take a t4 ∈ [t3 , ∞) such that τ2 (t) ≥ t3 , t ∈ [t4 , ∞). From here and from (11.40), we arrive to  Δ2 2 r(t)w Δ (t) = −q(t)G(x(τ2 (t)))   ≤ −q(t)G P −1 F − τ1−1 (t)   = −q(t)G(P −1 )G F − τ1−1 (t) ,

t ∈ [t4 , ∞).

Then    Δ2 2 q(t)G(P −1 )G F − τ1−1 (t) ≤ − r(t)w Δ (t) ,

t ∈ [t4 , ∞),

and G(P −1 )

∞ t4

  q(t)G F − τ1−1 (t) Δt ≤ −

t

 Δ2 2 r(t)w Δ (t) Δt

t4

< ∞.

This is a contradiction.

648

11 Oscillations of Fourth-Order Functional Dynamic Equations

(ii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) < 0,

2

 Δ2 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞),

2

 Δ2 2 r(t)w Δ (t) < 0,

t ∈ [t3 , ∞).

w Δ (t) < 0,

or w Δ (t) < 0,

w Δ (t) < 0,

Since w(t) < 0, t ∈ [t4 , ∞), we have that w is unbounded on [t4 , ∞) and lim w(t) = −∞.

t→∞

Hence from P x(τ1 (t)) ≤ x(t) + P x(τ1 (t)) < w(t) + F (t),

t ∈ [t3 , ∞),

we get lim sup P x(τ1 (t)) ≤ lim w(t) + lim F (t) t→∞

t→∞

t→∞

= −∞. Therefore lim x(t) = ∞.

t→∞

(iii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) < 0,

2

w Δ (t) > 0,

 Δ2 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞).

Since w < 0 on [t3 , ∞), we have that w is bounded on [t3 , ∞). Then there are a t4 ∈ [t3 , ∞) and a constant c > 0 such that 2

r(t)w Δ (t) ≥ c,

t ∈ [t4 , ∞).

Hence, c

t 2 ≤ tw Δ (t), r(t)

t ∈ [t4 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

649

Then t

c t4

s Δs ≤ r(s)

t

2

sw Δ (s)Δs

t4

!s=t ! = sw Δ (s)! − s=t4

t

w Δσ (s)Δs

t4 t

≤ tw Δ (t) − t4 w Δ (t4 ) −

w Δ (s)Δs

t4

≤ −t4 w Δ (t4 ) + w(t) − w(t4 ),

t ∈ [t4 , ∞).

Because w is bounded on [t4 , ∞), we conclude that ∞ t4

s Δs < ∞. r(s)

This is a contradiction. 2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain a contradiction. This completes the proof. Corollary 11.5. Suppose that all the conditions of Theorem 11.17 are satisfied. Then every bounded solution of the equation (11.39) oscillates. Theorem 11.18. Let 0 ≤ p(t) ≤ P < 1, t ∈ [t0 , ∞), P is a constant, and (F 1)– (F 3) hold, 1. there exists a λ > 0 such that G(u) + G(v) ≥ λG(u + v), 2. G(u)G(v) ≥ G(uv), u > 0, v > 0, 3. G(−u) = −G(u),

u > 0,

v > 0,

650

11 Oscillations of Fourth-Order Functional Dynamic Equations

2 ([t , ∞)) such that F does not change sign, lim 4. there exists F ∈ Crd 0 t→∞ F (t) = 2  Δ 2 2 ([t , ∞)), rF Δ2 = f. 0, rF Δ ∈ Crd 0

If ∞

Q(t)G (|F (τ2 (t))|) Δt = ∞,

t0

then every bounded solution of the equation (11.39) oscillates or tends to zero as t → ∞. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞). 1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set t ∈ [t1 , ∞).

z(t) = x(t) + p(t)x(τ1 (t)), Then 0 < z(t) ≤ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then, using (11.40), we have that  Δ2 2 r(t)w Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). Take a t3 ∈ [t2 , ∞) such that τ2 (t) ≥ t2 , t ∈ [t3 , ∞). Then w(τ2 (t)) > 0, t ∈ [t3 , ∞). Hence, w(t) = z(t) − F (t) > 0,

t ∈ [t3 , ∞),

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

651

i.e., t ∈ [t3 , ∞).

z(t) > F (t),

(i) Assume that there is a t4 ∈ [t3 , ∞) such that F (t) > 0, t ∈ [t4 , ∞). Take a t5 ∈ [t4 , ∞) so that τ2 (t) > t4 , t ∈ [t5 , ∞). From (11.41), we get Δ2  Δ2  2 2 + G(P ) r(τ1 (t))w Δ (τ1 (t)) 0 ≥ r(t)w Δ (t) +λQ(t)G(z(τ2 (t)))  Δ2  Δ2 2 2 ≥ r(t)w Δ (t) + G(P ) r(τ1 (t))w Δ (τ1 (t)) +λQ(t)G(F (τ2 (t))),

t ∈ [t5 , ∞),

or  Δ2  Δ2 2 2 λQ(t)G(F (τ2 (t))) ≤ − r(t)w Δ (t) − G(P ) r(τ1 (t))w Δ (τ1 (t)) ,

t ∈ [t5 , ∞), and ∞

∞

λQ(t)G(F (τ2 (t)))Δt ≤ −

2

r(t)w Δ (t)

t5

Δ2 Δt

t5 ∞

−G(P )

2

r(τ1 (t))w Δ (τ1 (t))

Δ2 Δt

t5

< ∞. This is a contradiction. (ii) Let t4 ∈ [t3 , ∞) be such that F (t) < 0, t ∈ [t4 , ∞). By (11.40), we get ∞

q(t)G(x(τ2 (t)))Δt ≤ −

t4

∞

2

r(t)w Δ (t)

Δ2 Δt

t4

< ∞. Hence, lim inf x(t) = 0. t→∞

Because w is bounded and monotonic, we conclude that there exists limt→∞ w(t). From here, there exists limt→∞ z(t). Hence from Lemma 11.7, we conclude that lim z(t) = 0.

t→∞

652

11 Oscillations of Fourth-Order Functional Dynamic Equations

Since t ∈ [t4 , ∞),

x(t) ≤ z(t), we get

lim x(t) = 0.

t→∞

Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0, t ∈ [t2 , ∞). Then x(t) ≤ z(t) < F (t) → 0 as

t → ∞.

Consequently lim x(t) = 0.

t→∞

2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain our claim. This completes the proof. Theorem 11.19. Let −1 < P ≤ p(t) ≤ 0, t ∈ [t0 , ∞), P is a constant, and (F 1)– 2 ([t , ∞)) such that F does not change (F 3) hold, there exists a function F ∈ Crd 0 Δ2  2 2 ([t , ∞)), rF Δ2 = f . If sign, limt→∞ F (t) = 0, rF Δ ∈ Crd 0 ∞

q(t)Δt = ∞,

t0

then every solution of the equation (11.39) oscillates or tends to zero as t → ∞. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

653

1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

Then z(t) ≥ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then, using (11.40), we have that  Δ2 2 r(t)w Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). By (11.40), we get  Δ2 2 q(t)G(x(τ2 (t))) ≤ − r(t)w Δ (t) ,

t ∈ [t2 , ∞),

and #∞ t2

q(t)G(x(τ2 (t)))Δt ≤ −

Δ2 #∞ Δ2 (t) r(t)w Δt t2

< ∞.

(11.42)

Hence, lim inf x(t) = 0. t→∞

(i) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

2

w Δ (t) > 0,

 Δ2 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞).

654

11 Oscillations of Fourth-Order Functional Dynamic Equations

Then lim w(t) = ∞.

t→∞

Because lim F (t) = 0,

t→∞

we conclude that lim z(t) = ∞.

t→∞

On the other hand, z(t) = x(t) + p(t)x(τ1 (t)) ≤ x(t),

t ∈ [t3 , ∞).

Then lim x(t) = ∞.

t→∞

This is a contradiction. (ii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

2

w Δ (t) < 0,

 Δ2 2 r(t)w Δ (t) > 0,

t ∈ [t3 , ∞).

Let lim w(t) = l.

t→∞

We have 0 < l ≤ ∞. If l = ∞, then lim z(t) = ∞ and

t→∞

lim x(t) = ∞.

t→∞

This is a contradiction. Then 0 < l < ∞. By Lemma 11.7, it follows that l = 0. This is a contradiction.

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

655

(b) Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0, t ∈ [t2 , ∞). Assume that x is unbounded on [t2 , ∞). Then there is a sequence {sn }n∈N ⊂ [t2 , ∞) such that sn → ∞,

x(sn ) → ∞,

as

n → ∞,

and x(sn ) = max{x(t) : t2 ≤ t ≤ sn }. We have w(sn ) ≥ x(sn ) + P x(τ1 (sn )) − F (sn ) ≥ (1 + P )x(sn ) − F (sn ) >0 for each n ∈ N large enough. This is a contradiction. Therefore x is bounded and w is bounded. Then there is a t3 ∈ [t2 , ∞) such that 2



2



w Δ (t) > 0,

w Δ (t) < 0,

w Δ (t) < 0,

w Δ (t) > 0,

2

Δ2

2

Δ2

r(t)w Δ (t)

> 0,

t ∈ [t3 , ∞),

> 0,

t ∈ [t3 , ∞).

or r(t)w Δ (t)

Hence from (11.40), we obtain (11.42). Then lim inf x(t) = 0 t→∞

and limt→∞ w(t) exists. Therefore limt→∞ z(t) exists. From here and from Lemma 11.7, it follows that lim z(t) = 0.

t→∞

Then 0 = lim z(t) t→∞

= lim sup (x(t) + p(t)x(τ1 (t))) t→∞

≥ lim sup x(t) + lim inf P x(τ1 (t)) t→∞

t→∞

= (1 + P ) lim sup x(t). t→∞

656

11 Oscillations of Fourth-Order Functional Dynamic Equations

Since 1 + P > 0, we get lim x(t) = 0.

t→∞

2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain our claim. This completes the proof. Example 11.13. Let T = 2N0 . Consider the equation % t x(t) −

1 x t +1

1 

Δ2 &Δ2 63 t t = +t 3x , 2 4 256t 4

Here σ (t) = 2t, r(t) = t, p(t) = −

1 , t +1

q(t) = t, 63 , 256t 4 t τ1 (t) = , 2 t τ2 (t) = , 4 √ G(u) = 3 u, u ∈ R. f (t) =

t ∈ [4, ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

We have ∞ t0

∞ t Δt = Δt r(t) 4 = ∞,

r(t) > 0, t ∈ [4, ∞),

q(t) > 0, ∞

∞

q(t)Δt =

tΔt 4

t0

= ∞. Let F (t) =

1 , t

t ∈ [4, ∞).

We have F (t) > 0,

t ∈ [4, ∞),

lim F (t) = 0,

t→∞

F Δ (t) = −

1 tσ (t)

1 , 2t 2 σ (t) + t 2 F Δ (t) = 2 2t (σ (t))2 =−

=

2t + t 2t 2 (2t)2

3t 8t 4 3 = 3, 8t 

3 2 r(t)F Δ (t) = t 8t 3 =

3 , 8t 2  Δ 3 σ (t) + t 2 r(t)F Δ (t) = − 2 8 t (σ (t))2 =

657

658

11 Oscillations of Fourth-Order Functional Dynamic Equations

=−

3 2t + t 8 t 2 (4t 2 )

9t 32t 4 9 = − 3, 32t =−



2

r(t)F Δ (t)

Δ2

=

9 (σ (t))2 + tσ (t) + t 2 32 t 3 (σ (t))3

=

9 4t 2 + 2t 2 + t 2 32 t 3 (8t 3 )

63t 2 256t 6 63 = 256t 4 = f (t), t ∈ [4, ∞). =

Therefore any solution of the considered equation oscillates or tends to zero as t → ∞. Exercise 11.12. Let T = 4N0 . Prove that any solution of the equation 1 

Δ4 1 t 1 t 3 5 x(t) − x = 5, +t x 2t 4 16 t

t ∈ [16, ∞),

oscillates or tends to zero as t → ∞. Theorem 11.20. Let −∞ < p1 ≤ p(t) ≤ p2 < −1, t ∈ [t0 , ∞), p1 , and p2 are 2 ([t , ∞)) such that F constants, (F 1)–(F 3) hold, there exists a function F ∈ Crd 0 Δ2  2 2 ([t , ∞)), rF Δ2 = f . If does not change sign, limt→∞ F (t) = 0, rF Δ ∈ Crd 0 ∞

q(t)Δt = ∞,

t0

then every bounded solution of the equation (11.39) oscillates or tends to zero as t → ∞. Proof. Suppose that the equation (11.39) has a nonoscillatory solution x on [t0 , ∞).

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

659

1. Let x is an eventually positive solution of the equation (11.39) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

x(τ1 (τ2 (t))) > 0,

t ∈ [t1 , ∞). Set z(t) = x(t) + p(t)x(τ1 (t)),

t ∈ [t1 , ∞).

Then z(t) ≥ x(t) + P x(τ1 (t)),

t ∈ [t1 , ∞).

Let w(t) = z(t) − F (t),

t ∈ [t1 , ∞).

Then, using (11.40), we have that  Δ2 2 r(t)w Δ (t) ≤ 0,

t ∈ [t1 , ∞).

(a) Let there is a t2 ∈ [t1 , ∞) such that w(t) > 0, t ∈ [t2 , ∞). By (11.40), we get  Δ2 2 q(t)G(x(τ2 (t))) ≤ − r(t)w Δ (t) ,

t ∈ [t2 , ∞),

and we get (11.42). Hence, lim inf x(t) = 0. t→∞

(i) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

2

w Δ (t) > 0,

 Δ2 2 r(t)w Δ (t) > 0,

Then lim w(t) = ∞.

t→∞

t ∈ [t3 , ∞).

660

11 Oscillations of Fourth-Order Functional Dynamic Equations

Because lim F (t) = 0,

t→∞

we conclude that lim z(t) = ∞.

t→∞

On the other hand, z(t) = x(t) + p(t)x(τ1 (t)) ≤ x(t),

t ∈ [t3 , ∞).

Then lim x(t) = ∞.

t→∞

This is a contradiction. (ii) Let there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

 Δ2 2 r(t)w Δ (t) > 0,

2

w Δ (t) < 0,

t ∈ [t3 , ∞).

Let lim w(t) = l.

t→∞

We have 0 < l ≤ ∞. If l = ∞, then lim z(t) = ∞ and

lim x(t) = ∞.

t→∞

t→∞

This is a contradiction. Then 0 < l < ∞. By Lemma 11.7, it follows that l = 0. This is a contradiction. (b) Let there is a t2 ∈ [t1 , ∞) such that w(t) < 0, t ∈ [t2 , ∞). Since x is bounded on [t2 , ∞) we conclude that w is bounded on [t2 , ∞). Then there is a t3 ∈ [t2 , ∞) such that w Δ (t) > 0,

2

w Δ (t) < 0,



2

r(t)w Δ (t)

Δ2

> 0,

t ∈ [t3 , ∞),

11.3 Oscillations of Fourth-Order Nonlinear Neutral Delay Dynamic Equations

661

or w Δ (t) < 0,

2

w Δ (t) > 0,



2

r(t)w Δ (t)

Δ2

> 0,

t ∈ [t3 , ∞).

Hence from (11.40), we obtain (11.42). Then lim inf x(t) = 0 t→∞

and limt→∞ w(t) exists. Therefore limt→∞ z(t) exists. From here and from Lemma 11.7, it follows that lim z(t) = 0.

t→∞

Then 0 = lim z(t) t→∞

= lim sup (x(t) + p(t)x(τ1 (t))) t→∞

≤ lim sup x(t) + lim inf p2 x(τ1 (t)) t→∞

t→∞

= (1 + p2 ) lim sup x(t). t→∞

Since 1 + p2 < 0, we get lim x(t) = 0.

t→∞

2. Let x is an eventually negative solution of the equation (11.39) on [t0 , ∞). Set y(t) = −x(t), f˜(t) = −f (t), F˜ (t) = −F (t),

t ∈ [t0 , ∞).

Then y satisfies the equation 2  2 Δ r(t) (y(t) + p(t)y(τ1 (t)))Δ + q(t)G(y(τ2 (t))) = f˜(t),

t ∈ [t0 , ∞).

Proceeding as above, we obtain our claim. This completes the proof.

662

11 Oscillations of Fourth-Order Functional Dynamic Equations

Exercise 11.13. Let T = 2N0 . Prove that every bounded solution of the equation 1  %

Δ2 &Δ2 t 3 12t + 1 t x = 4, +t 3x t x(t) − t +1 2 4 t

t ∈ [4, ∞),

oscillates or tends to zero as t → ∞. Corollary 11.6. Suppose that all the conditions of Theorem 11.20 are satisfied. Then every nonoscillatory solution of the equation (11.39) which does not tend to zero as t → ∞ is unbounded.

11.4 Advanced Practical Problems Problem 11.1. Let T = 2N0 . Write an analogue of Theorem 11.1 in the following cases. 1. 2. 3. 4. 5. 6. 7.

α(t) = 1, β(t) = t, t ∈ T, α(t) = t, β(t) = 1, t ∈ T, α(t) = β(t) = t, t ∈ T, α(t) = t 2 , β(t) = t, t ∈ T, α(t) = 1, β(t) = t 2 , t ∈ T, α(t) = t, β(t) = t 2 , t ∈ T, α(t) = β(t) = t 2 , t ∈ T.

Problem 11.2. Let T = 4N0 . Prove that the equation 4

x Δ (t) +

1 t3

5 t x = 0, 4

t ∈ [4, ∞),

is oscillatory. Problem 11.3. Write an analogue of Theorem 11.2 in the following cases. 1. T = hZ, h > 0, 2. T = q N0 , q > 1. Problem 11.4. Let T = R+ . Prove that the equation  √ 3 = 0, x (4) (t) + x 4 t is oscillatory.

t ∈ R+ ,

11.4 Advanced Practical Problems

663

Problem 11.5. Let T = 4N0 . Write an analogue of Theorem 11.4 in the following cases. 1. 2. 3. 4. 5.

α(t) = 1, β(t) = t, t ∈ T, α(t) = β(t) = t, t ∈ T, α(t) = t, β(t) = 1, t ∈ T, α(t) = t, β(t) = t 2 , t ∈ T, α(t) = β(t) = t 2 , t ∈ T.

Problem 11.6. Let T = 3N0 . Investigate if the equation x

Δ4

t  17 (t) + t + 3t + 2 x = 0, 3 

4

2

t ∈ [3, ∞),

is oscillatory. Problem 11.7. Let T = 2N0 . Prove that any bounded solution x of the equation

 3 Δ x Δ (t) φ + 20 5

1 t

 t t 40 2 φ3 x Δ 250000 2

 t = 0, +t 20 φ3 x 4

t ∈ [4, ∞),

2

either x or x Δ is oscillatory. Problem 11.8. Let T = 2N0 . Prove that any bounded solution x of the equation



 1 9 Δ2 t t φ3 x 150000 2 t

 t t 3 16 3 Δτ = 0, t ∈ [16, ∞), + τ φ3 x 27 4 τ  3 Δ φ x Δ (t) + 90 5

1

2

either x or x Δ is oscillatory. Problem 11.9. Let T = 4N0 . Prove that any solution of the equation % t 3 x(t) + t ∈ [16, ∞), oscillates.

1 x 3 t +1

1 

Δ2 &Δ2 t t = 0, + t9 7 x 4 16

664

11 Oscillations of Fourth-Order Functional Dynamic Equations

Problem 11.10. Let T = 4N0 . Prove that any solution of the equation % √ t x(t) −

1 x 12(t 4 + t 2 + 1)

1 

Δ2 &Δ2 t t = 0, + t3 7 4 16

t ∈ [16, ∞), oscillates or tends to zero. Problem 11.11. Let T = 3N0 . Prove that every solution of the equation %

3t x(t) −

1 x 3t 2 + t + 1

1 

Δ2 &Δ2 t t 1 4 15 +t x = 4, 3 9 t

t ∈ [9, ∞),

oscillates or tends to zero. Problem 11.12. Let T = 3N0 . Prove that every bounded solution of the equation ⎞ 2 1  2 Δ

6 + t 2 + 1 t Δ 29t ⎠ + t 10 11 x t = 1 , ⎝17t x(t) − x 2 3 9 3t + t + 1 t4 ⎛

oscillates or tends to zero.

t ∈ [9, ∞),

Chapter 12

Oscillations of Higher-Order Functional Dynamic Equations

The material of this chapter are based on results of the papers and monographs [80, 83, 84, 94, 154, 160, 174, 225–227, 229, 241, 250, 254]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

12.1 Oscillations of Higher-Order Delay Dynamic Equations In this section we investigate the following higher-order delay dynamic equation. %

% an (t)



 Δ Δ an−1 (t) . . . a1 (t)x Δ (t) . . .

Δ &α &Δ + g(t, x(τ (t))) = 0,

t ∈ [t0 , ∞),

(12.1) where n ≥ 2, n ∈ N, and (G1) (G2) (G3)

ak ∈ Crd (T), k ∈ {1, . . . , n}, are positive functions, τ : T → T is an increasing differentiable function with τ (t) ≤ t, τ (σ (t)) = σ (τ (t)), t ∈ T, limt→∞ τ (t) = ∞, g ∈ C (T × R) with g(t, x) ≥ q(t) xβ for some positive function q ∈ Crd (T), when x = 0, α ≥ 1, β ≥ 1 are two quotients of odd positive integers.

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_12

665

666

12 Oscillations of Higher-Order Functional Dynamic Equations

Write ⎧ ⎨ x(t) if k = 0 Δ (t, x(t)) if Sk (t, x(t)) = ak (t)Sk−1  Δ α ⎩ an (t) Sn−1 (t, x(t))

k ∈ {1, . . . , n − 1} if k = n,

t ∈ [t0 , ∞). Then the equation (12.1) can be written in the following form SnΔ (t, x(t)) + g(t, x(τ (t))) = 0,

t ∈ [t0 , ∞).

(12.2)

l ∈ {1, . . . , n − 1},

(12.3)

We will start with the following useful lemmas. Lemma 12.1. Assume ∞ t0

1 an (s)

1

α

Δs =

∞ t0

Δs = ∞, al (s)

and m ∈ {1, . . . , n}. 1. If lim inf Sm (t, x(t)) > 0, t→∞

then lim Sl (t, x(t)) = ∞,

l ∈ {0, . . . , m − 1}.

t→∞

2. If lim sup Sm (t, x(t)) < 0, t→∞

then lim Sl (t, x(t)) = −∞,

t→∞

l ∈ {0, . . . , m − 1}.

Proof. 1. Let c1 = lim inf Sm (t, x(t)). t→∞

(a) Suppose that 0 < c1 < ∞. Then there exist an  > 0 and a t1 ∈ [t0 , ∞) such that Sm (t, x(t)) ≥

c1 , 2

t ∈ [t1 , ∞).

12.1 Oscillations of Higher-Order Delay Dynamic Equations

667

(i) Let m ∈ {1, . . . , n − 1}. Then c1 , 2

t ∈ [t1 , ∞),

c1 1 , 2 am (t)

t ∈ [t1 , ∞).

Δ (t, x(t)) ≥ am (t)Sm−1

and Δ Sm−1 (t, x(t)) ≥

Hence, Sm−1 (t, x(t)) − Sm−1 (s, x(s)) ≥

t

c1 2

s

1 Δy, am (y)

s, t ∈ [t1 , ∞), s < t. Therefore lim Sm−1 (t, x(t)) = ∞.

t→∞

By induction, lim Sl (t, x(t)) = ∞,

t→∞

l ∈ {0, . . . , m − 1}.

(ii) Let m = n. Then  Δ α c1 an (t) Sn−1 (t, x(t)) ≥ , 2

t ∈ [t1 , ∞),

and Δ Sn−1 (t, x(t))

 c 1 1  α1 1 α ≥ , 2 an (t)

t ∈ [t1 , ∞).

Hence, Sn−1 (t, x(t)) − Sn−1 (s, x(s)) ≥

c 1 1

2

t

α

s

1 an (y)

s, t ∈ [t1 , ∞), s < t. Therefore lim Sn−1 (t, x(t)) = ∞.

t→∞

By induction, lim Sl (t, x(t)) = ∞,

t→∞

l ∈ {0, . . . , n − 1}.

1

α

Δy,

668

12 Oscillations of Higher-Order Functional Dynamic Equations

(b) Now we suppose that c1 = ∞. Then for any c2 > 0 there is a t2 ∈ [t0 , ∞) such that 1 , c2

Sm (t, x(t)) ≥

t ∈ [t2 , ∞).

(i) Let m ∈ {1, . . . , n − 1}. Then 1 , c2

t ∈ [t2 , ∞),

1 1 , c2 am (t)

t ∈ [t2 , ∞).

Δ am (t)Sm−1 (t, x(t)) ≥

and Δ Sm−1 (t, x(t)) ≥

Hence, Sm−1 (t, x(t)) − Sm−1 (s, x(s)) ≥

1 c2

t s

1 Δy, am (y)

s, t ∈ [t2 , ∞), s < t. Therefore lim Sm−1 (t, x(t)) = ∞.

t→∞

By induction, lim Sl (t, x(t)) = ∞,

l ∈ {0, . . . , m − 1}.

t→∞

(ii) Let m = n. Then  Δ α 1 (t, x(t)) ≥ , an (t) Sn−1 c2

t ∈ [t2 , ∞),

and

Δ Sn−1 (t, x(t))

≥

1 c2

1 α

1 an (t)

1

α

t ∈ [t2 , ∞).

,

Hence,

Sn−1 (t, x(t)) − Sn−1 (s, x(s)) ≥

1 c2

1

t

α

s

1 an (y)

1

α

Δy,

12.1 Oscillations of Higher-Order Delay Dynamic Equations

669

s, t ∈ [t2 , ∞), s < t. Therefore lim Sn−1 (t, x(t)) = ∞.

t→∞

By induction, lim Sl (t, x(t)) = ∞,

t→∞

l ∈ {0, . . . , n − 1}.

2. Let c3 = lim sup Sm (t, x(t)). t→∞

(a) Suppose that −∞ < c3 < 0. Then there exists a t3 ∈ [t0 , ∞) such that Sm (t, x(t)) ≤

c3 , 2

t ∈ [t3 , ∞).

