173 16 2MB
English Pages 578 Year 2019
Table of contents :
Preface......Page 5
Contents......Page 7
1.1 Banach Space Valued Functions......Page 11
1.2 Unbounded Operators......Page 13
1.3 C0 Semigroup......Page 14
1.4 Evolution Equations......Page 16
1.5 C0 Groups......Page 17
1.6 Notation......Page 18
1.7 Admissible Control Operators......Page 19
1.8 Examples......Page 25
1.9 Turnpike Property for Variational Problems......Page 30
2.1 Preliminaries and Main Results......Page 35
2.2 Lower Semicontinuity Property......Page 41
2.3 Perturbed Problems......Page 45
2.4 A Subclass of Problems......Page 46
2.5 Structure of Solutions in the Regions Close to the Endpoints......Page 52
2.6 Examples......Page 55
2.7 Auxiliary Results for Theorem 2.1......Page 59
2.8 Proofs of Theorems 2.1–2.4......Page 62
2.9 Proof of Theorem 2.9......Page 65
2.10 An Auxiliary Result for Theorem 2.10......Page 73
2.11 Proof of Theorem 2.10......Page 75
2.12 Proof of Proposition 2.12......Page 80
2.13 Auxiliary Results for Theorems 2.13 and 2.14......Page 81
2.14 Proofs of Theorems 2.13 and 2.14......Page 87
2.15 Proofs of Theorems 2.15–2.17......Page 90
2.16 Proof of Theorem 2.21......Page 95
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31......Page 96
2.18 Proof of Proposition 2.37......Page 104
2.19 Auxiliary Results for Theorems 2.38 and 2.39......Page 106
2.20 Proofs of Theorems 2.38 and 2.39......Page 107
2.21 Proofs of Theorem 2.40 and 2.41......Page 112
2.22 The First Bolza Problem......Page 114
2.23 Proof of Theorem 2.63......Page 118
2.24 Proof of Theorem 2.66......Page 120
2.25 Proof of Proposition 2.69......Page 125
2.26 The Second Bolza Problem......Page 127
2.27 Proof of Theorem 2.70......Page 129
2.28 Proof of Theorem 2.71......Page 130
2.29 Proof of Proposition 2.73......Page 137
3.1 Preliminaries and Main Results......Page 141
3.2 The Boundedness Results......Page 144
3.3 Turnpike Properties......Page 146
3.4 Lower Semicontinuity Property and Infinite Horizon Problems......Page 152
3.5 Perturbed Problems......Page 155
3.6 Examples......Page 158
3.7 Auxiliary Results for Theorems 3.1 and 3.2......Page 163
3.8 Proofs of Theorems 3.1–3.5......Page 166
3.9 An Auxiliary Result for Theorem 3.17......Page 168
3.10 Proof of Theorem 3.17......Page 169
3.11 An Auxiliary Result......Page 175
3.12 Proof of Theorem 3.18......Page 178
3.13 Auxiliary Results for Theorems 3.19......Page 182
3.14 Proofs of Theorems 3.19 and 3.20......Page 189
3.15 An Auxiliary Result for Theorem 3.24......Page 193
3.16 Proof of Theorem 3.24......Page 194
3.17 Proof of Theorem 3.26......Page 201
4.1 Preliminaries and Main Results......Page 206
4.2 Boundedness Results......Page 209
4.3 Turnpike Results......Page 211
4.4 Perturbed Problems......Page 216
4.5 Auxiliary Results for Theorems 4.1 and 4.2......Page 217
4.6 Proof of Theorems 4.1 and 4.2......Page 220
4.7 Proofs of Propositions 4.10, 4.12 and Theorems 4.3–4.5......Page 222
4.8 Auxiliary Results for Theorem 4.7......Page 224
4.9 STP Implies (P1)–(P3)......Page 228
4.10 (P1)–(P3) Imply STP......Page 238
4.11 Proof of Theorem 4.8......Page 262
4.12 Proof of Theorem 4.13......Page 264
4.13 Proof of Theorem 4.16......Page 271
5.1 Preliminaries......Page 277
5.2 Boundedness Results......Page 283
5.3 Turnpike Results......Page 285
5.4 Lower Semicontinuity Property......Page 288
5.5 Perturbed Problems......Page 290
5.6 The Triplet (f,A,G)......Page 291
5.7 Auxiliary Results for Theorem 5.1......Page 296
5.8 Proofs of Theorems 5.1–5.4......Page 302
5.9 Proofs of Theorems 5.6–5.9......Page 307
5.10 Proofs of Theorems 5.11 and 5.12......Page 311
5.11 Proof of Theorem 5.13......Page 312
5.12 An Auxiliary Result......Page 317
5.13 Proof of Theorem 5.14......Page 319
5.14 Proof of Proposition 5.16......Page 323
5.15 Proof of Proposition 5.18......Page 324
5.16 Auxiliary Results for Theorems 5.19 and 5.20......Page 326
5.17 Proofs of Theorems 5.19 and 5.20......Page 330
5.18 Proofs of Theorems 5.21–5.23......Page 333
5.19 Proof of Proposition 5.27......Page 337
5.20 Proof of Theorem 5.28......Page 344
5.21 Auxiliary Results......Page 345
5.22 Structure of Solutions in the Regions Close to the Endpoints......Page 356
5.23 Auxiliary Results for Theorems 5.58–5.61......Page 358
5.24 Proofs of Theorems 5.58 and 5.59......Page 359
5.25 Proofs of Theorems 5.60 and 5.61......Page 363
5.26 The First Bolza Problem......Page 365
5.27 Proof of Theorem 5.68......Page 369
5.28 Proof of Theorem 5.71......Page 370
5.29 Proof of Proposition 5.74......Page 375
5.30 The Second Bolza Problem......Page 377
5.31 Proof of Theorem 5.75......Page 379
5.32 Proof of Theorem 5.76......Page 380
5.33 Proof of Proposition 5.78......Page 386
5.34 Examples......Page 389
5.35 Exercises for Chapter 5......Page 392
6.1 Preliminaries......Page 393
6.2 Boundedness Results......Page 399
6.3 Turnpike Results......Page 401
6.4 Lower Semicontinuity Property......Page 408
6.5 Perturbed Problems......Page 410
6.6 Auxiliary Results for Theorems 6.2 and 6.3......Page 411
6.7 Proofs of Theorems 6.2 and 6.3......Page 418
6.8 Proofs of Theorems 6.4–6.10......Page 421
6.9 Proof of Theorem 6.22......Page 427
6.10 An Auxiliary Result for Theorem 6.23......Page 432
6.11 Proof of Theorem 6.23......Page 434
6.12 An Auxiliary Result......Page 439
6.13 Proof of Proposition 6.27......Page 440
6.14 Auxiliary Results for Theorems 6.28 and 6.29......Page 448
6.15 Proofs of Theorems 6.28 and 6.29......Page 452
6.16 Auxiliary Results for Theorem 6.33......Page 455
6.17 Proof of Theorem 6.33......Page 462
6.18 Proof of Theorem 6.35......Page 477
6.19 Examples......Page 479
6.20 Exercises for Chapter 6......Page 486
7.1 Preliminaries......Page 487
7.2 Boundedness Results......Page 490
7.3 Turnpike Results......Page 491
7.4 Perturbed Problems......Page 495
7.5 Auxiliary Results......Page 496
7.6 Proofs of Theorems 7.2–7.5......Page 505
7.7 Auxiliary Results......Page 510
7.8 STP Implies (P1), (P2), and (P3)......Page 513
7.9 An Auxiliary Result......Page 521
7.10 The Main Lemma......Page 524
7.11 Completion of the Proof of Theorem 7.8......Page 544
7.12 An Auxiliary Result for Theorem 7.9......Page 545
7.13 Proofs of Propositions 7.11 and 7.13......Page 547
7.14 Proofs of Theorems 7.9 and 7.14......Page 548
7.15 Auxiliary Results for Theorem 7.15......Page 556
7.16 Proof of Theorems 7.15......Page 560
7.17 Proof of Theorem 7.16......Page 563
7.18 An Example......Page 569
References......Page 571
Index......Page 577
Springer Optimization and Its Applications 148
Alexander J. Zaslavski
Turnpike Conditions in Infinite Dimensional Optimal Control
Springer Optimization and Its Applications Volume 148 Managing Editor Panos M. Pardalos
, University of Florida
EditorCombinatorial Optimization DingZhu Du, University of Texas at Dallas Advisory Board J. Birge, University of Chicago S. Butenko, Texas A&M University F. Giannessi, University of Pisa S. Rebennack, Karlsruhe Institute of Technology T. Terlaky, Lehigh University Y. Ye, Stanford University
Aims and Scope Optimization has been expanding in all directions at an astonishing rate during the last few decades. New algorithmic and theoretical techniques have been developed, the diffusion into other disciplines has proceeded at a rapid pace, and our knowledge of all aspects of the field has grown even more profound. At the same time, one of the most striking trends in optimization is the constantly increasing emphasis on the interdisciplinary nature of the field. Optimization has been a basic tool in all areas of applied mathematics, engineering, medicine, economics and other sciences. The series Springer Optimization and Its Applications publishes undergraduate and graduate textbooks, monographs and stateoftheart expository works that focus on algorithms for solving optimization problems and also study applications involving such problems. Some of the topics covered include nonlinear optimization (convex and nonconvex), network flow problems, stochastic optimization, optimal control, discrete optimization, multiobjective programming, description of software packages, approximation techniques and heuristic approaches.
More information about this series at http://www.springer.com/series/7393
Alexander J. Zaslavski
Turnpike Conditions in Infinite Dimensional Optimal Control
123
Alexander J. Zaslavski Department of Mathematics The Technion – Israel Institute of Technology Haifa, Israel
ISSN 19316828 ISSN 19316836 (electronic) Springer Optimization and Its Applications ISBN 9783030201777 ISBN 9783030201784 (eBook) https://doi.org/10.1007/9783030201784 Mathematics Subject Classification: 49J20, 49J99, 49K20, 49K27 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
Preface
The book is devoted to the study of the turnpike phenomenon arising in optimal control theory. The term was first coined by P. Samuelson in 1948 when he showed that an efficient expanding economy would spend most of the time in the vicinity of a balanced equilibrium path (also called a von Neumann path). To have the turnpike property means, roughly speaking, that the approximate solutions of the problems are determined mainly by the objective function and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints. The turnpike property discovered by P. Samuelson is well known in the economic literature, where it was studied for various models of economic growth. Usually for these models a turnpike is a singleton. Now it is well known that the turnpike property is a general phenomenon which holds for large classes of finitedimensional variational and optimal control problems. In our research, using the Baire category (generic) approach, it was shown that the turnpike property holds for a generic (typical) variational problem [104] and for a generic optimal control problem [117]. According to the generic approach, we say that a property holds for a generic (typical) element of a complete metric space (or the property holds generically) if the set of all elements of the metric space possessing this property contains a Gδ everywhere dense subset of the metric space which is a countable intersection of open everywhere dense sets. This means that the property holds for most elements of the metric space. In this book we are interested in individual (nongeneric) turnpike results and in sufficient and necessary conditions for the turnpike phenomenon which are of great interest because of their numerous applications in engineering and the economic theory. Sufficient and necessary conditions for the turnpike phenomenon were obtained in our previous research [102, 103, 105, 118] for finitedimensional variational problems and for discretetime optimal control problems in compact metric space. In the present book we study sufficient and necessary conditions for the turnpike phenomenon, using the approach developed in [102, 103, 105, 118], for discretetime optimal control problems in metric spaces, which are not necessarily compact (see Chapters 2–4), and for continuoustime infinite dimensional optimal control problems (see Chapters 5–7). All the results obtained in the book are new. v
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The monograph contains seven chapters. Chapter 1 is an introduction. In Chapter 2 we study discretetime autonomous problems. Chapter 3 is devoted to the study of discretetime nonautonomous problems on subintervals of halfaxis. Discretetime nonautonomous problems on subintervals of axis are analyzed in Chapter 4. Continuoustime autonomous problems are studied in Chapter 5. In Chapter 6 we consider continuoustime nonautonomous problems on subintervals of halfaxis. Chapter 7 is devoted to the study of continuoustime nonautonomous problems on subintervals of axis. Rishon LeZion, Israel October 19, 2018
Alexander J. Zaslavski
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 Banach Space Valued Functions . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Unbounded Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 C0 Semigroup .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.4 Evolution Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.5 C0 Groups .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.6 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.7 Admissible Control Operators . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.8 Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.9 Turnpike Property for Variational Problems.. . . . .. . . . . . . . . . . . . . . . . . . .
1 1 3 4 6 7 8 9 15 20
2 DiscreteTime Autonomous Problems . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1 Preliminaries and Main Results . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2 Lower Semicontinuity Property .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.4 A Subclass of Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.5 Structure of Solutions in the Regions Close to the Endpoints .. . . . . . 2.6 Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.7 Auxiliary Results for Theorem 2.1.. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.8 Proofs of Theorems 2.1–2.4 .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.9 Proof of Theorem 2.9 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.10 An Auxiliary Result for Theorem 2.10 . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.11 Proof of Theorem 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.12 Proof of Proposition 2.12.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.13 Auxiliary Results for Theorems 2.13 and 2.14 .. .. . . . . . . . . . . . . . . . . . . . 2.14 Proofs of Theorems 2.13 and 2.14 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.15 Proofs of Theorems 2.15–2.17 . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.16 Proof of Theorem 2.21 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31 . . . . . . . . . . . . . . . . . . . . 2.18 Proof of Proposition 2.37.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.19 Auxiliary Results for Theorems 2.38 and 2.39 .. .. . . . . . . . . . . . . . . . . . . .
25 25 31 35 36 42 45 49 52 55 63 65 70 71 77 80 85 86 94 96 vii
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2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29
Proofs of Theorems 2.38 and 2.39 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proofs of Theorem 2.40 and 2.41 . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The First Bolza Problem . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 2.63 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 2.66 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Proposition 2.69.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Second Bolza Problem . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 2.70 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 2.71 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Proposition 2.73.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
97 102 104 108 110 115 117 119 120 127
3 DiscreteTime Nonautonomous Problems on the HalfAxis .. . . . . . . . . . . . 3.1 Preliminaries and Main Results . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2 The Boundedness Results . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3 Turnpike Properties .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4 Lower Semicontinuity Property and Infinite Horizon Problems . . . . 3.5 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.6 Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.7 Auxiliary Results for Theorems 3.1 and 3.2. . . . . .. . . . . . . . . . . . . . . . . . . . 3.8 Proofs of Theorems 3.1–3.5 .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.9 An Auxiliary Result for Theorem 3.17 . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.10 Proof of Theorem 3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.11 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.12 Proof of Theorem 3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.13 Auxiliary Results for Theorems 3.19 . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.14 Proofs of Theorems 3.19 and 3.20 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.15 An Auxiliary Result for Theorem 3.24 . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.16 Proof of Theorem 3.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.17 Proof of Theorem 3.26 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
131 131 134 136 142 145 148 153 156 158 159 165 168 172 179 183 184 191
4 DiscreteTime Nonautonomous Problems on Axis . . .. . . . . . . . . . . . . . . . . . . . 4.1 Preliminaries and Main Results . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2 Boundedness Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 Turnpike Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.4 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.5 Auxiliary Results for Theorems 4.1 and 4.2. . . . . .. . . . . . . . . . . . . . . . . . . . 4.6 Proof of Theorems 4.1 and 4.2 .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.7 Proofs of Propositions 4.10, 4.12 and Theorems 4.3–4.5 .. . . . . . . . . . . 4.8 Auxiliary Results for Theorem 4.7.. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.9 STP Implies (P1)–(P3) . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.10 (P1)–(P3) Imply STP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.11 Proof of Theorem 4.8 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.12 Proof of Theorem 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.13 Proof of Theorem 4.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
197 197 200 202 207 208 211 213 215 219 229 253 255 262
Contents
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5 ContinuousTime Autonomous Problems . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Boundedness Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Turnpike Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4 Lower Semicontinuity Property .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.5 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.6 The Triplet (f, −A, −G). . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.7 Auxiliary Results for Theorem 5.1.. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.8 Proofs of Theorems 5.1–5.4 .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.9 Proofs of Theorems 5.6–5.9 .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.10 Proofs of Theorems 5.11 and 5.12 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.11 Proof of Theorem 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.12 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.13 Proof of Theorem 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.14 Proof of Proposition 5.16.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.15 Proof of Proposition 5.18.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.16 Auxiliary Results for Theorems 5.19 and 5.20 .. .. . . . . . . . . . . . . . . . . . . . 5.17 Proofs of Theorems 5.19 and 5.20 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.18 Proofs of Theorems 5.21–5.23 . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.19 Proof of Proposition 5.27.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.20 Proof of Theorem 5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.21 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.22 Structure of Solutions in the Regions Close to the Endpoints .. . . . . . 5.23 Auxiliary Results for Theorems 5.58–5.61 .. . . . . .. . . . . . . . . . . . . . . . . . . . 5.24 Proofs of Theorems 5.58 and 5.59 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.25 Proofs of Theorems 5.60 and 5.61 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.26 The First Bolza Problem . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.27 Proof of Theorem 5.68 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.28 Proof of Theorem 5.71 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.29 Proof of Proposition 5.74.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.30 The Second Bolza Problem . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.31 Proof of Theorem 5.75 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.32 Proof of Theorem 5.76 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.33 Proof of Proposition 5.78.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.34 Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.35 Exercises for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
269 269 275 277 280 282 283 288 294 299 303 304 309 311 315 316 318 322 325 329 336 337 348 350 351 355 357 361 362 367 369 371 372 378 381 384
6 ContinuousTime Nonautonomous Problems on the HalfAxis.. . . . . . . . 6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.2 Boundedness Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.3 Turnpike Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.4 Lower Semicontinuity Property .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.5 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.6 Auxiliary Results for Theorems 6.2 and 6.3. . . . . .. . . . . . . . . . . . . . . . . . . . 6.7 Proofs of Theorems 6.2 and 6.3 .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
385 385 391 393 400 402 403 410
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6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20
Proofs of Theorems 6.4–6.10 . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 6.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . An Auxiliary Result for Theorem 6.23 . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 6.23 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Proposition 6.27.. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Auxiliary Results for Theorems 6.28 and 6.29 .. .. . . . . . . . . . . . . . . . . . . . Proofs of Theorems 6.28 and 6.29 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Auxiliary Results for Theorem 6.33 . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 6.33 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Proof of Theorem 6.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
413 419 424 426 431 432 440 444 447 454 469 471 478
7 ContinuousTime Nonautonomous Problems on Axis . . . . . . . . . . . . . . . . . . . 7.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2 Boundedness Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.3 Turnpike Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.4 Perturbed Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.5 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.6 Proofs of Theorems 7.2–7.5 .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.7 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.8 STP Implies (P1), (P2), and (P3).. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.9 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.10 The Main Lemma .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.11 Completion of the Proof of Theorem 7.8 .. . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.12 An Auxiliary Result for Theorem 7.9.. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.13 Proofs of Propositions 7.11 and 7.13 . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.14 Proofs of Theorems 7.9 and 7.14 . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.15 Auxiliary Results for Theorem 7.15 . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.16 Proof of Theorems 7.15 . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.17 Proof of Theorem 7.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.18 An Example .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .
479 479 482 483 487 488 497 502 505 513 516 536 537 539 540 548 552 555 561
References .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 563 Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 569
Chapter 1
Introduction
The study of infinitedimensional optimal control has been a rapidly growing area of research [1–4, 13–18, 34, 46, 52, 56, 58, 59, 64, 76, 77, 81, 88–91, 100]. In this chapter we present preliminaries which we need in order to study turnpike properties of infinitedimensional optimal control problems. We discuss Banach space valued functions, unbounded operators, C0 semigroups, evolution equations, admissible control operators, and a turnpike property for variational problems.
1.1 Banach Space Valued Functions Let (X, · ) be a Banach space and a < b be real numbers. For any set E ⊂ R 1 define χE (t) = 1 for all t ∈ E and χE (t) = 0 for all t ∈ R 1 \ E. If a set E ⊂ R 1 is Lebesgue measurable, then its Lebesgue measure is denoted by E or by mes(E). A function f : [a, b] → X is called a simple function if there exists a finite collection of Lebesgue measurable sets Ei ⊂ [a, b], i ∈ I , mutually disjoint, and xi ∈ X, i ∈ I such that f (t) =
χEi (t)xi , t ∈ [a, b].
i∈I
A function f : [a, b] → X is strongly measurable if there exists a sequence of simple functions φk : [a, b] → X, k = 1, 2, . . . such that
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_1
1
2
1 Introduction
lim φk (t) − f (t) = 0, t ∈ [a, b] almost everywhere (a. e.).
k→∞
For every simple function f (·) = Bochner integral by
b
i∈I
(1.1)
χEi (·)xi , where the set I is finite, define its
f (t)dt =
a
Ei xi .
i∈I
Let f : [a, b] → X be a strongly measurable function. We say that f is Bochner integrable if there exists a sequence of simple functions φk : [a, b] → X, k = b 1, 2, . . . such that (1.1) holds and the sequence { a φk (t)dt}∞ k=1 strongly converges in X. In this case we define the Bochner integral of the function f by
b
f (t)dt = lim
b
k→∞ a
a
φk (t)dt.
It is known that the integral defined above is independent of the choice of the sequence {φk }∞ k=1 [64]. Similar to the Lebesgue integral, for any measurable set E ⊂ [a, b], the Bochner integral of f over E is defined by
b
f (t)dt =
χE (t)f (t)dt. a
E
The following result is true (see Proposition 3.4, Chapter 2 of [64]). Proposition 1.1. Let f : [a, b] → X be a strongly measurable function. Then f is Bochner integrable if and only if the function f (·) is Lebesgue integrable. Moreover, in this case
b
b
f (t)dt ≤
a
f (t)dt.
a
The Bochner integral possesses almost the same properties as the Lebesgue integral. If f : [a, b] → X is strongly measurable and f (·) ∈ Lp (a, b), for some p ∈ [1, ∞), then we say that f (·) is Lp Bochner integrable. For every p ≥ 1, the set of all Lp Bochner integrable functions is denoted by Lp (a, b; X) and for every f ∈ Lp (a, b; X),
b
f Lp (a,b;X) = (
f (t)p dt)1/p .
a
Clearly, the set of all Bochner integrable functions on [a, b] is L1 (a, b; X). Let a < b be real numbers. A function x : [a, b] → X is absolutely continuous (a. c.) on [a, b] if for each > 0 there exists δ > 0 such that for each pair of q q sequences {tn }n=1 , {sn }n=1 ⊂ [a, b] satisfying
1.2 Unbounded Operators
3
tn < sn , n = 1, . . . , q,
q
(sn − tn ) ≤ δ,
n=1
(tn , sn ) ∩ (tm , sm ) = ∅ for all m, n ∈ {1, . . . , q} such that m = n we have q
x(tn ) − x(sn ) ≤ .
n=1
The following result is true (see Theorem 1.124 of [18]). Proposition 1.2. Let X be a reflexive Banach space. Then every a. c. function x : [a, b] → X is a. e. differentiable on [a, b] and x(t) = x(a) +
t
(dx/dt)(s)ds, t ∈ [a, b]
a
where dx/dt ∈ L1 (a, b; X) is the strong derivative of x.
1.2 Unbounded Operators Let (X, · ) and (Y, · ) be Banach spaces. Denote by L(X, Y ) the set of all linear continuous operators from X to Y . For every A ∈ L(X, Y ), define A = sup{Ax : x ∈ X, x ≤ 1}. Set L(X) = L(X, X), X∗ = L(X, R 1 ), and Y ∗ = L(Y, R 1 ). For all x ∈ X, set I x = x. In the space X, we always consider the norm convergence. Let D(A) be a linear subspace of X (not necessarily closed), and let A : D(A) → Y be a linear operator. We say that A is densely defined if D(A) is dense in X. We say that A is closed if the graph G(A) = {(x, y) ∈ X × Y : x ∈ D(A), y = Ax} of A is closed in X × Y . We say that A is closable if there exists a closed operator ¯ → Y such that A¯ : D(A) ¯ ⊂ X, Ax ¯ = Ax, x ∈ D(A). D(A) ⊂ D(A) Let the symbol ·, · be referred to as the duality pairing between X∗ and X.
4
1 Introduction
Similar to a bounded operator, for any linear operator (not necessarily bounded) A : D(A) ⊂ X → X we still define the resolvent ρ(A) = {λ ∈ C 1 : (λI − A)−1 ∈ L(X)}, the spectrum σ (A) = C 1 \ ρ(A), the point spectrum (or the set of eigenvalues) σp (A) of A and set Ran(A) = {Ax : x ∈ D(A)}. Let A : D(A) ⊂ X → X be densely defined. Clearly, the map x → Ax, y, x ∈ D(A), y ∈ X∗ is welldefined. Suppose that y ∈ X∗ and  Ax, y ≤ cy x for all x ∈ D(A). Then the functional fy (x) = Ax, y, x ∈ D(A) can be extended linearly and continuously to the whole X which is the closure of D(A). Such an extension (still denoted by itself) fy is in X∗ and unique. Hence we obtain
Ax, y = fy (x) = x, fy for all x ∈ D(A). Define D(A∗ ) = {y ∈ X∗ : ∃cy ≥ 0 such that  Ax, y ≤ cy x ∀x ∈ D(A)}, A∗ y = fy , y ∈ D(A∗ ). Clearly, A∗ : D(A∗ ) ⊂ X∗ → X∗ is a linear operator satisfying
Ax, y = x, A∗ y for all x ∈ D(A) and all y ∈ D(A∗ ). The mapping A∗ is called the adjoint operator of A.
1.3 C0 Semigroup Let (X, · ) be a Banach space and let {T (t) : t ∈ [0, ∞)} ⊂ L(X). We call T (·) a C0 semigroup (or a strongly continuous semigroup of operators) on X if the following properties hold: T (0) = I,
(1.2)
T (t + s) = T (s)T (t) for all s, t ≥ 0,
(1.3)
lim T (s)x − x = 0, x ∈ X.
(1.4)
s→0
1.3 C0 Semigroup
5
In the case when T (t) is defined for all t ∈ R 1 and (1.3) holds for all t, s ∈ R 1 , we call T (·) as C0 group or a strongly continuous group of operators. Equations (1.3) and (1.4) are usually referred to as the semigroup property and the strong continuity, respectively. The following result is true (see Proposition 4.7, Chapter 2 of [64]). Proposition 1.3. Let T (·) be a C0 semigroup on X. Then there exist constants M ≥ 1 and ω ∈ R 1 such that T (t) ≤ Meωt , t ∈ [0, ∞). Let T (·) be a C0 semigroup on X, and let D(A) = {x ∈ X : lim t −1 (T (t) − I )x exists}, t →0
Ax = lim t −1 (T (t) − I )x, x ∈ D(A). t →0
The operator A : D(A) → X is called the generator of the semigroup T (·). We also say that A generates the C0 semigroup T (·). In general, the operator A is not bounded. The following two results hold. Proposition 1.4 (Proposition 4.10, Chapter 2 of [64]). Let T (·) be a C0 semigroup on X. Then the generator A of T (·) is densely defined closed operator and n ∩∞ n=1 D(A ) is dense in X. Furthermore, is S(·) is another C0 semigroup on X with the same generator A as T (·), then S(·) = T (·). Theorem 1.5 ( [50, 93] and Theorem 4.11, Chapter 2 of [64]). Let A : D(A) ⊂ X → X be a linear operator. The following properties are equivalent. (i) A generates a C0 semigroup T (·) on X such that T (t) ≤ Meωt for all t ≥ 0 with some M ≥ 1 and ω ∈ R 1 . (ii) A is a densely defined and closed operator such that for the above M ≥ 1 and ω ∈ R 1 , {λ ∈ C 1 : Re λ > ω} ⊂ ρ(A) and (λI − A)−n ≤ M(Re λ − ω)−n for all integers n ≥ 0 and all λ ∈ C 1 with Re λ > ω. (iii) A is a densely defined and closed operator such that for the above M ≥ 1 and ω ∈ R 1 , there exists a sequence of positive numbers λk → ∞ as k → ∞ such that for all integers k ≥ 1, λk ∈ ρ(A) and (λk I − A)−n ≤ M(λk − ω)−n for all integers n ≥ 0 and all integers k ≥ 1.
6
1 Introduction
Because the generator A determines the C0 semigroup T (·) uniquely and in the case A ∈ L(X), the C0 semigroup has an explicit expression eAt = ∞ −1 n n At the C 0 n=0 (n!) t A (Proposition 4.9, Chapter 2 of [64]), we denote by e semigroup generated by A.
1.4 Evolution Equations Let (X, · ) be a Banach space, A : D(A) ⊂ X → X generate a C0 semigroup eAt on X, T > 0, f : [0, T ] → X be a Bochner integrable function and y0 ∈ X. We consider the following evolution equation y (t) = Ay(t) + f (t), t ∈ [0, T ],
(P0 )
y(0) = y0 . A continuous function y : [0, T ] → X is called a (mild) solution of (P0 ) if
t
y(t) = eAt y0 +
eA(t −s)f (s)ds, t ∈ [0, T ].
0
A continuous function y : [0, T ] → X is called a weak solution of (P0 ) if for any x ∗ ∈ D(A∗ ), y(·), x ∗ is an absolutely continuous function on [0, T ] and that for all t ∈ [0, T ], ∗
∗
t
y(t), x = y0 , x +
[ y(s), A∗ x ∗ + f (s), x ∗ ]ds.
0
We have the following result. Proposition 1.6 ( [15], Proposition 5.2, Chapter 2 of [64]). A continuous function y : [0, T ] → X is a solution of (P0 ) if and only if it is a weak solution of (P0 ). The following useful result is also valid (see Proposition 4.14, Chapter 2 of [64]). Proposition 1.7. For any x ∈ D(A), eAt x ∈ D(A) for all t ≥ 0 and (d/dt)(eAt x) = AeAt x = eAt Ax for all t ≥ 0,
t
e x−e x = At
As
s
eAr Axdr for all t > s ≥ 0.
1.5 C0 Groups
7
1.5 C0 Groups Let (H, ·, ·) be a Hilbert space equipped with an inner product ·, · which induces the norm · . A oneparametric family S(t), t ∈ R 1 of continuous linear operators from H onto H is a strongly continuous group of continuous linear operators on H or C0 group on H if S(0)x = x for all x ∈ H, S(t1 + t2 ) = S(t1 )S(t2 ) for all t1 , t2 ∈ R 1 , lim S(t)x = x, x ∈ H.
t →0
Let S(t), t ∈ R 1 be a C0 group on H . Then the generator of S is the linear operator A : D(A) ⊂ H → H defined by D(A) = {x ∈ H : lim t −1 (S(t) − I )x exists}, t →0
Ax = lim t −1 (S(t) − I )x, x ∈ D(A). t →0
The following result is true. Theorem 1.8 ( [34]). Let S(·) be a C0 group on H . Then the following assertions hold. (i) D(A) is dense in H and A is a closed linear operator. (ii) for every x 0 ∈ D(A), there exists a unique function x ∈ C 1 (R 1 ; H ) ∩ C 0 (R 1 ; D(A)) such that x(0) = x 0 , x(t) ∈ D(A) for all t ∈ R 1 and (dx/dt)(t) = Ax(t) for all t ∈ R 1 ; moreover, this solution satisfies x(t) = S(t)x 0 for all t ∈ R 1 . (iii) S(t)∗ , t ∈ R 1 is a C0 group on H , and its generator is the adjoint operator A∗ of A. It is clear that if S(·) is a C0 group with the generator A : D(A) → H , then S(−t), t ∈ R 1 is also a C0 group with the generator −A : D(A) → H . Let A : D(A) ⊂ H → H . It is known that A and −A are both generators of C0 semigroups if and only if A is a generator of a C0 group [91]. Note that a C0 group is determined by its generator A uniquely and is denoted by eAt , t ∈ R 1 . Clearly, for all t ∈ R 1 , e−At = eA(−t ).
8
1 Introduction
1.6 Notation In this section we collect the notation which will be used in the book. Let (X, ρX ) be a metric space equipped with the metric ρX which induces the topology in X. For each x ∈ X and each r > 0 set BX (x, r) = {y ∈ X : ρ(x, y) ≤ r}. Let (Xi , ρXi ), i = 1, 2 be metric spaces. The set X1 × X2 is equipped with the metric ρX1 ×X2 ((x1 , x2 ), (y1 , y2 )) = ρX1 (x1 , y1 ) + ρX2 (x2 , y2 ), (x1 , x2 ), (y1 , y2 ) ∈ X1 × X2 . We denote by mes(Ω) the Lebesgue measure of a Lebesgue measurable set Ω ⊂ R 1 and define χΩ (x) = 1 for all x ∈ Ω, χΩ (x) = 0 for all x ∈ R 1 \ Ω. For each function f : X → [−∞, ∞], where X is nonempty, we set inf(f ) = inf{f (x) : x ∈ X}. For each s ∈ R 1 set s+ = max{s, 0}, s− = min{s, 0}, s = max{z : z is an integer and z ≤ s}. If (X, · ) is a normed space, then X is equipped with the metric ρX (x, y) = x − y, x, y ∈ X. Let (X, ρX ) be a metric space and T be a Lebesgue measurable subset of R 1 . A function f : T → X is called Lebesgue measurable if for any Borel set D ⊂ X the set f −1 (D) is a Lebesgue measurable set. In the sequel we denote by Card(D) the cardinality of a set D, suppose that the sum over an empty set is zero and that the infimum over an empty set is ∞. Let (X, ·, ·X ) be a Hilbert space equipped with an inner product ·, ·X which induces the norm · X . If the space X is understood, we use the notation ·, · =
·, ·X and · = · X . Let X be a Banach space equipped with the norm · X and X∗ is its dual space with the norm · X∗ . If the space X is understood, we use the notation · = · X . For x ∈ X and l ∈ X∗ , we set l(x) = l, xX∗ ,X . The symbol ·, ·X∗ ,X is referred to as the duality pairing between X∗ and X. When the pair X, X∗ is understood, we use the notation ·, · = ·, ·X∗ ,X .
1.7 Admissible Control Operators
9
1.7 Admissible Control Operators Let J be an open interval and U be a Hilbert space. Denote by H 1 (J ; U ) the Sobolev space of all locally absolutely continuous functions z : J → U for which dz/dt ∈ L2 (J ; U ). The space H 2 (J ; U ) is the set of all locally absolutely continuous functions z : J → U for which dz/dt ∈ H 1 (J ; U ). The space H01 (J ; U ) is the set of all functions in H 1 (J ; U ) which have limits equal to zero at the endpoints of J . Define H02 (J ; U ) = {h ∈ H 2 (J ; U ) ∩ H01 (J ; U ) : dh/dx ∈ H01 (J ; U )}. In this section we collect several useful results of [91]. Proposition 1.9 (Proposition 2.1.2 of [91]). Let eAt , t ≥ 0 be a C0 semigroup on a Hilbert space E. Then the mapping (t, z) → eAt z, t ∈ [0, ∞), z ∈ E is continuous on [0, ∞) × E in the product topology. Let X, Z be Hilbert spaces and D(A) be a linear subspace of X. A linear operator A : D(A) → Z is called closed if its graph G(A) = {(f, Af ) : f ∈ D(A)} is closed in X × Z. If A is closed, then D(A) is a Hilbert space with the graph norm · gr : z2gr = z2X + Az2Z , z ∈ D(A). Proposition 1.10 (Proposition 2.10.1 of [91]). Let X be a Hilbert space and A : D(A) → X be a densely defined linear operator such that ρ(A) = ∅. Then for every β ∈ ρ(A) ∩ R 1 , the space D(A) with the norm z1 = (βI − A)(z), z ∈ D(A) is a Hilbert space denoted by X1 . The norms generated as above for different β ∈ ρ(A) ∩ R 1 are equivalent to the graph norm, and the embedding X1 ⊂ X is continuous. If L ∈ L(X) satisfies LD(A) ⊂ D(A), then L ∈ L(X1 ). Note that the relation ρ(A) = ∅ implies that the operator A is closed. Let A be as in Proposition 1.10. It is clear that A∗ has the same property. We define a Hilbert space X1d = D(A∗ ) equipped with the norm zd1 = (βI − A∗ )z, z ∈ D(A∗ ), where β ∈ ρ(A) ∩ R 1 , or equivalently, β ∈ ρ(A∗ ) ∩ R 1 . Proposition 1.11 (Proposition 2.10.2 of [91]). Let A be as in Proposition 1.10 and let β ∈ ρ(A) ∩ R 1 .We denote by X−1 the completion of X with respect to the norm z−1 = (βI − A)−1 z, z ∈ X.
(1.5)
The norms generated as above for different β ∈ ρ(A) ∩ R 1 are equivalent (in particular, X−1 is independent of the choice of β). Moreover, X−1 is the dual of X1d with respect to the pivot space X. If L ∈ L(X) satisfies L∗ D(A∗ ) ⊂ D∗ (A), then L has a unique extension to an operator L˜ ∈ L(X−1 ).
10
1 Introduction
Proposition 1.12 (Proposition 2.10.3 of [91]). Let A : D(A) → X be a densely defined linear operator such that ρ(A) = ∅, β ∈ ρ(A) ∩ R 1 , X1 be as in Proposition 1.10 and X−1 be as in Proposition 1.11. Then A ∈ L(X1 , X) and A has a unique extension A˜ ∈ L(X, X−1 ). Moreover, the operators (βI − A)−1 ∈ ˜ −1 ∈ L(X−1 , X) are unitary. L(X, X1 ) and (βI − A) Proposition 1.13 (Proposition 2.10.4 of [91]). We use the notation from Proposition 1.12 and assume that A generates a C0 semigroup S(·) on X. Then for every ˜ ˜ t ≥ 0, S(t) has a unique extension S(t) ∈ L(X−1 ), S(t), t ≥ 0 is a C0 semigroup on X−1 , and A˜ is its generator. Let U, X be Hilbert spaces, S(·) be a C0 semigroup on X and A : D(A) → X be its generator. We use the notation above and denote by A and S(·) also the extension of the generator to X and the extension of the semigroup to X−1 , respectively. Consider the differential equation z (t) = Az(t) + f (t)
(1.6)
where f ∈ L1loc ([0, ∞); X−1 ). A solution of (1.6) in X−1 is a function z ∈ L1loc ([0, ∞); X) ∩ C([0, ∞); X−1 ) which satisfies the following equations in X−1 :
t
z(t) − z(0) =
[Az(s) + f (s)]ds for all t ∈ [0, ∞).
(1.7)
0
It is also called a “strong” solution of (1.6) in X−1 . Equation (1.7) implies that z is an a. c. function with values in X−1 and (1.6) holds for almost every t ≥ 0, with the derivative computed with respect to the norm of X−1 . We can also define the concept of a “weak” solution of (1.6) in X−1 by regarding instead of (1.7) that for every φ ∈ X1d and every t ≥ 0,
z(t) − z(0), φX−1 ,Xd = 1
t 0
[ z(s), A∗ φX + f (s), φX−1 ,Xd ]ds. 1
These two concepts are equivalent. Proposition 1.14 (Proposition 4.1.4 of [91]). Suppose that z is a solution of (1.6) in X−1 and let z0 = z(0). Then z is given by
t
z(t) = S(t)z0 +
S(t − σ )f (σ )dσ, t ≥ 0.
(1.8)
0
In particular, for every z0 ∈ X, there exists at most one solution in X−1 of (1.6) which satisfies initial condition z(0) = z0 . Note that z satisfying (1.8) is called the mild solution of (1.6) with the initial state z0 ∈ X. It is not difficult to see that the following result is valid.
1.7 Admissible Control Operators
11
Proposition 1.15. Let f ∈ L1loc ([0, ∞); X−1 ), T > 0 and z ∈ C 0 ([0, T ]; X−1 ). The z satisfies (1.8) for all t ∈ [0, T ] in X−1 if and only if for all ξ ∈ X1d = D(A∗ ) and all t ∈ [0, T ],
z(t), ξ X−1 ,Xd = S(t)z0 , ξ X−1 ,Xd + 1
1
= S(t)z0 , ξ X−1 ,Xd + 1
t
0
t 0
[ S(t − σ )f (σ ), ξ X−1 ,Xd ]dσ 1
[ f (σ ), S ∗ (t − σ )ξ X−1 ,Xd ]dσ. 1
Proposition 1.16 (Proposition 12.1.2 of [91]). Assume that Z1 , Z2 and Z3 are Hilbert spaces, F ∈ L(Z1 , Z3 ), G ∈ L(Z2 , Z3 ) and Ran(F ) = {F (z) : z ∈ Z1 } ⊂ Ran(G) = {G(z) : z ∈ Z2 }. Then there exists an operator L ∈ L(Z1 , Z2 ) such that F = GL. Corollary 1.17. Assume that Z1 , Z2 are Hilbert spaces and G ∈ L(Z1 , Z2 ) satisfies Ran(G) = Z2 . Then there exists an operator L ∈ L(Z2 , Z1 ) such that GL = I  the identity operator in Z2 . Let B ∈ L(U, X−1 ) and τ ≥ 0. Define Φτ ∈ L(L2 (0, ∞; U ), X−1 ) by
τ
Φτ u =
S(τ − σ )Bu(σ )dσ.
(1.9)
0
The operator B ∈ L(U, X−1 ) is called an admissible control operator for S(·) if for some τ > 0, Ran(Φτ ) = {Φτ u : u ∈ L2 (0, ∞; U )} ⊂ X. Proposition 1.18 (Proposition 4.2.2 and (4.2.5) of [91]). Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every t ≥ 0, Φt ∈ L(L2 (0, ∞; U ), X) and for all T > t > 0, Φt ≤ ΦT . Proposition 1.19 (Proposition 4.2.5 of [91]). Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every z0 ∈ X and every u ∈ L2loc (0, ∞; U ), the initial value problem z (t) = Az(t) + Bu(t), z(0) = z0
12
1 Introduction
has a unique solution in X−1 . This solution is given by z(t) = S(t)z0 + Φt u = S(t)z0 +
t
S(t − σ )Bu(σ )dσ
0
and it satisfies 1 ((0, ∞); X−1 ). z ∈ C([0, ∞); X) ∩ Hloc
Proposition 1.20 (Proposition 4.2.6 of [91]). Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every z0 ∈ X and every u ∈ L2loc (0, ∞; U ), there exists a unique function z ∈ C([0, ∞); X) such that for every t ≥ 0 and every ψ ∈ D(A∗ ),
t
z(t) − z0 , ψX =
[ z(σ ), A∗ ψX + u(σ ), B ∗ ψU ]dσ.
0
Propositions 1.18 and 1.19 and Theorem 1.5 imply the following result. Proposition 1.21. Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·) and T > 0. Then there exists a constant cT > 0 such that for every z0 ∈ X and every u ∈ L2 (0, T ; U ) the unique solution z of the initial value problem z (t) = Az(t) + Bu(t), t ∈ (0, T ) a. e., z(0) = z0 satisfies z ∈ C([0, T ]; X) and that for all t ∈ [0, T ], z(t) ≤ cT (z0 + u). Proposition 1.18 and Corollary 1.17 imply the following result. Proposition 1.22. Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·), T > 0 and Ran(ΦT ) = X. Then there exists L ∈ L(X, L2 (0, T ; U )) such that ΦT Lx = x for all x ∈ X. Theorem 1.23. Assume that B ∈ L(U, X−1 ) is an admissible control operator for a C0 semigroup S(·), T > 0, Ran(ΦT ) = X, xf ∈ X, uf ∈ U , x ∗ (t) = xf , u∗ (t) = uf , t ∈ [0, ∞), and x ∗ (·) is a unique solution in X−1 of the initial value problem z (t) = Az(t) + Bu∗ (t), z(0) = xf .
(1.10)
Then there exists a constant c > 0 such that for each z0 , z1 ∈ X, there exist u ∈ L2 (0, T ; U ) and z ∈ C 0 ([0, T ]; X) which is a solution of the problem z (t) = Az(t) + Bu(t), t ∈ [0, T ] a. e., z(0) = z0
1.7 Admissible Control Operators
13
in X−1 and satisfy z(T ) = z1 , u(t) − uf ≤ c(z1 − xf + z0 − xf ), t ∈ [0, T ], z(t) − xf ≤ c(z1 − xf + z0 − xf ), t ∈ [0, T ]. Proof. Let L ∈ L(X, L2 (0, T ; U )) be as guaranteed by Proposition 1.22. Therefore ΦT Lx = x for all x ∈ X.
(1.11)
Let cT > 0 be as guaranteed by Proposition 1.21. Set c = cT + cT L.
(1.12)
Let z0 , z1 ∈ X. By Proposition 1.19, there exists z˜ ∈ C 0 ([0, T ]; X) which is a solution of the initial value problem z˜ = A˜z + Buf , t ∈ (0, T ),
(1.13)
z˜ (0) = z0
(1.14)
in X−1 . Proposition 1.19 implies that there exists a unique function z ∈ C 0 ([0, T ]; X) such that z + B(L(z1 − z˜ (T )), t ∈ (0, T ), z = A
(1.15)
z(0) = 0
(1.16)
in X−1 . In view of (1.11) and Proposition 1.19, z(T ) = S(T ) z(0)+ΦT (L(z1 −˜z(T ))) = ΦT (L(z1 −˜z(T ))) = z1 −˜z(T ).
(1.17)
Set z = z˜ + z.
(1.18)
z(0) = z˜ (0) + z(0) = z0 ,
(1.19)
z(T ) = z˜ (T ) + z(T ) = z1 .
(1.20)
By (1.14) and (1.16)–(1.18),
14
1 Introduction
It follows from (1.13), (1.15), (1.17), and (1.18) that z = z˜ + z = A(˜z + z) + B(L(z1 − z˜ (T )) + uf ) = Az + B(L(z1 − z˜ (T )) + uf ), t ∈ (0, T )
(1.21)
in X−1 . In view of (1.10), (1.13) and (1.14), (˜z − x ∗ ) = A(˜z − x ∗ ), t ∈ (0, T )
(1.22)
(˜z − x ∗ )(0) = z0 − xf .
(1.23)
in X−1 and
It follows from (1.14), (1.22), (1.23), the choice of cT and Proposition 1.21 that for each τ ∈ [0, T ], ˜z(τ ) − xf ≤ cT ˜z(0) − xf = cT z0 − xf .
(1.24)
In view of (1.10), (1.19), and (1.21), (z − x ∗ ) = A(z − x ∗ ) + B((L(z1 − z˜ (T )))), t ∈ (0, T )
(1.25)
in X−1 , (z − x ∗ )(0) = z0 − xf .
(1.26)
By (1.24), L(z1 − z˜ (T )) ≤ Lz1 − z˜ (T ) ≤ L(z1 − x f + xf − z˜ (T )) ≤ L(z1 − xf + cT z0 − xf ).
(1.27)
Proposition 1.21, the choice of cT , (1.12) and (1.24)–(1.27) imply that for all t ∈ [0, T ], z(t) − xf ≤ cT (z0 − xf + L(z1 − z˜ (T ))) ≤ cT (z0 − xf + L(z1 − xf + cT z0 − xf )) = z0 − xf (cT + cT L) + cT Lz1 − xf ≤ c(z0 − xf + z1 − xf ). Theorem 1.23 is proved.
1.8 Examples
15
1.8 Examples Let U, X be Hilbert spaces, S(·) be a C0 semigroup on X, A : D(A) → X be its generator and B ∈ L(U, X−1 ) be an admissible control operator for S(·). We say that the pair (A, B) is exactly controllable in a time τ > 0 if Ran(Φτ ) = X [34, 91]. It is known that the exactly controllability in time τ > 0 is equivalent to the following property: for each pair z0 , z1 ∈ X, there exists u ∈ L2 (0, τ ; U ) such that the solution z of the initial value problem z = Az + Bu, z(0) = z0
(1.28)
satisfies z(τ ) = z1 . In this section we consider examples of the pairs (A, B) which are exactly controllable in some time τ > 0. All of these examples were discussed in [91] and [34]. It should be mentioned that in Chapters 5 and 6 our turnpike results will be established for large and general classes of infinitedimensional optimal control problems. One of this classes contains problems which are defined by an integrand and a pair of operators (A, B) introduced above, which is exactly controllable in some time τ > 0. Therefore, each example of pairs (A, B) considered below gives us a subclass of infinitedimensional optimal control problems, for which the results of Chapters 5 and 6 hold. Example 1.24 (Example 11.2.2 of [91]). We consider the problem of controlling the vibrations of an elastic membrane by a force field acting on a part of this membrane. More precisely, let n be a natural number and let Ω ⊂ R n be a bounded open set with ∂Ω of class C 2 or Ω be a rectangular domain. The physical problem described above can be modeled by the equations ∂ 2w − Δw = u in Ω × (0, ∞), ∂t 2 w = 0 on ∂Ω × (0, ∞), w(x, 0) = f (x),
∂w (x, 0) = g(x) for x ∈ Ω, ∂t
where f is the initial displacement and g is the initial velocity. Let O be a nonempty open subset of Ω and u ∈ L2 (0, ∞; L2 (O)) be the input function. For any such u, we assume that u(x, t) = 0 for all x ∈ Ω \ O. The equation above can be written in the form (1.28) using the following spaces and operators: X = H01 (Ω) × L2 (Ω), D(A) = (H 2 (Ω) ∩ H01 (Ω)) × H01 (Ω), U = L2 (O) ⊂ L2 (Ω), A
f g f 0 = for all ∈ D(A), Bu = for all u ∈ U. g Δf g u
16
1 Introduction
Here B ∈ L(U, X). It was shown in Example 11.2.2 of [91] that B is an admissible control operator for the C0 semigroup eAt and that the pair (A, B) is exactly controllable in time τ > 0 if Γ and O satisfy the assumptions of Theorem 7.4.1 of [91] where Γ is a relatively open subset of ∂Ω. In particular, (A, B) is exactly controllable in time τ > 0 if there exist x0 ∈ R n and > 0 such that N ({x ∈ ∂Ω : (x − x0 ) · ν(x) > 0}) ⊂ closO, τ > 2r(x0 ), where r(x0 ) = sup{x − x0 : x ∈ Ω},  ·  is the Euclidean norm in R n , ν is the unit outward normal vector field in ∂Ω and N (D) = {x ∈ Ω : d(x, D) < } for any D ⊂ Ω with d(x, D) = inf{x − y : y ∈ D}. Example 1.25 (Example 11.2.4 of [91]). Let n be a natural number, let Ω ⊂ R n be a bounded open set with ∂Ω of class C 2 or let Ω be a rectangular domain, and let O be a nonempty open subset of Ω. We consider the problem of controlling the vibrations of an elastic plate occupying the domain Ω by a force field acting on O. More precisely, we consider the following initial and boundary initial value problem ∂ 2w + Δ2 w = u in Ω × (0, ∞), ∂t 2 w = Δw = 0 on ∂Ω × (0, ∞), w(x, 0) = 0,
∂w (x, 0) = 0 for x ∈ Ω, ∂t
where u ∈ L2 (0, ∞; L2 (O)) is the input function. As usual we assume that u(x, t) = 0 for all x ∈ Ω \ O. The equations above determines a system with the state space X = (H 2 (Ω)∩H01 (Ω))×L2 (Ω) and the input space U = L2 (Ω) which is exactly controllable in any time τ > 0, as it was shown in Example 11.2.4 of [91], if the pair (Ω, O) satisfies one of the assumptions (A1) or (A2) in Example 11.2.3 of [91]. More precisely, let H = L2 (Ω) and D(A0 ) = H1 be the Sobolev space H 2 (Ω) ∩ H01 (Ω), A0 : D(A0 ) → H be defined by A0 φ = −Δφ, φ ∈ D(A0 ), and let H2 = D(A20 ) be endowed with the graph norm. Let χ be the Hilbert space H1 × H , consider the dense subset of χ defined by D(A) = H2 × H1 and let the linear operator A : D(A) → χ be defined by
0 I A= . −A20 0 Then the equations above can be written in the form z = Az + Bu, z(0) = 0
1.8 Examples
17
0 where B ∈ L(U, χ) is defined by Bu = for all u ∈ U . It was shown in u Example 11.2.4 of [91] that the pair (A, B) is exactly controllable in any time τ > 0 if the pair (Ω, O) satisfies one of the assumptions (A1) or (A2) in Example 11.2.3 of [91]. Example 1.26 (Example 11.2.6 of [91]). We consider the problem of controlling the vibrations of a string occupying the interval [0, π] by means of a force u(t) acting at its left end. The equations describing this problem are formulated as a wellposed boundary control system in subsection 10.2.2 of [91]: ∂ 2w ∂ 2w (x, t) = (x, t), 0 < x < π, t ≥ 0, ∂t 2 ∂x 2 ∂w (0, t) = u(t), t ≥ 0, ∂x
w(π, t) = 0, w(x, 0) = f (x),
∂w (x, 0) = g(x), 0 < x < π. ∂t
For this problem X = HR1 (0, π) × L2 (0, π), A : D(A) → X is defined by D(A) = 1 {f ∈ H 2 (0, π) ∩ HR1 (0, π) : ∂f ∂x (0) = 0} × HR (0, π),
g f f A ∈ D(A). = d 2 f for all g g 2 dx
f f = −g(0) for all ∈ D(A). It was g g shown in Example 10.2.6 of [91] that the pair (A, B) is exactly controllable in any time τ ≥ 2π.
The control operator B satisfies B ∗
Example 1.27 (Example 11.2.7 of [91]). We consider the boundary control of the nonhomogeneous elastic string. The model is described by the equation ∂w ∂ ∂ 2w (a(x) (x, t)) − b(x)w(x, t), 0 < x < π, t > 0, (x, t) = 2 ∂t ∂x ∂x w(0, t) = u(t), w(π, t) = 0, w(·, 0) = f,
∂w (·, 0) = g. ∂t
Here a ∈ C 2 [0, π], b ∈ L∞ (0, π), a(x) ≥ m > 0, b(x) ≥ 0 for all x ∈ [0, π]. These equations correspond to a wellposed boundary control system with state space X = L2 (0, π) × H −1 (0, π). The generator A is defined by
18
1 Introduction
f f g A for all = ∈ D(A) = H01 (0, π) × L2 (0, π), g g −A0 f where A0 ∈ L(H01 (0, π), H −1 (0, π)) is defined by A0 f =
df d (a ) + bf for all f ∈ H01 (0, π). dx dx
The control operator B of this system is determined by d φ φ −1 B = a(0) (A0 ψ)x=0 for all ∈ D(A∗ ) = D(A). ψ ψ dx ∗
It was shown in Example 11.2.7 of [91] that the pair (A, B) is exactly controllable π in any time τ ≥ 2 0 (a(x))−1/2dx. Example 1.28 (Example 11.2.8 of [91]). We consider the problem of controlling the vibrations of a beam occupying the interval [0, π] by means of a torque u(t) acting at its left end. The model is described by the initial and boundary value problem ∂ 4w ∂ 2w (x, t) = − (x, t), 0 < x < π, t > 0, ∂t 2 ∂x 4 w(0, t) = 0, w(π, t) = 0,
∂ 2w ∂ 2w (0, t) = u(t), (π, t) = 0, ∂x 2 ∂x 2
w(·, 0) = f,
∂w (·, 0) = g. ∂t
These equations correspond to a control system where H = L2 (0, π), H1 = 2 H 2 (0, π) ∩ H01 (0, π), A0 : H1 → H is defined by A0 f = − ddxf2 for all f ∈ H1 , H1/2 = H01 (0, π), H−1/2 = H0−1 (0, π). The unique extensions of A0 to unitary operators from H1/2 onto H−1/2 and from H onto H−1 are still denoted by A0 . The space H3/2 = A−1 0 H1/2 is H3/2 = {g ∈ H 3 (0, π) ∩ H01 (0, π) :
d 2g d 2g (0) = (π) = 0}. dx 2 dx 2
We set X = H1/2 × H−1/2 , D(A) = H3/2 × H1/2 , f g f A = for all ∈ D(A). g −A20 f g
1.8 Examples
19
The control operator B of this system is determined by B∗
d f f g) for all ∈ D(A∗ ) = D(A). = − (A−1 x=0 g g dx 0
It was shown in Example 11.2.8 of [91] that the pair (A, B) is exactly controllable in any time τ > 0. Example 1.29 (Example 11.2.9 of [91]). We consider the problem of controlling the vibrations of a beam occupying the interval [0, 1] by means of an angular velocity u(t) applied at its left end. The equations describing this problem have been formulated as a wellposed boundary control system in Section 10.5 of [91] as follows: ∂ 2w ∂ 4w (x, t) = − (x, t), 0 < x < 1, t > 0, ∂t 2 ∂x 4 w(0, t) = 0, w(1, t) = 0,
∂w ∂w (0, t) = u(t), (1, t) = 0, ∂x ∂x
w(·, 0) = f,
∂f (·, 0) = g. ∂t
We denote X = V ×L2 (0, 1), where V = {h ∈ H 2 (0, 1) : h(0) = h(1) = 0}. The norm on X is defined by z = 2
z1 2V
+ z2 2L2 ,
where
z1 2V
1
= 0

dh dx (1)
=
d 2 z1 2  dx. dx 2
Let Z ⊂ X be defined by Z = (V ∩ H 4 (0, 1)) × V , L : Z → X, G : Z → R 1 be defined by L
0 4
d − dx 4
I dz2 z (0), , G 1 = z2 dx 0
KerG = {z ∈ Z : G(z) = 0} = (V ∩ H 4 (0, 1)) × H02 (0, 1), A = LKerG and a control operator B be defined by d 2 ψ1 ψ1 ψ1 =− ∈ D(A∗ ) = D(A). (0), for all B 2 ψ2 ψ2 dx ∗
It was shown in Example 11.2.9 of [91] that the pair (A, B) is exactly controllable in any time τ > 0. The next two examples are discussed in [34].
20
1 Introduction
Example 1.30. The transport equation. Let L > 0. We consider the linear control system yt + yx = 0, t ∈ (0, T ), x ∈ (0, L), y(t, 0) = u(t), t ∈ (0, T ), where u(t) ∈ R 1 , y(t, ·) : (0, L) → R 1 , with X = L2 (0, L), D(A) = {f ∈ H 1 (0, L) : f (0) = 0}, Af = −fx , f ∈ D(A), U = R 1 and B : R 1 → D(A∗ ) defined by (Bu)z = uz(0) for all u ∈ R 1 and all z ∈ D(A∗ ). It was shown in [34] that the pair (A, B) is controllable. Example 1.31. The Kortewegde Vries equation. Let L > 0. We consider the linear control system yt + yx + yxxx = 0, t ∈ (0, T ), x ∈ (0, L), y(t, 0) = y(t, L) = 0, yx (t, L) = u(t), t ∈ (0, T ), where u(t) ∈ R 1 , y(t, ·) : (0, L) → R 1 , with X = L2 (0, L), D(A) = {f ∈ H 3 (0, L) : f (0) = f (L) = fx (L) = 0}, Af = −fx − fxxx , f ∈ D(A), U = R 1 and B : R 1 → D(A∗ ) is defined by (Bu)z = uzx (L) for all u ∈ R 1 and all z ∈ D(A∗ ). It was shown in [34] that the pair (A, B) is controllable.
1.9 Turnpike Property for Variational Problems The study of the existence and the structure of solutions of optimal control problems and dynamic games defined on infinite intervals and on sufficiently large intervals has been a rapidly growing area of research [7, 8, 21, 29, 36, 38– 40, 51, 54, 55, 67, 78, 104, 106, 108, 111, 116, 120, 125, 126, 129, 131, 134] which has various applications in engineering [5, 29, 61, 104], in models of economic growth [6, 10, 26–29, 37, 42, 53, 60, 66, 72, 79, 84–86, 92, 104, 114, 121, 130, 131], in infinite discrete models of solidstate physics related to dislocations in onedimensional crystals [11, 12, 83, 94], in model predictive control [35, 45], and in the theory of thermodynamical equilibrium for materials [33, 62, 69–71]. Discretetime problems optimal control problems were considered in [9, 13, 14, 24, 41, 48, 95, 96, 101, 103, 107, 110, 112, 114, 118, 119, 122, 127, 128, 130], finitedimensional continuoustime problems were analyzed in [20, 22, 23, 25, 30, 60, 63, 65, 68, 80, 99, 102, 105, 117, 132, 133], infinitedimensional optimal control was studied in [28, 29, 46, 74, 76, 77, 81, 88, 89, 98, 100, 109], while solutions of dynamic games were discussed in [19, 43, 44, 47, 49, 57, 82, 113, 115, 123, 124]. Sufficient and necessary conditions for the turnpike phenomenon were obtained in our previous research [102, 103, 105, 118] for finitedimensional variational problems and for discretetime optimal control problems in compact metric space.
1.9 Turnpike Property for Variational Problems
21
In this section, which is based on [102], we discuss the structure of approximate solutions of variational problems with continuous integrands f : [0, ∞) × R n × R n → R 1 which belong to a complete metric space of functions. We do not impose any convexity assumption. The main result of this section obtained in [102] deals with the turnpike property of variational problems. We consider the variational problems
T2
f (t, z(t), z (t))dt → min, z(T1 ) = x, z(T2 ) = y,
(P )
T1
z : [T1 , T2 ] → R n is an absolutely continuous function, where T1 ≥ 0, T2 > T1 , x, y ∈ R n and f : [0, ∞) × R n × R n → R 1 belongs to a space of integrands described below. It is well known that the solutions of the problems (P) exist for integrands f which satisfy two fundamental hypotheses concerning the behavior of the integrand as a function of the last argument (derivative): one that the integrand should grow superlinearly at infinity and the other that it should be convex [87]. Moreover, certain convexity assumptions are also necessary for properties of lower semicontinuity of integral functionals which are crucial in most of the existence proofs, although there are some interesting theorems without convexity [31, 73, 75]. For integrands f which do not satisfy the convexity assumption, the existence of solutions of the problems (P) is not guaranteed, and in this situation we consider δapproximate solutions. Let T1 ≥ 0, T2 > T1 , x, y ∈ R n , f : [0, ∞) × R n × R n → R 1 be an integrand, and let δ be a positive number. We say that an absolutely continuous (a.c.) function u : [T1 , T2 ] → R n satisfying u(T1 ) = x, u(T2 ) = y is a δapproximate solution of the problem (P) if
T2 T1
f (t, u(t), u (t))dt ≤
T2
f (t, z(t), z (t))dt + δ
T1
for each a.c. function z : [T1 , T2 ] → R n satisfying z(T1 ) = x, z(T2 ) = y. The main result of [102] deals with the turnpike property of the variational problems (P). As usual, to have this property means, roughly speaking, that the approximate solutions of the problems (P) are determined mainly by the integrand and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints. In the classical turnpike theory, it was assumed that a cost function (integrand) is convex. The convexity of the cost function played a crucial role there. In [102] we get rid of convexity of integrands and establish necessary and sufficient conditions for the turnpike property for a space of nonconvex integrands M described below. Let us now define the space of integrands. Denote by  ·  the Euclidean norm in R n . Let a be a positive constant, and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → +∞ as t → ∞. Denote by M the set of all continuous functions f : [0, ∞) × R n × R n → R 1 which satisfy the following assumptions:
22
1 Introduction
A(i) the function f is bounded on [0, ∞) × E for any bounded set E ⊂ R n × R n ; A(ii) f (t, x, u) ≥ max{ψ(x), ψ(u)u} − a for each (t, x, u) ∈ [0, ∞) × R n × Rn ; A(iii) for each M, > 0, there exist Γ, δ > 0 such that f (t, x1 , u) − f (t, x2 , u) ≤ max{f (t, x1 , u), f (t, x2 , u)} for each t ∈ [0, ∞) and each u, x1 , x2 ∈ R n which satisfy xi  ≤ M, i = 1, 2, u ≥ Γ,
x1 − x2  ≤ δ;
A(iv) for each M, > 0 there exists δ > 0 such that f (t, x1 , u1 ) − f (t, x2 , u2 ) ≤ for each t ∈ [0, ∞) and each u1 , u2 , x1 , x2 ∈ R n which satisfy xi , ui  ≤ M, i = 1, 2,
max{x1 − x2 , u1 − u2 } ≤ δ.
It is easy to show that an integrand f = f (t, x, u) ∈ C 1 ([0, ∞) × R n × R n ) belongs to M if f satisfies assumption A(ii), and if sup{f (t, 0, 0) : t ∈ [0, ∞)} < ∞ and also there exists an increasing function ψ0 : [0, ∞) → [0, ∞) such that sup{∂f/∂x(t, x, u), ∂f/∂u(t, x, u)} ≤ ψ0 (x)(1 + ψ(u)u) for each t ∈ [0, ∞) and each x, u ∈ R n . For the set M we consider the uniformity which is determined by the following base: E(N, , λ) = {(f, g) ∈ M × M : f (t, x, u) − g(t, x, u) ≤ for each t ∈ [0, ∞) and each x, u ∈ R n satisfying x, u ≤ N and (f (t, x, u) + 1)(g(t, x, u) + 1)−1 ∈ [λ−1 , λ] for each t ∈ [0, ∞) and each x, u ∈ R n satisfying x ≤ N}, where N > 0, > 0, λ > 1. It is not difficult to show that the space M with this uniformity is metrizable (by a metric ρw ). It is known (see [102]) that the metric space (M, ρw ) is complete. The metric ρw induces in M a topology. We consider functionals of the form I (T1 , T2 , x) = f
T2
f (t, x(t), x (t))dt
T1
where f ∈ M, 0 ≤ T1 < T2 < ∞ and x : [T1 , T2 ] → R n is an a.c. function.
1.9 Turnpike Property for Variational Problems
23
For f ∈ M, y, z ∈ R n and numbers T1 , T2 satisfying 0 ≤ T1 < T2 , we set U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x) : x : [T1 , T2 ] → R n is an a.c. function satisfying x(T1 ) = y, x(T2 ) = z}. It is easy to see that −∞ < U f (T1 , T2 , y, z) < ∞ for each f ∈ M, each y, z ∈ R n and all numbers T1 , T2 satisfying 0 ≤ T1 < T2 . Let f ∈ M. A locally absolutely continuous (a.c.) function x : [0, ∞) → R n is called an (f )good function [104] if for any a.c function y : [0, ∞) → R n there is a number My such that I f (0, T , y) ≥ My + I f (0, T , x) for each T ∈ (0, ∞). The following result was proved in [102]. Proposition 1.32. Let f ∈ M and let x : [0, ∞) → R n be a bounded a.c. function. Then the function x is (f )good if and only if there is M > 0 such that I f (0, T , x) ≤ U f (0, T , x(0), x(T )) + M for any T > 0. Let us now give the precise definition of the turnpike property. Assume that f ∈ M. We say that f has the turnpike property, or briefly TP, if there exists a bounded continuous function Xf : [0, ∞) → R n which satisfies the following condition: For each K, > 0, there exist constants δ, L > 0 such that for each x, y ∈ R n satisfying x, y ≤ K, each T1 ≥ 0, T2 ≥ T1 + 2L, and each a.c. function v : [T1 , T2 ] → R n which satisfies v(T1 ) = x, v(T2 ) = y, I f (T1 , T2 , v) ≤ U f (T1 , T2 , x, y) + δ the inequality v(t) − Xf (t) ≤ holds for all t ∈ [T1 + L, T2 − L]. The function Xf is called the turnpike of f . Assume that f ∈ M and X : [0, ∞) → R n is a bounded continuous function. How to verify if the integrand f has TP and X is its turnpike? In [102] we introduced two properties (P1) and (P2) and show that f has TP if and only if f possesses the properties (P1) and (P2). The property (P2) means that all (f )good functions have the same asymptotic behavior while the property (P1) means that if an a.c. function v : [0, T ] → R n is an approximate solution and T is large enough, then there is τ ∈ [0, T ] such that v(τ ) is close to X(τ ). The next theorem is the main result [102]. Theorem 1.33. Let f ∈ M and Xf : [0, ∞) → R n be a bounded absolutely continuous function. Then f has the turnpike property with Xf being the turnpike if and only if the following two properties hold:
24
1 Introduction
(P1) For each K, > 0, there exist γ , l > 0 such that for each T ≥ 0 and each a.c. function w : [T , T + l] → R n which satisfies w(T ), w(T +l) ≤ K, I f (T , T +l, w) ≤ U f (T , T +l, w(T ), w(T +l))+γ there is τ ∈ [T , T + l] for which Xf (τ ) − v(τ ) ≤ . (P2) For each (f )good function v : [0, ∞) → R n , v(t) − Xf (t) → 0 as t → ∞. In [102] we proved the following theorem which is an extension of Theorem 1.33. Theorem 1.34. Let f ∈ M, Xf : [0, ∞) → R n be an (f )good function. Assume that the properties (P1), (P2) hold. Then for each K, > 0, there exist δ, L > 0 and a neighborhood U of f in M such that for each g ∈ U, each T1 ≥ 0, T2 ≥ T1 + 2L and each a.c. function v : [T1 , T2 ] → R n which satisfies v(T1 ), v(T2 ) ≤ K, I g (T1 , T2 , v) ≤ U g (T1 , T2 , v(T1 ), v(T2 )) + δ the inequality v(t) − Xf (t) ≤ holds for all t ∈ [T1 + L, T2 − L]. In the present book, we study sufficient and necessary conditions for the turnpike phenomenon, using the approach developed in [102, 103, 105, 118]; for discretetime optimal control problems in metric spaces, which are not necessarily compact (see Chapters 2–4); and for continuoustime infinitedimensional optimal control problems (see Chapters 5–7). Our main results have Theorem 1.33 as their prototype. Since the discovery of the turnpike phenomenon by Paul Samuelson in 1948, different versions of the turnpike property were considered in the literature. In this book as well as in [102, 103, 105, 118], we study the turnpike property introduced and used in our previous research [99, 104, 117, 118, 129]. This turnpike property differs from other versions and has important features. Our turnpike property is a property of approximate solutions, and the turnpike is a nonstationary trajectory. As it was shown in [99, 117], our turnpike property holds for most problems belonging to large classes of variational and optimal control problems.
Chapter 2
DiscreteTime Autonomous Problems
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for discretetime optimal control problems in metric spaces, which are not necessarily compact. For these optimal control problems, the turnpike is a singleton. We also study the structure of approximate solutions on large intervals in the regions close to the endpoints and the existence of solutions of the corresponding infinite horizon optimal control problems.
2.1 Preliminaries and Main Results Let (E, ρE ) and (F, ρF ) be metric spaces. We suppose that A is a nonempty subset of E, U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U(x)}, G : M → E and f : M → R 1 . Let 0 ≤ T1 < T2 be integers. We denote by X(T1 , T2 , M, G) the set of all pairs 2 2 −1 , {ut }Tt =T ) such that for each integer t ∈ {T1 , . . . , T2 − 1}, of sequences ({xt }Tt =T 1 1 xt ∈ A, ut ∈ U(xt ), xt +1 = G(xt , ut ) and which are called trajectorycontrol pairs. For simplicity we use the notation X(T1 , T2 ) = X(T1 , T2 , M, G) if the pair (M, G) is understood. Let T1 ≥ 0 be an integer. Denote by X(T1 , ∞, M, G) (or X(T1 , ∞) if the pair ∞ (M, G) is understood) the set of all pairs of sequences {xt }∞ t =T1 ⊂ A, {ut }t =T1 ⊂ F 2 2 −1 such that for each integer T2 > T1 , ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 , M, G). The 1 1 elements of X(T1 , ∞, M, G) are called trajectorycontrol pairs. 2 Let 0 ≤ T1 < T2 be integers. A sequence {xt }Tt =T ⊂ A ({xt }∞ t =T1 ⊂ A, 1
2 −1 ⊂ F respectively) is called a trajectory if there exists a sequence {ut }Tt =T 1 ∞ ({ut }t =T1 ⊂ F , respectively) referred to as a control such that
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_2
25
26
2 DiscreteTime Autonomous Problems 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
∞ (({xt }∞ t =T1 , {ut }t =T1 ) ∈ X(T1 , ∞), respectively). Let θ0 ∈ E, θ1 ∈ F , a0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → ∞ as t → ∞. We suppose that the function f satisfies
f (x, u) ≥ ψ(ρE (x, θ0 )) − a0 for each (x, u) ∈ M.
(2.1)
For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we consider the following problems: T 2 −1 t =T1
2 2 −1 f (xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, xT2 = z, 1 1
T 2 −1 t =T1
(P1 ) 2 2 −1 f (xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, 1 1
T 2 −1 t =T1
(P2 )
2 2 −1 f (xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). 1 1
(P3 )
For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we define U f (T1 , T2 , y, z) = inf{
T 2 −1 t =T1
2 2 −1 f (xt , ut ) : ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
xT1 = y, xT2 = z},
(2.2)
σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ A},
(2.3)
σ f (T1 , T2 , y) = inf{U f (T1 , T2 , h, y) : h ∈ A},
(2.4)
σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, y) : h, y ∈ A}.
(2.5)
We assume that xf ∈ A, uf ∈ U(xf ) and xf = G(xf , uf ).
(2.6)
This implies that ({x¯t }∞ ¯ t }∞ ¯ t = uf for all t =0 , {u t =0 ) ∈ X(0, ∞), where x¯ t = xf , u integers t ≥ 0. We suppose that the following assumptions hold.
2.1 Preliminaries and Main Results
27
(A1) For each S1 > 0, there exist S2 > 0 and an integer c > 0 such that (T2 − T1 )f (xf , uf ) ≤
T 2 −1
f (xt , ut ) + S2
t =T1 2 2 −1 for each pair of integers T1 ≥ 0, T2 ≥ T1 + c and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 − 1. (A2) There exists an integer bf > 0, and for each > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ δ, i = 1, 2, there exist an −1 integer τ ∈ (0, bf ] and ({xt }τt=0, {ut }τt =0 ) ∈ X(0, τ ) which satisfies x0 = z1 , xτ = z2 and
τ −1
f (xt , ut ) ≤ τf (xf , uf ) + .
t =0
Section 2.6 contains examples of optimal control problems satisfying assumptions (A1) and (A2). Many examples can also be found in [106– 108, 118, 124, 125, 134]. The following result is proved in Section 2.8. Theorem 2.1. 1. There exists S > 0 such that for each pair of integers T2 > T1 ≥ 0 and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 T 2 −1
f (xt , ut ) + S ≥ (T2 − T1 )f (xf , uf ).
t =T1 ∞ 2. For each ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) either T −1
f (xt , ut ) − Tf (xf , uf ) → ∞ as T → ∞
t =0
or sup{
T −1
f (xt , ut ) − Tf (xf , uf ) : T ∈ {1, 2, . . . }} < ∞.
t =0
Moreover, if (2.7) holds, then the sequence {xt }∞ t =0 is bounded.
(2.7)
28
2 DiscreteTime Autonomous Problems
∞ We say that ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f, M, G)good ((f )good if the pair (M, G) is understood; (f, M)good if G is understood) if (2.7) holds [29, 104, 116, 120, 129, 134]. The next boundedness result is proved in Section 2.8. It has a prototype in [104].
Theorem 2.2. Let M0 > 0. Then there exists M1 > 0 such that for each pair of 2 2 −1 integers T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Let L > 0 be an integer. Denote by AL the set of all z ∈ A for which there exist −1 an integer τ ∈ (0, L] and ({xt }τt=0 , {ut }τt =0 ) ∈ X(0, τ ) such that x0 = z, xτ = xf ,
τ −1
f (xt , ut ) ≤ L.
t =0
L the set of all z ∈ A for which there exist an integer τ ∈ (0, L] and Denote by A −1 ({xt }τt=0 , {ut }τt =0 ) ∈ X(0, τ ) such that x0 = xf , xτ = z,
τ −1
f (xt , ut ) ≤ L.
t =0
The following Theorems 2.3–2.5 are also boundedness results. They are proved in Section 2.8. Theorem 2.3. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 L , xT1 ∈ AL , xT2 ∈ A T 2 −1
f (xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 2.4. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + L, and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1
2.1 Preliminaries and Main Results
xT1 ∈ AL ,
T 2 −1
29
f (xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 2.5. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + L, and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 L , xT2 ∈ A
T 2 −1
f (xt , ut ) ≤ σ f (T1 , T2 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. We say that the triplet (f, M, G) (or f if the pair (M, G) is understood; (f, M) if G is understood) possesses the turnpike property (or TP for short) if for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such that for each integer 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) T1 ≥ 0, each integer T2 ≥ T1 + 2L, and each ({xt }Tt =T 1 1 which satisfies T 2 −1
f (xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
we have ρE (xt , xf ) ≤ , t = T1 + L, . . . , T2 − L. Moreover, if ρE (xT1 , xf ) ≤ δ, then ρE (xt , xf ) ≤ for all t = T1 , . . . , T2 − L, and if ρE (xT2 , xf ) ≤ δ, then ρE (xt , xf ) ≤ for all t = T1 + L, . . . , T2 . Theorem 2.1 implies the following result. Proposition 2.6. Assume that f has TP and that , M > 0. Then there exist δ > 0 and an integer L > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L, 2 2 −1 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 T 2 −1
f (xt , ut ) ≤ min{(T2 − T1 )f (xf , uf ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . The following turnpike results are proved in Section 2.8.
30
2 DiscreteTime Autonomous Problems
Theorem 2.7. Assume that f has TP, that L > 0 is an integer, and that > 0. Then there exist δ > 0 and an integer L0 > L such that for each integer T1 ≥ 0, each 2 2 −1 integer T2 ≥ T1 + 2L0 , and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 L , xT1 ∈ AL , xT2 ∈ A T 2 −1
f (xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Theorem 2.8. Assume that f has TP, that L > 0 is an integer, and that M, > 0. Then there exist δ > 0 and an integer L0 > L such that for each integer T1 ≥ 0, 2 2 −1 each integer T2 ≥ T1 + 2L0 , and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which 1 1 satisfies xT1 ∈ AL , T 2 −1
f (xt , ut ) ≤ min{σ f (T1 , T2 , xT1 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . The next theorem is our main result in this chapter. It is proved in Section 2.9. Theorem 2.9. f has TP if and only if the following properties hold: ∞ (P1) for each (f )good pair of sequences ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞),
lim ρ(xt , xf ) = 0;
t →∞
(P2) for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such L−1 that for each ({xt }L t =0 , {ut }t =0 ) ∈ X(0, L) which satisfies L−1
f (xt , ut ) ≤ min{U f (0, L, x0 , xL ) + δ, Lf (xf , uf ) + M},
t =0
there exists an integer s ∈ [0, L] such that ρE (xs , xf ) ≤ .
2.2 Lower Semicontinuity Property
31
We say that the triplet (f, M, G) (or f if the pair (M, G) is understood; (f, M) if G is understood) possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0, there exist natural numbers Q, l such that for 2 2 −1 each integer T1 ≥ 0, each integer T2 l ≥ T1 + lQ, and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) which satisfies T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M,
t =T1 q
q
there exist finite sequences of natural numbers {ai }i=1 , {bi }i=1 ⊂ {T1 , . . . , T2 } such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
ρE (xt , xf ) ≤ for all integers t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. WTP was studies in [95, 96] for discretetime unconstrained problems, in [97] for variational problems and in [114] for discretetime constrained problems. The next result is proved in Section 2.11. Theorem 2.10. f has WTP if and only if f has TP.
2.2 Lower Semicontinuity Property We say that the triplet (f, M, G) (or f if the pair (M, G) is understood; (f, M) if G is understood) possesses the lower semicontinuity property (LSC property for short) if for each pair of integers T2 > T1 ≥ 0 and each sequence 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), j = 1, 2, . . . 1 1
(j )
(j )
which satisfies sup{
T 2 −1
(j )
(j )
f (xt , ut ) : j = 1, 2, . . . } < ∞,
t =T1
there exist a strictly increasing sequence of natural numbers {jk }∞ k=1 and T2 −1 2 ({xt }Tt =T , {u } ) ∈ X(T , T ) such that for any t ∈ {T , . . . , T − 1}, t t =T1 1 2 1 2 1 (jk )
xt
→ xt as k → ∞,
32
2 DiscreteTime Autonomous Problems T 2 −1
f (xt , ut ) ≤ lim inf j →∞
t =T1
T 2 −1
(j )
(j )
f (xt , ut ).
t =T1
LSC property plays an important role in the calculus of variations and optimal control theory [32]. Theorem 2.11. Assume that f possesses LSC property. Then f has TP if and only if f has (P1). Theorem 2.11 follows from Theorem 2.9 and the next proposition which is proved in Section 2.12. Proposition 2.12. Assume that f has LSC property and (P1). Then f has (P2). ∞ A pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is called (f, M, G)overtaking optimal (or (f )overtaking optimal if the pair (M, G) is understood; (f, M)overtaking ∞ optimal if G is understood) [29, 104, 120] if for every ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) satisfying x0 = y0 ,
lim sup[
T −1
T →∞
f (xt , ut ) −
t =0
T −1
f (yt , vt )] ≤ 0.
t =0
∞ A pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is called (f, M, G)weakly optimal (or (f )weakly optimal if the pair (M, G) is understood; (f, M)weakly optimal if G ∞ is understood) [29, 104, 120] if for every ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) satisfying x0 = y0 ,
lim inf[ T →∞
T −1 t =0
f (xt , ut ) −
T −1
f (yt , vt )] ≤ 0.
t =0
∞ A pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is called (f, M, G)minimal (or (f )minimal if the pair (M, G) is understood; (f, M)minimal if G is understood) [12, 94] if for every integer T ≥ 1, T −1
f (xt , ut ) = U f (0, T , x0 , xT ).
t =0
In infinite horizon optimal control, the main goal is to show the existence of solutions using the optimality criterions above. The next result is proved in Section 2.14. Theorem 2.13. Assume that f has (P1) and LSC property and that ∞ ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞)
2.2 Lower Semicontinuity Property
33
is (f )good. Then there exists an (f )overtaking optimal pair ∗ ∞ ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞)
such that x0∗ = x0 . The following theorem is also proved in Section 2.14. Theorem 2.14. Assume that f has (P1) and LSC property, ({x˜t }∞ ˜ t }∞ t =0 , {u t =0 ) ∈ X(0, ∞) ∗ ∞ ∗ is (f )good and that ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) satisfies x0 = x˜ 0 . Then the following conditions are equivalent:
(i) (ii) (iii) (iv) (v)
∗ ∞ ({xt∗ }∞ t =0 , {ut }t =0 ) is (f )overtaking optimal; ∞ ∗ ({xt }t =0 , {u∗t }∞ t =0 ) is (f )weakly optimal; ∗ }∞ ) is (f )minimal and (f )good; ({xt∗ }∞ , {u t t =0 t =0 ∗ ∞ ({xt∗ }∞ t =0 , {ut }t =0 ) is (f )minimal and satisfies limt →∞ ρE (xt , xf ) = 0; ∞ ∗ ({xt }t =0 , {u∗t }∞ t =0 ) is (f )minimal and satisfies lim inft →∞ ρE (xt , xf ) = 0.
The following three results are proved in Section 2.15. Theorem 2.15. Assume that f has (P1) and LSC property and that > 0. Then there exists δ > 0 such that the following assertions hold. (i) For every z ∈ A satisfying ρE (z, xf ) ≤ δ, there exists an (f )overtaking ∞ optimal pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) which satisfies x0 = z. ∞ (ii) If an (f )overtaking optimal pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) satisfies ρE (x0 , xf ) ≤ δ, then ρE (xt , xf ) ≤ for all integers t ≥ 0. Theorem 2.16. Assume that f has (P1) and LSC property, that > 0, and that L > 0 is an integer. Then there exists an integer τ0 > 0 such that for every (f )∞ overtaking optimal pair ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) which satisfies x0 ∈ AL , the inequality ρE (xt , xf ) ≤ holds for all integers t ≥ τ0 . Theorem 2.17. Assume that f has (P1) and LSC property, ∞ ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞)
is an (f )overtaking optimal and (f )good pair and that integers t2 > t1 ≥ 0 satisfy xt1 = xt2 . Then xt = xf for all integers t ≥ t1 . In the sequel we use the following result which easily follows from (P1) and (A2).
34
2 DiscreteTime Autonomous Problems
Proposition 2.18. Let f have (P1) and z ∈ A. There exists an (f )good pair ∞ ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞)
satisfying x0 = z if and only if z ∈ ∪{AL : L = 1, 2, . . . }. LSC property implies the following result. Proposition 2.19. Let f have LSC property, T2 > T1 ≥ 0 and L > 0 be integers. 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) such that 1. There exists ({xt }Tt =T 1 1
T 2 −1 t =T1
f (xt , ut ) ≤
T 2 −1 t =T1
2 2 −1 f (yt , vt ) for all ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ). 1 1
L , then there exists 2. If T2 − T1 ≥ 2L, z1 ∈ AL , z2 ∈ A 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that xTi = zi , i = 1, 2 and T 2 −1
f (xt , ut ) = U f (T1 , T2 , z1 , z2 ).
t =T1 2 2 −1 3. If T2 − T1 ≥ L, z ∈ AL , then there exists ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) such 1 1 that xT1 = z and
T 2 −1
f (xt , ut ) = σ f (T1 , T2 , z).
t =T1
L , then there exists ({xt }T2 , {ut }T2 −1 ) ∈ X(T1 , T2 ) such 4. If T2 − T1 ≥ L, z ∈ A t =T1 t =T1 that xT2 = z and T 2 −1
f (xt , ut ) = σ f (T1 , T2 , z).
t =T1
Theorem 2.20. f has (P1) and (P2) if and only if the following property holds: (i) for each > 0 and each M > 0, there exists an integer L > 0 such that for each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) pair of integers T1 ≥ 0, T2 ≥ T1 + L and each ({xt }Tt =T 1 1 which satisfies
2.3 Perturbed Problems
35 T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M,
t =T1
the inequality Card({t ∈ {T1 , . . . , T2 } : ρE (xt , xf ) > }) ≤ L holds. Proof. In view of (2.10) and Theorem 2.9, (P1) and (P2) imply (i). Clearly, (i) implies (P2). In order to show that (i) implies (P1), it is sufficient to note that if ∞ ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )good, then for some M > 0, T −1
f (xt , ut ) ≤ Tf (xf , uf ) + M for all integers T ≥ 1
t =0
and Card({t ∈ {0, 1, . . . } : ρE (xt , xf ) > }) < ∞.
2.3 Perturbed Problems In this section we use the following assumption. (A3) There exists an integer bf > 0, and for each > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ δ, i = 1, 2, there exist an −1 integer τ ∈ (0, bf ] and ({xt }τt=0, {ut }τt =0 ) ∈ X(0, τ ) which satisfies x0 = z1 , xτ = z2 and τ −1
f (xt , ut ) ≤ τf (xf , uf ) + , ρ(xt , xf ) ≤ , t = 0, . . . , τ.
t =0
Clearly, (A3) implies (A2). Assume that φ : E × E → [0, 1] is a continuous function satisfying φ(x, x) = 0 for all x ∈ E and such that the following property holds: (a) for each > 0, there exists δ > 0 such that if x, y ∈ E and ρE (x, y) ≤ δ, then φ(x, y) ≤ , and if x, y ∈ E satisfies φ(x, y) ≤ δ, then ρE (x, y) ≤ . For each r ∈ (0, 1) set fr (x, u) = f (x, u) + rφ(x, xf ), (x, u) ∈ M.
(2.8)
36
2 DiscreteTime Autonomous Problems
Clearly, for any r ∈ (0, 1], if (A3) holds, then (A1) and (A3) hold for fr with (xfr , ufr ) = (xf , uf ). Theorem 2.21. Let (A3) hold and r ∈ (0, 1]. Then fr has (P1) and (P2). Theorem 2.21 is proved in Section 2.16.
2.4 A Subclass of Problems For each x ∈ A define = {y ∈ A : there exists u ∈ U(x) such that y = G(x, u)}, U(x) = {(x, y) ∈ A × A : y ∈ U(x)}, y) = y, (x, y) ∈ M M G(x, define and for every (x, y) ∈ M f(x, y) = inf{f (x, u) : u ∈ U(x), y = G(x, u)}. G) G) satisfies (A1) and (A2), and (f, M, has TP if and only if Clearly, (f, M, (f, M, G) has TP. Suppose that (F, ρF ) = (E, ρE ) and G(x, u) = u, (x, u) ∈ M.
(2.9)
We continue to use the assumptions and definitions introduced in Section 2.1 but with some modifications which are given below. We assume that x f = uf ,
(2.10)
2 ⊂ A for all integers T2 > T1 ≥ 0, X(T1 , T2 ) is the set of all sequences {xt }Tt =T 1 satisfying (xt , xt +1 ) ∈ M for all t = T1 , . . . , T2 − 1 and X(T1 , ∞) is the set of all sequences {xt }∞ t =T1 ⊂ A satisfying (xt , xt +1 ) ∈ M for all integers t ≥ T1 . For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we define
U (T1 , T2 , y, z) = inf{ f
T 2 −1 t =T1
2 f (xt , ut ) : {xt }Tt =T ∈ X(T1 , T2 ), 1
xT1 = y, xT2 = z}, σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ A}, σ f (T1 , T2 , y) = inf{U f (T1 , T2 , h, y) : h ∈ A},
2.4 A Subclass of Problems
37
σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, y) : h, y ∈ A}. Define ¯ = {(y, x) ∈ A × A : (x, y) ∈ M} M
(2.11)
¯ set and for all (y, x) ∈ M f¯(y, x) = f (x, y).
(2.12)
2 Let T2 > T1 ≥ 0 be integers and {xt }Tt =T ∈ X(T1 , T2 ). Set 1
x¯t = xT2 −t +T1 , t = T1 , . . . , T2 .
(2.13)
By (2.12) and (2.13), for all t = T1 , . . . , T2 − 1, ¯ (x¯t , x¯t +1 ) = (xT2 −t +T1 , xT2 −t +T1 −1 ) ∈ M, T 2 −1
f¯(x¯t , x¯t +1 ) =
t =T1
=
T 2 −1
T 2 −1
(2.14)
f¯(xT2 −t +T1 , xT2 −t +T1 −1 )
t =T1
f (xT2 −t +T1 −1 , xT2 −t +T1 ) =
t =T1
T 2 −1
f (xt , xt +1 ).
(2.15)
t =T1
¯ 1 , T2 ) the set of all sequences Let T2 > T1 ≥ 0 be integers. Denote by X(T T2 ¯ for all t = T1 , . . . , T2 −1 and by X(T ¯ 1 , ∞) {xt }t =T1 ⊂ A satisfying (xt , xt +1 ) ∈ M ∞ ¯ the set of all sequences {xt }t =T1 ⊂ A satisfying (xt , xt +1 ) ∈ M for all integers t ≥ T1 . Suppose that for all (x, y) ∈ M, f (x, y) ≥ max{ψ(ρE (x, θ0 )), ψ(ρE (y, θ0 ))} − a0 .
(2.16)
For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we define f
U− (T1 , T2 , y, z) = inf{
T 2 −1 t =T1
2 ¯ 1 , T2 ), ∈ X(T f¯(xt , xt +1 ) : {xt }Tt =T 1
xT1 = y, xT2 = z}, f
f
f
f
σ− (T1 , T2 , y) = inf{U− (T1 , T2 , y, h) : h ∈ A}, σ− (T1 , T2 , y) = inf{U− (T1 , T2 , h, y) : h ∈ A},
38
2 DiscreteTime Autonomous Problems f
f
σ− (T1 , T2 ) = inf{U− (T1 , T2 , h, y) : h, y ∈ A}. Relations (2.12), (2.13), and (2.15) imply the following result. 2 Proposition 2.22. Let S2 > S1 ≥ 0 be integers, M ≥ 0 and {xt(i) }St =S ∈ X(S1 , S2 ), 1 i = 1, 2. Then
S 2 −1 t =S1
(1)
(1)
f (xt , xt +1 ) ≥
if and only if
S 2 −1 t =S1
S 2 −1 t =S1
(2)
(2)
f (xt , xt +1 ) − M
f¯(x¯t(1) , x¯t(1) +1 ) ≥
S 2 −1 t =S1
f¯(x¯t(2) , x¯t(2) +1 ) − M.
Proposition 2.22 implies the following result. 2 ∈ X(S1 , S2 ). Proposition 2.23. Let S2 > S1 ≥ 0 be integers, M ≥ 0 and {xt }St =S 1 Then the following assertions hold:
S 2 −1
f (xt , xt +1 ) ≤ σ f (S1 , S2 ) + M if and only if
t =S1 S 2 −1 t =S1 S 2 −1
f f¯(x¯t , x¯t +1 ) ≤ σ− (S1 , S2 ) + M;
f (xt , xt +1 ) ≤ U f (S1 , S2 , xS1 , xS2 ) + M
t =S1
if and only if
S 2 −1 t =S1
S 2 −1
f f¯(x¯t , x¯t +1 ) ≤ U− (S1 , S2 , x¯S1 , x¯S2 ) + M;
f (xt , xt +1 ) ≤ σ f (S1 , S2 , xS1 ) + M if and only if
t =S1 S 2 −1 t =S1
f σ− (S1 , S2 , x¯S2 ) + M; f¯(x¯t , x¯t +1 ) ≤
2.4 A Subclass of Problems S 2 −1
39
f (xt , xt +1 ) ≤ σ f (S1 , S2 , xS2 ) + M if and only if
t =S1 S 2 −1 t =S1
f f¯(x¯t , x¯t +1 ) ≤ σ− (S1 , S2 , x¯S1 ) + M.
¯ In view of Assertion 1 of Theorem 2.1 and ProposiClearly, (xf , xf ) ∈ M. ¯ and if (A3) holds for (f, M), then (A3) tion 2.22, (A1) and (A2) hold for (f¯, M), ¯ ¯ holds for (f , M) too. We suppose that (f, M) has (P1) and (P2). Then by Theorem 2.9, (f, M) has ¯ has TP too. Clearly, (P1) and (P2) hold for TP. Proposition 2.23 implies that (f¯, M) ¯ ¯ ¯ (f , M). Clearly, if (f, M) possesses LSC property, then it holds also for (f¯, M). By Theorem 2.1, there exists S∗ > 0 such that for each pair of integers T2 > 2 ∈ X(T1 , T2 ), T1 ≥ 0 and each {xt }Tt =T 1 T 2 −1
f (xt , xt +1 ) ≥ (T2 − T1 )f (xf , uf ) − S∗ .
(2.17)
t =T1
For all z ∈ A \ ∪{AL : L = 1, 2, . . . } set π f (z) = ∞.
(2.18)
z ∈ ∪{AL : L = 1, 2, . . . }.
(2.19)
Let
Define π (z) = inf{lim inf[ f
T →∞
T −1
f (xt , xt +1 ) − Tf (xf , xf )] :
t =0
{xt }∞ t =0 ∈ X(0, ∞), x0 = z}.
(2.20)
− S∗ ≤ π f (z) < ∞.
(2.21)
By (2.17) and (2.19),
There exists an integer L > 0 such that z ∈ AL .
(2.22)
40
2 DiscreteTime Autonomous Problems
In view of (2.22), there exist {x˜t }∞ t =0 ∈ X(0, ∞) and an integer τ ∈ {1, . . . , L} such that x˜0 = z, x˜t = xf for all integers t ≥ τ,
τ −1
f (x˜t , x˜t +1 ) ≤ L.
t =0
Then for all integers T > τ , T −1
f (x˜t , x˜t +1) − Tf (xf , uf ) ≤ L + (T − τ )f (xf , xf ) − Tf (xf , xf )
t =0
≤ L(1 + f (xf , xf )), π f (z) ≤ L(1 + f (xf , xf )) for all z ∈ AL .
(2.23)
Definition (2.20) implies the following result. Proposition 2.24. 1. Let S > T ≥ 0 be integers and {xt }St=T ∈ X(T , S) satisfy π f (xT ), π f (xS ) < ∞. Then π f (xT ) ≤
S−1
f (xt , xt +1 ) − (S − T )f (xf , uf ) + π f (xS ).
(2.24)
t =T
2. Let {xt }∞ t =0 ∈ X(0, ∞) be (f )good. Then for each pair of integers S > T ≥ 0, (2.24) holds. The following three propositions are proved in Section 2.17. Proposition 2.25. π f (xf ) = 0. Proposition 2.26. The function π f : A → R 1 ∪ {∞} is finite in a neighborhood of xf and continuous at xf . Proposition 2.27. For each M > 0, the set {z ∈ A : π f (z) ≤ M} is bounded, Corollary 2.28. π f (z) → ∞ as z ∈ A and ρE (z, θ0 ) → ∞.
(2.25)
2 ∈ X(T1 , T2 ) satisfy π f (xT1 ) < ∞. Let T2 > T1 ≥ 0 be integers and {xt }Tt =T 1 Define
2 )= Γ f ({xt }Tt =T 1
T 2 −1
f (xt , xt +1 ) − (T2 − T1 )f (xf , xf ) − π f (xT1 ) + π f (xT2 ).
t =T1
(2.26)
2.4 A Subclass of Problems
41
Proposition 2.24 implies that 2 0 ≤ Γ f ({xt }Tt =T ) ≤ ∞. 1
(2.27)
The next result is proved in Section 2.17. Proposition 2.29. Let {xt }∞ t =0 ∈ X(0, ∞) and x0 ∈ ∪{AL : L = 1, 2, . . . }. Then {xt }∞ t =0 is (f )good if and only if sup{Γ f ({xt }Tt=0 ) : T = 1, 2, . . . } = lim Γ f ({xt }Tt=0 ) < ∞. T →∞
If {xt }∞ t =0 is (f )good, then lim [
T →∞
T −1 t =0
f (xt , xt +1 ) − Tf (xf , xf )] = π f (x0 ) + lim Γ f ({xt }Tt=0). T →∞
Corollary 2.30. Let z ∈ A satisfy π f (z) < ∞ and > 0. Then there exists an (f )good {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z and lim Γ f ({xt }Tt=0 ) ≤ .
T →∞
Set inf(π f ) = inf{π f (z) : z ∈ A}. Proposition 2.25 and (2.21) imply that inf(π f ) is finite. Set Af = {z ∈ A : π f (z) ≤ inf(π f ) + 1}.
(2.28)
The next result is proved in Section 2.17. Proposition 2.31. There exists an integer M0 > 0 such that Af ⊂ AM0 . Corollary 2.30 implies the following result. Proposition 2.32. Let z ∈ ∪{AL : L = 1, 2, . . . } and > 0. Then there exists {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z and Γ f ({xt }Tt=0 ) ≤ for all integers T > 0. The next result if proved in Section 2.17. Proposition 2.33. Let M0 > 0 be an integer. Then the set AM0 is bounded.
42
2 DiscreteTime Autonomous Problems
2.5 Structure of Solutions in the Regions Close to the Endpoints We continue to study the problem introduced in Section 2.4 and assume that f possesses (P1) and LSC property. Then f has (P2) and TP. Let z ∈ ∪{AL : L = 1, 2, . . . }.
(2.29)
Denote by Λ(f, z) the set of all (f )overtaking optimal sequences {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z. In view of Theorem 2.13 and (2.29), Λ(f, z) = ∅. Theorem 2.1 implies that any element of Λ(f, z) is (f )good. Equation (2.20) implies the following result. Proposition 2.34. For every z ∈ ∪{AL : Λ(f, z), π (z) = lim inf[ f
T −1
T →∞
L = 1, 2, . . . } and every {xt }∞ t =0 ∈
f (xt , xt +1 ) − Tf (xf , xf )].
t =0
The next result follows from (P1) and Propositions 2.25, 2.26, and 2.34. Proposition 2.35. Let {xt }∞ t =0 ∈ X(0, ∞) be (f )good. Then for each pair of integers S > T ≥ 0, π f (xT ) =
S−1
f (xt , xt +1 ) − (S − T )f (xf , xf ) + π f (xS )
t =T
if and only if {xt }∞ t =0 is (f )overtaking optimal. Proposition 2.36. Let {xt }∞ t =0 ∈ X(0, ∞) be (f )overtaking optimal and (f )good. Then π f (x0 ) = lim [ T →∞
T −1
f (xt , xt +1 ) − Tf (xf , xf )].
t =0
Proof. Propositions 2.25, 2.26, and 2.35 and (P1) imply that for all integers T > 0, T −1
f (xt , xt +1 ) − Tf (xf , xf ) = π f (x0 ) − π f (xT ) → π f (x0 ) as T → ∞.
t =0
Proposition 2.36 is proved. The next result is proved in Section 2.18.
2.5 Structure of Solutions in the Regions Close to the Endpoints
43
Proposition 2.37. π f : A → R 1 ∪ {∞} is a lower semicontinuous function on A. It is clear that all the results obtained for the pair (f, M) also hold for the pair ¯ (f¯, M) L : L = 1, 2, . . . } set For all z ∈ A \ ∪{A f
π− (z) = ∞
(2.30)
L : L = 1, 2, . . . } define and for all z ∈ ∪{A f
π− (z) = inf{lim inf[ T →∞
T −1
f¯(xt , xt +1 ) − Tf (xf , xf )] :
t =0
¯ {xt }∞ t =0 ∈ X(0, ∞), x0 = z}.
(2.31)
2 ¯ 1 , T2 ). Define ∈ X(T Let T2 > T1 ≥ 0 be integers and {xt }Tt =T 1
f
2 Γ− ({xt }Tt =T )= 1
T 2 −1 t =T1
f f f¯(xt , xt +1 ) − (T2 − T1 )f (xf , xf ) − π− (xT1 ) + π− (xT2 ).
(2.32) In view of (2.27), f
2 ) ≤ ∞. 0 ≤ Γ− ({xt }Tt =T 1
(2.33)
The following two results are proved in Section 2.20. They describe the structure of approximate solutions in the regions close to the endpoints. Theorem 2.38. Let L0 , M be natural numbers and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies x0 ∈ AM ,
T −1
f (xt , xt +1 ) ≤ σ f (0, T , x0 ) + δ,
t =0
the inequalities f
f
f
L
0 π− (xT ) ≤ inf(π− ) + , Γ− ({x¯t }t =0 )≤
hold, where x¯ t = xT −t , t = 0, . . . , L0 . Theorem 2.39. Let L0 > 0 be an integer and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies
44
2 DiscreteTime Autonomous Problems T −1
f (xt , xt +1 ) ≤ σ f (0, T ) + δ,
t =0
the inequalities 0 π f (x0 ) ≤ inf(π f ) + , Γ f ({xt }L t =0 ) ≤ ,
f
f
f
0 π− (xT ) ≤ inf(π− ) + , Γ− ({x¯t }L t =0 ) ≤
hold, where x¯ t = xT −t , t = 0, . . . , T . The following two results are proved in Section 2.21. They also describe the structure of approximate solutions in the regions close to the endpoints. Theorem 2.40. Let (f, M) have LSC property, L0 , M > 0 be integers and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies x0 ∈ AM ,
T −1
f (xt , xt +1 ) ≤ σ f (0, T , x0 ) + δ,
t =0
¯ ¯ there exists an (f¯, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that π− (x0∗ ) = inf(π− ), ρE (xt∗ , xT −t ) ≤ , t = 0, . . . , L0 . f
f
Theorem 2.41. Let (f, M) have LSC property, L0 > 0 be an integer and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies T −1
f (xt , xt +1 ) ≤ σ f (0, T ) + δ,
t =0
¯ ¯ there exist an (f, M)overtaking optimal {xt∗,1 }∞ t =0 ∈ X(0, ∞) and an (f , M)∗,2 ∞ ¯ overtaking optimal {xt }t =0 ∈ X(0, ∞) such that π f (x0∗,1 ) = inf(π f ), π− (x0∗,2 ) = inf(π− ), f
f
ρE (xt , xt∗,1 ) ≤ , ρE (xT −t , xt∗,2 ) ≤ , t = 0, . . . , L0 .
2.6 Examples
45
2.6 Examples Example 2.42. We consider an example which is a particular case of the problem introduced in Section 2.1 and continue to assume that xf ∈ A, uf ∈ U(xf ), xf = G(xf , uf ). We suppose that G is continuous at (xf , uf ) and that the following assumption holds. (B1) For each > 0, there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ δ, i = 1, 2, there exists u ∈ U(z1 ) such that ρF (u, uf ) ≤ and z2 = G(z1 , u). Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ ∈ R 1 , a function π : E → R 1 be continuous at xf , a function L : M → [0, ∞) be upper semicontinuous at (xf , uf ) and for all (x, u) ∈ M, L(x, u) = 0 if and only if x = xf , u = uf , f (x, u) = μ + L(x, u) + π(x) − π(G(x, u)),
(2.34)
π(x) ≥ −a1 for all x ∈ E,
(2.35)
L(x, u) − π(G(x, u)) ≥ −a1 + ψ1 (ρE (θ0 , x)), (x, u) ∈ M.
(2.36)
Clearly, μ = f (xf , uf ).
(2.37)
2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ). By (2.34), Let T2 > T1 ≥ 0 be integers and ({xt }Tt =T 1 1
T 2 −1
f (xt , xt +1 ) = (T2 −T1 )μ(f )+
t =T1
T 2 −1
L(xt , ut )+π(xT1 )−π(G(xT2 −1 , uT2 −1 )).
t =T1
(2.38) It is easy to see that (2.1) holds for some a0 > 0 and an increasing ψ : [0, ∞) → [0, ∞) with limt →∞ ψ(t) = ∞. By (2.35)–(2.38), (A1) holds. It follows from (B1) that (A2) and (A3) hold. Example 2.43. We consider an example which is a particular case of the problem introduced in Section 2.1 and continue to assume that xf ∈ A, uf ∈ U(xf ), xf = G(xf , uf ). We suppose that G is continuous at (xf , uf ) and that the following assumption holds. (B1) There exists an integer df > 0, and for each > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ δ, i = 1, 2, there d
d −1
f f exists ({xt }t =0 , {ut }t =0 ) ∈ X(0, df ) which satisfies x0 = z1 , xdf = z2 and ρ(ut , uf ) ≤ , t = 0, . . . , df − 1.
46
2 DiscreteTime Autonomous Problems
Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ ∈ R 1 , a function π : E → R 1 be continuous at xf , a function L : M → [0, ∞) be upper semicontinuous at (xf , uf ) and that for all (x, u) ∈ M, L(x, u) = 0 if and only if x = xf , u = uf , f (x, u) = μ + L(x, u) + π(x) − π(G(x, u)),
(2.39)
π(x) ≥ −a1 for all x ∈ E,
(2.40)
L(x, u) − π(G(x, u)) ≥ −a1 + ψ1 (ρE (θ0 , x)), (x, u) ∈ M.
(2.41)
Clearly, μ = f (xf , uf ).
(2.42)
2 2 −1 Let T2 > T1 ≥ 0 be integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). By (2.39), 1 1
T 2 −1
f (xt , ut ) = (T2 − T1 )μ +
t =T1
T 2 −1
L(xt , ut ) + π(xT1 ) − π(G(xT2 −1 , uT2 −1 )).
t =T1
(2.43) It follows from (2.39)–(2.41) that (2.1) holds with an appropriate choice of a0 , ψ. In view of (2.40), (2.42), and (2.43), (A1) holds. By (2.41), (B1), and the continuity of G at (xf , uf ), (A2) and (A3) hold. Assume now that π is bounded on bounded subsets of E and that the following property holds: (B2) for each M, > 0 there exists δ > 0 such that for each (x, u) ∈ M which satisfies ρE (θ0 , x) ≤ M and L(x, u) ≤ δ we have ρE (x, xf ) ≤ . We claim that f has TP. In view of Theorems 2.9 and 2.20, it is sufficient to show that property (i) of Theorem 2.20 holds. Let , M > 0 and M1 > 0 be as guaranteed by Theorem 2.2 with M0 = M. Since π is bounded on bounded sets, there exists M2 > 0 such that π(z) ≤ M2 for all z ∈ BE (θ0 , M1 ). Let δ ∈ (0, 1) be as guaranteed by (B2) with M = M1 and L = 3 + δ −1 (2M2 + M + a0 + f (xf , uf )).
2.6 Examples
47
2 2 −1 Assume that T1 ≥ 0, T2 ≥ T1 + L are integers and that ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfies
T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M.
t =T1
Combined with Theorem 2.2 and the choice of M1 , this implies that ρE (θ0 , xt ) ≤ M1 , t = T1 , . . . , T2 − 1. It is not difficult to see that T 2 −2
f (xt , ut ) ≤
t =T1
T 2 −1
f (xt , ut )−f (xT2 −1 , uT2 −1 ) ≤ (T2 −T1 )f (xf , uf )+M +a0 .
t =T1
Together with (2.39), (2.42), the choice of M2 and δ, and (B2), this implies that M + a0 + f (xf , uf ) ≥
T 2 −2
(f (xt , ut ) − f (xf , uf ))
t =T1
=
T 2 −2
L(xt , ut ) + π(xT1 ) − π(xT2 −1 ) ≥
t =T1
T 2 −2
L(xt , ut ) − 2M2
t =T1
and 2M2 + M + a0 + f (xf , uf ) ≥
T 2 −2
L(xt , ut )
t =T1
≥ δCard({t ∈ {T1 , . . . , T2 − 2} : L(xt , ut ) > δ}) ≥ δCard({t ∈ {T1 , . . . , T2 − 2} : ρE (xt , xf ) > }). This implies that Card({t ∈ {T1 , . . . , T2 − 1} : ρE (xt , xf ) > }) ≤ 1 + δ −1 (2M2 + M + a0 + f (xf , uf )) < L and that property (i) of Theorem 2.20 holds. Therefore f has TP.
48
2 DiscreteTime Autonomous Problems
Example 2.44. We consider a particular case of the problem introduced in Section 2.1. Let A be a closed compact set, M be a closed compact set, G be continuous, and f be lower semicontinuous. Then LSC property holds. Example 2.45. We consider a particular case of the problem introduced in Section 2.1. Assume that for each M > 0, the set {(x, u) ∈ M : ρE (x, θ0 ) ≤ M} is closed and compact, G is continuous, and f is lower semicontinuous. It is not difficult to see that LSC property holds. Example 2.46. We consider an example which is a particular case of the problem introduced in Section 2.4 and continue to assume that (E, ρE ) = (F, ρF ), G(x, u) = u, (x, u) ∈ M, uf = xf , and (xf , uf ) ∈ M and assume that the following assumption holds. (B1) There exists 0 > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ 0 , i = 1, 2 we have (z1 , z2 ) ∈ M. Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ ∈ R 1 , a function π : E → R 1 be continuous at xf , a function L : M → [0, ∞) be upper semicontinuous at (xf , xf ) and that for all (x, u) ∈ M, L(x, u) = 0 if and only if x = xf , u = xf , f (x, u) = μ + L(x, u) + π(x) − π(u),
(2.44)
π(x) ≥ −a1 for all x ∈ E,
(2.45)
L(x, u) − π(u) ≥ −a1 + ψ1 (ρE (θ0 , x)), (x, u) ∈ M.
(2.46)
2 ∈ X(T1 , T2 ). By (2.44), Let T2 > T1 ≥ 0 be integers and {xt }Tt =T 1
T 2 −1
f (xt , xt +1 ) = (T2 − T1 )μ +
t =T1
T 2 −1
L(xt , ut ) + π(xT1 ) − π(xT2 ).
(2.47)
t =T1
It follows from (2.44)–(2.46) that (2.1) holds for an appropriate choice of a0 > 0 and ψ. By (2.45)–(2.47), (A1) holds. It follows from (B1) and upper semicontinuity of L at (xf , xf ) that (A2) and (A3) hold. Example 2.47. We consider an example which is a particular case of the problem introduced in Section 2.4 and continue to assume that (E, ρE ) = (F, ρF ), G(x, u) = u, (x, u) ∈ M, uf = xf , (xf , xf ) ∈ M and assume that the following assumption holds. (B1) There exists an integer df > 0, and for each > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ δ, i = 1, 2, there exists d
f ∈ X(0, df ) which satisfies x0 = z1 , xdf = z2 and ρ(xt , xf ) ≤ , {xt }t =0 t = 0, . . . , df .
2.7 Auxiliary Results for Theorem 2.1
49
Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ ∈ R 1 , a function π : E → R 1 be continuous at xf , a function L : M → [0, ∞) be upper semicontinuous at (xf , xf ) and for all (x, u) ∈ M, L(x, u) = 0 if and only if x = xf , u = xf , f (x, u) = μ + L(x, u) + π(x) − π(u),
(2.48)
π(x) ≥ −a1 for all x ∈ E, L(x, u) − π(u) ≥ −a1 + ψ1 (ρE (θ0 , x)), (x, u) ∈ M.
(2.49)
2 ∈ X(T1 , T2 ). By (2.48), equation Let T2 > T1 ≥ 0 be integers and {xt }Tt =T 1 (2.47) holds. It follows from (2.48) and (2.49) that (2.1) holds with an appropriate choice of a0 , ψ. In view of (2.47) and (2.49), (A1) holds. By (B1), (2.48), the upper semicontinuity of L at (xf , xf ) and continuity of π at xf , (A2) and (A3) hold.
2.7 Auxiliary Results for Theorem 2.1 Lemma 2.48. There exist integers S, c0 > 0 such that for each pair of integers 2 2 −1 T1 ≥ 0, T2 ≥ T1 + c0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 (T2 − T1 )f (xf , uf ) ≤
T 2 −1
f (xt , ut ) + S.
(2.50)
t =T1
Proof. There exists an integer S1 > 0 such that ψ(S1 ) > f (xf , uf ) + a0 + 1.
(2.51)
By (A1), there exist integers S2 > 0, c0 > 0 such that (T2 − T1 )f (xf , uf ) ≤
T 2 −1
f (xt , ut ) + S2
t =T1 2 2 −1 for each pair of integers T1 ≥ 0, T2 ≥ T1 + c0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 − 1. Fix an integer
S > S2 + (c0 + 2)(f (xf , uf ) + a0 + 2).
(2.52)
50
2 DiscreteTime Autonomous Problems
2 2 −1 Assume that T1 ≥ 0, T2 ≥ T1 + c0 are integers and that ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ). We show that (2.50) is true. Assume that
ρE (θ0 , xt ) ≥ S1 , t = T1 , . . . , T2 − 1.
(2.53)
By (2.1), (2.51), and (2.53), for all t = T1 , . . . , T2 − 1, f (xt , ut ) ≥ −a0 + ψ(ρE (θ0 , xt )) ≥ −a0 + ψ(S1 ) > f (xf , uf ) + 1, T 2 −1
f (xt , ut ) ≥ (T2 − T1 )(f (xf , uf ) + 1) ≥ (T2 − T1 )f (xf , uf )
t =T1
and (2.50) holds. Assume that min{ρE (θ0 , xt ) : t = T1 , . . . , T2 − 1} < S1 .
(2.54)
Set τ1 = min{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 },
(2.55)
τ2 = max{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 }.
(2.56)
By (2.54)–(2.56), τ1 , τ2 are welldefined, τ1 ≤ τ2 and ρE (θ0 , xτi ) ≤ S1 , i = 1, 2.
(2.57)
τ2 − τ1 ≥ c 0 ;
(2.58)
τ2 − τ1 < c 0 .
(2.59)
There are two cases:
Assume that (2.58) holds. It follows from (2.57), (2.58), and the choice of S2 and c0 that (τ2 − τ1 + 1)f (xf , uf ) ≤
τ 2 −1
f (xt , ut ) + S2 .
(2.60)
t =τ1
By (2.1), (2.51), (2.55), and (2.56), for each t ∈ ({T1 , . . . , τ1 } \ {τ1 }) ∪ ({τ2 + 1, . . . , T2 } \ {T2 }), f (xt , ut ) ≥ −a0 + ψ(ρE (θ0 , xt )) ≥ −a0 + ψ(S1 ) > f (xf , uf ) + 1.
(2.61)
2.7 Auxiliary Results for Theorem 2.1
51
In view of (2.52), (2.60), and (2.61), (T2 − T1 )f (xf , uf ) ≤ (τ1 − T1 )f (xf , uf ) +(τ2 − τ1 + 1)f (xf , uf ) + (T2 − τ2 − 1)f (xf , uf ) ≤ +
τ2
{f (xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
f (xt , ut ) + S2 +
{f (xt , ut ) : t ∈ {τ2 + 1, . . . , T2 } \ {T2 }}
t =τ1
=
T 2 −1
f (xt , ut ) + S2 ≤
t =T1
T 2 −1
f (xt , ut ) + S.
t =T1
Assume that (2.59) holds. Note that (2.61) is true. By (2.1), (2.52), (2.59), and (2.61), T 2 −1
f (xt , ut ) =
{f (xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
t =T1
+
τ2
f (xt , ut ) +
{f (xt , ut ) : t ∈ {τ2 + 1, . . . , T2 } \ {T2 }}
t =τ1
≥ (f (xf , uf ) + 1)Card({t ∈ {T1 , . . . , τ1 } \ {τ1 }}) − a0 (τ2 − τ1 + 1) +(f (xf , uf ) + 1)Card({t ∈ {τ2 + 1, . . . , T2 } \ {T2 }}) ≥ −a0 (c0 + 1) + (f (xf , uf ) + 1)(T2 − T1 − c0 − 2) ≥ f (xf , uf )(T2 −T1 )−(c0 +2)(a0 +f (xf , uf )+2) ≥ (T2 −T1 )f (xf , uf )−S. Lemma 2.48 is proved. The next auxiliary result easily follows from (2.1). Lemma 2.49. Let M0 > 0 and τ0 be natural number. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each T2 ∈ {T1 , . . . , T1 + τ0 }, and each T2 −1 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies t =T1 f (xt , ut ) ≤ M0 , the 1 1 inequality ρE (xt , θ0 ) ≤ M1 holds for all t = T1 , . . . , T2 − 1,
52
2 DiscreteTime Autonomous Problems
2.8 Proofs of Theorems 2.1–2.4 Proof of Theorem 2.1. Assertion 1 follows from Lemma 2.48 and (2.1). Let us prove Assertion 2. Let S > 0 be as guaranteed by Assertion 1. Assume that (2.7) does not hold and that Q > 0. Since (2.7) does not hold, Assertion 1 implies that there exists an integer TQ > 0 such that TQ −1
f (xt , ut ) − TQ f (xf , uf ) > Q.
(2.62)
t =0
By (2.62) and the choice of S, for each integer T > TQ , T −1
TQ −1
f (xt , ut ) =
t =0
f (xt , ut ) +
t =0
T −1
f (xt , ut )
t =TQ
≥ TQ f (xf , uf ) + Q + (T − TQ )f (xf , uf ) − S = Tf (xf , uf ) + Q − S. −1 Since Q is any positive number, we obtain that Tt =0 f (xt , ut )−Tf (xf , uf ) → ∞ as T → ∞. Assume that (2.7) holds. Then there exists S1 > 0 such that for each integer T > 0, 
T −1
f (xt , ut ) − Tf (xf , uf ) < S1 .
(2.63)
t =0
By (2.63), for each pair of integers T2 > T1 > 0, 
T 2 −1
f (xt , ut ) − (T2 − T1 )f (xf , uf ) < 2S1 .
(2.64)
t =T1
In view of (2.64), for each integer T ≥ 0, T +9
f (xt , ut ) ≤ 10f (xf , uf ) + 2S1 .
(2.65)
t =T
Lemma 2.49 and (2.65) imply that the sequence {xt }∞ t =0 is bounded. Thus Assertion 2 holds. Theorem 2.1 is proved. Proof of Theorem 2.2. By Theorem 2.1, there exists S0 > 0 such that the following property holds:
2.8 Proofs of Theorems 2.1–2.4
53
2 2 −1 (i) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
T 2 −1
f (xt , ut ) + S0 ≥ (T2 − T1 )f (xf , uf ).
t =T1
There exists M1 > 0 such that ψ(M1 ) > a0 + 2S0 + f (xf , uf ) + 1 + M0 .
(2.66)
2 2 −1 Let T2 > T1 ≥ 0 be integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfy 1 1
T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M0 .
(2.67)
t =T1
We show that ρE (θ0 , xt ) ≤ M1 , t = T1 , . . . , T2 − 1. Assume the contrary. Then there exists j ∈ {T1 , . . . , T2 − 1} such that ρE (θ0 , xj ) > M1 .
(2.68)
By (2.1) and (2.68), f (xj , uj ) ≥ ψ(ρE (θ0 , xj )) − a0 ≥ ψ(M1 ) − a0 . Property (i), (2.66), and (2.69) imply that T 2 −1
f (xt , ut ) =
{f (xt , ut ) : t ∈ {T1 , . . . , j } \ {j }} + f (xj , uj )
t =T1
+{f (xt , ut ) : t ∈ {j, . . . , T2 − 1} \ {j }} ≥ f (xf , uf )(Card({T1 , . . . , j } \ {j }) − S0 + ψ(M1 ) − a0 +f (xf , uf )(Card({j, . . . , T2 − 1} \ {j }) − S0 = f (xf , uf )((j − T1 ) + (T2 − 1 − j )) − a0 − 2S0 + ψ(M1 ) = f (xf , uf )(T2 − T1 ) + ψ(M1 ) − a0 − 2S0 − f (xf , uf ) > M0 + 1 + f (xf , uf )(T2 − T1 ).
(2.69)
54
2 DiscreteTime Autonomous Problems
This contradicts (2.67). The contradiction we have reached proves Theorem 2.2. Theorem 2.3 follows from Theorem 2.2 and the following proposition. Proposition 2.50. Let L > 0 be an integer. Then for each pair of integers T1 ≥ 0, L , T2 ≥ T1 + 2L and each pair of points z1 ∈ AL , z2 ∈ A U f (T1 , T2 , z1 , z2 ) ≤ (T2 − T1 )f (xf , uf ) + (2L + 2)(2 + a0 ). L . By the definition Proof. Let T1 ≥ 0, T2 ≥ T1 + 2L be integers, z1 ∈ AL , z2 ∈ A L , there exist integers τ1 ∈ (0, L], τ2 ∈ (0, L] and ({xt }T2 , {ut }T2 −1 ) ∈ of AL , A t =T1 t =T1 X(T1 , T2 ) such that xT1 = z1 , xt = xf , t = T1 + τ1 , . . . , T2 − τ2 , ut = uf , t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }, xT2 = z2 , T1 +τ 1 −1
f (xt , ut ) ≤ L,
t =T1
T 2 −1
f (xt , ut ) ≤ L.
t =T2 −τ2
In view of the relations above and (2.1), U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f (xt , ut )
t =T1
≤ 2L +
{f (xt , ut ) : t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }} ≤ 2L + (T2 − T1 )f (xf , uf ) + (2L + 2)a0.
Proposition 2.50 is proved. Theorem 2.4 follows from Theorem 2.2 and the following proposition. Proposition 2.51. Let L > 0 be an integer, T1 ≥ 0, T2 ≥ T1 + L be integers, z1 ∈ AL . Then σ f (T1 , T2 , z1 ) ≤ (T2 − T1 )f (xf , uf ) + (L + 1)(1 + a0 ). Proof. By the definition of AL , there exist an integer τ1 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) such that ({xt }Tt =T 1 1
∈
(0, L] and
xT1 = z1 , xt = xf , t = T1 + τ1 , . . . , T2 , ut = uf , t ∈ {T1 + τ1 , . . . , T2 } \ {T2 }, T1 +τ 1 −1 t =T1
f (xt , ut ) ≤ L.
2.9 Proof of Theorem 2.9
55
In view of the relations above and (2.1), σ f (T1 , T2 , z1 ) ≤
T 2 −1
f (xt , ut )
t =T1
≤ L + (T2 − T1 )f (xf , uf ) + (L + 1)a0 . Proposition 2.51 is proved. Theorem 2.5 follows from Theorem 2.2 and the following proposition. Proposition 2.52. Let L > 0 be an integer, T1 ≥ 0, T2 ≥ T1 + L be integers, L . Then z1 ∈ A σ f (T1 , T2 , z1 ) ≤ (T2 − T1 )f (xf , uf ) + (L + 1)(1 + a0 ). L , there exist an integer τ ∈ (0, L] and Proof. By the definition of A 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that xt = xf , t = T1 , . . . , T2 − τ, ut = uf , t = T1 , . . . , T2 − τ − 1, xT2 = z1 ,
T 2 −1
f (xt , ut ) ≤ L.
t =T2 −τ
In view of the relations above and (2.1), σ f (T1 , T2 , z1 ) ≤
T 2 −1
f (xt , ut )
t =T1
≤ L + (T2 − T1 )f (xf , uf ) + (L + 1)a0 . Proposition 2.52 is proved. Theorem 2.7 follows from Propositions 2.6 and 2.50, while Theorem 2.8 follows from Propositions 2.6 and 2.51.
2.9 Proof of Theorem 2.9 In view of Theorem 2.1, TP implies (P2). We show that TP implies (P1). Assume ∞ that TP holds, ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )good and > 0. There exists S > 0 such that for each integer T > 0,
56
2 DiscreteTime Autonomous Problems

T −1
f (xt , ut ) − Tf (xf , uf ) < S.
t =0
This implies that for each pair of integers T2 > T1 ≥ 0, 
T 2 −1
f (xt , ut ) − (T2 − T1 )f (xf , uf ) < 2S.
(2.70)
t =T1
Let δ > 0. We show that there exists an integer Tδ > 0 such that for each integer T > Tδ , T −1
f (xt , ut ) ≤ U f (Tδ , T , xTδ , xT ) + δ.
t =Tδ
Assume the contrary. Then for each integer T ≥ 0, there exists an integer S > T such that S−1
f (xt , ut ) > U f (T , S, xT , xS ) + δ.
t =T
This implies that there exists a strictly increasing sequence of integers {Ti }∞ i=0 such that T0 = 0
(2.71)
and for every integer i ≥ 0, Ti+1 −1
f (xt , ut ) > U f (Ti , Ti+1 , xTi , xTi+1 ) + δ.
(2.72)
t =Ti ∞ By (2.71) and (2.72), there exists ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that for every integer i ≥ 0,
yTi = xTi , Ti+1 −1
t =Ti
(2.73)
Ti+1 −1
f (xt , ut ) >
t =Ti
f (yt , vt ) + δ.
(2.74)
2.9 Proof of Theorem 2.9
57
In view of (2.70), (2.73), and (2.74), for each integer k ≥ 1, T k −1
f (yt , vt ) − Tk f (xf , uf ) =
t =0
+
T k −1
T k −1
f (yt , vt ) −
t =0
T k −1
f (xt , ut )
t =0
f (xt , ut ) − Tk f (xf , uf ) ≤ −kδ + 2S → −∞ as k → ∞.
t =0
This contradicts Theorem 2.1. The contradiction we have reached proves that the following property holds: (i) for each δ > 0 there exists an integer Tδ > 0 such that for each integer T > Tδ , T −1
f (xt , ut ) ≤ U f (Tδ , T , xTδ , xT ) + δ.
t =Tδ
Proposition 2.6 implies that there exist δ > 0 and an integer L > 0 such that the following property holds: (ii) for each integer S1 ≥ 0, each integer S2 ≥ S1 + 2L, and each 2 2 −1 , {ξt }St =S ) ∈ X(S1 , S2 ) which satisfies ({zt }St =S 1 1 S 2 −1
f (zt , ξt ) ≤ min{(S2 − S1 )f (xf , uf ) + 2S, U f (S1 , S2 , zS1 , zS2 ) + δ},
t =S1
we have ρE (zt , xf ) ≤ , t = S1 + L, . . . , S2 − L. Let an integer Tδ be as guaranteed by (i). Properties (i) and (ii), (2.70), and the choice of Tδ imply that for each integer T ≥ Tδ + 2L, ρE (xt , xf ) ≤ for all integers t ∈ [Tδ + L, T − L]. Therefore
ρE (xt , xf ) ≤ for all integers t ≥ Tδ + L
and (P1) holds. Thus TP implies (P1). Lemma 2.53. Assume that (P1) holds and that > 0. Then there exists δ > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2bf , and each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({xt }Tt =T 1 1 ρE (xTi , xf ) ≤ δ, i = 1, 2,
58
2 DiscreteTime Autonomous Problems T 2 −1
f (xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
the inequality ρE (xt , xf ) ≤ holds for all t = T1 , . . . , T2 . Proof. By (A2), for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: (iii) for each z ∈ A satisfying ρE (z, xf ) ≤ δk , there exist integers τ1 , τ2 ∈ (0, bf ] and 1 1 −1 2 2 −1 ({xt }τt =0 , {ut }τt =0 ) ∈ X(0, τ1 ), ({xt }τt =0 , {ut }τt =0 ) ∈ X(0, τ2 )
(1)
(1)
(2)
(2)
which satisfy (1)
(2)
= xf , x0 = xf , xτ(2) = z, x0 = z, xτ(1) 1 2 τ i −1
−k f (xt(i) , u(i) t ) ≤ τi f (xf , uf ) + 2 , i = 1, 2.
t =0
Assume that the lemma does not hold. Then for each integer k ≥ 1, there exist an integer Tk ≥ 2bf and Tk −1 k ({xt(k) }Tt =0 , {u(k) t }t =0 ) ∈ X(0, Tk )
such that (k)
(k)
ρE (x0 , xf ) ≤ δk , ρE (xTk , xf ) ≤ δk , T k −1 t =0
(k)
(k)
(k)
(k)
f (xt , ut ) ≤ U f (0, Tk , x0 , xTk ) + δk , (k)
sup{ρE (xt , xf ) : t ∈ {0, . . . , Tk }) > .
