Fuzzy Dynamic Equations, Dynamic Inclusions, and Optimal Control Problems on Time Scales 3030761312, 9783030761318

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Svetlin G. Georgiev

Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales

Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales

Svetlin G. Georgiev

Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales

Svetlin G. Georgiev Faculty of Mathematics & Info Sofia University St Kliment Ohridski Sofia, Bulgaria

ISBN 978-3-030-76131-8 ISBN 978-3-030-76132-5 (eBook) https://doi.org/10.1007/978-3-030-76132-5 Mathematics Subject Classification: 34A60, 34N05, 34A37, 34B37 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The theory of dynamic equations has many interesting applications in control theory, mathematical economics, mathematical biology, engineering, and technology. In some cases, there exists uncertainty, ambiguity, or vague factors in such problems, and fuzzy theory and interval analysis are powerful tools for modeling these equations on time scales. The aim of this book is to present a systematic account of recent developments; describe the current state of the useful theory; show the essential unity achieved in the theory fuzzy dynamic equations, dynamic inclusions, and optimal control problems on time scales; and initiate several new extensions to other types of fuzzy dynamic systems and dynamic inclusions. The book contains 16 chapters. Chapter 1 deals with fuzzy functions on time scales. We focus on a new class derivative and a new class integral of such fuzzy functions. The corresponding fundamental properties of the introduced derivative and integral are studied and discussed. In this chapter, we introduce the concept for shifts operators. They are defined shift almost periodic fuzzy functions and some of their properties are deduced. In Chap. 2, certain classes linear fuzzy dynamic equations are considered, and we give formulas for their solutions. They are proved some results for existence and uniqueness of the solutions of some classes nonlinear fuzzy dynamic equations. The continuous dependence of the solutions of first order fuzzy dynamic equations on the initial data is investigated. Some comparison results are provided. The Lyapunov functional is introduced, and criteria for stability and exponential stability of the trivial solution of some classes nonlinear fuzzy dynamic equations are proved by using it. In Chap. 3, some classes linear second order fuzzy dynamic equations are investigated and formulas for their solutions are deduced. They are formulateds and proved some existence and uniqueness results. The continuous dependence of the solutions of second order fuzzy dynamic equations on the initial data is investigated. Some classes of BVPs for second order fuzzy dynamic equations are investigated and formulas for their solutions are given. In Chap. 4, some basic results for periodic functions on the periodic time scales are formulated and proved. A phase space is built for fuzzy dynamic equations with infinite delays on the periodic time scales. By using the monotone methods, we systematically consider the existence of periodic v

vi

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solutions for the fuzzy dynamic equations with infinite delays on the periodic time scales, which generalize and incorporate, as special cases, some known results for fuzzy differential equations and fuzzy difference equations. In Chap. 5, linear first order impulsive fuzzy dynamic equations on time scales are considered, and we deduce formulas for their solutions. In the chapter, some classes nonlinear first order impulsive fuzzy dynamic equations are investigated, and existence and uniqueness of their solutions are proved. We introduce the conception for stable trivial solution of some classes nonlinear first order impulsive fuzzy dynamic equations and some stability criteria are deduced. In Chap. 6, the concept for Δmeasurable functions on time scales and Δ-Lebesgue integration on time scales is introduced. In the chapter, some of the properties of the Δ-measurable functions and the Δ-Lebesgue integral are deduced. The spaces L1 (T) and the Sobolev spaces are introduced. The concept for absolutely continuous functions on time scales is established and a characterization of such functions is provided. In this chapter, we introduce weak solutions and the Euler solutions for first order dynamic equations on time scales and an analogue of the Gronwall inequality is formulated and proved. In Chap. 7, some IVPs for first order dynamic inclusions are investigated. Some classes first order dynamic inclusions with nonlocal initial conditions are investigated for existence of solutions. In the chapter, BVPs for some classes first order dynamic inclusions with general boundary conditions and periodic boundary conditions are studied. Also, the dual time scales are introduced and a method for investigations for existence of solutions of first order dynamic inclusions via duality is proposed. In Chap. 8, a periodic boundary value problem for second order dynamic inclusions for existence of solutions is investigated. In the chapter, m-point boundary value problem for second order dynamic inclusions is considered and some existence results are deduced. In Chap. 9, some classes first order impulsive dynamic inclusions are investigated. They are introduced lower and upper solutions for first order impulsive dynamic inclusions. They are deducted some existence results for some classes boundary value problems for different classes first order impulsive dynamic inclusions. In Chap. 10, the controllability and observability of linear equations and the bang-bang principle are investigated. Chapter 11 is devoted on linear time optimal control. The existence of time-optimal controls and the maximum principle for linear time-optimal control are proven. Chapter 12 deals with the Pontryagin maximum principle. Adjoint linear dynamics, variations of the control, simple control variations, and the free endpoint problem are investigated. Dynamic programming is introduced in Chap. 13. The Hamilton-Jacobi-Bellman partial dynamic equations are deduced and the dynamic programming method is deduced. The dynamic games are introduced and the Isaacs equations are deduced. Chapter 14 is devoted on weak solutions and optimal control problems for some classes linear first order dynamic systems. Nonlinear dynamic equations and optimal control problems are investigated in Chap. 15 and some necessary conditions for optimality are provided. In Chap. 16, nonlinear integro-dynamic equations are investigated and some necessary conditions for optimality are deduced.

Preface

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The text material of this book is presented in highly readable, mathematically solid format. Many practical problems are illustrated, displaying a wide variety of solution techniques. The author welcomes suggestions for the improvement of the text. Paris, France March 2021

Svetlin G. Georgiev

Contents

1

2

3

Calculus of Fuzzy Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 First Type Fuzzy Delta Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Second Type Fuzzy Delta Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 First Type Fuzzy Delta Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Second Type Fuzzy Delta Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Shift Operators—Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Complete-Closed Time Scales under Non-translational Shifts. . . . . 1.8 Shift Almost Periodic Fuzzy Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 63 85 93 108 112 120 124 131 135

First Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Linear First Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . 2.2 Existence and Uniqueness of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Continuous Dependence of the Solutions of First Order Fuzzy Dynamic Equations on the Initial Data . . . . . . . . . . . . . . . . . . . . . . 2.4 Comparison Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Exponential Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 137 167

Second Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Linear Second Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . 3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Continuous Dependence of the Solutions of Second Order Fuzzy Dynamic Equations on the Initial Data . . . . . . . . . . . . . . .

205 205

175 176 184 192 201 203

226 239 248 ix

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3.5 3.6

Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

4

Functional Fuzzy Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Periodic Properties of Time Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Phase Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Periodic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

257 257 274 290 306 308

5

Impulsive Fuzzy Dynamic Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Linear First Order Impulsive Fuzzy Dynamic Equations . . . . . . . . . . 5.2 Existence of Solutions for First Order Nonlinear Impulsive Fuzzy Dynamic Equations-I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Existence of Solutions for First Order Nonlinear Impulsive Fuzzy Dynamic Equations-II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Stability of the Solutions of First Order Impulsive Fuzzy Dynamic Equations-I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Stability of the Solutions of First Order Impulsive Fuzzy Dynamic Equations-II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

309 310

The Lebesgue Integration. Lp -Spaces. Sobolev Spaces . . . . . . . . . . . . . . . . 6.1 The Lebesgue Delta-Integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Absolutely Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Alternative Way for Defining of Lebesgue Type Measure and Integration over T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Spaces Lp (T) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Sobolev Type Spaces and Generalized Derivatives . . . . . . . . . . . . . . . . . 6.7 Weak Solutions of Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Euler Solutions for Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 The Gronwall Type Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Δ × B-Measurable Set-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

361 361 377

6

7

First Order Dynamic Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Existence and Approximations of Solutions of First Order Dynamic Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Existence of Solutions of First Order Dynamic Inclusions with General Boundary Conditions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

324 337 350 353 357 359

383 386 390 393 408 419 423 427 433 434 435 435 455 471 489

Contents

7.5 7.6 7.7 7.8 8

9

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The Dual Time Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Existence of Solutions of First Order Dynamic Inclusions via Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Second Order Dynamic Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Fixed Point Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Existence Results for Second Order Dynamic Inclusions . . . . . . . . . . 8.3 Existence Results for Second Order Dynamic Inclusions with m-Point Boundary Value Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Boundary Value Problems for First Order Impulsive Dynamic Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Periodic Boundary Value Problems for First Order Linear Dynamic Inclusions with Impulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Periodic Boundary Value Problems for First Order Nonlinear Dynamic Inclusions with Impulses-I . . . . . . . . . . . . . . . . . . . . 9.4 Periodic Boundary Value Problems for First Order Nonlinear Dynamic Inclusions with Impulses-II . . . . . . . . . . . . . . . . . . . 9.5 Extremal Solutions of Periodic Boundary Value Problems for First Order Impulsive Integro-Dynamic Inclusions of Mixed Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Multiple Positive Solutions for First Order Impulsive Integral Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

500 510 511 514 515 515 522 537 548 550 551 551 563 574 601

609 631 658 660

10

Controllability, Bang–Bang Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Controllability of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Observability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Bang–Bang Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

661 661 668 684 695 698 699

11

Linear Time-Optimal Control. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Existence of Time-Optimal Controls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Maximum Principle for Linear Time-Optimal Control . . . . . . . . 11.3 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

701 701 702 714 715

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12

The Pontryagin Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Adjoint Linear Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Variations of the Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Simple Control Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 The Free Endpoint Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 The Pontryagin Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

717 717 718 720 722 723 726 728

13

Dynamic Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 The Hamilton–Jacobi–Bellman Partial Dynamic Equations . . . . . . . 13.2 The Dynamic Programming Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 The Linear Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Dynamic Games: Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 The Isaacs Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

729 729 735 746 754 756 761 763

14

Weak Solutions and Optimal Control Problems for Some Classes of Linear First Order Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Optimal Control Governed by a Class of First Order Linear Dynamic Systems-I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Optimal Control Governed by a Class of First Order Linear Dynamic Systems-II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

765 765 773 780 781

15

Nonlinear Dynamic Equations and Optimal Control Problems . . . . . . . 15.1 Existence of Optimal Controls. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Necessary Conditions of Optimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

783 783 792 801 803

16

Nonlinear Integro-Dynamic Equations and Optimal Control Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Existence of Optimal Controls. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Necessary Conditions of Optimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

805 806 810 821 823

Fuzzy Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Basic Operations with Fuzzy Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3 Fuzzy Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.4 Multiplication of Compact Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.5 Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

825 825 828 831 841 856

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xiii

B

Set-Valued Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.1 Limits of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Definition of Set-Valued Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.3 Continuity Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

857 857 861 863

C

Alaoglu’s Theorem. Krein–Milman Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 871

D

Mazur’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879

Chapter 1

Calculus of Fuzzy Functions

In this chapter are introduced new classes of delta derivatives called first type fuzzy delta derivative and second type fuzzy delta derivative. Their existence and uniqueness are proved. Some of their basic properties such as fuzzy delta differentiation of a sum of two fuzzy functions and fuzzy delta differentiation of a multiplication of a fuzzy function with a constant are deducted. α-levels of fuzzy functions are defined, and formulae for their fuzzy delta derivatives are given. In the chapter are defined six types of multiplication of fuzzy functions, and for each type a formula for its fuzzy delta derivative is deducted. First type fuzzy delta integration and second type fuzzy delta integration of fuzzy functions are defined, and some of their basic properties are deducted. Shift operators are investigated, and complete-closed time scales under non-translational shifts are introduced. Shift almost periodic fuzzy functions are defined, and some of their properties are given. Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. With F (R) we will denote the space of the real fuzzy numbers and with D(·, ·) we will denote the Hausdorff distance between the real fuzzy numbers. For more details about fuzzy numbers and Hausdorff distance between the real fuzzy numbers, we refer the reader to the appendix of this book.

1.1 First Type Fuzzy Delta Differentiation Definition 1.1 Assume that f : T → F (R) is a fuzzy function and t ∈ Tκ . Then f is said to be first type right fuzzy delta differentiable at t, shortly right δH + differentiable at t, if there exists an element δH f (t) ∈ F (R) with the property  that, for any given  > 0, there exists a neighborhood UT of t, i.e., UT = (t −δ, t +δ) T for some δ > 0, such that for all t + h ∈ UT the H -difference f (t + h) H f (σ (t)) exists and

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_1

1

2

1 Calculus of Fuzzy Functions

  + D f (t + h) H f (σ (t)), δH f (t)(h − μ(t)) ≤ |h − μ(t)| + with 0 ≤ h < δ. In this case, δH f (t) is said to be first type right fuzzy delta derivative of f at t, shortly right δH -derivative of f at t.

Definition 1.2 Assume that f : T → F (R) is a fuzzy function and t ∈ Tκ . Then f is said to be first type left fuzzy delta differentiable at t, shortly left δH -differentiable − at t, if there exists an element δH f (t) ∈ F (R) with the property that, for any given  > 0, there exists a neighborhood UT of t, i.e., UT = (t − δ, t + δ) T for some δ > 0, such that for all t − h ∈ UT the H -difference f (σ (t)) H f (t − h) exists and   − f (t)(h + μ(t)) ≤ (h + μ(t)) D f (σ (t)) H f (t − h), δH with 0 ≤ h < δ. Definition 1.3 Let f : T → F (R) be a fuzzy function and t ∈ Tκ . Then f is said to be first type fuzzy delta differentiable at t, shortly δH -differentiable at t, if f is both − + first type left and right fuzzy delta differentiable at t ∈ Tκ and δH f (t) = δH f (t), and we will denote it by δH f (t). We call δH f (t) the first type fuzzy delta derivative of f at t, shortly δH -derivative of f at t. We say that f is first type fuzzy delta differentiable at t, shortly δH -differentiable at t, if its δH -derivative exists at t. We say that f is first type fuzzy delta differentiable on Tκ , shortly δH -differentiable on Tκ , if its δH -derivative exists at each t ∈ Tκ . The fuzzy function δH f : Tκ → F (R) is then called first type fuzzy delta derivative, shortly δH -derivative of f on Tκ . Theorem 1.1 If the δH -derivative of f at t ∈ Tκ exists, then it is unique. Hence, δH -derivative is well-defined. 1 f (t) and δ 2 f (t) are the δ -derivative of f at t ∈ Tκ . Take Proof Assume that δH H H  > 0 arbitrarily. Then there exists a δ = δ() > 0 such that for any t + h ∈ (t − δ, t + δ) the H -difference f (t + h) H f (σ (t)) exists and

  1 D δH f (t)(h − μ(t)), f (t + h) H f (σ (t)) ≤ |h − μ(t)|,   2 D δH f (t)(h − μ(t)), f (t + h) H f (σ (t)) ≤ |h − μ(t)| with 0 ≤ h < δ. Then,   1 2 D δH f (t), δH f (t) =

  1 1 2 D δH f (t)(h − μ(t)), δH f (t)(h − μ(t)) |h − μ(t)|

 1 1 D δH f (t)(h − μ(t)) + f (t + h) H f (σ (t)), = |h − μ(t)|  2 δH f (t)(h − μ(t)) + f (t + h) H f (σ (t))

1.1 First Type Fuzzy Delta Differentiation

3

  1 1 f (t)(h − μ(t)), f (t + h) H f (σ (t)) D δH |h − μ(t)|   1 2 D f (t + h) H f (σ (t)), δH + f (t)(h − μ(t)) |h − μ(t)| ≤+

≤

= 2 for all h − μ(t) = 0. Since  > 0 was arbitrarily chosen, we conclude that   1 2 f (t), δH f (t) = 0. D δH 1 f (t) = δ 2 f (t). This completes the proof. Therefore, δH H

Theorem 1.2 Assume that f : T → F (R) is a continuous function at t1 ∈ Tκ and t1 is right-scattered. Then f is δH -differentiable at t1 and δH f (t1 ) =

f (σ (t1 )) H f (t1 ) . μ(t1 )

Proof We have lim

h→0+

f (t1 ) H f (σ (t1 )) f (t1 + h) H f (σ (t1 )) = h − μ(t1 ) −μ(t1 ) =

f (σ (t1 )) H f (t1 ) μ(t1 )

and f (σ (t1 )) H f (t1 ) f (σ (t1 )) H f (t1 − h) = . h→0+ h + μ(t1 ) μ(t1 ) lim

Hence, 

f (t1 + h) H f (σ (t1 )) f (σ (t1 )) H f (t1 ) , h→0+ h − μ(t1 ) μ(t1 )   f (σ (t1 )) H f (t1 ) f (σ (t1 )) H f (t1 ) , =D μ(t1 ) μ(t1 ) lim D

=0



4

1 Calculus of Fuzzy Functions

and 

f (σ (t1 )) H f (t1 − h) f (σ (t1 )) H f (t1 ) , lim D h→0+ h + μ(t1 ) μ(t1 )   f (σ (t1 )) H f (t1 ) f (σ (t1 )) H f (t1 ) =D , μ(t1 ) μ(t1 )



= 0. This completes the proof. Example 1.1 Let T = N. Consider   f (t) = t, t + 1, t 2 + 1 ,

t ∈ T.

Here σ (t) = t + 1, μ(t) = 1,   f (σ (t)) = σ (t), σ (t) + 1, (σ (t))2 + 1   = t + 1, t + 1 + 1, (t + 1)2 + 1   = t + 1, t + 2, t 2 + 2t + 2 , f (σ (t)) H f (t) = (1, 1, 2t + 1), f (σ (t)) H f (t) = (1, 1, 2t + 1), μ(t)

t ∈ T.

δH f (t) = (1, 1, 2t + 1),

t ∈ T.

Therefore,

Example 1.2 Let T = 2N0 . Consider   f (t) = t 2 − t, t 2 + 1, t 3 + 1 ,

t ∈ T.

We have σ (t) = 2t, μ(t) = t,   f (σ (t)) = (σ (t))2 − σ (t), (σ (t))2 + 1, (σ (t))3 + 1

1.1 First Type Fuzzy Delta Differentiation

5

  = 4t 2 − 2t, 4t 2 + 1, 8t 3 + 1 ,   f (σ (t)) H f (t) = 3t 2 − t, 3t 2 , 7t 3 ,  f (σ (t)) H f (t)  = 3t − 1, 3t, 7t 2 , μ(t)

t ∈ T.

Therefore,   δH f (t) = 3t − 1, 3t, 7t 2 , Example 1.3 Let T = [0, 1]



t ∈ T.

{3, 7, 8}. Consider

f (t) = (−3t + 2, t, 3t + 5) ,

t ∈ Tκ .

We will find δH f (1). We have σ (1) = 3, μ(1) = σ (1) − 1 = 3−1 = 2, f (1) = (−1, 1, 8), f (σ (1)) = (−3σ (1) + 2, σ (1), 3σ (1) + 5) = (2 − 9, 3, 9 + 5) = (−7, 3, 14), f (σ (1)) H f (1) = (−6, 2, 6), f (σ (1)) H f (1) = (−3, 1, 3) . μ(1) Consequently, δH f (1) = (−3, 1, 3) . 



[1, 4] {11, 15}, where 0, 12 and [1, 4] are the Exercise 1.1 Let T = 0, 12 real-valued intervals. Let also,   t +1 2 t +1 3 t +1 f (t) = ,t + , t + t2 + t + 1 + , t ∈ T. t +2 t +2 t +2

6

1 Calculus of Fuzzy Functions

Find   1 , δH f 2

δH f (4),

δH f (11).

Theorem 1.3 Assume that f : T → F (R) is δH -differentiable at t ∈ Tκ . Then f is continuous at t. Proof Take  > 0 arbitrarily. Since f is δH -differentiable at t, there exists a δ > 0 such that for any t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, 0 ≤ h < , the H -differences f (t + h) H f (σ (t)) and f (σ (t)) H f (t − h) exist and D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) ≤ |h − μ(t)|, D (f (σ (t)) H f (t − h), δH f (t)(h + μ(t))) ≤ (h + μ(t)). In particular, we have D (f (σ (t)) H f (t), δH f (t)μ(t)) ≤ μ(t). Hence, for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, we get D(f (t + h), f (t)) = D (f (t + h) H f (σ (t)), f (t) H f (σ (t)))  = D f (t + h) H f (σ (t)) + δH f (t)(h − μ(t)),  f (t) H f (σ (t)) + δH f (t)(h − μ(t)) ≤ D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) +D (δH f (t)(h − μ(t)), f (t) H f (σ (t)))   ≤ (h − μ(t)) + D δH f (t)h, 0 +D (δH f (t)(−μ(t)), f (t) H f (σ (t))) = (h − μ(t)) + hD(δH f (t), 0) +D (f (σ (t)) H f (t), δH f (t)μ(t)) ≤ (h − μ(t)) + hδH f (t) + μ(t) = h ( + δH f (t)) <  ( + δH f (t)) and D(f (t), f (t − h)) = D (f (t) H f (σ (t)), f (t − h) H f (σ (t)))

1.1 First Type Fuzzy Delta Differentiation

7

= D (f (σ (t)) H f (t), f (σ (t)) H f (t − h))  = D f (σ (t)) H f (t) + δH f (t)(h + μ(t)),  f (σ (t)) H f (t − h) + δH f (t)(h + μ(t)) ≤ D (δH f (t)(h + μ(t)), f (σ (t)) H f (t − h)) +D (f (σ (t)) H f (t), δH f (t)(h + μ(t))) ≤ (h + μ(t)) + D (f (σ (t)) H f (t), δH f (t)μ(t))   +D δH f (t)h, 0   0 ≤ (h + μ(t)) + μ(t) + hD δH f (t), < ( + μ(t)) + μ(t) + δH f (t) =  ( + 2μ(t) + δH f (t)) . This completes the proof. Theorem 1.4 Let f : T → F (R) be a fuzzy function and let t ∈ Tκ be right-dense. Then f is δH -differentiable at t if and only if the limits f (t + h) H f (t) h→0+ h lim

and

f (t) H f (t − h) h→0+ h lim

(1.1)

exist and satisfy the relations lim

h→0+

f (t + h) H f (t) f (t) H f (t − h) = lim = δH f (t). h→0+ h h

Proof 1. Assume that f is δH -differentiable at t. Take  > 0 arbitrarily. Since f is δH differentiable at t, there exists a δ > 0 such that for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, the H -differences f (t + h) H f (σ (t)) and f (σ (t)) H f (t − h) exist and D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) ≤ |h − μ(t)|, D (f (σ (t)) H f (t − h), δH f (t)(μ(t) + h)) ≤ (μ(t) + h), and using that σ (t) = t, we arrive at the following inequalities: D (f (t + h) H f (t), δH f (t)h) ≤ h, D (f (t) H f (t − h), δH f (t)h) ≤ h.

(1.2)

8

1 Calculus of Fuzzy Functions

Thus, 



f (t+h)H f (t) , δH f (t) h   D f (t)Hhf (t−h) , δH f (t)

D

≤ , ≤

(1.3)

for all t − h, t + h ∈ (t − δ, t + δ), 0 < h < δ. Because  > 0 was arbitrarily chosen, we obtain the following relations: δH f (t) = lim

h→0+

f (t + h) H f (t) f (t) H f (t − h) = lim . h→0+ h h

2. Suppose that t is right-dense and the limits (1.1) exist. Then, for any  > 0, there is a δ > 0 such that for t − h, t + h ∈ (t − δ, t + δ), 0 < h < δ, we have (1.3), whereupon we get (1.2). This completes the proof.  Example 1.4 Let T = [0, 1] 1 + 21k , where [0, 1] is the real-valued k∈N interval. Let also,   f (t) = t, 3t + 7, 7 + 3t + t 2 , t ∈ T. We will find δH f (1). Note that σ (1) = 1. Consequently, t = 1 is a right-dense point. Take h > 0 arbitrarily. We have f (1) = (1, 10, 11),   f (1 + h) = 1 + h, 3(1 + h) + 7, 7 + 3(1 + h) + (1 + h)2   = 1 + h, 10 + 3h, 10 + 3h + 1 + 2h + h2   = 1 + h, 10 + 3h, 11 + 5h + h2 ,   f (1 − h) = 1 − h, 3(1 − h) + 7, 7 + 3(1 − h) + (1 − h)2   = 1 − h, 10 − 3h, 10 − 3h + 1 − 2h + h2   = 1 − h, 10 − 3h, 11 − 5h + h2 ,   f (1 + h) H f (1) = h, 3h, 5h + h2 ,   f (1) H f (1 − h) = h, 3h, 5h − h2 , f (1 + h) H f (1) = (1, 3, 5 + h), h

1.1 First Type Fuzzy Delta Differentiation

9

f (1) H f (1 − h) = (1, 3, 5 − h), h f (1 + h) H f (1) lim = (1, 3, 5), h→0+ h f (1) H f (1 − h) = (1, 3, 5). h→0+ h lim

Therefore, δH f (1) = (1, 3, 5). Exercise 1.2 Let T =



1 {0} 2k k∈N

and

  t +4 t +4 , 11 + 5t + 7t 2 + , f (t) = 1 + t 2 , 1 + 3t + 4t 2 + t +1 t +1 Find δH f (0). Theorem 1.5 Let f : T → F (R) be δH -differentiable at t ∈ Tκ . Then, f (σ (t)) = f (t) + μ(t) · δH f (t) or f (t) = f (σ (t)) + (−1) · (μ(t) · δH f (t)) . Proof 1. Let σ (t) = t. Then μ(t) = 0 and f (σ (t)) = f (t) + μ(t) · δH f (t) or f (t) = f (σ (t)) + (−1)μ(t) · δH f (t). 2. Let σ (t) > t. Then, by Theorem 1.2, we have f (σ (t)) H f (t) = μ(t) · δH f (t). Hence, using the definition of H , we obtain f (σ (t)) = f (t) + μ(t) · δH f (t)

t ∈ T.

10

1 Calculus of Fuzzy Functions

or f (t) = f (σ (t)) + (−1) · (μ(t) · δH f (t)) . This completes the proof. Example 1.5 Let T = 2N0 and  f (t) =

 t +1 2 , t, t + 5t + 1 , t +2

t ∈ T.

We have σ (t) = t + 2, μ(t) = 2,   σ (t) + 1 f (σ (t)) = , σ (t), (σ (t))2 + 5σ (t) + 1 σ (t) + 2   t +2+1 2 , t + 2, (t + 2) + 5(t + 2) + 1 = t +2+2   t +3 2 = , t + 2, t + 4t + 4 + 5t + 11 t +4   t +3 , t + 2, t 2 + 9t + 15 , = t +4   t +3 t +1 − , t + 2 − t, t 2 + 9t + 15 − t 2 − 5t − 1 f (σ (t)) H f (t) = t +4 t +2   (t + 3)(t + 2) − (t + 1)(t + 4) , 2, 4t + 14 = (t + 2)(t + 4)   2 t + 5t + 6 − t 2 − 5t − 4 , 2, 4t + 14 = (t + 2)(t + 4)   2 , 2, 4t + 14 , = (t + 2)(t + 4) f (σ (t)) H f (t) μ(t)   1 = , 1, 2t + 7 , (t + 2)(t + 4)   t +1 , t, t 2 + 5t + 1 f (t) + μ(t) · δH f (t) = t +2 δH f (t) =

1.1 First Type Fuzzy Delta Differentiation

11

 2 , 2, 4t + 14 + (t + 2)(t + 4)   2 t +1 + , t + 2, t 2 + 5t + 1 + 4t + 14 t + 2 (t + 2)(t + 4)   (t + 1)(t + 4) + 2 , t + 2, t 2 + 9t + 15 (t + 2)(t + 4)   2 t + 5t + 6 , t + 2, t 2 + 9t + 15 (t + 2)(t + 4)   (t + 2)(t + 3) 2 , t + 2, t + 9t + 15 (t + 2)(t + 4)   t +3 2 , t + 2, t + 9t + 15 , t ∈ T. t +4 

= = = = = Thus,

f (σ (t)) = f (t) + μ(t) · δH f (t),

t ∈ T.

Theorem 1.6 Let f, g : T → F (R) be δH -differentiable at t ∈ Tκ . Then f + g : T → F (R) is δH -differentiable at t ∈ Tκ and δH (f + g)(t) = δH f (t) + δH g(t). Proof Take  > 0 arbitrarily. Since f, g : T → F (R) are δH -differentiable at t ∈ Tκ , there exists a δ > 0 such that for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, the H -differences f (t + h) H f (σ (t)) and f (σ (t)) H f (t − h) exist and  |h − μ(t)|, 2  D (f (σ (t)) H f (t − h), δH f (t)(h + μ(t))) ≤ (h + μ(t)), 2  D (g(t + h) H g(σ (t)), δH g(t)(h − μ(t))) ≤ |h − μ(t)|, 2  D (g(σ (t)) H g(t − h), δH g(t)(h + μ(t))) ≤ (h + μ(t)). 2

D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) ≤

Hence, for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, we have D ((f + g)(t + h) H (f + g)(σ (t)), (δH f (t) + δH g(t)) (h − μ(t))) = D ((f (t + h) H f (σ (t))) + (g(t + h) H g(σ (t))) , δH f (t)(h − μ(t)) + δH g(t)(h − μ(t)))

12

1 Calculus of Fuzzy Functions

≤ D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) +D (g(t + h) H g(σ (t)), δH g(t)(h − μ(t)))   ≤ |h − μ(t)| + |h − μ(t)| 2 2 = |h − μ(t)| and D ((f + g)(σ (t)) H (f + g)(t − h), (δH f (t) + δH g(t)) (h + μ(t))) = D ((f (σ (t)) H f (t − h)) + (g(σ (t)) H g(t − h)) , δH f (t)(h + μ(t)) + δH g(t)(h + μ(t))) ≤ D (f (σ (t)) H f (t − h), δH f (t)(h + μ(t))) +D (g(σ (t)) H g(t − h), δH g(t)(h + μ(t)))   ≤ (h + μ(t)) + (h + μ(t)) 2 2 = (h + μ(t)). This completes the proof.

 Example 1.6 Let T = 31k

k∈N

{0} and 

f (t) =

 t +1 , 1, 2 , t +2

g(t) = (t, 3, 4) ,

t ∈ N.

Note that σ (0) = 0, i.e., t = 0 is a right-dense point. We have  f (0) =

 1 , 1, 2 , 2

g(0) = (0, 3, 4). Take h > 0 arbitrarily. Then, 

 h+1 , 1, 2 , h+2   1−h , 1, 2 , f (−h) = 2−h   h+1 1 − , 0, 0 f (h) H f (0) = h+2 2 f (h) =

1.1 First Type Fuzzy Delta Differentiation

13

 2h + 2 − h − 2 , 0, 0 2(h + 2)   h , 0, 0 , 2(h + 2)   1 1−h − , 0, 0 2 2−h   2 − h − 2 + 2h , 0, 0 2(2 − h)   h , 0, 0 , 2(2 − h)   1 , 0, 0 , 2(h + 2)   1 , 0, 0 , 2(2 − h)   1 , 0, 0 , 4   1 , 0, 0 . 4 

= = f (0) H f (−h) = = = f (h) H f (0) = h f (0) H f (−h) = h lim

h→0

lim

h→0

f (h) H f (0) = h

f (0) H f (−h) = h

Therefore, f (h) H f (0) f (0) H f (−h) = lim h→0 h→0 h h = δH f (0)   1 , 0, 0 . = 4 lim

Next, g(h) = (h, 3, 4) , g(−h) = (−h, 3, 4) , g(h) H g(0) = (h, 0, 0) , g(0) H g(−h) = (h, 0, 0) , g(h) H g(0) = (1, 0, 0) , h g(0) H g(−h) = (1, 0, 0) , h

14

1 Calculus of Fuzzy Functions

g(h) H g(0) = (1, 0, 0), h g(0) H g(−h) lim = (1, 0, 0). h→0 h lim

h→0

Thus, lim

h→0

g(h) H g(0) g(0) H g(−h) = lim h→0 h h = δH g(0) = (1, 0, 0).

Then, δH f (0) + δH g(0) = =





1 + 1, 0, 0  4 5 , 0, 0 . 4

Also,  (f + g)(t) =

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

 t + 1 + t 2 + 2t , 4, 6 t +2   2 t + 3t + 1 , 4, 6 , t ∈ T, = t +2   1 , 4, 6 , (f + g)(0) = 2   2 h + 3h + 1 , 4, 6 , (f + g)(h) = h+2   2 h − 3h + 1 , 4, 6 , (f + g)(−h) = 2−h   2 h + 3h + 1 1 − , 0, 0 (f + g)(h) H (f + g)(0) = h+2 2   2 2h + 6h + 2 − h − 2 , 0, 0 = 2(h + 2)   2 2h + 5h , 0, 0 , = 2(h + 2) 

=

(1.4)

1.1 First Type Fuzzy Delta Differentiation

15

 2h + 5 , 0, 0 , 2(h + 2)   5 (f + g)(h) H (f + g)(0) = , 0, 0 , lim h→0 h 4   1 h2 − 3h + 1 − , 0, 0 (f + g)(0) H (f + g)(−h) = 2 2−h   2 − h − 2h2 + 6h − 2 , 0, 0 = 2(2 − h)   −2h2 + 5h , 0, 0 , = 2(2 − h)   −2h + 5 (f + g)(0) H (f + g)(−h) = , 0, 0 , h 2(2 − h)   5 (f + g)(h) H (f + g)(0) = , 0, 0 , lim h→0 h 4   5 (f + g)(0) H (f + g)(−h) = , 0, 0 . lim h→0 h 4 (f + g)(h) H (f + g)(0) = h



Consequently, (f + g)(h) H (f + g)(0) (f + g)(0) H (f + g)(−h) = lim h→0 h→0 h h = δH (f + g)(0)   5 , 0, 0 . = 4 lim

From here and from (1.4), we arrive at δH f (0) + δH g(0) = δH (f + g)(0). Theorem 1.7 Let f : T → F (R) be δH -differentiable at t ∈ Tκ . Then for any λ ∈ R the function λ · f : T → F (R) is δH -differentiable at t ∈ Tκ and δH (λ · f )(t) = λ · δH f (t). Proof If λ = 0, then the assertion is evident. Let λ = 0. Take  > 0 arbitrarily. Since f : T → F (R) is δH -differentiable at t ∈ Tκ , there exists a δ > 0 such that for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, the H -differences f (t + h) H f (σ (t)) and f (σ (t)) H f (t − h) exist and

16

1 Calculus of Fuzzy Functions

 |h − μ(t)|, |λ|  (h + μ(t)). D (f (σ (t)) H f (t − h), δH f (t)(h + μ(t))) ≤ |λ|

D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t))) ≤

Hence, for t − h, t + h ∈ (t − δ, t + δ), 0 ≤ h < δ, we get D ((λ · f )(t + h) H (λ · f )(σ (t)), λ · δH f (t)(h − μ(t))) = D (λ · (f (t + h) H f (σ (t))), λ · δH f (t)(h − μ(t))) = |λ|D (f (t + h) H f (σ (t)), δH f (t)(h − μ(t)))  ≤ |λ| |h − μ(t)| |λ| = |h − μ(t)|, D ((λ · f )(σ (t)) H (λ · f )(t − h), λ · δH f (t)(h + μ(t))) = D (λ · (f (σ (t)) H f (t − h)), λ · δH f (t)(h + μ(t))) = |λ|D (f (σ (t)) H f (t − h), δH f (t)(h + μ(t)))  ≤ |λ| (h + μ(t)) |λ| = (h + μ(t)). This completes the proof. Theorem 1.8 Let t ∈ Tκ , f : T → F (R) and fα (t) = [f (t)]α , α ∈ [0, 1]. If f is δH -differentiable at t, then fα is δH -differentiable at t and δH [f (t)]α = δH fα (t),

α ∈ [0, 1].

Proof Let f be δH -differentiable at t and let α ∈ [0, 1] be arbitrarily chosen. 1. Let t be right-dense. Take h > 0 arbitrarily. Since f is δH -differentiable at t, there exist f (t + h) H f (t) h→0 h lim

and

f (t) H f (t − h) h→0 h lim

and f (t + h) H f (t) h f (t) H f (t − h) . = lim h→0 h

δH f (t) = lim

h→0

1.1 First Type Fuzzy Delta Differentiation

17

Next, [f (t + h) H f (t)]α = fα (t + h) H fα (t), [f (t) H f (t − h)]α = fα (t) H fα (t − h), fα (t + h) H fα (t) [f (t + h) H f (t)]α = , h h fα (t) H fα (t − h) [f (t) H f (t − h)]α = . h h Hence, [f (t + h) H f (t)]α h→0 h fα (t + h) H fα (t) = lim h→0 h

[δH f (t)]α = lim

[f (t) H f (t + h)]α h→0 h fα (t) H fα (t − h) . = lim h→0 h

= lim

Thus, the limits fα (t + h) H fα (t) h→0 h lim

and

fα (t) H fα (t − h) h→0 h lim

exist and fα (t + h) H fα (t) fα (t) H fα (t − h) = lim h→0 h→0 h h = δH fα (t) lim

= [δH f (t)]α . 2. Let t be right-scattered. Since f is δH -differentiable at t, we have δH f (t) =

f (σ (t)) H f (t) . μ(t)

From here, [δH f (t)]α =

fα (σ (t)) H fα (t) μ(t)

= δH fα (t). This completes the proof.

18

1 Calculus of Fuzzy Functions

Theorem 1.9 Let t ∈ Tκ , f : T → F (R) is δH -differentiable at t. Let also,

 α [f (t)]α = f α (t), f (t) ,

α ∈ [0, 1].

α

Then f α and f are Δ-differentiable at t and

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

α ∈ [0, 1].

Proof Let α ∈ [0, 1] be arbitrarily chosen. 1. Suppose that t is right-dense. Then,  [δH f (t)] = α

lim

f α (t + h) − f α (t)

h

 αΔ = f αΔ (t), f (t) . h→0

α

α

f (t + h) − f (t) , lim h→0 h



2. Suppose that t is right-scattered. Then, 

f α (σ (t)) − f α (t) f α (σ (t)) − f α (t) , [δH f (t)] = μ(t) μ(t)

 αΔ = f αΔ (t), f (t) .



α

This completes the proof.

Example 1.7 Let T = [−1, 1] 2N , where [−1, 1] is the real-valued interval. Consider the function f : T → F (R) whose α-levels are given by [f (t)]α = g(t) · [−1, 1],

t ∈ T,

α ∈ [0, 1],

where  g(t) =

1 + sin t if t = 0 1 if t = 0.

We will find [δH f (0)]α , α ∈ [0, 1]. Here t = 0 is a right-dense point. For h > 0 sufficiently small, we have f (h) = g(h) · [−1, 1] = (1 + sin h) · [−1, 1] = [−1 − sin h, 1 + sin h],

1.1 First Type Fuzzy Delta Differentiation

19

f (0) = g(0) · [−1, 1] = 1 · [−1, 1] = [−1, 1],  α [f (h) H f (0)] = min{−1 − sinh + 1, 1 + sin h − 1},  max{−1 − sin h + 1,

1 + sin h − 1}

 =

 min{− sin h, sin h}, max{− sin h, sin h}

= [− sin h, sin h], 1 1 [f (h) H f (0)]α = · [− sin h, sin h] h h   sin h sin h , = − , h h   sin h sin h 1 , lim [f (h) H f (0)]α = lim − h→0 h h→0 h h = [−1, 1]. Example 1.8 Let T = [−1, 3] {5, 6, 9}, where [−1, 3] is the real-valued interval. Consider the function f : T → F (R) whose α-levels are given by [f (t)]α = g(t) · [−1, 2],

t ∈ T,

α ∈ [0, 1],

where  g(t) =

t (t+1) t+4

t

if t ∈ [−1, 3] if t ∈ {5, 6, 9}.

We will find [δH f (0)]α

and

[δH f (3)]α .

Note that 0 is a right-dense point and 3 is a right-scattered point. We have, for h ∈ (0, 3), h(h + 1) , h+4 f (h) = g(h) · [−1, 2] g(h) =

20

1 Calculus of Fuzzy Functions

h(h + 1) · [−1, 2] h+4   h(h + 1) 2h(h + 1) = − , , h+4 h+4

=

g(0) = 0, f (0) = g(0) · [−1, 2] = 0 · [−1, 2] = [0, 0], 12 , 7 f (3) = g(3) · [−1, 2] g(3) =

12 · [−1, 2] 7   12 24 = − , , 7 7

=

σ (3) = 5, μ(3) = σ (3) − 3 = 5−3 = 2, g(σ (3)) = g(5) = 5, f (σ (3)) = g(σ (3)) · [−1, 2] = 5 · [−1, 2] = [−5, 10]. Then,  h(h + 1) 2h(h + 1) [f (h) H f (0)] = min − , , h+4 h+4   h(h + 1) 2h(h + 1) , max − h+4 h+4   h(h + 1) 2h(h + 1) , , = − h+4 h+4   1 h(h + 1) 2h(h + 1) 1 α · [f (h) H f (0)] = · − , h h h+4 h+4 

α



1.1 First Type Fuzzy Delta Differentiation

21

  h + 1 2(h + 1) , , = − h+4 h+4 1 · [f (h) H f (0)]α h→0+ h   h + 1 2(h + 1) , lim − h→0+ h+4 h+4   1 1 , − , 4 2    24 12 , min −5 + , 10 − 7 7   24 12 max −5 + , 10 − 7 7      23 46 23 46 , max − , min − , 7 7 7 7   23 46 − , , 7 7

[δH f (0)]α = lim = = [f (σ (3)) H f (3)]α =

= =

1 · [f (σ (3)) H f (3)]α σ (3)   23 46 1 = · − , 5 7 7   23 46 . = − , 35 35

[δH f (3)]α =

Exercise 1.3 Let T = [−1, 4] {5, 11, 21}, where [−1, 4] is the real-valued interval. Consider the function f : T → F (R) whose α-levels are given by [f (t)]α = g(t) · [−1, 7], where g(t) =

⎧ 2 +t+1) ⎪ ⎨ t2t(t2 +3t+4 ⎪ ⎩ t 2 + 1 if

if

t ∈ [−1, 4]

t ∈ {5, 11, 21}.

Find [δH f (0)]α ,

[δH f (4)]α ,

[δH f (5)]α .

22

1 Calculus of Fuzzy Functions

Example 1.9 Let T = 2N0 and f : [1, 8] → F (R) be a fuzzy function whose α-levels are given by [f (t)]α = g(t) · [1, 8],

α ∈ [0, 1],

t ∈ [1, 8],

where g(t) = 1 + t 2 , t ∈ T. We have σ (t) = 2t,

t ∈ T,

and [f (t)]α = g(t) · [1, 8] = [1 + t 2 , 8 + 8t 2 ], [f (σ (t))]α = [1 + (σ (t))2 , 8 + 8(σ (t))2 ] = [1 + 4t 2 , 8 + 32t 2 ],

t ∈ [1, 8],

α ∈ [0, 1].

Then, [f (σ (t)) H f (t)]α = [min{1 + 4t 2 − 1 − t 2 , 8 + 32t 2 − 8 − 8t 2 }, max{1 + 4t 2 − 1 − t 2 , 8 + 32t 2 − 8 − 8t 2 }] = [min{3t 2 , 24t 2 }, max{3t 2 , 24t 2 }] = [3t 2 , 24t 2 ],

t ∈ [1, 8],

α ∈ [0, 1].

Thus, [δH f (t)]α =

1 · [f (σ (t)) H f (t)]α μ(t)

1 · [3t 2 , 24t 2 ] t = [3t, 24t], t ∈ [1, 8], =

α ∈ [0, 1].

Example 1.10 Let T = 2Z and f : [0, 10] → F (R) be a fuzzy function whose α-levels are given by [f (t)]α = [−t, 1 + 3t],

t ∈ [0, 10],

We will find δH f (t), t ∈ [0, 10]. Here σ (t) = t + 2,

t ∈ T,

[f (σ (t))] = [−σ (t), 1 + 3σ (t)] α

α ∈ [0, 1].

1.1 First Type Fuzzy Delta Differentiation

23

= [−t − 2, 1 + 3(t + 2)] = [−2 − t, 7 + 3t],

t ∈ [0, 10],

α ∈ [0, 1].

Hence,  [f (σ (t)) H f (t)] =

min{−2 − t + t, 7 + 3t − 1 − 3t},

α

 max{−2 − t + t, 7 + 3t − 1 − 3t} = [min{−2, 6}, max{−2, 6}] = [−2, 6], 1 1 · [f (σ (t)) H f (t)]α = · [−2, 6] μ(t) 2 = [−1, 3],

t ∈ [0, 10],

α ∈ [0, 1].

Therefore, [δH f (t)]α = [−1, 3],

t ∈ [0, 10],

α ∈ [0, 1].

Example 1.11 Let T = 3N0 and f : T → F (R) be a fuzzy function whose α-levels are given by [f (t)]α = g(t) · [1, 2],

t ∈ [1, 27],

where  g(t) =

t+1 t+2

4 if

if t = 3 t = 3.

We will find [δH f (1)]α . Here σ (t) = 3t, μ(t) = 2t,

t ∈ T,

f (1) = g(1) · [1, 2] 2 · [1, 2] 3   2 4 = , , 3 3

=

f (σ (1)) = f (3)

α ∈ [0, 1],

24

1 Calculus of Fuzzy Functions

= g(3) · [1, 2] = 4 · [1, 2] = [4, 8],  2 4 α [f (σ (1)) H f (1)] = min{4 − , 8 − }, 3 3  4 2 max{4 − , 8 − } 3 3   10 20 , . = 3 3 Exercise 1.4 Find [δH f (t)]α , where 1. T = Z, [f (t)]α = g(t) · [−1, 4], g(t) = t + t 2 , t ∈ T, α ∈ [0, 1], 2. T = 2Z, [f (t)]α = g(t) · [1, 7], g(t) = 1 + t, t ∈ T, α ∈ [0, 1], 2 α ∈ [0, 1]. 3. T = N20 , [f (t)]α = g(t) · [1, 30], g(t) = 1+t+t 1+3t , t ∈ T, Example 1.12 Let T = 3N0 , f, g : T → F (R) be fuzzy functions whose α-levels are given by [f (t)]α = [−3t, 4t 2 ], [g(t)]α = [t + 1, t 2 + t + 1],

t ∈ T,

α ∈ [0, 1].

We have σ (t) = t + 3, f α (t) = −3t, α

f (t) = 4t 2 , g α (t) = t + 1, g α (t) = t 2 + t + 1,

t ∈ T,

Then, f αΔ (t) = −3, f

αΔ

(t) = 4(σ (t) + t) = 4(t + 3 + t) = 4(2t + 3) = 8t + 12,

α ∈ [0, 1].

1.1 First Type Fuzzy Delta Differentiation

25

g αΔ (t) = 1, g αΔ (t) = σ (t) + t + 1 = t +3+t +1 = 2t + 4,

t ∈ T,

α ∈ [0, 1],

and [δH f (t)]α = [−3, 8t + 12], [δH g(t)]α = [1, 2t + 4],

t ∈ T,

α ∈ [0, 1].

Therefore, f and g are δH -differentiable in T. Next, [δH f (t)]α + [δH g(t)]α = [−2, 10t + 16], [(f + g)(t)]α = [−2t + 1, 5t 2 + t + 1], (f + g)α (t) = −2t + 1, α

(f + g) (t) = 5t 2 + t + 1, (f + g)αΔ (t) = −2, αΔ

(f + g)

(t) = 5(σ (t) + t) + 1 = 5(t + 3 + t) + 1 = 5(2t + 3) + 1 = 10t + 15 + 1 = 10t + 16,

t ∈ T,

α ∈ [0, 1].

Thus, [δH (f + g)(t)]α = [−2, 10t + 16] = [δH f (t)]α + [δH g(t)]α ,

t ∈ T,

α ∈ [0, 1].

[δH (f + g)(t)]α = [δH f (t)]α + [δH g(t)]α ,

t ∈ T,

α ∈ [0, 1],

Exercise 1.5 Check if

where 1. T = Z and

 [f (t)]α = t 2 + t − 1, t 4 + t 2 + t ,

26

1 Calculus of Fuzzy Functions



[g(t)]α = t 2 , t 8 + t 6 + t 4 + t 2 , 2. T = 3Z and



[f (t)]α = t 2 − 1, t 6 + t 4 + t 2 ,

 [g(t)]α = t 2 − 3t − 1, t 8 + 10t 6 + t 2 − 3t − 1 ,

3. T = 2N0 and  t 2 + 3t − 4 2t 4 + t 2 + 3t − 4 , , [f (t)] = t2 + t + 1 t2 + t + 1

 [g(t)]α = t 2 − 10, t 8 + 7t 6 + 6t 4 + 8t 2 − 10 . 

α

Example 1.13 Let T = 3N0 , f : T → F (R) be a fuzzy function whose α-levels are given by [f (t)]α = g(t) · [−1, 3],

α ∈ [0, 1],

t ∈ T,

where g(t) =

t2 + 1 , t +3

t ∈ T.

 α We will find δH f (t) , α ∈ [0, 1], t ∈ T. Here σ (t) = 3t, f α (t) = − α

f (t) =

t2 + 1 , t +3

3t 2 + 3 , t +3

t ∈ T,

α ∈ [0, 1].

We have g Δ (t) =

(σ (t) + t)(t + 3) − (t 2 + 1) (t + 3)(σ (t) + 3)

=

(3t + t)(t + 3) − t 2 − 1 (t + 3)(3t + 3)

=

4t (t + 3) − t 2 − 1 3(t + 1)(t + 3)

1.1 First Type Fuzzy Delta Differentiation

27

=

4t 2 + 12t − t 2 − 1 3(t + 1)(t + 3)

=

3t 2 + 12t − 1 , 3(t + 1)(t + 3)

t ∈ T.

Therefore,  α δH f (t) = g Δ (t) · [−1, 3],

t ∈ T,

α ∈ [0, 1].

Exercise 1.6 Let T = 4N0 , f : T → F (R) be a fuzzy function whose α-levels are given by [f (t)]α = g(t) · [−10, 31],

α ∈ [0, 1],

t ∈ T,

where g(t) =

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

t ∈ T.

 α Find δH f (t) , α ∈ [0, 1], t ∈ T. Definition 1.4 Let f, g : T → F (R) be fuzzy functions with α-levels given by

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , t ∈ T, Define: α,1 If,g (t)

=

α,2 If,g (t) =

α,3 If,g (t) =

α,4 If,g (t) =

   f α (t) f α (t)    ,  α  g (t) g α (t)    α  f (t) f α (t)    ,  α  g (t) g α (t)     f α (t) f α (t)    ,  α  g (t) g α (t)    α α    f (t) f (t)  ,  α α  g (t) g (t) 

α ∈ [0, 1].

28

1 Calculus of Fuzzy Functions

   f α (t) f α (t)    = α ,  g (t) g α (t)    α   α  f (t) f (t)  α,6 If,g (t) =  α , α  g (t) g (t) 

α,5 If,g (t)

t ∈ T,

α ∈ [0, 1].

Definition 1.5 Let f, g : T → F (R) be fuzzy functions with α-levels given by

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , t ∈ T,

α ∈ [0, 1].

Define: 1. f ◦ g : T → F (R) be a fuzzy function whose α-levels are given by [(f ◦ g)(t)]α = [f (t)]α ◦ [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

2. f  g : T → F (R) be a fuzzy function whose α-levels are given by [(f  g)(t)]α = [f (t)]α  [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

3. f  g : T → F (R) be a fuzzy function whose α-levels are given by [(f  g)(t)]α = [f (t)]α  [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

4. f  g : T → F (R) be a fuzzy function whose α-levels are given by [(f  g)(t)]α = [f (t)]α  [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

5. f ⊗ g : T → F (R) be a fuzzy function whose α-levels are given by [(f ⊗ g)(t)]α = [f (t)]α ⊗ [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

6. f g : T → F (R) be a fuzzy function whose α-levels are given by [(f g)(t)]α = [f (t)]α [g(t)]α ,

t ∈ T,

α ∈ [0, 1].

Theorem 1.10 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≤ 0, If,g

Ifα,1 σ ,δ g (t) ≤ 0, H

f ◦ g is δH -differentiable at t. Then,

Iδα,1 (t) ≤ 0, H f,g

1.1 First Type Fuzzy Delta Differentiation

29

δH (f ◦ g)(t) = f σ (t) ◦ δH g(t) + δH f (t) ◦ g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , t ∈ T,

α ∈ [0, 1].

Since f and g are δH -differentiable at t, we get

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) . α,1 α,1 (t) ≤ 0, Ifα,1 Now, using that If,g σ ,δ g (t) ≤ 0, Iδ f,g (t) ≤ 0, we obtain H H

α,1 (t) If,g

   f α (t) f α (t)    = α   g (t) g α (t)  α

= f α (t)g α (t) − g α (t)f (t) ≤ 0,    f ασ (t) f ασ (t)    α,1 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≤ 0,   αΔ  αΔ   f (t) f (t)  α,1 IδH f,g (t) =  α   g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f

αΔ

(t)

≤ 0. α,1 Because If,g (t) ≤ 0, we have

 α [(f ◦ g)(t)]α = f α (t)g α (t), f (t)g α (t) . Since f ◦ g is δH -differentiable at t, we obtain

(1.5)

30

1 Calculus of Fuzzy Functions



Δ



Δ α f g α (t)



(t), [δH (f ◦ g)(t)] = 

ασ ασ = f ασ (t)g αΔ (t) + g α (t)f αΔ (t), f (t)g αΔ (t) + g α (t)f (t) . α

f α gα

α,1 Because Ifα,1 σ ,δ g (t) ≤ 0 and Iδ f,g (t) ≤ 0, we have H H

  σ  α

ασ f ◦ δH g (t) = f ασ (t)g αΔ (t), g αΔ (t)f (t) and

 αΔ [(δH f ◦ g) (t)]α = f αΔ (t)g α (t), g α (t)f (t) . From here,   σ  α α f ◦ δH g (t) + [(δH f ◦ g) (t)] = f ασ (t)g αΔ (t) + f αΔ (t)g α (t), g

αΔ

(t)f

ασ

(t) + g (t)f α

αΔ

 (t) .

From the last equality and from (1.6), we get (1.5). This completes the proof. Exercise 1.7 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≤ 0, If,g

α,1 If,δ (t) ≤ 0, Hg

Iδα,1 σ (t) ≤ 0, H f,g

f ◦ g is δH -differentiable at t. Prove that δH (f ◦ g)(t) = f (t) ◦ δH g(t) + δH f (t) ◦ g σ (t). Theorem 1.11 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≥ 0, If,g

Ifα,1 σ ,δ g (t) ≥ 0, H

Iδα,1 (t) ≥ 0, H f,g

f ◦ g is δH -differentiable at t. Then, δH (f ◦ g)(t) = f σ (t) ◦ δH g(t) + δH f (t) ◦ g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) .

(1.6)

1.1 First Type Fuzzy Delta Differentiation

31

Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , t ∈ Tκ ,

α ∈ [0, 1].

α,1 α,1 (t) ≥ 0, Ifα,1 Since If,g σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we get H H

α,1 (t) If,g

   f α (t) f α (t)    = α   g (t) g α (t)  α

= f α (t)g α (t) − g α (t)f (t) ≥ 0,    f ασ (t) f ασ (t)    Ifα,1  σ ,δ g (t) =  αΔ H  g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,   αΔ  αΔ   f (t) f (t)  α,1 IδH f,g (t) =  α   g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f

αΔ

(t)

≥ 0. α,1 Now, using that If,g (t) ≥ 0, we have

α  [(f ◦ g)(t)]α = f (t)g α (t), f α (t)g α (t) . Since f ◦ g is δH -differentiable at t, we obtain  Δ Δ   α [δH (f ◦ g)(t)]α = f g α (t), f α g α (t) 

ασ αΔ = f (t)g αΔ (t) + g α (t)f (t), f ασ (t)g αΔ (t) + g α (t)f ασ (t) . α,1 Because Ifα,1 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α ασ f ◦ δH g (t) = f (t)g αΔ (t), g αΔ (t)f ασ (t)

(1.7)

32

1 Calculus of Fuzzy Functions

and

αΔ  [(δH f ◦ g) (t)]α = f (t)g α (t), g α (t)f αΔ (t) .

From here,   α  σ ασ αΔ f ◦ δH g (t) + [(δH f ◦ g) (t)]α = f (t)g αΔ (t) + f (t)g α (t),  g αΔ (t)f ασ (t) + g α (t)f αΔ (t) .s From the last equality and from (1.7), we get the desired result. This completes the proof. Exercise 1.8 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≥ 0, If,g

α,1 If,δ (t) ≥ 0, Hg

Iδα,1 σ (t) ≥ 0, H f,g

f ◦ g is δH -differentiable at t. Prove that δH (f ◦ g)(t) = f σ (t) ◦ δH g(t) + δH f (t) ◦ g(t). Example 1.14 Let T = 2N0 and f, g : T → F (R) be fuzzy functions whose αlevels are [f (t)]α = [1, 2t] , 

[g(t)]α = t, t 2 ,

t ∈ T,

We have σ (t) = 2t, f α (t) = 1, α

f (t) = 2t, g α (t) = t, g α (t) = t 2 , f ασ (t) = 1, f

ασ

(t) = 2σ (t) = 4t,

[δH f (t)] = [0, 2], α

[δH g(t)]α = [1, σ (t) + t]

α ∈ [0, 1].

1.1 First Type Fuzzy Delta Differentiation

33

= [1, 2t + t] = [1, 3t] ,

t ∈ T,

α ∈ [0, 1],

i.e., f and g are δH -differentiable at t. Next, α,1 If,g (t)

   f α (t) f α (t)    = α   g (t) g α (t)     1 2t  =  2  t t = t 2 − 2t 2 = −t 2

< 0,    f ασ (t) f ασ (t)    α,1 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)     1 4t    = 1 3t  = 3t − 4t = −t < 0,   αΔ   αΔ  f (t) f (t)  α,1 IδH f,g (t) =  α  α  g (t) g (t)    0 2  =  2  t t = −2t < 0,

t ∈ T,

α ∈ [0, 1].

Thus, 

[(f ◦ g) (t)]α = t 2 , 2t 2 , [δH (f ◦ g)(t)]α = [σ (t) + t, 2(σ (t) + t)] = [2t + t, 2(2t + t)] 





= [3t, 6t],

f σ ◦ δH g (t) = [3t, 4t] ,

34

1 Calculus of Fuzzy Functions

[(δH f ◦ g) (t)]α = [0, 2t] ,  σ   f ◦ δH g (t) + [(δH f ◦ g) (t)]α = [3t, 6t],

t ∈ T,

α ∈ [0, 1].

Therefore, [δH (f ◦ g) (t)]α =

 σ  α f ◦ δH g (t) + [(δH f ◦ g) (t)]α ,

t ∈ T,

Theorem 1.12 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,2 (t) ≥ 0, If,g

Ifα,2 σ ,δ g (t) ≥ 0, H

Iδα,2 (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Then, δH (f  g)(t) = f σ (t)  δH g(t) + δH f (t)  g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) . Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , t ∈ T,

α ∈ [0, 1].

α,2 α,2 (t) ≥ 0, Ifα,2 Because If,g σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we obtain H H

α,1 (t) If,g

  α  f (t) f α (t)    = α   g (t) g α (t)  α

= f (t)g α (t) − g α (t)f α (t) ≥ 0,  ασ   f (t) f ασ (t)    α,2 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  =f

ασ

≥ 0,

(t)g αΔ (t) − g αΔ (t)f ασ (t)

α ∈ [0, 1].

1.1 First Type Fuzzy Delta Differentiation

Iδα,2 (t) H f,g

35

    αΔ  f (t) f αΔ (t)  = α  α  g (t) g (t)  =f

αΔ

(t)g α (t) − g α (t)f αΔ (t)

≥ 0. α,2 We will use that If,g (t) ≥ 0 to get

 α [(f  g)(t)]α = f α (t)g α (t), f (t)g α (t) . Since f  g is δH -differentiable at t, we obtain  Δ  α Δ  α α (t), f g α (t) [δH (f  g)(t)] = f g 

ασ ασ = f ασ (t)g αΔ (t) + g α (t)f αΔ (t), f (t)g αΔ (t) + g α (t)f (t) . α

(1.8)

α,2 Because Ifα,2 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α

ασ f  δH g (t) = f ασ (t)g αΔ (t), g αΔ (t)f (t) and

 αΔ [(δH f  g) (t)]α = f αΔ (t)g α (t), g α (t)f (t) . Hence,   σ  α α f  δH g (t) + [(δH f  g) (t)] = f ασ (t)g αΔ (t) + f αΔ (t)g α (t), g

αΔ

(t)f

ασ

(t) + g (t)f α

αΔ

 (t) .

From the last equality and from (1.8), we get the desired result. This completes the proof. Exercise 1.9 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,2 (t) ≥ 0, If,g

α,2 If,δ (t) ≥ 0, Hg

Iδα,2 σ (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Prove that δH (f  g)(t) = f (t)  δH g(t) + δH f (t)  g σ (t).

36

1 Calculus of Fuzzy Functions

Theorem 1.13 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,3 (t) ≥ 0, If,g

Ifα,3 σ ,δ g (t) ≥ 0, H

Iδα,3 (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Then, δH (f  g)(t) = f σ (t)  δH g(t) + δH f (t)  g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) . Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , t ∈ T,

α ∈ [0, 1].

α,3 α,3 Because If,g (t) ≥ 0, Ifα,3 σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we obtain H H

α,3 (t) If,g

   f α (t) f α (t)    = α   g (t) g α (t)  = f α (t)g α (t) − g α (t)f α (t)

≥ 0,    f ασ (t) f ασ (t)    α,3 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f ασ (t) ≥ 0,    f αΔ (t) f αΔ (t)    Iδα,3 (t) =  α  H f,g  g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f αΔ (t) ≥ 0. α,3 We will use that If,g (t) ≥ 0 to get

1.1 First Type Fuzzy Delta Differentiation

37

 [(f  g)(t)]α = f α (t)g α (t), f α (t)g α (t) . Since f  g is δH -differentiable at t, we obtain  Δ Δ   α α α α (t), f g (t) [δH (f  g)(t)] = f g 

= f ασ (t)g αΔ (t) + g α (t)f αΔ (t), f ασ (t)g αΔ (t) + g α (t)f ασ (t) . α

(1.9)

α,3 Since Ifα,3 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α

f  δH g (t) = f ασ (t)g αΔ (t), g αΔ (t)f ασ (t) and

 [(δH f  g) (t)]α = f αΔ (t)g α (t), g α (t)f αΔ (t) . Hence,   σ  α α f  δH g (t) + [(δH f  g) (t)] = f ασ (t)g αΔ (t) + f αΔ (t)g α (t), g

αΔ

(t)f

ασ

(t) + g (t)f α

αΔ

 (t) .

From the last equality and from (1.9), we get the desired result. This completes the proof. Exercise 1.10 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,3 (t) ≥ 0, If,g

α,3 If,δ (t) ≥ 0, Hg

Iδα,3 σ (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Then, δH (f  g)(t) = f (t)  δH g(t) + δH f (t)  g σ (t). Theorem 1.14 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,4 (t) ≥ 0, If,g

Ifα,4 σ ,δ g (t) ≥ 0, H

Iδα,4 (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Then, δH (f  g)(t) = f σ (t)  δH g(t) + δH f (t)  g(t).

38

1 Calculus of Fuzzy Functions

Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) . Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , t ∈ T,

α ∈ [0, 1].

α,4 α,4 (t) ≥ 0, Ifα,4 Because If,g σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we obtain H H

α,4 (t) If,g

  α α    f (t) f (t)  = α   g (t) g α (t)  α

α

= f (t)g α (t) − g α (t)f (t) ≥ 0,   ασ ασ    f (t) f (t)  α,4 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  =f

ασ

(t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,   αΔ   αΔ  f (t) f (t)  α,4 IδH f,g (t) =  α   g (t) g α (t)  =f

αΔ

(t)g α (t) − g α (t)f

αΔ

(t)

≥ 0. α,4 We will use that If,g (t) ≥ 0 to get

α  α [(f  g)(t)]α = f (t)g α (t), f (t)g α (t) . Since f  g is δH -differentiable at t, we obtain    α Δ  α α Δ (t), f g α (t) [δH (f  g)(t)] = f g 

ασ αΔ ασ ασ = f (t)g αΔ (t) + g α (t)f (t), f (t)g αΔ (t) + g α (t)f (t) . α

(1.10)

1.1 First Type Fuzzy Delta Differentiation

39

α,4 Since Ifα,4 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α ασ ασ f  δH g (t) = f (t)g αΔ (t), g αΔ (t)f (t) and

αΔ  αΔ [(δH f  g) (t)]α = f (t)g α (t), g α (t)f (t) . Hence,   σ  α ασ αΔ f  δH g (t) + [(δH f  g) (t)]α = f (t)g αΔ (t) + f (t)g α (t), g

αΔ

(t)f

ασ

(t) + g (t)f α

αΔ

 (t) .

From the last equality and from (1.10), we get the desired result. This completes the proof. Exercise 1.11 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,4 If,g (t) ≥ 0,

α,4 If,δ (t) ≥ 0, Hg

Iδα,4 σ (t) ≥ 0, H f,g

f  g is δH -differentiable at t. Then, δH (f  g)(t) = f (t)  δH g(t) + δH f (t)  g σ (t). Theorem 1.15 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,5 (t) ≥ 0, If,g

Ifα,5 σ ,δ g (t) ≥ 0, H

Iδα,5 (t) ≥ 0, H f,g

f ⊗ g is δH -differentiable at t. Then, δH (f ⊗ g)(t) = f σ (t) ⊗ δH g(t) + δH f (t) ⊗ g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) . Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

40

1 Calculus of Fuzzy Functions

 [δH g(t)]α = g αΔ (t), g αΔ (t) ,

t ∈ T,

α ∈ [0, 1].

α,5 α,5 (t) ≥ 0, Ifα,5 Because If,g σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we obtain H H

α,5 (t) If,g

   f α (t) f α (t)    = α   g (t) g α (t)  α

= f α (t)g α (t) − g α (t)f (t) ≥ 0,    f ασ (t) f ασ (t)    α,5 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,   αΔ  αΔ   f (t) f (t)  α,5 IδH f,g (t) =  α   g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f

αΔ

(t)

≥ 0. α,5 We will use that If,g (t) ≥ 0 to get

α  [(f ⊗ g)(t)]α = f (t)g α (t), f α (t)g α (t) . Since f ⊗ g is δH -differentiable at t, we obtain   Δ   α α Δ α α (t), f g (t) [δH (f ⊗ g)(t)] = f g 

ασ αΔ = f (t)g αΔ (t) + g α (t)f (t), f ασ (t)g αΔ (t) + g α (t)f ασ (t) . α

α,5 Since Ifα,5 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α ασ f ⊗ δH g (t) = f (t)g αΔ (t), g αΔ (t)f ασ (t) and

αΔ  [(δH f ⊗ g) (t)]α = f (t)g α (t), g α (t)f αΔ (t) . Hence,

(1.11)

1.1 First Type Fuzzy Delta Differentiation

41

  σ  α ασ αΔ α f ⊗ δH g (t) + [(δH f ⊗ g) (t)] = f (t)g αΔ (t) + f (t)g α (t),  g αΔ (t)f ασ (t) + g α (t)f αΔ (t) . From the last equality and from (1.11), we get the desired result. This completes the proof. Exercise 1.12 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,5 (t) ≥ 0, If,g

α,5 If,δ (t) ≥ 0, Hg

Iδα,5 σ (t) ≥ 0, H f,g

f ⊗ g is δH -differentiable at t. Then, δH (f ⊗ g)(t) = f (t) ⊗ δH g(t) + δH f (t) ⊗ g σ (t). Theorem 1.16 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,6 (t) ≥ 0, If,g

Ifα,6 σ ,δ g (t) ≥ 0, H

Iδα,6 (t) ≥ 0, H f,g

f g is δH -differentiable at t. Then, δH (f g)(t) = f σ (t) δH g(t) + δH f (t) g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) . Then,

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , t ∈ T,

α ∈ [0, 1].

α,6 α,6 (t) ≥ 0, Ifα,6 Because If,g σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, we obtain H H

α,6 (t) If,g

  α   α  f (t) f (t)  = α   g (t) g α (t)  α

= f α (t)g α (t) − g α (t)f (t) ≥ 0,

42

1 Calculus of Fuzzy Functions

Ifα,6 σ ,δ g (t) H

  ασ   ασ  f (t) f (t)  =  αΔ  αΔ  g (t) g (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,   αΔ   αΔ  f (t) f (t)  α,6 IδH f,g (t) =  α   g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f

αΔ

(t)

≥ 0. α,6 We will use that If,g (t) ≥ 0 to get

α  [(f g)(t)]α = f (t)g α (t), f α (t)g α (t) . Since f g is δH -differentiable at t, we obtain   Δ   α α Δ α α (t), f g (t) [δH (f g)(t)] = f g 

ασ αΔ = f (t)g αΔ (t) + g α (t)f (t), f ασ (t)g αΔ (t) + g α (t)f ασ (t) . α

(1.12)

α,6 Since Ifα,6 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, we have H H

  σ  α ασ f δH g (t) = f (t)g αΔ (t), g αΔ (t)f ασ (t) and

αΔ  [(δH f g) (t)]α = f (t)g α (t), g α (t)f αΔ (t) . Hence,   σ  α ασ αΔ α f δH g (t) + [(δH f g) (t)] = f (t)g αΔ (t) + f (t)g α (t), g

αΔ

(t)f

ασ

(t) + g (t)f α

αΔ

 (t) .

From the last equality and from (1.12), we get the desired result. This completes the proof.

1.1 First Type Fuzzy Delta Differentiation

43

Exercise 1.13 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,6 (t) ≥ 0, If,g

α,6 If,δ (t) ≥ 0, Hg

Iδα,6 σ (t) ≥ 0, H f,g

f g is δH -differentiable at t. Then, δH (f g)(t) = f (t) δH g(t) + δH f (t) g σ (t). Before formulating and proving the next properties of the δH -derivative, we have a need of the following useful lemmas. Lemma 1.1 Let a, b, c, d, w, r ∈ F (R) and c H d, w H d exist. Then, D (c H d, a + b) ≤ D (w H d, a) + D(c, w + r) + D(b, r). Proof We have D (c H d, a + b) = D (c H d + w, a + b + w) ≤ D (w H d, a) + D(c, b + w) = D (w H d, a) + D (c + r, b + w + r) ≤ D (w H d, a) + D(c, w + r) + D(r, b). This completes the proof. Lemma 1.2 Let a, b, c, d, w, r ∈ F (R) and c H d, c H w exist. Then, D (c H d, a + b) ≤ D (c H w, a) + D(w, d + r) + D(b, r). Proof We have D (c H d, a + b) ≤ D(c + w, a + b + d + w) ≤ D(c, w + a) + D(w, b + d) = D(c, w + a) + D(w + r, b + d + r) ≤ D (c H w, a) + D(w, d + r) + D(r, b). This completes the proof. Theorem 1.17 Let f : T → R be Δ-differentiable at t ∈ Tκ and g : T → F (R) be δH -differentiable at t. Let also, for any h > 0 sufficiently small there exist u1 (σ (t), h), u2 (σ (t), h) ∈ F (R) such that g(t + h) = g(σ (t)) + u1 (σ (t), h), g(σ (t)) = g(t − h) + u2 (σ (t), h).

44

1 Calculus of Fuzzy Functions

If f (t)f Δ (t) > 0, then δH (f · g) (t) = f Δ (t) · g(t) + f (σ (t)) · δH g(t). Proof Firstly, we note that   f (σ (t))f Δ (t) = f (t) + μ(t)f Δ (t) f Δ (t)  2 = f (t)f Δ (t) + μ(t) f Δ (t) > 0. Let  ∈ (0, 1). Define  −1   ∗ =  1 + |f (σ (t))| + g(t) + f Δ (t) . We have  ∗ ∈ (0, 1). Since f is Δ-differentiable at t and g is δH -differentiable at t, there exist neighborhoods U1 (δ, t) and U2 (δ, t) of t for some positive δ such that for all 0 < h < δ with t + h ∈ U1 (t, δ), we have   f (σ (t)) − f (t + h) − f Δ (t)(μ(t) − h) ≤  ∗ (h − μ(t))

(1.13)

and for all 0 < h < δ with t − h, t + h ∈ U2 (t, δ), we have D (g(t + h) H g(σ (t)), (h − μ(t)) · δH g(t)) ≤  ∗ (h − μ(t)), D (g(σ (t)) H g(t − h), (h + μ(t)) · δH g(t)) ≤  ∗ (h + μ(t)).

(1.14)

Since f is Δ-differentiable at t, then f is continuous at t. Thus, for h > 0 sufficiently small, f (σ (t)) and f (t + h) have the same sign and we have f (t + h) = f (σ (t)) + v1 (σ (t), h). Now, using that f (t)f Δ (t) > 0, we conclude that v1 (σ (t), h) has the same sign as f (σ (t)) and f (t + h) for h > 0 sufficiently small. From here, we get (f · g)(t + h) = f (σ (t)) · g(σ (t)) +f (σ (t)) · u1 (σ (t), h) +v1 (σ (t), h) · g(σ (t)) +(v1 · u1 )(σ (t), h), i.e., the H -difference (f (t + h) · g(t + h)) H (f (σ (t)) · g(σ (t)))

1.1 First Type Fuzzy Delta Differentiation

45

exists. Next, we have f (σ (t)) = f (t − h) + v2 (σ (t), h), where v2 (σ (t), h) has the same sign as f (σ (t)) and f (t − h) for h > 0 sufficiently small. Therefore, (f · g)(σ (t)) = f (t − h) · g(t − h) +f (σ (t)) · u2 (σ (t), h) +v2 (σ (t), h) · g(σ (t)) +(v2 · u2 )(σ (t), h), for h > 0 sufficiently small, and the H -difference (f (σ (t)) · g(σ (t))) H (f (t − h) · g(t − h)) exists for h > 0 sufficiently small. Because g is δH -differentiable at t, we have that g is continuous at t. Then there exists a neighborhood U3 (t, δ) of t such that for all 0 < h < δ with t − h, t + h ∈ U3 (t, δ), we have D(g(t + h), g(t)) ≤  ∗ ,

Let U (t, δ) = U1 (t, δ)



  g(t + h) = D g(t + h), 0   ≤ D g(t), 0 +1 = g(t) + 1,

(1.15)

g(t − h) = D(g(t − h), 0) ≤ D(g(t), 0) + 1 = g(t) + 1,

(1.16)

D(g(t − h), g(t)) ≤  ∗ .

(1.17)

U2 (t, δ)



U3 (t, δ). For

c = f (t + h) · g(t + h), d = f (σ (t)) · g(σ (t)), a = ((h − μ(t))f (σ (t))) · δH g(t),   b = f Δ (t)(h − μ(t)) · g(t), w = f (σ (t)) · g(t + h),   r = f Δ (t)(h − μ(t)) · g(t + h)

46

1 Calculus of Fuzzy Functions

and for t − h, t + h ∈ U (t, δ), we apply Lemma 1.1 and we get    D (f · g)(t + h) H (f · g)(σ (t)), (h − μ(t)) · f (σ (t)) · δH g(t) + f Δ (t) · g(t) ≤ D (f(σ (t)) · (g(t + h) H g(σ (t))) , ((h − μ(t))f  (σ (t))) · δH g(t)) 

 +D f (t + h) · g(t + h), f (σ (t)) · g(t + h) + (h − μ(t))f Δ (t) · g(t + h)      +D (h − μ(t))f Δ (t) · g(t), (h − μ(t))f Δ (t) · g(t + h) . (1.18)

Since h − μ(t) > 0, f (t)f Δ (t) > 0, we have that (h − μ(t))f Δ (t) has the same sign as f (σ (t)). Therefore,   f (σ (t)) · g(t + h) + (h − μ(t))f Δ (t) · g(t + h)   = f (σ (t)) + f Δ (t)(h − μ(t)) · g(t + h). Thus,     D f (t + h) · g(t + h), f (σ (t)) · g(t + h) + (h − μ(t))f Δ (t) · g(t + h) = f (t + h) − f (σ (t)) − f Δ (t)(h − μ(t)) g(t + h). (1.19) Hence by (1.13)–(1.19), we arrive at D ((f · g)(t + h) H (f · g)(σ (t)), (h − μ(t))   · f (σ (t)) · δH g(t) + f Δ (t) · g(t) ≤  ∗ |f (σ (t))|(h − μ(t)) −μ(t)) + ∗ g(t + h)(h  +(h − μ(t)) f Δ (t) D (g(t + h), g(t))    ≤  ∗ (h − μ(t)) |f (σ (t))| + 1 + g(t) + f Δ (t) ≤ (h − μ(t)).

(1.20)

Next, there exists a neighborhood U4 (t, δ) of t such that for all 0 < h < δ with t − h ∈ U4 (t, δ), we have |f (σ (t)) − f (t − h) − f Δ (t)(h + μ(t))| ≤  ∗ (h + μ(t)). For d = f (t − h) · g(t − h), c = f (σ (t)) · g(σ (t)), a = ((h + μ(t))f (σ (t))) · δH g(t),   b = f Δ (t)(h + μ(t)) · g(t), w = f (σ (t)) · g(t − h),   r = f Δ (t)(h + μ(t)) · g(t − h),

(1.21)

1.1 First Type Fuzzy Delta Differentiation

47

we apply Lemma 1.2 and we get    D (f · g)(σ (t)) H (f · g)(t−h), (h + μ(t)) · f (σ (t)) · δH g(t)+f Δ (t) · g(t) ≤ D ((f (σ (t))·g(σ (t))) H (f (σ (t))·g(t−h)) , ((h+μ(t))f (σ (t))) ·δH g(t))   +D f (σ (t)) · g(t−h), f (t−h) · g(t − h)+f Δ (t)(h+μ(t)) · g(t − h)      +D (h + μ(t))f Δ (t) · g(t), (h + μ(t))f Δ (t) · g(t − h) . (1.22) Since h + μ(t) > 0 and f (t)f Δ (t) > 0, we have that f Δ (t)(h + μ(t)) has the same sign as f (σ (t)). Also, we have   f (t − h) · g(t − h) + (h + μ(t))f Δ (t) · g(t − h)   = f (t − h) + f Δ (t)(h + μ(t)) · g(t − h). Therefore,     D f (σ (t)) · g(t − h), f (t − h) · g(t − h) + (h + μ(t))f Δ (t) · g(t − h) = f (σ (t)) − f (t − h) − f Δ (t)(h + μ(t)) g(t − h). (1.23) Hence, using (1.14), (1.21), (1.22), (1.23), (1.16), and (1.17), we obtain D ((f · g)(σ (t)) H (f · g)(t − h), f (σ (t)) · δH g(t)    + (h + μ(t))f Δ (t) · g(t) ≤  ∗ (h + μ(t))|f (σ (t))| + ∗ g(t − h)(h + μ(t)) +(h + μ(t))|f Δ (t)|D(g(t − h), g(t))   ≤  ∗ (h + μ(t)) |f (σ (t))| + 1 + g(t) + |f Δ (t)| ≤ (h + μ(t)). By (1.20) and (1.24), we get the desired result. This completes the proof. Example 1.15 Let T = 2N0 and f (t) = t 2 + t + 1,   g(t) = t, t + 1, t 2 + 1 , Here σ (t) = 2t, μ(t) = t,

t ∈ T.

t ∈ T.

(1.24)

48

1 Calculus of Fuzzy Functions

Then,   f (t) · g(t) = t (t 2 + t + 1), (t + 1)(t 2 + t + 1), (t 2 + 1)(t 2 + t + 1)   = t 3 + t 2 + t, t 3 + t 2 + t + t 2 + t + 1, t 4 + t 3 + t 2 + t 2 + t + 1   = t 3 + t 2 + t, t 3 + 2t 2 + 2t + 1, t 4 + t 3 + 2t 2 + t + 1 , t ∈ T. Let h1 (t) = t 4 + t 3 + 2t 2 + t + 1, h2 (t) = t 3 + t 2 + t, h3 (t) = t 3 + 2t 2 + 2t + 1,

t ∈ T.

Then, 3 2 2 3 hΔ 1 (t) = (σ (t)) + t (σ (t)) + t σ (t) + t

+(σ (t))2 + tσ (t) + t 2 + 2(σ (t) + t) + 1 = 8t 3 + 4t 3 + 2t 3 + t 3 + 4t 2 + 2t 2 + t 2 + 6t + 1 = 15t 3 + 7t 2 + 6t + 1, 2 2 hΔ 2 (t) = (σ (t)) + tσ (t) + t + σ (t) + t + 1

= 4t 2 + 2t 2 + t 2 + 2t + t + 1 = 7t 2 + 3t + 1, 2 2 hΔ 3 (t) = (σ (t)) + tσ (t) + t + 2(σ (t) + t) + 2

= 4t 2 + 2t 2 + t 2 + 6t + 2 = 7t 2 + 6t + 2,

t ∈ T.

Hence,   Δ Δ δH (f · g)(t) = hΔ 2 (t), h3 (t), h1 (t)   = 7t 2 + 3t + 1, 7t 2 + 6t + 2, 15t 3 + 7t 2 + 6t + 1 , Next, f Δ (t) = σ (t) + t + 1 = 3t + 1,

t ∈ T.

1.1 First Type Fuzzy Delta Differentiation

49

  f Δ (t) · g(t) = (3t + 1)t, (3t + 1)(t + 1), (3t + 1)(t 2 + 1)   = 3t 2 + t, 3t 2 + 3t + t + 1, 3t 3 + 3t + t 2 + 1   = 3t 2 + t, 3t 2 + 4t + 1, 3t 3 + t 2 + 3t + 1 , t ∈ T. Let g1 (t) = t 2 + 1, g2 (t) = t, g3 (t) = t + 1,

t ∈ T.

Then, g1Δ (t) = σ (t) + t = 3t, g2Δ (t) = 1, g3Δ (t) = 1,   δH g(t) = g2Δ (t), g3Δ (t), g1Δ (t) = (1, 1, 3t),

t ∈ T.

Also, f (σ (t)) = (σ (t))2 + σ (t) + 1 = 4t 2 + 2t + 1,

t ∈ T.

We get   f (σ (t)) · δH g(t) = 4t 2 + 2t + 1, 4t 2 + 2t + 1, 3t (4t 2 + 2t + 1)   = 4t 2 + 2t + 1, 4t 2 + 2t + 1, 12t 3 + 6t 2 + 3t ,

t ∈ T,

and f Δ (t) · g(t) + f (σ (t)) · δH g(t)   = 3t 2 + t, 3t 2 + 4t + 1, 3t 3 + t 2 + 3t + 1   + 4t 2 + 2t + 1, 4t 2 + 2t + 1, 12t 3 + 6t 2 + 3t   = 7t 2 + 3t + 1, 7t 2 + 6t + 2, 15t 3 + 7t 2 + 6t + 1 ,

t ∈ T.

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1 Calculus of Fuzzy Functions

Therefore, δH (f · g)(t) = f Δ (t) · g(t) + f (σ (t)) · δH g(t).

Exercise 1.14 Let T = 2 +



1 {2}, 4k k∈N

f (t) = t 3 + 3t 2 + 4t + 7,   t +1 2 2 3 g(t) = ,t ,t + t , t +2

t ∈ T.

Prove that δH (f · g)(2) = f Δ (2) · g(2) + f (σ (2)) · δH g(2). In fact, Theorem 1.17 is equivalent to the following theorem. Theorem 1.18 Let f : T → R be Δ-differentiable at t ∈ Tκ with f (t)f Δ (t) > 0. Let also, g : T → F (R) be δH -differentiable at t with α-levels

 [g(t)]α = g α (t), g α (t) ,

α ∈ [0, 1].

Then f · g : T → F (R) is δH -differentiable at t and  α [δH (f · g)(t)]α = f Δ (t) · g(t) + [f (σ (t)) · δH g(t)]α ,

α ∈ [0, 1].

Proof 1. Let f (t) > 0, f Δ (t) > 0. Then, [f (t) · g(t)]α = f (t) · [g(t)]α

 = f (t) · g α (t), g α (t) 

= f (t)g α (t), f (t)g α (t) ,

 [δH (f · g)(t)]α = (f g α )Δ (t), (f g α )Δ (t)  = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

αΔ

 (t) ,

(1.25)

1.1 First Type Fuzzy Delta Differentiation



α

f Δ (t) · g(t)

51

= f Δ (t) · [g(t)]α

 = f Δ (t) · g α (t), g α (t) 

= f Δ (t)g α (t), f Δ (t)g α (t) ,

[f (σ (t)) · δH g(t)]α = f (σ (t)) · [δH g(t)]α

 = f (σ (t)) · g αΔ (t), g αΔ (t) 

= f (σ (t))g αΔ (t), f (σ (t))g αΔ (t) . Hence,  α  Δ α f (t) · g(t) + [f (σ (t)) · δH g(t)] = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

αΔ

Therefore, we get (1.25). 2. Suppose that f (t) < 0, f Δ (t) < 0. Then f (σ (t)) < 0 and [f (t) · g(t)]α = f (t) · [g(t)]α

 = f (t) · g α (t), g α (t) 

= f (t)g α (t), f (t)g α (t) ,

 [δH (f · g)(t)]α = (f g α )Δ (t), (f g α )Δ (t)  = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ



α

f Δ (t) · g(t)

α

αΔ

 (t) ,

= f Δ (t) · [g(t)]α

 = f Δ (t) · g α (t), g α (t) 

= f Δ (t)g α (t), f Δ (t)g α (t) ,

[f (σ (t)) · δH g(t)]α = f (σ (t)) · [δH g(t)]α

 = f (σ (t)) · g αΔ (t), g αΔ (t) 

= f (σ (t))g αΔ (t), f (σ (t))g αΔ (t) .

 (t) .

52

1 Calculus of Fuzzy Functions

From here,   Δ α f (t) · g(t) + [f (σ (t)) · δH g(t)]α = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

Consequently, we obtain (1.25). This completes the proof. Example 1.16 Let T = 2N0 , f : T → R be given by f (t) =

t +1 , t +2

t ∈ T,

and g : T → F (R) be with α-levels

 [g(t)]α = t + 2, t 2 + t + 2 ,

t ∈ T,

α ∈ [0, 1].

We have f (t) > 0, σ (t) = 2t, f Δ (t) = =

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

> 0, f (σ (t)) =

σ (t) + 1 σ (t) + 2

=

2t + 1 , 2(t + 1)

g α (t) = t + 2, g α (t) = t 2 + t + 2, g αΔ (t) = 1, g αΔ (t) = σ (t) + t + 1 = 3t + 1, [δH g(t)] = [1, 3t + 1] , α

t ∈ T,

α ∈ [0, 1].

αΔ

 (t) .

1.1 First Type Fuzzy Delta Differentiation

53

Therefore, 

α f Δ (t) · g(t) = f Δ (t) · [g(t)]α 

1 · t + 2, t 2 + t + 2 2(t + 1)(t + 2)   t2 + t + 2 1 , , = 2(t + 1) 2(t + 1)(t + 2)

=

[f (σ (t)) · δH g(t)]α = f (σ (t)) · [δH g(t)]α 2t + 1 · [1, 3t + 1] 2(t + 1)   2t + 1 (2t + 1)(3t + 1) = , 2(t + 1) 2(t + 1)   2 2t + 1 6t + 5t + 1 , , t ∈ T, = 2(t + 1) 2(t + 1)

=

α ∈ [0, 1].

Consequently,  α [δH (f · g)(t)]α = f Δ (t) · g(t) + [f (σ (t)) · δH g(t)]α   t2 + t + 2 1 , = 2(t + 1) 2(t + 1)(t + 2)   2t + 1 6t 2 + 5t + 1 , + 2(t + 1) 2(t + 1)   2 t +t +2 6t 2 + 5t + 1 2t + 2 , + = 2(t + 1) 2(t + 1)(t + 2) 2(t + 1)   2 2 t + t + 2 + (6t + 5t + 1)(t + 2) = 1, 2(t + 1)(t + 2)   2 t + t + 2 + 6t 3 + 12t 2 + 5t 2 + 10t + t + 2 = 1, 2(t + 1)(t + 2)   3 2 6t + 18t + 12t + 4 = 1, 2(t + 1)(t + 2)   3 3t + 9t 2 + 6t + 2 , t ∈ T, α ∈ [0, 1]. = 1, (t + 1)(t + 2) Exercise 1.15 Let T = 3N0 , f : T → R be given by f (t) =

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

t ∈ T,

54

1 Calculus of Fuzzy Functions

and g : T → F (R) be with α-levels  [g(t)]α = t + 2 +

 1 2 , t2 + t + 2 + , t +1 t +1

t ∈ T,

α ∈ [0, 1].

Find [δH (f · g)(t)]α ,

t ∈ T,

α ∈ [0, 1].

Theorem 1.19 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (σ (t))f Δ (t) < 0, g : T → F (R) be δH -differentiable at t and the H -differences (f (t + h) · g(t + h)) H (f (σ (t)) · g(σ (t))) and (f (σ (t)) · g(σ (t))) H (f (t − h) · g(t − h)) exist. Let also, for any h > 0 sufficiently small, there exist u1 (σ (t), h), u2 (σ (t), h) ∈ F (R) such that g(t + h) = g(σ (t)) + u1 (σ (t), h), g(σ (t)) = g(t − h) + u2 (σ (t), h). Then, δH (f · g)(t) = (f (σ (t)) · δH g(t)) H

 Δ   −f (t) · g(t) .