(i) Let m ∈ {1, . . . , n − 1}. Then c3 , 2

t ∈ [t3 , ∞),

c3 1 , 2 am (t)

t ∈ [t3 , ∞).

Δ am (t)Sm−1 (t, x(t)) ≤

and Δ Sm−1 (t, x(t)) ≤

Hence, Sm−1 (t, x(t)) − Sm−1 (s, x(s)) ≤

c3 2

t s

1 Δy, am (y)

s, t ∈ [t3 , ∞), s < t. Therefore lim Sm−1 (t, x(t)) = −∞.

t→∞

By induction, lim Sl (t, x(t)) = −∞,

t→∞

l ∈ {0, . . . , m − 1}.

(ii) Let m = n. Then  Δ α c3 an (t) Sn−1 (t, x(t)) ≤ , 2

t ∈ [t3 , ∞),

670

12 Oscillations of Higher-Order Functional Dynamic Equations

and Δ (t, x(t)) Sn−1

 c 1 1  α1 3 α ≤ , 2 an (t)

t ∈ [t3 , ∞).

Hence, Sn−1 (t, x(t)) − Sn−1 (s, x(s)) ≤

c 1 3

t

α

2

s

1 an (y)

1

α

Δy,

s, t ∈ [t3 , ∞), s < t. Therefore lim Sn−1 (t, x(t)) = −∞.

t→∞

By induction, lim Sl (t, x(t)) = −∞,

t→∞

l ∈ {0, . . . , n − 1}.

(b) Now we suppose that c3 = −∞. Then for any c4 < 0 there is a t4 ∈ [t0 , ∞) such that Sm (t, x(t)) ≤ c4 ,

t ∈ [t4 , ∞).

(i) Let m ∈ {1, . . . , n − 1}. Then Δ (t, x(t)) ≤ c4 , am (t)Sm−1

t ∈ [t4 , ∞),

1 , am (t)

t ∈ [t4 , ∞).

and Δ Sm−1 (t, x(t)) ≤ c4

Hence, t

Sm−1 (t, x(t)) − Sm−1 (s, x(s)) ≤ c4 s

1 Δy, am (y)

s, t ∈ [t4 , ∞), s < t. Therefore lim Sm−1 (t, x(t)) = −∞.

t→∞

By induction, lim Sl (t, x(t)) = −∞,

t→∞

l ∈ {0, . . . , m − 1}.

12.1 Oscillations of Higher-Order Delay Dynamic Equations

671

(ii) Let m = n. Then  Δ α (t, x(t)) ≤ c4 , an (t) Sn−1

t ∈ [t4 , ∞),

and Δ (t, x(t)) Sn−1

≤ (c4 )

1 α

1 an (t)

1

α

t ∈ [t4 , ∞).

,

Hence, t

1

Sn−1 (t, x(t)) − Sn−1 (s, x(s)) ≤ (c4 ) α

s

1 an (y)

1

α

Δy,

s, t ∈ [t4 , ∞), s < t. Therefore lim Sn−1 (t, x(t)) = −∞.

t→∞

By induction, lim Sl (t, x(t)) = −∞,

t→∞

l ∈ {0, . . . , n − 1}.

This completes the proof. Lemma 12.2. Assume (12.3). If SnΔ (t, x(t)) < 0 and x(t) > 0, t ∈ [t0 , ∞), then there exists an m ∈ {0, . . . , n} such that 1. m + n is even, 2. (−1)m+l Sl (t, x(t)) > 0, t ∈ [t0 , ∞), l ∈ {m, . . . , n}, 3. if m > 1, then there exists a t1 ∈ [t0 , ∞) such that Sl (t, x(t)) > 0, t ∈ [t1 , ∞), l ∈ {1, . . . , m − 1}. Proof. Assume that there is a t1 ∈ [t0 , ∞) such that Sn (t1 , x(t1 )) < 0. Since SnΔ (t, x(t)) < 0, t ∈ [t0 , ∞), we conclude that Sn (t, x(t)) < 0,

t ∈ [t1 , ∞),

and Sn (t, x(t)) < Sn (t1 , x(t1 )),

t ∈ (t1 , ∞).

672

12 Oscillations of Higher-Order Functional Dynamic Equations

Hence,  Δ α (t, x(t)) < Sn (t1 , x(t1 )), an (t) Sn−1  Δ α Sn (t1 , x(t1 )) Sn−1 (t, x(t)) < , an (t) 1

Δ (t, x(t)) < Sn−1

(Sn (t1 , x(t1 ))) α 1

,

(an (t)) α

t ∈ (t1 , ∞),

and 1

Sn−1 (t, x(t)) ≤ Sn−1 (t1 , x(t1 )) + (Sn (t1 , x(t1 ))) α

t t1

→ −∞

as

Δs

t → ∞.

Therefore there exists a t2 ∈ [t1 , ∞) such that t ∈ [t2 , ∞).

Sn−1 (t, x(t)) < 0,

By induction, we get points t1 , t2 , . . ., tn+1 such that t ∈ [tn−l+1 , ∞).

Sl (t, x(t)) < 0, In particular,

x(t) = S0 (t, x(t)) < 0,

t ∈ [tn+1 , ∞).

This is a contradiction. Therefore Sn (t, x(t)) > 0,

t ∈ [t0 , ∞).

We have two possibilities. Case 1. Case 2.

Sl (t) > 0 for any l ∈ {0, . . . , n − 1}, t ∈ [t0 , ∞), Sk (t) < 0 for some k ∈ {0, . . . , n − 1}, t ∈ [t0 , ∞).

1. Suppose Case 1. Then m = n. 2. Suppose Case 2. Let

M = j ∈ N0 : (−1)l+j Sl (t, x(t)) > 0, l ∈ {j + 1, . . . , n − 1},

j +l

t ∈ [t0 , ∞),

 even .

1

(an (s)) α

12.1 Oscillations of Higher-Order Delay Dynamic Equations

673

Note that n − 1 ∈ M and there exists m = min M. Assume that m > 1. Then Δ (t, x(t)) = Sm−1

=

Sm (t, x(t)) am (t) (−1)m+m Sm (t, x(t)) am (t)

> 0,

t ∈ [t0 , ∞).

(a) Assume that there is a t1 ∈ [t0 , ∞) such that Sm−1 (t, x(t)) > Sm−1 (t1 , x(t1 )),

t ∈ (t1 , ∞).

Then Δ (t, x(t)) = Sm−2

≥

Sm−1 (t, x(t)) am−1 (t) Sm−1 (t1 , x(t1 )) , am−1 (t)

t ∈ [t1 , ∞),

and Sm−2 (t, x(t)) ≥ Sm−2 (t1 , x(t1 )) t

+Sm−1 (t1 , x(t1 ))

t1

→ ∞ as

Δs am−1 (s)

t → ∞.

By induction, Sl (t, x(t)) → ∞,

as

t → ∞,

l ∈ {0, . . . , m − 2}.

(b) Assume Sm−1 (t, x(t)) < 0, t ∈ [t0 , ∞). Suppose that there is a t1 ∈ [t0 , ∞) such that Sm−2 (t1 , x(t1 )). Since Δ (t, x(t)) = Sm−2

Sm−1 (t, x(t)) am−1 (t)

< 0,

t ∈ [t1 , ∞),

674

12 Oscillations of Higher-Order Functional Dynamic Equations

we get Sm−2 (t, x(t)) < Sm−2 (t1 , x(t1 )) < 0,

t ∈ (t1 , ∞).

Hence, Δ (t, x(t)) = Sm−3


0,

t ∈ [t0 , ∞),

which contradicts with the definition of m. This completes the proof. Lemma 12.3. Assume (12.3) and ∞ t0

∞

1 an−1 (u)

u

1 an (s)

∞

1 α q(u)Δu ΔsΔu = ∞.

(12.4)

s

If x is an eventually positive solution of the equation (12.1), then there exists a t1 ∈ [t0 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t1 , ∞), 2. either limt→∞ x(t) = 0 or Sl (t, x(t)) > 0, t ∈ [t1 , ∞), l ∈ {0, . . . , n}. Proof. Since x is an eventually positive solution of the equation (12.1), then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

12.1 Oscillations of Higher-Order Delay Dynamic Equations

675

1. By the equation (12.2), we get SnΔ (t, x(t)) = −g(t, x(τ (t))) ≤ −q(t)(x(τ (t)))β t ∈ [t1 , ∞).

< 0,

2. Assume that there is a t2 ∈ [t1 , ∞) such that Sn (t2 , x(t2 )) < 0. Since SnΔ (t, x(t)) < 0, t ∈ [t2 , ∞), we conclude that Sn (t, x(t)) < Sn (t2 , x(t2 )) t ∈ (t2 , ∞).

< 0, Then

1

Δ Sn−1 (t, x(t))

=

(Sn (t, x(t))) α 1

(an (t)) α 1


0,

t ∈ [t3 , ∞).

676

12 Oscillations of Higher-Order Functional Dynamic Equations

By Lemma 12.2, it follows that there exists an m ∈ {0, . . . , n} such that m + n is even, (−1)m+l Sl (t, x(t)) > 0, t ∈ [t0 , ∞), l ∈ {m, . . . , n}, and if m > 1, then there exists a t4 ∈ [t0 , ∞) such that Sl (t, x(t)) > 0, t ∈ [t4 , ∞), l ∈ {1, . . . , m − 1}. If m = n, then Sl (t, x(t)) > 0, t ∈ [t4 , ∞), l ∈ {0, . . . , n}. Let m = n. Then Sn−1 (t, x(t)) < 0

and

Sn−2 (t, x(t)) > 0,

t ∈ [t4 , ∞).

Hence, there exists lim x(t) = l.

t→∞

We have 0 ≤ l < ∞. Assume that l > 0. Then there exists a t5 ∈ [t4 , ∞) such that x(τ (t)) ≥

l , 2

t ∈ [t5 , ∞).

By the equation (12.2), we get SnΔ (t, x(t)) = −g(t, x(τ (t))) ≤ −q(t)(x(τ (t)))β

β l ≤− q(t), t ∈ [t5 , ∞). 2 Hence,

β t6 l −Sn (t, x(t)) ≤ −Sn (t6 , x(t6 )) − q(s)Δs 2 t

β t6 l q(s)Δs, t, t6 ∈ [t5 , ∞), ≤− 2 t

t6 > t.

Therefore

β l −Sn (t, x(t)) ≤ − 2

∞

q(s)Δs,

t ∈ [t5 , ∞),

t

and Sn (t, x(t)) ≥

β l 2

∞

q(s)Δs, t

t ∈ [t5 , ∞).

12.1 Oscillations of Higher-Order Delay Dynamic Equations

677

From here, Δ (t, x(t)) Sn−1

β 1 l α ≥ 2 an (t)

1

∞

α

q(s)Δs

t ∈ [t5 , ∞),

,

t

and −Sn−1 (t, x(t)) ≥ −Sn−1 (t6 , x(t6 )) 1

 β t6 ∞ α 1 l α q(v)Δv Δs + 2 an (s) s t

 β t6 1 ∞ α 1 l α ≥ q(v)Δv Δs, 2 an (s) s t t, t6 ∈ [t5 , ∞), t < t6 . Therefore

β l α −Sn−1 (t, x(t)) ≥ 2

∞ t

1 an (s)

1

∞

α

q(v)Δv

Δs,

s

t ∈ [t5 , ∞), and Δ (t, x(t)) Sn−2

β l α ≤− an−1 (t) 2

∞

1

t

1 an (s)

1

∞

α

q(v)Δv

Δs,

s

t ∈ [t5 , ∞). Hence, Sn−2 (t, x(t)) ≤ Sn−2 (t5 , x(t5 )) −

t t5

β l α an−1 (u) 2

→ −∞ as

∞

1

u

1 an (s)

1

∞

α

q(v)Δv

ΔsΔu

s

t → ∞.

This is a contradiction. Therefore l = 0 or limt→∞ x(t) = 0. This completes the proof. Lemma 12.4. Assume that x is an eventually positive solution of the equation (12.1). If there exists a t1 ∈ [t0 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t1 , ∞), 2. Sl (t, x(t)) > 0, t ∈ [t1 , ∞), l ∈ {0, . . . , n}, then 1

Sl (t, x(t)) ≥ (Sn (t, x(t))) α Bl+1 (t, t1 ),

l ∈ {0, . . . , n − 1},

678

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t1 , ∞), and there exist a t2 ∈ [t1 , ∞) and a constant c > 0 such that t ∈ [t2 , ∞).

x(t) ≤ cB1 (t, t1 ), Here

⎧  1 ⎨#t α 1 Δs if l = n Bl (t, t1 ) = #t1 an (s) ⎩ t Bl+1 (s,t1 ) Δs if l ∈ {1, . . . , n − 1}. t1 al (s) Proof. Since SnΔ (t, x(t)) < 0, t ∈ [t1 , ∞), we have that 0 < Sn (t, x(t)) < Sn (t1 , x(t1 )),

t ∈ [t1 , ∞).

Hence, Δ (t, x(t)) Sn−1

= (Sn (t, x(t)))

1 α

1 an (t)

1

α

t ∈ [t1 , ∞),

,

and Sn−1 (t, x(t)) = Sn−1 (t1 , x(t1 ))

1 α 1 Δs an (s) t1

1 t α 1 1 ≥ Δs (Sn (s, x(s))) α an (s) t1 1 t 1 α 1 ≥ (Sn (t, x(t))) α Δs an (s) t1 +

t

1

(Sn (s, x(s))) α

1

= (Sn (t, x(t))) α Bn (t, t1 ),

t ∈ [t1 , ∞).

By induction, we get 1

Sl (t, x(t)) ≥ (Sn (t, x(t))) α Bl+1 (t, t1 ),

t ∈ [t1 , ∞),

l ∈ {0, . . . , n − 1}. On the other hand, Sn−1 (t, x(t)) = Sn−1 (t1 , x(t1 )) +

t

(Sn (s, x(s))) t1 1

1 α

1 an (s)

≤ Sn−1 (t1 , x(t1 )) + (Sn (t1 , x(t1 ))) α Bn (t, t1 ),

1

α

Δs

12.1 Oscillations of Higher-Order Delay Dynamic Equations

679

t ∈ [t1 , ∞). Hence, there exist a t2 ∈ [t1 , ∞) and a constant b1 > 0 so that Sn−1 (t, x(t)) ≤ b1 Bn (t, t1 ),

t ∈ [t2 , ∞).

Then Δ (t, x(t)) = Sn−2

Sn−1 (t, x(t)) an−1 (t)

≤ b1

Bn (t, t1 ) , an−1 (t)

t ∈ [t2 , ∞),

and t

Sn−2 (t, x(t)) ≤ Sn−2 (t2 , x(t2 )) + b1

t2 t

≤ Sn−2 (t2 , x(t2 )) + b1

t1

Bn (s, t1 ) Δs an−1 (s) Bn (s, t1 ) Δs an−1 (s)

= Sn−2 (t2 , x(t2 )) + b1 Bn−1 (t, t1 ),

t ∈ [t2 , ∞).

Hence, there exist a t3 ∈ [t2 , ∞) and a constant b2 > 0 such that Sn−2 (t, x(t)) ≤ b2 Bn−1 (t, t1 ),

t ∈ [t3 , ∞).

By induction, we get that there exist points t2 ≤ t3 ≤ . . . ≤ tn+1 and constants bl > 0, l ∈ {1, . . . , n}, such that x(t) = S0 (t, x(t)) ≤ bn B1 (t, t1 ),

t ∈ [tn+1 , ∞).

This completes the proof. For functions Φ : T → (0, ∞), φ : T → [0, ∞) such that Φ and an φ are differentiable, write h1 (t, t1 ) =

B2 (t, t1 ) , a1 (t1 )

h2 (t, t1 ) = h1 (t, t1 ) (B1 (σ (t), t1 ))α−1 , ⎧ ⎨ c1 if α < β δ1 (t, t1 , c1 , c2 ) = 1 if α = β ⎩ c2 (B1 (σ (t), t1 ))β−α if α > β,

680

12 Oscillations of Higher-Order Functional Dynamic Equations

c1 and c2 are any positive constants, g1 (t, t1 , c1 , c2 ) = Φ Δ (t) + 2βΦ(t)τ Δ (t)h2 (τ (t), t1 ) ×δ1 (τ (t), t1 , c1 , c2 )(an (t)φ(t))σ , G1 (t, t1 , c1 , c2 ) = Φ(t)q(t) − Φ(t)(an (t)φ(t))Δ

 2 +βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 ) (an (t)φ(t))σ , % % &&σ  Δ (t, x(t)) α Sn−1 X(t) = an (t) + φ(t) , (x(τ (t)))β

g+ (t) = max{0, g(t)}, g− (t) = min{0, −g(t)},

t ∈ [t0 , ∞).

Theorem 12.1. Suppose (G1)–(G3), (12.3), and (12.4). If there exist differentiable functions Φ : T → (0, ∞), φ : T → [0, ∞) with an φ being differentiable such that for any sufficiently large t1 ∈ [t0 , ∞) and for any positive constants c1 , c2 , there is a t2 ∈ [t1 , ∞) with τ (t2 ) > t1 so that t

lim sup t→∞

G1 (s, t1 , c1 , c2 )

t2

(g1 (s,t1 ,c1 ,c2 ))2 − 4βΦ(s)τ Δ (s)h 2 (τ (s),t1 )δ1 (τ (s),t1 ,c1 ,c2 )

 Δs = ∞,

(12.5)

then every solution of the equation (12.1) is either oscillatory or tends to 0 as t → ∞. Proof. Assume that the equation (12.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Lemma 12.3, it follows that there exists sufficiently large t2 ∈ [t1 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t2 , ∞), 2. either limt→∞ x(t) = 0 or Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Assume that Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Consider the function % w(t) = Φ(t)an (t)

Δ (t, x(t)) Sn−1

(x(τ (t)))β

α

& + φ(t) ,

t ∈ [t2 , ∞).

(12.6)

12.1 Oscillations of Higher-Order Delay Dynamic Equations

681

Then X(t) =

w σ (t) , Φ σ (t)

w(t) > 0,

t ∈ [t2 , ∞),

and

X(t) =

σ Sn (t, x(t)) + an (t)φ(t) , (x(τ (t)))β

t ∈ [t2 , ∞).

We have w(t) =

Φ(t)Sn (t, x(t)) + Φ(t)an (t)φ(t), (x(τ (t)))β

t ∈ [t2 , ∞),

and w Δ (t) =

Φ(t) S Δ (t, x(t)) + (x(τ (t)))β n

Φ(t) (x(τ (t)))β

Δ Snσ (t, x(t))

+Φ(t)(an (t)φ(t))Δ + Φ Δ (t)(an (t)φ(t))σ Δ

Φ(t) Φ(t) Δ = S (t, x(t)) + Snσ (t, x(t)) (x(τ (t)))β n (x(τ (t)))β +Φ(t)(an (t)φ(t))Δ + Φ Δ (t)(an (t)φ(t))σ ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ Φ Δ (t)Snσ (t, x(t)) (x(τ (t)))βσ Δ  (x(τ (t)))β Snσ (t, x(t)) −Φ(t) (x(τ (t)))β (x(τ (t)))βσ +

+Φ Δ (t)(an (t)φ(t))σ ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ +Φ Δ (t)X(t) Δ  (x(τ (t)))β Snσ (t, x(t)) −Φ(t) , (x(τ (t)))β (x(τ (t)))βσ

t ∈ [t2 , ∞).

By the chain rule, we get 

(x(τ (t)))β

Δ

≥ β(x(τ (t)))β−1 x Δ (τ (t))τ Δ (t),

t ∈ [t2 , ∞).

682

12 Oscillations of Higher-Order Functional Dynamic Equations

Therefore w Δ (t) ≤ −Φ(t)q(t) + Φ(t)(an(t)φ(t))Δ + Φ Δ (t)X(t) Δ

(t)) −βΦ(t)τ Δ (t) xx(τ(τ(t))

Sn (t,x(t)) (x(τ (t)))β

σ

(12.7)

,

t ∈ [t2 , ∞). We choose a t3 ∈ [t2 , ∞) such that τ (t) ≥ t2 , t ∈ [t3 , ∞). Then, using Lemma 12.4, we obtain a1 (τ (t))x Δ (τ (t)) = S1 (τ (t), x(τ (t))) 1

≥ (Sn (τ (t), x(τ (t)))) α B2 (τ (t), t1 ), t ∈ [t3 , ∞), or 1

1) x Δ (τ (t)) ≥ (Sn (τ (t), x(τ (t)))) α Ba2 1(τ(τ(t),t (t))  σ 1 ≥ (Sn (t, x(t))) α h1 (τ (t), t1 ) σ  1 Δ (t, x(t)) h1 (τ (t), t1 ) = (an (t)) α Sn−1 σ  Sn (t,x(t)) (x(τ (t)))β ≥ h1 (τ (t), t1 ) (x(τ (t)))β

(12.8) σ

(Sn (τ (t),x(τ (t))))

α−1 α

,

t ∈ [t3 , ∞). By Lemma 12.4, we have 1

x(t) ≥ (Sn (t, x(t))) α B1 (t, t1 ),

t ∈ [t2 , ∞).

Therefore (Sn (τ (t), x(τ (t))))

α−1 α

≤

(x(τ (t)))α−1 , (B1 (τ (t), t1 ))α−1

t ∈ [t3 , ∞),

and hence,  Sn (t, x(t)) σ x (τ (t)) ≥ h1 (τ (t), t1 ) (x(τ (t)))β σ

(x(τ (t)))β (B1 (τ (t), t1 ))α−1 × (x(τ (t)))α−1  σ

(x(τ (t)))β Sn (t, x(t)) σ = h2 (τ (t), t1 ) , (x(τ (t)))β (x(τ (t)))α−1

Δ

12.1 Oscillations of Higher-Order Delay Dynamic Equations

683

t ∈ [t3 , ∞), and w Δ (t) ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t)   σ

Sn (t, x(t)) σ 2 (x(τ (t)))β , −βτ Δ (t)Φ(t)h2 (τ (t), t1 ) (x(τ (t)))β (x(τ (t)))α t ∈ [t3 , ∞). Consider the following cases. 1. Let α < β. Then x Δ (t) > 0, t ∈ [t3 , ∞), and x(t) ≥ x(t3 ) = b1 ,

t ∈ [t1 , ∞).

Then β−α

(x(τ (t)))β−α ≥ b1 = c1 > 0,

t ∈ [t3 , ∞).

2. Let α = β. Then σ  (x(τ (t)))β−α = 1,

t ∈ [t3 , ∞).

3. Let α > β. Then, using Lemma 12.4, we get that there exists a constant c > 0 such that x(t) ≤ cB1 (t, t1 ),

t ∈ [t3 , ∞).

Thus 

(x(τ (t)))β−α

σ

≥ c2 (B1 (τ (t), t1 ))σ ,

t ∈ [t3 , ∞),

where c2 = cβ−α > 0. Consequently

(x(τ (t)))β (x(τ (t)))α

σ ≥ δ1 (τ (t), t1 , c1 , c2 ),

t ∈ [t3 , ∞),

684

12 Oscillations of Higher-Order Functional Dynamic Equations

and w Δ (t) ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t) σ 2  Sn (t,x(t)) −βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 ) (x(τ β (t))) = −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t) −βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 ) × (X(t) − (an (t)φ(t))σ )2 = −G1 (t, t1 , c1 , c2 ) + g1 (t, t1 , c1 , c2 )X(t) −βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 )(X(t))2 ≤ −G1 (t, t1 , c1 , c2 ) +

(12.9)

(g1 (t, t1 , c1 , c2 ))2 , 4βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 )

t ∈ [t3 , ∞), or G1 (t, t1 , c1 , c2 ) −

(g1 (t, t1 , c1 , c2 ))2 ≤ −w Δ (t), 4βΦ(t)τ Δ (t)h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 )

t ∈ [t3 , ∞), and t

% G1 (s, t1 , c1 , c2 ) −

t3

& (g1 (s, t1 , c1 , c2 ))2 Δs ≤ −w(t) + w(t3 ) 4βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 ) ≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. This completes the proof. Example 12.1. Let α < β, Φ(t) = φ(t) = 1,

t ∈ T.

Take c1 = 1. Then g1 (t, t1 , c1 , c2 ) = 2βτ Δ (t)h2 (τ (t), t1 )anσ (t), G1 (t, t1 , c1 , c2 ) = q(t) − anΔ (t)

 2 +βτ Δ (t)h2 (τ (t), t1 ) anσ (t) ,

12.1 Oscillations of Higher-Order Delay Dynamic Equations

685

t ∈ [t1 , ∞). Then (12.5) takes the form t

lim sup t→∞

 −β

t2

 2 q(s) − anΔ (s) + βτ Δ (s)h2 (τ (s), t1 ) anσ (s)

τ Δ (s)

= lim sup t→∞

t t2

2



 2  h2 (τ (s), t1 ) anσ (s) Δs τ Δ (s)

 q(s) − anΔ (s) Δs

= ∞. Example 12.2. Let T = {0}

  1 : k ∈ N {2k : k ∈ N}. 2k

Consider the following higher-order dynamic equation SnΔ (t, x(t)) +

γ t β+1

β+1 t x = 0, 2

t ∈ [4, ∞),

where an (t) = t α , al (t) = 1, l ∈ {1, . . . , n − 1}, γ q(t) = β+1 , t 1 γ > 1−β , 0 < β < 1. 2 Here σ (t) = 2t, t τ (t) = , 2 σ (t) τ (σ (t)) = 2 2t = 2 = t,

686

12 Oscillations of Higher-Order Functional Dynamic Equations

 t σ (τ (t)) = σ 2 t =2 2 = t, t ∈ T, 1 ∞ 1 α ∞ Δs Δs = an (s) s t0 t0 = ∞, ∞ t0

Δs = al (s)

∞

Δs t0

= ∞,

l ∈ {1, . . . , n − 1},

and ∞ t0

=

%

1

∞

an−1 (t) t % ∞ ∞ t0

1 γ α ≥ β

1 γ α = β

1 γ α = β

1 γ α ≥ β

1 γ α = β = ∞.

t ∞ t0 ∞

⎜ ⎝ %

t0

∞ t

∞ t0

∞

∞ t0

∞

1 sα

s

1 α s sβ

1 β

∞ t

1 Δt t

Δs ss σ



Δ uβ &

α

Δs Δt 

Δs Δt  Δt

& α1

 σ Δu uβ uβ

1

ss α

t

∞ t0

%

t

s

γ

s ∞

& 1 α q(u)Δu Δs Δt

& 1 α Δu Δs Δt uβ+1

∞

1 sα ⎛

∞

1 an (s)

⎞ ⎟ Δs ⎠ Δt

12.1 Oscillations of Higher-Order Delay Dynamic Equations

687

Also, lim Bn (t, t1 ) = lim

t→∞

t

t→∞ t 1

= lim

t

t→∞ t 1

1 an (s)

1

α

Δs

1 Δs s

=∞ and hence, lim B2 (t, t1 ) = lim B1 (σ (t), t1 )

t→∞

t→∞

= ∞, lim h2 (t, t1 ) = lim h1 (σ (t), t1 )

t→∞

t→∞

= ∞. Take c1 > 0, c2 > 0 arbitrarily. Let t2 ∈ [t1 , ∞) be so that B1 (τ (t), t1 ) ≥ 1, h1 (τ (t), t1 ) ≥

1 , c2



1 h2 (τ (t), t1 ) ≥ max 1, , c1

t ∈ [t2 , ∞).