(2.75)
(2.76)
(2.77)
Let k ≥ 1 be an integer. Property (iii) and (2.75) imply that there exist integers τk,1 , τk,2 ∈ (0, bf ] and T +τk,1 +τk,2
k ({x˜t(k)}t =0
T +τk,1 +τk,2 −1
k , {u˜ (k) t }t =0
) ∈ X(0, Tk + τk,1 + τk,2 )
such that (k)
(k)
x˜0 = xf , x˜τ(k) = x0 , k,1
(2.78)
2.9 Proof of Theorem 2.9
59
τk,1 −1
−k f (x˜t(k), u˜ (k) t ) ≤ τk,1 f (xf , uf ) + 2 ,
(2.79)
t =0 (k)
(k)
(k)
(k)
x˜τk,1 +t = xt , t = 0, . . . , Tk , u˜ τk,1 +t = ut , t = 0, . . . , Tk − 1, (k)
x˜Tk +τk,1 +τk,2 = xf , Tk +τk,1 +τk,2 −1
(2.80) (2.81)
−k f (x˜t(k), u˜ (k) t ) ≤ τk,2 f (xf , uf ) + 2 .
(2.82)
t =Tk +τk,1
By (2.76), (2.79), (2.80), and (2.82), Tk +τk,1 +τk,2 −1
(k)
(k)
f (x˜t , u˜ t )
t =0
≤ (τk,1 + τk,2 )f (xf , uf ) + 2
−k+1
+
T k −1
f (xt(k), u(k) t )
t =0
) + δk + 2−k+1 . ≤ (τk,1 + τk,2 )f (xf , uf ) + U f (0, Tk , x0(k) , xT(k) k
(2.83)
Property (iii) and (2.75) imply that there exist integers τk,3 , τk,4 ∈ (0, bf ] and Tk −1 k ({ xt(k)}Tt =0 , { u(k) t }t =0 ) ∈ X(0, Tk ) such that x0(k) = x0(k) , xT(k) = xT(k) , k k
(2.84)
xt(k) = xf , t = τk,3 , . . . , Tk − τk,4 ,
(2.85)
u(k) t = uf , t ∈ {τk,3 , . . . , T − τk,4 } \ {T − τk,4 }, τk,3 −1
−k f ( xt(k), u(k) t ) ≤ τk,3 f (xf , uf ) + 2 ,
(2.86)
(2.87)
t =0 T k −1 t =Tk −τk,4
In view of (2.84)–(2.88),
−k f ( xt(k) , u(k) t ) ≤ τk,4 f (xf , uf ) + 2 .
(2.88)
60
2 DiscreteTime Autonomous Problems
U f (0, Tk , x0(k) , xT(k) )≤ k
T k −1
−k+1 f ( xt(k), u(k) . t ) ≤ Tk f (xf , uf ) + 2
(2.89)
t =0
By (2.83) and (2.89), Tk +τk,1 +τk,2 −1
f (x˜t , u˜ t ) ≤ (Tk + τk,1 + τk,2 )f (xf , uf ) + 3 · 2−k+1 . (k)
(k)
(2.90)
t =0 ∞ By (2.78) and (2.81), there exists ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) such that (1)
xt = x˜t , t = 0, . . . , T1 + τ1,1 + τ1,2 ,
(2.91)
ut = u˜ (1) t , t = 0, . . . , T1 + τ1,1 + τ1,2 − 1
(2.92)
and for each integer k ≥ 1, = x˜t(k+1), t = 0, . . . , Tk+1 + τk+1,1 + τk+1,2 ,
(2.93)
= u˜ (k+1) , t = 0, . . . , Tk+1 + τk+1,1 + τk+1,2 − 1. t
(2.94)
x k
i=1 (Ti +τi,1 +τi,2 )+t
uk
i=1 (Ti +τi,1 +τi,2 )+t
In view of (2.91)–(2.94), for each integer k ≥ 1,
{f (xt , ut ) : t = 0, . . . ,
k (Ti + τi,1 + τi,2 ) − 1} i=1
=
+τi,2 −1 k Ti +τi,1 i=1
≤
f (x˜t(i), u˜ (i) t )
t =0
k k (Ti + τi,1 + τi,2 )f (xf , uf ) + 3 2−i+1 i=1
≤
i=1 k
(Ti + τi,1 + τi,2 )f (xf , uf ) + 6.
(2.95)
i=1 ∞ Theorem 2.1 and (2.95) imply that ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )good. Together with (P1) this implies that limt →∞ ρE (xt , xf ) = 0. On the other hand in view of (2.77) and (2.93), lim supt →∞ ρE (xt , xf ) > . The contradiction we have reached completes the proof of Lemma 2.53.
2.9 Proof of Theorem 2.9
61
Completion of the Proof of Theorem 2.9. Assume that properties (P1) and (P2) hold. Let , M > 0. By Lemma 2.53, there exists δ0 > 0 such that the following property holds: (iv) for each integer T1 ≥ 0, each integer T2 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({xt }Tt =T 1 1
≥ T1 + 2bf , and each
ρE (xTi , xf ) ≤ δ0 , i = 1, 2, T 2 −1
f (xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ0 ,
t =T1
the inequality ρE (xt , xf ) ≤ holds for all t = T1 , . . . , T2 . By Theorem 2.1, there exists S0 > 0 such that the following property holds: 2 2 −1 (v) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ), we have T 2 −1
f (xt , ut ) + S0 ≥ (T2 − T1 )f (xf , uf ).
t =T1
In view of (P2), there exist δ ∈ (0, δ0 ) and an integer L0 > 0 such that the following property holds: L0 L0 −1 (vi) for each ({xt }t =0 , {ut }t =0 ) ∈ X(0, L0 ) which satisfies L 0 −1
f (xt , ut ) ≤ min{U f (0, L0 , x0 , xL0 ) + δ, L0 f (xf , uf ) + 2S0 + M},
t =0
there exists an integer t0 ∈ [0, L0 ] such that ρE (xt0 , xf ) ≤ δ0 . Set L = L0 + bf .
(2.96)
Assume that T1 ≥ 0 and T2 ≥ T1 + 2L are integers and that 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
satisfies T 2 −1 t =T1
f (xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ}.
(2.97)
62
2 DiscreteTime Autonomous Problems
Theorem 2.1, properties (iv) and (v), and (2.97) imply that for each pair of integers Q1 , Q2 ∈ [T1 , T2 ] satisfying Q1 < Q2 , Q 2 −1
f (xt , ut ) =
t =Q1
T 2 −1
f (xt , ut ) −
{f (xt , ut ) : t ∈ {T1 , . . . , Q1 } \ {Q1 }}
t =T1
−
{f (xt , ut ) : t ∈ {Q2 , . . . , T2 } \ {T2 }}
≤ (T2 − T1 )f (xf , uf ) + M − (Q1 − T1 )f (xf , uf ) +S0 − (T2 − Q2 )f (xf , uf ) + S0 = (Q2 − Q1 )f (xf , uf ) + M + 2S0 . In particular, L 0 −1
f (xt , ut ) ≤ L0 f (xf , uf ) + 2S0 + M,
(2.98)
t =0 T 2 −1
f (xt , ut ) ≤ L0 f (xf , uf ) + 2S0 + M.
(2.99)
t =T2 −L0
By (2.97)–(2.99) and property (vi), there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xτi , xf ) ≤ δ0 , i = 1, 2.
(2.100)
If ρE (xT1 , xf ) ≤ δ, then we may assume that τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then we may assume that τ2 = T2 . In view of (2.97), τ 2 −1
f (xt , ut ) ≤ U f (τ1 , τ2 , xτ1 , xτ2 ) + δ.
(2.101)
t =τ1
It follows from (2.96) that τ2 − τ1 ≥ T2 − T1 − 2L0 ≥ 2L − 2L0 ≥ 2bf . Property (iv) and (2.100)–(2.102) imply that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Theorem 2.9 is proved.
(2.102)
2.10 An Auxiliary Result for Theorem 2.10
63
2.10 An Auxiliary Result for Theorem 2.10 Lemma 2.54. Assume that (P2) holds and that M, > 0. Then there exists a L−1 natural number L such that for each ({xt }L t =0 , {ut }t =0 ) ∈ X(0, L) which satisfies L−1
f (xt , ut ) ≤ Lf (xf , uf ) + M,
t =0
the following inequality holds: min{ρE (xt , xf ) : t = 0, . . . , L} ≤ . Proof. By Theorem 2.1, there exists S0 > 0 such that the following property holds: 2 2 −1 , {ut }Tt =T ) ∈ (i) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T 1 1 X(T1 , T2 ),
T 2 −1
f (xt , ut ) + S0 ≥ (T2 − T1 )f (xf , uf ).
t =T1
By (P2), there exist δ0 ∈ (0, ) and an integer L0 > 0 such that the following property holds: L0 L0 −1 , {ut }t =0 ) ∈ X(0, L0 ) which satisfies (ii) for each ({xt }t =0 L 0 −1
f (xt , ut ) ≤ min{U f (0, L0 , x0 , xL0 ) + δ0 ,
t =0
L0 f (xf , uf ) + 2S0 + 2M + f (xf , uf )}, we have min{ρE (xt , xf ) : t = 0, . . . , L0 } ≤ . Choose an integer q0 > (M + S0 )δ −1
(2.103)
L = q0 L0 .
(2.104)
and set
64
2 DiscreteTime Autonomous Problems
L−1 Let ({xt }L t =0 , {ut }t =0 ) ∈ X(0, L) satisfy L−1
f (xt , ut ) ≤ Lf (xf , uf ) + M.
(2.105)
t =0
We show that min{ρE (xt , xf ) : t = 0, . . . , L} ≤ . Assume the contrary. Then ρE (xt , xf ) > , t = 0, . . . , L.
(2.106)
Let an integer i ∈ {0, . . . , q0 − 1}. Property (i) and (2.105) imply that (i+1)L 0−1
f (xt , ut ) =
t =iL0
L−1
f (xt , ut ) −
{f (xt , ut ) : t ∈ {0, . . . , iL0 } \ {iL0 }}
t =0
−
{f (xt , ut ) : t ∈ {(i + 1)L0 , . . . , L} \ {L}}
≤ Lf (xf , uf ) + M − iL0 f (xf , uf ) + S0 − (L − (i + 1)L0 )f (xf , uf ) + S0 = L0 f (xf , uf ) + M + 2S0 .
(2.107)
By (2.106), (2.107), and property (ii), (i+1)L 0−1
f (xt , ut ) > U f (iL0 , (i + 1)L0 , xiL0 , x(i+1)L0 ) + δ0 .
(2.108)
t =iL0 L−1 It follows from (2.104) and (2.108) that there exists ({yt }L t =0 , {vt }t =0 ) ∈ X(0, L) such that yiL0 = xiL0 , i = 0, . . . , q0 and for all i = 0, . . . , q0 − 1, (i+1)L 0 −1 t =iL0
f (xt , ut ) >
(i+1)L 0 −1
f (yt , vt ) + δ0 .
t =iL0
In view of (2.105), (2.109), and property (i), Lf (xf , uf ) + M ≥
L−1 t =0
f (xt , ut )
(2.109)
2.11 Proof of Theorem 2.10
≥
L−1
65
f (yt , vt ) + q0 δ ≥ Lf (xf , uf ) − S0 + q0 δ,
t =0
q0 ≤ (M + S0 )δ −1 . This contradicts (2.103). The contradiction we have reached proves Lemma 2.54.
2.11 Proof of Theorem 2.10 Assume that f has TP. In view of Theorem 2.9, (P1) and (P2) hold. We show that WTP holds. Let , M > 0. By Theorem 2.1, there exists S0 > 0 such that the following property holds: 2 2 −1 , {vt }τt =τ ) ∈ X(τ1 , τ2 ), (i) for each pair of integers τ2 > τ1 ≥ 0 and each ({yt }τt =τ 1 1
τ 2 −1
f (yt , vt ) + S0 ≥ (τ2 − τ1 )f (xf , uf ).
t =τ1
TP implies that there exist δ ∈ (0, ) and an integer L0 > 0 such that the following property holds: (ii) for each integer τ1 ≥ 0, each integer τ2 ≥ τ1 + 2L0 , and each 2 2 −1 ({yt }τt =τ , {vt }τt =τ ) ∈ X(τ1 , τ2 ) which satisfies 1 1 τ 2 −1
f (yt , vt ) ≤ min{σ f (τ1 , τ2 ) + M + 3S0 , U f (τ1 , τ2 , yτ1 , yτ2 ) + δ},
t =τ1
we have ρE (yt , xf ) ≤ , t = τ1 + L0 , . . . , τ2 − L0 . Set l = 2L0 + 4.
(2.110)
Q ≥ 6 + 6(M + S0 )δ −1 .
(2.111)
Choose a natural number
66
2 DiscreteTime Autonomous Problems
2 2 −1 Assume that T1 ≥ 0, T2 ≥ T1 +lQ are integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 satisfies
T 2 −1
f (xt , ut ) ≤ (T2 − T1 )f (xf , uf ) + M.
(2.112)
t =T1
Property (i) and (2.112) imply that for each pair of integers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , τ 2 −1 t =τ1
f (xt , ut ) =
T 2 −1
f (xt , ut ) −
{f (xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
t =T1
−
{f (xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {T2 }}
≤ M+(T2 −T1 )f (xf , uf )−(τ1 −T1 )f (xf , uf )+S0 − (T2 − τ2 )f (xf , uf ) + S0 ≤ M + 2S0 + (τ2 − τ1 )f (xf , uf ).
(2.113)
t0 = T1 .
(2.114)
Set
If T 2 −1
f (xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
then we set t1 = T2 . Assume that T 2 −1
f (xt , ut ) > U f (T1 , T2 , xT1 , xT2 ) + δ.
(2.115)
t =T1
Set t1 = min{t ∈ {T1 + 1, . . . , T2 } :
t −1
f (xs , us ) > U f (T1 , t, xT1 , xt ) + δ}.
s=T1
Assume that k ≥ 1 is an integer and we defined a sequence of integers {ti }ki=0 ⊂ [T1 , T2 ] such that t0 < t1 · · · < tk ,
2.11 Proof of Theorem 2.10
67
for each integer i = 1, . . . , k if ti < T2 , then t i −1
f (xs , us ) − U f (ti−1 , ti , xti−1 , xti ) > δ,
(2.116)
s=ti−1
and if ti − 1 > ti−1 , then t i −2
f (xs , us ) − U f (ti−1 , ti − 1, xti−1 , xti −1 ) ≤ δ.
(2.117)
s=ti−1
(Note that in view of (2.115) and the choice of t1 , our assumption holds for k = 1.) 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) such that By (2.116), there exists ({yt }Tt =T 1 1 yti = xti , i = 1, . . . , k, t i −1
f (xs , us ) −
t i −1
(2.118)
f (ys , vs ) > δ, i ∈ {1, . . . , k} \ {k},
(2.119)
yt = xt , t ∈ {tk , . . . , T2 }, vt = ut , t ∈ {tk , . . . , T2 } \ {T2 }.
(2.120)
s=ti−1
s=ti−1
Property (i), (2.112), and (2.118)–(2.120) imply that (T2 − T1 )f (xf , uf ) + M ≥
T 2 −1 t =T1
f (xt , ut ) ≥
T 2 −1
f (yt , vt ) + δ(k − 1)
t =T1
≥ (T2 − T1 )f (xf , uf ) − S0 + δ(k − 1), k ≤ 1 + δ −1 (M + S0 ).
(2.121)
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If T 2 −1
f (xs , us ) ≤ U f (tk , T2 , xtk , xT2 ) + δ,
s=tk
then we set tk+1 = T2 , and the construction of the sequence is completed. Assume that T 2 −1 s=tk
f (xs , us ) > U f (tk , T2 , xtk , xT2 ) + δ.
68
2 DiscreteTime Autonomous Problems
Set tk+1 = min{t ∈ {tk + 1, . . . , T2 } :
t −1
f (xs , us ) > U f (tk , t, xtk , xt ) + δ}.
s=tk
It is not difficult to see that the assumption made for k also holds for k + 1. By q induction we constructed a finite sequence {ti }i=0 ⊂ [T1 , T2 ] such that T1 = t0 < t1 · · · < tq = T2 , for each integer i = 1, . . . , q if ti < T2 , then (2.116) holds and if ti − 1 > ti−1 , then (2.117) holds. In view of (2.121), q ≤ 1 + δ −1 (M + S0 ).
(2.122)
i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 2L0 + 1.
(2.123)
Assume that
Then (2.117) holds. Property (i) and (2.113) imply that ti+1 −1
f (xs , us ) ≤ M + 2S0 + (ti+1 − ti − 1)f (xf , uf )
s=ti
≤ M + 3S0 + σ f (ti , ti+1 − 1).
(2.124)
It follows from (2.117), (2.123), (2.124), and property (ii) that ρE (xt , xf ) ≤ , t = ti + L0 , . . . , ti+1 − 1 − L0 .
(2.125)
In view of (2.125), {t ∈ {T1 , . . . , T2 } : ρE (xt , xf ) > } ⊂ ∪{{ti , . . . , ti + L0 } ∪ {ti+1 − 1 − L0 , . . . , ti+1 } : i ∈ {0, . . . , q − 1} and ti+1 − ti ≥ 2L0 + 1} ∪{{ti , . . . , ti+1 } : i ∈ {0, . . . , q − 1} and ti+1 − ti < 2L0 + 1}. In view of (2.111) and (2.122), the number of intervals in the righthand side of the relation above does not exceed 3q ≤ 3(1 + δ −1 (M + S0 )) < Q, and their maximal length does not exceed 2L0 + 2 < l. Thus WTP holds. Assume that WTP holds. We show that TP holds. In view of Theorem 2.9, it is sufficient to show that (P1) and (P2) hold. It is clear that (P2) holds. Let us show ∞ that (P1) holds. Let ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) be (f )good. There exists M0 > 0 such that for each integer T ≥ 1,
2.11 Proof of Theorem 2.10

69
T −1
f (xt , ut ) − Tf (xf , uf ) ≤ M0 .
(2.126)
t =0
By (2.126), for each pair of integers T2 > T1 ≥ 0, T 2 −1
f (xt , ut ) − (T2 − T1 )f (xf , uf ) =
t =T1
−
T 2 −1
f (xt , ut ) − T2 f (xf , uf )
t =0
{f (xt , ut ) : t ∈ {0, . . . , T1 } \ {T1 }} − T1 f (xf , uf ) ≤ 2M0 .
(2.127)
Let > 0 and natural numbers l, Q be as guaranteed by WTP with and M = 2M0 . Set Ω = {t ∈ {0, 1, . . . } : ρE (xt , xf ) > }. We show that Ω is bounded. Assume that contrary. Then there exists a sequence of natural numbers {ti }∞ i=1 such that for all integers i ≥ 1, ti+1 − ti ≥ 2l + 2, ρE (xti , xf ) > .
(2.128)
Let k ≥ 1 + Q be an integer. It follows from (2.128) that tk ≥ tQ+1 ≥ Q(2l + 2).
(2.129)
By (2.127), (2.129), WTP, and the choice of Q, l, there exist an integer q ≤ Q and q q finite sequences of integers {ai }i=1 , {bi }i=1 ⊂ [0, tk ] such that 0 ≤ bi − ai ≤ l, i = 1, . . . , q,
(2.130)
bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
ρE (xt , xf ) ≤ for all integers t ∈ {0, . . . , tk } \ ∪i=1 [ai , bi ].
(2.131)
In view of (2.128) and (2.130), q
{ti }ki=1 ⊂ ∪i=1 [ai , bi ].
(2.132)
By (2.128), (2.130), and (2.132), k ≤ q ≤ Q, a contradiction. The contradiction we have reached proves that Ω is bounded. Since is an arbitrary positive number, we conclude that (P1) holds. This completes the proof of Theorem 2.10.
70
2 DiscreteTime Autonomous Problems
2.12 Proof of Proposition 2.12 Assume that (P2) does not hold. Then there exist , M > 0 such that for each natural (k) (k) number k, there exists ({xt }kt=0 , {ut }k−1 t =0 ) ∈ X(0, k) which satisfies k−1
f (xt(k) , u(k) t ) ≤ kf (xf , uf ) + M,
(2.133)
t =0
ρE (xt(k) , xf ) > , t = 0, . . . , k.
(2.134)
Let S0 > 0 be as guaranteed by Assertion 1 of Theorem 2.1. By the choice of S0 and (2.133), for each integer k ≥ 1 and each pair of integers τ1 , τ2 ∈ [0, k] satisfying τ1 < τ2 , τ 2 −1
f (xt(k), u(k) t ) − (τ2 − τ1 )f (xf , uf ) =
t =τ1
k−1
f (xt(k) , u(k) t ) − kf (xf , uf )
t =0
(k) (k) − {f (xt , ut ) : t ∈ {0, . . . , τ1 } \ {τ1 }} + τ1 f (xf , uf )
{f (xt(k) , u(k) t ) : t ∈ {τ2 , . . . , k} \ {k}} + (k − τ2 )f (xf , uf ) ≤ M + 2S0 . (2.135) By (2.1), (2.135), and LSC property, extracting subsequences, using the diagonalization process, and reindexing, we may assume that for each integer i ≥ 1, there (k) (k) (i) (i) (i) exists limk→∞ f (xi−1 , ui−1 ) and there exists ({yi−1 , yi }, {vi−1 }) ∈ X(i − 1, i) such that −
(k) (i) → yi−1 , xi(k) → yi(i) as k → ∞, xi−1 (i)
(i)
(k)
(2.136)
(k)
f (yi−1 , vi−1 ) ≤ lim f (xi−1 , ui−1 ).
(2.137)
k→∞
∞ In view of (2.136), there exists ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that for each integer i ≥ 1, (i) (i) , vi−1 ). (yi−1 , vi−1 ) = (yi−1
(2.138)
It follows from (2.137), (2.138), and (2.140) that for every integer m ≥ 1 m−1 i=0
f (yi , vi ) =
m i=1
(i)
(i)
f (yi−1 , vi−1 ) =
m i=1
(k)
(k)
lim f (xi−1 , ui−1 )
k→∞
2.13 Auxiliary Results for Theorems 2.13 and 2.14
= lim
k→∞
m
(k)
71
(k)
f (xi−1 , ui−1 ) ≤ M + 2S0 + mf (xf , uf ).
t =1
∞ Together with Theorem 2.1, this implies that ({yt }∞ t =0 , {vt }t =0 ) is (f )good. Property (P1) implies that there exists an integer τ0 > 0 such that
ρE (yt , xf ) ≤ /4 for all integers t ≥ τ0 − 1.
(2.139)
It follows from (2.136), (2.138), and (2.139) that (τ0 )
ρE (yt
, xf ) ≤ /4, t = τ0 − 1, τ0 .
(2.140)
By (2.136), there exists an integer k0 ≥ 1 + τ0 such that for all integers k ≥ k0 , (τ0 )
ρE (yt
(k)
, xt ) ≤ /4, t = τ0 − 1, τ0 .
(2.141)
In view of (2.140) and (2.141), for all integers k ≥ k0 and all t ∈ {τ0 − 1, τ0 }, ρE (xt(k), xf ) ≤ ρE (xt(k), yt(τ0 ) ) + ρE (yt(τ0 ) , xf ) ≤ /2. This contradicts (2.134). The contradiction we have reached proves that (P2) holds and completes the proof of Proposition 2.12.
2.13 Auxiliary Results for Theorems 2.13 and 2.14 Lemma 2.55. Let > 0. Then there exists δ > 0 such that for each bf bf −1 ({xt }t =0 , {ut }t =0 ) ∈ X(0, bf ) which satisfies ρE (x0 , xf ) ≤ δ, ρE (xbf , xf ) ≤ δ, the following inequality holds: bf −1
f (xt , ut ) ≥ bf f (xf , uf ) − .
t =0
Proof. By (A2), there exists δ > 0 such that the following property holds: (i) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 ∈ (0, bf ], τ1 −1 (2) τ2 τ2 −1 1 τ2 ∈ (0, bf ] and ({x˜t(1)}τt =0 , {u˜ (1) ˜ (2) t }t =0 ) ∈ X(0, τ1 ), ({x˜ t }t =0 , {u t }t =0 ) ∈ X(0, τ2 ) such that (1)
x˜ 0 = z , x˜τ(1) = xf , 1
τ 1 −1 t =0
(1)
(1)
f (x˜t , u˜ t ) ≤ τ1 f (xf , uf ) + /4,
72
2 DiscreteTime Autonomous Problems
(2)
x˜0 = xf , x˜τ(2) = z, 2
τ 2 −1
(2)
(2)
f (x˜t , u˜ t ) ≤ τ2 f (xf , uf ) + /4.
t =0 b −1
b
f f Assume that ({xt }t =0 , {ut }t =0 ) ∈ X(0, bf ) and
ρE (x0 , xf ) ≤ δ, ρE (xbf , xf ) ≤ δ.
(2.142)
∞ By property (i), there exists ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that
yt +3bf = yt , vt +3bf = vt for all integers t ≥ 0, bf −1
y0 = xf , ybf = x0 ,
f (yt , vt ) ≤ bf f (xf , uf ) + /4,
t =0
yt = xt −bf , t ∈ {bf , . . . , 2bf }, vt = ut −bf , t ∈ {bf , . . . , 2bf − 1}, 3bf −1
y3bf = xf ,
f (yt , vt ) ≤ bf f (xf , uf ) + /4.
t =2bf
In view of the relations above and Theorem 2.1, 3bf −1
3bf f (xf , uf ) ≤
f (yt , vt )
t =0 bf −1
=
bf −1
f (yt , vt ) +
t =0
3bf −1
f (xt , ut ) +
t =0
f (yt , vt ),
t =2bf
bf −1
f (xt , ut ) ≥ 3bf f (xf , uf ) − 2bf f (xf , uf ) − /2 ≥ bf f (xf , uf ) − .
t =0
Lemma 2.55 is proved. Proposition 2.56. Assume that f has LSC property and (P1) and that z ∈ ∪{AL : L = 1, 2, . . . }. Then there exists an (f )good and (f )minimal pair ∗ ∗ ∞ ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) such that x0 = z. Proof. Theorem 2.1 implies that f possesses (P2) and TP. There exists an integer L0 > 0 such that z ∈ AL0 .
(2.143)
2.13 Auxiliary Results for Theorems 2.13 and 2.14
73
It follows from Theorem 2.1 that there exists S0 > 0 such that for each pair of 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), integers T2 > T1 ≥ 0 and each ({xt }Tt =T 1 1 T 2 −1
f (xt , ut ) + S0 ≥ (T2 − T1 )f (xf , uf ).
(2.144)
t =T1
LSC property and (2.143) imply that for each integer k ≥ L0 there exists (k) (k) ({xt }kt=0 , {ut }k−1 t =0 ) ∈ X(0, k) satisfying (k)
x0 = z, k−1
(2.145)
f f (xt(k) , u(k) t ) = σ (0, k, z).
(2.146)
t =0
In view of (2.143), for each integer k ≥ L0 , σ f (0, k, z) ≤ L0 + f (xf , uf )(k − L0 ).
(2.147)
By (2.144), (2.146), and (2.147), for each integer k ≥ L0 and each pair of integers T1 , T2 ∈ [0, k] satisfying T1 < T2 , T 2 −1
(k)
(k)
f (xt , ut ) =
t =T1
−
k−1
(k)
(k)
f (xt , ut )
t =0
(k) (k) {f (xt , ut ) : t ∈ {0, . . . , T1 } \ {T1 }}
−
(k) (k) {f (xt , ut ) : t ∈ {T2 , . . . , k} \ {k}}
≤ L + f (xf , uf )(k − L0 ) − T1 f (xf , uf ) + S0 − (k − T2 )f (xf , uf ) + S0 ≤ (T2 − T1 )f (xf , uf ) + L0 (1 − f (xf , uf )) + 2S0 .
(2.148)
By (2.1), (2.148), and LSC property, extracting subsequences, using the diagonalization process, and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 such that k1 ≥ L0 and for each integer (kp )
i ≥ 0, there exists limp→∞ f (xi X(i, i + 1) such that (kp )
xi
(kp )
, ui
(k )
(i) ) and there exists ({yi(i) , yi+1 }, {vi(i) }) ∈
p (i) → yi(i) , xi+1 → yi+1 as p → ∞,
(2.149)
74
2 DiscreteTime Autonomous Problems (kp )
f (yi(i) , vi(i) ) ≤ lim f (xi p→∞
(kp )
, ui
).
(2.150)
∗ ∞ In view of (2.149), there exists ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) such that for each integer i ≥ 1,
(xi∗ , u∗i ) = (yi , vi ). (i)
(i)
(2.151)
It follows from (2.148), (2.150), and (2.151) that for every integer q ≥ 1 q−1
f (xt∗ , u∗t ) = lim
q−1
p→∞
i=0
(kp )
f (xt
(kp )
, ut
) ≤ qf (xf , uf )+L0 (1−f (xf , uf ))+2S0 .
t =0
(2.152) ∗ }∞ ) is (f )good. In view of , {u Theorem 2.1 and (2.152) imply that ({xt∗ }∞ t t =0 t =0 (2.145), (2.149), and (2.151), x0∗ = z.
(2.153)
In order to complete the proof of the proposition, it is sufficient to show that ∗ ∞ ({xt∗ }∞ t =0 , {ut }t =0 ) is (f )minimal. Assume the contrary. Then there exist Δ > 0, 0 0 −1 an integer τ0 ≥ 1, and ({yt }τt =0 , {vt }τt =0 ) ∈ X(0, τ0 ) such that y0 = x0∗ , yτ0 = xτ∗0 , τ 0 −1
f (xt∗ , u∗t ) >
t =0
τ 0 −1
f (yt , vt ) + 2Δ.
(2.154)
(2.155)
t =0
By (A2) and Lemma 2.55, there exists δ > 0 such that the following properties hold: (ii) for each ξ ∈ A satisfying ρE (ξ, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ] and τ1 −1 (2) τ2 τ2 −1 1 ({x˜t(1)}τt =0 , {u˜ (1) ˜ (2) t }t =0 ) ∈ X(0, τ1 ), ({x˜ t }t =0 , {u t }t =0 ) ∈ X(0, τ2 )
such that (1)
= xf , x˜0 = ξ , x˜τ(1) 1
τ 1 −1
(1)
(1)
(2)
(2)
f (x˜t , u˜ t ) ≤ τ1 f (xf , uf ) + Δ/8,
t =0
(2)
x˜0 = xf , x˜τ(2) = ξ, 2
τ 2 −1 t =0
f (x˜t , u˜ t ) ≤ τ2 f (xf , uf ) + Δ/8;
2.13 Auxiliary Results for Theorems 2.13 and 2.14
75
b −1
b
f f (iii) for each ({xt }t =0 , {ut }t =0 ) ∈ X(0, bf ) which satisfies ρE (x0 , xf ) ≤ δ, ρE (xbf , xf ) ≤ δ, we have
bf −1
f (xt , ut ) ≥ bf f (xf , uf ) − Δ/8.
t =0
Theorems 2.1 and 2.11, TP, (P1), LSC property, (2.146), and (2.147) imply that there exists an integer L1 > L0 such that for each integer k ≥ L0 + 2L1 , (k)
ρE (xt , xf ) ≤ δ, t ∈ {L1 , . . . , k − L1 }.
(2.156)
In view of (2.149) and (2.151), ρE (xt∗ , xf ) ≤ δ for all integers t ≥ L1 .
(2.157)
By (2.150) and (2.151), there exists a natural number p0 such that kp0 > L0 + 2L1 + 2τ0 + 2 + 2bf , τ0 +L 1 −1
f (xt∗ , u∗t )
≤
τ0 +L 1 −1
t =0
(kp0 )
f (xt
(kp0 )
, ut
(2.158)
) + Δ/2.
(2.159)
t =0 kp
kp −1
0 0 , {ut }t =0 ) ∈ Property (ii), (2.154), and (2.157) imply that there is ({xt }t =0 X(0, kp0 ) such that
xt = yt , t = 0, . . . , τ0 , ut = vt , t = 0, . . . , τ0 − 1,
(2.160)
xt = xt∗ , t = τ0 , . . . , τ0 + L1 , ut = u∗t , t = τ0 , . . . , τ0 + L1 − 1,
(2.161)
τ0 +L1 +bf −1
xτ0 +L1 +bf = xf ,
f (xt , ut ) ≤ bf f (xf , uf ) + Δ/8,
(2.162)
t =τ0 +L1
(kp0 )
xt = xt (kp0 )
ut = ut
, t = τ0 + L1 + 2bf , . . . , kp0 ,
, t = τ0 + L1 + 2bf , . . . , kp0 − 1,
(2.163)
τ0 +L1 +2bf −1
t =τ0 +L1 +bf
f (xt , ut ) ≤ bf f (xf , uf ) + Δ/8.
(2.164)
76
2 DiscreteTime Autonomous Problems
By (2.145), (2.146), (2.153), (2.154), and (2.160), kp0 −1
kp0 −1
f (xt , ut ) ≥
t =0
(kp0 )
f (xt
(kp0 )
, ut
(2.165)
).
t =0
It follows from (2.155), (2.156), (2.158)–(2.160), (2.163), (2.165), and property (iii) that kp0 −1
0≤
kp0 −1
f (xt , ut ) −
t =0
τ0 +L1 +2bf −1
f (xt , ut ) −
t =0
≤
τ 0 −1
(kp0 )
f (xt
)
(kp0 )
, ut
)
t =0
f (yt , vt ) + 2bf f (xf , uf ) + Δ/4 +
τ0 +L 1 −1
f (xt∗ , u∗t )
t =τ0
t =0
−
(kp0 )
, ut
t =0
τ0 +L1 +2bf −1
=
(kp0 )
f (xt
τ0 +L 1 −1
(kp0 )
f (xt
(kp0 )
, ut
τ0 +L1 +bf −1
)−
t =0
(kp0 )
f (xt
(kp0 )
, ut
)
t =τ0 +L1 τ0 +L1 +2bf −1
−
(kp0 )
f (xt
(kp0 )
, ut
)
t =τ0 +L1 +bf
≤
τ 0 −1
f (yt , vt ) +
τ0 +L 1 −1 t =τ0
t =0
−
f (xt∗ , u∗t ) + 2bf f (xf , uf ) + Δ/4
τ0 +L 1 −1
(kp0 )
f (xt
(kp0 )
, ut
) − 2bf f (xf , uf ) + Δ/4
t =0
≤
τ 0 −1
f (xt∗ , u∗t ) − 2Δ +
τ0 +L 1 −1 t =τ0
t =0
−
τ0 +L 1 −1
f (xt∗ , u∗t )
(kp0 )
f (xt
(kp0 )
, ut
) + Δ/2 ≤ −Δ,
t =0
a contradiction. The contradiction we have reached completes the proof of Proposition 2.56.
2.14 Proofs of Theorems 2.13 and 2.14
77
2.14 Proofs of Theorems 2.13 and 2.14 Proof of Theorem 2.14. Clearly, (i) implies (ii). In view of Theorem 2.1, (ii) implies (iii). By (P1), (iii) implies (iv). Evidently (iv) implies (v). We show that (v) implies ∗ ∞ (iii). Assume that ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )minimal and satisfies lim inf ρE (xf , xt∗ ) = 0.
(2.166)
lim ρE (x˜t , xf ) = 0.
(2.167)
t →∞
(P1) implies that t →∞
In view of Theorem 2.1, there exists S0 > 0 such that for all natural numbers T , 
T −1
f (x˜t , u˜ t ) − Tf (xf , uf ) ≤ S0 .
(2.168)
t =0
By (A2), there exists δ > 0 such that the following property holds: for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ] and (i) τi (i) τi −1 ({xt }t =0 , {ut }t =0 ) ∈ X(0, τi ), i = 1, 2 which satisfy (1)
(2)
x0 = z, xτ(1) = xf , x0 = xf , xτ(2) = z, 1 2 τ i −1
f (xt(i), u(i) t ) ≤ τi f (xf , uf ) + 1, i = 1, 2.
t =0
In view of (2.166) and (2.167), there exists an increasing sequence of natural numbers {tk }∞ k=1 such that ρE (x˜t , xf ) ≤ δ for all integers t ≥ t0 ,
(2.169)
lim tk = ∞,
k→∞
ρE (xt∗k +2bf , xf ) ≤ δ, k = 1, 2, . . . .
(2.170)
Let k ≥ 1 be an integer. By (A2), the choice of δ, (2.169), and (2.170), there exists t +2b t +2b −1 ({yt }tk=0 f , {vt }tk=0 f ) ∈ X(0, tk + 2bf ) such that yt = x˜t , t = 0, . . . , tk , vt = u˜ t , t = 0, . . . , tk − 1 ytk +bf = xf ,
(2.171) (2.172)
78
2 DiscreteTime Autonomous Problems tk +bf −1
f (yt , vt ) ≤ bf f (xf , uf ) + 1,
(2.173)
t =tk
ytk +2bf = xt∗k +2bf ,
(2.174)
tk +2bf −1
f (yt , vt ) ≤ bf f (xf , uf ) + 1.
(2.175)
t =tk +bf
Property (v), (2.168), and (2.171)–(2.175) imply that tk +2bf −1
f (xt∗ , u∗t ) ≤
tk +2bf −1
t =0
f (yt , vt ) ≤
t =0
t k −1
f (x˜t , u˜ t ) + 2bf f (xf , uf ) + 2
t =0
≤ tk f (xf , uf ) + S0 + 2bf f (xf , uf ) + 2 = (tk + 2bf )f (xf , uf ) + 2 + S0 . ∗ ∞ Together with Theorem 2.1, this implies that ({xt∗ }∞ t =0 , {ut }t =0 ) is (f )good and (iii) holds. ∗ ∞ We show that (iii) implies (i). Assume that ({xt∗ }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )minimal and (f )good. (P1) implies that
lim ρE (xt∗ , xf ) = 0.
t →∞
(2.176)
In view of Theorem 2.1, there exists S0 > 0 such that 
T −1
f (xt∗ , u∗t ) − Tf (xf , uf ) ≤ S0 , T ∈ {1, 2, . . . }.
t =0 ∞ Let ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) satisfy
x0 = x0∗ .
(2.177)
We show that lim sup[ T →∞
T −1
f (xt∗ , u∗t ) −
t =0
T −1
f (xt , ut )] ≤ 0.
(2.178)
t =0
∞ In view of Theorem 2.1 and (iii), we may assume that ({xt }∞ t =0 , {ut }t =0 ) is (f )good. (P1) implies that
lim ρE (xt , xf ) = 0.
t →∞
(2.179)
2.14 Proofs of Theorems 2.13 and 2.14
79
Let > 0. By (A2) and Lemma 2.55, there exists δ ∈ (0, ) such that the following property holds: (a) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ] τi −1 i and ({xt(i) }τt =0 , {u(i) t }t =0 ) ∈ X(0, τi ), i = 1, 2 satisfying x0(1) = z , xτ(1) = xf , x0(2) = xf , xτ(2) = z, 1 2 τ i −1
(i)
(i)
f (xt , ut ) ≤ τi f (xf , uf ) + /8, i = 1, 2;
t =0 b −1
b
f f (b) for each ({yt }t =0 , {vt }t =0 ) ∈ X(0, bf ) which satisfies ρE (y0 , xf ) ≤ δ, ρE (ybf , xf ) ≤ δ, we have
bf −1
f (yt , vt ) ≥ bf f (xf , uf ) − /8.
t =0
It follows from (2.176) and (2.179) that there exists an integer τ0 > 0 such that ρE (xt , xf ) ≤ δ, ρE (xt∗ , xf ) ≤ δ for all integers t ≥ τ0 .
(2.180)
Let T ≥ τ0 be an integer. Property (a) and (2.180) imply that there exists T +2b T +2b −1 ({yt }t =0 f , {vt }t =0 f ) ∈ X(0, T + 2bf ) which satisfies yt = xt , t = 0, . . . , T , vt = ut , t = 0, . . . , T − 1, yT +bf = xf ,
(2.181)
T +bf −1
f (yt , vt ) ≤ bf f (xf , uf ) + /8,
(2.182)
t =T T +2bf −1
yT +2bf = xT∗ +2bf ,
f (yt , vt ) ≤ bf f (xf , uf ) + /8.
(2.183)
t =T +bf
By property (b) and (2.180), T +2bf −1
t =T +bf
T +2bf −1
f (xt∗ , u∗t ),
t =T +bf
f (xt , ut ) ≥ bf f (xf , uf ) − /8.
(2.184)
80
2 DiscreteTime Autonomous Problems
It follows from property (iii), (2.177), and (2.181)–(2.184) that T −1
T +2bf −1
f (xt∗ , u∗t ) + 2bf f (xf , uf ) − /4
≤
t =0
f (xt∗ , u∗t )
t =0
T +2bf −1
≤
f (yt , vt ) ≤
t =0
T −1
f (xt , ut ) + 2bf f (xf , uf ) + /4
t =0
and T −1
f (xt∗ , u∗t )
≤
t =0
T −1
f (xt , ut ) + /2 for all integers T ≥ τ0 .
t =0
Since is any positive number, we conclude that (2.178) holds and that the ∗ ∞ pair ({xt∗ }∞ t =0 , {ut }t =0 ) is (f )overtaking optimal. Thus (iii) implies (i) and Theorem 2.14 is proved. Theorem 2.14 and Proposition 2.56 imply Theorem 2.13.
2.15 Proofs of Theorems 2.15–2.17 Proof of Theorem 2.15. Theorem 2.11 implies TP. Lemma 2.53, TP, (P1), and (A2) imply that there exists δ ∈ (0, ) such that the following properties hold: (a) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist an integer τ0 ∈ (0, bf ] and (0) τ0 (0) τ0 −1 , {ut }t =0 ) ∈ X(0, τ0 ) such that ({xt }t =0 = xf , x0(0) = z , xτ(0) 0
τ 0 −1
f (xt(0) , u(0) t ) ≤ τ0 f (xf , uf ) + 1;
t =0
(b) for each integer T1 ≥ 0, each integer T2 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({yt }Tt =T 1 1
≥
T1 + 2bf , and each
ρE (yTi , xf ) ≤ δ, i = 1, 2, T 2 −1
f (yt , vt ) ≤ U f (T1 , T2 , yT1 , yT2 ) + δ,
t =T1
we have ρE (yt , xf ) ≤ , t = T1 , . . . , T2 .
2.15 Proofs of Theorems 2.15–2.17
81
Assertion (i) follows from (a) and Theorem 2.13. We prove Assertion (ii). ∞ Assume that ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )overtaking optimal and that ρE (x0 , xf ) ≤ δ.
(2.185)
Theorem 2.14 and (P1) imply that lim ρE (xt , xf ) = 0.
t →∞
There exists an integer t0 > 2bf such that for each integer t ≥ t0 , ρE (xt , xf ) ≤ δ.
(2.186)
Let T ≥ t0 be an integer. It follows from property (b), (2.185), and (2.186) that ρE (xt , xf ) ≤ , t = 0, . . . , T . Therefore Assertion (ii) holds and Theorem 2.15 is proved. Proof of Theorem 2.16. In view of Theorem 2.11, TP holds. Together with Proposition 2.6, this implies that there exists an integer τ0 > 0 such that the following property holds: −1 ) ∈ X(0, T ) which satisfies (c) for each integer T ≥ 2τ0 and each ({xt }Tt=0 , {ut }Tt =0 T −1
f (xt , ut ) = U f (0, T , x0 , xT ),
t =0 T −1
f (xt , ut ) ≤ Tf (xf , xf ) + L(1 + f (xf , uf )) + 2,
t =0
we have ρE (xt , xf ) ≤ , t = τ0 , . . . , T − τ0 . ∞ Let ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) be (f )overtaking optimal and
x0 ∈ AL .
(2.187)
Theorem 2.13 and (2.187) imply that lim ρE (xt , xf ) = 0.
t →∞
(2.188)
82
2 DiscreteTime Autonomous Problems
∞ In view of (2.188), there exists an integer s0 ∈ (0, L] and ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that
y0 = x0 , ys0 = xf ,
s 0 −1
f (yt , vt ) ≤ L,
(2.189)
yt = xf , ut = uf for all integers t ≥ s0 .
(2.190)
t =0
By (2.189) and (2.190), for all sufficiently large integers T > L + 2τ0 , T −1
f (xt , ut ) ≤
t =0
T −1
f (yt , vt ) + 1 ≤ 1 + L +
t =0
T −1
f (yt , vt )
t =s0
≤ 1 + L + (T − s0 )f (xf , uf ) ≤ Tf (xf , uf ) + 1 + L(1 + f (xf , uf )) and in view of property (c), ρE (xt , xf ) ≤ , t = τ0 , . . . , T − τ0 . Since the relation above holds for all sufficiently large natural numbers T , we conclude that ρE (xt , xf ) ≤ for all integers t ≥ τ0 . This completes the proof of Theorem 2.16. Proof of Theorem 2.17. We may assume without loss of generality that t1 = 0. ∞ Clearly, there exists ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that yt = xt , t = 0, . . . , t2 , vt = ut , t = 0, . . . , t2 − 1, yt +t2 = yt , vt +t2 = vt for all integers t ≥ 0.
(2.191)
Theorem 2.1 and (2.191) imply that t 2 −1
f (yt , vt ) ≥ t2 f (xf , uf ).
(2.192)
f (yt , vt ) = t2 f (xf , uf ).
(2.193)
t =0
We show that t 2 −1 t =0
2.15 Proofs of Theorems 2.15–2.17
83
Assume the contrary. Then in view of (2.191) and (2.192), t 2 −1
f (xt , ut ) =
t =0
t 2 −1
f (yt , vt ) > t2 f (xf , uf ).
(2.194)
t =0
Set Δ=
t 2 −1
f (xt , ut ) − t2 f (xf , uf ).
(2.195)
t =0
By (A2) and Lemma 2.55, there exists δ ∈ (0, 1) such that the following properties hold: (d) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ] τi −1 i and ({xt(i) }τt =0 , {u(i) t }t =0 ) ∈ X(0, τi ), i = 1, 2 satisfying x0(1) = z , xτ(1) = xf , x0(2) = xf , xτ(2) = z, 1 2 τ i −1
(i)
(i)
f (xt , ut ) ≤ τi f (xf , uf ) + Δ/8, i = 1, 2;
t =0 b
b −1
f f (e) for each ({zt }t =0 , {ξt }t =0 ) ∈ X(0, bf ) which satisfies ρE (z0 , xf ) ≤ δ, ρE (zbf , xf ) ≤ δ, we have
bf −1
f (zt , ξt ) ≥ bf f (xf , uf ) − Δ/8.
t =0
It follows from Theorem 2.14 and (P1) that there exists an integer T0 > 0 such that ρE (xt , xf ) ≤ δ for all integers t ≥ T0 .
(2.196)
T1 > 2T0 + 2t2 + 2bf + 4.
(2.197)
Choose an integer
ut } ∞ By (d) and (2.196), there exists ({ x t }∞ t =0 , { t =0 ) ∈ X(0, ∞) such that xt = xt +t2 , t = 0, . . . , T0 + t2 + 2,
(2.198)
ut = ut +t2 , t = 0, . . . , T0 + t2 + 1,
(2.199)
84
2 DiscreteTime Autonomous Problems T0 +t2 +bf +1
f ( xt , ut ) ≤ bf f (xf , uf ) + Δ/8,
(2.200)
t =T0 +t2 +2
xT0 +t2 +bf +2 = xf ,
(2.201)
xt = xf , t = T0 + t2 + 2 + bf , . . . , T0 + 2t2 + 2 + bf ,
(2.202)
ut = uf , t = T0 + t2 + 2 + bf , . . . , T0 + 2t2 + 1 + bf ,
(2.203)
xT0 +2t2 +2bf +2 = xT0 +2t2 +2bf +2 ,
(2.204)
T0 +2t2 +2bf +1
f ( xt , ut ) ≤ bf f (xf , uf ) + Δ/8,
(2.205)
t =T0 +2t2 +2+bf
ut = ut for all integers t ≥ T0 + 2t2 + 2 + 2bf . xt = xt ,
(2.206)
In view of (2.198), (2.206), and the relation x0 = xt2 , T0 +2t2 +2bf +1
T0 +2t2 +2bf +1
f (xt , ut ) ≤
t =0
(2.207)
f ( xt , ut ).
t =0
It follows from (2.198)–(2.200), (2.202), (2.203), and (2.205) that T0 +2t2 +2bf +1
f ( xt , ut ) ≤
T0 +2t 2 +1
f (xt , ut ) + bf f (xf , uf ) + Δ/8
t =t2
t =0
+t2 f (xf , uf ) + bf f (xf , uf ) + Δ/8 =
T0 +2t 2 +1
f (xt , ut ) + f (xf , uf )(t2 + 2bf ) + Δ/4.
t =t2
Property (e), (2.195), and (2.196) imply that T0 +2t2 +2bf +1
f (xt , ut ) =
t =0
t 2 −1
f (xt , ut ) +
t =T0 +2t2 +2
f (xt , ut )
t =t2
t =0
T0 +2t2 +bf +1
+
T0 +2t 2 +1
T0 +2t2 +2bf +1
f (xt , ut ) +
t =T0 +2t2 +2+bf
f (xt , ut )
(2.208)
2.16 Proof of Theorem 2.21
85
≥ t2 f (xf , uf ) + Δ +
T0 +2t 2 +1
f (xt , ut ) + 2bf f (xf , uf ) − Δ/4.
(2.209)
t =t2
By (2.207)–(2.209), T0 +2t2 +2bf +1
0≤
T0 +2t2 +1+2bf
f ( xt , ut ) −
t =0
≤ −Δ + Δ/8,
t =0
a contradiction. The contradiction we have reached proves that t 2 −1
f (yt , vt ) = t2 f (xf , uf ).
(2.210)
t =0 ∞ Theorem 2.1, (2.191), and (2.210) imply that ({yt }∞ t =0 , {vt }t =0 ) is (f )good. By (P1), limt →∞ ρE (yt , xf ) = 0. Together with (2.191) this implies that yt = xf for all integers t ≥ 0 and that xt = xf for all t ∈ {0, . . . , t2 }. Combined with Theorem 2.15, this implies that xt = xf for all integers t ≥ 0. Theorem 2.17 is proved.
2.16 Proof of Theorem 2.21 ∞ We show that (P1) holds. Let ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) be (fr )good. There exists S0 > 0 such that for all integers T ≥ 1,
S0 > 
T −1
fr (xt , ut ) − Tff (xf , uf )
t =0
=
T −1
f (xt , ut ) − Tf (xf , uf ) + r
t =0
T −1
φ(xt , xf ).
(2.211)
t =0
Theorem 2.1 and (2.211) imply that Δ := lim
T →∞
T −1
φ(xt , xf ) < ∞.
(2.212)
t =0
Property (a) and (2.212) imply (P1). We show that (P2) holds. Let M, > 0. By property (a) there exists γ ∈ (0, 1) such that if ξ ∈ A, φ(ξ, xf ) ≤ γ , then ρE (ξ, xf ) ≤ .
(2.213)
86
2 DiscreteTime Autonomous Problems
By Theorem 2.1, there exists S0 > 0 such that for each pair of integers T2 > T1 ≥ 0 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), we have and each ({xt }Tt =T 1 1 T 2 −1
f (xt , ut ) + S0 ≥ (T2 − T1 )f (xf , uf ).
(2.214)
t =T1
Choose a natural number L > (rγ )−1 (M + S0 ).
(2.215)
L−1 Let ({xt }L t =0 , {ut }t =0 ) ∈ X(0, L) satisfy L−1
fr (xt , ut ) ≤ Lf (xf , uf ) + M.
(2.216)
t =0
We show that there exists an integer s ∈ {0, . . . , L} such that ρE (xs , xf ) ≤ . Assume the contrary. By (2.213), for all t = 0, . . . , L, ρE (xs , xf ) > , φ(xs , xf ) ≥ γ .
(2.217)
It follows from (2.8), (2.214), (2.216), and (2.217) that M + Lf (xf , uf ) ≥
L−1
fr (xt , ut ) =
t =0
L−1
f (xt , ut ) + r
t =0
L−1
φ(xt , xf )
t =0
≥ Lf (xf , xf ) − S0 + rLγ , L ≤ (γ r)−1 (M + S0 ). This contradicts (2.215). The contradiction we have reached proves that there exists an integer s ∈ {0, . . . , L} such that ρE (xs , xf ) ≤ . Therefore (P2) holds for fr and Theorem 2.21 is proved.
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31 Proof of Proposition 2.25. In view of (2.20), π f (xf ) ≤ 0. Assume that π f (xf ) < (0) 0. By (2.20), there exists {xt }∞ t =0 ∈ X(0, ∞) such that (0)
x0 = xf ,
(2.218)
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31
lim inf[ T →∞
T −1 t =0
−1 f f (xt(0), xt(0) +1 ) − Tf (xf , xf )] < 2 π (xf ).
87
(2.219)
Theorem 2.1 and (2.219) imply that {xt(0)}∞ t =0 is (f )good. In view of (P1), lim ρE (xt(0), xf ) = 0.
t →∞
(2.220)
It follows from (2.219) that there exists a strictly increasing sequence of natural number {Tk }∞ k=1 such that T k −1 t =0
−1 f f (xt(0), xt(0) +1 ) − Tk f (xf , xf ) < 2 π (xf ), k = 1, 2, . . . .
(2.221)
By (A2), there exists δ ∈ (0, 1) such that the following property holds: (i) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ] (i) τi ∈ X(0, τi ), i = 1, 2 which satisfy and {yt }t =0 (1)
(2)
y0 = z, yτ(1) = xf , y0 = xf , yτ(2) = z, 1 2 τ i −1 t =0
(i)
(i)
f (yt , yt +1 ) ≤ τi f (xf , xf ) − π f (xf )/8, i = 1, 2.
In view of (2.220), there exists an integer τ∗ > 0 such that (0)
ρE (xt , xf ) ≤ δ for all integers t ≥ τ∗ .
(2.222)
Let k be a natural number such that Tk > τ∗ . Property (i), (2.218), and (2.222) imply that there exists {yt }∞ t =0 ∈ X(0, ∞) such that yt +Tk +bf = yt for all integers t ≥ 0, (0)
(2.223)
yt = xt , t = 0, . . . , Tk , yTk +bf = xf ,
(2.224)
f (yt , yt +1 ) ≤ bf f (xf , uf ) − π f (xf )/8.
(2.225)
Tk +bf −1
t =Tk
88
2 DiscreteTime Autonomous Problems
By (2.221), (2.223), and (2.225), Tk +bf −1
f (yt , yt +1 ) =
T k −1
t =0
Tk +bf −1
f (yt , yt +1 ) +
f (yt , yt +1 )
t =Tk
t =0
< Tk f (xf , xf ) + 2−1 π f (xf ) + bf f (xf , xf ) − π f (xf )/8 ≤ (Tk + bf )f (xf , xf ) + π f (xf )/4, lim inf[ T →∞
T −1
f (yt , yt +1 ) − Tf (xf , xf )] = −∞,
t =0
a contradiction which proves that π f (xf ) = 0 and Proposition 2.25 itself. Proof of Proposition 2.26. Let > 0. By (A2), there exists δ > 0 such that the following property holds: (ii) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist integers τ1 , τ2 ∈ [1, bf ] i and {xt(i)}τt =0 ∈ X(0, τi ), i = 1, 2 which satisfy x0(1) = z, xτ(1) = xf , x0(2) = xf , xτ(2) = z, 1 2 τ i −1 t =0
(i)
(i)
f (xt , xt +1 ) ≤ τi f (xf , uf ) + /8, i = 1, 2.
Let z ∈ A satisfy ρE (z, xf ) ≤ δ
(2.226)
i and let integers τ1 , τ2 ∈ (0, bf ] and {xt(i) }τt =0 ∈ X(0, τi ), i = 1, 2 be as guaranteed by property (ii). Set
(1)
x˜t
(1)
(1)
= xt , t = 0, . . . , τ1 − 1, x˜t
= xf for all integers t ≥ τ1 .
(2.227)
By (2.20) and (2.227), π f (z) ≤ lim inf[ T →∞
=
τ 1 −1 t =0
T −1 t =0
(1)
(1)
f (x˜t , x˜t +1 ) − Tf (xf , xf )]
f (xt(1), xt(1) +1 ) − τ1 f (xf , xf ) ≤ /8.
(2.228)
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31
89
Propositions 2.24 and 2.25, property (ii), and (2.228) imply that 0 = π f (xf ) ≤
τ 2 −1 t =0
(2)
(2)
f (xt , xt +1 ) − τ2 f (xf , xf ) + π f (z) ≤ /8 + π f (z), π f (z) ≥ −/8.
Thus π f (z) ≤ /8. Proposition 2.26 is proved. Proof of Proposition 2.27. Let M > 0 and suppose that the set {z ∈ A : π f (z) ≤ M} is nonempty. Propositions 2.25 and 2.26 imply that there exists δ > 0 such that for each z ∈ A satisfying ρE (z, xf ) ≤ δ, π f (z) is finite and π f (z) ≤ 1.
(2.229)
By Theorem 2.20, there exists an integer L0 > 0 such that the following property holds: (iii) for each integer T ≥ L0 and each {xt }Tt=0 ∈ X(0, T ) which satisfies T −1
f (xt , xt +1 ) ≤ Tf (xf , uf ) + M + 4
t =0
and each S ∈ {0, . . . , T − L0 }, we have min{ρE (xt , xf ) : t = S, . . . , S + L0 } ≤ δ. Lemma 2.49 implies that there exists M1 > 0 such that the following property holds: (iv) for each S ∈ {1, . . . , L0 + 1} and each {xt }St=0 ∈ X(0, S) which satisfies S−1
f (xt , xt +1 ) ≤ (1 + L0 )f (xf , uf ) + M + 4,
t =0
we have ρE (θ0 , xt ) ≤ M1 , t = 0, . . . , S − 1. Assume that z ∈ A satisfies π f (z) ≤ M.
(2.230)
90
2 DiscreteTime Autonomous Problems
In view of (2.20) and (2.230), there exists {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z, lim inf[ T →∞
T −1
f (xt , xt +1 ) − Tf (xf , xf )] < π f (z) + 1 ≤ M + 1.
(2.231) (2.232)
t =0
By property (iii) and (2.232), there exists t0 ∈ {1, . . . , L0 + 1}
(2.233)
ρE (xt0 , xf ) ≤ δ.
(2.234)
such that
It follows from (2.229) and (2.234) that π f (xt0 ) ≤ 1.
(2.235)
Proposition 2.24, (2.230)–(2.232), and (2.235) imply that π f (x0 ) ≤
t 0 −1
f (xt , xt +1 ) − t0 f (xf , xf ) + π f (xt0 )
t =0
≤ lim inf[ T →∞
T −1
f (xt , xt +1 ) − Tf (xf , xf )] < M + 1.
t =0
Together with (2.235) this implies that t 0 −1
f (xt , xt +1 ) ≤ t0 f (xf , xf ) + M + 2.
(2.236)
t =0
By (2.233) and (2.236), ρE (z, θ0 ) = ρE (x0 , θ0 ) ≤ M1 . Proposition 2.27 is proved. Proof of Proposition 2.29. Assume that {xt }∞ t =0 ∈ X(0, ∞) is (f )good. Propositions 2.25 and 2.26, (P1), and (2.20) imply that lim ρE (xt , xf ) = 0, lim π f (xt ) = 0
t →∞
t →∞
(2.237)
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31
91
and π f (xt ) < ∞ for all integers t ≥ 0. It follows from (2.26) and (2.27) that sup{Γ f ({xt }Tt=0 ) : T = 1, 2, . . . } = lim Γ f ({xt }Tt=0 ) T →∞
= lim (
T −1
T →∞
f (xt , xt +1 ) − Tf (xf , xf ) − π f (x0 )) < ∞.
t =0
Assume that Δ := lim Γ f ({xt }Tt=0 ) < ∞. T →∞
(2.238)
It follows from (2.21), (2.26), (2.27), and (2.238) that for all integers T > 0, π f (xT ) < ∞, and T −1 t =0
f (xt , xt +1 ) − Tf (xf , xf ) = Γ f ({xt }Tt=0 )
+π f (x0 ) − π f (xT ) ≤ Δ + π f (x0 ) + S∗ . In view of Theorem 2.1, {xt }∞ t =0 is (f )good. Assume that {xt }∞ t =0 is (f )good. By (2.26), (2.27), and (2.237), for all integers T ≥ 1, T −1 t =0
f (xt , xt +1 ) − Tf (xf , xf ) = Γ f ({xt }Tt=0 ) + π f (x0 ) − π f (xT ) → lim Γ f ({xt }Tt=0 ) + π f (x0 ). T →∞
Proposition 2.29 is proved. Proof of Proposition 2.31. Proposition 2.27 and (2.28) imply that there exists M1 > 0 such that Af ⊂ {z ∈ A : ρE (z, θ0 ) ≤ M1 }.
(2.239)
By Theorem 2.2, there exists M2 > 0 such that the following property holds: (v) for each integer T > 0 and each {xt }Tt=0 ∈ X(0, T ) satisfying T −1
f (xt , xt +1 ) ≤ Tf (xf , xf ) +  inf(π f ) + 3,
t =0
the inequality ρE (xf , xt ) ≤ M2 holds for all t = 0, . . . , T − 1.
92
2 DiscreteTime Autonomous Problems
Theorem 2.1 implies that there exists c1 > 0 such that for each integer T > 0 and each {xt }Tt=0 ∈ X(0, T ), T −1
f (xt , xt +1 ) ≥ Tf (xf , uf ) − c1 .
(2.240)
t =0
In view of (A2), there exists δ ∈ (0, 1) such that the following property holds: (vi) for each z ∈ A satisfying ρE (z, xf ) ≤ δ, there exist an integer τ ∈ [1, bf ] and {xt }τt=0 ∈ X(0, τ ) which satisfies x0 = z, xτ = xf ,
τ −1
f (xt , xt +1 ) ≤ τf (xf , uf ) + 1.
t =0
By Theorem 2.20, there exists an integer L0 ≥ 1 such that the following property holds: 2 (vii) for each integer T1 ≥ 0, each integer T2 ≥ T1 + L0 , and each {xt }Tt =T ∈ 1 X(T1 , T2 ) which satisfies T 2 −1
f (xt , xt +1 ) − (T2 − T1 )f (xf , uf ) ≤  inf(π f ) + 3 + c1
t =T1
there exists t0 ∈ {0, . . . , L0 } such that ρE (xt0 , xf ) ≤ δ. Set M0 = (L0 + bf )(1 + f (xf , xf )) +  inf(π f ) + 4 + c1 .
(2.241)
Let z ∈ Af .
(2.242)
π f (z) ≤ inf(π f ) + 1.
(2.243)
In view of (2.28) and (2.242),
Proposition 2.29, (2.20), and (2.243) imply that there exists {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z, lim inf[ T →∞
T −1 t =0
f (xt , xt +1 ) − Tf (xf , xf )]
(2.244)
2.17 Proofs of Propositions 2.25–2.27, 2.29, and 2.31
93
= lim Γ f ({xt }Tt=0 ) + π f (z) ≤ π f (z) + 1 ≤ inf(π f ) + 2. T →∞
(2.245)
By (2.245), there exists a natural number T0 such that for all integers T ≥ T0 , T −1
f (xt , xt +1) − Tf (xf , xf ) ≤ inf(π f ) + 3.
(2.246)
t =0
Property (v) and (2.246) imply that ρE (xt , xf ) ≤ M2 for all integers t ≥ 0.
(2.247)
It follows from (2.240) and (2.246) that for every integer T ∈ (0, T0 ), T −1
f (xt , xt +1 ) − Tf (xf , xf ) =
t =0
−(
T 0 −1
f (xt , xt +1 ) − T0 f (xf , xf )
t =0
{f (xt , xt +1 ) : t ∈ {T , . . . , T0 } \ {T0 }} − (T − T0 )f (xf , xf )) ≤ inf(π f ) + 3 + c1 .
Together with (2.246) this implies that for all integers T > 0, T −1
f (xt , xt +1 ) − Tf (xf , xf ) ≤ inf(π f ) + 3 + c1 .
(2.248)
t =0
Property (vii) and (2.248) imply that there exists an integer t0 ∈ [0, L0 ]
(2.249)
ρE (xt0 , xf ) ≤ δ.
(2.250)
such that
t +b
It follows from property (vi) and (2.250) that there exists {yt }t0=0 f ∈ X(0, t0 + bf ) such that yt = xt , t = 0, . . . , t0 , yt0 +bf = xf ,
(2.251)
t0 +bf −1
t =t0
f (yt , yt +1 ) ≤ bf f (xf , uf ) + 1.
(2.252)
94
2 DiscreteTime Autonomous Problems
In view of (2.241) and (2.249), t0 + bf ≤ L0 + bf ≤ M0 .
(2.253)
By (2.241), (2.248), (2.249), (2.251), and (2.252), t0 +bf −1
f (yt , yt +1 ) ≤ t0 f (xf , xf ) + inf(π f ) + 3 + c1 + bf f (xf , uf ) + 1
t =0
≤ f (xf , xf )(L0 + bf ) +  inf(π f ) + 4 + c1 ≤ M0 . Together with (2.244), (2.251), and (2.253), this implies that z ∈ AM0 . Proposition 2.31 is proved. Proof of Proposition 2.33. Let z ∈ AM0 . There exists {xt }∞ t =0 ∈ X(0, ∞) such that x0 = z, xt = xf for all integers t ≥ M0 , M 0 −1
f (xt , xt +1 ) ≤ M0 + M0 f (xf , uf ).
t =0
By the relations above, π (z) ≤ lim inf[ f
T →∞
T −1
f (xt , xt +1 ) − Tf (xf , xf )]
t =0
≤ M0 (1 + f (xf , xf )) + M0 f (xf , xf ). Thus AM0 ⊂ {z ∈ A : π f (z) ≤ 2M0 (1 + f (xf , uf ))} which is bounded by Proposition 2.27. This completes the proof of Proposition 2.33.
2.18 Proof of Proposition 2.37 Let {x (k)}∞ k=1 ⊂ A, x ∈ A and x = lim x (k). k→∞
(2.254)
We show that π f (x) ≤ lim infk→∞ π f (x (k) ). Extracting subsequences and reindexing, we may assume without loss of generality that there exists lim π f (x (k) ) < ∞
k→∞
(2.255)
2.18 Proof of Proposition 2.37
95
and that π f (x (k)) < ∞ for all integers k ≥ 1. For each integer k ≥ 1, there exists (k) an (f, M)overtaking optimal {yt }∞ t =0 ∈ X(0, ∞) such that (k)
y0 = x (k).
(2.256)
Proposition 2.36 implies that for every integer k ≥ 1, π f (x (k) ) = lim [ T →∞
T −1 t =0
(k)
(k)
f (yt , yt +1 ) − Tf (xf , xf )].
(2.257)
It follows from Proposition 2.35, (2.21), and (2.255) that for every integer T ≥ −1 (k) (k) f (yt , yt +1 )}∞ 1, the sequence { Tt =0 k=1 is bounded. By LSC property, extracting subsequences, using the diagonalization process, and reindexing, we may assume (ik ) ∞ that there exists {yt }∞ t =0 ∈ X(0, ∞) and a subsequence {yt }t =0 , k = 1, 2, . . . such that for every integer t ≥ 0, (ik )
yt
→ yt as k → ∞,
(2.258)
for every integer T ≥ 1, there exists lim
T −1
k→∞ T −1
(ik )
f (yt
t =0
f (yt , yt +1 ) ≤ lim
k→∞
t =0
(i )
k , yt +1 ),
T −1
(ik )
f (yt
t =0
(2.259)
(i )
k , yt +1 ).
(2.260)
Let > 0. By Propositions 2.25 and 2.26, there exists δ > 0 such that for each ξ ∈ A satisfying ρE (ξ, xf ) ≤ δ, we have π f (ξ ) ≤ /2.
(2.261)
Since the sequences {yt }∞ t =0 , k = 1, 2, . . . are (f, M)overtaking optimal, it follows from (2.255), (2.257), and Proposition 2.6 that there exists an integer L0 > 0 such that for each integer k ≥ 1 and each integer t ≥ L0 , (k)
(k)
ρE (yt , xf ) ≤ δ.
(2.262)
Proposition 2.35, (2.261), and (2.262) imply that for each integer k ≥ 1 and each integer T ≥ L0 , π f (yT(k) ) ≤ /2,
(2.263)
96
2 DiscreteTime Autonomous Problems T −1 t =0
f (k) f (k) f (yt(k) , yt(k) +1 ) = Tf (xf , xf ) + π (y0 ) − π (yT )
≤ Tf (xf , xf ) + π f (y0(k)) + /2.
(2.264)
In view of (2.255), (2.256), (2.260), and (2.264), for each integer T ≥ L0 , T −1
f (yt , yt +1 ) ≤ Tf (xf , xf ) + lim π f (y0(k)) + /2, k→∞
t =0 T −1 t =0
f (yt , yt +1 ) − Tf (xf , xf ) ≤ lim π f (x (k) ) + /2. k→∞
(2.265)
By (2.254), (2.256), and (2.258), y0 = x. Together with (2.265) this implies that π f (x) ≤ limk→∞ π f (x (k) ) + /2. Since is any positive number, this completes the proof of Proposition 2.37.
2.19 Auxiliary Results for Theorems 2.38 and 2.39 ¯ in view Since the results obtained for the pair (f, M) also hold for the pair (f¯, M), of Theorem 2.8, we have the following result. Proposition 2.57. Assume that L, M > 0 are integers and that > 0. Then there exist δ > 0 and an integer L0 > L such that for each integer T1 ≥ 0, each integer 2 ¯ 1 , T2 ) which satisfies T2 ≥ T1 + 2L0 , and each {xt }Tt =T ∈ X(T 1 L , xT1 ∈ A T 2 −1 t =T1
f f f¯(xt , xt +1 ) ≤ min{σ− (T1 , T2 , xT1 ) + M, U− (T1 , T2 , xT1 , xT2 ) + δ},
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Proposition 2.57 implies the following result. Proposition 2.58. Assume that L, M > 0 are integers and that > 0. Then there exist δ > 0 and an integer L0 > 0 such that for each integer T1 ≥ 0, each integer 2 ∈ X(T1 , T2 ) which satisfies T2 ≥ T1 + 2L0 , and each {xt }Tt =T 1
2.20 Proofs of Theorems 2.38 and 2.39
97
L , xT2 ∈ A T 2 −1
f (xt , xt +1 ) ≤ min{ σ f (T1 , T2 , xT2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 .
2.20 Proofs of Theorems 2.38 and 2.39 We prove the following result. Theorem 2.59. Let L0 , M be natural numbers and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies M , xT ∈ A
T −1
f (xt , xt +1 ) ≤ σ f (0, T , xT ) + δ,
t =0
the inequalities 0 π f (x0 ) ≤ inf(π f ) + , Γ f ({xt }L t =0 ) ≤
hold. Proof. By Propositions 2.25 and 2.26, Lemma 2.55, and (A2), there exists δ1 ∈ (0, /4) such that: (i) for each z ∈ A satisfying ρE (z, xf ) ≤ 2δ1 , we have π f (z) ≤ /16; b
f (ii) for each {xt }t =0 ∈ X(0, bf ) which satisfies
ρE (x0 , xf ) ≤ 2δ1 , ρE (xbf , xf ) ≤ 2δ1 , we have bf −1
t =0
f (xt , xt +1 ) ≥ bf f (xf , xf ) − /16;
98
2 DiscreteTime Autonomous Problems
(iii) for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ 2δ1 , i = 1, 2, there exist an integer τ ∈ (0, bf ] and {xt }τt=0 ∈ X(0, τ ) which satisfies x0 = z1 , xτ = z2 and τ −1
f (xt , xt +1 ) ≤ τf (xf , xf ) + /16.
t =0
By Proposition 2.58, there exist δ2 ∈ (0, δ1 /8) and an integer l0 > 0 such that the following property holds: 2 (iv) for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2l0 , and each {xt }Tt =T ∈ 1 X(T1 , T2 ) which satisfies
M , xT2 ∈ A T 2 −1
f (xt , xt +1 ) ≤ σ f (T1 , T2 , xT2 ) + δ2 ,
t =T1
we have ρE (xt , xf ) ≤ δ1 , t = T1 + l0 , . . . , T2 − l0 . Corollary 2.30 and (2.21) imply that there exists z∗ ∈ A such that π f (z∗ ) ≤ inf(π f ) + δ2 /8,
(2.266)
and {xt∗ }∞ t =0 ∈ X(0, ∞) for which x0∗ = z∗ , lim Γ f ({xt∗ }Tt=0 ) ≤ δ2 /8. T →∞
(2.267)
By Proposition 2.29, (P1), (2.266), and (2.267), there exists an integer l1 > 0 such that ρE (xt∗ , xf ) ≤ δ2 /8 for all integers t ≥ l1 .
(2.268)
δ ∈ (0, δ2 /4)
(2.269)
L1 > 2L0 + 2l0 + 2l1 + 2bf + 8.
(2.270)
Choose
and an integer
2.20 Proofs of Theorems 2.38 and 2.39
99
Assume that T ≥ L1 is an integer and that {xt }Tt=0 ∈ X(0, T ) satisfies M , xT ∈ A T −1
(2.271)
f (xt , xt +1 ) ≤ σ f (0, T , xT ) + δ.
(2.272)
ρE (xt , xf ) ≤ δ1 , t = l0 , . . . , T − l0 .
(2.273)
t =0
By (2.269)–(2.272),
In view of (2.270) and (2.273), [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8] ⊂ [l0 , T − l0 − l1 − L0 ],
(2.274)
ρE (xt , xf ) ≤ δ1 , t = l0 + l1 + L0 , . . . , l0 + l1 + L0 + 2bf + 8.
(2.275)
Property (iii), (2.268), and (2.275) imply that there exist integers τ1 , τ2 ∈ (0, bf ] and {xt(1)}Tt=0 ∈ X(0, T ) such that xt(1) = xt∗ , t = 0, . . . , l0 + l1 + L0 + 4, xl(1) = xf , 0 +l1 +L0 +4+τ1 l0 +l1 +L 0 +3+τ1 t =l0 +l1 +L0 +4
(1)
(1)
f (xt , xt +1 ) ≤ τ1 f (xf , uf ) + /16, (1)
xl0 +l1 +L0 +4+2bf −τ2 = xf , xl(1) = xl0 +l1 +L0 +4+2bf , 0 +l1 +L0 +4+2bf l0 +l1 +L0 +3+2bf
t =l0 +l1 +L0 +4+2bf −τ2 (1)
xt
f (xt(1), xt(1) +1 ) ≤ τ2 f (xf , uf ) + /16,
= xf , t = l0 + l1 + L0 + 4 + τ1 , . . . , l0 + l1 + L0 + 4 + 2bf − τ2 , (1)
xt
= xt , t = l0 + l1 + L0 + 4 + 2bf , . . . , T .
(2.276)
100
2 DiscreteTime Autonomous Problems
It follows from (2.272) and (2.276) that T −1
δ≥
f (xt , xt +1 ) −
T −1
t =0
t =0
l0 +l1 +L0 +3+2bf
=
l0 +l1 +L0 +3+2bf
f (xt , xt +1 ) −
t =0
=
f (xt(1) , xt(1) +1 )
t =0
l0 +l 1 +L0 +3
f (xt(1) , xt(1) +1 )
l0 +l1 +L0 +3+bf
f (xt , xt +1 ) +
t =0
f (xt , xt +1 )
t =l0 +l1 +L0 +4 l0 +l1 +L0 +3+2bf
+
f (xt , xt +1 )
t =l0 +l1 +L0 +4+bf
−
l0 +l 1 +L0 +3 t =0
f (xt∗ , xt∗+1 ) − τ1 f (xf , xf ) − /16
− τ2 f (xf , xf ) − /16 − f (xf , xf )(2bf − τ1 − τ2 ).
(2.277)
Property (iii) and (2.275) imply that l0 +l1 +L0 +3+bf
f (xt , xt +1 ) ≥ bf f (xf , xf ) − /16,
(2.278)
t =l0 +l1 +L0 +4 l0 +l1 +L0 +3+2bf
f (xt , xt +1 ) ≥ bf f (xf , xf ) − /16.
(2.279)
t =l0 +l1 +L0 +4+bf
In view of (2.277)–(2.279), δ≥
l0 +l1 +L0 +3
f (xt , xt +1 ) + 2bf f (xf , xf ) − /8
t =0
−
l0 +l1 +L0 +3 t =0
f (xt∗ , xt∗+1 ) − 2bf f (xf , xf ) − /8.
(2.280)
2.20 Proofs of Theorems 2.38 and 2.39
101
By property (ii), (2.26), (2.266), (2.267), (2.275), and (2.280), l0 +l1 +L0 +3
f (xt , xt +1 ) ≤ /4 + /16 +
t =0
l0 +l1 +L0 +3 t =0
l +l +L0 +4
= 5/16 + Γ f ({xt∗ }t0=0 1
f (xt∗ , xt∗+1 )
) + (l0 + l1 + L0 + 4)f (xf , xf )
+π f (x0∗ ) − π f (xl∗0 +l1 +L0 +4 ) l +l +L0 +4
≤ Γ f ({xt∗ }t0=0 1
) + (l0 + l1 + L0 + 4)f (xf , xf ) + π f (x0∗ ) + 6/16
≤ δ2 /8 + (l0 + l1 + L0 + 4)f (xf , xf ) + π f (x0∗ ) + 3/8 ≤ inf(π f ) + δ2 /4 + (l0 + l1 + L0 + 4)f (xf , xf ) + 3/8.
(2.281)
It follows from (2.26), (2.275), (2.281), and property (i) that inf(π f ) + (l0 + l1 + L0 + 4)f (xf , xf ) ≥ −3/8 − δ2 /4 +
l0 +l1 +L0 +3
f (xt , xt +1 )
t =0 l +l +L0 +4
≥ −/2 + Γ f ({xt }t0=0 1
) + (l0 + l1 + L0 + 4)f (xf , xf )
+π f (x0 ) − π f (xl0 +l1 +L0 +4 ) l +l +L0 +4
≥ −/2 + Γ f ({xt }t0=0 1
) + (l0 + l1 + L0 + 4)f (xf , xf ) + π f (x0 ) − /16, l +l +L0 +4
π f (x0 ) + Γ f ({xt }t0=0 1
) ≤ 9/16 + inf(π f ).
The inequality above implies that +l1 +L0 +4 π f (x0 ) ≤ inf(π f ) + , Γ f ({xt }lt0=0 ) ≤ .
Theorem 2.59 is proved. ¯ and ProposiTheorem 2.38 follows from Theorem 2.59, applied for (f¯, M), tion 2.23. Theorem 2.40 follows from Proposition 2.26 and Theorem 2.38 and Theorem 2.59.
102
2 DiscreteTime Autonomous Problems
2.21 Proofs of Theorem 2.40 and 2.41 We prove the following result. Theorem 2.60. Let (f, M) have LSC property, L0 , M > 0 be integers and ∈ (0, 1). Then there exist δ > 0 and an integer L1 > L0 such that for each integer T ≥ L1 and each {xt }Tt=0 ∈ X(0, T ) which satisfies M , xT ∈ A
T −1
f (xt , xt +1 ) ≤ σ f (0, T , xT ) + δ,
t =0
there exists an (f, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that π f (x0∗ ) = inf(π f ), ρE (xt∗ , xt ) ≤ , t = 0, . . . , L0 . ¯ Note that Theorem 2.40 follows from Theorem 2.60, applied for (f¯, M), and Proposition 2.23. Theorem 2.41 follows from Theorems 2.40 and 2.60 and Proposition 2.6. Proposition 2.61. Let (f, M) have LSC property, L0 > 0 be an integer, and ∈ 0 (0, 1). Then there exist δ ∈ (0, ) such that for each {xt }L t =0 ∈ X(0, L0 ) which satisfies 0 π f (x0 ) ≤ inf(π f ) + δ, Γ f ({xt }L t =0 ) ≤ δ,
there exists an (f, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that π f (x0∗ ) = inf(π f ), ρE (xt∗ , xt ) ≤ , t = 0, . . . , L0 . Proof. Assume that the proposition does not hold. Then there exist a sequence (k) L0 {δk }∞ k=1 ⊂ (0, 1] and a sequence {xt }t =0 ∈ X(0, L0 ), k = 1, 2, . . . such that lim δk = 0
(2.282)
π f (x0 ) ≤ inf(π f ) + δk ,
(2.283)
k→∞
and that for all integers k ≥ 1, (k)
(k) L
0 Γ f ({xt }t =0 ) ≤ δk
(2.284)
and the following property holds: f (i) for each (f, M)overtaking optimal {yt }∞ t =0 ∈ X(0, ∞) satisfying π (y0 ) = inf(π f ), we have max{ρE (xt(k) , yt ) : t = 0, . . . , L0 } >
2.21 Proofs of Theorem 2.40 and 2.41
103
In view of (2.26), (2.282)–(2.284), and boundedness from below of the 0the (k) (k) −1 function π f (see (2.21)), the sequence { L f (xt , xt +1 )}∞ k=1 is bounded. By t =0 LSC property, extracting a subsequence and reindexing if necessary, we may 0 assume without loss of generality that there exists {xt }L t =0 ∈ X(0, L0 ) such that xt(k) → xt as k → ∞ for all t = 0, . . . , L0 , L 0 −1
f (xt , xt +1 ) ≤ lim inf k→∞
t =0
L 0 −1 t =0
(2.285)
f (xt(k), xt(k) +1 ).
(2.286)
The lower semicontinuity of π f (see Proposition 2.37) and (2.285) imply that (k)
π f (x0 ) ≤ lim inf π f (x0 ) = inf(π f ),
(2.287)
π f (x0 ) = inf(π f ).
(2.288)
k→∞
By the lower semicontinuity of π f (see Proposition 2.37) and (2.285), (k)
π f (xL0 ) ≤ lim inf π f (xL0 ).
(2.289)
k→∞
It follows from (2.26), (2.27), (2.282)–(2.284), and (2.286)–(2.288) that L 0 −1
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 )
t =0
≤ lim inf(
L 0 −1
k→∞
t =0
≤ lim inf( k→∞
(k)
(k)
(k)
(k)
f (xt , xt +1 ) − L0 f (xf , xf )) − lim π f (x0 ) + lim inf π f (xL0 ) k→∞
L 0 −1 t =0
(k)
(k)
k→∞
(k)
(k)
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 )) ≤ 0. (2.290)
In view of (2.26), (2.27), and (2.290), L 0 −1
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 ) = 0.
(2.291)
t =0
Theorem 2.13 and the relation π f (xT0 ) < ∞ imply that there exists an (f, M)overtaking optimal {x˜t }∞ t =0 ∈ X(0, ∞) such that x˜0 = xL0 .
(2.292)
104
2 DiscreteTime Autonomous Problems
For all integers t > L0 set xt = x˜t −L0 .
(2.293)
It is clear that {xt }∞ t =0 ∈ X(0, ∞) is (f )good. By Proposition 2.35, (2.26), (2.27), (2.291), and (2.292), for all integers S > 0, S−1
f (xt , xt +1 ) − Sf (xf , xf ) − π f (x0 ) + π f (xS ) = 0.
(2.294)
t =0
In view of (2.288) and (2.294), {xt }∞ t =0 ∈ X(0, ∞) is (f, M)overtaking optimal and (2.288) holds. By (2.285), for all sufficiently large natural numbers k, (k) ρE (xt , xt ) ≤ /2 for all t = 0, . . . , L0 . This contradicts property (i). The contradiction we have reached proves Proposition 2.61. Theorem 2.60 follows from Proposition 2.61 and Theorem 2.59.
2.22 The First Bolza Problem We consider the control system introduced in Section 2.4 and use the notation, definitions, and assumptions used there and in Section 2.5. Let a1 > 0. Denote by A(A) the set of all lower semicontinuous functions h : A → R 1 which are bounded on bounded subsets of A and such that h(z) ≥ −a1 for all z ∈ A.
(2.295)
The set A(A) is equipped with the uniformity which is determined by the base E(N, ) = {(h1 , h2 ) : A(A) × A(A) : h1 (z) − h2 (z) ≤ for all z ∈ A satisfying ρE (z, θ0 ) ≤ N},
(2.296)
where N, > 0. It is not difficult to see that the uniform space A(A) is metrizable and complete. For each pair of integers T2 > T1 ≥ 0, each y ∈ A, and each h ∈ A(A), we define σ f,h (T1 , T2 , y) = inf{
T 2 −1
f (xt , xt +1 ) + h(xT2 ) :
t =T1 2 {xt }Tt =T 1
∈ X(T1 , T2 ), xT1 = y},
(2.297)
2.22 The First Bolza Problem
σ f,h (T1 , T2 , y) = inf{
105 T 2 −1
f (xt , xt +1 ) + h(xT1 ) :
t =T1 2 ∈ X(T1 , T2 ), xT2 = y}, {xt }Tt =T 1
f,h
σ− (T1 , T2 , y) = inf{
T 2 −1
f¯(xt , xt +1 ) + h(xT2 ) :
t =T1
2 ¯ 1 , T2 ), xT1 = y}, {xt }Tt =T ∈ X(T 1
f,h
σ− (T1 , T2 , y) = inf{
(2.298)
T 2 −1
(2.299)
f¯(xt , xt +1 ) + h(xT1 ) :
t =T1
2 ¯ 1 , T2 ), xT2 = y}. ∈ X(T {xt }Tt =T 1
(2.300)
The next result follows from (2.12), (2.13), and (2.15). 2 Proposition 2.62. Let S2 > S1 ≥ 0 be integers, M ≥ 0, h ∈ A(A) and {xt }St =S ∈ 1 X(S1 , S2 ). Then the following assertions hold:
S 2 −1
f (xt , xt +1 ) + h(xS2 ) ≤ σ f,h (S1 , S2 , xS1 ) + M if and only if
t =S1 S 2 −1 t =S1 S 2 −1
f,h f¯(x¯t , x¯t +1 ) + h(x¯S1 ) ≤ σ− (S1 , S2 , x¯S2 ) + M;
f (xt , xt +1 ) + h(xS1 ) ≤ σ f,h (S1 , S2 , xS2 ) + M if and only if
t =S1 S 2 −1 t =S1
f,h f¯(x¯t , x¯t +1 ) + h(x¯S2 ) ≤ σ− (S1 , S2 , x¯S1 ) + M.
The next result is proved in Section 2.23. Theorem 2.63. Assume that (f, M) has TP, that M, L > 0 are integers, and that > 0. Then there exist δ > 0 and an integer L0 > L such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L0 , each h ∈ A(A) satisfying h(xf ) ≤ M, and 2 each {xt }Tt =T ∈ X(T1 , T2 ) which satisfies 1
106
2 DiscreteTime Autonomous Problems
xT1 ∈ AL , T 2 −1
f (xt , ut ) + h(xT2 ) ≤ σ f,h (T1 , T2 , xT1 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ then τ1 = T1 and if ρE (xT2 , xf ) ≤ δ then τ2 = T2 . ¯ too, and combined with Note that Theorem 2.63 is valid for the triplet (f¯, h, M) Proposition 2.62, this implies the following result. Theorem 2.64. Assume that (f, M) has TP, that M, L > 0 are integers, and that > 0. Then there exist δ > 0 and an integer L0 > L such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L0 , each h ∈ A(A) satisfying h(xf ) ≤ M, and 2 each {xt }Tt =T ∈ X(T1 , T2 ) which satisfies 1 L , xT2 ∈ A T 2 −1
f (xt , xt +1 ) + h(xT1 ) ≤ σ f,h (T1 , T2 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Let g ∈ A(A) be given. By (2.21), (2.25), and (2.295), π f + g : A → R 1 ∪ {∞} is bounded from below and satisfies (π f + g)(z) → ∞ as z ∈ A, ρE (z, θ0 ) → ∞.
(2.301)
We prove the following results. Theorem 2.65. Let (f, M) have TP, L0 , M be natural numbers and ∈ (0, 1). Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A(A) such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies x0 ∈ AM ,
T −1 t =0
f (xt , xt +1 ) + h(xT ) ≤ σ f,h (0, T , x0 ) + δ,
2.22 The First Bolza Problem
107
the inequalities f
f
f
0 (g + π− )(xT ) ≤ inf(g + π− ) + , Γ− ({x¯t }L t =0 ) ≤
hold where x¯t = xT −t , t = 0, . . . , T . Theorem 2.66. Let (f, M) have TP, L0 , M be natural numbers and ∈ (0, 1). Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A(A) such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies M , xT ∈ A
T −1
f (xt , xt +1 ) + h(x0 ) ≤ σ f,h (0, T , xT ) + δ,
t =0
the inequalities 0 (g + π f )(x0 ) ≤ inf(g + π f ) + , Γ f ({xt }L t =0 ) ≤
hold. ¯ Together Theorem 2.66 is proved in Section 2.24. It is also valid for (f¯, M). with Proposition 2.62, this implies Theorem 2.65. Theorem 2.67. Let (f, M) have TP and LSC property, L0 , M > 0 be integers and ∈ (0, 1). Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A(A) such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies x0 ∈ AM ,
T −1
f (xt , xt +1 ) + h(xT ) ≤ σ f,h (0, T , x0 ) + δ,
t =0
¯ there exists an (f¯, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that (π− + g)(x0∗ ) = inf(π− + g), ρE (xt∗ , xT −t ) ≤ , t = 0, . . . , L0 . f
f
Theorem 2.68. Let (f, M) have TP and LSC property, M, L0 > 0 be integers and ∈ (0, 1). Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A(A) such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies M , xT ∈ A
T −1 t =0
f (xt , xt +1 ) + h(x0 ) ≤ σ f,h (0, T , xT ) + δ,
108
2 DiscreteTime Autonomous Problems
there exist an (f, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that (π f + g)(x0∗ ) = inf(π f + g), ρE (xt , xt∗ ) ≤ , t = 0, . . . , L0 . ¯ Together with PropoWe prove only Theorem 2.68. It is also valid for (f¯, M). sition 2.62, this implies Theorem 2.67. Theorem 2.68 follows from Theorem 2.66 and the following result which is proved in Section 2.25. Proposition 2.69. Let (f, M) have TP and LSC property, L0 > 0 be an integer, 0 and ∈ (0, 1). Then there exist δ ∈ (0, ) such that for each {xt }L t =0 ∈ X(0, L0 ) which satisfies L
0 (π f + g)(x0 ) ≤ inf(π f + g) + δ, Γ f ({xt }t =0 ) ≤ δ,
there exists an (f, M)overtaking optimal {xt∗ }∞ t =0 ∈ X(0, ∞) such that (π f + g)(x0∗ ) = inf(π f + g), ρE (xt∗ , xt ) ≤ , t = 0, . . . , L0 .
2.23 Proof of Theorem 2.63 By Proposition 2.6, there exist δ ∈ (0, 1) and an integer L0 > L such that the following property holds: 2 ∈ (i) for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L0 , and each {xt }Tt =T 1 X(T1 , T2 ) which satisfies
T 2 −1
f (xt , ut ) ≤ min{(T2 −T1 )f (xf , xf )+a1 +M +a0 +1+2(1+f (xf , xf )),
t =T1
U f (T1 , T2 , xT1 , xT2 ) + δ}, there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T 2 . Assume that T1 ≥ 0, T2 ≥ T1 + 2L0 are integers, h ∈ A(A) satisfies h(xf ) ≤ M,
(2.302)
2.23 Proof of Theorem 2.63
109
2 and {xt }Tt =T ∈ X(T1 , T2 ) satisfies 1
xT1 ∈ AL , T 2 −1
(2.303)
f (xt , xt +1 ) + h(xT2 ) ≤ σ f,h (T1 , T2 , xT1 ) + δ.
(2.304)
t =T1
By (2.304), T 2 −1
f (xt , xt +1 ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ.
(2.305)
t =T1 2 In view of (2.303), there exists {yt }Tt =T ∈ X(T1 , T2 ) such that 1
yT1 = xT1 , yt = xf for all integers t ∈ [L, T2 ], T1 +L−1
f (yt , yt +1 ) ≤ L + Lf (xf , xf ).
(2.306)
(2.307)
t =T1
It follows from (2.295), (2.304), and (2.306) that −a1 +
T 2 −1
f (xt , xt +1 ) ≤
t =T1
T 2 −1 t =T1
f (xt , xt +1 )+h(xT2 ) ≤
T 2 −1
f (yt , yt +1 )+h(xf )+1
t =T1
≤ h(xf ) + 1 + L(1 + f (xf , xf )) + (T2 − T1 − L)f (xf , xf ).
(2.308)
By (2.302) and (2.308), T 2 −1
f (xt , xt +1 ) ≤ a1 + 1 + M + 2L(1 + f (xf , xf )) + (T2 − T1 )f (xf , xf ).
t =T1
(2.309) In view of (2.305) and (2.309), there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Theorem 2.63 is proved.
110
2 DiscreteTime Autonomous Problems
2.24 Proof of Theorem 2.66 By Propositions 2.25 and 2.26, Lemma 2.55, and (A2), there exists δ1 ∈ (0, /4) such that: (i) for each z ∈ A satisfying ρE (z, xf ) ≤ 2δ1 , we have π f (z) ≤ /16; b
f (ii) for each {xt }t =0 ∈ X(0, bf ) which satisfies ρE (x0 , xf ) ≤ 2δ1 , ρE (xbf , xf ) ≤ 2δ1 , we have
bf −1
f (xt , xt +1 ) ≥ bf f (xf , uf ) − /16;
t =0
(iii) for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ 2δ1 , i = 1, 2, there exist an integer τ ∈ (0, bf ] and {xt }τt=0 ∈ X(0, τ ) which satisfies x0 = z1 , xτ = z2 and τ −1
f (xt , xt +1 ) ≤ τf (xf , xf ) + /16.
t =0
By (2.25), (2.295), and (2.296), there exists a neighborhood U1 of g in A(A) such that for each h ∈ U1 ,  inf(π f + g) − inf(π f + h) ≤ δ1 /16.
(2.310)
Theorem 2.64 implies that there exist δ2 ∈ (0, δ1 /8), an integer l0 > M, and a neighborhood U2 of g in A(A) such that the following property holds: (iv) for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2l0 , each h ∈ U2 , and each 2 ∈ X(T1 , T2 ) which satisfies {xt }Tt =T 1 M , xT2 ∈ A T 2 −1
f (xt , xt +1 ) + h(xT1 ) ≤ σ f h (T1 , T2 , xT2 ) + δ2 ,
t =T1
we have ρE (xt , xf ) ≤ δ1 , t = T1 + l0 , . . . , T2 − l0 . There exists z∗ ∈ A such that (π f + g)(z∗ ) ≤ inf(π f + g) + δ2 /8.
(2.311)
2.24 Proof of Theorem 2.66
111
Corollary 2.30 implies that there exists {xt∗ }∞ t =0 ∈ X(0, ∞) for which x0∗ = z∗ , lim Γ f ({xt∗ }Tt=0 ) ≤ δ2 /8. T →∞
(2.312)
In view of (P1) and (2.311), lim ρE (xt∗ , xf ) = 0.
t →∞
(2.313)
M are bounded. Equation (2.313) In view of Proposition 2.33, the sets AM and A implies that there exists an integer l1 > 0 such that ρE (xt∗ , xf ) ≤ δ2 /8 for all integers t ≥ l1 .
(2.314)
+l1 +L0 +4 ∈ X(0, l0 + By Theorem 3.2, there exists M˜ > 0 such that for each {xt }lt0=0 l1 + L0 + 4) satisfying l0 +l 1 +L0 +3
f (xt , xt +1 ) ≤ (l0 + l1 + L0 + 4)f (xf , xf ) + a1 + 2 + inf(π f + g),
t =0
we have ˜ t = 0, . . . , l0 + l1 + L0 + 3. ρE (xt , θ0 ) ≤ M,
(2.315)
In view of (2.296), there exists a neighborhood U of g in A(A) such that U ⊂ U1 ∩ U2 ,
(2.316)
h(x0∗ ) − g(x0∗ ) ≤ δ1 /16 for all h ∈ U,
(2.317)
˜ h(z)−g(z) ≤ δ1 /16 for all h ∈ U and all z ∈ AM ∪(A∩BE (θ0 , M)).
(2.318)
Choose δ ∈ (0, δ2 /4)
(2.319)
L1 > 2L0 + 2l0 + 2l1 + 2bf + 8.
(2.320)
and an integer
Assume that T ≥ L1 is an integer, h ∈ U,
(2.321)
112
2 DiscreteTime Autonomous Problems
{xt }Tt=0 ∈ X(0, T ) satisfies M , xT ∈ A T −1
f (xt , xt +1 ) + h(x0 ) ≤ σ f,h (0, T , xT ) + δ.
(2.322) (2.323)
t =0
Property (iv), (2.316), and (2.319)–(2.323) imply that ρE (xt , xf ) ≤ δ1 , t = l0 , . . . , T − l0 .
(2.324)
In view of (2.320), [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8] ⊂ [l0 , T − l0 − l1 − L0 ].
(2.325)
It follows from (2.324) and (2.325) that ρE (xt , xf ) ≤ δ1 , t = l0 + l1 + L0 , . . . , l0 + l1 + L0 + 2bf + 8.
(2.326)
Property (iii), (2.314), (2.320), and (2.326) imply that there exist integers τ1 , τ2 ∈ (0, bf ] and {xt(1)}Tt=0 ∈ X(0, T ) such that (1)
xt
= xt∗ , t = 0, . . . , l0 + l1 + L0 + 4, (1)
xl0 +l1 +L0 +4+τ1 = xf , l0 +l1 +L 0 +3+τ1 t =l0 +l1 +L0 +4
f (xt(1), xt(1) +1 ) ≤ τ1 f (xf , uf ) + /16,
xl(1) = xl0 +l1 +L0 +4+2bf , 0 +l1 +L0 +4+2bf (1)
xl0 +l1 +L0 +4+2bf −τ2 = xf ,
(2.327) (2.328)
(2.329)
(2.330) (2.331)
l0 +l1 +L0 +3+2bf
t =l0 +l1 +L0 +4+2bf −τ2
(1)
(1)
f (xt , xt +1 ) ≤ τ2 f (xf , uf ) + /16,
(2.332)
xt(1) = xf , t = l0 + l1 + L0 + 4 + τ1 , . . . , l0 + l1 + L0 + 4 + 2bf − τ2 ,
(2.333)
xt(1) = xt , t = l0 + l1 + L0 + 4 + 2bf , . . . , T .
(2.334)
2.24 Proof of Theorem 2.66
113
By (2.329) and (2.334), T −1
f (xt , xt +1 ) + h(x0 ) ≤
t =0
T −1 t =0
(1) f (xt(1), xt(1) +1 ) + h(x0 ) + δ.
(2.335)
It follows from (2.333)–(2.335) that δ≥
T −1
f (xt , xt +1 ) + h(x0 ) −
t =0
t =0
l0 +l1 +L0 +3+2bf
(1)
f (xt , xt +1 ) + h(x0 ) −
t =0
=
(1)
(1)
l0 +l1 +L0 +3
(1)
(1)
f (xt , xt +1) − h(x0 )
t =0 l0 +l1 +L0 +3+2bf
f (xt , xt +1) +
t =0
−
(1)
f (xt , xt +1 ) − h(x0 )
l0 +l1 +L0 +3+2bf
=
T −1
l0 +l1 +L0 +3 t =0
f (xt , xt +1 ) + h(x0 )
t =l0 +l1 +L0 +4
(1)
(1)
f (xt , xt +1 ) −
l0 +l1 +L 0 +3+τ1 t =l0 +l1 +L0 +4
(1)
(1)
f (xt , xt +1 )
l0 +l1 +L0 +3+2bf
− f (xf , xf )(2bf − τ1 − τ2 ) −
t =l0 +l1 +L0 +4+2bf −τ2
(1) f (xt(1), xt(1) +1 ) − h(x0 ).
(2.336) Property (ii) and (2.326) imply that l0 +l1 +L0 +3+2bf
f (xt , xt +1 ) ≥ 2bf f (xf , xf ) − /8.
(2.337)
t =l0 +l1 +L0 +4
By (2.26), (2.311), (2.312), (2.317), (2.319), (2.326), (2.327), (2.329), (2.332), (2.333), (2.336), (2.337), and property (i), l0 +l1 +L0 +3
f (xt , xt +1 ) + h(x0 ) ≤ δ + /4 +
t =0
l0 +l1 +L0 +3 t =0
= δ + /4 +
l0 +l 1 +L0 +3 t =0
f (xt , xt +1 ) + h(x0∗ ) (1)
(1)
f (xt∗ , xt∗+1 ) + h(x0∗ )
+l1 +L0 +3 = δ + /4 + Γ f ({xt∗ }lt0=0 ) + (l0 + l1 + L0 + 4)f (xf , xf )
114
2 DiscreteTime Autonomous Problems
+π f (x0∗ ) − π f (xl∗0 +l1 +L0 +4 ) + h(x0∗ ) l +l +L0 +3
≤ Γ f ({xt∗ }t0=0 1
)+(l0 +l1 +L0 +4)f (xf , xf )+(π f +h)(x0∗ )+δ+/4+/16
≤ (l0 + l1 + L0 + 4)f (xf , xf ) + δ2 /8 + δ + /4 + /16 + (π f + h)(x0∗ ) ≤ (l0 + l1 + L0 + 4)f (xf , xf ) + /64 + /64 + 5/16 + (π f + g)(x0∗ ) + δ1 /16 ≤ (l0 + l1 + L0 + 4)f (xf , xf ) + inf(π f + g) + δ2 /8 + δ1 /16 + /32 + 5/16 ≤ (l0 + l1 + L0 + 4)f (xf , xf ) + inf(π f + g) + 3/8.
(2.338)
In view of (2.295) and (2.338), l0 +l1 +L0 +3
f (xt , xt +1) ≤ (l0 + l1 + L0 + 4)f (xf , xf ) + inf(π f + g) + a1 + 2.
t =0
(2.339)
It follows from (2.315) and (2.339) that ˜ t = 0, . . . , l0 + l1 + L0 + 3. ρE (xt , θ0 ) ≤ M,
(2.340)
It follows from (2.318), (2.321), and (2.340) that h(x0 ) − g(x0 ) ≤ δ1 /16 ≤ /16.
(2.341)
By (2.338) and (2.341), l0 +l 1 +L0 +3
f (xt , xt +1 ) + g(x0 )
t =0
≤ (l0 + l1 + L0 + 4)f (xf , xf ) + inf(π f + g) + 3/8 + /16.
(2.342)
In view of (2.26) and (2.340), +l1 +L0 +4 ) + π f (x0 ) − π f (xl0 +l1 +L0 +4 ) + g(x0 ) Γ f ({xt }lt0=0
≤ inf(π f + g) + 3/8 + /16.
(2.343)
Property (i) and (2.326) imply that π f (xl0 +l1 +L0 +4 ) ≤ /16.
(2.344)
2.25 Proof of Proposition 2.69
115
It follows from (2.343) and (2.344) that l +l +L0 +4
Γ f ({xt }t0=0 1
) + (π f + g)(x0 ) ≤ inf(π f + g) + /2.
(2.345)
By (2.27) and (2.345), +l1 +L0 +4 (π f + g)(x0 ) ≤ inf(π f + g) + , Γ f ({xt }lt0=0 ) ≤ .
Theorem 2.66 is proved.
2.25 Proof of Proposition 2.69 Assume that the proposition does not hold. Then there exist a sequence {δk }∞ k=1 ⊂ (k) L0 (0, 1] and a sequence {xt }t =0 ∈ X(0, L0 ), k = 1, 2, . . . such that lim δk = 0
(2.346)
k→∞
and that for all integers k ≥ 1, (k)
(π f + g)(x0 ) ≤ inf(π f + g) + δk , (k) L
0 Γ f ({xt }t =0 ) ≤ δk
(2.347) (2.348)
and that the following property holds: f (i) for each (f, M)overtaking optimal {yt }∞ t =0 ∈ X(0, ∞) satisfying (π + (k) g)(y0 ) = inf(π f + g), we have max{ρE (xt , yt ) : t = 0, . . . , L0 } > .
By (2.21), (2.295), and (2.347), the sequence {π f (x0(k) )}∞ k=1 is bounded.
(2.349)
Proposition 2.27 and (2.349) imply that the sequence {x0 }∞ k=1 is bounded. In view of (2.20), (2.26), (2.348), and (2.349), the sequence (k)
{
L 0 −1 t =0
∞ f (xt(k) , xt(k) +1 )}k=1
is bounded. By LSC property, extracting a subsequence and reindexing if necessary, L0 we may assume without loss of generality that there exists {xt }t =0 ∈ X(0, L0 ) such that xt(k) → xt as k → ∞ for all t = 0, . . . , L0 ,
(2.350)
116
2 DiscreteTime Autonomous Problems L 0 −1
f (xt , xt +1 ) ≤ lim inf k→∞
t =0
L 0 −1 t =0
f (xt(k), xt(k) +1 ).
(2.351)
By Proposition 2.37, (2.347), (2.350), and the lower semicontinuity of π f , π f + g, (k)
(k)
π f (x0 ) ≤ lim inf π f (x0 ), g(x0 ) ≤ lim inf g(x0 ), k→∞
(2.352)
k→∞
(k)
(π f + g)(x0 ) ≤ lim inf(π f + g)(x0 ) ≤ inf(π f + g),
(2.353)
k→∞
(k)
(π f + g)(x0 ) = inf(π f + g) = lim (π f + g)(x0 ).
(2.354)
k→∞
In view of (2.352) and (2.354), (k)
(k)
g(x0 ) = lim g(x0 ), π f (x0 ) = lim π f (x0 ). k→∞
(2.355)
k→∞
By (2.350) and the lower semicontinuity of π f + g, (k)
(π f + g)(xL0 ) ≤ lim inf(π f + g)(xL0 ).
(2.356)
k→∞
It follows from (2.26), (2.27), (2.346), (2.348), (2.351), and the lower semicontinuity of π f that L 0 −1
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 )
t =0
≤ lim inf(
L 0 −1
k→∞
t =0
≤ lim inf( k→∞
L 0 −1
f (k) f (k) f (xt(k) , xt(k) +1 ) − L0 f (xf , xf )) − lim π (x0 ) + lim inf π (xL0 ) k→∞
L 0 −1 t =0
(k)
(k)
k→∞
(k)
(k)
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 )) ≤ 0,
f (xt , xt +1 ) − L0 f (xf , xf ) − π f (x0 ) + π f (xL0 ) = 0.
(2.357)
t =0
Theorem 2.13 and (2.357) imply that there exists an (f, M)overtaking optimal {x˜t }∞ t =0 ∈ X(0, ∞) such that x˜0 = xL0 .
(2.358)
2.26 The Second Bolza Problem
117
For all integers t > L0 set xt = x˜t −L0 .
(2.359)
It is clear that {xt }∞ t =0 ∈ X(0, ∞) is (f )good. By Proposition 2.25, (2.26), (2.27), (2.359), and (2.365), for all integers S > 0, S−1
f (xt , xt +1) − Sf (xf , xf ) − π f (x0 ) + π f (xS ) = 0
t =0
and {xt }∞ t =0 ∈ X(0, ∞) is (f, M)overtaking optimal. It follows from (2.346), (2.347) and (2.356) that (π f + g)(x0 ) = inf(π f + g). By (2.350), for all sufficiently (k) large natural numbers k, ρE (xt , xt ) ≤ /2 for all t = 0, . . . , L0 . This contradicts property (i). The contradiction we have reached proves Proposition 2.69.
2.26 The Second Bolza Problem We consider the control system introduced and studied in Section 2.4 and in Section 2.5 and use the notation, definitions, and assumptions used there. Let a1 > 0. Denote by A the set of all lower semicontinuous functions h : A × A → R 1 which are bounded on bounded subsets of A × A and such that h(z1 , z2 ) ≥ −a1 for all z1 , z2 ∈ A.
(2.360)
The set A is equipped with the uniformity which is determined by the base E(N, ) = {(h1 , h2 ) : A × A : h1 (z1 , z2 ) − h2 (z1 , z2 ) ≤ for all z1 , z2 ∈ A satisfying ρE (zi , θ0 ) ≤ N, i = 1, 2},
(2.361)
where N, > 0. It is not difficult to see that the uniform space A is metrizable and complete. For each pair of integers T2 > T1 ≥ 0 and each h ∈ A, we define σ f,h (T1 , T2 ) = inf{
T 2 −1 t =T1
2 f (xt , xt +1 ) + h(xT1 , xT2 ) : {xt }Tt =T ∈ X(T1 , T2 )}. 1
(2.362) The next result is proved in Section 2.27. Theorem 2.70. Assume that (f, M) has TP, M, > 0. Then there exist δ > 0 and an integer L > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 +2L, each 2 ∈ X(T1 , T2 ) which satisfies h ∈ A satisfying h(xf , xf ) ≤ M, and each {xt }Tt =T 1
118
2 DiscreteTime Autonomous Problems T 2 −1
f (xt , xt +1 ) + h(xT1 , xT2 ) ≤ σ f,h (T1 , T2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Let h ∈ A. Define f
ψh (z1 , z2 ) = π f (z1 ) + π− (z2 ) + h(z1 , z2 ), z1 , z2 ∈ A.
(2.363)
By (2.21), (2.25), (2.360), and Proposition 2.37, ψh : A × A → R 1 ∪ {∞} is bounded from below and satisfies ψh (z1 , z2 ) → ∞ as z1 , z2 ∈ A, ρE (z1 , θ0 ) + ρE (z2 , θ0 ) → ∞.
(2.364)
The next theorem is proved in Section 2.28. Theorem 2.71. Let (f, M) have TP, g ∈ A, L0 be a natural number and ∈ (0, 1). Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies T −1
f (xt , xt +1 ) + h(x0 , xT ) ≤ σ f,h (0, T ) + δ,
t =0
the inequalities L
f
L
0 0 ψg (x0 , xT ) ≤ inf(ψg ) + , Γ f ({xt }t =0 ) ≤ , Γ− ({x¯t }t =0 )≤
hold where x¯t = xT −t , t = 0, . . . , T . Theorem 2.72. Let (f, M) have TP and LSC property, g ∈ A, L0 > 0 be an integer and > 0. Then there exist δ > 0, an integer L1 > L0 , and a neighborhood U of g in A such that for each integer T ≥ L1 , each h ∈ U, and each {xt }Tt=0 ∈ X(0, T ) which satisfies T −1
f (xt , xt +1 ) + h(x0 , xT ) ≤ σ f,h (0, T ) + δ,
t =0
¯ ¯ there exist an (f, M)overtaking optimal {xt∗,1 }∞ t =0 ∈ X(0, ∞) and an (f , M)∗,2 ∞ ¯ overtaking optimal {xt }t =0 ∈ X(0, ∞) such that
2.27 Proof of Theorem 2.70
119
ψg (x0∗,1 , x0∗,2 ) = inf(ψg ), ρE (xt∗,1 , xt ) ≤ , ρE (xt∗,2, xT −t ) ≤ , t = 0, . . . , L0 . Theorem 2.72 follows from Theorem 2.71 and the following result which is proved in Section 2.29. Proposition 2.73. Let (f, M) have TP and LSC property, g ∈ A, L0 > 0 be an (1) L0 integer and ∈ (0, 1). Then there exist δ > 0 such that for each {xt }t =0 ∈ (2) L0 ¯ X(0, L0 ) and each {xt }t =0 ∈ X(0, L0 ) which satisfy (1)
(2)
(1) L
f
(2) L
0 0 ψg (x0 , x0 ) ≤ inf(ψg ) + δ, Γ f ({xt }t =0 ) ≤ δ, Γ− ({xt }t =0 ) ≤ δ,
¯ ¯ there exist an (f, M)overtaking optimal {yt }∞ t =0 ∈ X(0, ∞) and an (f , M)(2) ∞ ¯ overtaking optimal {yt }t =0 ∈ X(0, ∞) such that (1)
ψg (y0(1), y0(2) ) = inf(ψg ), ρE (xt(i) , yt(i) ) ≤ , i = 1, 2, t = 0, . . . , L0 .
2.27 Proof of Theorem 2.70 By Proposition 2.6, there exist δ ∈ (0, 1) and an integer L > 0 such that the following property holds: 2 ∈ (i) for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L, and each {xt }Tt =T 1 X(T1 , T2 ) which satisfies
T 2 −1
f (xt , ut ) ≤ min{(T2 − T1 )f (xf , uf ) + a1 + M + a0 + 1,
t =T1
U f (T1 , T2 , xT1 , xT2 ) + δ}, there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T 2 . Assume that T1 ≥ 0, T2 ≥ T1 + 2L are integers, h ∈ A satisfies h(xf , xf ) ≤ M,
(2.365)
120
2 DiscreteTime Autonomous Problems
2 and {xt }Tt =T ∈ X(T1 , T2 ) satisfies 1
T 2 −1
f (xt , xt +1 ) + h(xT1 , xT2 ) ≤ σ f,h (T1 , T2 ) + δ.
(2.366)
t =T1
By (2.366), T 2 −1
f (xt , xt +1 ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ.
(2.367)
t =T1
It follows from (2.360), (2.365), and (2.366) that −a1 +
T 2 −1
f (xt , xt +1 ) ≤
t =T1
T 2 −1
f (xt , xt +1 ) + h(xT1 , xT2 )
t =T1
≤ h(xf , xf ) + 1 + (T2 − T1 )f (xf , xf ), T 2 −1
f (xt , xt +1 ) ≤ (T2 − T1 )f (xf , xf ) + M + a1 + 1.
(2.368)
t =T1
In view of (2.367), (2.368), and property (i), there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . Moreover, if ρE (xT1 , xf ) ≤ δ, then τ1 = T1 , and if ρE (xT2 , xf ) ≤ δ, then τ2 = T2 . Theorem 2.70 is proved.
2.28 Proof of Theorem 2.71 By Propositions 2.25 and 2.26, Lemma 2.55, and (A2), there exists δ1 ∈ (0, /16) such that: (i) for each z ∈ A satisfying ρE (z, xf ) ≤ 2δ1 , we have π f (z) ≤ /16, f π− (z) ≤ /16; b
f ∈ X(0, bf ) which satisfies ρE (x0 , xf ) ≤ 2δ1 , ρE (xbf , xf ) ≤ (ii) for each {xt }t =0 2δ1 , we have
2.28 Proof of Theorem 2.71
121
bf −1
f (xt , xt +1 ) ≥ bf f (xf , uf ) − /16;
t =0
(iii) for each zi ∈ A, i = 1, 2 satisfying ρE (zi , xf ) ≤ 2δ1 , i = 1, 2, there exist an integer τ ∈ (0, bf ] and {xt }τt=0 ∈ X(0, τ ) which satisfies x0 = z1 , xτ = z2 and τ −1
f (xt , xt +1 ) ≤ τf (xf , uf ) + /16.
t =0
By (2.21), (2.360), (2.361), (2.363), and (2.364), there exists a neighborhood U1 of g in A such that for each h ∈ U1 ,  inf(ψh ) − inf(ψg ) ≤ δ1 /16.
(2.369)
Theorem 2.70 implies that there exist δ2 ∈ (0, δ1 /8), an integer l0 > 0, and a neighborhood U2 of g in A such that the following property holds: (iv) for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2l0 , each h ∈ U2 , and each 2 {xt }Tt =T ∈ X(T1 , T2 ) which satisfies 1 T 2 −1
f (xt , xt +1 ) + h(xT1 , xT2 ) ≤ σ f,h (T1 , T2 ) + δ2 ,
t =T1
we have ρE (xt , xf ) ≤ δ1 , t = T1 + l0 , . . . , T2 − l0 . There exist z∗,1 , z∗,2 ∈ A such that ψg (z∗,1 , z∗,2 ) ≤ inf(ψg ) + δ2 /8.
(2.370)
Corollary 2.30, (2.363), and (2.370) imply that there exist {xt∗,1}∞ t =0 ∈ X(0, ∞), ¯ ∈ X(0, ∞) for which {xt∗,2}∞ t =0 x0∗,i = z∗,i , i = 1, 2,
(2.371)
lim Γ f ({xt∗,1 }Tt=0 ) ≤ δ2 /8,
(2.372)
lim Γ− ({xt∗,2 }Tt=0 ) ≤ δ2 /8.
(2.373)
T →∞
f
T →∞
122
2 DiscreteTime Autonomous Problems
In view of (P1), Proposition 2.29, (2.372), and (2.373), lim ρE (xt∗,i , xf ) = 0, i = 1, 2.
t →∞
(2.374)
Equation (2.374) implies that there exists an integer l1 > 0 such that ρE (xt∗,i , xf ) ≤ δ2 /8, i = 1, 2 for all integers t ≥ l1 .
(2.375)
In view of (2.25), (2.361), and (2.363), there is a neighborhood U of g in A such that U ⊂ U1 ∩ U2 , h(x0∗,1 , x0∗,2 ) − g(x0∗,1 x0∗,2 ) ≤ δ1 /16 for all h ∈ U,
(2.376) (2.377)
and for all h ∈ U and all z1 , z2 ∈ A satisfying ψh (z1 , z2 ) ≤ inf(ψg ) + 4, we have ψg (z1 , z2 ) − ψh (z1 , z2 ) ≤ /64.
(2.378)
δ ∈ (0, δ2 /8)
(2.379)
L1 > 4L0 + 4l0 + 4l1 + 4bf + 16.
(2.380)
Choose
and an integer
Assume that T ≥ L1 is an integer, h ∈ U,
(2.381)
{xt }Tt=0 ∈ X(0, T ) satisfies T −1
f (xt , xt +1 ) + h(x0 , xT ) ≤ σ f,h (0, T ) + δ.
(2.382)
t =0
Property (iv), (2.376), and (2.379)–(2.382) imply that ρE (xt , xf ) ≤ δ1 , t = l0 , . . . , T − l0 .
(2.383)
2.28 Proof of Theorem 2.71
123
Set x¯t = xT −t , t = 0, . . . , T .
(2.384)
Property (iii), (2.375), (2.380), (2.381), and (2.383) imply that there exist integers τ1 , τ2 , τ3 , τ4 ∈ (0, bf ] and {xt(1)}Tt=0 ∈ X(0, T ) such that xt(1) = xt∗,1 , t = 0, . . . , l0 + l1 + L0 + 4, l0 +l1 +L 0 +3+τ1
(1)
xl0 +l1 +L0 +4+τ1 = xf ,
t =l0 +l1 +L0 +4
(1)
(2.385)
(1)
f (xt , xt +1 ) ≤ τ1 f (xf , xf ) + /16,
xl(1) = xl0 +l1 +L0 +4+2bf , xl(1) = xf , 0 +l1 +L0 +4+2bf 0 +l1 +L0 +4+2bf −τ2
(2.386) (2.387)
l0 +l1 +L0 +3+2bf
(1)
(1)
(2.388)
= xf , t = l0 + l1 + L0 + 4 + τ1 , . . . , l0 + l1 + L0 + 4 + 2bf − τ2 ,
(2.389)
t =l0 +l1 +L0 +4+2bf −τ2 (1)
xt
f (xt , xt +1 ) ≤ τ2 f (xf , xf ) + /16,
xT −t = xt∗,2 , t = 0, . . . , l0 + l1 + L0 + 4, (1)
(1)
xT −l0 −l1 −L0 −4−τ3 = xf , T −l0 −l 1 −L0 −5
(1)
t =T −l0 −l1 −L0 −4−τ3
(1)
f (xt , xt +1 ) ≤ τ3 f (xf , xf ) + /16,
(1)
xT −l0 −l1 −L0 −4−2bf = xT −l0 −l1 −L0 −4−2bf ,
(2.390) (2.391)
(2.392)
(2.393)
(1)
(2.394)
f (xt(1), xt(1) +1 ) ≤ τ4 f (xf , xf ) + /16,
(2.395)
xT −l0 −l1 −L0 −4−2bf +τ4 = xf , T −l0 −l1 −L0 −5−2bf +τ4
t =T −l0 −l1 −L0 −4−2bf
xt(1) = xf , t = T − l0 − l1 − L0 − 4 − 2bf + τ4 , . . . , T − l0 − l1 − L0 − 4 − τ3 , (2.396) xt(1) = xt , t = l0 + l1 + L0 + 4 + 2bf , . . . , T − l0 − l1 − L0 − 4 − 2bf . (2.397)
124
2 DiscreteTime Autonomous Problems
By (2.382) and (2.397), T −1
f (xt , xt +1 ) + h(x0 , xT ) ≤
t =0
T −1 t =0
(1) (1) f (xt(1), xt(1) +1 ) + h(x0 , xT ) + δ.
(2.398)
It follows from (2.397) and (2.398) that δ≥
T −1
T −1
f (xt , xt +1 ) + h(x0 , xT ) −
t =0
t =0
=
l0 +l 1 +L0 +3
(1)
(1)
(1)
(1)
f (xt , xt +1 ) − h(x0 , xT )
l0 +l1 +L0 +3+bf
f (xt , xt +1 ) +
t =0
f (xt , xt +1 )
t =l0 +l1 +L0 +4 l0 +l1 +L0 +3+2bf
+
f (xt , xt +1 )
t =l0 +l1 +L0 +4+bf T −l0 −l1 −L0 −5−bf
+
T −l0 −l 1 −L0 −5
f (xt , xt +1 ) +
t =T −l0 −l1 −L0 −4−2bf T −l
+
f (xt , xt +1 )
t =T −l0 −l1 −L0 −4−bf
f (xt , xt +1 ) − h(x0 , xT )
t =T −l0 −l1 −L0 −4
−
l0 +l1 +L0 +3
f (xt(1), xt(1) +1 )
t =0
−
l0 +l1 +L0 +2bf −τ2 +3
−
T −l0 −l1 −L0 −5−2bf +τ4
−
(1)
t =T −l0 −l1 −L0 −4−2bf
−
T −l0 −l 1 −L0 −5 t =T −l0 −l1 −L0 −4−τ3
t =l0 +l1 +L0 +4
f (xt(1), xt(1) +1 )
l0 +l1 +L0 +3+2bf
f (xt(1), xt(1) +1 ) −
t =l0 +l1 +L0 +4+τ1
l0 +l1 +L 0 +3+τ1
(1)
f (xt , xt +1 ) −
(1)
(1)
f (xt , xt +1 ) −
t =l0 +l1 +L0 +4+2bf −τ2
f (xt(1), xt(1) +1 )
T −l0 −l1 −L0 −5−τ3
(1)
t =T −l0 −l1 −L0 −4−2bf +τ4 T −l
t =T −l0 −l1 −L0 −4
(1)
(1)
(1)
f (xt , xt +1 )
(1)
(1)
f (xt , xt +1 ) − h(x0 , xT ). (2.399)
2.28 Proof of Theorem 2.71
125
Property (ii), (2.15), (2.380), (2.381), (2.383)–(2.386), (2.397), and (2.399) imply that δ≥
l0 +l 1 +L0 +3
f (xt , xt +1 ) + 2(bf f (xf , xf ) − /16) + 2(bf f (xf , xf ) − /16)
t =0
+
l0 +l1 +L0 +3
f¯(x¯t , x¯t +1 ) + h(x0 , x¯0 )
t =0
−
l0 +l 1 +L0 +3 t =0
f (xt∗,1 , xt∗,1 +1 ) − τ1 f (xf , xf ) − /16
−(2bf − τ1 − τ2 )f (xf , xf ) − τ2 f (xf , xf ) − /16 −τ4 f (xf , xf ) − /16 − (2bf − τ3 − τ4 )f (xf , xf ) − τ3 f (xf , xf ) − /16 −
l0 +l 1 +L0 +3 t =0
≥
l0 +l1 +L0 +3
∗,1 ∗,2 f¯(xt∗,2 , xt∗,2 +1 ) − h(x0 , x0 )
f (xt , xt +1 ) +
l0 +l1 +L0 +3
t =0
−
l0 +l 1 +L0 +3 t =0
f¯(x¯t , x¯t +1 ) + h(x0 , x¯0 )
t =0
f (xt∗,1 , xt∗,1 +1 ) −
l0 +l 1 +L0 +3 t =0
∗,1 ∗,2 f¯(xt∗,2 , xt∗,2 +1 ) − h(x0 , x0 ) − /2.
(2.400) By property (i), (2.363), (2.372), (2.373), (2.375), (2.377), (2.380), (2.381), (2.383), (2.288), and (2.400), l +l +L0 +4
δ + /2 ≥ Γ f ({xt }t0=0 1
)
+(l0 + l1 + L0 + 4)f (xf , xf ) + π f (x0 ) − π f (xl0 +l1 +L0 +4 ) +l1 +L0 +4 ) + (l0 + l1 + L0 + 4)f (xf , xf ) +Γ− ({x¯t }lt0=0 f
f
f
+π− (x¯0 ) − π− (x¯l0 +l1 +L0 +4 ) + h(x0 , x¯0 ) l +l +L0 +4
−Γ f ({xt∗,1 }t0=0 1
) − (l0 + l1 + L0 + 4)f (xf , xf )
−π f (x0∗,1 ) + π f (xl∗,1 ) 0 +l1 +L0 +4
126
2 DiscreteTime Autonomous Problems +l1 +L0 +4 −Γ− ({xt∗,2 }lt0=0 ) − (l0 + l1 + L0 + 4)f (xf , xf ) f
−π− (x0∗,2) + π− (xl∗,2 ) − h(x0∗,1 , x0∗,2 ) 0 +l1 +L0 +4 f
f
l +l +L0 +4
≥ Γ f ({xt }t0=0 1
l +l +L0 +4
f
) + Γ− ({x¯t }t0=0 1
)
f
+π f (x0 ) + π− (x¯0 ) + h(x0 , x¯0 ) − /8 l +l +L0 +4
−Γ f ({xt∗,1}t0=0 1
l +l +L0 +4
) − Γ− ({xt∗,2}t0=0 1 f
)
−π f (x0∗,1 ) − π− (x0∗,2 ) − h(x0∗,1 , x0∗,2 ) − /8 f
+l1 +L0 +4 +l1 +L0 +4 ≥ Γ f ({xt }lt0=0 ) + Γ− ({x¯t }lt0=0 ) + ψh (x0 , x¯0 ) f
−/4 − δ1 /4 − ψh (x0∗,1 , x0∗,2 ) l +l +L0 +4
≥ Γ f ({xt }t0=0 1
l +l +L0 +4
f
) + Γ− ({x¯t }t0=0 1
) + ψh (x0 , x¯0 )
− /4 − δ1 /4 − δ1 /16 − ψg (x0∗,1 , x0∗,2 ).
(2.401)
It follows from (2.370), (2.371), and (2.401) that +l1 +L0 +4 +l1 +L0 +4 ) + Γ− ({x¯t }lt0=0 ) + ψh (x0 , x¯0 ) Γ f ({xt }lt0=0 f
≤ δ + /2 + δ1 /4 + δ1 /16 + ψg (x0∗,1 , x0∗,2 ) + /4 ≤ 3/4 + /8 + inf(ψg ).
(2.402)
By (2.369), (2.381), and (2.402), +l1 +L0 +4 +l1 +L0 +4 Γ f ({xt }lt0=0 ) + Γ− ({x¯t }lt0=0 ) + ψh (x0 , x¯0 ) f
≤ 3/4 + /8 + inf(ψh ) + δ1 /16 ≤ inf(φh ) + 3/4 + /8 + /64.
(2.403)
In view of (2.26), (2.27), (2.369), and (2.403), l +l +L0 +4
Γ f ({xt }t0=0 1
f
l +l +L0 +4
), Γ− ({x¯t }t0=0 1
) ≤ ,
ψh (x0 , x¯0 ) ≤ inf(ψh ) + (57/64) ≤ inf(ψg ) + (58/64).
(2.404)
2.29 Proof of Proposition 2.73
127
By (2.378), (2.381), and (2.404), ψg (x0 , x¯0 ) ≤ ψh (x0 , x¯0 ) + /64 ≤ inf(ψg ) + . Theorem 2.71 is proved.
2.29 Proof of Proposition 2.73 Assume that the proposition does not hold. Then there exist a sequence {δk }∞ k=1 ⊂ (k,2) L0 0 (0, 1] and sequences {xt(k,1)}L ∈ X(0, L ), k = 1, 2, . . . and {x } 0 t t =0 t =0 ∈ ¯ X(0, L0 ), k = 1, 2, . . . such that lim δk = 0
(2.405)
k→∞
and that for all integers k ≥ 1, (k,1) L0 }t =0 )
Γ f ({xt
f
(k,2) L0 }t =0 )
≤ δk , Γ− ({xt (k,1)
ψg (x0
(k,2)
, x0
≤ δk ,
) ≤ inf(ψg ) + δk ,
(2.406) (2.407)
and that the following property holds: ¯ ¯ (i) for each (f, M)overtaking optimal {yt }∞ t =0 ∈ X(0, ∞) and each (f , M)(2) ∞ (1) (2) ¯ overtaking optimal {yt }t =0 ∈ X(0, ∞) satisfying ψg (y0 , y0 ) = inf(ψg ), we have (1)
(k,1)
max{ρE (xt
(1)
(k,2)
, yt ) + ρE (xt
(2)
, yt ) : t = 0, . . . , L0 } > .
By (2.21), (2.360), and (2.407), the sequences )}∞ k=1 are bounded.
(2.408)
∞ {π f (xL0 )}∞ k=1 , {π− (xL0 )}k=1 are bounded.
(2.409)
(k,1)
{π f (x0
)}∞ k=1 , {π− (x0 f
(k,2)
It follows from (2.25) and (2.408) that (k,1)
f
(k,2)
In view of (2.20), (2.26), (2.405), (2.406), and (2.409), the sequences {
L 0 −1 t =0
(k,1)
f (xt
, xt +1 )}∞ k=1 , { (k,1)
L 0 −1 t =0
(k,2) (k,2) f¯(xt , xt +1 )}∞ k=1 are bounded.