Proof Firstly, note that   f (σ (t))f Δ (t) = f (t) + μ(t)f Δ (t) f Δ (t)  2 = f (t)f Δ (t) + μ(t) f Δ (t) < 0. Therefore, f (t)f Δ (t) < 0. Without loss of generality, suppose that f (t) > 0, f (σ (t)) > 0, and f Δ (t) < 0. Take  ∈ (0, 1) and  −1   ∗ =  1 + |f (σ (t))| + g(t) + f Δ (t) .

1.1 First Type Fuzzy Delta Differentiation

55

Note that  ∗ ∈ (0, 1). Because f is Δ-differentiable at t and g is δH -differentiable at t, there exist neighborhoods U1 (t, δ) and U2 (t, δ) of t for some positive δ such that for all 0 < h < δ with t + h ∈ U1 (t, δ), we have   f (σ (t)) − f (t + h) − f Δ (t)(μ(t) − h) ≤  ∗ (h − μ(t)),

(1.26)

and for all 0 < h < δ with t − h, t + h ∈ U2 (t, δ), we have D (g(t + h) H g(σ (t)), (h − μ(t)) · δH g(t)) ≤  ∗ (h − μ(t)), D (g(σ (t)) H g(t − h), (h + μ(t)) · δH g(t)) ≤  ∗ (h + μ(t)).

(1.27)

Because g is δH -differentiable at t, we have that g is continuous at t. Then there exists a neighborhood U3 (t, δ) of t such that for all 0 < h < δ with t − h, t + h ∈ U3 (t, δ), we have D(g(t + h), g(t)) ≤  ∗ ,   g(t + h) = D g(t + h), 0   ≤ D g(t), 0 +1 = g(t) + 1, and D (g(t − h), g(t)) ≤  ∗ ,   g(t − h) = D g(t − h), 0   ≤ D g(t), 0 +1 = g(t) + 1. Let U (t, δ) = U1 (t, δ)



U2 (t, δ)



U3 (t, δ). For

c = f (t + h) · g(t + h), d = f (σ (t)) · g(σ (t)), a = ((h − μ(t))f (σ (t))) · δH g(t),   b = f Δ (t)(h − μ(t)) · g(t), w = f (σ (t)) · g(t + h),   r = f Δ (t)(h − μ(t)) · g(t + h). and for t − h, t + h ∈ U (t, δ) we apply Lemma 1.1 and we get

56

1 Calculus of Fuzzy Functions     D (f · g)(t + h) H (f · g)(σ (t)), (h − μ(t)) · f (σ (t)) · δH g(t) H (−f Δ (t)) · g(t) ≤ D (f (σ (t)) · (g(t + h) H g(σ (t))) , ((h − μ(t))f (σ (t))) · δH g(t))   +D f (t + h) · g(t + h), f (σ (t)) · g(t + h) + ((h − μ(t))f Δ (t)) · g(t + h)   +D ((h − μ(t))f Δ (t)) · g(t), ((h − μ(t))f Δ (t)) · g(t + h) ≤  ∗ (h − μ(t))|f (σ (t))| +  ∗ (h − μ(t))g(t + h) + ∗ (h − μ(t))|f Δ (t)|   =  ∗ (h − μ(t)) 1 + |f (σ (t))| + |f Δ (t)| + g(t) = (h − μ(t)).

Next, there exists a neighborhood U4 (t, δ) of t such that for all 0 < h < δ with t − h ∈ U4 (t, δ), we have |f (σ (t)) − f (t − h) − f Δ (t)(h + μ(t))| ≤  ∗ (h + μ(t)). For d = f (t − h) · g(t − h), c = f (σ (t)) · g(σ (t)), a = (h + μ(t))f (σ (t)) · δH g(t), b = (f Δ (t)(h + μ(t))) · g(t), w = f (σ (t)) · g(t − h), r = (f Δ (t)(h + μ(t))) · g(t − h), we apply Lemma 1.2 and we get    D (f · g)(σ (t)) H (f · g)(t − h), (h + μ(t)) · f (σ (t)) · δH g(t) H ((−f Δ (t)) · g(t)) ≤ D ((f (σ (t)) · g(σ (t))) H (f (σ (t)) · g(t − h)) , (h + μ(t))f (σ (t)) · δH g(t))   +D f (σ (t)) · g(t − h), f (t − h) · g(t − h) + (f Δ (t)(h + μ(t))) · g(t − h)   +D ((h + μ(t))f Δ (t)) · g(t), ((h + μ(t))f Δ (t)) · g(t − h) ≤ |f (σ (t))| ∗ (h + μ(t)) +  ∗ (h + μ(t))g(t − h) + ∗ (h + μ(t))|f Δ (t)|   =  ∗ 1 + |f Δ (t)| + |f (σ (t))| + g(t) (h + μ(t)) = (h + μ(t)).

This completes the proof. Example 1.17 Let T = N and f (t) = g(t) =

1 , t 

 t 2 , t, t , t +1

t ∈ N.

1.1 First Type Fuzzy Delta Differentiation

57

We have σ (t) = t + 1, μ(t) = 1,

t ∈ T.

Let t , t +1 g2 (t) = t,

g1 (t) =

g3 (t) = t 2 ,

t ∈ T.

Then, g1Δ (t) = =

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

g2Δ (t) = 1, g3Δ (t) = σ (t) + t = t +1+t = 2t + 1, f Δ (t) = − =− f (σ (t)) = =

1 tσ (t) 1 , t (t + 1)

1 σ (t) 1 , t +1

t ∈ T.

From here,   δH g(t) = g1Δ (t), g2Δ (t), g3Δ (t)   1 = , 1, 2t + 1 , (t + 1)(t + 2)

58

1 Calculus of Fuzzy Functions

 2t + 1 1 1 , , , f (σ (t)) · δH g(t) = (t + 1)2 (t + 2) t + 1 t + 1    Δ  t 1 1 , , t ∈ T, −f (t) · g(t) = , (t + 1)2 t + 1 t + 1 

and    (f (σ (t)) · δH g(t)) H −f Δ (t) · g(t)   1 t 1 2t + 1 = − − , 0, t +1 t +1 (t + 1)2 (t + 2) (t + 1)2   1 , 0, 1 , t ∈ T. = − (t + 1)(t + 2) Next,  (f · g)(t) =

 1 , 1, t , t +1

t ∈ T.

Let 1 , t +1 h2 (t) = 1,

h1 (t) =

h3 (t) = t,

t ∈ T.

Then, hΔ 1 (t) = − =−

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

hΔ 2 (t) = 0, hΔ 3 (t) = 1,

t ∈ T.

Therefore,   Δ Δ δH (f · g)(t) = hΔ 1 (t), h2 (t), h3 (t)   1 = − , 0, 1 , (t + 1)(t + 2)

t ∈ T.

1.1 First Type Fuzzy Delta Differentiation

59

Consequently, δH (f · g)(t) = (f (σ (t)) · δH g(t)) H

 Δ   −f (t) · g(t) .

Exercise 1.16 Let T = 3N0 , 1 , t +1   g(t) = t 2 , t 4 , t 4 + 1 ,

f (t) =

t ∈ T.

Prove that δH (f · g)(t) = (f (σ (t)) · δH g(t)) H

 Δ   −f (t) · g(t) ,

t ∈ T.

Exercise 1.17 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (σ (t))f Δ (t) < 0, g : T → F (R) be δH -differentiable at t and the H -differences (f (t + h) · g(t + h)) H (f (σ (t)) · g(σ (t))) and (f (σ (t)) · g(σ (t))) H (f (t − h) · g(t − h)) exist. Let also, for any h > 0 sufficiently small, there exist u1 (σ (t), h), u2 (σ (t), h) ∈ F (R) such that g(t + h) = g(σ (t)) + u1 (σ (t), h), g(σ (t)) = g(t − h) + u2 (σ (t), h). Prove that   δH (f · g)(t) = f Δ (t) · δH g(t) H ((−f (σ (t))) · g(t)) . Theorem 1.20 Let f : T → R be Δ-differentiable at t ∈ Tκ with f (σ (t))f Δ (t) < 0. Let also, g : T → F (R) be δH -differentiable at t with α-levels

 [g(t)]α = g α (t), g α (t) ,

α ∈ [0, 1].

Then f · g : T → F (R) is δH -differentiable at t and  α [δH (f · g)(t)]α = f Δ (t) · g(t) H [(−f (σ (t))) · δH g(t)]α ,

α ∈ [0, 1]. (1.28)

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1 Calculus of Fuzzy Functions

Proof 1. Assume f (σ (t)) > 0, f Δ (t) < 0. Then f (t) > 0 and [f (t) · g(t)]α = f (t) · [g(t)]α

 = f (t) · g α (t), g α (t) 

= f (t)g α (t), f (t)g α (t) ,

 [δH (f · g)(t)]α = (f g α )Δ (t), (f g α )Δ (t)  = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

αΔ

 (t) ,

 Δ α f (t) · g(t) = f Δ (t) · [g(t)]α

 = f Δ (t) · g α (t), g α (t) 

= f Δ (t)g α (t), f Δ (t)g α (t) , [(−f (σ (t))) · δH g(t)]α = (−f (σ (t))) · [δH g(t)]α

 = (−f (σ (t))) · g αΔ (t), g αΔ (t) 

= −f (σ (t))g αΔ (t), −f (σ (t))g αΔ (t) . From here,   Δ α f (t) · g(t) H [(−f (σ (t))) · δH g(t)]α = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

Therefore, we get (1.28). 2. Let f (σ (t)) < 0, f Δ (t) > 0. Then f (t) < 0 and [f (t) · g(t)]α = f (t) · [g(t)]α

 = f (t) · g α (t), g α (t) 

= f (t)g α (t), f (t)g α (t) ,

αΔ

 (t) .

1.1 First Type Fuzzy Delta Differentiation

61

 [δH (f · g)(t)]α = (f g α )Δ (t), (f g α )Δ (t)  = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

αΔ

 (t) ,

 Δ α f (t) · g(t) = f Δ (t) · [g(t)]α

 = f Δ (t) · g α (t), g α (t) 

= f Δ (t)g α (t), f Δ (t)g α (t) , [(−f (σ (t))) · δH g(t)]α = (−f (σ (t))) · [δH g(t)]α

 = (−f (σ (t))) · g αΔ (t), g αΔ (t) 

= −f (σ (t))g αΔ (t), −f (σ (t))g αΔ (t) . Hence,   Δ α f (t) · g(t) H [(−f (σ (t))) · δH g(t)]α = f Δ (t)g α (t) + f (σ (t))g αΔ (t), f (t)g (t) + f (σ (t))g Δ

α

Thus, we obtain (1.28). This completes the proof. Example 1.18 Let T = 3N0 , f : T → R be given by f (t) =

1 , t

t ∈ T,

and g : T → F (R) be with α-levels α

[g(t)]α = t, t + t 2 , We have σ (t) = 3t, f (t) > 0, f (σ (t)) =

1 σ (t)

t ∈ T,

α ∈ [0, 1].

αΔ

 (t) .

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1 Calculus of Fuzzy Functions

1 3t > 0, =

f Δ (t) = − =−

1 tσ (t) 1 3t 2

< 0, g (t) = t, α

g α (t) = t + t 2 , g αΔ (t) = 1, g αΔ (t) = 1 + σ (t) + t = 1 + 4t,

t ∈ T,

α ∈ [0, 1].

Then,  Δ α f (t) · g(t) = f Δ (t) · [g(t)]α 

  1 = − 2 · t, t + t 2 3t   1+t 1 , = − ,− 3t 3t [(−f (σ (t))) · δH g(t)]α = (−f (σ (t))) · [δH g(t)]α   1 · [1, 1 + 4t] = − 3t   1 1 + 4t ,− , t ∈ T, = − 3t 3t

α ∈ [0, 1].

Consequently,   1 1+t ,− [δH (f · g)(t)] = − 3t 3t   1 + 4t 1 ,− H − 3t 3t   1 1+t 1 + 4t 1 + = − + ,− 3t 3t 3t 3t α

= [0, 1],

t ∈ T,

α ∈ [0, 1].

1.2 Second Type Fuzzy Delta Differentiation

63

Exercise 1.18 Let T = 2N0 , f : T → R be given by f (t) =

t2

t , +t +1

t ∈ T,

and g : T → F (R) be with α-levels 

1 [g(t)] = t + t , 1 + 2t + 3t + t +1 α

2

2

α ,

t ∈ T,

α ∈ [0, 1].

Find [δH (f · g)(t)]α ,

t ∈ T,

α ∈ [0, 1].

1.2 Second Type Fuzzy Delta Differentiation Definition 1.6 Assume that f : T → F (R) is a fuzzy function and t ∈ Tκ . Then f is said to be second type right fuzzy delta differentiable at t, shortly right δSH + differentiable at t, if there exists an element δSH f (t) ∈ F (R) with the property that for any given  > 0 there exists a neighborhood UT of t, i.e., UT = (t −δ, t +δ) T for some δ > 0 such that for all t + h ∈ UT the H -difference f (σ (t)) H f (t + h) exists and   + D f (σ (t)) H f (t + h), δSH f (t)(−(h − μ(t))) ≤ |h − μ(t)| + with 0 ≤ h < δ. In this case, δSH f (t) is said to be second type right fuzzy delta derivative of f at t, shortly right δSH -derivative of f at t.

Definition 1.7 Assume that f : T → F (R) is a fuzzy function and t ∈ Tκ . Then f is said to be second type left fuzzy delta differentiable at t, shortly left δSH − differentiable at t, if there exists an element δSH f (t) ∈ F (R) with the property that for any given  > 0 there exists a neighborhood UT of t, i.e., UT = (t −δ, t +δ) T for some δ > 0 such that for all t − h ∈ UT the H -difference f (t − h) H f (σ (t)) exists and   − f (t)(−(h + μ(t))) ≤ (h + μ(t)) D f (t − h) H f (σ (t)), δSH with 0 ≤ h < δ. Definition 1.8 Let f : T → F (R) be a fuzzy function and t ∈ Tκ . Then f is said to be second type fuzzy delta differentiable at t, shortly δSH -differentiable at t, if f is both second type left and right fuzzy delta differentiable at t ∈ Tκ and − + f (t) = δSH f (t), and we will denote it by δSH f (t). We call δSH f (t) the second δSH

64

1 Calculus of Fuzzy Functions

type fuzzy delta derivative of f at t, shortly δSH -derivative of f at t. We say that f is second type fuzzy delta differentiable at t, shortly δSH -differentiable at t, if its δSH -derivative exists at t. We say that f is second type fuzzy delta differentiable on Tκ , shortly δSH -differentiable on Tκ , if its δSH -derivative exists at each t ∈ Tκ . The fuzzy function δSH f : Tκ → F (R) is then called second type fuzzy delta derivative, shortly δSH -derivative of f on Tκ . The proofs of the following properties of the δSH -derivative are similar to the proofs of the corresponding properties of the δH -derivative. Therefore, we leave them to the reader as exercises. Theorem 1.21 If the δSH -derivative of f at t ∈ Tκ exists, then it is unique. Therefore, δSH -derivative is well-defined. Theorem 1.22 Assume that f : T → F (R) is a continuous function at t1 ∈ Tκ and t1 is right-scattered. Then f is δSH -differentiable at t1 and δSH f (t1 ) =

f (t1 ) H f (σ (t1 )) . −μ(t1 )

Example 1.19 Let T = 2N0 ,  f (t) =

 1 1 1 , , , t +2 t +1 t

t ∈ T.

Here σ (t) = 2t, μ(t) = t,  f (σ (t)) =

 1 1 1 , , σ (t) + 2 σ (t) + 1 σ (t)   1 1 1 , , , t ∈ T. = 2t + 2 2t + 1 2t

Then,  1 1 1 1 1 1 − , − , − t + 2 2t + 2 t + 1 2t + 1 t 2t   2t + 2 − t − 2 2t + 1 − t − 1 1 , , = 2(t + 1)(t + 2) (t + 1)(2t + 1) 2t   t 1 t , , , t ∈ T, = 2(t + 1)(t + 2) (t + 1)(2t + 1) 2t 

f (t) H f (σ (t)) =

1.2 Second Type Fuzzy Delta Differentiation

and

65

    t t 1 1 · , , δSH f (t) = − t 2(t + 1)(t + 2) (t + 1)(2t + 1) 2t   1 1 1 ,− , t ∈ T. = − 2,− (t + 1)(2t + 1) 2(t + 1)(t + 2) 2t

Exercise 1.19 Let T = 2Z,  f (t) =

 1 1 1 , , , t2 + 3 t2 + 2 t2 + 1

t ∈ T.

Find δSH f (t), t ∈ T. Theorem 1.23 Assume that f : T → F (R) is δSH -differentiable at t ∈ Tκ . Then f is continuous at t. Theorem 1.24 Let f : T → F (R) be a function and let t ∈ Tκ be right-dense. Then f is δSH -differentiable at t if and only if the limits f (t) H f (t + h) h→0+ −h lim

and

f (t − h) H f (t) h→0+ −h lim

exist and satisfy the relations lim

h→0+

f (t) H f (t + h) f (t − h) H f (t) = lim = δSH f (t). h→0+ −h −h

Example 1.20 Let T = {1}



f (t) =

1+



1 , 12k k∈N

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

t ∈ T.

Note that t = 1 is right-dense. Take h > 0 arbitrarily. Then,   1 1 1 , f (1) = , , 5 4 3   1 1 1 , , , f (1 + h) = 5+h 4+h 3+h   1 1 1 , , , f (1 − h) = 5−h 4−h 3−h   1 1 1 1 1 1 − , − , − f (1) H f (1 + h) = 5 5+h 4 4+h 3 3+h   h h h , , , = 5(5 + h) 4(4 + h) 3(3 + h)

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1 Calculus of Fuzzy Functions

 1 1 1 1 1 1 − , − , − f (1 − h) H f (1) = 5−h 5 4−h 4 3−h 3   h h h , , , = 5(5 − h) 4(4 − h) 3(3 − h)   1 1 1 f (1) H f (1 + h) = − ,− ,− , −h 3(3 + h) 4(4 + h) 5(5 + h)   1 1 1 f (1 − h) H f (1) = − ,− ,− . −h 3(3 − h) 4(4 − h) 5(5 − h) 

Hence,   1 1 1 f (1) H f (1 + h) = − ,− ,− , h→0 −h 9 16 25   1 1 1 f (1 − h) H f (1) = − ,− ,− . lim h→0 −h 9 16 25 lim

Therefore, f (1) H f (1 + h) −h f (1 − h) H f (1) = lim h→0 −h   1 1 1 . = − ,− ,− 9 16 25

δSH f (1) = lim

h→0

Exercise 1.20 Let T = {0}

1 

 f (t) =

3k k∈N

,

 1 1 1 , , , t2 + 4 t2 + 2 t2 + 1

t ∈ T.

Find δSH f (0),

δSH f

  1 , 3

δSH f

  1 . 9

Theorem 1.25 Let f : T → F (R) is δSH -differentiable at t ∈ Tκ . Then, f (σ (t)) = f (t) + (−μ(t)) · δSH f (t) or f (t) = f (σ (t)) + (−μ(t)) · δSH f (t),

t ∈ Tκ .

1.2 Second Type Fuzzy Delta Differentiation

67

Theorem 1.26 Let f, g : T → F (R) be δSH -differentiable at t ∈ Tκ . Then f + g : T → F (R) is δSH -differentiable at t ∈ Tκ and δSH (f + g)(t) = δSH f (t) + δSH g(t). Example 1.21 Let T = 2Z,  f (t) =  g(t) =

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

 1 1 1 , , , 2t + 5 2t + 3 2t + 1

t ∈ T.

We have σ (t) = t + 2, μ(t) = 2,

t ∈ T,

and  f (0) =

 1 1 , ,1 , 3 2

f (σ (0)) = f (2)   1 1 1 = , , , 5 4 3   1 1 , ,1 , g(0) = 5 3 g(σ (0)) = g(2)   1 1 1 = , , , 9 7 5  1 1 1 1 − , − ,1 − f (0) H f (σ (0)) = 3 5 2 4   2 1 2 , , , = 15 4 3  1 1 1 1 − , − ,1 − g(0) H g(σ (0)) = 5 9 3 7   4 4 4 , , , = 45 21 5 δSH f (0) =

f (0) H f (σ (0)) −2

1 3

1 5





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1 Calculus of Fuzzy Functions

  1 1 1 , = − ,− ,− 3 8 15 g(0) H g(σ (0)) −2   2 2 2 , = − ,− ,− 5 21 45   1 2 1 2 1 2 δSH f (0) + δSH g(0) = − − , − − , − − 3 5 8 21 15 45   37 1 11 ,− , = − ,− 15 168 9   1 1 1 1 + , + ,1 + 1 (f + g)(0) = 3 5 2 3   8 5 , ,2 , = 15 6   1 1 1 1 1 1 + , + , + (f + g)(σ (0)) = 5 9 4 7 3 5   14 11 8 , , , = 45 28 15   14 5 11 8 8 − , − ,2 − (f + g)(0) H (f + g)(σ (0)) = 15 45 6 28 15   24 − 14 70 − 33 30 − 8 , , = 45 84 15   2 37 22 , , , = 9 84 15 δSH g(0) =

(f + g)(0) H (f + g)(σ (0)) −2   11 37 1 = − ,− ,− . 15 168 9

δSH (f + g)(0) =

Consequently, δSH (f + g)(0) = δSH f (0) + δSH g(0). Exercise 1.21 Let T = {0}

1  2k k∈N

 f (t) =

,

 1 1 1 , , , t +8 t +6 t +4

1.2 Second Type Fuzzy Delta Differentiation

 g(t) =

69

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

t ∈ T.

Prove that δSH (f + g)(0) = δSH f (0) + δSH g(0). Theorem 1.27 Let f : T → F (R) be δSH -differentiable at t ∈ Tκ . Then for any λ ∈ R the function λ · f : T → F (R) is δSH -differentiable at t ∈ Tκ and δSH (λ · f )(t) = λ · δSH f (t). Theorem 1.28 Let t ∈ Tκ , f : T → F (R) and fα (t) = [f (t)]α , t ∈ T, α ∈ [0, 1]. If f is δSH -differentiable at t, then fα is δSH -differentiable at t and [δSH f (t)]α = δSH fα (t),

t ∈ T,

α ∈ [0, 1].

Theorem 1.29 Let t ∈ Tκ , f : T → F (R) is δSH -differentiable at t. Let also,

 α [f (t)]α = f α (t), f (t) ,

α ∈ [0, 1].

α

Then f α and f are Δ-differentiable at t and

αΔ  [δSH f (t)]α = f (t), f αΔ (t) , Example 1.22 Let T =



1 , 4k k∈N

α ∈ [0, 1].

f : T → F (R) and

[f (t)]α = [e−1 (t, 1) + t − 10, e−1 (t, 1) − t + 10] ,

t ∈ T.

Set g1 (t) = e−1 (t, 1) + t − 10, g2 (t) = e−1 (t, 1) − t + 10,

t ∈ T.

Then, g1Δ (t) = −e−1 (t, 1) + 1, g2Δ (t) = −e−1 (t, 1) − 1,

t ∈ T,

and [δSH f (t)]α = [−e−1 (t, 1) − 1, −e−1 (t, 1) + 1] ,

t ∈ T.

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1 Calculus of Fuzzy Functions

Exercise 1.22 Let T = 2N0 , f : T → F (R) and  1 1 , , [f (t)] = t +1 t 

t ∈ T.

α

Find [δSH f (t)]α . Example 1.23 Let T = 2N0 , f, g : T → F (R) be fuzzy functions whose α-levels are given by [f (t)]α = [e−1 (t, 0) − t + 2, e−1 (t, 0) + t + 5], [g(t)]α = [−e−1 (t, 0) + t − 10, −e−1 (t, 0) − t + 10],

t ∈ T,

α ∈ [0, 1].

Here σ (t) = 2t, f α (t) = e−1 (t, 0) − t + 2, α

f (t) = e−1 (t, 0) + t + 5, g α (t) = −e−1 (t, 0) + t − 10, g α (t) = −e−1 (t, 0) − t + 10, f αΔ (t) = −e−1 (t, 0) − 1, f g

αΔ

(t) = −e−1 (t, 0) + 1,

αΔ

(t) = e−1 (t, 0) + 1,

g αΔ (t) = e−1 (t, 0) − 1,

 αΔ [δH f (t)]α = f αΔ (t), f (t) = [−e−1 (t, 0) − 1, −e−1 (t, 0) + 1], [δSH g(t)] = [e−1 (t, 0) − 1, e−1 (t, 0) + 1], α

t ∈ T,

α ∈ [0, 1].

Thus, f is δH -differentiable in T and g is δSH -differentiable in T. We have [δH f (t)]α + [δSH g(t)]α = [−2, 2], [f (t) + g(t)]α = [−8, 15], [δH (f + g)(t)]α = [0, 0],

t ∈ T,

α ∈ [0, 1].

Note that [δH (f + g)(t)]α = [δH f (t)]α + [δSH g(t)]α ,

t ∈ T,

α ∈ [0, 1].

1.2 Second Type Fuzzy Delta Differentiation

71

Theorem 1.30 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (σ (t))f Δ (t) < 0, g : T → F (R) be δSH -differentiable at t. Let also, there exist u1 (σ (t), h), u2 (σ (t), h) ∈ F (R) such that g(σ (t)) = g(t + h) + u1 (σ (t), h), g(t − h) = g(σ (t)) + u2 (σ (t), h) for h > 0 sufficiently small. Then f · g is δSH -differentiable at t and δSH (f · g)(t) = f Δ (t) · g(t) + f (σ (t)) · δSH g(t). Proof Let  ∈ (0, 1) and  −1 .  ∗ =  1 + |f (σ (t))| + g(t) + |f Δ (t)| We have  ∗ ∈ (0, 1). Since f is Δ-differentiable at t, there exists a neighborhood UT1 (t, δ) of t such that for 0 < h < δ with t + h, t − h ∈ UT1 (t, δ) we have |f (σ (t)) − f (t + h) − f Δ (t)(μ(t) − h)| ≤  ∗ (h − μ(t)),

(1.29)

|f (σ (t)) − f (t − h) − f Δ (t)(h + μ(t))| ≤  ∗ (h + μ(t)).

(1.30)

Because g is δSH -differentiable at t, there exists a neighborhood UT2 (t, δ) such that for all 0 < h < δ with t − h, t + h ∈ UT2 (t, δ), we have that the H -differences g(σ (t)) H g(t + h) and

g(t − h) H g(σ (t))

exist and D (g(σ (t)) H g(t + h), δSH g(t)(μ(t) − h)) ≤  ∗ (h − μ(t)),

(1.31)

D (g(t − h) H g(σ (t)), δSH g(t)(−(h + μ(t)))) ≤  ∗ (h + μ(t)).

(1.32)

Also, g(t − h) ≤ 1 + g(t),

(1.33)

g(t + h) ≤ 1 + g(t),

(1.34)

D(g(t), g(t − h)) ≤  ∗ ,

(1.35)

D(g(t), g(t + h)) ≤  ∗ ,

(1.36)

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1 Calculus of Fuzzy Functions

for h > 0 sufficiently small. Because f is continuous at t and f (σ (t))f Δ (t) < 0, we have that f (t), f (t + h), and f (t − h) have the same sign for h > 0 sufficiently small and f (σ (t)) = f (t + h) + v1 (σ (t), h), where v1 (σ (t), h) has the same sign as f (σ (t)) and f (t + h). Consequently, f (σ (t)) · g(σ (t)) = (f (t + h) + v1 (σ (t), h)) · (g(t + h) + u1 (σ (t), h)) = f (t + h) · g(t + h) + f (t + h) · u1 (σ (t), h) +v1 (σ (t), h) · g(t + h) + v1 (σ (t), h) · u1 (σ (t), h). Therefore, the H -difference (f (σ (t)) · g(σ (t))) H (f (t + h) · g(t + h)) exists. Next, f (t − h) = f (σ (t)) + v2 (σ (t), h), where v2 (σ (t), h) has the same sign as f (t − h) and f (σ (t)) for h > 0 sufficiently small. From here, f (t − h) · g(t − h) = (f (σ (t)) + v2 (σ (t), h)) · (g(σ (t)) + u2 (σ (t), h)) = f (σ (t)) · g(σ (t)) + f (σ (t)) · u2 (σ (t), h) +v2 (σ (t), h) · g(σ (t)) + v2 (σ (t), h) · u2 (σ (t), h) and thus, the H -difference (f (t − h) · g(t − h)) H (f (σ (t)) · g(σ (t))) exists for h > 0 sufficiently small. Let c = f (t − h) · g(t − h), d = f (σ (t)) · g(σ (t)), a = f (σ (t)) · δSH g(t)(−(h + μ(t))), b = f Δ (t) · g(t)(−(h + μ(t))), w = f (σ (t)) · g(t − h), r = f Δ (t) · g(t − h)(−(h + μ(t))).

1.2 Second Type Fuzzy Delta Differentiation

73

Now we apply Lemma 1.1 for a, b, c, d, w, r and using (1.30), (1.32), (1.33), (1.35), we obtain     D (f ·g)(t−h) H (f ·g)(σ (t)), f (σ (t))·δSH g(t)+f Δ (t)·g(t) (−(h+μ(t))) ≤ D ((f (σ (t))·g(t−h)) H (f (σ (t))·g(σ (t))) , f (σ (t))·δSH g(t) (−(h+μ(t))))   +D f (t − h)·g(t−h), f (σ (t))·g(t−h)+f Δ (t)·g(t−h) (−(h+μ(t)))   +D f Δ (t)·g(t) (−(h+μ(t))) , f Δ (t)·g(t − h) (−(h+μ(t))) ≤ |f (σ (t))|D (g(t − h) H g(σ (t)), δSH g(t) (−(h+μ(t))))   + f (t − h) − f (σ (t)) − f Δ (t) (−(h+μ(t))) g(t − h)   + f Δ (t) D(g(t), g(t − h))(h+μ(t)) ≤  ∗ (h+μ(t))|f (σ (t))|+ ∗ (h+μ(t))(g(t)+1)   + ∗ f Δ (t) (h+μ(t))    =  ∗ 1+|f (σ (t))|+g(t)+ f Δ (t) (h+μ(t)) = (h+μ(t)) for h > 0 sufficiently small. Let c = f (σ (t)) · g(σ (t)), d = f (t + h) · g(t + h), a = f (σ (t)) · δSH g(t) (μ(t) − h) , b = f Δ (t) · g(t) (μ(t) − h) , w = f (σ (t)) · g(t + h), r = f Δ (t) · g(t + h) (μ(t) − h) . We apply Lemma 1.2 and using (1.29), (1.31), (1.33), (1.36), we get     D (f · g)(σ (t)) H (f · g)(t+h), f (σ (t)) · δSH g(t)+f Δ (t) · g(t) (μ(t)−h) ≤ D ((f · g)(σ (t)) H (f (σ (t)) · g(t+h)) , f (σ (t)) · δSH g(t)(μ(t)−h))   +D f (σ (t)) · g(t+h), f (t+h) · g(t+h)+f Δ (t) · g(t+h)(μ(t)−h)   +D f Δ (t) · g(t)(μ(t)−h), f Δ (t) · g(t+h)(μ(t)−h) = |f (σ (t))|D (g(σ (t)) H g(t+h), δSH g(t)(μ(t)−h))   + f (σ (t))−f (t+h)−f Δ (t)(μ(t)−h) g(t+h)   + f Δ (t) (h−μ(t))D(g(t), g(t+h))

74

1 Calculus of Fuzzy Functions

≤  ∗ |f (σ (t))|(h−μ(t))+ ∗ (1+g(t)) (h−μ(t))   + ∗ f Δ (t) (h−μ(t))     =  ∗ 1+|f (σ (t))|+ f Δ (t) +g(t) (h−μ(t)) = (h−μ(t)) for h > 0 sufficiently small. This completes the proof. Example 1.24 Let T = 2N0 , 1 , t   1 1 1 g(t) = , , , 3t 2t t

f (t) =

t ∈ T.

We have σ (t) = 2t,   1 1 1 , , , f (t) · g(t) = 3t 2 2t 2 t 2

t ∈ T.

Let h1 (t) =

1 , t2

t ∈ T.

Then, hΔ 1 (t) = − =−

σ (t) + t t 2 (σ (t))2 2t + t t 2 (2t)2

3t 4t 4 3 = − 3, 4t   3 3 1 δSH (f · g)(t) = − 3 , − 3 , − 3 , 4t 8t 4t =−

Next, f Δ (t) = −

1 tσ (t)

t ∈ T.

1.2 Second Type Fuzzy Delta Differentiation

75

1 = − 2, 2t   1 1 1 δSH g(t) = − 2 , − 2 , − 2 , 2t 4t 6t f (σ (t)) =

1 σ (t)

1 , 2t   1 1 1 f (σ (t)) · δSH g(t) = − 3 , − 3 , − 3 , 4t 8t 12t   1 1 1 f Δ (t) · g(t) = − 3 , − 3 , − 3 , t ∈ T, 2t 4t 6t =

and   1 1 1 f (t) · g(t) + f (σ (t)) · δSH g(t) = − 3 , − 3 , − 3 2t 4t 6t   1 1 1 + − 3,− 3,− 3 4t 8t 12t   1 1 1 1 1 1 = − 3 − 3,− 3 − 3,− 3 − 2t 4t 4t 8t 6t 12t 3   3 3 1 = − 3 , − 3 , − 3 , t ∈ T. 4t 8t 4t Δ

Consequently, δSH (f · g)(t) = f Δ (t) · g(t) + f (σ (t)) · δSH g(t), Exercise 1.23 Let T =

t ∈ T.



1 {0}, 2k k∈N

1 , t +1   1 1 1 , g(t) = , , t +7 t +6 t +5

f (t) =

t ∈ T.

Prove that δSH (f · g)(0) = f Δ (0) · g(0) + f (σ (0)) · δSH g(0).

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1 Calculus of Fuzzy Functions

Lemma 1.3 Let a, b, c, d, e, w, r ∈ F (R) and the H -differences c H d, a H b, r H c exist. Then, D (c H d, a H b) ≤ D(r, d + w) + D(a, w) + D(e, r H c) + D(e, b). Proof We have D(c H d, a H b) = D(c H d + r, a H b + r) = D(b + r + c, a + d + r) = D(b + r, a + d + r H c) ≤ D(r, a + d) + D(b, r H c) = D(r + w, a + d + w) + D(b + e, r H c + e) ≤ D(r, d + w) + D(w, a) +D(e, r H c) + D(b, e). This completes the proof. Lemma 1.4 Let a, b, c, d, w, r ∈ F (R) and c H d, a H b, c H w exist. Then, D(c H d, a H b) ≤ D(c H w, a) + D(w + r, d) + D(r, b). Proof We have D(c H d, a H b) = D(c + b, a + d) = D(c + b + w, a + d + w) ≤ D(c, a + w) + D(b + w, d) = D(c H w, a) + D(b + w + r, d + r) ≤ D(c H w, a) + D(w + r, d) + D(r, b). This completes the proof. Theorem 1.31 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (t)f Δ (t) > 0, g : T → F (R) be δSH -differentiable at t, and the H -differences ((f · g)(t + h)) H ((f · g)(σ (t)))

and

((f · g)(σ (t))) H ((f · g)(t − h))

exist for h > 0 sufficiently small. Then f · g is δH -differentiable at t and   δH (f · g)(t) = f Δ (t) · g(t) H ((−f (σ (t))) · δSH g(t)) .

1.2 Second Type Fuzzy Delta Differentiation

77

Proof Let  ∈ (0, 1) and  ∗ =  min



   −1 2 + g(t) + f Δ (t) + |f (t)| + δSH g(t) ,

 −1  Δ . 2 + 2g(t) + |f (t)| Note that  ∗ ∈ (0, 1). Since f is Δ-differentiable at t, there exists a neighborhood UT1 (t, δ) of t such that for 0 < h < δ with t + h, t − h ∈ UT1 (t, δ), we have |f (t + h) − f (σ (t)) − f Δ (t)(h − μ(t))| ≤  ∗ (h − μ(t)),

(1.37)

|f (σ (t)) − f (t − h) − f Δ (t)(h + μ(t))| ≤  ∗ (h + μ(t)).

(1.38)

and

Because g is δSH -differentiable at t, there exists a neighborhood UT2 (t, δ) of t such that for 0 < h < δ with t − h, t + h ∈ UT2 (t, δ), we have that the H -differences g(σ (t)) H g(t + h) and

g(t − h) H g(σ (t))

exist and D (g(σ (t)) H g(t + h), δSH g(t)(h − μ(t))) ≤  ∗ (h − μ(t)),

(1.39)

D (g(t − h) H g(σ (t)), δSH g(t)(−(h + μ(t)))) ≤  ∗ (h + μ(t)).

(1.40)

Since g is δSH -differentiable at t, then it is continuous at t. So, there exists a neighborhood UT3 (t, δ) of t such that

and

D(g(σ (t)), g(t)) ≤  ∗ ,

(1.41)

D(g(σ (t)), g(t + h)) ≤  ∗

(1.42)

  g(σ (t)) = D g(σ (t)), 0 ≤ D (g(σ (t)), g(t)) + D(g(t), 0) ≤  ∗ + g(t) < 1 + g(t).

(1.43)

Because f is Δ-differentiable at t, then it is continuous at t. Then there exists a neighborhood UT4 (t, δ) of t so that t + h ∈ UT4 (t, δ) and |f (σ (t)) − f (t + h)| ≤  ∗ ,

0 < h < δ.

(1.44)

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1 Calculus of Fuzzy Functions

Let UT (t, δ) = UT1 (t, δ)



UT2 (t, δ)



UT3 (t, δ)



UT4 (t, δ).

Set c = f (t + h) · g(t + h), d = f (σ (t)) · g(σ (t)), a = f Δ (t) · g(t)(h − μ(t)), b = −f (σ (t)) · δSH g(t)(h − μ(t)), r = f (t + h) · g(σ (t)), w = f Δ (t) · g(σ (t))(h − μ(t)), e = −f (t + h) · δSH g(t)(h − μ(t)). We apply Lemma 1.3 for a, b, c, d, e, w, r and using (1.37), (1.39), (1.41), (1.43), (1.44), we get     D (f ·g)(t+h) H (f ·g)(σ (t)), f Δ (t)·g(t) H ((−f (σ (t))·δSH g(t))) (h−μ(t))   ≤ D f (t+h)·g(σ (t)), f (σ (t))·g(σ (t))+f Δ (t)·g(σ (t))(h−μ(t))   +D f Δ (t)·g(t)(h−μ(t)), f Δ (t)·g(σ (t))(h−μ(t)) +D (−f (t+h)·δSH g(t)(h−μ(t)), (f (t+h)·g(σ (t))) H (f (t+h)·g(t+h))) +D (−f (t+h)·δSH g(t)(h−μ(t)), f (σ (t))·δSH g(t)(h−μ(t)))     ≤ f (t+h)−f (σ (t))−f Δ (t)(h−μ(t)) g(σ (t))     + f Δ (t) (h−μ(t))D(g(t), g(σ (t))) +|f (t+h)|D (δSH g(t)(h−μ(t)), g(σ (t)) H g(t+h)) +δSH g(t)|f (σ (t))−f (t+h)|(h−μ(t)) ≤  ∗ g(σ (t))(h−μ(t))+ ∗ |f Δ (t)|(h−μ(t)) + ∗ |f (t+h)|(h−μ(t))+ ∗ δSH g(t)(h−μ(t))   ≤  ∗ 1+|f (t)|+|f Δ (t)|+g(σ (t))+δSH g(t) (h−μ(t))   ≤  ∗ 2+|f (t)|+|f Δ (t)|+g(t)+δSH g(t) (h−μ(t)) ≤ (h−μ(t)).

1.2 Second Type Fuzzy Delta Differentiation

79

Let c=d = f (σ (t)) · g(σ (t)), a = f Δ (t) · g(σ (t))(h + μ(t)), b = f Δ (t) · g(t + h)(h + μ(t)), r = f Δ (t) · g(σ (t))(h + μ(t)), w = f (t − h) · g(σ (t)). Now we apply Lemma 1.4 for a, b, c, d, r, w and using (1.38), (1.40), (1.41), (1.42), (1.43), we obtain     D (f ·g)(σ (t)) H (f ·g)(t−h), f Δ (t)·g(t) H ((−f (σ (t)))·δSH g(t)(h+μ(t)))   ≤ D (f ·g)(σ (t)) H (f (t−h)·g(σ (t))), f Δ (t)·g(σ (t))(h+μ(t))   +D f (t−h)·g(σ (t))+f Δ (t)·g(σ (t))(h+μ(t)), f (σ (t))·g(σ (t))   +D f Δ (t)·g(σ (t))(h+μ(t)), f Δ (t)·g(t+h)(h+μ(t)) ≤ 2|f (σ (t))−f (t−h)−f Δ (t)(h+μ(t))|g(σ (t)) +|f Δ (t)|D(g(σ (t)), g(t+h))(h+μ(t)) ≤ 2 ∗ (h+μ(t))g(σ (t))+ ∗ |f Δ (t)|(h+μ(t))   =  ∗ 2g(σ (t))+|f Δ (t)| (h+μ(t))   ≤  ∗ 2+2g(t)+|f Δ (t)| (h+μ(t)) ≤ (h+μ(t)).

This completes the proof. Example 1.25 Let T = 2N0 , f (t) = t,   1 1 1 , , , g(t) = 4t 3t 2t Here σ (t) = 2t,

t ∈ T.

t ∈ T.

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1 Calculus of Fuzzy Functions

Let h(t) =

1 , t

t ∈ T.

Then, hΔ (t) = − =−

1 tσ (t) 1 , 2t 2

t ∈ T.

Hence,   1 1 1 δSH g(t) = − 2 , − 2 , − 2 , 4t 6t 8t f (σ (t)) = σ (t) = 2t, −f (σ (t)) = −2t,   1 1 1 (−f (σ (t)) · δSH g(t) = , , , 4t 3t 2t f Δ (t) = 1,   1 1 1 Δ , , , f (t) · g(t) = 4t 3t 2t

t ∈ T.

Therefore,  Δ  f (t) · g(t) H (−f (σ (t)) · δSH g(t)) = (0, 0, 0),

t ∈ T.

Next,  f (t) · g(t) =

 1 1 1 , , , 4 3 2

δH (f · g)(t) = (0, 0, 0),

t ∈ T.

Consequently,   δH (f · g)(t) = f Δ (t) · g(t) H ((−f (σ (t))) · δSH g(t)) . Exercise 1.24 Let T = 3N0 , f (t) = t + 1,   1 1 1 , 2 , 2 , g(t) = 2 t + 4t t + 3t t + 2t

t ∈ T.

1.2 Second Type Fuzzy Delta Differentiation

81

Check if   δH (f · g)(t) = f Δ (t) · g(t) H ((−f (σ (t))) · δSH g(t)) ,

t ∈ Tκ .

Exercise 1.25 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (t)f Δ (t) > 0, g : T → F (R) be δSH -differentiable at t, and the H -differences ((f · g)(σ (t))) H ((f · g)(t + h))

((f · g)(t − h)) H ((f · g)(σ (t)))

and

exist. Prove that f · g is δSH -differentiable at t and δSH (f · g)(t) = (f (σ (t)) · δSH g(t)) H

 Δ   −f (t) · g(t) .

Example 1.26 Let T = 2N0 , f (t) = t,   1 1 1 , , , g(t) = 3t 2 2t 2 t 2

t ∈ T.

Here σ (t) = 2t,

t ∈ T.

Let 1 , t2 1 h2 (t) = , t ∈ T. t h1 (t) =

We have hΔ 1 (t) = −

σ (t) + t t 2 (σ (t))2

3t 4t 4 3 = − 3, 4t 1 hΔ 2 (t) = − tσ (t) =−

=−

1 , 2t 2

t ∈ T.

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1 Calculus of Fuzzy Functions

Then,   3 3 1 δSH g(t) = − 3 , − 3 − 3 , 4t 8t 4t f (σ (t)) = 2t,   3 3 1 f (σ (t)) · δSH g(t) = − 2 , − 2 , − 2 , 2t 4t 2t f Δ (t) = 1,   1 1 1 Δ −f (t) · g(t) = − 2 , − 2 , − 2 , t 2t 3t

t ∈ T.

Hence,    (f (σ (t)) · δSH g(t)) H −f Δ (t) · g(t)     1 3 3 1 1 1 = − 2 , − 2 , − 2 H − 2 , − 2 , − 2 2t 4t 2t t 2t 3t   1 1 1 = − 2 , − 2 , − 2 , t ∈ T. 2t 4t 6t Next,  f (t) · g(t) =

 1 1 1 , , , 3t 2t t

  1 1 1 δSH (f · g)(t) = − 2 , − 2 , − 2 , 2t 4t 6t

t ∈ T.

Consequently, δSH (f · g)(t) = (f (σ (t)) · δSH g(t)) H

 Δ   −f (t) · g(t) .

Exercise 1.26 Let t ∈ Tκ , f : T → R be Δ-differentiable at t, f (σ (t))f Δ (t) < 0, g : T → F (R) be δH -differentiable at t, and the H -differences ((f · g)(σ (t))) H ((f · g)(t + h))

and

((f · g)(t − h)) H ((f · g)(σ (t)))

exist. Prove that f · g is δSH -differentiable at t and   δSH (f · g)(t) = f Δ (t) · g(t) H ((−f (σ (t))) · δH g(t)) .

1.2 Second Type Fuzzy Delta Differentiation

Example 1.27 Let T =

 N0 1 3

83

,

1 , t2   g(t) = t, t 2 + 1, t 2 + t + 2 ,

f (t) =

t ∈ T.

We have σ (t) = 3t,

t ∈ T.

Then, f Δ (t) = − =−

σ (t) + t t 2 (σ (t))2 3t + t t 2 (3t)2

4t 9t 4 4 = − 3, 9t δH g(t) = (1, σ (t) + t, σ (t) + t + 1) =−

= (1, 3t + t, 3t + t + 1) = (1, 4t, 4t + 1), f (σ (t)) =

1 (σ (t))2

1 , 9t 2 1 −f (σ (t)) = − 2 , 9t   1 4t + 1 4 ,− ,− 2 , (−f (σ (t))) · δH g(t) = − 9t 9t 2 9t    2 4 t + t + 2 −4(t 2 + 1) 4 Δ f (t) · g(t) = − , ,− 2 , 9t 3 9t 3 9t =

Hence,  Δ  f (t) · g(t) H ((−f (σ (t))) · δH g(t))

t ∈ T.

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1 Calculus of Fuzzy Functions



     4 t 2 + t + 2 −4(t 2 + 1) 4t + 1 1 4 4 − , − = − , , − , −  H 9t 9t 3 9t 3 9t 2 9t 2 9t 2   3t + 8 4 1 = − , − 3 , − 2 , t ∈ T. 9t 3 9t 3t Next,  f (t) · g(t) =

1 1 2 1 ,1 + 2,1 + + 2 t t t t

 t ∈ T.

,

Let h1 (t) =

1 , t

1 , t2 2 1 h3 (t) = 1 + + 2 , t t

h2 (t) = 1 +

t ∈ T.

We have hΔ 1 (t) = −

1 tσ (t)

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

4t 9t 4 4 = − 3, 9t =−

Δ Δ hΔ 3 (t) = h1 (t) + 2h2 (t)

1 8 − 3 3t 2 9t 3t + 8 =− , t ∈ T, 9t 3

=−

and 

3t + 8 4 1 δSH (f · g)(t) = − ,− 3,− 2 3 9t 9t 3t

 ,

t ∈ T.

1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation

85

Consequently,   δSH (f · g)(t) = f Δ (t) · g(t) H (−f (σ (t)) · δH g(t)) .

1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation Definition 1.9 Let f : [a, b] → F (R) be δH -differentiable on A ⊆ [a, b] and δSH -differentiable on [a, b]\A fuzzy function whose α-levels are given by

 α [f (t)]α = f α (t), f (t) ,

t ∈ [a, b],

α ∈ [0, 1].

Let also, t0 ∈ (a, b). We say that t0 is a switching point for f , if in any neighborhood U of t0 there exist points t1 < t0 < t2 such that 1. (type 1 switch) f is δH -differentiable at t1 and δSH -differentiable at t2 . 2. (type 2 switch) f is δSH -differentiable at t1 and δH -differentiable at t2 . Example 1.28 Let T = Z, f : T → R be a fuzzy function whose α-levels are defined by  [f (t)]α =

 1 1 2 , t , + t2 + 1 t2 + 1

t ∈ T,

α ∈ [0, 1].

Here σ (t) = t + 1, f α (t) = α

t2

1 , +1

f (t) = t 2 +

1 , t2 + 1

t ∈ T,

Then, f αΔ (t) = − =− =−

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

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

α ∈ [0, 1].

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1 Calculus of Fuzzy Functions

f

αΔ

(t) = σ (t) + t + f αΔ (t) = t +1+t − = 2t + 1 −

(t 2

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

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

t ∈ T,

α ∈ [0, 1].

We have 3 , 10 3 αΔ f (1) = 3 − 10 27 , = 10 1 f αΔ (−1) = , 2 1 αΔ f (−1) = −1 + 2 1 =− . 2 f αΔ (1) = −

Thus, f is δH -differentiable at t = 1 and δSH -differentiable at t = −1. Therefore, t = 0 is a type 2 switching point for f . Theorem 1.32 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≤ 0, If,g

Ifα,1 σ ,δ g (t) ≥ 0, H

Iδα,1 (t) ≥ 0, H f,g

α ∈ [0, 1],

f ◦ g is δSH -differentiable at t. Then, δSH (f ◦ g)(t) = f σ (t) ◦ δH g(t) + δH f (t) ◦ g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , α ∈ [0, 1]. Since f and g are δH -differentiable at t, we get

 αΔ [δH f (t)]α = f αΔ (t), f (t) ,

 [δH g(t)]α = g αΔ (t), g αΔ (t) , α ∈ [0, 1].

(1.45)

1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation

87

α,1 α,1 Now, using that If,g (t) ≤ 0, Ifα,1 σ ,δ g (t) ≥ 0, Iδ f,g (t) ≥ 0, α ∈ [0, 1], we obtain H H

α,1 (t) If,g

   f α (t) f α (t)    = α   g (t) g α (t)  α

= f α (t)g α (t) − g α (t)f (t) ≤ 0,    f ασ (t) f ασ (t)    α,1 If σ ,δH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,   αΔ  αΔ   f (t) f (t)  α,1 IδH f,g (t) =  α   g (t) g α (t)  = f αΔ (t)g α (t) − g α (t)f ≥ 0,

αΔ

(t)

α ∈ [0, 1].

α,1 Because If,g (t) ≤ 0, α ∈ [0, 1], we have

 α [(f ◦ g)(t)]α = f α (t)g α (t), f (t)g α (t) ,

α ∈ [0, 1].

Since f ◦ g is δSH -differentiable at t, we obtain 

Δ Δ  α f g α (t), f α g α (t)



[δSH (f ◦ g)(t)] = 

ασ αΔ = f (t)g αΔ (t) + g α (t)f (t), f ασ (t)g αΔ (t) + g α (t)f αΔ (t) , α ∈ [0, 1]. (1.46) α,1 Because Ifα,1 σ ,δ g (t) ≥ 0 and Iδ f,g (t) ≥ 0, α ∈ [0, 1], we have H H α

  σ  α

ασ f ◦ δH g (t) = g αΔ (t)f (t), f ασ (t)g αΔ (t) ,

α ∈ [0, 1],

and

 αΔ [(δH f ◦ g) (t)]α = g α (t)f (t), f αΔ (t)g α (t) ,

α ∈ [0, 1].

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1 Calculus of Fuzzy Functions

From here,  α  σ f ◦ δH g (t) + [(δH f ◦ g) (t)]α   ασ αΔ = g αΔ (t)f (t) + g α (t)f (t), f ασ (t)g αΔ (t) + f αΔ (t)g α (t) ,

α ∈ [0, 1].

From the last equality and from (1.46), we get (1.45). This completes the proof. Exercise 1.27 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≤ 0, If,g

α,1 If,δ (t) ≥ 0, Hg

Iδα,1 σ (t) ≥ 0, H f,g

α ∈ [0, 1],

f ◦ g is δSH -differentiable at t. Prove that δSH (f ◦ g)(t) = f (t) ◦ δH g(t) + δH f (t) ◦ g σ (t). Exercise 1.28 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≥ 0, If,g

Ifα,1 σ ,δ g (t) ≤ 0, H

Iδα,1 (t) ≤ 0, H f,g

α ∈ [0, 1],

f ◦ g is δSH -differentiable at t. Prove that δSH (f ◦ g)(t) = f σ (t) ◦ δH g(t) + δH f (t) ◦ g(t). Exercise 1.29 Let t ∈ Tκ and f , g be δH -differentiable at t, and α,1 (t) ≥ 0, If,g

α,1 If,δ (t) ≤ 0, Hg

Iδα,1 σ (t) ≤ 0, H f,g

α ∈ [0, 1],

f ◦ g is δSH -differentiable at t. Prove that δSH (f ◦ g)(t) = f (t) ◦ δH g(t) + δH f (t) ◦ g σ (t), Theorem 1.33 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≥ 0, If,g

Ifα,1 (t) ≥ 0, σ ,δ SH g

Iδα,1 (t) ≤ 0, SH f,g

α ∈ [0, 1],

f ◦ g is δH -differentiable at t. Then, δSH (f  g)(t) = f σ (t) ◦ δSH g(t) + δSH f (t) ◦ g(t). Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , α ∈ [0, 1].

(1.47)

1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation

89

Since f and g are δSH -differentiable at t, we get

αΔ  [δSH f (t)]α = f (t), f αΔ (t) ,

 [δSH g(t)]α = g αΔ (t), g αΔ (t) , α ∈ [0, 1]. α,2 (t) ≥ 0, Ifα,1 (t) ≥ 0, Iδα,1 (t) ≤ 0, α ∈ [0, 1], we obtain Now, using that If,g σ ,δ SH g SH f,g

  α  f (t) f α (t)    α,2 (t) =  α If,g   g (t) g α (t)  α

= −f α (t)g α (t) + g α (t)f (t) ≥ 0,    f ασ (t) f ασ (t)    α,1 If σ ,δSH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≥ 0,     αΔ  f (t) f αΔ (t)  α,1 IδSH f,g (t) =  α   g (t) g α (t)  =f

αΔ

≤ 0,

(t)g α (t) − g α (t)f αΔ (t) α ∈ [0, 1].

α,2 Because If,g (t) ≥ 0, α ∈ [0, 1], we have

 α [(f  g)(t)]α = f α (t)g α (t), f (t)g α (t) ,

α ∈ [0, 1].

Since f  g is δSH -differentiable at t, we obtain  Δ Δ   α [δSH (f  g)(t)]α = f g α (t), f α g α (t) 

ασ αΔ = f (t)g αΔ (t) + g α (t)f (t), f ασ (t)g αΔ (t)+g α (t)f αΔ (t) , α ∈ [0, 1]. (1.48) α,1 Because Ifα,1 (t) ≥ 0 and I (t) ≤ 0, α ∈ [0, 1], we have σ ,δ δSH f,g SH g   σ  α

ασ f ◦ δSH g (t) = g αΔ (t)f (t), f ασ (t)g αΔ (t)

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1 Calculus of Fuzzy Functions

and

αΔ  [(δSH f ◦ g) (t)]α = f (t)g α (t), g α (t)f αΔ (t) ,

α ∈ [0, 1].

From here,  σ  α f ◦ δSH g (t) + [(δSH f ◦ g) (t)]α   ασ αΔ αΔ α αΔ ασ α αΔ = f (t)g (t) + f (t)g (t), g (t)f (t) + g (t)f (t) ,

α ∈ [0, 1].

From the last equality and from (1.48), we get (1.47). This completes the proof. Exercise 1.30 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≥ 0, If,g

α,1 If,δ (t) ≥ 0, SH g

Iδα,1 σ (t) ≤ 0, SH f,g

α ∈ [0, 1],

f  g is δSH -differentiable at t. Prove that δSH (f  g)(t) = f (t) ◦ δSH g(t) + δSH f (t) ◦ g σ (t). Theorem 1.34 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≤ 0, If,g

Ifα,1 (t) ≤ 0, σ ,δ SH g

Iδα,1 (t) ≥ 0, SH f,g

α ∈ [0, 1],

f  g is δSH -differentiable at t. Then, δSH (f  g)(t) = f σ (t) ◦ δSH g(t) + δSH f (t) ◦ g(t).

(1.49)

Proof Let

 α [f (t)]α = f α (t), f (t) ,

 [g(t)]α = g α (t), g α (t) , α ∈ [0, 1]. Since f and g are δSH -differentiable at t, we get

αΔ  [δSH f (t)]α = f (t), f αΔ (t) ,

 [δSH g(t)]α = g αΔ (t), g αΔ (t) , α ∈ [0, 1]. α,2 (t) ≤ 0, Ifα,1 (t) ≤ 0, Iδα,1 (t) ≥ 0, α ∈ [0, 1], we obtain Now, using that If,g σ ,δ SH g SH f,g

1.3 Other Properties of First Type and Second Type Fuzzy Delta Differentiation

α,2 If,g (t)

91

  α  f (t) f α (t)    = α   g (t) g α (t)  α

= f (t)g α (t) − g α (t)f α (t) ≤ 0,    f ασ (t) f ασ (t)    α,1 If σ ,δSH g (t) =  αΔ   g (t) g αΔ (t)  = f ασ (t)g αΔ (t) − g αΔ (t)f

ασ

(t)

≤ 0,    αΔ αΔ (t)  f (t) f   Iδα,1 (t) =  α  SH f,g  g (t) g α (t)  =f

αΔ

≥ 0,

(t)g α (t) − g α (t)f αΔ (t) α ∈ [0, 1].

α,2 Because If,g (t) ≤ 0, α ∈ [0, 1], we have

α  [(f  g)(t)]α = f (t)g α (t), f α (t)g α (t) ,

α ∈ [0, 1].

Since f  g is δSH -differentiable at t, we obtain  Δ  α Δ  α α α (t), f g α (t) [δSH (f ◦ g)(t)] = f g 

ασ αΔ = f ασ (t)g αΔ (t) + g α (t)f αΔ (t), f (t)g αΔ (t) + g α (t)f (t) , Because Ifα,1 (t) ≤ 0 and Iδα,1 (t) ≥ 0, α ∈ [0, 1], we have σ ,δ SH g SH f,g   σ  α

ασ f ◦ δSH g (t) = f ασ (t)g αΔ (t), g αΔ (t)f (t) ,

α ∈ [0, 1]. (1.50)

α ∈ [0, 1],

and

 αΔ [(δSH f ◦ g) (t)]α = g α (t)f αΔ (t), f (t)g α (t) ,

α ∈ [0, 1].

From here,  σ  α f ◦ δSH g (t) + [(δSH f ◦ g) (t)]α   ασ αΔ αΔ ασ α αΔ αΔ α = g (t)f (t)+g (t)f (t), f (t)g (t)+f (t)g (t) ,

α ∈ [0, 1].

From the last equality and from (1.50), we get (1.49). This completes the proof.

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1 Calculus of Fuzzy Functions

Exercise 1.31 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≤ 0, If,g

α,1 If,δ (t) ≤ 0, SH g

Iδα,1 σ (t) ≥ 0, SH f,g

α ∈ [0, 1],

f  g is δSH -differentiable at t. Prove that δSH (f g)(t) = f (t) ◦ δSH g(t) + δSH f (t) ◦ g σ (t). Exercise 1.32 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≥ 0, If,g

Ifα,1 (t) ≥ 0, σ ,δ SH g

Iδα,1 (t) ≤ 0, SH f,g

α ∈ [0, 1],

f  g is δSH -differentiable at t. Prove that δSH (f  g)(t) = f σ (t) ◦ δSH g(t) + δSH f (t) ◦ g(t). Exercise 1.33 Let t ∈ Tκ and f , g be δSH -differentiable at t, and α,2 (t) ≥ 0, If,g

α,1 If,δ (t) ≥ 0, SH g

Iδα,1 σ (t) ≤ 0, SH f,g

α ∈ [0, 1],

f  g is δSH -differentiable at t. Prove that δSH (f  g)(t) = f σ (t) ◦ δSH g(t) + δSH f (t) ◦ g(t). Remark 1.1 As in above, one can get the corresponding assertions for , , , ⊗,

, , respectively.   Example 1.29 Let T = R and f, g : 12 , 1 → R be fuzzy functions whose α-levels are [f (t)]α = [−2, 1], [g(t)]α = [t 2 , t],

 t∈

 1 ,1 , 2

α ∈ [0, 1].

Then, [δSH f (t)]α = [0, 0], [δSH g(t)] = [1, 2t], [f (σ (t))]α = [−2, 1],  [f (σ (t))]α [δSH g(t)]α = min{−2, −4t, 1, 2t},  max{−2, −4t, 1, 2t} = [−4t, 2t] ,

1.4 First Type Fuzzy Delta Integration

93

 [g(t)] [δSH f (t)] = α

α

 min{0, 0},

max{0, 0} 

= [0, 0],

t∈

 1 ,1 , 2

α ∈ [0, 1],

whereupon [f (σ (t))]α [δSH g(t)]α + [g(t)]α [δSH f (t)]α = [−4t, 2t] + [0, 0]   1 ,1 , = [−4t, 2t], t ∈ 2

α ∈ [0, 1].

Next,  [f (t)]α [g(t)]α =

min{−2t 2 , −2t, t 2 , t},  max{−2t , −2t, t , t} 2

2

= [−2t, t], [δH (f g)(t)]α = [−2, 1],

 t∈

 1 ,1 , 2

α ∈ [0, 1].

Therefore, [f (σ (t))]α [δSH g(t)]α + [g(t)]α [δSH f (t)]α = [δH (f g)(t)]α , t∈





1 2, 1

,

α ∈ [0, 1].

Remark 1.2 By Example 1.29, for the traditional multiplication of fuzzy functions f and g we cannot obtain δH (f g) = f σ δSH g + δSH fg.

1.4 First Type Fuzzy Delta Integration In this section, we introduce the conception for the first type fuzzy delta integration on time scales. We deduct some of its properties. Let I ⊂ T. Definition 1.10 A function f : T → R is called a sector of the fuzzy function F : I → F (R) if f (t) ∈ F (t) for all t ∈ I . The set of all rd-continuous sectors of F on I is denoted by SH F (I ).

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1 Calculus of Fuzzy Functions

Example 1.30 Let T = 2N0 , I = [1, 64],

 F (t) = t, t 2 + 1, 2, 4, 16, t 3 − 1, 64, t 4 − 10 , f (t) = t,

t ∈ I.

Then f (t) ∈ F (t) for all t ∈ I . Therefore, f is a sector of F . Example 1.31 Let T = Z, I = [0, 16],  F (t) =

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

f (t) = t 2 ,

t ∈ I.

We have that f (t) ∈ / F (t) for all t ∈ I . Therefore, f is not a sector of F . Definition 1.11 A fuzzy function F : I → F (R) is said to be regulated provided its regulated sectors exist. Definition 1.12 A fuzzy function F : I → F (R) is said to be rd-continuous provided its rd-continuous sectors exist. The set of all rd-continuous fuzzy functions F : I → F (R) is denoted by Cf rd = Cf rd (I ) = Cf rd (I, F (R)). The set of fuzzy functions F : I → F (R) which are δH -differentiable whose δH derivatives are rd-continuous is denoted by 1 1 1 CHf rd = CHf rd (I ) = CHf rd (I, F (R)).

Definition 1.13 A continuous fuzzy function F : I → F (R) is said to be δH pre-differentiable with (region of differentiation) D provided D ⊂ Tκ , Tκ \D is countable and contains no right-scattered elements of T, and F is δH -differentiable at each t ∈ T. If F is rd-continuous, then F is regulated and there exists a fuzzy function f which is pre-differentiable with region of differentiation D such that δH f (t) = F (t),

t ∈ D.

Definition 1.14 A fuzzy function F : T → F (R) is said to be δH -integrable on I if its sectors are rd-continuous on I . In this case, we define the δH -integral of F on I by 



F (s)δH s = I

f (s)Δs : f ∈ SH F (I ) I

1.4 First Type Fuzzy Delta Integration

95

and defined level-wise by α

 F (s)δH s

 =

 α f (s)Δs ,

f α (s)Δs,

I

I

I

where

 α [F (s)]α = f α (s), f (s) ,

s ∈ I,

α ∈ [0, 1].

Example 1.32 Let T = N,   F (t) = 2t + 2, 2t + 4, 3t 2 + 3t + 5 ,

t ∈ T.

We will find 4

F (s)δH s. 1

We have σ (t) = t + 1,

t ∈ T.

Also, 4 1



4

F (s)δH s =

4

(2s + 2)Δs,

1

  3s + 3s + 5 Δs .

4

(2s + 4)Δs,

1

1

Let h1 (t) = t 2 + t, h2 (t) = t 2 + 3t + 4, h3 (t) = t 3 + 4t + 7, Then, hΔ 1 (t) = σ (t) + t + 1 = t +1+t +1 = 2t + 2, hΔ 2 (t)

= σ (t) + t + 3 = t +1+t +3 = 2t + 4,

t ∈ T.

2

96

1 Calculus of Fuzzy Functions 2 2 hΔ 3 (t) = (σ (t)) + tσ (t) + t + 4

= (t + 1)2 + t (t + 1) + t 2 + 4 = t 2 + 2t + 1 + t 2 + t + t 2 + 4 = 3t 2 + 3t + 5,

t ∈ T.

4

4

Therefore, 4



4

F (s)δH s =

1

1

hΔ 1 (s)Δs,

1

hΔ 2 (s)Δs,

1

 hΔ 3 (s)Δs

 s=4 s=4 s=4     = h1 (s) , h2 (s) , h3 (s) s=1

s=1

s=1

= (16 + 4 − 1 − 1, 16 + 12 + 4 − 1 − 3 − 4, 64 + 16 + 7 − 1 − 4 − 7) = (18, 24, 75). Exercise 1.34 Let T = 3N0 and  F (t) =

1 1 1 , , t4 t3 t2

 ,

t ∈ T.

Find 81

F (s)δH s. 1

Example 1.33 Let T = 2N0 , f : T → F (R) be a fuzzy function whose α-levels are given by α

[f (t)]α = 3t + 1, 7t 2 ,

t ∈ T,

α ∈ [0, 1].

We will find 

α

4

f (t)δH t

,

α ∈ [0, 1].

1

Here σ (t) = 2t, f α (t) = 3t + 1, α

f (t) = 7t 2 ,

t ∈ T,

α ∈ [0, 1].

1.4 First Type Fuzzy Delta Integration

97

Let f1 (t) = t 2 + t, f2 (t) = t 3 ,

t ∈ T.

Then, f1Δ (t) = σ (t) + t + 1 = 2t + t + 1 = 3t + 1, f2Δ (t)

= (σ (t))2 + tσ (t) + t 2 = (2t)2 + 2t 2 + t 2 = 4t 2 + 3t 2 = 7t 2 ,

t ∈ T.

Hence, 4

4

f α (t)Δt =

1

(3t + 1)Δt

1 4

= 1

f1Δ (t)Δt

t=4  = f1 (t) t=1

= f1 (4) − f1 (1) = (16 + 4) − (1 + 1) = 20 − 2 = 18, 4

4

α

f (t)Δt = 7

1

t 2 Δt

1 4

= 1

f2Δ (t)Δt

t=4  = f2 (t) t=1

= f2 (4) − f2 (1) = 64 − 1 = 63.

98

1 Calculus of Fuzzy Functions

Consequently, 

α

4



4

=

f (t)δH t

4

α

f (t)Δt,

1

1

α



f (t)Δt 1

= [18, 63]. Exercise 1.35 Let T = 3Z. Find 

α

9 −3

f (t)δH t

α ∈ [0, 1],

,

where

α [f (t)]α = t − 3, t 2 + t ,

t ∈ T,

α ∈ [0, 1].

Theorem 1.35 Let t0 , T ∈ T, t0 < T , F, G : [t0 , T ] → F (R) be δH -integrable. Then F + G : [t0 , T ] → F (R) is δH -integrable and T

(F (s) + G(s))δH s =

t0

T

T

F (s)δH s +

t0

G(s)δH s.

(1.51)

t0

Proof Since F and G are δH -integrable on [t0 , T ], then their sectors are rdcontinuous on [t0 , T ]. Hence, the sectors of F + G are rd-continuous on [t0 , T ]. Let u, v, w be rd-continuous sectors of F , G, and F + G, respectively. Then there exist f ∈ SH F ([t0 , T ]), g ∈ SH G ([t0 , T ]), and h ∈ SH (F +G) ([t0 , T ]) such that T

u(T ) =

f (s)Δs, t0 T

v(T ) =

g(s)Δs, t0 T

w(T ) =

h(s)Δs. t0

Hence, u(T ) + v(T ) =

T t0

=

T

f (s)Δs +

T

g(s)Δs t0

(f (s) + g(s))Δs

t0

∈ SH (F +G) ([t0 , T ]).

1.4 First Type Fuzzy Delta Integration

99

Thus,  !  T f (s)Δs : f ∈ SH F ([t0 , T ]) + t0 g(s)Δs : g ∈ SH G ([t0 , T ]) 

! T ⊂ t0 h(s)Δs : h ∈ SH (F +G) ([t0 , T ]) . (1.52) Next, there exist h1 ∈ SH F ([t0 , T ]) and h2 ∈ SH G ([t0 , T ]) such that

! T t0

h(s) = h1 (s) + h2 (s),

s ∈ [t0 , T ].

Therefore, T

w(T ) =

h(s)Δs t0 T

=

(h1 (s) + h2 (s))Δs

t0 T

=

h1 (s)Δs +

t0

T

h2 (s)Δs t0



T

∈



f (s)Δs : f ∈ SH F ([t0 , T ])

t0

 +

T

 g(s)Δs : g ∈ SH G ([t0 , T ]) .

t0

Consequently, 

T t0

 h(s)Δs : h ∈ SH (F +G) ([t0 , T ]) 

⊂

T

  f (s)Δs : f ∈ SH F ([t0 , T ]) +

t0

T

 g(s)Δs : g ∈ SH G ([t0 , T ]) .

t0

From the last relation and from (1.52), we get 

T t0

=

 h(s)Δs : h ∈ SH (F +G) ([t0 , T ]) 

T

 f (s)Δs : f ∈ SH F ([t0 , T ]) +

t0

This completes the proof.



T t0

 g(s)Δs : g ∈ SH G ([t0 , T ]) .

100

1 Calculus of Fuzzy Functions

Example 1.34 Let T = 2N0 , F (t) = {1, t 2 }, G(t) = {1, t},

t ∈ T.

Then, F (t) + G(t) = {2, 1 + t, 1 + t 2 , t + t 2 },

t ∈ T.

We will compute 4

4

F (t)δH t,

4

G(t)δH t,

1

1

(F (t) + G(t))δH t.

1

Let h1 (t) =

t2 , 3

h2 (t) =

t3 , 7

h3 (t) = t +

t2 , 3

h4 (t) = t +

t3 , 7

h5 (t) =

t3 t2 + , 3 7

t ∈ T.

We have σ (t) = 2t, 1 (σ (t) + t) 3 1 = (2t + t) 3 = t,  1 (σ (t))2 + tσ (t) + t 2 hΔ 2 (t) = 7 1 = (4t 2 + 2t 2 + t 2 ) 7

hΔ 1 (t) =

= t 2, 1 hΔ 3 (t) = 1 + (σ (t) + t) 3

1.4 First Type Fuzzy Delta Integration

101

1 = 1 + (2t + t) 3 = 1 + t,  1 2 2 (σ (t)) (t) = 1 + + tσ (t) + t hΔ 4 7  1 2 4t + 2t 2 + t 2 = 1+ 7 = 1 + t 2,

 1 1 (σ (t) + t) + (σ (t))2 + tσ (t) + t 2 3 7 1 1 = (2t + t) + (4t 2 + 2t 2 + t 2 ) 3 7

hΔ 5 (t) =

t ∈ T.

= t + t 2, Hence, 4 1



4

F (t)δH t =



4

Δt, 1

2

t Δt 1

 = 3,

4 1



hΔ 2 (t)Δt

 t=4   = 3, h2 (t) t=1

 3  t t=4 = 3,  7 t=1   64 1 = 3, − 7 7 = {3, 9},  4 4 G(t)δH t = Δt, 1

1



4

tΔt 1

 = 3,

4 1

hΔ 1 (t)Δt

 t=4   = 3, h1 (t) t=1

 2  t t=4 = 3,  3 t=1



102

1 Calculus of Fuzzy Functions

  16 1 − = 3, 3 3 = {3, 5},  4 4 (F (t) + G(t))δH t = 2Δt, 1

1

4

(1 + t)Δt,

1

 = 6,

4 1

4

(1 + t )Δt,

1

hΔ 3 (t)Δt,

4 1

4

2

1

hΔ 4 (t)Δt,

4 1

 (t + t )Δt 2



hΔ 5 (t)Δt

 t=4 t=4 t=4     = 6, h3 (t) , h4 (t) , h5 (t) t=1

t=1

t=1

        t 3 t=4 t 2 t=4 t 3 t=4 t 2 + = 6, t + , t + ,    3 t=1 7 t=1 3 7 t=1   16 1 64 1 16 64 1 1 = 6, 4 + − 1 − , 4 + −1− , + − − 3 3 7 7 3 7 3 7 = {6, 8, 12, 14}. Therefore, 4

4

F (t)δH t +

1

4

G(t)δH t =

1

(F (t) + G(t))δH t.

1

Theorem 1.36 Let t0 , T ∈ T, t0 < T , F : [t0 , T ] → F (R) be δH -integrable. Then λ · F : [t0 , T ] → F (R) is δH -integrable and T

λ · F (s)δH s = λ ·

t0

T

F (s)δH s t0

for any λ ∈ R. Proof Let f ∈ SH F ([t0 , T ]) and T

u(T ) =

f (s)Δs. t0

Then, T

λu(T ) = λ

f (s)Δs t0 T

=

λf (s)Δs t0

 ∈

T t0

 λf (s)Δs : f ∈ SH F ([t0 , T ]) .

1.4 First Type Fuzzy Delta Integration

103

Therefore,  λ·

T

  f (s)Δs : f ∈ SH F ([t0 , T ]) ⊂

 g(s)Δs : g ∈ SH (λ·F ) ([t0 , T ]) .

T

t0

t0

(1.53) Next, let g ∈ SH (λ·F ) ([t0 , T ]). Then there exists an h ∈ SH F ([t0 , T ]) such that s ∈ [t0 , T ].

g(s) = λh(s), Hence, T

T

λh(s)Δs = λ

h(s)Δs

t0

t0

and thus,  T   g(s)Δs : g ∈ SH (λ·F ) ([t0 , T ]) ⊂ λ ·

T

t0

 f (s)Δs : f ∈ SH F ([t0 , T ]) .

t0

From the last relation and from (1.53), it follows that  T   T  λ· f (s)Δs : f ∈ SH F ([t0 , T ]) = g(s)Δs : g ∈ SH (λ·F ) ([t0 , T ]) t0

t0

and λ·

T

T

F (s)δH s =

t0

λ · F (s)δH s.

t0

This completes the proof. Theorem 1.37 Let t0 , T ∈ T, t0 < T , and F : [t0 , T ] → F (R) be δH -integrable. Then, T

t

F (s)δH s =

t0

T

F (s)δH s +

t0

F (s)δH s t

for any t ∈ [t0 , T ]. Proof Let f ∈ SH F ([t0 , T ]) and t ∈ [t0 , T ] be fixed. Then, T

t

f (s)Δs =

t0

T

f (s)Δs +

t0

f (s)Δs t



t

⊂



f (s)Δs : f ∈ SH F ([t0 , t])

t0



T

+ t

 f (s)Δs : f ∈ SH F ([t, T ]) .

104

1 Calculus of Fuzzy Functions

Consequently,

!

 !  t f (s)Δs : f ∈ SH F ([t0 , T ]) ⊂ t0 f (s)Δs : f ∈ SH F ([t0 , t]) 

! T + t f (s)Δs : f ∈ SH F ([t, T ]) . (1.54) Let now g ∈ SH F ([t0 , t]), h ∈ SH F ([t, T ]) and T t0

z=

t

T

g(s)Δs +

h(s)Δs.

t0

t

Define  h1 (s) =

g(s) if t0 ≤ s ≤ t h(s) if t ≤ s ≤ T .

Then, T

t

h1 (s)Δs =

t0

T

g(s)Δs +

t0

h(s)Δs t

= z. Thus, 

t t0





T

f (s)Δs : f ∈ SH F ([t0 , t]) + 

⊂

 f (s)Δs : f ∈ SH F ([t, T ])

t

T

 f (s)Δs : f ∈ SH F ([t0 , T ]) .

t0

From the last relation and from (1.54), we get 

t t0

=

  f (s)Δs : f ∈ SH F ([t0 , t]) + 

T

 f (s)Δs : f ∈ SH F ([t, T ])

t

T

 f (s)Δs : f ∈ SH F ([t0 , T ])

t0

and T

F (s)δH s =

t0

This completes the proof.

t t0

T

F (s)δH s +

F (s)δH s. t

1.4 First Type Fuzzy Delta Integration

105

Theorem 1.38 Let t0 , T ∈ T, t0 < T , F : [t0 , T ] → F (R) is rd-continuous. If X0 ∈ F (R) and t

f (t) = X0 +

t ∈ [t0 , T ],

F (s)δH s, t0

then f is δH -differentiable and δH f (t) = F (t),

t ∈ [t0 , T ].

Proof Since F : [t0 , T ] → F (R) is rd-continuous, there exists a pre-differentiable function f with region of differentiation D, such that δH f (t), t ∈ [t0 , T ], exists and δH f (t) = F (t),

t ∈ D.

Let t ∈ I κ \D. Then t is a right-dense point of T. Hence, for 0 < h < δ with t − h, t + h ∈ [t0 , T ], we have f (t + h) H f (σ (t)) =

t+h t0

=

σ (t)

F (s)δH s H

F (s)δH s t0

t+h

F (s)δH s. σ (t)

Thus, for  > 0 sufficiently small, using the rd-continuity of F , we have  D

f (t+h) H f (σ (t)) , F (t) h−μ(t)



1 D (f (t+h) H f (σ (t)), (h−μ(t)) · F (t)) h−μ(t)   t+h t+h 1 D F (s)δH s, F (t)δH s = h−μ(t) σ (t) σ (t) =

t+h 1 D(F (s), F (t))δH s h−μ(t) σ (t) 0 arbitrarily. By the rd-continuity of F , we get  D

 −1 · (G(σ (t)) H G(t + h)), F (t) h − μ(t)

1 D ((−1) · (G(σ (t)) H G(t + h)) , (h − μ(t)) · F (t)) h − μ(t)  t+h  t+h 1 = D F (s)δSH s, F (t)δSH s h − μ(t) σ (t) σ (t) =

1 h − μ(t) 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. (P4) 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 1.37 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 (s, t), (s, u) ∈ D± and t < u. Then,

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1 Calculus of Fuzzy Functions

δ+ (s, t) = st < su = δ+ (s, u), t δ− (s, t) = s u < s = δ− (s, u). Therefore, (P 1) holds. 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

1.6 Shift Operators—Properties

115

and (s, δ+ (s, t)) ∈ D− . Next, t s = t ∈ T∗ .

sδ− (t, s) = s

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

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1 Calculus of Fuzzy Functions

and (s, δ− (u, t)) ∈ D+ . 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 1.36 Let T = T∗ = Z, t0 = 0. Prove that δ+ (s, t) = t + s,

δ− (s, t) = t − s

satisfy (P 1)-(P 5). Theorem 1.46 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

1.6 Shift Operators—Properties

117

(u, t) ∈ ([t0 , ∞) × [t0 , ∞))



D+ ,

(vi) δ+ (s, t) ∈ [t0 , ∞) for all (s, t) ∈ D+ with t ≥ t0 , (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 by (P 4), we get δ− (t, t) = δ− (t, δ+ (t, t0 )) = t0 . (ii) By (P 3), we have δ+ (t0 , t) = t. Hence by (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 by (P 4), δ− (s, u) = δ− (s, δ+ (s, t)) = t.

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Let (s, u) ∈ D− . Then, (s, δ− (s, u)) ∈ D+ 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 obtain 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 get 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 obtain δ+ (u, t) = δ+ (u, δ− (u, δ+ (t, u))) = δ+ (t, u).

1.6 Shift Operators—Properties

119

(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) (a) Let μ(t) > 0. Then, Δt δ+ (s, t) =

δ+ (s, σ (t)) − δ+ (s, t) μ(t)

> 0. (b) 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)

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1 Calculus of Fuzzy Functions

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.

1.7 Complete-Closed Time Scales under Non-translational Shifts In this section, we will introduce a concept of complete-closed time scales. Suppose that T is an unbounded above time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let t0 ∈ T and T∗ be the largest subset of the time scale T, i.e., T∗ = T. If δ± (s, t) are Δ-differentiable with respect Δ (s, t) we will denote the Δ-derivative of to their second argument, then with δ± δ± (s, t) to their second argument. Definition 1.18 The time scale T is said to be periodic in shifts δ± if there exists a p ∈ (t0 , ∞) such that (p, t) ∈ D∓ for any t ∈ T∗ . Moreover, if P = inf{p ∈ (t0 , ∞) : (p, t) ∈ D∓

for

t ∈ T∗ } = t0 ,

any

then P is called the period of the time scale T. Example 1.38 Let T = −q Z

" {1} = {−q n : q > 1,

n ∈ Z}

"

{1}.

We have T∗ = {−q n : q > 1,

n ∈ Z}

" {1}.

Take t0 = 1 and δ− (s, t) = −st,

t δ+ (s, t) = − , s

(s, t) ∈ [1, ∞) × T∗ .

(1.60)

1.7 Complete-Closed Time Scales under Non-translational Shifts

121

Denote Π − = {−q n : q > 1,

n ∈ Z+ }.

We have Π − ⊂ T∗ . Fix s = −q n1 ∈ Π − arbitrarily, where n1 ∈ Z+ . Then, δ− (s, t) = q n1 t ∈ T∗ , δ+ (s, t) = q −n1 t ∈ T∗ ,

t ∈ T∗ .

Thus, δ± (s, t) ∈ T∗ for any s ∈ Π − and for any t ∈ T∗ . Note that there is no P that satisfies (1.60). Therefore, the time scale T cannot be regarded as a periodic time scale under the shifts δ± . Introduce the following notations: ± = {(s, t) ∈ T∗ × T∗ : δ± (s, t) ∈ T∗ }. D For any s ∈ T∗ , denote − , T∗s = δ− (s, T∗ ) = {δ− (s, t) : (s, t) ∈ D

∀t ∈ T∗ },

+ , T∗s = δ+ (s, T∗ ) = {δ+ (s, t) : (s, t) ∈ D

∀t ∈ T∗ }.

δ

δ

−

+

Definition 1.19 Let T be a time scale with the shift operators δ± associated with the initial point t0 ∈ T∗ . The time scale T is said to be bi-direction shift completeclosed time scale, shortly S-CCTS, in the shifts δ± if ± , Π = {p ∈ T∗ : (p, t) ∈ D

∀t ∈ T∗ } ∈ / {{t0 }, ∅}.

Denote δ

+

δ

−

Π + = {p ∈ T∗ : T∗p ⊆ T∗ }, Π − = {p ∈ T∗ : T∗p ⊆ T∗ }.

Definition 1.20 Let T be a S-CCTS. Then, 1. We say S-CCTS with positive direction if Π + ∈ / {{t0 }, ∅}. 2. We say S-CCTS with negative direction if Π − ∈ / {{t0 }, ∅}. 3. We say S-CCTS with bi-direction if Π ± ∈ / {{t0 }, ∅}.

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1 Calculus of Fuzzy Functions

Example 1.39 Let T = (−q)Z {1} = {(−q)n : q > 1, n ∈ Z} {0, 1}. Take t0 = 1 and for any t ∈ T∗ , s ∈ [1, ∞), consider the following shift operators:  δ+ (s, t) =  δ− (s, t) =

st t s t s

st

if t > 0 if t < 0, if t > 0 if t < 0.

We have δ

−

T∗ p = δ− (p, T∗ ) − , ∀t ∈ T∗ } = {δ− (p, t) : (p, t) ∈ D   t ∗ = : (p, t) ∈ D− , ∀t ∈ T , t > 0 p "# $ − , ∀t ∈ T∗ , t < 0 . pt : (p, t) ∈ D δ

−

Then T∗p ⊆ T∗ if and only if 

" # t ∗ − , : (p, t) ∈ D− , ∀t ∈ T , t > 0 pt : (p, t) ∈ D p $" # {1}. ⊆ (−q)n : q > 1, n ∈ Z

∀t ∈ T∗ ,

$ t 0 = tp : (p, t) ∈ D  " t ∗ : (p, t) ∈ D− , ∀t ∈ T , t < 0 p δ

+

and T∗p ⊆ T∗ if and only if #

$" + , ∀t ∈ T∗ , t > 0 tp : (p, t) ∈ D # ⊆ (−q)n : q > 1,

n∈Z

$" {1}.



t + , : (p, t) ∈ D p

∗

∀t ∈ T ,

 t 1, n ∈ Z+ } {1}. Take t0 = 1 and consider the shift operators δ+ (s, t) = st,

t δ− (s, t) = , s

t ∈ T∗ .

Then δ+ (s, t) ∈ T∗ for any t ∈ T∗ and if t1 = q, s1 = q 2 , we have t1 s1 q = 2 q

δ− (s1 , t1 ) =

=

1 ∈ / T∗ . q

/ {{1}, ∅} and then T is a S-CCTS with positive direction. Thus, Π + ∈ Example 1.41 Let T = {q n : q > 1, n ∈ Z− } {0, 1}. Take t0 = 1 and consider the shift operators t δ+ (s, t) = , s

δ− (s, t) = st,

t ∈ T∗ .

We have that δ− (s, t) ∈ T∗ for any s ∈ Π − , t ∈ T∗ . Let t1 = q1 , s1 = δ+ (s1 , t1 ) =

t1 s1

=

1 q 1 q2

1 . q2

Then,

=q∈ / T∗ . / {{1}, ∅}. Therefore, T is S-CCTS with negative direction. Note that Π − ∈ 1 √ Exercise 1.37 Consider N±2 = {± n : n ∈ N}. Take t0 = 0 and

√ s 2 + t 2 if t √ − t 2 − s 2 if √ t 2 − s 2 if t √ δ− (s, t) = − t 2 + s 2 if δ+ (s, t) =

>0 t < 0, >0 t < 0.

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1 Calculus of Fuzzy Functions

Prove

√ 1 1. Π ± = N 2 = { n : n ∈ N}. 2. T is S-CCTS with bi-direction. Definition 1.21 If T is a bi-direction S-CCTS and t0 ∈ T is the initial point, for any s ∈ Π , we define a function A : Π → Π as follows:  A(s) =

if if

δ+ (s, t0 ) δ− (s, t0 )

s > t0 s < t0 .

We have A(s) > t0

and

A(s) ≥ s.

1.8 Shift Almost Periodic Fuzzy Functions Assume that D ⊂ F (R) is an open set. Definition 1.22 Let T be a bi-direction S-CCTS and f : T × D → F (R) be a continuous fuzzy function on T × D. 1. The function f is called a shift almost periodic fuzzy function, shortly S-almost periodic fuzzy function, in t ∈ T uniformly for x ∈ D with shift operators if the -shift number set of f  E(, f, S0 ) = τ ∈ Π : D (f (δ+ (τ, t), x) , f (t, x)) <  for

all

t ∈T

∗

 and

x ∈ S0

is relatively dense with respect to the pair (Π, δ± ) for all  > 0 and for each compact subset S0 of D, that is, for any given  > 0 and each compact subset S0 of D, there exists a constant l(, S0 ) > 0 such that each interval of length l(, S0 ) contains a τ (, S0 ) ∈ E(, f, S0 ) so that D (f (δ± (τ, t), x) , f (t, x)) <  for all t ∈ T∗ and x ∈ S0 . Here τ is called the -shift number of f and l(, S0 ) is called the inclusion length of E(, f, S0 ). 2. The function f is called shift-normal fuzzy function, shortly S-normal fuzzy function, if for any sequence {Fn }n∈N , Fn : T × D → F (R), n ∈ N, of the form Fn (t, x) = f (δ+ (hn , t), x) ,

n ∈ N,

1.8 Shift Almost Periodic Fuzzy Functions

125

where {hn }n∈N ⊂ Π is a sequence of real numbers, one can extract a subsequence {Fnk }k∈N of {Fn }n∈N , such that   D Fnk (t, x), F (t, x) → 0,

as

k → ∞,

uniformly with respect to (t, x) ∈ T × D. 3. Let δ± (s, t) be Δ-differentiable to its second argument. The function f is called shift Δ-almost periodic fuzzy function, shortly S Δ-almost periodic fuzzy function, in t ∈ T uniformly for x ∈ D with shift operators if the -shift number set of f    Δ E(, f, S0 ) = τ ∈ Π : D f (δ+ (τ, t), x) δ± (τ, t), f (t, x) <  for all

t ∈T

∗

 and

x ∈ S0

is a relatively dense set with respect to the pair (Π, δ± ) for all  > 0 and for each compact subset S0 of D. 4. Let δ± (s, t) be Δ-differentiable to its second argument. The function f is called shift Δ-normal fuzzy function, shortly S Δ-normal fuzzy function, if for any sequence {Fn }n∈N , Fn : T × D → F (R), n ∈ N, of the form Δ (hn , t), Fn (t, x) = f (δ+ (hn , t), x) δ+

n ∈ N,

where {hn }n∈N ⊂ Π is a sequence of real numbers, one can extract a subsequence {Fnk }k∈N of {Fn }n∈N , such that   D Fnk (t, x), F (t, x) → 0,

as

k → ∞,

uniformly with respect to (t, x) ∈ T × D. We denote APS (T) the set of all shift almost periodic fuzzy functions in shifts on T. Let α = {αn }n∈N , β = {βn }n∈N ⊂ Π be two sequences. Then β ⊂ α means that β is a subsequence of α, δ± (α, β) = {δ± (αn , βn )}n∈N ,

δ− (α, t0 ) = {δ− (αn , t0 )}n∈N .