Then ⎧ ⎪ ⎨ c1 h2 (τ (t), t1 ) ≥ 1 if α < β + 1 h2 (τ (t), t1 )δ1 (τ (t), t1 , c1 , c2 ) = h2 (τ (t), t1 ) ≥ 1 if α = β + 1 ⎪ ⎩ c h (τ (t), t ) (B (σ (t), t ))β σ if 2 1 1 1 1

α > β + 1.

For φ(t) = 0 and Φ(t) = t, we obtain t

lim sup t→∞

−

G1 (s, t1 , c1 , c2 )

t2

4(β

 (g1 (s, t1 , c1 , c2 ))2 Δs Δ + 1)Φ(s)τ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )

≥ lim sup t→∞

t t2

≥ lim sup t→∞

t t2

 1 γ 1 Δs − β s 2(β + 1) s

1 Δs 12s

= ∞.

Therefore every solution of the considered equation is either oscillatory or tends to 0 as t → ∞.

688

12 Oscillations of Higher-Order Functional Dynamic Equations

Exercise 12.1. Let   1 T = {0} :k∈N {2k : k ∈ N}, 2k and α = 5,

an (t) = t 5 ,

al (t) = 1,

t ∈ T,

l ∈ {1, . . . , n − 1}.

Prove that any solution of the equation SnΔ (t, x(t)) +

 5 3 t x = 0, 5 2 t3 2

t ∈ [2, ∞),

is either oscillatory or tends to 0 as t → ∞. Exercise 12.2. Let T = hZ, h > 0. Write an analogue of Theorem 12.1. For functions Φ : T → (0, ∞) and φ : T → [0, ∞) such that Φ and an φ are differentiable, we will write ⎧ ⎪ ⎨ c1 if α < β δ2 (t, t1 , c1 , c2 ) − 1 if α = β ⎪ β ⎩c −1 if α > β, 2 (B1 (σ (t), t1 )) α where c1 and c2 are any positive constants, 

1 βΦ(t)τ Δ (t)h1 (τ (t), t1 ) g2 (t, t1 , c1 , c2 ) = Φ Δ (t) + 1 + α 1  ×δ2 (τ (t), t1 , c1 , c2 ) (an (t)φ(t))σ α , G2 (t, t1 , c1 , c2 ) = Φ(t)q(t) − Φ(t)(an (t)φ(t))Δ β + Φ(t)τ Δ (t)h1 (τ (t), t1 ) α 1+ 1  α , ×δ2 (t, t1 , c1 , c2 ) (an (t)φ(t))σ t, t1 ∈ [t0 , ∞), t > t1 . Theorem 12.2. Suppose (G1)–(G3), (12.3), and (12.4). If there exist differentiable functions Φ : T → (0, ∞), φ : T → [0, ∞) with an φ being differentiable such that for any positive constants c1 and c2 there is a t2 ∈ [t1 , ∞), t1 ∈ [t0 , ∞), with τ (t2 ) > t1 , so that

12.1 Oscillations of Higher-Order Delay Dynamic Equations

lim supt→∞

t2 G2 (s, t1 , c1 , c2 )

689

#t

− (1+α)1+α

α α ((g2 (s,t1 ,c1 ,c2 ))+ )1+α (βΦ(s)τ Δ (s)h1 (τ (s),t1 )δ2 (τ (s),t1 ,c1 ,c2 ))α

 Δs = ∞,

(12.10) then every solution of the equation (12.1) is either oscillatory or tends to 0 as t → ∞. Proof. Assume that the equation (12.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that τ (t) > t1 ,

t ∈ [t2 , ∞),

and x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Lemma 12.3, it follows that there exists sufficiently large t2 ∈ [t1 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t2 , ∞), 2. either limt→∞ x(t) = 0 or Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Assume that Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. By (12.8), we have  1 σ x Δ (τ (t)) ≥ (Sn (t, x(t))) α h1 (τ (t), t1 ) σ

S (t, x(t)) σ  α1 n , = h1 (τ (t), t1 ) (x(τ (t))) (x(τ (t)))β 

β α

t ∈ [t2 , ∞). Define the function w by (12.6). Then, using (12.7) and the inequality x(τ (t)) ≤ (x(τ (t)))σ ,

t ∈ [t2 , ∞),

we get wΔ (t) ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t)  β σ 1 h1 (τ (t), t1 ) (x(τ (t))) α −βΦ(t)τ Δ (t) x(τ (t))  1 

Sn (t, x(t)) σ α Sn (t, x(t)) σ × (x(τ (t)))β (x(τ (t)))β

690

12 Oscillations of Higher-Order Functional Dynamic Equations

= −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t)  β σ (x(τ (t))) α −βΦ(t)τ Δ (t)h1 (τ (t), t1 ) x(τ (t)) σ 1+ α1

Sn (t, x(t)) × (x(τ (t)))β ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t)  β−α σ −βΦ(t)τ Δ (t)h1 (τ (t), t1 ) (x(τ (t))) α

×

Sn (t, x(t)) (x(τ (t)))β

σ 1+ α1 ,

t ∈ [t2 , ∞). Consider the following cases. 1. Let α < β. Then  β−α σ β−α (x(τ (t))) α ≥ (x(τ (t))) α ≥ (x(t1 ))

β−α α

= c1 > 0,

t ∈ [t2 , ∞).

= 1,

t ∈ [t2 , ∞).

2. Let α = β. Then  (x(τ (t)))

β−α α

σ

3. Let α > β. Then, using Lemma 12.4, we obtain that there exist a t3 ∈ [t2 , ∞) and a constant c > 0 such that x(τ (t)) ≤ cB1 (τ (t), t1 ),

t ∈ [t3 , ∞).

Therefore  (x(τ (t)))

β−α α

σ

 β−α σ ≥ c2 (B1 (τ (t), t1 )) α ,

t ∈ [t3 , ∞), where c2 = c

β−α α

> 0.

12.1 Oscillations of Higher-Order Delay Dynamic Equations

691

Consequently  β−α σ (x(τ (t))) α ≥ δ2 (τ (t), t1 , c1 , c2 ), t ∈ [t3 , ∞). Then w Δ (t) ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t) −βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )   1

Sn (t, x(t)) σ 1+ α × , (x(τ (t)))β t ∈ [t3 , ∞). By the inequality 1+ γ1

(U − V )

≥U

1+ γ1

+



1 1 1+ γ1 1 V V γ U, − 1+ γ γ

where γ ≥ 1 is a quotient of odd positive integers, we get

Sn (t, x(t)) (x(τ (t)))β

σ 1+ α1

 1+ 1 α = X(t) − (an (t)φ(t))σ 1+ 1 1 1 α (an (t)φ(t))σ ≥ (X(t))1+ α + α 

1 1  (an (t)φ(t))σ α X(t), − 1+ α

t ∈ [t3 , ∞). Then w Δ (t) ≤ −Φ(t)q(t) + Φ(t)(an (t)φ(t))Δ + Φ Δ (t)X(t) Δ (t)h (τ (t), t )δ (τ (t), t , c , c ) −βΦ(t)τ 1 1 2 1 1 2

1

1

× (X(t))1+ α + α1 ((an (t)φ(t))σ )1+ α   1 1 σ α − 1 + α ((an (t)φ(t)) ) X(t)

(12.11)

≤ −G2 (t, t1 , c1 , c2 ) + (g2 (t, t1 , c1 , c2 ))+ X(t) 1 −βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α , t ∈ [t3 , ∞). Let 1

1

A1+ α = βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α , 1

Bα =



α(g2 (t, t1 , c1 , c2 ))+

(1 + α) βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )



α α+1

,

692

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t3 , ∞). Using the inequality Aλ − λAB λ−1 + (λ − 1)B λ ≥ 0, for λ =

α+1 α ,

we get 1

βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α α 1 α+1 α+1 βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α − X(t) α α(g2 (t, t1 , c1 , c2 ))+ × α  1 α+1 (1 + α) βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α + ≥ 0,

α α+1 (g2 (t, t1 , c1 , c2 ))α+1 1 +   α (1 + α)1+α βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 ) α t ∈ [t3 , ∞),

or 1

(g2 (t, t1 , c1 , c2 ))+ X(t) − βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )(X(t))1+ α ≤

α α (g2 (t, t1 , c1 , c2 ))α+1 +  α , (1 + α)1+α βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )

t ∈ [t3 , ∞). Therefore w Δ (t) ≤ −G2 (t, t1 , c1 , c2 ) +



α+1 α α (g2 (t, t1 , c1 , c2 ))+

α ,

(1 + α)1+α βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )

t ∈ [t3 , ∞), or G2 (t, t1 , c1 , c2 ) −



α+1 α α (g2 (t, t1 , c1 , c2 ))+

α ≤ −w Δ (t),

(1 + α)1+α βΦ(t)τ Δ (t)h1 (τ (t), t1 )δ2 (τ (t), t1 , c1 , c2 )

t ∈ [t3 , ∞). Hence, t t3

% G2 (s, t1 , c1 , c2 ) −

(1 + α)1+α



α+1 α α (g2 (s, t1 , c1 , c2 ))+

α βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )

& Δs ≤ w(t3 ) − w(t) ≤ w(t3 ),

This is a contradiction. This completes the proof.

t ∈ [t3 , ∞).

12.1 Oscillations of Higher-Order Delay Dynamic Equations

693

Example 12.3. Let α < β and Φ(t) = t,

φ(t) = 0,

t ∈ T.

Take c1 = 1. Then δ2 (t, t1 , c1 , c2 ) = 1, g2 (t, t1 , c1 , c2 ) = 1, G2 (t, t1 , c1 , c2 ) = tq(t),

t ∈ T.

Then (12.10) takes the form t

lim sup t→∞

sq(s) −

t2

 αα  α Δs = ∞. (1 + α)1+α βsτ Δ (s)h1 (τ (s), t1 )

Exercise 12.3. Write an analogue of Theorem 12.2 in the following cases. 1. Φ(t) = 1, φ(t) = 0, t ∈ T, 2. Φ(t) = φ(t) = t, t ∈ T, 3. Φ(t) = t 2 , φ(t) = t 2 + 1, t ∈ T. Let D = {(t, s) ∈ T × T : t ≥ s ≥ t0 ,

∗ S = G ∈ Crd (D) : G(s, s) = 0, ∃G (t, s) ∈ Crd (D),

t, s ∈ [t0 , ∞)}, G(t, s) > 0,

G (t, s) ≤ 0,

Δs

Δs

t, s ∈ [t0 , ∞),

 t > s ≥ t0 .

Theorem 12.3. Suppose (G1)–(G3), (12.3), and (12.4). If there exist functions r, R ∈ S ∗ and differentiable functions Φ : T → (0, ∞), φ : T → [0, ∞) with an φ being differentiable such that for all sufficiently large t1 ∈ [t0 , ∞) and for any positive constants c1 and c2 there is a t2 ∈ [t1 , ∞) with τ (t2 ) > t1 so that R Δs (t, s) +

1 r(t, s) R(t, s)g1 (s, t1 , c1 , c2 ) = σ (R(t, s)) 2 σ Φ (s) Φ (s)

and lim sup t→∞

1 R(t, t2 )

t t2

R(t, s)G1 (s, t1 , c1 , c2 )

 (r(t, s))2 − Δs = ∞, 4βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )

694

12 Oscillations of Higher-Order Functional Dynamic Equations

then every solution of the equation (12.1) is either oscillatory or tends to 0 as t → ∞. Proof. Assume that the equation (12.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Lemma 12.3, it follows that there exists sufficiently large t2 ∈ [t1 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t2 , ∞), 2. either limt→∞ x(t) = 0 or Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Assume that Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Consider the function w defined by (12.6). Take t2 ∈ [t1 , ∞) such that τ (t2 ) > t1 and (12.9) holds for any t ∈ [t2 , ∞). Then G1 (s, t1 , c1 , c2 ) ≤ −w Δ (s) + g1 (s, t1 , c1 , c2 )X(s) −βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 , s ∈ [t2 , ∞), and hence, R(t, s)G1 (s, t1 , c1 , c2 ) ≤ −w Δ (s)R(t, s) + g1 (s, t1 , c1 , c2 )X(s)R(t, s) −βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 R(t, s), t, s ∈ [t2 , ∞), t ≥ s. Therefore t

G1 (s, t1 , c1 , c2 )R(t, s)Δs ≤ −

t2

t

R(t, s)w Δ (s)Δs

t2

+

t

g1 (s, t1 , c1 , c2 )X(s)R(t, s)Δs t2

−

t

βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 R(t, s)Δs

t2

≤ R(t, t2 )w(t2 ) +

t

R Δs (t, s)w σ (s)Δs

t2

+

t

g1 (s, t1 , c1 , c2 )X(s)R(t, s)Δs t2

−

t t2

βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 R(t, s)Δs

12.1 Oscillations of Higher-Order Delay Dynamic Equations t

= R(t, t2 )w(t2 ) +

695

R Δs (t, s)X(s)Φ σ (s)Δs

t2 t

+

g1 (s, t1 , c1 , c2 )X(s)R(t, s)Δs t2 t

−

βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 R(t, s)Δs

t2 t

= R(t, t2 )w(t2 ) +

1

r(t, s)(R(t, s)) 2 X(s)Δs t2

t

−

βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )(X(s))2 R(t, s)Δs

t2 t

≤ R(t, t2 )w(t2 ) +

t2

(r(t, s))2 Δs, 4βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )

t ∈ [t2 , ∞). Hence, 1 R(t, t2 )

t

R(t, s)G1 (s, t1 , c1 , c2 ) −

t2

4βΦ(s)τ Δ (s)h

 (r(t, s))2 Δs ≤ w(t2 ), 2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )

t ∈ [t2 , ∞). This is a contradiction. This completes the proof. Exercise 12.4. Let T = {0}



  1 2 :k∈N :k∈N . 2k k

Write an analogue of Theorem 12.3. Theorem 12.4. Suppose (G1)–(G3), (12.3), and (12.4). If there exist functions r, R ∈ S ∗ and differentiable functions Φ : T → (0, ∞), φ : T → [0, ∞) with an φ being differentiable such that for sufficiently large t1 ∈ [t0 , ∞) and for any positive constants c1 and c2 there is a t2 ∈ (t1 , ∞) with τ (t2 ) > t1 so that R Δs (t, s) +

α r(t, s) R(t, s)g2 (s, t1 , c1 , c2 ) = σ (R(t, s)) α+1 σ Φ (s) Φ (s)

and lim sup t→∞

−

1 R(t, t2 )

t t2

R(t, s)G2 (s, t1 , c1 , c2 )

 α α (r(t, s)+ )α+1  α Δs = ∞, (1 + α)1+α βΦ(s)τ Δ (s)h2 (τ (s), t1 )δ1 (τ (s), t1 , c1 , c2 )

696

12 Oscillations of Higher-Order Functional Dynamic Equations

then every solution of the equation (12.1) is either oscillatory or tends to 0 as t → ∞. Proof. Assume that the equation (12.1) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is eventually positive on [t0 , ∞). Then there exists a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ (t)) > 0,

t ∈ [t1 , ∞).

By Lemma 12.3, it follows that there exists sufficiently large t2 ∈ [t1 , ∞) such that 1. SnΔ (t, x(t)) < 0, t ∈ [t2 , ∞), 2. either limt→∞ x(t) = 0 or Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Assume that Sl (t, x(t)) > 0, t ∈ [t2 , ∞), l ∈ {0, . . . , n}. Consider the function w defined by (12.6). Take t2 ∈ [t1 , ∞) such that τ (t2 ) > t1 and (12.11) holds for any t ∈ [t2 , ∞). Then G2 (s, t1 , c1 , c2 ) ≤ −w Δ (s) + (g2 (s, t1 , c1 , c2 ))+ X(s) 1

−βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α (g2 (s, t1 , c1 , c2 ))+ σ w (s) Φ σ (s)

= −w Δ (s) +

1

−βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α , s ∈ [t2 , ∞), and R(t, s)G2 (s, t1 , c1 , c2 ) ≤ −R(t, s)w Δ (s) + R(t, s)

(g2 (s, t1 , c1 , c2 ))+ σ w (s) Φ σ (s) 1

−βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α R(t, s),

t, s ∈ [t2 , ∞), t ≥ s. Hence, t

R(t, s)G2 (s, t1 , c1 , c2 )Δs ≤ −

t2

t

R(t, s)w Δ (s)Δs

t2

+

t

R(t, s) t2

−

t

(g2 (s, t1 , c1 , c2 ))+ σ w (s)Δs Φ σ (s) 1

βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α R(t, s)Δs

t2

= R(t, t2 )w(t2 ) +

t t2

R Δs (t, s)w σ (s)Δs

12.1 Oscillations of Higher-Order Delay Dynamic Equations t

+

R(t, s) t2 t

−

697

(g2 (s, t1 , c1 , c2 ))+ σ w (s)Δs Φ σ (s) 1

βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α R(t, s)Δs

t2 t

≤ R(t, t2 )w(t2 ) +

t2 t

−

α

r(t, s)+ (R(t, s)) α+1 X(s)Δs 1

βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))1+ α R(t, s)Δs,

t2

t ∈ [t2 , ∞). Let A

α+1 α

= βR(t, s)Φ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s))

1

Bα =

(1 + α)



αr(t, s)+ βΦ(s)τ Δ (s)h

1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )



α α+1

α+1 α

,

.

Using the inequality Aλ − λAB λ−1 + (λ − 1)B λ ≥ 0, for λ =

α+1 α ,

we get α+1

βR(t, s)Φ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s)) α α α+1 α+1 α+1 βR(t, s)Φ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s)) α − α αr(t, s)+ ×   α Δ (1 + α) βΦ(s)τ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 ) α+1 + ≥ 0,

α α+1 (r(t, s)+ )α+1 1   α (1 + α)1+α βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 ) α t, s ∈ [t2 , ∞),

t ≥ s,

or α

r(t, s)+ (R(t, s)) α+1 X(s) −βR(t, s)Φ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )(X(s)) ≤

α+1 α

α α (r(t, s)+ )α+1  α , (1 + α)1+α βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )

698

12 Oscillations of Higher-Order Functional Dynamic Equations

t, s ∈ [t2 , ∞), t ≥ s. Therefore t

R(t, s)G2 (s, t1 , c1 , c2 )Δs ≤ R(t, t2 )w(t2 )

t2 t

+

t2

α α (r(t, s)+ )α+1  α Δs, (1 + α)1+α βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )

t ∈ [t2 , ∞), whereupon 1 R(t, t2 ) −

t

R(t, s)G2 (s, t1 , c1 , c2 )

t2

 α α (r(t, s)+ )α+1  α Δs (1 + α)1+α βΦ(s)τ Δ (s)h1 (τ (s), t1 )δ2 (τ (s), t1 , c1 , c2 )

≤ w(t2 ),

t ∈ [t2 , ∞).

This is a contradiction. This completes the proof. Exercise 12.5. Let T = hZ, h > 0. Write an analogue of Theorem 12.4.

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations In this section we investigate the following even order nonlinear delay dynamic equation   n−1 α Δ  β φ(t) x Δ (t) + q(t) x σ (τ (t)) = 0,

t ∈ [t0 , ∞),

(12.12)

where (H1) (H2)

n ≥ 4 is even, α and β are quotients of odd positive integers, 1 ([t , ∞)), φ Δ (t) ≥ 0, φ, q ∈ Crd ([t0 , ∞)) are positive functions, φ ∈ Crd 0 t ∈ [t0 , ∞), and ∞

1

(φ(t))− α Δt = ∞,

t0

(H3)

τ : T → T, τ ∈ Crd (T), τ (t) ≤ t, t ∈ T, limt→∞ τ (t) = ∞.

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

699

Definition 12.1. By a solution of the equation (12.12) we mean a nontrivial real 1 ([t , ∞)) with the following property function x ∈ Crd 1  n−1 α 1 ∈ Crd ([t1 , ∞)) φ xΔ for some t1 ∈ [t0 , ∞) and satisfies (12.12) on [t1 , ∞). Lemma 12.5. Let x be an eventually positive solution of the equation (12.12) and assume (H 1)–(H 3), and #∞

= ∞, t0 q(u)Δu 1 #∞ 1 #∞ α −α Δs < ∞, t0 (φ(s)) s q(u)Δu  1 #∞ #∞ #  ∞ (φ(s))−1 s q(u)Δu α Δs Δv = ∞. t0 v

(12.13)

Then there exists a t1 ∈ [t0 , ∞) such that 

 n−1 α Δ φ(t) x Δ (t) < 0,  n−1 α Δ x Δ (t) < 0,

l

x Δ (t) > 0, l ∈ {0, . . . , n − 1},

t ∈ [t1 , ∞),

and n−1

x Δ (t) > x Δ

(t)hn−2 (t, t1 ),

t ∈ [t1 , ∞).

Proof. Let t1 ∈ [t0 , ∞) be such that x(t) > 0,

x σ (τ (t)) > 0,

t ∈ [t1 , ∞).

Hence from (12.12), we get 

 n−1 α Δ  β = −q(t) x σ (τ (t)) φ(t) x Δ (t) < 0,

t ∈ [t1 , ∞).

 n−1 α Therefore φ(t) x Δ (t) is strictly decreasing on [t1 , ∞) and eventually of one sign. Assume that there is a t2 ∈ [t1 , ∞) such that n−1

xΔ

(t) < 0,

t ∈ [t2 , ∞).

Then  α  n−1 α n−1 −φ(t) x Δ (t) = φ(t) −x Δ (t)

700

12 Oscillations of Higher-Order Functional Dynamic Equations

is positive and strictly increasing on [t2 , ∞) and  1 n−1 (φ(t)) α −x Δ (t) is positive and strictly increasing on [t2 , ∞). Hence, n−2

xΔ

n−2

(t) = x Δ

(t2 ) +

t

n−1

xΔ

(s)Δs

t2 n−2

= xΔ

(t2 ) −

t

 1 1 n−1 (φ(s)) α −x Δ (s) (φ(s))− α Δs

t2 n−2

≤ xΔ

 1 n−1 (t2 ) − (φ(t2 )) α −x Δ (t2 )

t

1

(φ(s))− α Δs

t2

→ −∞ as

t → ∞.

By induction, we get l

x Δ (t) → −∞,

as

t → ∞,

l ∈ {0, . . . , n − 2}.

This is a contradiction. Consequently there exists a t2 ∈ [t1 , ∞) such that n−1

xΔ

(t) > 0,

t ∈ [t2 , ∞).

By Knesser’s theorem (see the appendix), it follows that there exists a t3 ∈ [t1 , ∞) k and an odd integer l1 ∈ {1, . . . , n − 1} such that if l1 > 0 we have x Δ (t) > 0, t ∈ [t3 , ∞), k ∈ {0, . . . , l1 } and l1 ≤ n − 1 implies (−1)l1 +k x Δ (t) > 0, k

t ∈ [t3 , ∞),

k ∈ {l1 , . . . , n − 2}.

Therefore x Δ (t) > 0,

t ∈ [t3 , ∞).

Take t4 ∈ [t3 , ∞) such that x σ (τ (t)) ≥ x(t1 ) > 0,

t ∈ [t4 , ∞).

Then, using (12.12), we get 

 n−1 α Δ ≤ −q(t)(x(t1 ))β φ(t) x Δ (t) = −c1 q(t),

t ∈ [t4 , ∞),

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

701

where c1 = (x(t1 ))β > 0. Then  n−1 α  n−1 α #v φ(t) x Δ (t) ≥ φ(v) x Δ (v) + c1 t q(u)Δu #v ≥ c1 t q(u)Δu,

(12.14)

v, t ∈ [t4 , ∞), v ≥ t. Letting v → ∞, we obtain  n−2 α φ(t) x Δ (t) ≥ c1

∞

q(u)Δu,

t ∈ [t4 , ∞),

t

or x

Δn−1

(t) ≥ c1 (φ(t))−1

1 α q(u)Δu ,

∞

1 α

t ∈ [t4 , ∞).

t

Assume that l1 ≤ n − 3. Then n−1

xΔ

(t) < 0,

t ∈ [t4 , ∞),

(t) > 0,

t ∈ [t4 , ∞).

and n−3

xΔ

Hence from (12.15), we obtain n−2

−x Δ

n−2

(t) > x Δ

v

=

n−2

(v) − x Δ n−1

xΔ

(t)

(s)Δs

t v

1 α

≥ c1

(φ(s))−1

t

∞

1 α q(u)Δu Δs,

s

v, t ∈ [t4 , ∞), v ≥ t. Letting v → ∞, we arrive to −x

Δn−2

∞

1 α

(t) ≥ c1

(φ(s)) t

∞

−1 s

1 α q(u)Δu Δs,

(12.15)

702

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t4 , ∞), whereupon n−3

xΔ

n−3

(t4 ) > −x Δ

t

=−

n−3

(t) + x Δ n−2

xΔ

(t4 )

(s)Δs

t4 ∞

t

1

≥ c1α

t4

∞

(φ(s))−1

v

1 α q(u)Δu ΔsΔv

s

and letting t → ∞, we obtain ∞

∞

(φ(s)) t4

∞

−1

v

s

1 α n−3 −1 q(u)Δu ΔsΔv ≤ c1 α x Δ (t4 ) < ∞.