(2.410)
128
2 DiscreteTime Autonomous Problems
By LSC property and (2.410), extracting a subsequence and reindexing if nec0 essary, we may assume without loss of generality that there exist {yt(1)}L t =0 ∈ (2) L0 ¯ ∈ X(0, L0 ) such that X(0, L0 ), {yt }t =0 (k,i)
xt
(i)
→ yt as k → ∞ in E for all t = 0, . . . , L0 , i = 1, 2, L 0 −1 t =0 L 0 −1 t =0
(1)
(1)
f (yt , yt +1 ) ≤ lim inf k→∞
(2) (2) f¯(yt , yt +1 ) ≤ lim inf k→∞
L 0 −1
(k,1)
, xt +1 ),
(2.412)
(k,2) (k,2) f¯(xt , xt +1 ).
(2.413)
f (xt
t =0 L 0 −1 t =0
(k,1)
(2.411)
In view of (2.363), (2.405), (2.407), and (2.411), ψg (y0(1), y0(2) ) ≤ lim inf ψg (x0(k,1), x0(k,2)) = inf(ψg ), k→∞
ψg (y0(1), y0(2) ) = inf(ψg ).
(2.414) (2.415)
Proposition 2.37 and (2.411) imply that f
f
π f (y0(1)) ≤ lim inf π f (x0(k,1)), π− (y0(2) ) ≤ lim inf π− (x0(k,2)), k→∞
k→∞
(1)
(2)
(k,1)
g(y0 , y0 ) ≤ lim inf g(x0 k→∞
(k,2)
, x0
(2.416) (2.417)
).
It follows from (2.363), (2.407), and (2.415)–(2.417) that f
f
π f (y0(1)) = lim π f (x0(k,1)), π− (y0(2) ) = lim π− (x0(k,2)), k→∞
k→∞
g(y0(1) , y0(2)) = lim g(x0(k,1), x0(k,2)).
(2.418) (2.419)
k→∞
Proposition 2.37 and (2.411) imply that (1)
(k,1)
f
f
(2)
(k,2)
π f (yL0 ) ≤ lim inf π f (xL0 ), π− (yL0 ) ≤ lim inf π− (xL0 ), k→∞
k→∞
It follows from (2.26), (2.406), (2.412), (2.418), and (2.420) that L 0 −1 t =0
f (1) f (1) f (yt(1) , yt(1) +1 ) − L0 f (xf , xf ) − π (y0 ) + π (yL0 )
(2.420)
2.29 Proof of Proposition 2.73
≤ lim inf
129 L 0 −1
k→∞
t =0
f (xt(k,1), xt(k,1) +1 ) − L0 f (xf , xf )
) − lim π f (x0(k,1)) + lim inf π f (xL(k,1) 0 k→∞
≤ lim inf( k→∞
k→∞
L 0 −1
(k,1)
f (xt
t =0 (k,1)
−π f (x0
(k,1)
, xt +1 ) − L0 f (xf , xf ) (k,1)
) + π f (xL0 )) ≤ 0.
(2.421)
It follows from (2.26), (2.405), (2.406), (2.413), (2.418), and (2.420) that L 0 −1 t =0
f (2) f (2) (2) (2) f¯(yt , yt +1 ) − L0 f (xf , xf ) − π− (y0 ) + π− (yL0 )
≤ lim inf
L 0 −1
k→∞
t =0
(k,2) (k,2) f¯(xt , xt +1 ) − L0 f (xf , xf )
f
f
) − lim π− (x0(k,2)) + lim inf π− (xL(k,2) 0 k→∞
≤ lim inf( k→∞
k→∞
L 0 −1 t =0 f
(k,2) (k,2) f¯(xt , xt +1 ) − L0 f (xf , xf ) (k,2)
−π− (x0
f
(k,2)
) + π− (xL0 )) ≤ 0.
(2.422)
By (2.26), (2.27), (2.421), and (2.422), L 0 −1 t =0 L 0 −1 t =0
(1)
(1)
(1)
(1)
f (yt , yt +1 ) − L0 f (xf , xf ) − π f (y0 ) + π f (yL0 ) = 0,
(2.423)
f (2) f (2) f¯(yt(2), yt(2) +1 ) − L0 f (xf , xf ) − π− (y0 ) + π− (yL0 ) = 0.
(2.424)
In view of (2.415), (2.423), and (2.424), (1)
f
(2)
π f (yL0 ), π− (yL0 ) are finite.
(2.425)
130
2 DiscreteTime Autonomous Problems
Theorem 2.13 and (2.425) imply that there exist an (f, M)overtaking optimal (2) ∞ ¯ ¯ ¯ {y˜t(1)}∞ t =0 ∈ X(0, ∞) and (f , M)overtaking optimal {y˜t }t =0 ∈ X(0, ∞) such that (i)
(i)
y˜0 = yL0 , i = 1, 2.
(2.426)
For all integers t > L0 and i = 1, 2 set yt(i) = y˜t(i) −L0 . Proposition 2.25, (2.26), (2.27), and (2.423)–(2.426) imply that for all integers T2 > T1 ≥ 0, T 2 −1 t =T1 T 2 −1 t =T1
f (1) f (1) f (yt(1), yt(1) +1 ) − (T2 − T1 )f (xf , xf ) − π (yT1 ) + π (yT2 ) = 0,
(2.427)
f (2) f (2) (2) (2) f¯(yt , yt +1 ) − (T2 − T1 )f (xf , xf ) − π− (yT1 ) + π− (yT2 ) = 0.
(2.428)
Proposition 2.35, (2.425), (2.427), and (2.428) imply that {yt }∞ t =0 ∈ X(0, ∞) is (2) ∞ ¯ ¯ ¯ (f, M)overtaking optimal and {yt }t =0 ∈ X(0, ∞) is (f , M)overtaking optimal. It follows from (2.411) that for all sufficiently large natural numbers k and i = 1, 2, ρE (xt(k,i), yt(i) ) ≤ /4 for all t = 0, . . . , L0 . Combined with (2.415) this contradicts property (i). The contradiction we have reached proves Proposition 2.73. (1)
Chapter 3
DiscreteTime Nonautonomous Problems on the HalfAxis
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for discretetime nonautonomous problems on subintervals of halfaxis in metric spaces, which are not necessarily compact. For these optimal control problems, the turnpike is not a singleton. We also study the existence of solutions of the corresponding infinite horizon optimal control problems.
3.1 Preliminaries and Main Results Let (E, ρE ) and (F, ρF ) be metric spaces. We suppose that A is a nonempty subset of {0, 1, . . . , } × E, U : A → 2F is a point to set mapping with a graph M = {(t, x, u) : (t, x) ∈ A, u ∈ U(t, x)},
(3.1)
G : M → E and f : M → R 1 . Let 0 ≤ T1 < T2 be integers. We denote by X(T1 , T2 ) the set of all pairs of 2 2 −1 , {ut }Tt =T ) such that for each integer t ∈ {T1 , . . . , T2 }, sequences ({xt }Tt =T 1 1 (t, xt ) ∈ A,
(3.2)
for each integer t ∈ {T1 , . . . , T2 − 1}, ut ∈ U(t, xt ),
(3.3)
xt +1 = G(t, xt , ut )
(3.4)
and which are called trajectorycontrol pairs.
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_3
131
132
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Let T1 ≥ 0 be an integer. Denote by X(T1 , ∞) the set of all pairs of ∞ sequences {xt }∞ t =T1 ⊂ E, {ut }t =T1 ⊂ F such that for each integer T2 > T1 ,
2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). Elements of X(T1 , ∞) are called trajectory1 1
2 control pairs. Let 0 ≤ T1 < T2 be integers. A sequence {xt }Tt =T ⊂ E ({xt }∞ t =T1 ⊂ 1 2 −1 E, respectively) is called a trajectory if there exists a sequence {ut }Tt =T ⊂ F 1
T2 T2 −1 ({ut }∞ t =T1 ⊂ F , respectively) referred to as a control such that ({xt }t =T1 , {ut }t =T1 ) ∈ ∞ ∞ X(T1 , T2 ) (({xt }t =T1 , {ut }t =T1 ) ∈ X(T1 , ∞), respectively). Let θ0 ∈ E, θ1 ∈ F , a0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that
ψ(t) → ∞ as t → ∞.
(3.5)
We suppose that the function f satisfies f (t, x, u) ≥ ψ(ρE (x, θ0 )) − a0 for each (t, x, u) ∈ M.
(3.6)
For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ E satisfying (T1 , y), (T2 , z) ∈ A, we consider the following problems: T 2 −1 t =T1
2 2 −1 f (t, xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, xT2 = z, 1 1
T 2 −1 t =T1
(P1 ) 2 2 −1 f (t, xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, 1 1
T 2 −1 t =T1
2 2 −1 f (t, xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). 1 1
(P2 )
(P3 )
For each pair of integers T2 > T1 ≥ 0 and each pair of points y, z ∈ E satisfying (T1 , y), (T2 , z) ∈ A, we define U (T1 , T2 , y, z) = inf{ f
T 2 −1 t =T1
2 2 −1 f (t, xt , ut ) : ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
xT1 = y, xT2 = z},
(3.7)
σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ E, (T2 , h) ∈ A},
(3.8)
σ f (T1 , T2 , z) = inf{U f (T1 , T2 , h, z) : h ∈ E, (T1 , h) ∈ A},
(3.9)
σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, ξ ) : h, ξ ∈ E, (T1 , h), (T2 , ξ ) ∈ A}.
(3.10)
3.1 Preliminaries and Main Results
133
∞ We suppose that bf > 0 is an integer, ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞), f
f
f
{xt : t = 0, 1, . . . } is bounded , f
(3.11)
f
Δf := sup{f (t, xt , ut ) : t = 0, 1, . . . } < ∞
(3.12)
and the the following assumptions hold. (A1)
For each S1 > 0, there exist S2 > 0 and an integer c > 0 such that T 2 −1
f
f
f (t, xt , ut ) ≤
t =T1
(A2)
T 2 −1
f (t, xt , ut ) + S2
t =T1
2 2 −1 for each pair of integers T1 ≥ 0, T2 ≥ T1 +c and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 . For each > 0, there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = f 1, 2 satisfying ρE (zi , xTi ) ≤ δ, i = 1, 2 and T2 ≥ bf , there exist integers τ1 , τ2 ∈ (0, bf ],
T1 +τ1 −1 1 +τ1 ({xt(1)}Tt =T , {u(1) ) ∈ X(T1 , T1 + τ1 ), t }t =T1 1 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T2 − τ2 , T2 ) 2 −τ2 2 −τ2
(2)
(2)
such that (1)
f
(1)
xT1 = z1 , xT1 +τ1 = xT1 +τ1 , T1 +τ 1 −1
f (t, xt(1), u(1) t )≤
t =T1
T1 +τ 1 −1
f
f
f
f
f (t, xt , ut ) + ,
t =T1 f
xT(2) = z2 , xT(2) = xT2 −τ2 , 2 2 −τ2 T 2 −1 t =T2 −τ2
(2) (2) f (t, xt , ut )
≤
T 2 −1
f (t, xt , ut ) + .
t =T2 −τ2
Section 3.6 contains examples of optimal control problems satisfying assumptions (A1) and (A2). Many examples can also be found in [106–108, 118, 124, 125, 134].
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3 DiscreteTime Nonautonomous Problems on the HalfAxis
3.2 The Boundedness Results The following result is proved in Section 3.8. Theorem 3.1. 1. There exists S > 0 such that for each pair of integers T2 > T1 ≥ 0 and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 T 2 −1
f (t, xt , ut ) + S ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
∞ 2. For each ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) either T −1
f (t, xt , ut ) −
t =0
T −1
f
f
f (t, xt , ut ) → ∞ as T → ∞
t =0
or sup{
T −1
f (t, xt , ut ) −
t =0
T −1
f
f
f (t, xt , ut ) : T ∈ {1, 2, . . . }} < ∞.
(3.13)
t =0
Moreover, if (3.13) holds, then sup{ρE (xt , θ0 ) : t = 0, 1, . . . } < ∞. ∞ We say that ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )goof if (3.13) holds [29, 104, 116, 120, 129, 134]. The next boundedness result is proved in Section 3.8. It has a prototype in [104].
Theorem 3.2. Let c > 0 be an integer and M0 > 0. Then there exists M1 > 0 such 2 2 −1 that for each pair of integers T1 ≥ 0, T2 ≥ T1 + c and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Let L > 0 be an integer. Denote by AL the set of all (S, z) ∈ A for which there S+τ −1 exist an integer τ ∈ (0, L] and ({xt }S+τ ) ∈ X(S, S + τ ) such that t =S , {ut }t =S f
xS = z, xS+τ = xS+τ ,
S+τ −1 t =S
f (t, xt , ut ) ≤ L.
3.2 The Boundedness Results
135
L the set of all (S, z) ∈ A such that S ≥ L and there exist an integer Denote by A τ ∈ (0, L] and ({xt }St=S−τ , {ut }S−1 t =S−τ ) ∈ X(S − τ, S) satisfying f
xS−τ = xS−τ , xS = z,
S−1
f (t, xt , ut ) ≤ L.
t =S−τ
The following Theorems 3.3–3.5 are also boundedness results. They are proved in Section 3.8. Theorem 3.3. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 L , (T1 , xT1 ) ∈ AL , (T2 , xT2 ) ∈ A T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 3.4. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + L, and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 (T1 , xT1 ) ∈ AL ,
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 3.5. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + L and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying 1 1 L , (T2 , xT2 ) ∈ A
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1.
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3 DiscreteTime Nonautonomous Problems on the HalfAxis
3.3 Turnpike Properties We say that f possesses the turnpike property (or TP for short) if for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such that for each integer 2 2 −1 T1 ≥ 0, each integer T2 ≥ T1 + 2L, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 which satisfies T 2 −1
f (t, xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, T1 ≥ L, then τ1 = T 1 . We say that f possesses the strong turnpike property (or STP for short) if for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such that for 2 2 −1 , {ut }Tt =T )∈ each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L, and each ({xt }Tt =T 1 1 X(T1 , T2 ) which satisfies T 2 −1
f (t, xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . Theorem 3.1 implies the following two results. Theorem 3.6. Assume that f has TP and that , M > 0. Then there exist δ > 0 and an integer L > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L, 2 2 −1 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 T 2 −1 t =T1
f (t, xt , ut ) ≤ min{
T 2 −1 t =T1
f
f
f (t, xt , ut ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
3.3 Turnpike Properties
137
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, T1 ≥ L, then τ1 = T 1 . Theorem 3.7. Assume that f has STP and that , M > 0. Then there exist δ > 0 and an integer L > 0 such that for each integer T1 ≥ 0, each integer T2 ≥ T1 + 2L, 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies and each ({xt }Tt =T 1 1 T 2 −1
f (t, xt , ut ) ≤ min{
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ},
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . Proposition 3.8. Let L > 0 be an integer, T1 ≥ 0, T2 ≥ T1 + 2L be integers, L . Then (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A U (T1 , T2 , z1 , z2 ) ≤ f
T 2 −1
f
f
f (t, xt , ut ) + 2L(1 + a0 ).
t =T1
L , there exist integers τ1 ∈ (0, L], τ2 ∈ (0, L] and Proof. By the definition of AL , A T2 T2 −1 ({yt }t =T1 , {vt }t =T1 ) ∈ X(T1 , T2 ) such that f
yT1 = z1 , yT2 = z2 , yt = xt , t = T1 + τ1 , . . . , T2 − τ2 , f
vt = ut , t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }, T1 +τ 1 −1
T 2 −1
f (t, yt , vt ) ≤ L,
t =T1
f (t, yt , vt ) ≤ L.
t =T2 −τ2
In view of the relations above, U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1 t =T1
f (t, yt , vt )
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3 DiscreteTime Nonautonomous Problems on the HalfAxis
≤ 2L +
f
f
{f (t, xt , ut ) : t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }} ≤ 2L + 2a0 L +
T 2 −1
f
f
f (t, xt , ut ).
t =T1
Proposition 3.8 is proved. Proposition 3.8 and Theorems 3.6 and 3.7 imply the following two results. Theorem 3.9. Assume that f has TP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each integer 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies T2 ≥ T1 + 2L, and each ({xt }Tt =T 1 1 L0 , (T1 , xT1 ) ∈ AL0 , (T2 , xT2 ) ∈ A T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, T1 ≥ L, then τ1 = T 1 . Theorem 3.10. Assume that f has STP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each 2 2 −1 integer T2 ≥ T1 + 2L, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 L0 , (T1 , xT1 ) ∈ AL0 , (T2 , xT2 ) ∈ A T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 .
3.3 Turnpike Properties
139
Proposition 3.11. Let L > 0 be an integer, (T1 , z) ∈ AL , T2 ≥ T1 + L be an integer. Then σ f (T1 , T2 , z) ≤
T 2 −1
f
f
f (t, xt , ut ) + L(1 + a0 ).
t =T1
Proof. By the definition of AL , there exist an integer τ ∈ (0, L] and 2 2 −1 ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that f
f
yT1 = z, yt = xt , t = T1 + τ, . . . , T2 , vt = ut , t ∈ {T1 + τ, . . . , T2 } \ {T2 }, T1 +τ −1
f (t, yt , vt ) ≤ L.
t =T1
In view of the relations above, σ f (T1 , T2 , z) ≤
T 2 −1
f (t, yt , vt )
t =T1
≤L+
f
f
{f (t, xt , ut ) : t ∈ {T1 + τ, . . . , T2 } \ {T2 }} ≤ L + a0 L +
T 2 −1
f
f
f (t, xt , ut ).
t =T1
Proposition 3.11 is proved. Proposition 3.11 and Theorems 3.6, 3.7 imply the following two results. Theorem 3.12. Assume that f has TP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each integer 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies T2 ≥ T1 + 2L, and each ({xt }Tt =T 1 1 (T1 , xT1 ) ∈ AL0 ,
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, T1 ≥ L, then τ1 = T 1 .
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3 DiscreteTime Nonautonomous Problems on the HalfAxis
Theorem 3.13. Assume that f has STP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each 2 2 −1 integer T2 ≥ T1 + 2L, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 (T1 , xT1 ) ∈ AL ,
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . L Proposition 3.14. Let L > 0 be an integer, T1 ≥ 0 be an integer, (T2 , z) ∈ A satisfy T2 ≥ T1 + L. Then σ f (T1 , T2 , z) ≤
T 2 −1
f
f
f (t, xt , ut ) + L(1 + a0 ).
t =T1
L , there exist an integer τ ∈ (0, L] and Proof. By the definition of A 2 2 −1 ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that f
f
yT2 = z, yt = xt , t = T1 , . . . , T2 − τ, vt = ut , t ∈ {T1 , . . . , T2 − τ } \ {T2 − τ }, T 2 −1
f (t, yt , vt ) ≤ L.
t =T2 −τ
In view of the relations above, σ f (T1 , T2 , z) ≤
T 2 −1
f (t, yt , vt )
t =T1
≤L+
f
f
{f (t, xt , ut ) : t ∈ {T1 , . . . , T2 − τ } \ {T2 − τ }} ≤ L + a0 L +
T 2 −1 t =T1
Proposition 3.14 is proved.
f
f
f (t, xt , ut ).
3.3 Turnpike Properties
141
Proposition 3.14 and Theorems 3.6 and 3.7 imply the following two results. Theorem 3.15. Assume that f has TP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each integer 2 2 −1 T2 ≥ T1 + 2L, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 L0 , (T2 , xT2 ) ∈ A
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, T1 ≥ L, then τ1 = T 1 . Theorem 3.16. Assume that f has STP, L0 > 0 is an integer, and > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 ≥ 0, each 2 2 −1 integer T2 ≥ T1 + 2L, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 L0 , (T2 , xT2 ) ∈ A
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT2 ) + δ,
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . The next theorem is our main result in this chapter. It is proved in Section 3.10. Theorem 3.17. f has TP if and only if the following properties hold: (P1)
∞ for each (f )good pair of sequences ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞), f
lim ρE (xt , xt ) = 0;
t →∞
(P2)
for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such +L +L−1 , {ut }Tt =T ) ∈ X(T , T + L) that for each integer T ≥ 0 and each ({xt }Tt =T which satisfies
142
3 DiscreteTime Nonautonomous Problems on the HalfAxis
T +L−1
f (t, xt , ut ) ≤ min{U f (T , T + L, xT , xT +L )
t =T
+δ,
T +L−1
f
f
f (t, xt , ut ) + M},
t =T f
there exists an integer s ∈ [T , T + L] such that ρE (xs , xs ) ≤ . We say that f possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0, there exist natural numbers Q, l such that for each integer 2 2 −1 T1 ≥ 0, each integer T2 ≥ T1 + lQ, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 which satisfies T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M,
t =T1 q
q
there exist finite sequences of integers {ai }i=1 , {bi }i=1 ⊂ {T1 , . . . , T2 } such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, f
q
ρE (xt , xt ) ≤ for all integers t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. WTP was studies in [95, 96] for discretetime unconstrained problems, in [97] for variational problems, and in [114] for discretetime constrained problems. The next result is proved in Section 3.12. Theorem 3.18. f has WTP if and only if f has (P1) and (P2).
3.4 Lower Semicontinuity Property and Infinite Horizon Problems We say that f possesses lower semicontinuity property (or LSC property for short) if (j ) 2 (j ) 2 −1 , {ut }Tt =T )∈ for each pair of integers T2 > T1 ≥ 0 and each sequence ({xt }Tt =T 1 1 X(T1 , T2 ), j = 1, 2, . . . which satisfies sup{
T 2 −1 t =T1
(j )
(j )
f (t, xt , ut ) : j = 1, 2, . . . } < ∞,
3.4 Lower Semicontinuity Property and Infinite Horizon Problems (jk ) T2 (j ) 2 −1 }t =T1 , {ut k }Tt =T ), 1
there exist a subsequence ({xt
143
k = 1, 2, . . . and
2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) ({xt }Tt =T 1 1
such that for any t ∈ {T1 , . . . , T2 }, (jk )
xt T 2 −1
→ xt as k → ∞,
f (t, xt , ut ) ≤ lim inf j →∞
t =T1
T 2 −1
(j )
(j )
f (t, xt , ut ).
t =T1
LSC property plays an important role in the calculus of variations and optimal control theory [32]. ∞ Let S ≥ 0 be an integer. A pair ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is called (f )∞ overtaking optimal [29, 104, 120] if for every ({yt }∞ t =S , {vt }t =S ) ∈ X(S, ∞) satisfying xS = yS , lim sup[ T →∞
T −1
f (t, xt , ut ) −
t =S
T −1
f (t, yt , vt )] ≤ 0.
t =S
∞ A pair ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is called (f )weakly optimal [29, 104, 120] ∞ if for every ({yt }t =S , {vt }∞ t =S ) ∈ X(S, ∞) satisfying xS = yS ,
lim inf[ T →∞
T −1 t =S
f (t, xt , ut ) −
T −1
f (t, yt , vt )] ≤ 0.
t =S
∞ A pair ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is called (f )minimal [12, 94] if for every integer T > S, T −1
f (t, xt , ut ) = U f (S, T , xS , xT ).
t =S
In infinite horizon optimal control, the main goal is to show the existence of solutions using the optimality criterions above. The next result is proved in Section 3.14. Theorem 3.19. Assume that f has (P1), (P2), and LSC property, S ≥ 0 is an ∞ integer, and ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is (f )good. Then there exists an (f )∗ ∞ ∗ overtaking optimal pair ({xt∗ }∞ t =S , {ut }t =S ) ∈ X(S, ∞) such that xS = xS . The following theorem is also proved in Section 3.14.
144
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Theorem 3.20. Assume that f has (P1), (P2), and LSC property, S ≥ 0 is an integer, ({x˜t }∞ ˜ t }∞ t =S , {u t =S ) ∈ X(S, ∞) is (f )good, and ∗ ∞ ({xt∗ }∞ t =S , {ut }t =S ) ∈ X(S, ∞)
satisfies xS∗ = x˜S . Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)
∗ ∞ ({xt∗ }∞ t =S , {ut }t =S ) is (f )overtaking optimal; ∗ ∞ ({xt }t =S , {u∗t }∞ t =S ) is (f )weakly optimal; ∗ }∞ ) is (f )minimal and (f )good; ({xt∗ }∞ , {u t t =S t =S f ∗ }∞ ) is (f )minimal and satisfies lim , {u ({xt∗ }∞ t →∞ ρE (xt , xt ) = 0; t t =S t =S ∞ ∞ ∗ ∗ ({xt }t =S , {ut }t =S ) is (f )minimal and satisfies f
lim inf ρE (xt , xt ) = 0. t →∞
The next result easily follows from Theorems 3.9 and 3.17. Theorem 3.21. Assume that f has (P1), (P2), and LSC property, L > 0 be an integer, and > 0. Then there exists an integer τ0 > 0 such that for each integer ∞ T0 ≥ 0 and each (f )overtaking optimal pair ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) satisfying (T0 , xT0 ) ∈ AL , f
ρE (xt , xt ) ≤ for all integers t ≥ T0 + τ0 . In the sequel we use the following result which easily follows from (P1) and (A2). Proposition 3.22. Let f have (P1) and (T0 , z0 ) ∈ A. There exists an (f )good ∞ pair ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) satisfying xT0 = z0 if and only if (T0 , z0 ) ∈ ∪{AL : L = 1, 2, . . . }. LSC property implies the following result. Proposition 3.23. Let f have LSC property, T2 > T2 ≥ 0, and L > 0 be integers. 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) such that 1. If X(T1 , T2 ) = ∅, then there exists ({xt }Tt =T 1 1
T 2 −1 t =T1
f (t, xt , ut ) ≤
T 2 −1 t =T1
2 2 −1 f (t, yt , vt ) for all ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ). 1 1
L , then there exists 2. If T2 − T1 ≥ 2L, (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A T2 T2 −1 ({xt }t =T1 , {ut }t =T1 ) ∈ X(T1 , T2 ) such that xTi = zi , i = 1, 2 and T 2 −1 t =T1
f (t, xt , ut ) = U f (T1 , T2 , z1 , z2 ).
3.5 Perturbed Problems
145
2 2 −1 3. If T2 − T1 ≥ L, (T1 , z) ∈ AL , then there exists ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 such that xT1 = z and
T 2 −1
f (t, xt , ut ) = σ f (T1 , T2 , z).
t =T1
L , then there exists ({xt }T2 , {ut }T2 −1 ) ∈ X(T1 , T2 ) 4. If T2 − T1 ≥ L, (T2 , z) ∈ A t =T1 t =T1 such that xT2 = z and T 2 −1
f (t, xt , ut ) = σ f (T1 , T2 , z).
t =T1
The following result is proved in Section 3.16. It provides necessary and sufficient conditions for STP. Theorem 3.24. Let f have LSC property. If f has STP, then (P1), (P2), and the following property hold: (P3)
∞ there exists {u˜ t }∞ ˜ t }∞ t =0 ⊂ F such that ({xt }t =0 , {u t =0 ) ∈ X(0, ∞) is (f )overtaking optimal and for each (f )overtaking optimal f
f
f
∞ ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) f
f
satisfying y0 = x0 , the equality yt = xt holds for all integers t ≥ 0. f
f
If (P1), (P2), and (P3) hold and u˜ t = ut for all integers t ≥ 0, then f has STP. The next result follows from (P1) and STP. Theorem 3.25. Assume that f has STP and > 0. Then there exists δ > 0 such that for every integer T1 ≥ 0 and every (f )overtaking optimal pair f ∞ ({xt }∞ t =T1 , {ut }t =T1 ) ∈ X(T1 , ∞) satisfying ρE (xT1 , xT1 ) ≤ δ, the inequality f
ρE (xt , xt ) ≤ holds for all integers t ≥ T1 .
3.5 Perturbed Problems In this section we suppose that the following assumption holds: (A3) For each > 0, there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 f satisfying ρE (zi , xTi ) ≤ δ, i = 1, 2 and T2 ≥ bf , there exist integers τ1 , τ2 ∈ (0, bf ], T1 +τ1 −1 1 +τ1 ({xt(1)}Tt =T , {u(1) ) ∈ X(T1 , T1 + τ1 ), t }t =T1 1
146
3 DiscreteTime Nonautonomous Problems on the HalfAxis 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T2 − τ2 , T2 ) 2 −τ2 2 −τ2
(2)
(2)
such that (1)
f
(1)
xT1 = z1 , xT1 +τ1 = xT1 +τ1 , T1 +τ 1 −1
f (t, xt(1) , u(1) t )≤
T1 +τ 1 −1
t =T1
f
f
f (t, xt , ut ) + ,
t =T1 f
ρ(xt(1), xt ) ≤ , t = T1 , . . . , T1 + τ1 , (2)
f
(2)
xT2 = z2 , xT2 −τ2 = xT2 −τ2 , T 2 −1
(2)
(2)
f (t, xt , ut ) ≤
t =T2 −τ2
T 2 −1
f
f
f (t, xt , ut ) + ,
t =T2 −τ2 (2)
f
ρ(xt , xt ) ≤ , t = T2 − τ2 , . . . , T2 . Clearly, (A3) implies (A2). Assume that φ : E×E → [0, 1] is a continuous function satisfying φ(x, x) = 0 for all x ∈ E and such that the following property holds: (i) for each > 0, there exists δ > 0 such that if x, y ∈ E and φ(x, y) ≤ δ, then ρE (x, y) ≤ . For each r ∈ (0, 1) set f
fr (t, x, u) = f (t, x, u) + rφ(x, xt ), (t, x, u) ∈ M. f
(3.14) f
f
f
Clearly, for any r ∈ (0, 1), (A1), (A3) hold for fr with (xt r , ut r ) = (xt , ut ), t = 0, 1, . . . . The next result is proved in Section 3.17. Theorem 3.26. Let r ∈ (0, 1). Then fr has TP, (P1), and (P2). If ∞ ({xt }∞ t =0 , {ut }t =0 ) f
f
∞ is (f )minimal, then f has (P3) and {u˜ t }∞ t =0 = {ut }t =0 . f
f
In the sequel we use the following result for which we assume only that (A2) holds and does not need (A3). Proposition 3.27. Let γ > 0. Then there exists δ > 0 such that if (T , z1 ) ∈ A, (T + 2bf , z2 ) ∈ A satisfy f
f
ρE (z1 , xT ) ≤ δ, ρE (z2 , xT +2bf ) ≤ δ,
(3.15)
3.5 Perturbed Problems
147
then there exists T +2bf
({xt }t =T
T +2bf −1
, {ut }t =T
) ∈ X(T , T + 2bf )
such that T +2bf −1
xT = z1 , xT +2bf = z2 ,
T +2bf −1
f (t, xt , ut ) ≤
t =T
f
f
f (t, xt , ut ) + γ .
t =T
Proof. Set = γ /2. Let δ > 0 be as guaranteed by (A2). Let (T , z1 ) ∈ A, (T + 2bf , z2 ) ∈ A and (3.15) hold. By (A2), there exist integers τ1 , τ2 ∈ (0, bf ], +τ1 +τ1 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + τ1 ), (1)
T +2b
(1)
T +2b −1
f f ({xt(2)}t =T +2b , {u(2) t }t =T +2bf −τ2 ) ∈ X(T + 2bf − τ2 , T + 2bf ) f −τ2
such that f
xT(1) = z1 , xT(1)+τ1 = xT +τ1 , f
xT(2)+2bf = z2 , xT(2)+2bf −τ2 = xT +2bf −τ2 , T +τ 1 −1
(1) (1) f (t, xt , ut )
≤
t =T
f
f
f (t, xt , ut ) + γ /2,
t =T
T +2bf −1
T +τ 1 −1
T +2bf −1
f (t, xt(2), u(2) t )≤
t =T +2bf −τ2
f
f
f (t, xt , ut ) + γ /2.
t =T +2bf −τ2
Define xt = xt(1), t = T . . . , T + τ1 , ut = u(1) t , t = T . . . , T + τ1 − 1, f
f
xt = xt , t = T + τ1 . . . , T + 2bf − τ2 , ut = ut , t ∈ {T + τ1 . . . , T + 2bf − τ2 } \ {T + 2bf − τ2 }, xt = xt(2), t = T + 2bf − τ2 , . . . , T , ut = u(2) t , t = T + 2b − τ2 , . . . , T − 1.
148
3 DiscreteTime Nonautonomous Problems on the HalfAxis
By the relations above, T +2bf −1
T +2bf −1
f (t, xt , ut ) ≤
t =T
f
f
f (t, xt , ut ) + γ .
t =T
Proposition 3.27 is proved.
3.6 Examples Example 3.28. We consider an example which is a particular case of the problem introduced in Section 3.1. Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ : {0, 1, . . . } → R 1 , π : {0, 1, . . . } × E → R 1 , L : M → [0, ∞), π(t, x) ≥ −a1 for all (t, x) ∈ {0, 1, . . . } × E, π be bounded on bounded subsets of {0, 1, . . . } × E, μ(t) ≥ −a1 for all t ∈ {0, 1, . . . }, and for all (t, x, u) ∈ M, L(t, x, u) − π(t, G(t, x, u)) ≥ −a1 + ψ1 (ρE (θ0 , x)), f
f
L(t, x, u) = 0 if and only if x = xt , u = ut , f (t, x, u) = μ(t) + L(t, x, u) + π(t, x) − π(t + 1, G(t, x, u)).
(3.16) (3.17) (3.18)
By the relation above, for all (t, x, u) ∈ M. f (t, x, u) ≥ −3a1 + ψ1 (ρE (θ0 , x))
(3.19)
and (3.6) holds with a0 = 3a1 , ψ0 = ψ1 . 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ). It is not Let T2 > T1 ≥ 0 be integers and ({xt }Tt =T 1 1 difficult to see that T 2 −1 t =T1
f (t, xt , ut ) =
T 2 −1 t =T1
μ(t) +
T 2 −1 t =T1
L(t, xt , ut ) +
T 2 −1 t =T1
(π(t, xt ) − π(t + 1, xt +1 ))
3.6 Examples
=
149
T 2 −1 t =T1
T 2 −1
μ(t) +
T 2 −1
L(t, xt , ut ) + π(T1 , xT1 ) − π(T2 , xT2 ),
(3.20)
t =T1
f f f (t, xt , ut )
T 2 −1
f
(3.21)
≥ π(T1 , xT1 ) − π(T2 , xT2 ) − π(T1 , xT1 ) + π(T2 , xT2 ).
(3.22)
=
t =T1
t =T1
f
μ(t) + π(T1 , xT1 ) − π(T2 , xT2 )
and T 2 −1
f (t, xt , ut ) −
t =T1
T 2 −1
f
f
f (t, xt , ut )
t =T1 f
f
Since π is bounded on bounded subsets of {0, 1, . . . , } × E, we conclude that (A1) holds. We suppose that the following assumptions hold: (B1)
for each > 0, there exists δ > 0 such that for each integer T ≥ 0, each f f pair (T , z1 ), (T + 1, z2 ) ∈ A satisfying ρE (z1 , xT ), ρE (z2 , xT +1 ) ≤ δ, there exists u ∈ U(T , z1 ) such that f
ρF (u, uT ) ≤ , z2 = G(T , z1 , u); (B2)
for each > 0, there exists δ > 0 such that for each (T , z, ξ ) ∈ M satisfying f f f f ρE (z, xT ) ≤ δ, ρF (ξ, uT ) ≤ δ, we have f (T , z, ξ ) ≤ f (T , xT , uT ) + . It is clear that (B1) and (B2) imply (A2) and (A3).
Example 3.29. We consider an example which is a particular case of the problem introduced in Section 3.1. Let a1 > 0, ψ1 : [0, ∞) → [0, ∞) be an increasing function such that ψ1 (t) → ∞ as t → ∞, μ : {0, 1, . . . } → R 1 , π : {0, 1, . . . } × E → R 1 , L : M → [0, ∞), π(t, x) ≥ −a1 for all (t, x) ∈ {0, 1, . . . } × E, π be bounded on bounded subsets of {0, 1, . . . } × E, μ(t) ≥ −a1 for all t ∈ {0, 1, . . . }, and for all (t, x, u) ∈ M, (3.16)–(3.18) hold. Then as it was shown in Example 2.28, (A1) and (3.6) hold with a0 = 3a1, ψ0 = ψ1 . We suppose that there exists a natural number df such that the following assumptions hold:
150
(B3)
3 DiscreteTime Nonautonomous Problems on the HalfAxis
for each > 0, there exists δ > 0 such that for each integer T ≥ 0, each f f pair (T , z1 ), (T + df , z2 ) ∈ A satisfying ρE (z1 , xT ), ρE (z2 , xT +df ) ≤ δ, T +d
T +d −1
there exists ({xt }t =T f , {ut }t =T f
) ∈ X(T , T + df ) such that f
xT = z1 , xT +df = z2 , ρ(ut , ut ) ≤ , t = T , . . . , T + df − 1; (B4)
for each > 0, there exists δ > 0 such that for each (T , z) ∈ A satisfying f f ρE (z, xT ) ≤ δ and each u ∈ U(T , z) satisfying ρF (u, uT ) ≤ δ, we have f
f
f
ρE (xT +1 , G(T , z, u)) ≤ , f (T , z, u) ≤ f (T , xT , uT ) + . (B3) and (B4) imply (A2) and (A3). Assume now that for any nonempty bounded set Ω ⊂ E, the function π is bounded on {0, 1, . . . , } × Ω, the function μ is bounded, and the following property holds: (B5)
for each M, > 0, there exists δ > 0 such that for each (t, x, u) ∈ M which satisfies ρE (θ0 , x) ≤ M and L(t, x, u) ≤ δ f
we have ρE (x, xt ) ≤ . We claim that f has TP. In view of Theorem 3.17, it is sufficient to show that f possesses (P1) and (P2). Let us show that f has (P1). Assume that ∞ ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )good. Theorem 3.1 and the boundedness assumption on π imply that sup{ρE (θ0 , xt ) : t = 0, 1, . . . } < ∞, sup{π(xt ) : t = 0, 1, . . . } < ∞, sup{
T −1
f (t, xt , ut ) −
t =0
T
f
f
f (t, xt , ut ) : T ∈ {1, 2, . . . }} < ∞,
t =0 f
sup{π(xt ) : t = 0, 1, . . . } < ∞. Combined with (3.17) and (3.18), these relations imply that sup{
T −1
L(t, xt , ut ) : T ∈ {1, 2, . . . }}
t =0
= sup{
T −1 t =0
f (t, xt , ut ) −
T −1 t =0
f
f
f (t, xt , ut )
3.6 Examples
151 f
f
−π(0, x0) + π(T , xT ) + π(0, x0 ) − π(T , xT ) : T ∈ {1, 2, . . . }} < ∞. f
Together with (B5) this implies that limt →∞ ρE (xt , xt ) = 0 and (P1) holds. We claim that f has (P2). Let , M > 0 and M1 > 0 be as guaranteed by Theorem 3.2 with c = 1 and M0 = M. Since π satisfies the boundedness assumption, there exists M2 > 0 such that π(t, z) ≤ M2 for all integers t ≥ 0 and all z ∈ BE (θ0 , M1 ). Let δ ∈ (0, 1) be as guaranteed by (B5) with M = M1 . Choose an integer L > 3 + δ −1 (M + 3a1 + sup{μ(t) : t = 0, 1, . . . } +6 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)}). +L +L−1 Assume that T ≥ 0 is an integer and that ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L) satisfies T +L−1
f (t, xt , ut ) ≤
t =T
T +L−1
f
f
f (t, xt , ut ) + M.
t =T
Combined with Theorem 3.2 and the choice of M1 , this implies that ρE (θ0 , xt ) ≤ M1 , t = T , . . . , T + L − 1, f
ρE (θ0 , xt ) ≤ M1 , t = 0, 1, . . . . It is not difficult to see that T +L−2
f (t, xt , ut ) ≤
T +L−1
t =T
f (t, xt , ut ) − f (T + L − 1, xT +L−1 , uT +L−1 )
t =T
≤
T +L−1
f
f
f (t, xt , ut ) + M + 3a1
t =T
≤
T +L−2
f
f
f
f
f (t, xt , ut ) + M + 3a1 + sup{f (s, xs , us ) : s = 0, 1, . . . }
t =T
≤
T +L−2
f
f
f (t, xt , ut ) + M + 3a1 + sup{μ(t) : t = 0, 1, . . . }
t =T
+2 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)}.
152
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Combined with (3.17) and (3.18), this implies that M + 3a1 + sup{μ(t) : t = 0, 1, . . . } +2 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)} ≥
T +L−2
f (t, xt , ut ) −
t =T
=
T +L−2
T +L−2
f
f
f (t, xt , ut )
t =T
L(t, xt , ut ) + π(T , xT ) − π(T + L − 1, xT +L−1 )
t =T f
f
−π(T , xT ) − π(T + L − 1, xT +L−1 ) ≥
T +L−2
L(t, xt , ut ) − 4 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)}.
t =T
Together with the choice of δ and (B5), this implies that M + 3a1 + sup{μ(t) : t = 0, 1, . . . } +6 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)} ≥
T +L−2
L(t, xt , ut )
t =T
≥ δCard({t ∈ {T , . . . , T + L − 2} : L(t, xt , ut ) > δ}) f
≥ δCard({t ∈ {T , . . . , T + L − 2} : ρE (xt , xt ) > }) and f
Card({t ∈ {T , . . . , T + L − 1} : ρE (xt , xt ) > }) ≤ 1 + δ −1 (M + 3a1 + sup{μ(t) : t = 0, 1, . . . } +6 sup{π(t, z) : t ∈ {0, 1, . . . }, z ∈ BE (θ0 , M)}) < L − 2. Therefore (P2) holds and f has TP. Example 3.30. We consider a particular case of the problem introduced in Section 3.1. Assume that for each integer t ≥ 0, the set {x ∈ E : (t, x) ∈ A} is
3.7 Auxiliary Results for Theorems 3.1 and 3.2
153
a closed compact set, {(x, u) ∈ E × F : (t, x, u) ∈ M} is a closed compact set, G(t, ·, ·) : {(x, u) ∈ E × F : (t, x, u) ∈ M} → E is continuous and that f (t, ·, ·) : {(x, u) ∈ E × F : (t, x, u) ∈ M} → R 1 is lower semicontinuous. Then LSC property holds. Example 3.31. We consider a particular case of the problem introduced in Section 3.1. Assume that for each M > 0 and each integer t ≥ 0, the set {(x, u) ∈ E × F : (t, x, u) ∈ M, ρE (x, θ0 ) ≤ M} is a closed and compact, G(t, ·, ·) : {(x, u) ∈ E × F : (t, x, u) ∈ M} → E is continuous and f (t, ·, ·) : {(x, u) ∈ E × F : (t, x, u) ∈ M} → R 1 is lower semicontinuous. Then LSC property holds.
3.7 Auxiliary Results for Theorems 3.1 and 3.2 Lemma 3.32. There exist S > 0 and an integer c0 ≥ 1 such that for each pair of 2 2 −1 integers T1 ≥ 0, T2 ≥ T1 + c0 , and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 T 2 −1
f f f (t, xt , ut )
≤
t =T1
T 2 −1
f (t, xt , ut ) + S.
(3.23)
t =T1
Proof. In view of (3.5), there exists S1 > 0 such that ψ(S1 ) > a0 + 1 + Δf .
(3.24)
By (A1), there exist S2 > 0 and an integer c0 > 0 such that T 2 −1
f
f
f (t, xt , ut ) ≤
t =T1
T 2 −1
f (t, xt , ut ) + S2
t =T1
2 2 −1 for each pair of integers T1 ≥ 0, T2 ≥ T1 + c0 , and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 . Fix an integer
S ≥ S2 + 2 + (c0 + 2)(2a0 + Δf ).
(3.25)
Assume that T1 ≥ 0, T2 ≥ T1 + c0 are integers and that 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ). ({xt }Tt =T 1 1
We show that (3.23) is true. Assume that ρE (θ0 , xt ) ≥ S1 , t = T1 , . . . , T2 − 1.
(3.26)
154
3 DiscreteTime Nonautonomous Problems on the HalfAxis
By (3.6) and (3.26), for all t = T1 , . . . , T2 − 1, f (t, xt , ut ) ≥ −a0 + ψ(ρE (θ0 , xt )) ≥ −a0 + ψ(S1 ).
(3.27)
It follows from (3.12), (3.24), and (3.27), T 2 −1
f (t, xt , ut ) ≥ (T2 − T1 )(ψ(S1 ) − a0 )
t =T1
≥ (T2 − T1 )(Δf + 1) ≥
T 2 −1
f
f
f (t, xt , ut ) + 1
t =T1
and (3.23) holds. Assume that min{ρE (θ0 , xt ) : t = T1 , . . . , T2 − 1} < S1 .
(3.28)
Set τ1 = min{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 }, τ2 = max{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 }.
(3.29)
Clearly, τ1 , τ2 are welldefined, τ1 ≤ τ2 and ρE (θ0 , xτi ) ≤ S1 , i = 1, 2.
(3.30)
τ2 − τ1 ≥ c 0 ;
(3.31)
τ2 − τ1 < c 0 .
(3.32)
There are two cases:
Assume that (3.31) holds. It follows from (3.29), (3.31), and the choice of S2 and c0 that τ 2 −1 t =τ1
f
f
f (t, xt , ut ) ≤
τ 2 −1
f (t, xt , ut ) + S2 .
(3.33)
t =τ1
By (3.6), (3.12), (3.24), and (3.29), for each integer t ∈ ([T1 , τ1 ] ∪ [τ2 , T2 ]) \ {τ1 , τ2 , T2 }, ρE (θ0 , xt ) > S1 , f (t, xt , ut ) ≥ −a0 + ψ(ρE (θ0 , xt )) ≥ −a0 + ψ(S1 ).
(3.34)
3.7 Auxiliary Results for Theorems 3.1 and 3.2
155
It follows from (3.34) that
{f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }} ≥ (τ1 − T1 )(ψ(S1 ) − a0 ) ≥
f f {f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }},
≥
(3.35)
{f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {τ2 , T2 }} f
f
{f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {τ2 , T2 }}.
(3.36)
In view of (3.6), (3.12), (3.25), (3.33), (3.35), and (3.36), T 2 −1
f (t, xt , ut ) ≥
t =T1
T 2 −1
f f f (t, xt , ut )
− S2 − Δf − a0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ) − S
t =T1
and (3.23) holds. Assume that (3.32) holds. By (3.6), (3.24), (3.25), (3.29), and (3.32), T 2 −1
f (t, xt , ut ) =
{f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
t =T1
+ +
{f (t, xt , ut ) : t ∈ {τ1 , . . . , τ2 }}
{f (t, xt , ut ) : t ∈ {τ2 + 1, . . . , T2 + 1} \ {T2 , T2 + 1}}
≥ (ψ(S1 ) − a0 )(τ1 − T1 ) − (c0 + 1)a0 + (ψ(S1 ) − a0 )(T2 − τ2 − 1) ≥ Δf (τ1 − T1 ) + Δf (τ2 − τ1 ) − Δf c0 − (c0 + 1)a0 + Δf (T2 − τ2 ) − Δf  = Δf (T2 − T1 ) − (c0 + 1)a0 − (c0 + 1)Δf  ≥
T 2 −1
f
f
f (t, xt , ut ) − S
t =T1
and (3.23) is true. Lemma 3.32 is proved. The next auxiliary result easily follows from (3.5) and (3.6). Lemma 3.33. Let M0 > 0 and τ0 be a natural number. Then there exists M1 > M0 such that for each integer T1 ≥ 0, each integer T2 ∈ (T1 , T1 + τ0 ], and each 2 −1 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies Tt =T f (t, xt , ut ) ≤ M0 , the ({xt }Tt =T 1 1 1 inequality ρE (xt , θ0 ) ≤ M1 holds for all t = T1 , . . . , T2 − 1,
156
3 DiscreteTime Nonautonomous Problems on the HalfAxis
3.8 Proofs of Theorems 3.1–3.5 Assertion 1 of Theorem 3.1 follows from Lemmas 3.32, (3.6), and (3.12). Let us prove Assertion 2. Assume that there exists a sequence of natural numbers {tk }∞ k=1 such that tk → ∞ as k → ∞, t k −1
f (t, xt , ut ) −
t =0
t k −1
f
f
f (t, xt , ut ) → ∞ as k → ∞.
(3.37)
t =0
Let a number S be as guaranteed by Assertion 1 and let Q > 0. In view of (3.37), there exists a natural number k0 such that for each integer k ≥ k0 , t k −1
f (t, xt , ut ) −
t k −1
t =0
f
f
f (t, xt , ut ) > Q + S.
(3.38)
t =0
Let T > tk0 be an integer. By (3.38), the choice of S and Assertion 1 of Theorem 3.1, T −1
f (t, xt , ut ) −
T −1
t =0
tk0 −1 f f f (t, xt , ut )
=
t =0
+
T −1
tk0 −1
f (t, xt , ut ) −
t =0
f (t, xt , ut ) −
t =tk0
T −1
f
f
f
f (t, xt , ut )
t =0
f
f (t, xt , ut ) ≥ Q + S − S = Q.
t =tk0
Since Q is any positive number, we obtain that lim [
T →∞
T −1
f (t, xt , ut ) −
t =0
T −1
f
f
f (t, xt , ut )] = ∞.
t =0
Assume that sup{
T −1
f (t, xt , ut ) −
t =0
T −1
f
f
f (t, xt , ut ) : t = 1, 2, . . . } < ∞.
t =0
This implies that there exists an integer S1 > 0 such that for each pair of integers T2 > T1 ≥ 0, 
T 2 −1 t =T1
f (t, xt , ut ) −
T 2 −1 t =T1
f
f
f (t, xt , ut ) < S1 .
3.8 Proofs of Theorems 3.1–3.5
157
Together with (3.12) the inequality above implies that for all integers t ≥ 0, we have f (t, xt , ut ) ≤ Δf + S1 . Combined with (3.5) and (3.6), this implies that sup{ρE (xt , θ0 ) : t = 0, 1, 2, . . . } < ∞. Theorem 3.1 is proved. Proof of Theorem 3.2. We may assume that c = 1. By Theorem 3.1, there exists S0 > 0 such that the following property holds: 2 2 −1 (i) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
In view of (3.5) and (3.6), there exists M1 > 0 such that for each (t, x, u) ∈ M satisfying f (t, x, u) ≤ 2S0 + M0 + 2 + Δf , we have ρE (x, θ0 ) ≤ M1 .
(3.39)
2 2 −1 Let T2 > T1 ≥ 0 be integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfy 1 1
T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M0 .
(3.40)
t =T1
We show that ρE (θ0 , xt ) ≤ M1 , t = T1 , . . . , T2 − 1. Assume the contrary. Then there exists t0 ∈ {T1 , . . . , T2 − 1} such that (3.41)
ρE (θ0 , xt0 ) > M1 . Property (i), (3.12), and (3.40) imply that f (t0 , xt0 , ut0 ) =
T 2 −1
f (t, xt , ut ) −
{f (t, xt , ut ) : t ∈ {T1 , . . . , t0 } \ {t0 }}
t =T1 T 2 −1 f f − {f (t, xt , ut ) : t ∈ {t0 , . . . , T2 − 1} \ {t0 }} ≤ f (t, xt , ut ) + M0 t =T1
−
f f {f (t, xt , ut ) : t ∈ {T1 , . . . , t0 } \ {t0 }} + S0
158
3 DiscreteTime Nonautonomous Problems on the HalfAxis
−
f f {f (t, xt , ut ) : t ∈ {t0 , . . . , T2 − 1} \ {t0 }} + S0 ≤ Δf + M0 + 2S0 .
In view of the relation above and (3.39), ρE (θ0 , xt0 ) ≤ M1 . This contradicts (3.41). The contradiction we have reached proves Theorem 3.2. Theorems 3.3–3.5 follow from Theorem 3.2 and Propositions 3.8, 3.11, and 3.14, respectively.
3.9 An Auxiliary Result for Theorem 3.17 ∞ Lemma 3.34. Let ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) be (f )good and δ > 0. Then there exists an integer Tδ > 0 such that for each integer T > Tδ , T −1
f (t, xt , ut ) ≤ U f (Tδ , T , xTδ , xT ) + δ.
t =Tδ
Proof. Assume the contrary. Then for each integer T ≥ 0, there exists an integer S > T such that S−1
f (t, xt , ut ) > U f (T , S, xT , xS ) + δ.
t =T
This implies that there exists a strictly increasing sequence of integers {Ti }∞ i=0 such that T0 = 0 and for every integer i ≥ 0, Ti+1 −1
f (t, xt , ut ) > U f (Ti , Ti+1 , xTi , xTi+1 ) + δ.
(3.42)
t =Ti ∞ By (3.42), there exists ({yt }∞ t =0 , {vt }t =0 ) ∈ X(0, ∞) such that for every integer i ≥ 0, yTi = xTi and Ti+1 −1
Ti+1 −1
f (t, xt , ut ) >
t =Ti
f (t, yt , vt ) + δ.
(3.43)
t =Ti
∞ Since ({xt }∞ t =0 , {ut }t =0 ) is (f )good there exists S > 0 such that for each pair of integers T2 > T1 ≥ 0,

T 2 −1 t =T1
f (t, xt , ut ) −
T 2 −1 t =T1
f
f
f (t, xt , ut ) < S.
(3.44)
3.10 Proof of Theorem 3.17
159
In view of (3.43) and (3.44), for every integer k ≥ 1, T k −1
f (t, yt , vt ) −
t =0
+
T k −1
f
f
f (t, xt , ut ) =
t =T1 T k −1
f (t, xt , ut ) −
t =0
T k −1
f (t, yt , vt ) −
T k −1
t =0
T k −1
f
f (t, xt , ut )
t =T1
f
f (t, xt , ut ) ≤ −kδ + S → −∞ as k → ∞.
t =T1
This contradicts Theorem 3.1. The contradiction we have reached proves Lemma 3.34.
3.10 Proof of Theorem 3.17 First we show that TP implies (P1) and (P2). In view of Theorem 3.1, TP implies ∞ (P2). We show that TP implies (P1). Assume that TP holds, ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )good and > 0. There exists S > 0 such that for each integer T > 0, 
T −1
f (t, xt , ut ) −
t =0
T −1
f
f
f (t, xt , ut ) < S.
t =0
This implies that for each pair of integers T2 > T1 ≥ 0, 
T 2 −1
f (t, xt , ut ) −
t =T1
T 2 −1
f
f
f (t, xt , ut ) < 2S.
(3.45)
t =T1
Theorem 3.6 implies that there exist δ > 0 and an integer L > 0 such that the following property holds: (i) for each integer S1 ≥ 0, each integer S2 2 2 −1 , {ξt }St =S ) ∈ X(S1 , S2 ) which satisfies ({zt }St =S 1 1 S 2 −1
f (t, zt , ξt ) ≤ min{
t =S1
S 2 −1
f
≥
S1 + 2L and each
f
f (t, xt , ut ) + 2S, U f (S1 , S2 , zS1 , zS2 ) + δ}
t =S1
we have f
ρE (zt , xt ) ≤ , t = S1 + L, . . . , S2 − L.
160
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Let an integer Tδ > 0 be as guaranteed by Lemma 3.34. Equation (3.45), Lemma 3.34, and the choice of Tδ imply that for each integer T ≥ Tδ + 2L, f
ρE (xt , xt ) ≤ for all integers t ≥ Tδ + L. Thus TP implies (P1). Lemma 3.35. Assume that (P1) holds and that > 0. Then there exist δ > 0 and an integer L > 0 such that for each integer T1 ≥ L, each integer T2 ≥ T1 + 2bf 2 2 −1 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 f
ρE (xTi , xTi ) ≤ δ, i = 1, 2, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 . Proof. By (A2), for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: f
(ii) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying ρE (zi , xTi ) ≤ δk , i = 1, 2 and T2 ≥ bf there exist integers τ1 , τ2 ∈ (0, bf ], T1 +τ1 −1 1 +τ1 ({xt(1)}Tt =T , {u(1) ) ∈ X(T1 , T1 + τ1 ), t }t =T1 1 T2 −1 2 ({xt(2)}Tt =T , {u(2) t }t =T2 −τ2 ) ∈ X(T2 − τ2 , T2 ) 2 −τ2
such that (1)
(1)
f
(2)
f
(2)
xT1 = z1 , xT1 +τ1 = xT1 +τ1 , xT2 = z2 , xT2 −τ2 = xT2 −τ2 , T1 +τ 1 −1
(1) (1) f (t, xt , ut )
≤
t =T1 T 2 −1 t =T2 −τ2
T1 +τ 1 −1
f (t, xt , ut ) + 2−k , f
f
t =T1
f (t, xt(2), u(2) t )≤
T 2 −1
f (t, xt , ut ) + 2−k . f
f
t =T2 −τ2
Assume that the lemma does not hold. Then for each integer k ≥ 1, there Tk,2 Tk,2 −1 exist integers Tk,1 ≥ k + bf , Tk,2 ≥ Tk,1 + 2bf and ({xt(k) }t =T , {u(k) t }t =Tk,1 ) ∈ k,1 X(Tk,1 , Tk,2 ) such that
3.10 Proof of Theorem 3.17
161 f
ρE (xT(k) , xTi ) ≤ δk , i = 1, 2, i
(3.46)
Tk,2 −1
t =Tk,1
(k) (k) f f (t, xt(k), u(k) t ) ≤ U (Tk,1 , Tk,2 , xTk,1 , xTk,2 ) + δk ,
f
max({ρE (xt(k), xt ) : t ∈ {Tk,1 , . . . , Tk,2 }) > .
(3.47)
(3.48)
Extracting a subsequence and reindexing, we may assume without loss of generality that for each integer k ≥ 1, Tk+1,1 ≥ Tk,2 + 4bf .
(3.49)
Let k ≥ 1 be an integer. Property (ii) and (3,46) imply that there exist integers Tk,2 +τk,2 Tk,2 +τk,2 −1 , {u˜ (k) τk,1 , τk,2 ∈ (0, bf ] and ({x˜t(k)}t =T t }t =Tk,1 −τk,1 ) ∈ X(Tk,1 − τk,1 , Tk,2 + k,1 −τk,1 τk,2 ) such that (k) x˜t(k) = xt(k), t = Tk,1 , . . . , Tk,2 , u˜ (k) t = ut , t = Tk,1 , . . . , Tk,2 − 1, f
f
= xTk,1 −τk,1 , x˜T(k) = xTk,2 +τk,2 , x˜T(k) k,1 −τk,1 k,2 +τk,2 Tk,1 −1
Tk,1 −1 (k) (k) f (t, x˜t , u˜ t )
≤
t =Tk,1 −τk,1
(3.51)
f (t, xt , ut ) + 2−k ,
(3.52)
f (t, xt , ut ) + 2−k .
(3.53)
f
f
t =Tk,1 −τk,1
Tk,2 +τk,2 −1
(3.50)
Tk,2 +τk,2 −1
f (t, x˜t(k) , u˜ (k) t )≤
t =Tk,2
f
f
t =Tk,2
Property (ii) and (3.46) imply that there exist integers τk,3 , τk,4 ∈ (0, bf ] and (k) T
−1
(k) T
k,2 k,2 ({ xt }t =T , { ut }t =T ) ∈ X(Tk,1 , Tk,2 ) such that k,1 k,1
(k)
(k)
(k)
(k)
xTk,2 = xTk,2 , xTk,1 = xTk,1 , (k)
xt (k)
ut
(3.54)
f
= xt , t = Tk,1 + τk,3 . . . , Tk,2 − τk,4 ,
f
= ut , t ∈ {Tk,1 + τk,3 , . . . , Tk,2 − τk,4 } \ {Tk,2 − τk,4 },
Tk,1 +τk,3 −1
t =Tk,1
Tk,1 +τk,3 −1
f (t, xt(k) , u(k) t )≤
t =Tk,1
f (t, xt , ut ) + 2−k , f
f
(3.55)
(3.56)
162
3 DiscreteTime Nonautonomous Problems on the HalfAxis Tk,2 −1
Tk,2 −1
f (t, xt(k) , u(k) t )≤
t =Tk,2 −τk,4
f (t, xt , ut ) + 2−k . f
f
(3.57)
t =Tk,2 −τk,4
By (3.54)–(3.57), Tk,2 −1
U
f
(Tk,1 , Tk,2 , xT(k) , xT(k) ) k,1 k,2 Tk,2 −1
≤
≤
f (t, xt(k), u(k) t )
t =Tk,1
f (t, xt , ut ) + 2−k+1 . f
f
(3.58)
t =Tk,1
It follows from (3.47) and (3.58) that Tk,2 −1
Tk,2 −1
f (t, xt(k) , u(k) t )≤
t =Tk,1
f (t, xt , ut ) + 2−k+2 . f
f
(3.59)
t =Tk,1
By (3.52), (3.53), (3.56), and (3,59), Tk,2 +τk,2 −1
Tk,1 −1 (k) (k) f (t, x˜t , u˜ t )
t =Tk,1 −τk,1 Tk,2 +τk,2 −1
+
Tk,2 −1
≤
f f f (t, xt , ut )
+
t =Tk,1 −τk,1
f f f (t, xt , ut ) + 2−k+1
t =Tk,2
(k)
(k)
f (t, xt , ut )
t =Tk,1
Tk,2 +τk,2 −1
≤
f (t, xt , ut ) + 2−k+3 . f
f
(3.60)
t =Tk,1 −τk,1
∞ By (3.49) and (3.51), there exists ({xt }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) such that for every integer k ≥ 1,
xt = x˜t(k), t = Tk,1 − τk,1 , . . . , Tk,2 + τk,2 , ut = u˜ (k) t , t = Tk,1 − τk,1 , . . . , Tk,2 + τk,2 − 1,
(3.61)
xt = xt , t ∈ {0, 1, . . . } \ ∪∞ k=0 {Tk,1 − τk,1 , . . . , Tk,2 + τk,2 },
(3.62)
ut = ut , t ∈ {0, 1, . . . } \ ∪∞ k=0 [Tk,1 − τk,1 , . . . , Tk,2 + τk,2 ).
(3.63)
f
f
It follows from (3.60)–(3.63) that {f (t, xt , ut ) : t = 0, . . . , Tk,2 + τk,2 − 1}
3.10 Proof of Theorem 3.17
−
=
k Ti,2 +τ i,2 −1
163
f f {f (t, xt , ut ) : t = 0, . . . , Tk,2 + τk,2 − 1} (i)
(i)
f (t, x˜t , u˜ t ) −
i=1 t =Ti,1 −τi,1
k Ti,2 +τ i,2 −1
f
f
f (t, xt , ut ) ≤
i=1 t =Ti,1 −τi,1
k
2−i+3 < ∞.
i=1
∞ Theorem 3.1 implies that ({xt }∞ t =0 , {ut }t =0 ) is (f )good. By (P1), f
lim ρE (xt , xt ) = 0.
t →∞
f
On the other hand, in view of (3.48)–(3.50), lim supt →∞ ρE (xt , xt ) ≥ . The contradiction we have reached completes the proof of Lemma 3.35. Completion of the Proof of Theorem 3.17. Assume that properties (P1) and (P2) hold. Let , M > 0. By Lemma 3.35, there exist δ0 > 0 and an integer L0 > 0 such that the following property holds: (iii) for each integer T1 ≥ L0 , each integer T2 ≥ T1 + 2bf and each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({xt }Tt =T 1 1 f
ρE (xTi , xTi ) ≤ δ0 , i = 1, 2, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ0
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 . By Theorem 3.1, there exists S0 > 0 such that the following property holds: 2 2 −1 , {ut }Tt =T ) ∈ (iv) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T 1 1 X(T1 , T2 ), we have
T 2 −1 t =T1
f (t, xt , ut ) + S0 ≥
T 2 −1
f
f
f (t, xt , ut ).
t =T1
In view of (P2), there exist δ ∈ (0, δ0 ) and an integer L1 > 0 such that the following property holds: +L1 +L1 −1 (v) for each integer T ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T +L1 ) which satisfies T +L 1 −1 t =T
f (t, xt , ut ) ≤ min{U f (T , T + L1 , xT , xT +L1 ) + δ,
164
3 DiscreteTime Nonautonomous Problems on the HalfAxis T +L 1 −1
f
f
f (t, xt , ut ) + 2S0 + M},
t =T f
there exists an integer s ∈ [0, L1 ] such that ρE (xT +s , xT +s ) ≤ δ0 . Set L = L0 + L1 + bf .
(3.64)
Assume that T1 ≥ 0 and T2 ≥ T1 + 2L are integers and that 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
(3.65)
satisfies T 2 −1
f (t, xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ}.
(3.66)
t =T1
In view of (3.66), T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M.
(3.67)
t =T1
Properties (iv) and (3.67) imply that for each pair of integers Q1 , Q2 ∈ [T1 , T2 ] satisfying Q1 < Q2 , Q 2 −1
f (t, xt , ut ) =
t =Q1
T 2 −1
T 2 −1
{f (t, xt , ut ) : t ∈ {T1 , . . . , Q1 } \ {Q1 }}
t =T1
− ≤
f (t, xt , ut ) −
f
f
{f (t, xt , ut ) : t ∈ {Q2 , . . . , T2 } \ {T2 }}
f (t, xt , ut ) + M −
f
f
{f (t, xt , ut ) : t ∈ {T1 , . . . , Q1 } \ {Q1 }} + S0
t =T1 Q 2 −1 f f f f − {f (t, xt , ut ) : t ∈ {Q2 , . . . , T2 }\{T2 }}+S0 = f (t, xt , ut )+M +2S0 . t =Q1
(3.68)
3.11 An Auxiliary Result
165
By (3.68),
≤
{f (t, xt , ut ) : t = max{T1 , L0 }, . . . , {T1 , L0 } + L1 − 1} f
f
{f (t, xt , ut ) : t = max{T1 , L0 }, . . . , {T1 , L0 } + L1 − 1} + M + 2S0 , (3.69) T T 2 −1 2 −1 f f f (t, xt , ut ) ≤ f (t, xt , ut ) + M + 2S0 . (3.70) t =T2 −L1
t =T2 −L1
It follows from (3.64), (3.66), (3.69), (3.70), and property (v) that there exist integers τ1 ∈ [max{T1 , L0 }, max{T1 , L0 } + L1 ], τ2 ∈ [T2 − L1 , T2 ] such that ρE (xτi , xτfi ) ≤ δ0 , i = 1, 2.
(3.71)
f
If ρE (xT2 , xT2 ) ≤ δ, then we may assume that τ2 = T2 and if T1 ≥ L0 and f
ρE (xT1 , xT1 ) ≤ δ, then we may assume that τ1 = T1 . In view of (3.64), τ2 − τ1 ≥ T2 − T1 − 2L1 − L0 ≥ L0 + 2bf .
(3.72)
Property (iii), (3.66), (3.71), and (3.72) imply that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . Theorem 3.17 is proved.
3.11 An Auxiliary Result Lemma 3.36. Assume that (P2) holds and that M, > 0. Then there exists a natural number L such that for each integer T ≥ 0 and each +L +L−1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L)
which satisfies T +L−1
f (t, xt , ut ) ≤
t =T
T +L−1
f
f
f (t, xt , ut ) + M
t =T
the following inequality holds: f
min{ρE (xt , xt ) : t = T , . . . , T + L} ≤ .
166
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Proof. By Theorem 3.1, there exists S0 > 0 such that the following property holds: 2 2 −1 (i) for each pair of integers T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
By (P2), there exist δ0 ∈ (0, ) and an integer L0 > 0 such that the following property holds: T +L
T +L0 −1
(ii) for each integer T ≥ 0 and each ({xt }t =T 0 , {ut }t =T which satisfies T +L 0 −1
) ∈ X(T , L0 + T )
f (t, xt , ut ) ≤ min{U f (T , T + L0 , xT , xT +L0 ) + δ0 ,
t =T T +L 0 −1
f
f
f (t, xt , ut ) + 2S0 + M},
t =T
we have f
min{ρE (xt , xt ) : t = T , . . . , T + L0 } ≤ . Choose an integer q0 > (M + S0 )δ0−1
(3.73)
L = q0 L0 .
(3.74)
and set
+L +L−1 Let T ≥ 0 be an integer and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L) satisfy T +L−1
f (t, xt , ut ) ≤
t =T
T +L−1
f
f
f (t, xt , ut ) + M.
(3.75)
t =T
We show that f
min{ρE (xt , xt ) : t = T , . . . , T + L} ≤ . Assume the contrary. Then f
ρE (xt , xt ) > , t = T , . . . , T + L.
(3.76)
3.11 An Auxiliary Result
167
Let an integer i ∈ {0, . . . , q0 − 1}. Property (i), (3.74), and (3.75) imply that T +(i+1)L 0 −1
f (t, xt , ut ) =
T +L−1
t =T +iL0
− −
f (t, xt , ut )
t =T
{f (t, xt , ut ) : t ∈ {T , . . . , T + iL0 } \ {T + iL0 }}
{f (t, xt , ut ) : t ∈ {T + (i + 1)L0 , . . . , T + L} \ {T + L}} ≤
T +L−1
f
f
f (t, xt , ut ) + M
t =T
f f − {f (t, xt , ut ) : t ∈ {T , . . . , T + iL0 } \ {T + iL0 }} + S0 −
f f {f (t, xt , ut ) : t ∈ {T + (i + 1)L0 , . . . , T + L} \ {T + L}} + S0 =
T +(i+1)L 0 −1
f
f
f (t, xt , ut ) + M + 2S0 .
(3.77)
t =T +iL0
By (3.76), (3.77), and property (ii), T +(i+1)L 0 −1
f (t, xt , ut ) > U f (T + iL0 , T + (i + 1)L0 , xT +iL0 , xT +(i+1)L0 ) + δ0 .
t =T +iL0
(3.78) +L +L−1 It follows from (3.78) that there exists ({yt }Tt =T , {vt }Tt =T ) ∈ X(T , T + L) such that yT +iL0 = xT +iL0 , i = 0, . . . , q0
(3.79)
and for all i = 0, . . . , q0 − 1, T +(i+1)L 0 −1
f (t, xt , ut ) >
T +(i+1)L 0 −1
t =T +iL0
f (t, yt , vt ) + δ0 .
(3.80)
t =T +iL0
In view of (3.74), (3.75), (3.80), and property (i), T +L−1 t =T
f
f
f (t, xt , ut ) + M ≥
T +L−1 t =T
f (t, xt , ut ) ≥
T +L−1 t =T
f (t, yt , vt ) + q0 δ0
168
3 DiscreteTime Nonautonomous Problems on the HalfAxis
≥
T +L−1
f
f
f (t, xt , ut ) − S0 + q0 δ0 ,
t =T
q0 ≤ (M + S0 )δ0−1 . This contradicts (3.73). The contradiction we have reached proves Lemma 3.36.
3.12 Proof of Theorem 3.18 Assume that f has (P1) and (P2). In view of Theorem 3.17, f has TP. We show that WTP holds. Let , M > 0. By Theorem 3.1, there exists S0 > 0 such that the following property holds: 2 2 −1 , {vt }τt =τ ) ∈ X(τ1 , τ2 ), (i) for each pair of integers τ2 > τ1 ≥ 0 and each ({yt }τt =τ 1 1
τ 2 −1
f (t, yt , vt ) + S0 ≥
t =τ1
τ 2 −1
f
f
f (t, xt , ut ).
t =τ1
Theorem 3.17 and TP imply that there exist δ ∈ (0, ) and an integer L0 > 0 such that the following property holds: (ii) for each integer τ1 ≥ 0, each integer τ2 2 2 −1 ({yt }τt =τ , {vt }τt =τ ) ∈ X(τ1 , τ2 ) which satisfies 1 1 τ 2 −1
≥
τ1 + 2L0 , and each
f (t, yt , vt ) ≤ min{σ f (τ1 , τ2 ) + M + 3S0 , U f (τ1 , τ2 , yτ1 , yτ2 ) + δ},
t =τ1
we have f
ρE (yt , xt ) ≤ , t = τ1 + L0 , . . . , τ2 − L0 . Set l = 2L0 + 1.
(3.81)
Q ≥ 2 + 2(M + S0 )δ −1 .
(3.82)
Choose a natural number
3.12 Proof of Theorem 3.18
169
2 2 −1 Assume that T1 ≥ 0, T2 ≥ T1 +lQ are integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 satisfies
T 2 −1
T 2 −1
f (t, xt , ut ) ≤
t =T1
f
f
f (t, xt , ut ) + M.
(3.83)
t =T1
Property (i) and (3.83) imply that for each pair of integers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , τ 2 −1
f (t, xt , ut ) =
t =τ1
T 2 −1
{f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
t =T1
− ≤
f (t, xt , ut ) −
T 2 −1
f
{f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {T2 }}
f
f (t, xt , ut ) + M −
f f {f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }} + S0
t =T1
−
f f {f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {T2 }} + S0 =
τ 2 −1
f
f
f (t, xt , ut ) + M + 2S0 .
(3.84)
t =τ1
Set t0 = T1 .
(3.85)
If T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
then we set t1 = T2 . Assume that T 2 −1
f (t, xt , ut ) > U f (T1 , T2 , xT1 , xT2 ) + δ.
t =T1
Set t1 = min{t ∈ {T1 + 1, . . . , T2 } :
t −1
f (τ, xτ , uτ ) − U f (T1 , t, xT1 , xt ) > δ}.
τ =T1
(3.86) Clearly, t1 is welldefined.
170
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Assume that k ≥ 1 is an integer and we defined a sequence of integers {ti }ki=0 ⊂ [T1 , T2 ] such that t0 < t1 · · · < tk , for each integer i = 1, . . . , k if ti − ti−1 ≥ 2, then t i −2
f (t, xt , ut ) − U f (ti−1 , ti − 1, xti−1 , xti −1 ) ≤ δ
(3.87)
t =ti−1
and if ti < T2 , then t i −1
f (t, xt , ut ) − U f (ti−1 , ti , xti−1 , xti ) > δ.
(3.88)
t =ti−1
(Note that in view of (3.86), our assumption holds for k = 1.) 2 2 −1 By (3.88), there exists ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) such that 1 1 yti = xti , i = 0, . . . , k, t i −1
f (t, xt , ut ) −
t i −1
(3.89)
f (t, yt , vt ) > δ, i ∈ {1, . . . , k} \ {k},
(3.90)
yt = xt , t ∈ {tk−1 , . . . , T2 }, vt = ut , t ∈ {tk−1 , . . . , T2 } \ {T2 }.
(3.91)
t =ti−1
t =ti−1
Property (i), (3.83), and (3.89)–(3.91) imply that T 2 −1
f
f
f (t, xt , ut ) + M ≥
t =T1
T 2 −1
f (t, xt , ut ) ≥
t =T1
≥
T 2 −1
T 2 −1
f (t, yt , vt ) + δ(k − 1)
t =T1
f
f
f (t, xt , ut ) − S0 + δ(k − 1),
t =T1
k ≤ 1 + δ −1 (M + S0 ).
(3.92)
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If T 2 −1 t =tk
f (t, xt , ut ) ≤ U f (tk , T2 , xtk , xT2 ) + δ,
3.12 Proof of Theorem 3.18
171
then we set tk+1 = T2 and the construction of the sequence is completed. Assume that T 2 −1
f (t, xt , ut ) > U f (tk , T2 , xtk , xT2 ) + δ.
t =tk
Set tk+1 = min{t ∈ {tk + 1, . . . , T2 } :
t −1
f (τ, xτ , uτ ) − U f (tk , t, xtk , xt ) > δ}.
τ =tk
Clearly, tk+1 is welldefined, and the assumption made for k also holds for k + 1. q By induction (see (3.87) and (3.88)), we constructed a finite sequence {ti }i=0 ⊂ [T1 , T2 ] such that q ≤ 1 + δ −1 (M + S0 ),
(3.93)
T1 = t0 < t1 · · · < tq = T2 , for each integer i = 1, . . . , q if ti − ti−1 ≥ 2, then t i −2
f (t, xt , ut ) − U f (ti−1 , ti − 1, xti−1 , xti −1 ) ≤ δ
(3.94)
t =ti−1
and if ti < T2 , then t i −1
f (t, xt , ut ) − U f (ti−1 , ti , xti−1 , xti ) > δ.
(3.95)
t =ti−1
Assume that i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 2L0 + 1.
(3.96)
By (3.94) and (3.96), ti+1 −2
f (t, xt , ut ) − U f (ti , ti+1 − 1, xti , xti+1 −1 ) ≤ δ.
(3.97)
t =ti
Property (i), (3.84), and (3.96) imply that ti+1 −2
t =ti
ti+1 −2
f (t, xt , ut ) ≤
f
f
f (t, xt , ut ) + M + 2S0
t =ti
≤ M + 3S0 + σ f (ti , ti+1 − 1).
(3.98)
172
3 DiscreteTime Nonautonomous Problems on the HalfAxis
It follows from (3.96)–(3.98) and property (ii) that f
ρE (xt , xt ) ≤ , t = ti + L0 , . . . , ti+1 − 1 − L0 .
(3.99)
In view of (3.99), f
{t ∈ {T1 , . . . , T2 } : ρE (xt , xt ) > } ⊂ ∪{{ti , . . . , ti + L0 − 1} ∪ {ti+1 − L0 , . . . , ti+1 } : i ∈ {0, . . . , q − 1} and ti+1 − ti ≥ 2L0 + 1} ∪{{ti , . . . , ti+1 } : i ∈ {0, . . . , q − 1} and ti+1 − ti < 2L0 + 1}. In view of (3.81), (3.82), and (3.93), WTP holds. Assume that WTP holds. It is easy to see that (P1) and (P2) hold. This completes the proof of Theorem 3.18.
3.13 Auxiliary Results for Theorems 3.19 Lemma 3.37. Let > 0. Then there exist δ > 0 and an integer L > 0 such that 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) for each pair of integers T2 > T1 ≥ L and each ({xt }Tt =T 1 1 f
f
which satisfies ρE (xT1 , xT1 ) ≤ δ, ρE (xT2 , xT2 ) ≤ δ, the following inequality holds: T 2 −1
f (t, xt , ut ) ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ) − .
t =T1
Proof. By (A2), there exists δ > 0 such that the following property holds: f
(i) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ, there exist an integer τ1 ∈ (0, bf ], +τ1 T +τ1 −1 and ({x˜t(1)}Tt =T , {u˜ (1) ) ∈ X(T , T + τ1 ), such that t }t =T (1)
f
(1)
x˜T = z, x˜T +τ1 = xT +τ1 , T +τ 1 −1
f (t, x˜t(1), u˜ (1) t )≤
t =T
T +τ 1 −1
f
f
f (t, xt , ut ) + /4
t =T
and if T ≥ bf , there exist an integer τ2 ∈ (0, bf ] and −1 ({x˜t }Tt=T −τ2 , {u˜ t }Tt =T −τ2 ) ∈ X(T − τ2 , T ) (2)
(2)
3.13 Auxiliary Results for Theorems 3.19
173
such that f
(2)
(2)
x˜T −τ2 = xT −τ2 , x˜T = z, T −1
T −1
f (t, x˜t(2), u˜ (2) t )≤
t =T −τ2
f
f
f (t, xt , ut ) + /4.
t =T −τ2
Lemma 3.34 implies that there exists an integer L > 2bf such that for each pair of integers T2 > T1 ≥ L − bf , T 2 −1 t =T1
f
f
f
f
f (t, xt , ut ) < U f (T1 , T2 , xT1 , xT2 ) + /4.
(3.100)
2 2 −1 Assume that T2 > T1 ≥ L are integers, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) and 1 1
f
ρE (xTi , xTi ) ≤ δ, i = 1, 2.
(3.101)
By property (i) and (3.101), there exist integers τ1 , τ2 ∈ (0, bf ] and 2 +τ2 2 +τ2 −1 , {vt }Tt =T ) ∈ X(T1 − τ1 , T2 + τ2 ) ({yt }Tt =T 1 −τ1 1 −τ1
such that f
yT1 −τ1 = xT1 −τ1 , yt = xt , t = T1 , . . . , T2 , vt = ut , t = T1 , . . . , T2 − 1, (3.102) T T 1 −1 1 −1 f f f (t, yt , vt ) ≤ f (t, xt , ut ) + /4, (3.103) t =T1 −τ1
f
yT2 +τ2 = xT2 +τ2 ,
t =T1 −τ1 T2 +τ 2 −1
f (t, yt , vt ) ≤
T2 +τ 2 −1
t =T2
f
f
(3.104)
f (t, yt , vt ) + /4.
(3.105)
f (t, xt , ut ) + /4.
t =T2
In view of (3.100)–(3.102) and (3.104), T2 +τ 2 −1
f
f
f (t, xt , ut )
0 such that (T0 , z0 ) ∈ AL0 .
(3.106)
It follows from Theorem 3.1 that there exists S0 > 0 such that for each pair of 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), integers T2 > T1 ≥ 0 and each ({xt }Tt =T 1 1 T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
(3.107)
t =T1
Fix an integer k0 ≥ L0 . LSC property and (3.106) imply that for each integer k ≥ k0 T0 +k−1 0 +k there exists ({xt(k)}Tt =T , {u(k) ) ∈ X(T0 , T0 + k) satisfying t }t =T0 0 (k)
xT0 = z0 , T0 +k−1
(3.108)
f f (t, xt(k), u(k) t ) = σ (T0 , T0 + k, z0 ).
(3.109)
t =T0
In view of (3.6) and (3.106), for each integer k ≥ k0 , σ (T0 , T0 + k, z0 ) ≤ L0 + f
T0 +k−1
f
f
f (t, xt , ut ) + a0 L0 .
(3.110)
t =T0
By (3.107), (3.109), and (3.110), for each integer k ≥ k0 and each pair of integers T1 , T2 ∈ [T0 , T0 + k] satisfying T1 < T2 , T 2 −1 t =T1
(k)
(k)
f (t, xt , ut ) =
T0 +k−1 t =T0
(k)
(k)
f (t, xt , ut )
3.13 Auxiliary Results for Theorems 3.19
− −
175
{f (t, xt(k) , u(k) t ) : t ∈ {T0 , . . . , T1 } \ {T1 }}
(k) (k) {f (t, xt , ut ) : t ∈ {T2 , . . . , T0 + k} \ {T0 + k}} ≤
T0 +k−1
f
f
f (t, xt , ut ) + L0 (1 + a0 )
t =T0
− −
f f {f (t, xt , ut ) : t ∈ {T0 , . . . , T1 } \ {T1 }} + S0
f f {f (t, xt , ut ) : t ∈ {T2 , . . . , T0 + k} \ {T0 + k}} + S0 =
T 2 −1
f
f
f (t, xt , ut ) + 2S0 + L0 (1 + a0 ).
(3.111)
t =T1
By (3.12), (3.111), and LSC property, extracting subsequences, using the diagonalization process, and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 such that k1 ≥ k0 , and for each integer t ≥ (k )
(k )
∗ ∞ T0 , there exists limp→∞ f (t, xt p , ut p ), and there exists ({xt∗ }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) such that for each integer t ≥ T0 , (kp )
xt
→ xt∗ as p → ∞,
f (t, xt∗ , u∗t ) ≤ lim f (t, xt
(kp )
p→∞
(3.112) (kp )
, ut
(3.113)
).
In view of (3.111) and (3.113) for every integer q ≥ 1 T0 +q−1
f (t, xt∗ , u∗t ) ≤ lim
p→∞
t =T0
≤
T0 +q−1
f
+q−1 T0
(kp )
f (t, xt
(kp )
, ut
)
t =T0
f
f (t, xt , ut ) + 2S0 + L0 (1 + a0 ).
(3.114)
t =T0 ∗ ∞ Theorem 3.1 and (3.114) imply that ({xt∗ }∞ t =T0 , {ut }t =T0 ) is (f )good. In view of (3.108) and (3.112),
xT∗0 = z0 .
(3.115)
176
3 DiscreteTime Nonautonomous Problems on the HalfAxis
In order to complete the proof of the proposition, it is sufficient to show that ∗ ∞ ({xt∗ }∞ t =T0 , {ut }t =T0 ) is (f )minimal. Assume the contrary. Then there exist Δ > 0, 0 +τ0 0 +τ0 −1 an integer τ0 ≥ 1 and ({yt }Tt =T , {vt }Tt =T ) ∈ X(T0 , T0 + τ0 ) such that 0 0
yT0 = xT∗0 , yT0 +τ0 = xT∗0 +τ0 , T0 +τ 0 −1
f (t, xt∗ , u∗t ) >
t =T0
T0 +τ 0 −1
(3.116)
f (t, yt , vt ) + 2Δ.
(3.117)
t =T0
By (A2) and Lemma 3.37, there exist δ > 0 and an integer L1 > 0 such that the following properties hold: f
(ii) for each (T , ξ ) ∈ A satisfying ρE (ξ, xT ) ≤ δ, there exist an integer τ1 ∈ +τ1 T +τ1 −1 (0, bf ] and ({x˜t(1)}Tt =T , {u˜ (1) ) ∈ X(T , T + τ1 ), such that t }t =T (1)
f
(1)
x˜T = ξ, x˜T +τ1 = xT +τ1 , T +τ 1 −1
(1)
T +τ 1 −1
(1)
f (t, x˜t , u˜ t ) ≤
t =T
f
f
f (t, xt , ut ) + Δ/8
t =T
and if T ≥ bf , then there exist an integer τ2 ∈ (0, bf ] and −1 ({x˜t }Tt=T −τ2 , {u˜ t }Tt =T −τ2 ) ∈ X(T − τ2 , T ) (2)
(2)
such that f
x˜T(2)−τ2 = xT −τ2 , x˜T(2) = ξ, T −1
(2)
(2)
f (t, x˜t , u˜ t ) ≤
t =T −τ2
T −1
f
f
f (t, xt , ut ) + Δ/8;
t =T −τ2
2 2 −1 (iii) for each pair of integers T2 > T1 ≥ L1 and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1
f
f
X(T1 , T2 ) which satisfies ρE (xT1 , xT1 ) ≤ δ, ρE (xT2 , xT2 ) ≤ δ, the following inequality holds: T 2 −1 t =T1
f (t, xt , ut ) ≥
T 2 −1
f
f
f (t, xt , ut ) − Δ/8.
t =T1
Theorems 3.1 and 3.17, (P1), (P2), (3.106), (3.108), and (3.110) imply that there exists an integer L2 > L0 + L1 such that for each integer k ≥ k0 + 2L2 ,
3.13 Auxiliary Results for Theorems 3.19
177
f
ρE (xt(k), xt ) ≤ δ, t ∈ {T0 + L2 , . . . , T0 + k − L2 },
(3.118)
ρE (xt∗ , xt ) ≤ δ for all integers t ≥ T0 + L2 .
(3.119)
f
By (3.113), there exists a natural number p0 such that kp0 > k0 + 2L2 + 2L1 + 2τ0 + 2 + 2bf + 2T0 , T0 +τ 0 +L2 −1
f (t, xt∗ , u∗t ) ≤
T0 +τ 0 +L2 −1
t =T0
(kp0 )
f (t, xt
(kp0 )
, ut
(3.120)
) + Δ/2.
(3.121)
t =T0
Property (ii), (3.116), and (3.118)–(3.120) imply that there exists T0 +kp
T0 +kp0 −1
({xt }t =T0 0 , {ut }t =T0
) ∈ X(T0 , T0 + kp0 )
such that xt = yt , t = T0 , . . . , T0 + τ0 , ut = vt , t = T0 , . . . , T0 + τ0 − 1,
(3.122)
xt = xt∗ , t = T0 +τ0 , . . . , T0 +τ0 +L2 , ut = u∗t , t = T0 +τ0 , . . . , T0 +τ0 +L2 −1, (3.123) f (3.124) xT0 +τ0 +L2 +bf = xT0 +τ0 +L2 +bf , T0 +τ0 +L2 +bf −1
T0 +τ0 +L2 +bf −1
f (t, xt , ut ) ≤
t =T0 +τ0 +L2
f
f
f (t, xt , ut ) + Δ/8,
(3.125)
t =T0 +τ0 +L2 (kp0 )
xt = xt
(kp0 )
ut = ut
, t = T0 + τ0 + L2 + 2bf , . . . , kp0 + T0 ,
(3.126)
, t = T0 + τ0 + L2 + 2bf , . . . , kp0 + T0 − 1,
(3.127)
T0 +τ0 +L2 +2bf −1
T0 +τ0 +L2 +2bf −1
f (t, xt , ut ) ≤
t =T0 +τ0 +L2 +bf
f
f
f (t, xt , ut ) + Δ/8.
(3.128)
t =T0 +τ0 +L2 +bf
It follows from (3.108), (3.109), (3.115), (3.116) and (3.122), T0 +kp0 −1
t =T0
T0 +kp0 −1
f (t, xt , ut ) ≥
t =T0
(kp0 )
f (t, xt
(kp0 )
, ut
).
(3.129)
178
3 DiscreteTime Nonautonomous Problems on the HalfAxis
By (3.117), (3.118), (3.120), (3.122), (3.123), (3.125)–(3.129) and property (iii), T0 +kp0 −1
0≤
T0 +kp0 −1
f (t, xt , ut ) −
t =T0
(kp0 )
, ut
)
t =T0
T0 +τ 0 −1
=
(kp0 )
f (t, xt
T0 +τ 0 +L2 −1
f (t, yt , vt ) +
t =T0
f (t, xt∗ , u∗t )
t =T0 +τ0
T0 +τ0 +L2 +bf −1
+
f
f
f (t, xt , ut ) + Δ/8
t =T0 +τ0 +L2 T0 +τ0 +L2 +2bf −1
+
f
f
f (t, xt , ut ) + Δ/8 −
T0 +τ 0 +L2 −1
t =T0 +τ0 +L2 +bf T0 +τ0 +L2 +bf −1
−
(kp ) (kp ) f (t, xt 0 , ut 0 )
T0 +τ0 +L2 +2bf −1
T0 +τ 0 −1
T0 +τ 0 +L2 −1
f (t, yt , vt ) +
)
f (t, xt∗ , u∗t )
Δ/8 +
f
f
f (t, xt , ut ) + Δ/8
t =T0 +τ0 +L2 +bf
(kp0 )
f (t, xt
(kp0 )
, ut
T0 +τ0 +L2 +bf −1
)−
t =T0
f
T0 +τ0 +L2 +2bf −1
−
f
f
f (t, xt , ut ) + Δ/8
t =T0 +τ0 +L2 +bf T0 +τ 0 −1
T0 +τ 0 +L2 −1
f (t, yt , vt ) +
t =T0
−
f (t, xt∗ , u∗t )
t =T0 +τ0
T0 +τ 0 +L2 −1 t =T0
(kp0 )
f (t, xt
(kp0 )
, ut
f
f (t, xt , ut ) + Δ/8
t =T0 +τ0 +L2 +1
=
(kp0 )
, ut
T0 +τ0 +L2 +2bf −1 f f f (t, xt , ut ) +
T0 +τ 0 +L2 −1
(kp0 )
f (t, xt
t =T0 +τ0
t =T0 +τ0 +L2
−
)
t =T0 +τ0 +L2 +bf
T0 +τ0 +L2 +bf −1
+
−
t =T0
(kp0 )
, ut
t =T0
t =T0 +τ0 +L2
≤
(kp0 )
f (t, xt
) + Δ/2
3.14 Proofs of Theorems 3.19 and 3.20
0 such that for all natural numbers T > S, 
T −1
f (t, x˜t , u˜ t ) −
t =S
T −1
f
f
f (t, xt , ut ) ≤ S0 .
(3.132)
t =S
By (A2), there exists δ > 0 such that the following property holds: f
(a) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ there exist an integer τ1 ∈ (0, bf ] (1) +τ1 (1) +τ1 −1 and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + τ1 ) satisfying f
xT(1) = z, xT(1)+τ1 = xT +τ1 , T +τ 1 −1
(1)
(1)
f (t, xt , ut ) ≤
t =T
T +τ 1 −1
f
f
f (t, xt , ut ) + 1
t =T
and if T ≥ bf , then there exist an integer τ2 ∈ (0, bf ] and −1 ({xt }Tt=T −τ2 , {ut }Tt =T −τ2 ) ∈ X(T − τ2 , T ) (2)
(2)
180
3 DiscreteTime Nonautonomous Problems on the HalfAxis
satisfying f
xT(2)−τ2 = xT −τ2 , xT(2) = z, T −1
T −1
f (t, xt(2), u(2) t )≤
t =T −τ2
f (t, xf , uf ) + 1.
t =T −τ2
In view of (3.130) and (3.131), there exists a strictly increasing sequence of natural numbers {tk }∞ k=1 such that ρE (xt∗k +2bf , xtk +2bf ) ≤ δ, k = 1, 2, . . . . f
(3.133)
f
ρE (x˜t , xt ) < δ for all integers t ≥ t0 .
(3.134)
Let k ≥ 1 be an integer. By property (a) and (3.134), there exists t +2bf
({yt }tk=S
t +2bf −1
, {vt }tk=S
) ∈ X(S, tk + 2bf )
such that yt = x˜t , t = S, . . . , tk , vt = u˜ t , t = S, . . . , tk − 1 tk +bf −1 f
ytk +bf = xt +bf ,
ytk +2bf =
tk +bf −1
f (t, yt , vt ) ≤
t =tk
xt∗k +2bf ,
(3.135)
f
f
f (t, xt , ut ) + 1,
(3.136)
t =tk
tk +2bf −1
tk +2bf −1
f (t, yt , vt ) ≤
t =tk +bf
f
f
f (t, xt , ut ) + 1.
t =tk +bf
(3.137) It follows from (3.132) and (3.135)–(3.137) that tk +2bf −1
f (t, xt∗ , u∗t ) ≤
tk +2bf −1
t =S
≤
t k −1
f (t, yt , vt )
t =S tk +2bf −1
f (t, x˜t , u˜ t ) +
f
f
f (t, xt , ut ) + 2
t =tk
t =S tk +2bf −1
≤
t =S
f
f
f (t, xt , ut ) + S0 + 2.
3.14 Proofs of Theorems 3.19 and 3.20
181
∗ ∞ Together with Theorem 3.1, this implies that ({xt∗ }∞ t =S , {ut }t =S ) is (f )good and (iii) holds. ∗ ∞ We show that (iii) implies (i). Assume that ({xt∗ }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is (f )minimal and (f )good. (P1) implies that
lim ρE (xt∗ , xt ) = 0. f
(3.138)
t →∞
There exists S0 > 0 such that 
T −1
f (t, xt∗ , u∗t ) −
t =S
T −1
f
f
f (t, xt , ut ) ≤ S0 for all integers T > S.
(3.139)
t =S
∞ Let ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) satisfy
xS = xS∗ .
(3.140)
We show that lim sup[ T →∞
T −1
f (t, xt∗ , u∗t )
−
t =S
T −1
f (t, xt , ut )] ≤ 0.
t =S
∞ In view of Theorem 3.1, we may assume that ({xt }∞ t =S , {ut }t =S ) is (f )good. (P1) implies that f
lim ρE (xt , xt ) = 0.
(3.141)
t →∞
Let > 0. By (A2) and Lemma 3.37, there exist δ ∈ (0, ) and an integer L1 > 0 such that the following properties hold: f (b) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ, there exist an integer τ1 ∈ (0, bf ] (1) +τ1 (1) +τ1 −1 and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + τ1 ) satisfying (1)
f
(1)
xT = z , xT +τ1 = xT +τ1 , T +τ 1 −1 t =T
f (t, xt(1), u(1) t )≤
T +τ 1 −1
f
f
f (t, xt , ut ) + /8
t =T
and if T ≥ bf , then there exist an integer τ2 ∈ (0, bf ] and T −1 ({xt(2)}Tt=T −τ2 , {u(2) t }t =T −τ2 ) ∈ X(T − τ2 , T )
182
3 DiscreteTime Nonautonomous Problems on the HalfAxis
satisfying f
xT(2)−τ2 = xT −τ2 , xT(2) = z, T −1
T −1
f (t, xt(2) , u(2) t )≤
t =T −τ2
f
f
f (t, xt , ut ) + /8;
t =T −τ2
2 2 −1 (c) for each pair of integers T2 > T1 ≥ L1 and each ({yt }Tt =T , {vt }Tt =T ) ∈ 1 1
f
X(T1 , T2 ) which satisfies ρE (yTi , xTi ) ≤ δ, i = 1, 2, we have T 2 −1
f (t, yt , vt ) ≥
t =T1
T 2 −1
f
f
f (t, xt , yt ) − /8.
t =T1
It follows from (3.138) and (3.141) that there exists an integer τ0 > S such that ρE (xt , xt ) ≤ δ, ρE (xt∗ , xt ) ≤ δ for all integers t ≥ τ0 . f
f
(3.142)
Let T ≥ τ0 + L1 be an integer. Property (b) and (3.142) imply that there exists T +2b T +2b −1 ({yt }t =S f , {vt }t =S f ) ∈ X(S, T + 2bf ) which satisfies f
yt = xt , t = S, . . . , T , vt = ut , t = S, . . . , T − 1, yT +bf = xT +bf , T +bf −1
T +bf −1
f (t, yt , vt ) ≤
t =T
f
f
f (t, xt , ut ) + /8,
(3.144)
t =T
yT +2bf = xT∗ +2bf , T +2bf −1
(3.143)
(3.145)
T +2bf −1
f (t, yt , vt ) ≤
t =T +bf
f
f
(3.146)
f
f
(3.147)
f (t, xt , ut ) + /8.
t =T +bf
By property (c) and (3.142), T +2bf −1
T +2bf −1
f (t, xt∗ , u∗t ) ≥
t =T
f (t, xt , ut ) − /8.
t =T
It follows from (3.140), (3.143), and (3.145)–(3.147) that T −1 t =S
T +2bf −1
f (t, xt∗ , u∗t )
+
t =T
T +2bf −1 f f f (t, xt , ut ) −
/8 ≤
t =S
f (t, xt∗ , u∗t )
3.15 An Auxiliary Result for Theorem 3.24 T +2bf −1
≤
f (t, yt , vt ) =
t =S
T −1
183 T +2bf −1
f (t, xt , ut ) +
t =S
f
f
f (t, xt , ut ) + /4
t =T
and T −1
T −1
f (t, xt∗ , u∗t ) ≤
t =S
f (t, xt , ut ) +
t =S
for all integers T ≥ τ0 + L1 . Since is any positive number, we conclude that ∗ ∞ ({xt∗ }∞ t =S , {ut }t =S ) is (f )overtaking optimal and (i) holds. Theorem 3.20 is proved. Theorem 3.19 follows from Proposition 3.38 and Theorem 3.20.
3.15 An Auxiliary Result for Theorem 3.24 Lemma 3.39. Let Δ > 0. Then there exists δ > 0 such that for each (T1 , z1 ), (T2 , z2 ) ∈ A satisfying T2 ≥ T1 + 2bf ,
(3.148)
f
ρE (zi , xTi ) ≤ δ, i = 1, 2
(3.149)
the following inequality holds: U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f
f
f (t, xt , ut ) + Δ.
t =T1
Proof. Let δ > 0 be as guaranteed by (A2) with = Δ/4. Let (T1 , z1 ), (T2 , z2 ) ∈ A satisfy (3.148) and (3.149). By (3.148), (3.149), and (A2) with = Δ/4, there 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) such that exists ({yt }Tt =T 1 1 yT1 = z1 , yT2 = z2 , f
(3.150)
f
yt = xt , t = T1 +bf , . . . , T2 −bf , vt = ut , t ∈ {T1 +bf , . . . , T2 −bf }\{T2 −bf }, (3.151) T1 +bf −1 T1 +bf −1 f f f (t, yt , vt ) ≤ f (t, xt , ut ) + Δ/4, (3.152) t =T1
t =T1
T 2 −1
T 2 −1
t =T2 −bf
f (t, yt , vt ) ≤
t =T2 −bf
f
f
f (t, xt , ut ) + Δ/4.
(3.153)
184
3 DiscreteTime Nonautonomous Problems on the HalfAxis
In view of (3.150)–(3.153), U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f (t, yt , vt ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + Δ/2.
t =T1
Lemma 3.39 is proved.
3.16 Proof of Theorem 3.24 Assume that STP holds. Theorem 3.17 implies that (P1) and (P2) hold. Let us show that (P3) holds. By Theorem 3.19, there exists an (f )overtaking optimal f ∗ ∞ ∗ ∞ ({xt∗ }∞ ˜ t }∞ t =0 , {ut }t =0 ) ∈ X(0, ∞) such that x0 = x0 . Assume that ({x˜ t }t =0 , {u t =0 ) ∈ f X(0, ∞) is (f )overtaking optimal and that x˜0 = x0 . By STP, (P1) and Theof rem 3.1, x˜t = xt for all integers t ≥ 0. f Assume that (P1), (P2), and (P3) hold and that u˜ t = ut for all integers t ≥ 0. f ∞ f ∞ Thus, ({xt }t =0 , {ut }t =0 ) ∈ X(0, ∞) is (f )overtaking optimal. Lemma 3.40. Let ∈ (0, 1). Then there exists δ > 0 such that for each pair 2 2 −1 of integers T1 ≥ 0, T2 ≥ T1 + 3bf and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 satisfying f
ρE (xTi , xTi ) ≤ δ, i = 1, 2 T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 . Proof. Lemma 3.35 and (A2) imply that there exist δ0 ∈ (0, /4) and an integer L0 > 0 such that (i) (A2) holds with = 1 and δ = δ0 ; (ii) for each pair of integers T1 ≥ L0 , T2 ≥ T1 + 2bf and each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) ({xt }Tt =T 1 1
satisfying f
ρE (xTi , xTi ) ≤ δ0 , i = 1, 2
3.16 Proof of Theorem 3.24 T 2 −1
185
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ0
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 . Theorem 3.1 implies that there exists S0 > 0 such that for each pair of integers 2 2 −1 T2 > T1 ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
(3.154)
t =T1
It follows from (P2) and Lemma 3.36 that there exist a natural number L1 such that the following property holds: +L1 +L1 −1 (iii) for each integer T ≥ 0 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L1 ) satisfying T +L 1 −1
f (t, xt , ut ) ≤
t =T
T +L 1 −1
f
f
f (t, xt , ut ) + 2S0 + 3
t =T
we have f
min{ρE (xt , xt ) : t = T , . . . , T + L1 } ≤ δ0 . Consider a sequence {δi }∞ i=1 ⊂ (0, 1) such that δi < 2−1 δi−1 , i = 2, 3, . . . .
(3.155)
Assume that the lemma does not hold. Then for each integer i ≥ 1, there exist Ti,2 Ti,2 −1 integers Ti,1 ≥ 0, Ti,2 ≥ Ti,1 + 3bf , and ({xt(i)}t =T , {u(i) t }t =Ti,1 ) ∈ X(Ti,1 , Ti,2 ) i,1 such that (i)
f
ρE (xTi,j , xTi,j ) ≤ δi , j = 1, 2,
(3.156)
Ti,2 −1
t =Ti,1
(i) (i) f f (t, xt(i) , u(i) t ) ≤ U (Ti,1 , Ti,2 , xTi,1 , xTi,2 ) + δi
(3.157)
and ti ∈ {Ti,1 , . . . , Ti,2 } for which f
ρE (xt(i) , xti ) > . i
(3.158)
186
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Let i be a natural number. Property (ii) and (3.155)–(3.158) imply that Ti,1 < L0 .
(3.159)
ti ≤ 2bf + 1 + L1 + L0 .
(3.160)
We show that
Property (i), (A2) with = 1, δ = δ0 , (3.155), and (3.156) imply that there exists Ti,2 Ti,2 −1 ({yt(i)}t =T , {vt(i) }t =T ) ∈ X(Ti,1 , Ti,2 ) such that i,1 i,1 (i)
(i)
yt
(i)
(i)
(i)
yTi,1 = xTi,1 , yTi,2 = xTi,2 ,
(3.161)
f
(3.162)
= xt , t ∈ {Ti,1 + bf , . . . , Ti,2 − bf },
f
vt(i) = ut , t ∈ {Ti,1 + bf , . . . , Ti,2 − bf − 1}, Ti,1 +bf −1
(3.163)
Ti,1 +bf −1
f (t, yt(i) , vt(i) ) ≤
t =Ti,1
f
f
(3.164)
f
f
(3.165)
f (t, xt , ut ) + 1,
t =Ti,1
Ti,2 −1
Ti,2 −1
(i)
(i)
f (t, yt , vt ) ≤
t =Ti,2 −bf
f (t, xt , ut ) + 1.
t =Ti,2 −bf
It follows from (3.155), (3.157), and (3.161)–(3.165) that Ti,2 −1
Ti,2 −1 (i) (i) f (t, xt , ut )
t =Ti,1
≤
(i)
(i)
f (t, yt , vt ) + 1
t =Ti,1 Ti,2 −1
≤
f
f
f (t, xt , ut ) + 3.
(3.166)
t =Ti,1
In view of (3.154), (3.166), for each pair of integers S1 , S2 ∈ [Ti,1 , Ti,2 ] satisfying S1 < S2 , S 2 −1 t =S1
−
Ti,2 −1
f (t, xt(i), u(i) t )=
f (t, xt(i) , u(i) t )
t =Ti,1
(i) (i) {f (t, xt , ut ) : t ∈ {Ti,1 , . . . , S1 } \ {S1 }}
3.16 Proof of Theorem 3.24
−
187
{f (t, xt(i) , u(i) t ) : t ∈ {S2 , . . . , Ti,2 } \ {Ti,2 }}
Ti,2 −1
≤
f
f
f (t, xt , ut ) + 3 −
f f {f (t, xt , ut ) : t ∈ {Ti,1 , . . . , S1 } \ {S1 }} + S0
t =Ti,1
−
S 2 −1 f f f f {f (t, xt , ut ) : t ∈ {S2 , . . . , Ti,2 }\{Ti,2 }}+S0 ≤ f (t, xt , ut )+3+2S0. t =S1
(3.167) Assume that (3.160) does not hold. Then ti > 2bf + 1 + L1 + L0 .
(3.168)
In view of (3.168), [ti − 2bf − 1 − L1 , ti − 1 − 2bf ] ⊂ [L0 , ∞). t −2b −1
(3.169)
t −2b −2
f i f , {u(i) Consider the pair ({xt(i) }ti=ti −L t }t =ti −L1 −2bf −1 ) ∈ X(ti − L1 − 2bf − 1 −2bf −1 1, ti − 2bf − 1). Property (iii) and (3.167) imply that there exists
t˜ ∈ {ti − L1 − 2bf − 1, . . . , ti − 2bf − 1}
(3.170)
f
(3.171)
such that ρE (xt˜, xt˜ ) ≤ δ0 . By property (ii), (3.155)–(3.157), and (3.169)–(3.171), ρE (xt , xt ) ≤ , t = t˜, . . . , Ti,2 f
f
and in particular, ρE (xti , xti ) ≤ . This contradicts (3.158). The contradiction we have reached proves (3.160). In view of (3.12) and (3.167), there exists M1 > 0 such that (i)
(i)
f (t, xt , ut ) ≤ M1 for each integer i ≥ 1 each t ∈ {Ti,1 , . . . , Ti,2 − 1}. (3.172) It follows from (3.5), (3.6), and (3.172) that there exists M2 > 0 such that ρE (xt(i), θ0 ) ≤ M2 , i = 1, 2, . . . , t ∈ {Ti,1 , . . . , Ti,2 − 1}.
(3.173)
188
3 DiscreteTime Nonautonomous Problems on the HalfAxis
Extracting a subsequence and reindexing, we may assume without loss of generality that Ti,1 = T1,1 for all integers i ≥ 1,
(3.174)
ti = t1 for all integers i ≥ 1,
(3.175)
there exists lim Ti,2 .
(3.176)
lim Ti,2 < ∞,
(3.177)
lim Ti,2 = ∞.
(3.178)
i→∞
There are two cases: i→∞
i→∞
Assume that (3.177) holds. Then extracting a subsequence and reindexing we may assume that Ti,2 = T1,2 for all integers i ≥ 1.
(3.179)
In view of (3.174), (3.176), (3.179), and LSC property, extracting a subsequence and reindexing, we may assume without loss of generality that there exists T1,2 T1,2 −1 , { ut }t =T ) ∈ X(T1,1 , T1,2 ) such that ({ xt }t =T 1,1 1,1 xt as i → ∞ for all t ∈ {T1,1 , . . . , T1,2 }, xt(i) → there exists limi→∞
T1,2 −1 t =T1,1
f (t, xt(i), u(i) t ) and
T1,2 −1
(3.180)
T1,2 −1
f (t, xt , ut ) ≤ lim
i→∞
t =T1,1
(i)
(i)
(3.181)
f (t, xt , ut ).
t =T1,1
It follows from (3.155), (3.156), (3.158), (3.175), and (3.180) that f
f
f
xT1,1 = xT1,1 , xT1,2 = xT1,2 , ρE ( xt1 , xt1 ) ≥ .
(3.182)
Lemmas 3.39, (3.155)–(3.157), and (3.181) imply that T1,2 −1
t =T1,1
T1,2 −1
f (t, xt , ut ) ≤ lim inf U f (T1,1 , T1,2 , xT(i) , xT(i) )≤ 1,1 1,2 i→∞
f
f
f (t, xt , ut ).
t =T1,1
(3.183)
3.16 Proof of Theorem 3.24
189
It is clear that (3.182) and (3.183) contradict (P3). The contradiction we have reached proves that (3.177) does not hold. Therefore (3.178) holds. In view of (3.172) and LSC property, extracting a subsequence, using the diagonalization process, and reindexing, we may assume without loss of generality that for every integer t ≥ T1,1 , there exists limi→∞ f (t, xt(i) , u(i) t ) and there exists ∞ ({ x t }∞ , { u } ) ∈ X(T , ∞) such that for every integer t ≥ T1,1 , t t =T1,1 1,1 t =T1,1 xt = lim xt(i) ,
(3.184)
f (t, xt , ut ) ≤ lim f (t, xt(i) , u(i) t ).
(3.185)
i→∞
i→∞
By (3.155), (3.156), (3.158), (3.174), (3.175), and (3.184), f
f
xT ,1 = xT ,1, ρE (xt1 , xt1 ) ≥ .
(3.186)
For all nonnegative integers t < T1,1 , set f
xt = xt . Let Δ > 0. Lemma 3.39 implies that there exists δ > 0 such that the following property holds: (iv) for each pair of integers τ1 ≥ 0, τ2 ≥ τ1 + 2bf and each 2 2 −1 , {ut }τt =τ ) ∈ X(τ1 , τ2 ) ({xt }τt =τ 1 1
satisfying ρE (xτi , xτfi ) ≤ δ, i = 1, 2, τ 2 −1
f (t, xt , ut ) ≤ U f (τ1 , τ2 , xτ1 , xτ2 ) + δ
t =τ1
we have τ 2 −1 t =τ1
f (t, xt , ut ) ≤
τ 2 −1
f
f
f (t, xt , ut ) + Δ/2.
t =τ1
In view of (3.155), there exists an integer k0 ≥ 1 such that δk < δ for all integers k ≥ k0 .
(3.187)
190
3 DiscreteTime Nonautonomous Problems on the HalfAxis
By (3.156), (3.157), (3.174), and (3.187), the following property holds: (v) for each integer k ≥ k0 , Tk,2 −1
Tk,2 −1 (k)
(k)
f (t, xt , ut ) ≤
t =T1,1
f
f
f (t, xt , ut ) + Δ/2.
t =T1,1
It follows from (3.167) and (3.185) that for each pair of integers S1 , S2 ∈ [T1,1, ∞) satisfying S1 < S2 , S 2 −1
f (t, xt , ut ) ≤
t =S1
S 2 −1
f
f
f (t, xt , ut ) + 3 + S0 .
(3.188)
t =S1
Theorem 3.1 and (3.188) imply that ({ x t }∞ ut } ∞ t =0 , { t =0 ) is (f )good. (P1) implies that f
lim ρE ( xt , xt ) = 0.
t →∞
(3.189)
In view of (3.189), there exists an integer τ0 > 0 such that f
ρE ( xt , xt ) ≤ δ/4 for all integers t ≥ τ0 .
(3.190)
Let q ≥ 1 be an integer satisfying (3.191)
Tq,2 > τ0 .
By (3.184), there exists an integer k1 ≥ k0 such that for each integer k ≥ k1 , (k)
Tk,2 > Tq,2 , ρE ( xTq,2 , xTq,2 ) ≤ δ/4.
(3.192)
Assume that an integer k ≥ k1 . Then (3.192) holds. By (3.156), (3.157), (3.174), (3.187), and (3.190)–(3.192), Tq,2 −1
t =T1,1
(k) (k) f f (t, xt(k) , u(k) t ) ≤ U (T1,1 , Tq,2 , xT1,1 , xTq,2 ) + δ,
(3.193)
f
(3.194)
ρE (xT(k) , xT1,1 ) ≤ δk ≤ δ, 1,1 f
f
ρE (xT(k) , xTq,2 ) ≤ ρE (xT(k) , xTq,2 ) + ρE ( xTq,2 , xTq,2 ) ≤ δ/4 + δ/4. q,2 q,2
(3.195)
3.17 Proof of Theorem 3.26
191
Property (iv), (3.174) and (3.193)–(3.195) imply that Tq,2 −1
Tq,2 −1
f (t, xt(k), u(k) t )
t =T1,1
≤
f
f
f (t, xt , ut ) + Δ/2.
(3.196)
t =T1,1
In view of (3.185), Tq,2 −1
Tq,2 −1
f (t, xt , ut ) ≤
t =T1,1
f
f
f (t, xt , ut ) + Δ/2.
t =T1,1
Since the relation above holds for every integer q ≥ 1 satisfying Tq,2 > τ0 , we have T −1
lim inf( T →∞
T −1
f (t, xt , ut ) −
t =T1,1
f
f
f (t, xt , ut )) ≤ Δ.
t =T1,1
Since Δ is an arbitrary positive number, the inequality above implies that lim inf( T →∞
T −1
f (t, xt , ut ) −
t =T1,1
T −1
f
f
f (t, xt , ut )) ≤ 0
t =T1,1
ut } ∞ and ({ x t }∞ t =0 , { t =0 ) is (f )weakly optimal. Theorem 3.20 implies that f ∞ ∞ ({ xt }t =0 , { ut }t =0) is (f )overtaking optimal. In view of (P3), xt = xt for all integers t ≥ 0. This contradicts (3.186). The contradiction we have reached completes the proof of Lemma 3.40. Completion of the Proof of Theorem 3.24. Theorem 3.17 implies TP. Lemma 3.40 and TP imply STP.
3.17 Proof of Theorem 3.26 We show that fr possesses (P2). Let , M > 0. Theorem 3.1 implies that there exists M0 > 0 such that for each pair of integers S2 > S1 ≥ 0 and each 2 2 −1 ({xt }St =S , {ut }St =S ) ∈ X(S1 , S2 ), we have 1 1 S 2 −1 t =S1
f (t, xt , ut ) + M0 ≥
S 2 −1
f
f
f (t, xt , ut ).
(3.197)
t =S1
Property (i) (see Section 3.5) implies that there exists δ > 0 such that the following property holds:
192
3 DiscreteTime Nonautonomous Problems on the HalfAxis
(a) for each x1 , x2 ∈ E satisfying φ(x1 , x2 ) ≤ δ, we have ρE (x1 , x2 ) ≤ . Choose a natural number L > 2(rδ)−1 (M + M0 ).
(3.198)
+L +L−1 Assume that a integer T ≥ 0 and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L) satisfy T +L−1
T +L−1
fr (t, xt , ut ) ≤
t =T
f
f
fr (t, xt , ut ) + M.
(3.199)
t =T
It follows from (3.14) and (3.197)–(3.199) that T +L−1
f (t, xt , ut ) + r
T +L−1
t =T
≤
f
φ(xt , xt ) =
t =T T +L−1
f
T +L−1
fr (t, xt , ut )
t =T
f
fr (t, xt , ut ) + M =
T +L−1
t =T
f
f
f (t, xt , ut ) + M
t =T
≤
T +L−1
f (t, xt , ut ) + M + M0 ,
t =T
r
T +L−1
f
φ(xt , xt ) ≤ M0 + M,
t =T
min{φ(xt , xt ) : t ∈ {T , . . . , T + L − 1}} ≤ (Lr)−1 (M0 + M) < δ f
f
and there exists an integer τ ∈ [T , T + L − 1] such that φ(xτ , xτ ) < δ. Property f (a) implies that ρE (xτ , xτ ) < . Therefore fr has (P2). We show that fr has (P1). Let T0 ≥ 0 be an integer and let ∞ ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞)
be (fr )good. We claim that f
lim ρE (xt , xt ) = 0.
t →∞
(3.200)
Assume that (3.200) does not hold. Then there exist > 0 and a sequence of natural numbers {tk }∞ k=1 ⊂ [T0 , ∞) such that for each integer k ≥ 1, f
tk+1 ≥ tk + 8, ρE (xtk , xtk ) > .
(3.201)
3.17 Proof of Theorem 3.26
193
There exists M0 > 0 such that for each integer T > T0 , M0 > 
T −1
fr (t, xt , ut ) −
t =T0
=
T −1
f (t, xt , ut ) −
t =T0
T −1
f
f
fr (t, xt , ut )
t =T0
T −1
f
f
f (t, xt , ut ) + r
t =T0
T −1
f
φ(xt , xt ).
(3.202)
t =T0
Set ∞
Δ=
f
(3.203)
φ(xt , xt ).
t =T0
Theorem 3.1, (3.202), and (3.203) imply that Δ < ∞.
(3.204) f
In view of (3.204), for every δ > 0 the set {t is an integer : t ≥ T0 , φ(xt , xt ) ≥ δ} is finite. Together with property (a), this implies that the set f
{t is an integer : t ≥ T0 , ρE (xt , xt ) > } f
is finite for any > 0. This implies that limt →∞ ρE (xt , xt ) = 0. Thus fr has (P1). f f ∞ Assume that ({xt }∞ t =0 , {ut }t =0 ) is (f )minimal. Theorem 3.20 and (3.14) imply f ∞ f ∞ that ({xt }t =0 , {ut }t =0 ) is (fr )minimal and (fr )overtaking optimal. ∞ Let T0 ≥ 0 be an integer and ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) be (fr )overtaking optimal and f
(3.205)
f
(3.206)
xT0 = xT0 . Property (P1) and (3.205) imply that lim ρE (xt , xt ) = 0.
t →∞ f
We show that xt = xt for all integers t ≥ T0 . Assume the contrary. Then there exists an integer t0 > T0 such that f
γ := ρ(xt0 , xt0 ) > 0.
(3.207)
Property (a) implies that there exists δ0 > 0 such that the following property holds: if z1 , z2 ∈ E, φ(z1 , z2 ) ≤ δ0 , then ρE (z1 , z2 ) ≤ γ /4.
(3.208)
194
3 DiscreteTime Nonautonomous Problems on the HalfAxis
In view of (3.207) and (3.208), f
(3.209)
φ(xt0 , xt0 ) > δ0 . By (3.14), (3.205) and (3.209), 0 = lim [ T →∞
= lim [ T →∞
fr (t, xt , ut ) −
t =T0
T −1
f (t, xt , ut ) + r
T →∞
f
f (t, xt , ut ) −
T −1
f
fr (t, xt , ut )]
T −1
f
φ(xt , xt ) −
t =T0
t =T0
≥ lim [
T −1 t =T0
t =T0
= lim [ T →∞
T −1
T −1
T −1
T −1
t =T0
f
t =T0
f
f
f (t, xt , ut )] + r
t =T0
∞
f
φ(xt , xt )
t =T0
T −1
f (t, xt , ut ) −
f
f (t, xt , ut )]
f
f
f (t, xt , ut )] + δ0 .
(3.210)
t =T0
By (3.210), there exists an integer T1 > T0 such that for each integer T ≥ T1 , T −1
f (t, xt , ut ) −
t =T0
T −1
f
f
f (t, xt , ut ) ≤ −δ0 /2.
(3.211)
t =T0
It follows from (A2) that there exists δ1 > 0 such that the following property holds: f
(b) for each (S, z) ∈ A satisfying ρE (z, xS ) ≤ δ1 , there exist an integer τ ∈ (0, bf ], S+τ −1 ({yt }S+τ ) ∈ X(S, S + τ ) t =S , {vt }t =S
such that f
yS = z, yS+τ = xS+τ , S+τ −1 t =S
f (t, yt , vt ) ≤
S+τ −1 t =S
f (t, xt , ut ) + 8−1 δ0 . f
f
3.17 Proof of Theorem 3.26
195
In view of (3.206), there exists an integer T2 > T1 such that f
ρE (xt , xt ) ≤ δ1 for all integers t ≥ T2 .
(3.212)
Property (b) and (3.212) imply that there exists T +t +bf
2 0 ({yt }t =T 0
T +t +bf −1
2 0 , {vt }t =T 0
) ∈ X(T0 , T2 + t0 + bf )
such that yt = xt , t = T0 , . . . , T2 + t0 , vt = ut , t = T0 , . . . , T2 + t0 − 1,
(3.213)
f
yT2 +t0 +bf = xT2 +t0 +bf , T2 +t0 +bf −1
(3.214)
T2 +t0 +bf −1
f (t, yt , vt ) ≤
t =T2 +t0
f (t, xt , ut ) + 8−1 δ0 . f
f
(3.215)
t =T2 +t0
In view of (3.205) and (3.213), f
yT0 = xT0 .
(3.216)
It follows from (3.211), (3.213), and (3.215) that T2 +t0 +bf −1
f (t, yt , vt ) ≤
t =T0
≤
T2 +t0 −1
T2 +t0 +bf −1
t =T0 T2 +t0 −1
f (t, xt , ut ) +
f (t, xt , ut ) + 8−1 δ0 f
f
t =T2 +t0 T2 +t0 +bf −1
f
f
f (t, xt , ut ) − δ0 /2 +
t =T0
f (t, xt , ut ) + 8−1 δ0 f
f
t =T2 +L0 T2 +t0 +bf −1
=
f
f
f (t, xt , ut ) − 3δ0 /8.
t =T0 ∞ In view of (3.214) and (3.216), ({xt }∞ t =0 , {ut }t =T ) is not (f )minimal, a contradicf tion. The contradiction we have reached proves that xt = xt for all integers t ≥ T0 and (P3) holds for fr . This completes the proof of Theorem 3.26. f
f
Chapter 4
DiscreteTime Nonautonomous Problems on Axis
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for discretetime nonautonomous problems on subintervals of axis in metric spaces, which are not necessarily compact. For these optimal control problems the turnpike is not a singleton. We also study the existence of solutions of the corresponding infinite horizon optimal control problems.
4.1 Preliminaries and Main Results Denote by Z the set of all integers. Let (E, ρE ) and (F, ρF ) be metric spaces. We suppose that A is a nonempty subset of Z × E, U : A → 2F is a point to set mapping with a graph M = {(t, x, u) : (t, x) ∈ A, u ∈ U(t, x)}, G : M → E and f : M → R 1 . Let T1 < T2 be integers. We denote by X(T1 , T2 ) the set of all pairs of sequences 2 2 −1 , {ut }Tt =T ) such that for each integer t ∈ {T1 , . . . , T2 }, ({xt }Tt =T 1 1 (t, xt ) ∈ A, for each integer t ∈ {T1 , . . . , T2 − 1}, ut ∈ U(t, xt ),
(4.1)
xt +1 = G(t, xt , ut )
(4.2)
and which are called trajectorycontrol pairs. © Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_4
197
198
4 DiscreteTime Nonautonomous Problems on Axis
Let T1 be an integer. Denote by X(T1 , ∞) the set of all pairs of sequences ∞ {xt }∞ t =T1 ⊂ E, {ut }t =T1 ⊂ F , which are called trajectorycontrol pairs, such 2 2 −1 that for each integer T2 > T1 , ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). We denote by 1 1 ∞ X(−∞, ∞) the set of all pairs of sequences {xt }∞ t =−∞ ⊂ E, {ut }t =−∞ ⊂ F , which are called trajectorycontrol pairs, such that for each pair of integers T2 > T1 , 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ). ({xt }Tt =T 1 1
2 Let T1 < T2 be integers. A sequence {xt }Tt =T ⊂ E ({xt }∞ t =T1 ⊂ E, 1 ∞ {xt }t =−∞ ⊂ E, respectively) is called a trajectory if there exists a sequence 2 −1 ∞ ⊂ F ({ut }∞ {ut }Tt =T t =T1 ⊂ F , {ut }t =−∞ ⊂ F , respectively) referred to as a control 1
2 2 −1 ∞ such that ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) (({xt }∞ t =T1 , {ut }t =T1 ) ∈ X(T1 , ∞), 1 1 ∞ ∞ ({xt }t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞), respectively). Let θ0 ∈ E and a0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that
ψ(t) → ∞ as t → ∞.
(4.3)
We suppose that the function f satisfies f (t, x, u) ≥ ψ(ρE (x, θ0 )) − a0 for each (t, x, u) ∈ M.
(4.4)
For each pair of integers T2 > T1 and each pair of points y, z ∈ E satisfying (T1 , y), (T2 , z) ∈ A we consider the following problems: T 2 −1 t =T1
2 2 −1 f (t, xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, xT2 = z, 1 1
T 2 −1 t =T1
(P1 ) 2 2 −1 f (t, xt , ut ) → min, ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), xT1 = y, 1 1
T 2 −1 t =T1
f (t, xt , ut ) → min, ({xt }TT21 , {ut }TT21 −1 ) ∈ X(T1 , T2 ).
(P2 )
(P3 )
For each pair of integers T2 > T1 and each pair of points y, z ∈ E satisfying (T1 , y), (T2 , z) ∈ A we define U (T1 , T2 , y, z) = inf{ f
T 2 −1 t =T1
2 2 −1 f (t, xt , ut ) : ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
xT1 = y, xT2 = z}, σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ E, (T2 , h) ∈ A},
(4.5) (4.6)
4.1 Preliminaries and Main Results
199
σ f (T1 , T2 , z) = inf{U f (T1 , T2 , h, z) : h ∈ E, (T1 , h) ∈ A},
(4.7)
σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, ξ ) : h, ξ ∈ E, (T1 , h), (T2 , ξ ) ∈ A}.
(4.8)
We suppose that bf > 0 is an integer, ∞ ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞), f
f
f
{xt : t ∈ Z} is bounded, f
(4.9)
f
Δf := sup{f (t, xt , ut ) : t ∈ Z} < ∞
(4.10)
and that the following assumptions hold. (A1) For each S1 > 0 there exist S2 > 0 and an integer c > 0 such that T 2 −1
f
f
f (t, xt , ut ) ≤
t =T1
T 2 −1
f (t, xt , ut ) + S2
t =T1
2 2 −1 for each pair of integers T1 , T2 ≥ T1 + c and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 . (A2) For each > 0 there exists δ > 0 such that for each (T , z) ∈ A, satisfying f ρE (z, xT ) ≤ δ there exist integers τ1 , τ2 ∈ (0, bf ],
+τ1 +τ1 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + τ1 ), (1)
(1)
T −1 ({xt(2)}Tt=T −τ2 , {u(2) t }t =T −τ2 ) ∈ X(T − τ2 , T )
such that (1)
f
(1)
xT = z, xT +τ1 = xT +τ1 , T +τ 1 −1
(1)
(1)
f (t, xt , ut ) ≤
t =T
T +τ 1 −1
f
f
f
f
f (t, xt , ut ) + ,
t =T f
xT(2) = z, xT(2)−τ2 = xT −τ2 , T −1 t =T −τ2
f (t, xt(2), u(2) t )≤
T −1 t =T −τ2
f (t, xt , ut ) + ,
200
4 DiscreteTime Nonautonomous Problems on Axis
Many examples of optimal control problems satisfying assumptions (A1) and (A2) can be found in [106–108, 118, 124, 125, 134]. The following result is proved in Section 4.6. Theorem 4.1. 1. There exists S > 0 such that for each pair of integers T2 > T1 and each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), ({xt }Tt =T 1 1 T 2 −1
f (t, xt , ut ) + S ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
∞ 2. For each ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) either T −1
f (t, xt , ut ) −
t =−T
T −1
f
f
f (t, xt , ut ) → ∞ as T → ∞
t =−T
or sup{
T −1
f (t, xt , ut )−
t =−T
T −1
f
f
f (t, xt , ut ) : T ∈ {1, 2, . . . }} < ∞.