We introduce the moving operator TαS , TαS f (t, x) = g(t, x) by g(t, x) = lim f (δ+ (αn , t), x) . n→∞

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Theorem 1.47 Let T be a bi-direction S-CCTS with shifts δ± and f ∈ C (T × D) be S-almost periodic in t uniformly for x ∈ D, where δ+ (τ, t) is continuous in t. Then it is uniformly continuous and bounded on T∗ × S. Proof Take  ∈ (0, 1] and a compact set S0 ⊂ D arbitrarily. Since f is S-almost periodic in t uniformly for x ∈ D, there exists a constant l(, S0 ) such that in any interval of length l(, S0 ), there is a τ ∈ E(, t, S0 ) so that D (f (δ+ (τ, t), x) , f (t, x)) <  ≤ 1 for all (t, x) ∈ T∗ × S0 . Now, using that f ∈ C (T × D), if t0 ∈ T∗ is the initial point, for any (t, x) ∈ [t0 , δ+ (l, t0 )] × S0 there exists a constant M > 0 such that f (t, x) < M. For any given t ∈ T∗ we choose τ ∈ E(, f, S0 )



[δ− (t, t0 ), δ+ (l, δ− (t, t0 ))] .

We have that δ+ (τ, t) ∈ [t0 , δ+ (l, t0 )]. Hence, for x ∈ S0 , we have f (δ+ (τ, t), x)  < M and D (f (δ+ (τ, t), x) , f (t, x)) < 1. Thus, for all (t, x) ∈ T∗ × S0 , we get   f (t, x) = D f (t, x), 0

  0 ≤ D (f (δ+ (τ, t), x) , f (t, x)) + D f (δ+ (τ, t), x) ,

< M + 1.     Let now l1 3 , S0 be an inclusion length of E 3 , f, S0 . Note that f (t, x) is uniformly continuous on   [t0 , δ+ (l1 , t0 )] × S0 . Therefore, there exists a positive constant δ ∗ = δ ∗ 3 , S0 , for any t1 , t2 ∈ [t0 , δ+ (l1 , t0 )] and |t1 − t2 | < δ ∗ , D (f (t1 , x), f (t2 , x))
0 such that |t − v| < δ ∗∗ implies |δ+ (τ, t) − δ+ (τ, v)| < δ ∗ . Let δ∗∗ = min{δ ∗ , δ ∗∗ }. Then the inequality |t − v| < δ∗∗ implies D (f (δ+ (τ, t), x) , f (δ+ (τ, v), x))
0 and a compact set S0 ⊂ D arbitrarily. Let l = l 4 , S0 be   an inclusion length of E 4 , f, S0 . Note that for any given subsequence α  = {αn }n∈N ⊂ Π , there exists γ  = {γn }n∈N ⊂ Π and t0 ≤ γn ≤ l, n ∈ N, such that   αn = δ+ τn , γn

and

  γn = δ− τn , αn ,

  where τn ∈ E 4 , f, S0 , n ∈ N. Therefore, there exists a subsequence γ = {γn }n∈N ⊂ γ  = {γn }n∈N such that γn → s, as n → ∞, where t0 ≤ s ≤ l. By Theorem 1.47, it follows that f (t, x) is uniformly continuous on T∗ × S0 . Hence, there exists a δ ∗ = δ ∗ (, S0 ) > 0 such that |t1 − t2 | < δ ∗ , for x ∈ S0 , implies D (f (δ+ (τ, t1 ), x) , f (δ+ (τ, t2 ), x))
N implies |γp − γm | < δ ∗ . Take α ⊂ α  , τ ⊂ τ  such that α, τ common with γ . Then, for p, m ≥ N , we can obtain

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1 Calculus of Fuzzy Functions

      D f δ+ δ− (τm , τp ), t , x , f (t, x)        ≤ D f δ− τm , δ+ (τp , t) , x , f δ+ (τp , t), x     +D f δ+ (τp , t), x , f (t, x)   < + 4 4  = 2 and     δ+ δ− (αm , αp ), δ− (γp , γm ) = δ+ δ− (γp , αp ), δ− (αm , γm )   = δ+ τp , δ− (τm , t0 ) = δ− (τm , τp )   ∈E , f, S0 . 2 Hence, for p, m > N, we obtain     D f δ+ (αp , t), x , f (δ+ (αm , t), x)     ≤ sup D f δ+ (αp , t), x , f (δ+ (αm , t), x) (t,x)∈T∗ ×S0

≤ ≤

sup

      D f δ+ δ− (αm , αp ), t , x , f (t, x)

sup

         D f δ+ δ− (αm , αp ), t , x , f δ+ δ− (γm , γp ), t , x

(t,x)∈T∗ ×S0 (t,x)∈T∗ ×S

+

sup

0

(t,x)∈T∗ ×S0

      D f δ+ δ− (γm , γp ), t , x , f (t, x)

  + 2 2 = .


0 and sequences δm , m∈N ∗ ∗ {tm }m∈N , {xm }m∈N , where δm > 0, δm → 0, as m → ∞, ∗ |t0 − tm | + |x0 − xm | < δm

and D (g(t0 , x0 ), g(tm , xm )) ≥ 0 .

(1.61)

Let X = {xm }m∈N {x0 }. We have that X is a compact subset of D. Then there exists a positive integer N = N(0 , X) such that n > N implies D (f (δ+ (βm , tm ), xm ) , g(tm , xm ))
2|α1 |, where a1 , b1 ∈ Π such that there are no 0 -shift numbers of f (t, x) in this interval. Now, we take        / E(0 , f, S0 ). α2 ∈ δ+ α1 , a1 , δ+ α1 , b1 , δ− α1 , α2 ∈ (a1 , b1 ), δ− (α1 , α2 ) ∈ Then we take an interval (a2 , b2 ) with   b2 − a2 > 2 |α1 | + |α2 | , where a2 , b2 ∈ Π such that there are no 0 -shift numbers of f (t, x) in this interval. Select     α3 ∈ δ+ (α2 , a2 ), δ+ (α2 , b2 ) , α3 ∈ δ+ (α1 , a2 ), δ+ (α1 , b2 ) , / E(0 , f, S0 ). δ− (α2 , α3 ), δ− (α1 , α3 ) ∈ Repeating this process, we get numbers α4 , α5 , . . . such that / E(0 , f, S0 ), δ− (αj , αi ) ∈

i > j.

Hence, for any i = j , i, j ∈ N, without loss of generality, let i > j , for x ∈ S0 we obtain      D f δ+ (αi , t), x , f δ+ (αj , t), x sup (t,x)∈T∗ ×S0

1.9 Advanced Practical Problems

=

sup

(t,x)∈T∗ ×S0

131

      D f δ+ δ− (αj , αi ), t , x , f (t, x)

≥ 0 . Therefore, there$ is no uniformly convergent subsequence of the sequence #  f δ+ (αn , t), x n∈N , (t, x) ∈ T∗ × S0 . This is a contradiction. Thus, f (t, x) is S-almost periodic in t uniformly for x ∈ D under shifts δ± . This completes the proof.

1.9 Advanced Practical Problems Problem 1.1 Let T = [−1, 3] Let also,



{4k }k∈N , where [−1, 3] is the real-valued interval.

  t2 + 1 t2 + 1 , t + 2t 2 + t 4 + 3 2 f (t) = t, t + t 2 + 2 , t +4 t +4

t ∈ T.

Find δH f (3),

δH f (4),

δH f (16).

 Problem 1.2 Let T = [0, 2] 2 + 23k , where [0, 2] is the real-valued interval. Let also,   t +1 f (t) = 2 , t + 3, t 3 + t + 3 , t ∈ T. t +t +1 Find δH f (2). Problem 1.3 Let T = {0} f (t) =

t3

 g(t) =

1  2k k∈N

,

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

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

t ∈ T.

Prove that δH (f · g)(0) = f Δ (0) · g(0) + f (σ (0)) · δH g(0).

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1 Calculus of Fuzzy Functions

Problem 1.4 Let T = 2N0 and 1 , t2   t +1 g(t) = , t, t 2 + t , t +2

f (t) =

t ∈ T.

Prove that δH (f · g)(t) = (f (σ (t)) · δH g(t)) H

 Δ   −f (t) · g(t) ,

t ∈ T.

Problem 1.5 Find [δH f (t)]α , where 1. T = 7Z, [f (t)]α = g(t) · [−10, 41], g(t) = 3 + 4t + t 2 , t ∈ T, α ∈ [0, 1], 1+t 2. T = 3Z, [f (t)]α = g(t) · [−1, 14], g(t) = 1+2t , t ∈ T, α ∈ [0, 1], 1+t 2 , t ∈ T, 1+3t 4 1+t 8 α Z, [f (t)] = g(t) · [−7, 12], g(t) = 1+3t , t ∈ T, 1+t , t ∈ T, 2N0 , [f (t)]α = g(t) · [−10, 3], g(t) = 1+7t

3. T = N40 , [f (t)]α = g(t) · [−10, 3], g(t) = 4. T = 5. T =

α ∈ [0, 1], α ∈ [0, 1], α ∈ [0, 1].

Problem 1.6 Let T = [−2, 1] {3, 4, 7}, where [−2, 1] is the real-valued interval. Consider the function f : T → F (R) whose α-levels are given by [f (t)]α = g(t) · [1, 4], where  g(t) =

t (t+2) 2t+5 t2 − 1

if t ∈ [−2, 1] if t ∈ {3, 4, 7}.

Find [δH f (1)]α ,

[δH f (0)]α ,

[δH f (4)]α .

Problem 1.7 Let T = 4N0 and  F (t) =

1 1 1 , , t7 t5 t4

 ,

Find 256

F (s)δH s. 1

t ∈ T.

1.9 Advanced Practical Problems

133

Problem 1.8 Check if [δH (f + g)(t)]α = [δH f (t)]α + [δH g(t)]α ,

t ∈ T,

α ∈ [0, 1],

where 1. T = 7Z and

 [f (t)]α = −e−2 (t, 0) + t 2 , e−2 (t, 0) + t 2 + 7 , 

[g(t)]α = t 4 , t 4 + t 2 , 2. T = Z and 

[f (t)]α = t 2 , t 6 + t 4 + t 2 , 

[g(t)]α = t 2 , t 8 + 10t 6 + t 2 , 3. T = 3N0 and [f (t)]α = [− sin−3 (t, 0) − 2, sin−3 (t, 0) + 2] , 

[g(t)]α = t 2 , t 10 + t 6 + t 4 + t 2 . Problem 1.9 Let T = 3N0 . Find 

α

27

f (t)δH t

α ∈ [0, 1],

,

1

where α

[f (t)]α = 2t + 8, t 3 + 2t 2 ,

t ∈ T,

α ∈ [0, 1].

Problem 1.10 Let T = N,  f (t) =

 1 1 1 , , , t +7 t +6 t +5

t ∈ T.

Find δSH f (t), t ∈ T. Problem 1.11 Let f, g : [a, b] → F (R) be δH -differentiable and f ◦ g is also δH -differentiable on [a, b]. If there is no switching point in [a, b] and α,1 (t) ≥ 0, If,g

Ifα,1 σ ,δ g (t) ≤ 0, H

Iδα,1 (t) ≥ 0, H f,g

t ∈ [a, b]κ ,

134

1 Calculus of Fuzzy Functions

prove that !b a

f (σ (t)) ◦ δH g(t)δH t = (f (b) ◦ g(b)) H (f (a) ◦ g(a)) !b H a δH f (t) ◦ g(t)δH t.

Problem 1.12 Let T = {2}



f (t) =

2+



1 , 4k k∈N

 1 1 1 , , , t + 11 t + 7 t + 1

t ∈ T.

Find δSH f (2), Problem 1.13 Let T = {1}

δSH f



1+

  9 , 4

 δSH f

 33 . 16



1 , 4k k∈N

 1 1 1 , , , t2 + 4 t2 + 3 t2 + 2   1 1 1 , , , t ∈ T. g(t) = t + 11 t + 9 t + 7 

f (t) =

Prove that δSH (f + g)(1) = δSH f (1) + δSH g(1). Problem 1.14 Let T = 3N0 , f : T → F (R) and  1 1 , , [f (t)] = 2t + 1 t 

α

t ∈ T.

Find [δSH f (t)]α . Problem 1.15 Let T = 3N0 , 1 , t2 + t + 1   1 1 1 g(t) = , , , 4t 3 3t 3 t 3

f (t) =

t ∈ T.

Prove that δSH (f · g)(t) = f Δ (t) · g(t) + f (σ (t)) · δSH g(t),

t ∈ T.

1.10 Notes and References

135

Problem 1.16 Let T = 2N, f (t) = t 2 ,   1 1 1 , g(t) = , , 7t 4 5t 4 3t 4

t ∈ T.

Prove that   δSH (f · g)(t) = f Δ (t) · g(t) H ((−f (σ (t))) · δSH g(t)) ,

t ∈ Tκ .

Problem 1.17 Consider T = {q n : q > 1,

n ∈ Z}

" {−q n : q > 1,

n ∈ Z}

" {0}.

Take t0 = 1 and  δ+ (s, t) =  δ− (s, t) =

st t s

if t > 0 if t < 0,

t s

if t > 0 st if t < 0.

Prove 1. δ± (s, t) satisfy all axioms for shift operators. 2. Π ± = {q n : q > 1, n ∈ Z+ }. 3. T is S-CCTS with bi-direction. Problem 1.18 Let T be a bi-direction S-CCTS with shifts δ± and f : T × D → F (R), f ∈ C (T × D), be S-almost periodic in t uniformly for x ∈ D under shifts δ± , where δ+ (τ, t) is continuous in t. Let also, α ∈ R. Prove that α · f is S-almost periodic in t uniformly for x ∈ D under shifts δ± .

1.10 Notes and References In this chapter, we deal with fuzzy functions on time scales. We focus on a new class derivative and a new class integral of such fuzzy functions. The corresponding fundamental properties of the introduced derivative and integral are studied and discussed. In this chapter, we introduce the concept for shifts operators. The shift almost periodic fuzzy functions are defined, and some of their properties are deducted. The results contained in this chapter can be found in the papers [1, 16, 43].

Chapter 2

First Order Fuzzy Dynamic Equations

This chapter is devoted to a qualitative analysis of first order fuzzy dynamic equations. First, we deducted formulae for the solutions of linear first order fuzzy dynamic equations and then investigated the Cauchy problem for first order fuzzy dynamic equations for existence and uniqueness. For this aim we introduced Lipschitz fuzzy functions and determined some of their properties. In this chapter we investigated the solutions of first order fuzzy dynamic equations for continuous dependence on the initial data. Some criteria as to when the trivial solution of first order fuzzy dynamic equations is equi-stable, uniformly stable, uniformly asymptotically stable, equi-asymptotically stable, exponentially stable, uniformly exponentially stable, and uniformly asymptotically stable are discussed. Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

2.1 Linear First Order Fuzzy Dynamic Equations Consider the following IVP: δH y(t) = a(t) · y(t) + b(t),

t ∈ (t0 , t1 ],

(2.1)

y(t0 ) = y0 ,

(2.2)

where t0 , t1 ∈ T, t0 < t1 , a : [t0 , t1 ] → R, a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R). We will start with the following useful result. Theorem 2.1 Let a > 0 on [t0 , t1 ] and   t y1 (t) = ea (t, t0 ) · y0 + ea (σ (τ ), t0 ) · b(τ )δH τ ,

t ∈ [t0 , t1 ].

t0

Then y1 is δH -differentiable and it is a solution of the IVPs (2.1), (2.2). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_2

137

138

2 First Order Fuzzy Dynamic Equations

Proof Let f (t) = ea (t, t0 ), g(t) = y0 +

t t0

ea (σ (τ ), t0 ) · b(τ )δH τ,

t ∈ [t0 , t1 ].

We have f Δ (t) = a(t)ea (t, t0 ),

t ∈ [t0 , t1 ].

Hence, f (t)f Δ (t) > 0,

f (σ (t))f Δ (t) > 0,

t ∈ [t0 , t1 ].

Therefore all conditions of Theorem 1.17 are satisfied. Consequently y1 is δH differentiable on [t0 , t1 ] and δH y1 (t) = f Δ (t) · g(t) + f (σ (t)) · δH g(t) = a(t) · (f (t) · g(t)) + ea (σ (t), t0 ) · (ea (σ (t), t0 ) · b(t)) = a(t) · y(t) + (ea (σ (t), t0 )ea (t0 , σ (t))) · b(t) = a(t) · y(t) + b(t),

t ∈ [t0 , t1 ].

Also, 0) y1 (t0 ) = 1 · (y0 + = 1 · y0 = y0 . Consequently y1 is a solution of the IVPs (2.1), (2.2). This completes the proof. Example 2.1 Let T = R. Consider the following IVP: δH y(t) = y(t) + (t, t, t),

t > 0,

y(0) = (1, 2, 3). Here a(t) = 1, ea (t, 0) = et , ea (t, 0) = e−t ,

t > 0.

2.1 Linear First Order Fuzzy Dynamic Equations

139

We have   y1 (t) = e · (1, 2, 3) +

t

t

e

−s

t

sds,

0

e

−s

t

sds,

0

e

−s

 sds

,

t ≥ 0.

0

Note that t 0

s=t  e−s sds = −e−s s  + s=0

t

= −e−t t − e−s  −t

= −e t − e

−t

e−s ds

0 s=t  s=0

+1 −t

= 1 − (1 + t)e ,

t ≥ 0.

Therefore    y1 (t) = et · (1, 2, 3) + 1 − (1 + t)e−t , 1 − (1 + t)e−t , 1 − (1 + t)e−t = (et , 2et , 3et ) + (et − 1 − t, et − 1 − t, et − 1 − t) = (2et − 1 − t, 3et − 1 − t, 4et − 1 − t),

t ≥ 0,

is a solution of the considered IVP. The solution is shown in Fig. 2.1 Example 2.2 Let T = N0 . Consider the IVP δH y(t) = t · y + (t, t 2 , t 2 + t),

t > 0,

y(0) = (−1, 0, 1). Here σ (t) = t + 1, a(t) = 1, b(t) = (t, t 2 , t 2 + t), Fig. 2.1 δH -differentiable solution of Example 2.1

t ∈ T.

400 300 200 100 1

2

3

4

5

140

2 First Order Fuzzy Dynamic Equations

We have ea (t, 0) = e =e =e =

!t

1 0 μ(τ )

!t

log(1+μ(τ )a(τ ))Δτ

log(1+τ )Δτ

0

%t−1

s=0 log(1+s)

t−1 &

(1 + s),

s=0

a(t) 1 + μ(t)a(t) t , =− 1+t t 1 + μ(t)(a)(t) = 1 − 1+t 1 , = 1+t a(t) = −

ea (σ (t), 0) = e =e

! σ (t) 0

! σ (t) 0

= e−

1 μ(τ )

log

log(1+μ(τ )(a)(τ ))Δτ

1 1+τ

Δτ

%t

s=0 log(1+s)

1 , s=0 (1 + s)

= 't

1 · (t, t 2 , t 2 + t) s=0 (1 + s)

ea (σ (t), 0) · b(t) = 't  = t 0

t−1 ( τ τ 'τ Δτ = , s=0 (1 + s) s=0 (1 + s)

'τ

τ =0

t 0

t−1 ( τ2 τ2 'τ Δτ = , s=0 (1 + s) s=0 (1 + s)

'τ

τ =0

t 0

 t2 t2 + t t , 't , 't , 't s=0 (1 + s) s=0 (1 + s) s=0 (1 + s)

t−1 ( τ2 + τ τ2 + τ 'τ 'τ Δτ = , s=0 (1 + s) s=0 (1 + s) τ =0

t > 0,

2.1 Linear First Order Fuzzy Dynamic Equations

141

and t

y0 + 0

 t−1 ( τ 'τ ea (σ (τ ), 0) · b(τ )Δτ = −1+ , s=0 (1+s) τ =0

t−1 (

 t−1 ( τ2 τ 2 +τ ' , 1+ , τ s=0 (1+s) s=0 (1+s)

'τ

τ =0

t > 0.

τ =0

Consequently  y1 (t) = ea (t, 0) · y0 +

t 0

=

t−1 &

 ea (σ (τ ), 0) · b(τ )Δτ

 (1 + s) ·

−1+

t−1 (

( τ2 τ2 + τ 'τ ,1 + s=0 (1 + s) s=0 (1 + s) t−1

τ =0

τ =0

 (1 + s) −1 +

s=0

s=0



'τ

& t−1

t−1 &

τ , s=0 (1 + s)

'τ

τ =0

s=0

=

t−1 (

(1 + s)

 t−1 ( τ =0

t−1 ( τ =0

 τ 'τ , s=0 (1 + s)

  t−1  t−1 & ( τ2 τ2 + τ 'τ 'τ , , (1 + s) 1 + s=0 (1 + s) s=0 (1 + s) τ =0

s=0

t > 0, is a solution of the considered IVP. Example 2.3 Let T = 2N0 . Consider the following IVP: δH y = (1 + t) · y + (1, t, t 2 ),

t > 1,

y(1) = (1, 1, 1). Here σ (t) = 2t, a(t) = 1 + t, b(t) = (1, t, t 2 ),

t ∈ T.

We have ea (t, 1) = e =e

!t

1 1 μ(τ )

!t

1 1 τ

log(1+μ(τ )a(τ ))Δτ

log(1+τ +τ 2 )Δτ

142

2 First Order Fuzzy Dynamic Equations

=e

% 2t

s=1 log(1+s+s

2)

t

=

2 &

(1 + s + s 2 ),

s=1

(a)(t) = −

a(t) 1 + μ(t)a(t)

=−

1+t 1 + t (1 + t)

1+t , 1 + t + t2 t (1 + t) 1 + μ(t)(a)(t) = 1 − 1 + t + t2 =−

1 + t + t2 − t − t2 1 + t + t2 1 = , 1 + t + t2 =

ea (σ (t), 1) = e =e =e =

! 2t 1

! 2t 1

1 μ(τ )

log(1+μ(τ )(a)(τ ))Δτ

1 μ(τ )

log

1 Δτ 1+τ +τ 2

%t

1 s=1 log 1+s+s 2

t & s=1

1 , 1 + s + s2

t

t 1

ea (σ (τ ), 1)Δτ =

2 (

μ(s)ea (σ (s), 1)

s=1 t

=

2 (

 s

s & τ =1

s=1

1 1 + τ + τ2

 ,

t

t 1

τ ea (σ (τ ), 1)Δτ =

2 (

sμ(s)ea (σ (s), 1)

s=1 t

=

2 (

s=1

 s

2

s & τ =1

1 1 + τ + τ2

 ,

2.1 Linear First Order Fuzzy Dynamic Equations

143

t

t 1

τ ea (σ (τ ), 1)Δτ = 2

2 (

s 2 μ(s)ea (σ (s), 1)

s=1



t

=

2 (

s

s &

3

1 1 + τ + τ2

τ =1

s=1

 t > 1.

,

Hence, 

t

t

ea (σ (τ ), 1) · b(τ )δH τ =

1

ea (σ (τ ), 1)Δτ,

1 t

τ ea (σ (τ ), 1)Δτ,

1 t 1

 τ 2 ea (σ (τ ), 1)Δτ

t  ( s 2 & s =

τ =1

s=1 t

2 (

 s

τ =1

s=1 t

2 (

s &

2



s &

3

s

τ =1

s=1

1 1 + τ + τ2

1 1 + τ + τ2 1 1 + τ + τ2

 ,

 ,  t > 1,

,

and t

y0 + 1

t   s 2 ( & ea (σ (τ ), 1) · b(τ )δH τ = 1 + s

τ =1

s=1 t

1+

2 (

 s2

τ =1

s=1 t

1+

2 (

s=1

s &

 s

3

s & τ =1

1 1 + τ + τ2 1 1 + τ + τ2 1 1 + τ + τ2

 ,

 ,  ,

t > 1,

144

2 First Order Fuzzy Dynamic Equations

and  ea (t, 1) · y0 +

t 1

t

2 &

=

 ea (σ (τ ), 1) · b(τ )δH τ

t   s 2 ( & (1 + s + s ) · 1 + s

s=1 2 (

1+

 s

2

t

2 (

1+

s=1

& t 2

s &

 s

3

s &

⎛

s=1 t

(1 + s + s 2 ) ⎝1 +

s=1

 s

2 (

t

(1 + s + s 2 ) ⎝1 +

s=1

2 (

⎞ 1 ⎠, 1 + τ + τ2

s & τ =1

 s2

s & τ =1

s=1

⎛

t

2 &

2 (

s=1

⎛

t



t

(1 + s + s 2 ) ⎝1 +

,

,

1 1 + τ + τ2

τ =1





1 1 + τ + τ2

τ =1

s=1

2 &

τ =1

s=1 t

=

1 1 + τ + τ2

2

 s3

s & τ =1

s=1

⎞ 1 ⎠, 1 + τ + τ2 ⎞  1 ⎠ , 1 + τ + τ2

t > 1.

Consequently  y1 (t) = ea (t, 1) · y0 +

t 1

& t 2

=

 ea (σ (τ ), 1) · b(τ )δH τ

⎛

(1 + s + s 2 ) ⎝1 +

s=1 t

2 &

t

s

⎛ (1 + s + s 2 ) ⎝1 +

t

2 (

 s2

(1 + s + s 2 ) ⎝1 +

s=1

is a solution of the considered IVP.

t

2 (

s=1

s & τ =1

s=1

⎛

s & τ =1

s=1

s=1 2 &



t

2 (

 s3

s & τ =1

⎞ 1 ⎠, 1 + τ + τ2

⎞ 1 ⎠, 1 + τ + τ2 ⎞  1 ⎠ , 1 + τ + τ2

t > 1,

2.1 Linear First Order Fuzzy Dynamic Equations

145

Example 2.4 Let T = 2N0 . Consider the IVP δH y = {1, t, 1 + t},

t > 1,

y(1) = {−1, 0, 1}. Here σ (t) = 2t,

t ∈ T.

Let h1 (t) =

1 2 t , 3

t ∈ T.

Then 1 (σ (t) + t) 3 1 = (2t + t) 3 = t, t ∈ T.

hΔ 1 (t) =

Note that t 1



t

{1, s, 1 + s}δH s =

t

Δs, 1

t

sΔs, 1

 (1 + s)Δs

1

 s=t s=t    = t − 1, h1 (s) , (s + h1 (s)) s=1

s=1

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

y(t) = y(1) +

{1, s, 1 + s}δH s

1

  1 1 4 = {−1, 0, 1} + t − 1, (t 2 − 1), t 2 + t − 3 3 3  1 1 7 = t − 2, (t 2 − 4), t 2 + t − , 3 3 3 1 1 4 t − 1, (t 2 − 1), t 2 + t − , 3 3 3  1 1 1 t, (t 2 + 2), t 2 + t − , t ≥ 1, 3 3 3 is a solution of the considered IVP.

t > 1.

146

2 First Order Fuzzy Dynamic Equations

Exercise 2.1 Let T = 2Z. Find a solution of the following IVP: δH y = (1 + t 2 ) · y + (1 + t 2 , 1 + t 2 + t 4 , 3 + t 2 + t 4 ),

t > 0,

y(0) = (−3, −2, 2). Now we consider the following IVP: δH y + (−a(t)) · y = b(t),

t ∈ (t0 , t1 ],

(2.3)

y(t0 ) = y0 ,

(2.4)

where t0 , t1 ∈ T, t0 < t1 , a : [t0 , t1 ] → R, a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R). Theorem 2.2 Let a < 0 on [t0 , t1 ], a ∈ R+ , and  y1 (t) = ea (t, t0 ) · y0 +

t t0

 ea (σ (τ ), t0 ) · b(τ )δH τ ,

t ∈ [t0 , t1 ].

Suppose that there exist the H -differences y1 (t + h) H y1 (σ (t)) and

y1 (σ (t)) H y1 (t − h),

t ∈ [t0 , t1 ],

for h > 0 sufficiently small. Then y1 is a solution of the IVPs (2.3), (2.4). Proof Let f (t) = ea (t, t0 ), g(t) = y0 +

t t0

ea (σ (τ ), t0 ) · b(τ )δH τ,

t ∈ [t0 , t1 ].

We have y1 (t) = f (t) · g(t), f Δ (t) = a(t)ea (t, t0 ), f (t)f Δ (t) < 0, f (σ (t))f Δ (t) < 0,

t ∈ [t0 , t1 ].

Therefore f · g satisfies all conditions of Theorem 1.19. Consequently δH y1 (t) = δH (f · g)(t) = (f (σ (t)) · δH g(t)) H

 Δ   −f (t) · g(t)

2.1 Linear First Order Fuzzy Dynamic Equations

147

= ((ea (σ (t), t0 )) · (ea (σ (t), t0 ) · b(t))) H ((−a(t)ea (t, t0 )) · g(t)) = ((ea (σ (t), t0 )ea (t0 , σ (t))) · b(t)) H ((−a(t)ea (t, t0 )) · g(t)) = b(t) H ((−a(t)ea (t, t0 )) · g(t)) ,

t ∈ (t0 , t1 ].

Hence, δH y1 (t) + (−a(t)) · y1 (t) = b(t),

t ∈ (t0 , t1 ].

Also, y1 (t0 ) = y0 . Therefore y1 is a solution of the IVPs (2.3), (2.4). This completes the proof. Example 2.5 Let T = 2N0 . Consider the following IVP: δH y +

  1 · y = 22t , 22t+2 , 22t+4 , 4

t > 0,

y(0) = (0, 1, 1). Here σ (t) = t + 2, 1 a(t) = − , 4   b(t) = 22t , 22t+2 , 22t+4 ,

t ∈ T.

We have ea (t, 0) = e =e

!t

1 0 μ(τ )

!t

1 0 2

log(1+μ(τ )a(τ ))Δτ

  log 1− 24 Δτ

%t−2

1

= e s=0 log 2 1 = t, 22 a(t) (a)(t) = − 1 + μ(t)a(t) =−

− 14 1−

1 4

·2

148

2 First Order Fuzzy Dynamic Equations

=

1 4

1−

=

1 4 1 2

=

1 , 2

ea (σ (t), 0) = e =e =e

1 2

! σ (t) 0

! t+2 0

1 μ(τ ) 1 2

log(1+μ(τ )(a)(τ ))Δτ

  log 1+2· 21 Δτ

%t

s=0 log 2

t

= 2 2 +1 , t 0

ea (σ (τ ), 0)22τ Δτ =

t

τ

22τ · 2 2 +1 Δτ

0 t

=

5

2 2 τ +1 Δτ

0

=

t−2 (

5

2 2 τ +2

s=0 5

22t − 1 =4 5 2 −1  4  5t 22 − 1 , = 31

t ∈ T.

Hence, 

 16  5  4  5t 22 − 1 , 22t − 1 , 31 31 0    64 5 t 22 − 1 , 31     t 5 4 16  5 t 22t − 1 , 1 + 22 − 1 , y0 + ea (σ (τ ), 0) · b(τ )Δτ = 31 31 0    64 5 t 2 2 − 1 , t ∈ T. 1+ 31 t

ea (σ (τ ), 0) · b(τ )Δτ =

2.1 Linear First Order Fuzzy Dynamic Equations

149

Consequently  y1 (t) = ea (t, 0) · y0 +

t 0



 ea (σ (τ ), 0) · b(τ )Δτ

  4  5t 16  5 t 22 − 1 , 1 + 22 − 1 , = t 31 2 2 31    64 5 t 22 − 1 1+ 31    5   5t 1 16 2 2 t − 1 4 22 − 1 , , + = t t t 31 31 22 22 22   5 1 64 2 2 t − 1 , t > 0. t + t 31 22 22 1

Exercise 2.2 Let T = 2N0 . Find a solution of the following IVP: δH y + t · y = (1, 2, 1 + t + t 2 ),

t > 1,

y(1) = (−3, −1, 0). Now we will consider the following IVP: δSH y = a(t) · y + b(t),

t ∈ (t0 , t1 ],

(2.5)

y(t0 ) = y0 ,

(2.6)

where t0 , t1 ∈ T, t0 < t1 , a : [t0 , t1 ] → R, a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R). Theorem 2.3 Let a < 0 on [t0 , t1 ] and the H -difference t

y0 H

t0

(−ea (σ (τ ), t0 )) · b(τ )δSH τ

exists for any t ∈ [t0 , t1 ]. Then  y2 (t) = ea (t, t0 ) · y0 H

t t0

 (−ea (σ (τ ), t0 )) · b(τ )δSH τ ,

is δSH -differentiable and it is a solution of the IVPs (2.5), (2.6).

t ∈ [t0 , t1 ],

150

2 First Order Fuzzy Dynamic Equations

Proof Let f (t) = ea (t, t0 ), t

g(t) = y0 H

t0

(−ea (σ (τ ), t0 )) · b(τ )δSH τ,

t ∈ [t0 , t1 ].

Then y2 (t) = f (t) · g(t),

t ∈ [t0 , t1 ].

Note that g is δSH -differentiable on [t0 , t1 ] and δSH g(t) = ea (σ (t), t0 ) · b(t),

t ∈ [t0 , t1 ].

Since a < 0 on [t0 , t1 ], we have f Δ (t) = a(t)ea (t, t0 ) t ∈ [t0 , t1 ].

< 0, Therefore f (t)f Δ (t) < 0

and

f (σ (t))f Δ (t) < 0,

t ∈ [t0 , t1 ].

Thus, the conditions of Theorem 1.30 are satisfied. From here, we conclude that the function y2 is δSH -differentiable on [t0 , t1 ] and δSH y2 (t) = f Δ (t) · g(t) + f (σ (t)) · δSH g(t) = a(t) · (f (t) · g(t)) + (ea (σ (t), t0 )) · (ea (σ (t), t0 ) · b(t)) = a(t) · y2 (t) + (ea (σ (t), t0 )ea (t0 , σ (t))) · b(t) = a(t) · y2 (t) + b(t),

t ∈ [t0 , t1 ].

Next, using that t0 t0

(−ea (σ (τ ), t0 )) · b(τ )δSH τ = 0,

we conclude that g(t0 ) = y0 and y2 (t0 ) = f (t0 ) · g(t0 ) = 1 · y0 = y0 . Consequently y2 is a solution of the IVPs (2.5), (2.6). This completes the proof.

2.1 Linear First Order Fuzzy Dynamic Equations

151

Example 2.6 Let T = R. Consider the IVP δSH y = (−1) · y + (t, t, t),

t > 0,

y(0) = (1, 2, 3). Here a(t) = −1 < 0, b(t) = (t, t, t), ea (t, 0) = e−t , ea (σ (t), 0) = et ,

t ≥ 0.

Next, t 0

ea (σ (s), 0)sds =

t

ses ds

0

s=t  = ses  − s=0

t

es ds

0 s=t 

= tet − es 

s=0

= te − e + 1 t

t

= (t − 1)et + 1,

t ≥ 0.

Hence, t 0

 (−ea (σ (τ ), 0)) · b(τ )δSH τ =

t

− 0 t

−

ea (σ (s), 0)sds

0

=



t

− 

ea (σ (s), 0)sds,

0

ea (σ (s), 0)sds

− (t − 1)et − 1, −(t − 1)et − 1  −(t − 1)e − 1 , t

t ≥ 0,

152

2 First Order Fuzzy Dynamic Equations

and t

y0 H 0

(−ea (σ (τ ), 0)) · b(τ )δSH τ

= (1, 2, 3) H (−(t − 1)et − 1, −(t − 1)et − 1, −(t − 1)et − 1) = (2 + (t − 1)et , 3 + (t − 1)et , 4 + (t − 1)et ), and

 y2 (t) = ea (t, 0) · y0 H

t 0

=e

−t

t ≥ 0, 

(−ea (σ (τ ), 0)) · b(τ )δSH τ

· (2 + (t − 1)e , 3 + (t − 1)et , 4 + (t − 1)et ) t

= (2e−t + t − 1, 3e−t + t − 1, 4e−t + t − 1), is a solution of the considered IVP and it is shown in Fig. 2.2. Example 2.7 Let T = 3N0 . Consider the following IVP:   t · y + (t, t, t), t > 1, δSH y = − 4 y(1) = (1, 1, 1). Here σ (t) = 3t, t a(t) = − 4 < 0, b(t) = (t, t, t),

t ∈ T,

y0 = (1, 1, 1). 3.0 2.5

Out[11]=

2.0 1.5 1.0 1

2

Fig. 2.2 δSH -differentiable solution of Example 2.6

3

4

t ≥ 0,

2.1 Linear First Order Fuzzy Dynamic Equations

153

We have (a)(t) = −

a(t) 1 + μ(t)a(t)

=−

− 4t   1 + 2t − 4t

=

t 4 t2 2

1−

t , 2(2 − t 2 ) & (1 + 2s(a)(s)) ea (σ (t), 1) = =

s∈[1,t]

&  1+ = s∈[1,t]

=

& s∈[1,t]

ea (t, 1) = =

&

2s 2 2(2 − s 2 )



2 , 2 − s2 (1 + 2sa(s))

  s∈ 1, 3t

 s  &  1 + 2s − 4  t

s∈ 1, 3

 &  s2 1− , = 2  t s∈ 1, 3

 (−ea (σ (t), 1)) · b(t) =

−t

& s∈[1,t]

−t

& s∈[1,t]

−t

& s∈[1,t]

2 , 2 − s2

2 , 2 − s2  2 , 2 − s2

t > 1.

154

2 First Order Fuzzy Dynamic Equations

Note that

⎛ t

1

t

(−ea (σ (τ ), 1)) τ Δτ = −

τ⎝

1

s∈[1,τ ]

⎛

t

= −2

&

3 (

⎞

&

τ2 ⎝

τ =1

⎞ 2 ⎠ Δτ 2 − s2

s∈[1,τ ]

2 ⎠ , 2 − s2

t > 1.

Hence, t 1

 (−ea (σ (τ ), 1)) · b(τ )δSH τ =

⎛

t

−2

3 (

τ =1 t

−2

3 (

⎛

t

−2 t 1

⎞ 2 ⎠ , 2 − s2

⎞ 2 ⎠ , 2 − s2

& s∈[1,τ ]

⎞  2 ⎠ , 2 − s2

&

τ2 ⎝

τ =1

y0 H

s∈[1,τ ]

⎛

τ2 ⎝

τ =1 3 (

τ2 ⎝

&

s∈[1,τ ]

(−ea (σ (τ ), 1)) · b(τ )δSH τ = (1, 1, 1)  −2

H

⎛

t

3 (

τ2 ⎝

τ =1

⎛

t

−2

3 (

τ2 ⎝

τ =1

⎛

t

−2

3 (

τ2 ⎝

τ =1

&

s∈[1,τ ]

& s∈[1,τ ]

& s∈[1,τ ]

⎞ 2 ⎠ , 2 − s2 ⎞  2 ⎠ 2 − s2

⎛ t  3 & ( = 1+2 τ2 ⎝ τ =1 t

1+2

3 (

τ =1 t

1+2

3 (

τ =1

⎛

τ2 ⎝

s∈[1,τ ]

&

s∈[1,τ ]

⎛ τ2 ⎝

⎞ 2 ⎠ , 2 − s2

&

s∈[1,τ ]

⎞ 2 ⎠ , 2 − s2 ⎞

2 ⎠ , 2 − s2 ⎞

 2 ⎠ , 2 − s2

t > 1,

2.1 Linear First Order Fuzzy Dynamic Equations

155

and 

t

y2 (t) = ea (t, 1) · y0 H 1

 (−ea (σ (τ ), 1)) · b(τ )δSH τ

⎞ ⎛ ⎞ t    3 2 & ( & s ⎟ 2 ⎜ ⎠, 1− τ2 ⎝ =⎝ ⎠· 1+2 2 2 − s2  t ⎛

τ =1

s∈ 1, 3

t

1+2

3 (

τ =1 t 3

1+2 ⎛

( τ =1

⎛ τ2 ⎝ ⎛ τ2 ⎝

& s∈[1,τ ]

& s∈[1,τ ]

s∈[1,τ ]

⎞ 2 ⎠ , 2 − s2 ⎞  2 ⎠ 2 − s2

⎞⎛ ⎛ ⎞⎞ t   3 2 & & ( s 2 ⎜ ⎟⎝ ⎠⎠ , 1− τ2 ⎝ ⎝ ⎠ 1+2 2 2 − s2  t

 =

τ =1

s∈ 1, 3

s∈[1,τ ]

⎞⎛ ⎛ ⎞⎞ t   3 2 & ( & s ⎟⎝ 2 ⎠⎠ ⎜ 1− τ2 ⎝ , ⎠ 1+2 ⎝ 2 2 − s2  t ⎛

τ =1

s∈ 1, 3

s∈[1,τ ]

⎞⎛ ⎛ ⎞⎞ t    3 2 & ( & s ⎟⎝ 2 ⎠⎠ ⎜ 2⎝ , 1 − τ 1 + 2 ⎠ ⎝ 2 2 − s2  t ⎛

τ =1

s∈ 1, 3

t > 1,

s∈[1,τ ]

is a solution of the considered IVP. Exercise 2.3 Let T = 2Z. Find a solution of the following IVP:   t + t2 δSH y = − · y + (1, 1 + 2t, 1 + 2t 2 ), 2

t > 1,

y(1) = (−1, 0, 1). Now we consider the following IVP: δSH y + (−a(t)) · y = b(t), y(t0 ) = y0 ,

t ∈ (t0 , t1 ],

(2.7) (2.8)

where t0 , t1 ∈ T, t0 < t1 , a : [t0 , t1 ] → R, a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R).

156

2 First Order Fuzzy Dynamic Equations

Theorem 2.4 Let a > 0 on [t0 , t1 ] and  y2 (t) = ea (t, t0 ) · y0 H

t t0

 (−ea (σ (τ ), t0 )) · b(τ )δSH τ ,

t ∈ [t0 , t1 ].

Suppose that the H -differences y2 (σ (t)) H y(t + h) and

y2 (t − h) H y(σ (t))

exist for t ∈ [t0 , t1 ] and h > 0 sufficiently small. Then y2 is δSH -differentiable on [t0 , t1 ] and it is a solution of the IVPs (2.7), (2.8). Proof Let f (t) = ea (t, t0 ), g(t) = y0 H

t t0

(−ea (σ (τ ), t0 )) · b(τ )δSH τ,

t ∈ [t0 , t1 ].

Then y2 (t) = f (t) · g(t),

t ∈ [t0 , t1 ].

Note that f Δ (t) = a(t)ea (t, t0 ) > 0, Δ

f (t)f (t) > 0, f (σ (t))f Δ (t) > 0,

t ∈ [t0 , t1 ].

Therefore the conditions of Exercise 1.25 are satisfied. We have that g is δSH differentiable on [t0 , t1 ] and δSH g(t) = (−ea (σ (t), t0 )) · b(t),

t ∈ [t0 , t1 ].

By Exercise 1.25, it follows that the function y2 is δSH -differentiable on [t0 , t1 ] and δSH y2 (t) = δSH (f · g)(t) = (f (σ (t)) · δSH g(t)) H

 Δ   −f (t) · g(t)

= (ea (σ (t), t0 )) · ((ea (σ (t), t0 )) · b(t)) H ((−a(t)) · (ea (t, t0 ) · g(t))) = (ea (σ (t), t0 )ea (t0 , σ (t))) · b(t) H ((−a(t)) · (f (t) · g(t)))

2.1 Linear First Order Fuzzy Dynamic Equations

157

= 1 · b(t) H (−a(t) · y2 (t)) = b(t) H ((−a(t)) · y2 (t)) ,

t ∈ (t0 , t1 ],

whereupon δSH y2 (t) + (−a(t)) · y2 (t) = b(t),

t ∈ (t0 , t1 ].

Also, y2 (t0 ) = f (t0 ) · g(t0 )

 = (ea (t0 , t0 )) · y0 H

t0 t0

 (−ea (σ (τ ), t0 )) · b(τ )δSH τ

= 1 · y0 = y0 . Consequently y2 is a solution of the IVPs (2.7), (2.8). This completes the proof. Example 2.8 Let T = R. Consider the following IVP: δSH y + (−t) · y = (t + t 3 , t + t 3 , t + t 3 ),

t > 0,

y(0) = (−2, −1, 1). Here a(t) = t, b(t) = (t + t 2 , t + t 2 , t + t 2 ),

t ≥ 0,

y0 = (−2, −1, 1). We have t2

ea (t, 0) = e 2 , t2

ea (σ (t), 0) = e− 2 , t 0

ea (σ (τ ), 0)(τ + τ 3 )dτ =

t

τ2

e− 2 (τ + τ 3 )dτ

0 t

= 0

t

=



2

e 0

t

τ2

e− 2 τ dτ + − τ2

d

 τ2 2

τ2

τ 3 e− 2 dτ

0 t

+

τ e 0



2

2 − τ2

d

τ2 2



158

2 First Order Fuzzy Dynamic Equations

  τ 2 τ =t τ 2 τ =t = −e− 2  −τ 2 e− 2  τ =0

t

+2

τe

τ =0

2 − τ2

dτ

0 t2

t2

= −e− 2 + 1 − t 2 e− 2  τ 2 τ =t −2e− 2  τ =0

t2

t2

= −(1 + t 2 )e− 2 + 1 − 2e− 2 + 2 t2

= −(3 + t 2 )e− 2 + 3,

t ≥ 0.

Hence, t 0

 t2 (−ea (σ (τ ), 0)) · b(τ )δSH τ = (3 + t 2 )e− 2 − 3, t2

(3 + t 2 )e− 2 − 3,  t2 (3 + t 2 )e− 2 − 3 ,

t ≥ 0,

and t

y0 H 0

(−ea (σ (τ ), 0)) · b(τ )δSH τ = (−2, −1, 1)  t2 H (3 + t 2 )e− 2 − 3, t2

(3 + t 2 )e− 2 − 3,  t2 (3 + t 2 )e− 2 − 3  t2 = 1 − (3 + t 2 )e− 2 , t2

2 − (3 + t 2 )e− 2 ,  t2 4 − (3 + t 2 )e− 2 , and

t ≥ 0,

2.1 Linear First Order Fuzzy Dynamic Equations

 y2 (t) = ea (t, 0) · y0 H

t 0

=e

t2 2

t2

= e

 (−ea (σ (τ ), 0)) · b(τ )δSH τ

 t2 t2 · 1 − (3 + t 2 )e− 2 , 2 − (3 + t 2 )e− 2 ,

4 − (3 + t 2 )e− 2 

159

t2 2



− 3 − t , 2e 2

t2 2

− 3 − t , 4e 2

t2 2

 −3−t , 2

is a solution of the considered IVP and it is shown in Fig. 2.3. Example 2.9 Let T = Z. Consider the following IVP:   1 δSH y + − · y = (1, 2, 3), 2

t > 0,

y(0) = (1, 2, 4). Here 1 , 2 b(t) = (1, 2, 3),

a(t) =

t ∈ T,

y0 = (1, 2, 4). We have   1 t ea (t, 0) = 1 + 2

3500 3000 2500 Out[12]= 2000

1500 1000 500 1

2

Fig. 2.3 δSH -differentiable solution of Example 2.8

3

4

t ≥ 0,

160

2 First Order Fuzzy Dynamic Equations

=

 t 3 , 2

(a)(t) = − =−

a(t) 1 + μ(t)a(t) 1 2

1+

1 2

1 =− , 3   1 t+1 ea (σ (t), 0) = 1 − 3  t+1 2 = , 3 t t  2 τ +1 ea (σ (τ ), 0)Δτ = Δτ 3 0 0 t−1  τ +1 ( 2 = 3 τ =0 ⎛  t ⎞ 2 − 1 2⎜ 3 ⎟ = ⎝ 2 ⎠, 3 − 1 3   t  2 , = 2 1− 3

t > 0.

Then t 0

 t  2 , (−ea (σ (τ ), 0)) · b(τ )δSH τ = −2 1 − 3 



 t  2 , −6 1 − 3



 t  2 −4 1 − , 3



t > 0,

and y0 H

t 0

 (−ea (σ (τ ), 0)) · b(τ )δSH τ =



 t  2 1+2 1− , 3 

4+6 1−

 t  2 2+4 1− , 3

 t  2 , t > 0. 3



2.1 Linear First Order Fuzzy Dynamic Equations

161

Therefore  y2 (t) = ea (t, 0) · y0 H

t 0

 (−ea (σ (τ ), 0)) · b(τ )δSH τ

  t   t    t   2 3 2 · 1+2 1− = , 2+4 1− , 2 3 3   t  2 4+6 1− 3   t   t    t   t  2 3 2 3 1+2 1− , 2+4 1− , = 2 3 2 3  t    t  3 2 4+6 1− , t > 0, 2 3 is a solution of the considered IVP and it is shown in Fig. 2.4. Exercise 2.4 Let T = 4Z. Find a solution of the following IVP:     1 + t2 δSH y + − · y = 1 + t + t 2, 2 + t + t 2 + t 4, 3 + t + t 2 + t 8 , 2

t > 0,

y(0) = (−1, 0, 4). Now we consider the following IVP: 6

5

4 Out[109]=

3

2

0

1

2

3

Fig. 2.4 δSH -differentiable solution of Example 2.9

4

5

6

162

2 First Order Fuzzy Dynamic Equations

δH y + (−b(t)) = a(t) · y,

t ∈ (t0 , t1 ],

(2.9)

y(t0 ) = y0 ,

(2.10)

where t0 , t1 ∈ T, t0 < t1 , a : [t0 , t1 ] → R, a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]). Theorem 2.5 Let a > 0 on [t0 , t1 ], and the H -difference t

y0 H

t0

ea (σ (τ ), t0 ) · (−b(τ ))δSH τ

exists for t ∈ [t0 , t1 ]. Let also,  y2 (t) = ea (t, t0 ) · y0 H

t t0

 ea (σ (τ ), t0 ) · (−b(τ ))δSH τ ,

t ∈ [t0 , t1 ],

and the H -differences f (t + h) H f (σ (t)) and

f (σ (t)) H f (t − h)

exist for t ∈ [t0 , t1 ] and h > 0 sufficiently small. Then y2 is δH -differentiable on [t0 , t1 ] and it is a solution of the IVPs (2.9), (2.10). Proof Let f (t) = ea (t, t0 ), g(t) = y0 H

t t0

ea (σ (τ ), t0 ) · (−b(τ ))δSH τ,

t ∈ [t0 , t1 ].

Then y2 (t) = f (t) · g(t),

t ∈ [t0 , t1 ].

Since a > 0 on [t0 , t1 ], we get f Δ (t) = a(t)ea (t, t0 ), f (t)f Δ (t) > 0, f (σ (t))f Δ (t) > 0,

t ∈ [t0 , t1 ].

Consequently y2 satisfies all conditions of Theorem 1.31. Hence, y2 is δH differentiable and   δH y2 (t) = f Δ (t) · g(t)

2.1 Linear First Order Fuzzy Dynamic Equations

163

H ((−f (σ (t))) · δSH g(t)) = (a(t)f (t)) · g(t) H ((−ea (σ (t), t0 )) · (ea (σ (t), t0 ) · b(t))) = a(t) · y2 (t) H ((−(ea (σ (t), t0 )ea (t0 , σ (t)))) · b(t)) = a(t) · y2 (t) H (−b(t)),

t ∈ [t0 , t1 ],

whereupon δH y2 (t) + (−b(t)) = a(t) · y(t),

t ∈ [t0 , t1 ].

Next, y2 (t0 ) = y0 . This completes the proof. Example 2.10 Let T = 2N0 . Consider the IVP δH y + (−t 2 , −t, −1) = y,

t > 1,

y(1) = (1, 2, 3). Here σ (t) = 2t, a(t) = 1, b(t) = (1, t, t 2 ), (a)(t) = −

a(t) 1 + μ(t)a(t)

1 , 1+t &  1− ea (σ (t), 1) = =−

s∈[1,t]

=

& s∈[1,t]

&

ea (t, 1) =

s∈[

 ea (σ (t), 1) · b(t) =

1, 2t

s 1+s



1 , 1+s (1 + s), ] &

− t2

s∈[1,t]

& 1 1 , −t , 1+s 1+s

 & 1 − , 1+s s∈[1,t]

s∈[1,t]

t > 1.

164

2 First Order Fuzzy Dynamic Equations

We have ⎛ ⎞ & ( 1 1 ⎠, Δτ = − τ3 ⎝ τ2 − 1+s 1+s 1 t s∈[1,τ ] s∈[1,τ ] τ ∈[1, 2 ] ⎛ ⎞ t & ( & 1 1 ⎠ Δτ = − τ2 ⎝ τ − , 1 + s 1 + s 1 t s∈[1,τ ] s∈[1,τ ] τ ∈[1, 2 ] ⎛ ⎞ t & ( & 1 1 ⎠, Δτ = − τ⎝ − 1+s 1 s∈[1,τ ] 1 + s t s∈[1,τ ] τ ∈[1, 2 ] t

&

t > 1. Hence, t 1

 (ea (σ (τ ), 1)) · (−b(τ ))δSH τ =

−

τ ∈[

−

1, 2t

( τ ∈[

−

⎛

(

1, 2t

τ3 ⎝ ]

s∈[1,τ ]

⎛

τ2 ⎝

& s∈[1,τ ]

]

(

&

⎛ τ⎝

& s∈[1,τ ]

τ ∈[1, 2t ]

⎞ 1 ⎠ , 1+s

⎞ 1 ⎠ , 1+s

⎞  1 ⎠ , 1+s

and t

y0 H 1

(ea (σ (τ ), 1)) · (−b(τ ))δSH τ 

= (1, 2, 3) H

−

τ

τ ∈[1, 2t ]

−

( τ ∈[

1, 2t

− ⎛

(

2⎝

⎛ τ3 ⎝

τ ∈[1, 2t ]

&

s∈[1,τ ]

⎛ τ⎝

]

(

&

s∈[1,τ ]

s∈[1,τ ]

⎞ 1 ⎠ , 1+s

⎞  1 ⎠ 1+s

&

⎞ 1 ⎠ , 1+s

t > 1,

2.1 Linear First Order Fuzzy Dynamic Equations

 = 1+

τ ∈[

2+

⎛

(

(

]

( τ ∈[

1, 2t

s∈[1,τ ]

⎛

τ2 ⎝

&

s∈[1,τ ]

τ ∈[1, 2t ]

3+

&

τ⎝

1, 2t

⎛ τ3 ⎝

]

165

&

s∈[1,τ ]

⎞ 1 ⎠ , 1+s ⎞

1 ⎠ , 1+s ⎞  1 ⎠ , 1+s

t > 1.

Therefore  y2 (t) = ea (t, 1) · y0 H

t 1

 (ea (σ (τ ), 1)) · (−b(τ ))δSH τ

⎛ ⎞  ( & 1 ⎠, (1 + s) · 1 + τ⎝ = 1+s s∈[1,τ ] s∈[1, 2t ] τ ∈[1, 2t ] ⎛ ⎞ & ( 1 ⎠ τ2 ⎝ 2+ , 1 + s t s∈[1,τ ] τ ∈[1, 2 ] ⎛ ⎞  & ( 1 ⎠ τ3 ⎝ 3+ 1+s s∈[1,τ ] τ ∈[1, 2t ] ⎛ ⎛ ⎞⎞  & ( & 1 ⎠⎟ ⎜ (1 + s) ⎝1 + τ⎝ = ⎠, 1 + s t t s∈[1,τ ] s∈[1, 2 ] τ ∈[1, 2 ] ⎛ ⎛ ⎞⎞ & ( & 1 ⎠⎟ ⎜ (1 + s) ⎝2 + τ2 ⎝ ⎠, 1 + s s∈[1,τ ] s∈[1, 2t ] τ ∈[1, 2t ] ⎛ ⎛ ⎞⎞  ( & & 1 ⎠⎟ ⎜ 3⎝ (1 + s) ⎝3 + τ ⎠ , 1+s s∈[1,τ ] s∈[1, 2t ] τ ∈[1, 2t ] &

t > 1, is a solution of the considered IVP. Exercise 2.5 Let T = 3Z. Find a solution of the following IVP: δH y + (−t 2 − t − 1, −t − 1, −1) = t · y, y(0) = (−1, 1, 3).

t > 0,

166

2 First Order Fuzzy Dynamic Equations

Exercise 2.6 Let a : [t0 , t1 ] → (−∞, 0), a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R), and  y1 (t) = ea (t, t0 ) · y0 +

t t0

 ea (σ (τ ), t0 ) · b(τ )δH τ ,

t ∈ [t0 , t1 ].

Let also, the H -differences y1 (σ (t)) H y1 (t + h) and

y1 (t − h) H y1 (σ (t))

exist for t ∈ [t0 , t1 ] and for h > 0 sufficiently small. Prove that y1 is δSH differentiable on [t0 , t1 ] and it is a solution of the following IVP: δSH y + (−b(t)) = a(t) · y,

t ∈ (t0 , t1 ],

y(t0 ) = y0 .

(2.11) (2.12)

Hint. Use Exercise 1.26. Exercise 2.7 Let T = Z. Find a solution of the following IVP: δSH y + (−t, −t + 1, −t + 2) = (−1) · y,

t > 0,

y(0) = (−3, −2, −1). Remark 2.1 Let us remark that the IVPs 0, δSH y + (−b(t)) + (−a(t)) · y =

t ∈ (t0 , t1 ],

y(t0 ) = y0 , and 0, δH y + (−b(t)) + (−a(t)) · y =

t ∈ (t0 , t1 ],

y(t0 ) = y0 , have no fuzzy solutions. Here a : [t0 , t1 ] → (−∞, 0), a ∈ Crd ([t0 , t1 ]), b : [t0 , t1 ] → F (R), b ∈ Cf rd ([t0 , t1 ]), y0 ∈ F (R). Remark 2.2 1. Note that the IVPs (2.1), (2.2) and the IVPs (2.3), (2.4) and (2.9), (2.10) are not equivalent. Also, in the general case, the IVPs (2.3), (2.4) and (2.9), (2.10) are not equivalent.

2.2 Existence and Uniqueness of Solutions

167

2. Note that the IVPs (2.5), (2.6) and the IVPs (2.7), (2.8) and (2.11), (2.12) are not equivalent. Moreover, in the general case, the IVPs (2.7), (2.8) and (2.11), (2.12) are not equivalent.

2.2 Existence and Uniqueness of Solutions Let t0 , t1 ∈ T, t0 < t1 , and y0 ∈ F (R). Consider the following IVP: δH y = f (t, y),

t ∈ (t0 , t1 ],

(2.13)

y(t0 ) = y0 ,

(2.14)

where (A1)

f : Tκ × F (R) → F (R), f ∈ Cf rd (Tκ × F (R)).

Definition 2.1 We say that a fuzzy function F : Tκ × F (R) → F (R) satisfies the Lipschitz condition with a constant K > 0 if D(F (t, x), F (t, y)) ≤ KD(x, y),

t ∈ T,

x, y ∈ F (R).

Theorem 2.6 Assume (A1) and f satisfy the Lipschitz condition with a constant K > 0 such that K(t1 − t0 ) < 1. Then the IVPs (2.13), (2.14) has a unique solution on [t0 , t1 ]. Proof Define a metric H on Cf rd ([t0 , t1 ]) as follows: H (ξ, ψ) = sup D(ξ(t), ψ(t)) t∈[t0 ,t1 ]

(2.15)

for ξ, ψ ∈ Cf rd ([t0 , t1 ]). Since (F (R), D) is a complete metric space, we have that (Cf rd ([t0 , t1 ]), H ) is a complete metric space. For ξ ∈ Cf rd ([t0 , t1 ]), define the operator Gξ(t) = y0 +

t

f (s, ξ(s))δH s,

t ∈ [t0 , t1 ].

t0

Let ξ ∈ Cf rd ([t0 , t1 ]) be arbitrarily chosen. Then there exists a positive real number M such that f (t, ξ(t)) ≤ M,

t ∈ [t0 , t1 ].

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2 First Order Fuzzy Dynamic Equations

Let t, t2 ∈ [t0 , t1 ]. Without loss of generality, assume that t2 < t. Then  D(Gξ(t), Gξ(t2 )) = D

t

 =D



t2

f (s, ξ(s))δH s,

f (s, ξ(s))δH s t0

t0 t

 f (s, ξ(s))δH s, 0

t2 t

≤

  D f (s, ξ(s)), 0 Δs

t2 t

=

f (s, ξ(s))Δs

t2

≤ M(t − t2 ). Thus, Gξ ∈ Cf rd ([t0 , t1 ]). Let now ξ, ψ ∈ Cf rd ([t0 , t1 ]) be arbitrarily chosen. Then  t  t F (s, ξ(s))Δs, F (s, ψ(s))Δs H (Gξ, Gψ) = sup D t∈[t0 ,t1 ]

≤ sup

t0

t∈[t0 ,t1 ] t0

≤ K sup

t0

t

D(F (s, ξ(s)), F (s, ψ(s)))Δs t

t∈[t0 ,t1 ] t0

D(ξ(s), ψ(s))Δs

≤ K(t − t0 )H (ξ, ψ) ≤ K(t1 − t0 )H (ξ, ψ) < H (ξ, ψ). From here, using the Banach contraction principle, we conclude that the operator G has a unique fixed point in Cf rd ([t0 , t1 ]), which is a unique solution of the IVPs (2.13), (2.14). This completes the proof. As in above, one can prove the following result. Theorem 2.7 Assume (A1) and f satisfy the Lipschitz condition with a constant K > 0 such that K(t1 − t0 ) < 1. Then the IVP δSH y = f (t, y), y(t0 ) = y0 , has a unique solution on [t0 , t1 ].

t ∈ (t0 , t1 ],

2.2 Existence and Uniqueness of Solutions

169

Example 2.11 Consider the following IVP:  δH y(t) = g(t) H

 1 · y(t) , 2

t ∈ (t0 , t1 ],

y(t0 ) = y0 , where g : [t0 , t1 ] → F (R), g ∈ Cf rd ([t0 , t1 ]) is a given fuzzy function. Here,  f (t, y(t)) = g(t) H

 1 · y(t) , 2

t ∈ [t0 , t1 ],

y ∈ Cf rd ([t0 , t1 ]).

For y, z ∈ Cf rd ([t0 , t1 ]), we have      1 1 · y(t) , g(t) H · z(t) D(f (t, y(t)), f (t, z(t))) = D g(t) H 2 2   1 1 · y(t), · z(t) =D 2 2 =

1 D(y(t), z(t)), 2

t ∈ [t0 , t1 ].

Therefore, if 12 (t1 − t0 ) < 1, the considered IVP has a unique solution. Exercise 2.8 Let T = 2Z, t0 , t1 ∈ T, t0 < t1 . Find conditions for t0 and t1 so that the following IVP: 1 t · y, t ∈ (t0 , t1 ], 2 y(t0 ) = (−1, 2, 3), δH y =

has a unique solution. Exercise 2.9 Let T = 3N0 , t0 , t1 ∈ T, t0 < t1 , and a ∈ R, a = 0. Find conditions for a, t0 , and t1 so that the following IVP:   t +1 · y, δSH y = a 2 t +1

t ∈ (t0 , t1 ],

y(t0 ) = (0, 1, 2), has a unique solution. Suppose that γ > 0. For x, y ∈ Cf rd ([t0 , t1 ]), define D(x(t), y(t)) . eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup

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2 First Order Fuzzy Dynamic Equations

In addition, since e0 (t, s) = 1 for any t, s ∈ T, d0 is defined by d0 (x, y) = sup D(x(t), y(t)), t∈[t0 ,t1 ]

i.e., d0 (x, y) = H (x, y), where H is defined by (2.15). Consider xγ = supt∈[t0 ,t1 ]

D(x(t), 0) eγ (t,t0 ) ,

x0 = supt∈[t0 ,t1 ] D(x(t), 0).

Theorem 2.8 We have: 1. dγ is a metric equivalent to the metric d0 .  2. Cf rd ([t0 , t1 ]), dγ is a complete metric space. Proof 1. Firstly, we will prove that dγ is a metric. Let x, y, z ∈ Cf rd ([t0 , t1 ]). (a) We have D(x(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup ≥ 0, dγ (x, y) = 0

⇐⇒

D(x(t), y(t)) =0 eγ (t, t0 ) t∈[t0 ,t1 ]

⇐⇒

sup

D(x(t), y(t)) = 0,

t ∈ [t0 , t1 ].

Thus, dγ (x, y) = 0 if and only if x(t) = y(t),

t ∈ [t0 , t1 ].

(b) We have D(x(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup

D(y(t), x(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

= sup

= dγ (y, x). (c) We have

(2.16)

2.2 Existence and Uniqueness of Solutions

171

D(x(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup

D(x(t), z(t)) + D(z(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

≤ sup

D(x(t), z(t)) D(z(t), y(t)) + sup e (t, t ) eγ (t, t0 ) γ 0 t∈[t0 ,t1 ] t∈[t0 ,t1 ]

= sup

= dγ (x, z) + dγ (z, y). Therefore dγ satisfies all axioms for a metric. Next, since eγΔ (t, t0 ) = γ eγ (t, t0 ),

t ∈ [t0 , t1 ],

and γ > 0, we have that eγ (·, t0 ) is an increasing function on [t0 , t1 ]. Therefore 1 ≤ eγ (t, t0 ) ≤ eγ (t1 , t0 ),

t ∈ [t0 , t1 ],

and 1 1 ≤ eγ (t1 , t0 ) eγ (t, t0 ) ≤ 1,

t ∈ [t0 , t1 ].

Hence, D(x(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup

≤ sup D(x(t), y(t)) t∈[t0 ,t1 ]

= d0 (x, y) and D(x(t), y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]

dγ (x, y) = sup

D(x(t), y(t)) eγ (t1 , t0 ) t∈[t0 ,t1 ]

≥ sup =

1 sup D(x(t), y(t)) eγ (t1 , t0 ) t∈[t0 ,t1 ]

=

1 d0 (x, y). eγ (t1 , t0 )

Consequently the metrics dγ and d0 are equivalent.

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2 First Order Fuzzy Dynamic Equations

  2. Let {xn }n∈N be a Cauchy sequence in Cf rd ([t0 , t1 ]), dγ . Take  > 0 arbitrarily. Then there exists an N = N() > 0 so that dγ (xm , xn )
N, m, n ∈ N, and then D(xm (t), xn (t)) ≤    for any m, n ≥ N , m, n ∈ N, and for any t ∈ [t0 , t1 ]. Because Cf rd ([t0 , t1 ]), d0 is a complete metric space, there exists an x ∈ Cf rd ([t0 , t1 ]) so that lim D(xm (t), x(t)) = 0,

m→∞

t ∈ [t0 , t1 ].

In particular, sup D(xm (t), x(t)) = 0.

lim

m→∞ t∈[t ,t ] 0 1

Consequently lim dγ (xm , x) ≤

m→∞

1 lim sup D(xm (t), x(t)) eγ (t1 , t0 ) m→∞ t∈[t0 ,t1 ]

= 0, i.e., lim dγ (xm , x) = 0.

m→∞

This completes the proof. Exercise 2.10 Prove that the map  · γ : Cf rd ([t0 , t1 ]) → [0, ∞), defined with (2.16), satisfies all axioms for a norm. Consider the following IVP: δH y = f (t, y σ ),

t ∈ (t0 , t1 ],

y(t0 ) = y0 ,

(2.17) (2.18)

2.2 Existence and Uniqueness of Solutions

173

where (A2)

f ∈ C ([t0 , t1 ] × F (R)), y0 ∈ R.

Theorem 2.9 Suppose (A2). If there exists a positive constant L such that and

L γ

0 on [t0 , ∞) × F (R), and g(·, ·) is a nondecreasing function with respect to the second argument on [t0 , ∞), and D (f (t, u(t)), f (t, v(t))) ≤ g (t, D(u(t), v(t))) ,

t ≥ t0 .

Suppose that there exists the maximal solution r(t, t0 , w0 ) of the IVP w Δ (t) = g(t, w(t)), w(t0 ) = w0 .

t ∈ (t0 , t1 ],

(2.31)

Then, if D(u0 , v0 ) ≤ w0 , we have D (u(t), v(t)) ≤ r(t, t0 , w0 ),

t ∈ [t0 , t1 ].

Proof Since u and v are solutions of the IVPs (2.29) and (2.30), respectively, the H -differences u(s)H u(t) and v(s)H v(t) exist for small s −t > 0, s, t ∈ [t0 , t1 ]. Let m(t) = D(u(t), v(t)),

t ∈ (t0 , t1 ].

For s, t ∈ [t0 , t1 ], s − t > 0 small enough, we get m(s) − m(t) = D(u(s), v(s)) − D(u(t), v(t)) ≤ D (u(s), u(t) + (s − t)f (t, u(t))) +D (u(t) + (s − t)f (t, u(t)), v(t) + (s − t)f (t, v(t))) +D (v(t) + (s − t)f (t, v(t)), v(s)) − D(u(t), v(t)) ≤ D (u(s), u(t) + (s − t)f (t, u(t))) +D (v(t) + (s − t)f (t, v(t)), v(s)) +(s − t)D (f (t, u(t)), f (t, v(t)))

178

2 First Order Fuzzy Dynamic Equations

= D (u(s) H u(t), (s − t)f (t, u(t))) +D ((s − t)f (t, v(t)), v(s) H v(t)) +(s − t)D (f (t, u(t)), f (t, v(t))) and  u(s) H u(t) , f (t, u(t)) s−t   v(s) H v(t) +D f (t, v(t)), s−t

m(s) − m(t) ≤D s−t



+D (f (t, u(t)), f (t, v(t))) .

1. Let t ∈ [t0 , t1 ] is right-dense. Then m(s) − m(t) s−t   u(s) H u(t) ≤ lim D , f (t, u(t)) s−t s→t +   v(s) H v(t) + lim D f (t, v(t)), s−t s→t +

mΔ (t) = lim

s→t +

+D (f (t, u(t)), f (t, v(t))) = D (δH u(t), f (t, u(t))) + D (δH v(t), f (t, v(t))) +D (f (t, u(t)), f (t, v(t))) = D (f (t, u(t)), f (t, v(t))) ≤ g (t, D(u(t), v(t))) = g(t, m(t)). 2. Let t ∈ [t0 , t1 ] is a right-scattered point. Then m(σ (t)) − m(t) μ(t)   u(σ (t)) H u(t) ≤D , f (t, u(t)) μ(t)   v(σ (t)) H v(t) +D f (t, v(t)), μ(t)

mΔ (t) =

+D (f (t, u(t)), f (t, v(t)))

2.4 Comparison Results

179

= D (f (t, u(t)), f (t, v(t))) ≤ g (t, D(u(t), v(t))) = g(t, m(t)). Next, m(t0 ) = D (u(t0 ), v(t0 )) = D(u0 , v0 ) ≤ w0 . Therefore we get mΔ (t) ≤ g(t, m(t)),

t ∈ (t0 , t1 ],

m(t0 ) ≤ w0 . Consequently m(t) ≤ r(t, t0 , w0 ),

t ∈ [t0 , t1 ],

i.e., D(u(t), v(t)) ≤ r(t, t0 , w0 ),

t ∈ [t0 , t1 ].

This completes the proof. Theorem 2.14 Assume f ∈ Crd (T × F (R)) satisfies f (t, φ) ≤ g (t, φ) for each (t, φ) ∈ T × F (R), where g : T × R+ → R+ , g ∈ Crd (T × R+ ) and g(·, ·) is nondecreasing with respect to the second argument on T. Let also the solution of the IVP w Δ (t) = g(t, w(t)),

t > t0 ,

w(t0 ) ≥ w0 , exists on [t0 , ∞). If f is smooth enough to ensure the local existence, then the largest interval of existence of any solution u(t, t0 , u0 ) of the IVP (2.29) with u0  ≤ w0 is [t0 , ∞). Proof Assume that u(t) = u(t, t0 , u0 ) is a solution of the IVP (2.29) with u0  ≤ w0 and it exists on the largest interval [t0 , τ ). We will show that τ = ∞. Suppose

180

2 First Order Fuzzy Dynamic Equations

the contrary, i.e., suppose that τ < ∞. Then there exists a β ≤ τ so that u exists on [t0 , β] and β cannot be increased. For t ∈ [t0 , β], set m(t) = D(u(t), 0). As in the proof of Theorem 2.13, we have mΔ (t) ≤ g(t, m(t)),

t ∈ [t0 , β],

and   D u(t), 0 ≤ r(t, t0 , w0 ),

t ∈ [t0 , β].

Take t1 , t2 ∈ [t0 , β] such that t1 < t2 . Then  D(u(t1 ), u(t2 )) = D u(t1 ), u(t1 ) +  =D 0,



t2

f (s, u(s))δH s t1

t2



f (s, u(s))δH s t1

≤

t2

  D 0, f (s, u(s)) Δs

t1

≤

t2

   g s, D u(s), 0 Δs

t1

≤

t2

g(s, r(s, t0 , w0 ))Δs t1

=

t2

r Δ (s, t0 , w0 )Δs

t1

= r(t2 , t0 , w0 ) − r(t1 , t0 , w0 ). Let β is left-dense. Then, taking t1 , t2 → β and using that lim r(t, t0 , w0 )

t→β −

exists and it is finite, we get that there exists u(β) = lim u(t). t→β −

2.4 Comparison Results

181

If β is right-scattered, then u(β) exists. Next, by the assumption for local existence, there exists a solution v(t, β, u(β)) of Eq. (2.29) defined on the interval [β, γ ] with γ > β. Therefore the solution u can be continued beyond β. This is a contradiction because β cannot be increased. Consequently τ = ∞. This completes the proof. As in above, one can prove the following results. Theorem 2.15 Assume (B1) and consider the IVPs δSH u(t) = f (t, u(t)), = u0 , u(t0 )

t ∈ (t0 , t1 ],

(2.32)

δSH v(t) = f (t, v(t)), = v0 , v(t0 )

t ∈ (t0 , t1 ],

(2.33)

and let u(t) = u(t, t0 , u0 ) and v(t) = v(t, t0 , v0 ) be solutions of the IVPs (2.32) and (2.33), respectively. Suppose that g, w satisfy all conditions of Theorem 2.13. Then, if D(u0 , v0 ) ≤ w0 , we have D(u(t), v(t)) ≤ r(t, t0 , w0 ),

t ∈ (t0 , t1 ].

Theorem 2.16 Assume that f , g, and w satisfy all conditions of Theorem 2.14. If f is smooth enough to ensure the local existence, then the largest interval of existence of any solution u(t, t0 , u0 ) of the IVP (2.32) with u0  ≤ w0 is [t0 , ∞). Definition 2.2 Let A : T → F (R), A ∈ Cf rd (T), V : T × F (R) → R, V ∈ Crd (T × F (R)). We call Δr V (t, A) and Δr V (t, A) the right upper and right lower derivatives of the function V at (t, A(t)) along the solutions of the fuzzy dynamic equation (2.29), if

Δr V (t, A(t)) =

⎧ V (σ (t),A(σ (t)))−V (t,A(t)) ⎨ μ(t) ⎩

lim sups→t +

if σ (t) > t,

V (s,A(t)+(s−t)f (t,A(t)))−V (t,A(t)) s−t

if σ (t) = t,

and Δr V (t, A(t)) =

⎧ V (σ (t),A(σ (t)))−V (t,A(t)) ⎨ μ(t) ⎩

lim infs→t +

if σ (t) > t,

V (s,A(t)+(s−t)f (t,A(t)))−V (t,A(t)) s−t

if σ (t) = t,

respectively. Similarly, we call Δl V (t, A(t)) and Δl V (t, A(t)) the left upper and left lower derivatives of the function V at (t, A(t)) along the solutions of the fuzzy dynamic equation (2.29), if

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2 First Order Fuzzy Dynamic Equations

Δl V (t, A(t)) =

⎧ V (t,A(t))−V (ρ(t),A(ρ(t))) ⎨ t−ρ(t) ⎩

lim sups→t −

if t > ρ(t),

V (s,A(t)+(s−t)f (t,A(t)))−V (t,A(t)) s−t

if

ρ(t) = t,

and Δl V (t, A(t)) =

⎧ V (t,A(t))−V (ρ(t),A(ρ(t))) ⎨ t−ρ(t) ⎩

lim infs→t −

if t > ρ(t),

V (s,A(t)+(s−t)f (t,A(t)))−V (t,A(t)) s−t

if t = ρ(t),

respectively. Theorem 2.17 Assume that V is given as in Definition 2.2 and satisfies Δr V (t, u(t)) ≤ g(t, V (t, u(t))),

t ∈ T,

|V (t, u(t)) − V (t, v(t))| ≤ LD(u(t), v(t)),

L ≥ 0,

t ∈ T,

where g : T × R+ → R, g ∈ Crd (T × R+ ), and u(t) = u(t, t0 , u0 ), v(t) = v(t, t0 , v0 ) are solutions of the IVPs (2.29) and (2.30), respectively. Let also r(t, t0 , w0 ) is the maximal solution of the IVP (2.31) existing on [t0 , ∞). If V (t0 , u0 ) ≤ w0 and u is a solution of the IVP (2.29) defined on [t0 , ∞), then V (t, u(t)) ≤ r(t, t0 , w0 ),

t ∈ [t0 , ∞).

Proof Let m(t) = V (t, u(t)),

t ∈ [t0 , ∞).

We have m(t0 ) = V (t0 , u0 ) ≤ w0 . Take s, t ∈ T, s > t. Then m(s) − m(t) = V (s, u(s)) − V (t, u(t)) = V (s, u(s)) − V (s, u(t) + (s − t)f (t, u(t))) +V (s, u(t) + (s − t)f (t, u(t))) − V (t, u(t)) ≤ LD(u(s), u(t) + (s − t)f (t, u(t))) +V (s, u(t) + (s − t)f (t, u(t))) − V (t, u(t)). Because u is a solution of the IVP (2.29), we have that the H -difference z(t) = u(s) H u(t)

(2.34)

2.4 Comparison Results

183

exists for s − t small enough. Then D(u(s), u(t) + (s − t)f (t, u(t))) = D(u(s) H u(t), (s − t)f (t, u(t))), whereupon 1 D(u(s), u(t) + (s − t)f (t, u(t))) = D s−t



 u(s) H u(t) , f (t, u(t)) . s−t

Hence and (2.34), we arrive at m(s)−m(t) s−t





u(s)H u(t) , f (t, u(t)) s−t V (s,u(t)+(s−t)f (t,u(t)))−V (t,u(t)) . + s−t

≤ LD

(2.35)

Next, lim sup s→t +

1 D(u(s), u(t)+(s−t)f (t, u(t))) = lim sup D s−t s→t +



u(s) H u(t) , f (t, u(t)) s−t

= D (δH u(t), f (t, u(t))) = 0.

From the last relation and from (2.35), we obtain m(s) − m(t) s−t   u(s) H u(t) ≤ L lim D , f (t, u(t)) s−t s→t +

mΔ + (t) = lim

s→t +

+ lim

s→t +

V (s, u(t) + (s − t)f (t, u(t))) − V (t, u(t)) s−t

= Δ V (t, u(t)) r

≤ g(t, V (t, u(t))) = g(t, m(t)),

t ∈ [t0 , ∞).

Consequently m(t) ≤ r(t, t0 , w0 ),

t ∈ [t0 , ∞),

or V (t, u(t)) ≤ r(t, t0 , w0 ), This completes the proof.

t ∈ [t0 , ∞).



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2 First Order Fuzzy Dynamic Equations

2.5 Stability Criteria Suppose that T is a time scale with minimal element t0 and it has no maximal element. Let u0 , v0 ∈ F (R) be such that u0 H v0 = w0 exists. Consider the equation δH y = f (t, y),

t > t0 ,

(2.36)

where (C1)

f : T × F (R) → F (R), f (t, 0) = 0.

In this section, suppose that the solutions of the IVP for Eq. (2.36) exist and are unique for any t ≥ t0 . For z ∈ F (R), with u(t, t0 , u0 ), t ≥ t0 , we will denote the solution of Eq. (2.36) for which u(t0 , t0 , z) = z. Definition 2.3 The trivial solution u(t) = 0, t ≥ t0 , is said to be: (i) Equi-stable if for each  > 0 and t0 ∈ T, there exists a δ = δ(t0 , ) such that u0  < δ implies that u(t) < ,

t ∈ [t0 , ∞)

(2.37)

(ii) Uniformly stable if in (i) the δ = δ() > 0 is independent of t0 (iii) Equi-asymptotically stable if (i) holds and for any  > 0 there exists a T > 0 such that (2.37) holds for all t ∈ [t0 + T , ∞) (iv) Uniformly asymptotically stable if (ii) and (iii) hold simultaneously For b > 0, denote S(b) = {u ∈ F (R) : u ≤ b}. With K we will denote the class of all continuous functions φ : [a, b) → [0, ∞) such that φ(0) = 0 and φ is increasing. Definition 2.4 The function V ∈ Crd (T×F (R)) is said to be positive definite if: (i) V (t, 0) = 0 for all t ∈ T. (ii) There exists a function φ ∈ K such that V (t, A) ≥ φ(A) for each (t, A) ∈ T × S(b). V is said to be negative definite if −V is positive definite. Theorem 2.18 Suppose (C1) and f ∈ Cf rd (T × F (R)), g : T × R → R, g ∈ Crd (T × R+ ) with g(t, 0) = 0 for all t ∈ T. Moreover:

2.5 Stability Criteria

185

(i) There exists V , given in Definition 2.2, such that V is positive definite and Δr V (t, u(t)) ≤ g(t, V (t, u(t))), |V (t, u(t)) − V (t, v(t))| ≤ LD(u(t), v(t)),

L ≥ 0,

t ∈ T,

where u(t) = u(t, t0 , u0 ), v(t) = v(t, t0 , v0 ). (ii) g is nondecreasing with respect to the second argument on T. Then the equi-stability properties of the trivial solution of Eq. (2.36) imply the corresponding stability properties of the trivial solution of (2.36). Proof Since V (t0 , 0) = 0 and V (t0 , A) is continuous with respect to A, for any  > 0, there exists a δ1 = δ1 (t0 , ) > 0 such that u0  ≤ δ1 implies V (t0 , u0 ) ≤ . Let φ ∈ K be as in Definition 2.4. Suppose that the trivial solution of (2.31) is equi-stable. Then, for any  > 0 and t0 ≥ 0, there exists a δ2 = δ2 (t0 , ) > 0 such that the inequalities 0 ≤ w0 < δ2 imply w(t) ≤ φ(),

t ∈ T,

where w(t) = w(t, t0 , w0 ) is any solution of (2.31). Assume that with these  and δ = min{δ1 , δ2 } the trivial solution of (2.36) is not equi-stable. Then there exists a solution u(t) = u(t, t0 , u0 ) of (2.36) with u0  < δ and t1 ∈ T, t1 > t0 , such that φ() ≤ V (t1 , u(t1 )) and

V (t, u(t)) < φ(),

t ∈ [t0 , t1 ).

By Theorem 2.17, we have V (t, u(t)) ≤ r(t, t0 , w0 ),

t ∈ [t0 , ∞).

In particular, we have V (t1 , u(t1 )) ≤ r(t1 , t0 , w0 ). Hence, φ() ≤ V (t1 , u(t1 )) ≤ r(t1 , t0 , w0 ) < φ(), which is a contradiction. This completes the proof.

186

2 First Order Fuzzy Dynamic Equations

Theorem 2.19 Suppose (C1) and f ∈ Cf rd (T × F (R)). Moreover, there exists a function V as in Definition 2.2 such that V is positive definite and Δr V (t, u(t)) ≤ 0,

t ∈ T,

|V (t, u(t)) − V (t, w(t))| ≤ LD(u(t), w(t)),

L ≥ 0,

t ∈ T,

where u(t) = u(t, t0 , u0 ), w(t) = w(t, t0 , w0 ). Then the trivial solution of (2.36) is equi-stable. Proof Define m(t) = V (t, u(t)), t ∈ T. Because V is positive definite, there exists a φ ∈ K such that m(t) = V (t, u(t)) ≥ φ(u(t)). Since V (t0 , 0) = 0 and V (t0 , A) is continuous with respect to A, for any  > 0 there exists δ = δ(t0 , ) > 0, δ ≤ φ(), such that u0  < δ implies V (t0 , u0 ) < φ(). Therefore m(t0 ) = V (t0 , u0 ) < φ(). On the other hand, the IVP w Δ = 0,

t > t0 ,

w(t0 ) = V (t0 , u0 ), has a unique solution w(t) = V (t0 , u0 ). By Theorem 2.18, we obtain m(t) = V (t, u(t)) ≤ V (t0 , u0 ),

t ∈ [t0 , ∞).

Therefore, for all t ∈ [t0 , ∞), we have φ(u(t)) ≤ V (t, u(t)) ≤ V (t0 , u0 ) < φ(),

t ∈ [t0 , ∞).

Since φ is increasing, we conclude that u(t) < ,

t ∈ [t0 , ∞).

Thus, the trivial solution of (2.36) is equi-stable. This completes the proof.

2.5 Stability Criteria

187

Theorem 2.20 Suppose that all assumptions of Theorem 2.19 are satisfied except that the estimate Δr V (t, A) ≤ 0 be strengthened to Δr V (t, A) ≤ −φ(A)

(2.38)

for (t, A) ∈ T × F (R), where φ ∈ K is a given function. Then the trivial solution of (2.36) is equi-asymptotically stable. Proof By Theorem 2.19, we have that the trivial solution of (2.36) is equi-stable. Then for any  > 0, there exists a δ = δ(t0 , ) > 0 such that u0  < δ implies that t ∈ [t0 , ∞).

u(t) < , Next, t

t

Δr V (s, u(s))Δs ≤ −

t0

φ(u(s))Δs,

t ∈ [t0 , ∞),

t0

or t

V (t, u(t)) − V (t0 , u(t0 )) ≤ −

φ(u(s))Δs,

t ∈ [t0 , ∞),

t0

or V (t, u(t)) +

t

φ(u(s))Δs ≤ V (t0 , u0 ),

t ∈ [t0 , ∞).

t0

Since V (t, u(t)) ≥ 0, t ∈ T, and Δr V (t, A) < 0, (t, A) ∈ T × F (R), we have that V (t, u(t)) is decreasing in t ∈ T. Then there exists lim V (t, u(t)) = α.

t→∞

We have that α ≥ 0. Assume that α > 0. Because V (t, u(t)) is decreasing in t, we get V (t, u(t)) ≥ α,

t ∈ [t0 , ∞).

Since V (t, u(t)) is continuous and V (t, 0) = 0, we conclude that there exists a positive constant ξ such that u(t) > ξ,

t ∈ [t0 , ∞).

188

2 First Order Fuzzy Dynamic Equations

Therefore V (t, u(t)) ≤ V (t0 , u0 ) −

t

φ(u(s))Δs t0

≤ V (t0 , u0 ) − φ(ξ )(t − t0 ),

t ∈ [t0 , ∞).

Take t1 ∈ [t0 , ∞) large enough so that V (t0 , u0 ) − φ(ξ )(t1 − t0 ) < 0. Hence, V (t1 , u(t1 )) < 0. This is a contradiction. Therefore α = 0 and lim V (t, u(t)) = 0.

t→∞

Assume that lim u(t) = 0.

t→∞

Then there exists 0 > 0 such that for any k ∈ N there is a tk ∈ T, with tk > k, so that u(tk ) > 0 . Now, using that V is positive definite, there exists a φ1 ∈ K so that V (t, A) ≥ φ1 (A) for any (t, A) ∈ T × S(b), where b > 0 is large enough. From here, V (tk , u(tk )) ≥ φ1 (u(tk )) > φ1 (0 ),

k ∈ N,

and lim V (tk , u(tk )) > 0.

k→∞

This is a contradiction. Consequently lim u(t) = 0

t→∞

and the trivial solution of (2.36) is equi-asymptotically stable. This completes the proof.

2.5 Stability Criteria

189

Theorem 2.21 Suppose that the assumptions of Theorem 2.19 are satisfied except the estimate Δr V (t, A) ≤ 0, (t, A) ∈ T × S(b), be strengthened by (2.38). Then the trivial solution of (2.36) is uniformly asymptotically stable. Proof By Theorem 2.20, it follows that the trivial solution of (2.36) is uniformly stable. Next, there exist φ1 , φ2 , ψ ∈ K so that φ1 (A) ≤ V (t, A) ≤ ψ(A),

(t, A) ∈ T × S(b),

Δr V (t, u(t)) ≤ −φ2 (u(t))   ≤ −φ2 ψ −1 (V (t, u(t))) t ∈ [t0 , ∞).