This is a contradiction. Therefore l1 = n − 1 and k

k ∈ {0, . . . , n − 1},

x Δ (t) > 0,

t ∈ [t4 , ∞).

Since φ Δ (t) ≥ 0, t ∈ [t4 , ∞), we have   n−1 α Δ  n−1 α φ(t) x Δ (t) = φ Δ (t) x Δ (t) +φ σ (t) < 0,

 n−1 α Δ x Δ (t)

t ∈ [t4 , ∞).

Therefore  n−1 α Δ x Δ (t) < 0,

t ∈ [t4 , ∞).

Assume that there is a t5 ∈ [t4 , ∞) such that n

x Δ (t) > 0,

t ∈ [t5 , ∞).

Then, if 0 < α < 1, we have 0>

 n−1 α Δ x Δ (t)

 n−1σ α−1 n ≥ α xΔ (t) x Δ (t),

t ∈ [t5 , ∞),

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

703

and if α ≥ 1, we have 0>

 n−1 α Δ x Δ (t)

 n−1 α−1 n ≥ α x Δ (t) x Δ (t),

t ∈ [t5 , ∞).

This is a contradiction. Therefore n

t ∈ [t4 , ∞),

x Δ (t) < 0, n−1

and x Δ

is strictly decreasing function on [t4 , ∞). Next, n−2

xΔ

n−2

(t) = x Δ

(t4 ) +

t

n−1

xΔ

(s)Δs

t4 t

≥

n−1

xΔ

(s)Δs

t4 n−1

≥ xΔ n−3

xΔ

n−3

(t) = x Δ

t ∈ [t4 , ∞),

(t)h1 (t, t4 ), (t4 ) +

t

n−2

xΔ

(s)Δs

t4 t

≥

n−2

xΔ

(s)Δs

t4 t

≥

n−1

xΔ

(s)h1 (s, t4 )Δs

t4 n−1

≥ xΔ

(t)h2 (t, t4 ),

t ∈ [t4 , ∞).

Repeating the above argument, we have n−1

x Δ (t) ≥ x Δ

(t)hn−2 (t, t4 ),

t ∈ [t4 , ∞).

This completes the proof. Theorem 12.5. Suppose (H 1)–(H 3) and ∞

q(u)Δu = ∞.

(12.16)

t0

Then the equation (12.12) is oscillatory. Proof. Assume that the equation (12.12) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the

704

12 Oscillations of Higher-Order Functional Dynamic Equations

equation (12.12) on [t0 , ∞). Then, by the proof of Lemma 12.5, it follows that there exists a t4 ∈ [t0 , ∞) such that (12.14) holds. Then v t4

 n−1 α q(u)Δu ≤ c1−1 φ(t4 ) x Δ (t4 ) ,

v ∈ [t4 , ∞).

Hence, letting v → ∞, we get ∞ t4

 n−1 α q(u)Δu ≤ c1−1 φ(t4 ) x Δ (t4 ) < ∞.

This is a contradiction. This completes the proof. Example 12.4. Let T = {0}



2k : k ∈ N0

  1 : k ∈ N . 2k

Consider the equation



3 5 Δ t 3 t 2 x Δ (t) + t3 x = 0, 2 Here σ (t) = 2t, φ(t) = t 2 , q(t) = t 3 , t τ (t) = , 2 n = 4,

t ∈ T,

α = 5, β = 3. Then φ Δ (t) = σ (t) + t = 2t + t = 3t ≥ 0,

t ∈ T,

t ∈ [4, ∞).

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations ∞

∞

1

(φ(t))− α Δt =

2

Δt

t5

4

t0

1

705

= ∞, ∞

∞

q(t)Δt =

t 3 Δt

4

t0

= ∞. Therefore the considered equation is oscillatory. Exercise 12.6. Let T = {0}



  1 3 : k ∈ N0 :k∈N . 3k k

Prove that the equation



5 7 Δ t 3 Δ5 11 t x (t) x +t = 0, 3

t ∈ [9, ∞),

is oscillatory. Theorem 12.6. Suppose (H 1)–(H 3) and #∞

t0 q(u)Δu < ∞, #∞ # 1  −1 ∞ q(u)Δu α Δs = ∞. t0 (φ(s)) s

(12.17)

Then every bounded solution of the equation (12.12) is oscillatory. Proof. Assume that the equation (12.12) has a bounded nonoscillatory solution x on [t0 , ∞). Without loss of generality, suppose that x is an eventually positive solution of the equation (12.12) on [t0 , ∞). Then, by the proof of Lemma 12.5, it follows that there exists a t4 ∈ [t0 , ∞) such that (12.15) holds on [t4 , ∞). Hence from (12.15), we get x

Δn−2

(t) ≥ x

Δn−2

t

1 α

(t4 ) + c1

(φ(s))−1

t4

∞

1 α q(u)Δu Δs,

s

t ∈ [t4 , ∞). Therefore n−2

lim x Δ

t→∞

(t) = ∞.

Take a constant c3 > 0 arbitrarily. Then there exists a t5 ∈ [t4 , ∞) such that n−2

xΔ

(t) ≥ c3 ,

t ∈ [t5 , ∞).

706

12 Oscillations of Higher-Order Functional Dynamic Equations

Hence, n−3

xΔ

n−3

(t) ≥ x Δ

(t5 ) + c3 (t − t5 ),

t ∈ [t5 , ∞),

and n−3

lim x Δ

t→∞

(t) = ∞.

By induction, we get lim x(t) = ∞.

t→∞

This is a contradiction. This completes the proof. Example 12.5. Let T = {0}

  1 {2k : k ∈ N}. : k ∈ N 0 2k

Consider the equation



7 7 Δ t 1 2 Δ3 t x (t) + 2 x = 0, 2 t Here σ (t) = 2t, φ(t) = t 2 , 1 , t2 t τ (t) = , t ∈ [4, ∞), 2 n = 4,

q(t) =

α = 7, β = 7, t0 = 4.

t ∈ [4, ∞).

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

707

Then ∞

∞

1

(φ(t))− α Δt =

2

Δt

t7

4

t0

1

= ∞, ∞

4

t0

∞

(φ(s))−1

t0

∞

q(t)Δt =

1 Δt t2

2 !!t=∞ =− ! t t=4 1 = 2 < ∞, 1  1 ∞ ∞ 1 ∞ 1 α 7 q(u)Δu Δs = Δu Δs s 2 s u2 s 4 1 ∞ 1 27 = Δs s2 s 4 ∞ 1 √ 7 Δs = 2 3 4 s7 = ∞.

Therefore every bounded solution of the considered equation is oscillatory. Exercise 12.7. Let T = {0}

  1 : k ∈ N {3k : k ∈ N}. 0 3k

Prove that every bounded solution of the equation



7 7 Δ t 1 5 t 5 x Δ (t) + 3 x = 0, 9 t

t ∈ [27, ∞),

is oscillatory. Theorem 12.7. Suppose (H 1)–(H 3), τ Δ (t) > 0, t ∈ T, τ σ (t) = σ (τ (t)), t ∈ T, 1 ([t , ∞)) such that and (12.13) holds. Let there exists a positive function f ∈ Crd 0 ⎛ t

lim sup t→∞

t2

⎜ ⎝f (s)q(s) −

 α α β

(α

+ 1)α+1



 α+1 φ(τ (s)) f Δ (s)+

Φ(s, t1

)f (s)τ Δ (s)h

⎞

⎟ α ⎠ Δs = ∞ n−2 (τ (s), t1 )

(12.18)

708

12 Oscillations of Higher-Order Functional Dynamic Equations

for sufficiently large t1 , t2 ∈ [t0 , ∞), where ⎧ ⎪ ⎪  if β > α ⎪ ⎨ 1 if β = α Φ(s, t1 ) = β ⎪ (0 + 1 h1 (τ σ (s), t1 ) + · · · + n−1 hn−1 (τ σ (s), t1 )) α −1 ⎪ ⎪ ⎩ if β < α, , l , l ∈ {0, . . . , n − 1}, are arbitrary positive constants, t2 satisfies τ (s) > t1 , s ∈ [t2 , ∞). Then the equation (12.12) is oscillatory. Proof. Let x be a nonoscillatory solution of the equation (12.12) on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.12) on [t0 , ∞). Take t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞),

x σ (τ (t)) > 0,

and   n−1 α Δ φ(t) x Δ (t) < 0,

l

x Δ (t) > 0,

l ∈ {0, . . . , n − 1},

 n−1 α Δ x Δ (t) < 0, n−1

x Δ (t) ≥ x Δ

(t)hn−2 (t, t1 ),

t ∈ [t1 , ∞).

Take t2 ∈ [t1 , ∞) such that τ (t) ≥ t1 , t ∈ [t2 , ∞). Then x σ (τ (t)) ≥ x(τ (t)) ≥ x(t1 ),

t ∈ [t2 , ∞).

Define the function w by the generalized Riccati substitution

w(t) = f (t)

 n−1 α φ(t) x Δ (t) (x(τ (t))β

,

t ∈ [t2 , ∞).

We have that w(t) > 0, t ∈ [t2 , ∞), and   n−1 α Δ w Δ (t) = φ(t) x Δ (t)

f (t) (x(τ (t)))β   n−1 α σ f (t) Δ + φ(t) x Δ (t) (x(τ (t)))β   n−1 α Δ f (t) = φ(t) x Δ (t) (x(τ (t)))β

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

709

  n−1 α σ + φ(t) x Δ (t)

f Δ (t) (x(τ (t)))βσ   n−1 α σ f (t) (x(τ (t)))β Δ − φ(t) x Δ (t) (x(τ (t)))β (x(τ (t)))βσ   n−1 α Δ f (t) (x σ (τ (t)))β = φ(t) x Δ (t) β (x σ (τ (t))) (x(τ (t)))β +

f Δ (t) σ w (t) f σ (t)

 n−1 α σ − φ(t) x Δ (t) 

≤ −q(t)f (t) +

Δ  f (t) (x(τ (t)))β (x(τ (t)))β (x(τ (t)))βσ

f Δ (t)+ σ w (t) f σ (t)

 n−1 α σ − φ(t) x Δ (t) 

Δ  f (t) (x(τ (t)))β , (x(τ (t)))β (x(τ (t)))βσ

t ∈ [t3 , ∞). If 0 < β < 1, we have  Δ  β−1 Δ (x(τ (t)))β ≥ β x σ (τ (t)) x (τ (t))τ Δ (t), t ∈ [t2 , ∞). If β ≥ 1, we have 

(x(τ (t)))β

Δ

≥ β(x(τ (t)))β−1 x Δ (τ (t))τ Δ (t),

t ∈ [t2 , ∞).

Then, if 0 < β < 1, we have f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) (x σ (τ (t)))β−1 x Δ (τ (t))τ Δ (t)

w Δ (t) ≤ −f (t)q(t) +

−βf (t)

(x(τ (t)))β (x(τ (t)))βσ f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) (x σ (τ (t)))β x Δ (τ (t))τ Δ (t)

≤ −f (t)q(t) +

−βf (t)

(x σ (τ (t)))β+1 (x(τ (t)))β

710

12 Oscillations of Higher-Order Functional Dynamic Equations

f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) x Δ (τ (t))τ Δ (t)

≤ −f (t)q(t) +

−βf (t)

,

(x σ (τ (t)))β+1

t ∈ [t2 , ∞), and if β ≥ 1, we have f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) (x(τ (t)))β−1 x Δ (τ (t))τ Δ (t)

w Δ (t) ≤ −f (t)q(t) +

−βf (t)

(x(τ (t)))β (x(τ (t)))βσ f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) x Δ (τ (t))τ Δ (t)

= −f (t)q(t) +

−βf (t)

x(τ (t))(x(τ (t)))βσ f Δ (t)+ σ w (t) f σ (t)   n−1 α σ φ(t) x Δ (t) x Δ (τ (t))τ Δ (t)

≤ −f (t)q(t) +

−βf (t)

,

(x σ (τ (t)))β+1

 n−1 α t ∈ [t2 , ∞). Note that φ(t) x Δ (t) is decreasing on [t2 , ∞) and τ (t) ≤ t ≤ σ (t), t ∈ [t2 , ∞). Then  n−1 α   n−1 α σ φ(τ (t)) x Δ (τ (t)) ≥ φ(t) x Δ (t) , t ∈ [t2 , ∞), and

n−1

xΔ

(τ (t)) ≥

  n−1 α σ α1 φ(t) x Δ (t) 1

(φ(τ (t))) α

,

t ∈ [t2 , ∞).

Therefore n−1

x Δ (τ (t)) ≥ x Δ (τ (t))hn−1 (τ (t), t1 )   n−1 α σ α1 φ(t) x Δ (t) ≥ hn−2 (τ (t), t1 ), 1 (φ(τ (t))) α

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

711

t ∈ [t2 , ∞), and w Δ (t) ≤ −f (t)q(t) + −βτ Δ (t)f (t)

f Δ (t)+ σ w (t) f σ (t)

  n−1 α σ 1+ α1 (t) φ(t) x Δ

1

(φ(τ (t))) α

f Δ (t)+ σ f σ (t) w (t)

= −f (t)q(t) +

hn−2 (τ (t),t1 )

(x σ (τ (t)))β+1

−βτ Δ (t)f (t) hn−2 (τ (t),t1 )(x

β σ (τ (t))) α −1



1

(φ(τ (t))) α

wσ (t) f σ (t)

1+ 1

α

,

t ∈ [t2 , ∞). Now we consider the following cases. 1. Let β > α. Then  σ  β −1 β x (τ (t)) α ≥ (x(τ (t2 ))) α −1 :=  > 0,

t ∈ [t2 , ∞).

2. Let α = β. Then  σ  β −1 x (τ (t)) α = 1,

t ∈ [t2 , ∞).

3. Let β < α. Take t3 ∈ [t2 , ∞) such that τ (t) ≥ t2 , t ∈ [t3 , ∞). Then τ σ (t) ≥ τ (t) ≥ t2 ,

t ∈ [t3 , ∞),

and n−1

xΔ

n−1

(t) ≤ x Δ

t ∈ [t3 , ∞).

(t3 ),

Hence, n−2

xΔ

n−2

(t3 ) + x Δ

n−2

(t3 ) + x Δ

(t) ≤ x Δ = xΔ

n−1 n−1

(t3 )(t − t3 ) (t3 )h1 (t, t3 ),

t ∈ [t3 , ∞).

From here, n−3

xΔ

n−3

(t) ≤ x Δ

+x

n−2

(t3 ) + x Δ

Δn−1

(t3 )h1 (t, t3 )

(t3 )h2 (t, t3 ),

t ∈ [t3 , ∞).

712

12 Oscillations of Higher-Order Functional Dynamic Equations

By induction, we get 2

x(t) ≤ x(t3 ) + x Δ (t3 )h1 (t, t3 ) + x Δ (t3 )h2 (t, t3 ) n−1

+ · · · + xΔ

(t3 )hn−1 (t, t3 )

:= 0 + 1 h1 (t, t3 ) + 2 h2 (t, t3 ) + · · · + n−1 hn−1 (t, t3 ),

t ∈ [t3 , ∞),

where l

l = x Δ (t3 ),

l ∈ {0, . . . , n − 1}.

Hence, x σ (τ (t)) ≤ 0 + 1 h1 (τ σ (t), t3 ) + 2 h2 (τ σ (t), t3 ) + · · · + n−1 hn−1 (τ σ (t), t3 ),

t ∈ [t3 , ∞),

and

 β −1  σ α ≥ 0 + 1 h1 (τ σ (t), t3 ) + 2 h2 (τ σ (t), t3 ) x (τ (t)) + · · · + n−1 hn−1 (τ (t), t3 ) σ

 βα −1 ,

t ∈ [t3 , ∞).

Consequently  β −1  σ x (τ (t)) α ≥ Φ(t, t3 ),

t ∈ [t3 , ∞),

and w Δ (t) ≤ −f (t)q(t) +

f Δ (t)+ σ w (t) f σ (t)  σ

3) −βτ Δ (t)f (t) hn−2 (τ (t),t1 )Φ(t,t 1

≤ −f (t)q(t) +

(φ(τ (t))) α f Δ (t)+ σ σ f (t) w (t)

3) −βτ Δ (t)f (t) hn−2 (τ (t),t3 )Φ(t,t 1

(φ(τ (t))) α

w (t) f σ (t)



wσ (t) f σ (t)

1+ 1

α

(12.19) 1+ 1

α

,

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

713

t ∈ [t3 , ∞). Let %

hn−2 (τ (t), t3 )Φ(t, t3 )

X = βτ Δ (t)f (t) 

f Δ (t)+ Y = λα t ∈ [t3 , ∞), for λ =

α+1 α .

α+1 α ,

λ

1

(φ(τ (t))) α

α % βτ Δ (t)f (t)

w σ (t) , f σ (t)

hn−2 (τ (t), t3 )Φ(t, t3 )

&− α

1

λ

,

(φ(τ (t))) α Now, using the inequality

λXY λ−1 − Xλ ≤ (λ − 1)Y λ , for λ =

&1

λ > 1,

we get

& α % α+1 hn−2 (τ (t), t3 )Φ(t, t3 ) α+1 w σ (t) Δ βτ (t)f (t) 1 α f σ (t) (φ(τ (t))) α αf Δ (t)+ × α+1

% Δ

βτ (t)f (t)

hn−2 (τ (t), t3 )Φ(t, t3 )

&−

α α+1

1

(φ(τ (t))) α

 α+1 hn−2 (τ (t), t3 )Φ(t, t3 ) w σ (t) α −βτ Δ (t)f (t) 1 f σ (t) (φ(τ (t))) α % &−α α+1 α α+1 1 Δ hn−2 (τ (t), t3 )Φ(t, t3 ) Δ f (t)+ ≤ , βτ (t)f (t) 1 α (α + 1)α+1 (φ(τ (t))) α t ∈ [t3 , ∞), or

 α+1 f Δ (t)+ σ hn−2 (τ (t), t3 )Φ(t, t3 ) w σ (t) α Δ w (t) − βτ (t)f (t) 1 f σ (t) f σ (t) (φ(τ (t))) α  α  α+1 α f Δ (t)+ φ(τ (t)) β  α , ≤ (α + 1)α+1 τ Δ (t)f (t)hn−2 (τ (t), t3 )Φ(t, t3 ) t ∈ [t3 , ∞), and w Δ (t) ≤ −f (t)q(t)  α  α+1 α φ(τ (t)) f Δ (t)+ β  α , + (α + 1)α+1 τ Δ (t)f (t)hn−2 (τ (t), t3 )Φ(t, t3 )

714

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t3 , ∞), or  α  α+1 α f Δ (t)+ φ(τ (t)) β  α ≤ −w Δ (t), f (t)q(t) − (α + 1)α+1 τ Δ (t)f (t)hn−2 (τ (t), t3 )Φ(t, t3 ) t ∈ [t3 , ∞). Then ⎛

⎞  α  α+1 α Δ (s) f φ(τ (s)) + β ⎜ ⎟  α ⎠ Δs ≤ −w(t) + w(t3 ) ⎝f (s)q(s) − (α + 1)α+1 τ Δ (s)f (s)hn−2 (τ (s), t3 )Φ(s, t3 ) t3 t

≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. This completes the proof. Exercise 12.8. Let T = Z, τ (t) = t − 1, f (t) = t 2 + t + 1, t ∈ T. Write an analogue of Theorem 12.7. Theorem 12.8. Assume (H 1)–(H 3), τ Δ (t) > 0, τ σ (t) = σ (τ (t)), t ∈ T, 1 ([t , ∞)) and a and (12.13) holds and there exist a positive function f ∈ Crd 0 function G ∈ Crd (D), D = {(t, s) ∈ T × T : t ≥ s ≥ t0 }, such that G(t, t) = 0,

t ∈ [t0 , ∞),

G(t, s) > 0,

(t, s) ∈ D,

t > s,

G has a rd-continuous delta derivative GΔs (t, s) on D with respect to the second variable, and there exists a function g ∈ Crd (D) such that GΔs (t, s) + G(t, s)

α g(t, s) f Δ (s)+ = σ (G(t, s)) α+1 , σ f (s) f (s)

(t, s) ∈ D,

and 1 lim sup t→∞ G(t, t2 ) ×

∞

 α

G(t, s)f (s)q(s) −

t2

(g(t, s)+ )α+1  (α + 1)α+1 τ Δ (s)f (s)hn−2 (τ (s), t1 )

α β

φ(τ (s))

(Φ(s, t1 ))α  α Δs = ∞

for all sufficiently large t1 , t2 ∈ [t0 , ∞), where t2 satisfies τ (s) > t1 , s ∈ [t2 , ∞), Φ is defined as in Theorem 12.7. Then the equation (12.12) is oscillatory.

12.2 Oscillations of Even Order Nonlinear Delay Dynamic Equations

715

Proof. Let x be a nonoscillatory solution of the equation (12.12) on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.12) on [t0 , ∞). Take t1 ∈ [t0 , ∞) such that x(t) > 0,

t ∈ [t1 , ∞).

x σ (τ (t)) > 0,

Take t3 ∈ [t1 , ∞) such that (12.19) holds. Then, by (12.19), we obtain f Δ (s)+ σ w (s) f σ (s)

σ 1+ 1 α w (s) −ξ(s, t3 ) , s ∈ [t3 , ∞), σ f (s)

f (s)q(s) ≤ −w Δ (s) +

where ξ(s, t3 ) = βτ Δ (s)f (s)

hn−2 (τ (s), t3 )Φ(s, t3 ) 1

(φ(τ (s))) α

,

s ∈ [t3 , ∞).

Hence, f Δ (s)+ σ w (s) f σ (s)

σ 1+ 1 α w (s) −G(t, s)ξ(s, t3 ) , s ∈ [t3 , ∞), σ f (s)

G(t, s)f (s)q(s) ≤ −G(t, s)w Δ (s) + G(t, s)

and t

G(t, s)f (s)q(s)Δs ≤ −

t3

t

G(t, s)w Δ (s)Δs

t3 t

f Δ (s)+ σ w (s)Δs f σ (s) t3

σ 1+ 1 t α w (s) − G(t, s)ξ(s, t3 ) Δs f σ (s) t3 +

G(t, s)

= G(t, t3 )w(t3 ) +

t

GΔs (t, s)w σ (s)Δs

t3 t

f Δ (s)+ σ w (s)Δs f σ (s) t3

σ 1+ 1 t α w (s) − G(t, s)ξ(s, t3 ) Δs σ f (s) t3

+

G(t, s)

716

12 Oscillations of Higher-Order Functional Dynamic Equations t

α g(t, s) (G(t, s)) α+1 w σ (s)Δs σ t3 f (s)

σ 1+ 1 t α w (s) − G(t, s)ξ(s, t3 ) Δs σ f (s) t3

= G(t, t3 )w(t3 ) +

t

α g(t, s)+ (G(t, s)) α+1 w σ (s)Δs σ t3 f (s)

σ 1+ 1 t α w (s) − G(t, s)ξ(s, t3 ) Δs, σ (s) f t3

≤ G(t, t3 )w(t3 ) +

t ∈ [t3 , ∞). Let w σ (s) , f σ (s)  α α g(t, s)+ (G(t, s)) α+1 Y = , α λα (G(t, s)ξ(s, t3 )) λ 1

X = (G(t, s)ξ(s, t3 )) λ

λ=

α+1 . α

Now, using the inequality λXY λ−1 − Xλ ≤ (λ − 1)Y λ for λ =

α+1 α ,

we get α

σ α w (s) αg(t, s)+ (G(t, s)) α+1 α+1 (G(t, s)ξ(s, t3 )) α+1 σ α α f (s) (α + 1)(G(t, s)ξ(s, t3 )) α+1

 α+1 w σ (s) α −G(t, s)ξ(s, t3 ) f σ (s)  α+1  1 g(t, s)+ (G(t, s))α+1 α α α+1 ≤ , α (α + 1)α+1 (G(t, s)ξ(s, t3 ))α

t, s ∈ [t3 , ∞), t > s, or

σ 1+ 1 α α g(t, s)+ w (s) σ α+1 w (s) − G(t, s)ξ(s, t ) (G(t, s)) 3 f σ (s) f σ (s) ≤

αα (g(t, s)+ )α+1 , α+1 (α + 1) (ξ(s, t3 ))α

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

717

t, s ∈ [t3 , ∞), t > s. Therefore t

G(t, s)f (s)q(s)Δs ≤ G(t, t3 )w(t3 )

t3

+

t t3

αα (g(t, s)+ )α+1 Δs, (α + 1)α+1 (ξ(s, t3 ))α

t ∈ [t3 , ∞), whereupon 1 G(t, t3 )

t t3

%

αα (g(t, s)+ )α+1 G(t, s)f (s)q(s) − (α + 1)α+1 (ξ(s, t3 ))α

& Δs ≤ w(t3 ),

t ∈ [t3 , ∞). This is a contradiction. This completes the proof. Exercise 12.9. Let T = 2Z, τ (t) = t − 2, f (t) = t 2 + 1, t ∈ T. Write an analogue of Theorem 12.8.

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations In this section we investigate the following nth order nonlinear neutral delay dynamic equation % an (t)

% & &   Δ Δ αn Δ α an−1 (t) . . . a1 (t) (x(t) − p(t)x(τ1 (t)))Δ 1 . . . + f (t, x(τ2 (t))) = 0,

(12.20) t ∈ [t0 , ∞), where n ≥ 2, αk , k ∈ {1, . . . , n}, are quotients of odd positive integers, (I1)

ak ∈ Crd (T), k ∈ {1, . . . , n}, are positive functions, p ∈ Crd (T), limt→∞ p(t) = p0 , |p0 | < 1, ∞ t0

(I2)

1 ak (t)



1 αk

Δt = ∞,

k ∈ {1, . . . , n},

τ1 , τ2 ∈ Crd (T), τ1 , τ2 : T → T, τ1 (t) ≤ t, t ∈ T, lim τ1 (t) = lim τ2 (t) = ∞,

t→∞

t→∞

718

12 Oscillations of Higher-Order Functional Dynamic Equations

(I3)

f ∈ C (T × R), uf (u) > 0, u = 0, t ∈ T, there exists a positive function q ∈ Crd (T) such that |f (t, u)| ≥ q(t)|u|,

u = 0,

t ∈ T.