(4.11)
t =−T
Moreover, if (4.11) holds, then sup{ρE (xt , θ0 ) : t ∈ Z} < ∞. ∞ We say that ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) is (f )good [29, 104, 116, 120, 129, 134] if
sup{
T −1 t =−T
f (t, xt , ut ) −
T −1
f
f
f (t, xt , ut ) : T ∈ {1, 2, . . . }} < ∞.
t =−T
4.2 Boundedness Results The next result is proved in Section 4.6. Theorem 4.2. Let M0 > 0. Then there exists M1 > 0 such that for each pair of 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying integers T2 > T1 and each ({xt }Tt =T 1 1
4.2 Boundedness Results
201
T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Let L > 0 be an integer. Denote by AL the set of all (S, z) ∈ A for which there S+τ −1 exist an integer τ ∈ (0, L] and ({xt }S+τ ) ∈ X(S, S + τ ) such that t =S , {ut }t =S f
xS = z, xS+τ = xS+τ ,
S+τ −1
f (t, xt , ut ) ≤ L.
t =S
L the set of all (S, z) ∈ A such that there exist an integer τ ∈ (0, L] Denote by A S and ({xt }t =S−τ , {ut }S−1 t =S−τ ) ∈ X(S − τ, S) satisfying f
xS−τ = xS−τ , xS = z,
S−1
f (t, xt , ut ) ≤ L.
t =S−τ
The next three results are proved in Section 4.7. Theorem 4.3. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such 2 2 −1 that for each integer T1 , each integer T2 ≥ T1 + 2L and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying L , (T1 , xT1 ) ∈ AL , (T2 , xT2 ) ∈ A T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 4.4. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such 2 2 −1 that for each integer T1 , each integer T2 ≥ T1 + L and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying (T1 , xT1 ) ∈ AL ,
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. Theorem 4.5. Let L > 0 be an integer and M0 > 0. Then there exists M1 > 0 such 2 2 −1 that for each integer T1 , each integer T2 ≥ T1 + L and each ({xt }Tt =T , {ut }Tt =T )∈ 1 1 X(T1 , T2 ) satisfying
202
4 DiscreteTime Nonautonomous Problems on Axis
L , (T2 , xT2 ) ∈ A
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT2 ) + M0
t =T1
the inequality ρE (θ0 , xt ) ≤ M1 holds for all t = T1 , . . . , T2 − 1. ∞ Theorem 4.6. For each integer T0 and each ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) either T −1
f (t, xt , ut ) −
t =T0
T −1
f
f
f (t, xt , ut ) → ∞ as T → ∞
t =T0
or sup{
T −1
f (t, xt , ut ) −
t =T0
T −1
f
f
f (t, xt , ut ) : T > T0 is an integer} < ∞.
t =T0
(4.12) Moreover, if (4.12) holds, then {xt : t ∈ Z, t ≥ T0 } is bounded. Theorem 4.6 easily follows from Assertion 1 of Theorem 4.1 and Theorem 4.2. ∞ Let T0 be an integer. We say that ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) is (f )good [29, 104, 116, 120, 129, 134] if (4.12) holds. ∞ Let T0 be an integer. We say that ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) is (f )minimal if for each integer T > T0 , T −1
f (t, xt , ut ) = U f (T0 , T , xT0 , xT ).
t =T0 ∞ We say that ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) is (f )minimal [12, 94] if for each pair of integers T2 > T1 , T 2 −1
f (t, xt , ut ) = U f (T1 , T2 , xT1 , xT2 ).
t =T1
4.3 Turnpike Results We say that f possesses the strong turnpike property (or STP for short) if for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 such that for each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) integer T1 , each integer T2 ≥ T1 + 2L, and each ({xt }Tt =T 1 1 which satisfies
4.3 Turnpike Results T 2 −1
203
f (t, xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ}
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ, then τ2 = T2 , and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . We say that f possesses lower semicontinuity property (or LSC property for short) if for each pair of integers T2 > T1 and each sequence (j ) 2 (j ) 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), j = 1, 2, . . . which satisfies ({xt }Tt =T 1 1 sup{
T 2 −1
(j )
(j )
f (t, xt , ut ) : j = 1, 2, . . . } < ∞
t =T1 (jk ) T2 (j ) 2 −1 }t =T1 , {ut k }Tt =T ), 1
there exist a subsequence ({xt
k = 1, 2, . . . and
2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that for any t ∈ {T1 , . . . , T2 }, (jk )
xt T 2 −1
→ xt as k → ∞,
f (t, xt , ut ) ≤ lim inf j →∞
t =T1
T 2 −1
(j )
(j )
f (t, xt , ut ).
t =T1
LSC property plays an important role in the calculus of variations and optimal control theory [32]. The next theorem is the main result of this chapter. It is proved in Sections 4.9 and 4.10. Theorem 4.7. Let f have LSC property. If f has STP, then the following properties hold: (P1)
∞ for each (f )good pair ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞), f
lim ρ(xt , xt ) = 0,
t →∞
(P2)
f
lim ρ(xt , xt ) = 0;
t →−∞
for each > 0 and each M > 0, there exist δ > 0 and an integer L > 0 +L +L−1 such that for each integer T and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + L) which satisfies
204
4 DiscreteTime Nonautonomous Problems on Axis T +L−1
f (t, xt , ut ) ≤ min{U f (T , T + L, xT , xT +L )
t =T
+δ,
T +L−1 t =T
f
f
f (t, xt , ut ) + M}, f
(P3)
there exists an integer s ∈ [T , T + L] such that ρE (xs , xs ) ≤ . f f ∞ f there exists {u˜ t }∞ ˜ t }∞ t =−∞ ⊂ F such that ({xt }t =−∞ , {u t =−∞ ) ∈ X(−∞, ∞) is (f )good and (f )minimal and for each (f )good and f ∞ (f )minimal ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞), the equality xt = xt holds for all integers t. f
f
If (P1), (P2), and (P3) hold and u˜ t = ut for all integers t ≥ 0, then f has STP. The next result is proved in Section 4.11. Theorem 4.8. f has properties (P1) and (P2) if and only if the following property holds: (P4)
for each > 0 and each M > 0 there exist δ > 0 and an integer L > 0 such that for each integer T1 , each integer T2 > T1 and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 T 2 −1
f (t, xt , ut ) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , xT1 , xT2 ) + δ}
t =T1 f
if T2 ≥ (T1 )+ + 2L, then ρE (xt , xt ) ≤ , t = (T1 )+ + L, . . . , T2 − L, and f if T1 ≤ (T2 )− − 2L, then ρE (xt , xt ) ≤ , t = T1 + L, . . . , (T2 )− − L. Theorem 4.9. Assume that f has STP, that L0 > 0 is an integer, and that > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 , each 2 2 −1 integer T2 ≥ T1 + 2L and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 L0 , (T1 , xT1 ) ∈ AL0 , (T2 , xT2 ) ∈ A T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that ρE (xt , xf ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . Theorem 4.9 follows from STP and the following result which is proved in Section 4.7.
4.3 Turnpike Results
205
Proposition 4.10. Let L > 0 be an integer. There exists M0 > 0 such that for each L satisfying T2 ≥ T1 + 2L, (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A U f (T1 , T2 , z1 , z2 ) ≤ σ f (T1 , T2 ) + M0 . Theorem 4.11. Assume that f has STP, L0 > 0 is an integer and that > 0. Then there exist δ > 0 and an integer L > L0 such that for each integer T1 , each integer 2 2 −1 T2 ≥ T1 + 2L and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies 1 1 (T1 , xT1 ) ∈ AL0 ,
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 , xT1 ) + δ
t =T1
there exist integers τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . f
f
Moreover, if ρE (xT2 , xT2 ) ≤ δ then τ2 = T2 and if ρE (xT1 , xT1 ) ≤ δ, then τ1 = T1 . Theorem 4.11 follows from STP and the next result which is proved in Section 4.7. Proposition 4.12. Let L0 > 0 be an integer. Then there exists M > 0 such that for each (T1 , z1 ) ∈ AL0 , each integer T2 ≥ T1 + L0 , σ f (T1 , T2 , z1 ) ≤ σ f (T1 , T2 ) + M. We say that f possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0, there exist natural numbers Q, l such that for each 2 2 −1 integer T1 , each integer T2 ≥ T1 + lQ, and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 which satisfies T 2 −1 t =T1
f (t, xt , ut ) ≤
T 2 −1
f
f
f (t, xt , ut ) + M
t =T1 q
q
there exist finite sequences of integers {ai }i=1 , {bi }i=1 ⊂ {T1 , . . . , T2 } such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, f
q
ρE (xt , xt ) ≤ for all integers t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. The next result is proved in Section 4.12.
206
4 DiscreteTime Nonautonomous Problems on Axis
Theorem 4.13. f has WTP if and only if f has (P1), (P2). ∞ Let T0 be an integer. A pair ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) is called (f )∞ ∞ overtaking optimal if ({xt }t =T0 , {ut }t =T0 ) is (f )good and for every ∞ ({yt }∞ t =T0 , {vt }t =T0 ) ∈ X(T0 , ∞)
satisfying yT0 = xT0 , lim sup[
T −1
T →∞ t =T 0
f (t, xt , ut ) −
T −1
f (t, yt , vt )] ≤ 0.
t =T0
∞ A pair ({xt }∞ t =T0 , {ut }t =T0 ) ∈ X(T0 , ∞) is called (f )weakly optimal if ∞ ∞ ∞ ({xt }t =T0 , {ut }t =T0 ) is (f )good and for every ({yt }∞ t =T0 , {vt }t =T0 ) ∈ X(T0 , ∞) satisfying yT0 = xT0 ,
lim inf[ T →∞
T −1
f (t, xt , ut ) −
t =T0
T −1
f (t, yt , vt )] ≤ 0.
t =T0
The next result is proved by the same scheme as Theorem 3.19. Its proof is omitted. Theorem 4.14. Assume that f has (P1), (P2), and LSC property, that S is an ∞ integer, and that ({xt }∞ t =S , {ut }t =S ) ∈ X(S, ∞) is (f )good. Then there exists an ∗ ∗ ∞ (f )overtaking optimal pair ({xt∗ }∞ t =S , {ut }t =S ) ∈ X(S, ∞) such that xS = xS . The following theorem is proved by the same scheme as Theorem 3.20. Its proof is omitted. Theorem 4.15. Assume that f has (P1), (P2), and LSC property, that S is an integer, and that ({x˜t }∞ ˜ t }∞ t =S , {u t =S ) ∈ X(S, ∞) is (f )good and that ∗ ∞ ∗ ∞ ({xt }t =S , {ut }t =S ) ∈ X(S, ∞) satisfies xS∗ = x˜S . Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)
∗ ∞ ({xt∗ }∞ t =S , {ut }t =S ) is (f )overtaking optimal; ∗ ∞ ({xt }t =S , {u∗t }∞ t =S ) is (f )weakly optimal; ∗ }∞ ) is (f )minimal and (f )good; ({xt∗ }∞ , {u t t =S t =S ∗ }∞ ) is (f )minimal and satisfies lim ∗ f , {u ({xt∗ }∞ t →∞ ρE (xt , xt ) = 0, t t =S t =S ∗ ∞ ∗ ∞ ({xt }t =S , {ut }t =S ) is (f )minimal and satisfies
lim inf ρE (xt∗ , xt ) = 0. f
t →∞
4.4 Perturbed Problems
207
4.4 Perturbed Problems In this section we suppose that the following assumption holds. (A3) For each > 0, there exists δ > 0 such that for each (T , z) ∈ A satisfying f ρE (z, xT ) ≤ δ, there exist integers τ1 , τ2 ∈ (0, bf ], +τ1 T +τ1 −1 ({xt(1)}Tt =T , {u(1) ) ∈ X(T , T + τ1 ), t }t =T −1 ({xt }Tt=T −τ2 , {ut }Tt =T −τ2 ) ∈ X(T − τ2 , T ) (2)
(2)
such that f
xT(1) = z, xT(1)+τ1 = xT +τ1 , T +τ 1 −1
(1)
(1)
f (t, xt , ut ) ≤
t =T
T +τ 1 −1
f
f
f (t, xt , ut ) + ,
t =T f
ρE (xt(1), xt ) ≤ , t = T , . . . , T + τ1 , f
xT(2) = z, xT(2)−τ2 = xT −τ2 , T −1
f (t, xt(2) , u(2) t )≤
t =T −τ2
T −1
f
f
f (t, xt , ut ) + ,
t =T −τ2 (2)
f
ρE (xt , xt ) ≤ , t = T − τ2 , . . . , T . Clearly, (A3) implies (A2). Assume that f has LSC property and that φ : E × E → [0, 1] is a continuous function satisfying φ(x, x) = 0 for all x ∈ E and such that the following property holds: (i) for each > 0, there exists δ > 0 such that if x, y ∈ E and φ(x, y) ≤ δ, then ρE (x, y) ≤ . For each r ∈ (0, 1) set f
fr (t, x, u) = f (t, x, u) + rφ(x, xt ), (t, x, u) ∈ M.
(4.13)
Clearly, for any r ∈ (0, 1), (4.4), (A1), (A3), LSC property hold for fr with f f f f (xt r , ut r ) = (xt , ut ), t = 0, 1, . . . . The next result is proved in Section 4.13. ∞ Theorem 4.16. Let r ∈ (0, 1) and ({xt }∞ t =−∞ , {ut }t =−∞ ) be (f )minimal. Then fr has STP. f
f
208
4 DiscreteTime Nonautonomous Problems on Axis
4.5 Auxiliary Results for Theorems 4.1 and 4.2 (A2) easily implies the following result. Proposition 4.17. Let γ > 0. Then there exists δ > 0 such that if (T , z1 ) ∈ A, (T + 2bf , z2 ) ∈ A satisfy f
f
ρE (z1 , xT ) ≤ δ, ρE (z2 , xT +2bf ) ≤ δ, then there exists T +2bf
({xt }t =T
T +2bf −1
, {ut }t =T
) ∈ X(T , T + 2bf )
such that T +2bf −1
xT = z1 , xT +2bf = z2 ,
T +2bf −1
f (t, xt , ut ) ≤
t =T
f
f
f (t, xt , ut ) + γ .
t =T
Lemma 4.18. There exists S > 0 such that for each pair of integers T2 > T1 and 2 2 −1 each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1 T 2 −1
f f f (t, xt , ut )
t =T1
≤
T 2 −1
f (t, xt , ut ) + S.
(4.14)
t =T1
Proof. In view of (4.3), there exists S1 > 0 such that ψ(S1 ) > a0 + 1 + Δf .
(4.15)
By (A1), there exist S2 > 0 and an integer c0 > 0 such that the following property holds: (i) T 2 −1 t =T1
f
f
f (t, xt , ut ) ≤
T 2 −1
f (t, xt , ut ) + S2
t =T1
2 2 −1 for each pair of integers T1 , T2 ≥ T1 + c0 , and each ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfying ρE (θ0 , xj ) ≤ S1 , j = T1 , T2 .
Fix S > S2 + 2a0 + 2(c0 + 1)(a0 + 1 + Δf ) + 2.
(4.16)
4.5 Auxiliary Results for Theorems 4.1 and 4.2
209
2 2 −1 Assume that T2 > T1 are integers and that ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ). 1 1 We show that (4.14) is true. In view of (4.4), (4.10), and (4.16), if T2 ≤ T1 + c0 , then (4.14) holds. Therefore we may assume without loss of generality that
T2 > T1 + c0 .
(4.17)
ρE (θ0 , xt ) ≥ S1 , t = T1 , . . . , T2 − 1.
(4.18)
Assume that
By (4.4), (4.15), and (4.18), for all t = T1 , . . . , T2 − 1, f (t, xt , ut ) ≥ −a0 + ψ(ρE (θ0 , xt )) ≥ −a0 + ψ(S1 ).
(4.19)
It follows from (4.10), (4.15), and (4.19), T 2 −1
f (t, xt , ut ) ≥ (T2 − T1 )(ψ(S1 ) − a0 )
t =T1
≥ (T2 − T1 )Δf  ≥
T 2 −1
f
f
f (t, xt , ut )
t =T1
and (4.14) holds. Assume that min{ρE (θ0 , xt ) : t = T1 , . . . , T2 − 1} < S1 .
(4.20)
Set τ1 = min{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 },
(4.21)
τ2 = max{t ∈ {T1 , . . . , T2 − 1} : ρE (θ0 , xt ) ≤ S1 }.
(4.22)
There are two cases: τ2 − τ1 ≥ c 0 ;
(4.23)
τ2 − τ1 < c 0 .
(4.24)
Assume that (4.13) holds. It follows from property (i) and (4.21)–(4.23) that τ 2 −1 t =τ1
f
f
f (t, xt , ut ) ≤
τ 2 −1
f (t, xt , ut ) + S2 .
t =τ1
By (4.4), (4.15), (4.21), and (4.22), for each integer t ∈ ([T1 , τ1 ] \ {τ1 }) ∪ ([τ2 , T2 − 1] \ {τ2 }),
(4.25)
210
4 DiscreteTime Nonautonomous Problems on Axis
we have ρE (θ0 , xt ) > S1 , f (t, xt , ut ) ≥ −a0 + ψ(S1 ) > Δf .
(4.26)
It follows from (4.20) and (4.26) that
{f (t, xt , ut ) : t ∈ ({T1 , . . . , τ1 } \ {τ1 }) ∪ ({τ2 , . . . , T2 − 1})} + a0 + Δf f f ≥ {f (t, xt , ut ) : t ∈ ({T1 , . . . , τ1 } \ {τ1 }) ∪ ({τ2 , . . . , T2 − 1})}. (4.27) In view of (4.16), (4.25), and (4.27), T 2 −1
f f f (t, xt , ut )
≤
T 2 −1
t =T1
f (t, xt , ut ) + S2 + Δf + a0 ≤
t =T1
T 2 −1
f (t, xt , ut ) + S.
t =T1
Assume that (4.24) holds. By (4.4), (4.15), (4.21), and (4.22), for all t ∈ ({T1 , . . . , τ1 } \ {τ1 }) ∪ ({τ2 , . . . , T2 − 1} \ {τ2 }), (4.26) holds. It follows from (4.4), (4.10), and (4.26) that (4.27) is true. In view of (4.4), (4.10), and (4.24), τ 2 −1
f f f (t, xt , ut ) −
t =τ1
τ 2 −1
f (t, xt , ut ) ≤ (τ2 − τ1 )(Δf  + a0 ) ≤ c0 (Δf  + a0 ).
t =τ1
(4.28) Equations (4.27) and (4.28) imply that T 2 −1
f (t, xt , ut ) + (c0 + 1)(Δf  + a0 ) ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
Lemma 4.18 is proved. The next auxiliary result easily follows from (4.3) and (4.4). Lemma 4.19. Let M1 > 0 and τ0 be a natural number. Then there exists M2 > M1 such that for each integer T1 , each T2 ∈ (T1 , T1 + τ0 ], and each 2 −1 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies Tt =T f (t, xt , ut ) ≤ M1 the 1 1 1 inequality ρE (xt , θ0 ) ≤ M2 holds for all t = T1 , . . . , T2 − 1. Lemma 4.20. Let Δ > 0. Then there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 satisfying f
T2 ≥ T1 + 2bf , ρE (zi , xTi ) ≤ δ, i = 1, 2 the following inequality holds: U (T1 , T2 , z1 , z2 ) ≤ f
T 2 −1 t =T1
f
f
f (t, xt , ut ) + Δ.
(4.29)
4.6 Proof of Theorems 4.1 and 4.2
211
Proof. Let δ > 0 be as guaranteed by (A2) with = Δ/4 and let (Ti , zi ) ∈ A, i = 1, 2 satisfy (4.29). By (A2) with = Δ/4 and (4.29), there exists 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) such that ({yt }Tt =T 1 1 f
yT1 = z1 , yT2 = z2 , yt = xt , t = T1 + bf , . . . , T2 − bf , f
vt = ut , t ∈ {T1 + bf , . . . , T2 − bf } \ {T2 − bf }, T1 +bf −1
T1 +bf −1
f (t, yt , vt ) ≤
t =T1
f
f
f
f
f (t, xt , ut ) + Δ/4,
t =T1
T 2 −1
f (t, yt , vt ) ≤
t =T2 −bf
T 2 −1
f (t, xt , ut ) + Δ/4.
t =T2 −bf
By the relations above U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f (t, yt , vt ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + Δ/2.
t =T1
Lemma 4.20 is proved.
4.6 Proof of Theorems 4.1 and 4.2 Proof of Theorem 4.1. Assertion 1 of Theorem 4.1 follows from Lemma 4.18. Let us prove assertion 2. Assume that there exists a sequence of natural numbers {Tk }∞ k=1 such that Tk → ∞ as k → ∞, T k −1
f (t, xt , ut ) −
t =−Tk
T k −1
f
f
f (t, xt , ut ) → ∞ as k → ∞.
(4.30)
t =−Tk
Let a number S > 0 be as guaranteed by Assertion 1. Let k ≥ 1 be an integer and let τ > Tk be an integer. Then τ −1 t =−τ
f (t, xt , ut ) −
τ −1 t =−τ
f
f
f (t, xt , ut )
212
=
4 DiscreteTime Nonautonomous Problems on Axis T k −1
f (t, xt , ut ) −
t =−Tk
T k −1
f
f
f (t, xt , ut ) +
τ −1
f (t, xt , ut ) −
t =Tk
≥
T k −1
f (t, xt , ut ) −
t =−Tk
f (t, xt , ut ) −
t =−τ
t =−Tk
+
−T k −1
τ −1
−T k −1
f
f
f (t, xt , ut )
t =−τ
f
f
f (t, xt , ut )
t =Tk T k −1
f
f
f (t, xt , ut ) − 2S → ∞ as k → ∞.
t =−Tk
Assertion 2 is proved. Assume that sup{
T −1
f (t, xt , ut ) −
t =−T
T −1
f
f
f (t, xt , ut ) : T = 1, 2, . . . } < ∞.
t =−T
By the relation above and (4.10), sup{f (t, xt , ut ) : t ∈ Z} < ∞. Together with Lemma 4.19 the inequality above implies that {xt : bounded. Theorem 4.1 is proved.
t ∈ Z} is
Proof of Theorem 4.2. By Theorem 4.1, there exists S0 > 0 such that the following property holds: 2 2 −1 (i) for each pair of integers T2 > T1 and each ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ), 1 1
T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ).
t =T1
In view of (4.3), there exists M1 > 0 such that ψ(M1 ) > a0 + Δf  + 3S0 + M0 .
(4.31)
2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfy Let T2 > T1 be integers and ({xt }Tt =T 1 1
T 2 −1 t =T1
f (t, xt , ut ) ≤
T 2 −1 t =T1
f
f
f (t, xt , ut ) + M0 .
(4.32)
4.7 Proofs of Propositions 4.10, 4.12 and Theorems 4.3–4.5
213
We show that ρE (θ0 , xt ) ≤ M1 , t = T1 , . . . , T2 − 1. Assume the contrary. Then there exists j0 ∈ {T1 , . . . , T2 − 1} such that (4.33)
ρE (θ0 , xj0 ) > M1 . Property (i), (4.4), (4.10), (4.32), and (4.33) imply that T 2 −1
f
f
f (t, xt , ut ) + M0 ≥
t =T1
≥
f (t, xt , ut )
t =T1
{f (t, xt , ut ) : t ∈ {T1 , . . . , j0 } \ {j0 }}
+f (j0 , xj0 , uj0 ) + ≥ ψ(M1 ) − a0 + −
T 2 −1
{f (t, xt , ut ) : t ∈ {j0 , . . . , T2 − 1} \ {j0 }}
f f {f (t, xt , ut ) : t ∈ {T1 , . . . , j0 } \ {j0 }} − S0
f f {f (t, xt , ut ) : t ∈ {j0 , . . . , T2 − 1} \ {j0 }} − S0 ≥ ψ(M1 ) − a0 − 2S0 − Δf  +
T 2 −1
f
f
f (t, xt , ut ),
t =T1
ψ(M1 ) ≤ a0 + 2S0 + Δf  + M0 . This contradicts (4.31). The contradiction we have reached proves Theorem 4.2.
4.7 Proofs of Propositions 4.10, 4.12 and Theorems 4.3–4.5 Proof of Proposition 4.10. Let S > 0 be as guaranteed by Theorem 4.1. Let L0 , T2 ≥ T1 + 2L0 . There exist integers τ1 ∈ (0, L0 ], (T1 , z1 ) ∈ AL0 , (T2 , z2 ) ∈ A T2 2 −1 ) ∈ X(T1 , T2 ) such that τ2 ∈ (0, L0 ] and ({yt }t =T1 , {vt }Tt =T 1 f
yT1 = z1 , yT2 = z2 , yt = xt , t = T1 + τ1 , . . . , T2 − τ2 , f
vt = ut , t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }, T1 +τ 1 −1 t =T1
f (t, yt , vt ) ≤ L0 ,
T 2 −1 t =T2 −τ2
f (t, yt , vt ) ≤ L0 .
214
4 DiscreteTime Nonautonomous Problems on Axis
In view of the relations above, (4.4) and the choice of S0 , U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f (t, yt , vt )
t =T1
≤ 2L +
f
f
{f (t, xt , ut ) : t ∈ {T1 + τ1 , . . . , T2 − τ2 } \ {T2 − τ2 }}
≤ 2L + 2a0 L +
T 2 −1
f
f
f (t, xt , ut ) ≤ 4L(1 + a0 ) + σ f (T1 , T2 ) + S.
t =T1
Proposition 4.10 is proved. Theorem 4.3 follows from Theorem 4.2 and Proposition 4.10. Proof of Proposition 4.12. Let S > 0 be as guaranteed by Theorem 4.1. Let (T1 , z1 ) ∈ AL0 , T2 ≥ T1 + L0 be an integer. Clearly, there exist an integer 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) such that τ ∈ (0, L0 ] and ({yt }Tt =T 1 1 f
f
yT1 = z1 , yt = xt , t = T1 + τ, . . . , T2 , vt = ut , t ∈ {T1 + τ, . . . , T2 } \ {T2 }, T1 +τ −1
f (t, yt , vt ) ≤ L.
t =T1
In view of the relations above, the choice of S and (4.32), σ f (T1 , T2 , z1 ) ≤
T 2 −1
f (t, yt , vt )
t =T1
≤ L+ ≤L+
{f (t, yt , vt ) : t ∈ {T1 + τ, . . . , T2 } \ {T2 }}
≤ L + a0 L +
f
f
{f (t, xt , ut ) : t ∈ {T1 + τ, . . . , T2 } \ {T2 }}
T 2 −1
f
f
f (t, xt , ut ) ≤ σ f (T1 , T2 ) + S + L(1 + a0 ).
t =T1
Proposition 4.12 is proved. Theorem 4.4 follows from Theorem 4.2 and Proposition 4.12. Theorem 4.5 follows from Theorem 4.2 and the next result.
4.8 Auxiliary Results for Theorem 4.7
215
L , Proposition 4.21. Let L > 0 be an integer, T2 > T1 be integers, (T2 , z) ∈ A T2 ≥ T1 + L. Then σ (T1 , T2 , z) ≤ f
T 2 −1
f
f
f (t, xt , ut ) + a0 (1 + L).
t =T1
L , there exist an integer τ ∈ (0, L] and Proof. By the definition of A 2 2 −1 ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) 1 1
such that f
f
yT2 = z, yt = xt , t = T1 , . . . , T2 − τ, vt = ut , t ∈ {T1 , . . . , T2 − τ } \ {T2 − τ }, T 2 −1
f (t, yt , vt ) ≤ L.
t =T2 −τ
In view of the relations above, σ (T1 , T2 , z) ≤ f
T 2 −1
f (t, yt , vt )
t =T1
≤L+
f
f
{f (t, xt , ut ) : t ∈ {T1 , . . . , T2 − τ } \ {T2 − τ }} ≤ L + a0 L +
T 2 −1
f
f
f (t, xt , ut ).
t =T1
Proposition 4.21 is proved.
4.8 Auxiliary Results for Theorem 4.7 Lemma 4.22. Let > 0. Then there exists an integer L > 0 such that the following properties hold: (i) for each pair of integers T2 > T1 ≥ L, T 2 −1 t =T1
f
f
f
f
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + ;
(ii) for each pair of integers T1 < T2 ≤ −L, (4.34) is true.
(4.34)
216
4 DiscreteTime Nonautonomous Problems on Axis
Proof. Assume the contrary. Then there exists a sequence of closed intervals [ai , bi ], i = 1, 2, . . . such that ai < bi , i = 1, 2, . . . are integers, [ai , bi ] ∩ [aj , bj ] = ∅ for each pair of natural numbers j > i, b i −1 t =ai
f
f
f (t, xt , ut ) > U f (ai , bi , xai , xbi ) + , i = 1, 2, . . . .
(4.35)
(4.36)
∞ By (4.35) and (4.36), there exists ({yt }∞ t =−∞ , {vt }t =−∞ ) ∈ X(−∞, ∞) such that
yt = xt for all integers t ∈ ∪∞ p=1 (ap , bp ),
(4.37)
vt = ut for all integers t ∈ ∪∞ p=1 [ap , bp − 1],
(4.38)
f
f
and for each natural number i, b i −1
f
f
f (t, xt , ut ) >
t =ai
b i −1
f (t, yt , vt ) + .
(4.39)
t =ai
Let q ≥ 1 be an integer. Then for every integer T > 0 such that [ai , bi ] ⊂ [−T , T ], i = 1, . . . , q, it follows from (4.37)–(4.39) that T −1
f
T −1
f
f (t, xt , ut ) −
t =−T
≥
f (t, yt , vt )
t =−T
q b b i −1 i −1 f f ( f (t, xt , ut ) − f (t, yt , vt )) ≥ q, i=1 t =ai T −1
t =ai
f (t, yt , vt ) ≤
t =−T
T −1
f
f
f (t, xt , ut ) − q.
t =−T
Since q is an arbitrary natural number, we conclude that T −1 t =−T
f (t, yt , vt ) −
T −1
f
f
f (t, xt , ut ) → −∞ as T → ∞.
t =−T
This contradicts Theorem 4.1. The contradiction we have reached proves Lemma 4.22.
4.8 Auxiliary Results for Theorem 4.7
217
Lemma 4.23. Assume that (P1) holds and that > 0. Then there exists δ > 0 and an integer T > 0 such that the following properties hold: 2 2 −1 , {vt }Tt =T ) ∈ (iii) for each pair of integers T2 > T1 ≥ T and each ({yt }Tt =T 1 1 X(T1 , T2 ) which satisfies
f
ρE (yTi , xTi ) ≤ δ, i = 1, 2,
(4.40)
we have T 2 −1
f (t, yt , vt ) ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ) − ;
(4.41)
t =T1
(iv) for each pair of integers T1 < T2 ≤ −T and each 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) ({yt }Tt =T 1 1
which satisfies (4.40) inequality (4.41) holds. Proof. By (A2), there exists δ ∈ (0, /4) such that the following property holds: f
(v) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ there exist integers τ1 , τ2 ∈ (0, bf ], +τ1 T +τ1 −1 ({x˜t(1)}Tt =T , {u˜ (1) ) ∈ X(T , T + τ1 ), t }t =T −1 ({x˜t }Tt=T −τ2 , {u˜ t }Tt =T −τ2 ) ∈ X(T − τ2 , T ) (2)
(2)
such that f
= z, x˜T(1)+τ1 = xT +τ1 , x˜T(1) 1 T +τ 1 −1
(1)
(1)
f (t, x˜t , u˜ t ) ≤
t =T
T +τ 1 −1
f
f
f (t, xt , ut ) + /4,
t =T (2)
(2)
f
x˜T = z, x˜T −τ2 = xT −τ2 , T −1 t =T −τ2
(2) (2) f (t, x˜t , u˜ t )
≤
T −1
f
f
f (t, xt , ut ) + /4.
t =T −τ2
Lemma 4.22 implies that there exists an integer L > bf such that the following properties hold:
218
4 DiscreteTime Nonautonomous Problems on Axis
(vi) for each pair of integers T2 > T1 ≥ L, T 2 −1 t =T1
f
f
f
f
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + /4;
(4.42)
(vii) for each pair of integers T1 < T2 ≤ −L, (4.42) is true. Set T = L + b f .
(4.43)
either T1 ≥ L + bf or T2 ≤ −L − bf ,
(4.44)
Assume that T2 > T1 are integers,
2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ), ({yt }Tt =T 1 1
f
ρE (yTi , xTi ) ≤ δ, i = 1, 2.
(4.45)
We show that (4.41) holds. Assume the contrary. Then T 2 −1
f (t, yt , vt )
0 such that ρE (xt , θ0 ) ≤ M0 for all integers t,
(4.52)
for each integer T > 0, 
T −1 t =−T
f (t, xt , ut ) −
T −1 t =−T
f
f
f (t, xt , ut ) ≤ M0 .
(4.53)
220
4 DiscreteTime Nonautonomous Problems on Axis
By Theorem 4.1, there exists M1 > 0 such that the following property holds: 2 2 −1 (i) for each pair of integers T2 > T1 and each ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ), 1 1
T 2 −1
T 2 −1
f (t, yt , vt ) + M1 ≥
t =T1
f
f
f (t, xt , ut ).
t =T1
Let S1 < S2 be integers. Choose a natural number T such that − T < S1 < S2 < T .
(4.54)
Property (i), (4.53), and (4.54) imply that S 2 −1
S 1 −1
f (t, xt , ut ) = −
t =S1
f (t, xt , ut ) −
T 2 −1
t =−T T −1
≤
t =S2
f f f (t, xt , ut )
T −1
S 1 −1
+ M0 −
t =−T
−
f (t, xt , ut ) +
T −1
f (t, xt , ut )
t =−T
f
f
f (t, xt , ut ) + M1
t =−T
f
f
f (t, xt , ut ) + M1 ≤
t =S2
S 2 −1
f
f
f (t, xt , ut ) + M0 + 2M1 .
t =S1
Thus for each pair of integers S2 > S1 , S 2 −1
f (t, xt , ut ) ≤
t =S1
S 2 −1
f
f
f (t, xt , ut ) + M0 + 2M1 .
(4.55)
t =S1
Let γ > 0. We show that there exists an integer Tγ > 0 such that the following properties hold: (ii) for each pair of integers S1 < S2 ≤ −Tγ , S 2 −1
f (t, xt , ut ) ≤ U f (S1 , S2 , xS1 , xS2 ) + γ ;
(4.56)
t =S1
(iii) for each pair of integers S2 > S1 ≥ Tγ , (4.56) is true. Assume the contrary. Then there exists a sequence of closed intervals [ai , bi ], i = 1, 2, . . . such that ai < bi , i = 1, 2, . . . are integers, [ai , bi ] ∩ [aj , bj ] = ∅ for each pair of natural numbers j > i,
(4.57)
4.9 STP Implies (P1)–(P3) b i −1 t =ai
221
f
f
f (t, xt , ut ) > U f (ai , bi , xai , xbi ) + γ , i = 1, 2, . . . .
(4.58)
∞ By (4.57) and (4.58), there exists ({yt }∞ t =−∞ , {vt }t =−∞ ) ∈ X(−∞, ∞) such that
yt = xt for all integers t ∈ ∪∞ i=1 (ai , bi ),
(4.59)
vt = ut for all integers t ∈ ∪∞ i=1 [ai , bi ),
(4.60)
and for each natural number i, b i −1
f (t, yt , vt )
0 there exists an integer Tγ > 0 such that properties (ii) and (iii) hold. Let > 0. By STP, there exist δ > 0 and an integer L > 0 such that the following property holds: (iv) for each pair of integers T1 , T2 ≥ T1 + 2L and each 2 2 −1 ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) 1 1
which satisfies T 2 −1
f (t, yt , vt ) ≤ min{σ f (T1 , T2 ) + 2M0 + 3M1 , U f (T1 , T2 , xT1 , xT2 ) + δ}
t =T1
we have f
ρE (yt , xt ) ≤ , t = T1 + L, . . . , T2 − L. Let a natural number Tδ be as guaranteed by properties (ii), (iii) with γ = δ and let k ≥ 1 be an integer. By properties (ii), (iii) with γ = δ, property (i), and (4.55), Tδ +2L+k−1
f (t, xt , ut ) ≤ U f (Tδ , Tδ + 2L + k, xTδ , xTδ +2L+k ) + δ,
t =Tδ Tδ +2L+k−1
f (t, xt , ut ) ≤
t =Tδ
Tδ +2L+k−1
f
f
f (t, xt , ut ) + M0 + 2M1
t =Tδ
≤ σ f (Tδ , Tδ + 2L + k) + M0 + 3M1 , −T δ −1
f (t, xt , ut ) ≤ U f (−Tδ − 2L − k, −Tδ , x−Tδ −2L−k , x−Tδ ) + δ,
t =−Tδ −2L−k −T δ −1
f (t, xt , ut ) ≤
t =−Tδ −2L−k
−T δ −1
f
f
f (t, xt , ut ) + M0 + 2M1
t =−Tδ −2L−k
≤ σ f (−Tδ − 2L − k, −Tδ ) + M0 + 3M1 . The relations above and property (iv) imply that f
ρE (xt , xt ) ≤ for all integers t ∈ [Tδ + L, Tδ + L + k] ∪ [−Tδ − L − k, −Tδ − L].
4.9 STP Implies (P1)–(P3)
223
Since k is an arbitrary natural number, we conclude that f
ρE (xt , xt ) ≤ for all integers t ∈ [Tδ + L, ∞) ∪ (−∞, −Tδ − L]. Since is any positive number, (P1) holds. We show that (P3) holds. In view of LSC property, for each integer k ≥ 1 there k−1 exists ({xt(k) }kt=−k , {u(k) t }t =−k ) ∈ X(−k, k) such that (k)
f
f
(k)
x−k = x−k , xk = xk ,
k−1
(k)
t =−k
f
(k)
f
f (t, xt , ut ) = U f (−k, k, x−k , xk ).
(4.63)
It follows from (A2) that for each integer k ≥ 1 there exists δk ∈ (0, 2−k ) such that the following property holds: f
(v) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δk there exist integers τ1 , τ2 ∈ (0, bf ], +τ1 +τ1 −1 ({yt(1)}Tt =T , {vt(1) }Tt =T ) ∈ X(T , T + τ1 ), 1 −1 ({yt(2)}Tt=T −τ2 , {vt(2) }Tt =T −τ2 ) ∈ X(T − τ2 , T2 )
such that (1)
f
(1)
yT = z, yT +τ1 = xT +τ1 , T +τ 1 −1
(1) (1) f (t, yt , vt )
≤
T +τ 1 −1
t =T
f (t, xt ut ) + k −1 , f f
t =T f
yT(2) = z, yT(2)−τ2 = xT −τ2 , T −1
T −1
f (t, yt(2), vt(2) ) ≤
t =T −τ2
f (t, xt ut ) + k −1 . f f
t =T −τ2
Theorem 4.1 implies that there exists M0 > 0 such that for each pair of integers 2 2 −1 , {vt }St =S ) ∈ X(S1 , S2 ), S2 > S1 and each ({yt }St =S 1 1 S 2 −1 t =S1
f (t, yt , vt ) ≥
S 2 −1
f
f
f (t, xt , ut ) − M0 .
(4.64)
t =S1
By (4.63) and (4.64), for each pair of integers S2 > S1 and each natural number k > max{S2 , −S1 },
224
4 DiscreteTime Nonautonomous Problems on Axis S 2 −1
(k)
(k)
f (t, xt , ut ) =
t =S1
−
k−1
k−1
S 1 −1
f
(k)
t =−k
f (t, xt(k), u(k) t )
−
t =−k
≤
(k)
f (t, xt , ut )
k−1
f (t, xt(k), u(k) t )
t =S2
f
f (t, xt , ut ) −
t =−k
S 1 −1
f
t =−k
=
S 2 −1
k−1
f
f (t, xt , ut ) −
f
f
f (t, xt , ut ) + 2M0
t =S2
f
f
f (t, xt , ut ) + 2M0 .
(4.65)
t =S1
It follows from STP, (4.63), and (4.65) that the following property holds: (vi) for each > 0 there exists an integer L > 0 such that for each integer k > L, (k)
f
ρE (xt , xt ) ≤ , t = −k + L, . . . , k − L.
(4.66)
In view of (4.65), for each integer i and each integer k > i + 1, (k)
f
(k)
f
f (i, xi , ui ) ≤ f (i, xi , ui ) + 2M0 .
(4.67)
By LSC property and (4.67), extracting subsequences and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 and there ∞ exists ({x˜t }∞ , { u ˜ } ) ∈ X(−∞, ∞) such that for each integer t, t t =−∞ t =−∞ (kp )
f (t, x˜t , u˜ t ) ≤ lim inf f (t, xt p→∞
(kp )
xt
(kp )
, ut
(4.68)
),
→ x˜t as p → ∞.
(4.69)
It follows from (4.66) and (4.69) that that for all integers t, f
x˜t = xt .
(4.70)
Let S be a natural number. By (4.65), (4.68), and (4.70), S−1 t =−S
f (t, x˜t , u˜ t ) ≤ lim inf p→∞
S−1 t =−S
(kp )
f (t, xt
(kp )
, ut
)≤
S−1 t =−S
f
f
f (t, xt , ut ) + 2M0 . (4.71)
4.9 STP Implies (P1)–(P3)
225
Theorem 4.1 and (4.71) imply that ({xt }∞ ˜ t }∞ t =−∞ , {u t =−∞ ) is (f )good. We show that it is (f )minimal. Assume the contrary. Then there exist γ > 0 and an integer S0 ≥ 1 such that f
S 0 −1 t =−S0
f
f
f
f
f
f (t, xt , ut ) > U f (−S0 , S0 , x−S0 , xS0 ) + γ .
(4.72)
S −1
0 0 , {vt∗ }t =−S ) ∈ X(−S0 , S0 ) such that In view of (4.72), there exists ({yt∗ }t =−S 0 0
S
S 0 −1
f
S 0 −1
f
f (t, xt , u˜ t ) >
t =−S0
f (t, yt∗ , vt∗ ) + γ ,
(4.73)
t =−S0 ∗ y−S = x−S0 , yS∗0 = xS0 , 0 f
f
(4.74)
Lemma 4.23 implies that there exist δ∗ > 0 and an integer T∗ > 0 such that the following properties hold: 2 2 −1 (vii) for each pair of integers T2 > T1 ≥ T∗ and each ({yt }Tt =T , {vt }Tt =T ) ∈ 1 1
f
X(T1 , T2 ) which satisfies ρE (yTi , xTi ) ≤ δ∗ , i = 1, 2 we have: T 2 −1
f (t, yt , vt ) ≥
t =T1
T 2 −1
f
f
f (t, xt , ut ) − γ /16;
(4.75)
t =T1
(viii) for each pair of integers T1 < T2 ≤ −T∗ and each 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) ({yt }Tt =T 1 1
f
which satisfies ρE (yTi , xTi ) ≤ δ∗ , i = 1, 2, (4.75) holds. Choose a natural number k0 > 8γ −1 + S0 + 8.
(4.76)
It follows from (4.68)–(4.70) that there exists a natural number q such that kq > k0 + T∗ + bf + 4,
(4.77)
for each integer t ∈ [−k0 − T∗ − bf − 4, k0 + T∗ + bf + 4], (kq )
ρE (xt
f
, xt ) ≤ min{δk0 , δ∗ },
(4.78)
226
4 DiscreteTime Nonautonomous Problems on Axis k0 +T ∗ −1
f
k0 +T ∗ −1
f
f (t, xt , u˜ t ) ≤
t =−k0 −T∗
(kq )
f (t, xt
(kq )
, ut
) + γ /16.
(4.79)
t =−k0 −T∗
For all integers t ∈ [−k0 − T∗ , k0 + T∗ ] \ [−S0 , S0 ] set yt∗ = xt , f
(4.80)
and for all integers t ∈ [−k0 − T∗ , k0 + T∗ − 1] \ [−S0 , S0 − 1] set vt∗ = u˜ t . f
(4.81)
0 +T∗ 0 +T∗ −1 , {vt∗ }kt =−k ) ∈ In view of (4.73), (4.74), (4.80), and (4.81), ({yt∗ }kt =−k 0 −T∗ 0 −T∗ X(−k0 − T∗ , k0 + T∗ ),
∗ = x−k0 −T∗ , yk∗0 +T∗ = xk0 +T∗ , y−k 0 −T∗ f
k0 +T ∗ −1
f
f
f (t, xt , u˜ t ) >
t =−k0 −T∗
f
k0 +T ∗ −1
(4.82)
f (t, yt∗ , vt∗ ) + γ .
(4.83)
t =−k0 −T∗
Property (v) and (4.78) imply that there exist integers τ1 , τ2 ∈ (0, bf ], (1) k +T∗ +b
(1) k +T∗ +b −1
0 f 0 f ({yt }t =k , {vt }t =k ) ∈ X(k0 + T∗ + bf − τ1 , k0 + T∗ + bf ), 0 +T∗ +bf −τ1 0 +T∗ +bf −τ1
−k −T∗ −b +τ
−k −T∗ −b +τ −1
0 f 2 0 f 2 , {vt(2) }t =−k ({yt(2)}t =−k 0 −T∗ −bf 0 −T∗ −bf
)
∈ X(−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 ) such that f
(1)
k
(1)
yk0 +T∗ +bf −τ1 = xk0 +T∗ +bf −τ1 , yk0 +T∗ +bf = xk0q+T∗ +bf , k0 +T∗ +bf −1
k0 +T∗ +bf −1 (1) (1) f (t, yt , vt )
≤
t =k0 +T∗ +bf −τ1
f (t, xt , ut ) + k0−1 , f
f
k
f
−k0 −T∗ −bf +τ2 −1 t =−k0 −T∗ −bf
(4.85)
t =k0 +T∗ +bf −τ1
(2) q (2) y−k = x−k , y−k = x−k0 −T∗ −bf +τ2 , 0 −T∗ −bf 0 −T∗ −bf 0 −T∗ −bf +τ2
(4.84)
−k0 −T∗ −bf +τ2 −1
f (t, yt(2), vt(2) )
≤
t =−k0 −T∗ −bf
f (t, xt , ut ) + k0−1 . f
f
(4.86)
(4.87)
4.9 STP Implies (P1)–(P3)
227
Define (2)
(2)
yt = yt , vt = vt
for all integers t ∈ [−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 ), (4.88) f f vt = ut for all integers t ∈ [−k0 − T∗ − bf + τ2 , −k0 − T∗ ] \ {−k0 − T∗ }, yt = xt , (4.89) yt = yt∗ , vt = vt∗ for all integers t ∈ [−k0 − T∗ , k0 + T∗ ), (4.90) f
yt = xt for all integers t = k0 + T∗ , . . . , k0 + T∗ + bf − τ1 ,
(4.91)
f
vt = ut for all integers t ∈ [k0 + T∗ , k0 + T∗ + bf − τ1 ] \ {k0 + T∗ + bf − τ1 }, (4.92) (1) yt = yt for all integers t = k0 + T∗ + bf − τ1 , . . . , k0 + T∗ + bf , (4.93) (1)
vt = vt
for all integers t ∈ [k0 + T∗ + bf − τ1 , k0 + T∗ + bf ).
(4.94)
By (4.82), (4.84), (4.86), and (4.88)–(4.94), k +T∗ +b
k +T∗ +b −1
0 f 0 f , { vt }t =−k ) ∈ X(−k0 − T∗ − bf , k0 + T∗ + bf ), ({ yt }t =−k 0 −T∗ −bf 0 −T∗ −bf
(k )
(k )
yk0 +T∗ +bf = xk0 q+T∗ +bf . y−k0 −T∗ −bf = x−kq0 −T∗ −bf ,
(4.95)
It follows from (4.83), (4.85), and (4.87–4.94) that k0 +T∗ +bf −1
−k0 −T∗ −bf +τ2 −1
f (t, yt , vt ) =
t =−k0 −T∗ −bf
+
f (t, yt , vt )
t =−k0 −T∗ −bf
{f (t, yt , vt ) : t ∈ {−k0 − T∗ − bf + τ2 , . . . , −k0 − T∗ } \ {−k0 − T∗ }} +
k0 +T ∗ −1
f (t, yt∗ , vt∗ )
t =−k0 −T∗
+
{f (t, yt , vt ) : t ∈ {k0 + T∗ , . . . , k0 + T∗ + bf − τ1 } \ {k0 + T ∗ + bf − τ1 }} k0 +T∗ +bf −1
+
(1)
(1)
f (t, yt , vt )
t =k0 +T∗ +bf −τ1 −k0 −T∗ −bf +τ2 −1
≤
t =−k0 −T∗ −bf
f (t, xt , ut ) + k0−1 f
f
228
4 DiscreteTime Nonautonomous Problems on Axis
+
f
f
{f (t, xt , ut ) : t ∈ {−k0 − T∗ − bf + τ2 , . . . , −k0 − T∗ } \ {−k0 − T∗ }} k0 +T ∗ −1
+
f (t, yt∗ , vt∗ )
t =−k0 −T∗
+
f f {f (t, xt , ut ) : t ∈ {k0 + T∗ , . . . , k0 + T∗ + bf − τ1 } \ {k0 + T∗ + bf − τ1 }} k0 +T∗ +bf −1
+
f (t, xt , ut ) + k0−1 f
f
t =k0 +T∗ +bf −τ1 −k0 −T∗ −1
≤
f
k0 +T ∗ −1
f
f (t, xt , ut ) +
t =−k0 −T∗ −bf
f
f
f (t, xt , u˜ t ) − γ
t =−k0 −T∗
k0 +T∗ +bf −1
+
f (t, xt , ut ) + 2k0−1 . f
f
(4.96)
t =k0 +T∗
In view of (4.79) and (4.96), k0 +T∗ +bf −1
f (t, yt , vt ) ≤ 2k0−1 − γ
t =−k0 −T∗ −bf −k0 −T∗ −1
+
f
f
f (t, xt , ut ) +
t =−k0 −T∗ −bf
k0 +T ∗ −1
(kq )
f (t, xt
(kq )
, ut
)
t =−k0 −T∗
k0 +T∗ +bf −1
+
f
f
f (t, xt , ut ) + γ /16.
(4.97)
t =k0 +T∗
Properties (vii) and (viii) and (4.75) and (4.78) imply that −k0 −T∗ −1
(kq )
f (t, xt
(kq )
, ut
)≥
t =−k0 −T∗ −bf k0 +T∗ +bf −1
t =k0 +T∗
−k0 −T∗ −1
f
f
f (t, xt , ut ) − γ /16,
(4.98)
t =−k0 −T∗ −bf
(kq ) (kq ) f (t, xt , ut )
k0 +T∗ +bf −1
≥
t =k0 +T∗
f
f
f (t, xt , ut ) − γ /16.
(4.99)
4.10 (P1)–(P3) Imply STP
229
By (4.76) and (4.97)–(4.99), k0 +T∗ +bf −1
k0 +T∗ +bf −1
f (t, yt , vt ) −
t =−k0 −T∗ −bf
(kq )
f (t, xt
(kq )
, ut
)
t =−k0 −T∗ −bf
≤ 2k0−1 − γ + γ /16 + γ /8 ≤ −γ /4. This contradicts (4.63) and (4.95). The contradiction we have reached proves that f f ({xt }∞ ˜ t }∞ t =−∞ , {u t =−∞ ) is (f )minimal. ∞ Assume that ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) is (f )good and (f )minimal. Then sup{
T −1
f (t, xt , ut ) −
t =−T
T −1
f
f
f (t, xt , ut ) : T ∈ {1, 2, . . . }} < ∞.
t =−T
Property (P1) implies that f
lim ρE (xt , xt ) = 0,
t →∞
f
lim ρE (xt , xt ) = 0.
t →−∞ f
STP and the relations above imply that xt = xt for all integers t. Thus (P3) holds.
4.10 (P1)–(P3) Imply STP Lemma 4.24. Assume that (P1) holds and that > 0. Then there exists δ > 0 and an integer L > 0 such that the following properties hold: (i) for each integer T1 ≥ L, each integer T2 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({xt }Tt =T 1 1
≥ T1 + 2bf , and each
f
ρE (xTi , xTi ) ≤ δ, i = 1, 2, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ
(4.100)
(4.101)
t =T1
the inequality f
ρE (xt , xt ) ≤ , t = T1 , . . . , T2 is true;
(4.102)
230
4 DiscreteTime Nonautonomous Problems on Axis
(ii) for each integer T2 ≤ −L, each integer T1 ≤ T2 − 2bf , and each 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) which satisfies (4.100) and (4.101) the ({xt }Tt =T 1 1 relation (4.102) is true. Proof. By (A2), for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: f
(iii) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δk there exist integers τ1 , τ2 ∈ (0, bf ], +τ1 +τ1 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T , T + τ1 ), (1)
(1)
T −1 ({xt(2)}Tt=T −τ2 , {u(2) t }t =T −τ2 ) ∈ X(T − τ2 , T )
such that (1)
f
(1)
(2)
f
(2)
xT = z, xT +τ1 = xT +τ1 , xT = z, xT −τ2 = xT −τ2 , T +τ 1 −1
(1)
(1)
f (t, xt , ut ) ≤
t =T T −1
T +τ 1 −1
f (t, xt , ut ) + 2−k , f
f
t =T
f (t, xt(2), u(2) t )≤
t =T −τ2
T −1
f (t, xt , ut ) + 2−k . f
f
t =T −τ2
Assume that the lemma does not hold. Then for each integer k ≥ 1, there exist Tk,2 Tk,2 −1 integers Tk,1 , Tk,2 ≥ Tk,1 + 2bf , and ({xt(k) }t =T , {u(k) t }t =Tk,1 ) ∈ X(Tk,1 , Tk,2 ) k,1 such that f
(k)
ρE (xTk,i , xTk,i ) ≤ δk , i = 1, 2,
(4.103)
Tk,2 −1
t =Tk,1
(k) (k) f f (t, xt(k) , u(k) t ) ≤ U (Tk,1 , Tk,2 , xTk,1 , xTk,2 ) + δk ,
f
max{ρE (xt(k), xt ) : t ∈ {Tk,1 , . . . , Tk,2 }} >
(4.104)
(4.105)
and that either for all integers k ≥ 1, Tk,1 ≥ k + bf ,
(4.106)
Tk,2 ≤ −k − bf .
(4.107)
or for all integers k ≥ 1,
4.10 (P1)–(P3) Imply STP
231
Extracting a subsequence and reindexing, we may assume without loss of generality that for each integer k ≥ 1, min{Tk+1,1, Tk+1,2 } ≥ max{Tk,1 , Tk,2 } + 4bf .
(4.108)
Let k ≥ 1 be an integer. Property (iii) and (4.103) imply that there exist integers Tk,2 +τk,2 Tk,2 +τk,2 −1 , {u˜ (k) τk,1 , τk,2 ∈ (0, bf ] and ({x˜t(k)}t =T t }t =Tk,1 −τk,1 ) ∈ X(Tk,1 − τk,1 , Tk,2 + k,1 −τk,1 τk,2 ) such that (k) x˜t(k) = xt(k), t = Tk,1 , . . . , Tk,2 , u˜ (k) t = ut , t = Tk,1 , . . . , Tk,2 − 1, f
f
= xTk,1 −τk,1 , x˜T(k) = xTk,2 +τk,2 , x˜T(k) k,1 −τk,1 k,2 +τk,2 Tk,1 −1
Tk,1 −1
(k)
(k)
f (t, x˜t , u˜ t ) ≤
t =Tk,1 −τk,1
(4.110)
f (t, xt , ut ) + 2−k ,
(4.111)
f (t, xt , ut ) + 2−k .
(4.112)
f
f
t =Tk,1 −τk,1
Tk,2 +τk,2 −1
(4.109)
Tk,2 +τk,2 −1
f (t, x˜t(k), u˜ (k) t )≤
t =Tk,2
f
f
t =Tk,2
Property (iii) and (4.103) imply that there exist integers τk,3 , τk,4 ∈ (0, bf ] and (k) T
(k) T
−1
k,2 k,2 ({ xt }t =T , { ut }t =T ) ∈ X(Tk,1 , Tk,2 ) such that k,1 k,1
(k)
(k)
(k)
(k)
xTk,2 = xTk,2 , xTk,1 = xTk,1 , (k)
xt (k)
ut
(4.113)
f
= xt , t = Tk,1 + τk,3 , . . . , Tk,2 − τk,4 ,
(4.114)
= ut , t ∈ {Tk,1 + τk,3 , . . . , Tk,2 − τk,4 } \ {Tk,2 − τk,4 },
(4.115)
f
Tk,1 +τk,3 −1
Tk,1 +τk,3 −1
f (t, xt(k) , u(k) t )≤
t =Tk,1
f (t, xt , ut ) + 2−k ,
(4.116)
f (t, xt , ut ) + 2−k .
(4.117)
f
f
t =Tk,1
Tk,2 −1
Tk,2 −1
(k)
(k)
f (t, xt , ut ) ≤
t =Tk,2 −τk,4
f
f
t =Tk,2 −τk,4
By (4.104) and (4.113)–(4.117), Tk,2 −1
U
f
(Tk,1 , Tk,2 , xT(k) , xT(k) ) k,1 k,2
≤
t =Tk,1
f (t, xt(k), u(k) t )
232
4 DiscreteTime Nonautonomous Problems on Axis Tk,2 −1
≤ U f (Tk,1 , Tk,2 , xT(k) , xT(k) ) + δk ≤ k,1 k,2 Tk,2 −1
≤
f (t, xt(k) , u(k) t ) + δk
t =Tk,1
f (t, xt , ut ) + 3 · 2−k . f
f
(4.118)
t =Tk,1
By (4.109)–(4.112) and (4.118), Tk,2 +τk,2 −1
Tk,1 −1 (k) (k) f (t, x˜t , u˜ t )
t =Tk,1 −τk,1 Tk,2 +τk,2 −1
+
Tk,2 −1
≤
f f f (t, xt , ut )
+
t =Tk,1 −τk,1
f f f (t, xt , ut ) + 2−k+1
t =Tk,2
(k)
(k)
f (t, xt , ut )
t =Tk,1
Tk,2 +τk,2 −1
≤
f (t, xt , ut ) + 2−k+3 . f
f
(4.119)
t =Tk,1 −τk,1
∞ By (4.108) and (4.110), there exists ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) such that for every integer k ≥ 1,
xt = x˜t(k), t = Tk,1 − τk,1 , . . . , Tk,2 + τk,2 , ut = u˜ (k) t , t = Tk,1 − τk,1 , . . . , Tk,2 + τk,2 − 1,
(4.120)
xt = xt , ut = ut for all integers t ∈ ∪∞ k=1 (Tk,1 − τk,1 , Tk,2 + τk,2 ). f
f
(4.121)
It follows from (4.119)–(4.121) that for each integer k ≥ 1, in the case of (4.106), {f (t, xt , ut ) : t = 0, . . . , Tk,2 + τk,2 − 1} −
=
k
f f {f (t, xt , ut ) : t = 0, . . . , Tk,2 + τk,2 − 1}
Ti,2 +τi,2 −1
(
Ti,2 +τi,2 −1 (i) (i) f (t, x˜t , u˜ t )
i=1 t =Ti,1 −τi,1
−
f
f
f (t, xt , ut )) ≤
t =Ti,1 −τi,1
k
2−i+3 ≤ 8
i=1
(4.122) and in the case of (4.107), {f (t, xt , ut ) : t = Tk,1 − τk,1 , . . . , −1} −
f f {f (t, xt , ut ) : t = Tk,1 − τk,1 , . . . , −1}
4.10 (P1)–(P3) Imply STP
=
233
Ti,2 +τi,2 −1 k Ti,2 +τ k i,2 −1 f f ( f (t, x˜t(i) , u˜ (i) ) − f (t, x , u )) ≤ 2−i+3 ≤ 8. t t t i=1 t =Ti,1 −τi,1
t =Ti,1 −τi,1
i=1
(4.123) Theorem 4.1 and (4.120)–(4.123) imply that in the both cases ∞ ({xt }∞ t =−∞ , {ut }t =−∞ ) f
f
is (f )good. In view of (P1), limt →∞ ρE (xt , xt ) = 0 and limt →−∞ ρE (xt , xt ) = 0. On the other hand in view of (4.105), (4.109), (4.120), and (4.121), at f least one of the following relations holds: lim supt →∞ ρE (xt , xt ) ≥ ; f lim supt →−∞ ρE (xt , xt ) ≥ . The contradiction we have reached completes the proof of Lemma 4.24. Lemma 4.25. Assume that (P1), (P2), (P3) hold, f
u˜ t = ut for all integers t
(4.124)
and that ∈ (0, 1). Then there exists δ > 0 such that for each pair of integers T1 , 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying T2 ≥ T1 + 3bf and each ({xt }Tt =T 1 1 f
ρE (xTi , xTi ) ≤ δ, i = 1, 2 T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ
t =T1 f
the inequality ρE (xt , xt ) ≤ for all t = T1 , . . . , T2 . Proof. Lemma 4.24 and (A2) imply that there exist δ0 ∈ (0, /4) and an integer L0 > 0 such that: (iv) (A2) holds with = 1 and δ = δ0 ; (v) for each pair of integers T1 , T2 ≥ T1 + 2bf such that either T1 ≥ L0 or 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ) satisfying T2 ≤ −L0 and each ({xt }Tt =T 1 1 f
ρE (xTi , xTi ) ≤ δ0 , i = 1, 2 T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ0
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 .
234
4 DiscreteTime Nonautonomous Problems on Axis
Theorem 4.1 implies that there exists S0 > 0 such that for each pair of integers 2 2 −1 , {ut }Tt =T ) ∈ X(T1 , T2 ), T2 > T1 and each ({xt }Tt =T 1 1 T 2 −1
f (t, xt , ut ) + S0 ≥
t =T1
T 2 −1
f
f
(4.125)
f (t, xt , ut ).
t =T1
It follows from (P2) that there exist δ1 ∈ (0, δ0 ) and a natural number L1 such that the following property holds: +L1 +L1 −1 , {ut }Tt =T ) ∈ X(T , T +L1 ) satisfying (vi) for each integer T and each ({xt }Tt =T T +L 1 −1
f (t, xt , ut ) ≤ min{
T +L 1 −1
t =T
f
f
f (t, xt , ut ) + 2S0 + 3,
t =T
U f (T , T + L1 , xT , xT +L1 ) + δ1 } we have f
min{ρE (xt , xt ) : t = T , . . . , T + L1 } ≤ δ0 . By (A2), there exists a sequence {δi }∞ i=1 ⊂ (0, 1) such that δi < 2−1 δi−1 , i = 2, 3, . . . .
(4.126)
and that for each integer i ≥ 1 the following property holds: f
(vii) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δi there exist (1) T +b
(1) T +b −1
({yt }t =T f , {vt }t =T f
) ∈ X(T , T + bf ),
−1 ({yt }Tt=T −bf , {vt }Tt =T −bf ) ∈ X(T − bf , T ) (2)
(2)
such that (1)
f
(1)
(2)
f
(2)
yT = z, yT +bf = xT +bf , yT = z, yT −bf = xT −bf , T +bf −1
T +bf −1 (1)
(1)
f (t, yt , vt ) ≤
t =T T −1 t =T −bf
f
f
f
f
f (t, xt , ut ) + 1/i,
t =T
(2)
(2)
f (t, yt , vt ) ≤
T −1 t =T −bf
f (t, xt , ut ) + 1/i.
4.10 (P1)–(P3) Imply STP
235
Assume that the lemma does not hold. Then for each integer i ≥ 1 there exist (i) Ti,2 (i) Ti,2 −1 , {ut }t =T ) ∈ X(Ti,1 , Ti,2 ) such integers Ti,1 , Ti,2 ≥ Ti,1 + 3bf and ({xt }t =T i,1 i,1 that f
, xTi,j ) ≤ δi , j = 1, 2, ρE (xT(i) i,j
(4.127)
Ti,2 −1
t =Ti,1
(i)
(i)
(i)
(i)
f (t, xt , ut ) ≤ U f (Ti,1 , Ti,2 , xTi,1 , xTi,2 ) + δi
(4.128)
and ti ∈ {Ti,1 , . . . , Ti,2 } for which (i)
f
(4.129)
ρE (xti , xti ) > . Let i be a natural number. Property (v) and (4.126)–(4.129) imply that Ti,1 < L0 , Ti,2 > −L0 .
(4.130)
We show that − 2bf − 1 − L1 − L0 ≤ ti ≤ 2bf + 1 + L1 + L0 .
(4.131)
Property (iv), (A2) with = 1, δ = δ0 , (4.126) and (4.127) imply that there exists (i) Ti,2 (i) Ti,2 −1 , {vt }t =T ) ∈ X(Ti,1 , Ti,2 ) such that ({yt }t =T i,1 i,1 = xT(i) , yT(i) = xT(i) , yT(i) i,1 i,1 i,2 i,2
(4.132)
f
yt(i) = xt , t ∈ {Ti,1 + bf , . . . , Ti,2 − bf }, (i)
vt
f
= ut , t ∈ {Ti,1 + bf , . . . , Ti,2 − bf − 1},
Ti,1 +bf −1
Ti,1 +bf −1 (i) (i) f (t, yt , vt )
≤
t =Ti,1
t =Ti,2 −bf
f
f
(4.134)
f
f
(4.135)
f (t, xt , ut ) + 1,
t =Ti,1
Ti,2 −1
(4.133)
Ti,2 −1
f (t, yt(i) , vt(i) )
≤
t =Ti,2 −bf
f (t, xt , ut ) + 1.
236
4 DiscreteTime Nonautonomous Problems on Axis
It follows from (4.128) and (4.132)–(4.135) that Ti,2 −1
Ti,2 −1 (i)
(i)
f (t, xt , ut ) ≤
t =Ti,1
(i)
(i)
f (t, yt , vt ) + 1
t =Ti,1 Ti,2 −1
≤
f
f
f (t, xt , ut ) + 3.
(4.136)
t =Ti,1
In view of (4.125), (4.136), for each pair of integers S1 , S2 ∈ [Ti,1 , Ti,2 ] satisfying S1 < S2 , S 2 −1
Ti,2 −1
f (t, xt(i), u(i) t )=
t =S1
− −
f
f (t, xt(i) , u(i) t )
t =Ti,1
(i) (i) {f (t, xt , ut ) : t ∈ {Ti,1 , . . . , S1 } \ {S1 }}
(i) (i) {f (t, xt , ut ) : t ∈ {S2 , . . . , Ti,2 } \ {Ti,2 }}
Ti,2 −1
≤
f
f (t, xt , ut ) + 3 −
f f {f (t, xt , ut ) : t ∈ {Ti,1 , . . . , S1 } \ {S1 }} + S0
t =Ti,1
−
f f {f (t, xt , ut ) : t ∈ {S2 , . . . , Ti,2 } \ {Ti,2 }} + S0 ≤
S 2 −1
f
f
f (t, xt , ut ) + 3 + 2S0 .
(4.137)
t =S1
Assume that ti > 2bf + 1 + L1 + L0 .
(4.138)
In view of (4.138), [ti − 2bf − 1 − L1 , ti − 1 − 2bf ] ⊂ [L0 , ∞). t −2b −1
(4.139)
t −2b −2
f i f , {u(i) Consider the pair ({xt(i) }ti=ti −L t }t =ti −2bf −L1 −1 ) ∈ X(ti − L1 − 2bf − 1 −2bf −1 1, ti − 2bf − 1). Property (vi), (4.126), (4.128), and (4.137) imply that there exists
t˜ ∈ {ti − L1 − 2bf − 1, . . . , ti − 2bf − 1}
(4.140)
4.10 (P1)–(P3) Imply STP
237
such that f
ρE (xt(i) ˜ , xt˜ ) ≤ δ0 .
(4.141)
By property (v), (4.126)–(4.128) and (4.139)–(4.141), (i) f ρE (xt , xt ) ≤ , t = t˜, . . . , Ti,2
and in particular, f
, xti ) ≤ . ρE (xt(i) i
(4.142)
This contradicts (4.129). The contradiction we have reached proves that ti ≤ 2bf + 1 + L1 + L0 .
(4.143)
ti < −2bf − 1 − L1 − L0 .
(4.144)
Assume that
In view of (4.144), [ti + 2bf + 1, ti + 1 + 2bf + L1 ] ⊂ (−∞, −L0 ]. t +2b +L +1
(4.145)
t +2b +L
f 1 i f 1 , {u(i) Consider the pair ({xt(i)}ti=ti +1+2b t }t =ti +1+2bf ) ∈ X(ti + 1 + 2bf , ti + 2bf + f 1+L1 ). Property (vi), (4.126), (4.128), (4.130), (4.137), and (4.145) imply that there exists
t˜ ∈ {ti + 1 + 2bf , . . . , ti + 2bf + 1 + L1 }
(4.146)
such that (i)
f
ρE (xt˜ , xt˜ ) ≤ δ0 . By property (v), (4.126)–(4.128), (4.130), (4.145), and (4.147), ρE (xt , xt ) ≤ , t = Ti,1 . . . , t˜, (i)
f
and in particular, (i)
f
ρE (xti , xti ) ≤ . This contradicts (4.129). The contradiction we have reached proves that ti ≥ −2bf − 1 − L1 − L0 .
(4.147)
238
4 DiscreteTime Nonautonomous Problems on Axis
Together with (4.143) this implies (4.131). In view of (4.10) and (4.137), there exists M1 > 0 such that (i)
(i)
f (t, xt , ut ) ≤ M1 for each integer i ≥ 1 each t ∈ {Ti,1 , . . . , Ti,2 − 1}. (4.148) It follows from (4.3), (4.4), (4.9), (4.127), and (4.148) that there exists M2 > 0 such that ρE (xt(i) , θ0 ) ≤ M2 , i = 1, 2, . . . , t ∈ {Ti,1 , . . . , Ti,2 }.
(4.149)
Extracting a subsequence and reindexing, we may assume without loss of generality that either Ti,1 = T1,1 for all integers i ≥ 1 or lim Ti,1 = −∞,
(4.150)
either Ti,2 = T1,2 for all integers i ≥ 1 or lim Ti,2 = ∞,
(4.151)
i→∞
i→∞
ti = t1 for all integers i ≥ 1.
(4.152)
There are four cases: (a) (b) (c) (d)
Ti,1 = T1,1 , Ti,2 = T1,2 for all integers i ≥ 1; Ti,1 = T1,1 for all integers i ≥ 1 and {Ti,2 }∞ i=1 is a strictly increasing sequence; Ti,2 = T1,2 for all integers i ≥ 1 and {Ti,1 }∞ i=1 is a strictly decreasing sequence; {Ti,1 }∞ is a strictly decreasing sequence and {Ti,2 }∞ i=1 i=1 is a strictly increasing sequence.
Assume that the case (a) holds. In view of (4.48) and LSC property, extracting a subsequence and reindexing we may assume without loss of generality that there T1,2 T1,2 −1 exists ({ xt }t =T , { ut }t =T ) ∈ X(T1,1, T1,2 ) such that 1,1 1,1 xt as i → ∞ for all t ∈ {T1,1 , . . . , T1,2 } xt(i) → (i)
(i)
ut ) ≤ lim inf f (t, xt , ut ), t ∈ {T1,1 , . . . , T1,2 − 1}. f (t, xt , i→∞
(4.153) (4.154)
It follows from (4.126), (4.127), (4.129), and (4.153) that f
f
f
xT1,2 = xT1,2 , ρE ( xt1 , xt1 ) ≥ . xT1,1 = xT1,1 ,
(4.155)
Property (P3), (4.124), and (4.155) imply that there exists γ > 0 such that T1,2 −1
t =T1,1
T1,2 −1 f f f (t, xt , ut )
S1 ≥ T1,1 , S 2 −1
f (t, xt , ut ) ≤
t =S1
S 2 −1
f
f
f (t, xt , ut ) + 3 + 2S0 .
(4.159)
f
(4.160)
t =S1
Set f
xt = xt , ut = ut for all integers t < T1,1 . Theorem 4.1, (4.127), (4.157), (4.159), and (4.160) imply that ({ x t }∞ ut } ∞ t =−∞ , { t =−∞ ) ∈ X(−∞, ∞) is (f )good. (P1) implies that f
lim ρE ( xt , xt ) = 0.
t →∞
(4.161)
By (4.129) and (4.157), f
xt1 ) ≥ . ρE (xt1 ,
(4.162)
240
4 DiscreteTime Nonautonomous Problems on Axis
Property (P3), (4.102), and (4.124) imply that ({ x t }∞ ut } ∞ t =−∞ , { t =−∞ ) is not (f )minimal. There exist Δ > 0 and an integer S∗ > T1,1  + 1 + bf such that S ∗ −1
f (t, xt , ut ) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ.
(4.163)
t =−S∗
Lemma 4.20 and (A2) imply that there exists δ(Δ) ∈ (0, Δ/8) such that the following properties hold: (viii) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying f
T2 ≥ T1 + 2bf , ρE (zi , xTi ) ≤ δ(Δ), i = 1, 2 we have U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f
f
f (t, xt , ut ) + Δ/8;
t =T1 f
(ix) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ(Δ) there exist (1) T +b
(1) T +b −1
({yt }t =T f , {vt }t =T f
) ∈ X(T , T + bf ),
−1 ({yt }Tt=T −bf , {vt }Tt =T −bf ) ∈ X(T − bf , T ) (2)
(2)
such that (1)
f
(1)
(2)
f
(2)
yT = z, yT +bf = xT +bf , yT = z, yT −bf = xT −bf , T +bf −1
T +bf −1 (1)
(1)
f (t, yt , vt ) ≤
t =T
f
f
f
f
f (t, xt , ut ) + Δ/16,
t =T
T −1 t =T −bf
(2)
(2)
f (t, yt , vt ) ≤
T −1
f (t, xt , ut ) + Δ/16.
t =T −bf
In view of (4.126), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ)/4 for all integers k ≥ k(Δ).
(4.164)
Property (viii), (4.64), (4.127), and (4.128) imply that the following property holds:
4.10 (P1)–(P3) Imply STP
241
(x) for each integer k ≥ k(Δ), Tk,2 −1
Tk,2 −1
f (t, xt(k), u(k) t )
≤
t =T1,1
f
f
f (t, xt , ut ) + Δ/2.
t =T1,1
Property (P1) implies that there exists an integer τ0 > 1 such that f
ρE ( xt , xt ) ≤ δ(Δ)/4 for all integers t ≥ τ0 .
(4.165)
Let q ≥ 1 be an integer satisfying Tq,2 > τ0 + S∗ .
(4.166)
By (4.157), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , ρE ( xTq,2 , xT(k) ) ≤ δ(Δ)/4. q,2
(4.167)
Assume that an integer k ≥ k1 . Then (4.167) holds. By (4.128) and (4.164), Tq,2 −1
(k)
t =T1,1
(k)
(k)
(k)
f (t, xt , ut ) ≤ U f (T1,1 , Tq,2 , xT1,1 , xTq,2 ) + δ(Δ)/4.
(4.168)
It follows from (4.127) and (4.164) that f
(k)
ρE (xT1,1 , xT1,1 ) ≤ δk ≤ δ(Δ)/4.
(4.169)
In view of (4.165) and (4.167), (k)
f
f
(k)
xTq,2 ) + ρE ( xTq,2 , xTq,2 ) ≤ δ(Δ)/4 + δ(Δ)/4. ρE (xTq,2 , xTq,2 ) ≤ ρE (xTq,2 , (4.170) Property (viii), (4.118), (4.164), (4.169), and (4.170) imply that Tq,2 −1
Tq,2 −1
f (t, xt(k), u(k) t )≤
t =T1,1
f
f
f (t, xt , ut ) + Δ/2.
(4.171)
t =T1,1
In view of (4.158), (4.170), and (4.171), Tq,2 −1
t =T1,1
Tq,2 −1
f (t, xt , ut ) ≤
f
f
f (t, xt , ut ) + Δ/2,
(4.172)
f
(4.173)
t =T1,1
ρE ( xTq,2 , xTq,2 ) ≤ δ(Δ)/2 for each integer q ≥ 1 satisfying (4.166).
242
4 DiscreteTime Nonautonomous Problems on Axis
Let an integer q ≥ 1 satisfy (4.166) and q −1 < Δ/4.
(4.174)
Property (ix), (4.163), and (4.173) imply that there exists ∞ ({ξt }∞ t =−∞ , {ηt }t =−∞ ) ∈ X(−∞, ∞)
such that xt , ηt = ut for all integers t < −S∗ , ξ−S∗ = x−S∗ , ξS∗ = x S∗ , ξt = S ∗ −1
f (t, ξt , ηt )
T1,2 . xt = xt , It follows from (4.127), (4.182), (4.184), and (4.185) that ut } ∞ ({ x t }∞ t =−∞ , { t =−∞ ) ∈ X(−∞, ∞) is (f )good. (P1) implies that f
lim ρE ( xt , xt ) = 0.
t →−∞
(4.186)
244
4 DiscreteTime Nonautonomous Problems on Axis
By (4.129), (4.152), and (4.182), f
ρE (xt1 , xt1 ) ≥ .
(4.187)
Property (P3), (4.124), and (4.187) imply that ({ x t }∞ ut } ∞ t =−∞ , { t =−∞ ) is not (f )minimal. Therefore there exist Δ > 0 and an integer S∗ > 0 such that S∗ > T1,2  + 1 + bf , S ∗ −1
(4.188)
f (t, xt , ut ) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ.
(4.189)
t =−S∗
Lemma 4.20 and (A2) imply that there exists δ(Δ) ∈ (0, Δ/8) such that the following properties hold: (xi) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying f
T2 ≥ T1 + 2bf , ρE (zi , xTi ) ≤ δ(Δ), i = 1, 2 we have U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f
f
f (t, xt , ut ) + Δ/8;
t =T1 f
(xii) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ(Δ) there exist T +b
T +b −1
({yt(1)}t =T f , {vt(1) }t =T f
) ∈ X(T , T + bf ),
−1 ({yt }Tt=T −bf , {vt }Tt =T −bf ) ∈ X(T − bf , T ) (2)
(2)
such that (1)
f
(1)
(2)
f
(2)
yT = z, yT +bf = xT +bf , yT = z, yT −bf = xT −bf , T +bf −1
T +bf −1 (1)
(1)
f (t, yt , vt ) ≤
t =T T −1 t =T −bf
f
f
f
f
f (t, xt , ut ) + Δ/16,
t =T
f (t, yt(2), vt(2) )
≤
T −1 t =T −bf
f (t, xt , ut ) + Δ/16.
4.10 (P1)–(P3) Imply STP
245
In view of (4.126), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ)/4 for all integers k ≥ k(Δ).
(4.190)
Property (xi), (4.127), (4.128), and (4.190) imply that for each integer k ≥ k(Δ), Tk,2 −1
Tk,2 −1
f (t, xt(k), u(k) t )
≤
t =T1,1
f
f
f (t, xt , ut ) + Δ/2.
(4.191)
t =T1,1
By (4.186), there exists an integer τ0 < −1 such that f
ρE ( xt , xt ) ≤ δ(Δ)/4 for all integers t ≤ τ0 .
(4.192)
Let q ≥ 1 be an integer satisfying Tq,1 < −τ0 − S∗ .
(4.193)
By (4.182), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , (k)
ρE ( xTq,1 , xTq,1 ) ≤ δ(Δ)/4.
(4.194)
Assume that an integer k ≥ k1 . Then (4.194) holds. By (4.127), (4.128), and (4.190), T1,2 −1
t =Tq,1
(k) (k) f f (t, xt(k) , u(k) t ) ≤ U (Tq,1 , T1,2 , xTq,1 , xT1,2 ) + δ(Δ)/4,
f
, xT1,2 ) ≤ δk ≤ δ(Δ)/4. ρE (xT(k) 1,2
(4.195)
(4.196)
In view of (4.192)–(4.194), (k)
f
f
(k)
xTq,1 ) + ρE ( xTq,1 , xTq,1 ) ≤ δ(Δ)/4 + δ(Δ)/4. ρE (xTq,1 , xTq,1 ) ≤ ρE (xTq,1 , (4.197) Property (xi), (4.128), (4.196), and (4.197) imply that T1,2 −1
T1,2 −1 (k)
(k)
f (t, xt , ut ) ≤
t =Tq,1
f
f
f (t, xt , ut ) + Δ/2.
(4.198)
t =Tq,1
In view of (4.182), (4.183), (4.197), and (4.198), T1,2 −1
t =Tq,1
T1,2 −1
f (t, xt , ut ) ≤
t =Tq,1
f
f
f (t, xt , ut ) + Δ/2,
(4.199)
246
4 DiscreteTime Nonautonomous Problems on Axis f
ρE ( xTq,1 , xTq,1 ) ≤ δ(Δ)/2
(4.200)
for each integer q ≥ 1 satisfying (4.193). Let an integer q ≥ 1 satisfy (4.193) and q −1 < Δ/4.
(4.201)
Property (xii), (4.189), and (4.200) imply that there exists ∗ ∗ −1 , {ηt }St =T ) ∈ X(Tq,1 − bf , S∗ ) ({ξt }St =T q,1 −bf q,1 −bf
such that ξ−S∗ = x−S∗ , ξS∗ = x S∗ , S ∗ −1
f (t, ξt , ηt )
0 and an integer S∗ such that
248
4 DiscreteTime Nonautonomous Problems on Axis
S∗ > 1 + bf , S ∗ −1
(4.214)
f (t, xt , ut ) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ.
(4.215)
t =−S∗
Lemma 4.20 and (A2) imply that there exists δ(Δ) ∈ (0, Δ/8) such that the following properties hold: (xiii) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying f
T2 ≥ T1 + 2bf , ρE (zi , xTi ) ≤ δ(Δ), i = 1, 2 we have U f (T1 , T2 , z1 , z2 ) ≤
T 2 −1
f
f
f (t, xt , ut ) + Δ/8;
t =T1 f
(xiv) for each (T , z) ∈ A satisfying ρE (z, xT ) ≤ δ(Δ) there exist T +b
T +b −1
({yt(1)}t =T f , {vt(1) }t =T f
) ∈ X(T , T + bf ),
−1 ({yt(2)}Tt=T −bf , {vt(2) }Tt =T −bf ) ∈ X(T − bf , T )
such that f
f
yT(1) = z, yT(1)+bf = xT +bf , yT(2) = z, yT(2)−bf = xT −bf , T +bf −1
T +bf −1
f (t, yt(1) , vt(1) )
≤
t =T
f
f
f
f
f (t, xt , ut ) + Δ/16,
t =T
T −1
f (t, yt(2), vt(2) ) ≤
t =T −bf
T −1
f (t, xt , ut ) + Δ/16.
t =T −bf
In view of (4.126), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ)/4 for all integers k ≥ k(Δ).
(4.216)
Property (xiii), (4.127), (4.128), and (4.216) imply that for each integer k ≥ k(Δ), Tk,2 −1
t =Tk,1
Tk,2 −1 (k)
(k)
f (t, xt , ut ) ≤
t =Tk,1
f
f
f (t, xt , ut ) + Δ/2.
(4.217)
4.10 (P1)–(P3) Imply STP
249
By (4.212), there exists an integer τ0 > 1 such that f
ρE ( xt , xt ) ≤ δ(Δ)/4 for all integers t ∈ [τ0 , ∞) ∪ (−∞, −τ0 ].
(4.218)
Let q ≥ 1 be an integer satisfying Tq,1 < −τ0 − S∗ , Tq,2 > S∗ + τ0 .
(4.219)
By (4.209), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , ρE ( xTq,1 , xT(k) ) ≤ δ(Δ)/4, ρE ( xTq,2 , xT(k) ) ≤ δ(Δ)/4. q,1 q,2
(4.220)
Assume that an integer k ≥ k1 . Then (4.220) holds. By (4.128) and (4.216), Tq,2 −1
(k)
t =Tq,1
(k)
(k)
(k)
f (t, xt , ut ) ≤ U f (Tq,1 , Tq,2 , xTq,1 , xTq,2 ) + δ(Δ)/4.
(4.221)
In view of (4.218)–(4.220), for j = 1, 2, f
f
, xTq,j ) ≤ ρE (xT(k) , xTq,j ) + ρE ( xTq,j , xTq,j ) ≤ δ(Δ)/4 + δ(Δ)/4. ρE (xT(k) q,j q,j (4.222) Property (xiii), (4.221), and (4.222) imply that Tq,2 −1
Tq,2 −1 (k)
(k)
f (t, xt , ut ) ≤
t =Tq,1
f
f
f (t, xt , ut ) + Δ/2.
(4.223)
t =Tq,1
In view of (4.210) and (4.223), Tq,2 −1
Tq,2 −1
f (t, xt , ut ) ≤
f
f
f (t, xt , ut ) + Δ/2.
(4.224)
ρE ( xTq,j , xTq,j ) ≤ δ(Δ)/4, j = 1, 2.
(4.225)
t =Tq,1
t =Tq,1
By (4.209) and (4.222), f
Let an integer q ≥ 1 satisfy (4.219). Property (xiv), (4.215), (4.219), and (4.225) Tq,2 +bf Tq,2 +bf −1 imply that there exists ({ξt }t =T , {ηt }t =T ) ∈ X(Tq,1 −bf , Tq,2 +bf ) such q,1 −bf q,1 −bf that ξ−S∗ = x−S∗ , ξS∗ = x S∗ ,
(4.226)
250
4 DiscreteTime Nonautonomous Problems on Axis S ∗ −1
S ∗ −1
f (t, ξt , ηt )
0. By Lemma 4.25, there exist δ0 ∈ (0, ) such that the following property holds: (xv) for each integer T1 , each integer T2 ≥ T1 + 3bf and each 2 2 −1 ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1
which satisfies f
ρE (xTi , xTi ) ≤ δ0 , i = 1, 2, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ0
t =T1 f
the inequality ρE (xt , xt ) ≤ holds for all t = T1 , . . . , T2 . By Theorem 4.1, there exists S0 > 0 such that for each pair of integers T2 > T1 2 2 −1 and each ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ), we have 1 1 T 2 −1
f (t, yt , vt ) + S0 ≥
t =T1
T 2 −1
f
f
(4.233)
f (t, xt , ut ).
t =T1
In view of (P2), there exist δ ∈ (0, δ0 ) and an integer L0 > 0 such that the following property holds: +L0 +L0 −1 (xvi) for each integer T and each ({yt }Tt =T , {vt }Tt =T ) ∈ X(T , T + L0 ) which satisfies T +L 0 −1
f (t, yt , vt )
t =T
≤ min{U f (T , T + L0 , yT , yT +L0 ) + δ,
T +L 0 −1
f
f
f (t, xt , ut ) + 2S0 + M}
t =T f
there exists an integer s ∈ [T , T + L0 ] such that ρE (ys , xs ) ≤ δ0 . Set L = 2L0 + 2bf + 2.
(4.234)
252
4 DiscreteTime Nonautonomous Problems on Axis
2 2 −1 Assume that T1 and T2 ≥ T1 + 2L are integers and that ({xt }Tt =T , {ut }Tt =T ) ∈ 1 1 X(T1 , T2 ) satisfies
T 2 −1
f (t, xt , ut ) ≤ σ f (T1 , T2 ) + M,
(4.235)
t =T1 T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ.
(4.236)
t =T1
In view of (4.236), T1 +L 0 −1
f (t, xt , ut ) ≤ U f (T1 , T1 + L0 , xT1 , xT1 +L0 ) + δ,
(4.237)
f (t, xt , ut ) ≤ U f (T2 − L0 , T2 , xT2 −L0 , xT2 ) + δ.
(4.238)
t =T1 T 2 −1 t =T2 −L0
By (4.235), T 2 −1
f (t, xt , ut ) ≤
t =T1
T 2 −1
f
f
f (t, xt , ut ) + M.
(4.239)
t =T1
It follows from (4.233) and (4.239) that for each pair of integers S1 , S2 ∈ [T1 , T2 ] satisfying S1 < S2 , S 2 −1
f (t, xt , ut ) =
t =S1
T 2 −1
{f (t, xt , ut ) : t ∈ {T1 , . . . , S1 } \ {S1 }}
t =T1
− ≤
f (t, xt , ut ) −
T 2 −1
f
f
{f (t, xt , ut ) : t ∈ {S2 , . . . , T2 } \ {T2 }}
f (t, xt , ut ) + M −
f
f
{f (t, xt , ut ) : t ∈ {T1 , . . . , S1 } \ {S1 }} + S0
t =T1
−
S 2 −1 f f f f {f (t, xt , ut ) : t ∈ {S2 , . . . , T2 } \ {T2 }} + S0 = f (t, xt , ut ) + M + 2S0 . t =S1
(4.240) It follows from (4.237), (4.238), (4.240), and property (xvi) that there exist integers τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that
4.11 Proof of Theorem 4.8
253
ρE (xτi , xτfi ) ≤ δ0 , i = 1, 2.
(4.241)
f
f
If ρE (xT1 , xT1 ) ≤ δ, then we may assume that τ1 = T1 and if ρE (xT2 , xT2 ) ≤ δ, then we may assume that τ2 = T2 . Property (xv), (4.234), (4.236), and (4.241) imply that f
ρE (xt , xt ) ≤ , t = τ1 , . . . , τ2 . Theorem 4.7 is proved.
4.11 Proof of Theorem 4.8 ∞ Lemma 4.26. Let ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) be (f )good and > 0. Then there exists an integer L > 0 such that for each pair of integers T2 > T1 ≥ L, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) +
t =T1
for each pair of integers T1 < T2 ≤ −L, T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + .
t =T1
Proof. There exists M0 > 0 such that for each integer T > 0, 
T −1
f (t, xt , ut ) −
t =−T
T −1
f
f
(4.242)
f (t, xt , ut ) < M0 .
t =−T
Theorem 4.1 implies that there exists M1 > 0 such that for each pair of integers 2 2 −1 S2 > S1 and each ({yt }St =S , {vt }St =S ) ∈ X(S1 , S2 ) 1 1 M1 +
S 2 −1 t =S1
f (t, yt , vt ) ≥
S 2 −1
f
f
f (t, xt , ut ).
(4.243)
t =S1
Assume that the lemma does not hold. Then there exists a sequence of intervals [ai , bi ], i = 1, 2, . . . such that for each integer i ≥ 1, ai < bi are integers, for each pair of integers i < j , [ai , bi ] ∩ [aj , bj ] = ∅,
(4.244)
254
4 DiscreteTime Nonautonomous Problems on Axis b i −1
f (t, xt , ut ) > U f (ai , bi , xai , xbi ) + , i = 1, 2, . . . .
(4.245)
t =ai ∞ By (4.244) and (4.245), there exists ({yt }∞ t =−∞ , {vt }t =−∞ ) ∈ X(−∞, ∞) such that
yt = xt for all integers t ∈ ∪∞ i=1 (ai , bi ),
(4.246)
vt = ut for all integers t ∈ ∪∞ i=1 [ai , bi ),
(4.247)
for all integer i ≥ 1, b i −1
f (t, yt , vt )
(M0 + M1 ) −1 .
(4.249)
Choose an integer T > 0 such that [ai , bi ] ⊂ [−T , T ], i = 1, . . . , q.
(4.250)
By (4.246)–(4.248) and (4.250), T −1
f (t, yt , vt ) −
t =−T
≤
T −1
f (t, xt , ut )
t =−T
q b b i −1 i −1 ( f (t, yt , vt ) − f (t, xt , ut )) ≤ −q. i=1 t =ai
(4.251)
t =ai
In view of (4.242), (4.243), and (4.251), T −1
f
f
f (t, xt , ut ) − M1 ≤
t =−T
T −1
f (t, yt , vt ) ≤
t =−T
≤
T −1
T −1
f (t, xt , ut ) − q
t =−T
f
f
f (t, xt , ut ) + M0 − q,
t =−T
q ≤ M0 + M1 . This contradicts (4.249). The contradiction we have reached proves Lemma 4.26.
4.12 Proof of Theorem 4.13
255
Proof of Theorem 4.8. Assume that (P4) holds. Clearly, (P4) implies (P2) and (P1) follows from (P4) and Lemma 4.26. Assume that (P1) and (P2) hold. Then (P4) follows from (P2) and Lemma 4.24. Theorem 4.8 is proved.
4.12 Proof of Theorem 4.13 Clearly, WTP implies (P1) and (P2). Assume that f has (P1) and (P2). We show that WTP holds. Let , M > 0. By Theorem 4.1, there exists S0 > 0 such that the following property holds: 2 2 −1 , {vt }τt =τ ) ∈ X(τ1 , τ2 ), (i) for each pair of integers τ2 > τ1 and each ({yt }τt =τ 1 1
τ 2 −1
f (t, yt , vt ) + S0 ≥
t =τ1
τ 2 −1
f
f
f (t, xt , ut ).
t =τ1
Lemma 4.24 implies that there exist δ0 > 0 and an integer L0 > 2bf such that the following properties hold: (ii) for each integer T1 ≥ L0 , each integer T2 ≥ T1 + 2bf , and each 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) which satisfies ({yt }Tt =T 1 1 f
ρE (yTi , xTi ) ≤ δ0 , i = 1, 2, T 2 −1
f (t, yt , vt ) ≤ U f (T1 , T2 , yT1 , yT2 ) + δ0
(4.252)
(4.253)
t =T1
we have f
ρE (yt , xt ) ≤ , t = T1 , . . . , T2 ;
(4.254)
(iii) for each integer T2 ≤ −L0 , each integer T1 ≤ T2 − 2bf , and each 2 2 −1 , {vt }Tt =T ) ∈ X(T1 , T2 ) which satisfies (4.252) and (4.253) rela({yt }Tt =T 1 1 tion (4.254) holds. Property (P2) implies that there exist δ ∈ (0, δ0 ) and an integer L1 > 0 such that the following property holds: +L1 +L1 −1 , {vt }Tt =T ) ∈ X(T , T + L1 ) which (iv) for each integer T and each ({yt }Tt =T satisfies
256
4 DiscreteTime Nonautonomous Problems on Axis T +L 1 −1
f (t, yt , vt )
t =T
≤ min{U f (T , T + L1 , yT , yT +L1 ) + δ,
T +L 1 −1
f
f
f (t, xt , ut ) + M + 2S0 }
t =T f
there exists an integer s ∈ [T , T + L1 ] such that ρE (xs , xs ) ≤ δ0 . Set l = 8L0 + 8L1 + 8 + 8bf .
(4.255)
Q ≥ 3(2 + 2(M + S0 )δ −1 ).
(4.256)
Choose a natural number
2 2 −1 Assume that T1 , T2 ≥ T1 + lQ are integers and ({xt }Tt =T , {ut }Tt =T ) ∈ X(T1 , T2 ) 1 1 satisfies
T 2 −1
T 2 −1
f (t, xt , ut ) ≤
t =T1
f
f
f (t, xt , ut ) + M.
(4.257)
t =T1
Property (i) and (4.257) imply that for each pair of integers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , τ 2 −1
f (t, xt , ut ) =
t =τ1
T 2 −1
{f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }}
t =T1
− ≤
f (t, xt , ut ) −
T 2 −1
f
{f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {T2 }}
f
f (t, xt , ut ) + M −
f f {f (t, xt , ut ) : t ∈ {T1 , . . . , τ1 } \ {τ1 }} + S0
t =T1
−
f f {f (t, xt , ut ) : t ∈ {τ2 , . . . , T2 } \ {T2 }} + S0 =
τ 2 −1
f
f
f (t, xt , ut ) + M + 2S0 .
(4.258)
t =τ1
Set t0 = T1 .
(4.259)
4.12 Proof of Theorem 4.13
257
If T 2 −1
f (t, xt , ut ) ≤ U f (T1 , T2 , xT1 , xT2 ) + δ,
t =T1
then we set t1 = T2 . Assume that T 2 −1
f (t, xt , ut ) > U f (T1 , T2 , xT1 , xT2 ) + δ.
(4.260)
t =T1
Set t1 = min{t ∈ {T1 + 1, . . . , T2 } :
t −1
f (τ, xτ , uτ ) > U f (T1 , t, xT1 , xt ) + δ}.
τ =T1
(4.261) Assume that k ≥ 1 is an integer and we defined a sequence of integers {ti }ki=0 ⊂ [T1 , T2 ] such that t0 < t1 · · · < tk ,
(4.262)
for each integer i = 1, . . . , k, if ti < T2 , then t i −1
f (t, xt , ut ) − U f (ti−1 , ti , xti−1 , xti ) > δ,
(4.263)
t =ti−1
and if ti − ti−1 ≥ 2, then t i −2
f (t, xt , ut ) − U f (ti−1 , ti − 1, xti−1 , xti −1 ) ≤ δ.
(4.264)
t =ti−1
(Note that in view of (4.261), our assumption holds for k = 1.) 2 2 −1 By (4.262)–(4.264), there exists ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) such that 1 1 yti = xti , i = 1, . . . , k, t i −1 t =ti−1
f (t, xt , ut ) −
t i −1
f (t, yt , vt ) > δ, i ∈ {1, . . . , k} \ {k},
(4.265)
(4.266)
t =ti−1
yt = xt , t ∈ {tk , . . . , T2 }, vt = ut , t ∈ {tk , . . . , T2 } \ {T2 }.
(4.267)
258
4 DiscreteTime Nonautonomous Problems on Axis
Property (i), (4.257), and (4.265)–(4.267) imply that M+
T 2 −1
f
f
f (t, xt , ut ) ≥
t =T1
T 2 −1
f (t, xt , ut ) ≥
t =T1 T 2 −1
≥
T 2 −1
f (t, yt , vt ) + δ(k − 1)
t =T1
f
f
f (t, xt , ut ) − S0 + δ(k − 1),
t =T1
k ≤ 1 + δ −1 (M + S0 ).
(4.268)
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If T 2 −1
f (t, xt , ut ) ≤ U f (tk , T2 , xtk , xT2 ) + δ,
t =tk
then we set tk+1 = T2 and the construction of the sequence is completed. Assume that T 2 −1
f (t, xt , ut ) > U f (tk , T2 , xtk , xT2 ) + δ.
t =tk
Set tk+1 = min{t ∈ {tk + 1, . . . , T2 } :
t −1
f (τ, xτ , uτ ) − U f (tk , t, xtk , xt ) > δ}.
τ =tk
Clearly, tk+1 is welldefined, and the assumption made for k also holds for k + 1. By induction and by (4.263), (4.264), and (4.268), we constructed a finite sequence q of integers {ti }i=0 ⊂ [T1 , T2 ] such that T1 = t0 < t1 · · · < tq = T2 , q ≤ 2 + δ −1 (M + S0 ),
(4.269)
for each integer i = 1, . . . , q if ti < T2 , then t i −1
f (t, xt , ut ) − U f (ti−1 , ti , xti−1 , xti ) > δ,
t =ti−1
and if ti − ti−1 ≥ 2, then t i −2 t =ti−1
f (t, xt , ut ) − U f (ti−1 , ti − 1, xti−1 , xti −1 ) ≤ δ.
(4.270)
4.12 Proof of Theorem 4.13
259
Let i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 8 + 8L1 + 8L0 + 8bf .
(4.271)
By (4.270) and (4.271), ti+1 −2
f (t, xt , ut ) ≤ U f (ti , ti+1 − 1, xti , xti+1 −1 ) + δ.
(4.272)
t =ti
Assume that ti ≥ −2L0 − 2L1 .
(4.273)
Property (iv), (4.258), (4,271), and (4.272) imply that there exist ti,1 ∈ {ti + 3L0 + 2L1 , . . . , ti + 3L0 + 3L1 }
(4.274)
such that f
ρE (xti,1 , xti,1 ) ≤ δ0
(4.275)
ti,2 ∈ {ti+1 − L1 − 1, . . . , ti+1 − 1}
(4.276)
and
such that f
ρE (xti,2 , xti,2 ) ≤ δ0 .
(4.277)
It follows from (4.271), (4.272), (4.274)–(4.277), and property (ii) that f
ρE (xt , xt ) ≤ , t = ti,1 , . . . , ti,2 . Together with (4.274) and (4.276), this implies that f
ρE (xt , xt ) ≤ , t = ti + 3L0 + 3L1 , . . . , ti+1 − L1 − 1.
(4.278)
Assume that ti+1 ≤ 2L0 + 2L1 + 1.
(4.279)
Property (iv), (4.258), (4,271), and (4.272) imply that there exist ti,1 ∈ {ti , . . . , ti + L1 }
(4.280)
260
4 DiscreteTime Nonautonomous Problems on Axis
such that f
ρE (xti,1 , xti,1 ) ≤ δ0
(4.281)
and ti,2 ∈ {ti+1 − 1 − 3L0 − 3L1 , . . . , ti+1 − 1 − 3L0 − 2L1 }
(4.282)
such that f
ρE (xti,2 , xti,2 ) ≤ δ0 .
(4.283)
It follows from (4.271), (4.272), (4.281), (4.283), and property (ii) that f
ρE (xt , xt ) ≤ , t = ti,1 , . . . , ti,2 . Together with (4.280) and (4.282), this implies that f
ρE (xt , xt ) ≤ , t = ti + L1 , . . . , ti+1 − 1 − 3L1 − 3L0 . Therefore (see (4.278)) in both cases f
ρE (xt , xt ) ≤ , t = ti + 3L0 + 3L1 , . . . , ti+1 − 2 − 3L1 − 3L0 .
(4.284)
Assume that ti < −2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1.
(4.285)
Property (iv), (4.258), (4.271), and (4.272) imply that there exist ti,1 ∈ {L0 , . . . , L0 + L1 }
(4.286)
such that f
ρE (xti,1 , xti,1 ) ≤ δ0
(4.287)
ti,2 ∈ {ti+1 − 1 − L1 , . . . , ti+1 − 1]
(4.288)
and
such that f
ρE (xti,2 , xti,2 ) ≤ δ0 ,
(4.289)
4.12 Proof of Theorem 4.13
261
ti,3 ∈ {ti , . . . , ti + L1 }
(4.290)
such that f
ρE (xti,3 , xti,3 ) ≤ δ0
(4.291)
ti,4 ∈ {−L0 − L1 , . . . , −L0 }
(4.292)
and
such that f
ρE (xti,4 , xti,4 ) ≤ δ0 .
(4.293)
It follows from property (ii), (4.272), and (4.285)–(4.293) that f
ρE (xt , xt ) ≤ , t ∈ {ti,1 , . . . , ti,2 } ∪ {ti,3 . . . , ti,4 }, f
ρE (xt , xt ) ≤ , t ∈ {L0 + L1 , . . . , ti+1 − 1 − L1 } ∪ {ti + L1 , . . . , −L1 − L0 .}. (4.294) In view of (4.294), f
ρE (xt , xt ) ≤ , t ∈ {ti +L1 , . . . , ti+1 −1−L1 }\(−L0 −L1 , L0 +L1 ).
(4.295)
In view of (4.284) and (4.295), f
{t ∈ {T1 , . . . , T2 } : ρE (xt , xt ) ≤ } ⊃ ∪{{ti + 3L0 + 3L1 , . . . , ti+1 − 2 − 3L0 − 3L1 } : i ∈ {0, . . . , q − 1} and ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8, ti ≥ −2L0 − 2L1 or ti+1 ≤ 2L0 + 2L1 + 1} ∪{{ti + L1 , . . . , ti+1 − 1 − L1 } \ (−L0 − L1 , L0 + L1 ) : i ∈ {0, . . . , q − 1} and ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8, ti < −2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1}. By (4.296), f
{t ∈ {T1 , . . . , T2 } : ρE (xt , xt ) > }
(4.296)
262
4 DiscreteTime Nonautonomous Problems on Axis
⊂ ∪{{ti , . . . ti+1 } : i ∈ {0, . . . , q − 1}, ti+1 − ti < 8L0 + 8L1 + 8bf + 8} ∪{{ti , . . . , ti + 3L0 + 3L1 } ∪ {ti+1 − 3L0 − 3L1 − 2, . . . ti+1 } : i ∈ {0, . . . , q − 1}, ti+1 − ti > 8L1 + 8L0 + 8 + 8bf } ∪{{ti , . . . , ti + L1 } ∪ {ti+1 − 1 − L1 , . . . , ti+1 } ∪ {−L0 − L1 , . . . , L0 + L1 } : i ∈ {0, . . . , q − 1} : ti < −2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1}.
(4.297)
The righthand side of equation (4.297) is the union of intervals. By (4.256) and (4.269), the number of the intervals does not exceed 3q ≤ 3(M + S0 )δ −1 + 6 < Q.
(4.298)
In view of (4.255), the maximal length of these intervals does not exceed l. Together with (4.298), this completes the proof of Theorem 4.13.
4.13 Proof of Theorem 4.16 By Theorem 4.7, it is sufficient to show that fr possesses (P1), (P2), and (P3) with u˜ t = ut for all integers t. Theorem 4.1 implies that there exists M0 > 0 such that for 2 2 −1 each pair of integers S2 > S1 and each ({xt }St =S , {ut }St =S ) ∈ X(S1 , S2 ), we have 1 1 S 2 −1
f (t, xt , ut ) + M0 ≥
t =S1
S 2 −1
f
f
f (t, xt , ut ).
(4.299)
t =S1
We show that (P2) holds. Let , M > 0. Property (i) implies that there exists δ > 0 such that the following property holds: (a) for each x1 , x2 ∈ E satisfying φ(x1 , x2 ) ≤ δ, we have ρE (x1 , x2 ) ≤ . Choose a natural number L > (rδ)−1 (M + M0 ).
(4.300)
+L +L−1 , {ut }Tt =T ) ∈ X(T , T + L) satisfy Assume that T in a integer and ({xt }Tt =T T +L−1 t =T
fr (t, xt , ut ) ≤
T +L−1 t =T
f
f
fr (t, xt , ut ) + M.
(4.301)
4.13 Proof of Theorem 4.16
263
It follows from (4.13) and (4.299)–(4.301) that T +L−1
f (t, xt , ut ) + r
T +L−1
t =T
≤
f
φ(xt , xt ) =
t =T
T +L−1
f f fr (t, xt , ut )
T +L−1
fr (t, xt , ut )
t =T
+M ≤
t =T
T +L−1
f (t, xt , ut ) + M + M0 ,
t =T T +L−1
φ(xt , xt ) ≤ (M0 + M)r −1 , f
t =T
min{φ(xt , xt ) : t ∈ {T , . . . , T + L − 1}} ≤ (Lr)−1 (M0 + M) < δ f
f
and there exists an integer τ ∈ [T , T + L − 1] such that φ(xτ , xτ ) < δ. Property f (a) implies that ρE (xτ , xτ ) ≤ . Therefore fr has (P2). ∞ We show that fr has (P1). Let ({xt }∞ t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) be (fr )good. There exists M1 > 0 such that for each integer T > 0, 
T −1
fr (t, xt , ut ) −
t =−T
T −1
f
f
(4.302)
f (t, xt , ut ) < M1 .
t =−T
We claim that f
lim ρE (xt , xt ) = 0,
t →∞
f
lim ρE (xt , xt ) = 0.
t →−∞
Assume the contrary. Then there exist > 0 such that at least one of the following properties hold: (b) there exists a sequence of natural numbers {tk }∞ k=1 such that for each integer k ≥ 1, f
tk+1 ≥ tk + 8, ρE (xtk , xtk ) > ; (c) there exists a sequence of negative integers {tk }∞ k=1 such that for each integer k ≥ 1, f
tk+1 < tk − 8, ρE (xtk , xtk ) > . In view of (4.13), (4.299), and (4.302), for each integer T > 0, M1 ≥
T −1 t =−T
fr (t, xt , ut ) −
T −1 t =−T
f
f
f (t, xt , ut )
264
4 DiscreteTime Nonautonomous Problems on Axis T −1
=
f (t, xt , ut ) −
t =−T
T −1
f
t =−T T −1
≥r
T −1
f
f (t, xt , ut ) + r
f
φ(xt , xt )
t =−T
f
φ(xt , xt ) − M0 ,
t =−T
Δ :=
∞
f
φ(xt , xt ) = lim
T →∞
t =−∞
T −1
φ(xt , xt ) < (M0 + M1 )r −1 . f
t =−T f
Properties (a), (b), and (c) imply that for each integer k ≥ 1, φ(xtk , xtk ) > δ, Δ≥
∞
f
φ(xtk , xtk ) = ∞,
k=1
a contradiction. The contradiction we have reached shows that (P1) holds. f f ∞ Let us show that (P3) holds. It is clear that ({xt }∞ t =−∞ , {ut }t =−∞ ) is (fr )∞ ∞ minimal. Assume that ({xt }t =−∞ , {ut }t =−∞ ) ∈ X(−∞, ∞) is (fr )minimal and (fr )good. Property (P1) implies that f
f
lim ρE (xt , xt ) = 0, lim ρE (xt , xt ) = 0.
t →∞
(4.303)
t →−∞
f
We show that xt = xt for all integers t. Assume the contrary. Then there exists an integer t0 such that f
γ := ρ(xt0 , xt0 ) > 0.
(4.304)
Lemma 4.23 and (A3) imply that there exist δ1 > 0 and an integer L1 > 0 such that the following properties hold: f
(d) for each (S, z) ∈ A satisfying ρE (z, xT ) ≤ δ1 there exist an integer τ1 ∈ (0, bf ] (1) S+τ1 −1 1 and ({yt(1)}S+τ ) ∈ X(S, S + τ1 ) satisfying t =S , {vt }t =S f
(1) yS(1) = z , yS+τ = xS+τ1 , 1 S+τ 1 −1
(1)
(1)
fr (t, yt , vt ) ≤
S+τ 1 −1
t =S
f
f
fr (t, xt , ut ) + δ0 r/8
t =S (2)
(2)
and an integer τ2 ∈ (0, bf ] and ({yt }St=S−τ2 , {vt }S−1 t =S−τ2 ) ∈ X(S − τ2 , S) satisfying
4.13 Proof of Theorem 4.16
265 f
(2) yS−τ = xS−τ2 , yS(2) = z, 2 S−1
S−1
fr (t, yt(2), vt(2) ) ≤
t =S−τ2
f
f
fr (t, xt , ut ) + δ0 r/8,
t =S−τ2 f
ρE (yt(1), xt ) ≤ δ0 r/8, t = S, . . . , S + τ1 , f
ρE (yt(2), xt ) ≤ δ0 r/8, t = S − τ2 , . . . , S; (e) for each pair of integers T2 > T1 such that either T2 ≤ −L1 or T1 ≥ L1 f 2 2 −1 and each ({yt }Tt =T , {vt }Tt =T ) ∈ X(T1 , T2 ) which satisfies ρE (yTi , xTi ) ≤ δ1 , 1 1 i = 1, 2, we have T 2 −1
fr (t, yt , vt ) ≥
t =T1
T 2 −1
f
f
fr (t, xt , yt ) − δ0 r/8.
(4.305)
t =T1
In view of (4.303), there exists an integer T0 > 0 such that [−L1 − t0  − 2bf − 2, t0  + 2bf + 2 + L1 ] ⊂ [−T0 , T0 ], f
ρE (xt , xt ) ≤ δ1 for all integers t ∈ (−∞, −T0 ] ∪ [T0 , ∞).
(4.306) (4.307)
Property (d) and (4.307) imply that there exists T +b
T +b −1
0 f 0 f , {vt }t =−T ) ∈ X(−T0 − bf , T0 + bf ) ({yt }t =−T 0 −bf 0 −bf
such that y−T0 −bf = x−T0 −bf , f
(4.308)
f
yt = xt , t = −T0 , . . . , T0 , vt = ut , t = −T0 , . . . , T0 − 1, −T 0 −1
fr (t, yt , vt ) ≤
t =−T0 −bf
−T 0 −1
fr (t, xt , ut ) + 8−1 δ0 r, f
f
T0 +bf −1 t =T0
(4.310)
t =−T0 −bf
yT0 +bf = xT0 +bf ,
(4.309)
(4.311)
T0 +bf −1
fr (t, yt , vt ) ≤
t =T0
fr (t, xt ut ) + 8−1 δ0 r. f f
(4.312)
266
4 DiscreteTime Nonautonomous Problems on Axis
In view of (4.308) and (4.311), T0 +bf −1
T0 +bf −1
fr (t, yt , vt ) ≥
t =−T0 −bf
(4.313)
fr (t, xt , ut ).
t =−T0 −bf
It follows from (4.306), (4.309), (4.310), (4.312), (4.313), and property (e) that −T 0 −1
fr (t, xt , ut )−δ0 r/8+
t =−T0 −bf
fr (t, xt , ut )+
t =−T0 −T 0 −1
≤
T0 +bf −1
T 0 −1
fr (t, xt , ut ) +
t =−T0 −bf
T0 +bf −1
T 0 −1
T 0 −1
−T 0 −1
f
f
f (t, xt , ut ) + δ0 r/8
t =−T0 −bf T0 +bf −1 f
f
f (t, xt , ut ) +
t =−T0 T 0 −1
fr (t, xt , ut )
t =T0
fr (t, yt , vt ) ≤
t =−T0 −bf
+
fr (t, xt , ut ) +
t =−T0
f
t =T0
T0 +bf −1
≤
f
fr (t, xt , ut )−δ0 r/8
fr (t, xt , ut ) ≤
t =−T0
f
f
f (t, xt , ut ) + δ0 r/8,
t =T0 T 0 −1
f
f
f (t, xt , ut ) + δ0 r/2.
(4.314)
t =−T0
By property (d), (4.306), and (4.307), there exists T +b
T +b −1
0 f 0 f , {ηt }t =−T ) ∈ X(−T0 − bf , T0 + bf ) ({ξt }t =−T 0 −bf 0 −bf
such that f
ξ−T0 −bf = x−T0 −bf , ξt = xt , t = −T0 , . . . , T0 , ηt = ut , t = −T0 , . . . , T0 −1, (4.315) f ξT0 +bf = xT0 +bf , (4.316) −T 0 −1 t =−T0 −bf
fr (t, ξt , ηt ) ≤
−T 0 −1 t =−T0 −bf
fr (t, xt , ut ) + 8−1 δ0 , f
f
(4.317)
4.13 Proof of Theorem 4.16
267
T0 +bf −1
T0 +bf −1
fr (t, ξt , ηt ) ≤
t =T0
fr (t, xt , ut ) + 8−1 δ0 r. f
f
(4.318)
t =T0
It follows from (4.13), and (4.315)–(4.318) that T0 +bf −1
T0 +bf −1 f
t =−T0 −bf −T 0 −1
=
T 0 −1
f (t, ξt , ηt ) +
T0 +bf −1
T 0 −1
T0 +bf −1
−T 0 −1
f
f (t, ξt , ηt )
t =T0
f
f
f (t, xt , ut ) +
t =−T0 −bf T 0 −1
f (t, xt , ut ) +
t =−T0
f (t, xt , ut ) +
t =−T0
f (t, ξt , ηt )
t =−T0 −bf
t =−T0 −bf
≤
f
f (t, xt , ut ) ≤
f
f
f (t, xt , ut ) + δ0 r/4,
t =T0
f
f (t, xt , ut ) ≤
t =−T0
T 0 −1
f (t, xt , ut ) + δ0 r/4.
t =−T0
By (4.13), (4.19), (4.305), and (4.314), T 0 −1
f
t =−T0
≥
T 0 −1 t =−T0
T 0 −1
f
f (t, xt , ut ) + δ0 r/2 ≥
fr (t, xt , ut )
t =−T0
f
f (t, xt , ut ) + rφ(xt0 , xt0 ) ≥
T 0 −1
f
f
f (t, xt , ut ) + rδ0 − rδ0 /4,
t =−T0
a contradiction. The contradiction we have reached proves (P3). This completes the proof of Theorem 4.16.
Chapter 5
ContinuousTime Autonomous Problems
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for continuoustime optimal control problems in infinite dimensional spaces. For these optimal control problems the turnpike is a singleton. We also study the structure of approximate solutions on large intervals in the regions close to the endpoints and the existence of solutions of the corresponding infinite horizon optimal control problems. The results of this chapter will be obtained for two large classes of problems which will be treated simultaneously.
5.1 Preliminaries We begin with the description of the first class of problems. Let (E, ·) be a Banach space and (F, ρF ) be a metric space. We suppose that A is a nonempty subset of E, U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U(x)}.
(5.1)
We suppose that M is a Borel measurable subset of E × F , G : M → E is a Borelian function and a linear operator A : D(A) → E generates a C0 semigroup eAt on E. Let f : M → R 1 be a bounded from below Borelian function. Let 0 ≤ T1 < T2 . We consider the following equation: x (t) = Ax(t) + G(x(t), u(t)), t ∈ [T1 , T2 ].
(5.2)
A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (5.2) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function,
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_5
269
270
5 ContinuousTime Autonomous Problems
x(t) ∈ A, t ∈ [T1 , T2 ], u(t) ∈ U(x(t)), t ∈ [T1 , T2 ] almost everywhere (a.e.),
(5.3) (5.4)
G(x(s), u(s)), s ∈ [T1 , T2 ] is Bochner integrable, and x(t) = eA(t −T1) x(T1 ) +
t
eA(t −s)G(x(s), u(s))ds, t ∈ [T1 , T2 ].
(5.5)
T1
The set of all pairs (x, u) which are solutions of (5.2) is denoted by X(T1 , T2 , A, G). In the sequel for simplicity, we use the notation X(T1 , T2 ) = X(T1 , T2 , A, G) if the pair (A, G) is understood. Let 0 ≤ T1 < T2 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F . Set y(t) = x(t + T1 ), v(t) = u(t + T1 ), t ∈ [0, T2 − T1 ]. Assume that (y, v) ∈ X(0, T2 − T1 ). We show that (x, u) ∈ X(T1 , T2 ). Indeed, we have t At y(t) = e y(0) + eA(t −s)G(y(s), v(s))ds, t ∈ [0, T2 − T1 ]. 0
In view of the relations above, for all t ∈ [T1 , T2 ], x(t) = y(t − T1 ) = eA(t −T1) x(T1 ) +
t −T1
eA(t −T1 −s) G(y(s), v(s))ds
0
= eA(t −T1 ) x(T1 ) +
t −T1
eA(t −T1−s) G(x(s + T1 ), u(s + T1 ))ds
0
= eA(t −T1 ) x(T1 ) +
t
eA(t −s)G(x(s), u(s))ds
T1
and (x, u) ∈ X(T1 , T2 ). Assume that (x, u) ∈ X(T1 , T2 ). We show that (x, u) ∈ X(τ, T2 ) for all τ ∈ (T1 , T2 ]. Let τ ∈ (T1 , T2 ). By (5.5), for all t ∈ [τ, T2 ], x(t) = eA(t −T1 ) x(T1 ) +
t
eA(t −s)G(x(s), u(s))ds
T1
= eA(t −τ )eA(τ −T1 ) x(T1 ) +
τ T1
eA(t −τ )eA(τ −s)G(x(s), u(s))ds
5.1 Preliminaries
271
t
+
eA(t −τ )eA(τ −s)G(x(s), u(s))ds
τ
=e
A(t −τ )
(e
A(τ −T1 )
x(T1 ) +
τ
eA(τ −s)G(x(s), u(s))ds)
T1
+
t
eA(t −s)G(x(s), u(s))ds
τ
=e
A(t −τ )
T2
x(τ ) +
eA(t −s)G(x(s), u(s))ds.
τ
Thus (x, u) ∈ X(τ, T2 ) for all τ ∈ (T1 , T2 ). Let 0 ≤ T1 < T2 < T3 , (x1 , u1 ) ∈ X(T1 , T2 ), (x2 , u2 ) ∈ X(T2 , T3 ), x1 (T2 ) = x2 (T2 ). For each t ∈ [T1 , T2 ] set x(t) = x1 (t), u(t) = u1 (t) and for each t ∈ (T2 , T3 ] set x(t) = x2 (t), u(t) = u2 (t). We show that (x, u) ∈ X(T1 , T3 ). It is not difficult to see that for each t ∈ [T1 , T2 ], eA(t −s)G(x(s), u(s)) = eA(t −s)G(x1 (s), u1 (s)), s ∈ [T1 , t], x(t) = x1 (t) = eA(t −T1) x1 (T1 ) +
t
eA(t −s)G(x1 (s), u1 (s))ds
T1
=e
A(t −T1 )
x(T1 ) +
t
eA(t −s)G(x(s), u(s))ds.
T1
Let t ∈ (T2 , T3 ]. For every s ∈ [T1 , t], if s ≤ T2 , then eA(t −s)G(x(s), u(s)) = eA(t −T2 ) eA(T2 −s) G(x1 (s), u1 (s)) and if s > T2 , then eA(t −s)G(x(s), u(s)) = eA(t −s)G(x2 (s), u2 (s)), e
A(t −T1 )
x(T1 ) +
t T1
eA(t −s)G(x(s), u(s))ds = eA(t −T2 ) eA(T2 −T1 ) x(T1 )
272
5 ContinuousTime Autonomous Problems
+
T2
eA(t −T2 ) eA(T2 −s) G(x(s), u(s))ds +
t
eA(t −s)G(x(s), u(s))ds
T2
T1
=e
A(t −T2 )
[e
A(T2 −T1 )
x(T1 ) +
T2
eA(T2 −s) G(x1 (s), u2 (s))ds]
T1
t
+
eA(t −s)G(x2 (s), u2 (s))ds
T2
=e
A(t −T2 )
x2 (T1 ) +
t
eA(t −s)G(x2 (s), u2 (s))ds = x2 (t) = x(t).
T2
Thus (x, u) ∈ X(T1 , T3 ). Now we describe the second class of problems. Let (E, ·, ·)E be a Hilbert space equipped with an inner product ·, ·E which induces the norm ·E , and let (F, ·, ·F ) be a Hilbert space equipped with an inner product ·, ·F which induces the norm · F . For simplicity, we set ·, ·E = ·, ·, · E = · , ·, ·F = ·, ·, · F = · , if E, F are understood. We suppose that A is a nonempty subset of E, U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U(x)}. We suppose that M is a Borel measurable subset of E × F and that a linear operator A : D(A) → E generates a C0 semigroup S(t) = eAt , t ≥ 0 on E. As usual, we denote by S(t)∗ the adjoint operator of S(t). Then S(t)∗ , t ∈ [0, ∞) is C0 semigroup, and its generator is the adjoint operator A∗ of A. The domain D(A∗ ) is a Hilbert space equipped with the graph norm · D(A∗ ) of the operator A∗ : z2D(A∗ ) := z2E + A∗ z2E , z ∈ D(A∗ ). Let D(A∗ ) be the dual space of D(A∗ ) with the pivot space E. In particular, E1d := D(A∗ ) ⊂ E ⊂ D(A∗ ) = E−1 . (Here we use the notation of Section 1.7.) Let G : M → D(A∗ ) = E−1 and f : M → R 1 be Borelian functions, B ∈ L(F, E−1 ) is an admissible control operator for eAt , t ≥ 0, G(x, u) = Bu, (x, u) ∈ M. Let 0 ≤ T1 < T2 . We consider the following equation x (t) = Ax(t) + Bu(t), t ∈ [T1 , T2 ] a. e. .
(5.6)
5.1 Preliminaries
273
A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (5.6) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, u ∈ L2 ([T1 , T2 ]; F ), x(t) ∈ A, t ∈ [T1 , T2 ], u(t) ∈ U(x(t)), t ∈ [T1 , T2 ] a.e., and for each t ∈ [T1 , T2 ], x(t) = eA(t −T1) x(T1 ) +
t
eA(t −s)Bu(s)ds
(5.7)
T1
in E−1 . The set of all pairs (x, u) which are solutions of (5.6) is denoted by X(T1 , T2 , A, G). In the sequel for simplicity, we use the notation X(T1 , T2 ) = X(T1 , T2 , A, G) if the pair (A, G) is understood. From now these two problems will be treated simultaneously. Let T1 ≥ 0. A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a (mild) solution of the system x (t) = Ax(t) + G(x(t), u(t)), t ∈ [T1 , ∞) if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (5.2) for the first problem and is a solution of (5.6) for the second problem. The set of all such pairs (x, u) which are solutions of the equation above is denoted by X(T1 , ∞). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) (0 ≤ T1 < T2 ) is called a trajectory if there exists a Lebesgue measurable function u : I → F (referred to as a control) such that (x, u) ∈ X(T1 , T2 ) or (x, u) ∈ X(T1 , ∞), respectively). Let T2 > T1 ≥ 0, (x, u) ∈ X(T1 , T2 ). Define I f (T1 , T2 , x, u) =
T2
f (x(t), u(t))dt T1
which is welldefined but can be ∞. Proposition 1.3 implies that there exist M∗ ≥ 1, ω∗ ∈ R 1 such that eAt ≤ M∗ eω∗ t , t ∈ [0, ∞).
(5.8)
Let a0 , K0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → ∞ as t → ∞. We suppose that for the first problem the function f satisfies
(5.9)
274
5 ContinuousTime Autonomous Problems
f (x, u) ≥ −a0 + max{ψ(x), ψ((G(x, u) − a0 x)+ )(G(x, u) − a0x)+ } (5.10) for each (x, u) ∈ M and for the second problem f satisfies f (x, u) ≥ −a0 + max{ψ(x), K0 u2 }, (x, u) ∈ M.
(5.11)
Let T2 > T1 ≥ 0 and y, z ∈ A. We consider the following problems: I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z,
(P1 )
I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ), x(T1 ) = y,
(P2 )
I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ).
(P3 )
We consider functionals of the form I f (T1 , T2 , x, u), where 0 ≤ T1 < T2 and (x, u) ∈ X(T1 , T2 ). For each pair of numbers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we define U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z},
(5.12)
σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ A},
(5.13)
σ f (T1 , T2 , y) = inf{U f (T1 , T2 , h, y) : h ∈ A},
(5.14)
σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, y) : h, y ∈ A}.
(5.15)
We suppose that xf ∈ E, uf ∈ F, (xf , uf ) ∈ X(0, ∞).
(5.16)
This means that x(t) ¯ = xf , u(t) ¯ = uf , t ≥ 0 is a solution of the equation x (t) = Ax(t) + G(x(t), u(t)), t ∈ [0, ∞). For the first class of problems, Proposition 1.6 implies that this is equivalent to the following property: for every x ∗ ∈ D(A∗ ), x(t), ¯ x ∗ is an absolutely continuous ( a. c. ) function on any bounded subinterval of [0, ∞) and for a. e. t ∈ [0, ∞), 0 = (d/dt) x(t), ¯ x ∗ = xf , A∗ x ∗ + G(xf , uf ), x ∗ .
(5.17)
Thus for the first class of problems, (5.16) holds if and only if for every x ∗ ∈ D(A∗ ),
5.2 Boundedness Results
275
0 = xf , A∗ x ∗ + G(xf , uf ), x ∗ .
(5.18)
It is not difficult to see that for the second class of problems, (5.16) holds if and only if (5.18) is true for every x ∗ ∈ D(A∗ ) (see Section 1.7). We suppose that there exists a number bf > 0 and the following assumptions hold. (A1)
For each S1 > 0 there exist S2 > 0 and an integer c > 0 such that (T2 − T1 )f (xf , uf ) ≤ I f (T1 , T2 , x, u) + S2
(A2)
for each T1 ≥ 0, each T2 ≥ T1 + c and each (x, u) ∈ X(T1 , T2 ) satisfying x(Tj ) ≤ S1 , j = 1, 2. For each > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying zi − xf ≤ δ, i = 1, 2 there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ ) which satisfies x(0) = z1 , x(τ ) = z2 and I f (0, τ, x, u) ≤ τf (xf , uf ) + .
Section 5.34 contains examples of optimal control problems satisfying assumptions (A1) and (A2). Many examples can also be found in [106–108, 118, 124, 125, 134].
5.2 Boundedness Results The following result is proved in Section 5.8. Theorem 5.1. 1. There exists S > 0 such that for each pair of numbers T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S ≥ (T2 − T1 )f (xf , uf ). 2. For each (x, u) ∈ X(0, ∞) either I f (0, t, x, u) − tf (xf , uf ) → ∞ as t → ∞ or sup{I f (0, t, x, u) − tf (xf , uf ) : t ∈ (0, ∞)} < ∞. Moreover, if (5.19) holds, then sup{x(t) : t ∈ (0, ∞)} < ∞.
(5.19)
276
5 ContinuousTime Autonomous Problems
We say that (x, u) ∈ X(0, ∞) is (f, A, G)good (or (f )good if A, G are understood) if (5.19) holds [29, 104, 116, 120, 129, 134]. The next boundedness result is proved in Section 5.8. Theorem 5.2. Let M0 > 0, c > 0, c0 ∈ (0, c). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c0 , T2 ]. Let L > 0. Denote by AL the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ X(0, τ ) such that x(0) = z, x(τ ) = xf , I f (0, τ, x, u) ≤ L. L the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ Denote by A X(0, τ ) such that x(0) = xf , x(τ ) = z, I f (0, τ, x, u) ≤ L. The following Theorems 5.3–5.5 are also boundedness results. They are proved in Section 5.8. Theorem 5.3. Let L > 0, M0 > 0, and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , x(T1) ∈ AL , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Theorem 5.4. Let L > 0, M0 > 0, and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Theorem 5.5. Let L > 0, M0 > 0, and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , I f (T1 , T2 , x, u) ≤ x(T2 ) ∈ A σ f (T1 , T2 , xT2 ) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [c + T1 , T2 ]. The following Theorems 5.6–5.9 are proved in Section 5.9.
5.3 Turnpike Results
277
Theorem 5.6. Let M0 > 0 and c > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 5.7. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , x(T1 ) ≤ M0 , x(T1 ) ∈ AL , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 5.8. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 5.9. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , xT2 ) + M0 , x(T1 ) ≤ M0 , x(T2 ) ∈ A the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ].
5.3 Turnpike Results We say that the triplet (f, A, G) (or f if the pair (A, G) is understood) possesses the turnpike property (or TP for short) if for each > 0 and each M > 0 there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, we have x(t) − xf ≤ for all t ∈ [T1 + L, T2 − L].
278
5 ContinuousTime Autonomous Problems
Moreover, if x(T1 ) − xf ≤ δ then x(t) − xf ≤ for all t ∈ [T1 , T2 − L] and if x(T2 ) − xf ≤ δ then x(t) − xf ≤ for all t ∈ [T1 + L, T2 ]. Theorem 5.1 implies the following result. Proposition 5.10. Assume that f has TP and that , M > 0. Then there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{(T2 − T1 )f (xf , uf ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ then τ2 = T2 . The next two results are proved in Section 5.10. Theorem 5.11. Assume that f has TP and that , L > 0. Then there exist δ > 0 and L0 > L such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , and each (x, u) ∈ X(T1 , T2 ) which satisfies L , x(T1) ∈ AL , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ) ≤ δ then τ2 = T2 . Theorem 5.12. Assume that f has TP and that , L, M > 0. Then there exist δ > 0 and L0 > L such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , and each (x, u) ∈ X(T1 , T2 ) which satisfies x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 , x(T1 )) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ then τ2 = T2 . The next theorem is the main result of this chapter. It is proved in Section 5.11.