< 0, Therefore

Δr V (t, u(t))   ≤ −1, φ2 ψ −1 (V (t, u(t)))

t ∈ [t0 , ∞).

From here, V (t,u(t)) V (t0 ,u0 )

φ2



ΔV (s, u(s))  ≤− ψ −1 (V (s, u(s)))

t

Δs, t0

or V (t,u(t)) V (t0 ,u0 )

ΔV (s, u(s))   ≤ −(t − t0 ), φ2 ψ −1 (V (s, u(s)))

i.e., V (t0 ,u0 ) V (t,u(t))

ΔV (s, u(s))   ≥ t − t0 , φ2 ψ −1 (V (s, u(s)))

t ∈ [t0 , ∞).

Next, using that V (t, u(t)) ≥ φ1 (u(t)),

t ∈ [t0 , ∞),

and V (t0 , u0 ) ≤ ψ(u0 ) ≤ ψ(b), for any  > 0,  < b, we get

(2.39)

190

2 First Order Fuzzy Dynamic Equations ψ(b) φ1 (u(t))

ΔV (s, u(s))   = φ2 ψ −1 (V (s, u(s)))

φ1 () φ1 (u(t))

+

ψ(b) φ1 () V (t0 ,u0 )

≥

V (t,u(t))

φ2



ΔV (s, u(s))  ψ −1 (V (s, u(s)))

ΔV (s, u(s))   φ2 ψ −1 (V (s, u(s))) ΔV (s, u(s))   φ2 ψ −1 (V (s, u(s))) t ∈ [t0 , ∞).

≥ t − t0 , Take ψ(b)

T > φ1 ()

ΔV (s, u(s))  . φ2 ψ −1 (V (s, u(s)))

Then T is independent of t0 and u0 and ψ(b) φ1 (u(t))

ΔV (s, u(s))   ≥ t − t0 − φ2 ψ −1 (V (s, u(s)))

ψ(b) φ1 ()

ΔV (s, u(s))   φ2 ψ −1 (V (s, u(s)))

> t − t0 − T > 0,

t ∈ (t0 + T , ∞).

φ1 (u(t)) < φ1 (),

t ∈ (t0 + T , ∞).

Hence,

Because φ1 is increasing, we obtain u(t) < ,

t ∈ (t0 + T , ∞).

Therefore the trivial solution is uniformly asymptotically stable. This completes the proof. Theorem 2.22 Suppose (C1) and V is given as in Definition 2.2 and V (t, 0) = 0 and for any c > 0 there exists an A ∈ S(b) such that V (t, A) > 0, t ∈ T. If Δr V (t, A), (t, A) ∈ T × F (R), is positive definite and there exists a ψ ∈ K so that V (t, A) ≤ ψ(A),

(t, A) ∈ T × S(b),

then the trivial solution of (2.36) is unstable. Proof Suppose that the trivial solution of (2.36) is stable. Then for any  > 0,  < b, there exists a δ > 0 such that u0  < δ implies u(t) < , t ∈ [t0 , ∞), where u(t) = u(t, t0 , u0 ). Since Δr V (t, u(t)) is positive definite, then V (t, u(t)) is increasing. Thus,

2.5 Stability Criteria

191

V (t, u(t)) ≥ V (t0 , u0 ) t ∈ (t0 , ∞).

> 0, Hence,

ψ(u(t)) ≥ V (t, u(t)) ≥ V (t0 , u0 ) t ∈ (t0 , ∞),

> 0, and

u(t) ≥ ψ −1 (V (t0 , u0 )) =α t ∈ (t0 , ∞).

> 0,

Because Δr V (t, u(t)) is positive definite, there exists a φ ∈ K such that Δr V (t, u(t)) ≥ φ(u(t)),

(t, u(t)) ∈ T × S(b).

Hence, t t0

t

Δr V (s, u(s))Δs ≥

t ∈ [t0 , ∞),

φ(u(s))Δs, t0

or V (t, u(t)) − V (t0 , u0 ) ≥

t

φ(u(s))Δs t0

≥ φ(α)(t − t0 ),

t ∈ [t0 , ∞).

Since u(t) < , t ∈ [t0 , ∞), we have ψ() ≥ ψ(u(t)) ≥ V (t, u(t)) ≥ V (t0 , u0 ) + φ(α)(t − t0 ), and taking t → ∞, we get ψ() = ∞. This is a contradiction. Therefore the trivial solution is unstable. This completes the proof. Example 2.12 Consider the following IVP: δH u = 3 · u,

t > t0 ,

u(t0 ) = (1, 2, 3).

192

2 First Order Fuzzy Dynamic Equations

Let V (t, A) = A,

t ∈ [t0 , ∞),

A ∈ F (R).

We have that u(t) = (e3 (t, t0 ), 2e3 (t, t0 ), 3e3 (t, t0 )),

t ∈ [t0 , ∞),

is the solution of the considered IVP. We have (1, 2, 3) = 3, V (t, u(t)) = 3e3 (t, t0 ),

t ∈ [t0 , ∞),

V (t, 0) = 0, Δr V (t, u(t)) = 9e3 (t, t0 ) ≥ 9,

t ∈ [t0 , ∞).

Thus, for any c > 0 there exists A ∈ S(c) so that V (t, A) > 0, t ∈ [t0 , ∞). Also, Δr V (t, A), (t, A) ∈ [t0 , ∞) × F (R), is positive definite. Take ψ = I d. Then all conditions of Theorem 2.22 are satisfied. Consequently the trivial solution of the considered IVP is unstable. Exercise 2.11 Let T = 2Z. Prove that the trivial solution of the IVP δH y = −4 · y,

t > 0,

y(0) = (−3, −2, −1), is uniformly stable.

2.6 Exponential Stability Suppose that T is a time scale with minimal element t0 and it has no maximal element. Let u0 , v0 ∈ F (R) be such that u0 H v0 = w0 exists. Consider Eq. (2.36) and assume that the solutions of the IVP for Eq. (2.36) exist and are unique for any t ≥ t0 . Definition 2.5 Let u(t) = u(t, t0 , u0 ), t ∈ [t0 , ∞). The trivial solution of Eq. (2.36) is said to be: (i) Exponentially stable on T, if there exist positive constants d, M, and a function C : R+ × T → R+ such that u(t, t0 , u0 ) ≤ C (u0 , t0 ) (eM (t, t0 ))2 ,

t ∈ [t0 , ∞)

(ii) Uniformly exponentially stable if in (i) the function C is independent of t0

2.6 Exponential Stability

193

(iii) Exponentially asymptotically stable if (i) holds and for any  > 0 there exists a positive real number T such that u(t, t0 , u0 ) < ,

t ∈ [t0 + T , ∞)

(iv) Uniformly exponentially asymptotically stable if (ii) and (iii) hold simultaneously. 1 (T × F (R)) and Theorem 2.23 Assume (C1) and let V ∈ Crd

(i) There exist strictly increasing continuous functions w, φ : [t0 , ∞) → [t0 , ∞) such that w(u(t)) ≤ V (t, u(t)) ≤ φ(u(t)),

t ∈ [t0 , ∞)

(ii) There exists a nonincreasing continuous function ψ : [0, ∞) → (−∞, 0] and constants L ≥ 0, δ ≥ M ≥ 0 such that V Δ (t, u(t)) ≤

ψ(u(t)) − L(M  δ)(t)eδ (t, 0) , 1 + Mμ(t)

t ∈ [t0 , ∞), and   ψ φ −1 (V (t, u(t))) + MV (t, u(t)) ≤ 0,

t ∈ [t0 , ∞),

where φ is given in (i). Then all solutions of (2.36) satisfy u(t) ≤ w−1 ((V (t0 , u0 ) + L)eM (t, t0 )) ,

t ∈ [t0 , ∞).

Proof We have (V (·, u(·))eM (·, t0 ))Δ (t) = V Δ (t, u(t))eM (σ (t), 0) + MV (t, u(t))eM (t, 0) ≤

(ψ(u(t)) − L(M  δ)(t)eδ (t, 0)) eM (σ (t), 0) 1 + Mμ(t) +MV (t, u(t))eM (t, 0)

= (ψ(u(t)) − L(M  δ)(t)eδ (t, 0)) eM (t, 0) +MV (t, u(t))eM (t, 0)     ≤ ψ φ −1 (V (t, u(t))) + MV (t, u(t)) eM (t, 0) −L(M  δ)(t)eδ (t, 0)eM (t, 0) ≤ −L(M  δ)(t)eδ (t, 0)eM (t, 0) = −L(M  δ)(t)eMδ (t, 0),

t ∈ [t0 , ∞).

194

2 First Order Fuzzy Dynamic Equations

Hence, t

V (t, u(t))eM (t, 0) − V (t0 , u0 )eM (t0 , 0) ≤ −L

(M  δ)(s)eMδ (s, 0)Δs

t0

s=t  = −LeMδ (s, 0)

s=t0

= −LeMδ (t, 0) + LeMδ (t0 , 0) ≤ LeMδ (t0 , 0),

t ∈ [t0 , ∞).

Then V (t, u(t))eM (t, 0) ≤ V (t0 , u0 )eM (t0 , 0) + LeMδ (t0 , 0) ≤ (L + V (t0 , u0 ))eM (t0 , 0),

t ∈ [t0 , ∞),

and V (t, u(t)) ≤ (L + V (t0 , u0 ))eM (t0 , 0)eM (0, t) = (L + V (t0 , u0 ))eM (t0 , t) = (L + V (t0 , u0 ))eM (t, t0 ),

t ∈ [t0 , ∞).

From the last inequality and from the condition (i), it follows that u(t) ≤ w−1 (V (t, u(t))) ≤ w −1 ((L + V (t0 , u0 )eM (t, t0 )) ,

t ∈ [t0 , ∞).

This completes the proof. 1 (T×F (R)), V : T×F (R) → [0, ∞) Theorem 2.24 Assume (C1) and let V ∈ Crd and satisfies the following conditions:

(i) There exist positive functions λ1 , λ2 and positive constants p, q such that λ1 (t)u(t)p ≤ V (t, u(t)) ≤ λ2 (t)u(t)q , λ1 (t0 ) ≤ λ1 (t),

t ∈ [t0 , ∞),

t ∈ [t0 , ∞).

(ii) There exist a positive function λ3 , a nonnegative constant L, a positive constant r, and a constant δ > M = inf

t≥t0

λ3 (t) r

(λ2 (t)) q

>0

2.6 Exponential Stability

195

such that V Δ (t, u(t)) ≤

−λ3 (t)u(t)r − L(M  δ)(t)eδ (t, 0) , 1 + Mμ(t)

t ∈ [t0 , ∞).

(iii) r

V (t, u(t)) − (V (t, u(t))) q ≤ 0,

t ∈ [t0 , ∞),

where q and r are constants given as in (i) and (ii), respectively. Then the trivial solution of (2.36) is exponentially stable on T. Proof We have (V (·, u(·))eM (·, 0))Δ (t) = V Δ (t, u(t))eM (σ (t), 0) + MV (t, u(t))eM (t, 0) ≤

−λ3 (t)u(t)r − L(M  δ)(t)eδ (t, 0) eM (σ (t), 0) 1 + Mμ(t)

+MV (t, u(t))eM (t, 0)   = −λ3 (t)u(t)r − L(M  δ)(t)eδ (t, 0) eM (t, 0) +MV (t, u(t))eM (t, 0)   r λ3 (t) q ≤ − r (V (t, u(t))) − L(M  δ)(t)eδ (t, 0) eM (t, 0) (λ2 (t)) q +MV (t, u(t))eM (t, 0)   r ≤ −M(V (t, u(t))) q − L(M  δ)(t)eδ (t, 0) + MV (t, u(t)) eM (t, 0) ≤ −L(M  δ)(t)eδ (t, 0)eM (t, 0) = −L(M  δ)(t)eMδ (t, 0),

t ∈ [t0 , ∞).

Integrating both sides of the last inequality from t0 to t, we get V (t, u(t))eM (t, 0) − V (t0 , u0 )eM (t0 , 0) ≤ −L

t

(M  δ)(s)eMδ (s, 0)Δs

t0

s=t  = −LeMδ (s, 0)

s=t0

= −LeMδ (t, 0) + LeMδ (t0 , 0) ≤ LeMδ (t0 , 0),

t ∈ [t0 , ∞),

whereupon V (t, u(t))eM (t, 0) ≤ V (t0 , u0 )eM (t0 , 0) + LeMδ (t0 , 0) ≤ (V (t0 , u0 ) + L)eM (t0 , 0),

t ∈ [t0 , ∞).

196

2 First Order Fuzzy Dynamic Equations

This yields V (t, u(t)) ≤ (V (t0 , u0 ) + L)eM (t0 , 0)eM (0, t) = (V (t0 , u0 ) + L)eM (t0 , t) = (V (t0 , u0 ) + L)eM (t, t0 ),

t ∈ [t0 , ∞).

Hence and (i), we get 1

u(t) ≤

(V (t, u(t))) p 1

(λ1 (t)) p ≤ (λ1 (t))

− p1

≤ (λ1 (t0 ))

− p1

1

((V (t0 , u0 ) + L)eM (t, t0 )) p 1

((V (t0 , u0 ) + L)eM (t, t0 )) p ,

t ∈ [t0 , ∞). This completes the proof. 1 (T × F (R)) Theorem 2.25 Assume (C1), and V : T × F (R) → R+ , V ∈ Crd satisfies the following conditions:

(i) There exist nonnegative constants k1 and p such that k1 u(t)p ≤ V (t, u(t)),

t ∈ [t0 , ∞).

(ii) There exist nonnegative constants k2 and L, positive constants  and δ, such that 0 <  ≤ min{k2 , δ} and V Δ (t, u(t)) ≤

−k2 V (t, u(t)) − L(  δ)(t)eδ (t, 0) , 1 + μ(t)

t ∈ [t0 , ∞). Then the trivial solution of (2.36) is uniformly exponentially stable. Proof We have (V (·, u(·))e (·, 0))Δ (t) = V Δ (t, u(t))e (σ (t), 0) + V (t, u(t))e (t, 0) ≤

−k2 V (t, u(t)) − L(  δ)(t)eδ (t, 0) e (σ (t), 0) 1 + μ(t) +V (t, u(t))e (t, 0)

2.6 Exponential Stability

197

= (−k2 V (t, u(t)) − L(  δ)(t)eδ (t, 0)) e (t, 0) +V (t, u(t))e (t, 0) ≤ −L(  δ)(t)eδ (t, 0)e (t, 0) = −L(  δ)(t)eδ (t, 0),

t ∈ [t0 , ∞).

Integrating both sides of this inequality from t0 and t, we obtain V (t, u(t))e (t, 0) − V (t0 , u0 )e (t0 , 0) ≤ −L

t t0

(  δ)(s)eδ (s, 0)Δs

s=t  = −Leδ (s, 0)

s=t0

= −Leδ (t, 0) + Leδ (t0 , 0),

t ∈ [t0 , ∞),

or V (t, u(t))e (t, 0) ≤ V (t0 , u0 )e (t0 , 0) − Leδ (t, 0) +Leδ (t0 , 0) ≤ V (t0 , u0 )e (t0 , 0) + Leδ (t0 , 0) ≤ (V (t0 , u0 ) + L)e (t0 , 0),

t ∈ [t0 , ∞).

By the last inequality, we arrive at V (t, u(t)) ≤ (V (t0 , u0 ) + L) e (t0 , 0)e (0, t) = (V (t0 , u0 ) + L)e (t0 , t) = (V (t0 , u0 ) + L)e (t, t0 ),

t ∈ [t0 , ∞).

Now we apply (i) and we obtain u(t) ≤

1 1 p

1

(V (t, u(t))) p

k1

1

≤

(V (t0 , u0 ) + L) p 1 p

1

(e (t, t0 )) p ,

t ∈ [t0 , ∞).

k1 This completes the proof.

Theorem 2.26 Assume that all conditions of Theorem 2.24 are satisfied except the estimate (ii) is strengthened to V Δ (t, u(t)) ≤ −λ3 (t)u(t)r ,

t ∈ [t0 , ∞),

(2.40)

198

2 First Order Fuzzy Dynamic Equations

where λ3 is positive and nondecreasing. In addition, assume that λ1 in (i) is nondecreasing and V (t, 0) = 0 for t ∈ T. Then the trivial solution of (2.36) is exponentially asymptotically stable. Proof By Theorem 2.24, it follows that the trivial solution of (2.36) is exponentially stable. Then there exist positive constants d, M and a function C : R+ × T → R+ such that u(t, t0 , u0 ) ≤ C(u0 , t0 ) (eM (t, t0 ))d ,

t ∈ [t0 , ∞).

We have that lim C(u0 , t0 ) (eM (t, t0 ))d = 0.

t→∞

By (2.40), we get V (t, u(t)) − V (t0 , u0 ) ≤ −

t

λ3 (s)u(s)r Δs,

t ∈ [t0 , ∞),

(2.41)

t0

and V is a decreasing function on T. Therefore there exists β = lim V (t, u(t)). t→∞

We have that β ≥ 0. Assume that β > 0. Then V (t, u(t)) ≥ β > 0,

t ∈ [t0 , ∞).

On the other hand, by (2.41), we arrive at a contradiction lim V (t, u(t)) = −∞.

t→∞

Thus, β = 0. Assume that limt→∞ u(t) = 0. Then there exists an  > 0 such that for any n ∈ N there is a tn ∈ T, with tn ≥ n, so that u(tn ) ≥ . Hence, using (i) of Theorem 2.24 and using that λ1 is positive and nondecreasing, we obtain V (tn , u(tn )) ≥ λ1 (tn )u(tn )p ≥ λ1 (t0 ) p > 0,

n ∈ N,

2.6 Exponential Stability

199

and lim V (tn , u(tn )) > 0,

n→∞

which is a contradiction. Therefore limt→∞ u(t) = 0. This completes the proof. Theorem 2.27 Assume that all conditions of Theorem 2.25 are satisfied except the estimate of (ii) is strengthened to V Δ (t, u(t)) ≤ −k2 V (t, u(t)),

t ∈ [t0 , ∞).

(2.42)

In addition, assume that V (t, 0) = 0, t ∈ T. Then the trivial solution of (2.36) is uniformly exponentially asymptotically stable. Proof By Theorem 2.25, it follows that the trivial solution of (2.36) is uniformly exponentially stable. Therefore there exist positive constants d, M and a function C : R+ × T → R+ such that u(t, t0 , u0 ) ≤ C(u0 ) (eM (t, t0 ))d ,

t ∈ [t0 , ∞).

Also, we have lim C(u0 ) (eM (t, t0 ))d = 0.

t→∞

Integrating both sides of (2.42) from t0 to t, we get !t V (t, u(t)) ≤ V (t0 , u0 ) − k2 t0 V (s, u(s))Δs !t ≤ V (t0 , u0 ) − k2 k1 t0 u(s)p Δs,

t ∈ [t0 , ∞).

(2.43)

By (2.42), we have that V (·, u(·)) is a decreasing function on T. Therefore there exists α = lim V (t, u(t)). t→∞

We have that α ≥ 0. Assume that α > 0. We have V (t, u(t)) ≥ α > 0,

t ∈ [t0 , ∞).

On the other hand, by (2.43), we get lim V (t, u(t)) = −∞,

t→∞

200

2 First Order Fuzzy Dynamic Equations

which is a contradiction. Consequently α = 0. Now, we assume that limt→∞ u(t) = 0. Then there exists an  > 0 such that for any n ∈ N there is a tn ∈ T, with tn ≥ n, so that u(tn ) ≥ . Then V (tn , u(tn )) ≥ k1 u(tn )p ≥ k1  p n ∈ N,

> 0, whereupon

lim V (tn , u(tn )) > 0,

n→∞

which is a contradiction. Therefore limt→∞ u(t) = 0. This completes the proof. Example 2.13 Let T = R. Consider the equation δH y = (−2) · y,

t ∈ [t0 , ∞).

Take V (t, A) = A,

(t, A) ∈ T × F (R).

We have V Δ (t, u(t)) = −2u(t),

t ∈ [t0 , ∞).

Let k1 =

1 , 2

p = 1,

k2 =

1 , 2

L = 0,

δ = 1,

=

1 . 2

Then all conditions of Theorem 2.25 are satisfied. Therefore the trivial solution of the considered equation is uniformly exponentially stable. Exercise 2.12 Let T = 2Z. Investigate the trivial solution of the equation δH y = (−7) · y, for: 1. 2. 3. 4.

Exponential stability Uniform exponential stability Exponential asymptotic stability Uniform exponential asymptotic stability

t > 0,

2.7 Advanced Practical Problems

201

2.7 Advanced Practical Problems Problem 2.1 Let T = N0 . Check if  y(t) = 1 + t, 3 + t + t 2 , 1 + 3t + t 2 +

3 1 + t2

 ,

t > 0,

δH y = (t + 1) · y + (1 + t, 11 + 4t, 15 + 4t + 3t 2 ),

t > 0,

is a solution of the following IVP:

y(0) = (1, 3, 4). Problem 2.2 Let T = 4Z. Find a solution of the following IVP: δH y = 2 · y + {1 + t, 1 + 3t, 1 + 3t + t 2 },

t > 0,

y(0) = {−1, 0, 1}. Problem 2.3 Let T = 3N0 . Find a solution of the following IVP: δH y = (1 + t + t 2 ) · y + {1, 3, 1 + t + t 2 },

t > 1,

y(1) = {−3, 0, 4}. Problem 2.4 Let T = 4N0 . Find a solution of the following IVP:     1 1 2 δH y = 1 + t + 4 · y + −1, 2 , 1 + t + 2 , t t t

t > 1,

y(1) = {−3, 0, 4}. Problem 2.5 Let T = 3N0 . Find a solution of the following IVP: δH y + (1 + 2t) · y = {t, t + 2, 2 + 3t + 4t 2 },

t > 1,

y(1) = {1, 2, 3}. Problem 2.6 Let T = 2N0 . Find a solution of the following IVP:   δSH y = −3 − t 2 · y + {−1 − t, 2t, 2 + 4t + 4t 2 }, y(1) = {1, 3, 5}.

t > 1,

202

2 First Order Fuzzy Dynamic Equations

Problem 2.7 Let T = 4N0 . Find a solution of the following IVP:  

 1 + t + 2t 2 · y = 1, 2 + t, 3 + t + t 2 , δSH y + − 12

t > 0,

y(0) = {−3, 1, 4}. Problem 2.8 Let T = 3N0 . Find a solution of the following IVP: δH y + {−t 3 , −t 2 , −1} = t 2 · y,

t > 0,

y(0) = {1, 3, 5}. Problem 2.9 Let T = 2N0 . Find a solution of the following IVP: δSH y + {−t 4 , −t 3 , −t 2 , −t} = (−t) · y,

t > 1,

y(1) = {1, 5, 9}. Problem 2.10 Let T = 2N0 . Find conditions for t0 and t1 so that the following IVP: 1 2 t · y + {1, 2, 3}, 8 y(t0 ) = {−1, 0, 5}, δH y =

has a unique solution. Problem 2.11 Let T = 2N0 . Prove that the trivial solution of the IVP δH y = (−4t − 3) · y,

t > 1,

y(1) = {−7, −6, −5}, is uniformly asymptotically stable. Problem 2.12 Let T = 2N0 . Investigate the trivial solution of the equation δH y = (−e1 (t, 1)) · y, for: 1. 2. 3. 4.

Exponential stability Uniform exponential stability Exponential asymptotic stability Uniform exponential asymptotic stability

t > 1,

2.8 Notes and References

203

2.8 Notes and References In this chapter we considered some classes of linear fuzzy dynamic equations and gave formulas for their solutions. We proved some results for existence and uniqueness of the solutions of some classes of nonlinear fuzzy dynamic equations. We investigated the continuous dependence of the solutions of first order fuzzy dynamic equations on the initial data and gave some comparison results. We introduced the Lyapunov functional, and, using it, proved criteria for stability and exponential stability of the trivial solution of some classes of nonlinear fuzzy dynamic equations. The material in this chapter is based on the results in the papers [18, 26, 29, 32, 42].

Chapter 3

Second Order Fuzzy Dynamic Equations

This chapter is devoted on a qualitative analysis of second order fuzzy dynamic equations. Firstly, the α-levels of the solutions of linear second order fuzzy dynamic equations are investigated. Then, it is introduced a class of BVPs for linear second order fuzzy dynamic equations and for it is given the corresponding Green function. In the chapter, it is investigated the Cauchy problem for a class of second order fuzzy dynamic equations for existence and uniqueness of the solutions and for the continuous dependence of the solutions on the initial data. For this aim, it is defined a suitable Banach space. Suppose that T is a time scale with forward jump operator and delta differentiation operator, σ and Δ, respectively.

3.1 Linear Second Order Fuzzy Dynamic Equations Let t0 , t1 ∈ T, t0 < t1 . With J , we will denote the interval (t0 , t1 ] or (t0 , t1 ). In this section, we will investigate the following IVP: 2 δH y = a(t) · δH y + b(t) · y + c(t),

δH y(t0 ) = a0 ,

t ∈ J,

y(t0 ) = a1 ,

(3.1) (3.2)

where a, b : [t0 , t1 ] → R, a, b ∈ Crd ([t0 , t1 ]), a0 , a1 ∈ F (R) with α-levels   [a0 ]α = a α0 , a α0 ,

α ∈ [0, 1],

  [a1 ]α = a α1 , a α1 ,

α ∈ [0, 1],

and

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_3

205

206

3 Second Order Fuzzy Dynamic Equations

respectively, c : [t0 , t1 ] → F (R), c ∈ Cf rd ([t0 , t1 ]) with α-levels   [c(t)]α = cα (t), cα (t) ,

t ∈ [t0 , t1 ],

α ∈ [0, 1],

and y : [t0 , t1 ] → F (R) is unknown fuzzy function. We will denote the α-levels of the unknown function y as follows:

 [y(t)]α = y α (t), y α (t) ,

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Then,

 [δH y(t)]α = y αΔ (t), y αΔ (t) ,

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) ,

t ∈ [t0 , t1 ],

α ∈ [0, 1].

We consider the following cases: 1. Suppose that a ≥ 0, b ≥ 0 on [t0 , t1 ]. Then,

 [a(t) · δH y(t)]α = a(t)y αΔ (t), a(t)y αΔ (t) ,

 [b(t) · y(t)]α = b(t)y α (t), b(t)y α (t) , t ∈ (t0 , t1 ],

α ∈ [0, 1].

Hence, [a(t) · δH y(t)]α + [b(t) · y(t)]α + c(t) 



= a(t)y αΔ (t), a(t)y αΔ (t) + b(t)y α (t), b(t)y α (t)   + cα (t), cα (t)

 = a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) , t ∈ [t0 , t1 ], α ∈ [0, 1]. From here and from (3.1), we get

 2 2 y αΔ (t), y αΔ (t) 

= a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) ,

t ∈ [t0 , t1 ], α ∈ [0, 1], and 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t),

3.1 Linear Second Order Fuzzy Dynamic Equations

207

t ∈ [t0 , t1 ], α ∈ [0, 1]. By the initial conditions (3.2), we find

   y αΔ (t0 ), y αΔ (t0 ) = a α0 , a α0 ,

   y α (t0 ), y α (t0 ) = a α1 , a α1 , α ∈ [0, 1], whereupon y αΔ (t0 ) = a α0 , y αΔ (t0 ) = a α0 , y α (t0 ) = a α1 , y α (t0 ) = a α1 ,

α ∈ [0, 1].

Consequently, we arrive at the following IVPs: 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), y αΔ (t0 ) = a α0 ,

y α (t0 ) = a α1 ,

t ∈ (t0 , t1 ],

α ∈ [0, 1],

t ∈ (t0 , t1 ],

α ∈ [0, 1],

α ∈ [0, 1],

and 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), y

αΔ

(t0 ) =

a α0 ,

y (t0 ) = α

a α1 ,

α ∈ [0, 1].

We find their solutions, and then we obtain the α-levels of the unknown function y. 2. Let a ≥ 0, b ≤ 0 on [t0 , t1 ]. Then,

 [a(t) · δH y(t)]α = a(t)y αΔ (t), a(t)y αΔ (t) ,

 [b(t) · y(t)]α = b(t)y α (t), b(t)y α (t) , t ∈ [t0 , t1 ],

α ∈ [0, 1],

and hence, [a(t) · δH y(t) + b(t) · y(t) + c(t)]α

 = a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) , t ∈ [t0 , t1 ],

α ∈ [0, 1], and

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) 

= a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) ,

208

3 Second Order Fuzzy Dynamic Equations

t ∈ [t0 , t1 ],

α ∈ [0, 1]. Thus, we get

2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

By the initial conditions, we arrive at

 [δH y(t0 ]α = y αΔ (t0 ), y αΔ (t0 ) 

= y α1 , y α1 ,

 [y(t0 )]α = y α (t0 ), y α (t0 ) 

= y α0 , y α0 , α ∈ [0, 1], whereupon y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 ,

α ∈ [0, 1].

Hence, we obtain the following IVP: 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 , α ∈ [0, 1].

t ∈ [t0 , t1 ],

α ∈ [0, 1],

(3.3)

Let z(t) = y α (t) − y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Then, 2

zΔ (t) = a(t)zΔ (t) − b(t)z(t) + cα (t) − cα (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

3.1 Linear Second Order Fuzzy Dynamic Equations

209

and z(t0 ) = y α (t0 ) − y α (t0 ) = y α0 − y α0 , zΔ (t) = y αΔ (t) − y αΔ (t), zΔ (t0 ) = y αΔ (t0 ) − y αΔ (t0 ) = y α1 − y α1 ,

α ∈ [0, 1],

i.e., we arrive at the following IVP: 2

zΔ (t) = a(t)zΔ (t) − b(t)z(t) + cα (t) − cα (t), z (t0 ) = Δ

z(t0 ) =

− y α1 , y α0 − y α0 ,

t ∈ (t0 , t1 ],

y α1

α ∈ [0, 1].

By the last IVP, we find z, and then, using that y α (t) = z(t) + y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

we get y αΔ (t) = zΔ (t) + y αΔ (t), 2

2

2

y αΔ (t) = zΔ (t) + y αΔ (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Hence, by the second equation of (3.3), we obtain   2 2 zΔ (t) + y αΔ (t) = a(t) zΔ (t) + y αΔ (t) + b(t)y α (t) + cα (t) = a(t)y αΔ (t) + b(t)y α (t) + a(t)zΔ (t) +cα (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

or 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + a(t)zΔ (t) 2

+cα (t) − zΔ (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Therefore, y α is a solution of the IVP 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + a(t)zΔ (t) 2

+cα (t) − zΔ (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

(3.4)

210

3 Second Order Fuzzy Dynamic Equations

y α (t0 ) = y 0 , y αΔ (t0 ) = y 1 ,

α ∈ [0, 1].

From here, we find y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1], and using (3.4), we obtain y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1]. 3. Let a ≤ 0, b ≥ 0 on [t0 , t1 ]. Then,

 [a(t) · δH y(t)]α = a(t)y αΔ (t), a(t)y αΔ (t) ,

 [b(t) · y(t)]α = b(t)y α (t), b(t)y α (t) , t ∈ [t0 , t1 ],

α ∈ [0, 1],

and hence, [a(t) · δH y(t) + b(t) · y(t) + c(t)]α

 = a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) , t ∈ [t0 , t1 ],

α ∈ [0, 1], and

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) 

= a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) ,

t ∈ [t0 , t1 ],

α ∈ [0, 1]. Thus, we get

2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t),

t ∈ [t0 , t1 ],

By the initial conditions, we arrive at

 [δH y(t0 ]α = y αΔ (t0 ), y αΔ (t0 ) 

= y α1 , y α1 ,

 [y(t0 )]α = y α (t0 ), y α (t0 ) 

= y α0 , y α0 , α ∈ [0, 1],

α ∈ [0, 1].

3.1 Linear Second Order Fuzzy Dynamic Equations

211

whereupon y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 ,

α ∈ [0, 1].

Hence, we obtain the following IVP: 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 , α ∈ [0, 1].

t ∈ [t0 , t1 ],

α ∈ [0, 1],

(3.5)

Let z(t) = y α (t) − y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

zΔ (t) = −a(t)zΔ (t) + b(t)z(t) + cα (t) − cα (t),

t ∈ [t0 , t1 ],

Then, 2

α ∈ [0, 1],

and z(t0 ) = y α (t0 ) − y α (t0 ) = y α0 − y α0 , zΔ (t) = y αΔ (t) − y αΔ (t), zΔ (t0 ) = y αΔ (t0 ) − y αΔ (t0 ) = y α1 − y α1 ,

α ∈ [0, 1],

i.e., we arrive at the following IVP: 2

zΔ (t) = −a(t)zΔ (t) + b(t)z(t) + cα (t) − cα (t), z (t0 ) = Δ

z(t0 ) =

− y α1 , y α0 − y α0 , y α1

α ∈ [0, 1].

t ∈ (t0 , t1 ],

212

3 Second Order Fuzzy Dynamic Equations

By the last IVP, we find z, and then, using that y α (t) = z(t) + y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

(3.6)

we get y αΔ (t) = zΔ (t) + y αΔ (t), 2

2

2

y αΔ (t) = zΔ (t) + y αΔ (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Hence, by the second equation of (3.5), we obtain 2

2

zΔ (t) + y αΔ (t) = a(t)y αΔ (t)   +b(t) z(t) + y α (t) + cα (t) = a(t)y αΔ (t) + b(t)y α (t) +b(t)z(t) + cα (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

2

t ∈ [t0 , t1 ],

α ∈ [0, 1].

2

t ∈ [t0 , t1 ],

α ∈ [0, 1],

or 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) +b(t)z(t) + cα (t) − zΔ (t), Therefore, y α is a solution of the IVP 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) +b(t)z(t) + cα (t) − zΔ (t), y (t0 ) = α

y α0 ,

y αΔ (t0 ) = y α1 ,

α ∈ [0, 1].

Hence, we find y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1], and using (3.6), we get y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1]. 4. Let a ≤ 0, b ≤ 0 on [t0 , t1 ]. Then,

 [a(t) · δH y(t)]α = a(t)y αΔ (t), a(t)y αΔ (t) ,

 [b(t) · y(t)]α = b(t)y α (t), b(t)y α (t) , t ∈ [t0 , t1 ], α ∈ [0, 1], and hence, [a(t) · δH y(t) + b(t) · y(t) + c(t)]α

 = a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) ,

3.1 Linear Second Order Fuzzy Dynamic Equations

t ∈ [t0 , t1 ],

213

α ∈ [0, 1], and

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) 

= a(t)y αΔ (t) + b(t)y α (t) + cα (t), a(t)y αΔ (t) + b(t)y α (t) + cα (t) ,

t ∈ [t0 , t1 ],

α ∈ [0, 1]. Thus, we get

2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

By the initial conditions, we arrive at

 [δH y(t0 ]α = y αΔ (t0 ), y αΔ (t0 ) 

= y α1 , y α1 ,

 [y(t0 )]α = y α (t0 ), y α (t0 ) 

= y α0 , y α0 , α ∈ [0, 1], whereupon y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 ,

α ∈ [0, 1].

Hence, we obtain the following IVP: 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), 2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), y αΔ (t0 ) = y α1 , y αΔ (t0 ) = y α1 , y α (t0 ) = y α0 , y α (t0 ) = y α0 , α ∈ [0, 1].

t ∈ [t0 , t1 ],

α ∈ [0, 1],

(3.7)

214

3 Second Order Fuzzy Dynamic Equations

Let z(t) = y α (t) − y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

zΔ (t) = −a(t)zΔ (t) − b(t)z(t) + cα (t) − cα (t),

t ∈ [t0 , t1 ],

Then, 2

α ∈ [0, 1],

and z(t0 ) = y α (t0 ) − y α (t0 ) = y α0 − y α0 , zΔ (t) = y αΔ (t) − y αΔ (t), zΔ (t0 ) = y αΔ (t0 ) − y αΔ (t0 ) = y α1 − y α1 ,

α ∈ [0, 1],

i.e., we arrive at the following IVP: 2

zΔ (t) = −a(t)zΔ (t) − b(t)z(t) + cα (t) − cα (t), z (t0 ) = Δ

z(t0 ) =

− y α1 , y α0 − y α0 ,

t ∈ (t0 , t1 ],

y α1

α ∈ [0, 1].

By the last IVP, we find z, and then, using that y α (t) = z(t) + y α (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1],

we get y αΔ (t) = zΔ (t) + y αΔ (t), 2

2

2

y αΔ (t) = zΔ (t) + y αΔ (t),

t ∈ [t0 , t1 ],

α ∈ [0, 1].

Hence, by the second equation of (3.7), we obtain 2

2

zΔ (t) + y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t), t ∈ [t0 , t1 ], α ∈ [0, 1], or 2

2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t) − zΔ (t), t ∈ [t0 , t1 ], α ∈ [0, 1]. Therefore, y α is a solution of the following IVP:

(3.8)

3.1 Linear Second Order Fuzzy Dynamic Equations

215

2

2

y αΔ (t) = a(t)y αΔ (t) + b(t)y α (t) + cα (t) − zΔ (t), t ∈ (t0 , t1 ],

α ∈ [0, 1],

y α (t0 ) = y α0 , y αΔ (t0 ) = y α1 ,

α ∈ [0, 1].

From here, we find y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1], and using (3.8), we obtain y α (t), t ∈ [t0 , t1 ], α ∈ [0, 1]. Example 3.1 Let T = 12 Z. Consider the IVP 2 δH y = 1 · δH y + 2 · y,

t > 0,

δH y(0) = y(0) = y0 , where y0 ∈ F (R) and [y0 ]α = [0, 1]. Let

 [y(t)]α = y α (t), y α (t) , Then,

t ∈ [t0 , t1 ],

 [δH y(t)]α = y αΔ (t), y αΔ (t) ,

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) ,

Next,

α ∈ [0, 1].

t ∈ [t0 , t1 ],

 [1 · δH y(t)]α = y αΔ (t), y αΔ (t) ,

 [2 · y(t)]α = 2y α (t), 2y α (t) , t ≥ 0,

α ∈ [0, 1].

α ∈ [0, 1],

and

 [1 · δH y(t) + 2 · y(t)] = y αΔ (t) + 2y α (t), y αΔ (t) + 2y α (t) , t > 0, α ∈ [0, 1]. Therefore,



 2 2 y αΔ (t), y αΔ (t) = y αΔ (t) + 2y α (t), y αΔ (t) + 2y α (t) ,

t > 0, α ∈ [0, 1]. Hence, 2

y αΔ (t) = y αΔ (t) + 2y α (t), 2

y αΔ (t) = y αΔ (t) + 2y α (t),

t > 0,

α ∈ [0, 1].

216

3 Second Order Fuzzy Dynamic Equations

By the initial conditions, we find y αΔ (0) = y α (0) = 0, y αΔ (0) = y α (0) = 1,

α ∈ [0, 1].

Thus, we get the following IVPs: 2

y αΔ (t) = y αΔ (t) + 2y α (t), y αΔ (0) = y α (0) = 0,

t > 0,

α ∈ [0, 1],

α ∈ [0, 1],

and 2

y αΔ (t) = y αΔ (t) + 2y α (t), y αΔ (0) = y α (0) = 1,

t > 0,

α ∈ [0, 1],

α ∈ [0, 1].

For their solutions, we have the following representations: y α (t) = c1 e−1 (t, 0) + c2 e2 (t, 0), y α (t) = c3 e−1 (t, 0) + c4 e2 (t, 0),

t ∈ [0, ∞),

α ∈ [0, 1],

where c1 , c2 , c3 , and c4 are real constants. Since, y αΔ (t) = −c1 e−1 (t, 0) + 2c2 e2 (t, 0),

t ∈ [0, ∞),

α ∈ [0, 1],

y αΔ (t) = −c3 e−1 (t, 0) + 2c4 e2 (t, 0),

t ∈ [0, ∞),

α ∈ [0, 1],

we get 0 = y α (0) = c1 + c2 , 0 = y αΔ (0) = −c1 + 2c2 , 1 = y α (0) = c3 + c4 , 1 = y αΔ (0) = −c3 + 2c4 ,

α ∈ [0, 1].

3.1 Linear Second Order Fuzzy Dynamic Equations

217

In this way, for the constants c1 , c2 , c3 , and c4 , we obtain the systems c1 + c2 = 0 −c1 + 2c2 = 0 and c3 + c4 = 1 −c3 + 2c4 = 1. Then, c1 = c2 = 0, 1 , 3 2 c4 = . 3 c3 =

Consequently, y α (t) = 0, y α (t) =

1 2 e−1 (t, 0) + e2 (t, 0), 3 3

t ∈ [0, ∞),

α ∈ [0, 1],

and   1 2 [y(t)] = 0, e−1 (t, 0) + e2 (t, 0) , 3 3 α

Exercise 3.1 Let T =

 N0 1 5

t ∈ [0, ∞),

. Find a solution of the IVP

2 δH y = 1 · δH y + 6 · y,

t > 0,

δH y(0) = y1 , y(0) = y0 , where y0 , y1 ∈ F (R) and [y0 ]α = [−1, 4] , [y1 ]α = [0, 5] ,

α ∈ [0, 1].

α ∈ [0, 1].

218

3 Second Order Fuzzy Dynamic Equations

Example 3.2 Let T = 15 Z. Consider the IVP 2 y = (−1) · δH y + 2 · y, δH

t ∈ (0, ∞),

δH y(0) = y1 , y(0) = y0 , where [y0 ]α = [0, 1], [y1 ]α = [−1, 1],

α ∈ [0, 1].

Here, 1 σ (t) = t + , 5 a(t) = −1, b(t) = 2, c(t) = 0,

t ∈ [0, ∞),

α ∈ [0, 1].

Let

 [y(t)]α = y α (t), y α (t) ,

t ∈ [0, ∞),

α ∈ [0, 1].

Then,

 [δH y(t)]α = y αΔ (t), y αΔ (t) , α



2 2 2 y(t) = y αΔ (t), y αΔ (t) , δH

 [(−1) · δH y(t)]α = −y αΔ (t), −y αΔ (t) , 

[2 · y(t)] = 2y α (t), 2y α (t) , t ∈ [0, ∞),

α ∈ [0, 1].

Hence,

 [(−1) · δH y(t) + 2 · y(t)]α = −y αΔ (t) + 2y α (t), −y αΔ (t) + 2y α (t) , t ∈ [0, ∞), α ∈ [0, 1], and

α

 2 2 2 δH y(t) = y αΔ (t), y αΔ (t) 

= −y αΔ (t) + 2y α (t), −y αΔ (t) + 2y α (t) ,

3.1 Linear Second Order Fuzzy Dynamic Equations

219

t ∈ [0, ∞), α ∈ [0, 1]. Therefore, we get the system 2

y αΔ (t) = −y αΔ (t) + 2y α (t) 2

y αΔ (t) = −y αΔ (t) + 2y α (t), t ∈ [0, ∞), α ∈ [0, 1]. By the initial conditions, we find

 [δH y(0)]α = y αΔ (0), y αΔ (0) = [−1, 1],

 [y(0)] = y α (0), y α (0) α

= [0, 1],

α ∈ [0, 1],

and y αΔ (0) = −1, y α (0) = 0, y αΔ (0) = 1, y α (0) = 1. Consequently, we obtain the following IVP: 2

y αΔ (t) = −y αΔ (t) + 2y α (t) 2

y αΔ (t) = y αΔ (0) = y α (0) = y αΔ (0) = y α (0) =

−y αΔ (t) + 2y α (t), −1 0 1 1, α ∈ [0, 1].

t ∈ (0, ∞),

z(t) = y α (t) − y α (t),

t ∈ [0, ∞),

α ∈ [0, 1], (3.9)

Let

Then, zΔ (t) = y αΔ (t) − y αΔ (t), 2

2

2

zΔ (t) = y αΔ (t) − y αΔ (t), z(0) = y α (0) − y α (0)

α ∈ [0, 1].

220

3 Second Order Fuzzy Dynamic Equations

= 1−0 = 1, zΔ (0) = y αΔ (0) − y αΔ (0) = 1 − (−1) = 2,

t ∈ [0, ∞),

α ∈ [0, 1].

Hence, by the first two equations of (3.9), we obtain the following IVP: 2

zΔ (t) = −zΔ (t) + 2z(t),

t ∈ (0, ∞),

z(0) = 1, z (0) = 2. Δ

Consider the homogeneous equation 2

zΔ (t) + zΔ (t) − 2z(t) = 0,

t ∈ [0, ∞).

Its characteristic equation is r 2 + r − 2 = 0. We have r1 = −2,

r2 = 1,

and z1 (t) = e−2 (t, 0)  t 2 5 = 1− 5 t   3 5 = , 5 z2 (t) = e1 (t, 0) t  1 5 = 1+ 5 t   6 5 = , t ∈ [0, ∞) 5

3.1 Linear Second Order Fuzzy Dynamic Equations

221

form a fundamental system of solutions. Next, z1Δ (t) = −2e−2 (t, 0), z1Δ (0) = −2, z2Δ (t) = e1 (t, 0), z2Δ (0) = 1,    z1 (t) z2 (t)    W (z1 , z2 )(t) =  Δ z1 (t) z2Δ (t)  = z1 (t)z2Δ (t) − z1Δ (t)z2 (t) = e−2 (t, 0)e1 (t, 0) + 2e−2 (t, 0)e1 (t, 0) = 3e−2 (t, 0)e1 (t, 0) = 3e1⊕(−2) (t, 0),

t ∈ [0, ∞).

Note that (1 ⊕ (−2)) (t) = 1 + (−2) + μ(t)(−2) = −1 −

2 5

7 =− , 5

t ∈ [0, ∞).

Therefore, W (z1 , z2 )(t) = 3e− 7 (t, 0), 5

W (z1 , z2 )(0) = 3,

t ∈ [0, ∞).

Also, α0 =

z2Δ (0)z(0) − z2 (0)zΔ (0) W (z1 , z2 )(0)

1−2 3 1 =− , 3 =

β0 =

z1 (0)zΔ (0) − z1Δ (0)z(0) W (z1 , z2 )(0)

2 − (−2) 3 4 = . 3 =

222

3 Second Order Fuzzy Dynamic Equations

By the variation of parameters, we arrive at z(t) = α0 z1 (t) + β0 z2 (t) 1 4 = − e−2 (t, 0) + e1 (t, 0) 3 3  t  t 1 3 5 4 6 5 =− + , t ∈ [0, ∞). 3 5 3 5 Hence, y α (t) = z(t) + y α (t) 1 4 = y α (t) − e−2 (t, 0) + e1 (t, 0), 3 3

t ∈ [0, ∞),

α ∈ [0, 1].

From here, 2 4 y αΔ (t) = y αΔ (t) + e−2 (t, 0) + e1 (t, 0), 3 3 4 4 2 2 y αΔ (t) = y αΔ (t) − e−2 (t, 0) + e1 (t, 0), 3 3 t ∈ [0, ∞), α ∈ [0, 1]. By the second equation of (3.9), we find 4 4 2 y αΔ (t) − e−2 (t, 0) + e1 (t, 0) = −y αΔ (t) 3 3   1 4 α +2 y (t) − e−2 (t, 0) + e1 (t, 0) 3 3 = −y αΔ (t) + 2y α (t) 8 2 − e−2 (t, 0) + e1 (t, 0), 3 3 t ∈ [0, ∞), α ∈ [0, 1], whereupon 2 2 y αΔ (t) = −y αΔ (t) + 2y α (t) + e−2 (t, 0) 3 4 + e1 (t, 0), t ∈ [0, ∞), α ∈ [0, 1]. 3 We get the following IVP:

3.1 Linear Second Order Fuzzy Dynamic Equations

223

2 2 y αΔ (t) = −y αΔ (t) + 2y α (t) + e−2 (t, 0) 3 4 + e1 (t, 0), t ∈ [0, ∞), α ∈ [0, 1], 3 y αΔ (0) = −1, y α (0) = 0. We have δ0 =

z2Δ (0)y α (0) − z2 (0)y αΔ (0) W (z1 , z2 )(0)

0 − (−1) 3 1 = , 3 =

γ0 =

z1 (0)y αΔ (0) − z1Δ (0)y α (0) W (z1 , z2 )(0)

1 =− . 3 Then, by the variation of parameters, we get y α (t) = δ0 z1 (t) + γ0 z2 (t) t

+ 0

z1 (σ (τ ))z2 (t) − z2 (σ (τ ))z1 (t) W (z1 , z2 )(σ (τ ))



 2 4 e−2 (τ, 0) + e1 (τ, 0) Δτ 3 3

=

1 1 e−2 (t, 0) − e1 (t, 0) 3 3 t e (σ (τ ), 0)e (t, 0) − e (σ (τ ), 0)e (t, 0) −2 1 1 −2 + 3e−2 (σ (τ ), 0)e1 (σ (τ ), 0) 0   2 4 e−2 (τ, 0) + e1 (τ, 0) Δτ × 3 3

=

1 1 e−2 (t, 0) − e1 (t, 0) 3 3 t   1 e1 (σ (τ ), 0)e1 (t, 0) − e(−2) (σ (τ ), 0)e−2 (t, 0) + 3 0   2 4 e−2 (τ, 0) + e1 (τ, 0) Δτ, t ∈ [0, ∞). × 3 3

224

3 Second Order Fuzzy Dynamic Equations

Note that (1)(t) = −

1 1+

1 5

5 =− , 6 2 ((−2))(t) = 1 − 25 10 , 3 e1 (t, 0) = e− 5 (t, 0) =

6

 t 1 5 = 1− 6  t 5 5 = , 6 t  2 5 e(−2) (t, 0) = 1 + 3  t 5 5 = , t ∈ [0, ∞). 3 Therefore,  t  t 3 5 1 6 5 − 5 3 5   5τ +1   t   5τ +1   t  ( 5 25 6 5 3 5 1 5 25 + − 15

6 5 3 5 

1 y (t) = 3 α

τ ∈ 0,t− 15

  τ  τ  2 3 5 4 6 5 + × , 3 5 3 5 and y α (t) = z(t) + y α (t) 1 =− 3

 t  t 3 5 4 6 5 + 5 3 5

t ∈ [0, ∞),

3.1 Linear Second Order Fuzzy Dynamic Equations

 t  t 3 5 1 6 5 − 5 3 5   5τ +1   t   5τ +1   t  ( 5 25 6 5 3 5 1 5 25 + − 15

6 5 3 5  1 + 3

τ ∈ 0,t− 15

  τ  τ  2 3 5 4 6 5 + × 3 5 3 5 =

 t 6 5 5 1 + 15

(

τ ∈ 0,t− 15

  5τ +1   t   5τ +1   t  5 25 6 5 3 5 5 25 − 6 5 3 5 

  τ  τ  2 3 5 4 6 5 × + , 3 5 3 5 t ∈ [0, ∞), and

  α α [y(t)] = y (t), y (t) α

 t   t 1 6 5 1 3 5 − = 3 5 3 5   5τ +1   t   5τ +1   t  ( 5 25 5 25 6 5 3 5 1 + − 15

6 5 3 5  τ ∈ 0,t− 51

  τ  τ  2 3 5 4 6 5 + × , 3 5 3 5  t 6 5 5 1 + 15

(

τ ∈ 0,t− 51

  5τ +1   t   5τ +1   t  5 25 6 5 3 5 5 25 − 6 5 3 5 

  τ  τ   2 3 5 4 6 5 , + × 3 5 3 5 t ∈ [0, ∞), α ∈ [0, 1].

225

226

3 Second Order Fuzzy Dynamic Equations

Exercise 3.2 Let T =

1 27 N0 .

Find a solution of the IVP

2 y = (−2) · δH y + 3 · y + c(t), δH

t > 0,

δH y(0) = y1 , y(0) = y0 , where [c(t)]α = [0, t], [y0 ]α = [−1, 0] , [y1 ]α = [0, 2] ,

t ∈ [0, ∞),

α ∈ [0, 1].

Exercise 3.3 Let T = 16 N0 . Find a solution of the IVP 2 y = (−6) · δH y + 7 · y + c(t), δH

t > 0,

δH y(0) = y1 , y(0) = y0 , where [c(t)]α = [0, 1 + 2t + t 2 ], [y0 ]α = [−3, −1] , [y1 ]α = [0, 1] ,

t ∈ [0, ∞),

α ∈ [0, 1].

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations Let a, b ∈ T, a < b. Consider the following BVP: δH (p · δH y) (t) = h(t),

t ∈ (a, b),

  y(a) = y σ 2 (b) = 0,

(3.10) (3.11)

1 (T), p > 0 on T, and h : T → F (R), h ∈ C where p : T → R, p ∈ Crd f rd (T) has α-levels

 α [h(t)]α = hα (t), h (t) , t ∈ T, α ∈ [0, 1].

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

227

Let

 [y(t)]α = y α (t), y α (t) ,

t ∈ [a, b],

α ∈ [0, 1].

Then,

 [δH y(t)]α = y αΔ (t), y αΔ (t) ,

 [p(t) · δH y(t)]α = p(t)y αΔ (t), p(t)y αΔ (t) ,   Δ Δ  [δH (p · δH y) (t)]α = py αΔ (t), py αΔ (t) ,

 [y(a)]α = y α (a), y α (a) ,

 α

 y σ 2 (b) = y α (σ 2 (b)), y α (σ 2 (b)) ,

t ∈ [a, b],

α ∈ [0, 1].

Thus, we arrive at the following BVPs:  Δ py αΔ (t) = hα (t), t ∈ (a, b), α ∈ [0, 1],   = y α σ 2 (b) = 0, α ∈ [0, 1], y α (a)

(3.12)

 αΔ Δ α py (t) = h (t), t ∈ (a, b), α ∈ [0, 1],   y α (a) = y α σ 2 (b) = 0, α ∈ [0, 1].

(3.13)

and

Note that the Green function for the BVP  Δ Δ px (t) = 0, t ∈ (a, b),   x(a) = x σ 2 (b) = 0 is given by

G(t, s) =

⎧ !t ⎪ a ⎪ ⎪ ⎨− ⎪ ⎪ ⎪ ⎩−

! σ 2 (b) 1 1 p(τ ) Δτ σ (s) p(τ ) Δτ ! σ 2 (b) 1 a p(τ ) Δτ ! σ (s) 1 ! σ 2 (b) 1 a p(τ ) Δτ t p(τ ) Δτ ! σ 2 (b) 1 Δτ a p(τ )

if t ≤ s (3.14) if t ≥ σ (s).

Then, the solutions of the BVPs (3.12), (3.13) are given by

228

3 Second Order Fuzzy Dynamic Equations



b

y (t) = min α

b

α

G(t, s)h (s)Δs, 

a

G(t, s)h (s)Δs , a

b

y α (t) = max

b

G(t, s)hα (s)Δs,

a



α

 α G(t, s)h (s)Δs ,

t ∈ [a, b], α ∈ [0, 1],

a

respectively. Example 3.3 Let T = 2N0 . Consider the BVP (3.10), (3.11) for 1 , 2t 2 

[h(t)]α = t, t 2 , p(t) =

t ∈ T,

α ∈ [0, 1],

a = 1, b = 4. Here σ (t) = 2t, hα (t) = t, α

h (t) = t 2 ,

t ∈ T,

α ∈ [0, 1].

Firstly, we will find the Green function given by (3.14). Let 1 g(t) = − , t

t ∈ T.

Then, g Δ (t) =

1 tσ (t)

1 2t 2 = p(t),

=

t ∈ T.

Therefore, t 1

1 Δτ = p(τ )

t 1

g Δ (τ )Δτ

τ =t  = g(τ ) τ =1   1 τ =t = −  τ τ =1

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

1 = − + 1, t σ 2 (4) σ (s)

1 Δτ = p(τ )

16 2s

τ =16  = g(τ ) τ =2s   1 τ =16 = −  τ τ =2s =−

σ (s) a

1 Δτ = p(τ )

t

g Δ (τ )Δτ

1

τ =2s  = g(τ ) τ =1   1 τ =2s = −  τ τ =1

1 Δτ = p(τ )

1 + 1, 2s 16

g Δ (τ )Δτ

t

τ =16  = g(τ ) τ −t   1 τ =16 = −  τ τ =t =−

σ 2 (b) a

1 1 + , 16 2s 2s

=− σ 2 (b)

g Δ (τ )Δτ

1 Δτ = p(τ )

1 1 + , 16 t 16

g Δ (τ )Δτ

1

τ =16  = g(τ ) τ =1   1 τ =16 = −  τ τ =1 1 +1 16 15 = . 16

=−

229

230

3 Second Order Fuzzy Dynamic Equations

Thus, ⎧    ⎨ − 16 − 1 + 1 − 1 + 1 15  t 16 2s   G(t, s) = ⎩ − 16 − 1 + 1 − 1 + 1 15 2s 16 t

if

t ≤s

if

t ≥ 2s,

t, s ∈ [1, 4]. Let 1 2 1 t + t, 48 2 1 1 f2 (t) = − t + t 2 , 2 3 1 3 1 2 f3 (t) = − t + t , 112 6 1 3 1 2 f4 (t) = t − t , t ∈ T. 7 6 f1 (t) = −

Then, 1 1 (σ (t) + t) + 48 2 1 1 − (2t + t) + 48 2 1 1 − t+ , 16 2 1 1 − + (σ (t) + t) 2 3 1 1 − + (2t + t) 2 3 1 − + t, 2  1  (σ (t))2 + tσ (t) + t 2 − 112 1 + (σ (t) + t) 6  1 1  2 − 4t + 2t 2 + t 2 + (2t + t) 112 6 1 1 − t 2 + t, 16 2  1 1 (σ (t))2 + tσ (t) + t 2 − (σ (t) + t) 7 6

f1Δ (t) = − = = f2Δ (t) = = = f3Δ (t) =

= = f4Δ (t) =

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

1 2 1 (4t + 2t 2 + t 2 ) − (2t + t) 7 6 1 = t 2 − t, t ∈ T. 2

=

Therefore, 4

f5 (t) =

G(t, s)hα (s)Δs

1 t

=

4

G(t, s)hα (s)Δs +

1

G(t, s)hα (s)Δs

t

  t   1 1 1 16 − + s − + 1 Δs 15 16 t 2s 1   4  1 1 1 16 − + − +1 sΔs − 15 t 16 2s t   t − 16 t 1 = s− Δs 15t 2 1   s 16(t − 1) 4 1 − Δs − 15t 2 16 t =−

=

t − 16 15t −

t 1

f2Δ (s)Δs

16(t − 1) 15t

t 1

f1Δ (s)Δs

s=t t − 16  f2 (s) s=1 15t s=4 16(t − 1)  f1 (s) − s=t 15t   1 1 t − 16 s=t − s + s2  = s=1 15t 2 3   1 16(t − 1) 1 s=4 − s2 + s  − s=t 15t 48 2   2 t t 1 1 t − 16 − + + − = 15t 2 3 2 3   16 16(t − 1) t2 t − +2+ − − 15t 48 48 2 =

231

232

3 Second Order Fuzzy Dynamic Equations

t − 16 = 15t



2t 2 − 3t + 1 6



16(t − 1) − 15t



t 2 − 24t + 80 48

(t − 16)(t − 1)(2t − 1) (t − 1)(t − 4)(t − 20) − 90t 45t t −1 = ((t − 16)(2t − 1) − 2(t − 4)(t − 20)) 90t  t −1 2 2t − t − 32t + 16 − 2t 2 + 40t + 8t − 160 = 90t (t − 1)(15t − 144) , t ∈ [1, 4], α ∈ [0, 1], = 90t =

and 4

f6 (t) =

α

G(t, s)h (s)Δs 1 t

=

4

α

G(t, s)h (s)Δs +

1

α

G(t, s)h (s)Δs t

  t  1 1 16 1 − + 1 s 2 Δs − + =− 15 16 t 2s 1   4  1 1 1 16 − + − +1 s 2 Δs − 15 t 16 2s t 16 − t t  2 s  =− Δs s − 15t 2 1   s2 16(t − 1) 4 s − Δs − 15t 2 16 t =− −

16 − t 15t

t 1

16(t − 1) 15t

f4Δ (s)Δs 4 t

f3Δ (s)Δs

s=t 16 − t  f4 (s) s=1 15t s=4 16(t − 1)  f3 (s) − s=t 15t   16 − t 1 3 1 2 s=t s − s  =− s=1 15t 7 6   1 3 1 2 s=4 16(t − 1) − s + s  − s=t 15t 112 6 =−



3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

233

 1 3 1 2 1 1 t − t − + 7 6 7 6   64 16 t3 t2 16(t − 1) − + + − − 15t 112 6 112 6   16 − t 1 3 1 2 1 t − t + =− 15t 7 6 42   4 8 t3 t2 16t − 16 − + + − − 15t 7 3 112 6   16 − t =− 6t 3 − 7t 2 + 1 630t   16t − 16 −192 + 896 + 3t 3 − 56t 2 − 15t 336   3 (16 − t) 6t − 7t 2 + 1 =− 630t

16 − t =− 15t



(t − 1)(704 + 3t 3 − 56t 2 ) 315t  1 96t 3 − 112t 2 + 16 − 6t 4 + 7t 3 − t =− 630t −



+1408t + 6t 4 − 112t 3 − 1408 − 6t 3 + 112t 2 1 =− 630t



 − 15t + 1407t − 1392 , 3

t ∈ [1, 4],

α ∈ [0, 1].

Consequently, [y(t)]α = [min{f5 (t), f6 (t)}, max{f5 (t), f6 (t)}] ,

t ∈ [1, 4],

α ∈ [0, 1].

Exercise 3.4 Let T = 3Z. Find a solution of the BVP (3.10), (3.11) when 1. p(t) = 1,

a = 0,

b = 6,

[h(t)]α = [1, 1 + t] ,

a=3,

b=30,



[h(t)]α = −1, t+t 2 ,

t ∈ [0, 6],

α ∈ [0, 1],

2. p(t)=t+1,

t∈[3, 30], α ∈ [0, 1],

3. p(t) = 4,

a = −3,

b = 9,

[h(t)]α = [0, 1] , t ∈ [−3, 9], α ∈ [0, 1],

234

3 Second Order Fuzzy Dynamic Equations

4. p(t)=

1+t , 1+t 2

a=−6, b=3,

[h(t)]α = [0, 3] , t ∈ [−6, 3],

α ∈ [0, 1].

Now, we consider Eq. (3.10) subject to the following boundary conditions: 0. y(a) = δH y(σ (b)) =

(3.15)

As in above, we arrive at the following BVPs:  Δ py αΔ (t) = hα (t), =

y α (a)

t ∈ (a, b),

y αΔ (σ (b))

α ∈ [0, 1],

(3.16)

α ∈ [0, 1],

(3.17)

= 0,

and  αΔ Δ α py (t) = h (t), t ∈ (a, b), y α (a) = y αΔ (σ (b)) = 0. Note that the Green function for the BVP  Δ Δ px (t) = 0,

t ∈ (a, b),

x(a) = x (σ (b)) = 0 Δ

is given by  G(t, s) =

−

!t

1 p(τ ) Δτ if t ≤ s !aσ (s) 1 − a p(τ ) Δτ if t ≥ σ (s),

t, s ∈ [a, b]. Then, 

b

y α (t) = min 

G(t, s)hα (s)Δs,

a

 α G(t, s)h (s)Δs ,

b

 α G(t, s)h (s)Δs ,

a b

y α (t) = max

b

G(t, s)hα (s)Δs,

a

a

t ∈ [a, b], α ∈ [0, 1]. Example 3.4 Let T = 3N0 . Consider the BVP (3.10), (3.15) for 1 , 3t 2 a = 1,

p(t) =

(3.18)

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

b = 9, 

[h(t)]α = t, t 2 ,

t ∈ T,

α ∈ [0, 1].

Here, σ (t) = 2t, hα (t) = t, α

h (t) = t 2 ,

t ∈ T,

α ∈ [0, 1].

Firstly, we will compute the Green function G, defined by (3.18). Let g(t) =

1 , t

t ∈ T.

Then, g Δ (t) = − =−

1 tσ (t) 1 , 3t 2

t ∈ T.

Hence, t a

1 Δτ = p(τ )

t 1

1 Δτ 3τ 2 t

=−

g Δ (τ )Δτ

1

τ =t  = −g(τ )

τ =1

1 τ =t =−  τ τ =1 1 = − + 1, t σ (s) a

1 Δτ = p(τ )

3s 1

=−

1 Δτ p(τ )

1 + 1, 3s

t, s ∈ T.

235

236

3 Second Order Fuzzy Dynamic Equations

Therefore, G(t, s) =

1

t −1 1 3s − 1

if t ≤ s if t ≥ 3s.

Let 1 2 t , 4 1 1 f2 (t) = t − t 2 , 3 4 1 3 f3 (t) = t , 13 1 2 1 f4 (t) = t − t 3, 12 13 f1 (t) =

t ∈ T.

We have 1 (σ (t) + t) 4 1 = (3t + t) 4 = t,

f1Δ (t) =

f2Δ (t) = = = f3Δ (t) = =

1 1 − (σ (t) + t) 3 4 1 1 − (3t + t) 3 4 1 − t, 3  1  (σ (t))2 + tσ (t) + t 2 13  1  2 9t + 3t 2 + t 2 13

= t 2,

 1  1 (σ (t) + t) − (σ (t))2 + tσ (t) + t 2 12 13   1 4t − 9t 2 + 3t 2 + t 2 = 12 13 t = − t 2 , t ∈ T. 3

f4Δ (t) =

3.2 Boundary Value Problems for Second Order Fuzzy Dynamic Equations

Hence, 9

f5 (t) =

G(t, s)hα (s)Δs

1 t

= 1

= = = = = =

9

G(t, s)hα (s)Δs +

G(t, s)hα (s)Δs

t

  9 1 1 − 1 sΔs + − 1 sΔs 3s t 1 t     t 1 9 1 − s Δs + −1 sΔs 3 t 1 t   9 t 1 f2Δ (s)Δs + f1Δ (s)Δs −1 t 1 t s=t 1 − t s=9   f2 (s) + f1 (s) s=1 s=t t   1 2 s=t 1 − t 2 s=9 1 s− s  + s  s=1 s=t 3 4 4t    1 1−t  1 1 1 t − t2 − + + 81 − t 2 3 4 3 4 4t t



4t − 3t 2 − 1 (1 − t)(81 − t 2 ) + 12 4t   4t 2 − 3t 3 − t + 3 81 − t 2 − 81t + t 3 = 12t

=

=

4t 2 − 3t 3 − t + 243 − 3t 2 − 243t + 3t 3 12t

t 2 − 244t + 243 12t (t − 1)(t − 243) , = 12t

=

9

f6 (t) =

α

G(t, s)h (s)Δs 1 t

= 1

t

= 1

9

α

G(t, s)h (s)Δs + 

 1 − 1 s 2 Δs + 3s

α

G(t, s)h (s)Δs t

t

9 1

t

 − 1 s 2 Δs

237

238

3 Second Order Fuzzy Dynamic Equations t

= 1

t

= 1



 1 1−t s − s 2 Δs + 3 t

f4Δ (s)Δs +

1−t t

9 t

9

s 2 Δs

t

f3Δ (s)Δs

s=t 1 − t s=9   f3 (s) = f4 (s) + s=1 s=t t    1 2 1 s=t 1 − t 3 s=9 s − s3  + s  = s=1 s=t 12 13 13t   2  t3 1 1 1−t  t − − + + 729 − t 3 = 12 13 12 13 13t   13t 2 − 12t 3 − 1 (1 − t) 729 − t 3 + = 156 13t   3 4 13t − 12t − t + (12 − 12t) 729 − t 3 = 156t =

13t 3 − 12t 4 − t + 8748 − 12t 3 − 8748t + 12t 4 156t

t 3 − 8749t + 8748 156t   (t − 1) t 2 + t − 8748 , = 156t =

t ∈ T.

We have f5 (1) = 0, f6 (1) = 0, 2(3 − 243) 12 · 3 240 =− 18 40 =− , 3 2 · (9 + 3 − 8748) f6 (3) = 156 · 3 8736 =− 78 · 3 2912 =− 78 f5 (3) =

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic. . .

239

112 , 3 8 · (9 − 243) f5 (9) = 12 · 9 8(−234) = 12 · 9 52 =− , 3 8(81 + 9 − 8748) f6 (9) = 156 · 9 2(−8658) = 39 · 9 2(−666) = 27 444 . =− 9 =−

Thus, [y(1)]α = [0, 0] ,   40 112 ,− , [y(3)]α = − 3 3   52 444 α ,− . [y(9)] = − 9 3 Exercise 3.5 Let T = 2Z. Find a solution of the BVP (3.10), (3.15) for p(t) = 1+t, a = 0,

b = 8,



[h(t)]α = 0, 1 + t 2 ,

t ∈ [0, 8],

α ∈ [0, 1].

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic Equations Let t0 , t1 ∈ T, t0 < t1 . With CHk ([t0 , t1 ]), k ∈ N0 , we denote the space of all l g(t) exists, and it is continuous functions g : [t0 , t1 ] → F (R) that are continuous, δH at every point t ∈ [t0 , t1 ] for any l ∈ {0, . . . , k}. In CH1 ([t0 , t1 ]), we introduce the following metric: dγ1 (x, y) = dγ (x, y) + dγ (δH x, δH y), where dγ (·, ·) is defined as in Sect. 2.2.

x, y ∈ CH1 ([t0 , t1 ]),

240

3 Second Order Fuzzy Dynamic Equations

  Theorem 3.1 The couple CH1 ([t0 , t1 ]), dγ1 is a complete metric space. Proof Let {xn }n∈N be a Cauchy sequence of elements of CHk ([t0 , t1 ]) with respect to the metric dγ1 . Then, dγ1 (xn , xm ) = dγ (xn , xm ) + dγ (δH xn , δH xm ) → 0,

as

m, n → ∞.

Hence, dγ (xn , xm ) → 0,

as

m, n → ∞,

dγ (δH xn , δH xm ) → 0,

as

m, n → ∞.

Therefore, the sequences {xn }n∈N and {δH xn }n∈N are Cauchy sequences in CH ([t0 , t1 ]) with respect to the metric dγ . Because the couple (CH ([t0 , t1 ]), dγ ) is a complete metric space, there exist x, y ∈ CH ([t0 , t1 ]) such that dγ (xn , x) → 0,

as

n → ∞,

dγ (δH xn , y) → 0,

as

n → ∞.

We will prove that there exists δH x(t), t ∈ [t0 , t1 ], that and δH x(t) = y(t), t ∈ [t0 , t1 ]. Let t

ψ(t) = x(t0 ) +

y(s)δH s,

t ∈ [t0 , t1 ].

t0

Then, dγ (x, ψ) =

sup

  !t D x(t), x(t0 ) + t0 y(s)δH s

t∈[t0 ,t1 ]

 ≤

sup

t∈[t0 ,t1 ]

 =

sup

t∈[t0 ,t1 ]

1 eγ (t, t0 ) 1 eγ (t, t0 )

eγ (t, t0 )   D(x(t), xn (t)) + D xn (t), x(t0 ) +

y(s)δH s t0



 D(x(t), xn (t)) + D xn (t0 ) +

≤ dγ (x, xn ) + sup

t∈[t0 ,t1 ]

 ≤ dγ (x, xn ) + sup

t∈[t0 ,t1 ]

 ≤ dγ (x, xn ) + sup

t∈[t0 ,t1 ]

 t∈[t0 ,t1 ]

t

1 eγ (t, t0 ) 1 eγ (t, t0 ) 1 eγ (t, t0 ) 1 eγ (t, t0 )

 D xn (t0 ) +

t

y(s)δH s t0

δH xn (s), x(t0 ) +

t0



t

y(s)δH s t0

  D(xn (t0 ), x(t0 )) + D

t



t

δH xn (s), t0

 D(xn (t0 ), x(t0 )) +



t

δH xn (s), x(t0 ) +

t0



= dγ (x, xn ) + sup



t

y(s)δH s t0



t

D (δH xn (s), y(s)) Δs t0



D(xn (t0 ), x(t0 )) + eγ (t0 , t0 )



t

D (δH xn (s), y(s)) Δs t0

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic. . . 

 dγ (xn , x) +

t

D (δH xn (s), y(s)) eγ (s, t0 ) Δs eγ (s, t0 ) t0 t∈[t0 ,t1 ]   t 1 eγ (s, t0 )Δs ≤ dγ (x, xn ) + dγ (x, xn ) + dγ (δH xn , y) sup t∈[t0 ,t1 ] eγ (t, t0 ) t0 ≤ dγ (x, xn ) + sup

1 eγ (t, t0 )

= 2dγ (x, xn ) +

dγ (δH xn , y) eγ (t, t0 ) − 1 sup γ eγ (t, t0 ) t∈[t0 ,t1 ]

≤ 2dγ (x, xn ) +

dγ (δH xn , y) γ

→ 0,

241 

as n → ∞.

Thus, x(t) = ψ(t),

t ∈ [t0 , t1 ].

Hence, using that ψ ∈ CHk ([t0 , t1 ]), we get x ∈ CHk ([t0 , t1 ]). Then, x(t) = x(t0 ) +

t

δH x(s) t0

= x(t0 ) +

t

y(s)δH s,

t ∈ [t0 , t1 ].

t0

Consequently, δH x(t) = y(t),

t ∈ [t0 , t1 ].

This completes the proof. Now, we consider the following Cauchy problem: 2 δH y = f (t, y, δH y),

t ∈ (t0 , t1 ],

(3.19)

y(t0 ) = k1 ,

(3.20)

δH y(t0 ) = k2 ,

(3.21)

where (D1) (D2)

f : T × F (R) × F (R) → F (R), f ∈ CH (T × F (R) × F (R)), k1 , k2 ∈ F (R).

Theorem 3.2 Suppose (D1) and (D2), and assume that there exist positive constants L1 and L2 such that D (f (t, x1 , x2 ), f (t, y1 , y2 )) ≤ L1 D(x1 , y1 ) + L2 D(x2 , y2 )

242

3 Second Order Fuzzy Dynamic Equations

for any t ∈ [t0 , t1 ], x1 , x2 , y1 , y2 ∈ F (R). If L(1 + γ ) < 1, γ2

L = max{L1 , L2 },

then the IVP (3.19)–(3.21) has a unique solution on [t0 , t1 ]. Proof For x ∈ CH ([t0 , t1 ]), define the operator t

T x(t) = k1 + (t − t0 ) · k2 +

z

f (s, x(s), δH x(s))δH sδH z, t0

t0

t ∈ [t0 , t1 ]. We have T x ∈ CH1 ([t0 , t1 ]) for any x ∈ CH ([t0 , t1 ]), and t

δH T x(t) = k2 +

f (s, x(s), δH x(s))δH s,

t ∈ [t0 , t1 ].

t0

Now, we will make an estimation of dγ (T x, T y),

x, y ∈ CH1 ([t0 , t1 ]).

For x, y ∈ CH1 ([t0 , t1 ]), we have D(T x(t), T y(t)) eγ (t, t0 ) t∈[t0 ,t1 ]   1 D k1 + (t − t0 ) · k2 + = sup t∈[t0 ,t1 ] eγ (t, t0 )

dγ (T x, T y) = sup

t

k1 + (t − t0 ) · k2 +  = sup

t∈[t0 ,t1 ] t

z

f (s, x(s), δH x(s))δH sδH z, t0

t0

z



f (s, y(s), δH y(s))δH sδH z t0

1 D eγ (t, t0 )

t

t0



t

z

f (s, x(s), δH x(s))δH sδH z, t0

t0

z



f (s, y(s), δH y(s))δH sδH z t0

t0



≤ sup

t∈[t0 ,t1 ]

≤ sup

t∈[t0 ,t1 ]



1 eγ (t, t0 ) 1 eγ (t, t0 )

t



z

D (f (s, x(s), δH x(s)), f (s, y(s), δH y(s))) ΔsΔz t0 t

t0

z

L1 D(x(s), y(s)) t0

t0

  +L2 D (δH x(s), δH y(s)) δH sδH z

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic. . .

 ≤ L sup

1 eγ (t, t0 )

t

243

z  D(x(s), y(s))

eγ (s, t0 ) t0   D (δH x(s), δH y(s)) eγ (s, t0 )ΔsΔz + eγ (s, t0 )   ≤ L dγ (x, y) + dγ (δH x, δH y)   t z 1 × sup eγ (s, t0 )ΔsΔz t∈[t0 ,t1 ] eγ (t, t0 ) t0 t0   t  1 L eγ (z, t0 ) − 1 Δz = dγ1 (x, y) sup γ t∈[t0 ,t1 ] eγ (t, t0 ) t0    eγ (t, t0 ) − 1 1 L − (t − t0 ) = dγ1 (x, y) sup γ γ t∈[t0 ,t1 ] eγ (t, t0 ) t∈[t0 ,t1 ]

t0

≤

eγ (t, t0 ) − 1 L 1 d (x, y) sup eγ (t, t0 ) γ2 γ t∈[t0 ,t1 ]

≤

L 1 d (x, y), γ2 γ

and D (δH T x(t), δH T y(t)) eγ (t, t0 )   t 1 D k2 + f (s, x(s), δH x(s))δH s, = sup t0 t∈[t0 ,t1 ] eγ (t, t0 )  t k2 + f (s, y(s), δH y(s))δH s

dγ (δH T x, δH T y) =

sup

t∈[t0 ,t1 ]

t0



≤

sup

t∈[t0 ,t1 ]

 ≤

sup

t∈[t0 ,t1 ]

1 D eγ (t, t0 ) 1 eγ (t, t0 )



t



t

f (s, x(s), δH x(s))δH s, t0

f (s, y(s), δH y(s))δH s t0



t

D (f (s, x(s), δH x(s)), f (s, y(s), δH y(s))) Δs t0



 t 1 (L1 D(x(s), y(s)) + L2 D (δH x(s), δH y(s))) Δs t∈[t0 ,t1 ] eγ (t, t0 ) t0    t  D(x(s), y(s)) D (δH x(s), δH y(s)) 1 + eγ (s, t0 )Δs ≤ L sup eγ (s, t0 ) eγ (s, t0 ) t∈[t0 ,t1 ] eγ (t, t0 ) t0   t   1 eγ (s, t0 )Δs ≤ L dγ (x, y) + dγ (δH x, δH y) sup e (t, t ) γ 0 t0 t∈[t0 ,t1 ] ≤

sup

=

eγ (t, t0 ) − 1 L 1 d (x, y) sup γ γ eγ (t, t0 ) t∈[t0 ,t1 ]

≤

L 1 d (x, y). γ γ

244

3 Second Order Fuzzy Dynamic Equations

Therefore, dγ1 (T x, T y) = dγ (T x, T y) + dγ (δH T x, δH T y) L 1 L dγ (x, y) + dγ1 (x, y) 2 γ γ   L L dγ1 (x, y) + = γ γ2 ≤

< dγ1 (x, y).     Thus, T : CH1 ([t0 , t1 ]), dγ1 → CH1 ([t0 , t1 ]), dγ1 is a contraction map. Hence, from the Banach fixed point theorem, it follows that the operator T has a unique fixed point x ∈ CH1 ([t0 , t1 ]). We have t

x(t) = k1 + (t − t0 ) · k2 +

z

f (s, x(s), δH x(s))δH sδH z, t0

t0

t ∈ [t0 , t1 ]. Note that x ∈ CH2 ([t0 , t1 ]) and x(t0 ) = k1 , δH x(t) = k2 +

t

f (s, x(s), δH x(s))δH s, t0

δH x(t0 ) = k2 , 2 δH x(t) = f (t, x(t), δH x(t)),

t ∈ [t0 , t1 ].

Consequently, x ∈ CH2 ([t0 , t1 ]) is a solution of the IVP (3.19)–(3.21). This completes the proof. k ([t , t ]), k ∈ N , we denote the space of all functions g : [t , t ] → F (R) With CSH 0 1 0 0 1 l g(t) exists, and it is continuous at every point t ∈ [t , t ] that are continuous, δSH 0 1 1 ([t , t ]), we introduce the following metric: for any l ∈ {0, . . . , k}. In CSH 0 1 1 (x, y) = dγ (x, y) + dγ (δSH x, δSH y), dSγ

1 x, y ∈ CSH ([t0 , t1 ]),

as in Sect. 2.2. As in above, one can prove that the couple where dγ (·, ·) is defined   1 1 CSH ([t0 , t1 ]), dSγ is a complete metric space. Now, we consider the IVP 2 δSH y = f (t, y, δSH y),

y(t0 ) = k1 ,

t ∈ (t0 , t1 ],

(3.22) (3.23)

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic. . .

δSH y(t0 ) = k2 ,

245

(3.24)

1 ([t , t ]), where f and k1 , k2 satisfy (D1) and (D2), respectively. For x ∈ CSH 0 1 define the operator

 T x(t)=k1 H (−1) · (t−t0 ) · k2 H (−1) ·

t t0

z t0

 f (s, x(s), δSH x(s))δSH sδSH z ,

1 ([t , t ]) be a fixed point of the operator T , then it t ∈ [t0 , t1 ]. Note that if x ∈ CSH 0 1 1 ([t , t ]) is a fixed point is a solution of the IVP (3.22)–(3.24). Really, let x ∈ CSH 0 1 of the operator T . Then,

 x(t) = k1 H (−1) · (t − t0 ) · k2 H (−1) ·

t t0

z t0

 f (s, x(s), δSH x(s))δSH sδSH z ,

t ∈ [t0 , t1 ], and x(t0 ) = k1 , δSH x(t) = k2 H (−1) ·

t

f (s, x(s), δSH x(s))δSH s, t0

δSH x(t0 ) = k2 , 2 δSH x(t) = f (t, x(t), δSH x(t)) ,

t ∈ [t0 , t1 ].

As we have proved Theorem 3.2, one can prove the following result. Theorem 3.3 Suppose (D1) and (D2), and assume that there exist positive constants L1 and L2 such that D (f (t, x1 , x2 ), f (t, y1 , y2 )) ≤ L1 D(x1 , y1 ) + L2 D(x2 , y2 ) for any t ∈ [t0 , t1 ], x1 , x2 , y1 , y2 ∈ F (R). If L(1 + γ ) < 1, γ2

L = max{L1 , L2 },

then the IVP (3.22)–(3.24) has a unique solution on [t0 , t1 ]. Now, we consider the IVP δSH (δH y) = f (t, y, δH y),

t ∈ (t0 , t1 ],

(3.25)

y(t0 ) = k1 ,

(3.26)

δH y(t0 ) = k2 ,

(3.27)

246

3 Second Order Fuzzy Dynamic Equations

1 ([t , t ]) where f and k1 , k2 satisfy (D1) and (D2), respectively. For x ∈ CSH 0 1  1 CH ([t0 , t1 ]), define the operator

 T x(t) = k1 + (t − t0 ) · k2 H (−1) ·

t

z

t0

 f (s, x(s), δH x(s))δSH sδH z ,

t0

1 ([t , t ]) ∩ C 1 ([t , t ]) be a fixed point of t ∈ [t0 , t1 ]. Note that if x ∈ CSH 0 1 H 0 1 the operator  T , then it is a solution of the IVP (3.25)–(3.27). Really, let x ∈ 1 ([t , t ]) CSH CH1 ([t0 , t1 ]) is a fixed point of the operator T . Then, 0 1

 x(t) = k1 + (t − t0 ) · k2 H (−1) ·

t

z

t0

 f (s, x(s), δH x(s))δSH sδH z ,

t0

t ∈ [t0 , t1 ], and x(t0 ) = k1 , δH x(t) = k2 H (−1) ·

t

f (s, x(s), δH x(s))δSH s, t0

δH x(t0 ) = k2 , δSH (δH x) (t) = f (t, x(t), δH x(t)) ,

t ∈ [t0 , t1 ].

As we have proved Theorem 3.2, one can prove the following result. Theorem 3.4 Suppose (D1) and (D2), and assume that there exist positive constants L1 and L2 such that D (f (t, x1 , x2 ), f (t, y1 , y2 )) ≤ L1 D(x1 , y1 ) + L2 D(x2 , y2 ) for any t ∈ [t0 , t1 ], x1 , x2 , y1 , y2 ∈ F (R). If L(1 + γ ) < 1, γ2

L = max{L1 , L2 },

then the IVP (3.25)–(3.27) has a unique solution on [t0 , t1 ]. Now, we consider the IVP δH (δSH y) = f (t, y, δSH y),

t ∈ (t0 , t1 ],

(3.28)

y(t0 ) = k1 ,

(3.29)

δSH y(t0 ) = k2 ,

(3.30)

3.3 Existence and Uniqueness of Solutions of Second Order Fuzzy Dynamic. . .

247

1 ([t , t ]) where f and k1 , k2 satisfy (D1) and (D2), respectively. For x ∈ CSH 0 1  1 CH ([t0 , t1 ]), define the operator

 T x(t) = k1 H (−1) · (t − t0 ) · k2 +

t t0

z

 f (s, x(s), δSH x(s))δH sδSH z ,

t0

1 ([t , t ]) ∩ C 1 ([t , t ]) is a fixed point of t ∈ [t0 , t1 ]. Note that if x ∈ CSH 0 1 H 0 1 the operator  T , then it is a solution of the IVP (3.28)–(3.30). Really, let x ∈ 1 ([t , t ]) CSH CH1 ([t0 , t1 ]) be a fixed point of the operator T . Then, 0 1

 x(t) = k1 H (−1) · (t − t0 ) · k2 +

t t0

z

 f (s, x(s), δSH x(s))δH sδSH z ,

t0

t ∈ [t0 , t1 ], and x(t0 ) = k1 , δSH x(t) = k2 +

t

f (s, x(s), δSH x(s))δH s, t0

δSH x(t0 ) = k2 , δH (δSH x) (t) = f (t, x(t), δSH x(t)) ,

t ∈ [t0 , t1 ].

As we have proved Theorem 3.2, one can prove the following result. Theorem 3.5 Suppose (D1) and (D2), and assume that there exist positive constants L1 and L2 such that D (f (t, x1 , x2 ), f (t, y1 , y2 )) ≤ L1 D(x1 , y1 ) + L2 D(x2 , y2 ) for any t ∈ [t0 , t1 ], x1 , x2 , y1 , y2 ∈ F (R). If L(1 + γ ) < 1, γ2

L = max{L1 , L2 },

then the IVP (3.28)–(3.30) has a unique solution on [t0 , t1 ]. Example 3.5 Consider the following IVP:     1 2 · y(t) , δH y(t) = (3 + t 2 ) · g(t) H 3

t ∈ (t0 , t1 ],

y(t0 ) = y0 , δH y(t0 ) = y1 , where g : [t0 , t1 ] → F (R), g ∈ Cf rd ([t0 , t1 ]) is a given fuzzy function, y0 , y1 ∈ F (R). Here,

248

3 Second Order Fuzzy Dynamic Equations

    1 · y(t) , f (t, y(t), δH y(t)) = (3 + t 2 ) · g(t) H 3

t ∈ [t0 , t1 ],

y ∈ Cf rd ([t0 , t1 ]).

For y, z ∈ Cf rd ([t0 , t1 ]), we have D(f (t, y(t), δH y(t)), f (t, z(t), δH z(t)))         1 1 2 2 = D (3 + t ) · g(t) H · y(t) , (3 + t ) · g(t) H · z(t) 3 3   1 1 · y(t), · z(t) =D 3 3 =

1 D(y(t), z(t)), 3

Therefore, if

1+γ 3γ 2

t ∈ [t0 , t1 ].

< 1, the considered IVP has a unique solution.

Exercise 3.6 Let T = 7Z. Find conditions for t0 and t1 so that the following IVP 1 t · y, t ∈ (t0 , t1 ], 9 y(t0 ) = {1, 2, 8}, 2 δH y=

δH y(t0 ) = {0, 1, 3} has a unique solution.

3.4 Continuous Dependence of the Solutions of Second Order Fuzzy Dynamic Equations on the Initial Data Suppose that t0 , t1 ∈ T, t0 < t1 . Consider the following IVPs: 2 y = f (t, y), δH

t ∈ (t0 , t1 ],

(3.31)

y(t0 ) = y1 ,

(3.32)

δH y(t0 ) = y2 ,

(3.33)

and 2 x = f (t, x), δH

t ∈ (t0 , t1 ],

(3.34)

x(t0 ) = k1 ,

(3.35)

δH x(t0 ) = k2 ,

(3.36)

3.4 Continuous Dependence of the Solutions of Second Order Fuzzy Dynamic. . .

249

where (D3) (D4)

f : T × F (R) → F (R), f ∈ Cf rd (T × F (R)), y1 , y2 , k1 , k2 ∈ F (R).