Introduce

Sk (t, x(t)) =

x(t) − p(t)x(τ1 (t)) if k = 0  Δ α (t, x(t)) k if k ∈ {1, . . . , n}. ak (t) Sk−1

Let S0 (t) = S0 (t, x(t)). Then the equation (12.20) takes the form SnΔ (t, x(t)) + f (t, x(τ2 (t))) = 0,

t ∈ [t0 , ∞).

Lemma 12.6. Let x be an eventually positive solution of the equation (12.20). If there exists a constant l ≥ 0 such that lim S0 (t) = l,

t→∞

then lim x(t) =

t→∞

l . 1 − p0

Proof. Since x is an eventually positive solution of the equation (12.20) and limt→∞ p(t) = p0 , |p0 | < 1, there exist a t1 ∈ [t0 , ∞) and a constant p1 ∈ (0, 1) such that x(t) > 0,

x(τ1 (t)) > 0,

t ∈ [t1 , ∞),

and |p(t)| ≤ p1 ,

t ∈ [t1 , ∞).

Assume that x is not bounded on [t0 , ∞). Then there exists a sequence {tn }n∈N ⊂ [t1 , ∞) such that tn → ∞, as n → ∞, and x(tn ) = max x(t), 0≤t≤tn

lim x(tn ) = ∞.

n→∞

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

719

Because τ1 (t) ≤ t, t ∈ T, we have x(τ1 (tn )) ≤ x(tn ),

n ∈ N,

and then S0 (tn ) = x(tn ) − p(tn )x(τ1 (tn )) ≥ x(tn ) − p(tn )x(tn ) ≥ x(tn ) − p1 x(tn ) = (1 − p1 )x(tn ) → ∞,

as

n → ∞.

This is a contradiction. Therefore x is bounded on [t0 , ∞). Let x1 = lim sup x(t), t→∞

x2 = lim inf x(t). t→∞

We have the following possibilities. 1. Let 0 ≤ p0 < 1. Then x1 − p0 x2 ≤ l ≤ x2 − p0 x1 , whereupon x1 ≤ x2 . 2. Let −1 < p0 ≤ 0. Then x1 − p0 x1 ≤ l ≤ x2 − p0 x2 , whereupon x1 ≤ x2 . Therefore limt→∞ x(t) exists and lim x(t) = x1 = x2 =

t→∞

l . 1 − p0

This completes the proof. Lemma 12.7. If SnΔ (t, x(t)) < 0 and x(t) > 0, t ∈ [t0 , ∞), then there exists an m ∈ {0, . . . , n} such that 1. m + n is even, 2. (−1)m+l Sl (t, x(t)) > 0, t ∈ [t0 , ∞), l ∈ {m, . . . , n},

720

12 Oscillations of Higher-Order Functional Dynamic Equations

3. if m > 1, then there exists a t1 ∈ [t0 , ∞) such that Sl (t, x(t)) > 0, t ∈ [t1 , ∞), l ∈ {1, . . . , m − 1}. Proof. Let there is a t1 ∈ [t0 , ∞) such that Sn (t1 , x(t1 )) < 0. Because SnΔ (t, x(t)) < 0, t ∈ [t0 , ∞), we get that t ∈ [t1 , ∞),

Sn (t, x(t)) < 0, and

t ∈ (t1 , ∞).

Sn (t, x(t)) < Sn (t1 , x(t1 )), Then

 Δ αn (t, x(t)) < Sn (t1 , x(t1 )), an (t) Sn−1  Δ αn Sn (t1 , x(t1 )) Sn−1 (t, x(t)) < , an (t) 1

Δ (t, x(t)) Sn−1


0, Consider the following cases. Case 1. Case 2.

Sl (t) > 0 for any l ∈ {0, . . . , n − 1}, t ∈ [t0 , ∞), Sk (t) < 0 for some k ∈ {0, . . . , n − 2}, t ∈ [t0 , ∞).

1. Suppose Case 1. Then m = n. 2. Suppose Case 2. Let

M = j ∈ N0 : (−1)l+j Sl (t, x(t)) > 0, l ∈ {j + 1, . . . , n − 1},

t ∈ [t0 , ∞),

 even .

j +l

Note that n − 1 ∈ M and there exists m = min M. Assume that m > 1. Then

Δ Sm−1 (t, x(t)) =

Sm (t, x(t)) am (t)



1 αm

1

=

(−1)m+m (Sm (t, x(t))) αm 1

(am (t)) αm > 0,

t ∈ [t0 , ∞).

(a) Assume that there is a t1 ∈ [t0 , ∞) such that t ∈ [t1 , ∞).

Sm−1 (t, x(t)) > 0, Then

Δ (t, x(t)) Sm−2

=

≥

Sm−1 (t, x(t)) am−1 (t)



Sm−1 (t1 , x(t1 )) am−1 (t)

1 αm−1



1 αm−1

t ∈ [t1 , ∞),

,

and Sm−2 (t, x(t)) ≥ Sm−2 (t1 , x(t1 )) 1

+ (Sm−1 (t1 , x(t1 ))) αm−1

t t1

→∞

as

t → ∞.

Δs 1

(am−1 (s)) αm−1

722

12 Oscillations of Higher-Order Functional Dynamic Equations

By induction, Sl (t, x(t)) → ∞,

as

t → ∞,

l ∈ {0, . . . , m − 2}.

(b) Assume Sm−1 (t, x(t)) < 0, t ∈ [t0 , ∞). Suppose that there is a t1 ∈ [t0 , ∞) such that t ∈ [t1 , ∞).

Sm−2 (t, x(t)) < 0, Since

Δ (t, x(t)) Sm−2

=

Sm−1 (t, x(t)) am−1 (t)

< 0,



1 αm−1

t ∈ [t1 , ∞),

we get Sm−2 (t, x(t)) < Sm−2 (t1 , x(t1 )) < 0,

t ∈ [t1 , ∞).

Hence,

Δ (t, x(t)) = Sm−3


0,

t ∈ [t0 , ∞),

which contradicts with the definition of m. This completes the proof. Lemma 12.8. Suppose (I 1)–(I 3) and ∞

An−1 (s)Δs = ∞,

(12.21)

t0

where ⎧ 1 #∞ αn ⎪ ⎨ 1 q(s)Δs if l = n a (t) t Al (t) =  n 1 ⎪ ⎩ 1 # ∞ A (s)Δs αl if l ∈ {1, . . . , n − 1}. l+1 al (t) t Let also, x is an eventually positive solution of the equation (12.20). Then there exists a t1 ∈ [t0 , ∞) sufficiently large such that SnΔ (t, x(t)) < 0, t ∈ [t1 , ∞) and 1. if n is odd, we have l ∈ {1, . . . , n},

Sl (t, x(t)) > 0,

(12.22)

2. if n is even, we have either (12.22) holds or (−1)l Sl (t, x(t)) > 0,

l ∈ {1, . . . , n},

(12.23)

and limt→∞ x(t) = 0. Proof. Note that there exist a t1 ∈ [t0 , ∞) and a constant p1 ∈ (0, 1) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

and |p(t)| ≤ p1 ,

t ∈ [t1 , ∞).

Then, using the equation (12.20) and (I 3), we arrive to SnΔ (t, x(t)) = −f (t, x(τ2 (t))) ≤ −q(t)x(τ2 (t)) < 0,

t ∈ [t1 , ∞).

t ∈ [t1 , ∞),

724

12 Oscillations of Higher-Order Functional Dynamic Equations

1. Let n is odd and m be as in Lemma 12.7. Then m is odd and t ∈ [t1 , ∞).

S1 (t, x(t)) > 0, Hence,

S0Δ (t)

S1 (t, x(t)) a1 (t)

=



1 α1

t ∈ [t1 , ∞).

> 0,

Therefore limt→∞ S0 (t) exists and it is positive or limt→∞ S0 (t, x(t)) = ∞. Consequently there exist a t2 ∈ [t1 , ∞) and a constant b > 0 such that S0 (t) ≥ b

b , 2(1 − p0 )

x(τ2 (t)) ≥

and

t ∈ [t2 , ∞).

Assume that m = n. Then, by Lemma 12.7, it follows that Sn−1 (t, x(t)) < 0

and

t ∈ [t2 , ∞).

Sn−2 (t, x(t)) > 0,

Next, ∞

Sn (t, x(t)) ≥

q(s)x(τ2 (s))Δs,

t ∈ [t2 , ∞).

t

From here,

Δ (t, x(t)) Sn−1

=

≥

≥

=

Sn (t, x(t)) an (t) 1 an (t)



1 αn



∞

q(s)x(τ2 (s))Δs

1 αn

t

b 2(1 − p0 ) b 2(1 − p0 )

 

1 αn

1 αn

1 an (t)

An (t),



∞

q(s)Δs

1 αn

t

t ∈ [t2 , ∞),

and

−Sn−1 (t, x(t)) ≥

b 2(1 − p0 )



1 αn

∞

An (s)Δs, t

t ∈ [t2 , ∞).

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

725

Then

Δ (t, x(t)) −Sn−2

 1 Sn−1 (t, x(t)) αn−1 = − an−1 (t)  1

 1 ∞ αn αn−1 αn−1 1 b ≥ An (s)Δs 2(1 − p0 ) an−1 (t) t  1

αn αn−1 b = An−1 (t), t ∈ [t2 , ∞), 2(1 − p0 )

and ∞ > Sn−2 (t2 , x(t2 ))  1

αn αn−1 b ≥ 2(1 − p0 )

∞

An−1 (s)Δs. t2

This is a contradiction. Therefore m = n. 2. Let n is even and m be as in Lemma 12.7. Then m is even. We have either S0Δ (t) > 0 or S0Δ (t) < 0, t ∈ [t1 , ∞). Hence, there exists r = lim S0 (t). t→∞

Since limt→∞ p(t) = p0 , |p0 | < 1, we have r ≥ 0. Assume that r > 0. Then there exists a t3 ∈ [t2 , ∞) such that x(τ2 (t)) ≥

r , 2

t ∈ [t3 , ∞).

From here and from (12.20), we obtain SnΔ (t, x(t)) = −f (t, x(τ2 (t))) ≤ −q(t)x(τ2 (t)) r ≤ − q(t), t ∈ [t3 , ∞), 2 and −Sn (t, x(t)) ≤ −Sn (t4 , x(t4 )) − ≤−

r 2

t4

q(s)Δs, t

r 2

t4

q(s)Δs t

t, t4 ∈ [t3 , ∞),

t < t4 .

726

12 Oscillations of Higher-Order Functional Dynamic Equations

Therefore −Sn (t, x(t)) ≤ −

∞

r 2

t ∈ [t3 , ∞),

q(s)Δs, t

and r 2

Sn (t, x(t)) ≥

∞

q(s)Δs,

t ∈ [t3 , ∞).

t

From here,

Δ (t, x(t)) Sn−1

1 αn 1 Sn (t, x(t)) = an (t) 1 r 1 1 ∞ αn αn ≥ q(s)Δs 2 an (t) t r 1 αn = An (t), t ∈ [t3 , ∞), 2

and −Sn−1 (t, x(t)) ≥ −Sn−1 (t4 , x(t4 )) r 1 t4 αn + An (s)Δs 2 t r 1 t4 αn ≥ An (s)Δs, t, t4 ∈ [t3 , ∞), 2 t

t < t4 .

Then Sn−1 (t, x(t)) ≤ −

r

∞

1 αn

2

An (s)Δs,

t ∈ [t3 , ∞),

t

and

Δ (t, x(t)) Sn−2

= ≤− =−

1 an−1 (t) r

 1 αn−1 Sn−1 (t, x(t))

1 αn αn−1

2 r 2

an−1 (t) 1 αn αn−1

An−1 (t),



∞

1

An (s)Δs t

t ∈ [t3 , ∞),

1 αn−1

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

727

and Sn−2 (t, x(t)) ≤ Sn−2 (t3 , x(t3 )) r 1 t αn αn−1 − An−1 (s)Δs 2 t3 → −∞ as

t → ∞.

This is a contradiction. This completes the proof. Lemma 12.9. Let x be an eventually positive solution of the equation (12.20) which satisfies Sl (t, x(t)) > 0,

l ∈ {0, . . . , n},

eventually. Then there exists a t1 ∈ [t0 , ∞) such that for t ∈ [t1 , ∞) and l ∈ {0, 1, . . . , n − 1} we have Sl (t, x(t)) ≥ (Sn (t, x(t)))

/n

1 k=l+1 αk

Bl+1 (t, t1 )

and S0Δ (t)

≥ (Sn (σ (t), x(σ (t))))

/n

1 k=l+1 αk

B2 (t, t1 ) a1 (t)



1 α1

and there exists a t2 ∈ [t1 , ∞) and a constant c > 0 such that S0 (t) ≤ cB1 (t, t1 ),

t ∈ [t1 , ∞),

where ⎧  1 # αn ⎪ 1 ⎨ t Δs if l = n t1 an (s) Bl (t, t1 ) =  1 # ⎪ ⎩ t Bl+1 (s,t) αl Δs if l ∈ {1, . . . , n − 1}. t1 al (s) Proof. Take t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

Sl (t, x(t)) > 0,

728

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t1 , ∞), l ∈ {1, . . . , n}. Then Sn (t, x(t)) is decreasing on [t1 , ∞) and t

Sn (s, x(s)) an (s) t1 1 t 1  αn Δs an (s) t1

Sn−1 (t, x(t)) = Sn−1 (t1 , x(t1 )) + 1

≥ (Sn (t, x(t))) αn



1 αn

Δs

1

= (Sn (t, x(t))) αn Bn (t, t1 ),  1 t S n−1 (s, x(s)) αn−1 Δs Sn−2 (t, x(t)) = Sn−2 (t1 , x(t1 )) + an−1 (s) t1 ≥

t t1

%

1

(Sn (s, x(s))) αn Bn (s, t1 ) an−1 (s)

≥ (Sn (t, x(t)))

t

1 αn αn−1

t1

&α 1

n−1

Δs

Bn (s, t1 ) an−1 (s)

1

= (Sn (t, x(t))) αn αn−1 Bn−1 (t, t1 ),



1 αn−1

Δs

t ∈ [t1 , ∞).

By induction, we have S1 (t, x(t)) ≥ (Sn (t, x(t))) S0 (t, x(t)) ≥ (Sn (t, x(t)))

/n

1 k=2 αk

B2 (t, t1 ),

/n

1 k=1 αk

B1 (t, t1 ),

t ∈ [t1 , ∞),

and

S0Δ (t, x(t))

=

S1 (t, x(t)) a1 (t)

≥ (Sn (t, x(t)))



1 α1

/n

1 k=1 αk

B2 (t, t1 ) a1 (t)



1 α1

,

t ∈ [t1 , ∞). On the other hand, for t ∈ [t1 , ∞), we have Sn−1 (t, x(t)) = Sn−1 (t1 , x(t1 )) +

t t1

Sn (s, x(s)) an (s) 1



1 αn

Δs

≤ Sn−1 (t1 , x(t1 )) + (Sn (t1 , x(t1 ))) αn Bn (t, t1 ).

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

729

There exists a t2 ∈ [t1 , ∞) and a constant b1 > 0 such that Sn−1 (t, x(t)) ≤ b1 Bn (t, t1 ),

t ∈ [t2 , ∞).

Next, t

Sn−2 (t, x(t)) = Sn−2 (t1 , x(t1 )) +

t2

≤ Sn−2 (t1 , x(t1 )) + b1 ≤ Sn−2 (t1 , x(t1 )) + b1

 1 Sn−1 (s, x(s)) αn−1 Δs an−1 (s) 1 t B (s, t )  αn−1 n 1 Δs an−1 (s) t2 1 t B (s, t )  αn−1 n 1 Δs an−1 (s) t1

≤ Sn−2 (t1 , x(t1 )) + b1 Bn−1 (t, t1 ),

t ∈ [t2 , ∞).

Then there exist a t3 ∈ [t2 , ∞) and a constant b2 > 0 such that Sn−2 (t, x(t)) ≤ b2 Bn−1 (t, t2 ),

t ∈ [t3 , ∞).

By induction, we get points t2 ≤ . . . ≤ tn and constants bl ≥ 0, l ∈ {1, . . . , n}, such that S0 (t) ≤ bn B1 (t, t1 ),

t ∈ [tn+1 , ∞).

This completes the proof. Theorem τ2 (t) > t, t ∈ T, / 12.9. Suppose that (12.21) holds, and p0 ∈ (0, 1), 1 (T) such that for all and nk=1 αk ≥ 1. If there exists a positive function z ∈ Crd sufficiently large t1 ∈ [t0 , ∞) t

lim sup t→∞

% z(s)q(s) −

t1

 Δ 2 z (s) 4Mz(s)τ2Δ (s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )

1& α1

Δs = ∞,

(12.24)

where M is a positive constant, then 1. every solution of the equation (12.20) is either oscillatory or tends to zero when n is even, 2. every solution of the equation (12.20) is oscillatory when n is odd. Proof. Suppose that the equation (12.20) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.20) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

p(t) > 0,

t ∈ [t1 , ∞).

730

12 Oscillations of Higher-Order Functional Dynamic Equations

By Lemma 12.8, it follows that (12.22) holds when n is odd, and either (12.22) holds or limt→∞ x(t) = 0 when n is even. 1. Let n is odd. Define the function w as follows. w(t) =

z(t)Sn (t, x(t)) , S0 (τ2 (t))

t ∈ [t1 , ∞).

We have that w(t) > 0, t ∈ [t1 , ∞), and

w (t) = Δ

Snσ (t, x(t))

≤ Snσ (t, x(t)) −z(t)q(t)

z(t) S0 (τ2 (t))

Δ + (Sn (t, x(t)))Δ

z(t) S0 (τ2 (t))

zΔ (t)S0 (τ2 (t)) − z(t) (S0 (τ2 (t)))Δ   S0 (τ2 (t))S0 τ2σ (t)

S0 (τ2 (t)) S0 (τ2 (t))

= −z(t)q(t) +

zΔ (t) σ w (t) zσ (t)

− (Sn (t, x(t)))σ z(t)

S0Δ (τ2 (t))τ2Δ (t)  , S0 (τ2 (t))S0 τ2σ (t)

t ∈ [t1 , ∞). Since Sn (t, x(t)) is decreasing on [t1 , ∞), there exists a constant d > 0 such that Snσ (t, x(t)) ≤ Sn (t, x(t)) ≤ d,

t ∈ [t1 , ∞).

On the other hand, we have

1 B2 (t, t1 ) α1 a1 (t) 1

/n 1 B2 (t, t1 ) α1 k=1 αk −1 ≥d Sn (σ (t), x(σ (t))) , a1 (t) /n

S0Δ (t) ≥ (Sn (σ (t), x(σ (t))))

1 k=1 αn

t ∈ [t1 , ∞). Let M=d

/n

1 k=1 αk

−1

.

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

731

Then zΔ (t) σ w (t) zσ (t)

w Δ (t) ≤ −z(t)q(t) +

1

−

Mz(t) (B2 (τ2 (t), t1 )) α1 τ2Δ (t) 1

(w(σ (t)))2

(zσ (t))2 (a1 (τ2 (t))) α1  Δ 2 z (t)

≤ −z(t)q(t) +

4Mz(t)τ2Δ (t)

a1 (τ2 (t)) B2 (τ2 (t), t1 )



1 α1

,

t ∈ [t1 , ∞), or z(t)q(t) −

 Δ 2 z (t)

4Mz(t)τ2Δ (t)

a1 (τ2 (t)) B2 (τ2 (t), t1 )



1 α1

≤ −w Δ (t),

t ∈ [t1 , ∞). Hence, t

%

 z(s)q(s) −

t1

zΔ (s)

2

4Mz(s)τ2Δ (s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )

1& α1

Δt ≤ −w(t) + w(t1 ) ≤ w(t1 ),

t ∈ [t1 , ∞).

This is a contradiction. 2. Let n is even. (a) If (12.22) holds, as in above we arrive to a contradiction. (b) Otherwise, using Lemma 12.8, we have (12.23) and limt→∞ x(t) = 0. This completes the proof. Example 12.6. Let z(t) = 1, t ∈ T. Then (12.24) takes the form t

lim sup t→∞

q(s)Δs = ∞.

t1

Example 12.7. Let z(t) = t, t ∈ T. Then zΔ (t) = 1,

t ∈ T,

and (12.24) takes the form t

lim sup t→∞

t1

%

1 sq(s) − 4Msτ2Δ (s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )

1& α1

Δs = ∞.

732

12 Oscillations of Higher-Order Functional Dynamic Equations

Example 12.8. Consider the equation ⎛

⎞Δ 1 ⎛⎛ ⎛ ⎞Δ ⎞Δ ⎞ n−1 & % Δ n+1 ⎜1 ⎟   1 1 ⎜ ⎜⎜ 1 ⎝ ⎟ ⎟ ⎟ x(t) − x(τ1 (t)) . . .⎠ ⎠ ⎠ ⎟ + t n x t n = 0, ... ⎜ ⎝⎝ ⎝t ⎠ t t 3 t ∈ [1, ∞). Here 1 , k ∈ {1, . . . , n}, t α1 = n + 1,

ak (t) =

αk = 1,

k ∈ {2, . . . , n − 1},

1 , n−1 1 p(t) = , 3 q(t) = t n , αn =

τ2 (t) = t n ,

t ∈ T.

Then ∞ t0

1 a1 (t)



1 α1

∞

Δt =

1

t n+1 Δt 1

= ∞, ∞ t0

1 an (t)



1 αn

∞

Δt =

Δt 1

= ∞, ∞ t0

1 ak (t)



1 αk

∞

Δt =

tΔt 1

= ∞, k ∈ {2, . . . , n − 1},

1 ∞ αn 1 An (t) = q(s)Δs an (t) t

n−1 ∞ n = t s Δs t

=t

∞

n−1

n−1 n

s Δs t

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

= ∞,

1 An−1 (t) = an−1 (t)



∞

An (s)Δs

733

1 αn−1

t

∞

=t

An (s)Δs t

= ∞, ∞

∞

An−1 (s)Δs =

An−1 (s)Δs 1

t0

= ∞. Take z(t) = 1. Then t

lim sup t→∞

%



z(s)q(s) −

t1

zΔ (s)

2

4Mz(s)τ2Δ (s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )

1& α1

Δs = lim sup t→∞

t

s n Δs

t1

= ∞.

Hence from Theorem 12.9, we conclude that every solution of the considered equation is either oscillatory or tends to zero. Theorem 12.10. Suppose that (12.21) holds, and p0 ∈ (0, 1), τ2 (t) ≥ t, and /n α ≥ 1. If there exist positive functions g, h ∈ Crd (D), D = {(t, s) ∈ k k=1 T × T, t ≥ s ≥ t0 }, such that h(t, t) = 0,

h(t, s) > 0 and

g(t, s) = hΔs (t, s) + h(t, s)

zΔ (s) zσ (s)

hΔs (t, s) ≤ 0,

t > s ≥ t0 ,

,

and for sufficiently large t1 , ⎛ 1 lim sup h(t, t1 ) t→∞

t

⎝h(t, s)z(s)q(s) −

t1

⎞

1

(g(t, s))2 (z(σ (s)))2 (a1 (τ2 (s))) α1 1

4Mh(t, s)z(s)τ2Δ (s) (B2 (τ2 (s), t1 )) α1

⎠ Δs = ∞,

where z and M are defined as in Theorem 12.9. Then 1. every solution of the equation (12.20) is either oscillatory or tends to zero when n is even. 2. every solution of the equation (12.20) is oscillatory when n is odd. Proof. Suppose that the equation (12.20) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.20) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

p(t) > 0,

t ∈ [t1 , ∞).

734

12 Oscillations of Higher-Order Functional Dynamic Equations

By Lemma 12.8, it follows that (12.22) holds when n is odd, and either (12.22) holds or limt→∞ x(t) = 0 when n is even. As in the proof of Theorem 12.9, we have that z(s)q(s) ≤ −w Δ (s) +

zΔ (s) w(σ (s)) z(σ (s))

− (Sn (s, x(s)))σ z(s) h(t, s)z(s)q(s) ≤ −h(t, s)wΔ (s) +

S0Δ (τ2 (s))τ2Δ (s) , S0 (τ2 (s))S0 (τ2 (σ (s)))

zΔ (s) w(σ (s))h(t, s) z(σ (s))

−h(t, s) (Sn (s, x(s)))σ z(s)

S0Δ (τ2 (s))τ2Δ (s) , S0 (τ2 (s))S0 (τ2 (σ (s)))

s ∈ [t1 , ∞). Then t

h(t, s)z(s)q(s)Δs ≤ −

t1

t t1

−

t

t

h(t, s)w Δ (s)Δs +

t1

zΔ (s) w(σ (s))h(t, s)Δs z(σ (s))

h(t, s) (Sn (s, x(s)))σ z(s)

t1

= h(t, t1 )w(t1 ) +

t

S0Δ (τ2 (s))τ2Δ (s) Δs S0 (τ2 (s))S0 (τ2 (σ (s)))

hΔs (t, s)w σ (s)Δs

t1

+

t t1

−

t

zΔ (s) z(σ (s))

w(σ (s))h(t, s)Δs

h(t, s) (Sn (s, x(s)))σ z(s)

t1

= h(t, t1 )w(t1 ) +

S0Δ (τ2 (s))τ2Δ (s) Δs S0 (τ2 (s))S0 (τ2 (σ (s)))

t

g(t, s)w(σ (s))Δs t1

−

t

h(t, s) (Sn (s, x(s)))σ z(s)

t1

≤ h(t, t1 )w(t1 ) +

S0Δ (τ2 (s))τ2Δ (s) Δs S0 (τ2 (s))S0 (τ2 (σ (s)))

t

g(t, s)w(σ (s))Δs t1

−

1

t

h(t, s)

Mz(s) (B2 (τ2 (s), t1 )) α1 τ2Δ (s) (zσ (s))2 (a1 (τ2 (s)))

t1

1 α1

(w(σ (s)))2 Δs

≤ h(t, t1 )w(t1 ) +

t t1

(g(t, s))2 zσ (s) 4Mz(s)τ2Δ (s)h(t, s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )



1 α1

Δs,

12.3 Oscillations of Higher-Order Nonlinear Neutral Delay Dynamic Equations

735

t ∈ [t1 , ∞). Therefore 1 h(t, t1 )

t t1

%

(g(t, s))2 zσ (s) h(t, s)z(s)q(s) − 4Mz(s)τ2Δ (s)h(t, s)

a1 (τ2 (s)) B2 (τ2 (s), t1 )



1 α1

& Δs ≤ w(t1 ),

t ∈ [t1 , ∞). This is a contradiction. This completes the proof. Theorem 12.11. Suppose that (12.21) holds, and p0 ∈ (0, 1) and τ2 (t) > t. If for all sufficiently large t1 ∈ [t0 , ∞) there are positive constants d1 and d2 such that lim sup (B1 (τ2 (t), t1 ))

/n

k=1 αk

t→∞

∞

q(s)Δs > 1,

γ (τ2 (t), t1 , d1 , d2 ) t

where ⎧ /n αk = 1 ⎨ 1 if /k=1 n γ (τ2 (t), t1 , d1 , d2 ) = d1 if k=1 αk 1,

then 1. every solution of the equation (12.20) is either oscillatory or tends to zero when n is even. 2. every solution of the equation (12.20) is oscillatory when n is odd. Proof. Suppose that the equation (12.20) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.20) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

p(t) > 0,

t ∈ [t1 , ∞).