5.3 Turnpike Results
279
Theorem 5.13. f has TP if and only if the following properties hold: (P1)
for each (f )good pair (x, u) ∈ X(0, ∞), lim x(t) − xf = 0;
t →∞
(P2)
for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each (x, u) ∈ X(0, L) which satisfies I f (0, L, x, u) ≤ min{U f (0, L, x(0), x(L)) + δ, Lf (xf , uf ) + M}, there exists s ∈ [0, L] such that x(s) − xf ≤ .
The next result is proved in Section 5.13. Theorem 5.14. Assume that f has properties (P1) and (P2). Let , M > 0. Then there exist a natural number Q and l > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + lQ, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M, q
q
there exist finite sequences {ai }i=1 , {bi }i=1 ⊂ [T1 , T2 ] such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf ≤ for all t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. Theorem 5.14 leads to the following definition. We say that the triplet (f, A, G) (or f if the pair (A, G) is understood) possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0, there exist a natural number Q and l > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + lQ, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M, q
q
there exist finite sequences {ai }i=1 , {bi }i=1 ⊂ [T1 , T2 ] such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf ≤ for all t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. WTP was studied in [95, 96] for discretetime unconstrained problems, in [97] for variational problems, and in [114] for discretetime constrained problems.
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5 ContinuousTime Autonomous Problems
Theorem 5.15. f has WTP if and only if f has (P1) and (P2). In view of Theorem 5.14, (P1) and (P2) imply WTP. Clearly, WTP implies (P2). Therefore Theorem 5.15 follows from the next result which is proved in Section 5.14. Proposition 5.16. Let f have WTP. Then f has (P1).
5.4 Lower Semicontinuity Property We say that (f, A, G) (or f if the pair (A, G) is understood) possesses lower semicontinuity property (LSC property for short) if for each T2 > T1 ≥ 0 and each sequence (xj , uj ) ∈ X(T1 , T2 ), j = 1, 2, . . . which satisfies sup{I f (T1 , T2 , xj , uj ) : j = 1, 2, . . . } < ∞, there exist a subsequence {(xjk , ujk )}∞ k=1 and (x, u) ∈ X(T1 , T2 ) such that, xjk (t) → x(t) as k → ∞ for every t ∈ [T1 , T2 ], I f (T1 , T2 , x, u) ≤ lim inf I f (T1 , T2 , xj , uj ). j →∞
LSC property plays an important role in the calculus of variations and optimal control theory [32]. Theorem 5.17. Assume that f possesses LSC property. Then f has TP if and only if f has (P1). Theorem 5.17 follows from Theorem 5.13 and the next proposition which is proved in Section 5.15. Proposition 5.18. Assume that f has LSC property and (P1). Then f has (P2). A pair (x, u) ∈ X(0, ∞) is called (f, A, G)overtaking optimal (or (f )overtaking optimal if the pair (A, G) is understood) [29, 104, 120] if for every (y, v) ∈ X(0, ∞) satisfying x(0) = y(0), lim sup[I f (0, T , x, u) − I f (0, T , y, v)] ≤ 0. T →∞
A pair (x, u) ∈ X(0, ∞) is called (f, A, G)weakly optimal (or (f )weakly optimal if the pair (A, G) is understood) [29, 104, 120] if for every (y, v) ∈ X(0, ∞) satisfying x(0) = y(0), lim inf[I f (0, T , x, u) − I f (0, T , y, v)] ≤ 0. T →∞
5.4 Lower Semicontinuity Property
281
A pair (x, u) ∈ X(0, ∞) is called (f, A, G)minimal (or (f )minimal optimal if the pair (A, G) is understood) [12, 94] if for every T > 0 I f (0, T , x, u) = U f (0, T , x(0), x(T )). In infinite horizon optimal control the main goal is to show the existence of solutions using the optimality criterions above. The next result is proved in Section 5.17. Theorem 5.19. Assume that f has (P1) and LSC property and that (x, u) ∈ X(0, ∞) is (f )good. Then there is an (f )overtaking optimal pair (x∗ , u∗ ) ∈ X(0, ∞) such that x∗ (0) = x(0). The following theorem is also proved in Section 5.17. Theorem 5.20. Assume that f has (P1) and LSC property, (x, ˜ u) ˜ ∈ X(0, ∞) is (f )good and that (x∗ , u∗ ) ∈ X(0, ∞) satisfies x∗ (0) = x(0). ˜ Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)
(x∗ , u∗ ) is (f )overtaking optimal. (x∗ , u∗ ) is (f )weakly optimal. (x∗ , u∗ ) is (f )minimal and (f )good. (x∗ , u∗ ) is (f )minimal and satisfies limt →∞ x∗ (t) = xf . (x∗ , u∗ ) is (f )minimal and satisfies lim inft →∞ x∗ (t) − xf = 0.
The following three results are proved in Section 5.18. Theorem 5.21. Assume that f has (P1) and LSC property and > 0. Then there exists δ > 0 such that the following assertions hold. (i) For every z ∈ A satisfying z−xf ≤ δ, there exists an (f )overtaking optimal pair (x, u) ∈ X(0, ∞) which satisfies x(0) = z. (ii) If an (f )overtaking optimal pair (x, u) ∈ X(0, ∞) satisfies x(0) − xf ≤ δ, then x(t) − xf ≤ for all t ≥ 0. Theorem 5.22. Assume that f has (P1) and LSC property, > 0, and L > 0. Then there exists τ0 > 0 such that for every (f )overtaking optimal pair (x, u) ∈ X(0, ∞) which satisfies x(0) ∈ AL , the inequality x(t) − xf ≤ holds for all t ≥ τ0 . Theorem 5.23. Assume that f has (P1) and LSC property, (x, u) ∈ X(0, ∞) is an (f )overtaking optimal and (f )good pair, and numbers t2 > t1 ≥ 0 satisfy x(t1 ) = x(t2 ). Then x(t) = xf for all t ≥ t1 . In the sequel we use the following result which easily follows the definitions. Proposition 5.24. Let f have (P1) and z ∈ A. There exists an (f )good pair (x, u) ∈ X(0, ∞) satisfying x(0) = z if and only if z ∈ ∪{AL : L > 0}. LSC property implies the following result.
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5 ContinuousTime Autonomous Problems
Proposition 5.25. Let f have LSC property, T2 > T2 ≥ 0, and L > 0. Then 1. There exists (x, u) ∈ X(T1 , T2 ) such that I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , y, v) for all (y, v) ∈ X(T1 , T2 ). L , then there exists (x, u) ∈ X(T1 , T2 ) such 2. If T2 − T1 ≥ 2L, z1 ∈ AL , z2 ∈ A that x(Ti ) = zi , i = 1, 2 and I f (T1 , T2 , x, u) = U f (T1 , T2 , z1 , z2 ). 3. If T2 − T1 ≥ L, z ∈ AL , then there exists (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = z and I f (T1 , T2 , x, u) = σ f (T1 , T2 , z). L , then there exists (x, u) ∈ X(T1 , T2 ) such that x(T2 ) = z 4. If T2 − T1 ≥ L, z ∈ A and σ f (T1 , T2 , z). I f (T1 , T2 , x, u) = Theorem 5.26. f has (P1) and (P2) if and only if the following property holds (P3)
for each > 0 and each M > 0, there exists L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M the inequality mes({t ∈ [T1 , T2 ] : x(t) − xf > }) ≤ L holds.
In view of Theorem 5.14, (P1) and (P2) imply (P3). Clearly, (P3) implies (P2). Therefore Theorem 5.26 follows from the next result which is proved in Section 5.19. Proposition 5.27. Let f have (P3). Then f has (P1).
5.5 Perturbed Problems In this section we use the following assumption. (A3) For each > 0, there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying zi − xf ≤ δ, i = 1, 2, there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ ) which satisfies x(0) = z1 , x(τ ) = z2 and
5.6 The Triplet (f, −A, −G)
283
x(t) − xf ≤ , t ∈ [0, τ ], I f (0, τ, x, u) ≤ τf (xf , uf ) + . Clearly, (A3) implies (A2). Assume that φ : E → [0, 1] is a continuous function satisfying φ(0) = 0 and such that the following property holds: (i) for each > 0 there exists δ > 0 such that for each x ∈ E satisfying φ(x) ≤ δ we have x ≤ . For each r ∈ (0, 1) set, fr (x, u) = f (x, u) + rφ(x − xf ), (x, u) ∈ M. Clearly, for any r ∈ (0, 1), fr is a Borelian function, if (A3) holds, then (A1) and (A3) hold for fr with (xfr , ufr ) = (xf , uf ). Theorem 5.28. Let (A3) hold and r ∈ (0, 1). Then fr has (P1) and (P2). Theorem 5.28 is proved in Section 5.20.
5.6 The Triplet (f, −A, −G) We use the notation, definitions, and assumptions introduced in Sections 5.1–5.4. We assume that −A : D(A) → E also generates a C0 semigroup e−At , t ≥ 0 on E. Clearly, D(−A) = D(A), (−A)∗ = −A∗ . Let 0 ≤ T1 < T2 . We consider the following equation x (t) = −Ax(t) − G(x(t), u(t)), t ∈ [T1 , T2 ].
(5.20)
Let us consider our first class of problems. A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (5.20) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, x(t) ∈ A, t ∈ [T1 , T2 ],
(5.21)
u(t) ∈ U(x(t)), t ∈ [T1 , T2 ] a. e.,
(5.22)
G(x(s), u(s)), s ∈ [T1 , T2 ] is Bochner integrable, and x(t) = e−A(t −T1 ) x(T1 ) −
t T1
e−A(t −s)G(x(s), u(s))ds, t ∈ [T1 , T2 ].
(5.23)
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5 ContinuousTime Autonomous Problems
For the second class of problems, a pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (5.20) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, u ∈ L2 (T1 , T2 ; F ), (5.21) and (5.22) hold and for each t ∈ [T1 , T2 ], x(t) = e−A(t −T1 ) x(T1 ) +
t
e−A(t −s)(−Bu(s))ds
(5.24)
T1
in E−1 . It is not difficult to show that −B is an admissible control operator for e−At , t ≥ 0. The set of all pairs (x, u) which are solutions of (5.20) is denoted by X(T1 , T2 , −A, −G). Let T1 ≥ 0. A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a (mild) solution of the system x (t) = −Ax(t) − G(x(t), u(t)), t ∈ [T1 , ∞) if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (5.20). The set of all such pairs (x, u), which are solutions of the equation above, is denoted by X(T1 , ∞, −A − G). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) (0 ≤ T1 < T2 ), is called a trajectory if there exists a Lebesgue measurable function u : I → F (referred to as a control) such that (x, u) ∈ X(T1 , T2 , −A, −G) or (x, u) ∈ X(T1 , ∞, −A, −G), respectively. Let T2 > T1 ≥ 0 and (x, u) ∈ X(T1 , T2 , −A, −G). Define I (T1 , T2 , x, u) = f
T2
f (x(t), u(t))dt
(5.25)
T1
which is welldefined but can be ∞. Proposition 1.3 implies that there exists M¯ ∗ ≥ 1, ω¯ ∗ ∈ R 1 such that e−At ≤ M¯ ∗ eω¯ ∗ t , t ∈ [0, ∞).
(5.26)
We consider functionals of the form I f (T1 , T2 , x, u), where 0 ≤ T1 < T2 and (x, u) ∈ X(T1 , T2 , −A, −G). For each pair of numbers T2 > T1 ≥ 0 and each pair of points y, z ∈ A, we define f
U− (T1 , T2 , y, z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 , −A, −G), x(T1 ) = y, x(T2 ) = z}, f
f
σ− (T1 , T2 , y) = inf{U− (T1 , T2 , y, h) : h ∈ A},
(5.27) (5.28)
5.6 The Triplet (f, −A, −G)
285
f
f
(5.29)
σ− (T1 , T2 ) = inf{U− (T1 , T2 , h, y) : h, y ∈ A}.
(5.30)
σ− (T1 , T2 , y) = inf{U− (T1 , T2 , h, y) : h ∈ A}, f
f
Assume that S2 > S1 ≥ 0, (x, u) ∈ X(S1 , S2 , A, G). For every t ∈ [T1 , T2 ] set x(t) ¯ = x(S2 − t + S1 ), u(t) ¯ = u(S2 − t + S1 ).
(5.31)
We show that (x, ¯ u) ¯ ∈ X(S1 , S2 , −A, −G). First we consider the first class of problems. Proposition 1.6 implies that for every x ∗ ∈ D(A∗ ), x(·), x ∗ is an a. c. function on [S1 , S2 ] and for all t ∈ [S1 , S2 ], ∗
∗
t
x(t), x = x(S1 ), x +
[ x(s), A∗ x ∗ + G(x(s), u(s)), x ∗ ]ds.
(5.32)
S1
Let x ∗ ∈ D(A∗ ). By (5.31) and (5.32), for every t ∈ [S1 , S2 ],
x(t), ¯ x ∗ = x(S2 − t + S1 ), x ∗
∗
S2 −t +S1
= x(S1 ), x +
[ x(s), A∗ x ∗ + G(x(s), u(s)), x ∗ ]ds
S1
= x(S2 ), x ∗ −
S2
[ x(s), A∗ x ∗ + G(x(s), u(s)), x ∗ ]ds
S1
+
S2 −t +S1
[ x(s), A∗ x ∗ + G(x(s), u(s)), x ∗ ]ds
S1
= x(S ¯ 1 ), x ∗ −
S2 S2 −t +S1
[ x(s), A∗ x ∗ + G(x(s), u(s)), x ∗ ]ds = x(S ¯ 1 ), x ∗
−
t
[ x(S2 − s + S1 ), A∗ x ∗ + G(x(S2 − s + S1 ), u(S2 − s + S1 )), x ∗ ]ds
S1 ∗
= x(S ¯ 1 ), x +
t
S1
and for every t ∈ [S1 , S2 ],
[ x(s), ¯ −A∗ x ∗ + −G(x(s), ¯ u(s)), ¯ x ∗ ]ds
286
5 ContinuousTime Autonomous Problems ∗
∗
x(t), ¯ x = x(S ¯ 1 ), x +
t
[ x(s), ¯ −A∗ x ∗ + −G(x(s), ¯ u(s)), ¯ x ∗ ]ds.
S1
(5.33) Proposition 1.6 and (5.33) imply that (x, ¯ u) ¯ is a solution of (5.20) on [S1 , S2 ] and satisfies (5.23). This implies that (x, ¯ u) ¯ ∈ X(S1 , S2 , −A, −G). For the second class of problems (x, u) ∈ X(S1 , S2 , −A, −G) because the second problem can be considered as a special case of the first class of problems when E is replaced by E−1 . By (5.31), ¯ u) ¯ = I f (S1 , S2 , x,
S2
f (x(t), ¯ u(t))dt ¯
S1
=
S2
f (x(S2 − t + S1 ), u(S2 − t + S1 ))dt =
S1
S2
f (x(t), u(t))dt.
(5.34)
S1
Thus we showed that for each S2 > S1 ≥ 0 and each (x, u) ∈ X(S1 , S2 , A, G), ¯ u) ¯ = I f (S1 , S2 , x, u), (x, ¯ u) ¯ ∈ X(S1 , S2 , −A, −G), I f (S1 , S2 , x,
(5.35)
and moreover, if (x, u) ∈ X(S1 , S2 , −A, −G), then (x, ¯ u) ¯ ∈ X(S1 , S2 , A, G). This implies the following result. Proposition 5.29. Let S2 > S1 ≥ 0, M ≥ 0 and (xi , ui ) ∈ X(S1 , S2 , A, G), i = 1, 2. Then I f (S1 , S2 , x1 , u1 ) ≥ I f (S1 , S2 , x2 , u2 ) − M if and only if I f (S1 , S2 , x¯1 , u¯ 1 ) ≥ I f (S1 , S2 , x¯2 , u¯ 2 ) − M.
(5.36) (5.37)
Proposition 5.29 implies the following result. Proposition 5.30. Let S2 > S1 ≥ 0, M ≥ 0 and (x, u) ∈ X(S1 , S2 , A, G). Then the following assertions hold: I f (S1 , S2 , x, u) ≤ σ f (S1 , S2 ) + M if and only if f
I f (S1 , S2 , x, ¯ u) ¯ ≤ σ− (S1 , S2 ) + M; I f (S1 , S2 , x, u) ≤ U f (S1 , S2 , x(S1 ), x(S2 )) + M f
¯ u) ¯ ≤ U− (S1 , S2 , x(S ¯ 1 ), x(S ¯ 2 )) + M; if and only if I f (S1 , S2 , x, I f (S1 , S2 , x, u) ≤ σ f (S1 , S2 , x(S1 )) + M
5.6 The Triplet (f, −A, −G)
287 f
if and only if I f (S1 , S2 , x, ¯ u) ¯ ≤ σ− (S1 , S2 , x(S ¯ 2 )) + M; I f (S1 , S2 , x, u) ≤ σ f (S1 , S2 , x(S2 )) + M f
¯ u) ¯ ≤ σ− (S1 , S2 , x(S ¯ 1 ) + M. if and only if I f (S1 , S2 , x, Clearly, (xf , uf ) ∈ X(0, ∞, −A, −G).
(5.38)
In view of (5.35), (A1) and (A2) hold for the triplet (f, −A, −G), and if (A3) holds for (f, A, G), then (A3) holds for (f, −A, −G) too. Therefore the results obtained for (f, A, G) also hold for the triplet (f, −A, −G). Theorem 5.31. Assume that (f, A, G) has properties (P1) and (P2). Then (f, −A, −G) possesses properties (P1) and (P2). Proof. By Theorem 5.13, (f, A, G) has TP. Proposition 5.30, Theorems 5.13 and (5.35) imply that (f, −A, −G) has TP, (P1), and (P2). We can now improve our boundedness results. Proposition 5.30 and Theorems 5.2 and (5.35) for (f, A, G) and (f, −A, −G) imply the following result. Theorem 5.32. Let M0 > 0 and c > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Proposition 5.30, (5.35) and Theorem 5.3 for (f, A, G) and (f, −A, −G) imply the following result. Theorem 5.33. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , x(T1) ∈ AL , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Propositions 5.30 and (5.35) and Theorem 5.4 for (f, A, G) and Theorem 5.5 for (f, −A, −G) imply the following result. Theorem 5.34. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying
288
5 ContinuousTime Autonomous Problems
x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Proposition 5.30, (5.35), Theorem 5.5 for (f, A, G), and Theorem 5.4 (f, −A, −G) imply the following result. Theorem 5.35. Let L > 0 and M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , xT2 ) + M0 , x(T2 ) ∈ A the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ].
5.7 Auxiliary Results for Theorem 5.1 Lemma 5.36. There exist numbers S > 0 and c0 ≥ 1 such that for each T1 ≥ 0, each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ), (T2 − T1 )f (xf , uf ) ≤ I f (T1 , T2 , x, u) + S2 .
(5.39)
Proof. By (5.9), there exists S1 > 0 such that ψ(S1 ) > f (xf , uf ) + a0 + 1.
(5.40)
By (A1), there exist S2 > 0, c0 ≥ 1 such that (T2 − T1 )f (xf , uf ) ≤ I f (T1 , T2 , x, u) + S2
(5.41)
for each T1 ≥ 0, each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ), x(T2 ) ≤ S1 .
(5.42)
S = S2 + c0 (f (xf , uf ) + a0 + 1).
(5.43)
T1 ≥ 0, T2 ≥ T1 + c0 , (x, u) ∈ X(T1 , T2 ).
(5.44)
Set
Assume that
We show that (5.39) is true. Assume that
5.7 Auxiliary Results for Theorem 5.1
289
x(t) ≥ S1 , t ∈ [T1 , T2 ].
(5.45)
By (5.10), (5.11), (5.40), and (5.45), f (x(t), u(t)) ≥ −a0 + ψ(x(t)) ≥ −a0 + ψ(S1 ) > f (xf , uf ) + 1.
(5.46)
In view of (5.46), I f (T1 , T2 , x, u) ≥ (T2 − T1 )(f (xf , uf ) + 1) ≥ (T2 − T1 )f (xf , uf ) and (5.39) holds. Assume that inf{x(t) : t ∈ [T1 , T2 ]} < S1 .
(5.47)
τ1 = inf{t ∈ [T1 , T2 ] : x(t) ≤ S1 },
(5.48)
τ2 = sup{t ∈ [T1 , T2 ] : x(t) ≤ S1 }.
(5.49)
Set
By (5.47)–(5.49), τ1 , τ2 are welldefined, τ1 ≤ τ2 and x(τi ) ≤ S1 , i = 1, 2.
(5.50)
It follows from (5.10), (5.11), (5.40), (5.48), and (5.49) that for all t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], (5.46) is true. There are two cases: τ2 − τ1 ≥ c 0 ;
(5.51)
τ2 − τ1 < c 0 .
(5.52)
Assume that (5.51) holds. It follows from (5.41), (5.50), and (5.51) and the choice of S2 and c0 that (τ2 − τ1 )f (xf , uf ) ≤ I f (τ1 , τ2 , x, u) + S2 .
(5.53)
By (5.43), (5.46) holding for each t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], (5.48), (5.49), and (5.53), (T2 −T1 )f (xf , uf ) ≤ (τ1 −T1 )f (xf , uf )+(τ2 −τ1 )f (xf , uf )+(T2 −τ2 )f (xf , uf ) ≤ I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + S2 + I f (τ2 , T2 , x, u) ≤ I f (T1 , T2 , x, u) + S2 ≤ I f (T1 , T2 , x, u) + S and (5.39) holds.
290
5 ContinuousTime Autonomous Problems
Assume that (5.52) holds. By (5.46) holding for each t [τ2 , T2 ], (5.10), (5.11), (5.43), and (5.52),
∈ [T1 , τ1 ] ∪
I f (T1 , T2 , x, u) = I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + I f (τ2 , T2 , x, u) ≥ (f (xf , uf ) + 1)(τ1 − T1 ) − a0 (τ2 − τ1 ) + (f (xf , uf ) + 1)(T2 − τ2 ) ≥ (f (xf , uf ) + 1)(T2 − T1 ) − (a0 + f (xf , uf ) + 1)(τ2 − τ1 ) ≥ (f (xf , uf ) + 1)(T2 − T1 ) − c0 (f (xf , uf ) + 1 + a0 ), (T2 − T1 )f (xf , uf ) ≤ I f (T1 , T2 , x, u) +c0 (f (xf , uf ) + 1 + a0 ) ≤ I f (T1 , T2 , x, u) + S. Lemma 5.36 is proved. Lemma 5.37. Let M0 , M1 , τ0 > 0. Then there exists M2 > M1 such that for each T1 ≥ 0, each T2 ∈ (T1 , T1 + τ0 ], and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M1 ,
(5.54)
I f (T1 , T2 , x, u) ≤ M0 ,
(5.55)
the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ].
(5.56)
Proof. Let us consider the first class of problems. Fix a positive number δ < 2−1 min{1, 16−1τ0 , 8−1 (a0 + 1)−1 e−ω∗ δ (a0 + 1)−1 (M∗ + 1)−1 }.
(5.57)
By (5.9) and (5.10), there exist h0 > M1 + 1 and γ0 > 0 such that f (x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (x, u) ∈ M satisfying x ≥ h0 ,
(5.58)
f (x, u) ≥ 8(G(x, u) − a0 x)+ for each (x, u) ∈ M satisfying G(x, u) − a0 x ≥ γ0 .
(5.59)
Choose a number M2 > (2(h0 + γ0 δ + M0 + a0 τ0 ) + 8M1 )(M∗ + 1)eω∗  δ −1 . Let
(5.60)
5.7 Auxiliary Results for Theorem 5.1
291
T1 ≥ 0, T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (5.54) and (5.55) hold. We show that (5.56) hold. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that x(t0 ) > M2 .
(5.61)
By the choice of h0 , (5.54), (5.55), and (5.58), there exists t1 ∈ [T1 , t0 ] such that x(t1 ) ≤ h0 , t1 − t0  ≤ δ.
(5.62)
There exists t2 ∈ [t1 , t0 ] such that x(t2 ) ≥ x(t) for all t ∈ [t1 , t0 ].
(5.63)
In view of (5.61)–(5.63), (5.64)
t2 > t1 . By the definition of X(T1 , T2 ), x(t2 ) − eA(t2 −t1 ) x(t1 ) =
t2
eA(t2−s) G(x(s), u(s))ds.
(5.65)
t1
It follows from (5.8), (5.62), (5.64), and (5.65) that x(t2 ) − M∗ eω∗ δ h0 ≤ x(t2 ) − eA(t2 −t1 ) x(t1 )
≤ M∗ eω∗ δ
t2
G(x(s), u(s))ds.
(5.66)
t1
Define E1 = {t ∈ [t1 , t2 ] : G(x(t), u(t)) ≥ a0 x(t) + γ0 }, E2 = [t1 , t2 ] \ E1 . By (5.10), (5.55), (5.59), (5.62), (5.63), (5.66)–(5.68), x(t2 )M∗−1 e−ω∗ δ − h0 ≤ ≤ a0
t2 t1
x(t)dt +
t2 t1
t2
G(x(s), u(s))ds
t1
(G(x(s), u(s)) − a0 x(s))+ ds
(5.67) (5.68)
292
5 ContinuousTime Autonomous Problems
≤ a0
t2
x(t)dt +
t1
E1
(G(x(t), u(t)) − a0 x(t))+ dt
+ E2
(G(x(t), u(t)) − a0 x(t))+ dt
≤ a0 x(t2 )δ + γ0 δ + E1
(G(x(t), u(t)) − a0 x(t))+ dt
≤ a0 x(t2 )δ + γ0 δ + 8−1
f (x(t), u(t))dt E1
≤ a0 x(t2 )δ + γ0 δ + 8−1 (M0 + a0 τ0 ).
(5.69)
It follows from (5.57), (5.61), (5.63) and (5.69) that 2−1 M∗−1 e−ω∗ δ M2 ≤ M2 (M∗−1 e−ω∗ δ − a0 δ) ≤ x(t2 )(M∗−1 e−ω∗ δ − a0 δ) ≤ h0 + γ0 δ + 8−1 (M0 + a0 τ0 ). This contradicts (5.60). The contradiction we have reached proves (5.56) and Lemma 5.37 for the first class of problems. Consider the second class of problems. Recall (see (5.8)) that eAt ≤ M∗ eω∗ t , t ∈ [0, ∞)
(5.70)
and that for every τ ≥ 0,
τ
Φτ u =
eA(τ −s) Bu(s)ds,
(5.71)
0
where Φτ ∈ L(L2 (0, τ ; U ), F ) (see (1.9)). Choose a number δ ∈ (0, 1).
(5.72)
h0 > M1 + 1
(5.73)
By (5.11), there exists
and γ > 1 such that f (x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (x, u) ∈ M satisfying x ≥ h0 ,
(5.74)
f (x, u) ≥ K0 u2 /2 for each u ∈ F satisfying u ≥ γ .
(5.75)
5.7 Auxiliary Results for Theorem 5.1
293
Choose M2 > h0 M∗ eω∗  + Φ1 γ + Φ1 (2K0−1 (M0 + a0 τ0 ) + 1) + 1.
(5.76)
Let T1 ≥ 0, T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (5.54) and (5.55) hold. We show that (5.56) holds. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that (5.61) holds. By the choice of h0 , (5.54), (5.55), and (5.74), there exists t1 ∈ [T1 , t0 ]
(5.77)
t2 ∈ [t1 , t0 ]
(5.78)
satisfying (5.62). There exists
satisfying (5.63) and (5.64). Clearly, (x, u) ∈ X(t1 , t2 ), x(t2 ) = eA(t2 −t1 ) x(t1 ) +
t2
eA(t2 −s) Bu(s)ds
(5.79) (5.80)
t1
in E−1 = D(A∗ ) . Proposition 1.18, (5.62), (5.70)–(5.72), (5.78), and (5.80) imply that t2 ω∗ δ x(t2 ) ≤ x(t1 )M∗ e + eA(t2−s) Bu(s)ds t1
≤ M∗ eω∗ δ h0 +
t2 −t1
eA(t2 −t1 −s) Bu(t1 + s)ds
0
≤ M∗ eω∗ δ h0 + Φt2 −t1 u(t1 + ·)L2 (0,t2 −t1 ;F ) ≤ M∗ eω∗ δ h0 + Φ1 (
t2
u(s)2 ds)1/2 .
(5.81)
t1
Define Ω1 = {t ∈ [t1 , t2 ] : u(t) ≥ γ },
(5.82)
Ω2 = [t1 , t2 ] \ Ω1 .
(5.83)
By (5.55), (5.63), (5.75), (5.76), (5.81), and (5.82), x(t0 ) ≤ x(t2 ) ≤ M∗ eω∗ δ h0
294
5 ContinuousTime Autonomous Problems
+Φ1 [(
u(s)2 ds)1/2 + ( Ω1
u(s)2 ds)1/2 ] Ω2
≤ M∗ eω∗ δ h0 + Φ1 (2K0−1
f (x(s), u(s))ds)1/2 + Φ1 γ δ 1/2 Ω1
≤ M∗ eω∗  h0 + Φ1 γ + Φ1 (2K0−1 (M0 + a0 τ0 ))1/2 < M2 .
(5.84)
This contradicts (5.61). The contradiction we have reached proves (5.56) and Lemma 5.37 for the second class of problems. Lemma 5.37, (5.9) and (5.10) imply the following result. Lemma 5.38. Let M1 > 0, 0 < τ < τ0 < τ1 . Then there exists M2 > 0 such that for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 , the inequality x(t) ≤ M2 for all t ∈ [T1 + τ, T2 ] holds.
5.8 Proofs of Theorems 5.1–5.4 Proof of Theorem 5.1. Assertion 1 follows from Lemma 5.36, (5.10), and (5.11). We will prove Assertion 2. Let (x, u) ∈ X(0, ∞). Assume that there exists a sequence of numbers {tk }∞ k=1 such that tk → ∞ as k → ∞,
(5.85)
I f (0, tk , x, u) − tk f (xf , uf ) → ∞ as k → ∞.
(5.86)
I f (0, t, x, u) − tf (xf , uf ) → ∞ as t → ∞.
(5.87)
We show that
In view of (5.9), there exists S0 > 0 such that S0 > 2x(0) + 8,
(5.88)
5.8 Proofs of Theorems 5.1–5.4
295
ψ(S0 − 4) > a0 + 8 + 8f (xf , uf ).
(5.89)
Let S > 0 be as guaranteed by Assertion 1. By (5.10), (5.11), and (5.89), we may assume that lim inf x(t) ≤ S0 − 1. t →∞
For each integer k ≥ 1 set, τk = inf{t ∈ [tk , ∞) : x(t) ≤ S0 }.
(5.90)
Let k ≥ 1 be an integer and t ≥ τk . It follows from (5.10), (5.11), (5.86), (5.89), and (5.90) the choice of S and Assertion 1 that I f (0, t, x, u) − tf (xf , uf ) ≥ I f (0, tk , x, u) − tk f (xf , uf ) +I f (tk , τk , x, u) − (τk − tk )f (xf , uf ) − S ≥ I f (0, tk , x, u) − tk f (xf , uf ) − S → ∞ as k → ∞. Thus Assertion 2 holds. Assume that sup{I f (0, t, x, u) − tf (xf , uf ) : t ∈ [0, ∞)} < ∞. Therefore there exists S1 > 0 such that for each T2 > T1 ≥ 0, I f (T1 , T2 , x, u) − (T2 − T1 )f (xf , uf ) ≤ S1 .
(5.91)
Let S0 > 0 be such that ψ(S0 ) > f (xf , uf ) + 2 + a0 .
(5.92)
It follows from (5.10), (5.11), (5.91) and (5.92) that for each T ≥ 0, min{x(t) : t ∈ [T , T + S1 ]} ≤ S0 . By the relation above, there exists a sequence {Ti }∞ i=1 ⊂ [0, ∞) such that for each integer i ≥ 1, Ti+1 − Ti ∈ [1, 1 + S1 ], x(Ti ) ≤ S0 .
296
5 ContinuousTime Autonomous Problems
This relation, Lemma 5.37 and (5.91) imply that sup{x(t) : t ∈ [0, ∞)} < ∞. Theorem 5.1 is proved. Proof of Theorem 5.2. We may assume that c0 < c/4. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (i) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). Lemma 5.38 implies that there exists M1 > 0 such that the following property holds: (ii) for each T ≥ 0 and each (x, u) ∈ X(T , T + c/2) satisfying I f (T , T + c/2, x, u) ≤ M0 + 2S0 + 1 + cf (xf , uf ), we have x(t) ≤ M1 for all t ∈ [c0 + T , T + c/2]. Assume that T1 ≥ 0, T2 ≥ T1 + c, (x, u) ∈ X(T1 , T2 ) and I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 .
(5.93)
In order to prove the theorem, it is sufficient to show that x(t) ≤ M1 , t ∈ [T1 + c0 , T2 ]. Assume the contrary. Then there exists t0 ∈ [T1 + c0 , T2 ]
(5.94)
x(t0 ) > M1 .
(5.95)
such that
By (5.94), there exists a closed interval [a, b] ⊂ [T1 , T2 ] such that t0 ∈ [a + c0 , b], b − a = 2−1 c. Property (ii), (5.95) and (5.96) imply that
(5.96)
5.8 Proofs of Theorems 5.1–5.4
297
I f (a, b, x, u) > M0 + 2S0 + 1 + cf (xf , uf ).
(5.97)
Property (i), (5.96) and (5.97) imply that I f (T1 , T2 , x, u) = I f (T1 , a, x, u) + I f (a, b, x, u) + I f (b, T2 , x, u) ≥ (a − T1 )f (xf , uf ) − S0 + cf (xf , uf ) + M0 + 2S0 + 1 + (T2 − b)f (xf , uf ) − S0 > (T2 − T1 )f (xf , uf ) + M0 + 1. This contradicts (5.93). The contradiction we have reached proves Theorem 5.2. Proof of Theorem 5.3. Theorem 5.2 implies that there exists M1 > 0 such that the following property holds: (iii) for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + 2L + M0 + 2Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + 2L, and (x, u) ∈ X(T1 , T2 ) satisfy L , x(T1) ∈ AL , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 .
(5.98) (5.99)
In view of (5.98), there exist τ1 ∈ (0, L], τ2 ∈ (0, L], and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), y(T2 ) = x(T2 ),
(5.100)
(y(t), v(t)) = (xf , uf ), t ∈ [τ1 + T1 , T2 − τ2 ],
(5.101)
I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L.
(5.102)
By (5.99)–(5.102), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ1 , y, v) + I f (T1 + τ1 , T2 − τ2 , y, v) + I f (T2 − τ2 , T2 , y, v) + M0 ≤ 2L + M0 + (T2 − T1 − τ2 − τ1 )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + 2L + M0 + 2Lf (xf , uf ).
298
5 ContinuousTime Autonomous Problems
Together with property (iii), this implies that x(t) ≤ M1 for all t ∈ [T1 + c, T2 ]. Theorem 5.3 is proved. Proof of Theorem 5.4. Theorem 5.2 implies that there exists M1 > 0 such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + L + M0 + Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L, and (x, u) ∈ X(T1 , T2 ) satisfy x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 .
(5.103) (5.104)
In view of (5.103), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), (y(t), v(t)) = (xf , uf ), t ∈ [τ + T1 , T2 ], I f (T1 , T1 + τ, y, v) ≤ L.
(5.105) (5.106)
By (5.104)–(5.106), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ, y, v) + I f (T1 + τ, T2 , y, v) + M0 ≤ L + M0 + (T2 − T1 − τ )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + L + M0 + Lf (xf , uf ). Together with property (iv), this implies that x(t) ≤ M1 for all t ∈ [T1 + c, T2 ]. Theorem 5.4 is proved. Proof of Theorem 5.5. Theorem 5.2 implies that there exists M1 > 0 such that the following property holds: (v) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + L + M0 + Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L, and (x, u) ∈ X(T1 , T2 ) satisfy
5.9 Proofs of Theorems 5.6–5.9
299
L , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + M0 .
(5.107) (5.108)
In view of (5.107), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that (y(t), v(t)) = (xf , uf ), t ∈ [T1 , T2 − τ ], y(T2 ) = x(T2 ), I f (T2 − τ, T2 , y, v) ≤ L.
(5.109) (5.110)
By (5.108)–(5.110), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T2 − τ, y, v) + I f (T2 − τ, T2 , y, v) + M0 ≤ L + M0 + (T2 − T1 − τ )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + L + M0 + Lf (xf , uf ). Together with property (v), this implies that x(t) ≤ M1 for all t ∈ [T1 + c, T2 ]. Theorem 5.5 is proved.
5.9 Proofs of Theorems 5.6–5.9 Proof of Theorem 5.6. Fix c0 = c/4. Theorem 5.2 implies that there exists M2 > 0 such that the following property holds: (i) for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 , the inequality x(t) ≤ M2 holds for all t ∈ [T1 + c0 , T2 ]. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (ii) for each T2 > T1 ≥ 0, and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). Lemma 5.37 implies that there exists M1 > M2 such that the following property holds:
300
5 ContinuousTime Autonomous Problems
(iii) for each T ≥ 0 and each (x, u) ∈ X(T , T + c0 ) satisfying x(T1 ) ≤ M0 , I f (T , T + c0 , x, u) ≤ M0 + S0 + 2c0 f (xf , uf ), we have x(t) ≤ M0 for all t ∈ [T , T + c0 ]. Assume that T1 ≥ 0, T2 ≥ T1 + c, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M0 .
(5.111) (5.112)
Property (i) and (5.112) imply that x(t) ≤ M2 < M1 , t ∈ [T1 + c0 , T2 ].
(5.113)
It follows from (5.112) and property (ii) that I f (T1 , T1 + c0 , x, u) = I f (T1 , T2 , x, u) − I f (T1 + c0 , T2 , x, u) ≤ (T2 −T1 )f (xf , uf )+M0 −(T2 −T1 −c0 )f (xf , uf )+S0 = c0 f (xf , uf )+M0 +S0 . By the relation above, (5.111) and property (iii), x(t) ≤ M1 , t ∈ [T1 , T1 + c0 ]. Together with (5.113), this completes the proof of Theorem 5.6. Proof of Theorem 5.7. Theorem 5.6 implies that there exists M1 > 0 such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + 2L + 2M0 + 2Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (v) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ).
5.9 Proofs of Theorems 5.6–5.9
301
Assume that T1 ≥ 0, T2 ≥ T1 + 2L, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 ,
(5.114)
L , x(T1) ∈ AL , x(T2 ) ∈ A
(5.115)
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 .
(5.116)
In view of (5.115), there exist τ1 ∈ (0, L], τ2 ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), y(T2 ) = x(T2 ),
(5.117)
(y(t), v(t)) = (xf , uf ), t ∈ [τ1 + T1 , T2 − τ2 ],
(5.118)
I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L.
(5.119)
By (5.116)–(5.119), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ1 , y, v) + I f (T1 + τ1 , T2 − τ2 , y, v) + I f (T2 − τ2 , T2 , y, v) + M0 ≤ 2L + M0 + (T2 − T1 − τ2 − τ1 )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + 2L + M0 + 2Lf (xf , uf ).
(5.120)
Property (iv), (5.114) and (5.120) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 5.7 is proved. Proof of Theorem 5.8. Theorem 5.6 implies that there exists M1 > 0 such that the following property holds: (vi) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + L + 2M0 + Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (vii) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ).
302
5 ContinuousTime Autonomous Problems
Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 , x(T1 ) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 .
(5.121) (5.122)
In view of (5.121), there exist τ ∈ (0, L], (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), (y(t), v(t)) = (xf , uf ), t ∈ [τ + T1 , T2 ], I f (T1 , T1 + τ, y, v) ≤ L.
(5.123) (5.124)
By (5.122)–(5.124), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ, y, v) + I f (T1 + τ, T2 , y, v) + M0 ≤ L + M0 + (T2 − T1 − τ )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + L + M0 + Lf (xf , uf ).
(5.125)
Together with property (vi), (5.121) and (5.125), this implies that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 5.8 is proved. Proof of Theorem 5.9. Theorem 5.6 implies that there exists M1 > 0 such that the following property holds: (viii) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + L + 2M0 + Lf (xf , uf ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (ix) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ), L , x(T1 ) ≤ M0 , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + M0 .
(5.126) (5.127)
5.10 Proofs of Theorems 5.11 and 5.12
303
In view of (5.126), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that (y(t), v(t)) = (xf , uf ), t ∈ [T1 , T2 − τ ],
(5.128)
y(T2 ) = x(T2 ),
(5.129)
I f (T2 − τ, T2 , y, v) ≤ L.
(5.130)
By (5.127)–(5.130), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) ≤ L+M0 +(T2 −T1 −τ )f (xf , uf ) ≤ (T2 −T1 )f (xf , uf )+L+M0 +Lf (xf , uf ). (5.131) Property (viii), (5.126) and (5.131) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 5.9 is proved.
5.10 Proofs of Theorems 5.11 and 5.12 Theorem 5.11 follows from Proposition 5.10 and the following result. Proposition 5.39. Let L > 0. Then for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each L , pair of points y ∈ AL , z ∈ A U f (T1 , T2 , y, z) ≤ (T2 − T1 )f (xf , uf ) + 2L(1 + f (xf , uf )). L . By the definition of AL , A L , Proof. Let T1 ≥ 0, T2 ≥ T1 + 2L, y ∈ AL , z ∈ A there exist τ1 ∈ (0, L], τ2 ∈ (0, L], and (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = y, (x(t), u(t)) = (xf , uf ), t ∈ [T1 + τ1 , T2 − τ2 ], x(T2 ) = z2 , I f (T1 , T1 + τ1 , x, u) ≤ L, I f (T2 − τ2 , T2 , x, u) ≤ L. In view of the relations above, U f (T1 , T2 , y, z) ≤ I f (T1 , T2 , x, u) = I f (T1 , T1 + τ1 , x, u) + I f (T1 + τ1 , T2 − τ2 , x, u) + I f (T2 − τ2 , T2 , x, u) ≤ 2L + (T2 − T1 − τ2 − τ1 )f (xf , uf ) ≤ 2L + (T2 − T1 )f (xf , uf ) + 2Lf (xf , uf ). Proposition 5.39 is proved. Theorem 5.12 follows from Proposition 5.10 and the following result.
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5 ContinuousTime Autonomous Problems
Proposition 5.40. Let L > 0. Then for each T1 ≥ 0, each T2 ≥ T1 + L, and each y ∈ AL , σ f (T1 , T2 , y) ≤ (T2 − T1 )f (xf , uf ) + L(1 + f (xf , uf )). Proof. Let T1 ≥ 0, T2 ≥ T1 + L, y ∈ AL . By the definition of AL , there exist τ ∈ (0, L] and (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = y, (x(t), u(t)) = (xf , uf ), t ∈ [T1 + τ, T2 ], I f (T1 , T1 + τ, x, u) ≤ L. In view of the relations above, σ f (T1 , T2 , y) ≤ I f (T1 , T2 , x, u) = I f (T1 , T1 + τ, x, u) + I f (T1 + τ, T2 , x, u) ≤ L + (T2 − T1 − τ )f (xf , uf ) ≤ (T2 − T1 )f (xf , uf ) + L(1 + f (xf , uf )). Proposition 5.40 is proved.
5.11 Proof of Theorem 5.13 First we show that TP implies (P1) and (P2). In view of Theorem 5.1, TP implies (P2). We show that TP implies (P1). Assume that TP holds, (x, u) ∈ X(0, ∞) is (f )good. There exists S > 0 such that for each T > 0, I f (0, T , x, u) − Tf (xf , uf ) < S. This implies that for each pair of numbers T2 > T1 ≥ 0, I f (T1 , T2 , x, u) − (T2 − T1 )f (xf , uf ) < 2S. Let δ > 0. We show that there exists Tδ > 0 such that for each T > Tδ , I f (Tδ , T , x, u) ≤ U f (Tδ , T , x(Tδ ), x(T )) + δ. Assume the contrary. Then for each T ≥ 0, there exists S > T such that I f (T , S, x, u) > U f (T , S, x(T ), x(S)) + δ.
(5.132)
5.11 Proof of Theorem 5.13
305
This implies that there exists a strictly increasing sequence of numbers {Ti }∞ i=0 such that T0 = 0 and for every integer i ≥ 0, I f (Ti , Ti+1 , x, u) > U f (Ti , Ti+1 , x(Ti ), x(Ti+1 )) + δ.
(5.133)
By (5.133), there exists (y, v) ∈ X(0, ∞) such that for every integer i ≥ 0, y(Ti ) = x(Ti ), I f (Ti , Ti+1 , x, u) > I f (Ti , Ti+1 , y, v) + δ.
(5.134) (5.135)
In view of (5.132), (5.134) and (5.135), for each integer k ≥ 1, I f (0, Tk , y, v) − Tk f (xf , uf ) = I f (0, Tk , y, v) − I f (0, Tk , x, u) + I f (0, Tk , x, u) −Tk f (xf , uf ) ≤ −kδ + 2S → −∞ as k → ∞. This contradicts Theorem 5.1. The contradiction we have reached proves that the following property holds: (i) for each δ > 0 there exists Tδ > 0 such that for each integer T > Tδ , I f (Tδ , T , x, u) ≤ U f (Tδ , T , x(Tδ ), x(T )) + δ. Let > 0. Proposition 5.10 implies that there exist δ > 0, L > 0 such that the following property holds: (ii) for each S1 ≥ 0, each S2 ≥ S1 +2L, and each (z, ξ ) ∈ X(S1 , S2 ) which satisfies I f (S1 , S2 , z, ξ ) ≤ min{(S2 −S1 )f (xf , uf )+2S, U f (S1 , S2 , z(S1 ), z(S2 ))+δ}, we have z(t) − xf ≤ , t ∈ [S1 + L, S2 − L]. Let Tδ > 0 be as guaranteed by (i). Properties (i) and (ii) (5.132) and the choice of Tδ imply that for each T ≥ Tδ + 2L, x(t) − xf ≤ , t ∈ [Tδ + L, T − L].
306
5 ContinuousTime Autonomous Problems
Therefore x(t) − xf ≤ for all t ≥ Tδ + L and (P1) holds. Thus TP implies (P1). Lemma 5.41. Assume that (P1) holds and that > 0. Then there exists δ > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf ≤ δ, i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, the inequality x(t) − xf ≤ holds for all t ∈ [T1 , T2 ]. Proof. By (A2), for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: (iii) for each z ∈ A satisfying z − xf ≤ δk , there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(0, τ1 ), (x2 , u2 ) ∈ X(0, τ2 ) which satisfy x1 (0) = z, x1 (τ1 ) = xf , x2 (0) = xf , x2 (τ2 ) = z, I f (0, τi , xi , ui ) ≤ τi f (xf , uf ) + 2−k , i = 1, 2. Assume that the lemma does not hold. Then for each integer k ≥ 1, there exist Tk ≥ 2bf and (xk , uk ) ∈ X(0, Tk ) such that xk (0) − xf ≤ δk , xk (Tk ) − xf ≤ δk ,
(5.136)
I f (0, Tk , xk , uk ) ≤ U f (0, Tk , xk (0), xk (Tk )) + δk ,
(5.137)
sup{xk (t) − xf : t ∈ [0, Tk ]} > .
(5.138)
Let k ≥ 1 be an integer. Property (iii) and (5.136) imply that there exist τk,1 , τk,2 ∈ (0, bf ] and (x˜k , u˜ k ) ∈ X(0, Tk + τk,1 + τk,2 ) such that x˜k (0) = xf , x˜k (τk,1 ) = xk (0),
(5.139)
I f (0, τk,1 , x˜k , u˜ k ) ≤ τk,1 f (xf , uf ) + 2−k ,
(5.140)
(x˜k (τk,1 + t), u˜ k (τk,1 + t)) = (xk (t), uk (t)), t ∈ [0, Tk ], x˜k (Tk + τk,1 + τk,2 ) = xf ,
(5.141) (5.142)
5.11 Proof of Theorem 5.13
307
I f (Tk + τk,1 , Tk + τk,1 + τk,2 , x˜k , u˜ k ) ≤ τk,2 f (xf , uf ) + 2−k .
(5.143)
By (5.137), (5.140), (5.141), and (5.143), I f (0, Tk + τk,1 + τk,2, x˜k , u˜ k ) ≤ (τk,1 + τk,2 )f (xf , uf ) + 2−k+1 + I f (0, Tk , xk , uk ) ≤ (τk,1 + τk,2 )f (xf , uf ) + U f (0, Tk , xk (0), xk (Tk )) + δk + 2−k+1 .
(5.144)
xk , uk ) ∈ Property (iii) and (5.136) imply that there exist τk,3 , τk,4 ∈ (0, bf ] and ( X(0, Tk ) such that xk (0) = xk (0), xk (Tk ) = xk (Tk ), ( xk (t), uk (t)) = (xf , uf ), t ∈ [τk,3 , Tk − τk,4 ], I f (0, τk,3 , xk , uk ) ≤ τk,3 f (xf , uf ) + 2−k , xk , uk ) ≤ τk,4 f (xf , uf ) + 2−k . I f (Tk − τk,4 , Tk ,
(5.145) (5.146) (5.147) (5.148)
In view of (5.145)–(5.148), U f (0, Tk , xk (0), xk (Tk )) ≤ I f (0, Tk , xk , uk ) ≤ Tk f (xf , uf ) + 2−k+1 .
(5.149)
By (5.144) and (5.149), I f (0, Tk +τk,1 +τk,2 , x˜k , u˜ k ) ≤ (Tk +τk,1 +τk,2 )f (xf , uf )+3·2−k+1.
(5.150)
By (5.139) and (5.142), there exists (x, u) ∈ X(0, ∞) such that (x(t), u(t)) = (x˜1 (t), u˜ 1 (t)), t = [0, T1 + τ1,1 + τ1,2 ],
(5.151)
and for each integer k ≥ 1 and each t ∈ [0, Tk+1 + τk+1,1 + τk+1,2 ], (x(
k k (Ti + τi,1 + τi,2 + t)), u( (Ti + τi,1 + τi,2 + t))) = (x˜k+1 (t), u˜ k+1 (t)). i=1
i=1
(5.152) In view of (5.150)–(5.152), for each integer k ≥ 1, I f (0,
k k (Ti + τi,1 + τi,2 ), x, ˜ u) ˜ = I f (0, Ti + τi,1 + τi,2 , x˜i , u˜ i ) i=1
i=1
308
≤
5 ContinuousTime Autonomous Problems
k
(Ti +τi,1 +τi,2 )f (xf , uf )+3
i=1
k
2−i−1 ≤
i=1
k
(Ti +τi,1 +τi,2 )f (xf , uf )+6.
i=1
Theorem 5.1 implies that (x, ˜ u) ˜ is (f )good. Together with (P1), this implies that limt →∞ x(t) ˜ − xf = 0. On the other hand, in view of (5.138), (5.141), and (5.152), lim supt →∞ x(t) ˜ − xf > . The contradiction we have reached completes the proof of Lemma 5.41. Completion of the Proof of Theorem 5.13. Assume that properties (P1) and (P2) hold. Let , M > 0. By Lemma 5.41, there exists δ0 > 0 such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf ≤ δ0 , i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ0 , the inequality x(t) − xf ≤ holds for all t ∈ [T1 , T2 ]. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (v) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), we have I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). Set S1 = 2S0 + M.
(5.153)
In view of (P2), there exist δ ∈ (0, δ0 ) and L0 > 0 such that the following property holds: (vi) for each (x, u) ∈ X(0, L0 ) which satisfies I f (0, L0 , x, u) ≤ min{U f (0, L0 , x(0), x(L0 )) + δ, L0 f (xf , uf ) + S1 }, there exists t0 ∈ [0, L0 ] such that x(t0 ) − xf ≤ δ0 . Set L = L0 + bf .
(5.154)
Assume that T1 ≥ 0 and T2 ≥ T1 + 2L and that (x, u) ∈ X(T1 , T2 ) satisfy I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 )+M, U f (T1 , T2 , x(T1 ), x(T2 ))+δ}.
(5.155)
5.12 An Auxiliary Result
309
Property (v) and (5.155) imply that for each pair of numbers Q1 , Q2 ∈ [T1 , T2 ] satisfying Q1 < Q2 , I f (Q1 , Q2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , Q1 , x, u) − I f (Q2 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M − (Q1 − T1 )f (xf , uf ) + S0 − (T2 − Q2 )f (xf , uf ) + S0 = (Q2 − Q1 )f (xf , uf ) + M + 2S0 .
(5.156)
It follows from (5.156) that I f (T1 , T1 + L0 , x, u) ≤ L0 f (xf , uf ) + 2S0 + M,
(5.157)
I f (T2 − L0 , T2 , x, u) ≤ L0 f (xf , uf ) + 2S0 + M.
(5.158)
By (5.153), (5.155), (5.157), (5.158), and property (vi), there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ]
(5.159)
x(τi ) − xf ≤ δ0 , i = 1, 2.
(5.160)
such that
If x(T1 ) − xf ≤ δ, then we may assume that τ1 = T1 and if x(T2 ) − xf ≤ δ, then we may assume that τ2 = T2 . In view of (5.155) and (5.159), I f (τ1 , τ2 , x, u) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ.
(5.161)
It follows from (5.154) and (5.159) that τ2 − τ1 ≥ T2 − T1 − 2L0 ≥ 2L − 2L0 ≥ 2bf .
(5.162)
Property (iv) and (5.160)–(5.162) imply that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Theorem 5.13 is proved.
5.12 An Auxiliary Result Lemma 5.42. Assume that (P2) holds and that M, > 0. Then there exists L > 0 such that for each (x, u) ∈ X(0, L) which satisfies
310
5 ContinuousTime Autonomous Problems
I f (0, L, x, u) ≤ Lf (xf , uf ) + M, the following inequality holds: inf{x(t) − xf : t ∈ [0, L]} ≤ . Proof. By Theorem 5.1, there exists S0 > 0 such that the following property holds: (i) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). By (P2), there exist δ0 ∈ (0, ) and L0 > 0 such that the following property holds: (ii) for each (x, u) ∈ X(0, L0 ) which satisfies I f (0, L0 , x, u) ≤ min{U f (0, L0 , x(0), x(L0 )) + δ0 , L0 f (xf , uf ) + 2S0 + M + f (xf , uf )}, we have inf{x(t) − xf : t ∈ [0, L0 ]} ≤ . Choose an integer q0 > (M + S0 )δ −1
(5.163)
L = q0 L0 .
(5.164)
I f (0, L, x, u) ≤ Lf (xf , uf ) + M.
(5.165)
and set
Let (x, u) ∈ X(0, L) satisfy
We show that inf{x(t) − xf : t ∈ [0, L]} ≤ . Assume the contrary. Then x(t) − xf > , t ∈ [0, L].
(5.166)
5.13 Proof of Theorem 5.14
311
Let an integer i ∈ {0, . . . , q0 − 1}. Property (i) and (5.165) imply that I f (iL0 , (i+1)L0 , x, u) = I f (0, L, x, u)−I f (0, iL0 , x, u)−I f ((i+1)L0, L, x, u) ≤ Lf (xf , uf ) + M − iL0 f (xf , uf ) + S0 − (L − (i + 1)L0 )f (xf , uf ) + S0 = L0 f (xf , uf ) + M + 2S0 .
(5.167)
By (5.166), (5.167), and property (ii), I f (iL0 , (i + 1)L0 , x, u) > U f (iL0 , (i + 1)L0 , x(iL0 ), x((i + 1)L0 )) + δ0 . (5.168) It follows from (5.168) that there exists (y, v) ∈ X(0, L) such that y(iL0 ) = x(iL0 ), i = 0, . . . , q0
(5.169)
and for all i = 0, . . . , q0 − 1, I f (iL0 , (i + 1)L0 , x, u) > I f (iL0 , (i + 1)L0 , y, v) + δ0 .
(5.170)
In view of (5.165), (5.169), (5.170) and property (i), Lf (xf , uf )+M ≥ I f (0, L, x, u) > I f (0, L, y, v)+q0 δ ≥ Lf (xf , uf )−S0 +q0 δ, q0 ≤ (M + S0 )δ −1 . This contradicts (5.163). The contradiction we have reached proves Lemma 5.42.
5.13 Proof of Theorem 5.14 By Theorem 5.1, there exists S0 > 0 such that the following property holds: (i) for each τ2 > τ1 ≥ 0 and each (y, v) ∈ X(τ1 , τ2 ), I f (τ1 , τ2 , y, v) + S0 ≥ (τ2 − τ1 )f (xf , uf ). Theorem 5.13 and STP imply that there exist δ ∈ (0, ) and L0 > 0 such that the following property holds: (ii) for each τ1 ≥ 0, each τ2 ≥ τ1 +2L0 , and each (y, v) ∈ X(τ1 , τ2 ) which satisfies I f (τ1 , τ2 , y, v) ≤ min{σ f (τ1 , τ2 ) + M + 3S0 , U f (τ1 , τ2 , y(τ1 ), y(τ2 )) + δ}, we have
312
5 ContinuousTime Autonomous Problems
y(t) − xf ≤ , t ∈ [τ1 + L0 , τ2 − L0 ]. Set l = 2L0 + 1.
(5.171)
Q ≥ 2 + 2(M + S0 )δ −1 .
(5.172)
Choose a natural number
Assume that T1 ≥ 0, T2 ≥ T1 + lQ and (x, u) ∈ X(T1 , T2 ) satisfy I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M.
(5.173)
Property (i), (5.172), and (5.173) imply that for each pair of numbers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , I f (τ1 , τ2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , τ1 , x, u) − I f (τ2 , T2 , x, u) ≤ M +(T2 −T1 )f (xf , uf )−(τ1 −T1 )f (xf , uf )+S0 −(T2 −τ2 )f (xf , uf )+S0 ≤ M + 2S0 + (τ2 − τ1 )f (xf , uf ).
(5.174)
Set t0 = T1 . If I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, then we set t1 = T2 , s1 = T2 . Assume that I f (T1 , T2 , x, u) > U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(5.175)
It is easy to see that for each t ∈ (T1 , T2 ) such that t − T1 is sufficiently small, we have I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) < δ.
(5.176)
Set t˜1 = inf{t ∈ (T1 , T2 ] : I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) > δ}. (5.177)
5.13 Proof of Theorem 5.14
313
Clearly, t˜1 ∈ (T1 , T2 ] is welldefined. By (5.175) and (5.176), there exist s1 , t1 ∈ [T1 , T2 ] such that t0 < s1 < t˜1 , s1 ≥ t˜1 − 1/4, I f (t0 , s1 , x, u) − U f (t0 , s1 , x(t0 ), x(s1 )) ≤ δ, t˜1 ≤ t1 < t˜1 + 1/4, t1 ≤ T2 , I f (t0 , t1 , x, u) − U f (t0 , t1 , x(t0 ), x(t1 )) > δ.
(5.178) (5.179) (5.180) (5.181)
Assume that k ≥ 1 is an integer and we defined finite sequences of numbers {ti }ki=0 ⊂ [T1 , T2 ], {si }ki=1 ⊂ [T1 , T2 ] such that t0 < t1 · · · < tk ,
(5.182)
ti−1 < si ≤ ti , ti − si ≤ 2−1 ,
(5.183)
for each integer i = 1, . . . , k
I f (ti−1 , si , x, u) − U f (ti−1 , si , x(ti−1 ), x(si )) ≤ δ,
(5.184)
and if ti < T2 , then I f (ti−1 , ti , x, u) − U f (ti−1 , ti , x(ti−1 ), x(ti )) > δ.
(5.185)
(Note that in view of (5.175), (5.178)–(5.181), our assumption holds for k = 1.) By (5.182) and (5.185), there exists (y, v) ∈ X(T1 , T2 ) such that y(ti ) = x(ti ), i = 1, . . . , k, I f (ti−1 , ti , x, u)−I f (ti−1 , ti , y, v) > δ for all intergers i ∈ [1, k]\{k}, (y(t), v(t)) = (x(t), u(t)), t ∈ [tk−1 , T2 ].
(5.186) (5.187) (5.188)
Property (i), (5.173), and (5.187) imply that (T2 − T1 )f (xf , uf ) + M ≥ I f (T1 , T2 , x, u) ≥ I f (T1 , T2 , y, v) + δ(k − 1) ≥ (T2 − T1 )f (xf , uf ) − S0 + δ(k − 1), k ≤ 1 + δ −1 (M + S0 ).
(5.189)
314
5 ContinuousTime Autonomous Problems
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If I f (tk , T2 , x, u) ≤ U f (tk , T2 , x(tk ), x(T2 )) + δ, then we set tk+1 = T2 , sk+1 = T2 , and the construction of the sequence is completed. Assume that I f (tk , T2 , x, u) > U f (tk , T2 , x(tk ), x(T2 )) + δ.
(5.190)
It is easy to see that for each t ∈ (tk , T2 ) such that t − tk is sufficiently small, we have I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) < δ.
(5.191)
Set t˜ = inf{t ∈ (tk , T2 ] : I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) > δ}.
(5.192)
By (5.190) and (5.191), t˜ is welldefined and t˜ ∈ (tk , T2 ].
(5.193)
There exist sk+1 , tk+1 ∈ [T1 , T2 ] such that tk < sk+1 < t˜, sk+1 > t˜ − 1/4, t˜ + 4−1 > tk+1 ≥ t˜, I f (tk , sk+1 , x, u) − U f (tk , sk+1 , x(tk ), x(sk+1 )) ≤ δ, I f (tk , tk+1 , x, u) − U f (tk , tk+1 , x(tk ), x(tk+1 )) > δ. It is not difficult to see that the assumption made for k also holds for k + 1. In q view of (5.189), by induction we constructed finite sequences {ti }i=0 ⊂ [T1 , T2 ], q {si }i=0 ⊂ [T1 , T2 ] such that q ≤ 1 + δ −1 (M + S0 ),
(5.194)
T1 = t0 < t1 · · · < tq = T2 , for each integer i = 1, . . . , q, (5.183) and (5.184) hold and if ti < T2 , then (5.185) holds. Assume that i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 2L0 + 1.
(5.195)
5.14 Proof of Proposition 5.16
315
By (5.183) and (5.195), si+1 − ti ≥ 2L0 .
(5.196)
In view of (5.184), I f (ti , si+1 , x, u) − U f (ti , si+1 , x(ti ), x(si+1 )) ≤ δ.
(5.197)
Property (i) and (5.174) imply that I f (ti , si+1 , x, u) ≤ M + 2S0 + (si+1 − ti )f (xf , uf ) ≤ M + 3S0 + σ f (ti , si+1 ). (5.198) It follows from (5.196)–(5.198)) and property (ii) that x(t) − xf ≤ , t ∈ [ti + L0 , si+1 − L0 ], and in view of (5.183), x(t) − xf ≤ , t ∈ [ti + L0 , ti+1 − L0 − 1].
(5.199)
By (5.199), {t ∈ [T1 , T2 ] : x(t) − xf > } ⊂ ∪{[ti , ti +L0 ]∪[ti+1 −1−L0 , ti+1 ] : i ∈ {0, . . . , q −1} and ti+1 −ti ≥ 2L0 +1} ∪ {[ti , ti+1 ] : i ∈ {0, . . . , q − 1} and ti+1 − ti < 2L0 + 1}.
(5.200)
In view of (5.171) and (5.193), the maximal length of intervals in the righthand side of (5.200) does not exceed l, and their number does not exceed 2q ≤ 2(1 + δ −1 (M + S0 )) ≤ Q. Theorem 5.14 is proved.
5.14 Proof of Proposition 5.16 Let us show that (P1) holds. Let (x, u) ∈ X(0, ∞) be (f )good. There exists M0 > 0 such that for each T > 0, I f (0, T , x, u) − Tf (xf , uf ) ≤ M0 . This implies that for each pair of numbers T2 > T1 ≥ 0, I f (T1 , T2 , x, u) − (T2 − T1 )f (xf , uf ) = I f (0, T2 , x, u) − T2 f (xf , uf )
316
5 ContinuousTime Autonomous Problems
− I f (0, T1 , x, u) + T1 f (xf , uf ) ≤ 2M0 .
(5.201)
Let > 0, and let l > 0 and a natural number Q be as guaranteed by WTP with and M = 2M0 . Set Ω = {t ∈ [0, ∞) : x(t) − xf > }. We show that Ω is bounded. Assume that contrary. Then there exists a sequence of positive numbers {ti }∞ i=1 such that for all integers i ≥ 1, ti+1 − ti ≥ 2l + 2, x(ti ) − xf > .
(5.202)
Let k ≥ 1 + Q be an integer. It follows from (5.202) that tQ ≥ Q(2l + 2).
(5.203)
By WTP and the choice of Q, l, there exist an integer q ≤ Q and finite sequences q q {ai }i=1 , {bi }i=1 ⊂ [0, tk ] such that 0 ≤ bi − ai ≤ l, i = 1, . . . , q,
(5.204)
bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf ≤ for all t ∈ [0, tk ] \ ∪i=1 [ai , bi ].
(5.205)
In view of (5.202) and (5.205), q
{ti }ki=1 ⊂ ∪i=1 [ai , bi ].
(5.206)
By (5.202), (5.204), and (5.206), k ≤ q ≤ Q, a contradiction. The contradiction we have reached proves that Ω is bounded. Since is an arbitrary positive number, this completes the proof of Proposition 5.16.
5.15 Proof of Proposition 5.18 Assume that (P2) does not hold. Then there exist , M > 0 such that for each natural number k, there exists an integer nk ≥ k and (xk , uk ) ∈ X(0, nk ) which satisfies I f (0, nk , xk , uk ) ≤ nk f (xf , uf ) + M,
(5.207)
xk (t) − xf > , t ∈ [0, nk ].
(5.208)
5.15 Proof of Proposition 5.18
317
Let S0 > 0 be as guaranteed by Theorem 5.1. By the choice of S0 , Theorem 5.1 and (5.207), for each integer k ≥ 1 and each pair of numbers τ1 , τ2 ∈ [0, k] satisfying τ1 < τ2 , I f (τ1 , τ2 , xk , uk ) − (τ2 − τ1 )f (xf , uf ) = I f (0, nk , xk , uk ) − nk f (xf , uf ) −I f (0, τ1 , xk , uk ) + τ1 f (xf , uf ) − I f (τ2 , nk , xk , uk ) + (nk − τ2 )f (xf , uf ) ≤ M + 2S0 .
(5.209)
By the relation above and LSC property, extracting subsequences, using the diagonalization process and reindexing, we may assume that for each integer i ≥ 1, there exists limk→∞ I f (i − 1, i, xk , uk ), and there exists (yi , vi ) ∈ X(i − 1, i) such that xk (t) → yi (t) as k → ∞ for every t ∈ [i − 1, i],
(5.210)
I f (i − 1, i, yi , vi ) ≤ lim I f (i − 1, i, xk , uk ).
(5.211)
k→∞
In view of (5.210), there exists (y, v) ∈ X(0, ∞) such that for each integer i ≥ 1, (y(t), v(t)) = (yi (t), vi (t)), t ∈ [i − 1, i]. It follows from the relation above, (5.209) and (5.211), that for every integer m ≥ 1 I f (0, m, y, v) =
m
I f (i − 1, i, y, v) =
i=1
≤
m i=1
m
I f (i − 1, i, yi , vi )
i=1
( lim I f (i −1, i, xk , uk )) = lim I f (0, m, xk , uk ) ≤ mf (xf , uf )+M +2S0 . k→∞
k→∞
Together with Theorem 5.1, this implies that (y, v) is (f )good. Property (P1) implies that there exists an integer τ0 > 0 such that y(t) − xf ≤ /4 for all t ≥ τ0 − 1, and in view of the definition of (y, v), yτ0 (t) − xf ≤ /4 for all t ∈ [τ0 − 1, τ0 ]. By (5.210), there exists an integer k0 > τ0 such that for all integers k ≥ k0 , yτ0 (τ0 ) − xk (τ0 ) ≤ /4.
(5.212)
318
5 ContinuousTime Autonomous Problems
In view of (5.212) and the relation above, for all integers k ≥ k0 , xk (τ0 ) − xf ≤ xk (τ0 ) − yτ0 (τ0 ) + yτ0 (τ0 ) − xf ≤ /2. This contradicts (5.208). The contradiction we have reached proves Proposition 5.18.
5.16 Auxiliary Results for Theorems 5.19 and 5.20 Lemma 5.43. Let > 0. Then there exists δ > 0 such that for each T ≥ bf and each (x, u) ∈ X(0, T ) which satisfies x(0) − xf ≤ δ, x(T ) − xf ≤ δ, the following inequality holds: I f (0, T , x, u) ≥ Tf (xf , uf ) − . Proof. By (A2), there exists δ > 0 such that the following property holds: (i) for each z ∈ A satisfying z − xf ≤ δ, there exist τ1 ∈ (0, bf ], τ2 ∈ (0, bf ] and (x˜1 , u˜ 1 ) ∈ X(0, τ1 ), (x˜2 , u˜ 2 ) ∈ X(0, τ2 ) such that x˜1 (0) = z , x˜1 (τ1 ) = xf , I f (0, τ1 , x˜1 , u˜ 1 ) ≤ τ1 f (xf , uf ) + /4, x˜2 (0) = xf , x˜2 (τ2 ) = z, I f (0, τ2 , x˜2 , u˜ 2 ) ≤ τ2 f (xf , uf ) + /4. Assume that T ≥ bf , (x, u) ∈ X(0, T ) and x(0) − xf ≤ δ, x(T ) − xf ≤ δ. By the relation above and property (i), there exists (y, v) ∈ X(0, ∞) such that (y(t + 2bf + T ), v(t + 2bf + T )) = (y(t), v(t)) for all t ≥ 0,
(5.213)
y(0) = xf , y(bf ) = x(0), y(2bf + T ) = xf ,
(5.214)
I f (0, bf , y, v) ≤ bf f (xf , uf ) + /4,
(5.215)
(y(t), v(t)) = (x(t − bf ), u(t − bf )), t ∈ [bf , bf + T ],
(5.216)
I f (bf + T , 2bf + T , y, v) ≤ bf f (xf , uf ) + /4.
(5.217)
5.16 Auxiliary Results for Theorems 5.19 and 5.20
319
Theorem 5.1, (5.213), (5.214), and (5.216) imply that (T + 2bf )f (xf , uf ) ≤ I f (0, T + 2bf , y, v) = I f (0, bf , y, v) +I f (0, T , x, u) + I f (T + bf , T + 2bf , y, v). By the relation above, (5.215) and (5.217), I f (0, T , x, u) ≥ (T + 2bf )f (xf , uf ) − 2bf f (xf , uf ) + /4 ≥ Tf (xf , uf ) − /2. Lemma 5.43 is proved. Assume that f has LSC property and (P1). Theorem 5.13 and Proposition 5.18 imply that f has (P2) and TP. Proposition 5.44. Assume that z ∈ ∪{AL : L > 0}. Then there exists an (f )good and (f )minimal pair (x∗ , u∗ ) ∈ X(0, ∞) such that x∗ (0) = z. Proof. There exists L0 > 0 such that z ∈ AL0 .
(5.218)
It follows from Theorem 5.1 that there exists S0 > 0 such that for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ (T2 − T1 )f (xf , uf ). Fix an integer k0 ≥ L0 . LSC property implies that for each integer k ≥ k0 , there exists (xk , uk ) ∈ X(0, k) satisfying xk (0) = z, I f (0, k, xk , uk ) = σ f (0, k, z).
(5.219)
In view of (5.218), for each integer k ≥ k0 , σ f (0, k, z) ≤ L0 + f (xf , uf )(k − L0 ) + L0 f (xf , uf ).
(5.220)
By (5.219), (5.220) and the choice of S0 , for each integer k ≥ k0 and each pair of numbers T1 , T2 ∈ [0, k] satisfying T1 < T2 , I f (T1 , T2 , xk , uk ) = I f (0, k, xk , uk ) − I f (0, T1 , xk , uk ) − I f (T2 , k, xk , uk ) ≤ kf (xf , uf )+L0 (1+2f (xf , uf ))−T1 f (xf , uf )+S0 −(k −T2)f (xf , uf )+S0 ≤ (T2 − T1 )f (xf , uf ) + L0 (1 + 2f (xf , uf )) + 2S0 .
(5.221)
320
5 ContinuousTime Autonomous Problems
By (5.221) and LSC property, extracting subsequences, using the diagonalization process and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 such that k1 ≥ k0 and for each integer i ≥ 0, there exists limp→∞ I f (i, i + 1, xkp , ukp ), and there exists (yi , vi ) ∈ X(i, i + 1) such that xkp (t) → yi (t) as p → ∞ for all t ∈ [i, i + 1],
(5.222)
I f (i, i + 1, yi , vi ) ≤ lim I f (i, i + 1, xkp , ukp ).
(5.223)
p→∞
In view of (5.222), there exists (x∗ , u∗ ) ∈ X(0, ∞) such that for each integer i ≥ 0, (x∗ (t), u∗ (t)) = (yi (t), vi (t)), t ∈ [i, i + 1]. It follows from (5.221), (5.223), and the definition of (x∗ , u∗ ) that for every integer q ≥ 1, I f (0, q, x∗ , u∗ ) ≤ lim I f (0, q, xkp , ukp ) p→∞
≤ qf (xf , uf ) + L0 (1 + 2f (xf , uf )) + 2S0 .
(5.224)
Theorem 5.1, the definition of (x∗ , u∗ ), (5.219), (5.222), and (5.224), imply that (x∗ , u∗ ) is (f )good and x∗ (0) = z.
(5.225)
In order to complete the proof of the proposition, it is sufficient to show that (x∗ , u∗ ) is (f )minimal. Assume the contrary. Then there exist Δ > 0, an integer τ0 ≥ 1 and (y, v) ∈ X(0, τ0 ) such that y(0) = x∗ (0), y(τ0 ) = x∗ (τ0 ), I f (0, τ0 , x∗ , u∗ ) > I f (0, τ0 , y, v) + 2Δ.
(5.226)
By (A2) and Lemma 5.43, there exists δ > 0 such that the following properties hold: (ii) for each ξ ∈ A satisfying ξ − xf ≤ δ, there exist τ1 , τ2 ∈ (0, bf ] and (x˜1 , u˜ 1 ) ∈ X(0, τ1 ), (x˜2 , u˜ 2 ) ∈ X(0, τ2 ) such that x˜1 (0) = ξ , x˜1 (τ1 ) = xf , I f (0, τ1 , x˜1 , u˜ 1 ) ≤ τ1 f (xf , uf ) + Δ/8, x˜ 2 (0) = xf , x˜2 (τ2 ) = ξ, I f (0, τ2 , x˜2 , u˜ 2 ) ≤ τ2 f (xf , uf ) + Δ/8;
5.16 Auxiliary Results for Theorems 5.19 and 5.20
321
(iii) for each (x, u) ∈ X(0, bf ) which satisfies x(0)−xf ≤ δ, x(bf )−xf ≤ δ, we have I f (0, bf , x, u) ≥ bf f (xf , uf ) − Δ/8. Theorem 5.1, (TP), (5.219) and (5.220) imply that there exists an integer L1 > L0 such that for each integer k ≥ k0 + 2L1 , xk (t) − xf ≤ δ, t ∈ [L1 , k − L1 ].
(5.227)
In view of (5.222), (5.227) and the definition of x∗ , x∗ (t) − xf ≤ δ for all t ≥ L1 .
(5.228)
By (5.223) and the definition of (x∗ , u∗ ), there exists a natural number p0 such that kp0 > k0 + 2L1 + 2τ0 + 2 + 2bf , I f (0, τ0 + L1 , x∗ , u∗ ) ≤ I f (0, τ0 + L1 , xkp0 , ukp0 ) + Δ/2.
(5.229) (5.230)
Property (ii) and (5.226)–(5.229) imply that there exists (x, u) ∈ X(0, kp0 ) such that (x(t), u(t)) = (y(t), v(t)), t ∈ [0, τ0 ], (x(t), u(t)) = (x∗ (t), u∗ (t)), t ∈ (τ0 , τ0 + L1 ], x(τ0 +L1 +bf ) = xf , I f (τ0 +L1 , τ0 +L1 +bf , x, u) ≤ bf f (xf , uf )+Δ/8, (x(t), u(t)) = (xkp0 (t), ukp0 (t)), t ∈ [τ0 + L1 + 2bf , kp0 ], I f (τ0 + L1 + bf , τ0 + L1 + 2bf , x, u) ≤ bf f (xf , uf ) + Δ/8. It follows from the relations above, (5.219), (5.225), (5.226), and property (iii) that I f (0, kp0 , x, u) ≥ I f (0, kp0 , xkp0 , ukp0 ). By the relation above, (5.226), (5.227), (5.229), (5.230), property (iii), and the choice of (x, u),
322
5 ContinuousTime Autonomous Problems
0 ≤ I f (0, kp0 , x, u) − I f (0, kp0 , xkp0 , ukp0 ) = I f (0, τ0 + L1 + 2bf , x, u) − I f (0, τ0 + L1 + 2bf , xkp0 , ukp0 ) = I f (0, τ0 , y, v) + I f (τ0 , τ0 + L1 , x∗ , u∗ ) + 2bf f (xf , uf ) + Δ/4 −I f (0, τ0 + L1 , xkp0 , ukp0 ) − I f (τ0 + L1 , τ0 + L1 + bf , xkp0 , ukp0 ) −I f (τ0 + L1 + bf , τ0 + L1 + 2bf , xkp0 , ukp0 ) ≤ I f (0, τ0 , y, v) + I f (τ0 , τ0 + L1 , x∗ , u∗ ) + 2bf f (xf , uf ) + Δ/4 −I f (0, τ0 + L1 , xkp0 , ukp0 ) − 2bf f (xf , uf ) + Δ/4 ≤ I f (0, τ0 , x∗ , u∗ ) − 2Δ + I f (τ0 , τ0 + L1 , x∗ , u∗ ) +Δ/2 − I f (0, τ0 + L1 , xkp0 , ukp0 ) ≤ −Δ, a contradiction. The contradiction we have reached completes the proof of Proposition 5.44.
5.17 Proofs of Theorems 5.19 and 5.20 Proof of Theorem 5.20. Clearly, (i) implies (ii). In view of Theorem 5.1, (ii) implies (iii). By (P1), (iii) implies (iv). Evidently (iv) implies (v). We show that (v) implies (iii). Assume that (x∗ , u∗ ) ∈ X(0, ∞) is (f )minimal and satisfies lim inf x∗ (t) − xf = 0.
(5.231)
lim x(t) ˜ − xf = 0.
(5.232)
t →∞
(P1) implies that t →∞
In view of Theorem 5.1, there exists S0 > 0 such that for all numbers T > 0, ˜ u) ˜ − Tf (xf , uf ) ≤ S0 . I f (0, T , x,
(5.233)
By (A2), there exists δ > 0 such that the following property holds: (a) for each z ∈ A satisfying z − xf ≤ δ, there exist τ1 , τ2 ∈ (0, bf ] and (xi , ui ) ∈ X(0, τi ), i = 1, 2 which satisfy
5.17 Proofs of Theorems 5.19 and 5.20
323
x1 (0) = z, x1 (τ1 ) = xf , x2 (0) = xf , x2 (τ2 ) = z, I f (0, τi , xi , ui ) ≤ τi f (xf , uf ) + 1, i = 1, 2. In view of (5.231) and (5.232), there exists an increasing sequence {tk }∞ k=1 ⊂ (0, ∞) such that lim tk = ∞, x∗ (tk + 2bf ) − xf < δ for all integers k ≥ 1,
k→∞
x(t) ˜ − xf ≤ δ for all t ≥ t0 .
(5.234) (5.235)
Let k ≥ 1 be an integer. By (a), (5.234) and (5.235), there exists (y, v) ∈ X(0, tk + 2bf ) such that (y(t), v(t)) = (x(t), ˜ u(t)), ˜ t ∈ [0, tk ], y(tk + bf ) = xf , I f (tk , tk + bf , y, v) ≤ bf f (xf , uf ) + 1, y(tk + 2bf ) = x∗ (tk + 2bf ), I f (tk + bf , tk + 2bf , y, v) ≤ bf f (xf , uf ) + 1. The relations above and (5.233) imply that I f (0, tk + 2bf , x∗ , u∗ ) ≤ I f (0, tk + 2bf , y, v) ≤ I f (0, tk , x, ˜ u) ˜ + 2bf f (xf , uf ) + 2 ≤ tk f (xf , uf ) + S0 + 2bf f (xf , uf ) + 2 = (tk + 2bf )f (xf , uf ) + 2S0 . Together with Theorem 5.1, this implies that (x∗ , u∗ ) is (f )good and (iii) holds. We show that (iii) implies (i). Assume that (x∗ , u∗ ) ∈ X(0, ∞) is (f )minimal and (f )good. (P1) implies that lim x∗ (t) − xf = 0.
t →∞
(5.236)
Clearly, there exists S0 > 0 such that I f (0, T , x∗ , u∗ ) − Tf (xf , uf ) ≤ S0 for all T > 0.
(5.237)
Let (x, u) ∈ X(0, ∞) satisfy x(0) = x∗ (0).
(5.238)
324
5 ContinuousTime Autonomous Problems
We show that lim sup[I f (0, T , x∗ , u∗ ) − I f (0, T , x, u)] ≤ 0. T →∞
(5.239)
In view of Theorem 5.1, we may assume that (x, u) is (f )good. (P1) implies that lim x(t) − xf = 0.
t →∞
(5.240)
Let > 0. By (A2) and Lemma 5.43, there exists δ ∈ (0, ) such that the following properties hold: (b) for each z ∈ A satisfying z − xf ≤ δ, there exist τ1 , τ2 ∈ (0, bf ] and (xi , ui ) ∈ X(0, τi ), i = 1, 2 satisfying x1 (0) = z , x1 (τ1 ) = xf , x2 (0) = xf , x2 (τ2 ) = z, I f (0, τi , xi , ui ) ≤ τi f (xf , uf ) + /8, i = 1, 2; (c) for each (y, v) ∈ X(0, bf ) which satisfies y(0) − xf ≤ δ, y(bf ) − xf ≤ δ, we have I f (0, bf , y, v) ≥ bf f (xf , uf ) − /8. It follows from (5.236) and (5.240) that there exists τ0 > 0 such that x(t) − xf ≤ δ, x∗ (t) − xf ≤ δ for all t ≥ τ0 .
(5.241)
Let T ≥ τ0 . Property (b) and (5.241) imply that there exists (y, v) ∈ X(0, T + 2bf ) which satisfies (y(t), v(t)) = (x(t), u(t)), t ∈ [0, T ],
(5.242)
y(T + bf ) = xf ,
(5.243)
I f (T , T + bf , y, v) ≤ bf f (xf , uf ) + /8, y(T + 2bf ) = x∗ (T + 2bf ), I f (T + bf , T + 2bf , y, v) ≤ bf f (xf , uf ) + /8.
(5.244) (5.245) (5.246)
By property (c) and (5.241), I f (T , T + bf , x∗ , u∗ ), I f (T + bf , T + 2bf , x∗ , u∗ ) ≥ bf f (xf , uf ) − /8. (5.247)
5.18 Proofs of Theorems 5.21–5.23
325
It follows from (5.238), (5.242), and (5.244)–(5.246) that I f (0, T , x∗ , u∗ ) + 2bf f (xf , uf ) − /4 ≤ I f (0, T + 2bf , x∗ , u∗ ) ≤ I f (0, T + 2bf , y, v) ≤ I f (0, T , x, u) + 2bf f (xf , uf ) + /4, I f (0, T , x∗ , u∗ ) ≤ I f (0, T , x, u) + /2 for all T ≥ τ0 . Since is any positive number, we conclude that (5.239) holds and (x∗ , u∗ ) is (f )overtaking optimal. Thus (iii) implies (i) and Theorem 5.20 is proved. Theorem 5.20 and Proposition 5.44 imply Theorem 5.19.
5.18 Proofs of Theorems 5.21–5.23 Proof of Theorem 5.21. Proposition 5.18, Lemma 5.41, TP, and (A2) imply that there exists δ ∈ (0, ) such that the following properties hold: (a) for each z ∈ A satisfying z − xf ≤ δ, there exist τ0 ∈ (0, bf ] and (x0 , u0 ) ∈ X(0, τ0 ) such that x0 (0) = z, x0 (τ0 ) = xf , I f (0, τ0 , x0 , u0 ) ≤ τ0 f (xf , uf ) + 1; (b) for each T1 ≥ 0, each T2 ≥ T1 + 2bf , and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf ≤ δ, i = 1, 2, I f (T1 , T2 , y, v) ≤ U f (T1 , T2 , y(T1 ), y(T2 )) + δ, we have y(t) − xf ≤ , t ∈ [T1 , T2 ]. Assertion (i) follows from (a) and Theorem 5.19. We prove assertion (ii). Assume that (x, u) ∈ X(0, ∞) is (f )overtaking optimal and x(0) − xf ≤ δ.
(5.248)
Properties (P1) and (a) and (5.248) imply that lim x(t) − xf = 0.
t →∞
(5.249)
326
5 ContinuousTime Autonomous Problems
By (5.249), there exists τ0 > 0 such that for each t ≥ τ0 , x(t) − xf ≤ δ.
(5.250)
Let T ≥ τ0 + 2bf . It follows from property (b), (5.248), and (5.250) that x(t) − xf ≤ , t ∈ [0, T ]. Theorem 5.21 is proved. Proof of Theorem 5.22. By Proposition 5.18, f has (P2). In view of Theorem 5.13, TP holds. Together with Proposition 5.10, this implies that there exists τ0 > 0 such that the following property holds: (c) for each T ≥ 2τ0 and each (x, u) ∈ X(0, T ) which satisfies I f (0, T , x, u) = U f (0, T , x(0), x(T )), I f (0, T , x, u) ≤ Tf (xf , uf ) + L(1 + f (xf , uf )) + 2 we have x(t) − xf ≤ , t ∈ [τ0 , T − τ0 ]. Let (x, u) ∈ X(0, ∞) be (f )overtaking optimal and x(0) ∈ AL .
(5.251)
Property (P1) and (5.251) imply that lim x(t) − xf = 0.
t →∞
(5.252)
In view of (5.251), there exist S0 ∈ (0, L] and (y, v) ∈ X(0, ∞) such that I f (0, S0 , y, v) ≤ L, y(0) = x(0), y(S0 ) = xf , y(t) = xf , t ∈ [S0 , ∞), v(t) = uf , t ∈ (S0 , ∞). By the relations above, for all sufficiently large T > L, I f (0, T , x, u) ≤ I f (0, T , y, v) + 1 ≤ 1 + L + I f (S0 , T , y, v) ≤ 1 + L + (T − S0 )f (xf , uf ) ≤ Tf (xf , uf ) + 1 + L(1 + f (xf , uf )). (5.253)
5.18 Proofs of Theorems 5.21–5.23
327
It follows from (5.253) and property (c) that for all sufficiently large T > L + 2τ0 , x(t) − xf ≤ , t ∈ [τ0 , T − τ0 ]. Since the relation above holds for all sufficiently large numbers T , we conclude that x(t) − xf ≤ , t ∈ [τ0 , ∞). This completes the proof of Theorem 5.22. Proof of Theorem 5.23. We may assume without loss of generality that t1 = 0. Clearly, there exists (y, v) ∈ X(0, ∞) such that (y(t), v(t)) = (x(t), u(t)), t ∈ [0, t2 ],
(5.254)
(y(t + t2 ), v(t + t2 )) = (y(t), v(t)) for all t ≥ 0.
(5.255)
Theorem 5.1, (5.254) and (5.255) imply that I f (0, t2 , y, v) ≥ t2 f (xf , uf ).
(5.256)
We show that I f (0, t2 , y, v) = t2 f (xf , uf ). Assume the contrary. Then in view of (5.254), I f (0, t2 , x, u) = I f (0, t2 , y, v) > t2 f (xf , uf ).
(5.257)
Set Δ = I f (0, t2 , x, u) − t2 f (xf , uf ) > 0.
(5.258)
By (A2) and Lemma 5.43, there exists δ ∈ (0, 1) such that the following properties hold: (d) for each z ∈ A satisfying z − xf ≤ δ, there exist τ1 , τ2 ∈ (0, bf ], and (xi , ui ) ∈ X(0, τi ), i = 1, 2 satisfying x1 (0) = z, x1 (τ1 ) = xf , x2 (0) = xf , x2 (τ2 ) = z, I f (0, τi , xi , ui ) ≤ τi f (xf , uf ) + Δ/8, i = 1, 2;
328
5 ContinuousTime Autonomous Problems
(e) for each (z, ξ ) ∈ X(0, bf ) which satisfies z(0) − xf ≤ δ, z(bf ) − xf ≤ δ, we have I f (0, bf , z, ξ ) ≥ bf f (xf , uf ) − Δ/8. It follows from (P1) that there exists T0 > 0 such that x(t) − xf ≤ δ for all t ≥ τ0 .
(5.259)
T1 > 2T0 + 2t2 + 2bf + 4.
(5.260)
Choose
By (d), (5.259), and (5.260), there exists ( x , u) ∈ X(0, ∞) such that ( x (t), u(t)) = (x(t + t2 ), u(t + t2 )), t ∈ [0, T0 + t2 + 2],
(5.261)
I f (T0 + t2 + 2, T0 + t2 + 2 + bf , x , u) ≤ bf f (xf , uf ) + Δ/8,
(5.262)
x (T0 + t2 + bf + 2) = xf ,
(5.263)
( x (t), u(t)) = (xf , uf ), t ∈ [T0 + t2 + 2 + bf , T0 + 2t2 + 2 + bf ],
(5.264)
x (T0 + 2t2 + 2bf + 2) = x(T0 + 2t2 + 2bf + 2),
(5.265)
I f (T0 + 2t2 + 2 + bf , T0 + 2t2 + 2 + 2bf , x , u) ≤ bf f (xf , uf ) + Δ/8, (5.266) ( x (t), u(t)) = (x(t), u(t)) for all t ≥ T0 + 2t2 + 2 + 2bf . (5.267) In view of (5.261) and (5.265), x , u). I f (0, T0 + 2t2 + 2 + 2bf , x, u) ≤ I f (0, T0 + 2t2 + 2 + 2bf ,
(5.268)
It follows from (5.261), (5.262), (5.264), and (5.266) that x , u) ≤ I f (t2 , T0 +2t2 +2, x, u)+t2f (xf , uf )+Δ/8 I f (0, T0 +2t2 +2+2bf , +f (xf , uf )bf + bf f (xf , uf ) + Δ/8 = I f (t2 , T0 + 2t2 + 2, x, u) + (t2 + 2bf )f (xf , uf ) + Δ/4. Property (e), (5.258), and (5.259) imply that
(5.269)
5.19 Proof of Proposition 5.27
329
I f (0, T0 + 2t2 + 2 + 2bf , x, u) = I f (0, t2 , x, u) + I f (t2 , T0 + 2t2 + 2, x, u) +I f (T0 +2t2 +2, T0 +2t2 +2+bf , x, u)+I f (T0 +2t2 +2+bf , T0 +2t2 +2+2bf , x, u) ≥ t2 f (xf , uf ) + Δ + I f (t2 , T0 + 2t2 + 2, x, u) + 2bf f (xf , uf ) − Δ/4.
By (5.269) and the relations above, I f (0, T0 +2t2 +2bf +2, x , u)−I f (0, T0 +2t2 +2bf +2, x, u) ≤ Δ/4−Δ+Δ/4. This contradicts (5.268). The contradiction we have reached proves that I f (0, t2 , y, v) = t2 f (xf , uf ). Theorem 5.1, (5.255), and the relation above imply that (y, v) is (f )good. By (P1), lim y(t) = xf .
t →∞
Together with (5.254) and (5.255), this implies that y(t) = xf for all t ≥ 0 and that x(t) = xf for all t ∈ [0, t2 ]. Combined with Theorem 5.21, this implies that x(t) = xf for all t ≥ 0. Theorem 5.23 is proved.
5.19 Proof of Proposition 5.27 Assume that (x, u) ∈ X(0, ∞) is (f )good. Theorem 5.1 implies that there exists M1 > 0 such that x(t) ≤ M1 , t ∈ [0, ∞),
(5.270)
I f (0, t, x, u) − tf (xf , uf ) < M1 , t ∈ (0, ∞). By the relation above, for all S > T ≥ 0, I f (T , S, x, u) − (S − T )f (xf , uf ) ≤ 2M1 .
(5.271)
Proposition 1.3 imply that there exist M ≥ 1, ω ∈ R 1 such that eAt ≤ Meωt , t ∈ [0, ∞) ∗
and in the case of the second class of problems eA t ≤ Meωt , t ∈ [0, ∞). (5.272)
330
5 ContinuousTime Autonomous Problems
Set 1 = (Meω )−1 .
(5.273)
We show that lim x(t) = xf .
t →∞
Assume the contrary. Then there exists ∈ (0, 1) and a sequence of numbers {tk }∞ k=1 such that t1 ≥ 8, tk+1 ≥ tk + 8 for all integers k ≥ 1, x(tk ) − xf > for all integers k ≥ 1.
(5.274) (5.275)
From now we consider the two classes of problems separately and begin with the first class of problems. By (5.9) and (5.10), there exists h0 > 0 such that f (x, u) ≥ 41−1 (M1 + a0 + 8)(G(x, u) − a0 x+ )
(5.276)
for each (x, u) ∈ M satisfying G(x, u) − a0 x ≥ h0 . There exists δ > 0 such that δ(f (xf , uf ) + a0 + 1) < 1,
(5.277)
δ ∈ (0, 1 (4a0 + 4h0 + 4 + 4a0 M1 )−1 ),
(5.278)
Meω δ < 1 /16,
(5.279)
eAt xf − xf < 1 /16 for all t ∈ [0, δ].
(5.280)
Let k ≥ 1 be an integer. We show that for all t ∈ [tk − δ, tk ], x(t) − xf ≥ δ.
(5.281)
Assume the contrary. Then there exists τ ∈ [tk − δ, tk ]
(5.282)
x(τ ) − xf < δ.
(5.283)
such that
5.19 Proof of Proposition 5.27
331
By (5.5),
x(tk ) = eA(tk −τ ) x(τ ) +
tk
eA(tk −s) G(x(s), u(s))ds.
(5.284)
τ
It follows from (5.272), (5.279), (5.280), and (5.282)–(5.284) that x(tk ) − xf = e
A(tk −τ )
tk
x(τ ) +
eA(tk −s) G(x(s), u(s))ds − xf
τ
≤ eA(tk −τ ) x(τ ) − eA(tk −τ ) xf + eA(tk −τ ) xf − xf
+Meωδ
tk
G(x(s), u(s))ds
τ
≤ Me
ωδ
x(τ ) − xf + e
A(tk −τ )
xf − xf + Me
ωδ
tk
G(x(s), u(s))ds
τ
≤ Me
ω
δ + e
A(tk −τ )
xf − xf + Me
ω
tk
G(x(s), u(s))ds
τ
< /16 + /16 + Meω
tk
G(x(s), u(s))ds.
(5.285)
τ
Set Ω1 = {t ∈ [τ, tk ] : G(x(t), u(t)) − a0 x(t) ≥ h0 }, Ω2 = [τ, tk ] \ Ω1 .
(5.287)
By (5.270), (5.271), (5.276)–(5.278), (5.282), (5.286) and (5.287),
tk
tk
G(x(s), u(s))ds ≤ a0
τ
x(s)ds
τ
tk
+ τ
(G(x(s), u(s)) − a0 x(s))+ ds
≤ a0 δM1 + h0 δ +
(5.286)
(G(x(s), u(s)) − a0 x(s))ds Ω1
≤ a0 δM1 + h0 δ + 4−1 1 (M1 + a0 + 8)−1
f (x(s), u(s)ds Ω1
332
5 ContinuousTime Autonomous Problems
≤ a0 δM1 + h0 δ + (I f (τ, tk , x, u) + a0 δ)(4−1 1 (M1 + a0 + 8)−1 ) ≤ a0 δM1 + h0 δ + (4−1 1 (M1 + a0 + 8)−1 )(2M1 + δf (xf , uf ) + a0 δ) ≤ δ(a0 M1 + h0 ) + 4−1 1 (M1 + a0 + 8)−1 (2M1 + 2) ≤ 1 /4 + 1 /2.
(5.288)
In view of (5.279), (5.285), and (5.288), x(tk ) − xf < /16 + /16 + Meω 31 /4 ≤ /8 + 3/4 < . This contradicts (5.275). The contradiction we have reached proves that (5.281) holds for all t ∈ [tk − δ, tk ]. This implies that mes({t ∈ [0, T ] : x(t) − xf ≥ δ}) → ∞ as T → ∞. On the other hand, property (P3) and (5.271) imply that there exists L > 0 such that for all T > 0, mes({t ∈ [0, T ] : x(t) − xf ≥ δ}) ≤ L. The contradiction we have reached proves that limt →∞ x(t) = xf and (P1) holds. Therefore for the first class of problems Proposition 5.27 is proved. Consider the second class of problems. Recall that for each τ ≥ 0, Φτ ∈ L(L2 (0, τ ; F ), E) is defined by Φτ u =
τ
eA(τ −s) Bu(s)ds, u ∈ L2 (0, τ ; F ).
(5.289)
0
Lemma 5.43 and (A2) imply that there exists δ ∈ (0, 1) such that M∗ eω∗  δ < /8, 32(a0 + f (xf , uf ) + 1)δ < K0 2 (Φ1 + 1)−1 , eAs xf − xf < /8 for all s ∈ [0, δ],
(5.290) (5.291) (5.292)
and the following properties hold: (a) for each zi ∈ A, i = 1, 2 satisfying zi − xf ≤ δ, i = 1, 2 there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ ) satisfying x(0) = z1 , x(τ ) = z2 , I f (0, τ, x, u) ≤ τf (xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 ;
5.19 Proof of Proposition 5.27
333
(b) for each T ≥ bf and each (y, v) ∈ X(0, T ) which satisfies y(0) − xf ≤ δ, y(T ) − xf ≤ δ, we have I f (0, T , y, v) ≥ Tf (xf , uf ) − 32−1 2 K0 (Φ1 + 1)−2 (recall K0 in (5.11)). Property (P3) and (5.271) imply that there exists L > 0 such that for all T > 0, mes({t ∈ [0, T ] : x(t) − xf ≥ δ}) ≤ L. Therefore there exists T0 > 0 such that for each S > T0 , mes({t ∈ [T0 , S] : x(t) − xf > δ}) ≤ δ/2.
(5.293)
Since the pair (x, u) is (f )good, it follows from Theorem 5.1 that there exists T1 > 0 such that for each S > T1 , I f (T1 , S, x, u) ≤ U f (T1 , S, x(T1 ), x(S)) + 16−1 2 K0 (Φ1 + 1)−2 .
(5.294)
Let k ≥ 1 be an integer such that tk > T0 + T1 + 1.
(5.295)
inf{x(t) − xf : t ∈ [tk − δ, tk ]} < δ.
(5.296)
τ ∈ [tk − δ, tk ]
(5.297)
x(τ ) − xf < δ.
(5.298)
Assume that
Therefore there exists
such that
It is clear that (x, u) ∈ X(τ, tk ) and in view of (5.7), x(tk ) = e
A(tk −τ )
x(τ ) +
tk
eA(tk −s) Bu(s)ds
τ
= eA(tk −τ ) x(τ ) + Φtk −τ u(τ + ·)
(5.299)
in E−1 = D(A∗ ) . It follows from (5.272), (5.275), (5.290), (5.292), (5.297)– (5.299), and Proposition 1.18 that
334
5 ContinuousTime Autonomous Problems
≤ x(tk ) − xf = eA(tk −τ ) x(τ ) + Φtk −τ u(τ + ·) − xf ≤ eA(tk −τ ) x(τ ) − eA(tk −τ ) xf + eA(tk −τ ) xf − xf + Φtk −τ u(τ + ·) ≤ eA(tk −τ ) x(τ ) − xf + eA(tk −τ ) xf − xf + Φtk −τ u(τ + ·)L2 (0,tk −τ ;F ) ≤ M∗ eω∗  δ + /8 + Φ1 u(τ + ·)L2 (0,tk −τ ;F )
tk
≤ /4 + Φ1 (
u(s)2 ds)1/2 .
(5.300)
τ
In view of (5.300),
tk
u(s)2 ds ≥ (2−1 (Φ1 + 1)−1 )2 .
(5.301)
τ
By (5.11), (5.297), and (5.301),
tk
tk
f (x(s), u(s))ds + a0 (tk − τ ) ≥ K0
τ
u(s)2 ds ≥ 4−1 2 K0 (Φ1 + 1)−2 ,
τ
tk
f (x(s), u(s))ds ≥ 4−1 K0 2 (Φ1 + 1)−2 − a0 δ.
(5.302)
τ
In view of (5.293) and (5.295), for each S > tk , mes({t ∈ [tk , S] : x(t) − xf > δ}) ≤ δ/2.
(5.303)
Inequality (5.303) implies that there exists S1 ∈ [tk , tk + δ]
(5.304)
x(S1 ) − xf ≤ δ.
(5.305)
such that
By (5.303) and (5.304), there exists S2 ∈ [S1 + 2bf , S1 + 2bf + δ]
(5.306)
x(S2 ) − xf ≤ δ.
(5.307)
such that
5.19 Proof of Proposition 5.27
335
It follows from (5.11), (5.297), (5.302), and (5.304) that I f (τ, S1 , x, u) = I f (τ, tk , x, u) + I f (tk , S1 , x, u) ≥ 4−1 K0 2 (Φ1 + 1)−2 − a0 δ − a0 δ.
(5.308)
Property (b) and (5.305)–(5.307) imply that I f (S1 , S2 , x, u) ≥ (S2 − S1 )f (xf , uf ) − 32−1 K0 2 (Φ1 + 1)−2 .
(5.309)
By (5.304), (5.306), (5.308), and (5.309), I f (τ, S2 , x, u) ≥ (S2 − S1 )f (xf , uf ) − 32−1 K0 2 (Φ1 + 1)−2 +4−1 K0 2 (Φ1 + 1)−2 − 2a0 δ ≥ K0 2 (Φ1 +1)−2 (4−1 −32−1 )−2a0 δ+(S2 −τ )f (xf , uf )−(S1 −τ )f (xf , uf ) ≥ K0 2 (Φ1 + 1)−2 (4−1 − 32−1 ) + (S2 − τ )f (xf , uf ) − 2δ(a0 + f (xf , uf )). (5.310) Property (a), (5.298), (5.304), (5.306), and (5.307) imply that that there exists (y, v) ∈ X(τ, S2 ) such that y(τ ) = x(τ ), y(τ + bf ) = xf , I f (τ, τ + bf , y, v) ≤ bf f (xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 ,
(5.311)
y(S2 ) = x(S2 ), y(S2 − bf ) = xf , I f (S2 − bf , S2 , y, v) ≤ bf f (xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 ,
(5.312)
(y(t), v(t)) = (xf , uf ), t ∈ [τ + bf , S2 − bf ].
(5.313)
In view of (5.311)–(5.313), I f (τ, S2 , y, v) ≤ (S2 − τ )f (xf , uf ) + 16−1 2 K0 (Φ1 + 1)−2 . Together with (5.291) and (5.310), the relation above implies that I f (τ, S2 , x, u) − I f (τ, S2 , y, v) ≥ 2 K0 (Φ1 + 1)−2 (4−1 − 16−1 − 32−1 ) − 2δ(a0 + f (xf , uf )) ≥ 2 K0 (Φ1 + 1)−2 (4−1 − 8−1 ) = 8−1 2 K0 (Φ1 + 1)−2 .
(5.314)
336
5 ContinuousTime Autonomous Problems
Since tk > T1 + 1 it follows from (5.295) and (5.297) that τ > T1 . In view of the choice of T1 (see (5.294)), I f (τ, S2 , x, u) ≤ U f (τ, S2 , x(τ ), x(S2 )) + 16−1 2 K0 (Φ1 + 1)−2 . This contradicts (5.311), (5.312), and (5.314). The contradiction we have reached proves that (5.296) is not true and inf{x(t) − xf : t ∈ [tk − δ, tk ]} ≥ δ for each natural number k satisfying tk ≥ T0 + T1 + 1. This implies that mes({t ∈ [0, T ] : x(t) − xf ≥ δ}) → ∞ as T → ∞. This contradicts (5.293). The contradiction we have reached proves that limt →∞ x(t) = xf and (P1) holds for the second class of problems too. Proposition 5.27 is proved.
5.20 Proof of Theorem 5.28 In view of Theorem 5.26, it is sufficient to show that fr has (P3). Let , M > 0. Theorem 5.1 implies that there exists M0 > 0 such that for each S2 > S1 ≥ 0 and each (x, u) ∈ X(S1 , S2 ), I f (S1 , S2 , x, u) ≥ (S2 − S1 )f (xf , uf ) − M0 .
(5.315)
There exists δ > 0 such that if z ∈ E satisfies φ(z) ≤ δ, then z ≤ .
(5.316)
L = δ −1 r −1 (M0 + M).
(5.317)
Set
Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ) satisfies I fr (T1 , T2 , x, u) ≤ (T2 − T1 )fr (xf , uf ) + M.
(5.318)
By (5.318) and the definition of fr , I f (T1 , T2 , x, u) + r
T2 T1
φ(x(t) − xf ))dt ≤ (T2 − T1 )f (xf , uf ) + M.
(5.319)
5.21 Auxiliary Results
337
In view of (5.315) and (5.319), (T2 − T1 )f (xf , uf ) − M0 + r
T2
φ(x(t) − xf )dt ≤ (T2 − T1 )f (xf , uf ) + M,
T1
T2
φ(x(t) − xf )dt ≤ r −1 (M0 + M).
(5.320)
T1
It follows from the choice of δ, (5.316), (5.317), and (5.320) that mes({t ∈ [T1 , T2 ] : x(t) − xf ≥ }) ≤ mes({t ∈ [T1 , T2 ] : φ(x(t) − xf ) > δ}) ≤δ
−1
T2
φ(x(t) − xf )dt ≤ δ −1 r −1 (M0 + M) ≤ L.
T1
Thus fr has (P3). Theorem 5.28 is proved.
5.21 Auxiliary Results We suppose that all the assumptions used in Section 5.6 hold and that (P1) and (P2) hold. By Theorem 5.1, there exists S∗ > 0 such that for each pair of numbers T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) ≥ (T2 − T1 )f (xf , uf ) − S∗ .
(5.321)
For all z ∈ A \ ∪{AL : L ∈ (0, ∞)}, set π f (z) = ∞.
(5.322)
z ∈ ∪{AL : L ∈ (0, ∞)}.
(5.323)
Let
Define π f (z) = inf{lim inf(I f (0, T , x, u)−Tf (xf , uf )) : (x, u) ∈ X(0, ∞), x(0) = z}. T →∞
(5.324)
By (5.321), (5.323) and (5.324), − S∗ ≤ π f (z) < ∞. There exists a L > 0 such that
(5.325)
338
5 ContinuousTime Autonomous Problems
z ∈ AL .
(5.326)
In view of (5.326), there exist (x, ˜ u) ˜ ∈ X(0, ∞) and τ ∈ (0, L] such that ˜ u) ˜ ≤ L, x(0) ˜ = z, x(τ ˜ ) = xf , I f (0, τ, x, (x(t), ˜ u(t)) ˜ = (xf , uf ) for all t ≥ τ. By the relations above, for all T ≥ τ , ˜ u) ˜ − Tf (xf , uf ) I f (0, T , x, ≤ τ + (T − τ )f (xf , uf ) − Tf (xf , uf ) ≤ L(1 + f (xf , uf )) and π f (z) ≤ L(1 + f (xf , uf )) for all z ∈ AL .
(5.327)
Definition (5.324) implies the following result. Proposition 5.45. 1. Let S > T ≥ 0 and (x, u) ∈ X(T , S) satisfy π f (x(T )), π f (x(S)) < ∞. Then π f (x(T )) ≤ I f (T , S, x, u) − (S − T )f (xf , uf ) + π f (x(S)).
(5.328)
2. Let (x, u) ∈ X(0, ∞) be an (f )good pair. Then for each pair of numbers S > T > 0, (5.328) holds. Proposition 5.46. π f (xf ) = 0. Proof. In view of (5.324), π f (xf ) ≤ 0.
(5.329)
π f (xf ) < 0.
(5.330)
Assume that
There exists (x0 , u0 ) ∈ X(0, ∞) such that x0 (0) = xf ,
(5.331)
5.21 Auxiliary Results
339
lim inf[I f (0, T , x0 , u0 ) − Tf (xf , uf )] < 2−1 π f (xf ). T →∞
(5.332)
It follows from (5.332) that there exists a strictly increasing sequence of positive numbers {Tk }∞ k=1 such that Tk → ∞ as k → ∞, I f (0, Tk , x0 , u0 ) − Tk f (xf , uf ) < 2−1 π f (xf ), k = 1, 2, . . . .
(5.333)
By (A2), there exists δ ∈ (0, 1) such that the following property holds: (i) for each z ∈ A, satisfying z − xf ≤ δ, there exist τ1 , τ2 ∈ (0, bf ] and (yi , vi ) ∈ X(0, τi ), i = 1, 2 which satisfy y1 (0) = z, y1 (τ1 ) = xf , I f (0, τ1 , y1 , v1 ) ≤ τ1 f (xf , uf ) − π f (xf )/8, y2 (0) = xf , y2 (τ2 ) = z, I f (0, τ2 , y2 , v2 ) ≤ τ2 f (xf , uf ) − π f (xf )/8. Theorem 5.1 and (5.333) imply that (x0 , u0 ) is (f )good. In view of (P1), lim x0 (t) − xf = 0.
t →∞
Choose a natural number k such that x0 (Tk ) − xf < δ.
(5.334)
Property (i), (5.331), and (5.334) imply that there exists (y, v) ∈ X(0, ∞) such that (y(t + Tk + bf ), v(t + Tk + bf )) = (y(t), v(t)) for all numbers t ≥ 0,
(5.335)
(y(t), v(t)) = (x0 (t), u0 (t)), t ∈ [0, Tk ],
(5.336)
y(Tk + bf ) = xf ,
(5.337)
I f (Tk , Tk + bf , y, v) ≤ bf f (xf , uf ) − π f (xf )/8. By (5.333), (5.336), and (5.338), I f (0, Tk + bf , y, v) = I f (0, Tk , y, v) + I f (Tk , Tk + bf , y, v) < Tk f (xf , uf ) + 2−1 π f (xf ) + bf f (xf , uf ) − π f (xf )/8 ≤ (Tk + bf )f (xf , uf ) + 4−1 π f (xf ).
(5.338)
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5 ContinuousTime Autonomous Problems
Combined with (5.335) this implies that lim inf[I f (0, T , y, v) − Tf (xf , uf )] = −∞, T →∞
a contradiction which proves that π f (xf ) = 0 and Proposition 5.46 itself. Proposition 5.47. The function π f : A → R 1 ∪ {∞} is finite in a neighborhood of xf and continuous at xf . Proof. Let > 0. By (A2), there exists δ > 0 such that the following property holds: (ii) for each z ∈ A satisfying z−xf ≤ δ there exist τ1 , τ2 ∈ (0, bf ] and (xi , ui ) ∈ X(0, τi ), i = 1, 2 which satisfy x1 (0) = z, x1 (τ1 ) = xf , x2 (0) = xf , x2 (τ2 ) = z, I f (0, τi , xi , ui ) ≤ τi f (xf , uf ) + /8, i = 1, 2. Let z ∈ A satisfy z − xf ≤ δ,
(5.339)
and let τ1 , τ2 ∈ (0, bf ] and (xi , ui ) ∈ X(0, τi ), i = 1, 2 be as guaranteed by property (ii). Set (x˜1 (t), u˜ 1 (t)) = (x1 (t), u1 (t)) for all t ∈ [0, τ1 ],
(5.340)
(x˜1 (t), u˜ 1 (t)) = (xf , uf ) for all t > τ1 .
(5.341)
Clearly, (x˜1 , u˜ 1 ) ∈ X(0, ∞). Property (ii), (5.340), and (5.341) imply that π f (z) ≤ lim inf[I f (0, T , x˜1 , u˜ 1 ) − Tf (xf , uf )] T →∞
= I f (0, τ1 , x1 , u1 ) − τ1 f (xf , uf ) ≤ /8. Propositions 5.45 and 5.46 and property (ii) imply that 0 = π f (xf ) ≤ I f (0, τ2 , x2 , u2 ) − τ2 f (xf , uf ) + π f (z) ≤ /8 + π f (z), π f (z) ≥ −/8. Proposition 5.47 is proved.
5.21 Auxiliary Results
341
Proposition 5.48. For each M > 0 the set {z ∈ A : π f (z) ≤ M} is bounded, Proof. Let M > 0 and suppose that the set {z ∈ A : π f (z) ≤ M} is nonempty. Propositions 5.46 and 5.47 imply that there exists δ > 0 such that for each z ∈ A satisfying z − xf ≤ δ, π f (z) is finite and π f (z) ≤ 1.
(5.342)
By Theorem 5.14, there exists L0 > 0 such that the following property holds: (iii) for each T ≥ L0 , each (x, u) ∈ X(0, T ) which satisfies I f (0, T , x, u) ≤ Tf (xf , uf ) + M + 4, and each S ∈ [0, T − L0 ], we have min{x(t) − xf : t ∈ [S, S + L0 ]} ≤ δ. Theorem 5.32 implies that there exists M1 > 0 such that the following property holds: (iv) for each S ∈ [1, L0 + 1] and each (x, u) ∈ X(0, S) which satisfies I f (0, S, x, u) ≤ (1 + L0 )f (xf , uf ) + M + 4, we have x(t) ≤ M1 , t ∈ [0, S]. Assume that z ∈ A satisfies π f (z) ≤ M.
(5.343)
In view of (5.324) and (5.343), there exists (x, u) ∈ X(0, ∞) such that x(0) = z, lim inf[I f (0, T , x, u) − Tf (xf , uf )] < π f (z) + 1 ≤ M + 1. T →∞
(5.344) (5.345)
By property (iii) and (5.345), there exists t0 ∈ [1, L0 + 1] such that x(t0 ) − xf ≤ δ.
(5.346)
342
5 ContinuousTime Autonomous Problems
It follows from (5.346) and the choice of δ (see (5.342)) that π f (x(t0 )) ≤ 1.
(5.347)
Proposition 5.45, (5.343)–(5.345), and (5.347) imply that π f (x(0)) ≤ I f (0, t0 , x, u) − t0 f (xf , xf ) + π f (x(t0 )) ≤ lim inf[I f (0, T , x, u) − Tf (xf , uf )] < π f (z) + 1 ≤ M + 1. T →∞
(5.348)
In view of (5.347) and (5.348), I f (0, t0 , x, u) ≤ t0 f (xf , uf ) + M + 2. Together with property (iv), this implies that x(t) ≤ M1 for all t ∈ [0, t0 ].
(5.349)
By (5.344) and (5.349), z = x(0) ≤ M1 . Proposition 5.48 is proved. Corollary 5.49. π f (z) → ∞ as z ∈ A and z → ∞. Let T2 > T1 ≥ 0 and (x, u) ∈ X(T1 , T2 ) satisfy π f (x(T1 )) < ∞. Define Γ f (T1 , T2 , x, u) = I f (T1 , T2 , x, u)−(T2 −T1 )f (xf , uf )−π f (x(T1 ))+π f (x(T2 )). (5.350) Proposition 5.45, (5.322), (5.325), and (5.350) implies that 0 ≤ Γ f (T1 , T2 , x, u) ≤ ∞.
(5.351)
Proposition 5.50. Let (x, u) ∈ X(0, ∞) and x(0) ∈ ∪{AL : L ∈ (0, ∞)}. Then (x, u) is (f )good if and only if sup{Γ f (0, T , x, u) : T ∈ (0, ∞)} = lim Γ f (0, T , x, u) < ∞. T →∞
If (x, u) is (f )good, then lim [I f (0, T , x, u) − Tf (xf , uf )] = π f (x(0)) + lim Γ f (0, T , x, u).
T →∞
T →∞
Proof. Assume that (x, u) ∈ X(0, ∞) is (f )good. Propositions 5.46, 5.47, (P1), (5.350), and (5.351) imply that
5.21 Auxiliary Results
343
lim x(t) = xf , lim π f (x(t)) = 0,
(5.352)
π f (x(t)) < ∞ for all t ≥ 0
(5.353)
t →∞
t →∞
and sup{Γ f (0, T , x, u) : T ∈ (0, ∞)} = lim Γ f (0, T , x, u) T →∞
≤ lim (I f (0, T , x, u) − Tf (xf , uf ) − π f (x(0)) < ∞. T →∞
Assume that Δ := lim Γ f (0, T , x, u) < ∞. T →∞
It follows from (5.325), (5.327), (5.350), (5.351), and (5.353) that for all T > 0, π f (x(T )) < ∞ and I f (0, T , x, u) − Tf (xf , uf ) = Γ f (0, T , x, u) +π f (x(0)) − π f (x(T )) ≤ Δ + π f (x(0)) + S∗ and (x, u) is (f )good. Assume that (x, u) is (f )good. By Propositions 5.46 and 5.47, (P1), (5.350) and (5.352), for all T > 0, I f (0, T , x, u) − Tf (xf , uf ) = Γ f (0, T , x, u) + π f (x(0)) − π f (x(T )) → lim Γ f (0, T , x, u) + π f (x(0)). T →∞
Proposition 5.50 is proved. Corollary 5.51. Let z ∈ A satisfy π f (z) < ∞ and > 0. Then there exists an (f )good (x, u) ∈ X(0, ∞) such that x(0) = z and lim Γ f (0, T , x, u) ≤ .
T →∞
Set inf(π f ) = inf{π f (z) : z ∈ A}.
(5.354)
Proposition 5.46, (5.325), and (5.354) imply that inf(π f ) is finite. Set Af = {z ∈ A : π f (z) ≤ inf(π f ) + 1}.
(5.355)
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5 ContinuousTime Autonomous Problems
Proposition 5.52. There exists M0 > 0 such that Af ⊂ AM0 . Proof. Proposition 5.48, (5.354), and (5.355) imply that there exists M1 > 0 such that Af ⊂ {z ∈ A : z ≤ M1 }.
(5.356)
By Theorem 5.6 and Lemma 5.37, there exists M2 > 0 such that for each T > 0 and each (x, u) ∈ X(0, T ) satisfying x(0) ≤ M1 , I f (0, T , x, u) ≤ Tf (xf , uf ) +  inf(π f ) + 3, we have x(t) ≤ M2 , t ∈ [0, T ].
(5.357)
Theorem 5.1 implies that there exists c1 > 0 such that for each T > 0 and each (x, u) ∈ X(0, T ), I f (0, T , x, u) ≥ Tf (xf , uf ) − c1 .
(5.358)
In view of (A2), there exists δ ∈ (0, 1) such that the following property holds: (v) for each z ∈ A satisfying z − xf ≤ δ, there exist τ1 ∈ (0, bf ] and (x1 , u1 ) ∈ X(0, τ1 ) which satisfies x1 (0) = z, x1 (τ1 ) = xf , I f (0, τ1 , x1 , u1 ) ≤ τ1 f (xf , uf ) + 1. By Theorem 5.14, there exists L0 > 0 such that the following property holds: (vi) for each T1 ≥ 0, each T2 ≥ T1 + L0 , and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) − (T2 − T1 )f (xf , uf ) ≤  inf(π f ) + 3 + c1 , there exists t0 ∈ [0, L0 ] such that x(t0 ) − xf ≤ δ. Set M0 = (L0 + bf )(1 + f (xf , uf )) +  inf(π f ) + 4 + c1 . Let z ∈ Af . In view of (5.355) and (5.356),
(5.359)
5.21 Auxiliary Results
345
π f (z) ≤ inf(π f ) + 1, z ≤ M1 ,
(5.360)
and by Proposition 5.50, there exists (x, u) ∈ X(0, ∞) such that x(0) = z,
(5.361)
lim [I f (0, T , x, u) − Tf (xf , uf )]
T →∞
= lim Γ f (0, T , x, u) + π f (z) ≤ π f (z) + 1 ≤ inf(π f ) + 2. T →∞
(5.362)
By the relation above, there exists a number T0 such that for all T ≥ T0 , I f (0, T , x, u) − Tf (xf , uf ) ≤ inf(π f ) + 3.
(5.363)
It follows from the choice of M2 (see (5.357)), (5.360), (5.361), and (5.363) that x(t) ≤ M1 for all t ≥ 0.
(5.364)
It follows from (5.358) and (5.363) that for any T ∈ (0, T0 ), I f (0, T , x, u) − Tf (xf , uf ) = I f (0, T0 , x, u) − T0 f (xf , uf ) −(I f (T , T0 , x, u) − (T0 − T )f (xf , uf )) ≤ inf(π f ) + 3 + c1 . Therefore for all T > 0, I f (0, T , x, u) − Tf (xf , uf ) ≤ inf(π f ) + 3 + c1 .
(5.365)
Property (vi) and (5.365) imply that there exists a number t0 such that t0 ∈ [0, L0 ], (x(t0 ) − xf ≤ δ.
(5.366)
It follows from property (v) and (5.366) that there exists (y, v) ∈ X(0, t0 + bf ) such that (y(t), v(t)) = (x(t), u(t)), t ∈ [0, t0 ], y(t0 + bf ) = xf , I f (t0 , t0 + bf , y, v) ≤ bf f (xf , uf ) + 1. In view of (5.359) and (5.366), t0 + bf ≤ L0 + bf ≤ M0 . By (5.359) and (5.365)–(5.368),
(5.367) (5.368)
346
5 ContinuousTime Autonomous Problems
I f (0, t0 + bf , y, v) ≤ t0 f (xf , uf ) + inf(π f ) + 3 + c1 + bf f (xf , uf ) + 1 ≤ f (xf , uf )(L0 + bf ) +  inf(π f ) + 4 + c1 ≤ M0 . Proposition 5.52 is proved. Proposition 5.53. Let M0 > 0 be an integer. Then the set AM0 is bounded. Proof. Let z ∈ AM0 . There exists (x, u) ∈ X(0, ∞) such that x(0) = z, x(t) = xf , u(t) = uf for all t ≥ M0 , I f (0, M0 , x, u) ≤ M0 + M0 f (xf , uf ). By the relations above, π f (z) ≤ lim inf[I f (0, T , x, u) − Tf (xf , uf )] T →∞
≤ M0 (1 + f (xf , uf )) + M0 f (xf , uf ). Thus AM0 ⊂ {z ∈ A : π f (z) ≤ 2M0 (1 + f (xf , uf ))} which is bounded by Proposition 5.48. This completes the proof of Proposition 5.53. Let z ∈ ∪{AL : L ∈ (0, ∞)}. Denote by Λ(f, z) the set of all (f )overtaking optimal pairs (x, u) ∈ X(0, ∞) such that x(0) = z. If f has LSC property, then in view of Theorem 5.19, Λ(f, z) = ∅, and by Theorem 5.1, any element of Λ(f, z) is (f )good. Equation (5.324) implies the following result. Proposition 5.54. Let f have LSC property. Then for every z ∈ ∪{AL : L ∈ (0, ∞)} and every (x, u) ∈ Λ(f, z), π f (z) = lim inf[I f (0, T , x, u) − Tf (xf , uf )]. T →∞
The next result follows from Proposition 5.54. Proposition 5.55. Let f have LSC property and (x, u) ∈ X(0, ∞) be (f )overtaking optimal and (f )good. Then for each pair of numbers S > T ≥ 0, π f (x(T )) = I f (T , S, x, u) − (S − T )f (xf , uf ) + π f (x(S)). Proposition 5.56. Let f have LSC property and (x, u) ∈ X(0, ∞) be (f )overtaking optimal and (f )good. Then
5.21 Auxiliary Results
347
π f (x0 ) = lim [I f (0, T , x, u) − Tf (xf , uf )]. T →∞
Proof. Propositions 5.46, 5.47, 5.55, and (P1) imply that for all T > 0, I f (0, T , x, u) − Tf (xf , uf ) = π f (x(0)) − π f (x(T )) → π f (x(0)) as T → ∞. Proposition 5.56 is proved. Proposition 5.57. Let LSC property holds. Then π f : A → R 1 ∪ {∞} is a lower semicontinuous function on A. Proof. Let {xk }∞ k=1 ⊂ A, x ∈ A and x = lim xk . k→∞
We show that π f (x) ≤ lim infk→∞ π f (xk ). Extracting subsequences and reindexing, we may assume without loss of generality that there exists lim π f (xk ) < ∞
k→∞
(5.369)
and that π f (xk ) < ∞ for all integers k ≥ 1. By Theorem 5.19, for each integer k ≥ 1, there exists (f, A, G)overtaking optimal and (f, A, G)good (yk , vk ) ∈ X(0, ∞, A, G) such that yk (0) = xk .
(5.370)
Proposition 5.56 implies that for every integer k ≥ 1, π f (xk ) = lim [I f (0, T , yk , vk ) − Tf (xf , uf )]. T →∞
(5.371)
It follows from Proposition 5.55, (5.325), and (5.369) that for every integer T ≥ 1, the sequence {I f (0, T , yk , vk )}∞ k=1 is bounded. By LSC property, extracting subsequences, using the diagonalization process and reindexing, we may assume that there exists (y, v) ∈ X(0, ∞, A, G), and a subsequence (yik , vik ), k = 1, 2, . . . such that for every integer t ≥ 0, yk (t) → y(t) as k → ∞, and for every natural number T , there exists lim I f (T − 1, T , yik , vik )
k→∞
and
(5.372)
348
5 ContinuousTime Autonomous Problems
I f (T − 1, T , y, v) ≤ lim I f (T − 1, T , yik , vik ). k→∞
(5.373)
Let > 0. By Propositions 5.46 and 5.47, there exists δ > 0 such that for each ξ ∈ A satisfying ξ − xf ≤ δ, we have π f (ξ ) ≤ /2.
(5.374)
Since the pairs (yk , vk ), k = 1, 2, . . . are (f, A, G)overtaking optimal, it follows from (5.325), (5.369), (5.371), and Proposition 5.10 that there exists L0 > 0 such that for each integer k ≥ 1 and each t ≥ L0 , yk (t) − xf ≤ δ.
(5.375)
It follows from (5.375) and the choice of δ (see (5.374)) that for each integer k ≥ 1 and each integer T ≥ L0 , π f (yk (T )) ≤ /2
(5.376)
and in view of Proposition 5.55, I f (0, T , yk , vk ) = Tf (xf , uf ) + π f (yk (0)) − π f (yk (T )) ≤ Tf (xf , uf ) + π f (yk (0)) + /2.
(5.377)
In view of (5.369), (5.370), (5.372), (5.373), and (5.377), for each integer T ≥ L0 , I f (0, T , y, v) ≤ Tf (xf , uf ) + lim π f (yk (0)) + /2, k→∞
I f (0, T , y, v) − Tf (xf , uf ) ≤ lim π f (xk ) + /2. k→∞
(5.378)
By (5.370), (5.372), and (5.378), y(0) = x, π f (x) ≤ lim inf[I f (0, T , y, v) − Tf (xf , uf )] ≤ lim π f (xk ) + /2. T →∞
k→∞
Since is any positive number, this completes the proof of Proposition 5.57.
5.22 Structure of Solutions in the Regions Close to the Endpoints We suppose that all the assumptions of Sections 5.6 and 5.21 hold and use all the notations and definitions introduced in these sections. As we have already
5.22 Structure of Solutions in the Regions Close to the Endpoints
349
mentioned in Section 5.6, all the results obtained for the triplet (f, A, G) also hold for the triplet (f, −A, −G). Assume that (f, A, G) has properties (P1) and (P2). By Theorem 5.31, (f, −A − G) has (P1) and (P2) too. L : L ∈ (0, ∞)}, set For all z ∈ A \ ∪{A f
π− (z) = ∞, L : L ∈ (0, ∞)}, define and for all z ∈ ∪{A f
π− (z) = inf{lim inf(I f (0, T , x, u) − Tf (xf , uf )) : T →∞
(x, u) ∈ X(0, ∞, −A, −G), x(0) = z}.
(5.379)
f
Let T2 > T1 ≥ 0 and (x, u) ∈ X(T1 , T2 , −A, −G) satisfy π− (x(T1 )) < ∞. Define f
f
f
Γ− (T1 , T2 , x, u) = I f (T1 , T2 , x, u)−(T2 −T1 )f (xf , uf )−π− (x(T1 ))+π− (x(T2 )). (5.380) Analogously to (5.351) we have, f
0 ≤ Γ− (T1 , T2 , x, u) ≤ ∞.
(5.381)
In this section we state the results which describe the asymptotic behavior of approximate solutions in the regions close to the endpoints. The following two results are proved in Section 5.24. Theorem 5.58. Let L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies x(0) ∈ AM , I f (0, T , x, u) ≤ σ f (0, T , x(0)) + δ, the inequalities f
f
f
π− (x(T )) ≤ inf(π− ) + , Γ− (0, L0 , x, ¯ u) ¯ ≤ hold where x(t) ¯ = x(T − t), u(t) ¯ = u(T − t), t ∈ [0, L0 ]. Theorem 5.59. Let L0 > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies I f (0, T , x, u) ≤ σ f (0, T ) + δ, the inequalities π f (x(0)) ≤ inf(π f ) + , Γ f (0, L0 , x, u) ≤ ,
350
5 ContinuousTime Autonomous Problems f
f
f
π− (x(T )) ≤ inf(π− ) + , Γ− (0, L0 , x, ¯ u) ¯ ≤ hold where x(t) ¯ = x(T − t), u(t) ¯ = u(T − t), t ∈ [0, T ]. The following two results are proved in Section 5.25. Theorem 5.60. Let (f, A, G) have LSC property, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies x(0) ∈ AM , I f (0, T , x, u) ≤ σ f (0, T , x(0)) + δ, there exists an (f, −A, −G)overtaking optimal (x∗ , u∗ ) ∈ X(0, ∞, −A, −G) such that f
f
π− (x∗ (0)) = inf(π− ), x∗ (t) − x(T − t) ≤ , t ∈ [0, L0 ]. Theorem 5.61. Let (f, A, G) have LSC property, L0 > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies I f (0, T , x, u) ≤ σ f (0, T ) + δ, there exist an (f, A, G)overtaking optimal (x∗,1 , u∗,1 ) ∈ X(0, ∞, A, G) and an (f, −A, −G)overtaking optimal (x∗,2 , u∗,2 ) ∈ X(0, ∞, −A, −G) such that f
f
π f (x∗,1 (0)) = inf(π f ), π− (x∗,2 (0)) = inf(π− ), x(t) − x∗,1 (t) ≤ , x(T − t) − x∗,2 (t) ≤ , t ∈ [0, L0 ].