Theorem 3.6 Suppose (D3) and (D4), and assume that there exists a positive constant L such that D(f (t, x1 ), f (t, x2 )) ≤ LD(x1 , x2 )

(3.37)

for any t ∈ [t0 , t1 ] and for any x1 , x2 ∈ F (R). If y and x are solutions of the IVPs (3.31)–(3.33) and (3.34)–(3.36), respectively, then D(x(t), y(t)) ≤ (D(k1 , y1 ) + (t1 − t0 )D(k2 , y2 )) eL(t1 −t0 ) (t, t0 ),

(3.38)

t ∈ [t0 , t1 ]. Proof By Theorem 3.2, it follows that y and x are the unique solutions of the IVPs (3.31)–(3.33) and (3.34)–(3.36), respectively. Also, they satisfy the following integral equations: t

y(t) = y1 + (t − t0 ) · y2 +

z

f (s, y(s))δH sδH z,

x(t) = k1 + (t − t0 ) · k2 +

t0

t0

t

z

f (s, x(s))δH sδH z, t0

t0

t ∈ [t0 , t1 ], respectively. Then,  D(x(t), y(t)) = D k1 + (t − t0 ) · k2 +

t

f (s, x(s))δH sδH z, t0

y1 + (t − t0 ) · y2 +

t

z t0



z

f (s, y(s))δH sδH z t0

t0

≤ D(k1 , y1 ) + D ((t − t0 ) · k2 , (t − t0 ) · y2 )  t z  t z +D f (s, x(s))δH sδH z, f (s, y(s))δH sδH z t0

t0

t0

t0

≤ D(k1 , y1 ) + (t1 − t0 )D(k2 , y2 ) +

t

z

D(f (s, x(s)), f (s, z(s)))ΔsΔz t0

t0

≤ D(k1 , y1 ) + (t1 − t0 )D(k2 , y2 ) +L

t

z

D(x(s), y(s))ΔsΔz t0

t0

= D(k1 , y1 ) + (t1 − t0 )D(k2 , y2 )

250

3 Second Order Fuzzy Dynamic Equations

+L

t

(t − σ (s))D(x(s), y(s))Δs

t0

≤ D(k1 , y1 ) + (t1 − t0 )D(k2 , y2 ) +L(t1 − t0 )

t

D(x(s), y(s))Δs, t0

t ∈ [t0 , t1 ]. From the last inequality and from the Gronwall inequality, we get (3.38). This completes the proof. Now, we consider the following IVPs: 2 δSH y = f (t, y),

t ∈ (t0 , t1 ],

(3.39)

y(t0 ) = y1 ,

(3.40)

δSH y(t0 ) = y2 ,

(3.41)

and 2 x = f (t, x), δSH

t ∈ (t0 , t1 ],

(3.42)

x(t0 ) = k1 ,

(3.43)

δSH x(t0 ) = k2 ,

(3.44)

where f and k1 , k2 , y1 , y2 satisfy (D3) and (D4), respectively. As we have proved Theorem 3.6, one can prove the following result. Theorem 3.7 Suppose (D3), (D4), and (3.37). If y and x are solutions of the IVPs (3.39)–(3.41) and (3.42)–(3.44), respectively, then they satisfy (3.38). Let us consider the following IVPs: δSH (δH y) = f (t, y),

t ∈ (t0 , t1 ],

(3.45)

y(t0 ) = y1 ,

(3.46)

δH y(t0 ) = y2 ,

(3.47)

and δSH (δH x) = f (t, x),

t ∈ (t0 , t1 ],

(3.48)

x(t0 ) = k1 ,

(3.49)

δH x(t0 ) = k2 ,

(3.50)

3.4 Continuous Dependence of the Solutions of Second Order Fuzzy Dynamic. . .

251

where f and k1 , k2 , y1 , y2 satisfy (D3) and (D4), respectively. As we have proved Theorem 3.6, one can prove the following result. Theorem 3.8 Suppose (D3), (D4), and (3.37). If y and x are solutions of the IVPs (3.45)–(3.47) and (3.48)–(3.50), respectively, then they satisfy (3.38). Now, consider the following IVPs: δH (δSH y) = f (t, y),

t ∈ (t0 , t1 ],

(3.51)

y(t0 ) = y1 ,

(3.52)

δSH y(t0 ) = y2 ,

(3.53)

and δH (δSH x) = f (t, x),

t ∈ (t0 , t1 ],

(3.54)

x(t0 ) = k1 ,

(3.55)

δSH x(t0 ) = k2 ,

(3.56)

where f and k1 , k2 , y1 , y2 satisfy (D3) and (D4), respectively. As we have proved Theorem 3.6, one can prove the following result. Theorem 3.9 Suppose (D3), (D4), and (3.37). If y and x are solutions of the IVPs (3.51)–(3.53) and (3.54)–(3.56), respectively, then they satisfy (3.38). Exercise 3.7 Let T = Z, and x and y are the solutions of the following IVPs: 2 δH y = 2 · y,

t ∈ (0, 4],

y(0) = {0, 1, 2}, δH y(0) = {−1, 1, 2}, and 2 y = 2 · y, δH

t ∈ (0, 4],

y(0) = {0, 2, 3}, δH y(0) = {−3, 4, 5}, respectively. Prove that D(x(t), y(t)) ≤ 1000e8 (t, 0),

t ∈ [0, 4].

252

3 Second Order Fuzzy Dynamic Equations

3.5 Advanced Practical Problems Problem 3.1 Let T = Z. Find a solution of the following IVP: 2 y = {1, 1 + t, 1 + t + t 2 }, δH

t > 0,

y(0) = {0, 1, 3}, δH y(0) = {−1, 0, 1}. Problem 3.2 Let T = 2N0 . Find a solution of the following IVP: 2 y = {t, 3 + t, 4 + 3t + 5t 2 }, δSH

t > 1,

y(1) = {−3, −2, −1}, δSH y(1) = {−1, 1, 3}. Problem 3.3 Let T = 3N0 . Find a solution of the following IVP: δSH (δH y) = {1, 3, t 2 },

t > 1,

y(1) = {1, 2, 7}, δH y(1) = {0, 1, 3}. Problem 3.4 Let T = 2N0 . Find a solution of the following IVP: δH (δSH y) = {−1, 0, 1 + 4t 2 },

t > 1,

y(1) = {−1, 0, 3}, δSH y(1) = {1, 2, 4}. Problem 3.5 Let T =

 N0 1 9

. Find a solution of the IVP

2 δH y = 2 · δH y + 8 · y,

t > 0,

δH y(0) = y1 , y(0) = y0 , where y0 , y1 ∈ F (R) and [y0 ]α = [1, 3] , [y1 ]α = [4, 10] ,

α ∈ [0, 1].

3.5 Advanced Practical Problems

Problem 3.6 Let T =

 N0 1 9

253

. Find a solution of the IVP

2 δH y = 9 · δH y + 10 · y,

t > 0,

δH y(0) = y1 , y(0) = y0 , where y0 , y1 ∈ F (R) and [y0 ]α = [−1, 5] , [y1 ]α = [0, 3] ,

α ∈ [0, 1].

Problem 3.7 Let T = 18 Z. Find a solution of the IVP 2 y = (−4) · δH y + 5 · y + c(t), δH

t > 0,

δH y(0) = y1 , y(0) = y0 , where [c(t)]α = [0, 1 + 2t], [y0 ]α = [−2, 1] , [y1 ]α = [0, 1] , Problem 3.8 Let T =

1 11 Z.

t ∈ [0, ∞),

α ∈ [0, 1].

Find a solution of the IVP

2 δSH y = (−10) · δH y + 11 · y + c(t),

t > 0,

δH y(0) = y1 , y(0) = y0 , where [c(t)]α = [−3, 1 + 4t], [y0 ]α = [−2, 1] , [y1 ]α = [−1, 1] ,

t ∈ [0, ∞),

α ∈ [0, 1].

254

3 Second Order Fuzzy Dynamic Equations

Problem 3.9 Let T = 2N0 . Find a solution of the BVP (3.10), (3.11) when 1. 

p(t)=t+t 2 , a=1, b=16, [h(t)]α = 0, t+2t 2 , t ∈ [1, 16], α ∈ [0, 1], 2. p(t) = 1,

a = 1,

b = 8,

[h(t)]α = [1, 3t] ,

t ∈ [1, 8],

b = 32,

[h(t)]α = [t, 3t] ,

t ∈ [1, 32],

α ∈ [0, 1],

3. p(t) = 2,

a = 1,

α ∈ [0, 1],

4. p(t)=1+t 2 ,

a=1,

b=64,

[h(t)]α = [0, t] ,

t ∈ [1, 64],

α ∈ [0, 1].

Problem 3.10 Let T = 2N0 . Find a solution of the BVP (3.10), (3.15) for p(t) = t 4 ,

a = 1,

b = 16,



[h(t)]α = t, t 2 + t 3 .

Problem 3.11 Let T = 2N0 . Find conditions for t0 and t1 so that the following IVP 1 t · y + {1, 1 + t, 1 + t + t 2 }, 9 y(t0 ) = {0, 1, 2},

2 δSH y=

δSH y(t0 ) = {−1, 0, 1} has a unique solution. Problem 3.12 Let T = 2Z, and x and y are the solutions of the following IVPs: 2 δSH y = 5 · y,

t ∈ (0, 4],

y(0) = {−1, 0, 1}, δSH y(0) = {−2, −1, 0}, and 2 y = 2 · y, δSH

t ∈ (0, 8],

y(0) = {−5, 1, 4}, δSH y(0) = {7, 8, 9},

3.6 Notes and References

255

respectively. Prove that D(x(t), y(t)) ≤ 10000e40 (t, 0),

t ∈ [0, 8].

3.6 Notes and References In this chapter are investigated some classes of linear second order fuzzy dynamic equations, and they are deducted formulas for their solutions. They are formulated and proved some existence and uniqueness results. These results can be found in [17]. It is investigated the continuous dependence of the solutions of second order fuzzy dynamic equations on the initial data. Some classes of BVPs for second order fuzzy dynamic equations are investigated, and formulas for their solutions are given.

Chapter 4

Functional Fuzzy Dynamic Equations

This chapter is devoted to a qualitative analysis of functional fuzzy dynamic equations. First, we deducted chain rules for fuzzy functions. These chain rules can be considered as analogues of the well known chain rules in the dynamic calculus on time scales. Then we investigated the periodic properties of time scales. In this chapter we deducted the periodic solutions of linear first order functional fuzzy dynamic equations. We considered the lower solutions and upper solutions for first order functional fuzzy dynamic equations. Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

4.1 Periodic Properties of Time Scales Suppose that v : T → R is a strictly increasing function such that T = v(T) is a time scale with forward jump operator and delta differentiation operator σ respectively. With and Δ, δH and δSH we will denote the first type fuzzy delta differentiation operator and the second type fuzzy delta differentiation operator on T, respectively. Assume that v is Δ-differentiable on T and v(σ (t)) = σ (v(t)),

t ∈ T.

  δH f (v(t)) Theorem 4.1 (The Chain Rule) Let f : T → F (R). If v Δ (t) and exist for t ∈ Tκ , then     δH f (v(t)). (δH f (v)) (t) = v Δ (t) ·

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_4

257

258

4 Functional Fuzzy Dynamic Equations

Proof Let  ∈ (0, 1) and t ∈ T be arbitrarily chosen. Set ∗ =

   . 1 + |v Δ (t)| +  δH f (v(t))

There exist neighborhoods U and V of t and v(t), respectively, such that |v(σ (t)) − v(s) − (σ (t) − s)v Δ (t)| ≤  ∗ |σ (t) − s|,

s ∈ U,

and     D f (v(t + h)) − f ( σ (v(t))), (v(t + h) − σ (v(t))) · δH f (v(t)) ≤  ∗ (v(t + h) − σ (v(t))), for v(t + h) ∈ V , and     D f ( σ (v(t))) − f (v(t − h)), ( σ (v(t)) − v(t − h)) · δH f (v(t)) ≤  ∗ ( σ (v(t)) − v(t − h)) for v(t − h) ∈ V . Take h > 0 such that v(t − h), v(t + h) ∈ V and t − h, t + h ∈ U . Then     D f (v(t + h)) − f (v(σ (t))), δH f (v(t))v Δ (t)(t + h − σ (t)) |    = D f (v(t + h)) − f ( σ (v(t))) + (v(t + h) − σ (v(t))) · δH f (v(t)),     δH f (v(t)) (v(t + h) − σ (v(t))) · δH f (v(t)) + (t + h − σ (t))v Δ (t) ·



    ≤ D f (v(t + h)) − f ( σ (v(t))), (v(t) − σ (v(t))) · δH f (v(t))       +D (v(t + h) − σ (v(t))) · δH f (v(t)), (t + h − σ (t))v Δ (t) · δH f (v(t))   ≤  ∗ (v(t + h) − σ (v(t))) +  ∗ |σ (t) − t − h| δH f (v(t)) ≤  ∗ |v(t + h) − σ (v(t)) − (t + h − σ (t))v Δ (t)|   + ∗ (t + h − σ (t))v Δ (t)   + ∗ |μ(t) − h| δH f (v(t)) ≤  ∗ (h − μ(t)) +  ∗ (h − μ(t))|v Δ (t)|   + ∗ (h − μ(t)) δH f (v(t))     =  ∗ 1 + |v Δ (t)| +  δH f (v(t)) (h − μ(t)) < (h − μ(t))

4.1 Periodic Properties of Time Scales

259

and     D f (v(σ (t))) − f (v(t − h)), δH f (v(t))v Δ (t)(h + μ(t))    = D f (v(σ (t))) − f (v(t − h)) + ( σ (v(t)) − v(t − h)) · δH f (v(t)),     ( σ (v(t)) − v(t − h)) · δH f (v(t)) + δH f (v(t))v Δ (t)(h + μ(t))



    ≤ D f ( σ (v(t))) − f (v(t + h)), ( σ (v(t)) − v(t − h)) · δH f (v(t))       +D ( σ (v(t)) − v(t − h)) · δH f (v(t)), δH f (v(t))v Δ (t)(h + μ(t))   ≤  ∗ ( δH f (v(t)) σ (v(t)) − v(t − h)) +  ∗ (h + μ(t)) ≤  ∗ |v(σ (t)) − v(t − h) + (μ(t) + h)v Δ (t)|   + ∗ (μ(t) + h)|v Δ (t)| +  ∗ (h + μ(t)) δH f (v(t))

  ≤  ∗ (μ(t) + h) +  ∗ (μ(t) + h)|v Δ (t)| +  ∗ (h + μ(t)) δH f (v(t))     =  ∗ 1 + |v Δ (t)| +  δH f (v(t)) (h + μ(t))

< (h + μ(t)). Thus,     δH f (v(t)). (δH f (v)) (t) = v Δ (t) · This completes the proof. Example 4.1 Let T = 2N0 , v(t) = t 2 , 

f (t) = 1, t, t 2 ,

t ∈ T.

We have T = v(T) = 4N0 , σ (t) = 2t, σ (t1 ) = 4t1 ,

t ∈ T, t1 ∈ T,

v(σ (t)) = (σ (t))2 = (2t)2 = 4t 2 ,

260

4 Functional Fuzzy Dynamic Equations

σ (v(t)) = 4v(t) = 4t 2 ,

t ∈ T.

Thus, v(σ (t)) = σ (v(t)),

t ∈ T.

Next, v Δ (t) = σ (t) + t = 2t + t = 3t,

 f (v(t)) = 1, v(t), (v(t))2 

= 1, t 2 , t 4 , 

(δH f (v)) (t) = 0, t + σ (t), (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 = {0, t + 2t, 8t 3 + 2t 3 + 4t 3 + t 3 } = {0, 3t, 15t 3 }, δH f (t1 ) = {0, 1, σ (t1 ) + t1 } = {0, 1, 5t1 },   δH f (v(t)) = {0, 1, 5t 2 },   Δ   δH f (v(t)) = (3t) · {0, 1, 5t 2 } v (t) · = {0, 3t, 15t 3 },

t ∈ T.

Consequently     δH f (v(t)), (δH f (v)) (t) = v Δ (t) ·

t ∈ T.

Exercise 4.1 Let T = 3N0 , v(t) = t 3 , f (t) = {t, 1 + t, 1 + t + t 2 },

t ∈ T.

Prove that     δH f (v(t)), (δH f (v)) (t) = v Δ (t) · As in the above, one can prove the following result.

t ∈ T.

4.1 Periodic Properties of Time Scales

261

  Theorem 4.2 (The Chain Rule) Let f : T → F (R). If v Δ (t) and δSH f (v(t)) exist for t ∈ Tκ , then     δSH f (v(t)). (δSH f (v)) (t) = v Δ (t) · Theorem 4.3 Assume that a, b ∈ T, a < b, f : T → F (R), f ∈ Cf rd (T), 1 (T). Then v ∈ Crd b

v(b)

v Δ (t) · f (t)δH t =

a

(f (v −1 ))(s) δH s.

v(a)

Proof Let F be an anti-derivative of v Δ · f . Then !b a

!b v Δ (t) · f (t)δH t = a δH F (t) = F (b) − F (a)     = F (v −1 ) (v(b)) − F (v −1 ) (v(a)) ! v(b) δH (F (v −1 ))(t). = v(a)

By Theorem 4.1, we get v(b)

δH (F (v −1 ))(t) =

v(a)

v(b)



v −1Δ (t) · (δH F ) (v −1 (t))

v(a)

=

v(b)

  v −1Δ (t) · v Δ · f (v −1 (t)) δH t

v(a)

=

v(b) 

 v Δ (v −1 (t))v −1Δ (t) · f (v −1 (t)) δH t

v(a)

=

v(b)

f (v −1 (t)) δH t.

v(a)

Hence by (4.1), we get b

v Δ (t) · f (t)δH t =

a

v(b)

(f (v −1 ))(s) δH s.

v(a)

This completes the proof. Example 4.2 Let T = N0 , v(t) = 2t, f (t) = {0, t, 1 + t},

t ∈ T,

(4.1)

262

4 Functional Fuzzy Dynamic Equations

a = 0, b = 2. Then σ (t) = t + 1,

t ∈ T,

T = v(T) = 2N0 , σ (t1 ) = t1 + 2,

t1 ∈ T,

v(σ (t)) = 2σ (t) = 2(t + 1) = 2t + 2 = σ (v(t)),

t ∈ T.

Let f1 (t) = t 2 − t, f2 (t) = t 2 + t,

t ∈ T.

Then f1Δ (t) = σ (t) + t − 1 = t +1+t −1 = 2t, f2Δ (t) = σ (t) + t + 1 = t +1+t +1 = 2t + 2,

t ∈ T.

Next, v Δ (t) = 2, v Δ (t) · f (t) = 2 · {0, t, 1 + t} 2 0

= {0, 2t, 2 + 2t},  2 Δ v (t) · f (t)δH t = 0, 2tΔt, 0

2 0

 (2 + 2t)Δt

4.1 Periodic Properties of Time Scales

263

 = 0,

2 0

f1Δ (t)Δt,

2 0

 f2Δ (t)Δt

 t=2 t=2    = 0, f1 (t) , f2 (t) t=0

t=0

 t=2 t=2    = 0, (t 2 − t) , (t 2 + t) t=0

= {0, 4 − 2, 4 + 2} = {0, 2, 6}. Also, v −1 (t) =

1 t, 2

f (v −1 )(t) = {0, v −1 (t), 1 + v −1 (t)}   1 1 T. = 0, t, 1 + t , t ∈ 2 2 Let 1 2 t − 4 1 g2 (t) = t 2 + 4

g1 (t) =

1 t, 2 1 t, 2

t ∈ T.

Then

g1Δ (t) = = =

g2Δ (t) = = =

1 1 σ (t) + t) − ( 4 2 1 1 (t + 2 + t) − 4 2 1 t, 2 1 1 σ (t) + t) + ( 4 2 1 1 (t + 2 + t) + 4 2 1 t + 1, t ∈ T, 2

t=0

264

4 Functional Fuzzy Dynamic Equations

and v(2)

4

(f (v −1 ))(s) δH s =

(f (v −1 ))(s) δH s

0

v(0)

 = 0,  = 0,

4

1 s Δs, 2

0 4 0

4 0



g1Δ (s)Δs,

0

  1 1 + s Δs 2  4 g2Δ (s)Δs

 s=4 s=4    = 0, g1 (s) , g2 (s) s=0

s=0

      1 2 1 s=4 1 2 1 s=4 = 0, s − s  , s + s  s=0 s=0 4 2 4 2 = {0, 4 − 2, 4 + 2} = {0, 2, 6}. Consequently b

v Δ (t) · f (t)δH t =

a

v(b)

(f (v −1 ))(s) δH s.

v(a)

Exercise 4.2 Let T = 2N0 , v(t) = t 3 , f (t) = {1 + t, 1 + 2t + t 2 , 1 + 3t + t 4 },

t ∈ T,

a = 1, b = 8. Prove that b

v Δ (t) · f (t)δH t =

a

v(b)

(f (v −1 ))(s) δH s.

v(a)

As in the above, one can prove the following result. Theorem 4.4 Assume that a, b ∈ T, a < b, f : T → F (R), f ∈ Cf rd (T), 1 (T). Then v ∈ Crd b a

v Δ (t) · f (t)δSH t =

v(b) v(a)

(f (v −1 ))(s) δSH s.

4.1 Periodic Properties of Time Scales

265

Exercise 4.3 Let T = 4N0 , v(t) = 5t + 3, f (t) = {−3, −1 + t, 1 + t + t 2 },

t ∈ T,

a = 0, b = 16. Prove that b

v Δ (t) · f (t)δSH t =

a

v(b)

(f (v −1 ))(s) δSH s.

v(a)

Definition 4.1 Let w > 0. A time scale T is called w-periodic if t + w ∈ T for any t ∈ T. Definition 4.2 A (fuzzy) function f , defined on T, is called w-periodic, if f (t + w) = f (t), t ∈ T. Theorem 4.5 Suppose that T is w-periodic. Let P : T → F (R), P ∈ Cf rd (T) be w-periodic, p : T → R, p ∈ R, and a, b ∈ T. Then: σ (t + w) = σ (t), t ∈ T. ρ(t + w) = ρ(t), t ∈ T. μ(t + w) = μ(t), t ∈ T. ep (t + w, s + w) = ep (t, s), s, t ∈ T. ep (t + w, t) is independent of t ∈ T. ! b+w !b P (t)δH t = a P (t)δH t. a+w ! b+w !b 7. a+w P (t)δSH t = a P (t)δSH t.

1. 2. 3. 4. 5. 6.

Proof Let v(t) = t + w, t ∈ T. Then v(T) = T + w = T.

1. We have σ (t + w) = = ≥ =

inf{s ∈ T : s > t + w} t1 + w inf{s ∈ T : s > t} + w σ (t) + w, t ∈ T.

(4.2)

266

4 Functional Fuzzy Dynamic Equations

Also, σ (t) + w = inf{s : s > t} + w ≥ inf{s : s > t + w} = σ (t + w),

t ∈ T.

By the last inequality and (4.2), we get σ (t + w) = σ (t) + w,

t ∈ T.

2. We have ρ(t + w) = sup{s ∈ T : s < t + w} = t2 + w ≤ sup{s ∈ T : s < t} + w = ρ(t) + w, t ∈ T. Also, ρ(t) + w = sup{s ∈ T : s < t} + w ≤ sup{s ∈ T : s < t + w} = ρ(t + w),

t ∈ T.

By the last inequality and from (4.3), we get ρ(t + w) = ρ(t) + w,

t ∈ T.

3. By (1), we have μ(t + w) = σ (t + w) − (t + w) = σ (t) + w − t − w = σ (t) − t = μ(t),

t ∈ T.

4. Using the chain rule, we get t+w s+w

1 log(1 + p(τ )μ(τ ))Δτ μ(τ ) t+w

=

s+w t

= s

1 log (1 + p(τ + w)μ(τ + w)) Δτ μ(τ + w)

1 log(1 + p(τ )μ(τ ))Δτ, μ(τ )

s, t ∈ T,

(4.3)

4.1 Periodic Properties of Time Scales

267

and hence, ep (t + w, s + w) = ep (t, s),

s, t ∈ T.

5. By (4), we obtain t+w t

0

1 log(1 + p(τ )μ(τ ))Δτ = μ(τ )

1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

t

w

+ 0

1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

t+w

+

w 0

=

1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

t

w

+ 0

t

+ 0 w

=

1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

0

1 log(1 + p(τ )μ(τ ))Δτ μ(τ ) 1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

1 log(1 + p(τ )μ(τ ))Δτ, μ(τ )

t ∈ T.

Consequently t+w t

1 log(1 + p(τ )μ(τ ))Δτ μ(τ )

is independent of t, t ∈ T. Hence, ep (t + w, t), t ∈ T, is independent of t, t ∈ T. 6. We apply the chain rule and we get b

b

P (t)δH t =

a

v Δ (t) · P (t)δH t

a

=

v(b)

  P v −1 (t) δH t

v(a)

=

b+w

P (t − w) δH t

a+w

=

b+w

P (t) δH t

a+w

=

b+w

P (t)δH t. a+w

268

4 Functional Fuzzy Dynamic Equations

7. The idea of the proof repeats the idea of the proof of (6). Therefore we leave it to the reader as an exercise. This completes the proof. Theorem 4.6 Let T be w-periodic, f : Tκ × Tκ → F (R) is continuous at (t, t), t f (t, ·) exists and it is rd-continuous on [a, σ (t)], and for each  > 0 t ∈ Tκ , δH there exists a neighborhood U of t, independent of τ ∈ [a, σ (t)], such that   t f (t, τ )(h − μ(t)) ≤ (h − μ(t)), D f (t + h, τ ) − f (σ (t), τ ), δH   t D f (σ (t), τ ) − f (t − h, τ ), δH f (t, τ )(h + μ(t)) ≤ (h + μ(t)), for all t − h, t + h ∈ U . Then t+w

G(t) =

f (t, τ )δH τ t

implies t+w

δH G(t) = t

t δH f (t, τ )δH τ + f (σ (t), t + w) H f (σ (t), t),

t ∈ T. Proof Let t ∈ T and take a ∈ (t, t + w). Then a

G(t) =

t+w

f (t, τ )δH τ +

t

f (t, τ )δH τ. a

Therefore it is enough to prove that the following assertions hold, for a, b, c ∈ Tκ with a < t, b > t: t

M(t) =

f (t, τ )δH τ

(4.4)

a

implies t

δH M(t) = a

t δH f (t, τ )δH τ + f (σ (t), t),

(4.5)

and b

M(t) =

f (t, τ )δH τ

(4.6)

t

implies b

δH M(t) = t

t δH f (t, τ )δH τ H f (σ (t), t).

(4.7)

4.1 Periodic Properties of Time Scales

269

We will prove the assertion (4.4) and (4.5). Take  > 0 arbitrarily. Then there exists a neighborhood U of t such that   t D f (t + h, τ ) − f (σ (t), τ ), δH f (t, τ )(h − μ(t)) ≤

 (h − μ(t)) 2(σ (t) − a)

for t + h ∈ U , and   t D f (σ (t), τ ) − f (t − h, τ ), δH f (t, τ )(h + μ(t)) ≤

 (h + μ(t)) 2(σ (t) − a)

for t − h ∈ U . Since F is continuous at (t, t), there is a neighborhood V of t such that D (F (s, τ ), F (t, t)) ≤ Let t + h, t − h ∈ U



 , 2

V . Then

  D M(t + h) H M(σ (t)), (h − μ(t)) ·

t a



t+h

=D



t a

σ (t)

=D

f (σ (t), τ )δH τ,

 t δH f (t, τ )δH τ + f (σ (t), t)

(f (t + h, τ ) H f (σ (t), τ )) δH τ +

a σ (t) a

·

σ (t)



t+h

f (t + h, τ )δH τ,

σ (t)

(h − μ(t)) · t

t δH f (t, τ )δH τ + (h − μ(t))

t δH f (t, τ )δH τ

 + (h − μ(t)) · f (σ (t), t)

σ (t)

σ (t)

(f (t + h, τ ) H f (σ (t), τ )) δH τ, (h−μ(t))·

≤D a

 +D

+ f (σ (t), t)

a

(h − μ(t)) · 

 t δH f (t, τ )δH τ

σ (t)

f (t + h, τ )δH τ H

a

s, τ ∈ V .

a t+h σ (t)

t δH f (t, τ )δH τ

t f (t + h, τ )δH τ, (−(h − μ(t))μ(t)) · δH f (t, t)

 +(h − μ(t)) · f (σ (t), t) σ (t)

≤ a

  t D f (t + h, τ ) H f (σ (t), τ ), (h − μ(t)) · δH f (t, τ ) Δτ



270

4 Functional Fuzzy Dynamic Equations



t+h

+D

 f (t + h, τ )δH τ, (h − μ(t)) · f (t, t)

σ (t) σ (t)

≤ a



 (h − μ(t))Δτ 2(σ (t) − a) t+h

+D



t+h

f (t + h, τ )δH τ,

f (t, t)δH τ

σ (t)

σ (t)

t+h  (h − μ(t)) + D (f (t + h, τ ), f (t, t)) Δτ 2 σ (t)   ≤ (h − μ(t)) + (h − μ(t)) 2 2 = (h − μ(t))

≤

and   t  t f (t, τ )δ τ + f (σ (t), t) D M(σ (t)) H M(t − h), (h + μ(t)) · δH H a



σ (t)

=D

a

f (σ (t), τ )δH τ H

(h + μ(t)) · 

σ (t)

=D

a σ (t)

+  ·

t−h σ (t) a

 ≤D

σ (t) a

≤

a

a

f (t − h, τ )δH τ, 

t f (t, τ )δ τ + f (σ (t), t) δH H

(f (σ (t), τ ) H f (t − h, τ )) δH τ

t f (t, τ )δ τ  δH H H

σ (t) t

 t f (t, τ )δ τ δH H

 + (h + μ(t)) · f (σ (t), t)

(f (σ (t), τ ) H f (t − h, τ )) δH τ, 

t f (t, τ )δ τ (h + μ(τ )) · δH H

(h + μ(t)) · σ (t)

 t

a

f (t − h, τ )δH τ, (h + μ(t))

σ (t) a

t−h

σ (t) t

 + D (−1) ·

σ (t) t−h

f (t − h, τ )δH τ, 

t f (t, τ )δ τ + (−h − μ(t)) · f (σ (t), t) δH H

 t f (t, τ ) Δτ D f (σ (t), τ ) H f (t − h, τ ), (h + μ(t)) · δH

 +D (−1) ·

σ (t) t−h

f (t − h, τ )δH τ,

4.1 Periodic Properties of Time Scales

271

 t f (t, τ ) + (−h − μ(t)) · f (σ (t), t) ((h + μ(t))μ(t)) · δH

≤

σ (t) a



 (h + μ(t))Δτ 2(σ (t) − a)

+D (−1) ·

σ (t) t−h

 = (h + μ(t)) + D 2  (h + μ(t)) + 2  ≤ (h + μ(t)) + 2 = (h + μ(t)).

≤



f (t − h, τ )δH τ, (−h − μ(t)) · f (t, t)



σ (t) t−h

σ (t) t−h



σ (t)

f (t − h, τ )δH τ,

f (t, t)δH τ

t−h

D (f (t − h, τ ), f (t, t)) Δτ

 (h + μ(t)) 2

Thus, we have proved the assertion (4.4) and (4.5). As in the above, one can prove the assertion (4.6) and (4.7). Then, for c ∈ T∗ , c < t, c + w > t, c+w

G(t) = t

f (t, τ )δH τ c+w

c+w

=

t+w

f (t, τ )δH τ +

t

f (t, τ )δH τ +

t

f (t, τ + w)δH τ,

c

whereupon  δH G(t) =

c+w

t δH





c+w

f (t, τ )δH τ t

c+w

=

f (t, τ )δH τ +

t

t = δH

t t

+

t

 f (t, τ + w)δH τ

c



t + δH



t

f (t, τ + w)δH τ

c

t δH f (t, τ )δH τ H f (σ (t), t)

δH f (t, τ + w)δH τ + f (σ (t), t + w)

c t+w

= t

t δH f (t, τ )δH τ

+f (σ (t), t + w) H f (σ (t), t), This completes the proof.

t ∈ Tκ .

272

4 Functional Fuzzy Dynamic Equations

Theorem 4.7 Let T be w-periodic, f : Tκ × Tκ → F (R) is continuous at (t, t), t f (t, ·) exists and it is rd-continuous on [a, σ (t)], and for each  > 0 t ∈ Tκ , δSH there exists a neighborhood U of t, independent of τ ∈ [a, σ (t)], such that   t f (t, τ )(−(h + μ(t))) ≤ (h + μ(t)), D f (t − h, τ ) − f (σ (t), τ ), δSH   t D f (σ (t), τ ) − f (t + h, τ ), δSH f (t, τ )(−(h − μ(t))) ≤ (h − μ(t)), for all t − h, t + h ∈ U . Then t+w

G(t) =

f (t, τ )δSH τ t

implies t+w

δSH G(t) = t

t δSH f (t, τ )δSH τ + f (σ (t), t + w) H f (σ (t), t),

t ∈ T. Example 4.3 Let T = N0 , f (t, τ ) = (1, τ, t + τ ),

t, τ ∈ T.

Take w = 1. Let 1 2 1 t − t, 2 2

g(t) =

t ∈ T.

Then 1 1 (σ (t) + t) − 2 2 1 1 = (t + 1 + t) − 2 2 1 1 =t+ − 2 2 = t, t ∈ T.

g Δ (t) =

We have t+w

G(t) =

f (t, τ )δH τ t



t+w

=

t+w

Δτ, t

t



t+w

τ Δτ,

(t+τ )Δτ t

4.1 Periodic Properties of Time Scales

273

 = w,

t+w

g (τ )Δτ, tw+ t



t+w

Δ

Δ

g (τ )Δτ t

 τ=t+w  τ=t+w   , tw+g(τ ) = w, g(τ ) τ=t

τ=t

      1 2 1 1 2 1 τ=t+w τ=t+w = w, , tw+ τ − τ  τ − τ  τ=t τ=t 2 2 2 2  1 1 1 1 = w, (t+w)2 − (t+w)− t 2 + t, 2 2 2 2  1 1 2 1 1 2 1 tw+ (t+w) − t− w− t + t 2 2 2 2 2  1 1 w2 1 1 1 = w, t 2 +tw+ − t− w− t 2 + t, 2 2 2 2 2 2  2 1 2 1 1 2 w 1 1 tw+ t +tw+ − t− w− t + t 2 2 2 2 2 2   2 w w 1 , = w, tw− w, 2tw− + 2 2 2 δH G(t) = (0, w, 2w), f (σ (t), t+w) = (1, t+w, σ (t)+t+w) = (1, t+w, t+1+t+w) = (1, t+w, 2t+1+w), f (σ (t), t) = (1, t, σ (t)+t) = (1, t, 2t+1), f (σ (t), t+w) H f (σ (t), t) = (0, w, w), t δH f (t, τ ) = (0, 0, 1),

and t+w t

t δH f (t, τ )δH τ + f (σ (t), t + w) H f (σ (t), t) = (0, 0, w) + (0, w, w)

= (0, w, 2w). Therefore t+w

δH G(t) = t

t ∈ T.

t δH f (t, τ )δH τ + f (σ (t), t + w) H f (σ (t), t),

274

4 Functional Fuzzy Dynamic Equations

Exercise 4.4 Let T = 3N0 , f (t, τ ) = (t + τ, 2t + τ + τ 2 , 2t + τ + tτ + τ 3 ), (t, τ ) ∈ T × T. Prove that t+w

δH G(t) = t

t δH f (t, τ )δH τ + f (σ (t), t + w) H f (σ (t), t),

t ∈ T.

4.2 The Phase Space Suppose that inf T = −∞ and t1 , t2 ∈ T imply t1 + t2 ∈ T. Denote T− = T T+ = T

 

(−∞, 0], [0, ∞).

Let (E1)

h : T− → [0, ∞), h(s) > 0, s ∈ T− and 0 −∞

h(s)Δs = 1.

Definition 4.3 Suppose (E1). The phase space Ch is defined by  Ch = φ ∈ Crd (T) :

0 −∞

 h(s)|φ|[s,0] Δs < ∞ ,

where |φ|[s,0] = sup |φ(t)|, t∈[s,0]

and endowed with the norm |φ|h =

0 −∞

h(s)|φ|[s,0] Δs.

Let I be a bounded closed subset of T. For F, G : I → F (R), F, G ∈ Cf rd (I ), define

4.2 The Phase Space

275

D(F, G) = sup D(F (t), G(t)). t∈I

Theorem 4.8 For any F ∈ Cf rd (I ) and f ∈ SH F there exists a positive constant η = η(F ) < 1 such that for each G ∈ Cf rd (I ) we can find g ∈ SH G satisfying ηf − g ≤ D(F, G), where f − g = sup |f (t) − g(t)|. t∈I

Proof Note that for any F, G ∈ Cf rd (I ), there exists a μ ∈ (0, 1) such that μD(SH F , SH G ) ≤ D(F, G). On the other hand, for any given t0 ∈ I and f ∈ SH F , there exists a gt0 ∈ SH G such that μ|f (t0 ) − gt0 (t0 )| ≤ D(SH F (t0 ), SH G (t0 )),

(4.8)

where SH F (t) = {f (t) : f ∈ SH F }. By the continuity of the metric, one can find a δ0 > 0 such that (4.8) holds on (t0 − δ0 , t0 + δ0 ). Since I is a bounded closed interval, there exist a finite number of intervals, say Ii = (ti − δi , ti + δi ),

i ∈ {1, . . . , n},

such that I⊂

n "

Ii ,

i=1

and μ|f (t) − gi (t)| ≤ D(SH F (t), SH G (t)), where gi ∈ SH Gi , i ∈ {1, . . . , n}. Let

t ∈ Ii ,

276

4 Functional Fuzzy Dynamic Equations

⎧ ⎪ ⎨ g1 (t) t ∈ I1 g (t) = ... ⎪ ⎩ gn (t) t ∈ In . We have g ∈ SH G and for any t ∈ I we have t ∈ Ii for some i ∈ {1, . . . , n} such that μ|f (t) − g (t)| = μ|f (t) − gi (t)| ≤ D(SH F (t), SH G (t)). Because t was arbitrarily chosen, it follows that μf − g  ≤ D(SH F , SH G ). Let η = μ2 . Then 0 < η < 1 and g ηf − g  = μ2 f − ≤ μD(SH F , SH G ) ≤ D(F, G). This completes the proof. With 2T we will denote the collection of bounded nonempty subsets of T. With Kc (R) we will denote the collection of nonempty compact and convex subsets of R. The space of all rd-continuous functions f : I → Kc (R) will be denoted by Ccf rd (I ). Definition 4.4 The function γ : 2T → [0, ∞) defined by  γ (I ) = inf η(F ) ∈ (0, 1) : F ∈ Ccf rd (I ), F : I → Kc (R) such exists

that

for any

G ∈ Ccf rd (I )

admits

f ∈ SH F

there

g ∈ SH G

satisfying  η(F )f − g ≤ D(F, G)

is called the measure of continuity on I . By Theorem 4.8, it follows that γ ∈ [0, 1). Theorem 4.9 Let I ∈ 2T be a closed subset of T. If γ (I ) > 0, then the space Ccf rd (I ), endowed with the metric D, is a complete metric space.

4.2 The Phase Space

277

Proof Let {Fn }n∈N ⊂ Ccf rd (T− ) be a Cauchy sequence. Then, for any  > 0 there exists a natural number N such that D(Fm , Fn ) <  for any m, n ≥ N , and t ∈ I . Then D(Fm (t), Fn (t)) < . Consequently {Fn (t)}n∈N is a Cauchy sequence in Kc (R). Because Kc (R) is a complete metric space, we get that the sequence {Fn (t)}n∈N is a convergent sequence to F (t) ∈ Kc (R). Take N 3 ∈ N large enough so that D(Fn (t), F (t))
0. Hence, for t ∈ (t0 − δ, t0 + δ), we get

278

4 Functional Fuzzy Dynamic Equations

  D(F (t), A) ≤ D F (t), FN  (t) 3   +D FN  (t), A 3

 2 + 3 3 = .


0, with IT , we denote IT = [−T , 0], and by C C (IT ) we mean the space Ccf rd (IT ) to satisfy that the measure of continuity on IT is positive. Define  C Ch = Φ ∈ Ccf rd (T− ),   Φ  ∈ C C (IT ) IT

LDh (Φ, 0) < ∞, for

 all T > 0 ,

where Dh (F, G) =

0 −∞

h(s)D [s,0] (F, G)Δs.

If F, G ∈ C Ch , we have that Dh (F, G) < ∞.

4.2 The Phase Space

281

Theorem 4.10 Suppose (E1). The following results are true: (i) For any  > 0 and T > 0, there exists a δ = δ(, T ) > 0 such that, for any Φ1 , Φ2 ∈ C Ch , if Dh (Φ1 , Φ2 ) < δ, then D IT (Φ1 , Φ2 ) ≤ . (ii) Suppose that {Fn }n∈N ⊂ Ccf rd (T− ) satisfies D Is (Fn , 0) ≤ φ(s) for any s > 0, where φ is a positive function and satisfies 0 −∞

h(s)φ(s)Δs < ∞.

Then lim Dh (Fn , F ) = 0

n→∞

if and only if for any n0 ∈ N, one has lim D In0 (Fn , F ) = 0.

n→∞

(iii) (C Ch , Dh ) is a complete metric space. Proof (i) Suppose the contrary. Then there exist an 0 > 0 and a T0 > 0 such that, for any δ > 0 there exist Φ1 , Φ2 ∈ C Ch such that Dh (Φ1 , Φ2 ) ≤ δ and D [−T0 ,0] (Φ1 , Φ2 ) > 0 . Take δ= Then

0 2

−T0 −∞

h(s)Δs.

282

4 Functional Fuzzy Dynamic Equations

δ ≥ Dh (Φ1 , Φ2 ) = ≥

0 −∞

h(s)D [s,0] (Φ1 , Φ2 )Δs

−T0 −∞

h(s)D [−T0 ,0] (Φ1 , Φ2 )Δs

−T0

> 0

−∞

h(s)Δs

= 2δ. This is a contradiction. (ii) Suppose that lim Dh (Fn , F ) = 0.

n→∞

Hence, using the definition of Dh (·, ·) and (E1), we get 0

lim

n→∞ −∞

h(s)D [s,0] (Fn , F )Δs = 0.

From here, we conclude that for any n0 ∈ N, one has lim D In0 (Fn , F ) = 0.

n→∞

Now, we assume that {Fn }n∈N ⊂ C C (T− ) satisfies 0) ≤ φ(s) D Is (Fn , for any s ≥ 0, where φ is a positive function for which 0 −∞

h(s)φ(s)Δs < ∞,

and for any n0 ∈ N, we have lim D In0 (Fn , F ) = 0.

n→∞

Since we have (4.10), for any  > 0 there exists an n0 ∈ N such that −n0 −∞

h(s)φ(s)Δs < .

(4.10)

4.2 The Phase Space

283

Hence, using the Lebesgue dominated convergence theorem, we get lim Dh (Fn , F ) = lim

n→∞

0

n→∞ −∞

= lim

h(s)D [s,0] (Fn , F )Δs

−n0

n→∞ −∞

h(s)D [s,0] (Fn , F )Δs

0

+ lim

n→∞ −n 0

≤ lim

−n0

n→∞ −∞

h(s)D [s,0] (Fn , F )Δs

h(s)D [s,0] (Fn , 0)Δs

−n0

+ lim

n→∞ −∞ 0

+ lim

n→∞ −n 0

≤ lim

−n0

n→∞ −∞

h(s)D [s,0] (Fn , F )Δs

h(s)φ(s)Δs + lim

n→∞ −n 0

n→∞ −∞

< 2 +

0 −n0

h(s)φ(s)Δs

h(s)D [s,0] (Fn , F )Δs

−n0

= 2 lim

−n0

n→∞ −∞

0

+ lim

h(s)D [s,0] (F, 0)Δs

h(s)φ(s)Δs + lim

0

n→∞ −n 0

h(s)D [s,0] (Fn , F )Δs

h(s) lim D [s,0] (Fn , F )Δs n→∞

= 2. Because  > 0 was arbitrarily chosen, we get lim Dh (Fn , F ) = 0.

n→∞

(iii) Let {Fn }n∈N be a Cauchy sequence of C Ch . Then, for any given s > 0, using (i), there exist a δ > 0 and an n0 ∈ N such that Dh (Fn , Fn0 ) ≤ δ and D Is (Fn , Fn0 ) < 1 for any n ≥ n0 . Thus,

284

4 Functional Fuzzy Dynamic Equations

D Is (Fn , 0) ≤ D Is (Fn , Fn0 ) + D Is (Fn0 , 0) < 1 + D Is (Fn0 , 0). Let 0). φ(s) = 1 + D Is (Fn0 , We have φ(s) > 0 and 0 −∞

0

h(s)φ(s)Δs =

−∞

h(s)Δs +

0 −∞

h(s)D Is (Fn0 , 0)Δs

0) = 1 + Dh (Fn0 , < ∞. By (ii), we have lim D IT (Fm , Fn ) = 0,

m,n→∞

whenever lim Dh (Fm , Fn ) = 0,

m,n→∞

where T > 0. Thus, {Fn }n∈N is a Cauchy sequence in Ccf rd (IT ). Hence by Theorem 4.9, we conclude that there exists F ∈ Ccf rd (IT ) such that lim D IT (Fn , F ) = 0.

n→∞

By (ii), we obtain lim D(Fn , F ) = 0.

n→∞

Then, for any  > 0, there exists an N ∈ N such that  , 2  D IT (Fn , F ) < 2 Dh (Fn , F )
0 was arbitrarily chosen, it follows that f is rd-continuous on T− . Also, d(f (t), F (t)) ≤ |f (t) − fn (t)| + d(fn (t), F (t)) ≤ |f (t) − fn (t)| + D(Fn (t), F (t))   < + 2 2 = for any n ∈ N large enough. Because  > 0 was arbitrarily chosen, we get d(f (t), F (t)) = 0. Hence, using that F (t) is a compact set, we get that  f (t) ∈ F (t). Thus, f ∈  SH F . By the completeness of C C (IT ), we get F  ∈ C C (IT ) for all T > 0. Therefore F ∈ C Ch . This completes the proof.

IT

286

4 Functional Fuzzy Dynamic Equations

Theorem 4.11 Suppose (E1), Φ ∈ C Ch , and Ut (ξ ) = U (t + ξ ) for ξ ∈ T− :   ∈ C C ([0, T ]), U0 = (i) Let T > 0. Assume that U : (−∞, T ) → Kc (R), U  [0,T ]

Φ. Then for any t ∈ [0, T ] we have Ut ∈ C Ch and V (t) = Ut ∈ C C ([0, T ]). (ii) There exists a nonnegative constant M such that 0). Φ(0) ≤ KMDh (Φ, Proof (i) We have Dh (Ut , 0) = =

0 −∞ −t −∞

−t −t

−∞

−t 0

−∞

+2 ≤

0 −∞

+2

h(s)D [s,0] (Ut , 0)Δs

 h(s) max D [s+t,0] (U, 0), D [−t,0] (Ut , 0) Δs

0

+ ≤

h(s)D [s,0] (Ut , 0)Δs 0

+ ≤

h(s)D [s,0] (Ut , 0)Δs

h(s)D [s,0] (Ut , 0)Δs

h(s)D [s+t,0] (Ut , 0)Δs 0 −∞

   h(s)D [0,t] U 

[0,T ]

 , 0 Δs

h(s)D [s,0] (Φ, 0)Δs 0 −∞

h(s)D

[0,T ]

   U

[0,T ]

   = Dh (Φ, 0) + 2D [0,T ] U 

 , 0 Δs

[0,T ]

 , 0

< ∞.

  ∈ C C ([−T + t, t]), which is We have that Ut ∈ Ccf rd (T− ) and U  [−T +t,t]   equivalent to Ut  ∈ C C (IT ). Hence, Ut ∈ C Ch , t ∈ [0, T ]. Now, we will IT

show that V is rd-continuous on [0, T ]. Let t0 ∈ [0, T ] be a right-dense point.

4.2 The Phase Space

287

Note that Ut0 ∈ C Ch . Then U is bounded on [0, T ]. Hence, there exists a constant L > 0 such that 0) ≤ L. D [0,T ] (U, Note that for any  > 0 there exists M = M(t0 , ) > 0 such that −M −∞

h(s)D [s,0] (Ut0 , 0)Δs < −M −∞

h(s)Δs
0 such that for any t ∈ (t0 − δ1 , t0 + δ1 ), we have D(Ut (r), Ut0 (r)) = D(Ut (r), Ut (r − t + t0 ))  < . 4 If r is right-scattered and left-dense, then Ut0 has a finite left-sided limit at r, say A. Then there exists a δ2 > 0 such that    D Ut0 (s), A < , 8

s ∈ (r − δ2 , r + δ2 ).

Let |t − t0 | < δ2 . We have r + t0 − t ∈ (r − δ2 , r + δ2 ) and   D Ut (r), Ut0 (r) = D (Ut (r), Ut (r + t0 − t)) ≤ D (Ut (r), A) + D (A, Ut (r + t0 − t))   < + 8 8  = . 4 If r is isolated, then there is a δ3 > 0 such that s = r whenever s ∈ (r − δ3 , r + δ3 ). Thus, r + t0 − t = r if |t − t0 | < δ. Hence,   D Ut (r), Ut0 (r) = D (Ut (r), Ut (r + t0 − t)) =0  < . 4

288

4 Functional Fuzzy Dynamic Equations

Therefore, for each r ∈ [−M, 0], we have    D [−M,0] Ut , Ut0 < 4 if t ∈ (t0 − δ, t0 + δ) with δ = min{δ1 , δ2 , δ3 }. Next, Dh (Ut , Ut0 ) = = ≤

0 −∞

−M −∞ −M −∞

+ ≤

−∞

≤L = .

h(s)D [s,0] (Ut , Ut0 )Δs

   ˜ + D [s,0] (Ut0 , h(s) max D [t0 ,t] (Ut , 0) 0)

0 −M

−∞

−M

h(s)D [s,0] (Ut , Ut0 )Δs

[s,0]

−M

0

  ˜ + D [s,0] (Ut0 , h(s) D [s,0] (Ut , 0) 0) Δs

−M −M

+

h(s)D [s,0] (Ut , Ut0 )Δs +

0

+D

≤

h(s)D [s,0] (Ut , Ut0 )Δs

 (Ut0 , 0) Δs

h(s)D [s,0] (Ut , Ut0 )Δs

  h(s) L + 2D [s,0] (Ut0 , 0) Δs + D [−M,0] (Ut , Ut0 )

   +2 + 4L 4 4

Therefore Ut is continuous at the right-dense point t0 with respect to t ∈ [0, T ]. If t0 is left-dense, we have that Ut has a finite limit at t0 with respect to t0 . Since Ut ∈ C Ch , there exists a ut ∈ SH Ut for each t ∈ [0, T ]. Because ut is rd-continuous on T− , we have that v ∈ SH V with v(t) = ut , t ∈ [0, T ]. Thus, V ∈ Ccf rd ([0, T ]). Now, we will prove that V ∈ C C ([0, T ]). Let = G(t)



G(t) if t ∈ (0, T ] Φ(t) if t ∈ T− .

t ∈ C Ch and G 0 = Φ. Also, G(t) = G t (0) and V (t) = Ut (0). Note We have G that for any fixed t ∈ [0, T ], if v(t) = ut ∈ SH Ut , there exists gt ∈ SH G t such that

4.2 The Phase Space

289

  t , Ut gt (s) − ut (s)| ≤ D IM G γ (M) max | s∈IM

for all M > 0. Since γ (·) is a decreasing function, we have 0 0, C C w = {U ∈ Ccf rd (T) : U (t + w) = U (t), and Ω = [0, w].

t ∈ T}

290

4 Functional Fuzzy Dynamic Equations

Theorem 4.12 Suppose that {Un }n∈N ⊂ C C w , U ∈ C C w , and lim D Ω (Un , U ) = 0.

n→∞

Then {Utn }n∈N converges uniformly to Ut ∈ C Ch with respect to t ∈ T. Proof Let  > 0 be arbitrarily chosen. Then there exists an N ∈ N such that D Ω (Un , U ) <  for any n ≥ N . Thus, for any n ≥ N, we have 0

Dh (Utn , Ut ) =

−∞ 0

=

−∞

h(s)D [s,0] (Utn , Ut )Δs h(s)D [s+t,t] (Un , U )Δs

≤ D Ω (Un , U )

0 −∞

h(s)Δs

≤ D Ω (Un , U ) 0 be arbitrarily chosen. Define 

W (t) = W (t) + eLt ,

t ∈ T.

We have V (0) ≤ W (0) ≤ W (0) + {} = W (0). Let tδ ∈ T+ be the supremum of all positive numbers δ such that V (0) ≤ W (0) implies V (t) ≤ W (t), t ∈ [0, δ]. We have that tδ ≥ 0. Because V and W are continuous, we conclude that tδ > 0. Assume that tδ < ∞. Then V (tδ ) ≤ W (tδ ) and σ (tδ ) ≤ a for some a ∈ T+ . Note that, if tδ is right-dense, then 

δH W (tδ ) = δH W (tδ ) + LeLtδ , and if tδ is right-scattered, then 

eLσ (tδ ) − eLtδ δH W (tδ ) = δH W (tδ ) +  μ(tδ )

/ .

Since V is a lower solution of Eq. (4.11), we have δH V (tδ ) ≤ (−a(tδ , V (tδ ))) · V (σ (tδ )) + F (tδ , Vtδ )   ≤ (−a(tδ , W (tδ ))) · W (σ (tδ )) + F tδ , (W )tδ ≤ (−a(tδ , W (tδ ))) · W (σ (tδ )) + F (tδ , Wtδ ) +L (W (tδ ) − W (tδ )) = (−a(tδ , W (tδ ))) · W (σ (tδ )) + F (tδ , Wtδ ) +LeLtδ ≤ δH W (tδ ) + LeLtδ .

(4.15)

296

4 Functional Fuzzy Dynamic Equations

Note that LeLtδ ≤

eLσ (tδ ) − eLtδ . μ(tδ )

Then, using (4.15), we get δH V (tδ ) ≤ δH W (tδ ) + L e = δH W (tδ ).

Lσ (tδ ) −eLtδ

μ(tδ )

(4.16)

If tδ is right-dense, using (4.16), we get W (tδ ) − V (tδ ) ≤ W (t) − V (t),

tδ < t < tδ + η.

This is a contradiction because tδ > 0 is the supremum of all t ∈ T+ for which V (t) ≤ W (t). Next, if tδ is right-scattered, then V (σ (tδ )) − V (tδ ) W (σ (tδ )) − W (tδ ) ≤ , μ(tδ ) μ(tδ ) i.e., W (tδ ) − V (tδ ) ≤ W (σ (tδ )) − V (σ (tδ )). We have that tδ < σ (tδ ) ≤ a and W (σ (tδ )) ≥ V (σ (tδ )), which is a contradiction with the definition of tδ . Consequently V (t) < W (t),

t ∈ T+ .

V (t) ≤ W (t),

t ∈ T+ .

Taking  → 0, we obtain

Let sδ ∈ T− be the infimum of all positive numbers δ such that V (0) ≤ W (0) implies that V (t) ≤ W (t), t ∈ [−δ, 0]. We have that sδ ≤ 0. Since V and W are continuous on T, we get sδ < 0 if 0 is a left-dense point. In addition, we have ρ(σ (sδ )) = sδ and δH V (ρ(sδ )) ≤ (−a (ρ(sδ ), V (ρ(sδ )))) · V (sδ )   +F ρ(sδ ), Vρ(sδ ) , δH W (ρ(sδ )) ≥ (−a (ρ(sδ ), W (ρ(sδ )))) · W (sδ )   +F ρ(sδ ), Wρ(sδ ) .

4.3 Periodic Solutions

297

Assume that sδ > −∞. Then ρ(sδ ) ≥ b for some b ∈ T− and V (sδ ) ≤ W (sδ ). Then δH V (ρ(sδ )) ≤ (−a (ρ(sδ ), V (ρ(sδ )))) · V (sδ )   +F ρ(sδ ), Vρ(sδ ) ≤ (−a (ρ(sδ ), W (ρ(sδ )))) · W (sδ )   +F ρ(sδ ), (W )ρ(sδ )

  ≤ (−a (ρ(sδ ), W (ρ(sδ )))) · W (sδ ) + F ρ(sδ ), Wρ(sδ ) +L (W (ρ(sδ )) − W (ρ(sδ )))

  = (−a (ρ(sδ ), W (ρ(sδ )))) · W (sδ ) + F ρ(sδ ), Wρ(sδ ) + LeLρ(sδ ) ≤ δH W (ρ(sδ )) + LeLρ(sδ ) ≤ δH W (ρ(sδ )). Hence, sδ must be a left-dense point and δH V (sδ ) ≤ δH W (sδ ), and we can choose η > 0 such that W (sδ ) − V (sδ ) ≤ W (t) − V (t),

sδ − η < t < sδ .

This is a contradiction with the definition of sδ . Therefore V (t) < W (t),

t ∈ T− .

V (t) ≤ W (t),

t ∈ T− .

Taking  → 0, we obtain

This completes the proof. Corollary 4.1 Let V , W ∈ C CT and δH V (t) ≤ δH W (t),

t ∈ T.

Then V (0) ≤ W (0) implies that V (t) ≤ W (t), t ∈ T. Theorem 4.16 Assume (E2), (E3), (4.14), and that: (i) Equation (4.11) has the lower solution V and the upper solution W with V , W ∈ C CT .

298

4 Functional Fuzzy Dynamic Equations

(ii) F (t, Φ) is rd-continuous in t and continuous in Φ ∈ C Ch , F (t + w, Φ) = F (t, Φ),

t ∈ T,

Φ ∈ C Ch ,

and it maps bounded sets into bounded sets in C Ch . (iii) a(t, U ) is rd-continuous and w-periodic in t and it is continuous and nondecreasing in U , and a(t, U (t)) ≥ α(t) for all U ∈ C CT , where α ∈ R + with eα (t + w, t) − 1 > 0, t ∈ T. Then Eq. (4.11) has the minimal solution Φ and the maximal solution Ψ in C CT . Moreover, there exist monotone sequences {Vn }n∈N and {Wn }n∈N in C CT such that lim Vn (t) = V (t),

n→∞

lim Wn (t) = W (t)

n→∞

uniformly in C Ch . Proof Construct the following sequences. t+w

Vn+1 (t) =

vn (s, t) · F (s, Vns )δH s,

t t+w

Wn+1 (t) =

wn (s, t) · F (s, Wns )δH s,

n ∈ N0 .

t

We stipulate that V0 (t) = V (t), W0 (t) = W (t), vn (s, t) =

ean (s, t) , ean (t + w, t) − 1

wn (s, t) =

ebn (s, t) , ebn (t + w, t) − 1

an (t) = a(t, Vn (t)), bn (t) = a(t, Wn (t)),

t ∈ T,

n ∈ N0 .

We have that Vn , Wn ∈ Ccf rd (T), n ∈ N, and δH Vn+1 (t) = (−a(t, Vn (t))) · Vn+1 (σ (t)) + F (t, Vnt ), δH Wn+1 (t) = (−a(t, Wn (t))) · Wn+1 (σ (t)) + F (t, Wnt ),

t ∈ T,

n ∈ N0 . (4.17)

4.3 Periodic Solutions

299

Since V is the lower solution of Eq. (4.11), we have   δH ea0 (t, 0) · V (t) = ea0 (t, 0) · δH V (t) + a0 (t)ea0 (t, 0) · V (σ (t)) = ea0 (t, 0) · (δH V (t) + a0 (t) · V (σ (t))) ≤ ea0 (t, 0) · F (t, Vt ),

t ∈ T.

Integrating both sides of the last inequality from 0 to w, we have w 0

w

ea0 (s, 0) · F (s, Vs )δH s ≥

0

  δH ea0 (s, 0) · V (s) δH s

= ea0 (w, 0) · V (w) H V (0) = (ea (w, 0) − 1) · V (0). Thus, w

V1 (0) =

v0 (s, 0) · F (s, Vs )δH s

0

=

1 · ea0 (w, 0) − 1

w 0

ea0 (s, 0) · F (s, Vs )δH s

≥ V (0). Let U (t) = V1 (t) H V0 (t),

t ∈ T.

Then δH U (t) = δH V1 (t) H δH V (t) ≥ ((−a(t, V0 (t))) · V1 (σ (t)) + F (t, V0t )) × H ((−a(t, V0 (t))) · V (σ (t)) + F (t, V0t )) = (−a(t, V (t))) · (V1 (σ (t)) H V0 (σ (t))) = (−a0 (t)) · U (σ (t)),

t ∈ T.

Hence,   δH ea0 (t, 0) · U (t) = a0 (t)ea0 (t, 0) · U (σ (t)) +ea0 (t, 0) · δH U (t) = ea0 (t, 0) · (a0 (t) · U (σ (t)) + δH U (t)) ≥ 0,

t ∈ T.

300

4 Functional Fuzzy Dynamic Equations

Therefore     δH ea0 (t, 0) · V (t) ≥ δH ea0 (t, 0) · V1 (t) ,

t ∈ T.

From the last inequality and from Corollary 4.1, it follows that ea0 (t, 0) · V (t) ≤ ea0 (t, 0) · V1 (t),

t ∈ T,

and V (t) ≤ V1 (t),

t ∈ T.

Let Z(t) = W0 (t) H W1 (t),

t ∈ T.

We have δH Z(t) = δH W0 (t) H δH W1 (t) ≥ ((−a(t, W (t))) · W (σ (t)) + F (t, Wt )) × H ((−a(t, W0 (t))) · W1 (σ (t)) + F (t, Wt )) = (−a(t, W (t))) · (W (σ (t)) H W1 (σ (t))) = (−a(t, W (t))) · Z(σ (t)),

t ∈ T.

Hence,   δH eb0 (t, 0) · Z(t) = b0 (t)eb0 (t, 0) · Z(σ (t)) +eb0 (t, 0) · δH Z(t) = eb0 (t, 0) · (b0 (t) · Z(σ (t)) + δH Z(t)) ≥ 0,

t ∈ T.

Therefore     δH eb0 (t, 0) · W0 (t) ≥ δH eb0 (t, 0) · W1 (t) ,

t ∈ T.

Hence by Corollary 4.1, we get eb0 (t, 0) · W0 (t) ≥ eb0 (t, 0) · W1 (t), whereupon W0 (t) ≥ W1 (t),

t ∈ T.

t ∈ T,

4.3 Periodic Solutions

301

Then, using that v0 (s, 0) ≤ w0 (s, 0), s ∈ [0, w], we get w

V1 (0) =

v0 (s, 0) · F (s, V0s )δH s

0 w

≤

w0 (s, 0) · F (s, W0s )δH s

0

= W1 (0). Let U1 (t) = W1 (t) H V1 (t),

t ∈ T.

Then δH U1 (t) = δH (W1 (t) H V1 (t)) = δH W1 (t) H δH V1 (t) = ((−a(t, W0 (t))) · W1 (σ (t)) + F (t, W0t )) × H ((−a(t, V0 (t))) · V1 (σ (t)) + F (t, V0t )) ≥ (−a(t, W0 (t))) · (W1 (σ (t)) H V1 (σ (t))) +F (t, W0t ) H F (t, V0t ) ≥ (−a(t, W0 (t))) · U1 (σ (t)),

t ∈ T.

This yields that   δH eb0 (t, 0) · U1 (t) = eb0 (t, 0) · δH U1 (t) +b0 (t)eb0 (t, 0) · U1 (σ (t)) = eb0 (t, 0) · (δH U1 (t) + b0 (t) · U1 (σ (t))) ≥ 0,

t ∈ T.

Therefore     δH eb0 (t, 0) · W1 (t) ≥ δH eb0 (t, 0) · V1 (t) ,

t ∈ T.

Hence by Corollary 4.1, we obtain eb0 (t, 0) · W1 (t) ≥ eb0 (t, 0) · V1 (t), whereupon W1 (t) ≥ V1 (t),

t ∈ T.

t ∈ T,

302

4 Functional Fuzzy Dynamic Equations

Consequently V0 (t) ≤ V1 (t) ≤ W1 (t) ≤ W0 (t),

t ∈ T.

By induction, V0 (t) ≤ V1 (t) ≤ · · · ≤ Vn (t) ≤ · · · ≤ Wn (t) ≤ · · · ≤ W1 (t) ≤ W0 (t), t ∈ T. For −s ∈ T− , denote 0). φ(s) = D Is (W0 , V0 ) + D Is (V0 ,   Since V0 

T

  , W 0 −

0 −∞

T−

∈ C Ch , we have

h(s) · φ(s)δH s =

0 −∞

  h(s) · D Is (W0 , V0 ) + D Is (V0 , 0) δH s

< ∞. Next, 0) = D Is (Vn H V0 , V0 ) D Is (Vn ,   ≤ D Is Vn H V0 , 0 + D Is (V0 , 0) 0) + D Is (V0 , 0) ≤ D Is (W0 H V0 , = φ(s),

n ∈ N.

Therefore the sequence {Vn (t)}n∈N is uniformly bounded on Is . Also, t+w

δH Vn+1 (t) =

(an )(t)vn (s, t) · F (s, Vns )δH s

t

+vn (t + w, σ (t)) · F (t, Vnt ) × H vn (t, σ (t)) · F (t, Vnt ) = (an )(t) · Vn (t) +

1 · F (t, Vnt ), 1 + μ(t)an (t)

t ∈ T.

By (ii), we have that F is bounded on {(t, Vn ) : t ∈ Is ,

n ∈ N0 }.

Also, using (iii), we have that {an }n∈N is bounded. Therefore there exist positive constants M1 and M2 such that

4.3 Periodic Solutions

303

  D F (t, Vnt ), 0 ≤ M1 , |an (t)| ≤ M2 ,

t ∈ Is ,

n ∈ N0 .

Hence,   0 ≤ M 2 M 1 + M1 , D δH Vn+1 (t),

t ∈ T.

Therefore {Vn }n∈N is equi-continuous on Is . Then, by the Arzela–Ascoli theorem, it follows that {Vn (t)}n∈N converges uniformly to Φ(t) on Is , i.e., lim D Is (Vn , Φ) = 0.

n→∞

Hence, lim Dh (Vn , Φ) = 0

n→∞

and

Φ ∈ C Ch .

Similarly, there exists Ψ ∈ C Ch such that lim Dh (Wn , Ψ ) = 0.

n→∞

Also, we have that lim D Ω (Vn , Φ) = 0 and

n→∞

lim D Ω (Wn , Ψ ) = 0,

n→∞

and if = Φt , Φ(t)

(t) = Ψt , Ψ

Ψ ∈ Ccf rd (Ω). By the definitions of the we have Φt , Ψt ∈ C Ch , T ∈ Ω, and Φ, sequences {Vn }n∈N , {Wn }n∈N , we have t+w

Φ(t) =

w(t, s) · F (s, Φs )δH s,

(4.18)

t t+w

Ψ (t) =

w(t, s) · F (s, Ψs )δH s,

t ∈ T−

t

where v(s, t) =

ea (s, t) , ea (t + w, t) − 1

a(t) = a(t, Φ(t)),



Ω,

(4.19)

304

4 Functional Fuzzy Dynamic Equations

w(s, t) =

eb (s, t) , eb (t + w, t) − 1

b(t) = a(t, Ψ (t)),

t ∈ T.

If t > w, we write t = t1 + nw, t1 ∈ Ω, n ∈ N, and define Φ(t) = Φ(t1 ),

Ψ (t) = Ψ (t1 ).

Thus, Φ and Ψ are w-periodic, lim Dh (Vn , Φ) = 0,

n→∞

lim Dh (Wn , Ψ ) = 0

n→∞

uniformly, and (4.18), (4.19) hold on T. Also, Φ, Ψ ∈ C CT and lim Dh (Vnt , Φt ) = 0,

n→∞

lim Dh (Wnt , Ψt ) = 0

n→∞

uniformly with respect to t ∈ T. Letting n → ∞ in (4.17), we find that Φ and Ψ are the solutions of Eq. (4.11) and V0 (t) ≤ Φ(t) ≤ Ψ (t) ≤ W0 (t),

t ∈ T.

Assume that U is a solution of Eq. (4.11) satisfying V0 (t) ≤ U (t) ≤ W0 (t),

t ∈ T.

We have w

U (0) =

va (s, 0) · F (s, Us )δH s

0 w

≥

v0 (s, 0) · F (s, V0s )δH s

0

= V1 (0). Let P (t) = U (t) H V1 (t),

t ∈ T,

and then δH P (t) = δH U (t) H δH V1 (t) = ((−a(t, U (t))) · U (σ (t)) + F (t, Ut )) × H ((−a(t, V0 (t))) · V1 (σ (t)) + F (t, V0t ))

4.3 Periodic Solutions

305

≥ (−a (t, U (t))) · (U (σ (t)) H V1 (σ (t))) +F (t, Ut ) H F (t, V0t ) ≥ (−a(t, U (t))) · P (σ (t)),

t ∈ T.

Thus, δH (ea (t, 0) · P (t)) = ea (t, 0) · δH P (t) +a(t)ea (t, 0) · P (σ (t)) = ea (t, 0) · (δH P (t) + a(t) · P (σ (t))) ≥ 0,

t ∈ T.

Consequently δH (ea (t, 0) · U (t)) ≥ δH (ea (t, 0) · V1 (t)) ,

t ∈ T.

Hence by Corollary 4.1, we obtain ea (t, 0) · U (t) ≥ ea (t, 0) · V1 (t),

t ∈ T,

whereupon U (t) ≥ V1 (t),

t ∈ T.

U (t) ≤ W1 (t),

t ∈ T.

As in the above,

By induction, we get Vn (t) ≤ U (t) ≤ Wn (t),

t ∈ T,

n ∈ N0 .

Taking limit n → ∞ in the last inequality, we get V0 (t) ≤ Φ(t) ≤ U (t) ≤ Ψ (t) ≤ W0 (t),

t ∈ T.

This completes the proof. Exercise 4.5 Let T = Z. Find 1-periodic solution to the equation δH y(t) = (−3) · y(t + 1) + {sin(2π t), 2 + cos(6π t), 4},

t ∈ T.

306

4 Functional Fuzzy Dynamic Equations

Exercise 4.6 Let T = 11Z. Find 11-periodic solution to the system δH y(t) = (−2) · z(t + 11) + {−3, −2, −1} δH z(t) = (−8) · y(t + 11) + {sin(6π t), 5, 10},

t ∈ T.

4.4 Advanced Practical Problems Problem 4.1 Let T = 4N0 , v(t) = t 2 , f (t) = {t 2 , 1 + t + t 2 , 2 + 3t + t 2 },

t ∈ T.

Prove that     δH f (v(t)), (δH f (v)) (t) = v Δ (t) ·

t ∈ T.

Problem 4.2 Let T = 3N0 , v(t) = t 3 , f (t) = {t, 1 + t, 2t 3 },

t ∈ T.

Prove that     δSH f (v(t)), (δSH f (v)) (t) = v Δ (t) · Problem 4.3 Let T =

t ∈ T.

 N0 1 {0}, 3

v(t) = t + t 2 , f (t) = {1, 1 + t + t 2 , 1 + t + 3t 2 + 4t 4 },

t ∈ T,

a = 0, b=

1 . 3

Prove that b a

v Δ (t) · f (t)δH t =

v(b) v(a)

(f (v −1 ))(s) δH s.

4.4 Advanced Practical Problems

307

Problem 4.4 Let T = 4N0 , v(t) = t 2 + 3, f (t) = {1, 3 + 2t, 5 + 4t 2 },

t ∈ T,

a = 1, b = 256. Prove that b

v Δ (t) · f (t)δSH t =

a

v(b)

(f (v −1 ))(s) δSH s.

v(a)

Problem 4.5 Let T = 4N0 , f (t, τ ) = {t 2 , t 2 + τ + tτ, t 2 + τ + t 2 τ 2 }, (t, τ ) ∈ T × T. Prove that t+w

δH G(t) = t

t δH f (t, τ )δH τ + f (σ (t), t + w) H f (σ (t), t),

t ∈ T. Problem 4.6 Let T = Z. Find 1-periodic solution to the equation δH y(t) = (−1) · y(t + 1) + {1, 2, 3},

t ∈ T.

Problem 4.7 Let T = 2Z. Find 2-periodic solution to the equation δH y(t) = (−2) · y(t + 2) + {1 + sin(2π t), 3, 7 + cos(4π t)},

t ∈ T.

Problem 4.8 Let T = 3Z. Find 3-periodic solution to the equation δSH y(t) = (− sin(4π t) − 8) · y(t + 3)   sin(8π t) , 2 + sin(2π t), 11 + cos(6π t) , + 1+ 10 + cos(2π t)

t ∈ T.

Problem 4.9 Let T = 5Z. Find 5-periodic solution to the equation δSH y(t) = (−4) · y(t + 5)   7 + 2 sin(4π t) , 4, 11 + sin(8π t) , + 1+ 10 + (cos(2π t) + sin(4π t))4

t ∈ T.

308

4 Functional Fuzzy Dynamic Equations

Problem 4.10 Let T = 7Z. Find 7-periodic solution to the system δH y(t) = (−2) · z(t + 7) + {1, 4, 9} δH z(t) = (−1) · y(t + 7) + {1 + sin(4π t), 4, 11},

t ∈ T.

4.5 Notes and References In this chapter we formulated and proved some basic results for periodic functions on the periodic time scales. A phase space is built for fuzzy dynamic equations with infinite delays on the periodic time scales. By using the monotone methods, we systematically considered the existence of periodic solutions for the fuzzy dynamic equations with infinite delays on the periodic time scales, which generalize and incorporate as special cases some known results for fuzzy differential equations and fuzzy difference equations. Some of the results in this chapter can be found in [27].

Chapter 5

Impulsive Fuzzy Dynamic Equations

This chapter is devoted to a qualitative analysis for impulsive fuzzy dynamic equations. First, we deducted the solutions of linear first order impulsive dynamic equations. Then we investigated some problems for first order impulsive fuzzy dynamic equations for existence and uniqueness of the solutions, using the Krasnosel’skii fixed point theorem and Sadovskii fixed point theorem. In this chapter we give criteria as to when the trivial solution of first order impulsive fuzzy dynamic equations is stable. Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. 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}. Such time scales exist.

 Example 5.1 Let U = 21n : n ∈ N {0} and T=U

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

Note that σ (n) = n for

1. Consider U . Let t =

1 2n ,

any

n ∈ N0 ∩ U.

n ∈ N. Then σ (t) =

1 2n−1

2 2n = 2t. =

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_5

309

310

5 Impulsive Fuzzy Dynamic Equations

2. Consider 1 − U . Let t = 1 −

1 2n .

Then

1 2n

= 1 − t and 1

σ (t) = 1 −

2n+1

1 1 · 2 2n 1 = 1 − (1 − t) 2 1 = (1 + t). 2 = 1−

3. Consider 1 + U . Let t = 1 +

1 2n .

Then

1 2n

= t − 1 and

1 2n−1 2 = 1+ n 2 = 1 + 2(t − 1)

σ (t) = 1 +

= 2t − 1. 4. For n − U , n ∈ N0 , we have σ (t) = 12 (n + t), t = n. 5. For n + U , n ∈ N0 , we have σ (t) = 2t − n, t = n.

5.1 Linear First Order Impulsive Fuzzy Dynamic Equations Consider the following impulsive fuzzy dynamic equation: δH y(t) = p(t) · y(t) + q(t), t ∈ [0, T ], y(0) = y0 ,   y(ti+ ) = y ti− + I (ti , y(ti )), i ∈ {1, . . . , n},

(5.1)

where   y ti± = lim y(t), t→ti±

  y(ti ) = y ti− , (F1) (F2)

i ∈ {1, . . . , n},

p : T → R, p ∈ R, y0 ∈ F (R), q : T → F (R), q ∈ Cf (T), I : T × F (R) → F (R), I ∈ Cf (T × F (R)).

5.1 Linear First Order Impulsive Fuzzy Dynamic Equations

311

Define tn+1 = T , J0 = [0, t1 ], Jk = (tk , tk+1 ], k ∈ {1, . . . , n},  P Cf = y : [0, T ] → F (R), y ∈ C (Jk ),   y tk− = y(tk ),

  ∃y tk± ,

 k ∈ {1, . . . , n} ,

P Cf1 H = y : [0, T ] → F (R),

y ∈ CH1 (Jk ),

 k ∈ {1, . . . , n} .

Theorem 5.1 Suppose (F 1) and (F 2). The function y ∈ P Cf1 H is a solution of Eq. (5.1) if and only if y ∈ P Cf satisfies the following integral equation: y(t) =

!t

ep (t, σ (τ )) · q(τ )δH τ + ep (t, 0) · y0 0% {k:tk t0 , k ∈ N, and tk → t0 , as k → ∞. We have [t0 , t1 ) ⊃ [t0 , t2 ) ⊃ . . . [t0 , tk ) ⊃ . . . and ∞ 

{t0 } =

[t0 , tk ).

k=1

Hence, {t0 } is Δ-measurable as a countable intersection of Δ-measurable sets. By the continuity property of μΔ , we get μΔ ({t0 }) = lim μΔ ([t0 , tk )) k→∞

= lim (tk − t0 ) k→∞

= 0. This completes the proof. Theorem 6.3 Let a, b ∈ T, a ≤ b. Then μΔ ([a, b)) = b − a, μΔ ((a, b)) = b − σ (a). Proof We have μΔ ([a, b)) = m1 ([a, b)) = b − a. Next, [a, b] = {a}

" (a, b).

Since {a}

 (a, b) = ∅,

we get b − a = μΔ ([a, b))

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

366

= μΔ ({a}

" (a, b))

= μΔ ({a}) + μΔ ((a, b)) = σ (a) − a + μΔ ((a, b)). Hence, μΔ ((a, b)) = σ (a) − a − b + a = σ (a) − b. This completes the proof. Example 6.1 Let T = 2Z. We will find μΔ ([0, 10))

and

μΔ ((0, 10)).

We have σ (t) = t + 2,

t ∈ T.

Then μΔ ([0, 10)) = 10 − 0 = 10, μΔ ((0, 10)) = 10 − σ (0) = 10 − 2 = 8. Example 6.2 Let T = 2N0



{0}. We will find

μΔ ([1, 16))

and

μΔ ((2, 32)).

We have σ (0) = 1,

σ (t) = 2t,

t ∈ 2N0 .

Then μΔ ([1, 16)) = 16 − 1 = 15, μΔ ((2, 32)) = 32 − σ (2)

6.1 The Lebesgue Delta-Integral

367

= 32 − 4 = 28. Exercise 6.3 Let T = 3N0 1. 2. 3. 4. 5.



{0}. Find:

μΔ ([0, 27)) μΔ ([1, 81)) μΔ ((0, 81)) μΔ ((0, 243)) μΔ ([1, 243))

Theorem 6.4 Let a, b ∈ T\{max T}, a ≤ b. Then μΔ ((a, b]) = σ (b) − σ (a),

μΔ ([a, b]) = σ (b) − a.

Proof We have (a, b] = (a, b)

"

{b}.

Then, using that (a, b)



{b} = ∅,

we obtain μΔ ((a, b]) = μΔ ((a, b)

"

{b})

= μΔ ((a, b)) + μΔ ({b}) = b − σ (a) + σ (b) − b = σ (b) − σ (a). Next, [a, b] = {a}

" (a, b].

Since {a}

 (a, b] = ∅,

we obtain μΔ ([a, b]) = μΔ ({a}

" (a, b])

= μΔ ({a}) + μΔ ((a, b])

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

368

= σ (a) − a + σ (b) − σ (a) = σ (b) − a. This completes the proof. Example 6.3 Let T = Z. We will find μΔ ((0, 10])

and

μΔ ([−2, 8]).

We have σ (t) = t + 1,

t ∈ T.

Then μΔ ((0, 10]) = σ (10) − σ (0) = 11 − 1 = 10, μΔ ([−2, 8]) = σ (8) − (−2) = 9+2 = 11. Example 6.4 Let T = 4N0



{0}. We will find

μΔ ((0, 4])

and

μΔ ([1, 64]).

We have σ (0) = 1, σ (t) = 4t,

t ∈ 4N0 .

Then μΔ ((0, 4]) = σ (4) − σ (0) = 16 − 1 = 15, μΔ ([1, 64]) = σ (64) − 1 = 256 − 1 = 255.

6.1 The Lebesgue Delta-Integral

369

Exercise 6.4 Let T = 8Z. Find: 1. 2. 3. 4. 5.

μΔ ((0, 8]) μΔ ([−8, 8]) μΔ ([−16, 32]) μΔ ([−16, 0]) μΔ ((−16, 8])

For a set E ⊂ [a, b], define IE = {i ∈ I : ti ∈ E



R},

where I ⊂ N and R is given by (6.1). Theorem 6.5 If E ⊂ [a, b], then the following properties are satisfied: 1. μΔ (E) ≤ m∗1 (E). 2. If b ∈ / E and E has no right-scattered points, then μΔ (E) = m∗1 (E). 3. If R is defined by (6.1) and I \R is Lebesgue measurable, then μ(R) = 0, where μ(·) is the classical Lebesgue measure. 4. μΔ (E



R) =

(

(σ (ti ) − ti )

i∈IE ≤

b−a

=μΔ ([a, b). 5. If b ∈ / E, then m∗1 (E) =

(

(σ (ti ) − ti ) + μ(E).

i∈IE

/ E and E has no right-scattered points. 6. m∗1 (E) = μΔ (E) if and only if b ∈ Proof 1. The assertions (1), (2), (3), and (4) follow directly from the definitions of m∗1 and μΔ . 2. Now, we will prove (5). Suppose that b ∈ / E. We have

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

370



μΔ (E) = μΔ (E



T)

" (T\R)))   = μΔ (E R) + μΔ (E (T\R))  = μΔ (E (T\R)). = μΔ (E

Because b ∈ /E have



(T\R) and E

(R

 (T\R) has no right-scattered points, by (2), we  (T\R))  = m∗1 (E (T\R)).

μΔ (E) = μΔ (E

Thus,   m∗1 (E) = m∗1 (E R) + m∗1 (E (T\R)) ( = (σ (ti ) − ti ) + μΔ (E). i∈IE

3. Now, we will prove (6). a. Let b ∈ / E and E has no right-scattered points. Then (

(σ (ti ) − ti ) = 0

i∈IE

and m∗1 (E) = μΔ (E). b. Let m∗1 (E) = μΔ (E). Then, using (5), we get (

(σ (ti ) − ti ) = 0.

i∈IE

This completes the proof. Theorem 6.6 Let A ⊂ [a, b]. Then A is Δ-measurable if and only if A is Lebesgue measurable. In such a case, the following properties hold for any Δ-measurable set A. (i) If b ∈ / A, then

6.1 The Lebesgue Delta-Integral

371

μΔ (A) =

(

(σ (ti ) − ti ) + μ(A).

i∈IA

(ii) μΔ (A) = μ(A) if and only if b ∈ / A and A has no right-scattered points. Proof 1. Let A be Δ-measurable. a. Let b ∈ / A. Take E ⊂ [a, b] arbitrarily. i. Suppose that b ∈ / E. Then, using that [a, b]\A = (T\A)

" ([a, b]\T)

and A is Δ-measurable and T is Lebesgue measurable, we obtain μ(E) ≤ μ(E



A) + μ(E



([a, b]\A))

 A) + μ(E (T\A))  +μ(E ([a, b]\T))   = m∗1 (E A) + m∗1 (E (T\A)) (  − (σ (ti ) − ti ) + μ(E ([a, b]\T)) ≤ μ(E



i∈IE  T

= m∗1 (E



T) −

(

(σ (ti ) − ti )

i∈IE  T

 +μ(E ([a, b]\T))   = μ(E T) + μ(E ([a, b]\T)) = μ(E). Thus, μ(E) = μ(E



A) + μ(E

ii. Let b ∈ E. Then μ({b}) = 0 and

 ([a, b]\A)).

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

372



 A) + μ(E ([a, b]\A))   ≤ μ((E [a, b)) A)   +μ((E [a, b)) ([a, b]\A))  = μ(E [a, b))

μ(E) ≤ μ(E

≤ μ(E), i.e., μ(E) = μ(E



A) + μ(E

 ([a, b]\A)).

Consequently A is Lebesgue measurable. b. Let b ∈ A. Then A\{b} is Δ-measurable, and by the previous case, it follows that A\{b} is Lebesgue measurable. Since {b} is Lebesgue measurable and the union of two Lebesgue measurable sets is a Lebesgue measurable set, we conclude that the set A is a Lebesgue measurable set. 2. The fact that if A is a Lebesgue measurable set, then it is a Δ-measurable set follows similarly and we leave its proof to the reader as an exercise. Note that (i) and (ii) follow by (5) and (6) by Theorem 6.5. This completes the proof. Definition 6.4 The Lebesgue integral associated with the measure μΔ we call the Lebesgue Δ-integral on T. For a set E ⊆ T and a (measurable) function f : E → R, the corresponding integral of f over E we denote by f (t)Δt. E

So, all theorems of the general Lebesgue integration theory, including the Lebesgue dominated convergence theorem, will be held for the Lebesgue deltaintegral on T. Below is a comparison of the Lebesgue Δ-integral with the Riemann Δ-integral. Theorem 6.7 Let [a, b) be a half-closed bounded interval in T and let f be a bounded real-valued function on [a, b). If f is Riemann Δ-integrable from a to b, then f is Lebesgue Δ-integrable on [a, b), and b

R a

b

f (t)Δt = L

f (t)Δt, a

where R and L indicate the Riemann and Lebesgue integrals, respectively.

(6.2)

6.1 The Lebesgue Delta-Integral

373

Proof Let f be a Riemann Δ-integrable from a to b. Then for each positive integer k we can choose a δk > 0, δk → 0, as k → ∞, and a partition (k)

(k)

k =b Pk : a = t0 < t1 < . . . < tn(k)

of the interval [a, b) such that Pk ∈ Pδk and U (f, Pk ) − L(f, Pk )
δ and ρ(tj ) = tj −1 , U (f, Pk ) and L(f, Pk ) are the upper Darboux Δ-sum and the lower Darboux Δ-sum of f with respect to Pk . Then b

lim L(f, Pk ) = lim U (f, Pk ) = R

k→∞

k→∞

f (t)Δt. a

By replacing the partitions Pk with finer partitions if necessary, we can assume that for each k the partition Pk+1 is a refinement of the partition Pk . Set (k)

(k)

(k)

mj = inf{f (t) : t ∈ [tj −1 , tj )}, (k) Mj(k) = sup{f (t) : t ∈ [tj(k) −1 , tj )},

j = 1, 2, . . . , n(k).

Define the sequences {φk } and {Φk } of functions on [a, b) such that φk (t) = m(k) j

and

Φk (t) = Mj(k) ,

(k) t ∈ [tj(k) −1 , tj ),

j = 1, 2, . . . , n(k). We have that {φk } is a nondecreasing sequence and {Φk } is a nonincreasing sequence. Also, for each positive integer k, we have φk ≤ φk+1 , Φk ≥ Φk+1 , φk ≤ f ≤ Φk , L L

[a,b)

[a,b)

φk (t)Δt = L(f, Pk ), Φk (t)Δt = U (f, Pk ).

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

374

Because f is bounded, we have that the sequences {φk } and {Φk } are bounded. Therefore φ(t) = lim φk (t), k→∞

Φ(t) = lim Φk (t),

t ∈ [a, b).

φ(t) ≤ f (t) ≤ Φ(t),

t ∈ [a, b),

k→∞

We have

and φ and Φ are Δ-measurable functions on [a, b). By the Lebesgue dominated theorem, we obtain lim L

k→∞

φk (t)Δt = L

[a,b)

lim L

k→∞

[a,b)

Φk (t)Δt = L

[a,b)

[a,b)

φ(t)Δt, Φ(t)Δt.

Therefore L

[a,b)

φ(t)Δt = lim

k→∞ [a,b)

φk (t)Δt

= lim L(f, Pk ) k→∞

b

=R

f (t)Δt a

= lim U (f, Pk ) k→∞

= lim L k→∞

=L

[a,b)

[a,b)

Φk (t)Δt

Φ(t)Δt.

Hence, L

[a,b)

(Φ(t) − φ(t))Δt = 0

and using that φ(t) = Φ(t),

t ∈ [a, b),

6.1 The Lebesgue Delta-Integral

375

we get φ(t) = Φ(t) for Δ-almost every t ∈ [a, b). Consequently φ(t) = f (t) = Φ(t) for Δ-almost every t ∈ [a, b). Therefore f is Lebesgue Δ-integrable and (6.2) holds. This completes the proof. Theorem 6.8 Let f be a bounded function defined on the half-closed bounded interval [a, b) of T. Then f is Riemann Δ-integrable from a to b if and only if the set of all right-dense points of [a, b) at which f is discontinuous is a set of Δ-measure zeros. Proof 1. Suppose that f is Riemann Δ-integrable from a to b. For each positive integer k let Pk , φk , Φk , φ, and Φ be defined as in the proof of Theorem 6.7. Let Λ=

∞ "

Pk ,

k=1

Λrd = {t ∈ [a, b) : t ∈ Λ and G = {t ∈ [a, b) : f Grd = {t ∈ G : t

t

is right-dense},

is discontinuous

at

t},

is right-dense},

A = {t ∈ [a, b) : φ(t) = Φ(t)}. Let t ∈ [a, b) be such that φ(t) = f (t) = Φ(t) and t ∈ / Λ. Assume that f is not continuous at t. Then there exist an  > 0 and a sequence {tj }j ∈N such that tj → t, as j → ∞, and |f (tj ) − f (t)| >  for any j ∈ N. Hence, f (tj ) >  + f (t) and Φ(t) ≥  + φ(t),

376

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

which is a contradiction. Therefore f is continuous at t. Observe that all rightscattered points of [a, b) belong to Λ. Hence, for each right-scattered point t of [a, b) and all sufficiently large k, we have φk (t) = Φk (t) = f (t) and from here, φ(t) = Φ(t) = f (t). Therefore Grd ⊂ A

"

Λrd .