By Lemma 12.8, it follows that (12.22) holds when n is odd, and either (12.22) holds or limt→∞ x(t) = 0 when n is even. 1. Let n is odd. Then, using Lemma 12.9 and SnΔ (t, x(t)) ≤ −q(t)x(τ2 (t)),

t ∈ [t1 , ∞),

we get ∞

q(s)S0 (τ2 (s))Δs ≤ Sn (t, x(t))

t

≤

S0 (t) B1 (t, t1 )

/nk=1 αk ,

t ∈ [t1 , ∞).

736

12 Oscillations of Higher-Order Functional Dynamic Equations

Since S0Δ (t) > 0 and τ2 (t) > t, t ∈ [t1 , ∞), we get ∞

q(s)Δs ≤ Sn (t, x(t))

S0 (τ2 (t)) t

≤

S0 (τ2 (t)) B1 (t, t1 )

/nk=1 αk ,

t ∈ [t1 , ∞), and (B1 (t, t1 ))

/n

k=1 αk

∞

/n

(S0 (τ2 (t)))1−

k=1 αk

q(s)Δs ≤ 1,

t

t ∈ [t1 , ∞). Consider the following cases. / (a) If nk=1 αk = 1, then /n

(S0 (τ2 (t)))1− (b) If

/n

k=1 αk

= 1,

k=1 αk

t ∈ [t1 , ∞).

< 1, then S0 (τ2 (t)) ≥ S0 (τ2 (t1 )),

t ∈ [t1 , ∞),

and /n

(S0 (τ2 (t)))1−

k=1 αk

/n

≥ d1 (S0 (τ2 (t1 )))1−

k=1 αk

,

t ∈/[t1 , ∞). (c) If nk=1 αk > 1, then, using Lemma 12.9, we get S0 (τ2 (t)) ≤ cB1 (τ2 (t), t1 ),

t ∈ [t1 , ∞).

Therefore /n

(S0 (τ2 (t)))1−

k=1 αk

/n

≥ c1−

k=1 αk

/n

(B1 (τ2 (t), t1 ))1−

k=1 αk

t ∈ [t1 , ∞). Let /n

d2 = c1−

k=1 αk

.

Then /n

(S0 (τ2 (t)))1− t ∈ [t1 , ∞).

k=1 αk

/n

≥ d2 (B1 (τ2 (t), t1 ))1−

k=1 αk

,

,

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay. . .

737

Consequently /n

(B1 (τ2 (t), t1 ))

k=1 αk

∞

γ (τ2 (t), t1 , d1 , d2 )

q(s)Δs ≤ 1,

t

t ∈ [t1 , ∞). This is a contradiction. 2. Let n is even. (a) If (12.22) holds, as in above we arrive to a contradiction. (b) Otherwise, using Lemma 12.8, we have (12.23) and limt→∞ x(t) = 0. This completes the proof.

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay Dynamic Equations In this section the following neutral delay dynamic equation is investigated: n

(x(t) + a(t)x(τ1 (t)))Δ + b(t)x(τ2 (t)) = 0,

t ∈ [t0 , ∞),

(12.25)

where n ∈ N, (J1) (J2)

a ∈ Crd ([t0 , ∞)), b ∈ Crd ([t0 , ∞)) is a positive function, 1 ([t , ∞)) are strictly increasing such that τ , τ : τ1 ∈ Crd ([t0 , ∞)), τ2 ∈ Crd 0 1 2 T → T and lim τ1 (t) = lim τ2 (t) = ∞,

t→∞

t→∞

τ1 (t) ≤ t,

τ2 (t) ≤ t,

t ∈ [t0 , ∞).

Set

t−1 = min

 inf

t∈[t0 ,∞)

τ1 (t),

inf

t∈[t0 ,∞)

τ2 (t) .

We will start with the following useful lemma. n

n (T), n ≥ 2, f Δ Lemma 12.10. Let f ∈ Crd ≤ 0 on T, there exists an n m ∈ {1, . . . , n − 1} such that (−1)n−m f Δ (t) ≥ 0 for sufficiently large t, k k ∈ {0, . . . , m − 1} implies f Δ (t) > 0 for sufficiently large t, k ∈ {m, . . . , n − 1} k implies (−1)m−k f Δ (t) > 0 for sufficiently large t. Then there exists a sufficiently large t1 ∈ [t0 , ∞) such that m

f Δ (t) ≥ hm−1 (t, t1 )f Δ (t),

t ∈ [t1 , ∞).

(12.26)

738

12 Oscillations of Higher-Order Functional Dynamic Equations m

Proof. Note that f Δ is nonincreasing on T. If m = 1, then (12.26) holds. Let m ∈ {2, . . . , n − 1}. By Taylor’s formula applied for f Δ at t1 ∈ [t0 , ∞), we have m−2 

f Δ (t) =

k+1

hk (t, t1 )f Δ

(t1 ) +

t

m

hm−2 (t, σ (η))f Δ (η)Δη

t1

≥f

m

hm−2 (t, σ (η))f Δ (η)Δη

t1

k=0

≥

t

Δm



t

(t)

hm−2 (t, σ (η))Δη t1

m

= f Δ (t)hm−1 (t, t1 ),

t ∈ [t1 , ∞).

This completes the proof. Introduce  (1) bm

=

b(t) (1 − a(τ1 (t))) if m = n − 1 #∞ (1) (−1)n−m−2 t hn−m−2 (t, σ (η))bn−1 (η)Δη

otherwise,

t ∈ [t0 , ∞). Theorem 12.12 (Kamenev-Type Oscillation Criteria). Assume that n ∈ 2N, (J 1) and (J 2) hold, 0 ≤ a(t) ≤ 1, t ∈ [t0 , ∞). Let also, for every k ∈ 1 ([t , ∞)) {1, 3, . . . , n−1} there exist a fixed nk ∈ N and a positive function φk ∈ Crd 0 such that lim sup t→∞

1 hnk (t, r)

t r

(1) hnk (t, σ (η)) bk (η)φk (η) −

 Δ 2 φk (η) 4φk (η)τ2Δ (η)hk−1 (τ2 (η), s)

 Δη = ∞

for all sufficiently large r, s ∈ [t0 , ∞) with τ2 (r) > s. Then every solution of the equation (12.25) is oscillatory. Proof. Suppose that the equation (12.25) has a nonoscillatory solution x on [t0 , ∞). Without loss of generality, assume that x is an eventually positive solution of the equation (12.25) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞).

Let y(t) = x(t) + a(t)x(τ1 (t)),

t ∈ [t1 , ∞).

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay. . .

739

Then n

y Δ (t) = −b(t)x(τ2 (t)),

t ∈ [t1 , ∞).

By Knesser’s theorem, it follows that there are a t2 ∈ [t1 , ∞) and an odd integer m ∈ {1, . . . , n − 1} such that k

t ∈ [t2 , ∞),

y Δ (t) > 0,

k ∈ {0, . . . , m − 1},

and k

k ∈ {m, . . . , n − 1},

(−1)m−k y Δ (t) > 0,

t ∈ [t2 , ∞).

We have y(t) ≥ x(t) > 0, t ∈ [t2 , ∞).

Δ

y (t) > 0, Now, using that

τ1 (τ2 (t)) ≤ τ2 (t),

t ∈ [t2 , ∞),

we get x(τ2 (t)) = y(τ2 (t)) − a(τ2 (t))x(τ1 (τ2 (t))) ≥ y(τ2 (t)) − a(τ2 (t))y(τ1 (τ2 (t))) ≥ y(τ2 (t)) − a(τ2 (t))y(τ2 (t)) = (1 − a(τ2 (t)))y(τ2 (t)),

t ∈ [t2 , ∞).

Integrating (12.25) over [t, ∞) ⊂ [t2 , ∞) for a total (n − m − 1) times and changing the order of integration from the innermost integral to the outermost integral, we get m+1

(−1)n−m−1 y Δ

∞

(t) = −

∞

∞

... t

ηn−m−2

b(η1 )x(τ2 (η1 ))Δη1 . . . Δηn−m−2 Δηn−m−1 η2

∞

=−

hn−m−2 (t, σ (η))b(η)x(τ2 (η))Δη t ∞

≤− t

hn−m−2 (t, σ (η))b(η)(1 − a(τ2 (η)))y(τ2 (η))Δη

740

12 Oscillations of Higher-Order Functional Dynamic Equations ∞

≤ −y(τ2 (t))

hn−m−2 (t, σ (η))b(η)(1 − a(τ2 (η)))Δη

t (1) = −y(τ2 (t))bm (t),

t ∈ [t2 , ∞),

or m+1

yΔ

(1) (t) + bm (t)y(τ2 (t)) ≤ 0,

t ∈ [t2 , ∞).

Define m

Ym (t) =

φm (t)y Δ (t) , y(τ2 (t))

t ∈ [t2 , ∞).

Then m

φm (t)y Δ (τ2 (t)) ≥ Ym (t) y(τ2 (t)) m

φm (t)y Δ σ (t) ≥ y(τ2 (σ (t))) =

φm (t)Ymσ (t) , σ (t) φm

t ∈ [t2 , ∞).

Hence, m+1

YmΔ (t)

φm (t)y Δ (t) = y(τ2 (t))   Δ m (t)y(τ2 (t)) − φm (t) (y(τ2 (t)))Δ y Δ σ (t) φm + y(τ2 (t))y(τ2 (σ (t))) (1) (t)φm (t) ≤ −bm % & φm (t)τ2Δ (t)(y Δ (τ2 (t))) Ymσ (t) Δ , + φm (t) − σ (t) y(τ2 (t)) φm

t ∈ [t2 , ∞). Hence from Lemma 12.10, it follows that there exists a t3 ∈ [t2 , ∞) such that m

y Δ (t) ≥ hm−1 (t, t3 )y Δ (t),

t ∈ [t3 , ∞).

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay. . .

741

Then (1)

YmΔ (t) ≤ −bm (t)φm (t) m y Δ (τ2 (t)) Ymσ (t) Δ (t) − φ (t)τ Δ (t)h + φm m m−1 (τ2 (t), t3 ) y(τ2 (t)) 2 φ σ (t) m

(1)

≤ −bm (t)φm (t) m y Δ (τ2 (σ (t))) Ymσ (t) Δ (t) − φ (t)τ Δ (t)h (τ (t), t ) + φm m m−1 2 3 2 y(τ2 (t)) φ σ (t)

(12.27)

m

(1)

≤ −bm (t)φm (t) Ymσ (t) Ymσ (t) Δ (t) − φ (t)τ Δ (t)h + φm (τ (t), t ) σ m m−1 2 3 2 φ (t) φ σ (t)  Δ 2 φm (t) (1) −bm (t)φm (t) + 4φ (t)τ Δ (t)h , m m−1 (τ2 (t),t3 ) 2

≤

m

m

t ∈ [t3 , ∞). Note that ∞ t3

!η=t ! YmΔ (η)hnm (t, σ (η))Δη = hnm (t, η)Ym (η)!

η=t3

−

t t3

Δ

Ym (η)hnmη (t, η)Δη

= −hnm (t, t3 )Ym (t3 ) +

t t3

Ym (η)hnm −1 (t, σ (η))Δη

≥ −hnm (t, t3 )Ym (t3 ),

t ∈ [t3 , ∞).

Hence from (12.27), we get (1) (t)φm (t) − bm

 Δ 2 φm (t) 4φm (t)τ2Δ (t)hm−1 (τ2 (t), t3 )

≤ −YmΔ (t),

t ∈ [t3 , ∞), and % (1) bm (η)φm (η) −

 Δ 2 φm (η) 4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

≤ −YmΔ (η)hnm (t, σ (η)),

& hnm (t, σ (η))

742

12 Oscillations of Higher-Order Functional Dynamic Equations

t, η ∈ [t3 , ∞), t > η. Therefore t

% (1) (η)φm (η) − bm

t3

t

≤−

 Δ 2 φm (η)

& hnm (t, σ (η))Δη

4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

YmΔ (η)hnm (t, σ (η))Δη

t3

≤ hnm (t, t3 )Ym (t3 ),

t ∈ [t3 , ∞).

Consequently 1 hnm (t, t3 )

t

% (1) (η)φm (η) − bm

t3

≤ Ym (t3 ),

 Δ 2 φm (η) 4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

& hnm (t, σ (η))Δη

t ∈ [t3 , ∞).

This is a contradiction. This completes the proof. Corollary 12.1. Assume that n ∈ 2N, (J 1) and (J 2) hold, 0 ≤ a(t) ≤ 1, t ∈ [t0 , ∞). Let also, for every k ∈ {1, 3, . . . , n − 1} there exists a fixed nk ∈ N such that lim sup t→∞

1 hnk (t, r)

t r

(1) hnk (t, σ (η)) bk (η)hk (η, s) −

 (hk−1 (η, s))2 Δη = ∞ 4hk (η, s)τ2Δ (η)hk−1 (τ2 (η), s)

for all sufficiently large r, s ∈ [t0 , ∞) with τ2 (r) > s. Then every solution of the equation (12.25) is oscillatory. Exercise 12.10. Write an analogue of Theorem 12.12 in the following cases. 1. φk (t) = t, t ∈ T, 2. φ(t) = t 2 , t ∈ T, 3. φ(t) = t 4 + t 2 + 1,

t ∈ T.

Introduce  (2) bm (t)

=

b(t) if m = n − 1 #∞ (2) (η)Δη, (−1)n−m−2 t hn−m−2 (t, σ (η))bn−1

t ∈ [t0 , ∞). Theorem 12.13 (Kamenev-Type Oscillation Criteria). Assume that n ∈ 2N, (J 1), and (J 2) hold, and 0 ≥ a(t) > −1 and t ∈ [t0 , ∞). Let also, for every k ∈ {1, 3, . . . , n − 1} there exist a fixed nk ∈ N and a positive function 1 ([t , ∞)) such that φk ∈ Crd 0

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay. . .

743

 2 

φkΔ (η) t 1 (2) Δη = ∞ lim sup hnk (t, σ (η)) bk (η)φk (η) − Δ 4τ2 (η)φk (η)hk−1 (τ2 (η), s) t→∞ hnk (t, r) r

for all sufficiently large r, s ∈ [t0 , ∞) with τ2 (r) > s. Then every solution of the equation (12.25) is oscillatory or tends to zero asymptotically. Proof. Suppose that the equation (12.25) has a nonoscillatory solution x on [t0 , ∞) which does not tend to zero asymptotically. Without loss of generality, assume that x is an eventually positive solution of the equation (12.25) on [t0 , ∞). Then there is a t1 ∈ [t0 , ∞) such that x(t) > 0,

x(τ1 (t)) > 0,

x(τ2 (t)) > 0,

t ∈ [t1 , ∞)

and 0 ≥ a(t) ≥ −a0 ,

t ∈ [t1 , ∞), k

where a0 ∈ (0, 1). Note that for each k ∈ {0, . . . , n − 1}, y Δ is of fixed sign on [t , ∞) for some t ∈ [t1 , ∞). Then there exists lim y(t) = l,

t→∞

and l ∈ [0, ∞). Assume that l > 0. 1. Let x is unbounded. Take an increasing divergent sequence {ξk }k∈N ⊂ [t , ∞) such that x(ξk ) = max x(t), t∈(t0 ,ξk ]

k ∈ N,

and {x(ξk )}k∈N is increasing and divergent. We have y(ξk ) = x(ξk ) + a(ξk )x(τ1 (ξk )) ≥ (1 − a0 )x(ξk ),

k ∈ N.

Therefore l = ∞. 2. Let x is bounded. We pick an increasing divergent sequence {ξk }k∈N ⊂ [t , ∞) such that lim x(ξk ) = p,

k→∞

744

12 Oscillations of Higher-Order Functional Dynamic Equations

where p = lim sup x(t). t→∞

We have x(τ1 (t)) ≤ p,

t ∈ [t , ∞).

p > 0,

Then y(ξk ) ≥ x(ξk ) − a0 x(τ1 (ξk )),

k ∈ N.

Hence, l ≥ (1 − a0 )p > 0. Consequently there exists a t2 ∈ [t1 , ∞) such that y(t) > 0, t ∈ [t2 , ∞). By Knesser’s theorem, it follows that there are a t2 ∈ [t1 , ∞) and an odd integer m ∈ {1, . . . , n − 1} such that k

y Δ (t) > 0,

t ∈ [t2 , ∞),

k ∈ {0, . . . , m − 1},

and k

(−1)m−k y Δ (t) > 0,

t ∈ [t2 , ∞),

k ∈ {m, . . . , n − 1}.

Observe that x(τ2 (t)) = y(τ2 (t)) − a(τ2 (t))x(τ1 (τ2 (t))) ≥ y(τ2 (t)),

t ∈ [t2 , ∞),

and m+1

(−1)n−m−1 y Δ

∞

(t) = −

hn−m−2 (t, σ (η))b(η)x(τ2 (η))Δη t ∞

≤−

hn−m−2 (t, σ (η))b(η)y(τ2 (η))Δη t ∞

≤ −y(τ2 (t))

hn−m−2 (t, σ (η))b(η)Δη, t

12.4 The Kamenev-Type Oscillation Criteria for Higher-Order Neutral Delay. . .

745

t ∈ [t2 , ∞). Hence, m+1

yΔ

(2) (t) + bm (t)y(τ2 (t)) ≤ 0,

t ∈ [t2 , ∞).

Define m

Ym (t) =

φm (t)y Δ (t) , y(τ2 (t))

t ∈ [t2 , ∞).

Then m

φm (t)y Δ (τ2 (t)) ≥ Ym (t) y(τ2 (t)) m

≥

φm (t)y Δ σ (t) y(τ2 (σ (t)))

=

φm (t)Ymσ (t) , σ (t) φm

t ∈ [t2 , ∞).

Hence, m+1

YmΔ (t)

φm (t)y Δ (t) = y(τ2 (t))  Δ  m (t)y(τ2 (t)) − φm (t) (y(τ2 (t)))Δ y Δ σ (t) φm + y(τ2 (t))y(τ2 (σ (t))) (2) (t)φm (t) ≤ −bm % & φm (t)τ2Δ (t)y Δ (τ2 (t)) Ymσ (t) Δ , + φm (t) − σ (t) y(τ2 (t)) φm

t ∈ [t2 , ∞). Hence from Lemma 12.10, it follows that there exists a t3 ∈ [t2 , ∞) such that m

y Δ (t) ≥ hm−1 (t, t3 )y Δ (t),

t ∈ [t3 , ∞).

Then (2)

YmΔ (t) ≤ −bm (t)φm (t)

Δm

y (τ2 (t)) Δ (t) − φ (t)τ Δ (t)h + φm m m−1 (τ2 (t), t3 ) y(τ2 (t)) 2 (2)

≤ −bm (t)φm (t)

Y σ (t)

Δ (t) − φ (t)τ Δ (t)h m + φm m m−1 (τ2 (t), t3 ) φ σ (t) 2

(2) ≤ −bm (t)φm (t) +

 Δ 2 φm (t) , 4φm (t)τ2Δ (t)hm−1 (τ2 (t),t3 )

m





Ymσ (t) σ (t) φm

Ymσ (t) σ (t) φm

746

12 Oscillations of Higher-Order Functional Dynamic Equations

t ∈ [t3 , ∞), or (2) bm (t)φm (t) −

 Δ 2 φm (t) 4φm (t)τ2Δ (t)hm−1 (τ2 (t), t3 )

≤ −YmΔ (t),

t ∈ [t3 , ∞), and % (2) bm (η)φm (η) −

&

 Δ 2 φm (η) 4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

hnm (t, σ (η))

≤ −YmΔ (η)hnm (t, σ (η)), t, η ∈ [t3 , ∞), t > η. Therefore %

t

(2) (η)φm (η) − bm

t3

≤−

t t3

 Δ 2 φm (η)

& hnm (t, σ (η))Δη

4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

YmΔ (η)hnm (t, σ (η))Δη

≤ hnm (t, t3 )Ym (t3 ),

t ∈ [t3 , ∞).

Consequently 1 hnm (t, t3 ) ≤ Ym (t3 ),

t

% (2) (η)φm (η) − bm

t3

 Δ 2 φm (η) 4φm (η)τ2Δ (η)hm−1 (τ2 (η), t3 )

& hnm (t, σ (η))Δη

t ∈ [t3 , ∞).

This is a contradiction. This completes the proof. Corollary 12.2. Assume that n ∈ 2N, (J 1), and (J 2) hold, and 0 ≥ a(t) > −1 and t ∈ [t0 , ∞). Let also, for every k ∈ {1, 3, . . . , n − 1} there exists a fixed nk ∈ N such that lim sup t→∞

1 hnk (t, r)

t r

(2) hnk (t, σ (η)) bk (η)hk (η, s) −

 (hk−1 (η, s))2 Δη = ∞ 4hk (η, s)τ2Δ (η)hk−1 (τ2 (η), s)

for all sufficiently large r, s ∈ [t0 , ∞) with τ2 (r) > s. Then every solution of the equation (12.25) is oscillatory. Exercise 12.11. Write an analogue of Theorem 12.13 in the following cases. 1. φk (t) = t + 3, t ∈ T, 2. φk (t) = t 2 + t + 1, t ∈ T, 3. φk (t) = t 4 , t ∈ T.

12.5 Advanced Practical Exercises

747

Remark 12.1. The results in this section can be extended for odd order neutral delay dynamic equations and we leave to the reader as an exercise.

12.5 Advanced Practical Exercises Problem 12.1. Let T = 2N0 . Write an analogue of Theorem 12.1 in the following cases. 1. 2. 3. 4. 5. 6. 7.

Φ(t) = φ(t) = 1, t ∈ T, Φ(t) = 1, φ(t) = t, t ∈ T, Φ(t) = φ(t) = t, t ∈ T, Φ(t) = t 2 , φ(t) = 1, t ∈ T, Φ(t) = 1, φ(t) = t 2 , t ∈ T, Φ(t) = φ(t) = t 2 , t ∈ T, Φ(t) = t 3 , φ(t) = 1, t ∈ T.

Problem 12.2. Let   1 T = {0} {3k : k ∈ N}, : k ∈ N0 3k and α = 7,

an (t) = t 7 ,

al (t) = 1,

t ∈ T,

l ∈ {1, . . . , n − 1}.

Prove that any solution of the equation SnΔ (t, x(t)) +

 7 5 t x = 0, 7 3 t5

23

t ∈ [2, ∞),

is either oscillatory or tends to 0 as t → ∞. Problem 12.3. Let T = {0}

  1 : k ∈ N {2k : k ∈ N}, 0 2k

and α = 11,

an (t) = t 11 ,

al (t) = 1,

t ∈ T,

l ∈ {1, . . . , n − 1}.

748

12 Oscillations of Higher-Order Functional Dynamic Equations

Prove that any solution of the equation SnΔ (t, x(t)) +

201 9

t7

 9 7 t x = 0, 2

t ∈ [2, ∞),

is either oscillatory or tends to 0 as t → ∞. Problem 12.4. Let T = {0}



4k : k ∈ N0

  1 : k ∈ N . 4k

Prove that the equation



9 9 Δ t 7 t 7 x Δ (t) + t 13 x = 0, 4

t ∈ [16, ∞),

is oscillatory. Problem 12.5. Let T = Z. Prove that the equation

  15 9 Δ  t 12 + t 6 + 1 x Δ (t) + t 4 + 1 (x (t − 4))9 = 0,

t ∈ [5, ∞),

is oscillatory. Problem 12.6. Let   1 T = {0} {4k : k ∈ N}. : k ∈ N0 4k Prove that every bounded solution of the equation

t

11

15  9 13 Δ t 1 Δ x (t) + 2 x = 0, 4 t

t ∈ [16, ∞),

is oscillatory. Problem 12.7. Let T = 2N0 . Prove that every solution of the equation ⎛

⎛⎛

⎛

%

⎜1 1 ⎜ ⎜⎜ 1 ⎝ ⎜ 2 ⎝⎝ 2 . . . 2 ⎝t t t

⎞Δ 1 ⎞Δ ⎞Δ ⎞ n−1

Δ &n+1 ⎟  t t ⎟ ⎟ ⎟ . . .⎠ ⎠ ⎠ ⎟ + t 3n x t 5n = 0, x(t) − 2 x ⎠ 2 t +3

t ∈ [4, ∞), is oscillatory or tends to zero.

Chapter 13

Shift Operators

The results contained in this chapter can be found in the papers [2, 210]. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T.