5.23 Auxiliary Results for Theorems 5.58–5.61 Since the results obtained for the triplet (f, A, G) also hold for the triplet (f, −A, −G), in view of Theorem 5.12, we have the following result. Proposition 5.62. Assume that L, M > 0 and that > 0. Then there exist δ > 0 and L0 > L such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , and each (x, u) ∈ X(T1 , T2 , −A, −G) which satisfies L , x(T1 ) ∈ A f
f
I f (T1 , T2 , x, u) ≤ min{σ− (T1 , T2 , x(T1 )) + M, U− (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that
5.24 Proofs of Theorems 5.58 and 5.59
351
x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ then τ2 = T2 . Propositions 5.30 and 5.62 imply the following result. Proposition 5.63. Assume that L, M > 0 and that > 0. Then there exist δ > 0 and L0 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies L , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ min{( σ f (T1 , T2 , x(T2 )) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ, then τ2 = T2 .
5.24 Proofs of Theorems 5.58 and 5.59 We prove the following result. Theorem 5.64. Let L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies M , I f (0, T , x, u) ≤ σ f (0, T , x(T )) + δ, x(T ) ∈ A the inequalities π f (x(0)) ≤ inf(π f ) + , Γ f (0, L0 , x, u) ≤ hold. Applying Theorem 5.64 with the triplet (f, −A, −G) and using Proposition 5.30, we obtain Theorem 5.58. Theorems 5.58 and 5.64 and TP imply Theorem 5.59. Proof of Theorem 5.64. By Propositions 5.46, 5.47 and Lemma 5.43 and (A2), there exists δ1 ∈ (0, /4) such that: (i) for each z ∈ A satisfying z − xf ≤ 2δ1 , we have π f (z) ≤ /16; (ii) for each (x, u) ∈ X(0, bf , A, G) which satisfies x(0) − xf ≤ 2δ1 , x(bf ) − xf ≤ 2δ1 , we have I f (0, bf , x, u) ≥ bf f (xf , uf ) − /16;
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5 ContinuousTime Autonomous Problems
(iii) for each zi ∈ A, i = 1, 2 satisfying zi − xf ≤ 2δ1 , i = 1, 2, there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ, A, G) which satisfies x(0) = z1 , x(τ ) = z2 and I f (0, τ, x, u) ≤ τf (xf , uf ) + /16. By Proposition 5.63, there exist δ2 ∈ (0, δ1 /8) and l0 > 0 such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + 2l0 , and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies M , x(T2 ) ∈ A I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + δ2 , we have x(t) − xf ≤ δ1 , t ∈ [T1 + l0 , T2 − l0 ]. Clearly, there exists z∗ ∈ A such that π f (z∗ ) ≤ inf(π f ) + δ2 /8.
(5.382)
Corollary 5.51 and (5.382) imply that there exists an (f, A, G)good (x∗ , u∗ ) ∈ X(0, ∞, A, G) for which x∗ (0) = z∗ ,
(5.383)
lim Γ f (0, T , x∗ , u∗ ) ≤ δ2 /8.
(5.384)
T →∞
By (P1), there exists l1 > 0 such that x∗ (t) − xf ≤ δ2 /8 for all numbers t ≥ l1 .
(5.385)
Choose δ ∈ (0, δ2 /4)
(5.386)
L1 > 2L0 + 2l0 + 2l1 + 2bf + 8.
(5.387)
and
Assume that T ≥ L1 and that (x, u) ∈ X(0, T , A, G) satisfies
5.24 Proofs of Theorems 5.58 and 5.59
353
M , x(T ) ∈ A
(5.388)
I f (0, T , x, u) ≤ σ f (0, T , x(T )) + δ.
(5.389)
Property (iv), (5.382), (5.386), (5.387), and (5.389) imply that x(t) − xf ≤ δ1 , t ∈ [l0 , T − l0 ].
(5.390)
In view of (5.387), [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8] ⊂ [l0 , T − l0 − l1 − L0 ].
(5.391)
It follows from (5.390) and (5.391) that x(t) − xf ≤ δ1 , t ∈ [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8].
(5.392)
Property (iii), (5.385), and (5.392) imply that there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(0, T , A, G) such that x1 (t) = x∗ (t), u1 (t) = u∗ (t), t ∈ [0, l0 + l1 + L0 + 4], x1 (l0 + l1 + L0 + 4 + τ1 ) = xf ,
(5.393) (5.394)
I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + τ1 , x1 , u1 ) ≤ τ1 f (xf , uf ) + /16, (5.395) x1 (l0 + l1 + L0 + 4 + 2bf ) = x(l0 + l1 + L0 + 4 + 2bf ), (5.396) x1 (l0 + l1 + L0 + 4 + 2bf − τ2 ) = xf ,
(5.397)
I f (l0 +l1 +L0 +4+2bf −τ2 , l0 +l1 +L0 +4+2bf , x1 , u1 ) ≤ τ2 f (xf , uf )+/16, (5.398) x1 (t) = xf , u1 (t) = uf , t ∈ [l0 +l1 +L0 +4+τ1, l0 +l1 +L0 +4+2bf −τ2], (5.399) x1 (t) = x(t), u1 (t) = u(t), t ∈ [l0 + l1 + L0 + 4 + 2bf , T ]. (5.400)
It follows from (5.389) and (5.400) that I f (0, T , x, u) ≤ I f (0, T , x1 , u1 ) + δ.
(5.401)
By (5.393), (5.395), and (5.398)–(5.401), δ ≥ I f (0, T , x, u) − I f (0, T , x1 , u1 ) = I f (0, l0 + l1 + L0 + 4 + 2bf , x, u) − I f (0, l0 + l1 + L0 + 4 + 2bf , x1 , u1 )
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5 ContinuousTime Autonomous Problems
≥ I f (0, l0 + l1 + L0 + 4, x, u) + I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + bf , x, u) +I f (l0 +l1 +L0 +4+bf , l0 +l1 +L0 +4+2bf , x, u)−I f (0, l0 +l1 +L0 +4, x∗ , u∗ ) −τ1 f (xf , uf )−/16−τ2 f (xf , uf )−/16−f (xf , uf )(2bf −τ1 −τ2 ).
(5.402)
Property (ii) and (5.392) imply that I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + bf , x, u), I f (l0 + l1 + L0 + 4 + bf , l0 + l1 + L0 + 4 + 2bf , x, u) ≥ bf f (xf , uf ) − /16. (5.403) In view of (5.402) and (5.403), δ ≥ I f (0, l0 + l1 + L0 + 4, x, u) + 2bf f (xf , uf ) − /8 − I f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) − 2bf f (xf , uf ) − /8.
(5.404)
By property (i), (5.350), (5.382)–(5.386), (5.392), and (5.404), I f (0, l0 + l1 + L0 + 4, x, u) ≤ /4 + /16 + I f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) = 5/16 + Γ f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) + (l0 + l1 + L0 + 4)f (xf , uf ) +π f (x∗ (0)) − π f (x∗ (l0 + l1 + L0 + 4)) ≤ Γ f (0, l0 +l1 +L0 +4, x∗ , u∗ )+(l0 +l1 +L0 +4)f (xf , uf )+π f (x∗ (0))+6/16 ≤ δ2 /8 + (l0 + l1 + L0 + 4)f (xf , uf ) + π f (x∗ (0)) + 6/16 ≤ δ2 /8 + (l0 + l1 + L0 + 4)f (xf , uf ) + inf(π f ) + δ2 /8 + 6/16.
(5.405)
It follows from (5.350), (5.392), (5.405), and property (i) that inf(π f )+(l0 +l1 +L0 +4)f (xf , uf ) ≥ −3/8−δ2 /4+I f (0, l0 +l1 +L0 +4, x, u) ≥ −7/16 + Γ f (0, l0 + l1 + L0 + 4, x, u) + (l0 + l1 + L0 + 4)f (xf , uf ) +π f (x(0)) + π f (x(l0 + l1 + L0 + 4)) ≥ −7/16+Γ f (0, l0 +l1 +L0 +4, x, u)+(l0 +l1 +L0 +4)f (xf , uf )+π f (x0 )−/16. (5.406) By (5.351) and (5.406), π f (x0 ) + Γ f (0, l0 + l1 + L0 + 4, x, u) ≤ inf(π f ) + /2.
5.25 Proofs of Theorems 5.60 and 5.61
355
The inequality above implies that π f (x(0)) ≤ inf(π f ) + , Γ f (0, l0 + l1 + L0 + 4, x, u) ≤ .
Theorem 5.64 is proved.
5.25 Proofs of Theorems 5.60 and 5.61 We prove the following result. Theorem 5.65. Let (f, A, G) have LSC property, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0 and L1 > L0 such that for each T ≥ L1 and each (x, u) ∈ X(0, T , A, G) which satisfies M , I f (0, T , x, u) ≤ x(T ) ∈ A σ f (0, T , x(T )) + δ, there exists an (f, A, G)overtaking optimal (x∗ , u∗ ) ∈ X(0, ∞, A, G) such that π f (x∗ (0)) = inf(π f ), x(t) − x∗ (t) ≤ , t ∈ [0, L0 ]. Note that Theorem 5.60 follows from Theorem 5.65, applied for the triplet (f, −A, −G) and Proposition 5.30. Theorem 5.61 follows from Theorems 5.60, 5.65 and TP. Theorem 5.65 follows from Theorem 5.64 and the following result. Proposition 5.66. Let (f, A, G) have LSC property, L0 > 0 and ∈ (0, 1). Then there exist δ ∈ (0, ) such that for each (x, u) ∈ X(0, L0 , A, G) which satisfies π f (x(0)) ≤ inf(π f ) + δ, Γ f (0, L0 , x, u) ≤ δ, there exists an (f, A, G)overtaking optimal (x∗ , u∗ ) ∈ X(0, ∞, A, G) such that π f (x∗ (0)) = inf(π f ), x∗ (t) − x(t) ≤ , t ∈ [0, L0 ]. Proof. Assume that the proposition does not hold. Then there exist a sequence ∞ {δk }∞ k=1 ⊂ (0, 1] and a sequence {(xk , uk )}k=1 ⊂ X(0, L0 , A, G) such that lim δk = 0
(5.407)
π f (xk (0)) ≤ inf(π f ) + δk ,
(5.408)
Γ f (0, L0 , xk , uk ) ≤ δk ,
(5.409)
k→∞
and that for all integers k ≥ 1,
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5 ContinuousTime Autonomous Problems
and the following property holds: (i) for each (f, A, G)overtaking optimal pair (y, v) ∈ X(0, ∞, A, G) satisfying π f (y(0)) = inf(π f ), we have max{xk (t) − y(t) : t ∈ [0, L0 ]} > . In view of (5.407)–(5.409) and the boundedness from below of the function π f , the sequence {I f (0, L0 , xk , uk )}∞ k=1 is bounded. By LSC property, extracting a subsequence and reindexing if necessary, we may assume without loss of generality that there exists (x, u) ∈ X(0, L0 , A, G) such that xk (t) → x(t) as k → ∞ for all t ∈ [0, L0 ],
(5.410)
I f (0, L0 , x, u) ≤ lim inf I f (0, L0 , xk , uk ).
(5.411)
k→∞
The lower semicontinuity of π f , (5.408) and (5.410) implies that π f (x(0)) ≤ lim inf π f (xk (0)) = inf(π f ), k→∞
π f (x(0)) = inf(π f ).
(5.412)
By the lower semicontinuity of π f and (5.410), π f (x(L0 )) ≤ lim inf π f (xk (L0 )). k→∞
(5.413)
It follows from (5.350), (5.407)–(5.409), and (5.411)–(5.413) that I f (0, L0 , x, u) − L0 f (xf , uf ) − π f (x(0)) + π f (x(L0 )) ≤ lim inf[I f (0, L0 , xk , uk ) − L0 f (xf , uf )] k→∞
− lim π f (xk (0)) + lim inf π f (xk (L0 )) k→∞
k→∞
≤ lim inf[I f (0, L0 , xk , uk ) − L0 f (xf , uf ) − π f (xk (0)) + π f (xk (L0 ))] ≤ 0. k→∞
(5.414) In view of (5.350), (5.351), and (5.414), I f (0, L0 , x, u) − L0 f (xf , uf ) − π f (x(0)) + π f (x(L0 )) = 0.
(5.415)
Theorem 5.19, (5.412), and (5.415) imply that there exists an (f, A, G)overtaking optimal pair (x, ˜ u) ˜ ∈ X(0, ∞, A, G) such that x(0) ˜ = x(L0 ).
(5.416)
5.26 The First Bolza Problem
357
For all numbers t > L0 , set x(t) = x(t ˜ − L0 ), u(t) = u(t ˜ − L0 ).
(5.417)
It is clear that (x, u) ∈ X(0, ∞, A, G) is (f, A, G)good. By Proposition 5.55, (5.350), (5.351), and (5.415), for all S > 0, I f (0, S, x, u) − Sf (xf , uf ) − π f (x(0)) + π f (x(S)) = 0.
(5.418)
In view of (5.412), (5.418), and Theorem 5.20, (x, u) is (f, A, G)overtaking optimal pair satisfying π f (x(0)) = inf(π f ). By (5.410), for all sufficiently large natural numbers k, xk (t) − x(t) ≤ /2, t ∈ [0, L0 ]. This contradicts property (i). The contradiction we have reached proves Proposition 5.66.
5.26 The First Bolza Problem We use the notation, definitions, and assumptions introduced and used in Sections 5.6, 5.21, and 5.22. Let a1 > 0. Denote by A(E)) the set of all lower semicontinuous functions h : E → R 1 which are bounded on bounded subsets of E and such that h(z) ≥ −a1 for all z ∈ E.
(5.419)
The set A(E) is equipped with the uniformity which is determined by the base E(N, ) = {(h1 , h2 ) ∈ A(E) × A(E) : h1 (z) − h2 (z) ≤ for all z ∈ E satisfying z ≤ N},
(5.420)
where N, > 0. It is not difficult to see that the uniform space A(E) is metrizable and complete. For each pair of numbers T2 > T1 ≥ 0, each y ∈ A, and each h ∈ A(E), we define
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5 ContinuousTime Autonomous Problems
σ f,h (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) + h(x(T2 )) : (x, u) ∈ X(T1 , T2 , A, G), x(T1 ) = y},
(5.421)
σ f,h (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) + h(x(T1 )) : (x, u) ∈ X(T1 , T2 , A, G), x(T2 ) = y},
(5.422)
f,h
σ− (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) + h(x(T2 )) : (x, u) ∈ X(T1 , T2 , −A, −G), x(T1 ) = y},
(5.423)
f,h
σ− (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) + h(x(T1 )) : (x, u) ∈ X(T1 , T2 , −A − G), x(T2 ) = y}.
(5.424)
The next result follows from Proposition 5.29 and (5.35). Proposition 5.67. Let S2 > S1 ≥ 0, M ≥ 0, h ∈ A(E) and (x, u) ∈ X(S1 , S2 , A, G). Then the following assertions hold: I f (S1 , S2 , x, u) + h(x(S2 )) ≤ σ f,h (S1 , S2 , x(S1 )) + M if and only if f,h
I f (S1 , S2 , x, ¯ u) ¯ + h(x(S ¯ 1 )) ≤ σ− (S1 , S2 , x(S ¯ 2 )) + M; I f (S1 , S2 , x, u) + h(x(S1 )) ≤ σ f,h (S1 , S2 , x(S2 )) + M if and only if f,h
I f (S1 , S2 , x, ¯ u) ¯ + h(x(S ¯ 2 )) ≤ σ− (S1 , S2 , x(S ¯ 1 )) + M. The next result is proved in Section 5.27. Theorem 5.68. Assume that (f, A, G) has TP, M, L > 0 and that > 0. Then there exist δ > 0 and L0 > L such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , each h ∈ A(E) satisfying h(xf ) ≤ M and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies x(T1 ) ∈ AL , I f (T1 , T2 , x, u) + h(x(T2 )) ≤ σ f,h (T1 , T2 , x(T1 )) + δ there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ then τ2 = T2 .
5.26 The First Bolza Problem
359
Note that Theorem 5.68 is valid for the (f, h, −A, −G) too and combined with Proposition 5.67 this implies the following result. Theorem 5.69. Assume that (f, A, G) has TP, M, L > 0, and > 0. Then there exist δ > 0 and L0 > L such that for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , each h ∈ A(E) satisfying h(xf ) ≤ M, and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies L , x(T2 ) ∈ A I f (T1 , T2 , x, u) + h(x(T1 )) ≤ σ f,h (T1 , T2 , x(T2 )) + δ, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ, then τ1 = T1 , and if x(T2 ) − xf ≤ δ, then τ2 = T 2 . R1
Let g ∈ A(E) be given. By Corollary 5.49, (5.325), and (5.419), π f + g : A → ∪ {∞} is bounded from below and satisfies (π f + g)(z) → ∞ as z ∈ A, z → ∞.
(5.425)
We prove the following results which describe the asymptotic behavior of approximate solutions in the regions close to the endpoints. Theorem 5.70. Let (f, A, G) have TP, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies x(0) ∈ AM , I f (0, T , x, u) + h(x(T )) ≤ σ f,h (0, T , x(0)) + δ, the inequalities f
f
f
(g + π− )(x(T )) ≤ inf(g + π− ) + , Γ− (0, L0 , x, ¯ u) ¯ ≤ hold where x(t) ¯ = x(T − t), u(t) ¯ = u(T − t), t ∈ [0, T ]. Theorem 5.71. Let (f, A, G) have TP, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies M , I f (0, T , x, u) + h(x(0)) ≤ x(T ) ∈ A σ f,h (0, T , x(T )) + δ, the inequalities (g + π f )(x(0)) ≤ inf(g + π f ) + , Γ f (0, L0 , x, u) ≤ .
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5 ContinuousTime Autonomous Problems
Theorem 5.71 is proved in Section 5.28. It is also valid for (f, −A, −G). Together with Proposition 5.67, this implies Theorem 5.70. Theorem 5.72. Let (f, A, G) have TP and LSC property, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies x(0) ∈ AM , I f (0, T , x, u) + h(x(T )) ≤ σ f,h (0, T , x(0)) + δ, there exists an (f, −A, −G)overtaking optimal pair (x∗ , u∗ ) ∈ X(0, ∞, −A, −G) such that f
f
(π− + g)(x∗ (0)) = inf(π− + g), x∗ (t) − x(T − t) ≤ , t ∈ [0, L0 ]. Theorem 5.73. Let (f, A, G) have TP and LSC property, L0 , M > 0 and ∈ (0, 1). Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies M , I f (0, T , x, u) + h(x(0)) ≤ σ f,h (0, T , x(T )) + δ, x(T ) ∈ A there exists an (f, A, G)overtaking optimal pair (x∗ , u∗ ) ∈ X(0, ∞, A, G) such that (π f + g)(x∗ (0)) = inf(π f + g), x∗ (t) − x(t) ≤ , t ∈ [0, L0 ]. We prove Theorem 5.73. It is also valid for (f, −A, −G). Together with Proposition 5.67, this implies Theorem 5.72. Theorem 5.73 follows from Theorem 5.71 and the following result which is proved in Section 5.29. Proposition 5.74. Let (f, A, G) have TP and LSC property, L0 > 0 and ∈ (0, 1). Then there exist δ ∈ (0, ) such that for each and each (x, u) ∈ X(0, L0 , A, G) which satisfies (g + π f )(x(0)) ≤ inf(π f + g) + δ, Γ f (0, L0 , x, u) ≤ δ, there exists an (f, A, G)overtaking optimal pair (x∗ , u∗ ) ∈ X(0, ∞, A, G) such that (π f + g)(x∗ (0)) = inf(π f + g), x∗ (t) − x(t) ≤ , t ∈ [0, L0 ].
5.27 Proof of Theorem 5.68
361
5.27 Proof of Theorem 5.68 By Proposition 5.10, there exist δ ∈ (0, 1) and L0 > L such that the following property holds: (i) for each T1 ≥ 0, each T2 ≥ T1 + 2L0 , and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies I f (T1 , T2 , x, u) ≤ min{(T2 −T1 )f (xf , uf )+a1 +M+a0 +2+2L(1+f (xf , uf )),
U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]; moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 , and if x(T2 ) − xf ≤ δ, then τ2 = T 2 . Assume that T1 ≥ 0, T2 ≥ T1 + 2L0 , h ∈ A(E) satisfies h(xf ) ≤ M,
(5.426)
and (x, u) ∈ X(T1 , T2 , A, G) satisfies x(T1 ) ∈ AL , I f (T1 , T2 , x, u) + h(x(T2 )) ≤ σ f,h (T1 , T2 , x(T1 )) + δ.
(5.427) (5.428)
By (5.428), I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(5.429)
In view of (5.427), there exists (y, v) ∈ X(T1 , T2 , A, G) such that y(T1 ) = x(T1 ),
(5.430)
y(t) = xf , v(t) = uf , t ∈ [T1 + L, T2 ],
(5.431)
I f (T1 , T1 + L, y, v) ≤ L + Lf (xf , uf ).
(5.432)
It follows from (5.29), (5.419), (5.428), (5.430), and (5.432) that −a1 +I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , x, u)+h(x(T2 )) ≤ 1+I f (T1 , T2 , y, v)+h(xf ) ≤ h(xf ) + 1 + L(1 + f (xf , uf )) + (T2 − T1 − L)f (xf , uf ),
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5 ContinuousTime Autonomous Problems
and together with (5.426), this implies that I f (T1 , T2 , x, u) ≤ a1 + 1 + M + 2L(1 + f (xf , uf )) + (T2 − T1 )f (xf , uf ). (5.433) In view of (5.429), (5.433), and property (i), there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ, then τ1 = T1 , and if x(T2 ) − xf ≤ δ, then τ2 = T2 . Theorem 5.68 is proved.
5.28 Proof of Theorem 5.71 By Propositions 5.46, 5.47 and Lemma 5.43 and (A2), there exists δ1 ∈ (0, /4) such that: (i) for each z ∈ A satisfying z − xf ≤ 2δ1 , we have π f (z) ≤ /16; (ii) for each (x, u) ∈ X(0, bf , A, G) which satisfies x(0) − xf ≤ 2δ1 , x(bf ) − xf ≤ 2δ1 , we have I f (0, bf , x, u) ≥ bf f (xf , uf ) − /16; (iii) for each zi ∈ A, i = 1, 2 satisfying zi − xf ≤ 2δ1 , i = 1, 2, there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ, A, G) which satisfies x(0) = z1 , x(τ ) = z2 and I f (0, τ, x, u) ≤ τf (xf , uf ) + /16. By (5.419), (5.420), and Corollary 5.49, there exists a neighborhood U1 of g in A(E) such that for each h ∈ U1 ,  inf(π f + g) − inf(π f + h) ≤ δ1 /16.
(5.434)
Theorem 5.69 implies that there exist δ2 ∈ (0, δ1 /8), l0 > M, and a neighborhood U2 of g in A(E) such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + 2l0 , each h ∈ U2 , and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies M , x(T2 ) ∈ A σ f,h (T1 , T2 , x(T2 )) + δ2 , I f (T1 , T2 , x, u) + h(x(T1 )) ≤
5.28 Proof of Theorem 5.71
363
we have x(t) − xf ≤ δ1 , t ∈ [T1 + l0 , T2 − l0 ]. Clearly, there exists z∗ ∈ A such that (π f + g)(z∗ ) ≤ inf(π f + g) + δ2 /8.
(5.435)
Corollary 5.51 implies that there exists an (f, A, G)good (x∗ , u∗ ) ∈ X(0, ∞, A, G) for which x∗ (0) = z∗ , lim Γ f (0, T , x∗ , u∗ ) ≤ δ2 /8. T →∞
(5.436)
Property (P1) implies that lim x∗ (t) − xf = 0.
t →∞
(5.437)
By (5.437), there exists l1 > 0 such that x∗ (t) − xf ≤ δ2 /8 for all numbers t ≥ l1 .
(5.438)
Theorem 5.32 implies that there exists M˜ > 0 such that the following property holds: (v) for each (x, u) ∈ X(0, l0 + l1 + L0 + 4, A, G) satisfying I f (0, l0 +l1 +L0 +4, x, u) ≤ (l0 +l1 +L0 +4)f (xf , uf )+a1 +2+a0 +inf(π f +g),
we have ˜ t ∈ [0, l0 + l1 + L0 + 4]. x(t) ≤ M, Proposition 5.53 implies that the set AM is bounded. In view of (5.420), there exists a neighborhood U of g in A(E) such that U ⊂ U1 ∩ U2 ,
(5.439)
h(x∗ (0)) − g(x∗ (0)) ≤ δ1 /16 for all h ∈ U,
(5.440)
h(z) − g(z) ≤ δ1 /16 for all h ∈ U and all z ∈ AM ,
(5.441)
˜ h(z) − g(z) ≤ δ1 /16 for all h ∈ U and all z ∈ E satisfying z ≤ M.
(5.442)
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5 ContinuousTime Autonomous Problems
Choose δ ∈ (0, δ2 /4),
(5.443)
L1 > 2L0 + 2l0 + 2l1 + 2bf + 8.
(5.444)
T ≥ L1 , h ∈ U,
(5.445)
(x, u) ∈ X(0, T , A, G),
(5.446)
M , x(T ) ∈ A
(5.447)
Assume that
I f (0, T , x, u) + h(x(0)) ≤ σ f,h (0, T , x(T )) + δ.
(5.448)
Property (iv), (5.439), (5.443)–(5.446), and (5.448) imply that x(t) − xf ≤ δ1 , t ∈ [l0 , T − l0 ].
(5.449)
In view of (5.444) and (5.445), [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8] ⊂ [l0 , T − l0 − l1 − L0 ].
(5.450)
It follows from (5.449) and (5.450) that x(t) − xf ≤ δ1 , t ∈ [l0 + l1 + L0 , l0 + l1 + L0 + 2bf + 8].
(5.451)
Property (iii), (5.438), (5.443), and (5.444), (5.451) imply that there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(0, T , A, G) such that x1 (t) = x∗ (t), u1 (t) = u∗ (t), t ∈ [0, l0 + l1 + L0 + 4], x1 (l0 + l1 + L0 + 4 + τ1 ) = xf ,
(5.452) (5.453)
I f (l0 +l1 +L0 +4, l0 +l1 +L0 +4+τ1, x1 , u1 ) ≤ τ1 f (xf , uf )+/16,
(5.454)
x1 (l0 + l1 + L0 + 4 + 2bf ) = x(l0 + l1 + L0 + 4 + 2bf ),
(5.455)
x1 (l0 + l1 + L0 + 4 + 2bf − τ2 ) = xf ,
(5.456)
I f (l0 +l1 +L0 +4+2bf −τ2 , l0 +l1 +L0 +4+2bf , x1 , u1 ) ≤ τ2 f (xf , uf )+/16, (5.457)
5.28 Proof of Theorem 5.71
365
x1 (t) = xf , u1 (t) = uf , t ∈ [l0 + l1 + L0 + 4 + τ1 , l0 + l1 + L0 + 4 + 2bf − τ2 ], (5.458) x1 (t) = x(t), u1 (t) = u(t), t ∈ [l0 + l1 + L0 + 4 + 2bf , T ]. (5.459) It follows from (5.448) and (5.459) that I f (0, T , x, u) + h(x(0)) ≤ I f (0, T , x1 , u1 ) + h(x1 (0)) + δ.
(5.460)
By (5.458)–(5.460), δ ≥ I f (0, T , x, u) + h(x(0)) − I f (0, T , x1 , u1 ) − h(x1 (0)) = I f (0, l0 + l1 + L0 + 4 + 2bf , x, u) + h(x(0)) −I f (0, l0 + l1 + L0 + 4 + 2bf , x1 , u1 ) − h(x1 (0)) = I f (0, l0 +l1 +L0 +4, x, u)+I f (l0 +l1 +L0 +4, l0 +l1 +L0 +4+2bf , x, u)+h(x(0))
−I f (0, l0 + l1 + L0 + 4, x1 , u1 ) − I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + τ1, x1 , u1 ) −f (xf , uf )(2bf − τ1 − τ2 ) −I f (l0 +l1 +L0 +4+2bf −τ2 , l0 +l1 +L0 +4+2bf , x1 , u1 )−h(x1 (0)).
(5.461)
Property (ii) and (5.451) imply that I f (l0 +l1 +L0 +4, l0 +l1 +L0 +4+2bf , x, u) ≥ 2bf f (xf , uf )−/8.
(5.462)
By property (i), (5.350), (5.435), (5.436), (5.438), (5.452), (5.454), (5.457), (5.461), and (5.462), I f (0, l0 + l1 + L0 + 4, x, u) + h(x(0)) ≤ δ + /4 + I f (0, l0 + l1 + L0 + 4, x1 , u1 ) + h(x1 (0)) = δ + /4 + I f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) + h(x∗ (0)) = δ + /4 + Γ f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) + (l0 + l1 + L0 + 4)f (xf , uf ) +π f (x∗ (0)) − π f (x∗ (l0 + l1 + L0 + 4)) + h(x∗ (0)) ≤ Γ f (0, l0 + l1 + L0 + 4, x∗ , u∗ ) + (l0 + l1 + L0 + 4)f (xf , uf )
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5 ContinuousTime Autonomous Problems
+π f (x∗ (0)) + /16 + h(x∗ (0)) + δ + /4 ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + δ2 /8 + δ + /4 + (π f + h)(x∗ (0)) + /16 ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + /64 + /64 + /4 + (π f + g)(x∗ (0)) + /16 ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + inf(π f + g) + δ2 /8 + /16 + /4 + /32 ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + inf(π f + g) + /4 + /16 + /16.
(5.463)
In view of (5.419) and (5.463), I f (0, l0 + l1 + L0 + 4, x, u) ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + inf(π f + g) + a1 + 1. (5.464) Property (v) and (5.464) imply that ˜ x(0) ≤ M. Together with (5.442) and (5.444), this implies that h(x(0)) − g(x(0)) ≤ δ1 /16 ≤ /16.
(5.465)
It follows from (5.340), (5.463), and (5.465) that I f (0, l0 + l1 + L0 + 4, x, u) + g(x(0)) ≤ (l0 + l1 + L0 + 4)f (xf , uf ) + inf(π f + g) + /4 + /8 + /16.
(5.466)
By (5.350) and (5.466), Γ f (0, l0 + l1 + L0 + 4, x, u) + π f (x(0)) + π f (x(l0 + l1 + L0 + 4)) + g(x(0)) ≤ inf(π f + g) + /4 + /4.
(5.467)
It follows from (5.451) and property (i) that π f (x(l0 + l1 + L0 + 4)) ≤ /16.
(5.468)
By (5.467) and (5.468), Γ f (0, l0 + l1 + L0 + 4, x, u) + (π f + g)(x(0)) ≤ inf(π f + g) + /2 + /16. This implies that
5.29 Proof of Proposition 5.74
367
Γ f (0, l0 + l1 + L0 + 4, x, u) ≤ , (π f + g)(x(0)) ≤ inf(π f + g) + 3/4. Theorem 5.71 is proved.
5.29 Proof of Proposition 5.74 Assume that the proposition does not hold. Then there exist a sequence {δk }∞ k=1 ⊂ (0, 1] and a sequence {(xk , uk )}∞ ⊂ X(0, L , A, G) such that 0 k=1 lim δk = 0
(5.469)
(π f + g)(xk (0)) ≤ inf(π f + g) + δk ,
(5.470)
Γ f (0, L0 , xk , uk ) ≤ δk
(5.471)
k→∞
and that for all integers k ≥ 1,
and the following property holds: (i) for each (f, A, G)overtaking optimal pair (y, v) ∈ X(0, ∞, A, G) satisfying (π f + g)(y(0)) = inf(π f + g), we have sup{xk (t) − y(t) : t ∈ [0, L0 ]} > . In view of (5.419), (5.469), and (5.470), the sequence {π f (xk (0))}∞ k=1 is bounded. Together with Corollary 5.49, this implies that the sequence {xk (0)}∞ k=1 is bounded. By (5.325), (5.350), (5.419), (5.469), and (5.471), the sequence {I f (0, L0 , xk , uk )}∞ k=1 is bounded. By LSC property, extracting a subsequence and reindexing if necessary, we may assume without loss of generality that there exists (x, u) ∈ X(0, L0 , A, G) such that xk (t) → x(t) as k → ∞ for all t ∈ [0, L0 ],
(5.472)
I f (0, L0 , x, u) ≤ lim inf I f (0, L0 , xk , uk ).
(5.473)
k→∞
Proposition 5.57, (5.470), and (5.472) imply that the function π f + g is lower semicontinuous and that π f (x(0)) ≤ lim inf π f (xk (0)),
(5.474)
g(x(0)) ≤ lim inf g(xk (0)),
(5.475)
k→∞
k→∞
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5 ContinuousTime Autonomous Problems
(π f + g)(x(0)) ≤ lim inf(π f + g)(xk (0)) ≤ inf(π f + g), k→∞
(π f + g)(x(0)) = inf(π f + g) = lim (π f + g)(xk (0)). k→∞
(5.476)
In view of (5.474)–(5.476), g(x(0)) = lim g(xk (0)),
(5.477)
π f (x(0)) = lim π f (xk (0)).
(5.478)
k→∞
k→∞
By the lower semicontinuity of π f and (5.472), π f (x(L0 )) ≤ lim inf π f (xk (L0 )).
(5.479)
k→∞
It follows from (5.350), (5.351), (5.469), (5.471), (5.473), (5.478), and (5.479) that I f (0, L0 , x, u) − L0 f (xf , uf ) − π f (x(0)) + π f (x(L0 )) ≤ lim inf[I f (0, L0 , xk , uk )−L0 f (xf , uf )]− lim π f (xk (0))+lim inf π f (xk (L0 )) k→∞
k→∞
k→∞
≤ lim inf[I f (0, L0 , xk , uk ) − L0 f (xf , uf ) − π f (xk (0)) + π f (xk (L0 ))] ≤ 0. k→∞
In view of the relation above and (5.351), I f (0, L0 , x, u) − L0 f (xf , uf ) − π f (x(0)) + π f (x(L0 )) = 0.
(5.480)
Theorem 5.19 and (5.480) imply that there exists an (f, A, G)overtaking optimal pair (x, ˜ u) ˜ ∈ X(0, ∞, A, G) such that x(0) ˜ = x(L0 ).
(5.481)
˜ − L0 ). x(t) = x(t ˜ − L0 ), u(t) = u(t
(5.482)
For all numbers t > L0 , set
It is clear that (x, u) ∈ X(0, ∞, A, G) is (f, A, G)good. By Propositions 5.55, (5.350), (5.351), and (5.480), for all S > 0, I f (0, S, x, u) − Sf (xf , uf ) − π f (x(0)) + π f (x(S)) = 0.
5.30 The Second Bolza Problem
369
In view of the relation above and Theorem 5.20, (x, u) is (f, A, G)overtaking optimal pair satisfying (π f + g)(x(0)) = inf(π f + g). By (5.472), for all sufficiently large natural numbers k, xk (t) − x(t) ≤ /2, t ∈ [0, L0 ]. This contradicts property (i). The contradiction we have reached proves Proposition 5.74.
5.30 The Second Bolza Problem We use the notation, definitions, and assumptions introduced and used in Sections 5.6, 5.21, and 5.22. Let a1 > 0. Denote by A(E × E) the set of all lower semicontinuous functions h : E × E → R 1 which are bounded on bounded subsets of E × E and such that h(z1 , z2 ) ≥ −a1 for all z1 , z2 ∈ E.
(5.483)
The set A(E × E) is equipped with the uniformity which is determined by the base E(N, ) = {(h1 , h2 ) ∈ A(E × E) × A(E × E) : h1 (z1 , z2 ) − h2 (z1 , z2 ) ≤ for all z1 , z2 ∈ E satisfying zi ≤ N, i = 1, 2},
(5.484)
where N, > 0. It is not difficult to see that the uniform space A(E × E) is metrizable and complete. For each pair of numbers T2 > T1 ≥ 0 and each h ∈ A(E × E), we define σ f,h (T1 , T2 ) = inf{I f (T1 , T2 , x, u) + h(x(T1 ), x(T2 )) : (x, u) ∈ X(T1 , T2 , A, G)}. (5.485) The next result is proved in Section 5.31. Theorem 5.75. Assume that (f, A, G) has TP, M, > 0. Then there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, each h ∈ A(E × E) satisfying h(xf , xf ) ≤ M, and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies I f (T1 , T2 , x, u) + h(x(T1 ), x(T2 )) ≤ σ f,h (T1 , T2 ) + δ, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that
370
5 ContinuousTime Autonomous Problems
x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ then τ1 = T1 and if x(T2 ) − xf ≤ δ then τ2 = T2 . Let h ∈ A(E × E). Define f
ψh (z1 , z2 ) = π f (z1 ) + π− (z2 ) + h(z1 , z2 ), z1 , z2 ∈ A.
(5.486)
By (5.325), (5.483), and Corollary 5.49, ψh : A × A → R 1 ∪ {∞} is bounded from below and satisfies ψh (z1 , z2 ) → ∞ as z1 , z2 ∈ A, z1 + z2 → ∞.
(5.487)
Fix g ∈ A(E × E). In this section we state the following results which describe the asymptotic behavior of approximate solutions in the regions close to the endpoints. The next result is proved in Section 5.32. Theorem 5.76. Let (f, A, G) have TP, L0 > 0 and ∈ (0, 1). Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E × E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies I f (0, T , x, u) + h(x(0), x(T )) ≤ σ f,h (0, T ) + δ, the inequalities ψg (x(0), x(T )) ≤ inf(ψg ) + , f
¯ u) ¯ ≤ Γ f (0, L0 , x, u) ≤ , Γ− (0, L0 , x, hold where x(t) ¯ = x(T − t), u(t) ¯ = u(T − t), t ∈ [0, T ]. Theorem 5.77. Let (f, A, G) have TP and LSC property, L0 > 0 and > 0. Then there exist δ > 0, L1 > L0 , and a neighborhood U of g in A(E × E) such that for each T ≥ L1 , each h ∈ U, and each (x, u) ∈ X(0, T , A, G) which satisfies I f (0, T , x, u) + h(x(0), x(T ))) ≤ σ f,h (0, T ) + δ, there exists an (f, A, G)overtaking optimal pair (x∗,1 , u∗,1 ) ∈ X(0, ∞, A, G) and an (f, −A, −G)overtaking optimal pair (x∗,2 , u∗,2 ) ∈ X(0, ∞, −A, −G) such that ψg (x∗,1 (0), x∗,2 (0)) = inf(ψg ), x∗,1 (t) − x(t)) ≤ , t ∈ [0, L0 ], x∗,2 (t) − x(T − t) ≤ , t ∈ [0, L0 ].
5.31 Proof of Theorem 5.75
371
Theorem 5.77 follows from Theorem 5.76 and the following result which is proved in Section 5.33. Proposition 5.78. Let (f, A, G) have TP and LSC property, L0 > 0 and ∈ (0, 1). Then there exist δ > 0 such that for each (x1 , u1 ) ∈ X(0, L0 , A, G) and each (x2 , u2 ) ∈ X(0, L0 , −A, −G) which satisfy ψg (x1 (0), x2 (0)) ≤ inf(ψg ) + δ, f
Γ f (0, L0 , x1 , u1 ) ≤ δ, Γ− (0, L0 , x2 , u2 ) ≤ δ, there exists an (f, A, G)overtaking optimal pair (y1 , v1 ) ∈ X(0, ∞, A, G) and an (f, −A, −G)overtaking optimal pair (y2 , v2 ) ∈ X(0, ∞, −A, −G) such that ψg (y1 (0), y2 (0)) = inf(ψg ), xi (t) − yi (t) ≤ , t ∈ [0, L0 ], i = 1, 2.
5.31 Proof of Theorem 5.75 By Proposition 5.10, there exist δ ∈ (0, 1) and L > 0 such that the following property holds: (i) for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies I f (T1 , T2 , x, u) ≤ min{(T2 − T1 )f (xf , uf ) + a1 + M + a0 + 1, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]; moreover, if x(T1 ) − xf ≤ δ, then τ1 = T1 , and if x(T2 ) − xf ≤ δ, then τ2 = T 2 . Assume that T1 ≥ 0, T2 ≥ T1 + 2L, h ∈ A(E × E) satisfies h(xf , xf ) ≤ M,
(5.488)
and (x, u) ∈ X(T1 , T2 , A, G) satisfies I f (T1 , T2 , x, u) + h(x(T1 ), x(T2 )) ≤ σ f,h (T1 , T2 ) + δ.
(5.489)
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5 ContinuousTime Autonomous Problems
By (5.489), I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(5.490)
It follows from (5.483), (5.488), and (5.489) that −a1 + I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , x, u) + h(x(T1 ), x(T2 )) ≤ 1 + (T2 − T1 )f (xf , uf ) + h(xf , xf ), I f (T1 , T2 , x, u) ≤ a1 + 1 + M + (T2 − T1 )f (xf , uf ).
(5.491)
In view of (5.490), (5.491), and property (i), there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf ≤ δ, then τ1 = T1 , and if x(T2 ) − xf ≤ δ, then τ2 = T2 . Theorem 5.75. is proved.
5.32 Proof of Theorem 5.76 By Propositions 5.46 and 5.47, Lemma 5.43, and (A2), there exists δ1 ∈ (0, /4) such that: (i) for each z ∈ A satisfying z − xf ≤ 2δ1 , we have f
π f (z) ≤ /16, π− (z) ≤ /16; (ii) for each (x, u) ∈ X(0, bf , A, G) which satisfies x(0) − xf ≤ 2δ1 , x(bf ) − xf ≤ 2δ1 , we have I f (0, bf , x, u) ≥ bf f (xf , uf ) − /16; (iii) for each zi ∈ E, i = 1, 2 satisfying zi − xf ≤ 2δ1 , i = 1, 2, there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ, A, G) which satisfies x(0) = z1 , x(τ ) = z2 and I f (0, τ, x, u) ≤ τf (xf , uf ) + /16. By (5.483), (5.484), and Corollary 5.49, there exists a neighborhood U1 of g in A(E × E) such that for each h ∈ U1 ,
5.32 Proof of Theorem 5.76
373
 inf(ψg ) − inf(ψh ) ≤ δ1 /16.
(5.492)
Theorem 5.75 implies that there exist δ2 ∈ (0, δ1 /8), l0 > 0, and a neighborhood U2 of g in A(E × E) such that the following property holds: (iv) for each T1 ≥ 0, each T2 ≥ T1 + 2l0 , each h ∈ U2 and each (x, u) ∈ X(T1 , T2 , A, G) which satisfies I f (T1 , T2 , x, u) + h(x(T1 ), x(T2 )) ≤ σ f,h (T1 , T2 ) + δ2 , we have x(t) − xf ≤ δ1 , t ∈ [T1 + l0 , T2 − l0 ]. Clearly, there exists z∗,1 , z∗,2 ∈ A such that ψg (z∗,1 , z∗,2 ) ≤ inf(ψg ) + δ2 /8.
(5.493)
Corollary 5.51 and (5.493) imply that there exist an (f, A, G)good (x∗,1 , u∗,1 ) ∈ X(0, ∞, A, G) and an (f, −A, −G)good (x∗,2 , u∗,2 ) ∈ X(0, ∞, −A, −G) for which x∗,1 (0) = z∗,1 , x∗,2 (0) = z∗,2 ,
(5.494)
lim Γ f (0, T , x∗,1 , u∗,1 ) ≤ δ2 /8,
(5.495)
T →∞
f
lim Γ− (0, T , x∗,2 , u∗,2 ) ≤ δ2 /8,
T →∞
(5.496)
Property (P1), (5.495), and (5.496) imply that lim x∗,i (t) − xf = 0, i = 1, 2.
t →∞
(5.497)
By (5.497), there exists l1 > 0 such that x∗,i (t) − xf ≤ δ2 /8, i = 1, 2 for all numbers t ≥ l1 .
(5.498)
It follows from Corollary 5.49, (5.483), (5.484), and (5.487) that there exists a neighborhood U of g in A(E × E) such that U ⊂ U1 ∩ U2 , h(x∗,1 (0), x∗,2 (0)) − g(x∗,1 (0), x∗,2 (0)) ≤ δ1 /16 for all h ∈ U,
(5.499) (5.500)
374
5 ContinuousTime Autonomous Problems
and that for all h ∈ U and all (z1 , z2 ) ∈ A × A satisfying ψh (z1 , z2 ) ≤ inf(ψg ) + 4, we have ψh (z1 , z2 ) − ψg (z1 , z2 ) ≤ /64.
(5.501)
δ ∈ (0, δ2 /8),
(5.502)
L1 > 4L0 + 4l0 + 4l1 + 4bf + 16.
(5.503)
T ≥ L1 , h ∈ U,
(5.504)
Choose
Assume that
(x, u) ∈ X(0, T , A, G), I f (0, T , x, u) + h(x(0), x(T )) ≤ σ f,h (0, T ) + δ.
(5.505)
Property (iv), (5.499), (5.502)–(5.505), and (5.511) imply that x(t) − xf ≤ δ1 , t ∈ [T1 + l0 , T2 − l0 ].
(5.506)
Set x(t) ¯ = x(T − t), u(t) ¯ = u(T − t), t ∈ [0, T ].
(5.507)
Property (iii), (5.498), (5.503), (5.504), and (5.506) imply that there exist τ1 , τ2 , τ3 , τ4 ∈ (0, bf ] and (x1 , u1 ) ∈ X(0, T , A, G) such that x1 (t) = x∗,1 (t), u1 (t) = u∗,1 (t), t ∈ [0, l0 + l1 + L0 + 4], x1 (l0 + l1 + L0 + 4 + τ1 ) = xf ,
(5.508) (5.509)
I f (l0 +l1 +L0 +4, l0 +l1 +L0 +4+τ1, x1 , u1 ) ≤ τ1 f (xf , uf )+/16,
(5.510)
x1 (l0 + l1 + L0 + 4 + 2bf ) = x(l0 + l1 + L0 + 4 + 2bf ),
(5.511)
x1 (l0 + l1 + L0 + 4 + 2bf − τ2 ) = xf ,
(5.512)
5.32 Proof of Theorem 5.76
375
I f (l0 +l1 +L0 +4+2bf −τ2 , l0 +l1 +L0 +4+2bf , x1 , u1 ) ≤ τ2 f (xf , uf )+/16, (5.513) x1 (t) = xf , u1 (t) = uf , t ∈ [l0 + l1 + L0 + 4 + τ1 , l0 + l1 + L0 + 4 + 2bf − τ2 ], (5.514) x1 (T − t) = x∗,2 (t), u1 (T − t) = u∗,2 (t), t ∈ [0, l0 + l1 + L0 + 4], (5.515) x1 (T − l0 − l1 − L0 − 4 − τ3 ) = xf ,
(5.516)
I f (T − l0 − l1 − L0 − 4 − τ3 , T − l0 − l1 − L0 − 4, x1 , u1 ) ≤ τ3 f (xf , uf ) + /16, (5.517) x1 (T − l0 − l1 − L0 − 4 − 2bf ) = x(T − l0 − l1 − L0 − 4 − 2bf ), (5.518) x1 (T − l0 − l1 − L0 − 4 − 2bf + τ4 ) = xf ,
(5.519)
I f (T − l0 − l1 − L0 − 4 − 2bf , T − l0 − l1 − L0 − 4 − 2bf + τ4 , x1 , u1 ) ≤ τ4 f (xf , uf ) + /16,
(5.520)
x1 (t) = xf , u1 (t) = uf , t ∈ [T −l0 −l1 −L0 −4−2bf +τ4 , T −l0 −l1 −L0 −4−τ3], (5.521) x1 (t) = x(t), u1 (t) = u(t), t ∈ [l0 + l1 + L0 + 4 + 2bf , T − l0 − l1 − L0 − 4 − 2bf ]. (5.522) It follows from (5.505) that I f (0, T , x, u) + h(x(0), x(T )) ≤ I f (0, T , x1 , u1 ) + h(x1 (0), x1 (T )) + δ. (5.523) By (5.522) and (5.523), δ ≥ I f (0, T , x, u) + h(x(0), x(T )) − I f (0, T , x1 , u1 ) − h(x1 (0), x1 (T )) = I f (0, l0 + l1 + L0 + 4, x, u) + I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + bf , x, u) +I f (l0 + l1 + L0 + 4 + bf , l0 + l1 + L0 + 4 + 2bf , x, u) +I f (T − l0 − l1 − L0 − 4 − 2bf , T − l0 − l1 − L0 − 4 − bf , x, u) +I f (T − l0 − l1 − L0 − 4 − bf , T − l0 − l1 − L0 − 4, x, u) +I f (T − l0 − l1 − L0 − 4, T , x, u) + h(x(0), x(T )) −I f (0, l0 + l1 + L0 + 4, x1 , u1 ) − I f (l0 + l1 + L0 + 4, l0 + l1 + L0 + 4 + τ1, x1 , u1 ) −I f (l0 + l1 + L0 + 4 + τ1 , l0 + l1 + L0 + 4 + 2bf − τ2 , x1 , u1 )
376
5 ContinuousTime Autonomous Problems
−I f (l0 + l1 + L0 + 4 + 2bf − τ2 , l0 + l1 + L0 + 4 + 2bf , x1 , u1 ) −I f (T − l0 − l1 − L0 − 4 − 2bf , T − l0 − l1 − L0 − 4 − 2bf + τ4 , x1 , u1 ) −I f (T − l0 − l1 − L0 − 4 − 2bf + τ4 , T − l0 − l1 − L0 − 4 − τ3 , x1 , u1 ) −I f (T − l0 − l1 − L0 − 4 − τ3 , T − l0 − l1 − L0 − 4, x1 , u1 ) − I f (T − l0 − l1 − L0 − 4, T , x1 , u1 ) − h(x1 (0), x1 (T )).
(5.524)
Proposition 5.28, property (ii), (5.503), (5.504), (5.506), (5.508), (5.510), (5.513)– (5.515), (5.517), (5.520), (5.521), and (5.524) imply that δ ≥ I f (0, l0 + l1 + L0 + 4, x, u) + 2bf f (xf , uf ) − /8 + 2bf f (xf , uf ) − /8 +I f (0, l0 + l1 + L0 + 4, x, ¯ u) ¯ + h(x(0), x(0)) ¯ −I f (0, l0 +l1 +L0 +4, x∗,1, u∗,1 )−τ1 f (xf , uf )−/16−(2bf −τ1 −τ2 )f (xf , uf ) −τ2 f (xf , uf ) − /16 − τ4 f (xf , uf ) − /16 − (2bf − τ3 − τ4 )f (xf , uf ) −τ3 f (xf , uf ) − /16 − I f (0, l0 + l1 + L0 + 4, x∗,2 , u∗,2 ) − h(x∗,1 (0), x∗,2(0)) ≥ I f (0, l0 + l1 + L0 + 4, x, u) + I f (0, l0 + l1 + L0 + 4, x, ¯ u) ¯ + h(x(0), x(0)) ¯ −I f (0, l0 + l1 + L0 + 4, x∗,1 , u∗,1 ) − I f (0, l0 + l1 + L0 + 4, x∗,2 , u∗,2 ) − h(x∗,1 (0), x∗,2 (0)) − /2.
(5.525)
Property (i), (5.350), (5.351), (5.408), (5.410), (5.495)–(5.498), (5.500), (5.503), (5.504), (5.506), (5.517), and (5.525) imply that δ + /2 ≥ Γ f (0, l0 + l1 + L0 + 4, x, u) +(l0 + l1 + L0 + 4)f (xf , uf ) + π f (x(0)) − π f (x(l0 + l1 + L0 + 4)) f
¯ u) ¯ + (l0 + l1 + L0 + 4)f (xf , uf ) +Γ− (0, l0 + l1 + L0 + 4, x, f
f
+π− (x(0)) ¯ − π− (x(l ¯ 0 + l1 + L0 + 4)) + h(x(0), x(0)) ¯ −Γ f (0, l0 + l1 + L0 + 4, x∗,1 , u∗,1 ) − (l0 + l1 + L0 + 4)f (xf , uf ) −π f (x∗,1 (0)) + π f (x∗,1 (l0 + l1 + L0 + 4)) − h(x∗,1 (0), x∗,2(0))
5.32 Proof of Theorem 5.76
377
f
−Γ− (0, l0 + l1 + L0 + 4, x∗,2 , u∗,2 ) − (l0 + l1 + L0 + 4)f (xf , uf ) f
f
−π− (x∗,2 (0)) + π− (x∗,2 (l0 + l1 + L0 + 4)) ≥ Γ f (0, l0 + l1 + L0 + 4, x, u) + π f (x(0)) f
f
¯ u) ¯ + π− (x(0)) ¯ + h(x(0), x(0)) ¯ − /8 +Γ− (0, l0 + l1 + L0 + 4, x, −Γ f (0, l0 + l1 + L0 + 4, x∗,1 , u∗,1 ) − π f (x∗,1 (0)) f
f
−Γ− (0, l0 + l1 + L0 + 4, x∗,2 , u∗,2 ) − π− (x∗,2 (0)) − h(x∗,1 (0), x∗,2 (0)) − /8 f
≥ Γ f (0, l0 + l1 + L0 + 4, x, u) + Γ− (0, l0 + l1 + L0 + 4, x, ¯ u) ¯ + ψh (x(0), x(0)) ¯ −δ2 /4 − ψh (x∗,1 (0)x∗,2 (0)) − /4 f
¯ u) ¯ + ψh (x(0), x(0)) ¯ ≥ Γ f (0, l0 + l1 + L0 + 4, x, u) + Γ− (0, l0 + l1 + L0 + 4, x, − δ2 /4 − /4 − ψg (x∗,1 (0), x∗,2(0)) − δ1 /16.
(5.526)
By (5.493), (5.494), (5.502), and (5.526), f
Γ f (0, l0 + l1 + L0 + 4, x, u) + Γ− (0, l0 + l1 + L0 + 4, x, ¯ u) ¯ + ψh (x(0), x(0)) ¯ ≤ δ + /2 + /4 + δ2 /4 + δ1 /16 + ψg (x∗,1 (0)x∗,2(0)) ≤ 3/4 + 3/64 + inf(ψg ) + /64 ≤ 3/4 + /8 + inf(ψg ).
(5.527)
In view of (5.492), (5.504), and (5.527), f
¯ u) ¯ + ψh (x(0), x(0)) ¯ Γ f (0, l0 + l1 + L0 + 4, x, u) + Γ− (0, l0 + l1 + L0 + 4, x, ≤ 3/4 + /8 + inf(ψh ) + δ1 /16 ≤ inf(ψh ) + 3/4 + /8 + /64. The relation above, (5.492), and (5.504) imply that f
Γ f (0, l0 + l1 + L0 + 4, x, u), Γ− (0, l0 + l1 + L0 + 4, x, ¯ u) ¯ ≤ , ψh (x(0), x(0)) ¯ ≤ inf(ψh ) + 57/64 ≤ inf(ψg ) + 58/64. By (5.501), (5.504), and (5.528),
(5.528)
378
5 ContinuousTime Autonomous Problems
ψg (x(0), x(0)) ¯ ≤ ψh (x(0), x(0)) ¯ + /64 ≤ inf(ψg ) + (59/64). Theorem 5.76 is proved.
5.33 Proof of Proposition 5.78 Assume that the proposition does not hold. Then there exist a sequence {δk }∞ k=1 ⊂ (0, 1] and sequences ∞ {(xk,1 , uk,1 )}∞ k=1 ⊂ X(0, L0 , A, G), {(xk,2 , uk,2 )}k=1 ⊂ X(0, L0 , −A, −G)
such that lim δk = 0
(5.529)
Γ f (0, L0 , xk,1 , uk,1 ) ≤ δk ,
(5.530)
k→∞
and that for all integers k ≥ 1,
f
Γ− (0, L0 , xk,2 , uk,2 ) ≤ δk ,
(5.531)
ψg (xk,1 (0), xk,2(0)) ≤ inf(ψg ) + δk ,
(5.532)
and the following property holds: (i) for each (f, A, G)overtaking optimal pair (y1 , v1 ) ∈ X(0, ∞, A, G) and each (f, −A, −G)overtaking optimal pair (y2 , v2 ) ∈ X(0, ∞, −A, −G) satisfying ψg (y1 (0), y2 (0)) = inf(ψg ), we have sup{xk,1 (t) − y1 (t) + xk,2 (t) − y2 (t) : t ∈ [0, L0 ]} > . In view of (5.325), (5.529), (5.531) , and (5.532), the sequences ∞ {π f (xk,1 (0))}∞ k=1 , {π− (xk,2 (0))}k=1 are bounded. f
(5.533)
By (5.350), (5.530), and (5.531), the sequences ∞ {π f (xk,1 (L0 ))}∞ k=1 , {π− (xk,2 (L0 ))}k=1 are bounded. f
(5.534)
5.33 Proof of Proposition 5.78
379
It follows from (5.350) and (5.529)–(5.534) that the sequences f ∞ {I f (0, L0 , xk,1 , uk,1 )}∞ k=1 , {I (0, L0 , xk,2 , uk,1 )}k=1
are bounded. By LSC property, extracting a subsequence and reindexing if necessary, we may assume without loss of generality that there exist (y1 , v1 ) ∈ X(0, L0 , A, G), (y2 , v2 ) ∈ X(0, L0 , −A, −G) such that for i = 1, 2, xk,i (t) → yi (t) as k → ∞ for all t ∈ [0, L0 ],
(5.535)
I f (0, L0 , yi , vi ) ≤ lim inf I f (0, L0 , xk,i , uk,i ).
(5.536)
k→∞
Proposition 5.57, (5.529), (5.532), and (5.535) imply that ψg (y1 (0), y2 (0)) ≤ lim inf ψg (xk,1 (0), xk,2(0)) = inf(ψg ),
(5.537)
ψg (y1 (0), y2 (0)) = inf(ψg ) = lim ψg (xk,1(0), xk,2 (0)).
(5.538)
k→∞
k→∞
Proposition 5.57 and (5.535) imply that π f (y1 (0)) ≤ lim inf π f (xk,1 (0)), k→∞
f
f
(5.539)
π− (y2 (0)) ≤ lim inf π− (xk,2 (0)),
(5.540)
g(y1 (0), y2 (0)) ≤ lim inf g(xk,1 (0), xk,2 (0)).
(5.541)
k→∞
k→∞
By (5.537)–(5.541), f
f
π f (y1 (0)) = lim π f (xk,1 (0)), π− (y2 (0)) = lim π− (xk,2 (0)), k→∞
k→∞
g(y1 (0), y2 (0)) = lim g(xk,1 (0), xk,2 (0)). k→∞
(5.542) (5.543)
Proposition 5.57 and (5.535) imply that π f (y1 (L0 )) ≤ lim inf π f (xk,1 (L0 )), k→∞
f
f
π− (y2 (L0 )) ≤ lim inf π− (xk,2 (L0 )). k→∞
It follows from (5.529), (5.530), (5.536), (5.542), and (5.544) that I f (0, L0 , y1 , v1 ) − L0 f (xf , uf ) − π f (y1 (0)) + π f (y1 (L0 ))
(5.544) (5.545)
380
5 ContinuousTime Autonomous Problems
≤ lim inf I f (0, L0 , xk,1 , uk,1 ) − L0 f (xf , uf ) k→∞
− lim π f (xk,1(0)) + lim inf π f (xk,1 (L0 )) k→∞
k→∞
≤ lim inf[I f (0, L0 , xk,1 , uk,1 )−L0 f (xf , uf )−π f (xk,1 (0))+π f (xk,1 (L0 ))] ≤ 0. k→∞
(5.546) It follows from (5.529), (5.531), (5.536), (5.542), and (5.545) that f
f
I f (0, L0 , y2 , v2 ) − L0 f (xf , uf ) − π− (y2 (0)) + π− (y2 (L0 )) ≤ lim inf I f (0, L0 , xk,2 , uk,2 ) − L0 f (xf , uf ) k→∞
f
f
− lim π− (xk,2(0)) + lim inf π− (xk,2 (L0 )) k→∞
k→∞
f
f
≤ lim inf[I f (0, L0 , xk,2 , uk,2 )−L0 f (xf , uf )−π− (xk,2 (0))+π− (xk,2 (L0 ))] ≤ 0. k→∞
(5.547) In view of (5.351), (5.546), and (5.547), I f (0, L0 , y1 , v1 ) − L0 f (xf , uf ) − π f (y1 (0)) + π f (y1 (L0 )) = 0, f
f
I f (0, L0 , y2 , v2 ) − L0 f (xf , uf ) − π− (y2 (0)) + π− (y2 (L0 )) = 0.
(5.548) (5.549)
f
By (5.548) and (5.549), π f (y1 (L0 )), π− (y2 (L0 )) are finite. Theorem 5.19 implies that there exist an (f, A, G)overtaking optimal pair (y˜1 , v˜1 ) ∈ X(0, ∞, A, G) and an (f, −A, −G)overtaking optimal pair (y˜2 , v˜2 ) ∈ X(0, ∞, −A, −G) such that y˜i (0) = yi (L0 ), i = 1, 2.
(5.550)
For i = 1, 2 and all numbers t > L0 , set yi (t) = y˜i (t − L0 ), vi (t) = v˜i (t − L0 ).
(5.551)
It is clear that (y1 , v1 ) ∈ X(0, ∞, A, G) is (f, A, G)good and (y2 , v2 ) ∈ X(0, ∞, −A, −G) is (f, −A, −G)good. Proposition 5.55, (5.350), (5.351), (5.548), and (5.549) imply that for all T2 > T1 ≥ 0,
5.34 Examples
381
I f (T1 , T2 , y1 , v1 ) − (T2 − T1 )f (xf , uf ) − π f (y1 (T1 )) + π f (y1 (T2 )) = 0, f
f
I f (T1 , T2 , y2 , v2 ) − (T2 − T1 )f (xf , uf ) − π− (y2 (T1 )) + π− (y2 (T2 )) = 0. In view of the relations above and Theorem 5.20, (y1 , v1 ) is (f, A, G)overtaking optimal pair and (y2 , v2 ) is (f, −A, −G)overtaking optimal pair. By (5.545), for all sufficiently large natural numbers k and i = 1, 2, xk,i (t) − yi (t) ≤ /4, t ∈ [0, L0 ]. This contradicts property (i). The contradiction we have reached proves Proposition 5.78.
5.34 Examples In this section we present a family of problems which belong to the second class of problems and for which the results of this chapter hold. Let (E, ·, ·E ) be a Hilbert space equipped with an inner product ·, ·E which induces the norm ·E , and let (F, ·, ·F ) be a Hilbert space equipped with an inner product ·, ·F which induces the norm · F . For simplicity, we set ·, ·E = ·, ·, · E = · , ·, ·F = ·, ·, · F = · , if E and F are understood. We suppose that A is a nonempty subset of E, and U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U(x)}. We suppose that M is a Borel measurable subset of E × F and a linear operator A : D(A) → E generates a C0 semigroup S(t) = eAt , t ≥ 0 on E. Let B ∈ L(F, E−1 ) be an admissible control operator for eAt , t ≥ 0, f : M → R 1 be a borelian function. For τ ≥ 0, τ Φτ u = eA(τ −s) Bu(s)ds, u ∈ L2 (0, τ ; F ). (5.552) 0
Since B is an admissible control operator for eAt , t ≥ 0, Φτ ∈ L(L2 (0, τ ; F ), E), τ ≥ 0
(5.553)
(see Proposition 1.18). We consider the control system x = Ax + Bu.
(5.554)
382
5 ContinuousTime Autonomous Problems
Assume that there exists Tf > 0 such that RanΦTf = E.
(5.555)
In other words the pair (A, B) is exactly controllable. We suppose that xf ∈ E, uf ∈ F , the pair x(t) ¯ = xf , u(t) ¯ = uf , t ≥ 0 is a solution of (5.554) and that (xf , uf ) is an interior point of M in E × F . Assume that L : E × F → [0, ∞) is a borelian function which is upper semicontinuous at (xf , uf ), L(xf , uf ) = 0,
(5.556)
ψ0 : [0, ∞) → [0, ∞) is an increasing function, K1 , a1 > 0 such that lim ψ0 (t) = ∞,
t →∞
L(x, u) ≥ −a1 + max{ψ0 (x)x, K1 u2 }
(5.557)
for all (x, u) ∈ E × F , μ ∈ R 1 , p¯ ∈ D(A∗ ). Let for all (x, u) ∈ E × F ¯ + Bu, p. ¯ f (x, u) = L(x, u) + μ + x, A∗ p
(5.558)
Clearly, f is a Borelian function. It is not difficult to see that there exist a0 , K0 > 0 and an increasing function ψ : [0, ∞) → [0, ∞) such that limt →∞ ψ(t) = ∞, f (x, u) ≥ −a0 + max{ψ(x), K0 u2 }, x ∈ E, u ∈ F, f is upper semicontinuous at (xf , uf ) and (5.21) holds. It follows from (5.555), Theorem 1.23, and the upper semicontinuity of f at (xf , uf ) that (A3) holds. Let 0 ≤ T1 < T2 , (x, u) ∈ X(T1 , T2 ). It is not difficult to see that
T2
I (T1 , T2 , x, u) = f
f (x(t), u(t))dt =
T1
+
T2
x(t), A∗ pdt ¯ +
T1
T2
L(x(t), u(t))dt + μ(T2 − T1 )
T1
T2
Bu(t), pdt ¯ ≥ μ(T2 − T1 ) + x(T2 ) − x(T1 ).p. ¯
T1
(5.559) This implies that (A1) holds. Assume now that the following property holds: (a) for each M, > 0 there exists δ > 0 such that for each (x, u) ∈ E × F which satisfies x ≤ M and L(x, u) ≤ δ, we have x − xf ≤ .
5.34 Examples
383
We claim that f has TP. In view of Theorems 5.13 and 5.26, it is sufficient to show that property (P3) of Theorem 5.26 holds. Let , M > 0 and M1 > 0 be as guaranteed by Theorem 5.2 with M0 = M, c0 = 1, c = 2. Let S > 0 be as guaranteed by Theorem 5.1, and let δ ∈ (0, 1) be as guaranteed by property (a) with M = M1 . Choose L > 3 + δ −1 (M + S + 2M1 p). ¯ Assume that T1 ≥ 0, T2 ≥ T1 + L and that (x, u) ∈ X(T1 , T2 ) satisfies I f (T1 , T2 , x, u) ≤ (T2 − T1 )f (xf , uf ) + M. Combined with Theorem 5.2 and the choice of M1 , this implies that x(t) ≤ M1 , t ∈ [T1 + 1, T2 ]. It is not difficult to see that, in view of the choice of S and Theorem 5.1, I f (T1 + 1, T2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , T1 + 1, x, u) ≤ (T2 − T1 )f (xf , uf ) + M − f (xf , uf ) + S = (T2 − T1 − 1)f (xf , uf ) + M + S. Combined with (5.559), this implies that
T2 T1 +1
L(x(t), u(t))dt
= I f (T1 + 1, T2 , x, u) − μ(T2 − T1 − 1) − x(T2 ) − x(T1 + 1), p ¯ ≤ M + S + 2M1 p. ¯ Together with the choice of δ and property (a) with M = M1 , this implies that ¯ ≥ M + S + 2M1 p
T2 T1 +1
L(x(t), u(t))dt
≥ δmes({t ∈ [T1 + 1, T2 ] : L(x(t), u(t)) > δ}) ≥ δmes({t ∈ [T1 + 1, T2 ] : x(t) − xf > }) and
384
5 ContinuousTime Autonomous Problems
mes({t ∈ [T1 , T2 ] : x(t) − xf > }) ¯ < L. ≤ 1 + δ −1 (M + S + 2M1 p) Thus (P3) holds and f has TP. We can easily obtain particular cases of this example with the pairs (A, B) considered in Section 1.8.
5.35 Exercises for Chapter 5 Exercise 5.79. Use the control problem considered in Example 1.24, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.80. Use the control problem considered in Example 1.25, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.81. Use the control problem considered in Example 1.26, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.82. Use the control problem considered in Example 1.27, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.83. Use the control problem considered in Example 1.28, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.84. Use the control problem considered in Example 1.29, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.85. Use the control problem considered in Example 1.30, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34. Exercise 5.86. Use the control problem considered in Example 1.31, and construct an optimal control problem which is a special case of the example analyzed in Section 5.34.
Chapter 6
ContinuousTime Nonautonomous Problems on the HalfAxis
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for continuoustime optimal control problems on subintervals of halfaxis in infinite dimensional spaces. For these optimal control problems the turnpike is not a singleton. We also study the existence of solutions of the corresponding infinite horizon optimal control problems. The results of this chapter will be obtained for two large classes of problems which will be treated simultaneously.
6.1 Preliminaries We begin with the description of the first class of problems. Let (E, ·) be a Banach space and E ∗ be its dual. Let {A(t) : t ∈ [0, ∞)} be the family of closed densely defined linear operators with the domain and range in the Banach space E. Let (F, ρF ) be a metric space. We suppose that A is a nonempty subset of [0, ∞) × E, U : A → 2F is a point to set mapping with a graph M = {(t, x, u) : (t, x) ∈ A, u ∈ U(t, x)}. We suppose that M is a Borel measurable subset of [0, ∞)×E ×F , G : M → E is a Borelian function. Let f : M → R 1 be a Borelian function bounded from below. We consider the homogeneous Cauchy problem x (t) = A(t)x(t), t ∈ [0, ∞).
(6.1)
We assume that there exists a function U : {(t, s) ∈ R 2 : 0 ≤ s ≤ t < ∞} → L(E) which has the following properties [18]:
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_6
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(i) for each x0 ∈ E the function (t, s) → U (t, s)x0 is continuous on the set {(t, s) ∈ R 2 : 0 ≤ s ≤ t < ∞}; (ii) U (s, s) = I d for all s ∈ [0, ∞), where I d is the identity operator; (iii) U (t, s)U (s, τ ) = U (t, τ ) for all numbers t ≥ s ≥ τ ≥ 0; (iv) for each s ≥ 0 there exists a densely linear subspace Es of E such that for each x0 ∈ Es the function t → U (t, s)x0 is continuously differentiable on [s, ∞) and (∂/∂t)U (t, s)x0 = A(t)U (t, s)x0 , t ∈ [s, ∞);
(6.2)
(v) there exists an increasing function τ → Δτ > 0, τ > 0 such that for each τ > 0, each s ≥ 0 and each t ∈ [s, s + τ ], U (t, s) ≤ Δτ . In this case problem (6.1) is called wellposed [18]. Let 0 ≤ T1 < T2 and consider the following equation x (t) = A(t)x(t) + f (t), t ∈ [T1 , T2 ], x(0) = xT1 ,
(6.3)
where xT1 ∈ E and f ∈ L1 (T1 , T2 ; E). A continuous function x : [T1 , T2 ] → E is a solution of (6.3) if x(t) = U (t, T1 )x(T1 ) +
t
U (t, s)f (s)ds, t ∈ [T1 , T2 ].
(6.4)
T1
Assume that (6.4) holds and τ ∈ [T1 , T2 ]. In view of (6.4), x(τ ) = U (τ, T1 )x(T1 ) +
τ
(6.5)
U (τ, s)f (s)ds. T1
Property (iii), (6.4), and (6.5) imply that for all t ∈ [τ, T2 ], x(t) = U (t, τ )U (τ, T1 )x(T1 ) +
τ
T1
= U (t, τ )U (τ, T1 )x(T1 ) + U (t, τ )
t
U (t, s)f (s)ds +
U (t, s)f (s)ds τ
τ
t
U (τ, s)f (s)ds +
T1
U (t, s)f (s)ds τ
t
= U (t, τ )x(τ ) +
U (t, s)f (s)ds. τ
Thus x : [τ, T2 ] → E is a solution of the equation y (t) = A(t)y(t) + f (t), t ∈ [τ, T2 ], y(τ ) = x(τ ).
6.1 Preliminaries
387
Let 0 ≤ T1 < T2 < T3 , z0 ∈ E, f ∈ L1 (T1 , T3 ; E), a continuous function x1 : [T1 , T2 ] → E is a solution of the equation x (t) = A(t)x(t) + f (t), t ∈ [T1 , T2 ], x(T1 ) = z0 , a continuous function x2 : [T2 , T3 ] → E is a solution of the equation x (t) = A(t)x(t) + f (t), t ∈ [T2 , T3 ], x(T2 ) = x1 (T2 ). Set x(t) = x1 (t), t ∈ [T1 , T2 ], x(t) = x2 (t), t ∈ [T2 , T3 ].
(6.6)
Clearly, the function x : [T1 , T3 ] → E is continuous. In view of (6.4) and (6.6), for all t ∈ [T1 , T2 ],
t
x(t) = x1 (t) = U (t, T1 )z0 +
U (t, s)f (s)ds.
(6.7)
T1
Property (iii), (6.4), (6.6), and (6.7) imply that for all t ∈ [T2 , T3 ],
t
x(t) = x2 (t) = U (t, T2 )x1 (T2 ) +
U (t, s)f (s)ds T2
= U (t, T2 )(U (T2 , T1 )z0 +
T2
U (T2 , s)f (s)ds) +
T2
U (t, s)f (s)ds +
t
U (t, s)f (s)ds T2
T1
= U (t, T1 )z0 +
U (t, s)f (s)ds T2
T1
= U (t, T1 )z0 +
t
t
U (t, s)f (s)ds. T1
Thus x(·) is a solution of the equation x (t) = A(t)x(t) + f (t), t ∈ [T1 , T3 ], x(T1 ) = z0 . Let 0 ≤ T1 < T2 . We consider the following equation x (t) = A(t)x(t) + G(t, x(t), u(t)), t ∈ [T1 , T2 ].
(6.8)
A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (6.8) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function,
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(t, x(t)) ∈ A, t ∈ [T1 , T2 ], u(t) ∈ U(t, x(t)), t ∈ [T1 , T2 ] almost everywhere (a.e.),
(6.9) (6.10)
G(s, x(s), u(s)), s ∈ [T1 , T2 ] is Bochner integrable and for every t ∈ [T1 , T2 ], x(t) = U (t, T1 )x(T1 ) +
t
U (t, s)G(s, x(s), u(s))ds.
(6.11)
T1
The set of all pairs (x, u) which are solutions of (6.8) is denoted by X(T1 , T2 ). Let T1 ≥ 0. A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a (mild) solution of the system x (t) = A(t)x(t) + G(t, x(t), u(t)), t ∈ [T1 , ∞)
(6.12)
if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (6.8). The set of all such pairs (x, u) which are solutions of the equation above, is denoted by X(T1 , ∞). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) (0 ≤ T1 < T2 ) is called a trajectory if there exists a Lebesgue measurable function u : I → F (referred to as a control) such that (x, u) ∈ X(T1 , T2 ) or (x, u) ∈ X(T1 , ∞), respectively). Now we describe the second class of problems. Let (E, ·, ·)E be a Hilbert space equipped with an inner product ·, ·E which induces the norm ·E , and let (F, ·, ·F ) be a Hilbert space equipped with an inner product ·, ·F which induces the norm · F . For simplicity, we set ·, ·E = ·, ·, · E = · , ·, ·F = ·, ·, · F = · , if E, F are understood. We suppose that A0 is a nonempty subset of E, U0 : A0 → 2F is a point to set mapping with a graph M0 = {(x, u) : x ∈ A0 , u ∈ U0 (x)}. We suppose that M0 is a Borel measurable subset of E × F . Define A = [0, ∞) × A0 , U : A → 2F by U(t, x) = U0 (x), (t, x) ∈ A, M = [0, ∞) × M0 . Let a linear operator A : D(A) → E generates a C0 semigroup S(t) = eAt , t ∈ [0, ∞) on E. As usual, we denote by S(t)∗ the adjoint of S(t). Then S(t)∗ , t ∈
6.1 Preliminaries
389
[0, ∞) is C0 semigroup and its generator is the adjoint A∗ of A. The domain D(A∗ ) is a Hilbert space equipped with the graph norm · D(A∗ ) : z2D(A∗ ) = z2E + A∗ z2E , z ∈ D(A∗ ).
(6.13)
Let D(A∗ ) be the dual of D(A∗ ) with the pivot space E. In particular, E1d := D(A∗ ) ⊂ E ⊂ D(A∗ ) = E−1 . (Here we use the notation of Section 1.7.) Let G : M → D(A∗ ) = E−1 and f : M → R 1 be Borelian functions, B ∈ L(F, E−1 ) is an admissible control operator for eAt , t ≥ 0, for all (t, x, u) ∈ M, G(t, x, u) = Bu. Let 0 ≤ T1 < T2 . We consider the following equation x (t) = Ax(t) + Bu(t), t ∈ [T1 , T2 ] a. e. .
(6.14)
A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (6.14) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, u ∈ L2 (T1 , T2 ; F ), (t, x(t)) ∈ A, t ∈ [T1 , T2 ],
(6.15)
u(t) ∈ U(t, x(t)), t ∈ [T1 , T2 ] a.e.,
(6.16)
and for each t ∈ [T1 , T2 ], x(t) = e
A(t −T1 )
x(T1 ) +
t
eA(t −s)Bu(s)ds
(6.17)
T1
in E−1 . The set of all pairs (x, u) which are solutions of (6.14) is denoted by X(T1 , T2 , A, G). In the sequel for simplicity, we use the notation X(T1 , T2 ) = X(T1 , T2 , A, G) if the pair (A, G) is understood. Let T1 ≥ 0. A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a (mild) solution of the system x (t) = Ax(t) + Bu(t), t ∈ [T1 , ∞)
(6.18)
if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (6.14). The set of all such pairs (x, u) which are solutions of (6.18) is denoted by X(T1 , ∞). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) (0 ≤ T1 < T2 ) is called a trajectory if there exists a Lebesgue measurable function u : I → F
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(referred to as a control) such that (x, u) ∈ X(T1 , T2 ) or (x, u) ∈ X(T1 , ∞), respectively. We treat the two problems simultaneously. Let T2 > T1 ≥ 0, (x, u) ∈ X(T1 , T2 ). Define I f (T1 , T2 , x, u) =
T2
f (t, x(t), u(t))dt T1
which is welldefined but can be ∞. Let a0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → ∞ as t → ∞.
(6.19)
We suppose that for the first class of problems the function f satisfies f (t, x, u) ≥ −a0 + max{ψ(x), ψ((G(t, x, u) − a0 x)+ )(G(t, x, u) − a0 x)+ }
(6.20)
for each (t, x, u) ∈ M and for the second class of problems f satisfies f (t, x, u) ≥ −a0 + max{ψ(x), K0 u2 }, (t, x, u) ∈ M.
(6.21)
We consider functionals of the form I f (T1 , T2 , x, u), where 0 ≤ T1 < T2 and (x, u) ∈ X(T1 , T2 ). For each pair of numbers T2 > T1 ≥ 0 and each pair of points (T1 , y), (T2 , z) ∈ A we define U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z}, σ f (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y}, σ f (T1 , T2 , z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T2 ) = z}, σ f (T1 , T2 ) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 )}.
(6.22)
We suppose that (xf , uf ) ∈ X(0, ∞) satisfies sup{xf (t) : t ∈ [0, ∞)} < ∞, Δf := sup{I f (j, j + 1, xf , uf ) : j = 0, 1, . . . } < ∞.
(6.23) (6.24)
We suppose that there exists a number bf > 0 and the following assumptions hold.
6.2 Boundedness Results
391
(A1) For each S1 > 0 there exist S2 > 0 and c > 0 such that I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S2 for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying x(Tj ) ≤ S1 , j = 1, 2. (A2) For each > 0 there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi −xf (Ti ) ≤ δ, i = 1, 2 and T2 ≥ bf there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfies x1 (T1 ) = z1 , x1 (T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + , x2 (T2 ) = z2 , x2 (T2 − τ2 ) = xf (T2 − τ2 ), I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + . Relations (6.20), (6.21), and (6.24) imply the following result. Lemma 6.1. Let c > 0. Then Δf (c) = sup{I f (T1 , T2 , xf , uf ) : T1 ≥ 0, T2 ∈ (T1 , T1 + c]} < ∞. Section 6.19 contains examples of optimal control problems satisfying assumptions (A1) and (A2). Many examples can also be found in [106–108, 118, 124, 125, 134].
6.2 Boundedness Results The following result is proved in Section 6.7. Theorem 6.2. 1. There exists S > 0 such that for each pair of numbers T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S ≥ I f (T1 , T2 , xf , uf ). 2. For each (x, u) ∈ X(0, ∞) either I f (0, T , x, u) − I f (0, T , xf , uf ) → ∞ as T → ∞ or
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sup{I f (0, T , x, u) − I f (0, T , xf , uf ) : T ∈ (0, ∞)} < ∞.
(6.25)
Moreover, if (6.25) holds, then sup{x(t) : t ∈ (0, ∞)} < ∞. We say that (x, u) ∈ X(0, ∞) is (f )good if (6.25) holds. The next boundedness result is proved in Section 6.7. Theorem 6.3. Let M0 > 0, c > 0, c0 ∈ (0, c). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c0 , T2 ]. Let L > 0. Denote by AL the set of all (s, z) ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ X(s, s + τ ) such that x(s) = z, x(s + τ ) = xf (s + τ ), I f (s, s + τ, x, u) ≤ L. L the set of all (s, z) ∈ A for which s ≥ L and there exist τ ∈ (0, L] Denote by A and (x, u) ∈ X(s − τ, s) such that x(s − τ ) = xf (s − τ ), x(s) = z, I f (s − τ, s, x, u) ≤ L. Theorems 6.4–6.10 stated below are also boundedness results. They are proved in Section 6.8. Theorem 6.4. Let L > 0, M0 > 0 and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Theorem 6.5. Let L > 0, M0 > 0 and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying (T1 , x(T1 )) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Theorem 6.6. Let L > 0, M0 > 0 and c ∈ (0, L). Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying
6.3 Turnpike Results
393
L , I f (T1 , T2 , x, u) ≤ (T2 , x(T2 )) ∈ A σ f (T1 , T2 , x(T2 )) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [c + T1 , T2 ]. Theorem 6.7. Let M0 > 0 and c > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 6.8. Let L > 0, M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , x(T1 ) ≤ M0 , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 6.9. Let L > 0, M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , (T1 , x(T1 )) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 6.10. Let L > 0, M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying L , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + M0 x(T1 ) ≤ M0 , (T2 , x(T2 )) ∈ A the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ].
6.3 Turnpike Results We say that the integrand f possesses the turnpike property (or TP for short) if for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ} there exist τ1 , τ2 ∈ [0, L] such that x(t) − xf (t) ≤ for all t ∈ [T1 + τ1 , T2 − τ2 ].
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Moreover, if x(T2 ) − xf (T2 ) ≤ δ then τ2 = 0 and if T1 ≥ L and x(T1 ) − xf (T1 ) ≤ δ, then τ1 = 0. We say that the integrand f possesses the strong turnpike property (or STP for short) if for each > 0 and each M > 0 there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ} there exist τ1 , τ2 ∈ [0, L] such that x(t) − xf (t) ≤ for all t ∈ [T1 + τ1 , T2 − τ2 ]. Moreover, if x(T1 ) − xf (T1 ) ≤ δ then τ1 = 0 and if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = 0. Theorem 6.2 implies the following result. Theorem 6.11. Assume that f has TP and that , M > 0. Then there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{I f (T1 , T2 , xf , uf ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 , and if T1 ≥ L and x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 . Theorem 6.12. Assume that f has STP and that , M > 0. Then there exist δ > 0 and L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{I f (T1 , T2 , xf , uf ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, for i = 1, 2, if x(Ti ) − xf (Ti ) ≤ δ, then τi = Ti . Proposition 6.13. Let L > 0. Then for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each L , pair of points (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , xf , uf ) + 2L(1 + a0 ).
6.3 Turnpike Results
395
L . By the definition Proof. Let T1 ≥ 0, T2 ≥ T1 + 2L, (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A L , there exist τ1 ∈ (0, L], τ2 ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that of AL , A y(T1 ) = z1 , y(T2 ) = z2 , (y(t), v(t)) = (xf (t), uf (t)), t ∈ [T1 + τ1 , T2 − τ2 ], I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L. In view of the relations above, (6.20) and (6.21), U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , y, v) ≤ 2L + I f (T1 + τ1 , T2 − τ2 , xf , uf ) ≤ 2L + 2a0 L + I f (T1 , T2 , xf , uf ). Proposition 6.13 is proved. Proposition 6.13 and Theorems 6.11 and 6.12 imply the following two results. Theorem 6.14. Assume that f has TP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies L0 , (T1 , x(T1 )) ∈ AL0 , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ then τ2 = T2 and if T1 ≥ L and x(T1 ) − xf (T1 ) ≤ δ then τ1 = T1 . Theorem 6.15. Assume that f has STP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies L0 , (T1 , x(T1 )) ∈ AL0 , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 , and if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 .
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Proposition 6.16. Let L > 0. Then for each (T1 , z) ∈ AL , each T2 ≥ T1 + L, σ f (T1 , T2 , z) ≤ I f (T1 , T2 , xf , uf ) + L(1 + a0 ). Proof. Let (T , z) ∈ AL , T2 ≥ T1 +L. By the definition of AL , there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = z, (y(t), v(t)) = (xf (t), uf (t)), t ∈ [T1 + τ, T2 ], I f (T1 , T1 + τ, y, v) ≤ L. In view of the relations above, (6.20) and (6.21), σ f (T1 , T2 , z) ≤ I f (T1 , T2 , y, v) ≤ L + I f (T1 + τ, T2 , xf , uf ) ≤ L + a0 L + I f (T1 , T2 , xf , uf ). Proposition 6.16 is proved. Proposition 6.16 and Theorems 6.11 and 6.12 imply the following two results. Theorem 6.17. Assume that f has TP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies (T1 , x(T1 )) ∈ AL0 , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + δ there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 , and if T1 ≥ L and x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 . Theorem 6.18. Assume that f has STP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies (T1 , x(T1 )) ∈ AL0 , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + δ, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that
6.3 Turnpike Results
397
x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 , and if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 . L Proposition 6.19. Let L > 0. Then for each T1 ≥ 0 and each (T2 , z) ∈ A satisfying T2 ≥ T1 + L, σ f (T1 , T2 , z) ≤ I f (T1 , T2 , xf , uf ) + L(1 + a0 ). L , T2 ≥ T1 + L. Then there exist τ ∈ (0, L] and Proof. Let T1 ≥ 0, (T2 , z) ∈ A (y, v) ∈ X(T1 , T2 ) such that y(T2 ) = z, (y(t), v(t)) = (xf (t), uf (t)), t ∈ [T1 , T2 − τ ], I f (T2 − τ, T2 , y, v) ≤ L. In view of the relations above, (6.21) and (6.26), σ f (T1 , T2 , z) ≤ I f (T1 , T2 , y, v) ≤ L + I f (T1 , T2 − τ, xf , uf ) ≤ L + a0 L + I f (T1 , T2 , xf , uf ). Proposition 6.19 is proved. Proposition 6.19 and Theorems 6.11 and 6.12 imply the following two results. Theorem 6.20. Assume that f has TP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies L0 , (T2 , x(T2 )) ∈ A σ f (T1 , T2 , x(T2 )) + δ, I f (T1 , T2 , x, u) ≤ there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 , and if T1 ≥ L and x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 . Theorem 6.21. Assume that f has STP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies
398
6 ContinuousTime Nonautonomous Problems on the HalfAxis
L0 , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + δ, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 , and if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 . The next theorem is the main result of this chapter. It is proved in Section 6.9. Theorem 6.22. f has TP if and only if the following properties hold: (P1) for each (f )good pair (x, u) ∈ X(0, ∞), lim x(t) − xf (t) = 0;
t →∞
(P2) for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each T ≥ 0 and each (x, u) ∈ X(T , T + L) which satisfies I f (T , T + L, x, u) ≤ min{U f (T , T + L, x(T ), x(T + L)) + δ, I f (T , T + L, xf , uf ) + M} there exists s ∈ [T , T + L] such that x(s) − xf (s) ≤ . The next result is proved in Section 6.11. Theorem 6.23. Assume that f has properties (P1) and (P2). Let , M > 0. Then there exist a natural number Q and l > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + lQ, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M, q
q
there exist finite sequences {ai }i=1 , {bi }i=1 ⊂ [T1 , T2 ] such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf (t) ≤ for all t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. Theorem 6.23 leads to the following definition.
6.3 Turnpike Results
399
We say that f possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0 there exist a natural number Q and l > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + lQ, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M, q
q
there exist finite sequences {ai }i=1 , {bi }i=1 ⊂ [T1 , T2 ] such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf (t) ≤ for all t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. Theorem 6.24. f has WTP if and only if f has (P1) and (P2). In view of Theorem 6.23, (P1) and (P2) imply WTP. Clearly, WTP implies (P2). Therefore Theorem 6.24 follows from the next result which can be easily proved. Proposition 6.25. Let f have WTP. Then f has (P1). Theorem 6.26. For the second class of problems, we assume that the following assumption holds: (A3) for each > 0 there exist δ, L > 0 such that for each S1 ≥ L and each S2 ∈ [S1 , S1 + δ] the inequality I f (S1 , S2 , xf , uf ) ≤ . The integrand f has (P1) and (P2) if and only if the following property holds: (P3) for each > 0 and each M > 0 there exists L > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M the inequality mes({t ∈ [T1 , T2 ] : x(t) − xf (t) > }) ≤ L is valid. In view of Theorem 6.23, (P1) and (P2) imply (P3). Clearly, (P3) implies (P2). Therefore Theorem 6.26 follows from the next result which is proved in Section 6.13. Proposition 6.27. For the second class of problems, we assume that assumption (A3) holds. Let f have (P3). Then f has (P1).
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
6.4 Lower Semicontinuity Property We say that f possesses lower semicontinuity property (or LSC property for short) if for each T2 > T1 ≥ 0 and each sequence (xj , uj ) ∈ X(T1 , T2 ), j = 1, 2, . . . which satisfies sup{I f (T1 , T2 , xj , uj ) : j = 1, 2, . . . } < ∞, there exist a subsequence {(xjk , ujk )}∞ k=1 and (x, u) ∈ X(T1 , T2 ) such that xjk (t) → x(t) as k → ∞ for every t ∈ [T1 , T2 ], I f (T1 , T2 , x, u) ≤ lim inf I f (T1 , T2 , xj , uj ). j →∞
LSC property plays an important role in the calculus of variations and optimal control theory [32]. Let S ≥ 0. A pair (x, u) ∈ X(S, ∞) is called (f )overtaking optimal if for every (y, v) ∈ X(S, ∞) satisfying x(S) = y(S), lim sup[I f (S, T , x, u) − I f (S, T , y, v)] ≤ 0. T →∞
A pair (x, u) ∈ X(S, ∞) is called f weakly optimal if for every (y, v) ∈ X(S, ∞) satisfying x(S) = y(S), lim inf[I f (S, T , x, u) − I f (S, T , y, v)] ≤ 0. T →∞
A pair (x, u) ∈ X(S, ∞) is called (f )minimal if for every T > S I f (S, T , x, u) = U f (S, T , x(S), x(T )). The next result is proved in Section 6.15. Theorem 6.28. Assume that f has (P1), (P2), and (LSC) property, S ≥ 0, and that (x, u) ∈ X(S, ∞) is (f )good. Then there exists an (f )overtaking optimal pair (x∗ , u∗ ) ∈ X(S, ∞) such that x∗ (S) = x(S). The following theorem is also proved in Section 6.15. Theorem 6.29. Assume that f has (P1), (P2), and LSC property, S ≥ 0, (x, ˜ u) ˜ ∈ ˜ Then X(S, ∞) is (f )good, and that (x∗ , u∗ ) ∈ X(S, ∞) satisfies x∗ (S) = x(S). the following conditions are equivalent: (i) (x∗ , u∗ ) is (f )overtaking optimal; (ii) (x∗ , u∗ ) is (f )weakly optimal; (iii) (x∗ , u∗ ) is (f )minimal and (f )good;
6.4 Lower Semicontinuity Property
401
(iv) (x∗ , u∗ ) is (f )minimal and satisfies limt →∞ (x∗ (t) − xf (t)) = 0; (v) (x∗ , u∗ ) is (f )minimal and satisfies lim inft →∞ x∗ (t) − xf (t) = 0. The next result easily follows from Theorems 6.14 and 6.23. Theorem 6.30. Assume that f has (P1), (P2), and LSC property and , L > 0. Then there exists τ0 > 0 such that for every T0 ≥ 0 and every (f )overtaking optimal pair (x, u) ∈ X(T0 , ∞) which satisfies (T0 , x(T0 )) ∈ AL the inequality x(t) − xf (t) ≤ holds for all t ≥ T0 + τ0 . In the sequel we use the following result which can be easily proved. Proposition 6.31. Let f have (P1) and (T0 , z0 ) ∈ A. There exists an (f )good pair (x, u) ∈ X(T0 , ∞) satisfying x(T0 ) = z0 if and only if (T0 , z0 ) ∈ ∪{AL : L > 0}. LSC property implies the following result. Proposition 6.32. Let f have LSC property, T2 > T2 ≥ 0, L > 0. Then the following assertions hold. 1. There exists (x, u) ∈ X(T1 , T2 ) such that I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , y, v) for all (y, v) ∈ X(T1 , T2 ). L , then there exists (x, u) ∈ 2. If T2 − T1 ≥ 2L, (T1 , z1 ) ∈ AL , (T2 , z2 ) ∈ A X(T1 , T2 ) such that x(Ti ) = zi , i = 1, 2 and I f (T1 , T2 , x, u) = U f (T1 , T2 , z1 , z2 ). 3. If T2 − T1 ≥ L, (T1 , z) ∈ AL , then there exists (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = z and I f (T1 , T2 , x, u) = σ f (T1 , T2 , z). L , then there exists (x, u) ∈ X(T1 , T2 ) such that 4. If T2 − T1 ≥ L, (T2 , z) ∈ A x(T2 ) = z and I f (T1 , T2 , x, u) = σ f (T1 , T2 , z). The next result is proved in Section 6.17. Theorem 6.33. For the first class of problems, assume that A(t) = 0 for all t ≥ 0, and for the second class of problems, assume that A = 0. Let f have LSC property and (xf , uf ) be (f )overtaking optimal. Then f has STP if and only if (P1) and (P2) hold and the following property holds: (P3) for each (f )overtaking optimal pair of sequences (y, v) ∈ X(0, ∞) satisfying y(0) = xf (0) the equality y(t) = xf (t) holds for all t ≥ 0. The next result follows from (A2), (P1), and STP.
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
Theorem 6.34. Assume that f has STP and > 0. Then there exists δ > 0 such that for every T1 ≥ 0 and every (f )overtaking optimal pair (x, u) ∈ X(T1 , ∞) satisfying x(T1 ) − xf (T1 ) ≤ δ, the inequality x(t) − xf (t) ≤ holds for all t ≥ T1 .
6.5 Perturbed Problems In this section we use the following assumption. (A4) For each > 0 there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi −xf (Ti ) ≤ δ, i = 1, 2 and T2 ≥ bf , there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfy x1 (T1 ) = z1 , x1 (T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + , x1 (t) − xf (t) ≤ , t ∈ [T1 , T1 + τ1 ], x2 (T2 ) = z2 , x2 (T2 − τ2 ) = xf (T2 − τ2 ), I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + , x2 (t) − xf (t) ≤ , t ∈ [T2 − τ2 , T2 ]. Clearly, (A4) implies (A2). Assume that φ : E → [0, 1] is a continuous function satisfying φ(0) = 0 and such that the following property holds: (i) for each > 0 there exists δ > 0 such that for each x ∈ E satisfying φ(x) ≤ δ we have x ≤ . For each r ∈ (0, 1) set fr (t, x, u) = f (t, x, u) + rφ(x − xf (t)), (t, x, u) ∈ M. Clearly, for any r ∈ (0, 1), fr is a Borealian function, if (A4) holds, then (A1) and (A4) hold for fr with (xfr , ufr ) = (xf , uf ). Theorem 6.35. Let (A4) hold and r ∈ (0, 1). Then fr has (P1) and (P2). Moreover, if (xf , uf ) is (f )minimal, then fr has (P4) and (xf , uf ) is (fr )overtaking optimal. Theorem 6.35 is proved in Section 6.18.
6.6 Auxiliary Results for Theorems 6.2 and 6.3
403
6.6 Auxiliary Results for Theorems 6.2 and 6.3 Proposition 6.36. Let γ > 0. Then there exists δ > 0 such that for each (T , z1 ), (T + 2bf , z2 ) ∈ A satisfying z1 − xf (T ) ≤ δ, z2 − xf (T + 2bf ) ≤ δ,
(6.26)
there exists (x, u) ∈ X(T , T + 2bf ) which satisfies x(T ) = z1 , x(T + 2bf ) = z2 , I f (T , T + 2bf , x, u) ≤ I f (T , T + 2bf , xf , uf ) + γ . Proof. Set = γ /2 and let δ > 0 be as guaranteed by (A2). Let (T , z1 ), (T + 2bf , z2 ) ∈ A satisfy (6.26). By (A2), there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T , T + τ1 ), (x2 , u2 ) ∈ X(T + 2bf − τ2 , T + 2bf ) which satisfy x1 (T ) = z1 , x1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x1 , u1 ) ≤ I f (T , T + τ1 , xf , uf ) + γ /2, x2 (T + 2bf ) = z2 , x2 (T + 2bf − τ2 ) = xf (T + 2bf − τ2 ), I f (T + 2bf − τ2 , T + 2bf , x2 , u2 ) ≤ I f (T + 2bf − τ2 , T + 2bf , xf , uf ) + γ /2. Define x(t) = x1 (t), u(t) = u1 (t), t ∈ [T , T + τ1 ], x(t) = x2 (t), u(t) = u2 (t), t ∈ (T + 2bf − τ2 , T + 2bf ], x(t) = xf (t), u(t) = uf (t), t ∈ (T + τ1 , T + 2bf − τ2 ]. It is clear that (x, u) ∈ X(T , T + 2bf ), x(T ) = z1 , x(T + 2bf ) = z2 , I f (T , T + 2bf , x, u) ≤ I f (T , T + 2bf , xf , uf ) + γ . Proposition 6.36 is proved. Lemma 6.37. There exist numbers S > 0, c0 ≥ 1 such that for each T1 ≥ 0, each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S.
(6.27)
404
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Proof. In view of (6.19)–(6.21) and (6.24), there exists S1 > 0 such that ψ(S1 ) > a0 + 1 + sup{I f (j, j + 1, xf , uf ) : j = 0, 1, . . . }.
(6.28)
By (A1), there exist S2 > 0, c0 > 1 such that I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S2
(6.29)
for each T1 ≥ 0, each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ), x(T2 ) ≤ S1 . Fix a number S ≥ S2 + 2 + 2a0 (4 + 2c0 ) + 4(c0 + 1) sup{I f (j, j + 1, xf , uf ) : j = 0, 1, . . . }. (6.30) Assume that T1 ≥ 0, T2 ≥ T1 + c0 , (x, u) ∈ X(T1 , T2 ). We show that (6.27) is true. Assume that x(t) ≥ S1 , t ∈ [T1 , T2 ].
(6.31)
By (6.20), (6.21), (6.24), (6.28), (6.30), and (6.31), for all t ∈ [T1 , T2 ], f (t, x(t), u(t)) ≥ −a0 + ψ(x(t)) ≥ −a0 + ψ(S1 ) and I f (T1 , T2 , x, u) ≥ (T2 − T1 )(ψ(S1 ) − a0 ) ≥ (T2 − T1 )(sup{I f (j, j + 1, xf , uf ) : j = 0, 1, . . . } + 1) ≥ I f (T1 , T2 , xf , uf ) − 2Δf − 2a0 ≥ I f (T1 , T2 , xf , uf ) − S and (6.27) holds. Assume that inf{x(t) : t ∈ [T1 , T2 ]} < S1 .
(6.32)
τ1 = inf{t ∈ [T1 , T2 ] : x(t) ≤ S1 },
(6.33)
Set
6.6 Auxiliary Results for Theorems 6.2 and 6.3
τ2 = sup{t ∈ [T1 , T2 ] : x(t) ≤ S1 }.
405
(6.34)
Clearly, τ1 , τ2 are welldefined, τ1 ≤ τ2 and x(τi ) ≤ S1 , i = 1, 2.
(6.35)
τ2 − τ1 ≥ c 0 ;
(6.36)
τ2 − τ1 < c 0 .
(6.37)
There are two cases:
Assume that (6.36) holds. It follows from (6.29), (6.35), (6.36), and the choice of S2 and c0 that I f (τ1 , τ2 , xf , uf ) ≤ I f (τ1 , τ2 , x, u) + S2 .
(6.38)
By (6.20), (6.21), (6.33), and (6.34), for each t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], x(t) ≥ S1 , f (t, x(t), u(t)) ≥ −a0 + ψ(x(t)) ≥ −a0 + ψ(S1 ).
(6.39)
It follows from (6.24), (6.28), and (6.39) that I f (T1 , τ1 , x, u) ≥ (τ1 − T1 )(ψ(S1 ) − a0 ) ≥ (τ1 − T1 )(Δf + 1) ≥ I f (T1 , τ1 , xf , uf ) − 2Δf − 2a0 ,
(6.40)
I f (τ2 , T2 , x, u) ≥ (T2 − τ2 )(ψ(S1 ) − a0 ) ≥ (T2 − τ2 )(Δf + 1) ≥ I f (τ2 , T2 , xf , uf ) − 2Δf − 2a0 .
(6.41)
In view of (6.30), (6.38), (6.40), and (6.41), I f (T1 , T2 , x, u) = I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + I f (τ2 , T2 , x, u) ≥ I f (T1 , τ1 , xf , uf ) − 2Δf − 2a0 + I f (τ1 , τ2 , xf , uf ) −S2 + I f (τ2 , T2 , xf , uf ) − 2Δf − 2a0 ≥ I f (T1 , T2 , xf , uf ) − S. Assume that (6.37) holds. By (6.20), (6.21), (6.33), and (6.34), for each t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], (6.39) is true. In view of (6.24), (6.28), and (6.39), (6.40) and (6.41) are true.
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
It follows from (6.20), (6.21), (6.30), (6.38), (6.40), and (6.41) that I f (T1 , T2 , x, u) = I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + I f (τ2 , T2 , x, u) ≥ I f (T1 , τ1 , xf , uf ) − 2Δf − 2a0 − a0 (τ2 − τ1 ) + I f (τ2 , T2 , xf , uf ) − 2Δf − 2a0 ≥ I f (T1 , T2 , xf , uf ) − 4Δf − 4a0 − a0 c0 − Δf c0 − 2Δf − 2a0 ≥ I f (T1 , T2 , xf , uf ) − S. Lemma 6.37 is proved. Lemma 6.38. Let M0 , M1 , τ0 > 0. Then there exists M2 > M1 such that for each T1 ≥ 0, each T2 ∈ (T1 , T1 + τ0 ], and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M1 ,
(6.42)
I f (T1 , T2 , x, u) ≤ M0
(6.43)
the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ].
(6.44)
Proof. Let us consider the first class of problems. Fix a positive number δ < min{16−1 τ0 , 8−1 (a0 + 1)−1 , 8−1 (a0 + 1)−1 Δ−1 δ }.
(6.45)
By (6.19) and (6.20), there exist h0 > M1 + 1 and γ0 > 0 such that f (t, x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (t, x, u) ∈ M satisfying x ≥ h0 , (6.46) f (t, x, u) ≥ 8(G(t, x, u) − a0 x)+ for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ γ0 .
(6.47)
Choose a number M2 > (8M1 + 8 + 2h0 + 2γ0 δ + M0 + a0 τ0 )(Δδ + 1).
(6.48)
Let T1 ≥ 0, T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (6.42) and (6.43) hold. We show that (6.44) hold. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that
6.6 Auxiliary Results for Theorems 6.2 and 6.3
407
x(t0 ) > M2 .
(6.49)
By the choice of h0 (see (6.46)), there exists t1 ∈ [T1 , t0 ] such that x(t1 ) ≤ h0 , t1 − t0  ≤ δ.
(6.50)
There exists t2 ∈ [t1 , t0 ] such that x(t2 ) ≥ x(t) for all t ∈ [t1 , t0 ].
(6.51)
In view of (6.48)–(6.51), (6.52)
t2 > t1 . By (6.11), (6.51), and (6.52),
t2
x(t2 ) − U (t2 , t1 )x(t1 ) =
U (t2 , s)G(x(s), u(s))ds.
(6.53)
t1
It follows from property (i) from Section 6.1, (6.50), and (6.53) that x(t2 )−Δδ h0 ≤ x(t2 )−Δδ x(t1 ) ≤ Δδ
t2
G(s, x(s), u(s))ds.
(6.54)
t1
Define E1 = {t ∈ [t1 , t2 ] : G(t, x(t), u(t)) ≥ a0 x(t) + γ0 }, E2 = [t1 , t2 ] \ E1 . By (6.43), (6.47), (6.50), (6.51) and (6.54)–(6.56), Δ−1 δ x(t2 ) ≤ h0 + ≤ h0 + a0
t2
t2
x(t)dt +
t1
t1
t2
G(s, x(s), u(s))ds
t1
(G(s, x(s), u(s)) − a0 x(s))+ ds
≤ h0 + a0 δx(t2 ) + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
+ E2
(G(t, x(t), u(t)) − a0 x(t))+ dt
(6.55) (6.56)
408
6 ContinuousTime Nonautonomous Problems on the HalfAxis
≤ h0 + a0 x(t2 )δ + γ0 δ + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ h0 + a0 x(t2 )δ + γ0 δ + 8
−1
f (t, x(t), u(t))dt E1
≤ h0 + a0 x(t2 )δ + γ0 δ + 8−1 (M0 + a0 τ0 ).
(6.57)
It follows from (6.45), (6.49), (6.51), and (6.57) that −1 −1 −1 2−1 M2 Δ−1 δ ≤ 2 Δδ x(t2 ) ≤ h0 + γ0 δ + 8 (M0 + a0 τ0 )
and M2 ≤ (2h0 + 2γ0 δ + 4−1 (M0 + a0 τ0 ))Δδ . This contradicts (6.48). The contradiction we have reached proves Lemma 6.38 for the first class of problems. Consider the second class of problems. Recall (see Proposition 1.3) that there exists M∗ ≥ 1 and ω∗ ∈ R 1 such that eAt ≤ M∗ eω∗ t , t ∈ [0, ∞).
(6.58)
and that for every τ ≥ 0,
τ
Φτ u =
eA(τ −s)Bu(s)ds, u ∈ L2 (0, τ ; F ),
(6.59)
0
where Φτ ∈ L(L2 (0, τ ; F ), E) (see Proposition 1.18). Choose a number δ ∈ (0, 1). By (6.19) and (6.21), there exists h0 > M1 + 1
(6.60)
and γ > 1 such that f (t, x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (t, x, u) ∈ M satisfying x ≥ h0 , (6.61) f (t, x, u) ≥ K0 u2 /2 for each u ∈ F satisfying u ≥ γ . (6.62) Choose M2 > M1 + M∗ eω∗ δ h0 + Φ1 γ δ 1/2 + Φ1 (2K0−1 (M0 + a0 τ0 ))1/2 .
(6.63)
6.6 Auxiliary Results for Theorems 6.2 and 6.3
409
Let T1 ≥ 0, T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (6.42) and (6.43) hold. We show that (6.44) hold. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that (6.49) holds. By the choice of h0 , (6.60), (6.61), there exists t1 ∈ [T1 , t0 ] satisfying (6.50). There exists t2 ∈ [t1 , t0 ] such that x(t2 ) ≥ x(t) for all t ∈ [t1 , t0 ].
(6.64)
Clearly, (x, u) ∈ X(t1 , t2 ), x(t2 ) = eA(t2 −t1 ) x(t1 ) +
t2
eA(t2 −s) Bu(s)ds
(6.65)
t1
in E−1 = D(A∗ ). Proposition 1.18, (6.50), (6.58), (6.59), and (6.65) imply that x(t2 ) ≤ x(t1 )M∗ eω∗ δ + ≤ M∗ eω∗ δ h0 +
t2 −t1
t2
eA(t2−s) Bu(s)ds
t1
eA(t2 −t1 −s) Bu(t1 + s)ds
0
≤ M∗ eω∗ δ h0 + Φt2 −t1 u(t1 + ·)L2 (0,t2 −t1 ;F ) ≤ M∗ e
ω∗ δ
h0 + Φ1 (
t2
u(s)2 ds)1/2 .
(6.66)
t1
Define Ω1 = {t ∈ [t1 , t2 ] : u(t) ≥ γ }, Ω2 = [t1 , t2 ] \ Ω1 . By (6.21), (6.43), (6.49), (6.50), (6.62)–(6.64), (6.60), and (6.67), M2 < x(t0 ) ≤ x(t2 ) ≤ M∗ eω∗ δ h0 +Φ1 [( u(s)2 ds)1/2 + ( u(s)2 ds)1/2 ] Ω1
≤ M∗ eω∗ δ h0 + Φ1 (2K0−1
Ω2
f (s, x(s), u(s))ds)1/2 + Φ1 γ δ 1/2 Ω1
≤ M∗ eω∗ δ h0 + Φ1 γ δ 1/2 + Φ1 (2K0−1 (M0 + a0 τ0 ))1/2 < M2 .
(6.67)
410
6 ContinuousTime Nonautonomous Problems on the HalfAxis
The contradiction we have reached proves Lemma 6.38 for the second class of problems. Lemma 6.38, (6.20), and (6.21) imply the following result. Lemma 6.39. Let M1 > 0, 0 < τ < τ0 < τ1 . Then there exists M2 > 0 such that for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 + τ, T2 ].
6.7 Proofs of Theorems 6.2 and 6.3 Proof of Theorem 6.2. Assertion 1 follows from Lemma 6.37, (6.20), (6.21), and (6.24). We will prove Assertion 2. Let (x, u) ∈ X(0, ∞). Assume that there exists a sequence of numbers {tk }∞ k=1 such that tk → ∞ as k → ∞, I f (0, tk , x, u) − I f (0, tk , xf , uf ) → ∞ as k → ∞.
(6.68)
Let a number S > 0 be as guaranteed by Assertion 1. Let k be a natural number and t ≥ tk . In view of (6.68) and the choice of S, I f (0, t, x, u) − I f (0, t, xf , uf ) = I f (0, tk , x, u) − I f (0, tk , xf , uf ) + I f (tk , t, x, u) − I f (tk , t, xf , uf ) ≥ I f (0, tk , x, u) − I f (0, tk , xf , uf ) − S → ∞ as k → ∞. Assume that sup{I f (0, t, x, u) − I f (0, t, xf , uf ) : t ∈ [0, ∞)} < ∞. Therefore there exists an integer S1 > 2 + a0 such that for each T2 > T1 ≥ 0, I f (T1 , T2 , x, u) − I f (T1 , T2 , xf , uf ) ≤ S1 .
(6.69)
6.7 Proofs of Theorems 6.2 and 6.3
411
Let S0 > 0 be such that ψ(S0 ) > 8Δf  + 32 + 4a0 .
(6.70)
We show that for each T ≥ 0, min{x(t) : t ∈ [T , T + S1 ]} ≤ S0 . Assume the contrary. Then there exists T ≥ 0 such that x(t) > S0 , t ∈ [T , T + S1 ].
(6.71)
By (6.20), (6.21), (6.24), (6.70), and (6.71), I f (T , T + S1 , x, u) ≥ −2a0 + I f (T + 1, T + S1 , x, u) ≥ −2a0 + (ψ(S0 ) − a0 )(S1 − 1) ≥ −2a0 + 4−1 ψ(S0 )S1 .
(6.72)
In view of (6.20), (6.21), and (6.24), I f (T , T + S1 , xf , uf ) ≤ I f (T , T + S1 + 1, xf , uf ) + 2a0 ≤ Δf (S1 + 1) + 2a0 ≤ 2Δf S1 + 2a0 .
(6.73)
It follows from (6.72) and (6.73) that I f (T , T + S1 , x, u) − I f (T , T + S1 , xf , uf ) ≥ 4−1 ψ(S0 )S1 − 4a0 − 2Δf S1 ≥ S1 (4−1 ψ(S0 ) − 2Δf ) − 4a0 ≥ −4a0 + 8S1 ≥ 4S1 . This contradicts (6.69). The contradiction we have reached proves that for each T ≥ 0, min{x(t) : t ∈ [T , T + S1 ]} ≤ S0 .
(6.74)
By (6.74), there exists a sequence {Ti }∞ i=1 ⊂ (0, ∞) such that for each integer i ≥ 1, Ti+1 − Ti ∈ [1, 1 + S1 ], x(Ti ) ≤ S0 . Lemma 6.38, (6.20), (6.21), (6.24), (6.69), and (6.75) imply that sup{x(t) : t ∈ [0, ∞)} < ∞. Theorem 6.2 is proved.
(6.75)
412
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Proof of Theorem 6.3. We may assume that c < 1/2 and c0 < c/4. By Theorem 6.2, there exists S0 > 0 such that the following property holds: (i) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ). Lemma 6.39 implies that there exists M1 > 0 such that the following property holds: (ii) for each T ≥ 0 and each (x, u) ∈ X(T , T + c/2) satisfying I f (T , T + c/2, x, u) ≤ M0 + 2S0 + 2 + 2a0 + 2Δf , we have x(t) ≤ M1 for all t ∈ [c0 + T , T + c/2]. Assume that T1 ≥ 0, T2 ≥ T1 + c, (x, u) ∈ X(T1 , T2 ) and I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 .
(6.76)
We show that x(t) ≤ M1 , t ∈ [T1 + c0 , T2 ]. Assume the contrary. Then there exists t0 ∈ [T1 + c0 , T2 ]
(6.77)
x(t0 ) > M1 .
(6.78)
such that
By (6.77), there exists a closed interval [a, b] ⊂ [T1 , T2 ] such that t0 ∈ [a + c0 , b], b − a = 2−1 c.
(6.79)
Property (ii), (6.78), and (6.79) imply that I f (a, b, x, u) > M0 + 2S0 + 2 + 2a0 + 2Δf . Property (i), (6.19)–(6.21), (6.24), (6.76), and (6.79) imply that I f (a, b, x, u) = I f (T1 , T2 , x, u) − I f (T1 , a, x, u) − I f (b, T2 , x, u)
(6.80)
6.8 Proofs of Theorems 6.4–6.10
413
≤ I f (T1 , T2 , xf , uf ) + M0 − I f (T1 , a, xf , uf ) + S0 − I f (b, T2 , xf , uf ) + S0 = I f (a, b, xf , uf ) + 2S0 + M0 ≤ 2Δf + 2a + 2S0 + M0 . This contradicts (6.86). The contradiction we have reached proves Theorem 6.3.
6.8 Proofs of Theorems 6.4–6.10 Proof of Theorem 6.4. Theorem 6.2 implies that there exists M1 > 0 such that the following property holds: (i) for each T1 ≥ 0, each T2 ≥ T1 + 2L and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + 2L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + 2L and (x, u) ∈ X(T1 , T2 ) satisfies L , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 .
(6.81) (6.82)
In view of (6.81), there exist τ1 ∈ (0, L], τ2 ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), y(T2 ) = x(T2 ), (y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ1 + T1 , T2 − τ2 ], I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.82), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) ≤ M0 + I f (T1 , T1 + τ1 , y, v) + I f (T1 + τ1 , T2 − τ2 , y, v) + I f (T2 − τ2 , T2 , y, v) ≤ 2L + M0 + I f (T1 + τ1 , T2 − τ2 , xf , uf ) ≤ 2L + M0 + I f (T1 , T2 , xf , uf ) + 2La0 .
(6.83)
414
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Property (i) and (6.83) imply that x(t) ≤ M1 for all t ∈ [T1 + c, T2 ]. Theorem 6.4 is proved. Proof of Theorem 6.5. Theorem 6.3 implies that there exists M1 > 0 such that the following property holds: (ii) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L and (x, u) ∈ X(T1 , T2 ) satisfies (T1 , x(T1 )) ∈ AL ,
(6.84)
I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 .
(6.85)
In view of (6.84), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), (y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ + T1 , T2 ], I f (T1 , T1 + τ, y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.85), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ, y, v) + I f (T1 + τ, T2 , xf , uf ) + M0 ≤ L + M0 + I f (T1 , T2 , xf , uf ) + La0 .
(6.86)
Property (ii) and (6.86) imply that x(t) ≤ M1 for all t ∈ [T1 +c, T2 ]. Theorem 6.5 is proved. Proof of Theorem 6.6. Theorem 6.3 implies that there exists M1 > 0 such that the following property holds: (iii) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 + c, T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L and (x, u) ∈ X(T1 , T2 ) satisfies L , (T2 , x(T2 )) ∈ A
(6.87)
6.8 Proofs of Theorems 6.4–6.10
I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + M0 .
415
(6.88)
In view of (6.87), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that (y(t), v(t)) = (xf (t), uf (t)), t ∈ [T1 , T2 − τ ], y(T2 ) = x(T2 ), I f (T2 − τ, T2 , y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.88), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T2 − τ, y, v) + I f (T2 − τ, T2 , y, v) + M0 ≤ L + M0 + I f (T1 , T2 − τ, xf , uf ) ≤ L + M0 + I f (T1 , T2 , xf , uf ) + La0 .
(6.89)
Property (iii) and (6.89) imply that x(t) ≤ M1 for all t ∈ [T1 +c, T2 ]. Theorem 6.6 is proved. Proof of Theorem 6.7. We may assume that c < 1. Theorem 6.2 implies that there exists S0 > 0 such that the following property holds: (iv) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), we have I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ). By Theorem 6.3, there exists M2 > 0 such that the following property holds: (v) for each T1 ≥ 0, each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ), satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 the inequality x(t) ≤ M2 holds for all t ∈ [T1 + c/4, T2 ]. Lemma 6.38 implies that there exists M1 > M2 such that the following property holds: (vi) for each T ≥ 0 and each (x, u) ∈ X(T , T + c/4) satisfying x(T ) ≤ M0 , I f (T , T + c/4, x, u) ≤ M0 + S0 + a0 c + 2Δf  + 2a0 , we have x(t) ≤ M1 for all t ∈ [T , T + c/4].
416
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Assume that T1 ≥ 0, T2 ≥ T1 + c, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 ,
(6.90)
I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 .
(6.91)
Property (v) and (6.91) imply that x(t) ≤ M2 < M1 , t ∈ [T1 + c/4, T2 ].
(6.92)
It follows from (6.20), (6.21), (6.24), (6.91), and property (iv) that I f (T1 , T1 + c/4, x, u) = I f (T1 , T2 , x, u) − I f (T1 + c/4, T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 − I f (T1 + c/4, T2 , x, u) + S0 ≤ S0 + M0 + I f (T1 , T1 + c/4, xf , uf ) ≤ M0 + S0 + 2Δf  + 2a0 . By the relation above, (6.90) and property (vi), x(t) ≤ M1 , t ∈ [T1 , T1 + c/4]. Theorem 6.7 is proved. Proof of Theorem 6.8. Theorem 6.7 implies that there exists M1 > 0 such that the following property holds: (vii) for each T1 ≥ 0, each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + 2L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + 2L, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 ,
(6.93)
L , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A
(6.94)
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 .
(6.95)
In view of (6.94), there exist τ1 ∈ (0, L], τ2 ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that
6.8 Proofs of Theorems 6.4–6.10
417
y(T1 ) = x(T1 ), y(T2 ) = x(T2 ), (y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ1 + T1 , T2 − τ2 ], I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.95), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ1 , y, v) + I f (T1 + τ1 , T2 − τ2 , y, v) + I f (T2 − τ2 , T2 , y, v) + M0 ≤ 2L + M0 + I f (T1 + τ1 , T2 − τ2 , xf , uf ) ≤ I f (T1 , T2 , xf , uf ) + 2L + M0 + 2La0 . Property (vii), (6.93), and the relation above imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 6.8 is proved. Proof of Theorem 6.9. Theorem 6.7 implies that there exists M1 > 0 such that the following property holds: (viii) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 ,
(6.96)
(T1 , x(T1 )) ∈ AL ,
(6.97)
I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 .
(6.98)
In view of (6.97), there exist τ ∈ (0, L], (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), (y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ + T1 , T2 ], I f (T1 , T1 + τ, y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.98),
418
6 ContinuousTime Nonautonomous Problems on the HalfAxis
I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T1 + τ, y, v) + I f (T1 + τ, T2 , xf , uf ) + M0 ≤ L + M0 + I f (T1 , T2 , xf , uf ) + La0 . The relations above, property (viii), and (6.96) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 6.9 is proved. Proof of Theorem 6.10. Theorem 6.7 implies that there exists M1 > 0 such that the following property holds: (ix) for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ) ≤ M0 , I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + L(1 + a0 ) the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ), x(T1 ) ≤ M0 ,
(6.99)
L , (T2 , x(T2 )) ∈ A
(6.100)
I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T2 )) + M0 .
(6.101)
In view of (6.100), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that (y(t), v(t)) = (xf (t), uf (t)), t ∈ [T1 , T2 − τ ], y(T2 ) = x(T2 ), I f (T2 − τ, T2 , y, v) ≤ L. By the relations above, (6.20), (6.21), and (6.101), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) = I f (T1 , T2 − τ, y, v) + I f (T2 − τ, T2 , y, v) + M0 ≤ L + M0 + I f (T1 , T2 − τ, xf , uf ) ≤ L + M0 + La0 + I f (T1 , T2 , xf , uf ). The relations above, property (ix) and (6.99) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 6.10 is proved.
6.9 Proof of Theorem 6.22
419
6.9 Proof of Theorem 6.22 First we show that TP implies (P1) and (P2). In view of Theorem 6.11, TP implies (P2). We show that TP implies (P1). Assume that TP holds, (x, u) ∈ X(0, ∞) is (f )good. Let > 0. There exists S > 0 such that for each T > 0, I f (0, T , x, u) − I f (0, T , xf , uf ) < S. This implies that for each pair of numbers T2 > T1 ≥ 0, I f (T1 , T2 , x, u) − I f (T1 , T2 , xf , uf ) < 2S.
(6.102)
Let δ > 0. We show that there exists Tδ > 0 such that for each T > Tδ , I f (Tδ , T , x, u) ≤ U f (Tδ , T , x(Tδ ), x(T )) + δ. Assume the contrary. Then for each T ≥ 0 there exists S > T such that I f (T , S, x, u) > U f (T , S, x(T ), x(S)) + δ. This implies that there exists a strictly increasing sequence of numbers {Ti }∞ i=0 such that T0 = 0 and for every integer i ≥ 0, I f (Ti , Ti+1 , x, u) > U f (Ti , Ti+1 , x(Ti ), x(Ti+1 )) + δ. By the relation above, there exists (y, v) ∈ X(0, ∞) such that for every integer i ≥ 0, y(Ti ) = x(Ti ), I f (Ti , Ti+1 , x, u) > I f (Ti , Ti+1 , y, v) + δ. Combined with (6.102) this implies that for each integer k ≥ 1, I f (0, Tk , y, v) − I f (0, Tk , xf , uf ) = I f (0, Tk , y, v) − I f (0, Tk , x, u) + I f (0, Tk , x, u) − I f (0, Tk , xf , uf ) ≤ −kδ + 2S → −∞ as k → ∞.
420
6 ContinuousTime Nonautonomous Problems on the HalfAxis
This contradicts Theorem 6.1. The contradiction we have reached proves that the following property holds: (i) for each δ > 0 there exists Tδ > 0 such that for each T > Tδ , I f (Tδ , T , x, u) ≤ U f (Tδ , T , x(Tδ ), x(T )) + δ. Theorem 6.11 implies that there exist δ > 0, L > 0 such that the following property holds: (ii) for each S1 ≥ 0, each S2 ≥ S1 +2L, and each (z, ξ ) ∈ X(S1 , S2 ) which satisfies I f (S1 , S2 , z, ξ ) ≤ min{I f (S1 , S2 , xf , uf )+2S, U f (S1 , S2 , z(S1 ), z(S2 ))+δ} we have z(t) − xf (t) ≤ , t ∈ [S1 + L, S2 − L]. Let Tδ > 0 be as guaranteed by (i). Properties (i) and (ii), (6.102), and the choice of Tδ imply that for each T ≥ Tδ + 2L, x(t) − xf (t) ≤ and (P1) holds. Thus TP implies (P1). Lemma 6.40. Assume that (P1) holds and that > 0. Then there exists δ, L > 0 such that for each T1 ≥ L, each T2 ≥ T1 + 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ the inequality x(t) − xf (t) ≤ holds for all t ∈ [T1 , T2 ]. Proof. By (A2), for each integer k ≥ 1 there exists δk ∈ (0, 2−k ) such that the following property holds: (iii) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δk , i = 1, 2 and T2 ≥ bf , there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfy x1 (T1 ) = z1 , x1 (T1 +τ1 ) = xf (T1 +τ1 ), x2 (T2 ) = z2 , x2 (T2 −τ2 ) = xf (T2 −τ2 ),
I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + 2−k , I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + 2−k .
6.9 Proof of Theorem 6.22
421
Assume that the lemma does not hold. Then for each integer k ≥ 1 there exist Tk,1 ≥ k + bf , Tk,2 ≥ Tk,1 + 2bf ,
(6.103)
and (xk , uk ) ∈ X(Tk,1 , Tk,2 ) such that xk (Tk,i ) − xf (Tk,i ) ≤ δk , i = 1, 2,
(6.104)
I f (Tk,1 , Tk,2 , xk , uk ) ≤ U f (Tk,1 , Tk,2 , xk (Tk,1 ), xk (Tk,2 )) + δk ,
(6.105)
sup{xk (t) − xf (t) : t ∈ [Tk,1 , Tk,2 ]} > .
(6.106)
Extracting a subsequence and reindexing, we may assume without loss of generality that for each integer k ≥ 1, Tk+1,1 ≥ Tk,2 + 4bf .
(6.107)
Let k ≥ 1 be an integer. Property (iii), (6.103), and (6.104) imply that there exist τk,1 , τk,2 ∈ (0, bf ] and (x˜k , u˜ k ) ∈ X(Tk,1 − τk,1 , Tk,2 + τk,2 ) such that (x˜k (t), u˜ k (t)) = (xk (t), uk (t)), t ∈ [Tk,1, Tk,2 ],
(6.108)
x˜k (Tk,1 − τk,1 ) = xf (Tk,1 − τk,1 ), x˜k (Tk,2 + τk,2 ) = xf (Tk,2 + τk,2 ),
(6.109)
I f (Tk,1 − τk,1 , Tk,1 , x˜k , u˜ k ) ≤ I f (Tk,1 − τk,1 , Tk,1 , xf , uf ) + 2−k ,
(6.110)
I f (Tk,2 , Tk,2 + τk,2 , x˜k , u˜ k ) ≤ I f (Tk,2 , Tk,2 + τk,2 , xf , uf ) + 2−k .
(6.111)
Property (iii), (6.103), and (6.104) imply that there exist τk,3 , τk,4 ∈ (0, bf ] and ( xk , uk ) ∈ X(Tk,1 , Tk,2 ) such that xk (Tk,1 ) = xk (Tk,1 ), xk (Tk,2 ) = xk (Tk,2 ),
(6.112)
( xk (t), uk (t)) = (xf (t), uf (t)), t ∈ [Tk,1 + τk,3 , Tk,2 − τk,4 ],
(6.113)
I f (Tk,1 , Tk,1 + τk,3 , xk , uk ) ≤ I f (Tk,1 , Tk,1 + τk,3 , xf , uf ) + 2−k ,
(6.114)
xk , uk ) ≤ I f (Tk,2 − τk,4 , Tk,2 , xf , uf ) + 2−k . I f (Tk,2 − τk,4 , Tk,2 ,
(6.115)
In view of (6.112), xk , uk ) U f (Tk,1 , Tk,2 , xk (Tk,1 ), xk (Tk,2 )) ≤ I f (Tk,1 , Tk,2 , ≤ I f (Tk,1 , Tk,2 , xf , uf ) + 2−k+1 .
(6.116)
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
By ((6.105) and (6.116), I f (Tk,1 , Tk,2 , xk , uk ) ≤ I f (Tk,1 , Tk,2 , xf , uf ) + 2−k+2 .
(6.117)
It follows from (6.108), (6.110), (6.111), and (6.117) that xk , uk ) I f (Tk,1 − τk,1 , Tk,2 + τk,2 , ≤ I f (Tk,1 − τk,1 , Tk,1 , xf , uf ) +I f (Tk,1 , Tk,2 , xk , uk ) + I f (Tk,2 , Tk,2 + τk,2 , xf , uf ) + 2−k+1 ≤ I f (Tk,1 − τk,1 , Tk,2 + τk,2 , xf , uf ) + 2−k+3 .