By the proof of Theorem 6.7, it follows that φ(t) = Φ(t) for Δ-almost every t in [a, b). Therefore μΔ (A) = μΔ (Λrd ) = 0 and hence, μΔ (Grd ) = 0. 2. Suppose that the set of all right-dense points of [a, b) at which f is discontinuous is of Δ-measure zero. Then μΔ (Grd ) = 0. For each positive integer k we choose δk > 0, δk → 0, as k → ∞, and a partition (k)

(k)

(k)

Pk : a = t0 < t1 < . . . < tn(k) = b of [a, b) such that Pk ∈ Pδk and Pk+1 is a refinement of Pk . Let φk , Φk , φ, and Φ be defined as in the proof of Theorem 6.7. Suppose that t ∈ [a, b) is rightdense and f is continuous at t. Then for a given  > 0, there exists a δ > 0 such that sup f − inf f < , where the supremum and the infimum are taken over the interval (t −δ, t +δ). For all k sufficiently large, a subinterval of Pk containing t will be lain in (t −δ, t +δ) and then Φk (t) − φk (t) < .

6.2 Absolutely Continuous Functions

377

Since  > 0 was arbitrarily chosen, we conclude that φ(t) = Φ(t). Next, at each right-scattered point t of [a, b), using the first part of the proof, we have φ(t) = Φ(t). Consequently A ⊂ Grd and using that μΔ (Grd ) = 0, we conclude that μΔ (A) = 0. From here, φ(t) = Φ(t) Δ-almost everywhere on [a, b) and L

[a,b)

φ(t)Δt = L

[a,b)

Φ(t)Δt.

Hence, using the proof of Theorem 6.7, we obtain lim L(f, Pk ) = lim U (f, Pk )

k→∞

k→∞

and thus, f is Riemann Δ-integrable on [a, b). This completes the proof.

6.2 Absolutely Continuous Functions Let a, b ∈ T, a < b. For a given function f : T → R, consider an auxiliary function f : [a, b] → R defined by

f (t) =

⎧ ⎪ ⎨ f (t)

if t ∈ T

⎪ ⎩ f (t ) + i

f (σ (ti ))−f (ti ) (t μ(ti )

− ti )

if t ∈ (ti , σ (ti )),

for some

i ∈ I, (6.3)

with I ⊂ N and {ti }i∈I given by (6.1). Theorem 6.9 Let f : T → R and f : [a, b] → R be the extension of f , defined by (6.3). Then the following statements are equivalent: (i) f maps every Δ-null subset of T into a null set. (ii) f maps every null subset of [a, b] into a null set.

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

378

Proof 1. Suppose (i). Let E ⊂ [a, b] be a null set. We will prove that f (E) is a null set. a. Let b ∈ / E. Then E=

 "

(E



 [ti , σ (ti ))

 "  E (T\R)

i∈I

and μ(f (E)) ≤

   μ f (E [ti , σ (ti )]) i∈I  +μ(f (E (T\R))). %

  Fix i ∈ I . By the definition of the function f , it follows that f  absolutely continuous. Hence, μ((E

(6.4)

[ti ,σ (ti )]

is

 [ti , σ (ti ))) = 0.

Now, applying the Banach–Zarecki theorem,1 we get μ(f (E

 [ti , σ (ti )))) = 0.

(6.5)

 Moreover,  since the set E (T\R) does not have any right-scattered points and b ∈ / E (T\R), using Theorem 6.5 (6), we get μΔ (E

  (T\R)) = μ(E (T\R))

= 0.   Hence by f  = f and our assumption (i), we obtain T

μ(f (E

  (T\R))) = μ(f (E (T\R)) = 0.

1

Theorem 6.10 (The Banach–Zarecki Theorem, [25]) A real-valued function g : X ⊂ R → R is an absolutely continuous function on the closed interval I if and only if the following conditions are satisfied: (i) g is continuous and of bounded variation. (ii) g maps every null subset of X into a null set.

6.2 Absolutely Continuous Functions

379

From the last relation and from (6.4), (6.5), we get μ(f (E)) = 0. b. Let b ∈ E. Then we consider both sets E\{b} and {b}. Since μ(f ({b})) = 0, we obtain μ(f (E)) ≤ μ(f (E\{b})) + μ(f ({b})) = 0, i.e., μ(f (E)) = 0. 2. Suppose (ii). Let E ⊂ T be a Δ-null set. a. Let b ∈ / E. Then, by Theorem 6.6 (i), we obtain 0 = μΔ (E) ( = (σ (ti ) − ti ) + μ(E) i∈IE

=

(

(σ (ti ) − ti ).

i∈IE

Therefore, the set E does not have any right-scattered points. So, f (E) = f (E) and (i) holds. b. Let b ∈ E. Consider both sets E\{b} and {b}. By the previous case, it follows that E\{b} does not have any right-scattered points. Then f (E\{b}) is a null set. Since μ({b}) = 0, then, using (ii), we have that f ({b}) is a null set. Consequently f (E) is a null set. This completes the proof. Definition 6.5 For a given function f {x0 , x1 , . . . , xn } of T, define

: T → R and a partition P

=

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

380

N (

V (P , f ) =

|f (xi ) − f (xi−1 )|.

k=1

We define the total variation of the function f as follows: b 1 (f ) = sup{V (P , f ) : P

partition of

T}.

a

If T.

b 2 (f ) < ∞, we say that the function f is of bounded variation on the time scale a

Note that if f : T → R is of bounded variation, then f : T → R, defined by (6.3), is of bounded variation. Definition 6.6 A function f : T → R is said to be absolutely continuous on T if  for every  > 0 there exists a δ > 0 such that if {(ak , bk ) T}nk=1 , with ak , bk ∈ T, is a finite partition disjoint family of subintervals of T satisfying n (

(bk − ak ) < δ,

k=1

then n (

|f (bk ) − f (ak )| < .

k=1

Theorem 6.11 A function f : T → R is absolutely continuous on T if and only if the following conditions hold: (i) f is continuous and of bounded variation on T. (ii) f maps every Δ-null subset of T into a null set. Proof Let f be the extension of the function f , defined by (6.3). 1. Let f be absolutely continuous on T. Then f is continuous on T. Now, we will prove that it is of bounded variation on T. Let δ > 0 correspond to  = 1 as in the definition of absolutely continuous function on T. Then there exists a partition P = {x0 , x1 , . . . , xn } such that for any k ∈ {1, . . . , n} either xk − xk−1 ≤

δ 2

6.2 Absolutely Continuous Functions

381

or δ 2

xk − xk−1 >

and

σ (xk−1 ) = xk .

If xk − xk−1 ≤ 2δ , then xk 1

(f ) ≤ 1.

xk−1

  Thus, f  then

[xk−1 ,xk ]



T

is of bounded variation on [xk−1 , xk ] xk 1



T. If xk −xk−1 > 2δ ,

(f ) = |f (xk ) − f (xk−1 )|

xk−1

   and f  is of bounded variation on [xk−1 , xk ] T. Therefore  [xk−1 ,xk ] T   f is of bounded variation on [xk−1 , xk ]. Consequently f is of bounded [xk−1 ,xk ]

variation on T. Let now E ⊂ T be a Δ-null set. We will prove that f (E) is a null set. Let  > 0 be arbitrarily chosen and δ > 0 corresponds to  as in the definition for absolutely continuous function. Because μΔ (E) = 0, we have that b ∈ / E and  it has not any right-scattered points. Then we may choose a family {[ak , bk ] T}∞ k=1 with ak , bk ∈ T of pairwise disjoint subintervals of T such that ∞ "

E⊂

([ak , bk ]



T)

k=1

and ∞ (

(bk − ak ) < δ.

k=1

Hence, f (E) ⊂ f

∞ "

([ak , bk )



 T)

k=1

=

∞ " k=1

f ([ak , bk )



T)

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and  ∞   " μ(f (E)) ⊂ μ f ([ak , bk ) T) k=1 ∞ "

=

 μ f ([ak , bk )



 T) .

k=1

Becausef is continuous on [ak , bk ] [ak , bk ] T, ck < dk , such that



T, it follows that there exist ck , dk ∈

      T) = μ f ([ak , bk ] T) μ f ([ak , bk ) ≤ |f (dk ) − f (ck )|. Since [ck , dk ] ⊂ [ak , bk ] and f is absolutely continuous on T, we have n (

|f (dk ) − f (ck )| < 

k=1

and ∞ (

|f (dk ) − f (ck )| ≤ .

k=1

From here, μ(f (E)) ≤

∞ (

   μ f ([ak , bk ) T)

k=1

≤

∞ (

|f (dk ) − f (ck )|

k=1

≤ . Because  > 0 was arbitrarily chosen, we get that μ(f (E)) = 0. 2. Suppose (i) and (ii). By Theorem 6.9 (ii), it follows that f maps every null subset of [a, b] into a null set. Hence by the Banach–Zarecki theorem, it follows   that f is absolutely continuous on [a, b]. By the equality f = f  , we conclude T that f is absolutely continuous on T. This completes the proof.

6.3 Alternative Way for Defining of Lebesgue Type Measure and Integration. . .

383

6.3 Alternative Way for Defining of Lebesgue Type Measure and Integration over T Let a = inf T and b = sup T. In this section, we assume that −∞ < a < ρ(b) ≤ b < ∞. Define the function E : [a, b) → R as follows: E(t) = sup {s ∈ T : s ≤ t} ,

t ∈ [a, b].

Suppose that u : Tκ → R is an arbitrary function. Then the function u◦E : [a, b) → R is an extension of u to [a, b) and it is constant on “gaps” in T, with value equal to the value of u at left-hand end of the gap. Definition 6.7 We say that u is measurable, respectively integrable, if u ◦ E is measurable, respectively integrable, on the real interval [a, b) in the usual Lebesgue sense. Let L1 (T) denote the set of such integrable functions on T. Definition 6.8 For any u ∈ L1 (T), we define the integral of u by t

t

uΔ =

s

u ◦ Edx,

s, t ∈ T,

s

where the integration on the right-hand side is the standard Lebesgue integral over the real interval [s, t] ⊂ [a, b]. Using the above construction, we can also define a measure and a class of measurable subsets of T, and integrals over such sets. For a set A ⊆ T, with χA , we will denote the characteristic function of A. Definition 6.9 We say that A is measurable if χA is measurable in the sense of Definition 6.8. Definition 6.10 If A ⊆ T is measurable, we define the integral of u ∈ L1 (T) over A to be b

uΔ = A

uχA Δ. a

We define the measure μT (A) of A to be μT (A) = From Definition 6.10, it follows    μT [s, t) T = t − s,

1Δ. A

s, t ∈ T,

s < t,

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and μT ({b}) = 0,

μT ({t}) = σ (t) − t,

t ∈ T.

Also, if t ∈ T is right-scattered and u ∈ L1 (T), then u(t) is well-defined and σ (t)

uΔ = u(t) (σ (t) − t) .

t

Remark 6.2 1. From Definition 6.8, it follows that the most of the standard Lebesgue integration results on [a, b] hold immediately for the above integral on T. In particular, Minkowsky’s type inequality and Hölder’s type inequality, and the dominated convergence theorem ! t hold. 2. A nabla integral s u∇ could be obtained in a similar manner to the above integral, by replacing the function E with the function E1 (t) = inf {s ∈ T : s ≥ t} . We will not consider this further here. For u ∈ Crd (T), we define |u|0,T = sup {|u(t)| : t ∈ T} . For u ∈ C1rd (T), we define   |u|1,T = |u|0,T + uΔ 0,Tκ . Theorem 6.12 If u ∈ L1 (T) and s ∈ T, then the function t

Us (t) =

t ∈ T,

uΔ, s

is well-defined and Us ∈ C(T) with |Us |0,T ≤

b

|u|Δ.

a

Proof By Definition 6.8, we get t

Us (t) =

uΔ s t

= s

u ◦ Edx,

t ∈ T.

6.3 Alternative Way for Defining of Lebesgue Type Measure and Integration. . .

385

!t Hence, using the standard Lebesgue integration theory, the integral s u ◦ Edx defines an absolutely continuous function of t ∈ [a, b]. Therefore Us is well-defined on T. Also,  t     |Us (t)| =  u ◦ Edx  s

t

≤

|u ◦ E| dx

s b

≤

|u ◦ E|dx

a b

≤

|u|Δ,

t ∈ T.

a

Hence, |Us |0,T = sup {|Us (t)| : t ∈ T} b

≤

|u|Δ.

a

This completes the proof. Theorem 6.13 If u ∈ C1 (T), then t

u(t) − u(s) =

uΔ Δ,

s, t ∈ T.

s

Proof Let s ∈ T be fixed. We define the function t

H (t) = u(t) − u(s) −

uΔ Δ,

t ∈ T.

s

!t Since uΔ ∈ C(T), by Theorem 6.12, it follows that s uΔ Δ is a continuous function of t ∈ T. Therefore H is a continuous function on T. Let t ∈ T be arbitrarily chosen and fixed. We take  > 0 arbitrarily. Then   σ (t)  |H (σ (t)) − H (t)| = u (σ (t)) − u(s) − uΔ Δ − u(t) + u(s) +  s     σ (t)  Δ  = u (σ (t)) − u(t) − u Δ   t   = u (σ (t)) − u(t) − uΔ (t) (σ (t) − t) ≤ (σ (t) − t).

t s

   uΔ Δ 

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

386

In the last inequality we have used that u ∈ C1 (T). Since H is a continuous function on T, there is a δ > 0 such that for any z ∈ (t − δ, t + δ) we have |H (t) − H (z)| < . Therefore |H (σ (t)) − H (z)| = |H (σ (t)) − H (t) + H (t) − H (z)| ≤ |H (σ (t)) − H (t)| + |H (t) − H (z)| < (σ (t) − t) +  for any z ∈ (t − δ, t + δ). Since t ∈ T was arbitrarily chosen, we conclude that H Δ (t) exists and H Δ (t) = 0 for any t ∈ T. Hence, using the mean value theorem for time scale derivatives (see [6, Theorem 2.41]), we conclude that H is a constant on T. Because H (s) = 0, we obtain H ≡ 0 on T. This completes the proof.

6.4 The Fundamental Theorem of Calculus First, we will show some properties about Δ-differentiability and differentiability. Theorem 6.14 Let I ⊂ N and R = {ti }i∈I be given as in (6.1). Assume that f : T → R and f : [a, b] → R is the extension of f to [a, b], defined by (6.3). Then we have the following statements: 1. f is differentiable on the set (ti , σ (ti )) and i∈I 

f (t) =

f (σ (ti )) − f (ti ) , μ(ti )

t ∈ (ti , σ (ti )), for some i ∈ I .  2. If f is differentiable at t ∈ [a, b) T, then f is Δ-differentiable at t and its Δ-derivative at t satisfies 

f Δ (t) = f (t).  3. If f is Δ-differentiable at ([a, b) T)\(R σ (R)), then f is differentiable at t and its derivative at t satisfies the equality 

f (t) = f Δ (t).

6.4 The Fundamental Theorem of Calculus

387

Proof 

1. By the definition of f , it follows that there exists f (t) at any t ∈ and 

f (t) = 2. Let t ∈ [a, b)





i∈I (ti , σ (ti ))

f (σ (ti )) − f (ti ) . μ(ti )

T be such that f is differentiable at t.

a. Let t be right-scattered. Then there is a ti ∈ R such that t = ti . Since f is continuous at ti , then f is continuous at ti and so, f is Δ-differentiable at ti , and f Δ (ti ) =

f (σ (ti )) − f (ti ) μ(ti ) 

= f (ti ). b. Let t be right-dense. We have 

f (t) = lim s→t

= =

f (t) − f (s) t −s

f (t) − f (s) s→t,s∈T t −s lim

lim

s→t,s∈T

f (t) − f (s) t −s

= f Δ (t). Therefore f is Δ-differentiable at t and 

f Δ (t) = f (t).  3. Let t ∈ ([a, b) T)\(R σ (R)) and f be Δ-differentiable at t. Take  > 0 arbitrarily. Then there exists a δ > 0 such that       f (t)−f (s)   f (t)−f (s)  t−s − f Δ (t) =  t−s − f Δ (t)

(6.6)

0 for which |u(t)| ≤ C, |un (t)| ≤ C for any n ∈ N and for any t ∈ T. Hence, b

p

u − un Lp (T) =

|u − un |p Δ

a b

=

|u − un |p−1 |u − un |Δ

a b

≤

(|u| + |un |)p−1 |u − un |Δ

a b

≤ (2C)p−1

|u − un |Δ

a

= (2C)p−1 u − un L1 (T) → 0, as n → ∞. This completes the proof. Theorem 6.18 Suppose that {un }n∈N is a sequence in Lp (T) for some p ≥ 1. If u−un Lp (T) → 0, as n → ∞, for some u ∈ Lp (T), and if t ∈ T is right-scattered, then un (t) → u(t), as n → ∞. Proof Let  · Lp (a,b) denote the standard Lp (a, b)-norm on the real interval [a, b]. Then

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

392

p

0 = lim (u − un ) ◦ ELp (a,b) n→∞

= lim

n→∞ a

≥ lim

b

|(u − un ) ◦ E|p dx

σ (t)

n→∞ t

= lim

σ (t)

n→∞ t

|(u − un ) ◦ E|p dx |u − un |p Δ

= lim |u(t) − un (t)|p (σ (t) − t) n→∞

≥ 0. Consequently un (t) → u(t), as n → ∞. This completes the proof. Theorem 6.19 If {un }n∈N is a Cauchy sequence in Lp (T), then there exists a unique u ∈ Lp (T) such that u − un Lp (T) → 0, as n → ∞. Proof Note that the sequence {un ◦ E}n∈N is a Cauchy sequence in Lp (a, b). Then, by the standard Lebesgue theory, it follows that there is a unique z ∈ Lp (a, b) such that z − un ◦ ELp ((a,b)) → 0, as n → ∞. Suppose that t ∈ T is right-scattered. Then b

p

 (un − um ) ◦ ELp (T) =

|(un − um ) ◦ E|p dx

a σ (t)

≥

|(un − um ) ◦ E|p dx

t σ (t)

=

|un − um |p Δ

t

= |un (t) − um (t)|p (σ (t) − t) . Therefore the sequence {un (t)}n∈N is a Cauchy sequence. Thus, it converges to c(t) and the function z must equal to c(t) almost everywhere on the interval [t, σ (t)), and so we may suppose that this equality holds everywhere on [t, σ (t)). Since the set of right-scattered points t ∈ T is at most countable, we may suppose that this is true for all right-scattered points t ∈ T. Defining u to be the restriction of z to T, it follows that u ∈ Lp (T), z = u ◦ E, and u − un Lp (T) → 0, as n → ∞. This completes the proof.

6.6 Sobolev Type Spaces and Generalized Derivatives

393

Remark 6.3 1. By Theorem 6.19, it follows that Lp (T), p ≥ 1, is a Banach space with respect to the norm  · Lp (T) . 2. We can define a natural inner product on L2 (T) by u, vT =

b

uvΔ,

u, v ∈ L2 (T).

(6.11)

a

With respect to this inner product, the space L2 (T) is a Hilbert space. 3. To simplify the notation, from now on we will use the notation ·T for the norm  · L2 (T) . 4. The notations  · T and ·, ·T for the above norm and inner product on L2 (T) indicate that their values depend on the entire time scale T, even if u is not defined at b. Exercise 6.6 Prove that (6.11) satisfies all axioms for an inner product.

6.6 Sobolev Type Spaces and Generalized Derivatives Definition 6.11 For u ∈ C1rd (T), we define u21,T = u2T + uΔ 2T

(6.12)

and define the space H1 (T) ⊂ L2 (T) to be the completion of C1 (T) with respect to the norm u1,T . The space H1 (T) will be called Sobolev space over T. Exercise 6.7 Prove that (6.12) satisfies all axioms for a norm. Theorem 6.20 u ∈ H1 (T) if and only if there exists a function uΔg ∈ L2 (T) such that the following condition holds: there exists a sequence {un }n∈N in C1 (T) such Δg 2 1 Δg is that un → u and uΔ n → u , as n → ∞, in L (T). If u ∈ H (T), then u 2 1 Δ Δ g unique in the L (T) sense. If u ∈ C (T), then u = u . Proof 1. Let u ∈ H1 (T). By the definition of the space H1 (T), it follows that there exists a sequence {un }n∈N of elements of C1 (T) such that un − u#1,T $→ 0, as n → ∞. Hence, un − um 1,T → 0, as m, n → ∞. Consequently uΔ n n∈N is a Cauchy sequence in L2 (T). Because L2 (T) is a Banach space, there exists a unique uΔg ∈ Δg 2 1 L2 (T) such that uΔ n → u , as n → ∞, in L (T). In particular, if u ∈ C (T), Δ Δ then u = u g . 2. Let {un }n∈N be a sequence of elements of C1 (T) such that un → u and uΔ n → uΔg , as n → ∞, in L2 (T), for some uΔg ∈ L2 (T). Hence, using (6.12), we

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

394

conclude that {un }n∈N is a convergent sequence in H1 (T). By the definition of H1 (T), it follows that there exists u1 ∈ H1 (T) such that un → u1 , as n → ∞, in H1 (T). Hence, un → u1 , as n → ∞, in L2 (T). Because un → u, as n → ∞, in L2 (T), we conclude that u = u1 and u ∈ H1 (T). This completes the proof. Definition 6.12 For any u ∈ H1 (T), the function uΔg in Theorem 6.20 will be called generalized derivative of u. Theorem 6.21 If u ∈ H1 (T), then u ∈ C(T). There exists a constant C > 0 such that |u|0,T ≤ Cu1,T ,

u ∈ H1 (T).

(6.13)

Furthermore, t

u(t) − u(s) =

s, t ∈ T.

uΔg Δ,

(6.14)

s

Proof Let u ∈ C1 (T) and s, t ∈ T, s ≤ t. By Theorem 6.13, we get t

u(t) − u(s) =

uΔ Δ.

(6.15)

s

Hence,   |u(t)| = u(s) +

  uΔ Δ

t s

  ≤ |u(s)| + 

t s t

≤ |u(s)| +

  u Δ Δ

 Δ u  Δ

s

≤ |u(s)| + (t − s)

1 2



 Δ 2 u  Δ

t

 12

s

0 0 ≤ |u(s)| + (b − a) 0uΔ 0T . 1 2

Now we integrate the last inequality over T with respect to s and we get b a

b

|u(t)|Δ ≤

b

1

|u(s)|Δ + (b − a) 2

a

≤ (b − a)

a 1 2



b

|u(s)| Δ 2

 12

a 1 2

3

uΔ T Δ 0 3 0 + (b − a) 2 0uΔ 0T

= (b − a) uT + (b − a) 2 uΔ T

6.6 Sobolev Type Spaces and Generalized Derivatives

395

or 0 1 3 0 (b − a)|u(t)| ≤ (b − a) 2 uT + (b − a) 2 0uΔ 0T . Hence, |u(t)| ≤

0 1 0 uT + (b − a) 2 0uΔ 0T

1 (b − a)

1 2

and sup |u(t)| ≤ t∈T

1 (b − a)

1 2

0 1 0 uT + (b − a) 2 0uΔ 0T ,

or |u|0,T ≤

1 (b − a)

1 2

0 1 0 uT + (b − a) 2 0uΔ 0T .

Let  C = max

/

1 1

(b − a) 2

, (b − a)

1 2

.

Then 0 0   |u|0,T ≤ C uT + 0uΔ 0T = Cu1,T . By Theorem 6.20, we have that uΔ = uΔg . Hence by (6.15), we get (6.14). Let now u ∈ H1 (T). Then there exists a sequence {un }n∈N of elements of C1 (T) such that un → u, as n → ∞, in H1 (T). For this sequence we have |un − um |0,T ≤ Cun − um 1,T ,

m, n ∈ N.

(6.16)

Note that un → u, as n → ∞, in L2 (T). Because un −um 1,T → 0, as m, n → ∞, using (6.16), we get |un − um |0,T → 0, as m, n → ∞. Therefore {un }n∈N is a Cauchy sequence in C(T). Because C(T) is a Banach space, we conclude that the sequence {un }n∈N is a convergent sequence in C(T). Let u1 ∈ C(T) be its limit, i.e., un → u, as n → ∞, in C(T). Then

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

396

 2   un − u1  Δ

b

un − u1 2T =

a

 2   ≤ (b − a) un − u1 

0,T

→ 0, as n → ∞. Therefore un → u1 , as n → ∞, in L2 (T). Because un → u, as n → ∞, in L2 (T), we conclude that u = u1 and u ∈ C(T). From here, using that |un |0,T ≤ Cun 1,T , we obtain that u satisfies the inequality (6.13). Since u ∈ H1 (T), using Theorem 6.20, there exists a unique function uΔg ∈ L2 (T) and there exists a sequence {vn }n∈N of elements of C1 (T) such that vn → u and vnΔ → uΔg , as n → ∞, in L2 (T). Also, we have t

vn (t) − vn (s) =

Δ

vn g Δ, s, t ∈ T,

s ≤ t.

(6.17)

s

Note that    

t

t

Δ

vn g Δ −

s

s

    uΔg Δ =    = 

t s t s t

≤ s

t

vnΔ Δ −

s

  uΔg Δ

  Δ   vn − uΔg Δ

 Δ  v − uΔg  Δ n

≤ (t − s)

1 2



t s

 Δ  v − uΔg 2 Δ n

 12

0 0 ≤ (b − a) 0vnΔ − uΔg 0T 1 2

→ 0,

s ≤ t,

as n → ∞. Hence, using that vn → u, as n → ∞, in C(T), and (6.17), we conclude that u satisfies (6.14). This completes the proof. Theorem 6.22 Let u ∈ H1 (T). If uΔg = 0, then u is a constant on T. Proof By (6.14), we get u(t) = u(s) for any s, t ∈ T. This completes the proof.

6.6 Sobolev Type Spaces and Generalized Derivatives

397

Theorem 6.23 Let u ∈ H1 (T). If t ∈ T is a right-scattered point, then uΔg (t) =

u (σ (t)) − u(t) . σ (t) − t

Proof Let t ∈ T be a right-scattered point. Since u ∈ H1 (T), we have that u ∈ L2 (T) and there exist a sequence {un }n∈N of elements of C1 (T) and an unique Δg 2 uΔg ∈ L2 (T) such that un → u and uΔ n → u , as n → ∞, in L (T). Hence Δ Δ g by Theorem 6.18, we get that un (t) → u(t) and un (t) → u (t), as n → ∞. By (6.14), we have σ (t)

u (σ (t)) = u(t) +

uΔg Δ

t

and σ (t)

un (σ (t)) = un (t) + t

uΔ nΔ

= un (t) + uΔ n (t) (σ (t) − t) . Hence,   lim un (σ (t)) = lim un (t) + uΔ n (t) (σ (t) − t)

n→∞

n→∞

σ (t)

= u(t) +

uΔg Δ

t

= u (σ (t)) . Therefore uΔg (t) = lim uΔ n (t) n→∞

= lim

n→∞

=

un (σ (t)) − un (t) σ (t) − t

u (σ (t)) − u(t) . σ (t) − t

This completes the proof. Theorem 6.24 Let u, v ∈ H1 (T) and α, β ∈ R. Then αu + βv ∈ H1 (T) and (αu + βv)Δg = αuΔg + βv Δg .

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

398

Proof Since u, v ∈ H1 (T), there exist sequences {un }n∈N , {vn }n∈N of elements of C1 (T) and unique uΔg , v Δg ∈ L2 (T) such that un → u,

vn → v,

Δg uΔ n →u ,

vnΔ → v Δg ,

as n → ∞, in L2 (T). Hence, αun + βvn → αu + βv,

Δ Δg αuΔ + βv Δg , n + βvn → αu

as n → ∞, in L2 (T). Hence by Theorem 6.20, it follows that αu + βv ∈ H1 (T). Applying again Theorem 6.20, we have that there exists a unique (αu + βv)Δg ∈ L2 (T) such that (αun + βvn )Δ → (αu + βv)Δg , as n → ∞, in L2 (T). Since Δ (αun + βvn )Δ = αuΔ n + βvn

→ αuΔg + βv Δg , as n → ∞, in L2 (T), we conclude that (αu + βv)Δg = αuΔg + βv Δg . This completes the proof. Theorem 6.25 Let u, v ∈ H1 (T). Then uv ∈ H1 (T) and (uv)Δg = uΔg v + uσ v Δg = uv Δg + uΔg v σ ,

(6.18)

and !t s

uΔg vΔ +

!t s

!t !t uσ v Δg Δ = s uv Δg Δ + s uΔg v σ Δ = (uv)(t) − (uv)(s), s, t ∈ T.

Proof Since u, v ∈ H1 (T), there exist sequences {un }n∈N , {vn }n∈N of elements of C1 (T) and unique uΔg , v Δg ∈ L2 (T) such that un → u,

vn → v,

Δg uΔ n →u ,

as n → ∞, in L2 (T). Then un vn → uv,

vnΔ → v Δg ,

6.6 Sobolev Type Spaces and Generalized Derivatives

399

as n → ∞, in L2 (T), uΔg v,

uσ v Δg ,

uΔg v σ ∈ L2 (T),

uv Δg ,

uσn → uσ ,

vnσ → v σ ,

as n → ∞, in L2 (T). Hence, σ Δ (un vn )Δ = uΔ n vn + un vn

(6.19) →

uΔg v

+ uσ v Δg ,

σ (un vn )Δ = un vnΔ + uΔ n vn → uv Δg + uΔg v σ ,

(6.20)

as n → ∞, in L2 (T). Therefore uv ∈ H1 (T). Hence, using Theorem 6.20, there exists a unique (uv)Δg ∈ L2 (T) such that (un vn )Δ → (uv)Δg , as n → ∞, in L2 (T). From here and from (6.19), (6.20), we obtain (6.18). Hence by (6.14), we get t

t

uΔg vΔ +

s

t

uσ v Δg Δ =

s

 Δ  u g v + uσ v Δg Δ

s t

=

(uv)Δg Δ

s

= (uv)(t) − (uv)(s) and t

t

uv Δg Δ +

s

t

uΔg v σ Δ =

s

 Δ  uv g + uΔg v σ Δ

s t

=

(uv)Δg Δ

s

= (uv)(t) − (uv)(s), This completes the proof.

s, t ∈ T.

!t Theorem 6.26 Let u ∈ L2 (T) and U (t) = a uΔ, t ∈ T. Then U ∈ H1 (T), U Δg = u, and if V ∈ H 1 (T) satisfies V Δg = u, then U − V is a constant. In addition, there exists a constant C > 0 such that

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

400

U 1,T ≤ CuT . Proof Let u ∈ L2 (T). By Theorem 6.17, it follows that there exists a sequence {un }n∈N of elements of C(T) such that un → u, as n → ∞, in L2 (T). Since un ∈ C(T), n ∈ N, we have that t

Un (t) =

un Δ ∈ C1 (T) and

UnΔ (t) = un (t),

a

t ∈ T.

For n ∈ N, we have Un − U 2T =

b a

   

t

t

un Δ −

a

a

2  uΔ Δ

 2 b t    =  (un − u)Δ Δ a a  b t 2 (un − u) Δ (t − a)Δ ≤ a

a b

≤ (b − a)



a

b

 (un − u) Δ Δ 2

a

= (b − a)2 un − u2T → 0, UnΔ

− uT = un − uT → 0,

as n → ∞, i.e., Un → U , UnΔ → u, as n → ∞, in L2 (T). Hence by Theorem 6.20, it follows that U ∈ H1 (T) and U Δg = u. Next, U 2T

   

2  = uΔ Δ a a  t  b 2 (t − a) u Δ Δ ≤ b

t

a b

≤ (b − a)



a

= (b − a)2 u2T ,

Hence,

 2

u Δ Δ a

U Δg 2T = u2T .

b a

6.6 Sobolev Type Spaces and Generalized Derivatives

401

1  2 U 1,T = U 2T + U Δg 2T 1  2 ≤ (b − a)2 u2T + u2T 1  2 = 1 + (b − a)2 uT . Let V ∈ H1 (T) satisfy V Δg = u. Then, using Theorem 6.24, we obtain (U − V )Δg = U Δg − V Δg = u−u = 0. Hence by Theorem 6.22, we conclude that U − V is a constant. This completes the proof. Definition 6.13 The function U in Theorem 6.26 will be called anti-derivative of u. Theorem 6.27 We have C1rd (T) ⊂ H1 (T).

(6.21)

Proof Let v ∈ C1rd (T) be arbitrarily chosen and u = v Δ ∈ Crd (T). By Theorem 6.16, it follows that u ∈ L2 (T). Hence by Theorem 6.26, we have t

U=

uΔ ∈ H1 (T) and

U Δg = u.

a

Because v Δ = v Δg = u, by Theorem 6.26, it follows that there exists a constant C such that v = U + C ∈ H1 (T). Because v ∈ C1rd (T) was arbitrarily chosen and for it we get that it is an element of H1 (T), we obtain the inclusion (6.21). This completes the proof. Definition 6.14 Define the space

  H2 (T) = u ∈ C1 (T) : uΔ ∈ H1 Tκ with the norm u22,T = u2T + uΔ 21,T . Exercise 6.8 Prove that (6.22) satisfies all axioms for a norm.

(6.22)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

402

Theorem 6.28 If {un }n∈N is a bounded sequence in H1 (T), then {un }n∈N has a subsequence that converges in C(T). Proof Since {un }n∈N is a bounded sequence in H1 (T), there exists a constant M > 0 such that un 1,T ≤ M for any n ∈ N. Hence by Theorem 6.21, it follows that |un |0,T ≤ Cun 1,T ≤ CM  with C = max

1

 1 2 , for any n ∈ N. Therefore the sequence {un }n∈N , (b − a) 1

(b−a) 2

is a bounded sequence in C(T). Also, using Cauchy–Schwarz inequality and (6.14), we have  t   Δg   |un (s) − un (t)| =  un Δ s

t

≤ s

   Δg  un  Δ 1



t

≤ (t − s) 2 s

   12  Δg 2 un  Δ

1 2

≤ (t − s) un 1,T 1

s, t ∈ T,

s ≤ t.

1

s, t ∈ T,

s ≥ t.

≤ M(t − s) 2 , As in the above, |un (t) − un (s)| ≤ M(s − t) 2 ,

Therefore the sequence {un }n∈N is equi-continuous $ on T. Hence, using the Arzela– # Ascoli theorem, there exists a subsequence unm m∈N of the sequence {un }n∈N that converges in C(T). This completes the proof. Theorem 6.29 The embedding Hi+1 (T) → C(T), is compact.

i = 0, 1,

6.6 Sobolev Type Spaces and Generalized Derivatives

403

Proof Let i = 0. By Theorem 6.21, we have |u|0,T ≤ Cu1,T ,  with C = max

1

 1 2 . Therefore , (b − a) 1

(b−a) 2

H1 (T) → C(T).

(6.23)

By Theorem 6.28, we have that every bounded sequence in H1 (T) has a subsequence that converges in C(T). Therefore the embedding (6.23) is compact. Let i = 1. Suppose that u ∈ H2 (T). By the definition of H2 (T), it follows that u ∈ C1 (T). Therefore H2 (T) → C1 (T).

(6.24)

Let {un }n∈N be a bounded sequence in H2 (T). Then {uΔ n }n∈N is a bounded sequence 1 (T). Hence by Theorem 6.28, it follows that there exists a subsequence in H # Δ$ unm m∈N of the sequence {uΔ n }n∈N that converges in C(T). From here, it follows # Δ$ that unm m∈N is a bounded sequence in C(T) and then there exists a positive constant M such that  Δ  u (t) ≤ M nm for any t ∈ T and for any m ∈ N. Note that     un (t) − un (s) =  m m 

t s t

≤ s

   uΔ Δ nm 

 Δ u  Δ nm

≤ M(t − s),

s, t ∈ T,

s ≤ t.

  un (t) − un (s) ≤ M(s − t), m m

s, t ∈ T,

s ≥ t.

As in the above,

# $ Therefore unm m∈N is equi-continuous on T. Since unm ∈ C1 (T), we have that unm ∈ H1 (T) for any m ∈ N. By Theorem 6.21, it follows that   un  m

0,T

0 0 ≤ C 0unm 01,T 0 0 ≤ C 0unm 02,T

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

404



1 2



, (b − a) , for any m ∈ N. Since {un }n∈N is a bounded $ $ # sequence in H2 (T) and unm m∈N is its subsequence, we conclude that unm m∈N is a bounded sequence in C(T). by the Arzela–Ascoli theorem, it follows that 

Hence $ # of the sequence unm m∈N that converges in there exists a subsequence unmk k∈N 

$ # Δ C(T). Because uΔ converges in C(T) we have that u converges nm m∈N n mk k∈N 

in C(T). Consequently the sequence unmk converges in C1 (T). Therefore the k∈N embedding (6.24) is compact. This completes the proof. with C = max

1

1

(b−a) 2 #

Theorem 6.30 For any  > 0 there exists C() > 0 such that 0 0   |u|0,T ≤ uΔg T + C() uT + 0uσ 0T ,

u ∈ H1 (T).

(6.25)

Proof Let u ∈ H1 (T). Then there exists a sequence {un }n∈N of elements of C1 (T) Δg 2 such that un → u, uΔ n → u , as n → ∞, in L (T). Take n ∈ N arbitrarily. Consider an arbitrary t0 ∈ T. Suppose that   2 2  T = {t0 }, T0, = t0 − , t0 + 2 2 i.e., t0 is not isolated in T. Let i0 = inf T0, ,

s0 = sup T0, .

We have 0 < s 0 − i0 ≤  2 . Then   |un (t)| = un (s) +

   uΔ Δ n 

t s

  ≤ |un (s)| + 

t s t

≤ |un (s)| + s

   uΔ Δ n 

 Δ u  Δ n

≤ |un (s)| + (t − s)

1 2



0 0 0 ≤ |un (s)| +  0uΔ n T,

t s

 Δ 2 u  Δ n

 12

s, t ∈ T0, ,

s ≤ t.

6.6 Sobolev Type Spaces and Generalized Derivatives

405

As in the above, 0 0 0 |un (t)| ≤ |un (s)| +  0uΔ n T,

s, t ∈ T0, ,

s ≥ t.

Integrating over T0, with respect to s, we get 2

(s0 − i0 )|un (t)| ≤

t0 + 2 2 t0 − 2

0 0 0 |un (s)|Δ + (s0 − i0 ) 0uΔ n T ⎛

≤ (s0 − i0 ) 2 ⎝ 1

2

t0 + 2 2

t0 − 2

⎞1 2

0 0 0 |un (s)|2 Δ⎠ + (s0 − i0 ) 0uΔ n T

0 0 1 0 ≤ (s0 − i0 ) 2 un T + (s0 − i0 ) 0uΔ n T,

t ∈ T0, .

Hence, 0 0 1 0 |un (t)| ≤ (s0 − i0 )− 2 un T +  0uΔ n T,

t ∈ T0, .

(6.26)

Now, we suppose that T0, = {t0 }, i.e., t0 is isolated in T. If σ (t0 ) > t0 , then |un (t0 )| = (σ (t0 ) − t0 )−1 ≤ (σ (t0 ) − t0 )

− 12

! σ (t0 )

|un |Δ t 0! 1 σ (t0 ) 2Δ 2 |u | n t0

(6.27)

1

≤ (σ (t0 ) − t0 )− 2 un T . If t0 > ρ(t0 ), then |un (t0 )| = (t0 − ρ(t0 ))−1 1

≤ (t0 − ρ(t0 ))− 2 ≤ (t0 − ρ(t0 ))

− 12

 σ u  Δ n ! 1 2 t0  σ 2 ρ(t0 ) un Δ 0 σ0 0u 0 . n T

! t0

ρ(t0 )

(6.28)

Suppose that there exists  > 0 such that for each m ∈ N and tm ∈ T we have 0 0 0 σ0   0 0 0 |un (tm )| >  0uΔ n T + m un T + un T ,

t m → t0 .

(6.29)

If t0 is isolated, (6.29) contradicts with (6.27) and (6.28). If t0 is not isolated, then (6.29) contradicts with (6.26). Consequently for each  > 0 there exists C() > 0 such that 0 σ0   0 0 |un |0,T ≤ uΔ n T + C() un T + un T .

(6.30)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

406

Because n ∈ N was arbitrarily chosen and |un |0,T → |u|0,T , un T → uT ,

0 Δ0 0 0 0u 0 → 0uΔg 0 , n T T 0 σ0 0 0 0u 0 → 0uσ 0 , n T T

as n → ∞, by (6.30), we get (6.25). This completes the proof. Theorem 6.31 For any  > 0 there exists C() > 0 such that 0 0 |u|0,Tκ ≤ uΔg T + C() 0uσ 0T ,

u ∈ H1 (T).

(6.31)

Proof Let u ∈ H1 (T). Then there exists a sequence {un }n∈N of elements of C1 (T) Δg 2 such that un → u, uΔ n → u , as n → ∞, in L (T). Take n ∈ N arbitrarily. Consider an arbitrary t0 ∈ Tκ . Suppose that   2 2  T0, = t0 − , t0 + Tκ = {t0 }, 2 2 i.e., t0 is not isolated in T. Let i0 = inf T0, ,

s0 = sup T0, .

We have 0 < s 0 − i0 ≤  2 . Then   |un (t)| = uσn (s) +

t σ (s)

    ≤ uσn (s) +    ≤ uσn (s) +

   uΔ Δ n 

t σ (s) t σ (s)

   uΔ Δ n 

 Δ u  Δ n

  1 ≤ uσn (s) + (t − σ (s)) 2



t σ (s)

 Δ 2 u  Δ

 12

n

0 0   0 ≤ uσn (s) +  0uΔ n T,

s, σ (s), t ∈ T0, ,

σ (s) ≤ t.

0 0   0 |un (t)| ≤ uσn (s) +  0uΔ n T,

s, σ (s), t ∈ T0, ,

σ (s) ≥ t.

As in the above,

6.6 Sobolev Type Spaces and Generalized Derivatives

407

Integrating over T0, with respect to s, we get 2

(s0 − i0 )|un (t)| ≤

t0 + 2 2 t0 − 2

0 0  σ  u (s) Δ + (s0 − i0 ) 0uΔ 0 n

⎛ ≤ (s0 − i0 ) 2 ⎝ 1

n

2

t0 + 2 2

t0 − 2

T

⎞1 2 0 0  σ 2 u (s) Δ⎠ + (s0 − i0 ) 0uΔ 0 n n T

0 0 1 0 ≤ (s0 − i0 ) 2 un T + (s0 − i0 ) 0uΔ n T,

t ∈ T0, .

Hence, 0 0 1 0 |un (t)| ≤ (s0 − i0 )− 2 uσn T +  0uΔ n T,

t ∈ T0, .

(6.32)

Now we suppose that T0, = {t0 }, i.e., t0 is isolated in T. Then |un (t0 )| = (t0 − ρ(t0 ))−1 1

≤ (t0 − ρ(t0 ))− 2 ≤ (t0 − ρ(t0 ))

− 12

 σ u  Δ n ! 1 2 t0  σ 2 ρ(t0 ) un Δ 0 σ0 0u 0 . n T

! t0

ρ(t0 )

(6.33)

Suppose that there exists  > 0 such that for each m ∈ N and tm ∈ Tκ we have 0 0 0 σ0 0 0 0 |un (tm )| >  0uΔ n T + m un T ,

tm → t0 .

(6.34)

If t0 is isolated, then (6.34) contradicts with (6.33). If t0 is not isolated, then (6.34) contradicts with (6.32). Consequently for each  > 0 there exists C() > 0 such that 0 σ0 0 0 |un |0,T ≤ uΔ n T + C() un T . Because n ∈ N was arbitrarily chosen and 0 0 0 Δ 0 0 g0 , 0 |un |0,T → |u|0,T , 0uΔ n T → u T 0 0 0 σ0 0u 0 → 0uσ 0 , n T T as n → ∞, by (6.35), we get (6.31). This completes the proof.

(6.35)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

408

6.7 Weak Solutions of Dynamic Systems Definition 6.15 Let f : T → R be a Lebesgue measurable function. If |f (t)|Δt < ∞ K

on all compact subsets K of T, then we say that f is locally integrable. The set of all such functions we will denote by L1loc (T). Define Γ1 (T) = {p ∈ L1loc (T) : 1 + μ(t)p(t) = 0}. Note that for any p ∈ Γ1 (T) and a, b ∈ R fixed, there exist positive constants a1 and a2 such that  T, a1 ≤ |1 + μ(t)p(t)| ≤ a2 , t ∈ [a, b] and the set {t ∈ T : 1 + μ(t)p(t) < 0} ⊂ T is finite. Definition 6.16 For p ∈ Γ1 (T) and s, t ∈ T, define  ξμ(t) (p(t)) =

Log(1+μ(t)p(t)) μ(t)

if μ(t) = 0 if μ(t) = 0,

p(t)

and ep (t, s) = e

!

[s,t) ξμ(τ ) (p(τ ))Δτ

Exercise 6.9 Let p ∈ Γ1 (T). Prove that ξμ(·) (p(·)) ∈ L1loc (T). Definition 6.17 For p, q ∈ Γ1 (T), define p ⊕ q = p + q + μpq, p , p = − 1 + μp

.

6.7 Weak Solutions of Dynamic Systems

409

pq =

p−q . 1 + μq

Exercise 6.10 Let p, q ∈ Γ1 (T). Prove that p ⊕ q,

p  q,

p ∈ Γ1 (T).

Exercise 6.11 Let p, q ∈ Γ1 (T) and s, t, r ∈ T. Prove that: e0 (t, s) = 1 ep (t, t) = 1 ep (t, s)ep (s, r) = ep (t, r) ep (σ (t), s) = (1 + μ(t)p(t))ep (t, s) 1 ep (t, s) = ep (s,t) = ep (s, t) ep (t, s)eq (t, s) = ep⊕q (t, s) ep (t,s) eq (t,s) = epq (t, s) ep (·, s) ∈ Crd (T) Δ  ep (·, s) = p(·)ep (·, s) Δ − a.e. on T  Δ 10. ep (s, ·) = −p(·)ep (s, σ (·)) Δ − a.e. on 1. 2. 3. 4. 5. 6. 7. 8. 9.

T

Theorem 6.32 Let x ∈ Crd (T), p, f ∈ L1 (T), p ≥ 0 on T, a ∈ T, and x Δ (t) ≤ p(t)x(t) + f (t) Δ − a.e.

on

T.

Then x(t) ≤ ep (t, a)x(a) +

[a,t)

ep (t, σ (τ ))f (τ )Δτ,

t ∈ T.

Proof Since p ∈ L1 (T) and p ≥ 0 on T, we have that p ∈ Γ1 (T) and 1 + μ(t)p(t) > 0,

t ∈ T.

Therefore, for any s, t ∈ T, we have ep (t, s) > 0,

ep (t, s) > 0.

Next, Δ  x(·)ep (·, s) (t) = x Δ (t)ep (σ (t), a) + x(t)(p)(t)ep (t, a) =

p(t)x(t) x Δ (t) ep (t, a) − ep (t, a) 1 + μ(t)p(t) 1 + μ(t)p(t)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

410

x Δ (t) − p(t)x(t) ep (t, a) 1 + μ(t)p(t)   = x Δ (t) − p(t)x(t) ep (σ (t), a)

=

Δ − a.e.

on

T.

Therefore x(t)ep (t, a) − x(a) = ≤

[a,t)

[a,t)

 Δ  x (τ ) − p(τ )x(τ ) ep (σ (τ ), a)Δτ f (τ )ep (σ (τ ), a)Δτ,

t ∈ T.

Hence, x(t)ep (t, a) ≤ x(a) +

t ∈ T,

f (τ )ep (σ (τ ), a)Δτ,

[a,t)

and x(t) ≤ x(a)ep (t, a) + = x(a)ep (t, a) + = x(a)ep (t, a) +

[a,t)

[a,t)

[a,t)

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

t ∈ T.

This completes the proof. Theorem 6.33 Let x ∈ Crd (T), p, g ∈ L1 T), x ≥ 0, g ≥ 0 on T, α ≥ 0, λ ∈ (0, 1), a, b ∈ T, a < b, x(t) ≤ α +

[a,t)

p(τ )x(τ )Δτ +

[a,t)

g(τ )(x(σ (τ )))λ Δτ,

t ∈ [a, b].

Then there exists a positive constant M such that x(t) ≤ M,

t ∈ [a, b].

Proof Let y(t) = α +

[a,t)

p(τ )x(τ )Δτ +

[a,t)

g(τ )(x(σ (τ )))λ Δτ,

t ∈ [a, b].

6.7 Weak Solutions of Dynamic Systems

411

Then x(t) ≤ y(t), and y is differentiable Δ − a.e.

on

t ∈ [a, b],

[a, b], y(a) = α. We have

y Δ (t) = p(t)x(t) + g(t)(x(σ (t)))λ ≤ p(t)y(t) + g(t)(y(σ (t)))λ

Δ − a.e.

on

[a, b].

Hence by Theorem 6.32, it follows that y(t) ≤ αep (t, a) +

[a,t)

ep (t, σ (τ ))g(τ )(y(σ (τ )))λ Δτ

≤ αep (b, a) + ep (b, a) ≤ αep (b, a) + ep (b, a) ≤ αep (b, a) + ep (b, a)

[a,t)

[a,t)

[a,b)

ep (a, σ (τ ))g(τ )(y(σ (τ )))λ Δτ g(τ )(y(σ (τ )))λ Δτ g(τ )(y(σ (τ )))λ Δτ,

t ∈ [a, b].

Define q(t) = αep (b, a) + ep (b, a) +ep (b, a)

[a,t)

[a,b)

g(τ )(y(σ (τ )))λ Δτ

g(τ )(y(σ (τ )))λ Δτ,

t ∈ [a, b].

Then q is monotone increasing on [a, b] and y(t) ≤ q(a), y(σ (t)) ≤ q(a),

t ∈ [a, b],

q(b) = αep (b, a) + 2ep (b, a)

[a,b)

g(τ )(y(σ (τ )))λ Δτ

= −αep (b, a) + 2q(a), and q Δ (t) = ep (b, a)g(t)(y(σ (t)))λ ≤ ep (b, a)g(t)(q(a))λ ≤ ep (b, a)g(t)(q(t))λ

Δ − a.e.

on [a, b].

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

412

Hence, q(t) − q(a) ≤ ep (b, a)

[a,t)

g(τ )(q(τ ))λ Δτ

≤ (q(t))λ ep (b, a)

[a,t)

t ∈ [a, b].

g(τ )Δτ,

From here, (q(t))1−λ − (q(a))1−λ ≤ (q(t))1−λ − (q(t))−λ q(a) ≤ ep (b, a)

[a,t)

g(τ )Δτ,

t ∈ [a, b].

Hence, (q(b))1−λ − (q(a))1−λ ≤ ep (b, a)

[a,b)

g(τ )Δτ

or (2q(a) − αep (b, a))1−λ − (q(a))1−λ ≤ ep (b, a)

[a,b)

g(τ )Δτ.

Let h(z) = (2z − αep (b, a))1−λ − z1−λ ,

z ∈ R.

We have lim h(z) = lim

z→∞

h(z)

z→∞ z1−λ

z1−λ



= lim

z→∞

αep (b, a) 2− z

1−λ

= ∞. Hence, there is a positive constant M such that q(a) ≤ M and x(t) ≤ y(t) ≤ q(a) ≤ M, This completes the proof.

t ∈ [a, b].

 − 1 z1−λ

6.7 Weak Solutions of Dynamic Systems

413

Consider the following IVP: x Δ (t) + p(t)x σ (t) = f (t) x(a) = x0 ,

(6.36)

where p ∈ Γ1 (T), f ∈ L1 (T), x0 ∈ R is given, a ∈ T. Note that the integral

[a,·)

ep (·, τ )f (τ )Δτ

is well-defined. Definition 6.18 The function x ∈ Crd (T) given by x(t) = ep (t, a)x0 +

[a,t)

ep (t, τ )f (τ )Δτ,

t ∈ T,

(6.37)

is said to be a weak solution of the IVP (6.36). If x ∈ Crd (T) is a weak solution of the IVP (6.36) and if it is given by the expression (6.37), we have x Δ (t) + p(t)x σ (t) = f (t)

Δ − a.e.

on

T.

Now we consider the IVP x Δ (t) + p(t)x σ (t) = f (t, x(t), x σ (t)), x(a) = x0 ,

a ≤ t < b,

where a, b ∈ T, a < b, p ∈ Γ1 (T), (H1) f : T × R × R → R is Δ-measurable in t ∈ T and |f (t, x1 , y1 ) − f (t, x2 , y2 )| ≤ L (|x1 − x2 | + |y1 − y2 |) for all t ∈ T, x1 , x2 , y1 , y2 ∈ R, and for some positive constant L. (H2) There exist a constant λ ∈ (0, 1) and a function q ∈ L1 (T), such that 1 − 2 sup |ep (t, s)| t,s∈[a,b)

[a,b)

q(τ )Δτ > 0,

and   |f (t, x, y)| ≤ q(t) 1 + |x| + |y|λ for all t ∈ T, x, y ∈ R.

(6.38)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

414

Definition 6.19 A function x ∈ Crd (T) is said to be a weak solution of the IVP (6.38), if x satisfies the following integral equation: x(t) = ep (t, a)x0 +

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ,

t ∈ T.

Theorem 6.34 p ∈ Γ1 (T). Suppose (H 1), (H 2). Then the IVP (6.38) has a unique weak solution x ∈ Crd ([a, b)) such that x Δ (t) + p(t)x σ (t) = f (t, x(t), x σ (t)) Δ − a.e.

on

T.

Proof For x ∈ Crd ([a, b)), define the operator (Qx)(t) = ep (t, a)x0 +

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ.

We have Qx ∈ Crd ([a, b)). Let ρ > 0 be arbitrarily chosen and M=

sup |ep (t, s)|.

t,s∈[a,b)

Take x, y ∈ Crd ([a, b)) such that xCrd ([a,b)) ≤ ρ,

yCrd ([a,b)) ≤ ρ,

where  · Crd ([a,b)) = sup | · |. [a,b)

We have   |(Qx)(t) − (Qy)(t)| = ep (t, a)x0 + −ep (t, a)x0 −   = 

[a,t)

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ   ep (t, τ )f (τ, y(τ ), y (τ ))Δτ  σ

[a,t)

 ep (t, τ ) f (τ, x(τ ), x σ (τ ))

   −f (τ, y(τ ), y (τ )) Δτ  σ

6.7 Weak Solutions of Dynamic Systems

≤

[a,t)

415

|ep (t, τ )||f (τ, x(τ ), x σ (τ ))

−f (τ, y(τ ), y σ (τ ))|Δτ ≤

[a,b)

|ep (t, τ )||f (τ, x(τ ), x σ (τ ))

−f (τ, y(τ ), y σ (τ ))|Δτ   |x(τ ) − y(τ )| + |x σ (τ ) − y σ (τ )| Δτ ≤ LM [a,b)

≤ 2LM(b − a)x − yCrd ([a,b)) ,

t ∈ [a, b).

Hence, Qx − QyCrd ([a,b)) ≤ 2LM(b − a)x − yCrd ([a,b)) . Therefore Q : Crd ([a, b)) → Crd ([a, b)) is a continuous operator. Define W = {x ∈ Crd ([a, b)) : xCrd ([a,b)) ≤ ρ}. Note that for x ∈ W , we have   |f (t, x(t), x σ (t))| ≤ q(t) 1 + |x(t)| + |x σ (t)|λ   ≤ q(t) 1 + xCrd ([a,b)) + xλCrd ([a,b))   ≤ q(t) 1 + ρ + ρ λ , t ∈ [a, b), and   |(Qx)(t)| = ep (t, a)x0 + ≤ |eP (t, a)x0 | +

  ep (t, τ )f (τ, x(τ ), x (τ ))Δτ  σ

[a,t)

[a,t)

|ep (t, τ )||f (τ, x(τ ), x σ (τ ))|Δτ

≤ M|x0 | + M(1 + ρ + ρ λ )  = M |x0 | + (1 + ρ + ρ λ )

[a,b)

[a,b)

q(τ )Δτ  q(τ )Δτ ,

t ∈ [a, b),

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

416

and QxCrd ([a,b))

 ≤ M |x0 | + (1 + ρ + ρ λ )

 [a,b)

q(τ )Δτ .

Therefore QW ⊆ Crd ([a, b)) is bounded. Let t1 , t2 ∈ [a, b), t2 > t1 . Then   |(Qx)(t2 ) − (Qx)(t1 )| = ep (t2 , a)x0 + ep (t2 , τ )f (τ, x(τ ), x σ (τ ))Δτ [a,t2 )

−ep (t1 , a)x0 −

[a,t1 )

  ep (t1 , τ )f (τ, x(τ ), x σ (τ ))Δτ 

≤ |ep (t2 , a) − ep (t1 , a)||x0 |   ep (t2 , τ )f (τ, x(τ ), x σ (τ ))Δτ +  [a,t2 )

−

[a,t1 )

  ep (t1 , τ )f (τ, x(τ ), x σ (τ ))Δτ 

≤ |ep (t1 , a)||ep (t2 , t1 ) − 1||x0 | + +

[a,t1 )

[t1 ,t2 )

|ep (t2 , τ ) − ep (t1 , τ )||f (τ, x(τ ), x σ (τ ))|Δτ |ep (t2 , τ )||f (τ, x(τ ), x σ (τ ))|Δτ

≤ M|x0 ||ep (t2 , t1 ) − 1| +

[a,t1 )

|ep (t2 , t1 ) − 1||ep (t1 , τ )|

×|f (τ, x(τ ), x σ (τ ))|Δτ +

[t1 ,t2 )

|ep (t2 , τ )||f (τ, x(τ ), x σ (τ ))|Δτ

≤ M|x0 ||ep (t2 , t1 ) − 1| +M(1 + ρ + ρ λ )|ep (t2 , t1 ) − 1| +M(1 + ρ + ρ λ )

[t1 ,t2 )

q(τ )Δτ a,t1 )

q(τ )Δτ

   ≤ M |x0 | + 1 + ρ + ρ λ + qL1 ([a,b)) ep (t2 , t1 ) − 1 +M(1 + ρ + ρ λ )

[t1 ,t2 )

q(τ )Δτ.

6.7 Weak Solutions of Dynamic Systems

417

Therefore QW is rd-equi-continuous. By the Arzela–Ascoli theorem, it follows that Q is a compact operator in Crd ([a, b)). Define Y = {x ∈ Crd ([a, b)) : x = δQx,

δ ∈ [0, 1]}.

Let y=

1 x, δ

δ = 0,

otherwise y = 0 for x ∈ Y . Then |y(t)| = |(Q(δy))(t)|   = ep (t, a)x0 +

[a,t)

≤ |ep (t, a)x0 | + ≤ M|x0 | + M

+Mδ λ

[a,t)

[a,t)

= M|x0 | + M

[a,t)

[a,t)

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

|ep (t, τ )||f (τ, δy(τ ), δy σ (τ ))|Δτ

  q(τ ) 1 + δ|y(τ )| + δ λ |y(τ )|λ Δτ q(τ )Δτ + Mδ

q(τ )|y σ (τ )|λ Δτ,

[a,t)

q(τ )|y(τ )|Δτ

t ∈ [a, b).

Hence by Theorem 6.33, it follows that there exists a positive constant r such that |y(t)| ≤ r,

t ∈ [a, b).

We take r > 0 large enough and b close enough to a so that M|x0 | + M(1 + r + r λ )

[a,b)

q(τ )Δτ ≤ r.

Thus, Y is a bounded set. Hence by the Leray–Schauder fixed point theorem, it follows that Q has a fixed point x ∈ Crd ([a, b)). We have x(t) = ep (t, a)x0 +

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ,

t ∈ [a, b).

Consequently x is a weak solution of the IVP (6.38) and x Δ (t) + p(t)x σ (t) = f (t, x(t), x σ (t)) Δ − a.e.

on

[a, b).

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

418

For x ∈ Crd ([a, b)), define |x(t)| , t∈[a,b) eβ (t, a)

xβ = sup

(6.39)

where β > 0 is chosen so that

[a,b)

1 Δτ ≤ . eβ (b, σ (τ )) 4LM

Note that Crd ([a, b)) is a Banach space with respect to the norm  · β . Define B = {x ∈ Crd ([a, b)) : xCrd ([a,b)) ≤ r}. Note that B is a Banach space with respect to the norm (6.39). For x ∈ B, define the operator H x(t) = ep (t, a)x0 +

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ,

t ∈ [a, b).

For x ∈ B, we get   |H x(t)| = ep (t, a)x0 +

[a,t)

≤ |ep (t, a)||x0 | + ≤ M|x0 | + M

[a,t)

  ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ 

[a,t)

|ep (t, τ )||f (τ, x(τ ), x σ (τ ))|Δτ

  q(τ ) 1 + |x(τ )| + |x σ (τ )|λ Δτ

≤ M|x0 | + M(1 + r + r λ ) ≤ M|x0 | + M(1 + r + r λ ) ≤ r,

[a,t)

[a,b)

q(τ )Δτ q(τ )Δτ

t ∈ [a, b).

Therefore H B ⊆ B. Next, for x, y ∈ B, we have  1  H x − Hyβ = sup ep (t, a)x0 + t∈[a,b) eβ (t, a) −ep (t, a)x0 −

[a,t)

ep (t, τ )f (τ, x(τ ), x σ (τ ))Δτ

  ep (t, τ )f (τ, y(τ ), y (τ ))Δτ  σ

[a,t)

6.8 Euler Solutions for Dynamic Equations

1 e (t, a) t∈[a,b) β

≤ sup

[a,t)

419

|ep (t, τ )||f (τ, x(τ ), x σ (τ ))

−f (τ, y(τ ), y (τ ))|Δτ   ML |x(τ ) − y(τ )| + |x σ (τ ) − y σ (τ )| Δτ ≤ sup e (t, a) [a,t) t∈[a,b) β     1 eβ (τ, a) + eβ (σ (τ ), a) Δτ ≤ x − yβ ML sup t∈[a,b) eβ (t, a) [a,t)     1 1 + Δτ = x − yβ ML sup eβ (t, σ (τ )) t∈[a,b) [a,t) eβ (t, τ )     1 1 = x − yβ ML sup 1+ Δτ 1 + βμ(τ ) t∈[a,b) [a,t) eβ (t, σ (τ ))   Δτ ≤ x − yβ 2ML sup t∈[a,b) [a,t) eβ (t, σ (τ )) σ

≤

1 x − yβ . 2

Therefore H has a unique weak solution x in (B,  · β ). Hence, the IVP (6.38) has a unique weak solution. This completes the proof.

6.8 Euler Solutions for Dynamic Equations Suppose that x0 ∈ R, a, b ∈ T, a < b. Consider the following IVP: x Δ (t) = f (t, x(t))

a.e.

t ∈ [a, b),

x(a) = x0 ,

(6.40) (6.41)

where f : [a, b) × R → R. Take δ > 0, N ∈ N, and π = {tk }N k=0 ∈ Pδ ([a, b)) arbitrarily. Define the function xπ : [a, b] → R as follows: xπ (ti ) = xi , xπ (t) = xi + (t − ti )f (ti , xi ),

t ∈ [ti , ti+1 ],

i ∈ {0, . . . , n − 1}.

(6.42)

Definition 6.20 The function xπ , defined by (6.42), will be called the Euler polygonal arc.

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420

Definition 6.21 A function x is said to be an Euler solution to the IVP (6.40), (6.41) if it is the uniform limit of a sequence of Euler polygonal arcs xπj related to the partition πj ∈ Pδj ([a, b]), for some decreasing sequence {δj }j ∈N , δj → 0, as j → ∞. Before proving the main result in this section, we will formulate and prove the following useful lemma. Lemma 6.2 Let n ∈ N. Consider the real numbers r 0 , r 1 , . . . , rn ≥ 0 obeying ri+1 ≤ (1 + βi )ri + ηi ,

i ∈ {0, . . . , n − 1},

(6.43)

where βi , ηi ≥ 0, i ∈ {0, . . . , n}, and r0 = 0. Then 

rn ≤ e

%n

i=0 βi

n−1  (

 ηi .

(6.44)

i=0

Proof We will use the principle of the mathematical induction. 1. Let n = 1. Then r1 ≤ (1 + β0 )r0 + η0 = η0 and n−1   %n−1  ( β e i=0 i ηi = eβ0 η0 i=0

≥ η0 ≥ r1 . 2. Assume that (6.44) holds for some n ∈ N. 3. We will prove that rn+1

  n  (  %n ≤ e i=0 βi ηi . i=1

Really, by (6.43), we have rn+1 ≤ (1 + βn )rn + ηn .

(6.45)

6.8 Euler Solutions for Dynamic Equations

421

Now, using (6.44), we arrive at rn+1

n−1   %n−1  ( β ≤ (1 + βn ) e i=0 i ηi + ηn

≤e

βn

 %n−1 e i=0 βi

n−1  (

i=0



ηi

+ ηn

i=0

n−1   (  %n βi i=0 = e ηi + ηn i=0

 %n ≤ e i=0 βi

n−1  (

 ηi

+ ηn e

i=0

 %n = e i=0 βi

n−1  (

%n

i=0 βi

 ηi + ηn

i=0

  n  %n  ( β i = e i=0 ηi , i=0

i.e., (6.45) holds. From here and from the principle of the mathematical induction, we conclude that (6.44) holds for any n ∈ N. This completes the proof. Theorem 6.35 Consider a bounded function c : T → R. Suppose that |f (t, x)| ≤ γ |x| + c(t),

(t, x) ∈ T × R,

where γ is a positive constant. Then the problem (6.40), (6.41) has an Euler solution. Proof Since c : T → R is a bounded function, then there exists a positive constant c1 such that c(t) ≤ c1 ,

t ∈ T.

Take δ > 0 and π = {t0 , t1 , . . . , tN } ∈ Pδ (T). Let xπ be the corresponding Euler polygonal arc. Fix i ∈ {0, 1, . . . , N − 1} and let t ∈ [ti , ti+1 ) be such that xπ is Δ-differentiable at t. If σ (t) > t, then xπΔ (t) =

xπ (σ (t)) − xπ (t) μ(t)

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

422

=

xi + (σ (t) − ti )f (ti , xi ) − (xi + (t − ti )f (ti , xi )) μ(t)

= f (ti , xi ). If σ (t) = t, then we take a sequence {sj }j ∈N ⊂ T decreasing and sj → t, as j → ∞. Since lim

j →∞



[ti , ti+1 ) such that it is

xπ (sj ) − xπ (t) = f (ti , xi ), sj − t

we get xπΔ (t) = f (ti , xi ). Thus, in both cases, we obtain |xπΔ (t)| = |f (ti , xi )| ≤ γ |xi | + c1 ≤ γ |xi − x0 | + γ |x0 | + c1 . For each i ∈ {0, 1, . . . , N − 1}, we have |xi+1 − x0 | ≤ |xi+1 − xi | + |xi − x0 | = |(ti+1 − ti )f (ti , xi )| + |xi − x0 | ≤ (ti+1 − ti )(γ |xi | + c1 ) + |xi − x0 | ≤ (ti+1 − ti ) (γ |xi − x0 | + γ |x0 | + c1 ) +|xi − x0 | ≤ (γ (ti+1 − ti ) + 1) |xi − x0 | +(ti+1 − ti )(γ |x0 | + c1 ). Set ri = |xi − x0 |, βi = (ti+1 − ti )γ , ηi = (ti+1 − ti )(γ |x0 | + c1 ),

i ∈ {0, . . . , N − 1}.

Thus, for any M ∈ {1, . . . , N }, we get rj +1 ≤ (1 + βj )rj + ηj ,

j ∈ {0, . . . , M − 1}.

6.9 The Gronwall Type Inequality

423

Hence by Lemma 6.2, it follows that

rM

⎛ ⎞  %M−1  M−1 ( β ≤ e j =0 j ⎝ ηj ⎠ j =0

= e(tM −t0 )γ (tM − t0 )(γ |x0 | + c1 ) ≤ (b − a)eγ (b−a) (γ |x0 | + c1 ) := K. So, |xπΔ (t)| ≤ γ K + γ |x0 | + c1 := L

a.e.

t ∈ [a, b).

Hence, for any t, s ∈ T, we have |xπ (t) − xπ (s)| ≤ L|t − s|. Now, consider a decreasing sequence {δJ }j ∈N of nonnegative numbers such that δj → 0, as j → ∞. For any j ∈ N, take a partition πj ∈ Pδj (T) and let xπj be the corresponding Euler polygonal arc. Since xπj (a) = x0 and |xπΔj (t)| ≤ L

a.e.

t ∈ [a, b),

then there is a subsequence {xπjk }k∈N of the sequence {xπj }j ∈N and an arc x : T → R such that xπjk → x,

as

k → ∞,

uniformly. Therefore x is an Euler solution of the problem (6.40), (6.41). This completes the proof.

6.9 The Gronwall Type Inequality Let a, b ∈ T, a < b. In this section, we state and prove a result that is a sort of the Gronwall inequality. Theorem 6.36 (The Gronwall Type Inequality) Let X : T → R and k : T → [0, ∞), k ∈ L1 ([a, b)). Suppose |x Δ (t)| ≤ γ1 |x(t)| + k(t) Δ − a.e.

t ∈ [a, b],

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

424

for some positive constant γ . Then 1. |x(t) − x(a)| ≤ γ1 |x(a)| +

[a,t)

[a,t)

eγ1 (t−s) Δs

k(s)eγ1 (t−s) Δs,

t ∈ T.

2.   |x(t)| ≤ γ1 eγ1 (b−a) (b − a) + 1 |x(a)| +eγ1 (b−a)

[a,b)

k(s)Δs,

t ∈ T.

Proof Define z : T → [0, ∞) as follows: z(t) = |x(t) − x(a)| and take t∗ ∈ [a, b) such that z and x are Δ-differentiable at t∗ . If σ (t∗ ) > t∗ , then zΔ (t∗ ) =

z(σ (t∗ )) − z(t∗ ) μ(t∗ )

=

|x(σ (t∗ )) − x(a)| − |x(t∗ ) − x(a)| μ(t∗ )

≤

|x(σ (t∗ )) − x(t∗ )| μ(t∗ )

= |x Δ (t∗ )|. If σ (t∗ ) = t∗ , let {sj }j ∈N ⊂ T be a decreasing sequence such that sj → t∗ , as j → ∞. Since |x(sj ) − x(a)| − |x(t∗ ) − x(a)| z(sj ) − z(t∗ ) = s j − t∗ s j − t∗ ≤

|x(sj ) − x(t∗ )| , s j − t∗

we arrive at the inequality zΔ (t∗ ) ≤ |x Δ (t∗ )|. Therefore

6.9 The Gronwall Type Inequality

425

zΔ (t) ≤ |x Δ (t)| Δ − a.e.

t ∈ [a, b).

Hence, zΔ (t) ≤ |x Δ (t)| ≤ γ1 |x(t)| + k(t) ≤ γ1 |x(t) − x(a)| + γ1 |x(a)| + k(t) = γ1 z(t) + γ1 |x(a)| + k(t) Δ − a.e.

(6.46) t ∈ [a, b).

Define the functions m : T → R and mc : R → R as follows: m(t) = e−γ1 t , mc (t) = e

−γ1 t

,

t ∈ T, t ∈ R.

Let ψ(t) = z(t)m(t),

t ∈ T.

Take t∗ ∈ [a, b) such that z and m are Δ-differentiable at t∗ and (6.46) holds. We have ψ Δ (t∗ ) = mΔ (t∗ )z(t∗ ) + m(σ (t∗ ))zΔ (t∗ ). If σ (t∗ ) = t∗ , then we take a sequence {sj }j ∈N ⊂ T such that sj → t∗ and mΔ (t∗ ) = lim

sj →t∗

= lim

sj →t∗

m(t∗ ) − m(sj ) t∗ − sj mc (t∗ ) − mc (sj ) t∗ − sj

= mc (t∗ ) = −γ1 e−γ1 t∗ . So, ψ Δ (t∗ ) = e−γ1 t∗ zΔ (t∗ ) − γ1 e−γ1 t∗ z(t∗ )   = zΔ (t∗ ) − γ1 z(t∗ ) e−γ1 t∗ . If σ (t∗ ) > t∗ , then, applying the mean value theorem, there exists a θ ∈ (t∗ , σ (t∗ )) such that mΔ (t∗ ) =

m(σ (t∗ )) − m(t∗ ) μ(t∗ )

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

426

=

mc (σ (t∗ )) − mc (t∗ ) μ(t∗ )

= mc (θ ) = −γ1 e−γ1 θ ≤ −γ1 e−γ1 σ (t∗ ) . Hence, ψ Δ (t∗ ) ≤ −γ1 z(t∗ )e−γ1 σ (t∗ ) + e−γ1 σ (t∗ ) zΔ (t∗ )   = zΔ (t∗ ) − γ1 z(t∗ ) e−γ1 σ (t∗ ) . Let λ(t∗ ) = zΔ (t∗ ) − γ1 z(t∗ ). If λ(t∗ ) ≤ 0, then ψ Δ (t∗ ) ≤ λ(t∗ )e−γ1 σ (t∗ ) ≤0 ≤ (γ1 |x(a)| + k(t∗ )) e−γ1 t∗ . If λ(t∗ ) > 0, then ψ Δ (t∗ ) ≤ λ(t∗ )e−γ1 σ (t∗ ) ≤ λ(t∗ )e−γ1 t∗   = zΔ (t∗ ) − γ1 z(t∗ ) e−γ1 t∗ ≤ (|γ1 ||x(a)| + k(t∗ )) e−γ1 t∗ . So, ψ Δ (t) ≤ (γ1 |x(a)| + k(t)) e−γ1 t

Δ − a.e.

t ∈ [a, b).

Now, for each t ∈ T, we have ! ψ(t) − ψ(a) = [a,t) ψ Δ (s)Δs ! ≤ [a,t) (γ1 |x(a)| + k(s)) e−γ1 s Δs Then

t ∈ T.

(6.47)

6.10 Δ × B-Measurable Set-Valued Functions

|x(t) − x(a)| ≤ eγ1 t

 [a,t)

= γ1 |x(a)|

427

 γ1 |x(a)|e−γ1 s + k(s)e−γ1 s Δs

[a,t)

eγ1 (t−s) Δs +

[a,t)

k(s)eγ1 (t−s) Δs,

t ∈ T. Now, by (6.47), we obtain ψ(t) ≤ γ1 |x(a)|e−γ1 a (b − a) + e−γ1 a

[a,b)

k(s)Δs,

t ∈ T.

Hence, |x(t)| − |x(a)| ≤ |x(t) − x(a)| ≤ γ1 |x(a)|eγ1 t e−γ1 a (b − a) +eγ1 t e−γ1 a

[a,b)

k(s)Δs

≤ γ1 |x(a)|eγ1 (b−a) (b − a) +eγ1 (b−a)

[a,b)

k(s)Δs,

t ∈ T,

whereupon   |x(t)| ≤ γ1 eγ1 (b−a) (b − a) + 1 |x(a)| +eγ1 (b−a)

[a,b)

k(s)Δs,

t ∈ T.

This completes the proof.

6.10 Δ × B-Measurable Set-Valued Functions With B we will denote the Borel σ -algebra of R and with Δ × B we will denote the least σ -algebra of T × R that contains all products A × B, where A ⊂ T is Δ-measurable and B ∈ B. Definition 6.22 A function φ : T×R → R is said to be a Δ-Caratheodory ´ function if it satisfies the following properties: 1. For each t ∈ T, the function x → φ(t, x) is continuous. 2. For each x ∈ R, the function t → φ(t, x) is Δ-measurable.

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

428

Theorem 6.37 Let F : T × R  R be a Δ × B-measurable set-valued function and u : T → R be a single-valued Δ-measurable function. Then the set-valued function G : T × R  R, defined by G(t) = F (t, u(t)),

t ∈ T,

is Δ-measurable. Proof Let D be the collection of subsets  D = E ⊂ T × R : {t ∈ T : (t, u(t)) ∈ E}

is

 Δ − measurable . We have that D is a σ -algebra in T × R. Let A ⊂ T be a Δ-measurable set and B ∈ B. Then  {t ∈ T : (t, u(t)) ∈ A × B} = A u−1 (B). (6.48) Really, let t1 ∈ {t ∈ T : (t, u(t)) ∈ A × B} be arbitrarily chosen. Then (t1 , u(t1 )) ∈ A × B  and t1 ∈ A, u(t1 ) ∈ B. Hence, t1 ∈ A and t1 ∈ u−1 (B). Thus, t1 ∈ A u−1 (B). Because t1 was arbitrarily chosen element of the set {t ∈ T : (t, u(t)) ∈ A × B}, we conclude that  (6.49) {t ∈ T : (t, u(t)) ∈ A × B} ⊂ A u−1 (B).  Let now t2 ∈ A u−1 (B) be arbitrarily chosen. Then t2 ∈ A and t2 ∈ u−1 (B), or t2 ∈ A and u(t2 ) ∈ B. Consequently (t2 , u(t2 )) ∈ A × B and t2 ∈ {t ∈ T : (t, u(t)) ∈ A × B}. Because t2 ∈ A



u−1 (B) was arbitrarily chosen, we conclude that A



u−1 (B) ⊂ {t ∈ T : (t, u(t)) ∈ A × B}.

 By the last relation and (6.49), we obtain the relation (6.48). Note that A u−1 (B) is a Δ-measurable set. Therefore Δ × B ⊂ D. Now, we take a compact set V ⊂ R arbitrarily. Then

6.10 Δ × B-Measurable Set-Valued Functions

429

F −1 (V ) = {(t, u) ∈ T × R : F (t, u)



V = ∅}

∈ Δ × B. Therefore G−1 (V ) = {t ∈ T : F (t, u(t))



V = ∅}

= {t ∈ T : (t, u(t)) ∈ F −1 (V )} is a Δ-measurable set. Consequently G is a Δ-measurable set-valued function. This completes the proof. Theorem 6.38 Take a Δ-Caratheodory ´ function φ : T × R → R and a Δmeasurable nonempty closed set-valued function H : T  R. Suppose that for each t ∈ T there exists a u ∈ H (t) such that φ(t, u) = 0. Then the set-valued function G : T → R given by G(t) = {u ∈ H (t) : φ(t, u) = 0} has a Δ-measurable selection. Proof Let a, b ∈ T, a < b. Define (t, u) = φ



φ(t, u) if t ∈ T φ(ti , u) if t ∈ (ti , σ (ti )) for

some

ti ∈ T.

: [a, b] × R → R is a Lebesgue Caratheodory Then φ ´ function. Define the setvalued function Φ : [a, b]  R as follows: (t, u) = 0}. Φ(t) = {u ∈ R : φ Hence, Graph(Φ) = {(t, γ ) ∈ [a, b] × R : γ ∈ Φ(t)} (t, γ ) = 0} = {(t, γ ) ∈ [a, b] × R : φ −1 ( =φ 0) is Lebesgue measurable. Then Φ is Lebesgue measurable because Φ is closed. Define the set-valued function N : [a, b]  R by (t) N(t) = H



Φ(t),

t ∈ [a, b],

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

430

where (t) = H



H (t) if t ∈ T H (ti ) if t ∈ (ti , σ (ti ))

for some

ti ∈ T.

is a closed and Lebesgue measurable set-valued function, we have that N Since H is Lebesgue measurable. Then, for any compact subset V ⊂ R, we have G−1 (V ) = {t ∈ T : G(t)



V = ∅}   = {t ∈ [a, b] : N(t) V = ∅} T

is Lebesgue measurable, i.e., G is Δ-measurable. Because G is nonempty and closed, we conclude that G has a Δ-measurable selection. This completes the proof. Theorem 6.39 Let F : T × R  R be a set-valued function such that: 1. F is a nonempty, closed, and Δ × B-measurable set-valued function. 2. There exists a function k : T → [0, ∞), k ∈ L1 ([a, b]) such that F (t, x) ⊂ F (t, y) + k(t)|y − x|{B}B∈B for all x, y ∈ R and for any B ∈ B. To the set-valued function F we associate the function ρ : T × R × R → [0, ∞) given by ρ(t, x, v) = inf{|v − y| : y ∈ F (t, x)}. Then: 1. The function t → ρ(t, x, v) is Δ-measurable for each (x, v) ∈ R × R. 2. For any t ∈ T and for any x1 , x2 , v1 , v2 ∈ R, one has |ρ(t, x1 , v1 ) − ρ(t, x2 , v2 )| ≤ k(t)|x1 − x2 | + |v1 − v2 |.

Proof Let α ∈ R be arbitrarily chosen. If α < 0, then {t ∈ T : ρ(t, x, v) ≤ α} = ∅, which is a Δ-measurable set. If α ≥ 0, then {t ∈ T : ρ(t, x, v) ≤ α} = {t ∈ T : F (t, x)



{v + αB}B∈B = ∅}.

(6.50)

Really, let t1 ∈ {t ∈ T : ρ(t, x, v) ≤ α} be arbitrarily chosen. Then t1 ∈ T and ρ(t1 , x, v) ≤ α

6.10 Δ × B-Measurable Set-Valued Functions

431

for any x, v ∈ R, and ρ(t1 , x, v) ≤ |v − y| for any y ∈ F (t1 , x) and for any x, v ∈ R. Hence, y ∈ {v + αB}B∈B and F (t1 , x)



{v + αB}B∈B = ∅.

Therefore {t ∈ T : ρ(t, x, v) ≤ α} ⊂ {t ∈ T : F (t, x) Let now t2 ∈ {t ∈ T : F (t, x)





{v + αB}B∈B = ∅}.

(6.51)

{v + αB}B∈B = ∅} be arbitrarily chosen. Then

F (t2 , x)



Therefore there exists a y1 ∈ F (t2 , x)

{v + αB}B∈B .



{v + αB}B∈B . For it we have

|v − y1 | ≤ α. On the other hand, by the definition of the function ρ, we get ρ(t2 , x, v) ≤ |v − y1 | ≤ α, whereupon t2 ∈ {t ∈ T : ρ(t, x, v) ≤ α}.  Because t2 ∈ {t ∈ T : F (t, x) {v + αB}B∈B = ∅} was arbitrarily chosen, we conclude that  {t ∈ T : F (t, x) {v + αB}B∈B = ∅} ⊂ {t ∈ T : ρ(t, x, v) ≤ α}. By the last relation and (6.51), we obtain (6.50). Since F is a Δ × B-measurable set-valued function, we have that the set  {t ∈ T : F (t, x) {v + αB}B∈B = ∅} is a Δ-measurable set. Hence by (6.50), we obtain that the set {t ∈ T : ρ(t, x, v) ≤ α} is a Δ-measurable set.

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

432

Let δ > 0 be arbitrarily chosen. By the definition of the function ρ, it follows that there exists an η2 ∈ F (t, x2 ) such that ρ(t, x2 , v2 ) > |v2 − η2 | − δ. By the assumptions for the set-valued function F , we have that there exists an η1 ∈ F (t, x1 ) such that |η1 − η2 | ≤ k(t)|x1 − x2 |. Then ρ(t, x1 , v1 ) ≤ |v1 − η1 | and ρ(t, x1 , v1 ) − ρ(t, x2 , v2 ) ≤ |v1 − η1 | − ρ(t, x2 , v2 ) < |v1 − η1 | − |v2 − η2 | + δ ≤ |v1 − v2 | + |v2 − η1 | − |v2 − η2 | + δ ≤ |v1 − v2 | + |η2 − η1 | + δ ≤ |v1 − v2 | + k(t)|x1 − x2 | + δ. Similarly, we obtain ρ(t, x2 , v2 ) − ρ(t, x1 , v1 ) < |v1 − v2 | + k(t)|x1 − x2 | + δ. Consequently |ρ(t, x2 , v2 ) − ρ(t, x1 , v1 )| ≤ |v1 − v2 | + k(t)|x1 − x2 | + δ. Because δ > 0 was arbitrarily chosen, by the last inequality, we arrive at the following inequality: ρ(t, x2 , v2 ) − ρ(t, x1 , v1 ) ≤ |v1 − v2 | + k(t)|x1 − x2 |. This completes the proof. Theorem 6.40 Let F and ρ be the same as in Theorem 6.39. If x : T → R is absolutely continuous, then the set-valued function # $ t  u ∈ F (t, x(t)) : |u − x Δ (t)| − ρ(t, x(t), x Δ (t)) = 0 has a Δ-measurable selection.

(6.52)

6.11 Advanced Practical Problems

433

Proof By Theorem 6.39, it follows that the function defined by t → ρ(t, x(t), x Δ (t)) is Δ-measurable. Next, the function defined with φ(t, x(t), x Δ (t)) = u − x Δ (t) − ρ(t, x(t), x Δ (t)) is a Δ-Caratheodory ´ function. Hence by Theorem 6.38, it follows that the set-valued function, defined by (6.52), has a Δ-measurable selection. This completes the proof.

6.11 Advanced Practical Problems Problem 6.1 Let T = 4Z. Find: 1. 2. 3. 4. 5.

μΔ ([−16, 0)) μΔ ([−32, 4)) μΔ ((0, 16)) μΔ ((4, 64)) μΔ ([4, 64))

Problem 6.2 Let T = 2N0 1. 2. 3. 4. 5.



{0}. Find:

μΔ ((0, 16]) μΔ ([0, 16]) μΔ ([1, 32]) μΔ ([1, 64]) μΔ ((0, 64])

Problem 6.3 Let T = hZ, h > 0, a = 0, b = 10h. Prove that the following functions: 1. 2. 3. 4. 5.

f (t) = t 2 1+1 f (t) = t 2 − 1 f (t) = 4e2 (t, 0) f (t) = cos2 (t, 0) f (t) = sin3 (t, 0)

belong to Lp ([a, b)), p ≥ 1. Problem 6.4 Let T = 2N0 and f (t) =

1 . t3

6 The Lebesgue Integration. Lp -Spaces. Sobolev Spaces

434

Find f 1,T . Problem 6.5 Let T = 3N0 and f (t) =

1 . t4 + t2 + 1

Find a constant C > 0 such that |f |0,T ≤ Cf 1,T .

6.12 Notes and References In this chapter we introduced the conception for Δ-measurable functions on time scales and Δ-Lebesgue integration on time scales and deducted some of the properties of the Δ-measurable functions and the Δ-Lebesgue integral. We introduced the spaces L1 (T) and the Sobolev spaces. We established the concept for absolutely continuous functions on time scales and gave a characterization of such functions. In this chapter we introduced weak solutions and the Euler solutions for first order dynamic equations on time scales and formulated and proved an analogue of the Gronwall inequality. The material in this chapter can be found in [6, 9, 10, 24], [25, 40] and [41].

Chapter 7

First Order Dynamic Inclusions

This chapter is devoted to a qualitative analysis of some classes of first order dynamic inclusions. They are investigated for existence of solutions of first order dynamic inclusions with local and nonlocal initial conditions, general boundary conditions, and periodic boundary conditions. Dual time scales are introduced and some of their basic properties are deducted. As their applications, the existence of solutions for some IVPs for first order dynamic inclusions is proved. Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also, a, b ∈ T, a < b.

7.1 Existence and Approximations of Solutions of First Order Dynamic Inclusions Consider the following problem: y Δ (t) ∈ F (t, y(t)) Δ − a.e.

t ∈ [a, b],

y(a) = x(a),

(7.1) (7.2)

where F : T × R  R is a set-valued function such that (H1) F is nonempty, closed, and Δ × B-measurable set-valued function, (H2) there exists a function k : T → [0, ∞), k ∈ L1 ([a, b]) such that F (t, x) ⊂ F (t, y) + k(t)|y − x|{B}B∈B for all x, y ∈ R,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_7

435

436

7 First Order Dynamic Inclusions

and (H3) x : T → R is an absolutely continuous single-valued function. To the set-valued function F we associate the function ρ : T × R × R → [0, ∞) given by ρ(t, x, v) = inf{|v − y| : y ∈ F (t, x)}. Definition 7.1 A single-valued function y ∈ AC([a, σ (b)]) is said to be a solution of the problem (7.1), (7.2) if there exists a function f ∈ L1 ([a, σ (b)]) such that f (t) ∈ F (t, y(t)), y Δ (t) = f (t) Δ − a.e.

t ∈ [a, b]

and the condition (7.2) holds. Example 7.1 Let T = N0 . Consider the problem (7.1), (7.2) for  F (t, y(t)) =

2 , (t + 2)(t + 3)

t + 1,

 t 4 + 7t 2 + 87t + 10 ,

 1 2 3t + 5t + 3 , 3 t ≥ 0,

x(0) = 1. We will prove that the functions t +1 , t +2  12 y2 (t) = t +t +2 , 2  13 t + t2 + t + 3 , y3 (t) = 3 y1 (t) = 2

are solutions of the considered problem. We have y1 (0) = 2 ·

0+1 0+2

= 1, 1 (0 + 0 + 2) 2 = 1,

y2 (0) =

t ≥ 0,

7.1 Existence and Approximations of Solutions of First Order Dynamic. . .

437

1 (0 + 0 + 0 + 3) 3 = 1,

y3 (0) =

and σ (t) = t + 1, y1Δ (t) = 2 =2 =

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

2 (t + 2)(t + 3)

∈ F (t, y(t)), 1 (σ (t) + t + 1) 2 1 = (t + 1 + t + 1) 2 1 = (2t + 2) 2 = t +1

y2Δ (t) =

∈ F (t, y(t)),  1 y3Δ (t) = (σ (t))2 + tσ (t) + t 2 + σ (t) + t + 1 3  1 (t + 1)2 + t (t + 1) + t 2 + t + 1 + t + 1 = 3  12 t + 2t + 1 + t 2 + t + t 2 + 2t + 2 = 3  1 2 3t + 5t + 3 = 3 ∈ F (t, y(t)), t ≥ 0. Example 7.2 Let T = 2N0 . Consider the problem (7.1), (7.2), where F (t, y(t)) = {7t 2 + 1, y(1) = 0.