13.1 Definition for Shift Operators. Properties Definition 13.1. Let T∗ be a nonempty subset of the time scale T. Define the operators δ± : [t0 , ∞) × T∗ → T∗ satisfying the following properties. (P1)

The functions δ± are strictly increasing with respect to their second arguments, i.e., if (T0 , t),

(T0 , u) ∈ D± = {(s, t) ∈ [t0 , ∞) × T∗ : δ± (s, t) ∈ T∗ }

and t < u, then δ± (T0 , t) < δ± (T0 , u). (P2)

If (T1 , u), (T2 , u) ∈ D± and T1 < T2 , then δ− (T1 , u) > δ− (T2 , u), δ+ (T1 , u) < δ+ (T2 , u).

(P3)

If t ∈ [t0 , ∞), then (t, t0 ) ∈ D+ and δ+ (t, t0 ) = t. If t ∈ T∗ , then (t0 , t) ∈ D+ and δ+ (t0 , t) = t.

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_13

749

750

(P4)

13 Shift Operators

If (s, t) ∈ D± , then (s, δ± (s, t)) ∈ D∓ and δ∓ (s, δ± (s, t)) = t.

(P5)

If (s, t) ∈ D± and (u, δ± (s, t)) ∈ D∓ , then (s, δ∓ (u, t)) ∈ D± and δ∓ (u, δ± (s, t)) = δ± (s, δ∓ (u, t)).

The operators δ+ and δ− associated with t0 ∈ T∗ (called the initial point) are said to be forward shift operator and backward shift operator on the set T∗ , respectively. The variable s ∈ [t0 , ∞) in δ± (s, t) is called the shift size. The sets D± are the domains of the shift operators δ± , respectively.  Example 13.1. Let T = q Z {0}, T∗ = q Z , q > 1, t0 = 1, δ+ (s, t) = st,

t δ− (s, t) = . s

Here D+ = {(s, t) ∈ [1, ∞) × T∗ : st ∈ T∗ }, 

∗ t ∗ D− = (s, t) ∈ [1, ∞) × T : ∈ T . s 1. Let (T0 , t), (T0 , u) ∈ D± , and t < u. Then δ+ (s, t) = st < su = δ+ (s, u), t δ− (s, t) = s u < s = δ− (s, u). Therefore (P 1) holds.

13.1 Definition for Shift Operators. Properties

751

2. Let (T1 , t), (T2 , t) ∈ D± , and T1 < T2 . Then δ+ (T1 , t) = T1 t < T2 t = δ+ (T2 , t), t δ− (T1 , t) = T1 t > T2 = δ− (T2 , t). Therefore (P 2) holds. 3. Let t ∈ [1, ∞). Then (t, 1) ∈ D+ and δ+ (t, 1) = t. If t ∈ T∗ , then (1, t) ∈ D+ and δ+ (1, t) = t. Therefore (P 3) holds. 4. Let (s, t) ∈ D± . Then (s, t) ∈ [1, ∞) × T∗ and ts ∈ T∗ . Hence, δ+ (s, t) ∈ T∗ , δ+ (s, t) ts = s s = t ∈ T∗ . Therefore δ− (s, δ+ (s, t)) =

δ+ (s, t) ∈ T∗ s

and (s, δ+ (s, t)) ∈ D− . Next, t s = t ∈ T∗ .

sδ− (t, s) = s

752

13 Shift Operators

Consequently δ− (s, δ+ (s, t)) = sδ− (s, t) ∈ T∗ and (s, δ− (t, s)) ∈ D+ . Also, δ+ (s, t) s ts = s = t,

δ− (s, δ+ (s, t)) =

δ+ (s, δ− (s, t)) = sδ− (s, t) t =s s = t. Consequently (P 4) holds. 5. Let (s, t) ∈ D± and (u, δ± (s, t)) ∈ D∓ . Then s, u ∈ [1, ∞), t, δ+ (s, t) ∈ T∗ , δ+ (s, t) ∈ T∗ , u or st ∈ T∗ . u Therefore δ+ (s, δ− (u, t)) = sδ− (u, t) st ∈ T∗ = u and (s, δ− (u, t)) ∈ D+ .

13.1 Definition for Shift Operators. Properties

753

Next, δ+ (u, t) s ut ∈ T∗ = s

δ− (s, δ+ (u, t)) =

and (s, δ+ (u, t)) ∈ D− . Also, δ+ (s, t) u st = u = δ+ (s, δ− (u, t)),

δ− (u, δ+ (s, t)) =

δ+ (u, δ− (s, t)) = uδ− (s, t) ut = s = δ− (s, δ+ (u, t)). Consequently (P 5) holds. Exercise 13.1. Let T = T∗ = Z, t0 = 0. Prove that δ+ (s, t) = t + s,

δ− (s, t) = t − s

satisfy (P 1)–(P 5). Theorem 13.1. Let δ± be the shift operators associated with the initial point t0 . Then we have. (i) δ− (t, t) = t0 for all t ∈ [t0 , ∞), (ii) δ− (t0 , t) = t for all t ∈ T∗ , (iii) If (s, t) ∈ D+ , then δ+ (s, t) = u implies δ− (s, u) = t. Conversely, if (s, u) ∈ D− , then δ− (s, u) = t implies δ+ (s, t) = u, (iv) δ+ (t, δ− (s, t0 )) = δ− (s, t) for all (s, t) ∈ D+ with t ≥ t0 , (v) δ+ (u, t) = δ+ (t, u) for all (u, t) ∈ ([t0 , ∞) × [t0 , ∞))

,

(vi) δ+ (s, t) ∈ [t0 , ∞) for all (s, t) ∈ D+ with t ≥ t0 ,

D+ ,

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13 Shift Operators

(vii) δ− (s, t) ∈ [t0 , ∞) for all (s, t) ∈ ([t0 , ∞) × [s, ∞))

,

D− ,

(viii) If δ+ (s, ·) is delta differentiable with respect to its second argument, then Δt (s, ·) > 0, δ+ (ix) δ+ (δ− (u, s), δ− (s, v)) = δ+ (u, v) for all (s, v) ∈ ([t0 , ∞) × [s, ∞)) (u, s) ∈ ([t0 , ∞) × [u, ∞))

, ,

D− , D− .

(x) If (s, t) ∈ D− and δ− (s, t) = t0 , then s = t. Proof.

(i) By (P 3), we have δ+ (t, t0 ) = t. Hence from (P 4), we get δ− (t, t) = δ− (t, δ+ (t, t0 )) = t0 .

(ii) By (P 3), we have δ+ (t0 , t) = t. Hence from (P 4), we get δ− (t0 , t) = δ− (t0 , δ+ (t0 , t)) = t. (iii) Let (s, t) ∈ D+ and u = δ+ (s, t). Then, by (P 4), we have (s, u) ∈ D− . Hence from (P 4), δ− (s, u) = δ− (s, δ+ (s, t)) = t. Let (s, u) ∈ D− . Then (s, δ− (s, u)) ∈ D+

13.1 Definition for Shift Operators. Properties

755

and δ+ (s, t) = δ+ (s, δ− (s, u)) = u. (iv) Let (s, t) ∈ D+ . By (P 5), we have δ+ (t, δ− (s, t0 )) = δ− (s, δ+ (t, t0 )) . By (P 3), we have δ+ (t, t0 ) = t. Therefore δ+ (t, δ− (s, t0 )) = δ− (s, t). (v) Let (u, t) ∈ ([t0 , ∞) × [t0 , ∞))

'

D+ . By (P 3), we have

t = δ+ (t, t0 ). By (i), we have δ− (u, u) = t0 . Therefore t = δ+ (t, t0 ) = δ+ (t, δ− (u, u)) . By (P 5), we have δ+ (t, δ− (u, u)) = δ− (u, δ+ (t, u)) . Therefore t = δ− (u, δ+ (t, u)) . By (iii), we have δ+ (u, t) = δ+ (u, δ− (u, δ+ (t, u))) = δ+ (t, u).

756

13 Shift Operators

(vi) Let (s, t) ∈ D+ . Since s ≥ t0 , using (P 1), we have δ+ (s, t) ≥ δ+ (t0 , t) =t ≥ t0 . (vii) Let (s, t) ∈ ([t0 , ∞) × [s, ∞))

'

D− . Then, using (P 1) and (i), we get

δ− (s, t) ≥ δ− (s, s) = t0 . (viii) 1. Let μ(t) > 0. Then Δt δ+ (s, t) =

δ+ (s, σ (t)) − δ+ (s, t) μ(t)

> 0. 2. Let μ(t) = 0. Then, for any h > 0, we have δ+ (s, t + h) − δ+ (s, t) > 0. For h < 0, t + h ∈ T∗ , we get δ+ (s, t + h) − δ+ (s, t) < 0. Therefore Δt (s, t) > 0. δ+

(ix) By (P 5) and (v), we have δ+ (δ− (u, s), δ− (s, v)) = δ− (s, δ+ (δ− (u, s), v)) = δ− (s, δ+ (v, δ− (u, s))) = δ− (s, δ− (u, δ+ (v, s))) = δ− (s, δ− (u, δ+ (s, v))) = δ− (s, δ+ (s, δ− (u, v))) = δ− (u, v)

13.2 The Delay Function

757

for all (s, v) ∈ ([t0 , ∞) × [s, ∞)) (u, s) ∈ ([t0 , ∞) × [u, ∞))

, ,

D− , D− .

(x) Let (s, t) ∈ D− and δ− (s, t) = t0 . Then, by (P 4), we have t = δ+ (s, δ− (s, t)) = δ+ (s, t0 ) = s. This completes the proof.

13.2 The Delay Function Definition 13.2. Let δ± : [t0 , ∞) × T∗ → T∗ be shift operators associated with t0 . Suppose that h ∈ (t0 , ∞) is a constant so that (h, t) ∈ D± for all t ∈ [t0 , ∞), the function δ− (h, ·) is differentiable with an rd-continuous derivative and δ− (h, ·) maps [t0 , ∞) onto [δ− (h, t0 ), ∞). The function δ− (h, ·) is called the delay function generated by the shift operator δ− on the time scale T. Below we suppose that δ− (h, ·) is a delay function on the time scale T. By (P 2), we have δ− (h, t) < δ− (t0 , t) = t,

t ∈ [t0 , ∞).

Also, by (P 1), we have that the delay function δ− (h, ·) is strictly increasing and it is invertible. By (P 4) and (P 5), we obtain −1 (h, t) = δ+ (h, t), δ−

t ∈ [t0 , ∞).

Let T1 = [t0 , ∞),

T2 = δ− (h, T1 ).

We have T2 = [δ− (h, t0 ), ∞).

758

13 Shift Operators

Note that T1 and T2 are time scales. With σ1 and σ2 , Δ1 and Δ2 we denote their forward jump operators and delta differentiation operators, respectively. We have T1 ⊂ T2 ⊂ T, σ (t) = σ2 (t),

t ∈ T2 ,

σ (t) = σ1 (t) = σ2 (t),

t ∈ T1 .

Therefore ! ! σ1 = σ ! , T1

! ! σ2 = σ ! . T2

Theorem 13.2. The delay function δ− (h, ·) preserves the structure of the points in T1 , i.e., σ1 (tˆ) = tˆ implies

σ2 (δ− (h, tˆ)) = δ− (h, tˆ),

σ1 (tˆ) > tˆ implies

σ2 (δ− (h, tˆ)) > δ− (h, tˆ).

Proof. Since σ1 (t) ≥ t,

t ∈ T1 ,

we have δ− (h, σ1 (t)) ≥ δ− (h, t),

t ∈ T1 .

Because σ2 (δ− (h, t)) ≥ δ− (h, t), we get δ− (h, σ1 (t)) ≥ σ2 (δ− (h, t)),

t ∈ T1 .

If σ1 (tˆ) = tˆ, then δ− (h, tˆ) = δ− (h, σ1 (tˆ)) ≥ σ2 (δ− (h, tˆ)). Hence from (13.1), we get σ2 (δ− (h, tˆ)) = δ− (h, tˆ).

(13.1)

13.2 The Delay Function

759

Let σ1 (tˆ) > tˆ. Then (tˆ, σ1 (tˆ))T1 = (tˆ, σ1 (tˆ))T = ∅ and δ− (h, σ1 (tˆ)) > δ− (h, tˆ). Suppose that δ− (h, tˆ) is right-dense. Then σ2 (δ− (h, tˆ)) = δ− (h, tˆ). Hence, (δ− (h, tˆ), δ− (h, σ1 (tˆ)))T2 = ∅. Let s ∈ (δ− (h, tˆ), δ− (h, σ1 (tˆ)))T2 . Since δ− (h, ·) is strictly increasing and invertible, there is a t ∈ (tˆ, σ1 (tˆ)) such that δ− (h, t) = s, which is a contradiction. Therefore δ− (h, tˆ) is right-scattered and σ2 (δ− (h, tˆ)) > δ− (h, tˆ). This completes the proof. Theorem 13.3. We have δ− (h, σ1 (t)) = σ2 (δ− (h, t)),

t ∈ T1 .

δ− (h, σ (t)) = σ (δ− (h, t)),

t ∈ T1 .

Thus

Proof. Let t ∈ T1 be arbitrarily chosen. 1. Let σ1 (t) = t. By Theorem 13.2, we get σ2 (δ− (h, t)) = δ− (h, t).

(13.2)

760

13 Shift Operators

2. Let σ1 (t) > t. Applying Theorem 13.2, we obtain σ2 (δ− (h, t)) > δ− (h, t). On the other hand, δ− (h, σ1 (t)) > δ− (h, t). Therefore σ2 (δ− (h, t)) = δ− (h, σ1 (t)). This completes the proof. Remark 13.1. By (13.2), we have δ− (h, σ (s)) = σ (δ− (h, s)),

s ∈ [t0 , ∞).

Let s = δ1 (h, t),

t ∈ [δ− (h, t0 ), ∞).

Then, for t ∈ [δ− (h, t0 ), ∞), we have δ− (h, σ (δ+ (h, t))) = σ (δ− (h, δ+ (h, t))) = σ (t). Since, for t ∈ [δ− (h, t0 ), ∞), we have δ− (h, δ+ (h, σ (t))) = σ (t), we get δ− (h, σ (δ+ (h, t))) = δ− (h, δ+ (h, σ (t))) . Therefore σ (δ+ (h, t)) = δ+ (h, σ (t)) ,

t ∈ [δ− (h, t0 ), ∞).

Theorem 13.4. Let g is an rd-continuous function on T. Then t δ− (h,t) s

t

g(u)ΔuΔs =

t δ− (h,t)

(σ (s) − δ− (h, t)) g(s)Δs,

t ∈ T.

13.3 Stability Analysis

761

Proof. Let s, t ∈ T. Define f1 (s) = s − δ− (h, t), t

f2 (s) =

g(u)Δu. s

Then f1Δ (s) = 1, f2Δ (s) = −g(s). Hence, using the formula for integration by parts, we obtain t

t

t

g(u)ΔuΔs =

δ− (h,t) s

δ− (h,t)

f1Δ (s)f2 (s)Δs

!s=t ! = f1 (s)f2 (s)!

s=δ− (h,t)

t

+

δ− (h,t)

=

t δ− (h,t)

f1 (σ (s))f2Δ (s)Δs

(σ (s) − δ− (h, t)) g(s)Δs.

This completes the proof.

13.3 Stability Analysis Let T be a time scale having a delay function δ− (h, t), where h ≥ t0 and t0 ∈ T is nonnegative and fixed. In this section we consider the equation Δ (h, t), x Δ (t) = a(t)x(t) + b(t)x (δ− (h, t)) δ−

(13.3)

t ∈ [t0 , ∞), where ! Δ ! (G1) !δ− (h, t)! ≤ M, t ∈ [t0 , ∞), M is some positive constant. (G2) a, b ∈ Crd (T). Let ψ : [δ− (h, t0 ), t0 ] → R be rd-continuous and x(t) = x(t, t0 , ψ) be the solution of the equation (13.3) on [t0 , ∞) with x(t) = ψ(t),

t ∈ [δ− (h, t0 ), t0 ].

762

13 Shift Operators

Let Q(t) = a(t) + b (δ+ (h, t)) ,

t ∈ [t0 , ∞).

Then

Δt

t

Q(t)x(t) −

δ− (h,t)

= (a(t) + (δ+ (h, t))) x(t)

b (δ+ (h, s)) x(s)Δs

−b (δ+ (h, t)) x(t) Δ +b (δ+ (h, δ− (h, t))) x (δ− (h, t)) δ− (h, t) Δ (h, t), = a(t)x(t) + b(t)x (δ− (h, t)) δ−

t ∈ [t0 , ∞).

Therefore the equation (13.3) takes the following form

Δt

t

x (t) = Q(t)x(t) − Δ

δ− (h,t)

b (δ+ (h, s)) x(s)Δs

,

(13.4)

t ∈ [t0 , ∞). Theorem 13.5. Suppose (G1) and (G2). Let t

A(t) = x(t) +

δ− (h,t)

b (δ+ (h, s)) x(s)Δs, t ∈ [t0 , ∞).

β(t) = t − δ− (h, t), Let also, there exists a λ > 0 such that −

Δ (h, t) λδ− ≤ Q(t) β(t) + λ(β(t) + μ(t))

≤ −λ(β(t) + μ(t)) (b (δ+ (h, t)))2 − μ(t) (Q(t))2 , t ∈ [t0 , ∞). If V (t) = (A(t))2 + λ

t δ− (h,t) s

t

(b (δ+ (h, u)))2 (x(u))2 Δu,

then, along the solutions of the equation (13.3), we have V Δ (t) ≤ Q(t)V (t),

t ∈ [t0 , ∞).

13.3 Stability Analysis

763

Proof. Note that AΔ (t) = x Δ (t) + b(δ+ (h, t))x(t) Δ −b(δ+ (h, δ− (h, t)))x(δ− (h, t))δ− (h, t) Δ (h, t) = x Δ (t) + b(δ+ (h, t))x(t) − b(t)x(δ− (h, t))δ− Δ (h, t) = a(t)x(t) + b(t)x(δ− (h, t))δ− Δ (h, t) +b(δ+ (h, t))x(t) − b(t)x(δ− (h, t))δ−

= (a(t) + b(δ+ (h, t)))x(t) = Q(t)x(t),

t ∈ [t0 , ∞).

Then V Δ (t) = (A(t) + A(σ (t))) AΔ (t) σ (t)

+λ t

(b (δ+ (h, u)))2 (x(u))2 Δu

Δ (h, t) −λδ−

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λ (t − δ− (h, t)) (b (δ+ (h, t)))2 (x(t))2   = 2A(t) + μ(t)AΔ (t) AΔ (t) σ (t)

+λ t

(b (δ+ (h, u)))2 (x(u))2 Δu

Δ (h, t) −λδ−

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λ (t − δ− (h, t)) (b (δ+ (h, t)))2 (x(t))2 = (2A(t) + μ(t)Q(t)x(t)) Q(t)x(t) Δ (h, t) −λδ−

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λμ(t) (b (δ+ (h, t)))2 (x(t))2 +λβ(t) (b (δ+ (h, t)))2 (x(t))2 = (2A(t) + μ(t)Q(t)x(t)) Q(t)x(t) Δ (h, t) −λδ−

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λ(β(t) + μ(t)) (b (δ+ (h, t)))2 (x(t))2 ,

764

13 Shift Operators

t ∈ [t0 , ∞). Note that 2A(t)x(t) = (x(t))2 + (A(t))2

t 2 − b(δ+ (h, u))x(u)Δu ,

t ∈ [t0 , ∞).

δ− (h,t)

Let

R(t) = −Q(t)

t δ− (h,t)

σ (t)

Δ −λδ− (h, t)

−λQ(t)

2 b (δ+ (h, u)) x(u)Δu

δ− (h,t)

t

t

δ− (h,t) s

(b (δ+ (h, u)))2 (x(u))2 Δu (b (δ+ (h, u)))2 (x(u))2 ΔuΔs,

t ∈ [t0 , ∞). Then V Δ (t) = 2A(t)Q(t)x(t) + μ(t) (Q(t)x(t))2 Δ −λδ− (h, t)

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λ(β(t) + μ(t)) (b (δ+ (h, t)))2 (x(t))2 = Q(t)(x(t))2 + Q(t)(A(t))2

t 2 − b (δ+ (h, u)) x(u)Δu Q(t) δ− (h,t)

+μ(t)(Q(t)x(t))2 Δ −λδ− (h, t)

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

+λ(β(t) + μ(t)) (b (δ+ (h, t)))2 (x(t))2  = (x(t))2 λ(β(t) + μ(t)) (b (δ+ (h, t)))2 + Q(t) + μ(t) (Q(t))2 +Q(t)V (t) −λQ(t)

−

t

t

δ− (h,t) s t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs

2 b (δ+ (h, u)) x(u)Δu Q(t)

13.3 Stability Analysis

765

Δ −λδ− (h, t)

σ (t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

 = (x(t))2 λ(β(t) + μ(t)) (b (δ+ (h, t)))2 + Q(t) + μ(t) (Q(t))2 +Q(t)V (t) + R(t) ≤ Q(t)V (t) + R(t),

t ∈ [t0 , ∞),

i.e., V Δ (t) ≤ Q(t)V (t) + R(t),

t ∈ [t0 , ∞).

(13.5)

Note that σ (t) δ− (h,t)

t

(b (δ+ (h, u)))2 (x(u))2 Δu ≥

δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [t0 , ∞), and

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu ≤ β(t)

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [t0 , ∞). On the other hand, t

t

δ− (h,t) s t

=

δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu (σ (u) − δ− (h, t)) (b (δ+ (h, u)))2 (x(u))2 Δu t

≤ (μ(t) + β(t))

δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [t0 , ∞). Hence, R(t) ≤ −Q(t)β(t)

t δ− (h,t)

Δ (h, t) −λδ−

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu (b (δ+ (h, u)))2 (x(u))2 Δu

−λQ(t)(β(t) + μ(t))

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

  Δ = − Q(t)β(t) + λδ− (h, t) + λQ(t)(β(t) + μ(t))

766

13 Shift Operators t

×

δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

t ∈ [t0 , ∞).

≤ 0,

Hence from (13.5), we obtain V Δ (t) ≤ Q(t)V (t),

t ∈ [t0 , ∞).

This completes the proof. Theorem 13.6. Suppose (G1) and (G2). Let 1 + μa > 0 on T, 1 + μQ = 0 on T and all hypotheses of Theorem 13.5. If there exists an α ∈ [t0 , h) such that t ∈ [t0 , ∞),

(a, t) ∈ D± , and δ− (h, t) ≤

δ− (α, t) + δ− (h, δ− (α, t)) , 2

t ∈ [α, ∞),

then any solution x of the equation (13.3) satisfies the inequalities %1 |x(t)| ≤

&

2 1−

1

1 ξ(t)

V (t0 ) e 2

# δ− (α,t) t0

Q(s)Δs

,

t ∈ [α, ∞),

and |x(t)| ≤ ψe

#t t0

a(s)Δs

1+M

t t0

! # !  ! − s a(u)Δu ! b(s) ! e t0 ! Δs , ! 1 + μ(s)a(s) !

t ∈ [t0 , α), where ξ(t) = 1 +

λΛ(t) , β(t)

Λ(t) = δ− (h, t) − δ− (h, δ− (α, t)) ,

t ∈ [t0 , ∞),

ψ = sup {|ψ(t)| : t ∈ [δ− (h, t0 ), t0 )} . Proof. Since t0 < α < h and δ− (α, t) < t,

t ∈ [α, ∞),

13.3 Stability Analysis

767

we get δ− (h, t) < δ− (α, t), δ− (h, t) > δ− (h, δ− (α, t)) ,

t ∈ [α, ∞).

Then 0 < Λ(t) = δ− (h, t) − δ− (h, δ− (α, t)) δ− (α, t) + δ− (h, δ− (α, t)) − δ− (h, δ− (α, t)) 2 δ− (α, t) − δ− (h, δ− (α, t)) = 2 δ− (α, t) − δ− (h, t) , t ∈ [α, ∞). < 2 ≤

Also, V (t) = (A(t))2 +λ

t

t

δ− (h,t) s

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs

= (A(t))2 +λ

t

(σ (u) − δ− (h, t)) (b (δ+ (h, u)))2 (x(u))2 Δu

δ− (h,t)

≥ (A(t))2 +λ (σ (δ− (α, t)) − δ− (h, t))

t δ− (α,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

≥ (A(t))2 +λ (δ− (α, t) − δ− (h, t))

t δ− (α,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

≥ (A(t))2 +λΛ(t)

t δ− (α,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

768

13 Shift Operators

t ∈ [α, ∞). Next, V (δ− (α, t)) = (A (δ− (α, t)))2 δ− (α,t)

+λ

δ− (α,t)

δ− (h,δ− (α,t)) s

≥λ

δ− (α,t)

δ− (α,t)

δ− (h,δ− (α,t)) s

=λ

δ− (α,t) δ− (h,δ− (α,t))

≥λ

δ− (α,t) δ− (h,t)

≥λ

δ− (α,t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs

(σ (u) − δ− (h, δ− (α, t))) (b (δ+ (h, u)))2 (x(u))2 Δu

(σ (u) − δ− (h, δ− (α, t))) (b (δ+ (h, u)))2 (x(u))2 Δu (u − δ− (h, δ− (α, t))) (b (δ+ (h, u)))2 (x(u))2 Δu

≥ λ (δ− (h, t) − δ− (h, δ− (α, t))) = λΛ(t)

δ− (α,t) δ− (h,t)

δ− (α,t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [α, ∞). Therefore V (t) + V (δ− (α, t)) ≥ (A(t))2 +λΛ(t)

t δ− (α,t)

+λΛ(t)

(b (δ+ (h, u)))2 (x(u))2 Δu

δ− (α,t) δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

= (A(t))2 +λΛ(t)

= x(t) +

t δ− (h,t) t δ− (h,t)

+λΛ(t)

t δ− (h,t)

≥ (x(t))2 + 2x(t)

(b (δ+ (h, u)))2 (x(u))2 Δu 2 b (δ+ (h, u)) x(u)Δu (b (δ+ (h, u)))2 (x(u))2 Δu t δ− (h,t)

b (δ+ (h, u)) x(u)Δu

13.3 Stability Analysis

769

t

+

δ− (h,t)

Λ(t) +λ β(t)

2 b (δ+ (h, u)) x(u)Δu t δ− (h,t)

t

= (x(t))2 + 2x(t)

+ξ(t)

2 b (δ+ (h, u)) x(u)Δu

δ− (h,t)

t δ− (h,t)

b (δ+ (h, u)) x(u)Δu

2 b (δ+ (h, u)) x(u)Δu



1 (x(t))2 = 1− ξ(t) x(t)  +2 √ ξ(t) ξ(t) (x(t))2 ξ(t)

+ξ(t)

t δ− (h,t)

b (δ+ (h, u)) x(u)Δu

+

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu



1 (x(t))2 = 1− ξ(t)

2 t  1 + √ b (δ+ (h, u)) x(u)Δu x(t) + ξ(t) ξ(t) δ− (h,t) 

1 (x(t))2 , t ∈ [α, ∞). ≥ 1− ξ(t) Consequently 

1−

1 ξ(t)



(x(t))2 ≤ V (t) + V (δ− (α, t)) ≤ 2V (δ− (α, t)) ,

t ∈ [α, ∞).