(6.118)
By (6.107) and (6.109), there exists (x, u) ∈ X(0, ∞) such that for each integer k ≥ 1, (x(t), u(t)) = (x˜k (t), u˜ k (t)), t ∈ [Tk,1 − τk,1 , Tk,2 + τk,2 ],
(6.119)
(x(t), u(t)) = (xf (t), uf (t)), t ∈ [0, ∞) \ ∪∞ k=1 [Tk,1 − τk,1 , Tk,2 + τk,2 ]. (6.120) It follows from (6.118)–(6.120) that for each integer k ≥ 1, I f (0, Tk,2 + τk,2 , x, u) − I f (0, Tk,2 + τk,2 , xf , uf ) =
k
(I f (Ti,1 − τi,1 , Ti,2 + τi,2 , x˜i , u˜ i ) − I f (Ti,1 − τi,1 , Ti,2 + τi,2 , xf , uf ))
i=1
≤
k
2−i+3 ≤ 8.
i=1
Theorem 6.1 implies that (x, u) is (f )good. Together with (P1) this implies that limt →∞ x(t) − xf (t) = 0. On the other hand in view of (6.106), (6.108), (6.117), and (6.119), lim supt →∞ x(t) − xf (t) ≥ . The contradiction we have reached completes the proof of Lemma 6.40. Completion of the Proof of Theorem 6.22. Assume that properties (P1) and (P2) hold. Let , M > 0. By Lemma 6.40, there exist δ0 , L0 > 0 such that the following property holds: (iv) for each T1 ≥ L0 , each T2 ≥ T1 + 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies
6.9 Proof of Theorem 6.22
423
x(Ti ) − xf (Ti ) ≤ δ0 , i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ0 the inequality x(t) − xf (t) ≤ holds for all t ∈ [T1 , T2 ]. By Theorem 6.1, there exists S0 > 0 such that the following property holds: (v) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), we have I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ). In view of (P2), there exist δ ∈ (0, δ0 ) and L1 > 0 such that the following property holds: (vi) for each T ≥ 0 and each (x, u) ∈ X(T , T + L1 ) which satisfies I f (T , T + L1 , x, u) ≤ min{U f (T , T + L1 , x(T ), x(T + L1 )) + δ, I f (T , T + L1 , xf , uf ) + 2S0 + M}, there exists S0 ∈ [0, L1 ] such that x(T + S0 ) − xf (T + S0 ) ≤ δ0 . Set L = L0 + L1 + bf .
(6.121)
Assume that T1 ≥ 0, T2 ≥ T1 + 2L and that (x, u) ∈ X(T1 , T2 ) satisfies I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 ) + M,
(6.122)
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(6.123)
I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M.
(6.124)
By (6.122),
Property (v) and (6.124) imply that for each pair of numbers Q1 , Q2 ∈ [T1 , T2 ] satisfying Q1 < Q2 , I f (Q1 , Q2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , Q1 , x, u) − I f (Q2 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M − I f (T1 , Q1 , xf , uf ) + S0 − I f (Q2 , T2 , xf , uf ) + S0 = I f (Q1 , Q2 , xf , uf ) + M + 2S0 .
(6.125)
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
It follows from (6.125) that I f (max{T1 , L0 }, max{T1 , L0 } + L1 , x, u) ≤ I f (max{T1 , L0 }, max{T1 , L0 } + L1 , xf , uf ) + 2S0 + M,
(6.126)
I f (T2 − L1 , T2 , x, u) ≤ I f (T2 − L1 , T2 , xf , uf ) + 2S0 + M.
(6.127)
By (6.123), (6.126), (6.127), and property (v), there exist τ1 ∈ [max{T1 , L0 }, max{T1 , L0 } + L1 ], τ2 ∈ [T2 − L1 , T2 ]
(6.128)
such that x(τi ) − xf (τi ) ≤ δ0 , i = 1, 2.
(6.129)
If x(T2 ) − xf (T2 ) ≤ δ, then we may assume that τ2 = T2 , and if T1 ≥ L0 and x(T1 ) − xf (T1 ) ≤ δ, then we may assume that τ1 = T1 . In view of (6.121) and (6.128), τ2 − τ1 ≥ T2 − T1 − L1 − L1 − L0 ≥ L0 + 2bf .
(6.130)
Property (iv), (6.123), and (6.128)–(6.130) imply that x(t) − xf (t) ≤ , t ∈ [τ1 , τ2 ]. Thus TP holds and Theorem 6.22 is proved.
6.10 An Auxiliary Result for Theorem 6.23 Lemma 6.41. Assume that (P2) holds and that M, > 0. Then there exists L > 0 such that for each T ≥ 0 and each (x, u) ∈ X(T , T + L) which satisfies I f (T , T + L, x, u) ≤ I f (T , T + L, xf , uf ) + M the following inequality holds: inf{x(t) − xf (t) : t ∈ [T , T + L]} ≤ . Proof. By Theorem 6.1, there exists S0 > 0 such that the following property holds:
6.10 An Auxiliary Result for Theorem 6.23
425
(i) for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ). By (P2), there exist δ0 ∈ (0, ) and L0 > 0 such that the following property holds: (ii) for each T ≥ 0 and each (x, u) ∈ X(T , T + L0 ) which satisfies I f (T , T + L0 , x, u) ≤ min{U f (T , T + L0 , x(T ), x(T + L0 )) + δ0 , I f (T , T + L0 , xf , uf ) + 2S0 + M} we have inf{x(t) − xf (t) : t ∈ [T , T + L0 ]} ≤ . Choose an integer q0 > (M + S0 )δ0−1
(6.131)
L = q0 L0 .
(6.132)
and set
Let T ≥ 0 and (x, u) ∈ X(T , T + L) satisfy I f (T , T + L, x, u) ≤ I f (T , T + L, xf , uf ) + M.
(6.133)
We show that inf{x(t) − xf (t) : t ∈ [T , T + L]} ≤ . Assume the contrary. Then x(t) − xf (t) > , t ∈ [T , T + L].
(6.134)
Let an integer i ∈ {0, . . . , q0 − 1}. Property (i), (6.132), and (6.133) imply that I f (T + iL0 , T + (i + 1)L0 , x, u) = I f (T , T + L, x, u) −I f (T , T + iL0 , x, u) − I f (T + (i + 1)L0 , T + L, x, u) ≤ I f (T , T + L, xf , uf ) + M − I f (T , T + iL0 , xf , uf ) + S0
426
6 ContinuousTime Nonautonomous Problems on the HalfAxis
−I f (T + (i + 1)L0 , T + L, xf , uf ) + S0 = I f (T + iL0 , T + (i + 1)L0 , xf , uf ) + M + 2S0 .
(6.135)
By (6.134), (6.135), and property (ii), I f (T + iL0 , T + (i + 1)L0 , x, u) > U f (T + iL0 , T + (i + 1)L0 , x(T + iL0 ), x(T + (i + 1)L0 )) + δ0 .
(6.136)
It follows from (6.136) that there exists (y, v) ∈ X(T , T + L) such that y(T + iL0 ) = x(T + iL0 ), i = 0, . . . , q0 and for all i = 0, . . . , q0 − 1, I f (T +iL0 , T +(i+1)L0, x, u) > I f (T +iL0 , T +(i+1)L0, y, v)+δ0 .
(6.137)
In view of (6.133), (6.137), and property (i), I f (T , T + L, xf , uf ) + M ≥ I f (T , T + L, x, u) ≥ I f (T , T + L, y, v) + q0 δ0 ≥ I f (T , T + L, xf , uf ) − S0 + q0 δ0 and q0 ≤ (M + S0 )δ0−1 . This contradicts (6.132). The contradiction we have reached proves Lemma 6.41.
6.11 Proof of Theorem 6.23 By Theorem 6.1, there exists S0 > 0 such that the following property holds: (i) for each τ2 > τ1 ≥ 0 and each (y, v) ∈ X(τ1 , τ2 ), I f (τ1 , τ2 , y, v) + S0 ≥ I f (τ1 , τ2 , xf , uf ). Theorem 6.22 implies that there exist δ ∈ (0, ) and L0 > 0 such that the following property holds:
6.11 Proof of Theorem 6.23
427
(ii) for each τ1 ≥ 0, each τ2 ≥ τ1 +2L0 , and each (y, v) ∈ X(τ1 , τ2 ) which satisfies I f (τ1 , τ2 , y, v) ≤ min{σ f (τ1 , τ2 ) + M + 3S0 , U f (τ1 , τ2 , y(τ1 ), y(τ2 )) + δ} we have y(t) − xf (t) ≤ , t ∈ [τ1 + L0 , τ2 − L0 ]. Set l = 2L0 + 1.
(6.138)
Q ≥ 2 + 2(M + S0 )δ −1 .
(6.139)
Choose a natural number
Assume that T1 ≥ 0, T2 ≥ T1 + lQ and (x, u) ∈ X(T1 , T2 ) satisfy I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M.
(6.140)
Property (i) and (6.140) imply that for each pair of numbers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , I f (τ1 , τ2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , τ1 , x, u) − I f (τ2 , T2 , x, u) ≤ M + I f (T1 , T2 , xf , uf ) − I f (T1 , τ1 , xf , uf ) + S0 − I f (τ2 , T2 , xf , uf ) + S0 ≤ I f (τ1 , τ2 , xf , uf ) + M + 2S0 .
(6.141)
t0 = T1 .
(6.142)
Set
If I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, then we set t1 = T2 , s1 = T2 .
(6.143)
Assume that I f (T1 , T2 , x, u) > U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(6.144)
428
6 ContinuousTime Nonautonomous Problems on the HalfAxis
It is easy to see that for each t ∈ (T1 , T2 ) such that t − T1 is sufficiently small, we have I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) < δ. Set t˜1 = inf{t ∈ (T1 , T2 ] : I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) > δ}. Clearly, t˜1 ∈ (T1 , T2 ] is welldefined. There exist s1 , t1 ∈ [T1 , T2 ] such that t0 < s1 < t˜1 , s1 ≥ t˜1 − 1/4, I f (t0 , s1 , x, u) − U f (t0 , s1 , x(t0 ), x(s1 )) ≤ δ, t˜1 ≤ t1 < t˜1 + 1/4, t1 ≤ T2 , I f (t0 , t1 , x, u) − U f (t0 , t1 , x(t0 ), x(t1 )) > δ.
(6.145) (6.146) (6.147) (6.148)
Assume that k ≥ 1 is an integer and we defined finite sequences of numbers {ti }ki=0 ⊂ [T1 , T2 ], {si }ki=1 ⊂ [T1 , T2 ] such that T1 = t0 < t1 · · · < tk ,
(6.149)
ti−1 < si ≤ ti , ti − si ≤ 2−1 ,
(6.150)
for each integer i = 1, . . . , k,
I f (ti−1 , si , x, u) − U f (ti−1 , si , x(ti−1 ), x(si )) ≤ δ,
(6.151)
and if ti < T2 , then I f (ti−1 , ti , x, u) − U f (ti−1 , ti , x(ti−1 ), x(ti )) > δ.
(6.152)
(Note that in view of (6.142), (6.143), and (6.145)–(6.148), our assumption holds for k = 1.) By (6.152), there exists (y, v) ∈ X(T1 , T2 ) such that y(ti ) = x(ti ), i = 1, . . . , k, I f (ti−1 , ti , x, u)−I f (ti−1 , ti , y, v) > δ for all integers i ∈ [1, k] \{k}, (y(t), v(t)) = (x(t), u(t)), t ∈ [tk−1 , T2 ].
(6.153) (6.154)
6.11 Proof of Theorem 6.23
429
Property (i), (6.140), (6.153), and (6.154) imply that I f (T1 , T2 , xf , uf ) + M ≥ I f (T1 , T2 , x, u) ≥ I f (T1 , T2 , y, v) + δ(k − 1) ≥ I f (T1 , T2 , xf , uf ) − S0 + δ(k − 1), k ≤ 1 + δ −1 (M + S0 ).
(6.155)
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If I f (tk , T2 , x, u) ≤ U f (tk , T2 , x(tk ), x(T2 )) + δ, then we set tk+1 = T2 , sk+1 = T2 and the construction of the sequence is completed. Assume that I f (tk , T2 , x, u) > U f (tk , T2 , x(tk ), x(T2 )) + δ. It is easy to see that for each t ∈ (tk , T2 ) such that t − tk is sufficiently small, we have I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) < δ. Set t˜ = inf{t ∈ (tk , T2 ] : I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) > δ}. Clearly, t˜ is welldefined and t˜ ∈ (tk , T2 ]. There exist sk+1 , tk+1 ∈ [T1 , T2 ] such that tk < sk+1 < t˜, sk+1 ≥ t˜ − 1/4, t˜ + 4−1 > tk+1 ≥ t˜, I f (tk , sk+1 , x, u) − U f (tk , sk+1 , x(tk ), x(sk+1 )) ≤ δ, I f (tk , tk+1 , x, u) − U f (tk , tk+1 , x(tk ), x(tk+1 )) > δ. It is not difficult to see that the assumption made for k also holds for k + 1. In q view of (6.155), by induction we constructed finite sequences {ti }i=0 ⊂ [T1 , T2 ], q {si }i=1 ⊂ [T1 , T2 ] such that q ≤ 1 + δ −1 (M + S0 ), T1 = t0 < t1 · · · < tq = T2 ,
(6.156)
430
6 ContinuousTime Nonautonomous Problems on the HalfAxis
for each integer i = 1, . . . , q, ti−1 < si ≤ ti , ti − si ≤ 2−1 , I f (ti−1 , si , x, u) − U f (ti−1 , si , x(ti−1 ), x(si )) ≤ δ,
(6.157) (6.158)
and if ti < T2 , then I f (ti−1 , ti , x, u) − U f (ti−1 , ti , x(ti−1 ), x(ti )) > δ.
(6.159)
Assume that i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 2L0 + 1.
(6.160)
By (6.157) and (6.160), si+1 − ti ≥ 2L0 .
(6.161)
In view of (6.158), I f (ti , si+1 , x, u) − U f (ti , si+1 , x(ti ), x(si+1 )) ≤ δ.
(6.162)
Property (i) and (6.140) imply that I f (ti , si+1 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , ti , x, u) − I f (si+1 , T2 , x, u) ≤ M + I f (T1 , T2 , xf , uf ) − I f (T1 , ti , xf , uf ) + S0 − I f (si+1 , T2 , xf , uf ) + S0 ≤ I f (ti , si+1 , xf , uf ) + M + 2S0 ≤ M + 3S0 + σ f (ti , si+1 ).
(6.163)
It follows from (6.161)–(6.163) and property (ii) that x(t) − xf (t) ≤ , t ∈ [ti + L0 , si+1 − L0 ] and together with (6.168) this implies that x(t) − xf (t) ≤ , t ∈ [ti + L0 , ti+1 − L0 − 1].
(6.164)
By (6.163), {t ∈ [T1 , T2 ] : x(t) − xf (t) > } ⊂ ∪{[ti , ti +L0 ]∪[ti+1 −1−L0 , ti+1 ] : i ∈ {0, . . . , q −1} and ti+1 −ti ≥ 2L0 +1} ∪{[ti , ti+1 ] : i ∈ {0, . . . , q − 1} and ti+1 − ti < 2L0 + 1}.
6.12 An Auxiliary Result
431
In view of (6.139) and (6.156), the maximal length of intervals in the righthand side of the relation above does not exceed 2L0 + 1 = l and their number does not exceed 2q ≤ Q. Theorem 6.23 is proved.
6.12 An Auxiliary Result Lemma 6.42. Let > 0. Then there exist δ, L > 0 such that for each T1 ≥ L, each T2 > T1 and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2 the following inequality holds: I f (T1 , T2 , T , x, u) ≥ I f (T1 , T2 , T , xf , uf ) − . Proof. By (A2), there exists δ > 0 such that the following property holds: (i) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ], (x˜1 , u˜ 1 ) ∈ X(T , T + τ1 ) such that x˜1 (T ) = z , x˜1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x˜1 , u˜ 1 ) ≤ I f (T , T + τ1 , xf , uf ) + /4, and if T ≥ bf , then there exist τ2 ∈ (0, bf ] and (x˜2 , u˜ 2 ) ∈ X(T − τ2 , T ) such that x˜2 (T − τ2 ) = xf (T − τ2 ) , x˜2 (T ) = z, I f (T − τ2 , T , x˜2 , u˜ 2 ) ≤ I f (T − τ2 , T , xf , uf ) + /4. Theorem 6.1 implies that there exists L > 2bf such that for each T1 ≥ L − bf and each T2 > T1 I f (T1 , T2 , xf , uf ) < U f (T1 , T2 , xf (T1 ), xf (T2 )) + /4.
(6.165)
Assume that T1 ≥ L, T2 > T1 , (x, u) ∈ X(T1 , T2 ), and x(T1 ) − xf (T1 ) ≤ δ, x(T2 ) − xf (T2 ) ≤ δ.
(6.166)
By (6.166) and property (i), there exists τ1 , τ2 ∈ (0, bf ] and (y, v) ∈ X(T1 − τ1 , T2 + τ2 ) such that
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
y(T1 − τ1 ) = xf (T1 − τ1 ),
(6.167)
(y(t), v(t)) = (x(t), u(t)) for all t ∈ [T1 , T2 ],
(6.168)
I f (T1 − τ1 , T1 , y, v) ≤ I f (T1 − τ1 , T1 , xf , uf ) + /4,
(6.169)
y(T2 + τ2 ) = xf (T2 + τ2 ), I f (T2 , T2 + τ2 , y, v) ≤ I f (T2 , T2 + τ2 , xf , uf ) + /4.
(6.170) (6.171)
It follows from (6.165), (6.167), and (6.170) that I f (T1 − τ1 , T2 + τ2 , xf , uf ) < I f (T1 − τ1 , T2 + τ2 , y, v) + /4.
(6.172)
Relations (6.168), (6.169), (6.171), and (6.172) imply that I f (T1 , T2 , x, u) = I f (T1 − τ1 , T2 + τ2 , y, v) −I f (T1 − τ1 , T1 , y, v) − I f (T2 , T2 + τ2 , y, v) ≥ I f (T1 − τ1 , T2 + τ2 , xf , uf ) − /4 − I f (T1 − τ1 , T1 , xf , uf ) −/4 − I f (T2 , T2 + τ2 , xf , uf ) − /4 = I f (T1 , T2 , xf , uf ) − 3/4. Lemma 6.42 is proved.
6.13 Proof of Proposition 6.27 Proposition 1.3 implies that there exist M∗ ≥ 1, ω∗ ∈ R 1 such that eAt ≤ M∗ eω∗ t , t ∈ [0, ∞), in the case of the second class of problems. We show that (P1) holds. Assume that (x, u) ∈ X(0, ∞) is (f )good. We show that limt →∞ x(t) − xf (t) = 0. Assume the contrary. Then there exists > 0 and a sequence of numbers {tk }∞ k=1 such that
6.13 Proof of Proposition 6.27
433
t1 ≥ 8, tk+1 ≥ tk + 8 for all integers k ≥ 1,
(6.173)
x(tk ) − xf (tk ) > for all integers k ≥ 1.
(6.174)
Theorem 6.2 implies that there exists M1 > 0 such that xf (t) ≤ M1 , x(t) ≤ M1 , t ∈ [0, ∞),
(6.175)
I f (0, t, x, u) − I f (0, t, xf , uf ) < M1 , t ∈ (0, ∞). By the relation above, for all S > T ≥ 0, I f (T , S, x, u) − I f (T , S, xf , uf ) ≤ 2M1 .
(6.176)
Lemma 6.1 implies that there exists M2 > 0 such that I f (S1 , S2 , xf , uf ) ≤ M2 for all S1 ≥ 0 and all S2 ∈ (S1 , S1 + 1].
(6.177)
Fix 1 = 4−1 (2Δ1 + 1)−1 .
(6.178)
Form now we consider the two classes of problems separately and first consider the first class of problems. By (6.19) and (6.20), there exists h0 > 0 such that f (t, x, u) ≥ 41−1 (M1 + M2 + a0 + 8)(G(t, x, u) − a0 x+ )
(6.179)
for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ h0 . There exists δ ∈ (0, 1) such that δ < /8, Δ1 δ < 1 /32, δ(a0 M1 + h0 ) < 1 /4.
(6.180)
Let k ≥ 1 be an integer. We show that for all t ∈ [tk − δ, tk ], x(t) − xf (t) ≥ δ.
(6.181)
Assume the contrary. Then there exists τ ∈ [tk − δ, tk ]
(6.182)
x(τ ) − xf (τ ) < δ.
(6.183)
such that
434
6 ContinuousTime Nonautonomous Problems on the HalfAxis
By (6.11) and (6.182),
tk
x(tk ) = U (tk , τ )x(τ ) +
U (tk , s)G(s, x(s), u(s))ds,
(6.184)
U (tk , s)G(s, xf (s), uf (s))ds.
(6.185)
τ
tk
xf (tk ) = U (tk , τ )xf (τ ) + τ
Property (v) (from Section 6.1), (6.181), (6.182), (6.184), and (6.185) imply that x(tk ) − xf (tk ) = U (tk , τ )xf (τ ) − x(τ )
tk
+
U (tk , s)G(s, x(s), u(s))ds
τ
tk
+
U (tk , s)G(s, xf (s), uf (s))ds
τ
tk
≤ Δδ xf (τ ) − x(τ ) + Δδ
G(s, x(s), u(s))ds
τ
tk
+Δδ
G(s, xf (s), ug (s))ds
τ
tk
≤ Δ1 δ + Δδ
tk
G(s, x(s), u(s))ds + Δδ
τ
G(s, xf (s), uf (s))ds.
τ
(6.186)
Let (y, v) ∈ {(x, u), (xf , uf )}.
(6.187)
Set Ω1 = {t ∈ [τ, tk ] : G(t, y(t), v(t)) − a0 y(t) ≥ h0 }, Ω2 = [τ, tk ] \ Ω1 . (6.188) By (6.20), (6.175)–(6.177), (6.180), (6.182), (6.187), and (6.188),
tk
G(s, y(s), v(s))ds ≤ a0
τ
tk
y(s)ds
τ
+ τ
tk
(G(s, y(s), v(s)) − a0 y(s))+ ds
6.13 Proof of Proposition 6.27
435
≤ a0 δM1 + h0 δ + Ω1
≤ a0 δM1 + h0 δ + 4
−1
(G(s, y(s), v(s)) − a0 y(s))+ ds
1 (M1 + M2 + a0 + 8)
−1
f (s, y(s), v(s))ds Ω1
≤ a0 δM1 + h0 δ + (4−1 1 (M1 + M2 + a0 + 8)−1 )(I f (τ, tk , y, v) + a0 δ) ≤ δ(a0 M1 + h0 ) + (4−1 1 (M1 + M2 + a0 + 8)−1 )(I f (τ, tk , xf , uf ) + 2M1 + a0 δ) ≤ δ(a0 M1 + h0 ) + (4−1 1 (M1 + M2 + a0 + 8)−1 )(M2 + 2M1 + a0 δ) < 1 /4 + 1 /2. In view of (6.174), (6.178), (6.180) and (6.186), ≤ x(tk ) − xf (tk ) ≤ Δ1 δ + 2Δδ (1 /4 + 1 /2) ≤ 1 /32 + 2Δ1 ≤ 1 (2Δ1 + 1) = 4−1 , a contradiction. The contradiction we have reached proves that (6.181) holds for all t ∈ [tk − δ, tk ]. Together with (6.173) and (6.174), this implies that mes({t ∈ [0, T ] : x(t) − xf (t) ≥ δ}) → ∞ as T → ∞. On the other hand, property (P3) and (6.176) imply that there exists L > 0 such that for all T > 0, mes({t ∈ [0, T ] : x(t) − xf (t) ≥ δ}) ≤ L. The contradiction we have reached proves that limt →∞ (x(t) − xf (t)) = 0 and (P1) holds for the first class of problems. Consider the second class of problems. Recall (see Proposition 1.18 and (1.9)) that for each τ ≥ 0, Φτ ∈ L(L2 (0, τ ; , F ), E) is defined by Φτ u =
τ
eA(τ −s) Bu(s)ds, u ∈ L2 (0, τ ; F ).
(6.189)
0
Lemma 6.42, (A2), and (A3) imply that there exist δ ∈ (0, 1), L¯ > 0 such that M∗ eω∗  δ < /8, δ < bf /16, 64a0δ < K0 2 (Φ1 + 1)−2
(6.190)
and the following properties hold: (i) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δ, i = 1, 2 and T2 ≥ bf there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) satisfying
436
6 ContinuousTime Nonautonomous Problems on the HalfAxis
x1 (T1 ) = z1 , x1 (T1 +τ1 ) = xf (T1 +τ1 ), x2 (T2 ) = z2 , x2 (T2 −τ2 ) = xf (T2 −τ2 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 , I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 ; ¯ each t2 ∈ (t1 , t1 + 4δ], (ii) for each t1 ≥ L, I f (t1 , t2 , xf , uf ) ≤ 32−1 2 K0 (Φ1 + 1)−2 ; ¯ each S2 > S1 , and each (ξ, η) ∈ X(S1 , S2 ) which satisfies (iii) for each S1 ≥ L, ξ(Si ) − xf (Si ) ≤ δ, i = 1, 2, we have I f (S1 , S2 , ξ, η) ≥ I f (S1 , S2 , xf , uf ) − 32−1 2 K0 (Φ1 + 1)−2 . Property (P3) and (6.176) imply that there exists L > 0 such that for all T > 0, mes({t ∈ [0, T ] : x(t) − xf (t) ≥ δ}) ≤ L. Therefore there exists T0 > 0 such that for each S > T0 , mes({t ∈ [T0 , S] : x(t) − xf (t) > δ}) ≤ δ/2.
(6.191)
Since the pair (x, u) is (f )good, it follows from Theorem 6.2 that there exists T1 > 0 such that for each S > T1 , I f (T1 , S, x, u) ≤ U f (T1 , S, x(T1 ), x(S)) + 32−1 2 K0 (Φ1 + 1)−2 ,
(6.192)
I f (T1 , S, xf , uf ) ≤ U f (T1 , S, xf (T1 ), xf (S)) + 32−1 2 K0 (Φ1 + 1)−2 . (6.193) Let k ≥ 1 be an integer such that ¯ tk > T0 + T1 + 1 + L.
(6.194)
inf{x(t) − xf (t) : t ∈ [tk − δ, tk ]} < δ.
(6.195)
Assume that
Therefore there exists τ ∈ [tk − δ, tk ] such that
(6.196)
6.13 Proof of Proposition 6.27
437
x(τ ) − xf (τ ) < δ.
(6.197)
It is clear that (x, u), (xf , uf ) ∈ X(τ, tk ). In view of (6.17), (6.189), and (6.196), x(tk ) = eA(tk −τ ) x(τ ) +
tk
τ
xf (tk ) = e
eA(tk −s) Bu(s)ds = eA(tk −τ ) x(τ ) + Φtk −τ u(τ + ·),
A(tk −τ )
(6.198)
tk
xf (τ ) +
e
A(tk −s)
Buf (s)ds
τ
= eA(tk −τ ) xf (τ ) + Φtk −τ uf (τ + ·),
(6.199)
Proposition 1.18, (6.174), (6.190), and (6.196)–(6.199) imply that ≤ x(tk ) − xf (tk ) = eA(tk −τ ) (x(τ ) − xf (τ )) + Φtk −τ u(τ + ·) + Φtk −τ uf (τ + ·) ≤ M∗ eω∗ δ x(τ ) − xf (τ )
tk
+Φ1 (
tk
u(s)2 ds)1/2 + Φ1 (
τ
uf (s)2 ds)1/2
τ
tk
< /8 + Φ1 (
tk
u(s)2 ds)1/2 + Φ1 (
τ
uf (s)2 ds)1/2
τ
and (4
−1
(Φ1 + 1)
−1 2
tk
) ≤ max{
tk
u(s) ds, 2
τ
uf (s)2 ds}.
(6.200)
τ
Let (y, v) ∈ {(x, u), (xf , uf )},
tk
tk
v(s)2 ds ≥ max{
τ
τ
tk
u(s)2 ds,
uf (s)2 ds}.
(6.201) (6.202)
τ
In view of (6.200)–(6.202),
tk τ
By (6.21) and (6.203),
v(s)2 ds ≥ (4−1 )2 (Φ1 + 1)−2 .
(6.203)
438
6 ContinuousTime Nonautonomous Problems on the HalfAxis
tk
tk
f (s, y(s), v(s))ds +a0 (tk −τ ) ≥ K0
τ
v(s)2 ds ≥ 4−1 2 K0 (Φ1 +1)−2 .
τ
(6.204)
It follows from (6.196) and (6.204) that
tk
f (s, y(s), v(s))ds ≥ 4−1 K0 2 (Φ1 + 1)−2 − a0 δ.
(6.205)
τ
In view of (6.191) and (6.194), for each S > tk , mes({t ∈ [tk , S] : y(t) − xf (t) > δ}) ≤ δ/2.
(6.206)
Inequality (6.206) implies that there exist S1 ∈ [tk , tk + δ]
(6.207)
y(S1 ) − xf (S1 ) ≤ δ
(6.208)
S2 ∈ [S1 + 2bf , S1 + 2bf + δ]
(6.209)
y(S2 ) − xf (S2 ) ≤ δ.
(6.210)
such that
and
such that
It follows from (6.205) and (6.207) that I f (τ, S1 , y, v) = I f (τ, tk , y, v) + I f (tk , S1 , y, v) ≥ 4−1 K0 2 (Φ1 + 1)−2 − a0 δ − a0 δ.
(6.211)
Property (iii), (6.194), and (6.207)–(6.210) imply that I f (S1 , S2 , y, v) ≥ I f (S1 , S2 , xf , uf ) − 32−1 K0 2 (Φ1 + 1)−2 . By (6.196), (6.197), (6.207), (6.209), (6.211), (6.212), and property (ii), I f (τ, S2 , y, v) = I f (τ, S1 , y, v) + I f (S1 , S2 , y, v) ≥ 4−1 K0 2 (Φ1 + 1)−2 − 2a0 δ +I f (S1 , S2 , xf , uf ) − 32−1 K0 2 (Φ1 + 1)−2
(6.212)
6.13 Proof of Proposition 6.27
439
≥ K0 2 (Φ1 + 1)−2 (4−1 − 32−1 ) −2a0 δ + I f (τ, S2 , xf , uf ) − I f (τ, S1 , xf , uf ) ≥ I f (τ, S2 , xf , uf ) + K0 2 (Φ1 + 1)−2 (4−1 − 32−1 ) − 2a0 δ − 16−1 K0 2 (Φ1 + 1)−2 .
(6.213)
Property (i), (6.196), (6.197), (6.207), and (6.209) imply that that there exists (x, ˜ u) ˜ ∈ X(τ, S2 ) such that x(τ ˜ ) = y(τ ), x(τ ˜ + bf ) = xf (τ + bf ),
(6.214)
˜ u) ˜ ≤ I f (τ, τ +bf , xf , uf )+32−1 2 K0 (Φ1 +1)−2 , I f (τ, τ +bf , x,
(6.215)
˜ 2 − bf ) = xf (S2 − bf ), x(S ˜ 2 ) = y(S2 ), x(S
(6.216)
I f (S2 − bf , S2 , x, ˜ u) ˜ ≤ I f (S2 − bf , S2 , xf , uf ) + 32−1 2 K0 (Φ1 + 1)−2 , (6.217) (x(t), ˜ u(t)) ˜ = (xf (t), uf (t)), t ∈ [τ + bf , S2 − bf ]. (6.218) In view of (6.215), (6.217), and (6.218), I f (τ, S2 , x, ˜ u) ˜ ≤ I f (τ, S2 , xf , uf ) + 16−1 2 K0 (Φ1 + 1)−2 .
(6.219)
It follows from (6.190), (6.213), and (6.219) that ˜ u) ˜ I f (τ, S2 , y, v) − I f (τ, S2 , x, ≥ 2 K0 (Φ1 + 1)−2 (4−1 − 32−1 − 8−1 ) − 2δa0 ≥ 2 K0 (Φ1 + 1)−2 16−1 .
(6.220)
By (6.220), I f (τ, S2 , y, v) ≥ U f (τ, S2 , y(τ ), y(S2 )) + 2 K0 (Φ1 + 1)−2 /16.
(6.221)
It follows from (6.192)–(6.194), (6.196), and (6.201) that τ > T1 and I f (τ, S2 , y, v) ≤ U f (τ, S2 , y(τ ), y(S2 )) + 32−1 2 K0 (Φ1 + 1)−2 . This contradicts (6.221). The contradiction we have reached proves that (6.195) is not true and x(t) − xf (t) ≥ δ, t ∈ [tk − δ, tk ]
440
6 ContinuousTime Nonautonomous Problems on the HalfAxis
¯ Together with (6.173) for each natural number k satisfying tk ≥ T0 + T1 + 1 + L. and (6.174), this implies that mes({t ∈ [0, T ] : x(t) − xf (t) ≥ δ}) → ∞ as T → ∞. This contradicts (6.191) holding for all S > T0 . The contradiction we have reached proves that limt →∞ (x(t) − xf (t) = 0 and (P1) holds for the second class of problems too. Proposition 6.27 is proved.
6.14 Auxiliary Results for Theorems 6.28 and 6.29 Recall that for each z ∈ R 1 , z = max{i ∈ R 1 : i is an integer, i ≤ z}. Proposition 6.43. Assume that f has (P1), (P2), and LSC property, T0 ≥ 0, z0 ∈ E, (T0 , z0 ) ∈ ∪{AL : L ∈ (0, ∞)}. Then there exists an (f )good and (f )minimal pair (x∗ , u∗ ) ∈ X(T0 , ∞) such that x∗ (T0 ) = z0 . Proof. There exists L0 > 0 such that (T0 , z0 ) ∈ AL0 .
(6.222)
It follows from Theorem 6.2 that there exists S0 > 0 such that for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(6.223)
Fix an integer k0 ≥ L0 . LSC property and (6.222) imply that for each integer k ≥ k0 , there exists (xk , uk ) ∈ X(T0 , T0 + k) satisfying xk (T0 ) = z0 , I f (T0 , T0 + k, xk , uk ) = σ f (T0 , T0 + k, z0 ).
(6.224) (6.225)
In view of (6.20), (6.21), and (6.222), for each integer k ≥ k0 , σ f (T0 , T0 + k, z0 ) ≤ L0 + I f (T0 , T0 + k, xf , uf ) + a0 L0 .
(6.226)
By (6.223), (6.225), and (6.226), for each integer k ≥ k0 and each pair of numbers T1 , T2 ∈ [T0 , T0 + k] satisfying T1 < T2 , I f (T1 , T2 , xk , uk ) = I f (T0 , T0 + k, xk , uk ) −I f (T0 , T1 , xk , uk ) − I f (T0 + k, T2 , xk , uk )
6.14 Auxiliary Results for Theorems 6.28 and 6.29
441
≤ I f (T0 , T0 + k, xf , uf ) + L0 (1 + a0 ) − I f (T0 , T1 , xf , uf ) +S0 − I f (T0 + k, T2 , xk , uk ) + S0 = I f (T1 , T2 , xf , uf ) + 2S0 + L0 (1 + a0 ).
(6.227)
By (6.227) and LSC property, extracting subsequences, using the diagonalization process and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 such that k1 ≥ k0 and for each integer i ≥ 0 there exists limp→∞ I f (T0 +i, T0 +i+1, xkp , ukp ) and there exists (yi , vi ) ∈ X(T0 +i, T0 +i+1) such that xkp (t) → yi (t) as p → ∞ for all t ∈ [T0 + i, T0 + i + 1],
(6.228)
I f (T0 + i, T0 + i + 1, yi , vi ) ≤ lim I f (T0 + i, T0 + i + 1, xkp , ukp ).
(6.229)
p→∞
In view of (6.228), there exists (x∗ , u∗ ) ∈ X(T0 , ∞) such that for each integer i ≥ 0, (x∗ (t), u∗ (t)) = (yi (t), vi (t)), t ∈ [T0 + i, T0 + i + 1].
(6.230)
It follows from (6.227), (6.229), and (6.230) that for every integer q ≥ 1, I f (T0 , T0 + q, x∗ , u∗ ) ≤ lim I f (T0 , T0 + q, xkp , ukp ) p→∞
≤ I f (T0 , T0 + q, xf , uf ) + 2S0 + L0 (1 + a0 ).
(6.231)
Theorem 6.2, (6.224), (6.228), (6.230), and (6.231) imply that (x∗ , u∗ ) is (f )good and x∗ (T0 ) = z0 .
(6.232)
In order to complete the proof of the proposition, it is sufficient to show that (x∗ , u∗ ) is (f )minimal. Assume the contrary. Then there exist Δ > 0, an integer τ0 ≥ 1 and (y, v) ∈ X(T0 , T0 + τ0 ) such that y(T0 ) = x∗ (T0 ), y(T0 + τ0 ) = x∗ (T0 + τ0 ),
(6.233)
I f (T0 , T0 + τ0 , x∗ , u∗ ) > I f (T0 , T0 + τ0 , y, v) + 2Δ.
(6.234)
By (A2) and Lemma 6.42, there exist L1 , δ > 0 such that the following properties hold:
442
6 ContinuousTime Nonautonomous Problems on the HalfAxis
(i) for each (T , ξ ) ∈ A satisfying ξ − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x˜1 , u˜ 1 ) ∈ X(T , T + τ1 ) such that x˜1 (T ) = ξ , x˜1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x˜1 , u˜ 1 ) ≤ I f (T , T + τ1 , xf , uf ) + Δ/8 and if T ≥ bf , then there exist τ2 ∈ (0, bf ] and (x˜2 , u˜ 2 ) ∈ X(T − τ2 , T ) such that x˜2 (T − τ2 ) = xf (T − τ2 ) , x˜2 (T ) = ξ, I f (T − τ2 , T , x˜2 , u˜ 2 ) ≤ I f (T − τ2 , T , xf , uf ) + Δ/8; (ii) if T2 > T1 ≥ L1 , (x, u) ∈ X(T1 , T2 ), x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, we have I f (T1 , T2 , x, u) ≥ I f (T1 , T2 , xf , uf ) − Δ/8. Theorems 6.17 and 6.22, (6.222), (6.224), and (6.225) imply that there exists an integer L2 > L0 + L1 such that for each integer k ≥ k0 + 2L2 , xk (t) − xf (t) ≤ δ, t ∈ [T0 + L2 , T0 + k − L2 ].
(6.235)
In view of (6.228), (6.230), and (6.235), x∗ (t) − xf (t) ≤ δ for all t ≥ T0 + L2 .
(6.236)
By (6.228) and (6.230), there exists a natural number p0 such that kp0 > k0 + 2L1 + 2L2 + 2τ0 + 2 + 2bf + 2T0 ,
(6.237)
I f (T0 , T0 + τ0 + L2 , x∗ , u∗ ) ≤ I f (T0 , T0 + τ0 + L2 , xkp0 , ukp0 ) + Δ/2.
(6.238)
Property (i), (6.233), and (6.235)–(6.237) imply that there is (x, u) ∈ X(T0 , T0 + kp0 ) such that (x(t), u(t)) = (y(t), v(t)), t ∈ [T0 , T0 + τ0 ],
(6.239)
(x(t), u(t)) = (x∗ (t), u∗ (t)), t ∈ (T0 + τ0 , T0 + τ0 + L2 ],
(6.240)
x(T0 + τ0 + L2 + bf ) = xf (T0 + τ0 + L2 + bf ),
(6.241)
I f (T0 + τ0 + L2 , T0 + τ0 + L2 + bf , x, u)
6.14 Auxiliary Results for Theorems 6.28 and 6.29
443
≤ I f (T0 + τ0 + L2 , T0 + τ0 + L2 + bf , xf , uf ) + Δ/8,
(6.242)
(x(t), u(t)) = (xkp0 (t), ukp0 (t)), t ∈ [T0 + τ0 + L2 + 2bf , kp0 + T0 ],
(6.243)
I f (T0 + τ0 + L2 + bf , T0 + τ0 + L2 + 2bf , x, u) ≤ I f (T0 + τ0 + L2 + bf , T0 + τ0 + L2 + 2bf , xf , uf ) + Δ/8.
(6.244)
It follows from (6.224), (6.225), (6.232), (6.233), and (6.238) that I f (T0 , T0 + kp0 , x, u) ≥ I f (T0 , T0 + kp0 , xkp0 , ukp0 ). By (6.234), (6.235), (6.237)–(6.240), (6.242)–(6.245), and property (ii), 0 ≤ I f (T0 , T0 + kp0 , x, u) − I f (T0 , T0 + kp0 , xkp0 , ukp0 ) = I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , T0 + τ0 + L2 , x∗ , u∗ ) +I f (T0 + τ0 + L2 , T0 + τ0 + L2 + bf , xf , uf ) + Δ/8 +I f (T0 + τ0 + L2 + bf , T0 + τ0 + L2 + 2bf , xf , uf ) + Δ/8 −I f (T0 , T0 + τ0 + L2 , xkp0 , ukp0 ) −I f (T0 + τ0 + L2 , T0 + τ0 + L2 + bf , xkp0 , ukp0 ) −I f (T0 + τ0 + L2 + bf , T0 + τ0 + L2 + 2bf , xkp0 , ukp0 ) ≤ I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , T0 + τ0 + L2 , x∗ , u∗ ) +I f (T0 + τ0 + L2 , T0 + τ0 + L2 + bf , xf , uf ) + Δ/8 +I f (T0 + τ0 + L2 + bf , T0 + τ0 + L2 + 2bf , xf , uf ) + Δ/8 −I f (T0 , T0 + τ0 + L2 , xkp0 , ukp0 ) −I f (T0 + τ0 + L2 , T0 + τ0 + L2 + 2bf , xkp0 , ukp0 ) + Δ/8 = I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , T0 + τ0 + L2 , x∗ , u∗ ) −I f (T0 , T0 + τ0 + L2 , xkp0 , ukp0 ) + Δ/2
(6.245)
444
6 ContinuousTime Nonautonomous Problems on the HalfAxis
< I f (T0 , T0 + τ0 , x∗ , u∗ ) − 2Δ + I f (T0 + τ0 , T0 + τ0 + L2 , x∗ , u∗ ) −I f (T0 , T0 + τ0 + L2 , xkp0 , ukp0 ) + Δ/2 ≤ −2Δ + Δ/2 + Δ/2, a contradiction. The contradiction we have reached completes the proof of Proposition 6.43.
6.15 Proofs of Theorems 6.28 and 6.29 Proof of Theorem 6.29. Clearly, (i) implies (ii). In view of Theorem 6.2, (ii) implies (iii). By (P1), (iii) implies (iv). Evidently (iv) implies (v). We show that (v) implies (iii). Assume that (x∗ , u∗ ) is (f )minimal and satisfies lim inf x(t) ˜ − xf (t) = 0. t →∞
(6.246)
Since (x, ˜ u) ˜ is (f )good, there exists S0 > 0 such that for all numbers T > S, ˜ u) ˜ − I f (S, T , xf , uf ) ≤ S0 . I f (S, T , x,
(6.247)
By (A2), there exists δ > 0 such that the following property holds: (a) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T , T + τ1 ) which satisfy x1 (T ) = z, x1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x1 , u1 ) ≤ I f (T , T + τ1 , xf , uf ) + 1, and if T ≥ bf , then there exist τ2 ∈ (0, bf ] and (x2 , u2 ) ∈ X(T − τ2 , T ) such that x2 (T − τ2 ) = xf (T − τ2 ), x2 (T ) = z, I f (T − τ2 , T , x2 , u2 ) ≤ I f (T − τ2 , T , xf , uf ) + 1. In view of (6.246) and (P1), there exists an increasing sequence {tk }∞ k=1 ⊂ (S, ∞) such that lim tk = ∞,
k→∞
x∗ (tk + 2bf ) − xf (tk + 2bf ) ≤ δ for all integers k ≥ 1,
(6.248) (6.249)
6.15 Proofs of Theorems 6.28 and 6.29
445
x(t) ˜ − xf (t) ≤ δ for all t ≥ t0 .
(6.250)
Let k ≥ 1 be an integer. By property (a), (6.249), and (6.250), there exists (y, v) ∈ X(S, tk + 2bf ) such that (y(t), v(t)) = (x(t), ˜ u(t)), ˜ t ∈ [S, tk ], y(tk + bf ) = xf (tk + bf ), I f (tk , tk + bf , y, v) ≤ I f (tk , tk + bf , xf , uf ) + 1, y(tk + 2bf ) = x∗ (tk + 2bf ), I f (tk + bf , tk + 2bf , y, v) ≤ I f (tk + bf , tk + 2bf , xf , uf ) + 1. The relations above and (6.247) imply that I f (S, tk + 2bf , x∗ , u∗ ) ≤ I f (S, tk + 2bf , y, v) ˜ u) ˜ + I f (tk , tk + 2bf , xf , uf ) + 2 ≤ I f (S, tk , x, ≤ I f (S, tk + 2bf , xf , uf ) + 2 + 2S0 . Together with Theorem 6.2, this implies that (x∗ , u∗ ) is (f )good and (iii) holds. We show that (iii) implies (i). Assume that (x∗ , u∗ ) is (f )minimal and (f )good. (P1) implies that lim x∗ (t) − xf (t) = 0.
t →∞
(6.251)
Since (x∗ , u∗ ) is (f )good, there exists S0 > 0 such that I f (S, T , x∗ , u∗ ) − I f (S, T , xf , uf ) ≤ S0 for all T > S.
(6.252)
Let (x, u) ∈ X(S, ∞) satisfy x(S) = x∗ (S).
(6.253)
We show that lim sup[I f (S, T , x∗ , u∗ ) − I f (S, T , x, u)] ≤ 0. T →∞
In view of Theorem 6.2 and (6.252), we may assume that (x, u) is (f )good. (P1) implies that
446
6 ContinuousTime Nonautonomous Problems on the HalfAxis
lim x(t) − xf (t) = 0.
t →∞
(6.254)
Let > 0. By (A2) and Lemma 6.42, there exist δ ∈ (0, ) and L1 > 0 such that the following properties hold: (b) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T , T + τ1 ) satisfying x1 (T ) = z , x1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x1 , u1 ) ≤ I f (T , T + τ1 , xf , uf ) + /8, and if T ≥ bf , then there exist τ2 ∈ (0, bf ] and (x2 , u2 ) ∈ X(T − τ2 , T ) satisfying x2 (T − τ2 ) = xf (T − τ2 ) , x2 (T ) = z, I f (T − τ2 , T , x2 , u2 ) ≤ I f (T − τ2 , T , xf , uf ) + /8; (c) for each T1 ≥ L1 , each T2 > T1 , and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, we have I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − /8. It follows from (6.251) and (6.254) that there exists τ0 > 0 such that x(t) − xf (t) ≤ δ, x∗ (t) − xf (t) ≤ δ for all t ≥ τ0 .
(6.255)
Let T ≥ τ0 + L1 .
(6.256)
Property (b), (6.255), and (6.256) imply that there exists (y, v) ∈ X(S, T + 2bf ) which satisfies (y(t), v(t)) = (x(t), u(t)), t ∈ [S, T ], y(T + bf ) = xf (T + bf ),
(6.257)
I f (T , T + bf , y, v) ≤ I f (T , T + bf , xf , uf ) + /8,
(6.258)
y(T + 2bf ) = x∗ (T + 2bf ), I f (T + bf , T + 2bf , y, v) ≤ I f (T + bf , T + 2bf , xf , uf ) + /8. By property (c) and (6.255),
(6.259) (6.260)
6.16 Auxiliary Results for Theorem 6.33
I f (T , T + 2bf , x∗ , u∗ ) ≥ I f (T , T + 2bf , xf , uf ) − /8.
447
(6.261)
It follows from (6.253) and (6.257)–(6.261) that I f (S, T , x∗ , u∗ ) + I f (T , T + 2bf , xf , uf ) − /8 ≤ I f (S, T + 2bf , x∗ , u∗ ) ≤ I f (S, T + 2bf , y, v) = I f (S, T , x, u) + I f (T , T + 2bf , xf , uf ) + /4, I f (S, T , x∗ , u∗ ) ≤ I f (S, T , x, u) + for all T ≥ τ0 + L1 . Since is any positive number, we conclude that lim sup[I f (S, T , x∗ , u∗ ) − I f (S, T , x, u)] ≤ 0, T →∞
(x∗ , u∗ ) is (f )overtaking optimal and (i) holds. Theorem 6.29 is proved. Theorem 6.29 and Proposition 6.43 imply Theorem 6.28.
6.16 Auxiliary Results for Theorem 6.33 In the next lemma, we consider only the first class of problems with A(t) = 0 for all t ≥ 0. Lemma 6.44. Let M0 , M1 , τ0 > 0. Then there exists M2 > M1 such that for each T1 ≥ 0, each T2 ∈ (T1 , T1 + τ0 ], and each (x, u) ∈ X(T1 , T2 ) satisfying inf{x(t) : t ∈ [T1 , T2 ]} ≤ M1 ,
(6.262)
I f (T1 , T2 , x, u) ≤ M0
(6.263)
the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ].
(6.264)
Proof. Fix a positive number δ < min{8−1 τ0 , 2−1 (a0 + 1)−1 }. By (6.19) and (6.20), there exist h0 > M1 + 1 and γ0 > 0 such that
(6.265)
448
6 ContinuousTime Nonautonomous Problems on the HalfAxis
f (t, x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (t, x, u) ∈ M satisfying x ≥ h0 , (6.266) f (t, x, u) ≥ 8(G(t, x, u) − a0 x)+ for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ γ0 .
(6.267)
Choose a number M2 > 2h0 + 2M1 + 2γ0 δ + M0 + a0 τ0 .
(6.268)
Let T1 ≥ 0, T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (6.262) and (6.263) hold. We show that (6.264) hold. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that x(t0 ) > M2 .
(6.269)
By the choice of h0 (see (6.266)), there exists t1 ∈ [T1 , T2 ] such that x(t1 ) ≤ h0 , t1 − t0  ≤ δ.
(6.270)
t2 ∈ [min{t1 , t0 }, max{t1 , t0 }]
(6.271)
There exists
such that x(t2 ) ≥ x(t) for all t ∈ [min{t1 , t0 }, max{t1 , t0 }].
(6.272)
In view of (6.268)–(6.270) and (6.272) t2 = t1 . By (6.11) (with U (t, s) = I d), x(t2 ) − x(t1 ) =
t2
G(x(s), u(s))ds.
(6.273)
t1
It follows from (6.270) and (6.273) that x(t2 ) − h0 ≤ 
t2 t1
G(s, x(s), u(s))ds.
(6.274)
6.16 Auxiliary Results for Theorem 6.33
449
Define E1 = {t ∈ [min{t1 , t2 }, max{t1 , t2 }] : G(t, x(t), u(t)) ≥ a0 x(t) + γ0 }, (6.275) E2 = [min{t1 , t2 }, max{t1 , t2 }] \ E1 . (6.276) By (6.263), (6.270), (6.272), and (6.274)–(6.276),
x(t2 ) − h0 ≤ a0
E1 ∪E2
x(t)dt +
E1 ∪E2
[G(s, x(s), u(s)) − a0 x(s)]+ ds
≤ a0 δx(t2 ) + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
+ E2
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 δx(t2 ) + γ0 δ + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 x(t2 )δ + γ0 δ + 8
−1
f (t, x(t), u(t))dt E1
≤ a0 x(t2 )δ + γ0 δ + 8−1 (M0 + a0 τ0 ). It follows from the relation above, (6.265), (6.269), and (6.272) that 2−1 M2 ≤ 2−1 x(t2 ) ≤ x(t2 )(1 − a0 δ) ≤ h0 + γ0 δ + 8−1 (M0 + a0 τ0 ) and M2 < 2h0 + 2γ0 δ + 4−1 (M0 + a0 τ0 ). This contradicts (6.268). The contradiction we have reached proves Lemma 6.44. Lemma 6.44, (6.19), and (6.20) imply the following result. Lemma 6.45. Let M1 > 0, 0 < τ0 < τ1 . Then there exists M2 > 0 such that for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ].
450
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Lemma 6.46. Let M1 > 0, ∈ (0, 1), 0 < τ0 < τ1 . Then there exists δ > 0 such that for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M0 , and each t1 , t2 ∈ [T1 , T2 ] satisfying t1 − t2  ≤ δ the inequality x(t1 ) − x(t2 ) ≤ holds. Proof. Lemma 6.45 implies that there exists M2 > 0 such that the following property holds: (i) for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 , we have x(t) ≤ M2 for all t ∈ [T1 , T2 ]. In view of (6.19) and (6.20), there exists h0 > 0 such that f (t, x, u) ≥ 4 −1 (M1 + a0 τ1 + 8)(G(t, x, u) − a0 x)+
(6.277)
for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ h0 . Choose a number δ ∈ (0, (4a0M2 + 4h0 + 4)−1 ).
(6.278)
Let T1 ≥ 0, T2 ∈ [T1 + τ0 , T1 + τ1 ], (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) ≤ M1 ,
(6.279)
t1 , t2 ∈ [T1 , T2 ], 0 < t2 − t2 ≤ δ.
(6.280)
Property (i) and (6.279) imply that x(t) ≤ M2 for all t ∈ [T1 , T2 ]. Define
(6.281)
6.16 Auxiliary Results for Theorem 6.33
451
Ω1 = {t ∈ [t1 , t2 ] : G(t, x(t), u(t)) − a0 x(t) ≥ h0 }, Ω2 = [t1 , t2 ] \ Ω1 .
(6.282) (6.283)
By (6.11) and (6.277)–(6.283), x(t2 ) − x(t1 ) =
t2
G(x(s), u(s))ds t1
≤ a0
t2
x(t)dt +
t1
t2 t1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 δM2 + δh0 + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 δM2 + δh0 + (4(M1 + a0 τ1 + 8))−1
f (t, x(t), u(t))dt E1
≤ a0 δM2 + δh0 + 4−1 < . Lemma 6.46 is proved. Lemma 6.47. Let Δ > 0. Then there exists δ > 0 such that for each (T1 , z1 ), (T2 , z2 ) ∈ A satisfying T2 ≥ T1 + 2bf , zi − xf (Ti ) ≤ δ, i = 1, 2
(6.284)
the following inequality holds: U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , xf , uf ) + Δ. Proof. Let δ > 0 be as guaranteed by (A2) with = Δ/4. Let (T1 , z1 ), (T2 , z2 ) ∈ A satisfy (6.284). By (6.284), the choice of δ and (A2) with = Δ/4, there exist (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = z1 , y(T2 ) = z2 , y(t) = xf (t), v(t) = uf (t), t ∈ [T1 + bf , T2 − bf ], I f (T1 , T1 + bf , y, v) ≤ I f (T1 , T1 + bf , xf , uf ) + Δ/4, I f (T2 − bf , T2 , y, v) ≤ I f (T2 − bf , T2 , xf , uf ) + Δ/4.
452
6 ContinuousTime Nonautonomous Problems on the HalfAxis
By the relations above, U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , y, v) ≤ I f (T1 , T2 , xf , uf ) + Δ/2. Lemma 6.47 is proved. Lemma 6.48. Let Δ ∈ (0, 1). Then there exists δ > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + 3bf , each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, and each τ1 ∈ [T1 , T1 + δ] and each τ2 ∈ [T2 − δ, T2 ] the inequality I f (τ1 , τ2 , x, u) ≤ I f (τ1 , τ2 , xf , uf ) + Δ holds. Proof. Lemma 6.47 implies that there exists δ0 ∈ (0, min{4−1 Δ, 2−1 bf }) such that the following property holds: (ii) for each (T1 , z1 ), (T2 , z2 ) ∈ A satisfying T2 ≥ T1 + 2bf , zi − xf (Ti ) ≤ δ0 , i = 1, 2 we have U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , xf , uf ) + Δ/4. Theorem 6.2 implies that exists M0 > 0 such that the following property holds: (iii) for each pair of numbers T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + M0 ≥ I f (T1 , T2 , xf , uf ). Lemma 6.46 implies that there exists δ1 ∈ (0, δ0 ) such that the following property holds: (iv) for each S ≥ 0, each (y, v) ∈ X(S, S + bf ) satisfying I f (S, S + bf , y, v) ≤ Δf (bf + 2) + 3a0 + 2M0 + 1
6.16 Auxiliary Results for Theorem 6.33
453
and each t1 , t2 ∈ [S, S +bf ] satisfying t1 −t2  ≤ δ1 we have y(t1 )−y(t2) ≤ δ0 /8. Set δ = δ1 /8.
(6.285)
T1 ≥ 0, T2 ≥ T1 + 3bf ,
(6.286)
x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2,
(6.287)
Assume that
(x, u) ∈ X(T1 , T2 ),
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, τ1 ∈ [T1 , T1 + δ], τ2 ∈ [T2 − δ, T2 ].
(6.288) (6.289)
Property (ii) and (6.286)–(6.288) imply that I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + 1. Property (iii) and the relation above imply that for each τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , I f (τ1 , τ2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , τ1 , x, u) − I f (τ2 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + 1 − I f (T1 , τ1 , xf , uf ) + M0 − I f (τ2 , T2 , xf , uf ) + M0 = I f (T1 , T2 , xf , uf ) + 2M0 + 1.
(6.290)
In view of (6.20) and (6.24), I f (T1 , T1 + bf , xf , uf ) ≤ I f (T1 , T1 + bf + 2, xf , uf ) + 3a0 ≤ (bf + 2)Δf + 3a0,
(6.291)
I f (T2 − bf , T2 , xf , uf ) ≤ I f (T2 − bf − 1, T2 + 1, xf , uf ) + 3a0 ≤ (bf + 2)Δf + 3a0,
(6.292)
It follows from (6.290)–(6.292) that I f (T1 , T1 + bf , x, u), I f (T2 − bf , T2 , x, u) ≤ Δf (bf + 2) + 3a0 + 2M0 + 1. (6.293)
454
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Property (iv), (6.285), (6.289), and (6.293) imply that xf (Ti ) − xf (τi ) ≤ δ0 /8, i = 1, 2,
(6.294)
x(Ti ) − x(τi ) ≤ δ0 /8, i = 1, 2.
(6.295)
By (6.285), (6.287), (6.294), and (6.295), for i = 1, 2, xf (τi ) − x(τi ) ≤ xf (τi ) − xf (Ti ) + xf (Ti ) − x(Ti ) + x(Ti ) − x(τi ) ≤ δ0 /8 + δ0 /8 + δ ≤ δ0 /2.
(6.296)
Property (i), (6.286), (6.289), and (6.296) imply that U f (τ1 , τ2 , x(τ1 ), x(τ2 )) ≤ I f (τ1 , τ2 , xf , uf ) + Δ/4.
(6.297)
By (6.288), (6.289), and (6.297), I f (τ1 , τ2 , xf , uf ) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ ≤ I f (τ1 , τ2 , xf , uf ) + Δ/4 + δ ≤ I f (τ1 , τ2 , xf , uf ) + Δ. Lemma 6.48 is proved.
6.17 Proof of Theorem 6.33 Assume that STP holds. Theorem 6.22 implies that (P1) and (P2) hold. Let us show that (P4) holds. Let (x, ˜ u) ˜ ∈ X(0, ∞) be (f )overtaking optimal and x(0) ˜ = xf (0).
(6.298)
(P1), STP, and (6.298) imply that x(t) ˜ = xf (t) for all t ≥ 0 and (P4) holds. Assume that (P1), (P2), and (P4) hold. Lemma 6.49. Let > 0. Then there exists δ > 0 such that for each pair of numbers T1 ≥ 0, T2 ≥ T1 + 3bf and each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2 I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ the inequality x(t) − xf (t) ≤ holds for all t ∈ [T1 , T2 ].
6.17 Proof of Theorem 6.33
455
Proof. Lemma 6.40 and (A2) imply that there exist δ0 ∈ (0, min{1, /4}) and L0 > 0 such that the following properties hold: (i) (A2) holds with = 1 and δ = δ0 ; (ii) for each pair of numbers T1 ≥ L0 , T2 ≥ T1 + 2bf and each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ0 , i = 1, 2 I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ0 we have x(t) − xf (t) ≤ , t ∈ [T1 , T2 ]. Theorem 6.2 implies that there exists S0 > 0 such that for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(6.299)
It follows from Theorem 6.23; there exists L1 > 0 such that the following property holds: (iii) for each T ≥ 0 and each (x, u) ∈ X(T , T + L1 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + 2S0 + 3 we have min{x(t) − xf (t) : t ∈ [T , T + L1 ]} ≤ δ0 . Consider a sequence {δi }∞ i=1 ⊂ (0, 1) such that δi < 2−1 δi−1 , i = 1, 2, . . . .
(6.300)
Assume that the lemma does not hold. Then for each integer i ≥ 1, there exist Ti,1 ≥ 0, Ti,2 ≥ Ti,1 + 3bf
(6.301)
and (xi , ui ) ∈ X(Ti,1 , Ti,2 ) such that xi (Ti,1 ) − xf (Ti,1 ) ≤ δi , xi (Ti,2 ) − xf (Ti,2 ) ≤ δi ,
(6.302)
I f (Ti,1 , Ti,2 , xi , ui ) ≤ U f (Ti,1 , Ti,2 , xi (Ti,1 ), xi (Ti,2 )) + δi
(6.303)
456
6 ContinuousTime Nonautonomous Problems on the HalfAxis
and ti ∈ [Ti,1 , Ti,2 ] for which x(ti ) − xf (ti ) > .
(6.304)
Let i be a natural number. Property (ii) and (6.300)–(6.304) imply that Ti,1 < L0 .
(6.305)
ti ≤ 2bf + 1 + L1 + L0 .
(6.306)
We show that
Property (i), (A2) with = 1, δ = δ0 , and (6.300)–(6.302) imply that there exists (yi , vi ) ∈ X(Ti,1 , Ti,2 ) such that yi (Ti,1 ) = xi (Ti,1 ), yi (Ti,2 ) = xi (Ti,2 ),
(6.307)
yi (t) = xf (t), vi (t) = uf (t), t ∈ [Ti,1 + bf , Ti,2 − bf ],
(6.308)
I f (Ti,1 , Ti,1 + bf , yi , vi ) ≤ I f (Ti,1 , Ti,1 + bf , xf , uf ) + 1,
(6.309)
I f (Ti,2 − bf , Ti,2 , yi , vi ) ≤ I f (Ti,2 − bf , Ti,2 , xf , uf ) + 1.
(6.310)
It follows from (6.303) and (6.307)–(6.310) that I f (Ti,1 , Ti,2 , xi , ui ) ≤ 1 + I f (Ti,1 , Ti,2 , yi , vi ) ≤ I f (Ti,1 , Ti,2 , xf , uf ) + 3. (6.311) In view of (6.299) and (6.311), for each pair of integers S1 , S2 ∈ [Ti,1 , Ti,2 ] satisfying S1 < S2 , I f (S1 , S2 , xi , ui ) = I f (Ti,1 , Ti,2 , xi , ui )−I f (Ti,1 , S1 , xi , ui )−I f (S2 , Ti,2 , xi , ui ) ≤ I f (Ti,1 , Ti,2 , xf , uf ) + 3 − I f (Ti,1 , S1 , xf , uf ) + S0 − I f (S2 , Ti,2 , xf , uf ) + S0 = I f (S1 , S2 , xf , uf ) + 3 + 2S0 .
(6.312)
Assume that (6.306) does not hold. Then ti > 2bf + 1 + L1 + L0 .
(6.313)
Consider the restriction of (xi , ui ) to the interval [ti − 2bf − 1 − L1 , ti − 1 − 2bf ] ⊂ [L0 , ∞).
(6.314)
6.17 Proof of Theorem 6.33
457
By property (iii) and (6.312), there exists t˜ ∈ [ti − 2bf − 1 − L1 , ti − 1 − 2bf ]
(6.315)
xf (t˜) − x(t˜) ≤ δ0 .
(6.316)
such that
Property (ii), (6.300), (6.302), (6.303), and (6.314)–(6.316) imply that x(t) − xf (t) ≤ , t ∈ [t˜, Ti,2 ] and in particular, x(ti ) − xf (ti ) ≤ . This contradicts (6.304). The contradiction we have reached proves (6.306). In view of (6.20), (6.24), and (6.312), there exists M1 > 0 such that the following property holds: (iv) for each integer i ≥ 1 and each S2 > S1 ≥ 0 satisfying S1 , S2 ∈ [Ti,1 , Ti,2 ], S2 − S1 ≤ 2bf , I f (S1 , S2 , xi , ui ) ≤ M1 . Lemma 6.45 and property (iv) imply that there exists M2 > 0 such that xi (t) ≤ M2 , t ∈ [Ti,1 , Ti,2 ], i = 1, 2, . . . .
(6.317)
In view of (6.305) and (6.306), extracting a subsequence and reindexing, we may assume without loss of generality that there exists T˜1 = lim Ti,1 ∈ [0, L0 ],
(6.318)
t˜ = lim ti ∈ [T˜1 , L0 + L1 + 2bf + 1],
(6.319)
T˜2 = lim Ti,2 ∈ [t˜, ∞].
(6.320)
i→∞
i→∞
i→∞
There exist strictly decreasing sequence {T1(k) }∞ k=1 and a strictly increasing sequence (k) ∞ {T2 }k=1 such that (k) (k) T˜1 = lim T1 , T˜2 = lim T2 , k→∞
k→∞
(6.321)
458
6 ContinuousTime Nonautonomous Problems on the HalfAxis
for all integers k ≥ 1, (k)
T1
(k)
(6.322)
< T2 .
If T˜2 = ∞, then we may assume that T2(k) = Tk,2 for all integers k ≥ 1.
(6.323)
By LSC property and property (iv), extracting a subsequence and reindexing we may assume without loss of generality that there exist x : (T˜1 , T˜2 ) → E, ˜ ˜ ˜ ˜ u : (T1 , T2 ) → F such that for each t ∈ (T1 , T2 ), x (t) = lim xi (t),
(6.324)
( x , u) ∈ X(T1(k) , T2(k) ),
(6.325)
i→∞
for each integer k ≥ 1,
(1)
(1)
(1)
(1)
I f (T1 , T2 , x , u) ≤ lim inf I f (T1 , T2 , xi , ui ), i→∞
(6.326)
for each integer k ≥ 1, (k+1)
I f (T1
(k)
(k)
(k+1)
, T1 , x , u) ≤ lim inf I f (T1 i→∞
(k+1)
I f (T2 , T2
(k)
(k)
, T1 , xi , ui ), (k+1)
, x , u) ≤ lim inf I f (T2 , T2 i→∞
, xi , ui ).
(6.327) (6.328)
By (6.317) and (6.324), x (t) ≤ M2 , t ∈ (T˜1 , T˜2 ).
(6.329)
xf (T˜1 ) = lim x (t).
(6.330)
We show that t →T˜1+
Let γ > 0. Lemma 6.46, property (iv), and (6.24) imply that there exists δ > 0 such that the following property holds: (v) for each integer i ≥ 1 and each S1 , S2 ∈ [Ti,1 , Ti,2 ] satisfying S2 − S1  ≤ 2δ, we have xi (S1 ) − xi (S2 ) ≤ γ /2;
6.17 Proof of Theorem 6.33
459
(vi) for each S1 , S2 ∈ [0, ∞) satisfying S2 − S1  ≤ 2δ, we have xf (S1 ) − xf (S2 ) ≤ γ /2. Let t ∈ (T˜1 , T˜1 + 2δ).
(6.331)
In view of property (v), (6.302), (6.321), and (6.331), for all sufficiently large natural numbers k, t ∈ (Tk,1 , Tk,1 + 2δ), xk (t) − xk (Tk,1 ) ≤ γ /2, xk (t) − xf (Tk,1 ) ≤ xk (t) − xk (Tk,1 ) + xk (Tk,1 ) − xf (Tk,1 ) ≤ γ /2 + δk . (6.332) Property (vi) and (6.321) imply that for all sufficiently large natural numbers k, xf (Tk,1 ) − xf (T˜1 ) ≤ γ /2.
(6.333)
It follows from (6.332)–(6.334) that for all sufficiently large natural numbers k, x (t) − xf (T˜1 ) = lim xk (t) − xf (T˜1 ) k→∞
≤ lim sup(xk (t) − xf (Tk,1 ) + xf (Tk,1 ) − xf (T˜1 )) ≤ γ . k→∞
Since γ is an arbitrary positive number, we conclude that lim x (t) − xf (T˜1 ) = 0
t →T˜1+
and (6.330) holds. Analogously we can show that if T˜2 < ∞, then xf (T˜2 ) = lim x (t).
(6.334)
u(T˜1 ) = uf (T˜1 ). x (T˜1 ) = xf (T˜1 ),
(6.335)
u(T˜2 ) = uf (T˜2 ). x (T˜2 ) = xf (T˜2 ),
(6.336)
t →T˜2−
Set
If T˜2 < ∞, then we set
460
6 ContinuousTime Nonautonomous Problems on the HalfAxis
We show that t˜ > T˜1 , x (t˜) − xf (t˜) ≥ .
(6.337)
Let γ ∈ (0, /4) and let δ > 0 be such that properties (v) and (vi) hold. Let k ≥ 1 be an integer. If tk − Tk,1 ≤ δ, then in view of properties (v), (vi), and the choice of δ, xf (tk ) − xf (Tk,1 ) ≤ γ /2, xk (tk ) − xk (Tk,1 ) ≤ γ /2 and together with (6.302) and (6.304) these relations imply that < xf (tk ) − xk (tk ) ≤ xf (tk ) − xf (Tk,1 ) + xf (Tk,1 ) − xk (Tk,1 ) +xk (Tk,1 ) − xk (tk ) ≤ γ /2 + δk + γ /2 ≤ γ + /4 < , a contradiction. The contradiction we have reached proves that tk − Tk,1 > δ.
(6.338)
If Tk,2 − tk ≤ δ, then in view of properties (v), (vi), and the choice of δ, xf (tk ) − xf (Tk,2 ) ≤ γ /2, xk (tk ) − xk (Tk,2 ) ≤ γ /2 and together with (6.302) and (6.304), these relations imply that < xf (tk ) − xk (tk ) ≤ xf (tk ) − xf (Tk,2 ) + xf (Tk,2 ) − xk (Tk,2 ) +xk (Tk,2 ) − xk (tk ) ≤ γ /2 + δk + γ /2 ≤ γ + /4 < , a contradiction. The contradiction we have reached proves that Tk,2 − tk ≥ δ.
(6.339)
By (6.338) and (6.339), for all natural numbers k, Tk,2 − tk ≥ δ, tk − Tk,1 ≥ δ.
(6.340)
It follows from (6.319) and (6.340) that T˜2 − t˜ ≥ δ, t˜ − T˜1 ≥ δ. Properties (v) and (vi), (6.306), (6.319), and (6.324) imply that
(6.341)
6.17 Proof of Theorem 6.33
461
x (t˜) − xf (t˜) = lim xk (t˜) − xf (t˜) = lim xk (tk ) − xf (tk ) ≥ k→∞
k→∞
and (6.337) holds. In view of (6.325), for every τ ∈ (T˜1 , T˜2 ), the function G(t, x (t), u(t)), t ∈ [T˜1 , τ ] is strongly measurable. Let k0 be a natural number and τ = T2(k0 ) .
(6.342)
We show that the function G(t, x (t), u(t)), t ∈ [T˜1 , τ ] is Bochner integrable. By (6.20), there exists γ0 > 0 such that f (t, x, u) ≥ 8(G(t, x, u) − a0 x)+ for all (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ γ0 . (k)
Let k ≥ 1 be an integer and T1
< τ . By (6.317),
τ (k)
(6.343)
G(t, x (t), u(t))dt ≤
T1
τ (k)
a0 x (t)dt
T1
+
τ (k)
T1
≤ a0 M2 (τ − T1(k) ) +
(G(t, x (t), u(t)) − a0 x (t))+ dt
τ (k)
T1
(G(t, x (t), u(t)) − a0 x (t))+ dt.
(6.344)
Set (k)
E1 = {t ∈ [T1 , τ ] : G(t, x (t), u(t)) ≥ a0 x (t) + γ0 }, (k)
E2 = [T1 , τ ] \ E1 .
(6.346)
It follows from (6.20), (6.312), (6.326), (6.328), and (6.342)–(6.346) that
τ
(k)
(k) T1
G(t, x (t), u(t))dt ≤ a0 M2 (τ − T1 )
+ E1
(G(t, x (t), u(t)) − a0 x (t))+ dt
+ E2
(6.345)
(G(t, x (t), u(t)) − a0 x (t))+ dt
462
6 ContinuousTime Nonautonomous Problems on the HalfAxis (k)
(k)
≤ a0 M2 (τ − T1 ) + γ0 (τ − T1 ) + E1
(G(t, x (t), u(t)) − a0 x (t))+ dt (k)
(k)
≤ a0 M2 (τ − T1 ) + γ0 (τ − T1 ) + (k)
(k)
f (t, x (t), u(t))dt E1 (k)
(k)
≤ a0 M2 (τ − T1 ) + γ0 (τ − T1 ) + I f (T1 , τ, x , u) + a0 (τ − T1 ) (k0 )
≤ (a0 M2 + γ0 + a0 )(T2
(k)
(k)
(k0 )
− T1 ) + I f (T1 , T2
, x , u)
≤ (T2(k0 ) − T1(k) )(a0 M2 + γ0 + a0 ) + lim inf I f (T1(k) , T2(k0 ) , xi , ui ) i→∞
≤ (T2(k0 ) − T1(k) )(a0 M2 + γ0 + a0 ) + I f (T1(k) , T2(k0 ) , xf , uf ) + 3 + 2S0.
(6.347)
By (6.321) and (6.347), lim
(k0 )
T2
k→∞ T (k) 1 (k0 )
≤ (T2
G(t, x (t), u(t))dt
(k ) − T˜1 )(a0 M2 + γ0 + a0 ) + I f (T˜1 , T2 0 , xf , uf ) + 3 + 2S0 .
Fatou’s lemma and (6.321) imply that
(k0 )
T2 T˜1
G(t, x (t), u(t))dt < ∞,
G(t, x (t), u(t)), t ∈ [T˜1 , T2(k0 ) ] is Bochner integrable for all integers k0 ≥ 1 and
(k0 )
T2 T˜1
G(t, x (t), u(t))dt ≤ I f (T˜1 , T2(k0 ) , xf , uf ) + 3 + 2S0 + (T2(k0 ) − T˜1 )(a0 M2 + γ0 + a0 ).
(6.348)
If T˜2 = ∞, then by (6.348), G(t, x (·), u(·)) is Bochner integrable on [T˜1 , τ ] for any τ > T˜1 .
(6.349)
If T˜2 < ∞, then by (6.321) and (6.348), G(·, x (·), u(·)) is strongly measurable on [T˜1 , T˜2 ] and
6.17 Proof of Theorem 6.33
lim
463
(k)
T2
k→∞ T˜1
G(t, x (t), u(t))dt ≤ I f (T˜1 , T˜2 , xf , uf ) + 3 + 2S0 +(T˜2 − T˜1 )(a0 M2 + γ0 + a0 )
and in view of Fatou’s lemma G(·, x (·), u(·)) is Bochner integrable on [T˜1 , T˜2 ].
(6.350)
Assume that T˜2 = ∞. We show that for every τ > T˜1 , ( x , u) ∈ X(T˜1 , τ ). Let ˜ ˜ τ > T1 . We claim that for every t ∈ (T1 , τ ], x (t) =
t T˜1
G(s, x (s), u(s))ds.
Let t ∈ (T˜1 , τ ], ¯ > 0. By (6.321), (6.330), (6.333), and (6.349), there exists an integer k0 ≥ 1 such that (k0 )
T1
(k0 )
< t ≤ τ < T2
(6.351)
,
(k) x (T1 ) − x (T˜1 ) ≤ ¯ /4 for all integers k ≥ k0 ,
(6.352)
(k)
T1 T˜1
G(s, x (s), u(s))dt < /4 ¯ for all integers k ≥ k0 . (k)
(6.353)
(k)
Let k > k0 be an integer. Since ( x , u) ∈ X(T1 , T2 ) (see (6.335)), it follows from (6.351)–(6.353) that x (t) = x (T1(k) ) +
t (k)
G(s, x (s), u(s))ds.
T1
In view of (6.354), x (t) − x (T˜1 ) −
≤
(k) x (T1 )
− x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
(k)
T1 T˜1
G(s, x (s), u(s))ds ≤ ¯ .
(6.354)
464
6 ContinuousTime Nonautonomous Problems on the HalfAxis
Since ¯ is any positive number, we conclude that x (t) = x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
for all t ∈ (T˜1 , τ ] and ( x , u) ∈ X(T˜1 , τ ) for any τ > T˜1 .
(6.355)
Assume that T˜2 < ∞ and let t ∈ (T˜1 , T˜2 ). We show that x (t) = x (T˜1 ) +
t
G(s, x (s), u(s))ds.
T˜1
Let ¯ > 0. By (6.320), (6.321), (6.330), and (6.335), there exists an integer k0 ≥ 1 such that T1(k) < t < T2(k) for all k ≥ k0 .
(6.356)
x (T1(k) ) − x (T˜1 ) ≤ ¯ /4 for all integers k ≥ k0 ,
(6.357)
(k)
T1 T˜1
G(s, x (s), u(s))ds ≤ ¯ /4 for all integers k ≥ k0 . (k)
(6.358)
(k)
Let k ≥ k0 be an integer. Since ( x , u) ∈ X(T1 , T2 ) it follows from (6.324) and (6.356) that
(k)
x (t) = x (T1 ) +
t (k)
G(s, x (s), u(s))ds.
(6.359)
T1
In view of (6.359), x (t) − x (T˜1 ) − = x (T1(k) ) +
≤
t (k)
t T˜1
G(s, x (s), u(s))ds
G(s, x (s), u(s))ds − x (T˜1 ) −
T1
(k) x (T1 )
− x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
(k)
T1 T˜1
G(s, x (s), u(s))ds < ¯ .
Since ¯ is any positive number, we conclude that
6.17 Proof of Theorem 6.33
465
x (t) = x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
(6.360)
for all t ∈ (T˜1 , T˜2 ). By (6.334), (6.336), (6.350), and (6.360), x (T˜2 ) = lim x (t) = x (T˜1 ) + lim t →T˜2−
t
t →T˜2− T˜1
= x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
G(s, x (s), u(s))ds.
Together with (6.360) this implies that ( x , u) ∈ X(T˜1 , T˜2 ).
(6.361)
It follows from (6.335)–(6.337) that x (T˜i ), i = 1, 2, x (t˜) − xf (t˜) ≥ . xf (T˜i ) =
(6.362)
Property (P4) and (6.362) imply that x , u). I f (T˜1 , T˜2 , xf , uf ) < I f (T˜1 , T˜2 ,
(6.363)
Let Δ > 0. Lemma 6.48 implies that there exists δ > 0 such that the following property holds: (vii) for each τ1 ≥ 0, each τ2 ≥ τ1 + 3bf , each (x, u) ∈ X(τ1 , τ2 ) satisfying x(τi ) − xf (τi ) ≤ δ, i = 1, 2, I f (τ1 , τ2 , x, u) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ, each t1 ∈ [τ1 , τ1 + δ], each t2 ∈ [τ2 − δ, τ2 ], I f (t1 , t2 , x, u) ≤ I f (t1 , t2 , xf , uf ) + Δ/2. In view of (6.300), there exists an integer k0 ≥ 1 such that δk < δ for all integers k ≥ k0 .
(6.364)
Let q ≥ 1 be an integer such that for all integers i ≥ q, T˜1 − T1 ≤ δ/2, T˜2 − T2 ≤ δ/2. (i)
(i)
(6.365)
466
6 ContinuousTime Nonautonomous Problems on the HalfAxis
By (6.318), (6.320), (6.321), and (6.365), there exists an integer k1 ≥ k0 such that for each integer k ≥ k1 , (q)
Tk,1 < T1
(q)
≤ Tk,1 + δ, Tk,2 − δ ≤ T2
< Tk,2 .
(6.366)
Property (vii), (6.301)–(6.303), (6.354), and (6.366) imply that for each integer k ≥ k1 , (q)
(q)
(q)
(q)
I f (T1 , T2 , xk , uk ) ≤ I f (T1 , T2 , xf , uf ) + Δ/2.
(6.367)
In view of (6.326)–(6.328) and (6.367), (q)
(q)
(q)
(q)
I f (T1 , T2 , x , u) ≤ lim inf I f (T1 , T2 , xk , uk ) k→∞
(q)
(q)
≤ I f (T1 , T2 , xf , uf ) + Δ/2
(6.368)
for all integers i ≥ q and for every integer q ≥ 1 satisfying (6.365). Fatou’s lemma, (6.20), and (6.321) imply that x , u) ≤ I f (T˜1 , T˜2 , xf , uf ) + Δ/2. I f (T˜1 , T˜2 , Since Δ is any positive number, we conclude that x , u) ≤ I f (T˜1 , T˜2 , xf , uf ). I f (T˜1 , T˜2 , This contradicts (6.363). The contradiction we have reached proves T˜2 = ∞. Recall (see (6.335) and (6.337)) that x (t˜) − xf (t˜) ≥ . x (T˜1 ) = xf (T˜1 ),
(6.369)
Let Δ > 0. Lemma 6.48 implies that there exists δ ∈ (0, bf /4) such that the following property holds: (viii) for each τ1 ≥ 0, each τ2 ≥ τ1 + 3bf , each (x, u) ∈ X(τ1 , τ2 ) satisfying x(τi ) − xf (τi ) ≤ δ, i = 1, 2, I f (τ1 , τ2 , x, u) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ, each t1 ∈ [τ1 , τ1 + δ], each t2 ∈ [τ2 − δ, τ2 ], I f (t1 , t2 , x, u) ≤ I f (t1 , t2 , xf , uf ) + Δ/2.
6.17 Proof of Theorem 6.33
467
Recall (see (6.323)) that for all integers k ≥ 1, (k)
T2
= Tk,2 .
(6.370)
In view of (6.300), there exists an integer k0 ≥ 1 such that δk < δ for all integers k ≥ k0 . Property (viii), (6.301), (6.303), (6.368), and (6.370) imply that the following property holds: (ix) for each integer k ≥ k0 , each τ1 ∈ [Tk,1 , Tk,1 + δ], each τ2 ∈ [Tk,2 − δ, Tk,2 ], I f (τ1 , τ2 , xk , uk ) ≤ I f (τ1 , τ2 , xf , uf ) + Δ/2. By (6.326)–(6.328) and (6.370), for each integer k ≥ 1 and each integer s ≥ k, (s)
I f (T1 , Tk,2 , x , u) (s)
(k)
(s)
(k)
= I f (T1 , T2 , x , u) ≤ lim inf I f (T1 , T2 , xi , ui ) i→∞ (s)
= lim inf I f (T1 , Tk,2 , xi , ui ). i→∞
(6.371)
In view of (6.312) and (6.371), for each integer k ≥ 1 and each integer s ≥ k, (s)
(s)
I f (T1 , Tk,2 , x , u) ≤ I f (T1 , Tk,2 , xf , uf ) + 3 + 2S0 .
(6.372)
Fatou’s lemma, (6.20), (6.321), and (6.372) imply that for all integers k ≥ 1, x , u) ≤ I f (T˜1 , Tk,2 , xf , uf ) + 4 + 2S0 . I f (T˜1 , Tk,2 , Together with Theorem 6.2 and (6.369), this implies that ( x , u) is (f )good. Property (P1) implies that x (t) − xf (t) = 0. lim
t →∞
Therefore there exists τ0 > 1 + T˜1 such that x (t) − xf (t) ≤ δ/4 for all t ≥ τ0 .
(6.373)
By (6.318)–(6.321) and (6.323), there exists an integer q ≥ 1 such that Tq,2 > τ0 , I f (T˜1 , T1 , x , u) < Δ/8, I f (T˜1 , T1 , xf , uf ) < Δ/8, (6.374) (q)
(q)
468
6 ContinuousTime Nonautonomous Problems on the HalfAxis
for all integers i ≥ q, T˜1 − T1  ≤ δ/2. (i)
(6.375)
It follows from (6.318), (6.321), (6.324), and (6.375) that there exists an integer k1 ≥ k0 + q such that for each integer k ≥ k1 , (q)
Tk,1 < T1
≤ Tk,1 + δ, x (Tq,2 ) − xk (Tq,2 ) ≤ δ/4.
(6.376)
Assume that an integer k ≥ k1 . Then (6.376) holds. By (6.301)–(6.303), (6.323), the choice of k0 , and (6.376), q
q
q
I f (T1 , Tq,2 , xk , uk ) ≤ U f (T1 , Tq,2 , xk (T1 ), xk (Tq,2 )) + δ, xk (Tk,1 ) − xf (Tk,1 ) ≤ δk ≤ δ.
(6.377) (6.378)
In view of (6.373)–(6.376), x (Tq,2 ) + x (Tq,2 ) − xf (Tq,2 ) ≤ δ/4 + δ/4. xk (Tq,2 ) − xf (Tq,2 ) ≤ xk (Tq,2 ) − (6.379) Properties (ix) and (vii) applied with τ1 = Tk,1 , τ2 = Tq,2 , (x, u) = (xk , uk ), (q) t1 = T1 , t2 = Tq,2 , (6.376)–(6.379) imply that (q)
(q)
I f (T1 , Tq,2 , xk , uk ) ≤ I f (T1 , Tq,2 , xf , uf ) + Δ/2.
(6.380)
In view of (6.380), (q)
(q)
I f (T1 , Tq,2 , x , u) ≤ lim inf I f (T1 , Tq,2 , xk , uk ) k→∞
(q)
≤ I f (T1 , Tq,2 , xf , uf ) + Δ/2.
(6.381)
By (6.375) and (6.381), x , u) ≤ I f (T˜1 , Tq,2 , xf , uf ) + Δ. I f (T˜1 , Tq,2 ,
(6.382)
Since the relation above holds for any integer q ≥ 1 satisfying (6.374) and (6.375), we have x , u) − I f (T˜1 , T , xf , uf )) ≤ Δ. lim inf(I f (T˜1 , T , T →∞
Since Δ is any positive, number we conclude that x , u) − I f (T˜1 , T , xf , uf )) ≤ 0. lim inf(I f (T˜1 , T , T →∞
(6.383)
6.18 Proof of Theorem 6.35
469
For all t ∈ [0, T˜1 ] \ {T˜1 }, set x (t) = xf (t), u(t) = uf (t). In view of (6.383), ( x , u) ∈ X(0, ∞) is (f )weakly optimal. Theorem 6.29 and (P4) imply that ( x , u) ∈ X(0, ∞) is (f )overtaking optimal and that x (t) = xf (t), t ≥ 0. This contradicts (6.362). The contradiction we have reached completes the proof of Lemma 6.49. Completion of Theorem 6.33. By Theorem 6.22, TP holds. Lemma 6.49 and TP imply STP. Theorem 6.33 is proved.
6.18 Proof of Theorem 6.35 In view of Theorem 6.26, in order to show that (P1), (P2) hold it is sufficient to show that fr has (P3). Let , M > 0. Theorem 6.2 implies that there exists S0 > 0 such that for each T2 > T1 ≥ 0 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(6.384)
Property (i) of Section 6.5 implies there exists δ ∈ (0, ) such that if z ∈ E satisfies φ(z) ≤ δ, then z ≤ .
(6.385)
L = δ −1 r −1 (S0 + M).
(6.386)
Set
Assume that T1 ≥ 0, T2 ≥ T1 + L, (x, u) ∈ X(T1 , T2 ) satisfies I fr (T1 , T2 , x, u) ≤ I fr (T1 , T2 , xf , uf ) + M. By (6.384), (6.385), the equality φ(0) = 0, and the definition of fr , I f (T1 , T2 , xf , uf ) + M = M + I fr (T1 , T2 , xf , uf ) ≥ I (T1 , T2 , x, u) + r f
T2 T1
φ(x(t) − xf (t))dt
(6.387)
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
≥ I f (T1 , T2 , xf , uf ) − S0 + r
T2
φ(x(t) − xf (t))dt,
T1
r
−1
(S0 + M) ≥
T2
φ(x(t) − xf (t))dt
T1
≥ δmes({t ∈ [T1 , T2 ] : φ(x(t) − xf (t)) ≥ δ}) ≥ δmes({t ∈ [T1 , T2 ] : x(t) − xf (t) > }) and mes({t ∈ [T1 , T2 ] : x(t) − xf (t) > }) ≤ δ −1 r −1 (S0 + M) ≤ L. Thus fr has (P3) and (P1), (P2) too. Assume that (xf , uf ) ∈ X(0, ∞) is (f )minimal. It is not difficult to see that (xf , uf ) is (fr )minimal too. Let (x, u) ∈ X(0, ∞) be (fr )good, x(0) = xf (0).
(6.388)
lim x(t) − xf (t) = 0.
(6.389)
Therefore t →∞
Let > 0. By (A3), there exists δ > 0 such that the following property holds: (i) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ there exist τ ∈ (0, bf ], (y, v) ∈ X(T , T + τ ) such that y(T ) = z, y(T + τ ) = xf (T + τ ), I fr (T , T + τ, y, v) ≤ I fr (T , T + τ, xf , uf ) + /8. In view of (6.389), there exists T0 > 0 such that x(t) − xf (t) ≤ δ for all t ≥ T0 .
(6.390)
Let T ≥ T0 . Property (i) and (6.390) imply that there exist τ ∈ (0, bf ], (y, v) ∈ X(T , T + τ ) such that y(T ) = x(T ), y(T + τ ) = xf (T + τ ), I fr (T , T + τ, y, v) ≤ I f (T , T + τ, xf , uf ) + /8. Set
(6.391)
6.19 Examples
471
y(t) = x(t), v(t) = u(t), t ∈ [0, T ].
(6.392)
Clearly, (y, v) ∈ X(0, T + τ ). Since (xf , uf ) is (f )minimal, it follows from (6.388), (6.391), and (6.392) that I f (0, T + τ, xf , uf ) ≤ I f (0, T + τ, y, v).
(6.393)
By the definition of fr and (6.391)–(6.393), I fr (0, T , xf , uf ) ≤ I f (0, T , y, v) + I f (T , T + τ, y, v) − I f (T , T + τ, xf , uf ) = I f (0, T , x, u) + I fr (T , T + τ, y, v)
T +τ
−r
φ(y(t) − xf (t))dt − I f (T , T + τ, xf , uf )
T
T +τ
≤ I f (0, T , x, u) − r
φ(y(t) − xf (t))dt + /8
T
T
= I (0, T , x, u) − r fr
T +τ
φ(x(t) − xf (t))dt − r
0
φ(y(t) − xf (t))dt + /8.
T
Since the relation above holds for any T ≥ T0 , we obtain lim sup(I fr (0, T , xf , uf )−I fr (0, T , x, u)) ≤ −r lim
T
φ(x(t)−xf (t))dt+/8.
T →∞ 0
T →∞
Since is any positive number, we have lim sup(I (0, T , xf , uf ) − I (0, T , x, u)) ≤ −r lim fr
T →∞
f
T →∞ 0
T
φ(x(t) − xf (t))dt.
(6.394) This implies that (xf , uf ) ∈ X(0, ∞) is (fr )overtaking optimal. Assume that (x, u) is (fr )overtaking optimal. Then (6.394) implies that x(t) = xf (t) for all t ≥ 0. Thus (P4) holds. Theorem 6.35 is proved.
6.19 Examples In this section we present a family of problems which belong to the second class of problems and for which the results of this chapter hold. We use the notation introduced in Section 6.1.
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
Let (E, ·, ·)E be a Hilbert space equipped with an inner product ·, ·E which induces the norm ·E , and let (F, ·, ·F ) be a Hilbert space equipped with an inner product ·, ·F which induces the norm · F . For simplicity, we set ·, ·E = ·, ·, · E = · , ·, ·F = ·, ·, · F = · , if E, F are understood. We suppose that A0 is a nonempty subset of E, U0 : A0 → 2F is a point to set mapping with a graph M0 = {(x, u) : x ∈ A0 , u ∈ U0 (x)}. We suppose that M0 is a Borel measurable subset of E × F . Define A = [0, ∞) × A0 , U : A → 2F by U(t, x) = U0 (x), (t, x) ∈ A, M = [0, ∞) × M0 . Let a linear operator A : D(A) → E generates a C0 semigroup S(t) = eAt , t ≥ 0 on E, E1d = D(A∗ ), E−1 = D(A∗ ) and let B ∈ L(F, E−1 ) is an admissible control operator for eAt , t ≥ 0. For T2 > T1 ≥ 0 we consider the following control system x (t) = Ax(t) + Bu(t), t ∈ [T1 , T2 ] a. e. . Assume that (xf , uf ) ∈ X(0, ∞) and sup{xf (t) : t ∈ [0, ∞)} < ∞. Recall that for every T > 0, ΦT ∈ L(L2 (0, ∞; F ), E) is defined by
T
ΦT u =
S(T − s)Bu(s)ds, u ∈ L2 (0, T ; F ).
(6.395)
0
Theorem 6.50. Assume that T0 > 0, Ran(ΦT0 ) = E. Then there exists a constant c > 0 such that for each T ≥ 0 and each z0 , z1 ∈ E, there exist u ∈ L2 (T , T + T0 ; F ) and z ∈ C 0 ([T , T + T0 ]); E) which is a solution of the initial value problem z (t) = Az(t) + Bu(t), t ∈ [T , T + T0 ] a. e. , z(T ) = z0
6.19 Examples
473
in E−1 and satisfies z(T + T0 ) = z1 and such that the following inequalities hold: z(t) − xf (t), u(t) − uf (t) ≤ c(z1 − xf (T + T0 ) + z0 − xf (T )), t ∈ [T , T + T0 ]. Proof. Let L ∈ L(E, L2 (0, T0 ; F )) be as guaranteed by Proposition 1.22. Therefore ΦT0 Lx = x for all x ∈ E.
(6.396)
Let cT0 > 1 be as guaranteed by Proposition 1.21 with T = T0 . Set c = cT0 + cT20 L.
(6.397)
Let z0 , z1 ∈ E. Define z1 (t) = S(t −T )(z0 −xf (T )) = eA(t −T ) (z0 −xf (T )), t ∈ [T , T +T0 ].
(6.398)
In view of (6.398), z1 (t) = Az1 (t), t ∈ [T , T + T0 ] a. e. ,
(6.399)
z1 (T ) = z0 − xf (T ),
(6.400)
(z1 + xf , uf ) ∈ X(T , T + T0 ), (z1 + xf ) (t) = A(z1 + xf )(t) + Buf (t), t ∈ [T , T + T0 ] a. e.
(6.401)
in E−1 , (z1 + xf )(T ) = z0 .
(6.402)
By Proposition 1.19, there exists a unique z2 ∈ C 0 ([0, T0 ]; E) such that z2 = Az2 + B(L(z1 − (z1 + xf )(T + T0 ))), t ∈ (0, T0 ) a. e. , z2 (0) = 0
(6.403) (6.404)
in E−1 . In view of (6.396), (6.403), and (6.404), z2 (T0 ) = S(T0 )z2 (0) + ΦT0 (L(z1 − (z1 + xf )(T + T0 ))) = z1 − (z1 + xf )(T + T0 ). (6.405)
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
Set z3 (t) = z2 (t − T ), t ∈ [T , T + T0 ],
(6.406)
z(t) = xf (t) + z1 (t) + z3 (t), t ∈ [T , T + T0 ].
(6.407)
By (6.398) and (6.404)–(6.407), z(T ) = xf (T ) + z1 (T ) + z3 (T ) = xf (T ) + z0 − xf (T ) + z2 (0) = z0 ,
(6.408)
z(T + T0 ) = xf (T + T0 ) + z1 (T + T0 ) + z2 (T0 ) = z1 .
(6.409)
It follows from (6.399), (6.403), (6.406), and (6.407) that for all t ∈ [T , T + T0 ], z (t) = xf (t) + z1 (t) + z3 (t) = Axf (t) + Buf (t) + Az1 (t) + z2 (t − T ) = Axf (t) + Buf (t) + Az1 (t) + Az3 (t) + B(L(z1 − (z1 + xf )(T + T0 ))) = Az(t) + B(uf (t) + L(z1 − (z1 + xf )(T + T0 )))
(6.410)
in E−1 . In view of (6.400), L(z1 −(z1 +xf )(T +T0 )) ≤ L(z1 −xf (T +T0 )+z1 (T +T0 )).
(6.411)
Proposition 1.21, the choice of cT0 , (6.399), (6.42), (6.404), and (6.406) imply that for all t ∈ [T , T + T0 ], z1 (t) ≤ cT0 z0 − xf (T ),
(6.412)
z3 (t) ≤ cT0 L(z1 − (z1 + xf )(T + T0 )) ≤ cT0 Lz1 − (z1 + xf )(T + T0 ) ≤ cT0 L(z1 − xf (T + T0 ) + cT0 z0 − xf (T )).
(6.413)
By (6.397) and (6.412), L(z1 − (z1 + xf )(T + T0 )) ≤ L(z1 − xf (T + T0 ) + cT0 z0 − xf (T )) ≤ c(z1 − xf (T + T0 ) + z0 − xf (T )). In view of (6.407), (6.412), and (6.413), for all t ∈ [T , T + L0 ],
6.19 Examples
475
z(t) − xf (t) ≤ z1 + z3 (t) ≤ cT0 z0 − xf (T ) + cT0 L(z1 − xf (T + T0 ) + cT0 z0 − xf (T )) ≤ c(z0 − xf (T ) + z1 − xf (T + T0 )). Theorem 6.50 is proved. Assume that there exists Tf > 0 such that RanΦTf = E. In other words the pair (A, B) is exactly controllable. We suppose that there exists r∗ > 0 such that for each t ≥ 0, {(x, u) ∈ E × F : x − xf (t) ≤ r∗ , u − uf (t) ≤ r∗ } ⊂ M0 . Assume that L : [0, ∞) × E × F → [0, ∞) is a Borelian function, L(t, xf (t), uf (t)) = 0, t ∈ [0, ∞), ψ0 : [0, ∞) → [0, ∞) is an increasing function, K1 , a1 > 0 such that lim ψ0 (t) = ∞,
t →∞
L(t, x, u) ≥ −a1 + max{ψ0 (x)x, K1 u2 } for all (t, x, u) ∈ M, μ ∈ R 1 , p¯ ∈ D(A∗ ). Let for all (t, x, u) ∈ M, f (t, x, u) = L(t, x, u) + μ + x, A∗ p ¯ + Bu, p ¯ E−1 ,E d . 1
(6.414)
It is not difficult to see that f is a Borelian function; there exist a0 , K0 > 0 and an increasing function ψ : [0, ∞) → [0, ∞) such that limt →∞ ψ(t) = ∞ and for all (t, x, u) ∈ M, f (t, x, u) ≥ −a0 + max{ψ(x), K0 u2 }. We suppose that the following property holds: (a) for each > 0 there exists δ > 0 such that for each (t, x, u) ∈ M satisfying x − xf (t) + u − uf (t) ≤ δ we have f (t, x, u) ≤ f (t, xf (t), uf (t)) + .
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
It follows from property (a) and Theorem 6.50 that (A3) holds. Let 0 ≤ T1 < T2 , (x, u) ∈ X(T1 , T2 ). It is not difficult to see that I f (T1 , T2 , x, u) − I f (T1 , T2 , xf , uf ) =
T2
L(t, x(t), u(t))dt +
T1
T2
∗
x(t), A pdt ¯ +
T1
−
T2
xf (t), A∗ pdt ¯ −
T1
T2 T1
T2 T1
Bu(t), p ¯ E−1 ,Ed dt
Buf (t), p ¯ E−1 ,Ed dt
≥ x(T2 ) − x(T1 ), p ¯ − xf (T2 ) − xf (T1 ), p. ¯
(6.415)
This implies that (A1) holds. Assume now that xf is uniformly continuous on [0, ∞) and that the following property holds: (a) for each M, > 0 there exists δ > 0 such that for each (t, x, u) ∈ [0, ∞) × E × F which satisfies x ≤ M and L(t, x, u) ≤ δ we have x − xf (t) ≤ . We claim that f has TP. In view of Theorems 6.22 and 6.26, it is sufficient to show that properties (A3) and (P3) of Theorem 6.26 hold. By (6.414), (A3) holds. Let us show that (P3) holds. Let , M > 0 and M1 > 0 be as guaranteed by Theorem 6.3 with M0 = M, c0 = 1, c = 2. Let S > 0 be as guaranteed by Theorem 6.2, and let δ ∈ (0, 1) be as guaranteed by property (a) with M = M1 . Choose L > 3 + δ −1 (M + S + 4M1 p). ¯ Assume that T1 ≥ 0, T2 ≥ T1 + L and that (x, u) ∈ X(T1 , T2 ) satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M. Combined with Theorem 6.3 and the choice of M1 , this implies that x(t) ≤ M1 , t ∈ [T1 + 1, T2 ]. It is not difficult to see that, in view of the choice of S and Theorem 6.2,
6.19 Examples
477
I f (T1 + 1, T2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , T1 + 1, x, u) ≤ I f (T1 , T2 , xf , uf ) + M − I f (T1 , T1 + 1, xf , uf ) + S = I f (T1 + 1, T2 , xf , uf ) + M + S. Combined with (6.415), this implies that M + S ≥ I f (T1 + 1, T2 , x, u) − I f (T1 + 1, T2 , xf , uf ) = + −
T2
T1 +1 T2 T1 +1
T2 T1 +1
L(t, x(t), u(t))dt
x(t), A∗ pdt ¯ +
xf (t), A∗ pdt ¯ − =
T2
T1 +1
T2 T1 +1
Bu(t), p ¯ E−1 ,Ed dt
Buf (t), p ¯ E−1 ,Ed dt
T2 T1 +1
L(t, x(t), u(t))dt
+ x(T1 + 1) − x(T2 ), p ¯ − xf (T1 + 1) − xf (T2 ), p ¯ ≥
T2 T1 +1
L(t, x(t), u(t))dt − 4M1 p. ¯
Together with the choice of δ and property (a) with M = M1 , this implies that ¯ ≥ M + S + 4M1 p
T2 T1 +1
L(t, x(t), u(t))dt
≥ δmes({t ∈ [T1 + 1, T2 ] : L(t, x(t), u(t)) > δ}) ≥ δmes({t ∈ [T1 + 1, T2 ] : x(t) − xf (t) > }) and mes({t ∈ [T1 , T2 ] : x(t) − xf (t) > }) ≤ 1 + δ −1 (M + S + 4M1 p) ¯ < L. Thus (P3) holds and f has TP.
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6 ContinuousTime Nonautonomous Problems on the HalfAxis
We can easily obtain particular cases of this example with the pairs (A, B) considered in Section 1.8.
6.20 Exercises for Chapter 6 Exercise 6.51. Use the control problem considered in Example 1.24, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.52. Use the control problem considered in Example 1.25, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.53. Use the control problem considered in Example 1.26, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.54. Use the control problem considered in Example 1.27, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.55. Use the control problem considered in Example 1.28, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.56. Use the control problem considered in Example 1.29, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.57. Use the control problem considered in Example 1.30, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19. Exercise 6.58. Use the control problem considered in Example 1.31, and construct an optimal control problem which is a special case of the example analyzed in Section 6.19.
Chapter 7
ContinuousTime Nonautonomous Problems on Axis
In this chapter we establish sufficient and necessary conditions for the turnpike phenomenon for continuoustime optimal control problems on subintervals of axis in infinite dimensional spaces. For these optimal control problems, the turnpike is not a singleton. We also study the existence of solutions of the corresponding infinite horizon optimal control problems.
7.1 Preliminaries Let (E, · ) be a Banach space and (F, ρF ) be a metric space. We suppose that A is a nonempty subset of R 1 × E, U : A → 2F is a point to set mapping with a graph M = {(t, x, u) : (t, x) ∈ A, u ∈ U(t, x)}. We suppose that M is a Borel measurable subset of R 1 × E × F , and G : M → E is a Borelian function. Let f : M → R 1 be a Borelian function bounded from below. Let −∞ < T1 < T2 < ∞ and consider the equation x (t) = G(t, x(t), u(t)), t ∈ [T1 , T2 ].
(7.1)
A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a solution of (7.1) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, (t, x(t)) ∈ A, t ∈ [T1 , T2 ],
(7.2)
u(t) ∈ U(t, x(t)), t ∈ [T1 , T2 ] almost everywhere (a.e.),
(7.3)
© Springer Nature Switzerland AG 2019 A. J. Zaslavski, Turnpike Conditions in Infinite Dimensional Optimal Control, Springer Optimization and Its Applications 148, https://doi.org/10.1007/9783030201784_7
479
480
7 ContinuousTime Nonautonomous Problems on Axis
G(s, x(s), u(s)), s ∈ [T1 , T2 ] is Bochner integrable and for every t ∈ [T1 , T2 ], x(t) = x(T1 ) +
t
G(s, x(s), u(s))ds, t ∈ [T1 , T2 ].
(7.4)
T1
The set of all pairs (x, u) which are solutions of (7.1) is denoted by X(T1 , T2 ). Let T1 ∈ R 1 . A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a solution of the system x (t) = G(t, x(t), u(t)), t ∈ [T1 , ∞)
(7.5)
if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (7.1). The set of all such pairs (x, u) which are solutions of the equation above is denoted by X(T1 , ∞). A pair of functions x : R 1 → E, u : R 1 → F is called a solution of the system x (t) = G(t, x(t), u(t)), t ∈ R 1
(7.6)
if for every pair T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (7.1). The set of all such pairs (x, u) which are solutions of (7.6) is denoted by X(−∞, ∞). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) or R 1 (T1 < T2 ), is called a trajectory if there exists a Lebesgue measurable function u : I → F (referred to as a control) such that (x, u) ∈ X(T1 , T2 ) or (x, u) ∈ X(T1 , ∞) or (x, u) ∈ X(−∞, ∞), respectively. Let T2 > T1 , (x, u) ∈ X(T1 , T2 ). Define I f (T1 , T2 , x, u) =
T2
f (t, x(t), u(t))dt T1
which is welldefined but can be ∞. Let a0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → ∞ as t → ∞.
(7.7)
We suppose that the function f satisfies f (t, x, u) ≥ −a0 + max{ψ(x), ψ((G(t, x, u) − a0 x)+ )(G(t, x, u) − a0 x)+ }
(7.8)
for each (t, x, u) ∈ M, and we consider functionals of the form I f (T1 , T2 , x, u), where T1 < T2 and (x, u) ∈ X(T1 , T2 ). For each pair of points (T1 , y), (T2 , z) ∈ A
7.1 Preliminaries
481
such that T1 < T2 , we define U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z},
(7.9)
σ f (T1 , T2 , y) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y},
(7.10)
σ f (T1 , T2 ) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 )}.
(7.11)
We suppose that (xf , uf ) ∈ X(−∞, ∞) satisfies sup{xf (t) : t ∈ R 1 } < ∞, Δf := sup{I f (j, j + 1, xf , uf ) : j is an integer } < ∞.
(7.12) (7.13)
We suppose that there exists a number bf > 0 and the following assumptions hold. (A1) For each S1 > 0, there exist S2 > 0 and c > 0 such that I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S2 for each T1 ∈ R 1 , each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying x(Tj ) ≤ S1 , j = 1, 2. (A2) For each > 0, there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δ, i = 1, 2 there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfy x1 (T1 ) = z1 , x1 (T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + , x2 (T2 ) = z2 , x2 (T2 − τ2 ) = xf (T2 − τ2 ), I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + . Relations (7.8) and (7.13) imply the following result. Lemma 7.1. Let c > 0. Then Δf (c) := sup{I f (T1 , T2 , xf , uf ) : T1 ∈ R 1 , T2 ∈ (T1 , T1 + c]} < ∞. Section 7.18 contains examples of optimal control problems satisfying assumptions (A1) and (A2). Many examples can also be found in [106–108, 118, 124, 125, 134]
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7 ContinuousTime Nonautonomous Problems on Axis
7.2 Boundedness Results The following result is proved in Section 7.6. Theorem 7.2. 1. There exists S > 0 such that for each pair of numbers T2 > T1 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S ≥ I f (T1 , T2 , xf , uf ). 2. For each (x, u) ∈ X(−∞, ∞) either I f (−T , T , x, u) − I f (−T , T , xf , uf ) → ∞ as T → ∞ or sup{I f (−T , T , x, u) − I f (−T , T , xf , uf ) : T ∈ (0, ∞)} < ∞.
(7.14)
Moreover, if (7.14) holds, then sup{x(t) : t ∈ R 1 } < ∞. We say that (x, u) ∈ X(−∞, ∞) is (f )good if (7.14) holds. The next boundedness result is proved in Section 7.6. Theorem 7.3. Let M0 > 0, c > 0. Then there exists M1 > 0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + c, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Let L > 0. Denote by AL the set of all (s, z) ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ X(s, s + τ ) such that x(s) = z, x(s + τ ) = xf (s + τ ), I f (s, s + τ, x, u) ≤ L. L the set of all (s, z) ∈ A such that there exist τ ∈ (0, L] and (x, u) ∈ Denote by A X(s − τ, s) such that x(s − τ ) = xf (s − τ ), x(s) = z, I f (s − τ, s, x, u) ≤ L. The following Theorems 7.4 and 7.5 are also boundedness results. They are proved in Section 7.6. Theorem 7.4. Let L > 0, M0 > 0. Then there exists M1 > 0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying
7.3 Turnpike Results
483
L , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 7.5. Let L > 0, M0 > 0. Then there exists M1 > 0 such that for each T1 ≥ 0, each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying (T1 , x(T1 )) ∈ AL , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 , the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Theorem 7.6. For each T0 ∈ R 1 and each (x, u) ∈ X(T0 , ∞) either I f (T0 , T , x, u) − I f (T0 , T , xf , uf ) → ∞ as T → ∞ or sup{I f (T0 , T , x, u) − I f (T0 , T , xf , uf ) : T ∈ (T0 , ∞)} < ∞.
(7.15)
Moreover, if (7.15) holds, then sup{x(t) : t ∈ [T0 , ∞)} < ∞. Theorem 7.6 easily follows from Assertion 1 of Theorem 7.2 and Theorem 7.3. Let T0 ∈ R 1 . We say that (x, u) ∈ X(T0 , ∞) is (f )good if (7.15) holds. Let T0 ∈ R 1 . We say that (x, u) ∈ X(T0 , ∞) is (f )minimal if for every T > T0 , I f (T0 , T , x, u) = U f (T0 , T , x(T0 ), x(T ))
(7.16)
and that (x, u) ∈ X(−∞, ∞) is called (f )minimal if for every pair T2 > T1 , I f (T1 , T2 , x, u) = U f (T1 , T2 , x(T1 ), x(T2 )).
(7.17)
7.3 Turnpike Results We say that the integrand f possesses the strong turnpike property (or STP for short) if for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ}, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ].
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7 ContinuousTime Nonautonomous Problems on Axis
Moreover, if x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 , and if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 . We say that f possesses lower semicontinuity property (or LSC property for short) if for each T2 > T1 and each sequence (xj , uj ) ∈ X(T1 , T2 ), j = 1, 2, . . . which satisfies sup{I f (T1 , T2 , xj , uj ) : j = 1, 2, . . . } < ∞, there exist a subsequence {(xjk , ujk )}∞ k=1 and (x, u) ∈ X(T1 , T2 ) such that xjk (t) → x(t) as k → ∞ for every t ∈ [T1 , T2 ], I f (T1 , T2 , x, u) ≤ lim inf I f (T1 , T2 , xj , uj ). j →∞
LSC property plays an important role in the calculus of variations and optimal control theory [32]. The next result follows from Lemma 7.21. Proposition 7.7. Let f have LSC property. Then for each T2 > T1 and each sequence (xj , uj ) ∈ X(T1 , T2 ), j = 1, 2, . . . which satisfies sup{I f (T1 , T2 , xj , uj ) : j = 1, 2, . . . } < ∞ there exist a subsequence {(xjk , ujk )}∞ k=1 and (x, u) ∈ X(T1 , T2 ) such that xjk (·) → x(·) as k → ∞ uniformly on [T1 , T2 ], I f (T1 , T2 , x, u) ≤ lim inf I f (T1 , T2 , xj , uj ). j →∞
The following result is proved in Section 7.11. Theorem 7.8. Let f have LSC property. If f has STP then the following three properties hold: (P1) for each (f )good pair (x, u) ∈ X(−∞, ∞), lim x(t) − xf (t) = 0,
t →∞
lim x(t) − xf (t) = 0;
t →−∞
(P2) for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each T ∈ R 1 and each (x, u) ∈ X(T , T + L) which satisfies I f (T , T + L, x, u) ≤ min{U f (T , T + L, x(T ), x(T + L)) + δ, I f (T , T + L, xf , uf ) + M}, there exists s ∈ [T , T + L] such that x(s) − xf (s) ≤ ;
7.3 Turnpike Results
485
(P3) there exists u˜ f : R 1 → F such that (xf , u˜ f ) ∈ X(−∞, ∞) is (f )good and (f )minimal and if (x, u) ∈ X(−∞, ∞) is (f )good and (f )minimal, then x(t) = xf (t) for all t ∈ R 1 . If (P1)–(P3) hold and u˜ f = uf , then f has STP. The following result is proved in Section 7.14. Theorem 7.9. f has (P1) and (P2) if and only if the following property holds: (P4) for each > 0 and each M > 0, there exist δ > 0 and L > 0 such that for each T1 ∈ R 1 , each T2 > T1 , and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ min{σ f (T1 , T2 ) + M, U f (T1 , T2 , x(T1 ), x(T2 )) + δ} if T2 ≥ 2L + (T1 )+ , then x(t) − xf (t) ≤ for all t ∈ [(T1 )+ + L, T2 − L] and if T1 ≤ (−T2 )+ − 2L, then x(t) − xf (t) ≤ for all t ∈ [T1 + L, −(−T2 )+ − L]. Theorem 7.10. Assume that f has STP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies L0 , (T1 , x(T1 )) ∈ AL0 , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 , and if and x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 . Theorem 7.10 follows from STP and the following result which is proved in Section 7.13. Proposition 7.11. Let L0 > 0. Then there exists M > 0 such that for each pair of L0 satisfying T2 ≥ T1 + 2L0 , points (T1 , z1 ) ∈ AL0 , (T2 , z2 ) ∈ A U f (T1 , T2 , z1 , z2 ) ≤ σ f (T1 , T2 ) + M.
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7 ContinuousTime Nonautonomous Problems on Axis
Theorem 7.12. Assume that f has STP and that , L0 > 0. Then there exist δ > 0 and L > L0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) which satisfies (T1 , x(T1 )) ∈ AL0 , I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + δ, there exist τ1 ∈ [T1 , T1 + L], τ2 ∈ [T2 − L, T2 ] such that x(t) − xf (t) ≤ for all t ∈ [τ1 , τ2 ]. Moreover, if x(T2 ) − xf (T2 ) ≤ δ, then τ2 = T2 , and if x(T1 ) − xf (T1 ) ≤ δ, then τ1 = T1 . Theorem 7.12 follows from STP and the following result which is proved in Section 7.13. Proposition 7.13. Let L0 > 0. Then there exists M > 0 such that for each (T1 , z1 ) ∈ AL0 and each T2 ≥ T1 + L0 , σ f (T1 , T2 , z1 ) ≤ σ f (T1 , T2 ) + M. We say that f possesses the weak turnpike property (or WTP for short) if for each > 0 and each M > 0, there exist a natural number Q and l > 0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + lQ, and each (x, u) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M, q
q
there exist finite sequences {ai }i=1 , {bi }i=1 ⊂ [T1 , T2 ] such that an integer q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, bi ≤ ai+1 for all integers i satisfying 1 ≤ i < q, q
x(t) − xf (t) ≤ for all t ∈ [T1 , T2 ] \ ∪i=1 [ai , bi ]. The next result is proved in Section 7.14. Theorem 7.14. f has WTP if and only if f has (P1) and (P2). Let T0 ∈ R 1 . A pair (x, u) ∈ X(T0 , ∞) is called (f )overtaking optimal if (x, u) is (f )good and if for every (y, v) ∈ X(T0 , ∞) satisfying x(T0 ) = y(T0 ), lim sup[I f (T0 , T , x, u) − I f (T0 , T , y, v)] ≤ 0. T →∞
7.4 Perturbed Problems
487
A pair (x, u) ∈ X(T0 , ∞) is called f weakly optimal if (x, u) is (f )good and if for every (y, v) ∈ X(T0 , ∞) satisfying x(T0 ) = y(T0 ), lim inf[I f (T0 , T , x, u) − I f (T0 , T , y, v)] ≤ 0. T →∞
The next result is proved in Section 7.16. Theorem 7.15. Assume that f has (P1), (P2) and LSC property, S ∈ R 1 , (x, ˜ u) ˜ ∈ X(S, ∞) is (f )good and that (x∗ , u∗ ) ∈ X(S, ∞) satisfies x∗ (S) = x(S). ˜ Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)
(x∗ , u∗ ) is (f )overtaking optimal; (x∗ , u∗ ) is (f )weakly optimal; (x∗ , u∗ ) is (f )minimal and (f )good; (x∗ , u∗ ) is (f )minimal and satisfies limt →∞ (x∗ (t) − xf (t)) = 0; (x∗ , u∗ ) is (f )minimal and satisfies lim inft →∞ x∗ (t) − xf (t) = 0.
Moreover, there exists an (f )overtaking optimal pair (y, v) ∈ X(S, ∞) such that x(S) ˜ = y(S).
7.4 Perturbed Problems Let f has LSC property. In this section we suppose that the following assumption holds. (A3) For each > 0, there exists δ > 0 such that for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δ, i = 1, 2, there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfy x1 (T1 ) = z1 , x1 (T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + , x1 (t) − xf (t) ≤ , t ∈ [T1 , T1 + τ1 ], x2 (T2 ) = z2 , x2 (T2 − τ2 ) = xf (T2 − τ2 ), I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + , x2 (t) − xf (t) ≤ , t ∈ [T2 − τ2 , T2 ]. Clearly, (A3) implies (A2). Assume that φ : E → [0, 1] is a continuous function satisfying φ(0) = 0 and such that the following property holds:
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7 ContinuousTime Nonautonomous Problems on Axis
(i) for each > 0 there exists δ > 0 such that for each x ∈ E satisfying φ(x) ≤ δ we have x ≤ . For each r ∈ (0, 1) set fr (t, x, u) = f (t, x, u) + rφ(x − xf (t)), (t, x, u) ∈ M. Clearly, for any r ∈ (0, 1), fr is a Borealian function and (A1) and (A3) hold for fr with (xfr , ufr ) = (xf , uf ). Theorem 7.16. Let r ∈ (0, 1), (xf , uf ) is (f )minimal . Then fr has STP. Theorem 7.16 is proved in Section 7.17.
7.5 Auxiliary Results Assumption (A2) easily implies the following result. Proposition 7.17. Let γ > 0. Then there exists δ > 0 such that for each (T , z1 ), (T + 2bf , z2 ) ∈ A satisfying z1 − xf (T ) ≤ δ, z2 − xf (T + 2bf ) ≤ δ there exists (x, u) ∈ X(T , T + 2bf ) which satisfies x(T ) = z1 , x(T + 2bf ) = z2 , I f (T , T + 2bf , x, u) ≤ I f (T , T + 2bf , xf , uf ) + γ . Lemma 7.18. There exist numbers S > 0, c0 ≥ 1 such that for each T1 ∈ R 1 , each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S.
(7.18)
Proof. In view of (7.7), there exists S1 > 0 such that ψ(S1 ) > a0 + 1 + Δf . By (A1), there exist S2 > 0, c0 > 1 such that I f (T1 , T2 , xf , uf ) ≤ I f (T1 , T2 , x, u) + S2 for each T1 ∈ R 1 , each T2 ≥ T1 + c0 , and each (x, u) ∈ X(T1 , T2 ) satisfying x(T1 ), x(T2 ) ≤ S1 .
(7.19)
7.5 Auxiliary Results
489
Fix S ≥ S2 + 2 + a0 (2 + 2c0 ) + 8(c0 + 1)Δf .
(7.20)
Assume that T1 ∈ R 1 , T2 ≥ T1 + c0 , (x, u) ∈ X(T1 , T2 ). We show that (7.18) is true. Assume that x(t) ≥ S1 , t ∈ [T1 , T2 ].
(7.21)
By (7.8), (7.13), (7.19), and (7.21), for all t ∈ [T1 , T2 ], f (t, x(t), u(t)) ≥ −a0 + ψ(x(t)) ≥ −a0 + ψ(S1 ) and I f (T1 , T2 , x, u) ≥ (T2 − T1 )(ψ(S1 ) − a0 ) ≥ (T2 − T1 )(Δf  + 1) ≥ I f (T1 , T2 , xf , uf ) − 2Δf  − 2a0 ≥ I f (T1 , T2 , xf , uf ) − S and (7.18) holds. Assume that inf{x(t) : t ∈ [T1 , T2 ]} < S1 .
(7.22)
τ1 = inf{t ∈ [T1 , T2 ] : x(t) ≤ S1 },
(7.23)
τ2 = sup{t ∈ [T1 , T2 ] : x(t) ≤ S1 }.
(7.24)
τ2 − τ1 ≥ c 0 ;
(7.25)
τ2 − τ1 < c 0 .
(7.26)
Set
There are two cases:
Assume that (7.25) holds. It follows from (7.23) to (7.25) and the choice of S2 and c0 that
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7 ContinuousTime Nonautonomous Problems on Axis
I f (τ1 , τ2 , xf , uf ) ≤ I f (τ1 , τ2 , x, u) + S2 .
(7.27)
By (7.8), (7.23), and (7.24), for each t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], x(t) > S1 , f (t, x(t), u(t)) ≥ −a0 + ψ(x(t)) ≥ −a0 + ψ(S1 ).
(7.28)
It follows from (7.13), (7.19), and (7.28) that I f (T1 , τ1 , x, u) ≥ (τ1 − T1 )(ψ(S1 ) − a0 ) ≥ (τ1 − T1 )(Δf  + 1) ≥ I f (T1 , τ1 , xf , uf ) − 2Δf  − 2a0 ,
(7.29)
I f (τ2 , T2 , x, u) ≥ (T2 − τ2 )(ψ(S1 ) − a0 ) ≥ (T2 − τ2 )(Δf  + 1) ≥ I f (τ2 , T2 , xf , uf ) − 2Δf  − 2a0.
(7.30)
In view of (7.27), (7.29), and (7.30), I f (T1 , T2 , x, u) = I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + I f (τ2 , T2 , x, u) ≥ I f (T1 , τ1 , xf , uf ) − 2Δf  − 2a0 + I f (τ1 , τ2 , xf , uf ) −S2 + I f (τ2 , T2 , xf , uf ) − 2Δf  − 2a0 ≥ I f (T1 , T2 , xf , uf ) − S and (7.18) holds. Assume that (7.26) holds. By (7.23) and (7.24), for each t ∈ [T1 , τ1 ] ∪ [τ2 , T2 ], (7.28)–(7.30) are true. It follows from (7.8), (7.13), (7.20), (7.29), and (7.30) that I f (T1 , T2 , x, u) = I f (T1 , τ1 , x, u) + I f (τ1 , τ2 , x, u) + I f (τ2 , T2 , x, u) ≥ I f (T1 , τ1 , xf , uf )−2Δf −2a0 −a0(τ2 −τ1 )+I f (τ2 , T2 , xf , uf )−2Δf −2a0 ≥ I f (T1 , T2 , xf , uf ) − a0 (c0 + 2) − Δf (c0 + 2) − 4a0 − 4Δf  ≥ I f (T1 , T2 , xf , uf ) − S. Lemma 7.18 is proved. Lemma 7.19. Let M0 , M1 , τ0 > 0. Then there exists M2 > M1 such that for each T1 ∈ R 1 , each T2 ∈ (T1 , T1 + τ0 ], and each (x, u) ∈ X(T1 , T2 ) satisfying
7.5 Auxiliary Results
491
inf{x(t) : t ∈ [T1 , T2 ]} ≤ M1 ,
(7.31)
I f (T1 , T2 , x, u) ≤ M0 ,
(7.32)
the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ].
(7.33)
Proof. Fix a positive number δ < 2−1 (a0 + 1)−1 . By (7.7) and (7.8), there exist h0 > M1 + 1 and γ0 > 0 such that f (t, x, u) ≥ 4(M0 + a0 τ0 )δ −1 for each (t, x, u) ∈ M satisfying x ≥ h0 , (7.34) f (t, x, u) ≥ 8(G(t, x, u) − a0 x)+ for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ γ0 .
(7.35)
Choose a number M2 > 2h0 + 2M1 + 2γ0 δ + M0 + a0 τ0 .
(7.36)
Let T1 ∈ R 1 , T2 ∈ (T1 , T1 + τ0 ], (x, u) ∈ X(T1 , T2 ), (7.31) and (7.32) hold. We show that (7.33) hold. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that x(t0 ) > M2 .
(7.37)
By (7.31) and (7.34), there exists t1 ∈ [T1 , T2 ] such that x(t1 ) ≤ h0 , t1 − t0  ≤ δ.
(7.38)
t2 ∈ [min{t1 , t0 }, max{t1 , t0 }]
(7.39)
There exists
such that x(t2 ) ≥ x(t) for all t ∈ [min{t1 , t0 }, max{t1 , t0 }].
(7.40)
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7 ContinuousTime Nonautonomous Problems on Axis
In view of (7.36)–(7.38) and (7.40) t2 = t1 .
(7.41)
By (7.4), x(t2 ) − x(t1 ) =
t2
G(x(s), u(s))ds.
(7.42)
G(s, x(s), u(s))ds.
(7.43)
t1
It follows from (7.38) and (7.42) that x(t2 ) − h0 ≤ 
t2
t1
Define E1 = {t ∈ [min{t1 , t0 }, max{t1 , t0 }] : G(t, x(t), u(t)) ≥ a0 x(t) + γ0 }, E2 = [min{t1 , t0 }, max{t1 , t0 }] \ E1 .
(7.44)
By (7.8), (7.32), (7.35), (7.38), (7.40), (7.43), and (7.44),
x(t2 ) − h0 ≤ a0
E1 ∪E2
x(t)dt +
E1 ∪E2
[G(s, x(s), u(s)) − a0 x(s)]+ ds
≤ a0 δx(t2 ) + E1
[G(s, x(s), u(s)) − a0 x(s)]+ ds
+ E2
[G(s, x(s), u(s)) − a0 x(s)]+ ds
≤ a0 x(t2 )δ + γ0 δ + 8−1
f (t, x(t), u(t))dt E1
≤ a0 x(t2 )δ + γ0 δ + 8−1 (M0 + a0 τ0 ).
(7.45)
It follows from (7.36) to (7.38), (7.40), (7.45), and the choice of δ that 2−1 M2 ≤ 2−1 x(t2 ) ≤ x(t2 (1 − a0 δ) ≤ h0 + γ0 δ + 8−1 (M0 + a0 τ0 ) and M2 < 2h0 + 2γ0 δ + 4−1 (M0 + a0 τ0 ).
7.5 Auxiliary Results
493
This contradicts (7.36). The contradiction we have reached proves Lemma 7.19. Lemma 7.19, (7.7) and (7.8) imply the following result. Lemma 7.20. Let M1 > 0, 0 < τ0 < τ1 . Then there exists M2 > 0 such that for each T1 ∈ R 1 , each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 , the following inequality holds: x(t) ≤ M2 for all t ∈ [T1 , T2 ]. Lemma 7.21. Let M1 > 0, ∈ (0, 1), 0 < τ0 < τ1 . Then there exists δ > 0 such that for each T1 ∈ R 1 , each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 and each t1 , t2 ∈ [T1 , T2 ] satisfying t1 − t2  ≤ δ, the inequality x(t1 ) − x(t2 ) ≤ holds. Proof. Lemma 7.20 implies that there exists M2 > 0 such that the following property holds: (i) for each T1 ≥ 0, each T2 ∈ [T1 + τ0 , T1 + τ1 ], and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ M1 , we have x(t) ≤ M2 for all t ∈ [T1 , T2 ]. In view of (7.7) and (7.8), there exists h0 > 0 such that f (t, x, u) ≥ 4 −1 (M1 + a0 τ1 + 8)(G(t, x, u) − a0 x)+ for each (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ h0 .
(7.46)
Choose a number δ ∈ (0, (4a0M2 + 4h0 + 4)−1 ). Let T1 ∈ R 1 , T2 ∈ [T1 + τ0 , T1 + τ1 ], (x, u) ∈ X(T1 , T2 ),
(7.47)
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7 ContinuousTime Nonautonomous Problems on Axis
I f (T1 , T2 , x, u) ≤ M1 ,
(7.48)
t1 , t2 ∈ [T1 , T2 ], 0 < t2 − t1 ≤ δ.
(7.49)
Property (i) and (7.48) imply that x(t) ≤ M2 for all t ∈ [T1 , T2 ].
(7.50)
Define Ω1 = {t ∈ [t1 , t2 ] : G(t, x(t), u(t)) − a0 x(t) ≥ h0 }, Ω2 = [t1 , t2 ] \ Ω1 .