7t 2 + 9t + 1,

15t 3 + 7t 2 + 12t + 1},

t ∈ T,

438

7 First Order Dynamic Inclusions

We reduce the problem (7.1), (7.2) to the following IVPs: y Δ (t) = 7t 2 + 1, y(1) = 0,

t ≥ 1,

y Δ (t) = 7t 2 + 9t + 1, y(1) = 0,

(7.3)

t ≥ 1,

(7.4)

and y Δ (t) = 15t 3 + 7t 2 + 12t + 1, y(1) = 0.

t ≥ 1,

Let f1 (t) = t 3 + t + 1, f2 (t) = t 3 + 3t 2 + t, f3 (t) = t 4 + t 3 + 4t 2 + t,

t ∈ T.

We have σ (t) = 2t, f1Δ (t) = (σ (t))2 + tσ (t) + t 2 + 1 = (2t)2 + t (2t) + t 2 + 1 = 4t 2 + 2t 2 + t 2 + 1 = 7t 2 + 1, f2Δ (t) = (σ (t))2 + tσ (t) + t 2 + 3(σ (t) + t) + 1 = (2t)2 + t (2t) + t 2 + 3(2t + t) + 1 = 4t 2 + 2t 2 + t 2 + 9t + 1 = 7t 2 + 9t + 1, f3Δ (t) = (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 +(σ (t))2 + tσ (t) + t 2 + 4(σ (t) + t) + 1 = (2t)3 + t (2t)2 + t 2 (2t) + t 3 +(2t)2 + t (2t) + t 2 + 4(2t + t) + 1 = 8t 3 + 4t 3 + 2t 3 + t 3 + 4t 2 + 2t 2 +t 2 + 12t + 1 = 15t 3 + 7t 2 + 12t + 1,

t ∈ T.

(7.5)

7.1 Existence and Approximations of Solutions of First Order Dynamic. . .

439

Consider the IVP (7.3). For its solution we have the representation. t

y(t) = y(1) +

(7s 2 + 1)Δs

1 t

=

(7s 2 + 1)Δs

1 t

= 1

f1Δ (s)Δs

s=t  = f1 (s)

s=1

s=t  = (s + s + 1) 3

s=1

= t +t +1−1−1−1 3

= t 3 + t − 2,

t ∈ T.

Now we consider the IVP (7.4). For its solution we have the following representation: t

y(t) = y(1) +

(7s 2 + 9s + 1)Δs

1 t

=

(7s 2 + 9s + 1)Δs

1 t

= 1

f2Δ (s)Δs

s=t  = f2 (s)

s=1

  s=t  = s 3 + 3s 2 + s 

s=1

= t + 3t + t − 1 − 3 − 1 3

2

= t 3 + 3t 2 + t − 5,

t ∈ T.

For the solution of the IVP (7.5) we have the following representation: t

y(t) = y(1) + t

= 1



1

  15s 3 + 7s 2 + 12s + 1 Δs

 15s 3 + 7s 2 + 12s + 1 Δs

440

7 First Order Dynamic Inclusions t

= 1

f3Δ (s)Δs

s=t  = f3 (s)

s=1

  s=t  = s 4 + s 3 + 4s 2 + s 

s=1

= t + t + 4t + t − 1 − 1 − 4 − 1 4

3

2

= t 4 + t 3 + 4t 2 + t − 7,

t ∈ T.

Consequently, y1 (t) = t 3 + t − 2, y2 (t) = t 3 + 3t 2 + t − 5, y3 (t) = t 4 + t 3 + 4t 2 + t − 7,

t ∈ T,

are solutions of the problem (7.1), (7.2). Example 7.3 Let T = 3N0 . Consider the problem (7.1), (7.2), where F (t, y(t)) = {y(t) + 1, y(t) + 1 + t, y(t) + 1 + t 2 },

t ∈ T,

y(1) = 1. We reduce the problem (7.1), (7.2) to the following IVPs: y Δ (t) = y(t) + 1, y(1) = 1,

t ∈ T,

(7.6)

y Δ (t) = y(t) + 1 + t, y(1) = 1,

t ∈ T,

(7.7)

y Δ (t) = y(t) + 1 + t 2 , y(1) = 1.

t ∈ T,

(7.8)

and

Note that σ (t) = 3t, & (1 + 2s), e1 (t, t0 ) = s∈[t0 ,t)

t, t0 ∈ T,

t ≥ t0 .

7.1 Existence and Approximations of Solutions of First Order Dynamic. . .

441

Now, we consider the IVP (7.6). For its solution we have the following representation: t

y(t) = e1 (t, 1) +

e1 (t, σ (τ ))Δτ 1

&

=

  s∈ 1, 3t

(1 + 2s) + 2

  s∈ 1, 3t

(1 + 2s)Δτ

  s∈ σ (τ ), 3t

1 t

&

=

&

t

(1 + 2s) +

3 (

τ

τ =1

&   s∈ σ (τ ), 3t

(1 + 2s),

t ∈ T.

Now, we consider the IVP (7.7). For its solution we have the following representation: t

y(t) = e1 (t, 1) +

e1 (t, σ (τ ))(1 + τ )Δτ

1

=

=

&

⎛ t

(1 + 2s) +

  s∈ 1, 3t

⎜ ⎝

⎞ &

  s∈ σ (τ ), 3t

1 t

&

(1 + 2s) + 2

  s∈ 1, 3t

3 (

⎟ (1 + 2s)⎠ (1 + τ )Δτ ⎛

⎜ τ (1 + τ ) ⎝

τ =1

⎞ &

⎟ (1 + 2s)⎠ ,

  s∈ σ (τ ), 3t

t ∈ T.

For the solution of the IVP (7.8), we have the representation. t

y(t) = e1 (t, 1) +

e1 (t, σ (τ ))(1 + τ 2 )Δτ

1

=

=

&

(1 + 2s) +

  s∈ 1, 3t

&

⎜ ⎝

(1 + 2s) + 2

⎞ &

  s∈ σ (τ ), 3t

1 t

  s∈ 1, 3t

Consequently,

⎛ t

3 (

τ =1

⎟ (1 + 2s)⎠ (1 + τ 2 )Δτ ⎛

⎜ τ (1 + τ 2 ) ⎝

⎞ &

⎟ (1 + 2s)⎠ ,

  s∈ σ (τ ), 3t

t ∈ T.

442

7 First Order Dynamic Inclusions

y1 (t) =

y2 (t) =

y3 (t) =

&

t

(1 + 2s) + 2

3 (

  s∈ 1, 3t

τ =1

&

3 (

τ

&   s∈ σ (τ ), 3t

⎛

t

(1 + 2s) + 2

  s∈ 1, 3t

τ =1

&

3 (

⎜ τ (1 + τ ) ⎝

  s∈ 1, 3t

⎞ &

⎟ (1 + 2s)⎠ ,

  s∈ σ (τ ), 3t

⎛

t

(1 + 2s) + 2

(1 + 2s),

⎜ τ (1 + τ 2 ) ⎝

τ =1

&

⎞ ⎟ (1 + 2s)⎠ ,

  s∈ σ (τ ), 3t

t ∈T

is a solution of the considered problem. Exercise 7.1 Let T = 3Z. Find a solution of the problem (7.1), (7.2), where F (t, y(t)) = {2y(t) + t 2 , 2y(t) + 1 + t 2 + t 4 , 2y(t) + 1 + 3t 2 + 4t 4 + t 6 },

t ∈ T,

y(1) = 0. Define K=e ρF (x) =

!

[a,b) k(t)Δt

[a,b)

,

ρ(t, x(t), x Δ (t))Δt.

Theorem 7.1 Suppose (H 1), (H 2), and (H 3). For any  > 0 satisfying ρF (x)
0 be chosen so that |yj (a)| ≤ C1 ,

j ∈ N.

Set k(t) = γ1 φ(t) + c∗ , t ∈ T,   K = γ1 eγ1 (b−a) (b − a) + 1 C1 +eγ1 (b−a)

[a,b)

k(s)Δs.

Then, by the Gronwall type inequality, Theorem 6.36, it follows that   |yj (t)| ≤ γ1 eγ1 (b−a) (b − a) + 1 |yj (a)| +eγ1 (b−a)

[a,b)

k(s)Δs

  ≤ γ1 eγ1 (b−a) (b − a) + 1 C1

7.1 Existence and Approximations of Solutions of First Order Dynamic. . .

+eγ1 (b−a) = K,

[a,b)

451

k(s)Δs

t ∈ T.

Thus, |yjΔ (t)| ≤ γ1 K + k(t) Δ − a.e.

t ∈ [a, b).

Therefore, there exists a subsequence {yjk }k∈N of the sequence {yj }j ∈N and an y : T → R such that yjk → y,

k → ∞,

as

uniformly, and for arbitrary c, d ∈ T, c < d, we have

[c,d)

yjΔk (s)Δs →

[c,d)

y Δ (s)Δs,

as

k → ∞.

Without loss of generality, we relabel {jk }k∈N by {k}k∈N . Take p ∈ R and consider the Hamiltonian function t → h(τk (t), xk (t) + yk (t), p) = lk (t). Define lk (t) = |p||xk (t) + yk (t)|. Note that lk is Δ-measurable for each k ∈ N. Also, |h(τk (t), xk (t) + yk (t), p)| ≤ |p||xk (t) + yk (t)| ≤ |p| (γ1 |xk (t) + yk (t)| + c∗ ) ≤ |p| (γ1 |xk (t)| + γ1 |yk (t)| + c∗ ) ≤ |p| (γ1 K + γ1 φ(t) + c∗ ) ,

t ∈ T.

Thus, lk ∈ L1 ([a, b)). Note that lim inf h(τk (s), xk (s) + yk (s), p)Δs ≤ lim inf

A k→∞

k→∞

h(τk (s), xk (s) + yk (s), p)Δs A

for each Δ-measurable set A ⊂ [a, b). Next, h(τk (s), xk (s) + yk (s), p) ≤ |p||ykΔ (s)|

Δ − a.e.

s ∈ [a, b).

452

7 First Order Dynamic Inclusions

Thus, if c, d ∈ T, c < d, then

[c,d)

  |p(s)||ykΔ (s)| − h(τk (s), xk (s) + yk (s), p) Δs ≥ 0.

Now, using that lim inf h(τk (s), xk (s) + yk (s), p)Δs

[c,d) k→∞

≤ lim inf k→∞

= lim

[c,d)

j →∞ [c,d)

h(τk (s), xk (s) + yk (s), p)Δs

h(τkj (s), xkj (s) + ykj (s), p)Δs,

we obtain

[c,d)

≥

|p(s)||y Δ (s)|Δs −

[c,d)

lim inf h(τk (s), xk (s) + yk (s), p)Δs

[c,d) k→∞

|p(s)||y Δ (s)|Δs − lim inf k→∞

[c,d)

h(τk (s), xk (s) + yk (s), p)Δs

= lim

|p(s)||ykΔj (s)|Δs − lim

= lim

  |p(s)||ykΔj (s)| − h(τkj (s), xkj (s) + ykj (s), p) Δs

j →∞ [c,d) j →∞ [c,d)

j →∞ [c,d)

h(τkj (s), xkj (s) + ykj (s), p)Δs

≥ 0. Since lim (τk (s), xk (s) + yk (s)) = (s, y(s))

Δ − a.e.

k→∞

s ∈ [a, b)

and h(t, x, p) is lower semi-continuous at (t, x), we get lim inf h(τk (s), xk (s) + yk (s), p) ≥ h(s, y(s), p)

Δ − a.e.

k→∞

s ∈ [a, b).

Hence,

[c,d)

  |p(s)||y Δ (s)| − h(s, y(s), p) Δs =

[c,d)

−

|p(s)||y Δ (s)|Δs

[c,d)

h(s, y(s), p)Δs

7.1 Existence and Approximations of Solutions of First Order Dynamic. . .

≥

[c,d)

−

453

|p(s)||y Δ (s)|Δs lim inf h(τk (s), xk (s)

[c,d) k→∞

+yk (s), p)Δs ≥ 0. Consequently, |p(s)||y Δ (s)| ≥ h(s, y(s), p)

Δ − a.e.

s ∈ [a, b).

Now, using the Hahn–Banach separation theorem, we conclude that y Δ (t) ∈ F (t, y(t)) Δ − a.e.

t ∈ [a, b).

This completes the proof. Theorem 7.3 Suppose (H 1) and (7.9). Then the problem (7.1), (7.2) has a solution. Proof Take an arbitrary selection f of F . Consider the IVP y Δ (t) = f (t, y(t)) y(a) = x(a).

Δ − a.e.

t ∈ [a, b),

(7.10)

Because F satisfies (7.9), we have that the function f satisfies the conditions of Theorem 6.35. Therefore, the IVP (7.10) has a Euler solution y. Let {yπj }j ∈N be a sequence of Euler solutions associated with the partitions πj ∈ Pδj (T) such that {δj }j ∈N is a decreasing sequence for which δj → 0, as j → ∞, and whose uniform limit is y. Without loss of generality, we take {yπj }j ∈N as {yj }j ∈N . Let j

j

j

πj = {t0 , t1 , . . . , tN (j ) } and t ∈ T. If t = b, then we define j

τj (b) = tN (j )−1 and if t ∈ [a, b), we define j

τj (t) = ti , j

i ∈ {0, 1, . . . , N(j ) − 1},

j

such that t ∈ [ti , ti+1 ). Then τj : T → R is Δ-measurable. Fix t ∈ [a, b). Let j

j

j

j

i ∈ {0, 1, . . . , N (j ) − 1} be such that t ∈ [ti , ti+1 ). If ρ(ti+1 ) = ti , it follows that

454

7 First Order Dynamic Inclusions j

t = ti

= τj (t) j

j

and if ti+1 − ti ≤ δj , we have j

|τj (t) − t| = t − ti j

j

< ti+1 − ti ≤ δj . Therefore, τj (t) → t,

as

j → ∞,

t ∈ [a, b).

Now, we define the function xj : T → R as follows: xj (t) = yj (τj (t)) − yj (t),

t ∈ T.

Hence by Theorem 6.35, there exists a constant C > 0 such that yj is Lipschitz continuous for each j . If t ∈ [a, b), it follows that |xj (t)| = |yj (τj (t)) − yj (t)| ≤ C|τj (t) − t| ≤ Cδj , and if t = b, we have |xj (b)| ≤ C|τj (b) − b| j

= C|tN (j )−1 − b| ≤ C(b − a). Consider t ∈ [a, b) such that yj is Δ- differentiable at t. Take i ∈ {0, 1, . . . , N (j ) − j j 1} such that t ∈ [ti , ti+1 ). From here,   j j yjΔ (t) = f ti , yj (ti )   j j ∈ F ti , yj (ti )   = F τj (t), yj (τj (t))   = F τj (t), xj (t) + yj (t) .

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

455

Since yj (a) = x(a), applying Theorem 7.2, we obtain a subsequence {yjm }m∈N of the sequence {yj }j ∈N and a z ∈ F such that yjm → z,

as

m → ∞,

yjm → y,

as

m → ∞,

uniformly. Because

uniformly, we conclude that y = z. This completes the proof.

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial Conditions Suppose that 0, b ∈ T, {tk }m k=1 ⊂ T, 0 < t1 < t2 < . . . < tm < b, y0 ∈ R, {ck }m ⊂ R, c =  0, k ∈ {1, . . . , m}. In this section, we investigate the following k k=1 IVP: y Δ (t) + p(t)y σ (t) ∈ F (t, y(t)) a.e. y(0) +

m (

t ∈ [a, b],

ck y(tk ) = y0 ,

(7.12)

k=1

where (H4) p is regressive and right-dense continuous on T. (H5) 0 = 1 +

m (

ck ep (tk , 0),

k=1



c = 1+

m (

−1 ck ep (tk , 0)

,

k=1

γ =

sup t∈[0,σ (b)]

δ = 1+

ep (t, 0)|c||y0 |,

sup t∈[0,σ (b)]

(7.11)

ep (t, 0)|c||y0 |.

(H6) F : [0, b] × R  R, t → F (t, y) is measurable for each y ∈ R. (H7) y → F (t, y) is u.s.c. for a.e. t ∈ [0, σ (b)]. (H8) F : [0, σ (b)] × R → Kc (R).

456

7 First Order Dynamic Inclusions

(H9) For each r > 0 there exists a nonnegative function φr ∈ L1 ([0, σ (b)]) such that F (t, y) = sup{|f | : f ∈ F (t, y)} ≤ φr (t) for each (t, y) ∈ [0, σ (b)] × R with |y| ≤ r, and lim inf r→∞

1 r

σ (b)

φr (t)Δt = β.

0

(H10) There exists a continuous nondecreasing function φ : [0, ∞) → (0, ∞), a nonnegative function q ∈ L1 ([0, σ (b)]) and a positive constant M such that F (t, y) ≤ q(t)φ (|y|) for each (t, y) ∈ [0, σ (b)] × R, and M γ + δ sup(t,s)∈[0,σ (b)]×[0,σ (b)] ep (t, s)φ(M)

! σ (b) 0

> 1. q(s)Δs

Definition 7.2 A single-valued function y ∈ AC([0, σ (b)]) is said to be a solution of the IVP (7.11), (7.12) if there exists a single-valued function f ∈ L1 ([0, σ (b)]) such that 1. f (t) ∈ F (t, y(t)) a.e. t ∈ [0, σ (b)], 2. y Δ (t) + p(t)y σ (t) = f (t) a.e. t ∈ [0, σ (b)], 3. The condition (7.12) holds. Example 7.5 Let T = 2Z. Consider the IVP  y Δ (t) ∈ 3t 2 + 8t + 7, 11t 2 + 10,

 1 4 3 + 12t + t + 5 , 1 + t2

t ∈ [0, 4],

y(0) + 2y(2) − y(4) = −56. Here  F (t, y(t)) = 3t 2 + 8t + 7, 11t 2 + 10,

 1 4 3 + 12t + t + 5 , 1 + t2

We will prove that y(t) = t 3 + t 2 + t,

t ∈ [0, 4]

is a solution of the considered problem. Really, we have

t ∈ [0, 4].

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

σ (t) = t + 2, y Δ (t) = (σ (t))2 + tσ (t) + t 2 + σ (t) + t + 1 = (t + 2)2 + t (t + 2) + t 2 + t + 2 + t + 1 = t 2 + 4t + 4 + t 2 + 2t + t 2 + 2t + 3 = 3t 2 + 8t + 7 ∈ F (t, y(t)),

t ∈ [0, 4],

y(0) = 0, y(2) = 23 + 22 + 2 =8+4+2 = 14, y(4) = 43 + 42 + 4 = 64 + 16 + 4 = 84 and y(0) + 2y(2) − y(4) = 0 + 28 − 84 = −56. Example 7.6 Let T = 2N0 . Consider the following IVP: 1 y Δ (t) + y σ (t) ∈ {7t + 3, 4 + t 2 + 11t 4 , 5t + 12t 5 + t 7 }, t

t ∈ [1, 8],

y(1) − 2y(2) + y(4) − y(8) = −62. Here F (t, y(t)) = {7t + 3, 4 + t 2 + 11t 4 , 5t + 12t 5 + t 7 }, We will prove that y(t) = t 2 + t,

t ∈ [1, 8],

is a solution of the considered IVP. Really, we have σ (t) = 2t,

t ∈ [1, 8].

457

458

7 First Order Dynamic Inclusions

y Δ (t) = σ (t) + t + 1 = 2t + t + 1 = 3t + 1, y σ (t) = (σ (t))2 + σ (t) = (2t)2 + 2t = 4t 2 + 2t, 1 1 σ y (t) = (4t 2 + 2t) t t = 4t + 2, 1 y Δ (t) + y σ (t) = 3t + 1 + 4t + 2 t = 7t + 3, t ∈ T, y(1) = 12 + 1 = 2, y(2) = 22 + 2 = 4+2 = 6, y(4) = 42 + 4 = 16 + 4 = 20, y(8) = 82 + 8 = 64 + 8 = 72 and y(1) − 2y(2) + y(4) − y(8) = 2 − 12 + 20 − 72 = 22 − 84 = −62. Exercise 7.3 Let T = Z. Prove that y(t) = t 2 − 2t + 1,

t ∈ T,

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

459

is a solution of the following IVP: y Δ (t) + ty σ (t) ∈ {t 3 + 2t − 1, t 3 + 2t + 1, t 3 + 2t + 7},

t ∈ [0, 3],

y(0) − y(1) − y(2) + y(3) = 4. Theorem 7.4 Suppose (H 4)–(H 9). Then the problem (7.11), (7.12) has at least one solution on [0, σ (b)], provided that sup (t,s)∈[0,σ (b)]×[0,σ (b)]

ep (t, s)βδ < 1.

Proof Let SF,y be the set of selections of F defined by

SF,y = f ∈ L1 ([0, σ (b)]) : f (t) ∈ F (t, y(t))

a.e.

 t ∈ [0, σ (b)] .

Define the operator N : C ([0, σ (b)]) → 2C ([0,σ (b)]) as follows:  N(y) = h ∈ C ([0, σ (b)]) :  h(t) = ep (t, 0)c y0 −

m (

ck 0

k=1 t

+ 0

ep (t, s)f (s)Δs,



tk

ep (tk , s)f (s)Δs

 f ∈ SF,y .

Note that any fixed point y ∈ C ([0, σ (b)]) of the operator N is a solution of the IVP (7.11), (7.12). 1. We will prove that N(y) is convex for any single-valued function y ∈ C ([a, b]). Let h1 , h2 ∈ N(y) be arbitrarily chosen. Then there exist f1 , f2 ∈ SF,y such that   m tk ( h1 (t) = ep (t, 0)c y0 − ck ep (tk , s)f1 (s)Δs k=1 t

+ 0

ep (t, s)f1 (s)Δs,

0

460

7 First Order Dynamic Inclusions

 h2 (t) = ep (t, 0)c y0 −

m (



tk

ck 0

k=1

ep (tk , s)f2 (s)Δs

t

+ 0

a.e.

ep (t, s)f2 (s)Δs,

t ∈ [0, σ (b)].

Take  ∈ [0, 1] arbitrarily. Since F : [0, σ (b)] × R → Kc (R), we have that f1 + (1 − )f2 ∈ SF,y . Next, for any t ∈ [0, σ (b)], we have h1 (t) + (1 − )h2 (t)  = ep (t, 0)c y0 −

m (



tk

ck 0

k=1

ep (tk , s)f1 (s)Δs

t

+ 0

ep (t, s)f1 (s)Δs, 

+(1 − )ep (t, 0)c y0 −

m (

ck 0

k=1

0

ep (t, s)f2 (s)Δs

 = ep (t, 0)c y0 −

m (

tk

ck

k=1

+ 0

ep (tk , s)f2 (s)Δs

t

+(1 − )

t



tk

0

 ep (tk , s) (f1 (s) + (1 − )f2 (s)) Δs

ep (t, s) (f1 (s) + (1 − )f2 (s)) Δs

a.e. t ∈ [0, σ (b)]. Thus, h1 + (1 − )h2 ∈ N(y). 2. Take  > 0 arbitrarily. Consider  Br = y ∈ C ([0, σ (b)]), y : [0, σ (b)] → R,  max |y(t)| = r .

t∈[0,σ (b)]

Suppose that for any r > 0 there exists an yr ∈ Br such that hr ∈ N(yr ),

sup t∈[0,σ (r)]

|hr (t)| > r

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

461

and  hr (t) = ep (t, 0)c y0 −

m (



tk

ck

ep (tk , s)fr (s)Δs

0

k=1 t

+ 0

ep (t, s)fr (s)Δs

t ∈ [0, σ (b)],

a.e.

for some fr ∈ SF,yr . Hence, |hr (t)|

sup

r
0, define φr (t) = a(t) + r θ d(t),

t ∈ [0, σ (b)].

Then (H 9) holds. Hence by Theorem 7.4, it follows that the problem (7.11), (7.12) has at least one solution on [0, σ (b)]. This completes the proof. Example 7.8 Let T = 2N0 {0}, p(t) = 1, 

4 4 4 F (t, y(t)) = t 2 + t 3 y(t), t + t 2 + t 3 y(t), t + t 2 + t 3 + 3 y(t) , a = 0, b = 8,

t ∈ [0, 16],

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

467

m = 3, t1 = 1, t2 = 2, t3 = 4, c1 = 1, c2 = 1, c3 = 4, y0 = 10. We have σ (0) = 1, σ (t) = 2t, (1)(t) = −

1 1 + μ(t)

1 , 1+t t 1 + μ(t)(1)(t) = 1 − 1+t 1 , t ∈ T, = 1+t =−

1+

m (

t = 0,

ck ep (tk , 0) = 1 + e1 (1, 0) + e1 (2, 0) + 4e1 (4, 0)

k=1

= 0, i.e., (H 4) and (H 5) hold. Also, (H 6)–(H 8) hold. Next, 4 F (t, y(t)) ≤ t + t 2 + t 3 + t 3 y(t), θ =

t ∈ [0, 16],

1 . 3

Hence by Corollary 7.1, it follows that the problem (7.11), (7.12) has at least one solution on [0, 16]. Exercise 7.5 Let T = 4Z, p(t) =

1 , 1 + t2

468

7 First Order Dynamic Inclusions

√ 3

√  3 y(t) y(t) , ,t + 1 + 1 + (y(t))2 1 + (y(t))2

 F (t, y(t)) = t +

t ∈ [0, 16],

m = 1, t1 = 4, c1 = 1, y0 = 3. Prove that the problem (7.11), (7.12) has at least one solution on [0, 16]. Theorem 7.6 Suppose (H 4)–(H 8) and (H 10). Then the problem (7.11), (7.12) has at least one solution on [0, σ (b)]. Proof Let the operator N : C ([0, σ (b)]) → 2C ([0,σ (b)]) be defined as in the proof of Theorem 7.4. Suppose that y ∈ λN(y) for some λ ∈ (0, 1). Then there exists a single-valued function f ∈ SF,y such that  y(t) = λep (t, 0)c y0 −

m (

ck

t 0

ep (tk , s)f (s)Δs

0

k=1

+λ



tk

t ∈ [0, σ (b)].

ep (t, s)f (s)Δs,

Then, using (H 10), we get |y(t)| ≤

sup t∈[0,σ (b)]

ep (t, 0)|c|



× |y0 | +

m (

|ck |

+ 

sup (t,s)∈[0,σ (b)]×[0,σ (b)] 0

k=1

ep (tk , s)|f (s)|Δs

t

sup 0 (t,s)∈[0,σ (b)]×[0,σ (b)]

≤ 1+

sup t∈[0,σ (b)]

ep (t, s)|f (s)|Δs

ep (t, 0)|c|

m (

 |ck |

k=1 σ (b)

×

sup (t,s)∈[0,σ (b)]×[0,σ (b)]

+



tk

sup t∈[0,σ (b)]

≤ γ +δ

ep (t, s)

|f (s)|Δs

0

ep (t, 0)|c||y0 |  sup

(t,s)∈[0,σ (b)]×[0,σ (b)]

ep (t, s)φ

 sup

t∈[0,σ (b)]

σ (b)

|y(t)|

q(s)Δs, 0

7.2 Existence Results for First Order Dynamic Inclusions with Nonlocal Initial. . .

469

t ∈ [0, σ (b)]. Hence, supt∈[0,σ (b)] |y(t)| ≤ 1.   ! σ (b) γ + δ sup(t,s)∈[0,σ (b)]×[0,σ (b)] ep (t, s)φ supt∈[0,σ (b)] |y(t)| 0 q(s)Δs This is a contradiction. Then, by (H 10), it follows that there exists a constant M such that sup

|y(t)| = M.

t∈[0,σ (b)]

Define 

/

V = y ∈ C ([0, σ (b)]) :

sup

|y(t)| < M .

t∈[0,σ (b)]

We have that N : V → 2C ([0,σ (b)]) and as in the proof in Theorem 7.4, we have that it is a compact multivalued map, u.s.c. with convex closed values. By the definition of V , it follows that there is no y ∈ ∂V such that y ∈ λN(y) for some λ ∈ (0, 1). From here and from Theorem B.9 (see the appendix of this book), we conclude that the operator N has a fixed point in V . This completes the proof. Example 7.9 Let T = R, p(t) = 0, a = 0, m = 2, t1 = 1, t2 = 2, c1 = 1, c2 = 2, y0 = 20, b = 3,  2  2 t +1 2 t +2 2 F (t, y(t)) = (y(t)) , (y(t)) , 800 800

t ∈ [0, 3].

We have that F (t, y(t)) ≤

t2 + 2 (y(t))2 , 800

t ∈ [0, 3].

470

7 First Order Dynamic Inclusions

Thus, φ(z) = z2 , z2 + 2 , 800

q(z) =

z ∈ R.

Note that σ (b)

q(s)Δs =

0

=

1 800 1 800

3

(s 2 + 2)ds

0



 1 3 s=3 s + 2s  s=0 3

1 (9 + 6) 800 15 = 800 3 , = 160 ep (t, s) = 1, =

c = (1 + 1 + 2)−1 1 . 4 1 γ = (20) 4 = 5,   1 δ = 1 + (1 + 2) 4 =

=

7 . 4

Then, M γ + δ sup(t,s)∈[0,σ (b)]×[0,σ (b)] ep (t, s)φ(M) M 5+

7 2 4M



3 160

 >1

! σ (b) 0

⇐⇒

>1 q(s)Δs

⇐⇒

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

M 5+

21 2 640 M

>1

471

⇐⇒

21 2 M − M + 5 < 0 ⇐⇒ 640   ⎛ 5 ⎞ 5  11 320 1 + 11 320 1 − 32 32 ⎟ ⎜ ⎟. , M∈⎜ ⎠ ⎝ 21 21 Therefore, if M belongs to the last interval, then the considered problem has at least one solution on [0, 3].

7.3 Existence of Solutions of First Order Dynamic Inclusions with General Boundary Conditions Let a, b ∈ T, a < b. In this section, we will investigate the following multivalued boundary value problem: y Δ (t) ∈ F (t, y(t)) Δ − a.e.

t ∈ [a, b]κ ,

(7.14)

L(y(a), y(b)) = 0,

(7.15)

where (H11) (H12) (H13) (H14)

t → F (t, y) is measurable for each y ∈ R. For every (t, y) ∈ [a, b]κ × R, F (t, y) is compact and convex. y → F (t, y) is u.s.c. for almost all t ∈ [a, b]κ . For every p > 0, there exists a single-valued function mp ∈ L1 ([a, b]κ ) such that F (·, y) = sup{|v| : v ∈ F (·, y)} ≤ mp (·)

a.e.

on

[a, b]κ ,

for all single-valued function y ∈ C ([a, b]) with sup |y(t)| ≤ p. t∈[a,b]

(H15) There exist single-valued functions α ∈ AC([a, b]), v1 ∈ L1 ([a, b]) such that v1 (·) ∈ F (·, α(·)) a.e. on [a, b]κ , α Δ (·) ≤ v1 (·) a.e. on [a, b]κ and L(α(a), α(b)) ≤ 0.

472

7 First Order Dynamic Inclusions

(H16) There exist single-valued functions β ∈ AC([a, b]), v2 ∈ L1 ([a, b]) such that v2 (·) ∈ F (·, β(·)) a.e. on [a, b]κ , β Δ (·) ≥ v2 (·) a.e. on [a, b]κ and L(β(a), β(b)) ≥ 0. (H17) L is a continuous single-valued map on [α(a), β(a)] × [α(b), β(b)], L is non-increasing with respect to the second argument. (H18) {v1 (t), v2 (t)} ∈ F (t, y) for all y ∈ [α(t), β(t)] and for all right-scattered t ∈ [a, b]κ . Definition 7.3 A single-valued function y ∈ AC([a, b]) is said to be a solution of (7.14), (7.15), if there exists a single-valued function f ∈ L1 ([a, b]) such that 1. f (t) ∈ F (t, y(t)) a.e. t ∈ [a, b]κ . 2. y Δ (t) = f (t) a.e. t ∈ [a, b]κ . 3. The condition (7.15) holds. Example 7.10 Let T = Z. Consider the following problem: y Δ (t) ∈ {3t 2 − 3t − 10, 3t 2 − 3t − 3, 3t 2 } a.e.

t ∈ [0, 4]κ ,

13y(0) − y(4) = 0. Here F (t, y(t)) = {3t 2 − 3t − 10, 3t 2 − 3t − 3, 3t 2 }, a = 0, b = 4, L(y(a), y(b)) = 13y(0) − y(4). We will prove that the function y(t) = t 3 − 3t 2 − t + 1,

t ∈T

is a solution of the considered problem. Really, we have σ (t) = t + 1, y Δ (t) = (σ (t))2 + tσ (t) + t 2 − 3(σ (t) + t) − 1 = (t + 1)2 + t (t + 1) + t 2 − 3(t + 1 + t) − 1 = t 2 + 2t + 1 + t 2 + t + t 2 − 6t − 3 − 1 = 3t 2 − 3t − 3 ∈ F (t, y(t)),

t ∈ [0, 4],

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

and y(0) = 1, y(4) = 43 − 3 · 42 − 4 + 1 = 64 − 48 − 3 = 13, 13y(0) − y(4) = 13 − 13 = 0. Example 7.11 Let T = 2N0 . Consider the following problem:  y (t) ∈ Δ

2 − 3t − 2t 2 −7 − 3t − t 2 , , 2(t 2 + 2)(2t 2 + 1) 2(t 2 + 2)(2t 2 + 1)  t2 a.e. t ∈ [1, 8], 2(t 2 + 2)(2t 2 + 1) 3y(1) + 22y(8) = 5.

Here  F (t, y(t)) =

2 − 3t − 2t 2 −7 − 3t − t 2 , , 2(t 2 + 2)(2t 2 + 1) 2(t 2 + 2)(2t 2 + 1)  t2 , t ∈ [1, 8], 2(t 2 + 2)(2t 2 + 1)

a = 1, b = 8, L(y(1), y(8)) = 3y(1) + 22y(8) − 5. We will prove that the function y(t) =

t +1 , t2 + 2

t ∈T

is a solution of the considered problem. Really, we have σ (t) = 2t, y Δ (t) =

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

473

474

7 First Order Dynamic Inclusions

=

t 2 + 2 − 3t (t + 1) (t 2 + 2)(4t 2 + 2)

=

t 2 + 2 − 3t 2 − 3t 2(t 2 + 2)(2t 2 + 1)

=

2 − 3t − 2t 2 2(t 2 + 2)(2t 2 + 1)

∈ F (t, y(t)),

t ∈ [1, 8],

and 1+1 1+2 2 = , 3 8+1 y(8) = 2 8 +2 9 = 66 3 , = 22     3 2 + 22 3y(1) + 22y(8) = 3 3 22 y(1) =

= 2+3 = 5. Exercise 7.6 Let T = 3Z. Prove that the function y(t) = t 2 + 2t + 4,

t ∈ T,

is a solution of the following problem: y Δ (t) ∈ {−4, 2t, 2t + 5},

t ∈ [0, 9],

103y(0) − 4y(9) = 0. Define A1,y = {t ∈ [a, b] : y(t) < α(t)}, A2,y = {t ∈ [a, b] : β(t) < y(t)},

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

475

A3,y = [a, b]\{A1,y , A2,y }. Theorem 7.7 Suppose (H 11)–(H 18) and α(t) ≤ β(t), t ∈ [a, b]. Then, the problem (7.14), (7.15) has at least one solution y such that α(t) ≤ y(t) ≤ β(t), t ∈ [a, b]. Proof Define the single-valued function τ : [a, b] × R → R as follows: τ (t, y) = max{α(t), min{y(t), β(t)}}. For a single-valued function y ∈ C ([a, b]), define y (t) = τ (t, y(t)),

t ∈ [a, b],

and 1 1 SF, y (t)) for a.e. t ∈ [a, b]κ }, y = {v ∈ L ([a, b]) : v(t) ∈ F (t,  1 1 for each t with σ (t) ∈ A1,y SF, y = v ∈ SF, y :

we have v(t) ≥ v1 (t) and for each t with σ (t) ∈ A2,y  we have v(t) ≤ v2 (t) . Let also, F1 (t, y) = F (t, τ (t, y)) + τ (t, y),

(t, y) ∈ [a, b] × R.

1. We will prove that for y ∈ C ([a, b]), we have 1 SF, y = ∅. 1 = ∅. For v ∈ S 1 , define vˆ : [a, b]κ → R by Note that SF, y F, y

v(t) ˆ = v1 (t)χA1,y (σ (t)) + v2 (t)χA2,y (σ (t)) +v(t)χA3,y (σ (t)). Since A1,y and A2,y are open sets, we have that A3,y is a Borel set. Then χA1,y , χA2,y , and χA3,y are Borel measurable functions. Hence, using that v1 , v2 , v, σ are Lebesgue measurable, it follows that vˆ ∈ L1 ([a, b]κ ). Let now t ∈ [a, b]κ and t is a right-dense point.

476

7 First Order Dynamic Inclusions

1 , we get a. Suppose that t ∈ A3,y . Then, using that v ∈ SF, y

v(t) ˆ = v(t) ∈ F (t, y (t)). b. Suppose that t ∈ A1,y . Then, v(t) ˆ = v1 (t) ∈ F (t, α(t)) = F (t, y (t)). c. Suppose that t ∈ A2,y . Then, v(t) ˆ = v2 (t) ∈ F (t, β(t)) = F (t, y (t)). Let now t ∈ [a, b]κ and t is a right-scattered point. 1 , we get a. Suppose that σ (t) ∈ A3,y . Then, using that v ∈ SF, y

v(t) ˆ = v(t) ∈ F (t, y (t)). b. Suppose that σ (t) ∈ A1,y . Then, using (H 18), we get v(t) ˆ = v1 (t) ∈ F (t, y (t)). c. Suppose that σ (t) ∈ A2,y . Then, again using (H 18), we obtain v(t) ˆ = v2 (t) ∈ F (t, y (t)). Define the operator T : C ([a, b]) → 2C ([a,b]) by  T y = w ∈ C ([a, b]) : w(t) = τ (a, y(a) − L( y (a), y (b))) t

+ a

y (s)) Δs (−y(s) + v(s) +

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

t ∈ [a, b],

for all

477

 1 v ∈ SF, y .

a. We will prove that T is convex-valued and compact-valued. Let y ∈ 1 C ([a, b]), w1 , w2 ∈ T y, λ ∈ [0, 1]. Hence, there exist u1 , u2 ∈ SF, y such that y (a), y (b))) w1 (t) = τ (a, y(a) − L( t

+

y (s)) Δs, (−y(s) + u1 (s) +

a

y (a), y (b))) w2 (t) = τ (a, y(a) − L( t

+

y (s)) Δs, (−y(s) + u2 (s) +

t ∈ [a, b].

a

Then, λw1 (t) + (1 − λ)w2 (t) = λτ (a, y(a) − L( y (a), y (b))) t

+λ

y (s)) Δs (−y(s) + u1 (s) +

a

+(1 − λ)τ (a, y(a) − L( y (a), y (b))) t

+(1 − λ)

y (s)) Δs (−y(s) + u2 (s) +

a

= τ (a, y(a) − L( y (a), y (b))) t

+

y (s)) Δs (−y(s) + λu1 (s) + (1 − λ)u2 (s) +

a

∈ T y(t),

a.e.

t ∈ [a, b]κ ,

1 . Now, we take a sequence {h } SF, because λu1 + (1 − λ)u2 ∈ n n∈N ⊂ T y y 1 SF, such that hn → h0 , as n → ∞. Then there exists a sequence {vn }n∈N ⊂ y such that

y (a), y (b))) hn (t) = τ (a, y(a) − L( t

+

y (s)) Δs, (−y(s) + vn (s) +

a 1 is closed, we get that h ∈ T y. Because SF, 0 y

n ∈ N,

a.e.

t ∈ [a, b]κ .

478

7 First Order Dynamic Inclusions

b. Now, we will prove that T is compact. Let   Br = y ∈ C ([a, b]) : max |y(t)| ≤ r . t∈[a,b]

1 such that Then, for each y ∈ Br and each w ∈ T y, there exists a v ∈ SF, y

w(t) = τ (a, y(a) − L( y (a), y (b))) t

+

y (s)) Δs, (−y(s) + v(s) +

t ∈ [a, b].

a

Then, |w(t)| ≤ max{|α(a)|, |β(a)|} t

+

y (s)|) Δs, (|y(s)| + |v(s)| + |

t ∈ [a, b].

0

Note that |y(s)| ≤ r,

s ∈ [a, b],

|v(s)| ≤ mr (s),

a.e.

s ∈ [a, b]κ ,

| y (s)| = |τ (s, y (s))| = |max{α(s), min{y(s), β(s)}, r}|   ≤ max max |α(s)|, max |β(s)|, r , s∈[a,b]

s∈[a,b]

s ∈ [a, b].

Hence, |w(t)| ≤ max{|α(a)|, |β(a)|} b

+r(b − a) +

mr (s)Δs a

 +(b − a) max

 max |α(s)|, max |β(s)|, r ,

s∈[a,b]

s∈[a,b]

a.e. t ∈ [a, b]. Thus, T maps bounded sets into bounded sets. Let now t1 , t2 ∈ 1 such that [a, b], t1 ≤ t2 and y ∈ Br . For each w ∈ T y, there exists a v ∈ SF, y w(t) = τ (a, y(a) − L( y (a), y (b))) t

+ a

y (s)) Δs, (−y(s) + v(s) +

t ∈ [a, b].

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

479

Then   y (a), y (b))) |w(t2 ) − w(t1 )| = τ (a, y(a) − L( t2

+

y (s)) Δs (−y(s) + v(s) +

a

−τ (a, y(a) − L( y (a), y (b))) t1

−   = 

a t2 t1 t2

≤

  y (s)) Δs  (−y(s) + v(s) +

  (−y(s) + v(s) + y (s))Δs 

y (s)|) Δs (|y(s)| + |v(s)| + |

t1 t2

≤ r(t2 − t1 ) +  + max

mr (s)Δs t1



max |α(s)|, max |β(s)|, r (t2 − t1 ).

s∈[a,b]

s∈[a,b]

Hence by the Arzela–Ascoli theorem, it follows that T maps bounded sets into equi-continuous sets. Thus, T is compact. c. Now, we will prove that T is u.s.c. For this aim, we will prove that T has a closed graph. Let yn → y0 , as n → ∞, hn ∈ T yn , n ∈ N, hn → h0 , as 1 such that n → ∞. Hence, for any n ∈ N0 , there exists a vn ∈ SF, y yn (a), yn (b))) hn (t) = τ (a, yn (a) − L( t

+

yn (s)) Δs, (−yn (s) + vn (s) +

a.e.

t ∈ [a, b].

a

Define Γ : L1 ([a, b]) → C ([a, b]) by t

(Γ v)(t) =

v(s)Δs. a

Note that Γ is a linear and continuous operator. Define yn (a), yn (b))) wn (t) = hn (t) − τ (a, yn (a) − L( t

−

yn (s)) Δs, (−yn (s) + vn (s) +

a

Note that wn → w0 in C ([a, b]),

t ∈ [a, b].

480

7 First Order Dynamic Inclusions t

wn (t) =

vn (s)Δs,

a.e.

t ∈ [a, b],

a

  1 SF, implies that wn ∈ Γ yn . Then, using Theorem B.8 (see the appendix of this book), we have w0 (t) = h0 (t) − τ (a, y0 (a) − L( y0 (a), y0 (b))) t

−

y0 (s)) Δs, (−y0 (s) + v0 (s) +

t ∈ [a, b],

a.e.

a 1 such that i.e., there exists a v0 ∈ SF, y t

v0 (s)Δs = h0 (t) − τ (a, y0 (a) − L( y0 (a), y0 (b)))

0 t

−

y0 (s)) Δs, (−y0 (s) + v0 (s) +

t ∈ [a, b].

a

Hence, h0 ∈ T y0 and T is u.s.c. d. We will prove that the set {v ∈ C ([a, b]), λv ∈ T v

for

some

λ > 1}

is bounded. By the boundedness condition on F , there exists a φ ∈ L1 ([a, b]) such that F (t, τ (t, y(t))) ≤ φ(t) a.e.

t ∈ [a, b]κ ,

∀y ∈ C ([a, b]).

Note that λy ∈ T y implies λy(t) = τ (a, y(a) − L( y (a), y (b))) t

+

y (s)) Δs, (−y(s) + v(s) +

a.e.

t ∈ [a, b],

a.e.

t ∈ [a, b].

a

whereupon y(t) =

1 τ (a, y(a) − L( y (a), y (b))) λ 1 t + y (s)) Δs, (−y(s) + v(s) + λ a

Also, there is a positive constant C such that

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

481

|τ (a, y(a) − L( y (a), y (b)))| ≤ C and t

t

|v(s)|Δs ≤

a

φ(s)Δs a b

≤

φ(s)Δs, a

t



| y (s)|Δs ≤ max

 max |α(s)|, max |β(s)|, r (b − a).

s∈[a,b]

a

s∈[a,b]

Then, b

|y(t)| ≤ C +

φ(s)Δs a



+ max t

+

 max |α(s)|, max |β(s)|, r

s∈[a,b]

|y(s)|Δs,

s∈[a,b]

t ∈ [a, b].

a

Let b

z0 = C +

φ(s)Δs a



+ max

 max |α(s)|, max |β(s)|, r .

s∈[a,b]

s∈[a,b]

Then t

|y(t)| ≤ z0 +

|y(s)|Δs,

t ∈ [a, b].

a

Hence by the Gronwall inequality, we arrive at the inequality |y(t)| ≤ z0 e1 (t, a),

t ∈ [a, b].

Now, we apply the Martelli theorem, Theorem B.10 (see the appendix of this book). We conclude that the operator T has a fixed point y. We thus obtain a solution y of the following problem: y (t)) a.e. y Δ (t) + y(t) ∈ F1 (t,

on

[a, b]κ ,

y(a) = τ (a, y(a) − L( y (a), y (b))) .

482

7 First Order Dynamic Inclusions

As in above, for each t with σ (t) ∈ A2,y , we have v(t) ˆ ≤ v2 (t). 2. We will prove that α(t) ≤ y(t) ≤ β(t),

t ∈ [a, b].

Assume the contrary. Since α(a) ≤ y(a) ≤ β(a), it follows that there exist t, tˆ ∈ [a, b], t < tˆ, α(t) ≤ y(t) ≤ β(t) and either y(t) < α(t), t ∈ (t, tˆ] or y(t)β(t), t ∈ (t, tˆ]. We assume the first of these. Then, y (t)) + y (t) y Δ (t) + y(t) ∈ F (t, = F (t, α(t)) + α(t),

a.e.

t ∈ (t, tˆ]



[a, b]κ ,

and y Δ (t) + y(t) ∈ F (t, y (t)) + y (t) = F (t, y(t)) + y(t), i.e., y Δ (t) ∈ F (t, y(t)), although this might not hold if {t} is a set of measure zero.  a. Assume that (t, tˆ) [a, b]κ = ∅. Then, there exists a v ∈ L1 ([a, b]) such that v(t) ∈ F (t, y (t)) and y (t), y Δ (t) + y(t) = v(t) +

a.e.

t ∈ [a, b]κ ,

and v(t) ≥ v1 (t), Let t ∈ (t, tˆ)



t ∈ [t, tˆ]

with σ (t) ∈ (t, tˆ].

[a, b]κ .

i. Let σ (t) = t. Hence, {t} is a set of measure zero. We have

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . . t

y(t) − y(t) =

y Δ (s)Δs

t t

=

(v(s) + α(s) − y(s))Δs

t t

=

t

v(s)Δs +

t

(α(s) − y(s))Δs

t

≥ α(t) − α(t), whereupon 0 > y(t) − α(t) ≥ y(t) − α(t) ≥ 0, which is a contradiction. ii. Let σ (t) > t. Then {t} is not a set of measure zero. Furthermore, t

y(t) − y(t) =

y Δ (s)Δs

t σ (t)

=

t

y Δ (s)Δs +

t

y Δ (s)Δs

σ (t) t

= y Δ (t)(σ (t) − t) +

(v(s) + α(s) − y(s))Δs

σ (t)

= v(t)(σ (t) − t) +

t

v(s)Δs σ (t)

t

+

(α(s) − y(s))Δs

σ (t)

≥ v1 (t)(σ (t) − t) +

t

v1 (s)Δs σ (t)

t

≥

v1 (s)Δs t

≥ α(t) − α(t), whereupon 0 > y(t) − α(t) ≥ y(t) − α(t) ≥ 0,

483

484

7 First Order Dynamic Inclusions

which is a contradiction.  b. Let (t, tˆ) [a, b]κ = ∅. Then we have tˆ = b, b is left-scattered, {ρ(b)} is not a set of measure zero, α(ρ(b)) ≤ y(ρ(b)),

α(b) > y(b).

Therefore, y (ρ(b)) y Δ (ρ(b)) + y(ρ(b)) = v(ρ(b)) + = v(ρ(b)) + y(ρ(b)). Also, y Δ (ρ(b)) =

y(b) − y(ρ(b)) . b − ρ(b)

Consequently, v(ρ(b)) =

y(b) − y(ρ(b)) . b − ρ(b)

We have v(ρ(b)) > v1 (ρ(b)). Hence, α(b) − α(ρ(b)) = α Δ (ρ(b)) b − ρ(b) ≤ v1 (ρ(b)) ≤ v(ρ(b)) =

y(b) − y(ρ(b)) , b − ρ(b)

i.e., α(b) − α(ρ(b)) ≤ y(b) − y(ρ(b)). From here, 0 < α(b) − y(b) ≤ α(ρ(b)) − y(ρ(b)) ≤ 0.

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

This is a contradiction. Therefore, y (t)), y Δ (t) + y(t) ∈ F1 (t,

a.e.

t ∈ [a, b]κ ,

implies that y Δ (t) ∈ F (t, y(t)),

a.e.

t ∈ [a, b]κ .

3. Now, we will prove that L(y(a), y(b)) = 0. Suppose that y(a) − L(y(a), y(b)) < α(a). By the definition of the function τ , it follows that y(a) = α(a). So, y(a) − L(α(a), y(b)) < α(a). Since L is non-increasing with respect to the second argument, we have α(a) ≤ α(a) − L(α(a), α(b)) ≤ α(a) − L(α(a), y(a)) < α(a), which is a contradiction. Therefore, y(a) − L(y(a), y(b)) ≥ α(a). As in above, y(a) − L(y(a), y(b)) ≤ β(a). Then, y(a) = τ (a, y(a) − L(y(a), y(b))) = y(a) − L(y(a), y(b)) and from here, L(y(a), y(b)) = 0.

485

486

7 First Order Dynamic Inclusions

Consequently, y is a solution of the problem (7.14), (7.15). This completes the proof. Example 7.12 Let T = Z. Consider the problem y Δ (t) = f (y(t)),

t ∈ [0, 2],

y(0) = 1, where ⎧ ⎨ 6 if y ≤ −1 f (y) = 2y(y − 2) if − 1 ≤ y ≤ 1 ⎩ 0 if y ≥ 2. Take ⎧ ⎨ −1 if α(t) = −1 if ⎩ −1 if ⎧ ⎨ 2 if t β(t) = 2 if t ⎩ 6 if t

t =0 t =1 t ≥ 2, =0 =1 ≥ 2.

We will prove that ⎧ ⎨ 1 if t = 0 y(t) = −1 if t = 1 ⎩ 5 if t ≥ 2 is its solution and α(t) ≤ y(t) ≤ β(t),

t ∈ [0, 2].

We have σ (t) = t + 1, y Δ (0) =

y(σ (0)) − y(0) σ (0) − 0

= y(1) − y(0) = −1 − 1 = −2, f (y(0)) = f (1)

7.3 Existence of Solutions of First Order Dynamic Inclusions with General. . .

= 2(1 − 2) = −2. Therefore, y Δ (0) = f (y(0)), α(0) ≤ y(0) ≤ β(0). Next, y Δ (1) =

y(σ (1)) − y(1) σ (1) − 1

= y(2) − y(1) = 5 − (−1) = 6, f (y(1)) = f (−1) = 2 · (−1) · (−1 − 2) = 6. Hence, y Δ (1) = f (y(1)), α(1) ≤ y(1) ≤ β(1). Also, y Δ (t) =

y(σ (t)) − y(t) σ (t) − t

= y(t + 1) − y(t) = 5−5 = 0, f (y(t)) = 0,

t ≥ 2.

Thus, y Δ (t) = f (y(t)), α(t) ≤ y(t) ≤ β(t),

t ≥ 2.

487

488

7 First Order Dynamic Inclusions

Example 7.13 Let T = Z, a = 0, b = 2, ⎧  1 ⎪ if y < −1 ⎪ ⎨ 2  1 F (0, y) = , −y + 2 if y ∈ [−1, 1] 2 ⎪ ⎪ ⎩ {1} if y > 1, ⎧  1 ⎪ if y < − 12 ⎪ 2 ⎨ 

F (1, y) = [y − 4, y + 1] if y ∈ − 1 , 2 2 ⎪ ⎪ ⎩ {−2} if y > 2, ⎧ ⎨ −1 if t = 0 α(t) = − 12 if t = 1 ⎩ 0 if t = 2, ⎧ ⎨ 1 if t = 0 β(t) = 2 if t = 1 ⎩ 0 if t = 2. Consider Eq. (7.14) subject to the boundary condition y(0) = y(2). We have that the functions F , L, α, and β satisfy (H 11)–(H 18). Hence by Theorem 7.7, it follows that the considered problem has at least one solution on [0, 2]κ . Now, we will prove that ⎧ ⎨ 0 if t = 0 y(t) = 1 if t = 1 ⎩ 0 if t = 2 is its solution. We have [0, 2]κ = [0, 1], σ (t) = t + 1,

t ∈ T,

y(σ (0)) − y(0) y Δ (0) = σ (0) − 0 = y(1) − y(0) =1 ∈ F (0, y(0)), y(1) = 1,

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .

y Δ (1) =

489

y(σ (1)) − y(1) σ (1) − 1

= y(2) − y(1) = 0−1 = −1 ∈ F (1, y(1)), and α(0) ≤ y(0) ≤ β(0), α(1) ≤ y(1) ≤ β(1).

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic Boundary Conditions Suppose that T is a compact time scale with a = min T < b = max T. Let also, x0 ∈ R and T0 = T\{b}. In this section, we will investigate the following periodic boundary value problem: x Δ (t) ∈ F (t, x(σ (t))) Δ − a.e. x(a) = x0

or

t ∈ T0 ,

(7.16)

x(a) = x(b),

(7.17)

where (H19) F : T0 × R  Kc (R) such that t → F (t, x) is Δ-measurable for every x ∈ R, x → F (t, x) is u.s.c. Δ-a.e. t ∈ T0 . (H20) For every r > 0, there exists a function hr : T0 → [−, ∞), hr ∈ L1 (T0 ) such that max{|y| : y ∈ F (t, x),

|x| ≤ r} ≤ hr (t) Δ − a.e.

t ∈ T0 .

Definition 7.4 Let v : T → R, M : T → [0, ∞), v, M ∈ H1 (T). We say that (v, M) is a solution-tube of (7.16) if 1. Δ-a.e. t ∈ T0 and for every x ∈ R such that |x − v(σ (t))| = M(σ (t)), there exists a δ > 0 so that, for any u ∈ R for which |u − x| < δ and

490

7 First Order Dynamic Inclusions

|u − v(σ (t))| ≥ M(σ (t)), there exists an y ∈ F (t, u) such that |y − v Δ (t)| ≤ M Δ (t). 2. v Δ (t) ∈ F (t, v(σ (t))) Δ-a.e. t ∈ T0 such that M(σ (t)) = 0. 3. M(t) = 0 for every t ∈ T0 such that M(σ (t)) = 0. 4. |v(b) − v(a)| ≤ M(a) − M(b). Denote T (v, M) = {x ∈ H1 (T) : |x(t) − v(t)| ≤ M(t),

t ∈ T}.

In addition, we suppose (H21) There exists a v : T → R and M : T → [0, ∞), v, M ∈ H1 (T) such that (v, M) is a solution-tube of (7.16). For t ∈ T0 and x ∈ R, define  −

x (t, x) =

M(t) |x−v(t)| (x

x

− v(t)) + v(t) otherwise,

if |x − v(t)| > M(t),

⎧ Δ v (t) if M(σ (t)) = 0, ⎪ ⎪ ⎨ R if M(σ (t)) > 0 and G(t, x) = ⎪ |x − v(σ (t))| ≤ M(σ (t)), ⎪ ⎩ {z : |z − v Δ (t)| ≤ M Δ (t)} otherwise,  F0 (t, x) = F (t, x − (σ (t), x)) G(t, x), x(σ (t)) = x − (σ (t), x(σ (t))). Note that if (H 19) and (H 20) hold, then G has nonempty, closed, convex values for all x ∈ R and for Δ-almost every t ∈ T0 . Theorem 7.8 Suppose (H 19) and (H 20). Then 1. x → G(t, x) has closed graph for Δ-almost every t ∈ T0 . 2. t → G(t, x) is Δ-measurable for every x ∈ R. Proof 1. For t ∈ T0 , define At = {(x, y) ∈ R2 : y ∈ G(t, x)}.

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .

491

Fix t ∈ T0 such that M(σ (t)) = 0. Take a sequence {(xk , yk )}k∈N of elements of the set At such that xk → x, yk → y, as k → ∞. If |x − v(σ (t))| ≤ M(σ (t)), then, G(t, x) = R and y ∈ G(t, x). So, (x, y) ∈ At . Suppose that |x − v(σ (t))| > M(σ (t)) and for k sufficiently large, let |xk − v(σ (t))| > M(σ (t)), |yk − v Δ (t)| ≤ M Δ (t). Then, |y − v Δ (t)| ≤ M Δ (t). 2. Let C be a nonempty, closed subset of R. Fix x ∈ R and let {ym }m∈N be a countable dense subset of C. Define  B1 = {t ∈ T0 : v Δ (t) ∈ C} {t ∈ T0 : M(σ (t)) = 0},  B2 = {t ∈ T0 : |x − v(σ (t))| − M(σ (t)) ≤ 0} {t ∈ T0 : M(σ (t)) > 0},  B3 = {t ∈ T0 : |x − σ (t)| − M(σ (t)) > 0} {t ∈ T0 : M(σ (t)) > 0},  "

1 {t ∈ T0 : |ym − v Δ (t)| ≤ M Δ (t) + }, k k∈N m∈N  Bx = {t ∈ T0 : G(t, x) C = ∅}. B4 =

We have that B x = B1

"

B2

"

(B3



B4 ).

Since the maps t → v(σ (t)), t → M(σ (t)), t → v Δ (t), and t → M Δ (t) are Δ-measurable, we get that Bx is Δ-measurable, and so is t → G(t, x). This completes the proof. Now, we define the multidimensional map F : C (T)  L1 (T) by F (x) = {w ∈ L1 (T0 ) : w(t) ∈ F0 (t, x(σ (t)))

Δ − a.e.

t ∈ T0 }.

492

7 First Order Dynamic Inclusions

Theorem 7.9 Assume (H 19)–(H 21). Then F has nonempty, convex values and there exists an h ∈ L1 (T0 ), h : T0 → [0, ∞), such that |w(t)| ≤ h(t) Δ − a.e.

t ∈ T0

for all w ∈ F (x) and all x ∈ C (T). Proof Let x ∈ C (T) be arbitrarily chosen. Then there exists a sequence {xm }m∈N of simple functions such that |xm (σ (t) − v(σ (t))| > M(σ (t))) Δ-a.e. on {t ∈ T0 : |x(σ (t)) − v(σ (t))| > M(σ (t))}, and such that xm → x in C (T). Because the multivalued maps t → F (t, y) and t → G(t, y) are Δ-measurable for every y ∈ R, we have that the maps t → F (t, xm (σ (t))) and

t → G(t, xm (σ (t)))

are Δ-measurable for every m ∈ N. Hence, the map t → F (t, xm (σ (t)))



G(t, xm (σ (t)))

is Δ-measurable for any m ∈ N, and for every k ∈ N, the map t→

  " F (t, xm (σ (t))) G(t, xm (σ (t))) m≥k

is Δ-measurable. From here, the map

t→

 k∈N

⎛ ⎝

"

F (t, xm (σ (t)))



⎞  G(t, xm (σ (t))) ⎠

m≥k

is Δ-measurable. Hence by (H 21), it follows that this map has nonempty values Δ-almost everywhere on {t ∈ T0 : M(σ (t)) = 0}. Thus, the multivalued map Γ : T0 → L1 (T0 ), defined by

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .

Γ (t) =

493

⎧   ⎪  ⎪    ⎪ ⎪ F (t, xm (σ (t))) G(t, xm (σ (t))) ⎪ ⎪ ⎪ ⎪ ⎨ k∈N m≥k ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

t ∈ {t ∈ T0 : M(σ (t)) = 0},

if

v Δ (t) if

t ∈ {t ∈ T0 : M(σ (t)) = 0},

is Δ-measurable and has nonempty and compact values. Hence, it has a Δmeasurable selection w. Because w(t) ∈ Γ (t) Δ-a.e., we have w(t) ∈

 "  F (t, xm (σ (t))) G(t, xm (σ (t))) m≥k

Δ-a.e. in {t : M(σ (t)) = 0}, for any k ∈ N. So, for Δ-almost every t ∈ {t ∈ T0 : M(σ (t)) = 0}, there exists a sequence {uml }l∈N such that uml (t) ∈ F (t, xml (σ (t)))



G(t, xml (σ (t)))

so that ul (t) → w(t). If |x(σ (t)) − v(σ (t))| ≤ M(σ (t)), using that y → F (t, y) and y → G(t, y) have closed graph and xml (σ (t)) → x(σ (t)) = x(σ (t)),

l → ∞,

as

we get w(t) ∈ F (t, x(σ (t)))



G(t, x(σ (t)))

= F0 (t, x(σ (t))). Next, if |x(σ (t)) − v(σ (t))| > M(σ (t)), using that xml (σ (t)) → x(σ (t)), as l → ∞, we have that there exists a sequence {yml }l∈N such that x − (σ (t), yml ) = x ml (σ (t)), yml → x(σ (t)),

as

l → ∞,

494

7 First Order Dynamic Inclusions

and xml (σ (t)) = θml x ml (σ (t)) + (1 − θml )yml for some θml ∈ [0, 1). Also, uml (t) ∈ F (t, xml (σ (t))) = F (t, xml (σ (t)))

 

G(t, xml (σ (t))) G(t, yml ).

Because y → F (t, y) and y → G(t, y) have closed graph and xml (σ (t)) → x(σ (t)), yml (σ (t)) → x(σ (t)),

as

l → ∞,

we conclude that w(t) ∈ F (t, x(σ (t)))



G(t, x(σ (t)))

= F0 (t, x(σ (t))). By (H 21), it follows that Δ-a.e. on {t ∈ T0 : M(σ (t)) = 0}, w(t) = v Δ (t) ∈ F (t, x(σ (t)))



G(t, x(σ (t)))

= F0 (t, x(σ (t))). Therefore, w ∈ F (x). Moreover, by the convexity of the values of F and G, it follows the convexity of F (x). By (H 20), we get that there exists an h = hr ∈ L1 (T0 ) with r = max{|v(t)| + M(t) : t ∈ T} such that |w(t)| ≤ h(t) for any w ∈ F (x) and for any x ∈ C (T). This completes the proof. Now, define the multivalued operator TI : C (T) → C (T) as follows:  TI (x) = u ∈ C (T), u : T → R :

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .





u(t) = e1 (a, t) x0 + where

495

 w ∈ F (x) .

[a,t)

e1 (s, a)(w(s) + x(σ (s)))Δs ,

Theorem 7.10 Assume (H 19)–(H 21). Then the operator TI is compact, u.s.c., with nonempty, convex and compact values. Proof By Theorem 7.9, it follows that the operator TI has nonempty, convex values and there exists a h ∈ L1 (T0 ), h : T0 → [0, ∞), such that w(t) ≤ h(t) Δ − a.e.

on

T0 .

Set r = max{|v(t)| + M(t) : t ∈ T}, c = max{|e1 (t, s)| : t, s ∈ T}. Note that for any u ∈ TI (C (T)), we have  |u(t)| ≤ c |x0 | +

[a,b)

 c(r + h(s))Δs ,

t ∈ T.

Next, for u ∈ TI (C (T)) and t, τ ∈ T, t > τ , we have     |u(t) − u(τ )| = e1 (a, t) x0 + e1 (s, a)(w(a) + x(σ (s)))Δs [a,t)

 −e1 (a, τ ) x0 +

[a,τ )

  e1 (s, a)(w(s) + x(σ (s)))Δs 

  = (e1 (a, t) − e1 (a, τ ))x0 +e1 (a, t) +e1 (a, t) −e1 (a, τ )

e1 (s, a)(w(s) + x(σ (s)))Δs

[a,τ )

[τ,t)

e1 (s, a)(w(s) + x(σ (s)))Δs

[a,τ )

  e1 (s, a)(w(s) + x(σ (s)))Δs 

  = (e1 (a, t) − e1 (a, τ ))x0

496

7 First Order Dynamic Inclusions

+(e1 (a, t) − e1 (a, τ )) −e1 (a, t)

[τ,t)

[a,τ )

e1 (s, a)(w(s) + x(σ (s)))Δs

  e1 (s, a)(w(s) + x(s))Δ

≤ |e1 (a, t) − e1 (a, τ )||x0 | +|e1 (a, t) − e1 (a, τ )| +|e1 (a, t)|

[τ,t)

[a,τ )

|e1 (s, a)|(|w(s)| + |x(σ (s))|)Δs

|e1 (s, a)|(|w(s)| + |x(σ (s))|)Δs 



≤ |e1 (a, t) − e1 (a, τ )| |x0 | + c +c2

[τ,t)

[a,τ )

(r + h(s))Δs

(r + h(s))Δs.

From here and from the fact that the maps t → e1 (a, t)

and

t→

[a,t)

(r + h(s))Δs

are continuous on T, we conclude that TI (C (T)) is equi-continuous. Then, by the Arzela–Ascoli theorem, we conclude that the operator TI (C (T)) is relatively compact in C (T). Now, we will prove that the operator TI has a closed graph. Let {xm }m∈N and {um }m∈N be convergent sequences in C (T) such that xm → x,

um → u,

as

m → ∞,

and um ∈ II (xm ). Let wm ∈ F (um ) be such that  um (t) = e1 (a, t) x0 +

 [a,t)

e1 (s, a)(um (s) + x m (σ (s)))Δs .

Let h be as in Theorem 7.9. Define  h(t) if t ∈ T ˆ h(t) = h(t ) if t ∈ (ti , σ (ti )), ti ∈ R,  i wm (t) if t ∈ T wˆ m (t) = wm (ti ) if t ∈ (ti , σ (ti )), ti ∈ R,

(7.18)

m ∈ N, where the set R is defined as in (6.1). We have wˆ m , hˆ ∈ L1 ([a, b]), m ∈ N, and

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .

497

ˆ |wˆ m (t)| ≤ h(t) for almost every t ∈ [a, b]. By the classical Dunford–Pettis theorem, it follows that there exists a g ∈ L1 ([a, b]) such that wˆ m → g, as m → ∞, in L1 ([a, b]) weakly. Then there exist zˆ m ∈ co{wˆ m , wˆ m+1 , . . .},

m ∈ N,

such that zˆ m → g, as m → ∞, in L1 ([a, b]). From here, there exists a subsequence again noted by {ˆzm }m∈N such that zˆ m (t) → g(t),

m → ∞,

as

for almost every t ∈ [a, b]. Therefore, for almost every t ∈ [a, b],  zˆ m (t) ∈ co ⊂ co

/



wˆ l (t)

l≥m

Fˆ (t, x l (σ (t)))



/ ˆ G(t, xl (σ (t))) ,

(7.19)

l≥m

ˆ are the extensions of the multivalued maps F and G, respectively, where Fˆ and G in the sense of (7.18). Now, taking the limit in (7.19), we arrive at g(t) ın co

⎧ ⎨" ⎩

Fˆ (t, x l (σ (t)))



ˆ xl (σ (t))) G(t,

l≥m

⊂ Fˆ (t, x(σ (t)))



⎫ ⎬ ⎭

ˆ x(σ (t))) G(t,

= Fˆ0 (t, x(σ (t))), ˆ y) have since xm → x, as m → ∞, in C (T) and the maps y → Fˆ (t, y), y → G(t, closed graphs and closed, convex values. Then there exists a function w : T0 → R such that g = w. ˆ So, w(t) ∈ Fˆ0 (t, x(σ (t))) = F0 (t, x(σ (t)))

Δ − a.e.

t ∈ T0 .

Therefore, w ∈ F (x). Because wˆ m → w, ˆ as m → ∞, in L1 ([a, b]), weakly, and xm → x, as m → ∞, in C (T), we have

[a,t)

e1 (s, a)(wm (s) + x m (σ (s)))Δs →

[a,t)

e1 (s, a)(w(s) + x(σ (s)))Δs,

498

7 First Order Dynamic Inclusions

as m → ∞. Hence by um → u, as m → ∞, in C (T), we get  u(t) = e1 (a, t) x0 +

[a,t)

 e1 (s, a)(w(s) + x(σ (s)))Δs ,

t ∈ T. Thus, u ∈ TI (x) and therefore TI has a closed graph. Because TI is compact and has a closed graph, we conclude that TI has compact values. Now, we will prove that TI is u.s.c. Let B ∈ C (T) be a closed set and A = {x ∈ C (T)LTI (x)



B = ∅}.

Let {xm }m∈N be a sequence in A converging to x in C (T). Then there exist um ∈ TI (xm ) B, m ∈ N. By the compactness of TI , it follows that there is a sequence again noted by {um }m∈N converging to uin C (T). Since B is closed and TI has a closed graph, we deduce that u ∈ TI (x) B. Consequently, x ∈ A. This completes the proof. Now, we define the multivalued operator TP : C (T) → C (T) as follows:  TP (x)(t) = v ∈ C (T) :  1 1 v(t) = e1 (t, a) e1 (b, a) − 1 +

[a,t)

where

[a,b)

(w(s) + x(σ (s)))e1 (s, a)Δs

 (w(s) + x(σ (s)))e1 (s, a)Δs ,  w ∈ F (x) .

As we have proved in Theorem 7.10, one can prove the following theorem. Theorem 7.11 Suppose (H 19)–(H 21). Then the operator TP is compact and u.s.c. with nonempty, convex and compact values. The main result in this section is as follows. Theorem 7.12 Assume  (H 19)–(H 21). Then the problem (7.16), (7.17) has a solution x ∈ H1 (T) T (v, M). Proof By Theorems 7.10 and 7.11, it follows that the operators TI and TP are compact and u.s.c. with nonempty, convex, and compact values. By the Kakutani fixed point theorem, Theorem B.11 (see the appendix of this book), it follows that the operators TI and TP have fixed points x and y, respectively, which satisfy the following problems: x Δ (t) + x(σ (t)) ∈ F0 (t, x(σ (t))) + x(σ (t)) Δ − a.e.

t ∈ T0 ,

7.4 Existence of Solutions of First Order Dynamic Inclusions with Periodic. . .

499

x(a) = x0 and y Δ (t) + y(σ (t)) ∈ F0 (t, y(σ (t))) + y(σ (t)) Δ − a.e.

t ∈ T0 ,

y(a) = y(b), respectively. We will show that x ∈ T (v, M). The fact that y ∈ T (v, M) we leave to the reader as an exercise. Define A = {t ∈ T0 : |x(σ (t)) − v(σ (t))| > M(σ (t))}. On A



R, μ({t}) > 0 and Δ-a.e. on A\R, we have |x(t) − v(t)|Δ ≤ |x Δ (t) − v Δ (t)|.

Let z(t) = x Δ (t) + x(σ (t)) − x(σ (t)) ∈ F0 (t, x(σ (t))) Δ-a.e. on T0 . Since (v, M) is a solution-tube of (7.16), we get (|x(t) − v(t)| − M(t))Δ < M Δ (t) − M Δ (t) =0 Δ-a.e. on {t ∈ A : M(σ (t)) > 0}. On the other hand, if M(σ (t)) = 0, then F0 (t, x(σ (t))) = {v Δ (t)} and Δ-a.e. on {t ∈ A : M(σ (t)) = 0} we have (|x(t) − v(t)| − M(t))Δ < −M Δ (t) < 0. Therefore, the function r(t) = |x(t) − v(t)| − M(t), satisfies r Δ (t) < 0

t ∈ T,

500

7 First Order Dynamic Inclusions

Δ-a.e. on A = {t ∈ T0 : r(σ (t)) > 0}. Since (v, M) is a solution-tube of (7.16), (7.17), then r(a) ≤ 0. Thus, A = ∅ and x ∈ T (v, M), i.e., x is a solution of the problem (7.16), (7.17). This completes the proof.

7.5 The Dual Time Scales Definition 7.5 Define the dual time scale T∗ as follows: T∗ = {s ∈ R : −s ∈ T}. ˆ and ∇ˆ we will denote the forward jump operator, the backward With σˆ , ρ, ˆ μ, ˆ νˆ , Δ, jump operator, the forward graininess operator, the backward graininess operator, the delta differentiation operator, and the nabla differentiation operator, respectively. Example 7.14 Let T = 2N0 . Then, T∗ = {−2n : n ∈ N}. Let t ∈ T∗ be arbitrarily chosen. Then there exists an n ∈ N0 such that t = −2n . Hence, σˆ (t) = σˆ (−2n ) = −2n−1 1 = (−2n ) 2 1 = t, 2 n ) ρ(t) ˆ = ρ(−2 ˆ = −2n+1 = 2(−2n ) = 2t, μ(t) ˆ = σˆ (t) − t 1 t −t 2 1 = − t, 2 νˆ (t) = t − ρ(t) ˆ =

= t − 2t = −t.

7.5 The Dual Time Scales

501

Example 7.15 Let T = hZ, h > 0. Then T = T∗ . Definition 7.6 For a given function f : T → R, define the dual function f ∗ : T∗ → R as follows: f ∗ (t) = f (−t),

t ∈ T∗ .

Example 7.16 Let T = 3N0 , f : T → R, be defined as follows: f (t) = (σ (t))2 + t 2 + t − ρ(t),

t ∈ T.

We have σ (t) = 3t, ρ(t) =

1 t, 3

t ∈ T.

Then, 1 f (t) = (3t)2 + t 2 + t − t 3 2 = 9t 2 + t 2 + t 3 2 = 10t 2 + t, t ∈ T. 3 Hence, f ∗ (t) = f (−t) 2 = 10(−t)2 + (−t) 3 2 = 10t 2 − t, t ∈ T∗ . 3 Exercise 7.7 Let T = 2N0 , f : T → R be defined as follows: f (t) = t 3 + (σ (t))2 t 2 + (σ (t))3 ρ(t) + t + 1,

t ∈ T.

Find f ∗ (t), t ∈ T∗ . By the definition for a dual function, we get the following properties of the dual time scales. Let a, b ∈ T, a < b and t ∈ T∗ . Then, 1. (Tκ )∗ = (T∗ )κ , 2. (Tκ )∗ = (T∗ )κ ,

502

3. 4. 5. 6. 7. 8.

7 First Order Dynamic Inclusions

[a, b]∗ = [−b, −a], σˆ (t) = −ρ ∗ (t), ρ(t) ˆ = −σ ∗ (t), νˆ (t) = μ∗ (t), μ(t) ˆ = ν ∗ (t), If f : T → R is rd-continuous(ld-continuous), then f ∗ : T∗ → R is ldcontinuous(rd-continuous) and the contrary.

Theorem 7.13 Let f : T → R be delta differentiable (resp. nabla differentiable) at t0 ∈ Tκ (resp. at t0 ∈ Tκ ), then f ∗ : T∗ → R is nabla differentiable (resp. delta differentiable) at −t0 ∈ (T∗ )κ (resp. at −t0 ∈ (T∗ )κ ) and the following identities are true: ˆ

f Δ (t0 ) = −(f ∗ )∇ (−t0 ), ˆ

f ∇ (t0 ) = −(f ∗ )Δ (−t0 ), )

(resp. or

  ˆ ∗ f Δ (t0 ) = − (f ∗ )∇ (t0 ),   ˆ ∗ (resp. f ∇ (t0 ) = − (f ∗ )Δ (t0 ), ) or 

(resp.

fΔ

∗

 ∇ˆ (−t0 ) = − f ∗ (−t0 ).

 ∇ˆ  Δ ∗ (−t0 ) = − f ∗ (−t0 ).) f

ˆ Proof Let f be Δ-differentiable at t0 ∈ Tκ . We will prove that f ∗ is ∇differentiable at −t0 ∈ (T∗ )κ . Take  > 0 arbitrarily. Then there exists a neighborhood U of t0 such that |f (σ (t0 )) − f (s) − f Δ (t0 )(σ (t0 ) − s)| < |σ (t0 ) − s| for all s ∈ U . Hence, using the properties of the dual functions, we get ˆ 0 ) − s)| < | − ρ(−t ˆ 0 ) − s| |f (−ρ(−t ˆ 0 )) − f ∗ (−s) − f Δ (t0 )(−ρ(−t for all s ∈ U . Let U ∗ be the dual of U . Then,  ∗  f (ρ(−t ˆ 0 ) − t| ˆ 0 )) − f ∗ (t) − (−f Δ (t0 ))(ρ(−t ˆ 0 ) − t) < |ρ(−t

7.5 The Dual Time Scales

503

ˆ for all t ∈ U ∗ . Therefore, f ∗ is ∇-differentiable at −t0 and ˆ

(f ∗ )∇ (−t0 ) = −f Δ (t0 ). Hence,   ˆ ∗ f Δ (t0 ) = − (f ∗ )∇ (t0 ) and 

fΔ

∗

 ∇ˆ (−t0 ) = − f ∗ (−t0 ).

Example 7.17 Let T = 2N0 , f : T → R be defined as follows: f (t) = 3t 2 − 2t + 4,

t ∈ T.

Here T∗ = {−2n : n ∈ N0 }, σ (t) = 2t,

t ∈ T,

ρ(t) ˆ = 2t,

t ∈ T∗ .

Let t ∈ T be arbitrarily chosen. Then, f Δ (t) = 3(σ (t) + t) − 2 = 3(2t + t) − 2 = 9t − 2, ∗

f (t) = f (−t) = 3(−t)2 − 2(−t) + 4 = 3t 2 + 2t + 4,  ∗ ∇ˆ   f (−t) = 3 ρ(−t) ˆ + (−t) + 2 = 3(−2t − t) + 2 = −9t + 2, 

− f

 ∗ ∇ˆ

(−t) = −(−9t + 2) = 9t − 2.

Consequently,

504

7 First Order Dynamic Inclusions

 ∇ˆ f ∗ (t) = − f ∗ (−t). Example 7.18 Let T = 3N0 and f : T → R be defined as follows: f (t) =

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

t ∈ T.

Here T∗ = {−3n : n ∈ N0 }, σ (t) = 3t, ρ(−t) ˆ = −3t,

t ∈ T.

Take t ∈ T arbitrarily. Then, f Δ (t) =

(σ (t) + t)(t + 4) − (t 2 + 2) + σ (t) + t (σ (t) + 4)(t + 4)

=

(3t + t)(t + 4) − t 2 − 2 + 3t + t (3t + 4)(t + 4)

=

4t 2 + 16t − t 2 − 2 + 4t (3t + 4)(t + 4)

=

3t 2 + 16t − 2 + 4t, (3t + 4)(t + 4)

f ∗ (t) = =

(−t)2 + 2 + (−t)2 + 1 −t + 4 t2 + 2 + t 2 + 1, −t + 4

 ∗ ∇ˆ (ρ(−t) ˆ + (−t))(t + 4) − (t 2 + 2)(−1) (−t) = f (−ρ(−t) ˆ + 4)(t + 4) +ρ(−t) ˆ + (−t) =

(−3t − t)(t + 4) + t 2 + 2 − 3t − t (3t + 4)(t + 4)

=

−4t 2 − 16t + t 2 + 2 − 4t (3t + 4)(t + 4)

=

−3t 2 − 16t + 2 − 4t, (3t + 4)(t + 4)

7.5 The Dual Time Scales

505

whereupon ˆ

f Δ (t) = −(f ∗ )∇ (−t). Exercise 7.8 Let T = 4N0 and f : T → R be defined as follows: f (t) =

t2 + t + 1 + t 2 + 2t + 7, t 2 + 3t + 4

t ∈ T.

Prove that ˆ

1. f Δ (t) = −  (f ∗ )∇ (−t),  ˆ ∗

2. f ∇ (t) = − (f ∗ )Δ (t),  ∗ ˆ 3. f Δ (−t) = − (f ∗ )∇ (−t), t ∈ T. f∗

1 (T)(resp. f ∈ C 1 (T)), then f ∗ : T∗ → R, Note that if f : T → R, f ∈ Crd ld 1 1 ∗ ∗ ∗ ∈ Cld (T ), (resp. f ∈ Crd (T ),) and the converse.

Example 7.19 Let f : T → R be delta differentiable at t ∈ Tκ . Then, f σ (t) = f (t) + μ(t)f Δ (t). Hence, using that μ(t) = ν(−t), ˆ  ∇ˆ f Δ (t) = − f ∗ (−t), f (t) = f ∗ (−t),   f (σ (t)) = f −ρ(−t) ˆ = f ∗ρˆ (−t), we get    ∇ˆ f ∗ρˆ (−t) = f ∗ (−t) + νˆ (−t) − f ∗ (−t) ˆ

= f ∗ (−t) − νˆ (−t)(f ∗ )∇ (−t), or, if s = −t, ˆ

f ∗ρˆ (s) = f ∗ (s) − νˆ (s)(f ∗ )∇ (s). Theorem 7.14 Let a, b ∈ T, a < b and f ∈ Crd ([a, b]). Then,

506

7 First Order Dynamic Inclusions b

−a

f (t)Δt =

−b

a

ˆ f ∗ (s)∇s.

Proof Let F be an antiderivative of f . Then, b

f (t)Δt = F (b) − F (a),

a

F Δ (t) = f (t),

t ∈ T.

By Theorem 7.13, it follows that  ∇ˆ F Δ (t) = − F ∗ (−t) = f (t),

t ∈ T,

whereupon ˆ

(F ∗ )∇ (s) = −f (−s) = −f ∗ (s),

s ∈ T∗ .

Hence, −a −b

s=−a ˆ = −(F ∗ )(s) f ∗ (s)∇s s=−b

∗

= −F (−a) + F ∗ (−b) = −F (a) + F (b) b

=

f (t)Δt. a

This completes the proof. Example 7.20 Let T = 2N0 . Consider 4

(7t 2 + 3t + 1)Δt.

1

Let f (t) = 7t 2 + 3t + 1, g(t) = t 3 + t 2 + t, We have

t ∈ T.

7.5 The Dual Time Scales

507

T∗ = {−2n : n ∈ N0 }, σ (t) = 2t,

t ∈ T,

ρ(t) ˆ = 2t,

t ∈ T∗ ,

and g Δ (t) = (σ (t))2 + tσ (t) + t 2 + σ (t) + t + 1 = (2t)2 + 2t 2 + t 2 + 2t + t + 1 = 7t 2 + 3t + 1,

t ∈ T.

Then, 4 1

4

f (t)Δt =

g Δ (t)Δt

1

t=4  = g(t) t=1

t=4  = (t 3 + t 2 + t) t=1

= 64 + 16 + 4 − 1 − 1 − 1 = 81. Next, f ∗ (s) = 7s 2 − 3s + 1,

s ∈ T∗ .

Let h(s) = s 3 − s 2 + s,

s ∈ T∗ .

Then, ˆ

2 ˆ + s ρ(s) ˆ + s 2 − (ρ(s) ˆ + s) + 1 h∇ (s) = (ρ(s))

= (2s)2 + 2s 2 + s 2 − (2s + s) + 1 s ∈ T∗ .

= 7s 2 − 3s + 1, Hence, −1 −4

ˆ = f ∗ (s)∇s

−1 −4

ˆ

ˆ h∇ (s)∇s

508

7 First Order Dynamic Inclusions

s=−1  = h(s) s=−4

s=−1  = (s 3 − s 2 + s) s=−4

= −1 − 1 − 1 − (−64 − 16 − 4) = −3 + 84 = 81. Consequently, 4

−1

f (t)Δt =

−4

1

ˆ f ∗ (s)∇s.

Exercise 7.9 Let T = 3N0 and f (t) = t 2 − 7t + 11,

t ∈ T.

Prove that 9

−1

f (t)Δt =

−9

1

ˆ f ∗ (t)∇t.

1 (T). By the Example 7.21 Let a, b ∈ T, a < b and f, g : T → R, f, g ∈ Crd integration by parts, we have

!b a

!b f (t)g Δ (t)Δt = f (b)g(b) − f (a)g(a) − a f Δ (t)g σ (t)Δt = f ∗ (−b)g ∗ (−b) − f ∗ (−a)g ∗ (−a) ! −a  Δ ∗ ˆ − f (s) (g σ )∗ (s)∇s −b

∗ (−b) − f ∗ (−a)g ∗ (−a) = f ∗ (−b)g  ! −a  ˆ ˆ − −b − (f ∗ )∇ (s) (g ∗ )ρˆ (s)∇s

= f ∗ (−b)g ∗ (−b) − f ∗ (−a)g ∗ (−a) ! −a ˆ ˆ + −b (f ∗ )∇ (s)(g ∗ )ρˆ (s)∇s. Also, b

f (t)g Δ (t)Δt =

a

−a −b

=− Hence by (7.20), we obtain

 ∗ ˆ f ∗ (s) g Δ (s)∇s

−a −b

 ∇ˆ ˆ f ∗ (s) g ∗ (s)∇s.

(7.20)

7.5 The Dual Time Scales

−

−a −b

509

 ∇ˆ ˆ = f ∗ (−b)g ∗ (−b) − f ∗ (−a)g ∗ (−a) f ∗ (s) g ∗ (s)∇s −a

+

−b

ˆ ˆ (f ∗ )∇ (s)(g ∗ )ρˆ (s)∇s,

or −a −b

 ∇ˆ ˆ = f ∗ (−a)g ∗ (−a) − f ∗ (−b)g ∗ (−b) f ∗ (s) g ∗ (s)∇s −

−a −b

ˆ

ˆ (f ∗ )∇ (s)(g ∗ )ρˆ (s)∇s.

We will finish this section with the duality principle. The Duality Principle For any statement true in the nabla(resp. delta) calculus in the time scale T, there is an equivalent dual statement in the delta(resp. nabla) calculus for the dual time scale T∗ . Example 7.22 Let T = 2N0 and f : T → R be defined as follows: f (t) = 4t 3 − 2t + 1,

t ∈ T.

Then, T∗ = {−2n : n ∈ N0 }, σ (t) = 2t,   f Δ (t) = 4 (σ (t))2 + tσ (t) + t 2 − 2 = 4((2t)2 + 2t 2 + t 2 ) − 2 = 28t 2 − 2, 2

f Δ (t) = 28(σ (t) + t) = 28(2t + t) = 84t,

t ∈ T.

Next, f ∗ (t) = f (−t) = 4(−t)3 − 2(−t) + 1 = −4t 3 + 2t + 1,

t ∈ T∗ ,

510

7 First Order Dynamic Inclusions

and ρ(t) ˆ = 2t, ˆ

2 ˆ + t ρ(t) ˆ + t 2) + 2 f ∗∇ (t) = −4((ρ(t))

= −4((2t)2 + 2t 2 + t 2 ) + 2 = −28t 2 + 2, ˆ2

f ∗∇ (t) = −28(ρ(t) ˆ + t) = −28(2t + t) = −84t,

t ∈ T∗ .

Therefore, ˆ2

f Δ (t) = f ∗∇ (−t), 2

t ∈ T.