In the last inequality we have used that V Δ (t) ≤ 0,

t ∈ [t0 , ∞).

Also, eQ (σ (t), t0 )V Δ (t) ≤ eQ (σ (t), t0 )Q(t)V (t),

t ∈ [t0 , ∞),

(13.6)

770

13 Shift Operators

or   Δ V (t) − Q(t)V (t) eQ (σ (t), t0 ) ≤ 0,

t ∈ [t0 , ∞).

Therefore t

0≥

t0 t

=

t0

 Δ  V (s) − Q(s)V (s) eQ (σ (s), t0 )Δs  Δ V (s)eQ (s, t0 ) Δs

= V (t)eQ (t, t0 ) − V (t0 ),

t ∈ [t0 , ∞).

Thus t ∈ [t0 , ∞).

V (t) ≤ V (t0 )eQ (t, t0 ), Hence from (13.6), we get

 1 1− (x(t))2 ≤ 2V (t0 )eQ (δ− (α, t), t0 ), ξ(t)

t ∈ [α, ∞),

or (x(t))2 ≤

2 1−

1 ξ(t)

t ∈ [α, ∞).

V (t0 )eQ (δ− (α, t), t0 ),

Therefore 1 |x(t)| ≤

2 1−

1 ξ(t)

1

V (t0 )e 2

# δ− (α,t) t0

Q(s)Δs

,

t ∈ [α, ∞). Now we multiply both sides of the equation (13.3) by ea (σ (t), t0 ) and we get x Δ (t)ea (σ (t), t0 ) − a(t)x(t)ea (σ (t), t0 ) Δ = b(t)x (δ− (h, t)) δ− (h, t)ea (σ (t), t0 ),

t ∈ [t0 , ∞),

or Δ (h, t)ea (σ (t), t0 ), (x(t)ea (t, t0 ))Δ = b(t)x (δ− (h, t)) δ−

13.3 Stability Analysis

771

t ∈ [t0 , ∞). Hence, t

x(t)ea (t, t0 ) = x(t0 ) +

t0 t

= x(t0 ) +

t0

Δ b(u)x (δ− (h, u)) δ− (h, u)ea (t0 , σ (u))Δu

b(u) Δ x (δ− (h, u)) δ− (h, u)ea (t0 , u)Δu, 1 + μ(u)a(u)

t ∈ [t0 , ∞), whereupon x(t) = x(t0 )ea (t, t0 ) t

+

b(u) Δ x (δ− (h, u)) δ− (h, u)ea (t0 , u)ea (t, t0 )Δu 1 + μ(u)a(u)

t0

= x(t0 )ea (t, t0 ) t

+

b(u) Δ x (δ− (h, u)) δ− (h, u)ea (t, u)Δu 1 + μ(u)a(u)

t0

= ψ(t0 )ea (t, t0 ) t

+

b(u) Δ x (δ− (h, u)) δ− (h, u)ea (t, u)Δu, 1 + μ(u)a(u)

t0

t ∈ [t0 , α). Hence,

#t a(u)Δu |x(t)| ≤ ψ e t0 t

+

t0

! ! ! ! ! Δ ! b(u) ! ! ! ! ! 1 + μ(u)a(u) ! |ψ (δ− (h, u))| δ− (h, u) ea (t, u)Δu

#t a(u)Δu ≤ ψ e t0 +M

t t0

! # !  ! t a(s)Δs ! b(u) !e u ! Δu , ! 1 + μ(u)a(u) !

t ∈ [t0 , α). In the last inequality we have used that δ− (h, t) < δ− (α, t) < δ− (α, α) = t0 , This completes the proof.

t ∈ [t0 , α).

772

13 Shift Operators

Theorem 13.7. If (t0 , h) = ∅, then the time scale T is isolated. Moreover, σ (t) = δ+ (h, t),

t ∈ [δ− (h, t0 ), ∞),

or equivalently σ (δ− (h, t)) = t,

t ∈ [t0 , ∞).

Proof. Define 0 (h, t0 ) = t0 , δ+

 k−1 k δ+ (h, t0 ) = δ+ h, δ+ (h, t0 ) ,

k ∈ N.

Since δ+ (h, t) is surjective and strictly increasing, we have    k−1 k−2 k−1 k δ+ (h, t0 ), δ+ (h, t0 ) = δ− h, δ+ (h, t0 ), δ+ (h, t0 ) , k ∈ N, k ≥ 2. Because (t0 , h) = ∅, we conclude that 

k−1 k δ+ (h, t0 ), δ+ (h, t0 ) = ∅,

k ∈ N.

 k−1 k σ δ+ (h, t0 ) = δ+ (h, t0 ),

k ∈ N,

Therefore (13.7)

and [t0 , ∞) =

∞ * 

k−1 k δ+ (h, t0 ), δ+ (h, t0 ) .

k=1

Consequently, for any t ∈ [t0 , ∞) there is a k0 ∈ N so that * k0 −1 k0 (h, t0 ), δ+ (h, t0 ) . t ∈ δ+ Hence from (13.7), it follows that for any t ∈ [t0 , ∞) there is a k0 ∈ N such that k0 −1 t = δ+ (h, t0 ).

Then  k0 −1 (h, t0 ) σ (t) = σ δ+ k0 = δ+ (h, t0 )

13.3 Stability Analysis

773

 k0 −1 = δ+ h, δ+ (h, t0 ) = δ+ (h, t),

t ∈ [δ− (h, t0 ), ∞) .

Note that σ (δ− (h, t)) = δ− (h, σ (t)) = δ− (h, δ+ (h, t)) = t,

t ∈ [t0 , ∞).

This completes the proof. Theorem 13.8. Suppose (G1), (G2), and the hypothesis of Theorem 13.5. If (t0 , h) = ∅, then any solution x of the equation (13.3) satisfies the exponential inequality 1  # 1 t 1 Q(s)Δs 1+ V (t0 )e 2 t0 |x(t)| ≤ , λ

t ∈ [t0 , ∞).

Proof. Define t

H (t) =

t

δ− (h,t) s

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs,

t ∈ [t0 , ∞). Then H (t) =

t δ− (h,t)

(σ (u) − δ− (h, t)) (b (δ+ (h, u)))2 (x(u))2 Δu

≥ (σ (δ− (h, t)) − δ− (h, t))

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

t 2 1 b (δ+ (h, u)) x(u)Δu ≥ (t − δ− (h, t)) β(t) δ− (h,t)

t 2 = b (δ+ (h, u)) x(u)Δu , δ− (h,t)

t ∈ [t0 , ∞). Hence, V (t) = (A(t))2 +λ

t δ− (h,t) s

t

(b (δ+ (h, u)))2 (x(u))2 ΔuΔs

774

13 Shift Operators

= (A(t))2 + λH (t) 2

t b (δ+ (h, u)) x(u)Δu ≥ x(t) + δ− (h,t)

t

+λ

δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

= (x(t))2 + 2x(t)

t δ− (h,t)

t

+

δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

t

+λ

b (δ+ (h, u)) x(u)Δu

δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

2 

x(t) 1 (x(t))2 + √ 1+λ 1+λ

t  x(t) √ +2 √ 1+λ b (δ+ (h, u)) x(u)Δu 1+λ δ− (h,t) 2

t b (δ+ (h, u)) x(u)Δu +(1 + λ)

= 1−

δ− (h,t)

=

≥

λ (x(t))2 1+λ

√ x(t) + √ + 1+λ 1+λ λ (x(t))2 , 1+λ

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

t ∈ [t0 , ∞).

From here, λ+1 V (t) λ 

1 V (t0 )eQ (t, t0 ) ≤ 1+ λ 

#t 1 Q(u)Δu V (t0 )e t0 , ≤ 1+ λ

(x(t))2 ≤

t ∈ [t0 , ∞).

13.4 A Criterion for Instability

775

Therefore 1

 # 1 t 1 Q(u)Δu , 1+ V (t0 )e 2 t0 λ

|x(t)| ≤

t ∈ [t0 , ∞).

This completes the proof.

13.4 A Criterion for Instability Theorem 13.9. Suppose (G1), (G2), and there exists a positive constant D such that β(t) < D ≤

Q(t) (b (δ+ (h, t)))2

,

t ∈ [t0 , ∞).

If V (t) = (A(t))2 − D

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [t0 , ∞),

then along the solutions of the equation (13.3), we have V Δ (t) ≥ Q(t)V (t),

t ∈ [t0 , ∞).

Proof. Since

2A(t)x(t) = (x(t))2 + (A(t))2 −

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu ,

t ∈ [t0 , ∞), and

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu ≤ β(t)

t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu,

t ∈ [t0 , ∞), we have V Δ (t) = (A(t) + A(σ (t))) AΔ (t) −D (b (δ+ (h, t)))2 (x(t))2 Δ +D (b (δ+ (h, δ− (h, t))))2 (x(δ− (h, t)))2 δ− (h, t)

776

13 Shift Operators

  = 2A(t) + μ(t)AΔ (t) AΔ (t) −D (b (δ+ (h, t)))2 (x(t))2 Δ +D(b(t))2 (x(δ− (h, t)))2 δ− (h, t)

≥ (2A(t) + μ(t)Q(t)x(t)) Q(t)x(t) −D (b (δ+ (h, t)))2 (x(t))2 ≥ 2A(t)Q(t)x(t) − D (b (δ+ (h, t)))2 (x(t))2 %

t

= Q(t) (x(t)) + (A(t)) − 2

2

δ− (h,t)

−D (b (δ+ (h, t)))2 (x(t))2

≥ Q(t) (x(t))2 + (A(t))2 − β(t)

2 & b (δ+ (h, u)) x(u)Δu

t δ− (h,t)

−D (b (δ+ (h, t)))2 (x(t))2

≥ Q(t) (x(t))2 + (A(t))2 − D

t

 (b (δ+ (h, u)))2 (x(u))2 Δu

 (b (δ+ (h, u))) (x(u)) Δu 2

δ− (h,t)

2

−D (b (δ+ (h, t))) (x(t))  = Q(t)V (t) + Q(t) − D (b (δ+ (h, t)))2 (x(t))2 2

≥ Q(t)V (t),

2

t ∈ [t0 , ∞).

This completes the proof. Theorem 13.10. Suppose that all hypotheses of Theorem 13.9 hold. Let β is bounded above by β0 with 0 < β0 < D. Then the zero solution of the equation (13.3) is unstable provided that t

lim

t→∞ t 0

(b (δ+ (h, u)))2 Δu = ∞.

Proof. By Theorem 13.9, we have V Δ (t) ≥ Q(t)V (t),

t ∈ [t0 , ∞).

Hence, V Δ (t)eQ (σ (t), t0 ) ≥ Q(t)V (t)eQ (σ (t), t0 ),

t ∈ [t0 , ∞),

whereupon Δ  V (t)eQ (t, t0 ) ≥ 0,

t ∈ [t0 , ∞).

13.4 A Criterion for Instability

777

Hence, V (t)eQ (t, t0 ) ≥ V (t0 ),

t ∈ [t0 , ∞).

Then t ∈ [t0 , ∞).

V (t) ≥ V (t0 )eQ (t, t0 ),

(13.8)

By the definitions of the functions V and A, we have V (t) = (x(t))2 + 2x(t)

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

t

+

δ− (h,t) t

−D

b (δ+ (h, u)) x(u)Δu

δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu.

(13.9)

Let C = D − β0 , K = |x(t)|, L=

t δ− (h,t)

|b (δ+ (h, u))| |x(u)|Δu,

t ∈ [t0 , ∞).

Using the inequality %

1 β0 K− C

C L β0

&2 ≥ 0,

t ∈ [t0 , ∞),

we get 2KL ≤

β0 2 C K + L2 . C β0

Thus, we get t

2|x(t)| δ− (h,t)

|b (δ+ (h, u))| |x(u)|Δu ≤

β0 (x(t))2 C

t 2 C + b (δ+ (h, u)) x(u)Δu , β0 δ− (h,t)

778

13 Shift Operators

t ∈ [t0 , ∞). Hence from (13.9), we find V (t) ≤ (x(t))2 +

C + β0

+

β0 (x(t))2 C

t δ− (h,t)

2 b (δ+ (h, u)) x(u)Δu

t δ− (h,t)

−D

t δ− (h,t)

= 1+

β0 C

(x(t))2

C + 1+ β0



t δ− (h,t)

(b (δ+ (h, u)))2 (x(u))2 Δu

(b (δ+ (h, u)))2 (x(u))2 Δu

D (x(t))2 C +

D β(t) β0

−D

t δ− (h,t)

t δ− (h,t)

≤

2 b (δ+ (h, u)) x(u)Δu

D (x(t))2 C

t 2 D + b (δ+ (h, u)) x(u)Δu β0 δ− (h,t) −D

≤

t δ− (h,t)

t δ− (h,t)

=

(b (δ+ (h, u)))2 (x(u))2 Δu



−D

2 b (δ+ (h, u)) x(u)Δu

D (x(t))2 , C

(b (δ+ (h, u)))2 (x(u))2 Δu

(b (δ+ (h, u)))2 (x(u))2 Δu t ∈ [t0 , ∞),

i.e., V (t) ≤

D (x(t))2 , C

t ∈ [t0 , ∞).

13.5 Advanced Practical Problems

779

Hence from (13.8), we get D (x(t))2 ≥ V (t0 )eQ (t, t0 ), C

t ∈ [t0 , ∞),

whereupon C V (t0 )eQ (t, t0 ) D

 t C ≥ V (t0 ) 1 + Q(s)Δs D t0

 t C ≥ V (t0 ) 1 + D (b (δ+ (h, u)))2 Δu , D t0

(x(t))2 ≥

t ∈ [t0 , ∞). Then 1 |x(t)| ≥

C V (t0 ) 1 + D D

t t0

 (b (δ+ (h, u)))2 Δu ,

t ∈ [t0 , ∞).

This completes the proof.

13.5 Advanced Practical Problems 1

Problem 13.1. Let T = T∗ = N 2 , t0 = 1. Prove that δ+ (s, t) =



t 2 + s2,

δ− (s, t) =



t 2 − s2

satisfy (P 1)–(P 5). Problem 13.2. Let T = Z. Investigate for stability and instability of the zero solution of the equation Δ x Δ (t) = tx(t) + t 2 x (δ− (1, t)) δ− (1, t),

t ∈ [0, ∞).

Problem 13.3. Let T = 2N0 . Investigate for stability and instability of the zero solution of the equation  Δ (4, t), x Δ (t) = t 3 + t 2 + t + 1 x(t) + (t − 2)x (δ− (4, t)) δ−

t ∈ [1, ∞).

780

13 Shift Operators

Problem 13.4. Let T = 3N0 . Investigate for stability and instability of the zero solution of the equation x Δ (t) =

t +1 Δ x(t) + tx (δ− (9, t)) δ− (9, t), t2 + 2

t ∈ [1, ∞).

1

Problem 13.5. Let T = N02 . Investigate for stability and instability of the zero solution of the equation x Δ (t) =

t4 + t2 + 1 Δ x(t) + tx (δ− (2, t)) δ− (2, t), t 10 + 1

t ∈ [1, ∞).

Chapter 14

Impulsive Functional Dynamic Equations

The material of this chapter are based on results of the papers and monographs [15, 16, 23, 32, 33, 37–43, 54, 70, 73, 81, 112, 114, 123–125, 130–132, 143, 156, 162, 163, 165, 171, 179, 181, 194, 195, 199, 202, 207, 210, 218]. Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

14.1 Existence of Solutions for First-Order Nonlinear Impulsive Dynamic Equations Suppose that 0, ti , T ∈ T, i ∈ {1, . . . , n}, 0 < ti < ti+1 , i ∈ {1, . . . , n − 1}, and ti is dense in T, i ∈ {1, . . . , n}. Consider the following nonlinear impulsive dynamic equation y Δ (t) = −a(t)y σ (t) + f (t, y(t)), t ∈ [0, T ], y(0) = 0, y(ti+ ) = y(ti− ) + I (ti , y(ti )), i ∈ {1, . . . , n},

(14.1)

where y(ti± ) = lim y(t), t→ti±

y(ti ) = y(ti− ), (H1) (H2)

i ∈ {1, . . . , n},

a ∈ R, f ∈ C (T × R),

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2_14

781

782

14 Impulsive Functional Dynamic Equations

(H3)

there exist g and h such that t

α = max

t∈[0,T ] 0 t

β = max

t∈[0,T ] 0

|ea (t, s)|g(s)Δs < ∞, |ea (t, s)|h(s)Δs < ∞,

|f (t, x) − f (t, y)| ≤ h(t)|x − y|, |f (t, y)| ≤ g(t) + h(t)|y|, (H4)

t ∈ T,

y ∈ R,

there exists a positive constant E such that |I (t, x) − I (t, y)| ≤ E|x − y|,

x, y ∈ R.

Define tn+1 = T , J0 = [0, t1 ], Jk = (tk , tk+1 ], k ∈ {1, . . . , n},

P C = y : [0, T ] → R, y ∈ C (Jk ), y(tk− )

∃y(tk± ), 

= y(tk ),

k ∈ {1, . . . , n} ,

 P C 1 = y : [0, T ] → R,

y ∈ C 1 (Jk ),

 k ∈ {1, . . . , n} .

The set P C is a Banach space endowed with the supremum norm u =

max uk ,

k∈{0,...,n}

where uk = sup |u(t)|,

k ∈ {0, . . . , n}.

t∈Jk

Theorem 14.1. The function y ∈ P C 1 is a solution of the equation (14.1) if and only if y ∈ P C satisfies the following integral equation. t

y(t) = 0

ea (t, s)f (s, y(s))Δs +

 {i:ti t1 , such that n−1

fΔ

n−1

(t) ≤ f Δ

n−1

(t2 ) < f Δ

(t1 ),

t ∈ L (∞),

t > t2 .

Hence, n−1

lim sup f Δ

(t) < 0

t→∞

and by Lemma E.1, we conclude that lim f (t) = −∞.

t→∞

This is a contradiction. Therefore n−1

fΔ

(t) > 0,

t ∈ L (∞).

Define

j M = m ∈ N0 : (−1)m+j f Δ (t) > 0, j ∈ (m, n − 1]

,

t ∈ L (∞), 

N0 ,

n+m

odd .

Then n − 1 ∈ M and there exists p = min M. If p ∈ {0, 1} we are done. Assume p−1 p−2 that p ≥ 2. Suppose that f Δ (t) < 0, t ∈ L (∞). Suppose that f Δ (t1 ) ≤ 0 for some t1 ∈ L (∞). Then there exists a t2 ∈ L (∞), t2 > t1 , such that p−2

fΔ

p−2

(t) ≤ f Δ

p−2

(t2 ) < f Δ

(t1 ),

t ∈ L (∞),

t > t2 .

E Knesser’s Theorem

869

Hence from Lemma E.1, we obtain that lim f (t) = −∞.

t→∞

p−2

This is a contradiction. Therefore f Δ j

(−1)p−2+j f Δ (t) > 0,

(t) > 0, t ∈ L (∞). Also, we have that

t ∈ L (∞),

j ∈ [p − 2, n − 1]

,

N0 ,

which is a contradiction by the definition of p. Therefore there exists a t3 ∈ L (∞) so that p−1

fΔ

(t3 ) ≥ 0.

p

Because f Δ (t) > 0, t ∈ L (∞), we conclude that p−1

fΔ

(t) > 0,

t ∈ L (∞),

t > t3 .

Hence, p−1

lim inf f Δ t→∞

(t) > 0

and by Lemma E.1, we obtain j

lim f Δ (t) = ∞,

0 ≤ j ≤ p − 2.

t→∞

n−1

2. Let s = 1. Assume that there exists a t4 ∈ L (∞) such that f Δ there exists a t5 ∈ [t4 , ∞), t5 > t4 , such that n−1

fΔ

n−1

(t) ≥ f Δ

n−1

(t5 ) > f Δ

(t4 ),

t ∈ L (∞),

Hence, n−1

lim inf f Δ t→∞

(t) > 0

and by Lemma E.1, we conclude that j

lim inf f Δ (t) = ∞, t→∞

So, p = n. This completes the proof.

j ∈ {1, . . . , n − 2}.

(t4 ) ≥ 0. Then

t > t5 .

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Index

A Abel’s Equation, 202 AC i , 801 Additive Inverse Under ⊕, 22 Asymptotic Stability, 222

E Eventually Negative Solution, 103 Eventually Positive Solution, 103 Exponential Stability of Linear Impulsive Dynamic Systems, 829

B Backward Jump Operator, 1 Backward Shift Operator, 750 Beverton-Holt Model, 92

F First Fundamental Solution, 376 Forward Jump Operator, 1 Forward Shift Operator, 750 Fréchet Derivative, 840 fσ, 4

C C 1 (U, Y ), 841 C 2 (U, Y ), 841 Cauchy Operator, 233 Chain Rule, 10, 12 Circle Dot Multiplication, 21 Circle Minus Substraction, 24 Circle Plus Addition, 21, 24 Complete Poset, 855 Cylindrical Transformation, 24

D Delay Function, 757 Delta Derivative, 5 Delta Differentiable Function, 5 Dense Point, 2 Derivative of the Inverse, 13 Cp1 (T), 15 1 (T), 15 Crd Crd (T), 15 Dual Version of Induction Principle, 16

G Gâteaux Derivative, 847 Generalized Exponential Function, 25 Generalized Food Limited Population Model, 97 Generalized Square, 22 Gilpin-Ayala single species growth model, 96 Global Asymptotical Stability, 198 Global Attractivity, 198 Graininess Function, 3, 850

H Harmonic Numbers, 3 Higher Order Delta Derivative, 9 Higher Order Hilger Derivative, 9 Higher Order Linear Impulsive Dynamic Equation, 808 Hilger Alternative Axis, 20 Hilger Complex Number, 19

© Springer Nature Switzerland AG 2019 S. G. Georgiev, Functional Dynamic Equations on Time Scales, https://doi.org/10.1007/978-3-030-15420-2

883

884 Hilger Derivative, 5 Hilger Differentiable Function, 5 Hilger Imaginary Circle, 20 Hilger Purely Imaginary Number, 20 Hilger Real Axis, 20 Hilger Real Part, 20 Homogeneous Linear Functional Dynamic Equation of nth Order, 141 Homogeneous Linear Impulsive Dynamic Systems, 819 Hyperbolic Functions, 28

I Improper Integral, 19 Impulsive Functional Dynamic Equation, 781 Impulsive Transition Matrix, 822 Induction Principle, 15 Isolated Point, 2

J Jump Operator, 850

K Kamenev-Type Oscillation Criteria, 738, 743 Knaster-Tarski Fixed Point Theorem, 855 Knesser’s Theorem, 868 Krasnosel’skii Fixed Point Theorem, 788

L Left-Dense Point, 2 Left-Scattered Point, 2 Leibniz Formula, 9 Leslie-Gower Competition Model, 94 Linear Functional Dynamic Equation of nth Order, 141 Linear Impulsive Dynamic System, 819 l ∞ (Rn ), 827 Liouville’s Formula, 67 Logistic Model, 84 Lower Solution of a First Order Impulsive Dynamic Equations, 796

M Matrix Exponential Function, 60 Measure Chain, 849

Index N Negative Definite Lyapunov Functional, 227 Nonhomogeneous Linear Impulsive Dynamic Systems, 827 Nonoscillatory Solution, 103

O Oscillatory Equation, 103 Oscillatory Solution, 103

P P, 34 P C, 782, 801 P C 1 , 782 Perron Property, 241 Poset, 855 Power Series, 34 Pre-Antiderivative, 17 Pre-Differentiable Function, 14 Predator Prey Model, 89 Putzer’s Algorithm, 79

R R, 24, 53 rd-Continuous Function, 15 rd-continuous Matrix, 53 Regressive Dynamic Equation, 38 Regressive Equation, 37 Regressive Function, 24 Regressive Group, 24 Regressive Matrix, 53 Regulated Function, 14 Remainder, 837 ρ, 1 Riccati Inequality, 387 Riccati Substitution, 370 Ricker Competition Model, 94 Ricker Model, 91 Right-Dense Point, 2 Right-Scattered Point, 2 R(T), 53 R(T, Rn×n ), 53 S Second Delta Derivative, 9 Second Fundamental Solution, 376 Second Hilger Derivative, 9 Second Order Impulsive Dynamic Equation, 800 Semigroup Property, 25

Index Shift Size, 750 σ, 1 Single species growth model, 95 Solution, 100, 161, 375 Solution of a First Order Impulsive Dynamic Equations, 796 Solution of a Homogeneous Linear Impulsive Dynamic System, 820 Solution of Linea Impulsive Dynamic Equation, 809 Solution of Second Order Impulsive Dynamic Equation, 801 Stable Solution, 792 T T, 1 Taylor’s Formula, 30, 31 Time Scale, 1

885 Totally Ordered Poset, 855 Trigonometric Functions, 29 Tumor Growth Model, 86, 87

U Uniform Exponential Stability, 233 Uniformly Asymptotically Stable Solution, 177 Uniformly Attractive Solution, 177 Uniform Stability, 198 Uniform Stable Solution, 162 Upper Solution of a First Order Impulsive Dynamic Equations, 796

Z Zero Solution, 161