(7.51) (7.52)
By (7.4), (7.8), and (7.47)–(7.52), x(t2 ) − x(t1 ) =
t2
G(s, x(s), u(s))ds t1
≤ a0
t2
x(t)dt +
t1
t2 t1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 δM2 + δh0 + E1
(G(t, x(t), u(t)) − a0 x(t))+ dt
≤ a0 δM2 + δh0 + (4(M1 + a0 τ1 + 8))
−1
f (t, x(t), u(t))dt E1
≤ a0 δM2 + δh0 + 4−1 < . Lemma 7.21 is proved. Lemma 7.22. Let Δ > 0. Then there exists δ > 0 such that for each (T1 , z1 ), (T2 , z2 ) ∈ A satisfying T2 ≥ T1 + 2bf , zi − xf (Ti ) ≤ δ, i = 1, 2 the following inequality holds: U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , xf , uf ) + Δ. Proof. Let δ > 0 be as guaranteed by (A2) with = Δ/4. Let (T1 , z1 ), (T2 , z2 ) ∈ A
(7.53)
7.5 Auxiliary Results
495
satisfy (7.53). By (7.53), the choice of δ and (A2) with = Δ/4, there exist (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = z1 , y(T2 ) = z2 , y(t) = xf (t), v(t) = uf (t), t ∈ [T1 + bf , T2 − bf ], I f (T1 , T1 + bf , y, v) ≤ I f (T1 , T1 + bf , xf , uf ) + Δ/4, I f (T2 − bf , T2 , y, v) ≤ I f (T2 − bf , T2 , xf , uf ) + Δ/4. By the relations above, U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , y, v) ≤ I f (T1 , T2 , xf , uf ) + Δ/2. Lemma 7.22 is proved. Lemma 7.23. Let Δ ∈ (0, 1). Then there exists δ > 0 such that for each T1 ∈ R 1 , each T2 ≥ T1 + 3bf , each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, each τ1 ∈ [T1 , T1 + δ] and each τ2 ∈ [T2 − δ, T2 ] the inequality I f (τ1 , τ2 , x, u) ≤ I f (τ1 , τ2 , xf , uf ) + Δ holds. Proof. Lemma 7.22 implies that there exists δ0 ∈ (0, min{4−1 Δ, 2−1 bf }) such that the following property holds: (ii) for each (T1 , z1 ), (T2 , z2 ) ∈ A satisfying T2 ≥ T1 + 2bf , zi − xf (Ti ) ≤ δ0 , i = 1, 2 we have U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , xf , uf ) + Δ/4. Lemmas 7.1 and 7.18, (7.8), and (7.13) imply that there exists M0 > 0 such that the following property holds:
496
7 ContinuousTime Nonautonomous Problems on Axis
(iii) for each pair of numbers T2 > T1 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + M0 ≥ I f (T1 , T2 , xf , uf ). Lemma 7.21 implies that there exists δ1 ∈ (0, δ0 ) such that the following property holds: (iv) for each S ∈ R 1 and each (y, v) ∈ X(S, S + bf ) satisfying I f (S, S + bf , y, v) ≤ Δf (bf + 2) + 3a0 + 2M0 + 1 and each t1 , t2 ∈ [S, S +bf ] satisfying t1 −t2  ≤ δ1 , we have y(t1 )−y(t2 ) ≤ δ0 /8. Set δ = δ1 /8.
(7.54)
T1 ∈ R 1 , T2 ≥ T1 + 3bf ,
(7.55)
x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2,
(7.56)
Assume that
(x, u) ∈ X(T1 , T2 ),
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, τ1 ∈ [T1 , T1 + δ], τ2 ∈ [T2 − δ, T2 ].
(7.57) (7.58)
Property (ii), (7.54), and (7.55) imply that I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + 1.
(7.59)
Property (iii) and (7.59) imply that for each t1 , t2 ∈ [T1 , T2 ] satisfying t1 < t2 , I f (t1 , t2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , t1 , x, u) − I f (t2 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + 1 − I f (T1 , t1 , xf , uf ) + M0 − I f (t2 , T2 , xf , uf ) + M0 = I f (t1 , t2 , xf , uf ) + 2M0 + 1. In view of (7.8), (7.13), and (7.55), I f (T1 , T1 + bf , xf , uf )
(7.60)
7.6 Proofs of Theorems 7.2–7.5
497
≤ I f (T1 , T1 + bf + 2, xf , uf ) + 3a0 ≤ (bf + 2)Δf + 3a0 ,
(7.61)
I f (T2 − bf , T2 , xf , uf ) ≤ I f (T2 − bf − 1, T2 + 1, xf , uf ) + 3a0 ≤ (bf + 2)Δf + 3a0 .
(7.62)
It follows from (7.60) and the relations above that I f (T1 , T1 +bf , xf , uf ), I f (T2 −bf , T2 , xf , uf ) ≤ Δf (bf +2)+3a0 +2M0 +1. (7.63) Property (iv), (7.54), (7.58), and (7.61)–(7.63) imply that xf (Ti ) − xf (τi ) ≤ δ0 /8, i = 1, 2,
(7.64)
x(Ti ) − x(τi ) ≤ δ0 /8, i = 1, 2,
(7.65)
By (7.54), (7.56), (7.64), and (7.65), for i = 1, 2, xf (τi ) − x(τi ) ≤ xf (τi ) − xf (Ti ) + xf (Ti ) − x(Ti ) + x(Ti ) − x(τi ) ≤ δ0 /8 + δ0 /8 + δ ≤ δ0 /2.
(7.66)
Property (ii), (7.55), and (7.66) imply that U f (τ1 , τ2 , x(τ1 ), x(τ2 )) ≤ I f (τ1 , τ2 , xf , uf ) + Δ/4.
(7.67)
By (7.54), (7.57), (7.58), and (7.67), I f (τ1 , τ2 , x, u) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ ≤ I f (τ1 , τ2 , xf , uf ) + Δ/4 + δ ≤ I f (τ1 , τ2 , xf , uf ) + Δ. Lemma 7.23 is proved.
7.6 Proofs of Theorems 7.2–7.5 Proof of Theorem 7.2. Assertion 1 of Theorem 7.2 follows from Lemma 7.18, (7.8), and (7.13). We will prove Assertion 2. Let (x, u) ∈ X(−∞, ∞). Assume that there exists a sequence of positive numbers {Tk }∞ k=1 such that Tk → ∞ as k → ∞,
(7.68)
498
7 ContinuousTime Nonautonomous Problems on Axis
I f (−Tk , Tk , x, u) − I f (−Tk , Tk , xf , uf ) → ∞ as k → ∞.
(7.69)
Let a number S > 0 be as guaranteed by Assertion 1. Let k be a natural number and T ≥ Tk . In view of (7.68), (7.69), Assertion 1, and the choice of S, I f (−T , T , x, u) − I f (−T , T , xf , uf ) = I f (−Tk , Tk , x, u) − I f (−Tk , Tk , xf , uf ) + I f (−T , −Tk , x, u) −I f (−T , −Tk , xf , uf ) +I f (Tk , T , x, u) − I f (Tk , T , xf , uf ) ≥ I f (−Tk , Tk , x, u) − I f (−Tk , Tk , xf , uf ) − 2S → ∞ as k → ∞. Thus Assertion 2 is proved. Assume that sup{I f (−T , T , x, u) − I f (−T , T , xf , uf ) : T ∈ [0, ∞)} < ∞. Therefore there exists an integer S1 > 4 + a0 such that for each T1 < T2 , I f (T1 , T2 , x, u) − I f (T1 , T2 , xf , uf ) ≤ S1 .
(7.70)
Let S0 > 0 be such that ψ(S0 ) > 4Δf  + 32.
(7.71)
We show that for each T ∈ R 1 , min{x(t) : t ∈ [T , T + S1 ]} ≤ S0 .
(7.72)
Assume the contrary. Then there exists T ∈ R 1 such that x(t) > S0 , t ∈ [T , T + S1 ].
(7.73)
By (7.8) and (7.73), I f (T , T + S1 , x, u) ≥ −2a0 + I f (T + 1, T + S1 , x, u) ≥ −2a0 + ψ(S0 )(S1 − 2) ≥ −2a0 + 2−1 ψ(S0 )S1 . In view of (7.7) and (7.13),
(7.74)
7.6 Proofs of Theorems 7.2–7.5
499
I f (T , T + S1 , xf , uf ) ≤ I f (T , T + S1 + 1, xf , uf ) + 2a0 ≤ Δf (S1 + 1) + 2a0 ≤ 2Δf S1 + 2a0 .
(7.75)
It follows from (7.71), (7.74), and (7.75) that I f (T , T + S1 , x, u) − I f (T , T + S1 , xf , uf ) ≥ 2−1 ψ(S0 )S1 − 4a0 − 2Δf S1 ≥ S1 (2−1 ψ(S0 ) − 2Δf ) − 4a0 ≥ −4a0 + 8S1 ≥ 4S1 . This contradicts (7.70). The contradiction we have reached proves that for each 1 T ∈ R 1 , (7.72) is true. By (7.72), there exists a sequence {Ti }∞ i=−∞ ⊂ R such that for each integer i ≥ 1, Ti+1 − Ti ∈ [1, 1 + S1 ], x(Ti ) ≤ S0 .
(7.76)
Lemma 7.19, (7.7), (7.13), (7.70), and (7.76) imply that sup{x(t) : t ∈ R 1 }. Theorem 7.2 is proved. Proof of Theorem 7.3. We may assume that c < 1/2. By Theorem 7.2, there exists S0 > 0 such that the following property holds: (i) for each T2 > T1 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ). Lemma 7.20 implies that there exists M1 > 0 such that the following property holds: (ii) for each T ∈ R 1 and each (x, u) ∈ X(T , T + c/2) satisfying I f (T , T + c/2, x, u) ≤ M0 + 2S0 + 2 + 2a0 + 2Δf  we have x(t) ≤ M1 for all t ∈ [T , T + c/2]. Assume that T1 ∈ R 1 , T2 ≥ T1 + c, (x, u) ∈ X(T1 , T2 ) and I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 .
(7.77)
500
7 ContinuousTime Nonautonomous Problems on Axis
We show that x(t) ≤ M1 , t ∈ [T1 , T2 ]. Assume the contrary. Then there exists t0 ∈ [T1 , T2 ] such that x(t0 ) > M1 .
(7.78)
Clearly, there exists a closed interval [a, b] ⊂ [T1 , T2 ] such that t0 ∈ [a, b], b − a = 2−1 c.
(7.79)
Property (ii), (7.78), and (7.79) imply that I f (a, b, x, u) > M0 + 2S0 + 2 + 2a0 + 2Δf .
(7.80)
Property (i), (7.7), (7.13), (7.77), and (7.79) imply that I f (a, b, x, u) = I f (T1 , T2 , x, u) − I f (T1 , a, x, u) − I f (b, T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 − I f (T1 , a, xf , uf ) + S0 − I f (b, T2 , xf , uf ) + S0 = I f (a, b, xf , uf ) + 2S0 + M0 ≤ 2Δf  + 2a0 + 2S0 + M0 . This contradicts (7.80). The contradiction we have reached proves Theorem 7.3. Proof of Theorem 7.4. Theorem 7.3 implies that there exists M1 > 0 such that the following property holds: (iii) for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + 2L(1 + a0 ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Assume that T1 ∈ R 1 , T2 ≥ T1 + 2L, and (x, u) ∈ X(T1 , T2 ) satisfy L , (T1 , x(T1 )) ∈ AL , (T2 , x(T2 )) ∈ A I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + M0 .
(7.81) (7.82)
7.6 Proofs of Theorems 7.2–7.5
501
In view of (7.81), there exist τ1 ∈ (0, L], τ2 ∈ (0, L], and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), y(T2 ) = x(T2 ),
(7.83)
(y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ1 + T1 , T2 − τ2 ],
(7.84)
I f (T1 , T1 + τ1 , y, v) ≤ L, I f (T2 − τ2 , T2 , y, v) ≤ L.
(7.85)
By (7.7) and (7.82)–(7.85), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v) ≤ M0 + I f (T1 , T1 + τ1 , y, v) + I f (T1 + τ1 , T2 − τ2 , y, v) + I f (T2 − τ2 , T2 , y, v) ≤ M0 + 2L + I f (T1 + τ1 , T2 − τ2 , xf , uf ) ≤ 2L + M0 + I f (T1 , T2 , xf , uf ) + 2La0 .
(7.86)
Property (iii), (7.85), and (7.86) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 7.4 is proved. Proof of Theorem 7.5. Theorem 7.3 implies that there exists M1 > 0 such that the following property holds: (iv) for each T1 ∈ R 1 , each T2 ≥ T1 + L, and each (x, u) ∈ X(T1 , T2 ) satisfying I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M0 + L(1 + a0 ), the inequality x(t) ≤ M1 holds for all t ∈ [T1 , T2 ]. Assume that T1 ∈ R 1 , T2 ≥ T1 + L and (x, u) ∈ X(T1 , T2 ) satisfies (T1 , x(T1 )) ∈ AL ,
(7.87)
I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 , x(T1 )) + M0 .
(7.88)
In view of (7.87), there exist τ ∈ (0, L] and (y, v) ∈ X(T1 , T2 ) such that y(T1 ) = x(T1 ), (y(t), v(t)) = (xf (t), uf (t)), t ∈ [τ + T1 , T2 ], I f (T1 , T1 + τ, y, v) ≤ L. By (7.7) and (7.88)–(7.90), I f (T1 , T2 , x, u) ≤ M0 + I f (T1 , T2 , y, v)
(7.89) (7.90)
502
7 ContinuousTime Nonautonomous Problems on Axis
= I f (T1 , T1 + τ, y, v) + I f (T1 + τ, T2 , xf , uf ) + M0 ≤ L + M0 + I f (T1 , T2 , xf , uf ) + La0 .
(7.91)
Property (iv) and (7.91) imply that x(t) ≤ M1 for all t ∈ [T1 , T2 ]. Theorem 7.5 is proved.
7.7 Auxiliary Results Lemma 7.24. Let > 0. Then there exist L > 0 such that the following properties hold: (i) for each T2 ≥ T1 ≥ L, I f (T1 , T2 , xf , uf ) ≤ U f (T1 , T2 , xf (T1 ), xf (T2 )) + ; (ii) for each T1 < T2 ≤ −L, I f (T1 , T2 , xf , uf ) ≤ U f (T1 , T2 , xf (T1 ), xf (T2 )) + . Proof. Assume the contrary. Then there exists a sequence of closed intervals {[ai , bi ]}∞ i=1 such that for each integer i ≥ 1 and each integer j > i, [ai , bi ] ∩ [aj , bj ] = ∅, I f (ai , bi , xf , uf ) > U f (ai , bi , xf (ai ), xf (bi )) + . By the relations above, there exists (y, v) ∈ X(−∞, ∞) such that y(t) = xf (t), v(t) = uf (t), t ∈ R 1 \ ∪∞ i=1 [ai , bi ],
(7.92)
for every integer i ≥ 1, I f (ai , bi , xf , uf ) > I f (ai , bi , y, v) + . Let q ≥ 1 be an integer. It follows from (7.92) and the relation above that for each T > 0 satisfying [ai , bi ] ⊂ [−T , T ], i = 1, . . . , q, we have I f (−T , T , xf , uf ) − I f (−T , T , y, v)
7.7 Auxiliary Results
≥
503
q (I f (ai , bi , xf , uf ) − I f (ai , bi , y, v)) ≥ q, i=1
I f (−T , T , y, v) ≤ I f (−T , T , xf , uf ) − q. Since q is any natural number, this implies that I f (−T , T , y, v) − I f (−T , T , xf , uf ) → −∞ as k → ∞. This contradicts Theorem 7.2. The contradiction we have reached proves Lemma 7.24. Lemma 7.25. Let (P1) hold and > 0. Then there exist δ, T > 0 such that the following properties hold: (iii) for each T1 ≥ T , each T2 > T1 , and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, the inequality I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − holds; (iv) for each T1 < T2 ≤ −T and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, the inequality I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − holds. Proof. By (A2), there exists δ ∈ (0, /4) such that the following property holds: (v) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 , τ2 ∈ (0, bf ], (x˜1 , u˜ 1 ) ∈ X(T , T + τ1 ), (x˜2 , u˜ 2 ) ∈ X(T − τ2 , T ) such that x˜1 (T ) = z , x˜1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x˜1 , u˜ 1 ) ≤ I f (T , T + τ1 , xf , uf ) + /4, x˜2 (T − τ2 ) = xf (T − τ2 ) , x˜2 (T ) = z, I f (T − τ2 , T , x˜2 , u˜ 2 ) ≤ I f (T − τ2 , T , xf , uf ) + /4.
504
7 ContinuousTime Nonautonomous Problems on Axis
Lemma 7.24 implies that there exists L > bf such that the following property holds: (vi) for each T2 > T1 ≥ L, I f (T1 , T2 , xf , uf ) < U f (T1 , T2 , xf (T1 ), xf (T2 )) + /4 and for each T1 < T2 ≤ −L, I f (T1 , T2 , xf , uf ) ≤ U f (T1 , T2 , xf (T1 ), xf (T2 )) + /4. Set T = L + b f . Let T2 > T1 , either T1 > L + bf or T2 < −L − bf ,
(7.93)
(y, v) ∈ X(T1 , T2 ) and y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2.
(7.94)
We show that I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , T , xf , uf ) − Assume the contrary. Then I f (T1 , T2 , y, v) < I f (T1 , T2 , xf , uf ) − .
(7.95)
By (7.94) and property (v), there exist τ1 , τ2 ∈ (0, bf ] and (x, u) ∈ X(T1 − τ1 , T2 + τ2 ) such that (x(t), u(t)) = (y(t), v(t)) for all t ∈ [T1 , T2 ],
(7.96)
x(T1 − τ1 ) = xf (T1 − τ1 ), x(T2 + τ2 ) = xf (T2 + τ2 ),
(7.97)
I f (T1 − τ1 , T1 , x, u) ≤ I f (T1 − τ1 , T1 , xf , uf ) + /4,
(7.98)
I f (T2 , T2 + τ2 , x, u) ≤ I f (T2 , T2 + τ2 , xf , uf ) + /4.
(7.99)
It follows from (7.95), (7.96), (7.98), and (7.99) that
7.8 STP Implies (P1), (P2), and (P3)
505
I f (T1 − τ1 , T2 + τ2 , x, u) = I f (T1 − τ1 , T1 , x, u) + I f (T1 , T2 , y, v) + I f (T2 , T2 + τ2 , x, u) ≤ I f (T1 , T2 , xf , uf ) − + I f (T1 − τ1 , T1 , xf , uf ) + /4 +I f (T2 , T2 + τ2 , xf , uf ) + /4 = I f (T1 − τ1 , T2 + τ2 , xf , uf ) − /2. Combined with (7.93) and (7.97), this contradicts (vi). The contradiction we have reached completes the proof of Lemma 7.25.
7.8 STP Implies (P1), (P2), and (P3) Assume that STP and LSC property hold. Theorem 7.2 and STP imply (P2). We show that (P1) holds. Assume that (x, u) ∈ X(−∞, ∞) is (f )good. Theorem 7.2 implies that there exists M0 > 0 such that x(t) ≤ M0 , t ∈ R 1 ,
(7.100)
for each T > 0, I f (−T , T , x, u) − I f (−T , T , xf , uf ) ≤ M0 .
(7.101)
Theorem 7.2 implies that there exists M1 > 0 such that the following property holds: (i) for each T1 < T2 and each (y, v) ∈ X(T1 , T2 ), M1 + I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ). Let S1 < S2 . Choose a positive number T such that − T < S1 < S2 < T .
(7.102)
By (7.101) and (7.102), I f (S1 , S2 , x, u) = I f (−T , T , x, u) − I f (−T , S1 , x, u) − I f (S2 , T , x, u) ≤ I f (−T , T , xf , uf ) + M0 − I f (T , S1 , xf , uf ) + M1 − I f (S2 , T , xf , uf ) + M1 = I f (S1 , S2 , xf , uf ) + M0 + 2M1 .
506
7 ContinuousTime Nonautonomous Problems on Axis
Thus for each pair of numbers S2 > S1 , I f (S1 , S2 , x, u) ≤ I f (S1 , S2 , xf , uf ) + M0 + 2M1 .
(7.103)
Let γ > 0. We show that there exists Tγ > 0 such that the following properties hold: (ii) for each S1 < S2 ≤ −Tγ , I f (S1 , S2 , x, u) ≤ U f (S1 , S2 , x(S1 ), x(S2 )) + γ ; (iii) for each S2 > S1 ≥ Tγ , I f (S1 , S2 , x, u) ≤ U f (S1 , S2 , x(S1 ), x(S2 )) + γ . Assume the contrary. Then there exists a sequence of closed intervals {[ai , bi ]}∞ i=1 such that for each integer i ≥ 1 and each integer j > i, [ai , bi ] ∩ [aj , bj ] = ∅, I f (ai , bi , x, u) > U f (ai , bi , x(ai ), x(bi )) + γ . By the relations above, there exists (y, v) ∈ X(−∞, ∞) such that y(t) = x(t), v(t) = u(t), t ∈ R 1 \ ∪∞ i=1 [ai , bi ],
(7.104)
for every integer i ≥ 1, I f (ai , bi , y, v) < I f (ai , bi , x, u) − γ .
(7.105)
Let q ≥ 1 be an integer and T > 0 satisfy [ai , bi ] ⊂ [−T , T ], i = 1, . . . , q.
(7.106)
By (7.103)–(7.106), we have I f (−T , T , x, u) − I f (−T , T , y, v) ≥
q (I f (ai , bi , x, u) − I f (ai , bi , y, v)) ≥ qγ , i=1
I f (−T , T , y, v) ≤ I f (−T , T , x, u)−qγ ≤ I f (−T , T , xf , uf )+M0 +2M1 −γ q. Thus for every T > 0 satisfying (7.106),
7.8 STP Implies (P1), (P2), and (P3)
507
I f (−T , T , y, v) − I f (−T , T , xf , uf ) ≤ M0 + 2M1 − γ q. Since q is any natural number, this implies that I f (−T , T , y, v) − I f (−T , T , xf , uf ) → −∞ as k → ∞. This contradicts Theorem 7.2. The contradiction we have reached proves that for each γ > 0, there exists Tγ > 0 such that properties (ii) and (iii) hold. Let > 0. By STP, there exist δ > 0, L > 0 such that the following property holds: (iv) for each T1 ∈ R 1 , each T2 ≥ T1 + 2L, and each (y, v) ∈ X(T1 , T2 ) which satisfies I f (T1 , T2 , y, v) ≤ min{σ f (T1 , T2 )+2M0 +2M1 , U f (T1 , T2 , y(T1 ), y(T2 ))+δ}, we have y(t) − xf (t) ≤ , t ∈ [T1 + L, T2 − L]. Let Tδ > 0 be such that (ii), (iii) hold with γ = δ, and let k ≥ 1 be an integer. Properties (i)–(iii), (7.103), and the choice of Tδ imply that I f (Tδ , Tδ + 2L + k, x, u) ≤ U f (Tδ , Tδ + 2L + k, x(Tδ ), x(Tδ + 2L + k)) + δ, I f (Tδ , Tδ + 2L + k, x, u) ≤ I f (Tδ , Tδ + 2L + k, xf , uf ) + M0 + 2M1 ≤ σ f (Tδ , Tδ + 2L + k) + 2M0 + 2M1 and I f (−Tδ − 2L − k, −Tδ , x, u) ≤ U f (−Tδ − 2L − k, −Tδ , x(−Tδ − 2L − k), x(−Tδ )) + δ, I f (−Tδ − 2L − k, −Tδ , x, u) ≤ I f (−Tδ − 2L − k, −Tδ , xf , uf ) + M0 + 2M1 ≤ σ f (−Tδ − 2L − k, −Tδ ) + 2M0 + 2M1 Property (iv) and the relations above imply that x(t) − xf (t) ≤ , t ∈ [Tδ + L, Tδ + L + k] ∪ [−Tδ − L − k, −Tδ − L].
508
7 ContinuousTime Nonautonomous Problems on Axis
Since k is any natural number, we conclude that x(t) − xf (t) ≤ , t ∈ [Tδ + L, ∞) ∪ (−∞, −Tδ − L]. Since is any positive number, (P1) holds. We show that (P3) holds. In view of LSC property, for each integer k ≥ 1, there exists (xk , uk ) ∈ X(−k, k) such that xk (−k) = xf (−k), xk (k) = xf (k), I f (−k, k, xk , uk ) = U f (−k, k, xf (−k), xf (k)).
(7.107) (7.108)
It follows from (A2) that for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: (v) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δk , i = 1, 2 there exist τ1 , τ2 ∈ (0, bf ], (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) such that x1 (T1 ) = z1 , x1 (T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + k −1 , x2 (T2 ) = z2 , x2 (T2 − τ2 ) = xf (T2 − τ2 ), I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + k −1 . Theorem 7.2 implies that there exists M0 > 0 such that for each pair of numbers S2 > S1 and each (y, v) ∈ X(S1 , S2 ), I f (S1 , S2 , y, v) ≥ I f (S1 , S2 , xf , uf ) − M0 .
(7.109)
By (7.108) and (7.109), for each pair of numbers S2 > S1 and each natural number k > max{S2 , −S1 }, I f (S1 , S2 , xk , uk ) = I f (−k, k, xk , uk ) − I f (−k, S1 , xk , uk ) − I f (S2 , k, xk , uk ) ≤ I f (−k, k, xf , uf ) − I f (−k, S1 , xf , uf ) + M0 − I f (S2 , k, xf , uf ) + M0 = I f (S1 , S2 , xf , uf ) + 2M0 .
(7.110)
7.8 STP Implies (P1), (P2), and (P3)
509
It follows from STP, (7.108), and (7.109) that the following property holds: (vi) for each > 0, there exists L > 0 such that for each integer k > L, xk (t) − xf (t) ≤ , t ∈ [−k + L, k − L]. Fix an integer i. In view of (7.110), for each integer k > i + 1, I f (i, i + 1, xk , uk ) ≤ I f (i, i + 1, xf , uf ) + 2M0 .
(7.111)
By LSC property, extracting subsequences and reindexing, we obtain that there exists a subsequence {(xkp , ukp )}∞ p=1 and that for each integer i ≥ 1, there exists (x˜i , u˜ i ) ∈ X(i, i + 1) such that I f (i, i + 1, x˜i , u˜ i ) ≤ lim inf I f (i, i + 1, xkp , ukp ),
(7.112)
xkp (t) → x˜i (t) as p → ∞ for all t ∈ [i, i + 1].
(7.113)
p→∞
It follows from (7.113) and (vi) that for all integers i, x˜i (t) = xf (t) for all t ∈ [i, i + 1].
(7.114)
For all integers i and all t ∈ [i, i + 1] set u˜ f (t) = u˜ i (t).
(7.115)
Clearly, (xf , u˜ f ) ∈ X(−∞, ∞). Let S be a natural number. By (7.110) and (7.112), I f (−S, S, xf , u˜ f ) ≤ lim inf I f (−S, S, xkp , ukp ) ≤ I f (−S, S, xf , uf ) + 2M0 . p→∞
(7.116) Theorem 7.2 and (7.116) imply that (xf , u˜ f ) is (f )good. We show that (xf , u˜ f ) is (f )minimal. Assume the contrary. Then there exist γ > 0 and an integer S0 ≥ 1 such that I f (−S0 , S0 , xf , u˜ f ) > U f (−S0 , S0 , xf (−S0 ), xf (S0 )) + γ . In view of the relation above, there exists (y∗ , v∗ ) ∈ X(−S0 , S0 ) such that I f (−S0 , S0 , xf , u˜ f ) > I f (−S0 , S0 , y∗ , v∗ ) + γ ,
(7.117)
y∗ (−S0 ) = xf (−S0 ), y∗ (S0 ) = xf (S0 ).
(7.118)
Lemma 7.25 implies that there exist δ∗ > 0 and T∗ > 0 such that the following property holds:
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7 ContinuousTime Nonautonomous Problems on Axis
(vii) for each T1 ≥ T∗ , each T2 > T1 , and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ∗ , i = 1, 2, the inequality I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − γ /16 holds and for each T1 < T2 ≤ −T∗ and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, the inequality I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − γ /16 holds. Choose k0 > 8γ −1 + S0 + 8.
(7.119)
It follows from (7.112) to (7.114) that there exists a natural number q such that kq > k0 + T∗ + bf + 4,
(7.120)
for each t ∈ [−k0 − T∗ − bf − 4, k0 + T∗ + bf + 4], xkq (t) − xf (t) ≤ min{δk0 , δ∗ },
(7.121)
I f (−k0 − T∗ , k0 + T∗ , xf , u˜ f ) ≤ I f (−k0 − T∗ , k0 + T∗ , xk,q , uk,q ) + γ /16. (7.122) For all t ∈ [−k0 − T∗ , k0 + T∗ ] \ [−S0 , S0 ] set y∗ (t) = xf (t), v∗ (t) = u˜ f (t).
(7.123)
In view of (7.118), (7.119), and (7.123), (y∗ , v∗ ) ∈ X(−k0 − T∗ , k0 + T∗ ), y∗ (−k0 − T∗ ) = xf (−k0 − T∗ ), y∗ (k0 + T∗ ) = xf (k0 + T∗ ),
(7.124)
I f (−k0 − T∗ , k0 + T∗ , xf , u˜ f ) > I f (−k0 − T∗ , k0 + T∗ , y∗ , v∗ ) + γ .
(7.125)
Property (v) and (7.121) imply that there exist τ1 , τ2 ∈ (0, bf ],
7.8 STP Implies (P1), (P2), and (P3)
511
(y1 , v1 ) ∈ X(k0 + T∗ + bf − τ1 , k0 + T∗ + bf ), (y2 , v2 ) ∈ X(−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 ) such that y1 (k0 +T∗ +bf −τ1 ) = xf (k0 +T∗ +bf −τ1 ), y1 (k0 +T∗ +bf ) = xkq (k0 +T∗ +bf ), (7.126) I f (k0 + T∗ + bf − τ1 , k0 + T∗ + bf , y1 , v1 ) ≤ I f (k0 + T∗ + bf − τ1 , k0 + T∗ + bf , xf , uf ) + k0−1 ,
(7.127)
y2 (−k0 − T∗ − bf ) = xkq (−k0 − T∗ − bf ), y2 (−k0 − T∗ − bf + τ2 ) = xf (−kq − T∗ − bf + τ2 ),
(7.128)
I f (−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 , y2 , v2 ) ≤ I f (−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 , xf , uf ) + k0−1 .
(7.129)
v (t) = v2 (t), t ∈ [−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 ), y (t) = y2 (t),
(7.130)
y (t) = xf (t), v (t) = uf (t), t ∈ [−k0 − T∗ − bf + τ2 , −k0 − T∗ ),
(7.131)
v (t) = v∗ (t), t ∈ [−k0 − T∗ , k0 + T∗ ], y (t) = y∗ (t),
(7.132)
Define
y (t) = xf (t), v (t) = uf (t), t ∈ (k0 + T∗ , k0 + T∗ + bf − τ1 ], y (t) = y1 (t), v (t) = v1 (t), t ∈ (k0 + T∗ + bf − τ1 , k0 + T∗ + bf ].
(7.133)
By (7.124), (7.126), (7.128), and (7.130)–(7.133), ( y , v ) ∈ X(−k0 − T∗ − bf , k0 + T∗ + bf ), y (k0 + T∗ + bf ) = xkq (k0 + T∗ + bf ). y (−k0 − T∗ − bf ) = xkq (−k0 − T∗ − bf ), (7.134) It follows from (7.125), (7.127), and (7.129)–(7.133) that I f (−k0 −T∗ −bf , k0 +T∗ +bf , y , v ) = I f (−k0 −T∗ −bf , −k0 −T∗ −bf +τ2 , y , v) +I f (−k0 − T∗ − bf + τ2 , −k0 − T∗ , y , v ) + I f (−k0 − T∗ , k0 + T∗ , y∗ , v∗ )
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7 ContinuousTime Nonautonomous Problems on Axis
+I f (k0 +T∗ , k0 +T∗ +bf −τ1 , xf , uf )+I f (k0 +T∗ +bf −τ1 , k0 +T∗ +bf , y1 , v1 ) ≤ I f (−k0 − T∗ − bf , −k0 − T∗ − bf + τ2 , xf , uf ) + k0−1 +I f (−k0 − T∗ − bf + τ2 , −k0 − T∗ , xf , uf ) +I f (−k0 − T∗ , k0 + T∗ , y∗ , v∗ ) + I f (k0 + T∗ , k0 + T∗ + bf − τ1 , xf , uf ) +I f (k0 + T∗ + bf − τ1 , k0 + T∗ + bf , xf , uf ) + k0−1 ≤ I f (−k0 − T∗ − bf , −k0 − T∗ , xf , uf ) + k0−1 + I f (−k0 − T∗ , k0 + T∗ , y∗ , v∗ ) +I f (k0 + T∗ , k0 + T∗ + bf , xf , uf ) + k0−1 ≤ 2k0−1 +I f (−k0 −T∗ −bf , −k0 −T∗ , xf , uf )+I f (k0 +T∗ , k0 +T∗ +bf , xf , uf ) + I f (−k0 − T∗ , k0 + T∗ , xf , u˜ f ) − γ .
(7.135)
By (7.122) and (7.135), y , v) I f (−k0 − T∗ − bf , k0 + T∗ + bf , ≤ 2k0−1 +I f (−k0 −T∗ −bf , −k0 −T∗ , xf , uf )+I f (k0 +T∗ , k0 +T∗ +bf , xf , uf ) +I f (−k0 − T∗ , k0 + T∗ , xf , u˜ f ) − γ ≤ 2k0−1 +I f (−k0 −T∗ −bf , −k0 −T∗ , xf , uf )+I f (k0 +T∗ , k0 +T∗ +bf , xf , uf ) − γ + I f (−k0 − T∗ , k0 + T∗ , xkq , ukq ) + γ /16.
(7.136)
Property (vii) and (7.121) imply that I f (−k0 −T∗ −bf , −k0 −T∗ , xkq , ukq ) ≥ I f (−k0 −T∗ −bf , −k0 −T∗ , xf , uf )−γ /16, (7.137) I f (k0 + T∗ , k0 + T∗ + bf , xkq , ukq ) ≥ I f (k0 + T∗ , k0 + T∗ + bf , xf , uf ) − γ /16. (7.138) In view of (7.119) and (7.136)–(7.138), I f (−k0 − T∗ − bf , k0 + T∗ + bf , y , v ) − I f (−k0 − T∗ − bf , k0 + T∗ + bf , xkq , ukq ) ≤ 2k0−1 − γ + γ /16 + γ /8 ≤ 2k0−1 − γ /2 < −γ /4.
7.9 An Auxiliary Result
513
Together with (7.134), this contradicts (7.108). The contradiction we have reached proves that (xf , u˜ f ) is (f )minimal. Assume that (x, u) ∈ X(−∞, ∞) is (f )good and (f )minimal. Then sup{I f (−T , T , x, u) − I f (−T , T , xf , uf ) : T ∈ (0, ∞)} < ∞. Property (P1) implies that lim x(t) − xf (t) = 0,
t →∞
lim x(t) − xf (t) = 0.
t →−∞
Combined with STP this implies that x(t) = xf (t) for all t ∈ R 1 . Thus (P3) holds.
7.9 An Auxiliary Result Lemma 7.26. Assume that (P1) holds and that > 0. Then there exist δ, L > 0 such that the following properties hold: (i) for each T1 ≥ L, each T2 ≥ T1 + 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ,
(7.139) (7.140)
the inequality x(t) − xf (t) ≤ , t ∈ [T1 , T2 ]
(7.141)
holds; (ii) for each T2 ≤ −L, each T1 ≤ T2 − 2bf , and each (x, u) ∈ X(T1 , T2 ) which satisfies (7.139) and (7.140), the inequality (7.141) holds. Proof. By (A2), for each integer k ≥ 1, there exists δk ∈ (0, 2−k ) such that the following property holds: (iii) for each (Ti , zi ) ∈ A, i = 1, 2 satisfying zi − xf (Ti ) ≤ δk , i = 1, 2 there exist τ1 , τ2 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T1 , T1 + τ1 ), (x2 , u2 ) ∈ X(T2 − τ2 , T2 ) which satisfy x1 (T1 ) = z1 , x1 (T1 +τ1 ) = xf (T1 +τ1 ), x2 (T2 ) = z2 , x2 (T2 −τ2 ) = xf (T2 −τ2 ), I f (T1 , T1 + τ1 , x1 , u1 ) ≤ I f (T1 , T1 + τ1 , xf , uf ) + 2−k , I f (T2 − τ2 , T2 , x2 , u2 ) ≤ I f (T2 − τ2 , T2 , xf , uf ) + 2−k .
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7 ContinuousTime Nonautonomous Problems on Axis
Assume that the lemma does not hold. Then for each integer k ≥ 1 there exist Tk,1 ∈ R 1 , Tk,2 ≥ Tk,1 + 2bf , (xk , uk ) ∈ X(Tk,1 , Tk,2 ) such that xk (Tk,i ) − xf (Tk,i ) ≤ δk , i = 1, 2,
(7.142)
I f (Tk,1 , Tk,2 , xk , uk ) ≤ U f (Tk,1 , Tk,2 , xk (Tk,1 ), xk (Tk,2 )) + δk ,
(7.143)
sup{xk (t) − xf (t) : t ∈ [Tk,1 , Tk,2 ]} > ,
(7.144)
either for all integers k ≥ 1, Tk,1 ≥ k + bf or for all integers k ≥ 1, Tk,2 ≤ −k.
(7.145)
Extracting a subsequence and reindexing, we may assume without loss of generality that for each integer k ≥ 1, min{Tk+1,2, Tk+1,1 } ≥ max{Tk,1 , Tk,2 } + 4bf .
(7.146)
Let k ≥ 1 be an integer. Property (iii) and (7.142) imply that there exist τk,1 , τk,2 ∈ (0, bf ] and (x˜k , u˜ k ) ∈ X(Tk,1 − τk,1 , Tk,2 + τk,2 ) such that (x˜k (t), u˜ k (t)) = (xk (t), uk (t)), t ∈ [Tk,1, Tk,2 ],
(7.147)
x˜k (s) = xf (s), s = Tk,1 − τk,1 , Tk,2 + τk,2 ,
(7.148)
I f (Tk,1 − τk,1 , Tk,1 , x˜k , u˜ k ) ≤ I f (Tk,1 − τk,1 , Tk,1 , xf , uf ) + 2−k ,
(7.149)
I f (Tk,2 , Tk,2 + τk,2 , x˜k , u˜ k ) ≤ I f (Tk,2 , Tk,2 + τk,2 , xf , uf ) + 2−k .
(7.150)
Property (iii) and (7.142) imply that there exist τk,3 , τk,4 ∈ (0, bf ] and ( xk , uk ) ∈ X(Tk,1 , Tk,2 ) such that xk (Tk,2 ) = xk (Tk,2 ), xk (Tk,1 ) = xk (Tk,1 ),
(7.151)
( xk (t), uk (t)) = (xf (t), uf (t)), t ∈ [Tk,1 + τk,3 , Tk,2 − τk,4 ],
(7.152)
I f (Tk,1 , Tk,1 + τk,3 , xk , uk ) ≤ I f (Tk,1 , Tk,1 + τk,3 , xf , uf ) + 2−k ,
(7.153)
I f (Tk,2 − τk,4 , Tk,2 , xk , uk ) ≤ I f (Tk,2 − τk,4 , Tk,2 , xf , uf ) + 2−k .
(7.154)
7.9 An Auxiliary Result
515
In view of (7.151)–(7.153), U f (Tk,1 , Tk,2 , xk (Tk,1 ), xk (Tk,2 )) xk , uk ) ≤ I f (Tk,1 , Tk,2 , xf , uf ) + 2−k+1 . ≤ I f (Tk,1 , Tk,2 ,
(7.155)
By (7.143) and (7.155), I f (Tk,1 , Tk,2 , xk , uk ) ≤ I f (Tk,1 , Tk,2 , xf , uf ) + 2−k+2 .
(7.156)
It follows from (7.147), (7.149), (7.150), and (7.156) that I f (Tk,1 − τk,1 , Tk,2 + τk,2 , x˜k , u˜ k ) ≤ I f (Tk,1 − τk,1 , Tk,1 , xf , uf ) + I f (Tk,1 , Tk,2 , xk , uk ) +I f (Tk,2 , Tk,2 + τk,2 , xf , uf ) + 2−k+1 ≤ I f (Tk,1 − τk,1 , Tk,2 + τk,2 , xf , uf ) + 2−k+3 .
(7.157)
By (7.146) and (7.148), there exists (x, u) ∈ X(−∞, ∞) such that for every integer k ≥ 1, (x(t), u(t)) = (x˜k (t), u˜ k (t)), t ∈ [Tk,1 − τk,1 , Tk,2 + τk,2 ],
(7.158)
(x(t), u(t)) = (xf (t), uf (t)), t ∈ R 1 \ ∪∞ k=1 [Tk,1 − τk,1 , Tk,2 + τk,2 ].
(7.159)
It follows from (7.145), (7.146), and (7.157)–(7.159) that for each integer k ≥ 1, if Ti,1 → ∞ as i → ∞, then I f (0, Tk,2 + τk,2 , x, u) − I f (0, Tk,2 + τk,2 , xf , uf ) =
k
(I f (Ti,1 − τi,1 , Ti,2 + τi,2 , x˜i , u˜ i ) − I f (Ti,1 − τi,1 , Ti,2 + τi,2 , xf , uf ))
i=1
≤
k i=1
and if Ti,2 → −∞ as i → ∞, then
2−i+3 ≤ 8
516
7 ContinuousTime Nonautonomous Problems on Axis
I f (Tk,1 − τk,1 , 0, x, u) − I f (Tk,1 − τk,1 ), 0, xf , uf ) =
k
(I f (Ti,1 − τi,1 , Ti,2 + τi,2 , x˜i , u˜ i ) − I f (Ti,1 − τi,1 , Ti,2 + τi,2 , xf , uf ))
i=1
≤
k
2−i+3 ≤ 8.
i=1
Theorem 7.2, the relations above, (7.145), (7.158), and (7.159) imply that (x, u) is (f )good. Together with (P1) this implies that limt →∞ x(t) − xf (t) = 0 and limt →−∞ x(t) − xf (t) = 0. On the other hand, in view of (7.144), (7.145), (7.147), and (7.158), at least one of the relation below holds: lim sup x(t) − xf (t) > 0, lim sup x(t) − xf (t) > 0. t →∞
t →−∞
The contradiction we have reached completes the proof of Lemma 7.26.
7.10 The Main Lemma Lemma 7.27. Let LSC properties, (P1), (P2), and (P3), hold, u˜ f = u, ∈ (0, 1). Then there exists δ > 0 such that for each pair of numbers T1 ∈ R 1 , T2 ≥ T1 + 3bf and each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2 I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, the inequality x(t) − xf (t) ≤ for all t ∈ [T1 , T2 ]. Proof. Lemma 7.26 and (A2) imply that there exist δ0 ∈ (0, min{1, /4}) and L0 > 0 such that the following properties hold: (i) (A2) holds with = 1 and δ = δ0 ; (ii) for each pair of numbers T1 ∈ R 1 , T2 ≥ T1 + 2bf such that either T1 ≥ L0 or T2 ≤ −L0 and each (x, u) ∈ X(T1 , T2 ) satisfying x(Ti ) − xf (Ti ) ≤ δ0 , i = 1, 2 I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ0
7.10 The Main Lemma
517
we have x(t) − xf (t) ≤ , t ∈ [T1 , T2 ]. Theorem 7.2 implies that there exists S0 > 0 such that for each pair of numbers T2 > T1 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(7.160)
It follows from property (P2) that there exist L1 > 0 and δ1 ∈ (0, δ0 ) such that the following property holds: (iii) for each T ∈ R 1 and each (x, u) ∈ X(T , T + L1 ) satisfying I f (T , T + L1 , x, u) ≤ min{I f (T , T +L1 , xf , uf )+2S0 +3, U f (T , T +L1 , x(T ), x(T +L1 ))+δ1 } we have min{x(t) − xf (t) : t ∈ [T , T + L1 ]} ≤ δ0 . (A2) implies that there exists a sequence {δi }∞ i=1 ⊂ (0, 1) such that δi < 2−1 δi−1 , i = 2, 3, . . .
(7.161)
and that the following property holds: (iv) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δi , there exist (y1 , v1 ) ∈ X(T , T + bf ), (y2 , v2 ) ∈ X(T − bf , T ) which satisfy y1 (T ) = z, y1 (T + bf ) = xf (T + bf ), I f (T , T + bf , y1 , v1 ) ≤ I f (T , T + bf , xf , uf ) + i −1 , y2 (T − bf ) = xf (T − bf ), y2 (T ) = z, I f (T − bf , T , y2 , v2 ) ≤ I f (T − bf , T , xf , uf ) + i −1 . Assume that the lemma does not hold. Then for each integer i ≥ 1, there exist Ti,1 ∈ R 1 , Ti,2 ≥ Ti,1 + 3bf and (xi , ui ) ∈ X(Ti,1 , Ti,2 ) such that
518
7 ContinuousTime Nonautonomous Problems on Axis
xi (Ti,1 ) − xf (Ti,1 ) ≤ δi , xi (Ti,2 ) − xf (Ti,2 ) ≤ δi ,
(7.162)
I f (Ti,1 , Ti,2 , xi , ui ) ≤ U f (Ti,1 , Ti,2 , xi (Ti,1 ), xi (Ti,2 )) + δi ,
(7.163)
and ti ∈ [Ti,1 , Ti,2 ] for which xi (ti ) − xf (ti ) > .
(7.164)
Let i be a natural number. Property (ii) and (7.161)–(7.164) imply that Ti,1 < L0 , Ti,2 > −L0 .
(7.165)
We show that −2bf − 1 − L1 − L0 ≤ ti ≤ 2bf + 1 + L1 + L0 . Property (i), (A2) with = 1, δ = δ0 , (7.161), and (7.163) imply that there exists (yi , vi ) ∈ X(Ti,1 , Ti,2 ) such that yi (Ti,1 ) = xi (Ti,1 ), yi (Ti,2 ) = xi (Ti,2 ),
(7.166)
yi (t) = xf (t), vi (t) = uf (t), t ∈ [Ti,1 + bf , Ti,2 − bf ],
(7.167)
I f (Ti,1 , Ti,1 + bf , yi , vi ) ≤ I f (Ti,1 , Ti,1 + bf , xf , uf ) + 1,
(7.168)
I f (Ti,2 − bf , Ti,2 , yi , vi ) ≤ I f (Ti,2 − bf , Ti,2 , xf , uf ) + 1.
(7.169)
It follows from (7.161), (7.163), and (7.166)–(7.169) that I f (Ti,1 , Ti,2 , xi , ui ) ≤ 1 + I f (Ti,1 , Ti,2 , yi , vi ) ≤ I f (Ti,1 , Ti,2 , xf , uf ) + 3. (7.170) In view of (7.160) and (7.170), for each pair of numbers S1 , S2 ∈ [Ti,1 , Ti,2 ] satisfying S1 < S2 , I f (S1 , S2 , xi , ui ) = I f (Ti,1 , Ti,2 , xi , ui )−I f (Ti,1 , S1 , xi , ui )−I f (S2 , Ti,2 , xi , ui ) ≤ I f (Ti,1 , Ti,2 , xf , uf ) + 3 − I f (Ti,1 , S1 , xf , uf ) + S0 − I f (S2 , Ti,2 , xf , uf ) + S0 = I f (S1 , S2 , xf , uf ) + 3 + 2S0 .
(7.171)
ti > 2bf + 1 + L1 + L0 .
(7.172)
Assume that
7.10 The Main Lemma
519
Consider the restriction of (xi , ui ) to the interval [ti − 2bf − 1 − L1 , ti − 1 − 2bf ] ⊂ [L0 , ∞).
(7.173)
Property (iii), (7.161), (7.163), (7.165), (7.171), and (7.173) imply that there exists t˜ ∈ [ti − 2bf − 1 − L1 , ti − 1 − 2bf ]
(7.174)
xf (t˜) − xi (t˜) ≤ δ0 .
(7.175)
such that
Property (ii), (7.157), (7.162), (7.163), and (7.173)–(7.175) imply that xi (t) − xf (t) ≤ , t ∈ [t˜, Ti,2 ] and in particular, xi (ti ) − xf (ti ) ≤ . This contradicts (7.164). The contradiction we have reached proves that ti ≤ 2bf + 1 + L1 + L0 .
(7.176)
ti < −2bf − 1 − L1 − L0 .
(7.177)
Assume that
Consider the restriction of (xi , ui ) to the interval [ti + 2bf + 1, ti + 1 + 2bf + L1 ] ⊂ (−∞, −L0 ].
(7.178)
Property (iii), (7.161), (7.163), (7.165), (7.171), and (7.178) imply that there exists t˜ ∈ [ti + 1 + 2bf , ti + 1 + 2bf + L1 ]
(7.179)
xf (t˜) − xi (t˜) ≤ δ0 .
(7.180)
such that
Property (ii), (7.162), (7.163), and (7.178)–(7.180) imply that xi (t) − xf (t) ≤ , t ∈ [Ti,1 , t˜]
520
7 ContinuousTime Nonautonomous Problems on Axis
and in particular, xi (ti ) − xf (ti ) ≤ . This contradicts (7.164). The contradiction we have reached proves that ti ≥ −2bf − 1 − L1 − L0 . Together with (7.176) this implies that − 2bf − 1 − L1 − L0 ≤ ti ≤ 2bf + 1 + L1 + L0 .
(7.181)
In view of (7.8), (7.13), and (7.171), there exists M1 > 0 such that the following property holds: (v) for each integer i ≥ 1 and each S2 > S1 satisfying S1 , S2 ∈ [Ti,1 , Ti,2 ], S2 − S1 ≤ 2bf , I f (S1 , S2 , xi , ui ) ≤ M1 . Lemma 7.20 and property (v) imply that there exists M2 > 0 such that xi (t) ≤ M2 , t ∈ [Ti,1 , Ti,2 ], i = 1, 2, . . . .
(7.182)
In view of (7.165) and (7.181), extracting a subsequence and reindexing, we may assume without loss of generality that there exists T˜1 = lim Ti,1 ∈ [−∞, L0 ], T˜2 = lim Ti,2 ∈ [−L0 , ∞]
(7.183)
t˜ = lim ti ∈ [−L1 − L0 − 2bf − 1, L0 + L1 + 2bf + 1].
(7.184)
i→∞
i→∞
i→∞
There exist a strictly decreasing sequence {T1 }∞ k=1 and a strictly increasing sequence {T2(k) }∞ such that k=1 (k)
T˜1 = lim T1(k) , T˜2 = lim T2(k) ,
(7.185)
T1(k) < T2(k) .
(7.186)
k→∞
k→∞
for all integers k ≥ 1,
If T˜2 = ∞, then we may assume that (k)
T2
= Tk,2 for all integers k ≥ 1
(7.187)
7.10 The Main Lemma
521
and if T˜1 = −∞, then we may assume that T1(k) = Tk,1 for all integers k ≥ 1.
(7.188)
By LSC property, property (v), and (7.171), extracting a subsequence and reindexing, we may assume without loss of generality that there exist x : (T˜1 , T˜2 ) → ˜ ˜ ˜ ˜ E, u : (T1 , T2 ) → F such that for each t ∈ (T1 , T2 ), x (t) = lim xi (t),
(7.189)
( x , u) ∈ X(T1(k) , T2(k) ),
(7.190)
i→∞
for each integer k ≥ 1,
I f (T1(1) , T2(1) , x , u) ≤ lim inf I f (T1(1) , T2(1), xi , ui ), i→∞
(7.191)
for each integer k ≥ 1, I f (T1(k+1) , T1(k) , x , u) ≤ lim inf I f (T1(k+1) , T1(k) , xi , ui ),
(7.192)
I f (T2(k) , T2(k+1) , x , u) ≤ lim inf I f (T2(k) , T2(k+1) , xi , ui ).
(7.193)
i→∞
i→∞
By (7.171), (7.182), (7.189), and (7.191)–(7.193), x (t) ≤ M2 , t ∈ (T˜1 , T˜2 )
(7.194)
and for each pair of integers k1 , k2 ≥ 1, (k1 )
I f (T1
(k2 )
, T2
(k1 )
, x , u) ≤ I f (T1
(k2 )
, T2
, xf , uf ) + 3 + 2S0 .
(7.195)
Lemma 7.21 and property (v) imply that the following property holds: (vi) for every ˜ > 0, there exists δ > 0 such that for each integer i ≥ 1 and each t1 , t2 ∈ [Ti,1 , Ti,2 ] satisfying t2 − t1  ≤ δ, we have xi (t1 ) − xi (t2 ) ≤ ˜ and that for each t1 , t2 ∈ R 1 satisfying t2 − t1  ≤ δ, we have xf (t1 ) − xf (t2 ) ≤ ˜ . Let ˜ > 0 and δ > 0 be as guaranteed by property (vi). It follows from property (vi), the choice of δ, and (7.189) that for each t1 , t2 ∈ (T˜1 , T˜2 ) satisfying t1 −t2  ≤ δ,
522
7 ContinuousTime Nonautonomous Problems on Axis
x (t1 ) − x (t2 ) ≤ .
(7.196)
Property (vi), (7.161), (7.162), (7.189), and (7.196) imply that if T˜1 > −∞, then there exist x (t) = lim xi (Ti,1 ) = lim xf (Ti,1 ) = xf (T˜1 ) lim
t →T˜1+
i→∞
i→∞
(7.197)
and we set x (T˜1 ) = lim x (t) t →T˜1+
(7.198)
and if T˜2 < ∞, then there exist lim x (t) = lim xi (Ti,2 ) = lim xf (Ti,2 ) = xf (T˜2 )
t →T˜2−
i→∞
i→∞
(7.199)
and we set x (t). x (T˜2 ) = lim t →T˜2−
(7.200)
Property (vi), (7.162), and (7.164) imply that T˜1 < t˜ < T˜2 .
(7.201)
Property (vi), (7.164), (7.184), (7.196), and (7.201) imply that x (t˜) = lim xf (tk ) − x (tk ) = lim xf (tk ) − xk (tk ) ≥ . xf (t˜) − k→∞
k→∞
(7.202)
In view of (7.190), for every pair of numbers τ1 , τ2 ∈ (T˜1 , T˜2 ) satisfying τ1 < τ2 , G(t, x (t), u(t)), t ∈ [τ1 , τ2 ] is strongly measurable; if T˜1 > −∞, then for every τ ∈ (T˜1 , T˜2 ), G(·, x (·), u(·)) is strongly measurable on [T˜1 , τ ); if T˜2 < ∞, then for every τ ∈ (T˜1 , T˜2 ), G(·, x (·), u(·)) is strongly measurable on [τ, T˜2 ]. Let k1 andk2 be natural numbers. By (7.7) and (7.8), there exists γ0 > 0 such that f (t, x, u) ≥ 8(G(t, x, u) − a0 x)+ for all (t, x, u) ∈ M satisfying G(t, x, u) − a0 x ≥ γ0 . In view of (7.194),
(7.203)
7.10 The Main Lemma
523
(k2 )
T2
G(t, x (t), u(t))dt
(k1 )
T1
≤
(k2 )
T2
a0 x (t)dt +
(k1 )
≤
(G(t, x (t), u(t)) − a0 x(t))+ dt
(k1 )
T1
a0 M2 (T2(k2 )
(k2 )
T2 T1
− T1(k1 ) )
(k2 )
T2
+
(k1 )
T1
(G(t, x (t), u(t)) − a0 x(t))+ dt.
(7.204)
Set E1 = {t ∈ [T1(k1 ) , T2(k2 ) ] : G(t, x (t), u(t)) ≥ a0 x(t) + γ0 }, (k1 )
E2 = [T1
(k2 )
, T2
(7.205)
] \ E1 .
It follows from (7.8), (7.195), and (7.203)–(7.205) that
(k2 )
T2
G(t, x (t), u(t))dt ≤ a0 M2 (T2(k2 ) − T1(k1 ) )
(k1 )
T1
+ E1
(G(t, x (t), u(t)) − a0 x (t))+ dt
+ E2
(G(t, x (t), u(t)) − a0 x (t))+ dt (k2 )
≤ a0 M2 (T2
(k1 )
− T1
(k2 )
) + γ0 (T2
(k1 )
− T1
)
+ E1 (k2 )
≤ a0 M2 (T2
(k1 )
(k2 )
(k1 )
−T1
(k2 )
(k1 )
− T1
) + γ0 (T2
− T1
)+
(k2 )
(k1 )
(k2 )
(k1 )
≤ (a0 M2 + γ0 )(T2 ≤ (T2
(G(t, x (t), u(t)) − a0 x (t))+ dt
− T1
) + a0 (T2
− T1
(k1 )
)(a0 M2 +γ0 +a0 )+I f (T1
f (t, x (t), u(t))dt E1
(k2 )
, T2
(k1 )
) + I f (T1
(k2 )
, T2
, xf , uf )+3+2S0 .
If T˜1 > −∞, T˜2 < ∞, then in view of (7.206), there exists
, x , u) (7.206)
524
7 ContinuousTime Nonautonomous Problems on Axis
lim
(k2 )
T2
k→∞ T (k1 ) 1
G(t, x (t), u(t))dt
≤ (T˜2 − T˜1 )(a0 M2 + γ0 + a0 ) + I f (T˜1 , T˜2 , xf , uf ) + 3 + 2S0 and in view of Fatou’s lemma
T˜2 T˜1
G(t, x (t), u(t))dt < ∞,
G(·, x (·), u(·)) is Bochner integrable on [T˜1 , T˜2 ]. If T˜1 > −∞, T˜2 = ∞, in view of (7.206), then for every integer i ≥ 1, there exists lim
(i)
T2
k→∞ T (k) 1
G(t, x (t), u(t))dt
≤ (T2 − T˜1 )(a0 M2 + γ0 + a0 ) + I f (T˜1 , T2 , xf , uf ) + 3 + 2S0 (i)
(i)
and in view of Fatou’s lemma
(i)
T2
T˜1
G(t, x (t), u(t))dt < ∞,
G(·, x (·), u(·)) is Bochner integrable on [T˜1 , T2 ]. (i)
(7.207)
If T˜1 = −∞, T˜2 < ∞, then for every integer i ≥ 1, there exists lim
T2(k)
k→∞ T (i) 1
G(t, x (t), u(t))dt
(i) (i) ≤ (T˜2 − T1 )(a0 M2 + γ0 + a0 ) + I f (T1 , T˜2 , xf , uf ) + 3 + 2S0
and in view of Fatou’s lemma,
T˜2 (i)
G(t, x (t), u(t))dt < ∞,
T1
(i) G(·, x (·), u(·)) is Bochner integrable on [T1 , T˜2 ].
Let
(7.208)
7.10 The Main Lemma
525
τ¯ ∈ (T˜1 , T˜2 ). Assume that T˜2 < ∞. Then for each sufficiently large natural number k, (k)
T2
(k)
> τ¯ > T1 ,
(k)
(k)
in view of (7.190), ( x , u) ∈ X(τ¯ , T2 ), for all t ∈ [τ¯ , T2 ], x (t) = x (τ¯ ) +
t
G(s, x (s), u(s))ds
τ¯
and this implies that for all t ∈ [τ¯ , T˜2 ), x (t) = x (τ¯ ) +
τ¯
t
G(s, x (s), u(s))ds → x (τ¯ ) +
T˜2
τ¯
G(s, x (s), u(s))ds as t → T˜2
and ( x , u) ∈ X(τ¯ , T˜2 ).
(7.209)
x , u) ∈ X(T˜1 , τ¯ ). It is sufficient to show Assume that T˜1 > −∞. We show that ( that for all t ∈ (T˜1 , τ¯ ], x (t) =
t T˜1
G(s, x (s), u(s))ds + x (T˜1 ).
Let t ∈ (T˜1 , τ¯ ] and > 0. Then there exists an integer k0 ≥ 1 such that (k0 )
T1
(k0 )
< t ≤ τ¯ < T2
(7.210)
,
(k) x (T1 ) − x (T˜1 ) ≤ ˜ /4
(7.211)
for all integers k ≥ k0 ,
T1(k) T˜1
G(s, x(s), u(s))ds < ˜ /4 for all integers k ≥ k0 . (k)
(7.212)
(k)
Let k ≥ k0 be an integer. Since ( x , u) ∈ X(T1 , T2 ), it follows from (7.190) and (7.210) that x (t)
= x (T1(k) ) +
t (k)
T1
G(s, x (s), u(s))ds.
(7.213)
526
7 ContinuousTime Nonautonomous Problems on Axis
By (7.211)–(7.213), x (t) − x (T˜1 ) −
≤
x (T1(k) )
− x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
(k)
T1 T˜1
G(s, x (s), u(s))ds ≤ ˜ .
Since ˜ is any positive number, we conclude that x (t) = x (T˜1 ) +
t T˜1
G(s, x (s), u(s))ds
and ( x , u) ∈ X(T˜1 , τ¯ ). Together with (7.209), this implies that if T˜2 < ∞, then ( x , u) ∈ X(τ¯ , T˜2 ) and if T˜1 > −∞ then ( x , u) ∈ X(T˜1 , τ¯ ). (7.214) There are the following cases: T˜1 > −∞, T˜2 < −∞;
(7.215)
T˜1 > −∞, T˜2 = ∞;
(7.216)
T˜1 = −∞, T˜2 < ∞;
(7.217)
T˜1 = −∞, T˜2 = ∞.
(7.218)
Lemma 7.23 implies that the following property holds: (vii) for each Δ ∈ (0, 1), there exists δ(Δ) > 0 such that for each τ1 ∈ R 1 , each τ2 ≥ τ1 + 3bf , each (y, v) ∈ X(τ1 , τ2 ) satisfying y(τi ) − xf (τi ) ≤ δ(Δ), i = 1, 2, I f (τ1 , τ2 , y, v) ≤ U f (τ1 , τ2 , y(τ1 ), y(τ2 )) + δ(Δ), each t1 ∈ [τ1 , τ1 + δ(Δ)], each t2 ∈ [τ2 − δ(Δ), τ2 ], I f (t1 , t2 , y, v) ≤ I f (t1 , t2 , xf , uf ) + Δ/2. Assume that (7.215) holds. It follows from (7.197) to (7.200), (7.202), (7.214), and (7.215) that x (T˜i ), i = 1, 2, x (t˜) − xf (t˜) ≥ . xf (T˜i ) =
(7.219)
7.10 The Main Lemma
527
Property (P3), the equality u˜ f = uf , and (7.219) imply that x , u). I f (T˜1 , T˜2 , xf , uf ) < I f (T˜1 , T˜2 ,
(7.220)
Let Δ ∈ (0, min{1, bf /8}), δ(Δ) be as guaranteed by property (vii) and let the following property hold: (viii) (A2) holds with δ = δ(Δ), = Δ/16. In view of (7.161), there exists an integer k0 ≥ 1 such that δk < δ(Δ) for all integers k ≥ k0 .
(7.221)
Let q be a natural number such that for all integers i ≥ q, T˜1 − T1 , T˜2 − T2  ≤ δ(Δ)/4. (i)
(i)
(7.222)
By (7.222), there exists an integer k1 ≥ k0 such that for each integer k ≥ k1 , (q)
Tk,1 < T1
(q)
≤ Tk,1 + δ(Δ), Tk,2 − δ(Δ) ≤ T2
< Tk,2 .
(7.223)
Property (vii), (7.162), (7.163), (7.221), and (7.223) imply that for each integer k ≥ k1 , (q)
(q)
(q)
(q)
I f (T1 , T2 , xk , uk ) ≤ I f (T1 , T2 , xf , uf ) + Δ/2.
(7.224)
By (7.191)–(7.193), (q)
(q)
I f (T1 , T2 , x , u) ≤ lim inf I f (T1(k) , T2(k) , xk , uk ) k→∞
(q)
(q)
≤ I f (T1 , T2 , xf , uf ) + Δ/2. Since the relation above holds for all integers q ≥ 1 satisfying (7.222), we conclude that x , u) ≤ I f (T˜1 , T˜2 , xf , uf ) + Δ/2. I f (T˜1 , T˜2 , Since Δ is any positive number, we conclude that I f (T˜1 , T˜2 , x , u) ≤ I f (T˜1 , T˜2 , xf , uf ). This contradicts (7.220). The contradiction we have reached proves that (7.215) does not hold.
528
7 ContinuousTime Nonautonomous Problems on Axis
Consider case (7.216) with T1 > −∞, T2 = ∞. We may assume (see (7.187)) that (k)
T2
= Tk,2 for all integers k ≥ 1.
(7.225)
By (7.173), (7.191)–(7.193), and (7.225), for each integer k ≥ 1, x , u) = I f (T1(k) , T2(k) , x , u) I f (T1(k) , Tk,2 , ≤ lim inf I f (T1(k) , T2(k) , xi , ui ) ≤ I f (T1(k) , Tk,2 , xf , uf ) + 3 + 2S0 . i→∞
(7.226)
In view of (7.214) and (7.226), for all sufficiently large natural numbers k, I f (T˜1 , Tk,2 , x , u) ≤ I f (T˜1 , Tk,2 , xf , uf ) + 4 + 2S0 .
(7.227)
u(t) = uf (t), t ∈ (−∞, T˜1 ). x (t) = xf (t),
(7.228)
Set
Theorem 7.2, (7.197), (7.198), (7.214), (7.227), and (7.228) imply that ( x , u) ∈ X(−∞, ∞) is (f )good. By (P1), x (t) − xf (t) = 0. lim
t →∞
(7.229)
Property (P3) and (7.202) imply that ( x , u) is not (f )minimal. Therefore there exist S∗ > 0 and Δ ∈ (0, 1) such that S∗ > T˜1  + 1 + bf , I f (−S∗ , S∗ , x , u) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ.
(7.230) (7.231)
Let δ(Δ) ∈ (0, min{Δ, bf /8}) be such that properties (vii) and (viii) hold. In view of (7.161), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ) for all integers k ≥ k(Δ).
(7.232)
Property (vii), (7.162), (7.163), and (7.232) imply that the following property holds: (ix) for each integer k ≥ k(Δ), each τ1 ∈ [Tk,1 , Tk,1 + δ(Δ)], and each τ2 ∈ [Tk,2 − δ(Δ), Tk,2 ], I f (τ1 , τ2 , xk , uk ) ≤ I f (τ1 , τ2 , xf , uf ) + Δ/2.
(7.233)
7.10 The Main Lemma
529
In view of (7.229), there exists τ0 > 1 such that x (t) − xf (t) ≤ δ(Δ)/4 for all t ≥ τ0 .
(7.234)
Let q be a natural number such that Tq,2 > τ0 + S∗ ,
(7.235)
I f (T˜1 , T1 , x , u) < Δ/8, I f (T˜1 , T1 , xf , uf ) < Δ/8, (q)
(q)
(7.236)
for all integers i ≥ q, T˜1 − T1(i)  ≤ δ(Δ)/2.
(7.237)
By (7.183), (7.189), and (7.234), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , (q)
Tk,1 < T1
≤ Tk,1 + δ(Δ),
x (Tq,2 ) − xk (Tq,2 ) ≤ δ(Δ)/4.
(7.238) (7.239)
Assume that an integer k ≥ k1 . Then (7.238) and (7.239) hold. It follows from (7.162), (7.225), (7.232), (7.234), (7.235), (7.238), and (7.239) that (q)
(q)
(q)
I f (T1 , Tq,2 , xk , uk ) ≤ U f (T1 , Tq,2 , xk (T1 ), xk (Tq,2 )) + δ(Δ), xk (Tk,1 ) − xf (Tk,1 ) ≤ δk ≤ δ(Δ),
(7.240) (7.241)
x (Tq,2 ) + x (Tq,2 ) − xf (Tq,2 ) xk (Tq,2 ) − xf (Tq,2 ) ≤ xk (Tq,2 ) − ≤ δ(Δ)/4 + δ(Δ)/4.
(7.242)
Property (vii), (7.138), (7.163), (7.238), (7.241), and (7.242) imply that (q)
(q)
I f (T1 , Tq,2 , xk , uk ) ≤ I f (T1 , Tq,2 , xf , uf ) + Δ/2.
(7.243)
By (7.191)–(7.193) and (7.243), (q)
(q)
I f (T1 , Tq,2 , x , u) ≤ lim inf I f (T1 , Tq,2 , xk , uk ) k→∞
(q)
≤ I f (T1 , Tq,2 , xf , uf ) + Δ/2.
(7.244)
In view of (7.236) and (7.244), x , u) ≤ I f (T˜1 , Tq,2 , xf , uf ) + 3Δ/4 I f (T˜1 , Tq,2 ,
(7.245)
530
7 ContinuousTime Nonautonomous Problems on Axis
for all integers q ≥ 1 satisfying (7.235)–(7.237). Let a natural number q satisfy (7.235)–(7.237) and q −1 < Δ/4. Property (viii), (7.231), and (7.234) imply that there exists (ξ, η) ∈ X(−∞, ∞) such that ξ(t) = x (t), η(t) = u(t), t ∈ (−∞, −S∗ ],
(7.246)
ξ(S∗ ) = x (S∗ ),
(7.247)
x , u) − 2Δ, I f (−S∗ , S∗ , ξ, η) < I f (−S∗ , S∗ ,
(7.248)
ξ(t) = x (t), η(t) = u(t), t ∈ (S∗ , Tq,2 ],
(7.249)
ξ(Tq,2 + bf ) = xf (Tq,2 + bf ),
(7.250)
I f (Tq,2 , Tq,2 + bf , ξ, η) ≤ I f (Tq,2 , Tq,2 + bf , xf , uf ) + Δ/16.
(7.251)
In view of (7.228), (7.230), (7.246), and (7.250), x (−S∗ ) = ξ(−S∗ ), ξ(Tq,2 + bf ) = xf (Tq,2 + bf ). xf (−S∗ ) =
(7.252)
Property (P3), equality u˜ f = uf , (7.228), (7.235), (7.245), (7.248), (7.249), (7.251), and (7.252) imply that 0 ≤ I f (−S∗ , Tq,2 + bf , ξ, η) − I f (−S∗ , Tq,2 + bf , xf , uf ) = I f (−S∗ , S∗ , ξ, η) + I f (S∗ , Tq,2 , x , u) +I f (Tq,2 , Tq,2 + bf , ξ, η) − I f (−S∗ , Tq,2 + bf , xf , uf ) < I f (−S∗ , S∗ , x , u) − 2Δ + I f (S∗ , Tq,2 , x , u) +I f (Tq,2 , Tq,2 + bf , xf , uf ) + Δ/16 − I f (−S∗ , Tq,2 + bf , xf , uf ) ≤ I f (−S∗ , Tq,2 , x , u) − 2Δ + Δ/16 − I f (−S∗ , Tq,2 , xf , uf ) = I f (T˜1 , Tq,2 , x , u) − 2Δ + Δ/16 − I f (T˜1 , Tq,2 , xf , uf ) ≤ −2Δ + Δ/16 + 3Δ/4 ≤ −Δ,
7.10 The Main Lemma
531
a contradiction. The contradiction we have reached proves that case (7.216) does not hold. Consider case (7.217) with T1 = −∞, T2 < ∞. We may assume that (k)
T1
= Tk,1 for all integers k ≥ 1.
(7.253)
By (7.171), (7.191)–(7.193), and (7.253), for each integer k ≥ 1, x , u) = I f (T1(k) , T2(k) , x , u) I f (Tk,1 , T2(k) , ≤ lim inf I f (T1(k) , T2(k) , xi , ui ) ≤ I f (Tk,1 , T2(k) , xf , uf ) + 3 + 2S0 . i→∞
(7.254)
In view of (7.117), (7.214), and (7.254), for all sufficiently large natural numbers k, I f (Tk,1 , T˜2 , , x , u) ≤ I f (Tk,1 , T˜2 , xf , uf ) + 4 + 2S0 .
(7.255)
Set u(t) = uf (t), t ∈ (T˜2 , ∞). x (t) = xf (t),
(7.256)
Theorem 7.2, (7.199), (7.200), (7.214), (7.217), (7.255), and (7.256) imply that ( x , u) ∈ X(−∞, ∞) is (f )good. By (P1), lim x (t) − xf (t) = 0.
t →−∞
(7.257)
Property (P3) and (7.202) imply that ( x , u) is not (f )minimal. Therefore there exist S∗ > 0 and Δ > 0 such that S∗ > T˜2  + 1 + bf , I f (−S∗ , S∗ , x , u) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ.
(7.258) (7.259)
Let δ(Δ) ∈ (0, min{Δ, bf /8}) be such that properties (vii) and (viii) hold. In view of (7.161), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ) for all integers k ≥ k(Δ).
(7.260)
Property (vii), (7.162), (7.163), and (7.260) imply that the following property holds: (x) for each integer k ≥ k(Δ), each τ1 ∈ [Tk,1, Tk,1 + δ(Δ)], each τ2 ∈ [Tk,2 − δ(Δ), Tk,2 ], I f (τ1 , τ2 , xk , uk ) ≤ I f (τ1 , τ2 , xf , uf ) + Δ/2.
532
7 ContinuousTime Nonautonomous Problems on Axis
In view of (7.257), there exists τ0 > 1 such that x (t) − xf (t) ≤ δ(Δ)/4 for all t ≤ −τ0 .
(7.261)
Let q be a natural number such that Tq,1 < −τ0 − S∗ ,
(7.262)
I f (T2 , T˜2 , x , u) < Δ/8, I f (T2 , T˜2 , xf , uf ) < Δ/8, (q)
(q)
(7.263)
for all integers i ≥ q, T˜2 − T2(i)  ≤ δ(Δ)/2.
(7.264)
By (7.189) and (7.264), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , (q)
Tk,2 − δ(Δ) ≤ T2
(7.265)
< Tk,2 ,
x (Tq,i ) − xk (Tq,i ) ≤ δ(Δ)/4, i = 1, 2.
(7.266)
Assume that an integer k ≥ k1 + q. Then (7.265) and (7.266) hold. It follows from (7.162), (7.163), (7.260)–(7.262), and (7.266) that I f (Tq,1 , Tk,2 , xk , uk ) ≤ U f (Tq,1 , Tk,2 , xk (Tq,1 ), xk (Tq,2 )) + δ(Δ), xk (Tk,2 ) − xf (Tk,2 ) ≤ δk ≤ δ(Δ),
(7.267) (7.268)
x (Tq,1 ) + x (Tq,1 ) − xf (Tq,1 ) xk (Tq,1 ) − xf (Tq,1 ) ≤ xk (Tq,1 ) − ≤ δ(Δ)/4 + δ(Δ)/4.
(7.269)
Property (vii), (7265), and (7.267)–(7.269) imply that (q)
(q)
I f (Tq,1 , T2 , xk , uk ) ≤ I f (Tq,1 , T2 , xf , uf ) + Δ/2
(7.270)
for all integers k ≥ k1 + q. By (7.191)–(7.193) and (7.270), (q)
(q)
I f (Tq,1 , T2 , x , u) ≤ lim inf I f (Tq,1 , T2 , xk , uk ) k→∞
(q)
≤ I f (Tq,1 , T2 , xf , uf ) + Δ/2.
(7.271)
In view of (7.263) and (7.271), I f (Tq,1 T˜2 , x , u) ≤ I f (Tq,1 , T˜2 , xf , uf ) + 3Δ/4
(7.272)
7.10 The Main Lemma
533
for all integers q ≥ 1 satisfying (7.262)–(7.264). Let a natural number q satisfy (7.262)–(7.264). Property (viii), the choice of δ(Δ), (7.259), and (7.261) imply that there exists (ξ, η) ∈ X(−∞, ∞) such that ξ(t) = x (t), η(t) = u(t), t ∈ [S∗ , ∞), x (−S∗ ), I f (−S∗ , S∗ , ξ, η) < I f (−S∗ , S∗ , x , u) − 2Δ, ξ(−S∗ ) =
(7.273) (7.274)
ξ(t) = x (t), η(t) = u(t), t ∈ [Tq,1 , −S∗ ),
(7.275)
ξ(Tq,1 − bf ) = xf (Tq,1 − bf ),
(7.276)
I f (Tq,1 − bf , Tq,1 , ξ, η) ≤ I f (Tq,1 − bf , Tq,1 , xf , uf ) + Δ/16.
(7.277)
ξ(t) = xf (t), η(t) = uf (t), t ∈ (−∞, Tq,1 − bf ).
(7.278)
In view of (7.256), (7.258), (7.273), and (7.276), xf (S∗ ) = x (S∗ ) = ξ(S∗ ), ξ(Tq,1 − bf ) = xf (Tq,1 − bf ).
(7.279)
Property (P3), equality u˜ f = uf , (7.258), (7.274), (7.275), and (7.277)–(7.279) imply that 0 ≤ I f (Tq,1 − bf , S∗ , ξ, η) − I f (Tq,1 − bf , S∗ , xf , uf ) = I f (Tq,1 − bf , Tq,1 , ξ, η) + I f (Tq,1 , −S∗ , ξ, η) +I f (−S∗ , S∗ , ξ, η) − I f (Tq,1 − bf , S∗ , xf , uf ) ≤ I f (Tq,1 − bf , Tq,1 , xf , uf ) + Δ/16 + I f (Tq,1 , −S∗ , x , u) +I f (−S∗ , S∗ , x , u) − 2Δ − I f (Tq,1 − bf , S∗ , xf , uf ) ≤ I f (Tq,1 , S∗ , x , u) − 2Δ + Δ/16 − I f (Tq,1 , S∗ , xf , uf ) x , u) − 2Δ + Δ/16 − I f (Tq,1 , T˜2 , xf , uf ) = I f (Tq,1 , T˜2 , ≤ −2Δ + Δ/16 + 3Δ/4 ≤ −Δ, a contradiction. The contradiction we have reached proves that case (7.217) does not hold. Consider case (7.218) with T1 = −∞, T2 = ∞. We may assume that T1(k) = Tk,1 , T2(k) = Tk,2 for all integers k ≥ 1.
(7.280)
534
7 ContinuousTime Nonautonomous Problems on Axis
By (7.171), (7.191)–(7.193), and (7.280), for each integer k ≥ 1, (k)
(k)
I f (Tk,1 , Tk,2 , x , u) = I f (T1 , T2 , x , u) (k)
(k)
≤ lim inf I f (T1 , T2 , xi , ui ) ≤ I f (Tk,1 , Tk,2 , xf , uf ) + 3 + 2S0 . i→∞
(7.281)
Theorem 7.2 and (7.281) imply that ( x , u) ∈ X(−∞, ∞) is (f )good. By (P1), lim x (t) − xf (t) = 0,
t →∞
lim x (t) − xf (t) = 0.
t →−∞
(7.282)
Property (P3) and (7.202) imply that ( x , u) is not (f )minimal. Therefore there exist S∗ > 0 and Δ > 0 such that S∗ > 1 + bf , x , u) > U f (−S∗ , S∗ , x (−S∗ ), x (S∗ )) + 2Δ. I f (−S∗ , S∗ ,
(7.283) (7.284)
Let δ(Δ) > 0 be such that properties (vii) and (viii) hold. In view of (7.161), there exists an integer k(Δ) ≥ 1 such that δk < δ(Δ) for all integers k ≥ k(Δ).
(7.285)
In view of (7.282), there exists τ0 > bf + 1 such that x (t) − xf (t) ≤ δ(Δ)/4 for all t ∈ (−∞, −τ0 ] ∪ [τ0 , ∞).
(7.286)
Let q be a natural number such that Tq,1 < −τ0 − S∗ , Tq,2 > τ0 + S∗ .
(7.287)
By (7.189), there exists an integer k1 ≥ k(Δ) + q such that for each integer k ≥ k1 , x (Tq,i ) − xk (Tq,i ) ≤ δ(Δ)/4, i = 1, 2.
(7.288)
Assume that an integer k ≥ k1 + q. Then (7.288) holds. It follows from (7.162) and (7.285)–(7.288) that I f (Tq,1 , Tq,2 , xk , uk ) ≤ U f (Tq,1 , Tq,2 , xk (Tq,1 ), xk (Tq,2 )) + δ(Δ)
(7.289)
and for i = 1, 2, x (Tq,i ) + x (Tq,i ) − xf (Tq,i ) xk (Tq,i ) − xf (Tq,i ) ≤ xk (Tq,i ) − ≤ δ(Δ)/4 + δ(Δ)/4.
(7.290)
7.10 The Main Lemma
535
Property (vii), the choice of δ(Δ), (7.289), and (7.290) imply that I f (Tq,1 , Tq,2 , xk , uk ) ≤ I f (Tq,1 , Tq,2 , xf , uf ) + Δ/2.
(7.291)
In view of (7.191)–(7.193) and (7.291), x , u) ≤ lim inf I f (Tq,1 , Tq,2 , xk , uk ) ≤ I f (Tq,1 , Tq,2 , xf , uf )+Δ/2. I f (Tq,1 , Tq,2 , k→∞
(7.292) Property (viii), the choice of δ(Δ), (7.283), (7.284), (7.286), and (7.287) imply that there exists (ξ, η) ∈ X(−∞, ∞) such that ξ(−S∗ ) = x (−S∗ ), ξ(S∗ ) = x (S∗ ),
(7.293)
I f (−S∗ , S∗ , ξ, η) < I f (−S∗ , S∗ , x , u) − 2Δ,
(7.294)
ξ(t) = x (t), η(t) = u(t), t ∈ [Tq,1 , −S∗ ) ∪ (S∗ , Tq,2 ],
(7.295)
ξ(Tq,1 − bf ) = xf (Tq,1 − bf ), I f (Tq,1 − bf , Tq,1 , ξ, η) ≤ I f (Tq,1 − bf , Tq,1 , xf , uf ) + Δ/16, ξ(Tq,2 + bf ) = xf (Tq,2 + bf ),
(7.296) (7.297) (7.298)
I f (Tq,2 , bf + Tq,2 , ξ, η) ≤ I f (Tq,2 , bf + Tq,2 , xf , uf ) + Δ/16,
(7.299)
ξ(t) = xf (t), η(t) = uf (t), t ∈ (−∞, Tq,1 − bf ] ∪ (Tq,2 + bf , ∞).
(7.300)
Property (P3), equality u˜ f = uf , (7.287), (7.292), and (7.294)–(7.299) imply that 0 ≤ I f (Tq,1 − bf , Tq,2 + bf , ξ, η) − I f (Tq,1 − bf , Tq,2 + bf , xf , uf ) = I f (Tq,1 − bf , Tq,1 , ξ, η) + I f (Tq,1 , −S∗ , ξ, η) +I f (−S∗ , S∗ , ξ, η) + I f (S∗ , Tq,2 , ξ, η) +I f (Tq,2 , Tq,2 + bf , ξ, η) − I f (Tq,1 − bf , Tq,2 + bf , xf , uf ) x , u) ≤ I f (Tq,1 − bf , Tq,1 , xf , uf ) + Δ/16 + I f (Tq,1 , −S∗ , +I f (−S∗ , S∗ , x , u)−2Δ+I f (S∗ , Tq,2 , x , u)+I f (Tq,2 , Tq,2 +bf , xf , uf )+Δ/16 −I f (Tq,1 − bf , Tq,2 + bf , xf , uf )
536
7 ContinuousTime Nonautonomous Problems on Axis
= Δ/16 − 2Δ + I f (Tq,1 , Tq,2 , x , u) − I f (Tq,1 , Tq,2 , xf , uf ) ≤ −2Δ + Δ/16 + Δ/2 < −Δ, a contradiction. The contradiction we have reached proves that case (7.218) does not hold and completes the proof of Lemma 7.27.
7.11 Completion of the Proof of Theorem 7.8 Assume that properties (P1), (P2), and (P3) hold and u˜ f = uf . Let , M > 0. By Lemma 7.27, there exist δ0 ∈ (0, ) such that the following property holds: (i) for each T1 ∈ R 1 , each T2 ≥ T1 + 3bf and each (x, u) ∈ X(T1 , T2 ) which satisfies x(Ti ) − xf (Ti ) ≤ δ0 , i = 1, 2, I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ0 the inequality x(t) − xf (t) ≤ holds for all t ∈ [T1 , T2 ]. By Theorem 7.2, there exists S0 > 0 such that for each T2 > T1 and each (x, u) ∈ X(T1 , T2 ), we have I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(7.301)
In view of (P2), there exist δ ∈ (0, δ0 ) and L0 > 0 such that the following property holds: (ii) for each T ∈ R 1 and each (y, v) ∈ X(T , T + L0 ) which satisfies I f (T , T + L0 , y, v) ≤ min{U f (T , T +L0 , y(T ), y(T + L0 ))+δ, I f (T , T +L0 , xf , uf )+M+2S0 } there exists S ∈ [T , T + L0 ] such that y(S) − xf (S) ≤ δ0 . Set L = 2L0 + 2bf + 2.
(7.302)
Assume that T1 ∈ R 1 , T2 ≥ T1 + 2L and that (x, u) ∈ X(T1 , T2 ) satisfies I f (T1 , T2 , x, u) ≤ σ f (T1 , T2 ) + M,
(7.303)
7.12 An Auxiliary Result for Theorem 7.9
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
537
(7.304)
By (7.302) and (7.304), I f (T1 , T1 + L0 , x, u) ≤ U f (T1 , T1 + L0 , x(T1 ), x(T1 + L0 )) + δ,
(7.305)
I f (T2 − L0 , T2 , x, u) ≤ U f (T2 − L0 , T2 , x(T2 − L0 ), x(T2 )) + δ.
(7.306)
In view of (7.303), I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M.
(7.307)
It follows from (7.301) and (7.306) that for each pair of numbers S1 , S2 ∈ [T1 , T2 ] satisfying S1 < S2 , I f (S1 , S2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , S1 , x, u) − I f (S2 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M − I f (T1 , S1 , xf , uf ) + S0 − I f (S2 , T2 , xf , uf ) + S0 = I f (S1 , S2 , xf , uf ) + M + 2S0 .
(7.308)
Property (ii), (7.305), (7.306), and (7.308) imply that there exist τ1 ∈ [T1 , T1 + L0 ], τ2 ∈ [T2 − L0 , T2 ]
(7.309)
x(τi ) − xf (τi ) ≤ δ0 , i = 1, 2.
(7.310)
such that
If x(T2 ) − xf (T2 ) ≤ δ, then we may assume that τ2 = T2 , and if x(T1 ) − xf (T1 ) ≤ δ, then we may assume that τ1 = T1 . It follows from (7.302), (7.304), (7.310), and property (i) that x(t) − xf (t) ≤ , t ∈ [τ1 , τ2 ]. Thus STP holds and Theorem 7.8 is proved.
7.12 An Auxiliary Result for Theorem 7.9 Lemma 7.28. Let (x, u) ∈ X(−∞, ∞) be (f )good and > 0. Then there exist L > 0 such that for each T2 > T1 satisfying either T1 ≥ L or T2 ≤ −L the following inequality holds:
538
7 ContinuousTime Nonautonomous Problems on Axis
I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + .
(7.311)
Proof. There exists M0 > 0 such that for each T > 0, I f (−T , T , x, u) − I f (−T , T , xf , uf ) < M0 .
(7.312)
By Theorem 7.2, there exists M1 > 0 such that for each S2 > S1 and each (y, v) ∈ X(S1 , S2 ), we have I f (S1 , S2 , y, v) + M1 ≥ I f (S1 , S2 , xf , uf ).
(7.313)
Assume that the lemma does not hold. Then there exists a sequence of closed intervals {[ai , bi ]}∞ i=1 such that for each integer i ≥ 1 and each integer j > i, [ai , bi ] ∩ [aj , bj ] = ∅, I f (ai , bi , x, u) > U f (ai , bi , x(ai ), x(bi )) + .
(7.314) (7.315)
By (7.314) and (7.315), there exists (y, v) ∈ X(−∞, ∞) such that y(t) = x(t), v(t) = u(t), t ∈ R 1 \ ∪∞ i=1 (ai , bi ),
(7.316)
for every integer i ≥ 1, I f (ai , bi , y, v) < I f (ai , bi , x, u) − .
(7.317)
Let q ≥ 1 be an integer such that q > (M0 + M1 ) −1 .
(7.318)
[ai , bi ] ⊂ [−T , T ], i = 1, . . . , q.
(7.319)
Choose T > 0 such that
In view of (7.316) and (7.319), I f (−T , T , y, v) − I f (−T , T , x, u) q ≤ (I f (ai , bi , y, v) − I f (ai , bi , x, u)) ≤ −q. i=1
By (7.312), (7.313) and the relation above,
7.13 Proofs of Propositions 7.11 and 7.13
539
I f (−T , T , xf , uf ) − M1 ≤ I f (−T , T , y, v) ≤ I f (−T , T , x, u) − q ≤ I f (−T , T , xf , uf ) + M0 − q, q ≤ M0 + M1 . This contradicts (7.318). The contradiction we have reached proves Lemma 7.28.
7.13 Proofs of Propositions 7.11 and 7.13 Proof of Proposition 7.11. By Theorem 7.2, there exists S0 > 0 such that the following property holds: (i) for each T2 > T1 and each (y, v) ∈ X(T1 , T2 ), we have I f (T1 , T2 , y, v) + S0 ≥ I f (S1 , S2 , xf , uf ). Set M = 2L0 + 2a0 L0 + S0 .
(7.320)
Assume that L0 , T2 ≥ T1 + 2L0 . (T1 , z1 ) ∈ AL0 , (T2 , z2 ) ∈ A Clearly, there exist τ1 ∈ (0, L0 ], τ2 ∈ (0, L0 ] and (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = z1 , x(T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x, u) ≤ L0 , (x(t), u(t)) = (xf (t), uf (t)), t ∈ [T1 + τ1 , T2 − τ2 ], x(T2 ) = z2 , I f (T2 − τ2 , T2 , x, u) ≤ L0 . In view of the relations above, property (i) and (7.320), U f (T1 , T2 , z1 , z2 ) ≤ I f (T1 , T2 , x, u) ≤ 2L0 + I f (T1 + τ1 , T2 − τ2 , xf , uf ) ≤ 2L0 + 2a0L0 + I f (T1 , T2 , xf , uf ) ≤ 2L0 + 2a0 L0 + S0 + σ f (T1 , T2 ) = M + σ f (T1 , T2 ). Proposition 7.11 is proved.
540
7 ContinuousTime Nonautonomous Problems on Axis
Proof of Proposition 7.13. By Theorem 7.2, there exists S0 > 0 such that property (i) of the proof of Proposition 7.11 holds. Set M = L0 + a0 L0 + S0 .
(7.321)
Assume that (T1 , z1 ) ∈ AL0 , T2 ≥ T1 + L0 . Clearly, there exist τ1 ∈ (0, L0 ] and (x, u) ∈ X(T1 , T2 ) such that x(T1 ) = z1 , x(T1 + τ1 ) = xf (T1 + τ1 ), I f (T1 , T1 + τ1 , x, u) ≤ L0 , (x(t), u(t)) = (xf (t), uf (t)), t ∈ [T1 + τ1 , T2 ]. In view of the relations above, property (i) and (7.321), σ f (T1 , T2 , z1 ) ≤ I f (T1 , T2 , x, u) ≤ L0 + I f (T1 + τ1 , T2 , xf , uf ) ≤ L0 + a0 L0 + I f (T1 , T2 , xf , uf ) ≤ L0 + a0 L0 + S0 + σ f (T1 , T2 ) = M + σ f (T1 , T2 ). Proposition 7.13 is proved.
7.14 Proofs of Theorems 7.9 and 7.14 Proof of Theorem 7.9. Assume that property (P4) holds. Then (P1) follows from Theorem 7.2, Lemma 7.28, and (P4), while (P2) follows from (P4) and Theorem 7.2. Assume that (P1) and (P2) hold. Then (P4) follows from Theorem 7.2, Lemma 7.26, and property (P2). Proof of Theorem 7.14. Assume that f has WTP. Then (P1) and (P2) hold. Assume that (P1) and (P2) hold. We show that f has WTP. Let , M > 0. By Theorem 7.2, there exists S0 > 0 such that the following property holds: (i) for each τ2 > τ1 and each (y, v) ∈ X(τ1 , τ2 ), I f (τ1 , τ2 , y, v) + S0 ≥ I f (τ1 , τ2 , xf , uf ). Lemma 7.26 implies that there exist δ0 ∈ (0, 1) and L0 > 2bf such that the following property holds:
7.14 Proofs of Theorems 7.9 and 7.14
541
(ii) for each T1 ≥ L0 , each T2 ≥ T1 + 2bf and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ0 , i = 1, 2, I f (T1 , T2 , y, v) ≤ U f (T1 , T2 , y(T1 ), y(T2 )) + δ0
(7.322) (7.323)
we have y(t) − xf (t) ≤ , t ∈ [T1 , T2 ]
(7.324)
and for each T2 ≤ −L0 , each T1 ≤ T2 − 2bf and each (y, v) ∈ X(T1 , T2 ) which satisfies (7.322) and (7.323) inequality (7.324) hold. Property (P2) implies that there exist δ ∈ (0, δ0 ) and L1 > 0 such that the following property holds: (iii) for each T ∈ R 1 and each (y, v) ∈ X(T , T + L1 ) which satisfies I f (T , T + L1 , y, v) ≤ min{U f (T , T + L1 , y(T ), y(T + L1 )) + δ, I f (T , T + L1 , xf , uf ) + M + 2S0 } there exists s ∈ [T , T + L1 ] such that y(s) − xf (s) ≤ δ0 . Set l = 8L0 + 8L1 + 8bf + 8.
(7.325)
Q ≥ 6 + 6(M + S0 )δ −1 .
(7.326)
Choose a natural number
Assume that T1 ∈ R 1 , T2 ≥ T1 + lQ and (x, u) ∈ X(T1 , T2 ) satisfies I f (T1 , T2 , x, u) ≤ I f (T1 , T2 , xf , uf ) + M.
(7.327)
Property (i) and (7.327) imply that for each pair of numbers τ1 , τ2 ∈ [T1 , T2 ] satisfying τ1 < τ2 , I f (τ1 , τ2 , x, u) = I f (T1 , T2 , x, u) − I f (T1 , τ1 , x, u) − I f (τ2 , T2 , x, u) ≤ M + I f (T1 , T2 , xf , uf ) − I f (T1 , τ1 , xf , uf ) + S0 − I f (τ2 , T2 , xf , uf ) + S0
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7 ContinuousTime Nonautonomous Problems on Axis
= I f (τ1 , τ2 , xf , uf ) + M + 2S0 .
(7.328)
t0 = T1 .
(7.329)
Set
If I f (T1 , T2 , x, u) ≤ U f (T1 , T2 , x(T1 ), x(T2 )) + δ, then we set t1 = T2 , s1 = T2 .
(7.330)
Assume that I f (T1 , T2 , x, u) > U f (T1 , T2 , x(T1 ), x(T2 )) + δ.
(7.331)
It is easy to see that for each t ∈ (T1 , T2 ) such that t − T1 is sufficiently small, we have I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) < δ. Set t˜1 = inf{t ∈ (T1 , T2 ] : I f (T1 , t, x, u) − U f (T1 , t, x(T1 ), x(t)) > δ}.
(7.332)
Clearly, t˜1 ∈ (T1 , T2 ] is welldefined, and there exist s1 , t1 ∈ [T1 , T2 ] such that t0 < s1 < t˜1 , s1 ≥ t˜1 − 1/4, I f (t0 , s1 , x, u) − U f (t0 , s1 , x(t0 ), x(s1 )) ≤ δ, t1 ≤ T2 , t˜1 ≤ t1 < t˜1 + 1/4, I f (t0 , t1 , x, u) − U f (t0 , t1 , x(t0 ), x(t1 )) > δ.
(7.333) (7.334) (7.335) (7.336)
Assume that k ≥ 1 is an integer, and we defined finite sequences of numbers {ti }ki=0 ⊂ [T1 , T2 ], {si }ki=1 ⊂ [T1 , T2 ] such that T1 = t0 < t1 · · · < tk , for each integer i = 1, . . . , k,
(7.337)
7.14 Proofs of Theorems 7.9 and 7.14
ti−1 < si ≤ ti , ti − si ≤ 2−1 , I f (ti−1 , si , x, u) − U f (ti−1 , si , x(ti−1 ), x(si )) ≤ δ,
543
(7.338) (7.339)
and if ti < T2 , then I f (ti−1 , ti , x, u) − U f (ti−1 , ti , x(ti−1 ), x(ti )) > δ.
(7.340)
(Note that in view of (7.329)–(7.336), our assumption holds for k = 1.) By (7.337)–(7.340), there exists (y, v) ∈ X(T1 , T2 ) such that y(ti ) = x(ti ), i = 1, . . . , k, I f (ti−1 , ti , x, u) − I f (ti−1 , ti , y, v) > δ for all integers i ∈ [1, k] \ {k}, (y(t), v(t)) = (x(t), u(t)), t ∈ [tk−1 , T2 ]. Property (i), the relations above, and (7.327) imply that I f (T1 , T2 , xf , uf ) + M ≥ I f (T1 , T2 , x, u) ≥ I f (T1 , T2 , y, v) + δ(k − 1) ≥ I f (T1 , T2 , xf , uf ) − S0 + δ(k − 1), k ≤ 1 + δ −1 (M + S0 ).
(7.341)
If tk = T2 , then the construction of the sequence is completed. Assume that tk < T2 . If I f (tk , T2 , x, u) ≤ U f (tk , T2 , x(tk ), x(T2 )) + δ, then we set tk+1 = T2 , sk+1 = T2 and the construction of the sequence is completed. Assume that I f (tk , T2 , x, u) > U f (tk , T2 , x(tk ), x(T2 )) + δ.
(7.342)
It is easy to see that for each t ∈ (tk , T2 ) such that t − tk is sufficiently small, we have I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) < δ. Set t˜ = inf{t ∈ (tk , T2 ] : I f (tk , t, x, u) − U f (tk , t, x(tk ), x(t)) > δ}.
544
7 ContinuousTime Nonautonomous Problems on Axis
Clearly, t˜ is welldefined, t˜ ∈ (tk , T2 ], and there exist sk+1 , tk+1 ∈ [T1 , T2 ] such that tk < sk+1 < t˜, sk+1 > t˜ − 1/4, t˜ + 4−1 > tk+1 ≥ t˜, I f (tk , sk+1 , x, u) − U f (tk , sk+1 , x(tk ), x(sk+1 )) ≤ δ, I f (tk , tk+1 , x, u) − U f (tk , tk+1 , x(tk ), x(tk+1 )) > δ. It is not difficult to see that the assumption made for k also holds for k + 1. In q view of (7.341), by induction we constructed finite sequences {ti }i=0 ⊂ [T1 , T2 ], q {si }i=1 ⊂ [T1 , T2 ] such that q ≤ 2 + δ −1 (M + S0 ),
(7.343)
T1 = t0 < t1 · · · < tq = T2 ,
(7.344)
for each integer i = 1, . . . , q, ti−1 < si ≤ ti , ti − si ≤ 2−1 , I f (ti−1 , si , x, u) − U f (ti−1 , si , x(ti−1 ), x(si )) ≤ δ,
(7.345) (7.346)
and if ti < T2 , then I f (ti−1 , ti , x, u) − U f (ti−1 , ti , x(ti−1 ), x(ti )) > δ.
(7.347)
Assume that i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8 = l.
(7.348)
By (7.345) and (7.348), si+1 − ti ≥ 8L0 + 8L1 + 8bf + 7.
(7.349)
ti ≥ −2L0 − 2L1 .
(7.350)
Assume that
Property (iii), (7.328), (7.346), and (7.349) imply that there exist ti,1 ∈ [ti + 3L0 + 2L1 , ti + 3L0 + 3L1 ] satisfying
(7.351)
7.14 Proofs of Theorems 7.9 and 7.14
545
x(ti,1 ) − xf (ti,1 ) ≤ δ0
(7.352)
ti,2 ∈ [si+1 − L1 , si+1 ]
(7.353)
x(ti,2 ) − xf (ti,2 ) ≤ δ0 .
(7.354)
and
satisfying
It follows from (7.346), (7.349)–(7.354), and property (ii) that x(t) − xf (t) ≤ , t ∈ [ti,1 , ti,2 ] and x(t) − xf (t) ≤ , t ∈ [ti + 3L0 + 3L1 , ti+1 − L1 − 1].
(7.355)
Assume that ti+1 ≤ 2L0 + 2L1 + 1.
(7.356)
Property (iii), (7.328), (7.346), and (7.349) imply that there exist ti,1 ∈ [ti , ti + L1 ]
(7.357)
x(ti,1 ) − xf (ti,1 ) ≤ δ0
(7.358)
satisfying
and ti,2 ∈ [si+1 − 3L1 − 3L0 − 1, si+1 − 3L0 − 2L1 − 1]
(7.359)
satisfying x(ti,2 ) − xf (ti,2 ) ≤ δ0 . It follows from (7.346), (7.356)–(7.360), and property (ii) that x(t) − xf (t) ≤ , t ∈ [ti,1 , ti,2 ] and x(t) − xf (t) ≤ , t ∈ [ti + L1 , ti+1 − 3L1 − 3L0 − 2].
(7.360)
546
7 ContinuousTime Nonautonomous Problems on Axis
In both cases, we have x(t) − xf (t) ≤ , t ∈ [ti + 3L0 + 3L1 , ti+1 − 3L1 − 3L0 − 2].
(7.361)
Assume that ti < −2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1.
(7.362)
In view of (7.335) and (7.362), si+1 > 2L0 + 2L1 .
(7.363)
Property (iii), (7.328), (7.346), (7.362), and (7.363) imply that there exist ti,1 ∈ [L0 , L0 + L1 ]
(7.364)
x(ti,1 ) − xf (ti,1 ) ≤ δ0
(7.365)
ti,2 ∈ [si+1 − L1 , si+1 ]
(7.366)
x(ti,2 ) − xf (ti,2 ) ≤ δ0 ,
(7.367)
ti,3 ∈ [ti , ti + L1 ]
(7.368)
x(ti,3 ) − xf (ti,3 ) ≤ δ0
(7.369)
ti,4 ∈ [−L0 − L1 , −L0 ]
(7.370)
x(ti,4 ) − xf (ti,4 ) ≤ δ0 .
(7.371)
satisfying
and
satisfying
satisfying
and
satisfying
It follows from (7.328), (7.346), (7.349), (7.362), (7.363), (7.364), (7.371), and property (ii) that
7.14 Proofs of Theorems 7.9 and 7.14
547
x(t) − xf (t) ≤ , t ∈ [ti,1 , ti,2 ] ∪ [ti,3 , ti,4 ], x(t) − xf (t) ≤ , t ∈ [L0 + L1 , ti+1 − 1 − L1 ] ∪ [ti + L1 , −L0 − L1 ]
(7.372)
and x(t)−xf (t) ≤ , t ∈ [ti +L1 , ti+1 −1 −L1 ] \(−L0 −L1 , L0 +L1 ).
(7.373)
By (7.361), (7.372), and (7.373), {t ∈ [T1 , T2 ] : x(t) − xf (t) ≤ } ⊃ ∪{[ti + 3L0 + 3L1 , ti+1 − 3L0 − 3L1 − 2] : i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8, ti ≥ −2L0 − 2L1 or ti+1 ≤ 2L0 + 2L1 + 1} ∪{[ti + L1 , ti+1 − L1 − 1] \ (−L0 − L1 , L0 + L1 ) : i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8, ti < −2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1}. This implies that {t ∈ [T1 , T2 ] : x(t) − xf (t) > } ⊂ ∪{[ti , ti+1 ] : i ∈ {0, . . . , q − 1}, ti+1 − ti < 8L0 + 8L1 + 8bf + 8} ∪{[ti , ti + 3L0 + 3L1 ] ∪ [ti+1 − 3L0 − 3L1 − 2, ti+1 ] : i ∈ {0, . . . , q − 1}, ti+1 − ti ≥ 8L0 + 8L1 + 8bf + 8} ∪{[ti , ti + L1 ] ∪ [ti+1 − L1 − 1, ti+1 ] ∪ [−L0 − L1 , L0 + L1 ] : i ∈ {0, . . . , q − 1}, ti < 2L0 − 2L1 , ti+1 > 2L0 + 2L1 + 1]}.
(7.374)
Clearly, the righthand side of (7.374) is a finite union of intervals. In view of (7.326) and (7.343), their number does not exceed 3q < Q. By (7.325), the maximal length of intervals in the righthand side of the relation above does not exceed l. Theorem 7.14 is proved.
548
7 ContinuousTime Nonautonomous Problems on Axis
7.15 Auxiliary Results for Theorem 7.15 Recall that for each z ∈ R 1 , z = max{i ∈ R 1 : i is an integer, i ≤ z}. Proposition 7.29. Assume that f has (P1), (P2) and LSC property and (T0 , z0 ) ∈ ∪{AL : L > 0}. Then there exists an (f )good and (f )minimal pair (x∗ , u∗ ) ∈ X(T0 , ∞) such that x∗ (T0 ) = z0 . Proof. There exists L0 > 0 such that (T0 , z0 ) ∈ AL0 .
(7.375)
It follows from Theorem 7.2 that there exists S0 > 0 such that for each T2 > T1 and each (x, u) ∈ X(T1 , T2 ), I f (T1 , T2 , x, u) + S0 ≥ I f (T1 , T2 , xf , uf ).
(7.376)
Fix an integer k0 ≥ L0 . LSC property and (7.375) imply that for each integer k ≥ k0 , there exists (xk , uk ) ∈ X(T0 , T0 + k) satisfying xk (T0 ) = z0 , I f (T0 , T0 + k, xk , uk ) = σ f (T0 , T0 + k, z0 ).
(7.377) (7.378)
In view of (7.8) and (7.375), for each integer k ≥ k0 , σ f (T0 , T0 + k, z0 ) ≤ L0 + I f (T0 , T0 + k, xf , uf ) + a0 L0 .
(7.379)
By (7.376), (7.378), and (7.379), for each integer k ≥ k0 and each pair of numbers s T1 , T2 ∈ [T0 , T0 + k] satisfying T1 < T2 , I f (T1 , T2 , xk , uk ) = I f (T0 , T0 + k, xk , uk ) −I f (T0 , T1 , xk , uk ) − I f (T2 , T0 + k, xk , uk ) ≤ I f (T0 , T0 + k, xf , uf ) + L0 (1 + a0 ) −I f (T0 , T1 , xf , uf ) + S0 − I f (T2 , T0 + k, xf , uf ) + S0 = I f (T1 , T2 , xf , uf ) + 2S0 + L0 (1 + a0 ).
(7.380)
By (7.370) and LSC property, extracting subsequences, using the diagonalization process, and reindexing, we obtain that there exists a strictly increasing sequence of natural numbers {kp }∞ p=1 such that k1 ≥ k0 and for each integer i ≥ 0, there exists
7.15 Auxiliary Results for Theorem 7.15
549
limp→∞ I f (T0 +i, T0 +i+1, xkp , ukp ) and there exists (yi , vi ) ∈ X(T0 +i, T0 +i+1) such that xkp (t) → yi (t) as p → ∞ for all t ∈ [T0 + i, T0 + i + 1],
(7.381)
I f (T0 + i, T0 + i + 1, yi , vi ) ≤ lim I f (T0 + i, T0 + i + 1, xkp , ukp ).
(7.382)
p→∞
In view of (7.381), there exists (x∗ , u∗ ) ∈ X(T0 , ∞) such that for each integer i ≥ 0, (x∗ (t), u∗ (t)) = (yi (t), vi (t)), t ∈ [T0 + i, T0 + i + 1].
(7.383)
It follows from (7.380), (7.382), and (7.383) that for every integer q ≥ 1, I f (T0 , T0 + q, x∗ , u∗ ) ≤ lim I f (T0 , T0 + q, xkp , ukp ) p→∞
≤ I f (T0 , T0 + q, xf , uf ) + 2S0 + L0 (1 + a0 ).
(7.384)
Theorem 7.6, (7.377), (7.381), and (7.384) imply that (x∗ , u∗ ) is (f )good and x∗ (T0 ) = z0 .
(7.385)
In order to complete the proof of the proposition, it is sufficient to show that (x∗ , u∗ ) is (f )minimal. Assume the contrary. Then there exist Δ > 0, an integer τ0 ≥ 1, and (y, v) ∈ X(T0 , T0 + τ0 ) such that y(T0 ) = x∗ (T0 ), y(T0 + τ0 ) = x∗ (T0 + τ0 ),
(7.386)
I f (T0 , T0 + τ0 , x∗ , u∗ ) > I f (T0 , T0 + τ0 , y, v) + 2Δ.
(7.387)
By (A2) and Lemma 7.25, there exists L1 , δ > 0 such that the following property holds: (i) for each (T , ξ ) ∈ A satisfying ξ − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x˜1 , u˜ 1 ) ∈ X(T , T + τ1 ) such that x˜1 (T ) = ξ , x˜1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x˜1 , u˜ 1 ) ≤ I f (T , T + τ1 , xf , uf ) + Δ/8 and there exist τ2 ∈ (0, bf ] and (x˜2 , u˜ 2 ) ∈ X(T − τ2 , T ) such that x˜2 (T − τ2 ) = xf (T − τ2 ) , x˜2 (T ) = ξ,
550
7 ContinuousTime Nonautonomous Problems on Axis
I f (T − τ2 , T , x˜2 , u˜ 2 ) ≤ I f (T − τ2 , T , xf , uf ) + Δ/8; if T2 > T1 ≥ L1 , (x, u) ∈ X(T1 , T2 ), x(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, we have I f (T1 , T2 , x, u) ≥ I f (T1 , T2 , xf , uf ) − Δ/8. Theorems 7.2 and 7.9, (7.375), and (7.378) imply that there exists an integer L2 > L0 + L1 + 2T0  such that for each integer k ≥ k0 + 2L2 , xk (t) − xf (t) ≤ δ, t ∈ [T0  + L2 , T0 + k − L2 ].
(7.388)
Set q0 = T0  + 1.
(7.389)
In view of (7.375), (7.381), (7.383), and (7.388), x∗ (t) − xf (t) ≤ δ for all t ≥ T0  + L2 .
(7.390)
By (7.382) and (7.383), there exists a natural number p0 such that kp0 > k0 + 2L1 + 4L2 + 2τ0 + 2 + 2bf + q0 ,
(7.391)
I f (T0 , T0 + τ0 + 2L2 + q0 , x∗ , u∗ ) ≤ I f (T0 , T0 + τ0 + 2L2 + q0 , xkp0 , ukp0 ) + Δ/2. (7.392) Property (i), (7.385), (7.388), and (7.390) imply that there exists (x, u) ∈ X(T0 , T0 + kp0 ) such that (x(t), u(t)) = (y(t), v(t)), t ∈ [T0 , T0 + τ0 ],
(7.393)
(x(t), u(t)) = (x∗ (t), u∗ (t)), t ∈ (T0 + τ0 , T0 + τ0 + 2L2 + q0 ],
(7.394)
x(T0 + τ0 + 2L2 + bf + q0 ) = xf (T0 + τ0 + 2L2 + bf + q0 ),
(7.395)
I f (T0 + τ0 + 2L2 + q0 , T0 + τ0 + 2L2 + bf + q0 , x, u) ≤ I f (T0 + τ0 + 2L2 + q0 , T0 + τ0 + 2L2 + bf + q0 , xf , uf ) + Δ/8,
(7.396)
(x(t), u(t)) = (xkp0 (t), ukp0 (t)), t ∈ [T0 + τ0 + 2L2 + 2bf + q0 , kp0 + T0 ], I f (T0 + τ0 + 2L2 + bf + q0 , T0 + τ0 + 2L2 + 2bf + q0 , x, u) ≤ I f (T0 +τ0 +2L2 +bf +q0 , T0 +τ0 +2L2 +2bf +q0 , xf , uf )+Δ/8.
(7.397)
7.15 Auxiliary Results for Theorem 7.15
551
It follows from (7.377), (7.378), (7.381), (7.383), (7.385), and (7.393) that I f (T0 , T0 + kp0 , x, u) ≥ I f (T0 , T0 + kp0 , xkp0 , ukp0 ).
(7.398)
By (7.327), (7.387), (7.392)–(7.396), (7.398), and property (i), 0 ≤ I f (T0 , T0 + kp0 , x, u) − I f (T0 , T0 + kp0 , xkp0 , ukp0 ) = I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , q0 + T0 + τ0 + 2L2 , x∗ , u∗ ) +I f (T0 + τ0 + 2L2 + q0 , q0 + T0 + τ0 + 2L2 + bf , xf , uf ) + Δ/8 +I f (T0 + τ0 + 2L2 + bf + q0 , q0 + T0 + τ0 + 2L2 + 2bf , xf , uf ) + Δ/8 +I f (T0 + τ0 + 2L2 + 2bf + q0 , kp0 + T0 , xkp0 , ukp0 ) −I f (T0 , T0 + τ0 + 2L2 + q0 , xkp0 , ukp0 ) −I f (T0 + τ0 + 2L2 + q0 , T0 + τ0 + 2L2 + bf + q0 , xkp0 , ukp0 ) −I f (T0 + τ0 + 2L2 + q0 + bf , T0 + τ0 + 2L2 + q0 + 2bf , xkp0 , ukp0 ) −I f (T0 + τ0 + 2L2 + 2bf + q0 , T0 + kp0 , xkp0 , ukp0 ) ≤ I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , T0 + τ0 + 2L2 + q0 , x∗ , u∗ ) +I f (T0 + τ0 + 2L2 + q0 , T0 + τ0 + 2L2 + bf + q0 , xf , uf ) +I f (T0 + τ0 + 2L2 + q0 + bf , T0 + τ0 + 2L2 + 2bf + q0 , xf , uf ) + Δ/4 −I f (T0 , T0 + τ0 + 2L2 + q0 , xkp0 , ukp0 ) −I f (T0 + τ0 + 2L2 + q0 , T0 + τ0 + 2L2 + bf + q0 , xf , uf ) + Δ/8 −I f (T0 + τ0 + 2L2 + bf + q0 , T0 + τ0 + 2L2 + 2bf + q0 , xf , uf ) + Δ/8 ≤ I f (T0 , T0 + τ0 , y, v) + I f (T0 + τ0 , T0 + τ0 + 2L2 + q0 , x∗ , u∗ ) −I f (T0 , T0 + τ0 + 2L2 + q0 , xkp0 , ukp0 ) + Δ/2 < I f (T0 , T0 + τ0 + q0 + 2L2 , x∗ , u∗ ) − 2Δ
552
7 ContinuousTime Nonautonomous Problems on Axis
−I f (T0 , T0 + τ0 + 2L2 + q0 , xkp0 , ukp0 ) + Δ/4 < −Δ, a contradiction. The contradiction we have reached completes the proof of Proposition 7.29.
7.16 Proof of Theorems 7.15 Clearly, (i) implies (ii) and (ii) implies (iii). By Theorem 7.9, (iii) implies (iv). Evidently (iv) implies (v). We show that (v) implies (iii). Assume that (x∗ , u∗ ) ∈ X(S, ∞) is (f )minimal and satisfies lim inf x∗ (t) − xf (t) = 0. t →∞
(7.399)
Since (x, ˜ u) ˜ is (f )good, there exists S0 > 0 such that for all numbers T > S, I f (S, T , x, ˜ u) ˜ − I f (S, T , xf , uf ) ≤ S0 .
(7.400)
By (A2), there exists δ > 0 such that the following property holds: (a) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T , T + τ1 ) which satisfy x1 (T ) = z, x1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x1 , u1 ) ≤ I f (T , T + τ1 , xf , uf ) + 1 and there exist τ2 ∈ (0, bf ] and (x2 , u2 ) ∈ X(T − τ2 , T ) such that x2 (T − τ2 ) = xf (T − τ2 ), x2 (T ) = z, I f (T − τ2 , T , x2 , u2 ) ≤ I f (T − τ2 , T , xf , uf ) + 1. In view of (7.399) and Theorem 7.9, there exists an increasing sequence {tk }∞ k=1 ⊂ (S, ∞) such that lim tk = ∞,
k→∞
x∗ (tk + 2bf ) − xf (tk + 2bf ) ≤ δ for all integers k ≥ 1, x(t) ˜ − xf (t) ≤ δ for all t ≥ t0 .
(7.401) (7.402)
7.16 Proof of Theorems 7.15
553
Let k ≥ 1 be an integer. By property (a), (7.401) and (7.402), there exists (y, v) ∈ X(S, tk + 2bf ) such that (y(t), v(t)) = (x(t), ˜ u(t)), ˜ t ∈ [S, tk ], y(tk + bf ) = xf (tk + bf ), I f (tk , tk + bf , y, v) ≤ I f (tk , tk + bf , xf , uf ) + 1, y(tk + 2bf ) = x∗ (tk + 2bf ), I f (tk + bf , tk + 2bf , y, v) ≤ I f (tk + bf , tk + 2bf , xf , uf ) + 1. The relations above and (7.400) imply that I f (S, tk + 2bf , x∗ , u∗ ) ≤ I f (S, tk + 2bf , y, v) ≤ I f (S, tk , x, ˜ u) ˜ + I f (tk + bf , tk + 2bf , xf , uf ) + 2 ≤ I f (S, tk + 2bf , xf , uf ) + 2 + S0 . This implies that (x∗ , u∗ ) is (f )good and (iii) holds. We show that (iii) implies (i). Assume that (x∗ , u∗ ) is (f )minimal and (f )good. Theorem 7.9 implies that lim x∗ (t) − xf (t) = 0.
t →∞
(7.403)
There exists S0 > 0 such that I f (S, T , x∗ , u∗ ) − I f (S, T , xf , uf ) ≤ S0 for all T > S.
(7.404)
Let (x, u) ∈ X(S, ∞) satisfy x(S) = x∗ (S).
(7.405)
We show that lim sup[I f (S, T , x∗ , u∗ ) − I f (S, T , x, u)] ≤ 0. T →∞
(7.406)
In view of Theorem 7.6, we may assume that (x, u) is (f )good. Theorem 7.9 implies that lim x(t) − xf (t) = 0.
t →∞
(7.407)
554
7 ContinuousTime Nonautonomous Problems on Axis
Let > 0. By (A2) and Lemma 7.25, there exist δ ∈ (0, ) and L1 > 0 such that the following property holds: (b) for each (T , z) ∈ A satisfying z − xf (T ) ≤ δ, there exist τ1 ∈ (0, bf ] and (x1 , u1 ) ∈ X(T , T + τ1 ) satisfying x1 (T ) = z , x1 (T + τ1 ) = xf (T + τ1 ), I f (T , T + τ1 , x1 , u1 ) ≤ I f (T , T + τ1 , xf , uf ) + /8 and then there exist τ2 ∈ (0, bf ] and (x2 , u2 ) ∈ X(T − τ2 , T ) satisfying x2 (T − τ2 ) = xf (T − τ2 ) , x2 (T ) = z, I f (T − τ2 , T , x2 , u2 ) ≤ I f (T − τ2 , T , xf , uf ) + /8; for each T1 ≥ L1 , each T2 > T1 and each (y, v) ∈ X(T1 , T2 ) which satisfies y(Ti ) − xf (Ti ) ≤ δ, i = 1, 2, we have I f (T1 , T2 , y, v) ≥ I f (T1 , T2 , xf , uf ) − /8. It follows from (7.403) and (7.407) that there exists τ0 > S0  such that x(t) − xf (t) ≤ δ, x∗ (t) − xf (t) ≤ δ for all t ≥ τ0 .
(7.408)
Let T ≥ τ0 + L1 . Property (b) and (7.408) imply that there exists (y, v) ∈ X(S, T + 2bf ) which satisfies (y(t), v(t)) = (x(t), u(t)), t ∈ [S, T ], y(T + bf ) = xf (T + bf ),
(7.409)
I f (T , T + bf , y, v) ≤ I f (T , T + bf , xf , uf ) + /8,
(7.410)
y(T + 2bf ) = x∗ (T + 2bf ), I f (T + bf , T + 2bf , y, v) ≤ I f (T + bf , T + 2bf , xf , uf ) + /8.
(7.411) (7.412)
By property (b) and (7.408), I f (T , T + 2bf , x∗ , u∗ ) ≥ I f (T , T + 2bf , xf , uf ) − /8. It follows from (7.405) and (7.409)–(7.413) that I f (S, T , x∗ , u∗ ) + I f (T , T + 2bf , x∗ , u∗ ) − /8
(7.413)
7.17 Proof of Theorem 7.16
555
≤ I f (S, T + 2bf , x∗ , u∗ ) ≤ I f (S, T + 2bf , y, v) = I f (S, T , x, u) + I f (T , T + 2bf , xf , uf ) + /4, I f (S, T , x∗ , u∗ ) ≤ I f (S, T , x, u) + for all T ≥ τ0 + L. Since is any positive number, we conclude that (7.406) holds and (x∗ , u∗ ) is (f )overtaking optimal and (i) holds. In order to complete the proof, we need only to apply Proposition 7.29.
7.17 Proof of Theorem 7.16 In view of Theorem 7.8, it is sufficient to show that fr has (P1), (P2), and (P3) with u˜ fr = uf . Theorem 7.2 implies that there exists M0 > 0 such that the following property holds: (i) for each S2 > S1 and each (x, u) ∈ X(S1 , S2 ), I f (S1 , S2 , x, u) + M0 ≥ I f (S1 , S2 , xf , uf ). We show that (P2) holds. Let , M > 0. Property (i) of Section 7.4 implies there exists δ > 0 such that the following property holds: (ii) if z ∈ E satisfies φ(z) ≤ δ, then z ≤ . Choose L > δ −1 r −1 (M0 + M).
(7.414)
Assume that T ∈ R 1 , (x, u) ∈ X(T , T + L) satisfies I fr (T , T + L, x, u) ≤ I fr (T , T + L, xf , uf ) + M.
(7.415)
By (7.415), property (i) and the definition of fr , I f (T , T + L, x, u) + r
T +L
φ(x(t) − xf (t))dt
T
= I fr (T , T +L, x, u) ≤ I fr (T , T +L, xf , uf )+M ≤ I f (T , T +L, x, u)+M+M0 ,
T +L T
φ(x(t) − xf (t))dt ≤ r −1 (M0 + M).
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7 ContinuousTime Nonautonomous Problems on Axis
Together with (7.414), this implies that inf{φ(x(t) − xf (t)) : t ∈ [T , T + L]} ≤ L−1 (M0 + M)r −1 < δ. Therefore there exists τ ∈ [T , T + L] such that φ(x(τ ) − xf (τ )) < δ. Together with property (ii) this implies that x(τ ) − xf (τ ) ≤ . Therefore (P2) holds. Let us show that fr has (P1). Let (x, u) ∈ X(−∞, ∞) be (fr )good. There exists M1 > 0 such that I fr (−T , T , x, u) − I fr (−T , T , xf , uf ) < M1 for all T > 0.
(7.416)
We show that lim (x(t) − xf (t)) = 0,
t →∞
lim (x(t) − xf (t)) = 0.
t →−∞
Assume the contrary. Then at least one of the following properties holds: (iii) there exists a sequence {tk }∞ k=1 ⊂ (0, ∞) such that for each integer k ≥ 1, tk+1 > tk + 8 and x(tk ) − xf (tk ) > ; (iv) there exists a sequence {tk }∞ k=1 ⊂ (−∞, 0) such that for each integer k ≥ 1, tk+1 < tk − 8 and x(tk ) − xf (tk ) > . Property (i), the definition of fr and (7.416) imply that for each T > 0, M1 > I fr (−T , T , x, u) − I fr (−T , T , xf , uf ) =r
T −T
φ(x(t) − xf (t))dt + I f (−T , T , x, u) − I f (−T , T , xf , uf ) ≥r
Δ := lim
T
T →∞ −T
T −T
φ(x(t) − xf (t))d − M0 ,
φ(x(t) − xf (t))d < (M0 + M1 )r −1 .
(7.417)
Property (i) and (7.416) imply that for each T2 > T1 and each T > 0 satisfying [T1 , T2 ] ⊂ [−T , T ], I f (−T , T , x, u) < I f (−T , T , xf , uf ) + M1 , I f (T1 , T2 , x, u) = I f (−T , T , x, u) − I f (−T , T1 , x, u) − I f (T2 , T , x, u) ≤ I f (−T , T , xf , uf ) + M1 − I f (−T , T1 , xf , uf ) + M0 − I f (T2 , T , x, u) + M0 = I f (T1 , T2 , xf , uf ) + M1 + 2M0 .
(7.418)
7.17 Proof of Theorem 7.16
557
In view of (7.8), (7.13), and (7.418), for each T ∈ R 1 , I f (T , T +1, x, u) ≤ I f (T , T +1, xf , uf )+2M0 +M1 ≤ 2M0 +M1 +2Δf +2a0 . (7.419) Lemma 7.21 implies that there exists δ ∈ (0, 4−1 ) such that the following property holds: (v) for each T ∈ R 1 , each (y, v) ∈ X(T , T + 1) satisfying I f (T , T + 1, y, v) ≤ 2M0 + M1 + 2Δf + 2a0 and each t1 , t2 ∈ [T , T + 1] satisfying t1 − t2  ≤ δ the inequality y(t1 ) − y(t2 ) ≤ holds. Property (i) of Section 7.4 implies that there exists γ > 0 such that the following property holds: (vi) for each ξ ∈ E satisfying ξ ≥ /2, then φ(ξ ) ≥ γ . Let k ≥ 1 be an integer. It follows from (7.8), (7.13), and (7.419) that I f (tk , tk+1 , xf , uf ) ≤ 2Δf + 2a0 , I f (tk , tk+1 , x, u) ≤ 2M0 + M1 + 2Δf + 2a0. Properties (iii)–(vi) and the relations above imply that for all t ∈ [tk , tk + δ], x(t) − x(tk ) ≤ /8, xf (t) − xf (tk ) ≤ /8, xf (t) − x(t) ≥ xf (tk ) − x(tk ) − xf (t) − xf (tk ) − x(t) − x(tk ) ≥ /2, φ(x(t) − xf (t)) ≥ γ . Therefore
tk +δ
φ(x(t) − xf (t))dt ≥ γ δ.
tk
This implies that lim
T
T →∞ −T
φ(x(t) − xf (t))dt = ∞.
This contradicts (7.417). The contradiction we have reached proves that (P1) holds. Let us show that (P3) holds. Clearly, (xf , uf ) is (fr )minimal. Assume that (x, u) ∈ X(−∞, ∞) is (fr )good and (fr )minimal. Property (P1) implies that
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7 ContinuousTime Nonautonomous Problems on Axis
lim x(t) − xf (t) = 0,
t →∞
lim x(t) − xf (t) = 0.
t →−∞
(7.420)
We show that x(t) = xf (t) for all t ∈ R 1 . Assume the contrary. Then there exists t0 ∈ R 1 such that γ := x(t0 ) − xf (t0 ) > 0.
(7.421)
By (7.421), there exists 0 ∈ (0, 1) such that x(t) − xf (t) > γ /2 for all t ∈ [t0 , t0 + 0 ].
(7.422)
Property (i) of Section 7.4 implies that there exists δ0 > 0 such that if z ∈ E satisfies φ(z) ≤ δ0 , then z ≤ γ /4.
(7.423)
It follows from (7.422) and (7.423) that φ(x(t) − xf (t))) > δ0 for allˆt ∈ [t0 , t0 + 0 ].
(7.424)
By (A3) and Lemma 7.25, there exists L1 > 0, δ1 ∈ (0, min{δ0 , 0 }) such that the following property holds: (vii) for each (S, z) ∈ A satisfying z − xf (S) ≤ δ1 there exist τ1 ∈ (0, bf ] and (y1 , v1 ) ∈ X(S, S + τ1 ), τ2 ∈ (0, bf ], (y2 , v2 ) ∈ X(S − τ2 , S2 ) such that y1 (S) = z , y1 (S + τ1 ) = xf (S + τ1 ), I fr (S, S + τ1 , y1 , v1 ) ≤ I fr (S, S + τ1 , xf , uf ) + 8−1 0 δ0 r, y2 (S − τ2 ) = xf (S − τ2 ) , y2 (S) = z, I fr (S − τ2 , S, y2 , v2 ) ≤ I fr (S − τ2 , S, xf , uf ) + 8−1 0 δ0 r, y1 (t) − xf (t) ≤ 8−1 0 δ0 , t ∈ [S, S + τ1 ], y2 (t) − xf (t) ≤ 8−1 0 δ0 , t ∈ [S − τ2 , S2 ]; for each T2 > T1 such that either T2 ≤ −L1 or T1 ≥ L1 and each (y, v) ∈ X(T1 , T2 ) satisfying y(Ti ) − xf (Ti ) ≤ δ1 , i = 1, 2, we have I fr (T1 , T2 , y, v) ≥ I fr (T1 , T2 , xf , uf ) − 8−1 0 δ0 r.
7.17 Proof of Theorem 7.16
559
In view of (7.422), there exists T0 > L1 such that [−L1 + L0 − 2bf − 2, L0 + 1 + 2bf + 2 + L1 ] ⊂ [−T0 , T0 ],
(7.425)
x(t) − xf (t) ≤ δ1 for all t ∈ (−∞, −T0 ] ∪ [T0 , ∞).
(7.426)
Property (vii) and (7.426) imply that there exists (y, v) ∈ X(−T0 − bf , T0 + bf ) such that y(−T0 − bf ) = x(−T0 − bf ),
(7.427)
(y(t), v(t)) = (xf (t), uf (t)), t ∈ [−T0 , T0 ],
(7.428)
I fr (−T0 − bf , −T0 , y, v) ≤ I fr (−T0 − bf , −T0 , xf , uf ) + 8−1 0 δ0 r,
(7.429)
y(T0 + bf ) = x(T0 + bf ), I fr (T0