7.6 Existence of Solutions of First Order Dynamic Inclusions via Duality Consider the following problem: y ∇ (t) ∈ G(t, y(t))

∇ − a.e.

t ∈ [a, b],

y(a) = x(a),

(7.21) (7.22)

where G : T × R  R is a set-valued function such that (H22) G is nonempty, closed, and ∇ × B-measurable set-valued function. (H23) There exists a function k : T → [0, ∞), k ∈ L1 ([a, b]) such that G(t, x) ⊂ G(t, y) + k(t)|y − x|{B}B∈B for all x, y ∈ R, and (H24) x : T → R is an absolutely continuous single-valued function. To the set-valued function G we associate the function ρ : T × R × R → [0, ∞) given by ρ(t, x, v) = inf{|v − y| : y ∈ G(t, x)}.

7.7 Advanced Practical Problems

511

Definition 7.7 A single-valued function y ∈ AC([a, b]) is said to be a solution of the problem (7.21), (7.22) if there exists a function g ∈ L1 ([ρ(a), b]) such that g(t) ∈ G(t, y(t)), ∇

y (t) = g(t) ∇ − a.e.

t ∈ [a, b]

and the condition (7.22) holds. Theorem 7.15 Suppose (H 22)–(H 24). For any  > 0 satisfying  , K

ρG (x)
0. In this section, we will investigate the dynamic inclusion T

y(t) ∈

g(t, s)F (s, y σ (s))Δs,

t ∈ [0, T ],

(8.1)

0

where (I1) F : T × R  Kc (R), F (t, ·) is u.s.c. for any t ∈ [0, T ], F (y) = ∅ for any y ∈ C (T), F (y) = {u ∈ C (T) : u(t) ∈ F (t, y σ (t)) for all

t ∈ T},

(I2) g : [0, T ] × [0, T ] → R is continuous. Definition 8.1 A single-valued function y ∈ C ([0, T ]) is said to be a solution of the dynamic inclusion (8.1) if there exists a single-valued function f ∈ C ([0, T ]) such that

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_8

515

516

8 Second Order Dynamic Inclusions

1. f (t) ∈ F (t, y σ (t)), t ∈ [0, T ], !T 2. y(t) = 0 g(t, s)f (s)Δs, t ∈ [0, T ]. Example 8.1 Let T = Z, T = 3, and g(t, s) = t 2 + 3t + 4,  10(s 2 + 5s + 3) y σ (s), F (s, y σ (s)) = 2 21(s + 5s + 6)(s 2 + 5s + 8) 10(s 2 + 5s + 3) y σ (s) + 1, + 5s + 6)(s 2 + 5s + 8)  10(s 2 + 5s + 3) σ y (s) + 2 , 21(s 2 + 5s + 6)(s 2 + 5s + 8) 21(s 2

t, s ∈ [0, 3].

We will show that the function y(t) =

21 2 (t + 3t + 4), 10

t ∈ [0, 3]

is a solution of the dynamic inclusion (8.1). We have σ (t) = t + 1, 21 ((σ (t))2 + 3σ (t) + 4) 10 21 = ((t + 1)2 + 3(t + 1) + 4) 10 21 2 = (t + 2t + 1 + 3t + 3 + 4) 10 21 2 = (t + 5t + 8), 10

y σ (t) =

10(s 2 + 5s + 3) 10(s 2 + 5s + 3) σ y (s) = 21(s 2 + 5s + 6)(s 2 + 5s + 8) 21(s 2 + 5s + 6)(s 2 + 5s + 8)   21 2 (s + 5s + 8) × 10 =

s 2 + 5s + 3 , s 2 + 5s + 6

10(s 2 + 5s + 3) s 2 + 5s + 3 σ y +1 (s) + 1 = 21(s 2 + 5s + 6)(s 2 + 5s + 8) s 2 + 5s + 6 =

s 2 + 5s + 3 + s 2 + 5s + 6 s 2 + 5s + 6

8.1 Fixed Point Results

517

=

2s 2 + 10s + 9 , s 2 + 5s + 6

10(s 2 + 5s + 3) s 2 + 5s + 3 σ y +2 (s) + 2 = 21(s 2 + 5s + 6)(s 2 + 5s + 8) s 2 + 5s + 6 =

s 2 + 5s + 3 + 2s 2 + 10s + 12 s 2 + 5s + 6

=

3s 2 + 15s + 15 , s 2 + 5s + 6

t, s ∈ [0, 3].

Let h(t) =

t2 + t + 1 , t +2

t ∈ [0, 3].

Then, hΔ (t) = =

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

=

2(t + 1)(t + 2) − t 2 − t − 1 (t + 2)(t + 3)

=

2t 2 + 6t + 4 − t 2 − t − 1 t 2 + 5t + 6

= h(3) = = h(0) = h(3) − h(0) = = Therefore,

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

t 2 + 5t + 3 , t 2 + 5t + 6 9+3+1 3+2 13 , 5 1 , 2 13 1 − 5 2 21 . 10

t ∈ [0, 3],

518

8 Second Order Dynamic Inclusions

F (s, y σ (s)) = {hΔ (s), hΔ (s) + 1, hΔ (s) + 2},  g(t, s)F (s, y σ (s)) = (t 2 + 3t + 4)hΔ (s), (t 2 + 3t + 4)(hΔ (s) + 1),  (t + 3t + 4)(h (s) + 2) , 2

Δ

t, s ∈ [0, 3].

Hence, 3 0

 g(t, s)F (s, y σ (s))Δs = (t 2 + 3t + 4) 3 0

3 0

hΔ (s)Δs, (t 2 + 3t + 4)

(hΔ (s) + 1)Δs, (t 2 + 3t + 4)



3 0

 (hΔ (s) + 2)Δs

s=3 s=3   , (t 2 + 3t + 4)(h(s) + s) , = (t 2 + 3t + 4)h(s) s=0

s=0

s=3   (t 2 + 3t + 4)(h(s) + 2s) s=0



= (t 2 + 3t + 4)(h(3) − h(0)), (t 2 + 3t + 4)(h(3) − h(0) + 3),  (t 2 + 3t + 4)(h(3) − h(0) + 6)   21 21 2 (t + 3t + 4), + 3 (t 2 + 3t + 4), 10 10    21 2 + 6 (t + 3t + 4) 10  51 21 2 = (t + 3t + 4), (t 2 + 3t + 4), 10 10  81 2 (t + 3t + 4) , t ∈ [0, 3]. 10 

=

Consequently, 3

y(t) ∈

g(t, s)F (s, y σ (s))Δs,

t ∈ [0, 3],

0

i.e., y is a solution of the considered dynamic inclusion. Exercise 8.1 Let T = 2N0 {0} and g(t, s) = 1,

8.1 Fixed Point Results

519

 F (s, y (s)) = σ

 5 σ 5 σ 5 σ y (s), y (s) + 1, y (s) + 3 , 18 18 18

(t, s) ∈ [0, 4]. Check if y(t) = 63t,

t ∈ [0, 4]

is a solution of the dynamic inclusion (8.1). Theorem 8.1 Suppose (I 1) and (I 2). If for all r > 0 there exists a nonnegative function hr ∈ C ([0, T ]) with t ∈ [0, T ],

|F (t, u)| ≤ hr (t),

|u| ≤ r,

and if there exists a constant M with max |y(t)| = M

t∈[0,T ]

for all solutions of the dynamic inclusion T

y(t) ∈ λ

g(t, s)F (s, y σ (s))Δs,

t ∈ [0, T ],

(8.2)

0

for all λ ∈ (0, 1), then the dynamic inclusion (8.1) has a solution. Proof Define a linear and continuous operator T : C ([0, T ]) → C ([0, T ]) as follows: T

T y(t) =

g(t, s)y(s)Δs,

t ∈ [0, T ],

0

so that (8.2) is equivalent to the fixed point problem y ∈ λ(T ◦ F )(y). We also define E = C ([0, T ]),   U = u ∈ E : max |u(t)| < M . t∈[0,T ]

Assume that there exists an u ∈ ∂U and a λ ∈ (0, 1) with u ∈ λ(T ◦ F )(u). Then max |u(t)| = M.

t∈[0,T ]

520

8 Second Order Dynamic Inclusions

Therefore, the second possibility of the Leray–Schauder nonlinear alternative, Theorem B.9(see the appendix of this book), is ruled out. Now, we will show that the function T ◦ F : U  Kc (E) is u.s.c and compact. Let {uk }k∈N , {wk }k∈N ⊂ R be such that uk → u0 , wk → w0 ,

as

k → ∞,

wk ∈ (T ◦ F )(uk ),

k ∈ N.

Then there is a vk ∈ F (uk ), k ∈ N, such that wk = T vk ,

k ∈ N.

Because uk ∈ U , k ∈ N, and |F (t, uk (t))| ≤ hM (t),

t ∈ [0, T ],

there exists a compact set Ω ⊂ E with {vk }k∈N ⊂ Ω. Hence, there exists a convergent subsequence {vkl }l∈N of the sequence {vk }k∈N , say vkl → v0 , as l → ∞. We have vkl → v0 , ukl → u0 ,

as

l → ∞,

vkl ∈ F (t, uσkl (t)),

t ∈ T,

l ∈ N.

From here, using that F (t, ·) is u.s.c. for any t ∈ [0, T ], we conclude that v0 (t) ∈ F (t, uσ0 (t)),

t ∈ [0, T ].

Therefore, v0 ∈ F (u0 ). Since T : E → E is continuous, we have wkl → w0 , wkl = T vkl → T v0 ,

as

l → ∞.

Consequently, w0 = T v0 and so, T ◦ F : U  Kc (E) is u.s.c. By the Arzela– Ascoli theorem, we have that T ◦ F : U  Kc (E) is compact. Now, applying the Leray–Schauder nonlinear alternative, we conclude that T ◦ F has a fixed point in E. This completes the proof. Theorem 8.2 Assume (I 1) and (I 2). If there exists a nonnegative function h ∈ C ([0, T ]) with

8.1 Fixed Point Results

521

|F (t, u)| ≤ h(t),

t ∈ [0, T ],

u ∈ R,

then the dynamic inclusion (8.1) has a solution. Proof Let T , F , E, and U be as in the proof of Theorem 8.1. By the proof of Theorem 8.1, it follows that T ◦ F : U  Kc (E) is u.s.c and compact. Hence by the Kakutani fixed point theorem, Theorem B.11(see the appendix of this book), it follows that the operator T ◦ F has a fixed point in E. This completes the proof. Example 8.2 Let T = 2Z, T = 8, g(t, s) = t 2 + s 2 ,  2 s + (y σ (s))2 s 2 + (y σ (s))2 σ , + 2, F (s, y (s)) = 1 + s 2 + (y σ (s))2 1 + s 2 + (y σ (s))2  s 2 + (y σ (s))2 + 3 , t, s ∈ [0, 8]. 1 + s 2 + (y σ (s))2 Note that  2   s + (y σ (s))2     1 + s 2 + (y σ (s))2  ≤ 1,  2    2  s + (y σ (s))2   s + (y σ (s))2       1 + s 2 + (y σ (s))2 + 2 ≤  1 + s 2 + (y σ (s))2  + 2 ≤ 3,  2     s + (y σ (s))2   s 2 + (y σ (s))2    ≤ + 3  1 + s 2 + (y σ (s))2   1 + s 2 + (y σ (s))2  + 3 ≤ 4,

s ∈ [0, 8].

Next, F is u.s.c. By the theory of nonlinear integral equations(see [21]), it follows that F (y) = ∅ for any y ∈ C ([0, 8]). Hence by Theorem 8.2, it follows that the dynamic inclusion (8.1) has a solution. Exercise 8.2 Let T = 2N0 {0}, T = 16 and g(t, s) = ts,  (y σ (s))2 (y σ (s))2 σ , + 1, F (s, y (s)) = 4 + (y σ (s))4 4 + (y σ (s))4  (y σ (s))2 + 3 , t, s ∈ [0, 16]. 4 + (y σ (s))4 Prove that the dynamic inclusion (8.1) has a solution.

522

8 Second Order Dynamic Inclusions

8.2 Existence Results for Second Order Dynamic Inclusions Suppose that 0, T ∈ T, 0 < T , σ (T ) = 0. In this section, we will investigate the following periodic BVP for the following second order dynamic inclusion: 2

y Δ (t) ∈ F (t, y σ (t)),

t ∈ [0, T ],

(8.3)

y(0) = y(σ (T )) = 0,

(8.4)

where F satisfies (I 1). Definition 8.2 A single-valued function y ∈ C 2 ([0, T ]) is said to be a solution of the problem (8.3), (8.4), if 1. There exists a single-valued function f F (t, y σ (t)), t ∈ [0, T ], and 2

y Δ (t) = f (t),

∈ C ([0, T ]) such that f (t) ∈

t ∈ [0, T ].

2. y(0) = y(σ (T )) = 0. Example 8.3 Let T = Z, T = 3, and F (t, y σ (t)) = {12t 2 + 6t − 10, 12t 2 + 6t, 12t 2 + 6t + 2},

t ∈ [0, 3].

We will prove that the function y(t) = t (t − 4)(t 2 + t + 1),

t ∈ [0, 4]

is a solution of the periodic BVP (8.3), (8.4). Indeed, we have σ (t) = t + 1, y(t) = (t 2 − 4t)(t 2 + t + 1) = t 4 + t 3 + t 2 − 4t 3 − 4t 2 − 4t = t 4 − 3t 3 − 3t 2 − 4t, y Δ (t) = (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3   −3 (σ (t))2 + tσ (t) + t 2 − 3(σ (t) + t) − 4 = (t + 1)3 + t (t + 1)2 + t 2 (t + 1) + t 3   −3 (t + 1)2 + t (t + 1) + t 2 − 3(t + 1 + t) − 4 −3(t + 1 + t)

8.2 Existence Results for Second Order Dynamic Inclusions

= t 3 + 3t 2 + 3t + 1 + t (t 2 + 2t + 1) + t 3 + t 2 + t 3 −3(t 2 + 2t + 1 + t 2 + t + t 2 ) − 4 − 3(2t + 1) = 3t 3 + 4t 2 + 3t + 1 + t 3 + 2t 2 + t −3(3t 2 + 3t + 1) − 3(2t + 1) − 4 = 4t 3 + 6t 2 + 4t + 1 − 9t 2 − 9t − 3 − 6t − 3 − 4 = 4t 3 − 3t 2 − 11t − 9,   2 y Δ (t) = 4 (σ (t))2 + tσ (t) + t 2 − 3(σ (t) + t) − 11   = 4 (t + 1)2 + t (t + 1) + t 2 − 3(t + 1 + t) − 11 = 4(t 2 + 2t + 1 + t 2 + t + t 2 ) − 3(2t + 1) − 11 = 4(3t 2 + 3t + 1) − 6t − 3 − 11 = 12t 2 + 12t + 4 − 6t − 14 = 12t 2 + 6t − 10 ∈ F (t, y σ (t)),

t ∈ [0, 3],

y(0) = 0, y(σ (3)) = y(4) = 0. Exercise 8.3 Let T = 3Z, T = 9, and F (t, y σ (t)) = {2, 3, 5},

t ∈ [0, 9].

Prove that y(t) = t 2 − 12t,

t ∈ [0, 12]

is a solution to the problem (8.3), (8.4). Theorem 8.3 Assume (I 1). Define a function g : [0, T ] × [0, T ] → R by  g(t, s) =

tσ (s) σ (T ) tσ (s) σ (T )

− t if t ≤ s − σ (s) if t ≥ σ (s).

Then y solves (8.1) if and only if y solves (8.3), (8.4). Proof 1. Assume that y solves (8.1). Then there exists a function τ ∈ F (y) such that

523

524

8 Second Order Dynamic Inclusions T

y(t) =

g(t, s)τ (s)Δs 0 t

=

T

g(t, s)τ (s)Δs +

0

g(t, s)τ (s)Δs t



 tσ (s) − σ (s) τ (s)Δs = σ (T ) 0  t tσ (s) + t− τ (s)Δs σ (T ) T t

=

t

t σ (T )

τ (s)Δs −

T

=

T

t σ (T )

t σ (T )

0 t

σ (s)τ (s)Δs T t

σ (s)τ (s)Δs −

σ (s)τ (s)Δs

0

0

t

+t

σ (s)τ (s)Δs

0 t

+t

t

σ (s)τ (s)Δs −

t ∈ [0, T ].

τ (s)Δs, T

Hence, y Δ (t) =

1 σ (T )

T

t

σ (s)τ (s)Δs − σ (t)τ (t) +

0

τ (s)Δs T

+σ (t)τ (t) =

1 σ (T )

2

y Δ (t) = τ (t),

T

t

σ (s)τ (s)Δs +

0

τ (s)Δs, T

t ∈ (0, T ).

Next, T

y(0) =

g(0, s)τ (s)Δs 0

= 0, T

y(σ (T )) =

σ (T )

σ (s)τ (s)Δs −

0

σ (s)τ (s)Δs 0

σ (T )

+σ (T )

τ (s)Δs T

8.2 Existence Results for Second Order Dynamic Inclusions σ (T )

=−

525 σ (T )

σ (s)τ (s)Δs + σ (T )

T

τ (s)Δs T

= 0. So, y solves (8.3), (8.4). 2. Let y solves (8.3), (8.4). Then, using integrating by parts, we get T

2

g(t, s)y Δ (s)Δs =

0

T

t σ (T )

0 t

+t

y

Δ2

t

2

σ (s)y Δ (s)Δs −

2

σ (s)y Δ (s)Δs

0

(s)Δs

T

=

T t t  Δ  s=T sy (s)  − y Δ (s)Δs s=0 σ (T ) σ (T ) 0 t  s=t  + y Δ (s)Δs − sy Δ (s)  s=0

0

  +t y Δ (t) − y Δ (T ) =

 t  Δ T y (T ) − y(T ) + y(0) σ (T ) −ty Δ (t) + y(t) − y(0)   +t y Δ (t) − y Δ (T )

=

tT Δ t y (T ) − y(T ) σ (T ) σ (T ) +y(t) − ty Δ (T )

t tμ(T ) Δ y (T ) − y(T ) + y(t) σ (T ) σ (T ) t t =− (y(σ (T )) − y(T )) − y(T ) + y(t) σ (T ) σ (T ) t =− y(σ (T )) + y(t) σ (T ) =−

= y(t),

t ∈ [0, T ].

Hence, t

y(t) =

2

g(t, s)y Δ (s)Δs

0 T

∈ 0

g(t, s)F (s, y σ (s))Δs,

t ∈ [0, T ].

526

8 Second Order Dynamic Inclusions

This completes the proof. Theorem 8.4 Assume (I 1). Suppose that there exists a continuous and nondecreasing function Ψ : [0, ∞) → [0, ∞) with Ψ (u) > 0 for u > 0, and a function q : T → [0, ∞) such that |F (t, u)| ≤ q(t)Ψ (|u|),

u ∈ R,

t ∈ T.

Let T

Q0 = max t∈T

|g(t, s)|q(s)Δs

0

and sup c>0

c > Q0 , Ψ (c)

where the function g is defined as in Theorem 8.3. Then the problem (8.3), (8.4) has a solution. Proof Let M > 0 be such that M > Q0 . Ψ (M) Consider the problem 2

y Δ (t) ∈ λF (t, y σ (t)),

t ∈ [0, T ],

y(0) = y(σ (T )) = 0, where λ ∈ (0, 1). By Theorem 8.3, it follows that the last problem is equivalent to the dynamic inclusion (8.2). Let y be any solution of the dynamic inclusion (8.2). Then, T

|y(t)| ≤ λ

|g(t, s)|q(s)Ψ (|y(s)|)Δs

0



 ≤Ψ

s∈[0,T ]



≤ Q0 ψ

t

max |y(s)|

|g(t, s)|q(s)Δs

0



max |y(s)| ,

s∈[0,T ]

t ∈ [0, T ].

8.2 Existence Results for Second Order Dynamic Inclusions

527

So, maxs∈[0,T ] |y(s)|   ≤ Q0 . Ψ maxs∈[0,T ] |y(s)| If M = max |y(s)|, s∈[0,T ]

by the last inequality, we get M ≤ Q0 , Ψ (M) which is a contradiction. Hence by Theorem 8.1, it follows that the dynamic inclusion (8.1) has a solution. Now, applying Theorem 8.3, we conclude that the problem (8.3), (8.4) has a solution. This completes the proof. Definition 8.3 A function α ∈ C ([0, T ]) is said to be a lower solution of the dynamic inclusion (8.3) if F (t, α σ (t))

  2 −∞, α Δ (t) = ∅,

t ∈ [0, T ],

and α(0) ≤ 0,

α(σ (T )) ≤ 0.

Definition 8.4 A function β ∈ C ([0, T ]) is said to be an upper solution of the dynamic inclusion (8.3) if F (t, β σ (t))

  2 β Δ (t), ∞ = ∅,

t ∈ [0, T ],

and β(0) ≥ 0,

β(σ (T )) ≥ 0.

Theorem 8.5 Let F satisfy all conditions of Theorem 8.1. If there exist lower and upper solutions α and β of (8.3) with α(t) ≤ β(t),

t ∈ [0, T ],

then the dynamic inclusion (8.3) has a solution y such that

528

8 Second Order Dynamic Inclusions

α(t) ≤ y(t) ≤ β(t),

t ∈ [0, T ].

Proof Define ⎧ σ ⎨ α (t) if x < α σ (t) h(t, x) = β σ (t) if x > β σ (t) ⎩ x otherwise, ⎧ ⎨ p(x − α σ (t)) if x < α σ (t) r(t, x) = p(x − β σ (t)) if x > β σ (t) ⎩ 0 otherwise,  x if |x| ≤ 1 p(x) = x |x| if |x| > 1,  ⎧ Δ2 (t) ⎪ −∞, α if x < α σ (t) ⎪ ⎨

 2 Γ+ (t, x) = β Δ (t), ∞ if x > β σ (t) ⎪ ⎪ ⎩ R otherwise,  Γ+ (t, x), F+ (t, x) = F (t, h(t, x)) F+∗ (t, x) = F+ (t, x) + r(t, x). We have that Γ+ (t, ·) is u.s.c. for each t ∈ [0, T ] and hence, so are F+ (t, ·) and F+∗ (t, ·) for any t ∈ [0, T ]. By Theorem 8.3, the problem for the dynamic inclusion y Δ (t) ∈ F+∗ (t, y σ (t)), 2

t ∈ [0, T ],

(8.5)

subject to the periodic boundary conditions (8.4) is equivalent to the problem t

y(t) ∈ 0

g(t, s)F+∗ (s, y σ (s))Δs,

t ∈ T,

(8.6)

where g is given as in Theorem 8.3. Since |F+∗ (t, u)| ≤ |F+ (t, u)| + |r(t, u)| ≤ hmaxt∈[0,T ] |β(t)| + 1, by Theorem 8.2, it follows that (8.6) has a solution y ∈ C (T). By the equivalence of (8.6) and (8.5), (8.4), we get that the problem (8.5), (8.4) has a solution y ∈ C (T). Now, we assume that there exists an m ∈ [0, T ] such that y(m) > β(m).

8.2 Existence Results for Second Order Dynamic Inclusions

529

Define u = y − β. There exists a θ ∈ [0, T ] such that u(θ ) = max u(t). t∈[0,T ]

We have u(θ ) > 0, Δ2

u

(ρ(θ )) ≤ 0,

σ (ρ(θ )) = θ. Next, y Δ (ρ(θ )) ∈ F+∗ (ρ(θ ), y σ (ρ(θ )) 2

= F+ (ρ(θ ), y(ρ(θ ))) + r(ρ(θ ), y(θ )) = F+ (ρ(θ ), y(ρ(θ ))) + p(y(θ ) − β(θ )).

2  Then there exists w(θ ) ∈ β Δ (ρ(θ )), ∞ with 2

y Δ (ρ(θ )) = w(θ ) + p(y(θ ) − β(θ )) 2

≥ β Δ (ρ(θ )) + p(y(θ ) − β(θ )). So, 2

2

2

uΔ (ρ(θ )) = y Δ (ρ(θ )) − β Δ (ρ(θ )) ≥ p(y(θ ) − β(θ )) > 0, which is a contradiction. Hence, y(t) ≤ β(t),

t ∈ [0, T ].

As in above, one can prove that α(t) ≤ y(t), This completes the proof.

t ∈ [0, T ].

530

8 Second Order Dynamic Inclusions

Below we suppose (I3) σ is delta differentiable. Before formulating and proving our next results in this section, we have a need of the following auxiliary result. Lemma 8.1 Assume (I 3). If y : T → R is delta differentiable, then so is y σ : Tκ → R and we have the following relation: y σ Δ (t) = σ Δ (t)y Δσ (t),

t ∈ Tκ .

Proof We have, for t ∈ Tκ , y σ (t) = y(t) + μ(t)y Δ (t) = y(t) + (σ (t) − t)y Δ (t), 2

y Δσ (t) = y Δ (t) + μ(t)y Δ (t). Then, y σ Δ (t) = y Δ (t) + (σ (t) − t)Δ y Δσ (t) 2

+(σ (t) − t)y Δ (t) 2

= y Δ (t) + μ(t)y Δ (t) +(σ Δ (t) − 1)y Δσ (t) = y Δσ (t) + (σ Δ (t) − 1)y Δσ (t) = σ Δ (t)y Δσ (t),

t ∈ Tκ .

This completes the proof. 2

Lemma 8.2 Assume (I 3). If y : T → R is twice delta differentiable on Tκ , then so is y 2 : T → R and we have the following relation:  Δ2  2 2 y2 (t) = 2y σ (t)y Δ (t) + y Δ (t)  2 +σ Δ (t) y Δσ (t) ,

2

t ∈ Tκ .

2

Proof Let t ∈ Tκ . We have  Δ y 2 (t) = y σ (t)y Δ (t) + y(t)y Δ (t).

8.2 Existence Results for Second Order Dynamic Inclusions

531

Hence, applying Lemma 8.1, we obtain  Δ2 2 y2 (t) = y σ Δ (t)y Δσ (t) + y σ (t)y Δ (t) 2  2 + y Δ (t) + y σ (t)y Δ (t)  2 2 = 2y σ (t)y Δ (t) + y Δ (t) +y σ Δ (t)y Δσ (t)

 2 2 = 2y σ (t)y Δ (t) + y Δ (t)  2 +σ Δ (t) y Δσ (t) . This completes the proof. Theorem 8.6 Assume (I 1) and (I 3). If there exist constants L, K ≥ 0 with |F (t, u)| ≤

inf

w∈F (t,u)

(Luw + K)

(8.7)

for all (t, u) ∈ T × R, then any solution y of the problem (8.3), (8.4) satisfies max |y(t)| ≤ K max

t∈[0,T ]

T

t∈[0,T ] 0

|g(t, s)|Δs,

where the function g is given as in Theorem 8.3. Proof Let y be a solution of the problem (8.3), (8.4). By Theorem 8.3, it follows that y solves (8.1). Then there is a τ ∈ F (y) such that T

y(t) =

g(t, s)τ (s)Δs,

t ∈ [0, T ].

0

By Lemma 8.2, using that σ is an increasing function, it follows that 2

y σ (t)y Δ (t) ≤

1 2 Δ (y ) (t), 2

t ∈ [0, T ].

Hence, using that τ ∈ F (y), we have |τ (s)| ≤

inf

w∈F (t,u)

 σ  Ly (s)w + K 2

≤ Ly σ (s)y Δ (s) + K ≤

L 2 Δ2 (y ) (s) + K, 2

s ∈ [0, T ].

532

8 Second Order Dynamic Inclusions

Thus,  !   T |y(t)| =  0 g(t, s)τ (s)Δs  !T ≤ 0 |g(t, s)||τ  (s)|Δs 2  !T ≤ 0 |g(t, s)| L2 (y 2 )Δ (s) + K Δs !T !T 2 = L2 0 |g(t, s)|(y 2 )Δ (s)Δs + K 0 |g(t, s)|Δs, t ∈ [0, T ]. Observe that T 0

t

2

|g(t, s)|(y 2 )Δ (s)Δs =

2

|g(t, s)|(y 2 )Δ (s)Δs

0 T

+

2

|g(t, s)|(y 2 )Δ (s)Δs

t

 t t 2 σ (s)(y 2 )Δ (s)Δs σ (T ) 0  T  σ (s) 2 1− (y 2 )Δ (s)Δs +t σ (T ) t   s=t t  s(y 2 )Δ (s) = 1− s=0 σ (T )  t  t − 1− (y 2 )Δ (s)Δs σ (T ) 0   s=T s  (y 2 )Δ (s) +t 1 − s=t σ (T )

 = 1−

T t (y 2 )Δ (s)Δs σ (T ) t   t = t 1− (y 2 )Δ (t) σ (T )   s=t t  (y(s))2  − 1− s=0 σ (T )   T (y 2 )Δ (T ) +t 1 − σ (T )   t (y 2 )Δ (t) −t 1 − σ (T ) s=T t  + (y(s))2  s=t σ (T )

+

(8.8)

8.2 Existence Results for Second Order Dynamic Inclusions

533

   t T 2 (y(t)) + t 1 − y(T )y Δ (T ) = − 1− σ (T ) σ (T ) t t + (y(T ))2 − (y(t))2 σ (T ) σ (T ) 

= −(y(t))2 + t

μ(T ) y(T )y Δ (T ) σ (T )

t (y(T ))2 σ (T ) t = −(y(t))2 + y(T )(y σ (T ) − y(T )) σ (T ) t (y(T ))2 + σ (T ) +

= −(y(t))2 ,

t ∈ [0, T ].

Hence by (8.8), we obtain L |y(t)| ≤ − (y(t))2 + K 2 T

≤K

T

|g(t, s)|Δs

0

|g(t, s)|Δs

0

≤ K max

T

t∈[0,T ] 0

|g(t, s)|Δs,

t ∈ [0, T ].

Consequently, max |y(t)| ≤ K max

t∈[0,T ]

T

|g(t, s)|Δs.

t∈[0,T ] 0

This completes the proof. Theorem 8.7 Assume (I 1), (I 3), and (8.7). Then the problem (8.3), (8.4) has a solution. Proof By (8.7), we get |λF (t, u)| ≤

inf

w∈λF (t,u)

(Luw + K),

(t, u) ∈ T × R.

Then Theorem 8.6 is applicable to the problem for the dynamic inclusion 2

y Δ (t) ∈ λF (t, y σ (t)),

t ∈ [0, T ],

(8.9)

534

8 Second Order Dynamic Inclusions

subject to the boundary conditions (8.4). Let R = K max

T

t∈[0,T ] 0

|g(t, s)|Δs.

Thus, for any solution y of the problem (8.9), (8.4), we have max |y(t)| ≤ R.

t∈[0,T ]

Define U = {y ∈ C ([0, T ]) : max |y(t)| < R + 1}. t∈[0,T ]

(8.10)

Note that there cannot be any solution y of (8.9), (8.4) with max |y(t)| = R + 1.

t∈[0,T ]

By Theorem 8.3, we have that the problem (8.3), (8.4) is equivalent to the dynamic inclusion (8.1). Let the operator T be defined as in the proof of Theorem 8.1. Then, as in the proof of Theorem 8.1, the operator T ◦ F : U  Kc (R) is u.s.c. and compact, where U is defined by (8.10). Now, applying the Leray–Schauder nonlinear alternative, we get that the operator T ◦ F has a fixed point in U , whereupon we conclude that the problem (8.3), (8.4) has a solution. This completes the proof. Example 8.4 Let T = Z, T = 4, and  F (t, u) =

 u u u , + 1, + 2 , 1 + u2 1 + u2 1 + u2

Since    u     1 + u2  ≤ 1,      u   u    ≤ + 1  1 + u2   1 + u2  + 1 ≤ 1+1 = 2,      u   u    ≤ + 2  1 + u2   1 + u2  + 2 ≤ 1+2 = 3,

t ∈ [0, 4].

8.2 Existence Results for Second Order Dynamic Inclusions

535

we take L = 1, K = 4. Here σ (t) = t + 1, σ (4) = 5,  g(t, s) =  = =

t (s+1) 5 t (s+1) 5

− t if t ≤ s − (s + 1) if t ≥ s + 1

s 5t − 45 t if t ≤ s s+1 5 (t − 5) if t ≥ s + 1

 t (s−4) 5 s+1 5 (t

if t ≤ s − 5) if t ≥ s + 1,

t ∈ [0, 4]. Let 1 2 (s − 9s), 2 1 h2 (s) = (s 2 + s), s ∈ T. 2 h1 (s) =

Then, 1 (σ (s) + s − 9) 2 1 = (s + 1 + s − 9) 2 = s − 4,

hΔ 1 (s) =

1 (σ (s) + s + 1) 2 1 = (s + 1 + s + 1) 2 = s + 1, s ∈ T.

hΔ 2 (s) =

Hence, T

t

|g(t, s)|Δs =

0

0

= =

4

|g(t, s)|Δs +

t −5 5 t −5 5

|g(t, s)|Δs

t t 0 t 0

4

1 (s + 1)Δs + t 5 1 hΔ 2 (s)Δs − t 5

(4 − s)Δs

t 4 t

hΔ 1 (s)Δs

536

8 Second Order Dynamic Inclusions

= = = = =

s=t s=4 t −5 1   − th1 (s) h2 (s) s=0 s=t 5 5  s=4 s=t t −5 2 1   − t (s 2 − 9s) (s + s) s=0 s=t 10 10 t −5 2 1 (t + t) − t (16 − 36 − t 2 + 9t) 10 10  1 3 t + t 2 − 5t 2 − 5t + 20t + t 3 − 9t 2 10  1  3 2t − 13t 2 + 15t , t ∈ [0, 4]. 10

and 4

max

t∈[0,4] 0

 1  3 2t − 13t 2 + 15t t∈[0,4] 10

|g(t, s)|Δs = max

1 (128 + 208 + 60) 10 < 40. ≤

By Theorem 8.7, it follows that the problem (8.3), (8.4) has a solution. By Theorem 8.6, it follows that any solution y of the problem (8.3), (8.4) satisfies the following estimate: max |y(t)| ≤ K max

t∈[0,4]

t∈[0,4] 0

4

|g(t, s)|Δs

≤ 4 · 40 = 160. Exercise 8.4 Let T = 2N0  F (t, u) =



{0}, T = 8, and

 u3 u3 u3 2 , + t, +1+t +t . 1 + 3u4 1 + 3u4 1 + 3u4

Prove that the problem (8.3), (8.4) has a solution and find an estimate of the solutions of the problem (8.3), (8.4).

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . .

537

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point Boundary Value Conditions Suppose that 0, b ∈ T, b > 0. Consider the following BVP: y Δ∇ (t) ∈ F (y(t)), y Δ (0) =

m−2 (

t ∈ [0, b],

(8.11)

aj y Δ (ξj ),

(8.12)

bj y(ξj ),

(8.13)

j =1

y(b) =

m−2 ( j =1

where (I4)

m−2 % j =1

aj = 1,

m−2 % j =1

bj = 1,

(I5) F  Kc (R) is l.s.c., (I6) there exists a continuous nondecreasing function φ : [0, ∞) → (0, ∞) such that F (u) = sup{|v| : v ∈ F (u)} ≤ φ(|u|),

u ∈ R,

(I7) lim infq→∞ φ(q) q = λ < ∞, (I8) 0 < ξ1 < ξ2 < . . . < ξm−2 < ρ(b). Definition 8.5 A function y : [0, b] → R is said to be a solution of the BVP (8.11)– (8.13) if 1. y Δ and y Δ∇ exist on [a, b], y Δ∇ : [0, b] → R is continuous, 2. there exists a function f (t) ∈ F (y(t)), t ∈ [0, b], such that y Δ∇ (t) = f (t), 3. y Δ satisfies (8.12), 4. y satisfies (8.13). Example 8.5 Let T = 2Z, b = 4, m = 3, a1 =

1 , 5

t ∈ [a, b],

538

8 Second Order Dynamic Inclusions

17 , 3 F (t, u) = {6t, 6t + 1, 6t + 2, 6t + 3}, b1 =

t ∈ [0, 4].

We will show that the function y(t) = t 3 + t 2 + t + 1,

t ∈ [0, 4],

is a solution of the BVP (8.11)–(8.13). Really, we have σ (t) = t + 2, ρ(t) = t − 2, y (t) = (σ (t))2 + tσ (t) + t 2 + σ (t) + t + 1 Δ

= (t + 2)2 + t (t + 2) + t 2 + t + 2 + t + 1 = t 2 + 4t + 4 + t 2 + 2t + t 2 + 2t + 3 = 3t 2 + 8t + 7, y Δ∇ (t) = 3(ρ(t) + t) + 8 = 3(t − 2 + t) + 8 = 6t − 6 + 8 = 6t + 2 ∈ F (y(t)),

t ∈ [0, 4],

and y Δ (0) = 7, y(4) = 43 + 42 + 4 + 1 = 64 + 16 + 4 + 1 = 85, y(2) = 23 + 22 + 2 + 1 = 8+4+3 = 15, y Δ (2) = 12 + 16 + 7 = 35,

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . .

539

1 Δ y (2) 5 1 = · 35 5 =7

a1 y Δ (ξ1 ) =

= y Δ (0), 17 y(2) 3 17 = · 15 3 = 85

b1 y(ξ1 ) =

= y(4). Before formulating and proving the existence results for the problem (8.11)–(8.13), we consider the following second order dynamic equation y Δ∇ (t) = g(t),

t ∈ [0, b],

(8.14)

subject to the boundary conditions (8.12) and (8.13), where (I9) g ∈ C (T). Theorem 8.8 Suppose (I 4) and (I 9). Then y is a solution of the BVP (8.14), (8.12), (8.13) if and only if it is a solution of the following integral equation: y(t) =

!t

%m−2

aj

! ξj

g(s)∇s

j =1 0 % 0 (t − s)g(s)∇s + t 1− m−2 j =1 aj  m−2 !ξ % 1 + %m−2 bj 0 j (ξj − s)g(s)∇s 1− j =1 bj j =1  %  ! ξj m−2 m−2 !b % j =1 aj 0 g(s)∇s %m−2 − 0 (b − s)g(s)∇s + , bj ξj − b

1−

j =1

j =1

(8.15)

aj

t ∈ [0, b]. Proof 1. Let y be a solution of the BVP (8.14), (8.12), (8.13). By nabla integrating of (8.14) from 0 to t, we get t

y Δ (t) = y Δ (0) +

g(s)∇s,

t ∈ [0, b].

0

Now, we delta integrate the Eq. (8.16) from 0 to t and we get

(8.16)

540

8 Second Order Dynamic Inclusions t

y(t) = y(0) + ty Δ (0) +

(t − s)g(s)∇s,

t ∈ [0, b].

0

By (8.16), we have ξj

y Δ (ξj ) = y Δ (0) +

j ∈ {1, . . . , m − 2}.

g(s)∇s, 0

Hence by (8.12), we obtain y Δ (0) =

m−2 (

aj y Δ (ξj )

j =1

=

m−2 (

 aj y Δ (0) +



ξj

g(s)∇s 0

j =1

= y Δ (0)

m−2 (

aj +

m−2 (

j =1

ξj

aj

g(s)∇s, 0

j =1

whereupon ⎛ y Δ (0) ⎝1 −

m−2 (

⎞ aj ⎠ =

j =1

m−2 (

ξj

aj

g(s)∇s, 0

j =1

or !ξ aj 0 j g(s)∇s . % 1 − jm−2 =1 aj

%m−2 y Δ (0) =

j =1

By (8.17), we have ξj

y(ξj ) = y(0) + ξj y Δ (0) +

(ξj − s)g(s)∇s,

0

and using (8.13), we get m−2 (

bj y(ξj ) = y(b)

j =1 b

= y(0) + by Δ (0) + 0

whereupon

(b − s)g(s)∇s,

(8.17)

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . . m−2 (

 bj y(0) + ξj y Δ (0) +

ξj

541

 (ξj − s)g(s)∇s

0

j =1 b

= y(0) + by Δ (0) +

(b − s)g(s)∇s,

0

or ⎛ ⎝1 −

m−2 (

⎞ bj ⎠ y(0) =

j =1

m−2 (

bj ξj y Δ (0) +

j =1

=⎝

ξj

bj

b 0

m−2 (

(b − s)g(s)∇s

⎞

bj ξj − b⎠ y Δ (0) +

j =1 b

− ⎛ =⎝

⎞%

m−2 (

bj ξj − b⎠

m−2 ( j =1 b

ξj

bj

(ξj − s)g(s)∇s

0

(b − s)g(s)∇s

0

−

m−2 ( j =1

j =1

+

(ξj − s)g(s)∇s

0

j =1

−by Δ (0) − ⎛

m−2 (

ξj

bj

! ξj m−2 j =1 aj 0 g(s)∇s % 1 − jm−2 =1 aj

(ξj − s)g(s)∇s

0

(b − s)g(s)∇s,

0

or ⎛⎛ y(0) = 

1−

+

1 %m−2 j =1

m−2 ( j =1 b

−

 ⎝⎝ bj

ξj

bj

m−2 (

⎞% bj ξj − b⎠

j =1

(ξj − s)g(s)∇s

0

 (b − s)g(s)∇s .

0

Substituting y(0) and y Δ (0) in (8.17), we get (8.15).

! ξj m−2 j =1 aj 0 g(s)∇s % 1 − jm−2 =1 aj

542

8 Second Order Dynamic Inclusions

2. Let y be a solution of the integral equation (8.15). Then, !ξ aj 0 j g(s)∇s + y (t) = % 1 − jm−2 =1 aj %m−2 ! ξj j =1 aj 0 g(s)∇s y Δ (0) = , % 1 − jm−2 =1 aj %m−2

t

j =1

Δ

y Δ∇ (t) = g(t),

g(s)∇s,

t ∈ [a, b],

0

t ∈ [0, b].

By (8.15), we get ξl

bl y(ξl ) = bl

(ξl − s)g(s)∇s + bl ξl

0

⎛ +

1− b

−

bl %m−2 j =1

m−2 (

⎝ bj

+⎝

ξj

bj

l=1

bl y(ξl ) =

m−2 (

m−2 (

⎞% bj ξj − b⎠ ξl

bl

(ξl − s)g(s)∇s

m−2 (

 %m−2 bl ξl

%m−2 1− b

−

l=1 bl %m−2

+⎝

bj

j =1

0

(b − s)g(s)∇s

0

⎛

j =1

!ξ aj 0 j g(s)∇s % 1 − jm−2 =1 aj ⎛ m−2 ξj ( ⎝ bj (ξj − s)g(s)∇s j =1

l=1

+

⎞

! ξj m−2 j =1 aj 0 g(s)∇s ⎠ , % 1 − jm−2 =1 aj

0

l=1

+

(ξj − s)g(s)∇s

0

j =1

j =1 m−2 (

j =1

(b − s)g(s)∇s

0

⎛

!ξ aj 0 j g(s)∇s % 1 − jm−2 =1 aj

%m−2

m−2 ( j =1

⎞% bj ξj − b⎠

⎞

! ξj m−2 j =1 aj 0 g(s)∇s ⎠ % 1 − jm−2 =1 aj

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . .

⎛ =

1−

1 %m−2

⎝

m−2 (

bj

j =1

j =1 bj %m−2

b

j =1

%m−2 −

543

⎞ ξj

bj

(ξj − s)g(s)∇s ⎠

0

(b − s)g(s)∇s 1 − j =1 aj 0 ⎛ ⎞ % ⎞ ⎛ m−2 m−2 m−2 ( ( bj j =1 bj ξj + ⎝ bj ξj − b⎠ +⎝ %m−2 ⎠ 1 − j =1 bj j =1 j =1 !ξ aj 0 j g(s)∇s × % 1 − jm−2 =1 aj ⎛ m−2 ( 1 ⎝ = bj %m−2 1 − j =1 bj j =1 %m−2 j −1

%m−2 −

1−

j =1 bj %m−2 j =1

b

aj

⎞ ξj

(ξj − s)g(s)∇s ⎠

0

(b − s)g(s)∇s

0

⎞  ⎛ %m−2 ! ξj % bj ξj − b jm−2 =1 bj ⎝ j −1 aj 0 g(s)∇s ⎠ , % % 1 − jm−2 1 − jm−2 =1 bj =1 aj

 %m−2 j =1

+ and b

y(b) =

!ξ aj 0 j g(s)∇s % 1 − jm−2 =1 aj

%m−2 j =1

(b − s)g(s)∇s + b

0

⎛ +

1− b

−

1 %m−2 j =1

m−2 (

⎝ bj

+⎝

m−2 (

⎞% bj ξj − b⎠

j =1

=

1−

(ξj − s)g(s)∇s

0

j =1

(b − s)g(s)∇s

0

⎛

ξj

bj

1 %m−2 j =1

⎛ ⎝ bj

m−2 ( j =1

⎞

! ξj m−2 j =1 aj 0 g(s)∇s ⎠ % 1 − jm−2 =1 aj

⎞

ξj

bj 0

(ξj − s)g(s)∇s ⎠

544

8 Second Order Dynamic Inclusions

%m−2 −

1−

j =1 bj %m−2 j =1

b

aj

(b − s)g(s)∇s

0

⎞  ⎛ %m−2 ! ξj % bj ξj − b jm−2 =1 bj ⎝ j −1 aj 0 g(s)∇s ⎠ . % % 1 − jm−2 1 − jm−2 =1 bj =1 aj

 %m−2 +

j =1

Thus, y(b) =

m−2 (

bj y(ξj ).

j =1

This completes the proof. Let %m−2

j =1 |aj |ξj α=  %m−2  ,  1 − j =1 aj 

%m−2

2 j =1 |bj |ξj  β= %m−2  ,  1 − j =1 bj 

b2 , γ =  %   m−2 1 − j =1 bj  %   m−2   j =1 bj ξj − b η=  %m−2  .  1 − j =1 bj  Theorem 8.9 Assume (I 4)–(I 8). Then the BVP (8.11)–(8.13) has a solution provided that 

 b2 + bα + β + γ + ηα λ < 1.

(8.18)

Proof By (I 5), we have that F is l.s.c. Then there exists a continuous single-valued function f : R → R such that f (y) ∈ F (y) for any y ∈ R. Consider the dynamic equation y Δ∇ (t) = f (y(t)),

t ∈ [0, b],

(8.19)

subject to the boundary conditions (8.12), (8.13). For y ∈ C ([0, b]), define the operator

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . .

t

Ny(t) =

%m−2 (t − s)f (y(s))∇s + t

0

⎛ +

1− b

−

1 %m−2 j =1

⎝ bj

+⎝

ξj

bj

(ξj − s)f (y(s))∇s

0

j =1

(b − s)f (y(s))∇s

0

⎛

m−2 (

! ξj

0 f (y(s))∇s %m−2 1 − j =1 aj

j =1

aj

545

m−2 (

⎞% bj ξj − b⎠

j =1

⎞

! ξj m−2 j =1 aj 0 f (y(s))∇s ⎠ , % 1 − jm−2 =1 aj

t ∈ [0, b].

By Theorem 8.8, it follows that any fixed point y ∈ C ([0, b]) of the operator N is a solution of the problem (8.19), (8.12), (8.13). 1. Since f is continuous, we have that N : C ([0, b]) → C ([0, b]) is a continuous operator. 2. For each q > 0, define Bq = {y ∈ C ([0, b]) : max |y(t)| ≤ q}. t∈[0,b]

We have that Bq is a bounded, closed, and convex set. Now, we will prove that N (Bq ) ⊆ Bq . Assume that there is an yq ∈ Bq such that |Nyq (t)| > q for some t ∈ [0, b]. Hence, q < |Nyq (t)|  %m−2 ! ξj  t  j =1 aj 0 f (yq (s))∇s =  (t − s)f (yq (s))∇s + t % 1 − jm−2  0 =1 aj ⎛ m−2 ξj ( 1 ⎝ + bj (ξj − s)f (yq (s))∇s %m−2 0 1 − j =1 bj j =1 ⎛

−

≤

t 0

b 0

m−2 (

(b − s)f (yq (s))∇s + ⎝

j =1

bj ξj − b⎠

! ξj j =1 |aj | 0 |f (yq (s))|∇s   %   1 − jm−2 =1 aj 

%m−2 (t − s)|f (yq (s))|∇s + t

⎞ ! ξj  a f (y (s))∇s  q j 0 j =1 ⎠ %m−2  1 − j =1 aj 

⎞% m−2

546

8 Second Order Dynamic Inclusions

⎛

m−2 ξj ( ⎝ |bj | (ξj − s)|f (yq (s))|∇s  0 j =1 bj  j =1  % ⎞ m−2  m−2 |a | ! ξj |f (y (s))|∇s b (  j =1 j 0 q ⎠   (b − s)|f (yq (s))|∇s +  bj ξj − b + %   0  j =1  1 − jm−2 =1 aj 

+  1 −

t

≤

0

1 %m−2

! ξj j =1 |aj | 0 φ(|yq (s)|)∇s   %   a 1 − jm−2  j =1

%m−2 (t − s)φ(|yq (s)|)∇s + t ⎛

m−2 ξj ( ⎝ |bj | (ξj − s)φ(|yq (s)|)∇s  0 j =1 bj  j =1  % ⎞ m−2  m−2 |a | ! ξj φ(|y (s)|)∇s b (  j =1 j 0 q ⎠   (b − s)φ(|yq (s)|)∇s +  bj ξj − b + %   0  j =1  1 − jm−2 =1 aj 

+  1 −

≤b

b 0

1 %m−2

φ(q)∇s + b ⎛

+  1 − +b

! ξj j =1 |aj | 0 φ(q)∇s   %   a 1 − jm−2  j =1

%m−2

m−2 ξj ( ⎝ |bj | ξj φ(q)∇s  0 j =1 bj  j =1

1 %m−2

b

φ(q)∇s 0

%  ⎞  m−2 |a | ! ξj φ(q)∇s m−2  j =1 j 0 ( ⎠   bj ξj − b +  %     j =1 a 1 − jm−2  j =1   ≤ b2 + bα + β + γ + ηα φ(q),

whereupon  φ(q)  1 < b2 + bα + β + γ + ηα q and   φ(q) 1 ≤ b2 + bα + β + γ + ηα lim inf q→∞ q   = λ b2 + bα + β + γ + ηα . This is a contradiction. Therefore, N(Bq ) ⊆ Bq .

8.3 Existence Results for Second Order Dynamic Inclusions with m-Point. . .

547

3. Let t1 , t2 ∈ [0, b], t1 < t2 . Then, for y ∈ Bq , we have |Ny(t2 ) − Ny(t1 )| %m−2 ! ξj  t  2 j =1 aj 0 f (y(s))∇s  (t2 − s)f (y(s))∇s + t2 =  % 1 − jm−2 0 =1 aj ⎛ m−2 ξj ( 1 + bj (ξj − s)f (y(s))∇s %m−2 ⎝ 1 − j =1 bj j =1 0 ⎛ ⎞% ⎞ !ξ m−2 m−2 b ( aj 0 j f (y(s))∇s j =1 ⎠ (b − s)f (y(s))∇s + ⎝ bj ξj − b⎠ − %m−2 1 − a 0 j j =1 j =1 t1

−

%m−2 (t1 − s)f (y(s))∇s − t1

0

⎛ −

1− b

−

1 %m−2 j =1

m−2 (

⎝ bj

ξj

bj

j =1

  = 

t2

(ξj − s)f (y(s))∇s

0

⎛

(b − s)f (y(s))∇s + ⎝

0

m−2 (

⎞% bj ξj − b⎠

j =1

%m−2 (t2 − s)f (y(s))∇s + (t2 − t1 )

t1 0

  = 

t1

  (t1 − s)f (y(s))∇s 

%m−2 (t2 − t1 )f (y(s))∇s + (t2 − t1 )

0 t2

+

t1 t1

≤ 0

+

aj

! ξj

  (t2 − s)f (y(s))∇s 

aj

! ξj

0 f (y(s))∇s %m−2 1 − j =1 aj

j =1

%   m−2 ! ξj   j =1 aj 0 f (y(s))∇s    (t2 − t1 )|f (y(s))|∇s + (t2 − t1 )  %  1 − jm−2 a   j =1 t2

(t2 − s)|f (y(s))|∇s

t1 t1

≤ (t2 − t1 ) 0

φ(q)∇s + (t2 − t1 )

t2

φ(q)∇s t1

⎞

! ξj  m−2 j =1 aj 0 f (y(s))∇s ⎠  %  1 − jm−2 =1 aj

0 f (y(s))∇s %m−2 1 − j =1 aj

j =1

0

−

aj

! ξj

0 f (y(s))∇s %m−2 1 − j =1 aj

j =1

548

8 Second Order Dynamic Inclusions

!ξ |aj | 0 j φ(q)∇s  %m−2   1 − j =1 aj 

%m−2 +(t2 − t1 )

j =1

≤ t1 (t2 − t1 )φ(q) + (t2 − t1 )2 φ(q) + α(t2 − t1 )φ(q) → 0 as

t1 → t2 .

Thus, the set {Ny : y ∈ Bq } is equi-continuous. Now, using the Arzela– Ascoli theorem, we conclude that the operator N : C ([0, b]) → C ([0, b]) is a completely continuous operator. Hence by the Sadovskii fixed point theorem, it follows that the operator N has a fixed point y ∈ C ([0, b]), which is a solution of the BVP (8.11)–(8.13). This completes the proof. Corollary 8.1 Assume (I 4), (I 5), (I 8) and there exist real numbers c, d > 0, 0 < τ < 1, such that F (u) ≤ c|u|τ + d,

u ∈ R.

Then the problem (8.11)–(8.13) has a solution. Proof Let φ(u) = c|u|τ + d,

u ∈ R.

Then F satisfies (I 6) and φ satisfies (I 7), and lim inf q→∞

φ(q) = 0. q

By the last relation, it follows that (8.18) holds. Hence by Theorem 8.9, it follows that the problem (8.11)–(8.13) has a solution. This completes the proof. Exercise 8.5 Let F (t, u) =

#√ 3

u,

√ 3

u + 1,

√ 3

$ u+2 .

Prove that the problem (8.11)–(8.13) has a solution.

8.4 Advanced Practical Problems Problem 8.1 Let T = 2N0 T = 4,



{0},

8.4 Advanced Practical Problems

549

g(t, s) = t 2 + t + 1,  s2 s2 σ σ y y σ (s) + s 2 , (s), F (s, y (s)) = 9(4s 2 + 2s + 1) 9(4s 2 + 2s + 1)  s2 σ 2 y (s) + s + s + 1 , t, s ∈ [0, 4]. 9(4s 2 + 2s + 1) Prove that y(t) = 63t,

t ∈ [0, 4]

is a solution of the dynamic inclusion (8.1). Problem 8.2 Let T = 4N0 {0}, T = 64, and g(t, s) = t 2 + 2ts + 3s 4 ,  s 4 + 3s 8 + 10(y σ (s))3 s 4 + 3s 8 + 10(y σ (s))3 σ , + 1, F (s, y (s)) = 4s 4 + 3s 8 + 15 + 18(y σ (s))6 4s 4 + 3s 8 + 15 + 18(y σ (s))6  s 4 + 3s 8 + 10(y σ (s))3 + 7 , t, s ∈ [0, 16]. 4s 4 + 3s 8 + 15 + 18(y σ (s))6 Prove that the dynamic inclusion (8.1) has a solution. Problem 8.3 Let T = 7Z, T = 14, and F (t, y σ (t)) = {2, 7, 11},

t ∈ [0, 14].

Prove that y(t) = t 2 − 21t,

t ∈ [0, 21],

is a solution to the problem (8.3), (8.4). Problem 8.4 Let T = 3N0 {0}, T = 8, and  F (t, u) =

 7u5 7u5 7u5 2 , + 2t, + 2 + 3t + 4t . 1 + u6 1 + u6 1 + u6

Prove that the problem (8.3), (8.4) has a solution and find an estimate of the solutions of the problem (8.3), (8.4).

550

8 Second Order Dynamic Inclusions

Problem 8.5 Let F (t, u) =



√ √ √ 5 u + t 2 , 5 u + t 2 + 1, 5 u + t 2 + t 4 + 2 .

Prove that the problem (8.11)–(8.13) has a solution.

8.5 Notes and References In this chapter is investigated a periodic boundary value problem for second order dynamic inclusions for existence of solutions. In the chapter is considered m-point boundary value problem for second order dynamic inclusions, and some existence results are deducted. The results in this chapter can be found in [7] and [13].

Chapter 9

Boundary Value Problems for First Order Impulsive Dynamic Inclusions

This chapter is devoted to a qualitative analysis of some classes of first order impulsive dynamic inclusions. They are investigated for existence of solutions of some periodic BVPs for first order impulsive dynamic inclusions. Some criteria for the existence of lower solutions and upper solutions for some classes of BVPs for first order impulsive dynamic inclusions are given. Next, extremal solutions for BVPs for first order impulsive integro-dynamic inclusions are considered. In the chapter, some BVPs for first order dynamic inclusions for existence of multiple positive solutions are investigated.

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions Suppose that T is a time scale with forward jump operator, backward jump operator, and delta differentiation operator σ , ρ, and Δ, respectively. Let also, 0, t1 , T ∈ T, 0 < t1 < T , t1 is right-dense. Set J = [0, T ] and consider the impulsive dynamic inclusion with a single impulse. y ∇ (t) ∈ G(t, y(t)), + = I (y(t1 )), y(t1 ) B(y(0), y(T )) = 0,

t ∈ J \{t1 }, (9.1)

where G : T × R  Kc (R) is l.s.c. Then there exists a continuous selection g(t) ∈ G(t), t ∈ J . Hence, to prove the existence of solutions of (9.1), we will prove that the following BVP has a solution:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. G. Georgiev, Fuzzy Dynamic Equations, Dynamic Inclusions and Optimal Control Problems on Time Scales, https://doi.org/10.1007/978-3-030-76132-5_9

551

552

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

y ∇ (t) = g(t, y(t)), + = I (y(t1 )), y(t1 ) B(y(0), y(T )) = 0.

t ∈ J \{t1 }, (9.2)

Let J1 = [0, t1 ],

J2 = (t1 , T ].

With yi , i ∈ {1, 2}, will be denoted the restriction of y : J → R on Ji , i ∈ {1, 2}. Define C (J1 ) = {y : J1 → R,

y ∈ C (J1 )},

C (J2 ) = {y : J2 → R,

y ∈ C (J2 ),

∃y(t1+ )},

A = {y : J → R,

y1 ∈ C (J1 ),

y2 ∈ C (J2 )}.

For y ∈ A, define y = sup{|y(t)| : t ∈ J }. Then (A,  · ) is a Banach space. For u, v ∈ A, define [u, v] = {w ∈ A : u(t) ≤ w(t) ≤ v(t),

t ∈ J }.

Definition 9.1 We say that y : T → R is a solution of (9.2) if 1. y ∈ A, 2. t

y(t) = y(0) +

g(s, y(s))∇s,

t ∈ J1 ,

0

y(t) = I (y(t1 )) +

t

g(s, y(s))∇s,

t ∈ J2 ,

t1

3. B(y(0), y(T )) = 0. Definition 9.2 We say that α : J → R is a lower solution of Eq. (9.2) if 1. α ∈ A, 2. b

α(b) − α(a) ≤

g(s, α(s))∇s a

for a ≤ b, a, b ∈ J1 or a, b ∈ J2 ,

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions

553

3. α(t1+ ) ≤ I (α(t1 )), 4. B(α(0), α(T )) ≤ 0. Definition 9.3 We say that β : J → R is an upper solution of Eq. (9.2) if 1. β ∈ A, 2. b

β(b) − β(a) ≤

g(s, β(s))∇s a

for a ≤ b, a, b ∈ J1 or a, b ∈ J2 , 3. β(t1+ ) ≥ I (β(t1 )), 4. B(β(0), β(T )) ≥ 0. Let p(t, x) = max {α(t),

min{x, β(t)}} .

Suppose (I1) g : J × R → R is continuous, (I2) There is an h : J → [0, ∞) such that T

h(s)∇s < ∞,

0

|g(t, p(t, x))| ≤ h(t),

(t, x) ∈ J × R,

(I3) I : R → R is continuous and nondecreasing, (I4) B : R × R → R is continuous and for each x ∈ [α(0), β(0)], B(x, ·) is nonincreasing. Define an operator G : A → A as follows: t

Gy(t) = p(0, y(0)) +

g(s, p(s, y(s)))∇s,

t ∈ J1 ,

(9.3)

0

Gy(t) = I (p(t1 , y(t1 ))) +

t

g(s, p(s, y(s)))∇s, t1

where y(0) = y(0) − B(y(0), y(T )).

t ∈ J2 ,

(9.4)

554

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

Lemma 9.1 Assume (I 1)–(I 4) and y ∈ A satisfy t

y(t) = p(0, y(0)) +

g(s, p(s, y(s)))∇s,

t ∈ J1 ,

(9.5)

0

y(t) = I (p(t1 , y(t1 ))) +

t

g(s, p(s, y(s)))∇s,

t ∈ J2 .

t1

Then, y(t) ∈ [α(t), β(t)],

t ∈ J.

Proof Take t = 0. From (9.5), we get y(0) = p(0, y(0)) = max {α(0),

min{y(0), β(0)}}

= max {α(0),

min{y(0) − B(y(0), y(T )), β(0)}} .

If y(0) = α(0), we get y(0) ∈ [α(0), β(0)]. If y(0) = min{y(0) − B(y(0), y(T )),

β(0)},

then α(0) ≤ y(0) ≤ β(0). Suppose that there is a t2 ∈ (0, t1 ] such that α(t2 ) > y(t2 ). Since α(0) ≤ y(0), there is a t3 ∈ [0, t2 ) so that α(t) ≤ y(t),

t ∈ [0, t3 ],

α(t) > y(t),

t ∈ (t3 , t2 ].

and

Then, p(t, y(t)) = max {α(t), = α(t),

min{y(t), β(t)}}

t ∈ (t3 , t2 ],

(9.6)

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions

555

and g(t, p(t, y(t))) = g(t, α(t)),

t ∈ (t3 , t2 ].

Hence, for t ∈ (t3 , t2 ], we obtain t

g(s, p(s, y(s)))∇s = y(t) − p(0, y(0))

0

= y(t3 ) − p(0, y(0)) + y(t) − y(t3 ) t3

=

g(s, p(s, y(s)))∇s + y(t) − y(t3 ),

0

and t

y(t) − y(t3 ) =

g(s, p(s, y(s)))∇s 0

0

=

t3

g(s, p(s, y(s)))∇s −

t

g(s, p(s, y(s)))∇s t3

=

t

g(s, α(s))∇s t3

≥ α(t) − α(t3 ). Since y(t3 ) ≥ α(t3 ), we get y(t) ≥ α(t), t ∈ (t3 , t2 ]. This is a contradiction. Therefore, y(t) ≥ α(t), t ∈ [0, t1 ]. Similarly, y(t) ≤ β(t), t ∈ [0, t1 ]. In particular, we have α(t1 ) ≤ y(t1 ) ≤ β(t1 ). Hence, α(t1+ ) ≤ I (α(t1 )) ≤ I (y(t1 )) ≤ I (β(t1 )) ≤ β(t1+ ). Since I (y(t1 )) = I (p(t1 , y(t1 ))) = y(t1+ ),

556

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

we obtain α(t1+ ) ≤ y(t1+ ) ≤ β(t1+ ). Proceeding as above, we arrive at α(t) ≤ y(t) ≤ β(t),

t ∈ J2 .

This completes the proof. Remark 9.1 Suppose that all conditions of Lemma 9.1 hold. Then, !t y(t) = y(0) + 0 g(s, ! t y(s))∇s, t ∈ J1 , y(t) = I (y(t1 )) + t1 g(s, y(s))∇s, t ∈ J2 .

(9.7)

Lemma 9.2 Assume (I 1)–(I 4) and y ∈ A satisfy (9.5), (9.6). Then, y(0) ∈ [α(0), β(0)]. Proof Suppose that α(0) > y(0). Thus, y(0) = p(0, y(0)) = max{α(0), min{y(0), β(0)}} = max{α(0), y(0)} = α(0). Hence, B(y(0), y(T )) > 0 and B(α(0), α(T )) = B(y(0), α(T )) ≥ B(y(0), y(T )) > 0, which is a contradiction. Therefore, α(0) ≤ y(0). Now, assume that y(0) > β(0). Then,

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions

557

y(0) − B(y(0), y(T )) = y(0) > β(0) and y(0) = p(0, y(0)) = max{α(0), min{y(0), β(0)}} = max{α(0), β(0)} = β(0), and B(y(0), y(T )) < 0. Hence, 0 > B(y(0), y(T )) = B(β(0), y(T )) ≥ B(β(0), β(T )), which is a contradiction. Consequently, y(0) ≤ β(0). This completes the proof. Lemma 9.3 Assume (I 1)–(I 4) and y ∈ A satisfy (9.5), (9.6). If y is a fixed point of the operator G, then y is a solution of (9.2). Proof By Lemma 9.1, we get y(t) ∈ [α(t), β(t)],

t ∈ J.

By Remark 9.1, it follows that y satisfies (9.7). By Lemma 9.2, we obtain y(0) ∈ [α(0), β(0)] and y(0) = p(0, y(0)) = max{α(0), min{y(0), β(0)}} = max{α(0), y(0)} = y(0) = y(0) − B(y(0), y(T )),

558

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

whereupon B(y(0), y(T )) = 0. Thus, y is a solution of Eq. (9.2). This completes the proof. Define K = α + β and w : J → R as follows: t

w(t) = K +

h(s)∇s. 0

Let  S = y ∈ A : |y(0)| ≤ K, on

0 ≤ a ≤ b ≤ t1

|y(t1+ )| ≤ K, or

|y(b) − y(a)| ≤ w(b) − w(a)

t1 < a ≤ b ≤ T ,

 a, b ∈ J .

Exercise 9.1 Prove that S is a convex and compact subset of (A,  · ). Lemma 9.4 Let (I 1)–(I 4) hold. Then G(S) ⊆ S. Proof Take y ∈ S arbitrarily. Then, |Gy(0)| = |p(0, y(0))| = |max{α(0), min{y(0), β(0)}}| ≤ max{|α(0)|, |β(0)|} ≤ K. Note that β(t1 ) ≥ p(t1 , y(t1 )) = max{α(t1 ), min{y(t1 ), β(t1 )}} ≥ α(t1 ). Hence, α(t1+ ) ≤ I (α(t1 )) ≤ I (p(t1 , y(t1 )))

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions

≤ I (β(t1 )) ≤ β(t1+ ) and |I (p(t1 , y(t1 )))| ≤ max{|α(t1+ )|, |β(t1+ )|} ≤ K. Let t → t1 , t > t1 , in (9.4) and we get |Gy(t1+ )| = |I (p(t1 , y(t1 )))| ≤ K. Let a, b ∈ J , 0 ≤ a ≤ b ≤ t1 . Then,   |Gy(b) − Gy(a)| = p(0, y(0)) +

b

g(s, p(s, y(s)))∇s 0

  g(s, p(s, y(s)))∇s 

a

−p(0, y(0)) − 0

  = 

b a b

≤

  g(s, p(s, y(s)))∇s 

|g(s, p(s, y(s)))|∇s

a b

≤

h(s)∇s a

= w(b) − w(a). Now, take t1 < a ≤ b ≤ T . Then,   |Gy(b) − Gy(a)| = I (p(t1 ), y(t1 )) + −I (p(t1 ), y(t1 )) −   = 

b

g(s, p(s, y(s)))∇s t1 a t1

b a b

≤ a

  g(s, p(s, y(s)))∇s 

  g(s, p(s, y(s)))∇s 

|g(s, p(s, y(s)))|∇s

559

560

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions b

≤

h(s)∇s a

= w(b) − w(a). Thus, G : S → S. This completes the proof. Lemma 9.5 Let (I 1)–(I 4) hold. Then G : S → S is continuous. Proof Let {yn }n∈N ⊆ S and yn → y, as n → ∞, y ∈ S, in the space (A,  · ). Note that yn → y, as n → ∞, uniformly on compact subsets of J . Then, for t ∈ J1 , we get   |Gy(t) − Gyn (t)| = Gy(t) − p(0, y(0)) − (Gyn (t) − p(0, y n (0)))   +p(0, y(0)) − p(0, y n (0))   = 

t 0

  +p(0, y(0)) − p(0, y n (0))   = 

t

t

g(s, p(s, y(s)))∇s −

g(s, p(s, yn (s)))∇s 0

(g(s, p(s, y(s))) − g(s, p(s, yn (s))))∇s

0

  +p(0, y(0)) − p(0, y n (0))  t     ≤  (g(s, p(s, y(s))) − g(s, p(s, yn (s))))∇s  0

+|p(0, y(0)) − p(0, y n (0))|   t    ≤ g(s, p(s, y(s))) − g(s, p(s, yn (s)))∇s 0

+|p(0, y(0)) − p(0, y n (0))| → 0,

as

n → ∞.

For t ∈ J2 , we obtain   |Gy(t) − Gyn (t)| = Gy(t) − I (p(t1 , y(t1 ))) − (Gyn (t) − I (p(t1 , yn (t1 ))))   +I (p(t1 , y(t1 ))) − I (p(t1 , yn (t1 )))

9.1 Lower and Upper Solutions for First Order Impulsive Dynamic Inclusions

  = 

t

g(s, p(s, y(s)))∇s −

t1

561

t

g(s, p(s, yn (s)))∇s t1

  +I (p(t1 , y(t1 ))) − I (p(t1 , yn (t1 )))  t  =  (g(s, p(s, y(s))) − g(s, p(s, yn (s))))∇s t1

  +I (p(t1 , y(t1 ))) − I (p(t1 , yn (t1 )))   ≤ 

t t1

  (g(s, p(s, y(s))) − g(s, p(s, yn (s))))∇s 

    +I (p(t1 , y(t1 ))) − I (p(t1 , yn (t1 )))   t    ≤ g(s, p(s, y(s))) − g(s, p(s, yn (s)))∇s t1

     +I (p(t1 , y(t1 ))) − I (p(t1 , yn (t1 ))) → 0,

as

n → ∞.

This completes the proof. Lemma 9.6 Suppose (I 1)–(I 4). Then G has a fixed point in S. Proof By Lemma 9.4, it follows that G : S → S. From Lemma 9.5, we have that G : S → S is continuous. Hence by Schauder’s fixed point theorem, we conclude that G has a fixed point in S. This completes the proof. Theorem 9.1 Suppose (I 1)–(I 4). Then Eq. (9.2) has at least one solution y ∈ S such that y(t) ∈ [α(t), β(t)], t ∈ J . Proof By Lemma 9.6, we have that the operator G has a fixed point y ∈ S. By Lemma 9.3, we obtain that y solves (9.2) and y(t) ∈ [α(t), β(t)], t ∈ J . This completes the proof. Example 9.1 Let T = [0, 1] [2, 3], where [0, 1] and [2, 3] are the real-valued intervals. Consider the following impulsive dynamic equation: y ∇ (t) = t 2 + (y(t))2 , y(2+ ) = y(2) + 1, B(y(0), y(3)) = 0.

t ∈ [0, 1]

"

(2, 3],

562

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

Note that α ≡ 0 is a lower solution for the considered equation. Let β be an upper solution of the considered equation. To construct β on [0, 1], consider the IVP β  = 1 + β 2,

t ∈ [0, 1],

β(0) = 1. We have β(t) = tan t,

t ∈ [0, 1].

We have 2

β(2) = β(0) +

β ∇ (s)∇s

0 1

= β(0) +

β ∇ (s)∇s + β ∇ (2)

0 1

= β(0) +

β  (s)ds +

0

β(2) − β(1) 2−1

= β(0) + β(1) − β(0) + β(2) − β(1) = β(2). Thus, β(2) is arbitrary. Take β(2) = 1. We have β(2+ ) = I (β(2)) = β(2) + 1 = 1+1 = 2. To construct β on [2, 3], consider the IVP β  = 9 + β 2,

t ∈ [2, 3],

β(2) = 2. For its solution we have     2 β(t) = 3 tan tan−1 + 3t − 6 , 3

t ∈ [2, 3].

9.2 Periodic Boundary Value Problems for First Order Linear Dynamic. . .

563

By Theorem 9.1, it follows that the considered equation has a solution y such that 0 ≤ y(t) ≤ β(t),

t ∈ T.

Exercise 9.2 Let T = [0, 2] [3, 4], where [0, 2] and [3, 4] are the real-valued intervals. Investigate for existence of solutions the following impulsive dynamic equation: y ∇ (t) = t 4 + 2(y(t)),

t ∈ [0, 2]

"

(3, 4],

y(3+ ) = y(3) − 1, B(y(0), y(4)) = 0.

9.2 Periodic Boundary Value Problems for First Order Linear Dynamic Inclusions with Impulses Let T be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also, 0, T ∈ T, T > 0, J = [0, T ]. Assume that 0 = t0 < t1 < . . . < tm < tm+1 = T , tk , k ∈ {1, . . . , m}, are right-dense. Set J0 = J \{tk }m k=1 . Consider the following linear impulsive dynamic inclusion: x Δ (t) + p(t)x σ (t) ∈ Q(t), t ∈ J0 , = x(tk− ) + Ik (x(tk )), x(tk+ )

(9.8)

with periodic boundary condition x(0) = x(σ (T )),

(9.9)

where Q : T  Kc (R) is u.s.c. Then there exists a continuous selection q(t) ∈ Q(t), t ∈ J . Thus, to prove the existence of solutions of the problem (9.8), (9.9), we will prove the existence of solutions of the following problem: x Δ (t) + p(t)x σ (t) = q(t), t ∈ J0 , = x(tk− ) + Ik (x(tk )), x(tk+ )

(9.10)

with periodic boundary condition x(0) = x(σ (T )),

(9.11)

564

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

(H1) p, q ∈ C ([0, σ (T )]), p is regressive on [0, σ (T )], (H2) Ik ∈ C (R), k ∈ {1, . . . , m}. Define  P C([0, σ (T )]) = x : [0, σ (T )] → R, on

J0 ,

is continuous

ld − continuous

∃ lim x(t), t→tk +0

x

at tk ,  k ∈ {1, . . . , m} .

P C([0, σ (T )]) is a Banach space with the norm x =

max

sup

j ∈{0,...,m} t∈[tj ,tj +1 ]

|x(t)|.

Now, we define another Banach space  P C Δ ([0, σ (T )]) = x : x ∈ P C([0, σ (T )]), it is

continuous

∃ lim x Δ (t) and t→tk +0

tk

on

∃x Δ

and

(tk , tk+1 ), lim x Δ (t)

t→tk −

is left − scattered,

if 

k ∈ {1, . . . , m}

with the norm   Δ    xΔ = max x, sup x (t) . 

t∈J0

Definition 9.4 As a solution of the problem (9.10), (9.11), we mean a function x ∈ P C Δ ([0, σ (T )]) such that (9.10) and (9.11) are satisfied. Lemma 9.7 Suppose ep (σ (T ), 0) = 1. If x ∈ P C Δ ([0, σ (T )]) is a solution of the problem (9.10), (9.11), then σ (T )

x(t) = 0

g(t, s)q(s)Δs +

m ( k=1

g(t, tj )Ij (x(tj )),

t ∈ [0, σ (T )],

(9.12)

9.2 Periodic Boundary Value Problems for First Order Linear Dynamic. . .

565

where 1 g(t, s) = 1 − ep (σ (T ), 0)



ep (t, s) 0 ≤ s < t ≤ σ (T ) ep (σ (T ), 0)ep (t, s) 0 ≤ t ≤ s ≤ σ (T ), (9.13)

t, s ∈ [0, σ (T )]. Proof Let y(t) = ep (t, 0)x(t),

t ∈ [0, σ (T )].

Then, y Δ (t) = p(t)ep (t, 0)x σ (t) + ep (t, 0)x Δ (t)   = ep (t, 0) p(t)x σ (t) + x Δ (t) = q(t)ep (t, 0),

t ∈ J0 ,

y(σ (T )) = ep (σ (T ), 0)x(σ (T )) = ep (σ (T ), 0)x(0) = ep (σ (T ), 0)y(0), y(tj+ )

= ep (tj+ , 0)x(tj+ ) = ep (tj , 0)(x(tj ) + Ij (x(tj ))) = ep (tj , 0)x(tj ) + ep (tj , 0)Ij (x(tj )) = ep (tj , 0)ep (0, tj )y(tj ) + ep (tj , 0)Ij (x(tj )) = y(tj ) + ep (tj , 0)Ij (x(tj )),

j ∈ {1, . . . , m},

i.e., y satisfies the impulsive periodic boundary value problem y Δ (t) = q(t)ep (t, 0), t ∈ J0 , y(tj+ ) = y(tj ) + ep (tj , 0)Ij (x(tj )),

j ∈ {1, . . . , m},

y(σ (T )) = ep (σ (T ), 0)y(0).

(9.14) (9.15)

By the first equation of (9.14), we derive t

y(t) = y(0) +

ep (s, 0)q(s)Δs,

t ∈ [0, t1 ],

0

and y(t) = y(tj+ ) +

t

ep (s, 0)q(s)Δs, tj

t ∈ (tj , tj +1 ],

j ∈ {1, . . . , m}.

566

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

Thus, tj +1

y(tj +1 ) = y(tj+ ) + tj +1

=

ep (s, 0)q(s)Δs tj

ep (s, 0)q(s)Δs + y(tj ) + ep (tj , 0)Ij (x(tj ))

tj tj +1

=

ep (s, 0)q(s)Δs +

tj

tj

ep (s, 0)q(s)Δs tj −1

+ep (tj −1 , 0)Ij −1 (x(tj −1 )) + ep (tj , 0)Ij (x(tj )) tj +1

=

tj −1

ep (s, 0)q(s)Δs + ep (tj −1 , 0)Ij −1 (x(tj −1 )) + ep (tj , 0)Ij (x(tj ))

= ··· tj +1

=

ep (s, 0)q(s)Δs +

0

j (

ep (tk , 0)Ik (x(tk )) + y(0).

k=1

Hence, t

y(t) = y(tj+ ) +

ep (s, 0)q(s)Δs tj t

= y(tj ) +

ep (s, 0)q(s)Δs + ep (tj , 0)Ij (x(tj ))

tj tj

=

ep (s, 0)q(s)Δs +

0

+

j −1 (

ep (tk , 0)Ik (x(tk )) + y(0)

k=1 t

ep (s, 0)q(s)Δs + ep (tj , 0)Ij (x(tj ))

tj t

= y(0) +

ep (s, 0)q(s)Δs +

0

t ∈ (tj , tj +1 ],

j (

ep (tj , 0)Ij (x(tj )),

k=1

j ∈ {1, . . . , m}.

Therefore, t

y(t) = y(0) + 0

ep (s, 0)q(s)Δs +

(

ep (tj , 0)Ij (x(tj )),

(9.16)

j :tj (Aσ (T )δ)−1 , we get that there exists a constant R > rδ such that f (t, x) ≥ Nx

for

x ≥ δR.

Define ΩR = {x ∈ X : x < R}. Then, for x ∈ K



∂ΩR , we have x(t) ≥ δx = δR,

and

t ∈ [0, σ (T )]

596

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions σ (T )

Ψ x(t) =

g(t, s)f (s, x(σ (s)))Δs +

0

m (

g(t, tk )Ik (x(tk ))

k=1



σ (T )

≥A

f (s, x(σ (s)))Δs +

0

m (



Ik (x(tk ))

k=1

σ (T )

≥A

f (s, x(σ (s)))Δs 0 σ (T )

≥ AN

x(σ (s))Δs 0

≥ ANδRσ (T ) ≥R = x,

t ∈ [0, σ (T )],

whereupon Ψ x ≥ x. Hence by Theorem 9.11, it follows that the problem (9.19), (9.20) has at least one positive solution. This completes the proof. Theorem 9.13 Suppose (H 4)–(H 6) and max f∞ = 0,

min f0 = ∞,

lim

x→∞

Ik (x) = 0, x

k ∈ {1, . . . , m}.

Then the problem (9.19), (9.20) has at least one positive solution. Proof Let Ψ be as in the proof of Theorem 9.6. As in the proof of Theorem 9.3, we have that Ψ : X → X is completely continuous. Also, as in the proof of Theorem 9.6, it follows that Ψ : K → K. Since max f∞ = 0 and

lim

x→∞

Ik (x) = 0, x

for  ≤ (B(σ (T ) + m))−1 , there exists a positive constant r such that f (t, x) ≤ x,

Ik (x) ≤ x,

k ∈ {1, . . . , m},

Let Ωr = {x ∈ X : x < r}.

x ≥ δr.

9.3 Periodic Boundary Value Problems for First Order Nonlinear Dynamic. . .

Then, for x ∈ K



597

∂Ωr , we have x = r and σ (T )

Ψ x(t) =

g(t, s)f (s, x(σ (s)))Δs +

0

m (

g(t, tk )Ik (x(tk ))

k=1



σ (T )

≤B

f (s, x(σ (s)))Δs +

0

m (

Ik (x(tk ))

k=1



σ (T )

≤ B

x(σ (s))Δs +

0

m (





x(tk )

k=1

≤ B(σ (T ) + m)x ≤ x,

t ∈ [0, σ (T )],

whereupon Ψ x ≤ x. Because min f0 = ∞, for N > (Aσ (T )δ)−1 , there exists a constant R < δr such that f (t, x) ≥ Nx

for

0 < x ≤ R.

Define ΩR = {x ∈ X : x < R}. Then, for x ∈ K



∂ΩR , we have x(t) ≥ δx = δR,

t ∈ [0, σ (T )],

and σ (T )

Ψ x(t) =

g(t, s)f (s, x(σ (s)))Δs +

0

k=1 σ (T )

≥

m (

g(t, s)f (s, x(σ (s)))Δs 0 σ (T )

≥ AN

x(σ (s))Δs 0

≥ ANσ (T )x

g(t, tk )Ik (x(tk ))

598

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

≥ x,

t ∈ [0, σ (T )],

whereupon Ψ x ≥ x. Hence by Theorem 9.11, it follows that the problem (9.19), (9.20) has at least one positive solution. This completes the proof. Theorem 9.14 Suppose (H 4)–(H 6) and min f0 = ∞, there exists a positive constant R such that f (t, x) ≤

R , 2Bσ (T )

0 < x ≤ R,

t ∈ [0, σ (T )],

and Ik (x) ≤

|x| , 2mB

k ∈ {1, . . . , m},

x ∈ R.

Then the problem (9.19), (9.20) has at least one positive solution. Proof Let Ψ be as in the proof of Theorem 9.6. As in the proof of Theorem 9.3, we have that Ψ : X → X is completely continuous. Also, as in the proof of Theorem 9.6, it follows that Ψ : K → K. Since min f0 = ∞, for N > (Aσ (T )δ)−1 , there exists a positive constant r < 2BNRσ (T ) such that f (t, x) ≥ Nx,

0 < x ≤ r,

t ∈ [0, σ (T )].

Define Ωr = {x ∈ X : x < r}. Then, for any x ∈ K



∂Ωr , we have x(t) ≥ δx = δr,

t ∈ [0, σ (T )],

and σ (T )

Ψ x(t) = 0

g(t, s)f (s, x(σ (s)))Δs +

m ( k=1

g(t, tk )Ik (x(tk ))

9.3 Periodic Boundary Value Problems for First Order Nonlinear Dynamic. . .

599

σ (T )

≥

g(t, s)f (s, x(σ (s)))Δs 0 σ (T )

≥A

f (s, x(σ (s)))Δs 0 σ (T )

≥ AN

x(σ (s))Δs 0

≥ ANδrσ (T ) ≥r = x,

t ∈ [0, σ (T )],

whereupon Ψ x ≥ x. Let ΩR = {x ∈ X : x < R}. Then, for x ∈ K Ψ x(t) =



∂ΩR , we obtain

! σ (T ) 0

m %

g(t, tk )Ik (x(tk ))

k=1

 ≤B

g(t, s)f (s, x(σ (s)))Δs +

! σ (T ) 0

f (s, x(σ (s)))Δs +

m %



(9.23)

Ik (x(tk ))

k=1



m R 1 ( σ (T ) + x(tk ) ≤B 2Bσ (T ) 2mB k=1   mR R + ≤B 2B 2mB



=R = x,

t ∈ [0, σ (T )]

whereupon Ψ x ≤ x. Hence by Theorem 9.11, it follows that the problem (9.19), (9.20) has at least one positive solution. This completes the proof.

600

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

Theorem 9.15 Suppose (H 4)–(H 6) and min f∞ = ∞, and there exists a positive constant R such that f (t, s) ≤

R , 2Bσ (T ) |x| , 2mB

Ik (x) ≤

0 < x ≤ R,

t ∈ [0, σ (T )],

k ∈ {1, . . . , m},

x ∈ R.

Then the problem (9.19), (9.20) has at least one positive solution. Proof Let Ψ be as in the proof of Theorem 9.6. As in the proof of Theorem 9.3, we have that Ψ : X → X is completely continuous. Also, as in the proof of Theorem 9.6, it follows that Ψ : K → K. Since min f∞ = ∞, for M ∗ ≥ (Aσ (T )δ)−1 , there exists a positive constant R ∗ > Rδ such that f (t, x) ≥ M ∗ x,

x ≥ δR ∗ ,

t ∈ [0, σ (T )].

Let ΩR ∗ = {x ∈ X : x < R ∗ }. Then, for any x ∈ K



∂ΩR ∗ , we have x(t) ≥ δx = δR ∗ ,

t ∈ [0, σ (T )],

and σ (T )

Ψ x(t) =

g(t, s)f (s, x(σ (s)))Δs +

0

k=1 σ (T )

≥

m (

g(t, s)f (s, x(σ (s)))Δs 0 σ (T )

≥A

f (s, x(σ (s)))Δs 0 σ (T )

≥ AM ∗

x(σ (s))Δs 0

≥ AM ∗ δR ∗ σ (T ) ≥ R∗

g(t, tk )Ik (x(tk ))

9.4 Periodic Boundary Value Problems for First Order Nonlinear Dynamic. . .

= x,

601

t ∈ [0, σ (T )],

whereupon Ψ x ≥ x. Hence, by (9.23) and Theorem 9.11, it follows that the problem (9.19), (9.20) has at least one positive solution. This completes the proof.

9.4 Periodic Boundary Value Problems for First Order Nonlinear Dynamic Inclusions with Impulses-II Suppose that T is a periodic time scale with period p such that 0, tj ∈ T, j ∈ {1, . . . , m}, where T = lp for some l ∈ N, 0 < tj < tj +1 and each tj is dense in T, j ∈ {1, . . . , m}. In this section, we will investigate the following BVP: y Δ (t) + a(t)y σ (t) ∈ F (t, y(t)), t ∈ T, = I (tj , y(tj )), j ∈ {1, . . . , m}, y(tj+ ) − y(tj− ) y(0) = y(T ),

(9.24)

where F : T × R  Kc (R) is u.s.c. Then there exists a continuous selection f (t, x(t)) ∈ F (t, x(t)), t ∈ T. To prove the existence of solutions of (9.24), we will investigate the following problem: y Δ (t) + a(t)y σ (t) = f (t, y(t)), t ∈ T, = I (tj , y(tj )), j ∈ {1, . . . , m}, y(tj+ ) − y(tj− ) y(0) = y(T ),

(9.25)

where (H1) a ∈ R, a(t + T ) = a(t), t ∈ T, η = ea (T , 0) < 1, (H2) f ∈ C (T × R), f (t + T , y(t + T )) = f (t, y(t)), t ∈ T. Define tm+1 = T , J0 = [0, t1 ], Jj = (tj , tj +1 ], j ∈ {1, . . . , m},  P Cp = y : T → R, ∃y(tj± ),

y(t ± T ) = y(t),

y(tj− ) = y(tj ),

P Cp1 = {y : T → R,

y ∈ C (Jj ),

 j ∈ {1, . . . , m} ,

y(t ± T ) = y(t),

y ∈ C 1 (Jj ),

The set P Cp is a Banach space endowed with the norm

j ∈ {1, . . . , m}}.

602

9 Boundary Value Problems for First Order Impulsive Dynamic Inclusions

y =

max {

j ∈{1,...,m}

(

|y(t)|}.

t∈Jj

Lemma 9.11 Let h ∈ P Cp . Then the function y ∈ P Cp1 is a solution of the BVP y Δ (t) + a(t)y σ (t) = h(t), t ∈ T, = I (tj , y(tj )), j ∈ {1, . . . , m}, y(tj+ ) − y(tj− ) y(0) = y(T )

(9.26)

if and only if y ∈ P Cp and y satisfy the integral equation T

y(t) =

G(t, s)h(s)Δs +

0

m (

G(t, tj )I (tj , y(tj )),

(9.27)

j =1

where G(t, s) =

1 1−η



ea (t, s) 0 ≤ s ≤ t ≤ T ηea (t, s) 0 ≤ t < s ≤ T .

Proof We have that y ∈ P Cp1 is a solution of the impulsive dynamic equation y Δ (t) + a(t)y σ (t) = h(t), t ∈ T, = I (tj , y(tj )), j ∈ {1, . . . , m}, y(tj+ ) − y(tj− )

(9.28)

if and only if y ∈ P Cp and it satisfies the integral equation t

y(t) = 0

ea (t, s)h(s)Δs + ea (t, 0)y(0) +

(

ea (t, tj )I (tj , y(tj )),

{j :tj